Reading through the proof of the Krein-Milman theorem
Suppose $X$ is a topological vector space on which $X^*$ separates points. If $K$ is a nonempty compact convex set in $X$, then $K$ is the closed convex hull of the set of its extreme points. In symbols $K \subset \bar{co}(E(K))$
Proof : Let $\mathcal{P}$ be the collection of all compact extreme sets of $K$. Since $K \in \mathcal{P}$, $\mathcal{P} \neq \emptyset$. We shall use the following two properties of $\mathcal{P}$:
a) The intersection $S$ of any nonempty subcollection of $\mathcal{P}$ is a member of $\mathcal{P}$, unless $S \neq \emptyset$.
b) If $S \in \mathcal{P}$, $\Lambda \in X^*$, $\mu$ is the maximum of $Re \; \Lambda$ on $S$ and $$ S_\Lambda = \left\{x \in S : Re \;\Lambda x = \mu \right\} $$ then $S_\Lambda \in \mathcal{P}$.
I struggle with b) a little bit
To prove (b), suppose $tx + (1-t)y = z \in S_\Lambda, x \in K, y \in K, 0 < t < 1$. Since $z \in S$ and $S \subset \mathcal{P}$, we have $x \in S$ and $y \in S$. Hence $Re \; \Lambda x \leq \mu,Re \; \Lambda y \leq \mu$. Since $Re \; \Lambda z = \mu$ and $\Lambda$ is linear, we conclude: $Re \;\Lambda = \mu = Re \; \Lambda y$ Hence $x \in S_\Lambda$ and $y \in S_\Lambda$.
I do not get why the linearity of $\Lambda$ implies $ Re \;\Lambda = \mu = Re \; \Lambda y$. The only clue I had was
$$ \mu = Re \; \Lambda z = Re \; \Lambda \left((1-t)x + t y\right) = (1 - t) Re \; \Lambda x + t Re \;\Lambda y $$ Also since $\mu = (1 - t) \mu + t \mu$ we get $$ (1 - t) \mu + t \mu = (1 - t) Re \; \Lambda x + t Re \;\Lambda y $$
And from here I would guess I can use continuity somehow to assess the result of the author.
Remember as $\mu$ is the maximum value of all $Re \;\Lambda v \quad (v\in S)$ we have: $$ \mu = Re \; \Lambda z = Re \; \Lambda \left((1-t)x + t y\right) = (1 - t) Re \; \Lambda x + t Re \;\Lambda y \le (1 - t) \mu + t \mu = \mu $$ so $$\mu = (1 - t) Re \; \Lambda x + t Re \;\Lambda y $$ Now since $0<t<1$ then in case of $Re \; \Lambda x \lt \mu $ the equality above could not hold. The same holds for $Re \; \Lambda y$ so it must hold $Re \; \Lambda x = \mu $ and $Re \; \Lambda y = \mu $