In this post I present some ideas which shed light on the question why one can expect the Fourier series to converge to the function (under certain assumptions).
Complete orthonormal systems
To do so let us first study a simpler case, one with which you are familiar with. (Throughout this section I will ignore convergence questions.)
Consider the vector space . For two vectors and we define the inner product
Assume we have given a set of orthonormal vectors that is, we have
Now the question arises whether for every vector there are such that
The answer is of course: it depends. If , then yes, if then no. (Notice that is not possible.)
Consider now equipped with the standard inner product (defined here). Then let us ask the same question. Assume we have given a set of orthonormal vectors that is, we have
Is it true that for every vector there are such that
The answer is again it depends. If , then certainly not because is infinite dimensional. If , then we still do not know. Why? To illustrate this, let be the vector whose entries are all except the entry at position is . More precisely, for let
where we set
Choose now an arbitrary element
In general, which vectors do we need to represent The answer is: all of them. Since we have
If we consider for instance the set of vectors
then the element cannot be represented using the elements in In other words, is missing from the set .
Hence, for an infinite dimensional vector space it is not enough to have infinitely many orthonormal vectors in order to be able to represent any element. Notice that, if we have an arbitrary infinite set of orthonormal vectors
and a vector which cannot be represented by the vectors in , then there exists a vector such that
Indeed, we can define the vector by
Since cannot be represented by the elements in , it follows that is not the zero vector. Further
since we have chosen the elements such that
Now let us consider Fourier series. Let us ignore questions concerning convergence in the following. For instance, say, let us only consider continuously differentiable functions. The same argument as for also applies in this case. If there would be a function for which we do not have
then there would have to be a (continuously differentiable), -periodic function such that
Therefore every (continuously differentiable), periodic function can be represented by its Fourier series if and only if the functions
are a complete orthonormal system, that is, there is no (continuously differentiable), periodic function which is orthogonal to all trigonometric functions given in line (1).
This does indeed hold, which follows from the convergence of the Fourier series to the function in the mean square sense. We prove this result in the next section.
Mean square convergence
We prove now that the Fourier series of converges to in the mean square sense. That is,
where is the th partial sum of the Fourier series and the -norm is defined by
To prove this we need some preparation. Instead of using for and for we use in the following the functions where . For complex valued functions , we define the inner product by
where stands for the complex conjugate of a complex number . Hence, we consider the Fourier series
where the Fourier coefficients are given by
Further, the th partial sum of the Fourier series is given by
Exercise Show that (2) holds if is a trigonometric polynomial .
Definition (Dirichlet kernel) The trigonometric polynomial defined for by
is called the th Dirichlet kernel.
Exercise Show that
Definition (Fejér kernel) The trigonometric polynomial defined for by
is called the th Fejér kernel.
Exercise Show that
The Fejér kernel satisfies the following properties:
- For all we have
- There exists an such that for all
- For every
For the first part we just note that for and for . The second part follows from the first, since .
To prove the third part, note that for there exists a such that for all . Hence for all and therefore
Hence the result follows.
Theorem Let be continuous and periodic with period . Let the convolution of and be given by
Since is continuous it follows that is uniformly continuous on any bounded interval. For choose such that for we have . Then, by the first property of the above lemma we have
where . Therefore
The result follows now by the third property of .
The function is a trigonometric polynomial. Indeed, using the variable transformation we obtain
where the are the Fourier coefficients of .
Hence the theorem above shows that a continuous, periodic function can be uniformly approximated by a trigonometric polynomial. The result now follows from the best approximation lemma.
Lemma (Best approximation lemma)
If is integrable with Fourier coefficients , then
for any complex numbers .
Let be Fourier coefficients of and set for . Then
Now we have
Since and , it follows that the inner product is . Hence we can use the Pythagorean theorem to obtain
Let be periodic and continuous. Then
The proof follows from the observation
The corollary states that trigonometric polynomials are dense in the space of continuous functions.
So far we have shown that the mean square convergence for Fourier series holds for continuous functions. To show that it also holds for merely integrable and bounded functions we need the following lemma.
Suppose that is integrable and bounded by . There there exists a sequence of continuous and periodic functions with so that
Given an integer , there is a partition of given by such that the upper and lower Riemann sums of differ by at most . Let
Thus we have and
We construct now functions in the following way. Let be large enough such that . Then, for construct by setting
For and let
Further, for let
and for let
For we set .
Hence is a sequence of continuous and periodic functions defined on Further, differs from only in the intervals , and in those intervals by at most . Therefore
By choosing large enough, say we obtain
Using the triangle inequality we obtain
Now we define
and extend periodically to . Then, are periodic and continuous, and, by the above construction,
which proves the result.
Let be periodic, integrable and bounded. Then
Let be periodic, bounded and integrable. Let . Then, by the above lemma, there exists a periodic, continuous function such that
Hence we get
By the corollary above, there exists a trigonometric polynomial such that and hence, using the triangle inequality,
By the best approximation lemma it follows that
where is the degree of . Hence the result follows.
For more information see for example E.M. Stein and R. Shakarchi, Princeton lectures in analysis I, Fourier analysis. Princeton University Press, Princeton, 2003. See also the Riesz-Fischer theorem and the Dini test.