Infinite Product of Analytic Functions

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Theorem

Let $D \subset \C$ be an open connected set.

Let $\left\langle{f_n}\right\rangle$ be a sequence of analytic functions $f_n: D \to \C$ that are not identically zero.

Let $\displaystyle \sum_{n \mathop = 1}^\infty \left({f_n - 1}\right)$ converge locally uniformly absolutely on $D$.


Then:

$(1): \quad f = \displaystyle \prod_{n \mathop = 1}^\infty f_n$ converges locally uniformly absolutely on $D$
$(2): \quad f$ is analytic
$(3): \quad$ For each $z \in D$, $f_n \left({z}\right) = 0$ for finitely many $n \in \N$
$(4): \quad$ For each $z \in D$, $\operatorname {mult}_z \left({f}\right) = \displaystyle \sum_{n \mathop = 1}^\infty \operatorname{mult}_z \left({f_n}\right)$

where $\operatorname{mult}$ denotes multiplicity.


Proof

Because $\displaystyle \sum_{n \mathop = 1}^\infty \left({f_n - 1}\right)$ converges locally uniformly absolutely on $D$, the series converges locally uniformly absolutely.

Thus $f = \displaystyle \prod_{n \mathop = 1}^\infty f_n$ converges locally uniformly absolutely.

By Uniform Limit of Analytic Functions is Analytic, $f$ is analytic

The last claim follows from Zeroes of Infinite Product of Analytic Functions

$\blacksquare$


Also see