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

## Definition

Let $X$ be a set.

Let $\left \langle {f_n} \right \rangle$ 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

### 2 implies 1

By the Monotone Convergence Theorem:

The sequence of partial products of $\displaystyle \prod_{n \mathop = n_0}^\infty \left({1 + \left\vert{f_n}\right\vert}\right)$ converges uniformly if ...

### 2 implies 3

By Terms in Uniformly Convergent Series Converge Uniformly to Zero, there exists $n_0\in\N$ such that $|f_n(x)|\leq\frac12$ for $n\geq n_0$.

Then $f_n(x)\neq-1$ for all $n\geq n_0$ and $x\in X$.

$|\log(1 + f_n(x))| \le \frac32 |f_n(x)|$

for $n \ge n_0$.

$\displaystyle \sum_{n \mathop = n_0}^\infty \log(1+f_n)$

$\Box$

### 3 implies 1

By Terms in Uniformly Convergent Series Converge Uniformly to Zero there exists $n_0>0$ such that $|\log(1 + f_n(x))|\leq\frac1{100}$ for $n\geq n_0$ and $x\in X$.

By Bounds for Complex Exponential $|f_n(x)| \leq \frac32|\log(1 + f_n(x))|\leq\frac12$

By Bounds for Complex Logarithm $|\log(1 + |f_n(x)|)|\leq \frac32 |f_n(x)|$

Thus $|\log(1 + |f_n(x)|)| \leq\frac94|\log(1 + f_n(x))|$.

$\displaystyle \sum_{n \mathop = n_0}^\infty \log(1 + |f_n(x)|)$ and $\displaystyle \sum_{n \mathop = n_0}^\infty |f_n(x)|$

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

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

By:

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

the sequence of partial products of $\displaystyle \prod_{n \mathop = 1}^\infty \left({1 + \left\vert{f_n}\right\vert}\right)$ converges uniformly.

$\Box$

### 3 implies 2

By Terms in Uniformly Convergent Series Converge Uniformly to Zero there exists $n_0>0$ such that $|\log(1 + f_n(x))|\leq\frac12$ for $n\geq n_0$ and $x\in X$.

$|f_n(x)| \leq \frac32|\log(1 + f_n(x))|$
$\displaystyle \sum_{n \mathop = n_0}^\infty f_n$

$\Box$