In Section 2 we considered partial sums of Fourier series and we asked what happens when . In this entry we take a more general approach by considering sequences of functions and studying their convergence behaviour.
Let be a vector space of functions equipped with a norm . We say that a sequence of functions converges to in if and
This means that for every there exists an such that
Exercise Consider the vector space of continuous functions equipped with the maximum norm. Show that the functions
Example Consider the vector space of bounded functions equipped with the supremum norm. Show that the functions
as . Show that does not converge to in the supremum norm however.
Note that the functions , in the above example are all continuous, but the limit is not.
Note (Interchanging limits) Limits cannot always be interchanged. For example let
Hence, in this example,
In particular, assume that is continuous and converges at every point to a function , we do not necessarily have that is continuous.
For the example above we have
There is, however, a type of convergence which ensures that if the sequence of functions are continuous then the limit is also continuous. This type of convergence is called uniform convergence.
Let be a sequence of functions and let We say that converges uniformly to if for every there is an integer such that implies that
for all .
In other words, let be the supremum norm Then uniform convergence just means that
The difference between uniform convergence and pointwise convergence is that for pointwise convergence the choice of depends on and on whereas for uniform convergence is only allowed to depend on (but not on ).
Note that uniform convergence implies pointwise convergence ( can you prove this?), whereas pointwise convergence does not imply uniform convergence.
Theorem (Uniform convergence theorem) If is a sequence of continuous functions which converge uniformly to a function , then is continuous.
A consequence of this is that if is continuous for all and as , then if has a discontinuity implies that the convergence of to cannot be uniform.
For Fourier series, this implies the following: Assume that is periodic with period and note that the partial sums are continuous. Then if is not continuous at some point , then cannot converge uniformly to as on any open interval containing .
There is a very convenient test for uniform convergence due to Weierstrass.
Theorem (Weierstrass test)
Suppose is a sequence of functions defined on a set and suppose
Then the series
converges uniformly on if converges.
Example Let (where ) and
where and . We must choose , such that
that is, we can choose Hence it remains to show that the sum
converges. This can be done using the ratio test
Exercise Show that the functions
converge uniformly to
Mean square convergence of Fourier series
An interesting property of the partial sums of a Fourier series is that among all trigonometric polynomials of degree , the partial sum yield the best approximation of in the -norm (or mean square sense). Note that the -norm is given by
Lemma (Best approximation lemma)
Assume is integrable and let denote the th partial sum of the Fourier series, then
for any real numbers and .
Exercise Give a geometrical interpretation of this result.
Let be periodic, bounded and integrable on . Then the Fourier series of converges in the mean square sense, that is,
You can find a proof of this result on this post.
Further, Parseval’s identity holds