M ATH 140 B - HW 8 S OLUTIONS
Problem 1 (WR Ch 8 #13).
Put f (x) = x if 0 ≤ x < 2π, and apply Parseval’s Theorem to conclude that
∞ 1
X
π2
=
.
2
6
n=1 n
Solution. First we calculate the Fourier coefficients of f (x) = x using integration by parts.
For n 6= 0,
Z
1 π
f (x)e −i nx d x
2π −π
Z
1 π
=
x e −i nx d x
2π −π
· −i nx ¸π
Z
1 π e −i nx
e
1
−
x
dx
=
2π
−i n −π 2π −π −i n
·
¸
·
¸π
1
(−1)n
1 e −i nx =
2π
−
2
2π
−i n
2π−n
−π
n
(−1)
=i
.
n
Z π
Z
1
1 π
c0 =
f (x) d x =
x d x = 0.
2π −π
2π −π
cn =
Parseval’s identity says that
1
2π
Z
π
−π
| f (x)|2 d x =
∞
X
|c n |2 .
−∞
Computing the right side we get
∞
X
−∞
And the left side is
1
2π
Z
π
−π
|c n |2 =
¯
¯
∞ 1
X ¯ (−1)n ¯2
X
¯i
¯ =2
.
¯ n ¯
2
n=1 n
n6=0
|x|2 d x =
· ¸π
1 x3
1 2π3 π2
=
·
=
.
2π 3 −π 2π 3
3
Putting this together,
∞ 1
X
π2
=2
2
3
n=1 n
=⇒
1
∞ 1
X
π2
=
.
2
6
n=1 n
Problem 2 (WR Ch 8 #19).
Suppose f is a continuous function on R1 , f (x + 2π) = f (x), and α/π is irrational. Prove
that
Z
N
1 π
1 X
f (x + nα) =
f (t ) d t
N →∞ N n=1
2π −π
lim
for every x.
Solution. First we prove it for f (x) = e i kx with k ∈ Z and k 6= 0.
1
2π
Z
π
e
−π
i kt
"
#π
1 e i kx
= 0.
dt =
2π i k
−π
Also,
¯
¯
¯
¯
¯1 X
¯ 1 ¯X
¯
N
N
¯
¯
i k(x+nα) ¯
i kx i knα ¯
e
e e
¯
¯= ¯
¯
¯ N n=1
¯ N ¯n=1
¯
¯
¯
¯
N
1 ¯¯ i kx X
i knα ¯
e
= ¯e
¯
¯
N¯
n=1
¯
¯
¯
N
1 ¯¯ i kx ¯¯ ¯¯ X
¯
= ¯e ¯ ¯
e i knα ¯
¯n=1
¯
N
¯
¯
¯
N
1 ¯¯ X
¯
e i knα ¯
= ¯
¯
N ¯n=1
¯
¯
´n ¯
N ³
1 ¯¯ X
i kα ¯
e
= ¯
¯,
¯
N ¯n=1
and since α/π is irrational, that means kα is not a multiple of π, so e i kα 6= 1, and we can use
the partial sum formula from page 61 of Rudin to get
¯
¯
¯ ¯
¯ ¯
¯
¯X
¯ ¯ 1 − ¡e i kα ¢N +1
´n ¯ ¯ X
´n
N ³
N ³
2
¯
¯
¯ ¯
i kα ¯ ¯
i kα
¡
¢ − 1¯ <
+1 = M,
e
− 1¯ = ¯
e
¯
¯=¯
¯ |1 − e i kα |
¯n=1
¯ ¯n=0
¯ ¯ 1 − e i kα
so the sum is bounded by M . Therefore,
¯
¯
¯
¯
¯
¯
¯1 X
¯
N
N
M
1 X
¯
¯
i k(x+nα) ¯
i k(x+nα) ¯
e
e
= 0.
¯ lim
¯ = lim ¯
¯ ≤ lim
¯N →∞ N n=1
¯ N →∞ ¯ N n=1
¯ N →∞ N
This establishes the equality for f (x) = e i kx with k 6= 0. If k = 0, we have f (x) = 1, so
Z
N
N
1 X
1 X
1
1 π
f (x + nα) = lim
1 = lim
N =1=
f (t ) d t .
N →∞ N n=1
N →∞ N n=1
N →∞ N
2π −π
lim
Now that we’ve shown it for all functions of the form e i kx , we set some ² > 0 and we use TheoP
rem 8.15 to get a trigonometric polynomial P (x) = Kk=−K c k e i kx such that |P (x) − f (x)| < ²/2
2
for all real x. Then
N
N ¡£
¤
¢
1 X
1 X
f (x + nα) = lim
f (x + nα) − P (x + nα) + P (x + nα)
N →∞ N n=1
N →∞ N n=1
lim
N £
N
¤
1 X
1 X
f (x + nα) − P (x + nα) + lim
P (x + nα)
N →∞ N n=1
N →∞ N n=1
= lim
1
2π
Z
π
Z
1 π
f (t ) d t =
[ f (t ) − P (t )] + P (t ) d t
2π −π
−π
Z π
Z
1
1 π
=
[ f (t ) − P (t )] +
P (t ) d t .
2π −π
2π −π
Z
N
1 π
1 X
P (x + nα) =
P (t ) d t
N →∞ N n=1
2π −π
lim
¯
¯
Z
¯
N
¯
1 π
1 X
¯
f (t ) d t ¯¯
f (x + nα) −
¯ lim
¯N →∞ N n=1
2π −π
¯
¯
Z
¯
¯
N
1 X
1 π
¯
¯
≤ ¯ lim
[ f (x + nα) − P (x + nα)] −
[ f (t ) − P (t )] d t ¯
¯N →∞ N n=1
¯
2π −π
¯ ¯
¯
¯
Z
¯ ¯ 1 π
¯
N
¯
1 X
¯
¯
[ f (x + nα) − P (x + nα)]¯ + ¯¯
[ f (t ) − P (t )] d t ¯¯
≤ ¯ lim
¯ 2π −π
¯N →∞ N n=1
Z
N
1 X
1 π
≤ lim
| f (x + nα) − P (x + nα)| +
| f (t ) − P (t )| d t
N →∞ N n=1
2π −π
Z
N
1 X
1 π
< lim
²/2 +
²/2 d t
N →∞ N n=1
2π −π
² ²
= + = ².
2 2
Since ² was arbitrary, this proves the equality.
Problem 3 (Supp-3 #1). Suppose f : [0, π] → R is continuous and
Rπ
0
f (x) sin nx = 0,
n = 1, 2, . . .. Does it follow that f ≡ 0? Proof or counterexample.
Solution. Let c =
R
1 π
π 0
f (x) d x and define g : [−π, π] → R by
(
f (x)
x ∈ [0, π],
g (x) =
− f (−x) x ∈ [−π, 0).
Then g is an odd function (i.e. g (−x) = −g (x)), and since cos nx is an even function, that
means that g (x) · cos nx is an odd function, and therefore
Z π
g (x) cos nx d x = 0,
−π
3
because integrating an odd function symmetrically around 0 gives 0. Also note that, (with a
change of variables in the first integral using u = −x)
Z
π
−π
Z
0
Z
π
g (x) sin nx d x =
− f (−x) sin nx d x +
f (x) sin nx d x
Z−π
Z π 0
π
=
f (u) sin nu d u +
f (x) sin nx d x
0Z
0
π
nx
= 2 f (x)sin
dx
0
= 0.
Calculating the Fourier coefficients for g for n ≥ 0:
Z
1 π
cn =
g (x) e −i nx d x
2π −π
Z
1 π
=
g (x) (cos nx − i sin nx) d x
2π −π
µZ π
¶
Z π
1
g
(x)
cos
nx
d
x
−
i
g
(x)
sin
nx
d
x
=
−π
2π −π
= 0.
Z
1 π
g (x) d x
2π −π
Z π
1
2
( f (x) − c) d x
=
2π 0
Z π
Z
1
1 π
=
f (x) d x −
c dx
π 0
π 0
Z π
1
=
f (x) d x − c
π 0
c0 =
= 0.
Now by Parseval’s identity,
1
2π
Z
π
−π
|g (x)|2 d x =
∞
X
|c n |2 = 0,
−∞
and since g 2 is continuous and g 2 ≥ 0, then we have g 2 ≡ 0 by Problem 6.2 proved on a
previous homework. Therefore, g ≡ 0, so f ≡ 0.
Problem 4 (Supp-3 #3). If f is real analytic in a neighborhood of x 0 and f (x 0 ) = 0, show
that f (x)/(x − x 0 ) is real analytic in the same neighborhood.
Solution. If f is real analytic in a neighborhood of x 0 and f (x 0 ) = 0 then there is some ² > 0
such that for x ∈ B (x 0 ; ²),
f (x) =
∞
X
c n (x − x 0 )n
and
0 = f (x 0 ) = c 0
n=0
=⇒
f (x) =
∞
X
n=1
4
c n (x − x 0 )n .
Letting f k =
k
P
n=1
c n (x − x 0 )n , we have f k â f on B (x 0 ; ²) because f is real analytic on B (x 0 ; ²).
Also, by Theorem 8.1, f k0 =
k
P
n=1
nc n (x − x 0 )n−1 is such that { f k0 } converges uniformly to f 0 on
B (x 0 ; ²) (this means the radius of convergence of f k0 is some number R 0 ≥ ²). We’re going to
prove that f k /(x − x 0 ) has the same radius of convergence as f k0 , and that will finish the proof.
Therefore,
k
k
X
X
f k (x)
= (x − x 0 )−1
c n (x − x 0 )n =
c n (x − x 0 )n−1 ,
x − x0
n=1
n=1
and checking the radius of convergence R of this last power series, we have
p
p
p
p
1
1
= lim sup n−1 |c n | = lim sup n−1 n n−1 |c n | = lim sup n−1 n|c n | = 0 ,
R
R
n→∞
n→∞
n→∞
p
since limn→∞ n−1 n = 1. Then R = R 0 ≥ ². That means f k /(x − x 0 ) converges uniformly on
B (x 0 ; ²), and thus f (x)/(x − x 0 ) is real analytic on B (x 0 ; ²).
Problem 5 (Supp #4). Prove that if f (x) is real analytic on (a, b) and c ∈ (a, b), then F (x) =
Rx
c f (t ) d t is also real analytic on (a, b).
b+a
2
(the point in the middle of a and b). If f is real analytic on (a, b) and
k
P
c ∈ (a, b), then letting f k (x) =
c n (x − x 0 )n we have f k â f on [a, b]. Let R 1 be the radius of
n=0
¯
¯
¯
¯
convergence of f k . What we have shown so far is that R 1 ≥ ¯ b−a
2 ¯. Now we set
Solution. Let x 0 =
Z
G k (x) =
x
x0
k
x X
Z
f k (t ) d t =
x 0 n=0
c n (t − x 0 )n d t =
k Z
X
x
n=0 x 0
c n (t − x 0 )n d t =
k
X
cn
(x − x 0 )n+1 ,
n
+
1
n=0
then the radius of convergence R 2 of this last power series is given by
1
= lim sup
R2
n→∞
s
n+1
|c n |
= lim sup
n +1
n→∞
1
p
n +1
p
p
1
|c n | = lim sup n+1 |c n | =
,
R1
n→∞
n+1
n+1
¯
¯
p
¯
¯
n + 1 = 1, and thus we have R 2 = R 1 ≥ ¯ b−a
2 ¯, so G k converges uniformly on
n→∞
Rx
[a, b]. If we let G(x) = x0 f (t ) d t , then by Theorem 7.16 we have G k â G on [a, b]. Lastly, if
Rx
we let C = c 0 f (t ) d t , we have
since lim
n+1
x
Z
F (x) =
c
x0
Z
f (t ) d t =
c
Z
f (t ) d t +
x
x0
f (t ) d t = G(x) +C ,
and similarly we define F k (x) = G k (x)+C , so that F k â F , and thus F is real analytic on [a, b].
5