Given two distinct pure state $\phi_1$ and $\phi_2$ in a commutative unital C$^*$-algebra, how do we show $\|\phi_1-\phi_2\|=2$?
Of course, we know the norm must be $\leq 2$ by the triangle inequality, so all we need to do is to find an element $a$ where $\|\phi_1(a)-\phi_2(a)\|=2$. I tried doing this but unfortunately, we don't have much information on the norm of $\|\phi_1(a)-\phi_2(a)\|$ given some arbitrary $a$. Most of the theorems I can currently use are for the converse: given some normal element $b$, we know there exists a state $\omega$ where $\omega(b)=\|b\|$. I could not find much of a way to use this though. Now I have previously proven that states are in fact multiplicative in a commutative unital C$^*$-algebra, but I can't think of a good way to apply that. Another approach is just proof by contradiction and assumes as element $a$ where $||\phi_1(a)-\phi_2(a)||=2$ does not exist and show that either $\phi_1$ or $\phi_2$ is not a pure state which would imply their corresponding GNS representation is not irreducible so I tried to find an invariant subspace in their GNS representation. Now that also seemed too hard to do so I am still stuck.
From the Gelfand-Naimark Theorem it follows that $A\cong C_0(X)$ for some locally compact Hausdorf space $X$. The state space of $C_0(X)$ can be identified with the convex set of (suitably regular) probability measures on $X$ and it is not hard to see that the pure states are precisely the point measures. To see that the norm difference of two distinct pure states $\delta_x$ and $\delta_y$ is two you only need a continuous function $f \in C_0(X)$ such that $f(x) =1$ and $f(y)=-1$ with $\|f\|_\infty =1$.