Theorem 3.5.5F3C6870EA9A8

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. $\square$