Proof of Comparison Principle with a Weighted Penalty for Operators Convex in the Value, Gradient and Matrix Variables

Each result cited is universally quantified over the data in its own statement. Fix $s,K$ as in hypothesis Comparison Principle with a Weighted Penalty for Operators Convex in the Value, Gradient and Matrix Variables §barrier, write $\varphi=sP-K$, a classical subsolution of $F$ on $D$, and put $a=w-s$, which is positive by claim 1 of Elementary Order Arithmetic in an Ordered Field. Fix $x\in D$; we show $u(x)\le v(x)$.

Step 1: the perturbed subsolution. Fix $t\in\mathbb{R}$ with $0<t<1$ and let $u_{t}=(1-t)u+t\varphi$. By hypothesis Comparison Principle with a Weighted Penalty for Operators Convex in the Value, Gradient and Matrix Variables §convex and A Convex Combination of a Viscosity Subsolution and a Classical Subsolution is a Viscosity Subsolution, for an Operator Convex in the Value, Gradient and Matrix Variables, $u_{t}$ is a viscosity subsolution of $F$ on $D$.

Step 2: $u_{t}<v$ outside a sublevel set. Put $\delta=ta/4$, positive by claims 5, 7 and 8 of Elementary Order Arithmetic in an Ordered Field. By Penalty-Subordinate Growth of a Function on an Open Subset of Euclidean Space §above applied to $u-wP$ and Penalty-Subordinate Growth of a Function on an Open Subset of Euclidean Space §below applied to $v-wP$, both with this $\delta$, there are $C_{1},C_{2}\in\mathbb{R}$ with

Multiplying the first by $1-t\ge0$ (claim 5 of Elementary Arithmetic in an Ordered Field), adding $t\varphi(y)=tsP(y)-tK$ and subtracting the second, we get for every $y\in D$

By the field axioms $c=-ta+(2-t)\delta=-\tfrac{ta}{2}-\tfrac{t^{2}a}{4}$, so with $b=\tfrac{ta}{2}>0$ we have $c+b=-\tfrac{t^{2}a}{4}\le0$. Hence for $y\in D$ with $0\le P(y)$, $(c+b)P(y)\le0$ (claim 5 of Elementary Arithmetic in an Ordered Field and claim 1 of Zero Products and Elementary Identities in a Field), that is $cP(y)\le-bP(y)$. Put $R=b^{-1}(|C_{3}|+1)$, which is positive. For every $y\in D$ with $R\le P(y)$ we have $0\le P(y)$, so, using $bR=|C_{3}|+1$ and $C_{3}\le|C_{3}|$ (claim 3 of Properties of the Absolute Value in an Ordered Field),

Step 3: comparison on $D_{R}$. If $D_{R}$ is empty, every $y\in D$ has $R\le P(y)$ and (1) gives $u_{t}<v$ on $D$. Otherwise $\Omega=D_{R}$ is nonempty, open and bounded, and $\overline{\Omega}$ is a subset of $D$, by Basic Properties of the Sublevel Sets of a Penalty §sublevel-sets; so $\Omega$ satisfies Bounded Open Domain in Euclidean Space §domain. We apply Comparison Principle for the Dirichlet Problem for Second-Order Equations on $\Omega$ to the operator $F|_{\Omega}$ of Restriction of a Second-Order Equation Operator to an Open Subset §operator, the constant $\gamma$, the modulus $\omega_{R}$ of hypothesis Comparison Principle with a Weighted Penalty for Operators Convex in the Value, Gradient and Matrix Variables §structure, and the restrictions $u_{t}|_{\overline{\Omega}}$ and $v|_{\overline{\Omega}}$.

Comparison Principle for the Dirichlet Problem for Second-Order Equations §continuity: $F|_{\Omega}$ is continuous by hypothesis Comparison Principle with a Weighted Penalty for Operators Convex in the Value, Gradient and Matrix Variables §continuity and Restriction of a Second-Order Equation Operator to an Open Subset §continuity.

Comparison Principle for the Dirichlet Problem for Second-Order Equations §strictly-proper: $F|_{\Omega}$ is degenerate elliptic by hypothesis Comparison Principle with a Weighted Penalty for Operators Convex in the Value, Gradient and Matrix Variables §strictly-proper and Restriction of a Second-Order Equation Operator to an Open Subset §elliptic, and condition 2 of Strictly Proper Second-Order Equation Operator holds for $F|_{\Omega}$ because it holds for $F$ at every point of $D\supseteq\Omega$ and the two operators agree on $\Omega$.

Comparison Principle for the Dirichlet Problem for Second-Order Equations §structure: this is hypothesis Comparison Principle with a Weighted Penalty for Operators Convex in the Value, Gradient and Matrix Variables §structure.

The functions: $u_{t}|_{\overline{\Omega}}$ is upper semicontinuous on $\overline{\Omega}$ by claim 4 of Semicontinuity and Continuity Under Composition with a Continuous Map, and its restriction to $\Omega$ is a viscosity subsolution of $F|_{\Omega}$ on $\Omega$ by Restriction of Viscosity Sub- and Supersolutions to an Open Subset §subsolution; so it is a viscosity subsolution of $F|_{\Omega}$ up to the boundary of $\Omega$. Likewise, by the lower-semicontinuous half of the same claim and Restriction of Viscosity Sub- and Supersolutions to an Open Subset §supersolution, $v|_{\overline{\Omega}}$ is a viscosity supersolution of $F|_{\Omega}$ up to the boundary of $\Omega$.

The boundary: every $y\in\partial\Omega$ lies in $D$ and satisfies $P(y)=R$ by Basic Properties of the Sublevel Sets of a Penalty §boundary, so $u_{t}(y)<v(y)$ by (1).

Hence $u_{t}\le v$ on $\overline{\Omega}\supseteq D_{R}$, and by (1) also at every $y\in D$ with $R\le P(y)$. Since every $y\in D$ satisfies $P(y)<R$ or $R\le P(y)$, $u_{t}(y)\le v(y)$ for every $y\in D$. In particular, at our fixed $x$,

by rearranging $(1-t)u(x)+t(sP(x)-K)\le v(x)$ with the field axioms.

Step 4: $t\to0$. Suppose $v(x)<u(x)$ and put $e=u(x)-v(x)>0$. Taking $t=\tfrac12$ in (2) gives $0<e\le\tfrac{M}{2}$, so $M$ is positive. Let $t$ be the least of $\tfrac12$ and $\tfrac{e}{2M}$ (claim 9 of Elementary Order Arithmetic in an Ordered Field); then $0<t<1$ and $tM\le\tfrac{e}{2}<e$, contradicting (2). Hence $u(x)\le v(x)$, and as $x\in D$ was arbitrary the theorem is proved.