Liouville property - a very basic question Let $\mathbb{F}_2$ be the free group on two generators. By a result of Kaimanovich and Vershik, for each measure $\mu$ on $\mathbb{F}_2$ such that the support of $\mu$ generates $\mathbb{F}_2$, we have that the random walk is not $\mu$-Liouville, i.e. there is a bounded $\mu$-harmonic function on $\mathbb{F}_2$ which is non-constant. Can one construct this function geometrically without involving KV result? I see this for finitely supported measures. Likely this has been clarified somewhere, I would like to have a citation in the latter case.
 A: If I understand correctly the question is a request for a formula expressing a specific non-constant bounded $\mu$-harmonic function on $\mathbb{F}_2$, where $\mu$ is a fixed generating probability measure. 
I assume below that $\mathbb{F}_2$ is freely generated by $\{a,b\}$. Let $A$ be the set consisting of all words starting with $a$ in $\mathbb{F}_2$. I claim that
$$ h(x)=\lim_{n\to\infty}\frac{1}{n+1} \sum_{k=0}^n\mu^k(xA), \quad x\in \mathbb{F}_2 $$
is such a $\mu$-harmonic, where $\mu^k$ stands for the $k$-th convolution power of $\mu$ ($\mu^0=\delta_e$).
In fact, it is not hard to see that the formula for $h$ above converges pointwise to a $[0,1]$-valued $\mu$-harmonic. With some work you can see that this function is non-constant.
A slightly more sophisticated way to see this, which is in line with my comments above, is as follows.
Consider the compact space $\bar{\mathbb{F}}_2=\mathbb{F}_2\cup \partial\mathbb{F}_2$. Then $\nu=\lim_{n\to\infty}\frac{1}{n+1} \sum_{k=0}^n\mu^n*\delta_e$ is a (in fact, the unique) stationary measure on $\bar{\mathbb{F}}_2$. It is supported on $\partial\mathbb{F}_2$, because $\mathbb{F}_2$ supports no stationary measure. It is fully supported there, by minimality. It is not hard to see that $(\partial \mathbb{F}_2,\nu)$ is a $\mu$-boundary in the sense of Furstenberg, thus the Poisson transform of any non constant $L^\infty$ function is non-constant. The expression for $h$ above is the Poisson transform for $\chi_\bar{A}$ wrt $\nu$ on $\bar{\mathbb{F}}_2$, which is the same as the Poisson transform for $\chi_{\partial A}$ wrt $\nu$ on $\partial \mathbb{F}_2$. The latter is non-constant by the fact that $\nu$ is fully supported.
Of course, the choice of $A$ in the construction above was quite arbitrary.
