In Dixmier's "C*-Algebras", lemma 3.4.1 states the following: if $A$ is a unital C*-algebra, $S(A)$ the set of states, and $Q\subseteq S(A)$ a discerning subset, then $\overline{\operatorname{conv}}^{w*}Q = S(A)$. Here a subset $Q$ of $S(A)$ is called discerning if whenever $x\in A$ is self-adjoint, and $f(x)\ge 0$ for all $f\in Q$, then $x$ is positive.
I'm having trouble following the proof. Let $x\in A$ be self-adjoint, and $\alpha\in\mathbb{R}$ be such that $f(x)\ge \alpha$ for all $f\in Q$. Then $x - \alpha\cdot 1\ge 0$, so $f(x)\ge \alpha$ for all $f\in S(A)$. Apparently, this is enough to conclude that $S(A)\subseteq\overline{\operatorname{conv}}^{w*}Q$, but I fail to see why. (The opposite inclusion is easy to see.)
The proof is so short on details that I imagine I'm missing something simple, but I've come back to this a couple times now and still can't see the correct thought process.
Ah, the key is the geometric Hahn-Banach. Suppose we had some $f\in S(A)$, but not $C:=\overline{\operatorname{conv}}^{w*}Q$. Assuming the latter set is absorbing in at least one place (which I haven't checked yet), we can exhibit $x\in A$ and $\alpha\in\mathbb{R}$ such that $\Re f(x)\ge \alpha \ge \Re g(x)$ for all $g\in C$. By passing to $\Re x = (x + x^*)/2$, we can exhibit a hermitian $x$ such that $f(x)\ge \alpha\ge g(x)$ for all $g\in C$ (since $f(\Re x) = \Re f(x)$).
By assumption that $Q$ is discerning, $g(\alpha\cdot1 - x)\ge 0$, so that $\alpha\cdot1 - x$ is positive, and so $h(\alpha\cdot1 - x)\ge 0$ for all $h\in S(A)$. Yet $f(\alpha\cdot1 - x)\le 0$, yielding a contradiction.
Also, I should note that this lemma occurs in a slightly different form in Brown and Ozawa (also labelled lemma 3.4.1).