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 $y=\infty$?



A question on minimum principle
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.


