Let $A$ be a separable (non-unital) C\*-algebra. Let $\tau: A\to\mathbb{C}$ a tracial state. Consider the strict topology on $A$, i.e. $a\_n\to a$ iff $a\_nx\to ax$ and $xa\_n\to xa$ for all $x\in A$. It is clear that this is the same as the usual norm-topology when $A$ has a unit.


My question is:



> 
> When is $\tau$, restricted on the unit ball of $A$, continuous in the strict topology?
> 
> 
> 


In other words: Under what conditions is it true that every norm-bounded sequence $a\_n\in A$, with $a\_nx~,~ xa\_n\to 0$ for all $x\in A$, satisfies $\tau(a\_n^\*a\_n)\to 0$?


Is this always possible? If not, I would be very glad to see a counterexample.
If it turns out to be wrong in general, I would also like to know about the case that $A$ has a unique tracial state and is algebraically simple. Looking at that case would be sufficient for my purposes.



When are tracial states continuous in the strict topology?
Go to the GNS representation. There is a cyclic vector $v$ such that $\tau(x) = \langle \pi(x)v,v\rangle$ for all $x \in A$. Now if $\|a\_nx\| \to 0$ in $A$ for every $x \in A$ then $\|\pi(a\_nx)v\|\to 0$ for all $x \in A$. Since $v$ is cyclic and $(a\_n)$ is bounded, this implies that $\pi(a\_n)v \to 0$, so $\tau(a\_n^\*a\_n) = \|\pi(a\_n)v\|^2 \to 0$.


