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 ${\sum_{n = 1}^{m} v_{n}}_{m}$ 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 $k \in N$ , there exists $N_{k} \in N$
such that for all $n, m \geq N_{k}$ , 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} + \dots + N_{k}

Then $n_{1} < n_{2} < \dots$ and for all $k$ , we have $n_{k} \geq N_{k}$ . Thus, for all $k \in N$ , we have

| | v_{n_{k + 1}} - v_{n_{k}} | | < \frac{1}{2^{k}}

Thus

\begin{aligned} \sum_{k} (v_{n_{k + 1}} - v_{n_{k}}) is absolutely summable \\ ⟹ & \sum_{k} (v_{n_{k + 1}} - v_{n_{k}}) is summable \\ ⟹ & {\sum_{k = 1}^{m} (v_{n_{k + 1}} - v_{n_{k}})}_{m = 1}^{\infty} = {v_{n_{m + 1} - v_{n_{1}}}}_{m = 1}^{\infty} converges \end{aligned}

Thus the (sub)sequence is ${v_{n_{m + 1}}}_{m = 1}^{\infty}$ converges in $V$ . Thus $V$ is Banach. $◼$

Mentions

File	Last Modified
bounded linear operator space is banach	2025-10-02
Functional Analysis Lecture 13	2025-10-02
Functional Analysis Lecture 2	2025-10-02

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