Let $\Gamma$ be a finitely generated group and let $\mu$ be a probability measure on $\Gamma$. Consider the Green function $G(x,y)=\sum\_{n\geq 0}\mu^{\*n}(x^{-1}y)$, where $\mu^{\*n}$ is the $n$th convolution power of $\mu$ and $x,y\in \Gamma$.
Denote by $e$ the neutral element of $\Gamma$.
The Martin kernel is then defined as
$K(x,y)=\frac{G(x,y)}{G(e,y)}$.
As a function of the variable $x$, $K(x,y)$ is $\mu$-harmonic everywhere except at $y$.


Now, the Martin compactification is defined as follows: it is the smallest compact metrizable space $M$, such that $\Gamma\subset M$ is open and dense and such that the Martin kernel $K(\cdot, \cdot)$ continuously extend to $\Gamma \times M$. In other words, a sequence $y\_n$ converges to a point $\tilde{y}$ in the Martin compactification if and only if the function $K(\cdot,y\_n)$ converges pointwise. In this case, denote by $K(\cdot,\tilde{y})$ the limit function. The Martin boundary is then defined as $\partial \Gamma=M\setminus \Gamma$.


Consider a point $\tilde{y}$ in the Martin boundary. Then, $K(\cdot,\tilde{y})$ is a limit a super-harmonic function and it is thus super-harmonic. Moreover, if $z\in \Gamma$ is fixed and if $y\_n$ converges to $\tilde{y}$, then for large enough $n$, $y\_n\neq z$, so $K(\cdot,\tilde{y})$ is a limit a harmonic functions at $z$. This does not ensure however that $K(\cdot, \tilde{y})$ is harmonic at $z$. It does when the measure $\mu$ is finitely supported, or when it has super-exponential moments (see for example the preprint Martin boundary covers Floyd boundary, Gekhtman, Gerasimov, Potyagailo, Yang, <https://arxiv.org/abs/1708.02133>).


In the paper harmonic measures versus quasiconformal measures for hyperbolic groups (<https://arxiv.org/abs/0806.3915>), Blachère, Haïssinsky and Mathieu claim that for every point $\tilde{y}\in \partial \Gamma$, the Martin kernel $K(\cdot,\tilde{y})$ is harmonic. They do not really use this property, so it's not a real issue, but I wonder if their claim is true.


**So my question is the following**: Is it true that Martin kernels $K(\cdot, \tilde{y})$ is always harmonic, when $\tilde{y}$ is in the Martin boundary ? If not, can we improve the assumption of super-exponential moments ?


Another question, more vague though: if $\tilde{y}$ is in the minimal-Martin boundary, then the function $K(\cdot,\tilde{y})$ is indeed harmonic. Is there a link between non-minimality of the Martin boundary and non-harmonicity of Martin kernels ? For example, is $K(\cdot,\tilde{y})$ harmonic if and only if $\tilde{y}$ is minimal ?



Harmonicity of the Martin kernels
There are explicit examples of radially symmetric random walks on free groups for which the Martin boundary contains potentials, see Theorems 2 and 4 in [Cartwright and Sawyer](https://mathscinet.ams.org/mathscinet-getitem?mr=1088396).


Concerning your second question, there are finitely generated groups $G$ for which the set of minimal harmonic functions is **not** closed in the Martin boundary for **any** random walk on $G$, see Remark 6.10 in [Kaimanovich](https://mathscinet.ams.org/mathscinet-getitem?mr=2191210) (the reason is that if the set of minimal harmonic functions is closed, then the group action on it is **topologically amenable**, whereas it is known that there are groups which do not have such actions). Thus, if you take a finitely supported random walk on a group like this, then the closure of minimal harmonic functions necessarily contains non-minimal harmonic functions as well.


