See that pure states are multiplicative linear functionals in a commutative $C^*$-algebra

c-star-algebrasfunctional-analysisoperator-algebrasoperator-theoryquantum-information

I use Zhu's An Introduction to Operator Algebras as my textbook. When doing exercises I encountered the following problem:

(Notations: $\mathcal{U}(\mathcal{A})$ is for the set of all unitary elements in $\mathcal{A}$. The pure states are the extreme points of a state space, which is known to be convex and weak*-closed.)

Problem. Let $\mathcal{A}$ be a $C^{*}$-algebra.

(i) Let $\varphi$ be a state on a $C^{*}$-algebra $\mathcal{A}$.
Suppose that $|\varphi(u)|=1$ for all unitary elements $u \in \mathcal{A}$. Show that $\varphi$ is a pure state. [Hint: $\operatorname{span} \mathcal{U}(\mathcal{A})=\mathcal{A}$ by Zhu Theorem 10.6.]

(ii) Let $\varphi$ be a multiplicative functional on a $C^{*}$-algebra
$\mathcal{A}$. Show that $\varphi$ is a pure state on $\mathcal{A}$.

(iii) Show that the pure and multiplicative states coincide for
commutative $\mathcal{A}$.

I managed to work out the first two problems but I have no idea about the last one. How to see from being an extreme element in the state space of a commutative $\mathcal{A}$ that the extreme element is multiplicative?

PS: There is already a post about this problem: prove that all pure states in a commutative C* algebra are multiplicative linear functionals however it used representations which I have not been taught and don't know anything about.

Any help is appreciated!

Best Answer

Suppose that $\varphi$ is an extreme point in the state space.

Let $a\geq0$ with $0<\varphi(a)<1$ (this exists unless $\varphi=0$). Define $$ \psi_a(x)=\frac{\varphi(ax)}{\varphi(a)}. $$ Because $A$ is abelian, for $x\geq0$ we have $ \psi_a(x)=\varphi(a^{1/2}xa^{1/2})\geq0. $ So $\psi_a$ is positive and $\psi(1)=1$, thus a state. We have, with $t=\varphi(a)$, $$ \varphi(x)=t\,\psi_a(x)+(1-t),\psi_{1-a}(x),\qquad x\in A. $$ As $\varphi$ is extreme, we have $\psi_a=\varphi$. This equality is $$ \varphi(ax)=\varphi(a)\varphi(x). $$ This works for all $x\in A$ and all $a\in A^+$ with $\varphi(a)<1$. As the positive elements span $A$, we get $$ \varphi(ax)=\varphi(a)\varphi(x),\qquad a,x\in A. $$

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