 Question:
----------


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.


 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.



A conjecture concerning symmetric convex sets
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.


