3
$\begingroup$

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 $t_i$?

$\endgroup$
4
  • $\begingroup$ 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. $\endgroup$ 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?? $\endgroup$ Jan 10, 2016 at 0:13
  • $\begingroup$ @NoamD.Elkies Ups, I forgot that the identity Theorem holds only in dimension 1 :) $\endgroup$
    – Nick S
    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
    Jan 10, 2016 at 0:26

2 Answers 2

5
$\begingroup$
  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.

$\endgroup$
4
  • $\begingroup$ But the support of positive definite functions always contains a neighborhood of zero. $\endgroup$
    – Nick S
    Jan 12, 2016 at 16:50
  • $\begingroup$ When $f$ is positive definite, your statement is of course correct. I edited the answer. $\endgroup$ 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
    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. $\endgroup$ Jan 12, 2016 at 21:29
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.