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  "body": "\n## Statement\n\nLet $f \\colon \\R \\to \\R$ be a [[continuously differentiable function|06SKCHDQT5]] that is [[periodic|089KWVYD69]] with period $2\\pi$ and satisfies $\\int_0^{2\\pi} f(x) \\, dx = 0$.\nThen\n$$\\int_0^{2\\pi} f(x)^2 \\, dx \\le \\int_0^{2\\pi} f'(x)^2 \\, dx,$$\nwith equality if and only if $f(x) = \\alpha \\cos x + \\beta \\sin x$ for some real numbers $\\alpha, \\beta$.\n\n## Remarks\n\nProof idea: expand $f$ in its Fourier series $\\sum_k c_k e^{i k x}$.\nThe mean zero condition gives $c_0 = 0$, and $f'$ has coefficients $i k c_k$.\nBy [[Parseval's identity|047RFCNVSK]] the two sides are $2\\pi \\sum_{k \\ne 0} \\abs{c_k}^2$ and $2\\pi \\sum_{k \\ne 0} k^2 \\abs{c_k}^2$, and $k^2 \\ge 1$ with equality only for $k = \\pm 1$.\n\nThe mean zero condition is needed: the constant function $f = 1$ gives $2\\pi \\le 0$, which is false.\n\nA related form holds on an interval with zero boundary values: if $f \\colon [0, \\pi] \\to \\R$ is continuously differentiable with $f(0) = f(\\pi) = 0$, then $\\int_0^\\pi f^2 \\le \\int_0^\\pi f'^2$.\n\n## Explanations\n\n### L1 human\n\nA periodic function with average zero cannot be large without also changing fast.\nWirtinger's inequality makes this precise: the total square of the function is at most the total square of its slope.\nThe extreme case is a single sine wave of the longest possible period.\n\n### L2 human\n\nThis is the simplest Poincaré-type inequality, and the constant $1$ is the smallest nonzero eigenvalue of $-d^2/dx^2$ on the circle of length $2\\pi$.\nIt gives a short proof of the isoperimetric inequality in the plane, by applying it to the coordinates of a closed curve parametrized by arc length.\n",
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