Proof of Sup- and Inf-Convolutions of Bounded Viscosity Sub- and Supersolutions of the Penalty-Drift Equation with a Convex Penalty

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.

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, and the rules for the absolute value) are justified by Elementary Order Arithmetic in an Ordered Field, Elementary Arithmetic in an Ordered Field and Properties of the Absolute Value in an Ordered Field and are not cited individually; the same applies to the vector-space identities in $\mathbb{R}^{n}$, such as $(x+h)-y=(x-y)+h$. 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 $p_{0}$, $r$, $\theta_{1}$, $\theta_{2}$, $g_{1}$, $g_{2}$, $F$, $F_{1}$, $F_{2}$ of the statement; in particular $p_{0}\le P(y)$ for every $y\in D$.

Step 0 (Preliminaries).

(0.1) Continuity. As recorded in the statement, $P$ is continuous at every point of $D$ and hence, by claim 2 of Semicontinuity Under Negation and Characterization of Continuity, both upper and lower semicontinuous on $D$. For a fixed $a\in\mathbb{R}^{n}$, the function $D\to\mathbb{R}$ with value $\tfrac{1}{2\tau}\lVert y-a\rVert^{2}=\tfrac{1}{2\tau}d_{E}(y,a)^{2}$ at $y$ 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 $c=\tfrac{1}{2\tau}$), so the function $e_{a}:D\to\mathbb{R}$, $e_{a}(y)=\eta P(y)+\tfrac{1}{2\tau}\lVert y-a\rVert^{2}$, 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; as for $P$, 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 $s'^{2}\le s's<ss$ (the strict step by claim 10 of Elementary Order Arithmetic in an Ordered Field), 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 claim 1 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$, hence, by Bilinearity and Symmetry of the Dot Product on $\mathbb{R}^n$ (symmetry, additivity and homogeneity in both arguments),

Moreover $\lVert\mu a\rVert=|\mu|\,\lVert a\rVert$ for $\mu\in\mathbb{R}$ by claim 5 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$; in particular $\lVert a-b\rVert=\lVert b-a\rVert$, and $\lVert a-a\rVert=0$ by claim 3 there.

(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 $|\omega(y)|\le M$, that is $-M\le\omega(y)\le M$, for every $y\in D$. Put $W(y)=\omega(y)-\eta P(y)$ for $y\in D$, and for $x,y\in D$ put

Since $p_{0}\le P(y)$ and $\omega(y)\le M$,

and $f_{x}(y)\le W(y)$ 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 $M-\eta p_{0}$, 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),

Step 2 (Existence and location of $\omega$-maximisers). Fix $x\in D$ and put $t=P(x)+\tfrac{2M}{\eta}$ and $L=\{y\in D:P(y)\le t\}$, which is compact by the sublevel property of the penalty $P$. Let $K=\{y\in D:f_{x}(x)\le f_{x}(y)\}$. If $y\in K$, then by (1.2) $W(x)=f_{x}(x)\le f_{x}(y)\le W(y)$, that is $\omega(x)-\eta P(x)\le\omega(y)-\eta P(y)$, so $\eta P(y)\le\omega(y)-\omega(x)+\eta P(x)\le2M+\eta P(x)$ and, dividing by $\eta>0$, $P(y)\le t$. 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), $W(x)=f_{x}(x)\le w_{\omega}(x)=f_{x}(z)=W(z)-\tfrac{1}{2\tau}\lVert x-z\rVert^{2}$, so by (1.1)

Multiplying by $2\tau>0$ gives $\lVert x-z\rVert^{2}\le r(x)^{2}$, and (0.2) gives

Step 3 (Bounds). For $x\in D$ let $z$ be an $\omega$-maximiser at $x$ (Step 2). By (1.2) and (1.1), $W(x)=f_{x}(x)\le w_{\omega}(x)=f_{x}(z)\le W(z)\le M-\eta p_{0}$, that is

Step 4 (Semiconvexity and continuity). 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 $A_{y}:\mathbb{R}^{n}\to\mathbb{R}$, $A_{y}(x)=W(y)-\tfrac{1}{2\tau}\lVert y\rVert^{2}+\tau^{-1}(x\cdot y)$. By (0.3), $\lVert x-y\rVert^{2}=\lVert x\rVert^{2}-2(x\cdot y)+\lVert y\rVert^{2}$, whence

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 $0\le1-t$,

Hence $w_{\omega}$ is semiconvex on $D$ with constant $\tau^{-1}$. Since $D$ is convex and open, the continuity clause for semiconvex functions (with $U=S=D$) shows: for every $x\in D$ and every positive $e\in\mathbb{R}$ there is a positive $\delta\in\mathbb{R}$ such that every $y\in D$ with $\lVert y-x\rVert<\delta$ satisfies $|w_{\omega}(y)-w_{\omega}(x)|<e$. We refer to this continuity property as (4.2).

Step 5 (Gradient). Let $x\in D$ be such that the partial derivative of $w_{\omega}$ with respect to every variable exists at $x$, and let $z$ be an $\omega$-maximiser at $x$. Let $\psi:D\to\mathbb{R}$, $\psi(x')=-\tfrac{1}{2\tau}\lVert x'-z\rVert^{2}=c\,d_{E}(x',z)^{2}$ with $c=-\tfrac{1}{2\tau}$. By claim 1 of A Scaled Squared Distance to a Point is of Class $C^2$, with Gradient and Hessian (with $U=D$, $a=z$), for each $i$ the partial derivative of $\psi$ with respect to the $i$th variable exists at $x$ and equals $2c(x_{i}-z_{i})=\tau^{-1}(z_{i}-x_{i})$. Let $\chi=w_{\omega}-\psi$ on $D$; by claim 1 of Constants, Coordinate Functions, Sums and Products of $C^k$ Functions on a Euclidean Open Set (applied to the multiple $(-1)\psi$ and then to the sum), $\partial_{i}\chi(x)$ exists and $\partial_{i}\chi(x)=\partial_{i}w_{\omega}(x)-\tau^{-1}(z_{i}-x_{i})$. For every $x'\in D$, (1.2) gives $W(z)+\psi(x')=f_{x'}(z)\le w_{\omega}(x')$, that is $\chi(x')\ge W(z)$, while $\chi(x)=w_{\omega}(x)-\psi(x)=f_{x}(z)-\psi(x)=W(z)$ because $z$ is an $\omega$-maximiser at $x$. Hence $\chi(x')-\chi(x)\ge0$ for every $x'\in D$.

We claim $\ell:=\partial_{i}\chi(x)=0$. For real $s$ write $x^{(s)}=(x_{1},\dots,x_{i-1},x_{i}+s,x_{i+1},\dots,x_{n})$. Suppose $0<\ell$. By Partial Derivative on a Euclidean Open Set, applied with the tolerance $\ell$, there is a positive $\delta$ such that every real $s$ with $0<|s|<\delta$ satisfies $x^{(s)}\in D$ and $\bigl|\tfrac{\chi(x^{(s)})-\chi(x)}{s}-\ell\bigr|<\ell$; taking $s=-\tfrac{\delta}{2}$ gives $0<\tfrac{\chi(x^{(s)})-\chi(x)}{s}$, whereas the numerator is nonnegative and $s<0$, so this quotient is at most $0$: a contradiction. Suppose $\ell<0$. With the tolerance $-\ell$ we obtain $\delta$ as before, and $s=\tfrac{\delta}{2}$ gives $\tfrac{\chi(x^{(s)})-\chi(x)}{s}<0$, whereas the quotient is nonnegative: a contradiction. As the order is total, $\ell=0$. Therefore $\partial_{i}w_{\omega}(x)=\tau^{-1}(z_{i}-x_{i})$ for every $i$, and by Gradient of a Real-Valued Function on a Euclidean Open Set together with the coordinatewise definitions of the difference and the scalar multiple,

the equality by (0.3) and the inequality by (2.1).

Step 6 (Transfer between points). Let $x,y\in D$ and let $z$ be an $\omega$-maximiser at $x$. By (1.2), $W(z)-\tfrac{1}{2\tau}\lVert y-z\rVert^{2}=f_{y}(z)\le w_{\omega}(y)$, and $w_{\omega}(x)=W(z)-\tfrac{1}{2\tau}\lVert x-z\rVert^{2}$, so

By claim 6 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ 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

Next, by (1.2), $W(y)-\tfrac{1}{2\tau}\lVert x-y\rVert^{2}=f_{x}(y)\le w_{\omega}(x)=W(z)-\tfrac{1}{2\tau}\lVert x-z\rVert^{2}$, so by (1.1)

which is at most $\bigl(r(y)+\lVert x-y\rVert\bigr)^{2}$ because $0\le2r(y)\lVert x-y\rVert$. By (0.2), $\lVert x-z\rVert\le r(y)+\lVert x-y\rVert$. Inserting this into (6.1), using $0\le2\lVert x-y\rVert$,

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 $\psi$ be as in Step 5 (with this $z$); 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. For $x'\in D$ with $\lVert x-x'\rVert<\beta_{1}$, (1.2) gives

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

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

the distance from $h$ to the origin used there being $\lVert h\rVert$ by claim 2 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$. 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

Let $\beta_{1}$ be as in (7a) and let $\beta$ be the lesser of $\beta_{1}$ and $\beta_{2}$ (claim 9 of Elementary Order Arithmetic in an Ordered Field). 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 $W(y)-\tfrac{1}{2\tau}\lVert x-z\rVert^{2}=f_{x'}(y)\le w_{\omega}(x')$. Together with $w_{\omega}(x)=f_{x}(z)=W(z)-\tfrac{1}{2\tau}\lVert x-z\rVert^{2}$ and (7.2) this yields

which is (7.1).

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

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 $V-b=\mathbb{R}^{n}$), $Q(y)=\hat{Q}(y-z)$ is of class $C^{2}$ on $\mathbb{R}^{n}$ with $DQ(y)=B(y-z)+p$ and $D^{2}Q(y)=B$; and by clause 2 of the calculus setting its restriction to $D$ is of class $C^{2}$ on $D$ with the same gradient and Hessian. As $P$ is of class $C^{2}$ on $D$, claim 3 of Constants, Coordinate Functions, Sums and Products of $C^k$ Functions on a Euclidean Open Set shows that $\Psi$ is of class $C^{2}$ on $D$. Claim 1 there, applied first to the partial derivatives of $Q$ and $\mu P$ at every point of $D$ and then to the partial derivatives of the functions $\partial_{j}Q$ and $\mu\,\partial_{j}P$ (which exist because $Q$ and $P$ are of class $C^{2}$), gives $\partial_{i}\Psi(z)=\partial_{i}Q(z)+\mu\,\partial_{i}P(z)$ and $\partial_{i}\partial_{j}\Psi(z)=\partial_{i}\partial_{j}Q(z)+\mu\,\partial_{i}\partial_{j}P(z)$ for all $i,j$. By Gradient of a Real-Valued Function on a Euclidean Open Set, Hessian Matrix of a C^2 Function and the entrywise definitions of sums and scalar multiples, $D\Psi(z)=B(z-z)+p+\mu DP(z)=p+\mu DP(z)$ (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 $D^{2}\Psi(z)=B+\mu D^{2}P(z)$; finally $\Psi(z)=\mu P(z)$ because every dot product with the origin vanishes.

Step 8 (Part A, clauses A1 to A5). By Viscosity Subsolution and Supersolution of a Second-Order Equation, $u$ is upper semicontinuous on $D$, and $|u(y)|\le M$ for $y\in D$; 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 (5.1), the parenthetical remark in A4 being claim 1 of A Derivative Matrix is the Jacobian Matrix, and is Unique applied to the single coordinate function $\overline{w}$; clause A5 is (6.2).

Step 9 (Part A, clause A6). Semicontinuity. Let $x\in D$ and let $e$ be positive; by (4.2) with $\omega=u$ there is a positive $\delta$ such that every $y\in D$ with $\lVert x-y\rVert<\delta$ satisfies $|\overline{w}(y)-\overline{w}(x)|<e$, 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 $p=D\varphi(x)$, $X=D^{2}\varphi(x)$, $\xi=DP(z)$ and $H=D^{2}P(z)$; fix a positive $\sigma\in\mathbb{R}$; and let $\beta$ be given by (7b) for this $\sigma$ (with $\omega=u$, so $W=u-\eta P$). By (7a), $p=\tau^{-1}(z-x)$. Let $\Psi$ be the function of (7c) with $B=X+\sigma I_{n}$ and $\mu=\eta$. For $y\in D$ with $\lVert y-z\rVert<\beta$, (7.1) reads $u(y)-\Psi(y)\le u(z)-\eta P(z)=u(z)-\Psi(z)$; 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 $D\Psi(z)=p+\eta\xi$ and $D^{2}\Psi(z)=X+\sigma I_{n}+\eta H$, Viscosity Subsolution and Supersolution of a Second-Order Equation and the formula for $F$ give

By claim 1 of Basic Properties of the Trace, $\operatorname{tr}(X+\sigma I_{n}+\eta H)=\operatorname{tr}X+\sigma\operatorname{tr}I_{n}+\eta\operatorname{tr}H$, and $\operatorname{tr}I_{n}=n$ (the sum of $n$ ones) by Trace of a Real Square Matrix and Identity Matrix. By (0.3) and Bilinearity and Symmetry of the Dot Product on $\mathbb{R}^n$, $\lVert p+\eta\xi\rVert^{2}=\lVert p\rVert^{2}+2\eta(p\cdot\xi)+\eta^{2}\lVert\xi\rVert^{2}$ and $\xi\cdot(p+\eta\xi)=\xi\cdot p+\eta\lVert\xi\rVert^{2}$. Hence

We bound three of these terms from below. First, the dissipation inequality at $z$, multiplied by $\eta>0$, gives $\tfrac{\kappa\eta}{2}\operatorname{tr}H\le\eta(1-\varepsilon)\lVert\xi\rVert^{2}+\eta\lambda P(z)+\eta C$. Second, $0\le\tfrac{\theta\eta^{2}}{2}\lVert\xi\rVert^{2}$. Third, the weighted Young inequality, applied to the points $-\theta p$ and $\xi$ with the positive number $t=(2\varepsilon)^{-1}$ (its remaining data $B$ and $\varepsilon$ play no role in that clause and may be taken to be $I_{n}$ and $1$), gives $2\bigl((-\theta p)\cdot\xi\bigr)\le\tfrac{1}{2\varepsilon}\lVert-\theta p\rVert^{2}+2\varepsilon\lVert\xi\rVert^{2}=\tfrac{\theta^{2}}{2\varepsilon}\lVert p\rVert^{2}+2\varepsilon\lVert\xi\rVert^{2}$ (using $\lVert-\theta p\rVert=\theta\lVert p\rVert$ from (0.3)); since $(-\theta p)\cdot\xi=-\theta\,(p\cdot\xi)$ by claim 4 of Bilinearity and Symmetry of the Dot Product on $\mathbb{R}^n$, halving and multiplying by $\eta$ gives $\theta\eta\,(p\cdot\xi)\ge-\tfrac{\theta^{2}\eta}{4\varepsilon}\lVert p\rVert^{2}-\eta\varepsilon\lVert\xi\rVert^{2}$. Inserting the three bounds into (9.1), and using $\eta\lVert\xi\rVert^{2}-\eta(1-\varepsilon)\lVert\xi\rVert^{2}-\eta\varepsilon\lVert\xi\rVert^{2}=0$ and $\tfrac{\theta}{2}-\tfrac{\theta^{2}\eta}{4\varepsilon}=\tfrac{\theta_{1}}{2}$,

Now we compare with the point $x$. (i) Since $z$ is a maximiser at $x$, $\overline{w}(x)=u(z)-\eta P(z)-\tfrac{1}{2\tau}\lVert x-z\rVert^{2}\le u(z)-\eta P(z)$, so $\lambda\overline{w}(x)\le\lambda(u(z)-\eta P(z))$. (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 $g$, $g(z)-g(x)\le|g(x)-g(z)|\le\rho(\lVert x-z\rVert)\le\rho(r(x))$, the last step because $\lVert x-z\rVert\le r(x)$ by clause A1 and $\rho$ is nondecreasing; so $g(z)+\eta C\le g_{1}(x)$. Combining (i), (ii), (iii) with (9.2),