Let $A$ be a simple unital $C^*$-algebra and denote $U(A)$ the group of unitaries in $A$.
For $u\in U(A)$ lets define the $^*$-automorphism $\text{Ad}u:A\to A$ given by $a\mapsto uau^*$.
It is a known fact that if $\varphi$ is a pure state of $A$ then the set $\{\varphi\circ\text{Ad}u:u\in U(A)\}$ is weak*-dense in the space of pure states of $A$. There exists some variants that also holds and it is not necessary to consider the set of $\textbf{all}$ unitaries in $A$. For example, the set $\{\varphi\circ\text{Ad}\exp(ia):0\leq a\leq\pi\}$ is also a weak*-dense subset of the pure state space.
I am wondering if it is possible to require the unitaries to be "not to far from 1", that is, if for all $\epsilon>0$ the set $\{\varphi\circ\text{Ad}u:\|1-u\|<\epsilon\}$ is weak*-dense in the pure state space.
Thank you very much!
Best Answer
My previous answer did not satisfy all assumptions from the question, as was pointed out by Aweygan in the comments, but one can fix this by considering the finite-dimensional case instead
Let $A=M_n(\mathbb{C})$. The states of the form $\phi_\xi=\langle\,\cdot\,\xi,\xi\rangle$ for $\xi\in \mathbb{C}^n$ with $\|\xi\|=1$ are pure states on $A$, and $\phi_\xi\circ \mathrm{Ad}(u)=\phi_{u^\ast\xi}$. Let $\xi,\eta\in H$ be orthogonal unit vectors and $p$ the orthogonal projection onto the span of $\eta$. If $u$ is unitary with $\|1-u\|<\epsilon$, then $$ |\langle pu^\ast\xi,u^\ast\xi\rangle|\leq \|p(u^\ast \xi-\xi)\|+\|p\xi\|<\epsilon, $$ while $\langle p\eta,\eta\rangle=1$. Thus $\phi_\eta$ does not lie in the weak$^\ast$ closure of $\{\phi_\xi\circ\mathrm{Ad}(u)\colon \|1-u\|<\epsilon\}$ if $\epsilon<1$.