I re-open this subject appeared the first time here Pure states of $C(K)$ are exactly evaluations
If $K$ is a compact Hausdorff space, how to prove that pure states are the functionals $\delta_t, t \in K$ definded by $\delta_t(f) = f(t)$ for all $f \in C(K)$.
By Riesz representation theorem, for each functional $\varphi \in C(K)^*$ there exists a Radon measure $\mu$ such that $\varphi(f) = \int_K f d\mu$ for all $f \in C(K)$.
Suppose that $S$ is the subspace of $C(K)$ states.
We can easily prove that $\delta_t \in S$ for all $t \in K$.
To prove that $\delta_t$ is pure, we suppose that there are $\varphi_1, \varphi_2 \in S$ and $0<\lambda < 1$ such that $\delta_t = (1-\lambda)\varphi_1 + \lambda \varphi_2$.
I want to show that $\varphi_1 = \varphi_2$.
Thanks for any help.
Best Answer
If $\mu_t$ is the probability measure concentrated at $t$, and if $\mu_1,\mu_2$ are probability measures, and if $\mu_t = \lambda \mu_1 + (1-\lambda)\mu_2$ for some $\lambda \in (0,1)$, then you can apply the measures to the singleton $\{t\}$ in order to conclude that $\mu_1\{t\}=1=\mu_2\{t\}$. Since they're probability measures, then $\mu_1=\mu_2=\mu_{t}$.