There exists a countable set of mutually orthogonal trigonometric functions which form a basis for $L^2(T)$. Proof?

Question

Evidently, this fact (for real or complex valued functions) is usually taken "for-granted" in derivations of Fourier series/transform, taking $\{e^{inx}|n\in\mathbf Z\}$ as the set of basis vectors.

My hope is to find proof for this. (preferably "simple", but any valid proof will do :).

Thanks.

Answer 1

I think that the Hilbert space you are referring to is $L^2(\mathbb{T})$, the space of square integrable and $2\pi$-periodic functions. In this case the trigonometric system $$\left\{\frac{e^{i n x}}{\sqrt{2\pi}}\ :\ n\in \mathbb{Z}\right\}$$ forms an orthonormal basis. The proof of this comprises of two parts: the first is showing that the system is orthonormal, which is a straightforward check, and the second is showing that it is complete$^{[1]}$, which is more involved. As far as I know, there are at least three ways of proving this:

1. You can prove this directly, as in Rudin R&CA, 3rd edition, §4.24. You do not need any particular prerequisite, but the proof is clearer if you know something about convolutions.
2. You can use Stone-Weierstrass's theorem which shows that the trigonometric system is complete in $C(\mathbb{T})$, from which you infer completeness in $L^2(\mathbb{T})$ by a simple embedding argument.
3. You can use Sturm-Liouville's theory or Hilbert-Schmidt's theorem on compact operators and observe that the trigonometric system is exactly the set of normalized eigenfunctions of the following boundary-value problem: $$\begin{cases} \frac{du}{dx}= i\lambda u, & x\in (0, 2\pi) \\ u(0)=u(2\pi) \end{cases}$$ (See Coddington-Levinson's book on ordinary differential equations, chapter "Self-adjoint problems on finite intervals").

Either approach has its advantages and disadvantages.

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$^{[1]}$ Definition: If $S$ is a subset of a Banach space $X$, then it is said to be complete if its linear span is dense in $X$. Warning: This terminology is not universally accepted.