A state $\omega$ on a unital $C^*$ algebra $A$ is called definite at $a\in A$ self-adjoint if $\omega(a^2)=\omega(a)^2$.
I proved that if we have such a definite state at $a$, then for all $b\in A$ we have: $$\omega(ab)=\omega(ba)= \omega(a)\omega(b).$$
Now I want to prove the following. Let $a\in A$ be self-adjoint and assume that for all $0\neq b\in A$, there exists a definite state $\omega$ at $a$ such that $\omega(b)\neq0$. Prove that $a\in Z(A)$, where $Z(A)$ is the center of $A$.
The problem is: we don't have any information which guarantee that $\omega$ is one-to-one, but I know that every state define a semi inner product on $A$ as follows : $\forall a,b\in A$: $\langle a,b\rangle =\omega(b^*a)$, does it help ??
Any idea is really appreciated. Thanks!
Let $I_a$ denote the set of all definite states $\omega$ at $a$. By what you have proved, which simply follows from Cauchy-Schwarz inequality applied to $\omega(x^*y)$ (see here, question a, for a proof), we get $$ ac-ca\in\bigcap_{\omega\in I_a}\;\mbox{Ker}\;\omega\qquad\forall c\in A. $$ Let $c$ in $A$. If $b=ac-ca\neq 0$, there exists $\omega \in I_a$ such that $\omega(b)=\omega(ac-ca)\neq 0$. So this contradicts what you have proved. Hence $ac-ca=0$ for every $c\in A$. That is $a$ belongs to the center of $A$. QED.
For short: the assumption means that $\bigcap_{\omega\in I_a}\;\mbox{Ker}\;\omega=\{0\}$. So every commutator $ac-ca$ must be $0$.