Proof of Lipschitz-Cone Comparison for the Quadratic Control Problem on Unitary Laws under Strong Free Noise

Each result cited is universally quantified over the data in its own statement.

Conventions. The conventions of Unitary Laws with Free Unitary Noise: Standing Data §spaces and Unitary Laws with Free Unitary Noise: Standing Data §test-functions are in force, and viscosity sub- and supersolutions are those of The Discounted Hamilton-Jacobi-Bellman Equation on Unitary Laws with Free Unitary Noise §equation with Hamiltonian $H_{Q}$, in the sense of Viscosity Sub- and Supersolutions of the Hamilton-Jacobi-Bellman Equation on Unitary Laws. By The Word Gauge Space Is a Real Hilbert Space, and the Unitary Laws Form a Compact Metric Space in It §inner-product and The Word Gauge Space Is a Real Hilbert Space, and the Unitary Laws Form a Compact Metric Space in It §hilbert, $E_{d}$ is a real Hilbert space with inner product $\langle\cdot,\cdot\rangle_{d}$ and norm $\lVert\cdot\rVert_{d}$; accordingly Elementary Identities in a Real Inner Product Space and the symmetry axiom of Real Inner Product Space §inner-product apply to $E_{d}$, and Affine and Quadratic Functions on a Real Hilbert Space are of Class $C^2$, in the setting of Real Hilbert Spaces: Standing Notation and Background, applies with $E_{d}$ as the Hilbert space and $\lVert\cdot\rVert_{d}$ as its norm. By The Classes $C^1$ and $C^2$ on an Open Subset of a Real Inner Product Space §c2, every member of $C^{2}(E_{d})$ belongs to $C^{1}(E_{d})$. For $\mu,\nu\in\mathcal{L}_{d}$, $d_{\mathcal{L}}(\mu,\nu)=\lVert\iota_{d}(\mu)-\iota_{d}(\nu)\rVert_{d}$ by The Embedding of Unitary Laws into the Word Gauge Space and the Gauge Distance §distance; $d_{\mathcal{L}}$ is a metric on $\mathcal{L}_{d}$ by The Word Gauge Space Is a Real Hilbert Space, and the Unitary Laws Form a Compact Metric Space in It §metric; $\mathcal{L}_{d}$ is nonempty by Unitary Laws: Adjoints, the Bound One, and Reduction to Cyclically Reduced Words §nonempty and compact in the topology induced by $d_{\mathcal{L}}$ by The Word Gauge Space Is a Real Hilbert Space, and the Unitary Laws Form a Compact Metric Space in It §compact. By Unitary Laws with Free Unitary Noise: Standing Data §constants, $\rho>0$ and $\beta\ge0$, and $K>0$ by The Trilinear Estimate for Cyclic Gradients of Differences of Unitary Laws §trilinear. Square roots $\sqrt{t}$ of reals $t\ge0$ are the nonnegative square roots of Existence and Uniqueness of the Nonnegative Square Root; comparisons of nonnegative reals through their squares use the weak form, claim 2, of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field; order arithmetic, absolute values $|t|$ (equal to $t$ or to $-t$) and limits of real sequences are as in The Real Numbers: Standing Notation and Background §numbers and The Real Numbers: Standing Notation and Background §sequences.

Three elementary facts. Let $\varepsilon>0$ be real.

(E1) For reals $A\ge0$ and $B>0$, $\sqrt{A}\le\frac{A+B^{2}}{2B}$, with equality when $A=B^{2}$. Indeed $0\le(\sqrt{A}-B)^{2}=A-2B\sqrt{A}+B^{2}$, so $2B\sqrt{A}\le A+B^{2}$, and we divide by $2B>0$. If $A=B^{2}$, then $\sqrt{A}=B$ by the uniqueness in Existence and Uniqueness of the Nonnegative Square Root, $B$ being nonnegative with square $A$, and $\frac{B^{2}+B^{2}}{2B}=B$.

(E2) For every real $D\ge0$, $D\le\sqrt{\varepsilon+D^{2}}\le\sqrt{\varepsilon}+D$ and $0<\sqrt{\varepsilon}\le\sqrt{\varepsilon+D^{2}}$. Indeed $D^{2}\le\varepsilon+D^{2}\le\varepsilon+2\sqrt{\varepsilon}\,D+D^{2}=(\sqrt{\varepsilon}+D)^{2}$ and $\varepsilon\le\varepsilon+D^{2}$, all quantities compared being squares of nonnegative reals, so claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field gives the inequalities; and $\sqrt{\varepsilon}\ne0$ since its square $\varepsilon$ is nonzero, so $\sqrt{\varepsilon}>0$.

(E3) For reals $D,D'\ge0$, $\bigl|\sqrt{\varepsilon+D^{2}}-\sqrt{\varepsilon+D'^{2}}\bigr|\le|D-D'|$. Indeed, put $a=\sqrt{\varepsilon+D^{2}}$ and $a'=\sqrt{\varepsilon+D'^{2}}$, both positive by (E2). Then $(a-a')(a+a')=a^{2}-a'^{2}=D^{2}-D'^{2}=(D-D')(D+D')$, and $0\le D+D'\le a+a'$ by (E2). Taking absolute values, which are multiplicative by claims 4 and 8 of Properties of Complex Conjugation and Modulus (reals being viewed as complex numbers), $|a-a'|\,(a+a')=|D-D'|\,(D+D')\le|D-D'|\,(a+a')$, and we divide by $a+a'>0$.

Part 1 (Lipschitz cone, clause 1). Let $L$ be real with $L_{f}/\rho\le L\le\beta^{2}/(4K)$, let $u$ be a viscosity subsolution and $v$ a viscosity supersolution. By Viscosity Sub- and Supersolutions of the Hamilton-Jacobi-Bellman Equation on Unitary Laws §subsolution and Viscosity Sub- and Supersolutions of the Hamilton-Jacobi-Bellman Equation on Unitary Laws §supersolution, $u$ is bounded and upper semicontinuous and $v$ is bounded and lower semicontinuous on $(\mathcal{L}_{d},d_{\mathcal{L}})$. Since $L_{f}\ge0$ and $\rho>0$, $L\ge L_{f}/\rho\ge0$; multiplying $L_{f}/\rho\le L$ by $\rho>0$ gives $L_{f}\le\rho L$, and multiplying $L\le\beta^{2}/(4K)$ by $K>0$ gives $LK\le\beta^{2}/4$. Fix a real $\varepsilon>0$.

Step 1 (Penalisation and a maximiser). Let $X=\mathcal{L}_{d}\times\mathcal{L}_{d}$ with the product metric $d_{X}\bigl((\mu,\nu),(\mu',\nu')\bigr)=\max\{d_{\mathcal{L}}(\mu,\mu'),d_{\mathcal{L}}(\nu,\nu')\}$, a metric by claim 1 of The Product Metric is a Metric. By A Product of Compact Subsets is Compact in the Product Metric, applied with both metric spaces equal to $(\mathcal{L}_{d},d_{\mathcal{L}})$ and both compact subsets equal to $\mathcal{L}_{d}$, the set $\mathcal{L}_{d}\times\mathcal{L}_{d}$ is compact in $(\mathcal{L}_{d}\times\mathcal{L}_{d},d_{X})$; it is nonempty since $\mathcal{L}_{d}$ is. Define $\varphi,\Phi,\Psi:X\to\mathbb{R}$ by

$$\varphi(\mu,\nu)=\sqrt{\varepsilon+d_{\mathcal{L}}(\mu,\nu)^{2}},\qquad\Phi(\mu,\nu)=u(\mu)-v(\nu),\qquad\Psi(\mu,\nu)=\Phi(\mu,\nu)-L\,\varphi(\mu,\nu).$$

$\Phi$ is upper semicontinuous on $X$ by claim 4 of Negation, Restriction, and Separated Differences of Semicontinuous Functions, applied with both metric spaces equal to $(\mathcal{L}_{d},d_{\mathcal{L}})$, $A=B=\mathcal{L}_{d}$, $f=u$ and $g=v$.

$L\varphi$ is lower semicontinuous on $X$. Let $z=(\mu,\nu)$ and $z'=(\mu',\nu')$ in $X$, and put $D=d_{\mathcal{L}}(\mu,\nu)$, $D'=d_{\mathcal{L}}(\mu',\nu')$. Both $d_{\mathcal{L}}(\mu,\mu')$ and $d_{\mathcal{L}}(\nu,\nu')$ are at most $d_{X}(z,z')$, the maximum dominating both entries (Product Metric on the Cartesian Product of Two Metric Spaces). By the triangle inequality and symmetry of the metric $d_{\mathcal{L}}$, $D\le d_{\mathcal{L}}(\mu,\mu')+D'+d_{\mathcal{L}}(\nu',\nu)$ and $D'\le d_{\mathcal{L}}(\mu',\mu)+D+d_{\mathcal{L}}(\nu,\nu')$, so $D-D'\le2\,d_{X}(z,z')$ and $D'-D\le2\,d_{X}(z,z')$; as $|D-D'|$ is one of $D-D'$ and $D'-D$, these two one-sided bounds give $|D-D'|\le2\,d_{X}(z,z')$, and (E3) gives $\varphi(z)-\varphi(z')\le2\,d_{X}(z,z')$. Now let $\varepsilon'>0$ and put $t=\varepsilon'/(2L+1)>0$. If $d_{X}(z,z')<t$, then, as $L\ge0$,

$$L\varphi(z')\ge L\varphi(z)-2L\,d_{X}(z,z')\ge L\varphi(z)-2Lt>L\varphi(z)-\varepsilon',$$

the last step because $2Lt=\varepsilon'\cdot\frac{2L}{2L+1}<\varepsilon'$. This is Lower Semicontinuous Function on a Subset of a Metric Space at $z$ relative to $X$.

Hence $\Psi=\Phi-L\varphi$ is upper semicontinuous on $X$, by claim 3 of Negation, Restriction, and Separated Differences of Semicontinuous Functions, applied at every point of $X$ with the metric space $(X,d_{X})$, $A=X$, $\Phi$ in the role of $u$ and $L\varphi$ in the role of $w$. By claim 1 of Semicontinuous Functions Attain Their Extrema on a Compact Set, applied with the metric space $(X,d_{X})$ and the nonempty compact set $X$, there is $(\hat\mu,\hat\nu)\in X$ with

$$\Psi(\mu,\nu)\le\Psi(\hat\mu,\hat\nu)\qquad(\mu,\nu\in\mathcal{L}_{d}).\tag{MX}$$

Put $x=\iota_{d}(\hat\mu)-\iota_{d}(\hat\nu)\in E_{d}$, so that $\lVert x\rVert_{d}=d_{\mathcal{L}}(\hat\mu,\hat\nu)$ and $\lVert x\rVert_{d,1}$ is defined by The Word Gauge Space Is a Real Hilbert Space, and the Unitary Laws Form a Compact Metric Space in It §sobolev; put $a=\varphi(\hat\mu,\hat\nu)$, $s=L/a$ and $p=s\,x\in E_{d}$. By (E2), $a>0$ and $\lVert x\rVert_{d}\le a$; moreover $s\ge0$ and $s\,a=L$.

Step 2 (The upper test function). Let $F:E_{d}\to\mathbb{R}$, $F(y)=\tfrac{s}{2}\lVert y-\iota_{d}(\hat\nu)\rVert_{d}^{2}$. By Affine and Quadratic Functions on a Real Hilbert Space are of Class $C^2$ §quadratic (with $s$ and $\iota_{d}(\hat\nu)$ in the roles of $\alpha$ and $y_{0}$), $F\in C^{2}(E_{d})\subseteq C^{1}(E_{d})$ and $DF(y)=s\,(y-\iota_{d}(\hat\nu))$, so $DF(\iota_{d}(\hat\mu))=p$. Let $\mu\in\mathcal{L}_{d}$. Then $F(\iota_{d}(\mu))=\frac{L}{2a}\,d_{\mathcal{L}}(\mu,\hat\nu)^{2}$, and (E1) with $A=\varepsilon+d_{\mathcal{L}}(\mu,\hat\nu)^{2}$ and $B=a$, multiplied by $L\ge0$, gives

$$L\,\varphi(\mu,\hat\nu)\le\frac{L}{2a}(\varepsilon+a^{2})+F(\iota_{d}(\mu)),$$

with equality for $\mu=\hat\mu$, where $A=\varepsilon+\lVert x\rVert_{d}^{2}=a^{2}$. By (MX) with $\nu=\hat\nu$, $u(\mu)-L\varphi(\mu,\hat\nu)\le u(\hat\mu)-L\varphi(\hat\mu,\hat\nu)$, after adding $v(\hat\nu)$ to both sides. Therefore

$$u(\mu)-F(\iota_{d}(\mu))\le u(\mu)-L\varphi(\mu,\hat\nu)+\frac{L}{2a}(\varepsilon+a^{2})\le u(\hat\mu)-L\varphi(\hat\mu,\hat\nu)+\frac{L}{2a}(\varepsilon+a^{2})=u(\hat\mu)-F(\iota_{d}(\hat\mu)).$$

Thus $u-F\circ\iota_{d}$ has a local maximum at $\hat\mu$ relative to $\mathcal{L}_{d}$ (with radius $1$), and Viscosity Sub- and Supersolutions of the Hamilton-Jacobi-Bellman Equation on Unitary Laws §subsolution gives

$$\rho\,u(\hat\mu)+H_{Q}(\hat\mu,p)-\frac{\beta^{2}}{2}\bigl\langle p,\widehat{\Theta}\hat\mu\bigr\rangle_{d}\le f(\hat\mu).\tag{S}$$

Step 3 (The lower test function). Let $G:E_{d}\to\mathbb{R}$, $G(y)=\tfrac{-s}{2}\lVert y-\iota_{d}(\hat\mu)\rVert_{d}^{2}$. By Affine and Quadratic Functions on a Real Hilbert Space are of Class $C^2$ §quadratic (with $-s$ and $\iota_{d}(\hat\mu)$ in the roles of $\alpha$ and $y_{0}$), $G\in C^{1}(E_{d})$ and $DG(y)=-s\,(y-\iota_{d}(\hat\mu))$, so $DG(\iota_{d}(\hat\nu))=-s\,(\iota_{d}(\hat\nu)-\iota_{d}(\hat\mu))=s\,x=p$ by the vector space axioms. Let $\nu\in\mathcal{L}_{d}$. By symmetry of the metric, $-G(\iota_{d}(\nu))=\frac{L}{2a}\,d_{\mathcal{L}}(\nu,\hat\mu)^{2}=\frac{L}{2a}\,d_{\mathcal{L}}(\hat\mu,\nu)^{2}$, and (E1) with $A=\varepsilon+d_{\mathcal{L}}(\hat\mu,\nu)^{2}$ and $B=a$, multiplied by $L\ge0$, gives

$$L\,\varphi(\hat\mu,\nu)\le\frac{L}{2a}(\varepsilon+a^{2})-G(\iota_{d}(\nu)),$$

with equality for $\nu=\hat\nu$. By (MX) with $\mu=\hat\mu$, after subtracting $u(\hat\mu)$ and rearranging, $v(\hat\nu)+L\varphi(\hat\mu,\hat\nu)\le v(\nu)+L\varphi(\hat\mu,\nu)$. Therefore

$$v(\hat\nu)-G(\iota_{d}(\hat\nu))=v(\hat\nu)+L\varphi(\hat\mu,\hat\nu)-\frac{L}{2a}(\varepsilon+a^{2})\le v(\nu)+L\varphi(\hat\mu,\nu)-\frac{L}{2a}(\varepsilon+a^{2})\le v(\nu)-G(\iota_{d}(\nu)).$$

Thus $v-G\circ\iota_{d}$ has a local minimum at $\hat\nu$ relative to $\mathcal{L}_{d}$, and Viscosity Sub- and Supersolutions of the Hamilton-Jacobi-Bellman Equation on Unitary Laws §supersolution gives

$$\rho\,v(\hat\nu)+H_{Q}(\hat\nu,p)-\frac{\beta^{2}}{2}\bigl\langle p,\widehat{\Theta}\hat\nu\bigr\rangle_{d}\ge f(\hat\nu).\tag{P}$$

Step 4 (Subtracting). Subtracting (P) from (S), and using $\langle p,\widehat{\Theta}\hat\mu\rangle_{d}-\langle p,\widehat{\Theta}\hat\nu\rangle_{d}=\langle p,\widehat{\Theta}\hat\mu-\widehat{\Theta}\hat\nu\rangle_{d}=s\,\langle x,\widehat{\Theta}\hat\mu-\widehat{\Theta}\hat\nu\rangle_{d}$ (by Elementary Identities in a Real Inner Product Space §bilinear and symmetry of the inner product),

$$\rho\bigl(u(\hat\mu)-v(\hat\nu)\bigr)\le\bigl(f(\hat\mu)-f(\hat\nu)\bigr)+\bigl(H_{Q}(\hat\nu,p)-H_{Q}(\hat\mu,p)\bigr)+\frac{\beta^{2}s}{2}\bigl\langle x,\widehat{\Theta}\hat\mu-\widehat{\Theta}\hat\nu\bigr\rangle_{d}.\tag{D}$$

Step 5 (Scaling of the truncated cyclic gradients). Let $\Xi^{i}_{q,n}$, $Z^{i}_{q,n}$ ($q\in E_{d}$, $n\in\mathbb{N}$, $i\in[d]$) be as in The Truncated Cyclic Gradient of a Gauge Vector §gradient. We show $\Xi^{i}_{p,n}=s\,\Xi^{i}_{x,n}$. For $w\in W^{\circ}_{d}$, $p(w)=s\,x(w)$, real multiples in $E_{d}$ being pointwise (The Word Gauge: Weights, the Space of Weighted Square-Summable Word Functions, Its Norm and Inner Product, and the Length-Weighted Gauge §space), and $\overline{s\,x(w)}=\overline{s}\;\overline{x(w)}=s\,\overline{x(w)}$ by claim 1 of Properties of Complex Conjugation and Modulus, $s$ being real. So each summand $c_{w}\overline{p(w)}D^{i}_{w}$ of $Z^{i}_{p,n}$ is $s$ times the corresponding summand of $Z^{i}_{x,n}$; the finite sum of word polynomials being evaluated pointwise (Word Polynomials in Unitary Letters: the Normed *-Algebra of Finitely Supported Word Functions §polynomials), claim 4 of Properties of a Sum over a Finite Index Set gives $Z^{i}_{p,n}=s\,Z^{i}_{x,n}$. By Word Polynomials under a Unitary Law: Linearity, Traciality, Positivity, the Cauchy-Schwarz Inequality and the l1 Bound §algebra, $(s\,Z^{i}_{x,n})^{*}=\overline{s}\,(Z^{i}_{x,n})^{*}=s\,(Z^{i}_{x,n})^{*}$, and hence, sums and multiples being pointwise (Word Polynomials in Unitary Letters: the Normed *-Algebra of Finitely Supported Word Functions §polynomials), $\Xi^{i}_{p,n}=\tfrac12\bigl(s\,Z^{i}_{x,n}+s\,(Z^{i}_{x,n})^{*}\bigr)=s\,\Xi^{i}_{x,n}$. By Word Polynomials under a Unitary Law: Linearity, Traciality, Positivity, the Cauchy-Schwarz Inequality and the l1 Bound §algebra again, $\Xi^{i}_{p,n}\Xi^{i}_{p,n}=s^{2}\,\Xi^{i}_{x,n}\Xi^{i}_{x,n}$, and by Word Polynomials under a Unitary Law: Linearity, Traciality, Positivity, the Cauchy-Schwarz Inequality and the l1 Bound §linear, $\lambda\bigl(\Xi^{i}_{p,n}\Xi^{i}_{p,n}\bigr)=s^{2}\lambda\bigl(\Xi^{i}_{x,n}\Xi^{i}_{x,n}\bigr)$ for every $\lambda\in\mathcal{L}_{d}$. Writing, for $q\in\{p,x\}$,

$$T_{n}(q)=\sum_{i\in[d]}\Bigl(\hat\mu\bigl(\Xi^{i}_{q,n}\Xi^{i}_{q,n}\bigr)-\hat\nu\bigl(\Xi^{i}_{q,n}\Xi^{i}_{q,n}\bigr)\Bigr),$$

real numbers by the preamble of The Quadratic Control Hamiltonian: Legendre Formula, Bounds, Differences in the Law and Admissibility, claims 3 and 4 of Properties of a Sum over a Finite Index Set give $T_{n}(p)=s^{2}\,T_{n}(x)$ for every $n\in\mathbb{N}$.

Step 6 (The Hamiltonian term). By The Quadratic Control Hamiltonian: Legendre Formula, Bounds, Differences in the Law and Admissibility §difference with $\hat\mu,\hat\nu$ and $p$, the sequence $(T_{n}(p))_{n\in\mathbb{N}}$ converges and $H_{Q}(\hat\nu,p)-H_{Q}(\hat\mu,p)=-\frac12\lim_{n\to\infty}T_{n}(p)$. By The Trilinear Estimate for Cyclic Gradients of Differences of Unitary Laws §trilinear with $\hat\mu,\hat\nu$ (so that its $x$ is our $x$), $|T_{n}(x)|\le K\lVert x\rVert_{d}\lVert x\rVert_{d,1}^{2}$, hence $T_{n}(x)\ge-|T_{n}(x)|\ge-K\lVert x\rVert_{d}\lVert x\rVert_{d,1}^{2}$ and, as $s^{2}\ge0$, $T_{n}(p)\ge-s^{2}K\lVert x\rVert_{d}\lVert x\rVert_{d,1}^{2}$ for every $n$. The constant sequence with this value converges to it (Limit of a Sequence of Real Numbers), so claim 1 of Order Properties of Limits of Real Sequences gives $\lim_{n\to\infty}T_{n}(p)\ge-s^{2}K\lVert x\rVert_{d}\lVert x\rVert_{d,1}^{2}$. Using $\lVert x\rVert_{d}\le a$, $\frac12s^{2}K\lVert x\rVert_{d,1}^{2}\ge0$ and $s\,a=L$,

$$H_{Q}(\hat\nu,p)-H_{Q}(\hat\mu,p)\le\frac{s^{2}K}{2}\lVert x\rVert_{d}\lVert x\rVert_{d,1}^{2}\le\frac{s^{2}K a}{2}\lVert x\rVert_{d,1}^{2}=\frac{s\,LK}{2}\lVert x\rVert_{d,1}^{2}.\tag{H}$$

Step 7 (The noise term). By The Free Unitary Heat Generator: Bound, Continuity and Strict Dissipation in the Length-Weighted Gauge §dissipation with $\hat\mu,\hat\nu$, $\langle x,\widehat{\Theta}\hat\mu-\widehat{\Theta}\hat\nu\rangle_{d}\le-\frac14\lVert x\rVert_{d,1}^{2}$; as $\beta^{2}s/2\ge0$,

$$\frac{\beta^{2}s}{2}\bigl\langle x,\widehat{\Theta}\hat\mu-\widehat{\Theta}\hat\nu\bigr\rangle_{d}\le-\frac{\beta^{2}s}{8}\lVert x\rVert_{d,1}^{2}.\tag{N}$$

Step 8 (The source term). By the hypothesis on $f$, $\lVert x\rVert_{d}\le a$, $L_{f}\ge0$, $L_{f}\le\rho L$ and $a>0$,

$$f(\hat\mu)-f(\hat\nu)\le|f(\hat\mu)-f(\hat\nu)|\le L_{f}\,\lVert x\rVert_{d}\le L_{f}\,a\le\rho L a.\tag{F}$$

Step 9 (The maximum of the penalised difference). Inserting (F), (H) and (N) into (D),

$$\rho\bigl(u(\hat\mu)-v(\hat\nu)\bigr)\le\rho L a+\frac{s}{2}\Bigl(LK-\frac{\beta^{2}}{4}\Bigr)\lVert x\rVert_{d,1}^{2}\le\rho L a,$$

the last step because $LK\le\beta^{2}/4$, $s/2\ge0$ and $\lVert x\rVert_{d,1}^{2}\ge0$. Dividing by $\rho>0$ gives $u(\hat\mu)-v(\hat\nu)\le L\varphi(\hat\mu,\hat\nu)$, that is $\Psi(\hat\mu,\hat\nu)\le0$. By (MX) and (E2), multiplied by $L\ge0$,

$$u(\mu)-v(\nu)\le L\,\varphi(\mu,\nu)\le L\,d_{\mathcal{L}}(\mu,\nu)+L\sqrt{\varepsilon}\qquad(\mu,\nu\in\mathcal{L}_{d}).\tag{C}$$

Step 10 (Removing $\varepsilon$). Steps 1 to 9 apply to every real $\varepsilon>0$, so (C) holds for every such $\varepsilon$. Let $\mu,\nu\in\mathcal{L}_{d}$ and let $\varepsilon''>0$ be real. Apply (C) with $\varepsilon=\bigl(\varepsilon''/(L+1)\bigr)^{2}>0$, whose square root is $\varepsilon''/(L+1)$ by the uniqueness in Existence and Uniqueness of the Nonnegative Square Root; since $0\le L/(L+1)\le1$, $L\sqrt{\varepsilon}=\varepsilon''\cdot\frac{L}{L+1}\le\varepsilon''$, and so $u(\mu)-v(\nu)\le L\,d_{\mathcal{L}}(\mu,\nu)+\varepsilon''$. By Comparison of Real Numbers with Arbitrary Positive Slack §slack-above, $u(\mu)-v(\nu)\le L\,d_{\mathcal{L}}(\mu,\nu)$. This proves clause 1.

Part 2 (Comparison, clause 2). Put $L_{0}=L_{f}/\rho$. Dividing the hypothesis $L_{f}\le\rho\beta^{2}/(4K)$ by $\rho>0$ gives $L_{f}/\rho\le L_{0}\le\beta^{2}/(4K)$, so clause 1 applies with $L=L_{0}$. Let $u,v$ be as in clause 1 and $\lambda\in\mathcal{L}_{d}$. Clause 1 with $\mu=\nu=\lambda$ gives $u(\lambda)-v(\lambda)\le L_{0}\,d_{\mathcal{L}}(\lambda,\lambda)=0$, the metric vanishing on the diagonal. Hence $u(\lambda)\le v(\lambda)$.

Part 3 (Uniqueness, clause 3). Let $V$ and $W$ be viscosity solutions. By Viscosity Sub- and Supersolutions of the Hamilton-Jacobi-Bellman Equation on Unitary Laws §solution each is a viscosity subsolution and a viscosity supersolution. Clause 2 with $(u,v)=(V,W)$ gives $V\le W$, and with $(u,v)=(W,V)$ gives $W\le V$; hence $V(\lambda)=W(\lambda)$ for every $\lambda\in\mathcal{L}_{d}$ by antisymmetry of the order, that is $V=W$.

Part 4 (Lipschitz bound, clause 4). Let $V$ be a viscosity solution, so a viscosity subsolution and a viscosity supersolution (Viscosity Sub- and Supersolutions of the Hamilton-Jacobi-Bellman Equation on Unitary Laws §solution), and let $\mu,\nu\in\mathcal{L}_{d}$. With $L_{0}=L_{f}/\rho$ admissible in clause 1 as shown in Part 2, clause 1 with $u=v=V$ gives $V(\mu)-V(\nu)\le L_{0}\,d_{\mathcal{L}}(\mu,\nu)$ and $V(\nu)-V(\mu)\le L_{0}\,d_{\mathcal{L}}(\nu,\mu)=L_{0}\,d_{\mathcal{L}}(\mu,\nu)$, by symmetry of the metric. As $|V(\mu)-V(\nu)|$ is one of $V(\mu)-V(\nu)$ and $V(\nu)-V(\mu)$, $|V(\mu)-V(\nu)|\le(L_{f}/\rho)\,d_{\mathcal{L}}(\mu,\nu)$.