In the proof of the above theorem, I was confused about the following statement:
Since $\rho$ takes its maximum on $(\mathscr{R}_h)_1$, there is an extreme point...
Can anyone explain to me how to conclude that $\rho$ takes its maximum at an extreme point?
$\def\R{\mathscr R}$
Since you are reading Kadison-Ringrose, I'll quote from that book.
On $(\R)_1$, the functional $\rho$ is bounded by $\|\rho\|$. By definition of the norm, this means that there exists a sequence $\{A_n\}\subset(\R)_1$ with $|\rho(A_n)|\to\|\rho\|$. Because $(\R)_1$ is weakly compact, there exists a subnet $\{A_{n_j}\}$ and $A\in(\R)_1$ such that $A_{n_j}\to A$ weakly. As $\rho$ is normal, $$ |\rho(A)|=\lim_j|\rho(A_{n_j})|=\|\rho\|. $$ Write $\rho(A)=|\rho(A)|\,e^{i\theta}$. Then $$ \rho(e^{-i\theta}A)=|\rho(A)|=\|\rho\|. $$ We now remane $e^{-i\theta}A$ to $A$ and we have $\rho(A)=\|\rho\|$. As we have $\rho(A^*)=\overline{\rho(A)}=\overline{\|\rho\|}=\|\rho\|$, $$ \rho\Big(\tfrac12(A+A^*)\Big)=\|\rho\|. $$ So we may replace $A$ with $\tfrac12(A+A^*)$ and thus assume that $A$ is selfadjoint.
Now apply Corollary 1.4.4 to obtain an extreme point $A$ in $(\R_h)_1$ with $\rho(A)=\|\rho\|$.