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Fourier Series
Ana Emilia de Orellana
Harmonic Analysis Course Notes

Separation of variables
Dirichlet’s problem in the disk D ⊂ R2 is
(
∆u = 0 in D
u=f
in S
In polar coordinates ∆u =
of the form

2u
∂2u
+ 1r ∂u
+ r12 partial

...
This gives us
1
1
∆u = v ′′ (r)ϕ(θ) + v ′ (r)ϕ(θ) + 2 v(r)ϕ′′ (θ) = 0
...
As the left hand side does not depend on
θ and the right hand side does not depend on r we obtain for λ ∈ R
r2 v ′′ (r) + rv ′ (r) = λv(r),
ϕ′′ (θ) = −λϕ(θ)
...

The independent solutions will be of the form v(r) = rn for n ≥ 1 and
v(r) = 1 or v(r) = log(r) for n = 0
...


And every linear combination will be a solution, thus the general solution is
X
u(r, θ) =
rn (an cos(nθ) + bn sin(nθ))
...

What if ρ is not a trigonometric polynomial? Fourier answered this by
proving that every function in the disk can be written as a trigonometric
polynomial
...
A function f is 2π-periodic if f (θ + 2π) = f (θ) for all θ ∈ R
...


Questions
1
...
Is there a series n∞ (an cos(nθ) + bn sin(nθ)) such that
X
rn (an cos(nθ) + bn sin(nθ)),
u(r, θ) =
n

converges to a solution of the Dirichlet problem?

Fourier Series
P
We want to write a function f on S as ∞
n (an cos(nθ) + bn sin(nθ))
...


and c0 = a0 , so

(an cos(nθ) + bn sin(nθ)) = c0 +

∞
X

(cn einθ + c−n e−inθ ) =

n

Note that the partial sums are sN (θ) =

2

∞
X
n=−∞

PN
−N

cn einθ
...


R 2π
R 2π
Suppose sN ⇒ f (converges uniformly), then 0 sN (θ)einθ dθ → 0 f (θ)e−inθ dθ
...


0

P∞

n=−∞ cn e

inθ

converges uniformly to f it must be
Z

1
cn =
2π

2π

f (θ)e−inθ dθ
...
For a Riemmann integrable function f on S, its Fourier coefficients are given by
Z 2π
1
ˆ
f (n) =
f (θ)e−inθ dθ
...

n∈Z

Observation 1
...

|f (n)| ≤
2π 0
2π

Examples
Sawtooth function
y
−T /2

A

T /2

T

3T /2

2T

5T /2

x
−A

A function f that is 2π-periodic and f (θ) = θ on [−π, π)
...

2π 0
2π −π

3

If n ̸= 0
Z 2π
1
ˆ
f (n) =
f (θ)e−inθ dθ
2π 0
Z π
1
=
θeinθ dθ
2π −π


Z
1 θe−inθ π
1 π −inθ
=
e
dθ
 +
2π −in −π in −π
1 πe−inπ + neinπ
=
2π
−in
n
i(−1)
=

...


• Converges at every θ by Dirichlet test
...

n=1
n
2
• At θ = π it converges to 0 but f (π) = −π
...

• The series converges uniformly by Weierstrass’ M -test
...

P
P∞ 1
1
π2
• At θ = 0 the series is π2 − π4 ∞
k=0 (2k+1)2
...


4

Flounders function
A function h that is 2π-periodic, h(θ) = (π 2 − θ2 )2 on [−π, π), its Fourier
coefficients are given by
(
8 4
π
n=0
ˆ
h(n)
= 15 (−1)n
−24 n4
n ̸= 0
...

4
4
15
n
15
n
n̸=0
n̸=0
• Faster convergence
...
Does the Fourier series of a function converge? In what sense? Does
regularity help?
4
...


Abel means
P
Definition 3
...

We say that

P

an is Abel-summable to s if
lim Ar = s
...
(Abel) If a series converges to s then it is Abel-summable to s
...
The converse is not true
P
n
• The series ∞
n=0 (−1) diverges, but
∞
X
Ar =
(−1)n rn =
n=0

• The series

P∞

n
n=0 (−1) (n

Ar =

1
1
→ , r → 1
...

n+r
4

Abel means of the Fourier series of f :

Ar f (θ) = fˆ(0) +

∞
X

rn (fˆ(n)einθ + fˆ(−n)e−inθ ) =

n=1

X

r|n| fˆ(n)einθ
...

2π 0
n∈Z
X
n∈Z

|n| inθ

r e

=1+

∞
X

n inθ

r e

+

n=1

∞
X

rn e−inθ =

n=1

1 − rn

...

Corollary 1
...
If
f ∈ C(S), then Ar f converges uniformly to f on S
...
If f is continuous at θ and its Fourier series converges at θ then it
converges to f (θ)
...
But
Ar f (θ) → f (θ)
...
If f is continuous at θ and fˆ(n) = 0 for all n ∈ Z, then f (θ) = 0
...

P ˆ
3
...

By M -test: Uniformly to f if f ∈ C(S)
...
If f ∈ C(S) and for 0 ≤ r < 1,
X
u(r, θ) =
r|n| einθ ,
n∈Z

then u is harmonic in D and converges uniformly to f as r → 1
...
The coefficients fˆ(n) are bounded, so for 0 < ρ < 1 the series (and its
term by term derivatives) converge uniformly for 0 ≤ r ≤ ρ and θ ∈ [0, 2π]
...
Since u(r, θ) = Af (θ)
then u(r, ·) → f uniformly as r → 1

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