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 $f_n: X \to \C$.
The following definitions of the concept of Uniform Absolute Convergence of Product are equivalent:
Definition 1
The infinite product $\ds \prod_{n \mathop = 1}^\infty \paren {1 + f_n}$ converges uniformly absolutely if and only if the sequence of partial products of $\ds \prod_{n \mathop = 1}^\infty \paren {1 + \size {f_n} }$ converges uniformly.
Definition 2
The infinite product $\ds \prod_{n \mathop = 1}^\infty \paren {1 + f_n}$ converges uniformly absolutely if and only if the series $\ds \sum_{n \mathop = 1}^\infty f_n$ converges uniformly absolutely.
Definition 3
The infinite product $\ds \prod_{n \mathop = 1}^\infty \paren {1 + f_n}$ converges uniformly absolutely if and only if there exists $n_0 \in \N$ such that:
- $(1): \quad \map {f_n} x \ne -1$ for $n \ge n_0$ and $x \in X$
and:
- $(2): \quad$ The series $\ds \sum_{n \mathop = n_0}^\infty \map \log {1 + f_n}$ is uniformly absolutely convergent.
Proof
1 implies 2
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2 implies 1
By Terms in Uniformly Convergent Series Converge Uniformly to Zero
By the Monotone Convergence Theorem:
- The sequence of partial products of $\ds \prod_{n \mathop = n_0}^\infty \paren {1 + \size {f_n} }$ converges uniformly if ...
Use Bounds for Finite Product of Real Numbers.
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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,
- $\ds \sum_{n \mathop = n_0}^\infty \ln {1 + \map {f_n} x}$
converges uniformly absolutely.
$\Box$
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:
- $\ds \sum_{n \mathop = n_0}^\infty \map \ln {1 + \size {\map {f_n} x} }$ and $\ds \sum_{n \mathop = n_0}^\infty \size {\map {f_n} x}$
By Bounds for Finite Product of Real Numbers the sequence of partial products of $\ds \prod_{n \mathop = n_0}^\infty \paren {1 + \size {f_n} }$ is bounded.
By Complex Exponential is Uniformly Continuous on Half-Planes, the sequence of partial products of $\ds \prod_{n \mathop = n_0}^\infty \paren {1 + \size {f_n} }$ converges uniformly.
By:
- Uniformly Convergent Sequence Multiplied with Function
- $f_1, \ldots, f_{n_0 - 1}$ are bounded
the sequence of partial products of $\ds \prod_{n \mathop = 1}^\infty \paren {1 + \size {f_n} }$ converges uniformly.
$\Box$
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:
- $\ds \sum_{n \mathop = n_0}^\infty f_n$
converges uniformly absolutely.
$\Box$
Sources
- 1973: John B. Conway: Functions of One Complex Variable ... (previous) ... (next) $\text {VII}$: Compact and Convergence in the Space of Analytic Functions: $\S 5$: Weierstrass Factorization Theorem: Lemma $5.8$