Let A be a C*-Algebra.
Prove the extreme points of $(A_{s.a.})_{1}$ are the self-adjoint unitaries in A.
I am reading the proof from notes on Von Neumann algebras by Peterson in page (14,15), but I have some problems in understanding details.
APPLICATIONS OF FUNCTIONAL CALCULUS
$\frac{1}{2}\left(\left(a-x_{-}\right)+\left(b-x_{-}\right)\right) .$ Since $x$ is an extreme point we conclude that $x=$ $a-x_{-}=b-x_{-}$ and hence $a=b=x_{+}$
We have shown now that $x_{+}$ is an extreme point in $\left(A_{+}\right)_{1}$ and thus by part (i) we conclude that $x_{+}$ is a projection. The same argument shows that $x_{-}$ is also a projection, and thus $x$ is a self-adjoint unitary.
(iii) If $x \in(A)_{1}$ such that $x^{*} x$ is not a projection then by applying functional calculus to $x^{*} x$ we can find an element $y \in A_{+}$ such that $x^{*} x y=y x^{*} x \neq 0$ and $\|x(1 \pm y)\|^{2}=\left\|x^{*} x(1 \pm y)^{2}\right\| \leq 1 .$ Since $x y \neq 0$ we conclude that $x=$ $\frac{1}{2}((x+x y)+(x-x y))$ is not an extreme point of $(A)_{1}$
1- why if $x_+$ and $x_-$ are projection, then we can conclude that x is unitary?( x is a self-adjoint element with norm one)
2- If $1= \frac{a+b}{2}$ where a and b are self adjoint element and $||a||=||b||=1$ , how by using functional calculus we can prove that $a=b=1$.
You have, since $x_+x_-=0$, that $$ x^*x=x^2=(x_+-x_-)^2=x_++x_-=|x|.$$ The assumption is that $|x|$ is a projection; let $q=I-|x|$. Then $$x=\frac12\,(x+q)+\frac12\,(x-q),$$ so by $x$ being an extreme point we get that $x=x+q$, so $q=0$. That is $|x|=I$.
I cannot image what functional calculus has to do with this. You have, for any state $f$, $$ 1=f(I)=\frac12\,(f(a)+f(b))\leq\frac12\,(|f(a)|+|f(b)|)\leq\frac12\,(\|a\|+\|b\|)=1. $$ It follows that $f(a)=f(b)=1$. As we can do this for any state $f$, we have that $f(I-a)=f(I-b)=0$ for all states. As states separate points, $I-a=I-b=0$.