3.5.4 Minimal Green functions on general open sets870251679082

Proof. The class $\mathcal B_ a$ is closed under minimum and lowering on balls away from $a$, since $\Phi _ a$ is harmonic there and lowering preserves superharmonicity of $w$. Its infimum $v$ is nonnegative and harmonic away from $a$ by Lemma 3.5.2. On a ball about $a$, every member dominates the ball Green function: their harmonic singularities cancel and the superharmonic difference has nonnegative boundary lower limits. Hence $v$ also dominates that ball Green function. The local Bôcher lemma gives $v=c\Phi _ a+h$ with $c\ge 0$; the preceding domination forces $c\ge 1$. This expression defines harmonic $h$ across $a$ and on all of $\Omega $. Now $v/c=\Phi _ a+h/c\ge 0$ belongs to $\mathcal B_ a$, so $v\le v/c\le v$. If $c>1$, this would force $v=0$ off $a$, contradicting the ball domination. Thus $c=1$, and $v$ is the required minimal kernel. Uniqueness and necessity follow at once from the definition.

For $d\ge 3$, $\Phi _ a\ge 0$ is an admissible member. For a bounded planar set, $\Phi _ a+C\ge 0$ for a sufficiently large constant. Restriction of an existing Green kernel supplies a member on any smaller open set. If a planar open set avoids a closed ball, translate and dilate so that it avoids $\overline{B(0,1)}$. For $a,y$ in it,

since $|1/a-1/y|<2$. The last two terms are harmonic in $y$ there. For a disconnected planar open set, each component avoids a ball in another component and is Greenian. For a pole in one component, extend its harmonic correction by $-\Phi _ a$ on all the others; the resulting admissible member is zero on those other components. This proves existence for the union.

A planar Green kernel would have a harmonic correction $h_ a\ge -\Phi _ a=\log |\cdot -a|$, which tends to infinity at infinity. On arbitrarily large circles the minimum principle would then force $h_ a(0)$ to exceed every constant, impossible. A minimal kernel is bounded outside every ball $B\Subset \Omega $ containing its pole. Indeed let $m=\max _{\partial B}G(a,\cdot )>0$, and replace it outside $B$ by $\min (G(a,\cdot ),2m)$. This agrees with $G$ near the sphere, is superharmonic, and after subtracting $\Phi _ a$ is superharmonic across the pole. Minimality bounds $G$ by this replacement. On a plane minus a closed discrete set, this bound and the Bôcher lemma remove every deleted point as a bounded harmonic singularity. The resulting nonnegative kernel has a harmonic correction on the whole plane, contradicting the preceding minimum-principle argument. Finally in dimension at least three on the whole space, $G=\Phi _ a+h_ a\le \Phi _ a$ makes $h_ a\le 0$ entire harmonic, hence constant by Liouville. Its constant must be zero because $G\ge 0$ and $\Phi _ a\to 0$ at infinity. $\square$

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. $\square$