2
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.