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The uniform approximation closure of the integer-coefficient polynomials on $[0,1]$ consists precisely of the continuous functions taking integer values at both endpoints. Necessity follows because evaluations at zero and one are integers and convergent integer sequences are eventually constant. For sufficiency, the rounded Bernstein polynomial has integer monomial coefficients and differs uniformly by a vanishing amount from the Bernstein polynomial, which converges uniformly to the target.

For each $k$, let ${\epsilon }_{n,k}=\left(\genfrac{}{}{0ex}{}{n}{k}\right)f(k/n)-\lfloor \left(\genfrac{}{}{0ex}{}{n}{k}\right)f(k/n)\rfloor$. The floor function ensures $0\le {\epsilon }_{n,k}<1$, even when the quantity being rounded is negative. The integer endpoint hypothesis gives ${\epsilon }_{n,0}={\epsilon }_{n,n}=0$. Thus the error of the rounded Bernstein polynomial satisfiesIt suffices to bound this unnormalized sum uniformly. If $0\le x\le 1/4$, put $q=x/(1-x)\le 1/3$ and sum a geometric series:The last inequality follows by maximizing $x(1-x{)}^n$, whose maximum occurs at $x=1/(n+1)$ and is at most $1/(n+1)$. Reflection gives the same bound for $3/4\le x\le 1$. In the middle interval both $x$ and $1-x$ are at most $3/4$, so each summand is at most $(3/4{)}^n$. Therefore, for $n\ge 2$,Endpoint integrality matters: at zero or one a noninteger value would leave a fixed nonzero rounding error.

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

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 $n$.