 Solution

For this Cartesian product, the requested $U^2$ norm is the box norm, with complex conjugates:Equivalently it is ${\mathbb{E}}_{y,y^{\prime }}|K(y,y^{\prime }){|}^2$, where $K(y,y^{\prime })={\mathbb{E}}_{x}f(x,y)\overline{f(x,y^{\prime })}$. This makes its nonnegativity explicit.

Apply the Cauchy-Schwarz inequality first in $x$:The right-hand expectation is nonnegative, since it came from ${\mathbb{E}}_x|{\mathbb{E}}_{y}f(x,y)v(y){|}^2$. A second application in $(y,y^{\prime })$ givesTaking fourth roots proves the bilinear correlation bound for the box norm