Let $\mathcal{A}$ be a unital $C^{*}$ algebra with closed unit ball $$ \bar{B}_{1}:=\{A \in \mathcal{A}:\|A\| \leq 1\} $$ Show that if $\mathcal{A}$ is commutative, then the set of extreme points of $\bar{B}_{1}$ coincides with the group of unitary elements of $\mathcal{A}$.
So from a previous result, I can show that if $A$ is a extreme point in $\bar{B}_{1}$, we know that $A^*A$ is a projection. Now from this, I was trying to use some Gelfand Naimark theorem or functional calculus to show that $A^*A$ is in fact the identity that would show that $A$ is unitary, but I was not able to make much progress. Any idea on how I should approach this?
Because $A$ is unital and commutative, $A=C(T)$ for some compact Hausdorff space $T$.
Let $f\in C(T)$. Suppose there exists $t_0\in T$ with $|f(t_0)|<1$. By continuity there exist $\delta\in(0,1)$ and an open set $t_0\ni V$ with $|f(t)|<1-\delta$ on $V$. Use Urysohn's Lemma to construct $g\in C(T)$ with $g(t_0)=\delta$, $\|g\|\leq\delta$ and $g|_{T\setminus V}=0$. Then $$ |f\pm g|\leq1, $$ and $$ f=\frac12\,(f+g)+\frac12\,(f-g). $$ It follows that if $f$ is extreme then $|f(t)|=1$ for all $t$. That is, $f$ is a unitary.