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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.

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  • $\begingroup$ What do the states of a commutative unital $C^\ast$-algebra look like in light of Gel'fand duality? $\endgroup$ Jun 5, 2022 at 18:26
  • $\begingroup$ I think we just call that the Gelfand Naimark theorem ….so we have a isomorphism with the space of continuous functions on the characters of our initial Calgebra $\endgroup$
    – Bill
    Jun 5, 2022 at 19:11

1 Answer 1

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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 $\|f\|_\infty =1$.

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    $\begingroup$ thank you. for the last part, the existence of those functions should come from Urysohns lemma right? $\endgroup$
    – Bill
    Jun 5, 2022 at 20:37
  • $\begingroup$ Yes, I think does. $\endgroup$
    – Lau
    Jun 5, 2022 at 20:40

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