# A Hahn-Banach separation theorem argument, claryfying the details

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.

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.