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Background

The n dimensional Euclidean ball of radius 1/2 has width 1 in every direction. Namely, when you consider a pair of parallel tangent hyperplanes in any direction the distance between them is 1.

There are other sets of constant width 1. A famous one is the Reuleaux triangle in the plane. The isoperimetric inequality implies that among all sets of constant width 1 the ball has largest volume. Lets denote the Volume of the n-ball of radius 1/2 by $V_n$.

The question

Is there some $\epsilon >0$ so that for every $n>1$ there exist a set $K_n$ of constant width 1 in dimension n whose volume satisfies $VOL(K_n) \le (1-\epsilon)^n V_n$.

This question was asked by Oded Schramm who also asked it for spherical sets of constant width r.

A proposed construction

Here is a proposed construction (also by Schramm). It will be interesting to examine what is the asymptotic behavior of the volume. (And also what is the volume in small dimensions 3,4,...)

Start with $K_1=[-1/2,1/2]$. Given $K_n$ consider it embedded in the hyperplane of all points in $R^{n+1}$ whose (n+1)th coordinate is zero.

Let $K^+_{n+1}$ be the set of all points $x$, with nonnegative (n+1)th coordinate, such that the ball of radius 1 with center at $x$ contains $K_n$.

Let $K^-_{n+1}$ be the set of all points $x$, with nonpositive (n+1)th coordinate, such that $x$ belongs to the intersection of all balls of radius 1 containing $K_n$.

Let $K_{n+1}$ be the union of these two sets $K^-_{n+1}$ and $K^+_{n+1}$.

Motivation

Sets of constant width (other than the ball) are not lucky enough to serve as norms of Banach spaces and to attract the powerful Banach space theorist to study their asymptotic properties for large dimensions. But they are very exciting and this looks like a very basic question.

References and additional motivation

In the paper: "On the volume of sets having constant width" Isr. J Math 63(1988) 178-182 Oded Schramm gives a lower bound on volumes of sets of constant width. Schramm wrote that a good way to present the volume of a set $K \subset R^n$ is to specify the radius of the ball having the same volume as $K$, called it the effective radius of the set $K$ and denote it by $er K$. Next he defined $r_n$ as the minimal effective radius of all sets having constant width two in $R^n$. Schramm proved that $r_n \ge \sqrt {3+2/(n+1)}-1$. He asked if the limit of $r_n$ exists and if $r_n$ is a monotone decreasing sequence.

Our question is essentially wheather $r_n$ tends to 1 as $n$ tends to infinity.

In the paper: O. Schramm, Illuminating sets of constant width. Mathematika 35 (1988), 180--189. Schramm proved a similar lower bound for the spherical case and deduced the best known upper bound for Borsuk's problem on covering sets with sets of smaller diameter.

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3  
Excellent question, Gil, not that I have any idea how to attack it. –  Bill Johnson Jun 22 '10 at 5:48
5  
The two 1988 papers of Oded Schramm mentioned in Gil's question were essentially Oded's MSc thesis written under the guidance of Gil. –  Gideon Schechtman Jul 1 '10 at 10:10
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I think Schramm's algorithm is the same as the one in lama.univ-savoie.fr/~lachand/pdfs/spheroforms.pdf ? Starting from the segment you generate Rouleaux, and then the Meissner body, and so on? According to springerlink.com/content/dn05ruk04k18l687/fulltext.pdf Rouleaux is about 10% smaller than the disk, and the Meissner body is a bit shy of 20% smaller than the ball. From the numerics in the first paper I linked to, the 4-d one is about 23, 24% smaller than its sphere. –  Willie Wong Jul 8 '10 at 15:02
3  
Maybe I'm missing a word like "convex" in the statement of the problem. As written, I could dig out larger and larger holes from the middle of the radius-$1/2$ balls, and make the volumes decay however I want. –  Theo Johnson-Freyd Jul 10 '10 at 21:10
4  
@Jonathan--really? I think that @Theo is right; indeed, the width of a convex body in any direction is the same as the width of its boundary in the same direction (e.g. the width of a unit sphere is 1 in every direction, just like for the unit ball). It doesn't hurt to add word "convex" to the formulation of the question anyway. –  Wlodzimierz Holsztynski Feb 27 '13 at 4:19
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