Proof of The Dissipative Hamilton-Jacobi Operator on a Hilbert Triple Satisfies the Comparison Hypotheses

Each result cited is universally quantified over the data in its own statement. Throughout, $|a+b|_{H}^{2}=|a|_{H}^{2}+2\langle a,b\rangle_{H}+|b|_{H}^{2}$ for $a,b\in H$ by Elementary Identities in a Real Inner Product Space (v), and $|\langle a,b\rangle_{H}|\le|a|_{H}|b|_{H}$ by The Cauchy–Schwarz Inequality in a Real Inner Product Space. We record one identity: for $x,y\in D(A)$,

$$\langle Ax-Ay,\,x-y\rangle_{H}=|x-y|_{V}^{2}, \tag{1}$$

since $x-y\in V$, so that $\langle Ax,x-y\rangle_{H}=\langle x,x-y\rangle_{V}$ and $\langle Ay,x-y\rangle_{H}=\langle y,x-y\rangle_{V}$ by Hilbert Triples: Standing Notation and Background (iii), and subtracting gives $\langle x-y,x-y\rangle_{V}$.

Claim 1. Each of $\lambda_{0}r$, $\tfrac12|p|_{H}^{2}$, $\langle Ax,p\rangle_{H}$ and $g(x)$ is a real number for $(x,r,p,X)\in D(A)\times\mathbb{R}\times H\times\mathrm{Sym}(V)$, so $F$ is a function from $D(A)\times\mathbb{R}\times H\times\mathrm{Sym}(V)$ to $\mathbb{R}$. Since $W=D(A)\cap H=D(A)$, this is exactly the shape required by Second-Order Equation Operator on an Open Subset of a Hilbert Triple and Its $\delta$-Shifts (i). The defining expression does not involve the fourth argument, so $F(x,r,p,X)=F(x,r,p,X')$ for all $X,X'\in\mathrm{Sym}(V)$, i.e. $F$ is first order; degenerate ellipticity follows from A First-Order Equation Operator is Degenerate Elliptic and Its $\delta$-Shifts Ignore the Form Argument (i).

Claim 2. Let $R$ be positive, let $x\in D(A)$, $p\in H$, $X\in\mathrm{Sym}(V)$ and let $r,s\in\mathbb{R}$ satisfy $-R\le s\le r\le R$. All terms of $F$ other than $\lambda_{0}r$ are unchanged when $r$ is replaced by $s$, so

$$F(x,r,p,X)-F(x,s,p,X)=\lambda_{0}(r-s),$$

and in particular $\lambda_{0}(r-s)\le F(x,r,p,X)-F(x,s,p,X)$. Thus $\lambda_{0}$ is a properness constant for $F$ at $R$, for every positive $R$, and $F$ is locally strictly proper.

Claim 3. For a real $\alpha>1$ the function $t\mapsto\omega_{2}(t,\alpha)=3\alpha^{2}t$ is the linear function with the nonnegative constant $3\alpha^{2}$, hence a modulus of continuity by Linear Moduli of Continuity (i). Two preliminary inequalities. First, for $x,y\in H$, since $|g(x)-g(y)|\le|g(x)|+|g(y)|\le2C_{g}$ (claim 5 of Properties of the Absolute Value in an Ordered Field) and $|g(x)-g(y)|\le\omega_{g}(|x-y|_{H})$, the last assertion of The Nondecreasing Envelope of a Truncated Modulus of Continuity (iv) gives $|g(x)-g(y)|\le\omega_{1}(|x-y|_{H})$. Secondly, for a real $\alpha>1$ and a nonnegative real $s$,

$$\alpha s^{2}+\tfrac{1}{\alpha}-s=\tfrac{1}{\alpha}\bigl(\alpha s-\tfrac12\bigr)^{2}+\tfrac{3}{4\alpha}\ \ge\ 0 ,$$

as one checks by expanding $\tfrac{1}{\alpha}(\alpha s-\tfrac12)^{2}=\alpha s^{2}-s+\tfrac{1}{4\alpha}$; so $s\le\alpha s^{2}+\tfrac{1}{\alpha}$, and by The Nondecreasing Envelope of a Truncated Modulus of Continuity (iii), $\omega_{1}(s)\le\omega_{1}\bigl(\alpha s^{2}+\tfrac{1}{\alpha}\bigr)$.

Let $R$ be positive, let $x,y\in D(A)$, $r\in\mathbb{R}$, $X,Y\in\mathrm{Sym}(H)$ and let $\alpha,\delta$ satisfy $1<\alpha$ and $0<\delta<1$; put $P=\alpha(x-y)$. By Second-Order Equation Operator on an Open Subset of a Hilbert Triple and Its $\delta$-Shifts (ii) and the expansion of the norm,

$$F^{-}_{\delta}(x,r,P,X)=\lambda_{0}r+\lambda_{0}\delta h(x)+\tfrac12|P|_{H}^{2}+\delta\langle P,Ax\rangle_{H}+\tfrac{\delta^{2}}{2}|Ax|_{H}^{2}+\langle Ax,P\rangle_{H}+\delta|Ax|_{H}^{2}-g(x),$$ $$F^{+}_{\delta}(y,r,P,Y)=\lambda_{0}r-\lambda_{0}\delta h(y)+\tfrac12|P|_{H}^{2}-\delta\langle P,Ay\rangle_{H}+\tfrac{\delta^{2}}{2}|Ay|_{H}^{2}+\langle Ay,P\rangle_{H}-\delta|Ay|_{H}^{2}-g(y).$$

Subtracting and using (1) in the form $\langle Ax,P\rangle_{H}-\langle Ay,P\rangle_{H}=\alpha|x-y|_{V}^{2}$,

$$F^{-}_{\delta}(x,r,P,X)-F^{+}_{\delta}(y,r,P,Y)=\lambda_{0}\delta\bigl(h(x)+h(y)\bigr)+\alpha|x-y|_{V}^{2}+T_{x}+T_{y}+g(y)-g(x),$$

where

$$T_{x}=\delta|Ax|_{H}^{2}+\tfrac{\delta^{2}}{2}|Ax|_{H}^{2}+\delta\langle P,Ax\rangle_{H},\qquad T_{y}=\delta|Ay|_{H}^{2}-\tfrac{\delta^{2}}{2}|Ay|_{H}^{2}+\delta\langle P,Ay\rangle_{H}.$$

Since $\tfrac{\delta^{2}}{2}|Ax|_{H}^{2}\ge0$ and $\delta\langle P,Ax\rangle_{H}\ge-\delta|P|_{H}|Ax|_{H}$, while $\delta\bigl(|Ax|_{H}-\tfrac{|P|_{H}}{2}\bigr)^{2}\ge0$ gives $\delta|Ax|_{H}^{2}-\delta|P|_{H}|Ax|_{H}\ge-\tfrac{\delta}{4}|P|_{H}^{2}$, we get $T_{x}\ge-\tfrac{\delta}{4}|P|_{H}^{2}$. Since $\delta\le1$ we have $\delta-\tfrac{\delta^{2}}{2}\ge\tfrac{\delta}{2}$, and $\tfrac{\delta}{2}\bigl(|Ay|_{H}-|P|_{H}\bigr)^{2}\ge0$ gives $\tfrac{\delta}{2}|Ay|_{H}^{2}-\delta|P|_{H}|Ay|_{H}\ge-\tfrac{\delta}{2}|P|_{H}^{2}$, so $T_{y}\ge-\tfrac{\delta}{2}|P|_{H}^{2}$. Moreover $h(x)+h(y)\ge0$ by The Penalty Function $h=\tfrac12|\cdot|_V^2$ of a Hilbert Triple: Expansion Identities, Closed Sublevel Sets, Density and Local Bounds (i) together with claim 2 of Elementary Arithmetic in an Ordered Field, $\alpha|x-y|_{V}^{2}\ge0$, and $g(y)-g(x)\ge-|g(x)-g(y)|$ by claim 3 of Properties of the Absolute Value in an Ordered Field. Since $|P|_{H}^{2}=\alpha^{2}|x-y|_{H}^{2}$, we conclude

$$F^{-}_{\delta}(x,r,P,X)-F^{+}_{\delta}(y,r,P,Y)\ \ge\ -|g(x)-g(y)|-\tfrac{3}{4}\delta\alpha^{2}|x-y|_{H}^{2}.$$

By the first preliminary inequality and then the second with $s=|x-y|_{H}$, $|g(x)-g(y)|\le\omega_{1}\bigl(\alpha|x-y|_{H}^{2}+\tfrac{1}{\alpha}\bigr)$. For the second term, $|x-y|_{H}\le|x|_{H}+|y|_{H}$ by The Norm Metric of a Real Inner Product Space: Triangle Inequalities, Limits and Continuity (i), $(a+b)^{2}\le2a^{2}+2b^{2}$ for real $a,b$ because $2a^{2}+2b^{2}-(a+b)^{2}=(a-b)^{2}\ge0$, and $|x|_{H}^{2}\le|x|_{V}^{2}=2h(x)$ by Hilbert Triples: Standing Notation and Background (i) and Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field; hence $|x-y|_{H}^{2}\le4\bigl(h(x)+h(y)\bigr)\le4\bigl(h(x)+h(y)+1\bigr)$, and multiplying by $\tfrac34\delta\alpha^{2}\ge0$,

$$\tfrac{3}{4}\delta\alpha^{2}|x-y|_{H}^{2}\le3\alpha^{2}\,\delta\bigl(h(x)+h(y)+1\bigr)=\omega_{2}\bigl(\delta(h(x)+h(y)+1),\alpha\bigr).$$

Therefore

$$F^{-}_{\delta}(x,r,P,X)-F^{+}_{\delta}(y,r,P,Y)\ \ge\ -\omega_{1}\Bigl(\alpha|x-y|_{H}^{2}+\tfrac{1}{\alpha}\Bigr)-\omega_{2}\bigl(\delta(h(x)+h(y)+1),\alpha\bigr).$$

No restriction on $r$ was used, so $(\omega_{1},\omega_{2})$ is a structure pair for $F$ at every positive $R$, by The First-Order Structure Condition for a Second-Order Equation Operator on a Hilbert Triple (i).

Claim 4. Let $\delta,R$ satisfy $0<\delta<1$ and $0<R$. Recall from Test Data for a Second-Order Equation Operator on a Hilbert Triple and the Admissible Sets (ii) that an $R$-bounded datum $(x,r,p,X)$ satisfies $h(x)<R$, $|r|<R$ and $|p|_{H}<R$.

(a) An upper bound for $F^{+}_{\delta}$. For $R$-bounded $\eta=(y,s,p',X')$, expanding as above,

$$F^{+}_{\delta}(\eta)=\lambda_{0}s-\lambda_{0}\delta h(y)+\tfrac12|p'|_{H}^{2}-\delta\langle p',Ay\rangle_{H}+\tfrac{\delta^{2}}{2}|Ay|_{H}^{2}+\langle Ay,p'\rangle_{H}-\delta|Ay|_{H}^{2}-g(y).$$

Here $\lambda_{0}s\le\lambda_{0}R$, $-\lambda_{0}\delta h(y)\le0$, $\tfrac12|p'|_{H}^{2}\le\tfrac12R^{2}$, $-\delta\langle p',Ay\rangle_{H}\le R|Ay|_{H}$, $\langle Ay,p'\rangle_{H}\le R|Ay|_{H}$, $\tfrac{\delta^{2}}{2}|Ay|_{H}^{2}-\delta|Ay|_{H}^{2}\le-\tfrac{\delta}{2}|Ay|_{H}^{2}$ and $-g(y)\le C_{g}$, so

$$F^{+}_{\delta}(\eta)\le\lambda_{0}R+\tfrac12R^{2}+C_{g}+2R|Ay|_{H}-\tfrac{\delta}{2}|Ay|_{H}^{2}. \tag{2}$$

Since $\tfrac{\delta}{2}\bigl(|Ay|_{H}-\tfrac{2R}{\delta}\bigr)^{2}\ge0$ gives $2R|Ay|_{H}-\tfrac{\delta}{2}|Ay|_{H}^{2}\le\tfrac{2R^{2}}{\delta}$, we obtain $F^{+}_{\delta}(\eta)\le\Sigma$, where $\Sigma=\lambda_{0}R+\tfrac12R^{2}+C_{g}+\tfrac{2R^{2}}{\delta}$.

(b) A lower bound for $F^{-}_{\delta}$. For $R$-bounded $\xi=(x,r,p,X)$, in the same way,

$$F^{-}_{\delta}(\xi)\ \ge\ -\lambda_{0}R-R|Ax|_{H}-R|Ax|_{H}+\delta|Ax|_{H}^{2}-C_{g}=\delta|Ax|_{H}^{2}-2R|Ax|_{H}-\lambda_{0}R-C_{g}, \tag{3}$$

using $\lambda_{0}\delta h(x)\ge0$, $\tfrac12|p|_{H}^{2}\ge0$ and $\tfrac{\delta^{2}}{2}|Ax|_{H}^{2}\ge0$. Since $\delta\bigl(|Ax|_{H}-\tfrac{R}{\delta}\bigr)^{2}\ge0$ gives $\delta|Ax|_{H}^{2}-2R|Ax|_{H}\ge-\tfrac{R^{2}}{\delta}$, we obtain $F^{-}_{\delta}(\xi)\ge\Xi$, where $\Xi=-\tfrac{R^{2}}{\delta}-\lambda_{0}R-C_{g}$.

(c) A bound on $|Ax|_{H}$ for admissible data. Let $\xi=(x,r,p,X)\in S^{-}_{\delta,R}$. By Test Data for a Second-Order Equation Operator on a Hilbert Triple and the Admissible Sets (iii) there is an $R$-bounded $\eta$ with $F^{-}_{\delta}(\xi)<R+F^{+}_{\delta}(\eta)\le R+\Sigma$, so by (3)

$$\delta|Ax|_{H}^{2}-2R|Ax|_{H}\le K,\qquad K=R+\Sigma+\lambda_{0}R+C_{g}\ \ (\ge0).$$

As $2R|Ax|_{H}\le\tfrac{\delta}{2}|Ax|_{H}^{2}+\tfrac{2R^{2}}{\delta}$, this gives $\tfrac{\delta}{2}|Ax|_{H}^{2}\le K+\tfrac{2R^{2}}{\delta}$, so $|Ax|_{H}\le\Lambda_{1}$, the nonnegative square root of $\tfrac{2}{\delta}\bigl(K+\tfrac{2R^{2}}{\delta}\bigr)$.

Let now $\eta=(y,s,p',X')\in S^{+}_{\delta,R}$. There is an $R$-bounded $\xi$ with $F^{+}_{\delta}(\eta)>F^{-}_{\delta}(\xi)-R\ge\Xi-R$, so by (2)

$$\tfrac{\delta}{2}|Ay|_{H}^{2}-2R|Ay|_{H}\le K',\qquad K'=\lambda_{0}R+\tfrac12R^{2}+C_{g}-\Xi+R\ \ (\ge0),$$

and since $2R|Ay|_{H}\le\tfrac{\delta}{4}|Ay|_{H}^{2}+\tfrac{4R^{2}}{\delta}$ we get $|Ay|_{H}\le\Lambda_{2}$, the nonnegative square root of $\tfrac{4}{\delta}\bigl(K'+\tfrac{4R^{2}}{\delta}\bigr)$.

(d) The shift modulus. Put $\Lambda=\Lambda_{1}+\Lambda_{2}$, $c=R+2\Lambda$ and let $\omega$ have value $ct+\tfrac12t^{2}$ at a nonnegative $t$. Then $\omega$ is a modulus of continuity: its values are nonnegative, and for positive $\epsilon$ every $t$ with $0\le t\le\tau$, where $\tau$ is the smaller of $1$ and $\tfrac{\epsilon}{c+1}$, satisfies $\tfrac12t^{2}\le\tfrac12 t$ and hence $\omega(t)\le(c+1)t\le\epsilon$. Note also that $\omega$ is nondecreasing on the nonnegative reals, since $c\ge0$.

Let $q\in H$ and $Y\in\mathrm{Sym}(H)$. For $\xi=(x,r,p,X)\in S^{-}_{\delta,R}$ we have $F^{-}_{\delta}(x,r,p+q,X+Y)=F^{-}_{\delta}(x,r,p+q,X)$ by A First-Order Equation Operator is Degenerate Elliptic and Its $\delta$-Shifts Ignore the Form Argument (ii) and claim 1, and, expanding the norm,

$$F^{-}_{\delta}(x,r,p+q,X)-F^{-}_{\delta}(x,r,p,X)=\langle p+\delta Ax,q\rangle_{H}+\tfrac12|q|_{H}^{2}+\langle Ax,q\rangle_{H}\ \le\ \bigl(R+\Lambda+\Lambda\bigr)|q|_{H}+\tfrac12|q|_{H}^{2}=\omega\bigl(|q|_{H}\bigr),$$

using $|p|_{H}<R$, $\delta<1$ and $|Ax|_{H}\le\Lambda_{1}\le\Lambda$. Since $\omega$ is nondecreasing, $\omega(|q|_{H})\le\omega\bigl(|q|_{H}+\lVert Y\rVert\bigr)$, which is the first condition of The Shift-Continuity Condition on Admissible Test Data (i).

For $\eta=(y,s,p',X')\in S^{+}_{\delta,R}$ we have $F^{+}_{\delta}(y,s,p'+q,X'+Y)=F^{+}_{\delta}(y,s,p'+q,X')$, again by A First-Order Equation Operator is Degenerate Elliptic and Its $\delta$-Shifts Ignore the Form Argument (ii) and claim 1, and the same computation gives

$$F^{+}_{\delta}(y,s,p'+q,X'+Y)-F^{+}_{\delta}(y,s,p',X')=\langle p'-\delta Ay,q\rangle_{H}+\tfrac12|q|_{H}^{2}+\langle Ay,q\rangle_{H}\ \ge\ -\bigl(R+\Lambda+\Lambda\bigr)|q|_{H}\ \ge\ -\omega\bigl(|q|_{H}+\lVert Y\rVert\bigr),$$

using $\tfrac12|q|_{H}^{2}\ge0$ and $|Ay|_{H}\le\Lambda_{2}\le\Lambda$. Hence $\omega$ is a shift modulus for $F$ at $(\delta,R)$, and as $\delta$ and $R$ were arbitrary, $F$ satisfies the shift-continuity condition.

Claim 5. By claims 1 to 4 the operator $F$ satisfies all hypotheses on the operator in A Comparison Principle on a Hilbert Triple under the First-Order Structure Condition, and $u$, $v$, $C$ satisfy the remaining ones, so A Comparison Principle on a Hilbert Triple under the First-Order Structure Condition (ii) gives $u(x)\le v(x)$ for every $x\in V$.