Assume $f$ is continuous on $[-1,1]$. If $\int_{-1}^{1}f(x)x^ndx=0$, for all nonnegative integer $n$, can we prove that $f(x)=0$?

Question

Assume $f$ is continuous. If $$\int_{-1}^{1}f(x)x^ndx=0$$ holds for every non-negative integer $n$ , can we prove $f(x)=0$ for all $|x|\le 1$ ? Is it possible to have $f(x)x^n$ odd?

Answer 1

Hint: use the Stone-Weierstrass approximation theorem (a continuous function $f\colon [a,b]\to\mathbb{R}$ can be uniformly approximated by a sequence of polynomials).

Answer 2

Let $M=\max|f|$. If $M=0$, then $f\equiv0$. Otherwise, fix $\varepsilon>0$ and let $P(x)$ be a polynomial with real coefficients such that $(\forall x\in[-1,1]):\bigl|f(x)-P(x)\bigr|<\frac\varepsilon{2M}$; such a polynomial exists by the Weierstrass approximation theorem. Then\begin{align*}\int_{-1}^1f(x)^2\,\mathrm dx&=\int_{1}^1f(x)P(x)\,\mathrm dx+\int_{-1}^1f(x)\bigl(f(x)-P(x)\bigr)\,\mathrm dx\\&=\int_{-1}^1f(x)\bigl(f(x)-P(x)\bigr)\,\mathrm dx,\end{align*}since $P(x)$ is a sum of monomials. But then$$\int_{-1}^1f(x)^2\,\mathrm dx=\left|\int_{-1}^1f(x)\bigl(f(x)-P(x)\bigr)\,\mathrm dx\right|<2M\frac\varepsilon{2M}=\varepsilon.$$Since this is true for every $\varepsilon>0$, $\int_{-1}^1f(x)^2\,\mathrm dx=0$ and the continuity of $f$ implies now that $f\equiv0$.