Suppose $\int_0^1 f(x)x^kdx = 0$ for any nonnegative integer $k$. Show $f\equiv 0$ on $[0,1]$.

Question

The problem is in the title with the added assumption that $f$ is continuous. My proof is that $p_n\to f$ uniformly for some sequence of polynomials $p_n$ by the Weierstrass Approximation Theorem. So, $$\int_0^1f(x)^2dx = \int_0^1f(x)(f(x)-p_n(x))dx$$ $$\leq\max |f|\int_0^1|f(x)-p_n(x)|dx\to 0$$ by uniform convergence of the $p_n$. Then $\int_0^1 f(x)^2dx = 0$ and $f$ continuous imply $f\equiv 0$. Does this seem correct to everyone?

Answer 1

This looks right. For the record, this was problem B7 [sic] in the 1958 Putnam exam (actually the question asked about functions on an arbitrary fixed interval $[a,b]$, but that's equivalent to $[0,1]$ by linear change of variable), and the published solution uses Weierstrass Approximation as you do.