Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 2 / 2F (Topics in Analysis)

The Weierstrass approximation theorem says that for every real continuous function $f$ on $[0,1]$ and every $\epsilon >0$, there is a real polynomial $P$ with ${\sup }_{[0,1]}|f-P|<\epsilon$. Equivalently, polynomials are dense in $C[0,1]$ in the uniform norm.

The given moments imply ${\int }_0^{1}f(x)P(x)dx=0$ for every polynomial $P$. Choose polynomials $P_k$ converging uniformly to $f$ by the Weierstrass approximation theorem. ThenA nonnegative continuous function with zero integral vanishes everywhere. Hence $f=0$.

This also forces $f=0$. Indeed, $h(t)=f(\sqrt{t})$ is a continuous function on $[0,1]$. Approximate $h$ uniformly by polynomials $P_k(t)$ using the Weierstrass approximation theorem. Then $P_k(x^2)\to f(x)$ uniformly, and the even moments give $\int f(x)P_k(x^2)dx=0$. The estimate from part (i), with $P_k(x^2)$ in place of $P_k(x)$, proves $\int f^2=0$ and hence the conclusion.

Again $f=0$. The change of variables $t=x^2$ givesApply part (i) to the continuous function $h(t)=f(\sqrt{t})$. It vanishes on the whole interval, so does $f$.