Question:
Suppose $A$ is a unital C*-algebra, and $a,b\in A_+$ be positive such that $\|a\|=\|b\|=1$. Show $\|a-b\|=1$ iff there is a pure state $\omega$ such that $\omega(a)\omega(b)=0$ and $\omega(a)+\omega(b)=1$.
It's been a few months since I started self-learning C*-algebras. But sometimes I still can't see how to go about a random question like this. The main result I know regarding pure states and the norm of positive elements is Thm 5.1.11 in Murphy(1990). But not sure how to use it here. Thanks in advance.
First assume that there exists a pure state $\omega$ such that $\omega(a)\omega(b)=0$ and $\omega(a+b)=1$. Then $$ \omega(a+b)^2-\omega(a-b)^2=4\omega(a)\omega(b)=0. $$ Thus $\omega(a-b)=\omega(a+b)=1$. It follows that $\lVert a-b\rVert\geq 1$. On the other hand, $-1\leq a-b\leq 1$ implies $\lVert a-b\rVert\leq 1$.
Now assume that $\lVert a-b\rVert=1$. Since $a-b$ is self-adjoint, there exists (see below) a pure state $\omega$ on $A$ such that $\omega(a-b)^2=1$. Since $0\leq \omega(a),\omega(b)\leq 1$, we have $\omega(a)-1\leq \omega(a-b)\leq 1-\omega(b)$, hence $\omega(a-b)^2\leq \max\{1-\omega(a),1-\omega(b)\}$. Therefore $(\omega(a)-\omega(b))^2=1$ is only possible if $\omega(a)=0$ or $\omega(b)=0$.
The existence of the pure state $\omega$ is justified as follows. By Gelfand-Naimark, the unital $C^\ast$-algebra $B$ generated by $a-b$ is $\ast$-isomorphic to $C(X)$ for some compact space $X$ and $a-b$ is sent to some real-valued function $f\in C(X)$. Since $X$ is compact, there exists $x\in X$ such that $\lvert f(x)\rvert=\lVert f\rVert_\infty$. The measure $\delta_x$ is a pure state on $C(X)$, which gives rise to a pure state $\omega_0$ on $B$ such that $\lvert\omega_0(a-b)\rvert=\lVert a-b\rVert$. Moreover, $\omega_0$ can be extended to a pure state $\omega$ on $A$ (see Murphy, Theorem 5.1.13 for example).