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  "body": "\n## Statement\n\nLet $f \\colon [0, 1] \\to \\R$ be continuous, and for $n \\ge 1$ let $B_n(f)$ be its $n$-th [[Bernstein polynomial|0C4VWBFY73]],\n$$B_n(f)(x) = \\sum_{k=0}^{n} f\\Big( \\frac{k}{n} \\Big) \\binom{n}{k} x^k (1 - x)^{n-k}.$$\nThen $B_n(f) \\to f$ [[uniformly|08FPABDQ4C]] on $[0, 1]$ as $n \\to \\infty$.\n\n## Remarks\n\nThe 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 $f(k/n) - f(x)$.\nBy [[uniform continuity|0Q4DR37DVE]], the terms with $\\abs{k/n - x} \u003c \\delta$ contribute less than $\\varepsilon$.\nThe 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 $2 \\norm{f}_\\infty / (4 n \\delta^2)$.\n\nExample: $B_n$ reproduces $1$ and $x$, and $B_n(x^2) = x^2 + x(1 - x)/n$, so $\\norm{B_n(x^2) - x^2}_\\infty = 1/(4n)$.\nThis shows that the convergence is slow: of order $1/n$ even for a polynomial.\nMore generally $\\norm{B_n(f) - f}_\\infty$ is bounded by a constant multiple of the [[modulus of continuity|AXCJBSQZWF]] $\\omega(f, n^{-1/2})$.\n\nContinuity 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.\nThe result gives a constructive proof of the [[Weierstrass approximation theorem|0JQE130DHW]].\n\n## Explanations\n\n### L1 human\n\nThe Bernstein polynomial of $f$ samples $f$ at the points $0, 1/n, 2/n, \\dots, 1$ and blends these values with smooth weights.\nThe theorem says that as you take more sample points, the blended curve hugs the graph of $f$ everywhere on $[0, 1]$.\nThis gives an explicit polynomial close to any continuous function.\n\n### L2 human\n\nThe proof is the model for the theory of positive linear operators.\nThe operator $B_n$ is linear and maps nonnegative functions to nonnegative functions.\nIt is exact on $1$ and $x$, with a small error on $x^2$.\nKorovkin's theorem shows that these three facts alone force uniform convergence.\nBernstein 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.\nThe price is the slow rate, which cannot beat order $1/n$ for nonlinear smooth $f$.\n",
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