I was reviewing for a test and I'm not sure how to approach this problem:
Suppose $f$ is a function continuous on $[0,1]$ with $$\int_0^1x^nf(x)dx=0$$ for all even integers $n\geq0$. Then $f(x)=0$ for all $x\in[0,1]$.
Is this true? And if so, how can it be proved? I believe that the Weierstrass approximation can be used but I'm not sure how to apply it exactly.
Let $g(x)=\sqrt x f(\sqrt x)$. Then $\int_0^{1} g(x) x^{n}\, dx =2\int_0^{1} f(y)y^{2n+2}\, dy=0$ for all $n \geq 0$. Hence $g(x)=0$ for all $x$ which implies $f(x)=0$ for all $x$.
Since $f(x)$ is continuous on $[0, 1]$, then it can be extended to an even continuous function $\tilde f(x)$ on $[-1, 1]$ by \begin{align} \tilde f(x) = \begin{cases} f(x) & \text{ if } 0\leq x\leq 1,\\ f(-x) & \text{ if } -1 \leq x <0 \end{cases}. \end{align} Note that since \begin{align} \int^1_0 f(x)x^{2k}\ dx = 0 \end{align} for any $k \in \mathbb{N}$, then \begin{align} \int^1_{-1} \tilde f(x) x^{2k}\ dx = \int^1_0 f(x)x^{2k}\ dx+\int^0_{-1} f(-x)x^{2k}\ dx = 0. \end{align}
Hence $\tilde f$ is an even function which is orthogonal to $x^{2k}$ for any $k$. In particular, $\tilde f$ is orthogonal to all the even Legendre polynomials. However, since even functions are generated by the even Legendre polynomials on $[-1, 1]$, then $\tilde f$ must be zero.
