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Let $f\in L^2(\mathbb R)$ be a not identically vanishing function that is compactly supported in the interval $(-\sigma,\sigma)$ for some fixed $\sigma>0$. For each $z\in \mathbb C$ let us define the function $$ F(z) = \int_{-\sigma}^{\sigma} f(x) e^{-izx}\,dx \quad z\in \mathbb C.$$ The above function is entire and of exponential type. Now suppose that $A=\{a_k\}_{k=1}^{\infty} \subset \mathbb R$ is a uniformly discrete set and assume that $$ F(a_k)=0 \quad \forall\, k \in \mathbb N.$$ What is the best universal constant $C$ to have the inequality $$ \limsup_{r\to \infty} \frac{|A \cap (0,r)|}{r} \leq C \sigma,$$ independent of $F$ and $\sigma$.

My own initial understanding was that based on interpolation results of Beurling, the constant $C= \frac{1}{\pi}$ works but I am not so sure anymore. It seems based on a reading of a paper of Eremenko and Yuditskii, the constant $C=\frac{2}{\pi}$ works for sure but that paper is in fact concerned with upper density of zeros on the entire complex plane and not just the positive real axis and also not under the additional assumption that the zeros are uniformly discrete.

Any comments on this is much appreciated.

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If you count positive zeros, then under your conditions $$\limsup \frac{|A\cap[0,r]|}{r}\leq \frac{1}{\pi},$$ and this is exact.

If instead of counting positive zeros you count those in an arbitrarily small sector $|\arg z|<\epsilon$ then the limit exists and we have equality. This implies your statement about real zeros.

The reason is that your condition $f\in L^2$ implies completely regular growth with indicator diagram $[-i\sigma,i\sigma]$. This is a deep theorem of N. Levinson.

Ref. B. Ya. Levin, Distribution of zeros of entire functions.

Remark. In my paper with Yuditskii that you mention we consider a more general class of functions which are of exponential type with indicator $[-\sigma i,\sigma i]$ but not necessarily of completely regular growth.

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    $\begingroup$ @Ali: No, this upper uniform density can be infinite. This follows from the Beurling-Malliavin theorem of the radius of completeness. $\endgroup$ Commented May 1 at 14:36
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    $\begingroup$ I am not sure what "uniformly discrete" exactly is, but you cannot estimate the "upper uniform density" in terms of the exponential type and $L^2$ norm. One can construct examples with "upper uniform density" greater than the usual density. The condition is Beurling-Malliavin theorem is essentially exact. $\endgroup$ Commented May 2 at 13:35
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    $\begingroup$ @Ali: tis is an interesting result, but it has much stronger conditions then your question: the function is absolutely continuous and $f(\sigma)f(-\sigma)\neq 0$. $\endgroup$ Commented May 3 at 23:41
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    $\begingroup$ @Ali: Yes. And before that replace $f$ by $f(z)\bar{f}(\bar{z})$ to make $f$ symmetric, which somewhat simplifies the argument. $\endgroup$ Commented May 5 at 13:01
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    $\begingroup$ Functions $f$ of exp type such that $f(x\pm ia)$ is bounded from above and below, for some $a>0$ are called "functions of the type of sine". Much more is known about their zero distribution, see for example, acadsci.fi/mathematica/Vol35/vol35pp23-46.pdf $\endgroup$ Commented May 5 at 13:08

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