In this blog entry you can find lecture notes for Math2111, several variable calculus. See also the table of contents for this course. This blog entry printed to pdf is available here.

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 .

The trigonometric polynomial defined for byDefinition(Dirichlet kernel)is called the

.th Dirichlet kernel

**Exercise** Show that

The trigonometric polynomial defined for byDefinition(Fejér kernel)is called the

.th Fejér kernel

**Exercise** Show that

Lemma

The Fejér kernel satisfies the following properties:

- For all we have
- There exists an such that for all
- For every

*Proof*

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.

Let be continuous and periodic with period . Let the convolution of and be given byTheoremThen

*Proof*

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

Therefore,

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 , thenfor any complex numbers .

*Proof*

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

Corollary

Let be periodic and continuous. Then

**Proof**

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.

Lemma

Suppose that is integrable and bounded by . There there exists a sequence of continuous and periodic functions with so thatand

**Proof**

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.

Theorem

Let be periodic, integrable and bounded. Then

**Proof**

Let be periodic, bounded and integrable. Let . Then, by the above lemma, there exists a periodic, continuous function such that

and

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.