Proposition 3.1.30DBDE3CC0301D

Proof. Choose $r>0$ with $\overline{B(\xi ,2r)}\subset W$ and $0<t<r$. The set $L=\overline\Omega \cap \{ r-t\leq |x-\xi |\leq r+t\} $ is compact, and $b$ is positive and continuous on it, so there is $c>0$ with $b\geq 2c$ on $L$ (if $L$ is empty, any $c>0$ will do). Define $\tilde b=\min \{ b,c\} $ on $\Omega \cap B(\xi ,r)$ and $\tilde b=c$ on $\Omega \setminus B(\xi ,r)$. Since $b\geq 2c$ on $L$, we have $\tilde b=\min \{ b,c\} $ on $\Omega \cap B(\xi ,r+t)$ and $\tilde b=c$ on $\Omega \setminus \overline{B(\xi ,r)}$. These two open sets cover $\Omega $, and on each of them $\tilde b$ is a continuous supersolution, being a minimum of two continuous supersolutions or a constant. By the locality of the subsolution property, noted after the definition of continuous subsolutions, $\tilde b$ is a continuous supersolution on $\Omega $.

The function $\tilde b$ is positive, and it tends to zero at $\xi $, because $0<\tilde b\leq b$ on $\Omega \cap B(\xi ,r)$ and the extension of $b$ vanishes at $\xi $. If $N$ is a neighborhood of $\xi $, then $\tilde b\geq c$ on $\Omega \setminus B(\xi ,r)$, while on $(\Omega \cap B(\xi ,r))\setminus N$ it is at least the minimum of $c$ and of $b$ over the compact set $(\overline\Omega \cap \overline{B(\xi ,r)})\setminus \operatorname{Int}N$, on which $b$ is positive. Hence $\tilde b$ is a barrier at $\xi $, and theorem 3.1.29 shows that $\xi $ is regular. $\square$