Still following the theorem 3.23 (Krein-Milman), follow up to this question.
The proof again continues
It now follows from (b) that every $\Lambda \in X^*$ is constant on $M$, since $X^*$ separates points on $X$, $M$ has only one point.
To get why $\Lambda$ is constant on $M$, I know from (b) that $M_{\Lambda}$ is in $\mathcal{P}$, which means is a compact extreme set, I don't understand why this might imply that $\Lambda$ is constant on $M$. My guess was using the definition of $M_\Lambda$, but I'm not sure how, and mainly the reason is since $\Lambda$ is in $X^*$ it can return a complex number in general, but the definition of $M_\Lambda$ would only use the real part of $\Lambda x$ not the imaginary.
Also the fact that $M$ contains only one point isn't clear to me.
Could you please elaborate?
Best Answer
Well, you know from (b) that $M_{\Lambda}$ is in $\mathcal{P}$. If $\Lambda$ would not be constant on $M_{\Lambda}$, then either the real part of $\Lambda$ is not constant, or the imaginary part is not constant. Without loss of generality we may assume the real part of $\Lambda$ not to be constant, otherwise use $\Lambda:=-i\Lambda\in X^*$
So, if the real part of $\Lambda$ is not constant let us define $Q:=\{x\in M_{\Lambda}: Re \;\Lambda x = \max_{y\in M_{\Lambda}} Re \;\Lambda y\}$. If $\Lambda$ is not a constant then $Q$ is a proper subset of $M_{\Lambda}$ and therefore a proper subset of $M$ - which is not possible by the design of $M$.
In case $M$ contains two different elements, then since $X^*$ separates points you can seperate the two elements with a continuous linear functional. But this functional has to be constant on $M$ as we just proved above, which is a contradiction. So $M$ may just contain one element.