Geometry problem about finite set in unit ball

Question

I think covering numbers (of the unit ball) is the right way but I think about this question for a few days now and a hint would by nice….

Let $B_{1}$ be the unit ball in $\mathbb R^d$ and let $A \subset B_{1}$ be a finite set.

Denote $\ell=\lvert A\rvert^{-1}\sum_{a \in A} R(a) $ where $R:A \to (0,\infty)$ is a function such that for all $a \in A$,

$$R(a) < \inf_{a' \in A \setminus \{a\}} \lVert a-a'\rVert.$$

We need to prove that

$$\lvert A\rvert \le \left(\frac{4}{\ell}\right)^d.$$

Just a hint will be great.

Answer 1

We must assume that $|A|>1$ which implies that $R(a) \le 2$ for all $a \in A$.

Let $V_d$ denote the volume of he unit ball $B_1=B(0,1)$. The open balls $\{B(a,R(a)/2)\}_{a \in A}$ are pairwise disjoint and contained in $B(0,2)$. Comparing the volume of their union to the volume of $B(0,2)$, we infer that
$$ 2^dV_d \ge \sum_{a \in A} R(a)^d \, 2^{-d}\, V_d\,.$$ Multiplying both sides by $2^d$ and dividing by $|A| V_d$, we obtain that $$|A|^{-1}4^d \ge |A|^{-1}\sum_{a \in A} R(a)^d \ge \Bigl(|A|^{-1}\sum_{a \in A} R(a)\Bigr)^d = \ell ^d \,,$$ where we used convexity of $x \mapsto x^d$ in the second inequality. This proves the claim $|A| \le (4/\ell)^d$.