Solutions to some exercises and problems
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Solutions to some exercises and problems Teck-Cheong Lim Department of Mathematical Sciences George Mason University 4400, University Drive Fairfax, VA 22030 U.S.A. e-mail address: [email protected] Abstract Solutions to some exercises and problems from Stein and Shakarchi’s Fourier Analysis. The book by Y. Ketznelson, ”An introduction of Har- monic Analysis” (2nd corrected edition) is referred to frequently. Chapter 1: The Genesis of Fourier Analysis Chapter 2: Basic Properties of Fourier Series Chapter 3: Convergence of Fourier Series Chapter 4: Some applications of Fourier Series Chapter 5: The Fourier transform on R Chapter 6: The Fourier transform on R d Chapter 7: Finite Fourier Analysis Chapter 8: Dirichlet’s Theorem Chapter 1 The Genesis of Fourier Analysis 1. (Exercise 8) Suppose F is a function on (a, b) with two continuous deriva- tives. Show that whenever x and x + h belong to (a, b), one may write F (x + h)= F (x)+ hF 0 (x)+ h 2 2 F 00 (x)+ h 2 φ(h), where φ(h) → 0 as h → 0. Deduce that F (x + h)+ F (x - h) - 2F (x) h 2 → F 00 (x) as h → 0. Proof. Firstly one has F (x + h) - F (x)= Z x+h x F 0 (y) dy. 1
Transcript of Solutions to some exercises and problems
Solutions to some exercises and problems
Teck-Cheong LimDepartment of Mathematical Sciences
George Mason University4400, University DriveFairfax, VA 22030
U.S.A.e-mail address: [email protected]
Abstract
Solutions to some exercises and problems from Stein and Shakarchi’sFourier Analysis. The book by Y. Ketznelson, ”An introduction of Har-monic Analysis” (2nd corrected edition) is referred to frequently.Chapter 1: The Genesis of Fourier AnalysisChapter 2: Basic Properties of Fourier SeriesChapter 3: Convergence of Fourier SeriesChapter 4: Some applications of Fourier SeriesChapter 5: The Fourier transform on RChapter 6: The Fourier transform on Rd
Chapter 7: Finite Fourier AnalysisChapter 8: Dirichlet’s Theorem
Chapter 1 The Genesis of Fourier Analysis
1. (Exercise 8) Suppose F is a function on (a, b) with two continuous deriva-tives. Show that whenever x and x+ h belong to (a, b), one may write
F (x+ h) = F (x) + hF ′(x) +h2
2F ′′(x) + h2φ(h),
where φ(h)→ 0 as h→ 0.Deduce that
F (x+ h) + F (x− h)− 2F (x)
h2→ F ′′(x) as h→ 0.
Proof.Firstly one has
F (x+ h)− F (x) =
∫ x+h
x
F ′(y) dy.
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Then F ′(y) = F ′(x)+(y−x)F ′′(x)+(y−x)ψ(y−x), where ψ is continuousand ψ(h)→ 0 as h→ 0. Therefore
F (x+ h)− F (x) = F ′(x)h+ F ′′(x)h2
2+
∫ h
0
uψ(u) du.
By mean-value theorem∫ h
0
uψ(u) du = ψ(ξ)
∫ h
0
u du = h2ψ(ξ)
2= h2φ(h)
where ξ is between 0 and h, and φ(h) → 0 as h → 0 by the continuity ofψ.From above, one easily gets
F (x+ h) + F (x− h)− 2F (x)
h2− F ′′(x) = φ(h) + φ(−h)→ 0 as h→ 0.
Remark. Suppose F is periodic. The function ζ(x, h) = F (x+h)+F (x−h)−2F (x)h2 −
F ′′(x) is continuous for all x, h. And there exists M such that
|F (x+ h) + F (x− h)− 2F (x)| ≤Mh2
for all x, h (since F, F ′′, ζ are all periodic in x).
2. (Exercise 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 .
Solution.From x = r cos θ, y = r sin θ, we get(
∂x∂r
∂x∂θ
∂y∂r
∂y∂θ
)=
(cos θ −r sin θsin θ r cos θ
),
hence (∂r∂x
∂r∂y
∂θ∂x
∂θ∂y
)=
(cos θ −r sin θsin θ r cos θ
)−1=
(cos θ sin θ− 1r sin θ 1
r cos θ
).
2
By Chain Rule,
∂u
∂x=∂u
∂r
∂r
∂x+∂u
∂θ
∂θ
∂x
=∂u
∂rcos θ − 1
r
∂u
∂θsin θ.
By Chain Rule and Product Rule,
∂
∂x
(∂u
∂rcos θ
)=
∂
∂r
(∂u
∂rcos θ
)∂r
∂x+
∂
∂θ
(∂u
∂rcos θ
)∂θ
∂x
=∂2u
∂r2cos2 θ +
1
r
∂u
∂rsin2 θ − 1
r
∂2u
∂θ∂rcos θ sin θ
Similarly,
∂
∂x
(−1
r
∂u
∂θsin θ
)=
1
r2∂2u
∂θ2sin2 θ +
2
r2+
2
r2∂u
∂θsin θ cos θ − 1
r
∂2u
∂r∂θcos θ sin θ.
Since ∂2u∂r∂θ = ∂2u
∂θ∂r , we get
∂2u
∂x2=∂2u
∂r2cos2 θ+
1
r
∂u
∂rsin2 θ+
1
r2∂2u
∂θ2sin2 θ+
2
r2∂u
∂θsin θ cos θ−2
r
∂2u
∂r∂θcos θ sin θ.
Similarly,∂u
∂y=∂u
∂rsin θ +
1
r
∂u
∂θcos θ,
∂2u
∂y2=∂2u
∂r2sin2 θ+
1
r
∂u
∂rcos2 θ+
1
r2∂2u
∂θ2cos2 θ− 2
r2∂u
∂θsin θ cos θ+
2
r
∂2u
∂r∂θcos θ sin θ.
Consequently,
∂2u
∂x2+∂2u
∂y2=∂2u
∂r2+
1
r
∂u
∂r+
1
r2∂2u
∂θ2.
Assume the general case that u is complex-valued. Then∣∣∣∣∂u∂x∣∣∣∣2 =
∣∣∣∣∂u∂r∣∣∣∣2 cos2 θ +
1
r2
∣∣∣∣∂u∂θ∣∣∣∣2 sin2 θ − 2
r<{∂u
∂r
∂u
∂θcos θ sin θ}
and ∣∣∣∣∂u∂y∣∣∣∣2 =
∣∣∣∣∂u∂r∣∣∣∣2 sin2 θ +
1
r2
∣∣∣∣∂u∂θ∣∣∣∣2 cos2 θ +
2
r<{∂u
∂r
∂u
∂θcos θ sin θ},
hence ∣∣∣∣∂u∂x∣∣∣∣2 +
∣∣∣∣∂u∂y∣∣∣∣2 =
∣∣∣∣∂u∂r∣∣∣∣2 +
1
r2
∣∣∣∣∂u∂θ∣∣∣∣2 .
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3. (Added in) Let f(x) be an odd 2π-periodic function. For n ∈ Z define
cn =1
2π
∫ π
−πf(t)e−int dt
Prove thatc−ne−inx + cne
inx = An sinnx
where
An =2
π
∫ π
0
f(t) sinnt dt.
Proof.
c−ne−inx + cneinx
=1
2π
∫ π
−πf(t)einte−inx + f(t)e−inteinx dt
=1
2π
∫ π
−πf(t)(e−in(x−t) + ein(x−t)) dt
=1
2π
∫ π
−πf(t)2 cosn(x− t) dt
=1
π
∫ π
−πf(t)(cosnx cosnt+ sinnx sinnt) dt
=1
π
∫ π
−πf(t) sinnx sinnt dt (since f(t) cosnt is odd)
= sinnx2
π
∫ π
0
f(t) sinnt dt (since f(t) sinnt is even)
= An sinnx
4. (Exercise 11) Show that if n ∈ Z the only solutions of the differentialequation
r2F ′′(r) + rF ′(r)− n2F (r) = 0,
which are twice differentiable when r > 0, are given by linear combina-tions of rn and r−n when n 6= 0, and 1 and log r when n = 0.Solution.If F (r) solves the equation, write g(r) = F (r)/rn. Then g is twice differ-entiable, F (r) = rng(r) and
F ′(r) = nrn−1g(r) + rng′(r),
F ′′(r) = n(n− 1)rn−2g(r) + nrn−1g′(r) + nrn−1g′(r) + rng′′(r).
Then
r2F ′′(r)+rF ′(r)−n2F (r) = rn+1(rg′′(r)+(2n+1)g′(r)) = rn+1(2ng′(r)+(rg′(r))′) = 0,
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hence2ng(r) + rg′(r) = c.
It follows that g(r) is a linear combination of r−2n and 1 if n 6= 0, andlog r and 1 if n = 0. The result follows.
5. (Problem 1) Consider the Dirichlet problem illustrated in Figure 11. Moreprecisely, we look for a solution of the steady-state heat equation 4u = 0in the rectangle R = {(x, y) : 0 ≤ x ≤ π, 0 ≤ y ≤ 1} that vanishes on thevertical 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 onthe horizontal sides of the rectangle.Use separation of variables to show that if f0 and f1 have Fourier expan-sions
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 kAk +
sinh ky
sinh kBk
)sin kx.
Compare this result with the solution of the Dirichlet problem in the stripobtained in Problem 3, Chapter 5.Solution.Consideration of basic solution of the form A(x)B(y) yields
A(x) = sin kx, B(y) = ck sinh ky + dk cosh ky.
Setting y = 0, y = 1 and comparing coefficients give
dk = Ak, ck sinh k + dk cosh k = Bk.
Solving for dk, ck and using the identity sinh(a − b) = sinh a cosh b −sinh b cosh a, we find
ck sinh ky + dk cosh ky =sinh k(1− y)
sinh kAk +
sinh ky
sinh kBk.
Chapter 2
1. (Exercise 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−2πf(x) dx.
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Also prove that∫ π
−πf(x+ a) dx =
∫ π
−πf(x) dx =
∫ π+a
−π+af(x) dx.
Proof.For the first two identities, substitute u = x− 2π and u = x+ 2π respec-tively, and note that f(x ± 2π) = f(x). For the second equalities, firstly
we have∫ π−π f(x+a) dx =
∫ π+a−π+a f(x) dx by substitution u = x+a. Then
by applying the first set of equalities, we get∫ −π−π+a
f(x) dx =
∫ π
π+a
f(x) dx,
so∫ π+a
−π+af(x) dx =
∫ −π−π+a
f(x) dx+
∫ π
−πf(x) dx+
∫ π+a
π
f(x) dx =
∫ π
−πf(x) dx.
2. (Exercise 2) In this exercise we show how the symmetries of a functionimply certain properties of its Fourier coefficients. Let f be a 2π-periodicRiemann integrable function on R.(a) Show that the Fourier series of the function f can be written as
f(θ) ∼ f(0) +∑n≥1
[f(n) + f(−n)] cosnθ + i[f(n)− f(−n)] sinnθ.
(b) Prove that if f is even, then f(n) = f(−n), and we get a cosine series.
(c) Prove that if f is odd, then f(n) = −f(−n), and we get a sine series.
(d) Suppose that f(θ + π) = f(θ) for all θ ∈ R. Show that f(n) = 0 forall odd n.(e) Show that f is real-valued if and only if f(n) = f(−n) for all n.Proof.
f(n)eint + f(−n)e−int
= (f(n) + f(−n))1
2(eint + e−int) + i(f(n)− f(−n))
1
2i(eint − e−int)
= An cosnt+Bn sinnt
whereAn = f(n)+f(−n) = 12π
∫ 2π
0f(s)(e−ins+eins) ds = 1
π
∫ 2π
0f(s) cosns ds
andBn = i(f(n)−f(−n)) = i 12π
∫ 2π
0f(s)(e−ins−eins) ds = 1
π
∫ 2π
0f(s) sinns ds.
If f is real, then An, Bn are real, which is true iff f(n) = f(−n) for all n.If f is odd, then f(s) cosns is odd, so An = 0. If f is even, then f(s) sinnsis odd, so Bn = 0.If f(θ + π) = f(θ) for all θ ∈ R, then∫ 2π
0
f(t)e−int dt =
∫ π
0
f(t)e−int dt+
∫ 2π
π
f(t)e−int dt
6
and∫ 2π
π
f(t)e−int dt =
∫ π
0
f(u+ π)e−inue−inπ du = −∫ π
0
f(t)e−int dt
for all odd n.
3. (Exercise 4) Consider the 2π-periodic odd function defined on [0, π] byf(θ) = θ(π − θ).(a) Draw the graph of f .(b) Computer the Fourier coefficients of f , and show that
f(θ) =8
π
∑k odd ≥1
sin kθ
k3
Solution.Checked with Maple and found that it is correct.
4. (Exercise 6) Let f be the function defined on [−π, π] by f(θ) = |θ|.(a) Draw the graph of f .(b) Calculate the Fourier coefficients of f , and show that ... O(1/n2).(c) What is the Fourier series of f in terms of sines and cosines?(d) Taking θ = 0, prove that
∑n odd ≥1
1
n2=π2
8and
∞∑n=1
1
n2=π2
6.
Solution.Since it is an even function, by Exercise 2, f(n) = f(−n). We find f(n) +
f(−n) = 0, if n is even, and − 2πn2 if n is odd. So the series is
π
2− 4
π
∑n odd ≥1
cosnθ
n2.
Assuming that the series converges at θ = 0, we find the first part of (d).It is easy to see that
∑n even ≥1
1n2 = 1
4
∑∞n=1
1n2 . Combining this with
what we got for the odd sum, we get the second part of (d).
5. (Exercise 7) Suppose {an}Nn=1 and {bn}Nn=1 are two finite sequences of
complex numbers. Let Bk =∑kn=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.
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(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.
Proof. (a)
N∑n=M
anbn
=
N∑n=M
an(Bn −Bn−1)
=
N∑n=M
anBn −N∑
n=M
anBn−1
=
N∑n=M
anBn −N−1∑
n=M−1an+1Bn
= anBN +
N−1∑n=M
anBn − aMBM−1 −N−1∑n=M
an+1Bn
= aNBN − aMBM−1 −N−1∑n=M
(an+1 − an)Bn
(b) Suppose |Bn| ≤ B for all n. Then for N > M
|N∑
n=M
anbn| ≤ B(aN + aM +
N−1∑n=M
(an − an+1)) = 2BaM
proving that the series is Cauchy and hence converges.
6. (Exercise 8) Verify that 12i
∑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 < 0π/2− x/2 if 0 < x < π.
Note that this function is not continuous. Show that nevertheless, theseries converges for every x (by which we mean, as usual, that the sym-metric partial sums of the series converge). In particular, the value of theseries at the origin, namely 0, is the average of the values of f(x) as xapproaches the origin from the left and the right.Solution.Checked with Maple and found it correct. bn = sinnx,
Bn =
n∑k=1
bk =sin(nx/2) sin((n+ 1)x/2)
sin(x/2)
8
(to prove write sinnx as 12i (e
inx − e−inx))) and |Bn| ≤ csc(|x|/2) for|x| < π, x 6= 0. 1/n is decreasing to 0. So series converges for x 6= 0. Atx = 0, all the symmetric sums are 0, so series converges to 0. It mustconverge to the function since its Cesaro means converge to the function.
7. (Exercise 9) Let f(x) = χ[a,b](x) be the characteristic function of theinterval [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− a2π
+∑n6=0
e−ina − e−inb
2πineinx.
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 doesnot converge absolutely for any x.(c) However, prove that Fourier series converges at every point x. Whathappens if a = −π and b = π?Solution.(a) Straightforward.(b) |e−ina − e−inb|2 = 1 + 1 − 2 cosn(b − a) = 4 sin2(n(b − a)/2), thus|e−ina − e−inb| = 2| sinnθ0|, where
0 < θ0 =b− a
2<π
2.
The series∞∑n=1
| sinnθ0|n
diverges because .......(c) an = 1/n, bn = sin(na) − sin(nb) which has bounded partial sums bythe solution of Exercise 8. So series converges by Exercise 7. If a = −πand b = π, then the Fourier series is 1.
8. (Exercise 10) Suppose f is a periodic function of period 2π which belongsto the class Ck. Show that
f(n) = O(1/|n|k) as |n| → ∞.
Solution.Integrate by parts k times as in Corollary 2.4.
9. (Exercise 11) Suppose that {fk}∞k=1 is a sequence of Riemann integrablefunctions on the interval [0, 1] such that∫ 1
0
|fk(x)− f(x)| dx→ 0 as k →∞.
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Show that fk(n)→ f(n) uniformly in n as k →∞.Proof.Use
|g(n)| ≤∫ 1
0
|g(x)| dx
for all n.
10. (Exercise 12) Prove that if a series of complex numbers∑cn converges to
s, then∑cn is Cesaro summable to s.
Proof.Let the sequence of partial sums be sn, n = 1, 2, · · · . First assume thats = 0. Let ε > 0. Choose N1 such that |sn| < ε
2 for all n ≥ N1. ChooseN > N1 such that ∑N1
k=1 |sk|n
<ε
2
for all n ≥ N . Then for all n ≥ N ,
|s1 + · · ·+ snn
|
≤ |s1|+ · · ·+ |sn|n
=|s1 + · · ·+ |sN1 |
n+|sN1+1 + · · ·+ |sn|
n
<ε
2+nε/2
n= ε
This proves that∑cn is Cesaro summable to 0. Notice that we prove
above that if an is any sequence converging to 0, then the sequence σn =a1+···+an
n converges to 0.Now suppose that s 6= 0. Then sequence tn = sn − s converges to 0. Soby the proof above (t1 + · · ·+ tn)/n = (s1 + · · ·+ sn)/n− s converges to0, i.e (s1 + · · ·+ sn)/n converges to s.
11. (Exercise 13) The purpose of this exercise is to prove that Abel summa-bility is more general than the standard or Cesaro 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.(b) However, show that there exist series which are Abel summable, butthat do not converge.(c) Argue similarly to prove that if a series
∑∞n=1 cn is Cesaro summable
to σ, then it is Abel summable to σ.(d) Give an example of a series that is Abel summable but not Cesarosummable.The results above can be summarized by the following implications aboutseries:convergent ⇒ Cesaro summable ⇒ Able summable,
10
and the fact that none of the arrows can be reversed.Solution:In what follows, 0 ≤ r < 1; sn denotes the sequence of partial sums of∑∞n=1 cn.
(a). First assume that s = 0. For 0 ≤ r < 1 consider tn =∑nk=1 skr
k.
We have rtn =∑nk=1 skr
k+1 =∑n+1k=2 sk−1r
k. Since s1 = c1 and sk −sk−1 = ck, we get (1 − r)tn =
∑nk=1 ckr
k − snrn+1, hence∑nk=1 ckr
k =(1− r)
∑nk=1 skr
k + snrn+1. Since ck, sk are bounded sequences (in fact,
they converge to 0) and 0 ≤ r < 1, both series∑∞k=1 ckr
k and∑∞k=1 skr
k
are absolutely convergent, and∑∞k=1 ckr
k = (1−r)∑∞k=1 skr
k. Let ε > 0.Choose K such that |sk| < ε for all k ≥ K. Then
|∞∑k=1
ckrk| ≤ (1− r)
K−1∑k=1
Mrk + (1− r)ε rK
1− r≤ (1− r)
K−1∑k=1
Mrk + ε
It follows that
lim supr→1−
|∞∑k=1
ckrk| ≤ ε.
Since ε > 0 is arbitrary, we have lim supr→1− |∑∞k=1 ckr
k| ≤ 0 which isequivalent to limr→1−
∑∞k=1 ckr
k = 0.Next assume that s 6= 0. Define dn = cn − s/2n. Then
∑∞n=1 dn = 0,
and∑∞n=1 dnr
n =∑∞n=1 cnr
n − sr/(2 − r). Thus by our proof above,limr→1−
∑∞n=1 cnr
n − s = 0, i.e. limr→1−∑∞n=1 cnr
n = s.(b). The series
∑∞n=1(−1)n does not converge, but limr→1−
∑∞n=1(−1)nrn =
limr→1−−r1+r = −1/2, i.e.
∑∞n=1(−1)n is Abel summable to −1/2.
(c). Write
σn =s1 + · · ·+ sn
n
and τn = nσn. Since τn − τn−1 = sn, by the same argument as in (a), wehave
(1− r)∞∑k=1
τkrk =
∞∑k=1
skrk,
hence
(1− r)2∞∑k=1
σkkrk =
∞∑k=1
ckrk
from the proof of (a). Assume that σ = 0. Let ε > 0. Choose N such that|σk| < ε for all n ≥ N , and B be a bound for |σk|, k = 1, 2, · · · . Then
|∞∑k=1
ckrk| ≤ (1−r)2
N−1∑k=1
Bkrk+(1−r)2εrN ((1− r)N + r)
(1− r)2≤ (1−r)2
N−1∑k=1
Bkrk+εrN ((1−r)N+r)
where we have use the fact that∑∞k=N kr
k = rN ((1−r)N+r)(1−r)2 . As in the
proof of (a), this implies that limr→1−∑∞k=1 ckr
k = 0. For the case σ 6= 0,
11
consider dn = cn − σ/2n. Since∑∞n=1 cn is Cesaro summable to σ and∑∞
n=1 σ/2n = σ (and hence Cesaro summable to σ), we have
∑∞n=1 dn
Cesaro summable to 0. Thus by our proof above and the same argumentas in the proof of (a), the result follows.(d). Consider the series
∑∞n=1(−1)nn. It is Abel summable to −1/4 since
∞∑n=1
(−1)nnrn =−r
(1 + r)2.
Note that
σn −n− 1
nσn−1 =
ann.
Thus for a Cesaro summable series∑∞n=1 an, limn→∞
ann must be 0. This
proves that∑∞n=1(−1)nn is not Cesaro summable.
12. (Exercise 14) This exercise deals with a theorem of Tauber which says thatunder an additional condition on the coefficients cn, the above arrows canbe reversed.(a) If
∑cn is Cesaro summable to σ and cn = o(1/n) (that is, ncn → 0),
then∑cn converges to σ.
(b) The above statement holds if we replace Cesaro summable by Abelsummable.Proof.(a). Denote (n − 1)cn by tn. Since tn = n−1
n ncn and ncn → 0 we havetn → 0. Now
sn−σn = sn−s1 + · · ·+ sn
n=
(sn − s1) + · · ·+ (sn − sn)
n=t2 + · · ·+ tn
n
and it is immediate that sn − σn → 0.(b). Let r = 1− 1
N . We have
|N∑n=1
cn −N∑n=1
cnrn| ≤
N∑n=1
|cn|(
1− (1− 1
N)n)≤
N∑n=1
|cn|n
N
where we have used (1− 1N )n ≥ 1− n
N for 1 ≤ n ≤ N , which can be proved
12
easily by induction on n. Also if |cnn| < ε for all n ≥ N , then
|∞∑n=1
cnrn −
N∑n=1
cnrn|
≤∞∑
n=N+1
|cn|(1−1
N)n
≤∞∑
n=N+1
n|cn|N
(1− 1
N)n
≤∞∑
n=N+1
ε
N(1− 1
N)n
= ε(1− 1
N)N+1 → ε
e
as N →∞.
13. (Exercise 15) Prove that the Fejer kernel is given by
FN (x) =1
N
sin2(Nx/2)
sin2(x/2).
Proof.Recall that NFN (x) = D0(x)+· · ·+DN−1(x) where Dn(x) is the Dirichletkernel. Write ω = eix. Then
Dn(x)
= ω−n + · · ·+ ω−1 + 1 + ω + · · ·+ ωn
= (ω−n + · · ·+ ω−1) + (1 + ω + · · ·+ ωn)
= ω−1ω−n − 1
ω−1 − 1+
1− ωn+1
1− ω
=ω−n − 1
1− ω+
1− ωn+1
1− ω
=ω−n − ωn+1
1− ω
13
So
NFN (x)
=
N−1∑n=0
ω−n − ωn+1
1− ω
=1
1− ω
(N−1∑n=0
ω−n −N−1∑n=0
ωn+1
)
=1
1− ω
(ω−N − 1
ω−1 − 1− ω 1− ωN
1− ω
)=
1
1− ω
(ω−N+1 − ω
1− ω− ω 1− ωN
1− ω
)= ω
ω−N − 2 + ωN
(1− ω)2
=1
(ω−1/2)2(ωN/2 − ω−N/2)2
(1− ω)2
=(ωN/2 − ω−N/2)2
(ω1/2 − ω−1/2)2
=−4 sin2(Nx/2)
−4 sin2(x/2)
=sin2(Nx/2)
sin2(x/2).
Therefore
FN (x) =1
N
sin2(Nx/2)
sin2(x/2).
14. (Exercise 16) The Weierstrass approximation theorem states: Let f be acontinuous function on the closed and bounded interval [a, b] ⊂ R. Then,for any ε > 0, there exists a polynomial P such that
supx∈[a,b]
|f(x)− P (x)| < ε.
Proof.Let ε > 0. We may extend f to a continuous (c−a)-periodic function whereb ≤ c. By Corollary 5.4 of Fejer’s theorem, there exists a trigonometricpolynomial Q =
∑Nn=M ane
inx such that |Q(x)−f(x)| < ε/2 for all x. Foreach n,M ≤ n ≤ N , there exists a polynomial pn(x) such that |aneinx −pn(x)| < ε/2N for all x ∈ [a, c]. Then P = pM + · · ·+ pN is a polynomialin x that satisfies the requirement.
15. (Exercise 17) In Section 5.4 we proved that the Abel means of f convergeto f at all points of continuity, that is,
limr→1
Ar(f)(θ) = limr→1
(Pr ∗ f)(θ) = f(θ), with 0 < r < 1,
14
whenever f is continuous at θ. In this exercise, we will study the behaviorof Ar(f)(θ) at certain points of discontinuity.An integrable function is said to have a jump discontinuity at θ if thetwo limits
limh→0,h>0
f(θ + h) = f(θ+) and limh→0,h<0
f(θ + h) = f(θ−)
exist.(a) Prove that if f has a jump discontinuity at θ, then
limr→1
Ar(f)(θ) =f(θ+) + f(θ−)
2, with 0 ≤ r < 1.
(b) Using a similar argument, show that if f has a jump discontinuity at
θ, the Fourier series of f at θ is Cesaro summmable to f(θ+)+f(θ−)2 .
Proof.Since Pr(θ) = Pr(−θ), and 1
2π
∫ π−π Pr(θ) dθ = 1, we have 1
2π
∫ 0
−π Pr(θ) dθ =12π
∫ π0Pr(θ) dθ = 1/2. Suppose f has a jump discontinuity at θ. Let ε > 0
be given. Choose δ > 0 so that 0 < h < δ implies |f(θ − h)− f(θ−)| < εand |f(θ + h) − f(θ+)| < ε. Let M be such that |f(y)| ≤ M for all y.Then ∣∣∣∣(f ∗ Pr)(θ)− f(θ+) + f(θ−)
2
∣∣∣∣=
∣∣∣∣( 1
2π
∫ π
−πPr(y)f(θ − y) dy
)− f(θ+) + f(θ−)
2
∣∣∣∣≤ 1
2π
∫ 0
−πPr(y)|f(θ − y)− f(θ+)| dy +
1
2π
∫ π
0
Pr(y)|f(θ − y)− f(θ−)| dy
≤ 1
2π
∫−δ<y<0
Pr(y)|f(θ − y)− f(θ+)| dy +1
2π
∫0<y<δ
Pr(y)|f(θ − y)− f(θ−)| dy
+1
2π2M
∫δ≤|y|≤π
Pr(y) dy
≤ ε
2+ε
2+M
π
∫δ≤|y|≤π
Pr(y) dy
Therefore, recalling that limr→1
∫δ≤|y|≤π Pr(y) dy = 0,
lim supr→1
∣∣∣∣(f ∗ Pr)(θ)− f(θ+) + f(θ−)
2
∣∣∣∣ ≤ εSince ε > 0 is arbitrary, we have
limr→1
(f ∗ Pr)(θ) =f(θ+) + f(θ−)
2.
(b) Since the Fejer kernel Fn(θ) is even and positive, we also have 12π
∫ 0
−π Fn(θ) dθ =12π
∫ π0Fn(θ) dθ = 1/2, for all n. Repeat the above argument.
15
Remark One can replace f(θ+)+f(θ−)2 by limh→0
f(θ+h)+f(θ−h)2 and obtain
a more general statement of Exercise 17. (See Katznelson’s book.)
16. (Exercise 18) If Pr(θ) denotes the Poisson kernel, show that the funciton
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.Solution.We have
u(r, θ) =
∞∑n=−∞
r|m|ineinθ.
Each summand is Laplacian (see chapter 1, separation of variable), and theseries and its derivatives converge uniformly on r < ρ for any 0 < ρ < 1,hence the infinite sum is also Laplacian.
u(r, θ) =2r(r2 − 1) sin θ
(1− 2r cos θ + r2)2
If θ 6= 0, then 1− 2 cos θ+ 1 6= 0, so the limit is 0 as r → 1. If θ = 0, thenu(r, 0) = 0 and the limit is trivially 0. Note that
u(x, y) =2y(x2 + y2 − 1)
((1− x)2 + y2)2
and u(1− ε, ε) → −∞ as ε → 0+. This implies that u does not convergeto 0 uniformly as r → 1.
17. (Exercise 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).
16
Express u as an integral involving f , analogous to the Poisson integralformula (6).Solution.By considering the odd extension of f and following the derivation ofPoisson’s kernel with e−πy and eiπt replacing r and eit, respectively, weobtain
u(x, y) =1
2
∫ 1
−1f(t)Qy(x− t) dt
where
Qy(t) =1− e−2πy
1− 2e−πy cosπt+ e−2πy.
or, using the fact that f is odd, we have the alternate form
u(x, y) =1
2
∫ 1
0
f(t)Q(x, t) dt
where
Q(x, t) =1− e−2πy
1− 2e−πy cosπ(x− t) + e−2πy− 1− e−2πy
1− 2e−πy cosπ(x+ t) + e−2πy.
18. (Exercise 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 isto solve
∂2
∂r2+
1
r
∂
∂r+
1
r2∂2
∂θ2= 0
subject to the boundary conditions{u(1, θ) = f(θ),u(ρ, θ) = g(θ),
where f and g are given continuous functions.Arguing as we have previously for the Dirichlet problem in the disc, wecan hope to write
u(r, θ) =∑
cn(r)einθ
with cn(r) = Anrn +Bnr
−n, n 6= 0. Set
f(θ) ∼∑
aneinθ 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.
17
Show that as a result we have
u(r, θ)− (Pr ∗ f)(θ)→ 0 as r → 1 uniformly in θ,
andu(r, θ)− (Pρ/r ∗ g)(θ)→ 0 as r → ρ uniformly in θ.
Solution.Note that rn−r−n
ρn−ρ−n = r2n−1ρ2n−1 (ρ/r)n → 0 and (ρ/r)n−(r/ρ)n
ρn−ρ−n = (ρ/r)2n−1ρ2n−1 rn →
0 as n→∞.The series converges because for instance
r2n − 1
ρ2n − 1(ρ/r)n ≤ 1 + r2
1− ρ2(ρ/r)n.
And the last result follows from, for instance,((ρ/r)2n − 1
ρ2n − 1− 1
)rn =
((ρ/r)2n − ρ2n
ρ2n − 1
)rn
≤ 2[(ρ/r)2n − ρ2n]rn
and∞∑n=1
[(ρ/r)2n − ρ2n]rn =ρ2/r
1− ρ2/r− ρ2r
1− ρ2r
which approaches 0 as r → 1. Details are to be given.
19. (Problem 2) Let DN denote the Dirichlet kernel
DN (θ) =
N∑n=N
eikθ =sin(N + 1/2)θ
sin(θ/2),
and define
LN =1
2π
∫ π
−π|DN (θ)| dθ.
(a) Prove thatLN ≥ c logN
for some constant c > 0. A more careful estimate gives
LN =4
π2logN +O(1).
(See Katznelson Chapter 2, Exercise 1.1 for section p.50.)(b) Prove the following as a consequence: for each n ≥ 1, there exists acontinuous function fn such that |fn| ≤ 1 and |Sn(fn)(0)| ≥ c′ log n.Solution.
18
(a) Note that xsin x ≥ 1 for x in the interval [−π/2, π/2]. It follows that
for θ ∈ [−π, π],
|DN (θ)| ≥ 2| sin(N + 1/2)θ|
θ.
Then ∫ π
−π|DN (θ)| dθ
≥ 4
∫ π
0
| sin(N + 1/2)θ|θ
dθ
= 4
∫ (N+1/2)π
0
| sin θ|θ
dθ
≥ 4
∫ Nπ
0
| sin θ|θ
dθ
= 4
N−1∑k=0
∫ (k+1)π
kπ
| sin θ|θ
dθ
≥ 4
N−1∑k=0
1
(k + 1)π
∫ (k+1)π
kπ
| sin θ| dθ
=8
π
N−1∑k=0
1
k + 1
=8
πlog(N + 1)
≥ 8
πlogN
Therefore LN ≥ c logN .(b) The function gn which is equal to 1 when Dn is positive and −1 whenDn is negative has the desired property but is not continuous. Approxi-mate gn in the integral norm (in the sense of Lemma 3.2) by continuousfunctions hk satisfying |hk| ≤ 1.
20. (Problem 3) Littlewood provided a refinement of Tauber’s theorem:(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 Cesaro 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 implythat if f is an integrable function that satisfies f(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−)
2as N →∞.
19
(iii) If f is continuous on [−π, π], then SN (f)→ f uniformly.For the simpler assertion (b), hence of proof of (i),(ii),and (iii), see Problem5 in Chapter 4.Solution.
Lemma 1 Let f be a C2 real function on [0, 1). Suppose as x → 1,
f(x) = o(1), f ′′(x) = O(
1(1−x)2
), then f ′(x) = o
(1
1−x
).
Proof. Choose C > 0 such that |(1 − x)2f ′′(x)| ≤ C for all x. Letε > 0. Choose a δ, 0 < δ < min{1/2, ε/(4C)}. Choose η < 1 such that|f(x)| < (1/4)δε for all x > η. We claim that |(1 − x)f ′(x)| < ε for allx > η. Let x > η. With x′ = x+ δ(1− x) we have
f(x′) = f(x) + δ(1− x)f ′(x) +1
2δ2(1− x)2f ′′(ζ)
for some x < ζ < x′.Since δ < 1/2, we have 1 − x ≤ 2(1 − ζ), so that |(1 − x)2f ′′(ζ)| ≤4|(1− ζ)2f ′′(ζ)| ≤ 4C. Therefore
|(1− x)f ′(x)|
= |f(x′)− f(x)
δ− 1
2δ(1− x)2f ′′(ζ)|
≤ |f(x′)|+ |f(x)|δ
+1
2δ(1− x)2|f ′′(ζ)|
<1
2ε+
1
2ε = ε
Lemma 2 Let∑∞n=0 anx
n be a real power series with an ≥ 0 for all n.Suppose
limx→1−
(1− x)
∞∑n=0
anxn = 1.
Then for any integrable function g(t) on [0, 1]
limx→1−
(1− x)
∞∑n=0
anxng(xn) =
∫ 1
0
g(t) dt.
Proof. First prove for functions of the type xk. Then for any polynomial;then for any continuous function; then for any integral function.
Corollary 1 Let∑∞n=0 anx
n be a real power series with an ≥ 0 for all n.Suppose
limx→1−
(1− x)
∞∑n=0
anxn = 1.
Then
limN→∞
∑Nn=0 anN
= 1.
20
Proof. Apply the above lemma to the function g(t) = 0 for t < e−1, 1/totherwise, and let x = e−1/N .Finishing the solution of Problem 3
We shall write f(x) ∼ g(x) to mean f(x)g(x) → 1 as x→ 1, and f(n) ∼ g(n)
to mean f(n)g(n) → 1 as n→∞.
(a) We may assume that s = 0 (see Exercise 13, Chapter 2), i.e. we assumethat f(x) =
∑∞n=0 anx
n → 0 as x → 1. Then f ′′(x) = O(1/(1 − x)2)because
f ′′(x) =
∞∑n=2
n(n− 1)anxn−2 = O(
∞∑n=2
(n− 1)xn−2) = O(1/(1− x)2).
So by Lemma 1,
f ′(x) = o
(1
1− x)
).
Suppose |nan| ≤ c. Then
∞∑n=1
(1− nanc
)xn−1 =1
1− x− f ′(x)
c∼ 1
1− x
Since 1− nanc ≥ 0, Corollary 1 implies that
n∑k=1
(1− kakc
) ∼ n
or what is the samen∑k=1
kak = o(n).
21
Write wn =∑nk=1 kak, w0 = 0. So wn/n→ 0 as n→∞. Then
f(x)− a0
=
∞∑n=1
anxn
=
∞∑n=1
wn − wn−1n
xn
=
∞∑n=1
wnnxn −
∞∑n=1
wn−1n
xn
=
∞∑n=1
wnnxn −
∞∑n=1
wnn+ 1
xn+1
=
∞∑n=1
wn
(xn
n− xn+1
n+ 1
)
=
∞∑n=1
wn
(xn − xn+1
n+ 1+
xn
n(n+ 1)
)
= (1− x)
∞∑n=1
wnxn
n+ 1+
∞∑n=1
wnn(n+ 1)
xn
Since f(x)→ 0 and the first term in the last sum approaches 0 as x→ 1,we get
limx→1
∞∑n=1
wnn(n+ 1)
xn = −a0
Since wnn(n+1) = o(1/n), by the regular Tauberian theorem
∞∑n=1
wnn(n+ 1)
= −a0.
Now
N∑n=1
wnn(n+ 1)
=
N∑n=1
wn
(1
n− 1
n+ 1
)
=
N∑n=1
wn − wn−1n
− wNN + 1
=
N∑n=1
an −wNN + 1
.
Letting N →∞ we get∑∞n=1 an = −a0, i.e.
∑∞n=0 an = 0.
22
Chapter 3
1. (Exercise 1) Show that the first two examples of inner product spaces,namely Rd and Cd, are complete.
2. (Exercise 2) Prove that the vector space `2(Z) is complete.
3. (Exercise 3) Construct a sequence of integrable functions {fk} on [0, 2π]such that
limk→∞
1
2π
∫ 2π
0
|fk(t)|2 dt = 0
but limk→∞ fk(t) fails to exist for any t.
4. (Exercise 4) (In (c), use the fact that f is continuous except possibly ona set of measure 0.)
5. (Exercise 5) Let
f(t) =
{0 for t = 0log(1/t) for 0 < t ≤ 2π,
and define a sequence of functions in R by
fn(t) =
{0 for 0 ≤ t ≤ 1/nf(t) for 1/n < t ≤ 2π.
Prove that {fn}∞n=1 is a Cauchy sequence in R. However, f does not be-long to R.Solution. ∫
(log t)2 dt = t(log t)2 − 2t log t+ 2t
and limt→0 t log t = 0, limt→0 t log2 t = 0.
6. (Exercise 6) Consider the sequence {ak}∞k=−∞ defined by
ak =
{1/k if k ≥ 10 if k ≤ 0.
Note that {ak} ∈ l2(Z), but that no Riemann integrable function has kth
Fourier coefficient equal to ak for all k.Solution.Let M = sup−π≤tπ |f(t)|. Recall that
Ar(f)(θ) =1
2π
∫ π
−πPr(θ − t)f(t) dt.
So
|Ar(f)(θ)| ≤M 1
2π
∫ π
−πPr(θ − t) dt = M
23
Also
Ar(f)(θ) =
∞∑n=−∞
f(n)einθr|n|.
So
Ar(f)(0) =
∞∑n=−∞
f(n)r|n|.
It follows that if f is a Riemann integrable function that has kth Fouriercoefficient equal to ak for all k, then
∞∑n=1
rn
n
is bounded for 0 ≤ r < 1. This is a contradiction since the sum of theseries is − log(1− r).
7. (Exercise 7) Show that the trigonometric series∑n≥2
sinnx
log n
converges for every x, yet it is not the Fourier series of a Riemann inte-grable function.The same is true for
∑sinnxnα for 0 < α < 1, but the case 1/2 < α < 1 is
more difficult. See Problem 1.Solution.Apply Parseval’s identity.(A Riemann integrable function is in L2.) Seriesconverges because the partial sums of
∑sinnx is bounded; see Exercise
9. Postpone the case 1/2 < α < 1 to Problem 1.
8. (Exercise 8) Exercise 6 in Chapter 2 dealt with the sums∑n odd ≥1
1
n2=π2
8and
∞∑n=1
1
n2=π2
6.
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’sidentity to find the sums of the following two series:
∞∑n=0
1
(2n+ 1)4and
∞∑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
960and
∞∑n=1
1
n6=
π6
945.
24
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 unre-
solved question.Solution. The Fourier series for f(θ) = |θ| is
π
2− 2
π
∑n odd ≥1
einx + e−inx
n2.
By Parseval’s identity,
π2
4+
4
π2
∑n odd ≥1
2
n4=
1
2π
∫ π
−π|θ|2 dθ =
π2
3
from which we get∞∑n=0
1
(2n+ 1)4=π4
96.
The sum∑∞n=1
1n4 is obtained from solving for x in the equation
x =π4
96+
x
16
which yields x = π4
90 .The Fourier series for the function f(θ) = θ(π − θ) is
8
π
∑k odd ≥1
sin kθ
k3= −i 4
π
∑k odd ≥1
eikθ − e−ikθ
k3.
By Parseval’s identity,
16
π2
∞∑n=0
2
(2n+ 1)6=
1
2π
∫ π
−πθ2(1− θ)2 dθ =
1
30π4,
from which we get∞∑n=0
1
(2n+ 1)6=
π6
960.
Solving x from
x =1
64x+
π6
960
we get x = π6
945 .
9. (Exercise 9) Show that for α not an integer, the Fourier series of
π
sinπαei(π−x)α
25
on [0, 2π) is given by∞∑−∞
einx
n+ α.
Apply Parseval’s formula to show that
∞∑−∞
1
(n+ α)2=
π2
(sinπα)2.
SolutionStraightforward checking.
10. (Exercise 10) Consider the example of a vibrating string which we analyzedin Chapter 1. The displacement u(x, t) of the string at time t satisfies thewave 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 totalenergy of the string by
E(t) =1
2ρ
∫ L
0
(∂u
∂t
)2
dx+1
2τ
∫ L
0
(∂u
∂x
)2
dx.
The first term corresponds to the ”kinetic energy” of the string (in analogywith (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 thatE(t) is constant. Therefore,
E(t) = E(0) =1
2ρ
∫ L
0
g(x)2 dx+1
2ρ
∫ L
0
f ′(x)2 dx.
Solution. We have
E′(t) = ρ
∫ L
0
∂u
∂t
∂2u
∂t2dx+ τ
∫ L
0
∂u
∂x
∂2u
∂x∂tdx
= ρ
∫ L
0
∂u
∂t
∂2u
∂t2dx− τ
∫ L
0
∂2u
∂2x
∂u
∂tdx
= 0
where we have used integration by parts, and that ∂u∂t (0, t) = ∂u
∂t (L, t) = 0for all t.
26
11. (Exercise 11) The inequalities of Wirtinger and Poincare establish a rela-tionship between the norm of a function and that of its derivative.
(a) If f is T -periodic, continuous, and piecewise C1 with∫ T0f(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 aa 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 func-tion 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 cannotbe improved.Solution.(a) The condition
∫ T0f(t) dt = 0 implies that f(0) = 0. The continuity
of f guarantees that f ′(n) = 2πinT f(n). Indeed, write τ for 2π
T . Then forn 6= 0,
f(n) =1
T
∫ T
0
f(t)e−inτt dt
=1
T(f(0+)− f(T−))
1
inτ+
1
inτ
1
T
∫ T
0
f ′(t)e−inτt dt
=T
2πinf ′(n)
Therefore, by Parseval’s identity,recalling that f(0) = 0,∫ T
0
|f(t)|2 dt = T∑|n|>0
|f(n)|2
=T 3
4π2
∑|n|>0
|f ′(n)|2
n2
≤ T 3
4π2
∑|n|>0
|f ′(n)|2
=T 3
4π2
1
T
∫ T
0
|f ′(t)|2 dt
=T 2
4π2
∫ T
0
|f ′(t)|2 dt
27
From the above inequalities, we see that equality holds if and only iff(n) = 0 for all n ≥ 2. This means that, writing an for f(n), f(x) =a1e
iτx + a−1e−iτx which simplifies to A sin(τx) +B cos(τx).
Remark. It is clear from the proof above that in the absence of the
condition∫ T0f(t) dt = 0, the inequality in (a) is∑
|n|>0
|an|2 ≤T 2
4π2
∑|n|>0
|bn|2
where an = f(n), bn = f ′(n).
(b) Let an = f(n), bn = g(n), cn = g′(n). Then∣∣∣∣∣∫ T
0
f(t)g(t) dt
∣∣∣∣∣2
= T |∑|n|≥0
anbn|2
= T |∑|n|>0
anbn|2
≤ (T∑|n|>0
|an|2)(∑|n|>0
|bn|2)
≤∫ T
0
|f(t)|2 dt T2
4π2
∫ T
0
|g′(t)| dt
(c) Extend f to a function on [a, 2b − a] such that f(b + h) = −f(b −h) for 0 ≤ h ≤ b − a and then extend it so that it is T = 2(b − a)-periodic. It is easy to see that now f(b+ h) = −f(b− h) and f ′(b+ h) =
f ′(b − h) for all h. Then∫ T0f(x) dx =
∫ 2b−aa
f(x) dx = 0,∫ T0|f(x)| dx =∫ 2b−a
a|f(x)| dx = 2
∫ ba|f(x)| dx, and
∫ T0|f ′(x)| dx = 2
∫ ba|f ′(x)| dx. Check
that f so extended is also C1 on R. Therefore by (a), we have∫ b
a
|f(x)| dx
=1
2
∫ T
0
|f(x)| dx
≤ 1
2
22(b− a)2
4π2
∫ T
0
|f ′(x)| dx
=(b− a)2
π2
1
2
∫ T
0
|f ′(x)| dx
=(b− a)2
π2
∫ b
a
|f ′(x)| dx
For the function f(x) that we define, its translation g(x) = f(x+ b) is an
odd function. So g(n) = einbf(n), and the Fourier series for g is of the
28
form∞∑n=1
An sin2πnx
2(b− a)=
∞∑n=1
An sinπnx
b− a.
Thus the Fourier series for f is of the form
∞∑n=1
An sinπn(x− b)b− a
.
According the inequality in (c) is an equality iff f(x) is of the form
A sin π(x−b)b−a . (The book saysA sin π(x−a)
b−a ; this is equivalent since sin π(x−b)b−a =
− sin π(x−a)b−a ).
12. (Exercise 12) Prove that∫∞0
sin xx dx = π
2 .Proof.We have
∫ π−πDN (t) dt = 2π. So∫ π
−π
sin(N + 1/2)x
sin(x/2)dx = 2π
Write csc(x/2) as csc(x/2)− 2/x+ 2/x. limx→0 csc(x/2)− 2/x = 0, so ithas a removable discontinuity at 0 on the interval [−π, π]. By Lebesgue-Riemann lemma we get∫ π
−π
2 sin(N + 1/2)x
xdx→ 2π as N →∞.
This yields ∫ π
0
sin(N + 1/2)x
xdx→ π
2as N →∞.
By change of variable, we get∫ (N+1/2)π
0
sinx
xdx→ π
2as N →∞.
Since∫ (N+1/2)π
Nπsin xx dx → 0 as N → ∞, (use Mean-Value Theorem), we
get ∫ Nπ
0
sinx
xdx→ π
2as N →∞.
By MVT, ∫ (N+1)π
Nπ
| sinx|x
dx ≤ 1
N,
so for any t > π there exists N > 0 such that∣∣∣∣∫ t
Nπ
sinx
xdx
∣∣∣∣ ≤ 1
N.
29
It follows that ∫ ∞0
sinx
xdx = lim
N→∞
∫ Nπ
0
sinx
xdx =
π
2.
13. (Exercise 13) Suppose that f is periodic and of class Ck. Show that
f(n) = o(1/|n|k),
that is, |n|kf(n) goes to 0 as |n| → ∞. This is an improvement overExercise 10 in Chapter 2.Solution.We have (see p.43)
2πf(n) =1
(in)k
∫ 2π
0
f (k)(θ)e−inθ dθ.
So
2π(in)kf(n) =
∫ 2π
0
f (k)(θ)e−inθ dθ.
Now use Lebesgue-Riemann lemma.
14. (Exercise 14) Prove that the Fourier series of a continuously differentiablefunction f on the circle is absolutely convergent.Proof.From f(n) = 1
in f′(n), we get, by Cauchy-Schwarz inequality and Parse-
val’s identity,
∞∑n=−∞
|f(n)|
=
∞∑n=−∞
1
|n||f ′(n)|
≤ (2π2
6)1/2(
∞∑n=−∞
|f ′(n)|2)1/2
= (2π2
6)1/2
1
2π
∫ π
−π|f ′(t)|2 dt <∞.
15. (Add in) Prove that the Fourier series of a 2π periodic absolutely contin-uous function whose derivative (exists a.e.) in [0, 2π] is square integrable(in particular, Riemann integrable), is absolutely convergent. (Note thatExercise 16 below shows that Lipschitz condition alone is enough. Butderivative of a Lipschitz function is bounded. And Lipschitz functionsare precisely functions representable as integral of a bounded measurablefunction.)Proof. Use the proof in Exercise 14. Note that the integration by partsformula is valid for absolutely continuous functions.
30
16. (Exercise 15) Let f be a 2π-periodic and Riemann integrable on [−π, π].(a) Show that
f(n) = − 1
2π
∫ π
−πf(x+
π
n)e−inx dx
hence
f(n) =1
4π
∫ π
−π[f(x)− f(x+
π
n)]e−inx dx.
(b) Now assume that f satisfies a Holder 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
f(n) = O(1/|n|α).
(c) Prove that the above result cannot be improved by showing that thefunction
f(x) =
∞∑k=0
2−kαei2kx,
where 0 < α < 1, satisfies
|f(x+ h)− f(x)| ≤ C|h|α,
and f(N) = 1/Nα whenever N = 2k.Solution.Note that for any real x, |1 − eix| ≤ |x| (to prove, just note that LHS is2| sin(x/2)| and | sin a| ≤ |a| for all real number a).(a).
1
2π
∫ π
−πf(x+
π
n)e−inx dx
=1
2π
∫ π+πn
−π+πn
f(u)e−inueiπ du
= − 1
2π
∫ π
−πf(u)e−inu du
= −f(n)
(b)
|f(n)|
≤ 1
4π
∫ π
−π|f(x)− f(x+
π
n)| dx
≤ 1
4π2πC
πα
|n|α
=C1
|n|α
31
(c).
|f(x+ h)− f(x)|
= |∞∑k=0
2−kαei2k(x+h) −
∞∑k=0
2−kαei2kx|
≤∞∑k=0
2−kα|ei2kh − 1|
≤∑
2k≤1/|h|
2−kα2k|h|+∑
2k>1/|h|
2−kα2
The second sum is easily seen to be less than 2|h|α. The first sum is 0 if|h| > 1. So assume |h| ≤ 1. Let l be the unique nonnegative integer suchthat 2−l−1 < |h| ≤ 2−l. Then the first sum is
l∑k=0
(2k|h|)1−α|h|α ≤ |h|αl∑
k=0
(2k−l)1−α ≤ 1
1− 2α−1|h|α.
Since the series converges uniformly, the coefficient of einx is f(n) for alln.
17. (Exercise 16) Let f be a 2π-periodic function which satisfies a Lipschitzcondition with constant K; that is
|f(x)− f(y)| ≤ K|x− y| for all x, y.
This is simply the Holder condition with α = 1, so by the previous exercise,we see that f(n) = O(1/|n|). Since the harmonic series
∑1/n diverges, we
cannot say anything (yet) about the absolute convergence of the Fourierseries of f . The outline below actually proves that the Fourier series of fconverges absolutely and uniformly.(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| sinnh|2|f(n)|2,
and show that∞∑
n=−∞| sinnh|2|f(n)|2 ≤ K2h2.
(b) Let p be a positive integer. By choosing h = π/2p+1, show that∑2p−1<|n|≤2p
|f(n)|2 ≤ K2π2
22p+1.
(c) Estimate∑
2p−1<|n|≤2p |f(n)|, and conclude that the Fourier series off converges absolutely, hence uniformly.
32
(d) In fact, modify the argument slightly to prove Bernstein’s theorem: Iff satisfies a Holder condition of order α > 1/2, then the Fourier series off converges absolutely.Solution.(a). The Fourier coefficients of the translated function f(x+h) is einhf(n).So
gh(n) = (einh − e−inh)f(n) = 2i sinnhf(n).
The first equation in (a) follows from Parseval’s identity. Since |gh(x)| ≤K[x+ h− (x− h)] = 2K|h|, the second inequality in (a) follows.(b). h = π/2p+1 and 2p−1 < |n| ≤ 2p imply π/4 < |n|h ≤ π/2, hence| sinnh|2 ≥ 1/2. Thus
1
2
∑2p−1<|n|≤2p
|f(n)|2 ≤∞∑
n=−∞| sinnh|2|f(n)|2 ≤ K2h2 =
K2π2
22(p+1)
from which it follows ∑2p−1<|n|≤2p
|f(n)|2 ≤ K2π2
22p+1.
(c). By Cauchy-Schwarz inequality, we have∑2p−1<|n|≤2p
|f(n)|
≤ (∑
2p−1<|n|≤2p12)1/2(
∑2p−1<|n|≤2p
|f(n)|2)1/2
≤ (2p − 2p−1)1/2Kπ
2p+1/2
≤ Kπ
(√
2)p
Therefore ∑1≤|n|<∞
|f(n)| ≤∑
1≤p<∞
Kπ
(√
2)p<∞
(d). Modifications are given below:
The Fourier coefficients of the translated function f(x+h) is einhf(n). So
gh(n) = (einh − e−inh)f(n) = 2i sinnhf(n).
The first equation in (a) follows from Parseval’s identity. Since |gh(x)| ≤K|x+ h− (x− h)|α = 2αK|h|α, the second inequality in (a) becomes
∞∑n=−∞
| sinnh|2|f(n)|2 ≤ 22(α−1)K2|h|2α.
33
h = π/2p+1 and 2p−1 < |n| ≤ 2p imply π/4 < |n|h ≤ π/2, hence| sinnh|2 ≥ 1/2. Thus
1
2
∑2p−1<|n|≤2p
|f(n)|2 ≤∞∑
n=−∞| sinnh|2|f(n)|2 ≤ 22(α−1)K2|h|2α =
K2π2α
22(αp+1)
from which it follows ∑2p−1<|n|≤2p
|f(n)|2 ≤ K2π2α
22αp+1.
By Cauchy-Schwarz inequality, we have∑2p−1<|n|≤2p
|f(n)|
≤ (∑
2p−1<|n|≤2p12)1/2(
∑2p−1<|n|≤2p
|f(n)|2)1/2
≤ (2p − 2p−1)1/2Kπα
2αp+1/2
≤ Kπ
(2α−1/2)p
Therefore, if α > 1/2, then∑1≤|n|<∞
|f(n)| ≤∑
1≤p<∞
Kπ
(2α−1/2)p<∞
Remark. The condition α > 1/2 is sharp. See Katznelson’s book, p. 32.
18. (Exercise 17) If f is a bounded monotonic function on [−π, π], then
f(n) = O(1/|n|).
19. (Exercise 18) Here are a few things we have learned about the decay ofFourier coefficients:(a) if f is of class Ck, then f(n) = o(1/|n|k);
(b) if f is Lipschitz, then f(n) = O(1/|n|);(c) if f is monotonic, then f(n) = O(1/|n|);(d) if f satisfies a Holder condition with exponent α where 0 < α < 1,
then f(n) = O(1/|n|α);
(e) if f is merely Riemann integrable, then∑|f(n)|2 < ∞ and therefore
f(n) = o(1).Nevertheless, show that the Fourier coefficients of a continuous functioncan tend to 0 arbitrarily slowly by proving that for every sequence ofnonnegative real numbers {εk} converging to 0, there exists a continuous
34
function f such that |f(n)| ≥ εn for infinitely many values of n.Solution.Choose a subsequence {εnk} such that
∑k εnk < ∞, e.g., εnk < 1/2k.
Then the function∑∞k=1 εnke
inkx, is such a function, because it is abso-lutely ( and hence uniformly) convergent.
20. (Exercise 19) Give another proof that the sum∑
0<|n|≤N einx/n is uni-
formly bounded in N and x ∈ [−π, π] by using the fact that
1
2i
∑0<|n|≤N
einx
n=
N∑n=1
sinnx
n=
1
2
∫ x
0
(DN (t)− 1) dt,
where DN is the Dirichlet kernel. Now use the fact that∫∞0
sin tt dt < ∞
which was proved in Exercise 12.
21. (Exercise 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 periodicityto all of R. The Fourier series of f is
f(x) ∼ 1
2i
∑n 6=0
einx
n=
∞∑n=1
sinnx
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
limN→∞
max0<x≤π/N
SN (f)(x)− π
2=
∫ π
0
sin t
tdt− π
2≈ 0.08949π,
which is roughly 9% of the jump π. This result is a manifestation ofGibbs’s phenomenon which states that near a jump discontinuity, theFourier series of a function overshoots (or undershoots) it by approxi-mately 9% of the jump.
22. (Problem 1) For each 0 < α < 1 the series
∞∑n=1
sinnx
nα
converges for every x but is not the Fourier series of a Riemann integrablefunction.(a) If the conjugate Dirichlet kernel is defined by
DN (x) =∑|n|≤N
sign(n)einx where sign(n) =
1 if n > 00 if n = 0−1 if n < 0,
35
then show that
DN (x) = icos(x/2)− cos((N + 1/2)x)
sin(x/2),
and ∫ π
−π|DN (x)| dx ≤ c logN, for N ≥ 2
(b) As a result, if f is Riemann integrable, then
(f ∗ D)(0) = O(logN).
(c) In the present case, this leads to
∞∑n=1
1
nα= O(logN),
which is a contradiction.Solution.Let ω = eix.
DN (x)
=
n∑k=1
ωk −n∑k=1
ω−k
= ω1− ωn
1− ω− ω−1 1− ω−n
1− ω−1
= ω1− ωn
1− ω+
1− ω−n
1− ω
=ω + 1− ωn+1 − ω−n
1− ω
=ω1/2 + ω−1/2 − ωn+1/2 − ω−n−1/2
ω−1/2 − ω1/2
=cos(x/2)− cos((N + 1/2)x)
sin(x/2)i
Note that by trigonometric identity, this is
2 sin(N+12 x) sin(N2 x)
sin(x/2)i.
Note that t/ sin t ≤ π/2 for t ∈ [−π/2, π/2], so for x ∈ [−π, π],
|DN (x)| ≤ 4| sin(N2 x)|
x
36
Then ∫ π
−π|DN (θ)| dθ
≤ 8
∫ π
0
| sin(N/2)θ|θ
dθ
= 8
∫ (N/2)π
0
| sin θ|θ
dθ
= 8
N−1∑k=0
∫ (k+1)π/2
kπ/2
| sin θ|θ
dθ
= 8
N−1∑k=1
∫ (k+1)π/2
kπ/2
| sin θ|θ
dθ +
∫ π/2
0
sin θ
θdθ
≤ 82
π
N−1∑k=1
1
k
∫ (k+1)π/2
kπ/2
| sin θ| dθ +
∫ π/2
0
sin θ
θdθ
=16
π
N−1∑k=1
1
k+
∫ π/2
0
sin θ
θdθ
=16
π
N−1∑k=2
1
k+
16
π+
∫ π/2
0
sin θ
θdθ
≤ 16
πlogN + c1
≤ c logN
where c is chosen such that c/2 ≥ 16π and (c/2) log 2 ≥ c1, so that
c logN = c/2 logN + c/2 logN ≥ 16π logN + c1 for all N ≥ 2.
(b),(c). If f is such a Riemann integrable function, then f(n) = 1/nα, f(−n) =
37
−1/nα for n > 0, and f(0) = 0. So
(f ∗ DN )(0) =1
2π
∫ π
−πf(t)DN (0− t) dt
= − 1
2π
∫ π
−πf(t)DN (t) dt
= − 1
2π
∫ π
−πf(t)
∑|n|≤N
sign(n)eint dt
= −∑|n|≤N
sign(n)1
2π
∫ π
−πf(t)eint dt
= −∑|n|≤N
sign(n)f(n)
= −2
N∑n=1
1
nα
Since f(t) is bounded, we also have by (a),
|∫ π
−πf(t)DN (t) dt| ≤M
∫ π
−π|DN (t)| dt = O(logN).
Thus∑Nn=1
1nα = O(logN); this is impossible (since
∑Nn=1
1nα grows like
N1−α, see my ApostolStudyNote, p. 25).
23. (Problem 2)
24. (Problem 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,
cotu =1
u+ 2
∞∑n=1
u
u2 − n2π2.
(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+ 1dt =
π
sin(απ).
38
Solution.From Maple, for all n,
f(n) = (−1)n+1 α sin(απ)
π(n2 − α2).
It follows that for −π ≤ x ≤ π,
cos(αx) =sin(απ)
απ+
2α sin(απ)
π
∞∑n=1
(−1)n+1 cosnx
n2 − α2.
Evaluating at x = π yields the first equation in part (b). The secondequation follows directly from the first by substituting u for απ.(c) follows by substituting x = 0 in the Fourier series.
(d). Split the integral as∫ 1
0+∫∞1
and change variables t = 1/u in thesecond integral, one gets∫ ∞
0
tα−1
t+ 1dt =
∫ 1
0
tα−1
t+ 1dt+
∫ 1
0
t(1−α)−1
t+ 1dt.
Claim: For 0 < γ < 1,∫ 1
0
tγ−1
t+ 1dt =
∞∑k=0
(−1)k1
k + γ.
Proof.For each 0 < r < 1, the series tγ−1
∑∞k=0(−1)ktk converges to tγ−1
t+1 uni-formly on [0, r]. So ∫ r
0
tγ−1
t+ 1dt =
∞∑k=0
(−1)krγ+k
k + γ
Since ∫ 1
0
tγ−1
t+ 1dt
= limr→1−
∫ r
0
tγ−1
t+ 1dt
= limr→1−
∞∑k=0
(−1)krγ+k
k + γ
= limr→1−
∞∑k=0
(−1)krk
k + γ
The series∑∞k=0(−1)k 1
k+γ is Abel summable and hence convergent to∫ 1
0tγ−1
t+1 dt, by Littlewood’s theorem since (−1)k1/(k+γ) = O(1/k). Q.E.D.
39
Therefore ∫ 1
0
tα−1
t+ 1dt+
∫ 1
0
t(1−α)−1
t+ 1dt
=
∞∑k=0
(−1)k
k + 1− α+
∞∑k=0
(−1)k
k + α
=
∞∑k=1
(−1)k−1
k − α+
1
α+
∞∑k=1
(−1)k
k + α
=1
α+ 2α
∞∑k=1
(−1)k−1
k2 − α2
=π
sin(απ)
by (c).
25. (Problem 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 numbers are expressed in terms of theBernoulli numbers; the related Bernoulli polynomials are discussed in thenext problem.Define the Bernoulli numbers Bn by the formula
z
ez − 1=
∞∑n=0
Bnn!zn.
(a) Show that B0 = 1, B1 = −1/2, B2 = 1/6, B3 = 0, B4 = −1/30, andB5 = 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
Bnn!zn,
show that Bn = 0 if n is odd and > 1. Also prove that
z cot z = 1 +
∞∑n=1
22nB2n
(2n)!z2n.
40
(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 cotx = 1− 2
∞∑m=1
ζ(2m)
π2mx2m.
(e) Conclude that
2ζ(2m) = (−1)m+1 (2π)2m
(2m)!B2m.
Solution.(a) Get these numbers directly from (1 + z/2! + z2/3! + · · · )−1 by longdivision.(b).
1 =z
ez − 1
ez − 1
z
=
( ∞∑n=0
Bnn!zn
)( ∞∑n=0
1
(n+ 1)!zn
)
=
∞∑n=0
(n∑k=0
Bkk!
1
(n− k + 1)!
)tn
from which it follows that B0 = 1, and for n ≥ 1,
n∑k=0
Bkk!
1
(n− k + 1)!= 0,
i.e.
Bnn!
= −n−1∑k=0
Bkk!
1
(n− k + 1)!
= − 1
(n+ 1)!
n−1∑k=0
(n+ 1
k
)Bk
(c). Since B0 = 1, B1 = − 12 , we have
z
ez − 1= 1− z
2+
∞∑n=2
Bnn!zn.
41
Now,
z
ez − 1+z
2
= z
(1
ez − 1+
1
2
)=
z
2
(ez + 1
ez − 1
)=
z
2
(ez/2 + e−z/2
ez/2 − e−z/2
)=
z
2coth
z
2
is an even function, so B2n+1 = 0 for all n ≥ 1. Note that
coth(iz) =eiz/2 + e−iz/2
eiz/2 − e−iz/2=
2 cos z
2i sin z
so cot z = i coth(iz). Therefore
z cot z = iz coth(iz) = 1 +
∞∑n=1
B2n
(2n)!(i2z)2n
=
∞∑n=0
(−1)n22nB2n
(2n)!z2n
From Problem 3 (b), we have for 0 < x < π,writing rn for xnπ ,
x cotx = 1− 2
∞∑n=1
x2
n2π2 − x2
= 1− 2
∞∑n=1
r2n1− r2n
= 1− 2∞∑n=1
∞∑m=1
r2mn
= 1− 2
∞∑m=1
( ∞∑m=1
1
n2m
)x2m
π2m
= 1− 2
∞∑m=1
ζ(2m)
π2mx2m
Comparing coefficients of x cotx in both formulas, we get (e).Remark. Ross Tang has discovered explicit formulas for Bernoulli num-bers and Euler numbers in
http://www.voofie.com/content/117/an-explicit-formula-for-the-euler-zigzag-numbers
-updown-numbers-from-power-series/#Explicit_Formula_for_Euler_number
42
His formula is
B2n =2n
22n − 42n
2n∑k=1
k∑j=0
(k
j
)(−1)j(k − 2j)2n
2kikk,
E2n = i
2n+1∑k=1
k∑j=0
(k
j
)(−1)j(k − 2j)2n+1
2kikk.
Update: Simpler formulas are already in
http://en.wikipedia.org/wiki/Bernoulli_number
Perhaps it is a good project to try to find explicit formulas forBernoulli polynomials.From the above solution, we get the following series expansion for cotxand cothx:
cot z =1
z+
∞∑n=1
(−1)n22nB2n
(2n)!z2n−1,
coth z =
∞∑n=1
22nB2n
(2n)!z2n−1.
I have yet to figure out the proofs of the next four formulas:
tan z =
∞∑n=1
(−1)n−14n(4n − 1)B2n
(2n)!z2n−1 = z +
z3
3+
2z5
15+
17z7
315+ · · · ,
tanh z =
∞∑n=1
4n(4n − 1)B2n
(2n)!z2n−1 = z − z3
3+
2z5
15− 17z7
315+ · · · ,
sec z =
∞∑n=0
(−1)nE2n
(2n)!z2n,
sech z =
∞∑n=0
E2n
(2n)!z2n,
where E2n are the Euler’s numbers. (The generating function for E2n issech z = 2
ez+e−z ).Update: First note that
2 coth(2z)− coth z = tanh z
from which we get the series for tanh z. Then use tan z = −i tanh(iz) toget tan z series. Other series that use Bernoulli numbers are:
z
sin z,
z
sinh z, log
sin z
z, log(cos z), log
tan z
z, · · · ,
43
but I have not looked into this assertion.Atkinson (American Math Monthly, vol. 93, no. 5,1986, p. 387-; thispaper in pdf form is in my computer under the name TangentSeries.pdf)has discovered that the above series for tangent and secant can be readoff from the sides of the following triangle (only seven rows are shown):
10 1
1 1 00 1 2 2
5 5 4 2 00 5 10 14 16 16
61 61 56 46 32 16 0· · · · · · · ·
The eighth row is 0, 61, 122, 178, 224, 256, 272, 272.
26. (Problem 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
k
)Bkx
n−k.
Show that B0(x) = 1, B1(x) = x−1/2, B2(x) = x2−x+1/6, and B3(x) =x3 − 3
2x2 + 1
2x.(b) If n ≥ 1, then
Bn(x+ 1)−Bn(x) = nxn−1,
and if n ≥ 2, thenBn(0) = Bn(1) = Bn.
(c) Define Sm(n) = 1m + 2m + · · ·+ (n− 1)m. Show that
(m+ 1)Sm(n) = Bm+1(n)−Bm+1.
(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
0Bn(x) dx =
0 for n ≥ 1, and show that from (b) one obtains∫ x+1
m
Bn(t) dt = xn.
(e) Calculate the Fourier series of B1(x) to conclude that for 0 < x < 1we have
B1(x) = x− 1
2=−1
π
∞∑k=1
sin(2πkx)
k.
44
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 (i.e.
requiring∫ 1
0Bn(x) dx = 0), successive integrals (antiderivatives) of the
sawtooth function x− 1/2, 0 < x < 1.Solution.
zexz
ez − 1=
z
ez − 1exz
=
∞∑n=0
Bnn!zn∞∑n=0
xn
n!zn
=
∞∑n=0
(n∑k=0
Bkk!
xn−k
(n− k)!
)zn
=
∞∑n=0
1
n!
(n∑k=0
(n
k
)Bkx
n−k
)zn
Therefore
Bn(x) =
n∑k=0
(n
k
)Bkx
n−k.
ze(x+1)z
ez − 1− zexz
ez − 1
=zexz
ez − 1(ez − 1)
= zexz
=∑n=0
xn
n!zn+1
=∑n=1
xn−1
(n− 1)!zn
45
Therefore comparing coefficients of zn, n ≥ 1 we get
Bn(x+ 1)
n!− Bn(x)
n!=
xn−1
(n− 1)!
and henceBn(x+ 1)−Bn(x) = nxn−1,
So for n ≥ 2, Bn(1) − Bn(0) = n0n−1 = 0,i.e. Bn(1) = Bn(0) = Bn.(Note that by definition of Bn(x), we have Bn(0) = Bn.)For m ≥ 1,
(m+ 1)
n−1∑k=1
km
=
n−1∑k=1
Bm+1(k + 1)−Bm+1(k)
= Bm+1(n)−Bm+1(1)
= Bm+1(n)−Bm+1
Note that this yields a formula for the sum
1m + 2m + · · ·+ nm =1
m+ 1(Bm+1(n+ 1)−Bm+1).
Write F (x, z) for zexz
ez−1 . Since
∂
∂xF (x, z) = zF (x, z)
, comparing coefficient of zn on both sides, we get
B′n(x)
n!=Bn−1(x)
(n− 1)!,
hence B′n(x) = nBn−1(x). Next,∫ 1
0
Bn(x) dx =1
n+ 1
∫ 1
0
B′n+1(x) dx =1
n+ 1(Bn+1(1)−Bn+1(0)) = 0.
From (b),∫ x+1
x
Bn(t) dt =1
n+ 1
∫ x+1
x
B′n+1(t) dt =1
n+ 1(Bn+1(x+1)−Bn+1(x)) = xn.
From Maple, one gets
B1(x) = x− 1
2=−1
π
∞∑k=1
sin(2πkx)
k.
46
B2(x) is an antiderivative of 2B1(x), so
B2(x) =−2
π
∞∑k=1
− cos(2πkx)
2πk2+ C
But∫ 1
0B2(x) dx = 0 and
∫ 1
0cos(2πkx) dx = 0, so C = 0, and
B2(x) =2
2π2
∞∑k=1
cos(2πkx)
2πk2=
2 · 2!
(2π)2
∞∑k=1
cos(2πkx)
k2.
Next, arguing the same manner, B3(x) is an antiderivative of 3B2(x), and
B3(x) =2 · 3!
(2π)3
∞∑k=1
sin(2πkx)
k3.
And then
B4(x) = − 2 · 4!
(2π)4
∞∑k=1
cos(2πkx)
k4.
and so on, and we get
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.
Writing sin, cos in terms of e powers, and noting that i2m = (−1)m, weget the last formula
Bn(x) = − n!
(2πi)n
∑k 6=0
e2πikx
kn.
Chapter 4
1. (Problem 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 = 4N ,
47
where 4N 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≤|j|≤N+K−1
(1− |j| −N
K
)f(j)eijθ.
From this last expression for σN,K conclude that
|σN,K − SM | ≤∑
N+1≤|j|≤N+K−1
|f(j)|
for all N ≤M < N +K.(b) Use one of the above formulas and Fejer’s theorem to show that withN = 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−)
2as 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 iscontinuous on [−π, π], show that σkn,n(f)→ f uniformly as n→∞.
(c) Using part (a), show that if f(j) = O(1/|j|) and kn ≤ m < (k + 1)n,we get
|σkn,n − Sm| ≤C
kfor some constant C > 0.
(d) Suppose that f(j) = O(1/|j|). 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−)
2as N →∞.
Also, show that if f is continuous on [−π, π], then SN (f)→ f uniformly.(e) The above arguments show if
∑cn is Cesaro 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).Solution.
48
(a). σN,K = 1K ((N + K)σN+K − NσN ) is straightforward. Write aj for
f(j)eijθ. We have
σN,K
=1
K[SN + (SN + aN+1 + a−N−1) + · · ·+ (SN +
∑N+1≤|j|≤N+K−1
aj)]
= SN +1
K
K−1∑l=1
∑N+1≤|j|≤N+l
aj
= SN +1
K
∑N+1≤|j|≤N+K−1
aj
K−1∑l=|j|−N
1
= SN +1
K
∑N+1≤|j|≤N+K−1
(K +N − |j|)aj
= SN +∑
N+1≤|j|≤N+K−1
(1− |j| −N
K
)aj
For N ≤ M < N +K, σN,K − SM =∑N+1≤|j|≤N+K−1 bjaj , where bj is
either 1− |j|−NK or − |j|−NK , hence |σN,K − SM | ≤∑N+1≤|j|≤N+K−1 |aj |.
(b). We have
σkn,n =1
n[(k + 1)nσ(k+1)n − knσkn] = (k + 1)σ(k+1)n − kσkn
Let A be either f(θ) or f(θ+)+f(θ−)2 . Since σ(k+1)n, σkn approach the
same value A, as n → ∞, we see that σkn,n → A. Uniform convergencestatement is also clear from σkn,n = (k+1)σ(k+1)n−kσkn. (I think I haveto elaborate this last sentence.)
(c). Suppose |f(j)| = |aj | ≤ C1/|j| for some constant C1 > 0. From thelast part of (a), for kn ≤ m < (k + 1)n,
|σkn,n − Sm|
≤ C1
∑kn+1≤|j|≤(k+1)n−1
1
|j|
≤ 2C1n− 1
kn+ 1
≤ 2C11
k=C
k
(d). Let ε > 0. Fix a positive integer k such that Ck < ε. Choose N such
that|σkn,n −A| < ε
49
for all n ≥ N . Then for all m ≥ kN , we have kn ≤ m < (k+ 1)n for somen ≥ N , and
|Sm −A| ≤ |Sm − σkn,n|+ |σkn,n −A| < 2ε.
This proves that Sm → A as m→∞.(e)Note that except for the uniform convergence part, the only propertiesof the series
∑j aj that we use in the above proof are that it is Cesaro
summable to A and that aj = O(1/|j|).
Chapter 5 The Fourier Transform on R
1. (Exercise 1) Corollary 2.3 in Chapter 2 leads to the following simplifiedversion of the Fourier inversion formula. Suppose f is a continuous func-tion supported on an interval [−M,M ], whose Fourier transform f is ofmoderate 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/2f(x)e−2πinx/L dx =
1
Lf(n/L).
Alternatively, we may write f(x) = δ∑∞n=−∞ f(nδ)e2πinδx with δ = 1/L.
(b) Prove that if F is continuous and of moderate decrease, then∫ ∞−∞
F (ξ) dξ = limδ→0,δ>0
δ
∞∑n=−∞
F (δn).
(c) Conclude that f(x) =∫∞−∞ f(ξ)e2πixξ dξ.
Solution.(a)
∞∑n=−∞
an(L)e2πinx/L =1
L
∞∑n=−∞
f(n
L)e2πinx/L
Let B,C be the constants defined as in Remark 1 below. Since |f(nL )| ≤Bn2 , the above series is absolutely convergent. Hence the series convergesuniformly (and absolutely) to f(x) by Corollary 2.3 in Chapter 2.(b). Given ε > 0, choose N > C such that
∫|x|>N |F (x)| dx < ε/4 and∑
|n|>NBn2 < ε/4. Choose 0 < δ1 < 1 such that for all 0 < δ < δ1,
|∫ N
−NF (x) dx− δ
∑|n|≤N/δ
F (nδ)| < ε
2.
This is possible because δ∑|n|≤N/δ F (nδ) is almost a Riemann sum. (Fill
in the details.) Since
δ|∑|n|>N/δ
F (δn)| ≤∑|n|>N
|F (δn)| ≤∑|n|>N
B
n2
50
the result follows.(c). Apply (b) to F (ξ) = f(ξ)e2πiξx and use (a).Remark 1. From the book’s definition, a function f defined on R is saidto be of moderate decrease if f is continuous and there exists a constantA > 0 so that
|f(x)| ≤ A
1 + x2for all x ∈ R.
Note that this is equivalent to saying that f is continuous and there arepositive constants B,C such that
|f(x)| ≤ B
x2for all |x| ≥ C.
Proof. Suppose f satisfies the first condition. Since for all x 6= 0
A
1 + x2=
A
1 + x2/2 + x2/2≤ 2A
x2
we can choose B = 2A,C = 1 in the second statement.Now suppose f satisfies the second condition. Let D = max{1, C}. Then
B
x2=
2B
x2 + x2≤ 2B
1 + x2for |x| ≥ D.
LetM = max{(1+x2)|f(x)| : |x| ≤ D}. Then we can letA = max{2B,M}in the first condition.Remark 2. Let f(x) = 1,−1 ≤ x ≤ 1, zero elsewhere. Then the Fourier
transform of f is sin(2πξπξ , which is not integrable. Modifying f , making it
continuous, we let fn be the even extension of the following function
g(x) = 1 for 0 ≤ x ≤ n− 1
n, and − n(x− 1) for
n− 1
n≤ x ≤ 1
and zero for x ≥ 1. The Fourier transform of fn, n ≥ 1, is
n(1− 2 cos2(πξ) + cos(n−1n 2πξ))
2π2ξ2
which is of moderate decrease. A continuous with compact support suchit Fourier transform is not of moderate decrease is in Exercise 3. Itremains to see an example of continuous function of compactsupport such that its Fourier transform is not integrable. Fromsci.math, cronopio said Terry Tao seems to say that the function (1 −log(1 − |x|))−a, |x| < 1, zero elsewhere, (0 < a < 1 is a fixed number) isan example. See http://mathoverflow.net/questions/43550/is-the-fourier-transform-of-1-1-log1-x2-supported-in-1-1-integrableExistence of such function also follows from closed graph theorem, see thelink in the above link.
51
2. (Exercise 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 transformstill make sense. Show that
f(ξ) =sin 2πξ
πξand g(ξ) =
(sinπξ
πξ
)2
,
with the understanding that f(0) = 2 and g(0) = 1.Solution. Check with Maple and found to agree.
3. (Exercise 3) The following exercise illustrates the principle that the decay
of f is related to the continuity properties of f .(a) Suppose that f is a function of moderate decrease on R whose Fourier
transform f is continuous and satisfies
f(ξ) = O
(1
|ξ|1+α
)as |ξ| → ∞
for some 0 < α < 1. Prove that f satisfies a Holder 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, withf(0) = 0, and which is equal to 1/ log(1/|x|) for all x in a neighborhood
of the origin. Prove that f is not of moderate decrease. In fact, there isno ε > 0 so that f(ξ) = O(1/|ξ|1+ε) as |ξ| → ∞.Solution.By remarks in section 1.7, since both f and f are of moderate decrease,the inverse Fourier transform of f is f . Thus
f(x+ h)− f(x) =
∫ ∞−∞
f(ξ)e2πiξx(e2πiξh − 1) dξ.
The condition on f is equivalent to
|f(ξ)| ≤ A
1 + |ξ|1+αfor all ξ ∈ R
52
for some A > 0. Thus (note that |e2πiξh − 1| = 2| sin(πξh)|)∣∣∣∣f(x+ h)− f(x)
hα
∣∣∣∣ ≤ 1
|h|α
∫ ∞−∞
A|e2πiξh − 1|1 + |ξ|1+α
dξ
≤ 4A
|h|α
∫ ∞0
| sin(πξh)|1 + ξ1+α
dξ
=4A
π
∫ ∞0
| sin(u)||h|1+α + u1+α
du
≤ 4A
π
(∫ 1
0
| sin(u)u |uα
du+
∫ ∞1
1
u1+αdu
)
≤ 4A
π
(∫ 1
0
1
uαdu+
∫ ∞1
1
u1+αdu
)<∞
(b). We have|f(h)− f(0)|
|h|ε=
1
−|h|ε log |h|→ ∞
as h→ 0 for any fixed ε > 0. So by (a), f is not of moderate decrease.
4. (Exercise 4) Examples of compactly supported functions in S(R) are veryhandy 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 orx ≥ 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 Rsuch that F (x) = 0 if x ≤ a, F (x) = 1 if x ≥ b, and F is strictly increasingon [a, b].(c) Let δ > 0 be so small that a + δ < b − δ. Show that there exists anindefinitely differentiable function g such that g is 0 if x ≤ a or x ≥ b, gis 1 on [a+ δ, b− δ], and g is strictly monotonic on [a, a+ δ] and [b− δ, b].Solution. Note that the graph of −1/(x − a) − 1/(b − x) is symmetricabout the line x = (a + b)/2 with maximum value of −4/(b − a) at thepoint x = (a+ b)/2.For (b) consider F (x) = c
∫ x−∞ f(t) dt where c is the reciprocal of
∫∞−∞ f(t) dt.
5. (Exercise 5) Suppose f is continuous and of moderate decrease.
(a) Prove that f is continuous and f(ξ)→ 0 as |ξ| → ∞.
(b) Show that if f(ξ) = 0 for all ξ, then f is identically 0.Solution.
53
|f(ξ + h)− f(ξ)| =∣∣∣∣∫ ∞−∞
f(x)e−2πiξx(e−2πihx − 1) dx
∣∣∣∣≤∫ ∞−∞|f(x)||e−2πihx − 1| dx
=
∫ ∞−∞|f(x)|2| sin 2πhx| dx
≤ 2A
∫ ∞−∞
| sin 2πhx|1 + x2
dx
Let ε > 0. Choose N > 0 such that 2A∫|x|>N
11+x2 dx < ε/2. Since
| sin 2πhx|1+x2 → 0 uniformly on [−N,N ] as h→ 0, there exists δ > 0 such that
2A∫ N−N
| sin 2πhx|1+x2 dx < ε/2 for |h| < δ. This proves that f is continuous.
Note: Lebesgue dominated convergence theorem could be applied to yielda simpler proof.(b).
f(ξ) =1
2
(∫ ∞−∞
f(x)e−2πiξx −∫ ∞−∞
f(x)e−2πi(ξx+1/2) dx
)=
1
2
∫ ∞−∞
(f(x)− f(x− 1
2ξ))e−2πiξx dx
Now the result follows from applying Lebesgue dominated convergencetheorem; without it, the proof is harder.(b). If f, g are of moderate decrease, then the function f(x)g(y)e−2πixy isintegrable over R2. Then it follows from Fubini’s theorem that∫
f(x)g(x) dx =
∫g(x)f(x) dx.
Suppose f(x) = 0 for all x. For any t and any δ > 0, Kδ(t−x) as a functionof x is in S(R), so by Corollary 1.10 it is equal to the Fourier transform gδfor some gδ in S(R). So
∫f(x)gδ(x) dx =
∫f(x)Kδ(t−x) dx = 0. Letting
δ → 0, we get f(t) = 0.
6. (Exercise 8) Prove that if f is continuous, of moderate decrease, and∫∞−∞ f(y)e−y
2
e2xy dy = 0 for all x ∈ R, then f = 0.Proof.Let g(x) = e−x
2
. Then (f∗g)(x) =∫∞−∞ f(y)e−(x−y)
2
dy = e−x2 ∫∞−∞ f(y)e−y
2
e2xy dy =
0 for all x. This implies that f ∗ g(ξ) = f(ξ)g(ξ) = f(ξ)√πe−π
2ξ2 = 0 for
all ξ. So f = 0. By Theorem 1.9, f = 0.
7. (Exercise 12) Show that the function defined
u(x, t) =x
tHt(x)
54
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. Proof.Maple shows that
∂u
∂t=∂2u
∂x2=x(x2 − 6t)
8√πt7/2
ex2/4t.
It is clear that limt→0+ u(x, t) = 0 for all x. If we approach the originalong the parabola x2 = t, we get
limx→0
1
x2√
4πe−1/4 =∞.
8. (Add in) Suppose f is an even function such that both f and f are of
moderate decrease. Then f is also even and the Fourier transform of f(x)is f(ξ).Proof.The
∫below denotes
∫∞−∞. We have
f(−ξ) =
∫f(x)e2πixξ dx =
∫f(−x)e−2πixξ dx =
∫f(x)e−2πixξ dx = f(ξ)
so f is even.Since
f(x) =
∫f(t)e2πixt dt
we have∫f(t)e−2πiξt dt =
∫f(−t)e2πiξt dt =
∫f(t)e2πiξt dt = f(ξ)
proving that the Fourier transform of f(x) is f(ξ).
9. (Exercise 15) This exercise provides another example of periodization.(a) Apply the Poisson summation formula to the function g(x) and itsFourier transform in Exercise 2 to obtain
∞∑n=−∞
1
(n+ α)2=
π2
(sinπα)2
whenever α is real, but not equal to an integer.(b) Prove as a consequence that
∞∑n=−∞
1
n+ α=
π
tanπα(1)
whenever α is real but not equal to an integer.Solution.
55
By the last item, the Fourier transform of h(x) = g(x) is g(x) in Exercise2. By Poisson summation formula
∞∑n=−∞
sin2(π(n+ α))
π2(n+ α)2=
∞∑n=−∞
g(n)e2πinα = 1.
(If α is an integer, then the equality is reduced to trivial 1 = 1.) Sincesin(nπ + πα) = (−1)n sin(πα), the result follows.(b). Assume that 0 < α < 1. We have from (a), for n 6= 0,∫ α
0
1
(−n+ x)2+
1
(n+ x)2dx = −
(1
−n+ α+
1
n+ α
)and since limx→0
π2
(sinπx)2 −1x2 = π2/3, it has a removable discontinuity at
x = 0 and limx→0− πtanπx + 1
x = 0, we have∫ α
0
π2
(sinπx)2− 1
x2dx = − π
tanπα+
1
α.
Therefore1
α+
∞∑n=1
(1
n+ α+
1
−n+ α
)=
π
tanπα.
Evaluating at α = 1/2 yields 0; this is correct since
∞∑n=1
4
4n2 − 1=
∞∑n=1
2
2n− 1− 2
2n+ 1
is a telescoping series whose sum is 2.Now we have to prove for any α, a noninteger. Let n1 = bαc. Thenα = n1 + α1 for some 0 < α1 < 1. Then
∞∑n=−∞
1
n+ α=
∞∑n=−∞
1
n+ α1=
π
tanπα1=
π
tanπα.
10. (Exercise 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
f(ξ) = e−2πt|ξ| where t > 0 in order to get
1
π
∞∑n=−∞
t
t2 + n2=
∞∑n=−∞
e−2πt|n|.
(b) Prove the following identity:
1
π
∞∑n=−∞
t
t2 + n2=
1
πt+
2
π
∞∑m=1
(−1)m+1ζ(2m)t2m−1, 0 < t < 1
56
as well as∞∑
n=−∞e−2πt|n| =
2
1− e−2πt− 1, 0 < t < 1.
(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.
Solution.(a) By Lemma 2.4, the Fourier transform of f(x) = t/(π(x2+t2)) is f(ξ) =e−2πt|ξ|. Then by Poisson summation formula with x = 0 ((x+n)2 = n2),we get (a).(b).
1
π
∞∑n=−∞
t
t2 + n2=
1
πt+
1
π
∑n 6=0
t
t2 + n2
=1
πt+
2
π
∞∑n=1
t
t2 + n2
=1
πt+
2
π
∞∑n=1
t/n2
1 + (t/n)2
=1
πt+
2
π
∞∑n=1
t
n2
∞∑m=0
(−1)mt2m
n2m
=1
πt+
2
π
∞∑n=1
∞∑m=0
(−1)mt2m+1
n2(m+1)
=1
πt+
2
π
∞∑n=1
∞∑m=1
(−1)m+1 t2m−1
n2m
=1
πt+
2
π
∞∑m=1
∞∑n=1
(−1)m+1 t2m−1
n2m
=1
πt+
2
π
∞∑m=1
(−1)m+1ζ(2m)t2m−1
57
Next,
∞∑n=−∞
e−2πt|n| = 1 + 2
∞∑n=1
e−2πtn
= 1 + 2e−2πt
1− e−2πt
=1 + e−2πt
1− e−2πt
=2− (1− e−2πt)
1− e−2πt
=2
1− e−2πt− 1
Dividing by 2 on both sides, we get
1
2πt+
1
π
∞∑m=1
(−1)m+1ζ(2m)t2m−1 =1
1− e−2πt− 1
2.
Letting z = −2πt and multiplying both sides by −z, we get
1 +1
π
∞∑m=1
(−1)m+1ζ(2m)2πz2m
(2π)2m=
z
ez − 1+z
2.
Comparing with
z
ez − 1= 1− z
2+
∞∑m=1
B2m
(2m)!z2m,
we get
2ζ(2m) = (−1)m+1 (2π)2m
(2m)!B2m.
11. (Exercise 23) The Heisenberg uncertainty principle can be formulated in
terms of the operator L = − d2
dx2 + x2, which acts on Schwartz functionsby the formula
L(f) = −d2f
dx2+ x2f.
This operator, sometimes called the Hermite operator, is the quantumanalogue of the harmonic oscillator. Consider the usual inner product onS given by
(f, g) =
∫ ∞−∞
f(x)g(x) dx whenever f, g ∈ S.
(a) Prove that the Heisenberg uncertainty principle implies
(Lf, f) ≥ (f, f) for all f ∈ S.
58
This is usually denoted by L ≥ I.(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.(c) Now for t ∈ R, let
At(f) =df
dx+ txf and A∗t (f) = − df
dx+ txf.
Use the fact that (A∗tAtf, f) ≥ 0 to give another proof of the Heisenberguncertainty principle which says that whenever
∫∞−∞ |f(x)|2 = 1 then(∫ ∞
−∞x2|f(x)|2
)(∫ ∞−∞
∣∣∣∣ dfdx∣∣∣∣2 dx
)≥ 1
4.
Solution.(a) By dividing f by
√(f, f) if necessary, we may assume that (f, f) = 1.
We need to prove that (Lf, f) ≥ 1. Now
(Lf, f) = −(f ′′, f) + (x2f, f) = (f ′, f ′) + (x2f, f) =
∫|f ′|2 +
∫x2|f |2 = 4π2
∫ξ2|f |2 +
∫x2|f |2
≥ 4π(
∫ξ2|f |2)1/2)(
∫x2|f |2)1/2 ≥ 4π
1
4π= 1
where we have used integration by parts and theorem 4.1 (Heisenberguncertainty principle).(b) Direct checking. Use integration by parts. Note: AA∗ = L+ I.(c) Similar to (b), we also have (A∗tAtf, f) ≥ 0 for all t. When this iswritten out, we get t2(x2f, f)−t(f, f)−(f ′′, f) = t2(x2f, f)−t+(f ′, f ′) ≥ 0for all t. It follows that
1− 4(f ′, f ′)(x2f, f) ≤ 0
from which the result follows.
12. (Problem 1) The equation
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 heatequation which occurs in a number of applications. To solve it, make the
59
change of variables 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,
with U(y, 0) = F (y). This can be solved like the usual heat equation (thecase a = 1) by taking the Fourier transform in the y variable. One mustthen compute the integral∫ ∞
−∞e(−4π
2ξ2+(1−a)2πiξ)te2πiξν dξ.
Show that the solution of the original problem is then given by
u(x, t) =1
(4πt)1/2
∫ ∞0
e(− log(ν/x)+(1−a)t)2/(4t)f(ν)dν
ν.
Solution.I have checked that what it says up to the new differential equation iscorrect.Taking the Fourier transform w.r.t. y of the new equation, we get
∂U
∂t= (−4π2ξ2 + (1− a)2πiξ)U .
Solve, using U(ξ, 0) = F (ξ), to get
U(ξ, t) = F (ξ)e(−4π2ξ2+(1−a)2πiξ)t.
It remains to find the inverse Fourier transform of e(−4π2ξ2+(1−a)2πiξ)t and
then write U(y, t) as a convolution. We have∫ ∞−∞
e(−4π2ξ2+(1−a)2πiξ)te2πiξy dξ =
∫ ∞−∞
e−4π2ξ2te((1−a)t+y)2πiξ dξ
which is equal to the inverse transform of e−4π2tξ2 evaluated at (1−a)t+y.
Using F(e−πy2
) = e−πξ2
and F(δf(δy)) = f(δ−1ξ) with δ = 1/√
4πt, wefind that the required inverse transform is
1√4πt
e−y2/4t
evaluating at (1− a)t+ y yields
1√4πt
e−(y+(1−a)t)2/(4t)
Therefore
U(y, t) =
∫ ∞−∞
F (µ)1√4πt
e−(y−µ+(1−a)t)2/(4t) dµ.
Make a substitution µ = − log ν, and substitute y = − log x, we get theresult.
60
13. (Problem 2) The Black-Scholes equation from finance theory is
∂V
∂t+ rs
∂V
∂s+σ2s2
2
∂2V
∂s2− rV = 0, 0 < t < T, (2)
subject to the ”final” boundary condition V (s, T ) = F (s). An appropri-ate change of variables reduces this to the equation in Problem 1. Al-ternatively, the substitution V (s, t) = eax+bτU(x, τ) where x = log s, τ =σ2
2 (T − t), a = 12 −
rσ2 , and b = −
(12 + r
σ2
)2reduces (2) to the one-
dimensional heat equation with the initial condition U(x, 0) = e−axF (ex).Thus a solution to the Black-Scholes equation 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∗
s∗.
Solution. Multiplying the equation by e−rt. Define V1 = e−rtV . Thenthe equation becomes
∂V1∂t
+ rs∂V1∂s
+σ2s2
2
∂2V1∂s2
= 0.
Let t1 = −(σ2/2)t. Then ∂V1
∂t = ∂V1
∂t1(−σ2/2), and upon dividing by −σ2/2
on both sides we get
∂V1∂t1
=2r
σ2s∂V1∂s
+ s2∂2V1∂s2
.
This is the equation in Problem 1.
For the alternative way, compute ∂V∂t ,
∂V∂s and ∂2V
∂s2 , replacing 1/s by ex
and canceling out eax+bτ , we get
−σ2
2(bU+
∂U
∂τ)+r(aU+
∂U
∂x)+
σ2
2(a(a−1)U+(2a−1)
∂U
∂x+∂2U
∂x2)−rU = 0.
Since r+ σ2
2 (2a− 1) = 0,−σ2
2 b+ ra+ σ2
2 a(a− 1)− r = 0, the coefficients
of U and ∂U∂x are zero. Upon canceling out σ2
2 , we get the heat equation∂U∂τ = ∂2U
∂x2 . Since F (ex) = F (s) = V (s, T ) = eaxU(x, 0), the initialcondition for the heat equation is U(x, 0) = e−axF (ex). From the formulafor solution of heat equation, we get
U(x, τ) =1√4πτ
∫ ∞−∞
e−ayF (ey)e−(x−y)2/(4τ) dy.
Making a change of variable y = log s∗ and substitute x = log s, we find
U(log s, τ) =1√4πτ
∫ ∞0
e−a log s∗F (s∗)e−(log(s/s∗)2/(4τ) ds
∗
s∗.
61
and
V (s, t) =1√4πτ
∫ ∞0
ea log s+bτe−a log s∗F (s∗)e−(log(s/s∗)2/(4τ) ds
∗
s∗.
Adding up the exponents of e, we get
a log(s/s∗) + bτ − (log(s/s∗)2/(4τ)
= − 1
4τ[log2(s/s∗)− 4aτ log(s/s∗)− 4bτ2]
= − 1
4τ(log(s/s∗)− 2aτ)2 + (a2 + b)τ
Since 4τ = 2σ2(T − t),−2aτ = (r− σ2
2 )(T − t) and (a2 + b)τ = −r(T − t),we are done.
14. (Problem 4) If g is a smooth function on R, define the formal power series
u(x, t) =
∞∑n=0
g(n)(t)x2n
(2n)!(3)
(a) Check formally that u solves the heat equation.(b) For a > 0, consider the function defined by
g(t) =
{e−t
−aif 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)ke−
12 t
−afor t > 0.
(c) As a result, for each x and t the series (3) converges; u solves the heatequation; u vanishes for t = 0; and u satisfies the estimate |u(x, t)| ≤Cec|x|
2a/(a−1)
for some constants C, c > 0.(d) Conclude that for every ε > 0 there exists a nonzero solution to theheat equation which is continuous for x ∈ R and t ≥ 0, which satisfiesu(x, 0) = 0 and |u(x, t)| ≤ Cec|x|2+ε .
15. (Problem 7) The Hermite functions hk(x) are defined by the generatingidentity
∞∑k=0
hk(x)tk
k!= e−(x
2/2−2tx+t2). (4)
(a) Show that an alternate definition of the Hermite functions is given bythe formula
hk(x) = (−1)kex2/2
(d
dx
)ke−x
2
. (5)
62
Conclude from the above expression that each hk(x) is of the form Pk(x)e−x2/2,
where Pk is a polynomial of degree k. In particular, the Hermite functionsbelong to the Schwartz space and h0(x) = e−x
2/2, h1(x) = 2xe−x2/2.
(b) Prove that the family {hk}∞k=0 is complete in the sense that if f is aSchwartz function, and
(f, hk) =
∫ ∞−∞
f(x)hk(x) dx = 0 for all k ≥ 0,
then f = 0.(c) Define h∗k(x) = hk((2π)1/2x). Then
h∗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 Exercise23, and in fact, prove that
Lhk = (2k + 1)hk.
In particular, we conclude that the functions hk are mutually orthogonalfor the L2 inner product on the Schwartz space.(e) Finally, show that
∫∞−∞ hk(x)2 dx = π1/22kk!.
Solution.(a) Note that e−(x
2/2−2tx+t2) = ex2/2e−(t−x)
2
. Note that(d
dx
)ke−x
2
= pk(x)e−x2
for some degree k polynomial pk(x) which is odd if k is odd, and even if k
is even. Therefore the k partial derivative of e−(t−x)2
w.r.t. t, evaluatedat t = 0 is pk(−x)e−x
2
= (−1)kpk(x)e−x2
. The formula then follows fromTaylor expansion.(b) Under the hypothesis, we have by equation (4)∫ ∞
−∞f(x)e−(x
2/2−2tx+t2) dx = e−t2
∫ ∞−∞
f(x)e−x2/2+2tx dx
for all t. By Exercise 8, f = 0.(c) From equation (4), we get
∞∑k=0
hk(√
2πx)tk
k!
= exp(−(πx2 − 2t√
2πx+ t2))
= exp(t2) exp(−π(x−√
2/πt)2)
63
Taking Fourier transform of both sides, we get
∞∑k=0
h∗k(ξ)tk
k!= exp(t2) exp(−2
√2πitξ) exp(−πξ2)
= exp(−(πξ2 − 2(−it)√
2πξ + (−it)2))
=
∞∑k=0
h∗k(ξ)(−it)k
k!
Comparing both sides, we get the result.(d) From the RHS of (4), we have
∞∑k=0
L(hk(x))tk
k!= L(e−(x
2/2−2tx+t2))
= − d2
dx2e−(x
2/2−2tx+t2) + x2e−(x2/2−2tx+t2)
= − d
dxe−(x
2/2−2tx+t2)(−x+ 2t) + x2e−(x2/2−2tx+t2)
= −e−(x2/2−2tx+t2)((−x+ 2t)2 − 1) + x2e−(x
2/2−2tx+t2)
= e−(x2/2−2tx+t2)(1− 4t2 + 4tx)
= e−(x2/2−2tx+t2) + (−4t2 + 4tx)e−(x
2/2−2tx+t2)
=
∞∑k=0
hk(x)tk
k!+ 2t
d
dte−(x
2/2−2tx+t2)
=
∞∑k=0
hk(x)tk
k!+
∞∑k=0
2khk(x)tk
k!
=
∞∑k=0
(2k + 1)hk(x)tk
k!
from which we get Lhk = (2k + 1)hk. By Exercise 23, L = I +A∗A, so itis Hermitian. Therefore the functions hk are mutually orthogonal.(e) Multiply both sides of equation (4) by hk(x) and integrate. On theLHS, using the orthogonality property, we get
tk
k!
∫ ∞−∞
[hk(x)]2 dx.
On the RHS, using equation (5) and integration by parts k times, we get
(−1)k(−1)k∫ ∞−∞
e−x2
(d
dx
)ke2tx−t
2
= (2t)k∫ ∞−∞
e−(x−t)2
dx = 2ktk√π.
Therefore result.
64
Chapter 6 The Fourier Transform on Rd
1. (Exercise 1) Suppose that R is a rotation in the plane R2, and let(a bc 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 equationsin a, b, c, d.(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→ ze−iφ,and if R is improper, then it takes the form z 7→ zeiφ, where z = x− iy.Solution.(a) Write D = det(R). Then D = ±1 and the conditions Rt = R−1
implies (a cb d
)=
(d/D −b/D−c/D a/D
),
i.e. we have D = ad− bc = ±1, a = dD, b = −cD, c = −bD, d = aD.(b) From (a) we have D = ad − bc = a2D − (−b2D) = (a2 + b2)D whichimplies that a2 + b2 = 1. Thus (b) follows.(c). Suppose D = 1. Then ze−iφ = (x+ iy)(a− ib) = (ax+ by) + i(−bx+ay) = (ax+ by) + i(cx+ dy).If D = −1, then zeiφ = (x − iy)(a + ib) = ax + by + i(bx − ay) =ax+ by + i(cx+ dy). This proves (c).
2. (Exercise 2) Suppose R : R3 → R3 is a proper rotation.(a) Show that p(t) = det(R − tI) is a polynomial of degree 3, and provethat there exists γ ∈ S2 (where S2 denotes the unit sphere in R3) with
R(γ) = γ.
(b) If P denotes the plane perpendicular to γ and passing through theorigin, show that
R : P → P,and that this linear map is a rotation.Solution.(a). p(t) is clearly a polynomial of degree 3. (The coefficient of t3 is−1.). p(0) = det(R) > 0. Since limt→∞ p(t) = −∞, we see that thereexists λ > 0 such that p(λ) = 0. So R − λI is singular, and its kernel isnontrivial.
Chapter 7 Finite Fourier Analysis
1. (Exercise 1.) Let f be a function on the circle. For each N ≥ 1 the discreteFourier coefficients of f are defined by
aN (n) =1
N
N−1∑k=0
f(e2πik/N )e−2πikn/N =1
N
N−1∑k=0
f(e2πik/N )en(k), for n ∈ Z.
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We also let
a(n) =
∫ 1
0
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 →∞.Proof.(a) is easy since e−2πi = 1.(b) Note that aN (n) is the Riemann sum of the integral a(n) with parti-tion: k/N, k = 0, · · · , N , and choice of points: k/N, k = 0, · · · , N − 1.
2. (Exercise 2) If f is a C1 function on the circle, prove that |aN (n)| ≤ c/|n|whenever 0 < |n| ≤ N/2.Proof. Let l be an integer. Then
aN (n)[1− e2πiln/N ]
=1
N
N−1∑k=0
f(e2πik/N )e−2πikn/N − 1
N
N−1∑k=0
f(e2πik/N )e−2πi(k−l)n/N
=1
N
N−1∑k=0
f(e2πik/N )e−2πikn/N − 1
N
N−l−1∑k=−l
f(e2πi(k+l)/N )e−2πikn/N
=1
N
N−1∑k=0
f(e2πik/N )e−2πikn/N − 1
N
N−1∑k=0
f(e2πi(k+l)/N )e−2πikn/N
=1
N
N−1∑k=0
[f(e2πik/N )− f(e2πi(k+l)/N )]e−2πikn/N
f ∈ C1 implies f is Lipschitz. Let M be its Lipschitz constant. Thenfrom the above equations,
|aN (n)||1− e2πiln/N | ≤M |1− e2πil/N |
Let l be the integer closest to N2n (in case of ambiguity, choose either
integer). Then |l −N/2n| ≤ 1/2, from which we get∣∣∣∣ lnN − 1
2
∣∣∣∣ ≤ |n|2N≤ 1
4.
Thus
|1− e2πil/N | ≤ |2πl/N | = |2πln/N |/|n| ≤ 3π
2|n|.
On the other hand, since ∣∣∣∣ lnN − 1
2
∣∣∣∣ ≤ 1
4
66
impliesπ
2≤ 2πln
N≤ 3π
2,
we have|1− e2πiln/N | ≥
√2.
Therefore for 0 < |n| ≤ N/2,
|aN (n)| ≤ 3πM
2√
2|n|.
3. (Exercise 3) By a similar method, show that if f is a C2 function on thecircle, then
|aN (n)| ≤ c
|n|2, whenever 0 < |n| ≤ N/2.
As a result, prove the inversion formula for f ∈ C2,
f(e2πix) =
∞∑n=−∞
a(n)e2πinx
from its finite version.Solution.We have
aN (n)[2− e2πiln/N − e−2πiln/N ]
=1
N
N−1∑k=0
[2f(e2πik/N )− f(e2πi(k+l)/N )− f(e2πi(k−l)/N )]e−2πikn/N
≤ 1
N
N−1∑k=0
M |e2πi(k+l)/N − e2πik/N |2
= M |e2πil/N − 1|2
Then do as we did in Exercise 2 to choose l to reach the conclusion.For the second part, let N be odd, then for a fixed k,∑
|n|<N/2
aN (n)e2πikn/N
=1
N
∑|n|<N/2
N−1∑j=0
f(e2πij/N )e−2πijn/Ne2πikn/N
=1
N
N−1∑j=0
f(e2πij/N )∑
|n|<N/2
e−2πijn/Ne2πikn/N
= f(e2πik/N )
where the last equality follows because∑|n|<N/2 e
−2πijn/Ne2πikn/N = 0if j 6= k and equal to N if j = k. Note that I have not used the condition|aN (n)| ≤ c/|n|2. (to be continued.....).
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4. (Exercise 4) Let e be a character on G = Z(N), the additive group ofintegers modulo N . Show that there exists a unique 0 ≤ ` ≤ N−1 so that
e(k) = e`(k) = e2πi`k/N for all k ∈ Z(N).
Conversely, every function of this type is a character on Z(N). Deducethat e` 7→ ` defines an isomorphism from G to G.Proof.Let the N roots of unit be 1, ζ, ζ2, cdots, ζN−1, where ζ = e2πi/N .Let e be a character in Z(N). We have 1 = e(0) = e(1+1+· · ·+1) = e(1)N ,so e(1) = ζl for some 0 ≤ l ≤ N − 1. Then for each n ∈ Z(N),
e(n) = e(1 + 1 + · · ·+ 1) = e(1)n = ζln
proving that e = el. elem = el+m 7→ l +m.
5. (Exercise 5) Show that all characters on S1 are given by
en(x) = einx with n ∈ Z,
and check that en 7→ n defines an isomorphism from S1 to Z.Solution.Since F (0) 6= 0 and F is continuous, there exists δ > 0 such that c =∫ δ0F (y) dy 6= 0. Then
cF (x) =
∫ δ
0
F (y)F (x) dy =
∫ δ
0
F (y + x) dy =
∫ δ+x
x
F (y) dy
for all x. Differentiating both sides,
cF ′(x) = F (δ + x)− F (x) = F (δ)F (x)− F (x) = (F (δ)− 1)F (x).
It follows that F (x) = eAx for some constant A. Since π + π = 0 in S1,we have 1 = F (0) = F (π + π) = e2πA. Therefore A = in for some integern.
6. (Exercise 6) Prove that all characters on R take the form
eξ(x) = e2πξx woth ξ ∈ R,
and that eξ 7→ ξ defines an isomorphism from R to R.Solution.Using the same argument in Exercise 5, F (x) = eAx for some complexnumber A. Since |F (x)| = 1, A must be pure imaginary. So A = 2πiξ forsome real number ξ. eξeη = eξ+η 7→ ξ + η.
7. (Exercise 8) Suppose that P (x) =∑Nn=1 ane
2πinx.(a) Show by using the Parseval identities for the circle and Z(N), that∫ 1
0
|P (x)|2 dx =1
N
N∑j=1
|P (j/N)|2.
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(b) Prove the reconstruction formula
P (x) =
N∑j=1
P (j/N)K(x− (j/N))
where
K(x) =e2πix
N
1− e2πiNx
1− e2πix=
1
N(e2πix + e2πi2x + · · ·+ e2πiNx).
Observe that P is completely determined by the values P (j/N) for 1 ≤j ≤ N . Note also that K(0) = 1, and K(j/N) = 0 whenever j is notcongruent to 0 modulo N .Solution.Considering P (x) as a 1-periodic function, we have by Parseval identity:∫ 1
0|P (x)|2 dx =
∑Nn=1 |an|2. On the other hand, considering f(j) =
P (j/N) =∑Nn=1 ane
2πinj/N as a function on Z(N), Parseval identity
yields∑Nn=1 |an|2 = ‖f‖2 = 1
N
∑Nj=1 |f(j)|2 = 1
N
∑Nj=1 |P (j/N)|2.
8. (Exercise 12) Suppose that G is a finite abelian group and e : G → C isa function that satisfies e(x + y) = e(x)e(y) for all x, y ∈ G. Prove thateither e is identically 0, or e never vanishes. In the second case, show thatfor each x, e(x) = e2πir(x) for some r = r(x) ∈ Q of the form p/q, whereq = |G|.Proof.Since e(0) = e(0 + 0) = e(0)e(0), we see that either e(0) = 0 or e(0) = 1.In the first case, e(x) = e(x + 0) = e(x)e(0) = 0 for all x ∈ G. In thesecond case, 1 = e(0) = e(x + (−x)) = e(x)e(−x), proving that e(x) 6= 0and e(−x) = e(x)−1.Let N = |G|. For any x ∈ G, Nx = x + x + · · · + x = 0, so 1 = e(0) =e(x)N , proving that e(x) is equal to ζk for some k = 0, · · · , N − 1, whereζ = e2πi/N .
9. (Exercise 13) In analogy with ordinary Fourier series, one may interpretfinite Fourier expansions using convolutions as follows. Suppose G is afinite abelian group, 1G its unit, and V the vector space of complex-valuedfunctions on G.(a) The convolution of two functions f and g in V is defined for each a ∈ Gby
(f ∗ g)(a) =1
|G|∑b∈G
f(b)g(a · b−1).
Show that for all e ∈ G one has (f ∗ g)(e) = f(e)g(e).(b) Use Theorem 2.5 to show that∑
e∈G
e(c) = 0 whenever c ∈ G and c 6= 1G.
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(c) As a result of (b), show that the Fourier series Sf(a) =∑e∈G f(e)e(a)
of a function f ∈ V take the form
Sf = f ∗D,
where D is defined by
D(c) =∑e∈G
e(c) =
{|G| if c = 1G,0 otherwise.
Since f ∗ D = f , we recover the fact that Sf = f . Loosely speaking, Dcorresponds 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. ThusD 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).Solution.Let N = |G|.
f ∗ g(e)
=1
N
∑x∈G
(f ∗ g)(x)e(x)
=1
N2
∑x∈G
∑b∈G
f(b)g(xb−1)e(x)
=1
N2
∑b∈G
f(b)∑x∈G
g(xb−1)e(x)
=1
N2
∑b∈G
f(b)∑y∈G
g(y)e(by)
=1
N2
∑b∈G
f(b)e(b)∑y∈G
g(y)e(y)
= f(e)g(e)
(b). If c 6= 1G, then there exists e′ ∈ G such that e′(c) 6= 1.[This can beproved in two ways. Suppose e(c) = 1 for all e ∈ G. Let H be the cyclicsubgroup generated by c. Since c 6= 1G, we have |H| > 1 so |G/H| < |G|.Each e ∈ G induces a character in G/H, and different e’s induce differentcharacters. This is impossible since there are exactly |G/H| characters onG/H.Another proof is to use the fact that G is isomorphic to a direct product ofcyclic groups G1×G2× · · ·×Gn where |G1|||G2|| · · · ||Gn|. Let for each i
70
between 1 and n gi be a generator of Gi, then c =∑i(ai ∗gi). Since c 6= 0
not all integers ai can be zero, let us say it’s k. Now the map defined bygi → 1 for i 6= k and gk → z, with z = e2πi/N where N = |Gk|, defines acharacter that is not 1 on c (it is zak).]Now back to the proof. We have e′
∑e∈G e =
∑e∈G e
′e =∑e∈G e since
G is a group. So e′(c)∑e∈G e(c) =
∑e∈G e(c). Since e′(c) 6= 1, we get∑
e∈G e(c) = 0.
Sf(a) =∑e∈G
f(e)e(a)
=∑e∈G
1
N
∑x∈G
f(x)e(x)e(a)
=1
N
∑x∈G
f(x)∑e∈G
e(a)e(x)−1
=1
N
∑x∈G
f(x)∑e∈G
e(a)e(x−1)
=1
N
∑x∈G
f(x)∑e∈G
e(ax−1)
= (f ∗D)(a)
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