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 $f_n: X \to \C$.

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.

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 $\displaystyle \sum_{n \mathop = n_0}^\infty \size {\map {f_n} x}$

converge uniformly.

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.

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.