Homework 2 Solutions
Proof of Riemann-Lebesgue
We first consider step functions f =
∑m
k=1 ck1Ik , where the Ik = [ak, bk) are
pairwise-disjoint subintervals of [−π, π], the ck are some real numbers, and 1A
denotes the indicator function of the set A (which means 1A(x) = 0 for x /∈ A
and 1A(x) = 1 for x ∈ A). In this case we compute
2πf̂(n) =
∫ π
−π
f(x)e−inxdx =
m∑
k=1
ck
∫ bk
ak
e−inxdx =
m∑
k=1
ck
e−inak − e−inbk
in
.
Noting that |einak − einbk | ≤ |einak |+ |einbk | = 2 and using the triangle inequal-
ity in the above finite sum, we find that |2πf̂(n)| ≤ 2
n
∑m
k=1 |ck|, which clearly
tends to 0 as |n| → ∞. Note here that m is not related to n in any way.
Next, we use the given fact to extend the convergence to all f which are in-
tegrable. Let f be integrable, and take some � > 0. Then there exists some step
function g� ≤ f such that 12π
∫ π
−π(f − g�) ≤ �. Consequently, we find that
|f̂(n)| =
∣∣∣∣ 12π
∫ π
−π
(f(x)− g�(x))e−inxdx+
1
2π
∫ π
−π
g�(x)dx
∣∣∣∣
≤
1
2π
∫
|f(x)− g�(x)|dx+ |ĝ�(n)|
≤ �+ |ĝ�(n)|.
Letting |n| → ∞ on both sides and noting that g� is a step function, we notice
that lim sup|n|→∞ |f̂(n)| ≤ �. Since � is arbitrary, this gives the desired result.
The proof extends to complex-valued f by considering real and imaginary parts
separately.
1
Chapter 2, Problem 2
Noting that | sin(x/2)| ≤ |x/2| and making the substitution u = (N + 1
2
)π,
we have ∫ π
−π
|DN (x)|dx = 2
∫ π
0
|DN (x)|dx
= 2
∫ π
0
| sin((N + 1
2
)x)|
| sin(x/2)|
dx
≥ 2
∫ π
0
| sin((N + 1
2
)x)|
|x/2|
dx
= 4
∫ (N+ 1
2
)π
0
| sin(u)|
u
du
≥ 4
N−1∑
k=0
∫ (k+1)π
kπ
| sinu|
u
du.
Now noting that 1
u
≤ 1
(k+1)π
for u ∈ [kπ, (k + 1)π] and then noting that∫ (k+1)π
kπ
| sinu|du = 2, we see that the last expression is bounded below by
4
N−1∑
k=0
1
(k + 1)π
∫ (k+1)π
kπ
| sinu|du =
8
π
N−1∑
k=0
1
k + 1
.
Now we take for granted that the partial sums of the harmonic series are asymp-
totic to logN , finishing the proof. If we multiply by 1
2π
then we get the constant
c = 4
π2
mentioned in the problem.
2
Chapter 3, Exercise 12
We have that
2π =
∫ π
−π
DN (t)dt =
∫ π
−π
sin((N+1/2)t)
(
1
sin(t/2)
−
1
t/2
)
dt+
∫ π
−π
sin((N + 1
2
)t)
t/2
dt.
Let us call these two terms on the right hand side as AN and BN , respec-
tively. Then letting f(t) = 1
sin(t/2)
− 1
t/2
, which is an odd function that extends
continuously to 0, we have
AN =
∫ π
−π
sin((N + 1/2)t)f(t)dt
= −i
∫ π
−π
ei(N+
1
2
)tf(t)dt
= −i
∫ π
−π
eiNt
[
e
1
2
itf(t)
]
dt
which tends to 0 as n → ∞ by the Riemann-Lebesgue lemma (applied to the
function t 7→ e
1
2
itf(t)).
Now for BN , we substitute u = (N +
1
2
)t to obtain that
BN = 2
∫ π
0
sin((N + 1
2
)t)
t/2
dt
= 4
∫ (N+ 1
2
)π
0
sinu
u
du.
Combining the results of the past several paragraphs, we find that
lim
N→∞
∫ (N+ 1
2
)π
0
sinu
u
du = lim
N→∞
1
4
BN = lim
N→∞
1
4
(2π −AN ) =
π
2
.
The proof is finished by noting that if |x− (N + 1
2
)π| ≤ π
2
, then x ≥ Nπ, so
∣∣∣∣
∫ (N+ 1
2
)π
x
sinu
u
du
∣∣∣∣ ≤ 12N .
3
Chapter 3, Exercise 13
Let f ∈ Ck(T), and let g := f (k). From Homework 1 (or directly by repeated
integration-by-parts), we know that ĝ(n) = iknkf̂(n). Since g is integrable (in
fact continuous) by assumption, we also know from Riemann-Lebesgue that
ĝ(n)→ 0, hence |nkf̂(n)| = |i−kĝ(n)| → 0 as |n| → ∞ (since |i−k| = 1).
Chapter 3, Exercise 15
Part a: Fixing n 6= 0 and making the substitution θ = x+ π
n
, we find that
2πf̂(n) =
∫ π
−π
f(θ)e−inθdθ =
∫ π+π/n
−π−π/n
f(x+ π/n)e−in(x+
π
n
)dx.
Now using the result of (the second part of) Exercise 1 of Chapter 2 with
a = −π/n, and then noting that e−in(x+
π
n
) = −e−inx, we find that the last
expression is just∫ π
−π
f(x+ π/n)e−in(x+
π
n
)dx = −
∫ π
−π
f(x+ π/n)e−inxdx,
proving the first part. For the second part, note that
4πf̂(n) = 2πf̂(n) + 2πf̂(n)
=
∫ π
−π
f(x)e−inxdx−
∫ π
−π
f(x+ π/n)e−inxdx
=
∫ π
−π
[
f(x)− f(x+ π/n)
]
e−inxdx.
Part b: If |f(x)−f(y)| ≤ C|x−y|α, then clearly |f(x)−f(x+π/n)| ≤ C(π/n)α =
C ′n−α, and therefore
2π|f̂(n)| ≤
∫ π
−π
∣∣[f(x)− f(x+ π/n)]e−inx|dx ≤ 2πC ′n−α.
Part c: We write
f(x)− f(y) =
∑
2k<|x−y|−1 2−kα(ei2 kx − ei2 ky) + ∑ 2k≥|x−y|−1 2−kα(ei2 kx − ei2 ky). Let’s call the terms on the right-hand side as A and B respectively. Note that these are functions of x and y. We define the integer quantity K := min{k ∈ N : 2k ≥ |x− y|−1} = d− log2 |x− y|e, 4 which is also a function of x and y. To solve the problem, the first observation we make is that if |z|, |w| ≤ 1 then |e2 kz − e2 kw| ≤ 2k|z − w|, which is given as a hint. In particular, applying this to A and applying a geometric sum identity gives A ≤ ∑ 2k<|x−y|−1 2−kα|ei2 kx − ei2 ky| ≤ ∑ 2k<|x−y|−1 2(1−α)k|x− y| = 2(1−α)K − 1 21−α − 1 |x− y|. Using the fact that 2K = 2 · 2K−1 ≤ 2|x − y|−1, we see that 2(1−α)K − 1 ≤ 2(1−α)K ≤ 21−α|x− y|α−1. So the last expression is bounded above by 21−α|x− y|α−1 21−α − 1 |x− y| = 21−α 21−α − 1 |x− y|α, which proves the desired bound for A. Now for B, we use that |ei2 kx−ei2 ky| ≤ 2 and 2−K ≤ |x− y| so that B ≤ ∑ 2k≥|x−y|−1 2−kα = 2−αK 1− 2−α ≤ 1 1− 2−α |x− y|α. which proves the claim. 5