2
$\begingroup$

This question already has an answer here:

Let $A$ be a normed space where every absolutely convergent series converges in $A$.

How do I prove that $A$ is Banach?

Let $\sum_{n=1}^\infty x_n$ be an absolutely convergent series in $A$, then we know that $\sum_{n=1}^\infty ||x_n||$ is convergent in $\mathbb{R}$ and that $\sum_{n=1}^\infty x_n$ is convergent in $A$.
Now using this, we want to show that all Cauchy sequences $\{x_n\}$ in $A$ converge. How would I do this?

|cite|improve this question
$\endgroup$

marked as duplicate by Watson, Stefan Mesken, Martin Sleziak, John B, Stahl Dec 19 '16 at 21:58

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

3
$\begingroup$

Let $(x_n)_{n\geq 1}$ be Cauchy in $A$. We may extract a subsequence $x_{n_k}$ for which $\|x_{n_{k+1}}-x_{n_{k}}\|\leq 2^{-k}$ for $k\geq 1$. Then $$ \sum_{k\geq 1} \|x_{n_{k+1}}-x_{n_{k}}\| \leq \sum_{k\geq 1} 2^{-k} =1$$ implies by the hypothesis on $A$ that the limit $$y= x_{n_1} + \lim_{p\rightarrow \infty}\sum_{k=1}^p (x_{n_{k+1}} - x_{n_k})\in A$$ exists (as the sum is abs convergent). We have $\|y-x_{n_k}\|\leq 2^{1-k}$ so $x_{n_k}$ converges to $y$ in $A$ and then so does $x_n$.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Do you mean $y_1=x_{n_1}$ and $y_{k+1} = x_{n_{k+1}} − x_{n_{k}}$? $\endgroup$ – user368886 Oct 2 '16 at 12:28
  • $\begingroup$ Yes, then the sum of $y_k$ will be convergent by hypothesis on $A$ and you just have to show that the limit is indeed the limit of $x_n$ as well. It is a standard trick to show e.g. that $L^1$ is complete. $\endgroup$ – H. H. Rugh Oct 2 '16 at 13:14
  • 1
    $\begingroup$ Would you mind completing the proof, if you have the time? I can't really see it $\endgroup$ – user368886 Oct 2 '16 at 13:33