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 $t\_i$?



Are the zeroes of the Fourier Transform of compactly supported functions isolated?
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). 


2. 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. 


