Let $c>0$ and $X$ the collection of all entire functions $f$ for which there exists $C_f > 0$ s.t. $$ |f(z)| \leq C_f e^{c|z|}. $$ Given a sequence $(z_n) \subset (0,\infty)$ s.t. $$ \frac {z_n}{n} \geq \frac \pi c, \quad \forall n $$ my question is whether $(z_n)$ is contained in the zero set of a non-trivial function in $X$. The question is motivated by theorems I read on completeness problems of complex exponentials. I have the feeling that the above claim is true and could be found in some book. Does someone have a reference for it or an easy proof/disproof?
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$\begingroup$ if say $c=\pi$ and $z_n=n$ then if $f$ has zeroes precisely at $z_n$ then $f(z) \Gamma (-z+1)$ is entire and has no zeroes so is $e^g$ and $1/\Gamma$ has maximal type so is not in $X$ hence $f$ is not in $X$; you need some symmetry for the zeroes to conclude a result of the type you want - see Levin Lectures on Entire Functions page $32$ $\endgroup$– ConradCommented Sep 29, 2023 at 15:13
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$\begingroup$ Thank you! Perhaps the notion of zero set I defined is a bit uncommon, for me it does not mean that $f$ has to vanish only on $Z$, but it is some nontrivial function with $Z$ a subset of the zero set (I edited the post). If ones assumes a symmetry in the zeros as you stated, is there some reference for a result of the above type? $\endgroup$– user975628Commented Sep 29, 2023 at 15:23
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$\begingroup$ The book by Levin (Lectures on Entire Functions) has results in this direction- here the result should be easy since the function that has zeroes at $\pm z_n$, has Hadamard product without any exponential term since one can group them in pairs and $f(z)=\Pi (1-z^2/z_n^2)$ converges absolutely as $\sum 1/z_n^2$ does; this should imply that the type of $f$ is equal to the upper density of the zeroes so is at most $c$ $\endgroup$– ConradCommented Sep 29, 2023 at 15:28
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$\begingroup$ No. Lindelof's theorem says that $\lim_{R\to\infty} \sum_{|z_n|\le R} 1/z_n$ exists for the zeros of an entire function of exponential type. Obviously, the condition can fail in your setting. See Koosis, The logarithmic integral I, Section III B. $\endgroup$– Christian RemlingCommented Sep 29, 2023 at 16:12
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$\begingroup$ But since I'm only interested in function whose zero set contains $(z_n)$, I could in principle consider functions with a symmetric zero set and hope to be in the desired class $\endgroup$– user975628Commented Sep 29, 2023 at 18:54
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As in Conrad's comment, let $$f(z):=\prod_1^\infty\Big(1-\frac{z^2}{z_n^2}\Big).$$ Then for complex $z\ne0$ $$|f(z)|=\prod_1^\infty\Big|1-\frac{z^2}{z_n^2}\Big| \le\prod_1^\infty\Big(1+\frac{|z|^2}{z_n^2}\Big)\le\prod_1^\infty\Big(1+\frac{c^2|z|^2}{\pi^2 n^2}\Big) =\frac{\sinh c|z|}{c|z|}\le e^{c|z|},$$ as desired.
answered Sep 29, 2023 at 19:59