counter-example: aboslute convergence => convergence in incomplete vector space

Question

Is the following statement true?

Let $X$ be a normed linear space, $x_k \in X$, $k \in \mathbb{N}$ and $\sum_{k=0}^\infty \lVert x_k\rVert$ convergent. Then $\sum_{k=0}^\infty x_k$ is also convergent.

I dont think so (for the incomplete case). However, I could not find a counterexample.

Thanks for your help!

Answer 1

It is slightly more interesting to show that for any incomplete space, there is a divergent series $\sum x_n$ with $\sum\|x_n\|<\infty$.

To see this, start with a non-convergent Cauchy sequence $(u_n)$. For each natural number $k$, pick $n_k$ so that $i,j\ge n_k$ implies $\|u_i-u_j\|<2^{-k}$. Also, make sure you pick them so $n_{k+1}>n_k$ for every $k$.

Now let $x_1=u_{n_1}$, and $x_k=u_{n_{k+1}}-u_{n_k}$ for $k\ge2$. Then $\|x_k\|<2^{-k}$, so $\sum\|x_n\|<\infty$. And yet $\sum_{j=1}^kx_j=u_{n_{k+1}}$, and that does not have a limit as $k\to\infty$.

(You need the fact that even in a non-complete space, if a Cauchy sequence has any convergent subsequence then the total sequence is convergent as well.)

Answer 2

Let

$$X:=\left\{\;\{x_n\}_{n\in\Bbb N}\subset\Bbb R\;;\; \exists N\in\Bbb N\;\;s.t.\;\;x_n=0\;\;\forall n>N\;\right\}$$

The above is just the long way to say $\;X\;$ is the vector space of all real sequences with finite support, and this is normed space wrt

$$||\{x_n\}||=\max_{n\in\Bbb N}|x_n|$$

Take now the sequence

$$\left\{h_m:=\left(0,0,...,\overbrace{\frac1{m^2}}^{m-th\;\;\text{place}},0,\ldots\right)\right\}_{m\in\Bbb N}$$

and observe that

$$\sum_{n=1}^\infty ||h_n||=\sum_{n=1}^\infty\frac1{n^2}$$

converges.

Now show that for any element $\;x\in X\; ,\;\;\; h_n\rlap{\;\;\;/}\longrightarrow x\;$