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Fourier series of function f: $$f(x)=\sum_{s=-\infty}^{\infty}f_{s}\exp(2\pi isx)$$

Suppose $f_{s}\sim\frac{1}{s^{p}}$. What can we say about $f(x)$? Can we find some bounds for $f(x)$ like $f(x)<c\sum_{s}\frac{\exp(-2\pi isx)}{s^{p}}$?

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You might try Parceval's Theorem's_theorem or Bessel's inequality's_inequality – Brian Mar 15 '11 at 15:36
up vote 3 down vote accepted

The size of the Fourier coefficients of a function is related to the smoothness of the function, not to its size. The smoother the function (i.e. the more derivatives it has), the faster the convergence to $0$ of its Fourier coefficients.

If $f$ is a periodic function of period $2\pi$, integrable on an interval of length $2\pi$, then, by the Riemann-Lebesgue theorem, $\hat f_s\to0$, where $\hat f_s$ is the $s$-th Fourier coefficient of $f$. If moreover $f$ is $k$ times differentiable and $f^{(k)}$ is integrable, then $|\hat f_s|\le C|s|^{-k}$ (even more, $|\hat f_s||s|^{k}\to0$.)

Conversely, if $|\hat f_s|\le C|s|^{-k}$ for some $k\ge2$, then $f$ is continuous with $k-2$ continuous derivatives. But there is nothing you can say about the size of $f$. For instance, if $f=M$ is constant, then $\hat f_s=0$ for all $s\ne0$.

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in my case $f_s\sim s^{-p}$. I mean $f_s=c_s s^{-p}$ with $c_{s}\neq 0$. I can also assume $c_s\in [a,b],a>0$. It doesn't change anything, does it? – Katja Mar 15 '11 at 14:16
@Katja No, it does not tell you nothing about the size of $f$. The only thing you can get if $p>1$ s the trivial bound $|f(x)|\le \max(|c_s|)\sum_{s=-\infty}^{+\infty}|s|^{-p}$. – Julián Aguirre Mar 15 '11 at 22:43

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