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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. –  Joel Cohen May 3 '12 at 16:14

2 Answers 2

up vote 5 down vote accepted

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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It seems to me that this answer is in contrast with @GregMartin 's one. For example, as of Greg's answer the series $$\sum_{n=1}^\infty \frac{1}{n^2}\sin(nx)$$ represents an analytic function, because it is uniformly convergent; however, its coefficients do not decrease as a geometric progression. Where's the problem? Am I wrong somewhere? –  Giuseppe Negro May 3 '12 at 17:34
    
@Giuseppe Negro I don't know what comment you are referring but it seems that it was meant continuous, not analytic. –  Andrew May 3 '12 at 17:45
    
I'm referring to this answer here: math.stackexchange.com/a/140474/8157 –  Giuseppe Negro May 3 '12 at 17:48
    
@Giuseppe Negro No, it is not analytic. The necessary condition is that the series is differentiable infinitely many times, and for that coefficients should decrease fast enougth. The first derivative of your example is equal to $\pi-x$ on $(0,2\pi)$ (and is $2\pi$-periodic). So this it is discontinuous at the origin. –  Andrew May 3 '12 at 17:59
    
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. –  Giuseppe Negro May 3 '12 at 23:53

[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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Would you please look at comments to this answer: math.stackexchange.com/a/140486/8157 ? –  Giuseppe Negro May 3 '12 at 17:49
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This is just plain false. A class of counter-examples being Weierstrass functions. –  D. Thomine May 3 '12 at 19:42
    
Damn. This is enough for differentiability, but not analyticity. I'm used to the first derivative being enough for analyticity, but here we're just considering behavior on the real axis, so it's not enough. –  Greg Martin May 3 '12 at 23:37
    
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. –  D. Thomine May 4 '12 at 9:09

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