Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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).

share|improve this question
add comment

2 Answers 2

up vote 11 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.

share|improve this answer
    
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
add comment

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.

share|improve this answer
    
I hope this is not confusing. Matt's answer appeared while I was typing, so I'm posting it anyway. –  Matt N. 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 $G$. –  Matt N. Aug 19 '12 at 21:06
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.