= Solution
For the unheaded existence request, use the <probabilistic method>. Choose every value $f(x,y)$ independently and uniformly from $\{-1,1\}$. For a rectangle with $x\ne x'$ and $y\ne y'$, its four entries are distinct, so independence and the zero mean of each sign give expectation zero for their product. If $x=x'$ or $y=y'$, the product is identically one. The number of ordered rectangles in this union of diagonal cases is
$$
mn^2+m^2n-mn.
$$
Therefore the <random sign rectangle fourth moment> has expectation
$$
\mathbb E Q(f)=\frac1m+\frac1n-\frac1{mn}.
$$
Some sign function has $Q(f)$ at most this expectation. Applying the bound in part (a) to that same function gives the simultaneous choices
$$
\boxed{c_1=\frac1m+\frac1n-\frac1{mn},\qquad c_2=\left(\frac1m+\frac1n-\frac1{mn}\right)^{1/4}.}
$$
Both constants tend to zero whenever both dimensions tend to infinity, without any assumption on their relative rates of growth.
Back to article page
