Let $f\in L^2[0,1]$. Show that $f(x)=x$ a.e. if and only if $\int_{0}^{1}x^{n}f(x)dx=\frac{1}{n+2}$ whenever $n\geq 0$.
What I did for one implication is: if $f(x)=x$ a.e, then $f(x)-x=0$ a.e. so that for each n $x^{n}(f(x)-x))=0$ a.e. and $x^{n}f(x)=x^{n+1}$ a.e. Integrating both sides gives the result $\int_{0}^{1}x^{n}f(x)dx=\frac{1}{n+2}$ whenever $n\geq 0$.
For the other implication, we need to use the fact that polynomials are dense in $L^2[0,1]$, that is, for any $f\in L^2[0,1]$ and any positive $\epsilon$, there is a sequence of polynomials $p_{n}$ such that $||f-p_{n}||<\epsilon$. My thought is to compute the integral $\int_{0}^{1}(f(x)-x)^{2}dx$ and get that it is zero and since the integrand is non-negative to conclude that $f(x)=x$ a.e I am stuck here and any help connecting my idea with approximation theorem is very appreciated!
I'll assume that all functions here are real-valued.
Let $g(x)=f(x)-x$. Then $\int_0^1 g(x)p(x)\,dx=0$ for all polynomials $p$. By Weierstrass, there is a sequence of polynomials $(p_n)$ converging uniformly to $g$ on $[0,1]$. Then $\int_0^1p_n(x)g(x)\,dx=0$. Deduce that $\int_0^1 g(x)^2\,dx=0$ and so that $g(x)=0$ identically.