Show that

A Normed Linear Space $X$ is a Banach Space iff every absolutely convergent series is convergent.

My try:

Let $X$ is a Banach Space .Let $\sum x_n$ be an absolutely convergent series .Consider $s_n=\sum_{i=1}^nx_i$. Now $\sum \|x_n\|<\infty \implies \exists N$ such that $\sum_{i=N}^ \infty \|x_i\|<\epsilon$ for any $\epsilon>0$

Then $\|s_n-s_m\|\le \sum _{i=m+1}^n \|x_i\|<\epsilon \forall n,m>N$

So $s_n$ is Cauchy in $X$ and hence converges $s_n\to s$ (say).

Thus $\sum x_i$ converges.

Conversely, let $x_n$ be a Cauchy Sequence in $X$. Here I can't proceed how to use the given fact.

Any help will be great.

  • $\begingroup$ How can you deduce that $\exists N$ such that $\sum_{i=N}^{\infty} ||x_{i}||<\epsilon$ $\endgroup$ – Soulostar Feb 2 '18 at 7:45

For the converse argument; let $X$ be a normed linear space in which every absolutely convergent series converges, and suppose that $\{x_n\}$ is a Cauchy sequence.

For each $k \in \mathbb{N}$, choose $n_k$ such that $||(x_m − x_n)|| < 2^{−k}$ for $m, n \geq n_k$. In particular, $||x_{n_{k+1}}−x_{n_k}||< 2^{-k}$. If we define $y_{1} = x_{n_{1}}$ and $y_{k+1} = x_{n_{k+1}} − x_{n_{k}}$ for $k \geq 1$, it follows that $\sum ||y_{n}|| ≤ ||x_{n_{1}} || + 1$ i. e., ($y_{n}$) is absolutely convergent, and hence convergent.

  • 1
    $\begingroup$ Hey, I am facing the same problem and your solution was really good! However, I don't know why proving $(y_n)$ is convergent finished the problem... Now, for every $\varepsilon >0$ there exists $N\in \mathbb{N}$ such that for all $n\geq N$ we have that $\sum_{n=N}^\infty y_n <\varepsilon$, but what does conclude the proof? $\endgroup$ – user326159 Feb 21 at 17:12

Hint: show that there is a subsequence $y_n$ such that $\|y_n - y_{n+1}\| < 2^{-n}$.


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.