TITLE: Polynomials form a Hilbert basis for $L^2$
QUESTION [3 upvotes]: 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 $L^2$.

REPLY [4 votes]: 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.
{"set_name": "stack_exchange", "score": 3, "question_id": 1598343}