For each $(i,j)\in \mathbb{N}^2$, $a(i,j)=1$ or $0$, and 1) $a(i,i)=0$ for all $i$; 2)for fixed $i$, there is at most one $j$ such that $a(i,j)=1$. Suppose we know that there is a finite $\kappa$ such that \begin{equation} \sum_{i\in S, j\in S^{C}}a(i,j)\le\kappa\end{equation} for any subset $S$ of $\mathbb{N}$,then one can show $\sum a(i,j)$ is bounded by $3\kappa$.
I am asking what is the largest possible value of $\sum a(i,j)$.
For people who are interested in the background, I am considering an operator $T\in\mathcal{L}(\ell^2)$. With respect to the natural basis, the entries of $T$ are $a(i,j)$. The existence of $\kappa$ in the first inequality is the same as \begin{equation}rank(TP-PTP)\le\kappa\end{equation} for all orthogonal projections $P$ in the masa (maximal abelian, self-adjoint subalgebra) of diagonal operators.
In a recent paper by Popov, Marcoux and Radjavi (Almost Invariant Halfspaces and Approximate Commutation), they have shown $\sum a(i,j)$ is bounded by $3\kappa$ (Thm 4.7), but in almost all examples I can think of, $\sum a(i,j)\le 2\kappa$, so I wonder whether the coefficient can be improved.
Again, I think the problem has been reduced to a purely algebraic or number theoretic problem in the first two paragraph of this post.
Thanks very much!