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 $0<A_f<\infty$ and $0<A_g<\infty$ (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 $$ \max\{A_f,A_g\}<A_{fg} $$ (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.


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 $1$: $$F(z)=\prod_n\left(1-\frac{z}{a_n}\right)e^{\displaystyle z/a_n}.$$ 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.

  • 1
    $\begingroup$ 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. $\endgroup$ – Bedovlat Nov 29 '18 at 20:51
  • $\begingroup$ 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. $\endgroup$ – Bedovlat Nov 29 '18 at 21:00
  • $\begingroup$ @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. $\endgroup$ – Alexandre Eremenko Nov 29 '18 at 21:50
  • $\begingroup$ Wow! What about the Weierstrass factorization of $\sin(\pi z)/(\pi z)$? $\endgroup$ – Bedovlat Nov 29 '18 at 21:54
  • $\begingroup$ $$\sin\pi z=\pi z\prod_{n\neq 0}\left(1-z/n\right)e^{z/n}=\pi z\prod_{n=1}^\infty\left(1-z^2/n^2\right)$$ are not of the kind you wrote! $\endgroup$ – Alexandre Eremenko Nov 29 '18 at 21:58

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