Assume if Odd moment of $h(x)$ is equal to zero, is $h(x)=0$? - Fourier moment problem, $x\in[0,1]$ - $\int _0^1\:h\left(x\right)\cdot x^{2n+1}dx$

Question

Showing that $ \int_0^1 x^{2n}f(x) dx = 0 $ implies $f = 0$
Hi, tried looking at this link.
I did not understand how to answer regarding odd function, I receive the answer is that it is not correct, The integral is indeed $0$, but only because the integral of multiplication of odd functions is $0$, it does not say that $h(x)$ must be constant 0.

Answer 1

Let $g(x)=xh(x)$. Then $\int_0^{1} x^{2n}g(x)dx=0$ for $n=0,1,2,\cdots$. Hence, $\int_0^{1}\phi (x)g(x)dx=0$ for any $\phi \in span \{1,x^{2},x^{4},\cdots\}$. By Stone-Wierstrass Theorem $span \{1,x^{2},x^{4},\cdots\}$ is dense in $C[0,1]$ (w.r.t. the sup norm). It follows now that $\int_0^{1}[g(x)]^{2}dx=0$ and so $g \equiv 0$ and hence, $h\equiv 0$.