Absolutely Convergent Generalized Sum Converges

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Let $V$ be a Banach space.

Let $\left\Vert{\cdot}\right\Vert$ denote the norm on $V$.

Let $d$ denote the corresponding induced metric.

Let $\left({v_i}\right)_{i \mathop \in I}$ be an indexed subset of $V$ such that the generalized sum $\displaystyle \sum_{i \mathop \in I} \left\{{v_i}\right\}$ converges absolutely.

Then the generalized sum $\displaystyle \sum \left\{{v_i: i \in I}\right\}$ converges.


The proof proceeds in two stages:

$(1):\quad$ Finding a candidate $v \in V$ where the sum might converge to
$(2):\quad$ Showing that the candidate is indeed sought limit.

That $\displaystyle \sum \left\{{v_i: i \mathop \in I}\right\}$ converges absolutely means that $\displaystyle \sum \left\{{\left\Vert{v_i}\right\Vert: i \mathop \in I}\right\}$ converges.

Now, for all $n \in \N$, let $F_n \subseteq I$ be finite such that:

$\displaystyle \sum_{i \mathop \in G} \left\Vert{v_i}\right\Vert > \sum \left\{{\left\Vert{v_i}\right\Vert: i \mathop \in I}\right\} - 2^{-n}$, for all finite $G$ with $F_n \subseteq G \subseteq I$

It may be arranged that $n \ge m \implies F_m \subseteq F_n$ by passing over to $\displaystyle F'_n = \bigcup_{m \mathop = 1}^n F_m$ if necessary.


$\displaystyle v_n = \sum_{i \mathop \in F_n} v_i$

Next, it is to be shown that the sequence $\left({v_n}\right)_{n \mathop \in \N}$ is Cauchy.

So let $\epsilon > 0$, and let $N \in \N$ be such that $2^{-N} < \epsilon$.

Then for $m \ge n \ge N$, have:

\(\displaystyle d \left({v_m, v_n}\right)\) \(=\) \(\displaystyle \left\Vert{\left({\sum_{i \mathop \in F_m} v_i}\right) - \left({\sum_{i \mathop \in F_n} v_i}\right)}\right\Vert\) $\quad$ Definition of Induced Metric $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left\Vert{\sum_{i \mathop \in F_m \setminus F_n} v_i}\right\Vert\) $\quad$ $F_n \subseteq F_m$ $\quad$
\(\displaystyle \) \(\le\) \(\displaystyle \sum_{i \mathop \in F_m \setminus F_n} \left\Vert{v_i}\right\Vert\) $\quad$ Triangle Inequality for $\left\Vert{\cdot}\right\Vert$ $\quad$

Now to estimate this last quantity, observe:

\(\displaystyle \sum \left\{ {\left\Vert{v_i}\right\Vert: i \mathop \in I}\right\} - 2^{-n} + \sum_{i \mathop \in F_m \setminus F_n} \left\Vert{v_i}\right\Vert\) \(<\) \(\displaystyle \sum_{i \mathop \in F_n} \left\Vert{v_i}\right\Vert + \sum_{i \mathop \in F_m \setminus F_n} \left\Vert{v_i}\right\Vert\) $\quad$ Defining property of $F_n$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{i \mathop \in F_m} \left\Vert{v_i}\right\Vert\) $\quad$ Union with Relative Complement, $F_n \subseteq F_m$ $\quad$
\(\displaystyle \) \(\le\) \(\displaystyle \sum \left\{ {\left\Vert{v_i}\right\Vert: i \mathop \in I}\right\}\) $\quad$ Generalized Sum is Monotone $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle \sum_{i \mathop \in F_m \setminus F_n} \left\Vert{v_i}\right\Vert\) \(<\) \(\displaystyle 2^{-n}\) $\quad$ $\quad$

Finally, by the defining property of $N$, as $n \ge N: : 2^{-n} < 2^{-N} < \epsilon$

Combining all of these estimates leads to the conclusion that:

$d \left({v_m, v_n}\right) < \epsilon$

It follows that $\left({v_n}\right)_{n \in \N}$ is a Cauchy sequence.

As $V$ is a Banach space:

$\displaystyle \exists v \in V: \lim_{n \to \infty} v_n = v$

Having identified a candidate $v$ for the sum $\displaystyle \sum \left\{{v_i: i \in I}\right\}$ to converge to, it remains to verify that this is indeed the case.

According to the definition of considered sum, the convergence is convergence of a net.

Next, Metric Induces Topology ensures that we can limit the choice of opens $U$ containing $v$ to open balls centered at $v$.

Now let $\epsilon > 0$.

We want to find a finite $F \subseteq I$ such that:

$d \left({\displaystyle \sum_{i \mathop \in G} v_i, v}\right) < \epsilon$

for all finite $G$ with $F \subseteq G \subseteq I$.

Now let $N \in \N$ such that:

$\forall n \ge N: d \left({v_n, v}\right) < \dfrac \epsilon 2$

with the $v_n$ as above.

By taking a larger $N$ if necessary, ensure that $2^{-N} < \dfrac \epsilon 2$ holds as well.

Let us verify that the set $F_N$ defined above has sought properties.

So let $G$ be finite with $F_N \subseteq G \subseteq I$.


\(\displaystyle d \left({\sum_{i \mathop \in G} v_i, v}\right)\) \(=\) \(\displaystyle \left\Vert{\left({\sum_{i \mathop \in G} v_i}\right) - v}\right\Vert\) $\quad$ Definition of Induced Metric $\quad$
\(\displaystyle \) \(\le\) \(\displaystyle \left\Vert{\left({\sum_{i \mathop \in G} v_i}\right) - \left({\sum_{i \mathop \in F_N} v_i}\right)}\right\Vert + \left\Vert{\left({\sum_{i \mathop \in F_N} v_i}\right) - v}\right\Vert\) $\quad$ Triangle inequality for $\left\Vert{\cdot}\right\Vert$ $\quad$
\(\displaystyle \) \(<\) \(\displaystyle \left\Vert{\sum_{i \mathop \in G \setminus F_N} v_i}\right\Vert + \frac \epsilon 2\) $\quad$ $F_N \subseteq G$, defining property of $N$ $\quad$
\(\displaystyle \) \(\le\) \(\displaystyle \sum_{i \mathop \in G \setminus F_N} \left\Vert{v_i}\right\Vert + \frac \epsilon 2\) $\quad$ Triangle inequality for $\left\Vert{\cdot}\right\Vert$ $\quad$

For the first of these terms, observe:

\(\displaystyle \sum \left\{ {\left\Vert{v_i}\right\Vert: i \in I}\right\} - 2^{-N} + \sum_{i \mathop \in G \setminus F_N} \left\Vert{v_i}\right\Vert\) \(<\) \(\displaystyle \sum_{i \mathop \in F_N} \left\Vert{v_i}\right\Vert + \sum_{i \mathop \in G \setminus F_N} \left\Vert{v_i}\right\Vert\) $\quad$ Defining property of $F_N$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{i \mathop \in G} \left\Vert{v_i}\right\Vert\) $\quad$ Union with Relative Complement, $F_N \subseteq G$ $\quad$
\(\displaystyle \) \(\le\) \(\displaystyle \sum \left\{ {\left\Vert{v_i}\right\Vert: i \in I}\right\}\) $\quad$ Generalized Sum is Monotone $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle \sum_{i \mathop \in G \setminus F_N} \left\Vert{v_i}\right\Vert\) \(<\) \(\displaystyle 2^{-N}\) $\quad$ $\quad$

Using that $2^{-N} < \dfrac \epsilon 2$, combine these inequalities to obtain:

$\displaystyle d \left({\sum_{i \mathop \in G} v_i, v}\right) < \frac \epsilon 2 + \frac \epsilon 2 = \epsilon$

By definition of convergence of a net, it follows that:

$\displaystyle \sum \left\{{v_i: i \in I}\right\} = v$


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