I'm considering a possible alternate proof of an elementary theorem in functional analysis:

Let $(X, \|\cdot\|)$ be a normed space over $\mathbb{R}$ or $\mathbb{C}$. $X$ is Banach if and only if every absolutely convergent series in $X$ converges in $X$.

  1. Say $X$ is not Banach. That means there exists a non-convergent Cauchy sequence $(x_n)_{n=1}^\infty$. Is there a way to explicitly construct a non-convergent but absolutely convergent series in $X$ using the Cauchy series $(x_n)_{n=1}^\infty$?
  2. Conversely, say we have an absolutely convergent, but non-convergent series $\sum_{n=1}^\infty y_n$ in $X$. Is there a simple way to get one explicit non-convergent Cauchy series using this series?

My attempt:

  1. Let's try with defining a sequence $(z_n)_{n=1}^\infty$ with: \[z_1=x_1\] \[z_n=x_n-x_{n-1}, \forall n \geq 2\] Notice that $\sum_{n=1}^N z_n = x_N$ so $\sum_{n=1}^\infty z_n$ does not converge. But is it absolutely convergent? I have no idea.
  2. Consider the sequence $\left(\sum_{n=1}^N y_n\right)_{N=1}^\infty$. This sequence does not converge. We show that it is Cauchy: let $M \geq N \in \mathbb{N}$ \[\left\Vert\sum_{n=1}^M y_n - \sum_{n=1}^N y_n\right\Vert = \left\Vert\sum_{n=N+1}^M y_n\right\Vert \leq \sum_{n=N+1}^M \left\Vert y_n \right\Vert \leq \sum_{n=N+1}^\infty \left\Vert y_n \right\Vert \xrightarrow{M, N \to \infty} 0\] It seems that it is Cauchy. Is this correct?

1 Answer 1

  1. is correct. ,1. Is not.

In case of one, you first need to extract a subsequence out of your Cauchy sequence, or you might end up with something that does not work.

Using the fact that $x_n$ is Cauchy, you extract a subsequence $x_{\phi (n)}$, which we shall call $y_n$, such that $||y_{n+1}-y_n||\leq \frac{1}{2^n}$ (or any other positive sequence whose series converges).

This new sequence still does not converge, since a Cauchy sequence that has a partial limit is convergent. And we have $y_{n+1}-y_0=\sum \limits_{k=0}^n y_{k+1}-y_{k}$, so the rest works just like you described it.


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