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For nonempty finite sets $X,Y$ and a function $f:X\times Y\to \mathbb{C}$ on their Cartesian product, using uniform expectations, defineThis equals ${\mathbb{E}}_{y,y^{\prime }}|{\mathbb{E}}_{x}f(x,y)\overline{f(x,y^{\prime })}{|}^2$. Hence it is nonnegative, and vanishing forces $f=0$ by taking $y=y^{\prime }$. Absolute homogeneity and the box Cauchy-Schwarz inequality establish that it is a norm. For a real-valued function the complex conjugates can be omitted. The bilinear correlation bound for the box norm also holds for complex functions.

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

Uniform expectations require nonempty finite sets; assume this throughout the analytic argument. For real-valued functions on $X\times Y$, define the box norm byAll four variables are sampled as independent random variables with replacement, so repeated coordinates are included. The defining fourth power equalsIf it vanishes, every summand vanishes. In particular, the terms $y_0=y_1=y$ give ${\mathbb{E}}_{x}f(x,y{)}^2=0$ for each $y$, so $f=0$. Absolute homogeneity of a norm follows directly: $\Vert af{\Vert }_{\square }=|a|\Vert f{\Vert }_{\square }$.

For the triangle inequality, prove the mixed box Cauchy-Schwarz inequality. WriteSeparating the two $y$ averages and applying the Cauchy-Schwarz inequality in $(x_0,x_1)$ givesExpanding the square and reversing the order of finite sums,by a second Cauchy-Schwarz inequality. Thus $|\Lambda |\le {\prod }_{i,j}\Vert f_{ij}{\Vert }_{\square }$. Expand $\Lambda (f+g,f+g,f+g,f+g)$ into its sixteen multilinear terms and apply this bound to each term. Their upper bounds sum to $(\Vert f{\Vert }_{\square }+\Vert g{\Vert }_{\square }{)}^4$. Taking fourth roots yields the triangle inequality. Together with definiteness of a norm and absolute homogeneity of a norm, this proves the box norm is a norm. Complex functions require conjugates and are outside the printed real-valued convention.

For the bilinear correlation bound for the box norm, use the normalized L2 norm: let $B={\mathbb{E}}_{x,y}f(x,y)u(x)v(y)$ and $\Vert u{\Vert }_2^2={\mathbb{E}}_{x}u(x{)}^2$, $\Vert v{\Vert }_2^2={\mathbb{E}}_{y}v(y{)}^2$. The Cauchy-Schwarz inequality in $y$ givesThe remaining factor is ${\mathbb{E}}_{x,x^{\prime }}u(x)u(x^{\prime }){\mathbb{E}}_{y}f(x,y)f(x^{\prime },y)$. By the Cauchy-Schwarz inequality in $(x,x^{\prime })$, its magnitude is at mostTaking square roots proves

For the tripartite graph, assume its parts are nonempty and write $g_{XY},g_{YZ},g_{XZ}$ for its adjacency indicator functions. Let $f=g_{XY}-\alpha$, so $\Vert f{\Vert }_{\square }\le c$. The normalized triangle count isThe constant-degree hypothesis is exactly ${\mathbb{E}}_y{g}_{YZ}(y,z)=\beta$ for each $z$. Therefore the contribution of the constant $\alpha$ isFor fixed $z$, apply the bilinear correlation bound for the box norm with $u_z(x)=g_{XZ}(x,z)$ and $v_z(y)=g_{YZ}(y,z)$. Since these are indicator functions, $\Vert v_z{\Vert }_2^2=\beta$ and $\Vert u_z{\Vert }_2^2=d_X(z)$, the fraction of $X$ adjacent to $z$. Averaging and applying the Cauchy-Schwarz inequality in $z$ gives the sharper triangle counting with one box-uniform pair and constant opposite degree estimateMultiplying by $|X||Y||Z|$ yieldsThe middle bound also handles $\beta =0$ or $\gamma =0$, when the triangle count is zero.

To connect explicitly with the suggested expansion, put $f_2=g_{YZ}-\beta$, $f_3=g_{XZ}-\gamma$. Both have overall mean zero, and ${\mathbb{E}}_y{f}_2(y,z)=0$ for every $z$. Consequently ${\mathbb{E}}_{x,y,z}{f}_2(y,z)f_3(x,z)=0$; all terms with constant $\alpha$ reduce to $\alpha \beta \gamma$. The terms containing $f$ combine into the single error just estimated. This is where the exact degree assumption is used. If any part is empty the triangle count itself is zero, but the printed uniform expectations and edge density of a bipartite graph values would be undefined; the nonempty-parts convention must therefore be stated rather than dividing by zero.

Let a tripartite graph have nonempty parts $X,Y,Z$, pair edge density of a bipartite graph values $\alpha ,\beta ,\gamma$ on $XY,YZ,XZ$, and $\Vert G(X,Y)-\alpha {\Vert }_{\square }\le c$. If every $z\in Z$ has exactly $\beta |Y|$ neighbors in $Y$, its normalized triangle count $\tau$ obeysThe constant-degree assumption makes the contribution of the constant $\alpha$ exactly $\alpha \beta \gamma$. For each fixed $z$, apply the bilinear correlation bound for the box norm to the indicator functions of its two vertex neighbourhoods. Their squared $L^2$ norms are $\beta$ and the relative $X$-degree of $z$. Average over $z$ and use the Cauchy-Schwarz inequality to bound the mean square root of that degree by $\sqrt{\gamma }$.