Uniform Absolute Convergence of Infinite Product of Complex Functions

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Theorem

Let $X$ be a compact metric space.

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

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

Then:

$\map f x = \ds \prod_{n \mathop = 1}^\infty \paren {1 + \map {f_n} x}$ converges uniformly absolutely on $X$
$f$ is continuous
there exists $n_0 \in \N$ such that $\ds \prod_{n \mathop = n_0}^\infty \paren {1 + \map {f_n} x}$ has no zeroes.

Proof

Lemma

Let $X$ be a set.

Let $\family {g_n}$ be a family of mappings $g_n : X \to \C$.

Let $g_n$ converge uniformly to $g: X \to \C$.

Let there be a constant $a \in \R$ such that $\map \Re {\map g x} \le a$ for all $x \in X$.

Then $\exp g_n$ converges uniformly to $\exp g$.

$\Box$

Because $\ds \sum_{n \mathop = 1}^\infty f_n$ converges uniformly, there exists $n_0 > 0$ such that $\size {\map {f_n} x} < \dfrac 1 2$ for $n \ge n_0$ and $x \in X$.

Then:

$\map \Re {1 + \map {f_n} x} > 0$
$\size {\map \log {1 + \map {f_n} x} } \le \dfrac 3 2 \size {\map {f_n} x}$
$\size {\map \log {1 + \size {\map {f_n} x} } } \le \dfrac 3 2 \size {\map {f_n} x}$

for $n \ge n_0$.

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

and:

$h = \ds \sum_{n \mathop = n_0}^\infty \map \log {1 + \size {f_n} }$

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

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

$\map \exp g = \ds \prod_{n \mathop = n_0}^\infty \paren {1 + f_n}$
$\map \exp h = \ds \prod_{n \mathop = n_0}^\infty \paren {1 + \size {f_n} }$

By the lemma, the products converge uniformly.

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

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

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

$\blacksquare$