Martin kernel

Remarks

For fixed $y$, the function $M(\cdot, y)$ is positive and harmonic on $\Omega \setminus \set{y}$, and $M(x_0, y) = 1$. By Harnack's inequality, when $y$ stays away from a compact set $K \ni x_0$, the functions $M(\cdot, y)$ are uniformly bounded on $K$. Let $(y_k)$ be a sequence that leaves each compact subset of $\Omega$. Then a subsequence of $M(\cdot, y_k)$ converges locally uniformly to a positive harmonic function $h$ with $h(x_0) = 1$. Example: let $\Omega$ be the unit ball of $\R^n$ and $x_0 = 0$. If $y_k \to \zeta$ with $\abs{\zeta} = 1$, then $M(x, y_k) \to (1 - \abs{x}^2) / \abs{x - \zeta}^n$, the Poisson kernel for the normalised surface measure. A change of reference point $x_0$ multiplies $M(\cdot, y)$ by the factor $G(x_0, y) / G(x_0', y)$.