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Hi, Let $\Lambda\subset\mathbb{R}$ be an infinite discrete set of finite density (for simplicity one may take the density equals 1) and $\delta_{\lambda}$ is a unit mass located at the point $\lambda\in\Lambda$. Define the tempered distribution $\delta_{\Lambda}=\underset{\lambda\in\Lambda}{\sum}\delta_{\lambda}$. It is known that if there exists a finite set $\Sigma$ such that for every two successive elements of $\Lambda$, $\lambda,\mu$ we have $\lambda-\mu\in\Sigma$ and $\delta_{\Lambda}$ has a spectral gap then $\Lambda$ must be periodic, i.e. a finite union of copies of a translated lattice.

I am trying to understand if I can drop the condition of having a finite set of differences. In other words I am trying to construct a set $\Lambda$ so that $\delta_{\Lambda}$ will have a spectral gap but $\Lambda$ will have an infinite set of differences. Obviously for periodicity of $\Lambda$ I cannot drop this condition altogether because $\Lambda$ cannot be periodic if its set of differences is inifinite, but constructing such a distribution will show that the two conditions are separate. And in a more general tone, How can one get intuition regarding whether a tempered distribution of the kind $\delta_{\Lambda}$ has a spectral gap at all?

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Could you please clarify what do you exactly mean by Spectral gap? –  Rami Feb 5 '12 at 1:04
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I believe he is referring to the recent result of Alex Iosevich, Mihail N. Kolountzakis ``Periodicity of the spectrum in dimension one'' arxiv.org/abs/1108.5689 where they prove such a result. In the context of that paper `spectral gap' would mean the existence of a > 0 such that supp (\widehat{\delta_\Lambda}) \cap (0,a) = \phi . –  Vagabond Feb 5 '12 at 6:02
    
see equation 10 page 6 of the above mentioned paper. –  Vagabond Feb 5 '12 at 6:05
    
That is exactly what I meant. –  Itay Feb 5 '12 at 6:36
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