banach spaces have all absolutely summable series are summable

[[concept]]

Theorem

A normed vector space $V$ is a Banach space if and only if every absolutely summable series is summable.

Proof

$(⟹)$ Suppose $V$ is a Banach space. If ${\sum }_n{v}_n$ is absolutely summable, then the sequence of partial sums is a Cauchy sequence in $V$ (see the previous theorem). Thus ${\left\{{\sum }_{n=1}^m{v}_n\right\}}_m$ converges in $V$. Thus by definition, the series is summable.

$(⟸)$ Suppose every absolutely summable series is summable. Let $\{v_n\}$ be a Cauchy sequence in $V$. We show that a subsequence of $\{v_n\}$ converges in $V$. Then $\{v_n{\}}_m$ converges by metric space theory.

$\{v_n{\}}_m$ is Cauchy $⟹$ for all $k\in \mathbb{N}$, there exists $N_k\in \mathbb{N}$ such that for all $n,m\ge N_k$, we have $||v_n-v_m||<\frac{1}{2^k}$ (we choose this expression because it is summable). Now, define $n_k=N_1+N_2+\cdots +N_k$ Then $n_1<n_2<\dots$ and for all $k$, we have $n_k\ge N_k$. Thus, for all $k\in \mathbb{N}$, we have $||v_{n_{k+1}}-v_{n_k}||<\frac{1}{2^k}$ Thus

&\sum_{k} (v_{n_{k+1}} - v_{n_{k}}) \text{ is absolutely summable} \\ \implies &\sum_{k}(v_{n_{k+1}} - v_{n_{k}}) \text{ is summable} \\ \implies &\left\{ \sum_{k=1}^m (v_{n_{k+1}} - v_{n_{k}}) \right\}_{m=1}^\infty = \{ v_{n_{m+1} - v_{n_{1}}} \}_{m=1}^\infty \text{ converges} \end{align}$$ Thus the (sub)sequence is $\{ v_{n_{m+1}} \}_{m=1}^\infty$ converges in $V$. Thus $V$ is [[Concept Wiki/Banach space\|Banach]]. $\blacksquare$

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