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For a study of the stability of certain maps taking values in a sphere I have the following question.

Let $A$ be a subset of $\mathbb{R}^n$. Suppose $A$ lies in a unit ball, but in no ball of smaller radius centered at any point. I'm interested in bounding the diameter of $A$ below. I conjecture that a regular $n$-dimensional simplex (inscribed in the unit sphere) minimizes this diameter. It's not hard to show that the diameter of such a simplex, when indexed by $n$, forms a strictly decreasing sequence converging to $\sqrt{2}$. Is $\sqrt{2}$ a strict lower bound for general $A$?

I wouldn't be surprised if this problem were solved long ago. I would like a reference I could cite in a paper.

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  • $\begingroup$ I think this a very non-trivial question. The problem is that $A$ does not necessarily contain orthonormal basis. $A$ can be "dusty" or $A$ can be like equilateral triangle. $\endgroup$ Jan 10, 2023 at 17:45
  • $\begingroup$ I don't think it's as bad as it sounds. I want to bound the diameter below. Discarding points from A can only reduce the diameter. In fact, i think it's sufficient to retain at most n+1 carefully chosen points of A. $\endgroup$
    – Steve
    Jan 12, 2023 at 17:31
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    $\begingroup$ I have no possibility to comment. But I think that taking taking a convex hull and using Jung's theorem would be a solution. $\endgroup$ Jan 12, 2023 at 20:11

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It is Jung's theorem. If the diameter would be less than $d=\sqrt{\frac{2(n+1)}{n}}$, there would be a smaller enclosing ball. https://en.wikipedia.org/wiki/Jung%27s_theorem

Supplement: In the plane, there is a remarkable strengthening of Jung: Every figure of unit diameter can be enclosed by a regular hexagon such that its opposite sides are at a distance of $1/\sqrt3$. http://gymarkiv.sdu.dk/MFM/kdvs/mfm%201-9/mfm-3-2.pdf. This usually is the main step in the solution of Borsuk's problem in the plane. https://en.wikipedia.org/wiki/Borsuk%27s_conjecture

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  • $\begingroup$ Thanks so much! I knew somebody must have already solved this problem. $\endgroup$
    – Steve
    Jan 13, 2023 at 15:29
  • $\begingroup$ You are welcome. It's a pleasure to be able to help. $\endgroup$
    – BerndM
    Jan 13, 2023 at 15:38
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$\sqrt{2}$ is indeed a strict lower bound. A fairly short proof goes as follows:

Working with closed sets, we can assume $A$ to be a compact set of the closed unit ball $B$ centered at $0$, perhaps after a suitable translation. No closed ball of radius strictly smaller than $1$ and arbitrary center contains $A$.

By compacity of $A$, it intersects $B$ in a point which we can assume to be the first coordinate $e_1$ of an orthonormal basis.

Suppose now that $A$ has exactly diameter $\sqrt{2}$. Suppose first that $A$ contains no unit vector orthogonal to $e_1$. By compacity of $A$, we can therefore assume that there exists $\epsilon>0$ such that every element of the intersection of $A$ with the unit sphere $S$ has scalar product $\geq \epsilon$ with $e_1$. This implies that the closed halfsphere defined by all unit-vectors making an obtuse angle with $e_1$ is at strictly positive distance from $A$. We can therefore move the center of the unit ball by a small amount in the direction of $e_1$ in order to get a unit ball containing $A$ in its interior.

We can therefore assume that $A$ contains a second element $e_2$ of an orthonormal basis. The same argument as above shows that we can find a smaller sphere if the $n-1$ dimensional halfsphere of unit vectors making obtuse angles with $e_1+e_2$ does not intersect $A$: Move the center of $U$ by a tiny amount in direction $e_1+e_2$. We can therefore assume the existence of a unit vector $e_3$ in $A$ making an obtuse angle with $e_1+e_2$. Since $A$ has diameter $\sqrt 2$, the element $e_3$ is orthogonal to $e_1$ and $e_2$.

Iteration of this argument shows that $A$ contains an orthonormal basis $e_1,\ldots,e_n$. But now we can apply our favourite argument once more and move the center of the unit ball by a tiny amount in the direction of $e_1+\ldots+e_n$ such that the moved unit ball contains $A$ in its interior.

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  • $\begingroup$ I don't find this argument convincing. $\endgroup$
    – Steve
    Jan 12, 2023 at 15:51
  • $\begingroup$ I rechecked it and I did not see problems but perhaps I am blind to an obvious mistake. $\endgroup$ Jan 13, 2023 at 11:29
  • $\begingroup$ In a comment, BerndM informed me that the result I'm looking for is "Jung's theorem". See Wikipedia. It turns out that Jung's proof is rather long, so maybe this problem is harder than it looks. For the case of a circle (n=1) see the reference to Rademacher and Toeplitz given in the Wikipedia article. Even that proof is rather involved. $\endgroup$
    – Steve
    Jan 16, 2023 at 17:54
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Your conjecture that a regular simplex minimizes the diameter is correct.

Let $M$ be the intersection of $A$ with the boundary of the minimum ball containing $A$. Then the convex hull of $M$ contains the center of the ball. By Caratheodory's theorem there are at most $n+1$ points in $M$ whose convex hull contains the center. It remains to prove that any inscribed simplex that contains the center in its interior has at least one edge longer than the edge of a regular inscribed simplex. (If less than $n+1$ points from $M$ were taken, then the dimension reduces; alternatively, just add copies of points.)

This should have been done somewhere. Here is a sketch of a proof. If all edges incident to a vertex of an inscribed simplex (containing the sphere center) are shorter than the edge of a regular inscribed simplex, then the opposite face is closer to the center than a face of a regular inscribed simplex. Then apply induction on the dimension.

Edit: for a more general statement on edge lengths of inscribed simplices see Ilya Bogdanov's answer to this question.

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  • $\begingroup$ The statement 'then the opposite face is closer to the center ...' is not true assuming only that the simplex is inscribed in a sphere: A very tiny simplex which is almost flat can be inscribed in a huge sphere and all its faces can be far away from the origin. One must therefore also use the assumptions. $\endgroup$ Jan 13, 2023 at 10:33
  • $\begingroup$ I assumed that the new simplex also contains the center of the sphere. Will add this to my answer. $\endgroup$ Jan 13, 2023 at 10:38
  • $\begingroup$ OK. Where do you treat the problem of $A$ reduced to $2$ points (and similar stuff given by large simplices of lower dimension) $\endgroup$ Jan 13, 2023 at 10:41
  • $\begingroup$ I do not understand your question. If you mean the induction base, then it is $n=1$, both the sphere and the simplex are segments, the lower bound is 2. $\endgroup$ Jan 13, 2023 at 10:46
  • $\begingroup$ The intersection of $A$ with the sphere can be a simplex of small dimension but $A$ is not necessarily contained in the affine hull of this simplex. Therefore the is perhaps a snag with applying Caratheodory directly. $\endgroup$ Jan 13, 2023 at 10:57

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