Proof of Elementary Properties of Local Slopes: Order, Negation, Bounds from Difference Quotients, Lipschitz Functions and Squared Distances

Notation and three general facts. For a function $\chi:\Omega\to\mathbb{R}$ locally Lipschitz on $\Omega$, a point $x\in\Omega$, one of the three maps $g$ and a real $r>0$, write $A^{g}_{r}(x;\chi)$, $\rho^{g}(x;\chi)$ and $S^{g}\chi(x)$ for the sets and the number formed in the preamble of Local Slope, Super-Slope, Sub-Slope and Upper Slope Envelope of a Locally Lipschitz Function on a Metric Space with $\chi$ in place of $\psi$; thus a slope pair $(\sigma,g)$ of $\chi$ at $x$ has $\sigma=S^{g}\chi(x)$, and $|\nabla\chi|(x)=S^{g}\chi(x)$ for $g(a)=|a|$ by Local Slope, Super-Slope, Sub-Slope and Upper Slope Envelope of a Locally Lipschitz Function on a Metric Space §slope. Recall from that preamble that $0\le g(a)\le|a|$ for every real $a$ and that $0\le S^{g}\chi(x)$.

(F1) If $r\in\rho^{g}(x;\chi)$, then $S^{g}\chi(x)\le\sup A^{g}_{r}(x;\chi)$, because an infimum is a lower bound.

(F2) If $r>0$ and $M$ is a real number with $a\le M$ for every $a\in A^{g}_{r}(x;\chi)$, then $A^{g}_{r}(x;\chi)$ is bounded above, so $r\in\rho^{g}(x;\chi)$, $\sup A^{g}_{r}(x;\chi)\le M$ because the supremum is the least upper bound, and $S^{g}\chi(x)\le M$ by (F1).

(F3) For every real $\varepsilon>0$ there is $r\in\rho^{|\cdot|}(x;\chi)$ with $|\chi(y)-\chi(x)|<(|\nabla\chi|(x)+\varepsilon)\,d(x,y)$ for all $y\in\Omega$ with $0<d(x,y)<r$. Indeed, the set $\{\sup A^{|\cdot|}_{r}(x;\chi):r\in\rho^{|\cdot|}(x;\chi)\}$ is nonempty and bounded below with infimum $|\nabla\chi|(x)$, so by claim 4 of Approximation Property of the Supremum and the Infimum in $\mathbb{R}$ some $r\in\rho^{|\cdot|}(x;\chi)$ has $\sup A^{|\cdot|}_{r}(x;\chi)<|\nabla\chi|(x)+\varepsilon$; for $y\in\Omega$ with $0<d(x,y)<r$ the quotient $|\chi(y)-\chi(x)|/d(x,y)$ belongs to $A^{|\cdot|}_{r}(x;\chi)$, hence is below $|\nabla\chi|(x)+\varepsilon$, and we multiply by $d(x,y)>0$.

Clause 1 (Order). Let $g(a)=[a]_{+}$ or $g(a)=[a]_{-}$, and let $r\in\rho^{|\cdot|}(x;\psi)$. The element $0$ of $A^{g}_{r}(x;\psi)$ is at most $\sup A^{|\cdot|}_{r}(x;\psi)$, which is nonnegative, and for $y\in\Omega$ with $0<d(x,y)<r$ we have $g(\psi(y)-\psi(x))/d(x,y)\le|\psi(y)-\psi(x)|/d(x,y)\le\sup A^{|\cdot|}_{r}(x;\psi)$, the middle term being an element of $A^{|\cdot|}_{r}(x;\psi)$. By (F2), $S^{g}\psi(x)\le\sup A^{|\cdot|}_{r}(x;\psi)$. As $r\in\rho^{|\cdot|}(x;\psi)$ was arbitrary, $S^{g}\psi(x)$ is a lower bound of the set whose infimum is $|\nabla\psi|(x)$, so $S^{g}\psi(x)\le|\nabla\psi|(x)$; by Local Slope, Super-Slope, Sub-Slope and Upper Slope Envelope of a Locally Lipschitz Function on a Metric Space §super-slope and Local Slope, Super-Slope, Sub-Slope and Upper Slope Envelope of a Locally Lipschitz Function on a Metric Space §sub-slope this gives $|\nabla^{+}\psi|(x)\le|\nabla\psi|(x)$ and $|\nabla^{-}\psi|(x)\le|\nabla\psi|(x)$.

For the envelope, let $E_{r}(x)$ and $\rho^{*}(x)$ be as in Local Slope, Super-Slope, Sub-Slope and Upper Slope Envelope of a Locally Lipschitz Function on a Metric Space §envelope, and let $r\in\rho^{*}(x)$. Since $x\in\Omega$ and $d(x,x)=0<r$, the number $|\nabla\psi|(x)$ belongs to $E_{r}(x)$, so $|\nabla\psi|(x)\le\sup E_{r}(x)$. Thus $|\nabla\psi|(x)$ is a lower bound of $\{\sup E_{r}(x):r\in\rho^{*}(x)\}$, and $|\nabla\psi|(x)\le|\nabla\psi|^{*}(x)$.

Clause 2 (Negation). For all $y,z\in\Omega$ we have $|(-\psi)(y)-(-\psi)(z)|=|-(\psi(y)-\psi(z))|=|\psi(y)-\psi(z)|$ by claim 2 of Properties of the Absolute Value in an Ordered Field, so every Lipschitz pair of $\psi$ at a point is one of $-\psi$, and $-\psi$ is locally Lipschitz on $\Omega$ by Locally Lipschitz Function on an Open Subset of a Metric Space §locally-lipschitz. For real $a$ we have $|-a|=|a|$, $[-a]_{+}=\max\{-a,0\}=[a]_{-}$ and $[-a]_{-}=\max\{a,0\}=[a]_{+}$. Applying this with $a=\psi(y)-\psi(x)$, so that $-a=(-\psi)(y)-(-\psi)(x)$, shows for every $r>0$ that $A^{|\cdot|}_{r}(x;-\psi)=A^{|\cdot|}_{r}(x;\psi)$, that $A^{g}_{r}(x;-\psi)=A^{h}_{r}(x;\psi)$ for $g(a)=[a]_{+}$ and $h(a)=[a]_{-}$, and that $A^{h}_{r}(x;-\psi)=A^{g}_{r}(x;\psi)$. Equal families of sets have the same radii of boundedness, the same suprema and hence the same infimum of suprema, which by Local Slope, Super-Slope, Sub-Slope and Upper Slope Envelope of a Locally Lipschitz Function on a Metric Space §slope, Local Slope, Super-Slope, Sub-Slope and Upper Slope Envelope of a Locally Lipschitz Function on a Metric Space §super-slope and Local Slope, Super-Slope, Sub-Slope and Upper Slope Envelope of a Locally Lipschitz Function on a Metric Space §sub-slope gives the three equalities.

Clause 3 (Upper bound). Fix a real $\varepsilon>0$. Choose $r>0$ as in the hypothesis for this $\varepsilon$, then $r'$ as in (F3) for $\chi=\varphi$ and this $\varepsilon$, and let $r''$ be the smaller of $r$ and $r'$. Put $M=c+|\nabla\varphi|(x)+2\varepsilon$, which is nonnegative because $c\ge0$ and $|\nabla\varphi|(x)\ge0$. For $y\in\Omega$ with $0<d(x,y)<r''$ the hypothesis and (F3) give

so the corresponding element of $A^{g}_{r''}(x;\psi)$ is below $M$; the element $0$ is at most $M$ as well. By (F2), $\sigma=S^{g}\psi(x)\le c+|\nabla\varphi|(x)+2\varepsilon$. Given any real $\varepsilon'>0$, this with $\varepsilon=\varepsilon'/2$ yields $\sigma\le c+|\nabla\varphi|(x)+\varepsilon'$, and slack above gives $\sigma\le c+|\nabla\varphi|(x)$.

Clause 4 (Lower bound). Fix $r\in\rho^{g}(x;\psi)$ and then a real $\varepsilon>0$. Choose $r'$ as in (F3) for $\chi=\varphi$ and this $\varepsilon$, and apply the hypothesis with $\varepsilon$ and with the smaller of $r$ and $r'$ as radius to obtain $y\in\Omega$ with $0<d(x,y)<r$, $d(x,y)<r'$ and

Dividing by $d(x,y)>0$, the element $g(\psi(y)-\psi(x))/d(x,y)$ of $A^{g}_{r}(x;\psi)$ exceeds $c-|\nabla\varphi|(x)-2\varepsilon$, hence so does $\sup A^{g}_{r}(x;\psi)$. As $\varepsilon>0$ was arbitrary (replace $\varepsilon$ by half a given positive number), slack below gives $c-|\nabla\varphi|(x)\le\sup A^{g}_{r}(x;\psi)$. As $r\in\rho^{g}(x;\psi)$ was arbitrary, $c-|\nabla\varphi|(x)$ is a lower bound of the set whose infimum is $\sigma$, so $c-|\nabla\varphi|(x)\le\sigma$.

Clause 5 (Lipschitz functions). Let $x\in\Omega$. As $\Omega$ is open, there is $r>0$ with $B_{d}(x,r)\subseteq\Omega$, and the hypothesis gives $|\psi(y)-\psi(z)|\le L\,d(y,z)$ for $y,z\in B_{d}(x,r)$; so $\psi$ is locally Lipschitz on $\Omega$ by Locally Lipschitz Function on an Open Subset of a Metric Space §locally-lipschitz. For every $r>0$, every element of $A^{|\cdot|}_{r}(x;\psi)$ is at most $L$: the element $0$ because $L\ge0$, and $|\psi(y)-\psi(x)|/d(x,y)\le L$ for $y\in\Omega$ with $0<d(x,y)<r$. By (F2) with $r=1$, $|\nabla\psi|(x)\le L$. Consequently, for every $r>0$ every element $|\nabla\psi|(y)$ of $E_{r}(x)$ (notation of Local Slope, Super-Slope, Sub-Slope and Upper Slope Envelope of a Locally Lipschitz Function on a Metric Space §envelope) is at most $L$, so $1\in\rho^{*}(x)$, $\sup E_{1}(x)\le L$, and $|\nabla\psi|^{*}(x)\le\sup E_{1}(x)\le L$.

Clause 6 (Squared distances). Write $D(y)=d(y,x_{0})$ for $y\in X$, so that $\varphi(y)=k\,D(y)^{2}+C$ on $\Omega$.

Step 1 (Basic estimate). For $y,z\in X$ the triangle inequality and symmetry of $d$ (Metric Space) give $D(y)\le d(y,z)+D(z)$ and $D(z)\le d(y,z)+D(y)$, so $|D(y)-D(z)|\le d(y,z)$ by claim 6 of Properties of the Absolute Value in an Ordered Field. For $y,z\in\Omega$, claim 4 of Zero Products and Elementary Identities in a Field gives $\varphi(y)-\varphi(z)=k\,(D(y)-D(z))(D(y)+D(z))$, and since $k\ge0$ and $D(y)+D(z)\ge0$, claim 4 of Properties of the Absolute Value in an Ordered Field yields

Step 2 (Local Lipschitz property). Let $x\in\Omega$ and choose $r>0$ with $B_{d}(x,r)\subseteq\Omega$ (openness of $\Omega$). For $y,z\in B_{d}(x,r)$ we have $D(y)\le d(y,x)+D(x)<D(x)+r$ and likewise $D(z)<D(x)+r$, so $(*)$ gives $|\varphi(y)-\varphi(z)|\le 2k(D(x)+r)\,d(y,z)$ with $2k(D(x)+r)\ge0$. Hence $\varphi$ is locally Lipschitz on $\Omega$ by Locally Lipschitz Function on an Open Subset of a Metric Space §locally-lipschitz.

Step 3 (Upper bound). Let $\omega:\Omega\to\mathbb{R}$ be the zero function. By Clause 5 with $L=0$ it is locally Lipschitz with $|\nabla\omega|(x)\le0$, and $|\nabla\omega|(x)\ge0$ by Local Slope, Super-Slope, Sub-Slope and Upper Slope Envelope of a Locally Lipschitz Function on a Metric Space, so $|\nabla\omega|(x)=0$ for every $x\in\Omega$. Fix $x\in\Omega$ and a real $\varepsilon>0$, and put $r=\varepsilon/(k+1)>0$. For $y\in\Omega$ with $0<d(x,y)<r$ we have $D(y)\le d(x,y)+D(x)$ and $k\,d(x,y)\le(k+1)\,d(x,y)<\varepsilon$, so $(*)$ with $z=x$ gives

Clause 3 (proved above), applied to $\varphi$ in the role of $\psi$, $\omega$ in the role of $\varphi$, the slope pair $(|\nabla\varphi|(x),|\cdot|)$ and $c=2k\,D(x)\ge0$, gives $|\nabla\varphi|(x)\le2k\,D(x)$.

Step 4 (Lower bound for the sub-slope). Fix $x\in\Omega$. If $D(x)=0$, then Clause 1 and Step 3 give $0\le|\nabla^{-}\varphi|(x)\le|\nabla\varphi|(x)\le0=2k\,D(x)$. Suppose $D(x)>0$, and let $r_{\Omega}>0$ satisfy $B_{d}(x,r_{\Omega})\subseteq\Omega$. We verify the hypothesis of Clause 4 for $\varphi$ in the role of $\psi$, $\omega$ in the role of $\varphi$, the slope pair $(|\nabla^{-}\varphi|(x),[\cdot]_{-})$ and $c=2k\,D(x)$. Let reals $\varepsilon>0$ and $r>0$ be given, let $s$ be the smaller of $r$ and $r_{\Omega}$, and let $t$ be the smallest of the three positive reals $\tfrac12$, $s/(2D(x))$ and $\varepsilon/((k+1)D(x))$. Then $0<t<1$, $t\,D(x)\le s/2<s$ and $k\,t\,D(x)\le(k+1)\,t\,D(x)\le\varepsilon$. Since $(X,d)$ has interpolation points, there is an interpolation point $z$ of $x$ and $x_{0}$ at parameter $t$:

By the triangle inequality $D(x)\le d(x,z)+D(z)\le d(x,z)+(1-t)D(x)$, so $d(x,z)\ge t\,D(x)$ and therefore $d(x,z)=t\,D(x)$. Hence $0<d(x,z)<s$, so $z\in B_{d}(x,r_{\Omega})\subseteq\Omega$ and $0<d(x,z)<r$. As $0\le D(z)\le(1-t)D(x)$, claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field gives $D(z)^{2}\le(1-t)^{2}D(x)^{2}$, and so

Since $[\varphi(z)-\varphi(x)]_{-}=\max\{\varphi(x)-\varphi(z),0\}\ge\varphi(x)-\varphi(z)$ by claim 1 of Elementary Properties of the Maximum of Two Elements, we obtain $[\varphi(z)-\varphi(x)]_{-}\ge(c-\varepsilon)\,d(x,z)-|\omega(z)-\omega(x)|$, which is the hypothesis of Clause 4 with the point $z$. Clause 4 (proved above) gives $2k\,D(x)=c-|\nabla\omega|(x)\le|\nabla^{-}\varphi|(x)$, and with Clause 1 and Step 3, $2k\,D(x)\le|\nabla^{-}\varphi|(x)\le|\nabla\varphi|(x)\le2k\,D(x)$.

In both cases $|\nabla\varphi|(x)=|\nabla^{-}\varphi|(x)=2k\,D(x)$ for every $x\in\Omega$.

Step 5 (Continuity and the test classes). For $x,y\in\Omega$, Step 1 gives $|2k\,D(y)-2k\,D(x)|=2k\,|D(y)-D(x)|\le2k\,d(x,y)$. Given a real $\varepsilon>0$, put $\delta=\varepsilon/(2k+1)>0$; then $d(x,y)<\delta$ implies $|2k\,D(y)-2k\,D(x)|\le(2k+1)\,d(x,y)<\varepsilon$. So $x\mapsto|\nabla\varphi|(x)=2k\,D(x)$ is continuous on $\Omega$, and together with Steps 2 and 4 this shows $\varphi\in\underline{\mathcal{C}}(\Omega)$ by Test Classes for Slope-Based Viscosity Solutions on a Metric Space §sub-class. By Clause 2, $-\varphi$ is locally Lipschitz on $\Omega$, $|\nabla(-\varphi)|(x)=|\nabla\varphi|(x)=2k\,D(x)$ and $|\nabla^{+}(-\varphi)|(x)=|\nabla^{-}\varphi|(x)=2k\,D(x)$ for every $x\in\Omega$; the map $x\mapsto|\nabla(-\varphi)|(x)$ is the continuous map just treated, so $-\varphi\in\overline{\mathcal{C}}(\Omega)$ by Test Classes for Slope-Based Viscosity Solutions on a Metric Space §super-class.