Theorem 3.1.278FE1B604CDA2

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\| $. $\square$