 # Let <mi class="MJX-tex-caligraphic" mathvariant="script">A be a C &#x2217;<!- Maliyah Robles 2022-07-07 Answered
Let $\mathcal{A}$ be a ${C}^{\ast }$-algebra.
(i) Let $\phi$ be a state on a $u\in \mathcal{A}$-algebra $\mathcal{A}$. Suppose that $|\phi \left(u\right)|=1$ for all unitary elements u∈A. Show that φ is a pure state. [Hint: $\mathrm{span}\mathcal{U}\left(\mathcal{A}\right)=\mathcal{A}$ ]
(ii) Let $\phi$ be a multiplicative functional on a ${C}^{\ast }$-algebra $\mathcal{A}$. Show that $\phi$ 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?
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Suppose that $\phi$ is an extreme point in the state space.
Let $a\ge 0$ with $0<\phi \left(a\right)<1$ (this exists unless $\phi =0$). Define
${\psi }_{a}\left(x\right)=\frac{\phi \left(ax\right)}{\phi \left(a\right)}.$Because $A$ is abelian, for $x\ge 0$ we have ${\psi }_{a}\left(x\right)=\phi \left({a}^{1/2}x{a}^{1/2}\right)\ge 0.$. So ${\psi }_{a}$ is positive and $\psi \left(1\right)=1$, thus a state. We have, with $t=\phi \left(a\right)$,
$\phi \left(x\right)=t\phantom{\rule{thinmathspace}{0ex}}{\psi }_{a}\left(x\right)+\left(1-t\right),{\psi }_{1-a}\left(x\right),\phantom{\rule{2em}{0ex}}x\in A.$
As $\phi$ is extreme, we have ${\psi }_{a}=\phi$. This equality is
$\phi \left(ax\right)=\phi \left(a\right)\phi \left(x\right).$
This works for all $x\in A$ and all $a\in {A}^{+}$ with $\phi \left(a\right)<1$. As the positive elements span $A$, we get
$\phi \left(ax\right)=\phi \left(a\right)\phi \left(x\right),\phantom{\rule{2em}{0ex}}a,x\in A.$