Distance from a state of a $C^*$-algebra to the state space Suppose $\tau$ is a state of a $C^*$-algebra $A$ and $S$ is a $\sigma(A^*, A)$ closed subspace of the state space $S(A)$. The disance forom $\tau$ to $S$ is defined as following:
$d(\tau, S)=inf_{\phi\in S(A)}\|\tau-\phi\|$.
Does there exist $\phi \in S(A)$ such that  $d(\tau,\phi)=d(\tau , S)$? Is the norm on the dual of $A$ lower semi-continuous?
 A: I can only answer for the separable and unital case. Under these conditions $S(A)$ is weak-* closed (because $A$ is unital), so $S$ is also weak-* closed in $A^*$, so $S$ is weak-* compact by Banach Alaoglu. Since $A$ is separable, the closed unit ball of $A$ endowed with the weak-* topology is metrizable. So $S$ endowed with the weak-* topology is homeomorphic to a compact metric space. We define $f:S\to[0,\infty)$ by $f(\phi)=\|\tau-\phi\|$. If we prove that $f$ is lower semi-continuous with respect to the weak-* topology, then since lower semi-continuous functions over compact metric spaces attain their minimum, we will have that there exists $\phi\in S$ so that $\|\tau-\phi\|=d(\tau,S)$.
Let $\phi_i\to\phi$ in $S$ in the weak-* topology. Then $\tau-\phi_i\to \tau-\phi$ in $A^*$ with the weak-* topology. It suffices to show that the norm is lower semi-continuous with respect to the weak-* topology on $A^*$. So let $\psi_i\to\psi$ in $A^*$. If $\varepsilon>0$, then we may find a unit element $a\in A$ so that $\|\psi\|\leq|\psi(a)|+\varepsilon$. Since $\psi_i(a)\to\psi(a)$, we find $i_o\in I$ such that for all $i\geq i_o$ we have $|\psi(a)-\psi_i(a)|<\varepsilon$. But then
$$\|\psi\|\leq|\psi(a)|+\varepsilon\leq|\psi_i(a)|+2\varepsilon\leq\|\psi_i\|+2\varepsilon$$
for all $i\geq i_o$. So the function is lower semi continuous.
Comment: I am not sure if lower semi-continuous functions attain their minima over arbitrary compact topological ($T_2$) spaces, so the separability assumption can be dropped. If anybody knows, please feel free to edit/comment.
