Let $A$ be a $C^*$-algebra and $\tau\in A^*$ a tracial state. Let $(\pi_\tau, H_\tau, \xi_\tau)$ be the associated GNS-representation. Is it true that $\pi_\tau(A)'\xi_\tau$ is dense in $H_\tau$?
Note that $\xi_\tau$ is separating for $\pi_\tau(A)$ by the traciality assumption.
In particular, if $\tau$ is normal, then $\pi_\tau(A)$ is a von Neumann algebra and the result follows.
If $\tau$ is faithful I can also finish.
Is this true for a general tracial state?
Context: This is used in the proof of theorem 6.1.4 in Brown Ozawa's book "C*-algebras and finite-dimensional approximations".
The reason this works is because you are doing GNS for the trace (it is not true in general, see here). When you do GNS for a trace, separating for $\pi_\tau(A)$ implies separating for $\pi_\tau(A)''$.
Let $x\in\pi_\tau(A)''$ with $x\xi_\tau=0$. Then there exists a net $\{a_j\}\subset A$ with $\pi_\tau(a_j)\to x$ sot. For any $b\in A$, \begin{align} \langle \pi_\tau(a_j)\pi_\tau( b)\xi_\tau,\pi_\tau( b)\xi_\tau\rangle &=\langle \pi_\tau(b^*a_jb)\xi_\tau,\xi_\tau\rangle=\tau(b^*a_jb)=\tau(bb^*a_j)\\[0.3cm] & =\langle \pi_\tau(a_j)\xi_\tau,\pi_\tau(bb^*)\xi_\tau\rangle\to\langle x\xi_\tau,\pi_\tau(bb^*)\xi_\tau\rangle=0. \end{align} As $\pi_\tau( A)\xi_\tau$ is dense, this shows that $a_j\to0$ wot. That is, $x=0$.