I have a continuous function $f$ on a locally compact Abelian group $G$ with compact support, and I would like to say that the zeroes of $f$ are sparse in some sense (isolated would be good, uniformly discrete would be great).

Now, if $G=\mathbb R^d$, then this is a consequence of the Paley-Wiener theorem, but $G$ is a general LCAG.

Now, for what I am doing I can replace $G$ with the subgroup generated by $\sup(f)$, thus by the structure theorem I can assume that $G$ has the form $\mathbb R^d \times \mathbb Z^n \times K$. Also, by a simple trick I am sure I can ignore the $\mathbb Z^n$ component. If $K$ was not there, I would be done, but I don't see any way of eliminating it.

Anyhow, since the dual of $K$ is discrete, intuitively anything in here is isolated and Paley-Wiener should solve the problem in $\mathbb R^n$. Unfortunately, it looks like this intuitive part becomes a proof only for functions $f: \mathbb R^d \times K \to \mathbb C$ of the form $$ f(s,t)=g(s)h(t) $$

So my questions are:

1) Is there any general result of the type I am seeking in the case of LCAG? I know few "uncertainty principle" type results, which unfortunately are not what I need but maybe there is a variant I am not familiar with which would work.

2) Is there any Paley -Wiener Theorem for the case $G= \mathbb R^d \times K$, where $K$ is any compact group? [ Note that $K$ need not be Lie].

Edit: As the comments already provide a counterexample, is the following weaker version true, at least in $\mathbb R^d$?

Question: Let $f$ be a continuous function with compact support, which is positive definite. Can we show that there exists $t_1,..,t_k$ such that $\sum T_{t_i} (\widehat{f})$ is nowhere vanishing, where $T_{t_i}$ denotes translation by $t_i$?

  • $\begingroup$ From the title I suppose you mean the zeros of $\hat f$ are discrete. That's true for $d=1$ (unless $f \equiv 0$) because $\hat f$ is analytic, but already for $d=2$ the zeros of an analytic function can easily contain curves, so I'm not sure how Paley-Wiener helps. $\endgroup$ Commented Jan 10, 2016 at 0:12
  • $\begingroup$ If $G=R^n$ Fourier transform is an entire function on $R^n$ by Paley-Wiener. So why zeros are isolated?? $\endgroup$ Commented Jan 10, 2016 at 0:13
  • $\begingroup$ @NoamD.Elkies Ups, I forgot that the identity Theorem holds only in dimension 1 :) $\endgroup$
    – Nick S
    Commented Jan 10, 2016 at 0:16
  • 2
    $\begingroup$ The two comments point counterexamples, I edited the question to replace sparseness by some type of "not too many". @AlexandreEremenko $\endgroup$
    – Nick S
    Commented Jan 10, 2016 at 0:26

2 Answers 2

  1. Without the condition that $f$ is positive definite, the answer to the modified question is "no", even when $G=R$, the real line. Suppose that the support of $f$ does not contain some neighborhood of $0$, say $[-1,-1/2]\cup[1/2,1]$, and $f(-x)=\overline{f(x)}$, so that Fourier transform $F$ is real. Then $\sum F(t-t_j)$ is the Fourier transform of $pf$, where $p$ is an exponential polynomial, and $pf$ has the same support at $f$. But there is a theorem which says that a real function $F$ whose (inverse) Fourier transform has a "spectral gap" (that is its support is disjoint from some neighborhood of zero), then $F$ has infinitely many real zeros.

For the case when $f$ has compact support, this is due to B. Logan, Properties of high-pass signals, (Thesis, Dept. Electrical Engineering, Columbia Univ. 1965), for the general case,

Eremenko and Novikov, Oscillation of Fourier integrals with a spectral gap, J. de Math. Pures et Appl., 8, 3 (2004), 313-365, http://www.math.purdue.edu/~eremenko/dvi/novik1011.pdf

which also reproduces Logan's elementary proof for the case of compact support. (If $f$ is a measure with finite support at the integers, this is a classical theorem of Ch. Sturm).

  1. Now if $f$ is positive definite, and with compact support, this means that $F=\hat{f}$ is non-negative and entire, of exponential type. So zeros of $F$ make a discrete sequence of isolated points. Then it is quite evident that you can make the sum $\sum_j F(t-t_j)$ positive: just take $t_0=0$ and $t_1$ different from all differences between zeros of $F$. Then $F(t-t_0)+F(t-t_1)>0$ for all real $t$.

To do this in $R^n$ you will need more than $2$ summands. You use the following lemma: if $Z$ is an analytic set in dimension $n$, not equal to the whole space, then you can find $n+1$ vectors $t_j$ such that the translations $Z+t_j$ are disjoint. This is proved by induction on $n$, using he Weierstrass Preparation Theorem.

  • $\begingroup$ But the support of positive definite functions always contains a neighborhood of zero. $\endgroup$
    – Nick S
    Commented Jan 12, 2016 at 16:50
  • $\begingroup$ When $f$ is positive definite, your statement is of course correct. I edited the answer. $\endgroup$ Commented Jan 12, 2016 at 21:09
  • $\begingroup$ As it was pointed in the comments, the zeroes of an entire type function are discrete only in $\mathbb R$, in higher dimensions the zeroes can contain curves... $\endgroup$
    – Nick S
    Commented Jan 12, 2016 at 21:26
  • $\begingroup$ Yes, of course. But if you add more than $n+1$ non-negative summands in dimension $n$ you can make your sum positive. $\endgroup$ Commented Jan 12, 2016 at 21:29

The 1976 paper by Liepins seems to do what you want.


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