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If $X$ is a Banach space, then the following theorem holds:

Let $\sum x_n$ be a series in $X$ which converges absolutely. Then every rearrangement $\sum x_{\sigma(n)}$ converges, and they all converge to the same value.

Proof: Let $(s_n')$ be the sequence of the partial sums of $\sum x_{\sigma(n)}$. Since $\sum x_n$ is absolutely convergent, given $\epsilon>0$ there is an integer $n_0$ such that $$\sum_{k=n_0}^m ||x_k|| <\epsilon$$ for all $m\geq n_0$. Let $$p=\max_{1\leq i< n_0}\sigma^{-1}(i).$$ If $n>p$, we have that $\{1,2,\dotsc,n_0-1\}$ is a subset of $\{\sigma(1),\sigma(2),\dotsc,\sigma(n)\}$. Hence all the $x_i$ for $i=1,2,\dots,n_0-1$ are cancelled in $s_n-s_n'$. So, $$||s_n-s_n'||\leq \sum_{k=n_0}^m ||x_k|| <\epsilon.$$ We conclude that $(s_n')$ converges to the same value as $(s_n)$.

However, if $X$ is a normed vector space which is not necessarily Banach, is any of the following true?

If $\sum x_n$ converges absolutely, then any rearrangement converges absolutely.

or

If $\sum x_n$ is a convergent series which converges absolutely, then any rearrangement converges absolutely.

or

If $\sum x_n$ is a convergent series which converges absolutely, then any rearrangement converges.

If so, it's possible to modify my proof to handle the more general result?

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Gabriel
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1 Answers1

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Every normed space $X$ can be thought of as a subspace of $\widetilde{X}$, the completion of $X$, which is Banach. All your propositions follow immediately.

Edit: Here is the more detailed version. Let $X$ be a normed space. In particular, $X$ is a metric space, and so, it has a metric completion $\widetilde{X}$, which is a Banach space. (The norm on $\widetilde{X}$ is defined to be the unique continuous extension of the one on $X$). Now, let $\sum x_n$ be an absolutely convergent series in $X$. In particular, it is in $\widetilde{X}$, which is Banach. It follows that every reordering of $\sum x_n$ converges to the same limit. We're done.

Amitai Yuval
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  • As I am starting to study normed spaces, I don't see why the propositions follow from this. Can you clarify it a bit? Isn't a simple change in my proof is enough to prove these results? – Gabriel May 11 '17 at 11:58
  • @GabrielRibeiro I edited my answer, have a look. – Amitai Yuval May 11 '17 at 12:49
  • Ok, I get that $\sum x_n$ is absolutely convergent in $\widetilde{X}$ and hence convergent in $\widetilde{X}$. This implies that every reordering of $\sum x_n$ converges to the same limit in $\widetilde{X}$. Does this imply that this reordering is absolutely convergent in $X$ or even convergent? Why? – Gabriel May 11 '17 at 12:54
  • @GabrielRibeiro To begin with, the limit $S$ of $\sum x_n$ in $\widetilde{X}$ is actually an element of $X$. Every reordering converges to the same limit $S$, which is still in $X$. – Amitai Yuval May 11 '17 at 13:01
  • But then why couldn't we take $\sigma(n)=n$ and conclude that every reordering of a absolutely convergent series in a normed space converges? (Which is not true if $X$ is not Banach, right?) – Gabriel May 11 '17 at 13:04
  • @GabrielRibeiro I see now. You and I use different terminology. I guess then that, in your language, your first statement is false and the other two are true. – Amitai Yuval May 11 '17 at 13:09