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 $0<A_f<\infty$ 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.
 A: 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.
