Let $A \subseteq \mathcal B(\mathcal H)$ be a unital C*-algebra in its universal representation. The GNS representation $\pi_\mu\colon A \rightarrow \mathcal B(\mathcal H_\mu)$ with base state $\mu$ extends uniquely to a normal $\ast$-homomorphism $\pi_\mu''\colon A'' \to \mathcal B(\mathcal H_\mu)$. Since $A''$ is a von Neumann algebra, there exists a unique projection $p\in A''$ such that $\mathbf{ker} \ \pi_\mu''\ = A''p$. Is $p$ the least upper bound of operators $a \in A$ such that $\mu(a) = 0$ and $0 \leq a \leq 1$?
EDIT: Jonas provided a simple counterexample. For non-commutative C*-algebras, $\mu(a)=0$ of course does not imply that $\pi_\mu(a)=0$. The naive question is therefore whether $p$ is the least upper bound of operators $a\in A$ such that $0 \leq a \leq 1$ and $\mu(c^\ast a c)=0$ for all $c \in A$. More generally, I would be grateful for any such "intrinsic" characterization of the projection $p$.
Let $A \subseteq \mathcal B(\mathcal H)$ be a unital C*-algebra in its universal representation. The GNS representation $\pi_\mu\colon A \rightarrow \mathcal B(\mathcal H_\mu)$ with base state $\mu$ extends uniquely to a normal $\ast$-homomorphism $\pi_\mu''\colon A'' \to \mathcal B(\mathcal H_\mu)$. Since $A''$ is a von Neumann algebra, there exists a unique projection $p\in A''$ such that $\mathbf{ker} \ \pi_\mu''\ = A''p$. Is $p$ the least upper bound of operators $a \in A$ such that $\mu(a) = 0$ and $0 \leq a \leq 1$?