Is there an L^p tauberian theorem? 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?
 A: OK, my first naive thoughts are as follows. What follows is incomplete, but I'm leaving it here in case it suggests a proper solution or jogs someone's memory.
Let's look at the case 1< p < 2, and let f\in Lp(R) be such that translates of f do not span a dense subspace. By duality, there exists a nonzero g \in Lq(R) such that the convolution of f with g is zero (this convolution is a continuous function on R, so we don't need an a.e. qualifier here.)
Now if f and g were known to have well-defined Fourier transforms, the hypothesis that g is not identically zero ought to imply that the FT of f has to vanish on a `visible' subset of R. This suggests trying to introduce some mollifier functions, i.e. some h and k whose Fourier transforms are compactly supported but are 1 on very large intervals (maybe de la Vallee Poussin kernels would be enough?) and then considering
h*f*g*k = 0
where we hope that h*f and g*k will be in L1(R), and that g*k will not be identically zero. Then applying the previous argument we know that the FT of h*f would have to vanish on some open interval, and then by varying h maybe we can eventually say something about f...
A: 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 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).
A: Regarding the argument of Yemon Choi below: If f is in $L^p$, $p\neq 1$, is there   any  mollifier  $h$ which can make $h\ast f$ to be $L^1$?  
