 Solution

First prove the pair-factor correlation bound for the three-dimensional box norm. For real $g$ and three functions $a(y,z),b(x,z),c(x,y)$ bounded in absolute value by one, letApply the Cauchy-Schwarz inequality in $(y,z)$, placing the $x$ average inside and removing $a$. This givesExpand this square, with independent $x_0,x_1$. Apply Cauchy-Schwarz in $(x_0,x_1,z)$, removing the product $b(x_0,z)b(x_1,z)$. The result isExpand with independent $y_0,y_1$ and apply Cauchy-Schwarz in $(x_0,x_1,y_0,y_1)$, removing the four $c$ factors. ThenTherefore $|T|\le \Vert g{\Vert }_{{\square }^3}$, with no independence assumption on the three pair functions.

In this multipartite hypergraph, a simplex is a three-uniform tetrahedron: one vertex of a hypergraph from each of $X,Y,Z,W$, with all four possible triples present. Write the corresponding hypergraph edge indicator functions as $h_1(x,y,z),h_2(x,y,w),h_3(x,z,w),h_4(y,z,w)$, of subset densities $p,q,r,s$. Its normalized count isTelescope the product exactly:For the first term fix $w$. The remaining three factors are bounded pair functions of $(x,y)$, $(x,z)$ and $(y,z)$, so the inequality just proved bounds its $(x,y,z)$ average by $\Vert h_1-p{\Vert }_{{\square }^3}\le {\alpha }^{1/8}$. Averaging over $w$ preserves that bound. For each other term fix the vertex of a hypergraph outside its balanced triple and apply the same argument, allowing an unused pair factor to be identically one. The subset density prefactors are at most one. HenceMultiplying by $|X||Y||Z||W|$ givesThus one may take the absolute constant $C=4$. This proves the complete-pair-support case of the tetrahedron counting lemma directly; no lower bound on $p,q,r,s$ is needed.