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: tlim@gmu.edu
Abstract
Solutions to some exercises and problems from Stein and Shakarchi’s
Fourier Analysis. The book by Y. Ketznelson, ”An introduction of Harmonic 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 Rd
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 derivatives. Show that whenever x and x + h belong to (a, b), one may write
h2 00
F (x) + h2 φ(h),
2
F (x + h) = F (x) + hF 0 (x) +
where φ(h) → 0 as h → 0.
Deduce that
F (x + h) + F (x − h) − 2F (x)
→ F 00 (x) as h → 0.
h2
Proof.
Firstly one has
Z
x+h
F (x + h) − F (x) =
x
1
F 0 (y) dy.
Then F 0 (y) = F 0 (x)+(y−x)F 00 (x)+(y−x)ψ(y−x), where ψ is continuous
and ψ(h) → 0 as h → 0. Therefore
F (x + h) − F (x) = F 0 (x)h + F 00 (x)
By mean-value theorem
Z h
Z
uψ(u) du = ψ(ξ)
0
h2
+
2
h
u du = h2
0
Z
h
uψ(u) du.
0
ψ(ξ)
= h2 φ(h)
2
where ξ is between 0 and h, and φ(h) → 0 as h → 0 by the continuity of
ψ.
From above, one easily gets
F (x + h) + F (x − h) − 2F (x)
− F 00 (x) = φ(h) + φ(−h) → 0 as h → 0.
h2
(x)
Remark. Suppose F is periodic. The function ζ(x, h) = F (x+h)+F (x−h)−2F
−
h2
00
F (x) is continuous for all x, h. And there exists M such that
|F (x + h) + F (x − h) − 2F (x)| ≤ M h2
for all x, h (since F, F 00 , ζ are all periodic in x).
2. (Exercise 10) Show that the expression of the Laplacian
4=
∂2
∂2
+
∂x2
∂y 2
is given in polar coordinates by the formula
4=
∂2
1 ∂
1 ∂2
+
+ 2 2.
2
∂r
r ∂r r ∂θ
Also prove that
2 2 2
2
∂u + ∂u = ∂u + 1 ∂u .
∂x ∂y ∂r 2
r ∂θ Solution.
From x = r cos θ, y = r sin θ, we get
∂x ∂x cos θ
∂r
∂θ
=
∂y
∂y
sin θ
∂r
∂θ
−r sin θ
,
r cos θ
hence
∂r
∂x
∂θ
∂x
∂r
∂y
∂θ
∂y
!
=
cos θ
sin θ
−r sin θ
r cos θ
2
−1
=
cos θ
− 1r sin θ
sin θ
.
1
r cos θ
By Chain Rule,
∂u
∂u ∂r
∂u ∂θ
=
+
∂x
∂r ∂x ∂θ ∂x
∂u
1 ∂u
=
cos θ −
sin θ.
∂r
r ∂θ
By Chain Rule and Product Rule,
∂ ∂u
cos θ
∂x ∂r
∂ ∂u
∂r
∂ ∂u
∂θ
=
cos θ
+
cos θ
∂r ∂r
∂x ∂θ ∂r
∂x
∂2u
1 ∂u
1 ∂2u
2
2
=
cos
θ
+
sin
θ
−
cos θ sin θ
∂r2
r ∂r
r ∂θ∂r
Similarly,
1 ∂u
∂
−
sin θ
∂x
r ∂θ
1 ∂2u
2
2 ∂u
1 ∂2u
2
=
sin
θ
+
+
sin
θ
cos
θ
−
cos θ sin θ.
r2 ∂θ2
r2
r2 ∂θ
r ∂r∂θ
Since
∂2u
∂r∂θ
=
∂2u
∂θ∂r ,
we get
∂2u
1 ∂u
1 ∂2u
2 ∂u
2 ∂2u
∂2u
2
2
2
=
cos
θ+
sin
θ+
sin
θ+
sin
θ
cos
θ−
cos θ sin θ.
∂x2
∂r2
r ∂r
r2 ∂θ2
r2 ∂θ
r ∂r∂θ
Similarly,
∂u
∂u
1 ∂u
=
sin θ +
cos θ,
∂y
∂r
r ∂θ
∂2u
∂2u
1 ∂u
1 ∂2u
2 ∂u
2 ∂2u
=
sin2 θ+
cos2 θ+ 2 2 cos2 θ− 2
sin θ cos θ+
cos θ sin θ.
2
2
∂y
∂r
r ∂r
r ∂θ
r ∂θ
r ∂r∂θ
Consequently,
∂2u ∂2u
∂ 2 u 1 ∂u
1 ∂2u
+ 2 =
+
+ 2 2.
2
2
∂x
∂y
∂r
r ∂r
r ∂θ
Assume the general case that u is complex-valued. Then
2 2
2
∂u = ∂u cos2 θ + 1 ∂u sin2 θ − 2 <{ ∂u ∂u cos θ sin θ}
∂x ∂r 2
r ∂θ r
∂r ∂θ
and
2 2
2
∂u = ∂u sin2 θ + 1 ∂u cos2 θ + 2 <{ ∂u ∂u cos θ sin θ},
∂y ∂r 2
r ∂θ r
∂r ∂θ
hence
2 2 2
2
∂u + ∂u = ∂u + 1 ∂u .
∂x ∂y ∂r 2
r ∂θ 3
3. (Added in) Let f (x) be an odd 2π-periodic function. For n ∈ Z define
Z π
1
cn =
f (t)e−int dt
2π −π
Prove that
c−ne−inx + cn einx = An sin nx
where
An =
2
π
Z
π
f (t) sin nt dt.
0
Proof.
=
=
=
=
=
=
=
c−ne−inx + cn einx
Z π
1
f (t)eint e−inx + f (t)e−int einx dt
2π −π
Z π
1
f (t)(e−in(x−t) + ein(x−t) ) dt
2π −π
Z π
1
f (t)2 cos n(x − t) dt
2π −π
Z
1 π
f (t)(cos nx cos nt + sin nx sin nt) dt
π −π
Z
1 π
f (t) sin nx sin nt dt (since f (t) cos nt is odd)
π −π
Z
2 π
sin nx
f (t) sin nt dt (since f (t) sin nt is even)
π 0
An sin nx
4. (Exercise 11) Show that if n ∈ Z the only solutions of the differential
equation
r2 F 00 (r) + rF 0 (r) − n2 F (r) = 0,
which are twice differentiable when r > 0, are given by linear combinations of rn and r−n when n 6= 0, and 1 and log r when n = 0.
Solution.
If F (r) solves the equation, write g(r) = F (r)/rn . Then g is twice differentiable, F (r) = rn g(r) and
F 0 (r) = nrn−1 g(r) + rn g 0 (r),
F 00 (r) = n(n − 1)rn−2 g(r) + nrn−1 g 0 (r) + nrn−1 g 0 (r) + rn g 00 (r).
Then
r2 F 00 (r)+rF 0 (r)−n2 F (r) = rn+1 (rg 00 (r)+(2n+1)g 0 (r)) = rn+1 (2ng 0 (r)+(rg 0 (r))0 ) = 0,
4
hence
2ng(r) + rg 0 (r) = c.
It follows that g(r) is a linear combination of r−2n and 1 if n 6= 0, and
log r and 1 if n = 0. The result follows.
5. (Problem 1) Consider the Dirichlet problem illustrated in Figure 11. More
precisely, we look for a solution of the steady-state heat equation 4u = 0
in the rectangle R = {(x, y) : 0 ≤ x ≤ π, 0 ≤ y ≤ 1} that vanishes on the
vertical sides of R, and so that
u(x, 0) = f0 (x) and u(x, 1) = f1 (x),
where f0 and f1 are initial data which fix the temperature distribution on
the horizontal sides of the rectangle.
Use separation of variables to show that if f0 and f1 have Fourier expansions
∞
∞
X
X
f0 (x) =
Ak sin kx and f1 (x) =
Bk sin kx,
k=1
then
u(x, y) =
k=1
∞ X
sinh k(1 − y)
sinh k
k=1
sinh ky
Ak +
Bk sin kx.
sinh k
Compare this result with the solution of the Dirichlet problem in the strip
obtained 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 ky
sinh k(1 − y)
Ak +
Bk .
sinh k
sinh k
Chapter 2
1. (Exercise 1) Suppose f is 2π-periodic and integrable on any finite interval.
Prove that if a, b ∈ R, then
Z
b
Z
b+2π
f (x) dx =
a
Z
b−2π
f (x) dx =
a+2π
f (x) dx.
a−2π
5
Also prove that
Z π
Z
π
f (x + a) dx =
−π
π+a
Z
f (x) dx =
−π
f (x) dx.
−π+a
Proof.
For the first two identities, substitute u = x − 2π and u = x + 2π respectively, and note that f (x ± 2π) = f (x). For the second equalities, firstly
Rπ
R π+a
we have −π f (x + a) dx = −π+a f (x) dx by substitution u = x + a. Then
by applying the first set of equalities, we get
Z −π
Z π
f (x) dx =
f (x) dx,
−π+a
so
Z π+a
Z
−π
f (x) dx =
−π+a
π+a
Z
π
f (x) dx+
−π+a
Z
π+a
f (x) dx+
−π
Z
π
f (x) dx =
π
f (x) dx.
−π
2. (Exercise 2) In this exercise we show how the symmetries of a function
imply certain properties of its Fourier coefficients. Let f be a 2π-periodic
Riemann integrable function on R.
(a) Show that the Fourier series of the function f can be written as
X
f (θ) ∼ fˆ(0) +
[fˆ(n) + fˆ(−n)] cos nθ + i[fˆ(n) − fˆ(−n)] sin nθ.
n≥1
(b) Prove that if f is even, then fˆ(n) = fˆ(−n), and we get a cosine series.
(c) Prove that if f is odd, then fˆ(n) = −fˆ(−n), and we get a sine series.
(d) Suppose that f (θ + π) = f (θ) for all θ ∈ R. Show that fˆ(n) = 0 for
all odd n.
(e) Show that f is real-valued if and only if fˆ(n) = fˆ(−n) for all n.
Proof.
fˆ(n)eint + fˆ(−n)e−int
1
1
= (fˆ(n) + fˆ(−n)) (eint + e−int ) + i(fˆ(n) − fˆ(−n)) (eint − e−int )
2
2i
= An cos nt + Bn sin nt
R 2π
R 2π
1
where An = fˆ(n)+fˆ(−n) = 2π
f (s)(e−ins +eins ) ds = π1 0 f (s) cos ns ds
0R
R 2π
2π
1
and Bn = i(fˆ(n)−fˆ(−n)) = i 2π
f (s)(e−ins −eins ) ds = π1 0 f (s) sin ns ds.
0
If f is real, then An , Bn are real, which is true iff fˆ(n) = fˆ(−n) for all n.
If f is odd, then f (s) cos ns is odd, so An = 0. If f is even, then f (s) sin ns
is odd, so Bn = 0.
If f (θ + π) = f (θ) for all θ ∈ R, then
Z 2π
Z π
Z 2π
−int
−int
f (t)e
dt =
f (t)e
dt +
f (t)e−int dt
0
0
π
6
and
Z
2π
f (t)e
−int
π
Z
dt =
π
f (u + π)e
−inu −inπ
e
Z
du = −
0
π
f (t)e−int dt
0
for all odd n.
3. (Exercise 4) Consider the 2π-periodic odd function defined on [0, π] by
f (θ) = θ(π − θ).
(a) Draw the graph of f .
(b) Computer the Fourier coefficients of f , and show that
f (θ) =
8
π
sin kθ
k3
X
k
odd
≥1
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
X
n
odd
≥1
∞
X
1
π2
1
π2
=
and
=
.
2
2
n
8
n
6
n=1
Solution.
Since it is an even function, by Exercise 2, fˆ(n) = fˆ(−n). We find fˆ(n) +
fˆ(−n) = 0, if n is even, and − πn2 2 if n is odd. So the series is
π
4
−
2
π
cos nθ
.
n2
X
n
odd
≥1
Assuming that the series
0, we find the first part of (d).
P converges at θ =P
∞
It is easy to see that n even ≥1 n12 = 14 n=1 n12 . Combining this with
what we got for the odd sum, we get the second part of (d).
5. (Exercise 7) Suppose {an }N
and {bn }N
n=1 are two finite sequences of
n=1 P
k
complex
numbers.
Let
B
=
b
k
n=1 n denote the partial sums of the
P
series
bn with the convention B0 = 0.
(a) Prove the summation by parts formula
N
X
an bn = aN BN − aM BM −1 −
n=M
N
−1
X
n=M
7
(an+1 − an )Bn .
(b) Deduce from this formula Dirichlet’s
test for convergence of a series :
P
if the partial sums of the series
bn are bounded, and {a
Pn } is a sequence
of real numbers that decreases monotonically to 0, then an bn converges.
Proof. (a)
N
X
an bn
n=M
=
N
X
an (Bn − Bn−1 )
n=M
=
N
X
an Bn −
n=M
=
N
X
N
X
an Bn−1
n=M
N
−1
X
an Bn −
an+1 Bn
n=M −1
n=M
= an BN +
N
−1
X
an Bn − aM BM −1 −
n=M
N
−1
X
an+1 Bn
n=M
= aN BN − aM BM −1 −
N
−1
X
(an+1 − an )Bn
n=M
(b) Suppose |Bn | ≤ B for all n. Then for N > M
|
N
X
an bn | ≤ B(aN + aM +
n=M
N
−1
X
(an − an+1 )) = 2BaM
n=M
proving that the series is Cauchy and hence converges.
P
einx
1
is the Fourier series of the 2π6. (Exercise 8) Verify that 2i
n6=0 n
periodic sawtooth function illustrated in Figure 6, defined by f (0) = 0,
and
−π/2 − x/2 if − π < x < 0
f (x) =
π/2 − x/2
if 0 < x < π.
Note that this function is not continuous. Show that nevertheless, the
series converges for every x (by which we mean, as usual, that the symmetric partial sums of the series converge). In particular, the value of the
series at the origin, namely 0, is the average of the values of f (x) as x
approaches the origin from the left and the right.
Solution.
Checked with Maple and found it correct. bn = sin nx,
Bn =
n
X
k=1
bk =
sin(nx/2) sin((n + 1)x/2)
sin(x/2)
8
1
(to prove write sin nx as 2i
(einx − 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. At
x = 0, all the symmetric sums are 0, so series converges to 0. It must
converge 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 the
interval [a, b] ⊂ [−π, π], that is,
1 if x ∈ [a, b]
χ[a,b] (x) =
0 otherwise.
(a) Show that the Fourier series of f is given by
f (x) ∼
b − a X e−ina − e−inb inx
+
e .
2π
2πin
n6=0
The sum extends over all positive and negative integers excluding 0.
(b) Show that if a 6= −π or b 6= π and a 6= b, then the Fourier series does
not converge absolutely for any x.
(c) However, prove that Fourier series converges at every point x. What
happens if a = −π and b = π?
Solution.
(a) Straightforward.
(b) |e−ina − e−inb |2 = 1 + 1 − 2 cos n(b − a) = 4 sin2 (n(b − a)/2), thus
|e−ina − e−inb | = 2| sin nθ0 |, where
0 < θ0 =
The series
b−a
π
< .
2
2
∞
X
| sin nθ0 |
n
n=1
diverges because .......
(c) an = 1/n, bn = sin(na) − sin(nb) which has bounded partial sums by
the solution of Exercise 8. So series converges by Exercise 7. If a = −π
and b = π, then the Fourier series is 1.
8. (Exercise 10) Suppose f is a periodic function of period 2π which belongs
to the class C k . Show that
fˆ(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 integrable
functions on the interval [0, 1] such that
Z 1
|fk (x) − f (x)| dx → 0 as k → ∞.
0
9
Show that fˆk (n) → fˆ(n) uniformly in n as k → ∞.
Proof.
Use
Z 1
|ĝ(n)| ≤
|g(x)| dx
0
for all n.
P
10. (Exercise
cn converges to
P12) Prove that if a series of complex numbers
s, then
cn is Cesaro summable to s.
Proof.
Let the sequence of partial sums be sn , n = 1, 2, · · · . First assume that
s = 0. Let > 0. Choose N1 such that |sn | < 2 for all n ≥ N1 . Choose
N > N1 such that
PN1
k=1 |sk |
<
n
2
for all n ≥ N . Then for all n ≥ N ,
s1 + · · · + sn
|
n
|s1 | + · · · + |sn |
n
|s1 + · · · + |sN1 | |sN1 +1 + · · · + |sn |
+
n
n
n/2
+
2
n
|
≤
=
<
=
P
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
converges to 0.
n
Now suppose that s 6= 0. Then sequence tn = sn − s converges to 0. So
by the proof above (t1 + · · · + tn )/n = (s1 + · · · + sn )/n − s converges to
0, i.e (s1 + · · · + sn )/n converges to s.
11. (Exercise 13) The purpose of this exercise is to prove that Abel summability is more general than the
standard or Cesaro methods of summation.
P∞
(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, but
that do not converge.
P∞
(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 Cesaro
summable.
The results above can be summarized by the following implications about
series:
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
P∞
n=1 cn .
Pn
(a). First assume that s = 0. For 0 ≤ r < 1 consider tn = k=1 sk rk .
Pn
Pn+1
We have rtn = k=1 sk rk+1 =P k=2 sk−1 rk . Since s1 = P
c1 and sk −
n
n
sk−1 = P
ck , we get (1 − r)tn = k=1 ck rk − sn rn+1 , hence k=1 ck rk =
n
(1 − r) k=1 sk rk + sn rn+1 . Since ck , sk are bounded
sequences
(in fact,
P∞
P∞
k
k
they converge to 0) and 0 ≤ r <
1,
both
series
c
r
and
k
k=1
k=1 sk r
P∞
P∞
are absolutely convergent, and k=1 ck rk = (1 − r) k=1 sk rk . Let > 0.
Choose K such that |sk | < for all k ≥ K. Then
|
∞
X
ck rk | ≤ (1 − r)
k=1
K−1
X
M rk + (1 − r)
K−1
X
rK
≤ (1 − r)
M rk + 1−r
k=1
k=1
It follows that
lim sup |
r→1−
∞
X
ck rk | ≤ .
k=1
P∞
Since > 0 is arbitrary,
lim supr→1− | k=1 ck rk | ≤ 0 which is
P∞we have
equivalent to limr→1− k=1 ck rk = 0.
P∞
NextP
assume that sP6= 0. Define dn = cn − s/2n . Then n=1 dn = 0,
∞
∞
n
n
and
r). Thus by our proof above,
n=1
Pd∞n r =n n=1 cn r − sr/(2 − P
∞
limr→1− n=1 P
cn r − s = 0, i.e. limr→1− n=1 cn rn = s. P
∞
∞
n
(b). The series n=1 (−1) Pdoes not converge, but limr→1− n=1 (−1)n rn =
∞
−r
n
limr→1− 1+r = −1/2, i.e.
n=1 (−1) is Abel summable to −1/2.
(c). Write
s1 + · · · + sn
σn =
n
and τn = nσn . Since τn − τn−1 = sn , by the same argument as in (a), we
have
∞
∞
X
X
(1 − r)
τk r k =
sk rk ,
k=1
hence
(1 − r)2
∞
X
k=1
σk krk =
k=1
∞
X
ck rk
k=1
from the proof of (a). Assume that σ = 0. Let > 0. Choose N such that
|σk | < for all n ≥ N , and B be a bound for |σk |, k = 1, 2, · · · . Then
|
∞
X
k=1
k
ck r | ≤ (1−r)
2
N
−1
X
k=1
N
−1
X
rN ((1 − r)N + r)
2
Bkr +(1−r) ≤ (1−r)
Bkrk +rN ((1−r)N +r)
(1 − r)2
k
2
k=1
N
P∞
+r)
where we have use the fact that k=N krk = r ((1−r)N
. As in the
(1−r)2
P∞
k
proof of (a), this implies that limr→1− k=1 ck r = 0. For the case σ 6= 0,
11
P∞
consider
dn = cn − σ/2n . Since n=1 cn is Cesaro summable to
σ and
P∞
P∞
n
σ/2
=
σ
(and
hence
Cesaro
summable
to
σ),
we
have
n=1
n=1 dn
Cesaro summable to 0. Thus by our proof above and the same argument
as in the proof of (a), the
follows.
Presult
∞
(d). Consider the series n=1 (−1)n n. It is Abel summable to −1/4 since
∞
X
(−1)n nrn =
n=1
−r
.
(1 + r)2
Note that
n−1
an
σn−1 =
.
n
n
P∞
Thus for a Cesaro
P∞ summable series n=1 an , limn→∞
proves that n=1 (−1)n n is not Cesaro summable.
σn −
an
n
must be 0. This
12. (Exercise 14) This exercise deals with a theorem of Tauber which says that
under an additional condition on the coefficients cn , the above arrows can
be reversed.
P
(a) IfP 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 Abel
summable.
Proof.
(a). Denote (n − 1)cn by tn . Since tn = n−1
n ncn and ncn → 0 we have
tn → 0. Now
sn −σn = sn −
s1 + · · · + sn
(sn − s1 ) + · · · + (sn − sn )
t2 + · · · + tn
=
=
n
n
n
and it is immediate that sn − σn → 0.
(b). Let r = 1 − N1 . We have
|
N
X
n=1
cn −
N
X
n=1
cn r n | ≤
X
N
n
1
|cn |
|cn | 1 − (1 − )n ≤
N
N
n=1
n=1
N
X
n
where we have used (1 − N1 )n ≥ 1 − N
for 1 ≤ n ≤ N , which can be proved
12
easily by induction on n. Also if |cn n| < for all n ≥ N , then
|
≤
≤
≤
∞
X
cn r n −
n=1
∞
X
N
X
cn r n |
n=1
|cn |(1 −
n=N +1
∞
X
n=N +1
∞
X
n=N +1
= (1 −
1 n
)
N
1
n|cn |
(1 − )n
N
N
1
(1 − )n
N
N
1 N +1
)
→
N
e
as N → ∞.
13. (Exercise 15) Prove that the Fejer kernel is given by
FN (x) =
1 sin2 (N x/2)
.
N sin2 (x/2)
Proof.
Recall that N FN (x) = D0 (x)+· · ·+DN −1 (x) where Dn (x) is the Dirichlet
kernel. Write ω = eix . Then
Dn (x)
=
ω −n + · · · + ω −1 + 1 + ω + · · · + ω n
(ω −n + · · · + ω −1 ) + (1 + ω + · · · + ω n )
ω −n − 1 1 − ω n+1
+
= ω −1 −1
ω −1
1−ω
n+1
−n
ω −1 1−ω
+
=
1−ω
1−ω
ω −n − ω n+1
=
1−ω
=
13
So
N FN (x)
=
N
−1
X
n=0
=
ω −n − ω n+1
1−ω
1
1−ω
=
1
1−ω
=
1
1−ω
= ω
N
−1
X
ω −n −
n=0
−N
N
−1
X
!
ω n+1
n=0
ω
−1
1 − ωN
−ω
−1
ω −1
1−ω
ω −N +1 − ω
1 − ωN
−ω
1−ω
1−ω
ω −N − 2 + ω N
(1 − ω)2
=
(ω N/2 − ω −N/2 )2
(1 − ω)2
(ω −1/2 )2
=
(ω N/2 − ω −N/2 )2
(ω 1/2 − ω −1/2 )2
=
−4 sin2 (N x/2)
−4 sin2 (x/2)
=
sin2 (N x/2)
.
sin2 (x/2)
1
Therefore
FN (x) =
1 sin2 (N x/2)
.
N sin2 (x/2)
14. (Exercise 16) The Weierstrass approximation theorem states: Let f be a
continuous function on the closed and bounded interval [a, b] ⊂ R. Then,
for any > 0, there exists a polynomial P such that
sup |f (x) − P (x)| < .
x∈[a,b]
Proof.
Let > 0. We may extend f to a continuous (c−a)-periodic function where
b ≤ c. By Corollary
PN 5.4 of Fejer’s theorem, there exists a trigonometric
polynomial Q = n=M an einx such that |Q(x)−f (x)| < /2 for all x. For
each n, M ≤ n ≤ N , there exists a polynomial pn (x) such that |an einx −
pn (x)| < /2N for all x ∈ [a, c]. Then P = pM + · · · + pN is a polynomial
in x that satisfies the requirement.
15. (Exercise 17) In Section 5.4 we proved that the Abel means of f converge
to f at all points of continuity, that is,
lim Ar (f )(θ) = lim (Pr ∗ f )(θ) = f (θ), with 0 < r < 1,
r→1
r→1
14
whenever f is continuous at θ. In this exercise, we will study the behavior
of Ar (f )(θ) at certain points of discontinuity.
An integrable function is said to have a jump discontinuity at θ if the
two limits
lim
h→0,h>0
f (θ + h) = f (θ+ ) and
lim
h→0,h<0
f (θ + h) = f (θ− )
exist.
(a) Prove that if f has a jump discontinuity at θ, then
lim Ar (f )(θ) =
r→1
f (θ+ ) + f (θ− )
, with 0 ≤ r < 1.
2
(b) Using a similar argument, show that if f has a jump discontinuity at
+
(θ − )
θ, the Fourier series of f at θ is Cesaro summmable to f (θ )+f
.
2
Proof.
Rπ
R0
1
1
Since Pr (θ) = Pr (−θ), and 2π
P (θ) dθ = 1, we have 2π
P (θ) dθ =
−π r
−π r
R
π
1
P
(θ)
dθ
=
1/2.
Suppose
f
has
a
jump
discontinuity
at
θ.
Let > 0
r
2π 0
be given. Choose δ > 0 so that 0 < h < δ implies |f (θ − h) − f (θ−)| < and |f (θ + h) − f (θ+)| < . Let M be such that |f (y)| ≤ M for all y.
Then
+
− (f ∗ Pr )(θ) − f (θ ) + f (θ ) 2
Z π
1
f (θ+ ) + f (θ− ) = Pr (y)f (θ − y) dy −
2π −π
2
Z 0
Z π
1
1
≤
Pr (y)|f (θ − y) − f (θ+ )| dy +
Pr (y)|f (θ − y) − f (θ− )| dy
2π −π
2π 0
Z
Z
1
1
≤
Pr (y)|f (θ − y) − f (θ+ )| dy +
Pr (y)|f (θ − y) − f (θ− )| dy
2π −δ<y<0
2π 0<y<δ
Z
1
+ 2M
Pr (y) dy
2π
δ≤|y|≤π
Z
M
≤
+ +
Pr (y) dy
2 2
π δ≤|y|≤π
R
Therefore, recalling that limr→1 δ≤|y|≤π Pr (y) dy = 0,
f (θ+ ) + f (θ− ) lim sup (f ∗ Pr )(θ) −
≤
2
r→1
Since > 0 is arbitrary, we have
lim (f ∗ Pr )(θ) =
r→1
f (θ+ ) + f (θ− )
.
2
(b) Since the Fejer kernel Fn (θ) is even and positive, we also have
Rπ
1
2π 0 Fn (θ) dθ = 1/2, for all n. Repeat the above argument.
15
1
2π
R0
−π
Fn (θ) dθ =
−
+
(θ−h)
(θ )
by limh→0 f (θ+h)+f
and obtain
Remark One can replace f (θ )+f
2
2
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
∞
X
u(r, θ) =
r|m| ineinθ .
n=−∞
Each summand is Laplacian (see chapter 1, separation of variable), and the
series 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, then
u(r, 0) = 0 and the limit is trivially 0. Note that
u(x, y) =
2y(x2 + y 2 − 1)
((1 − x)2 + y 2 )2
and u(1 − , ) → −∞ as → 0+. This implies that u does not converge
to 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

when 0 ≤ y,
 u(0, y) = 0
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) =
∞
X
an sin(nπx)
n=1
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 integral
formula (6).
Solution.
By considering the odd extension of f and following the derivation of
Poisson’s kernel with e−πy and eiπt replacing r and eit , respectively, we
obtain
Z
1 1
f (t)Qy (x − t) dt
u(x, y) =
2 −1
where
1 − e−2πy
.
1−
cos πt + e−2πy
or, using the fact that f is odd, we have the alternate form
Z
1 1
f (t)Q(x, t) dt
u(x, y) =
2 0
Qy (t) =
2e−πy
where
Q(x, t) =
1 − e−2πy
1 − e−2πy
−
.
1 − 2e−πy cos π(x − t) + e−2πy 1 − 2e−πy cos π(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 is
to solve
∂2
1 ∂
1 ∂2
+
+
=0
∂r2
r ∂r r2 ∂θ2
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, we
can hope to write
X
u(r, θ) =
cn (r)einθ
with cn (r) = An rn + Bn r−n , n 6= 0. Set
X
X
f (θ) ∼
an einθ and g(θ) ∼
bn einθ .
We want cn (1) = an and cn (ρ) = bn . This leads to the solution
X
1
u(r, θ) =
[((ρ/r)n − (r/ρ)n )an + (rn − r−n )bn ]einθ
ρn − ρ−n
n6=0
+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,
and
u(r, θ) − (Pρ/r ∗ g)(θ) → 0 as r → ρ uniformly in θ.
Solution.
n
−n
r 2n −1
n
Note that ρrn −r
−ρ−n = ρ2n −1 (ρ/r) → 0 and
0 as n → ∞.
The series converges because for instance
(ρ/r)n −(r/ρ)n
ρn −ρ−n
=
(ρ/r)2n −1 n
ρ2n −1 r
→
r2n − 1
1 + r2
n
(ρ/r)n .
(ρ/r)
≤
ρ2n − 1
1 − ρ2
And the last result follows from, for instance,
(ρ/r)2n − ρ2n
(ρ/r)2n − 1
n
−
1
r
=
rn
ρ2n − 1
ρ2n − 1
≤ 2[(ρ/r)2n − ρ2n ]rn
and
∞
X
ρ2 r
ρ2 /r
−
2
1 − ρ /r 1 − ρ2 r
[(ρ/r)2n − ρ2n ]rn =
n=1
which approaches 0 as r → 1. Details are to be given.
19. (Problem 2) Let DN denote the Dirichlet kernel
DN (θ) =
N
X
eikθ =
n=N
and define
LN =
1
2π
Z
sin(N + 1/2)θ
,
sin(θ/2)
π
|DN (θ)| dθ.
−π
(a) Prove that
LN ≥ c log N
for some constant c > 0. A more careful estimate gives
LN =
4
log N + O(1).
π2
(See Katznelson Chapter 2, Exercise 1.1 for section p.50.)
(b) Prove the following as a consequence: for each n ≥ 1, there exists a
continuous function fn such that |fn | ≤ 1 and |Sn (fn )(0)| ≥ c0 log n.
Solution.
18
(a) Note that sinx x ≥ 1 for x in the interval [−π/2, π/2]. It follows that
for θ ∈ [−π, π],
| sin(N + 1/2)θ|
.
|DN (θ)| ≥ 2
θ
Then
Z π
|DN (θ)| dθ
−π
Z π
| sin(N + 1/2)θ|
≥ 4
dθ
θ
0
Z (N +1/2)π
| sin θ|
= 4
dθ
θ
0
Z Nπ
| sin θ|
dθ
≥ 4
θ
0
N
−1 Z (k+1)π
X
| sin θ|
dθ
= 4
θ
k=0 kπ
Z (k+1)π
N
−1
X
1
≥ 4
| sin θ| dθ
(k + 1)π kπ
k=0
=
N −1
8 X 1
π
k+1
k=0
8
log(N + 1)
=
π
8
≥
log N
π
Therefore LN ≥ c log N .
(b) The function gn which is equal to 1 when Dn is positive and −1 when
Dn is negative has the desired property but is not continuous. Approximate gn in the integral norm (in the sense of Lemma 3.2) by continuous
functions hk satisfying |hk | ≤ 1.
20. (Problem
theorem:
P 3) Littlewood provided a refinement of Tauber’sP
(a) If
cn is Abel summable to s and cn = O(1/n), then
cn converges
to s.
P
(b) As
cn is Cesàro summable to s and cn = O(1/n),
Pa consequence, if
then
cn converges to s.
These results may be applied to Fourier series. By Exercise 17, they imply
that if f is an integrable function that satisfies fˆ(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− )
as N → ∞.
2
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 Problem
5 in Chapter 4.
Solution.
Lemma 1 Let f be a C 2 real function on [0, 1).
as x → 1,
Suppose
1
1
0
00
f (x) = o(1), f (x) = O (1−x)2 , then f (x) = o 1−x .
Proof. Choose C > 0 such that |(1 − x)2 f 00 (x)| ≤ C for all x. Let
> 0. Choose a δ, 0 < δ < min{1/2, /(4C)}. Choose η < 1 such that
|f (x)| < (1/4)δ for all x > η. We claim that |(1 − x)f 0 (x)| < for all
x > η. Let x > η. With x0 = x + δ(1 − x) we have
1
f (x0 ) = f (x) + δ(1 − x)f 0 (x) + δ 2 (1 − x)2 f 00 (ζ)
2
for some x < ζ < x0 .
Since δ < 1/2, we have 1 − x ≤ 2(1 − ζ), so that |(1 − x)2 f 00 (ζ)| ≤
4|(1 − ζ)2 f 00 (ζ)| ≤ 4C. Therefore
|(1 − x)f 0 (x)|
f (x0 ) − f (x) 1
= |
− δ(1 − x)2 f 00 (ζ)|
δ
2
|f (x0 )| + |f (x)| 1
≤
+ δ(1 − x)2 |f 00 (ζ)|
δ
2
1
1
+ =
<
2
2
P∞
n
Lemma 2 Let n=0 an x be a real power series with an ≥ 0 for all n.
Suppose
∞
X
an xn = 1.
lim− (1 − x)
x→1
n=0
Then for any integrable function g(t) on [0, 1]
lim− (1 − x)
x→1
∞
X
n
n
Z
an x g(x ) =
1
g(t) dt.
0
n=0
Proof. First prove for functions of the type xk . Then for any polynomial;
then for any continuous function; then for any integral function.
P∞
Corollary 1 Let n=0 an xn be a real power series with an ≥ 0 for all n.
Suppose
∞
X
lim− (1 − x)
an xn = 1.
x→1
n=0
Then
PN
n=0
lim
N
N →∞
20
an
= 1.
Proof. Apply the above lemma to the function g(t) = 0 for t < e−1 , 1/t
otherwise, and let x = e−1/N .
Finishing the solution of Problem 3
(x)
We shall write f (x) ∼ g(x) to mean fg(x)
→ 1 as x → 1, and f (n) ∼ g(n)
(n)
to mean fg(n)
→ 1 as n → ∞.
(a) We may assume
P∞ that s = 0 (see Exercise 13, Chapter 2), i.e. we assume
that f (x) = n=0 an xn → 0 as x → 1. Then f 00 (x) = O(1/(1 − x)2 )
because
f 00 (x) =
∞
X
n(n − 1)an xn−2 = O(
n=2
∞
X
(n − 1)xn−2 ) = O(1/(1 − x)2 ).
n=2
So by Lemma 1,
0
f (x) = o
1
1 − x)
.
Suppose |nan | ≤ c. Then
∞
X
(1 −
n=1
Since 1 −
nan
c
nan n−1
1
f 0 (x)
1
)x
=
−
∼
c
1−x
c
1−x
≥ 0, Corollary 1 implies that
n
X
(1 −
k=1
or what is the same
n
X
kak
)∼n
c
kak = o(n).
k=1
21
Write wn =
Pn
k=1
=
=
=
=
=
=
=
kak , w0 = 0. So wn /n → 0 as n → ∞. Then
f (x) − a0
∞
X
an xn
n=1
∞
X
wn − wn−1 n
x
n
n=1
∞
∞
X
wn n X wn−1 n
x −
x
n
n
n=1
n=1
∞
∞
X
wn n X wn n+1
x −
x
n
n+1
n=1
n=1
n
∞
X
xn+1
x
−
wn
n
n+1
n=1
∞
X
xn
xn − xn+1
+
wn
n+1
n(n + 1)
n=1
∞
X
(1 − x)
∞
wn
n=1
X
wn
xn
+
xn
n + 1 n=1 n(n + 1)
Since f (x) → 0 and the first term in the last sum approaches 0 as x → 1,
we get
∞
X
wn
lim
xn = −a0
x→1
n(n
+
1)
n=1
Since
wn
n(n+1)
= o(1/n), by the regular Tauberian theorem
∞
X
wn
= −a0 .
n(n
+ 1)
n=1
Now
N
X
=
=
=
wn
n(n
+ 1)
n=1
N
X
1
1
wn
−
n n+1
n=1
N
X
wn − wn−1
wN
−
n
N +1
n=1
N
X
an −
n=1
Letting N → ∞ we get
P∞
n=1
wN
.
N +1
an = −a0 , i.e.
22
P∞
n=0
an = 0.
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
Z 2π
1
lim
|fk (t)|2 dt = 0
k→∞ 2π 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 on
a set of measure 0.)
5. (Exercise 5) Let
f (t) =
0
log(1/t)
for t = 0
for 0 < t ≤ 2π,
and define a sequence of functions in R by
0
for 0 ≤ t ≤ 1/n
fn (t) =
f (t) for 1/n < t ≤ 2π.
Prove that {fn }∞
n=1 is a Cauchy sequence in R. However, f does not belong to R.
Solution.
Z
(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
1/k if k ≥ 1
ak =
0
if k ≤ 0.
Note that {ak } ∈ l2 (Z), but that no Riemann integrable function has k th
Fourier coefficient equal to ak for all k.
Solution.
Let M = sup−π≤tπ |f (t)|. Recall that
Z π
1
Ar (f )(θ) =
Pr (θ − t)f (t) dt.
2π −π
So
|Ar (f )(θ)| ≤ M
1
2π
23
Z
π
Pr (θ − t) dt = M
−π
Also
Ar (f )(θ) =
∞
X
fˆ(n)einθ r|n| .
n=−∞
So
Ar (f )(0) =
∞
X
fˆ(n)r|n| .
n=−∞
It follows that if f is a Riemann integrable function that has k th Fourier
coefficient equal to ak for all k, then
∞
X
rn
n
n=1
is bounded for 0 ≤ r < 1. This is a contradiction since the sum of the
series is − log(1 − r).
7. (Exercise 7) Show that the trigonometric series
X sin nx
n≥2
log n
converges for every x, yet it is not the Fourier series of a Riemann integrable function.
P sin nx
The same is true for
for 0 < α < 1, but the case 1/2 < α < 1 is
nα
more difficult. See Problem 1.
Solution.
Apply Parseval’s identity.(A Riemann integrable
function is in L2 .) Series
P
converges because the partial sums of
sin nx 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
X
n
odd
≥1
∞
X
1
π2
1
π2
=
and
=
.
n2
8
n2
6
n=1
Similar sums can be derived using the methods of this chapter.
(a) Let f be the function defined on [−π, π] by f (θ) = |θ|. Use Parseval’s
identity to find the sums of the following two series:
∞
X
∞
X
1
1
and
.
4
4
(2n
+
1)
n
n=0
n=1
4
4
In fact, they are π96 and π90 , respectively.
(b) Consider the 2π-periodic odd function defined on [0, π] by f (θ) =
θ(π − θ). Show that
∞
X
∞
X
π6
1
π6
1
=
and
=
.
(2n + 1)6
960
n6
945
n=1
n=0
24
P∞
Remark. The general expression when k is even for n=1 1/nk in terms
of
π k is given in Problem 4. P
However, finding a formula for the sum
P∞
∞
3
k
1/n
,
or
more
generally
n=1
n=1 1/n with k odd, is a famous unresolved question.
Solution. The Fourier series for f (θ) = |θ| is
π
2
−
2
π
einx + e−inx
.
n2
X
n
odd
≥1
By Parseval’s identity,
π2
4
+ 2
4
π
1
2
=
n4
2π
X
n
from which we get
odd
Z
≥1
π
|θ|2 dθ =
−π
π2
3
∞
X
π4
1
=
.
(2n + 1)4
96
n=0
The sum
P∞
1
n=1 n4
is obtained from solving for x in the equation
x=
π4
x
+
96 16
4
which yields x = π90 .
The Fourier series for the function f (θ) = θ(π − θ) is
8
π
4
sin kθ
= −i
k3
π
X
k
odd
≥1
eikθ − e−ikθ
.
k3
X
k
odd
≥1
By Parseval’s identity,
Z π
∞
16 X
2
1
1 4
=
π ,
θ2 (1 − θ)2 dθ =
2
6
π n=0 (2n + 1)
2π −π
30
from which we get
∞
X
1
π6
.
=
6
(2n + 1)
960
n=0
Solving x from
x=
we get x =
1
π6
x+
64
960
π6
945 .
9. (Exercise 9) Show that for α not an integer, the Fourier series of
π
ei(π−x)α
sin πα
25
on [0, 2π) is given by
∞
X
einx
.
n+α
−∞
Apply Parseval’s formula to show that
∞
X
π2
1
=
.
2
(n + α)
(sin πα)2
−∞
Solution
Straightforward checking.
10. (Exercise 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 ∂2u
∂2u 2
=
, c = τ /ρ.
2
2
c ∂t
∂x2
The string is subject to the initial conditions
u(x, 0) = f (x) and
∂u
(x, 0) = g(x),
∂t
where we assume that f ∈ C 1 and g is continuous. We define the total
energy of the string by
1
E(t) = ρ
2
L
Z
0
∂u
∂t
2
Z
1
dx + τ
2
L
0
∂u
∂x
2
dx.
The first term corresponds to the ”kinetic energy” of the string (in analogy
with (1/2)mv 2 , 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
ρ
2
Z
0
L
1
g(x)2 dx + ρ
2
Z
L
f 0 (x)2 dx.
0
Solution. We have
E 0 (t) = ρ
Z
L
∂u ∂ 2 u
dx + τ
∂t ∂t2
Z
L
∂u ∂ 2 u
dx − τ
∂t ∂t2
Z
0
Z
=ρ
0
L
∂u ∂ 2 u
dx
∂x ∂x∂t
L
∂ 2 u ∂u
dx
∂ 2 x ∂t
0
0
=0
where we have used integration by parts, and that
for all t.
26
∂u
∂t (0, t)
=
∂u
∂t (L, t)
=0
11. (Exercise 11) The inequalities of Wirtinger and Poincare establish a relationship between the norm of a function and that of its derivative.
RT
(a) If f is T -periodic, continuous, and piecewise C 1 with 0 f (t) dt = 0,
show that
Z T
Z T
T2
2
|f (t)| dt ≤
|f 0 (t)|2 dt,
4π 2 0
0
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 C 1 and T -periodic, prove that
Z
2
Z T
Z T
T
T2
2
≤
|f
(t)|
f
(t)g(t)
dt
dt
|g 0 (t)|2 dt.
0
4π 2 0
0
(c) For any compact interval [a, b] and any continuously differentiable function f with f (a) = f (b) = 0, show that
Z b
Z
(b − a)2 b 0 2
|f (t)|2 dt ≤
|f (t)| dt.
π2
a
a
Discuss the case of equality, and prove that the constant (b−a)2 /π 2 cannot
be improved.
Solution.
RT
(a) The condition 0 f (t) dt = 0 implies that fˆ(0) = 0. The continuity
2π
ˆ
of f guarantees that fˆ0 (n) = 2πin
T f (n). Indeed, write τ for T . Then for
n 6= 0,
Z
1 T
fˆ(n) =
f (t)e−inτ t dt
T 0
Z
1
1
1 1 T 0
=
(f (0+ ) − f (T − ))
+
f (t)e−inτ t dt
T
inτ
inτ T 0
T ˆ0
f (n)
=
2πin
Therefore, by Parseval’s identity,recalling that fˆ(0) = 0,
Z
T
|f (t)|2 dt = T
0
=
X
|n|>0
T 3 X |fˆ0 (n)|2
4π 2
n2
|n|>0
3
≤
=
=
T X ˆ0
|f (n)|2
4π 2
|n|>0
Z
3
T 1 T 0 2
|f (t)| dt
4π 2 T 0
Z T
T2
|f 0 (t)|2 dt
4π 2 0
27
|fˆ(n)|2
From the above inequalities, we see that equality holds if and only if
fˆ(n) = 0 for all n ≥ 2. This means that, writing an for fˆ(n), f (x) =
a1 eiτ x + a−1 e−iτ x which simplifies to A sin(τ x) + B cos(τ x).
Remark. It is clear from the proof above that in the absence of the
RT
condition 0 f (t) dt = 0, the inequality in (a) is
X
|an |2 ≤
|n|>0
T2 X
|bn |2
4π 2
|n|>0
where an = fˆ(n), bn = fˆ0 (n).
(b) Let an = fˆ(n), bn = ĝ(n), cn = gˆ0 (n). Then
2
Z
T
f (t)g(t) dt
0
X
= T|
an bn |2
|n|≥0
X
= T|
an bn |2
|n|>0
X
≤ (T
|an |2 )(
|n|>0
T
Z
X
|n|>0
T2
|f (t)| dt 2
4π
2
≤
0
|bn |2 )
Z
T
|g 0 (t)| dt
0
(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 0 (b + h) =
RT
R 2b−a
RT
f 0 (b − h) for all h. Then 0 f (x) dx = a
f (x) dx = 0, 0 |f (x)| dx =
R 2b−a
Rb
RT
Rb
|f (x)| dx = 2 a |f (x)| dx, and 0 |f 0 (x)| dx = 2 a |f 0 (x)| dx. Check
a
that f so extended is also C 1 on R. Therefore by (a), we have
Z b
|f (x)| dx
a
=
≤
=
=
1
2
Z
T
|f (x)| dx
0
2
Z
1 2 (b − a)2 T 0
|f (x)| dx
2
4π 2
0
Z
(b − a)2 1 T 0
|f (x)| dx
π2 2 0
Z
(b − a)2 b 0
|f (x)| dx
π2
a
For the function f (x) that we define, its translation g(x) = f (x + b) is an
odd function. So ĝ(n) = einb fˆ(n), and the Fourier series for g is of the
28
form
∞
X
∞
X
2πnx
πnx
An sin
=
An sin
.
2(b
−
a)
b
−a
n=1
n=1
Thus the Fourier series for f is of the form
∞
X
An sin
n=1
πn(x − b)
.
b−a
According the inequality in (c) is an equality iff f (x) is of the form
π(x−a)
π(x−b)
A sin π(x−b)
b−a . (The book says A sin b−a ; this is equivalent since sin b−a =
− sin π(x−a)
b−a ).
R∞
12. (Exercise 12) Prove that 0 sinx x dx =
Proof. R
π
We have −π DN (t) dt = 2π. So
Z
π
−π
π
2.
sin(N + 1/2)x
dx = 2π
sin(x/2)
Write csc(x/2) as csc(x/2) − 2/x + 2/x. limx→0 csc(x/2) − 2/x = 0, so it
has a removable discontinuity at 0 on the interval [−π, π]. By LebesgueRiemann lemma we get
Z π
2 sin(N + 1/2)x
dx → 2π as N → ∞.
x
−π
This yields
Z
π
0
sin(N + 1/2)x
π
dx → as N → ∞.
x
2
By change of variable, we get
Z
(N +1/2)π
0
Since
get
R (N +1/2)π
Nπ
sin x
x
sin x
π
dx → as N → ∞.
x
2
dx → 0 as N → ∞, (use Mean-Value Theorem), we
Z
Nπ
0
sin x
π
dx → as N → ∞.
x
2
By MVT,
Z
(N +1)π
Nπ
| sin x|
1
dx ≤ ,
x
N
so for any t > π there exists N > 0 such that
Z t
sin x 1
dx ≤ .
x
N
Nπ
29
It follows that
Z
0
∞
sin x
dx = lim
N →∞
x
Z
0
Nπ
sin x
π
dx = .
x
2
13. (Exercise 13) Suppose that f is periodic and of class C k . Show that
fˆ(n) = o(1/|n|k ),
that is, |n|k fˆ(n) goes to 0 as |n| → ∞. This is an improvement over
Exercise 10 in Chapter 2.
Solution.
We have (see p.43)
Z 2π
1
f (k) (θ)e−inθ dθ.
2π fˆ(n) =
(in)k 0
So
2π(in) fˆ(n) =
k
Z
2π
f (k) (θ)e−inθ dθ.
0
Now use Lebesgue-Riemann lemma.
14. (Exercise 14) Prove that the Fourier series of a continuously differentiable
function f on the circle is absolutely convergent.
Proof.
1 ˆ0
f (n), we get, by Cauchy-Schwarz inequality and ParseFrom fˆ(n) = in
val’s identity,
∞
X
=
|fˆ(n)|
n=−∞
∞
X
1 ˆ0
|f (n)|
|n|
n=−∞
∞
π 2 1/2 X ˆ0
) (
|f (n)|2 )1/2
6
n=−∞
Z π
2
1
π
|f 0 (t)|2 dt < ∞.
(2 )1/2
6
2π −π
≤ (2
=
15. (Add in) Prove that the Fourier series of a 2π periodic absolutely continuous function whose derivative (exists a.e.) in [0, 2π] is square integrable
(in particular, Riemann integrable), is absolutely convergent. (Note that
Exercise 16 below shows that Lipschitz condition alone is enough. But
derivative of a Lipschitz function is bounded. And Lipschitz functions
are precisely functions representable as integral of a bounded measurable
function.)
Proof. Use the proof in Exercise 14. Note that the integration by parts
formula 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
Z π
1
π
ˆ
f (n) = −
f (x + )e−inx dx
2π −π
n
hence
1
fˆ(n) =
4π
Z
π
[f (x) − f (x +
−π
π −inx
)]e
dx.
n
(b) Now assume that f satisfies a Hölder condition of order α, namely
|f (x + h) − f (x)| ≤ C|h|α
for some 0 < α ≤ 1, some C > 0, and all x, h. Use part (a) to show that
fˆ(n) = O(1/|n|α ).
(c) Prove that the above result cannot be improved by showing that the
function
∞
X
k
f (x) =
2−kα ei2 x ,
k=0
where 0 < α < 1, satisfies
|f (x + h) − f (x)| ≤ C|h|α ,
and fˆ(N ) = 1/N α whenever N = 2k .
Solution.
Note that for any real x, |1 − eix | ≤ |x| (to prove, just note that LHS is
2| sin(x/2)| and | sin a| ≤ |a| for all real number a).
(a).
Z π
π
1
f (x + )e−inx dx
2π −π
n
Z π+ nπ
1
f (u)e−inu eiπ du
=
2π −π+ nπ
Z π
1
= −
f (u)e−inu du
2π −π
= −fˆ(n)
(b)
|fˆ(n)|
Z π
1
π
≤
|f (x) − f (x + )| dx
4π −π
n
1
πα
≤
2πC α
4π
|n|
C1
=
|n|α
31
(c).
=
|f (x + h) − f (x)|
∞
∞
X
X
k
k
|
2−kα ei2 (x+h) −
2−kα ei2 x |
k=0
≤
∞
X
k=0
k
2−kα |ei2
h
− 1|
k=0
X
≤
X
2−kα 2k |h| +
2k ≤1/|h|
2−kα 2
2k >1/|h|
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 such
that 2−l−1 < |h| ≤ 2−l . Then the first sum is
l
X
k
1−α
(2 |h|)
α
α
|h| ≤ |h|
k=0
l
X
(2k−l )1−α ≤
k=0
1
|h|α .
1 − 2α−1
Since the series converges uniformly, the coefficient of einx is fˆ(n) for all
n.
17. (Exercise 16) Let f be a 2π-periodic function which satisfies a Lipschitz
condition with constant K; that is
|f (x) − f (y)| ≤ K|x − y| for all x, y.
This is simply the Hölder condition with α = 1, so by thePprevious exercise,
we see that fˆ(n) = O(1/|n|). Since the harmonic series 1/n diverges, we
cannot say anything (yet) about the absolute convergence of the Fourier
series of f . The outline below actually proves that the Fourier series of f
converges absolutely and uniformly.
(a) For every positive h we define gh (x) = f (x + h) − f (x − h). Prove that
Z 2π
∞
X
1
|gh (x)|2 dx =
4| sin nh|2 |fˆ(n)|2 ,
2π 0
n=−∞
and show that
∞
X
| sin nh|2 |fˆ(n)|2 ≤ K 2 h2 .
n=−∞
(b) Let p be a positive integer. By choosing h = π/2p+1 , show that
X
|fˆ(n)|2 ≤
2p−1 <|n|≤2p
K 2 π2
.
22p+1
P
(c) Estimate 2p−1 <|n|≤2p |fˆ(n)|, and conclude that the Fourier series of
f converges absolutely, hence uniformly.
32
(d) In fact, modify the argument slightly to prove Bernstein’s theorem: If
f satisfies a Hölder condition of order α > 1/2, then the Fourier series of
f converges absolutely.
Solution.
(a). The Fourier coefficients of the translated function f (x+h) is einh fˆ(n).
So
gˆh (n) = (einh − e−inh )fˆ(n) = 2i sin nhfˆ(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
| sin nh|2 ≥ 1/2. Thus
1
2
X
|fˆ(n)|2 ≤
2p−1 <|n|≤2p
∞
X
K 2 π2
| sin nh|2 |fˆ(n)|2 ≤ K 2 h2 = 2(p+1)
2
n=−∞
from which it follows
X
|fˆ(n)|2 ≤
2p−1 <|n|≤2p
K 2 π2
.
22p+1
(c). By Cauchy-Schwarz inequality, we have
X
|fˆ(n)|
2p−1 <|n|≤2p
≤
X
(
2p−1 <|n|≤2p
|fˆ(n)|2 )1/2
2p−1 <|n|≤2p
≤
(2p − 2p−1 )1/2
≤
Kπ
√
( 2)p
Kπ
2p+1/2
Therefore
X
X
12 )1/2 (
|fˆ(n)| ≤
1≤|n|<∞
Kπ
√
<∞
(
2)p
1≤p<∞
X
(d). Modifications are given below:
The Fourier coefficients of the translated function f (x + h) is einh fˆ(n). So
gˆh (n) = (einh − e−inh )fˆ(n) = 2i sin nhfˆ(n).
The first equation in (a) follows from Parseval’s identity. Since |gh (x)| ≤
K|x + h − (x − h)|α = 2α K|h|α , the second inequality in (a) becomes
∞
X
| sin nh|2 |fˆ(n)|2 ≤ 22(α−1) K 2 |h|2α .
n=−∞
33
h = π/2p+1 and 2p−1 < |n| ≤ 2p imply π/4 < |n|h ≤ π/2, hence
| sin nh|2 ≥ 1/2. Thus
1
2
X
|fˆ(n)|2 ≤
2p−1 <|n|≤2p
∞
X
K 2 π 2α
| sin nh|2 |fˆ(n)|2 ≤ 22(α−1) K 2 |h|2α = 2(αp+1)
2
n=−∞
from which it follows
X
|fˆ(n)|2 ≤
2p−1 <|n|≤2p
K 2 π 2α
.
22αp+1
By Cauchy-Schwarz inequality, we have
X
|fˆ(n)|
2p−1 <|n|≤2p
≤
(
X
2p−1 <|n|≤2p
≤
≤
X
12 )1/2 (
|fˆ(n)|2 )1/2
2p−1 <|n|≤2p
(2p − 2p−1 )1/2
Kπ α
2αp+1/2
Kπ
(2α−1/2 )p
Therefore, if α > 1/2, then
X
1≤|n|<∞
X
|fˆ(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
fˆ(n) = O(1/|n|).
19. (Exercise 18) Here are a few things we have learned about the decay of
Fourier coefficients:
(a) if f is of class C k , then fˆ(n) = o(1/|n|k );
(b) if f is Lipschitz, then fˆ(n) = O(1/|n|);
(c) if f is monotonic, then fˆ(n) = O(1/|n|);
(d) if f satisfies a Hölder condition with exponent α where 0 < α < 1,
then fˆ(n) = O(1/|n|α );
P ˆ
(e) if f is merely Riemann integrable, then
|f (n)|2 < ∞ and therefore
ˆ
f (n) = o(1).
Nevertheless, show that the Fourier coefficients of a continuous function
can tend to 0 arbitrarily slowly by proving that for every sequence of
nonnegative real numbers {k } converging to 0, there exists a continuous
34
function f such that |fˆ(n)| ≥ n for infinitely many values of n.
Solution.
P
k
Choose a subsequence
such that
k nk < ∞, e.g., nk < 1/2 .
P∞ {nk } in
x
Then the function k=1 nk e k , is such a function, because it is absolutely ( and hence uniformly) convergent.
P
20. (Exercise 19) Give another proof that the sum 0<|n|≤N einx /n is uniformly bounded in N and x ∈ [−π, π] by using the fact that
1
2i
X
0<|n|≤N
Z
N
X
einx
1 x
sin nx
(DN (t) − 1) dt,
=
=
n
n
2 0
n=1
where DN is the Dirichlet kernel. Now use the fact that
which was proved in Exercise 12.
R∞
0
sin t
t
dt < ∞
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 periodicity
to all of R. The Fourier series of f is
f (x) ∼
∞
1 X ei nx X sin nx
=
,
2i
n
n
n=1
n6=0
and f has a jump discontinuity at the origin with
f (0+ ) =
π
π
, f (0− ) = − , and hence f (0+ ) − f (0− ) = π.
2
2
Show that
π
lim
max SN (f )(x) − =
N →∞ 0<x≤π/N
2
Z
π
0
sin t
π
dt − ≈ 0.08949π,
t
2
which is roughly 9% of the jump π. This result is a manifestation of
Gibbs’s phenomenon which states that near a jump discontinuity, the
Fourier series of a function overshoots (or undershoots) it by approximately 9% of the jump.
22. (Problem 1) For each 0 < α < 1 the series
∞
X
sin nx
nα
n=1
converges for every x but is not the Fourier series of a Riemann integrable
function.
(a) If the conjugate Dirichlet kernel is defined by

if n > 0
 1
X
0
if n = 0
D̃N (x) =
sign(n)einx where sign(n) =

−1 if n < 0,
|n|≤N
35
then show that
D̃N (x) = i
and
Z
cos(x/2) − cos((N + 1/2)x)
,
sin(x/2)
π
|D̃N (x)| dx ≤ c log N, for N ≥ 2
−π
(b) As a result, if f is Riemann integrable, then
(f ∗ D̃)(0) = O(log N ).
(c) In the present case, this leads to
∞
X
1
= O(log N ),
α
n
n=1
which is a contradiction.
Solution.
Let ω = eix .
=
D̃N (x)
n
n
X
X
ωk −
ω −k
k=1
k=1
1−ω
1 − ω −n
− ω −1
1−ω
1 − ω −1
n
1−ω
1 − ω −n
ω
+
1−ω
1−ω
ω + 1 − ω n+1 − ω −n
1−ω
1/2
ω
+ ω −1/2 − ω n+1/2 − ω −n−1/2
ω −1/2 − ω 1/2
cos(x/2) − cos((N + 1/2)x)
i
sin(x/2)
n
=
=
=
=
=
ω
Note that by trigonometric identity, this is
2 sin( N2+1 x) sin( N2 x)
i.
sin(x/2)
Note that t/ sin t ≤ π/2 for t ∈ [−π/2, π/2], so for x ∈ [−π, π],
|D̃N (x)| ≤ 4
36
| sin( N2 x)|
x
Then
Z
π
|D̃N (θ)| dθ
−π
π
Z
≤
8
0
Z
=
=
| sin(N/2)θ|
dθ
θ
(N/2)π
8
8
0
N
−1 Z (k+1)π/2
X
k=0
=
8
=
=
8
kπ/2
N
−1 Z (k+1)π/2
X
k=1
≤
| sin θ|
dθ
θ
2
π
16
π
kπ/2
N
−1
X
k=1
N
−1
X
1
k
Z
| sin θ|
dθ
θ
| sin θ|
dθ +
θ
Z
π/2
0
(k+1)π/2
Z
| sin θ| dθ +
kπ/2
1
+
k
Z
0
sin θ
dθ
θ
π/2
sin θ
dθ
θ
π/2
sin θ
dθ
θ
0
k=1
Z π/2
N −1
sin θ
16 X 1 16
+
+
dθ
π
k
π
θ
0
k=2
≤
≤
16
log N + c1
π
c log N
where c is chosen such that c/2 ≥ 16
π and (c/2) log 2 ≥ c1 , so that
c log N = c/2 log N + c/2 log N ≥ 16
log
N + c1 for all N ≥ 2.
π
(b),(c). If f is such a Riemann integrable function, then fˆ(n) = 1/nα , fˆ(−n) =
37
−1/nα for n > 0, and fˆ(0) = 0. So
=
=
=
=
Z π
1
f (t)D̃N (0 − t) dt
(f ∗ D̃N )(0) =
2π −π
Z π
1
−
f (t)D̃N (t) dt
2π −π
Z π
X
1
f (t)
sign(n)eint dt
−
2π −π
|n|≤N
Z π
X
1
−
sign(n)
f (t)eint dt
2π −π
|n|≤N
X
−
sign(n)fˆ(n)
|n|≤N
= −2
N
X
1
α
n
n=1
Since f (t) is bounded, we also have by (a),
Z π
Z π
|
f (t)D̃N (t) dt| ≤ M
|D̃N (t)| dt = O(log N ).
−π
−π
PN
1
n=1 nα
= O(log N ); this is impossible (since
Thus
N 1−α , see my ApostolStudyNote, p. 25).
PN
1
n=1 nα
grows like
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:
∞
X
1
π
1
=
−
.
2 − α2
2
n
2α
2α
tan(απ)
n=1
For all u ∈ C − πZ,
cot u =
∞
X
1
u
+2
.
2 − n2 π 2
u
u
n=1
(c) Show that for all α ∈ C − Z we have
∞
X
απ
(−1)n−1
= 1 + 2α2
.
sin(απ)
n2 − α 2
n=1
(d) For all 0 < α < 1, show that
Z ∞ α−1
t
π
dt =
.
t
+
1
sin(απ)
0
38
Solution.
From Maple, for all n,
α sin(απ)
.
fˆ(n) = (−1)n+1
π(n2 − α2 )
It follows that for −π ≤ x ≤ π,
cos(αx) =
∞
sin(απ) 2α sin(απ) X
cos nx
.
+
(−1)n+1 2
απ
π
n − α2
n=1
Evaluating at x = π yields the first equation in part (b). The second
equation follows directly from the first by substituting u for απ.
(c) follows by substituting x = 0 in the Fourier series.
R1 R∞
(d). Split the integral as 0 + 1 and change variables t = 1/u in the
second integral, one gets
Z 1 α−1
Z 1 (1−α)−1
Z ∞ α−1
t
t
t
dt =
dt +
dt.
t+1
t+1
0 t+1
0
0
Claim: For 0 < γ < 1,
1
Z
0
∞
X
tγ−1
1
(−1)k
dt =
.
t+1
k+γ
k=0
Proof.
P∞
For each 0 < r < 1, the series tγ−1 k=0 (−1)k tk converges to
formly on [0, r]. So
Z
0
r
tγ−1
t+1
uni-
∞
X
tγ−1
rγ+k
dt =
(−1)k
t+1
k+γ
k=0
Since
=
=
=
1
tγ−1
dt
0 t+1
Z r γ−1
t
lim−
dt
r→1
0 t+1
∞
X
rγ+k
lim−
(−1)k
k+γ
r→1
Z
lim−
r→1
P∞
k=0
∞
X
(−1)k
k=0
rk
k+γ
k 1
The series
k=0 (−1) k+γ is Abel summable and hence convergent to
R 1 tγ−1
dt, by Littlewood’s theorem since (−1)k 1/(k+γ) = O(1/k). Q.E.D.
0 t+1
39
Therefore
Z
=
=
1
0
∞
X
k=0
∞
X
k=1
=
=
tα−1
dt +
t+1
Z
1 (1−α)−1
t
t+1
0
dt
∞
X (−1)k
(−1)k
+
k+1−α
k+α
k=0
∞
k−1
1 X (−1)k
(−1)
+ +
k−α
α
k+α
k=1
∞
X
1
(−1)k−1
+ 2α
α
k 2 − α2
k=1
π
sin(απ)
by (c).
25. (Problem 4) In this problem, we find the formula for the sum of the series
∞
X
1
k
n
n=1
where k is any even integer. These numbers are expressed in terms of the
Bernoulli numbers; the related Bernoulli polynomials are discussed in the
next problem.
Define the Bernoulli numbers Bn by the formula
∞
X
Bn n
z
=
z .
ez − 1 n=0 n!
(a) Show that B0 = 1, B1 = −1/2, B2 = 1/6, B3 = 0, B4 = −1/30, and
B5 = 0.
(b) Show that for n ≥ 1 we have
n−1 1 X n+1
Bn = −
Bk .
n+1
k
k=0
(c) By writing
∞
z
z X Bn n
=
1
−
+
z ,
ez − 1
2 n=2 n!
show that Bn = 0 if n is odd and > 1. Also prove that
z cot z = 1 +
40
∞
X
22n B2n 2n
z .
(2n)!
n=1
(d) The zeta function is defined by
ζ(s) =
∞
X
1
, for all s > 1.
s
n
n=1
Deduce from the result in (c), and the expression for the cotangent function obtained in the previous problem, that
x cot x = 1 − 2
∞
X
ζ(2m) 2m
x .
π 2m
m=1
(e) Conclude that
2ζ(2m) = (−1)m+1
(2π)2m
B2m .
(2m)!
Solution.
(a) Get these numbers directly from (1 + z/2! + z 2 /3! + · · · )−1 by long
division.
(b).
z ez − 1
ez − 1 z
! ∞
!
∞
X Bn
X
1
n
n
=
z
z
n!
(n + 1)!
n=0
n=0
!
∞
n
X
X
Bk
1
=
tn
k! (n − k + 1)!
n=0
1=
k=0
from which it follows that B0 = 1, and for n ≥ 1,
n
X
Bk
k=0
1
= 0,
k! (n − k + 1)!
i.e.
Bn
n!
n−1
X
Bk
1
k! (n − k + 1)!
k=0
n−1
X n + 1 1
= −
Bk
(n + 1)!
k
= −
k=0
(c). Since B0 = 1, B1 = − 21 , we have
∞
z
z X Bn n
=
1
−
+
z .
ez − 1
2 n=2 n!
41
Now,
=
=
=
=
z
z
+
ez − 1 2
1
1
z
+
ez − 1 2
z
z e +1
2 ez − 1
z ez/2 + e−z/2
2 ez/2 − e−z/2
z
z
coth
2
2
is an even function, so B2n+1 = 0 for all n ≥ 1. Note that
coth(iz) =
2 cos z
eiz/2 + e−iz/2
=
2i sin z
eiz/2 − e−iz/2
so cot z = i coth(iz). Therefore
z cot z = iz coth(iz)
=
=
∞
X
B2n
(i2z)2n
(2n)!
n=1
1+
∞
X
(−1)n
n=0
22n B2n 2n
z
(2n)!
From Problem 3 (b), we have for 0 < x < π,writing rn for
x cot x = 1 − 2
=
1−2
=
1−2
=
=
1−2
1−2
x
nπ ,
∞
X
x2
n2 π 2 − x2
n=1
∞
X
rn2
1 − rn2
n=1
∞ X
∞
X
rn2m
n=1 m=1
∞
∞
X
X
m=1
∞
X
m=1
1
n2m
!
x2m
π 2m
ζ(2m) 2m
x
π 2m
m=1
Comparing coefficients of x cot x in both formulas, we get (e).
Remark. Ross Tang has discovered explicit formulas for Bernoulli numbers 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 X
k X
2n
k (−1)j (k − 2j)2n
,
2n
2n
2 −4
j
2k ik k
j=0
k=1
E2n = i
2n+1
k XX
k=1 j=0
k (−1)j (k − 2j)2n+1
.
2k ik k
j
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 for
Bernoulli polynomials.
From the above solution, we get the following series expansion for cot x
and coth x:
∞
22n B2n 2n−1
1 X
(−1)n
z
,
cot z = +
z n=1
(2n)!
coth z =
∞
X
22n B2n 2n−1
z
.
(2n)!
n=1
I have yet to figure out the proofs of the next four formulas:
tan z =
∞
X
(−1)n−1
n=1
tanh z =
4n (4n − 1)B2n 2n−1
z3
2z 5
17z 7
z
=z+
+
+
+ ··· ,
(2n)!
3
15
315
∞
X
z3
2z 5
17z 7
4n (4n − 1)B2n 2n−1
z
=z−
+
−
+ ··· ,
(2n)!
3
15
315
n=1
sec z =
∞
X
(−1)n E2n 2n
z ,
(2n)!
n=0
sech z =
∞
X
E2n 2n
z ,
(2n)!
n=0
where E2n are the Euler’s numbers. (The generating function for E2n is
2
sech z = 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) to
get tan z series. Other series that use Bernoulli numbers are:
z
sin z
tan z
z
,
, log
, log(cos z), log
,··· ,
sin z sinh z
z
z
43
but I have not looked into this assertion.
Atkinson (American Math Monthly, vol. 93, no. 5,1986, p. 387-; this
paper in pdf form is in my computer under the name TangentSeries.pdf)
has discovered that the above series for tangent and secant can be read
off from the sides of the following triangle (only seven rows are shown):
1
0
1
1
1
0
5
0
61
·
1
5
4
5
61
·
0
2
10
56
·
2
2
14
46
·
0
16
32
·
16
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
∞
X
Bn (x) n
zexz
=
z .
ez − 1 n=0 n!
(a) The functions Bn (x) are polynomials in x and
n X
n
Bk xn−k .
Bn (x) =
k
k=0
Show that B0 (x) = 1, B1 (x) = x − 1/2, B2 (x) = x2 − x + 1/6, and B3 (x) =
x3 − 23 x2 + 12 x.
(b) If n ≥ 1, then
Bn (x + 1) − Bn (x) = nxn−1 ,
and if n ≥ 2, then
Bn (0) = Bn (1) = Bn .
m
(c) Define Sm (n) = 1 + 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
R1
satisfy (i) B0 (x) = 1, (ii) Bn0 (x) = nBn−1 (x) for n ≥ 1, (iii) 0 Bn (x) dx =
0 for n ≥ 1, and show that from (b) one obtains
Z x+1
Bn (t) dt = xn .
m
(e) Calculate the Fourier series of B1 (x) to conclude that for 0 < x < 1
we have
∞
1
−1 X sin(2πkx)
B1 (x) = x − =
.
2
π
k
k=1
44
Integrate and conclude that
n+1
B2n (x) = (−1)
∞
2(2n)! X cos(2πkx)
,
(2π)2n
k 2n
k=1
∞
B2n+1 (x) = (−1)n+1
2(2n + 1)! X sin(2πkx)
.
(2π)2n+1
k 2n+1
k=1
Finally, show that for 0 < x < 1,
Bn (x) = −
n! X e2πikx
.
(2πi)n
kn
k6=0
We observe that the Bernoulli polynomials are, up to normalization (i.e.
R1
requiring 0 Bn (x) dx = 0), successive integrals (antiderivatives) of the
sawtooth function x − 1/2, 0 < x < 1.
Solution.
zexz
z
= z
exz
z
e −1
e −1
∞
∞
X
Bn n X xn n
z
z
=
n!
n!
n=0
n=0
!
∞
n
X
X
Bk xn−k
=
zn
k!
(n
−
k)!
n=0 k=0
!
∞
n X
1 X n
n−k
=
Bk x
zn
k
n!
n=0
k=0
Therefore
Bn (x) =
n X
n
k=0
k
Bk xn−k .
ze(x+1)z
zexz
−
ez − 1
ez − 1
zexz z
=
(e − 1)
ez − 1
xz
= ze
X xn
=
z n+1
n!
n=0
=
X xn−1
zn
(n
−
1)!
n=1
45
Therefore comparing coefficients of z n , n ≥ 1 we get
xn−1
Bn (x + 1) Bn (x)
−
=
n!
n!
(n − 1)!
and hence
Bn (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
X
km
k=1
n−1
X
=
Bm+1 (k + 1) − Bm+1 (k)
k=1
= Bm+1 (n) − Bm+1 (1)
= Bm+1 (n) − Bm+1
Note that this yields a formula for the sum
1 m + 2 m + · · · + nm =
Write F (x, z) for
zexz
ez −1 .
1
(Bm+1 (n + 1) − Bm+1 ).
m+1
Since
∂
F (x, z) = zF (x, z)
∂x
, comparing coefficient of z n on both sides, we get
Bn0 (x)
Bn−1 (x)
=
,
n!
(n − 1)!
hence Bn0 (x) = nBn−1 (x). Next,
Z 1
Z 1
1
1
0
Bn (x) dx =
Bn+1
(x) dx =
(Bn+1 (1) − Bn+1 (0)) = 0.
n+1 0
n+1
0
From (b),
Z x+1
Bn (t) dt =
x
1
n+1
Z
x+1
0
Bn+1
(t) dt =
x
1
(Bn+1 (x+1)−Bn+1 (x)) = xn .
n+1
From Maple, one gets
∞
B1 (x) = x −
1
−1 X sin(2πkx)
=
.
2
π
k
k=1
46
B2 (x) is an antiderivative of 2B1 (x), so
∞
B2 (x) =
−2 X − cos(2πkx)
+C
π
2πk 2
k=1
But
R1
0
B2 (x) dx = 0 and
B2 (x) =
R1
0
cos(2πkx) dx = 0, so C = 0, and
∞
∞
2 X cos(2πkx)
2 · 2! X cos(2πkx)
=
.
2π 2
2πk 2
(2π)2
k2
k=1
k=1
Next, arguing the same manner, B3 (x) is an antiderivative of 3B2 (x), and
B3 (x) =
∞
2 · 3! X sin(2πkx)
.
(2π)3
k3
k=1
And then
B4 (x) = −
∞
2 · 4! X cos(2πkx)
.
(2π)4
k4
k=1
and so on, and we get
B2n (x) = (−1)n+1
∞
2(2n)! X cos(2πkx)
,
(2π)2n
k 2n
k=1
∞
B2n+1 (x) = (−1)n+1
2(2n + 1)! X sin(2πkx)
.
(2π)2n+1
k 2n+1
k=1
Writing sin, cos in terms of e powers, and noting that i2m = (−1)m , we
get the last formula
Bn (x) = −
n! X e2πikx
.
(2πi)n
kn
k6=0
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
1
σN,K = ((N + K)σN +K − N σN ),
K
and
X
|j| − N ˆ
σN,K = SN +
1−
f (j)eijθ .
K
N +1≤|j|≤N +K−1
From this last expression for σN,K conclude that
X
|σN,K − SM | ≤
|fˆ(j)|
N +1≤|j|≤N +K−1
for all N ≤ M < N + K.
(b) Use one of the above formulas and Fejer’s theorem to show that with
N = kn and K = n, then
σkn,n (f )(θ) → f (θ) as n → ∞
whenever f is continuous at θ, and also
σkn,n (f )(θ) →
f (θ+ ) + f (θ− )
as n → ∞
2
at a jump discontinuity (refer to the preceding chapters and their exercises for the appropriate definitions and results). In the case when f is
continuous on [−π, π], show that σkn,n (f ) → f uniformly as n → ∞.
(c) Using part (a), show that if fˆ(j) = O(1/|j|) and kn ≤ m < (k + 1)n,
we get
C
|σkn,n − Sm | ≤
for some constant C > 0.
k
(d) Suppose that fˆ(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− )
as N → ∞.
2
Also, show that if f is continuousP
on [−π, π], then SN (f ) → f uniformly.
(e) The aboveP
arguments show if
cn is Cesàro summable to s and cn =
O(1/n), then
cn converges to s. This is a weak version of Littlewood’s
theorem (Problem 3, Chapter 2).
Solution.
48
1
(a). σN,K = K
((N + K)σN +K − N σN ) is straightforward. Write aj for
ijθ
ˆ
f (j)e . We have
σN,K
1
[SN + (SN + aN +1 + a−N −1 ) + · · · + (SN +
=
K
= SN +
K−1
1 X
K
X
X
aj )]
N +1≤|j|≤N +K−1
aj
l=1 N +1≤|j|≤N +l
= SN +
= SN +
1
K
1
K
X
K−1
X
aj
N +1≤|j|≤N +K−1
X
(K + N − |j|)aj
N +1≤|j|≤N +K−1
X
= SN +
1−
N +1≤|j|≤N +K−1
|j| − N
K
For N ≤ M < N + K, σN,K − SM =
|j|−N
K
either 1 −
(b). We have
σkn,n =
1
l=|j|−N
or
− |j|−N
K ,
P
aj
N +1≤|j|≤N +K−1 bj aj ,
hence |σN,K − SM | ≤
where bj is
P
N +1≤|j|≤N +K−1
|aj |.
1
[(k + 1)nσ(k+1)n − knσkn ] = (k + 1)σ(k+1)n − kσkn
n
−
+
(θ )
Let A be either f (θ) or f (θ )+f
. Since σ(k+1)n , σkn approach the
2
same value A, as n → ∞, we see that σkn,n → A. Uniform convergence
statement is also clear from σkn,n = (k + 1)σ(k+1)n − kσkn . (I think I have
to elaborate this last sentence.)
(c). Suppose |fˆ(j)| = |aj | ≤ C1 /|j| for some constant C1 > 0. From the
last part of (a), for kn ≤ m < (k + 1)n,
≤
|σkn,n − Sm |
X
C1
kn+1≤|j|≤(k+1)n−1
≤
≤
1
|j|
n−1
kn + 1
1
C
2C1 =
k
k
2C1
(d). Let > 0. Fix a positive integer k such that
that
|σkn,n − A| < 49
C
k
< . Choose N such
for all n ≥ N . Then for all m ≥ kN , we have kn ≤ m < (k + 1)n for some
n ≥ N , and
|Sm − A| ≤ |Sm − σkn,n | + |σkn,n − A| < 2.
This proves that Sm → A as m → ∞.
(e)Note that P
except for the uniform convergence part, the only properties
of 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 simplified
version of the Fourier inversion formula. Suppose f is a continuous function supported on an interval [−M, M ], whose Fourier transform fˆ is of
moderate decrease.
P
(a) Fix L with L/2 > M , and show that f (x) = an (L)e2πinx/L where
Z
1
1 L/2
f (x)e−2πinx/L dx = fˆ(n/L).
an (L) =
L −L/2
L
P∞
Alternatively, we may write f (x) = δ n=−∞ fˆ(nδ)e2πinδx with δ = 1/L.
(b) Prove that if F is continuous and of moderate decrease, then
Z ∞
∞
X
F (ξ) dξ = lim δ
F (δn).
δ→0,δ>0
−∞
n=−∞
R∞
(c) Conclude that f (x) = −∞ fˆ(ξ)e2πixξ dξ.
Solution.
(a)
∞
∞
X
1 X ˆ n 2πinx/L
f ( )e
an (L)e2πinx/L =
L n=−∞ L
n=−∞
n
)| ≤
Let B, C be the constants defined as in Remark 1 below. Since |fˆ( L
B
,
the
above
series
is
absolutely
convergent.
Hence
the
series
converges
n2
uniformly (and absolutely) to f (x) by Corollary
R 2.3 in Chapter 2.
(b). Given > 0, choose N > C such that |x|>N |F (x)| dx < /4 and
P
B
|n|>N n2 < /4. Choose 0 < δ1 < 1 such that for all 0 < δ < δ1 ,
Z
N
|
F (x) dx − δ
−N
X
|n|>N/δ
F (nδ)| <
|n|≤N/δ
This is possible because δ
in the details.) Since
δ|
X
P
|n|≤N/δ
F (δn)| ≤
F (nδ) is almost a Riemann sum. (Fill
X
|n|>N
50
.
2
|F (δn)| ≤
X B
n2
|n|>N
the result follows.
(c). Apply (b) to F (ξ) = fˆ(ξ)e2πiξx and use (a).
Remark 1. From the book’s definition, a function f defined on R is said
to be of moderate decrease if f is continuous and there exists a constant
A > 0 so that
A
for all x ∈ R.
|f (x)| ≤
1 + x2
Note that this is equivalent to saying that f is continuous and there are
positive constants B, C such that
|f (x)| ≤
B
for all |x| ≥ C.
x2
Proof. Suppose f satisfies the first condition. Since for all x 6= 0
A
A
2A
=
≤ 2
1 + x2
1 + x2 /2 + x2 /2
x
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
2B
2B
= 2
≤
for |x| ≥ D.
x2
x + x2
1 + x2
Let M = max{(1+x2 )|f (x)| : |x| ≤ D}. Then we can let A = 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(2πξ
πξ , 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−1
, and − n(x − 1) for
≤x≤1
n
n
and zero for x ≥ 1. The Fourier transform of fn , n ≥ 1, is
n(1 − 2 cos2 (πξ) + cos( n−1
n 2πξ))
2
2
2π ξ
which is of moderate decrease. A continuous with compact support such
it Fourier transform is not of moderate decrease is in Exercise 3. It
remains to see an example of continuous function of compact
support such that its Fourier transform is not integrable. From
sci.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) is
an example. See http://mathoverflow.net/questions/43550/is-the-fouriertransform-of-1-1-log1-x2-supported-in-1-1-integrable
Existence of such function also follows from closed graph theorem, see the
link in the above link.
51
2. (Exercise 2) Let f and g be the functions defined by
1 if |x| ≤ 1,
1 − |x|
f (x) = χ[−1,1] (x) =
and g(x) =
0 otherwise,
0
if |x| ≤ 1,
otherwise.
Although f is not continuous, the integral defining its Fourier transform
still make sense. Show that
2
sin πξ
sin 2πξ
ˆ
,
and ĝ(ξ) =
f (ξ) =
πξ
πξ
with the understanding that fˆ(0) = 2 and ĝ(0) = 1.
Solution. Check with Maple and found to agree.
3. (Exercise 3) The following exercise illustrates the principle that the decay
of fˆ is related to the continuity properties of f .
(a) Suppose that f is a function of moderate decrease on R whose Fourier
transform fˆ is continuous and satisfies
1
ˆ
f (ξ) = O
as |ξ| → ∞
|ξ|1+α
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, with
f (0) = 0, and which is equal to 1/ log(1/|x|) for all x in a neighborhood
of the origin. Prove that fˆ is not of moderate decrease. In fact, there is
no > 0 so that fˆ(ξ) = O(1/|ξ|1+ ) as |ξ| → ∞.
Solution.
By remarks in section 1.7, since both f and fˆ are of moderate decrease,
the inverse Fourier transform of fˆ is f . Thus
Z ∞
f (x + h) − f (x) =
fˆ(ξ)e2πiξx (e2πiξh − 1) dξ.
−∞
The condition on fˆ is equivalent to
|fˆ(ξ)| ≤
A
for all ξ ∈ R
1 + |ξ|1+α
52
for some A > 0. Thus (note that |e2πiξh − 1| = 2| sin(πξh)|)
Z ∞
f (x + h) − f (x) A|e2πiξh − 1|
≤ 1
dξ
α
α
h
|h| −∞ 1 + |ξ|1+α
Z ∞
4A
| sin(πξh)|
≤
dξ
|h|α 0 1 + ξ 1+α
Z
4A ∞
| sin(u)|
=
du
π 0 |h|1+α + u1+α
!
Z 1 sin(u)
Z ∞
| u |
1
4A
≤
du +
du
π
uα
u1+α
0
1
Z 1
Z ∞
4A
1
1
≤
du +
du
α
π
u1+α
0 u
1
<∞
(b). We have
|f (h) − f (0)|
1
=
→∞
|h|
−|h| log |h|
as h → 0 for any fixed > 0. So by (a), fˆ is not of moderate decrease.
4. (Exercise 4) Examples of compactly supported functions in S(R) are very
handy in many applications in analysis. Some examples are:
(a) Suppose a < b, and f is the function such that f (x) = 0 if x ≤ a or
x ≥ b and
f (x) = e−1/(x−a) e−1/(b−x) if a < x < b.
Show that f is indefinitely differentiable on R.
(b) Prove that there exists an indefinitely differentiable function F on R
such that F (x) = 0 if x ≤ a, F (x) = 1 if x ≥ b, and F is strictly increasing
on [a, b].
(c) Let δ > 0 be so small that a + δ < b − δ. Show that there exists an
indefinitely differentiable function g such that g is 0 if x ≤ a or x ≥ b, g
is 1 on [a + δ, b − δ], and g is strictly monotonic on [a, a + δ] and [b − δ, b].
Solution. Note that the graph of −1/(x − a) − 1/(b − x) is symmetric
about the line x = (a + b)/2 with maximum value of −4/(b − a) at the
point x = (a + b)/2.
R∞
Rx
For (b) consider F (x) = c −∞ 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 fˆ is continuous and fˆ(ξ) → 0 as |ξ| → ∞.
(b) Show that if fˆ(ξ) = 0 for all ξ, then f is identically 0.
Solution.
53
Z ∞
f (x)e−2πiξx (e−2πihx − 1) dx
|fˆ(ξ + h) − fˆ(ξ)| = −∞
Z ∞
|f (x)||e−2πihx − 1| dx
≤
−∞
Z ∞
=
|f (x)|2| sin 2πhx| dx
−∞
Z ∞
| sin 2πhx|
≤ 2A
dx
1 + x2
−∞
R
1
Let > 0. Choose N > 0 such that 2A |x|>N 1+x
Since
2 dx < /2.
| sin 2πhx|
→ 0 uniformly on [−N, N ] as h →
1+x2
R N | sin 2πhx|
2A −N 1+x2 dx < /2 for |h| < δ. This
0, there exists δ > 0 such that
proves that fˆ is continuous.
Note: Lebesgue dominated convergence theorem could be applied to yield
a simpler proof.
(b).
Z ∞
Z ∞
1
−2πiξx
−2πi(ξx+1/2)
ˆ
f (ξ) =
f (x)e
−
f (x)e
dx
2
−∞
−∞
Z ∞
1
1
=
(f (x) − f (x − ))e−2πiξx dx
2 −∞
2ξ
Now the result follows from applying Lebesgue dominated convergence
theorem; without it, the proof is harder.
(b). If f, g are of moderate decrease, then the function f (x)g(y)e−2πixy is
integrable over R2 . Then it follows from Fubini’s theorem that
Z
Z
f (x)ĝ(x) dx = g(x)fˆ(x) dx.
Suppose fˆ(x) = 0 for all x. For any t and any δ > 0, Kδ (t−x) as a function
of x is in S(R), so by Corollary
1.10 it is equal
to the Fourier transform gˆδ
R
R
for some gδ in S(R). So f (x)gˆδ (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
R∞
2
f (y)e−y e2xy dy = 0 for all x ∈ R, then f = 0.
−∞
Proof.
R∞
2
2
2 R∞
2
Let g(x) = e−x . Then (f ∗g)(x) = −∞ f (y)e−(x−y) dy = e−x −∞ f (y)e−y e2xy dy =
√
2 2
0 for all x. This implies that f[
∗ g(ξ) = fˆ(ξ)ĝ(ξ) = fˆ(ξ) πe−π ξ = 0 for
all ξ. So fˆ = 0. By Theorem 1.9, f = 0.
7. (Exercise 12) Show that the function defined
u(x, t) =
54
x
Ht (x)
t
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
x(x2 − 6t) x2 /4t
∂u
∂2u
√
=
e
.
=
∂t
∂x2
8 πt7/2
It is clear that limt→0+ u(x, t) = 0 for all x. If we approach the origin
along the parabola x2 = t, we get
lim
x→0 x2
1
√
4π
e−1/4 = ∞.
8. (Add in) Suppose f is an even function such that both f and fˆ are of
moderate decrease. Then fˆ is also even and the Fourier transform of fˆ(x)
is f (ξ).
Proof.
R
R∞
The below denotes −∞ . We have
Z
Z
Z
fˆ(−ξ) = f (x)e2πixξ dx = f (−x)e−2πixξ dx = f (x)e−2πixξ dx = fˆ(ξ)
so fˆ is even.
Since
Z
f (x) =
we have
Z
fˆ(t)e−2πiξt dt =
Z
fˆ(t)e2πixt dt
fˆ(−t)e2πiξt dt =
Z
fˆ(t)e2πiξt dt = f (ξ)
proving that the Fourier transform of fˆ(x) is f (ξ).
9. (Exercise 15) This exercise provides another example of periodization.
(a) Apply the Poisson summation formula to the function ĝ(x) and its
Fourier transform in Exercise 2 to obtain
∞
X
1
π2
=
2
(n + α)
(sin πα)2
n=−∞
whenever α is real, but not equal to an integer.
(b) Prove as a consequence that
∞
X
1
π
=
n+α
tan πα
n=−∞
whenever α is real but not equal to an integer.
Solution.
55
(1)
By the last item, the Fourier transform of h(x) = ĝ(x) is g(x) in Exercise
2. By Poisson summation formula
∞
∞
X
X
sin2 (π(n + α))
=
g(n)e2πinα = 1.
2 (n + α)2
π
n=−∞
n=−∞
(If α is an integer, then the equality is reduced to trivial 1 = 1.) Since
sin(nπ + πα) = (−1)n sin(πα), the result follows.
(b). Assume that 0 < α < 1. We have from (a), for n 6= 0,
Z α
1
1
1
1
+
dx
=
−
+
2
(n + x)2
−n + α n + α
0 (−n + x)
2
and since limx→0 (sinππx)2 − x12 = π 2 /3, it has a removable discontinuity at
x = 0 and limx→0 − tanππx + x1 = 0, we have
Z α
π2
1
π
1
− 2 dx = −
+ .
2
(sin
πx)
x
tan
πα
α
0
Therefore
∞
1 X
+
α n=1
1
1
+
n + α −n + α
=
π
.
tan πα
Evaluating at α = 1/2 yields 0; this is correct since
∞
X
∞
X
4
2
2
=
−
2
4n − 1 n=1 2n − 1 2n + 1
n=1
is a telescoping series whose sum is 2.
Now we have to prove for any α, a noninteger. Let n1 = bαc. Then
α = n1 + α1 for some 0 < α1 < 1. Then
∞
X
∞
X
1
1
π
π
=
.
=
=
n + α n=−∞ n + α1
tan πα1
tan πα
n=−∞
10. P
(Exercise 19) The following is a variant of the calculation of ζ(2m) =
∞
2m
found in Problem 4, Chapter 3.
n=1 1/n
(a) Apply the Poisson summation formula to f (x) = t/(π(x2 + t2 )) and
fˆ(ξ) = e−2πt|ξ| where t > 0 in order to get
∞
∞
X
1 X
t
=
e−2πt|n| .
π n=−∞ t2 + n2
n=−∞
(b) Prove the following identity:
∞
∞
t
1
2 X
1 X
=
+
(−1)m+1 ζ(2m)t2m−1 , 0 < t < 1
π n=−∞ t2 + n2
πt π m=1
56
as well as
∞
X
e−2πt|n| =
n=−∞
2
− 1, 0 < t < 1.
1 − e−2πt
(c) Use the fact that
∞
X
z
z
B2m 2m
=
1
−
+
z ,
z
e −1
2 m=1 (2m)!
where Bk are the Bernoulli numbers to deduce from the above formula,
2ζ(2m) = (−1)m+1
(2π)2m
B2m .
(2m)!
Solution.
(a) By Lemma 2.4, the Fourier transform of f (x) = t/(π(x2 +t2 )) is fˆ(ξ) =
e−2πt|ξ| . Then by Poisson summation formula with x = 0 ((x + n)2 = n2 ),
we get (a).
(b).
∞
t
1
1X
t
1 X
=
+
π n=−∞ t2 + n2
πt π
t2 + n2
n6=0
∞
X
=
1
2
+
πt π
t
2 + n2
t
n=1
=
∞
t/n2
2X
1
+
πt π n=1 1 + (t/n)2
=
∞
∞
2m
1
2X t X
m t
+
(−1)
πt π n=1 n2 m=0
n2m
=
∞ ∞
2XX
t2m+1
1
+
(−1)m 2(m+1)
πt π n=1 m=0
n
∞ ∞
1
2XX
t2m−1
(−1)m+1 2m
=
+
πt π n=1 m=1
n
=
∞ ∞
1
2 XX
t2m−1
+
(−1)m+1 2m
πt π m=1 n=1
n
=
∞
1
2 X
+
(−1)m+1 ζ(2m)t2m−1
πt π m=1
57
Next,
∞
X
e−2πt|n| = 1 + 2
n=−∞
∞
X
e−2πtn
n=1
−2πt
=1+2
e
1 − e−2πt
1 + e−2πt
1 − e−2πt
2 − (1 − e−2πt )
=
1 − e−2πt
2
=
−1
1 − e−2πt
=
Dividing by 2 on both sides, we get
∞
1
1 X
1
1
+
(−1)m+1 ζ(2m)t2m−1 =
− .
2πt π m=1
1 − e−2πt
2
Letting z = −2πt and multiplying both sides by −z, we get
1+
∞
1 X
2πz 2m
z
z
(−1)m+1 ζ(2m)
= z
+ .
2m
π m=1
(2π)
e −1 2
Comparing with
∞
X
z
B2m 2m
z
=
1
−
+
z ,
ez − 1
2 m=1 (2m)!
we get
2ζ(2m) = (−1)m+1
(2π)2m
B2m .
(2m)!
11. (Exercise 23) The Heisenberg uncertainty principle can be formulated in
d2
2
terms of the operator L = − dx
2 + x , which acts on Schwartz functions
by the formula
d2 f
L(f ) = − 2 + x2 f.
dx
This operator, sometimes called the Hermite operator, is the quantum
analogue of the harmonic oscillator. Consider the usual inner product on
S given by
Z ∞
(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
df
+ xf and A∗ (f ) = −
+ xf.
dx
dx
The operators A and A∗ are sometimes called the annihilation and creation 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
df
+ txf and A∗t (f ) = −
+ txf.
dx
dx
Use the fact that (A∗t At f, f ) ≥ 0 to give anotherR proof of the Heisenberg
∞
uncertainty principle which says that whenever −∞ |f (x)|2 = 1 then
Z ∞
Z ∞ 2 !
df 2
2
dx ≥ 1 .
x |f (x)|
4
−∞
−∞ dx
Solution.
p
(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
Z
Z
Z
Z
(Lf, f ) = −(f 00 , f ) + (x2 f, f ) = (f 0 , f 0 ) + (x2 f, f ) = |f 0 |2 + x2 |f |2 = 4π 2 ξ 2 |fˆ|2 + x2 |f |2
Z
Z
1
2 ˆ 2 1/2
=1
≥ 4π( ξ |f | ) )( x2 |f |2 )1/2 ≥ 4π
4π
where we have used integration by parts and theorem 4.1 (Heisenberg
uncertainty principle).
(b) Direct checking. Use integration by parts. Note: AA∗ = L + I.
(c) Similar to (b), we also have (A∗t At f, f ) ≥ 0 for all t. When this is
written out, we get t2 (x2 f, f )−t(f, f )−(f 00 , f ) = t2 (x2 f, f )−t+(f 0 , f 0 ) ≥ 0
for all t. It follows that
1 − 4(f 0 , f 0 )(x2 f, f ) ≤ 0
from which the result follows.
12. (Problem 1) The equation
x2
∂2u
∂u
∂u
+ ax
=
∂x2
∂x
∂t
with u(x, 0) = f (x) for 0 < x < ∞ and t > 0 is a variant of the heat
equation which occurs in a number of applications. To solve 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
∂U
∂2U
∂U
+ (1 − a)
=
,
∂y 2
∂y
∂t
with U (y, 0) = F (y). This can be solved like the usual heat equation (the
case a = 1) by taking the Fourier transform in the y variable. One must
then compute the integral
Z ∞
2 2
e(−4π ξ +(1−a)2πiξ)t e2πiξν dξ.
−∞
Show that the solution of the original problem is then given by
Z ∞
2
dν
1
e(− log(ν/x)+(1−a)t) /(4t) f (ν) .
u(x, t) =
ν
(4πt)1/2 0
Solution.
I have checked that what it says up to the new differential equation is
correct.
Taking the Fourier transform w.r.t. y of the new equation, we get
∂ Û
= (−4π 2 ξ 2 + (1 − a)2πiξ)Û .
∂t
Solve, using Û (ξ, 0) = F̂ (ξ), to get
Û (ξ, t) = F̂ (ξ)e(−4π
2 2
ξ +(1−a)2πiξ)t
.
2 2
It remains to find the inverse Fourier transform of e(−4π ξ +(1−a)2πiξ)t and
then write U (y, t) as a convolution. We have
Z ∞
Z ∞
2 2
2 2
e(−4π ξ +(1−a)2πiξ)t e2πiξy dξ =
e−4π ξ t e((1−a)t+y)2πiξ dξ
−∞
−∞
2
2
which is equal to the inverse transform of e−4π tξ evaluated at (1−a)t+y.
√
2
2
Using F(e−πy ) = e−πξ and F(δf (δy)) = fˆ(δ −1 ξ) with δ = 1/ 4πt, we
find that the required inverse transform is
√
1 −y2 /4t
e
4πt
evaluating at (1 − a)t + y yields
√
1 −(y+(1−a)t)2 /(4t)
e
4πt
Therefore
Z
∞
U (y, t) =
−∞
F (µ) √
1 −(y−µ+(1−a)t)2 /(4t)
e
dµ.
4πt
Make a substitution µ = − log ν, and substitute y = − log x, we get the
result.
60
13. (Problem 2) The Black-Scholes equation from finance theory is
∂V
∂V
σ 2 s2 ∂ 2 V
− rV = 0, 0 < t < T,
+ rs
+
∂t
∂s
2 ∂s2
(2)
subject to the ”final” boundary condition V (s, T ) = F (s). An appropriate change of variables reduces this to the equation in Problem 1. Alternatively, the substitution V (s, t) = eax+bτ U (x, τ ) where x = log s, τ =
σ2
1
r
1
r 2
reduces (2) to the one2 (T − t), a = 2 − σ 2 , and b = − 2 + σ 2
dimensional heat equation with the initial condition U (x, 0) = e−ax F (ex ).
Thus a solution to the Black-Scholes equation is
Z ∞ (log(s/s∗ )+(r−σ2 /2)(T −t))2
ds∗
e−r(T −t)
−
2σ 2 (T −t)
e
F (s∗ ) ∗ .
V (s, t) = p
2
s
2πσ (T − t) 0
Solution. Multiplying the equation by e−rt . Define V1 = e−rt V . Then
the equation becomes
∂V1
∂V1
σ 2 s2 ∂ 2 V1
+ rs
+
= 0.
∂t
∂s
2 ∂s2
Let t1 = −(σ 2 /2)t. Then
on both sides we get
∂V1
∂t
=
∂V1
2
∂t1 (−σ /2),
and upon dividing by −σ 2 /2
∂V1
∂ 2 V1
2r ∂V1
+ s2
.
= 2s
∂t1
σ ∂s
∂s2
This is the equation in Problem 1.
For the alternative way, compute
and canceling out eax+bτ , we get
−
∂V
∂t
, ∂V
∂s and
∂2V
∂s2
, replacing 1/s by ex
∂U
∂U
σ2
∂U ∂ 2 U
σ2
(bU +
)+r(aU +
)+ (a(a−1)U +(2a−1)
+
)−rU = 0.
2
∂τ
∂x
2
∂x ∂x2
2
2
2
Since r + σ2 (2a − 1) = 0, − σ2 b + ra + σ2 a(a − 1) − r = 0, the coefficients
σ2
of U and ∂U
∂x are zero. Upon canceling out 2 , we get the heat equation
∂U
∂2U
x
ax
∂τ = ∂x2 . Since F (e ) = F (s) = V (s, T ) = e U (x, 0), the initial
−ax
F (ex ). From the formula
condition for the heat equation is U (x, 0) = e
for solution of heat equation, we get
Z ∞
2
1
U (x, τ ) = √
e−ay F (ey )e−(x−y) /(4τ ) dy.
4πτ −∞
Making a change of variable y = log s∗ and substitute x = log s, we find
Z ∞
∗
∗ 2
ds∗
1
e−a log s F (s∗ )e−(log(s/s ) /(4τ ) ∗ .
U (log s, τ ) = √
s
4πτ 0
61
and
V (s, t) = √
1
4πτ
Z
∞
∗
∗ 2
ea log s+bτ e−a log s F (s∗ )e−(log(s/s
) /(4τ )
0
ds∗
.
s∗
Adding up the exponents of e, we get
a log(s/s∗ ) + bτ − (log(s/s∗ )2 /(4τ )
1
= − [log2 (s/s∗ ) − 4aτ log(s/s∗ ) − 4bτ 2 ]
4τ
1
= − (log(s/s∗ ) − 2aτ )2 + (a2 + b)τ
4τ
2
Since 4τ = 2σ 2 (T − t), −2aτ = (r − σ2 )(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) =
∞
X
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
−a
e−t
if t > 0
g(t) =
0
if t ≤ 0.
One can show that there exists 0 < θ < 1 depending on a so that
|g (k) (t)| ≤
k! − 1 t−a
e 2
for t > 0.
(θt)k
(c) As a result, for each x and t the series (3) converges; u solves the heat
equation; u vanishes for t = 0; and u satisfies the estimate |u(x, t)| ≤
2a/(a−1)
Cec|x|
for some constants C, c > 0.
(d) Conclude that for every > 0 there exists a nonzero solution to the
heat equation which is continuous for x ∈ R and t ≥ 0, which satisfies
2+
u(x, 0) = 0 and |u(x, t)| ≤ Cec|x| .
15. (Problem 7) The Hermite functions hk (x) are defined by the generating
identity
∞
X
2
2
tk
hk (x) = e−(x /2−2tx+t ) .
(4)
k!
k=0
(a) Show that an alternate definition of the Hermite functions is given by
the formula
k
2
2
d
hk (x) = (−1)k ex /2
e−x .
(5)
dx
62
Conclude from the above expression that each hk (x) is of the form Pk (x)e−x
where Pk is a polynomial of degree k. In particular, the Hermite functions
2
2
belong to the Schwartz space and h0 (x) = e−x /2 , h1 (x) = 2xe−x /2 .
∞
(b) Prove that the family {hk }k=0 is complete in the sense that if f is a
Schwartz function, and
Z ∞
(f, hk ) =
f (x)hk (x) dx = 0 for all k ≥ 0,
−∞
then f = 0.
(c) Define h∗k (x) = hk ((2π)1/2 x). Then
c∗ (ξ) = (−i)k h∗ (ξ).
h
k
k
Therefore, each h∗k is an eigenfunction for the Fourier transform.
(d) Show that hk is an eigenfunction for the operator defined in Exercise
23, and in fact, prove that
Lhk = (2k + 1)hk .
In particular, we conclude that the functions hk are mutually orthogonal
for the L2 inner product
R ∞on the Schwartz space.
(e) Finally, show that −∞ hk (x)2 dx = π 1/2 2k k!.
Solution.
2
2
2
2
(a) Note that e−(x /2−2tx+t ) = ex /2 e−(t−x) . Note that
d
dx
k
2
e−x = pk (x)e−x
2
for some degree k polynomial pk (x) which is odd if k is odd, and even if k
2
is even. Therefore the k partial derivative of e−(t−x) w.r.t. t, evaluated
2
2
at t = 0 is pk (−x)e−x = (−1)k pk (x)e−x . The formula then follows from
Taylor expansion.
(b) Under the hypothesis, we have by equation (4)
Z ∞
Z ∞
2
2
2
2
f (x)e−(x /2−2tx+t ) dx = e−t
f (x)e−x /2+2tx dx
−∞
−∞
for all t. By Exercise 8, f = 0.
(c) From equation (4), we get
∞
X
√
tk
hk ( 2πx)
k!
k=0
√
= exp(−(πx2 − 2t 2πx + t2 ))
p
= exp(t2 ) exp(−π(x − 2/πt)2 )
63
2
/2
,
Taking Fourier transform of both sides, we get
∞
X
k
√
c∗ (ξ) t = exp(t2 ) exp(−2 2πitξ) exp(−πξ 2 )
h
k
k!
k=0
√
= exp(−(πξ 2 − 2(−it) 2πξ + (−it)2 ))
∞
X
(−it)k
=
h∗k (ξ)
k!
k=0
Comparing both sides, we get the result.
(d) From the RHS of (4), we have
∞
X
L(hk (x))
k=0
2
2
tk
= L(e−(x /2−2tx+t ) )
k!
2
2
d2 −(x2 /2−2tx+t2 )
e
+ x2 e−(x /2−2tx+t )
dx2
2
2
2
2
d
= − e−(x /2−2tx+t ) (−x + 2t) + x2 e−(x /2−2tx+t )
dx
2
2
2
2
= −e−(x /2−2tx+t ) ((−x + 2t)2 − 1) + x2 e−(x /2−2tx+t )
= −
= e−(x
2
/2−2tx+t2 )
2
(1 − 4t2 + 4tx)
2
= e−(x /2−2tx+t ) + (−4t2 + 4tx)e−(x
∞
X
2
2
d
tk
=
hk (x) + 2t e−(x /2−2tx+t )
k!
dt
=
=
k=0
∞
X
2
/2−2tx+t2 )
∞
hk (x)
k=0
∞
X
tk X
tk
+
2khk (x)
k!
k!
k=0
(2k + 1)hk (x)
k=0
tk
k!
from which we get Lhk = (2k + 1)hk . By Exercise 23, L = I + A∗ A, so it
is Hermitian. Therefore the functions hk are mutually orthogonal.
(e) Multiply both sides of equation (4) by hk (x) and integrate. On the
LHS, using the orthogonality property, we get
Z
tk ∞
[hk (x)]2 dx.
k! −∞
On the RHS, using equation (5) and integration by parts k times, we get
(−1)k (−1)k
Z
∞
−∞
2
e−x
d
dx
k
2
e2tx−t = (2t)k
Z
∞
−∞
Therefore result.
64
√
2
e−(x−t) dx = 2k tk π.
Chapter 6 The Fourier Transform on Rd
1. (Exercise 1) Suppose that R is a rotation in the plane R2 , and let
a b
c d
denote its matrix with respect to the standard basis vectors e1 = (1, 0)
and e2 = (0, 1).
(a) Write the conditions Rt = R−1 and det(R) = ±1 in terms of equations
in a, b, c, d.
(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→ z̄eiφ , where z̄ = x − iy.
Solution.
(a) Write D = det(R). Then D = ±1 and the conditions Rt = R−1
implies
a c
d/D −b/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 = a2 D − (−b2 D) = (a2 + b2 )D which
implies 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 z̄eiφ = (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 prove
that there exists γ ∈ S 2 (where S 2 denotes the unit sphere in R3 ) with
R(γ) = γ.
(b) If P denotes the plane perpendicular to γ and passing through the
origin, show that
R : P → P,
and that this linear map is a rotation.
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 there
exists λ > 0 such that p(λ) = 0. So R − λI is singular, and its kernel is
nontrivial.
Chapter 7 Finite Fourier Analysis
1. (Exercise 1.) Let f be a function on the circle. For each N ≥ 1 the discrete
Fourier coefficients of f are defined by
aN (n) =
N −1
N −1
1 X
1 X
f (e2πik/N )e−2πikn/N =
f (e2πik/N )en (k), for n ∈ Z.
N
N
k=0
k=0
65
We also let
Z
a(n) =
1
f (e2πix )e−2πinx dx
0
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 partition: k/N, k = 0, · · · , N , and choice of points: k/N, k = 0, · · · , N − 1.
2. (Exercise 2) If f is a C 1 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 ]
=
=
=
=
N −1
N −1
1 X
1 X
f (e2πik/N )e−2πikn/N −
f (e2πik/N )e−2πi(k−l)n/N
N
N
1
N
1
N
1
N
k=0
k=0
N
−1
X
NX
−l−1
f (e2πik/N )e−2πikn/N −
k=0
N
−1
X
f (e2πik/N )e−2πikn/N −
k=0
1
N
1
N
f (e2πi(k+l)/N )e−2πikn/N
k=−l
N
−1
X
f (e2πi(k+l)/N )e−2πikn/N
k=0
N
−1
X
[f (e2πik/N ) − f (e2πi(k+l)/N )]e−2πikn/N
k=0
f ∈ C 1 implies f is Lipschitz. Let M be its Lipschitz constant. Then
from the above equations,
|aN (n)||1 − e2πiln/N | ≤ M |1 − e2πil/N |
N
Let l be the integer closest to 2n
(in case of ambiguity, choose either
integer). Then |l − N/2n| ≤ 1/2, from which we get
ln 1 − ≤ |n| ≤ 1 .
N
2 2N
4
Thus
|1 − e2πil/N | ≤ |2πl/N | = |2πln/N |/|n| ≤
On the other hand, since
ln 1 1
− ≤
N
2 4
66
3π
.
2|n|
implies
π
2πln
3π
≤
≤
,
2
N
2
√
|1 − e2πiln/N | ≥ 2.
we have
Therefore for 0 < |n| ≤ N/2,
3πM
|aN (n)| ≤ √
.
2 2|n|
3. (Exercise 3) By a similar method, show that if f is a C 2 function on the
circle, then
c
|aN (n)| ≤
, whenever 0 < |n| ≤ N/2.
|n|2
As a result, prove the inversion formula for f ∈ C 2 ,
∞
X
f (e2πix ) =
a(n)e2πinx
n=−∞
from its finite version.
Solution.
We have
aN (n)[2 − e2πiln/N − e−2πiln/N ]
=
N −1
1 X
[2f (e2πik/N ) − f (e2πi(k+l)/N ) − f (e2πi(k−l)/N )]e−2πikn/N
N
k=0
≤
1
N
N
−1
X
M |e2πi(k+l)/N − e2πik/N |2
k=0
2πil/N
= M |e
− 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,
X
aN (n)e2πikn/N
|n|<N/2
=
=
1
N
X
N
−1
X
f (e2πij/N )e−2πijn/N e2πikn/N
|n|<N/2 j=0
N −1
X
1 X
f (e2πij/N )
e−2πijn/N e2πikn/N
N j=0
|n|<N/2
=
f (e
2πik/N
)
P
where the last equality follows because |n|<N/2 e−2πijn/N e2πikn/N = 0
if 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.....).
67
4. (Exercise 4) Let e be a character on G = Z(N ), the additive group of
integers 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 ). Deduce
that e` 7→ ` defines an isomorphism from Ĝ to G.
Proof.
Let the N roots of unit be 1, ζ, ζ 2 , cdots, ζ N −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) = ζ l for some 0 ≤ l ≤ N − 1. Then for each n ∈ Z(N ),
e(n) = e(1 + 1 + · · · + 1) = e(1)n = ζ ln
proving that e = el . el em = el+m 7→ l + m.
5. (Exercise 5) Show that all characters on S 1 are given by
en (x) = einx with n ∈ Z,
and check that en 7→ n defines an isomorphism from Sˆ1 to Z.
Solution.
Since F (0) 6= 0 and F is continuous, there exists δ > 0 such that c =
Rδ
F (y) dy 6= 0. Then
0
Z δ
Z δ
Z δ+x
cF (x) =
F (y)F (x) dy =
F (y + x) dy =
F (y) dy
0
0
x
for all x. Differentiating both sides,
cF 0 (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 S 1 ,
we have 1 = F (0) = F (π + π) = e2πA . Therefore A = in for some integer
n.
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 R̂ to R.
Solution.
Using the same argument in Exercise 5, F (x) = eAx for some complex
number A. Since |F (x)| = 1, A must be pure imaginary. So A = 2πiξ for
some real number ξ. eξ eη = eξ+η 7→ ξ + η.
PN
7. (Exercise 8) Suppose that P (x) = n=1 an e2πinx .
(a) Show by using the Parseval identities for the circle and Z(N ), that
Z
1
|P (x)|2 dx =
0
68
N
1 X
|P (j/N )|2 .
N j=1
(b) Prove the reconstruction formula
P (x) =
N
X
P (j/N )K(x − (j/N ))
j=1
where
K(x) =
1
e2πix 1 − e2πiN x
= (e2πix + e2πi2x + · · · + e2πiN x ).
N 1 − e2πix
N
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 not
congruent to 0 modulo N .
Solution.
Considering P (x) as a 1-periodic function, we have by Parseval identity:
R1
PN
2
|P (x)|2 dx =
n=1 |an | . On the other hand, considering f (j) =
0
PN
2πinj/N
P (j/N ) =
as a function on Z(N ), Parseval identity
n=1 an e
PN
PN
PN
2
yields n=1 |an | = kf k2 = N1 j=1 |f (j)|2 = N1 j=1 |P (j/N )|2 .
8. (Exercise 12) Suppose that G is a finite abelian group and e : G → C is
a function that satisfies e(x + y) = e(x)e(y) for all x, y ∈ G. Prove that
either e is identically 0, or e never vanishes. In the second case, show that
for each x, e(x) = e2πir(x) for some r = r(x) ∈ Q of the form p/q, where
q = |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 the
second case, 1 = e(0) = e(x + (−x)) = e(x)e(−x), proving that e(x) 6= 0
and e(−x) = e(x)−1 .
Let N = |G|. For any x ∈ G, N x = x + x + · · · + x = 0, so 1 = e(0) =
e(x)N , proving that e(x) is equal to ζ k for some k = 0, · · · , N − 1, where
ζ = e2πi/N .
9. (Exercise 13) In analogy with ordinary Fourier series, one may interpret
finite Fourier expansions using convolutions as follows. Suppose G is a
finite abelian group, 1G its unit, and V the vector space of complex-valued
functions on G.
(a) The convolution of two functions f and g in V is defined for each a ∈ G
by
1 X
(f ∗ g)(a) =
f (b)g(a · b−1 ).
|G|
b∈G
Show that for all e ∈ Ĝ one has (f[
∗ g)(e) = fˆ(e)ĝ(e).
(b) Use Theorem 2.5 to show that
X
e(c) = 0 whenever c ∈ G and c 6= 1G .
e∈Ĝ
69
(c) As a result of (b), show that the Fourier series Sf (a) =
of a function f ∈ V take the form
P
e∈Ĝ
fˆ(e)e(a)
Sf = f ∗ D,
where D is defined by
D(c) =
X
e(c) =
|G|
0
if c = 1G ,
otherwise.
e∈Ĝ
Since f ∗ D = f , we recover the fact that Sf = f . Loosely speaking, D
corresponds to a ”Dirac delta function”; it has unit mass
1 X
D(c) = 1,
|G|
c∈G
and (4) says that this mass is concentrated at the unit element in G. Thus
D has the same interpretation as the ”limit” of a family of good kernels.
(See Section 4, Chapter 2.)
Note. The function D reappears in the next chapter as δ1 (n).
Solution.
Let N = |G|.
=
=
=
=
f[
∗ g(e)
1 X
(f ∗ g)(x)e(x)
N
x∈G
1 XX
f (b)g(xb−1 )e(x)
N2
x∈G b∈G
X
1 X
f (b)
g(xb−1 )e(x)
2
N
x∈G
b∈G
X
1 X
f
(b)
g(y)e(by)
N2
b∈G
y∈G
=
X
1 X
f
(b)e(b)
g(y)e(y)
N2
=
fˆ(e)ĝ(e)
b∈G
y∈G
(b). If c 6= 1G , then there exists e0 ∈ Ĝ such that e0 (c) 6= 1.[This can be
proved in two ways. Suppose e(c) = 1 for all e ∈ Ĝ. Let H be the cyclic
subgroup generated by c. Since c 6= 1G , we have |H| > 1 so |G/H| < |G|.
Each e ∈ Ĝ induces a character in G/H, and different e’s induce different
characters. This is impossible since there are exactly |G/H| characters on
G/H.
Another proof is to use the fact that G is isomorphic to a direct product of
cyclic groups G1 × G2 × · · · × Gn where |G1|||G2|| · · · ||Gn|. Let for each i
70
P
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 by
gi → 1 for i 6= k and gk → z, with z = e2πi/N where N = |Gk |, defines a
character that is not 1 on c (it is z akP
).]
P
P
Now back to the proof. We have e0 e∈Ĝ e = e∈Ĝ e0 e = e∈Ĝ e since
P
P
Ĝ is a group. So e0 (c) e∈Ĝ e(c) = e∈Ĝ e(c). Since e0 (c) 6= 1, we get
P
e∈Ĝ e(c) = 0.
Sf (a)
=
X
fˆ(e)e(a)
e∈Ĝ
=
X 1 X
f (x)e(x)e(a)
N
e∈Ĝ
=
x∈G
X
1 X
f (x)
e(a)e(x)−1
N
x∈G
=
e∈Ĝ
X
1 X
f (x)
e(a)e(x−1 )
N
x∈G
e∈Ĝ
=
X
1 X
f (x)
e(ax−1 )
N
=
(f ∗ D)(a)
x∈G
71
e∈Ĝ