Equivalence of Definitions of Uniform Absolute Convergence of Product of Complex Functions

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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 $\displaystyle \prod_{n \mathop = 1}^\infty \left({1 + f_n}\right)$ converges uniformly absolutely if and only if the sequence of partial products of $\displaystyle \prod_{n \mathop = 1}^\infty \left({1 + \left\vert{f_n}\right\vert}\right)$ converges uniformly.

Definition 2

The infinite product $\displaystyle \prod_{n \mathop = 1}^\infty \left({1 + f_n}\right)$ converges uniformly absolutely if and only if the series $\displaystyle \sum_{n \mathop = 1}^\infty f_n$ converges uniformly absolutely.

Definition 3

The infinite product $\displaystyle \prod_{n \mathop = 1}^\infty \left({1 + f_n}\right)$ converges uniformly absolutely if and only if there exists $n_0 \in \N$ such that:

$(1): \quad f_n \left({x}\right) \ne -1$ for $n \ge n_0$ and $x \in X$

and:

$(2): \quad$ The series $\displaystyle \sum_{n \mathop = n_0}^\infty \log \left({1 + f_n}\right)$ is uniformly absolutely convergent.


Proof

1 implies 2


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.

$\Box$


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:

$\displaystyle \sum_{n \mathop = n_0}^\infty \map \ln {1 + \size {\map {f_n} x} }$ 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.

$\Box$


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:

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

converges uniformly absolutely.

$\Box$


Sources