Proof of Comparison and Uniqueness for Free-Energy-Penalised Viscosity Solutions of the Discounted HJB Equation with Free Langevin Noise in a Wall

Each result cited is applied with the data of its own statement. Throughout, $\rho$, $\sigma$ and $R$ are as in The Discounted Hamilton-Jacobi-Bellman Equation with Free Langevin Noise in a Wall on Noncommutative Laws: Standing Notation §parameters; $\mathcal{D}$, $\mathcal{E}$, $\mathcal{D}_{\Xi}$ and $\Xi$ as in The Discounted Hamilton-Jacobi-Bellman Equation with Free Langevin Noise in a Wall on Noncommutative Laws: Standing Notation §free-energy; and $\mathcal{H}$ as in The Discounted Hamilton-Jacobi-Bellman Equation with Free Langevin Noise in a Wall on Noncommutative Laws: Standing Notation §hamiltonian. By Square-Integrable Noncommutative Laws: the Wasserstein Completion of the Laws, Affine Push-Forwards, Moments, Couplings and Cost §laws, $\Sigma^{2}_{d}$ is the metric completion of $(\Sigma_{d},W_{2})$ with metric $\widehat{W}_{2}$ and canonical map $\kappa_{d}$, so by The Metric Completion is a Complete Metric Space with a Dense Isometric Copy of the Space, and Maps Preserving Cauchy Sequences Extend to It §isometry we have the isometry identity

$$\widehat{W}_{2}\bigl(\kappa_{d}(\lambda),\kappa_{d}(\lambda')\bigr)=W_{2}(\lambda,\lambda')\qquad\text{for all }\lambda,\lambda'\in\Sigma_{d}.$$

For $\lambda\in\Sigma_{d}$, $M(\lambda)=\sum_{j=1}^{d}\lambda(x_{j}x_{j})$ denotes the second moment as defined in Plan Jets of Squared Wasserstein Distances at an Optimal Coupling of Bounded Noncommutative Laws §realisation; it agrees with the $M$ of The Noncommutative Wasserstein Distance: Existence of Optimal Couplings, Symmetry, Separation, a Moment Bound, Weak-Star Lower Semicontinuity, and Displacement Interpolation, where $x_{j}^{2}=x_{j}x_{j}$. The tracial W*-probability space $(K,N,\Psi)$ provided by Plan Jets of Squared Wasserstein Distances at an Optimal Coupling of Bounded Noncommutative Laws §realisation is written $(H_{0},M_{0},\Omega_{0})$ below, as are the spaces quantified over in the structure condition (Step 3), to avoid clashes with the letters $M$, $K$ (Step 3) and $\Psi_{\delta,\alpha}$ (Step 1).

Proof of clause 1 (comparison). The objects are chosen in the following order: $\theta$, $b$, $e$, $(\delta_{u},\omega_{u})$, $(\delta_{v},\omega_{v})$ (Step 1); $r$, then $k$ and $\alpha$ (Step 3); $\delta_{1}$, then $\delta$, then the maximising pair $(\hat{\mu},\hat{\nu})$, then the optimal coupling $\gamma$ (Step 4); finally $(H_{0},M_{0},\Omega_{0})$, $X$ and $Y$ (Step 8).

Step 1 (Setup). If $\mathcal{D}=\varnothing$ there is nothing to prove, so assume $\mathcal{D}\ne\varnothing$. Suppose, for a contradiction, that there is $\mu_{0}\in\mathcal{D}$ with $u(\kappa_{d}(\mu_{0}))-v(\kappa_{d}(\mu_{0}))>0$, and let $\theta>0$ be half of this difference. Since $u$ and $v$ are bounded, fix a real $b\ge0$ with $|u(\lambda)|\le b$ and $|v(\lambda)|\le b$ for every $\lambda\in\Sigma^{2}_{d}$. By The Wall-Confined Free Energy: Norm Bound, Lower Bound, Weak-Star Compact Sublevel Sets and Displacement Monotonicity of the Score §bounds, applied with the free entropy penalty $(\mathcal{D}_{0},\mathcal{E}_{0})$ and the radius $R$ of The Discounted Hamilton-Jacobi-Bellman Equation with Free Langevin Noise in a Wall on Noncommutative Laws: Standing Notation §free-energy, $\mathcal{D}\subseteq\Sigma_{d,R}$, and we fix a real $e$ with $e\le\mathcal{E}(\lambda)$ for every $\lambda\in\mathcal{D}$. Since $u$ is a free-energy-penalised viscosity subsolution, fix a real $\delta_{u}>0$ and a nondecreasing $\omega_{u}:[0,\infty)\to[0,\infty)$ as there; since $v$ is a free-energy-penalised viscosity supersolution, fix a real $\delta_{v}>0$ and a nondecreasing $\omega_{v}:[0,\infty)\to[0,\infty)$ as there. The hypotheses of Maximisers of the Doubled Difference Penalised by the Wall-Confined Free Energy on Bounded Noncommutative Laws now hold with these $u$, $v$, $b$ and $e$ (the semicontinuity of $u$ and $v$ is assumed in clause 1); let $\Psi_{\delta,\alpha}$ and $S(\delta,\alpha)$ be as there.

Moment estimate. Let $\lambda\in\Sigma_{d,R}$ and $j\in[d]$. The variable $x_{j}$ is self-adjoint by Noncommutative Polynomials Form a Unital Complex Algebra with Involution: Linear Extension from Monomials, Products, Adjoints and Self-Adjoint Parts §adjoint, so $\lambda(x_{j}x_{j})=\lambda(x_{j}^{*}x_{j})$ is a nonnegative real number by condition (b) of Noncommutative Laws of Finitely Many Self-Adjoint Variables with a Norm Bound §tracial-state, and it is at most $R^{2}$ by Noncommutative Laws of Finitely Many Self-Adjoint Variables with a Norm Bound §norm-bound applied to the word $jj$ of length $2$, since $x_{j}x_{j}=x_{jj}$ by Noncommutative Polynomials Form a Unital Complex Algebra with Involution: Linear Extension from Monomials, Products, Adjoints and Self-Adjoint Parts §monomials. Hence

$$0\le M(\lambda)\le dR^{2}\qquad\text{for every }\lambda\in\Sigma_{d,R}.$$

Step 2 (The strength sequence). For $k\in\mathbb{N}$ let $s_{k}=S(0,2^{k})$, a real number by Maximisers of the Doubled Difference Penalised by the Wall-Confined Free Energy on Bounded Noncommutative Laws §bounds. By Maximisers of the Doubled Difference Penalised by the Wall-Confined Free Energy on Bounded Noncommutative Laws §diagonal with $\lambda=\mu_{0}$, $s_{k}\ge u(\kappa_{d}(\mu_{0}))-v(\kappa_{d}(\mu_{0}))=2\theta$ for every $k$. By Maximisers of the Doubled Difference Penalised by the Wall-Confined Free Energy on Bounded Noncommutative Laws §monotone, with $2^{k}$ in the role of $\alpha'$ and $2^{k+1}$ in the role of $\alpha$, $s_{k+1}\le s_{k}$ for every $k$. Thus $(s_{k})_{k\in\mathbb{N}}$ is nonincreasing and its set of terms is bounded below by $2\theta$, so by A Bounded Monotone Sequence of Real Numbers Converges §nonincreasing the sequence $(s_{k})_{k\in\mathbb{N}}$ converges to a real number $s$.

Step 3 (Choice of $r$ and of the strength). Since $\mathcal{H}$ satisfies the structure condition, applying it with $R\sqrt{d}$ in place of its radius and with $\eta=\rho\theta/2>0$ gives a real $r>0$ such that, for every tracial W*-probability space $(H_{0},M_{0},\Omega_{0})$, all $L^{2}$ $d$-tuples $X,Y$ of it with $\lVert X\rVert_{2}\le R\sqrt{d}$ and $\lVert Y\rVert_{2}\le R\sqrt{d}$, and every real $\alpha>0$ with $\alpha\lVert X-Y\rVert_{2}^{2}+\lVert X-Y\rVert_{2}<r$,

$$\mathcal{H}_{M_{0}}\bigl(Y,\alpha(X-Y)\bigr)-\mathcal{H}_{M_{0}}\bigl(X,\alpha(X-Y)\bigr)<\tfrac{\rho\theta}{2}.$$

Let $\tau=\min\{r/8,\,r^{2}/16\}>0$. By Step 2 there is $K\in\mathbb{N}$ with $|s_{j}-s|<\tau/4$ for every $j\ge K$. By the Archimedean property choose $k\in\mathbb{N}$ with $k\ge K$ and $k>\max\{4/\tau,\ 2(2\rho+1)/(3\rho\theta)\}$, and put $\alpha=2^{k+1}$. Then $\alpha>k$, $\alpha\ge2$ and $\alpha/2=2^{k}$, so $s_{k}=S(0,\alpha/2)$ and $s_{k+1}=S(0,\alpha)$. Put $\varepsilon=s_{k}-s_{k+1}+2/\alpha$. Since $k,k+1\ge K$, $s_{k}-s_{k+1}\le|s_{k}-s|+|s_{k+1}-s|<\tau/2$, and $2/\alpha<2/k<\tau/2$; together with Step 2,

$$0\le\varepsilon<\tau,\qquad\frac{2\rho+1}{\alpha}<\frac{3}{2}\rho\theta.$$

Step 4 (Choice of the weight, the maximising pair and the coupling). By Maximisers of the Doubled Difference Penalised by the Wall-Confined Free Energy on Bounded Noncommutative Laws §weight, applied with this $\alpha$ and with $1/\alpha$ in the role of $\varepsilon$, fix a real $\delta_{1}>0$ with $S(\delta',\alpha)\ge S(0,\alpha)-1/\alpha$ for every real $\delta'$ with $0<\delta'\le\delta_{1}$. Let $c=2\alpha R\sqrt{d}$ and

$$\delta=\min\Bigl\{\delta_{u},\ \delta_{v},\ \delta_{1},\ \frac{1}{\alpha\,(2|e|+\omega_{u}(c)+\omega_{v}(c)+1)}\Bigr\}.$$

Then $\delta>0$, $\delta\le\delta_{u}$, $\delta\le\delta_{v}$, and

$$S(\delta,\alpha)\ge s_{k+1}-\tfrac{1}{\alpha},\qquad 2\delta|e|\le\tfrac{1}{\alpha},\qquad\delta\bigl(\omega_{u}(c)+\omega_{v}(c)\bigr)\le\tfrac{1}{\alpha}.$$

By Maximisers of the Doubled Difference Penalised by the Wall-Confined Free Energy on Bounded Noncommutative Laws §maximiser fix a maximising pair $(\hat{\mu},\hat{\nu})\in\mathcal{D}\times\mathcal{D}$ of $\Psi_{\delta,\alpha}$, so $\Psi_{\delta,\alpha}(\hat{\mu},\hat{\nu})=S(\delta,\alpha)$, and put $W=W_{2}(\hat{\mu},\hat{\nu})$. Since $\hat{\mu},\hat{\nu}\in\Sigma_{d,R}$ by Step 1, The Noncommutative Wasserstein Distance: Existence of Optimal Couplings, Symmetry, Separation, a Moment Bound, Weak-Star Lower Semicontinuity, and Displacement Interpolation §attained (with radius $R$) gives an optimal coupling $\gamma\in\Pi(\hat{\mu},\hat{\nu})$; fix it.

Step 5 (Size of $W$). By The Noncommutative Wasserstein Distance: Existence of Optimal Couplings, Symmetry, Separation, a Moment Bound, Weak-Star Lower Semicontinuity, and Displacement Interpolation §moment-bound and the moment estimate, $W^{2}\le2M(\hat{\mu})+2M(\hat{\nu})\le4dR^{2}$, so $0\le W\le2R\sqrt{d}$ and $\alpha W\le c$. By Maximisers of the Doubled Difference Penalised by the Wall-Confined Free Energy on Bounded Noncommutative Laws §strength with $\alpha'=\alpha/2$,

$$S(\delta,\alpha)+\tfrac{\alpha}{4}W^{2}\le S(\delta,\alpha/2).$$

Here $-e\le|-e|=|e|$ by claims 3 and 2 of Properties of the Absolute Value in an Ordered Field, so, as $0\le2\delta$, claim 5 of Elementary Arithmetic in an Ordered Field and claim 2 of Zero Products and Elementary Identities in a Field give $-2\delta e=2\delta(-e)\le2\delta|e|$, and $2\delta|e|\le1/\alpha$ by Step 4. Hence, by Maximisers of the Doubled Difference Penalised by the Wall-Confined Free Energy on Bounded Noncommutative Laws §bounds, $S(\delta,\alpha/2)\le S(0,\alpha/2)-2\delta e\le s_{k}+2\delta|e|\le s_{k}+1/\alpha$, and by Step 4, $S(\delta,\alpha)\ge s_{k+1}-1/\alpha$. Hence $\frac{\alpha}{4}W^{2}\le\varepsilon$. Consequently $\alpha W^{2}\le4\varepsilon<4\tau\le r/2$, and, as $\alpha\ge2$, $W^{2}\le2\varepsilon<2\tau\le r^{2}/8<r^{2}/4$, so $W<r/2$. Therefore

$$\alpha W^{2}+W<r.$$

Step 6 (Subsolution test at $\hat{\mu}$). Apply Plan Jets of Squared Wasserstein Distances at an Optimal Coupling of Bounded Noncommutative Laws with $(\hat{\mu},\hat{\nu},\gamma,\alpha,1)$ in the roles of $(\mu,\nu,\gamma,\alpha,\beta)$, and let $\pi$, $\pi'$, $\varphi$ and $\psi$ be as there; thus

$$\varphi(\lambda)=\tfrac{\alpha}{2}\widehat{W}_{2}\bigl(\lambda,\kappa_{d}(\hat{\nu})\bigr)^{2}+\tfrac{1}{2}\widehat{W}_{2}\bigl(\lambda,\kappa_{d}(\hat{\mu})\bigr)^{2},\qquad\psi(\lambda)=-\tfrac{\alpha}{2}\widehat{W}_{2}\bigl(\kappa_{d}(\hat{\mu}),\lambda\bigr)^{2}-\tfrac{1}{2}\widehat{W}_{2}\bigl(\lambda,\kappa_{d}(\hat{\nu})\bigr)^{2}.$$

Let $\nu\in\mathcal{D}$. By the isometry identity, $\varphi(\kappa_{d}(\nu))=\frac{\alpha}{2}W_{2}(\nu,\hat{\nu})^{2}+\frac{1}{2}W_{2}(\nu,\hat{\mu})^{2}$, so, with $C=v(\kappa_{d}(\hat{\nu}))+\delta\mathcal{E}(\hat{\nu})$,

$$u\bigl(\kappa_{d}(\nu)\bigr)-\varphi\bigl(\kappa_{d}(\nu)\bigr)-\delta\mathcal{E}(\nu)=\Psi_{\delta,\alpha}(\nu,\hat{\nu})+C-\tfrac{1}{2}W_{2}(\nu,\hat{\mu})^{2}\le S(\delta,\alpha)+C-\tfrac{1}{2}W_{2}(\nu,\hat{\mu})^{2}.$$

At $\nu=\hat{\mu}$ the left-hand side equals $S(\delta,\alpha)+C$, because $\Psi_{\delta,\alpha}(\hat{\mu},\hat{\nu})=S(\delta,\alpha)$ and $W_{2}(\hat{\mu},\hat{\mu})=0$ by The Noncommutative Wasserstein Distance: Existence of Optimal Couplings, Symmetry, Separation, a Moment Bound, Weak-Star Lower Semicontinuity, and Displacement Interpolation §separation. If $\nu\ne\hat{\mu}$, then $W_{2}(\nu,\hat{\mu})\ne0$ by the same clause (as $\nu\in\mathcal{D}\subseteq\Sigma_{d,R}$ by Step 1), so $0<W_{2}(\nu,\hat{\mu})$ since $W_{2}$ is nonnegative by The Noncommutative Quadratic Wasserstein Distance and Optimal Couplings §distance, and $W_{2}(\nu,\hat{\mu})^{2}>0$ by claim 5 of Elementary Order Arithmetic in an Ordered Field and the left-hand side is strictly smaller than its value at $\hat{\mu}$. By Plan Jets of Squared Wasserstein Distances at an Optimal Coupling of Bounded Noncommutative Laws §plans, $\pi$ is a bounded plan at $\hat{\mu}$ with $|\pi|_{\mathrm{mom}}=\alpha W$, and by Plan Jets of Squared Wasserstein Distances at an Optimal Coupling of Bounded Noncommutative Laws §superjet, $\kappa_{2d}(\pi)\in J^{+}\varphi(\kappa_{d}(\hat{\mu}))$. Now Free-Energy-Penalised Viscosity Subsolutions, Supersolutions and Solutions of the Discounted HJB Equation with Free Langevin Noise in a Wall §sub, applied to $u$ with $(\delta_{u},\omega_{u})$, the test function $\varphi$, the weight $\delta$ (recall $0<\delta\le\delta_{u}$), the law $\hat{\mu}$ and the plan $\pi$, gives $\hat{\mu}\in\mathcal{D}_{\Xi}$ and, as $\omega_{u}$ is nondecreasing and $\alpha W\le c$,

$$\rho\,u\bigl(\kappa_{d}(\hat{\mu})\bigr)+\mathcal{H}\bigl(\kappa_{2d}(\pi)\bigr)+\tfrac{\sigma^{2}}{2}\mathcal{J}\bigl(\Xi(\hat{\mu}),\pi\bigr)\le\delta\,\omega_{u}(\alpha W)\le\delta\,\omega_{u}(c).$$

Step 7 (Supersolution test at $\hat{\nu}$). Let $\nu\in\mathcal{D}$. By the isometry identity, $\psi(\kappa_{d}(\nu))=-\frac{\alpha}{2}W_{2}(\hat{\mu},\nu)^{2}-\frac{1}{2}W_{2}(\nu,\hat{\nu})^{2}$, so, with $C'=u(\kappa_{d}(\hat{\mu}))-\delta\mathcal{E}(\hat{\mu})$,

$$v\bigl(\kappa_{d}(\nu)\bigr)-\psi\bigl(\kappa_{d}(\nu)\bigr)+\delta\mathcal{E}(\nu)=C'-\Psi_{\delta,\alpha}(\hat{\mu},\nu)+\tfrac{1}{2}W_{2}(\nu,\hat{\nu})^{2}\ge C'-S(\delta,\alpha)+\tfrac{1}{2}W_{2}(\nu,\hat{\nu})^{2}.$$

At $\nu=\hat{\nu}$ the left-hand side equals $C'-S(\delta,\alpha)$, and for $\nu\ne\hat{\nu}$ it is strictly larger, since then $W_{2}(\nu,\hat{\nu})^{2}>0$ by The Noncommutative Wasserstein Distance: Existence of Optimal Couplings, Symmetry, Separation, a Moment Bound, Weak-Star Lower Semicontinuity, and Displacement Interpolation §separation as in Step 6 (as $\nu\in\mathcal{D}\subseteq\Sigma_{d,R}$). By Plan Jets of Squared Wasserstein Distances at an Optimal Coupling of Bounded Noncommutative Laws §plans, $\pi'$ is a bounded plan at $\hat{\nu}$ with $|\pi'|_{\mathrm{mom}}=\alpha W$, and by Plan Jets of Squared Wasserstein Distances at an Optimal Coupling of Bounded Noncommutative Laws §subjet, $\kappa_{2d}(\pi')\in J^{-}\psi(\kappa_{d}(\hat{\nu}))$. Now Free-Energy-Penalised Viscosity Subsolutions, Supersolutions and Solutions of the Discounted HJB Equation with Free Langevin Noise in a Wall §super, applied to $v$ with $(\delta_{v},\omega_{v})$, the test function $\psi$, the weight $\delta$ (recall $0<\delta\le\delta_{v}$), the law $\hat{\nu}$ and the plan $\pi'$, gives $\hat{\nu}\in\mathcal{D}_{\Xi}$ and

$$\rho\,v\bigl(\kappa_{d}(\hat{\nu})\bigr)+\mathcal{H}\bigl(\kappa_{2d}(\pi')\bigr)+\tfrac{\sigma^{2}}{2}\mathcal{J}\bigl(\Xi(\hat{\nu}),\pi'\bigr)\ge-\delta\,\omega_{v}(\alpha W)\ge-\delta\,\omega_{v}(c).$$

Step 8 (Upper bound). Subtracting the inequality of Step 7 from that of Step 6 gives

$$\rho\bigl(u(\kappa_{d}(\hat{\mu}))-v(\kappa_{d}(\hat{\nu}))\bigr)\le\mathcal{H}\bigl(\kappa_{2d}(\pi')\bigr)-\mathcal{H}\bigl(\kappa_{2d}(\pi)\bigr)-\tfrac{\sigma^{2}}{2}\Bigl(\mathcal{J}\bigl(\Xi(\hat{\mu}),\pi\bigr)-\mathcal{J}\bigl(\Xi(\hat{\nu}),\pi'\bigr)\Bigr)+\delta\bigl(\omega_{u}(c)+\omega_{v}(c)\bigr).$$

By Plan Jets of Squared Wasserstein Distances at an Optimal Coupling of Bounded Noncommutative Laws §pairing, the bracket equals $\alpha\bigl(\mathcal{J}^{1}_{\gamma}(\Xi(\hat{\mu}))-\mathcal{J}^{2}_{\gamma}(\Xi(\hat{\nu}))\bigr)$, and this is nonnegative by The Wall-Confined Free Energy: Norm Bound, Lower Bound, Weak-Star Compact Sublevel Sets and Displacement Monotonicity of the Score §monotone, since $\hat{\mu},\hat{\nu}\in\mathcal{D}_{\Xi}$ (Steps 6 and 7) and $\gamma$ is optimal. With the last inequality of Step 4 this yields

$$\rho\bigl(u(\kappa_{d}(\hat{\mu}))-v(\kappa_{d}(\hat{\nu}))\bigr)\le\mathcal{H}\bigl(\kappa_{2d}(\pi')\bigr)-\mathcal{H}\bigl(\kappa_{2d}(\pi)\bigr)+\tfrac{1}{\alpha}.$$

By Plan Jets of Squared Wasserstein Distances at an Optimal Coupling of Bounded Noncommutative Laws §realisation, applied with the data of Step 6 and with $M$ the second moment defined in that clause, fix the tracial W*-probability space $(K,N,\Psi)$ of that clause, written here $(H_{0},M_{0},\Omega_{0})$, and $L^{2}$ $d$-tuples $X,Y$ of it with $\mathrm{law}(X,\alpha(X-Y))=\kappa_{2d}(\pi)$, $\mathrm{law}(Y,\alpha(X-Y))=\kappa_{2d}(\pi')$, $\lVert X-Y\rVert_{2}=W$, $\lVert X\rVert_{2}^{2}=M(\hat{\mu})$ and $\lVert Y\rVert_{2}^{2}=M(\hat{\nu})$. By the moment estimate, $\lVert X\rVert_{2}^{2}\le dR^{2}$ and $\lVert Y\rVert_{2}^{2}\le dR^{2}$, so $\lVert X\rVert_{2}\le R\sqrt{d}$ and $\lVert Y\rVert_{2}\le R\sqrt{d}$; and $\alpha\lVert X-Y\rVert_{2}^{2}+\lVert X-Y\rVert_{2}=\alpha W^{2}+W<r$ by Step 5. The choice of $r$ in Step 3 therefore gives $\mathcal{H}_{M_{0}}(Y,\alpha(X-Y))-\mathcal{H}_{M_{0}}(X,\alpha(X-Y))<\rho\theta/2$. By Plan Jets and Hamiltonians on Square-Integrable Noncommutative Laws: Standing Notation §lifts (the lift), $\mathcal{H}_{M_{0}}(X,\alpha(X-Y))=\mathcal{H}(\mathrm{law}(X,\alpha(X-Y)))=\mathcal{H}(\kappa_{2d}(\pi))$ and likewise $\mathcal{H}_{M_{0}}(Y,\alpha(X-Y))=\mathcal{H}(\kappa_{2d}(\pi'))$. Hence

$$\rho\bigl(u(\kappa_{d}(\hat{\mu}))-v(\kappa_{d}(\hat{\nu}))\bigr)<\tfrac{\rho\theta}{2}+\tfrac{1}{\alpha}.$$

Step 9 (Lower bound and contradiction). Since $\Psi_{\delta,\alpha}(\hat{\mu},\hat{\nu})=S(\delta,\alpha)$, $W^{2}\ge0$, and $\mathcal{E}(\hat{\mu}),\mathcal{E}(\hat{\nu})\ge e$ (Step 1),

$$u\bigl(\kappa_{d}(\hat{\mu})\bigr)-v\bigl(\kappa_{d}(\hat{\nu})\bigr)=S(\delta,\alpha)+\tfrac{\alpha}{2}W^{2}+\delta\mathcal{E}(\hat{\mu})+\delta\mathcal{E}(\hat{\nu})\ge S(\delta,\alpha)-2\delta|e|\ge s_{k+1}-\tfrac{2}{\alpha}\ge2\theta-\tfrac{2}{\alpha},$$

by Step 4 and Step 2. Multiplying by $\rho>0$ and combining with Step 8 gives $2\rho\theta-\frac{2\rho}{\alpha}<\frac{\rho\theta}{2}+\frac{1}{\alpha}$, that is, $\frac{3}{2}\rho\theta<\frac{2\rho+1}{\alpha}$, which contradicts the choice of $k$ in Step 3. Hence $u(\kappa_{d}(\mu))\le v(\kappa_{d}(\mu))$ for every $\mu\in\mathcal{D}$.

Proof of clause 2 (uniqueness). Step 10. By Weak-Star Semicontinuity on Bounded Laws of Functions on Square-Integrable Noncommutative Laws §continuous, $u$ and $v$ are both weak-star upper semicontinuous on bounded laws and weak-star lower semicontinuous on bounded laws, and by Free-Energy-Penalised Viscosity Subsolutions, Supersolutions and Solutions of the Discounted HJB Equation with Free Langevin Noise in a Wall §solution each is both a free-energy-penalised viscosity subsolution and supersolution of $(\mathrm{E})$. Clause 1 applied to the pair $(u,v)$ gives $u(\kappa_{d}(\mu))\le v(\kappa_{d}(\mu))$, and clause 1 applied to the pair $(v,u)$ gives $v(\kappa_{d}(\mu))\le u(\kappa_{d}(\mu))$, for every $\mu\in\mathcal{D}$. Hence $u(\kappa_{d}(\mu))=v(\kappa_{d}(\mu))$ for every $\mu\in\mathcal{D}$.