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If I have a bounded set $F$ in $N$ dimensional space and another set $G$ where every element $g$ in $G$ has $h'g=c$ and also must exist in $F$. $H$ is a vector in the $N$ dimensional space and $c$ is any constant $1\times 1$ matrix (scalar). $h$ is a vector of appropriate dimension.

How can I prove every extreme point of $G$ lies on the boundary of $F$?

That is to say if $x$ and $y$ are extreme points in $F$ then $xλ + (1-λ)y = g$

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    Best to define extreme points negatively. For a convex set $S$, $x$ is not an extreme point of $S$ if there exist $y,z \in S$ and 0 < \lambda < 1 such that $x=\lambda y + (1-\lambda) z$.2010-10-14

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I'll prove that every point of $G$ in the interior of $F$ is not an extreme point of $G$. I'll assume that N>1.

LEMMA. There is a vector $v\neq 0$ such that $h'v=0$.

Proof. Since N>1 there is a vector $u$ which is not a multiple of $h$. Let $v=u-\left({{h'u}\over{h'h}}\right)h$ Then $h'v=0$. Since $u$ is not a multiple of $h$, $v\neq 0$.

Answer. Suppose $x \in G$ and $x$ is also in the interior of $F$. Let $v$ be as in the lemma above. Then there is a small enough $\lambda$ such that $(x \pm \lambda v) \in F$. But $h'(x \pm \lambda v)=h'x$ so $(x \pm \lambda v) \in G$. But, $x= {(x + \lambda v) \over 2} + {(x - \lambda v) \over 2}$ So $x$ is not an extreme point of $G$.