Linked with this question and discussion (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 $\|f\|_K=\sup_{s\in K}|f_s|$)
  2. $\mathcal{H}(\Omega)$ (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 ?
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    $\begingroup$ $\mathcal{H}(\Omega)$ is a nuclear space, so absolute and unconditional convergence coincide, by a theorem from Grothendieck's PhD thesis. (You can also find it in Schäfer's Topological Vector Spaces and several other standard texts that touch on locally convex tensor products.) $\endgroup$ Dec 17, 2018 at 10:18
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    $\begingroup$ $C([0,1])$ is an infinite dimensional Banach space, so it is not nuclear and therefore has an unconditionally summable series that is not absolutely summable. However, you can get this special case more easily by taking your favourite non-absolutely but unconditionally summable series in a separable Banach space, such as $\ell^2$, and embedding it isometrically in $C([0,1])$. $\endgroup$ Dec 17, 2018 at 10:37
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    $\begingroup$ Could you arrange your two remarkable comments as an answer ? $\endgroup$ Dec 17, 2018 at 11:50
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    $\begingroup$ I'll do it when I can get to a computer, I'm travelling at the moment. $\endgroup$ Dec 17, 2018 at 12:49

1 Answer 1


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 $C([0,1])$.

  • $\begingroup$ Dear Robert, Thank you for your intelligent answer to my naïve question. This helped me a lot ! $\endgroup$ May 10, 2021 at 9:00
  • $\begingroup$ @DuchampGérardH.E. You're welcome, Gérard. $\endgroup$ May 15, 2021 at 6:55

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