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Let $\{a_k\}$ be an infinite sequence, with $\sum_{k=\infty}^\infty \vert a_k\vert^2<\infty$. Let $f(\omega)=\sum_{k=-\infty}^\infty a_k e^{ik\omega}$ be its Fourier series. By Plancheral's theorem, $f$ is bounded. How do I get a reasonable bound for $|f|$, i.e., if $|f|\leq g$, how do I find $g$?

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Any $f \in L^2$ (not necessarily bounded) has a Fourier series whose coefficients are square-summable. – Akhil Mathew Mar 8 '11 at 19:40
up vote 1 down vote accepted

The fact that $\sum_{k \in Z} |a_k|^2$ is finite just means that $f(\omega)$ is an $L^2$ function, and Plancherel's theorem says that any $L^2$ function arises as $f(\omega)$ for some such sequence $\{a_k\}_{k \in Z}$. If you want $f(\omega)$ to be bounded, you have to assume further conditions on $\{a_k\}_{k \in Z}$. For example, if you know that $\sum_k |a_k|$ is finite as well, then by summing the absolute values of the terms you get that $|f(\omega)| \leq \sum_k |a_k|$.

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yes, $\sum_k|a_k |$ finite would've made life easy, but unfortunately, it isn't finite. I guess I should've mentioned that in my question. It is bugging me as to how one would do it without this assumption. And there are functions like this. Take, for e.g., $a_k=sinc(\alpha k)$, where $\alpha$ is just some constant. This is not absolutely summable. However the Fourier transform is absolutely summable. – user7815 Mar 8 '11 at 22:39

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