Proof of Comparison and Uniqueness for the Linear-Quadratic Hamilton-Jacobi Equation on Square-Integrable Noncommutative Laws

Each result cited is universally quantified over the data in its own statement. By The Linear-Quadratic Hamilton-Jacobi Equation with Law-Dependent Affine Drift on Square-Integrable Noncommutative Laws §equation, $(\mathrm{LQ})$ is the discounted stationary Hamilton--Jacobi equation with discount rate $\rho$ and Hamiltonian $\mathcal{H}^{\mathrm{LQ}}$, so its plan-jet viscosity sub- and supersolutions are those of that equation in the sense of Plan-Jet Viscosity Subsolutions, Supersolutions and Solutions of the Discounted Hamilton-Jacobi Equation on Square-Integrable Noncommutative Laws. By The Linear-Quadratic Hamiltonian: Its Lift, the Structure Condition and Its Quadratic Structure §structure and The Linear-Quadratic Hamiltonian: Its Lift, the Structure Condition and Its Quadratic Structure §quadratic, $\mathcal{H}^{\mathrm{LQ}}$ satisfies the structure condition and is quadratic with a convex Lipschitz remainder.

Clause 1. The hypotheses of Comparison Principle for Plan-Jet Viscosity Solutions of the Discounted Hamilton-Jacobi Equation on Square-Integrable Noncommutative Laws §quadratic hold with $\mathcal{H}=\mathcal{H}^{\mathrm{LQ}}$, and it gives $u(\mu)\le v(\mu)$ for every $\mu\in\Sigma^{2}_{d}$.

Clause 2. A continuous function $w:\Sigma^{2}_{d}\to\mathbb{R}$ is upper and lower semicontinuous: given $x$ and $\varepsilon>0$, Continuous Map Between Metric Spaces yields $r>0$ with $|w(y)-w(x)|<\varepsilon$ whenever $\widehat{W}_{2}(x,y)<r$, which gives both $w(y)<w(x)+\varepsilon$ (Upper Semicontinuous Function on a Subset of a Metric Space) and $w(y)>w(x)-\varepsilon$ (Lower Semicontinuous Function on a Subset of a Metric Space). By Plan-Jet Viscosity Subsolutions, Supersolutions and Solutions of the Discounted Hamilton-Jacobi Equation on Square-Integrable Noncommutative Laws §solution, $u$ and $v$ are each both subsolutions and supersolutions of $(\mathrm{LQ})$. Clause 1 applied to the pair $(u,v)$ gives $u\le v$, and applied to the pair $(v,u)$ gives $v\le u$. Hence $u=v$.