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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).

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up vote 12 down vote accepted

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.

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Nice answer. I'd only like to underline what I think is the key in the correspondence: in a Hilbert space, you have a notion of ``projection''. If you know the Gram-Schmidt procedure, you actually know the following: given an orthonormal basis $\{\phi_n\}_{n \in I}$ for a Hilbert space, one can write a function as $f = \sum_{n\in I} \langle f, \phi_n\rangle \phi_n.$ Now take $\phi_n = \cos(nx)$ and $\phi_m = \cos(mx)$ and the inner product Matt wrote down -- the sum becomes a Fourier series, and the $\langle f, \phi_n \rangle$ are Fourier coefficients! – snarski Aug 19 '12 at 23:01

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.

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I hope this is not confusing. Matt's answer appeared while I was typing, so I'm posting it anyway. – Rudy the Reindeer Aug 19 '12 at 21:01
Basically, your two necessary conditions for a Fourier series are also two necessary conditions for a Hilbert space, right? – Tom Au Aug 19 '12 at 21:03
@TomAu The inner product is but the group $G$ is not. You do have a group structure in a Hilbert space but note that the addition in this case is addition of functions, $f + g \in L^2 (G)$, not elements of the group $G$. – Rudy the Reindeer Aug 19 '12 at 21:06

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