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Disclosure: I have asked this question on MSE (https://math.stackexchange.com/questions/4895621/maximal-sum-of-a-function-such-that-the-sum-is-bounded-on-product-sets) but received no comments in a week. If you know a better suited tag please let me know and I will attach it to the question.

The formal problem is listed below (copied from MSE):

Let $n \in \mathbb{N}$ and $f: [n] \times [n] \to \mathbb{R}$ such that

  1. $\sum_{x,y \in [n]} f(x,y) = 0$, and
  2. $\left|\sum_{x \in S, y \in S'} f(x,y) \right|\le 1$ for all $S,S' \subseteq [n]$.

What is the maximum value the expression $$ \sum_{(x,y) \in T} f(x,y) $$ can take for an arbitrary subset $T\subseteq [n]\times [n]$?

My intuition is that the optimal $T$ likely diagonal since it is far from product sets. For $n=1$ the answer is trivially $0$. For $n=2$, it is also easy to see that the answer is $2$, achieved by the function $$ f(x,y) = \begin{cases} 1, x=y \\ -1, x \ne y \end{cases}. $$ and $T = \{(0,0),(1,1)\}$. I also see that the answer is monotone in $n$. The example does not seem to generalize to $n>2$ in a trivial way, and I wasn't able to find good references for this problem. A non-trivial upper-bound would also be useful. Thank you.

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  • $\begingroup$ The order of magnitude is, of course, $\sqrt n$ but to find the exact maximum may be quite difficult. $\endgroup$
    – fedja
    Commented Apr 16 at 19:55

1 Answer 1

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I've addressed small cases with linear programming. For $n\leq 5$, the maximum happen to be just $n$, but for $n=6$ it's only $\frac{16}3$. Here are the constructed solutions:

Max: 3
[ 0 -1  1]
[-1  1  0]
[ 1  0 -1]

Max: 4
[ 1/4 -1/4 -3/4  3/4]
[ 3/4  1/4 -1/4 -3/4]
[-3/4  3/4  1/4 -1/4]
[-1/4 -3/4  3/4  1/4]

Max: 5
[ 1/2  1/2 -1/2 -1/2    0]
[   0 -1/2 -1/2  1/2  1/2]
[-1/2    0  1/2 -1/2  1/2]
[-1/2  1/2    0  1/2 -1/2]
[ 1/2 -1/2  1/2    0 -1/2]

Max: 16/3
[ -3/8  -1/3   3/8 11/24 -7/24   1/6]
[11/24  -1/3 -7/24  -3/8   3/8   1/6]
[  3/8   1/6 11/24 -7/24  -3/8  -1/3]
[-7/24   1/6  -3/8   3/8 11/24  -1/3]
[ -1/3   1/6   1/6  -1/3   1/6   1/6]
[  1/6   1/6  -1/3   1/6  -1/3   1/6]
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