Polynomials form a Hilbert basis for $L^2$ If you form a set of orthonormal polynomials on $[0,1]$, by applying the Gram-Schmidt process from monomials $\{1, x, x^2, \dots \}$ then what is required to show that this is a basis for $L^2[0,1]$?
$\text{Span}\{ e_n \}_{n \in \mathbb{N}} = L^2[0,1]$?
How can I use this to prove the above?
Edit:
I believe I need to show that the span of the vectors I got from Gram-Schmidt is orthonormal and complete. I can show it is orthonormal, to show completeness I need to use Stone-Weierstrass to show this for continuous functions on the domain, and then a density argument for $L^2$.
 A: An orthonormal set $\{ p_n \}$ in a Hilbert space $H$ is a complete orthonormal basis iff there is no non-zero element of $H$ that is orthogonal to every $p_n$.
If you know that $\{ e^{2\pi in x} \}_{n=-\infty}^{\infty}$ is a complete orthonormal basis, then you can use this to bootstrap to the normalized sequence of polynomials $\{ p_n \}$ obtained by applying the Gram-Schmidt process to $\{ 1,x,x^2,x^3,\cdots \}$. To do this, suppose $f\in H$ satisfies $(f,p_n)=0$ for all $n=0,1,2,3,\cdots$; it must be shown that $f=0$. Equivalently, suppose $(f,x^n)=0$ for all $n=0,1,2,3,\cdots$, and show that $f=0$.
If $f\in H$ and $(f,x^n)=0$ for all $n=0,1,2,3,\cdots$, then
$$
          (f,e^{2\pi ikx})=(f,\sum_{n=0}^{\infty}\frac{(2\pi ikx)^{n}}{n!})
          = \sum_{n=0}^{\infty}\frac{(-2\pi ik)^{n}}{n!}(f,x^{n})=0.
$$
Therefore, assuming $\{ e^{2\pi ik x} \}_{k=-\infty}^{\infty}$ is a complete orthonormal basis, it follows that $f=0$, which is what you wanted to prove.
A: A Schauder basis of a Hilbert space (or more generally a Banach space) is a linearly independent set of vectors such that every vector can be uniquely written as the sum of a norm convergent series where the individual terms of the series are multiples of the basis vectors.
Any polynomial is a linear combination of a finite number of our basis vectors.
Uniform convergence on a finite measure space (such as a bounded interval with Lebesgue measure) implies convergence in quadratic mean; in fact, for any two essentially bounded functions $f$ and $g$
$$\|f-g\|_2\leq\|f-g\|_\infty\cdot\sqrt{b-a}.$$
Finally note that the indicator function of any interval, and therefore any block function, can be approximated in $L^2$ norm by continuous functions.
From all of this it follows that the span of our normalised set of vectors is dense in $L^2.$ So an arbitrary $f\in L^2$ can be written as the limit of a sequence of polynomials $f_n.$ We shall now argue that it can even be written as the sum of the series $\sum_{i=1}^\infty\langle f,e_i\rangle e_i.$
If $f_n$ is a polynomial of degree $m(n),$ we obtain a better approximation replacing $f_n$ with the orthogonal projection of $f$ on the $m(n)$-dimensional subspace generated by the first $m(n)$ vectors of our Gram-Schmidt basis:
$$\overline f_n=\sum_{i=1}^{m(n)}\langle f,e_i\rangle e_i.$$
Thus at least some partial sums of the series $\sum_{i=1}^\infty\langle f,e_i\rangle e_i$ converge to $f.$ But the other partial sums are never worse approximations than the previous ones, i.e., $\sum_{i=1}^{k+1}\langle f,e_i\rangle e_i$ is always at least as close to $f$ as $\sum_{i=1}^{k}\langle f,e_i\rangle e_i$ is, by orthonormality.
The arbitrary nature of $f$ proves that $\{e_1,e_2,\ldots\}$ is a Schauder basis.
