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. Let's denote the volume of the $n$-ball of radius $1/2$ by $V_n$.

The question

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

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 $K^+_{n+1}$.


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 $\operatorname{er} K$. 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 whether $r_n$ 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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    $\begingroup$ Excellent question, Gil, not that I have any idea how to attack it. $\endgroup$ Jun 22, 2010 at 5:48
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    $\begingroup$ The two 1988 papers of Oded Schramm mentioned in Gil's question were essentially Oded's MSc thesis written under the guidance of Gil. $\endgroup$ Jul 1, 2010 at 10:10
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    $\begingroup$ 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. $\endgroup$ Jul 8, 2010 at 15:02
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    $\begingroup$ 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. $\endgroup$ Jul 10, 2010 at 21:10
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    $\begingroup$ @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. $\endgroup$ Feb 27, 2013 at 4:19

1 Answer 1


Warning: This is not an answer to the question as posed, just an explanation of my comment at Gil's request. The question itself remains as open as it was!

Let $L$ be the intersection of the unit (radius $1$) ball with the positive orthant $\{x\in\mathbb R^n: x_i\ge 0\ \forall i\}$. Note that the support function of $L$ with respect to the origin is $h_L(\theta)=|\theta_+|$ where $\theta$ is a unit vector and $\theta_+$ is its "positive part" (i.e., all negative coordinates get replaced by $0$). Take small $r>0$ and let $K$ be the convex hull of $L\cup (-rL)$.

Then $h_K(\theta)=\max(|\theta_+|,r|\theta_-|)$, so the width of $K$ in the direction $\theta$ is $h_K(\theta)+h_K(-\theta)|=\max(|\theta_-|,r|\theta_+|)+\max(|\theta_+|,r|\theta_-|)$. Since $|\theta_+|^2+|\theta_-|^2=1$, this is at least $\frac{1+r}{\sqrt{1+r^2}}\ge 1+\frac r2$ for small $r$.

Now let us estimate the volume of $K$ up to a polynomial in $n$ factor. $K$ is the union of the sets $K_{\rho,m}$ consisting of points $x$ with $m$ negative and $n-m$ positive coordinates such that $|x_-|\le\rho$, $|x_+|\le 1-\frac \rho r$. After some usual mumbo-jumbo about $n$ choices of $m$ and a polynomial net in $\rho$, we see that we just need to bound $\max_{m,\rho}|K_{\rho,m}|$.

Now, $|K_{\rho,m}|={n\choose m}\omega_m\omega_{n-m}2^{-n}\rho^m(1-\frac\rho r)^{n-m}$. We want to compare it to the volume of the ball of diameter $1+\frac r2$, which is $2^{-n}\omega_n(1+\frac r2)^n$ ($\omega_k$ is the volume of the $k$-dimensional unit ball). Note that $\frac{\omega_m\omega_{n-m}}{\omega_n}\approx {n\choose m}^{1/2}$ (up to a polynomial factor), so we want to bound $$ {n\choose m}^{3/2}\rho^m(1-\tfrac\rho r)^{n-m}\le\left[ {n\choose m}(\rho^{2/3})^m(1-\tfrac 23\tfrac\rho r)^{n-m} \right]^{3/2} \\ \le (1+\rho^{2/3}-\tfrac 23\tfrac{\rho}{r})^{\frac 32n}\le (1+r^2)^{\frac 32 n}\, $$ so the ratio is essentially $\left[\frac{(1+r^2)^{3/2}}{1+\frac r2}\right]^n$, which is exponentially small in $n$ for fixed small $r>0$.

I re-iterate that it, probably, doesn't say absolutely anything about the original question, but Gil was interested in details, so here they are. BTW, I'll not get surprised if an even simpler construction is possible. I just needed some bounds for self-dual cones and this particular example is merely a byproduct of one failed attempt to prove something decent in another problem...

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    $\begingroup$ Many thanks dear Fedja. It is very interesting. I wonder if this (or other) construction goes below Schramm's lower bound (roughly $(\sqrt 3 -1)^n$) or perhaps maybe his bound could be extended to your version of the problem. $\endgroup$
    – Gil Kalai
    Jan 3, 2022 at 20:18
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    $\begingroup$ @GilKalai This one, done carefully, gives $(2\sqrt 2-2)^n$, which is still above Schramm's lower bound. It would be interesting to learn how Schramm's proof uses that the width is exactly $1$, not just $\ge 1$. $\endgroup$
    – fedja
    Jan 3, 2022 at 20:34
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    $\begingroup$ Great thanks a lot! (Slightly more details on the mumbo-jumbo part could be helpful). I dont know if Schramm's proof (or any other lower bound) applies to your setting. Any improvement of his bound in the (quite similar) spherical case will improve the upper bound he gave (and later Bourgain and Lindenstrauss by a different way) for Borsuk's problem, on how many sets of smaller diameter are required to cover a set of diameter 1 in $\mathbb R^d$. (BTW, There was an easy argument of a lower bound of $(1/\sqrt 2)^n$ based on a theorem of Larman and Rogers.) $\endgroup$
    – Gil Kalai
    Jan 4, 2022 at 14:14

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