Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 2 / 11F (Topics in Analysis)

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.

True. For each $n$, the Weierstrass approximation theorem gives a polynomial $P_n$ such thatFor each fixed $x\in \mathbb{R}$, this estimate applies once $n\ge |x|$, so $P_n(x)\to f(x)$ pointwise.

False. Take the bounded nonconstant continuous function $f(x)=\sin x$. If polynomials $P_n$ converged uniformly to $f$ on $\mathbb{R}$, then each sufficiently late $P_n$ would be bounded on $\mathbb{R}$ and hence constant. A uniform limit of constants is constant, a contradiction. More generally, the globally uniform limit of real polynomials is itself a polynomial.

False. The bounded continuous function $f(x)=\sin (1/x)$ on $(0,1]$ has no limit at zero. If polynomials converged uniformly to it there, they would be uniformly Cauchy on $(0,1]$ and hence on $[0,1]$ by continuity at zero. Their uniform limit on $[0,1]$ would continuously extend $f$ to zero, which is impossible.

True. First choose polynomials $Q_n\to f$ uniformly by the Weierstrass approximation theorem. Let $L_1,\dots ,L_m$ be the Lagrange interpolation polynomials for the nodes $x_1,\dots ,x_m$, so $L_j(x_i)={\delta }_{ij}$. SetThen $P_n(x_j)=f(x_j)$ for every $j$, while

True. By the Weierstrass approximation theorem, choose polynomials $q_n$ with $q_n\to f^{(m)}$ uniformly. Define $P_n$ by integrating $q_n$ exactly $m$ times and choose the $m$ integration constants so thatThen $P_n^{(m)}=q_n$, and repeated use of the fundamental theorem of calculus givesEvery derivative therefore converges uniformly as required.