If $A$ is an abelian C*-algebra,and positive linear functional $\tau$ is pure then it is a character on $A$.
Murphy in his book(C*-algebras and operator theory) has below proof:
While I think we can do it more easier:
Let $\tau$ be pure. Since $A$ is abelian, $\phi_\tau(A)\subset \phi_\tau(A)'$. Also $\phi_\tau(A)'=\Bbb C 1$, so $\phi_\tau(A) \subset \Bbb C 1$ which shows that $$\tau(a)\tau(b)= \langle \phi_\tau(a)x_\tau,x_\tau\rangle\langle \phi_\tau(b)x_\tau,x_\tau\rangle =\langle \phi_\tau(ab)x_\tau,x_\tau\rangle =\tau(ab) $$for $a,b\in A$.
I do not know why murphy shows $B(H) = \Bbb C 1$ and then using whole $B(H)$ to show $\tau$ is multiplicative. Is my argument correct? or Is necessary to follow Murphy's argument? Thanks.
I have just copied and pasted from my own notes on Murphy (my comments in italic). Hopefully something here is helpful!
Assume now that $A$ is abelian.
If $\rho$ is pure, then $\varphi_\rho(A)'=\mathbb{C}1_{H_\rho}$. But $\varphi_\rho(A)\subset \varphi(A)'$, so $\varphi_\rho(A)$ consists of scalars, and therefore $ B(H_\rho)\subset \varphi_\rho(A)'$
I don't agree with this --- however, if $\rho$ is pure, then $(H_\rho,\varphi_\rho)$ is irreducible so that $\varphi_\rho$ has no invariant subspace of $H_\rho$. We want to prove that $B(H_\rho)=\mathbb{C}1_{H_\rho}$ --- i.e. that $H_\rho$ is one dimensional. Suppose on the contrary that $ H_\rho$ is not one dimensional --- in which case it has a two dimensional subspace with basis vectors $\{e_1,e_2\}$. However $\varphi_\rho(A)e_i\subset e_i$ as all of the $ \varphi_\rho(A)$ are scalars and this is a contradiction as $\varphi_\rho(A)$ acts irreducibly.
Hence, $B(H_\rho)=\mathbb{C}I_{H_\rho}$. Therefore, if $T,\,S\in B(H_\rho)$ they are scalars and (using the fact that $x_\rho$ is a unit vector and looking at $\langle \lambda \mu x_\rho,x_\rho\rangle=\langle \lambda x_\rho,x_\rho\rangle\langle\mu x_\rho,x_\rho\rangle$) it is clear $\rho(ab)=\rho(a)\rho(b)$ and therefore a character on $A$.
Now suppose conversely that $\rho$ is a character on $A$, and let $\tau$ be a positive linear functional on $A$ such that $\tau\leq\rho$. If $\rho(a)=0$, then $ \rho(a^*a)=0$, so $\tau(a^*a)=0$. Since $|\tau(a)|\leq\tau(a^*a)^{1/2}$, therefore $\tau(a)=0$ (via an application of Cauchy-Schwarz). Hence, $\text{ker }\rho\subset\text{ker }\tau$, and it follows from elementary linear algebra that there is a scalar such that $\tau=t\rho$
Choose a $a\not\in\text{ker }\rho$. Now for all $b\in A$, we have $$\rho\left(b-a\frac{\rho(b)}{\rho(a)}\right)=0.$$ Now we can write $$b=\left(b-a\frac{\rho(b)}{\rho(a)}\right)+a\frac{\rho(b)}{\rho(a)};$$ that is $A=\{c+\lambda a:c\in\text{ker }\rho,\lambda\in\mathbb{C}\}$ with $\rho(c+\lambda a)=\lambda \rho(a)$. Suppose $\text{ker }\rho\subsetneq\text{ker }\tau$. Now choose an $a\in\text{ker }\tau\backslash\text{ker }\rho$. Now any $b\in A$ is of the form $b=c+\lambda a$, with $c\in\text{ker }\rho$ and $a$ as above: $$\begin{align}\Rightarrow \tau(b)&=\tau(c)+\lambda(a)=0+\lambda 0=0 \\\Rightarrow\tau&=0.\end{align}$$ If $\tau$ is non-zero we therefore have that $\text{ker }\rho=\text{ker }\tau$. Now $$\begin{align}\tau(b)&=\tau(c)+\lambda \tau(a)=0+\lambda \tau(a) \\&=\lambda \rho(a)\frac{\tau(a)}{\rho(a)}=\underbrace{\frac{\tau(a)}{\rho(a)}}_{:=t} \rho(b).\end{align}$$ In other words, $\text{dim }A/\text{ker }\rho=1$.
Choose $a\in A$ such that $\rho(a)=1$. Then $\rho(a^*a)=1$, so $0\leq\tau(a^*a)=t\rho(a^*a)=t\leq\rho(a^*a)=1$, and therefore $t\in[0,1]$. This shows that $\rho$ is pure and the equivalence is proved $\bullet$