Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 69 / 1

For a continuous function $f$ on $[0,1]$, its Bernstein polynomial isThe weights are nonnegative and, by the binomial theorem, sum to $(x+1-x{)}^n=1$. Thus $|B_n(f,x)|\le {\sum }_j|f(j/n)|\left(\genfrac{}{}{0ex}{}{n}{j}\right)x^j(1-x{)}^{n-j}\le \Vert f{\Vert }_{\infty }$ at every $x\in [0,1]$, including the endpoints. Taking the supremum norm givesIn particular, $B_n$ is a linear operator of norm one, since it preserves the constant function one.

At a sampling point $j/n$, the given polynomial has value $f_{nm}(j/n)=n^{-m}{j}^{\underline{m}}$, where $j^{\underline{m}}=j(j-1)\cdots (j-m+1)$ is a falling factorial. It vanishes when $j<m$. The binomial coefficient identity $j^{\underline{m}}\left(\genfrac{}{}{0ex}{}{n}{j}\right)=n^{\underline{m}}\left(\genfrac{}{}{0ex}{}{n-m}{j-m}\right)$ therefore givesThe last step is the binomial theorem, and $n^{\underline{m}}/n^m={\prod }_{r=0}^{m-1}(1-r/n)=f_{nm}(1)$. For $m=0$ the empty product is one and the identity is simply $B_{n}1=1$. This is the Bernstein falling-factorial identity.

Fix $m$ and take $n\ge m$. Comparing the two products defining $f_{nm}$ and the monomial $g_m(x)=x^m$, change one factor at a time. Every factor has absolute value at most one on $[0,1]$, so the triangle inequality givesThe contraction proved in (a) and the Bernstein falling-factorial identity from (b) now giveThis proves uniform convergence for each fixed monomial; linearity then gives it for every fixed polynomial. The cases $m=0,1$ have zero error for every $n$.