4.4.4 Local barriers, cone regularity and regular exhaustions93318C47A3DE

Proof. The convex function $|x-a|$ is subharmonic by Jensen on harmonic coordinate functions, so lies in the lower class for $m$. Thus $H_ m\ge |x-a|>0$. Let $w$ be a weak local barrier, and choose a small sphere of radius $r$ centered at $a$ within its domain. Put $M=\max _{\partial D}|z-a|$. Choose compact $F\subset D\cap \partial B(a,r)$ whose complement in that open spherical set has normalized area less than $r/M$; if the open set is empty no $F$ is needed and comparison directly gives $H_ m\le r$ inside. Otherwise $k=\inf _ Fw>0$ (if $F$ is empty choose it nonempty by taking a smaller error). Let $b$ be the ball Poisson integral of boundary data $M$ on $(D\cap \partial B)\setminus F$ and zero elsewhere. It is nonnegative, continuous near the center, with $b(a)<r$. For every lower competitor $v$ for $m$, bounded comparison on $D\cap B$ gives

On $F$ the $w$ term dominates the bound $v\le M$; on the remaining open sphere the Poisson limit is $M$; on the original boundary $v$ has upper limit at most $m\le r$. These verify every boundary piece. Taking the supremum and then approaching $a$ gives $\limsup H_ m\le 2r$. Let $r\to 0$. The global barrier and its positive-away-from-$a$ lower bound now follow from $H_ m\ge |x-a|$. For bounded-above data with local limsup $L$, the function $L+\varepsilon +cH_ m$ belongs to the upper class when $c$ is large enough to control the data outside a small neighborhood. Its limit at $a$ is $L+\varepsilon $. The case $L=-\infty $ is handled with any finite upper target instead. Negation gives the lower inequality. Conversely, regularity applied to continuous $m$ makes $H_ m$ a weak barrier. $\square$

Proof. Restrict a weak barrier for inheritance. For an exterior ball tangent only at $a$ use minus its radial fundamental function after subtracting its value on the sphere; it is positive on the exterior and tends to zero at $a$. A smaller internally tangent ball removes other contact points. For a planar exterior segment choose the logarithm branch cut along its ray; on a sufficiently small slit ball $-\operatorname {Re}(1/\log (z-a))=-\log |z-a|/|\log (z-a)|^2$ is a positive harmonic barrier tending to zero.

For a cone, shrink it and enlarge the domain locally to the truncated cone complement $D_\rho =B(a,\rho )\setminus C$. All its boundary points other than $a$ have exterior balls, hence are regular. On $D_{2\rho }$ let $u$ be the Perron solution for $|z-a|$. On the outer sphere it equals $2\rho $; on lateral cone boundary it equals $|z-a|$. It is bounded between zero and $2\rho $ and is strictly less than $2\rho $ inside. On the compact radius-$\rho $ spherical section (including its regular lateral endpoints) its maximum is $b<2\rho $. Choose $q$ with $\max \{ 1/2,b/(2\rho )\} <q<1$. On the boundary of $D_\rho $ except $a$, $u(x)\le q u(a+2(x-a))$: on the outer section use $b<q2\rho $ and on the lateral section use the factor two. The difference is bounded harmonic; the polar boundary minimum principle permits the singleton exception at $a$. The inequality holds inside. Its finite nonnegative limsup $L$ at $a$ satisfies $L\le qL$, hence is zero. This is a weak local barrier and proves cone regularity.

For exhaustion we give details of the finite-ball construction. Cover any compact $K$ in the open set by finitely many balls of a common radius $r$, with centers in $K$ and closures still inside the open set; choose the cover so a small interval of larger common radii still covers $K$ and has closures inside. Avoid finitely many critical radii: at a boundary point of the union, if the active outward normals had zero in their convex hull, that point would be the circumcenter in the affine span of some affinely independent subset of at most $d+1$ centers, by a minimal convex representation. Each such subset has at most one circumradius, giving a finite forbidden list. At an allowed radius the convex hull of the active normals avoids zero. Its nearest point to zero gives a direction with strictly positive scalar product with every active normal. A small solid cone in this direction avoids all active balls; the other finitely many balls have a positive gap locally. Thus every boundary point of the union has an exterior cone, and the union is bounded regular. Apply this to a compact exhaustion, including the closure of the previously constructed set at each step, to obtain the desired increasing exhaustion. $\square$