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  "body": "\n## Statement\n\nLet $n \\ge 0$ and let $P$ be a polynomial with [[complex|09FTAET5Z4]] coefficients of [[degree|0DPF6WRZW6]] at most $n$.\nThen\n$$\\max_{\\abs{z} = 1} \\abs{P'(z)} \\le n \\max_{\\abs{z} = 1} \\abs{P(z)}.$$\nEquality holds if and only if $P(z) = a z^n$ for some $a \\in \\C$.\n\n## Remarks\n\nProof idea: let $M = \\max_{\\abs{z} = 1} \\abs{P(z)} \u003e 0$ and let $\\lambda \\in \\C$ with $\\abs{\\lambda} \u003e 1$.\nOn $\\abs{z} = 1$ one has $\\abs{\\lambda M z^n} \u003e \\abs{P(z)}$, so by Rouché's theorem all $n$ roots of $P(z) - \\lambda M z^n$ lie in $\\abs{z} \u003c 1$.\nBy the [[Gauss–Lucas theorem|A56086N8XQ]], so do the roots of $P'(z) - \\lambda n M z^{n-1}$.\nIf $\\abs{P'(z_0)} \u003e n M$ for some $\\abs{z_0} = 1$, then $\\lambda = P'(z_0) / (n M z_0^{n-1})$ has $\\abs{\\lambda} \u003e 1$ and makes $z_0$ such a root, a contradiction.\n\nThe inequality has an equivalent trigonometric form: a real or complex trigonometric polynomial $T$ of degree at most $n$ satisfies $\\max \\abs{T'} \\le n \\max \\abs{T}$.\nWith $x = \\cos \\theta$, the trigonometric form gives, for a real polynomial $P$ of degree at most $n$ and $-1 \u003c x \u003c 1$, the bound $\\abs{P'(x)} \\le \\frac{n}{\\sqrt{1 - x^2}} \\max_{[-1, 1]} \\abs{P}$.\n\nExample: $P(z) = z^n$ gives $\\abs{P'(z)} = n$ on the unit circle, so equality holds.\n\n## Explanations\n\n### L1 human\n\nA polynomial of degree $n$ that is at most $1$ in size on the unit circle has derivative at most $n$ there.\nThe pure power $z^n$, which turns around the circle $n$ times as fast, is the extreme case.\n\n### L2 human\n\nBernstein's inequality is the basic tool for inverse theorems in approximation theory: good approximation by polynomials of low degree forces smoothness.\nIts form on $[-1, 1]$ is sharp in the interior, while near the endpoints the [[Markov brothers' inequality|EHA5MF9VKE]] gives the correct bound $n^2$.\nThe proof through the Gauss–Lucas theorem shows how geometric facts about roots give analytic estimates.\n",
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