## Formalized Name Uniqueness of Bounded Uniformly Continuous Viscosity Solutions Under Spatial Infinity Compatibility ## Formalized Statement Let $T>0$, let $n\in\mathbb N$, set $Q=(0,T)\times\mathbb R^n$ and $\overline Q=[0,T]\times\mathbb R^n$, and let $\lambda>0$. Let $H:[0,T]\times\mathbb R^n\times\mathbb R^n\to\mathbb R$ be continuous. Assume that for every $R>0$ there exists a constant $L_R>0$ such that for every $t\in[0,T]$, every $x,y\in\overline{B}(0,R)$, and every $p,q\in\mathbb R^n$, \begin{align*} |H(t,x,p)-H(t,y,q)|\le L_R\bigl((1+|p|+|q|)|x-y|+|p-q|\bigr). \end{align*} Assume also that for every $R>0$ there exists a modulus of continuity $\omega_R:[0,\infty)\to[0,\infty)$, with $\omega_R(r)\to0$ as $r\to0$, such that for every $s,t\in[0,T]$, every $x\in\overline{B}(0,R)$, and every $p\in\mathbb R^n$, \begin{align*} |H(t,x,p)-H(s,x,p)|\le \omega_R(|t-s|). \end{align*} Let $g\in C_b(\mathbb R^n)$ be uniformly continuous. Let $\mathcal C$ be a class of functions $u:\overline Q\to\mathbb R$ such that every $u\in\mathcal C$ is bounded and uniformly continuous on $\overline Q$, is a viscosity solution in $Q$ of \begin{align*} \lambda u(t,x)-\partial_t u(t,x)+H(t,x,\nabla u(t,x))=0, \end{align*} and satisfies the terminal condition \begin{align*} u(T,x)=g(x) \end{align*} for every $x\in\mathbb R^n$. Assume moreover that $\mathcal C$ satisfies the following pairwise compatibility condition at spatial infinity: for every $u,v\in\mathcal C$, \begin{align*} \limsup_{R\to\infty}\sup\{u(t,x)-v(t,x):t\in[0,T],\ |x|\ge R\}\le 0 \end{align*} and \begin{align*} \limsup_{R\to\infty}\sup\{v(t,x)-u(t,x):t\in[0,T],\ |x|\ge R\}\le 0. \end{align*} Then $\mathcal C$ contains at most one function. Equivalently, if $u,v\in\mathcal C$, then \begin{align*} u(t,x)=v(t,x) \end{align*} for every $(t,x)\in\overline Q$. ## Proof [proofplan] Fix two functions $u,v\in\mathcal C$. We prove $u\le v$ by time reversal, reducing the terminal-value problem to an initial-value comparison problem for bounded uniformly continuous viscosity solutions. The spatial-infinity compatibility condition localizes any positive maximum of the difference to a bounded spatial ball. A doubling-of-variables argument, with a small penalization near the artificial terminal time in the reversed variable, gives viscosity inequalities at nearby points; the local Lipschitz hypothesis on $H$ then forces $\lambda\sup(u-v)\le 0$, a contradiction if the supremum is positive. Reversing the roles of $u$ and $v$ gives equality. [/proofplan] [step:Reverse time and rewrite the equation as an initial-value problem] Define the reversed time variable $\tau\in[0,T]$ by $\tau=T-t$. Define \begin{align*} U:[0,T]\times\mathbb R^n \to \mathbb R,\quad (\tau,x)\mapsto u(T-\tau,x) \end{align*} and \begin{align*} V:[0,T]\times\mathbb R^n \to \mathbb R,\quad (\tau,x)\mapsto v(T-\tau,x) \end{align*} Define also the reversed Hamiltonian \begin{align*} \widetilde H:[0,T]\times\mathbb R^n\times\mathbb R^n \to \mathbb R,\quad (\tau,x,p)\mapsto H(T-\tau,x,p) \end{align*} Since $u$ and $v$ are bounded and uniformly continuous on $\overline Q$, the functions $U$ and $V$ are bounded and uniformly continuous on $[0,T]\times\mathbb R^n$. The terminal condition gives \begin{align*} U(0,x)=V(0,x)=g(x) \end{align*} for every $x\in\mathbb R^n$. The viscosity equation for $u$ transforms into \begin{align*} \partial_\tau U(\tau,x)+\lambda U(\tau,x)+\widetilde H(\tau,x,\nabla U(\tau,x))=0 \end{align*} on $(0,T)\times\mathbb R^n$, and the same equation holds for $V$. Thus $U$ is a viscosity subsolution and $V$ is a viscosity supersolution of this initial-value Hamilton-Jacobi equation. The local Lipschitz hypothesis is inherited by $\widetilde H$: for every $R>0$ there is $L_R>0$ such that for every $\tau\in[0,T]$, every $x,y\in\overline{B}(0,R)$, and every $p,q\in\mathbb R^n$, \begin{align*} |\widetilde H(\tau,x,p)-\widetilde H(\tau,y,q)|\le L_R\bigl((1+|p|+|q|)|x-y|+|p-q|\bigr) \end{align*} The time-continuity hypothesis is also inherited: for every $R>0$ there is a modulus $\omega_R$ such that for every $\tau,\sigma\in[0,T]$, every $x\in\overline{B}(0,R)$, and every $p\in\mathbb R^n$, \begin{align*} |\widetilde H(\tau,x,p)-\widetilde H(\sigma,x,p)|\le \omega_R(|\tau-\sigma|) \end{align*} [/step] [step:Localize a hypothetical positive maximum in a bounded spatial region] Assume for contradiction that \begin{align*} M:=\sup_{(\tau,x)\in[0,T]\times\mathbb R^n}\bigl(U(\tau,x)-V(\tau,x)\bigr)>0. \end{align*} The spatial-infinity compatibility condition for $u$ and $v$ becomes \begin{align*} \limsup_{R\to\infty}\sup\{U(\tau,x)-V(\tau,x):\tau\in[0,T],\ |x|\ge R\}\le 0. \end{align*} Choose $R_0>0$ so large that \begin{align*} \sup\{U(\tau,x)-V(\tau,x):\tau\in[0,T],\ |x|\ge R_0\}\le \frac{M}{4}. \end{align*} Therefore every point at which $U-V$ is within $M/4$ of its supremum lies in the bounded cylinder $[0,T]\times B(0,R_0)$. Because $U(0,x)=V(0,x)$ for all $x\in\mathbb R^n$, the positive supremum $M$ cannot be attained or approached at the initial time alone. More precisely, there exist $\tau_0\in(0,T]$ and $x_0\in B(0,R_0)$ such that \begin{align*} U(\tau_0,x_0)-V(\tau_0,x_0)>\frac{3M}{4}. \end{align*} [/step] [step:Double the variables and force the maximum into the viscosity region] Fix positive parameters $\alpha,\beta,\delta>0$. Define the penalized function \begin{align*} \Phi_{\alpha,\beta,\delta}:[0,T)\times[0,T)\times\mathbb R^n\times\mathbb R^n \to \mathbb R, \quad (\tau,\sigma,x,y)\mapsto U(\tau,x)-V(\sigma,y)-\frac{|x-y|^2}{2\alpha}-\frac{|\tau-\sigma|^2}{2\alpha}-\beta(|x|^2+|y|^2)-\frac{\delta}{T-\tau}-\frac{\delta}{T-\sigma} \end{align*} The terms $-\delta/(T-\tau)$ and $-\delta/(T-\sigma)$ force maximizing points away from the artificial terminal face $\tau=T$ or $\sigma=T$. The quadratic spatial penalty forces existence of a maximum in space. Hence $\Phi_{\alpha,\beta,\delta}$ attains a maximum at some point \begin{align*} (\tau_{\alpha,\beta,\delta},\sigma_{\alpha,\beta,\delta},x_{\alpha,\beta,\delta},y_{\alpha,\beta,\delta})\in[0,T)\times[0,T)\times\mathbb R^n\times\mathbb R^n \end{align*} For $\alpha,\beta,\delta$ sufficiently small, the maximum value is positive, since it is bounded below by evaluating $\Phi_{\alpha,\beta,\delta}$ at $(\tau_0,\tau_0,x_0,x_0)$ and using $U(\tau_0,x_0)-V(\tau_0,x_0)>3M/4$. We now choose the parameters in the ordered limiting regime $\alpha\downarrow0$, then $\beta\downarrow0$, then $\delta\downarrow0$. For all sufficiently small parameters in this regime, neither $\tau_{\alpha,\beta,\delta}$ nor $\sigma_{\alpha,\beta,\delta}$ equals $0$. To justify this, suppose that there are parameters tending to $0$ in the ordered regime and maximizing points with $\tau_{\alpha,\beta,\delta}=0$. Maximality gives \begin{align*} \Phi_{\alpha,\beta,\delta}(0,\sigma_{\alpha,\beta,\delta},x_{\alpha,\beta,\delta},y_{\alpha,\beta,\delta})\ge \Phi_{\alpha,\beta,\delta}(\tau_0,\tau_0,x_0,x_0). \end{align*} The right-hand side tends to $U(\tau_0,x_0)-V(\tau_0,x_0)>3M/4$ after the ordered limits. Hence the left-hand side stays positive. Since the terms $-|x-y|^2/(2\alpha)$, $-|\tau-\sigma|^2/(2\alpha)$, $-\beta(|x|^2+|y|^2)$, $-\delta/(T-\tau)$, and $-\delta/(T-\sigma)$ are non-positive, positivity implies that \begin{align*} g(x_{\alpha,\beta,\delta})-V(\sigma_{\alpha,\beta,\delta},y_{\alpha,\beta,\delta})>0. \end{align*} The same maximality inequality also forces \begin{align*} \frac{|x_{\alpha,\beta,\delta}-y_{\alpha,\beta,\delta}|^2}{2\alpha}\to0 \end{align*} and \begin{align*} \frac{|\sigma_{\alpha,\beta,\delta}|^2}{2\alpha}\to0 \end{align*} in the ordered limit; otherwise the non-positive penalties would prevent the value from remaining above $3M/4$. Thus $|x_{\alpha,\beta,\delta}-y_{\alpha,\beta,\delta}|\to0$ and $\sigma_{\alpha,\beta,\delta}\to0$. Uniform continuity of $g$ on $\mathbb R^n$ and of $V$ on $[0,T]\times\mathbb R^n$, together with $V(0,y)=g(y)$, gives \begin{align*} g(x_{\alpha,\beta,\delta})-V(\sigma_{\alpha,\beta,\delta},y_{\alpha,\beta,\delta})\to0, \end{align*} contradicting the positive lower bound. The case $\sigma_{\alpha,\beta,\delta}=0$ is identical, using $U(0,x)=g(x)$. Therefore, for the small parameters used below, the maximizing points lie in $(0,T)\times(0,T)\times\mathbb R^n\times\mathbb R^n$, where the viscosity inequalities are available. [guided] The purpose of the penalized function is to manufacture a genuine maximum at points where the viscosity definitions may be used. On an unbounded spatial domain, $U-V$ need not attain its supremum. The term \begin{align*} -\beta(|x|^2+|y|^2) \end{align*} makes the expression tend to $-\infty$ as $|x|+|y|\to\infty$, so a maximum exists in the spatial variables. The terms \begin{align*} -\frac{\delta}{T-\tau} \end{align*} and \begin{align*} -\frac{\delta}{T-\sigma} \end{align*} make the expression tend to $-\infty$ as either reversed time variable approaches $T$. This is necessary because the reversed problem has no boundary condition at $\tau=T$, corresponding to the original time slice $t=0$. It remains to exclude the initial faces before applying viscosity inequalities. Suppose, for example, that a maximizing point had $\tau_{\alpha,\beta,\delta}=0$ along parameters tending to $0$ in the ordered regime $\alpha\downarrow0$, then $\beta\downarrow0$, then $\delta\downarrow0$. Evaluating the maximum against the interior point $(\tau_0,\tau_0,x_0,x_0)$ gives \begin{align*} \Phi_{\alpha,\beta,\delta}(0,\sigma_{\alpha,\beta,\delta},x_{\alpha,\beta,\delta},y_{\alpha,\beta,\delta})\ge \Phi_{\alpha,\beta,\delta}(\tau_0,\tau_0,x_0,x_0). \end{align*} The right-hand side tends to $U(\tau_0,x_0)-V(\tau_0,x_0)$, which is larger than $3M/4$. Thus the left-hand side must remain positive. But at $\tau=0$ the initial condition gives $U(0,x)=g(x)$, so the non-penalized part is \begin{align*} g(x_{\alpha,\beta,\delta})-V(\sigma_{\alpha,\beta,\delta},y_{\alpha,\beta,\delta}). \end{align*} Because every penalty in $\Phi_{\alpha,\beta,\delta}$ is non-positive, the value cannot stay positive unless this difference stays positive. The same lower bound also forces the quadratic penalties to vanish in the limit: \begin{align*} \frac{|x_{\alpha,\beta,\delta}-y_{\alpha,\beta,\delta}|^2}{2\alpha}\to0 \end{align*} and \begin{align*} \frac{|\sigma_{\alpha,\beta,\delta}|^2}{2\alpha}\to0. \end{align*} Hence $x_{\alpha,\beta,\delta}-y_{\alpha,\beta,\delta}\to0$ and $\sigma_{\alpha,\beta,\delta}\to0$. Since $g$ is uniformly continuous on $\mathbb R^n$, $V$ is uniformly continuous on $[0,T]\times\mathbb R^n$, and $V(0,y)=g(y)$, we obtain \begin{align*} g(x_{\alpha,\beta,\delta})-V(\sigma_{\alpha,\beta,\delta},y_{\alpha,\beta,\delta})\to0. \end{align*} This contradicts the positive lower bound. The same argument excludes $\sigma_{\alpha,\beta,\delta}=0$, using $U(0,x)=g(x)$. Thus the maximizing points used in the comparison are genuinely interior in time. [/guided] [/step] [step:Apply the viscosity inequalities at the doubled maximum] At the maximum point, define \begin{align*} p_{\alpha,\beta,\delta}:=\frac{x_{\alpha,\beta,\delta}-y_{\alpha,\beta,\delta}}{\alpha}+2\beta x_{\alpha,\beta,\delta},\qquad q_{\alpha,\beta,\delta}:=\frac{x_{\alpha,\beta,\delta}-y_{\alpha,\beta,\delta}}{\alpha}-2\beta y_{\alpha,\beta,\delta} \end{align*} Also define the time derivatives of the test functions by \begin{align*} a_{\alpha,\beta,\delta}:=\frac{\tau_{\alpha,\beta,\delta}-\sigma_{\alpha,\beta,\delta}}{\alpha}+\frac{\delta}{(T-\tau_{\alpha,\beta,\delta})^2},\qquad b_{\alpha,\beta,\delta}:=\frac{\tau_{\alpha,\beta,\delta}-\sigma_{\alpha,\beta,\delta}}{\alpha}-\frac{\delta}{(T-\sigma_{\alpha,\beta,\delta})^2} \end{align*} We now spell out the elementary doubling argument that produces the one-variable test functions. With $(\sigma_{\alpha,\beta,\delta},y_{\alpha,\beta,\delta})$ fixed, define a smooth function $\varphi_U:(0,T)\times\mathbb R^n\to\mathbb R$ by \begin{align*} \varphi_U(\tau,x):=V(\sigma_{\alpha,\beta,\delta},y_{\alpha,\beta,\delta})+\frac{|x-y_{\alpha,\beta,\delta}|^2}{2\alpha}+\frac{|\tau-\sigma_{\alpha,\beta,\delta}|^2}{2\alpha}+\beta(|x|^2+|y_{\alpha,\beta,\delta}|^2)+\frac{\delta}{T-\tau}+\frac{\delta}{T-\sigma_{\alpha,\beta,\delta}}. \end{align*} The maximality of $\Phi_{\alpha,\beta,\delta}$ says precisely that $U-\varphi_U$ has a local maximum at $(\tau_{\alpha,\beta,\delta},x_{\alpha,\beta,\delta})$. The time derivative of $\varphi_U$ there is $a_{\alpha,\beta,\delta}$ and the spatial gradient is $p_{\alpha,\beta,\delta}$. Since $U$ is a viscosity subsolution, \begin{align*} a_{\alpha,\beta,\delta}+\lambda U(\tau_{\alpha,\beta,\delta},x_{\alpha,\beta,\delta})+\widetilde H(\tau_{\alpha,\beta,\delta},x_{\alpha,\beta,\delta},p_{\alpha,\beta,\delta})\le 0. \end{align*} Similarly, with $(\tau_{\alpha,\beta,\delta},x_{\alpha,\beta,\delta})$ fixed, define $\varphi_V:(0,T)\times\mathbb R^n\to\mathbb R$ by \begin{align*} \varphi_V(\sigma,y):=U(\tau_{\alpha,\beta,\delta},x_{\alpha,\beta,\delta})-\frac{|x_{\alpha,\beta,\delta}-y|^2}{2\alpha}-\frac{|\tau_{\alpha,\beta,\delta}-\sigma|^2}{2\alpha}-\beta(|x_{\alpha,\beta,\delta}|^2+|y|^2)-\frac{\delta}{T-\tau_{\alpha,\beta,\delta}}-\frac{\delta}{T-\sigma}. \end{align*} The maximality of $\Phi_{\alpha,\beta,\delta}$ says that $V-\varphi_V$ has a local minimum at $(\sigma_{\alpha,\beta,\delta},y_{\alpha,\beta,\delta})$. The time derivative of $\varphi_V$ there is $b_{\alpha,\beta,\delta}$ and the spatial gradient is $q_{\alpha,\beta,\delta}$. Since $V$ is a viscosity supersolution, \begin{align*} b_{\alpha,\beta,\delta}+\lambda V(\sigma_{\alpha,\beta,\delta},y_{\alpha,\beta,\delta})+\widetilde H(\sigma_{\alpha,\beta,\delta},y_{\alpha,\beta,\delta},q_{\alpha,\beta,\delta})\ge 0. \end{align*} Subtracting the second inequality from the first gives \begin{align*} \lambda\bigl(U(\tau_{\alpha,\beta,\delta},x_{\alpha,\beta,\delta})-V(\sigma_{\alpha,\beta,\delta},y_{\alpha,\beta,\delta})\bigr) \le b_{\alpha,\beta,\delta}-a_{\alpha,\beta,\delta}+\widetilde H(\sigma_{\alpha,\beta,\delta},y_{\alpha,\beta,\delta},q_{\alpha,\beta,\delta})-\widetilde H(\tau_{\alpha,\beta,\delta},x_{\alpha,\beta,\delta},p_{\alpha,\beta,\delta}). \end{align*} By the definitions of $a_{\alpha,\beta,\delta}$ and $b_{\alpha,\beta,\delta}$, \begin{align*} b_{\alpha,\beta,\delta}-a_{\alpha,\beta,\delta}=-\frac{\delta}{(T-\sigma_{\alpha,\beta,\delta})^2}-\frac{\delta}{(T-\tau_{\alpha,\beta,\delta})^2}\le 0. \end{align*} Therefore \begin{align*} \lambda\bigl(U(\tau_{\alpha,\beta,\delta},x_{\alpha,\beta,\delta})-V(\sigma_{\alpha,\beta,\delta},y_{\alpha,\beta,\delta})\bigr) \le \widetilde H(\sigma_{\alpha,\beta,\delta},y_{\alpha,\beta,\delta},q_{\alpha,\beta,\delta})-\widetilde H(\tau_{\alpha,\beta,\delta},x_{\alpha,\beta,\delta},p_{\alpha,\beta,\delta}). \end{align*} [/step] [step:Prove the doubled-variable estimates and send the Hamiltonian error to zero] We record the limiting estimates in a precise ordered form. Let \begin{align*} Z_{\alpha,\beta,\delta}:=(\tau_{\alpha,\beta,\delta},\sigma_{\alpha,\beta,\delta},x_{\alpha,\beta,\delta},y_{\alpha,\beta,\delta}) \end{align*} denote the maximizing point, and write \begin{align*} P_{\alpha,\beta,\delta}:=\frac{|x_{\alpha,\beta,\delta}-y_{\alpha,\beta,\delta}|^2}{2\alpha}+\frac{|\tau_{\alpha,\beta,\delta}-\sigma_{\alpha,\beta,\delta}|^2}{2\alpha}. \end{align*} For each $\eta>0$, the definition of $M$ and continuity of $U-V$ on $[0,T]\times\mathbb R^n$ give a point $(\tau_\eta,x_\eta)\in[0,T)\times\mathbb R^n$ such that \begin{align*} U(\tau_\eta,x_\eta)-V(\tau_\eta,x_\eta)>M-\eta. \end{align*} If an approximating point initially lies at $\tau=T$, continuity in time allows replacing it by a nearby point with time coordinate strictly smaller than $T$. Evaluating the maximum at $(\tau_\eta,\tau_\eta,x_\eta,x_\eta)$ yields \begin{align*} \Phi_{\alpha,\beta,\delta}(Z_{\alpha,\beta,\delta})\ge U(\tau_\eta,x_\eta)-V(\tau_\eta,x_\eta)-2\beta |x_\eta|^2-\frac{2\delta}{T-\tau_\eta}. \end{align*} Taking first $\alpha\downarrow0$, then $\beta\downarrow0$, then $\delta\downarrow0$, and then $\eta\downarrow0$, gives \begin{align*} \liminf_{\delta\downarrow0}\liminf_{\beta\downarrow0}\liminf_{\alpha\downarrow0}\Phi_{\alpha,\beta,\delta}(Z_{\alpha,\beta,\delta})\ge M. \end{align*} Conversely, since the penalties in $\Phi_{\alpha,\beta,\delta}$ are non-positive and $M$ is the supremum of $U-V$ after identifying nearby doubled variables only by deletion of penalties, \begin{align*} \Phi_{\alpha,\beta,\delta}(Z_{\alpha,\beta,\delta})\le U(\tau_{\alpha,\beta,\delta},x_{\alpha,\beta,\delta})-V(\sigma_{\alpha,\beta,\delta},y_{\alpha,\beta,\delta}). \end{align*} To control the mismatch between $(\tau,x)$ and $(\sigma,y)$, compare $\Phi_{\alpha,\beta,\delta}(Z_{\alpha,\beta,\delta})$ with $\Phi_{\alpha,\beta,\delta}(\tau_{\alpha,\beta,\delta},\tau_{\alpha,\beta,\delta},x_{\alpha,\beta,\delta},x_{\alpha,\beta,\delta})$. Maximality gives \begin{align*} P_{\alpha,\beta,\delta}\le V(\tau_{\alpha,\beta,\delta},x_{\alpha,\beta,\delta})-V(\sigma_{\alpha,\beta,\delta},y_{\alpha,\beta,\delta})+\beta(|x_{\alpha,\beta,\delta}|^2-|y_{\alpha,\beta,\delta}|^2)+\frac{\delta}{T-\tau_{\alpha,\beta,\delta}}-\frac{\delta}{T-\sigma_{\alpha,\beta,\delta}}. \end{align*} Using instead the comparison point $(\sigma_{\alpha,\beta,\delta},\sigma_{\alpha,\beta,\delta},y_{\alpha,\beta,\delta},y_{\alpha,\beta,\delta})$ gives the symmetric bound with $U$ in place of $V$. Since $U$ and $V$ are bounded and uniformly continuous on $[0,T]\times\mathbb R^n$, these two maximality inequalities imply, in the ordered limiting regime, that \begin{align*} P_{\alpha,\beta,\delta}\to0. \end{align*} Hence \begin{align*} |x_{\alpha,\beta,\delta}-y_{\alpha,\beta,\delta}|\to0 \end{align*} and \begin{align*} |\tau_{\alpha,\beta,\delta}-\sigma_{\alpha,\beta,\delta}|\to0. \end{align*} Uniform continuity of $V$ then gives \begin{align*} \bigl(U(\tau_{\alpha,\beta,\delta},x_{\alpha,\beta,\delta})-V(\sigma_{\alpha,\beta,\delta},y_{\alpha,\beta,\delta})\bigr)-\bigl(U(\tau_{\alpha,\beta,\delta},x_{\alpha,\beta,\delta})-V(\tau_{\alpha,\beta,\delta},x_{\alpha,\beta,\delta})\bigr)\to0 \end{align*} Therefore the preceding lower bound for the penalized maxima implies \begin{align*} U(\tau_{\alpha,\beta,\delta},x_{\alpha,\beta,\delta})-V(\sigma_{\alpha,\beta,\delta},y_{\alpha,\beta,\delta})\to M \end{align*} in the same ordered limiting sense. We next justify the bounded spatial localization. Choose $R_1>R_0+1$ so large that the compatibility condition for the ordered pair $(u,v)$ gives \begin{align*} \sup\{U(\tau,x)-V(\tau,x):\tau\in[0,T],\ |x|\ge R_1-1\}\le \frac{M}{4}. \end{align*} If $|x_{\alpha,\beta,\delta}|\ge R_1$ along a sufficiently small ordered sequence, then the estimate $|x_{\alpha,\beta,\delta}-y_{\alpha,\beta,\delta}|\to0$ gives $|y_{\alpha,\beta,\delta}|\ge R_1-1$ eventually, and uniform continuity of $V$ gives \begin{align*} U(\tau_{\alpha,\beta,\delta},x_{\alpha,\beta,\delta})-V(\sigma_{\alpha,\beta,\delta},y_{\alpha,\beta,\delta})\le \frac{M}{4}+o(1), \end{align*} contradicting convergence of the unpenalized doubled difference to $M$. Hence, after discarding sufficiently small parameters, $x_{\alpha,\beta,\delta},y_{\alpha,\beta,\delta}\in\overline{B}(0,R_1)$. This is the only spatial-infinity condition needed for the comparison $u\le v$; the opposite condition will be used when the roles of $u$ and $v$ are interchanged. For the relevant points write $x=x_{\alpha,\beta,\delta}$, $y=y_{\alpha,\beta,\delta}$, $\tau=\tau_{\alpha,\beta,\delta}$, $\sigma=\sigma_{\alpha,\beta,\delta}$, $p=p_{\alpha,\beta,\delta}$, and $q=q_{\alpha,\beta,\delta}$. Split the Hamiltonian difference as \begin{align*} \widetilde H(\sigma,y,q)-\widetilde H(\tau,x,p)=\bigl(\widetilde H(\sigma,y,q)-\widetilde H(\sigma,x,p)\bigr)+\bigl(\widetilde H(\sigma,x,p)-\widetilde H(\tau,x,p)\bigr) \end{align*} On $\overline{B}(0,R_1)$, the spatial-gradient Lipschitz estimate gives \begin{align*} |\widetilde H(\sigma,y,q)-\widetilde H(\sigma,x,p)|\le L_{R_1}\bigl((1+|p|+|q|)|x-y|+|p-q|\bigr) \end{align*} By definition, \begin{align*} p-q=2\beta x+2\beta y \end{align*} so $|p-q|\to0$ as $\beta\downarrow0$ because $x,y\in\overline{B}(0,R_1)$. Also \begin{align*} (|p|+|q|)|x-y|\le 2\frac{|x-y|^2}{\alpha}+2\beta(|x|+|y|)|x-y| \end{align*} and the right-hand side tends to $0$ by the doubled-variable estimate and boundedness of $x,y$. Hence \begin{align*} \widetilde H(\sigma,y,q)-\widetilde H(\sigma,x,p)\to0 \end{align*} The time-continuity hypothesis gives the remaining estimate, uniformly in the possibly unbounded gradient $p$: \begin{align*} |\widetilde H(\sigma,x,p)-\widetilde H(\tau,x,p)|\le \omega_{R_1}(|\sigma-\tau|)\to0 \end{align*} Consequently \begin{align*} \widetilde H(\sigma_{\alpha,\beta,\delta},y_{\alpha,\beta,\delta},q_{\alpha,\beta,\delta})-\widetilde H(\tau_{\alpha,\beta,\delta},x_{\alpha,\beta,\delta},p_{\alpha,\beta,\delta})\to0 \end{align*} Passing to the ordered limit in the viscosity inequality yields \begin{align*} \lambda M\le 0 \end{align*} Since $\lambda>0$ and $M>0$, this is impossible. [/step] [step:Reverse the comparison and conclude uniqueness] The contradiction proves \begin{align*} U(\tau,x)\le V(\tau,x) \end{align*} for every $(\tau,x)\in[0,T]\times\mathbb R^n$. Returning to the original time variable $t=T-\tau$, this gives \begin{align*} u(t,x)\le v(t,x) \end{align*} for every $(t,x)\in\overline Q$. The hypotheses are symmetric in $u$ and $v$: both functions are viscosity solutions with the same terminal datum, and the compatibility assumption includes both spatial-infinity inequalities. Repeating the same argument with $u$ and $v$ interchanged gives \begin{align*} v(t,x)\le u(t,x) \end{align*} for every $(t,x)\in\overline Q$. Hence $u(t,x)=v(t,x)$ for all $(t,x)\in\overline Q$. Since $u,v\in\mathcal C$ were arbitrary, the class $\mathcal C$ contains at most one function. [/step]