[proofplan] We argue by contradiction and suppose that $u-v$ has a positive maximum somewhere before the terminal time. A doubled-variable penalization localizes two nearby space-time points while a small quadratic anchor prevents escape to spatial infinity and keeps the limiting contact point away from $t=T$. Applying the viscosity subsolution and supersolution inequalities at the two penalized contact points gives an inequality with the positive term $\lambda(u-v)$ on the left. The Crandall-Lions continuity condition controls the Hamiltonian difference from moving the base point, and local Lipschitz continuity in the gradient variable controls the small anchor-gradient error; sending the doubling parameter to zero and then the anchor strength to zero contradicts $\lambda > 0$ and the assumed positive gap. No external comparison or doubling lemma is invoked: the compactness, attainment, and collapse estimates are proved directly below. [/proofplan]

[step:Reduce the contradiction to a positive interior-in-time maximum of $u-v$] Define \begin{align*} M := \sup_{(t,x) \in [0,T] \times \mathbb{R}^n} \{u(t,x)-v(t,x)\}. \end{align*} Assume for contradiction that $M > 0$. Since $u$ is upper semicontinuous and $v$ is lower semicontinuous, the function \begin{align*} U: [0,T] \times \mathbb{R}^n \to \mathbb{R} \end{align*} defined by \begin{align*} U(t,x) := u(t,x)-v(t,x) \end{align*} is upper semicontinuous. The spatial-infinity ordering implies that no maximizing sequence for a positive value of $U$ can escape to spatial infinity. Hence there exist $t_0 \in [0,T]$ and $x_0 \in \mathbb{R}^n$ such that \begin{align*} U(t_0,x_0)=M. \end{align*} The terminal ordering gives $U(T,x) \leq 0$ for every $x \in \mathbb{R}^n$, so $t_0 < T$. Fix this point $(t_0,x_0)$ and set $m := U(t_0,x_0) = M > 0$. [/step]

[step:Introduce a coercive doubled-variable maximum problem]For $\beta > 0$ and $\varepsilon > 0$, define \begin{align*} \Phi_{\varepsilon,\beta}: [0,T]^2 \times \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R} \end{align*} by \begin{align*} \Phi_{\varepsilon,\beta}(t,s,x,y) := u(t,x)-v(s,y)-\frac{|x-y|^2}{2\varepsilon}-\frac{|t-s|^2}{2\varepsilon}-\beta |x-x_0|^2-\beta |y-x_0|^2-\beta |t-t_0|^2-\beta |s-t_0|^2. \end{align*} Because $u$ and $v$ are bounded and the two spatial quadratic terms are coercive, $\Phi_{\varepsilon,\beta}$ attains its maximum at some point \begin{align*} (t_{\varepsilon,\beta},s_{\varepsilon,\beta},x_{\varepsilon,\beta},y_{\varepsilon,\beta}) \in [0,T]^2 \times \mathbb{R}^n \times \mathbb{R}^n. \end{align*} Define also the diagonal anchored supremum \begin{align*} M_\beta := \sup_{(t,x) \in [0,T] \times \mathbb{R}^n} \{u(t,x)-v(t,x)-2\beta |x-x_0|^2-2\beta |t-t_0|^2\}. \end{align*} Since the point $(t_0,x_0)$ is admissible, \begin{align*} M_\beta \geq m. \end{align*} Conversely, $u(t,x)-v(t,x) \leq m$ for every $(t,x) \in [0,T] \times \mathbb{R}^n$ by the definition of $m=M$, and the two anchor terms in the definition of $M_\beta$ are non-negative. Hence $M_\beta \leq m$, so \begin{align*} M_\beta = m. \end{align*} The diagonal anchored function defining $M_\beta$ is upper semicontinuous because $u-v$ is upper semicontinuous and the anchor terms are continuous. It also tends to $-\infty$ as $|x| \to \infty$, uniformly in $t \in [0,T]$, because $u$ and $v$ are bounded and $\beta>0$. Since the time interval $[0,T]$ is compact, the supremum $M_\beta$ is attained at some point $(\tau_\beta,z_\beta) \in [0,T] \times \mathbb{R}^n$. Moreover, if $(\tau_\beta,z_\beta)$ maximizes the diagonal anchored function defining $M_\beta$, then \begin{align*} u(\tau_\beta,z_\beta)-v(\tau_\beta,z_\beta)-2\beta |z_\beta-x_0|^2-2\beta |\tau_\beta-t_0|^2=m. \end{align*} Since $u(\tau_\beta,z_\beta)-v(\tau_\beta,z_\beta) \leq m$, both non-negative anchor terms must vanish. Therefore \begin{align*} (\tau_\beta,z_\beta)=(t_0,x_0). \end{align*} For fixed $\beta > 0$, let $\varepsilon_j \downarrow 0$ and choose maximizers $(t_{\varepsilon_j,\beta},s_{\varepsilon_j,\beta},x_{\varepsilon_j,\beta},y_{\varepsilon_j,\beta})$. Coercivity of the spatial anchor terms and boundedness of $u$ and $v$ give a compact subset of $[0,T]^2 \times \mathbb{R}^n \times \mathbb{R}^n$ containing this sequence, after passing to a subsequence. Let its limit be $(\bar t,\bar s,\bar x,\bar y)$. Comparing the maximum value with the diagonal value at $(t_0,x_0)$ gives \begin{align*} u(t_{\varepsilon_j,\beta},x_{\varepsilon_j,\beta})-v(s_{\varepsilon_j,\beta},y_{\varepsilon_j,\beta})-\frac{|x_{\varepsilon_j,\beta}-y_{\varepsilon_j,\beta}|^2}{2\varepsilon_j}-\frac{|t_{\varepsilon_j,\beta}-s_{\varepsilon_j,\beta}|^2}{2\varepsilon_j}-\beta |x_{\varepsilon_j,\beta}-x_0|^2-\beta |y_{\varepsilon_j,\beta}-x_0|^2-\beta |t_{\varepsilon_j,\beta}-t_0|^2-\beta |s_{\varepsilon_j,\beta}-t_0|^2 \geq m. \end{align*} We now prove the doubled-variable collapse directly. Since $u$ is upper semicontinuous and $v$ is lower semicontinuous, the function $(t,s,x,y) \mapsto u(t,x)-v(s,y)$ is upper semicontinuous along the convergent subsequence. Taking the limsup in the preceding maximality inequality gives \begin{align*} m \leq u(\bar t,\bar x)-v(\bar s,\bar y)-\liminf_{j \to \infty}\left(\frac{|x_{\varepsilon_j,\beta}-y_{\varepsilon_j,\beta}|^2}{2\varepsilon_j}+\frac{|t_{\varepsilon_j,\beta}-s_{\varepsilon_j,\beta}|^2}{2\varepsilon_j}\right)-\beta |\bar x-x_0|^2-\beta |\bar y-x_0|^2-\beta |\bar t-t_0|^2-\beta |\bar s-t_0|^2. \end{align*} On the other hand, if either $\bar x \neq \bar y$ or $\bar t \neq \bar s$, then the corresponding quotient tends to $+\infty$, contradicting the finite lower bound $m$. Hence $\bar t=\bar s$ and $\bar x=\bar y$. With this diagonal limit, the same inequality becomes \begin{align*} m \leq u(\bar t,\bar x)-v(\bar t,\bar x)-2\beta |\bar x-x_0|^2-2\beta |\bar t-t_0|^2 \leq M_\beta=m. \end{align*} Therefore $(\bar t,\bar x)$ maximizes the diagonal anchored function, and the preceding paragraph identifies this diagonal maximizer as $(t_0,x_0)$. Returning to the displayed limsup inequality then forces \begin{align*} \lim_{j \to \infty}\left(\frac{|x_{\varepsilon_j,\beta}-y_{\varepsilon_j,\beta}|^2}{\varepsilon_j}+\frac{|t_{\varepsilon_j,\beta}-s_{\varepsilon_j,\beta}|^2}{\varepsilon_j}\right)=0. \end{align*} Thus, along every sequence $\varepsilon \downarrow 0$ and after subsequence extraction if necessary, \begin{align*} (t_{\varepsilon,\beta},s_{\varepsilon,\beta},x_{\varepsilon,\beta},y_{\varepsilon,\beta}) \to (t_0,t_0,x_0,x_0) \end{align*} and the displayed penalty quotient tends to $0$. Because $t_0<T$, for all sufficiently small $\varepsilon > 0$, both $t_{\varepsilon,\beta}$ and $s_{\varepsilon,\beta}$ lie in $[0,T)$.[/step]

[guided]The role of $\Phi_{\varepsilon,\beta}$ is to force two copies of the variables to meet while retaining compactness. The terms \begin{align*} \frac{|x-y|^2}{2\varepsilon} \end{align*} and \begin{align*} \frac{|t-s|^2}{2\varepsilon} \end{align*} penalize separation of the two space-time points. The terms multiplied by $\beta$ anchor both points near the fixed positive maximum point $(t_0,x_0)$ and prevent the spatial variables from escaping to infinity. For fixed $\beta > 0$, boundedness of $u$ and $v$ gives a constant $B > 0$ such that $|u| \leq B$ and $|v| \leq B$ on $[0,T] \times \mathbb{R}^n$. If a maximizing sequence for $\Phi_{\varepsilon,\beta}$ had $|x|+|y| \to \infty$, then the negative term \begin{align*} -\beta |x-x_0|^2-\beta |y-x_0|^2 \end{align*} would force $\Phi_{\varepsilon,\beta} \to -\infty$, contradicting the finite value obtained by evaluating at $(t_0,t_0,x_0,x_0)$. Thus the maximum is attained. Next, compare the maximum value with the diagonal value at any point $(t,x)$. Since \begin{align*} \Phi_{\varepsilon,\beta}(t_{\varepsilon,\beta},s_{\varepsilon,\beta},x_{\varepsilon,\beta},y_{\varepsilon,\beta}) \geq \Phi_{\varepsilon,\beta}(t,t,x,x), \end{align*} taking the supremum over diagonal points gives \begin{align*} \Phi_{\varepsilon,\beta}(t_{\varepsilon,\beta},s_{\varepsilon,\beta},x_{\varepsilon,\beta},y_{\varepsilon,\beta}) \geq M_\beta. \end{align*} The diagonal anchored supremum is actually exactly $m$. Indeed, $M_\beta \geq m$ by evaluating at $(t_0,x_0)$, while $u(t,x)-v(t,x) \leq m$ everywhere and the anchor terms are non-negative, so $M_\beta \leq m$. Thus $M_\beta=m$. The supremum is attained: the diagonal anchored function is upper semicontinuous, and because $u$ and $v$ are bounded while $\beta |x-x_0|^2 \to \infty$ as $|x| \to \infty$, its upper level sets are contained in compact subsets of $[0,T] \times \mathbb{R}^n$. Therefore there is a point $(\tau_\beta,z_\beta) \in [0,T] \times \mathbb{R}^n$ where this supremum is reached. If $(\tau_\beta,z_\beta)$ is such a diagonal maximizer, then \begin{align*} u(\tau_\beta,z_\beta)-v(\tau_\beta,z_\beta)-2\beta |z_\beta-x_0|^2-2\beta |\tau_\beta-t_0|^2=m. \end{align*} Because the first difference is at most $m$, the two non-negative anchor terms must both be zero. Hence $(\tau_\beta,z_\beta)=(t_0,x_0)$. Now take any sequence $\varepsilon_j \downarrow 0$ and a corresponding sequence of maximizers. The coercive anchor terms and boundedness of $u$ and $v$ put the spatial variables in a fixed compact ball, so a subsequence converges to some $(\bar t,\bar s,\bar x,\bar y)$. The maximality inequality against the diagonal point $(t_0,x_0)$ gives a lower bound by $m$. We make the semicontinuity step explicit. Upper semicontinuity of $u$ and lower semicontinuity of $v$ imply \begin{align*} \limsup_{j \to \infty}\{u(t_{\varepsilon_j,\beta},x_{\varepsilon_j,\beta})-v(s_{\varepsilon_j,\beta},y_{\varepsilon_j,\beta})\} \leq u(\bar t,\bar x)-v(\bar s,\bar y). \end{align*} If $\bar x \neq \bar y$ or $\bar t \neq \bar s$, then one of the quotients in the separation penalty tends to $+\infty$, while the remaining terms have finite limsup. This contradicts the lower bound by $m$. Hence the limit is diagonal: $\bar t=\bar s$ and $\bar x=\bar y$. The maximality inequality then says \begin{align*} m \leq u(\bar t,\bar x)-v(\bar t,\bar x)-2\beta |\bar x-x_0|^2-2\beta |\bar t-t_0|^2 \leq M_\beta=m, \end{align*} so $(\bar t,\bar x)$ maximizes the diagonal anchored function. The previous paragraph identifies the only possible diagonal maximizer as $(t_0,x_0)$. Returning to the lower bound now leaves no room for positive separation penalty, and therefore \begin{align*} \frac{|x_{\varepsilon_j,\beta}-y_{\varepsilon_j,\beta}|^2}{\varepsilon_j}+\frac{|t_{\varepsilon_j,\beta}-s_{\varepsilon_j,\beta}|^2}{\varepsilon_j} \to 0. \end{align*} Hence the penalized contact points converge to $(t_0,t_0,x_0,x_0)$. Since $t_0<T$, they lie in the viscosity domain $[0,T) \times \mathbb{R}^n$ once $\varepsilon$ is small.[/guided]

[step:Apply the viscosity inequalities at the two penalized contact points]Fix $\beta > 0$ and take $\varepsilon > 0$ sufficiently small so that $t_{\varepsilon,\beta},s_{\varepsilon,\beta} < T$. Set \begin{align*} r_{\varepsilon,\beta} := \frac{x_{\varepsilon,\beta}-y_{\varepsilon,\beta}}{\varepsilon}. \end{align*} Define \begin{align*} p_{\varepsilon,\beta} := r_{\varepsilon,\beta}+2\beta(x_{\varepsilon,\beta}-x_0) \end{align*} and \begin{align*} q_{\varepsilon,\beta} := r_{\varepsilon,\beta}-2\beta(y_{\varepsilon,\beta}-x_0). \end{align*} Fixing $(s_{\varepsilon,\beta},y_{\varepsilon,\beta})$ in the doubled-variable maximum, define the smooth [test function](/page/Test%20Function) \begin{align*} \varphi^u_{\varepsilon,\beta}: [0,T] \times \mathbb{R}^n \to \mathbb{R} \end{align*} by \begin{align*} \varphi^u_{\varepsilon,\beta}(t,x) := u(t_{\varepsilon,\beta},x_{\varepsilon,\beta})+\frac{|x-y_{\varepsilon,\beta}|^2-|x_{\varepsilon,\beta}-y_{\varepsilon,\beta}|^2}{2\varepsilon}+\frac{|t-s_{\varepsilon,\beta}|^2-|t_{\varepsilon,\beta}-s_{\varepsilon,\beta}|^2}{2\varepsilon}+\beta |x-x_0|^2-\beta |x_{\varepsilon,\beta}-x_0|^2+\beta |t-t_0|^2-\beta |t_{\varepsilon,\beta}-t_0|^2. \end{align*} Then $\varphi^u_{\varepsilon,\beta}(t_{\varepsilon,\beta},x_{\varepsilon,\beta})=u(t_{\varepsilon,\beta},x_{\varepsilon,\beta})$, and the maximality of $\Phi_{\varepsilon,\beta}$ implies that $u-\varphi^u_{\varepsilon,\beta}$ has a local maximum at $(t_{\varepsilon,\beta},x_{\varepsilon,\beta})$ relative to $[0,T) \times \mathbb{R}^n$. Similarly, fixing $(t_{\varepsilon,\beta},x_{\varepsilon,\beta})$, define \begin{align*} \varphi^v_{\varepsilon,\beta}: [0,T] \times \mathbb{R}^n \to \mathbb{R} \end{align*} by \begin{align*} \varphi^v_{\varepsilon,\beta}(s,y) := v(s_{\varepsilon,\beta},y_{\varepsilon,\beta})-\frac{|x_{\varepsilon,\beta}-y|^2-|x_{\varepsilon,\beta}-y_{\varepsilon,\beta}|^2}{2\varepsilon}-\frac{|t_{\varepsilon,\beta}-s|^2-|t_{\varepsilon,\beta}-s_{\varepsilon,\beta}|^2}{2\varepsilon}-\beta |y-x_0|^2+\beta |y_{\varepsilon,\beta}-x_0|^2-\beta |s-t_0|^2+\beta |s_{\varepsilon,\beta}-t_0|^2. \end{align*} Then $\varphi^v_{\varepsilon,\beta}(s_{\varepsilon,\beta},y_{\varepsilon,\beta})=v(s_{\varepsilon,\beta},y_{\varepsilon,\beta})$, and $v-\varphi^v_{\varepsilon,\beta}$ has a local minimum at $(s_{\varepsilon,\beta},y_{\varepsilon,\beta})$ relative to $[0,T) \times \mathbb{R}^n$. At $(t_{\varepsilon,\beta},x_{\varepsilon,\beta})$, the function touching $u$ from above has time derivative \begin{align*} a_{\varepsilon,\beta} := \frac{t_{\varepsilon,\beta}-s_{\varepsilon,\beta}}{\varepsilon}+2\beta(t_{\varepsilon,\beta}-t_0) \end{align*} and spatial gradient $p_{\varepsilon,\beta}$. The viscosity subsolution inequality gives \begin{align*} \lambda u(t_{\varepsilon,\beta},x_{\varepsilon,\beta})-a_{\varepsilon,\beta}+H(t_{\varepsilon,\beta},x_{\varepsilon,\beta},p_{\varepsilon,\beta}) \leq 0. \end{align*} At $(s_{\varepsilon,\beta},y_{\varepsilon,\beta})$, the function touching $v$ from below has time derivative \begin{align*} b_{\varepsilon,\beta} := \frac{t_{\varepsilon,\beta}-s_{\varepsilon,\beta}}{\varepsilon}-2\beta(s_{\varepsilon,\beta}-t_0) \end{align*} and spatial gradient $q_{\varepsilon,\beta}$. The viscosity supersolution inequality gives \begin{align*} \lambda v(s_{\varepsilon,\beta},y_{\varepsilon,\beta})-b_{\varepsilon,\beta}+H(s_{\varepsilon,\beta},y_{\varepsilon,\beta},q_{\varepsilon,\beta}) \geq 0. \end{align*} Subtracting the supersolution inequality from the subsolution inequality yields \begin{align*} \lambda\{u(t_{\varepsilon,\beta},x_{\varepsilon,\beta})-v(s_{\varepsilon,\beta},y_{\varepsilon,\beta})\} \leq a_{\varepsilon,\beta}-b_{\varepsilon,\beta}+H(s_{\varepsilon,\beta},y_{\varepsilon,\beta},q_{\varepsilon,\beta})-H(t_{\varepsilon,\beta},x_{\varepsilon,\beta},p_{\varepsilon,\beta}). \end{align*} Since \begin{align*} a_{\varepsilon,\beta}-b_{\varepsilon,\beta}=2\beta(t_{\varepsilon,\beta}-t_0)+2\beta(s_{\varepsilon,\beta}-t_0), \end{align*} we have \begin{align*} |a_{\varepsilon,\beta}-b_{\varepsilon,\beta}| \leq 4\beta T. \end{align*} The use of viscosity test functions is valid also when one of the times equals $0$, because the hypotheses explicitly interpret tests at $t=0$ relative to the topology of $[0,T) \times \mathbb{R}^n$.[/step]

[guided]At the maximum point of $\Phi_{\varepsilon,\beta}$, we freeze the variables belonging to $v$ and obtain a test function for $u$ from above. Its spatial gradient is the derivative with respect to $x$ of the two $x$-dependent penalty terms, namely \begin{align*} p_{\varepsilon,\beta}=\frac{x_{\varepsilon,\beta}-y_{\varepsilon,\beta}}{\varepsilon}+2\beta(x_{\varepsilon,\beta}-x_0). \end{align*} Its time derivative is similarly \begin{align*} a_{\varepsilon,\beta}=\frac{t_{\varepsilon,\beta}-s_{\varepsilon,\beta}}{\varepsilon}+2\beta(t_{\varepsilon,\beta}-t_0). \end{align*} Because $u$ is a viscosity subsolution at $(t_{\varepsilon,\beta},x_{\varepsilon,\beta}) \in [0,T) \times \mathbb{R}^n$, the defining inequality gives \begin{align*} \lambda u(t_{\varepsilon,\beta},x_{\varepsilon,\beta})-a_{\varepsilon,\beta}+H(t_{\varepsilon,\beta},x_{\varepsilon,\beta},p_{\varepsilon,\beta}) \leq 0. \end{align*} For $v$, we freeze the variables belonging to $u$ and obtain a test function from below. The derivative with respect to $y$ has the opposite sign for the doubled spatial term and the same sign pattern from differentiating the lower test function, giving \begin{align*} q_{\varepsilon,\beta}=\frac{x_{\varepsilon,\beta}-y_{\varepsilon,\beta}}{\varepsilon}-2\beta(y_{\varepsilon,\beta}-x_0) \end{align*} and \begin{align*} b_{\varepsilon,\beta}=\frac{t_{\varepsilon,\beta}-s_{\varepsilon,\beta}}{\varepsilon}-2\beta(s_{\varepsilon,\beta}-t_0). \end{align*} Since $v$ is a viscosity supersolution at $(s_{\varepsilon,\beta},y_{\varepsilon,\beta}) \in [0,T) \times \mathbb{R}^n$, we have \begin{align*} \lambda v(s_{\varepsilon,\beta},y_{\varepsilon,\beta})-b_{\varepsilon,\beta}+H(s_{\varepsilon,\beta},y_{\varepsilon,\beta},q_{\varepsilon,\beta}) \geq 0. \end{align*} Subtracting the supersolution inequality from the subsolution inequality isolates the positive properness term: \begin{align*} \lambda\{u(t_{\varepsilon,\beta},x_{\varepsilon,\beta})-v(s_{\varepsilon,\beta},y_{\varepsilon,\beta})\} \leq a_{\varepsilon,\beta}-b_{\varepsilon,\beta}+H(s_{\varepsilon,\beta},y_{\varepsilon,\beta},q_{\varepsilon,\beta})-H(t_{\varepsilon,\beta},x_{\varepsilon,\beta},p_{\varepsilon,\beta}). \end{align*} Finally, \begin{align*} a_{\varepsilon,\beta}-b_{\varepsilon,\beta}=2\beta(t_{\varepsilon,\beta}-t_0)+2\beta(s_{\varepsilon,\beta}-t_0), \end{align*} and $t_{\varepsilon,\beta},s_{\varepsilon,\beta},t_0 \in [0,T]$, so $|a_{\varepsilon,\beta}-b_{\varepsilon,\beta}| \leq 4\beta T$. The relative-topology convention at $t=0$ is exactly what allows this viscosity argument even if one contact time equals $0$.[/guided]

[step:Control the Hamiltonian difference by the doubled-variable continuity condition]For fixed $\beta > 0$, the penalized contact points remain in a compact set \begin{align*} K_\beta \subset [0,T] \times \mathbb{R}^n \end{align*} for all sufficiently small $\varepsilon > 0$. Let $L_{K_\beta} > 0$ be the local Lipschitz constant for $H$ on $K_\beta$, and let $\omega_{K_\beta}$ be the modulus from the Crandall-Lions doubled-variable continuity condition. Split the Hamiltonian difference as \begin{align*} H(s_{\varepsilon,\beta},y_{\varepsilon,\beta},q_{\varepsilon,\beta})-H(t_{\varepsilon,\beta},x_{\varepsilon,\beta},p_{\varepsilon,\beta}) = A_{\varepsilon,\beta}+B_{\varepsilon,\beta}+C_{\varepsilon,\beta}, \end{align*} where \begin{align*} A_{\varepsilon,\beta} := H(s_{\varepsilon,\beta},y_{\varepsilon,\beta},q_{\varepsilon,\beta})-H(s_{\varepsilon,\beta},y_{\varepsilon,\beta},r_{\varepsilon,\beta}), \end{align*} \begin{align*} B_{\varepsilon,\beta} := H(s_{\varepsilon,\beta},y_{\varepsilon,\beta},r_{\varepsilon,\beta})-H(t_{\varepsilon,\beta},x_{\varepsilon,\beta},r_{\varepsilon,\beta}), \end{align*} and \begin{align*} C_{\varepsilon,\beta} := H(t_{\varepsilon,\beta},x_{\varepsilon,\beta},r_{\varepsilon,\beta})-H(t_{\varepsilon,\beta},x_{\varepsilon,\beta},p_{\varepsilon,\beta}). \end{align*} The local Lipschitz condition in the gradient variable gives \begin{align*} |A_{\varepsilon,\beta}| \leq 2L_{K_\beta}\beta |y_{\varepsilon,\beta}-x_0| \end{align*} and \begin{align*} |C_{\varepsilon,\beta}| \leq 2L_{K_\beta}\beta |x_{\varepsilon,\beta}-x_0|. \end{align*} The doubled-variable continuity condition gives \begin{align*} |B_{\varepsilon,\beta}| \leq \omega_{K_\beta}\left(|t_{\varepsilon,\beta}-s_{\varepsilon,\beta}|+\frac{|x_{\varepsilon,\beta}-y_{\varepsilon,\beta}|^2}{\varepsilon}\right). \end{align*} Therefore \begin{align*} H(s_{\varepsilon,\beta},y_{\varepsilon,\beta},q_{\varepsilon,\beta})-H(t_{\varepsilon,\beta},x_{\varepsilon,\beta},p_{\varepsilon,\beta}) \leq 2L_{K_\beta}\beta |x_{\varepsilon,\beta}-x_0|+2L_{K_\beta}\beta |y_{\varepsilon,\beta}-x_0|+\omega_{K_\beta}\left(|t_{\varepsilon,\beta}-s_{\varepsilon,\beta}|+\frac{|x_{\varepsilon,\beta}-y_{\varepsilon,\beta}|^2}{\varepsilon}\right). \end{align*}[/step]

[guided]The Hamiltonian is evaluated at two different base points and two slightly different gradients. We separate these two sources of error. First set \begin{align*} r_{\varepsilon,\beta}=\frac{x_{\varepsilon,\beta}-y_{\varepsilon,\beta}}{\varepsilon}. \end{align*} The gradients $p_{\varepsilon,\beta}$ and $q_{\varepsilon,\beta}$ differ from this common doubled-variable gradient only by the anchor gradients: \begin{align*} |q_{\varepsilon,\beta}-r_{\varepsilon,\beta}|=2\beta |y_{\varepsilon,\beta}-x_0| \end{align*} and \begin{align*} |p_{\varepsilon,\beta}-r_{\varepsilon,\beta}|=2\beta |x_{\varepsilon,\beta}-x_0|. \end{align*} For fixed $\beta>0$, the contact points lie in the compact set $K_\beta$ for all sufficiently small $\varepsilon$. Therefore the local Lipschitz hypothesis in the gradient variable gives \begin{align*} |A_{\varepsilon,\beta}| \leq 2L_{K_\beta}\beta |y_{\varepsilon,\beta}-x_0| \end{align*} and \begin{align*} |C_{\varepsilon,\beta}| \leq 2L_{K_\beta}\beta |x_{\varepsilon,\beta}-x_0|. \end{align*} The remaining term $B_{\varepsilon,\beta}$ keeps the same gradient $r_{\varepsilon,\beta}$ and only changes the base point from $(t_{\varepsilon,\beta},x_{\varepsilon,\beta})$ to $(s_{\varepsilon,\beta},y_{\varepsilon,\beta})$. The Crandall-Lions doubled-variable continuity condition must be applied with the variables in the order $t=t_{\varepsilon,\beta}$, $x=x_{\varepsilon,\beta}$, $s=s_{\varepsilon,\beta}$, and $y=y_{\varepsilon,\beta}$, using the same parameter $\varepsilon$. With this substitution the gradient appearing in the hypothesis is exactly \begin{align*} \frac{x_{\varepsilon,\beta}-y_{\varepsilon,\beta}}{\varepsilon}=r_{\varepsilon,\beta}, \end{align*} so the hypothesis gives \begin{align*} \left|H\left(t_{\varepsilon,\beta},x_{\varepsilon,\beta},r_{\varepsilon,\beta}\right)-H\left(s_{\varepsilon,\beta},y_{\varepsilon,\beta},r_{\varepsilon,\beta}\right)\right| \leq \omega_{K_\beta}\left(|t_{\varepsilon,\beta}-s_{\varepsilon,\beta}|+\frac{|x_{\varepsilon,\beta}-y_{\varepsilon,\beta}|^2}{\varepsilon}\right). \end{align*} The left-hand side is $|B_{\varepsilon,\beta}|$, hence \begin{align*} |B_{\varepsilon,\beta}| \leq \omega_{K_\beta}\left(|t_{\varepsilon,\beta}-s_{\varepsilon,\beta}|+\frac{|x_{\varepsilon,\beta}-y_{\varepsilon,\beta}|^2}{\varepsilon}\right). \end{align*} Adding these three bounds gives the displayed Hamiltonian estimate.[/guided]

[step:Pass first to the doubled-variable limit]Combining the previous two steps gives \begin{align*} \lambda\{u(t_{\varepsilon,\beta},x_{\varepsilon,\beta})-v(s_{\varepsilon,\beta},y_{\varepsilon,\beta})\} \leq 4\beta T+2L_{K_\beta}\beta |x_{\varepsilon,\beta}-x_0|+2L_{K_\beta}\beta |y_{\varepsilon,\beta}-x_0|+\omega_{K_\beta}\left(|t_{\varepsilon,\beta}-s_{\varepsilon,\beta}|+\frac{|x_{\varepsilon,\beta}-y_{\varepsilon,\beta}|^2}{\varepsilon}\right). \end{align*} Here $K_\beta$ is fixed while $\varepsilon \downarrow 0$, so $L_{K_\beta}$ is a fixed finite constant during this limiting step. Since the maximum value of $\Phi_{\varepsilon,\beta}$ is at least the diagonal supremum $M_\beta$, and all penalty terms are non-negative, \begin{align*} u(t_{\varepsilon,\beta},x_{\varepsilon,\beta})-v(s_{\varepsilon,\beta},y_{\varepsilon,\beta}) \geq M_\beta. \end{align*} For fixed $\beta>0$, the compact set $K_\beta$ and the constant $L_{K_\beta}$ are fixed while $\varepsilon \downarrow 0$. Along any convergent subsequence, the doubled-variable construction gives \begin{align*} (t_{\varepsilon,\beta},s_{\varepsilon,\beta},x_{\varepsilon,\beta},y_{\varepsilon,\beta}) \to (t_0,t_0,x_0,x_0) \end{align*} and \begin{align*} \frac{|x_{\varepsilon,\beta}-y_{\varepsilon,\beta}|^2}{\varepsilon}+|t_{\varepsilon,\beta}-s_{\varepsilon,\beta}| \to 0. \end{align*} Hence the modulus term tends to $\omega_{K_\beta}(0)=0$, and the two local-Lipschitz anchor terms tend to $0$ because $x_{\varepsilon,\beta} \to x_0$ and $y_{\varepsilon,\beta} \to x_0$. Letting $\varepsilon \downarrow 0$ in the previous inequality therefore gives \begin{align*} \lambda M_\beta \leq 4\beta T. \end{align*}[/step]

[guided]The inequality just displayed is obtained by substituting the Hamiltonian estimate into the viscosity inequality. The important point is the order of limits: $\beta>0$ is fixed first, and only $\varepsilon$ tends to $0$. Therefore the compact set $K_\beta$ and the Lipschitz constant $L_{K_\beta}$ do not vary in this step. Since the doubled-variable contact points converge to $(t_0,t_0,x_0,x_0)$ for this fixed $\beta$, we have \begin{align*} |x_{\varepsilon,\beta}-x_0| \to 0 \end{align*} and \begin{align*} |y_{\varepsilon,\beta}-x_0| \to 0. \end{align*} Multiplying by the fixed number $2L_{K_\beta}\beta$ shows that the two anchor-gradient error terms tend to $0$. The collapse estimate also gives \begin{align*} |t_{\varepsilon,\beta}-s_{\varepsilon,\beta}|+\frac{|x_{\varepsilon,\beta}-y_{\varepsilon,\beta}|^2}{\varepsilon} \to 0. \end{align*} By the defining property of the modulus, $\omega_{K_\beta}(r) \to 0$ as $r \downarrow 0$, so the modulus term tends to $0$ as well.[/guided]

[step:Send the anchoring strength to zero and obtain the contradiction] The preceding construction gives $M_\beta=m$ for every $\beta>0$. Taking $\beta \downarrow 0$ in \begin{align*} \lambda M_\beta \leq 4\beta T \end{align*} therefore gives \begin{align*} \lambda m \leq 0. \end{align*} This contradicts $\lambda > 0$ and $m > 0$. Hence the assumption $M > 0$ is false, so \begin{align*} u(t,x)-v(t,x) \leq 0 \end{align*} for every $(t,x) \in [0,T] \times \mathbb{R}^n$. This is exactly the desired comparison estimate. [/step]