Uniform convergence of Bernstein approximations

Remarks

The idea of the proof: the weights $\binom{n}{k} x^k (1 - x)^{n-k}$ are nonnegative and sum to $1$, so $B_n(f)(x) - f(x)$ is an average of $f(k/n) - f(x)$. By uniform continuity, the terms with $\abs{k/n - x} < \delta$ contribute less than $\varepsilon$. The weights are binomial probabilities with mean $x$ and variance $x(1 - x)/n \le 1/(4n)$, so by Chebyshev's inequality the remaining terms carry total weight at most $1/(4 n \delta^2)$, and they contribute at most $2 \norm{f}_\infty / (4 n \delta^2)$.

Example: $B_n$ reproduces $1$ and $x$, and $B_n(x^2) = x^2 + x(1 - x)/n$, so $\norm{B_n(x^2) - x^2}_\infty = 1/(4n)$. This shows that the convergence is slow: of order $1/n$ even for a polynomial. More generally $\norm{B_n(f) - f}_\infty$ is bounded by a constant multiple of the modulus of continuity $\omega(f, n^{-1/2})$.

Continuity is needed: for the step function $f = 0$ on $[0, 1/2]$ and $f = 1$ on $(1/2, 1]$, each $B_n(f)$ is continuous, so the convergence cannot be uniform. The result gives a constructive proof of the Weierstrass approximation theorem.

Going deeper

For a student

The proof is the model for the theory of positive linear operators. The operator $B_n$ is linear and maps nonnegative functions to nonnegative functions. It is exact on $1$ and $x$, with a small error on $x^2$. Korovkin's theorem shows that these three facts alone force uniform convergence. Bernstein polynomials also preserve shape: if $f$ is monotone or convex, so is $B_n(f)$, which is why they appear in computer graphics as Bézier curves. The price is the slow rate, which cannot beat order $1/n$ for nonlinear smooth $f$.