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Let $\{x_{i,j}:i\in I,j \in J\}$ be a double indexed family of elements in a Banach space $X$.

I want to show the following result.

Proposition. If $\sum_{(i,j)\in I\times J}\|x_{i,j}\|<\infty$ then

$$\sum_{(i,j)\in I\times J}x_{i,j}=\sum_{i\in I} \sum_{j\in J} x_{i,j}=\sum_{j\in J} \sum_{i\in I} x_{i,j}$$

where the convergence of each series is part of the result and convergence is understood using nets as in here.

Proof. I will assume the results holds when the terms $x_{i,j}$ take values in $[0,\infty]$, since this can be seen as a special case of Fubini's Theorem for the Lebesgue integral with counting measure.

  1. Since $\sum_{(i,j)\in I\times J}\|x_{i,j}\|<\infty$ and $X$ is complete there series $\sum_{(i,j)\in I\times J}x_{i,j}$ converges (Cauchy nets converge in complete metric spaces).

  2. Since $\sum_{i\in I} \sum_{j\in J} \|x_{i,j}\|=\sum_{(i,j)\in I\times J}\|x_{i,j}\|<\infty$ we have $ \sum_{j\in J} \|x_{i,j}\|<\infty$ for each $i\in I$, and so arguing as in $1$ we get that $\sum_{j\in J} x_{i,j}$ converges for each $i\in I$. Also, $\|\sum_{j\in J} x_{i,j}\|\leq \sum_{j\in J} \|x_{i,j}\|$ for each $i\in I$. It follows that $\sum_{i\in I} \|\sum_{j\in J} x_{i,j}\|\leq \sum_{i\in I} \sum_{j\in J} \|x_{i,j}\|<\infty$ and so the double serie $\sum_{i\in I} \sum_{j\in J} x_{i,j}$ converges.

  3. A symmetric argument to 2 shows that $\sum_{i\in I} x_{i,j}$ converges for each $j\in J$ and that $\sum_{j\in J} \sum_{i\in I} x_{i,j}$ converges.

  4. It remains to show that all three series converge to the same element.

How can I complete step 4?

Alphie
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    Since the sum of norms is finite, there are at-most countably many non-zero terms, i.e you may as well assume $I,J$ are at-most countable. So, the question becomes extending Fubini-Tonelli from real/complex-valued doubly-indexed sequences in $\Bbb{N}^2$ to Banach-valued ones. This is now a standard trick which can be accomplished via Hahn-Banach (I explained this trick in a complex-analysis setting here, but it’s analogous). Or you could just revisit the original proof and show it follows through without much difficulty for Banach-valued guys. – peek-a-boo Jun 03 '25 at 17:12
  • @peek-a-boo Ok thank you for the suggestion. I tried to argue directly below. Do you think it works? – Alphie Jun 03 '25 at 17:39

1 Answers1

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Let $\epsilon>0$ be given and choose a finite subset $F\subset I\times J$ such that $\| \sum_{(i,j)\in F'}x_{i,j}-\sum_{(i,j)\in I\times J}x_{i,j}\|<\epsilon$ whenever $F\subset F'\subset I\times J$ with $F'$ finite. Let $F_I$ and $F_J$ denote $I$ and $J$ sections of $F$ respectively.

By definition we can choose a finite subset $I'\subset I$ such that $F_I\subset I'$ and

$$\|\sum_{i\in I'} \sum_{j\in J} x_{i,j} -\sum_{i\in I} \sum_{j\in J} x_{i,j}\|<\epsilon$$

Also, by definition we can choose a finite subset $J'\subset J$ such that $F_J\subset J'$ and

$$\|\sum_{i\in I'} \sum_{j\in J'} x_{i,j} -\sum_{i\in I'} \sum_{j\in J} x_{i,j}\|<\epsilon$$

Since $F\subset F_I\times F_J \subset I'\times J'$ it follows that $\|\sum_{i\in I} \sum_{j\in J} x_{i,j}-\sum_{(i,j)\in I\times J}x_{i,j}\|<3\epsilon$. Since $\epsilon>0$ was arbitrary we conclude that $\sum_{i\in I} \sum_{j\in J} x_{i,j}=\sum_{(i,j)\in I\times J}x_{i,j}$. By symmetry we also have $\sum_{j\in J} \sum_{i\in I} x_{i,j}=\sum_{(i,j)\in I\times J}x_{i,j}$.

Alphie
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