2Fourier series

IB Methods 2.2 Convergence of Fourier series
When Fourier first proposed the idea of a Fourier series, people didn’t really
believe in him. How can we be sure that the infinite series actually converges?
It turns out that in many cases, they don’t.
To investigate the convergence of the series, we define the partial Fourier
sum as
S
n
f =
n
X
m=n
ˆ
f
m
e
imθ
.
The question we want to answer is whether
S
n
f
“converges” to
f
. Here we
have to be careful with what we mean by convergence. As we (might) have seen
in Analysis, there are many ways of defining convergence of functions. If we
have a “norm” on the space of functions, we can define convergence to mean
lim
n→∞
kS
n
f fk = 0. Our “norm” can be defined as
kS
n
f fk =
1
2π
Z
π
π
|S
n
f(θ) f(θ)|
2
dθ.
However, this doesn’t really work if
S
n
f
and
f
can be arbitrary functions, as in
this does not necessarily have to be a norm. Indeed, if
lim
n→∞
S
n
f
differs from
f
on finitely or countably many points, the integral will still be zero. In particular,
lim S
n
f
and
f
can differ at all rational points, but this definition will say that
S
n
f converges to f.
Hence another possible definition of convergence is to require
lim
n→∞
S
n
f(θ) f(θ) = 0
for all
θ
. This is known as pointwise convergence. However, this is often too
weak a notion. We can ask for more, and require that the rate of convergent is
independent of θ. This is known as uniform convergence, and is defined by
lim
n→∞
sup
θ
|S
n
f(θ) f(θ)| = 0.
Of course, with different definitions of convergence, we can get different answers
to whether it converges. Unfortunately, even if we manage to get our definition
of convergence right, it is still difficult to come up with a criterion to decide if a
function has a convergent Fourier Series.
It is not difficult to prove a general criterion for convergence in norm, but
we will not do that, since that is analysis. If interested, one should take the IID
Analysis of Functions course. In this course, instead of trying to come up with
something general, let’s look at an example instead.
Example. Consider the sawtooth function f(θ) = θ.
3π
π
π
3π
Note that this function is discontinuous at odd multiples of π.
The Fourier coefficients (for n 6= 0) are
ˆ
f
n
=
1
2π
Z
π
π
e
inθ
θ dθ =
1
2πin
e
inθ
θ
π
π
+
1
2πin
Z
π
π
e
inθ
dθ =
(1)
n+1
in
.
We also have
ˆ
f
0
= 0.
Hence we have
θ =
X
n6=0
(1)
n+1
in
e
inθ
.
It turns out that this series converges to the sawtooth for all
θ 6
= (2
m
+ 1)
π
, i.e.
everywhere that the sawtooth is continuous.
Let’s look explicitly at the case where
θ
=
π
. Each term of the partial Fourier
series is zero. So we can say that the Fourier series converges to 0. This is the
average value of lim
ε0
f(π ± ε).
This is typical. At an isolated discontinuity, the Fourier series is the average
of the limiting values of the original function as we approach from either side.