At the beginning of Section 4.1 of the paper Limits of Dense Graph Sequences by Lovasz and Szegedy, the following is stated:
Let $U:[0, 1]^2\to \mathbf R$ be an integrable function. Define the rectangle norm of $U$ as $$\|U\|_{\blacksquare} = \sup_{A, B\subseteq [0, 1]}\left|\int_A\int_BU(x, y)\ dx dy\right|$$ (where $A$ and $B$ are implicitly assumed to be measurable).
Then the authors say that it is easy to see that
$$\|U\|_{\blacksquare}= \sup_{f, g:[0, 1]\to [0, 1]} \left| \int_0^1\int_0^1 U(x, y)f(x)g(y)\ dx dy \right|$$
(where $f$ and $g$ are implicitly assumed to be measurable.) (There is a misprint in the paper. The absolute value symbole above is missing.)
Question. I don't see how the second equation is true. Can somebody please provide a proof. Thanks.
Characteristic functions of measurable sets are measurable $[0,1]$-valued functions, this implies that $$\sup_{A,B\subseteq[0,1]}\left|\int_A\int_B U(x,y)\,dxdy\right|≤\sup_{f,g:[0,1]\to[0,1]}\left|\int_{[0,1]^2}U(x,y)f(x)g(y)\,dxdy\right|.$$ Now lets look at some specific $f,g:[0,1]\to[0,1]$ and show that we can make the integral larger by having these functions be only $\{0,1\}$-valued, ie by having them be characteristic functions. This would imply "$≥$".
Suppose for convenience that $\int_0^1\int_0^1 U(x,y)f(x)g(y)\,dxdy$ is positive. Look at the function $y\mapsto \int_0^1dx\, U(x,y)f(x)$ and let $\tilde g$ be the characteristic function of the set on which that function is positive (this is measurable). Since $g$ maps into $[0,1]$ we have: $$\tilde g(y)\int_0^1U(x,y)f(x)\,dx≥g(y)\int_0^1U(x,y)f(x)\,dx$$ and as such the integral can be made bigger by replacing $g$ with $\{0,1\}$-valued function. You can do the same now with $f$.