Can one detect a cyclic and separating vector for a concrete $C^*$-algebra using a dense subalgebra? Let $A$ be a $C^*$ algebra of operators acting on some Hilbert space $H$, and $A_0$ is a norm dense $*$-subalgebra of $A$. Suppose there exists some unit vector $\xi \in H$, such that (i) $A_0 \xi$ is dense in $H$; (ii) $a\xi = 0$ if and only if $a=0$ for all $a \in A_0$, i.e. $\xi$ is separating for $A_0$.
Question: is $\xi$ also separating for $A$, i.e. is it true that $a \xi = 0$ implies $a=0$ for all $a \in A$?
It is clear that if the vector state $\omega_\xi$ of $\xi$ is a trace on $A$, then the answer is affirmative by (i). More generally, if $\omega_\xi$ is a KMS state (so that we have sufficient control of $\omega_\xi$ from being away from a trace using suitable automorphisms of $A$), then the answer is still affirmative. But in general, I don't know the answer, and suspect that there is a counter-example.
 A: Here is another example, following a somewhat naive intuition that there could be a suitably chosen involutive algebra generated by operators with sufficiently large range, thus having a high chance of admitting a cyclic and separating vector, yet its norm closure contains sufficiently many rank-one projections, thus such a vector could not be separating for the closure.
Let $H=\ell^2(\mathbb{Z})$ and $(\delta_k)$ the standard orthonormal basis. Denote the corresponding matrix unit by $e_{i,j}$, i.e. $e_{i,j}(\sum_k \xi_k \delta_k) = \xi_j \delta_i$ and pose $e_k = e_{k,k}$. Consider two self-adjoint operators $S, T \in B(H)$ defined as follows:

*

*$S(\delta_0) = \delta_0$, and $S(\delta_k) = \frac{1}{2k}\delta_k$ for $k\ne 0$; or equivalently, $S = e_{0} + \sum_{k \ne 0} \frac{1}{2k}e_k$ with resepct to the norm topology.


*$T(\delta_0) = 0$, and $T(\delta_k) = - \delta_{-k}$; or equivalently, $T = -\sum_{k \ne 0} e_{k,-k}$ with respect to the strong topology.
We calculate
$$ ST = - \sum_{k \ne 0} \frac{1}{2k} e_{k,-k}
= \sum_{k \ne 0} \frac{1}{2k} e_{-k,k} = -TS. $$
Also note that $T^2 = 1 - e_0$, while $Te_0 = e_0 T = 0$ and $S^k e_0 = e_0 S^k = e_0$ for all $k \ge 0$, with the usual convention that $S^0=1$.
Let $\mathcal{A}_0$ be the unital (involutive since $S$, $T$ are self-adjoint) subalgebra generated by $S$ and $T$, and let $\mathcal{A}$ be the norm closure of $\mathcal{A}_0$. By the above calculation, we see that $\mathcal{A}_0$ is linearly spanned by $e_0$, $S^i$ with $i \ge 0$ and $S^i T$ with $i \ge 0$.
Let $\xi = \delta_0 + \sum_{k > 0}\frac{1}{k}(\delta_k + (-1)^k \delta_{-k}) = (\xi_k) \in H$. We claim that $\xi$ is separating for $\mathcal{A}_0$. Indeed, let $x = a e_0 + \sum_{i=0}^N (b_i S^i + c_i S^i T) \in \mathcal{A}_0$, with $a$, $b_i$, $c_i$ being coefficients. To prove the claim, we only need to show that $x \xi = 0$ implies $x = 0$. We calculate
$$ x \xi = \left(a+ \sum_{i=0}^N b_i\right) \delta_0
+ \sum_{k \ne 0}\sum_{i=0}^N (2k)^{-i} (b_i  \xi_k -  c_i \xi_{-k}) \delta_k. $$
Suppose $x \xi = 0$. For $k > 0$, we compare the coefficients of $\delta_k$ in $x \xi$.  If $k$ is odd, we have $\xi_k = - \xi_{-k} \ne 0$ and $\sum_{i=0}^N (2k)^{-i}(b_i + c_i)=0$. Take any $N+1$ different odd $k$ and note that the corresponding Vandermonde determinant does not vanish, we get $b_i + c_i = 0$ for all $0 \le i \le N$. Similarly, if $k$ is even, then $\xi_k = \xi_{-k} \ne 0$, and $\sum_{i=0}^N (2k)^{-i}(b_i - c_i)=0$. This time, taking $N+1$ different even $k$ shows that $b_i - c_i = 0$ for all $0 \le i \le N$. Hence $b_i = c_i = 0$ for all $0 \le i \le N$. Comparing the coefficient of $\delta_0$ now gives $a=0$ too. Thus $x=0$, and $\xi$ is indeed separating for $\mathcal{A}_0$.
Since $\mathcal{A}_0$ is norm dense in $\mathcal{A}$, cyclicity of $\xi$ for $\mathcal{A}_0$ is equivalent to that for $\mathcal{A}$, which we now establish. By the definition of $S$, we have $\| S - e_0 \| = \frac{1}{2}$ and $e_0 (S - e_0) = (S - e_0) e_0 = 0$. Hence for all positive integer $n$, we have $e_0 + (S - e_0)^n = S^n \in \mathcal{A}_0$, which converges to $e_0$ in norm. Thus $e_0 \in \mathcal{A}$. Similarly, $(2(S - e_0))^{2n+1} \in \mathcal{A}_0$ converges in norm to $(e_1 - e_{-1}) \in \mathcal{A}$, and $(2(S - e_0))^{2n} \in \mathcal{A}_0$ converges in norm to $(e_1 + e_{-1}) \in \mathcal{A}$. Thus both $e_1$ and $e_{-1}$ are in $\mathcal{A}$. Continue in this way (i.e. raising $4(S - e_0 - \sum_{0 < |k| \le 1} \frac{1}{2k}e_k)$ to odd powers and even powers then taking limits), we get $e_2, e_{-2} \in \mathcal{A}$. Just repeat the procedure, by induction, we see that every $e_k \in \mathcal{A}$. Since $\xi_k \ne 0$, we have $\delta_k = \xi_k^{-1}e_k \xi \in \mathcal{A}\xi$ for all $k \in \mathbb{Z}$, so $\xi$ is cyclic for $\mathcal{A}$, hence for $\mathcal{A}_0$.
Finally, by definition, $-Te_{-1} \delta_{-1} = \delta_1$ so $0 \ne e_1 - Te_{-1} = e_1 + e_{1,-1} \in \mathcal{A}$, and
$$ (e_1 + e_{1,-1})\xi = (\xi_1 + \xi_{-1})\delta_1 = 0 $$
showing $\xi$ is not separating for $\mathcal{A}$. Normalizing $\xi$ to a unit vector answers the question.
A: Here's a counter-example. Take $A_0:=\mathbb{C}[F(s,t)]\subset A:=\mathrm{C}^*_{\mathrm{r}}(F(s,t))$, where $F(s,t)$ is the free group on $\{s,t\}$, $E\colon A\to \mathrm{C}^*_{\mathrm{r}}(F(s))$ the canonical conditional expectation, and $\psi\colon \mathrm{C}^*_{\mathrm{r}}(F(s)) \cong C(\mathbb{R}/\mathbb{Z})\ni f\mapsto 2\int_0^{1/2} f(r)\,dr$.
Note that the conditional expectation $E$ is faithful and the state $\psi$ is faithful on the algebra $\mathbb{C}[F(s)]$ of trigonometric polynomials.
Put $\varphi=\psi\circ E$ and let $(\pi,H,\xi)$ denote the GNS-triplet.
Since $\varphi$ is faithful on $A_0$, the vector $\xi$ is separating on $\pi(A_0)$.
It is also cyclic because it is a GNS vector.
However $\xi$ is not separating for $\pi(A)$, since $\psi$ is supported on $[0,1/2]$ and $\pi$ is faithful.
