Relationship of Fourier series and Hilbert spaces?

I just read in a textbook that a Hilbert space can be defined or represented by an appropriate Fourier series. How might that be? Is it because a Fourier series is an infinite series that adequately "covers" a Hilbert space?

Apart from this I (a mathematical novice) have a hard time seeing the connection between a Hilbert space, a vector construct, and a Fourier series (of trigonometric functions).

The space of periodic $L^2$ functions (say with period $2\pi$) forms a Hilbert space. (Here $L^2$ means that $\int_0^{2\pi} f(x)^2 dx$ exists.)

The inner product of two functions is given by $\int_0^{2\pi} f(x)g(x) dx$. (Here and above I am thinking of real-valued functions; for complex valued functions the formulas are similar.)

Now we consider two facts, one about $L^2$-functions, and one about Hilbert space

• Every $L^2$-function can be expanded as a Fourier series.

• Every Hilbert space admits an orthonormal basis, and each vector in the Hilbert space can be expanded as a series in terms of this orthonormal basis.

It turns out that the first of these facts is a special case of the second: we can interpret the trigonometric functions as an orthonormal basis of the space of $L^2$-functions, and then the Fourier expansion of an arbitrary $L^2$-function is the same thing as its Hilbert space-theoretic expansion in terms of the orthonormal basis.

Summary/big picture: To see how a "vector construct" like Hilbert space relates to Fourier series, you don't consider a single function in isolation, but instead consider the entire vector space of $L^2$-functions, which is in fact a Hilbert space.

To define the Fourier series of a function you need two things:

(i) the domain of the function needs to be a compact Abelian topological group $G$

(ii) an inner product on your space of functions

If you have this, then you can say the Fourier series of an $f: G \to \mathbb R$ or $\mathbb C$ is the representation of $f$ in terms of the characters of $G$. The characters of a compact Abelian topological group are all continuous homomorphisms $G \to S^1$. This is why you need the domain to be a compact Abelian topological group.

Then you use that the characters of $G$ actually form a basis for the functions on $G$. But for this you need a way to define orthogonality and this is where the inner product comes in.

As it happens, if you take your function space to be $L^2 (G)$ with the usual inner product (as given in Matt E's answer) then this gives you a Hilbert space.