TITLE: Constant Martin kernel and amenability
QUESTION [8 upvotes]: Consider a finitely supported random walk on a discrete group G such that the support generates $G$ as a semigroup. The Martin kernels are then non-negative functions on the product $G \times M$ where $M$ denotes the Martin boundary of the random walk. Does anyone know an example with $G$ amenable such that there does not exist a point m in the Martin boundary for which the function $K( . , m)$ is constant $1$? And does anyone know an example of a non-amenable group for which there is an element $m$ in the Martin boundary such that the corresponding Martin kernel $K( . ,m)$ is constant $1$?

REPLY [3 votes]: This is a partial answer to your second question. If the function 1 is minimal, then every bounded harmonic function is constant. To reformulate, if there exists a point $m$ in the minimal Martin boundary such that $K(\cdot, m)$ is constant 1,then the Poisson boundary is trivial, which cannot happen if the group is non-amenable. So you have to look at non-minimal points.
EDIT (taking into account the comments of R W)
However, for finitely supported measures, in several classes of groups (such as hyperbolic groups and relatively hyperbolic groups with respect to virtually abelian subgroups), the Martin boundary is minimal.
So in particular, for such examples of groups, the answer is no. You thus have to look for groups such that the minimal boundary is not the whole Martin boundary (such as the examples given by R W).