Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 62 / 1

The Bernstein polynomial isIt has degree at most $n$, since cancellation can lower the degree. For $0\le x\le 1$, the Bernstein basis weights are nonnegative and their sum is $(x+(1-x){)}^n=1$ by the binomial theorem. ConsequentlyThus the Bernstein polynomial defines a positive linear operator on continuous functions satisfying $\Vert B_{n}f{\Vert }_{\infty }\le \Vert f{\Vert }_{\infty }$. Since $B_{n}1=1$, its operator norm in the supremum norm is exactly one.

Write $j^{\underline{m}}=j(j-1)\cdots (j-m+1)$ for the falling factorial. At the sampling points, $f_{nm}(j/n)=n^{-m}{j}^{\underline{m}}$, which vanishes for $j<m$. For $j\ge m$, cancellation of the factorials givesInsert this into the Bernstein polynomial and set $\ell =j-m$:The binomial theorem evaluates the remaining sum. For $m=0$, both sides equal one. This proves the Bernstein falling-factorial identity $B_n(f_{nm},x)=f_{nm}(1)x^m$.

Fix $m$ and take $n\ge m$. On $[0,1]$, every factor $x-r/n$ and every factor $x$ has absolute value at most one. Replacing the factors successively and applying the triangle inequality givesUsing the supremum norm contraction in (a) and the Bernstein falling-factorial identity in (b),For $m=0,1$, the monomials are reproduced exactly. For every fixed $m$, $B_n(x^m)\to x^m$ uniformly on $[0,1]$.

The Weierstrass approximation theorem states that the polynomials are a dense subspace of $C[0,1]$ in the supremum norm. By linearity and (c), $B_{n}p\to p$ uniformly for every polynomial $p$. Given $f\in C[0,1]$ and $\epsilon >0$, choose a polynomial $p$ with $\Vert f-p{\Vert }_{\infty }<\epsilon$. The operator norm estimate in (a) givesFirst let $n\to \infty$, then $\epsilon \downarrow 0$. Hence $\Vert B_{n}f-f{\Vert }_{\infty }\to 0$ for every continuous $f$. This is an instance of extension of approximation operators from a dense subspace.