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Hi, if the Fourier series development of $g(t)$ (periodic, $C^\infty$) is

$$ g(t)=\sum_{-\infty}^{+\infty}a_n e^{in\omega t} $$

does the series

$$ \sum_{-\infty}^{+\infty}\frac{a_n^2}{n^2}? $$ converges toward something known like average $g^2$ or something like that?

edited Nov 23 2010 at 14:26

asked Nov 23 2010 at 13:42

Mermoz
45●5

What do you know about $g(t)$? If it is merely $L^1$, the best you can say is given by the Riemann-Lebesgue lemma (which in particular tells you that the series you wrote down converges, since $a_n$ is better than bounded). Please modify the question to be more precise on what you are asking about. – Willie Wong Nov 23 2010 at 14:16

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Do you mean $a_n^2$ or $|a_n|^2$? Assume for now $a_0 = 0$. If it were not, then the sum diverges at $n = 0$. In this case, there exists $G(t)$ smooth periodic with $G' = g$. $G'$ has the property that its Fourier coefficients, $b_n$, are given by $b_n = a_n / (in\omega)$. So if you are looking at $|a_n|^2$ your expression is just the $L^2$ norm of $G$ (up to some constant factor due to the normalization of the Fourier transform), and if you are looking at $a_n^2$, your expression is some constant factor times $(g*g)(0)$, where the convolution is evaluated over the circle. – Willie Wong Nov 23 2010 at 15:46

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@Willie, I deleted my answer, since (a) your comment is better, and (b) as we established in the comments to my answer, this is a bit too elementary for MO. In fact, I think maybe we ought to close this one. – Harald Hanche-Olsen Nov 23 2010 at 15:59

anton · Accepted Answer · 2010-11-23 18:11:55Z UTC

Assume $a_0=0$, which easily can be arranged by adding a constant to $g$. Then the function $$ h(t)=\frac1{i\omega } \sum_{n}\frac{a_n}ne^{in\omega t} $$ is the primitive of $g$. Let $h^* (x)=\overline{h(-x)}$, then the sum you asked for equals the inner product $$ \langle h,h^*\rangle. $$

Series of squared Fourier coefficients

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