Unconditional and absolute convergence in non-Banach spaces

Question

I know that, by Dvoretzky-Rogers theorem, we know that in a Banach space $X$ the following are equivalent:

1. $X$ is of finite dimension.
2. Every unconditonally convergent series is absolutely convergent.

But, what if $X$ is a normed but not complete space. Is the previous theorem still true, or there are examples of infinite dimensional non-Banach spaces where every unconditionally convergent series is also absolutely convergent?

Thanks for the answers.

Answer 1

This is only a partial answer since it doesn't answer the question outright but provides a quite similar statement which is true and generalises the one in Banach spaces. I didn't want to leave it as only a comment, but I am aware that this somewhat "avoids" the question. This is not true anymore, it's an actual answer!

Firstly, a quick comment on what was my first approach: in their original paper (which I'll link here), Dvoretzky and Rogers did observe themselves that completeness of their Banach space $B$ was only used once, towards the end, hence the statement below holds, which is a consequence of their Theorem $3$ and the remark preceding it:

Let $X$ be a normed space. The following assertions are equivalent:

• $X$ is finite-dimensional;
• Every unconditionally Cauchy convergent series is absolutely convergent.

This is a simple generalisation of the original statement since unconditional Cauchy convergence is equivalent to unconditional convergence in Banach spaces. The problem is that I do not know whether their proof method can be adjusted to prove your claim.

However, the theory of nuclear spaces seems to come to the rescue: I'll refer to this answer on MathOverflow by Robert Fueber, itself based on Schaefer's Topological Vector Spaces. Disclaimer: I do not know that much myself on the theory presented, so I'll be more than happy to hear (well, I guess read) other users's opinions on my answer, whether I did put the pieces of this puzzle correctly, or if I missed or misunderstood something.

I'll quote the relevant parts of the answer for self-containment:

For $E$ a locally convex space, let $\ell^1[E]$ denote the set of absolutely summable ($\mathbb{N}$-indexed) series and $\ell^1(E)$ the set of unconditionally summable series. These spaces admit locally convex topologies such that the inclusion $\ell^1[E] \to \ell^1(E)$ is continuous. (...)
(...) It is then true that the inclusion mapping $\ell^1[E] \to \ell^1(E)$ is a linear homeomorphism iff $E$ is a nuclear space. This is actually Pietsch's sharpened version of Grothendieck's theorem.(...)
(...) If $E$ is infrabarrelled, for instance if $E$ is a normed space, then the mapping $\ell^1[E] \to \ell^1(E)$ is a linear homeomorphism iff it is surjective, so these spaces are nuclear iff unconditional and absolute summability coincide. Infinite-dimensional Banach spaces are not nuclear, so every infinite-dimensional Banach space contains an unconditionally convergent series that is not absolutely convergent. This fact was originally proven by Dvoretsky and Rogers, in quite a different way.(...)

I added the bolding to emphasize the key information: nuclear normed spaces are exactly the normed spaces where unconditional and absolute summability are one and the same, yet in Schaefer's book the definition of absolute summability already requires unconditional summability, thus this should be exactly saying that a locally convex space is nuclear iff every unconditionally convergent series is absolutely convergent, which would be exactly what we want. There only remains to see which normed spaces are nuclear.
Robert only mentions Banach spaces, but it's actually a fact that no infinite-dimensional normed space, even non-Banach, can be nuclear (see the third property listed here), hence the nuclear normed spaces are exactly the finite-dimensional normed spaces, which proves your claim, and we are done.

I don't know if it's a satisfying answer, I'd be glad to see someone with more knowledge on the matter write a more confident answer, but hopefully it helped.