When are positive elements mapped to $1$ by same pure states

Question

Let $\mathcal{A}$ be a unital C*-algebra, and $0\leq a\leq b\in\mathcal{A}$ be 2 positive elements with unit length, $\|a\|=\|b\|=1$. Show $\|b-(1-a)\|=1$.

I'm trying to prove by finding a pure state $f$ such that $f(a)=f(b)=1$. Does it always exist?

Answer 1

You get it for free. Just let $f$ be a pure state with $f(a)=1$. Since $$ 0\leq a\leq b\leq 1, $$ you get $$ 1=f(a)\leq f(b)\leq f(1)=1. $$ So $f(b)=1$.