Is this infinite product entire?

Question

Let $(z_i)$ be a square-summable sequence which is even summable but not absolute summable, i.e. $\sum_{i=1}^{\infty} \vert z_i \vert = \infty$,$\sum_{i=1}^{\infty} \vert z_i \vert^2 < \infty$ and $\sum_{i=1}^{\infty} z_i$ exists. I would like to ask if the following function $$f(\mu):=\prod_{i=1}^{\infty}(1+\mu^2 \vert z_i \vert^2 - 2 \mu \Re(z_i))$$ is necessarily entire?

I must say that I do not even know if this product exists away from the real axis. On the real axis it is clear that it exists by using that $(1+x) \le e^x$ such that $\vert f(\mu) \vert \le e^{\mu^2 \sum_i \vert z_i \vert^2 -2\mu \Re \sum_i z_i }.$

Assuming it was entire, does there exist a similar growth bound on $\vert f(\mu) \vert$ as the one I obtained on the real axis?

Answer 1

This function is (on the real line, at least) the product of $$ \exp( \mu^2 \sum_{i=1}^\infty |z_i|^2 - 2 \mu \Re(\sum_{i=1}^\infty z_i)) \quad (1)$$ and the Hadamard type product $$ \prod_{i=1}^\infty E_1( 2 \mu\Re z_i - \mu^2 |z_i|^2) \quad(2)$$ where $E_1$ is the first elementary factor $$ E_1(z) := (1-z) \exp(z).$$ The expression (1) is clearly entire in $\mu$; the product (2) is locally uniformly convergent (from the standard bound $E_1(z) = 1+O(|z|^2)$ when $|z| \leq 1$) and so is also entire. A refinement of this analysis (using for instance the upper bound $|E_1(z)| \leq \exp(O(|z|^2))$ for all $z$) also gives growth bounds comparable to the ones you already located in the real case.

An alternate factorisation is

$$ \exp( - 2 \mu \Re \sum_{i=1}^\infty z_i ) \prod_{i=1}^\infty E_1(\mu z_i) E_1(\mu \overline{z_i}).$$ Thus this function is an order two entire function with zeroes precisely at $1/z_i, 1/\overline{z_i}$ (counting multiplicity), which specifies the function uniquely up to quadratic exponential factors (such as (1)) by the Hadamard factorisation theorem.