# Given a positive element, $a$, of a $C$*-algebra, why does there exists a pure state, $p$, on $A$ such that $p(a)=||a||$? [duplicate]

I'm reading secondary literature where they make this claim, however, I cannot see why it holds true. This is a reformulation from a previous question that I didn't specify good enough.

• What is A? What does it mean to take p(a) where p is a pure state and a is an algebra element? How do you define ||a||? You need to be much more specific and rigorous. – oleg Sep 6 '19 at 15:36
• Reading which literature? – Qmechanic Sep 6 '19 at 16:12
• $A$ is the $C$*-algebra. And $||$ $||$ is the norm. That should be obvious. – Annonymus Sep 6 '19 at 17:30
• @Annonymus Please edit the original question, rather than reposting it as a new one. When you edit the original, it will put it into a queue for people to decide if it should be reopened or not. – tpg2114 Sep 8 '19 at 16:48