Let's suppose that $S \subset \mathbb{R}^n$ is convex and symmetric so:

\begin{equation} x \in S \iff -x \in S \tag{1} \end{equation}

Now, if we define the radius of $S$ as $R$ such that:

\begin{equation} R = \sup_{x \in S} \lVert x \rVert \tag{2} \end{equation}

and use (2) to define:

\begin{equation} V = \{x \in S: \lVert x \rVert = R\} \tag{3} \end{equation}

then I conjecture that:

\begin{equation} S = \text{conv}(V) \tag{*} \end{equation}

I have worked out special cases of this problem within the context of high-dimensional probability but I suspect that it's generally true.

Might there be a theorem which guarantees this result?

Special case:

  1. As some people are voting to close this question I'd like to share my intuition about a special case as I think it might clarify my perspective.

  2. I was thinking in particular about symmetric convex polytopes and my intuition was that all symmetric convex polytopes in $\mathbb{R}^n$ whose vertex set equalled $V$ in (3) were regular polytopes.


I consulted several texts on convexity in high dimensions and couldn't find an answer to this question. For this reason I decided to ask the question here.


closed as off-topic by Mateusz Kwaśnicki, user44191, Yoav Kallus, fedja, Aidan Rocke Jun 17 at 6:42

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Mateusz Kwaśnicki, user44191, Yoav Kallus, fedja, Aidan Rocke
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ It's even worse than in the answers given below: unless $S$ is assumed to be closed, $V$ can well be empty! $\endgroup$ – Mateusz Kwaśnicki Jun 16 at 20:33
  • $\begingroup$ I implicitly assumed we want to also assume that $S$ is closed and bounded (otherwise we can get $R=\infty$ or $V=\varnothing$). $\endgroup$ – user78375 Jun 16 at 22:35

Consider the set $$S:= \{ (x,y) \in \mathbb R^2 : x^2+4y^2 \leq 4 \}$$

This is convex and symmetric, and $R=2$.

But $V= \{ (2,0), (-2,0) \}$ and $\mbox{conv}(V)= \{ (x, 0) : -2 \leq x \leq 2 \} \neq S$.

P.S. About the new question.

Let $A=(0,1), C=(0,-1)$ and $B,D$ be the intersection between the circle $x^2+y^2=1$ and the line $y= \alpha$ where $0<\alpha<1$. Let $B'$ be the reflection of $B$ in the $x$-axis.

Then $AB'CD$ is convex, symmetric has $V= \{ A, B', C, D \}$ but it is not regular unless $\alpha=\frac{\sqrt{2}}{2}$, I think.

  • $\begingroup$ Has this obstruction anything to do with the fact that the isometry group of S is of order greater than 2? $\endgroup$ – Sylvain JULIEN Jun 16 at 19:48
  • $\begingroup$ @SylvainJULIEN I don't think so, you can take this example and expand the positive $x$ axix by a factor of $\alpha >1$ but keep the negative part unexpanded. Then $V$ becomes a single point, and there is no isometry. $\endgroup$ – Nick S Jun 17 at 6:40
  • $\begingroup$ @SylvainJULIEN Also see my PS about the new question. $\endgroup$ – Nick S Jun 17 at 6:45
  • $\begingroup$ @NickS Is $ABCD$ symmetric? $ABCD$ is symmetric about $x=0$ but not about $y=x$ and $y=-x$ as required. $\endgroup$ – Aidan Rocke Jun 17 at 8:14
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    $\begingroup$ @AidanRocke You are right, I need to replace $B$ by $-B$. See the edit, I think it is now. $\endgroup$ – Nick S Jun 17 at 10:29

Take any convex, symmetric, bounded set $T$ in $\mathbb{R}^n$. Choose any point $p\in \mathbb{R}^n$ such that $\|p\|>\sup_{x\in T}\|x\|$ and let $S$ be the convex hull of $T\cup \{\pm p\}$. This set is convex and symmetric, $V=\{\pm p\}$, and the convex hull of $V$ is just the line segment connecting $p$ and $-p$. It is easy to construct counterexmples in this way, for example if the original set $T$ has non-empty interior.


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