Uniform Absolute Convergence of Infinite Product of Complex Functions

Theorem
Let $X$ be a compact metric space.

Let $\left\langle{f_n}\right\rangle$ be a sequence of continuous functions $X \to \C$.

Let $\displaystyle \sum_{n \mathop = 1}^\infty f_n$ converge uniformly absolutely on $X$.

Then:
 * $f \left({x}\right) = \displaystyle \prod_{n \mathop = 1}^\infty \left({1 + f_n \left({x}\right)}\right)$ converges uniformly absolutely on $X$
 * $f$ is continuous
 * there exists $n_0 \in \N$ such that $\displaystyle \prod_{n \mathop = n_0}^\infty \left({1 + f_n \left({x}\right)}\right)$ has no zeroes.

Lemma
Because $\displaystyle \sum_{n \mathop = 1}^\infty f_n$ converges uniformly, there exists $n_0>0$ such that $|f_n(x)|<\frac12$ for $n\geq n_0$ and $x\in X$.

Then $\Re(1+f_n(x))>0$.

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

for $n\geq n_0$.

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

converge uniformly.

By Uniform Limit Theorem, $g$ and $h$ are continuous.

By Continuous Function on Compact Space is Bounded, $g$ and $h$ are bounded

By Exponential of Series Equals Infinite Product:
 * $\exp(g) = \displaystyle \prod_{n \mathop = n_0}^\infty (1+f_n)$
 * $\exp(h) = \displaystyle \prod_{n \mathop = n_0}^\infty (1+|f_n|)$

By the lemma, the products converge uniformly.

By Uniform Limit Theorem, $\exp(g)$ is continuous.

Because the second product converges uniformly, the first converges uniformly absolutely.

By Image of Complex Exponential Function, $\exp(g)$ has no zeroes.

Also see

 * Infinite Product of Analytic Functions