Shakarchi and Stein Measure Theory Solutions Chapter 3 Exercise 4
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Fourier Analysis, Stein and Shakarchi
Chapter 3 Convergence of Fourier Series
Yung-Hsiang Huang∗
2018.03.21
Abstract
T denotes [−π, π] or [−12 , 1 2 ]. Some estimates may differ a constant multiple from the real
situation because the author is familiar with the Fourier coefficients f̂(n) := ∫ 1
2
− 1 2
f(x)e−2πinx dx
which is different from this textbook.
Note that Exercise 16 (Bernstein's Theorem) contains an extended discussion (mainly on
the counterexamples for Hölder exponent α = 12) from page 8 to page 15. We also discuss a
Poincaré inequality on 2D rectangle in Exercise 11.
Exercises
1. We omit the proofs for showing that Rd and Cd are complete.
2. Prove that the vector space l2(Z) is complete.
Proof. Suppose Ak = {ak,n}n∈Z with k = 1, 2, · · · is a Cauchy sequence. For each n, {ak,n}k=1 is a Cauchy sequence of complex numbers since |ak,n − ak′,n| ≤ ‖Ak − A′k‖l2 , therefore it
converges to a limit, say bn. Given � > 0, there is N� ∈ N such that for all k, k′ ≥ N�,∑M n=1 |ak,n − ak′,n|2 ≤ ‖Ak − Ak′‖2l2 <
�2
2 for all M . For each k > N� and M , we let k
′ → ∞
to get ∑M
n=1 |ak,n − bn|2 < �2. Since M is arbitrary, one has shown that ‖Ak − B‖l2 < � as
k ≥ N�, where B = (· · · , b−1, b0, b1, · · · ). Also, B ∈ l2(Z) by Minkowski's inequality ‖B‖l2 ≤
‖B − AN�‖l2 + ‖AN�‖l2
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3. It's standard to 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. For example, take {χIk}k, the first interval
I1 be [0,1], the next two be the two halves of [0,1], and the next four be the four
quarters, and so on. Note that there is always a subsequence {fkj} such that fkj → 0
almost everywhere.
4. Recall the vector space R of Riemann-integrable functions, with its inner product
and norm
‖f‖ = ( 1
2π
∫ 2π 0
|f(x)|2 dx )1/2
.
(a) Then there exist non-zero integrable functions f for which ‖f‖ = 0, e.g. χ{π}.
(b) However, one can use the proof by contradiction to show if f ∈ R with ‖f‖ = 0,
then f(x) = 0 whenever f is continuous at x.
(c) Conversely, by using the Lebesgue criteria, one can show if f ∈ R vanishes at
all of its points of continuity, then the lower Riemann sum of |f |2 is 0 and hence
‖f‖ = 0.
5. Let f(θ) = log(1/θ) for 0 < θ ≤ 2π and f(0) = 0. Define a sequence of functions in
R by fn(θ) = χ( 1 n ,2π]f(θ). Then it's easy to show {fn}∞n=1 is a Cauchy sequence in R.
However, f does not belong to R since it's unbounded.
6. Let {ak}∞k=−∞ be a sequence defined by ak = k−1 if k ≥ 1 and ak = 0 if k ≤ 0. Note that
{ak} ∈ `2(Z), but one can show that no Riemann integrable function (or more generally L∞(T))
has k-th Fourier coefficient equal to ak for all k by the same proof as page 83-84. However, it's
known to be a Fourier coefficient of some L2(T) function by the L2 convergence of
the series and completeness of L2(T). Check Planchel's theorem and [13, Theorem
8.30]
7. By using the Abel-Dirichlet test and Planchel's theorem, one can show that the
trigonometric series ∑ n≥2
1
log n sinnx
converges for every x, yet it is not the Fourier series of a L2(T) function. However,
it's actually not he Fourier series of a L1(T) function, see [5, Section 14.I], especially
2
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(c)(iii) ⇒ (i), which we will provide a proof here. Also see its following remark for
cosine series.
The same method can be applied to ∑
sinnx nα
for 0 < α ≤ 1 2 . For the case 1/2 < α < 1
is more difficult to show it's not a Fourier series of a Riemann integrable function.
See Problem 1.
Proof. The most difficult part is to find a corrected start line. Suppose ∑∞
n=1 bn sinnx ∼ f(x)
for some f ∈ L1(T). We will prove that ∑∞
n=1 bn n
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If N ≤ 1 x
(which means x small and hence we don't want to use the cosectant estimate), then
we pick m = N and obtain SN(x) ≤ Nx ≤ 1.
If N > 1 x , then we choose m to be the unique integer satisfying 1
x − 1 < m ≤ 1
x . (m = 0 means
x is large, so we just use the cosectant estimate.) Then the upper bound becomes
|SN(x)| ≤ 1 + x2
2π ≤ 1 + π
2 .
Therefore we complete the proof.
8. Exercise 6 in Chapter 2 dealt with the sums
∞∑ n=0
1
(2n+ 1)2 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|. Using Parseval's identity one can
find the sums of the following two identity:
∞∑ n=0
1
(2n+ 1)4 = π4/96 and
∞∑ n=1
1
n4 = π4/90.
(b) Consider the 2π-periodic odd function defined on [0, π] by f(θ) = θ(π− θ). As (a), one can
find that ∞∑ n=0
1
(2n+ 1)6 =
π6
960 and
∞∑ n=1
1
n6 =
π6
945 .
Remark 1. . The general expression when k is even for ∑∞
n=1 1/n k in terms of πk is given in
Problem 4. However, finding a formula for the sum ∑∞
n=1 1/n 3, or more generally
∑∞ n=1 1/n
k
with k odd, is a famous unresolved question.
9. 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 one see that
∞∑ n=−∞
1
(n+ α)2 =
π2
(sinπα)2 .
Also one notes that Problem 2.3 (or Jordan's test) implies that ∑∞
n=−∞ 1
n+α = π
tanπα .
Other method to construct these identities are given in Exercise 5.15 and Book II's Exercise
3.12.
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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 ∈ C0. 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.
Proof. dE/dt ≡ 0 by wave equation. To passing d/dt inside the integral, maybe this problem
needs stronger assumptions on smoothness of f, g.
11. The inequalities of Wirtinger and Poincaŕe establish a relationship between the
norm of a function and that of its derivative.
(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 ).
(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,
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Discuss the case of equality, and prove that the constant (b−a) 2
π2 cannot 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)].
Remark 2. The Poincaré inequality, Sobolev inequality and isoperimetric inequality (intro-
duced in Section 4.1 and Book III's Section 3.4) are closely related.
Proof. (a) By hypothesis, f̂(0) = 0 and f̂ ′(n) exists for all n ∈ N and equals to 2πin T f̂(n) where
f̂(n) := 1 T
∫ T 0 f(x)e−
2πin T
x dx. Then Parseval's identity implies that∫ T 0
|f(t)|2 dt = T ∑ n6=0
|f̂(n)|2 = T ∑ n6=0
T 2
4π2n2 |f̂ ′(n)|2 ≤ T
∑ n6=0
T 2
4π2 |f̂ ′(n)|2 = T
2
4π2
∫ T 0
|f ′(x)|2 dx.
with equality if and only if f̂(n) = 0 whenever n 6= ±1, that is, f(x) is a linear combination o
Shakarchi and Stein Measure Theory Solutions Chapter 3 Exercise 4
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