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The classical Shannon sampling theorem states that a bandlimited function with $\mbox{supp } \hat f\subset [-1/2,1/2]$ can be uniquely determined by its samples $(f(i))_{i\in \mathbb{Z}}$ (The symbol $\hat f$ refers to the Fourier transform of $f$).

My question is as follows: Suppose that $f$ is not bandlimited but that we have an estimate of the form

$$ |\hat f (\omega)|\le C \exp(-|\omega|^2). $$

Can such $f$ still be determined by its samples at the integers?

Edit: The answer to the question as stated is `NO', by the example of Jean Duchon. But is the following true: Suppose that $\hat f$ satisfies the above growth estimate. Can $f$ be uniquely determined by its samples $(f(\alpha i))_{i\in\mathbb{Z}}$ for all $\alpha <1$?

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    $\begingroup$ I don't understand the update - one can write each real number as $\alpha i$ with $i\in\mathbb Z$ and $0<\alpha<1$. Do you mean for some $\alpha<1$? $\endgroup$ – Sebastian Goette Mar 3 '16 at 16:04
  • $\begingroup$ Why not? the lattice $\mathbb{Z}$ is not dense enough for sampling these functions, so the question is whether the denser lattices $\alpha \mathbb{Z}$, $\alpha <1$ are sampling sets. $\endgroup$ – conan Mar 3 '16 at 16:07
  • $\begingroup$ The edited question is unclear. Every number can be written as $\alpha i, i\in Z, 0<\alpha<1$. $\endgroup$ – Alexandre Eremenko Mar 3 '16 at 20:54
  • $\begingroup$ why don't you try to understand how works the sampling theorem, i.e. how sampling in the time domain corresponds to periodization in the frequency domain ? this is linked to the Poisson summation formula which says in reality that (in the sense of distributions) the Dirac comb $\sum_n \delta(x-n)$ ) is its own Fourier transform. hence the answer is no, the perioziation operator is invertible only for those functions compactly supported on less than one period. $\endgroup$ – reuns Mar 4 '16 at 1:07
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No as well to the edited question (if I understood it...): $f(x)=e^{-cx^2}\sin(\pi x/\alpha)$

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No. Example $f(x)=e^{-cx^2}\sin(\pi x)$, chosing $c$ in the proper range...

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