Proof of Comparison Principle for Slope-Based Solutions of the Eikonal Equation on a Complete Metric Space with Interpolation Points

Each result cited below is universally quantified over the data in its own statement.

Throughout, $H_{f}(x,r,p)=p-f(x)$ for $x\in\Omega$, $r\in\mathbb{R}$ and $p\in T$ is the Hamiltonian on $\Omega$ of The Eikonal Equation on an Open Subset of a Metric Space; by The Eikonal Equation on an Open Subset of a Metric Space §eikonal the hypotheses say that the restriction $u|_{\Omega}$ of $u$ to $\Omega$ is an s-subsolution and the restriction $v|_{\Omega}$ an s-supersolution of $H_{f}=0$ in $\Omega$, in the sense of Slope-Based Viscosity Subsolutions, Supersolutions and Solutions on a Metric Space §subsolution and Slope-Based Viscosity Subsolutions, Supersolutions and Solutions on a Metric Space §supersolution. The properties of $D$ used below are those of The Distance to a Set is Nonexpansive (with $A=K$): for all $x,y\in X$ and $k\in K$, $D(x)\le d(x,k)$ by its claim 2 and $D(x)\le d(x,y)+D(y)$ by its claim 3.

We argue by contradiction. Suppose that there is $z\in X$ with $v(z)<u(z)$, and put $\theta=u(z)-v(z)$, a positive real. The constants are chosen in the order $M$, $c$, $\beta$, $\sigma$, $\lambda$, $\sigma'$, $\gamma$, $\varepsilon$ of Step 2; the points $(x_{0},y_{0})$ and $(x_{\varepsilon},y_{\varepsilon})$ of Step 3 are chosen after $\varepsilon$.

Step 1 (the doubled space). Let $Z=X\times X$ and $\varrho\bigl((x,y),(x',y')\bigr)=d(x,x')+d(y,y')$. Conditions 1, 3 and 4 of Metric Space for $\varrho$ follow by adding the same conditions for $d$ in the two coordinates; and since both summands are nonnegative, $\varrho\bigl((x,y),(x',y')\bigr)=0$ holds exactly when $d(x,x')=0$ and $d(y,y')=0$, that is, by condition 2 for $d$, exactly when $(x,y)=(x',y')$. So $(Z,\varrho)$ is a metric space.

It is complete: let $\bigl((x_{m},y_{m})\bigr)$ be a Cauchy sequence in $(Z,\varrho)$. Since $d(x_{m},x_{\ell})\le\varrho\bigl((x_{m},y_{m}),(x_{\ell},y_{\ell})\bigr)$ and $d(y_{m},y_{\ell})\le\varrho\bigl((x_{m},y_{m}),(x_{\ell},y_{\ell})\bigr)$, the sequences $(x_{m})$ and $(y_{m})$ are Cauchy sequences in $(X,d)$, so by completeness of $(X,d)$ they converge to some $\bar{x}\in X$ and $\bar{y}\in X$. Given a real $\tau>0$, take $N_{1},N_{2}\in\mathbb{N}$ with $d(x_{m},\bar{x})<\tau/2$ for $m\ge N_{1}$ and $d(y_{m},\bar{y})<\tau/2$ for $m\ge N_{2}$; then $\varrho\bigl((x_{m},y_{m}),(\bar{x},\bar{y})\bigr)<\tau$ for $m\ge\max\{N_{1},N_{2}\}$, so $\bigl((x_{m},y_{m})\bigr)$ converges to $(\bar{x},\bar{y})$ in $(Z,\varrho)$.

Now let $c$ and $\varepsilon$ be positive reals and $\Phi(x,y)=c\,u(x)-v(y)-\frac{1}{\varepsilon}d(x,y)^{2}$ on $Z$. Then $\Phi$ is upper semicontinuous on $Z$ in the sense of Upper Semicontinuous Function on a Subset of a Metric Space (for the metric space $(Z,\varrho)$). Indeed, fix $(\bar{x},\bar{y})\in Z$ and a real $\tau>0$, and put $\ell=d(\bar{x},\bar{y})$. Since $u$ is upper semicontinuous at $\bar{x}$, there is a real $\delta_{1}>0$ with $u(x)<u(\bar{x})+\tau/(3c)$, hence $c\,u(x)<c\,u(\bar{x})+\tau/3$, for all $x\in X$ with $d(\bar{x},x)<\delta_{1}$; since $v$ is lower semicontinuous at $\bar{y}$, there is a real $\delta_{2}>0$ with $v(\bar{y})-\tau/3<v(y)$ for all $y\in X$ with $d(\bar{y},y)<\delta_{2}$. Let $\delta_{3}=\min\{1,\varepsilon\tau/(3(2\ell+1))\}$ and $\delta_{0}=\min\{\delta_{1},\delta_{2},\delta_{3}\}$, and let $(x,y)\in Z$ with $s=\varrho\bigl((x,y),(\bar{x},\bar{y})\bigr)<\delta_{0}$. Then $d(\bar{x},x)\le s$ and $d(\bar{y},y)\le s$, and the triangle inequality (condition 4 of Metric Space) gives $\ell\le d(\bar{x},x)+d(x,y)+d(y,\bar{y})$ and $d(x,y)\le d(x,\bar{x})+\ell+d(\bar{y},y)$, so $|d(x,y)-\ell|\le s$. If $d(x,y)\le\ell$ then

and if $\ell<d(x,y)$ then $\ell^{2}-d(x,y)^{2}<0<\varepsilon\tau/3$. Adding the three estimates, $\Phi(x,y)<\Phi(\bar{x},\bar{y})+\tau$.

Step 2 (constants). Since $u$ and $v$ are bounded above and below in the sense of The Real Numbers: Standing Notation and Background §bounds, there are reals $a_{1},b_{1},a_{2},b_{2}$ with $a_{1}\le u(x)\le b_{1}$ and $a_{2}\le v(x)\le b_{2}$ for every $x\in X$; with $M=1+|a_{1}|+|b_{1}|+|a_{2}|+|b_{2}|$, a positive real, we get $|u(x)|\le M$ and $|v(x)|\le M$ for every $x\in X$. In particular $\theta\le|u(z)|+|v(z)|\le2M$.

Let $c=1-\theta/(16M)$. Then $0<\theta/(16M)\le1/8$, so $7/8\le c<1$, and

Let $\beta=\theta/2$. By the boundary hypothesis of the statement, applied with the positive real $\beta/4$, there is a real $\sigma>0$ with $u(x)-v(y)<\beta/4$ for all $x,y\in X$ with $D(x)+D(y)+d(x,y)<\sigma$. For such $x,y$, (2.1) gives $-(1-c)\,u(x)\le\theta/16=\beta/8$, hence

Let $\lambda=(1-c)\,c_{0}$, a positive real. By uniform continuity of $f$ on $\Omega$ there is a real $\sigma'>0$ with $|f(x)-f(y)|<\lambda/2$ for all $x,y\in\Omega$ with $d(x,y)<\sigma'$. Let $\gamma=\min\{\sigma/4,\sigma'\}$, and finally

Then $0<\varepsilon<1$, so $\varepsilon^{2}\le\varepsilon\le\theta/4$; moreover $2M\varepsilon\le\gamma^{2}/2<\gamma^{2}$ and $2\varepsilon\le\lambda/2$.

Step 3 (doubling of variables and Ekeland's principle). Let $\Phi$ be the function of Step 1 for these $c$ and $\varepsilon$. Since $0<c\le1$, for all $(x,y)\in Z$ we have $\Phi(x,y)\le c\,u(x)-v(y)\le c\,M+M\le2M$, so the nonempty set $\Phi(Z)$ is bounded above and $S=\sup_{(x,y)\in Z}\Phi(x,y)$ exists by The Real Numbers: Standing Notation and Background §bounds. By (2.1),

By claim 3 of Approximation Property of the Supremum and the Infimum in $\mathbb{R}$, applied with the positive real $\varepsilon^{2}$, there is $(x_{0},y_{0})\in Z$ with $S-\varepsilon^{2}<\Phi(x_{0},y_{0})$. By Step 1, Ekeland's Variational Principle for Upper Semicontinuous Functions on a Complete Metric Space applies to $F=\Phi$ on the complete metric space $(Z,\varrho)$ with $\eta=\varepsilon^{2}$ and $\kappa=\varepsilon$, so that $\eta/\kappa=\varepsilon$; it gives $(x_{\varepsilon},y_{\varepsilon})\in Z$ such that, by Ekeland's Variational Principle for Upper Semicontinuous Functions on a Complete Metric Space §value and $\varepsilon^{2}\le\theta/4$,

and, by Ekeland's Variational Principle for Upper Semicontinuous Functions on a Complete Metric Space §perturbed, $\Phi(x,y)-\varepsilon\,d(x,x_{\varepsilon})-\varepsilon\,d(y,y_{\varepsilon})<\Phi(x_{\varepsilon},y_{\varepsilon})$ for every $(x,y)\ne(x_{\varepsilon},y_{\varepsilon})$. Since the left-hand side equals $\Phi(x_{\varepsilon},y_{\varepsilon})$ at $(x,y)=(x_{\varepsilon},y_{\varepsilon})$, we get

Put $\delta=d(x_{\varepsilon},y_{\varepsilon})$. From (3.1), $\beta<\Phi(x_{\varepsilon},y_{\varepsilon})\le c\,u(x_{\varepsilon})-v(y_{\varepsilon})$ and $\delta^{2}/\varepsilon=c\,u(x_{\varepsilon})-v(y_{\varepsilon})-\Phi(x_{\varepsilon},y_{\varepsilon})<2M$, so $\delta^{2}<2M\varepsilon<\gamma^{2}$; as $\delta$ and $\gamma$ are nonnegative, claim 1 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field gives

Step 4 (the points lie in $\Omega$). Suppose $D(x_{\varepsilon})\le\sigma/4$. Since $D(y_{\varepsilon})\le\delta+D(x_{\varepsilon})$, (3.3) gives $D(x_{\varepsilon})+D(y_{\varepsilon})+\delta\le2D(x_{\varepsilon})+2\delta<\sigma/2+\sigma/2=\sigma$, so $c\,u(x_{\varepsilon})-v(y_{\varepsilon})<\beta/2$ by (2.2), contradicting $\beta<c\,u(x_{\varepsilon})-v(y_{\varepsilon})$ from Step 3. Symmetrically, if $D(y_{\varepsilon})\le\sigma/4$ then $D(x_{\varepsilon})\le\delta+D(y_{\varepsilon})$ leads to the same contradiction. Hence $D(x_{\varepsilon})>\sigma/4>0$ and $D(y_{\varepsilon})>\sigma/4>0$. If $x_{\varepsilon}$ were in $K$, then $D(x_{\varepsilon})\le d(x_{\varepsilon},x_{\varepsilon})=0$; so $x_{\varepsilon}\in\Omega$, and likewise $y_{\varepsilon}\in\Omega$.

Step 5 (subsolution test at $x_{\varepsilon}$). Define $\psi_{1},\psi_{2}:\Omega\to\mathbb{R}$ by $\psi_{1}(x)=\frac{1}{c\varepsilon}d(x,y_{\varepsilon})^{2}+\frac{1}{c}v(y_{\varepsilon})$ and $\psi_{2}(x)=\frac{\varepsilon}{c}d(x,x_{\varepsilon})$. Since $(X,d)$ has interpolation points and $\Omega$ is open, Elementary Properties of Local Slopes: Order, Negation, Bounds from Difference Quotients, Lipschitz Functions and Squared Distances §squared-distance, applied with $x_{0}=y_{\varepsilon}$, $k=1/(c\varepsilon)$ and $C=v(y_{\varepsilon})/c$, shows that $\psi_{1}\in\underline{\mathcal{C}}(\Omega)$ and $|\nabla\psi_{1}|(x_{\varepsilon})=2\delta/(c\varepsilon)$. For $y,w\in\Omega$ the triangle inequality gives $|d(y,x_{\varepsilon})-d(w,x_{\varepsilon})|\le d(y,w)$, so $|\psi_{2}(y)-\psi_{2}(w)|\le\frac{\varepsilon}{c}d(y,w)$; by Elementary Properties of Local Slopes: Order, Negation, Bounds from Difference Quotients, Lipschitz Functions and Squared Distances §lipschitz with $L=\varepsilon/c$, $\psi_{2}$ is locally Lipschitz on $\Omega$ and $e_{1}=|\nabla\psi_{2}|^{*}(x_{\varepsilon})$ satisfies $e_{1}\le\varepsilon/c$. Taking $y=y_{\varepsilon}$ in (3.2), dividing by $c>0$ and noting $\psi_{2}(x_{\varepsilon})=0$, we get $u(x)-\psi_{1}(x)-\psi_{2}(x)\le u(x_{\varepsilon})-\psi_{1}(x_{\varepsilon})-\psi_{2}(x_{\varepsilon})$ for every $x\in\Omega$, so $u|_{\Omega}-\psi_{1}-\psi_{2}$ has a local maximum at $x_{\varepsilon}$ relative to $\Omega$ (with any radius). By Slope-Based Viscosity Subsolutions, Supersolutions and Solutions on a Metric Space §subsolution, $\max\{2\delta/(c\varepsilon)-e_{1},0\}-f(x_{\varepsilon})\le0$. By claim 1 of Elementary Properties of the Maximum of Two Elements and $e_{1}\le\varepsilon/c$,

Step 6 (supersolution test at $y_{\varepsilon}$). Define $\psi_{3},\psi_{4}:\Omega\to\mathbb{R}$ by $\psi_{3}(y)=-\frac{1}{\varepsilon}d(y,x_{\varepsilon})^{2}+c\,u(x_{\varepsilon})$ and $\psi_{4}(y)=-\varepsilon\,d(y,y_{\varepsilon})$. Then $\psi_{3}=-\varphi$ with $\varphi(y)=\frac{1}{\varepsilon}d(y,x_{\varepsilon})^{2}-c\,u(x_{\varepsilon})$, so the last sentence of Elementary Properties of Local Slopes: Order, Negation, Bounds from Difference Quotients, Lipschitz Functions and Squared Distances §squared-distance (with $x_{0}=x_{\varepsilon}$, $k=1/\varepsilon$, $C=-c\,u(x_{\varepsilon})$) shows that $\psi_{3}\in\overline{\mathcal{C}}(\Omega)$ and $|\nabla\psi_{3}|(y_{\varepsilon})=2\,d(y_{\varepsilon},x_{\varepsilon})/\varepsilon=2\delta/\varepsilon$. As in Step 5, $|\psi_{4}(y)-\psi_{4}(w)|\le\varepsilon\,d(y,w)$ for $y,w\in\Omega$, so by Elementary Properties of Local Slopes: Order, Negation, Bounds from Difference Quotients, Lipschitz Functions and Squared Distances §lipschitz $\psi_{4}$ is locally Lipschitz on $\Omega$ and $e_{2}=|\nabla\psi_{4}|^{*}(y_{\varepsilon})$ satisfies $e_{2}\le\varepsilon$. Taking $x=x_{\varepsilon}$ in (3.2) and using $d(x_{\varepsilon},y)=d(y,x_{\varepsilon})$, we get $c\,u(x_{\varepsilon})-v(y)-\frac{1}{\varepsilon}d(y,x_{\varepsilon})^{2}-\varepsilon\,d(y,y_{\varepsilon})\le c\,u(x_{\varepsilon})-v(y_{\varepsilon})-\frac{1}{\varepsilon}\delta^{2}$, which rearranges to $v(y_{\varepsilon})-\psi_{3}(y_{\varepsilon})-\psi_{4}(y_{\varepsilon})\le v(y)-\psi_{3}(y)-\psi_{4}(y)$ for every $y\in\Omega$. So $v|_{\Omega}-\psi_{3}-\psi_{4}$ has a local minimum at $y_{\varepsilon}$ relative to $\Omega$, and Slope-Based Viscosity Subsolutions, Supersolutions and Solutions on a Metric Space §supersolution gives $2\delta/\varepsilon+e_{2}-f(y_{\varepsilon})\ge0$, hence

Step 7 (contradiction). By (6.1) and (5.1), $f(y_{\varepsilon})\le c\,f(x_{\varepsilon})+2\varepsilon$, that is,

Since $x_{\varepsilon}\in\Omega$, $c_{0}\le f(x_{\varepsilon})$, and $0<1-c$, the left-hand side is at least $(1-c)\,c_{0}=\lambda$. Since $x_{\varepsilon},y_{\varepsilon}\in\Omega$ and $d(x_{\varepsilon},y_{\varepsilon})=\delta<\sigma'$ by (3.3), the choice of $\sigma'$ gives $f(x_{\varepsilon})-f(y_{\varepsilon})<\lambda/2$, and $2\varepsilon\le\lambda/2$ by Step 2. Hence $\lambda<\lambda/2+\lambda/2=\lambda$, which is impossible. Hence there is no $z\in X$ with $v(z)<u(z)$, that is, $u(x)\le v(x)$ for every $x\in X$.