If the Fourier transform of a function $f \in L^2$, whose frequency $\xi$ satisfies $|\xi| \leq \pi$, has compact support, it is famous that \begin{align*} f(x) = \sum_n f(n)\frac{\sin\pi(x-n)}{\pi(x-n)} \end{align*} holds. However, does this series always converge?
Daubechies' book "Ten Lectures on Wavelets" says in section 2.1 that this series absolutely converges independent of $x$, but I can't figure it out.

  • $\begingroup$ Do you know whether $\sum_n|f(n)|^2$ converges or not? $\endgroup$ – uniquesolution Sep 4 '15 at 8:43
  • $\begingroup$ Yes, it certainly converges. $\endgroup$ – dazaga Sep 4 '15 at 13:15

The $\operatorname{sinc}$ factor is $O(|n|^{-1})$ and is therefore in $\ell^2(\mathbb{Z})$. The first factor, $f(n)$, is also in $\ell^2$, and the conclusion follows from Cauchy-Schwarz.

Actually, $\sum |f(n)|^2<\infty$ for any function with compactly supported Fourier transform (it does not have to be within $|\xi|\le\pi$). Indeed, there is $N$ such that $\operatorname{supp} \hat f\subset I := [- \pi N, \pi N]$. The functions $\{e^{in\xi}:n\in\mathbb{Z}\}$ on $I$ are mutually orthogonal and have the same $L^2(I)$ norm. Bessel's inequality implies $\sum_n |\langle \hat f,e^{in\xi}\rangle|^2<\infty$ which is the desired conclusion.


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