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  "body": "\n## Statement\n\nLet $0 \u003c \\alpha \\le 1$ and let $f \\colon \\R \\to \\C$ be a $2\\pi$-periodic function that is [[Hölder continuous|012PBA0FMD]] with exponent $\\alpha$.\nSo there is $K \\ge 0$ with $\\abs{f(x) - f(y)} \\le K \\abs{x - y}^\\alpha$ for all $x, y \\in \\R$.\nLet $\\hat f(n)$, $n \\in \\Z$, be its [[Fourier coefficients|5NN2YZZR3V]].\nIf $\\alpha \u003e 1/2$, then $\\sum_{n \\in \\Z} \\abs{\\hat f(n)} \u003c \\infty$, so the Fourier series of $f$ is [[absolutely convergent|HHBGB5R5S2]].\n\n## Remarks\n\nProof idea: for $h \u003e 0$ the function $x \\mapsto f(x + h) - f(x - h)$ has Fourier coefficients $2i \\sin(nh) \\hat f(n)$.\nBy [[Parseval's identity|047RFCNVSK]] and the Hölder bound, $\\sum_n 4 \\sin^2(nh) \\abs{\\hat f(n)}^2 \\le K^2 (2h)^{2\\alpha}$.\nTake $h = \\pi / (3 \\cdot 2^j)$; for $2^j \\le \\abs{n} \u003c 2^{j+1}$, $\\sin^2(nh) \\ge 3/4$, so the sum of $\\abs{\\hat f(n)}^2$ over this block is $O(2^{-2j\\alpha})$.\nThe block has fewer than $2^{j+2}$ terms, so by Cauchy–Schwarz the sum of $\\abs{\\hat f(n)}$ over it is $O(2^{j(1/2 - \\alpha)})$.\nThese bounds form a convergent geometric series when $\\alpha \u003e 1/2$.\n\nExample: every continuously differentiable periodic function is Hölder continuous with exponent $1$, so its Fourier series converges absolutely.\n\nThe bound $\\alpha \u003e 1/2$ is sharp: there are Hölder continuous functions with exponent $1/2$ whose Fourier series does not converge absolutely.\nMere continuity is far from enough, since there are continuous functions whose Fourier series even diverges at a point.\n\n## Explanations\n\n### L1 human\n\nThe smoother a periodic function is, the faster its Fourier coefficients decay.\nBernstein's theorem gives a precise threshold: Hölder continuity with an exponent above one half already makes the coefficients absolutely summable.\nThen the Fourier series converges uniformly and without any delicate cancellation.\n\n### L2 human\n\nThe proof is a model of a standard technique: estimate the coefficients in dyadic blocks with Parseval's identity, then pass from $\\ell^2$ to $\\ell^1$ with Cauchy–Schwarz.\nThe loss of a factor $2^{j/2}$ in the last step explains the threshold $1/2$.\nThe theorem gives a large class of functions in the [[Wiener algebra|5H8XESYCHQ]].\nA related criterion of Zygmund says that a function of bounded variation that is Hölder continuous with some positive exponent also has an absolutely convergent Fourier series.\n",
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  "name": "Bernstein's theorem on absolutely convergent Fourier series",
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