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Consider a function $f:\mathbb{R}\to\mathbb{R}$ which is periodic with period $2\pi$. Let us impose the condition that $f$ is analytic. Now does that imply that $f$ has a finite Fourier series?

PS : Although this question seems to be related to this, I couldn't find anything that I can understand there

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    Being analytic is not sufficient for the Fourier series to be finite. For example $\theta \mapsto \frac{e^{e^{i\theta}} + e^{e^{-i\theta}}}{2} = \sum_{n = 0}^{\infty} \frac{\cos(n\theta)}{n!}$ is analytic and has an infinite Fourier series.2012-05-03

2 Answers 2

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Fourier series represents an analytic function if and only if its coefficients decrease at least as a geometric progression: $\limsup_{n\to\infty}\,(|a_n|+|b_n|)^{1/n}=q<1.$ This fact can be found in books on Forier series.

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    Perfectly clear. Thank you. Indeed, any $H^1(\mathbb{S}^1)$ function has a uniformly convergent Fourier series, but not any one of them is analytic.2012-05-03
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[This is false - do not believe it:]

Any uniformly convergent sum of analytic functions is again analytic. So you can construct as many counterexamples to your question as you want by taking sequences $\{ \dots, a_{-1}, a_0, a_1, \dots \}$ whose sum $\sum_{n=-\infty}^\infty a_n$ absolutely converges; the corresponding Fourier series $\sum_{n=-\infty}^\infty a_n e^{inx}$ will then be an analytic function.

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    See my comment: this is not enough for differentiability, or even any kind of Hölder regularity. Weierstrass functions with well-chosen parameters can get pretty irregular. For these functions, the sequence $(a_n)$ is lacunary, so that it can be summable while decreasing arbitrarily slowly.2012-05-04