程序代写代做代考 Homework 5 Solutions

Homework 5 Solutions

Chapter 4, Exercise 11

We will work over the interval [−π, π] rather than [− 1
2
, 1
2
]. This slightly changes

the formula for the heat kernel because e−4π
2n2t becomes e−n

2t and e2iπnx be-
comes einx (so the notation is more convenient but the constants change).

Note that Ht ∗ f − f has Fourier coefficients f̂(n)(e−n
2t − 1). Thus by Par-

seval
1

∫ π
−π

(
Ht ∗ f(x)− f(x)

)2
dx =


n∈Z
|e−n

2t − 1|2|f̂(n)|2. (1)

We want to show that the right-hand side tends to 0 as t → 0. For this, we
present two arguments.

As t → 0, it is clear that |e−n
2t − 1| → 0 for all n ∈ Z. Fixing � > 0, we can

truncate the series at some finite value N = N(�) such that

|n|≥N |f̂(n)|

2 < �, then take the limit as t → 0 for the finite number of remaining terms (see the proof of Exercise 5.1b below which is very similar and more explicit). Less explicitly, this is basically an application the dominated convergence theorem applied to counting measure on Z with dominating function g(n) = |f̂(n)|2. For the second proof, one may assume first that f is a trigonometric polynomial. Then using the identity |e−q − 1| ≤ q for q > 0, we see∑

n∈Z
|e−n

2t − 1|2|f̂(n)|2 ≤ t2

n∈Z

n4|f̂(n)|2 t→0−→ 0.

Now one may use the fact that any Riemann-integrable function can be approx-
imated arbitrarily closely in the mean-square-norm by trigonometric polynomi-
als, together with a standard �/3-trick and the fact that convolution with the
heat kernel contracts the mean-square-norm (by Parseval).

Remark: Assuming Corollary 5.3.4 in the next chapter: {Ht} is a family of
good kernels as t→ 0, from which the result may be deduced more directly.

1

Chapter 4, Exercise 13

Part a: Again, we work on [−π, π] rather than [−1/2, 1/2], so our constants
will be different than those in the book. Noting that Ĥt(n) = e

−n2t, we see by
Parseval that

1

∫ π
−π
|Ht(x)|2dx =


n∈Z

e−2n
2t.

Hence

t1/2
∫ π
−π
|Ht(x)|2dx = 2πt1/2


n∈Z

e−2n
2t = 2πδ


k∈δZ

e−2k
2

,

where δ := t1/2. Now by the result of Exercise 1b in Chapter 5 (see below), the

RHS converges as δ → 0 to some nonzero constant C = 2π

R e
−2u2du.

Part b: We use the given hint that x2 ≤ C sin2(x/2). Then we fix a t > 0, and we
define f(x) = sin2(x/2)Ht(x) = (1− cosx)Ht(x) and also define g(x) = Ht(x).
So we have∫ π

−π
x2Ht(x)

2dx ≤ C
∫ π
−π

sin2(x/2)Ht(x)
2dx = C

∫ π
−π

f(x)g(x)dx.

By Parseval, 1

∫ π
−π f(x)g(x)dx =


n∈Z f̂(n)ĝ(n). It is clear that ĝ(n) = e

−n2t.
Using the fact that the Fourier coefficients of a product of two functions is
the convolution of the respective Fourier coefficients, one may see that f̂(n) =

e−n
2t − 1

2
e−(n+1)

2t − 1
2
e−(n−1)

2t. Now recall summation by parts: if (an) is
square-summable then∑

n∈Z
an

(
an −

1

2
an+1 −

1

2
an−1

)
=

1

2


n∈Z

(
an+1 − an

)2
.

Applying this to an = e
−n2t, we find that∑

n∈Z
f̂(n)ĝ(n) =

1

2


n∈Z

(
e−(n+1)

2t − e−n
2t
)2
.

Letting F (x) = e−x
2t, we see that F ′(x) = −2xte−x

2t, and so by the mean value
theorem, we can always find constants xn,t ∈ [n, n + 1] such that F (n + 1) −
F (n) = F ′(xn,t) = −2xn,tte−x

2
n,tt. Summarizing, we have shown that for any

t > 0, there exist points xn,t ∈ [n, n+ 1] such that∫ π
−π

x2Ht(x)
2dx ≤ C


n∈Z

x2n,tt
2e−2x

2
n,tt.

Multiplying by t−1/2, we see that

t−1/2
∫ π
−π

x2Ht(x)
2 ≤ Ct1/2


n∈Z

(
t1/2xn,t

)2
e
−2
(
t1/2xn,t

)2
.

2

Since xn,t ∈ [n, n+ 1], a close inspection of the right-hand side shows that it is
basically a Riemann sum approximation of


R u

2e−2u
2

du, with mesh size t1/2.
Thus as t → 0, the right side converges to that integral (which can be made
rigorous by adapting the proof of Exercise 1b in Chapter 5 below, or just by
applying the dominated convergence theorem in a slightly clever way). This
proves the desired upper bound, but we remark that a lower bound may also be
proved in exactly the same way by noting that x2 ≥ c sin2(x/2) for x ∈ [−π, π].
Hence t1/2 is really the true order of this integral as t→ 0.

Chapter 5, Exercise 1

Part a: Since f̂ is of moderate decrease, it follows that

n∈Z |f̂(n/L)| <∞ for any L > 0. This means that

n |an(L)| < +∞. Noting that an(L) is just the Fourier coefficient of f viewed as a function on the bounded interval [−L/2, L/2], we can apply Corollary 2.3 in chapter 2 to see that ∑ n an(L)e 2πinx/L actually converges uniformly to f on [−L/2, L/2]. Part b: Suppose F is continuous and |F (x)| ≤ A(1 + x2)−1. Let � > 0, and set
M = M(�) := 4A�−1. Then we see that∫

{|x|>M}
|F (x)|dx ≤ 2A

∫ ∞
M

x−2dx = �/2,

and similarly, for any δ > 0,

δ

|n|>1+M/δ

|F (δn)| ≤ 2Aδ

n≥1+M/δ

δ−2n−2 ≤ 2Aδ−1
∫ ∞
M/δ

x−2dx = �/2.

Noting that F is uniformly continuous on [−M,M ], it holds that

lim
δ→0

δ

|n|≤1+M/δ

F (δn) =


{|x|≤M}

F (x)dx.

Now by the triangle inequality,∣∣∣∣δ∑
n∈Z

F (δn)−

R
F (x)dx

∣∣∣∣ ≤ �/2 + �/2 +
∣∣∣∣δ ∑
|n|≤1+M/δ

F (δn)−

{|x|≤M}

F (x)dx

∣∣∣∣.
Now letting δ → 0 on both sides (while keeping �,M fixed) shows that

lim sup
δ→0

∣∣∣∣δ∑
n∈Z

F (δn)−

R
F (x)dx

∣∣∣∣ ≤ �.
As � was arbitrary, we are done.

Part c: This follows immediately by letting δ → 0 (i.e., L → ∞) in the for-
mula from Part a, then applying the result of Part b.

3

Chapter 5, Exercise 2

We have

f̂(ξ) =


R
f(x)e−2πiξxdx =

∫ 1
−1
e−2πiξxdx = −

1

2πiξ
(e−2πiξ − e2πiξ) =

sin(2πξ)

πξ
.

Now a quick computation reveals that (f ∗f)(x) = 2g(x/2), which means g(x) =
1
2
(f ∗ f)(2x), hence

ĝ(ξ) =
1

2

[
1

2
f̂(ξ/2)2

]
=

(
sin(πξ)

πξ

)2
.

Chapter 5, Exercise 8

Let g(x) = e−x
2

. Under the given condition on f , we see that

(f ∗ g)(x) =

R
f(y)e−(x−y)

2

dy = e−x
2


R
f(y)e−y

2

e2xydy = e−x
2

· 0 = 0.

Thus,
1

2
e−ξ

2/4f̂(ξ) = f̂(ξ)ĝ(ξ) = (̂f ∗ g)(ξ) = 0.

Hence f̂(ξ) = 0 for all ξ, and we conclude that

f(x) =


R
f̂(ξ)e2πiξxdξ = 0.

4