= Solution
Use uniform <expectations> on the nonempty sets $X,Y$. Define the rectangle fourth moment and the normalized <cut norm> by
$$
Q(f)=\mathbb E_{x,x',y,y'}f(x,y)f(x,y')f(x',y)f(x',y'),\qquad D(f)=\max_{A\subseteq X,\,B\subseteq Y}|\mathbb E_{x,y}f(x,y)1_A(x)1_B(y)|.
$$
The first quantity is nonnegative, since
$$
Q(f)=\mathbb E_{x,x'}\left(\mathbb E_y f(x,y)f(x',y)\right)^2;
$$
its fourth root is the <box norm>. To bound the <cut norm>, fix $A,B$ and put $T=\mathbb E_{x,y}f(x,y)1_A(x)1_B(y)$. The <Cauchy-Schwarz inequality> in $y$ gives
$$
|T|^2\leq\mathbb E_y\left|\mathbb E_x1_A(x)f(x,y)\right|^2=\mathbb E_{x,x'}1_A(x)1_A(x')\mathbb E_y f(x,y)f(x',y).
$$
A second <Cauchy-Schwarz inequality>, now in $(x,x')$, yields
$$
|T|^4\leq\mathbb E_{x,x'}\left(\mathbb E_y f(x,y)f(x',y)\right)^2=Q(f),
$$
because the squared mean of $1_A(x)1_A(x')$ is at most one. Taking the maximum over rectangles gives the first direction of <cut norm and rectangle fourth-moment equivalence>:
$$
\boxed{D(f)\leq Q(f)^{1/4},\qquad c_2=c_1^{1/4}\text{ is admissible}.}
$$
The normalization divides the rectangle sum by $mn$ and the fourth-moment sum by $m^2n^2$, so the bound is independent of the two set sizes.
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