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  "body": "\n## Definition\n\nLet $f \\colon [0, 1] \\to \\R$ be Lebesgue integrable and let $n \\ge 0$ be an integer.\nThe *Kantorovich polynomial* of degree $n$ of $f$ is\n$$K_n(f)(x) = (n + 1) \\sum_{k=0}^{n} \\binom{n}{k} x^k (1 - x)^{n-k} \\int_{k/(n+1)}^{(k+1)/(n+1)} f(t) \\, dt, \\qquad x \\in [0, 1].$$\n\n## Remarks\n\nLet $F(x) = \\int_0^x f(t) \\, dt$ and let $B_{n+1}$ be the [[Bernstein polynomial|0JT92VK307]] operator of degree $n + 1$.\nThen $K_n(f) = (B_{n+1} F)'$, by the formula for the derivative of a Bernstein polynomial.\nEach $K_n$ is a positive linear operator.\nFor $f \\in L^p[0, 1]$ with $1 \\le p \u003c \\infty$, $K_n(f) \\to f$ in the [[$L^p$ norm|SASP14MA1Z]], and for continuous $f$ the convergence is uniform.\n\nExample: for $f(t) = t$, the integral over $[k/(n+1), (k+1)/(n+1)]$ is $\\frac{2k + 1}{2(n+1)^2}$.\nHence $K_n(f)(x) = \\sum_{k} \\binom{n}{k} x^k (1 - x)^{n-k} \\frac{2k + 1}{2(n + 1)} = \\frac{nx + 1/2}{n + 1}$, which tends to $x$.\n\nNon-example: the Bernstein polynomial $B_n f(x) = \\sum_k \\binom{n}{k} x^k (1 - x)^{n-k} f(k/n)$ is not a Kantorovich polynomial.\nIt uses the point values $f(k/n)$, so it changes when $f$ is changed on the null set $\\set{0, 1/n, \\dots, 1}$, while $K_n(f)$ does not.\n",
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