Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 2 / 11G (Topics in Analysis)

Let $X\sim \mathrm{Bin}(n,t)$ and $Y\sim \mathrm{Bin}(n,s)$ be independent. The displayed Bernstein polynomial isGiven $\epsilon >0$, uniform continuity of $f$ supplies $\delta >0$ such that changing both coordinates by less than $\delta$ changes $f$ by less than $\epsilon$. If $M=\Vert f{\Vert }_{\infty }$, then Chebyshev inequality and
$\mathrm{Var}(X/n),\mathrm{Var}(Y/n)\le 1/(4n)$ give, uniformly in $(t,s)$,Thus $B_{n}f\to f$ uniformly.

Each $B_{n}f$ is a polynomial, whose iterated integrals may be interchanged term by term. Uniform convergence permits passage to the limit in both iterated integrals, proving the asserted continuous-function form of Fubini's theorem.

Now suppose every monomial moment of $f$ vanishes. By linearity,
$\int \int pf=0$ for every two-variable polynomial $p$. The just-proved Bernstein approximation gives polynomials $p_n\to f$ uniformly. Thereforeso $\boxed{f=0}$. Statement (i) is true.

True. An affine change of variables carries $[a,b{]}^2$ to $[0,1{]}^2$. Equivalently, polynomials in $s,t$ are uniformly dense on $[a,b{]}^2$ by the rescaled Bernstein construction. The moment assumptions annihilate every polynomial, so approximating $f$ itself again gives $\int f^2=0$ and hence $f=0$.

False. Take $f(s,t)=s$. Then $s^{2n}{t}^{2m}f(s,t)$ is odd in $s$, so every stated integral over $[-1,1{]}^2$ vanishes, although $f$ is not zero.

True. The map $(s,t)\mapsto (s^2,t^2)$ is a homeomorphism of $[0,1{]}^2$. Hence every continuous $f(s,t)$ can be written as $g(s^2,t^2)$ for continuous
$g(u,v)=f(\sqrt{u},\sqrt{v})$. Bernstein polynomials approximate $g(u,v)$ uniformly, so polynomials in $s^2,t^2$ approximate $f$ uniformly. The assumed even moments annihilate all such polynomials; approximating $f$ and passing to $\int f^2$ proves $f=0$.