From Wiener's tauberian theorem we know that linear combinations of translates of f \in L^1(R) are dense in L^1(R) if and only if the Fourier transform of f never vanishes. It is also known that linear combinations of translates of f \in L^2(R) are dense in L^2 if and only if the Fourier transform of f is nonzero almost everywhere. Is there a characterization (in terms of the Fourier transform) of functions in L^p(R) with the property that linear combinations of its translates are dense in L^p?


If the answer is no can it be shown that no reasonable measure of the size of the zero set of the Fourier transform of f will suffice to give such a characterization?



Is there an L^p tauberian theorem?
Actually this is a well known question. N. Lev and A. Olevskii have shown the following theorem:


**Theorem (Lev, Olevskii)** Given any 1 < p < 2 one can find two vectors in $l^1(Z)$, such that one is cyclic in $l^p(Z)$ and the other is not, but their Fourier transforms have an identical set of zeros.


The same result follows for $L^p(R)$.


Look [here](http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6X1B-4SK0C3K-2&_user=10&_rdoc=1&_fmt=&_orig=search&_sort=d&_docanchor=&view=c&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=71a092e5dc0badfd237266b26a087a44) for example or on arxiv under Olevskii or Lev. This means more or less that for $p\neq 1,2$, there can be no characterization of $L^p$ generators in terms of the zero set of the Fourier transform. Hope this helps.


PS: Maybe I should add that I have the impression that this is a big open problem so you shouldn't expect an 'easy' answer. It is not clear in what terms one should seek for such a characterization. I would contact Nir Lev for more information (you can look for his e-mail on his web site).


