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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.

Edit: Question answered on mathoverflow here.

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