Im reading over the following theorem in C* algebras by Murphy, and I'm confused on two particular parts:
1: How is Hahn Banach being applied here exactly? What linear functional are we extending to somehow conclude there exists a $\rho \in C(\Omega,\mathbb{R})^{\sharp}$ (the dual over $\mathbb{R}$) with $\rho \circ \theta = \tau'$?
2: How is $\theta$ isometric? Theorem $3.3.6$ states for $a \in A$ a normal elemnent of a nonzero $C^{*}$ algebra, there exists a state $\tau \in A^{*}$ with $||a|| = |\tau(a)|$. From this theorem, clearly $||a|| \leq ||\theta(a)||_{\infty}$, but the other direction is confusing me.
My only guess is that if you apply the Gelfand representation to $C^{*}(a)$, then $\sup_{\Omega'}|\hat{a}(\tau)| = ||\theta(a)||_{\infty} \leq ||\hat{a}||_{\infty} = ||a||$ where $\Omega'$ is the set of all positive linear functionals on $C^{*}(a)$ with norm $\leq 1$. Then the restrictions of all the positive linear functionals on $A$ withh norm $\leq 1$ are contained in $\Omega'$, and since all of the functionals are acting only on a fixed $a$, $|\tau|_{C^{*}(a)}(a)| = |\tau(a)|$ for $\tau \in \Omega$, and hence $\sup_{\tau \in \Omega'} |\tau(a)| \geq ||\theta(a)|| \implies $ equality.
Any explanation would be appreciated.
You choose a sequence $\{a_k\}\subset A_{\rm sa}$ with $$\tag1\lim\tau(a_k)=\|\tau\|.$$ Let $M=\theta(A_{\rm sa})\subset C(\Omega,\mathbb R)$. Then consider $\rho_0:M:\to\mathbb R$ given by $\rho_0(\theta(a))=\tau(a)$. Then $$ |\rho_0(\theta(a))|=|\tau'(a)|\leq\|\tau'\|\,\|a\|=\|\tau'\|\,\|\theta(a)\|, $$ and from $(1)$ you get $\|\rho_0\|=\|\tau'\|$; now you apply Hahn-Banach to extend to all of $C(\Omega,\mathbb R)$.
By definition $\|\tau\|\leq1$ for all $\tau\in\Omega$, so $$|\theta(a)(\tau)|=|\tau(a)|\leq\|\tau\|\,\|a\|=\|a\|.$$