4.3.3 Boundary minimum principles with polar exceptions5F11D2AF64E9

Proof. For a bounded $\Omega $, a global polar function infinite on $Z$ can be made nonnegative on $\overline\Omega $ by adding a constant. Call it $v$. Then $u+\varepsilon v$ has boundary lower limits at least $\alpha $ at all points, including $Z$, since $u$ is bounded below. The bounded-domain minimum principle gives $u+\varepsilon v\ge \alpha $. Average over a small interior ball; its average of $v$ is finite. Let $\varepsilon \downarrow 0$ and then the radius decrease to zero to obtain $u\ge \alpha $ everywhere. The case with a condition at infinity follows by truncating with large balls, using $u\ge \alpha -\eta $ outside such a ball, and then letting $\eta \downarrow 0$.

We need one planar fact: the boundary of a nonempty Greenian open set is not polar. If $\Omega $ is not dense and its boundary were polar, take an entire superharmonic polar function $v$ infinite there and replace it by infinity on one component of $\Omega $. The result is lsc, locally supermean (the boundary values are infinity), finite somewhere in the open complement, and hence superharmonic on the plane. It cannot be infinity on a nonempty open set, a contradiction. If $\Omega $ is dense, its complement equals its boundary. A fixed-pole Green function is bounded away from its pole, so polar removability extends it harmonically across the complement outside a small pole ball. Its harmonic correction then extends to the whole plane, and the resulting nonnegative fundamental-plus-harmonic function contradicts the whole-plane nonexistence proved earlier. Thus the planar fact holds.

For unbounded planar $\Omega $, choose $z_0\in \partial \Omega \setminus Z$. For any $\eta >0$ choose $r>0$ with $u\ge \alpha -\eta $ on $\Omega \cap \overline{B(z_0,r)}$. On $D=\Omega \setminus \overline{B(z_0,r)}$ the function $v(x)=\log (|x-z_0|/r)$ is nonnegative harmonic and tends to infinity at infinity. Since $u$ is bounded below, $u+\varepsilon v$ satisfies the required boundary inequality $\alpha -\eta $ on a sufficiently large outer sphere and on the inner sphere, and on the original boundary outside $Z$. The bounded-domain result gives $u+\varepsilon v\ge \alpha -\eta $ on every large truncation of $D$. Let the truncation grow, then $\varepsilon \downarrow 0$, then $\eta \downarrow 0$. Together with the inner ball inequality this proves the assertion on $\Omega $. $\square$