Martin boundary

Remarks

Each $M(\cdot, \xi)$ with $\xi \in \Delta$ is a positive harmonic function on $\Omega$ with value $1$ at $x_0$. The set $\Omega \cup \Delta$ carries a compact metrisable topology that induces the usual topology on $\Omega$. In it, $\Omega$ is open and dense, and each fundamental sequence converges to its limit in $\Delta$. So it is a compactification of $\Omega$, and for each fixed $x \in \Omega$ the function $y \mapsto M(x, y)$ extends continuously to $\Delta$ by $\xi \mapsto M(x, \xi)$. It does not depend on the choice of $x_0$, up to a homeomorphism that is the identity on $\Omega$. A point $\xi \in \Delta$ is minimal if $M(\cdot, \xi)$ is a minimal harmonic function; the minimal points form the minimal Martin boundary $\Delta_1$. Example: let $B$ be the unit ball of $\R^n$ and $x_0 = 0$. A sequence is fundamental if and only if it converges to a point $\zeta$ of the sphere, and the limit is the Poisson kernel $P(\cdot, \zeta)$. Distinct $\zeta$ give distinct kernels, so $\Delta$ is homeomorphic to the unit sphere. By contrast, consider the slit disc $\set{\abs{z} < 1} \setminus [0, 1)$. Sequences converging to $1/2$ from above the slit and from below it give different limits, so the Martin boundary has two points over $1/2$.