I'm trying to understand the proof of this statement:
Let $G$ be a countable group, then: $$G \text{ amenable }\iff \exists \nu \in \text{Prob}(G) \text{ s.t }(G,\nu) \text{ is Liouville}$$
For that we first prove that for all group $G$ and $\nu \in \text{Prob}(G)$ we can find $P:l^{\infty}(G)\to l_{\nu}^{\infty}(G)$ such that $P$ is linear, $||P||=1$, $P$ is a $G$-map, $P\geq 0$, and its retriction to $l_{\nu}^{\infty}(G)$ is the identity. Where $l_{\nu}^{\infty}(G)$ is the set of $\nu$-harmonic function in $l^{\infty}(G)$ .
From that we imediatly say that if $(G,\nu)$ is Liouville then $G$ is amenable. This implication should be trivial but I don't see why.
Thank you for your help!
Suppose $(G, \nu)$ is Liouville. Then all harmonic functions are constant, so $\ell^\infty_\nu(G) = \mathbf R$ with trivial $G$-action. This means that $P$ can be identified with a nonnegative $G$-invariant linear map $\ell^\infty(G) \to \mathbf R$ of norm $1$, which is precisely a $G$-invariant mean.