Proof of Distance Functions Give Explicit Slope-Based Solutions of the Eikonal and Discounted Hopf-Lax Equations

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

For $x\in X$ let $S_{x}=\{d(x,k):k\in K\}$, so that $D(x)=\inf S_{x}$ by Distance from a Point to a Nonempty Subset of a Metric Space; $S_{x}$ is nonempty because $K$ is, and bounded below by $0$. For a real $a$ we write $[a]_{-}=\max\{-a,0\}$ as in Local Slope, Super-Slope, Sub-Slope and Upper Slope Envelope of a Locally Lipschitz Function on a Metric Space, so that $[\psi(y)-\psi(x)]_{-}=\max\{\psi(x)-\psi(y),0\}$; by claim 1 of Elementary Properties of the Maximum of Two Elements, $\psi(x)-\psi(y)\le[\psi(y)-\psi(x)]_{-}$. The slope pair of a locally Lipschitz $\psi$ at $x$ used below is always $(|\nabla^{-}\psi|(x),[\cdot]_{-})$, in the sense of Elementary Properties of Local Slopes: Order, Negation, Bounds from Difference Quotients, Lipschitz Functions and Squared Distances.

Step 1 (properties of $D$). By claims 1 to 4 of The Distance to a Set is Nonexpansive (with $A=K$), for all $x,y\in X$ and every $k\in K$,

Step 2 (points of almost steepest descent of $D$). We show: if $x\in X$ satisfies $0<D(x)$, then for all reals $\varepsilon'>0$ and $R>0$ there is $y\in X$ with

The choices are made in the order $t$, $k$, $y$. Let $t=\min\{1/2,\,R/(2(1+\varepsilon')D(x))\}$, so that $0<t\le1$, and let $\eta=\varepsilon' t\,D(x)$, a positive real. By claim 4 of Approximation Property of the Supremum and the Infimum in $\mathbb{R}$, applied to $S_{x}$ with $\eta$, there is $k\in K$ with $d(x,k)<D(x)+\eta$; by (1.1), $0<D(x)\le d(x,k)$. By Metric Space with Interpolation Points §interpolation there is an interpolation point $y$ of $x$ and $k$ at parameter $t$: $d(x,y)\le t\,d(x,k)$ and $d(y,k)\le(1-t)\,d(x,k)$. The triangle inequality gives $d(x,k)\le d(x,y)+d(y,k)\le d(x,y)+(1-t)\,d(x,k)$, so $t\,d(x,k)\le d(x,y)$, and therefore $d(x,y)=t\,d(x,k)>0$. Moreover $d(x,y)<t\,(1+\varepsilon' t)\,D(x)\le t\,(1+\varepsilon')\,D(x)\le R/2<R$. Finally, by (1.1), $D(y)\le d(y,k)\le(1-t)\,d(x,k)$, so

Part 1. Here $\Omega=X\setminus K$ is open, and by The Eikonal Equation on an Open Subset of a Metric Space the eikonal equation $|\nabla u|=\mathbf{1}$ in $\Omega$ is $H=0$ with $H(x,r,p)=p-1$. Write $D_{\Omega}$ for the restriction of $D$ to $\Omega$.

Step 3 (semicontinuity). Let $x\in\Omega$ and $\tau$ a positive real. For every $y\in\Omega$ with $d(x,y)<\tau$, (1.1) gives $|D(y)-D(x)|\le d(x,y)<\tau$, hence $D(y)<D(x)+\tau$ and $D(x)-\tau<D(y)$. So $D_{\Omega}$ is upper and lower semicontinuous on $\Omega$.

Step 4 (Part 1, subsolution). Let $\psi_{1}\in\underline{\mathcal{C}}(\Omega)$, let $\psi_{2}:\Omega\to\mathbb{R}$ be locally Lipschitz on $\Omega$, and let $x\in\Omega$ be a point at which $D_{\Omega}-\psi_{1}-\psi_{2}$ has a local maximum relative to $\Omega$, with a radius $r_{1}>0$. For $y\in\Omega$ with $d(x,y)<r_{1}$ we have $D(y)-\psi_{1}(y)-\psi_{2}(y)\le D(x)-\psi_{1}(x)-\psi_{2}(x)$, so by (1.1)

The right-hand side is nonnegative, so claim 3 of Elementary Properties of the Maximum of Two Elements gives $[\psi_{1}(y)-\psi_{1}(x)]_{-}\le d(x,y)+|\psi_{2}(y)-\psi_{2}(x)|\le(1+\varepsilon)\,d(x,y)+|\psi_{2}(y)-\psi_{2}(x)|$ for every real $\varepsilon>0$ and every $y\in\Omega$ with $0<d(x,y)<r_{1}$. Hence Elementary Properties of Local Slopes: Order, Negation, Bounds from Difference Quotients, Lipschitz Functions and Squared Distances §upper-bound, applied with $\psi=\psi_{1}$, $\varphi=\psi_{2}$, $c=1$ and $r=r_{1}$ for every $\varepsilon$, gives $|\nabla^{-}\psi_{1}|(x)\le1+|\nabla\psi_{2}|(x)$. Since $\psi_{1}\in\underline{\mathcal{C}}(\Omega)$, Test Classes for Slope-Based Viscosity Solutions on a Metric Space §sub-class gives $|\nabla\psi_{1}|(x)=|\nabla^{-}\psi_{1}|(x)$, and Elementary Properties of Local Slopes: Order, Negation, Bounds from Difference Quotients, Lipschitz Functions and Squared Distances §order gives $|\nabla\psi_{2}|(x)\le|\nabla\psi_{2}|^{*}(x)$; so $|\nabla\psi_{1}|(x)-|\nabla\psi_{2}|^{*}(x)\le1$. As also $0\le1$, claim 3 of Elementary Properties of the Maximum of Two Elements gives $\max\{|\nabla\psi_{1}|(x)-|\nabla\psi_{2}|^{*}(x),0\}\le1$, that is, $H\bigl(x,D(x),\max\{|\nabla\psi_{1}|(x)-|\nabla\psi_{2}|^{*}(x),0\}\bigr)\le0$. Together with Step 3, $D_{\Omega}$ is an s-subsolution of $H=0$ in $\Omega$ by Slope-Based Viscosity Subsolutions, Supersolutions and Solutions on a Metric Space §subsolution.

Step 5 (Part 1, supersolution). Let $\psi_{1}\in\overline{\mathcal{C}}(\Omega)$, let $\psi_{2}:\Omega\to\mathbb{R}$ be locally Lipschitz on $\Omega$, and let $x\in\Omega$ be a point at which $D_{\Omega}-\psi_{1}-\psi_{2}$ has a local minimum relative to $\Omega$, with a radius $r_{1}>0$. Since $\Omega$ is open, there is a real $r_{2}>0$ with $B_{d}(x,r_{2})\subseteq\Omega$. No $k\in K$ lies in $\Omega$, so no $k\in K$ lies in the open ball $B_{d}(x,r_{2})$, that is, $r_{2}\le d(x,k)$ for every $k\in K$; thus $r_{2}$ is a lower bound of $S_{x}$, and since $D(x)$ is the greatest lower bound of $S_{x}$, $0<r_{2}\le D(x)$.

We verify the hypothesis of Elementary Properties of Local Slopes: Order, Negation, Bounds from Difference Quotients, Lipschitz Functions and Squared Distances §lower-bound for $\psi=\psi_{1}$, $\varphi=\psi_{2}$ and $c=1$. Let $\varepsilon>0$ and $r>0$ be reals. Step 2, applied with $\varepsilon'=\varepsilon$ and $R=\min\{r,r_{1},r_{2}\}$, gives $y\in X$ with $0<d(x,y)<R$ and $(1-\varepsilon)\,d(x,y)\le D(x)-D(y)$; then $y\in B_{d}(x,r_{2})\subseteq\Omega$ and $d(x,y)<r_{1}$, so $D(x)-\psi_{1}(x)-\psi_{2}(x)\le D(y)-\psi_{1}(y)-\psi_{2}(y)$, and therefore

So Elementary Properties of Local Slopes: Order, Negation, Bounds from Difference Quotients, Lipschitz Functions and Squared Distances §lower-bound gives $1-|\nabla\psi_{2}|(x)\le|\nabla^{-}\psi_{1}|(x)$. By Elementary Properties of Local Slopes: Order, Negation, Bounds from Difference Quotients, Lipschitz Functions and Squared Distances §order, $|\nabla^{-}\psi_{1}|(x)\le|\nabla\psi_{1}|(x)$ and $|\nabla\psi_{2}|(x)\le|\nabla\psi_{2}|^{*}(x)$, so $1\le|\nabla\psi_{1}|(x)+|\nabla\psi_{2}|^{*}(x)$, that is, $H\bigl(x,D(x),|\nabla\psi_{1}|(x)+|\nabla\psi_{2}|^{*}(x)\bigr)\ge0$. (Only local Lipschitz continuity of $\psi_{1}$ was used.) Together with Step 3, $D_{\Omega}$ is an s-supersolution of $H=0$ in $\Omega$ by Slope-Based Viscosity Subsolutions, Supersolutions and Solutions on a Metric Space §supersolution, and with Step 4 an s-solution by Slope-Based Viscosity Subsolutions, Supersolutions and Solutions on a Metric Space §solution. This proves claim 1.

Part 2. Here $\Omega=X$, $u(x)=a\,D(x)^{2}$ and $f(x)=b\,D(x)^{2}$, and by The Discounted Stationary Hopf-Lax Equation on an Open Subset of a Metric Space the equation is $H=0$ with $H(x,r,p)=\rho\,r+\frac{1}{2}p^{2}-b\,D(x)^{2}$, a Hamiltonian on $X$; by Slope-Based Viscosity Subsolutions, Supersolutions and Solutions on a Metric Space §hamiltonian, $H(x,r,p)\le H(x,r,p')$ whenever $p,p'\in T$ and $p\le p'$.

Step 6 (the constant $a$). Since $0\le\rho^{2}$ and $0\le b$, the real $\rho^{2}+8b$ is nonnegative, and by Existence and Uniqueness of the Nonnegative Square Root $s=\sqrt{\rho^{2}+8b}$ is the unique real with $0\le s$ and $s^{2}=\rho^{2}+8b$. As $\rho^{2}\le s^{2}$ with $0\le\rho$ and $0\le s$, claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field gives $\rho\le s$, so $a=\frac{1}{4}(s-\rho)\ge0$. Moreover $\rho+2a=\frac{1}{2}(s+\rho)$, so

Consequently, for every $x\in X$,

Step 7 (two estimates for $u$). For $x,y\in X$ we have $u(x)-u(y)=a\bigl(D(x)-D(y)\bigr)\bigl(D(x)+D(y)\bigr)$, and by (1.1) $D(x)-D(y)\le d(x,y)$ and $D(x)+D(y)\le2D(x)+d(x,y)$, both right-hand sides being nonnegative. If $D(y)\le D(x)$, multiplying these bounds for nonnegative factors gives $\bigl(D(x)-D(y)\bigr)\bigl(D(x)+D(y)\bigr)\le d(x,y)\bigl(2D(x)+d(x,y)\bigr)$; if $D(x)<D(y)$ the left-hand side is negative. Since $0\le a$,

the second because $|u(x)-u(y)|=a\,|D(x)-D(y)|\,\bigl(D(x)+D(y)\bigr)$ and (1.1). Now let $x\in X$ and $\tau$ a positive real, and let $\delta_{0}=\min\{1,\tau/(a(2D(x)+1)+1)\}$. For $y\in X$ with $d(x,y)<\delta_{0}$, (7.1) gives $|u(x)-u(y)|\le a(2D(x)+1)\,d(x,y)\le a(2D(x)+1)\,\delta_{0}<\tau$. Hence $u(y)<u(x)+\tau$ and $u(x)-\tau<u(y)$, so $u$ is upper and lower semicontinuous on $X$.

Step 8 (Part 2, subsolution). Let $\psi_{1}\in\underline{\mathcal{C}}(X)$, let $\psi_{2}:X\to\mathbb{R}$ be locally Lipschitz on $X$, and let $x\in X$ be a point at which $u-\psi_{1}-\psi_{2}$ has a local maximum relative to $X$, with a radius $r_{1}>0$. For $y\in X$ with $d(x,y)<r_{1}$, $u(y)-\psi_{1}(y)-\psi_{2}(y)\le u(x)-\psi_{1}(x)-\psi_{2}(x)$, so by (7.1)

and since the right-hand side is nonnegative, claim 3 of Elementary Properties of the Maximum of Two Elements bounds $[\psi_{1}(y)-\psi_{1}(x)]_{-}$ by the same quantity. Given a real $\varepsilon>0$, let $r=\min\{r_{1},\varepsilon/(a+1)\}$; for $y\in X$ with $0<d(x,y)<r$ we have $a\,d(x,y)\le\varepsilon$, hence $[\psi_{1}(y)-\psi_{1}(x)]_{-}\le\bigl(2aD(x)+\varepsilon\bigr)\,d(x,y)+|\psi_{2}(y)-\psi_{2}(x)|$. So Elementary Properties of Local Slopes: Order, Negation, Bounds from Difference Quotients, Lipschitz Functions and Squared Distances §upper-bound, with $\psi=\psi_{1}$, $\varphi=\psi_{2}$ and the nonnegative real $c=2aD(x)$, gives $|\nabla^{-}\psi_{1}|(x)\le2aD(x)+|\nabla\psi_{2}|(x)$. By Test Classes for Slope-Based Viscosity Solutions on a Metric Space §sub-class, $|\nabla\psi_{1}|(x)=|\nabla^{-}\psi_{1}|(x)$, and by Elementary Properties of Local Slopes: Order, Negation, Bounds from Difference Quotients, Lipschitz Functions and Squared Distances §order, $|\nabla\psi_{2}|(x)\le|\nabla\psi_{2}|^{*}(x)$; so $|\nabla\psi_{1}|(x)-|\nabla\psi_{2}|^{*}(x)\le2aD(x)$, and as $0\le2aD(x)$, claim 3 of Elementary Properties of the Maximum of Two Elements gives $q=\max\{|\nabla\psi_{1}|(x)-|\nabla\psi_{2}|^{*}(x),0\}\le2aD(x)$. Both lie in $T$, so monotonicity of $H$ and (6.1) give $H(x,u(x),q)\le H\bigl(x,u(x),2aD(x)\bigr)=0$. With Step 7, $u$ is an s-subsolution of $H=0$ in $X$ by Slope-Based Viscosity Subsolutions, Supersolutions and Solutions on a Metric Space §subsolution.

Step 9 (Part 2, supersolution). Let $\psi_{1}\in\overline{\mathcal{C}}(X)$, let $\psi_{2}:X\to\mathbb{R}$ be locally Lipschitz on $X$, let $x\in X$ be a point at which $u-\psi_{1}-\psi_{2}$ has a local minimum relative to $X$, with a radius $r_{1}>0$, and let $q=|\nabla\psi_{1}|(x)+|\nabla\psi_{2}|^{*}(x)\in T$. If $D(x)=0$, then $u(x)=0$ and $H(x,u(x),q)=\frac{1}{2}q^{2}\ge0$.

Suppose $0<D(x)$. We verify the hypothesis of Elementary Properties of Local Slopes: Order, Negation, Bounds from Difference Quotients, Lipschitz Functions and Squared Distances §lower-bound for $\psi=\psi_{1}$, $\varphi=\psi_{2}$ and $c=2aD(x)$. Let $\varepsilon>0$ and $r>0$ be reals; choose, in this order, $\varepsilon'=\min\{1,\varepsilon/(2(2aD(x)+1))\}$ and $R=\min\{r,r_{1},D(x),\varepsilon/(2(a+1))\}$, and let $y$ be given by Step 2 for these $\varepsilon'$ and $R$, with $h=d(x,y)$. Then $0<h<R$ and $0\le(1-\varepsilon')h\le D(x)-D(y)$, while (1.1) gives $2D(x)-h\le D(x)+D(y)$ with $0<D(x)<2D(x)-h$. Multiplying these lower bounds of nonnegative factors and using $0\le a$,

Here $2aD(x)\,\varepsilon'\le\frac{\varepsilon}{2}\cdot\frac{2aD(x)}{2aD(x)+1}\le\frac{\varepsilon}{2}$ and $a\,h\le\frac{a}{a+1}\cdot\frac{\varepsilon}{2}\le\frac{\varepsilon}{2}$, so $u(x)-u(y)\ge\bigl(2aD(x)-\varepsilon\bigr)h$. Since $d(x,y)<r_{1}$, $u(x)-\psi_{1}(x)-\psi_{2}(x)\le u(y)-\psi_{1}(y)-\psi_{2}(y)$, and therefore

with $0<d(x,y)<r$. So Elementary Properties of Local Slopes: Order, Negation, Bounds from Difference Quotients, Lipschitz Functions and Squared Distances §lower-bound gives $2aD(x)-|\nabla\psi_{2}|(x)\le|\nabla^{-}\psi_{1}|(x)$, and Elementary Properties of Local Slopes: Order, Negation, Bounds from Difference Quotients, Lipschitz Functions and Squared Distances §order gives $|\nabla^{-}\psi_{1}|(x)\le|\nabla\psi_{1}|(x)$ and $|\nabla\psi_{2}|(x)\le|\nabla\psi_{2}|^{*}(x)$; hence $2aD(x)\le q$. Both lie in $T$, so monotonicity of $H$ and (6.1) give $H(x,u(x),q)\ge H\bigl(x,u(x),2aD(x)\bigr)=0$. (Only local Lipschitz continuity of $\psi_{1}$ was used.)

With Step 7, $u$ is an s-supersolution of $H=0$ in $X$ by Slope-Based Viscosity Subsolutions, Supersolutions and Solutions on a Metric Space §supersolution, and with Step 8 an s-solution by Slope-Based Viscosity Subsolutions, Supersolutions and Solutions on a Metric Space §solution. This proves claim 2.