Krdytkyu

I have a question regarding the proof of proposition 6.1. in https://arxiv.org/pdf/1509.01870.pdf, how exactly Hahn-Banach separation theorem has been used.

Proposition 6.1: A discrete group is $C^*$-simple iff for every bounded linear functional $phicolon C_r^*(G)tomathbb{C}$ and every $ain C_r^*(G)$ $$inf_{bin K}|phi(b)-phi(1)tau_{lambda}(a)|=0,$$ where $K$ denotes the norm closed conex hull of ${lambda_galambda_{g^{-1}}mid gin G}$.

Proof: By a result stated in the paper, $G$ is not $C^*$-simple if and only if there is a bounded linear functional $phicolon C_r^*(G)tomathbb{C}$ such that $phi(1)tau_lambda$ does not belong to the weak* closed convex hull of the orbit $Gphi$. By the Hahn-Banach separation theorem, this is equivalent to the existence of $ain C_r^*(G)$ such that $$inf_{bin K}|phi(b)-phi(1)tau_{lambda}(a)|>0,$$ where $K$ denotes the norm closed conex hull of ${lambda_galambda_{g^{-1}}mid gin G}$. (QED)

Note that ${lambda_galambda_{g^{-1}}mid gin G}$ is the G-orbit of $a$ in $C_r^*(G)$.
It's clear to me that if $phi(1)tau_lambda$ does not belong to the weak* closed convex hull of the orbit $Gphi$, that the functionals must have strictly positive disctance somewhere. But how precisly is that Hahn-Banach's separation theorem? I see it is applied to the convex disjoint sets A={the weak* closed convex hull of the orbit $Gphi$} and $B={phi(1)tau_lambda}$ right? But then there must be a functional which seperated $A$ and $B$. And here it doesn't look to me as an application of Hahn-Banach.

Let $A := C^*_r(G)$. Equip $A^*$, the topological dual of $A$, with the weak*-topology. Let us denote, as you already did,
$$
K := overline{mathrm{co}}^{w*}(Gphi), qquad F := {phi(1)tau_lambda}.
$$

Then, as you noted, by Hahn-Banach you find a continuous functional
$$
varphi in (A^*,text{weak*})^*,
$$
which separates $K$ and $F$. Now, the crucial point is that $varphi$ is given by some point evaluation, where we use that $A^*$ is equipped with the weak*-topology.

That gives you the required $a$, which does the job. Furthermore, one has to note that the weak* closure of a convex set in a normed space equals its norm-closure, which again invokes Hahn-Banach.