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 nonconstant. 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.

1$\begingroup$ What do you mean by "construct"? What do you mean by "this function"? One can always "find" a stationary measure on the geometric boundary, fix a nonconstant $L^\infty$ function on this boundary (say the characteristic function of the boundary of a half tree) and take its Poisson transform. This will give a nonconstant bounded harmonic function. Is this a satisfying "construction"? $\endgroup$ – Uri Bader Apr 2 '16 at 17:08

$\begingroup$ Let me add that one can "construct" the stationary measure as limit of averages of convolution operators applied to a given fixed measure (say the delta measure at the identity). Combining the two "constructions" above, one can find an actual limiting process that will converge to a bounded harmonic function. $\endgroup$ – Uri Bader Apr 2 '16 at 17:29
If I understand correctly the question is a request for a formula expressing a specific nonconstant 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 nonconstant.
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 nonconstant. 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 nonconstant by the fact that $\nu$ is fully supported.
Of course, the choice of $A$ in the construction above was quite arbitrary.

$\begingroup$ Actually, there is no need to take the Cesaro averages as the convolution powers themselves already converge to the stationary measure. $\endgroup$ – R W Apr 2 '16 at 20:27

$\begingroup$ Thanks for this. In fact I had in mind is something which does not really involve convolutions. For instance, if the measure is fin supported, then $f(x)=\mathbb{P}_x(\exists K \forall n\geq K\text{ } X_n \text{is in the left tree})$ is harmonic, nonconstant. $\endgroup$ – Kate Juschenko Apr 3 '16 at 15:05

$\begingroup$ @Kate Juschenko This is precisely the same harmonic function as the one user89334 is talking about, and there is no need to require finiteness of support of the step distribution $\mu$ for its construction. No matter what the step distribution is, $X_n$ almost surely converges to the boundary of the free group, and the arising hitting distribution is precisely the unique stationary measure evoked by user89334. $\endgroup$ – R W Apr 3 '16 at 18:22

$\begingroup$ I agree with both comments of R W. Kate, what exactly is it that you're looking for? the simplest expression? an easy to prove example? An exact citation? $\endgroup$ – Uri Bader Apr 3 '16 at 19:43

1$\begingroup$ @Kate Juschenko Honestly, don't see much difference. The averages are a vestige from what is used in the proof of the BogolyubovKrylov theorem on existence of invariant (in this context stationary) measures. What does make difference with the Schreier graph you have in mind is the very notion of a stationary measure  it does not really make sense for the graph itself, as a result of which the usual Furstenberg's technique for proving boundary convergence (hinged on using the martingale theorem) does not work. $\endgroup$ – R W Apr 4 '16 at 8:04