A boundary point is regular exactly when it admits a barrier

Suppose $b$ is a barrier at $\zeta$, and fix a continuous boundary datum $\phi$ and $\epsilon >0$. Choose a neighbourhood $V$ of $\zeta$ such that $\mathrm{\mid }\phi (\eta )-\phi (\zeta )\mathrm{\mid }<\epsilon$ for $\eta \in V\cap \mathrm{\partial }\mathrm{\Omega }$. By [L3] there is $c_V<0$ bounding the boundary limsup of $b$ on $\mathrm{\partial }\mathrm{\Omega }\setminus V$. Since $\phi$ is bounded on the compact boundary, choose $A>0$ large enough that, on this complement, both $\phi (\zeta )-\epsilon +A{c}_V\le \phi (\eta )$ and $\phi (\eta )+A{c}_V\le \phi (\zeta )+\epsilon$ hold. If the complement is empty, any $A>0$ suffices.

Conversely suppose $\zeta$ is regular. Set $\psi (\eta )=-\mathrm{\mid }\eta -\zeta {\mathrm{\mid }}^2$ on $\mathrm{\partial }\mathrm{\Omega }$ and $B=H_{\psi }$. By [L2], $B$ is subharmonic, and regularity gives $B(z)\to \psi (\zeta )=0$ as $z\to \zeta$. For any $v\in \mathcal{P}(\psi ,\mathrm{\Omega })$, [L4] and [L5] make $v(z)+q(z)$ subharmonic, with boundary limsup at most $\psi (\eta )+q(\eta )=0$ at every $\eta \in \mathrm{\partial }\mathrm{\Omega }$. The maximum principle in [L5] gives $v\le -q$. Taking the supremum and regularizing preserves this bound because $q$ is continuous: $B\le -q<0$ on $\mathrm{\Omega }$. For any neighbourhood $V$ of $\zeta$ with nonempty boundary complement, the compact set $\mathrm{\partial }\mathrm{\Omega }\setminus V$ has ${\delta }_V={\min }_{\eta \in \mathrm{\partial }\mathrm{\Omega }\setminus V}\mathrm{\mid }\eta -\zeta {\mathrm{\mid }}^2>0$, so the boundary limsup of $B$ there is at most $-{\delta }_V$. The empty-complement case is vacuous. Thus $B$ is a barrier at $\zeta$.

The function $\mathrm{\ell }(z)=\phi (\zeta )-\epsilon +Ab(z)$ is subharmonic by [L5]. Near $\zeta$ its boundary limsup is at most $\phi (\zeta )-\epsilon \le \phi (\eta )$; away from $\zeta$ the first inequality in step 1.1 gives the same bound. Thus $\mathrm{\ell }\in \mathcal{P}(\phi ,\mathrm{\Omega })$ by [L1], and $H_{\phi }\ge U_{\phi }\ge \mathrm{\ell }$. Since $b(z)\to 0$ at $\zeta$, this proves ${\mathrm{lim\hspace{0.17em}inf}}_{z\to \zeta }{H}_{\phi }(z)\ge \phi (\zeta )-\epsilon$.

Let $v\in \mathcal{P}(\phi ,\mathrm{\Omega })$ be arbitrary. The subharmonic function $s=v+Ab-\phi (\zeta )-\epsilon$ has boundary limsup at most $0$: on $V\cap \mathrm{\partial }\mathrm{\Omega }$ use $\mathrm{lim\hspace{0.17em}sup}v\le \phi (\eta )<\phi (\zeta )+\epsilon$ and $b<0$; on the complement use the second inequality in step 1.1. By [L5], $s\le 0$ throughout $\mathrm{\Omega }$. Taking the supremum over all $v$ gives $U_{\phi }(z)\le \phi (\zeta )+\epsilon -Ab(z)$. For any $\delta >0$, the barrier limit gives a neighbourhood $W$ of $\zeta$ on which $b>-\delta$, hence $U_{\phi }<\phi (\zeta )+\epsilon +A\delta$ on $W\cap \mathrm{\Omega }$. Its upper-semicontinuous regularization satisfies the same weak upper bound on a smaller neighbourhood of $\zeta$. Letting $\delta \downarrow 0$ yields ${\mathrm{lim\hspace{0.17em}sup}}_{z\to \zeta }{H}_{\phi }(z)\le \phi (\zeta )+\epsilon$. Together with step 2.1 and arbitrary $\epsilon$, this proves regularity.

Steps 1.1–3.1 establish both implications.