# Equivalence of Definitions of Uniform Absolute Convergence of Product of Complex Functions

## Contents

## Definition

Let $X$ be a set.

Let $\sequence {f_n}$ be a sequence of bounded mappings $f_n: X \to \C$.

The following definitions of the concept of **Uniform Absolute Convergence of Product** are equivalent:

### Definition 1

The infinite product $\displaystyle \prod_{n \mathop = 1}^\infty \left({1 + f_n}\right)$ **converges uniformly absolutely** if and only if the sequence of partial products of $\displaystyle \prod_{n \mathop = 1}^\infty \left({1 + \left\vert{f_n}\right\vert}\right)$ converges uniformly.

### Definition 2

The infinite product $\displaystyle \prod_{n \mathop = 1}^\infty \left({1 + f_n}\right)$ **converges uniformly absolutely** if and only if the series $\displaystyle \sum_{n \mathop = 1}^\infty f_n$ converges uniformly absolutely.

### Definition 3

The infinite product $\displaystyle \prod_{n \mathop = 1}^\infty \left({1 + f_n}\right)$ **converges uniformly absolutely** if and only if there exists $n_0 \in \N$ such that:

- $(1): \quad f_n \left({x}\right) \ne -1$ for $n \ge n_0$ and $x \in X$

and:

- $(2): \quad$ The series $\displaystyle \sum_{n \mathop = n_0}^\infty \log \left({1 + f_n}\right)$ is uniformly absolutely convergent.

## Proof

### 1 implies 2

### 2 implies 1

By Terms in Uniformly Convergent Series Converge Uniformly to Zero

By the Monotone Convergence Theorem:

- The sequence of partial products of $\displaystyle \prod_{n \mathop = n_0}^\infty \paren {1 + \size {f_n} }$ converges uniformly if ...

Use Bounds for Finite Product of Real Numbers.

### 2 implies 3

By Terms in Uniformly Convergent Series Converge Uniformly to Zero, there exists $n_0 \in \N$ such that $\size {\map {f_n} x} \le \dfrac 1 2$ for $n \ge n_0$.

Then $\map {f_n} x \ne -1$ for all $n \ge n_0$ and $x \in X$.

By Bounds for Complex Logarithm:

- $\size {\map \ln {1 + \map {f_n} x} } \le \dfrac 3 2 \size {\map {f_n} x}$

for $n \ge n_0$.

By Comparison Test for Uniformly Convergent Series,

- $\displaystyle \sum_{n \mathop = n_0}^\infty \ln {1 + \map {f_n} x}$

converges uniformly absolutely.

$\Box$

### 3 implies 1

By Terms in Uniformly Convergent Series Converge Uniformly to Zero there exists $n_0 > 0$ such that $\ln {1 + \map {f_n} x} \le \dfrac 1 {100}$ for $n \ge n_0$ and $x \in X$.

By Bounds for Complex Exponential:

- $\size {\map {f_n} x} \le \dfrac 3 2 \size {\ln {1 + \map {f_n} x} } \le \dfrac 1 2$

By Bounds for Complex Logarithm:

- $\size {\map \ln {1 + \size {\map {f_n} x} } } \le \dfrac 3 2 \size {\map {f_n} x}$

Thus:

- $\size {\map \ln {1 + \size {\map {f_n} x} } } \le \dfrac 9 4 \size {\ln {1 + \map {f_n} x} }$

By Comparison Test for Uniformly Convergent Series:

- $\displaystyle \sum_{n \mathop = n_0}^\infty \map \ln {1 + \size {\map {f_n} x} }$ and $\displaystyle \sum_{n \mathop = n_0}^\infty \size {\map {f_n} x}$

By Bounds for Finite Product of Real Numbers the sequence of partial products of $\displaystyle \prod_{n \mathop = n_0}^\infty \paren {1 + \size {f_n} }$ is bounded.

By Complex Exponential is Uniformly Continuous on Half-Planes, the sequence of partial products of $\displaystyle \prod_{n \mathop = n_0}^\infty \paren {1 + \size {f_n} }$ converges uniformly.

By:

- Uniformly Convergent Sequence Multiplied with Function
- $f_1, \ldots, f_{n_0 - 1}$ are bounded

the sequence of partial products of $\displaystyle \prod_{n \mathop = 1}^\infty \paren {1 + \size {f_n} }$ converges uniformly.

$\Box$

### 3 implies 2

By Terms in Uniformly Convergent Series Converge Uniformly to Zero there exists $n_0 > 0$ such that $\size {\ln {1 + \map {f_n} x} } \le \dfrac 1 2$ for $n \ge n_0$ and $x \in X$.

By Bounds for Complex Exponential:

- $\size {\map {f_n} x} \le \dfrac 3 2 \size {\ln {1 + \map {f_n} x} }$

By Comparison Test for Uniformly Convergent Series:

- $\displaystyle \sum_{n \mathop = n_0}^\infty f_n$

converges uniformly absolutely.

$\Box$

## Sources

- 1973: John B. Conway:
*Functions of One Complex Variable*... (previous) ... (next) $\text {VII}$: Compact and Convergence in the Space of Analytic Functions: $\S 5$: Weierstrass Factorization Theorem: Lemma $5.8$