Triangle Inequality for Generalized Sums

Theorem
Let $V$ be a Banach space, and let $\left\Vert{\cdot}\right\Vert$ denote its norm.

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

Then $\displaystyle \left\Vert{\sum \left\{{v_i: i \in I}\right\}}\right\Vert \le \sum \left\{{\left\Vert{v_i}\right\Vert: i \in I}\right\}$.

Proof
First of all, note that Absolutely Convergent Generalized Sum Converges assures us that the left expression in above equality is defined.

Suppose that there exists an $\epsilon > 0$ such that $\displaystyle \left\Vert{\sum \left\{{v_i: i \in I}\right\}}\right\Vert > \sum \left\{{\left\Vert{v_i}\right\Vert: i \in I}\right\} + \epsilon$.

This supposition is seen to be equivalent to $\displaystyle \left\Vert{\sum \left\{{v_i: i \in I}\right\}}\right\Vert > \sum \left\{{\left\Vert{v_i}\right\Vert: i \in I}\right\}$.

Then, by definition of a generalized sum, there necessarily exists a finite subset $F$ of $I$ with:


 * $\displaystyle \left\Vert{\sum_{i \in F} v_i}\right\Vert > \left\Vert{\sum \left\{{v_i: i \in I}\right\}}\right\Vert - \epsilon > \sum \left\{{\left\Vert{v_i}\right\Vert: i \in I}\right\}$

However, using the standard triangle equality on this finite sum (i.e., axiom N3 for a norm, repetitively), we also have:


 * $\displaystyle \left\Vert{\sum_{i \in F} v_i}\right\Vert \le \sum_{i \in F} \left\Vert{v_i}\right\Vert \le \sum \left\{{\left\Vert{v_i}\right\Vert: i \in I}\right\}$

Here the second inequality follows from Generalized Sum is Monotone.

These two estimates constitute a contradiction, and therefore such an $\epsilon$ cannot exist.

Hence $\displaystyle \left\Vert{\sum \left\{{v_i: i \in I}\right\}}\right\Vert \le \sum \left\{{\left\Vert{v_i}\right\Vert: i \in I}\right\}$.