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If $f$ is Lipschitz continuous on $[0,1]$ with Lipschitz constant $L$, then Cauchy-Schwarz inequality and the binomial distribution variance giveThis is an explicit error guarantee, rather than only an existence theorem. For the Bernstein approximation of a corner, choosing $n\ge 2500$ guarantees supremum norm error at most $0.01$; $n>2500$ guarantees error strictly less than $0.01$. The guarantee is sufficient, not optimal.

For this specific target, the error reaches its maximum at $x=1/2$. Indeed the derivative of $B_{n}f$ is a weighted average of the slopes $n[f((k+1)/n)-f(k/n)]$, all in $[-1,1]$: see derivatives of Bernstein polynomials. Subtracting the target makes the error nondecreasing to the left of the corner and nonincreasing to its right. For even $n\ge 2$, the symmetric binomial distribution at $x=1/2$ gives the exact valueThe last equality follows by pairing terms about $n/2$ and telescoping the binomial coefficients. It gives $1/4$ for $n=2$ and $3/16$ for $n=4$.

Write the Bernstein basis as $p_{n,j}(x)=\left(\genfrac{}{}{0ex}{}{n}{j}\right)x^j(1-x{)}^{n-j}$, with $p_{n,j}=0$ for out-of-range indices. The Bernstein polynomial isDifferentiation and the identities $j\left(\genfrac{}{}{0ex}{}{n}{j}\right)=n\left(\genfrac{}{}{0ex}{}{n-1}{j-1}\right)$ and $(n-j)\left(\genfrac{}{}{0ex}{}{n}{j}\right)=n\left(\genfrac{}{}{0ex}{}{n-1}{j}\right)$ give $p_{n,j}^{\prime }=n(p_{n-1,j-1}-p_{n-1,j})$. Shifting the first sum's index therefore yieldsRepeating this calculation takes another forward finite difference of the coefficients and lowers the degree by one. Induction proves the derivatives of Bernstein polynomials formulaFor $n>r$, define a continuous function on $[0,1]$ byAll sampled arguments lie in $[0,1]$, since the largest is $(n-r)t/n+r/n\le 1$. At the degree-$(n-r)$ sampling point $t=j/(n-r)$, this gives exactly the coefficient $(n{)}_r{\Delta }_{1/n}^{r}f(j/n)$, and hence $\boxed{B_n^{(r)}f=B_{n-r}{g}_{n,r}}$. The rescaling is essential: the sampling meshes are $1/n$ and $1/(n-r)$, and the function $g_{n,r}$ depends on $n$. For $n=r$, the derivative is the constant $n!{\Delta }_{1/n}^{n}f(0)$, interpreted as a degree-zero Bernstein polynomial.

Now fix $r$ and assume $f\in C^r[0,1]$. The supplied uniform limit of normalized finite differences, together with $(n{)}_r/n^r\to 1$ and uniform continuity of $f^{(r)}$, shows thatPositivity and ${\sum }_j{p}_{m,j}=1$ imply $\Vert B_{m}h{\Vert }_{\infty }\le \Vert h{\Vert }_{\infty }$. For completeness, if $X\sim \mathrm{Bin}(m,x)$ then $B_m(h,x)=\mathbb{E}[h(X/m)]$, and $\mathrm{Var}(X/m)=x(1-x)/m\le 1/(4m)$. Splitting according to whether $|X/m-x|\le \delta$ and using Chebyshev's inequality givesFirst let $m\to \infty$ and then $\delta \to 0$; this proves uniform convergence for each continuous $h$. Finally,Thus Bernstein polynomial derivatives converge uniformly to the corresponding derivatives of $f$, for each fixed $r$ with $f\in C^r[0,1]$. Continuity of $f$ alone is sufficient for the definition of $B_n$, but not for this derivative conclusion.