Suppose $C=\{ a_{i,j} \}_{i,j \in \Bbb{Z}}$ is a symmetric complex bisequence (which means $a_{i,j}=a_{j,i}$) such that $\sum_{i \le j} \lvert a_{i,j} \rvert$ diverges. Call $P \subseteq \Bbb{Z}^2$ square-symmetric if $P=F \times F$ for some finite set $F \subseteq \Bbb{Z}$. Denote by $S$ the collection of $\sum_{(i,j) \in P} a_{i,j}$ for all square-symmetric $P$'s. Must $S$ be unbounded? The square-symmetric restriction really annoyed me and I don't know where to start. Even a result about the case $a_{i,j}=\pm 1$ is of great help to me!

PS: The problem can be stated in a matrix-manner: Suppose the entry-wise $l^1$-norm of a countably-infinite symmetric complex matrix is unbounded, must the entry-sums of its finite principal minors be also unbounded?

  • $\begingroup$ Counterexample: $a_{i,j}=1/\max(i^2,j^2)$. $\endgroup$ Apr 30, 2022 at 17:47
  • $\begingroup$ @MikeEarnest This can not be counterexample since your numbers are positive. In this (and similar) case, the square-symmetric sums (indexed by $n$) correspond to $\{ (i,j)|1 \le i,j \le n \}$ are unbounded. $\endgroup$
    – Zerox
    Apr 30, 2022 at 17:55
  • $\begingroup$ $\{(i,j)\mid 1\le i\le n \text{ and }1\le j\le n\}$ is not a finite union of $\{(k,k)\}$ and $\{(i,j),(j,i),(i,i),(j,j)\}$. $\endgroup$ Apr 30, 2022 at 17:57
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    $\begingroup$ Never mind, I was confused, and thinking of "disjoint union" instead of union. Interesting question... $\endgroup$ Apr 30, 2022 at 18:14
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    $\begingroup$ Is $P\subseteq \mathbb{Z}^2$ a "square-symmetric set" if and only if $P=F \times F$ for some finite set $F \subset \mathbb{Z}$? $\endgroup$
    – Mike F
    May 1, 2022 at 0:28


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