Absolute Net Convergence Equivalent to Absolute Convergence

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Theorem

Let $V$ be a Banach space.

Let $\sequence {v_n}_{n \mathop \in \N}$ be a sequence of elements in $V$.

Let $r \in \R_{\mathop \ge 0}$


Then the following two statements are equivalent:

$(1): \quad$ the generalized sum $\ds \sum \set {v_n: n \in \N}$ is absolutely net convergent to $r$
$(2): \quad$ the series $\ds \sum_{n \mathop = 1}^\infty v_n$ is absolutely convergent to $r$


Proof

Statement $(1)$ implies Statement $(2)$

Let the generalized sum $\ds \sum \set {v_n: n \in \N}$ be absolutely net convergent to $r$.


Let $\epsilon \in \R_{\mathop > 0}$.


From Characterization of Convergent Net in Metric Space:

$(1) \quad \exists F \subset \N: F $ is finite $: \forall E \subseteq \N : E \supseteq F: E$ is finite $\implies \size{\ds \sum_{n \mathop \in E} \norm{v_n} - r} < \epsilon$


Let $N = \max \set{n : v_n \in F}$.

We have:

$F \subseteq \closedint 0 N$


Let $m \ge N$.

We have:

$\closedint 0 m \supseteq \closedint 0 N \supseteq F$


From $(1)$:

$\size{\ds \sum_{n \mathop \in \closedint 0 m} \norm{v_n} - r} < \epsilon$

By definition of summation over finite index:

$\size{\ds \sum_{n \mathop = 0}^m \norm{v_n} - r} < \epsilon$


Since $m \ge N$ was arbitrary, it follows that:

$\forall m \ge N : \size{\ds \sum_{n \mathop = 0}^m \norm{v_n} - r} < \epsilon$


Since $\epsilon$ was arbitrary, it follows that the series $\ds \sum_{n \mathop = 1}^\infty v_n$ is absolutely convergent to $r$ by definition.

$\Box$


Statement $(2)$ implies Statement $(1)$

Let the series $\ds \sum_{n \mathop = 1}^\infty v_n$ be absolutely convergent to $r$.


Let $\epsilon \in \R_{\mathop \ge 0}$.


By definition of absolutely convergent:

$(2) \quad \exists N \in \N : \forall m \ge N : \size{\ds \sum_{n \mathop = 0}^m \norm{v_m} - r} < \dfrac \epsilon 3$


Let:

$F = \closedint 0 N$


Let:

$E \subseteq \N : E \supseteq F : E$ is finite.


Let:

$m = \max \set{n : n \in E}$

Let:

$G = \closedint 0 m$


We have:

$F = \closedint 0 N \subseteq E \subseteq \closedint 0 m = G$


From Set Difference and Intersection form Partition:

$E = F \cup E \setminus F$

and

$G = F \cup G \setminus F$


From Set Difference Intersection with Second Set is Empty Set:

$F \cap E \setminus F = \O$

and

$F \cap G \setminus F = \O$


From Set Difference over Subset:

$E \setminus F \subseteq G \setminus F$


We have:

\(\ds \size {\sum_{n \mathop \in E} \norm {v_n} - r}\) \(=\) \(\ds \size{\sum_{n \mathop \in F} \norm {v_n} + \sum_{n \mathop \in E \setminus F} \norm {v_n} - r }\) Summation over Union of Disjoint Finite Index Sets
\(\ds \) \(\le\) \(\ds \size{\sum_{n \mathop \in F} \norm {v_n} - r } + \size{ \sum_{n \mathop \in E \setminus F} \norm {v_n} }\) Triangle Inequality for Real Numbers
\(\ds \) \(\le\) \(\ds \size{\sum_{n \mathop \in F} \norm {v_n} - r } + \size{ \sum_{n \mathop \in G \setminus F} \norm {v_n} }\)
\(\ds \) \(=\) \(\ds \size{\sum_{n \mathop \in F} \norm {v_n} - r } + \size{ \sum_{n \mathop \in G} \norm {v_n} - \sum_{n \mathop \in F} \norm {v_n} }\) Summation over Union of Disjoint Finite Index Sets
\(\ds \) \(=\) \(\ds \size{\sum_{n \mathop \in F} \norm {v_n} - r } + \size{ \sum_{n \mathop \in G} \norm {v_n} -r + r - \sum_{n \mathop \in F} \norm {v_n} }\)
\(\ds \) \(\le\) \(\ds \size{\sum_{n \mathop \in F} \norm {v_n} - r } + \size{ \sum_{n \mathop \in G} \norm {v_n} -r } + \size{ r - \sum_{n \mathop \in F} \norm {v_n} }\) Triangle Inequality for Real Numbers
\(\ds \) \(=\) \(\ds \size{\sum_{n \mathop = 0}^N \norm {v_n} - r } + \size{ \sum_{n \mathop = 0}^m \norm {v_n} -r } + \size{ r - \sum_{n \mathop = 0}^N \norm {v_n} }\) Definition of Summation over Finite Index
\(\ds \) \(<\) \(\ds \dfrac \epsilon 3 + \dfrac \epsilon 3 + \dfrac \epsilon 3\) from $(2)$
\(\ds \) \(=\) \(\ds \epsilon\)


Since $E$ was arbitrary, it follows:

$\exists F \subset \N: F $ is finite $: \forall E \subseteq \N : E \supseteq F: E$ is finite $\leadsto \size{\ds \sum_{n \mathop \in E} \norm{v_n} - r} < \epsilon$


Sine $\epsilon$ was arbitrary, it follows:

$\forall \epsilon \in \R_{\mathop > 0} : \exists F \subset \N: F $ is finite $: \forall E \subseteq \N : E \supseteq F: E$ is finite $\leadsto \size{\ds \sum_{n \mathop \in E} \norm{v_n} - r} < \epsilon$


From Characterization of Convergent Net in Metric Space:

the generalized sum $\ds \sum \set {v_n: n \in \N}$ is absolutely net convergent to $r$.

$\blacksquare$

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