2
$\begingroup$

Let $V$ be some `nice' vector space and let $T: V\to \mathbb{R}$ be a linear functional over $V$.

Define \begin{align} M= K \cap \bigcup_{i \in \mathbb{N} } \{ v \in V: T(v)=c_i \} \end{align} where $K$ is some compact and convex subset of $V$. Moreover, $K$ has at most $n$ extreme points.

That is, $M$ is an intersection of $K$ with countably many hyperplanes.

The question I have is, can we say something about the extreme points of $M$?

The general answer, I suspect, is that it is impossible to say something without extra assumptions on $T$. So, we would have to make some assumptions on $T$.

Some motivation: The following result can be shown when the intersection is finite.

Let $\tilde{M}=K \cap \bigcup_{i=1}^m \{ v \in V: T(v)=c_i \} $. Then, one can show, with little restriction on $T$, that the extreme points of $ \tilde{M}$ can be represented as a convex combination of at most $m$ extreme pints of $K$.

$\endgroup$
2
$\begingroup$

The last paragraph is wrong. Consider the case $m=1$. $\tilde{M}$ is the intersection of $K$ with one hyperplane. In general this will not contain any extreme point of $K$, so its extreme points will not be convex combinations of $m=1$ extreme points of $K$.

What is true is that every extreme point of $M$ is an extreme point of the intersection of $K$ with one hyperplane, and this is a convex combination of two extreme points of $K$. Namely, suppose $p = \sum_{i=1}^r t_i p_i$, $t_i \in (0,1)$, $\sum_i t_i = 1$, is a convex combination of $r > 2$ extreme points of $K$. Say $T(p) = c$. If any $T(p_j) = c$, then $p$ is a convex combination of $p_j$ and $(p - t_j p_j)/(1-t_j)$ which are both in $M$, so not an extreme point. Otherwise some $T(p_i) > c$ and some $< c$. Relabelling, suppose $T(p_1) > c$ and $T(p_2) < c$. Then $$q = \frac{T(p_1) - c}{T(p_1)-T(p_2)} p_2 + \frac{c - T(p_2)}{T(p_1)- T(p_2)} p_1$$ is a nontrivial convex combination of $p_1$ and $p_2$ which is in $M$, and $p$ is a convex combination of this and some other member of $M$, thus not an extreme point.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.