Does Murphy prove that states separate the points of a C*-algebra?

Question

Im currently reading chapter 3.3 of Murphy's book on C*-algebras. This chapter is about positive linear functionals. On the internet I read somewhere (I lost the source) that states (i.e. positive linear functionals with norm one) separate the points of a C*-algebra $A$. If I'm not mistaken, this means that for any $x,y\in A$ there exists a state $\tau$ on $A$ such that $\tau(x)\neq\tau(y)$. I can't find this result in Murphy's book, is this right? I need a source (preferably Murphy) that I can refer to. Or else, does it follow easily from the results Murphy presents?. Thanks in advance!

Answer 1

I don't know if it is stated explicitly (edit: it is). But here is an argument.

Assume first that $A$ is unital. You want to show that there is a state $\tau$ with $\tau(x-y)\ne0$. So, it is enough to prove that for any nonzero $a\in A$, there exists a state $\tau$ with $\tau(a)\ne0$. Suppose first that $a$ is selfadjoint. Via the Gelfand transform $\Gamma$, $C^*(a)\simeq C(\sigma(a))$, where $a$ is mapped to the identity function; take any nonzero $t\in\sigma(a)$, then $\tau(x)=\Gamma(x)(t)$ is a state on $C^*(a)$ with $\tau(a)\ne0$.

Now we have two problems: that $\tau$ is not defined on all of $A$, and that the above used that $a$ is selfadjoint.

For the first part, because $\tau$ is a state you have that $\|\tau\|=1=\tau(1)$. If you extend $\tau$ to all of $A$ by Hahn-Banach, you get a linear functional $\tilde\tau$ on $A$ with $\tilde\tau(a)=\tau(a)$, and $\|\tilde\tau\|=\|\tau\|$. So $$\|\tilde\tau\|=\|\tau\|=\tau(1)=\tilde\tau(1).$$ Now Corollary 3.3.4 in Murphy's book give you that $\tilde\tau\geq0$, so it is a state.

When $a$ is not selfadjoint, apply the above to $\operatorname{Re} a$.

When $A$ is not unital, apply the above to the unitization of $A$. The restriction of a state is a state.

And now that I wrote everything and I look at the next page, I see that this is Theorem 3.3.6 in Murphy's book.

Answer 2

This is an easy consequence of the Gelfand-Naimark theorem, which Murphy proves in section 3.4 (if I recall correctly).

Indeed, choose a Hilbert space $\mathcal{H}$ such that $A \subseteq B(\mathcal{H})$. Then the vector states $$\{\omega_{\xi}: A \to \mathbb{C}: a \mapsto \langle a\xi, \xi\rangle\mid \xi \in \mathcal{H}\}$$ separate the points of $A$, because $\langle a \xi, \xi\rangle = 0$ for all $\xi \in \mathcal{H}$ implies $a=0$.