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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?

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    $\begingroup$ This feels like it should be known (in the sense either of having a characterization, or a result showing no reasonable characterization is possible). Unfortunately I can't remember anything definite. Have you had any luck on MathSciNet? $\endgroup$
    – Yemon Choi
    Commented Nov 5, 2009 at 6:35
  • $\begingroup$ Not really. This problem came up in something I was thinking about about a long while ago. At the time I did some fairly intensive detective work trying to find a reference/answer. I remember finding passing references to the problem but I never was able to track down an answer. $\endgroup$
    – Mark Lewko
    Commented Nov 5, 2009 at 7:03
  • $\begingroup$ In $$ $$ math.tamu.edu/~thomas.schlumprecht/ossz.pdf $$ $$ Odell, Sari, Schlumprecht, and Zheng prove an interesting non Tauberian theorem: the translates of an $L_1$ function cannot contain a Schauder basis for $L_1$, nor can translates of an $L_p$ function contain an unconditional Schauder basis for $L_p$ when $p \le 4$. $\endgroup$
    – Bill Johnson
    Commented Mar 24, 2012 at 14:42
  • $\begingroup$ Is the reference that I give to this answer relevant to this question: mathoverflow.net/questions/333317/… ? $\endgroup$
    – Adrián González Pérez
    Commented Jun 6, 2019 at 11:03
  • $\begingroup$ ok, I saw that it is already referenced bellow, never mind. $\endgroup$
    – Adrián González Pérez
    Commented Jun 6, 2019 at 11:08

2 Answers 2

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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).

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  • $\begingroup$ Thanks! This explains why I thought the question "looked familiar" - I must have seen the abstract of something like this: arXiv:0908.0447 $\endgroup$
    – Yemon Choi
    Commented Nov 5, 2009 at 19:25
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    $\begingroup$ Thanks! Looking at the paper of N. Lev and A. Olevskii I learned that there is a partial generalization of Beurling (On a closure problem, Ark. Mat. 1 (1951), pp. 301–303.) This states that the translates of f \in L^{p} is dense in L^p (for 1<p<2) if the Hausdorff dimension of the zero set of the Fourier transform of f is less than 2(p-1)/p. The converse, as has be pointed out, is false. $\endgroup$
    – Mark Lewko
    Commented Nov 5, 2009 at 20:40
  • $\begingroup$ The link to sciencedirect.com is broken. I'm also unable to find any snapshot saved on the Wayback Machine. $\endgroup$
    – The Amplitwist
    Commented May 18, 2023 at 7:09
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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...

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  • $\begingroup$ If f is in $L^p$, $p\neq 1$, is there any mollifier $h$ which can make $h\ast f$ to be $L^1$? $\endgroup$
    – spr
    Commented Mar 24, 2012 at 10:25

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