Consider the following statement, which I suspect is false as written:

Let $E,F,G$ be (Hausdorff) topological vector spaces (over $\mathbb{R}$), let $\varphi\colon E\times F\to G$ be continuous and bilinear, and let $(x_i)_{i\in I}$ and $(y_j)_{j\in J}$ be summable families in $E$ and $F$ respectively with $\sum_{i\in I} x_i = x$ and $\sum_{j\in J} y_j = y$. Then $(\varphi(x_i,y_j))_{(i,j)\in I\times J}$ is summable with $\sum_{(i,j)\in I\times J} \varphi(x_i, y_j) = \varphi(x,y)$.

(Lest there be any doubt, “$(z_k)_{k\in K}$ summable with $\sum_{k\in K} z_k = z$” means that for every neighborhood $V$ of $z$ there is $K_0\subseteq K$ finite such that for any $K_0 \subseteq K_1 \subseteq K$ finite we have $\sum_{k\in K_1} z_k \in V$”.)

I am interested both in “nice” counterexamples to the statement above and in strengthenings of the hypotheses which would make it true—or basically any information regarding variations of this statement (I know essentially nothing except for the pretty much trivial fact that when $E,F,G$ are finite-dimensional it is correct). Since the rules of MO are to ask one specific question, and since I am mostly interested in counterexamples, let me ask:

Question: Is there a counterexample to the above statement with “nice” spaces $E,F,G$ (e.g., locally convex, complete, metrizable… or even Banach spaces)?

—but again, any information concerning it is welcome.

Comments (added 2018-01-04):

  • It is clear that, under the hypotheses of the statement above, $\sum_{(i,j)\in I_1\times J_1} \varphi(x_i, y_j)$ converges to $\varphi(x,y)$ where $I_1$, $J_1$ range over the finite subsets of $I$ and $J$ respectively; what is to be proven is that $\sum_{(i,j)\in K_1} \varphi(x_i, y_j)$ converges to $\varphi(x,y)$ where $K_1$ ranges over the finite subsets of $I\times J$. The subtlety, of course, is that $K_1$ can fail to be a rectangle.

  • The following result is found in Seth Warner's book Topological Rings (1993), theorem 10.15: if $E,F,G$ be are Hausdorff commutative topological groups, and $\varphi\colon E\times F\to G$ is continuous and $\mathbb{Z}$-bilinear, and $(x_i)_{i\in I}$ and $(y_j)_{j\in J}$ summable families in $E$ and $F$ respectively with $\sum_{i\in I} x_i = x$ and $\sum_{j\in J} y_j = y$, then provided $(\varphi(x_i,y_j))_{(i,j)\in I\times J}$ is summable, it sum is $\sum_{(i,j)\in I\times J} \varphi(x_i, y_j) = \varphi(x,y)$. So the crucial question is the summability of $(\varphi(x_i,y_j))$, not the equality with $\varphi(x,y)$. Even with the very weak hypothesis that $E,F,G$ are commutative topological groups, I still don't have a counterexample!

  • The following possibly related result is found in Kamal Kant Jha's 1972 paper “Analysis of Bounded Sets in Topological Tensor Products” (corollary 3.3): If $(x_i)$ is a totally summable family in a locally convex space $E$ [meaning that there exists $L\subseteq E$ closed, absolutely convex and bounded, such that $\{x_i\}\subseteq L$ and $\sum_i p_L(x_i) < +\infty$ for $p_L$ the gauge of $L$] and ditto for $(y_j)$ in $F$, then $(x_i \otimes y_j)$ is totally summable in $E \mathbin{\otimes_\varepsilon} F$.


I found an answer to my question in Raymond Ryan's book Introduction to Tensor Products of Banach Spaces (2002), example 4.30, which I reproduce here with only minor modifications.

Specifically, this is with $E = F = \ell^2$ and $G = \mathbb{R}$, and furthermore $(y_j) = (x_i)$, i.e., we construct a bilinear form $\varphi$ on $\ell^2$ and an unconditionally convergent ($\Leftrightarrow$ summable) series $\sum_{i=1}^{+\infty} x_i = x \in \ell^2$ such that $\sum_{i,j} \varphi(x_i,x_j)$, while convergent in $\mathbb{R}$ for a certain ordering of the pairs $(i,j)$, is not absolutely convergent (so, not summable in $\mathbb{R}$).

Let $(e_i)$ (numbered from $1$, say) be the standard Hilbert (orthonormal) basis of $\ell^2$. Let $I_n$ be the integer interval between $2^n$ and $2^{n+1}-1$ inclusive (so $\#I_n = 2^n$), let $u_n := 2^{-n/2} \sum_{i\in I_n} e_i$ so that $(u_n)$ is an orthonormal sequence, and let $x := \sum_{n=1}^{+\infty} \frac{1}{n} u_n$. So we have $x = \sum_{i=1}^{+\infty} x_i e_i$ where $x_i = \frac{1}{n\,2^{n/2}}$ when $i \in I_n$ (and $x_1 = 0$ but no matter): this sum is unconditionally convergent.

Now let $A_n$ be the $2^n \times 2^n$ real matrix defined inductively by $A_0 = (1)$ and $$A_{n+1} = \begin{pmatrix}A_n&A_n\\A_n&-A_n\end{pmatrix}$$ [Ryan puts a minus sign on the lower left entry instead of the lower right, I don't think this changes anything but the above sign convention will make $\varphi$ symmetric]. The (Euclidean operator) norm $\|A_n\|$ of $A_n$ is $2^{n/2}$ (because $2^{-n/2} A_n$ is orthogonal).

Construct the bilinear form on $\ell^2 \times \ell^2$ which is block diagonal with blocks $2^{-n/2} A_n$ in the obvious sense, i.e., $\varphi(\sum_{i=1}^{+\infty} z_i e_i, \sum_{j=0}^{+\infty} z'_j e_j)$ is defined as $\sum_{n=0}^{+\infty} 2^{-n/2} \sum_{(i,j)\in {I_n}^2} (A_n)_{i,j} z_i z'_j$ (where $A_n$ is considered as indexed by ${I_n}^2$). Then $\varphi$ is continuous with $\|\varphi\| \leq 1$ (in the sense that $|\varphi(z,z')|^2 \leq \|z\|^2\,\|z'\|^2$) because $\|2^{-n/2} A_n\| \leq 1$.

In particular, $\varphi(x,x)$ is well-defined (with value $\sum_{n=1}^{+\infty} \frac{1}{n^2\,2^{n/2}}$, but no matter), whereas $\varphi(x_i e_i, x_j e_j) = \frac{\pm 1}{n^2\,2^{3n/2}}$ gives $\sum_{(i,j)\in {I_n}^2} |\varphi(x_i e_i, x_j e_j)| = \frac{2^{n/2}}{n^2}$ which does not converge (so $\sum_{i,j} \varphi(x_i e_i,x_j e_j)$ is not absolutely convergent).

  • $\begingroup$ (+1,+1) I came to your counterexample and refered to it as an answer within the so useful and interesting big-list. Do you have counterexamples in Banach algebras with $\varphi$ as bilinear product ? $\endgroup$ Dec 16 '18 at 7:44
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    $\begingroup$ @DuchampGérardH.E. Yes, it's fairly easy: see proposition 5 in this note, which was written precisely to answer that question (that a colleague of mine raised in one of his papers). I was wondering whether I should send that note somewhere. $\endgroup$
    – Gro-Tsen
    Dec 16 '18 at 9:33
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    $\begingroup$ Oh, thank you very much (+1), I'll incorporate this in my zoo. Please, tell me when this will be published or quotable from somewhere (I got the "ping" from this discussion). $\endgroup$ Dec 16 '18 at 13:35
  • $\begingroup$ BTW, do you know a series $\sum_{n}\,f_n$ which is summable (i.e. unconditionally convergent) but not absolutely convergent in the Banach space $C(K)$ ($K$ is a compact domain in the complex plane) ? or, better, in $\mathcal{H}(\Omega)$ ($\Omega$ is a non-empty open subset in the complex plane) ? $\endgroup$ Dec 16 '18 at 17:16
  • $\begingroup$ @DuchampGérardH.E. Very good question. No, I don't know such a series. $\endgroup$
    – Gro-Tsen
    Dec 17 '18 at 7:58

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