MA342A (Harmonic Analysis 1) Tutorial sheet 8
[December 10, 2015]
Name: Solutions
√
1. Could there be F ∈ L2 (T) such that F̂ (n) = 1/ n for n ≥ 1 and F̂ (n) = 0 for n ≤ 0?
Solution:
According to the results on Fourier series of L2 functions, if F ∈ L2 , then
P
2
n∈Z |F̂ (n)| < ∞. (See Parseval’s identity 2.1.46.)
But
2 X
∞ ∞
X
1
1
√
=∞
=
n
n
n=1
n=1
So there cannot be such an F ∈ L2 (T).
2. Could there be F ∈ L1 (T) such that F̂ (n) = (−1)n for each n ∈ Z?
Solution: By the Riemann-Lebesgue Lemma, if F L1 (T), then lim|n|→∞ F̂ (n) = 0. But
limn→∞ (−1)n does not exist (or it is certainly not 0 as |(−1)n | = 1 always). Hence there
cannot be such an F ∈ L1 (T).
3. Show that there is F ∈ C(T) with F̂ (n) = 1/2n for n ≥ 1 and F̂ (n) = 0 for n ≤ 0.
P
1 n
Solution: Consider the series ∞
n=1 2n ζ with |ζ| = 1. It is absolutely convergent since
∞ ∞
X
1 n X
1
ζ =
<∞
2n n
2
n=1
n=1
(in fact the sum is (1/2)/(1 − 1/2) = 1 as it is a geometric series). It follows that it is
uniformly convergent, that is the partial sums
N
X
1 n
sN (ζ) =
ζ
2n
n=1
P∞ 1 n
converge uniformly (for ζ ∈ T) to some limit F (ζ) =
n=1 2n ζ . [See the proof of
Theorem
2.1.41
in the notes for more details. There the argument is that |F (ζ) − sN (ζ)| =
∞
∞
X
X
1 n 1
ζ ≤
→ 0 as N → ∞.]
n
2
2n
n=N +1
n=N +1
Uniform limits of continuous functions are continuous and so F ∈ C(T).
We need to be sure that F has the right Fourier coefficients.
Z
F̂ (n) =
F (ζ)ζ¯n dλ(ζ)F̂ (n)
T
Z =
lim sN (ζ) ζ¯n dλ(ζ)
N →∞
ZT
=
lim sN (ζ)ζ¯n dλ(ζ)
T N →∞
Since |F (ζ)ζ̄n − sN (ζ)ζ̄ n | = |F (ζ)−̄sN (ζ)| → 0 uniformly, we can exchange the order of
the integral and the limit, to get
Z
F̂ (n) = lim
sN (ζ)ζ¯n dλ(ζ) = lim sc
N (n)
N →∞
N →∞
T
But as a trigonometric polynomial, the Fourier coefficients of sN are the ones you see, that
n
is sc
c
N (n) = 0 if n ≤ 0 or n > N while s
N (n) = 1/2 if 0 ≤ n ≤ N .
So, in the limit we get F̂ (n) = 0 if n ≤ 0 and F̂ (n) = 1/2n if n > 0.
Note: Really all we used
If we had
P∞summable.
P∞here was that the coefficients are absolutely
n
number an ∈ C with n=−∞ |an | < ∞ we could define F (ζ) = n=−∞ an ζ and prove
that F ∈ C(T) and F̂ (n) = an for all n ∈ Z.
P
n
Just use essentially the same proof with sN (ζ) = N
n=−N an ζ .
Richard M. Timoney
2