Theorem 3.5.69E37095B7A8A

Proof. Restriction is admissible, proving domain monotonicity. Along an exhaustion the increasing limit is bounded above by $G_\Omega $ off the pole, hence harmonic there. The harmonic corrections $G_ j(a,\cdot )-\Phi _ a$ also increase and are bounded above by $h_ a$; harmonic monotone convergence across the pole makes their limit harmonic on all of $\Omega $. The limit is a nonnegative admissible kernel and is therefore $G_\Omega $ by minimality.

Every harmonic minorant $v$ of $G(a,\cdot )$ satisfies $v\le 0$, since $G-v$ is admissible and minimality gives $G\le G-v$. Thus its greatest harmonic minorant is zero. In the simultaneous lowering process of the preceding theorem, applied to $G(a,\cdot )=\Phi _ a+h_ a$, the harmonic correction is invariant, so

We verify that $q(a,y)=(\Phi _ a)_\infty (y)$ is harmonic in $a$ for each fixed $y$. Choose a locally finite cover of $\Omega $ by relatively compact balls, and enumerate it with repetitions. Such a cover is obtained by covering the compact layers of a nested compact exhaustion by finitely many small balls lying in adjacent exhaustion layers; only finitely many distinct balls meet each compact subset. Its union of boundary spheres $C$ is relatively closed. At a finite lowering stage, integration of $\Phi _ a(z)$ against the successive compact-sphere Poisson kernels is harmonic in $a$ away from those spheres and, until a ball containing $y$ has been used, possibly away from $y$. Differentiation is justified locally by positive distance from the finite union of spheres and bounded derivatives of the fundamental kernel; iterated finite integrals preserve those bounds. Once a ball containing $y$ has been used, there is no remaining point-pole exception. Consequently, on $\Omega \setminus C$ the decreasing stage limits are harmonic or $-\infty $. They are finite because $q(a,y)=-h_ a(y)$ for all $a,y$, as the preceding identity shows; hence they are harmonic there.

For any prescribed $a_0$, choose the locally finite covering balls so that none of their boundary spheres contains $a_0$. This is possible in each finite layer by perturbing radii while retaining the covering; the local finiteness construction is unchanged. Then $a_0\notin C$, and $q(\cdot ,y)$ is harmonic near $a_0$. The greatest-minorant characterization makes $q$ independent of the chosen cover. Thus it is harmonic on all of $\Omega $. For fixed $y$, $\Phi _ y-q(\cdot ,y)$ is nonnegative and admissible, so $G(a,y)\ge G(y,a)$; interchange $a,y$ for equality. On components not containing $a$, a competitor equal to zero there and unchanged on the pole component proves $G=0$ there. Within the pole component, the strong minimum principle on the punctured domain gives strict positivity.

Take disjoint compact balls around two distinct points, lying in slightly larger disjoint balls in $\Omega $. Harnack bounds in the first variable and the established bound away from a pole in the second variable bound $G$ uniformly on their product. Harmonic interior derivative estimates in each variable then give uniform equicontinuity there, hence joint continuity by the triangle inequality. Near a diagonal point, domain monotonicity bounds $G$ below by the Green function of one fixed ball containing both variables. Its explicit formula tends to infinity there.

Finally, away from the averaging sphere, joint continuity on a compact product gives continuity of the spherical integral. On the sphere itself, compare its normalized averages at radii $r+\eta $ and $r-\eta $. Superharmonic spherical-mean monotonicity places the radius-$r$ average between these two continuous functions of the pole $z$. At a fixed pole on the sphere, the radial averages converge as $\eta \to 0$: their continuity in radius follows from annular mean concavity, proved from the annular divergence formula and monotone smooth approximation. They are finite by local surface integrability of a superharmonic function on interior spheres. The squeeze gives continuity at the pole on the sphere. Multiply by the constant surface area to pass from means to the stated integrals. $\square$