most recent 30 from http://mathoverflow.net 2013-05-22T14:48:18Z http://mathoverflow.net/feeds/question/87541 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/87541/spectral-gap-of-tempered-distributions ET 2012-02-04T18:36:37Z 2012-02-04T18:36:37Z

<p>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. </p> <p>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?</p>