Let $G$ be a $k$-partite $k$-uniform hypergraph with at least $dn^k$ many edges. I want a lower bound on the number of $K\_{2, 2,\, \ldots\,,2}$ in $G$, preferably something like $\gamma n^{2k}$ for some constant $\gamma$ independent of $n$.


My ideas: I can get down to a minimum degree subhypergraph from the edge-density by cleaning up vertices with low degrees. Starting from the edge-density I can also get down to a hypergraph where all vertices have a partite minimum co-degree either $0$ or $\beta n^{k-1}$. But I have no idea of how to do both of that together. I have also been thinking of some kind of induction approach but have not been able to prove it with that.



Counting $K_{2, 2, \,\ldots\,,2}$ in a $k$-partite $k$-uniform hypergraph
You can use the doubling method (a.k.a. Cauchy-Schwarz)


For simplicity, suppose we are working in the graph case. So, let $G$ be a graph with bipartition $X\cup Y$ with $|X|=|Y|=n$ and suppose that $e(G)=\Omega(n^2)$. Let $f(x,y)$ be the indicator function of the edges of $G$ and let  

$$A=\sum\_{x\in X, y\in Y}f(x,y) $$
Using the Cauchy-Schwarz inequality, we have
$$A^2=\left(\sum\_{x\in X}\sum\_{y\in Y}f(x,y)\right)^2\le n\sum\_{x\in X}\left(\sum\_{y\in Y}f(x,y)\right)^2=n\sum\_{x\in X}\sum\_{y,y'\in Y}f(x,y)f(x,y'),$$
where the RHS is $n$ times the the number of "cherries" (copies of $K\_{1,2}$) in $G$. This operation is sometimes called "doubling" since we are duplicating the variable $y$. Now we double the variable $x$ by squaring again.


$$A^4\le n^2\left(\sum\_{y,y'\in Y}\sum\_{x\in X}f(x,y)f(x,y')\right)^2\le n^4\sum\_{x,x'\in X, y,y'\in Y}f(x,y)f(x,y')f(x',y)f(x',y'),$$
and thus the RHS is $n^4$ times the number of $K\_{2,2}$. Then the number of $K\_{2,2}$ is of order $n^{-4}A^4=\Omega(n^4)$. The same proof works for $k$-uniform hypergraphs but doing "doublings" $k$ times.


