= Solution
For the reverse direction, rectangular discrepancy also controls pairings with arbitrary bounded real test functions. If $u:X\to[-1,1]$ and $v:Y\to[-1,1]$, decompose each into its positive and negative parts. For example, the <layer cake representation> gives
$$
u_+(x)=\int_0^1 1_{\{u(x)>s\}}\,ds,\qquad u_-(x)=\int_0^1 1_{\{u(x)<-s\}}\,ds.
$$
Each of the four sign combinations in $u(x)v(y)$ is thus an integral of rectangular <indicator functions>. The <triangle inequality> and the definition of the <cut norm> show that
$$
|\mathbb E_{x,y}f(x,y)u(x)v(y)|\leq4D(f).
$$
Now fix $x',y'$ and take $u(x)=f(x,y')$, $v(y)=f(x',y)$. The hypothesis $|f|\leq1$ makes these admissible tests. Multiplying their pairing by $f(x',y')$ and averaging gives
$$
0\leq Q(f)\leq\mathbb E_{x',y'}\left|\mathbb E_{x,y}f(x,y)f(x,y')f(x',y)\right|\leq4D(f).
$$
Consequently
$$
\boxed{Q(f)\leq4D(f),\qquad c_1=4c_2\text{ is admissible}.}
$$
Together with part (a), this proves <cut norm and rectangle fourth-moment equivalence>. The intended equivalence is quantitative smallness: along any family of bounded functions, the normalized rectangle fourth moments tend to zero if and only if the normalized <cut norms> tend to zero. It does not assert equality of the two constants or norms.
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