4.4.7 Green boundary limits, Kellogg's theorem and planar Greenian setsBD2BAD3CFA50

Proof. For a bounded open set, continuous boundary data $\Phi _ a$ have a finite harmonic Perron solution, and $G(a,\cdot )=\Phi _ a-H_{\Phi _ a}$. Indeed $\Phi _ a$ is an upper competitor, so the right side is nonnegative and admissible, hence at least $G$. Writing $G=\Phi _ a+h_ a$, the function $-h_ a$ is a lower competitor: it is bounded above by $\Phi _ a$ and bounded on a pole neighborhood by harmonicity, so bounded above throughout the bounded set. Its boundary upper limits are at most $\Phi _ a$. Thus $-h_ a\le H_{\Phi _ a}$, giving the opposite inequality. At regular points the formula gives zero boundary limit; conversely such a Green function is positive harmonic near a boundary point and supplies a weak local barrier.

For arbitrary Greenian $\Omega $, exhaust by bounded regular open sets $D_ j\Subset \Omega $ containing a fixed pole $a$. Extend $G_{D_ j}(a,\cdot )$ by zero outside $D_ j$. On the complement of a small closed pole ball these are nonnegative continuous subharmonic functions: they are harmonic inside, zero outside, and at a boundary point their value zero is at most every sufficiently small sphere average. They increase to the zero extension $g$ of $G_\Omega (a,\cdot )$, bounded above there by the fixed-pole kernel bound. The checked subharmonic envelope/polar discrepancy theorem says $g=g^*$ outside a polar set. On $\partial \Omega $, $g=0$ and $g^*\ge \limsup _{\Omega \ni x\to z}G_\Omega (a,x)\ge 0$, so that limsup is zero off a polar set.

Choose one pole in each component; there are at most countably many components. The union of their exceptions is polar. Harnack chains in the pole variable make the zero limit uniform on any compact subset of poles in a component. A compact subset meets only finitely many components, so the limit is uniform on every compact subset of $\Omega $. Integration against a finite compactly supported measure proves admissibility with one common exceptional set. The bounded-density statement is then precisely the compact-tail estimate already proved. On a bounded connected set, regularity and the fixed-pole condition are equivalent by the first part, so the irregular set is contained in that pole's polar exception. These arguments remove the temporary admissibility hypothesis in the capacity theorem without using that theorem to prove the boundary discrepancy. $\square$

Proof. A Greenian set has nonpolar boundary by the planar fact in the polar boundary minimum proof, hence nonpolar complement. Conversely, fix $a\in \Omega $. Countably many open disks tangent at $a$ cover the plane minus $a$. If the closed complement is nonpolar, one of its compact intersections $K$ with the corresponding closed disks is nonpolar. Take a larger tangent disk $B$ with $K\Subset B$ and $a\in \partial B$. The ball capacitary potential $p$ of $K$ is nonzero, lies in $[0,1]$, is harmonic off $K$ and zero at $\partial B$. Extend it by zero outside $B$. This is continuous subharmonic on the complement of $K$, by the same boundary mean argument as above. Then $2-p$ is strictly positive superharmonic on $\Omega $ and nonconstant near $a$: it equals two on the outer side of $\partial B$ and is strictly smaller on the adjacent inner component. This formulation also works if $B\setminus K$ has other components where $p=1$.

It remains to obtain a Green kernel from a nonconstant positive superharmonic function $s$, first when $\Omega $ is connected. For any pole $a$, the strong minimum principle gives $b$ with $s(b)<s(a)$. Scale and truncate $s$ so it is at least one on a small closed ball $\overline{B(a,r)}$ and less than one at $b$. The regularized infimum $v$ of nonnegative superharmonic majorants of one on that ball lies in $[0,1]$, equals one inside, is harmonic outside, and is less than one at $b$; these use only saturated families and the checked polar discrepancy theorem, not a Green kernel on $\Omega $. It is continuous on the inner sphere: on a slightly larger concentric annulus its boundary values are continuous on the outer sphere, one on the inner sphere outside a polar exception; Perron comparison ignoring that exception and annulus regularity give inner limit one. Thus $w=1-v$ is continuous subharmonic, zero on the ball, and strictly positive on its complement. That complement is connected: any path crossing the ball may be detoured through a slightly larger sphere inside $\Omega $.

Choose $t>r$ with $\overline{B(a,t)}\subset \Omega $, and $2\alpha =\min _{|x-a|=t}w(x)>0$. Define a subharmonic $q$ by $\log (|x-a|/r)$ on the inner ball, by $\max \{ \log (|x-a|/r),(w(x)/\alpha )\log (t/r)\} $ on the annulus, and by the second term outside the radius-$t$ ball. At the inner sphere the extension dominates the harmonic logarithm with the same value; its sphere submeans follow. Near the outer sphere the second term dominates strictly, so locality proves subharmonicity. It is bounded above and $q+\Phi _ a=-\log r$ near $a$. If $C$ bounds $q$, the nonnegative superharmonic function $C-q$ is $\Phi _ a$ plus a superharmonic correction extending across $a$. Thus the admissible class $\mathcal B_ a$ is nonempty, and the minimal-kernel construction gives a Green function. A disconnected planar open set is already Greenian by the earlier component construction, so the connected argument suffices. $\square$