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Let $A$ be a countable subset of the open unit disk $\mathbb D$ centered at 0. For a point $x\in\mathbb D$ denote by $\nu_x$ the associated harmonic measure on the boundary circle $\partial\mathbb D$.

Question: When does there exist a probability measure $\alpha$ on $A$ which satisfies the following stationarity property: $ \nu_0 = \sum_{x\in A} \alpha(x) \nu_x ? $ If yes, what can one say about the collection of all measures $\alpha$ on $A$ with this property? - other than that it is a convex set :)

This question, of course, can be reformulated in terms of the hyperbolic plane $\mathbf H^2$: then $\nu_x$ is just the visual measure on $\partial \mathbf H^2$ as seen from the point $x$. In the same way one can ask it in numerous other situations when a natural boundary is present as well, for instance, for homogeneous trees, symmetric spaces, buildings, etc. Also, one can assume that $A$ is the orbit of a certain discrete group of isometries of the state space.

I am only aware of two results in this direction. One is a construction due to Furstenberg (later extended by Lyons and Sullivan). According to it, if $X$ is a Riemannian manifold of bounded geometry, and $A$ is an $r$-separated subset of $X$ such that the union of $r$-balls centered at points of $A$ is recurrent for the Brownian motion on $X$, then the Brownian motion can in some sense be approximated at infinity with a certain Markov chain on $X$. In particular, the transition probabilities of this chain satisfy the stationarity condition from my question.

The other result is due to Connell-Muchnik who proved that Gibbs measures on the boundary of a CAT(-1) space can always be stationarized in the above way with respect to the orbits of a discrete cocompact action.

Is there anything known about the purely analytic aspects of the problem, just for the unit disk (aka the hyperbolic plane)?

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  • $\begingroup$ Is your stationarity property equivalent with $u(0)=\sum_{x\in A}\alpha(x)u(x)$ for all harmonic functions $u$ in the disc? $\endgroup$ May 12, 2015 at 6:35
  • $\begingroup$ Yes - at least for positive ones. $\endgroup$
    – R W
    May 12, 2015 at 6:47

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