Let $A$ be a unital C*algebra, and let $\mathcal R$ be a separating family of irreducible representations of $A$. Each vector state of a representation in $\mathcal R$ is a pure state, and the span of the set $X$ of such states is weak* dense in $A^*$. Is the closed convex hull of $X$ the entire state space of $A$? Does the closure of $X$ contain every pure state of $A$? This is probably textbook material, so please suggest a reference.

$\begingroup$ If $A$ is separable, the answer to the second question should be yes by an application of Glimm's lemma. Are you interested in the nonseparable case? $\endgroup$ – Caleb Eckhardt Aug 25 '17 at 22:15

$\begingroup$ I find the question to be interesting and natural in the general case too, but I would accept an answer for separable $A$. $\endgroup$ – Andre Kornell Aug 26 '17 at 4:24
Good question. I don't know a reference, but I think the answer is yes. The direct sum $\pi$ of the irreps in $\mathcal{R}$ is faithful, so $\\pi(x)\ = \x\$ for all $x \in A$. But the norm of an element of a direct sum is the sup of its norms in the summands, so $\x\ = \sup \\pi_\alpha(x)\$ where $\pi_\alpha$ ranges over $\mathcal{R}$. If $x$ is selfadjoint then for any $\alpha$ there are unit vectors $v$ in the Hilbert space of $\pi_\alpha$ such that $\langle \pi_\alpha(x)v,v\rangle$ gets arbitrarily close to $\\pi_\alpha(x)\$. This means that there are pure states $f$ in your set $X$ for which $f(x)$ gets arbitrarily close to $\x\$. Now if the (weak*) closed convex hull of $X$ were not the entire state space then there would be a pure state $g$ outside it, and a selfadjoint element $x$ of $A$ and $a \in \mathbb{R}$ such that $g(x) > a \geq f(x)$ for all $f \in X$. By what I said above, this would force $g(x) > \x\$, a contradiction.

$\begingroup$ I had actually originally written the first part of the question as a statement, but I realized I couldn't remember the argument. It's really the second part that I'm after. $\endgroup$ – Andre Kornell Aug 24 '17 at 12:50

$\begingroup$ I didn't see that! Let me think some more. $\endgroup$ – Nik Weaver Aug 24 '17 at 13:10

$\begingroup$ (For future readers: this is the right argument for the first part of the question. Thanks Nik!) $\endgroup$ – Andre Kornell Aug 24 '17 at 14:24