Proof of Sup- and Inf-Convolutions of Bounded Viscosity Sub- and Supersolutions of the Penalty-Drift Equation on the Whole Space with a Monotone Drift

Each result cited below is universally quantified over the data in its own statement, and is applied to the data named at the point of use. This proof follows the scheme of the published proof for the penalised case, with the penalty term removed.

Elementary order and arithmetic manipulations of real numbers (rearranging finite sums, adding inequalities, multiplying an inequality by a nonnegative or by a positive number, reversing an inequality by negation, the rules for the absolute value, and the Archimedean property) are provided by The Real Numbers: Standing Notation and Background §background and are not cited individually; the same applies to the vector-space identities in $\mathbb{R}^{n}$, such as $(x+h)-y=(x-y)+h$. We write $D=\mathbb{R}^{n}$. By clause 2 of the setting the Euclidean distance is $d_{E}(x,y)=\lVert x-y\rVert$, so local extrema relative to $D$ and semicontinuity on $D$ are expressed below through $\lVert x-y\rVert$. We keep the notation $r$, $F$, $F_{+}$, $F_{-}$ of the statement; by Existence and Uniqueness of the Nonnegative Square Root of a Nonnegative Real Number, $0\le r$ and $r^{2}=4\tau M$.

Step 0 (Preliminaries).

(0.1) Continuity. For a fixed $a\in\mathbb{R}^{n}$, the function $e_{a}:D\to\mathbb{R}$, $e_{a}(y)=\tfrac{1}{2\tau}\lVert y-a\rVert^{2}=\tfrac{1}{2\tau}d_{E}(y,a)^{2}$, is of class $C^{2}$ on $D$ by claim 2 of A Scaled Squared Distance to a Point is of Class $C^2$, with Gradient and Hessian (with $U=D$ and $c=\tfrac{1}{2\tau}$); it is then continuous at every point of $D$ (claim 2 of Coordinate Functions, the $C^k$ Hierarchy, and Partial Derivatives of a Smooth Map and clause 2 of the calculus setting) and so lower semicontinuous on $D$ by claim 2 of Semicontinuity Under Negation and Characterization of Continuity.

(0.2) Squares. Let $s,s'\in\mathbb{R}$ with $0\le s$ and $0\le s'$. Then $s\le s'$ if and only if $s^{2}\le s'^{2}$. Indeed, if $s\le s'$ then $s^{2}\le ss'\le s'^{2}$; if instead $s'<s$, then $0<s$, and $s'^{2}\le s's<ss$, so $s^{2}\le s'^{2}$ fails.

(0.3) Expansions. For $a,b\in\mathbb{R}^{n}$ one has $\lVert a\rVert^{2}=a\cdot a$ and $0\le\lVert a\rVert$ by the square clause for the Euclidean norm, hence, by Bilinearity and Symmetry of the Dot Product on $\mathbb{R}^n$ (symmetry, additivity and homogeneity in both arguments),

$$\lVert a+b\rVert^{2}=\lVert a\rVert^{2}+2\,(a\cdot b)+\lVert b\rVert^{2},\qquad \lVert a-b\rVert^{2}=\lVert a\rVert^{2}-2\,(a\cdot b)+\lVert b\rVert^{2}.$$

Moreover $\lVert\mu a\rVert=|\mu|\,\lVert a\rVert$ for $\mu\in\mathbb{R}$ by Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ §homogeneity; in particular $\lVert a-b\rVert=\lVert b-a\rVert$, and $\lVert a-a\rVert=0$ by Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ §vanishing.

(0.4) A limiting fact. If $a,c\in\mathbb{R}$ with $0\le c$ satisfy $a\le c\sigma$ for every positive $\sigma\in\mathbb{R}$, then $a\le0$. Indeed, suppose $0<a$. If $c=0$ then $a\le0$, a contradiction; if $0<c$, then $\sigma=\tfrac{a}{2c}$ is positive and gives $a\le\tfrac{a}{2}<a$, again a contradiction.

Step 1 (A general construction). In Steps 1 to 7, $\omega:D\to\mathbb{R}$ denotes a function that is upper semicontinuous on $D$ and satisfies $|\omega(y)|\le M$ for every $y\in D$. For $x,y\in D$ put

$$f_{x}(y)=\omega(y)-\tfrac{1}{2\tau}\lVert x-y\rVert^{2}.$$

The bound on $\omega$ reads

$$-M\le\omega(y)\le M\qquad(y\in D),\tag{1.1}$$

and $f_{x}(y)\le\omega(y)$ because $0\le\tfrac{1}{2\tau}\lVert x-y\rVert^{2}$. Hence, exactly as in the statement, for each $x\in D$ the set $\{f_{x}(y):y\in D\}$ is nonempty and bounded above by $M$, and we let $w_{\omega}(x)$ be its least upper bound, which exists because the real numbers are Dedekind complete. We call $y\in D$ an $\omega$-maximiser at $x$ if $w_{\omega}(x)=f_{x}(y)$. For $\omega=u$ these are the function $\overline{w}$ and the maximisers of Part A. By the definition of a least upper bound, and by (0.3),

$$f_{x}(y)\le w_{\omega}(x)\quad\text{for all }x,y\in D,\qquad f_{x}(x)=\omega(x)\quad\text{for all }x\in D.\tag{1.2}$$

Step 2 (Existence and location of $\omega$-maximisers). Fix $x\in D$, let $L=\{y\in D:\lVert x-y\rVert\le r\}$, which is the closed ball in $(\mathbb{R}^{n},d_{E})$ with centre $x$ and radius $r$, and let $K=\{y\in D:f_{x}(x)\le f_{x}(y)\}$. By claims 2 and 3 of Elementary Properties of the Closed Ball in a Metric Space, $L$ is bounded in $(\mathbb{R}^{n},d_{E})$ and closed in $(\mathbb{R}^{n},\mathcal{T}_{d_{E}})$, so $L$ is compact by Heine-Borel Theorem in $\mathbb{R}^n$ (implication from 2 to 1). If $y\in K$, then $\omega(x)=f_{x}(x)\le f_{x}(y)=\omega(y)-\tfrac{1}{2\tau}\lVert x-y\rVert^{2}$, so by (1.1) $\tfrac{1}{2\tau}\lVert x-y\rVert^{2}\le\omega(y)-\omega(x)\le2M$; multiplying by $2\tau>0$ gives $\lVert x-y\rVert^{2}\le4\tau M=r^{2}$, and (0.2) gives $\lVert x-y\rVert\le r$. Hence $K\subseteq L$ and $K=\{y\in L:f_{x}(x)\le f_{x}(y)\}$.

The function $f_{x}$ equals $\omega-e_{x}$ on $D$ with $e_{x}$ from (0.1) (note $\lVert x-y\rVert=\lVert y-x\rVert$). Since $\omega$ is upper semicontinuous and $e_{x}$ is lower semicontinuous on $D$, claim 3 of Negation, Restriction, and Separated Differences of Semicontinuous Functions shows that $f_{x}$ is upper semicontinuous on $D$, and by claim 2 there its restrictions to $L$ and to $K$ are upper semicontinuous on $L$ and on $K$. By claim 2 of Compactness of Intersections with Closed Sets and of Level Sets of Semicontinuous Functions (with the compact set $L$ and $c=f_{x}(x)$), $K$ is compact. It is nonempty, as $x\in K$. By claim 1 of Semicontinuous Functions Attain Their Extrema on a Compact Set there is $z\in K$ with $f_{x}(y)\le f_{x}(z)$ for every $y\in K$; and every $y\in D\setminus K$ satisfies $f_{x}(y)<f_{x}(x)\le f_{x}(z)$. Thus $f_{x}(z)$ is an upper bound of $\{f_{x}(y):y\in D\}$ belonging to that set, hence its least upper bound: $w_{\omega}(x)=f_{x}(z)$, and $z$ is an $\omega$-maximiser at $x$.

Now let $z$ be any $\omega$-maximiser at $x$. By (1.2), $f_{x}(x)\le w_{\omega}(x)=f_{x}(z)$, so $z\in K\subseteq L$, that is

$$\lVert x-z\rVert\le r\qquad\text{for every }\omega\text{-maximiser }z\text{ at }x.\tag{2.1}$$

Step 3 (Bounds). For $x\in D$ let $z$ be an $\omega$-maximiser at $x$ (Step 2). By (1.2), Step 1 and (1.1),

$$\omega(x)=f_{x}(x)\le w_{\omega}(x)=f_{x}(z)\le\omega(z)\le M.\tag{3.1}$$

Step 4 (Semiconvexity). Let $G:D\to\mathbb{R}$, $G(x)=w_{\omega}(x)+\tfrac{1}{2\tau}\lVert x\rVert^{2}$; since $\tfrac{\tau^{-1}}{2}=\tfrac{1}{2\tau}$, it suffices by Semiconvex Function on a Convex Subset of $\mathbb{R}^n$ to show that $G$ is convex on $D$. For $y\in D$ let $A_{y}:\mathbb{R}^{n}\to\mathbb{R}$, $A_{y}(x)=\omega(y)-\tfrac{1}{2\tau}\lVert y\rVert^{2}+\tau^{-1}(x\cdot y)$. By (0.3), $\lVert x-y\rVert^{2}=\lVert x\rVert^{2}-2(x\cdot y)+\lVert y\rVert^{2}$, whence

$$f_{x}(y)+\tfrac{1}{2\tau}\lVert x\rVert^{2}=A_{y}(x)\qquad(x,y\in D).\tag{4.1}$$

By claims 1, 2 and 4 of Bilinearity and Symmetry of the Dot Product on $\mathbb{R}^n$, $A_{y}(t x_{1}+(1-t)x_{2})=t\,A_{y}(x_{1})+(1-t)\,A_{y}(x_{2})$ for all $x_{1},x_{2}\in\mathbb{R}^{n}$ and $t\in\mathbb{R}$. Now let $x_{1},x_{2}\in D$ and $t\in\mathbb{R}$ with $0\le t\le1$; the point $x_{t}=t x_{1}+(1-t)x_{2}$ lies in $D$ because $D$ is convex. Choose an $\omega$-maximiser $z_{t}$ at $x_{t}$ (Step 2). By (4.1) and (1.2), $G(x_{t})=f_{x_{t}}(z_{t})+\tfrac{1}{2\tau}\lVert x_{t}\rVert^{2}=A_{z_{t}}(x_{t})$ and $A_{z_{t}}(x_{i})=f_{x_{i}}(z_{t})+\tfrac{1}{2\tau}\lVert x_{i}\rVert^{2}\le G(x_{i})$ for $i=1,2$. As $0\le t$ and $0\le1-t$,

$$G(x_{t})=t\,A_{z_{t}}(x_{1})+(1-t)\,A_{z_{t}}(x_{2})\le t\,G(x_{1})+(1-t)\,G(x_{2}).$$

Hence $w_{\omega}$ is semiconvex on $D$ with constant $\tau^{-1}$.

Step 5 (A local bound). Let $x,y\in D$ and let $z$ be an $\omega$-maximiser at $x$. By (1.2), $\omega(z)-\tfrac{1}{2\tau}\lVert y-z\rVert^{2}=f_{y}(z)\le w_{\omega}(y)$, and $w_{\omega}(x)=\omega(z)-\tfrac{1}{2\tau}\lVert x-z\rVert^{2}$, so

$$w_{\omega}(x)-w_{\omega}(y)\le\tfrac{1}{2\tau}\bigl(\lVert y-z\rVert^{2}-\lVert x-z\rVert^{2}\bigr).$$

By Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ §triangle and (0.3), $\lVert y-z\rVert=\lVert(y-x)+(x-z)\rVert\le\lVert x-y\rVert+\lVert x-z\rVert$, so by (0.2) $\lVert y-z\rVert^{2}\le\lVert x-y\rVert^{2}+2\lVert x-y\rVert\lVert x-z\rVert+\lVert x-z\rVert^{2}$ and

$$w_{\omega}(x)-w_{\omega}(y)\le\tfrac{1}{2\tau}\bigl(\lVert x-y\rVert^{2}+2\lVert x-y\rVert\lVert x-z\rVert\bigr).\tag{5.1}$$

By (2.1), $\lVert x-z\rVert\le r$; inserting this into (5.1), using $0\le2\lVert x-y\rVert$, gives $w_{\omega}(x)-w_{\omega}(y)\le\tfrac{1}{2\tau}\lVert x-y\rVert\bigl(\lVert x-y\rVert+2r\bigr)$. Exchanging the roles of $x$ and $y$ (note $\lVert y-x\rVert=\lVert x-y\rVert$ by (0.3)) and combining the two inequalities,

$$|w_{\omega}(x)-w_{\omega}(y)|\le\tfrac{1}{2\tau}\lVert x-y\rVert\bigl(\lVert x-y\rVert+2r\bigr)\qquad(x,y\in D).\tag{5.2}$$

Step 6 (Lipschitz bound). Fix $x,y\in D$, put $\delta=\lVert x-y\rVert$, and let $k$ be a natural number, read in $\mathbb{R}$ as in clause 1 of the real-number setting, so that $k$ is positive. For each integer $j$ with $0\le j\le k$ put $x_{j}=x+\tfrac{j}{k}(y-x)\in D$; then $x_{0}=x$, $x_{k}=y$ and $x_{j+1}-x_{j}=\tfrac{1}{k}(y-x)$ for $0\le j<k$, so $\lVert x_{j+1}-x_{j}\rVert=\tfrac{\delta}{k}$ by (0.3). By (5.2), $|w_{\omega}(x_{j+1})-w_{\omega}(x_{j})|\le\tfrac{1}{2\tau}\tfrac{\delta}{k}\bigl(\tfrac{\delta}{k}+2r\bigr)$ for $0\le j<k$, and induction on $j$, using the triangle inequality for the absolute value, gives $|w_{\omega}(x_{j})-w_{\omega}(x)|\le j\cdot\tfrac{1}{2\tau}\tfrac{\delta}{k}\bigl(\tfrac{\delta}{k}+2r\bigr)$ for $0\le j\le k$. For $j=k$ this reads

$$|w_{\omega}(y)-w_{\omega}(x)|\le\tfrac{1}{2\tau}\delta\bigl(\tfrac{\delta}{k}+2r\bigr)=\tau^{-1}r\,\delta+\tfrac{\delta^{2}}{2\tau}\cdot\tfrac{1}{k}.$$

This holds for every natural number $k$. Given a positive $\sigma\in\mathbb{R}$, the Archimedean property provides a natural number $k$ with $\tfrac{1}{k}<\sigma$, so $|w_{\omega}(y)-w_{\omega}(x)|-\tau^{-1}r\,\delta\le\tfrac{\delta^{2}}{2\tau}\sigma$. As $0\le\tfrac{\delta^{2}}{2\tau}$, (0.4) gives

$$|w_{\omega}(x)-w_{\omega}(y)|\le\tau^{-1}r\,\lVert x-y\rVert\qquad(x,y\in D).\tag{6.1}$$

Step 7 (Transfer of test functions).

(7a) Let $\varphi:D\to\mathbb{R}$ be of class $C^{2}$ on $D$, let $x\in D$, suppose that $w_{\omega}-\varphi$ has a local maximum at $x$ relative to $D$, let $z$ be an $\omega$-maximiser at $x$, and put $p=D\varphi(x)$ and $X=D^{2}\varphi(x)\in\mathcal{S}(n)$. Then $p=\tau^{-1}(z-x)$.

Indeed, let $\beta_{1}$ be positive such that every $x'\in D$ with $\lVert x-x'\rVert<\beta_{1}$ satisfies $w_{\omega}(x')-\varphi(x')\le w_{\omega}(x)-\varphi(x)$. Let $\psi:D\to\mathbb{R}$, $\psi(x')=-\tfrac{1}{2\tau}\lVert x'-z\rVert^{2}=c\,d_{E}(x',z)^{2}$ with $c=-\tfrac{1}{2\tau}$; it is of class $C^{2}$ on $D$ with $D\psi(x)=(2c)(x-z)=\tau^{-1}(z-x)$, by claims 2 and 3 of A Scaled Squared Distance to a Point is of Class $C^2$, with Gradient and Hessian (with $U=D$ and $a=z$). For $x'\in D$ with $\lVert x-x'\rVert<\beta_{1}$, (1.2) gives

$$\omega(z)+\psi(x')=f_{x'}(z)\le w_{\omega}(x')\le w_{\omega}(x)+\varphi(x')-\varphi(x)=\omega(z)+\psi(x)+\varphi(x')-\varphi(x),$$

that is $\varphi(x)-\psi(x)\le\varphi(x')-\psi(x')$. So $\varphi-\psi$, which is of class $C^{2}$ on $D$ by claim 3 of Constants, Coordinate Functions, Sums and Products of $C^k$ Functions on a Euclidean Open Set, has a local minimum at $x$ relative to $D$, and its gradient at $x$ is the origin by claim 2 of First- and Second-Order Conditions at a Local Extremum of a Function of Class $C^2$. By claim 1 of Constants, Coordinate Functions, Sums and Products of $C^k$ Functions on a Euclidean Open Set that gradient is $D\varphi(x)-D\psi(x)$, so $p=D\psi(x)=\tau^{-1}(z-x)$.

(7b) In the situation of (7a), for every positive $\sigma\in\mathbb{R}$ there is a positive $\beta\in\mathbb{R}$ such that every $y\in D$ with $\lVert y-z\rVert<\beta$ satisfies

$$\omega(y)\le \omega(z)+p\cdot(y-z)+\tfrac{1}{2}\,(y-z)\cdot\bigl((X+\sigma I_{n})(y-z)\bigr).\tag{7.1}$$

Indeed, $X+\sigma I_{n}\in\mathcal{S}(n)$ by the clause on symmetric matrices. By Second-Order Taylor Expansion with Peano Remainder (with $U=D$, $f=\varphi$ and the tolerance $\tfrac{\sigma}{2}$) there is a positive $\beta_{2}$ such that every $h\in\mathbb{R}^{n}$ with $\lVert h\rVert<\beta_{2}$ satisfies $x+h\in D$ and

$$\Bigl|\varphi(x+h)-\varphi(x)-\sum_{i=1}^{n}\partial_{i}\varphi(x)\,h_{i}-\tfrac{1}{2}\sum_{i=1}^{n}\sum_{j=1}^{n}\partial_{j}\partial_{i}\varphi(x)\,h_{i}h_{j}\Bigr|\le\tfrac{\sigma}{2}\lVert h\rVert^{2},$$

the distance from $h$ to the origin used there being $\lVert h\rVert$ by Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ §distance. By Gradient of a Real-Valued Function on a Euclidean Open Set and the definition of the dot product, $\sum_{i}\partial_{i}\varphi(x)h_{i}=p\cdot h$. By Hessian Matrix of a C^2 Function, $\partial_{j}\partial_{i}\varphi(x)$ is the entry $X_{ji}$, which equals $X_{ij}$ as $X$ is symmetric, so claim 4 of Linearity of the Matrix-Vector Product and the Quadratic Form as a Double Sum gives $\sum_{i}\sum_{j}\partial_{j}\partial_{i}\varphi(x)h_{i}h_{j}=h\cdot(Xh)$. By claims 1 and 2 of Linearity of the Matrix-Vector Product and the Quadratic Form as a Double Sum, $(X+\sigma I_{n})h=Xh+\sigma h$, so $h\cdot((X+\sigma I_{n})h)=h\cdot(Xh)+\sigma\lVert h\rVert^{2}$ by Bilinearity and Symmetry of the Dot Product on $\mathbb{R}^n$ and (0.3). Consequently

$$\varphi(x+h)-\varphi(x)\le p\cdot h+\tfrac{1}{2}\,h\cdot\bigl((X+\sigma I_{n})h\bigr)\qquad\text{whenever }\lVert h\rVert<\beta_{2}.\tag{7.2}$$

Let $\beta_{1}$ be as in (7a) and let $\beta$ be the lesser of $\beta_{1}$ and $\beta_{2}$. Let $y\in D$ with $\lVert y-z\rVert<\beta$, and put $h=y-z$ and $x'=x+h$. Then $\lVert h\rVert<\beta_{2}$, so $x'\in D$ and (7.2) holds; and $\lVert x-x'\rVert=\lVert h\rVert<\beta_{1}$, so $w_{\omega}(x')\le w_{\omega}(x)+\varphi(x')-\varphi(x)$. Since $x'-y=x-z$, (1.2) gives $\omega(y)-\tfrac{1}{2\tau}\lVert x-z\rVert^{2}=f_{x'}(y)\le w_{\omega}(x')$. Together with $w_{\omega}(x)=f_{x}(z)=\omega(z)-\tfrac{1}{2\tau}\lVert x-z\rVert^{2}$ and (7.2) this yields

$$\omega(y)-\tfrac{1}{2\tau}\lVert x-z\rVert^{2}\le \omega(z)-\tfrac{1}{2\tau}\lVert x-z\rVert^{2}+p\cdot h+\tfrac{1}{2}\,h\cdot\bigl((X+\sigma I_{n})h\bigr),$$

which is (7.1).

(7c) Let $z\in D$, $p\in\mathbb{R}^{n}$ and $B\in\mathcal{S}(n)$, and let $\Psi:D\to\mathbb{R}$, $\Psi(y)=\tfrac{1}{2}(y-z)\cdot(B(y-z))+p\cdot(y-z)$. Then $\Psi$ is of class $C^{2}$ on $D$, and

$$\Psi(z)=0,\qquad D\Psi(z)=p,\qquad D^{2}\Psi(z)=B.$$

Indeed, by the clause on quadratic functions the function $\hat{Q}:\mathbb{R}^{n}\to\mathbb{R}$, $\hat{Q}(h)=\tfrac{1}{2}h\cdot(Bh)+p\cdot h+0$, is of class $C^{2}$ on $\mathbb{R}^{n}$ with $D\hat{Q}(h)=Bh+p$ and $D^{2}\hat{Q}(h)=B$; by the translation clause (with $V=\mathbb{R}^{n}$ and $b=-z$, so that $V-b=\mathbb{R}^{n}$), $\Psi(y)=\hat{Q}(y-z)$ is of class $C^{2}$ on $\mathbb{R}^{n}=D$ with $D\Psi(y)=B(y-z)+p$ and $D^{2}\Psi(y)=B$. In particular $D\Psi(z)=B(z-z)+p=p$ (as $B$ applied to the origin is the origin, by claim 3 of Linearity of the Matrix-Vector Product and the Quadratic Form as a Double Sum with the scalar $0$) and $D^{2}\Psi(z)=B$; finally $\Psi(z)=0$ because every dot product with the origin vanishes.

Step 8 (Part A, clauses A1 to A4). By Viscosity Subsolution and Supersolution of a Second-Order Equation, $u$ is upper semicontinuous on $D$, and $|u(y)|\le M$ for $y\in D$ by hypothesis; so Steps 1 to 7 apply with $\omega=u$, for which $w_{u}=\overline{w}$ and the $u$-maximisers at $x$ are the maximisers at $x$ of Part A. Clause A1 is Step 2 with (2.1); clause A2 is (3.1); clause A3 is Step 4; clause A4 is (6.1), since the distance of The Absolute Value Metric on the Real Line between $\overline{w}(x)$ and $\overline{w}(y)$ is $|\overline{w}(x)-\overline{w}(y)|$ and $d_{E}(x,y)=\lVert x-y\rVert$, so that (6.1) is the inequality of Lipschitz Map Between Metric Spaces with constant $\tau^{-1}r$, which is nonnegative.

Step 9 (Part A, clause A5). Semicontinuity. Let $x\in D$ and let $e$ be positive, and put $\delta=\tfrac{\tau e}{r+1}$, which is positive. Every $y\in D$ with $\lVert x-y\rVert<\delta$ satisfies, by clause A4, $\overline{w}(y)-\overline{w}(x)\le\tau^{-1}r\lVert x-y\rVert\le\tau^{-1}r\delta=\tfrac{re}{r+1}<e$, hence $\overline{w}(y)<\overline{w}(x)+e$. So $\overline{w}$ is upper semicontinuous on $D$.

The subsolution inequality. Let $\varphi:D\to\mathbb{R}$ be of class $C^{2}$ on $D$ and let $x\in D$ be such that $\overline{w}-\varphi$ has a local maximum at $x$ relative to $D$. Choose, in this order: a maximiser $z$ at $x$ (clause A1); put $p=D\varphi(x)$, $X=D^{2}\varphi(x)$ and $\xi=DP(z)$; fix a positive $\sigma\in\mathbb{R}$; and let $\beta$ be given by (7b) for this $\sigma$ (with $\omega=u$). By (7a), $p=\tau^{-1}(z-x)$. Let $\Psi$ be the function of (7c) with $B=X+\sigma I_{n}$. For $y\in D$ with $\lVert y-z\rVert<\beta$, (7.1) reads $u(y)-\Psi(y)\le u(z)=u(z)-\Psi(z)$; thus $u-\Psi$ has a local maximum at $z$ relative to $D$. Since $u$ is a viscosity subsolution of $F$ on $D$ and $\Psi$ is of class $C^{2}$ on $D$ with $D\Psi(z)=p$ and $D^{2}\Psi(z)=X+\sigma I_{n}$, Viscosity Subsolution and Supersolution of a Second-Order Equation and the formula for $F$ give

$$\lambda u(z)+\tfrac{\theta}{2}\lVert p\rVert^{2}+\xi\cdot p-\tfrac{\kappa}{2}\operatorname{tr}\bigl(X+\sigma I_{n}\bigr)-g(z)\le0 .$$

By claim 1 of Basic Properties of the Trace, $\operatorname{tr}(X+\sigma I_{n})=\operatorname{tr}X+\sigma\operatorname{tr}I_{n}$, and $\operatorname{tr}I_{n}=n$ (the sum of $n$ ones) by Trace of a Real Square Matrix and Identity Matrix. Hence

$$\lambda u(z)+\tfrac{\theta}{2}\lVert p\rVert^{2}+\xi\cdot p-\tfrac{\kappa}{2}\operatorname{tr}X-\tfrac{\kappa}{2}n\sigma\le g(z).\tag{9.1}$$

Now we compare with the point $x$. (i) Since $z$ is a maximiser at $x$, $\overline{w}(x)=u(z)-\tfrac{1}{2\tau}\lVert x-z\rVert^{2}\le u(z)$, so $\lambda\overline{w}(x)\le\lambda u(z)$. (ii) The monotonicity hypothesis with the points $x$ and $z$ gives $0\le(DP(x)-\xi)\cdot(x-z)$; as $p=-\tau^{-1}(x-z)$, Bilinearity and Symmetry of the Dot Product on $\mathbb{R}^n$ gives $\xi\cdot p-DP(x)\cdot p=\tau^{-1}\bigl((DP(x)-\xi)\cdot(x-z)\bigr)\ge0$, so $DP(x)\cdot p\le\xi\cdot p$. (iii) By the hypothesis on $g$, $g(z)-g(x)\le|g(x)-g(z)|\le\rho(\lVert x-z\rVert)\le\rho(r)$, the last step because $\lVert x-z\rVert\le r$ by clause A1 and $\rho$ is nondecreasing; so $g(z)\le g(x)+\rho(r)$. Combining (i), (ii), (iii) with (9.1), and using the formula for $F_{+}$, whose running cost is $g+\rho(r)$,

$$F_{+}\bigl(x,\overline{w}(x),D\varphi(x),D^{2}\varphi(x)\bigr)=\lambda\overline{w}(x)+\tfrac{\theta}{2}\lVert p\rVert^{2}+DP(x)\cdot p-\tfrac{\kappa}{2}\operatorname{tr}X-\bigl(g(x)+\rho(r)\bigr)\le\tfrac{\kappa}{2}n\,\sigma .$$

The left-hand side does not depend on $\sigma$, and $0\le\tfrac{\kappa}{2}n$; as $\sigma$ was an arbitrary positive number, (0.4) gives $F_{+}(x,\overline{w}(x),D\varphi(x),D^{2}\varphi(x))\le0$. Since $\varphi$ and $x$ were arbitrary and $\overline{w}$ is upper semicontinuous, $\overline{w}$ is a viscosity subsolution of $F_{+}$ on $D$ by Viscosity Subsolution and Supersolution of a Second-Order Equation; this is clause A5.

Step 10 (Part B, clauses B1 to B4). By Viscosity Subsolution and Supersolution of a Second-Order Equation, $v$ is lower semicontinuous on $D$, so $-v$ is upper semicontinuous on $D$ by claim 1 of Semicontinuity Under Negation and Characterization of Continuity; and $|-v(y)|=|v(y)|\le M$ by hypothesis. Hence Steps 1 to 7 apply with $\omega=-v$; then $f_{x}(y)=-\bigl(v(y)+\tfrac{1}{2\tau}\lVert x-y\rVert^{2}\bigr)$. For $x\in D$, the set of additive inverses of the elements of the set displayed in Part B is therefore $\{f_{x}(y):y\in D\}$, whose least upper bound is $w_{-v}(x)$; by the definition of $\underline{w}$ in the statement,

$$\underline{w}(x)=-w_{-v}(x)\qquad(x\in D),\tag{10.1}$$

and (by reversing inequalities under negation) this number is indeed the greatest lower bound of the set displayed in Part B. Moreover $y\in D$ is a minimiser at $x$ if and only if it is a $(-v)$-maximiser at $x$. Clause B1 is therefore Step 2 with (2.1). Clause B2 follows from (3.1), which reads $-v(x)\le-\underline{w}(x)\le M$, by negation. Clause B3 is Step 4, since $-\underline{w}=w_{-v}$. Clause B4 follows from (6.1) with $\omega=-v$, since $|\underline{w}(x)-\underline{w}(y)|=|w_{-v}(x)-w_{-v}(y)|$, exactly as for clause A4 in Step 8.

Step 11 (Part B, clause B5). Semicontinuity. Let $x\in D$ and let $e$ be positive; with $\delta=\tfrac{\tau e}{r+1}$ as in Step 9, clause B4 shows that every $y\in D$ with $\lVert x-y\rVert<\delta$ satisfies $\underline{w}(x)-\underline{w}(y)\le\tau^{-1}r\lVert x-y\rVert\le\tau^{-1}r\delta=\tfrac{re}{r+1}<e$, hence $\underline{w}(x)-e<\underline{w}(y)$. So $\underline{w}$ is lower semicontinuous on $D$.

The supersolution inequality. Let $\varphi:D\to\mathbb{R}$ be of class $C^{2}$ on $D$ and let $x\in D$ be such that $\underline{w}-\varphi$ has a local minimum at $x$ relative to $D$. By (10.1), $w_{-v}-(-\varphi)=-(\underline{w}-\varphi)$ has a local maximum at $x$ relative to $D$ (with the same radius), and $-\varphi=(-1)\varphi$ is of class $C^{2}$ on $D$ with $D(-\varphi)(x)=-D\varphi(x)$ and $D^{2}(-\varphi)(x)=-D^{2}\varphi(x)$, by claims 1 and 3 of Constants, Coordinate Functions, Sums and Products of $C^k$ Functions on a Euclidean Open Set (claim 1 applied first to $\varphi$ and then to the functions $\partial_{i}\varphi$, which exist because $\varphi$ is of class $C^{2}$, gives $\partial_{j}\partial_{i}(-\varphi)(x)=-\partial_{j}\partial_{i}\varphi(x)$; the entries of the gradient and Hessian are then those of $-D\varphi(x)$ and $-D^{2}\varphi(x)$ by Gradient of a Real-Valued Function on a Euclidean Open Set and Hessian Matrix of a C^2 Function). Choose, in this order: a minimiser $z$ at $x$ (clause B1), which is a $(-v)$-maximiser at $x$; put $p=D\varphi(x)$, $X=D^{2}\varphi(x)$ and $\xi=DP(z)$; fix a positive $\sigma\in\mathbb{R}$; and let $\beta$ be given by (7b) for $\omega=-v$, the test function $-\varphi$ (whose gradient and Hessian at $x$ are $-p$ and $-X$) and this $\sigma$. By (7a), $-p=\tau^{-1}(z-x)$, that is $p=\tau^{-1}(x-z)$. For $y\in D$ with $\lVert y-z\rVert<\beta$, (7.1) reads

$$-v(y)\le-v(z)+(-p)\cdot(y-z)+\tfrac{1}{2}(y-z)\cdot\bigl((-X+\sigma I_{n})(y-z)\bigr).$$

By claim 1 of Linearity of the Matrix-Vector Product and the Quadratic Form as a Double Sum and Bilinearity and Symmetry of the Dot Product on $\mathbb{R}^n$, $(-p)\cdot(y-z)=-\,p\cdot(y-z)$ and $(y-z)\cdot((-X+\sigma I_{n})(y-z))=-(y-z)\cdot((X-\sigma I_{n})(y-z))$, where $X-\sigma I_{n}\in\mathcal{S}(n)$ by the clause on symmetric matrices; negating,

$$v(y)\ge v(z)+p\cdot(y-z)+\tfrac{1}{2}(y-z)\cdot\bigl((X-\sigma I_{n})(y-z)\bigr).$$

Let $\Psi$ be the function of (7c) with $B=X-\sigma I_{n}$. The last inequality says $v(y)-\Psi(y)\ge v(z)=v(z)-\Psi(z)$ for $y\in D$ with $\lVert y-z\rVert<\beta$, so $v-\Psi$ has a local minimum at $z$ relative to $D$. Since $v$ is a viscosity supersolution of $F$ on $D$ and $D\Psi(z)=p$, $D^{2}\Psi(z)=X-\sigma I_{n}$, we obtain, with claim 1 of Basic Properties of the Trace and $\operatorname{tr}I_{n}=n$ as in Step 9,

$$g(z)\le\lambda v(z)+\tfrac{\theta}{2}\lVert p\rVert^{2}+\xi\cdot p-\tfrac{\kappa}{2}\operatorname{tr}X+\tfrac{\kappa}{2}n\sigma .\tag{11.1}$$

Now we compare with the point $x$. (i) Since $z$ is a minimiser at $x$, $\underline{w}(x)=v(z)+\tfrac{1}{2\tau}\lVert x-z\rVert^{2}\ge v(z)$, so $\lambda v(z)\le\lambda\underline{w}(x)$. (ii) As $p=\tau^{-1}(x-z)$, the monotonicity hypothesis and Bilinearity and Symmetry of the Dot Product on $\mathbb{R}^n$ give $DP(x)\cdot p-\xi\cdot p=\tau^{-1}\bigl((DP(x)-\xi)\cdot(x-z)\bigr)\ge0$, so $\xi\cdot p\le DP(x)\cdot p$. (iii) $g(x)-g(z)\le|g(x)-g(z)|\le\rho(\lVert x-z\rVert)\le\rho(r)$ by clause B1 and the monotonicity of $\rho$, so $g(x)-\rho(r)\le g(z)$. Combining (i), (ii), (iii) with (11.1), and using the formula for $F_{-}$, whose running cost is $g-\rho(r)$,

$$-F_{-}\bigl(x,\underline{w}(x),D\varphi(x),D^{2}\varphi(x)\bigr)=\bigl(g(x)-\rho(r)\bigr)-\lambda\underline{w}(x)-\tfrac{\theta}{2}\lVert p\rVert^{2}-DP(x)\cdot p+\tfrac{\kappa}{2}\operatorname{tr}X\le\tfrac{\kappa}{2}n\,\sigma .$$

As $\sigma$ was an arbitrary positive number, (0.4) gives $0\le F_{-}(x,\underline{w}(x),D\varphi(x),D^{2}\varphi(x))$. Since $\varphi$ and $x$ were arbitrary and $\underline{w}$ is lower semicontinuous, $\underline{w}$ is a viscosity supersolution of $F_{-}$ on $D$ by Viscosity Subsolution and Supersolution of a Second-Order Equation; this is clause B5.