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Let's say a sum of elements in a Banach space is absolutely convergent if even the sum of the norms converges, i.e. $\sum_{i=1}^\infty ||x_i|| < \infty$. This condition implies that the sum of the $x_i$ can be reordered, and in fact that it can be represented in this way that has no dependence on the ordering: You consider the directed set of finite subsets of $\mathbb{N}$ and the net formed on this directed subset is for each $J \subset \mathbb{N}$ with $J$ finite we assign the sum $\sum_{i \in J} x_i$. Do we have TFAE, or any more implications than what I've stated in the following:

(1) The sum converges absolutely

(2) The sum converges in the sense of the net

(3) The sum converges in the ordered sense, but any permutation on the naturals gives the same result

These are equivalent in $\mathbb{R}$ and $\mathbb{C}$. Otherwise I only see that (1) implies (2) implies (3). Actually also (3) implies (2) for general Banach spaces because one can check it against all continuous linear functionals and then it's back to familiar cases.

Jeff
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  • I do not know how to see that (3) implies (2) for general Banach spaces. Indeed, with (3), you get $$\sum_{i=1}^\infty |\varphi(x_i)|<+\infty$$ for any $\varphi$ in the strong dual but, what is the next step ? – Duchamp Gérard H. E. Mar 11 '19 at 14:43

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From http://en.wikipedia.org/wiki/Absolute_convergence:

A theorem of A. Dvoretzky and C. A. Rogers asserts that every infinite-dimensional Banach space admits an unconditionally convergent series that is not absolutely convergent.

Dvoretzky, A.; Rogers, C. A. (1950), "Absolute and unconditional convergence in normed linear spaces", Proc. Nat. Acad. Sci. U. S. A. 36:192–197.

Stephen Montgomery-Smith
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    Thanks! I even thought of the Hilbert space example but somehow thought it was not valid, forgetting that you don't take the square sum when you look at individual absolute values. – Jeff Nov 23 '13 at 21:56
  • I knew about it because I read about it many years ago in the book "Sequences and Series in Banach Spaces" by Joseph Diestel. – Stephen Montgomery-Smith Nov 23 '13 at 21:58