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

## 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 $\ds \prod_{n \mathop = 1}^\infty \paren {1 + f_n}$ converges uniformly absolutely if and only if the sequence of partial products of $\ds \prod_{n \mathop = 1}^\infty \paren {1 + \size {f_n} }$ converges uniformly.

### Definition 2

The infinite product $\ds \prod_{n \mathop = 1}^\infty \paren {1 + f_n}$ converges uniformly absolutely if and only if the series $\ds \sum_{n \mathop = 1}^\infty f_n$ converges uniformly absolutely.

### Definition 3

The infinite product $\ds \prod_{n \mathop = 1}^\infty \paren {1 + f_n}$ converges uniformly absolutely if and only if there exists $n_0 \in \N$ such that:

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

and:

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

## Proof

### 2 implies 1

By the Monotone Convergence Theorem:

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

### 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 $x \in X$.

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

for $n \ge n_0$.

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

$\Box$

### 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 $x \in X$.

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

Thus:

$\size {\map \ln {1 + \size {\map {f_n} x} } } \le \dfrac 9 4 \size {\ln {1 + \map {f_n} x} }$
$\ds \sum_{n \mathop = n_0}^\infty \map \ln {1 + \size {\map {f_n} x} }$ and $\ds \sum_{n \mathop = n_0}^\infty \size {\map {f_n} x}$

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

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

By:

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

the sequence of partial products of $\ds \prod_{n \mathop = 1}^\infty \paren {1 + \size {f_n} }$ 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 $x \in X$.

$\size {\map {f_n} x} \le \dfrac 3 2 \size {\ln {1 + \map {f_n} x} }$
$\ds \sum_{n \mathop = n_0}^\infty f_n$

$\Box$