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 $f_n: X \to \C$.

Then the following definitions of uniform absolute convergence of a product are equivalent:

2 implies 1
By Terms in Uniformly Convergent Series Converge Uniformly to Zero

By the Monotone Convergence Theorem:
 * The sequence of partial products of $\displaystyle \prod_{n \mathop = n_0}^\infty \left({1 + \left\vert{f_n}\right\vert}\right)$ 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 $|f_n(x)|\leq\frac12$ for $n\geq n_0$.

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

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

for $n\geq n_0$.

By Comparison Test for Uniformly Convergent Series,
 * $\displaystyle \sum_{n \mathop = n_0}^\infty \log(1+f_n)$

converges uniformly absolutely.

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

By Comparison Test for Uniformly Convergent Series:
 * $\displaystyle \sum_{n \mathop = n_0}^\infty \log(1 + |f_n(x)|)$ and $\displaystyle \sum_{n \mathop = n_0}^\infty |f_n(x)|$

converge uniformly.

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

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 $x\in X$.

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

By Comparison Test for Uniformly Convergent Series,
 * $\displaystyle \sum_{n \mathop = n_0}^\infty f_n$

converge uniformly absolutely.