Linked with this question and discussion
([Bilinear product of two summable families](https://mathoverflow.net/questions/289760/bilinear-product-of-two-summable-families)), I am very
interested in counterexamples/results about the following questions (cf the end).
First, I recall that a
family $(a\_i)\_{i\in I}$ in a topological abelian group $(G,+)$ is called *summable* with sum $S$ iff
for all neighbourhood of zero $W$ it exists $J\_W\subset\_{finite} I$ such that for all $J$ with
$J\_W\subset J\subset\_{finite} I$, $(S-\sum\_{i\in J}a\_i)\in W$. It amounts to the same to say that the
net $J\mapsto \sum\_{i\in J}a\_i$ (from $2^{(I)}$, the set of finite subsets of $I$, ordered by
inclusion, to $G$) converges to $S$.


It is known that, when $I=\mathbb{N}$ (series $\sum\_{n\geq 0}\,a\_n$) the series
$\sum\_{n\geq 0}\,a\_n$ is summable iff it is unconditionaly convergent, i.e. the sequence of partial
sums
$$
N\to \sum\_{n=0}^N\,a\_{\sigma(n)}
$$  

converges for all permutation $\sigma$ of $\mathbb{N}$.


**Question(s)** I am particularly interested in counterexamples/results about series
$\sum\_{n\geq 0}\,a\_n$ which are unconditionaly convergent but not absolutely convergent
in the following frameworks


1. $K=[0,1]\subset \mathbb{R}$ and a series of continuous real functions $\sum\_{n\geq 0}\,f\_n$ unconditionaly convergent but not absolutely convergent i.e. 
$$
\sum\_{n\geq 0}\,||f\_n||\_K<+\infty
$$
(where $\|f\|\_K=\sup\_{s\in K}|f\_s|$)
- $\mathcal{H}(\Omega)$ (space of holomorphic functions $\Omega\to \mathbb{C}$, where 
$\Omega\subset \mathbb{C}$ is not empty and open). In this context, absolutely convergent, 
for a series $\sum\_{n\geq 0}\,f\_n$, means that for all $K\subset\_{compact} \Omega$, one has 
$$
\sum\_{n\geq 0}\,||f\_n||\_K<+\infty
$$

 
are there (counter-)examples or general results in these directions ?

Unconditionally convergent series in some functional spaces
A good resource for these things is Section IV.10 of Schaefer's *Topological Vector Spaces*, so you should look there for the proofs of the following statements. 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 $\ell^1[E] \rightarrow \ell^1(E)$ is continuous, and in the case that $E$ is complete, $\ell^1[E] \cong \ell^1 \hat{\otimes} E$ and $\ell^1(E) \cong \ell^1 \check{\otimes} E$, where these are the projective and injective tensor products defined by Grothendieck.


It is then true that the inclusion mapping $\ell^1[E] \rightarrow \ell^1(E)$ is a linear homeomorphism iff $E$ is a nuclear space. This is actually Pietsch's sharpened version of Grothendieck's theorem. This answers (2), because $\mathcal{H}(\Omega)$ is a nuclear space for $\Omega$ open, so unconditional and absolute summability are the same.


If $E$ is infrabarrelled, for instance if $E$ is a normed space, then the mapping $\ell^1[E] \rightarrow \ell^1(E)$ 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.


Therefore $C([0,1])$ contains an unconditionally summable series that is not absolutely summable. However, in this special case, it can be proved more easily. Just take your favourite non-absolutely but unconditionally summable series in a separable Banach space, such as $\left(\frac{e\_n}{n}\right)\_n$ in $\ell^2$. Any separable Banach space can be isometrically embedded as a closed subspace in $C([0,1])$, so the image of this series defines a series with the same properties in $C([0,1])$.


