Measure, Integration & Real Analysis, 2021a

Section 11A Fourier Series and Poisson Integral 351
Now suppose n ∈ Z \{0}. Then
(
f
[k] )̂(n) =
∫ π
−π
˜f (k) (t)e −int dt
2π
= 1
2π ˜f (k−1) (t)e −int] t=π
t=−π + in ∫ π
−π
˜f (k−1) (t)e −int dt
2π
= in ( f [k−1])̂(n),
where the second equality above follows from integration by parts.
Iterating the equation above now produces the desired result.
Now we can prove the beautiful result that a twice continuously differentiable function
on ∂D equals its Fourier series, with uniform convergence of the Fourier series.
This conclusion holds with the weaker hypothesis that the function is continuously
differentiable, but the proof is easier with the hypothesis used here.
11.27 Fourier series of twice continuously differentiable functions converge
Suppose f : ∂D → C is twice continuously differentiable. Then
f (z) =
∞
∑
̂f (n)z n
n=−∞
for all z ∈ ∂D. Furthermore, the partial sums
on ∂D to f as K, M → ∞.
M
∑
̂f (n)z n converge uniformly
n=−K
Proof
If n ∈ Z \{0}, then
11.28 | ̂f (n)| = |( f [2])̂(n)|
n 2 ≤ ‖ f [2] ‖ 1
n 2 ,
where the equality above follows from 11.26 and the inequality above follows
from 11.9(c). Now 11.28 implies that
11.29
∞
∑ | ̂f (n)z n ∞
| = ∑ | ̂f (n)| < ∞
n=−∞
n=−∞
for all z ∈ ∂D. The inequality above implies that ∑ ∞ n=−∞ ̂f (n)z n converges and that
the partial sums converge uniformly on ∂D.
Furthermore, for each z ∈ ∂D we have
f (z) =lim
r↑1
∞
∑
n=−∞
r |n| ̂f (n)z n =
∞
∑
̂f (n)z n ,
n=−∞
where the first equality holds by 11.18 and 11.11, and the second equality holds by
the Dominated Convergence Theorem (use counting measure on Z) and 11.29.
Measure, Integration & Real Analysis, by Sheldon Axler