Lemma 3.1.243F83842AEB11

Proof. For the first assertion we may assume that $D\neq \varnothing $. Let $\varepsilon >0$, and put $w(y)=v(y)+\varepsilon |y|^2$ for $y\in D$. If $\overline{B(c,r)}\subset D$, then $v(c)\leq M_ v(c,r)$, and the identity $M_{|\cdot |^2}(c,r)=|c|^2+r^2$ gives $w(c)\leq M_ w(c,r)-\varepsilon r^2$. Consequently $w$ does not attain its supremum over $D$ at any point $c\in D$, for we would then have $M_ w(c,r)\leq w(c)$ for small $r$. Choose $x_ k\in D$ with $w(x_ k)\to \sup _ Dw$. Since $D$ is bounded, we may assume, after passing to a subsequence, that $x_ k$ converges to a point $p\in \overline D$. If $p$ were in $D$, the continuity of $w$ would give $w(p)=\sup _ Dw$, which we have excluded. Hence $p\in \partial D$. Since $\sup _ Dw=\lim _ k\bigl (v(x_ k)+\varepsilon |x_ k|^2\bigr )$, the hypothesis gives $\sup _ Dw\leq \varlimsup _{D\ni x\to p}v(x)+\varepsilon |p|^2\leq \varepsilon \max _{\overline D}|y|^2$. Thus $v\leq w\leq \varepsilon \max _{\overline D}|y|^2$ on $D$, and letting $\varepsilon \downarrow 0$ gives $v\leq 0$.

For the second assertion, $v^ B$ is harmonic on $B$ and equals $v$ on $D\setminus B$; it is continuous at the points of $\partial B$ because $H_ B(v|_{\partial B})$ extends continuously to $v$ on $\partial B$ (theorem 3.1.18). Moreover $v^ B\geq v$, because $v\leq H_ B(v|_{\partial B})$ on $B$. To verify the subsolution inequality, let $A$ be a ball with $\overline A\subset D$, and let $H=H_ A(v^ B|_{\partial A})$, regarded as a continuous function on $\overline A$. Since $v\leq v^ B$ and $H_ A$ preserves order, $v\leq H_ A(v|_{\partial A})\leq H$ on $A$; the same inequality holds on $\partial A$, where $H=v^ B\geq v$. On the open set $A\cap B$, the function $v^ B-H$ is harmonic, and it is continuous on the closure of $A\cap B$. A boundary point $\zeta $ of $A\cap B$ lies either on $\partial A$, where $v^ B(\zeta )=H(\zeta )$, or in $A\cap \partial B$, where $v^ B(\zeta )=v(\zeta )\leq H(\zeta )$. By part (2) of proposition 3.1.10, applied to $v^ B-H$ on the bounded open set $A\cap B$, we get $v^ B\leq H$ on $A\cap B$. On $A\setminus B$ we have $v^ B=v\leq H$. Hence $v^ B\leq H_ A(v^ B|_{\partial A})$ on $A$, which is the subsolution inequality for $v^ B$ on $A$. $\square$