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For $m\ge 1$ and $n\ge 2$, the Bernstein polynomial of the monomial $x^m$ satisfiesIt follows by applying $k\left(\genfrac{}{}{0ex}{}{n}{k}\right)=n\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)$ and expanding $k=(k-1)+1$ with the binomial theorem. The coefficients sum to one; only the first tends to one as $n\to \infty$. Induction on $m$ proves uniform convergence of these monomial approximants, starting from $B_n(1)=1$.

For the constant monomial $f_0=1$, the binomial theorem gives $B_n(f_0)(x)=(x+1-x{)}^n=1$. For $f_1$, the binomial coefficient identity $k\left(\genfrac{}{}{0ex}{}{n}{k}\right)=n\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)$ givesThus $\boxed{B_n(f_0)=1,B_n(f_1)=x}$, including $x=0,1$ by evaluation of the polynomials.

For $m\ge 1$ and $n\ge 2$, substitute the preceding coefficient identity into the Bernstein polynomial sum and write $j=k-1$. This yields the Bernstein monomial recurrenceThe coefficients are nonnegative and sum to $1$; $a_{n,m,0}=((n-1)/n{)}^{m-1}\to 1$ and the others tend to zero. The weights in any Bernstein polynomial sum are nonnegative and sum to $1$, so $0\le B_n(f_j)(x)\le 1$. Induction on $m$ therefore gives $B_n(f_m)(x)\to x{x}^{m-1}=x^m$: the $\ell =0$ term has that limit and the other finitely many terms vanish. In fact, the same estimate in the supremum norm proves uniform convergence on $[0,1]$.

As an independent quantitative bound, take a binomial distribution variable $K$ with parameters $(n,x)$. Then $B_n(f_m)(x)=\mathbb{E}[(K/n{)}^m]$. On $[0,1]$, factoring the difference of powers gives $|t^m-x^m|\le m|t-x|$; the Cauchy-Schwarz inequality and the variance of $K/n$ giveFor $m=0$ the error is exactly zero. Finally the requested sum equals $2^m{B}_{2n}(f_m)(1/2)$, so

First address the positivity request in part (a), which has no separate Solution heading in the stub. For $x\in [0,1]$, each $x^k(1-x{)}^{n-k}$ and each coefficient $c_k$ is nonnegative. Their sum is therefore nonnegative, including at the endpoints:This is the elementary direction of the positive Bernstein coefficients for a strictly positive polynomial criterion; the converse developed later needs strict positivity.

For completeness, the convergence fact listed as (i) also has a short proof. If $K\sim \mathrm{Bin}(n,x)$, the Bernstein polynomial is $B_n(f)(x)=\mathbb{E}[f(K/n)]$. Its mean argument is $x$ and $\mathrm{Var}(K/n)=x(1-x)/n\le 1/(4n)$. Given $ϵ>0$, uniform continuity of $f$ on the compact interval supplies $\delta >0$ such that $|f(s)-f(t)|<ϵ$ whenever $|s-t|\le \delta$. By Chebyshev's inequality, uniformly in $x$,Letting $n\to \infty$ and then $ϵ\downarrow 0$ proves $\boxed{\Vert B_n(f)-f{\Vert }_{\infty }\to 0}$. The two numbered items in the source are assumptions supplied for later parts, not additional exam questions; their stub sections contain these supporting derivations.

For $f\in C[0,1]$, define its Bernstein polynomialIf $X_{n,x}\sim \mathrm{Bin}(n,x)$, thenLet $M=\Vert f{\Vert }_{\infty }$. By uniform continuity, for any $\epsilon >0$ choose $\delta >0$ such that $|f(u)-f(v)|<\epsilon /2$ whenever $|u-v|<\delta$. Splitting the expectation according to that event and using Chebyshev's inequality gives, uniformly in $x$,For sufficiently large $n$ this is below $\epsilon$. Thus $B_{n}f\to f$ uniformly on $[0,1]$. An affine change of variable treats every compact interval, proving the Weierstrass approximation theorem.

Because $f$ is continuous on the compact interval $[0,1]$, it is uniformly continuous and bounded; write $|f|\le M$. Given $ϵ>0$, choose $\delta >0$ such thatThenThe bound is independent of $p$. Taking the supremum, then $n\to \infty$, gives a limit superior at most $ϵ$. Since $ϵ$ is arbitrary,This is the Bernstein polynomial proof of the Weierstrass approximation theorem.

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.