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Fourier Analysis: An Introduction (Princeton Lectures in Analysis, Volume 1)

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FOURIER ANALYSIS

Ibookroot October 20, 2007

Princeton Lectures in Analysis

I Fourier Analysis: An Introduction

II Complex Analysis

III Real Analysis:
Measure Theory, Integration, and

Hilbert Spaces

Ibookroot October 20, 2007

Princeton Lectures in Analysis

FOURIER ANALYSIS

an introduction

Elias M. Stein

Rami Shakarchi

PRINCETON UNIVERSITY PRESS

PRINCETON AND OXFORD

Princeton, New Jersey 08540
In the United Kingdom: Princeton University Press,

6 Oxford Street, Woodstock, Oxfordshire OX20 1TW

Library of Congress Control Number 2003103688
ISBN 978-0-691-11384-5

British Library Cataloging-in-Publication Data is available

The publisher would like to acknowledge the authors of this volume for
providing the camera-ready copy from which this book was printed

Printed on acid-free paper. ∞

press.princeton.edu

Printed in the United States of America

5 7 9 10 8 6

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To my grandchildren

Carolyn, Alison, Jason

E.M.S.

To my parents

Mohamed & Mireille

and my brother

Karim

R.S.

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Foreword

Beginning in the spring of 2000, a series of four one-semester courses
were taught at Princeton University whose purpose was to present, in
an integrated manner, the core areas of analysis. The objective was to
make plain the organic unity that exists between the various parts of the
subject, and to illustrate the wide applicability of ideas of analysis to
other fields of mathematics and science. The present series of books is
an elaboration of the lectures that were given.

While there are a number of excellent texts dealing with individual
parts of what we cover, our exposition aims at a different goal: pre-
senting the various sub-areas of analysis not as separate disciplines, but
rather as highly interconnected. It is our view that seeing these relations
and their resulting synergies will motivate the reader to attain a better
understanding of the subject as a whole. With this outcome in mind, we
have concentrated on the main ideas and theorems that have shaped the
field (sometimes sacrificing a more systematic approach), and we have
been sensitive to the historical order in which the logic of the subject
developed.

We have organized our exposition into four volumes, each reflecting
the material covered in a semester. Their contents may be broadly sum-
marized as follows:

I. Fourier series and integrals.

II. Complex analysis.

III. Measure theory, Lebesgue integration, and Hilbert spaces.

IV. A selection of further topics, including functional analysis, distri-
butions, and elements of probability theory.

However, this listing does not by itself give a complete picture of
the many interconnections that are presented, nor of the applications
to other branches that are highlighted. To give a few examples: the ele-
ments of (finite) Fourier series studied in Book I, which lead to Dirichlet
characters, and from there to the infinitude of primes in an arithmetic
progression; the X-ray and Radon transforms, which arise in a number of

Ziyang He

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FOREWORD

problems in Book I, and reappear in Book III to play an important role in
understanding Besicovitch-like sets in two and three dimensions; Fatou’s
theorem, which guarantees the existence of boundary values of bounded
holomorphic functions in the disc, and whose proof relies on ideas devel-
oped in each of the first three books; and the theta function, which first
occurs in Book I in the solution of the heat equation, and is then used
in Book II to find the number of ways an integer can be represented as
the sum of two or four squares, and in the analytic continuation of the
zeta function.

A few further words about the books and the courses on which they
were based. These courses where given at a rather intensive pace, with 48
lecture-hours a semester. The weekly problem sets played an indispens-
able part, and as a result exercises and problems have a similarly im-
portant role in our books. Each chapter has a series of “Exercises” that
are tied directly to the text, and while some are easy, others may require
more effort. However, the substantial number of hints that are given
should enable the reader to attack most exercises. There are also more
involved and challenging “Problems”; the ones that are most difficult, or
go beyond the scope of the text, are marked with an asterisk.

Despite the substantial connections that exist between the different
volumes, enough overlapping material has been provided so that each of
the first three books requires only minimal prerequisites: acquaintance
with elementary topics in analysis such as limits, series, differentiable
functions, and Riemann integration, together with some exposure to lin-
ear algebra. This makes these books accessible to students interested
in such diverse disciplines as mathematics, physics, engineering, and
finance, at both the undergraduate and graduate level.

It is with great pleasure that we express our appreciation to all who
have aided in this enterprise. We are particularly grateful to the stu-
dents who participated in the four courses. Their continuing interest,
enthusiasm, and dedication provided the encouragement that made this
project possible. We also wish to thank Adrian Banner and Jose Luis
Rodrigo for their special help in running the courses, and their efforts to
see that the students got the most from each class. In addition, Adrian
Banner also made valuable suggestions that are incorporated in the text.

viii

Ziyang He

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ixFOREWORD

We wish also to record a note of special thanks for the following in-
dividuals: Charles Fefferman, who taught the first week (successfully
launching the whole project!); Paul Hagelstein, who in addition to read-
ing part of the manuscript taught several weeks of one of the courses, and
has since taken over the teaching of the second round of the series; and
Daniel Levine, who gave valuable help in proof-reading. Last but not
least, our thanks go to Gerree Pecht, for her consummate skill in type-
setting and for the time and energy she spent in the preparation of all
aspects of the lectures, such as transparencies, notes, and the manuscript.

We are also happy to acknowledge our indebtedness for the support
we received from the 250th Anniversary Fund of Princeton University,
and the National Science Foundation’s VIGRE program.

Elias M. Stein

Rami Shakarchi

Princeton, New Jersey
August 2002

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Preface to Book I

Any effort to present an overall view of analysis must at its start deal
with the following questions: Where does one begin? What are the initial
subjects to be treated, and in what order are the relevant concepts and
basic techniques to be developed?

Our answers to these questions are guided by our view of the centrality
of Fourier analysis, both in the role it has played in the development of
the subject, and in the fact that its ideas permeate much of the present-
day analysis. For these reasons we have devoted this first volume to an
exposition of some basic facts about Fourier series, taken together with
a study of elements of Fourier transforms and finite Fourier analysis.
Starting this way allows one to see rather easily certain applications to
other sciences, together with the link to such topics as partial differential
equations and number theory. In later volumes several of these connec-
tions will be taken up from a more systematic point of view, and the ties
that exist with complex analysis, real analysis, Hilbert space theory, and
other areas will be explored further.

In the same spirit, we have been mindful not to overburden the begin-
ning student with some of the difficulties that are inherent in the subject:
a proper appreciation of the subtleties and technical complications that
arise can come only after one has mastered some of the initial ideas in-
volved. This point of view has led us to the following choice of material
in the present volume:

• Fourier series. At this early stage it is not appropriate to intro-
duce measure theory and Lebesgue integration. For this reason
our treatment of Fourier series in the first four chapters is carried
out in the context of Riemann integrable functions. Even with this
restriction, a substantial part of the theory can be developed, de-
tailing convergence and summability; also, a variety of connections
with other problems in mathematics can be illustrated.

• Fourier transform. For the same reasons, instead of undertaking
the theory in a general setting, we confine ourselves in Chapters 5
and 6 largely to the framework of test functions. Despite these lim-
itations, we can learn a number of basic and interesting facts about
Fourier analysis in Rd and its relation to other areas, including the
wave equation and the Radon transform.

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PREFACE TO BOOK I

• Finite Fourier analysis. This is an introductory subject par excel-
lence, because limits and integrals are not explicitly present. Nev-
ertheless, the subject has several striking applications, including
the proof of the infinitude of primes in arithmetic progression.

Taking into account the introductory nature of this first volume, we
have kept the prerequisites to a minimum. Although we suppose some
acquaintance with the notion of the Riemann integral, we provide an
appendix that contains most of the results about integration needed in
the text.

We hope that this approach will facilitate the goal that we have set
for ourselves: to inspire the interested reader to learn more about this
fascinating subject, and to discover how Fourier analysis affects decisively
other parts of mathematics and science.

xii

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Contents

Foreword

Preface

Chapter 1. The Genesis of Fourier Analysis 1

1 The vibrating string 2
1.1 Derivation of the wave equation 6
1.2 Solution to the wave equation 8
1.3 Example: the plucked string 16

2 The heat equation 18
2.1 Derivation of the heat equation 18
2.2 Steady-state heat equation in the disc 19

3 Exercises 22
4 Problem 27

Chapter 2. Basic Properties of Fourier Series 29

1 Examples and formulation of the problem 30
1.1 Main definitions and some examples 34

2 Uniqueness of Fourier series 39
3 Convolutions 44
4 Good kernels 48
5 Cesàro and Abel summability: applications to Fourier series 51

5.1 Cesàro means and summation 51
5.2 Fejér’s theorem 52
5.3 Abel means and summation 54
5.4 The Poisson kernel and Dirichlet’s problem in the

unit disc 55
6 Exercises 58
7 Problems 65

Chapter 3. Convergence of Fourier Series 69

1 Mean-square convergence of Fourier series 70
1.1 Vector spaces and inner products 70
1.2 Proof of mean-square convergence 76

2 Return to pointwise convergence 81
2.1 A local result 81
2.2 A continuous function with diverging Fourier series 83

vii

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CONTENTS

3 Exercises 87
4 Problems 95

Chapter 4. Some Applications of Fourier Series 100

1 The isoperimetric inequality 101
2 Weyl’s equidistribution theorem 105
3 A continuous but nowhere differentiable function 113
4 The heat equation on the circle 118
5 Exercises 120
6 Problems 125

Chapter 5. The Fourier Transform on R 129
1 Elementary theory of the Fourier transform 131

1.1 Integration of functions on the real line 131
1.2 Definition of the Fourier transform 134
1.3 The Schwartz space 134
1.4 The Fourier transform on S 136
1.5 The Fourier inversion 140
1.6 The Plancherel formula 142
1.7 Extension to functions of moderate decrease 144
1.8 The Weierstrass approximation theorem 144

2 Applications to some partial differential equations 145
2.1 The time-dependent heat equation on the real line 145
2.2 The steady-state heat equation in the upper half-

plane 149
3 The Poisson summation formula 153

3.1 Theta and zeta functions 155
3.2 Heat kernels 156
3.3 Poisson kernels 157

4 The Heisenberg uncertainty principle 158
5 Exercises 161
6 Problems 169

Chapter 6. The Fourier Transform on Rd 175
1 Preliminaries 176

1.1 Symmetries 176
1.2 Integration on Rd 178

2 Elementary theory of the Fourier transform 180
3 The wave equation in Rd × R 184

3.1 Solution in terms of Fourier transforms 184
3.2 The wave equation in R3 × R 189

xiv

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CONTENTS xv

3.3 The wave equation in R2 × R: descent 194
4 Radial symmetry and Bessel functions 196
5 The Radon transform and some of its applications 198

5.1 The X-ray transform in R2 199
5.2 The Radon transform in R3 201
5.3 A note about plane waves 207

6 Exercises 207
7 Problems 212

Chapter 7. Finite Fourier Analysis 218

1 Fourier analysis on Z(N) 219
1.1 The group Z(N) 219
1.2 Fourier inversion theorem and Plancherel identity

on Z(N) 221
1.3 The fast Fourier transform 224

2 Fourier analysis on finite abelian groups 226
2.1 Abelian groups 226
2.2 Characters 230
2.3 The orthogonality relations 232
2.4 Characters as a total family 233
2.5 Fourier inversion and Plancherel formula 235

3 Exercises 236
4 Problems 239

Chapter 8. Dirichlet’s Theorem 241

1 A little elementary number theory 241
1.1 The fundamental theorem of arithmetic 241
1.2 The infinitude of primes 244

2 Dirichlet’s theorem 252
2.1 Fourier analysis, Dirichlet characters, and reduc-

tion of the theorem 254
2.2 Dirichlet L-functions 255

3 Proof of the theorem 258
3.1 Logarithms 258
3.2 L-functions 261
3.3 Non-vanishing of the L-function 265

4 Exercises 275
5 Problems 279

Appendix: Integration 281

1 Definition of the Riemann integral 281

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x CONTENTS

1.1 Basic properties 282
1.2 Sets of measure zero and discontinuities of inte-

grable functions 286
2 Multiple integrals 289

2.1 The Riemann integral in Rd 289
2.2 Repeated integrals 291
2.3 The change of variables formula 292
2.4 Spherical coordinates 293

3 Improper integrals. Integration over Rd 294
3.1 Integration of functions of moderate decrease 294
3.2 Repeated integrals 295
3.3 Spherical coordinates 297

Notes and References 298

Bibliography 300

Symbol Glossary 303

Index 305

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1 The Genesis of Fourier
Analysis

Regarding the researches of d’Alembert and Euler could
one not add that if they knew this expansion, they
made but a very imperfect use of it. They were both
persuaded that an arbitrary and discontinuous func-
tion could never be resolved in series of this kind, and
it does not even seem that anyone had developed a
constant in cosines of multiple arcs, the first problem
which I had to solve in the theory of heat.

J. Fourier, 1808-9

In the beginning, it was the problem of the vibrating string, and the
later investigation of heat flow, that led to the development of Fourier
analysis. The laws governing these distinct physical phenomena were
expressed by two different partial differential equations, the wave and
heat equations, and these were solved in terms of Fourier series.

Here we want to start by describing in some detail the development
of these ideas. We will do this initially in the context of the problem of
the vibrating string, and we will proceed in three steps. First, we de-
scribe several physical (empirical) concepts which motivate correspond-
ing mathematical ideas of importance for our study. These are: the role
of the functions cos t, sin t, and eit suggested by simple harmonic mo-
tion; the use of separation of variables, derived from the phenomenon
of standing waves; and the related concept of linearity, connected to the
superposition of tones. Next, we derive the partial differential equation
which governs the motion of the vibrating string. Finally, we will use
what we learned about the physical nature of the problem (expressed
mathematically) to solve the equation. In the last section, we use the
same approach to study the problem of heat diffusion.

Given the introductory nature of this chapter and the subject matter
covered, our presentation cannot be based on purely mathematical rea-
soning. Rather, it proceeds by plausibility arguments and aims to provide
the motivation for the further rigorous analysis in the succeeding chap-
ters. The impatient reader who wishes to begin immediately with the
theorems of the subject may prefer to pass directly to the next chapter.

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2 Chapter 1. THE GENESIS OF FOURIER ANALYSIS

1 The vibrating string

The problem consists of the study of the motion of a string fixed at
its end points and allowed to vibrate freely. We have in mind physical
systems such as the strings of a musical instrument. As we mentioned
above, we begin with a brief description of several observable physical
phenomena on which our study is based. These are:

• simple harmonic motion,

• standing and traveling waves,

• harmonics and superposition of tones.
Understanding the empirical facts behind these phenomena will moti-

vate our mathematical approach to vibrating strings.

Simple harmonic motion

Simple harmonic motion describes the behavior of the most basic oscil-
latory system (called the simple harmonic oscillator), and is therefore
a natural place to start the study of vibrations. Consider a mass {m}
attached to a horizontal spring, which itself is attached to a fixed wall,
and assume that the system lies on a frictionless surface.

Choose an axis whose origin coincides with the center of the mass when
it is at rest (that is, the spring is neither stretched nor compressed), as
shown in Figure 1. When the mass is displaced from its initial equilibrium

0y y(t)y

Figure 1. Simple harmonic oscillator

position and then released, it will undergo simple harmonic motion.
This motion can be described mathematically once we have found the
differential equation that governs the movement of the mass.

Let y(t) denote the displacement of the mass at time t. We assume that
the spring is ideal, in the sense that it satisfies Hooke’s law: the restoring
force F exerted by the spring on the mass is given by F = −ky(t). Here

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1. The vibrating string 3

k > 0 is a given physical quantity called the spring constant. Applying
Newton’s law (force = mass × acceleration), we obtain

−ky(t) = my′′(t),

where we use the notation y′′ to denote the second derivative of y with
respect to t. With c =

√
k/m, this second order ordinary differential

equation becomes

(1) y′′(t) + c2y(t) = 0.

The general solution of equation (1) is given by

y(t) = a cos ct + b sin ct ,

where a and b are constants. Clearly, all functions of this form solve
equation (1), and Exercise 6 outlines a proof that these are the only
(twice differentiable) solutions of that differential equation.

In the above expression for y(t), the quantity c is given, but a and b
can be any real numbers. In order to determine the particular solution
of the equation, we must impose two initial conditions in view of the
two unknown constants a and b. For example, if we are given y(0) and
y′(0), the initial position and velocity of the mass, then the solution of
the physical problem is unique and given by

y(t) = y(0) cos ct +
y′(0)

c
sin ct .

One can easily verify that there exist constants A > 0 and ϕ ∈ R such
that

a cos ct + b sin ct = A cos(ct− ϕ).

Because of the physical interpretation given above, one calls A =
√

a2 + b2

the “amplitude” of the motion, c its “natural frequency,” ϕ its “phase”
(uniquely determined up to an integer multiple of 2π), and 2π/c the
“period” of the motion.

The typical graph of the function A cos(ct− ϕ), illustrated in
Figure 2, exhibits a wavelike pattern that is obtained from translating
and stretching (or shrinking) the usual graph of cos t.

We make two observations regarding our examination of simple har-
monic motion. The first is that the mathematical description of the most
elementary oscillatory system, namely simple harmonic motion, involves

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4 Chapter 1. THE GENESIS OF FOURIER ANALYSIS

Figure 2. The graph of A cos(ct− ϕ)

the most basic trigonometric functions cos t and sin t. It will be impor-
tant in what follows to recall the connection between these functions
and complex numbers, as given in Euler’s identity eit = cos t + i sin t.
The second observation is that simple harmonic motion is determined as
a function of time by two initial conditions, one determining the position,
and the other the velocity (specified, for example, at time t = 0). This
property is shared by more general oscillatory systems, as we shall see
below.

Standing and traveling waves

As it turns out, the vibrating string can be viewed in terms of one-
dimensional wave motions. Here we want to describe two kinds of mo-
tions that lend themselves to simple graphic representations.

• First, we consider standing waves. These are wavelike motions
described by the graphs y = u(x, t) developing in time t as shown
in Figure 3.

In other words, there is an initial profile y = ϕ(x) representing the
wave at time t = 0, and an amplifying factor ψ(t), depending on t,
so that y = u(x, t) with

u(x, t) = ϕ(x)ψ(t).

The nature of standing waves suggests the mathematical idea of
“separation of variables,” to which we will return later.

• A second type of wave motion that is often observed in nature is
that of a traveling wave. Its description is particularly simple:

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1. The vibrating string 5

u(x, 0) = ϕ(x)

u(x, t0)

Figure 3. A standing wave at different moments in time: t = 0 and
t = t0

there is an initial profile F (x) so that u(x, t) equals F (x) when
t = 0. As t evolves, this profile is displaced to the right by ct units,
where c is a positive constant, namely

u(x, t) = F (x− ct).

Graphically, the situation is depicted in Figure 4.

F (x) F (x− ct0)

Figure 4. A traveling wave at two different moments in time: t = 0 and
t = t0

Since the movement in t is at the rate c, that constant represents the
velocity of the wave. The function F (x− ct) is a one-dimensional
traveling wave moving to the right. Similarly, u(x, t) = F (x + ct)
is a one-dimensional traveling wave moving to the left.

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6 Chapter 1. THE GENESIS OF FOURIER ANALYSIS

Harmonics and superposition of tones

The final physical observation we want to mention (without going into
any details now) is one that musicians have been aware of since time
immemorial. It is the existence of harmonics, or overtones. The pure
tones are accompanied by combinations of overtones which are primar-
ily responsible for the timbre (or tone color) of the instrument. The idea
of combination or superposition of tones is implemented mathematically
by the basic concept of linearity, as we shall see below.

We now turn our attention to our main problem, that of describing the
motion of a vibrating string. First, we derive the wave equation, that is,
the partial differential equation that governs the motion of the string.

1.1 Derivation of the wave equation

Imagine a homogeneous string placed in the (x, y)-plane, and stretched
along the x-axis between x = 0 and x = L. If it is set to vibrate, its
displacement y = u(x, t) is then a function of x and t, and the goal is to
derive the differential equation which governs this function.

For this purpose, we consider the string as being subdivided into a
large number N of masses (which we think of as individual particles)
distributed uniformly along the x-axis, so that the nth particle has its
x-coordinate at xn = nL/N . We shall therefore conceive of the vibrat-
ing string as a complex system of N particles, each oscillating in the
vertical direction only ; however, unlike the simple harmonic oscillator we
considered previously, each particle will have its oscillation linked to its
immediate neighbor by the tension of the string.

yn−1
yn yn+1

xn−1 xn+1xn

Figure 5. A vibrating string as a discrete system of masses

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1. The vibrating string 7

We then set yn(t) = u(xn, t), and note that xn+1 − xn = h, with h =
L/N . If we assume that the string has constant density ρ > 0, it is
reasonable to assign mass equal to ρh to each particle. By Newton’s law,
ρhy′′n(t) equals the force acting on the n

th particle. We now make the
simple assumption that this force is due to the effect of the two nearby
particles, the ones with x-coordinates at xn−1 and xn+1 (see Figure 5).
We further assume that the force (or tension) coming from the right of
the nth particle is proportional to (yn+1 − yn)/h, where h is the distance
between xn+1 and xn; hence we can write the tension as

(τ
h

)
(yn+1 − yn),

where τ > 0 is a constant equal to the coefficient of tension of the string.
There is a similar force coming from the left, and it is

(τ
h

)
(yn−1 − yn).

Altogether, adding these forces gives us the desired relation between the
oscillators yn(t), namely

(2) ρhy′′n(t) =
τ

h
{yn+1(t) + yn−1(t)− 2yn(t)} .

On the one hand, with the notation chosen above, we see that

yn+1(t) + yn−1(t)− 2yn(t) = u(xn + h, t) + u(xn − h, t)− 2u(xn, t).

On the other hand, for any reasonable function F (x) (that is, one that
has continuous second derivatives) we have

F (x + h) + F (x− h)− 2F (x)
h2

→ F ′′(x) as h → 0.

Thus we may conclude, after dividing by h in (2) and letting h tend to
zero (that is, N goes to infinity), that

ρ
∂2u

∂t2
= τ

∂2u

∂x2
,

1
c2

∂2u

∂t2
=

∂2u

∂x2
, with c =

√
τ/ρ.

This relation is known as the one-dimensional wave equation, or
more simply as the wave equation. For reasons that will be apparent
later, the coefficient c > 0 is called the velocity of the motion.

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8 Chapter 1. THE GENESIS OF FOURIER ANALYSIS

In connection with this partial differential equation, we make an im-
portant simplifying mathematical remark. This has to do with scaling,
or in the language of physics, a “change of units.” That is, we can think of
the coordinate x as x = aX where a is an appropriate positive constant.
Now, in terms of the new coordinate X, the interval 0 ≤ x ≤ L becomes
0 ≤ X ≤ L/a. Similarly, we can replace the time coordinate t by t = bT ,
where b is another positive constant. If we set U(X,T ) = u(x, t), then

∂U

∂X
= a

∂u

∂x
,

∂2U

∂X2
= a2

∂2u

∂x2
,

and similarly for the derivatives in t. So if we choose a and b appropri-
ately, we can transform the one-dimensional wave equation into

∂2U

∂T 2
=

∂2U

∂X2
,

which has the effect of setting the velocity c equal to 1. Moreover, we have
the freedom to transform the interval 0 ≤ x ≤ L to 0 ≤ X ≤ π. (We shall
see that the choice of π is convenient in many circumstances.) All this
is accomplished by taking a = L/π and b = L/(cπ). Once we solve the
new equation, we can of course return to the original equation by making
the inverse change of variables. Hence, we do not sacrifice generality by
thinking of the wave equation as given on the interval [0, π] with velocity
c = 1.

1.2 Solution to the wave equation

Having derived the equation for the vibrating string, we now explain two
methods to solve it:

• using traveling waves,

• using the superposition of standing waves.

While the first approach is very simple and elegant, it does not directly
give full insight into the problem; the second method accomplishes that,
and moreover is of wide applicability. It was first believed that the second
method applied only in the simple cases where the initial position and
velocity of the string were themselves given as a superposition of standing
waves. However, as a consequence of Fourier’s ideas, it became clear that
the problem could be worked either way for all initial conditions.

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1. The vibrating string 9

Traveling waves

To simplify matters as before, we assume that c = 1 and L = π, so that
the equation we wish to solve becomes

∂2u

∂t2
=

∂2u

∂x2
on 0 ≤ x ≤ π.

The crucial observation is the following: if F is any twice differentiable
function, then u(x, t) = F (x + t) and u(x, t) = F (x− t) solve the wave
equation. The verification of this is a simple exercise in differentiation.
Note that the graph of u(x, t) = F (x− t) at time t = 0 is simply the
graph of F , and that at time t = 1 it becomes the graph of F translated
to the right by 1. Therefore, we recognize that F (x− t) is a traveling
wave which travels to the right with speed 1. Similarly, u(x, t) = F (x + t)
is a wave traveling to the left with speed 1. These motions are depicted
in Figure 6.

F (x + t) F (x) F (x− t)

Figure 6. Waves traveling in both directions

Our discussion of tones and their combinations leads us to observe
that the wave equation is linear. This means that if u(x, t) and v(x, t)
are particular solutions, then so is αu(x, t) + βv(x, t), where α and β
are any constants. Therefore, we may superpose two waves traveling in
opposite directions to find that whenever F and G are twice differentiable
functions, then

u(x, t) = F (x + t) + G(x− t)

is a solution of the wave equation. In fact, we now show that all solutions
take this form.

We drop for the moment the assumption that 0 ≤ x ≤ π, and suppose
that u is a twice differentiable function which solves the wave equation

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10 Chapter 1. THE GENESIS OF FOURIER ANALYSIS

for all real x and t. Consider the following new set of variables ξ = x + t,
η = x− t, and define v(ξ, η) = u(x, t). The change of variables formula
shows that v satisfies

∂2v

∂ξ∂η
= 0.

Integrating this relation twice gives v(ξ, η) = F (ξ) + G(η), which then
implies

u(x, t) = F (x + t) + G(x− t),

for some functions F and G.

We must now connect this result with our original problem, that is,
the physical motion of a string. There, we imposed the restrictions 0 ≤
x ≤ π, the initial shape of the string u(x, 0) = f(x), and also the fact
that the string has fixed end points, namely u(0, t) = u(π, t) = 0 for all
t. To use the simple observation above, we first extend f to all of R by
making it odd1 on [−π, π], and then periodic2 in x of period 2π, and
similarly for u(x, t), the solution of our problem. Then the extension u
solves the wave equation on all of R, and u(x, 0) = f(x) for all x ∈ R.
Therefore, u(x, t) = F (x + t) + G(x− t), and setting t = 0 we find that

F (x) + G(x) = f(x).

Since many choices of F and G will satisfy this identity, this suggests
imposing another initial condition on u (similar to the two initial condi-
tions in the case of simple harmonic motion), namely the initial velocity
of the string which we denote by g(x):

∂u

∂t
(x, 0) = g(x),

where of course g(0) = g(π) = 0. Again, we extend g to R first by mak-
ing it odd over [−π, π], and then periodic of period 2π. The two initial
conditions of position and velocity now translate into the following sys-
tem:

{
F (x) + G(x) = f(x) ,
F ′(x)−G′(x) = g(x) .

1A function f defined on a set U is odd if −x ∈ U whenever x ∈ U and f(−x) = −f(x),
and even if f(−x) = f(x).

2A function f on R is periodic of period ω if f(x + ω) = f(x) for all x.

Ibookroot October 20, 2007

1. The vibrating string 11

Differentiating the first equation and adding it to the second, we obtain

2F ′(x) = f ′(x) + g(x).

Similarly

2G′(x) = f ′(x)− g(x),

and hence there are constants C1 and C2 so that

F (x) =
1
2

[
f(x) +

∫ x
0

g(y) dy
]

+ C1

and

G(x) =
1
2

[
f(x)−

∫ x
0

g(y) dy
]

+ C2.

Since F (x) + G(x) = f(x) we conclude that C1 + C2 = 0, and therefore,
our final solution of the wave equation with the given initial conditions
takes the form

u(x, t) =
1
2

[f(x + t) + f(x− t)] + 1
2

∫ x+t
x−t

g(y) dy.

The form of this solution is known as d’Alembert’s formula. Observe
that the extensions we chose for f and g guarantee that the string always
has fixed ends, that is, u(0, t) = u(π, t) = 0 for all t.

A final remark is in order. The passage from t ≥ 0 to t ∈ R, and then
back to t ≥ 0, which was made above, exhibits the time reversal property
of the wave equation. In other words, a solution u to the wave equation
for t ≥ 0, leads to a solution u− defined for negative time t < 0 simply by setting u−(x, t) = u(x,−t), a fact which follows from the invariance of the wave equation under the transformation t 7→ −t. The situation is quite different in the case of the heat equation. Superposition of standing waves We turn to the second method of solving the wave equation, which is based on two fundamental conclusions from our previous physical obser- vations. By our considerations of standing waves, we are led to look for special solutions to the wave equation which are of the form ϕ(x)ψ(t). This procedure, which works equally well in other contexts (in the case of the heat equation, for instance), is called separation of variables and constructs solutions that are called pure tones. Then by the linearity Ibookroot October 20, 2007 12 Chapter 1. THE GENESIS OF FOURIER ANALYSIS of the wave equation, we can expect to combine these pure tones into a more complex combination of sound. Pushing this idea further, we can hope ultimately to express the general solution of the wave equation in terms of sums of these particular solutions. Note that one side of the wave equation involves only differentiation in x, while the other, only differentiation in t. This observation pro- vides another reason to look for solutions of the equation in the form u(x, t) = ϕ(x)ψ(t) (that is, to “separate variables”), the hope being to reduce a difficult partial differential equation into a system of simpler ordinary differential equations. In the case of the wave equation, with u of the above form, we get ϕ(x)ψ′′(t) = ϕ′′(x)ψ(t), and therefore ψ′′(t) ψ(t) = ϕ′′(x) ϕ(x) . The key observation here is that the left-hand side depends only on t, and the right-hand side only on x. This can happen only if both sides are equal to a constant, say λ. Therefore, the wave equation reduces to the following (3) { ψ′′(t)− λψ(t) = 0 ϕ′′(x)− λϕ(x) = 0. We focus our attention on the first equation in the above system. At this point, the reader will recognize the equation we obtained in the study of simple harmonic motion. Note that we need to consider only the case when λ < 0, since when λ ≥ 0 the solution ψ will not oscillate as time varies. Therefore, we may write λ = −m2, and the solution of the equation is then given by ψ(t) = A cosmt + B sinmt. Similarly, we find that the solution of the second equation in (3) is ϕ(x) = Ã cos mx + B̃ sinmx. Now we take into account that the string is attached at x = 0 and x = π. This translates into ϕ(0) = ϕ(π) = 0, which in turn gives Ã = 0, and if B̃ 6= 0, then m must be an integer. If m = 0, the solution vanishes identically, and if m ≤ −1, we may rename the constants and reduce to Ibookroot October 20, 2007 1. The vibrating string 13 the case m ≥ 1 since the function sin y is odd and cos y is even. Finally, we arrive at the guess that for each m ≥ 1, the function um(x, t) = (Am cos mt + Bm sin mt) sin mx, which we recognize as a standing wave, is a solution to the wave equa- tion. Note that in the above argument we divided by ϕ and ψ, which sometimes vanish, so one must actually check by hand that the standing wave um solves the equation. This straightforward calculation is left as an exercise to the reader. Before proceeding further with the analysis of the wave equation, we pause to discuss standing waves in more detail. The terminology comes from looking at the graph of um(x, t) for each fixed t. Suppose first that m = 1, and take u(x, t) = cos t sinx. Then, Figure 7 (a) gives the graph of u for different values of t. (b)(a) 0−π 2π 0 π−2π π 2π−π−2π Figure 7. Fundamental tone (a) and overtones (b) at different moments in time The case m = 1 corresponds to the fundamental tone or first har- monic of the vibrating string. We now take m = 2 and look at u(x, t) = cos 2t sin 2x. This corre- sponds to the first overtone or second harmonic, and this motion is described in Figure 7 (b). Note that u(π/2, t) = 0 for all t. Such points, which remain motionless in time, are called nodes, while points whose motion has maximum amplitude are named anti-nodes. For higher values of m we get more overtones or higher harmonics. Note that as m increases, the frequency increases, and the period 2π/m Ibookroot October 20, 2007 14 Chapter 1. THE GENESIS OF FOURIER ANALYSIS decreases. Therefore, the fundamental tone has a lower frequency than the overtones. We now return to the original problem. Recall that the wave equation is linear in the sense that if u and v solve the equation, so does αu + βv for any constants α and β. This allows us to construct more solutions by taking linear combinations of the standing waves um. This technique, called superposition, leads to our final guess for a solution of the wave equation (4) u(x, t) = ∞∑ m=1 (Am cos mt + Bm sin mt) sin mx. Note that the above sum is infinite, so that questions of convergence arise, but since most of our arguments so far are formal, we will not worry about this point now. Suppose the above expression gave all the solutions to the wave equa- tion. If we then require that the initial position of the string at time t = 0 is given by the shape of the graph of the function f on [0, π], with of course f(0) = f(π) = 0, we would have u(x, 0) = f(x), hence ∞∑ m=1 Am sin mx = f(x). Since the initial shape of the string can be any reasonable function f , we must ask the following basic question: Given a function f on [0, π] (with f(0) = f(π) = 0), can we find coefficients Am so that (5) f(x) = ∞∑ m=1 Am sin mx ? This question is stated loosely, but a lot of our effort in the next two chapters of this book will be to formulate the question precisely and attempt to answer it. This was the basic problem that initiated the study of Fourier analysis. A simple observation allows us to guess a formula giving Am if the expansion (5) were to hold. Indeed, we multiply both sides by sinnx Ibookroot October 20, 2007 1. The vibrating string 15 and integrate between [0, π]; working formally, we obtain ∫ π 0 f(x) sin nx dx = ∫ π 0 ( ∞∑ m=1 Am sinmx ) sinnx dx = ∞∑ m=1 Am ∫ π 0 sinmx sin nx dx = An · π 2 , where we have used the fact that ∫ π 0 sinmx sinnx dx = { 0 if m 6= n, π/2 if m = n. Therefore, the guess for An, called the nth Fourier sine coefficient of f , is (6) An = 2 π ∫ π 0 f(x) sin nx dx. We shall return to this formula, and other similar ones, later. One can transform the question about Fourier sine series on [0, π] to a more general question on the interval [−π, π]. If we could express f on [0, π] in terms of a sine series, then this expansion would also hold on [−π, π] if we extend f to this interval by making it odd. Similarly, one can ask if an even function g(x) on [−π, π] can be expressed as a cosine series, namely g(x) = ∞∑ m=0 A′m cosmx. More generally, since an arbitrary function F on [−π, π] can be expressed as f + g, where f is odd and g is even,3 we may ask if F can be written as F (x) = ∞∑ m=1 Am sin mx + ∞∑ m=0 A′m cosmx, or by applying Euler’s identity eix = cos x + i sinx, we could hope that F takes the form F (x) = ∞∑ m=−∞ ame imx. 3Take, for example, f(x) = [F (x)− F (−x)]/2 and g(x) = [F (x) + F (−x)]/2. Ibookroot October 20, 2007 16 Chapter 1. THE GENESIS OF FOURIER ANALYSIS By analogy with (6), we can use the fact that 1 2π ∫ π −π eimxe−inx dx = { 0 if n 6= m 1 if n = m, to see that one expects that an = 1 2π ∫ π −π F (x)e−inx dx. The quantity an is called the nth Fourier coefficient of F . We can now reformulate the problem raised above: Question: Given any reasonable function F on [−π, π], with Fourier coefficients defined above, is it true that (7) F (x) = ∞∑ m=−∞ ame imx ? This formulation of the problem, in terms of complex exponentials, is the form we shall use the most in what follows. Joseph Fourier (1768-1830) was the first to believe that an “arbitrary” function F could be given as a series (7). In other words, his idea was that any function is the linear combination (possibly infinite) of the most basic trigonometric functions sinmx and cos mx, where m ranges over the integers.4 Although this idea was implicit in earlier work, Fourier had the conviction that his predecessors lacked, and he used it in his study of heat diffusion; this began the subject of “Fourier analysis.” This discipline, which was first developed to solve certain physical problems, has proved to have many applications in mathematics and other fields as well, as we shall see later. We return to the wave equation. To formulate the problem correctly, we must impose two initial conditions, as our experience with simple harmonic motion and traveling waves indicated. The conditions assign the initial position and velocity of the string. That is, we require that u satisfy the differential equation and the two conditions u(x, 0) = f(x) and ∂u ∂t (x, 0) = g(x), 4The first proof that a general class of functions can be represented by Fourier series was given later by Dirichlet; see Problem 6, Chapter 4. Ibookroot October 20, 2007 1. The vibrating string 17 where f and g are pre-assigned functions. Note that this is consistent with (4) in that this requires that f and g be expressible as f(x) = ∞∑ m=1 Am sinmx and g(x) = ∞∑ m=1 mBm sinmx. 1.3 Example: the plucked string We now apply our reasoning to the particular problem of the plucked string. For simplicity we choose units so that the string is taken on the interval [0, π], and it satisfies the wave equation with c = 1. The string is assumed to be plucked to height h at the point p with 0 < p < π; this is the initial position. That is, we take as our initial position the triangular shape given by f(x) =    xh p for 0 ≤ x ≤ p h(π − x) π − p for p ≤ x ≤ π, which is depicted in Figure 8. 0 h p π Figure 8. Initial position of a plucked string We also choose an initial velocity g(x) identically equal to 0. Then, we can compute the Fourier coefficients of f (Exercise 9), and assuming that the answer to the question raised before (5) is positive, we obtain f(x) = ∞∑ m=1 Am sinmx with Am = 2h m2 sinmp p(π − p) . Ibookroot October 20, 2007 18 Chapter 1. THE GENESIS OF FOURIER ANALYSIS Thus (8) u(x, t) = ∞∑ m=1 Am cos mt sinmx, and note that this series converges absolutely. The solution can also be expressed in terms of traveling waves. In fact (9) u(x, t) = f(x + t) + f(x− t) 2 . Here f(x) is defined for all x as follows: first, f is extended to [−π, π] by making it odd, and then f is extended to the whole real line by making it periodic of period 2π, that is, f(x + 2πk) = f(x) for all integers k. Observe that (8) implies (9) in view of the trigonometric identity cos v sinu = 1 2 [sin(u + v) + sin(u− v)]. As a final remark, we should note an unsatisfactory aspect of the so- lution to this problem, which however is in the nature of things. Since the initial data f(x) for the plucked string is not twice continuously dif- ferentiable, neither is the function u (given by (9)). Hence u is not truly a solution of the wave equation: while u(x, t) does represent the position of the plucked string, it does not satisfy the partial differential equation we set out to solve! This state of affairs may be understood properly only if we realize that u does solve the equation, but in an appropriate generalized sense. A better understanding of this phenomenon requires ideas relevant to the study of “weak solutions” and the theory of “dis- tributions.” These topics we consider only later, in Books III and IV. 2 The heat equation We now discuss the problem of heat diffusion by following the same framework as for the wave equation. First, we derive the time-dependent heat equation, and then study the steady-state heat equation in the disc, which leads us back to the basic question (7). 2.1 Derivation of the heat equation Consider an infinite metal plate which we model as the plane R2, and suppose we are given an initial heat distribution at time t = 0. Let the temperature at the point (x, y) at time t be denoted by u(x, y, t). Ibookroot October 20, 2007 2. The heat equation 19 Consider a small square centered at (x0, y0) with sides parallel to the axis and of side length h, as shown in Figure 9. The amount of heat energy in S at time t is given by H(t) = σ ∫ ∫ S u(x, y, t) dx dy , where σ > 0 is a constant called the specific heat of the material. There-
fore, the heat flow into S is

∂H

∂t
= σ

∫ ∫

∂u

∂t
dx dy ,

which is approximately equal to

σh2
∂u

∂t
(x0, y0, t),

since the area of S is h2. Now we apply Newton’s law of cooling, which
states that heat flows from the higher to lower temperature at a rate
proportional to the difference, that is, the gradient.

(x0 + h/2, y0)(x0, y0)
h

Figure 9. Heat flow through a small square

The heat flow through the vertical side on the right is therefore

−κh ∂u
∂x

(x0 + h/2, y0, t) ,

where κ > 0 is the conductivity of the material. A similar argument for
the other sides shows that the total heat flow through the square S is

Ibookroot October 20, 2007

20 Chapter 1. THE GENESIS OF FOURIER ANALYSIS

given by

κh

[
∂u

∂x
(x0 + h/2, y0, t)−

∂u

∂x
(x0 − h/2, y0, t)

+
∂u

∂y
(x0, y0 + h/2, t)−

∂u

∂y
(x0, y0 − h/2, t)

]
.

Applying the mean value theorem and letting h tend to zero, we find
that

∂u

∂t
=

∂2u

∂x2
+

∂2u

∂y2
;

this is called the time-dependent heat equation, often abbreviated
to the heat equation.

2.2 Steady-state heat equation in the disc

After a long period of time, there is no more heat exchange, so that
the system reaches thermal equilibrium and ∂u/∂t = 0. In this case,
the time-dependent heat equation reduces to the steady-state heat
equation

(10)
∂2u

∂x2
+

∂2u

∂y2
= 0.

The operator ∂2/∂x2 + ∂2/∂y2 is of such importance in mathematics and
physics that it is often abbreviated as 4 and given a name: the Laplace
operator or Laplacian. So the steady-state heat equation is written as

4u = 0,

and solutions to this equation are called harmonic functions.

Consider the unit disc in the plane

D = {(x, y) ∈ R2 : x2 + y2 < 1}, whose boundary is the unit circle C. In polar coordinates (r, θ), with 0 ≤ r and 0 ≤ θ < 2π, we have D = {(r, θ) : 0 ≤ r < 1} and C = {(r, θ) : r = 1}. The problem, often called the Dirichlet problem (for the Laplacian on the unit disc), is to solve the steady-state heat equation in the unit Ibookroot October 20, 2007 2. The heat equation 21 disc subject to the boundary condition u = f on C. This corresponds to fixing a predetermined temperature distribution on the circle, waiting a long time, and then looking at the temperature distribution inside the disc. u(1, θ) = f(θ) x y 0 4u = 0 Figure 10. The Dirichlet problem for the disc While the method of separation of variables will turn out to be useful for equation (10), a difficulty comes from the fact that the boundary condition is not easily expressed in terms of rectangular coordinates. Since this boundary condition is best described by the coordinates (r, θ), namely u(1, θ) = f(θ), we rewrite the Laplacian in polar coordinates. An application of the chain rule gives (Exercise 10): 4u = ∂ 2u ∂r2 + 1 r ∂u ∂r + 1 r2 ∂2u ∂θ2 . We now multiply both sides by r2, and since 4u = 0, we get r2 ∂2u ∂r2 + r ∂u ∂r = −∂ 2u ∂θ2 . Separating these variables, and looking for a solution of the form u(r, θ) = F (r)G(θ), we find r2F ′′(r) + rF ′(r) F (r) = −G ′′(θ) G(θ) . Ibookroot October 20, 2007 22 Chapter 1. THE GENESIS OF FOURIER ANALYSIS Since the two sides depend on different variables, they must both be constant, say equal to λ. We therefore get the following equations: { G′′(θ) + λG(θ) = 0 , r2F ′′(r) + rF ′(r)− λF (r) = 0. Since G must be periodic of period 2π, this implies that λ ≥ 0 and (as we have seen before) that λ = m2 where m is an integer; hence G(θ) = Ã cos mθ + B̃ sinmθ. An application of Euler’s identity, eix = cos x + i sinx, allows one to rewrite G in terms of complex exponentials, G(θ) = Aeimθ + Be−imθ. With λ = m2 and m 6= 0, two simple solutions of the equation in F are F (r) = rm and F (r) = r−m (Exercise 11 gives further information about these solutions). If m = 0, then F (r) = 1 and F (r) = log r are two solu- tions. If m > 0, we note that r−m grows unboundedly large as r tends
to zero, so F (r)G(θ) is unbounded at the origin; the same occurs when
m = 0 and F (r) = log r. We reject these solutions as contrary to our
intuition. Therefore, we are left with the following special functions:

um(r, θ) = r
|m|eimθ, m ∈ Z.

We now make the important observation that (10) is linear , and so as
in the case of the vibrating string, we may superpose the above special
solutions to obtain the presumed general solution:

u(r, θ) =
∞∑

m=−∞
amr

|m|eimθ.

If this expression gave all the solutions to the steady-state heat equation,
then for a reasonable f we should have

u(1, θ) =
∞∑

m=−∞
ame

imθ = f(θ).

We therefore ask again in this context: given any reasonable function f
on [0, 2π] with f(0) = f(2π), can we find coefficients am so that

f(θ) =
∞∑

m=−∞
ame

imθ ?

Ibookroot October 20, 2007

3. Exercises 23

Historical Note: D’Alembert (in 1747) first solved the equation of the
vibrating string using the method of traveling waves. This solution was
elaborated by Euler a year later. In 1753, D. Bernoulli proposed the
solution which for all intents and purposes is the Fourier series given
by (4), but Euler was not entirely convinced of its full generality, since
this could hold only if an “arbitrary” function could be expanded in
Fourier series. D’Alembert and other mathematicians also had doubts.
This viewpoint was changed by Fourier (in 1807) in his study of the
heat equation, where his conviction and work eventually led others to a
complete proof that a general function could be represented as a Fourier
series.

3 Exercises

1. If z = x + iy is a complex number with x, y ∈ R, we define

|z| = (x2 + y2)1/2

and call this quantity the modulus or absolute value of z.

(a) What is the geometric interpretation of |z|?

(b) Show that if |z| = 0, then z = 0.

(d) If z1 and z2 are two complex numbers, prove that

|z1z2| = |z1||z2| and |z1 + z2| ≤ |z1|+ |z2|.

(e) Show that if z 6= 0, then |1/z| = 1/|z|.

2. If z = x + iy is a complex number with x, y ∈ R, we define the complex
conjugate of z by

z = x− iy.

(a) What is the geometric interpretation of z?

(b) Show that |z|2 = zz.

Ibookroot October 20, 2007

24 Chapter 1. THE GENESIS OF FOURIER ANALYSIS

3. A sequence of complex numbers {wn}∞n=1 is said to converge if there exists
w ∈ C such that

lim
n→∞

|wn − w| = 0,

and we say that w is a limit of the sequence.

(a) Show that a converging sequence of complex numbers has a unique limit.

The sequence {wn}∞n=1 is said to be a Cauchy sequence if for every ² > 0 there
exists a positive integer N such that

|wn − wm| < ² whenever n,m > N.

(b) Prove that a sequence of complex numbers converges if and only if it is a
Cauchy sequence. [Hint: A similar theorem exists for the convergence of a
sequence of real numbers. Why does it carry over to sequences of complex
numbers?]

A series
∑∞

n=1
zn of complex numbers is said to converge if the sequence formed

by the partial sums

SN =
N∑

n=1

converges. Let {an}∞n=1 be a sequence of non-negative real numbers such that
the series

∑
n

an converges.

∑
n

zn converges. [Hint: Use the Cauchy
criterion.]

4. For z ∈ C, we define the complex exponential by

ez =
∞∑

n=0

n!
.

(a) Prove that the above definition makes sense, by showing that the series
converges for every complex number z. Moreover, show that the conver-
gence is uniform5 on every bounded subset of C.

(b) If z1, z2 are two complex numbers, prove that ez1ez2 = ez1+z2 . [Hint: Use
the binomial theorem to expand (z1 + z2)n, as well as the formula for the
binomial coefficients.]

5A sequence of functions {fn(z)}∞n=1 is said to be uniformly convergent on a set S if
there exists a function f on S so that for every ² > 0 there is an integer N such that
|fn(z)− f(z)| < ² whenever n > N and z ∈ S.

Ibookroot October 20, 2007

3. Exercises 25

eiy = cos y + i sin y.

This is Euler’s identity. [Hint: Use power series.]

(d) More generally,

ex+iy = ex(cos y + i sin y)

whenever x, y ∈ R, and show that

|ex+iy| = ex.

(e) Prove that ez = 1 if and only if z = 2πki for some integer k.

(f) Show that every complex number z = x + iy can be written in the form

z = reiθ ,

where r is unique and in the range 0 ≤ r < ∞, and θ ∈ R is unique up to an integer multiple of 2π. Check that r = |z| and θ = arctan(y/x) whenever these formulas make sense. (g) In particular, i = eiπ/2. What is the geometric meaning of multiplying a complex number by i? Or by eiθ for any θ ∈ R? (h) Given θ ∈ R, show that cos θ = eiθ + e−iθ 2 and sin θ = eiθ − e−iθ 2i . These are also called Euler’s identities. (i) Use the complex exponential to derive trigonometric identities such as cos(θ + ϑ) = cos θ cosϑ− sin θ sinϑ, and then show that 2 sin θ sinϕ = cos(θ − ϕ)− cos(θ + ϕ) , 2 sin θ cos ϕ = sin(θ + ϕ) + sin(θ − ϕ). This calculation connects the solution given by d’Alembert in terms of traveling waves and the solution in terms of superposition of standing waves. Ibookroot October 20, 2007 26 Chapter 1. THE GENESIS OF FOURIER ANALYSIS 5. Verify that f(x) = einx is periodic with period 2π and that 1 2π ∫ π −π einx dx = { 1 if n = 0, 0 if n 6= 0. Use this fact to prove that if n,m ≥ 1 we have 1 π ∫ π −π cos nx cosmx dx = { 0 if n 6= m, 1 n = m, and similarly 1 π ∫ π −π sin nx sinmx dx = { 0 if n 6= m, 1 n = m. Finally, show that ∫ π −π sinnx cos mxdx = 0 for any n,m. [Hint: Calculate einxe−imx + einxeimx and einxe−imx − einxeimx.] 6. Prove that if f is a twice continuously differentiable function on R which is a solution of the equation f ′′(t) + c2f(t) = 0, then there exist constants a and b such that f(t) = a cos ct + b sin ct. This can be done by differentiating the two functions g(t) = f(t) cos ct− c−1f ′(t) sin ct and h(t) = f(t) sin ct + c−1f ′(t) cos ct. 7. Show that if a and b are real, then one can write a cos ct + b sin ct = A cos(ct− ϕ), where A = √ a2 + b2, and ϕ is chosen so that cosϕ = a√ a2 + b2 and sin ϕ = b√ a2 + b2 . 8. Suppose F is a function on (a, b) with two continuous derivatives. Show that whenever x and x + h belong to (a, b), one may write F (x + h) = F (x) + hF ′(x) + h2 2 F ′′(x) + h2ϕ(h) , Ibookroot October 20, 2007 4. Problem 27 where ϕ(h) → 0 as h → 0. Deduce that F (x + h) + F (x− h)− 2F (x) h2 → F ′′(x) as h → 0. [Hint: This is simply a Taylor expansion. It may be obtained by noting that F (x + h)− F (x) = ∫ x+h x F ′(y) dy, and then writing F ′(y) = F ′(x) + (y − x)F ′′(x) + (y − x)ψ(y − x), where ψ(h) → 0 as h → 0.] 9. In the case of the plucked string, use the formula for the Fourier sine coeffi- cients to show that Am = 2h m2 sinmp p(π − p) . For what position of p are the second, fourth, . . . harmonics missing? For what position of p are the third, sixth, . . . harmonics missing? 10. Show that the expression of the Laplacian 4 = ∂ 2 ∂x2 + ∂2 ∂y2 is given in polar coordinates by the formula 4 = ∂ 2 ∂r2 + 1 r ∂ ∂r + 1 r2 ∂2 ∂θ2 . Also, prove that ∣∣∣∣ ∂u ∂x ∣∣∣∣ 2 + ∣∣∣∣ ∂u ∂y ∣∣∣∣ 2 = ∣∣∣∣ ∂u ∂r ∣∣∣∣ 2 + 1 r2 ∣∣∣∣ ∂u ∂θ ∣∣∣∣ 2 . 11. Show that if n ∈ Z the only solutions of the differential equation r2F ′′(r) + rF ′(r)− n2F (r) = 0, which are twice differentiable when r > 0, are given by linear combinations of
rn and r−n when n 6= 0, and 1 and log r when n = 0.
[Hint: If F solves the equation, write F (r) = g(r)rn, find the equation satisfied
by g, and conclude that rg′(r) + 2ng(r) = c where c is a constant.]

Ibookroot October 20, 2007

28 Chapter 1. THE GENESIS OF FOURIER ANALYSIS

u = f1

u = 0

u = f0

u = 0

4u = 0

Figure 11. Dirichlet problem in a rectangle

4 Problem

1. Consider the Dirichlet problem illustrated in Figure 11.
More precisely, we look for a solution of the steady-state heat equation

4u = 0 in the rectangle R = {(x, y) : 0 ≤ x ≤ π, 0 ≤ y ≤ 1} that vanishes on
the vertical sides of R, and so that

u(x, 0) = f0(x) and u(x, 1) = f1(x) ,

where f0 and f1 are initial data which fix the temperature distribution on the
horizontal sides of the rectangle.

Use separation of variables to show that if f0 and f1 have Fourier expansions

f0(x) =
∞∑

k=1

Ak sin kx and f1(x) =
∞∑

k=1

Bk sin kx,

then

u(x, y) =
∞∑

k=1

(
sinh k(1− y)

sinh k
Ak +

sinh ky
sinh k

)
sin kx.

We recall the definitions of the hyperbolic sine and cosine functions:

sinh x =
ex − e−x

2
and cosh x =

ex + e−x

2
.

Compare this result with the solution of the Dirichlet problem in the strip ob-
tained in Problem 3, Chapter 5.

Ibookroot October 20, 2007

2 Basic Properties of Fourier
Series

Nearly fifty years had passed without any progress on
the question of analytic representation of an arbitrary
function, when an assertion of Fourier threw new light
on the subject. Thus a new era began for the de-
velopment of this part of Mathematics and this was
heralded in a stunning way by major developments in
mathematical Physics.

B. Riemann, 1854

In this chapter, we begin our rigorous study of Fourier series. We set
the stage by introducing the main objects in the subject, and then for-
mulate some basic problems which we have already touched upon earlier.

Our first result disposes of the question of uniqueness: Are two func-
tions with the same Fourier coefficients necessarily equal? Indeed, a
simple argument shows that if both functions are continuous, then in
fact they must agree.

Next, we take a closer look at the partial sums of a Fourier series. Using
the formula for the Fourier coefficients (which involves an integration),
we make the key observation that these sums can be written conveniently
as integrals:

1
2π

∫
DN (x− y)f(y) dy,

where {DN} is a family of functions called the Dirichlet kernels. The
above expression is the convolution of f with the function DN . Convo-
lutions will play a critical role in our analysis. In general, given a family
of functions {Kn}, we are led to investigate the limiting properties as n
tends to infinity of the convolutions

1
2π

∫
Kn(x− y)f(y) dy.

We find that if the family {Kn} satisfies the three important properties
of “good kernels,” then the convolutions above tend to f(x) as n →∞
(at least when f is continuous). In this sense, the family {Kn} is an

Ibookroot October 20, 2007

30 Chapter 2. BASIC PROPERTIES OF FOURIER SERIES

“approximation to the identity.” Unfortunately, the Dirichlet kernels
DN do not belong to the category of good kernels, which indicates that
the question of convergence of Fourier series is subtle.

Instead of pursuing at this stage the problem of convergence, we con-
sider various other methods of summing the Fourier series of a function.
The first method, which involves averages of partial sums, leads to con-
volutions with good kernels, and yields an important theorem of Fejér.
From this, we deduce the fact that a continuous function on the circle
can be approximated uniformly by trigonometric polynomials. Second,
we may also sum the Fourier series in the sense of Abel and again en-
counter a family of good kernels. In this case, the results about convo-
lutions and good kernels lead to a solution of the Dirichlet problem for
the steady-state heat equation in the disc, considered at the end of the
previous chapter.

1 Examples and formulation of the problem

We commence with a brief description of the types of functions with
which we shall be concerned. Since the Fourier coefficients of f are
defined by

an =
1
L

∫ L
0

f(x)e−2πinx/L dx, for n ∈ Z,

where f is complex-valued on [0, L], it will be necessary to place some in-
tegrability conditions on f . We shall therefore assume for the remainder
of this book that all functions are at least Riemann integrable.1 Some-
times it will be illuminating to focus our attention on functions that
are more “regular,” that is, functions that possess certain continuity or
differentiability properties. Below, we list several classes of functions in
increasing order of generality. We emphasize that we will not generally
restrict our attention to real-valued functions, contrary to what the fol-
lowing pictures may suggest; we will almost always allow functions that
take values in the complex numbers C. Furthermore, we sometimes think
of our functions as being defined on the circle rather than an interval.
We elaborate upon this below.

1Limiting ourselves to Riemann integrable functions is natural at this elementary stage
of study of the subject. The more advanced notion of Lebesgue integrability will be taken
up in Book III.

Ibookroot October 20, 2007

1. Examples and formulation of the problem 31

Everywhere continuous functions

These are the complex-valued functions f which are continuous at every
point of the segment [0, L]. A typical continuous function is sketched in
Figure 1 (a). We shall note later that continuous functions on the circle
satisfy the additional condition f(0) = f(L).

Piecewise continuous functions

These are bounded functions on [0, L] which have only finitely many
discontinuities. An example of such a function with simple discontinuities
is pictured in Figure 1 (b).

(a) (b)

0 x

L0 x

Figure 1. Functions on [0, L]: continuous and piecewise continuous

This class of functions is wide enough to illustrate many of the the-
orems in the next few chapters. However, for logical completeness we
consider also the more general class of Riemann integrable functions.
This more extended setting is natural since the formula for the Fourier
coefficients involves integration.

Riemann integrable functions

This is the most general class of functions we will be concerned with.
Such functions are bounded, but may have infinitely many discontinu-
ities. We recall the definition of integrability. A real-valued function f
defined on [0, L] is Riemann integrable (which we abbreviate as in-
tegrable2) if it is bounded, and if for every ² > 0, there is a subdivision
0 = x0 < x1 < · · · < xN−1 < xN = L of the interval [0, L], so that if U 2Starting in Book III, the term “integrable” will be used in the broader sense of Lebesgue theory. Ibookroot October 20, 2007 32 Chapter 2. BASIC PROPERTIES OF FOURIER SERIES and L are, respectively, the upper and lower sums of f for this subdivi- sion, namely U = N∑ j=1 [ sup xj−1≤x≤xj f(x)](xj − xj−1) and L = N∑ j=1 [ inf xj−1≤x≤xj f(x)](xj − xj−1) , then we have U − L < ². Finally, we say that a complex-valued function is integrable if its real and imaginary parts are integrable. It is worthwhile to remember at this point that the sum and product of two integrable functions are integrable. A simple example of an integrable function on [0, 1] with infinitely many discontinuities is given by f(x) =    1 if 1/(n + 1) < x ≤ 1/n and n is odd, 0 if 1/(n + 1) < x ≤ 1/n and n is even, 0 if x = 0. This example is illustrated in Figure 2. Note that f is discontinuous when x = 1/n and at x = 0. 1 3 1 2 1 5 1 4 0 1 1 Figure 2. A Riemann integrable function More elaborate examples of integrable functions whose discontinuities are dense in the interval [0, 1] are described in Problem 1. In general, while integrable functions may have infinitely many discontinuities, these Ibookroot October 20, 2007 1. Examples and formulation of the problem 33 functions are actually characterized by the fact that, in a precise sense, their discontinuities are not too numerous: they are “negligible,” that is, the set of points where an integrable function is discontinuous has “mea- sure 0.” The reader will find further details about Riemann integration in the appendix. From now on, we shall always assume that our functions are integrable, even if we do not state this requirement explicitly. Functions on the circle There is a natural connection between 2π-periodic functions on R like the exponentials einθ, functions on an interval of length 2π, and functions on the unit circle. This connection arises as follows. A point on the unit circle takes the form eiθ, where θ is a real number that is unique up to integer multiples of 2π. If F is a function on the circle, then we may define for each real number θ f(θ) = F (eiθ), and observe that with this definition, the function f is periodic on R of period 2π, that is, f(θ + 2π) = f(θ) for all θ. The integrability, continu- ity and other smoothness properties of F are determined by those of f . For instance, we say that F is integrable on the circle if f is integrable on every interval of length 2π. Also, F is continuous on the circle if f is continuous on R, which is the same as saying that f is continuous on any interval of length 2π. Moreover, F is continuously differentiable if f has a continuous derivative, and so forth. Since f has period 2π, we may restrict it to any interval of length 2π, say [0, 2π] or [−π, π], and still capture the initial function F on the circle. We note that f must take the same value at the end-points of the interval since they correspond to the same point on the circle. Conversely, any function on [0, 2π] for which f(0) = f(2π) can be extended to a periodic function on R which can then be identified as a function on the circle. In particular, a continuous function f on the interval [0, 2π] gives rise to a continuous function on the circle if and only if f(0) = f(2π). In conclusion, functions on R that 2π-periodic, and functions on an interval of length 2π that take on the same value at its end-points, are two equivalent descriptions of the same mathematical objects, namely, functions on the circle. In this connection, we mention an item of notational usage. When our functions are defined on an interval on the line, we often use x as the independent variable; however, when we consider these as functions Ibookroot October 20, 2007 34 Chapter 2. BASIC PROPERTIES OF FOURIER SERIES on the circle, we usually replace the variable x by θ. As the reader will note, we are not strictly bound by this rule since this practice is mostly a matter of convenience. 1.1 Main definitions and some examples We now begin our study of Fourier analysis with the precise definition of the Fourier series of a function. Here, it is important to pin down where our function is originally defined. If f is an integrable function given on an interval [a, b] of length L (that is, b− a = L), then the nth Fourier coefficient of f is defined by f̂(n) = 1 L ∫ b a f(x)e−2πinx/L dx, n ∈ Z. The Fourier series of f is given formally3 by ∞∑ n=−∞ f̂(n)e2πinx/L. We shall sometimes write an for the Fourier coefficients of f , and use the notation f(x) ∼ ∞∑ n=−∞ ane 2πinx/L to indicate that the series on the right-hand side is the Fourier series of f . For instance, if f is an integrable function on the interval [−π, π], then the nth Fourier coefficient of f is f̂(n) = an = 1 2π ∫ π −π f(θ)e−inθ dθ, n ∈ Z, and the Fourier series of f is f(θ) ∼ ∞∑ n=−∞ ane inθ. Here we use θ as a variable since we think of it as an angle ranging from −π to π. 3At this point, we do not say anything about the convergence of the series. Ibookroot October 20, 2007 1. Examples and formulation of the problem 35 Also, if f is defined on [0, 2π], then the formulas are the same as above, except that we integrate from 0 to 2π in the definition of the Fourier coefficients. We may also consider the Fourier coefficients and Fourier series for a function defined on the circle. By our previous discussion, we may think of a function on the circle as a function f on R which is 2π-periodic. We may restrict the function f to any interval of length 2π, for instance [0, 2π] or [−π, π], and compute its Fourier coefficients. Fortunately, f is periodic and Exercise 1 shows that the resulting integrals are independent of the chosen interval. Thus the Fourier coefficients of a function on the circle are well defined. Finally, we shall sometimes consider a function g given on [0, 1]. Then ĝ(n) = an = ∫ 1 0 g(x)e−2πinx dx and g(x) ∼ ∞∑ n=−∞ ane 2πinx. Here we use x for a variable ranging from 0 to 1. Of course, if f is initially given on [0, 2π], then g(x) = f(2πx) is defined on [0, 1] and a change of variables shows that the nth Fourier coefficient of f equals the nth Fourier coefficient of g. Fourier series are part of a larger family called the trigonometric se- ries which, by definition, are expressions of the form ∑∞ n=−∞ cne 2πinx/L where cn ∈ C. If a trigonometric series involves only finitely many non- zero terms, that is, cn = 0 for all large |n|, it is called a trigonometric polynomial; its degree is the largest value of |n| for which cn 6= 0. The N th partial sum of the Fourier series of f , for N a positive integer, is a particular example of a trigonometric polynomial. It is given by SN (f)(x) = N∑ n=−N f̂(n)e2πinx/L. Note that by definition, the above sum is symmetric since n ranges from −N to N , a choice that is natural because of the resulting decomposition of the Fourier series as sine and cosine series. As a consequence, the convergence of Fourier series will be understood (in this book) as the “limit” as N tends to infinity of these symmetric sums. In fact, using the partial sums of the Fourier series, we can reformulate the basic question raised in Chapter 1 as follows: Problem: In what sense does SN (f) converge to f as N →∞ ? Ibookroot October 20, 2007 36 Chapter 2. BASIC PROPERTIES OF FOURIER SERIES Before proceeding further with this question, we turn to some simple examples of Fourier series. Example 1. Let f(θ) = θ for −π ≤ θ ≤ π. The calculation of the Fourier coefficients requires a simple integration by parts. First, if n 6= 0, then f̂(n) = 1 2π ∫ π −π θe−inθ dθ = 1 2π [ − θ in e−inθ ]π −π + 1 2πin ∫ π −π e−inθ dθ = (−1)n+1 in , and if n = 0 we clearly have f̂(0) = 1 2π ∫ π −π θ dθ = 0. Hence, the Fourier series of f is given by f(θ) ∼ ∑ n 6=0 (−1)n+1 in einθ = 2 ∞∑ n=1 (−1)n+1 sin nθ n . The first sum is over all non-zero integers, and the second is obtained by an application of Euler’s identities. It is possible to prove by elementary means that the above series converges for every θ, but it is not obvious that it converges to f(θ). This will be proved later (Exercises 8 and 9 deal with a similar situation). Example 2. Define f(θ) = (π − θ)2/4 for 0 ≤ θ ≤ 2π. Then successive integration by parts similar to that performed in the previous example yield f(θ) ∼ π 2 12 + ∞∑ n=1 cos nθ n2 . Example 3. The Fourier series of the function f(θ) = π sinπα ei(π−θ)α on [0, 2π] is f(θ) ∼ ∞∑ n=−∞ einθ n + α , Ibookroot October 20, 2007 1. Examples and formulation of the problem 37 whenever α is not an integer. Example 4. The trigonometric polynomial defined for x ∈ [−π, π] by DN (x) = N∑ n=−N einx is called the N th Dirichlet kernel and is of fundamental importance in the theory (as we shall see later). Notice that its Fourier coefficients an have the property that an = 1 if |n| ≤ N and an = 0 otherwise. A closed form formula for the Dirichlet kernel is DN (x) = sin((N + 1 2 )x) sin(x/2) . This can be seen by summing the geometric progressions N∑ n=0 ωn and −1∑ n=−N ωn with ω = eix. These sums are, respectively, equal to 1− ωN+1 1− ω and ω−N − 1 1− ω . Their sum is then ω−N − ωN+1 1− ω = ω−N−1/2 − ωN+1/2 ω−1/2 − ω1/2 = sin((N + 1 2 )x) sin(x/2) , giving the desired result. Example 5. The function Pr(θ), called the Poisson kernel, is defined for θ ∈ [−π, π] and 0 ≤ r < 1 by the absolutely and uniformly convergent series Pr(θ) = ∞∑ n=−∞ r|n|einθ. This function arose implicitly in the solution of the steady-state heat equation on the unit disc discussed in Chapter 1. Note that in calcu- lating the Fourier coefficients of Pr(θ) we can interchange the order of integration and summation since the sum converges uniformly in θ for Ibookroot October 20, 2007 38 Chapter 2. BASIC PROPERTIES OF FOURIER SERIES each fixed r, and obtain that the nth Fourier coefficient equals r|n|. One can also sum the series for Pr(θ) and see that Pr(θ) = 1− r2 1− 2r cos θ + r2 . In fact, Pr(θ) = ∞∑ n=0 ωn + ∞∑ n=1 ωn with ω = reiθ, where both series converge absolutely. The first sum (an infinite geomet- ric progression) equals 1/(1− ω), and likewise, the second is ω/(1− ω). Together, they combine to give 1− ω + (1− ω)ω (1− ω)(1− ω) = 1− |ω|2 |1− ω|2 = 1− r2 1− 2r cos θ + r2 , as claimed. The Poisson kernel will reappear later in the context of Abel summability of the Fourier series of a function. Let us return to the problem formulated earlier. The definition of the Fourier series of f is purely formal, and it is not obvious whether it converges to f . In fact, the solution of this problem can be very hard, or relatively easy, depending on the sense in which we expect the series to converge, or on what additional restrictions we place on f . Let us be more precise. Suppose, for the sake of this discussion, that the function f (which is always assumed to be Riemann integrable) is defined on [−π, π]. The first question one might ask is whether the partial sums of the Fourier series of f converge to f pointwise. That is, do we have (1) lim N→∞ SN (f)(θ) = f(θ) for every θ? We see quite easily that in general we cannot expect this result to be true at every θ, since we can always change an integrable function at one point without changing its Fourier coefficients. As a result, we might ask the same question assuming that f is continuous and periodic. For a long time it was believed that under these additional assumptions the answer would be “yes.” It was a surprise when Du Bois-Reymond showed that there exists a continuous function whose Fourier series diverges at a point. We will give such an example in the next chapter. Despite this negative result, we might ask what happens if we add more smoothness conditions on f : for example, we might assume that f is continuously Ibookroot October 20, 2007 2. Uniqueness of Fourier series 39 differentiable, or twice continuously differentiable. We will see that then the Fourier series of f converges to f uniformly. We will also interpret the limit (1) by showing that the Fourier series sums, in the sense of Cesàro or Abel, to the function f at all of its points of continuity. This approach involves appropriate averages of the partial sums of the Fourier series of f . Finally, we can also define the limit (1) in the mean square sense. In the next chapter, we will show that if f is merely integrable, then 1 2π ∫ π −π |SN (f)(θ)− f(θ)|2 dθ → 0 as N →∞. It is of interest to know that the problem of pointwise convergence of Fourier series was settled in 1966 by L. Carleson, who showed, among other things, that if f is integrable in our sense,4 then the Fourier series of f converges to f except possibly on a set of “measure 0.” The proof of this theorem is difficult and beyond the scope of this book. 2 Uniqueness of Fourier series If we were to assume that the Fourier series of functions f converge to f in an appropriate sense, then we could infer that a function is uniquely determined by its Fourier coefficients. This would lead to the following statement: if f and g have the same Fourier coefficients, then f and g are necessarily equal. By taking the difference f − g, this proposition can be reformulated as: if f̂(n) = 0 for all n ∈ Z, then f = 0. As stated, this assertion cannot be correct without reservation, since calculating Fourier coefficients requires integration, and we see that, for example, any two functions which differ at finitely many points have the same Fourier series. However, we do have the following positive result. Theorem 2.1 Suppose that f is an integrable function on the circle with f̂(n) = 0 for all n ∈ Z. Then f(θ0) = 0 whenever f is continuous at the point θ0. Thus, in terms of what we know about the set of discontinuities of in- tegrable functions,5 we can conclude that f vanishes for “most” values of θ. Proof. We suppose first that f is real-valued, and argue by con- tradiction. Assume, without loss of generality, that f is defined on 4Carleson’s proof actually holds for the wider class of functions which are square inte- grable in the Lebesgue sense. 5See the appendix. Ibookroot October 20, 2007 40 Chapter 2. BASIC PROPERTIES OF FOURIER SERIES [−π, π], that θ0 = 0, and f(0) > 0. The idea now is to construct a fam-
ily of trigonometric polynomials {pk} that “peak” at 0, and so that∫

pk(θ)f(θ) dθ →∞ as k →∞. This will be our desired contradiction
since these integrals are equal to zero by assumption.

Since f is continuous at 0, we can choose 0 < δ ≤ π/2, so that f(θ) >
f(0)/2 whenever |θ| < δ. Let p(θ) = ² + cos θ, where ² > 0 is chosen so small that |p(θ)| < 1− ²/2, whenever δ ≤ |θ| ≤ π. Then, choose a positive η with η < δ, so that p(θ) ≥ 1 + ²/2, for |θ| < η. Finally, let pk(θ) = [p(θ)] k, and select B so that |f(θ)| ≤ B for all θ. This is possible since f is integrable, hence bounded. Figure 3 illustrates the family {pk}. By p p6 p15 Figure 3. The functions p, p6, and p15 when ² = 0.1 construction, each pk is a trigonometric polynomial, and since f̂(n) = 0 for all n, we must have ∫ π −π f(θ)pk(θ) dθ = 0 for all k. However, we have the estimate ∣∣∣∣∣ ∫ δ≤|θ| f(θ)pk(θ) dθ ∣∣∣∣∣ ≤ 2πB(1− ²/2) k. Ibookroot October 20, 2007 2. Uniqueness of Fourier series 41 Also, our choice of δ guarantees that p(θ) and f(θ) are non-negative whenever |θ| < δ, thus ∫ η≤|θ|<δ f(θ)pk(θ) dθ ≥ 0. Finally, ∫ |θ|<η f(θ)pk(θ) dθ ≥ 2η f(0) 2 (1 + ²/2)k. Therefore, ∫ pk(θ)f(θ) dθ →∞ as k →∞, and this concludes the proof when f is real-valued. In general, write f(θ) = u(θ) + iv(θ), where u and v are real-valued. If we define f(θ) = f(θ), then u(θ) = f(θ) + f(θ) 2 and v(θ) = f(θ)− f(θ) 2i , and since f̂(n) = f̂(−n), we conclude that the Fourier coefficients of u and v all vanish, hence f = 0 at its points of continuity. The idea of constructing a family of functions (trigonometric polynomials in this case) which peak at the origin, together with other nice properties, will play an important role in this book. Such families of functions will be taken up later in Section 4 in connection with the notion of convolution. For now, note that the above theorem implies the following. Corollary 2.2 If f is continuous on the circle and f̂(n) = 0 for all n ∈ Z, then f = 0. The next corollary shows that the problem (1) formulated earlier has a simple positive answer under the assumption that the series of Fourier coefficients converges absolutely. Corollary 2.3 Suppose that f is a continuous function on the circle and that the Fourier series of f is absolutely convergent, ∑∞ n=−∞ |f̂(n)| < ∞. Then, the Fourier series converges uniformly to f , that is, lim N→∞ SN (f)(θ) = f(θ) uniformly in θ. Proof. Recall that if a sequence of continuous functions converges uniformly, then the limit is also continuous. Now observe that the assumption ∑ |f̂(n)| < ∞ implies that the partial sums of the Fourier Ibookroot October 20, 2007 42 Chapter 2. BASIC PROPERTIES OF FOURIER SERIES series of f converge absolutely and uniformly, and therefore the function g defined by g(θ) = ∞∑ n=−∞ f̂(n)einθ = lim N→∞ N∑ n=−N f̂(n)einθ is continuous on the circle. Moreover, the Fourier coefficients of g are precisely f̂(n) since we can interchange the infinite sum with the integral (a consequence of the uniform convergence of the series). Therefore, the previous corollary applied to the function f − g yields f = g, as desired. What conditions on f would guarantee the absolute convergence of its Fourier series? As it turns out, the smoothness of f is directly related to the decay of the Fourier coefficients, and in general, the smoother the function, the faster this decay. As a result, we can expect that relatively smooth functions equal their Fourier series. This is in fact the case, as we now show. In order to state the result concisely we introduce the standard “O” notation, which we will use freely in the rest of this book. For exam- ple, the statement f̂(n) = O(1/|n|2) as |n| → ∞, means that the left- hand side is bounded by a constant multiple of the right-hand side; that is, there exists C > 0 with |f̂(n)| ≤ C/|n|2 for all large |n|. More
generally, f(x) = O(g(x)) as x → a means that for some constant C,
|f(x)| ≤ C|g(x)| as x approaches a. In particular, f(x) = O(1) means
that f is bounded.

Corollary 2.4 Suppose that f is a twice continuously differentiable func-
tion on the circle. Then

f̂(n) = O(1/|n|2) as |n| → ∞,

so that the Fourier series of f converges absolutely and uniformly to f .

Ibookroot October 20, 2007

2. Uniqueness of Fourier series 43

Proof. The estimate on the Fourier coefficients is proved by integrating
by parts twice for n 6= 0. We obtain

2πf̂(n) =
∫ 2π
0

f(θ)e−inθ dθ

[
f(θ) · −e

−inθ

]2π

+
1
in

∫ 2π
0

f ′(θ)e−inθ dθ

=
1
in

∫ 2π
0

f ′(θ)e−inθ dθ

=
1
in

[
f ′(θ) · −e

−inθ

]2π

+
1

(in)2

∫ 2π
0

f ′′(θ)e−inθ dθ

=
−1
n2

∫ 2π
0

f ′′(θ)e−inθ dθ.

The quantities in brackets vanish since f and f ′ are periodic. Therefore

2π|n|2|f̂(n)| ≤
∣∣∣∣
∫ 2π
0

f ′′(θ)e−inθ dθ
∣∣∣∣ ≤

∫ 2π
0

|f ′′(θ)| dθ ≤ C,

where the constant C is independent of n. (We can take C = 2πB where
B is a bound for f ′′.) Since

∑
1/n2 converges, the proof of the corollary

is complete.

Incidentally, we have also established the following important identity:

f̂ ′(n) = inf̂(n), for all n ∈ Z.

If n 6= 0 the proof is given above, and if n = 0 it is left as an exercise to the
reader. So if f is differentiable and f ∼ ∑ aneinθ, then f ′ ∼

∑
anine

inθ.
Also, if f is twice continuously differentiable, then f ′′ ∼ ∑ an(in)2einθ,
and so on. Further smoothness conditions on f imply even better decay
of the Fourier coefficients (Exercise 10).

There are also stronger versions of Corollary 2.4. It can be shown, for
example, that the Fourier series of f converges absolutely, assuming only
that f has one continuous derivative. Even more generally, the Fourier
series of f converges absolutely (and hence uniformly to f) if f satisfies
a Hölder condition of order α, with α > 1/2, that is,

sup
θ
|f(θ + t)− f(θ)| ≤ A|t|α for all t.

For more on these matters, see the exercises at the end of Chapter 3.

Ibookroot October 20, 2007

44 Chapter 2. BASIC PROPERTIES OF FOURIER SERIES

At this point it is worthwhile to introduce a common notation: we say
that f belongs to the class Ck if f is k times continuously differentiable.
Belonging to the class Ck or satisfying a Hölder condition are two possible
ways to describe the smoothness of a function.

3 Convolutions

The notion of convolution of two functions plays a fundamental role in
Fourier analysis; it appears naturally in the context of Fourier series but
also serves more generally in the analysis of functions in other settings.

Given two 2π-periodic integrable functions f and g on R, we define
their convolution f ∗ g on [−π, π] by

(2) (f ∗ g)(x) = 1
2π

∫ π
−π

f(y)g(x− y) dy.

The above integral makes sense for each x, since the product of two
integrable functions is again integrable. Also, since the functions are
periodic, we can change variables to see that

(f ∗ g)(x) = 1
2π

∫ π
−π

f(x− y)g(y) dy.

Loosely speaking, convolutions correspond to “weighted averages.” For
instance, if g = 1 in (2), then f ∗ g is constant and equal to 1

2π

∫ π
−π f(y) dy,

which we may interpret as the average value of f on the circle. Also, the
convolution (f ∗ g)(x) plays a role similar to, and in some sense replaces,
the pointwise product f(x)g(x) of the two functions f and g.

In the context of this chapter, our interest in convolutions originates
from the fact that the partial sums of the Fourier series of f can be
expressed as follows:

SN (f)(x) =
N∑

n=−N
f̂(n)einx

=
N∑

n=−N

(
1
2π

∫ π
−π

f(y)e−iny dy
)

einx

=
1
2π

∫ π
−π

f(y)

(
N∑

n=−N
ein(x−y)

)
dy

= (f ∗DN )(x),

Ibookroot October 20, 2007

3. Convolutions 45

where DN is the N th Dirichlet kernel (see Example 4) given by

DN (x) =
N∑

n=−N
einx.

So we observe that the problem of understanding SN (f) reduces to the
understanding of the convolution f ∗DN .

We begin by gathering some of the main properties of convolutions.

Proposition 3.1 Suppose that f , g, and h are 2π-periodic integrable
functions. Then:

(i) f ∗ (g + h) = (f ∗ g) + (f ∗ h).

(ii) (cf) ∗ g = c(f ∗ g) = f ∗ (cg) for any c ∈ C.

(iii) f ∗ g = g ∗ f .

(iv) (f ∗ g) ∗ h = f ∗ (g ∗ h).

(v) f ∗ g is continuous.

(vi) f̂ ∗ g(n) = f̂(n)ĝ(n).

The first four points describe the algebraic properties of convolutions:
linearity, commutativity, and associativity. Property (v) exhibits an im-
portant principle: the convolution of f ∗ g is “more regular” than f or g.
Here, f ∗ g is continuous while f and g are merely (Riemann) integrable.
Finally, (vi) is key in the study of Fourier series. In general, the Fourier
coefficients of the product fg are not the product of the Fourier coeffi-
cients of f and g. However, (vi) says that this relation holds if we replace
the product of the two functions f and g by their convolution f ∗ g.

Proof. Properties (i) and (ii) follow at once from the linearity of the
integral.

The other properties are easily deduced if we assume also that f and
g are continuous. In this case, we may freely interchange the order of

Ibookroot October 20, 2007

46 Chapter 2. BASIC PROPERTIES OF FOURIER SERIES

integration. For instance, to establish (vi) we write

f̂ ∗ g(n) = 1
2π

∫ π
−π

(f ∗ g)(x)e−inx dx

=
1
2π

∫ π
−π

1
2π

(∫ π
−π

f(y)g(x− y) dy
)

e−inx dx

=
1
2π

∫ π
−π

f(y)e−iny
(

1
2π

∫ π
−π

g(x− y)e−in(x−y) dx
)

=
1
2π

∫ π
−π

f(y)e−iny
(

1
2π

∫ π
−π

g(x)e−inx dx
)

= f̂(n)ĝ(n).

To prove (iii), one first notes that if F is continuous and 2π-periodic,
then

∫ π
−π

F (y) dy =
∫ π
−π

F (x− y) dy for any x ∈ R.

The verification of this identity consists of a change of variables y 7→ −y,
followed by a translation y 7→ y − x. Then, one takes F (y) = f(y)g(x− y).

Also, (iv) follows by interchanging two integral signs, and an appro-
priate change of variables.

Finally, we show that if f and g are continuous, then f ∗ g is continu-
ous. First, we may write

(f ∗ g)(x1)− (f ∗ g)(x2) =
1
2π

∫ π
−π

f(y) [g(x1 − y)− g(x2 − y)] dy.

Since g is continuous it must be uniformly continuous on any closed
and bounded interval. But g is also periodic, so it must be uniformly
continuous on all of R; given ² > 0 there exists δ > 0 so that |g(s)−
g(t)| < ² whenever |s− t| < δ. Then, |x1 − x2| < δ implies |(x1 − y)− (x2 − y)| < δ for any y, hence |(f ∗ g)(x1)− (f ∗ g)(x2)| ≤ 1 2π ∣∣∣∣ ∫ π −π f(y) [g(x1 − y)− g(x2 − y)] dy ∣∣∣∣ ≤ 1 2π ∫ π −π |f(y)| |g(x1 − y)− g(x2 − y)| dy ≤ ² 2π ∫ π −π |f(y)| dy ≤ ² 2π 2π B , Ibookroot October 20, 2007 3. Convolutions 47 where B is chosen so that |f(x)| ≤ B for all x. As a result, we conclude that f ∗ g is continuous, and the proposition is proved, at least when f and g are continuous. In general, when f and g are merely integrable, we may use the re- sults established so far (when f and g are continuous), together with the following approximation lemma, whose proof may be found in the appendix. Lemma 3.2 Suppose f is integrable on the circle and bounded by B. Then there exists a sequence {fk}∞k=1 of continuous functions on the circle so that sup x∈[−π,π] |fk(x)| ≤ B for all k = 1, 2, . . . , and ∫ π −π |f(x)− fk(x)| dx → 0 as k →∞. Using this result, we may complete the proof of the proposition as follows. Apply Lemma 3.2 to f and g to obtain sequences {fk} and {gk} of approximating continuous functions. Then f ∗ g − fk ∗ gk = (f − fk) ∗ g + fk ∗ (g − gk). By the properties of the sequence {fk}, |(f − fk) ∗ g(x)| ≤ 1 2π ∫ π −π |f(x− y)− fk(x− y)| |g(y)| dy ≤ 1 2π sup y |g(y)| ∫ π −π |f(y)− fk(y)| dy → 0 as k →∞. Hence (f − fk) ∗ g → 0 uniformly in x. Similarly, fk ∗ (g − gk) → 0 uni- formly, and therefore fk ∗ gk tends uniformly to f ∗ g. Since each fk ∗ gk is continuous, it follows that f ∗ g is also continuous, and we have (v). Next, we establish (vi). For each fixed integer n we must have f̂k ∗ gk(n) → f̂ ∗ g(n) as k tends to infinity since fk ∗ gk converges uni- formly to f ∗ g. However, we found earlier that f̂k(n)ĝk(n) = f̂k ∗ gk(n) because both fk and gk are continuous. Hence |f̂(n)− f̂k(n)|= 1 2π ∣∣∣∣ ∫ π −π (f(x)− fk(x))e−inx dx ∣∣∣∣ ≤ 1 2π ∫ π −π |f(x)− fk(x)| dx, Ibookroot October 20, 2007 48 Chapter 2. BASIC PROPERTIES OF FOURIER SERIES and as a result we find that f̂k(n) → f̂(n) as k goes to infinity. Similarly ĝk(n) → ĝ(n), and the desired property is established once we let k tend to infinity. Finally, properties (iii) and (iv) follow from the same kind of arguments. 4 Good kernels In the proof of Theorem 2.1 we constructed a sequence of trigonometric polynomials {pk} with the property that the functions pk peaked at the origin. As a result, we could isolate the behavior of f at the origin. In this section, we return to such families of functions, but this time in a more general setting. First, we define the notion of good kernel, and discuss the characteristic properties of such functions. Then, by the use of convolutions, we show how these kernels can be used to recover a given function. A family of kernels {Kn(x)}∞n=1 on the circle is said to be a family of good kernels if it satisfies the following properties: (a) For all n ≥ 1, 1 2π ∫ π −π Kn(x) dx = 1. (b) There exists M > 0 such that for all n ≥ 1,
∫ π
−π
|Kn(x)| dx ≤ M.

δ≤|x|≤π
|Kn(x)| dx → 0, as n →∞.

In practice we shall encounter families where Kn(x) ≥ 0, in which
case (b) is a consequence of (a). We may interpret the kernels Kn(x)
as weight distributions on the circle: property (a) says that Kn assigns
unit mass to the whole circle [−π, π], and (c) that this mass concentrates
near the origin as n becomes large.6 Figure 4 (a) illustrates the typical
character of a family of good kernels.

The importance of good kernels is highlighted by their use in connec-
tion with convolutions.

6In the limit, a family of good kernels represents the “Dirac delta function.” This
terminology comes from physics.

Ibookroot October 20, 2007

4. Good kernels 49

(a) (b)
y

Kn(y)

y = 0

f(x− y) f(x)

Figure 4. Good kernels

Theorem 4.1 Let {Kn}∞n=1 be a family of good kernels, and f an inte-
grable function on the circle. Then

lim
n→∞

(f ∗Kn)(x) = f(x)

whenever f is continuous at x. If f is continuous everywhere, then the
above limit is uniform.

Because of this result, the family {Kn} is sometimes referred to as an
approximation to the identity.

We have previously interpreted convolutions as weighted averages. In
this context, the convolution

(f ∗Kn)(x) =
1
2π

∫ π
−π

f(x− y)Kn(y) dy

is the average of f(x− y), where the weights are given by Kn(y). How-
ever, the weight distribution Kn concentrates its mass at y = 0 as n
becomes large. Hence in the integral, the value f(x) is assigned the full
mass as n →∞. Figure 4 (b) illustrates this point.

Proof of Theorem 4.1. If ² > 0 and f is continuous at x, choose δ so
that |y| < δ implies |f(x− y)− f(x)| < ². Then, by the first property of good kernels, we can write (f ∗Kn)(x)− f(x) = 1 2π ∫ π −π Kn(y)f(x− y) dy − f(x) = 1 2π ∫ π −π Kn(y)[f(x− y)− f(x)] dy. Ibookroot October 20, 2007 50 Chapter 2. BASIC PROPERTIES OF FOURIER SERIES Hence, |(f ∗Kn)(x)− f(x)| = ∣∣∣∣ 1 2π ∫ π −π Kn(y)[f(x− y)− f(x)] dy ∣∣∣∣ ≤ 1 2π ∫ |y|<δ |Kn(y)| |f(x− y)− f(x)| dy + 1 2π ∫ δ≤|y|≤π |Kn(y)| |f(x− y)− f(x)| dy ≤ ² 2π ∫ π −π |Kn(y)| dy + 2B 2π ∫ δ≤|y|≤π |Kn(y)| dy, where B is a bound for f . The first term is bounded by ²M/2π because of the second property of good kernels. By the third property we see that for all large n, the second term will be less than ². Therefore, for some constant C > 0 and all large n we have

|(f ∗Kn)(x)− f(x)| ≤ C²,

thereby proving the first assertion in the theorem. If f is continuous
everywhere, then it is uniformly continuous, and δ can be chosen in-
dependent of x. This provides the desired conclusion that f ∗Kn → f
uniformly.

Recall from the beginning of Section 3 that

SN (f)(x) = (f ∗DN )(x) ,

where DN (x) =
∑N

n=−N e
inx is the Dirichlet kernel. It is natural now for

us to ask whether DN is a good kernel, since if this were true, Theorem 4.1
would imply that the Fourier series of f converges to f(x) whenever f is
continuous at x. Unfortunately, this is not the case. Indeed, an estimate
shows that DN violates the second property; more precisely, one has (see
Problem 2)

∫ π
−π
|DN (x)| dx ≥ c log N, as N →∞.

However, we should note that the formula for DN as a sum of exponen-
tials immediately gives

1
2π

∫ π
−π

DN (x) dx = 1,

so the first property of good kernels is actually verified. The fact that the
mean value of DN is 1, while the integral of its absolute value is large,

Ibookroot October 20, 2007

5. Cesàro and Abel summability: applications to Fourier series 51

is a result of cancellations. Indeed, Figure 5 shows that the function
DN (x) takes on positive and negative values and oscillates very rapidly
as N gets large.

Figure 5. The Dirichlet kernel for large N

This observation suggests that the pointwise convergence of Fourier
series is intricate, and may even fail at points of continuity. This is
indeed the case, as we will see in the next chapter.

5 Cesàro and Abel summability: applications to Fourier

series

Since a Fourier series may fail to converge at individual points, we are
led to try to overcome this failure by interpreting the limit

lim
N→∞

SN (f) = f

in a different sense.

5.1 Cesàro means and summation

We begin by taking ordinary averages of the partial sums, a technique
which we now describe in more detail.

Ibookroot October 20, 2007

52 Chapter 2. BASIC PROPERTIES OF FOURIER SERIES

Suppose we are given a series of complex numbers

c0 + c1 + c2 + · · · =
∞∑

k=0

ck.

We define the nth partial sum sn by

sn =
n∑

k=0

ck,

and say that the series converges to s if limn→∞ sn = s. This is the
most natural and most commonly used type of “summability.” Consider,
however, the example of the series

(3) 1− 1 + 1− 1 + · · · =
∞∑

k=0

(−1)k.

Its partial sums form the sequence {1, 0, 1, 0, . . .} which has no limit.
Because these partial sums alternate evenly between 1 and 0, one might
therefore suggest that 1/2 is the “limit” of the sequence, and hence 1/2
equals the “sum” of that particular series. We give a precise meaning to
this by defining the average of the first N partial sums by

σN =
s0 + s1 + · · ·+ sN−1

N
.

The quantity σN is called the N th Cesàro mean7 of the sequence {sk}
or the N th Cesàro sum of the series

∑∞
k=0 ck.

If σN converges to a limit σ as N tends to infinity, we say that the
series

∑
cn is Cesàro summable to σ. In the case of series of functions,

we shall understand the limit in the sense of either pointwise or uniform
convergence, depending on the situation.

The reader will have no difficulty checking that in the above exam-
ple (3), the series is Cesàro summable to 1/2. Moreover, one can show
that Cesàro summation is a more inclusive process than convergence. In
fact, if a series is convergent to s, then it is also Cesàro summable to the
same limit s (Exercise 12).

5.2 Fejér’s theorem

An interesting application of Cesàro summability appears in the context
of Fourier series.

7Note that if the series
∑∞

k=1
ck begins with the term k = 1, then it is common prac-

tice to define σN = (s1 + · · ·+ sN )/N . This change of notation has little effect on what
follows.

Ibookroot October 20, 2007

5. Cesàro and Abel summability: applications to Fourier series 53

We mentioned earlier that the Dirichlet kernels fail to belong to the
family of good kernels. Quite surprisingly, their averages are very well
behaved functions, in the sense that they do form a family of good ker-
nels.

To see this, we form the N th Cesàro mean of the Fourier series, which
by definition is

σN (f)(x) =
S0(f)(x) + · · ·+ SN−1(f)(x)

N
.

Since Sn(f) = f ∗Dn, we find that

σN (f)(x) = (f ∗ FN )(x),

where FN (x) is the N -th Fejér kernel given by

FN (x) =
D0(x) + · · ·+ DN−1(x)

N
.

Lemma 5.1 We have

FN (x) =
1
N

sin2(Nx/2)
sin2(x/2)

and the Fejér kernel is a good kernel.

The proof of the formula for FN (a simple application of trigonometric
identities) is outlined in Exercise 15. To prove the rest of the lemma, note
that FN is positive and 12π

∫ π
−π FN (x) dx = 1, in view of the fact that a

similar identity holds for the Dirichlet kernels Dn. However, sin
2(x/2) ≥

cδ > 0, if δ ≤ |x| ≤ π, hence FN (x) ≤ 1/(Ncδ), from which it follows that
∫

δ≤|x|≤π
|FN (x)| dx → 0 as N →∞.

Applying Theorem 4.1 to this new family of good kernels yields the
following important result.

Theorem 5.2 If f is integrable on the circle, then the Fourier series of
f is Cesàro summable to f at every point of continuity of f .

Moreover, if f is continuous on the circle, then the Fourier series of
f is uniformly Cesàro summable to f .

We may now state two corollaries. The first is a result that we have
already established. The second is new, and of fundamental importance.

Ibookroot October 20, 2007

54 Chapter 2. BASIC PROPERTIES OF FOURIER SERIES

Corollary 5.3 If f is integrable on the circle and f̂(n) = 0 for all n,
then f = 0 at all points of continuity of f .

The proof is immediate since all the partial sums are 0, hence all the
Cesàro means are 0.

Corollary 5.4 Continuous functions on the circle can be uniformly ap-
proximated by trigonometric polynomials.

This means that if f is continuous on [−π, π] with f(−π) = f(π) and
² > 0, then there exists a trigonometric polynomial P such that

|f(x)− P (x)| < ² for all −π ≤ x ≤ π. This follows immediately from the theorem since the partial sums, hence the Cesàro means, are trigonometric polynomials. Corollary 5.4 is the periodic analogue of the Weierstrass approximation theorem for polyno- mials which can be found in Exercise 16. 5.3 Abel means and summation Another method of summation was first considered by Abel and actually predates the Cesàro method. A series of complex numbers ∑∞ k=0 ck is said to be Abel summable to s if for every 0 ≤ r < 1, the series A(r) = ∞∑ k=0 ckr k converges, and lim r→1 A(r) = s. The quantities A(r) are called the Abel means of the series. One can prove that if the series converges to s, then it is Abel summable to s. Moreover, the method of Abel summability is even more powerful than the Cesàro method: when the series is Cesàro summable, it is always Abel summable to the same sum. However, if we consider the series 1− 2 + 3− 4 + 5− · · · = ∞∑ k=0 (−1)k(k + 1), then one can show that it is Abel summable to 1/4 since A(r) = ∞∑ k=0 (−1)k(k + 1)rk = 1 (1 + r)2 , but this series is not Cesàro summable; see Exercise 13. Ibookroot October 20, 2007 5. Cesàro and Abel summability: applications to Fourier series 55 5.4 The Poisson kernel and Dirichlet’s problem in the unit disc To adapt Abel summability to the context of Fourier series, we define the Abel means of the function f(θ) ∼ ∑∞n=−∞ aneinθ by Ar(f)(θ) = ∞∑ n=−∞ r|n|ane inθ. Since the index n takes positive and negative values, it is natural to write c0 = a0, and cn = aneinθ + a−ne−inθ for n > 0, so that the Abel means
of the Fourier series correspond to the definition given in the previous
section for numerical series.

We note that since f is integrable, |an| is uniformly bounded in n, so
that Ar(f) converges absolutely and uniformly for each 0 ≤ r < 1. Just as in the case of Cesàro means, the key fact is that these Abel means can be written as convolutions Ar(f)(θ) = (f ∗ Pr)(θ), where Pr(θ) is the Poisson kernel given by (4) Pr(θ) = ∞∑ n=−∞ r|n|einθ. In fact, Ar(f)(θ) = ∞∑ n=−∞ r|n|ane inθ = ∞∑ n=−∞ r|n| ( 1 2π ∫ π −π f(ϕ)e−inϕ dϕ ) einθ = 1 2π ∫ π −π f(ϕ) ( ∞∑ n=−∞ r|n|e−in(ϕ−θ) ) dϕ, where the interchange of the integral and infinite sum is justified by the uniform convergence of the series. Lemma 5.5 If 0 ≤ r < 1, then Pr(θ) = 1− r2 1− 2r cos θ + r2 . Ibookroot October 20, 2007 56 Chapter 2. BASIC PROPERTIES OF FOURIER SERIES The Poisson kernel is a good kernel,8 as r tends to 1 from below. Proof. The identity Pr(θ) = 1−r 2 1−2r cos θ+r2 has already been derived in Section 1.1. Note that 1− 2r cos θ + r2 = (1− r)2 + 2r(1− cos θ). Hence if 1/2 ≤ r ≤ 1 and δ ≤ |θ| ≤ π, then 1− 2r cos θ + r2 ≥ cδ > 0.
Thus Pr(θ) ≤ (1− r2)/cδ when δ ≤ |θ| ≤ π, and the third property of
good kernels is verified. Clearly Pr(θ) ≥ 0, and integrating the expres-
sion (4) term by term (which is justified by the absolute convergence of
the series) yields

1
2π

∫ π
−π

Pr(θ) dθ = 1,

thereby concluding the proof that Pr is a good kernel.

Combining this lemma with Theorem 4.1, we obtain our next result.

Theorem 5.6 The Fourier series of an integrable function on the circle
is Abel summable to f at every point of continuity. Moreover, if f is
continuous on the circle, then the Fourier series of f is uniformly Abel
summable to f .

We now return to a problem discussed in Chapter 1, where we sketched
the solution of the steady-state heat equation 4u = 0 in the unit disc
with boundary condition u = f on the circle. We expressed the Laplacian
in terms of polar coordinates, separated variables, and expected that a
solution was given by

(5) u(r, θ) =
∞∑

m=−∞
amr

|m|eimθ,

where am was the mth Fourier coefficient of f . In other words, we were
led to take

u(r, θ) = Ar(f)(θ) =
1
2π

∫ π
−π

f(ϕ)Pr(θ − ϕ) dϕ.

We are now in a position to show that this is indeed the case.

8In this case, the family of kernels is indexed by a continuous parameter 0 ≤ r < 1, rather than the discrete n considered previously. In the definition of good kernels, we simply replace n by r and take the limit in property (c) appropriately, for example r → 1 in this case. Ibookroot October 20, 2007 5. Cesàro and Abel summability: applications to Fourier series 57 Theorem 5.7 Let f be an integrable function defined on the unit circle. Then the function u defined in the unit disc by the Poisson integral (6) u(r, θ) = (f ∗ Pr)(θ) has the following properties: (i) u has two continuous derivatives in the unit disc and satisfies 4u = 0. (ii) If θ is any point of continuity of f , then lim r→1 u(r, θ) = f(θ). If f is continuous everywhere, then this limit is uniform. (iii) If f is continuous, then u(r, θ) is the unique solution to the steady- state heat equation in the disc which satisfies conditions (i) and (ii). Proof. For (i), we recall that the function u is given by the series (5). Fix ρ < 1; inside each disc of radius r < ρ < 1 centered at the origin, the series for u can be differentiated term by term, and the differentiated se- ries is uniformly and absolutely convergent. Thus u can be differentiated twice (in fact infinitely many times), and since this holds for all ρ < 1, we conclude that u is twice differentiable inside the unit disc. Moreover, in polar coordinates, 4u = ∂ 2u ∂r2 + 1 r ∂u ∂r + 1 r2 ∂2u ∂θ2 , so term by term differentiation shows that 4u = 0. The proof of (ii) is a simple application of the previous theorem. To prove (iii) we argue as follows. Suppose v solves the steady-state heat equation in the disc and converges to f uniformly as r tends to 1 from below. For each fixed r with 0 < r < 1, the function v(r, θ) has a Fourier series ∞∑ n=−∞ an(r)e inθ where an(r) = 1 2π ∫ π −π v(r, θ)e−inθ dθ. Taking into account that v(r, θ) solves the equation (7) ∂2v ∂r2 + 1 r ∂v ∂r + 1 r2 ∂2v ∂θ2 = 0, Ibookroot October 20, 2007 58 Chapter 2. BASIC PROPERTIES OF FOURIER SERIES we find that (8) a′′n(r) + 1 r a′n(r)− n2 r2 an(r) = 0. Indeed, we may first multiply (7) by e−inθ and integrate in θ. Then, since v is periodic, two integrations by parts give 1 2π ∫ π −π ∂2v ∂θ2 (r, θ)e−inθ dθ = −n2an(r). Finally, we may interchange the order of differentiation and integra- tion, which is permissible since v has two continuous derivatives; this yields (8). Therefore, we must have an(r) = Anrn + Bnr−n for some constants An and Bn, when n 6= 0 (see Exercise 11 in Chapter 1). To evaluate the constants, we first observe that each term an(r) is bounded because v is bounded, therefore Bn = 0. To find An we let r → 1. Since v converges uniformly to f as r → 1 we find that An = 1 2π ∫ π −π f(θ)e−inθ dθ. By a similar argument, this formula also holds when n = 0. Our con- clusion is that for each 0 < r < 1, the Fourier series of v is given by the series of u(r, θ), so by the uniqueness of Fourier series for continuous functions, we must have u = v. Remark. By part (iii) of the theorem, we may conclude that if u solves 4u = 0 in the disc, and converges to 0 uniformly as r → 1, then u must be identically 0. However, if uniform convergence is replaced by pointwise convergence, this conclusion may fail; see Exercise 18. 6 Exercises 1. Suppose f is 2π-periodic and integrable on any finite interval. Prove that if a, b ∈ R, then ∫ b a f(x) dx = ∫ b+2π a+2π f(x) dx = ∫ b−2π a−2π f(x) dx. Also prove that ∫ π −π f(x + a) dx = ∫ π −π f(x) dx = ∫ π+a −π+a f(x) dx. Ibookroot October 20, 2007 6. Exercises 59 2. In this exercise we show how the symmetries of a function imply certain properties of its Fourier coefficients. Let f be a 2π-periodic Riemann integrable function defined on R. (a) Show that the Fourier series of the function f can be written as f(θ) ∼ f̂(0) + ∑ n≥1 [f̂(n) + f̂(−n)] cos nθ + i[f̂(n)− f̂(−n)] sin nθ. (b) Prove that if f is even, then f̂(n) = f̂(−n), and we get a cosine series. (c) Prove that if f is odd, then f̂(n) = −f̂(−n), and we get a sine series. (d) Suppose that f(θ + π) = f(θ) for all θ ∈ R. Show that f̂(n) = 0 for all odd n. (e) Show that f is real-valued if and only if f̂(n) = f̂(−n) for all n. 3. We return to the problem of the plucked string discussed in Chapter 1. Show that the initial condition f is equal to its Fourier sine series f(x) = ∞∑ m=1 Am sinmx with Am = 2h m2 sinmp p(π − p) . [Hint: Note that |Am| ≤ C/m2.] 4. Consider the 2π-periodic odd function defined on [0, π] by f(θ) = θ(π − θ). (a) Draw the graph of f . (b) Compute the Fourier coefficients of f , and show that f(θ) = 8 π ∑ k odd ≥ 1 sin kθ k3 . 5. On the interval [−π, π] consider the function f(θ) = { 0 if |θ| > δ,
1− |θ|/δ if |θ| ≤ δ.

Thus the graph of f has the shape of a triangular tent. Show that

f(θ) =
δ

2π
+ 2

∞∑
n=1

1− cos nδ
n2πδ

cos nθ.

6. Let f be the function defined on [−π, π] by f(θ) = |θ|.

Ibookroot October 20, 2007

60 Chapter 2. BASIC PROPERTIES OF FOURIER SERIES

(a) Draw the graph of f .

(b) Calculate the Fourier coefficients of f , and show that

f̂(n) =





2
if n = 0,

−1 + (−1)n
πn2

if n 6= 0.

(d) Taking θ = 0, prove that

∑

n odd ≥1

1
n2

=
π2

8
and

∞∑
n=1

1
n2

=
π2

6
.

See also Example 2 in Section 1.1.

7. Suppose {an}Nn=1 and {bn}Nn=1 are two finite sequences of complex numbers.
Let Bk =

∑k
n=1

bn denote the partial sums of the series
∑

bn with the convention
B0 = 0.

(a) Prove the summation by parts formula

N∑
n=M

anbn = aNBN − aMBM−1 −
N−1∑
n=M

(an+1 − an)Bn.

(b) Deduce from this formula Dirichlet’s test for convergence of a series: if the
partial sums of the series

∑
bn are bounded, and {an} is a sequence of

real numbers that decreases monotonically to 0, then
∑

anbn converges.

8. Verify that
1
2i

∑
n 6=0

einx

n
is the Fourier series of the 2π-periodic sawtooth

function illustrated in Figure 6, defined by f(0) = 0, and

f(x) =





−π
2
− x

2
if −π < x < 0, π 2 − x 2 if 0 < x < π. Note that this function is not continuous. Show that nevertheless, the series converges for every x (by which we mean, as usual, that the symmetric partial sums of the series converge). In particular, the value of the series at the origin, namely 0, is the average of the values of f(x) as x approaches the origin from the left and the right. Ibookroot October 20, 2007 6. Exercises 61 0 π 2 −π π −π 2 Figure 6. The sawtooth function [Hint: Use Dirichlet’s test for convergence of a series ∑ anbn.] 9. Let f(x) = χ[a,b](x) be the characteristic function of the interval [a, b] ⊂ [−π, π], that is, χ[a,b](x) = { 1 if x ∈ [a, b], 0 otherwise. (a) Show that the Fourier series of f is given by f(x) ∼ b− a 2π + ∑ n6=0 e−ina − e−inb 2πin einx. The sum extends over all positive and negative integers excluding 0. (b) Show that if a 6= −π or b 6= π and a 6= b, then the Fourier series does not converge absolutely for any x. [Hint: It suffices to prove that for many values of n one has | sin nθ0| ≥ c > 0 where θ0 = (b− a)/2.]

10. Suppose f is a periodic function of period 2π which belongs to the class Ck.
Show that

f̂(n) = O(1/|n|k) as |n| → ∞.
This notation means that there exists a constant C such |f̂(n)| ≤ C/|n|k. We
could also write this as |n|kf̂(n) = O(1), where O(1) means bounded.
[Hint: Integrate by parts.]

11. Suppose that {fk}∞k=1 is a sequence of Riemann integrable functions on the
interval [0, 1] such that

∫ 1
0

|fk(x)− f(x)| dx → 0 as k →∞.

Ibookroot October 20, 2007

62 Chapter 2. BASIC PROPERTIES OF FOURIER SERIES

Show that f̂k(n) → f̂(n) uniformly in n as k →∞.

12. Prove that if a series of complex numbers
∑

cn converges to s, then
∑

cn
is Cesàro summable to s.
[Hint: Assume sn → 0 as n →∞.]

13. The purpose of this exercise is to prove that Abel summability is stronger
than the standard or Cesàro methods of summation.

(a) Show that if the series
∑∞

n=1
cn of complex numbers converges to a finite

limit s, then the series is Abel summable to s. [Hint: Why is it enough to
prove the theorem when s = 0? Assuming s = 0, show that if sN = c1 +
· · ·+ cN , then

∑N
n=1

cnr
n = (1− r)

∑N
n=1

snr
n + sNrN+1. Let N →∞

to show that
∑

cnr
n = (1− r)

∑
snr

Finally, prove that the right-hand side converges to 0 as r → 1.]
(b) However, show that there exist series which are Abel summable, but that

do not converge. [Hint: Try cn = (−1)n. What is the Abel limit of
∑

cn?]

n=1
cn is Cesàro summable to

σ, then it is Abel summable to σ. [Hint: Note that

∞∑
n=1

cnr
n = (1− r)2

∞∑
n=1

nσnr
n,

and assume σ = 0.]

(d) Give an example of a series that is Abel summable but not Cesàro summable.
[Hint: Try cn = (−1)n−1n. Note that if

∑
cn is Cesàro summable, then

cn/n tends to 0.]

The results above can be summarized by the following implications about
series:

convergent =⇒ Cesàro summable =⇒ Abel summable,

and the fact that none of the arrows can be reversed.

14. This exercise deals with a theorem of Tauber which says that under an
additional condition on the coefficients cn, the above arrows can be reversed.

(a) If
∑

cn is Cesàro summable to σ and cn = o(1/n) (that is, ncn → 0), then∑
cn converges to σ. [Hint: sn − σn = [(n− 1)cn + · · ·+ c2]/n.]

(b) The above statement holds if we replace Cesàro summable by Abel summable.
[Hint: Estimate the difference between

∑N
n=1

cn and
∑N

n=1
cnr

n where
r = 1− 1/N .]

Ibookroot October 20, 2007

6. Exercises 63

15. Prove that the Fejér kernel is given by

FN (x) =
1
N

sin2(Nx/2)
sin2(x/2)

[Hint: Remember that NFN (x) = D0(x) + · · ·+ DN−1(x) where Dn(x) is the
Dirichlet kernel. Therefore, if ω = eix we have

NFN (x) =
N−1∑
n=0

ω−n − ωn+1
1− ω .]

16. The Weierstrass approximation theorem states: Let f be a continuous
function on the closed and bounded interval [a, b] ⊂ R. Then, for any ² > 0,
there exists a polynomial P such that

sup
x∈[a,b]

|f(x)− P (x)| < ². Prove this by applying Corollary 5.4 of Fejér’s theorem and using the fact that the exponential function eix can be approximated by polynomials uniformly on any interval. 17. In Section 5.4 we proved that the Abel means of f converge to f at all points of continuity, that is, lim r→1 Ar(f)(θ) = lim r→1 (Pr ∗ f)(θ) = f(θ), with 0 < r < 1, whenever f is continuous at θ. In this exercise, we will study the behavior of Ar(f)(θ) at certain points of discontinuity. An integrable function is said to have a jump discontinuity at θ if the two limits lim h → 0 h > 0

f(θ + h) = f(θ+) and lim
h → 0
h > 0

f(θ − h) = f(θ−)

exist.

(a) Prove that if f has a jump discontinuity at θ, then

lim
r→1

Ar(f)(θ) =
f(θ+) + f(θ−)

2
, with 0 ≤ r < 1. [Hint: Explain why 1 2π ∫ 0 −π Pr(θ) dθ = 1 2π ∫ π 0 Pr(θ) dθ = 12 , then modify the proof given in the text.] Ibookroot October 20, 2007 64 Chapter 2. BASIC PROPERTIES OF FOURIER SERIES (b) Using a similar argument, show that if f has a jump discontinuity at θ, the Fourier series of f at θ is Cesàro summable to f(θ +)+f(θ−) 2 . 18. If Pr(θ) denotes the Poisson kernel, show that the function u(r, θ) = ∂Pr ∂θ , defined for 0 ≤ r < 1 and θ ∈ R, satisfies: (i) 4u = 0 in the disc. (ii) limr→1 u(r, θ) = 0 for each θ. However, u is not identically zero. 19. Solve Laplace’s equation 4u = 0 in the semi infinite strip S = {(x, y) : 0 < x < 1, 0 < y}, subject to the following boundary conditions    u(0, y) = 0 when 0 ≤ y, u(1, y) = 0 when 0 ≤ y, u(x, 0) = f(x) when 0 ≤ x ≤ 1 where f is a given function, with of course f(0) = f(1) = 0. Write f(x) = ∞∑ n=1 an sin(nπx) and expand the general solution in terms of the special solutions given by un(x, y) = e −nπy sin(nπx). Express u as an integral involving f , analogous to the Poisson integral for- mula (6). 20. Consider the Dirichlet problem in the annulus defined by {(r, θ) : ρ < r < 1}, where 0 < ρ < 1 is the inner radius. The problem is to solve ∂2u ∂r2 + 1 r ∂u ∂r + 1 r2 ∂2u ∂θ2 = 0 subject to the boundary conditions { u(1, θ) = f(θ), u(ρ, θ) = g(θ), Ibookroot October 20, 2007 7. Problems 65 where f and g are given continuous functions. Arguing as we have previously for the Dirichlet problem in the disc, we can hope to write u(r, θ) = ∑ cn(r)e inθ with cn(r) = Anrn + Bnr−n, n 6= 0. Set f(θ) ∼ ∑ ane inθ and g(θ) ∼ ∑ bne inθ. We want cn(1) = an and cn(ρ) = bn. This leads to the solution u(r, θ)= ∑ n 6=0 ( 1 ρn − ρ−n )[ ((ρ/r)n − (r/ρ)n) an + (rn − r−n)bn ] einθ +a0 + (b0 − a0) log r log ρ . Show that as a result we have u(r, θ)− (Pr ∗ f)(θ) → 0 as r → 1 uniformly in θ, and u(r, θ)− (Pρ/r ∗ g)(θ) → 0 as r → ρ uniformly in θ. 7 Problems 1. One can construct Riemann integrable functions on [0, 1] that have a dense set of discontinuities as follows. (a) Let f(x) = 0 when x < 0, and f(x) = 1 if x ≥ 0. Choose a countable dense sequence {rn} in [0, 1]. Then, show that the function F (x) = ∞∑ n=1 1 n2 f(x− rn) is integrable and has discontinuities at all points of the sequence {rn}. [Hint: F is monotonic and bounded.] (b) Consider next F (x) = ∞∑ n=1 3−ng(x− rn), where g(x) = sin 1/x when x 6= 0, and g(0) = 0. Then F is integrable, discontinuous at each x = rn, and fails to be monotonic in any subinterval of [0, 1]. [Hint: Use the fact that 3−k >

∑
n>k

3−n.]

Ibookroot October 20, 2007

66 Chapter 2. BASIC PROPERTIES OF FOURIER SERIES

F (x) =
∞∑

n=1

(nx)
n2

where (x) = x for x ∈ (−1/2, 1/2] and (x) is continued to R by periodicity,
that is, (x + 1) = (x). It can be shown that F is discontinuous whenever
x = m/2n, where m,n ∈ Z with m odd and n 6= 0.

2. Let DN denote the Dirichlet kernel

DN (θ) =
N∑

k=−N
eikθ =

sin((N + 1/2)θ)
sin(θ/2)

and define

LN =
1
2π

∫ π
−π
|DN (θ)| dθ.

(a) Prove that

LN ≥ c log N

for some constant c > 0. [Hint: Show that |DN (θ)| ≥ c sin((N+1/2)θ)|θ| , change
variables, and prove that

LN ≥ c
∫ Nπ

| sin θ|
|θ| dθ + O(1).

Write the integral as a sum
∑N−1

k=1

∫ (k+1)π
kπ

. To conclude, use the fact that∑n
k=1

1/k ≥ c log n.] A more careful estimate gives

LN =
4
π2

log N + O(1).

(b) Prove the following as a consequence: for each n ≥ 1, there exists a contin-
uous function fn such that |fn| ≤ 1 and |Sn(fn)(0)| ≥ c′ log n. [Hint: The
function gn which is equal to 1 when Dn is positive and −1 when Dn is
negative has the desired property but is not continuous. Approximate gn
in the integral norm (in the sense of Lemma 3.2) by continuous functions
hk satisfying |hk| ≤ 1.]

3.∗ Littlewood provided a refinement of Tauber’s theorem:

Ibookroot October 20, 2007

7. Problems 67

(a) If
∑

cn is Abel summable to s and cn = O(1/n), then
∑

cn converges to
s.

(b) As a consequence, if
∑

cn is Cesàro summable to s and cn = O(1/n), then∑
cn converges to s.

These results may be applied to Fourier series. By Exercise 17, they imply that
if f is an integrable function that satisfies f̂(ν) = O(1/|ν|), then:

(i) If f is continuous at θ, then

SN (f)(θ) → f(θ) as N →∞.

(ii) If f has a jump discontinuity at θ, then

SN (f)(θ) →
f(θ+) + f(θ−)

2
as N →∞.

(iii) If f is continuous on [−π, π], then SN (f) → f uniformly.
For the simpler assertion (b), hence a proof of (i), (ii), and (iii), see Problem 5
in Chapter 4.

Ibookroot October 20, 2007

3 Convergence of Fourier Series

The sine and cosine series, by which one can repre-
sent an arbitrary function in a given interval, enjoy
among other remarkable properties that of being con-
vergent. This property did not escape the great ge-
ometer (Fourier) who began, through the introduc-
tion of the representation of functions just mentioned,
a new career for the applications of analysis; it was
stated in the Memoir which contains his first research
on heat. But no one so far, to my knowledge, gave a
general proof of it . . .

G. Dirichlet, 1829

In this chapter, we continue our study of the problem of convergence
of Fourier series. We approach the problem from two different points of
view.

The first is “global” and concerns the overall behavior of a function
f over the entire interval [0, 2π]. The result we have in mind is “mean-
square convergence”: if f is integrable on the circle, then

1
2π

∫ 2π
0

|f(θ)− SN (f)(θ)|2 dθ → 0 as N →∞.

At the heart of this result is the fundamental notion of “orthogonal-
ity”; this idea is expressed in terms of vector spaces with inner products,
and their related infinite dimensional variants, the Hilbert spaces. A
connected result is the Parseval identity which equates the mean-square
“norm” of the function with a corresponding norm of its Fourier coeffi-
cients. Orthogonality is a fundamental mathematical notion which has
many applications in analysis.

The second viewpoint is “local” and concerns the behavior of f near a
given point. The main question we consider is the problem of pointwise
convergence: does the Fourier series of f converge to the value f(θ)
for a given θ? We first show that this convergence does indeed hold
whenever f is differentiable at θ. As a corollary, we obtain the Riemann
localization principle, which states that the question of whether or not
SN (f)(θ) → f(θ) is completely determined by the behavior of f in an

Ibookroot October 20, 2007

70 Chapter 3. CONVERGENCE OF FOURIER SERIES

arbitrarily small interval about θ. This is a remarkable result since the
Fourier coefficients, hence the Fourier series, of f depend on the values
of f on the whole interval [0, 2π].

Even though convergence of the Fourier series holds at points where
f is differentiable, it may fail if f is merely continuous. The chapter
concludes with the presentation of a continuous function whose Fourier
series does not converge at a given point, as promised earlier.

1 Mean-square convergence of Fourier series

The aim of this section is the proof of the following theorem.

Theorem 1.1 Suppose f is integrable on the circle. Then

1
2π

∫ 2π
0

|f(θ)− SN (f)(θ)|2 dθ → 0 as N →∞.

As we remarked earlier, the key concept involved is that of orthogonal-
ity. The correct setting for orthogonality is in a vector space equipped
with an inner product.

1.1 Vector spaces and inner products

We now review the definitions of a vector space over R or C, an inner
product, and its associated norm. In addition to the familiar finite-
dimensional vector spaces Rd and Cd, we also examine two infinite-
dimensional examples which play a central role in the proof of Theo-
rem 1.1.

Preliminaries on vector spaces

A vector space V over the real numbers R is a set whose elements may be
“added” together, and “multiplied” by scalars. More precisely, we may
associate to any pair X, Y ∈ V an element in V called their sum and
denoted by X + Y . We require that this addition respects the usual laws
of arithmetic, such as commutativity X + Y = Y + X, and associativity
X + (Y + Z) = (X + Y ) + Z, etc. Also, given any X ∈ V and real num-
ber λ, we assign an element λX ∈ V called the product of X by λ. This
scalar multiplication must satisfy the standard properties, for instance
λ1(λ2X) = (λ1λ2)X and λ(X + Y ) = λX + λY . We may instead allow
scalar multiplication by numbers in C; we then say that V is a vector
space over the complex numbers.

Ibookroot October 20, 2007

1. Mean-square convergence of Fourier series 71

For example, the set Rd of d-tuples of real numbers (x1, x2, . . . , xd) is
a vector space over the reals. Addition is defined componentwise by

(x1, . . . , xd) + (y1, . . . , yd) = (x1 + y1, . . . , xd + yd),

and so is multiplication by a scalar λ ∈ R:

λ(x1, . . . , xd) = (λx1, . . . , λxd).

Similarly, the space Cd (the complex version of the previous example)
is the set of d-tuples of complex numbers (z1, z2, . . . , zd). It is a vector
space over C with addition defined componentwise by

(z1, . . . , zd) + (w1, . . . , wd) = (z1 + w1, . . . , zd + wd).

Multiplication by scalars λ ∈ C is given by

λ(z1, . . . , zd) = (λz1, . . . , λzd).

An inner product on a vector space V over R associates to any pair
X, Y of elements in V a real number which we denote by (X,Y ). In
particular, the inner product must be symmetric (X, Y ) = (Y, X) and
linear in both variables; that is,

(αX + βY,Z) = α(X, Z) + β(Y, Z)

whenever α, β ∈ R and X,Y, Z ∈ V . Also, we require that the inner prod-
uct be positive-definite, that is, (X, X) ≥ 0 for all X in V . In particular,
given an inner product (·, ·) we may define the norm of X by

‖X‖ = (X, X)1/2.

If in addition ‖X‖ = 0 implies X = 0, we say that the inner product is
strictly positive-definite.

For example, the space Rd is equipped with a (strictly positive-definite)
inner product defined by

(X, Y ) = x1y1 + · · ·+ xdyd

when X = (x1, . . . , xd) and Y = (y1, . . . , yd). Then

‖X‖ = (X, X)1/2 =
√

x21 + · · ·+ x2d,

Ibookroot October 20, 2007

72 Chapter 3. CONVERGENCE OF FOURIER SERIES

which is the usual Euclidean distance. One also uses the notation |X|
instead of ‖X‖.

For vector spaces over the complex numbers, the inner product of two
elements is a complex number. Moreover, these inner products are called
Hermitian (instead of symmetric) since they must satisfy
(X, Y ) = (Y, X). Hence the inner product is linear in the first variable,
but conjugate-linear in the second:

(αX + βY, Z) =α(X, Z) + β(Y, Z) and

(X,αY + βZ) =α(X, Y ) + β(X, Z).

Also, we must have (X,X) ≥ 0, and the norm of X is defined by
‖X‖ = (X,X)1/2 as before. Again, the inner product is strictly positive-
definite if ‖X‖ = 0 implies X = 0.

For example, the inner product of two vectors Z = (z1, . . . , zd) and
W = (w1, . . . , wd) in Cd is defined by

(Z,W ) = z1w1 + · · ·+ zdwd.

The norm of the vector Z is then given by

‖Z‖ = (Z, Z)1/2 =
√
|z1|2 + · · ·+ |zd|2.

The presence of an inner product on a vector space allows one to define
the geometric notion of “orthogonality.” Let V be a vector space (over R
or C) with inner product (·, ·) and associated norm ‖ · ‖. Two elements
X and Y are orthogonal if (X, Y ) = 0, and we write X ⊥ Y . Three
important results can be derived from this notion of orthogonality:

(i) The Pythagorean theorem: if X and Y are orthogonal, then

‖X + Y ‖2 = ‖X‖2 + ‖Y ‖2.

(ii) The Cauchy-Schwarz inequality: for any X,Y ∈ V we have

|(X,Y )| ≤ ‖X‖ ‖Y ‖.

(iii) The triangle inequality: for any X, Y ∈ V we have

‖X + Y ‖ ≤ ‖X‖+ ‖Y ‖.

Ibookroot October 20, 2007

1. Mean-square convergence of Fourier series 73

The proofs of these facts are simple. For (i) it suffices to expand
(X + Y, X + Y ) and use the assumption that (X, Y ) = 0.

For (ii), we first dispose of the case when ‖Y ‖ = 0 by showing that
this implies (X, Y ) = 0 for all X. Indeed, for all real t we have

0 ≤ ‖X + tY ‖2 = ‖X‖2 + 2t Re(X,Y )

and Re(X,Y ) 6= 0 contradicts the inequality if we take t to be large and
positive (or negative). Similarly, by considering ‖X + itY ‖2, we find that
Im(X, Y ) = 0.

If ‖Y ‖ 6= 0, we may set c = (X, Y )/(Y, Y ); then X − cY is orthogonal
to Y , and therefore also to cY . If we write X = X − cY + cY and apply
the Pythagorean theorem, we get

‖X‖2 = ‖X − cY ‖2 + ‖cY ‖2 ≥ |c|2‖Y ‖2.

Taking square roots on both sides gives the result. Note that we have
equality in the above precisely when X = cY .

Finally, for (iii) we first note that

‖X + Y ‖2 = (X,X) + (X,Y ) + (Y,X) + (Y, Y ).

But (X, X) = ‖X‖2, (Y, Y ) = ‖Y ‖2, and by the Cauchy-Schwarz inequal-
ity

|(X, Y ) + (Y, X)| ≤ 2 ‖X‖ ‖Y ‖,

therefore

‖X + Y ‖2 ≤ ‖X‖2 + 2 ‖X‖ ‖Y ‖+ ‖Y ‖2 = (‖X‖+ ‖Y ‖)2.

Two important examples

The vector spaces Rd and Cd are finite dimensional. In the context
of Fourier series, we need to work with two infinite-dimensional vector
spaces, which we now describe.

Example 1. The vector space `2(Z) over C is the set of all (two-sided)
infinite sequences of complex numbers

(. . . , a−n, . . . , a−1, a0, a1, . . . , an, . . .)

such that
∑

n∈Z
|an|2 < ∞; Ibookroot October 20, 2007 74 Chapter 3. CONVERGENCE OF FOURIER SERIES that is, the series converges. Addition is defined componentwise, and so is scalar multiplication. The inner product between the two vectors A = (. . . , a−1, a0, a1, . . .) and B = (. . . , b−1, b0, b1, . . .) is defined by the absolutely convergent series (A,B) = ∑ n∈Z anbn. The norm of A is then given by ‖A‖ = (A,A)1/2 = (∑ n∈Z |an|2 )1/2 . We must first check that `2(Z) is a vector space. This requires that if A and B are two elements in `2(Z), then so is the vector A + B. To see this, for each integer N > 0 we let AN denote the truncated element

AN = (. . . , 0, 0, a−N , . . . , a−1, a0, a1, . . . , aN , 0, 0, . . .),

where we have set an = 0 whenever |n| > N . We define the truncated
element BN similarly. Then, by the triangle inequality which holds in a
finite dimensional Euclidean space, we have

‖AN + BN‖ ≤ ‖AN‖+ ‖BN‖ ≤ ‖A‖+ ‖B‖.

Thus
∑

|n|≤N
|an + bn|2 ≤ (‖A‖+ ‖B‖)2,

and letting N tend to infinity gives
∑

n∈Z |an + bn|2 < ∞. It also fol- lows that ‖A + B‖ ≤ ‖A‖+ ‖B‖, which is the triangle inequality. The Cauchy-Schwarz inequality, which states that the sum ∑ n∈Z anbn con- verges absolutely and that |(A,B)| ≤ ‖A‖ ‖B‖, can be deduced in the same way from its finite analogue. In the three examples Rd, Cd, and `2(Z), the vector spaces with their inner products and norms satisfy two important properties: (i) The inner product is strictly positive-definite, that is, ‖X‖ = 0 implies X = 0. (ii) The vector space is complete, which by definition means that every Cauchy sequence in the norm converges to a limit in the vector space. Ibookroot October 20, 2007 1. Mean-square convergence of Fourier series 75 An inner product space with these two properties is called a Hilbert space. We see that Rd and Cd are examples of finite-dimensional Hilbert spaces, while `2(Z) is an example of an infinite-dimensional Hilbert space (see Exercises 1 and 2). If either of the conditions above fail, the space is called a pre-Hilbert space. We now give an important example of a pre-Hilbert space where both conditions (i) and (ii) fail. Example 2. Let R denote the set of complex-valued Riemann integrable functions on [0, 2π] (or equivalently, integrable functions on the circle). This is a vector space over C. Addition is defined pointwise by (f + g)(θ) = f(θ) + g(θ). Naturally, multiplication by a scalar λ ∈ C is given by (λf)(θ) = λ · f(θ). An inner product is defined on this vector space by (1) (f, g) = 1 2π ∫ 2π 0 f(θ)g(θ) dθ. The norm of f is then ‖f‖ = ( 1 2π ∫ 2π 0 |f(θ)|2 dθ )1/2 . One needs to check that the analogue of the Cauchy-Schwarz and tri- angle inequalities hold in this example; that is, |(f, g)| ≤ ‖f‖ ‖g‖ and ‖f + g‖ ≤ ‖f‖+ ‖g‖. While these facts can be obtained as consequences of the corresponding inequalities in the previous examples, the argument is a little elaborate and we prefer to proceed differently. We first observe that 2AB ≤ (A2 + B2) for any two real numbers A and B. If we set A = λ1/2|f(θ)| and B = λ−1/2|g(θ)| with λ > 0, we get

|f(θ)g(θ)| ≤ 1
2
(λ|f(θ)|2 + λ−1|g(θ)|2).

We then integrate this in θ to obtain

|(f, g)| ≤ 1
2π

∫ 2π
0

|f(θ)| |g(θ)| dθ ≤ 1
2
(λ‖f‖2 + λ−1‖g‖2).

Then, put λ = ‖g‖/‖f‖ to get the Cauchy-Schwarz inequality. The tri-
angle inequality is then a simple consequence, as we have seen above.

Ibookroot October 20, 2007

76 Chapter 3. CONVERGENCE OF FOURIER SERIES

Of course, in our choice of λ we must assume that ‖f‖ 6= 0 and ‖g‖ 6= 0,
which leads us to the following observation.

In R, condition (i) for a Hilbert space fails, since ‖f‖ = 0 implies only
that f vanishes at its points of continuity. This is not a very serious
problem since in the appendix we show that an integrable function is
continuous except for a “negligible” set, so that ‖f‖ = 0 implies that f
vanishes except on a set of “measure zero.” One can get around the
difficulty that f is not identically zero by adopting the convention that
such functions are actually the zero function, since for the purpose of
integration, f behaves precisely like the zero function.

A more essential difficulty is that the space R is not complete. One
way to see this is to start with the function

f(θ) =

{
0 for θ = 0,
log(1/θ) for 0 < θ ≤ 2π. Since f is not bounded, it does not belong to the space R. Moreover, the sequence of truncations fn defined by fn(θ) = { 0 for 0 ≤ θ ≤ 1/n, f(θ) for 1/n < θ ≤ 2π can easily be seen to form a Cauchy sequence inR (see Exercise 5). How- ever, this sequence cannot converge to an element in R, since that limit, if it existed, would have to be f ; for another example, see Exercise 7. This and more complicated examples motivate the search for the com- pletion of R, the class of Riemann integrable functions on [0, 2π]. The construction and identification of this completion, the Lebesgue class L2([0, 2π]), represents an important turning point in the development of analysis (somewhat akin to the much earlier completion of the rationals, that is, the passage from Q to R). A further discussion of these fun- damental ideas will be postponed until Book III, where we take up the Lebesgue theory of integration. We now turn to the proof of Theorem 1.1. 1.2 Proof of mean-square convergence Consider the space R of integrable functions on the circle with inner product (f, g) = 1 2π ∫ 2π 0 f(θ)g(θ) dθ Ibookroot October 20, 2007 1. Mean-square convergence of Fourier series 77 and norm ‖f‖ defined by ‖f‖2 = (f, f) = 1 2π ∫ 2π 0 |f(θ)|2 dθ. With this notation, we must prove that ‖f − SN (f)‖ → 0 as N tends to infinity. For each integer n, let en(θ) = einθ, and observe that the family {en}n∈Z is orthonormal; that is, (en, em) = { 1 if n = m 0 if n 6= m. Let f be an integrable function on the circle, and let an denote its Fourier coefficients. An important observation is that these Fourier coefficients are represented by inner products of f with the elements in the orthonor- mal set {en}n∈Z: (f, en) = 1 2π ∫ 2π 0 f(θ)e−inθ dθ = an. In particular, SN (f) = ∑ |n|≤N anen. Then the orthonormal property of the family {en} and the fact that an = (f, en) imply that the difference f −∑|n|≤N anen is orthogonal to en for all |n| ≤ N . Therefore, we must have (2) (f − ∑ |n|≤N anen) ⊥ ∑ |n|≤N bnen for any complex numbers bn. We draw two conclusions from this fact. First, we can apply the Pythagorean theorem to the decomposition f = f − ∑ |n|≤N anen + ∑ |n|≤N anen, where we now choose bn = an, to obtain ‖f‖2 = ‖f − ∑ |n|≤N anen‖2 + ‖ ∑ |n|≤N anen‖2. Since the orthonormal property of the family {en}n∈Z implies that ‖ ∑ |n|≤N anen‖2 = ∑ |n|≤N |an|2, Ibookroot October 20, 2007 78 Chapter 3. CONVERGENCE OF FOURIER SERIES we deduce that (3) ‖f‖2 = ‖f − SN (f)‖2 + ∑ |n|≤N |an|2. The second conclusion we may draw from (2) is the following simple lemma. Lemma 1.2 (Best approximation) If f is integrable on the circle with Fourier coefficients an, then ‖f − SN (f)‖ ≤ ‖f − ∑ |n|≤N cnen‖ for any complex numbers cn. Moreover, equality holds precisely when cn = an for all |n| ≤ N . Proof. This follows immediately by applying the Pythagorean theo- rem to f − ∑ |n|≤N cnen = f − SN (f) + ∑ |n|≤N bnen, where bn = an − cn. This lemma has a clear geometric interpretation. It says that the trigonometric polynomial of degree at most N which is closest to f in the norm ‖ · ‖ is the partial sum SN (f). This geometric property of the partial sums is depicted in Figure 1, where the orthogonal projection of f in the plane spanned by {e−N , . . . , e0, . . . , eN} is simply SN (f). f − SN (f) SN (f) f e0 e−N e1 eN Figure 1. The best approximation lemma We can now give the proof that ‖SN (f)− f‖ → 0 using the best ap- proximation lemma, as well as the important fact that trigonometric polynomials are dense in the space of continuous functions on the circle. Ibookroot October 20, 2007 1. Mean-square convergence of Fourier series 79 Suppose that f is continuous on the circle. Then, given ² > 0, there
exists (by Corollary 5.4 in Chapter 2) a trigonometric polynomial P , say
of degree M , such that

|f(θ)− P (θ)| < ² for all θ. In particular, taking squares and integrating this inequality yields ‖f − P‖ < ², and by the best approximation lemma we conclude that ‖f − SN (f)‖ < ² whenever N ≥ M . This proves Theorem 1.1 when f is continuous. If f is merely integrable, we can no longer approximate f uniformly by trigonometric polynomials. Instead, we apply the approximation Lemma 3.2 in Chapter 2 and choose a continuous function g on the circle which satisfies sup θ∈[0,2π] |g(θ)| ≤ sup θ∈[0,2π] |f(θ)| = B, and ∫ 2π 0 |f(θ)− g(θ)| dθ < ²2. Then we get ‖f − g‖2 = 1 2π ∫ 2π 0 |f(θ)− g(θ)|2 dθ = 1 2π ∫ 2π 0 |f(θ)− g(θ)| |f(θ)− g(θ)| dθ ≤ 2B 2π ∫ 2π 0 |f(θ)− g(θ)| dθ ≤ C²2. Now we may approximate g by a trigonometric polynomial P so that ‖g − P‖ < ². Then ‖f − P‖ < C ′², and we may again conclude by ap- plying the best approximation lemma. This completes the proof that the partial sums of the Fourier series of f converge to f in the mean square norm ‖ · ‖. Note that this result and the relation (3) imply that if an is the nth Fourier coefficient of an integrable function f , then the series ∑∞ n=−∞ |an|2 converges, and in fact we have Parseval’s identity ∞∑ n=−∞ |an|2 = ‖f‖2. Ibookroot October 20, 2007 80 Chapter 3. CONVERGENCE OF FOURIER SERIES This identity provides an important connection between the norms in the two vector spaces `2(Z) and R. We now summarize the results of this section. Theorem 1.3 Let f be an integrable function on the circle with f ∼ ∑∞n=−∞ aneinθ. Then we have: (i) Mean-square convergence of the Fourier series 1 2π ∫ 2π 0 |f(θ)− SN (f)(θ)|2 dθ → 0 as N →∞. (ii) Parseval’s identity ∞∑ n=−∞ |an|2 = 1 2π ∫ 2π 0 |f(θ)|2 dθ. Remark 1. If {en} is any orthonormal family of functions on the circle, and an = (f, en), then we may deduce from the relation (3) that ∞∑ n=−∞ |an|2 ≤ ‖f‖2. This is known as Bessel’s inequality. Equality holds (as in Parseval’s identity) precisely when the family {en} is also a “basis,” in the sense that ‖∑|n|≤N anen − f‖ → 0 as N →∞. Remark 2. We may associate to every integrable function the se- quence {an} formed by its Fourier coefficients. Parseval’s identity guar- antees that {an} ∈ `2(Z). Since `2(Z) is a Hilbert space, the failure of R to be complete, discussed earlier, may be understood as follows: there exist sequences {an}n∈Z such that ∑ n∈Z |an|2 < ∞, yet no Riemann in- tegrable function F has nth Fourier coefficient equal to an for all n. An example is given in Exercise 6. Since the terms of a converging series tend to 0, we deduce from Par- seval’s identity or Bessel’s inequality the following result. Theorem 1.4 (Riemann-Lebesgue lemma) If f is integrable on the circle, then f̂(n) → 0 as |n| → ∞. An equivalent reformulation of this proposition is that if f is integrable on [0, 2π], then ∫ 2π 0 f(θ) sin(Nθ) dθ → 0 as N →∞ Ibookroot October 20, 2007 2. Return to pointwise convergence 81 and ∫ 2π 0 f(θ) cos(Nθ) dθ → 0 as N →∞. To conclude this section, we give a more general version of the Parseval identity which we will use in the next chapter. Lemma 1.5 Suppose F and G are integrable on the circle with F ∼ ∑ ane inθ and G ∼ ∑ bne inθ. Then 1 2π ∫ 2π 0 F (θ)G(θ) dθ = ∞∑ n=−∞ anbn. Recall from the discussion in Example 1 that the series ∑∞ n=−∞ anbn converges absolutely. Proof. The proof follows from Parseval’s identity and the fact that (F, G) = 1 4 [ ‖F + G‖2 − ‖F −G‖2 + i ( ‖F + iG‖2 − ‖F − iG‖2 )] which holds in every Hermitian inner product space. The verification of this fact is left to the reader. 2 Return to pointwise convergence The mean-square convergence theorem does not provide further insight into the problem of pointwise convergence. Indeed, Theorem 1.1 by itself does not guarantee that the Fourier series converges for any θ. Exercise 3 helps to explain this statement. However, if a function is differentiable at a point θ0, then its Fourier series converges at θ0. After proving this result, we give an example of a continuous function with diverging Fourier series at one point. These phenomena are indicative of the intricate nature of the problem of pointwise convergence in the theory of Fourier series. 2.1 A local result Theorem 2.1 Let f be an integrable function on the circle which is dif- ferentiable at a point θ0. Then SN (f)(θ0) → f(θ0) as N tends to infinity. Ibookroot October 20, 2007 82 Chapter 3. CONVERGENCE OF FOURIER SERIES Proof. Define F (t) =    f(θ0 − t)− f(θ0) t if t 6= 0 and |t| < π −f ′(θ0) if t = 0. First, F is bounded near 0 since f is differentiable there. Second, for all small δ the function F is integrable on [−π,−δ] ∪ [δ, π] because f has this property and |t| > δ there. As a consequence of Proposition 1.4 in
the appendix, the function F is integrable on all of [−π, π]. We know
that SN (f)(θ0) = (f ∗DN )(θ0), where DN is the Dirichlet kernel. Since
1
2π

∫
DN = 1, we find that

SN (f)(θ0)− f(θ0) =
1
2π

∫ π
−π

f(θ0 − t)DN (t) dt− f(θ0)

=
1
2π

∫ π
−π

[f(θ0 − t)− f(θ0)]DN (t) dt

=
1
2π

∫ π
−π

F (t)tDN (t) dt.

We recall that

tDN (t) =
t

sin(t/2)
sin((N + 1/2)t),

where the quotient t
sin(t/2)

is continuous in the interval [−π, π]. Since we
can write

sin((N + 1/2)t) = sin(Nt) cos(t/2) + cos(Nt) sin(t/2),

we can apply the Riemann-Lebesgue lemma to the Riemann integrable
functions F (t)t cos(t/2)/ sin(t/2) and F (t)t to finish the proof of the the-
orem.

Observe that the conclusion of the theorem still holds if we only assume
that f satisfies a Lipschitz condition at θ0; that is,

|f(θ)− f(θ0)| ≤ M |θ − θ0|

for some M ≥ 0 and all θ. This is the same as saying that f satisfies a
Hölder condition of order α = 1.

A striking consequence of this theorem is the localization principle of
Riemann. This result states that the convergence of SN (f)(θ0) depends
only on the behavior of f near θ0. This is not clear at first, since forming
the Fourier series requires integrating f over the whole circle.

Ibookroot October 20, 2007

2. Return to pointwise convergence 83

Theorem 2.2 Suppose f and g are two integrable functions defined on
the circle, and for some θ0 there exists an open interval I containing θ0
such that

f(θ) = g(θ) for all θ ∈ I.
Then SN (f)(θ0)− SN (g)(θ0) → 0 as N tends to infinity.

Proof. The function f − g is 0 in I, so it is differentiable at θ0, and
we may apply the previous theorem to conclude the proof.

2.2 A continuous function with diverging Fourier series

We now turn our attention to an example of a continuous periodic func-
tion whose Fourier series diverges at a point. Thus, Theorem 2.1 fails
if the differentiability assumption is replaced by the weaker assumption
of continuity. Our counter-example shows that this hypothesis which
had appeared plausible, is in fact false; moreover, its construction also
illuminates an important principle of the theory.

The principle that is involved here will be referred to as “symmetry-
breaking.”1 The symmetry that we have in mind is the symmetry be-
tween the frequencies einθ and e−inθ which appear in the Fourier expan-
sion of a function. For example, the partial sum operator SN is defined
in a way that reflects this symmetry. Also, the Dirichlet, Fejèr, and
Poisson kernels are symmetric in this sense. When we break the symme-
try, that is, when we split the Fourier series

∑∞
n=−∞ ane

inθ into the two
pieces

∑
n≥0 ane

inθ and
∑

n<0 ane inθ, we introduce new and far-reaching phenomena. We give a simple example. Start with the sawtooth function f which is odd in θ and which equals i(π − θ) when 0 < θ < π. Then, by Exercise 8 in Chapter 2, we know that (4) f(θ) ∼ ∑ n 6=0 einθ n . Consider now the result of breaking the symmetry and the resulting series n=−1∑ n=−∞ einθ n . Then, unlike (4), the above is no longer the Fourier series of a Riemann integrable function. Indeed, suppose it were the Fourier series of an 1We have borrowed this terminology from physics, where it is used in a very different context. Ibookroot October 20, 2007 84 Chapter 3. CONVERGENCE OF FOURIER SERIES integrable function, say f̃ , where in particular f̃ is bounded. Using the Abel means, we then have |Ar(f̃)(0)| = ∞∑ n=1 rn n , which tends to infinity as r tends to 1, because ∑ 1/n diverges. This gives the desired contradiction since |Ar(f̃)(0)| ≤ 1 2π ∫ π −π |f̃(θ)|Pr(θ) dθ ≤ sup θ |f̃(θ)|, where Pr(θ) denotes the Poisson kernel discussed in the previous chapter. The sawtooth function is the object from which we will fashion our counter-example. We proceed as follows. For each N ≥ 1 we define the following two functions on [−π, π], fN (θ) = ∑ 1≤|n|≤N einθ n and f̃N (θ) = ∑ −N≤n≤−1 einθ n . We contend that: (i) |f̃N (0)| ≥ c log N . (ii) fN (θ) is uniformly bounded in N and θ. The first statement is a consequence of the fact that ∑N n=1 1/n ≥ log N , which is easily established (see also Figure 2): N∑ n=1 1 n ≥ N−1∑ n=1 ∫ n+1 n dx x = ∫ N 1 dx x = log N. To prove (ii), we argue in the same spirit as in the proof of Tauber’s theorem, which says that if the series ∑ cn is Abel summable to s and cn = o(1/n), then ∑ cn actually converges to s (see Exercise 14 in Chap- ter 2). In fact, the proof of Tauber’s theorem is quite similar to that of the lemma below. Lemma 2.3 Suppose that the Abel means Ar = ∑∞ n=1 r ncn of the series∑∞ n=1 cn are bounded as r tends to 1 (with r < 1). If cn = O(1/n), then the partial sums SN = ∑N n=1 cn are bounded. Ibookroot October 20, 2007 2. Return to pointwise convergence 85 y = 1 x n n + 1 1 n Figure 2. Comparing a sum with an integral Proof. Let r = 1− 1/N and choose M so that n|cn| ≤ M . We esti- mate the difference SN −Ar = N∑ n=1 (cn − rncn)− ∞∑ n=N+1 rncn as follows: |SN −Ar| ≤ N∑ n=1 |cn|(1− rn) + ∞∑ n=N+1 rn|cn| ≤ M N∑ n=1 (1− r) + M N ∞∑ n=N+1 rn ≤ MN(1− r) + M N 1 1− r = 2M, where we have used the simple observation that 1− rn = (1− r)(1 + r + · · ·+ rn−1) ≤ n(1− r). So we see that if M satisfies both |Ar| ≤ M and n|cn| ≤ M , then |SN | ≤ 3M . We apply the lemma to the series ∑ n 6=0 einθ n , Ibookroot October 20, 2007 86 Chapter 3. CONVERGENCE OF FOURIER SERIES which is the Fourier series of the sawtooth function f used above. Here cn = einθ/n + e−inθ/(−n) for n 6= 0, so clearly cn = O(1/|n|). Finally, the Abel means of this series are Ar(f)(θ) = (f ∗ Pr)(θ). But f is bounded and Pr is a good kernel, so SN (f)(θ) is uniformly bounded in N and θ, as was to be shown. We now come to the heart of the matter. Notice that fN and f̃N are trigonometric polynomials of degree N (that is, they have non-zero Fourier coefficients only when |n| ≤ N). From these, we form trigono- metric polynomials PN and P̃N , now of degrees 3N and 2N − 1, by displacing the frequencies of fN and f̃N by 2N units. In other words, we define PN (θ) = ei(2N)θfN (θ) and P̃N (θ) = ei(2N)θf̃N (θ). So while fN has non-vanishing Fourier coefficients when 0 < |n| ≤ N , now the coef- ficients of PN are non-vanishing for N ≤ n ≤ 3N , n 6= 2N . Moreover, while n = 0 is the center of symmetry of fN , now n = 2N is the center of symmetry of PN . We next consider the partial sums SM . Lemma 2.4 SM (PN ) =    PN if M ≥ 3N , P̃N if M = 2N , 0 if M < N. This is clear from what has been said above and from Figure 3. S2N (e i(2N)θfN )(θ) = ei(2N)θ f̃N (θ) −N N0 2N 3N0 N fN (θ) ei(2N)θfN (θ) = PN (θ) 2N 3N0 N Figure 3. Breaking symmetry in Lemma 2.4 The effect is that when M = 2N , the operator SM breaks the symme- try of PN , but in the other cases covered in the lemma, the action of SM Ibookroot October 20, 2007 3. Exercises 87 is relatively benign, since then the outcome is either PN or 0. Finally, we need to find a convergent series of positive terms ∑ αk and a sequence of integers {Nk} which increases rapidly enough so that: (i) Nk+1 > 3Nk,

(ii) αk log Nk →∞ as k →∞.
We choose (for example) αk = 1/k2 and Nk = 32

which are easily seen
to satisfy the above criteria.

Finally, we can write down our desired function. It is

f(θ) =
∞∑

k=1

αkPNk(θ).

Due to the uniform boundedness of the PN (recall that |PN (θ)| = |fN (θ)|),
the series above converges uniformly to a continuous periodic function.
However, by our lemma we get

|S2Nm(f)(0)| ≥ cαm log Nm + O(1) →∞ as m →∞.

3Nk+1Nk+13Nk−1Nk−1 3NkNk

2Nk

Figure 4. Symmetry broken in the middle interval (Nk, 3Nk)

Indeed, the terms that correspond to Nk with k < m or k > m con-
tribute O(1) or 0, respectively (because the PN ’s are uniformly bounded),
while the term that corresponds to Nm is in absolute value greater than
cαm log Nm because |P̃N (θ)| = |f̃N (θ)| ≥ c log N . So the partial sums of
the Fourier series of f at 0 are not bounded, and we are done since this
proves the divergence of the Fourier series of f at θ = 0. To produce a
function whose series diverges at any other preassigned θ = θ0, it suffices
to consider the function f(θ − θ0).

3 Exercises

1. Show that the first two examples of inner product spaces, namely Rd and Cd,
are complete.

Ibookroot October 20, 2007

88 Chapter 3. CONVERGENCE OF FOURIER SERIES

[Hint: Every Cauchy sequence in R has a limit.]

2. Prove that the vector space `2(Z) is complete.
[Hint: Suppose Ak = {ak,n}n∈Z with k = 1, 2, . . . is a Cauchy sequence. Show
that for each n, {ak,n}∞k=1 is a Cauchy sequence of complex numbers, therefore
it converges to a limit, say bn. By taking partial sums of ‖Ak −Ak′‖ and letting
k′ →∞, show that ‖Ak −B‖ → 0 as k →∞, where B = (. . . , b−1, b0, b1, . . .).
Finally, prove that B ∈ `2(Z).]

3. Construct a sequence of integrable functions {fk} on [0, 2π] such that

lim
k→∞

1
2π

∫ 2π
0

|fk(θ)|2 dθ = 0

but limk→∞ fk(θ) fails to exist for any θ.

[Hint: Choose a sequence of intervals Ik ⊂ [0, 2π] whose lengths tend to 0, and
so that each point belongs to infinitely many of them; then let fk = χIk .]

4. Recall the vector space R of integrable functions, with its inner product and
norm

‖f‖ =
(

1
2π

∫ 2π
0

|f(x)|2 dx
)1/2

(a) Show that there exist non-zero integrable functions f for which ‖f‖ = 0.

(b) However, show that if f ∈ R with ‖f‖ = 0, then f(x) = 0 whenever f is
continuous at x.

5. Let

f(θ) =

{
0 for θ = 0
log(1/θ) for 0 < θ ≤ 2π, and define a sequence of functions in R by fn(θ) = { 0 for 0 ≤ θ ≤ 1/n f(θ) for 1/n < θ ≤ 2π. Prove that {fn}∞n=1 is a Cauchy sequence in R. However, f does not belong to R. Ibookroot October 20, 2007 3. Exercises 89 [Hint: Show that ∫ b a (log θ)2 dθ → 0 if 0 < a < b and b → 0, by using the fact that the derivative of θ(log θ)2 − 2θ log θ + 2θ is equal to (log θ)2.] 6. Consider the sequence {ak}∞k=−∞ defined by ak = { 1/k if k ≥ 1 0 if k ≤ 0. Note that {ak} ∈ `2(Z), but that no Riemann integrable function has kth Fourier coefficient equal to ak for all k. 7. Show that the trigonometric series ∑ n≥2 1 log n sinnx converges for every x, yet it is not the Fourier series of a Riemann integrable function. The same is true for ∑ sin nx nα for 0 < α < 1, but the case 1/2 < α < 1 is more difficult. See Problem 1. 8. Exercise 6 in Chapter 2 dealt with the sums ∑ n odd ≥1 1 n2 and ∞∑ n=1 1 n2 . Similar sums can be derived using the methods of this chapter. (a) Let f be the function defined on [−π, π] by f(θ) = |θ|. Use Parseval’s identity to find the sums of the following two series: ∞∑ n=0 1 (2n + 1)4 and ∞∑ n=1 1 n4 . In fact, they are π4/96 and π4/90, respectively. (b) Consider the 2π-periodic odd function defined on [0, π] by f(θ) = θ(π − θ). Show that ∞∑ n=0 1 (2n + 1)6 = π6 960 and ∞∑ n=1 1 n6 = π6 945 . Remark. The general expression when k is even for ∑∞ n=1 1/nk in terms of πk is given in Problem 4. However, finding a formula for the sum ∑∞ n=1 1/n3, or more generally ∑∞ n=1 1/nk with k odd, is a famous unresolved question. Ibookroot October 20, 2007 90 Chapter 3. CONVERGENCE OF FOURIER SERIES 9. Show that for α not an integer, the Fourier series of π sinπα ei(π−x)α on [0, 2π] is given by ∞∑ n=−∞ einx n + α . Apply Parseval’s formula to show that ∞∑ n=−∞ 1 (n + α)2 = π2 (sin πα)2 . 10. Consider the example of a vibrating string which we analyzed in Chapter 1. The displacement u(x, t) of the string at time t satisfies the wave equation 1 c2 ∂2u ∂t2 = ∂2u ∂x2 , c2 = τ/ρ. The string is subject to the initial conditions u(x, 0) = f(x) and ∂u ∂t (x, 0) = g(x), where we assume that f ∈ C1 and g is continuous. We define the total energy of the string by E(t) = 1 2 ρ ∫ L 0 ( ∂u ∂t )2 dx + 1 2 τ ∫ L 0 ( ∂u ∂x )2 dx. The first term corresponds to the “kinetic energy” of the string (in analogy with (1/2)mv2, the kinetic energy of a particle of mass m and velocity v), and the second term corresponds to its “potential energy.” Show that the total energy of the string is conserved, in the sense that E(t) is constant. Therefore, E(t) = E(0) = 1 2 ρ ∫ L 0 g(x)2 dx + 1 2 τ ∫ L 0 f ′(x)2 dx. 11. The inequalities of Wirtinger and Poincaré establish a relationship between the norm of a function and that of its derivative. Ibookroot October 20, 2007 3. Exercises 91 (a) If f is T -periodic, continuous, and piecewise C1 with ∫ T 0 f(t) dt = 0, show that ∫ T 0 |f(t)|2 dt ≤ T 2 4π2 ∫ T 0 |f ′(t)|2 dt, with equality if and only if f(t) = A sin(2πt/T ) + B cos(2πt/T ). [Hint: Apply Parseval’s identity.] (b) If f is as above and g is just C1 and T -periodic, prove that ∣∣∣∣ ∫ T 0 f(t)g(t) dt ∣∣∣∣ 2 ≤ T 2 4π2 ∫ T 0 |f(t)|2 dt ∫ T 0 |g′(t)|2 dt. (c) For any compact interval [a, b] and any continuously differentiable function f with f(a) = f(b) = 0, show that ∫ b a |f(t)|2 dt ≤ (b− a) 2 π2 ∫ b a |f ′(t)|2 dt. Discuss the case of equality, and prove that the constant (b− a)2/π2 can- not be improved. [Hint: Extend f to be odd with respect to a and periodic of period T = 2(b− a) so that its integral over an interval of length T is 0. Apply part a) to get the inequality, and conclude that equality holds if and only if f(t) = A sin(π t−a b−a )]. 12. Prove that ∫ ∞ 0 sin x x dx = π 2 . [Hint: Start with the fact that the integral of DN (θ) equals 2π, and note that the difference (1/ sin(θ/2))− 2/θ is continuous on [−π, π]. Apply the Riemann- Lebesgue lemma.] 13. Suppose that f is periodic and of class Ck. Show that f̂(n) = o(1/|n|k), that is, |n|kf̂(n) goes to 0 as |n| → ∞. This is an improvement over Exercise 10 in Chapter 2. [Hint: Use the Riemann-Lebesgue lemma.] 14. Prove that the Fourier series of a continuously differentiable function f on the circle is absolutely convergent. [Hint: Use the Cauchy-Schwarz inequality and Parseval’s identity for f ′.] 15. Let f be 2π-periodic and Riemann integrable on [−π, π]. Ibookroot October 20, 2007 92 Chapter 3. CONVERGENCE OF FOURIER SERIES (a) Show that f̂(n) = − 1 2π ∫ π −π f(x + π/n)e−inx dx hence f̂(n) = 1 4π ∫ π −π [f(x)− f(x + π/n)]e−inx dx. (b) Now assume that f satisfies a Hölder condition of order α, namely |f(x + h)− f(x)| ≤ C|h|α for some 0 < α ≤ 1, some C > 0, and all x, h. Use part a) to show that

f̂(n) = O(1/|n|α).

f(x) =
∞∑

k=0

2−kαei2
kx,

where 0 < α < 1, satisfies |f(x + h)− f(x)| ≤ C|h|α, and f̂(N) = 1/Nα whenever N = 2k. [Hint: For (c), break up the sum as follows f(x + h)− f(x) = ∑ 2k≤1/|h|+∑ 2k>1/|h|. To estimate the first sum use the fact that |1− eiθ| ≤ |θ| whenever θ

is small. To estimate the second sum, use the obvious inequality |eix − eiy| ≤ 2.]

16. Let f be a 2π-periodic function which satisfies a Lipschitz condition with
constant K; that is,

|f(x)− f(y)| ≤ K|x− y| for all x, y.

This is simply the Hölder condition with α = 1, so by the previous exercise, we
see that f̂(n) = O(1/|n|). Since the harmonic series

∑
1/n diverges, we cannot

say anything (yet) about the absolute convergence of the Fourier series of f . The
outline below actually proves that the Fourier series of f converges absolutely
and uniformly.

Ibookroot October 20, 2007

3. Exercises 93

(a) For every positive h we define gh(x) = f(x + h)− f(x− h). Prove that

1
2π

∫ 2π
0

|gh(x)|2 dx =
∞∑

n=−∞
4| sin nh|2|f̂(n)|2,

and show that
∞∑

n=−∞
| sinnh|2|f̂(n)|2 ≤ K2h2.

(b) Let p be a positive integer. By choosing h = π/2p+1, show that

∑

2p−1<|n|≤2p |f̂(n)|2 ≤ K 2π2 22p+1 . (c) Estimate ∑ 2p−1<|n|≤2p |f̂(n)|, and conclude that the Fourier series of f converges absolutely, hence uniformly. [Hint: Use the Cauchy-Schwarz inequality to estimate the sum.] (d) In fact, modify the argument slightly to prove Bernstein’s theorem: If f satisfies a Hölder condition of order α > 1/2, then the Fourier series of f
converges absolutely.

17. If f is a bounded monotonic function on [−π, π], then

f̂(n) = O(1/|n|).

[Hint: One may assume that f is increasing, and say |f | ≤ M . First check that
the Fourier coefficients of the characteristic function of [a, b] satisfy O(1/|n|).
Now show that a sum of the form

N∑

k=1

αkχ[ak,ak+1](x)

with −π = a1 < a2 < · · · < aN < aN+1 = π and −M ≤ α1 ≤ · · · ≤ αN ≤ M has Fourier coefficients that are O(1/|n|) uniformly in N . Summing by parts one gets a telescopic sum ∑ (αk+1 − αk) which can be bounded by 2M . Now approximate f by functions of the above type.] 18. Here are a few things we have learned about the decay of Fourier coefficients: (a) if f is of class Ck, then f̂(n) = o(1/|n|k); (b) if f is Lipschitz, then f̂(n) = O(1/|n|); Ibookroot October 20, 2007 94 Chapter 3. CONVERGENCE OF FOURIER SERIES (c) if f is monotonic, then f̂(n) = O(1/|n|); (d) if f is satisfies a Hölder condition with exponent α where 0 < α < 1, then f̂(n) = O(1/|n|α); (e) if f is merely Riemann integrable, then ∑ |f̂(n)|2 < ∞ and therefore f̂(n) = o(1). Nevertheless, show that the Fourier coefficients of a continuous function can tend to 0 arbitrarily slowly by proving that for every sequence of nonnegative real numbers {²n} converging to 0, there exists a continuous function f such that |f̂(n)| ≥ ²n for infinitely many values of n. [Hint: Choose a subsequence {²nk} so that ∑ k ²nk < ∞.] 19. Give another proof that the sum ∑ 0<|n|≤N e inx/n is uniformly bounded in N and x ∈ [−π, π] by using the fact that 1 2i ∑ 0<|n|≤N einx n = N∑ n=1 sin nx n = 1 2 ∫ x 0 (DN (t)− 1) dt, where DN is the Dirichlet kernel. Now use the fact that ∫∞ 0 sin t t dt < ∞ which was proved in Exercise 12. 20. Let f(x) denote the sawtooth function defined by f(x) = (π − x)/2 on the interval (0, 2π) with f(0) = 0 and extended by periodicity to all of R. The Fourier series of f is f(x) ∼ 1 2i ∑ |n|6=0 einx n = ∞∑ n=1 sin nx n , and f has a jump discontinuity at the origin with f(0+) = π 2 , f(0−) = −π 2 , and hence f(0+)− f(0−) = π. Show that max 0 0
0 if n = 0
−1 if n < 0, then show that D̃N (x) = cos(x/2)− cos((N + 1/2)x) sin(x/2) , and ∫ π −π |D̃N (x)| dx ≤ c log N. (b) As a result, if f is Riemann integrable, then (f ∗ D̃N )(0) = O(log N). (c) In the present case, this leads to N∑ n=1 1 nα = O(log N), which is a contradiction. 2. An important fact we have proved is that the family {einx}n∈Z is orthonormal in R and it is also complete, in the sense that the Fourier series of f converges to f in the norm. In this exercise, we consider another family possessing these same properties. On [−1, 1] define Ln(x) = dn dxn (x2 − 1)n, n = 0, 1, 2, . . .. Then Ln is a polynomial of degree n which is called the nth Legendre poly- nomial. Ibookroot October 20, 2007 96 Chapter 3. CONVERGENCE OF FOURIER SERIES (a) Show that if f is indefinitely differentiable on [−1, 1], then ∫ 1 −1 Ln(x)f(x) dx = (−1)n ∫ 1 −1 (x2 − 1)nf (n)(x) dx. In particular, show that Ln is orthogonal to xm whenever m < n. Hence {Ln}∞n=0 is an orthogonal family. (b) Show that ‖Ln‖2 = ∫ 1 −1 |Ln(x)|2 dx = (n!)222n+1 2n + 1 . [Hint: First, note that ‖Ln‖2 = (−1)n(2n)! ∫ 1 −1(x 2 − 1)n dx. Write (x2 − 1)n = (x− 1)n(x + 1)n and integrate by parts n times to calculate this last integral.] (c) Prove that any polynomial of degree n that is orthogonal to 1, x, x2, . . . , xn−1 is a constant multiple of Ln. (d) Let Ln = Ln/‖Ln‖, which are the normalized Legendre polynomials. Prove that {Ln} is the family obtained by applying the “Gram-Schmidt process” to {1, x, . . . , xn, . . .}, and conclude that every Riemann integrable function f on [−1, 1] has a Legendre expansion ∞∑ n=0 〈f,Ln〉Ln which converges to f in the mean-square sense. 3. Let α be a complex number not equal to an integer. (a) Calculate the Fourier series of the 2π-periodic function defined on [−π, π] by f(x) = cos(αx). (b) Prove the following formulas due to Euler: ∞∑ n=1 1 n2 − α2 = 1 2α2 − π 2α tan(απ) . For all u ∈ C− πZ, cot u = 1 u + 2 ∞∑ n=1 u u2 − n2π2 . Ibookroot October 20, 2007 4. Problems 97 (c) Show that for all α ∈ C− Z we have απ sin(απ) = 1 + 2α2 ∞∑ n=1 (−1)n−1 n2 − α2 . (d) For all 0 < α < 1, show that ∫ ∞ 0 tα−1 t + 1 dt = π sin(απ) . [Hint: Split the integral as ∫ 1 0 + ∫∞ 1 and change variables t = 1/u in the second integral. Now both integrals are of the form ∫ 1 0 tγ−1 1 + t dt, 0 < γ < 1, which one can show is equal to ∑∞ k=0 (−1)k k+γ . Use part (c) to conclude the proof.] 4. In this problem, we find the formula for the sum of the series ∞∑ n=1 1 nk where k is any even integer. These sums are expressed in terms of the Bernoulli numbers; the related Bernoulli polynomials are discussed in the next problem. Define the Bernoulli numbers Bn by the formula z ez − 1 = ∞∑ n=0 Bn n! zn. (a) Show that B0 = 1, B1 = −1/2, B2 = 1/6, B3 = 0, B4 = −1/30, and B5 = 0. (b) Show that for n ≥ 1 we have Bn = − 1 n + 1 n−1∑ k=0 ( n + 1 k ) Bk. (c) By writing z ez − 1 = 1− z 2 + ∞∑ n=2 Bn n! zn, Ibookroot October 20, 2007 98 Chapter 3. CONVERGENCE OF FOURIER SERIES show that Bn = 0 if n is odd and > 1. Also prove that

z cot z = 1 +
∞∑

n=1

(−1)n 2
2nB2n
(2n)!

z2n.

(d) The zeta function is defined by

ζ(s) =
∞∑

n=1

1
ns

, for all s > 1.

Deduce from the result in (c), and the expression for the cotangent func-
tion obtained in the previous problem, that

x cot x = 1− 2
∞∑

m=1

ζ(2m)
π2m

x2m.

(e) Conclude that

2ζ(2m) = (−1)m+1 (2π)
2m

(2m)!
B2m.

5. Define the Bernoulli polynomials Bn(x) by the formula

zexz

ez − 1 =
∞∑

n=0

Bn(x)
n!

zn.

(a) The functions Bn(x) are polynomials in x and

Bn(x) =
n∑

k=0

(
n

)
Bkx

n−k.

Show that B0(x) = 1, B1(x) = x− 1/2, B2(x) = x2 − x + 1/6, and
B3(x) = x3 − 32×2 + 12x.

(b) If n ≥ 1, then
Bn(x + 1)−Bn(x) = nxn−1,

and if n ≥ 2, then
Bn(0) = Bn(1) = Bn.

(m + 1)Sm(n) = Bm+1(n)−Bm+1.

Ibookroot October 20, 2007

4. Problems 99

(d) Prove that the Bernoulli polynomials are the only polynomials that satisfy

(i) B0(x) = 1,

(ii) B′n(x) = nBn−1(x) for n ≥ 1,
(iii)

∫ 1
0

Bn(x) dx = 0 for n ≥ 1, and show that from (b) one obtains
∫ x+1

Bn(t) dt = x
n.

(e) Calculate the Fourier series of B1(x) to conclude that for 0 < x < 1 we have B1(x) = x− 1/2 = −1 π ∞∑ k=1 sin(2πkx) k . Integrate and conclude that B2n(x)= (−1)n+1 2(2n)! (2π)2n ∞∑ k=1 cos(2πkx) k2n , B2n+1(x)= (−1)n+1 2(2n + 1)! (2π)2n+1 ∞∑ k=1 sin(2πkx) k2n+1 . Finally, show that for 0 < x < 1, Bn(x) = − n! (2πi)n ∑ k 6=0 e2πikx kn . We observe that the Bernoulli polynomials are, up to normalization, successive integrals of the sawtooth function. Ibookroot October 20, 2007 4 Some Applications of Fourier Series Fourier series and analogous expansions intervene very naturally in the general theory of curves and surfaces. In effect, this theory, conceived from the point of view of analysis, deals obviously with the study of arbitrary functions. I was thus led to use Fourier series in sev- eral questions of geometry, and I have obtained in this direction a number of results which will be presented in this work. One notes that my considerations form only a beginning of a principal series of researches, which would without doubt give many new results. A. Hurwitz, 1902 In the previous chapters we introduced some basic facts about Fourier analysis, motivated by problems that arose in physics. The motion of a string and the diffusion of heat were two instances that led naturally to the expansion of a function in terms of a Fourier series. We propose next to give the reader a flavor of the broader impact of Fourier analysis, and illustrate how these ideas reach out to other areas of mathematics. In particular, consider the following three problems: I. Among all simple closed curves of length ` in the plane R2, which one encloses the largest area? II. Given an irrational number γ, what can be said about the distri- bution of the fractional parts of the sequence of numbers nγ, for n = 1, 2, 3, . . .? III. Does there exist a continuous function that is nowhere differen- tiable? The first problem is clearly geometric in nature, and at first sight, would seem to have little to do with Fourier series. The second question lies on the border between number theory and the study of dynamical systems, and gives us the simplest example of the idea of “ergodicity.” The third problem, while analytic in nature, resisted many attempts before the Ibookroot October 20, 2007 1. The isoperimetric inequality 101 solution was finally discovered. It is remarkable that all three questions can be resolved quite simply and directly by the use of Fourier series. In the last section of this chapter, we return to a problem that provided our initial motivation. We consider the time-dependent heat equation on the circle. Here our investigation will lead us to the important but enigmatic heat kernel for the circle. However, the mysteries surrounding its basic properties will not be fully understood until we can apply the Poisson summation formula, which we will do in the next chapter. 1 The isoperimetric inequality Let Γ denote a closed curve in the plane which does not intersect itself. Also, let ` denote the length of Γ, and A the area of the bounded region in R2 enclosed by Γ. The problem now is to determine for a given ` the curve Γ which maximizes A (if any such curve exists). Γ small A Γ large A Figure 1. The isoperimetric problem A little experimentation and reflection suggests that the solution should be a circle. This conclusion can be reached by the following heuristic con- siderations. The curve can be thought of as a closed piece of string lying flat on a table. If the region enclosed by the string is not convex (for ex- ample), one can deform part of the string and increase the area enclosed by it. Also, playing with some simple examples, one can convince oneself that the “flatter” the curve is in some portion, the less efficient it is in enclosing area. Therefore we want to maximize the “roundness” of the curve at each point. Although the circle is the correct guess, making the above ideas precise is a difficult matter. The key idea in the solution we give to the isoperimetric problem con- sists of an application of Parseval’s identity for Fourier series. However, before we can attempt a solution to this problem, we must define the Ibookroot October 20, 2007 102 Chapter 4. SOME APPLICATIONS OF FOURIER SERIES notion of a simple closed curve, its length, and what we mean by the area of the region enclosed by it. Curves, length and area A parametrized curve γ is a mapping γ : [a, b] → R2. The image of γ is a set of points in the plane which we call a curve and denote by Γ. The curve Γ is simple if it does not intersect itself, and closed if its two end-points coincide. In terms of the parametrization above, these two conditions translate into γ(s1) 6= γ(s2) unless s1 = a and s2 = b, in which case γ(a) = γ(b). We may extend γ to a periodic function on R of period b− a, and think of γ as a function on the circle. We also always impose some smoothness on our curves by assuming that γ is of class C1, and that its derivative γ′ satisfies γ′(s) 6= 0. Altogether, these conditions guarantee that Γ has a well-defined tangent at each point, which varies continuously as the point on the curve varies. More- over, the parametrization γ induces an orientation on Γ as the parameter s travels from a to b. Any C1 bijective mapping s : [c, d] → [a, b] gives rise to another parametrization of Γ by the formula η(t) = γ(s(t)). Clearly, the conditions that Γ be closed and simple are independent of the chosen parametrization. Also, we say that the two parametrizations γ and η are equivalent if s′(t) > 0 for all t; this means that η and γ
induce the same orientation on the curve Γ. If, however, s′(t) < 0, then η reverses the orientation. If Γ is parametrized by γ(s) = (x(s), y(s)), then the length of the curve Γ is defined by ` = ∫ b a |γ′(s)| ds = ∫ b a ( x′(s)2 + y′(s)2 )1/2 ds. The length of Γ is a notion intrinsic to the curve, and does not depend on its parametrization. To see that this is indeed the case, suppose that γ(s(t)) = η(t). Then, the change of variables formula and the chain rule imply that ∫ b a |γ′(s)| ds = ∫ d c |γ′(s(t))| |s′(t)| dt = ∫ d c |η′(t)| dt, Ibookroot October 20, 2007 1. The isoperimetric inequality 103 as desired. In the proof of the theorem below, we shall use a special type of parametrization for Γ. We say that γ is a parametrization by arc- length if |γ′(s)| = 1 for all s. This means that γ(s) travels at a constant speed, and as a consequence, the length of Γ is precisely b− a. Therefore, after a possible additional translation, a parametrization by arc-length will be defined on [0, `]. Any curve admits a parametrization by arc- length (Exercise 1). We now turn to the isoperimetric problem. The attempt to give a precise formulation of the area A of the region enclosed by a simple closed curve Γ raises a number of tricky questions. In a variety of simple situations, it is evident that the area is given by the following familiar formula of the calculus: A = 1 2 ∣∣∣∣ ∫ Γ (x dy − y dx) ∣∣∣∣(1) = 1 2 ∣∣∣∣∣ ∫ b a x(s)y′(s)− y(s)x′(s) ds ∣∣∣∣∣ ; see, for example, Exercise 3. Thus in formulating our result we shall adopt the easy expedient of taking (1) as our definition of area. This device allows us to give a quick and neat proof of the isoperimetric in- equality. A listing of issues this simplification leaves unresolved can be found after the proof of the theorem. Statement and proof of the isoperimetric inequality Theorem 1.1 Suppose that Γ is a simple closed curve in R2 of length `, and let A denote the area of the region enclosed by this curve. Then A ≤ ` 2 4π , with equality if and only if Γ is a circle. The first observation is that we can rescale the problem. This means that we can change the units of measurement by a factor of δ > 0 as
follows. Consider the mapping of the plane R2 to itself, which sends the
point (x, y) to (δx, δy). A look at the formula defining the length of a
curve shows that if Γ is of length `, then its image under this mapping
has length δ`. So this operation magnifies or contracts lengths by a
factor of δ depending on whether δ ≥ 1 or δ ≤ 1. Similarly, we see that

Ibookroot October 20, 2007

104 Chapter 4. SOME APPLICATIONS OF FOURIER SERIES

the mapping magnifies (or contracts) areas by a factor of δ2. By taking
δ = 2π/`, we see that it suffices to prove that if ` = 2π then A ≤ π, with
equality only if Γ is a circle.

Let γ : [0, 2π] → R2 with γ(s) = (x(s), y(s)) be a parametrization by
arc-length of the curve Γ, that is, x′(s)2 + y′(s)2 = 1 for all s ∈ [0, 2π].
This implies that

(2)
1
2π

∫ 2π
0

(x′(s)2 + y′(s)2) ds = 1.

Since the curve is closed, the functions x(s) and y(s) are 2π-periodic, so
we may consider their Fourier series

x(s) ∼
∑

ane
ins and y(s) ∼

∑
bne

ins.

Then, as we remarked in the later part of Section 2 of Chapter 2, we
have

x′(s) ∼
∑

anine
ins and y′(s) ∼

∑
bnine

ins.

Parseval’s identity applied to (2) gives

(3)
∞∑

n=−∞
|n|2

(
|an|2 + |bn|2

)
= 1.

We now apply the bilinear form of Parseval’s identity (Lemma 1.5, Chap-
ter 3) to the integral defining A. Since x(s) and y(s) are real-valued, we
have an = a−n and bn = b−n, so we find that

A = 1
2

∣∣∣∣
∫ 2π
0

x(s)y′(s)− y(s)x′(s) ds
∣∣∣∣ = π

∣∣∣∣∣
∞∑

n=−∞
n

(
anbn − bnan

)∣∣∣∣∣ .

We observe next that

(4) |anbn − bnan| ≤ 2 |an| |bn| ≤ |an|2 + |bn|2,

and since |n| ≤ |n|2, we may use (3) to get

A ≤ π
∞∑

n=−∞
|n|2

(
|an|2 + |bn|2

)

≤ π,

Ibookroot October 20, 2007

2. Weyl’s equidistribution theorem 105

as desired.
When A = π, we see from the above argument that

x(s) = a−1e
−is + a0 + a1e

is and y(s) = b−1e
−is + b0 + b1e

because |n| < |n|2 as soon as |n| ≥ 2. We know that x(s) and y(s) are real-valued, so a−1 = a1 and b−1 = b1. The identity (3) implies that 2 ( |a1|2 + |b1|2 ) = 1, and since we have equality in (4) we must have |a1| = |b1| = 1/2. We write a1 = 1 2 eiα and b1 = 1 2 eiβ . The fact that 1 = 2|a1b1 − a1b1| implies that | sin(α− β)| = 1, hence α− β = kπ/2 where k is an odd integer. From this we find that x(s) = a0 + cos(α + s) and y(s) = b0 ± sin(α + s), where the sign in y(s) depends on the parity of (k − 1)/2. In any case, we see that Γ is a circle, for which the case of equality obviously holds, and the proof of the theorem is complete. The solution given above (due to Hurwitz in 1901) is indeed very ele- gant, but clearly leaves some important issues unanswered. We list these as follows. Suppose Γ is a simple closed curve. (i) How is the “region enclosed by Γ” defined? (ii) What is the geometric definition of the “area” of this region? Does this definition accord with (1)? (iii) Can these results be extended to the most general class of sim- ple closed curves relevant to the problem—those curves which are “rectifiable”—that is, those to which we can ascribe a finite length? It turns out that the clarifications of the problems raised are connected to a number of other significant ideas in analysis. We shall return to these questions in succeeding books of this series. 2 Weyl’s equidistribution theorem We now apply ideas coming from Fourier series to a problem dealing with properties of irrational numbers. We begin with a brief discussion of congruences, a concept needed to understand our main theorem. Ibookroot October 20, 2007 106 Chapter 4. SOME APPLICATIONS OF FOURIER SERIES The reals modulo the integers If x is a real number, we let [x] denote the greatest integer less than or equal to x and call the quantity [x] the integer part of x. The fractional part of x is then defined by 〈x〉 = x− [x]. In particular, 〈x〉 ∈ [0, 1) for every x ∈ R. For example, the integer and fractional parts of 2.7 are 2 and 0.7, respectively, while the integer and fractional parts of −3.4 are −4 and 0.6, respectively. We may define a relation on R by saying that the two numbers x and y are equivalent, or congruent, if x− y ∈ Z. We then write x = y mod Z or x = y mod 1. This means that we identify two real numbers if they differ by an integer. Observe that any real number x is congruent to a unique number in [0, 1) which is precisely 〈x〉, the fractional part of x. In effect, reducing a real number modulo Z means looking only at its fractional part and disregarding its integer part. Now start with a real number γ 6= 0 and look at the sequence γ, 2γ, 3γ, . . . . An intriguing question is to ask what happens to this sequence if we reduce it modulo Z, that is, if we look at the sequence of fractional parts 〈γ〉, 〈2γ〉, 〈3γ〉, . . . . Here are some simple observations: (i) If γ is rational, then only finitely many numbers appearing in 〈nγ〉 are distinct. (ii) If γ is irrational, then the numbers 〈nγ〉 are all distinct. Indeed, for part (i), note that if γ = p/q, the first q terms in the sequence are 〈p/q〉, 〈2p/q〉, . . . , 〈(q − 1)p/q〉, 〈qp/q〉 = 0. The sequence then begins to repeat itself, since 〈(q + 1)p/q〉 = 〈1 + p/q〉 = 〈p/q〉, and so on. However, see Exercise 6 for a more refined result. Also, for part (ii) assume that not all numbers are distinct. We there- fore have 〈n1γ〉 = 〈n2γ〉 for some n1 6= n2; then n1γ − n2γ ∈ Z, hence γ is rational, a contradiction. Ibookroot October 20, 2007 2. Weyl’s equidistribution theorem 107 In fact, it can be shown that if γ is irrational, then 〈nγ〉 is dense in the interval [0, 1), a result originally proved by Kronecker. In other words, the sequence 〈nγ〉 hits every sub-interval of [0, 1) (and hence it does so infinitely many times). We will obtain this fact as a corollary of a deeper theorem dealing with the uniform distribution of the sequence 〈nγ〉. A sequence of numbers ξ1, ξ2, . . . , ξn, . . . in [0, 1) is said to be equidis- tributed if for every interval (a, b) ⊂ [0, 1), lim N→∞ #{1 ≤ n ≤ N : ξn ∈ (a, b)} N = b− a where #A denotes the cardinality of the finite set A. This means that for large N , the proportion of numbers ξn in (a, b) with n ≤ N is equal to the ratio of the length of the interval (a, b) to the length of the interval [0, 1). In other words, the sequence ξn sweeps out the whole interval evenly, and every sub-interval gets its fair share. Clearly, the ordering of the sequence is very important, as the next two examples illustrate. Example 1. The sequence 0, 1 2 , 0, 1 3 , 2 3 , 0, 1 4 , 2 4 , 3 4 , 0, 1 5 , 2 5 , · · · appears to be equidistributed since it passes over the interval [0, 1) very evenly. Of course this is not a proof, and the reader is invited to give one. For a somewhat related example, see Exercise 8 with σ = 1/2. Example 2. Let {rn}∞n=1 be any enumeration of the rationals in [0, 1). Then the sequence defined by ξn = { rn/2 if n is even, 0 if n is odd, is not equidistributed since “half” of the sequence is at 0. Nevertheless, this sequence is obviously dense. We now arrive at the main theorem of this section. Theorem 2.1 If γ is irrational, then the sequence of fractional parts 〈γ〉, 〈2γ〉, 〈3γ〉, . . . is equidistributed in [0, 1). In particular, 〈nγ〉 is dense in [0, 1), and we get Kronecker’s theo- rem as a corollary. In Figure 2 we illustrate the set of points 〈γ〉, 〈2γ〉, 〈3γ〉, . . . , 〈Nγ〉 for three different values of N when γ = √ 2. Ibookroot October 20, 2007 108 Chapter 4. SOME APPLICATIONS OF FOURIER SERIES N = 10 N = 30 N = 80 10 1 1 0 0 Figure 2. The sequence 〈γ〉, 〈2γ〉, 〈3γ〉, . . . , 〈Nγ〉 when γ = √ 2 Fix (a, b) ⊂ [0, 1) and let χ(a,b)(x) denote the characteristic function of the interval (a, b), that is, the function equal to 1 in (a, b) and 0 in [0, 1)− (a, b). We may extend this function to R by periodicity (pe- riod 1), and still denote this extension by χ(a,b)(x). Then, as a conse- quence of the definitions, we find that #{1 ≤ n ≤ N : 〈nγ〉 ∈ (a, b)} = N∑ n=1 χ(a,b)(nγ), and the theorem can be reformulated as the statement that 1 N N∑ n=1 χ(a,b)(nγ) → ∫ 1 0 χ(a,b)(x) dx, as N →∞. This step removes the difficulty of working with fractional parts and reduces the number theory to analysis. The heart of the matter lies in the following result. Lemma 2.2 If f is continuous and periodic of period 1, and γ is irra- tional, then 1 N N∑ n=1 f(nγ) → ∫ 1 0 f(x) dx as N →∞. The proof of the lemma is divided into three steps. Step 1. We first check the validity of the limit in the case when f is one of the exponentials 1, e2πix, . . . , e2πikx, . . . . If f = 1, the limit Ibookroot October 20, 2007 2. Weyl’s equidistribution theorem 109 surely holds. If f = e2πikx with k 6= 0, then the integral is 0. Since γ is irrational, we have e2πikγ 6= 1, therefore 1 N N∑ n=1 f(nγ) = e2πikγ N 1− e2πikNγ 1− e2πikγ , which goes to 0 as N →∞. Step 2. It is clear that if f and g satisfy the lemma, then so does Af + Bg for any A,B ∈ C. Therefore, the first step implies that the lemma is true for all trigonometric polynomials. Step 3. Let ² > 0. If f is any continuous periodic function of period 1,
choose a trigonometric polynomial P so that supx∈R |f(x)− P (x)| < ²/3 (this is possible by Corollary 5.4 in Chapter 2). Then, by step 1, for all large N we have ∣∣∣∣∣ 1 N N∑ n=1 P (nγ)− ∫ 1 0 P (x) dx ∣∣∣∣∣ < ²/3. Therefore ∣∣∣∣∣ 1 N N∑ n=1 f(nγ)− ∫ 1 0 f(x) dx ∣∣∣∣∣ ≤ 1 N N∑ n=1 |f(nγ)− P (nγ)|+ + ∣∣∣∣∣ 1 N N∑ n=1 P (nγ)− ∫ 1 0 P (x) dx ∣∣∣∣∣ + + ∫ 1 0 |P (x)− f(x)| dx < ², and the lemma is proved. Now we can finish the proof of the theorem. Choose two continuous periodic functions f+² and f − ² of period 1 which approximate χ(a,b)(x) on [0, 1) from above and below; both f+² and f − ² are bounded by 1 and agree with χ(a,b)(x) except in intervals of total length 2² (see Figure 3). In particular, f−² (x) ≤ χ(a,b)(x) ≤ f+² (x), and b− a− 2² ≤ ∫ 1 0 f−² (x) dx and ∫ 1 0 f+² (x) dx ≤ b− a + 2². If SN = 1N ∑N n=1 χ(a,b)(nγ), then we get 1 N N∑ n=1 f−² (nγ) ≤ SN ≤ 1 N N∑ n=1 f+² (nγ). Ibookroot October 20, 2007 110 Chapter 4. SOME APPLICATIONS OF FOURIER SERIES f−² f+² a0 b 1b + ²b− ²a + ²a− ² Figure 3. Approximations of χ(a,b)(x) Therefore b− a− 2² ≤ lim inf N→∞ SN and lim sup N→∞ SN ≤ b− a + 2². Since this is true for every ² > 0, the limit limN→∞ SN exists and must
equal b− a. This completes the proof of the equidistribution theorem.

This theorem has the following consequence.

Corollary 2.3 The conclusion of Lemma 2.2 holds for every function
f which is Riemann integrable in [0, 1], and periodic of period 1.

Proof. Assume f is real-valued, and consider a partition of the
interval [0, 1], say 0 = x0 < x1 < · · · < xN = 1. Next, define fU (x) = supxj−1≤y≤xj f(y) if x ∈ [xj−1, xj) and fL(x) = infxj−1≤y≤xj f(y) for x ∈ (xj−1, xj). Then clearly fL ≤ f ≤ fU and ∫ 1 0 fL(x) dx ≤ ∫ 1 0 f(x) dx ≤ ∫ 1 0 fU (x) dx. Moreover, by making the partition sufficiently fine we can guarantee that for a given ² > 0,

∫ 1
0

fU (x) dx−
∫ 1
0

fL(x) dx ≤ ².

However,

1
N

N∑

n=1

fL(nγ) →
∫ 1
0

fL(x) dx

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2. Weyl’s equidistribution theorem 111

by the theorem, because each fL is a finite linear combination of charac-
teristic functions of intervals; similarly we have

1
N

N∑

n=1

fU (nγ) →
∫ 1
0

fU (x) dx.

From these two assertions we can conclude the proof of the corollary by
using the previous approximation argument.

There is an interesting interpretation of the lemma and its corollary,
in terms of a simple dynamical system. In this example, the underlying
space is the circle parametrized by the angle θ. We also consider a
mapping of this space to itself: here, we choose a rotation ρ of the circle
by the angle 2πγ, that is, the transformation ρ : θ 7→ θ + 2πγ.

We want next to consider how this space, with its underlying action
ρ, evolves in time. In other words, we wish to consider the iterates of ρ,
namely ρ, ρ2, ρ3, . . ., ρn where

ρn = ρ ◦ ρ ◦ · · · ◦ ρ : θ 7→ θ + 2πnγ,

and where we think of the action ρn taking place at the time t = n.
To each Riemann integrable function f on the circle, we can also asso-

ciate the corresponding effects of the rotation ρ, and obtain a sequence
of functions

f(θ), f(ρ(θ)), f(ρ2(θ)), . . . , f(ρn(θ)), . . .

with f(ρn(θ)) = f(θ + 2πnγ). In this special context, the ergodicity of
this system is then the statement that the “time average”

lim
N→∞

1
N

N∑

n=1

f(ρn(θ))

exists for each θ and equals the “space average”

1
2π

∫ 2π
0

f(θ) dθ,

whenever γ is irrational. In fact, this assertion is merely a rephrasing of
Corollary 2.3, once we make the change of variables θ = 2πx.

Returning to the problem of equidistributed sequences, we observe that
the proof of Theorem 2.1 gives the following characterization.

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112 Chapter 4. SOME APPLICATIONS OF FOURIER SERIES

Weyl’s criterion. A sequence of real numbers ξ1, ξ2 . . . in
[0, 1) is equidistributed if and only if for all integers k 6= 0 one
has

1
N

N∑

n=1

e2πikξn → 0, as N →∞.

One direction of this theorem was in effect proved above, and the con-
verse can be found in Exercise 7. In particular, we find that to understand
the equidistributive properties of a sequence ξn, it suffices to estimate
the size of the corresponding “exponential sum”

∑N
n=1 e

2πikξn . For ex-
ample, it can be shown using Weyl’s criterion that the sequence 〈n2γ〉
is equidistributed whenever γ is irrational. This and other examples can
be found in Exercises 8, and 9; also Problems 2, and 3.

As a last remark, we mention a nice geometric interpretation of the
distribution properties of 〈nγ〉. Suppose that the sides of a square are
reflecting mirrors and that a ray of light leaves a point inside the square.
What kind of path will the light trace out?

Figure 4. Reflection of a ray of light in a square

To solve this problem, the main idea is to consider the grid of the
plane formed by successively reflecting the initial square across its sides.
With an appropriate choice of axis, the path traced by the light in the
square corresponds to the straight line P + (t, γt) in the plane. As a
result, the reader may observe that the path will be either closed and
periodic, or it will be dense in the square. The first of these situations

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3. A continuous but nowhere differentiable function 113

will happen if and only if the slope γ of the initial direction of the light
(determined with respect to one of the sides of the square) is rational.
In the second situation, when γ is irrational, the density follows from
Kronecker’s theorem. What stronger conclusion does one get from the
equidistribution theorem?

3 A continuous but nowhere differentiable function

There are many obvious examples of continuous functions that are not
differentiable at one point, say f(x) = |x|. It is almost as easy to con-
struct a continuous function that is not differentiable at any given finite
set of points, or even at appropriate sets containing countably many
points. A more subtle problem is whether there exists a continuous
function that is nowhere differentiable. In 1861, Riemann guessed that
the function defined by

(5) R(x) =
∞∑

n=1

sin(n2x)
n2

was nowhere differentiable. He was led to consider this function because
of its close connection to the theta function which will be introduced in
Chapter 5. Riemann never gave a proof, but mentioned this example in
one of his lectures. This triggered the interest of Weierstrass who, in an
attempt to find a proof, came across the first example of a continuous but
nowhere differentiable function. Say 0 < b < 1 and a is an integer > 1.
In 1872 he proved that if ab > 1 + 3π/2, then the function

W (x) =
∞∑

n=1

bn cos(anx)

is nowhere differentiable.
But the story is not complete without a final word about Riemann’s

original function. In 1916 Hardy showed that R is not differentiable at
all irrational multiples of π, and also at certain rational multiples of π.
However, it was not until much later, in 1969, that Gerver completely
settled the problem, first by proving that the function R is actually
differentiable at all the rational multiples of π of the form πp/q with p
and q odd integers, and then by showing that R is not differentiable in
all of the remaining cases.

In this section, we prove the following theorem.

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114 Chapter 4. SOME APPLICATIONS OF FOURIER SERIES

Theorem 3.1 If 0 < α < 1, then the function fα(x) = f(x) = ∞∑ n=0 2−nαei2 nx is continuous but nowhere differentiable. The continuity is clear because of the absolute convergence of the se- ries. The crucial property of f which we need is that it has many van- ishing Fourier coefficients. A Fourier series that skips many terms, like the one given above, or like W (x), is called a lacunary Fourier series. The proof of the theorem is really the story of three methods of sum- ming a Fourier series. First, there is the ordinary convergence in terms of the partial sums SN (g) = g ∗DN . Next, there is Cesàro summabil- ity σN (g) = g ∗ FN , with FN the Fejér kernel. A third method, clearly connected with the second, involves the delayed means defined by 4N (g) = 2σ2N (g)− σN (g). Hence 4N (g) = g ∗ [2F2N − FN ]. These methods can best be visualized as in Figure 5. Suppose g(x) ∼ ∑ aneinx. Then: • SN arises by multiplying the term aneinx by 1 if |n| ≤ N , and 0 if |n| > N .

• σN arises by multiplying aneinx by 1− |n|/N for |n| ≤ N and 0 for
|n| > N .

• 4N arises by multiplying aneinx by 1 if |n| ≤ N , by 2(1− |n|/(2N))
for N ≤ |n| ≤ 2N , and 0 for |n| > 2N .

For example, note that

σN (g)(x) =
S0(g)(x) + S1(g)(x) + · · ·+ SN−1(g)(x)

=
1
N

N−1∑

`=0

∑

|k|≤`
ake

ikx

=
1
N

∑

|n|≤N
(N − |n|)aneinx

=
∑

|n|≤N

(
1− |n|

)
ane

inx.

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3. A continuous but nowhere differentiable function 115

Partial sums

−N

−N 0 N

−2N

∆N (g)(x) = 2σ2N (g)(x)− σN (g)(x)
Delayed means

Cesàro means

σN (g)(x) =
∑
|n|≤N

(
1− |n|

)
ane

inx

SN (g)(x) =
∑
|n|≤N ane

inx

Figure 5. Three summation methods

Ibookroot October 20, 2007

116 Chapter 4. SOME APPLICATIONS OF FOURIER SERIES

The proof of the other assertion is similar.

The delayed means have two important features. On the one hand,
their properties are closely related to the (good) features of the Cesàro
means. On the other hand, for series that have lacunary properties like
those of f , the delayed means are essentially equal to the partial sums.
In particular, note that for our function f = fα

(6) SN (f) = 4N ′(f),

where N ′ is the largest integer of the form 2k with N ′ ≤ N . This is clear
by examining Figure 5 and the definition of f .

We turn to the proof of the theorem proper and argue by contradiction;
that is, we assume that f ′(x0) exists for some x0.

Lemma 3.2 Let g be any continuous function that is differentiable at
x0. Then, the Cesàro means satisfy σN (g)′(x0) = O(log N), therefore

4N (g)′(x0) = O(log N).

Proof. First we have

σN (g)
′(x0) =

∫ π
−π

F ′N (x0 − t)g(t) dt =
∫ π
−π

F ′N (t)g(x0 − t) dt,

where FN is the Fejér kernel. Since FN is periodic, we have
∫ π
−πF

′
N (t)dt=0

and this implies that

σN (g)
′(x0) =

∫ π
−π

F ′N (t)[g(x0 − t)− g(x0)] dt.

From the assumption that g is differentiable at x0 we get

|σN (g)′(x0)| ≤ C
∫ π
−π
|F ′N (t)| |t| dt.

Now observe that F ′N satisfies the two estimates

|F ′N (t)| ≤ AN2 and |F ′N (t)| ≤
A

|t|2 .

For the first inequality, recall that FN is a trigonometric polynomial
of degree N whose coefficients are bounded by 1. Therefore, F ′N is a
trigonometric polynomial of degree N whose coefficients are no bigger
than N . Hence |F ′(t)| ≤ (2N + 1)N ≤ AN2.

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3. A continuous but nowhere differentiable function 117

For the second inequality, we recall that

FN (t) =
1
N

sin2(Nt/2)
sin2(t/2)

Differentiating this expression, we get two terms:

sin(Nt/2) cos(Nt/2)
sin2(t/2)

− 1
N

cos(t/2) sin2(Nt/2)
sin3(t/2)

If we then use the facts that | sin(Nt/2)| ≤ CN |t| and | sin(t/2)| ≥ c|t|
(for |t| ≤ π), we get the desired estimates for F ′N (t).

Using all of these estimates we find that

|σN (g)′(x0)| ≤ C
∫

|t|≥1/N
|F ′N (t)| |t| dt + C

∫

|t|≤1/N
|F ′N (t)| |t| dt

≤ CA
∫

|t|≥1/N

|t| + CAN
∫

|t|≤1/N
dt

= O(log N) + O(1)
= O(log N).

The proof of the lemma is complete once we invoke the definition of
4N (g).

Lemma 3.3 If 2N = 2n, then

42N (f)−4N (f) = 2−nαei2
nx.

This follows from our previous observation (6) because 42N (f) =
S2N (f) and 4N (f) = SN (f).

Now, by the first lemma we have

42N (f)′(x0)−4N (f)′(x0) = O(log N),

and the second lemma also implies

|42N (f)′(x0)−4N (f)′(x0)| = 2n(1−α) ≥ cN1−α.

This is the desired contradiction since N1−α grows faster than log N .

A few additional remarks about our function fα(x) =
∑∞

n=0 2
−nαei2

are in order.

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118 Chapter 4. SOME APPLICATIONS OF FOURIER SERIES

This function is complex-valued as opposed to the examples R and W
above, and so the nowhere differentiability of fα does not imply the same
property for its real and imaginary parts. However, a small modification
of our proof shows that, in fact, the real part of fα,

∞∑

n=0

2−nα cos 2nx,

as well as its imaginary part, are both nowhere differentiable. To see
this, observe first that by the same proof, Lemma 3.2 has the following
generalization: if g is a continuous function which is differentiable at x0,
then

4N (g)′(x0 + h) = O(log N) whenever |h| ≤ c/N .

We then proceed with F (x) =
∑∞

n=0 2
−nα cos 2nx, noting as above that

42N (F )−4N (F ) = 2−nα cos 2nx; as a result, assuming that F is differ-
entiable at x0, we get that

|2n(1−α) sin(2n(x0 + h))| = O(log N)

when 2N = 2n, and |h| ≤ c/N . To get a contradiction, we need only
choose h so that | sin(2n(x0 + h))| = 1; this is accomplished by setting
δ equal to the distance from 2nx0 to the nearest number of the form
(k + 1/2)π, k ∈ Z (so δ ≤ π/2), and taking h = ±δ/2n.

Clearly, when α > 1 the function fα is continuously differentiable since
the series can be differentiated term by term. Finally, the nowhere dif-
ferentiability we have proved for α < 1 actually extends to α = 1 by a suitable refinement of the argument (see Problem 8 in Chapter 5). In fact, using these more elaborate methods one can also show that the Weierstrass function W is nowhere differentiable if ab ≥ 1. 4 The heat equation on the circle As a final illustration, we return to the original problem of heat diffusion considered by Fourier. Suppose we are given an initial temperature distribution at t = 0 on a ring and that we are asked to describe the temperature at points on the ring at times t > 0.

The ring is modeled by the unit circle. A point on this circle is de-
scribed by its angle θ = 2πx, where the variable x lies between 0 and 1.
If u(x, t) denotes the temperature at time t of a point described by the

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4. The heat equation on the circle 119

angle θ, then considerations similar to the ones given in Chapter 1 show
that u satisfies the differential equation

(7)
∂u

∂t
= c

∂2u

∂x2
.

The constant c is a positive physical constant which depends on the
material of which the ring is made (see Section 2.1 in Chapter 1). After
rescaling the time variable, we may assume that c = 1. If f is our initial
data, we impose the condition

u(x, 0) = f(x).

To solve the problem, we separate variables and look for special solutions
of the form

u(x, t) = A(x)B(t).

Then inserting this expression for u into the heat equation we get

B′(t)
B(t)

=
A′′(x)
A(x)

Both sides are therefore constant, say equal to λ. Since A must be
periodic of period 1, we see that the only possibility is λ = −4π2n2,
where n ∈ Z. Then A is a linear combination of the exponentials e2πinx
and e−2πinx, and B(t) is a multiple of e−4π

2n2t. By superposing these
solutions, we are led to

(8) u(x, t) =
∞∑

n=−∞
ane

−4π2n2te2πinx,

where, setting t = 0, we see that {an} are the Fourier coefficients of f .
Note that when f is Riemann integrable, the coefficients an are

bounded, and since the factor e−4π
2n2t tends to zero extremely fast, the

series defining u converges. In fact, in this case, u is twice differentiable
and solves equation (7).

The natural question with regard to the boundary condition is the
following: do we have u(x, t) → f(x) as t tends to 0, and in what sense?
A simple application of the Parseval identity shows that this limit holds
in the mean square sense (Exercise 11). For a better understanding of
the properties of our solution (8), we write it as

u(x, t) = (f ∗Ht)(x),

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120 Chapter 4. SOME APPLICATIONS OF FOURIER SERIES

where Ht is the heat kernel for the circle, given by

(9) Ht(x) =
∞∑

n=−∞
e−4π

2n2te2πinx,

and where the convolution for functions with period 1 is defined by

(f ∗ g)(x) =
∫ 1
0

f(x− y)g(y) dy.

An analogy between the heat kernel and the Poisson kernel (of Chapter 2)
is given in Exercise 12. However, unlike in the case of the Poisson kernel,
there is no elementary formula for the heat kernel. Nevertheless, it turns
out that it is a good kernel (in the sense of Chapter 2). The proof is
not obvious and requires the use of the celebrated Poisson summation
formula, which will be taken up in Chapter 5. As a corollary, we will
also find that Ht is everywhere positive, a fact that is also not obvious
from its defining expression (9). We can, however, give the following
heuristic argument for the positivity of Ht. Suppose that we begin with
an initial temperature distribution f which is everywhere ≤ 0. Then it
is physically reasonable to expect u(x, t) ≤ 0 for all t since heat travels
from hot to cold. Now

u(x, t) =
∫ 1
0

f(x− y)Ht(y) dy.

If Ht is negative for some x0, then we may choose f ≤ 0 supported near
x0, and this would imply u(x0, t) > 0, which is a contradiction.

5 Exercises

1. Let γ : [a, b] → R2 be a parametrization for the closed curve Γ.
(a) Prove that γ is a parametrization by arc-length if and only if the length

of the curve from γ(a) to γ(s) is precisely s− a, that is,
∫ s

|γ′(t)| dt = s− a.

(b) Prove that any curve Γ admits a parametrization by arc-length. [Hint: If
η is any parametrization, let h(s) =

∫ s
a
|η′(t)| dt and consider γ = η ◦ h−1.]

2. Suppose γ : [a, b] → R2 is a parametrization for a closed curve Γ, with
γ(t) = (x(t), y(t)).

Ibookroot October 20, 2007

5. Exercises 121

(a) Show that

1
2

∫ b
a

(x(s)y′(s)− y(s)x′(s)) ds =
∫ b

x(s)y′(s) ds = −
∫ b

y(s)x′(s) ds.

(b) Define the reverse parametrization of γ by γ− : [a, b] → R2 with
γ−(t) = γ(b + a− t). The image of γ− is precisely Γ, except that the
points γ−(t) and γ(t) travel in opposite directions. Thus γ− “reverses”
the orientation of the curve. Prove that

∫

(x dy − y dx) = −
∫

γ−
(x dy − y dx).

In particular, we may assume (after a possible change in orientation) that

A = 1
2

∫ b
a

(x(s)y′(s)− y(s)x′(s)) ds =
∫ b

x(s)y′(s) ds.

3. Suppose Γ is a curve in the plane, and that there exists a set of coordinates
x and y so that the x-axis divides the curve into the union of the graph of
two continuous functions y = f(x) and y = g(x) for 0 ≤ x ≤ 1, and with f(x) ≥
g(x) (see Figure 6). Let Ω denote the region between the graphs of these two
functions:

Ω = {(x, y) : 0 ≤ x ≤ 1 and g(x) ≤ y ≤ f(x)}.

0 1

y = g(x)

y = f(x)

Ω

Figure 6. Simple version of the area formula

With the familiar interpretation that the integral
∫

h(x) dx gives the area
under the graph of the function h, we see that the area of Ω is

∫ 1
0

f(x) dx−

Ibookroot October 20, 2007

122 Chapter 4. SOME APPLICATIONS OF FOURIER SERIES

∫ 1
0

g(x) dx. Show that this definition coincides with the area formula A given in
the text, that is,

∫ 1
0

f(x) dx−
∫ 1

g(x) dx =
∣∣∣∣−

∫

y dx

∣∣∣∣ = A.

Also, note that if the orientation of the curve is chosen so that Ω “lies to the
left” of Γ, then the above formula holds without the absolute value signs.

This formula generalizes to any set that can be written as a finite union of
domains like Ω above.

4. Observe that with the definition of ` and A given in the text, the isoperimetric
inequality continues to hold (with the same proof) even when Γ is not simple.

Show that this stronger version of the isoperimetric inequality is equivalent
to Wirtinger’s inequality, which says that if f is 2π-periodic, of class C1, and
satisfies

∫ 2π
0

f(t) dt = 0, then

∫ 2π
0

|f(t)|2 dt ≤
∫ 2π

|f ′(t)|2 dt

with equality if and only if f(t) = A sin t + B cos t (Exercise 11, Chapter 3).
[Hint: In one direction, note that if the length of the curve is 2π and γ is an
appropriate arc-length parametrization, then

2(π −A) =
∫ 2π

[x′(s) + y(s)]
2

ds +
∫ 2π

(y′(s)2 − y(s)2) ds.

A change of coordinates will guarantee
∫ 2π
0

y(s) ds = 0. For the other direction,
start with a real-valued f satisfying all the hypotheses of Wirtinger’s inequality,
and construct g, 2π-periodic and so that the term in brackets above vanishes.]

5. Prove that the sequence {γn}∞n=1, where γn is the fractional part of
(

1 +
√

5
2

)n
,

is not equidistributed in [0, 1].

[Hint: Show that Un =
(

1+
√

5
2

)n
+

(
1−
√

5
2

)n
is the solution of the difference

equation Ur+1 = Ur + Ur−1 with U0 = 2 and U1 = 1. The Un satisfy the same
difference equation as the Fibonacci numbers.]

6. Let θ = p/q be a rational number where p and q are relatively prime inte-
gers (that is, θ is in lowest form). We assume without loss of generality that
q > 0. Define a sequence of numbers in [0, 1) by ξn = 〈nθ〉 where 〈·〉 denotes the

Ibookroot October 20, 2007

5. Exercises 123

fractional part. Show that the sequence {ξ1, ξ2, . . .} is equidistributed on the
points of the form

0, 1/q, 2/q, . . . , (q − 1)/q.

In fact, prove that for any 0 ≤ a < q, one has #{n : 1 ≤ n ≤ N, 〈nθ〉 = a/q} N = 1 q + O ( 1 N ) . [Hint: For each integer k ≥ 0, there exists a unique integer n with kq ≤ n < (k + 1)q and so that 〈nθ〉 = a/q. Why can one assume k = 0? Prove the existence of n by using the fact1 that if p and q are relatively prime, there exist integers x, y such that xp + yq = 1. Next, divide N by q with remainder, that is, write N = `q + r where 0 ≤ ` and 0 ≤ r < q. Establish the inequalities ` ≤ #{n : 1 ≤ n ≤ N, 〈nθ〉 = a/q} ≤ ` + 1.] 7. Prove the second part of Weyl’s criterion: if a sequence of numbers ξ1, ξ2, . . . in [0, 1) is equidistributed, then for all k ∈ Z− {0} 1 N N∑ n=1 e2πikξn → 0 as N →∞. [Hint: It suffices to show that 1 N ∑N n=1 f(ξn) → ∫ 1 0 f(x) dx for all continuous f . Prove this first when f is the characteristic function of an interval.] 8. Show that for any a 6= 0, and σ with 0 < σ < 1, the sequence 〈anσ〉 is equidis- tributed in [0, 1). [Hint: Prove that ∑N n=1 e2πibn σ = O(Nσ) + O(N1−σ) if b 6= 0.] In fact, note the following N∑ n=1 e2πibn σ − ∫ N 1 e2πibx σ dx = O ( N∑ n=1 n−1+σ ) . 9. In contrast with the result in Exercise 8, prove that 〈a log n〉 is not equidis- tributed for any a. [Hint: Compare the sum ∑N n=1 e2πib log n with the corresponding integral.] 10. Suppose that f is a periodic function on R of period 1, and {ξn} is a sequence which is equidistributed in [0, 1). Prove that: 1The elementary results in arithmetic used in this exercise can be found at the begin- ning of Chapter 8. Ibookroot October 20, 2007 124 Chapter 4. SOME APPLICATIONS OF FOURIER SERIES (a) If f is continuous and satisfies ∫ 1 0 f(x) dx = 0, then lim N→∞ 1 N N∑ n=1 f(x + ξn) = 0 uniformly in x. [Hint: Establish this result first for trigonometric polynomials.] (b) If f is merely integrable on [0, 1] and satisfies ∫ 1 0 f(x) dx = 0, then lim N→∞ ∫ 1 0 ∣∣∣∣∣ 1 N N∑ n=1 f(x + ξn) ∣∣∣∣∣ 2 dx = 0. 11. Show that if u(x, t) = (f ∗Ht)(x) where Ht is the heat kernel, and f is Riemann integrable, then ∫ 1 0 |u(x, t)− f(x)|2 dx → 0 as t → 0. 12. A change of variables in (8) leads to the solution u(θ, τ) = ∑ ane −n2τeinθ = (f ∗ hτ )(θ) of the equation ∂u ∂τ = ∂2u ∂θ2 with 0 ≤ θ ≤ 2π and τ > 0,

with boundary condition u(θ, 0) = f(θ) ∼
∑

ane
inθ. Here hτ (θ) =∑∞

n=−∞ e
−n2τeinθ. This version of the heat kernel on [0, 2π] is the analogue

of the Poisson kernel, which can be written as Pr(θ) =
∑∞

n=−∞ e
−|n|τeinθ with

r = e−τ (and so 0 < r < 1 corresponds to τ > 0).

13. The fact that the kernel Ht(x) is a good kernel, hence u(x, t) → f(x) at
each point of continuity of f , is not easy to prove. This will be shown in the
next chapter. However, one can prove directly that Ht(x) is “peaked” at x = 0
as t → 0 in the following sense:

(a) Show that
∫ 1/2
−1/2 |Ht(x)|2 dx is of the order of magnitude of t−1/2 as t → 0.

More precisely, prove that t1/2
∫ 1/2
−1/2 |Ht(x)|2 dx converges to a non-zero

limit as t → 0.
(b) Prove that

∫ 1/2
−1/2 x

2|Ht(x)|2 dx = O(t1/2) as t → 0.

Ibookroot October 20, 2007

6. Problems 125

[Hint: For (a) compare the sum
∑∞
−∞ e

−cn2t with the integral
∫∞
−∞ e

−cx2t dx
where c > 0. For (b) use x2 ≤ C(sin πx)2 for −1/2 ≤ x ≤ 1/2, and apply the
mean value theorem to e−cx

2t.]

6 Problems

1.∗ This problem explores another relationship between the geometry of a curve
and Fourier series. The diameter of a closed curve Γ parametrized by
γ(t) = (x(t), y(t)) on [−π, π] is defined by

d = sup
P, Q∈Γ

|P −Q| = sup
t1, t2∈[−π,π]

|γ(t1)− γ(t2)|.

If an is the nth Fourier coefficient of γ(t) = x(t) + iy(t) and ` denotes the length
of Γ, then

(a) 2|an| ≤ d for all n 6= 0.
(b) ` ≤ πd, whenever Γ is convex.

Property (a) follows from the fact that 2an = 12π
∫ π
−π[γ(t)− γ(t + π/n)]e−int dt.

The equality ` = πd is satisfied when Γ is a circle, but surprisingly, this is
not the only case. In fact, one finds that ` = πd is equivalent to 2|a1| = d. We
re-parametrize γ so that for each t in [−π, π] the tangent to the curve makes an
angle t with the y-axis. Then, if a1 = 1 we have

γ′(t) = ieit(1 + r(t)),

where r is a real-valued function which satisfies r(t) + r(t + π) = 0, and
|r(t)| ≤ 1. Figure 7 (a) shows the curve obtained by setting r(t) = cos 5t. Also,
Figure 7 (b) consists of the curve where r(t) = h(3t), with h(s) = −1 if −π ≤
s ≤ 0 and h(s) = 1 if 0 < s < π. This curve (which is only piecewise of class C1) is known as the Reuleaux triangle and is the classical example of a convex curve of constant width which is not a circle. 2.∗ Here we present an estimate of Weyl which leads to some interesting results. (a) Let SN = ∑N n=1 e2πif(n). Show that for H ≤ N , one has |SN |2 ≤ c N H H∑ h=0 ∣∣∣∣∣ N−h∑ n=1 e2πi(f(n+h)−f(n)) ∣∣∣∣∣ , for some constant c > 0 independent of N , H, and f .

(b) Use this estimate to show that the sequence 〈n2γ〉 is equidistributed in
[0, 1) whenever γ is irrational.

Ibookroot October 20, 2007

126 Chapter 4. SOME APPLICATIONS OF FOURIER SERIES

(a) (b)

Figure 7. Some curves with maximal length for a given diameter

(c) More generally, show that if {ξn} is a sequence of real numbers so that
for all positive integers h the difference 〈ξn+h − ξn〉 is equidistributed in
[0, 1), then 〈ξn〉 is also equidistributed in [0, 1).

(d) Suppose that P (x) = cnxn + · · ·+ c0 is a polynomial with real coefficients,
where at least one of c1, . . . , cn is irrational. Then the sequence 〈P (n)〉 is
equidistributed in [0, 1).

[Hint: For (a), let an = e2πif(n) when 1 ≤ n ≤ N and 0 otherwise. Then write
H

∑
n

an =
∑H

k=1

∑
n

an+k and apply the Cauchy-Schwarz inequality. For (b),
note that (n + h)2γ − n2γ = 2nhγ + h2γ, and use the fact that for each integer
h, the sequence 〈2nhγ〉 is equidistributed. Finally, to prove (d), assume first that
P (x) = Q(x) + c1x + c0 where c1 is irrational, and estimate the exponential sum∑N

n=1
e2πikP (n). Then, argue by induction on the highest degree term which has

an irrational coefficient, and use part (c).]

3.∗ If σ > 0 is not an integer and a 6= 0, then 〈anσ〉 is equidistributed in [0, 1).
See also Exercise 8.

4. An elementary construction of a continuous but nowhere differentiable func-
tion is obtained by “piling up singularities,” as follows.

On [−1, 1] consider the function

ϕ(x) = |x|

and extend ϕ to R by requiring it to be periodic of period 2. Clearly, ϕ is
continuous on R and |ϕ(x)| ≤ 1 for all x so the function f defined by

f(x) =
∞∑

n=0

(
3
4

)n
ϕ(4nx)

is continuous on R.

Ibookroot October 20, 2007

6. Problems 127

(a) Fix x0 ∈ R. For every positive integer m, let δm = ± 124−m where the
sign is chosen so that no integer lies in between 4mx0 and 4m(x0 + δm).
Consider the quotient

γn =
ϕ(4n(x0 + δm))− ϕ(4nx0)

δm
.

Prove that if n > m, then γn = 0, and for 0 ≤ n ≤ m one has |γn| ≤ 4n
with |γm| = 4m.

(b) From the above observations prove the estimate

∣∣∣∣
f(x0 + δm)− f(x0)

δm

∣∣∣∣ ≥
1
2
(3m + 1),

and conclude that f is not differentiable at x0.

5. Let f be a Riemann integrable function on the interval [−π, π]. We define
the generalized delayed means of the Fourier series of f by

σN,K =
SN + · · ·+ SN+K−1

K
.

Note that in particular

σ0,N = σN , σN,1 = SN and σN,N = ∆N ,

where ∆N are the specific delayed means used in Section 3.

(a) Show that

σN,K =
1
K

((N + K)σN+K −NσN ) ,

and

σN,K = SN +
∑

N+1≤|ν|≤N+K−1

(
1− |ν| −N

)
f̂(ν)eiνθ.

From this last expression for σN,K conclude that

|σN,K − SM | ≤
∑

N+1≤|ν|≤N+K−1
|f̂(ν)|

for all N ≤ M < N + K. Ibookroot October 20, 2007 128 Chapter 4. SOME APPLICATIONS OF FOURIER SERIES (b) Use one of the above formulas and Fejér’s theorem to show that with N = kn and K = n, then σkn,n(f)(θ) → f(θ) as n →∞ whenever f is continuous at θ, and also σkn,n(f)(θ) → f(θ+) + f(θ−) 2 as n →∞ at a jump discontinuity (refer to the preceding chapters and their exer- cises for the appropriate definitions and results). In the case when f is continuous on [−π, π], show that σkn,n(f) → f uniformly as n →∞. (c) Using part (a), show that if f̂(ν) = O(1/|ν|) and kn ≤ m < (k + 1)n, we get |σkn,n − Sm| ≤ C k for some constant C > 0.

(d) Suppose that f̂(ν) = O(1/|ν|). Prove that if f is continuous at θ then

SN (f)(θ) → f(θ) as N →∞,

and if f has a jump discontinuity at θ then

SN (f)(θ) →
f(θ+) + f(θ−)

2
as N →∞.

Also, show that if f is continuous on [−π, π], then SN (f) → f uniformly.
(e) The above arguments show that if

∑
cn is Cesàro summable to s and cn =

O(1/n), then
∑

cn converges to s. This is a weak version of Littlewood’s
theorem (Problem 3, Chapter 2).

6. Dirichlet’s theorem states that the Fourier series of a real continuous peri-
odic function f which has only a finite number of relative maxima and minima
converges everywhere to f (and uniformly).

Prove this theorem by showing that such a function satisfies f̂(n) = O(1/|n|).
[Hint: Argue as in Exercise 17, Chapter 3; then use conclusion (d) in Problem 5
above.]

Ibookroot October 20, 2007

5 The Fourier Transform on R

The theory of Fourier series and integrals has always
had major difficulties and necessitated a large math-
ematical apparatus in dealing with questions of con-
vergence. It engendered the development of methods
of summation, although these did not lead to a com-
pletely satisfactory solution of the problem.. . . For the
Fourier transform, the introduction of distributions
(hence the space S) is inevitable either in an explicit
or hidden form.. . . As a result one may obtain all that
is desired from the point of view of the continuity and
inversion of the Fourier transform.

L. Schwartz, 1950

The theory of Fourier series applies to functions on the circle, or equiv-
alently, periodic functions on R. In this chapter, we develop an analogous
theory for functions on the entire real line which are non-periodic. The
functions we consider will be suitably “small” at infinity. There are sev-
eral ways of defining an appropriate notion of “smallness,” but it will
nevertheless be vital to assume some sort of vanishing at infinity.

On the one hand, recall that the Fourier series of a periodic function
associates a sequence of numbers, namely the Fourier coefficients, to
that function; on the other hand, given a suitable function f on R, the
analogous object associated to f will in fact be another function f̂ on R
which is called the Fourier transform of f . Since the Fourier transform
of a function on R is again a function on R, one can observe a symmetry
between a function and its Fourier transform, whose analogue is not as
apparent in the setting of Fourier series.

Roughly speaking, the Fourier transform is a continuous version of the
Fourier coefficients. Recall that the Fourier coefficients an of a function
f defined on the circle are given by

(1) an =
∫ 1
0

f(x)e−2πinx dx,

Ibookroot October 20, 2007

130 Chapter 5. THE FOURIER TRANSFORM ON R

and then in the appropriate sense we have

(2) f(x) =
∞∑

n=−∞
ane

2πinx.

Here we have replaced θ by 2πx, as we have frequently done previously.
Now, consider the following analogy where we replace all of the discrete

symbols (such as integers and sums) by their continuous counterparts
(such as real numbers and integrals). In other words, given a function f
on all of R, we define its Fourier transform by changing the domain of
integration from the circle to all of R, and by replacing n ∈ Z by ξ ∈ R
in (1), that is, by setting

(3) f̂(ξ) =
∫ ∞
−∞

f(x)e−2πixξ dx.

We push our analogy further, and consider the following continuous ver-
sion of (2): replacing the sum by an integral, and an by f̂(ξ), leads to
the Fourier inversion formula,

(4) f(x) =
∫ ∞
−∞

f̂(ξ)e2πixξ dξ.

Under a suitable hypotheses on f , the identity (4) actually holds, and
much of the theory in this chapter aims at proving and exploiting this
relation. The validity of the Fourier inversion formula is also suggested
by the following simple observation. Suppose f is supported in a finite
interval contained in I = [−L/2, L/2], and we expand f in a Fourier series
on I. Then, letting L tend to infinity, we are led to (4) (see Exercise 1).

The special properties of the Fourier transform make it an important
tool in the study of partial differential equations. For instance, we shall
see how the Fourier inversion formula allows us to analyze some equations
that are modeled on the real line. In particular, following the ideas
developed on the circle, we solve the time-dependent heat equation for
an infinite rod and the steady-state heat equation in the upper half-plane.

In the last part of the chapter we discuss further topics related to the
Poisson summation formula,

∑

n∈Z
f(n) =

∑

n∈Z
f̂(n),

which gives another remarkable connection between periodic functions
(and their Fourier series) and non-periodic functions on the line (and

Ibookroot October 20, 2007

1. Elementary theory of the Fourier transform 131

their Fourier transforms). This identity allows us to prove an assertion
made in the previous chapter, namely, that the heat kernel Ht(x) satisfies
the properties of a good kernel. In addition, the Poisson summation
formula arises in many other settings, in particular in parts of number
theory, as we shall see in Book II.

We make a final comment about the approach we have chosen. In our
study of Fourier series, we found it useful to consider Riemann integrable
functions on the circle. In particular, this generality assured us that even
functions that had certain discontinuities could be treated by the theory.
In contrast, our exposition of the elementary properties of the Fourier
transform is stated in terms of the Schwartz space S of testing functions.
These are functions that are indefinitely differentiable and that, together
with their derivatives, are rapidly decreasing at infinity. The reliance on
this space of functions is a device that allows us to come quickly to the
main conclusions, formulated in a direct and transparent fashion. Once
this is carried out, we point out some easy extensions to a somewhat
wider setting. The more general theory of Fourier transforms (which
must necessarily be based on Lebesgue integration) will be treated in
Book III.

1 Elementary theory of the Fourier transform

We begin by extending the notion of integration to functions that are
defined on the whole real line.

1.1 Integration of functions on the real line

Given the notion of the integral of a function on a closed and bounded
interval, the most natural extension of this definition to continuous func-
tions over R is

∫ ∞
−∞

f(x) dx = lim
N→∞

∫ N
−N

f(x) dx.

Of course, this limit may not exist. For example, it is clear that if
f(x) = 1, or even if f(x) = 1/(1 + |x|), then the above limit is infinite.
A moment’s reflection suggests that the limit will exist if we impose on
f enough decay as |x| tends to infinity. A useful condition is as follows.

A function f defined on R is said to be of moderate decrease if f
is continuous and there exists a constant A > 0 so that

|f(x)| ≤ A
1 + x2

for all x ∈ R.

Ibookroot October 20, 2007

132 Chapter 5. THE FOURIER TRANSFORM ON R

This inequality says that f is bounded (by A for instance), and also that
it decays at infinity at least as fast as 1/x2, since A/(1 + x2) ≤ A/x2.

For example, the function f(x) = 1/(1 + |x|n) is of moderate decrease
as long as n ≥ 2. Another example is given by the function e−a|x| for
a > 0.

We shall denote by M(R) the set of functions of moderate decrease
on R. As an exercise, the reader can check that under the usual addition
of functions and multiplication by scalars, M(R) forms a vector space
over C.

We next see that whenever f belongs to M(R), then we may define

(5)
∫ ∞
−∞

f(x) dx = lim
N→∞

∫ N
−N

f(x) dx,

where the limit now exists. Indeed, for each N the integral IN =∫ N
−N f(x) dx is well defined because f is continuous. It now suffices to

show that {IN} is a Cauchy sequence, and this follows because if M > N ,
then

|IM − IN | ≤
∣∣∣∣∣
∫

N≤|x|≤M
f(x) dx

∣∣∣∣∣

≤ A
∫

N≤|x|≤M

≤ 2A
N

→ 0 as N →∞.

Notice we have also proved that
∫
|x|≥N f(x) dx → 0 as N →∞. At this

point, we remark that we may replace the exponent 2 in the definition
of moderate decrease by 1 + ² where ² > 0; that is,

|f(x)| ≤ A
1 + |x|1+² for all x ∈ R.

This definition would work just as well for the purpose of the theory
developed in this chapter. We chose ² = 1 merely as a matter of conve-
nience.

We summarize some elementary properties of integration over R in a
proposition.

Proposition 1.1 The integral of a function of moderate decrease defined
by (5) satisfies the following properties:

Ibookroot October 20, 2007

1. Elementary theory of the Fourier transform 133

(i) Linearity: if f, g ∈M(R) and a, b ∈ C, then
∫ ∞
−∞

(af(x) + bg(x)) dx = a
∫ ∞
−∞

f(x) dx + b
∫ ∞
−∞

g(x) dx.

(ii) Translation invariance: for every h ∈ R we have
∫ ∞
−∞

f(x− h) dx =
∫ ∞
−∞

f(x) dx.

(iii) Scaling under dilations: if δ > 0, then

∫ ∞
−∞

f(δx) dx =
∫ ∞
−∞

f(x) dx.

(iv) Continuity: if f ∈M(R), then
∫ ∞
−∞

|f(x− h)− f(x)| dx → 0 as h → 0.

We say a few words about the proof. Property (i) is immediate. To
verify property (ii), it suffices to see that

∫ N
−N

f(x− h) dx−
∫ N
−N

f(x) dx → 0 as N →∞.

Since
∫ N
−N f(x− h) dx =

∫ N−h
−N−h f(x) dx, the above difference is majorized

by
∣∣∣∣∣
∫ −N
−N−h

f(x) dx

∣∣∣∣∣ +
∣∣∣∣∣
∫ N

N−h
f(x) dx

∣∣∣∣∣ ≤
A′

1 + N2

for large N , which tends to 0 as N tends to infinity.
The proof of property (iii) is similar once we observe that δ

∫ N
−N f(δx) dx =∫ δN

−δN f(x) dx.
To prove property (iv) it suffices to take |h| ≤ 1. For a preassigned ² > 0,

we first choose N so large that
∫

|x|≥N
|f(x)| dx ≤ ²/4 and

∫

|x|≥N
|f(x− h)| dx ≤ ²/4.

Now with N fixed, we use the fact that since f is continuous, it is uni-
formly continuous in the interval [−N − 1, N + 1]. Hence

Ibookroot October 20, 2007

134 Chapter 5. THE FOURIER TRANSFORM ON R

sup|x|≤N |f(x− h)− f(x)| → 0 as h tends to 0. So we can take h so
small that this supremum is less than ²/4N . Altogether, then,

∫ ∞
−∞

|f(x− h)− f(x)| dx ≤
∫ N
−N

|f(x− h)− f(x)| dx

+
∫

|x|≥N
|f(x− h)| dx

+
∫

|x|≥N
|f(x)| dx

≤ ²/2 + ²/4 + ²/4 = ²,

and thus conclusion (iv) follows.

1.2 Definition of the Fourier transform

If f ∈M(R), we define its Fourier transform for ξ ∈ R by

f̂(ξ) =
∫ ∞
−∞

f(x)e−2πixξ dx.

Of course, |e−2πixξ| = 1, so the integrand is of moderate decrease, and
the integral makes sense.

In fact, this last observation implies that f̂ is bounded, and moreover,
a simple argument shows that f̂ is continuous and tends to 0 as |ξ| → ∞
(Exercise 5). However, nothing in the definition above guarantees that
f̂ is of moderate decrease, or has a specific decay. In particular, it is not
clear in this context how to make sense of the integral

∫∞
−∞ f̂(ξ)e

2πixξ dξ
and the resulting Fourier inversion formula. To remedy this, we introduce
a more refined space of functions considered by Schwartz which is very
useful in establishing the initial properties of the Fourier transform.

The choice of the Schwartz space is motivated by an important prin-
ciple which ties the decay of f̂ to the continuity and differentiability
properties of f (and vice versa): the faster f̂(ξ) decreases as |ξ| → ∞,
the “smoother” f must be. An example that reflects this principle is
given in Exercise 3. We also note that this relationship between f and f̂
is reminiscent of a similar one between the smoothness of a function on
the circle and the decay of its Fourier coefficients; see the discussion of
Corollary 2.4 in Chapter 2.

1.3 The Schwartz space

The Schwartz space on R consists of the set of all indefinitely differ-
entiable functions f so that f and all its derivatives f ′, f ′′, . . . , f (`), . . .,

Ibookroot October 20, 2007

1. Elementary theory of the Fourier transform 135

are rapidly decreasing, in the sense that

sup
x∈R

|x|k|f (`)(x)| < ∞ for every k, ` ≥ 0. We denote this space by S = S(R), and again, the reader should verify that S(R) is a vector space over C. Moreover, if f ∈ S(R), we have f ′(x) = df dx ∈ S(R) and xf(x) ∈ S(R). This expresses the important fact that the Schwartz space is closed under differentiation and multiplication by polynomials. A simple example of a function in S(R) is the Gaussian defined by f(x) = e−x 2 , which plays a central role in the theory of the Fourier transform, as well as other fields (for example, probability theory and physics). The reader can check that the derivatives of f are of the form P (x)e−x 2 where P is a polynomial, and this immediately shows that f ∈ S(R). In fact, e−ax2 belongs to S(R) whenever a > 0. Later, we will normalize the Gaussian
by choosing a = π.

1−1

Figure 1. The Gaussian e−πx
2

An important class of other examples in S(R) are the “bump func-
tions” which vanish outside bounded intervals (Exercise 4).

As a final remark, note that although e−|x| decreases rapidly at infinity,
it is not differentiable at 0 and therefore does not belong to S(R).

Ibookroot October 20, 2007

136 Chapter 5. THE FOURIER TRANSFORM ON R

1.4 The Fourier transform on S
The Fourier transform of a function f ∈ S(R) is defined by

f̂(ξ) =
∫ ∞
−∞

f(x)e−2πixξ dx.

Some simple properties of the Fourier transform are gathered in the fol-
lowing proposition. We use the notation

f(x) −→ f̂(ξ)
to mean that f̂ denotes the Fourier transform of f .

Proposition 1.2 If f ∈ S(R) then:
(i) f(x + h) −→ f̂(ξ)e2πihξ whenever h ∈ R.
(ii) f(x)e−2πixh −→ f̂(ξ + h) whenever h ∈ R.
(iii) f(δx) −→ δ−1f̂(δ−1ξ) whenever δ > 0.
(iv) f ′(x) −→ 2πiξf̂(ξ).

(v) −2πixf(x) −→ d
dξ

f̂(ξ).

In particular, except for factors of 2πi, the Fourier transform inter-
changes differentiation and multiplication by x. This is the key property
that makes the Fourier transform a central object in the theory of differ-
ential equations. We shall return to this point later.

Proof. Property (i) is an immediate consequence of the translation
invariance of the integral, and property (ii) follows from the definition.
Also, the third property of Proposition 1.1 establishes (iii).

Integrating by parts gives
∫ N
−N

f ′(x)e−2πixξ dx =
[
f(x)e−2πixξ

]N
−N

+ 2πiξ
∫ N
−N

f(x)e−2πixξ dx,

so letting N tend to infinity gives (iv).
Finally, to prove property (v), we must show that f̂ is differentiable

and find its derivative. Let ² > 0 and consider

f̂(ξ + h)− f̂(ξ)
h

− ̂(−2πixf)(ξ)=
∫ ∞
−∞

f(x)e−2πixξ
[

e−2πixh − 1
h

+ 2πix

]
dx.

Ibookroot October 20, 2007

1. Elementary theory of the Fourier transform 137

Since f(x) and xf(x) are of rapid decrease, there exists an integer N
so that

∫
|x|≥N |f(x)| dx ≤ ² and

∫
|x|≥N |x| |f(x)| dx ≤ ². Moreover, for

|x| ≤ N , there exists h0 so that |h| < h0 implies ∣∣∣∣∣ e−2πixh − 1 h + 2πix ∣∣∣∣∣ ≤ ² N . Hence for |h| < h0 we have ∣∣∣∣∣ f̂(ξ + h)− f̂(ξ) h − ̂(−2πixf)(ξ) ∣∣∣∣∣ ≤ ∫ N −N ∣∣∣∣∣f(x)e −2πixξ [ e−2πixh − 1 h + 2πix ]∣∣∣∣∣ dx + C² ≤ C ′². Theorem 1.3 If f ∈ S(R), then f̂ ∈ S(R). The proof is an easy application of the fact that the Fourier transform interchanges differentiation and multiplication. In fact, note that if f ∈ S(R), its Fourier transform f̂ is bounded; then also, for each pair of non-negative integers ` and k, the expression ξk ( d dξ )` f̂(ξ) is bounded, since by the last proposition, it is the Fourier transform of 1 (2πi)k ( d dx )k [(−2πix)`f(x)]. The proof of the inversion formula f(x) = ∫ ∞ −∞ f̂(ξ)e2πixξ dξ for f ∈ S(R), which we give in the next section, is based on a careful study of the function e−ax 2 , which, as we have already observed, is in S(R) if a > 0.

Ibookroot October 20, 2007

138 Chapter 5. THE FOURIER TRANSFORM ON R

The Gaussians as good kernels

We begin by considering the case a = π because of the normalization:

(6)
∫ ∞
−∞

e−πx
2
dx = 1.

To see why (6) is true, we use the multiplicative property of the expo-
nential to reduce the calculation to a two-dimensional integral. More
precisely, we can argue as follows:

(∫ ∞
−∞

e−πx
2
dx

)2
=

∫ ∞
−∞

e−π(x
2+y2) dx dy

=
∫ 2π
0

∫ ∞
0

e−πr
2
r dr dθ

=
∫ ∞
0

2πre−πr
2
dr

=
[
−e−πr2

]∞
0

= 1,

where we have evaluated the two-dimensional integral using polar coor-
dinates.

The fundamental property of the Gaussian which is of interest to us,
and which actually follows from (6), is that e−πx

2
equals its Fourier

transform! We isolate this important result in a theorem.

Theorem 1.4 If f(x) = e−πx
2
, then f̂(ξ) = f(ξ).

Proof. Define

F (ξ) = f̂(ξ) =
∫ ∞
−∞

e−πx
2
e−2πixξ dx,

and observe that F (0) = 1, by our previous calculation. By property (v)
in Proposition 1.2, and the fact that f ′(x) = −2πxf(x), we obtain

F ′(ξ) =
∫ ∞
−∞

f(x)(−2πix)e−2πixξ dx = i
∫ ∞
−∞

f ′(x)e−2πixξ dx.

By (iv) of the same proposition, we find that

F ′(ξ) = i(2πiξ)f̂(ξ) = −2πξF (ξ).

Ibookroot October 20, 2007

1. Elementary theory of the Fourier transform 139

If we define G(ξ) = F (ξ)eπξ
2
, then from what we have seen above, it

follows that G′(ξ) = 0, hence G is constant. Since F (0) = 1, we conclude
that G is identically equal to 1, therefore F (ξ) = e−πξ

2
, as was to be

shown.

The scaling properties of the Fourier transform under dilations yield
the following important transformation law, which follows from (iii) in
Proposition 1.2 (with δ replaced by δ−1/2).

Corollary 1.5 If δ > 0 and Kδ(x) = δ−1/2e−πx
2/δ, then K̂δ(ξ) = e−πδξ

2
.

We pause to make an important observation. As δ tends to 0, the
function Kδ peaks at the origin, while its Fourier transform K̂δ gets
flatter. So in this particular example, we see that Kδ and K̂δ cannot both
be localized (that is, concentrated) at the origin. This is an example of a
general phenomenon called the Heisenberg uncertainty principle, which
we will discuss at the end of this chapter.

We have now constructed a family of good kernels on the real line,
analogous to those on the circle considered in Chapter 2. Indeed, with

Kδ(x) = δ
−1/2e−πx

2/δ,

we have:

(i)
∫∞
−∞Kδ(x) dx = 1.

(ii)
∫∞
−∞ |Kδ(x)| dx ≤ M .

(iii) For every η > 0, we have
∫
|x|>η |Kδ(x)| dx → 0 as δ → 0.

To prove (i), we may change variables and use (6), or note that the
integral equals K̂δ(0), which is 1 by Corollary 1.5. Since Kδ ≥ 0, it is
clear that property (ii) is also true. Finally we can again change variables
to get

∫

|x|>η
|Kδ(x)| dx =

∫

|y|>η/δ1/2
e−πy

2
dy → 0

as δ tends to 0. We have thus proved the following result.

Theorem 1.6 The collection {Kδ}δ>0 is a family of good kernels
as δ → 0.

We next apply these good kernels via the operation of convolution,
which is given as follows. If f, g ∈ S(R), their convolution is defined by

(7) (f ∗ g)(x) =
∫ ∞
−∞

f(x− t)g(t) dt.

Ibookroot October 20, 2007

140 Chapter 5. THE FOURIER TRANSFORM ON R

For a fixed value of x, the function f(x− t)g(t) is of rapid decrease in t,
hence the integral converges.

By the argument in Section 4 of Chapter 2 (with a slight modification),
we get the following corollary.

Corollary 1.7 If f ∈ S(R) , then

(f ∗Kδ)(x) → f(x) uniformly in x as δ → 0.

Proof. First, we claim that f is uniformly continuous on R. Indeed,
given ² > 0 there exists R > 0 so that |f(x)| < ²/4 whenever |x| ≥ R. Moreover, f is continuous, hence uniformly continuous on the compact interval [−R, R], and together with the previous observation, we can find η > 0 so that |f(x)− f(y)| < ² whenever |x− y| < η. Now we argue as usual. Using the first property of good kernels, we can write (f ∗Kδ)(x)− f(x) = ∫ ∞ −∞ Kδ(t) [f(x− t)− f(x)] dt, and since Kδ ≥ 0, we find |(f ∗Kδ)(x)− f(x)| ≤ ∫ |t|>η
+

∫

|t|≤η
Kδ(t) |f(x− t)− f(x)| dt.

The first integral is small by the third property of good kernels, and the
fact that f is bounded, while the second integral is also small since f
is uniformly continuous and

∫
Kδ = 1. This concludes the proof of the

corollary.

1.5 The Fourier inversion

The next result is an identity sometimes called the multiplication for-
mula.

Proposition 1.8 If f, g ∈ S(R), then
∫ ∞
−∞

f(x)ĝ(x) dx =
∫ ∞
−∞

f̂(y)g(y) dy.

To prove the proposition, we need to digress briefly to discuss the inter-
change of the order of integration for double integrals. Suppose F (x, y)
is a continuous function in the plane (x, y) ∈ R2. We will assume the
following decay condition on F :

|F (x, y)| ≤ A/(1 + x2)(1 + y2).

Ibookroot October 20, 2007

1. Elementary theory of the Fourier transform 141

Then, we can state that for each x the function F (x, y) is of moderate
decrease in y, and similarly for each fixed y the function F (x, y) is of
moderate decrease in x. Moreover, the function F1(x) =

∫∞
−∞ F (x, y) dy

is continuous and of moderate decrease; similarly for the function F2(y) =∫∞
−∞ F (x, y) dx. Finally

∫ ∞
−∞

F1(x) dx =
∫ ∞
−∞

F2(y) dy.

The proof of these facts may be found in the appendix.
We now apply this to F (x, y) = f(x)g(y)e−2πixy. Then F1(x) =

f(x)ĝ(x), and F2(y) = f̂(y)g(y) so
∫ ∞
−∞

f(x)ĝ(x) dx =
∫ ∞
−∞

f̂(y)g(y) dy,

which is the assertion of the proposition.
The multiplication formula and the fact that the Gaussian is its own

Fourier transform lead to a proof of the first major theorem.

Theorem 1.9 (Fourier inversion) If f ∈ S(R), then

f(x) =
∫ ∞
−∞

f̂(ξ)e2πixξ dξ.

Proof. We first claim that

f(0) =
∫ ∞
−∞

f̂(ξ) dξ.

Let Gδ(x) = e−πδx
2

so that Ĝδ(ξ) = Kδ(ξ). By the multiplication for-
mula we get

∫ ∞
−∞

f(x)Kδ(x) dx =
∫ ∞
−∞

f̂(ξ)Gδ(ξ) dξ.

Since Kδ is a good kernel, the first integral goes to f(0) as δ tends to 0.
Since the second integral clearly converges to

∫∞
−∞ f̂(ξ) dξ as δ tends to 0,

our claim is proved. In general, let F (y) = f(y + x) so that

f(x) = F (0) =
∫ ∞
−∞

F̂ (ξ) dξ =
∫ ∞
−∞

f̂(ξ)e2πixξ dξ.

As the name of Theorem 1.9 suggests, it provides a formula that inverts
the Fourier transform; in fact we see that the Fourier transform is its own

Ibookroot October 20, 2007

142 Chapter 5. THE FOURIER TRANSFORM ON R

inverse except for the change of x to −x. More precisely, we may define
two mappings F : S(R) → S(R) and F∗ : S(R) → S(R) by

F(f)(ξ) =
∫ ∞
−∞

f(x)e−2πixξ dx and F∗(g)(x) =
∫ ∞
−∞

g(ξ)e2πixξ dξ.

Thus F is the Fourier transform, and Theorem 1.9 guarantees that
F∗ ◦ F = I on S(R), where I is the identity mapping. Moreover, since
the definitions of F and F∗ differ only by a sign in the exponential, we
see that F(f)(y) = F∗(f)(−y), so we also have F ◦ F∗ = I. As a conse-
quence, we conclude that F∗ is the inverse of the Fourier transform on
S(R), and we get the following result.

Corollary 1.10 The Fourier transform is a bijective mapping on the
Schwartz space.

1.6 The Plancherel formula

We need a few further results about convolutions of Schwartz functions.
The key fact is that the Fourier transform interchanges convolutions with
pointwise products, a result analogous to the situation for Fourier series.

Proposition 1.11 If f, g ∈ S(R) then:
(i) f ∗ g ∈ S(R).
(ii) f ∗ g = g ∗ f .

(iii) (̂f ∗ g)(ξ) = f̂(ξ)ĝ(ξ).

Proof. To prove that f ∗ g is rapidly decreasing, observe first that for
any ` ≥ 0 we have supx |x|`|g(x− y)| ≤ A`(1 + |y|)`, because g is rapidly
decreasing (to check this assertion, consider separately the two cases
|x| ≤ 2|y| and |x| ≥ 2|y|). From this, we see that

sup
x
|x`(f ∗ g)(x)| ≤ A`

∫ ∞
−∞

|f(y)|(1 + |y|)` dy,

so that x`(f ∗ g)(x) is a bounded function for every ` ≥ 0. These esti-
mates carry over to the derivatives of f ∗ g, thereby proving that
f ∗ g ∈ S(R) because

(
d

)k
(f ∗ g)(x) = (f ∗

(
d

)k
g)(x) for k = 1, 2, . . ..

Ibookroot October 20, 2007

1. Elementary theory of the Fourier transform 143

This identity is proved first for k = 1 by differentiating under the inte-
gral defining f ∗ g. The interchange of differentiation and integration is
justified in this case by the rapid decrease of dg/dx. The identity then
follows for every k by iteration.

For fixed x, the change of variables x− y = u shows that

(f ∗ g)(x) =
∫ ∞
−∞

f(x− u)g(u) du = (g ∗ f)(x).

This change of variables is a composition of two changes, y 7→ −y and
y 7→ y − h (with h = x). For the first one we use the observation that∫∞
−∞ F (x) dx =

∫∞
−∞ F (−x) dx for any Schwartz function F , and for the

second, we apply (ii) of Proposition 1.1
Finally, consider F (x, y) = f(y)g(x− y)e−2πixξ. Since f and g are

rapidly decreasing, considering separately the two cases |x| ≤ 2|y| and
|x| ≥ 2|y|, we see that the discussion of the change of order of integration
after Proposition 1.8 applies to F . In this case F1(x) = (f ∗ g)(x)e−2πixξ,
and F2(y) = f(y)e−2πiyξ ĝ(ξ). Thus

∫∞
−∞ F1(x) dx =

∫∞
−∞ F2(y) dy, which

implies (iii). The proposition is therefore proved.

We now use the properties of convolutions of Schwartz functions to
prove the main result of this section. The result we have in mind is the
analogue for functions on R of Parseval’s identity for Fourier series.

The Schwartz space can be equipped with a Hermitian inner product

(f, g) =
∫ ∞
−∞

f(x)g(x) dx

whose associated norm is

‖f‖ =
(∫ ∞

−∞
|f(x)|2 dx

)1/2
.

The second major theorem in the theory states that the Fourier transform
is a unitary transformation on S(R).

Theorem 1.12 (Plancherel) If f ∈ S(R) then ‖f̂‖ = ‖f‖.

Proof. If f ∈ S(R) define f [(x) = f(−x). Then f̂ [(ξ) = f̂(ξ). Now
let h = f ∗ f [. Clearly, we have

ĥ(ξ) = |f̂(ξ)|2 and h(0) =
∫ ∞
−∞

|f(x)|2 dx.

Ibookroot October 20, 2007

144 Chapter 5. THE FOURIER TRANSFORM ON R

The theorem now follows from the inversion formula applied with x = 0,
that is,

∫ ∞
−∞

ĥ(ξ) dξ = h(0).

1.7 Extension to functions of moderate decrease

In the previous sections, we have limited our assertion of the Fourier
inversion and Plancherel formulas to the case when the function involved
belonged to the Schwartz space. It does not really involve further ideas to
extend these results to functions of moderate decrease, once we make the
additional assumption that the Fourier transform of the function under
consideration is also of moderate decrease. Indeed, the key observation,
which is easy to prove, is that the convolution f ∗ g of two functions f and
g of moderate decrease is again a function of moderate decrease (Exer-
cise 7); also f̂ ∗ g = f̂ ĝ. Moreover, the multiplication formula continues
to hold, and we deduce the Fourier inversion and Plancherel formulas
when f and f̂ are both of moderate decrease.

This generalization, although modest in scope, is nevertheless useful
in some circumstances.

1.8 The Weierstrass approximation theorem

We now digress briefly by further exploiting our good kernels to prove
the Weierstrass approximation theorem. This result was already alluded
to in Chapter 2.

Theorem 1.13 Let f be a continuous function on the closed and bounded
interval [a, b] ⊂ R. Then, for any ² > 0, there exists a polynomial P such
that

sup
x∈[a,b]

|f(x)− P (x)| < ². In other words, f can be uniformly approximated by polynomials. Proof. Let [−M, M ] denote any interval that contains [a, b] in its interior, and let g be a continuous function on R that equals 0 outside [−M,M ] and equals f in [a, b]. For example, extend f as follows: from b to M define g by a straight line segment going from f(b) to 0, and from a to −M by a straight line segment from f(a) also to 0. Let B be a Ibookroot October 20, 2007 2. Applications to some partial differential equations 145 bound for g, that is, |g(x)| ≤ B for all x. Then, since {Kδ} is a family of good kernels, and g is continuous with compact support, we may argue as in the proof of Corollary 1.7 to see that g ∗Kδ converges uniformly to g as δ tends to 0. In fact, we choose δ0 so that |g(x)− (g ∗Kδ0)(x)| < ²/2 for all x ∈ R. Now, we recall that ex is given by the power series expansion ex =∑∞ n=0 x n/n! which converges uniformly in every compact interval of R. Therefore, there exists an integer N so that |Kδ0(x)−R(x)| ≤ ² 4MB for all x ∈ [−2M, 2M ] where R(x) = δ−1/20 ∑N n=0 (−πx2/δ0)n n! . Then, recalling that g vanishes outside the interval [−M, M ], we have that for all x ∈ [−M, M ] |(g ∗Kδ0)(x)− (g ∗R)(x)| = ∣∣∣∣∣ ∫ M −M g(t) [Kδ0(x− t)−R(x− t)] dt ∣∣∣∣∣ ≤ ∫ M −M |g(t)| |Kδ0(x− t)−R(x− t)| dt ≤ 2MB sup z∈[−2M,2M ] |Kδ0(z)−R(z)| < ²/2. Therefore, the triangle inequality implies that |g(x)− (g ∗R)(x)| < ² whenever x ∈ [−M, M ], hence |f(x)− (g ∗R)(x)| < ² when x ∈ [a, b]. Finally, note that g ∗R is a polynomial in the x variable. Indeed, by definition we have (g ∗R)(x) = ∫ M −M g(t)R(x− t) dt, and R(x− t) is a polynomial in x since it can be expressed, after several expansions, as R(x− t) = ∑n an(t)xn where the sum is finite. This concludes the proof of the theorem. 2 Applications to some partial differential equations We mentioned earlier that a crucial property of the Fourier transform is that it interchanges differentiation and multiplication by polynomials. We now use this crucial fact together with the Fourier inversion theorem to solve some specific partial differential equations. 2.1 The time-dependent heat equation on the real line In Chapter 4 we considered the heat equation on the circle. Here we study the analogous problem on the real line. Ibookroot October 20, 2007 146 Chapter 5. THE FOURIER TRANSFORM ON R Consider an infinite rod, which we model by the real line, and suppose that we are given an initial temperature distribution f(x) on the rod at time t = 0. We wish now to determine the temperature u(x, t) at a point x at time t > 0. Considerations similar to the ones given in
Chapter 1 show that when u is appropriately normalized, it solves the
following partial differential equation:

(8)
∂u

∂t
=

∂2u

∂x2
,

called the heat equation. The initial condition we impose is
u(x, 0) = f(x).

Just as in the case of the circle, the solution is given in terms of a
convolution. Indeed, define the heat kernel of the line by

Ht(x) = Kδ(x), with δ = 4πt,

so that

Ht(x) =
1

(4πt)1/2
e−x

2/4t and Ĥt(ξ) = e−4π
2tξ2 .

Taking the Fourier transform of equation (8) in the x variable (for-
mally) leads to

∂û

∂t
(ξ, t) = −4π2ξ2û(ξ, t).

Fixing ξ, this is an ordinary differential equation in the variable t (with
unknown û(ξ, ·)), so there exists a constant A(ξ) so that

û(ξ, t) = A(ξ)e−4π
2ξ2t.

We may also take the Fourier transform of the initial condition and obtain
û(ξ, 0) = f̂(ξ), hence A(ξ) = f̂(ξ). This leads to the following theorem.

Theorem 2.1 Given f ∈ S(R), let

u(x, t) = (f ∗ Ht)(x) for t > 0

where Ht is the heat kernel. Then:
(i) The function u is C2 when x ∈ R and t > 0, and u solves the heat

equation.

Ibookroot October 20, 2007

2. Applications to some partial differential equations 147

(ii) u(x, t) → f(x) uniformly in x as t → 0. Hence if we set u(x, 0) =
f(x), then u is continuous on the closure of the upper half-plane

R2+ = {(x, t) : x ∈ R, t ≥ 0}.
(iii)

∫∞
−∞ |u(x, t)− f(x)|2 dx → 0 as t → 0.

Proof. Because u = f ∗ Ht, taking the Fourier transform in the x-
variable gives û = f̂Ĥt, and so û(ξ, t) = f̂(ξ)e−4π

2ξ2t. The Fourier inver-
sion formula gives

u(x, t) =
∫ ∞
−∞

f̂(ξ)e−4π
2tξ2e2πiξx dξ.

By differentiating under the integral sign, one verifies (i). In fact, one
observes that u is indefinitely differentiable. Note that (ii) is an imme-
diate consequence of Corollary 1.7. Finally, by Plancherel’s formula, we
have

∫ ∞
−∞

|u(x, t)− f(x)|2 dx =
∫ ∞
−∞

|û(ξ, t)− f̂(ξ)|2 dξ

=
∫ ∞
−∞

|f̂(ξ)|2 |e−4π2tξ2 − 1| dξ.

To see that this last integral goes to 0 as t → 0, we argue as follows:
since |e−4π2tξ2 − 1| ≤ 2 and f ∈ S(R), we can find N so that

∫

|ξ|≥N
|f̂(ξ)|2|e−4π2tξ2 − 1| dξ < ², and for all small t we have sup|ξ|≤N |f̂(ξ)|2|e−4π 2tξ2 − 1| < ²/2N since f̂ is bounded. Thus ∫ |ξ|≤N |f̂(ξ)|2 |e−4π2tξ2 − 1| dξ < ² for all small t. This completes the proof of the theorem. The above theorem guarantees the existence of a solution to the heat equation with initial data f . This solution is also unique, if uniqueness is formulated appropriately. In this regard, we note that u = f ∗ Ht, f ∈ S(R), satisfies the following additional property. Corollary 2.2 u(·, t) belongs to S(R) uniformly in t, in the sense that for any T > 0

(9) sup
x ∈ R

0 < t < T |x|k ∣∣∣∣∣ ∂` ∂x` u(x, t) ∣∣∣∣∣ < ∞ for each k, ` ≥ 0. Ibookroot October 20, 2007 148 Chapter 5. THE FOURIER TRANSFORM ON R Proof. This result is a consequence of the following estimate: |u(x, t)| ≤ ∫ |y|≤|x|/2 |f(x− y)|Ht(y) dy + ∫ |y|≥|x|/2 |f(x− y)|Ht(y) dy ≤ CN (1 + |x|)N + C√ t e−cx 2/t. Indeed, since f is rapidly decreasing, we have |f(x− y)| ≤ CN/(1 + |x|)N when |y| ≤ |x|/2. Also, if |y| ≥ |x|/2 thenHt(y) ≤ Ct−1/2e−cx 2/t, and we obtain the above inequality. Consequently, we see that u(x, t) is rapidly decreasing uniformly for 0 < t < T . The same argument can be applied to the derivatives of u in the x variable since we may differentiate under the integral sign and apply the above estimate with f replaced by f ′, and so on. This leads to the following uniqueness theorem. Theorem 2.3 Suppose u(x, t) satisfies the following conditions: (i) u is continuous on the closure of the upper half-plane. (ii) u satisfies the heat equation for t > 0.

(iii) u satisfies the boundary condition u(x, 0) = 0.

(iv) u(·, t) ∈ S(R) uniformly in t, as in (9).
Then, we conclude that u = 0.

Below we use the abbreviations ∂`xu and ∂tu to denote ∂
`u/∂x` and

∂u/∂t, respectively.

Proof. We define the energy at time t of the solution u(x, t) by

E(t) =
∫

R
|u(x, t)|2 dx.

Clearly E(t) ≥ 0. Since E(0) = 0 it suffices to show that E is a de-
creasing function, and this is achieved by proving that dE/dt ≤ 0. The
assumptions on u allow us to differentiate E(t) under the integral sign

dt
=

∫

R
[∂tu(x, t)u(x, t) + u(x, t)∂tu(x, t)] dx.

But u satisfies the heat equation, therefore ∂tu = ∂2xu and ∂tu = ∂
2
xu, so

that after an integration by parts, where we use the fact that u and its

Ibookroot October 20, 2007

2. Applications to some partial differential equations 149

x derivatives decrease rapidly as |x| → ∞, we find

dt
=

∫

[
∂2xu(x, t)u(x, t) + u(x, t)∂

2
xu(x, t)

]
dx

= −
∫

R
[∂xu(x, t)∂xu(x, t) + ∂xu(x, t)∂xu(x, t)] dx

= −2
∫

R
|∂xu(x, t)|2 dx

≤ 0,

as claimed. Thus E(t) = 0 for all t, hence u = 0.

Another uniqueness theorem for the heat equation, with a less restric-
tive assumption than (9), can be found in Problem 6. Examples when
uniqueness fails are given in Exercise 12 and Problem 4.

2.2 The steady-state heat equation in the upper half-plane

The equation we are now concerned with is

(10) 4u = ∂
2u

∂x2
+

∂2u

∂y2
= 0

in the upper half-plane R2+ = {(x, y) : x ∈ R, y > 0}. The boundary con-
dition we require is u(x, 0) = f(x). The operator 4 is the Laplacian and
the above partial differential equation describes the steady-state heat dis-
tribution in R2+ subject to u = f on the boundary. The kernel that solves
this problem is called the Poisson kernel for the upper half-plane, and
is given by

Py(x) =
1
π

x2 + y2
where x ∈ R and y > 0.

This is the analogue of the Poisson kernel for the disc discussed in Sec-
tion 5.4 of Chapter 2.

Note that for each fixed y the kernel Py is only of moderate decrease
as a function of x, so we will use the theory of the Fourier transform
appropriate for these types of functions (see Section 1.7).

We proceed as in the case of the time-dependent heat equation, by
taking the Fourier transform of equation (10) (formally) in the x variable,
thereby obtaining

−4π2ξ2û(ξ, y) + ∂
2û

∂y2
(ξ, y) = 0

Ibookroot October 20, 2007

150 Chapter 5. THE FOURIER TRANSFORM ON R

with the boundary condition û(ξ, 0) = f̂(ξ). The general solution of this
ordinary differential equation in y (with ξ fixed) takes the form

û(ξ, y) = A(ξ)e−2π|ξ|y + B(ξ)e2π|ξ|y.

If we disregard the second term because of its rapid exponential increase
we find, after setting y = 0, that

û(ξ, y) = f̂(ξ)e−2π|ξ|y.

Therefore u is given in terms of the convolution of f with a kernel whose
Fourier transform is e−2π|ξ|y. This is precisely the Poisson kernel given
above, as we prove next.

Lemma 2.4 The following two identities hold:
∫ ∞
−∞

e−2π|ξ|ye2πiξx dξ =Py(x),
∫ ∞
−∞

Py(x)e−2πixξ dx = e−2π|ξ|y.

Proof. The first formula is fairly straightforward since we can split
the integral from −∞ to 0 and 0 to ∞. Then, since y > 0 we have

∫ ∞
0

e−2πξye2πiξx dξ =
∫ ∞
0

e2πi(x+iy)ξ dξ =

[
e2πi(x+iy)ξ

2πi(x + iy)

]∞

− 1
2πi(x + iy)

and similarly,
∫ 0
−∞

e2πξye2πiξx dξ =
1

2πi(x− iy) .

Therefore
∫ ∞
−∞

e−2π|ξ|ye2πiξx dξ =
1

2πi(x− iy) −
1

2πi(x + iy)
=

π(x2 + y2)
.

The second formula is now a consequence of the Fourier inversion theorem
applied in the case when f and f̂ are of moderate decrease.

Lemma 2.5 The Poisson kernel is a good kernel on R as y → 0.

Ibookroot October 20, 2007

2. Applications to some partial differential equations 151

Proof. Setting ξ = 0 in the second formula of the lemma shows that∫∞
−∞ Py(x) dx = 1, and clearly Py(x) ≥ 0, so it remains to check the last

property of good kernels. Given a fixed δ > 0, we may change variables
u = x/y so that

∫ ∞
δ

x2 + y2
dx =

∫ ∞
δ/y

1 + u2
= [arctanu]∞δ/y = π/2− arctan(δ/y),

and this quantity goes to 0 as y → 0. Since Py(x) is an even function,
the proof is complete.

The following theorem establishes the existence of a solution to our
problem.

Theorem 2.6 Given f ∈ S(R), let u(x, y) = (f ∗ Py)(x). Then:
(i) u(x, y) is C2 in R2+ and 4u = 0.
(ii) u(x, y) → f(x) uniformly as y → 0.
(iii)

∫∞
−∞ |u(x, y)− f(x)|2 dx → 0 as y → 0.

(iv) If u(x, 0) = f(x), then u is continuous on the closure R2+ of the
upper half-plane, and vanishes at infinity in the sense that

u(x, y) → 0 as |x|+ y →∞.

Proof. The proofs of parts (i), (ii), and (iii) are similar to the case of
the heat equation, and so are left to the reader. Part (iv) is a consequence
of two easy estimates whenever f is of moderate decrease. First, we have

|(f ∗ Py)(x)| ≤ C
(

1
(1 + x2)

+
y

x2 + y2

)

which is proved (as in the case of the heat equation) by splitting the
integral

∫∞
−∞ f(x− t)Py(t) dt into the part where |t| ≤ |x|/2 and the part

where |t| ≥ |x|/2. Also, we have |(f ∗ Py)(x)| ≤ C/y, since supx Py(x) ≤
c/y.

Using the first estimate when |x| ≥ |y| and the second when |x| ≤ |y|
gives the desired decrease at infinity.

We next show that the solution is essentially unique.

Theorem 2.7 Suppose u is continuous on the closure of the upper half-
plane R2+, satisfies 4u = 0 for (x, y) ∈ R2+, u(x, 0) = 0, and u(x, y) van-
ishes at infinity. Then u = 0.

Ibookroot October 20, 2007

152 Chapter 5. THE FOURIER TRANSFORM ON R

A simple example shows that a condition concerning the decay of u at
infinity is needed: take u(x, y) = y. Clearly u satisfies the steady-state
heat equation and vanishes on the real line, yet u is not identically zero.

The proof of the theorem relies on a basic fact about harmonic func-
tions, which are functions satisfying 4u = 0. The fact is that the value
of a harmonic function at a point equals its average value around any
circle centered at that point.

Lemma 2.8 (Mean-value property) Suppose Ω is an open set in R2
and let u be a function of class C2 with 4u = 0 in Ω. If the closure of
the disc centered at (x, y) and of radius R is contained in Ω, then

u(x, y) =
1
2π

∫ 2π
0

u(x + r cos θ, y + r sin θ) dθ

for all 0 ≤ r ≤ R.
Proof. Let U(r, θ) = u(x + r cos θ, y + r sin θ). Expressing the Lapla-

cian in polar coordinates, the equation 4u = 0 then implies

0 =
∂2U

∂θ2
+ r

∂

∂r

(
r
∂U

∂r

)
.

If we define F (r) = 1
2π

∫ 2π
0

U(r, θ) dθ, the above gives

r
∂

∂r

(
r
∂F

∂r

)
=

1
2π

∫ 2π
0

−∂
2U

∂θ2
(r, θ) dθ.

The integral of ∂2U/∂θ2 over the circle vanishes since ∂U/∂θ is peri-
odic, hence r ∂

∂r

(
r ∂F

∂r

)
= 0, and consequently r∂F/∂r must be constant.

Evaluating this expression at r = 0 we find that ∂F/∂r = 0. Thus F is
constant, but since F (0) = u(x, y), we finally find that F (r) = u(x, y) for
all 0 ≤ r ≤ R, which is the mean-value property.

Finally, note that the argument above is implicit in the proof of The-
orem 5.7, Chapter 2.

To prove Theorem 2.7 we argue by contradiction. Considering sepa-
rately the real and imaginary parts of u, we may suppose that u itself
is real-valued, and is somewhere strictly positive, say u(x0, y0) > 0 for
some x0 ∈ R and y0 > 0. We shall see that this leads to a contradiction.
First, since u vanishes at infinity, we can find a large semi-disc of ra-
dius R, D+R = {(x, y) : x2 + y2 ≤ R, y ≥ 0} outside of which u(x, y) ≤
1
2
u(x0, y0). Next, since u is continuous in D

+
R , it attains its maximum

M there, so there exists a point (x1, y1) ∈ D+R with u(x1, y1) = M , while

Ibookroot October 20, 2007

3. The Poisson summation formula 153

u(x, y) ≤ M in the semi-disc; also, since u(x, y) ≤ 1
2
u(x0, y0) ≤ M/2 out-

side of the semi-disc, we have u(x, y) ≤ M throughout the entire upper
half-plane. Now the mean-value property for harmonic functions implies

u(x1, y1) =
1
2π

∫ 2π
0

u(x1 + ρ cos θ, y1 + ρ sin θ) dθ

whenever the circle of integration lies in the upper half-plane. In par-
ticular, this equation holds if 0 < ρ < y1. Since u(x1, y1) equals the maximum value M , and u(x1 + ρ cos θ, y1 + ρ sin θ) ≤ M , it follows by continuity that u(x1 + ρ cos θ, y1 + ρ sin θ) = M on the whole circle. For otherwise u(x, y) ≤ M − ², on an arc of length δ > 0 on the circle, and
this would give

1
2π

∫ 2π
0

u(x1 + ρ cos θ, y1 + ρ sin θ) dθ ≤ M −
²δ

2π
< M, contradicting the fact that u(x1, y1) = M . Now letting ρ → y1, and using the continuity of u again, we see that this implies u(x1, 0) = M > 0,
which contradicts the fact that u(x, 0) = 0 for all x.

3 The Poisson summation formula

The definition of the Fourier transform was motivated by the desire for
a continuous version of Fourier series, applicable to functions defined
on the real line. We now show that there exists a further remarkable
connection between the analysis of functions on the circle and related
functions on R.

Given a function f ∈ S(R) on the real line, we can construct a new
function on the circle by the recipe

F1(x) =
∞∑

n=−∞
f(x + n).

Since f is rapidly decreasing, the series converges absolutely and uni-
formly on every compact subset of R, so F1 is continuous. Note that
F1(x + 1) = F1(x) because passage from n to n + 1 in the above sum
merely shifts the terms on the series defining F1(x). Hence F1 is periodic
with period 1. The function F1 is called the periodization of f .

There is another way to arrive at a “periodic version” of f , this time
by Fourier analysis. Start with the identity

f(x) =
∫ ∞
−∞

f̂(ξ)e2πiξx dξ,

Ibookroot October 20, 2007

154 Chapter 5. THE FOURIER TRANSFORM ON R

and consider its discrete analogue, where the integral is replaced by a
sum

F2(x) =
∞∑

n=−∞
f̂(n)e2πinx.

Once again, the sum converges absolutely and uniformly since f̂ belongs
to the Schwartz space, hence F2 is continuous. Moreover, F2 is also
periodic of period 1 since this is the case for each one of the exponentials
e2πinx.

The fundamental fact is that these two approaches, which produce F1
and F2, actually lead to the same function.

Theorem 3.1 (Poisson summation formula) If f ∈ S(R), then
∞∑

n=−∞
f(x + n) =

∞∑
n=−∞

f̂(n)e2πinx.

In particular, setting x = 0 we have

∞∑
n=−∞

f(n) =
∞∑

n=−∞
f̂(n).

In other words, the Fourier coefficients of the periodization of f are
given precisely by the values of the Fourier transform of f on the integers.

Proof. To check the first formula it suffices, by Theorem 2.1 in
Chapter 2, to show that both sides (which are continuous) have the
same Fourier coefficients (viewed as functions on the circle). Clearly, the
mth Fourier coefficient of the right-hand side is f̂(m). For the left-hand
side we have

∫ 1
0

( ∞∑
n=−∞

f(x + n)

)
e−2πimx dx =

∞∑
n=−∞

∫ 1
0

f(x + n)e−2πimx dx

=
∞∑

n=−∞

∫ n+1
n

f(y)e−2πimy dy

=
∫ ∞
−∞

f(y)e−2πimy dy

= f̂(m),

where the interchange of the sum and integral is permissible since f is
rapidly decreasing. This completes the proof of the theorem.

Ibookroot October 20, 2007

3. The Poisson summation formula 155

We observe that the theorem extends to the case when we merely
assume that both f and f̂ are of moderate decrease; the proof is in fact
unchanged.

It turns out that the operation of periodization is important in a
number of questions, even when the Poisson summation formula does
not apply. We give an example by considering the elementary function
f(x) = 1/x, x 6= 0. The result is that ∑∞n=−∞ 1/(x + n), when summed
symmetrically, gives the partial fraction decomposition of the cotangent
function. In fact this sum equals π cot πx, when x is not an integer.
Similarly with f(x) = 1/x2, we get

∑∞
n=−∞ 1/(x + n)

2 = π2/(sinπx)2,
whenever x /∈ Z (see Exercise 15).

3.1 Theta and zeta functions

We define the theta function ϑ(s) for s > 0 by

ϑ(s) =
∞∑

n=−∞
e−πn

2s.

The condition on s ensures the absolute convergence of the series. A
crucial fact about this special function is that it satisfies the following
functional equation.

Theorem 3.2 s−1/2ϑ(1/s) = ϑ(s) whenever s > 0.

The proof of this identity consists of a simple application of the Poisson
summation formula to the pair

f(x) = e−πsx
2

and f̂(ξ) = s−1/2e−πξ
2/s.

The theta function ϑ(s) also extends to complex values of s when
Re(s) > 0, and the functional equation is still valid then. The theta
function is intimately connected with an important function in number
theory, the zeta function ζ(s) defined for Re(s) > 1 by

ζ(s) =
∞∑

n=1

1
ns

Later we will see that this function carries essential information about
the prime numbers (see Chapter 8).

It also turns out that ζ, ϑ, and another important function Γ are
related by the following identity:

π−s/2Γ(s/2)ζ(s) =
1
2

∫ ∞
0

ts/2−1(ϑ(t)− 1) dt,

Ibookroot October 20, 2007

156 Chapter 5. THE FOURIER TRANSFORM ON R

which is valid for s > 1 (Exercises 17 and 18).

Returning to the function ϑ, define the generalization Θ(z|τ) given by

Θ(z|τ) =
∞∑

n=−∞
eiπn

2τe2πinz

whenever Im(τ) > 0 and z ∈ C. Taking z = 0 and τ = is we get Θ(z|τ) =
ϑ(s).

3.2 Heat kernels

Another application related to the Poisson summation formula and the
theta function is the time-dependent heat equation on the circle. A
solution to the equation

∂u

∂t
=

∂2u

∂x2

subject to u(x, 0) = f(x), where f is periodic of period 1, was given in
the previous chapter by

u(x, t) = (f ∗Ht)(x)

where Ht(x) is the heat kernel on the circle, that is,

Ht(x) =
∞∑

n=−∞
e−4π

2n2te2πinx.

Note in particular that with our definition of the generalized theta func-
tion in the previous section, we have Θ(x|4πit) = Ht(x). Also, recall that
the heat equation on R gave rise to the heat kernel

Ht(x) =
1

(4πt)1/2
e−x

2/4t

where Ĥt(ξ) = e−4π
2ξ2t. The fundamental relation between these two

objects is an immediate consequence of the Poisson summation formula:

Theorem 3.3 The heat kernel on the circle is the periodization of the
heat kernel on the real line:

Ht(x) =
∞∑

n=−∞
Ht(x + n).

Ibookroot October 20, 2007

3. The Poisson summation formula 157

Although the proof that Ht is a good kernel on R was fairly straightfor-
ward, we left open the harder problem that Ht is a good kernel on the
circle. The above results allow us to resolve this matter.

Corollary 3.4 The kernel Ht(x) is a good kernel for t → 0.
Proof. We already observed that

∫
|x|≤1/2 Ht(x) dx = 1. Now note

that Ht ≥ 0, which is immediate from the above formula since Ht ≥ 0.
Finally, we claim that when |x| ≤ 1/2,

Ht(x) = Ht(x) + Et(x),

where the error satisfies |Et(x)| ≤ c1e−c2/t with c1, c2 > 0 and 0 < t ≤ 1. To see this, note again that the formula in the theorem gives Ht(x) = Ht(x) + ∑ |n|≥1 Ht(x + n); therefore, since |x| ≤ 1/2, Et(x) = 1√ 4πt ∑ |n|≥1 e−(x+n) 2/4t ≤ Ct−1/2 ∑ n≥1 e−cn 2/t. Note that n2/t ≥ n2 and n2/t ≥ 1/t whenever 0 < t ≤ 1, so e−cn2/t ≤ e− c 2n 2 e− c 2 1 t . Hence |Et(x)| ≤ Ct−1/2e− c 2 1 t ∑ n≥1 e− c 2n 2 ≤ c1e−c2/t. The proof of the claim is complete, and as a result ∫ |x|≤1/2 |Et(x)| dx → 0 as t → 0. It is now clear that Ht satisfies ∫ η<|x|≤1/2 |Ht(x)| dx → 0 as t → 0, because Ht does. 3.3 Poisson kernels In a similar manner to the discussion above about the heat kernels, we state the relation between the Poisson kernels for the disc and the upper half-plane where Pr(θ) = 1− r2 1− 2r cos θ + r2 and Py(x) = 1 π y y2 + x2 . Ibookroot October 20, 2007 158 Chapter 5. THE FOURIER TRANSFORM ON R Theorem 3.5 Pr(2πx) = ∑ n∈Z Py(x + n) where r = e−2πy. This is again an immediate corollary of the Poisson summation formula applied to f(x) = Py(x) and f̂(ξ) = e−2π|ξ|y. Of course, here we use the Poisson summation formula under the assumptions that f and f̂ are of moderate decrease. 4 The Heisenberg uncertainty principle The mathematical thrust of the principle can be formulated in terms of a relation between a function and its Fourier transform. The basic under- lying law, formulated in its vaguest and most general form, states that a function and its Fourier transform cannot both be essentially localized. Somewhat more precisely, if the “preponderance” of the mass of a func- tion is concentrated in an interval of length L, then the preponderance of the mass of its Fourier transform cannot lie in an interval of length essentially smaller than L−1. The exact statement is as follows. Theorem 4.1 Suppose ψ is a function in S(R) which satisfies the nor- malizing condition ∫∞ −∞ |ψ(x)|2 dx = 1. Then (∫ ∞ −∞ x2|ψ(x)|2 dx ) (∫ ∞ −∞ ξ2|ψ̂(ξ)|2 dξ ) ≥ 1 16π2 , and equality holds if and only if ψ(x) = Ae−Bx 2 where B > 0 and |A|2 =√
2B/π.
In fact, we have

(∫ ∞
−∞

(x− x0)2|ψ(x)|2 dx
) (∫ ∞

−∞
(ξ − ξ0)2|ψ̂(ξ)|2 dξ

)
≥ 1

16π2

for every x0, ξ0 ∈ R.
Proof. The second inequality actually follows from the first by re-

placing ψ(x) by e−2πixξ0ψ(x + x0) and changing variables. To prove the
first inequality, we argue as follows. Beginning with our normalizing as-
sumption

∫
|ψ|2 = 1, and recalling that ψ and ψ′ are rapidly decreasing,

an integration by parts gives

1 =
∫ ∞
−∞

|ψ(x)|2 dx

= −
∫ ∞
−∞

x
d

dx
|ψ(x)|2 dx

= −
∫ ∞
−∞

(
xψ′(x)ψ(x) + xψ′(x)ψ(x)

)
dx.

Ibookroot October 20, 2007

4. The Heisenberg uncertainty principle 159

The last identity follows because |ψ|2 = ψψ. Therefore

1 ≤ 2
∫ ∞
−∞

|x| |ψ(x)| |ψ′(x)| dx

≤ 2
(∫ ∞

−∞
x2|ψ(x)|2 dx

)1/2 (∫ ∞
−∞

|ψ′(x)|2 dx
)1/2

where we have used the Cauchy-Schwarz inequality. The identity
∫ ∞
−∞

|ψ′(x)|2 dx = 4π2
∫ ∞
−∞

ξ2|ψ̂(ξ)|2 dξ,

which holds because of the properties of the Fourier transform and the
Plancherel formula, concludes the proof of the inequality in the theorem.

If equality holds, then we must also have equality where we applied the
Cauchy-Schwarz inequality, and as a result we find that ψ′(x) = βxψ(x)
for some constant β. The solutions to this equation are ψ(x) = Aeβx

2/2,
where A is constant. Since we want ψ to be a Schwartz function, we must
take β = −2B < 0, and since we impose the condition ∫∞ −∞ |ψ(x)|2 dx = 1 we find that |A|2 = √ 2B/π, as was to be shown. The precise assertion contained in Theorem 4.1 first came to light in the study of quantum mechanics. It arose when one considered the extent to which one could simultaneously locate the position and momentum of a particle. Assuming we are dealing with (say) an electron that travels along the real line, then according to the laws of physics, matters are governed by a “state function” ψ, which we can assume to be in S(R), and which is normalized according to the requirement that (11) ∫ ∞ −∞ |ψ(x)|2 dx = 1. The position of the particle is then determined not as a definite point x; instead its probable location is given by the rules of quantum mechanics as follows: • The probability that the particle is located in the interval (a, b) is∫ b a |ψ(x)|2 dx. According to this law we can calculate the probable location of the particle with the aid of ψ: in fact, there may be only a small probability that the particle is located in a given interval (a′, b′), but nevertheless it is somewhere on the real line since ∫∞ −∞ |ψ(x)|2 dx = 1. Ibookroot October 20, 2007 160 Chapter 5. THE FOURIER TRANSFORM ON R In addition to the probability density |ψ(x)|2dx, there is the ex- pectation of where the particle might be. This expectation is the best guess of the position of the particle, given its probability distribution determined by |ψ(x)|2dx, and is the quantity defined by (12) x = ∫ ∞ −∞ x|ψ(x)|2 dx. Why is this our best guess? Consider the simpler (idealized) situation where we are given that the particle can be found at only finitely many different points, x1, x2, . . . , xN on the real axis, with pi the probability that the particle is at xi, and p1 + p2 + · · ·+ pN = 1. Then, if we knew nothing else, and were forced to make one choice as to the position of the particle, we would naturally take x = ∑N i=1 xipi, which is the appropriate weighted average of the possible positions. The quantity (12) is clearly the general (integral) version of this. We next come to the notion of variance, which in our terminology is the uncertainty attached to our expectation. Having determined that the expected position of the particle is x (given by (12)), the resulting uncertainty is the quantity (13) ∫ ∞ −∞ (x− x)2|ψ(x)|2 dx. Notice that if ψ is highly concentrated near x, it means that there is a high probability that x is near x, and so (13) is small, because most of the contribution to the integral takes place for values of x near x. Here we have a small uncertainty. On the other hand, if ψ(x) is rather flat (that is, the probability distribution |ψ(x)|2dx is not very concentrated), then the integral (13) is rather big, because large values of (x− x)2 will come into play, and as a result the uncertainty is relatively large. It is also worthwhile to observe that the expectation x is that choice for which the uncertainty ∫∞ −∞(x− x)2|ψ(x)|2 dx is the smallest. Indeed, if we try to minimize this quantity by equating to 0 its derivative with respect to x, we find that 2 ∫∞ −∞(x− x)|ψ(x)|2 dx = 0, which gives (12). So far, we have discussed the “expectation” and “uncertainty” related to the position of the particle. Of equal relevance are the corresponding notions regarding its momentum. The corresponding rule of quantum mechanics is: • The probability that the momentum ξ of the particle belongs to the interval (a, b) is ∫ b a |ψ̂(ξ)|2 dξ where ψ̂ is the Fourier transform of ψ. Ibookroot October 20, 2007 5. Exercises 161 Combining these two laws with Theorem 4.1 gives 1/16π2 as the lower bound for the product of the uncertainty of the position and the uncer- tainty of the momentum of a particle. So the more certain we are about the location of the particle, the less certain we can be about its mo- mentum, and vice versa. However, we have simplified the statement of the two laws by rescaling to change the units of measurement. Actually, there enters a fundamental but small physical number ~ called Planck’s constant. When properly taken into account, the physical conclusion is (uncertainty of position)×(uncertainty of momentum) ≥ ~/16π2. 5 Exercises 1. Corollary 2.3 in Chapter 2 leads to the following simplified version of the Fourier inversion formula. Suppose f is a continuous function supported on an interval [−M,M ], whose Fourier transform f̂ is of moderate decrease. (a) Fix L with L/2 > M , and show that f(x) =
∑

an(L)e2πinx/L where

an(L) =
1
L

∫ L/2
−L/2

f(x)e−2πinx/L dx =
1
L

f̂(n/L).

Alternatively, we may write f(x) = δ
∑∞

n=−∞ f̂(nδ)e
2πinδx with δ = 1/L.

(b) Prove that if F is continuous and of moderate decrease, then

∫ ∞
−∞

F (ξ) dξ = lim
δ → 0
δ > 0

∞∑
n=−∞

F (δn).

f̂(ξ)e2πixξ dξ.

[Hint: For (a), note that the Fourier series of f on [−L/2, L/2] converges ab-
solutely. For (b), first approximate the integral by

∫ N
−N F and the sum by

δ
∑
|n|≤N/δ F (nδ). Then approximate the second integral by Riemann sums.]

2. Let f and g be the functions defined by

f(x) = χ[−1,1](x) =
{

1 if |x| ≤ 1,
0 otherwise,

and g(x) =
{

1− |x| if |x| ≤ 1,
0 otherwise.

Although f is not continuous, the integral defining its Fourier transform still
makes sense. Show that

f̂(ξ) =
sin 2πξ

πξ
and ĝ(ξ) =

(
sinπξ

πξ

)2
,

Ibookroot October 20, 2007

162 Chapter 5. THE FOURIER TRANSFORM ON R

with the understanding that f̂(0) = 2 and ĝ(0) = 1.

3. The following exercise illustrates the principle that the decay of f̂ is related
to the continuity properties of f .

(a) Suppose that f is a function of moderate decrease on R whose Fourier
transform f̂ is continuous and satisfies

f̂(ξ) = O
(

1
|ξ|1+α

)
as |ξ| → ∞

for some 0 < α < 1. Prove that f satisfies a Hölder condition of order α, that is, that |f(x + h)− f(x)| ≤ M |h|α for some M > 0 and all x, h ∈ R.

(b) Let f be a continuous function on R which vanishes for |x| ≥ 1, with
f(0) = 0, and which is equal to 1/ log(1/|x|) for all x in a neighborhood
of the origin. Prove that f̂ is not of moderate decrease. In fact, there is
no ² > 0 so that f̂(ξ) = O(1/|ξ|1+²) as |ξ| → ∞.

[Hint: For part (a), use the Fourier inversion formula to express f(x + h)− f(x)
as an integral involving f̂ , and estimate this integral separately for ξ in the two
ranges |ξ| ≤ 1/|h| and |ξ| ≥ 1/|h|.]

4. Bump functions. Examples of compactly supported functions in S(R) are
very handy in many applications in analysis. Some examples are:

(a) Suppose a < b, and f is the function such that f(x) = 0 if x ≤ a or x ≥ b and f(x) = e−1/(x−a)e−1/(b−x) if a < x < b. Show that f is indefinitely differentiable on R. (b) Prove that there exists an indefinitely differentiable function F on R such that F (x) = 0 if x ≤ a, F (x) = 1 if x ≥ b, and F is strictly increasing on [a, b]. (c) Let δ > 0 be so small that a + δ < b− δ. Show that there exists an indef- initely differentiable function g such that g is 0 if x ≤ a or x ≥ b, g is 1 on [a + δ, b− δ], and g is strictly monotonic on [a, a + δ] and [b− δ, b]. [Hint: For (b) consider F (x) = c ∫ x −∞ f(t) dt where c is an appropriate constant.] 5. Suppose f is continuous and of moderate decrease. (a) Prove that f̂ is continuous and f̂(ξ) → 0 as |ξ| → ∞. Ibookroot October 20, 2007 5. Exercises 163 (b) Show that if f̂(ξ) = 0 for all ξ, then f is identically 0. [Hint: For part (a), show that f̂(ξ) = 1 2 ∫∞ −∞[f(x)− f(x− 1/(2ξ))]e−2πixξ dx. For part (b), verify that the multiplication formula ∫ f(x)ĝ(x) dx = ∫ f̂(y)g(y) dy still holds whenever g ∈ S(R).] 6. The function e−πx 2 is its own Fourier transform. Generate other functions that (up to a constant multiple) are their own Fourier transforms. What must the constant multiples be? To decide this, prove that F4 = I. Here F(f) = f̂ is the Fourier transform, F4 = F ◦ F ◦ F ◦ F , and I is the identity operator (If)(x) = f(x) (see also Problem 7). 7. Prove that the convolution of two functions of moderate decrease is a function of moderate decrease. [Hint: Write ∫ f(x− y)g(y) dy = ∫ |y|≤|x|/2 + ∫ |y|≥|x|/2 . In the first integral f(x− y) = O(1/(1 + x2)) while in the second integral g(y) = O(1/(1 + x2)).] 8. Prove that f is continuous, of moderate decrease, and ∫∞ −∞f(y)e −y2e2xydy =0 for all x ∈ R, then f = 0. [Hint: Consider f ∗ e−x2 .] 9. If f is of moderate decrease, then (14) ∫ R −R ( 1− |ξ| R ) f̂(ξ)e2πixξ dξ = (f ∗ FR)(x), where the Fejér kernel on the real line is defined by FR(t) =    R ( sinπtR πtR )2 if t 6= 0, R if t = 0. Show that {FR} is a family of good kernels as R →∞, and therefore (14) tends uniformly to f(x) as R →∞. This is the analogue of Fejér’s theorem for Fourier series in the context of the Fourier transform. 10. Below is an outline of a different proof of the Weierstrass approximation theorem. Ibookroot October 20, 2007 164 Chapter 5. THE FOURIER TRANSFORM ON R Define the Landau kernels by Ln(x) =    (1− x2)n cn if −1 ≤ x ≤ 1, 0 if |x| ≥ 1, where cn is chosen so that ∫∞ −∞ Ln(x) dx = 1. Prove that {Ln}n≥0 is a family of good kernels as n →∞. As a result, show that if f is a continuous func- tion supported in [−1/2, 1/2], then (f ∗ Ln)(x) is a sequence of polynomials on [−1/2, 1/2] which converges uniformly to f . [Hint: First show that cn ≥ 2/(n + 1).] 11. Suppose that u is the solution to the heat equation given by u = f ∗ Ht where f ∈ S(R). If we also set u(x, 0) = f(x), prove that u is continuous on the closure of the upper half-plane, and vanishes at infinity, that is, u(x, t) → 0 as |x|+ t →∞. [Hint: To prove that u vanishes at infinity, show that (i) |u(x, t)| ≤ C/ √ t and (ii) |u(x, t)| ≤ C/(1 + |x|2) + Ct−1/2e−cx2/t. Use (i) when |x| ≤ t, and (ii) other- wise.] 12. Show that the function defined by u(x, t) = x t Ht(x) satisfies the heat equation for t > 0 and limt→0 u(x, t) = 0 for every x, but u is
not continuous at the origin.
[Hint: Approach the origin with (x, t) on the parabola x2/4t = c where c is a
constant.]

13. Prove the following uniqueness theorem for harmonic functions in the strip
{(x, y) : 0 < y < 1, −∞ < x < ∞}: if u is harmonic in the strip, continuous on its closure with u(x, 0) = u(x, 1) = 0 for all x ∈ R, and u vanishes at infinity, then u = 0. 14. Prove that the periodization of the Fejér kernel FN on the real line (Exer- cise 9) is equal to the Fejér kernel for periodic functions of period 1. In other words, ∞∑ n=−∞ FN (x + n) = FN (x), when N ≥ 1 is an integer, and where FN (x) = N∑ n=−N ( 1− |n| N ) e2πinx = 1 N sin2(Nπx) sin2(πx) . Ibookroot October 20, 2007 5. Exercises 165 15. This exercise provides another example of periodization. (a) Apply the Poisson summation formula to the function g in Exercise 2 to obtain ∞∑ n=−∞ 1 (n + α)2 = π2 (sin πα)2 whenever α is real, but not equal to an integer. (b) Prove as a consequence that (15) ∞∑ n=−∞ 1 (n + α) = π tan πα whenever α is real but not equal to an integer. [Hint: First prove it when 0 < α < 1. To do so, integrate the formula in (b). What is the precise meaning of the series on the left-hand side of (15)? Evaluate at α = 1/2.] 16. The Dirichlet kernel on the real line is defined by ∫ R −R f̂(ξ)e2πixξ dξ = (f ∗ DR)(x) so that DR(x) = χ̂[−R,R](x) = sin(2πRx) πx . Also, the modified Dirichlet kernel for periodic functions of period 1 is defined by D∗N (x) = ∑ |n|≤N−1 e2πinx + 1 2 (e−2πiNx + e2πiNx). Show that the result in Exercise 15 gives ∞∑ n=−∞ DN (x + n) = D∗N (x), where N ≥ 1 is an integer, and the infinite series must be summed symmetrically. In other words, the periodization of DN is the modified Dirichlet kernel D∗N . 17. The gamma function is defined for s > 0 by

Γ(s) =
∫ ∞

e−xxs−1 dx.

(a) Show that for s > 0 the above integral makes sense, that is, that the
following two limits exist:

lim
δ → 0
δ > 0

∫ 1
δ

e−xxs−1 dx and lim
A→∞

∫ A
1

e−xxs−1 dx.

Ibookroot October 20, 2007

166 Chapter 5. THE FOURIER TRANSFORM ON R

(b) Prove that Γ(s + 1) = sΓ(s) whenever s > 0, and conclude that for every
integer n ≥ 1 we have Γ(n + 1) = n!.

Γ
(

1
2

)
=
√

π and Γ
(

3
2

)
=
√

2
.

[Hint: For (c), use
∫∞
−∞ e

−πx2 dx = 1.]

18. The zeta function is defined for s > 1 by ζ(s) =
∑∞

n=1
1/ns. Verify the

identity

π−s/2Γ(s/2)ζ(s) =
1
2

∫ ∞
0

t
s
2−1(ϑ(t)− 1) dt whenever s > 1

where Γ and ϑ are the gamma and theta functions, respectively:

Γ(s) =
∫ ∞

e−tts−1 dt and ϑ(s) =
∞∑

n=−∞
e−πn

2s.

More about the zeta function and its relation to the prime number theorem can
be found in Book II.

19. The following is a variant of the calculation of ζ(2m) =
∑∞

n=1
1/n2m found

in Problem 4, Chapter 3.

(a) Apply the Poisson summation formula to f(x) = t/(π(x2 + t2))
and f̂(ξ) = e−2πt|ξ| where t > 0 in order to get

1
π

∞∑
n=−∞

t2 + n2
=

∞∑
n=−∞

e−2πt|n|.

(b) Prove the following identity valid for 0 < t < 1: 1 π ∞∑ n=−∞ t t2 + n2 = 1 πt + 2 π ∞∑ m=1 (−1)m+1ζ(2m)t2m−1 as well as ∞∑ n=−∞ e−2πt|n| = 2 1− e−2πt − 1. Ibookroot October 20, 2007 5. Exercises 167 (c) Use the fact that z ez − 1 = 1− z 2 + ∞∑ m=1 B2m (2m)! z2m, where Bk are the Bernoulli numbers to deduce from the above formula, 2ζ(2m) = (−1)m+1 (2π) 2m (2m)! B2m. 20. The following results are relevant in information theory when one tries to recover a signal from its samples. Suppose f is of moderate decrease and that its Fourier transform f̂ is sup- ported in I = [−1/2, 1/2]. Then, f is entirely determined by its restriction to Z. This means that if g is another function of moderate decrease whose Fourier transform is supported in I and f(n) = g(n) for all n ∈ Z, then f = g. More precisely: (a) Prove that the following reconstruction formula holds: f(x) = ∞∑ n=−∞ f(n)K(x− n) where K(y) = sinπy πy . Note that K(y) = O(1/|y|) as |y| → ∞. (b) If λ > 1, then

f(x) =
∞∑

n=−∞

1
λ

f
(

)
Kλ

(
x− n

)
where Kλ(y) =

cos πy − cosπλy
π2(λ− 1)y2 .

Thus, if one samples f “more often,” the series in the reconstruction
formula converges faster since Kλ(y) = O(1/|y|2) as |y| → ∞. Note that
Kλ(y) → K(y) as λ → 1.

|f(x)|2 dx =
∞∑

n=−∞
|f(n)|2.

[Hint: For part (a) show that if χ is the characteristic function of I, then
f̂(ξ) = χ(ξ)

∑∞
n=−∞ f(n)e

−2πinξ. For (b) use the function in Figure 2 instead
of χ(ξ).]

21. Suppose that f is continuous on R. Show that f and f̂ cannot both be
compactly supported unless f = 0. This can be viewed in the same spirit as the
uncertainty principle.

Ibookroot October 20, 2007

168 Chapter 5. THE FOURIER TRANSFORM ON R

1
2

− 1
2

λ
2

−λ
2

Figure 2. The function in Exercise 20

[Hint: Assume f is supported in [0, 1/2]. Expand f in a Fourier series in the
interval [0, 1], and note that as a result, f is a trigonometric polynomial.]

22. The heuristic assertion stated before Theorem 4.1 can be made precise as
follows. If F is a function on R, then we say that the preponderance of its mass
is contained in an interval I (centered at the origin) if

(16)
∫

x2|F (x)|2 dx ≥ 1
2

∫

R
x2|F (x)|2 dx.

Now suppose f ∈ S, and (16) holds with F = f and I = I1; also with F = f̂ and
I = I2. Then if Lj denotes the length of Ij , we have

L1L2 ≥
1
2π

A similar conclusion holds if the intervals are not necessarily centered at the
origin.

23. The Heisenberg uncertainty principle can be formulated in terms of the
operator L = − d2

dx2
+ x2, which acts on Schwartz functions by the formula

L(f) = −d
2f

dx2
+ x2f.

This operator, sometimes called the Hermite operator, is the quantum ana-
logue of the harmonic oscillator. Consider the usual inner product on S given
by

(f, g) =
∫ ∞
−∞

f(x)g(x) dx whenever f, g ∈ S.

Ibookroot October 20, 2007

6. Problems 169

(a) Prove that the Heisenberg uncertainty principle implies

(Lf, f) ≥ (f, f) for all f ∈ S.

This is usually denoted by L ≥ I. [Hint: Integrate by parts.]
(b) Consider the operators A and A∗ defined on S by

A(f) =
df

dx
+ xf and A∗(f) = − df

dx
+ xf.

The operators A and A∗ are sometimes called the annihilation and cre-
ation operators, respectively. Prove that for all f, g ∈ S we have

(i) (Af, g) = (f, A∗g),

(ii) (Af, Af) = (A∗Af, f) ≥ 0,
(iii) A∗A = L− I.
In particular, this again shows that L ≥ I.

At(f) =
df

dx
+ txf and A∗t (f) = −

dx
+ txf.

Use the fact that (A∗t Atf, f) ≥ 0 to give another proof of the Heisenberg
uncertainty principle which says that whenever

∫∞
−∞ |f(x)|2 dx = 1 then

(∫ ∞
−∞

x2|f(x)|2 dx
)(∫ ∞

−∞

∣∣∣∣
df

∣∣∣∣
2

)
≥ 1/4.

[Hint: Think of (A∗t Atf, f) as a quadratic polynomial in t.]

6 Problems

1. The equation

(17) x2
∂2u

∂x2
+ ax

∂u

∂x
=

∂u

∂t

with u(x, 0) = f(x) for 0 < x < ∞ and t > 0 is a variant of the heat equation
which occurs in a number of applications. To solve (17), make the change of vari-
ables x = e−y so that −∞ < y < ∞. Set U(y, t) = u(e−y, t) and F (y) = f(e−y). Then the problem reduces to the equation ∂2U ∂y2 + (1− a)∂U ∂y = ∂U ∂t , Ibookroot October 20, 2007 170 Chapter 5. THE FOURIER TRANSFORM ON R with U(y, 0) = F (y). This can be solved like the usual heat equation (the case a = 1) by taking the Fourier transform in the y variable. One must then compute the integral ∫∞ −∞ e (−4π2ξ2+(1−a)2πiξ)te2πiξv dξ. Show that the solution of the original problem is then given by u(x, t) = 1 (4πt)1/2 ∫ ∞ 0 e−(log(v/x)+(1−a)t) 2/(4t)f(v) dv v . 2. The Black-Scholes equation from finance theory is (18) ∂V ∂t + rs ∂V ∂s + σ2s2 2 ∂2V ∂s2 − rV = 0, 0 < t < T , subject to the “final” boundary condition V (s, T ) = F (s). An appropriate change of variables reduces this to the equation in Problem 1. Alternatively, the substi- tution V (s, t) = eax+bτU(x, τ) where x = log s, τ = σ 2 2 (T − t), a = 1 2 − r σ2 , and b = − ( 1 2 + r σ2 )2 reduces (18) to the one-dimensional heat equation with the ini- tial condition U(x, 0) = e−axF (ex). Thus a solution to the Black-Scholes equa- tion is V (s, t) = e−r(T−t)√ 2πσ2(T − t) ∫ ∞ 0 e − (log(s/s ∗)+(r−σ2/2)(T−t))2 2σ2(T−t) F (s∗) ds∗. 3. ∗ The Dirichlet problem in a strip. Consider the equation 4u = 0 in the horizontal strip {(x, y) : 0 < y < 1, −∞ < x < ∞} with boundary conditions u(x, 0) = f0(x) and u(x, 1) = f1(x), where f0 and f1 are both in the Schwartz space. (a) Show (formally) that if u is a solution to this problem, then û(ξ, y) = A(ξ)e2πξy + B(ξ)e−2πξy. Express A and B in terms of f̂0 and f̂1, and show that û(ξ, y) = sinh(2π(1− y)ξ) sinh(2πξ) f̂0(ξ) + sinh(2πyξ) sinh(2πξ) f̂0(ξ). (b) Prove as a result that ∫ ∞ −∞ |u(x, y)− f0(x)|2 dx → 0 as y → 0 and ∫ ∞ −∞ |u(x, y)− f1(x)|2 dx → 0 as y → 1. Ibookroot October 20, 2007 6. Problems 171 (c) If Φ(ξ) = (sinh 2πaξ)/(sinh 2πξ), with 0 ≤ a < 1, then Φ is the Fourier transform of ϕ where ϕ(x) = sinπa 2 · 1 cosh πx + cos πa . This can be shown, for instance, by using contour integration and the residue formula from complex analysis (see Book II, Chapter 3). (d) Use this result to express u in terms of Poisson-like integrals involving f0 and f1 as follows: u(x, y) = sin πy 2 (∫ ∞ −∞ f0(x− t) cosh πt− cos πy dt + ∫ ∞ −∞ f1(x− t) cosh πt + cos πy dt ) . (e) Finally, one can check that the function u(x, y) defined by the above ex- pression is harmonic in the strip, and converges uniformly to f0(x) as y → 0, and to f1(x) as y → 1. Moreover, one sees that u(x, y) vanishes at infinity, that is, lim|x|→∞ u(x, y) = 0, uniformly in y. In Exercise 12, we gave an example of a function that satisfies the heat equation in the upper half-plane, with boundary value 0, but which was not identically 0. We observed in this case that u was in fact not continuous up to the boundary. In Problem 4 we exhibit examples illustrating non-uniqueness, but this time with continuity up to the boundary t = 0. These examples satisfy a growth condition at infinity, namely |u(x, t)| ≤ Cecx2+² , for any ² > 0. Problems 5 and 6 show
that under the more restrictive growth condition |u(x, t)| ≤ Cecx2 , uniqueness does
hold.

4.∗ If g is a smooth function on R, define the formal power series

(19) u(x, t) =
∞∑

n=0

g(n)(t)
x2n

(2n!)
.

(a) Check formally that u solves the heat equation.

(b) For a > 0, consider the function defined by

g(t) =
{

e−t
−a

if t > 0
0 if t ≤ 0.

One can show that there exists 0 < θ < 1 depending on a so that |g(k)(t)| ≤ k! (θt)k e− 1 2 t −a for t > 0.

(c) As a result, for each x and t the series (19) converges; u solves the heat
equation; u vanishes for t = 0; and u satisfies the estimate |u(x, t)| ≤
Cec|x|

2a/(a−1)
for some constants C, c > 0.

Ibookroot October 20, 2007

172 Chapter 5. THE FOURIER TRANSFORM ON R

(d) Conclude that for every ² > 0 there exists a non-zero solution to the heat
equation which is continuous for x ∈ R and t ≥ 0, which satisfies u(x, 0) =
0 and |u(x, t)| ≤ Cec|x|2+² .

5.∗ The following “maximum principle” for solutions of the heat equation will
be used in the next problem.

Theorem. Suppose that u(x, t) is a real-valued solution of the heat equation
in the upper half-plane, which is continuous on its closure. Let R denote the
rectangle

R = {(x, y) ∈ R2 : a ≤ x ≤ b, 0 ≤ t ≤ c}
and ∂′R be the part of the boundary of R which consists of the two vertical sides
and its base on the line t = 0 (see Figure 3). Then

min
(x,t)∈∂′R

u(x, t) = min
(x,t)∈R

u(x, t) and max
(x,t)∈∂′R

u(x, t) = max
(x,t)∈R

u(x, t).

∂′R

Figure 3. The rectangle R and part of its boundary ∂′R

The steps leading to a proof of this result are outlined below.

(a) Show that it suffices to prove that if u ≥ 0 on ∂′R, then u ≥ 0 in R.
(b) For ² > 0, let

v(x, t) = u(x, t) + ²t.

Then, v has a minimum on R, say at (x1, t1). Show that x1 = a or b,
or else t1 = 0. To do so, suppose on the contrary that a < x1 < b and 0 < t1 ≤ c, and prove that vxx(x1, t1)− vt(x1, t1) ≤ −². However, show also that the left-hand side must be non-negative. (c) Deduce from (b) that u(x, t) ≥ ²(t1 − t) for any (x, t) ∈ R and let ² → 0. 6.∗ The examples in Problem 4 are optimal in the sense of the following unique- ness theorem due to Tychonoff. Ibookroot October 20, 2007 6. Problems 173 Theorem. Suppose u(x, t) satisfies the following conditions: (i) u(x, t) solves the heat equation for all x ∈ R and and all t > 0.
(ii) u(x, t) is continuous for all x ∈ R and 0 ≤ t ≤ c.
(iii) u(x, 0) = 0.

(iv) |u(x, t)| ≤ Meax2 for some M , a, and all x ∈ R, 0 ≤ t < c. Then u is identically equal to 0. 7.∗ The Hermite functions hk(x) are defined by the generating identity ∞∑ k=0 hk(x) tk k! = e−(x 2/2−2tx+t2). (a) Show that an alternate definition of the Hermite functions is given by the formula hk(x) = (−1)kex 2/2 ( d dx )k e−x 2 . [Hint: Write e−(x 2/2−2tx+t2) = ex 2/2e−(x−t) 2 and use Taylor’s formula.] Conclude from the above expression that each hk(x) is of the form Pk(x)e−x 2/2, where Pk is a polynomial of degree k. In particular, the Her- mite functions belong to the Schwartz space and h0(x) = e−x 2/2, h1(x) = 2xe−x 2/2. (b) Prove that the family {hk}∞k=0 is complete in the sense that if f is a Schwartz function, and (f, hk) = ∫ ∞ −∞ f(x)hk(x) dx = 0 for all k ≥ 0, then f = 0. [Hint: Use Exercise 8.] (c) Define h∗k(x) = hk((2π) 1/2x). Then ĥ∗k(ξ) = (−i)kh∗k(ξ). Therefore, each h∗k is an eigenfunction for the Fourier transform. (d) Show that hk is an eigenfunction for the operator defined in Exercise 23, and in fact, prove that Lhk = (2k + 1)hk. In particular, we conclude that the functions hk are mutually orthogonal for the L2 inner product on the Schwartz space. (e) Finally, show that ∫∞ −∞[hk(x)] 2 dx = π1/22kk!. [Hint: Square the generat- ing relation.] Ibookroot October 20, 2007 174 Chapter 5. THE FOURIER TRANSFORM ON R 8.∗ To refine the results in Chapter 4, and to prove that fα(x) = ∞∑ n=0 2−nαe2πi2 nx is nowhere differentiable even in the case α = 1, we need to consider a variant of the delayed means4N , which in turn will be analyzed by the Poisson summation formula. (a) Fix an indefinitely differentiable function Φ satisfying Φ(ξ) = { 1 when |ξ| ≤ 1, 0 when |ξ| ≥ 2. By the Fourier inversion formula, there exists ϕ ∈ S so that ϕ̂(ξ) = Φ(ξ). Let ϕN (x) = Nϕ(Nx) so that ϕ̂N (ξ) = Φ(ξ/N). Finally, set 4̃N (x) = ∞∑ n=−∞ ϕN (x + n). Observe by the Poisson summation formula that 4̃N (x) =∑∞ n=−∞ Φ(n/N)e 2πinx, thus 4̃N is a trigonometric polynomial of degree ≤ 2N , with terms whose coefficients are 1 when |n| ≤ N . Let 4̃N (f) = f ∗ 4̃N . Note that SN (fα) = 4̃N ′(fα) where N ′ is the largest integer of the form 2k with N ′ ≤ N . (b) If we set 4̃N (x) = ϕN (x) + EN (x) where EN (x) = ∑ |n|≥1 ϕN (x + n), then one sees that: (i) sup|x|≤1/2 |E′N (x)| → 0 as N →∞. (ii) |4̃′N (x)| ≤ cN2. (iii) |4̃′N (x)| ≤ c/(N |x|3), for |x| ≤ 1/2. Moreover, ∫ |x|≤1/2 4̃′N (x) dx = 0, and − ∫ |x|≤1/2 x4̃′N (x) dx → 1 as N → ∞. (c) The above estimates imply that if f ′(x0) exists, then (f ∗ 4̃′N )(x0 + hN ) → f ′(x0) as N →∞, whenever |hN | ≤ C/N . Then, conclude that both the real and imaginary parts of f1 are nowhere differentiable, as in the proof given in Section 3, Chapter 4. Ibookroot October 20, 2007 6 The Fourier Transform on Rd It occurred to me that in order to improve treatment planning one had to know the distribution of the at- tenuation coefficient of tissues in the body. This in- formation would be useful for diagnostic purposes and would constitute a tomogram or series of tomograms. It was immediately evident that the problem was a mathematical one. If a fine beam of gamma rays of intensity I0 is incident on the body and the emerg- ing density is I, then the measurable quantity g equals log(I0/I) = ∫ L f ds, where f is the variable absorption coefficient along the line L. Hence if f is a function of two dimensions, and g is known for all lines intersect- ing the body, the question is, can f be determined if g is known? Fourteen years would elapse before I learned that Radon had solved this problem in 1917. A. M. Cormack, 1979 The previous chapter introduced the theory of the Fourier transform on R and illustrated some of its applications to partial differential equa- tions. Here, our aim is to present an analogous theory for functions of several variables. After a brief review of some relevant notions in Rd, we begin with some general facts about the Fourier transform on the Schwartz space S(Rd). Fortunately, the main ideas and techniques have already been considered in the one-dimensional case. In fact, with the appropriate notation, the statements (and proofs) of the key theorems, such as the Fourier inversion and Plancherel formulas, remain unchanged. Next, we highlight the connection to some higher dimensional prob- lems in mathematical physics, and in particular we investigate the wave equation in d dimensions, with a detailed analysis in the cases d = 3 and d = 2. At this stage, we discover a rich interplay between the Fourier transform and rotational symmetry, that arises only in Rd when d ≥ 2. Finally, the chapter ends with a discussion of the Radon transform. This topic is of substantial interest in its own right, but in addition has significant relevance in its application to the use of X-ray scans as well Ibookroot October 20, 2007 176 Chapter 6. THE FOURIER TRANSFORM ON Rd as to other parts of mathematics. 1 Preliminaries The setting in this chapter will be Rd, the vector space1 of all d-tuples of real numbers (x1, . . . , xd) with xi ∈ R. Addition of vectors is component- wise, and so is multiplication by real scalars. Given x = (x1, . . . , xd) ∈ Rd we define |x| = (x21 + · · ·+ x2d)1/2, so that |x| is simply the length of the vector x in the usual Euclidean norm. In fact, we equip Rd with the standard inner product defined by x · y = x1y1 + · · ·+ xdyd, so that |x|2 = x · x. We use the notation x · y in place of (x, y) of Chap- ter 3. Given a d-tuple α = (α1, . . . , αd) of non-negative integers (sometimes called a multi-index), the monomial xα is defined by xα = xα11 x α2 2 · · ·xαdd . Similarly, we define the differential operator (∂/∂x)α by ( ∂ ∂x )α = ( ∂ ∂x1 )α1 ( ∂ ∂x2 )α2 · · · ( ∂ ∂xd )αd = ∂|α| ∂xα11 · · · ∂xαdd , where |α| = α1 + · · ·+ αd is the order of the multi-index α. 1.1 Symmetries Analysis in Rd, and in particular the theory of the Fourier transform, is shaped by three important groups of symmetries of the underlying space: (i) Translations (ii) Dilations (iii) Rotations 1See Chapter 3 for a brief review of vector spaces and inner products. Here we find it convenient to use lower case letters such as x (as opposed to X) to designate points in Rd. Also, we use | · | instead of ‖ · ‖ to denote the Euclidean norm. Ibookroot October 20, 2007 1. Preliminaries 177 We have seen that translations x 7→ x + h, with h ∈ Rd fixed, and dila- tions x 7→ δx, with δ > 0, play an important role in the one-dimensional
theory. In R, the only two rotations are the identity and multiplica-
tion by −1. However, in Rd with d ≥ 2 there are more rotations, and
the understanding of the interaction between the Fourier transform and
rotations leads to fruitful insights regarding spherical symmetries.

A rotation in Rd is a linear transformation R : Rd → Rd which pre-
serves the inner product. In other words,

• R(ax + by) = aR(x) + bR(y) for all x, y ∈ Rd and a, b ∈ R.
• R(x) ·R(y) = x · y for all x, y ∈ Rd.

Equivalently, this last condition can be replaced by |R(x)| = |x| for all
x ∈ Rd, or Rt = R−1 where Rt and R−1 denote the transpose and inverse
of R, respectively.2 In particular, we have det(R) = ±1, where det(R) is
the determinant of R. If det(R) = 1 we say that R is a proper rotation;
otherwise, we say that R is an improper rotation.

Example 1. On the real line R, there are two rotations: the identity
which is proper, and multiplication by −1 which is improper.

Example 2. The rotations in the plane R2 can be described in terms of
complex numbers. We identify R2 with C by assigning the point (x, y)
to the complex number z = x + iy. Under this identification, all proper
rotations are of the form z 7→ zeiϕ for some ϕ ∈ R, and all improper rota-
tions are of the form z 7→ zeiϕ for some ϕ ∈ R (here, z = x− iy denotes
the complex conjugate of z). See Exercise 1 for the argument leading to
this result.

Example 3. Euler gave the following very simple geometric description
of rotations in R3. Given a proper rotation R, there exists a unit vector
γ so that:

(i) R fixes γ, that is, R(γ) = γ.

(ii) If P denotes the plane passing through the origin and perpendicular
to γ, then R : P → P, and the restriction of R to P is a rotation
in R2.

2Recall that the transpose of a linear operator A : Rd → Rd is the linear operator
B : Rd → Rd which satisfies A(x) · y = x ·B(y) for all x, y ∈ Rd. We write B = At. The
inverse of A (when it exists) is the linear operator C : Rd → Rd with A ◦ C = C ◦A = I
(where I is the identity), and we write C = A−1.

Ibookroot October 20, 2007

178 Chapter 6. THE FOURIER TRANSFORM ON Rd

Geometrically, the vector γ gives the direction of the axis of rotation. A
proof of this fact is given in Exercise 2. Finally, if R is improper, then
−R is proper (since in R3 det(−R) = − det(R)), so R is the composition
of a proper rotation and a symmetry with respect to the origin.

Example 4. Given two orthonormal bases {e1, . . . , ed} and {e′1, . . . , e′d}
in Rd, we can define a rotation R by letting R(ei) = e′i for i = 1, . . . , d.
Conversely, if R is a rotation and {e1, . . . , ed} is an orthonormal basis,
then {e′1, . . . , e′d}, where e′j = R(ej), is another orthonormal basis.

1.2 Integration on Rd

Since we shall be dealing with functions on Rd, we will need to discuss
some aspects of integration of such functions. A more detailed review of
integration on Rd is given in the appendix.

A continuous complex-valued function f on Rd is said to be rapidly
decreasing if for every multi-index α the function |xαf(x)| is bounded.
Equivalently, a continuous function is of rapid decrease if

sup
x∈Rd

|x|k |f(x)| < ∞ for every k = 0, 1, 2, . . .. Given a function of rapid decrease, we define ∫ Rd f(x) dx = lim N→∞ ∫ QN f(x) dx, where QN denotes the closed cube centered at the origin, with sides of length N parallel to the coordinate axis, that is, QN = {x ∈ Rd : |xi| ≤ N/2 for i = 1, . . . , d}. The integral over QN is a multiple integral in the usual sense of Riemann integration. That the limit exists follows from the fact that the integrals IN = ∫ QN f(x) dx form a Cauchy sequence as N tends to infinity. Two observations are in order. First, we may replace the square QN by the ball BN = {x ∈ Rd : |x| ≤ N} without changing the definition. Second, we do not need the full force of rapid decrease to show that the limit exists. In fact it suffices to assume that f is continuous and sup x∈Rd |x|d+² |f(x)| < ∞ for some ² > 0.

Ibookroot October 20, 2007

1. Preliminaries 179

For example, functions of moderate decrease on R correspond to ² = 1.
In keeping with this we define functions of moderate decrease on Rd
as those that are continuous and satisfy the above inequality with ² = 1.

The interaction of integration with the three important groups of sym-
metries is as follows: if f is of moderate decrease, then

(i)
∫

Rd
f(x + h) dx =

∫

Rd
f(x) dx for all h ∈ Rd,

(ii) δd
∫

Rd
f(δx) dx =

∫

Rd
f(x) dx for all δ > 0,

(iii)
∫

Rd
f(R(x)) dx =

∫

Rd
f(x) dx for every rotation R.

Polar coordinates

It will be convenient to introduce polar coordinates in Rd and find the
corresponding integration formula. We begin with two examples which
correspond to the case d = 2 and d = 3. (A more elaborate discussion
applying to all d is contained in the appendix.)

Example 1. In R2, polar coordinates are given by (r, θ) with r ≥ 0 and
0 ≤ θ < 2π. The Jacobian of the change of variables is equal to r, so that ∫ R2 f(x) dx = ∫ 2π 0 ∫ ∞ 0 f(r cos θ, r sin θ) r dr dθ. Now we may write a point on the unit circle S1 as γ = (cos θ, sin θ), and given a function g on the circle, we define its integral over S1 by ∫ S1 g(γ) dσ(γ) = ∫ 2π 0 g(cos θ, sin θ) dθ. With this notation we then have ∫ R2 f(x) dx = ∫ S1 ∫ ∞ 0 f(rγ) r dr dσ(γ). Example 2. In R3 one uses spherical coordinates given by    x1 = r sin θ cosϕ, x2 = r sin θ sinϕ, x3 = r cos θ, Ibookroot October 20, 2007 180 Chapter 6. THE FOURIER TRANSFORM ON Rd where 0 < r, 0 ≤ θ ≤ π and 0 ≤ ϕ ≤ 2π. The Jacobian of the change of variables is r2 sin θ so that ∫ R3 f(x) dx = ∫ 2π 0 ∫ π 0 ∫ ∞ 0 f(r sin θ cosϕ, r sin θ sinϕ, r cos θ)r2dr sin θ dθ dϕ. If g is a function on the unit sphere S2 = {x ∈ R3 : |x| = 1}, and γ = (sin θ cos ϕ, sin θ sinϕ, cos θ), we define the surface element dσ(γ) by ∫ S2 g(γ) dσ(γ) = ∫ 2π 0 ∫ π 0 g(γ) sin θ dθ dϕ. As a result, ∫ R3 f(x) dx = ∫ S2 ∫ ∞ 0 f(rγ) r2 dr dσ(γ). In general, it is possible to write any point in Rd − {0} uniquely as x = rγ where γ lies on the unit sphere Sd−1 ⊂ Rd and r > 0. Indeed, take r = |x|
and γ = x/|x|. Thus one may proceed as in the cases d = 2 or d = 3 to
define spherical coordinates. The formula we shall use is

∫

Rd
f(x) dx =

∫

Sd−1

∫ ∞
0

f(rγ) rd−1 dr dσ(γ),

whenever f is of moderate decrease. Here dσ(γ) denotes the surface
element on the sphere Sd−1 obtained from the spherical coordinates.

2 Elementary theory of the Fourier transform

The Schwartz space S(Rd) (sometimes abbreviated as S) consists of
all indefinitely differentiable functions f on Rd such that

sup
x∈Rd

∣∣∣∣∣x
α

(
∂

∂x

)β
f(x)

∣∣∣∣∣ < ∞, for every multi-index α and β. In other words, f and all its derivatives are required to be rapidly decreasing. Ibookroot October 20, 2007 2. Elementary theory of the Fourier transform 181 Example 1. An example of a function in S(Rd) is the d-dimensional Gaussian given by e−π|x| 2 . The theory in Chapter 5 already made clear the central role played by this function in the case d = 1. The Fourier transform of a Schwartz function f is defined by f̂(ξ) = ∫ Rd f(x)e−2πix·ξ dx, for ξ ∈ Rd. Note the resemblance with the formula in one-dimension, except that we are now integrating on Rd, and the product of x and ξ is replaced by the inner product of the two vectors. We now list some simple properties of the Fourier transform. In the next proposition the arrow indicates that we have taken the Fourier trans- form, so F (x) −→ G(ξ) means that G(ξ) = F̂ (ξ). Proposition 2.1 Let f ∈ S(Rd). (i) f(x + h) −→ f̂(ξ)e2πiξ·h whenever h ∈ Rd. (ii) f(x)e−2πixh −→ f̂(ξ + h) whenever h ∈ Rd. (iii) f(δx) −→ δ−df̂(δ−1ξ) whenever δ > 0.

(iv)
(

∂

∂x

)α
f(x) −→ (2πiξ)αf̂(ξ).

(v) (−2πix)αf(x) −→
(

∂

∂ξ

)α
f̂(ξ).

(vi) f(Rx) −→ f̂(Rξ) whenever R is a rotation.

The first five properties are proved in the same way as in the one-
dimensional case. To verify the last property, simply change variables
y = Rx in the integral. Then, recall that | det(R)| = 1, and
R−1y · ξ = y ·Rξ, because R is a rotation.

Properties (iv) and (v) in the proposition show that, up to factors of
2πi, the Fourier transform interchanges differentiation and multiplication
by monomials. This motivates the definition of the Schwartz space and
leads to the next corollary.

Corollary 2.2 The Fourier transform maps S(Rd) to itself.

At this point we disgress to observe a simple fact concerning the in-
terplay between the Fourier transform and rotations. We say that a

Ibookroot October 20, 2007

182 Chapter 6. THE FOURIER TRANSFORM ON Rd

function f is radial if it depends only on |x|; in other words, f is radial
if there is a function f0(u), defined for u ≥ 0, such that f(x) = f0(|x|).
We note that f is radial if and only if f(Rx) = f(x) for every rotation
R. In one direction, this is obvious since |Rx| = |x|. Conversely, suppose
that f(Rx) = f(x), for all rotations R. Now define f0 by

f0(u) =
{

f(0) if u = 0,
f(x) if |x| = u.

Note that f0 is well defined, since if x and x′ are points with |x| = |x′|
there is always a rotation R so that x′ = Rx.

Corollary 2.3 The Fourier transform of a radial function is radial.

This follows at once from property (vi) in the last proposition. Indeed,
the condition f(Rx) = f(x) for all R implies that f̂(Rξ) = f̂(ξ) for all
R, thus f̂ is radial whenever f is.

An example of a radial function in Rd is the Gaussian e−π|x|
2
. Also,

we observe that when d = 1, the radial functions are precisely the even
functions, that is, those for which f(x) = f(−x).

After these preliminaries, we retrace the steps taken in the previous
chapter to obtain the Fourier inversion formula and Plancherel theorem
for Rd.

Theorem 2.4 Suppose f ∈ S(Rd). Then

f(x) =
∫

Rd
f̂(ξ)e2πix·ξ dξ.

Moreover
∫

Rd
|f̂(ξ)|2 dξ =

∫

Rd
|f(x)|2 dx.

The proof proceeds in the following stages.

Step 1. The Fourier transform of e−π|x|
2

is e−π|ξ|
2
. To prove this,

notice that the properties of the exponential functions imply that

e−π|x|
2

= e−πx
2
1 · · · e−πx2d and e−2πix·ξ = e−2πix1·ξ1 · · · e−2πixd·ξd ,

so that the integrand in the Fourier transform is a product of d functions,
each depending on the variable xj (1 ≤ j ≤ d) only. Thus the assertion

Ibookroot October 20, 2007

2. Elementary theory of the Fourier transform 183

follows by writing the integral over Rd as a series of repeated integrals,
each taken over R. For example, when d = 2,
∫

R2
e−π|x|

2
e−2πix·ξ dx =

∫

R
e−πx

2
2e−2πix2·ξ2

(∫

R
e−πx

2
1e−2πix1·ξ1dx1

)
dx2

=
∫

R
e−πx

2
2e−2πix2·ξ2e−πξ

2
1 dx2

= e−πξ
2
1e−πξ

2
2

= e−π|ξ|
2
.

As a consequence of Proposition 2.1, applied with δ1/2 instead of δ, we
find that ̂(e−πδ|x|2) = δ−d/2e−π|ξ|

2/δ.
Step 2. The family Kδ(x) = δ−d/2e−π|x|

2/δ is a family of good kernels
in Rd. By this we mean that

(i)
∫

Rd
Kδ(x) dx = 1,

(ii)
∫

Rd
|Kδ(x)| dx ≤ M (in fact Kδ(x) ≥ 0),

(iii) For every η > 0,
∫

|x|≥η
|Kδ(x)| dx → 0 as δ → 0.

The proofs of these assertions are almost identical to the case d = 1. As
a result

∫

Rd
Kδ(x)F (x) dx → F (0) as δ → 0

when F is a Schwartz function, or more generally when F is bounded
and continuous at the origin.

Step 3. The multiplication formula
∫

Rd
f(x)ĝ(x) dx =

∫

Rd
f̂(y)g(y) dy

holds whenever f and g are in S. The proof requires the evaluation of the
integral of f(x)g(y)e−2πix·y over (x, y) ∈ R2d = Rd × Rd as a repeated
integral, with each separate integration taken over Rd. The justification
is similar to that in the proof of Proposition 1.8 in the previous chapter.
(See the appendix.)

The Fourier inversion is then a simple consequence of the multiplica-
tion formula and the family of good kernels Kδ, as in Chapter 5. It also

Ibookroot October 20, 2007

184 Chapter 6. THE FOURIER TRANSFORM ON Rd

follows that the Fourier transform F is a bijective map of S(Rd) to itself,
whose inverse is

F∗(g)(x) =
∫

Rd
g(ξ)e2πix·ξ dξ.

Step 4. Next we turn to the convolution, defined by

(f ∗ g)(x) =
∫

Rd
f(y)g(x− y) dy, f, g ∈ S.

We have that f ∗ g ∈ S(Rd), f ∗ g = g ∗ f , and (̂f ∗ g)(ξ) = f̂(ξ)ĝ(ξ).
The argument is similar to that in one-dimension. The calculation of the
Fourier transform of f ∗ g involves an integration of f(y)g(x− y)e−2πix·ξ
(over R2d = Rd × Rd) expressed as a repeated integral.

Then, following the same argument in the previous chapter, we obtain
the d-dimensional Plancherel formula, thereby concluding the proof of
Theorem 2.4.

3 The wave equation in Rd × R
Our next goal is to apply what we have learned about the Fourier trans-
form to the study of the wave equation. Here, we once again simplify
matters by restricting ourselves to functions in the Schwartz class S. We
note that in any further analysis of the wave equation it is important to
allow functions that have much more general behavior, and in particular
that may be discontinuous. However, what we lose in generality by only
considering Schwartz functions, we gain in transparency. Our study in
this restricted context will allow us to explain certain basic ideas in their
simplest form.

3.1 Solution in terms of Fourier transforms

The motion of a vibrating string satisfies the equation

∂2u

∂x2
=

1
c2

∂2u

∂t2
,

which we referred to as the one-dimensional wave equation.
A natural generalization of this equation to d space variables is

(1)
∂2u

∂x21
+ · · ·+ ∂

∂x2d
=

1
c2

∂2u

∂t2
.

In fact, it is known that in the case d = 3, this equation determines the
behavior of electromagnetic waves in vacuum (with c = speed of light).

Ibookroot October 20, 2007

3. The wave equation in Rd × R 185

Also, this equation describes the propagation of sound waves. Thus (1)
is called the d-dimensional wave equation.

Our first observation is that we may assume c = 1, since we can rescale
the variable t if necessary. Also, if we define the Laplacian in d dimen-
sions by

4 = ∂
2

∂x21
+ · · ·+ ∂

∂x2d
,

then the wave equation can be rewritten as

(2) 4u = ∂
2u

∂t2
.

The goal of this section is to find a solution to this equation, subject
to the initial conditions

u(x, 0) = f(x) and
∂u

∂t
(x, 0) = g(x),

where f, g ∈ S(Rd). This is called the Cauchy problem for the wave
equation.

Before solving this problem, we note that while we think of the variable
t as time, we do not restrict ourselves to t > 0. As we will see, the solution
we obtain makes sense for all t ∈ R. This is a manifestation of the fact
that the wave equation can be reversed in time (unlike the heat equation).

A formula for the solution of our problem is given in the next theorem.
The heuristic argument which leads to this formula is important since, as
we have already seen, it applies to some other boundary value problems
as well.

Suppose u solves the Cauchy problem for the wave equation. The
technique employed consists of taking the Fourier transform of the equa-
tion and of the initial conditions, with respect to the space variables
x1, . . . , xd. This reduces the problem to an ordinary differential equation
in the time variable. Indeed, recalling that differentiation with respect to
xj becomes multiplication by 2πiξj , and the differentiation with respect
to t commutes with the Fourier transform in the space variables, we find
that (2) becomes

−4π2|ξ|2û(ξ, t) = ∂
2û

∂t2
(ξ, t).

For each fixed ξ ∈ Rd, this is an ordinary differential equation in t whose
solution is given by

û(ξ, t) = A(ξ) cos(2π|ξ|t) + B(ξ) sin(2π|ξ|t),

Ibookroot October 20, 2007

186 Chapter 6. THE FOURIER TRANSFORM ON Rd

where for each ξ, A(ξ) and B(ξ) are unknown constants to be determined
by the initial conditions. In fact, taking the Fourier transform (in x) of
the initial conditions yields

û(ξ, 0) = f̂(ξ) and
∂û

∂t
(ξ, 0) = ĝ(ξ).

We may now solve for A(ξ) and B(ξ) to obtain

A(ξ) = f̂(ξ) and 2π|ξ|B(ξ) = ĝ(ξ).

Therefore, we find that

û(ξ, t) = f̂(ξ) cos(2π|ξ|t) + ĝ(ξ)sin(2π|ξ|t)
2π|ξ| ,

and the solution u is given by taking the inverse Fourier transform in
the ξ variables. This formal derivation then leads to a precise existence
theorem for our problem.

Theorem 3.1 A solution of the Cauchy problem for the wave equation
is

(3) u(x, t) =
∫

[
f̂(ξ) cos(2π|ξ|t) + ĝ(ξ)sin(2π|ξ|t)

2π|ξ|
]

e2πix·ξ dξ.

Proof. We first verify that u solves the wave equation. This is
straightforward once we note that we can differentiate in x and t un-
der the integral sign (because f and g are both Schwartz functions) and
therefore u is at least C2. On the one hand we differentiate the expo-
nential with respect to the x variables to get

4u(x, t) =
∫

[
f̂(ξ) cos(2π|ξ|t) + ĝ(ξ)sin(2π|ξ|t)

2π|ξ|
]

(−4π2|ξ|2)e2πix·ξ dξ,

while on the other hand we differentiate the terms in brackets with re-
spect to t twice to get

∂2u

∂t2
(x, t) =

∫

[
−4π2|ξ|2f̂(ξ) cos(2π|ξ|t)− 4π2|ξ|2ĝ(ξ)sin(2π|ξ|t)

2π|ξ|
]

e2πix·ξ dξ.

This shows that u solves equation (2). Setting t = 0 we get

u(x, 0) =
∫

Rd
f̂(ξ)e2πix·ξ dξ = f(x)

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3. The wave equation in Rd × R 187

by the Fourier inversion theorem. Finally, differentiating once with re-
spect to t, setting t = 0, and using the Fourier inversion shows that

∂u

∂t
(x, 0) = g(x).

Thus u also verifies the initial conditions, and the proof of the theorem
is complete.

As the reader will note, both f̂(ξ) cos(2π|ξ|t) and ĝ(ξ) sin(2π|ξ|t)
2π|ξ| are

functions in S, assuming as we do that f and g are in S. This is be-
cause both cosu and (sinu)/u are even functions that are indefinitely
differentiable.

Having proved the existence of a solution to the Cauchy problem for the
wave equation, we raise the question of uniqueness. Are there solutions
to the problem

4u = ∂
2u

∂t2
subject to u(x, 0) = f(x) and

∂u

∂t
(x, 0) = g(x),

other than the one given by the formula in the theorem? In fact the
answer is, as expected, no. The proof of this fact, which will not be
given here (but see Problem 3), can be based on a conservation of energy
argument. This is a local counterpart of a global conservation of energy
statement which we will now present.

We observed in Exercise 10, Chapter 3, that in the one-dimensional
case, the total energy of the vibrating string is conserved in time. The
analogue of this fact holds in higher dimensions as well. Define the
energy of a solution by

E(t) =
∫

∣∣∣∣
∂u

∂t

∣∣∣∣
2

+
∣∣∣∣
∂u

∂x1

∣∣∣∣
2

+ · · ·+
∣∣∣∣
∂u

∂xd

∣∣∣∣
2

dx.

Theorem 3.2 If u is the solution of the wave equation given by for-
mula (3), then E(t) is conserved, that is,

E(t) = E(0), for all t ∈ R.

The proof requires the following lemma.

Lemma 3.3 Suppose a and b are complex numbers and α is real. Then

|a cos α + b sinα|2 + | − a sin α + b cosα|2 = |a|2 + |b|2.

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188 Chapter 6. THE FOURIER TRANSFORM ON Rd

This follows directly because e1 = (cos α, sinα) and e2 = (− sinα, cos α)
are a pair of orthonormal vectors, hence with Z = (a, b) ∈ C2, we have

|Z|2 = |Z · e1|2 + |Z · e2|2,

where · represents the inner product in C2.
Now by Plancherel’s theorem,

∫

∣∣∣∣
∂u

∂t

∣∣∣∣
2

dx =
∫

∣∣∣−2π|ξ|f̂(ξ) sin(2π|ξ|t) + ĝ(ξ) cos(2π|ξ|t)
∣∣∣
2

dξ.

Similarly,

∫

d∑

j=1

∣∣∣∣
∂u

∂xj

∣∣∣∣
2

dx =
∫

∣∣∣2π|ξ|f̂(ξ) cos(2π|ξ|t) + ĝ(ξ) sin(2π|ξ|t)
∣∣∣
2

dξ.

We now apply the lemma with

a = 2π|ξ|f̂(ξ), b = ĝ(ξ) and α = 2π|ξ|t.

The result is that

E(t) =
∫

∣∣∣∣
∂u

∂t

∣∣∣∣
2

+
∣∣∣∣
∂u

∂x1

∣∣∣∣
2

+ · · ·+
∣∣∣∣
∂u

∂xd

∣∣∣∣
2

=
∫

Rd
(4π2|ξ|2|f̂(ξ)|2 + |ĝ(ξ)|2) dξ,

which is clearly independent of t. Thus Theorem 3.2 is proved.

The drawback with formula (3), which does give the solution of the
wave equation, is that it is quite indirect, involving the calculation of the
Fourier transforms of f and g, and then a further inverse Fourier trans-
form. However, for every dimension d there is a more explicit formula.
This formula is very simple when d = 1 and a little less so when d = 3.
More generally, the formula is “elementary” whenever d is odd, and more
complicated when d is even (see Problems 4 and 5).

In what follows we consider the cases d = 1, d = 3, and d = 2, which
together give a picture of the general situation. Recall that in Chapter 1,
when discussing the wave equation over the interval [0, L], we found that
the solution is given by d’Alembert’s formula

(4) u(x, t) =
f(x + t) + f(x− t)

2
+

1
2

∫ x+t
x−t

g(y) dy.

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3. The wave equation in Rd × R 189

with the interpretation that both f and g are extended outside [0, L] by
making them odd in [−L,L], and periodic on the real line, with period
2L. The same formula (4) holds for the solution of the wave equation
when d = 1 and when the initial data are functions in S(R). In fact, this
follows directly from (3) if we note that

cos(2π|ξ|t) = 1
2
(e2πi|ξ|t + e−2πi|ξ|t)

and

sin(2π|ξ|t)
2π|ξ| =

1
4πi|ξ|(e

2πi|ξ|t − e−2πi|ξ|t).

Finally, we note that the two terms that appear in d’Alembert’s for-
mula (4) consist of appropriate averages. Indeed, the first term is pre-
cisely the average of f over the two points that are the boundary of the
interval [x− t, x + t]; the second term is, up to a factor of t, the mean
value of g over this interval, that is, (1/2t)

∫ x+t
x−t g(y) dy. This suggests a

generalization to higher dimensions, where we might expect to write the
solution of our problem as averages of the initial data. This is in fact the
case, and we now treat in detail the particular situation d = 3.

3.2 The wave equation in R3 × R
If S2 denotes the unit sphere in R3, we define the spherical mean of
the function f over the sphere of radius t centered at x by

(5) Mt(f)(x) =
1
4π

∫

S2
f(x− tγ) dσ(γ),

where dσ(γ) is the element of surface area for S2. Since 4π is the area
of the unit sphere, we can interpret Mt(f) as the average value of f over
the sphere centered at x of radius t.

Lemma 3.4 If f ∈ S(R3) and t is fixed, then Mt(f) ∈ S(R3). Moreover,
Mt(f) is indefinitely differentiable in t, and each t-derivative also belongs
to S(R3).

Proof. Let F (x) = Mt(f)(x). To show that F is rapidly decreasing,
start with the inequality |f(x)| ≤ AN/(1 + |x|N ) which holds for every
fixed N ≥ 0. As a simple consequence, whenever t is fixed, we have

|f(x− γt)| ≤ A′N/(1 + |x|N ) for all γ ∈ S2.

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190 Chapter 6. THE FOURIER TRANSFORM ON Rd

To see this consider separately the cases when |x| ≤ 2|t|, and |x| > 2|t|.
Therefore, by integration

|F (x)| ≤ A′N/(1 + |x|N ),

and since this holds for every N , the function F is rapidly decreasing.
One next observes that F is indefinitely differentiable, and

(6)
(

∂

∂x

)α
F (x) = Mt(f

(α))(x)

where f (α)(x) = (∂/∂x)αf . It suffices to prove this when (∂/∂x)α =
∂/∂xk, and then proceed by induction to get the general case. Further-
more, it is enough to take k = 1. Now

F (x1 + h, x2, x3)− F (x1, x2, x3)
h

=
1
4π

∫

S2
gh(γ) dσ(γ)

where

gh(γ) =
f(x + e1h− γt)− f(x− γt)

h
,

and e1 = (1, 0, 0). Now, it suffices to observe that gh → ∂∂x1 f(x− γt)
as h → 0 uniformly in γ. As a result, we find that (6) holds, and by
the first argument, it follows that

(
∂
∂x

)α
F (x) is also rapidly decreasing,

hence F ∈ S. The same argument applies to each t-derivative of Mt(f).

The basic fact about integration on spheres that we shall need is the
following Fourier transform formula.

Lemma 3.5
1
4π

∫

S2
e−2πiξ·γ dσ(γ) =

sin(2π|ξ|)
2π|ξ| .

This formula, as we shall see in the following section, is connected to
the fact that the Fourier transform of a radial function is radial.

Proof. Note that the integral on the left is radial in ξ. Indeed, if R is
a rotation then

∫

S2
e−2πiR(ξ)·γ dσ(γ) =

∫

S2
e−2πiξ·R

−1(γ) dσ(γ) =
∫

S2
e−2πiξ·γ dσ(γ)

because we may change variables γ → R−1(γ). (For this, see formula (4)
in the appendix.) So if |ξ| = ρ, it suffices to prove the lemma with

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3. The wave equation in Rd × R 191

ξ = (0, 0, ρ). If ρ = 0, the lemma is obvious. If ρ > 0, we choose spherical
coordinates to find that the left-hand side is equal to

1
4π

∫ 2π
0

∫ π
0

e−2πiρ cos θ sin θ dθ dϕ.

The change of variables u = − cos θ gives
1
4π

∫ 2π
0

∫ π
0

e−2πiρ cos θ sin θ dθ dϕ =
1
2

∫ π
0

e−2πiρ cos θ sin θ dθ

=
1
2

∫ 1
−1

e2πiρu du

=
1

4πiρ
[
e2πiρu

]1
−1

=
sin(2πρ)

2πρ
,

and the formula is proved.

By the defining formula (5) we may interpret Mt(f) as a convolution
of the function f with the element dσ, and since the Fourier transform
interchanges convolutions with products, we are led to believe that M̂t(f)
is the product of the corresponding Fourier transforms. Indeed, we have
the identity

(7) M̂t(f)(ξ) = f̂(ξ)
sin(2π|ξ|t)

2π|ξ|t .

To see this, write

M̂t(f)(ξ) =
∫

R3
e−2πix·ξ

(
1
4π

∫

S2
f(x− γt) dσ(γ)

)
dx,

and note that we may interchange the order of integration and make a
simple change of variables to achieve the desired identity.

As a result, we find that the solution of our problem may be expressed
by using the spherical means of the initial data.

Theorem 3.6 The solution when d = 3 of the Cauchy problem for the
wave equation

4u = ∂
2u

∂t2
subject to u(x, 0) = f(x) and

∂u

∂t
(x, 0) = g(x)

is given by

u(x, t) =
∂

∂t
(tMt(f)(x)) + tMt(g)(x).

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192 Chapter 6. THE FOURIER TRANSFORM ON Rd

Proof. Consider first the problem

4u = ∂
2u

∂t2
subject to u(x, 0) = 0 and

∂u

∂t
(x, 0) = g(x).

Then by Theorem 3.1, we know that its solution u1 is given by

u1(x, t) =
∫

[
ĝ(ξ)

sin(2π|ξ|t)
2π|ξ|

]
e2πix·ξ dξ

= t
∫

[
ĝ(ξ)

sin(2π|ξ|t)
2π|ξ|t

]
e2πix·ξ dξ

= tMt(g)(x),

where we have used (7) applied to g, and the Fourier inversion formula.
According to Theorem 3.1 again, the solution to the problem

4u = ∂
2u

∂t2
subject to u(x, 0) = f(x) and

∂u

∂t
(x, 0) = 0

is given by

u2(x, t) =
∫

[
f̂(ξ) cos(2π|ξ|t)

]
e2πix·ξ dξ

=
∂

∂t

(
t

∫

[
f̂(ξ)

sin(2π|ξ|t)
2π|ξ|t

]
e2πix·ξ dξ

)

=
∂

∂t
(tMt(f)(x)).

We may now superpose these two solutions to obtain u = u1 + u2 as the
solution of our original problem.

Huygens principle

The solutions to the wave equation in one and three dimensions are given,
respectively, by

u(x, t) =
f(x + t) + f(x− t)

2
+

1
2

∫ x+t
x−t

g(y) dy

and

u(x, t) =
∂

∂t
(tMt(f)(x)) + tMt(g)(x).

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3. The wave equation in Rd × R 193

x + tx− t

(x, t)

0 x

Figure 1. Huygens principle, d = 1

We observe that in the one-dimensional problem, the value of the solution
at (x, t) depends only on the values of f and g in the interval centered
at x of length 2t, as shown in Figure 1.

If in addition g = 0, then the solution depends only on the data at the
two boundary points of this interval. In three dimensions, this boundary
dependence always holds. More precisely, the solution u(x, t) depends
only on the values of f and g in an immediate neighborhood of the sphere
centered at x and of radius t. This situation is depicted in Figure 2, where
we have drawn the cone originating at (x, t) and with its base the ball
centered at x of radius t. This cone is called the backward light cone
originating at (x, t).

x-space

(x, t)

Figure 2. Backward light cone originating at (x, t)

Alternatively, the data at a point x0 in the plane t = 0 influences the
solution only on the boundary of a cone originating at x0, called the
forward light cone and depicted in Figure 3.

This phenomenon, known as the Huygens principle, is immediate
from the formulas for u given above.

Another important aspect of the wave equation connected with these

Ibookroot October 20, 2007

194 Chapter 6. THE FOURIER TRANSFORM ON Rd

Figure 3. The forward light cone originating at x0

considerations is that of the finite speed of propagation. (In the
case where c = 1, the speed is 1.) This means that if we have an initial
disturbance localized at x = x0, then after a finite time t, its effects will
have propagated only inside the ball centered at x0 of radius |t|. To state
this precisely, suppose the initial conditions f and g are supported in the
ball of radius δ, centered at x0 (think of δ as small). Then u(x, t) is
supported in the ball of radius |t|+ δ centered at x0. This assertion is
clear from the above discussion.

3.3 The wave equation in R2 × R: descent
It is a remarkable fact that the solution of the wave equation in three
dimensions leads to a solution of the wave equation in two dimensions.
Define the corresponding means by

M̃t(F )(x) =
1
2π

∫

|y|≤1
F (x− ty)(1− |y|2)−1/2 dy.

Theorem 3.7 A solution of the Cauchy problem for the wave equation
in two dimensions with initial data f, g ∈ S(R2) is given by

(8) u(x, t) =
∂

∂t
(tM̃t(f)(x)) + tM̃t(g)(x).

Notice the difference between this case and the case d = 3. Here, u at
(x, t) depends on f and g in the whole disc (of radius |t| centered at x),
and not just on the values of the initial data near the boundary of that
disc.

Formally, the identity in the theorem arises as follows. If we start
with an initial pair of functions f and g in S(R2), we may consider the
corresponding functions f̃ and g̃ on R3 that are merely extensions of f
and g that are constant in the x3 variable, that is,

f̃(x1, x2, x3) = f(x1, x2) and g̃(x1, x2, x3) = g(x1, x2).

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3. The wave equation in Rd × R 195

Now, if ũ is the solution (given in the previous section) of the 3-dimensional
wave equation with initial data f̃ and g̃, then one can expect that ũ is
also constant in x3 so that ũ satisfies the 2-dimensional wave equation.
A difficulty with this argument is that f̃ and g̃ are not rapidly decreasing
since they are constant in x3, so that our previous methods do not apply.
However, it is easy to modify the argument so as to obtain a proof of
Theorem 3.7.

We fix T > 0 and consider a function η(x3) that is in S(R), such that
η(x3) = 1 if |x3| ≤ 3T . The trick is to truncate f̃ and g̃ in the x3-variable,
and consider instead

f̃ [(x1, x2, x3) = f(x1, x2)η(x3) and g̃
[(x1, x2, x3) = g(x1, x2)η(x3).

Now both f̃ [ and g̃[ are in S(R3), so Theorem 3.6 provides a solution
ũ[ of the wave equation with initial data f̃ [ and g̃[. It is easy to see
from the formula that ũ[(x, t) is independent of x3, whenever |x3| ≤ T
and |t| ≤ T . In particular, if we define u(x1, x2, t) = ũ[(x1, x2, 0, t), then
u satisfies the 2-dimensional wave equation when |t| ≤ T . Since T is
arbitrary, u is a solution to our problem, and it remains to see why u has
the desired form.

By definition of the spherical coordinates, we recall that the integral
of a function H over the sphere S2 is given by

1
4π

∫

S2
H(γ) dσ(γ) =

1
4π

∫ 2π
0

∫ π
0

H(sin θ cos ϕ, sin θ sinϕ, cos θ) sin θ dθ dϕ.

If H does not depend on the last variable, that is, H(x1, x2, x3) = h(x1, x2)
for some function h of two variables, then

Mt(H)(x1, x2, 0) =

1
4π

∫ 2π
0

∫ π
0

h(x1 − t sin θ cos ϕ, x2 − t sin θ sin ϕ) sin θ dθ dϕ.

To calculate this last integral, we split the θ-integral from 0 to π/2 and
then π/2 to π. By making the change of variables r = sin θ, we find, after
a final change to polar coordinates, that

Mt(H)(x1, x2, 0) =
1
2π

∫

|y|≤1
h(x− ty)(1− |y|2)−1/2 dy

= M̃t(h)(x1, x2).

Ibookroot October 20, 2007

196 Chapter 6. THE FOURIER TRANSFORM ON Rd

Applying this to H = f̃ [, h = f , and H = g̃[, h = g, we find that u is
given by the formula (8), and the proof of Theorem 3.7 is complete.

Remark. In the case of general d, the solution of the wave equation
shares many of the properties we have discussed in the special cases
d = 1, 2, and 3.

• At a given time t, the initial data at a point x only affects the solu-
tion u in a specific region. When d > 1 is odd, the data influences
only the points on the boundary of the forward light cone origi-
nating at x, while when d = 1 or d is even, it affects all points of
the forward light cone. Alternatively, the solution at a point (x, t)
depends only on the data at the base of the backward light cone
originating at (x, t). In fact, when d > 1 is odd, only the data in an
immediate neighborhood of the boundary of the base will influence
u(x, t).

• Waves propagate with finite speed: if the initial data is supported
in a bounded set, then the support of the solution u spreads with
velocity 1 (or more generally c, if the wave equation is not normal-
ized).

We can illustrate some of these facts by the following observation about
the different behavior of the propagation of waves in three and two dimen-
sions. Since the propagation of light is governed by the three-dimensional
wave equation, if at t = 0 a light flashes at the origin, the following hap-
pens: any observer will see the flash (after a finite amount of time) only
for an instant. In contrast, consider what happens in two dimensions. If
we drop a stone in a lake, any point on the surface will begin (after some
time) to undulate; although the amplitude of the oscillations will decrease
over time, the undulations will continue (in principle) indefinitely.

The difference in character of the formulas for the solutions of the
wave equation when d = 1 and d = 3 on the one hand, and d = 2 on
the other hand, illustrates a general principle in d-dimensional Fourier
analysis: a significant number of formulas that arise are simpler in the
case of odd dimensions, compared to the corresponding situations in even
dimensions. We will see several further examples of this below.

4 Radial symmetry and Bessel functions

We observed earlier that the Fourier transform of a radial function in
Rd is also radial. In other words, if f(x) = f0(|x|) for some f0, then

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4. Radial symmetry and Bessel functions 197

f̂(ξ) = F0(|ξ|) for some F0. A natural problem is to determine a relation
between f0 and F0.

This problem has a simple answer in dimensions one and three. If
d = 1 the relation we seek is

(9) F0(ρ) = 2
∫ ∞
0

cos(2πρr)f0(r) dr.

If we recall that R has only two rotations, the identity and multiplication
by −1, we find that a function is radial precisely when it is even. Having
made this observation it is easy to see that if f is radial, and |ξ| = ρ,
then

F0(ρ) = f̂(|ξ|) =
∫ ∞
−∞

f(x)e−2πix|ξ| dx

=
∫ ∞
0

f0(r)(e
−2πir|ξ| + e2πir|ξ|) dr

= 2
∫ ∞
0

cos(2πρr)f0(r) dr.

In the case d = 3, the relation between f0 and F0 is also quite simple
and given by the formula

(10) F0(ρ) = 2ρ
−1

∫ ∞
0

sin(2πρr)f0(r)r dr.

The proof of this identity is based on the formula for the Fourier trans-
form of the surface element dσ given in Lemma 3.5:

F0(ρ) = f̂(ξ) =
∫

R3
f(x)e−2πix·ξ dx

=
∫ ∞
0

f0(r)
∫

S2
e−2πirγ·ξdσ(γ)r2 dr

=
∫ ∞
0

f0(r)
2 sin(2πρr)

ρr
r2 dr

= 2ρ−1
∫ ∞
0

sin(2πρr)f0(r)r dr.

More generally, the relation between f0 and F0 has a nice description
in terms of a family of special functions that arise naturally in problems
that exhibit radial symmetry.

The Bessel function of order n ∈ Z, denoted Jn(ρ), is defined as the
nth Fourier coefficient of the function eiρ sin θ. So

Jn(ρ) =
1
2π

∫ 2π
0

eiρ sin θe−inθ dθ,

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198 Chapter 6. THE FOURIER TRANSFORM ON Rd

therefore

eiρ sin θ =
∞∑

n=−∞
Jn(ρ)e

inθ.

As a result of this definition, we find that when d = 2, the relation be-
tween f0 and F0 is

(11) F0(ρ) = 2π
∫ ∞
0

J0(2πrρ)f0(r)r dr.

Indeed, since f̂(ξ) is radial we take ξ = (0,−ρ) so that

f̂(ξ) =
∫

R2
f(x)e2πix·(0,ρ) dx

=
∫ 2π
0

∫ ∞
0

f0(r)e
2πirρ sin θr dr dθ

= 2π
∫ ∞
0

J0(2πrρ)f0(r)r dr,

as desired.

In general, there are corresponding formulas relating f0 and F0 in Rd
in terms of Bessel functions of order d/2− 1 (see Problem 2). In even
dimensions, these are the Bessel functions we have defined above. For
odd dimensions, we need a more general definition of Bessel functions to
encompass half-integral orders. Note that the formulas for the Fourier
transform of radial functions give another illustration of the differences
between odd and even dimensions. When d = 1 or d = 3 (as well as
d > 3, d odd) the formulas are in terms of elementary functions, but this
is not the case when d is even.

5 The Radon transform and some of its applications

Invented by Johann Radon in 1917, the integral transform we discuss
next has many applications in mathematics and other sciences, includ-
ing a significant achievement in medicine. To motivate the definitions
and the central problem of reconstruction, we first present the close con-
nection between the Radon transform and the development of X-ray
scans (or CAT scans) in the theory of medical imaging. The solution of
the reconstruction problem, and the introduction of new algorithms and
faster computers, all contributed to a rapid development of computerized
tomography. In practice, X-ray scans provide a “picture” of an internal
organ, one that helps to detect and locate many types of abnormalities.

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5. The Radon transform and some of its applications 199

After a brief description of X-ray scans in two dimensions, we define
the X-ray transform and formulate the basic problem of inverting this
mapping. Although this problem has an explicit solution in R2, it is
more complicated than the analogous problem in three dimensions, hence
we give a complete solution of the reconstruction problem only in R3.
Here we have another example where results are simpler in the odd-
dimensional case than in the even-dimensional situation.

5.1 The X-ray transform in R2

Consider a two-dimensional object O lying in the plane R2, which we
may think of as a planar cross section of a human organ.

First, we assume that O is homogeneous, and suppose that a very
narrow beam of X-ray photons traverses this object.

Figure 4. Attenuation of an X-ray beam

If I0 and I denote the intensity of the beam before and after passing
through O, respectively, the following relation holds:

I = I0e
−dρ.

Here d is the distance traveled by the beam in the object, and ρ denotes
the attenuation coefficient (or absorption coefficient), which depends on
the density and other physical characteristics of O. If the object is not
homogeneous, but consists of two materials with attenuation coefficients
ρ1 and ρ2, then the observed decrease in the intensity of the beam is

Ibookroot October 20, 2007

200 Chapter 6. THE FOURIER TRANSFORM ON Rd

given by

I = I0e
−d1ρ1−d2ρ2

where d1 and d2 denote the distances traveled by the beam in each ma-
terial. In the case of an arbitrary object whose density and physical
characteristics vary from point to point, the attenuation factor is a func-
tion ρ in R2, and the above relations become

I = I0e
∫

L
ρ
.

Here L is the line in R2 traced by the beam, and
∫
L

ρ denotes the line
integral of ρ over L. Since we observe I and I0, the data we gather
after sending the X-ray beam through the object along the line L is the
quantity

∫

L
ρ.

Since we may initially send the beam in any given direction, we may
calculate the above integral for every line in R2. We define the X-ray
transform (or Radon transform in R2) of ρ by

X(ρ)(L) =
∫

L
ρ.

Note that this transform assigns to each appropriate function ρ on R2
(for example, ρ ∈ S(R2)) another function X(ρ) whose domain is the set
of lines L in R2.

The unknown is ρ, and since our original interest lies precisely in the
composition of the object, the problem now becomes to reconstruct the
function ρ from the collected data, that is, its X-ray transform. We
therefore pose the following reconstruction problem: Find a formula for
ρ in terms of X(ρ).

Mathematically, the problem asks for a formula giving the inverse of
X. Does such an inverse even exist? As a first step, we pose the following
simpler uniqueness question: If X(ρ) = X(ρ′), can we conclude that ρ =
ρ′?

There is a reasonable a priori expectation that X(ρ) actually deter-
mines ρ, as one can see by counting the dimensionality (or degrees of
freedom) involved. A function ρ on R2 depends on two parameters (the
x1 and x2 coordinates, for example). Similarly, the function X(ρ), which
is a function of lines L, is also determined by two parameters (for ex-
ample, the slope of L and its x2-intercept). In this sense, ρ and X(ρ)

Ibookroot October 20, 2007

5. The Radon transform and some of its applications 201

convey an equivalent amount of information, so it is not unreasonable to
suppose that X(ρ) determines ρ.

While there is a satisfactory answer to the reconstruction problem,
and a positive answer to the uniqueness question in R2, we shall forego
giving them here. (However, see Exercise 13 and Problem 8.) Instead
we shall deal with the analogous but simpler situation in R3.

Let us finally remark that in fact, one can sample the X-ray trans-
form, and determine X(ρ)(L) for only finitely many lines. Therefore,
the reconstruction method implemented in practice is based not only on
the general theory, but also on sampling procedures, numerical approx-
imations, and computer algorithms. It turns out that a method used
in developing effective relevant algorithms is the fast Fourier transform,
which incidentally we take up in the next chapter.

5.2 The Radon transform in R3

The experiment described in the previous section applies in three dimen-
sions as well. If O is an object in R3 determined by a function ρ which
describes the density and physical characteristics of this object, sending
an X-ray beam through O determines the quantity

∫

L
ρ,

for every line in R3. In R2 this knowledge was enough to uniquely de-
termine ρ, but in R3 we do not need as much data. In fact, by using the
heuristic argument above of counting the number of degrees of freedom,
we see that for functions ρ in R3 the number is three, while the number
of parameters determining a line L in R3 is four (for example, two for
the intercept in the (x1, x2) plane, and two more for the direction of the
line). Thus in this sense, the problem is over-determined.

We turn instead to the natural mathematical generalization of the two-
dimensional problem. Here we wish to determine the function in R3 by
knowing its integral over all planes3 in R3. To be precise, when we speak
of a plane, we mean a plane not necessarily passing through the origin.
If P is such a plane, we define the Radon transform R(f) by

R(f)(P) =
∫

P
f.

To simplify our presentation, we shall follow our practice of assuming
that we are dealing with functions in the class S(R3). However, many

3Note that the dimensionality associated with points on R3, and that for planes in R3,
equals three in both cases.

Ibookroot October 20, 2007

202 Chapter 6. THE FOURIER TRANSFORM ON Rd

of the results obtained below can be shown to be valid for much larger
classes of functions.

First, we explain what we mean by the integral of f over a plane. The
description we use for planes in R3 is the following: given a unit vector
γ ∈ S2 and a number t ∈ R, we define the plane Pt,γ by

Pt,γ = {x ∈ R3 : x · γ = t}.

So we parametrize a plane by a unit vector γ orthogonal to it, and by its
“distance” t to the origin (see Figure 5). Note that Pt,γ = P−t,−γ , and
we allow t to take negative values.

Pt,γ

Figure 5. Description of a plane in R3

Given a function f ∈ S(Rd), we need to make sense of its integral over
Pt,γ . We proceed as follows. Choose unit vectors e1, e2 so that e1, e2, γ is
an orthonormal basis for R3. Then any x ∈ Pt,γ can be written uniquely
as

x = tγ + u where u = u1e1 + u2e2 with u1, u2 ∈ R.

If f ∈ S(R3), we define

(12)
∫

Pt,γ
f =

∫

R2
f(tγ + u1e1 + u2e2) du1 du2.

To be consistent, we must check that this definition is independent of
the choice of the vectors e1, e2.

Ibookroot October 20, 2007

5. The Radon transform and some of its applications 203

Proposition 5.1 If f ∈ S(R3), then for each γ the definition of
∫
Pt,γ f

is independent of the choice of e1 and e2. Moreover

∫ ∞
−∞

(∫

Pt,γ
f

)
dt =

∫

R3
f(x) dx.

Proof. If e′1, e
′
2 is another choice of basis vectors so that γ, e

′
1, e

′
2 is

orthonormal, consider the rotation R in R2 which takes e1 to e′1 and
e2 to e′2. Changing variables u

′ = R(u) in the integral proves that our
definition (12) is independent of the choice of basis.

To prove the formula, let R denote the rotation which takes the stan-
dard basis of unit vectors4 in R3 to γ, e1, and e2. Then

∫

R3
f(x) dx =

∫

R3
f(Rx) dx

=
∫

R3
f(x1γ + x2e1 + x3e2) dx1 dx2 dx3

=
∫ ∞
−∞

(∫

Pt,γ
f

)
dt.

Remark. We digress to point out that the X-ray transform deter-
mines the Radon transform, since two-dimensional integrals can be ex-
pressed as iterated one-dimensional integrals. In other words, the knowl-
edge of the integral of a function over all lines determines the integral of
that function over any plane.

Having disposed of these preliminary matters, we turn to the study of
the original problem. The Radon transform of a function f ∈ S(R3)
is defined by

R(f)(t, γ) =
∫

Pt,γ
f.

In particular, we see that the Radon transform is a function on the
set of planes in R3. From the parametrization given for a plane, we
may equivalently think of R(f) as a function on the product R× S2 =
{(t, γ) : t ∈ R, γ ∈ S2}, where S2 denotes the unit sphere in R3. The
relevant class of functions on R× S2 consists of those that satisfy the
Schwartz condition in t uniformly in γ. In other words, we define S(R×
S2) to be the space of all continuous functions F (t, γ) that are indefinitely

4Here we are referring to the vectors (1, 0, 0), (0, 1, 0), and (0, 0, 1).

Ibookroot October 20, 2007

204 Chapter 6. THE FOURIER TRANSFORM ON Rd

differentiable in t, and that satisfy

sup
t∈R, γ∈S2

|t|k
∣∣∣∣∣
d`F

∂t`
(t, γ)

∣∣∣∣∣ < ∞ for all integers k, ` ≥ 0. Our goal is to solve the following problems. Uniqueness problem: If R(f) = R(g), then f = g. Reconstruction problem: Express f in terms of R(f). The solutions will be obtained by using the Fourier transform. In fact, the key point is a very elegant and essential relation between the Radon and Fourier transforms. Lemma 5.2 If f ∈ S(R3), then R(f)(t, γ) ∈ S(R) for each fixed γ. More- over, R̂(f)(s, γ) = f̂(sγ). To be precise, f̂ denotes the (three-dimensional) Fourier transform of f , while R̂(f)(s, γ) denotes the one-dimensional Fourier transform of R(f)(t, γ) as a function of t, with γ fixed. Proof. Since f ∈ S(R3), for every positive integer N there is a con- stant AN < ∞ so that (1 + |t|)N (1 + |u|)N |f(tγ + u)| ≤ AN , if we recall that x = tγ + u, where γ is orthogonal to u. Therefore, as soon as N ≥ 3, we find (1 + |t|)NR(f)(t, γ) ≤ AN ∫ R2 du (1 + |u|)N < ∞. A similar argument for the derivatives shows that R(f)(t, γ) ∈ S(R) for each fixed γ. To establish the identity, we first note that R̂(f)(s, γ) = ∫ ∞ −∞ (∫ Pt,γ f ) e−2πist dt = ∫ ∞ −∞ ∫ R2 f(tγ + u1e1 + u2e2) du1 du2e −2πist dt. Ibookroot October 20, 2007 5. The Radon transform and some of its applications 205 However, since γ · u = 0 and |γ| = 1, we may write e−2πist = e−2πisγ·(tγ+u). As a result, we find that R̂(f)(s, γ) = ∫ ∞ −∞ ∫ R2 f(tγ + u1e1 + u2e2)e −2πisγ·(tγ+u) du1 du2 dt = ∫ ∞ −∞ ∫ R2 f(tγ + u)e−2πisγ·(tγ+u) du dt. A final rotation from γ, e1, e2 to the standard basis in R3 proves that R̂(f)(s, γ) = f̂(sγ), as desired. As a consequence of this identity, we can answer the uniqueness ques- tion for the Radon transform in R3 in the affirmative. Corollary 5.3 If f, g ∈ S(R3) and R(f) = R(g), then f = g. The proof of the corollary follows from an application of the lemma to the difference f − g and use of the Fourier inversion theorem. Our final task is to give the formula that allows us to recover f from its Radon transform. Since R(f) is a function on the set of planes in R3, and f is a function of the space variables x ∈ R3, to recover f we introduce the dual Radon transform, which passes from functions defined on planes to functions in R3. Given a function F on R× S2, we define its dual Radon transform by (13) R∗(F )(x) = ∫ S2 F (x · γ, γ) dσ(γ). Observe that a point x belongs to Pt,γ if and only if x · γ = t, so the idea here is that given x ∈ R3, we obtain R∗(F )(x) by integrating F over the subset of all planes passing through x, that is, R∗(F )(x) = ∫ {Pt,γ such that x∈Pt,γ} F, where the integral on the right is given the precise meaning in (13). We use the terminology “dual” because of the following observation. If V1 = S(R3) with the usual Hermitian inner product (f, g)1 = ∫ R3 f(x)g(x) dx, Ibookroot October 20, 2007 206 Chapter 6. THE FOURIER TRANSFORM ON Rd and V2 = S(R× S2) with the Hermitian inner product (F,G)2 = ∫ R ∫ S2 F (t, γ)G(t, γ) dσ(γ) dt, then R : V1 → V2, R∗ : V2 → V1, with (14) (Rf, F )2 = (f,R∗F )1. The validity of this identity is not needed in the argument below, and its verification is left as an exercise for the reader. We can now state the reconstruction theorem. Theorem 5.4 If f ∈ S(R3), then 4(R∗R(f)) = −8π2f. We recall that 4 = ∂2 ∂x21 + ∂ 2 ∂x22 + ∂ 2 ∂x23 is the Laplacian. Proof. By our previous lemma, we have R(f)(t, γ) = ∫ ∞ −∞ f̂(sγ)e2πits ds. Therefore R∗R(f)(x) = ∫ S2 ∫ ∞ −∞ f̂(sγ)e2πix·γs ds dσ(γ), hence 4(R∗R(f))(x) = ∫ S2 ∫ ∞ −∞ f̂(sγ)(−4π2s2)e2πix·γs ds dσ(γ) =− 4π2 ∫ S2 ∫ ∞ −∞ f̂(sγ)e2πix·γss2 ds dσ(γ) =− 4π2 ∫ S2 ∫ 0 −∞ f̂(sγ)e2πix·γss2 ds dσ(γ) − 4π2 ∫ S2 ∫ ∞ 0 f̂(sγ)e2πix·γss2 ds dσ(γ) =− 8π2 ∫ S2 ∫ ∞ 0 f̂(sγ)e2πix·γss2 ds dσ(γ) =− 8π2f(x). Ibookroot October 20, 2007 6. Exercises 207 In the first line, we have differentiated under the integral sign and used the fact 4(e2πix·γs) = (−4π2s2)e2πix·γs, since |γ| = 1. The last step fol- lows from the formula for polar coordinates in R3 and the Fourier inver- sion theorem. 5.3 A note about plane waves We conclude this chapter by briefly mentioning a nice connection between the Radon transform and solutions of the wave equation. This comes about in the following way. Recall that when d = 1, the solution of the wave equation can be expressed as the sum of traveling waves (see Chapter 1), and it is natural to ask if an analogue of such traveling waves exists in higher dimensions. The answer is as follows. Let F be a function of one variable, which we assume is sufficiently smooth (say C2), and consider u(x, t) defined by u(x, t) = F ((x · γ)− t), where x ∈ Rd and γ is a unit vector in Rd. It is easy to verify directly that u is a solution of the wave equation in Rd (with c = 1). Such a solution is called a plane wave; indeed, notice that u is constant on every plane perpendicular to the direction γ, and as time t increases, the wave travels in the γ direction. (It should be remarked that plane waves are never functions in S(Rd) when d > 1 because they are constant in
directions perpendicular to γ).5

The basic fact is that when d > 1, the solution of the wave equation
can be written as an integral (as opposed to sum, when d = 1) of plane
waves; this can in fact be done via the Radon transform of the initial
data f and g. For the relevant formulas when d = 3, see Problem 6.

6 Exercises

1. Suppose that R is a rotation in the plane R2, and let

R =
(

a b
c d

)

denote its matrix with respect to the standard basis vectors e1 = (1, 0) and
e2 = (0, 1).

(a) Write the conditions Rt = R−1 and det(R) = ±1 in terms of equations in
a, b, c, d.

5Incidentally, this observation is further indication that a fuller treatment of the wave
equation requires lifting the restriction that functions belong to S(Rd).

Ibookroot October 20, 2007

208 Chapter 6. THE FOURIER TRANSFORM ON Rd

(b) Show that there exists ϕ ∈ R such that a + ib = eiϕ.
(c) Conclude that if R is proper, then it can be expressed as z 7→ zeiϕ, and if

R is improper, then it takes the form z 7→ zeiϕ, where z = x− iy.

2. Suppose that R : R3 → R3 is a proper rotation.

(a) Show that p(t) = det(R− tI) is a polynomial of degree 3, and prove that
there exists γ ∈ S2 (where S2 denotes the unit sphere in R3) with

R(γ) = γ.

[Hint: Use the fact that p(0) > 0 to see that there is λ > 0 with p(λ) = 0.
Then R− λI is singular, so its kernel is non-trivial.]

(b) If P denotes the plane perpendicular to γ and passing through the origin,
show that

R : P → P,

and that this linear map is a rotation.

3. Recall the formula
∫

Rd
F (x) dx =

∫

Sd−1

∫ ∞
0

F (rγ)rd−1 dr dσ(γ).

Apply this to the special case when F (x) = g(r)f(γ), where x = rγ, to prove
that for any rotation R, one has

∫

Sd−1
f(R(γ)) dσ(γ) =

∫

Sd−1
f(γ) dσ(γ),

whenever f is a continuous function on the sphere Sd−1.

4. Let Ad and Vd denote the area and volume of the unit sphere and unit ball
in Rd, respectively.

(a) Prove the formula

Ad =
2πd/2

Γ(d/2)

so that A2 = 2π, A3 = 4π, A4 = 2π2, . . .. Here Γ(x) =
∫∞
0

e−ttx−1 dt is
the Gamma function. [Hint: Use polar coordinates and the fact that∫
Rd e

−π|x|2 dx = 1.]

Ibookroot October 20, 2007

6. Exercises 209

(b) Show that d Vd = Ad, hence

Vd =
πd/2

Γ(d/2 + 1)
.

In particular V2 = π, V3 = 4π/3, . . ..

5. Let A be a d× d positive definite symmetric matrix with real coefficients.
Show that ∫

Rd
e−π(x,A(x)) dx = (det(A))−1/2.

This generalizes the fact that
∫
Rd e

−π|x|2 dx = 1, which corresponds to the case
where A is the identity.
[Hint: Apply the spectral theorem to write A = RDR−1 where R is a rotation
and, D is diagonal with entries λ1, . . . , λd, where {λi} are the eigenvalues of A.]

6. Suppose ψ ∈ S(Rd) satisfies
∫
|ψ(x)|2 dx = 1. Show that

(∫

Rd
|x|2|ψ(x)|2 dx

)(∫

Rd
|ξ|2|ψ̂(ξ)|2 dξ

)
≥ d

16π2
.

This is the statement of the Heisenberg uncertainty principle in d dimensions.

7. Consider the time-dependent heat equation in Rd:

(15)
∂u

∂t
=

∂2u

∂x21
+ · · ·+ ∂

∂x2d
, where t > 0,

with boundary values u(x, 0) = f(x) ∈ S(Rd). If

H(d)t (x) =
1

(4πt)d/2
e−|x|

2/4t =
∫

Rd
e−4π

2t|ξ|2e2πix·ξ dξ

is the d-dimensional heat kernel, show that the convolution

u(x, t) = (f ∗ H(d)t )(x)
is indefinitely differentiable when x ∈ Rd and t > 0. Moreover, u solves (15), and
is continuous up to the boundary t = 0 with u(x, 0) = f(x).

The reader may also wish to formulate the d-dimensional analogues of Theo-
rem 2.1 and 2.3 in Chapter 5.

8. In Chapter 5, we found that a solution to the steady-state heat equation in the
upper half-plane with boundary values f is given by the convolution u = f ∗ Py
where the Poisson kernel is

Py(x) =
1
π

x2 + y2
where x ∈ R and y > 0.

Ibookroot October 20, 2007

210 Chapter 6. THE FOURIER TRANSFORM ON Rd

More generally, one can calculate the d-dimensional Poisson kernel using the
Fourier transform as follows.

(a) The subordination principle allows one to write expressions involv-
ing the function e−x in terms of corresponding expressions involving the
function e−x

2
. One form of this is the identity

e−β =
∫ ∞

e−u√
πu

e−β
2/4u du

when β ≥ 0. Prove this identity with β = 2π|x| by taking the Fourier
transform of both sides.

(b) Consider the steady-state heat equation in the upper half-space {(x, y) :
x ∈ Rd, y > 0}

d∑
j=1

∂2u

∂x2j
+

∂2u

∂y2
= 0

with the Dirichlet boundary condition u(x, 0) = f(x). A solution to this
problem is given by the convolution u(x, y) = (f ∗ P (d)y )(x) where P (d)y (x)
is the d-dimensional Poisson kernel

P (d)y (x) =
∫

Rd
e2πix·ξe−2π|ξ|y dξ.

Compute P (d)y (x) by using the subordination principle and the d-dimensional
heat kernel. (See Exercise 7.) Show that

P (d)y (x) =
Γ((d + 1)/2)

π(d+1)/2
y

(|x|2 + y2)(d+1)/2 .

9. A spherical wave is a solution u(x, t) of the Cauchy problem for the wave
equation in Rd, which as a function of x is radial. Prove that u is a spherical
wave if and only if the initial data f, g ∈ S are both radial.

10. Let u(x, t) be a solution of the wave equation, and let E(t) denote the energy
of this wave

E(t) =
∫

∣∣∣∣
∂u

∂t
(x, t)

∣∣∣∣
2

+
d∑

j=1

∫

∣∣∣∣
∂u

∂xj
(x, t)

∣∣∣∣
2

dx.

We have seen that E(t) is constant using Plancherel’s formula. Give an alternate
proof of this fact by differentiating the integral with respect to t and showing
that

dt
= 0.

Ibookroot October 20, 2007

6. Exercises 211

[Hint: Integrate by parts.]

11. Show that the solution of the wave equation

∂2u

∂t2
=

∂2u

∂x21
+

∂2u

∂x22
+

∂2u

∂x23

subject to u(x, 0) = f(x) and ∂u
∂t

(x, 0) = g(x), where f, g ∈ S(R3), is given by

u(x, t) =
1

|S(x, t)|
∫

S(x,t)

[tg(y) + f(y) +∇f(y) · (y − x)] dσ(y),

where S(x, t) denotes the sphere of center x and radius t, and |S(x, t)| its area.
This is an alternate expression for the solution of the wave equation given in
Theorem 3.6. It is sometimes called Kirchhoff’s formula.

12. Establish the identity (14) about the dual transform given in the text. In
other words, prove that

(16)
∫

∫

S2
R(f)(t, γ)F (t, γ)dσ(γ) dt =

∫

R3
f(x)R∗(F )(x) dx

where f ∈ S(R3), F ∈ S(R× S2), and

R(f) =
∫

Pt,γ
f and R∗(F )(x) =

∫

S2
F (x · γ, γ) dσ(γ).

[Hint: Consider the integral

∫ ∫ ∫
f(tγ + u1e2 + u2e2)F (t, γ) dt dσ(γ) du1 du2.

Integrating first in u gives the left-hand side of (16), while integrating in u and
t and setting x = tγ + u1e2 + u2e2 gives the right-hand side.]

13. For each (t, θ) with t ∈ R and |θ| ≤ π, let L = Lt,θ denote the line in the
(x, y)-plane given by

x cos θ + y sin θ = t.

This is the line perpendicular to the direction (cos θ, sin θ) at “distance” t from
the origin (we allow negative t). For f ∈ S(R2) the X-ray transform or two-
dimensional Radon transform of f is defined by

X(f)(t, θ) =
∫

Lt,θ

f =
∫ ∞
−∞

f(t cos θ + u sin θ, t sin θ − u cos θ) du.

Ibookroot October 20, 2007

212 Chapter 6. THE FOURIER TRANSFORM ON Rd

Calculate the X-ray transform of the function f(x, y) = e−π(x
2+y2).

14. Let X be the X-ray transform. Show that if f ∈ S and X(f) = 0, then
f = 0, by taking the Fourier transform in one variable.

15. For F ∈ S(R× S1), define the dual X-ray transform X∗(F ) by integrat-
ing F over all lines that pass through the point (x, y) (that is, those lines Lt,θ
with x cos θ + y sin θ = t):

X∗(F )(x, y) =
∫

F (x cos θ + y sin θ, θ) dθ.

Check that in this case, if f ∈ S(R2) and F ∈ S(R× S1), then
∫ ∫

X(f)(t, θ)F (t, θ) dt dθ =
∫ ∫

f(x, y)X∗(F )(x, y) dx dy.

7 Problems

1. Let Jn denote the nth order Bessel function, for n ∈ Z. Prove that
(a) Jn(ρ) is real for all real ρ.

(b) J−n(ρ) = (−1)nJn(ρ).
(c) 2J ′n(ρ) = Jn−1(ρ)− Jn+1(ρ).

(d)
(

2n
ρ

)
Jn(ρ) = Jn−1(ρ) + Jn+1(ρ).

(e) (ρ−nJn(ρ))′ = −ρ−nJn+1(ρ).
(f) (ρnJn(ρ))′ = ρnJn−1(ρ).

(g) Jn(ρ) satisfies the second order differential equation

J ′′n(ρ) + ρ
−1J ′n(ρ) + (1− n2/ρ2)Jn(ρ) = 0.

(h) Show that

Jn(ρ) =
(

)n ∞∑
m=0

(−1)m ρ
2m

22mm!(n + m)!
.

(i) Show that for all integers n and all real numbers a and b we have

Jn(a + b) =
∑

`∈Z
J`(a)Jn−`(b).

Ibookroot October 20, 2007

7. Problems 213

2. Another formula for Jn(ρ) that allows one to define Bessel functions for
non-integral values of n, (n > −1/2) is

Jn(ρ) =
(ρ/2)n

Γ(n + 1/2)
√

∫ 1
−1

eiρt(1− t2)n−(1/2) dt.

(a) Check that the above formula agrees with the definition of Jn(ρ) for in-
tegral n ≥ 0. [Hint: Verify it for n = 0 and then check that both sides
satisfy the recursion formula (e) in Problem 1.]

(b) Note that J1/2(ρ) =
√

2
π
ρ−1/2 sin ρ.

lim
n→−1/2

Jn(ρ) =

√
2
π

ρ−1/2 cos ρ.

(d) Observe that the formulas we have proved in the text giving F0 in terms
of f0 (when describing the Fourier transform of a radial function) take the
form

(17) F0(ρ) = 2πρ
−(d/2)+1

∫ ∞
0

J(d/2)−1(2πρr)f0(r)r
d/2 dr,

for d = 1, 2, and 3, if one uses the formulas above with the understanding
that J−1/2(ρ) = limn→−1/2 Jn(ρ). It turns out that the relation between
F0 and f0 given by (17) is valid in all dimensions d.

3. We observed that the solution u(x, t) of the Cauchy problem for the wave
equation given by formula (3) depends only on the initial data on the base on
the backward light cone. It is natural to ask if this property is shared by any
solution of the wave equation. An affirmative answer would imply uniqueness of
the solution.

Let B(x0, r0) denote the closed ball in the hyperplane t = 0 centered at x0
and of radius r0. The backward light cone with base B(x0, r0) is defined by

LB(x0,r0) = {(x, t) ∈ Rd × R : |x− x0| ≤ r0 − t, 0 ≤ t ≤ r0}.

Theorem Suppose that u(x, t) is a C2 function on the closed upper half-plane
{(x, t) : x ∈ Rd, t ≥ 0} that solves the wave equation

∂2u

∂t2
= 4u.

If u(x, 0) = ∂u
∂t

(x, 0) = 0 for all x ∈ B(x0, r0), then u(x, t) = 0 for all (x, t) ∈
LB(x0,r0).

Ibookroot October 20, 2007

214 Chapter 6. THE FOURIER TRANSFORM ON Rd

In words, if the initial data of the Cauchy problem for the wave equation
vanishes on a ball B, then any solution u of the problem vanishes in the backward
light cone with base B. The following steps outline a proof of the theorem.

(a) Assume that u is real. For 0 ≤ t ≤ r0 let Bt(x0, r0) = {x : |x− x0| ≤ r0 −
t}, and also define

∇u(x, t) =
(

∂u

∂x1
, . . . ,

∂u

∂xd
,
∂u

∂t

)
.

Now consider the energy integral

E(t) =
1
2

∫

Bt(x0,r0)

|∇u|2 dx

=
1
2

∫

Bt(x0,r0)

(
∂u

∂t

)2
+

d∑
j=1

(
∂u

∂xj

)2
dx.

Observe that E(t) ≥ 0 and E(0) = 0. Prove that

E′(t) =
∫

Bt(x0,r0)

∂u

∂t

∂2u

∂t2
+

d∑
j=1

∂u

∂xj

∂2u

∂xj∂t
dx− 1

∫

∂Bt(x0,r0)

|∇u|2 dσ(γ).

(b) Show that

∂

∂xj

[
∂u

∂xj

∂u

∂t

]
=

∂u

∂xj

∂2u

∂xj∂t
+

∂2u

∂x2j

∂u

∂t
.

E′(t) =
∫

∂Bt(x0,r0)

d∑
j=1

∂u

∂xj

∂u

∂t
νj dσ(γ)−

1
2

∫

∂Bt(x0,r0)

|∇u|2 dσ(γ),

where νj denotes the jth coordinate of the outward normal to Bt(x0, r0).

(d) Use the Cauchy-Schwarz inequality to conclude that

d∑
j=1

∂u

∂xj

∂u

∂t
νj ≤

1
2
|∇u|2,

and as a result, E′(t) ≤ 0. Deduce from this that E(t) = 0 and u = 0.

Ibookroot October 20, 2007

7. Problems 215

4.∗ There exist formulas for the solution of the Cauchy problem for the wave
equation

∂2u

∂t2
=

∂2u

∂x21
+ · · ·+ ∂

∂x2d
with u(x, 0) = f(x) and

∂u

∂t
(x, 0) = g(x)

in Rd × R in terms of spherical means which generalize the formula given in the
text for d = 3. In fact, the solution for even dimensions is deduced from that for
odd dimensions, so we discuss this case first.

Suppose that d > 1 is odd and let h ∈ S(Rd). The spherical mean of h on the
ball centered at x of radius t is defined by

Mrh(x) = Mh(x, r) =
1

∫

Sd−1
h(x− rγ) dσ(γ),

where Ad denotes the area of the unit sphere Sd−1 in Rd.

(a) Show that

4xMh(x, r) =
[
∂2r +

d− 1
r

]
Mh(x, r),

where 4x denotes the Laplacian in the space variables x, and ∂r = ∂/∂r.

(b) Show that a twice differentiable function u(x, t) satisfies the wave equation
if and only if

[
∂2r +

d− 1
r

]
Mu(x, r, t) = ∂2t Mu(x, r, t),

where Mu(x, r, t) denote the spherical means of the function u(x, t).

(c) If d = 2k + 1, define Tϕ(r) = (r−1∂r)k−1[r2k−1ϕ(r)], and let ũ = TMu.
Then this function solves the one-dimensional wave equation for each fixed
x:

∂2t ũ(x, r, t) = ∂
2
r ũ(x, r, t).

One can then use d’Alembert’s formula to find the solution ũ(x, r, t) of
this problem expressed in terms of the initial data.

(d) Now show that

u(x, t) = Mu(x, 0, t) = lim
r→0

ũ(x, r, t)
αr

where α = 1 · 3 · · · (d− 2).

Ibookroot October 20, 2007

216 Chapter 6. THE FOURIER TRANSFORM ON Rd

(e) Conclude that the solution of the Cauchy problem for the d-dimensional
wave equation, when d > 1 is odd, is

u(x, t) =
1

1 · 3 · · · (d− 2)
[
∂t(t

−1∂t)
(d−3)/2 (td−2Mtf(x)

)
+

(t−1∂t)
(d−3)/2 (td−2Mtg(x)

)]
.

5.∗ The method of descent can be used to prove that the solution of the Cauchy
problem for the wave equation in the case when d is even is given by the formula

u(x, t) =
1

1 · 3 · · · (d− 2)
[
∂t(t

−1∂t)
(d−3)/2

(
td−2M̃tf(x)

)
+

(t−1∂t)
(d−3)/2

(
td−2M̃tg(x)

)]
,

where M̃t denotes the modified spherical means defined by

M̃th(x) =
2

Ad+1

∫

f(x + ty)√
1− |y|2

dy.

6.∗ Given initial data f and g of the form

f(x) = F (x · γ) and g(x) = G(x · γ),

check that the plane wave given by

u(x, t) =
F (x · γ + t) + F (x · γ − t)

2
+

1
2

∫ x·γ+t
x·γ−t

G(s) ds

is a solution of the Cauchy problem for the d-dimensional wave equation.
In general, the solution is given as a superposition of plane waves. For the

case d = 3, this can be expressed in terms of the Radon transform as follows.
Let

R̃(f)(t, γ) = − 1
8π2

(
d

)2
R(f)(t, γ).

Then u(x, t) =

1
2

∫

[
R̃(f)(x · γ − t, γ) + R̃(f)(x · γ + t, γ) +

∫ x·γ+t
x·γ−t

R̃(g)(s, γ) ds
]

dσ(γ).

Ibookroot October 20, 2007

7. Problems 217

7. For every real number a > 0, define the operator (−4)a by the formula

(−4)af(x) =
∫

Rd
(2π|ξ|)2af̂(ξ)e2πiξ·x dξ

whenever f ∈ S(Rd).
(a) Check that (−4)a agrees with the usual definition of the ath power of

−4 (that is, a compositions of minus the Laplacian) when a is a positive
integer.

(b) Verify that (−4)a(f) is indefinitely differentiable.
(c) Prove that if a is not an integer, then in general (−4)a(f) is not rapidly

decreasing.

(d) Let u(x, y) be the solution of the steady-state heat equation

∂2u

∂y2
+

d∑
j=1

∂2u

∂x2j
= 0, with u(x, 0) = f(x),

given by convolving f with the Poisson kernel (see Exercise 8). Check
that

(−4)1/2f(x) = − lim
y→0

∂u

∂y
(x, y),

and more generally that

(−4)k/2f(x) = (−1)k lim
y→0

∂ku

∂yk
(x, y)

for any positive integer k.

8.∗ The reconstruction formula for the Radon transform in Rd is as follows:

(a) When d = 2,

(−4)1/2
4π

R∗(R(f)) = f,

where (−4)1/2 is defined in Problem 7.
(b) If the Radon transform and its dual are defined by analogy to the cases

d = 2 and d = 3, then for general d,

(2π)1−d

2
(−4)(d−1)/2R∗(R(f)) = f.

Ibookroot October 20, 2007

7 Finite Fourier Analysis

This past year has seen the birth, or rather the re-
birth, of an exciting revolution in computing Fourier
transforms. A class of algorithms known as the fast
Fourier transform or FFT, is forcing a complete re-
assessment of many computational paths, not only in
frequency analysis, but in any fields where problems
can be reduced to Fourier transforms and/or convolu-
tions…

C. Bingham and J. W. Tukey, 1966

In the previous chapters we studied the Fourier series of functions on
the circle and the Fourier transform of functions defined on the Euclidean
space Rd. The goal here is to introduce another version of Fourier analy-
sis, now for functions defined on finite sets, and more precisely, on finite
abelian groups. This theory is particularly elegant and simple since infi-
nite sums and integrals are replaced by finite sums, and thus questions
of convergence disappear.

In turning our attention to finite Fourier analysis, we begin with the
simplest example, Z(N), where the underlying space is the (multiplica-
tive) group of N th roots of unity on the circle. This group can also be
realized in additive form, as Z/NZ, the equivalence classes of integers
modulo N . The group Z(N) arises as the natural approximation to the
circle (as N tends to infinity) since in the first picture the points of Z(N)
correspond to N points on the circle which are uniformly distributed. For
this reason, in practical applications, the group Z(N) becomes a natural
candidate for the storage of information of a function on the circle, and
for the resulting numerical computations involving Fourier series. The
situation is particularly nice when N is large and of the form N = 2n.
The computations of the Fourier coefficients now lead to the “fast Fourier
transform,” which exploits the fact that an induction in n requires only
about log N steps to go from N = 1 to N = 2n. This yields a substantial
saving in time in practical applications.

In the second part of the chapter we undertake the more general the-
ory of Fourier analysis on finite abelian groups. Here the fundamental
example is the multiplicative group Z∗(q). The Fourier inversion formula

Ibookroot October 20, 2007

1. Fourier analysis on Z(N) 219

for Z∗(q) will be seen to be a key step in the proof of Dirichlet’s theorem
on primes in arithmetic progression, which we will take up in the next
chapter.

1 Fourier analysis on Z(N)
We turn to the group of N th roots of unity. This group arises naturally as
the simplest finite abelian group. It also gives a uniform partition of the
circle, and is therefore a good choice if one wishes to sample appropriate
functions on the circle. Moreover, this partition gets finer as N tends to
infinity, and one might expect that the discrete Fourier theory that we
discuss here tends to the continuous theory of Fourier series on the circle.
In a broad sense, this is the case, although this aspect of the problem is
not one that we develop.

1.1 The group Z(N)

Let N be a positive integer. A complex number z is an N th root of
unity if zN = 1. The set of N th roots of unity is precisely

{
1, e2πi/N , e2πi2/N , . . . , e2πi(N−1)/N

}
.

Indeed, suppose that zN = 1 with z = reiθ. Then we must have rNeiNθ =
1, and taking absolute values yields r = 1. Therefore eiNθ = 1, and this
means that Nθ = 2πk where k ∈ Z. So if ζ = e2πi/N we find that ζk
exhausts all the N th roots of unity. However, notice that ζN = 1 so if
n and m differ by an integer multiple of N , then ζn = ζm. In fact, it is
clear that

ζn = ζm if and only if n−m is divisible by N .

We denote the set of all N th roots of unity by Z(N). The fact that
this set gives a uniform partition of the circle is clear from its definition.
Note that the set Z(N) satisfies the following properties:

(i) If z, w ∈ Z(N), then zw ∈ Z(N) and zw = wz.
(ii) 1 ∈ Z(N).
(iii) If z ∈ Z(N), then z−1 = 1/z ∈ Z(N) and of course zz−1 = 1.

As a result we can conclude that Z(N) is an abelian group under complex
multiplication. The appropriate definitions are set out in detail later in
Section 2.1.

Ibookroot October 20, 2007

220 Chapter 7. FINITE FOURIER ANALYSIS

ζ8

ζ6

ζ4

ζ3
ζ2

Z(N), N = 26Z(9), ζ = e2πi/9

ζ5

ζ7

Figure 1. The group of N th roots of unity when N = 9 and N = 26 =
64

There is another way to visualize the group Z(N). This consists of
choosing the integer power of ζ that determines each root of unity. We
observed above that this integer is not unique since ζn = ζm whenever n
and m differ by an integer multiple of N . Naturally, we might select the
integer which satisfies 0 ≤ n ≤ N − 1. Although this choice is perfectly
reasonable in terms of “sets,” we ask what happens when we multiply
roots of unity. Clearly, we must add the corresponding integers since
ζnζm = ζn+m but nothing guarantees that 0 ≤ n + m ≤ N − 1. In fact,
if ζnζm = ζk with 0 ≤ k ≤ N − 1, then n + m and k differ by an integer
multiple of N . So, to find the integer in [0, N − 1] corresponding to the
root of unity ζnζm, we see that after adding the integers n and m we
must reduce modulo N , that is, find the unique integer 0 ≤ k ≤ N − 1
so that (n + m)− k is an integer multiple of N .

An equivalent approach is to associate to each root of unity ω the
class of integers n so that ζn = ω. Doing so for each root of unity we
obtain a partition of the integers in N disjoint infinite classes. To add
two of these classes, choose any integer in each one of them, say n and
m, respectively, and define the sum of the classes to be the class which
contains the integer n + m.

We formalize the above notions. Two integers x and y are congru-
ent modulo N if the difference x− y is divisible by N , and we write
x ≡ y mod N . In other words, this means that x and y differ by an
integer multiple of N . It is an easy exercise to check the following three
properties:

Ibookroot October 20, 2007

1. Fourier analysis on Z(N) 221

• x ≡ x mod N for all integers x.
• If x ≡ y mod N , then y ≡ x mod N .
• If x ≡ y mod N and y ≡ z mod N , then x ≡ z mod N .

The above defines an equivalence relation on Z. Let R(x) denote the
equivalence class, or residue class, of the integer x. Any integer of the
form x + kN with k ∈ Z is an element (or “representative”) of R(x).
In fact, there are precisely N equivalence classes, and each class has a
unique representative between 0 and N − 1. We may now add equiva-
lence classes by defining

R(x) + R(y) = R(x + y).

This definition is of course independent of the representatives x and y
because if x′ ∈ R(x) and y′ ∈ R(y), then one checks easily that x′ + y′ ∈
R(x + y). This turns the set of equivalence classes into an abelian group
called the group of integers modulo N , which is sometimes denoted
by Z/NZ. The association

R(k) ←→ e2πik/N

gives a correspondence between the two abelian groups, Z/NZ and Z(N).
Since the operations are respected, in the sense that addition of inte-
gers modulo N becomes multiplication of complex numbers, we shall
also denote the group of integers modulo N by Z(N). Observe that
0 ∈ Z/NZ corresponds to 1 on the unit circle.

Let V and W denote the vector spaces of complex-valued functions on
the group of integers modulo N and the N th roots of unity, respectively.
Then, the identification given above carries over to V and W as follows:

F (k) ←→ f(e2πik/N ),

where F is a function on the integers modulo N and f is a function on
the N th roots of unity.

From now on, we write Z(N) but think of either the group of integers
modulo N or the group of N th roots of unity.

1.2 Fourier inversion theorem and Plancherel identity on Z(N)

The first and most crucial step in developing Fourier analysis on Z(N) is
to find the functions which correspond to the exponentials en(x) = e2πinx

in the case of the circle. Some important properties of these exponentials
are:

Ibookroot October 20, 2007

222 Chapter 7. FINITE FOURIER ANALYSIS

(i) {en}n∈Z is an orthonormal set for the inner product (1) (in Chap-
ter 3) on the space of Riemann integrable functions on the circle.

(ii) Finite linear combinations of the en’s (the trigonometric polyno-
mials) are dense in the space of continuous functions on the circle.

(iii) en(x + y) = en(x)en(y).

On Z(N), the appropriate analogues are the N functions e0, . . . , eN−1
defined by

e`(k) = ζ
`k = e2πi`k/N for ` = 0, . . . , N − 1 and k = 0, . . . , N − 1,

where ζ = e2πi/N . To understand the parallel with (i) and (ii), we can
think of the complex-valued functions on Z(N) as a vector space V ,
endowed with the Hermitian inner product

(F, G) =
N−1∑

k=0

F (k)G(k)

and associated norm

‖F‖2 =
N−1∑

k=0

|F (k)|2.

Lemma 1.1 The family {e0, . . . , eN−1} is orthogonal. In fact,

(em, e`) =
{

N if m = `,
0 if m 6= `.

Proof. We have

(em, e`) =
N−1∑

k=0

ζmkζ−`k =
N−1∑

k=0

ζ(m−`)k.

If m = `, each term in the sum is equal to 1, and the sum equals N . If
m 6= `, then q = ζm−` is not equal to 1, and the usual formula

1 + q + q2 + · · ·+ qN−1 = 1− q
N

1− q

shows that (em, e`) = 0, because qN = 1.

Since the N functions e0, . . . , eN−1 are orthogonal, they must be lin-
early independent, and since the vector space V is N -dimensional, we

Ibookroot October 20, 2007

1. Fourier analysis on Z(N) 223

conclude that {e0, . . . , eN−1} is an orthogonal basis for V . Clearly, prop-
erty (iii) also holds, that is, e`(k + m) = e`(k)e`(m) for all `, and all
k,m ∈ Z(N).

By the lemma each vector e` has norm
√

N , so if we define

e∗` =
1√
N

e`,

then {e∗0, . . . , e∗N−1} is an orthonormal basis for V . Hence for any F ∈ V
we have

(1) F =
N−1∑

n=0

(F, e∗n)e
∗
n as well as ‖F‖2 =

N−1∑

n=0

|(F, e∗n)|2.

If we define the nth Fourier coefficient of F by

an =
1
N

N−1∑

k=0

F (k)e−2πikn/N ,

the above observations give the following fundamental theorem which is
the Z(N) version of the Fourier inversion and the Parseval-Plancherel
formulas.

Theorem 1.2 If F is a function on Z(N), then

F (k) =
N−1∑

n=0

ane
2πink/N .

Moreover,

N−1∑

n=0

|an|2 =
1
N

N−1∑

k=0

|F (k)|2.

The proof follows directly from (1) once we observe that

an =
1
N

(F, en) =
1√
N

(F, e∗n).

Remark. It is possible to recover the Fourier inversion on the circle
for sufficiently smooth functions (say C2) by letting N →∞ in the finite
model Z(N) (see Exercise 3).

Ibookroot October 20, 2007

224 Chapter 7. FINITE FOURIER ANALYSIS

1.3 The fast Fourier transform

The fast Fourier transform is a method that was developed as a means
of calculating efficiently the Fourier coefficients of a function F on Z(N).

The problem, which arises naturally in numerical analysis, is to deter-
mine an algorithm that minimizes the amount of time it takes a computer
to calculate the Fourier coefficients of a given function on Z(N). Since
this amount of time is roughly proportional to the number of operations
the computer must perform, our problem becomes that of minimizing
the number of operations necessary to obtain all the Fourier coefficients
{an} given the values of F on Z(N). By operations we mean either an
addition or a multiplication of complex numbers.

We begin with a naive approach to the problem. Fix N , and suppose
that we are given F (0), . . . , F (N − 1) and ωN = e−2πi/N . If we denote
by aNk (F ) the k

th Fourier coefficient of F on Z(N), then by definition

aNk (F ) =
1
N

N−1∑

r=0

F (r)ωkrN ,

and crude estimates show that the number of operations needed to cal-
culate all Fourier coefficients is ≤ 2N2 + N . Indeed, it takes at most
N − 2 multiplications to determine ω2N , . . . , ωN−1N , and each coefficient
aNk requires N + 1 multiplications and N − 1 additions.

We now present the fast Fourier transform, an algorithm that im-
proves the bound O(N2) obtained above. Such an improvement is possi-
ble if, for example, we restrict ourselves to the case where the partition
of the circle is dyadic, that is, N = 2n. (See also Exercise 9.)

Theorem 1.3 Given ωN = e−2πi/N with N = 2n, it is possible to calcu-
late the Fourier coefficients of a function on Z(N) with at most

4 · 2nn = 4N log2(N) = O(N log N)

operations.

The proof of the theorem consists of using the calculations for M
division points, to obtain the Fourier coefficients for 2M division points.
Since we choose N = 2n, we obtain the desired formula as a consequence
of a recurrence which involves n = O(log N) steps.

Let #(M) denote the minimum number of operations needed to cal-
culate all the Fourier coefficients of any function on Z(M). The key to
the proof of the theorem is contained in the following recursion step.

Ibookroot October 20, 2007

1. Fourier analysis on Z(N) 225

Lemma 1.4 If we are given ω2M = e−2πi/(2M), then

#(2M) ≤ 2#(M) + 8M.

Proof. The calculation of ω2M , . . . , ω2M2M requires no more than 2M
operations. Note that in particular we get ωM = e−2πi/M = ω22M . The
main idea is that for any given function F on Z(2M), we consider two
functions F0 and F1 on Z(M) defined by

F0(r) = F (2r) and F1(r) = F (2r + 1).

We assume that it is possible to calculate the Fourier coefficients of F0
and F1 in no more than #(M) operations each. If we denote the Fourier
coefficients corresponding to the groups Z(2M) and Z(M) by a2Mk and
aMk , respectively, then we have

a2Mk (F ) =
1
2

(
aMk (F0) + a

M
k (F1)ω

k
2M

)
.

To prove this, we sum over odd and even integers in the definition of the
Fourier coefficient a2Mk (F ), and find

a2Mk (F ) =
1

2M−1∑

r=0

F (r)ωkr2M

=
1
2

(
1
M

M−1∑

`=0

F (2`)ωk(2`)2M +
1
M

M−1∑

m=0

F (2m + 1)ωk(2m+1)2M

)

=
1
2

(
1
M

M−1∑

`=0

F0(`)ω
k`
M +

1
M

M−1∑

m=0

F1(m)ω
km
M ω

k
2M

)
,

which establishes our assertion.
As a result, knowing aMk (F0), a

M
k (F1), and ω

k
2M , we see that each

a2Mk (F ) can be computed using no more than three operations (one ad-
dition and two multiplications). So

#(2M) ≤ 2M + 2#(M) + 3× 2M = 2#(M) + 8M,

and the proof of the lemma is complete.

An induction on n, where N = 2n, will conclude the proof of the the-
orem. The initial step n = 1 is easy, since N = 2 and the two Fourier
coefficients are

aN0 (F ) =
1
2

(F (1) + F (−1)) and aN1 (F ) =
1
2

(F (1) + (−1)F (−1)) .

Ibookroot October 20, 2007

226 Chapter 7. FINITE FOURIER ANALYSIS

Calculating these Fourier coefficients requires no more than five opera-
tions, which is less than 4× 2 = 8. Suppose the theorem is true up to
N = 2n−1 so that #(N) ≤ 4 · 2n−1(n− 1). By the lemma we must have

#(2N) ≤ 2 · 4 · 2n−1(n− 1) + 8 · 2n−1 = 4 · 2nn,
which concludes the inductive step and the proof of the theorem.

2 Fourier analysis on finite abelian groups

The main goal in the rest of this chapter is to generalize the results about
Fourier series expansions obtained in the special case of Z(N).

After a brief introduction to some notions related to finite abelian
groups, we turn to the important concept of a character. In our set-
ting, we find that characters play the same role as the exponentials
e0, . . . , eN−1 on the group Z(N), and thus provide the key ingredient
in the development of the theory on arbitrary finite abelian groups. In
fact, it suffices to prove that a finite abelian group has “enough” charac-
ters, and this leads automatically to the desired Fourier theory.

2.1 Abelian groups

An abelian group (or commutative group) is a set G together with a
binary operation on pairs of elements of G, (a, b) 7→ a · b, that satisfies
the following conditions:

(i) Associativity : a · (b · c) = (a · b) · c for all a, b, c ∈ G.
(ii) Identity : There exists an element u ∈ G (often written as either 1

or 0) such that a · u = u · a = a for all a ∈ G.
(iii) Inverses: For every a ∈ G, there exists an element a−1 ∈ G such

that a · a−1 = a−1 · a = u.
(iv) Commutativity : For a, b ∈ G, we have a · b = b · a.

We leave as simple verifications the facts that the identity element and
inverses are unique.

Warning. In the definition of an abelian group, we used the “multi-
plicative” notation for the operation in G. Sometimes, one uses the “ad-
ditive” notation a + b and −a, instead of a · b and a−1. There are times
when one notation may be more appropriate than the other, and the
examples below illustrate this point. The same group may have different
interpretations, one where the multiplicative notation is more suggestive,
and another where it is natural to view the group with addition, as the
operation.

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2. Fourier analysis on finite abelian groups 227

Examples of abelian groups

• The set of real numbers R with the usual addition. The identity is
0 and the inverse of x is −x.
Also, R− {0} and R+ = {x ∈ R : x > 0} equipped, with the stan-
dard multiplication, are abelian groups. In both cases the unit is 1
and the inverse of x is 1/x.

• With the usual addition, the set of integers Z is an abelian group.
However, Z− {0} is not an abelian group with the standard mul-
tiplication, since, for example, 2 does not have a multiplicative
inverse in Z. In contrast, Q− {0} is an abelian group with the
standard multiplication.

• The unit circle S1 in the complex plane. If we view the circle as
the set of points {eiθ : θ ∈ R}, the group operation is the standard
multiplication of complex numbers. However, if we identify points
on S1 with their angle θ, then S1 becomes R modulo 2π, where the
operation is addition modulo 2π.

• Z(N) is an abelian group. Viewed as the N th roots of unity on the
circle, Z(N) is a group under multiplication of complex numbers.
However, if Z(N) is interpreted as Z/NZ, the integers modulo N ,
then it is an abelian group where the operation is addition modulo
N .

• The last example consists of Z∗(q). This group is defined as the set
of all integers modulo q that have multiplicative inverses, with the
group operation being multiplication modulo q. This important
example is discussed in more detail below.

A homomorphism between two abelian groups G and H is a map
f : G → H which satisfies the property

f(a · b) = f(a) · f(b),

where the dot on the left-hand side is the operation in G, and the dot
on the right-hand side the operation in H.

We say that two groups G and H are isomorphic, and write G ≈ H,
if there is a bijective homomorphism from G to H. Equivalently, G and
H are isomorphic if there exists another homomorphism f̃ : H → G, so
that for all a ∈ G and b ∈ H

(f̃ ◦ f)(a) = a and (f ◦ f̃)(b) = b.

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228 Chapter 7. FINITE FOURIER ANALYSIS

Roughly speaking, isomorphic groups describe the “same” object because
they have the same underlying group structure (which is really all that
matters); however, their particular notational representations might be
different.

Example 1. A pair of isomorphic abelian groups arose already when
we considered the group Z(N). In one representation it was given as
the multiplicative group of N th roots of unity in C. In a second repre-
sentation it was the additive group Z/NZ of residue classes of integers
modulo N . The mapping n 7→ R(n), which associates to a root of unity
z = e2πin/N = ζn the residue class in Z/NZ determined by n, provides
an isomorphism between the two different representations.

Example 2. In parallel with the previous example, we see that the circle
(with multiplication) is isomorphic to the real numbers modulo 2π (with
addition).

Example 3. The properties of the exponential and logarithm guarantee
that

exp : R→ R+ and log : R+ → R
are two homomorphisms that are inverses of each other. Thus R (with
addition) and R+ (with multiplication) are isomorphic.

In what follows, we are primarily interested in abelian groups that are
finite. In this case, we denote by |G| the number of elements in G, and
call |G| the order of the group. For example, the order of Z(N) is N .

A few additional remarks are in order:

• If G1 and G2 are two finite abelian groups, their direct product
G1 ×G2 is the group whose elements are pairs (g1, g2) with g1 ∈ G1
and g2 ∈ G2. The operation in G1 ×G2 is then defined by

(g1, g2) · (g′1, g′2) = (g1 · g′1, g2 · g′2).
Clearly, if G1 and G2 are finite abelian groups, then so is G1 ×G2.
The definition of direct product generalizes immediately to the case
of finitely many factors G1 ×G2 × · · · ×Gn.

• The structure theorem for finite abelian groups states that such a
group is isomorphic to a direct product of groups of the type Z(N);
see Problem 2. This is a nice result which gives us an overview of
the class of all finite abelian groups. However, since we shall not
use this theorem below, we omit its proof.

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2. Fourier analysis on finite abelian groups 229

We now discuss briefly the examples of abelian groups that play a
central role in the proof of Dirichlet’s theorem in the next chapter.

The group Z∗(q)

Let q be a positive integer. We see that multiplication in Z(q) can be
unambiguously defined, because if n is congruent to n′ and m is congruent
to m′ (both modulo q), then nm is congruent to n′m′ modulo q. An
integer n ∈ Z(q) is a unit if there exists an integer m ∈ Z(q) so that

nm ≡ 1 mod q.

The set of all units in Z(q) is denoted by Z∗(q), and it is clear from our
definition that Z∗(q) is an abelian group under multiplication modulo q.
Thus within the additive group Z(q) lies a set Z∗(q) that is a group under
multiplication. An alternative characterization of Z∗(q) will be given in
the next chapter, as those elements in Z(q) that are relatively prime to q.

Example 4. The group of units in Z(4) = {0, 1, 2, 3} is

Z∗(4) = {1, 3}.

This reflects the fact that odd integers are divided into two classes de-
pending on whether they are of the form 4k + 1 or 4k + 3. In fact, Z∗(4)
is isomorphic to Z(2). Indeed, we can make the following association:

Z∗(4) Z(2)
1 ←→ 0
3 ←→ 1

and then notice that multiplication in Z∗(4) corresponds to addition in
Z(2).

Example 5. The units in Z(5) are

Z∗(5) = {1, 2, 3, 4}.

Moreover, Z∗(5) is isomorphic to Z(4) with the following identification:

Z∗(5) Z(4)
1 ←→ 0
2 ←→ 1
3 ←→ 3
4 ←→ 2

Ibookroot October 20, 2007

230 Chapter 7. FINITE FOURIER ANALYSIS

Example 6. The units in Z(8) = {0, 1, 2, 3, 4, 5, 6, 7} are given by

Z∗(8) = {1, 3, 5, 7}.

In fact, Z∗(8) is isomorphic to the direct product Z(2)× Z(2). In this
case, an isomorphism between the groups is given by the identification

Z∗(8) Z(2)× Z(2)
1 ←→ (0, 0)
3 ←→ (1, 0)
5 ←→ (0, 1)
7 ←→ (1, 1)

2.2 Characters

Let G be a finite abelian group (with the multiplicative notation) and
S1 the unit circle in the complex plane. A character on G is a complex-
valued function e : G → S1 which satisfies the following condition:

(2) e(a · b) = e(a)e(b) for all a, b ∈ G.

In other words, a character is a homomorphism from G to the circle
group. The trivial or unit character is defined by e(a) = 1 for all
a ∈ G.

Characters play an important role in the context of finite Fourier se-
ries, primarily because the multiplicative property (2) generalizes the
analogous identity for the exponential functions on the circle and the
law

e`(k + m) = e`(k)e`(m),

which held for the exponentials e0, . . . , eN−1 used in the Fourier theory
on Z(N). There we had e`(k) = ζ`k = e2πi`k/N , with 0 ≤ ` ≤ N − 1 and
k ∈ Z(N), and in fact, the functions e0, . . . , eN−1 are precisely all the
characters of the group Z(N).

If G is a finite abelian group, we denote by Ĝ the set of all characters
of G, and observe next that this set inherits the structure of an abelian
group.

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2. Fourier analysis on finite abelian groups 231

Lemma 2.1 The set Ĝ is an abelian group under multiplication defined
by

(e1 · e2)(a) = e1(a)e2(a) for all a ∈ G.

The proof of this assertion is straightforward if one observes that the
trivial character plays the role of the unit. We call Ĝ the dual group
of G.

In light of the above analogy between characters for a general abelian
group and the exponentials on Z(N), we gather several more examples
of groups and their duals. This provides further evidence of the central
role played by characters. (See Exercises 4, 5, and 6.)

Example 1. If G = Z(N), all characters of G take the form e`(k) = ζ`k =
e2πi`k/N for some 0 ≤ ` ≤ N − 1, and it is easy to check that e` 7→ ` gives
an isomorphism from Ẑ(N) to Z(N).

Example 2. The dual group of the circle1 is precisely {en}n∈Z (where
en(x) = e2πinx). Moreover, en 7→ n gives an isomorphism between Ŝ1
and the integers Z.

Example 3. Characters on R are described by

eξ(x) = e
2πiξx where ξ ∈ R.

Thus eξ 7→ ξ is an isomorphism from R̂ to R.

Example 4. Since exp : R→ R+ is an isomorphism, we deduce from the
previous example that the characters on R+ are given by

eξ(x) = x
2πiξ = e2πiξ log x where ξ ∈ R,

and R̂+ is isomorphic to R (or R+).

The following lemma says that a nowhere vanishing multiplicative
function is a character, a result that will be useful later.

Lemma 2.2 Let G be a finite abelian group, and e : G → C− {0} a mul-
tiplicative function, namely e(a · b) = e(a)e(b) for all a, b ∈ G. Then e is
a character.

1In addition to (2), the definition of a character on an infinite abelian group requires
continuity. When G is the circle, R, or R+, the meaning of “continuous” refers to the
standard notion of limit.

Ibookroot October 20, 2007

232 Chapter 7. FINITE FOURIER ANALYSIS

Proof. The group G being finite, the absolute value of e(a) is bounded
above and below as a ranges over G. Since |e(bn)| = |e(b)|n, we conclude
that |e(b)| = 1 for all b ∈ G.

The next step is to verify that the characters form an orthonormal
basis of the vector space V of functions over the group G. This fact
was obtained directly in the special case G = Z(N) from the explicit
description of the characters e0, . . . , eN−1.

In the general case, we begin with the orthogonality relations; then we
prove that there are “enough” characters by showing that there are as
many as the order of the group.

2.3 The orthogonality relations

Let V denote the vector space of complex-valued functions defined on the
finite abelian group G. Note that the dimension of V is |G|, the order of
G. We define a Hermitian inner product on V by

(3) (f, g) =
1
|G|

∑

a∈G
f(a)g(a), whenever f, g ∈ V .

Here the sum is taken over the group and is therefore finite.

Theorem 2.3 The characters of G form an orthonormal family with
respect to the inner product defined above.

Since |e(a)| = 1 for any character, we find that

(e, e) =
1
|G|

∑

a∈G
e(a)e(a) =

1
|G|

∑

a∈G
|e(a)|2 = 1.

If e 6= e′ and both are characters, we must prove that (e, e′) = 0; we
isolate the key step in a lemma.

Lemma 2.4 If e is a non-trivial character of the group G, then∑
a∈G e(a) = 0.

Proof. Choose b ∈ G such that e(b) 6= 1. Then we have

e(b)
∑

a∈G
e(a) =

∑

a∈G
e(b)e(a) =

∑

a∈G
e(ab) =

∑

a∈G
e(a).

The last equality follows because as a ranges over the group, ab ranges
over G as well. Therefore

∑
a∈G e(a) = 0.

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2. Fourier analysis on finite abelian groups 233

We can now conclude the proof of the theorem. Suppose e′ is a char-
acter distinct from e. Because e(e′)−1 is non-trivial, the lemma implies
that

∑

a∈G
e(a)(e′(a))−1 = 0.

Since (e′(a))−1 = e′(a), the theorem is proved.

As a consequence of the theorem, we see that distinct characters are
linearly independent. Since the dimension of V over C is |G|, we conclude
that the order of Ĝ is finite and ≤ |G|. The main result to which we now
turn is that, in fact, |Ĝ| = |G|.

2.4 Characters as a total family

The following completes the analogy between characters and the complex
exponentials.

Theorem 2.5 The characters of a finite abelian group G form a basis
for the vector space of functions on G.

There are several proofs of this theorem. One consists of using the
structure theorem for finite abelian groups we have mentioned earlier,
which states that any such group is the direct product of cyclic groups,
that is, groups of the type Z(N). Since cyclic groups are self-dual, using
this fact we would conclude that |Ĝ| = |G|, and therefore the characters
form a basis for G. (See Problem 3.)

Here we shall prove the theorem directly without these considerations.

Suppose V is a vector space of dimension d with inner product (·, ·).
A linear transformation T : V → V is unitary if it preserves the inner
product, (Tv, Tw) = (v, w) for all v, w ∈ V . The spectral theorem from
linear algebra asserts that any unitary transformation is diagonalizable.
In other words, there exists a basis {v1, . . . , vd} (eigenvectors) of V such
that T (vi) = λivi, where λi ∈ C is the eigenvalue attached to vi.

The proof of Theorem 2.5 is based on the following extension of the
spectral theorem.

Lemma 2.6 Suppose {T1, . . . , Tk} is a commuting family of unitary trans-
formations on the finite-dimensional inner product space V ; that is,

TiTj = TjTi for all i, j.

Then T1, . . . , Tk are simultaneously diagonalizable. In other words, there
exists a basis for V which consists of eigenvectors for every
Ti, i = 1, . . . , k.

Ibookroot October 20, 2007

234 Chapter 7. FINITE FOURIER ANALYSIS

Proof. We use induction on k. The case k = 1 is simply the spec-
tral theorem. Suppose that the lemma is true for any family of k − 1
commuting unitary transformations. The spectral theorem applied to Tk
says that V is the direct sum of its eigenspaces

V = Vλ1 ⊕ · · · ⊕ Vλs ,

where Vλi denotes the subspace of all eigenvectors with eigenvalue λi.
We claim that each one of the T1, . . . , Tk−1 maps each eigenspace Vλi to
itself. Indeed, if v ∈ Vλi and 1 ≤ j ≤ k − 1, then

TkTj(v) = TjTk(v) = Tj(λiv) = λiTj(v)

so Tj(v) ∈ Vλi , and the claim is proved.
Since the restrictions to Vλi of T1, . . . , Tk−1 form a family of commut-

ing unitary linear transformations, the induction hypothesis guarantees
that these are simultaneously diagonalizable on each subspace Vλi . This
diagonalization provides us with the desired basis for each Vλi , and thus
for V .

We can now prove Theorem 2.5. Recall that the vector space V of
complex-valued functions defined on G has dimension |G|. For each
a ∈ G we define a linear transformation Ta : V → V by

(Taf)(x) = f(a · x) for x ∈ G.

Since G is abelian it is clear that TaTb = TbTa for all a, b ∈ G, and one
checks easily that Ta is unitary for the Hermitian inner product (3) de-
fined on V . By Lemma 2.6 the family {Ta}a∈G is simultaneously di-
agonalizable. This means there is a basis {vb(x)}b∈G for V such that
each vb(x) is an eigenfunction for Ta, for every a. Let v be one of these
basis elements and 1 the unit element in G. We must have v(1) 6= 0 for
otherwise

v(a) = v(a · 1) = (Tav)(1) = λav(1) = 0,

where λa is the eigenvalue of v for Ta. Hence v = 0, and this is a contra-
diction. We claim that the function defined by w(x) = λx = v(x)/v(1)
is a character of G. Arguing as above we find that w(x) 6= 0 for every x,
and

w(a · b) = v(a · b)
v(1)

=
λav(b)
v(1)

= λaλb
v(1)
v(1)

= λaλb = w(a)w(b).

We now invoke Lemma 2.2 to conclude the proof.

Ibookroot October 20, 2007

2. Fourier analysis on finite abelian groups 235

2.5 Fourier inversion and Plancherel formula

We now put together the results obtained in the previous sections to
discuss the Fourier expansion of a function on a finite abelian group G.
Given a function f on G and character e of G, we define the Fourier
coefficient of f with respect to e, by

f̂(e) = (f, e) =
1
|G|

∑

a∈G
f(a)e(a),

and the Fourier series of f as

f ∼
∑

e∈Ĝ
f̂(e)e.

Since the characters form a basis, we know that

f =
∑

e∈Ĝ
cee

for some set of constants ce. By the orthogonality relations satisfied by
the characters, we find that

(f, e) = ce,

so f is indeed equal to its Fourier series, namely,

f =
∑

e∈Ĝ
f̂(e)e.

We summarize our results.

Theorem 2.7 Let G be a finite abelian group. The characters of G form
an orthonormal basis for the vector space V of functions on G equipped
with the inner product

(f, g) =
1
|G|

∑

a∈G
f(a)g(a).

In particular, any function f on G is equal to its Fourier series

f =
∑

e∈Ĝ
f̂(e)e.

Finally, we have the Parseval-Plancherel formula for finite abelian
groups.

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236 Chapter 7. FINITE FOURIER ANALYSIS

Theorem 2.8 If f is a function on G, then ‖f‖2 =
∑

e∈Ĝ
|f̂(e)|2.

Proof. Since the characters of G form an orthonormal basis for the
vector space V , and (f, e) = f̂(e), we have that

‖f‖2 = (f, f) =
∑

e∈Ĝ
(f, e)f̂(e) =

∑

e∈Ĝ
|f̂(e)|2.

The apparent difference of this statement with that of Theorem 1.2
is due to the different normalizations of the Fourier coefficients that are
used.

3 Exercises

1. Let f be a function on the circle. For each N ≥ 1 the discrete Fourier
coefficients of f are defined by

aN (n) =
1
N

N∑

k=1

f(e2πik/N )e−2πikn/N , for n ∈ Z.

We also let

a(n) =
∫ 1

f(e2πix)e−2πinx dx

denote the ordinary Fourier coefficients of f .

(a) Show that aN (n) = aN (n + N).

(b) Prove that if f is continuous, then aN (n) → a(n) as N →∞.

2. If f is a C1 function on the circle, prove that |aN (n)| ≤ c/|n| whenever
0 < |n| ≤ N/2. [Hint: Write aN (n)[1− e2πi`n/N ] = 1 N N∑ k=1 [f(e2πik/N )− f(e2πi(k+`)/N )]e−2πikn/N , and choose ` so that `n/N is nearly 1/2.] 3. By a similar method, show that if f is a C2 function on the circle, then |aN (n)| ≤ c/|n|2, whenever 0 < |n| ≤ N/2. Ibookroot October 20, 2007 3. Exercises 237 As a result, prove the inversion formula for f ∈ C2, f(e2πix) = ∞∑ n=−∞ a(n)e2πinx from its finite version. [Hint: For the first part, use the second symmetric difference f(e2πi(k+`)/N ) + f(e2πi(k−`)/N )− 2f(e2πik/N ). For the second part, if N is odd (say), write the inversion formula as f(e2πik/N ) = ∑ |n| 1.

10. A group G is cyclic if there exists g ∈ G that generates all of G, that is,
if any element in G can be written as gn for some n ∈ Z. Prove that a finite
abelian group is cyclic if and only if it is isomorphic to Z(N) for some N .

11. Write down the multiplicative tables for the groups Z∗(3), Z∗(4), Z∗(5),
Z∗(6), Z∗(8), and Z∗(9). Which of these groups are cyclic?

12. Suppose that G is a finite abelian group and e : G → C is a function that
satisfies e(x · y) = e(x)e(y) for all x, y ∈ G. Prove that either e is identically 0,
or e never vanishes. In the second case, show that for each x, e(x) = e2πir for
some r ∈ Q of the form r = p/q, where q = |G|.

Ibookroot October 20, 2007

4. Problems 239

13. In analogy with ordinary Fourier series, one may interpret finite Fourier
expansions using convolutions as follows. Suppose G is a finite abelian group,
1G its unit, and V the vector space of complex-valued functions on G.

(a) The convolution of two functions f and g in V is defined for each a ∈ G
by

(f ∗ g)(a) = 1|G|
∑

b∈G
f(b)g(a · b−1).

Show that for all e ∈ Ĝ one has (̂f ∗ g)(e) = f̂(e)ĝ(e).
(b) Use Theorem 2.5 to show that if e is a character on G, then

∑

e∈Ĝ
e(c) = 0 whenever c ∈ G and c 6= 1G.

e∈Ĝ f̂(e)e(a) of
a function f ∈ V takes the form

Sf = f ∗D,

where D is defined by

(4) D(c) =
∑

e∈Ĝ
e(c) =

{
|G| if c = 1G,
0 otherwise.

Since f ∗D = f , we recover the fact that Sf = f . Loosely speaking, D
corresponds to a “Dirac delta function”; it has unit mass

1
|G|

∑
c∈G

D(c) = 1,

and (4) says that this mass is concentrated at the unit element in G. Thus
D has the same interpretation as the “limit” of a family of good kernels.
(See Section 4, Chapter 2.)

Note. The function D reappears in the next chapter as δ1(n).

4 Problems

1. Prove that if n and m are two positive integers that are relatively prime, then

Z(nm) ≈ Z(n)× Z(m).

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240 Chapter 7. FINITE FOURIER ANALYSIS

[Hint: Consider the map Z(nm) → Z(n)× Z(m) given by k 7→ (k mod n, k mod
m), and use the fact that there exist integers x and y such that xn + ym = 1.]

2.∗ Every finite abelian group G is isomorphic to a direct product of cyclic
groups. Here are two more precise formulations of this theorem.

• If p1, . . . , ps are the distinct primes appearing in the factorization of the
order of G, then

G ≈ G(p1)× · · · ×G(ps),

where each G(p) is of the form G(p) = Z(pr1)× · · · × Z(pr`), with 0 ≤
r1 ≤ · · · ≤ r` (this sequence of integers depends on p of course). This
decomposition is unique.

• There exist unique integers d1, . . . , dk such that

d1|d2, d2|d3, · · · , dk−1|dk

and

G ≈ Z(d1)× · · · × Z(dk).

Deduce the second formulation from the first.

3. Let Ĝ denote the collection of distinct characters of the finite abelian group
G.

(a) Note that if G = Z(N), then Ĝ is isomorphic to G.

(b) Prove that ̂G1 ×G2 = Ĝ1 × Ĝ2.
(c) Prove using Problem 2 that if G is a finite abelian group, then Ĝ is iso-

morphic to G.

4.∗ When p is prime the group Z∗(p) is cyclic, and Z∗(p) ≈ Z(p− 1).

Ibookroot October 20, 2007

8 Dirichlet’s Theorem

Dirichlet, Gustav Lejeune (Düren 1805-Göttingen 1859),
German mathematician. He was a number theorist at
heart. But, while studying in Paris, being a very like-
able person, he was befriended by Fourier and other
like-minded mathematicians, and he learned analysis
from them. Thus equipped, he was able to lay the
foundation for the application of Fourier analysis to
(analytic) theory of numbers.

S. Bochner, 1966

As a striking application of the theory of finite Fourier series, we now
prove Dirichlet’s theorem on primes in arithmetic progression. This the-
orem states that if q and ` are positive integers with no common factor,
then the progression

`, ` + q, ` + 2q, ` + 3q, . . . , ` + kq, . . .

contains infinitely many prime numbers. This change of subject matter
that we undertake illustrates the wide applicability of ideas from Fourier
analysis to various areas outside its seemingly narrower confines. In this
particular case, it is the theory of Fourier series on the finite abelian
group Z∗(q) that plays a key role in the solution of the problem.

1 A little elementary number theory

We begin by introducing the requisite background. This involves elemen-
tary ideas of divisibility of integers, and in particular properties regarding
prime numbers. Here the basic fact, called the fundamental theorem of
arithmetic, is that every integer is the product of primes in an essentially
unique way.

1.1 The fundamental theorem of arithmetic

The following theorem is a mathematical formulation of long division.

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242 Chapter 8. DIRICHLET’S THEOREM

Theorem 1.1 (Euclid’s algorithm) For any integers a and b with
b > 0, there exist unique integers q and r with 0 ≤ r < b such that a = qb + r. Here q denotes the quotient of a by b, and r is the remainder, which is smaller than b. Proof. First we prove the existence of q and r. Let S denote the set of all non-negative integers of the form a− qb with q ∈ Z. This set is non-empty and in fact S contains arbitrarily large positive integers since b 6= 0. Let r denote the smallest element in S, so that r = a− qb for some integer q. By construction 0 ≤ r, and we claim that r < b. If not, we may write r = b + s with 0 ≤ s < r, so b + s = a− qb, which then implies s = a− (q + 1)b. Hence s ∈ S with s < r, and this contradicts the minimality of r. So r < b, hence q and r satisfy the conditions of the theorem. To prove uniqueness, suppose we also had a = q1b + r1 where 0 ≤ r1 < b. By subtraction we find (q − q1)b = r1 − r. The left-hand side has absolute value 0 or ≥ b, while the right-hand side has absolute value < b. Hence both sides of the equation must be 0, which gives q = q1 and r = r1. An integer a divides b if there exists another integer c such that ac = b; we then write a|b and say that a is a divisor of b. Note that in particular 1 divides every integer, and a|a for all integers a. A prime number is a positive integer greater than 1 that has no positive divisors besides 1 and itself. The main theorem in this section says that any positive integer can be written uniquely as the product of prime numbers. The greatest common divisor of two positive integers a and b is the largest integer that divides both a and b. We usually denote the greatest common divisor by gcd(a, b). Two positive integers are relatively prime if their greatest common divisor is 1. In other words, 1 is the only positive divisor common to both a and b. Ibookroot October 20, 2007 1. A little elementary number theory 243 Theorem 1.2 If gcd(a, b) = d, then there exist integers x and y such that ax + by = d. Proof. Consider the set S of all positive integers of the form ax + by where x, y ∈ Z, and let s be the smallest element in S. We claim that s = d. By construction, there exist integers x and y such that ax + by = s. Clearly, any divisor of a and b divides s, so we must have d ≤ s. The proof will be complete if we can show that s|a and s|b. By Euclid’s algorithm, we can write a = qs + r with 0 ≤ r < s. Multiplying the above by q we find qax + qby = qs, and therefore qax + qby = a− r. Hence r = a(1− qx) + b(−qy). Since s was minimal in S and 0 ≤ r < s, we conclude that r = 0, therefore s divides a. A similar argument shows that s divides b, hence s = d as desired. In particular we record the following three consequences of the theo- rem. Corollary 1.3 Two positive integers a and b are relatively prime if and only if there exist integers x and y such that ax + by = 1. Proof. If a and b are relatively prime, two integers x and y with the desired property exist by Theorem 1.2. Conversely, if ax + by = 1 holds and d is positive and divides both a and b, then d divides 1, hence d = 1. Corollary 1.4 If a and c are relatively prime and c divides ab, then c divides b. In particular, if p is a prime that does not divide a and p divides ab, then p divides b. Proof. We can write 1 = ax + cy, so multiplying by b we find b = abx + cby. Hence c|b. Corollary 1.5 If p is prime and p divides the product a1 · · · ar, then p divides ai for some i. Proof. By the previous corollary, if p does not divide a1, then p divides a2 · · · ar, so eventually p|ai. We can now prove the main result of this section. Ibookroot October 20, 2007 244 Chapter 8. DIRICHLET’S THEOREM Theorem 1.6 Every positive integer greater than 1 can be factored uniquely into a product of primes. Proof. First, we show that such a factorization is possible. We do so by proving that the set S of positive integers > 1 which do not
have a factorization into primes is empty. Arguing by contradiction, we
assume that S 6= ∅. Let n be the smallest element of S. Since n cannot
be a prime, there exist integers a > 1 and b > 1 such that ab = n. But
then a < n and b < n, so a /∈ S as well as b /∈ S. Hence both a and b have prime factorizations and so does their product n. This implies n /∈ S, therefore S is empty, as desired. We now turn our attention to the uniqueness of the factorization. Sup- pose that n has two factorizations into primes n = p1p2 · · · pr = q1q2 · · · qs. So p1 divides q1q2 · · · qs, and we can apply Corollary 1.5 to conclude that p1|qi for some i. Since qi is prime, we must have p1 = qi. Continuing with this argument we find that the two factorizations of n are equal up to a permutation of the factors. We briefly digress to give an alternate definition of the group Z∗(q) which appeared in the previous chapter. According to our initial defini- tion, Z∗(q) is the multiplicative group of units in Z(q): those n ∈ Z(q) for which there exists an integer m so that (1) nm ≡ 1 mod q. Equivalently, Z∗(q) is the group under multiplication of all integers in Z(q) that are relatively prime to q. Indeed, notice that if (1) is satisfied, then automatically n and q are relatively prime. Conversely, suppose we assume that n and q are relatively prime. Then, if we put a = n and b = q in Corollary 1.3, we find nx + qy = 1. Hence nx ≡ 1 mod q, and we can take m = x to establish the equiva- lence. 1.2 The infinitude of primes The study of prime numbers has always been a central topic in arithmetic, and the first fundamental problem that arose was to determine whether Ibookroot October 20, 2007 1. A little elementary number theory 245 there are infinitely many primes or not. This problem was solved in Euclid’s Elements with a simple and very elegant argument. Theorem 1.7 There are infinitely many primes. Proof. Suppose not, and denote by p1, . . . , pn the complete set of primes. Define N = p1p2 · · · pn + 1. Since N is larger than any pi, the integer N cannot be prime. Therefore, N is divisible by a prime that belongs to our list. But this is also an absurdity since every prime divides the product, yet no prime divides 1. This contradiction concludes the proof. Euclid’s argument actually can be modified to deduce finer results about the infinitude of primes. To see this, consider the following prob- lem. Prime numbers (except for 2) can be divided into two classes de- pending on whether they are of the form 4k + 1 or 4k + 3, and the above theorem says that at least one of these classes has to be infinite. A natu- ral question is to ask whether both classes are infinite, and if not, which one is? In the case of primes of the form 4k + 3, the fact that the class is infinite has a proof that is similar to Euclid’s, but with a twist. If there are only finitely many such primes, enumerate them in increasing order omitting 3, p1 = 7, p2 = 11, . . . , pn, and let N = 4p1p2 · · · pn + 3. Clearly, N is of the form 4k + 3 and cannot be prime since N > pn.
Since the product of two numbers of the form 4m + 1 is again of the
form 4m + 1, one of the prime divisors of N , say p, must be of the form
4k + 3. We must have p 6= 3, since 3 does not divide the product in the
definition of N . Also, p cannot be one of the other primes of the form
4k + 3, that is, p 6= pi for i = 1, . . . n, because then p divides the product
p1 · · · pn but does not divide 3.

It remains to determine if the class of primes of the form 4k + 1 is
infinite. A simple-minded modification of the above argument does not
work since the product of two numbers of the form 4m + 3 is never of the
form 4m + 3. More generally, in an attempt to prove the law of quadratic
reciprocity, Legendre formulated the following statement:

Ibookroot October 20, 2007

246 Chapter 8. DIRICHLET’S THEOREM

If q and ` are relatively prime, then the sequence

` + kq, k ∈ Z

contains infinitely many primes (hence at least one prime!).

Of course, the condition that q and ` be relatively prime is necessary,
for otherwise ` + kq is never prime. In other words, this hypothesis says
that any arithmetic progression that could contain primes necessarily
contains infinitely many of them.

Legendre’s assertion was proved by Dirichlet. The key idea in his proof
is Euler’s analytical approach to prime numbers involving his product
formula, which gives a strengthened version of Theorem 1.7. This insight
of Euler led to a deep connection between the theory of primes and
analysis.

The zeta function and its Euler product

We begin with a rapid review of infinite products. If {An}∞n=1 is a se-
quence of real numbers, we define

∞∏

n=1

An = lim
N→∞

N∏

n=1

if the limit exists, in which case we say that the product converges. The
natural approach is to take logarithms and transform products into sums.
We gather in a lemma the properties we shall need of the function log x,
defined for positive real numbers.

Lemma 1.8 The exponential and logarithm functions satisfy the follow-
ing properties:

(i) elog x = x.

(ii) log(1 + x) = x + E(x) where |E(x)| ≤ x2 if |x| < 1/2. (iii) If log(1 + x) = y and |x| < 1/2, then |y| ≤ 2|x|. In terms of the O notation, property (ii) will be recorded as log(1 + x) = x + O(x2). Proof. Property (i) is standard. To prove property (ii) we use the power series expansion of log(1 + x) for |x| < 1, that is, (2) log(1 + x) = ∞∑ n=1 (−1)n+1 n xn. Ibookroot October 20, 2007 1. A little elementary number theory 247 Then we have E(x) = log(1 + x)− x = −x 2 2 + x3 3 − x 4 4 + · · · , and the triangle inequality implies |E(x)| ≤ x 2 2 ( 1 + |x|+ |x|2 + · · · ) . Therefore, if |x| ≤ 1/2 we can sum the geometric series on the right-hand side to find that |E(x)| ≤ x 2 2 ( 1 + 1 2 + 1 22 + · · · ) ≤ x 2 2 ( 1 1− 1/2 ) ≤ x2. The proof of property (iii) is now immediate; if x 6= 0 and |x| ≤ 1/2, then ∣∣∣∣ log(1 + x) x ∣∣∣∣ ≤ 1 + ∣∣∣∣ E(x) x ∣∣∣∣ ≤ 1 + |x| ≤ 2, and if x = 0, (iii) is clearly also true. We can now prove the main result on infinite products of real numbers. Proposition 1.9 If An = 1 + an and ∑ |an| converges, then the prod- uct ∏ n An converges, and this product vanishes if and only if one of its factors An vanishes. Also, if an 6= 1 for all n, then ∏ n 1/(1− an) converges. Proof. If ∑ |an| converges, then for all large n we must have |an| < 1/2. Disregarding finitely many terms if necessary, we may assume that this inequality holds for all n. Then we may write the partial products as follows: N∏ n=1 An = N∏ n=1 elog(1+an) = eBN , where BN = ∑N n=1 bn with bn = log(1 + an). By the lemma, we know that |bn| ≤ 2|an|, so that BN converges to a real number, say B. Since Ibookroot October 20, 2007 248 Chapter 8. DIRICHLET’S THEOREM the exponential function is continuous, we conclude that eBN converges to eB as N goes to infinity, proving the first assertion of the proposition. Observe also that if 1 + an 6= 0 for all n, the product converges to a non-zero limit since it is expressed as eB. Finally observe that the partial products of ∏ n 1/(1− an) are 1/ ∏N n=1(1− an), so the same argument as above proves that the product in the denominator converges to a non-zero limit. With these preliminaries behind us, we can now return to the heart of the matter. For s a real number (strictly) greater than 1, we define the zeta function by ζ(s) = ∞∑ n=1 1 ns . To see that the series defining ζ converges, we use the principle that whenever f is a decreasing function one can compare ∑ f(n) with∫ f(x) dx, as is suggested by Figure 1. Note also that a similar tech- nique was used in Chapter 3, that time bounding a sum from below by an integral. y = f(x) n− 1 n f(n) Figure 1. Comparing sums with integrals Here we take f(x) = 1/xs to see that ∞∑ n=1 1 ns ≤ 1 + ∞∑ n=2 ∫ n n−1 dx xs = 1 + ∫ ∞ 1 dx xs , Ibookroot October 20, 2007 1. A little elementary number theory 249 and therefore, (3) ζ(s) ≤ 1 + 1 s− 1 . Clearly, the series defining ζ converges uniformly on each half-line s > s0 > 1, hence ζ is continuous when s > 1. The zeta function was
already mentioned earlier in the discussion of the Poisson summation
formula and the theta function.

The key result is Euler’s product formula.

Theorem 1.10 For every s > 1, we have

ζ(s) =
∏
p

1
1− 1/ps ,

where the product is taken over all primes.

It is important to remark that this identity is an analytic expression
of the fundamental theorem of arithmetic. In fact, each factor of the
product 1/(1− p−s) can be written as a convergent geometric series

1 +
1
ps

+
1

p2s
+ · · ·+ 1

pMs
+ · · · .

So we consider

∏
pj

(
1 +

1
psj

+
1

p2sj
+ · · ·+ 1

pMsj
+ · · ·

)
,

where the product is taken over all primes, which we order in increasing
order p1 < p2 < · · · . Proceeding formally (these manipulations will be justified below), we calculate the product as a sum of terms, each term originating by picking out a term 1/pksj (in the sum corresponding to pj) with a k, which of course will depend on j, and with k = 0 for j sufficiently large. The product obtained this way is 1 (pk11 p k2 2 · · · pkmm )s = 1 ns , where the integer n is written as a product of primes n = pk11 p k2 2 · · · pkmm . By the fundamental theorem of arithmetic, each integer ≥ 1 occurs in this way uniquely, hence the product equals ∞∑ n=1 1 ns . Ibookroot October 20, 2007 250 Chapter 8. DIRICHLET’S THEOREM We now justify this heuristic argument. Proof. Suppose M and N are positive integers with M > N . Observe
now that any positive integer n ≤ N can be written uniquely as a product
of primes, and that each prime must be less than or equal to N and
repeated less than M times. Therefore

N∑

n=1

1
ns

≤
∏

p≤N

(
1 +

1
ps

+
1

p2s
+ · · ·+ 1

pMs

)

≤
∏

p≤N

(
1

1− p−s
)

≤
∏
p

(
1

1− p−s
)

Letting N tend to infinity now yields

∞∑

n=1

1
ns

≤
∏
p

(
1

1− p−s
)

For the reverse inequality, we argue as follows. Again, by the fundamen-
tal theorem of arithmetic, we find that

∏

p≤N

(
1 +

1
ps

+
1

p2s
+ · · ·+ 1

pMs

)
≤

∞∑

n=1

1
ns

Letting M tend to infinity gives

∏

p≤N

(
1

1− p−s
)
≤

∞∑

n=1

1
ns

Hence

∏
p

(
1

1− p−s
)
≤

∞∑

n=1

1
ns

and the proof of the product formula is complete.

We now come to Euler’s version of Theorem 1.7, which inspired Dirich-
let’s approach to the general problem of primes in arithmetic progression.
The point is the following proposition.

Ibookroot October 20, 2007

1. A little elementary number theory 251

Proposition 1.11 The series

∑
p

1/p

diverges, when the sum is taken over all primes p.

Of course, if there were only finitely many primes the series would
converge automatically.

Proof. We take logarithms of both sides of the Euler formula. Since
log x is continuous, we may write the logarithm of the infinite product
as the sum of the logarithms. Therefore, we obtain for s > 1

−
∑

log(1− 1/ps) = log ζ(s).

Since log(1 + x) = x + O(|x|2) whenever |x| ≤ 1/2, we get

−
∑

[
−1/ps + O(1/p2s)

]
= log ζ(s),

which gives
∑

1/ps + O(1) = log ζ(s).

The term O(1) appears because
∑

p 1/p
2s ≤ ∑∞n=1 1/n2. Now we let s

tend to 1 from above, namely s → 1+, and note that ζ(s) →∞ since∑∞
n=1 1/n

s ≥ ∑Mn=1 1/ns, and therefore

lim inf
s→1+

∞∑

n=1

1/ns ≥
M∑

n=1

1/n for every M.

We conclude that
∑

p 1/p
s →∞ as s → 1+, and since 1/p > 1/ps for all

s > 1, we finally have that

∑
p

1/p = ∞.

In the rest of this chapter we see how Dirichlet adapted Euler’s insight.

Ibookroot October 20, 2007

252 Chapter 8. DIRICHLET’S THEOREM

2 Dirichlet’s theorem

We remind the reader of our goal:

Theorem 2.1 If q and ` are relatively prime positive integers, then there
are infinitely many primes of the form ` + kq with k ∈ Z.

Following Euler’s argument, Dirichlet proved this theorem by showing
that the series

∑

p≡` mod q

1
p

diverges, where the sum is over all primes congruent to ` modulo q. Once
q is fixed and no confusion is possible, we write p ≡ ` to denote a prime
congruent to ` modulo q. The proof consists of several steps, one of
which requires Fourier analysis on the group Z∗(q). Before proceeding
with the theorem in its complete generality, we outline the solution to
the particular problem raised earlier: are there infinitely many primes of
the form 4k + 1? This example, which consists of the special case q = 4
and ` = 1, illustrates all the important steps in the proof of Dirichlet’s
theorem.

We begin with the character on Z∗(4) defined by χ(1) = 1 and
χ(3) = −1. We extend this character to all of Z as follows:

χ(n) =





0 if n is even,
1 if n = 4k + 1,

−1 if n = 4k + 3.

Note that this function is multiplicative, that is, χ(nm) = χ(n)χ(m) on
all of Z. Let L(s, χ) =

∑∞
n=1 χ(n)/n

s, so that

L(s, χ) = 1− 1
3s

+
1
5s
− 1

7s
+ · · · .

Then L(1, χ) is the convergent series given by

1− 1
3

+
1
5
− 1

7
+ · · · .

Since the terms in the series are alternating and their absolute values
decrease to zero we have L(1, χ) 6= 0. Because χ is multiplicative, the
Euler product generalizes (as we will prove later) to give

∞∑

n=1

χ(n)
ns

=
∏
p

1
1− χ(p)/ps .

Ibookroot October 20, 2007

2. Dirichlet’s theorem 253

Taking the logarithm of both sides, we find that

log L(s, χ) =
∑

χ(p)
ps

+ O(1).

Letting s → 1+, the observation that L(1, χ) 6= 0 shows that ∑p χ(p)/ps
remains bounded. Hence

∑

p≡1

1
ps
−

∑

p≡3

1
ps

is bounded as s → 1+. However, we know from Proposition 1.11 that
∑

1
ps

is unbounded as s → 1+, so putting these two facts together, we find
that

2
∑

p≡1

1
ps

is unbounded as s → 1+. Hence ∑p≡1 1/p diverges, and as a consequence
there are infinitely many primes of the form 4k + 1.

We digress briefly to show that in fact L(1, χ) = π/4. To see this, we
integrate the identity

1
1 + x2

= 1− x2 + x4 − x6 + · · · ,

and get
∫ y
0

1 + x2
= y − y

3
+

5
− · · · , 0 < y < 1. We then let y tend to 1. The integral can be calculated as ∫ 1 0 dx 1 + x2 = arctanu|10 = π 4 , so this proves that the series 1− 1/3 + 1/5− · · · is Abel summable to π/4. Since we know the series converges, its limit is the same as its Abel limit, hence 1− 1/3 + 1/5− · · · = π/4. The rest of this chapter gives the full proof of Dirichlet’s theorem. We begin with the Fourier analysis (which is actually the last step in the example given above), and reduce the theorem to the non-vanishing of L-functions. Ibookroot October 20, 2007 254 Chapter 8. DIRICHLET’S THEOREM 2.1 Fourier analysis, Dirichlet characters, and reduction of the theorem In what follows we take the abelian group G to be Z∗(q). Our formulas below involve the order of G, which is the number of integers 0 ≤ n < q that are relatively prime to q; this number defines the Euler phi- function ϕ(q), and |G| = ϕ(q). Consider the function δ` on G, which we think of as the characteristic function of `; if n ∈ Z∗(q), then δ`(n) = { 1 if n ≡ ` mod q, 0 otherwise. We can expand this function in a Fourier series as follows: δ`(n) = ∑ e∈Ĝ δ̂`(e)e(n), where δ̂`(e) = 1 |G| ∑ m∈G δ`(m)e(m) = 1 |G| e(`). Hence δ`(n) = 1 |G| ∑ e∈Ĝ e(`)e(n). We can extend the function δ` to all of Z by setting δ`(m) = 0 whenever m and q are not relatively prime. Similarly, the extensions of the characters e ∈ Ĝ to all of Z which are given by the recipe χ(m) = { e(m) if m and q are relatively prime 0 otherwise, are called the Dirichlet characters modulo q. We shall denote the extension to Z of the trivial character of G by χ0, so that χ0(m) = 1 if m and q are relatively prime, and 0 otherwise. Note that the Dirichlet characters modulo q are multiplicative on all of Z, in the sense that χ(nm) = χ(n)χ(m) for all n, m ∈ Z. Since the integer q is fixed, we may without fear of confusion, speak of “Dirichlet characters” omitting reference to q.1 With |G| = ϕ(q), we may restate the above results as follows: 1We use the notation χ instead of e to distinguish the Dirichlet characters (defined on Z) from the characters e (defined on Z∗(q)). Ibookroot October 20, 2007 2. Dirichlet’s theorem 255 Lemma 2.2 The Dirichlet characters are multiplicative. Moreover, δ`(m) = 1 ϕ(q) ∑ χ χ(`)χ(m), where the sum is over all Dirichlet characters. With the above lemma we have taken our first step towards a proof of the theorem, since this lemma shows that ∑ p≡` 1 ps = ∑ p δ`(p) ps = 1 ϕ(q) ∑ χ χ(`) ∑ p χ(p) ps . Thus it suffices to understand the behavior of ∑ p χ(p)p −s as s → 1+. In fact, we divide the above sum in two parts depending on whether or not χ is trivial. So we have ∑ p≡` 1 ps = 1 ϕ(q) ∑ p χ0(p) ps + 1 ϕ(q) ∑ χ 6=χ0 χ(`) ∑ p χ(p) ps = 1 ϕ(q) ∑ p not dividing q 1 ps + 1 ϕ(q) ∑ χ6=χ0 χ(`) ∑ p χ(p) ps .(4) Since there are only finitely many primes dividing q, Euler’s theorem (Proposition 1.11) implies that the first sum on the right-hand side di- verges when s tends to 1. These observations show that Dirichlet’s the- orem is a consequence of the following assertion. Theorem 2.3 If χ is a nontrivial Dirichlet character, then the sum ∑ p χ(p) ps remains bounded as s → 1+. The proof of Theorem 2.3 requires the introduction of the L-functions, to which we now turn. 2.2 Dirichlet L-functions We proved earlier that the zeta function ζ(s) = ∑ n 1/n s could be ex- pressed as a product, namely ∞∑ n=1 1 ns = ∏ p 1 (1− p−s) . Ibookroot October 20, 2007 256 Chapter 8. DIRICHLET’S THEOREM Dirichlet observed an analogue of this formula for the so-called L-functions defined for s > 1 by

L(s, χ) =
∞∑

n=1

χ(n)
ns

where χ is a Dirichlet character.

Theorem 2.4 If s > 1, then

∞∑

n=1

χ(n)
ns

=
∏
p

1
(1− χ(p)p−s) ,

where the product is over all primes.

Assuming this theorem for now, we can follow Euler’s argument for-
mally: taking the logarithm of the product and using the fact that
log(1 + x) = x + O(x2) whenever x is small, we would get

log L(s, χ) = −
∑

log(1− χ(p)/ps)

= −
∑

[
−χ(p)

ps
+ O

(
1

p2s

)]

=
∑

χ(p)
ps

+ O(1).

If L(1, χ) is finite and non-zero, then log L(s, χ) is bounded as s → 1+,
and we can conclude that the sum

∑
p

χ(p)
ps

is bounded as s → 1+. We now make several observations about the
above formal argument.

First, we must prove the product formula in Theorem 2.4. Since the
Dirichlet characters χ can be complex-valued we will extend the loga-
rithm to complex numbers w of the form w = 1/(1− z) with |z| < 1. (This will be done in terms of a power series.) Then we show that with this definition of the logarithm, the proof of Euler’s product formula given earlier carries over to L-functions. Second, we must make sense of taking the logarithm of both sides of the product formula. If the Dirichlet characters are real, this argument works Ibookroot October 20, 2007 2. Dirichlet’s theorem 257 and is precisely the one given in the example corresponding to primes of the form 4k + 1. In general, the difficulty lies in the fact that χ(p) is a complex number, and the complex logarithm is not single valued; in particular, the logarithm of a product is not the sum of the logarithms. Third, it remains to prove that whenever χ 6= χ0, then log L(s, χ) is bounded as s → 1+. If (as we shall see) L(s, χ) is continuous at s = 1, then it suffices to show that L(1, χ) 6= 0. This is the non-vanishing we mentioned earlier, which corresponds to the alternating series being non-zero in the previous example. The fact that L(1, χ) 6= 0 is the most difficult part of the argument. So we will focus on three points: 1. Complex logarithms and infinite products. 2. Study of L(s, χ). 3. Proof that L(1, χ) 6= 0 if χ is non-trivial. However, before we enter further into the details, we pause briefly to discuss some historical facts surrounding Dirichlet’s theorem. Historical digression In the following list, we have gathered the names of those mathematicians whose work dealt most closely with the series of achievements related to Dirichlet’s theorem. To give a better perspective, we attach the years in which they reached the age of 35: Euler 1742 Legendre 1787 Gauss 1812 Dirichlet 1840 Riemann 1861 As we mentioned earlier, Euler’s discovery of the product formula for the zeta function is the starting point in Dirichlet’s argument. Legendre in effect conjectured the theorem because he needed it in his proof of the law of quadratic reciprocity. However, this goal was first accomplished by Gauss who, while not knowing how to establish the theorem about primes in arithmetic progression, nevertheless found a number of different proofs of quadratic reciprocity. Later, Riemann extended the study of the zeta function to the complex plane and indicated how properties Ibookroot October 20, 2007 258 Chapter 8. DIRICHLET’S THEOREM related to the non-vanishing of that function were central in the further understanding of the distribution of prime numbers. Dirichlet proved his theorem in 1837. It should be noted that Fourier, who had befriended Dirichlet when the latter was a young mathematician visiting Paris, had died several years before. Besides the great activity in mathematics, that period was also a very fertile time in the arts, and in particular music. The era of Beethoven had ended only ten years earlier, and Schumann was now reaching the heights of his creativity. But the musician whose career was closest to Dirichlet was Felix Mendelssohn (four years his junior). It so happens that the latter began composing his famous violin concerto the year after Dirichlet succeeded in proving his theorem. 3 Proof of the theorem We return to the proof of Dirichlet’s theorem and to the three difficulties mentioned above. 3.1 Logarithms The device to deal with the first point is to define two logarithms, one for complex numbers of the form 1/(1− z) with |z| < 1 which we denote by log1, and one for the function L(s, χ) which we will denote by log2. For the first logarithm, we define log1 ( 1 1− z ) = ∞∑ k=1 zk k for |z| < 1. Note that log1 w is then defined if Re(w) > 1/2, and because of equa-
tion (2), log1 w gives an extension of the usual log x when x is a real
number > 1/2.

Proposition 3.1 The logarithm function log1 satisfies the following prop-
erties:

(i) If |z| < 1, then elog1( 1 1−z ) = 1 1− z . (ii) If |z| < 1, then log1 ( 1 1− z ) = z + E1(z), where the error E1 satisfies |E1(z)| ≤ |z|2 if |z| < 1/2. Ibookroot October 20, 2007 3. Proof of the theorem 259 (iii) If |z| < 1/2, then ∣∣∣∣log1 ( 1 1− z )∣∣∣∣ ≤ 2|z|. Proof. To establish the first property, let z = reiθ with 0 ≤ r < 1, and observe that it suffices to show that (5) (1− reiθ) e ∑∞ k=1 (reiθ)k/k = 1. To do so, we differentiate the left-hand side with respect to r, and this gives [ −eiθ + (1− reiθ) ( ∞∑ k=1 (reiθ)k/k )′ ] e ∑∞ k=1 (reiθ)k/k . The term in brackets equals −eiθ + (1− reiθ)eiθ ( ∞∑ k=1 (reiθ)k−1 ) = −eiθ + (1− reiθ)eiθ 1 1− reiθ = 0. Having found that the left-hand side of the equation (5) is constant, we set r = 0 and get the desired result. The proofs of the second and third properties are the same as their real counterparts given in Lemma 1.8. Using these results we can state a sufficient condition guaranteeing the convergence of infinite products of complex numbers. Its proof is the same as in the real case, except that we now use the logarithm log1. Proposition 3.2 If ∑ |an| converges, and an 6= 1 for all n, then ∞∏ n=1 ( 1 1− an ) converges. Moreover, this product is non-zero. Proof. For n large enough, |an| < 1/2, so we may assume without loss of generality that this inequality holds for all n ≥ 1. Then N∏ n=1 ( 1 1− an ) = N∏ n=1 elog1( 1 1−an ) = e ∑ N n=1 log1( 11−an ). Ibookroot October 20, 2007 260 Chapter 8. DIRICHLET’S THEOREM But we know from the previous proposition that ∣∣∣∣log1 ( 1 1− z )∣∣∣∣ ≤ 2|z|, so the fact that the series ∑ |an| converges, immediately implies that the limit lim N→∞ N∑ n=1 log1 ( 1 1− an ) = A exists. Since the exponential function is continuous, we conclude that the product converges to eA, which is clearly non-zero. We may now prove the promised Dirichlet product formula ∑ n χ(n) ns = ∏ p 1 (1− χ(p)p−s) . For simplicity of notation, let L denote the left-hand side of the above equation. Define SN = ∑ n≤N χ(n)n−s and ΠN = ∏ p≤N ( 1 1− χ(p)p−s ) . The infinite product Π = limN→∞ΠN = ∏ p ( 1 1−χ(p)p−s ) converges by the previous proposition. Indeed, if we set an = χ(pn)p−sn , where pn is the nth prime, we note that if s > 1, then

∑ |an| < ∞. Also, define ΠN,M = ∏ p≤N ( 1 + χ(p) ps + · · ·+ χ(p M ) pMs ) . Now fix ² > 0 and choose N so large that

|SN − L| < ² and |ΠN −Π| < ². We can next select M large enough so that |SN −ΠN,M | < ² and |ΠN,M −ΠN | < ². To see the first inequality, one uses the fundamental theorem of arith- metic and the fact that the Dirichlet characters are multiplicative. The Ibookroot October 20, 2007 3. Proof of the theorem 261 second inequality follows merely because each series ∑∞ n=1 χ(pn) pns con- verges. Therefore |L−Π| ≤ |L− SN |+ |SN −ΠN,M |+ |ΠN,M −ΠN |+ |ΠN −Π| < 4², as was to be shown. 3.2 L-functions The next step is a better understanding of the L-functions. Their behav- ior as functions of s (especially near s = 1) depends on whether or not χ is trivial. In the first case, L(s, χ0) is up to some simple factors just the zeta function. Proposition 3.3 Suppose χ0 is the trivial Dirichlet character, χ0(n) = { 1 if n and q are relatively prime, 0 otherwise, and q = pa11 · · · paNN is the prime factorization of q. Then L(s, χ0) = (1− p−s1 )(1− p−s2 ) · · · (1− p−sN )ζ(s). Therefore L(s, χ0) →∞ as s → 1+. Proof. The identity follows at once on comparing the Dirichlet and Euler product formulas. The final statement holds because ζ(s) →∞ as s → 1+. The behavior of the remaining L-functions, those for which χ 6= χ0, is more subtle. A remarkable property is that these functions are now defined and continuous for s > 0. In fact, more is true.

Proposition 3.4 If χ is a non-trivial Dirichlet character, then the series
∞∑

n=1

χ(n)/ns

converges for s > 0, and we denote its sum by L(s, χ). Moreover:

(i) The function L(s, χ) is continuously differentiable for 0 < s < ∞. (ii) There exists constants c, c′ > 0 so that

L(s, χ) = 1 + O(e−cs) as s →∞, and

L′(s, χ) = O(e−c
′s) as s →∞.

Ibookroot October 20, 2007

262 Chapter 8. DIRICHLET’S THEOREM

We first isolate the key cancellation property that non-trivial Dirichlet
characters possess, which accounts for the behavior of the L-function
described in the proposition.

Lemma 3.5 If χ is a non-trivial Dirichlet character, then

∣∣∣∣∣
k∑

n=1

χ(n)

∣∣∣∣∣ ≤ q, for any k.

Proof. First, we recall that

q∑

n=1

χ(n) = 0.

In fact, if S denotes the sum and a ∈ Z∗(q), then the multiplicative
property of the Dirichlet character χ gives

χ(a)S =
∑

χ(a)χ(n) =
∑

χ(an) =
∑

χ(n) = S.

Since χ is non-trivial, χ(a) 6= 1 for some a, hence S = 0. We now write
k = aq + b with 0 ≤ b < q, and note that k∑ n=1 χ(n) = aq∑ n=1 χ(n) + ∑ aq 1 by the series

∞∑

n=1

χ(n)
ns

which converges absolutely and uniformly for s > δ > 1. Moreover, the
differentiated series also converges absolutely and uniformly for s > δ >
1, which shows that L(s, χ) is continuously differentiable for s > 1. We

Ibookroot October 20, 2007

3. Proof of the theorem 263

sum by parts2 to extend this result to s > 0. Indeed, we have

N∑

k=1

χ(k)
ks

=
N∑

k=1

sk − sk−1
ks

=
N−1∑

k=1

[
1
ks
− 1

(k + 1)s

]
+

sN
Ns

=
N−1∑

k=1

fk(s) +
sN
Ns

where fk(s) = sk [k−s − (k + 1)−s]. If g(x) = x−s, then g′(x) = −sx−s−1,
so applying the mean-value theorem between x = k and x = k + 1, and
the fact that |sk| ≤ q, we find that

|fk(s)| ≤ qsk−s−1.

Therefore, the series
∑

fk(s) converges absolutely and uniformly for s >
δ > 0, and this proves that L(s, χ) is continuous for s > 0. To prove that
it is also continuously differentiable, we differentiate the series term by
term, obtaining

∑
(log n)

χ(n)
ns

Again, we rewrite this series using summation by parts as
∑

sk
[
−k−s log k + (k + 1)−s log(k + 1)

]
,

and an application of the mean-value theorem to the function g(x) =
x−s log x shows that the terms are O(k−δ/2−1), thus proving that the
differentiated series converges uniformly for s > δ > 0. Hence L(s, χ) is
continuously differentiable for s > 0.

Now, observe that for all s large,

|L(s, χ)− 1| ≤ 2q
∞∑

n=2

n−s

≤ 2−sO(1),

and we can take c = log 2, to see that L(s, χ) = 1 + O(e−cs) as s →∞.
A similar argument also shows that L′(s, χ) = O(e−c

′s) as s →∞ with
in fact c′ = c, and the proof of the proposition is complete.

2For the formula of summation by parts, see Exercise 7 in Chapter 2.

Ibookroot October 20, 2007

264 Chapter 8. DIRICHLET’S THEOREM

With the facts gathered so far about L(s, χ) we are in a position to
define the logarithm of the L-functions. This is done by integrating its
logarithmic derivative. In other words, if χ is a non-trivial Dirichlet
character and s > 1 we define3

log2 L(s, χ) = −
∫ ∞

L′(t, χ)
L(t, χ)

dt.

We know that L(t, χ) 6= 0 for every t > 1 since it is given by a product
(Proposition 3.2), and the integral is convergent because

L′(t, χ)
L(t, χ)

= O(e−ct),

which follows from the behavior at infinity of L(t, χ) and L′(t, χ) recorded
earlier.

The following links the two logarithms.

Proposition 3.6 If s > 1, then

elog2 L(s,χ) = L(s, χ).

Moreover

log2 L(s, χ) =
∑

log1

(
1

1− χ(p)/ps
)

Proof. Differentiating e− log2 L(s,χ)L(s, χ) with respect to s gives

−L
′(s, χ)

L(s, χ)
e− log2 L(s,χ)L(s, χ) + e− log2 L(s,χ)L′(s, χ) = 0.

So e− log2 L(s,χ)L(s, χ) is constant, and this constant can be seen to be 1
by letting s tend to infinity. This proves the first conclusion.

To prove the equality between the logarithms, we fix s and take the ex-
ponential of both sides. The left-hand side becomes elog2 L(s,χ) = L(s, χ),
and the right-hand side becomes

e
∑

p
log1

(
1

1−χ(p)/ps
)

=
∏
p

e
log1

(
1

1−χ(p)/ps
)

=
∏
p

(
1

1− χ(p)/ps
)

= L(s, χ),

3The notation log2 used in this context should not be confused with the logarithm to
the base 2.

Ibookroot October 20, 2007

3. Proof of the theorem 265

by (i) in Proposition 3.1 and the Dirichlet product formula. Therefore,
for each s there exists an integer M(s) so that

log2 L(s, χ)−
∑

log1

(
1

1− χ(p)/ps
)

= 2πiM(s).

As the reader may verify, the left-hand side is continuous in s, and this
implies the continuity of the function M(s). But M(s) is integer-valued
so we conclude that M(s) is constant, and this constant can be seen to
be 0 by letting s go to infinity.

Putting together the work we have done so far gives rigorous meaning
to the formal argument presented earlier. Indeed, the properties of log1
show that

∑
p

log1

(
1

1− χ(p)/ps
)

=
∑

χ(p)
ps

+ O

(∑
p

1
p2s

)

=
∑

χ(p)
ps

+ O(1).

Now if L(1, χ) 6= 0 for a non-trivial Dirichlet character, then by its in-
tegral representation log2 L(s, χ) remains bounded as s → 1+. Thus
the identity between the logarithms implies that

∑
p χ(p)p

−s remains
bounded as s → 1+, which is the desired result. Therefore, to finish the
proof of Dirichlet’s theorem, we need to see that L(1, χ) 6= 0 when χ is
non-trivial.

3.3 Non-vanishing of the L-function

We now turn to a proof of the following deep result:

Theorem 3.7 If χ 6= χ0, then L(1, χ) 6= 0.

There are several proofs of this fact, some involving algebraic number
theory (among them Dirichlet’s original argument), and others involving
complex analysis. Here we opt for a more elementary argument that
requires no special knowledge of either of these areas. The proof splits
in two cases, depending on whether χ is complex or real. A Dirich-
let character is said to be real if it takes on only real values (that is,
+1, −1, or 0) and complex otherwise. In other words, χ is real if and
only if χ(n) = χ(n) for all integers n.

Ibookroot October 20, 2007

266 Chapter 8. DIRICHLET’S THEOREM

Case I: complex Dirichlet characters

This is the easier of the two cases. The proof is by contradiction, and we
use two lemmas.

Lemma 3.8 If s > 1, then
∏
χ

L(s, χ) ≥ 1,

where the product is taken over all Dirichlet characters. In particular the
product is real-valued.

Proof. We have shown earlier that for s > 1

L(s, χ) = exp

(∑
p

log1

(
1

1− χ(p)p−s
))

Hence,

∏
χ

L(s, χ) = exp

(∑
χ

∑
p

log1

(
1

1− χ(p)p−s
))

= exp

(∑
χ

∑
p

∞∑

k=1

1
k

χ(pk)
pks

)

= exp

(∑
p

∞∑

k=1

∑
χ

1
k

χ(pk)
pks

)
.

Because of Lemma 2.2 (with ` = 1) we have
∑

χ χ(p
k) = ϕ(q)δ1(pk), and

hence

∏
χ

L(s, χ) = exp

(
ϕ(q)

∑
p

∞∑

k=1

1
k

δ1(pk)
pks

)
≥ 1,

since the term in the exponential is non-negative.

Lemma 3.9 The following three properties hold:

(i) If L(1, χ) = 0, then L(1, χ) = 0.

(ii) If χ is non-trivial and L(1, χ) = 0, then

|L(s, χ)| ≤ C|s− 1| when 1 ≤ s ≤ 2.

Ibookroot October 20, 2007

3. Proof of the theorem 267

(iii) For the trivial Dirichlet character χ0, we have

|L(s, χ0)| ≤
C

|s− 1| when 1 < s ≤ 2. Proof. The first statement is immediate because L(1, χ) = L(1, χ). The second statement follows from the mean-value theorem since L(s, χ) is continuously differentiable for s > 0 when χ is non-trivial. Finally, the
last statement follows because by Proposition 3.3

L(s, χ0) = (1− p−s1 )(1− p−s2 ) · · · (1− p−sN )ζ(s),

and ζ satisfies the similar estimate (3).

We can now conclude the proof that L(1, χ) 6= 0 for χ a non-trivial
complex Dirichlet character. If not, say L(1, χ) = 0, then we also have
L(1, χ) = 0. Since χ 6= χ, there are at least two terms in the product

∏
χ

L(s, χ),

that vanish like |s− 1| as s → 1+. Since only the trivial character con-
tributes a term that grows, and this growth is no worse than O(1/|s− 1|),
we find that the product goes to 0 as s → 1+, contradicting the fact that
it is ≥ 1 by Lemma 3.8.

Case II: real Dirichlet characters

The proof that L(1, χ) 6= 0 when χ is a non-trivial real Dirichlet character
is very different from the earlier complex case. The method we shall
exploit involves summation along hyperbolas. It is a curious fact that
this method was introduced by Dirichlet himself, twelve years after the
proof of his theorem on arithmetic progressions, to establish another
famous result of his: the average order of the divisor function. However,
he made no connection between the proofs of these two theorems. We will
instead proceed by proving first Dirichlet’s divisor theorem, as a simple
example of the method of summation along hyperbolas. Then, we shall
adapt these ideas to prove the fact that L(1, χ) 6= 0. As a preliminary
matter, we need to deal with some simple sums, and their corresponding
integral analogues.

Sums vs. Integrals

Here we use the idea of comparing a sum with its corresponding integral,
which already occurred in the estimate (3) for the zeta function.

Ibookroot October 20, 2007

268 Chapter 8. DIRICHLET’S THEOREM

Proposition 3.10 If N is a positive integer, then:

(i)
∑

1≤n≤N

1
n

=
∫ N
1

x
+ O(1) = log N + O(1).

(ii) More precisely, there exists a real number γ, called Euler’s constant,
so that

∑

1≤n≤N

1
n

= log N + γ + O(1/N).

Proof. It suffices to establish the more refined estimate given in
part (ii). Let

γn =
1
n
−

∫ n+1
n

x
.

Since 1/x is decreasing, we clearly have

0 ≤ γn ≤
1
n
− 1

n + 1
≤ 1

n2
,

so the series
∑∞

n=1 γn converges to a limit which we denote by γ. More-
over, if we estimate

∑
f(n) by

∫
f(x) dx, where f(x) = 1/x2, we find

∞∑

n=N+1

γn ≤
∞∑

n=N+1

1
n2

≤
∫ ∞

x2
= O(1/N).

Therefore

N∑

n=1

1
n
−

∫ N
1

x
= γ −

∞∑

n=N+1

γn +
∫ N+1

x
,

and this last integral is O(1/N) as N →∞.

Proposition 3.11 If N is a positive integer, then

∑

1≤n≤N

1
n1/2

=
∫ N
1

x1/2
+ c′ + O(1/N1/2)

= 2N1/2 + c + O(1/N1/2).

The proof is essentially a repetition of the proof of the previous proposi-
tion, this time using the fact that

∣∣∣∣
1

n1/2
− 1

(n + 1)1/2

∣∣∣∣ ≤
C

n3/2
.

Ibookroot October 20, 2007

3. Proof of the theorem 269

This last inequality follows from the mean-value theorem applied to
f(x) = x−1/2, between x = n and x = n + 1.

Hyperbolic sums

If F is a function defined on pairs of positive integers, there are three
ways to calculate

SN =
∑ ∑

F (m,n),

where the sum is taken over all pairs of positive integers (m,n) which
satisfy mn ≤ N .

We may carry out the summation in any one of the following three
ways. (See Figure 2.)

(a) Along hyperbolas:

SN =
∑

1≤k≤N

( ∑

nm=k

F (m,n)

)

(b) Vertically:

SN =
∑

1≤m≤N


 ∑

1≤n≤N/m
F (m,n)




SN =
∑

1≤n≤N


 ∑

1≤m≤N/n
F (m,n)




It is a remarkable fact that one can obtain interesting conclusions from
the obvious fact that these three methods of summation give the same
sum. We apply this idea first in the study of the divisor problem.

Intermezzo: the divisor problem

For a positive integer k, let d(k) denote the number of positive divisors
of k. For example,

k 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
d(k) 1 2 2 3 2 4 2 4 3 4 2 6 2 4 4 5 2

Ibookroot October 20, 2007

270 Chapter 8. DIRICHLET’S THEOREM

(c)(a) (b)

Figure 2. The three methods of summation

One observes that the behavior of d(k) as k tends to infinity is rather
irregular, and in fact, it does not seem possible to approximate d(k) by
a simple analytic expression in k. However, it is natural to inquire about
the average size of d(k). In other words, one might ask, what is the
behavior of

1
N

N∑

k=1

d(k) as N →∞?

The answer was provided by Dirichlet, who made use of hyperbolic sums.
Indeed, we observe that

d(k) =
∑

nm=k, 1≤n,m
1.

Theorem 3.12 If k is a positive integer, then

1
N

N∑

k=1

d(k) = log N + O(1).

More precisely,

1
N

N∑

k=1

d(k) = log N + (2γ − 1) + O(1/N1/2),

where γ is Euler’s constant.

Proof. Let SN =
∑N

k=1 d(k). We observed that summing F = 1 along
hyperbolas gives SN . Summing vertically, we find

SN =
∑

1≤m≤N

∑

1≤n≤N/m
1.

Ibookroot October 20, 2007

3. Proof of the theorem 271

But
∑

1≤n≤N/m 1 = [N/m] = N/m + O(1), where [x] denote the greatest
integer ≤ x. Therefore

SN =
∑

1≤m≤N
(N/m + O(1)) = N


 ∑

1≤m≤N
1/m


 + O(N).

Hence, by part (i) of Proposition 3.10,

SN
N

= log N + O(1)

which gives the first conclusion.
For the more refined estimate we proceed as follows. Consider the

three regions I, II, and III shown in Figure 3. These are defined by

I = {1 ≤ m < N1/2, N1/2 < n ≤ N/m}, II = {1 ≤ m ≤ N1/2, 1 ≤ n ≤ N1/2}, III = {N1/2 < m ≤ N/n, 1 ≤ n < N1/2}. N III N1/2 nm = NI N m n II N1/2 Figure 3. The three regions I, II, and III If SI , SII , and SIII denote the sums taken over the regions I, II, and III, respectively, then SN = SI + SII + SIII = 2(SI + SII)− SII , Ibookroot October 20, 2007 272 Chapter 8. DIRICHLET’S THEOREM since by symmetry SI = SIII . Now we sum vertically, and use (ii) of Proposition 3.10 to obtain SI + SII = ∑ 1≤m≤N1/2   ∑ 1≤n≤N/m 1   = ∑ 1≤m≤N1/2 [N/m] = ∑ 1≤m≤N1/2 (N/m + O(1)) = N   ∑ 1≤m≤N1/2 1/m   + O(N1/2) = N log N1/2 + Nγ + O(N1/2). Finally, SII corresponds to a square so SII = ∑ 1≤m≤N1/2 ∑ 1≤n≤N1/2 1 = [N1/2]2 = N + O(N1/2). Putting these estimates together and dividing by N yields the more re- fined statement in the theorem. Non-vanishing of the L-function Our essential application of summation along hyperbolas is to the main point of this section, namely that L(1, χ) 6= 0 for a non-trivial real Dirich- let character χ. Given such a character, let F (m,n) = χ(n) (nm)1/2 , and define SN = ∑ ∑ F (m,n), where the sum is over all integers m,n ≥ 1 that satisfy mn ≤ N . Proposition 3.13 The following statements are true: (i) SN ≥ c log N for some constant c > 0.
(ii) SN = 2N1/2L(1, χ) + O(1).

Ibookroot October 20, 2007

3. Proof of the theorem 273

It suffices to prove the proposition, since the assumption L(1, χ) = 0
would give an immediate contradiction.

We first sum along hyperbolas. Observe that

∑

nm=k

χ(n)
(nm)1/2

=
1

k1/2

∑

n|k
χ(n).

For conclusion (i) it will be enough to show the following lemma.

Lemma 3.14
∑

n|k
χ(n) ≥

{
0 for all k
1 if k = `2 for some ` ∈ Z.

From the lemma, we then get

SN ≥
∑

k=`2, `≤N1/2

1
k1/2

≥ c log N,

where the last inequality follows from (i) in Proposition 3.10.

The proof of the lemma is simple. If k is a power of a prime, say
k = pa, then the divisors of k are 1, p, p2, . . . , pa and

∑

n|k
χ(n) = χ(1) + χ(p) + χ(p2) + · · ·+ χ(pa)

= 1 + χ(p) + χ(p)2 + · · ·+ χ(p)a.

So this sum is equal to




a + 1 if χ(p) = 1,
1 if χ(p) = −1 and a is even,
0 if χ(p) = −1 and a is odd,
1 if χ(p) = 0, that is p|q.

In general, if k = pa11 · · · paNN , then any divisor of k is of the form pb11 · · · pbNN
where 0 ≤ bj ≤ aj for all j. Therefore, the multiplicative property of χ
gives

∑

n|k
χ(n) =

N∏

j=1

(
χ(1) + χ(pj) + χ(p

2
j) + · · ·+ χ(p

aj
j )

)
,

and the proof is complete.

To prove the second statement in the proposition, we write

SN = SI + (SII + SIII),

Ibookroot October 20, 2007

274 Chapter 8. DIRICHLET’S THEOREM

where the sums SI , SII , and SIII were defined earlier (see also Figure 3).
We evaluate SI by summing vertically, and SII + SIII by summing hor-
izontally. In order to carry this out we need the following simple results.

Lemma 3.15 For all integers 0 < a < b we have (i) b∑ n=a χ(n) n1/2 = O(a−1/2), (ii) b∑ n=a χ(n) n = O(a−1). Proof. This argument is similar to the proof of Proposition 3.4; we use summation by parts. Let sn = ∑ 1≤k≤n χ(k), and remember that |sn| ≤ q for all n. Then b∑ n=a χ(n) n1/2 = b−1∑ n=a sn [ n−1/2 − (n + 1)−1/2 ] + O(a−1/2) = O ( ∞∑ n=a n−3/2 ) + O(a−1/2). By comparing the sum ∑∞ n=a n −3/2 with the integral of f(x) = x−3/2, we find that the former is also O(a−1/2). A similar argument establishes (ii). We may now finish the proof of the proposition. Summing vertically we find SI = ∑ m 1/2.
(b) Show that if Re(w) > 1/2 and w = ρeiϕ with ρ > 0, |ϕ| < π, then log1 w = log ρ + iϕ. [Hint: If eζ = w, then the real part of ζ is uniquely determined and its imaginary part is determined modulo 2π.] 8. Let ζ denote the zeta function defined for s > 1.

(a) Compare ζ(s) with
∫∞
1

x−s dx to show that

ζ(s) =
1

s− 1 + O(1) as s → 1
+.

(b) Prove as a consequence that

∑
p

1
ps

= log
(

1
s− 1

)
+ O(1) as s → 1+.

9. Let χ0 denote the trivial Dirichlet character mod q, and p1, . . . , pk the distinct
prime divisors of q. Recall that L(s, χ0) = (1− p−s1 ) · · · (1− p−sk )ζ(s), and show
as a consequence

L(s, χ0) =
ϕ(q)

1
s− 1 + O(1) as s → 1

[Hint: Use the asymptotics for ζ in Exercise 8.]

10. Show that if ` is relatively prime to q, then

∑

p≡`

1
ps

=
1

ϕ(q)
log

(
1

s− 1

)
+ O(1) as s → 1+.

This is a quantitative version of Dirichlet’s theorem.

[Hint: Recall (4).]

11. Use the characters for Z∗(3), Z∗(4), Z∗(5), and Z∗(6) to verify directly that
L(1, χ) 6= 0 for all non-trivial Dirichlet characters modulo q when q = 3, 4, 5,
and 6.

[Hint: Consider in each case the appropriate alternating series.]

Ibookroot October 20, 2007

278 Chapter 8. DIRICHLET’S THEOREM

12. Suppose χ is real and non-trivial; assuming the theorem that L(1, χ) 6= 0,
show directly that L(1, χ) > 0.

[Hint: Use the product formula for L(s, χ).]

13. Let {an}∞n=−∞ be a sequence of complex numbers such that an = am if
n = m mod q. Show that the series

∞∑
n=1

an
n

converges if and only if
∑q

n=1
an = 0.

[Hint: Sum by parts.]

14. The series

F (θ) =
∑

|n|6=0

einθ

n
, for |θ| < π, converges for every θ and is the Fourier series of the function defined on [−π, π] by F (0) = 0 and F (θ) = { i(−π − θ) if −π ≤ θ < 0 i(π − θ) if 0 < θ ≤ π, and extended by periodicity (period 2π) to all of R (see Exercise 8 in Chapter 2). Show also that if θ 6= 0 mod 2π, then the series E(θ) = ∞∑ n=1 einθ n converges, and that E(θ) = 1 2 log ( 1 2− 2 cos θ ) + i 2 F (θ). 15. To sum the series ∑∞ n=1 an/n, with an = am if n = m mod q and ∑q n=1 an = 0, proceed as follows. (a) Define A(m) = q∑ n=1 anζ −mn where ζ = e2πi/q. Ibookroot October 20, 2007 5. Problems 279 Note that A(q) = 0. With the notation of the previous exercise, prove that ∞∑ n=1 an n = 1 q q−1∑ m=1 A(m)E(2πm/q). [Hint: Use Fourier inversion on Z(q).] (b) If {am} is odd, (a−m = −am) for m ∈ Z, observe that a0 = aq = 0 and show that A(m) = ∑ 1≤n 0 so that d(k) ≥
c(log k)N for infinitely many k. [Hint: Let p1, . . . , pN be N distinct primes,
and consider k of the form (p1p2 · · · pN )m for m = 1, 2, . . ..]

3. Show that if p is relatively prime to q, then

∏
χ

(
1− χ(p)

)
=

(
1

1− pfs
)g

where g = ϕ(q)/f , and f is the order of p in Z∗(q) (that is, the smallest n for
which pn ≡ 1 mod q). Here the product is taken over all Dirichlet characters
modulo q.

4. Prove as a consequence of the previous problem that

∏
χ

L(s, χ) =
∑
n≥1

an
ns

where an ≥ 0, and the product is over all Dirichlet characters modulo q.

Ibookroot October 20, 2007

Appendix : Integration

This appendix is meant as a quick review of the definition and main
properties of the Riemann integral on R, and integration of appropriate
continuous functions on Rd. Our exposition is brief since we assume that
the reader already has some familiarity with this material.

We begin with the theory of Riemann integration on a closed and
bounded interval on the real line. Besides the standard results about
the integral, we also discuss the notion of sets of measure 0, and give
a necessary and sufficient condition on the set of discontinuities of a
function that guarantee its integrability.

We also discuss multiple and repeated integrals. In particular, we
extend the notion of integration to the entire space Rd by restricting
ourselves to functions that decay fast enough at infinity.

1 Definition of the Riemann integral

Let f be a bounded real-valued function defined on the closed interval
[a, b] ⊂ R. By a partition P of [a, b] we mean a finite sequence of num-
bers x0, x1, . . . , xN with

a = x0 < x1 < · · · < xN−1 < xN = b. Given such a partition, we let Ij denote the interval [xj−1, xj ] and write |Ij | for its length, namely |Ij | = xj − xj−1. We define the upper and lower sums of f with respect to P by U(P, f) = N∑ j=1 [ sup x∈Ij f(x)] |Ij | and L(P, f) = N∑ j=1 [ inf x∈Ij f(x)] |Ij |. Note that the infimum and supremum exist because by assumption, f is bounded. Clearly U(P, f) ≥ L(P, f), and the function f is said to be Riemann integrable, or simply integrable, if for every ² > 0 there
exists a partition P such that

U(P, f)− L(P, f) < ². To define the value of the integral of f , we need to make a simple yet important observation. A partition P ′ is said to be a refinement of the partition P if P ′ is obtained from P by adding points. Then, adding one Ibookroot October 20, 2007 282 Appendix: INTEGRATION point at a time, it is easy to check that U(P ′, f) ≤ U(P, f) and L(P ′, f) ≥ L(P, f). From this, we see that if P1 and P2 are two partitions of [a, b], then U(P1, f) ≥ L(P2, f), since it is possible to take P ′ as a common refinement of both P1 and P2 to obtain U(P1, f) ≥ U(P ′, f) ≥ L(P ′, f) ≥ L(P2, f). Since f is bounded we see that both U = inf P U(P, f) and L = sup P L(P, f) exist (where the infimum and supremum are taken over all partitions of [a, b]), and also that U ≥ L . Moreover, if f is integrable we must have U = L, and we define ∫ b a f(x) dx to be this common value. Finally, a bounded complex-valued function f = u + iv is said to be integrable if its real and imaginary parts u and v are integrable, and we define ∫ b a f(x) dx = ∫ b a u(x) dx + i ∫ b a v(x) dx. For example, the constants are integrable functions and it is clear that if c ∈ C, then ∫ b a c dx = c(b− a). Also, continuous functions are inte- grable. This is because a continuous function on a closed and bounded interval [a, b] is uniformly continuous, that is, given ² > 0 there exists δ
such that if |x− y| < δ then |f(x)− f(y)| < ². So if we choose n with (b− a)/n < δ, then the partition P given by a, a + b− a n , . . . , a + k b− a n , . . . , a + (n− 1)b− a n , b satisfies U(P, f)− L(P, f) ≤ ²(b− a). 1.1 Basic properties Proposition 1.1 If f and g are integrable on [a, b], then: (i) f + g is integrable, and ∫ b a f(x) + g(x) dx = ∫ b a f(x) dx + ∫ b a g(x) dx. Ibookroot October 20, 2007 1. Definition of the Riemann integral 283 (ii) If c ∈ C, then ∫ b a cf(x) dx = c ∫ b a f(x) dx. (iii) If f and g are real-valued and f(x) ≤ g(x), then ∫ b a f(x) dx ≤ ∫ b a g(x) dx. (iv) If c ∈ [a, b], then ∫ b a f(x) dx = ∫ c a f(x) dx + ∫ b c f(x) dx. Proof. For property (i) we may assume that f and g are real-valued. If P is a partition of [a, b], then U(P, f + g) ≤ U(P, f) + U(P, g) and L(P, f + g) ≥ L(P, f) + L(P, g). Given ² > 0, there exist partitions P1 and P2 such that U(P1, f)− L(P1, f) < ² and U(P2, g)− L(P2, g) < ², so that if P0 is a common refinement of P1 and P2, we get U(P0, f + g)− L(P0, f + g) < 2². So f + g is integrable, and if we let I = infP U(P, f + g) = supP L(P, f + g), then we see that I ≤ U(P0, f + g) + 2² ≤ U(P0, f) + U(P0, g) + 2² ≤ ∫ b a f(x) dx + ∫ b a g(x) dx + 4². Similarly I ≥ ∫ b a f(x) dx + ∫ b a g(x) dx− 4², which proves that ∫ b a f(x) + g(x) dx = ∫ b a f(x) dx + ∫ b a g(x) dx. The second and third parts of the proposition are just as easy to prove. For the last property, simply refine partitions of [a, b] by adding the point c. Another important property we need to prove is that fg is integrable whenever f and g are integrable. Lemma 1.2 If f is real-valued integrable on [a, b] and ϕ is a real-valued continuous function on R, then ϕ ◦ f is also integrable on [a, b]. Proof. Let ² > 0 and remember that f is bounded, say |f | ≤ M . Since
ϕ is uniformly continuous on [−M, M ] we may choose δ > 0 so that if
s, t ∈ [−M, M ] and |s− t| < δ, then |ϕ(s)− ϕ(t)| < ². Now choose a par- tition P = {x0, . . . , xN} of [a, b] with U(P, f)− L(P, f) < δ2. Let Ij = [xj−1, xj ] and distinguish two classes: we write j ∈ Λ if supx∈Ij f(x)− infx∈Ij f(x) < δ so that by construction sup x∈Ij ϕ ◦ f(x)− inf x∈Ij ϕ ◦ f(x) < ². Ibookroot October 20, 2007 284 Appendix: INTEGRATION Otherwise, we write j ∈ Λ′ and note that δ ∑ j∈Λ′ |Ij | ≤ ∑ j∈Λ′ [ sup x∈Ij f(x)− inf x∈Ij f(x)] |Ij | ≤ δ2 so ∑ j∈Λ′ |Ij | < δ. Therefore, separating the cases j ∈ Λ and j ∈ Λ′ we find that U(P,ϕ ◦ f)− L(P, ϕ ◦ f) ≤ ²(b− a) + 2Bδ, where B is a bound for ϕ on [−M, M ]. Since we can also choose δ < ², we see that the proposition is proved. ¿From the lemma we get the following facts: • If f and g are integrable on [a, b], then the product fg is integrable on [a, b]. This follows from the lemma with ϕ(t) = t2, and the fact that fg = 1 4 ( [f + g]2 − [f − g]2 ) . • If f is integrable on [a, b], then the function |f | is integrable, and∣∣∣ ∫ b a f(x) dx ∣∣∣ ≤ ∫ b a |f(x)| dx. We can take ϕ(t) = |t| to see that |f | is integrable. Moreover, the in- equality follows from (iii) in Proposition 1.1. We record two results that imply integrability. Proposition 1.3 A bounded monotonic function f on an interval [a, b] is integrable. Proof. We may assume without loss of generality that a = 0, b = 1, and f is monotonically increasing. Then, for each N , we choose the uniform partition PN given by xj = j/N for all j = 0, . . . , N . If αj = f(xj), then we have U(PN , f) = 1 N N∑ j=1 αj and L(PN , f) = 1 N N∑ j=1 αj−1. Therefore, if |f(x)| ≤ B for all x we have U(PN , f)− L(PN , f) = αN − α0 N ≤ 2B N , and the proposition is proved. Ibookroot October 20, 2007 1. Definition of the Riemann integral 285 Proposition 1.4 Let f be a bounded function on the compact interval [a, b]. If c ∈ (a, b), and if for all small δ > 0 the function f is integrable
on the intervals [a, c− δ] and [c + δ, b], then f is integrable on [a, b].

Proof. Suppose |f | ≤ M and let ² > 0. Choose δ > 0 (small) so
that 4δM ≤ ²/3. Now let P1 and P2 be partitions of [a, c− δ] and [c +
δ, b] so that for each i = 1, 2 we have U(Pi, f)− L(Pi, f) < ²/3. This is possible since f is integrable on each one of the intervals. Then by taking as a partition P = P1 ∪ {c− δ} ∪ {c + δ} ∪ P2 we immediately see that U(P, f)− L(P, f) < ². We end this section with a useful approximation lemma. Recall that a function on the circle is the same as a 2π-periodic function on R. Lemma 1.5 Suppose f is integrable on the circle and f is bounded by B. Then there exists a sequence {fk}∞k=1 of continuous functions on the circle so that sup x∈[−π,π] |fk(x)| ≤ B for all k = 1, 2, . . ., and ∫ π −π |f(x)− fk(x)| dx → 0 as k →∞. Proof. Assume f is real-valued (in general apply the following argu- ment to the real and imaginary parts separately). Given ² > 0, we may
choose a partition −π = x0 < x1 < · · · < xN = π of the interval [−π, π] so that the upper and lower sums of f differ by at most ². Denote by f∗ the step function defined by f∗(x) = sup xj−1≤y≤xj f(y) if x ∈ [xj−1, xj) for 1 ≤ j ≤ N . By construction we have |f∗| ≤ B, and moreover (1) ∫ π −π |f∗(x)− f(x)| dx = ∫ π −π (f∗(x)− f(x)) dx < ². Now we can modify f∗ to make it continuous and periodic yet still ap- proximate f in the sense of the lemma. For small δ > 0, let f̃(x) = f∗(x)
when the distance of x from any of the division points x0, . . . , xN is
≥ δ. In the δ-neighborhood of xj for j = 1, . . . , N − 1, define f̃(x) to be
the linear function for which f̃(xj ± δ) = f∗(xj ± δ). Near x0 = −π, f̃

Ibookroot October 20, 2007

286 Appendix: INTEGRATION

f∗

x3x1 x2

f̃

x1 x2 x3

x2 − δ x2 + δ
x0 x0

Figure 1. Portions of the functions f∗ and f̃

is linear with f̃(−π) = 0 and f̃(−π + δ) = f∗(−π + δ). Similarly, near
xN = π the function f̃ is linear with f̃(π) = 0 and f̃(π − δ) = f∗(π − δ).
In Figure 1 we illustrate the situation near x0 = −π. In the second pic-
ture the graph of f̃ is shifted slightly below to clarify the situation.

Then, since f̃(−π) = f̃(π), we may extend f̃ to a continuous and 2π-
periodic function on R. The absolute value of this extension is also
bounded by B. Moreover, f̃ differs from f∗ only in the N intervals of
length 2δ surrounding the division points. Thus

∫ π
−π
|f∗(x)− f̃(x)| dx ≤ 2BN(2δ).

If we choose δ sufficiently small, we get

(2)
∫ π
−π
|f∗(x)− f̃(x)| dx < ². As a result, equations (1), (2), and the triangle inequality yield ∫ π −π |f(x)− f̃(x)| dx < 2². Denoting by fk the f̃ so constructed, when 2² = 1/k, we see that the sequence {fk} has the properties required by the lemma. 1.2 Sets of measure zero and discontinuities of integrable func- tions We observed that all continuous functions are integrable. By modifying the argument slightly, one can show that all piecewise continuous func- tions are also integrable. In fact, this is a consequence of Proposition 1.4 Ibookroot October 20, 2007 1. Definition of the Riemann integral 287 applied finitely many times. We now turn to a more careful study of the discontinuities of integrable functions. We start with a definition1: a subset E of R is said to have measure 0 if for every ² > 0 there exists a countable family of open intervals {Ik}∞k=1
such that

(i) E ⊂ ⋃∞k=1 Ik,
(ii)

∑∞
k=1 |Ik| < ², where |Ik| denotes the length of the interval Ik. The first condition says that the union of the intervals covers E, and the second that this union is small. The reader will have no difficulty proving that any finite set of points has measure 0. A more subtle argument is needed to prove that a countable set of points has measure 0. In fact, this result is contained in the following lemma. Lemma 1.6 The union of countably many sets of measure 0 has mea- sure 0. Proof. Say E1, E2, . . . are sets of measure 0, and let E = ∪∞i=1Ei. Let ² > 0, and for each i choose open interval Ii,1, Ii,2, . . . so that

Ei ⊂
∞⋃

k=1

Ii,k and
∞∑

k=1

|Ii,k| < ²/2i. Now clearly we have E ⊂ ⋃∞i,k=1 Ii,k, and ∞∑ i=1 ∞∑ k=1 |Ii,k| ≤ ∞∑ i=1 ² 2i ≤ ², as was to be shown. An important observation is that if E has measure 0 and is com- pact, then it is possible to find a finite number of open intervals Ik, k = 1, . . . , N , that satisfy the two conditions (i) and (ii) above. We can prove the characterization of Riemann integrable functions in terms of their discontinuities. Theorem 1.7 A bounded function f on [a, b] is integrable if and only if its set of discontinuities has measure 0. 1A systematic study of the measure of sets arises in the theory of Lebesgue integration, which is taken up in Book III. Ibookroot October 20, 2007 288 Appendix: INTEGRATION We write J = [a, b] and I(c, r) = (c− r, c + r) for the open interval centered at c of radius r > 0. Define the oscillation of f on I(c, r) by

osc(f, c, r) = sup |f(x)− f(y)|

where the supremum is taken over all x, y ∈ J ∩ I(c, r). This quantity
exists since f is bounded. Define the oscillation of f at c by

osc(f, c) = lim
r→0

osc(f, c, r).

This limit exists because osc(f, c, r) is ≥ 0 and a decreasing function of
r. The point is that f is continuous at c if and only if osc(f, c) = 0. This
is clear from the definitions. For each ² > 0 we define a set A² by

A² = {c ∈ J : osc(f, c) ≥ ²}.

Having done that, we see that the set of points in J where f is discon-
tinuous is simply

⋃
²>0 A². This is an important step in the proof of our

theorem.

Lemma 1.8 If ² > 0, then the set A² is closed and therefore compact.

Proof. The argument is simple. Suppose cn ∈ A² converges to c
and assume that c /∈ A². Write osc(f, c) = ²− δ where δ > 0. Select r
so that osc(f, c, r) < ²− δ/2, and choose n with |cn − c| < r/2. Then osc(f, cn, r/2) < ² which implies osc(f, cn) < ², a contradiction. We are now ready to prove the first part of the theorem. Suppose that the set D of discontinuities of f has measure 0, and let ² > 0.
Since A² ⊂ D, we can cover A² by a finite number of open intervals,
say I1, . . . , IN , whose total length is < ². The complement of this union I of intervals is compact, and around each point z in this complement we can find an interval Fz with supx,y∈Fz |f(x)− f(y)| ≤ ², simply because z /∈ A². We may now choose a finite subcovering of ∪z∈IcIz, which we denote by IN+1, . . . , IN ′ . Now, taking all the end points of the intervals I1, I2, . . . , IN ′ we obtain a partition P of [a, b] with U(P, f)− L(P, f) ≤ 2M N∑ j=1 |Ij |+ ²(b− a) ≤ C². Hence f is integrable on [a, b], as was to be shown. Conversely, suppose that f is integrable on [a, b], and let D be its set of discontinuities. Since D equals ∪∞n=1A1/n, it suffices to prove Ibookroot October 20, 2007 2. Multiple integrals 289 that each A1/n has measure 0. Let ² > 0 and choose a partition P =
{x0, x1, . . . , xN} so that U(P, f)− L(P, f) < ²/n. Then, if A1/n inter- sects Ij = (xj−1, xj) we must have supx∈Ij f(x)− infx∈Ij f(x) ≥ 1/n, and this shows that 1 n ∑ {j:Ij∩A1/n 6=∅} |Ij | ≤ U(P, f)− L(P, f) < ²/n. So by taking intervals intersecting A1/n and making them slightly larger, we can cover A1/n with open intervals of total length ≤ 2². Therefore A1/n has measure 0, and we are done. Note that incidentally, this gives another proof that fg is integrable whenever f and g are. 2 Multiple integrals We assume that the reader is familiar with the standard theory of multi- ple integrals of functions defined on bounded sets. Here, we give a quick review of the main definitions and results of this theory. Then, we de- scribe the notion of “improper” multiple integration where the range of integration is extended to all of Rd. This is relevant to our study of the Fourier transform. In the spirit of Chapters 5 and 6, we shall define the integral of functions that are continuous and satisfy an adequate decay condition at infinity. Recall that the vector space Rd consists of all d-tuples of real numbers x = (x1, x2, . . . , xd) with xj ∈ R, where addition and multiplication by scalars are defined componentwise. 2.1 The Riemann integral in Rd Definitions The notion of Riemann integration on a rectangle R ⊂ Rd is an imme- diate generalization of the notion of Riemann integration on an interval [a, b] ⊂ R. We restrict our attention to continuous functions; these are always integrable. By a closed rectangle in Rd, we mean a set of the form R = {aj ≤ xj ≤ bj : 1 ≤ j ≤ d} where aj , bj ∈ R for 1 ≤ j ≤ n. In other words, R is the product of the one-dimensional intervals [aj , bj ]: R = [a1, b1]× · · · × [ad, bd]. Ibookroot October 20, 2007 290 Appendix: INTEGRATION If Pj is a partition of the closed interval [aj , bj ], then we call P = (P1, . . . , Pd) a partition of R; and if Sj is a subinterval of the partition Pj , then S = S1 × · · · × Sd is a subrectangle of the partition P . The volume |S| of a subrectangle S is naturally given by the product of the length of its sides |S| = |S1| × · · · × |Sd|, where |Sj | denotes the length of the interval Sj . We are now ready to define the notion of integral over R. Given a bounded real-valued function f defined on R and a partition P , we define the upper and lower sums of f with respect to P by U(P, f) = ∑ [sup x∈S f(x)] |S| and L(P, f) = ∑ [ inf x∈S f(x)] |S|, where the sums are taken over all subrectangles of the partition P . These definitions are direct generalizations of the analogous notions in one di- mension. A partition P ′ = (P ′1, . . . , P ′ d) is a refinement of P = (P1, . . . , Pd) if each P ′j is a refinement of Pj . Arguing with these refinements as we did in the one-dimensional case, we see that if we define U = inf P U(P, f) and L = sup P L(P, f), then both U and L exist, are finite, and U ≥ L. We say that f is Rie- mann integrable on R if for every ² > 0 there exists a partition P so
that

U(P, f)− L(P, f) < ². This implies that U = L, and this common value, which we shall denote by either ∫ R f(x1, . . . , xd) dx1 · · · dxd, ∫ R f(x) dx, or ∫ R f, is by definition the integral of f over R. If f is complex-valued, say f(x) = u(x) + iv(x), where u and v are real-valued, we naturally define ∫ R f(x) dx = ∫ R u(x) dx + i ∫ R v(x) dx. In the results that follow, we are primarily interested in continuous functions. Clearly, if f is continuous on a closed rectangle R then f is integrable since it is uniformly continuous on R. Also, we note that if f is continuous on, say, a closed ball B, then we may define its integral Ibookroot October 20, 2007 2. Multiple integrals 291 over B in the following way: if g is the extension of f defined by g(x) = 0 if x /∈ B, then g is integrable on any rectangle R that contains B, and we may set ∫ B f(x) dx = ∫ R g(x) dx. 2.2 Repeated integrals The fundamental theorem of calculus allows us to compute many one dimensional integrals, since it is possible in many instances to find an antiderivative for the integrand. In Rd, this permits the calculation of multiple integrals, since a d-dimensional integral actually reduces to d one-dimensional integrals. A precise statement describing this fact is given by the following. Theorem 2.1 Let f be a continuous function defined on a closed rect- angle R ⊂ Rd. Suppose R = R1 ×R2 where R1 ⊂ Rd1 and R2 ⊂ Rd2 with d = d1 + d2. If we write x = (x1, x2) with xi ∈ Rdi , then F (x1) =∫ R2 f(x1, x2) dx2 is continuous on R1, and we have ∫ R f(x) dx = ∫ R1 (∫ R2 f(x1, x2) dx2 ) dx1. Proof. The continuity of F follows from the uniform continuity of f on R and the fact that |F (x1)− F (x′1)| ≤ ∫ R2 |f(x1, x2)− f(x′1, x2)| dx2. To prove the identity, let P1 and P2 be partitions of R1 and R2, respec- tively. If S and T are subrectangles in P1 and P2, respectively, then the key observation is that sup S×T f(x1, x2) ≥ sup x1∈S ( sup x2∈T f(x1, x2) ) and inf S×T f(x1, x2) ≤ inf x1∈S ( inf x2∈T f(x1, x2) ) . Ibookroot October 20, 2007 292 Appendix: INTEGRATION Then, U(P, f) = ∑ S,T [ sup S×T f(x1, x2)] |S × T | ≥ ∑ S ∑ T sup x1∈S [ sup x2∈T f(x1, x2)] |T | × |S| ≥ ∑ S sup x1∈S (∫ R2 f(x1, x2) dx2 ) |S| ≥ U ( P1, ∫ R2 f(x1, x2) dx2 ) . Arguing similarly for the lower sums, we find that L(P, f) ≤ L(P1, ∫ R2 f(x1, x2) dx2) ≤ U(P1, ∫ R2 f(x1, x2) dx2) ≤ U(P, f), and the theorem follows from these inequalities. Repeating this argument, we find as a corollary that if f is continuous on the rectangle R ⊂ Rd given by R = [a1, b1]× · · · [ad, bd], then ∫ R f(x) dx = ∫ b1 a1 (∫ b2 a2 · · · (∫ bd ad f(x1, . . . , xd) dxd ) . . . dx2 ) dx1, where the right-hand side denotes d-iterates of one-dimensional integrals. It is also clear from the theorem that we can interchange the order of integration in the repeated integral as desired. 2.3 The change of variables formula A diffeomorphism of class C1, g : A → B, is a mapping that is contin- uously differentiable, invertible, and whose inverse g−1 : B → A is also continuously differentiable. We denote by Dg the Jacobian or derivative of g. Then, the change of variables formula says the following. Theorem 2.2 Suppose A and B are compact subsets of Rd and g : A → B is a diffeomorphism of class C1. If f is continuous on B, then ∫ g(A) f(x) dx = ∫ A f(g(y)) |det(Dg)(y)| dy. The proof of this theorem consists first of an analysis of the special situation when g is a linear transformation L. In this case, if R is a rectangle, then |g(R)| = | det(L)| |R|, Ibookroot October 20, 2007 2. Multiple integrals 293 which explains the term | det(Dg)|. Indeed, this term corresponds to the new infinitesimal element of volume after the change of variables. 2.4 Spherical coordinates An important application of the change of variables formula is to the case of polar coordinates in R2, spherical coordinates in R3, and their general- ization in Rd. These are particularly important when the function, or set we are integrating over, exhibit some rotational (or spherical) symme- tries. The cases d = 2 and d = 3 were given in Chapter 6. More generally, the spherical coordinates system in Rd is given by x = g(r, θ1, . . . , θd−1) where    x1 = r sin θ1 sin θ2 · · · sin θd−2 cos θd−1, x2 = r sin θ1 sin θ2 · · · sin θd−2 sin θd−1, ... xd−1 = r sin θ1 sin θ2, xd = r cos θ1, with 0 ≤ θi ≤ π for 1 ≤ i ≤ d− 2 and 0 ≤ θd−1 ≤ 2π. The determinant of the Jacobian of this transformation is given by rd−1 sind−2 θ1 sin d−3 θ2 · · · sin θd−2. Any point in x ∈ Rd − {0} can be written uniquely as rγ with γ ∈ Sd−1 the unit sphere in Rd. If we define ∫ Sd−1 f(γ) dσ(γ) = ∫ π 0 ∫ π 0 · · · ∫ 2π 0 f(g(r, θ)) sind−2 θ1 sin d−3 θ2 · · · sin θd−2 dθd−1 · · · dθ1, then we see that if B(0, N) denotes the ball of radius N centered at the origin, then (3) ∫ B(0,N) f(x) dx = ∫ Sd−1 ∫ N 0 f(rγ) rd−1 dr dσ(γ). In fact, we define the area of the unit sphere Sd−1 ⊂ Rd as ωd = ∫ Sd−1 dσ(γ). An important application of spherical coordinates is to the calculation of the integral ∫ A(R1,R2) |x|λ dx, where A(R1, R2) denotes the annulus Ibookroot October 20, 2007 294 Appendix: INTEGRATION A(R1, R2) = {R1 ≤ |x| ≤ R2} and λ ∈ R. Applying polar coordinates, we find ∫ A(R1,R2) |x|λ dx = ∫ Sd−1 ∫ R2 R1 rλ+d−1 drdσ(γ). Therefore ∫ A(R1,R2) |x|λ dx = { ωd λ+d [Rλ+d2 −Rλ+d1 ] if λ 6= −d, ωd[log(R2)− log(R1)] if λ = −d. 3 Improper integrals. Integration over Rd Most of the theorems we just discussed extend to functions integrated over all of Rd once we impose some decay at infinity on the functions we integrate. 3.1 Integration of functions of moderate decrease For each fixed N > 0 consider the closed cube in Rd centered at the origin
with sides parallel to the axis, and of side length N : QN = {|xj | ≤ N/2 :
1 ≤ j ≤ d}. Let f be a continuous function on Rd. If the limit

lim
N→∞

∫

f(x) dx

exists, we denote it by
∫

Rd
f(x) dx.

We deal with a special class of functions whose integrals over Rd exist.
A continuous function f on Rd is said to be of moderate decrease if
there exists A > 0 such that

|f(x)| ≤ A
1 + |x|d+1 .

Note that if d = 1 we recover the definition given in Chapter 5. An
important example of a function of moderate decrease in R is the Poisson
kernel given by Py(x) = 1π

y
x2+y2

We claim that if f is of moderate decrease, then the above limit exists.
Let IN =

∫
QN

f(x) dx. Each IN exists because f is continuous hence
integrable. For M > N , we have

|IM − IN | ≤
∫

QM−QN
|f(x)| dx.

Ibookroot October 20, 2007

3. Improper integrals. Integration over Rd 295

Now observe that the set QM −QN is contained in the annulus
A(aN, bM) = {aN ≤ |x| ≤ bM}, where a and b are constants that de-
pend only on the dimension d. This is because the cube QN is contained
in the annulus N/2 ≤ |x| ≤ N

√
d/2, so that we can take a = 1/2 and

b =
√

d/2. Therefore, using the fact that f is of moderate decrease yields

|IM − IN | ≤ A
∫

aN≤|x|≤bM
|x|−d−1 dx.

Now putting λ = −d− 1 in the calculation of the integral of the previous
section, we find that

|IM − IN | ≤ C
(

1
aN

− 1
bM

)
.

So if f is of moderate decrease, we conclude that {IN}∞N=1 is a Cauchy
sequence, and therefore

∫
Rd f(x) dx exists.

Instead of the rectangles QN , we could have chosen the balls BN cen-
tered at the origin and of radius N . Then, if f is of moderate decrease,
the reader should have no difficulties proving that limN→∞

∫
BN

f(x) dx
exists, and that this limit equals limN→∞

∫
QN

f(x) dx.

Some elementary properties of the integrals of functions of moderate
decrease are summarized in Chapter 6.

3.2 Repeated integrals

In Chapters 5 and 6 we claimed that the multiplication formula held for
functions of moderate decrease. This required an appropriate interchange
of integration. Similarly for operators defined in terms of convolutions
(with the Poisson kernel for example).

We now justify the necessary formula for iterated integrals. We only
consider the case d = 2, although the reader will have no difficulty ex-
tending this result to arbitrary dimensions.

Theorem 3.1 Suppose f is continuous on R2 and of moderate decrease.
Then

F (x1) =
∫

R
f(x1, x2) dx2

is of moderate decrease on R, and
∫

R2
f(x) dx =

∫

(∫

R
f(x1, x2) dx2

)
dx1.

Ibookroot October 20, 2007

296 Appendix: INTEGRATION

Proof. To see why F is of moderate decrease, note first that

|F (x1)| ≤
∫

Adx2
1 + (x21 + x

2
2)3/2

≤
∫

|x2|≤|x1|
+

∫

|x2|≥|x1|
.

In the first integral, we observe that the integrand is ≤ A/(1 + |x1|3), so
∫

|x2|≤|x1|

Adx2
1 + (x21 + x

2
2)3/2

≤ A
1 + |x1|3

∫

|x2|≤|x1|
dx2 ≤

A′

1 + |x1|2
.

For the second integral, we have
∫

|x2|≥|x1|

Adx2
1 + (x21 + x

2
2)3/2

≤ A′′
∫

|x2|≥|x1|

dx2
1 + |x2|3

≤ A
′′′

|x1|2
,

thus F is of moderate decrease. In fact, this argument together with
Theorem 2.1 shows that F is the uniform limit of continuous functions,
thus is also continuous.

To establish the identity we simply use an approximation and Theo-
rem 2.1 over finite rectangles. Write Sc to denote the complement of a
set S. Given ² > 0 choose N so large that

∣∣∣∣
∫

R2
f(x1, x2) dx1dx2 −

∫

IN×IN
f(x1, x2) dx1dx2

∣∣∣∣ < ², where IN = [−N,N ]. Now we know that ∫ IN×IN f(x1, x2) dx1dx2 = ∫ IN (∫ IN f(x1, x2) dx2 ) dx1. But this last iterated integral can be written as = ∫ R (∫ R f(x1, x2) dx2 ) dx1− ∫ Ic N (∫ R f(x1, x2) dx2 ) dx1 − ∫ IN (∫ Ic N f(x1, x2) dx2 ) dx1. We can now estimate ∣∣∣∣∣ ∫ IN (∫ Ic N f(x1, x2) dx2 ) dx1 ∣∣∣∣∣ ≤ O ( 1 N2 ) + C ∫ 1≤|x1|≤N (∫ |x2|≥N dx2 (|x1|+ |y1|)3 ) dx1 ≤ O ( 1 N ) . Ibookroot October 20, 2007 3. Improper integrals. Integration over Rd 297 A similar argument shows that ∣∣∣∣∣ ∫ Ic N (∫ R f(x1, x2) dx2 ) dx1 ∣∣∣∣∣ ≤ C N . Therefore, we can find N so large that ∣∣∣∣ ∫ IN×IN f(x1, x2) dx1dx2 − ∫ R (∫ R f(x1, x2) dx2 ) dx1 ∣∣∣∣ < ², and we are done. 3.3 Spherical coordinates In Rd, spherical coordinates are given by x = rγ, where r ≥ 0 and γ belongs to the unit sphere Sd−1. If f is of moderate decrease, then for each fixed γ ∈ Sd−1, the function of f given by f(rγ)rd−1 is also of moderate decrease on R. Indeed, we have ∣∣∣f(rγ)rd−1 ∣∣∣ ≤ A r d−1 1 + |rγ|d+1 ≤ B 1 + r2 . As a result, by letting R →∞ in (3) we obtain the formula ∫ Rd f(x) dx = ∫ Sd−1 ∫ ∞ 0 f(rγ) rd−1 dr dσ(γ). As a consequence, if we combine the fact that ∫ Rd f(R(x)) dx = ∫ Rd f(x) dx, whenever R is a rotation, with the identity (3), then we obtain that (4) ∫ Sd−1 f(R(γ)) dσ(γ) = ∫ Sd−1 f(γ) dσ(γ) . Ibookroot October 20, 2007 Notes and References Seeley [29] gives an elegant and brief introduction to Fourier series and the Fourier transform. The authoritative text on Fourier series is Zygmund [36]. For further applications of Fourier analysis to a variety of other topics, see Dym and McKean [8] and Körner [21]. The reader should also consult the book by Kahane and Lemarié-Rieusset [20], which contains many historical facts and other results related to Fourier series. Chapter 1 The citation is taken from a letter of Fourier to an unknown correspondent (probably Lagrange), see Herivel [15]. More facts about the early history of Fourier series can be found in Sections I-III of Riemann’s memoir [27]. Chapter 2 The quote is a translation of an excerpt in Riemann’s paper [27]. For a proof of Littlewood’s theorem (Problem 3), as well as other related “Tauberian theorems,” see Chapter 7 in Titchmarsh [32]. Chapter 3 The citation is a translation of a passage in Dirichlet’s memoir [6]. Chapter 4 The quote is translated from Hurwitz [17]. The problem of a ray of light reflecting inside a square is discussed in Chap- ter 23 of Hardy and Wright [13]. The relationship between the diameter of a curve and Fourier coefficients (Problem 1) is explored in Pfluger [26]. Many topics concerning equidistribution of sequences, including the results in Problems 2 and 3, are taken up in Kuipers and Niederreiter [22]. Chapter 5 The citation is a free translation of a passage in Schwartz [28]. For topics in finance, see Duffie [7], and in particular Chapter 5 for the Black- Scholes theory (Problems 1 and 2). The results in Problems 4, 5, and 6 are worked out in John [19] and Wid- der [34]. For Problem 7, see Chapter 2 in Wiener [35]. The original proof of the nowhere differentiability of f1 (Problem 8) is in Hardy [12]. Chapter 6 The quote is an excerpt from Cormack’s Nobel Prize lecture [5]. More about the wave equation, as well as the results in Problems 3, 4, and 5 can be found in Chapter 5 of Folland [9]. 298 Ibookroot October 20, 2007 NOTES AND REFERENCES 299 A discussion of the relationship between rotational symmetry, the Fourier transform, and Bessel functions is in Chapter 4 of Stein and Weiss [31]. For more on the Radon transform, see Chapter 1 in John [18], Helgason [14], and Ludwig [25]. Chapter 7 The citation is taken from Bingham and Tukey [2]. Proofs of the structure theorem for finite abelian groups (Problem 2) can be found in Chapter 2 of Herstein [16], Chapter 2 in Lang [23], or Chapter 104 in Körner [21]. For Problem 4, see Andrews [1], which contains a short proof. Chapter 8 The citation is from Bochner [3]. For more on the divisor function, see Chapter 18 in Hardy and Wright [13]. Another “elementary” proof that L(1, χ) 6= 0 can be found in Chapter 3 of Gelfond and Linnik [11]. An alternate proof that L(1, χ) 6= 0 based on algebraic number theory is in Weyl [33]. Also, two other analytic variants of the proof that L(1, χ) 6= 0 can be found in Chapter 109 in Körner [21] and Chapter 6 in Serre [30]. See also the latter reference for Problems 3 and 4. Appendix Further details about the results on integration reviewed in the appendix can be found in Folland [10] (Chapter 4), Buck [4] (Chapter 4), or Lang [24] (Chap- ter 20). Ibookroot October 20, 2007 Bibliography [1] G. E. Andrews. Number theory. Dover Publications, New York, 1994. Corrected reprint of the 1971 originally published by W.B. Saunders Company. [2] C. Bingham and J.W. Tukey. Fourier methods in the frequency analysis of data. The Collected Works of John W Tukey, Volume II Time Series: 1965-1984(Wadsworth Advanced Books & Software), 1984. [3] S. Bochner. The role of Mathematics in the Rise of Science. Prince- ton University Press, Princeton, NJ, 1966. [4] R. C. Buck. Advanced Calculus. McGraw-Hill, New York, third edition, 1978. [5] A. M. Cormack. Nobel Prize in Physiology and Medicine Lecture, volume Volume 209. Science, 1980. [6] G. L. Dirichlet. Sur la convergence des séries trigonometriques qui servent à representer une fonction arbitraire entre des limites données. Crelle, Journal für die reine angewandte Mathematik, 4:157–169, 1829. [7] D. Duffie. Dynamic Asset Pricing Theory. Princeton University Press, Princeton, NJ, 2001. [8] H. Dym and H. P. McKean. Fourier Series and Integrals. Academic Press, New York, 1972. [9] G. B. Folland. Introduction to Partial Differential Equations. Princeton University Press, Princeton, NJ, 1995. [10] G. B. Folland. Advanced Calculus. Prentice Hall, Englewood Cliffs, NJ, 2002. [11] A. O. Gelfond and Yu. V. Linnik. Elementary Methods in Analytic Number Theory. Rand McNally & Compagny, Chicago, 1965. [12] G. H. Hardy. Weierstrass’s non-differentiable function. Transac- tions, American Mathematical Society, 17:301–325, 1916. [13] G. H. Hardy and E. M. Wright. An Introduction to the Theory of Numbers. Oxford University Press, London, fifth edition, 1979. 300 Ibookroot October 20, 2007 BIBLIOGRAPHY 301 [14] S. Helgason. The Radon transform on Euclidean spaces, com- pact two-point homogeneous spaces and Grassman manifolds. Acta. Math., 113:153–180, 1965. [15] J. Herivel. Joseph Fourier The Man and the Physicist. Clarendon Press, Oxford, 1975. [16] I. N. Herstein. Abstract Algebra. Macmillan, New York, second edition, 1990. [17] A. Hurwitz. Sur quelques applications géometriques des séries de Fourier. Annales de l’Ecole Normale Supérieure, 19(3):357–408, 1902. [18] F. John. Plane Waves and Spherical Mean Applied to Partial Dif- ferential Equations. Interscience Publishers, New York, 1955. [19] F. John. Partial Differential Equations. Springer-Verlag, New York, fourth edition, 1982. [20] J.P. Kahane and P. G. Lemarié-Rieusset. Séries de Fourier et on- delettes. Cassini, Paris, 1998. English version: Gordon & Breach, 1995. [21] T. W. Körner. Fourier Analysis. Cambridge University Press, Cam- bridge, UK, 1988. [22] L. Kuipers and H. Niederreiter. Uniform Distribution of Sequences. Wiley, New York, 1974. [23] S. Lang. Undergraduate Algebra. Springer-Verlag, New York, second edition, 1990. [24] S. Lang. Undergraduate Analysis. Springer-Verlag, New York, sec- ond edition, 1997. [25] D. Ludwig. The Radon transform on Euclidean space. Comm. Pure Appl. Math., 19:49–81, 1966. [26] A. Pfluger. On the diameter of planar curves and Fourier coeffi- cients. Colloquia Mathematica Societatis János Bolyai, Functions, series, operators, 35:957–965, 1983. [27] B. Riemann. Ueber die Darstellbarkeit einer Function durch eine trigonometrische Reihe. Habilitation an der Universität zu Göttingen, 1854. Collected Works, Springer Verlag, New York, 1990. Ibookroot October 20, 2007 302 BIBLIOGRAPHY [28] L. Schwartz. Théorie des distributions, volume Volume I. Hermann, Paris, 1950. [29] R. T. Seeley. An Introduction to Fourier Series and Integrals. W. A. Benjamin, New York, 1966. [30] J.P. Serre. A course in Arithmetic. GTM 7. Springer Verlag, New York, 1973. [31] E. M. Stein and G. Weiss. Introduction to Fourier Analysis on Euclidean Spaces. Princeton University Press, Princeton, NJ, 1971. [32] E. C. Titchmarsh. The Theory of Functions. Oxford University Press, London, second edition, 1939. [33] H. Weyl. Algebraic Theory of Numbers, volume Volume 1 of Annals of Mathematics Studies. Princeton University Press, Princeton, NJ, 1940. [34] D. V. Widder. The Heat Equation. Academic Press, New York, 1975. [35] N. Wiener. The Fourier Integral and Certain of its Applications. Cambridge University Press, Cambridge, UK, 1933. [36] A. Zygmund. Trigonometric Series, volume Volumes I and II. Cambridge University Press, Cambridge, UK, second edition, 1959. Reprinted 1993. Ibookroot October 20, 2007 Symbol Glossary The page numbers on the right indicate the first time the symbol or notation is defined or used. As usual, Z, Q, R and C denote the integers, the rationals, the reals, and the complex numbers respectively. 4 Laplacian 20, 185 |z|, z Absolute value and complex conjugate 22 ez Complex exponential 24 sinhx, coshx Hyperbolic sine and hyperbolic cosine 28 f̂(n), an Fourier coefficient 34 f(θ) ∼ ∑ aneinθ Fourier series 34 SN (f) Partial sum of a Fourier series 35 DN , DR, D̃N , D∗N Dirichlet kernel, conjugate, and modified 37, 95, 165 Pr, Py, P(d)y Poisson kernels 37, 149, 210 O, o Big O and little o notation 42, 62 Ck Functions that are k times dif- ferentiable 44 f ∗ g Convolution 44, 139, 184, 239 σN , σN (f) Cesàro mean 52, 53 FN , FR Fejér kernel 53, 163 A(r), Ar(f) Abel mean 54, 55 χ[a,b] Characteristic function 61 f(θ+), f(θ−) One-sided limits at jump dis- continuities 63 Rd, Cd Euclidean spaces 71 X ⊥ Y Orthogonal vectors 72 `2(Z) Square summable sequences 73 R Riemann integrable functions 75 ζ(s) Zeta function 98 [x], 〈x〉 Integer and fractional parts 106 4N , σN,K , 4̃N Delayed means 114, 127, 174 Ht, Ht, H(d)t Heat kernels 120, 146, 209 M(R) Space of functions of moderate decrease on R 131 303 Ibookroot October 20, 2007 304 SYMBOL GLOSSARY f̂(ξ) Fourier transform 134, 181 S,S(R),S(Rd) Schwartz space 134, 180 R2+, R2+ Upper half-plane and its closure 149 ϑ(s), Θ(z|τ) Theta functions 155, 156 Γ(s) Gamma function 165 ‖x‖, |x|; (x, y), x · y Norm and inner product in Rd 71, 176 xα, |α|, ( ∂ ∂x )α Monomial, its order, and differential operator 176 S1, S2, Sd−1 Unit circle in R2, and unit spheres in R3, Rd 179, 180 Mt, M̃t Spherical mean 194, 216 Jn Bessel function 197, 213 P, Pt,γ Plane 202 R, R∗ Radon and dual Radon transforms 203, 205 Ad, Vd Area and volume of the unit sphere in Rd 208 Z(N) Group of N th roots of unity 219 Z/NZ Group of integers modulo N 221 G, |G| Abelian group and its or- der 226, 228 G ≈ H Isomorphic groups 227 G1 ×G2 Direct product of groups 228 Z∗(q) Group of units modulo q 227, 229, 244 Ĝ Dual group of G 231 a|b a divides b 242 gcd(a, b) Greatest common divisor of a and b 242 ϕ(q) Number of integers rela- tively prime to q 254 χ, χ0 Dirichlet character, and trivial Dirichlet character 254 L(s, χ) Dirichlet L-function 256 log1 ( 1 1−z ) , log2 L(s, χ) Logarithms 258, 264 d(k) Number of positive divi- sors of k 269 Ibookroot October 20, 2007 Index Relevant items that also arise in Book I are listed in this index, preceeded by the numeral I. Abel means, 54 summable, 54 abelian group, 226 absolute value, 23 absorption coefficient, 199 amplitude, 3 annihilation operator, 169 approximation to the identity, 49 attenuation coefficient, 199 Bernoulli numbers, 97, 167 polynomials, 98 Bernstein’s theorem, 93 Bessel function, 197 Bessel’s inequality, 80 best approximation lemma, 78 Black-Scholes equation, 170 bump functions, 162 Cauchy problem (wave equa- tion), 185 Cauchy sequence, 24 Cauchy-Schwarz inequality, 72 Cesàro means, 52 sum, 52 summable, 52 character, 230 trivial (unit), 230 class Ck, 44 closed rectangle, 289 complete vector space, 74 complex conjugate, 23 exponential, 24 congruent integers, 220 conjugate Dirichlet kernel, 95 convolution, 44, 139, 239 coordinates spherical in Rd, 293 creation operator, 169 curve, 102 area enclosed, 103 closed, 102 diameter, 125 length, 102 simple, 102 d’Alembert’s formula, 11 delayed means, 114 generalized, 127 descent (method of), 194 dilations, 133, 177 direct product of groups, 228 Dirichlet characters, 254 complex, 265 real, 265 Dirichlet kernel conjugate (on the circle), 95 modified (on the real line), 165 on the circle, 37 Dirichlet problem 305 Ibookroot October 20, 2007 306 INDEX annulus, 64 rectangle, 28 strip, 170 unit disc, 20 Dirichlet product formula, 256 Dirichlet’s test, 60 Dirichlet’s theorem, 128 discontinuity jump, 63 of a Riemann integrable function, 286 divisibility of integers, 242 divisor, 242 greatest common, 242 divisor function, 269, 280 dual X-ray transform, 212 group, 231 Radon transform, 205 eigenvalues and eigenvectors, 233 energy, 148, 187 of a string, 90 equidistributed sequence, 107 ergodicity, 111 Euclid’s algorithm, 241 Euler constant γ, 268 identities, 25 phi-function, 254, 276 product formula, 249 even function, 10 expectation, 160 exponential function, 24 exponential sum, 112 fast Fourier transform, 224 Fejér kernel on the circle, 53 on the real line, 163 Fibonacci numbers, 122 Fourier coefficient (discrete), 236 coefficient, 16, 34 on Z(N), 223 on a finite abelian group, 235 series, 34 sine coefficient, 15 Fourier inversion finite abelian group, 235 on Z(N), 223 on R, 141 on Rd, 182 Fourier series, 34, 235 Abel means, 55 Cesàro mean, 53 delayed means, 114 generalized delayed means, 127 lacunary, 114 partial sum, 35 uniqueness, 39 Fourier series convergence mean square, 70 pointwise, 81, 128 Fourier series divergence, 83 Fourier transform, 134, 136, 181 fractional part, 106 function Bessel, 197 exponential, 24 gamma, 165 moderate decrease, 131, 179, 294 radial, 182 rapidly decreasing, 135, 178 sawtooth, 60, 83 Schwartz, 135, 180 theta, 155 zeta, 98 gamma function, 165 Ibookroot October 20, 2007 INDEX 307 Gaussian, 135, 181 Gibbs’s phenomenon, 94 good kernels, 48 greatest common divisor, 242 group (abelian), 226 cyclic, 238 dual, 231 homomorphism, 227 isomorphic, 227 of integers modulo N , 221 of units, 227, 229, 244 order, 228 Hölder condition, 43 harmonic function, 20 mean value property, 152 harmonics, 13 heat equation, 20 d-dimensions, 209 on the real line, 146 steady-state, 20 time-dependent, 20 heat kernel d-dimensions, 209 of the real line, 146, 156 of the circle, 120, 124, 156 Heisenberg uncertainty princi- ple, 158, 168, 209 Hermite functions, 173 operator, 168 Hermitian inner product, 72 Hilbert space, 75 homomorphism, 227 Hooke’s law, 2 Huygens’ principle, 193 hyperbolic cosine and sine func- tions, 28 hyperbolic sums, 269 inner product, 71 Hermitian, 72 strictly positive definite, 71 integer part, 106 integrable function (Riemann), 31, 281 inverse of a linear operator, 177 isoperimetric inequality, 103, 122 jump discontinuity, 63 Kirchhoff’s formula, 211 L-function, 256 Landau kernels, 164 Laplace operator, 20 Laplacian, 20, 149, 185 polar coordinates, 27 Legendre expansion, 96 polynomial, 95 light cone backward, 193, 213 forward, 193 linearity, 6, 22 Lipschitz condition, 82 logarithm log1, 258 log2, 264 mean value propery, 152 measure zero, 287 moderate decrease (function), 131 modulus, 23 monomial, 176 multi-index, 176 multiplication formula, 140, 183 natural frequency, 3 Newton’s law, 3 of cooling, 19 nowhere differentiable function, 113, 126 Ibookroot October 20, 2007 308 INDEX odd function, 10 order of a group, 228 orthogonal elements, 72 orthogonality relations, 232 orthonormal family, 77 oscillation (of a function), 288 overtones, 6, 13 parametrization arc-length, 103 reverse, 121 Parseval’s identity, 79 finite abelian group, 236 part fractional, 106 integer, 106 partition of a rectangle, 290 of an interval, 281 period, 3 periodic function, 10 periodization, 153 phase, 3 Plancherel formula on Z(N), 223 on R, 143 on Rd, 182 Planck’s constant, 161 plucked string, 17 Poincaré’s inequality, 90 Poisson integral formula, 57 Poisson kernel d-dimensions, 210 on the unit disc, 37, 55 on the upper half-plane, 149 Poisson kernels,comparison, 157 Poisson summation formula, 154–156, 165, 174 polar coordinates, 179 polynomials Bernoulli, 98 Legendre, 95 pre-Hilbert space, 75 prime number, 242 primes in arithmetic progres- sion, 245, 252, 275 probability density, 160 profile, 5 pure tones, 6 Pythagorean theorem, 72 Radon transform, 200, 203 rapid decrease, 135 refinement (partition), 281, 290 relatively prime integers, 242 repeated integrals, 295 Reuleaux triangle, 125 Riemann integrable function, 31, 281, 290 Riemann localization principle, 82 Riemann-Lebesgue lemma, 80 root of unity, 219 rotation, 177 improper, 177 proper, 177 sawtooth function, 60, 83, 84, 94, 99, 278 scaling, 8 Schwartz space, 134, 180 separation of variables, 4, 11 simple harmonic motion, 2 space variables, 185 spectral theorem, 233 commuting family, 233 speed of propagation (finite), 194 spherical coordinates in Rd, 293 mean, 189 wave, 210 spring constant, 3 standing wave, 4 Ibookroot October 20, 2007 INDEX 309 subordination principle, 210 subrectangle, 290 summable Abel, 54 Cesàro, 52 summation by parts, 60 superposition, 6, 14 symmetry-breaking, 83 theta function, 155 functional equation, 155 tone fundamental, 13 pure, 11 translations, 133, 177 transpose of a linear operator, 177 traveling wave, 4 trigonometric polynomial, 35 degree, 35 series, 35 Tychonoff’s uniqueness theo- rem, 172 uncertainty, 160 unit, 229 unitary transformation, 143, 233 variance, 160 vector space, 70 velocity of a wave, 5 velocity of the motion, 7 vibrating string, 90 wave standing, 4, 13 traveling, 4 velocity, 5 wave equation, 184 d-dimensional, 185 d’Alembert’s formula, 11 is linear, 9 one-dimensional, 7 time reversal, 11 Weierstrass approximation theo- rem, 54, 63, 144, 163 Weyl criterion, 112, 123 estimate, 125 theorem, 107 Wirtinger’s inequality, 90, 122 X-ray transform, 200 zeta function, 98, 155, 166, 248

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