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If polynomials with integer monomial coefficients converge uniformly to $f$, their values at zero and one are integers converging to $f(0)$ and $f(1)$. A convergent integer sequence is eventually constant: once all its terms lie in an interval of length less than one they must agree. Hence both endpoint values are integers.

Conversely assume those values are integers. The rounded Bernstein polynomial is an integer-coefficient polynomial, because its displayed scalar coefficients are integers and each $x^k(1-x{)}^{n-k}$ expands with integer coefficients. Part b gives its vanishing distance from $B_{n}f$. For completeness, the remaining uniform convergence $B_{n}f\to f$ follows directly from uniform continuity. Let $X$ have the binomial distribution with parameters $n,x$; then $B_n(f,x)=\mathbb{E}f(X/n)$, $\mathbb{E}(X/n)=x$, and $\mathrm{Var}(X/n)=x(1-x)/n\le 1/(4n)$. Splitting into $|X/n-x|\le \delta$ and its complement, and applying Chebyshev's inequality, giveswhere $\omega$ is the modulus of continuity. First choose $\delta$ to make the first term small uniformly, then take $n$ large. The triangle inequality now gives $\Vert B_n^{\ast }f-f{\Vert }_{\infty }\to 0$. We have proved the integer-coefficient polynomial approximation characterization

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.