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  "body": "\n## Statement\n\nLet $a \u003c b$, let $f \\colon [a, b] \\to \\R$ be continuous, let $n \\ge 0$, and let $p$ be a real [[polynomial function|1G0B73WPX8]] of degree at most $n$.\nLet $a \\le x_0 \u003c x_1 \u003c \\cdots \u003c x_{n+1} \\le b$ be points at which the values $f(x_i) - p(x_i)$ are nonzero and alternate in sign.\nThen the [[error of best approximation|YDN0T4MEX0]] of $f$ by polynomials of degree at most $n$ in the supremum norm satisfies\n$$E_n(f) \\ge \\min_{0 \\le i \\le n+1} \\abs{f(x_i) - p(x_i)}.$$\n\n## Remarks\n\nThe idea of the proof: let $\\mu$ be the minimum on the right and suppose some polynomial $q$ of degree at most $n$ has $\\norm{f - q}_\\infty \u003c \\mu$.\nAt each $x_i$, the difference $q - p = (f - p) - (f - q)$ has the sign of $f(x_i) - p(x_i)$, because $\\abs{f(x_i) - q(x_i)} \u003c \\mu \\le \\abs{f(x_i) - p(x_i)}$.\nSo $q - p$ alternates in sign at $n + 2$ points and has at least $n + 1$ zeros, hence $q = p$.\nBut then $\\norm{f - q}_\\infty \\ge \\mu$, a contradiction.\n\nExample: for $f(x) = x^2$ on $[0, 1]$, $n = 1$ and $p(x) = x - 1/10$, the error $x^2 - x + 1/10$ takes the values $1/10, -3/20, 1/10$ at $0, 1/2, 1$.\nSo $E_1(f) \\ge 1/10$, while $\\norm{f - p}_\\infty = 3/20$ gives $E_1(f) \\le 3/20$; the true value is $1/8$.\n\nThe number of points matters: with only $n + 1$ alternation points no lower bound follows.\nFor $f(x) = x$, $n = 1$ and $p = 1/2$, the error $x - 1/2$ takes the values $-1/2, 1/2$ at $0, 1$, but $E_1(f) = 0$.\n\n## Explanations\n\n### L1 human\n\nIf an approximation's error changes sign enough times, then no approximation of the same degree can be much better.\nThe theorem turns this into a number: the best possible error is at least the smallest of the alternating error values.\nTogether with the actual error of $p$, it gives a two-sided bound on the best achievable error.\n\n### L2 human\n\nThe bound is the easy half of the equioscillation theorem: when all the alternating values equal $\\norm{f - p}_\\infty$, it shows $p$ is best.\nIn the Remez algorithm it certifies progress, since each iteration produces a polynomial and a set of $n + 2$ alternation points, and the lower and upper bounds squeeze $E_n(f)$.\nThe same argument works for any Haar subspace of dimension $n + 1$.\n",
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