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  "body": "\n## Statement\n\nLet $f \\colon \\R \\to \\C$ be a [[continuous|0X3WGX95FE]] $2\\pi$-periodic function whose [[Fourier coefficients|5NN2YZZR3V]] satisfy\n$$\\sum_{n \\in \\Z} \\abs{\\hat f(n)} \u003c \\infty.$$\nThen the partial sums $S_N f(x) = \\sum_{n=-N}^{N} \\hat f(n) e^{inx}$ of its [[Fourier series|G12KXDHPPS]] converge to $f$ [[uniformly|08VZ57PT26]] on $\\R$.\n\n## Remarks\n\nSince $\\abs{\\hat f(n) e^{inx}} = \\abs{\\hat f(n)}$, the [[Weierstrass M-test|0NX5NWAFC9]] shows that $S_N f$ converges uniformly to a continuous $2\\pi$-periodic function $g$.\nUniform convergence allows integration term by term, so $\\hat g(n) = \\hat f(n)$ for all $n$.\nThen $f - g$ is continuous with all Fourier coefficients zero, and its Fejér means, which converge uniformly to $f - g$ by [[Fejér's theorem|TPJZCSXSVN]], are all zero; hence $f = g$.\nExample: for $f(x) = \\abs{x}$ on $[-\\pi, \\pi]$, extended periodically, $\\hat f(0) = \\frac{\\pi}{2}$ and $\\hat f(n) = \\frac{(-1)^n - 1}{\\pi n^2}$ for $n \\ne 0$, which is summable.\nWithout continuity the conclusion fails: if $f$ is only integrable with summable coefficients, the series converges uniformly to a continuous function that equals $f$ only almost everywhere.\nEvery $f \\in C^1$ periodic satisfies the hypothesis, by the Cauchy–Schwarz inequality and Parseval's identity applied to $f'$.\n\n## Explanations\n\n### L1 human\n\nIf the sizes of the Fourier coefficients add up to a finite number, the Fourier series converges as nicely as possible: uniformly, to the function itself.\nThe finite sum of sizes bounds the error of all the tails at once.\n\n### L2 human\n\nThis is the simplest sufficient condition for uniform convergence, and it is easy to check from decay of the coefficients, for instance $\\abs{\\hat f(n)} \\le C \\abs{n}^{-1-\\delta}$ with $\\delta \u003e 0$.\nThe functions with summable Fourier coefficients form the [[Wiener algebra|5H8XESYCHQ]] $A(\\mathbb{T})$.\nBernstein's theorem shows that Hölder continuity of order $\\alpha \u003e \\frac{1}{2}$ is enough to be in $A(\\mathbb{T})$.\nThe condition is far from necessary: there are continuous functions whose Fourier series converge uniformly but not absolutely.\n",
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