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Hello. I'd like to consider the open unit ball in an infinite dimensional Hilbert space and ask when can we fit infinitely many open balls of radius $r<1$ inside.

For example, when $r=1/(1+\sqrt2)$, we can pick an orthonormal basis $(x_1,...)$ for our Hilbert space and put the centers of the balls at $(1-r)x_i = \sqrt2/(1+\sqrt2)x_i$ for each $i$. The distance between any two centers is thus $\sqrt2/(1+\sqrt2)\sqrt2 = 2r$, so indeed the balls just kiss each other.

Can we fit any larger balls? What is the critical radius $r_\infty$ such that for $r>r_\infty$ we may only fit finitely many balls of radius $r$, but for smaller $r$ we may fit infinitely many?

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If one poses the problem in finite dimension $d$, we get a critical radius $r_d$. Because we may choose $d$ balls and project orthogonal onto the subspace spanned by their centers, we have $r_\infty\le r_d$. I guess that $r_\infty=\lim r_d$ as $d\rightarrow\infty$. Besides, looking at the packing of $d+1$ spheres in $d$ space dimension, whose centers form a $(d+1)$-simplex, I anticipate that $r_d\rightarrow 1/(1+\sqrt2)$. –  Denis Serre Apr 13 '11 at 9:40
    
Do you know an exact or asymptotic formula, or lower bound, for $M(n, r)$, the maximum number of $n$--dimensional balls of radius $r<1$ which can be packed inside an $n$--dimensional unit ball? In other words, what happens in $n$--dimensions, asymptotically, as $n$ goes to infinity? Intuitively, we would guess that if $M(n, r_n) \to \infty$ for some sequence $r_n \to 1$, then $r_{\infty} = 1$; and, presumably, it wouldn't be too hard to prove, IF we had some kind of algorithm ``uniform in $n$''. Sorry, no answers from me, just more stupid questions! –  Zen Harper Apr 13 '11 at 9:41
    
In the direction of Denis' comment, if $r_d$ denotes the critical radius for which there are exactly $d$ disjoint balls of radius $r$ in the unit ball of a Hilbert space, I believe that the argument in my answer might prove (by induction on $d$) that $r_d$ is attained for balls centered on a simplex. –  Mikael de la Salle Apr 13 '11 at 10:17
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2 Answers

up vote 9 down vote accepted

Your value of $r$ is the best.

Equivalently, $\rho=\sqrt 2$, where $\rho$ is the sup, over all infinite sequence $(x_i)$ in the unit ball of a Hilbert space, of $\inf_{i\neq j} |x_i-x_j|$.

Here is a proof, by contradiction. Assume that $\rho>\sqrt 2$ and pick a sequence $(x_i)$ such that $\inf_{i\neq j} |x_i-x_j|$ is almost $\rho$. Take $e$ the unit vector $x_1/|x_1|$. Then from the inequality $|x_i-x_1|>\sqrt 2$ we get that $\langle x_i,e\rangle<0$, and even that $\langle x_i,e\rangle<-\delta$ for some positive $\delta$ depending on $\rho$ only. In particular, every element in the sequence $(x_2,x_3,...)$ belongs to the ball of radius $1-\epsilon$ around $-\delta e$ for some $\epsilon>0$ depending on $\rho$ only. This implies that $\inf_{i\neq j >1} |x_i-x_j| \leq (1-\epsilon)\rho$. But $\inf_{i\neq j} |x_i-x_j|$ was arbitraly close to $\rho$. We thus get $\rho \leq (1-\epsilon)\rho$, a contradiction.

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Mikael, $\rho$ is called the Kottman constant of $\ell_2$. Elton and Odell proved in Colloq. Math. 44 (1981), no. 1, 105–109. 46B20 that the Kottman constant of any infinite dimensional space is larger than one. Papers that reference that paper contain finer results for many special spaces. –  Bill Johnson Apr 13 '11 at 17:28
    
Bill, thank you for these explanations. –  Mikael de la Salle Apr 13 '11 at 18:36
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The optimality of your configuration can be shown as a plain consequence of the Kirszbraun theorem.

(I happened to ask myself this problem too, and eventually added this short section in a wiki article, thinking that it could be useful one day --not completely true, since your question has been already answered by Mikael de la Salle).

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