Proof of The Viscous Hamilton-Jacobi Operator is Continuous, Strictly Proper and Satisfies the Structure Condition

Conventions. The order $\le$ and the arithmetic of $\mathbb{R}$ are those of the ordered field of real numbers. Multiplication by a nonnegative real number preserves $\le$: if $a\le b$ and $0\le\lambda$, then either $a=b$, and the two products are equal, or $a<b$, and then $\lambda a\le\lambda b$ by claim 10 of Elementary Order Arithmetic in an Ordered Field when $0<\lambda$, while $\lambda=0$ makes both products $0$. Adding a fixed real number to both sides of an inequality also preserves it, by the compatibility of $\le$ with addition. Finally, $0<2^{-1}$ by claims 8 and 7 of Elementary Order Arithmetic in an Ordered Field, so $\tfrac{\kappa}{2}=\kappa\cdot2^{-1}$ is nonnegative and $\tfrac{1}{2}=1\cdot 2^{-1}$ is positive.

Proof of claim 2. We first check degenerate ellipticity. Let $x\in\Omega$, $r\in\mathbb{R}$, $p\in\mathbb{R}^{n}$ and $X,Y\in\mathcal{S}(n)$ with $X\preceq Y$. By The Trace as a Sum of Quadratic Forms, its Monotonicity and a Norm Bound (iv), $\operatorname{tr}(X)\le\operatorname{tr}(Y)$. Multiplying by the nonnegative $\tfrac{\kappa}{2}$ gives $\tfrac{\kappa}{2}\operatorname{tr}(X)\le\tfrac{\kappa}{2}\operatorname{tr}(Y)$, and claim 4 of Elementary Order Arithmetic in an Ordered Field then gives

$$-\tfrac{\kappa}{2}\operatorname{tr}(Y)\le-\tfrac{\kappa}{2}\operatorname{tr}(X).$$

Adding $\gamma r+\tfrac{1}{2}\lVert p\rVert^{2}-f(x)$ to both sides yields $F(x,r,p,Y)\le F(x,r,p,X)$. As $x$, $r$, $p$, $X$ and $Y$ were arbitrary, $F$ is degenerate elliptic, which is condition 1 of Strictly Proper Second-Order Equation Operator.

For condition 2, let $x\in\Omega$, $p\in\mathbb{R}^{n}$, $X\in\mathcal{S}(n)$ and $r,s\in\mathbb{R}$ with $s\le r$. The three terms $-\tfrac{\kappa}{2}\operatorname{tr}(X)$, $\tfrac{1}{2}\lVert p\rVert^{2}$ and $-f(x)$ are the same in $F(x,r,p,X)$ and in $F(x,s,p,X)$, so

$$F(x,r,p,X)-F(x,s,p,X)=\gamma r-\gamma s=\gamma(r-s)$$

by distributivity. In particular $\gamma(r-s)\le F(x,r,p,X)-F(x,s,p,X)$, and claim 2 is proved.

Proof of claim 3. Define $G:\mathbb{R}\times\mathbb{R}^{n}\times\mathcal{S}(n)\to\mathbb{R}$ by

$$G(r,p,X)=\gamma r-\tfrac{\kappa}{2}\operatorname{tr}(X)+\tfrac{1}{2}\lVert p\rVert^{2},$$

so that $F(x,r,p,X)=G(r,p,X)-f(x)$ for every $x\in\Omega$, $r\in\mathbb{R}$, $p\in\mathbb{R}^{n}$ and $X\in\mathcal{S}(n)$. The computation carried out in the proof of claim 2, with the term $-f(x)$ omitted throughout, shows that $G(r,p,Y)\le G(r,p,X)$ whenever $r\in\mathbb{R}$, $p\in\mathbb{R}^{n}$ and $X,Y\in\mathcal{S}(n)$ satisfy $X\preceq Y$; that is, $G$ is degenerate elliptic in the matrix variable in the sense required by Structure Condition: a Degenerate Elliptic Operator with a Continuous Inhomogeneity. Since $f$ and $\omega$ are as required there, that example applies to this $G$, this $f$ and this $\omega$, and its conclusion Structure Condition: a Degenerate Elliptic Operator with a Continuous Inhomogeneity (i) is exactly claim 3.

Proof of claim 1. Fix $x_{0}\in\Omega$, $r_{0}\in\mathbb{R}$, $p_{0}\in\mathbb{R}^{n}$ and $X_{0}\in\mathcal{S}(n)$, and let $\varepsilon\in\mathbb{R}$ be positive. Put $4=2+2$; then $0<2$ by claim 8 of Elementary Order Arithmetic in an Ordered Field and hence $0<4$ by claim 3 there, so $4^{-1}$ exists and is positive by claim 7 there, and $\varepsilon'=\varepsilon\cdot4^{-1}$ is positive by claim 5 there. Moreover $\varepsilon'+\varepsilon'+\varepsilon'+\varepsilon'=(1+1+1+1)\varepsilon'=4\bigl(\varepsilon\cdot4^{-1}\bigr)=\varepsilon$ by distributivity, commutativity and the defining property of the multiplicative inverse.

Four bounds. We record four estimates.

(i) For $s\in\mathbb{R}$, distributivity and claim 4 of Properties of the Absolute Value in an Ordered Field give $|\gamma s-\gamma r_{0}|=|\gamma(s-r_{0})|=|\gamma||s-r_{0}|=\gamma|s-r_{0}|$, the last step because $|\gamma|=\gamma$ by the definition of the absolute value, $\gamma$ being nonnegative.

(ii) Let $Y\in\mathcal{S}(n)$. Then $Y-X_{0}\in\mathcal{S}(n)$ by Euclidean Space and Real Matrices: Standing Notation (iii), and since the entries of $Y-X_{0}$ are the differences of the corresponding entries by the definition of a matrix difference, claim 1 of Basic Properties of the Trace, applied with the coefficients $1$ and $-1$, gives $\operatorname{tr}(Y)-\operatorname{tr}(X_{0})=\operatorname{tr}(Y-X_{0})$. Hence, by The Trace as a Sum of Quadratic Forms, its Monotonicity and a Norm Bound (v) and Euclidean Space and Real Matrices: Standing Notation (v),

$$\bigl|\operatorname{tr}(Y)-\operatorname{tr}(X_{0})\bigr|\le\beta_{n}\lVert Y-X_{0}\rVert=\beta_{n}\,d_{\mathcal{S}(n)}(Y,X_{0}),$$

where $\beta_{n}=\sum_{k=1}^{n}1$ is as in The Trace as a Sum of Quadratic Forms, its Monotonicity and a Norm Bound.

(iii) Let $q\in\mathbb{R}^{n}$ and put $h=q-p_{0}$, so that $q=p_{0}+h$. By claim 1 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square and claims 1, 2 and 5 of The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite,

$$\lVert q\rVert^{2}=q\cdot q=p_{0}\cdot p_{0}+(p_{0}\cdot h)+(h\cdot p_{0})+h\cdot h=\lVert p_{0}\rVert^{2}+2(p_{0}\cdot h)+\lVert h\rVert^{2},$$

where $2(p_{0}\cdot h)$ abbreviates $(p_{0}\cdot h)+(p_{0}\cdot h)$, the two middle terms being equal by the symmetry of the dot product. Hence $\lVert q\rVert^{2}-\lVert p_{0}\rVert^{2}=2(p_{0}\cdot h)+\lVert h\rVert^{2}$, and by claims 4 and 5 of Properties of the Absolute Value in an Ordered Field, The Cauchy-Schwarz Inequality on $\mathbb{R}^n$ and the nonnegativity of norms recorded in claim 1 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square,

$$\bigl|\lVert q\rVert^{2}-\lVert p_{0}\rVert^{2}\bigr|\le2|p_{0}\cdot h|+\lVert h\rVert^{2}\le2\lVert p_{0}\rVert\lVert h\rVert+\lVert h\rVert^{2}.$$

If moreover $\lVert h\rVert\le1$, then multiplying by the nonnegative $\lVert h\rVert$ gives $\lVert h\rVert^{2}\le\lVert h\rVert$, so that by distributivity

$$\bigl|\lVert q\rVert^{2}-\lVert p_{0}\rVert^{2}\bigr|\le\bigl(2\lVert p_{0}\rVert+1\bigr)\lVert h\rVert .$$

(iv) Since $\Omega\subseteq\overline{\Omega}$ and $f$ is continuous at $x_{0}$ relative to $\overline{\Omega}$, there is a positive $\delta_{f}\in\mathbb{R}$ such that every $y\in\overline{\Omega}$ with $d_{E}(x_{0},y)<\delta_{f}$ satisfies $|f(y)-f(x_{0})|<\varepsilon'$; in particular this holds for every $y\in\Omega$.

Choice of $\delta$. Put

$$c_{1}=\gamma,\qquad c_{2}=\tfrac{\kappa}{2}\beta_{n}+1,\qquad c_{3}=\tfrac{1}{2}\bigl(2\lVert p_{0}\rVert+1\bigr)+1 .$$

Each $c_{i}$ is positive: $c_{1}=\gamma$ is positive by hypothesis, while $\tfrac{\kappa}{2}\beta_{n}$ and $\tfrac{1}{2}(2\lVert p_{0}\rVert+1)$ are nonnegative, being products of nonnegative numbers — for $\beta_{n}$ this is recorded in The Trace as a Sum of Quadratic Forms, its Monotonicity and a Norm Bound, where $1\le\beta_{n}$ — so $c_{2}$ and $c_{3}$ are at least $1$ and $0<1$ by claim 6 of Elementary Order Arithmetic in an Ordered Field. By claim 7 there each $c_{i}^{-1}$ exists and is positive, and by claim 5 there each $\varepsilon'c_{i}^{-1}$ is positive. Applying claim 9 of Elementary Order Arithmetic in an Ordered Field repeatedly to the five positive numbers $1$, $\delta_{f}$, $\varepsilon'c_{1}^{-1}$, $\varepsilon'c_{2}^{-1}$ and $\varepsilon'c_{3}^{-1}$ produces a real number $\delta$ that is at most each of them and equal to one of them, hence positive.

Conclusion. Let $y\in\Omega$, $s\in\mathbb{R}$, $q\in\mathbb{R}^{n}$ and $Y\in\mathcal{S}(n)$ satisfy

$$d_{E}(y,x_{0})<\delta,\qquad|s-r_{0}|<\delta,\qquad\lVert q-p_{0}\rVert<\delta,\qquad d_{\mathcal{S}(n)}(Y,X_{0})<\delta,$$

and put $h=q-p_{0}$, so that $\lVert h\rVert<\delta\le1$. Grouping the four terms of $F$,

$$F(y,s,q,Y)-F(x_{0},r_{0},p_{0},X_{0})=\bigl(\gamma s-\gamma r_{0}\bigr)-\tfrac{\kappa}{2}\bigl(\operatorname{tr}(Y)-\operatorname{tr}(X_{0})\bigr)+\tfrac{1}{2}\bigl(\lVert q\rVert^{2}-\lVert p_{0}\rVert^{2}\bigr)-\bigl(f(y)-f(x_{0})\bigr).$$

By claims 2, 4 and 5 of Properties of the Absolute Value in an Ordered Field, the absolute value of the left-hand side is at most the sum of

$$\gamma|s-r_{0}|,\qquad\tfrac{\kappa}{2}\bigl|\operatorname{tr}(Y)-\operatorname{tr}(X_{0})\bigr|,\qquad\tfrac{1}{2}\bigl|\lVert q\rVert^{2}-\lVert p_{0}\rVert^{2}\bigr|,\qquad|f(y)-f(x_{0})| .$$

We bound each summand by $\varepsilon'$. By (i) and claim 10 of Elementary Order Arithmetic in an Ordered Field, $\gamma|s-r_{0}|<\gamma\delta$, and multiplying $\delta\le\varepsilon'c_{1}^{-1}$ by the nonnegative $\gamma=c_{1}$ gives $\gamma\delta\le\varepsilon'$; hence the first summand is less than $\varepsilon'$. By (ii), $\tfrac{\kappa}{2}\beta_{n}\le c_{2}$, so multiplying by the nonnegative $d_{\mathcal{S}(n)}(Y,X_{0})$ and using (ii) again gives $\tfrac{\kappa}{2}|\operatorname{tr}(Y)-\operatorname{tr}(X_{0})|\le c_{2}\,d_{\mathcal{S}(n)}(Y,X_{0})<c_{2}\delta\le\varepsilon'$, the last two steps by claim 10 and by multiplying $\delta\le\varepsilon'c_{2}^{-1}$ by $c_{2}$. By (iii), and since $\lVert h\rVert\le1$, the same reasoning with $c_{3}$ gives $\tfrac{1}{2}|\lVert q\rVert^{2}-\lVert p_{0}\rVert^{2}|\le c_{3}\lVert h\rVert<c_{3}\delta\le\varepsilon'$. By (iv) and $d_{E}(y,x_{0})<\delta\le\delta_{f}$, the fourth summand is less than $\varepsilon'$.

Adding the four strict inequalities by claim 3 of Elementary Order Arithmetic in an Ordered Field, applied repeatedly, and using claim 2 there,

$$\bigl|F(y,s,q,Y)-F(x_{0},r_{0},p_{0},X_{0})\bigr|<\varepsilon'+\varepsilon'+\varepsilon'+\varepsilon'=\varepsilon .$$

Thus $F$ is continuous at $(x_{0},r_{0},p_{0},X_{0})$ in the sense of Continuity of a Second-Order Equation Operator (i). Since the quadruple was arbitrary, $F$ is continuous, which is claim 1.