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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.

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@Rasmus the C* algebra in which I'm working may not be unital. – Jeff Sep 17 '12 at 8:47
I don't really know what you are saying. Afterall, being a state automatically implies in the unital case that $\phi(1)=1$. – Jeff Sep 17 '12 at 8:59
I am following Takesaki's book so I would really like to stick with his definitions, which are given above. I really thought the solution would just be messing around with convex combinations and scalars, but maybe I'm wrong. – Jeff Sep 17 '12 at 9:12
What if you write $\psi$ as a linear combination of states $\psi = \| \phi\| \frac{\phi}{\| \phi \|} + \| \psi- \phi \| \frac{\psi - \phi}{\| \psi - \phi\|}$. Can you derive a contradiction by showing that $\| \phi\| + \| \psi - \phi \| = 1$? – Christopher A. Wong Sep 17 '12 at 19:10
@Christopher A. Wong I considered that. It didn't seem hopeful, but maybe I'll try that way again. Instead I was thinking about using either a Zorn's lemma argument to find somehow the "largest" $\phi$ dominated by $\psi$, or alternatively one may consider things like for a given $\phi$ and $\psi$, $c:=Sup\{\lambda: \lambda \phi \leq \psi\}$ – Jeff Sep 17 '12 at 19:16
up vote 2 down vote accepted

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 $$\rho(x+\lambda 1)=\rho(x)+\lambda=\psi(x)+\lambda=\tilde\psi(x+\lambda).$$

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Your solution presents the interesting fact that ways to extend that are compatible with my desires exist, and are actually unique hence extension itself preserves some basic operations like +, -. Actually, a different answer was suggested, and was something I was trying already myself yesterday. The important detail for any solution is that the norm is additive, not just subadditive, on positive linear functionals, and one can get this by making a very special choice of approximate identity $u_i$ (according to Takesaki for instance) and then using $||f||=lim|f(u_i)|$. – Jeff Sep 19 '12 at 4:06
Incidentally, the proof of the statement $|\psi(x)|^2 \leq \psi(x^*x)$ itself also requires an approximate identity of the C* type. But here the choice can be made with no special care. – Jeff Sep 19 '12 at 4:09
Jeff, you can also prove $|\psi(x)|^2 \leq \psi(x^*x)$ without using C$^*$ theory, by means of Stinespring's Dilation. – Martin Argerami Sep 19 '12 at 4:51

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