# Exponential type of a product of entire functions

Let $$\{a_n\}_{n=1}^\infty$$ and $$\{b_m\}_{m=1}^\infty$$ be two sequences of points in $$\mathbb{C}$$ such that $$f(z)=\prod_{n=1}^\infty\left(1-\frac{z}{a_n}\right)\quad\mbox{and}\quad g(z)=\prod_{m=1}^\infty\left(1-\frac{z}{b_m}\right)$$ are entire functions of finite exponential types $$0 and $$0 (growth orders are $$\rho_f=\rho_g=1$$), respectively. Is it true that the exponential type $$A_{fg}$$ of the entire function $$f(z)g(z)$$ satisfies $$\max\{A_f,A_g\} (strict inequality)?

More generally, given that the growth of $$f$$ as above is $$\rho_f=1$$, is there a way to establish the exponential type from the sequence $$\{a_n\}$$?

Thank you.

I expected this to be a university-level question, so I posted it on MSE (here). But it doesn't seem to even get any attention. So I am posting it here. Apologies if it sounds too elementary to some; I am not a specialist in complex analysis.

## 1 Answer

On the first question: Your inequality is incorrect for exponential type functions. Take $$f(z)=e^z,\; g(z)=e^{-z}$$, both have exponential type 1. These examples are of course not of the form of infinite product that you request.

But if $$f$$ is defined by an infinite product as you wrote, then (with usual understanding of an infinite product) the condition of convergence is $$\sum\frac{1}{|a_n|}<\infty,$$ and this implies that your functions are of exponential type $$0$$. (Levin, Ch. I, section 4, Lemma 3). Then $$A_f=A_g=A_{fg}=0$$, so your strict inequality is wrong and the non-strict inequality holds trivially.

In general, for functions of exponential type it is true that $$A_{fg}\leq A_f+A_g$$, and equality can hold.

On your second question. To tell the exponential type of the infinite product from its zeros is possible (but not simple in general). Let $$n(r)$$ be the counting function of zeros. For functions of exponential type, $$n(r)/r$$ has finite upper limit as $$r\to\infty$$. This is a consequence of Jensen's formula. But this upper limit is not simply related to the exponential type: the arguments of zeros also have a strong influence.

Simple relations can be obtained only in the (important) special cases, for example when the zeros are real, and limits $$\lim n^+(r)/r$$ and $$\lim n^-(r)/r$$ exist, where $$n_{\pm}$$ counts the number of positive and negative zeros.

For the details, see B. Levin, Distribution of zeros of entire functions, AMS, 1970.

EDIT. @Christian Remling suggested to use the following definition of the class of sequences: $$f(z)=\lim_{r\to\infty}\sum_{|a_n|\leq r}\left(1-\frac{z}{a_n}\right)$$ converges. Here is an example with $$\max\{A_f,A_g\}>A_{fg}$$ in this class. I use the facts and notation from Levin's book, English edition. Define $$a_{2k}=k,\; a_{2k+1}=-k^2/(k+b),$$ where $$b>0$$ is a constant to be chosen. Then define a canonical product of genus $$1$$: $$F(z)=\prod_n\left(1-\frac{z}{a_n}\right)e^z/a_n}.$$ To this function, Theorem 2 from section 1, Chapter II can be applied. In the notation of this section, we have $$\tau_F=0$$, and $$d\Delta$$ has two atoms of mass $$1$$: one at $$0$$ another at $$\pi$$. Then formula (2.06) gives $$H_F(\theta)=\pi\sin|\theta|.$$ It is not surprising, of course that $$F$$ behaves like $$\sin\pi z$$, and this can be proved by direct estimation, without a reference to Levin's book.

On the other hand, the product $$f$$ is conditionally convergent, so exponents in the product of $$F$$ can be taken out, and we obtain $$F(z)=f(z)\exp(-\pi^2b z/6).$$ Therefore $$H_f(\theta)=\pi\sin|\theta|+\frac{\pi^2b}{6}\cos\theta.$$ Now it is clear that when $$b$$ is very large, the function $$f$$ behaves like an exponential, and we will have a counterexample with $$g(z)=f(-z)$$.

To be specific, take $$b=7$$. Then $$A_f=\max_{\theta}H_f(\theta)> 3.5\pi.$$ Take $$g(z)=f(ze^{-3.4i})$$ and we obtain $$A_{fg}=\max_\theta\left(H_f(\theta)+H_f(\theta-3.4)\right)<1.5\pi,$$ where I used Maple to plot trigonometric functions and read their maxima from display.

Notice: In the Russian edition of Levin, formula (2.06) is printed with a misprint.

• If it is easy to construct counterexamples, would you, please, do me a favour and construct one for me? Thank you. Of course, I had exponenrials in mind when posting this question. But for Weierstrass products as above, the product has twice as many zeros (multiplicity counted), and this might force a faster growth at a first glance. – Bedovlat Nov 29 '18 at 20:51
• Thanks. I will definitely have a look into Levin's book, but I may still not be able to conclude without a specialist's help. Hence the enquiry. – Bedovlat Nov 29 '18 at 21:00
• @Bedovlat: Your question makes little sense, as stated, because the exponential type of your infinite products is always $0$, if the products are understood in the usual way. See my edited answer. – Alexandre Eremenko Nov 29 '18 at 21:50
• Wow! What about the Weierstrass factorization of $\sin(\pi z)/(\pi z)$? – Bedovlat Nov 29 '18 at 21:54
• $$\sin\pi z=\pi z\prod_{n\neq 0}\left(1-z/n\right)e^{z/n}=\pi z\prod_{n=1}^\infty\left(1-z^2/n^2\right)$$ are not of the kind you wrote! – Alexandre Eremenko Nov 29 '18 at 21:58