# A conjecture concerning symmetric convex sets [closed]

## Question:

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

$$$$x \in S \iff -x \in S \tag{1}$$$$

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

$$$$R = \sup_{x \in S} \lVert x \rVert \tag{2}$$$$

and use (2) to define:

$$$$V = \{x \in S: \lVert x \rVert = R\} \tag{3}$$$$

then I conjecture that:

$$$$S = \text{conv}(V) \tag{*}$$$$

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.

## Remark:

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 RockeJun 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.

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

• Has this obstruction anything to do with the fact that the isometry group of S is of order greater than 2? – Sylvain JULIEN Jun 16 at 19:48
• @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. – Nick S Jun 17 at 6:40
• @SylvainJULIEN Also see my PS about the new question. – Nick S Jun 17 at 6:45
• @NickS Is $ABCD$ symmetric? $ABCD$ is symmetric about $x=0$ but not about $y=x$ and $y=-x$ as required. – Aidan Rocke Jun 17 at 8:14
• @AidanRocke You are right, I need to replace $B$ by $-B$. See the edit, I think it is now. – 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.