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

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.