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 $\phi\colon C_r^*(G)\to\mathbb{C}$ and every $a\in C_r^*(G)$ $$\inf_{b\in K}|\phi(b)-\phi(1)\tau_{\lambda}(a)|=0,$$ where $K$ denotes the norm closed conex hull of $\{\lambda_ga\lambda_{g^{-1}}\mid g\in G\}$.
Proof: By a result stated in the paper, $G$ is not $C^*$-simple if and only if there is a bounded linear functional $\phi\colon C_r^*(G)\to\mathbb{C}$ such that $\phi(1)\tau_\lambda$ does not belong to the weak* closed convex hull of the orbit $G\phi$. By the Hahn-Banach separation theorem, this is equivalent to the existence of $a\in C_r^*(G)$ such that $$\inf_{b\in K}|\phi(b)-\phi(1)\tau_{\lambda}(a)|>0,$$ where $K$ denotes the norm closed conex hull of $\{\lambda_ga\lambda_{g^{-1}}\mid g\in G\}$. (QED)
Note that $\{\lambda_ga\lambda_{g^{-1}}\mid g\in G\}$ 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 $G\phi$, 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 $G\phi$} 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.
Best Answer
Let $A := C^*_r(G)$. Equip $A^*$, the topological dual of $A$, with the weak*-topology. Let us denote, as you already did, $$ K := \overline{\mathrm{co}}^{w*}(G\phi), \qquad F := \{\phi(1)\tau_\lambda\}. $$
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.