# Derivative of Infinite Product of Analytic Functions

## Contents

## Theorem

Let $D \subset \C$ be open.

Let $\left\langle{f_n}\right\rangle$ be a sequence of analytic functions $f_n: D \to \C$.

Let the product $\displaystyle \prod_{n \mathop = 1}^\infty f_n$ converge locally uniformly to $f$.

Then:

- $\displaystyle f' = \sum_{n \mathop = 1}^\infty f_n'\cdot \prod_{\substack{k \mathop = 1 \\ k\mathop \ne n} }^\infty f_k$

and the series converges locally uniformly in $D$.

## Outline of Proof

Using Logarithmic Derivative of Infinite Product of Analytic Functions we write $\displaystyle \frac {f'} f = \sum_{n \mathop = 1}^\infty \frac{f_n'}{f_n}$, and we multiply this by $f$.

## Proof

Note that by Infinite Product of Analytic Functions is Analytic, $f$ is analytic.

We may suppose none of the $f_n$ is identically zero on any open subset of $D$.

Let $E = D \setminus \left\{ {z \in D: f \left({z}\right) = 0}\right\}$.

By Logarithmic Derivative of Infinite Product of Analytic Functions, $\displaystyle \frac{f'} f = \sum_{n \mathop = 1}^\infty \frac {f_n'} {f_n}$ converges locally uniformly in $E$.

By Linear Combination of Convergent Series, $\displaystyle f' = \sum_{n \mathop = 1}^\infty f_n' \cdot \prod_{\substack {k \mathop = 1 \\ k \mathop \ne n}}^\infty f_k$ on $E$.

By Uniformly Convergent Sequence Multiplied with Function, the series converges locally uniformly in $E$.

By Uniformly Convergent Sequence on Dense Subset, the series converges locally uniformly in $D$.

Let $g$ denote its limit.

By Uniform Limit of Analytic Functions is Analytic, $g$ is analytic in $D$.

By Uniqueness of Analytic Continuation, $f' = g$.

$\blacksquare$