A planar barrier forces the regularized Perron envelope to have the prescribed boundary limit

The boundary $\mathrm{\partial }\mathrm{\Omega }$ is closed, and it is bounded because $\mathrm{\Omega }$ is bounded, so $\mathrm{\partial }\mathrm{\Omega }$ is compact by [F5]. Hence there is a neighbourhood $V$ of $\zeta$ with $\mathrm{\mid }\phi (\eta )-\phi (\zeta )\mathrm{\mid }<\epsilon$ for every $\eta \in \mathrm{\partial }\mathrm{\Omega }\cap V$, namely a disc around $\zeta$ whose intersection with the boundary lies inside the open set ${\phi }^{-1}(\phi (\zeta )-\epsilon ,\phi (\zeta )+\epsilon )$. By the barrier property [F2] there is $c_V<0$ with ${\mathrm{lim\hspace{0.17em}sup}}_{z\to \eta ,z\in \mathrm{\Omega }}b(z)\le c_V$ for every $\eta \in \mathrm{\partial }\mathrm{\Omega }\setminus V$. The set $S:=\mathrm{\partial }\mathrm{\Omega }\setminus V$ is a closed subset of the compact set $\mathrm{\partial }\mathrm{\Omega }$, hence compact by [F5]. Put $D=0$ if $S=\mathrm{\varnothing }$; otherwise, since the two displayed functions are continuous on the nonempty compact set $S$, put $D:=\max (0,{\max }_{\eta \in S}\max \{\phi (\zeta )-\epsilon -\phi (\eta ),\phi (\eta )-\phi (\zeta )-\epsilon \})\mathrm{.}$ This finite nonnegative number bounds both deviations that will be needed. Let $A$ be the least positive integer with $A\cdot (-c_V)>D$, which exists because $(-c_V)>0$. Then for every $\eta \in S$ both $\phi (\zeta )-\epsilon +A{c}_V\le \phi (\eta )$ and $\phi (\eta )+A{c}_V\le \phi (\zeta )+\epsilon$, since $A(-c_V)>D$ bounds respectively $\phi (\zeta )-\epsilon -\phi (\eta )$ and $\phi (\eta )-\phi (\zeta )-\epsilon$.

The function $\mathrm{\ell }(z):=\phi (\zeta )-\epsilon +Ab(z)$ is subharmonic on $\mathrm{\Omega }$ by [F3], since $b$ is subharmonic, $A\ge 0$ and constants are harmonic. It belongs to $\mathcal{P}(\phi ,\mathrm{\Omega })$: at a boundary point $\eta \in \mathrm{\partial }\mathrm{\Omega }\cap V$ its limsup is at most $\phi (\zeta )-\epsilon +0<\phi (\eta )$ by the choice of $V$ and $b<0$, and at $\eta \in \mathrm{\partial }\mathrm{\Omega }\setminus V$ it is at most $\phi (\zeta )-\epsilon +A{c}_V\le \phi (\eta )$ by step 1.1. Therefore $U_{\phi }\ge \mathrm{\ell }$ on $\mathrm{\Omega }$ by [F1], that is $U_{\phi }(z)\ge \phi (\zeta )-\epsilon +Ab(z)(z\in \mathrm{\Omega })\mathrm{.}$

Let $v\in \mathcal{P}(\phi ,\mathrm{\Omega })$ and put $w:=v+Ab$. Then $w$ is subharmonic on $\mathrm{\Omega }$ by [F3], and at every boundary point its limsup is at most $\phi (\zeta )+\epsilon$: at $\eta \in \mathrm{\partial }\mathrm{\Omega }\cap V$ we have $\mathrm{lim\hspace{0.17em}sup}w\le \phi (\eta )+0<\phi (\zeta )+\epsilon$ by step 1.1 and $b<0$, while at $\eta \in \mathrm{\partial }\mathrm{\Omega }\setminus V$ we have $\mathrm{lim\hspace{0.17em}sup}w\le \phi (\eta )+A{c}_V\le \phi (\zeta )+\epsilon$ by the upper-deviation bound in step 1.1. So $w\in \mathcal{P}(\phi (\zeta )+\epsilon ,\mathrm{\Omega })$ for the constant datum $\phi (\zeta )+\epsilon$, and [F4] gives the pointwise bound $v(z)\le \phi (\zeta )+\epsilon -Ab(z)(z\in \mathrm{\Omega })\mathrm{.}$

Fix $\delta >0$. Since $b(z)\to 0$ as $z\to \zeta$ inside $\mathrm{\Omega }$ [F2], there is a neighbourhood $N$ of $\zeta$ with $-\delta \le b\le 0$ on $N\cap \mathrm{\Omega }$. Steps 2.1 and 2.2 apply to every $z\in N\cap \mathrm{\Omega }$ and give, after taking suprema in $v$ and using that $b\le 0$ only strengthens the upper bound, $\phi (\zeta )-\epsilon -A\delta \le \mathrm{\ell }(z)\le U_{\phi }(z)\le \phi (\zeta )+\epsilon +A\delta (z\in N\cap \mathrm{\Omega })\mathrm{.}$

The regularization inherits the two bounds on a smaller neighbourhood. Indeed $H_{\phi }\ge U_{\phi }$ by the defining limit in [F1], so $H_{\phi }\ge \phi (\zeta )-\epsilon -A\delta$ on $N\cap \mathrm{\Omega }$; and if $w\in N^{\mathrm{\prime }}\cap \mathrm{\Omega }$ for a neighbourhood $N^{\mathrm{\prime }}$ of $\zeta$ with $\{\mathrm{\mid }w-y\mathrm{\mid }<{\rho }_0\}\subseteq N$ for some ${\rho }_0>0$, then every $y\in \mathrm{\Omega }$ with $\mathrm{\mid }y-w\mathrm{\mid }<{\rho }_0$ lies in $N\cap \mathrm{\Omega }$, so the supremum defining $H_{\phi }(w)$ is at most $\phi (\zeta )+\epsilon +A\delta$ and hence so is its limit $H_{\phi }(w)$ $\mathrm{\mid }{H}_{\phi }(w)-\phi (\zeta )\mathrm{\mid }\le \epsilon +A\delta (w\in N^{\mathrm{\prime }}\cap \mathrm{\Omega })\mathrm{.}$

Given $\eta >0$, apply step 1.1 with $\epsilon :=\eta \mathrm{/}2$ and step 3.1 with $\delta :=\eta \mathrm{/}(2A)$, where $A\ge 1$ is the positive integer of step 1.1; then step 4.1 yields a neighbourhood of $\zeta$ on which $\mathrm{\mid }{H}_{\phi }-\phi (\zeta )\mathrm{\mid }\le \eta$. Hence the limit exists and equals $\phi (\zeta )$, that is, $\zeta$ is regular in the sense of [F2]; the argument used only the barrier $b$ at $\zeta$, the continuity of $\phi$ at $\zeta$ and the compactness of $\mathrm{\partial }\mathrm{\Omega }$, and it made no use of boundary behaviour of $H_{\phi }$ at any other point.