Many antipodes implies many neighbors

Abstract

Suppose $\left\{x_1,\dots ,x_n\right\}\subset {\mathbb{R}}^2$ is a set of $n$ points in the plane with diameter $\le 1$, meaning $\Vert x_i-x_j\Vert \le 1$ for all $1\le i,j\le n$. We show that if there are many `antipodes', these are pairs of points of with distance $\ge 1-\epsilon$, then there are many neighbors, these are pairs of points that are distance $\le \epsilon$. More precisely, we prove that for some universal $c>0$, $$\#\left\{(i,j):\Vert x_i-x_j\Vert \le \epsilon \right\}\ge \frac{c\cdot {\epsilon }^{3/4}}{{\left(\log {\epsilon }^{-1}\right)}^{1/4}}\cdot \#\left\{(i,j):\Vert x_i-x_j\Vert \ge 1-\epsilon \right\}.$$ The inequality is very easy too prove with factor ${\epsilon }^2$ and easy with $\epsilon$. The optimal rate might be ${\epsilon }^{1/2}$ which is attained by several examples.