TITLE: What fraction of n-point sets in the unit ball have diameter smaller than 1?
QUESTION [13 upvotes]: This question is inspired by a recent talk by Matt Kahle on random geometric complexes. 
Some simple notation: let $\mathcal{B} \subset \mathbb{R}^d$ be the unit ball in $d$-dimensional Euclidean space with the usual norm $\|\cdot\|$ and let $\mathcal{B}^k$ denote the $k$-fold product of this ball with itself for any positive integer $k$. For each $n \in \mathbb{N}$, define $K_n \subset \mathcal{B}^n$ as follows:
$$ K_n = \lbrace(p_1,\ldots,p_n) \in \mathcal{B}^n \text{ so that } \|p_i - p_j\| < 1 \text{ for all } 1 \leq i,j \leq n\rbrace.$$
Thus, $K_n$ consists of those $n$-tuples of points in the unit ball whose diameter is bounded above by $1$. I would like to know what fraction of $\mathcal{B}^n$ lies in $K_n$ asymptotically as $n$ increases. More precisely, let $\mu$ be the usual $d$-dimensional Lebesgue measure, and define 
$$\chi(n) = \frac{\mu(K_n)}{\mu(\mathcal{B})^n}.$$

How does $\chi(n)$ behave for fixed dimension $d$ as $n \to \infty$?

While an answer to this specific question would be great, I would be much more interested in some insight regarding how one should tackle such problems. The Calculus approach of setting up some integral fails horribly -- at least, I should confess that I tried $n = 2 = d$ and got hopelessly stuck with what appears to be an elliptic integral of the murderous kind -- so I expect that there is no closed formula for $\chi(n)$. But surely someone has worked out asymptotic envelopes for such a basic quantity!

REPLY [15 votes]: I'm going to show that, for any $x>2^{-d}$, this is $O(x^{n})$. By Ricardo's lower bound, this is tight.
Given a set of $n$ points of diameter at most $1$, take all the points you get from those points by rounding the coordinates up or down to multiples of $\epsilon$. This new set of points will contain the old set of points in its convex hull, and since the new set of points will be of a distance no more than $\sqrt{d}\epsilon$ from the old set of points, will have a diameter at most $1+2\sqrt{d}\epsilon$. Thus each configuration of points of diameter at most $1$ in a sphere of radius $1$ is contained in the convex hull of some configuration of $\epsilon$-lattice points of diameter at most $1+2\sqrt{d}\epsilon$ in a sphere of radius $1+\sqrt{d}\epsilon$. Let $N$ be the (finite) number of such configurations of lattice points. Then since the sphere has the greatest volume of any convex body with a given diameter, the convex hulls each take up at most $(1+2\sqrt{d} \epsilon)^d/2^d$ of the volume of the sphere, so landing in any hull has probability
\[ N \left(\frac{ 1 + 2 \sqrt{d} \epsilon}{2}\right)^{dn} \]
If we let $\epsilon$ go to $0$ we have the desired result.
To be more specific, we can set $\epsilon = n^{-1/(d+1)}$, and get the upper bound
\[\ln \left(2^{nd} \chi(n)\right) = O\left( n^{d/(d+1)} \right) \]
using the trivial upper bound $N < 2^{O(\epsilon^{-d} )}$