Proof of Comparison Principle for First-Order Strictly Proper Equations by Doubling of Variables

Claims are cited by number from Elementary Order Arithmetic in an Ordered Field (below, the order arithmetic lemma), Elementary Arithmetic in an Ordered Field, and, for the maximum and minimum of two real numbers, Elementary Properties of the Maximum of Two Elements and Elementary Properties of the Minimum of Two Elements (below, the maximum lemma and the minimum lemma), whose claim 1 bounds and whose claim 2 identifies the value with one of the two arguments.

Step 0: setup. Write $K=\overline{\Omega}$. Since $\Omega$ is bounded in $(\mathbb{R}^n,d_E)$, the set $K$ is compact in $(\mathbb{R}^n,\mathcal{T}_{d_E})$ by The Closure of a Bounded Subset of $\mathbb{R}^n$ is Compact; and $\Omega\subseteq K$ by The Closure is the Smallest Closed Superset, so $K$ is nonempty. By Viscosity Subsolution and Supersolution up to the Boundary, $u$ is upper semicontinuous on $K$ and $v$ is lower semicontinuous on $K$, and the restrictions $u|_{\Omega}$ and $v|_{\Omega}$ are respectively a viscosity subsolution and a viscosity supersolution of $F$ on $\Omega$.

Equip $\mathbb{R}^n\times\mathbb{R}^n$ with the product metric $d_{E\times E}$ obtained from $d_E$ and $d_E$, a metric by claim 1 of The Product Metric is a Metric, and with the collection of its subsets open in that metric, a topology by Metric Open Sets Form a Topology; by A Product of Compact Subsets is Compact in the Product Metric the set $K\times K$ is compact.

Let $\psi:K\times K\to\mathbb{R}$ be $\psi(x,y)=d_E(x,y)^2$ and, for $\alpha\in\mathbb{R}$ with $0<\alpha$, let

$$\Phi_{\alpha}(x,y)=u(x)-v(y)-\alpha\,\psi(x,y)\qquad\bigl((x,y)\in K\times K\bigr).$$

By The Squared-Distance Penalization Limit on a Compact Set the function $\psi$ is lower semicontinuous on $K\times K$, satisfies $0\le\psi$, satisfies $\psi(x,y)=0$ if and only if $x=y$, and claims 1 to 6 of Limits of Penalized Maxima on a Compact Set hold for these functions. In particular claim 1 provides the maximum value $M$ of $u-v$ on $K$ and, for each $\alpha$ with $0<\alpha$, the maximum value $M_{\alpha}$ of $\Phi_{\alpha}$ on $K\times K$; for each such $\alpha$ fix a point $(x_{\alpha},y_{\alpha})\in K\times K$ with $\Phi_{\alpha}(x_{\alpha},y_{\alpha})=M_{\alpha}$, the remaining claims holding for any such choice.

Step 1: the contradiction hypothesis. Suppose, for contradiction, that there is $z\in K$ for which $u(z)\le v(z)$ fails. Since $\le$ is a total order, this means $v(z)<u(z)$, so adding $-v(z)$ and using claim 1 of the order arithmetic lemma gives $0<u(z)-v(z)$. By claim 1 of Limits of Penalized Maxima on a Compact Set we have $u(z)-v(z)\le M$, so by claim 2 of the order arithmetic lemma

$$0<M .$$

Write $\delta=M$. By claim 4 of Limits of Penalized Maxima on a Compact Set,

$$\delta\le u(x_{\alpha})-v(y_{\alpha})\qquad\text{for every }\alpha\text{ with }0<\alpha. \tag{$*$}$$

Step 2: for large $\alpha$ both $x_{\alpha}$ and $y_{\alpha}$ lie in $\Omega$. Let

$$B=\{(x,y)\in\mathbb{R}^n\times\mathbb{R}^n:\ x\in\partial_{\mathbb{R}^n}\Omega\ \text{or}\ y\in\partial_{\mathbb{R}^n}\Omega\}.$$

(a) $B$ is closed. By claim 2 of Decomposition of a Topological Space by the Boundary of a Subset the set $\partial_{\mathbb{R}^n}\Omega$ is closed in $(\mathbb{R}^n,\mathcal{T}_{d_E})$, so its complement $N=\mathbb{R}^n\setminus\partial_{\mathbb{R}^n}\Omega$ belongs to $\mathcal{T}_{d_E}$. Let $(a,b)$ lie in the complement of $B$, that is, $a\in N$ and $b\in N$. By Open Subset of a Metric Space there are $r_1,r_2\in\mathbb{R}$ with $0<r_1$, $0<r_2$ such that every $x$ with $d_E(a,x)<r_1$ lies in $N$ and every $y$ with $d_E(b,y)<r_2$ lies in $N$. Let $r$ be the minimum of $r_1$ and $r_2$; by claims 2 and 1 of the minimum lemma, $0<r$, $r\le r_1$ and $r\le r_2$. If $d_{E\times E}((a,b),(x,y))<r$, then by the definition of the product metric and claim 1 of the maximum lemma we have $d_E(a,x)\le d_{E\times E}((a,b),(x,y))$ and $d_E(b,y)\le d_{E\times E}((a,b),(x,y))$, so by claim 2 of the order arithmetic lemma $d_E(a,x)<r_1$ and $d_E(b,y)<r_2$; hence $x\in N$ and $y\in N$, that is, $(x,y)$ lies in the complement of $B$. Therefore the complement of $B$ is open in $d_{E\times E}$ and $B$ is closed.

(b) A compact set of bad pairs. By claim 1 of Compactness of Intersections with Closed Sets and of Level Sets of Semicontinuous Functions, applied in the metric space $(\mathbb{R}^n\times\mathbb{R}^n,d_{E\times E})$ to the compact set $K\times K$ and the closed set $B$, the set $C=(K\times K)\cap B$ is compact. Let $h:K\times K\to\mathbb{R}$ be $h(x,y)=u(x)-v(y)$; by claim 4 of Negation, Restriction, and Separated Differences of Semicontinuous Functions the function $h$ is upper semicontinuous on $K\times K$, and by claim 2 of that lemma its restriction to $C$ is upper semicontinuous on $C$. Hence, by claim 2 of Compactness of Intersections with Closed Sets and of Level Sets of Semicontinuous Functions applied to the compact set $C$, that restriction and the real number $\delta$, the set

$$D=\{(x,y)\in C:\ \delta\le h(x,y)\}$$

is compact.

(c) On $D$ the penalty is bounded below by a positive number. Suppose $D$ is nonempty. The function $\psi$ is lower semicontinuous on $K\times K$, hence, by claim 2 of Negation, Restriction, and Separated Differences of Semicontinuous Functions and $D\subseteq K\times K$, its restriction to $D$ is lower semicontinuous on $D$. By claim 2 of Semicontinuous Functions Attain Their Extrema on a Compact Set there is $(x^{*},y^{*})\in D$ with

$$\psi(x^{*},y^{*})\le\psi(x,y)\qquad\text{for every }(x,y)\in D .$$

We claim $\theta=\psi(x^{*},y^{*})$ satisfies $0<\theta$. Since $0\le\psi$, it suffices to exclude $\theta=0$. If $\theta=0$, then $x^{*}=y^{*}$ by the property of $\psi$ recorded in Step 0. As $(x^{*},y^{*})\in B$, either $x^{*}\in\partial_{\mathbb{R}^n}\Omega$ or $y^{*}\in\partial_{\mathbb{R}^n}\Omega$, and in both cases $x^{*}\in\partial_{\mathbb{R}^n}\Omega$. On the other hand $\delta\le h(x^{*},y^{*})=u(x^{*})-v(x^{*})$, and $0<\delta$, so $0<u(x^{*})-v(x^{*})$ by claim 2 of the order arithmetic lemma, whence $v(x^{*})<u(x^{*})$ by claim 1. This contradicts the boundary hypothesis $u(x^{*})\le v(x^{*})$ together with antisymmetry of $\le$. Hence $0<\theta$.

(d) Interiority. By claim 4 of Limits of Penalized Maxima on a Compact Set applied with $\varepsilon=1$ there is $\alpha_1\in\mathbb{R}$ with $0<\alpha_1$ such that

$$\alpha\,\psi(x_{\alpha},y_{\alpha})\le 1\qquad\text{whenever }\alpha_1\le\alpha .$$

Define $\alpha_3$ as follows: if $D$ is empty, let $\alpha_3=\alpha_1$; otherwise, with $\theta$ as in (c), let $\alpha_3$ be the maximum of $\alpha_1$ and $2\,\theta^{-1}$, which is positive because $\theta^{-1}$ is positive by claim 7 of the order arithmetic lemma and $0<2$.

Let $\alpha$ satisfy $\alpha_3\le\alpha$; then $\alpha_1\le\alpha$ and $0<\alpha$. We show $(x_{\alpha},y_{\alpha})\notin D$. If $D$ is empty this is trivial. Otherwise $2\,\theta^{-1}\le\alpha$, so multiplying by the nonnegative $\theta$ gives $2=(2\,\theta^{-1})\theta\le\alpha\,\theta$; since $0<1$ gives $1<1+1=2$ by claim 1 of the order arithmetic lemma, claim 2 yields $1<\alpha\,\theta$. If we had $\theta\le\psi(x_{\alpha},y_{\alpha})$, then multiplying by the nonnegative $\alpha$ would give $\alpha\,\theta\le\alpha\,\psi(x_{\alpha},y_{\alpha})\le1$, contradicting $1<\alpha\,\theta$. Hence $\psi(x_{\alpha},y_{\alpha})<\theta$, and since $\theta\le\psi(x,y)$ for every $(x,y)\in D$, we conclude $(x_{\alpha},y_{\alpha})\notin D$.

Now $(x_{\alpha},y_{\alpha})\in K\times K$ and, by $(*)$, $\delta\le h(x_{\alpha},y_{\alpha})$. Were $(x_{\alpha},y_{\alpha})\in B$, it would lie in $C$ and therefore in $D$, which we have excluded. Hence $x_{\alpha}\notin\partial_{\mathbb{R}^n}\Omega$ and $y_{\alpha}\notin\partial_{\mathbb{R}^n}\Omega$. 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 $K$, and $\operatorname{int}_{\mathbb{R}^n}(\Omega)=\Omega$ by The Interior is the Largest Open Subset because $\Omega\in\mathcal{T}_{d_E}$. Therefore

$$x_{\alpha}\in\Omega\quad\text{and}\quad y_{\alpha}\in\Omega\qquad\text{whenever }\alpha_3\le\alpha .$$

Step 3: the two viscosity inequalities. Fix $\alpha$ with $\alpha_3\le\alpha$ and write $\hat{x}=x_{\alpha}$, $\hat{y}=y_{\alpha}$ and $p_{\alpha}=(2\alpha)(\hat{x}-\hat{y})$.

(a) Subsolution. Let $q_1:\Omega\to\mathbb{R}$ be $q_1(x)=\alpha\,d_E(x,\hat{y})^2$ and let $k_1:\Omega\to\mathbb{R}$ be the constant function with value $-v(\hat{y})$. By A Scaled Squared Distance to a Point is of Class $C^2$, with Gradient and Hessian, applied with $U=\Omega$, $a=\hat{y}$ and $c=\alpha$, the function $q_1$ is of class $C^2$ on $\Omega$ with gradient $Dq_1(x)=(2\alpha)(x-\hat{y})$ at every $x\in\Omega$. By claim 2 of Differences and Constants for Functions of Class $C^2$ on a Euclidean Open Set the function $k_1$ is of class $C^2$ on $\Omega$ with gradient the origin of $\mathbb{R}^n$ at every point, and by claim 1 of that lemma the function $\varphi=q_1-k_1$ is of class $C^2$ on $\Omega$ with

$$D\varphi(x)=Dq_1(x)-0_{\mathbb{R}^n}=(2\alpha)(x-\hat{y}),$$

where the identity $z-0_{\mathbb{R}^n}=z$ holds because $\mathbb{R}^n$ is a real vector space by Euclidean Space $\mathbb{R}^n$ is a Real Vector Space. In particular $D\varphi(\hat{x})=p_{\alpha}$. Note $\varphi(x)=q_1(x)+v(\hat{y})=v(\hat{y})+\alpha\,\psi(x,\hat{y})$ for $x\in\Omega$.

Since $(\hat{x},\hat{y})$ maximizes $\Phi_{\alpha}$ on $K\times K$ and $\Omega\subseteq K$, for every $x\in\Omega$

$$\bigl(u|_{\Omega}-\varphi\bigr)(x)=u(x)-v(\hat{y})-\alpha\,\psi(x,\hat{y})=\Phi_{\alpha}(x,\hat{y})\le\Phi_{\alpha}(\hat{x},\hat{y})=\bigl(u|_{\Omega}-\varphi\bigr)(\hat{x}),$$

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 $\hat{x}$ relative to $\Omega$. Since $u|_{\Omega}$ is a viscosity subsolution of $F$ on $\Omega$, Viscosity Subsolution and Supersolution of a Second-Order Equation gives

$$F\bigl(\hat{x},u(\hat{x}),p_{\alpha},X_1\bigr)\le0,\qquad X_1=D^2\varphi(\hat{x}),$$

where $X_1\in\mathcal{S}(n)$ by Equality of Mixed Second Partial Derivatives and Symmetry of the Hessian.

(b) Supersolution. Let $q_2:\Omega\to\mathbb{R}$ be $q_2(y)=\alpha\,d_E(y,\hat{x})^2$ and let $k_2:\Omega\to\mathbb{R}$ be the constant function with value $u(\hat{x})$. By A Scaled Squared Distance to a Point is of Class $C^2$, with Gradient and Hessian, applied with $a=\hat{x}$ and $c=\alpha$, the function $q_2$ is of class $C^2$ on $\Omega$ with $Dq_2(y)=(2\alpha)(y-\hat{x})$, and by claims 1 and 2 of Differences and Constants for Functions of Class $C^2$ on a Euclidean Open Set the function $\chi=k_2-q_2$ is of class $C^2$ on $\Omega$ with

$$D\chi(y)=0_{\mathbb{R}^n}-(2\alpha)(y-\hat{x})=(2\alpha)(\hat{x}-y),$$

the last identity again holding in the real vector space $\mathbb{R}^n$ of Euclidean Space $\mathbb{R}^n$ is a Real Vector Space. In particular $D\chi(\hat{y})=p_{\alpha}$. Since $d_E$ is symmetric, $\chi(y)=u(\hat{x})-\alpha\,\psi(\hat{x},y)$ for $y\in\Omega$.

For every $y\in\Omega$ we have $\Phi_{\alpha}(\hat{x},y)\le\Phi_{\alpha}(\hat{x},\hat{y})$, that is,

$$\chi(y)-v(y)\le\chi(\hat{y})-v(\hat{y}),$$

and adding $v(y)+v(\hat{y})-\chi(y)-\chi(\hat{y})$ to both sides, which preserves the relation $\le$ by claim 3 (translation) of Elementary Arithmetic in an Ordered Field applied twice, gives

$$\bigl(v|_{\Omega}-\chi\bigr)(\hat{y})\le\bigl(v|_{\Omega}-\chi\bigr)(y).$$

Hence, taking $\delta=1$ in the definition of a local minimum, the function $v|_{\Omega}-\chi$ has a local minimum at $\hat{y}$ relative to $\Omega$. Since $v|_{\Omega}$ is a viscosity supersolution of $F$ on $\Omega$, Viscosity Subsolution and Supersolution of a Second-Order Equation gives

$$0\le F\bigl(\hat{y},v(\hat{y}),p_{\alpha},X_2\bigr),\qquad X_2=D^2\chi(\hat{y}),$$

with $X_2\in\mathcal{S}(n)$ by Equality of Mixed Second Partial Derivatives and Symmetry of the Hessian.

(c) Removing the matrix arguments. Set $X_{\alpha}=X_1$. By hypothesis 1 of the theorem, $F(\hat{y},v(\hat{y}),p_{\alpha},X_2)=F(\hat{y},v(\hat{y}),p_{\alpha},X_{\alpha})$, so

$$F\bigl(\hat{x},u(\hat{x}),p_{\alpha},X_{\alpha}\bigr)\le0\le F\bigl(\hat{y},v(\hat{y}),p_{\alpha},X_{\alpha}\bigr). \tag{$**$}$$

Step 4: strict properness and the structure condition. By $(*)$ we have $\delta\le u(\hat{x})-v(\hat{y})$, and $0<\delta$, so $0<u(\hat{x})-v(\hat{y})$ and hence $v(\hat{y})\le u(\hat{x})$ by claim 3 of Elementary Arithmetic in an Ordered Field. Condition 2 of Strictly Proper Second-Order Equation Operator, applied at the point $\hat{x}$ with $p=p_{\alpha}$, $X=X_{\alpha}$, $r=u(\hat{x})$ and $s=v(\hat{y})$, gives

$$\gamma\bigl(u(\hat{x})-v(\hat{y})\bigr)\le F\bigl(\hat{x},u(\hat{x}),p_{\alpha},X_{\alpha}\bigr)-F\bigl(\hat{x},v(\hat{y}),p_{\alpha},X_{\alpha}\bigr).$$

Multiplying $\delta\le u(\hat{x})-v(\hat{y})$ by the nonnegative $\gamma$ (claim 5 of Elementary Arithmetic in an Ordered Field) and using transitivity,

$$\gamma\,\delta\le F\bigl(\hat{x},u(\hat{x}),p_{\alpha},X_{\alpha}\bigr)-F\bigl(\hat{x},v(\hat{y}),p_{\alpha},X_{\alpha}\bigr).$$

Write the right-hand side as the sum

$$\Bigl(F\bigl(\hat{x},u(\hat{x}),p_{\alpha},X_{\alpha}\bigr)-F\bigl(\hat{y},v(\hat{y}),p_{\alpha},X_{\alpha}\bigr)\Bigr)+\Bigl(F\bigl(\hat{y},v(\hat{y}),p_{\alpha},X_{\alpha}\bigr)-F\bigl(\hat{x},v(\hat{y}),p_{\alpha},X_{\alpha}\bigr)\Bigr),$$

an identity of field arithmetic. By $(**)$ the first summand is at most $0$. Since $\hat{x},\hat{y}\in\Omega$ and $p_{\alpha}=(2\alpha)(\hat{x}-\hat{y})$ with $0<2\alpha$, hypothesis 3 of the theorem, applied with $x=\hat{x}$, $y=\hat{y}$, $r=v(\hat{y})$, $X=X_{\alpha}$ and $\beta=2\alpha$, bounds the second summand by $\omega(t_{\alpha})$, where

$$t_{\alpha}=2\alpha\,\psi(\hat{x},\hat{y})+d_E(\hat{x},\hat{y}),$$

which satisfies $0\le t_{\alpha}$ by claims 2 and 5 of Elementary Arithmetic in an Ordered Field. Adding the two bounds by claim 3 of the order arithmetic lemma and using transitivity,

$$\gamma\,\delta\le\omega(t_{\alpha})\qquad\text{whenever }\alpha_3\le\alpha. \tag{$*\!*\!*$}$$

Step 5: the contradiction. Since $0<\gamma$ and $0<\delta$, claim 5 of the order arithmetic lemma gives $0<\gamma\,\delta$. By claim 8 of that lemma there is $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$ and $\varepsilon+\varepsilon=\gamma\,\delta$; adding $\varepsilon$ to $0<\varepsilon$ gives $\varepsilon<\gamma\,\delta$. Since $\omega$ is a modulus of continuity, condition 2 of Modulus of Continuity provides $\sigma\in\mathbb{R}$ with $0<\sigma$ such that every $t$ with $0\le t\le\sigma$ satisfies $\omega(t)\le\varepsilon$.

Applying claim 8 of the order arithmetic lemma twice, choose $\eta\in\mathbb{R}$ with $0<\eta$ and $(\eta+\eta)+(\eta+\eta)=\sigma$. By claim 4 of Limits of Penalized Maxima on a Compact Set, applied with $\eta$ in the role of its $\varepsilon$, there is $\alpha_4$ with $0<\alpha_4$ such that $\alpha\,\psi(x_{\alpha},y_{\alpha})\le\eta$ whenever $\alpha_4\le\alpha$; multiplying by the nonnegative $2$ gives $2\alpha\,\psi(x_{\alpha},y_{\alpha})\le\eta+\eta$ for such $\alpha$. By claim 5 of Limits of Penalized Maxima on a Compact Set, applied with $\eta+\eta$ in the role of its $\eta$, there is $\alpha_5$ with $0<\alpha_5$ such that $d_E(x_{\alpha},y_{\alpha})<\eta+\eta$ whenever $\alpha_5\le\alpha$.

Let $\alpha$ be the maximum of $\alpha_3$ and the maximum of $\alpha_4$ and $\alpha_5$; then $0<\alpha$ and $\alpha_3\le\alpha$, $\alpha_4\le\alpha$, $\alpha_5\le\alpha$ by claim 1 of the maximum lemma and transitivity. For this $\alpha$, adding the two displayed bounds by claim 3 of the order arithmetic lemma gives

$$t_{\alpha}=2\alpha\,\psi(x_{\alpha},y_{\alpha})+d_E(x_{\alpha},y_{\alpha})<(\eta+\eta)+(\eta+\eta)=\sigma ,$$

so $0\le t_{\alpha}\le\sigma$ and therefore $\omega(t_{\alpha})\le\varepsilon$. Combining with $(*\!*\!*)$ and $\varepsilon<\gamma\,\delta$,

$$\gamma\,\delta\le\omega(t_{\alpha})\le\varepsilon<\gamma\,\delta ,$$

which contradicts antisymmetry of $\le$.

Therefore no $z\in\overline{\Omega}$ with $u(z)\le v(z)$ failing exists, that is, $u(x)\le v(x)$ for every $x\in\overline{\Omega}$.