 Solution

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.