I've been searching around for an answer to this question on the web for some time, but I keep coming up short (it may very well be that I don't have the right terms to be searching with).
In any case, here's what I'm trying to do:
I'm working with Fourier series, and I want to show that every piece-wise continuous function $f(x)$ has a Fourier series representation in $L^2([a,b])$. This requires that there be an orthonormal complete sequence upon which the Fourier series relies. It's known that the set of complex exponentials are orthonormal complete.
To show this, we can use Bessel's Inequality:
$\sum\limits_{k=1}^{\infty}|<x,e_k>|^2 \leq ||x||^2$
For the case of a complete orthonormal sequence, Bessel's Inequality becomes Parseval's Identity (Wikipedia Link):
$\sum\limits_{k=1}^{\infty}|<x,e_k>|^2 = ||x||^2$
My question, then, is two fold: (1), why does Bessel's Inequality only reach equivalence in the case of a complete orthonormal sequence, and (2) how can I show that the complex exponentials form a complete orthonormal sequence?
Any help would be appreciated.
1.) Take any complete orthonormal sequence and form the subsequence of every second element. Then this subsequence is still orthonormal and satisfies the Bessel inequality. But obviously, the missing odd-index elements of the original sequence are orthogonal to the subsequence and thus not representable.
2.) This is usually done using the Stone-Weierstraß approximation theorem. $\sin$ and $\cos$ as pair are separating on the interval $[-\pi,\pi)$ and thus any continuous function can be approximatied by a polynomial in them. Any $L^2$ function can be approximated by a continuous function. And any polynomial in $\sin$ and $\cos$ can be transformed into a trigonometric polynomial. Or just replace as usual $\sin x=(e^{ix}-e^{-ix})/(2i)$ etc.