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3

Yes, this is standard. Let $T\in A$ selfadjoint, i.e. with $T=T^*$. Note first that $\phi(T)$ is real: since $T+\|T\|\,\text{id}$ is positive, we have that $$\phi(T)+\|T||\in\mathbb R,$$ so $\phi(T)\in \mathbb R$. Now, as $-T+\|T\|\,\text{id}\geq0$, we get $-\phi(T)+\|T\|\geq0$, so $$\phi(T)\leq\|T\|.$$ Since $-T$ is also selfadjoint, we can also get ...

1

The spectrum of $a$ is a compact set that does not contain $0$. So there is a disk $D$ around $0$ with $D\cap\sigma(a)=\emptyset$. Thus, on $\sigma(a)$, $f:t\longmapsto 1/t$ is continuous, so $f\in C(\sigma(a))$. Then $f(a)\in C^*(a)$ via the Gelfand transform. Or, even easier, you could check that $f$ is analytic on $\sigma(a)$, and so $a^{-1}$ belongs ...

1

Let $\{\phi\}$ be a net of pure states on $A$ and assume that it is $w^*$-convergent to $\phi$. Clearly $\phi$ is a positive linear functional on $A$. To prove $\phi$ is a pure state, it is enough to show that support of $\phi$ is a minimal projection in $A^{**}$. Assume $q$ is a non-trivial projection majorized by the support of $\phi$. We have that ...

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0

Let $f\in A=C_0(0,1)$. Then $f(t)>0$ for all $t$ if and only if $f$ is strictly positive. Assume first that $f(t)>0$ for all $t$. Fix, initially, $g\in C_0(0,1)$ such that $g(t)=0$ for all $t\in (0,b)\cup(1-b,1)$ for some $b>0$. As $[b,1-b]$ is compact and $f>0$ on $[b,1-b]$, there exists $\delta>0$ with $f(t)\geq\delta$ for all $t\in ... 0 Let$a(t)>0$for all$t\in (0,1)$(eg$1-2|t-1/2|$). Then for each$n \in \mathbb{N}$, there exists a function$c_n \in C_0(0,1)$so that$c_n(t)a(t)=1$whenever$t \in I_n :=[\frac{1}{n},1-\frac{1}{n}]$. This consideration shows that for any$f \in C_0(0,1)(a c_n f c_n a)(t)=f(t)$whenever$t \in I_n$. The construction implies that$f \in ...

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