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Suppose that $(z_n) \subset \mathbb C$ is a sequence (repetitions allowed) such that $$ F(z) = \prod_n \left ( 1-\frac{z}{z_n} \right ) $$ defines an entire function of exponential type, that is, $|F(z)| \leq ae^{b|z|}$ for $a,b > 0$. Further, let $(p_n) \subseteq \mathbb C$ be a bounded sequence. Does $$ \tilde F(z) = \prod_n \left ( 1-\frac{z}{z_n+p_n} \right ) $$ still define an entire function of exponential type? Notice, that the zeros (z_n+p_n) of the new function $\tilde F$ arise as a "perturbation" of the original zeros $(z_n)$. I feel like the statement is true and should follow from the general theory of entire functions of exponential type but I don't have a reference.

I would appreciate any help.

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  • $\begingroup$ When you write the product without explanation, you must assume that it is convergent in the usual sense, which means $\sum|z_n|^{-1}$ is finite. Then your function is necessary of zero exponential type. And the perturbed function is of zero type as well. The posted answers are related to conditionally convergent products is various senses. The best source I can recommend is Levin, Distribution of zeros of entire functions. $\endgroup$
    – Alexandre Eremenko
    Commented Apr 19 at 12:27

2 Answers 2

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The answer is yes. Your function definies a function of finite exponential type if and only if the counting function of $z_n$ is bounded by a linear function (that is, if the number of $z_n$ in the disk $B(0, r)$ is at most $ar + b$ for some $a, b > 0$) and if $$\limsup \left|\sum_{z_n < r} \frac{1}{z_n}\right| < \infty.$$

For these facts see Lecture 5 in [1], specifically Theorem $4$.

Now, when we perturb $z_n$ by a bounded sequence the first condition is clearly preserved, we don't even have to change $a$, and then when we consider difference of the sums for a given $r$. The terms that were added or removed from the sum due to the change in absolute value each contribute $O(\frac{1}{r})$, and there are $O(r)$ of them by the first condition, so in total we change the sum by at most a finite amount. for the remaining ones we can bound the impact by $$\sum_n \left|\frac{1}{z_n} - \frac{1}{z_n + p_n}\right| = \sum_n \left|\frac{p_n}{z_n(z_n+p_n)}\right|$$ and this series converges absolutely since (if we reorder $z_n$ by increasing absolute value) this is majorated by the sum $\frac{1}{n^2}$, which is convergent.

[1]Levin, B. Ya., Lectures on entire functions. In collab. with Yu. Lyubarskii, M. Sodin, V. Tkachenko. Transl. by V. Tkachenko from an original Russian manuscr, Translations of Mathematical Monographs. 150. Providence, RI: American Mathematical Society (AMS). xv, 248 p. (1996). ZBL0856.30001.

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Yes, this works, with the understanding that we may have to renormalize the products as $\prod (1-z/z_n)e^{z/z_n}$, as is usually done anyway. [Or you could only consider $\prod (1-z/z_n)$ with the understanding that you are making the extra assumption that this product converges, but then you are not discussing zero sets of arbitrary functions of exponential type.]

If $z_n$ are the zeros of an entire function of exponential type, then the partial sums of $\sum 1/z_n$, taken over $|z_n|\le R$, stay bounded. This is Lindelof's theorem. It also has an (easier) converse, saying that if $z_n$ is such a sequence and $|z_n|\gtrsim n$, then the product will define an entire function of exponential type.

Finally, since this ($|z_n|\gtrsim n$) holds for a function of exponential type, bounded shifts of the $z_n$ will not affect these conditions.

Koosis, The logarithmic integral 1 discusses this in detail.

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    $\begingroup$ I believe $\sum \frac{1}{z_n}$ being convergent on the one hand is too strong (boundedness is enough), and on the other hand is too weak -- you also need an a priori linear bound on the counting function of $z_n$. $\endgroup$
    – Aleksei Kulikov
    Commented Apr 18 at 20:11
  • $\begingroup$ Also I did not see your answer when I posted mine, since we were writing at the same time -- my apologies. $\endgroup$
    – Aleksei Kulikov
    Commented Apr 18 at 20:14
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    $\begingroup$ @AlekseiKulikov: You are right of course. I misquoted Lindelof's theorem and its converse from faulty memory. Also, nothing wrong of course with writing a second (or first) answer. $\endgroup$
    – Christian Remling
    Commented Apr 18 at 20:54
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    $\begingroup$ @AlekseiKulikov: To my small credit, I was already wondering why anyone bothers to include the exponentials $e^{z/z_n}$ if they are not needed... This is now explained. $\endgroup$
    – Christian Remling
    Commented Apr 18 at 20:59
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    $\begingroup$ @J.Swail: Yes, that is correct. The assumption you then need is the one I originally had in the erroneous version of this answer, namely, $\sum 1/z_n$ must converge. Since always $\sum 1/|z_n|^2<\infty$ for the zeros of an entire function, we will also have $\sum 1/w_n$ convergent, with $w_n=z_n+p_n$. If we want to be extra cautious, we can then define $F(z)=\prod (1-z/w_n)e^{z/w_n}$, but now we see that the two products $\prod (1-z/w_n)$ and $\prod e^{z/w_n}$ converge separately, the latter to $e^{az}$ for some $a\in\mathbb C$, $\endgroup$
    – Christian Remling
    Commented Apr 19 at 13:56

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