Proof of Comparison of a Viscosity Subsolution with a Strict Classical Supersolution

Claims are cited by number from Elementary Order Arithmetic in an Ordered Field, below the order arithmetic lemma.

Step 1: $\overline{\Omega}$ is nonempty and compact. Since $\Omega$ is bounded in $(\mathbb{R}^n,d_E)$, its closure $\overline{\Omega}$ is compact in $(\mathbb{R}^n,\mathcal{T}_{d_E})$ by The Closure of a Bounded Subset of $\mathbb{R}^n$ is Compact. By The Closure is the Smallest Closed Superset we have $\Omega\subseteq\overline{\Omega}$, so $\overline{\Omega}$ is nonempty because $\Omega$ is.

Step 2: the difference attains a maximum. Since $u$ is a viscosity subsolution of $F$ up to the boundary of $\Omega$, it is upper semicontinuous on $\overline{\Omega}$, and $v$ is lower semicontinuous on $\overline{\Omega}$ by hypothesis. Let $w:\overline{\Omega}\to\mathbb{R}$ be the function $w(x)=u(x)-v(x)$. By claim 3 of Negation, Restriction, and Separated Differences of Semicontinuous Functions, applied at every point of $\overline{\Omega}$, the function $w$ is upper semicontinuous on $\overline{\Omega}$. By claim 1 of Semicontinuous Functions Attain Their Extrema on a Compact Set there exists $x_0\in\overline{\Omega}$ with

$$w(x)\le w(x_0)\qquad\text{for every }x\in\overline{\Omega}.$$

Step 3: assume the conclusion fails. Suppose, for contradiction, that there is $z\in\overline{\Omega}$ for which $u(z)\le v(z)$ fails. Since $\le$ is a total order on $\mathbb{R}$, this means $v(z)\le u(z)$ and $u(z)\ne v(z)$, that is, $v(z)<u(z)$. Adding $-v(z)$ to both sides and using claim 1 of the order arithmetic lemma gives $0<w(z)$, and then $w(z)\le w(x_0)$ together with claim 2 gives

$$0<w(x_0),$$

hence, adding $v(x_0)$ and using claim 1 again, $v(x_0)<u(x_0)$; in particular $v(x_0)\le u(x_0)$.

Step 4: the maximum point lies in $\Omega$. If $x_0$ belonged to $\partial_{\mathbb{R}^n}\Omega$, then hypothesis 2 would give $u(x_0)\le v(x_0)$, which together with $v(x_0)<u(x_0)$ contradicts antisymmetry of $\le$. Hence $x_0\notin\partial_{\mathbb{R}^n}\Omega$. By claim 5 of Decomposition of a Topological Space by the Boundary of a Subset, the union of $\operatorname{int}_{\mathbb{R}^n}(\Omega)$ and $\partial_{\mathbb{R}^n}\Omega$ is $\overline{\Omega}$, and since $\Omega$ belongs to $\mathcal{T}_{d_E}$ we have $\operatorname{int}_{\mathbb{R}^n}(\Omega)=\Omega$ by The Interior is the Largest Open Subset. As $x_0\in\overline{\Omega}$ and $x_0\notin\partial_{\mathbb{R}^n}\Omega$, we conclude $x_0\in\Omega$.

Step 5: $v|_{\Omega}$ is an admissible test function at $x_0$. Write $\varphi=v|_{\Omega}$, which is of class $C^2$ on $\Omega$ by hypothesis, and let $u|_{\Omega}:\Omega\to\mathbb{R}$ be the restriction of $u$, so that $u|_{\Omega}-\varphi$ is the function on $\Omega$ with value $u(y)-v(y)=w(y)$ at $y\in\Omega$. Since $\Omega\subseteq\overline{\Omega}$, Step 2 gives

$$(u|_{\Omega}-\varphi)(y)=w(y)\le w(x_0)=(u|_{\Omega}-\varphi)(x_0)\qquad\text{for every }y\in\Omega,$$

so, taking $\delta=1$ in Local Maximum of a Function Relative to a Subset of a Metric Space, the function $u|_{\Omega}-\varphi$ has a local maximum at $x_0$ relative to $\Omega$.

Step 6: the subsolution inequality. By Viscosity Subsolution and Supersolution up to the Boundary the restriction $u|_{\Omega}$ is a viscosity subsolution of $F$ on $\Omega$. Applying that definition with the test function $\varphi$ and the point $x_0$ furnished by Step 5, and using $u|_{\Omega}(x_0)=u(x_0)$, $D\varphi(x_0)=Dv(x_0)$ and $D^2\varphi(x_0)=D^2v(x_0)$, we obtain

$$F\bigl(x_0,u(x_0),Dv(x_0),D^2v(x_0)\bigr)\le 0 .$$

Step 7: monotonicity in the second argument and the contradiction. By Step 3 we have $v(x_0)\le u(x_0)$, so condition 2 of Proper Second-Order Equation Operator, applied at the point $x_0$ with $p=Dv(x_0)$ and $X=D^2v(x_0)$, gives

$$F\bigl(x_0,v(x_0),Dv(x_0),D^2v(x_0)\bigr)\le F\bigl(x_0,u(x_0),Dv(x_0),D^2v(x_0)\bigr)\le 0 .$$

Since $x_0\in\Omega$, hypothesis 1 gives $0<F(x_0,v(x_0),Dv(x_0),D^2v(x_0))$, and the two displayed relations contradict antisymmetry of $\le$.

Therefore no such $z$ exists, and $u(x)\le v(x)$ for every $x\in\overline{\Omega}$.