Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 70 / 2

The Bernstein polynomial is the positive weighted averageFor $0\le x\le 1$, the Bernstein basis weights are nonnegative and the binomial theorem gives ${\sum }_{k=0}^n{b}_{n,k}(x)=1$. Hence $|B_n(f,x)|\le {\sum }_k\Vert f{\Vert }_{\infty }{b}_{n,k}(x)=\Vert f{\Vert }_{\infty }$. In particular,This is the contraction property of the positive linear operator on continuous functions defined by the Bernstein polynomial.

The integer endpoint hypothesis is important. Let $r_{n,k}=\left(\genfrac{}{}{0ex}{}{n}{k}\right)f(k/n)-\lfloor \left(\genfrac{}{}{0ex}{}{n}{k}\right)f(k/n)\rfloor$. The floor function gives $0\le r_{n,k}<1$, while $r_{n,0}=r_{n,n}=0$ because $f(0)$ and $f(1)$ are integers. Therefore the rounded Bernstein polynomial satisfiesFor $0\le x\le 1/4$, the successive terms have ratio $x/(1-x)\le 1/3$, so the geometric series estimate givesThe last inequality follows by maximizing $x(1-x{)}^{n-1}$ at $x=1/n$. The same bound holds on $[3/4,1]$ by reflection. On $[1/4,3/4]$, each factor $x$ and $1-x$ is at most $3/4$, giving $S_n(x)\le (n-1)(3/4{)}^n$. ThusHere $n\ge 2$; for $n=1$ the difference vanishes. Without the integer endpoint hypothesis the claim would fail: at $x=0$ the error is $f(0)-\lfloor f(0)\rfloor$, independently of $n$.

First prove necessity. If polynomials with integer coefficients converge to $f$ in the supremum norm, their values at $0$ and $1$ are convergent sequences of integers. Each such sequence is eventually constant, since a convergent sequence is Cauchy and two distinct integers are at distance at least one. Its limit is an integer. Thus $f(0),f(1)\in \mathbb{Z}$.

For sufficiency we also need uniform convergence of the Bernstein polynomial to $f$. Here is a direct argument. For $X\sim \mathrm{Bin}(n,x)$, write $B_n(f,x)=\mathbb{E}f(X/n)$. The binomial distribution has $\mathbb{E}(X/n)=x$ and $\mathrm{Var}(X/n)=x(1-x)/n\le 1/(4n)$. For every $\delta >0$, uniform continuity of $f$ and the Chebyshev inequality giveFirst make $\delta$ small and then $n$ large; the bound is uniform in $x$. Hence $\Vert B_n(f)-f{\Vert }_{\infty }\to 0$.

If the endpoints are integers, part (b) gives $\Vert B_n^{\ast }(f)-B_n(f){\Vert }_{\infty }\to 0$. Each coefficient multiplying $x^k(1-x{)}^{n-k}$ in $B_n^{\ast }$ is an integer. Expanding $(1-x{)}^{n-k}$ by the binomial theorem therefore gives integer coefficients in the ordinary monomial basis. The triangle inequality proves $\Vert B_n^{\ast }(f)-f{\Vert }_{\infty }\to 0$. Thus the integer endpoint condition is necessary and sufficient for integer-coefficient polynomial approximation.