Counting intersections of hyperplanes

Question

This is a dublicate from stackexchange:

Consider two families of hyperplanes $F_1$ and $F_2$ in $\mathbb{R}^d$ both containing $n$ hyperplanes. We have that for all $f \in F_1$ and $g \in F_2$ that $f$ and $g$ intersect. Further we know that for each point $p \in \mathbb{R}^d$ we have that at most $c \cdot n$ hyperplanes from each family contain this point, $ c < 1$.

The second family induces $d-2$ dimensional planes on the hyperplanes of the first family. I want to show that there are at least $\epsilon \cdot n$ hyperplanes in the first family on which at least $\tau \cdot n$ planes are induced.

What I am able to show is: If there is a plane in which $k \cdot n$ hyperplanes form the second family intersect then, there this plane ist contained in at most $c \cdot n$ hyperplanes from the first family. Consider the $(1-c)n$ hyperplanes of the first family which do not contain the plane, than on each of them we get $k \cdot n$ induced planes.

On the other hand it is clear if at most a constant number, $k$, of hyperplanes intersect in one plane we get $\frac{n}{k}$ induced planes on each of the hyperplanes in the first family.

So somehow I can show the extremal cases but I'm stuck with the ones in between. Any help or suggestions?

Edit: Let $n_H$ denote the number of $d-2$ dimensional planes induced on the plane $H \in F_1$. Further let $f_{H,1},...,f_{H,n_H}$ denote this induced planes and $S(f_{H,i}):=\{K \in F_2 \mid f_{H,i} \subset K\}$. Then for $H, H' \in F_1$ we have:

\begin{equation} \frac{n}{d} = \sum_{i=1}^{n_H} \sum_{j=1}^{n_K} \vert S(f_{H,i}) \cap S(f_{H',j})\vert \leq c \cdot \frac{n}{d}+ n_{H} \cdot n_{H'} \end{equation}

Since $\vert S(f_{H,i}) \cap S(f_{H',j})\vert \leq 1$ for all $f_{H_i} \neq H \cap H'$ and $\vert S(f_{H,i}) \cap S(f_{H',j})\vert \leq c \cdot \frac{n}{d}$ if $f_{H,i} = f_{H',j}=H \cap H'$. Therefore $(1-c)\frac{n}{d} \leq n_{H'} \cdot n_H$. This holds for all $H, H' \in F_1$ so $n_H \gg \sqrt{n}$.

Answer 1

As by my comment above, it is enough to solve the problem for $d=2$.
Using point-line duality, we get the following:
Given $n$ red and $n$ blue points in the plane, such that no line contains more than $cn$ points, prove that through at least $\varepsilon n$ of the red points there are at least $\tau n$ lines that each go through some blue point.

This can be proved similarly to Beck's theorem that can be found here.
More precisely, there are $n^2$ red-blue point pairs.
At most $O(cn^2)$ of the lines of these pairs contains at least $1/c$ points (see the linked proof).
As there are no lines with $cn$ points, most pairs will contain at most $1/c$ points, which implies that there are at least $\varepsilon n$ red points with $\tau n$ lines through each with some blue point.

Answer 2

According to domotorp's comment, we only need to solve the problem for $d=2$. As $\mathbb R^2 \subset \mathbb {RP}^2$, by the well-known point-line duality, we can take the duality on \mathbb {RP}^2. Thus points will become lines, lines will become points, and the inclusion is reversed. The duality can be taken sufficiently generic to make sure the lines and points are all defined in $\mathbb R^2$.

Thus, the problem can be restated as follows:

Consider two families of points $F_1$ and $F_2$ in $\mathbb{R}^2$ both containing $n$ points, and the set of lines $L$ passing through an element of $F_1$ and an element of $F_2$. Further we know that each line in $L$ contains at most $cn$ points from either $F_1$ or $F_2$ for some fixed $c<1$.

The goal is to show that there are at least $\epsilon n$ points in $F_1$ such that each of them are on $τn$ lines in $L$.

The starting point is a "bichromatic Beck's theorem" (Theorem 2, paraphrased):

Let $p$ and $q$ be subsets of the points on the Euclidean plane such that $|p|=|q|=n$, and $B$ the set of lines that is incident with at least one point from $p$ and one point from $q$. If every element in $B$ is incident to at most $r$ points in $p \cup q$, then $|B|=\Omega(n(2n − r)).$ (The implied constant in $\Omega$ is absolute.)

Let $r=2cn$, and we will have $L=\Omega(2(1-c)n^2)$.

Since we have established $|L| \geq \delta n^2$ for some $\delta$ only depending on $c$, we can take $\epsilon=\tau=\delta/2$. If the conclusion were false, then the number of lines in $L$ would be less than $n^2\epsilon+n^2\tau=n^2\delta$, contradiction.