# Infinite Product of Analytic Functions

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## Theorem

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

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

Let $\ds \sum_{n \mathop = 1}^\infty \paren {f_n - 1}$ converge locally uniformly absolutely on $D$.

Then:

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

where $\operatorname {mult}$ denotes multiplicity.

## Proof

Because $\ds \sum_{n \mathop = 1}^\infty \paren {f_n - 1}$ converges locally uniformly absolutely on $D$, the series converges locally uniformly absolutely.

Thus $f = \ds \prod_{n \mathop = 1}^\infty f_n$ 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

$\blacksquare$