Preimage of Maximum of Bounded Linear Functional on Extreme Set in Convex Compact Set is Extreme Set

Theorem

Let $\struct {X, \norm \cdot}$ be a normed vector space over $\R$.

Let $\struct {X^\ast, \norm \cdot_{X^\ast} }$ be the normed dual space of $\struct {X, \norm \cdot}$.

Let $K$ be a compact convex subset of $X$.

Let $M \subseteq K$ be an extreme set in $K$.

Let $f \in X^\ast$.

Let:

$\ds M^f = \set {x \in M : \map f x = \max_{y \in M} \map f y}$

Then $M^f$ is an extreme set in $K$.

Proof

Note that from the extreme value theorem for normed vector spaces we have that:

$f$ attains its supremum

and so $M^f$ is well-defined.

From Mapping is Continuous iff Inverse Images of Open Sets are Open: Corollary, we have:

$M^f$ is closed.

Let $x, y \in K$ and $t \in \openint 0 1$ be such that:

$t x + \paren {1 - t} y \in M^f$

We show that $x, y \in M^f$.

Since $M^f \subseteq M$, we have:

$t x + \paren {1 - t} y \in M$

and so $x, y \in M$ since $M$ is an extreme set in $K$.

Since $t x + \paren {1 - t} y \in M^f$, we have:

$\ds \map f {t x + \paren {1 - t} y} = \max_{z \in M} \map f z$

Since $f \in X^\ast$, we have:

$\ds t \map f x + \paren {1 - t} \map f y = \max_{z \in M} \map f z$

On the other hand, since $x, y \in M$, we have:

$\ds \map f x \le \max_{z \in M} \map f z$

and:

$\ds \map f y \le \max_{z \in M} \map f z$

from the definition of maximum.

If:

$\ds \map f x < \max_{z \in M} \map f z$

or:

$\ds \map f y < \max_{z \in M} \map f z$

we have:

$\ds t \map f x + \paren {1 - t} \map f y < \max_{z \in M} \map f z$

So, we must have:

$\ds \map f x = \map f y = \max_{z \in M} \map f z$

So:

$x, y \in M^f$

So:

$M^f$ is an extreme set in $K$.

$\blacksquare$

Sources

• 2020: James C. Robinson: Introduction to Functional Analysis ... (previous) ... (next) $21.6$: The Krein-Milman Theorem