Extreme points in the set of positive linear functionals of norm $\leq 1$

Question

Let $A$ be a C* algebra, and $S$ the set of positive linear functionals on $A$ in the unit ball of $A^*$ (Which has the weak-* topology.) I am having difficulty seeing that all nonzero extreme points of S must be pure states. For me, a pure state $ \psi$ has norm 1 and if $\phi$ is a positive linear functional for which $ \phi \leq \psi$ then there exists $c \in [0, 1]$ for which $\phi=c*\psi$. So far, my attempts have been as follows. If $\psi$ is nonzero, then for it to be an extreme point, it's clear $\psi$ must have norm 1. So I just need to see it's pure now. Assume that $0 \leq \phi \leq \psi$, from which it follows that $||\phi|| \leq ||\psi||$ and $||\psi-\phi|| \leq ||\psi||$. Then $1/2\psi=1/2\phi+1/2(\psi-\phi)$. That's not good enough though, since $1/2\psi$ need not be extreme. I don't see how I can use a condition like "extreme" which relies on convex combinations when multiplying by scalars like $1/2$ is not permissible.

Answer 1

Let $\psi$ be extreme. We want to show it is pure.

Assume first that $A$ is unital. In this case, we know that $\psi(1)=1$. If $0\leq\phi\leq\psi$, then $\psi-\phi$ is a positive linear functional. Let $\lambda=\phi(1)$. If $\lambda=1$, then $(\psi-\phi)(1)=0$; as positive functionals achieve their norm at the identity, $\phi=\psi$. If $\lambda<1$, then $$ \psi=\lambda\,\frac\phi\lambda+(1-\lambda)\,\frac{\psi-\phi}{1-\lambda}, $$ a convex combination of states. As $\psi$ is extremal, we deduce that $\phi/\lambda=\psi$, i.e. $\phi=\lambda\psi$.

Now, for the case where $A$ is non-unital. Let $\tilde A$ be its unitization. Consider the following claim, which we prove at the end:

Claim: each state on $A$ has a unique extension to a state on $\tilde A$.

Using the claim, it is clear that $\tilde\psi$, the extension of $\psi$ to $\tilde A$ is extremal. Indeed, if $\tilde\psi=\alpha\tilde\varphi_1+(1-\alpha)\tilde\varphi_2$ then both need to be states for the equality to hold, and we get the same equality when restricting to $A$ (where again the equality forces both functionals to be states); there, we use that $\psi$ is extremal to deduce that $\varphi_1=\varphi_2=\psi$. As extensions are unique, we get $\tilde\varphi_1=\tilde\varphi_2=\tilde\psi$.

Now, since $\tilde\psi$ is extremal, by the first part $\tilde\psi$ is pure. Any extension $\phi'$ of $\phi$ will satisfy $\phi'\leq\tilde\psi$ (this is a computation similar to the one in the proof of the claim), so by the first part of the proof we get $\psi'=c\tilde\psi$ for some $c\in[0,1]$, and this of course still holds for the restrictions.

Proof of the Claim: Since $\tilde A=A+\mathbb{C}1$, we can define $\tilde\psi(x+\lambda 1)=\psi(x)+\lambda$. This is clearly unital. It is also positive, as $$ \tilde\psi((x+\lambda 1)^*(x+\lambda 1))=\tilde\psi(x^*x+2\mbox{Re}\lambda x+|\lambda|^2) =\psi(x^*x)+2\mbox{Re}\lambda\psi(x)+|\lambda|^2\geq|\psi(x)|^2+2\mbox{Re}\lambda\psi(x)+|\lambda|^2=|\psi(x)+\lambda|^2\geq0 $$ (here we are using that $\psi\geq0$ in two places: in that it preserves adjoints, and in the Cauchy-Schwarz inequality $|\psi(x)|^2\leq\psi(x^*x)$). It is a state, because it is positive and $\psi(1)=1$.

For the uniqueness, if $\rho$ is a state on $\tilde A$ such that $\rho|_A=\psi$, then $$\rho(x+\lambda 1)=\rho(x)+\lambda=\psi(x)+\lambda=\tilde\psi(x+\lambda).$$