Recall that a separable C*-algebra $A$ is *quasi-diagonal* if there are completely positive and contractive maps $\varphi_k \colon A \rightarrow M_{n(k)}$ such that $||\varphi_k(ab) - \varphi_k(a)\varphi_k(b)|| \rightarrow 0$ and $||\varphi_k(a)|| \rightarrow ||a||$ for every $a, b \in A$, where $M_{n(k)}$ denotes the algebra of $n(k) \times n(k)$ complex matrices.

The following ought to be known to experts, but I've had some issues trying to prove it.

**Question:** Let $A$ be a quasi-diagonal separable C*-algebra. Does $A$ admit a faithful trace?

Observe we do not assume that $A$ is simple. The idea would be to use an ultralimit of the cp maps $\varphi_k \colon A \rightarrow M_{n(k)}$ composed with the usual traces $tr_{n(k)} \colon M_{n(k)} \rightarrow \mathbb{C}$. This does not necessarily yield a faithful trace, though, as the traces $tr_{n(k)}(\varphi_k(a^*a))$ might be very small while $||\varphi_k(a^*a)||$ remains large. However, one could in some cases tweak $\varphi_k$ by adding a factor that enlarges the trace, maintains the asymptotic multiplicativity and the norm. The actual proof requires to do this everywhere at once, that is, for every possible element $a^*a$, and this seems impossible (or unreasonably hard).

Note, moreover, that there is no obvious K-theoretic obstruction, as every countable abelian group $K_0(A)$ embeds into $\mathbb{R}$.

Any help is greatly appreciated.