3.1.23 Perron envelopes and boundary barriersA3DA602C02CB

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 $A$. $\square$

Proof. The bounds for $U_ f$ were proved above. To prove harmonicity, fix a ball $B$ with $\overline B\subset \Omega $ and a point $x_0\in B$; we show that $U_ f$ agrees on $B$ with a harmonic function. If $v\in \mathcal S_ f$, its harmonic lifting $v^ B$ on $B$ is a continuous subsolution by lemma 3.1.24, and it coincides with $v$ near $\partial \Omega $ because $\overline B$ is a compact subset of $\Omega $; hence $v^ B\in \mathcal S_ f$.

Choose $v_ j\in \mathcal S_ f$ with $v_ j(x_0)\to U_ f(x_0)$, and put $w_ j=\max \{ v_1,\ldots ,v_ j,\min _{\partial \Omega }f\} $. Then $(w_ j)$ is an increasing sequence in $\mathcal S_ f$ with values in $[\min _{\partial \Omega }f,\max _{\partial \Omega }f]$, and $w_ j(x_0)\to U_ f(x_0)$ because $v_ j\leq w_ j\leq U_ f$. The liftings $w_ j^ B$ belong to $\mathcal S_ f$ and satisfy $w_ j\leq w_ j^ B\leq U_ f$. On $B$ they are harmonic, increase with $j$ because $H_ B$ preserves order, and are bounded by $\max _{\partial \Omega }f$. By part (2) of corollary 3.1.21, applied on the domain $B$ to this sequence, which is bounded above, they converge on $B$ to a harmonic function $h$, and $h\leq U_ f$ on $B$, while $h(x_0)=U_ f(x_0)$.

Now let $v\in \mathcal S_ f$ be arbitrary. The same argument applied to the increasing sequence $\max \{ w_ j,v\} \in \mathcal S_ f$ shows that the liftings $(\max \{ w_ j,v\} )^ B$ increase on $B$ to a harmonic function $k$. Since these liftings dominate $w_ j^ B$ and $v$ and lie in $\mathcal S_ f$, we have $h\leq k\leq U_ f$ and $k\geq v$ on $B$. In particular $k(x_0)\leq U_ f(x_0)=h(x_0)$, so the nonnegative harmonic function $k-h$ on $B$ vanishes at $x_0$ and therefore vanishes identically, by proposition 3.1.19(2). Hence $v\leq h$ on $B$. Taking the supremum over $v\in \mathcal S_ f$ gives $U_ f\leq h$, so $U_ f=h$ on $B$. Since $B$ was arbitrary, $U_ f$ is harmonic on $\Omega $.

Finally, let $H$ be harmonic on $\Omega $ and continuous on $\overline\Omega $ with $H=f$ on $\partial \Omega $. Then $H\in \mathcal S_ f$, so $H\leq U_ f$. Conversely, for $v\in \mathcal S_ f$ the function $v-H$ is a continuous subsolution with $\varlimsup _{x\to \zeta }(v-H)(x)\leq 0$ at every $\zeta \in \partial \Omega $, so $v\leq H$ by lemma 3.1.24. Hence $U_ f=H$. $\square$

Proof. We may assume that $K\neq \varnothing $. Let $\delta >0$. By compactness there are points $\zeta _1,\ldots ,\zeta _ N\in K$ with $K\subset \bigcup _ iB(\zeta _ i,\delta )$. Put $\chi _ i(x)=\rho ((x-\zeta _ i)/(2\delta ))$, where $\rho $ is the mollifier of section 2.5; this function is smooth, positive on $B(\zeta _ i,2\delta )$ and zero outside it. The function $\sigma =\sum _ i\chi _ i$ is positive on the open set $W=\bigcup _ iB(\zeta _ i,2\delta )\supset K$. Choose a smooth function $\eta $ with compact support in $W$ which equals one near $K$ (lemma 2.5.1(4)), and put

extended by zero outside $W$. This is a smooth function with compact support on $\mathbb R^ d$, because it is smooth on $W$, where $\sigma >0$, and vanishes on the open set $\mathbb R^ d\setminus \operatorname{Supp}\eta $; it is real-valued if $f$ is. For $\zeta \in K$ we have $\eta (\zeta )=1$, so $F(\zeta )$ is a convex combination of values $f(\zeta _ i)$ with $|\zeta _ i-\zeta |<2\delta $, and therefore $|F(\zeta )-f(\zeta )|\leq \sup \{ |f(y)-f(y')|:y,y'\in K,\ |y-y'|<2\delta \} $. By the uniform continuity of $f$ on the compact set $K$, this bound tends to zero as $\delta \to 0$. $\square$

Proof. Write $\| f\| =\max _{\partial \Omega }|f|$, and let $\mathcal R$ be the set of $f\in C(\partial \Omega )$ with $U_ f=V_ f$. We show that $\mathcal R$ is a closed linear subspace of $C(\partial \Omega )$ containing a dense subset.

Step 1: $\mathcal R$ is a closed linear subspace, and $f\mapsto U_ f$ is linear on it. If $v\in \mathcal S_ f$ and $w\in \mathcal S_ g$, then $v+w\in \mathcal S_{f+g}$; taking suprema over $v$ and $w$ at a fixed point gives $U_ f+U_ g\leq U_{f+g}$. Applying this to $-f$ and $-g$ and changing signs gives $V_{f+g}\leq V_ f+V_ g$. Together with the inequality $U_{f+g}\leq V_{f+g}$ proved above, we obtain

Moreover $\mathcal S_{tf}=t\, \mathcal S_ f$ for $t>0$, so $U_{tf}=tU_ f$ and $V_{tf}=tV_ f$, while $U_{-f}=-V_ f$ and $V_{-f}=-U_ f$ by definition. Also $U_0=0$, because $0\in \mathcal S_0$ and every member of $\mathcal S_0$ is nonpositive by lemma 3.1.24; hence $V_0=-U_0=0$ and $0\in \mathcal R$. Consequently, if $f,g\in \mathcal R$, the display forces $U_{f+g}=V_{f+g}=U_ f+U_ g$, and $tf\in \mathcal R$ for every real $t$. Thus $\mathcal R$ is a linear subspace, and $f\mapsto U_ f$ is linear on it.

To see that $\mathcal R$ is closed, we first show that $U_ f$ and $V_ f$ depend $1$-Lipschitz continuously on $f$ for the supremum norm. If $f\leq g$, then $\mathcal S_ f\subset \mathcal S_ g$ and hence $U_ f\leq U_ g$; and $\mathcal S_{f+t}=\mathcal S_ f+t$ for constants $t$, so $U_{f+t}=U_ f+t$. Since $f\leq g+\| f-g\| $, we get $U_ f\leq U_ g+\| f-g\| $, and by symmetry $|U_ f-U_ g|\leq \| f-g\| $; as $V_ f=-U_{-f}$ and $V_ g=-U_{-g}$, also $|V_ f-V_ g|\leq \| f-g\| $. If $f_ n\in \mathcal R$ converge uniformly to $f$, then $0\leq V_ f-U_ f\leq |V_ f-V_{f_ n}|+|U_{f_ n}-U_ f|\leq 2\| f-f_ n\| \to 0$, so $f\in \mathcal R$. Hence $\mathcal R$ is closed.

Step 2: $\mathcal R$ contains the restriction to $\partial \Omega $ of every real smooth function on a neighborhood of $\overline\Omega $. Suppose first that $F$ is a real smooth function on a neighborhood of $\overline\Omega $ with $\Delta F\geq 0$ there. Then $F$ is a continuous subsolution on $\Omega $, and $F(x)\to F(\zeta )$ as $\Omega \ni x\to \zeta \in \partial \Omega $, so $F\in \mathcal S_{F|_{\partial \Omega }}$ and $U_{F|_{\partial \Omega }}\geq F$. Consequently the harmonic function $U_{F|_{\partial \Omega }}$ satisfies $\varliminf _{x\to \zeta }U_{F|_{\partial \Omega }}(x)\geq F(\zeta )$ at every $\zeta \in \partial \Omega $, so it belongs to $-\mathcal S_{-F|_{\partial \Omega }}$ and $V_{F|_{\partial \Omega }}\leq U_{F|_{\partial \Omega }}$. Thus $F|_{\partial \Omega }\in \mathcal R$.

Now let $F$ be an arbitrary real smooth function on an open neighborhood $W$ of $\overline\Omega $, and choose $\delta >0$ such that the compact set $L=\{ x:\operatorname {dist}(x,\overline\Omega )\leq \delta \} $ lies in $W$. Since $\Delta F$ is continuous, hence bounded below, on $L$, there is a constant $A\geq 0$ with $\Delta F+2dA\geq 0$ on $L$. As $\Delta (A|x|^2)=2dA$, both $F+A|x|^2$ and $A|x|^2$ are smooth with nonnegative Laplacian on the open neighborhood $\{ x:\operatorname {dist}(x,\overline\Omega )<\delta \} $ of $\overline\Omega $. By the first case their restrictions to $\partial \Omega $ lie in $\mathcal R$, and by Step 1 so does the restriction of their difference $F$.

Step 3: these restrictions are dense in $C(\partial \Omega )$. This is lemma 3.1.26, applied to real-valued functions on the compact set $\partial \Omega $.

By Steps 1–3, $\mathcal R=C(\partial \Omega )$, and $f\mapsto U_ f$ is linear. By Step 1 it preserves order and satisfies $|U_ f-U_ g|\leq \| f-g\| $. Since $U_0=0$, it follows that $U_ f\geq 0$ whenever $f\geq 0$, and that $|U_ f|\leq \| f\| $. $\square$

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 $. $\square$

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. $\square$