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Let $\Gamma$ be a finitely generated group, and let $M$ be the Martin boundary of $\Gamma$. I was reading the article on Martin boundary on Encyclopedia of Math, and I have a few questions about what is written there. To fix notations, for $x, y \in \Gamma$, let $K(x, y) = \frac{G(x, y)}{G(e, y)}$ denote the Martin kernel, which is a ratio of Green functions. For the latter to make sense, we have assumed the existence of a non-degenerate measure on $\Gamma$ which generates a transient random walk. Let $\xi$ denote a typical point on $M$.

  1. It is mentioned there that the function $K(., \xi)$ is harmonic. I am not so sure whether this is true in general. I was under the impression that one needs some condition on the measure $\mu$ for harmonicity. For example, see Lemma 7.1 of this paper. Otherwise, the function $K(x, \xi)$ is only superharmonic. Am I missing something? If I am correct about this, how much is the failure of $K(x, \xi)$ to be a harmonic function?

  2. It is mentioned there that the function $K(x, \xi): \Gamma \times M \to [0, \infty]$ is jointly continuous in the variables $x, \xi$. I am trying to make sure what this means. Since we are on a discrete group, does this mean that $\sup_{x \in \Gamma} |K(x, \xi) - K(x, \xi_0)| \to 0$ as $\xi \to \xi_0$ in the Martin boundary metric?

  3. Is there some literature as to when the minimal Martin boundary and the Martin boundary are the same (and not up to measure $0$)?

Thanks in advance!

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  1. The Martin kernels are not always harmonic. You need to assume something, for instance that the measure $\mu$ to be finitely supported. In such case, harmonicity is easy to prove : let $y_n$ converge to a point in the Martin boundary, whose Martin kernel I denote by $K(\cdot, \xi)$. Then, for any $x$, for large enough $n$, $G(\cdot,y_n)/G(e,y_n)$ is harmonic at $x$, so $P G(\cdot, y_n)/G(e,y_n)(x)=G(x,y_n)/G(e,y_n)$. Because $\mu$ is finitely supported $P G(\cdot,y_n)/G(e,y_n)$ converges to $P K(\cdot, \xi)$, so $K(\cdot,\xi)$ is indeed harmonic at $x$ and this is true for all $x$. Note that you can relax a bit the assumption and assume that $\mu$ has finite super-exponential moments. However, in general, Martin kernels may fail to be harmonic, see the paper The Martin boundary for general isotropic random walks in a tree by Cartwright and Sawyer.
  2. The group $\Gamma$ being discrete, continuity means that for every fixed $x$, $K(x,\xi_n)$ converges to $K(x,\xi)$ as $\xi_n$ converges to $\xi$. In general the Martin compactification is metrizable and can be defined as the completion of $\Gamma$ endowed with the metric $$d(x,y)=\sum_{z\in \Gamma}\alpha(z)\frac{|K(z,x)-K(z,y)|+|\delta_{x}(z)-\delta_{y}(z)|}{D_z+1},$$ where $\alpha$ satisfies that $\sum_{z\in \Gamma}\alpha(z)$ is finite and $D_z$ satisfies that $K(z,y)\leq D_z$ for every $z$ and $y$ in $\Gamma$. See Sawyer's survey for more details on this.
  3. Not much is known about the difference between the minimal Martin boundary and the whole Martin boundary. For finitely supported random walks on abelian groups (Ney and Spitzer) and on hyperbolic groups (Gouëzel and references therein for random walks on co-compact Fuchsian groups and random walks on trees), the Martin boundary is minimal. But there are also examples where it is not. First of all, minimal points in the Martin boundary are given by harmonic Martin kernels, so if there are non-harmonic limit Martin kernels, those are not in the minimal boundary. Second, assuming the minimal Martin boundary to be compact, the action of $\Gamma$ on it would be topologically amenable. So any group which is not amenable at infinity cannot have a compact minimal Martin boundary and in particular, the Martin boundary is bigger than its minimal part. See Section 6 in Amenability and the Liouville property by Kaimanovich.
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    $\begingroup$ Thanks for the answer, very helpful! A quick clarification, if I may: I was looking for a statement of the kind "on a hyperbolic group with finitely supported measure, all $K(x, \xi)$ correspond to minimal harmonic functions", but I am unable to find such a statement in Gouëzel's paper. A little pointer, please? $\endgroup$ Commented Jun 15, 2024 at 14:15
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    $\begingroup$ @SMS You're absolutely right, I forgot that this is not stated in Gouëzel's paper, although this follows from the results in the cited paper. Also, Gouëzel is not really the best citation. What he did in the cited paper in terms of Martin boundary was to generalize Ancona's results for the $r$-Martin boundary, for any $r$ up to the radius of convergence of the Green function. The result for the usual boundary, which corresponds to $r=1$, should be attributed to Ancona. $\endgroup$ Commented Jun 16, 2024 at 6:40
  • $\begingroup$ @SMS If you want proper references : Ancona proved that the Martin boundary coincides with the Gromov boundary and that it is minimal in the 80s, see for instance Negatively curved manifolds, elliptic operators, and the Martin bound ary in Ann. Math. Minimality follows from the so-called Ancona inequalities. This is summarized in Woess's book random walks on infinite graphs and groups, see Theorem 27.1 in there. Finally, Gouêzel proved that Ancona inequalities hold uniformly for $r$ up to the radius of convergence of the Green function, this is Theorem 2.3 in the cited paper in my answer $\endgroup$ Commented Jun 16, 2024 at 6:43

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