= Solution
Fix $\varepsilon>0$ and choose a boundary neighbourhood $U$ of $z$ on which $|\varphi-\varphi(z)|<\varepsilon$. The positive continuous barrier $w$ has a positive minimum $m$ on the compact set $\partial\Omega\setminus U$. Choosing $A$ large enough makes
$$
\varphi(z)-\varepsilon-Aw\leq\varphi
\leq\varphi(z)+\varepsilon+Aw
$$
on all of $\partial\Omega$. The left function is subharmonic and belongs to the Perron family; the right function is superharmonic and dominates every family member by part (ii). Therefore
$$
\varphi(z)-\varepsilon-Aw(x)leq\overline u(x)
\leq\varphi(z)+\varepsilon+Aw(x).
$$
As $x\to z$, continuity gives $w(x)\to0$. Letting $\varepsilon\downarrow0$ proves $\overline u(x)\to\varphi(z)$. Such a $w$ is a <barrier for the Dirichlet problem>, and $z$ is a <regular boundary point>.
Back to article page
