A question on minimum principle

Question

Suppose $D$ be an unbounded domain of $\mathbb{R}^m$ for $m\geq3$, and $u$ is superharmonic on $D$. We know that if $\liminf_{x\to y}u(x)\geq0$ for all $y$ in $\partial^\infty D$ (the boundary of $D$ union the point at infinity), then $ u$ in nonnegative in $D$. Is there any condition (s) that allows us to skip the case $y=\infty$?

Answer 1

Question: For which unbounded domains $D$ does the condition $\liminf_{x\to y}u(x)\geq0$ for all $y$ in $\partial D$ (the finite boundary of $D$), imply that $ u$ in nonnegative in $D$ for bounded subharmonic functions in $D$?

(Remark: The stipulation that $u$ must be bounded was missing in the first version of this answer, thanks to Mateusz Kwaśnicki for pointing that out.)

Answer: The exact criterion is

$(*)$ The point at infinity should have zero harmonic measure in $\partial^\infty D$, i.e., Brownian motion $W_t$ started at a point in $D$ should hit $\partial D$ almost surely.

Indeed, if $(*)$ holds, then the required inequality follows from the supermartingale property of $u(W_t)$ for superharmonic $u$. And if $(*)$ does, not hold define $-u(x0$ to be the harmonic measure of the point at infinity for Brownian motion started at $x$.

There are several other equivalent criteria:

(1) The Martin capacity of $\partial D \setminus B(0,R)$ should be be bounded away from zero as $R \to \infty$;

(2) A Wiener test in terms of Greenian capacity:

Consider the shells $S_k=B(0,2^k) \setminus B(0,2^{k-1})$. Then the requirement is $$\sum_{k \ge 1} 2^{k(2-d)} \cdot \text{Cap}_G (\partial D \cap S_k) =\infty \,.$$

[1] Benjamini, Itai, Robin Pemantle, and Yuval Peres. "Martin capacity for Markov chains." The Annals of Probability (1995): 1332-1346.

[2] Lamperti, John. "Wiener's test and Markov chains." Journal of Mathematical Analysis and Applications 6, no. 1 (1963): 58-66.

Answer 2

The question can be phrased equivalently as follows: for what $D$ there is no infinite Martin boundary point.

This will not be the case for most "typical" domains.

A simple example of an unbounded domain $D$ in $\mathbb R^2 = \mathbb C$ with no infinite Martin boundary point is a "snake" that goes back to the unit ball infinitely many times as it becomes unbounded: a very thin open set that follows the curve $$ x(t) = (1 + t \sin^2 t) e^{i/t} , \qquad t \geqslant \pi . $$ Then $D \setminus B(0,2)$ only contains bounded components, and this easily leads to the desired property.

To some extent, all counterexample must follow the same pattern. I do not have a reference, but I am rather sure this has been studied, perhaps also for finite boundary points (which is an equivalent problem by means of the Kelvin transform).