Green's function of a Markov process

Remarks

For each $x$ the function $u(x, \cdot)$ is determined only $m$-almost everywhere. One usually fixes a version for which $x \mapsto u(x, y)$ is excessive for each $y$; under standard duality hypotheses such a version exists and is unique. Example: for Brownian motion in $\R^3$ with Lebesgue measure, $u(x, y) = 1/(2\pi\abs{x - y})$, because this is the density of $U(x, \cdot)$. For Brownian motion killed on leaving a planar domain $D$, the Green's function is $u = g_D/\pi$. Here $g_D$ is the classical Green's function with $\Delta g_D(\cdot, y) = -2\pi\delta_y$, and the factor comes from the generator $\frac12 \Delta$. For a Markov chain on a countable set with counting measure, $u(x, y) = \sum_{n \ge 0} P^n(x, y)$ is the expected number of visits to $y$ starting from $x$. Non-example: Brownian motion in $\R^2$ on the whole plane has no Green's function with finite values, since $U(x, A) = \infty$ for every nonempty open $A$.