 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 givesEach 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 thatNow fix $x^{\prime },y^{\prime }$ and take $u(x)=f(x,y^{\prime })$, $v(y)=f(x^{\prime },y)$. The hypothesis $|f|\le 1$ makes these admissible tests. Multiplying their pairing by $f(x^{\prime },y^{\prime })$ and averaging givesConsequentlyTogether 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.