Suppose I have a convex set $C\subset\mathbb{R}^n$ such that $0\in C$ and every Cauchy sequence in $C$ converges in $C$, but $C$ need not be bounded. (Actually I want unbounded $C$). Consider the set \begin{equation}\mathfrak{L}=\lbrace T:\mathbb{R}^n\rightarrow \mathbb{R}^n, \text{ $T$ is linear}, T(C)\subseteq C\rbrace\end{equation} Is there any relation between extremal points (as well as exposed points) of $C$ and extremal points (exposed points) of $\mathfrak{L}$?

I do not work in convex geometry, and so do not know whether the statement is making sense. I have asked this question here, but did not get any response (and so I decided to ask it here). Please give suggestions and feel free to correct (and edit as well), if I am wrong and/or there is some ambiguity. Advanced thanks for any help.

  • $\begingroup$ You mean that $C$ is closed but not bounded. $\endgroup$ – Denis Serre Sep 13 '12 at 16:00
  • $\begingroup$ @Denis Serre Yes. Example $\lbrace (x,y,z)\in\mathbb{R}^3: ~x^2+y^2\leq z^2\rbrace$ $\endgroup$ – RSG Sep 13 '12 at 16:37

I'm not sure what counts as a "relation" here. It is tempting to think that an extreme point of $\mathfrak L$ would map at least some extreme points of $C$ to extreme points of $C$, but this is not true. For example, in ${\mathbb R}^2$ let $C$ be the convex hull of $(-1,-2)$, $(-1,2)$, $(2,-1)$ and $(2,1)$. Then one of the extreme points of $\mathfrak L$ is $\pmatrix{-1/2 & 0\cr 0 &3/4\cr}$, which maps the extreme points of $C$ to $(1/2, -3/2)$, $(1/2, 3/2)$, $(-1,-3/4)$ and $(-1,3/4)$.

If you want an example with $C$ unbounded, take the cartesian product of this with $[0,\infty)$.


If $C$ is a closed, convex, pointed, solid cone, then there are some results. For example, see

  1. B. S. Tam, A geometric treatment of generalized inverses and semigroups of nonnegative matrices, Linear Algebra Appl., 41 (1981), 225-272.

  2. R. Loewy and H. Schneider, Positive operators on the $n$-dimensional ice-cream cone, J. Math. Anal. Appl., 49 (1975), 375-392.


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