Comparison Principle for the Dirichlet Problem for Second-Order Equations

In the setting of Second-Order Equations on Euclidean Open Sets and of Bounded Open Domain in Euclidean Space, let $F$ be a second-order equation operator on $\Omega$, let $\gamma\in\mathbb{R}$ be positive and let $\omega$ be a modulus of continuity, with domain the set $T$ of nonnegative real numbers. We abbreviate $\lVert z\rVert^{2}=\lVert z\rVert\lVert z\rVert$ and write $3=1+1+1$. Assume the following three conditions.

1. (Continuity)¶ $F$ is continuous.

2. (Strict properness)¶ $F$ is strictly proper with constant $\gamma$.

3. (Structure condition)¶ For all $x,y\in\Omega$, every $r\in\mathbb{R}$, every positive $\alpha\in\mathbb{R}$ and all $X,Y\in\mathcal{S}(n)$ satisfying

one has

where $\alpha(x-y)$ is the scalar multiple by $\alpha$ of the difference $x-y$ in $\mathbb{R}^{n}$. The argument of $\omega$ lies in $T$: both summands are nonnegative, since $\lVert x-y\rVert$ is nonnegative by claim 1 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$, hence so is $\lVert x-y\rVert^{2}$ by claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field, and $\alpha\lVert x-y\rVert^{2}$ is then either $0$, if $\lVert x-y\rVert^{2}=0$, or positive by claim 5 of Elementary Order Arithmetic in an Ordered Field; and adding $\lVert x-y\rVert$ to both sides of $0\le\alpha\lVert x-y\rVert^{2}$, which the compatibility of $\le$ with addition in the ordered field axioms permits, gives $\lVert x-y\rVert\le\alpha\lVert x-y\rVert^{2}+\lVert x-y\rVert$, so that the argument of $\omega$ is nonnegative by transitivity.

Let $u:\overline{\Omega}\to\mathbb{R}$ be a viscosity subsolution of $F$ up to the boundary of $\Omega$, let $v:\overline{\Omega}\to\mathbb{R}$ be a viscosity supersolution of $F$ up to the boundary of $\Omega$, and assume

Then $u(x)\le v(x)$ for every $x\in\overline{\Omega}$.