Theorem 3.1.29C1B72E118338

Proof. Suppose that $b$ is a barrier at $\xi $, and let $f\in C(\partial \Omega )$ and $\varepsilon >0$. We show that $|U_ f-f(\xi )|\leq \varepsilon +Cb$ on $\Omega $ for a suitable constant $C$. Choose $\delta >0$ such that $|f(\zeta )-f(\xi )|<\varepsilon $ for all $\zeta \in \partial \Omega $ with $|\zeta -\xi |<2\delta $, and put $c=\inf _{\Omega \setminus B(\xi ,\delta )}b>0$; if $\Omega \subset B(\xi ,\delta )$, then $|\zeta -\xi |\leq \delta $ for every $\zeta \in \partial \Omega $, and we put $c=1$. If $\zeta \in \partial \Omega $ and $|\zeta -\xi |\geq 2\delta $, then $\Omega \cap B(\zeta ,\delta )$ lies outside $B(\xi ,\delta )$, so $\varliminf _{x\to \zeta }b(x)\geq c$. Let $M=\max _{\partial \Omega }|f|$ and fix $C\geq 2M/c$, which depends on $f$, $\varepsilon $ and $b$. We claim that, for every $\zeta \in \partial \Omega $,

If $|\zeta -\xi |<2\delta $, this holds because $b>0$ and $f(\xi )-\varepsilon <f(\zeta )<f(\xi )+\varepsilon $. If $|\zeta -\xi |\geq 2\delta $, it holds because $f(\xi )-Cc\leq M-2M\leq f(\zeta )$ and $f(\xi )+Cc\geq -M+2M\geq f(\zeta )$. Hence the continuous subsolution $f(\xi )-\varepsilon -Cb$ belongs to $\mathcal S_ f$, and the continuous supersolution $f(\xi )+\varepsilon +Cb$ belongs to $-\mathcal S_{-f}$, so that it dominates every member of $\mathcal S_ f$, as shown after theorem 3.1.25. Therefore $|U_ f-f(\xi )|\leq \varepsilon +Cb$ on $\Omega $. Letting $x\to \xi $ and then $\varepsilon \downarrow 0$ shows that $U_ f(x)\to f(\xi )$. Thus $\xi $ is regular.

Conversely, let $\xi $ be regular, and take the boundary function $f(\zeta )=|\zeta -\xi |$. The smooth function $v(x)=|x-\xi |^2/\operatorname {diam}\Omega $ satisfies $\Delta v\geq 0$, and $v\leq f$ on $\partial \Omega $ because $|\zeta -\xi |\leq \operatorname {diam}\overline\Omega =\operatorname {diam}\Omega $ for $\zeta \in \partial \Omega $; hence $v\in \mathcal S_ f$. Thus $b=U_ f$ is harmonic by theorem 3.1.25, and hence a continuous supersolution, with $b\geq |x-\xi |^2/\operatorname {diam}\Omega $. In particular $b$ is positive on $\Omega $, and if $N$ is a neighborhood of $\xi $ containing a ball $B(\xi ,t)$, then $\inf _{\Omega \setminus N}b\geq t^2/\operatorname {diam}\Omega >0$. By regularity, $b(x)\to f(\xi )=0$ as $x\to \xi $. Hence $b$ is a barrier.

Now suppose that every boundary point is regular, and let $f\in C(\partial \Omega )$. Then $U_ f$ is harmonic on $\Omega $ by theorem 3.1.25. By regularity and the continuity of $f$ on $\partial \Omega $, the function equal to $U_ f$ on $\Omega $ and to $f$ on $\partial \Omega $ is continuous on $\overline\Omega $. Uniqueness follows from proposition 3.1.10.

Finally, suppose that $\overline{B(a,r)}\cap \overline\Omega =\{ \xi \} $. Then $|\xi -a|=r$, since otherwise the open ball $B(a,r)$ would be a neighborhood of $\xi $ and would therefore contain points of $\Omega $, which differ from $\xi \notin \Omega $. In particular $a\notin \overline\Omega $, and every point of $\overline\Omega \setminus \{ \xi \} $ satisfies $|x-a|>r$. Put

For $d\geq 2$ this is harmonic on $\mathbb R^ d\setminus \{ a\} $ by proposition 3.1.13, being a positive multiple of $E_ d(x-a)$ plus a constant. In dimension one, each component of $\Omega $ is an open interval disjoint from $[a-r,a+r]$, so it lies on one side of this interval; hence $|x-a|$ is affine on each component, and $b$ is harmonic on $\Omega $ as well. In all dimensions $b$ is an increasing function of $|x-a|$ which vanishes when $|x-a|=r$, so it is continuous on the compact set $\overline\Omega $, positive on $\overline\Omega \setminus \{ \xi \} $ and zero at $\xi $. Hence $b(x)\to 0$ as $x\to \xi $, and if $N$ is a neighborhood of $\xi $ with $\Omega \setminus N\neq \varnothing $, the infimum of $b$ over the nonempty compact set $\overline\Omega \setminus \operatorname{Int}N$ is attained and positive. Thus $b$ is a barrier at $\xi $. $\square$