Let $f:[0,1]\rightarrow\mathbb{R}$ be continuous such that $\int_0^1 x^{2k}f(x)dx=0$, $\forall k\geq 1$, integers.
Show that $f(x)=0$, $\forall x\in[0,1]$.
*Let $y=x^2$, then this function can be written as $\int_0^1 yy^nf(\sqrt y)dx$, where $n\geq 0$.
Since $yf(\sqrt y)$ is continuous on a closed interval, by Weierstrass Approximation theorem, we can find a polynomial $p(x)$ converging to it uniformly.
Thus, we have $\int_0^1yf(\sqrt y)p(x)dx=0$, for any polynomial.Then we can get $f(\sqrt y)=0$ for all $y\in [0,1]$.
Therefore, $f(x)=0$, for all $y\in [0,1]$, i.e. for all $x\in[0,1]$.
Is this correct?