Comparison for first-order Hamilton--Jacobi equations

Case (a): setup and uniform localisation estimates. Assume $\sigma :={\sup }_Z(u-v)>0$ and fix $(x_1,t_1)\in Z$ with $u(x_1,t_1)-v(x_1,t_1)>3\sigma \mathrm{/}4$. Choose $\eta ,{\eta }^{\mathrm{\prime }}>0$ with $(\eta +{\eta }^{\mathrm{\prime }})\mathrm{/}(T-t_1)<\sigma \mathrm{/}4$ and put $B:=\eta +{\eta }^{\mathrm{\prime }}$; then ${\tilde{u}}_{\eta }(x_1,t_1)-{\tilde{v}}_{{\eta }^{\mathrm{\prime }}}(x_1,t_1)>\sigma \mathrm{/}2$. Fix $\rho >0$ with $2\rho \mathrm{\mid }{x}_1{\mathrm{\mid }}^2<\sigma \mathrm{/}8$; then for the doubling function of case (a) with these penalties, $M_{\alpha }\ge {\mathrm{\Phi }}_{\alpha }(x_1,x_1,t_1,t_1)>\sigma \mathrm{/}2-\sigma \mathrm{/}8=3\sigma \mathrm{/}8$ for every $\alpha$. Every maximiser $(x,y,t,s)$ of ${\mathrm{\Phi }}_{\alpha }$ has $t,s<T$, and by [F2] the weight bound gives $\rho (\mathrm{\mid }x{\mathrm{\mid }}^2+\mathrm{\mid }y{\mathrm{\mid }}^2)\le \sup u-\inf v-3\sigma \mathrm{/}8=:K_{\rho }$, so $\mathrm{\mid }x\mathrm{\mid },\mathrm{\mid }y\mathrm{\mid }\le C_{\rho }:=\sqrt{K_{\rho }\mathrm{/}\rho }$. Writing $d^2:=\mathrm{\mid }x-y{\mathrm{\mid }}^2+\mathrm{\mid }t-s{\mathrm{\mid }}^2$ and evaluating ${\mathrm{\Phi }}_{\alpha \mathrm{/}2}$ at any maximiser of ${\mathrm{\Phi }}_{\alpha }$ (the penalty difference is $\frac{\alpha }{4}{d}^2$) gives $M_{\alpha \mathrm{/}2}\ge M_{\alpha }+\frac{\alpha }{4}{d}^2$, hence $d\le {\delta }_{\alpha }:=(4(M_{\alpha \mathrm{/}2}-M_{\alpha })\mathrm{/}\alpha {)}^{1\mathrm{/}2}$ for every maximiser; since $M_{\alpha }\downarrow M_{\mathrm{\infty }}$ boundedly as $\alpha \to \mathrm{\infty }$, ${\delta }_{\alpha }\to 0$. With $p=\alpha (x-y)+2\rho x$ as in [F2] we get $\mathrm{\mid }p\mathrm{\mid }\le \alpha d+2\rho C_{\rho }$ and therefore $(1+\mathrm{\mid }p\mathrm{\mid })d\le d+\alpha d^2+2\rho C_{\rho }d\le {\delta }_{\alpha }+4(M_{\alpha \mathrm{/}2}-M_{\alpha })+2\rho C_{\rho }{\delta }_{\alpha }\to 0$ and $\mathrm{\mid }q-p\mathrm{\mid }=2\rho \mathrm{\mid }x+y\mathrm{\mid }\le 4\rho C_{\rho }$ uniformly over all maximisers, both limits being as $\alpha \to \mathrm{\infty }$ with $\rho$ fixed.

Case (b): setup and uniform localisation estimates. Let $\sigma :={\max }_{\bar{Z}}(u-v)>0$; the maximum is attained by compactness and continuity, and if it were attained on $\mathrm{\Gamma }$ it would be $\le 0$, then continuity supplies a point $(x_1,t_1)\in Z$ with $u(x_1,t_1)-v(x_1,t_1)>3\sigma \mathrm{/}4$, even if the maximum occurs at $t=T$. Choose $\eta ,{\eta }^{\mathrm{\prime }}>0$ with $B:=\eta +{\eta }^{\mathrm{\prime }}<\sigma (T-t_1)\mathrm{/}4$; then ${\tilde{u}}_{\eta }(x_1,t_1)-{\tilde{v}}_{{\eta }^{\mathrm{\prime }}}(x_1,t_1)>\sigma \mathrm{/}2$, and the doubling function of case (b) satisfies $M_{\alpha }\ge \sigma \mathrm{/}2$ for every $\alpha$. It is upper semicontinuous on the compact box $\bar{O}\times \bar{O}\times [0,T{]}^2$ (the penalties tend to $-\mathrm{\infty }$ at the terminal faces, where the value is declared $-\mathrm{\infty }$); a nonempty compact superlevel set and [F6] give a finite attained maximum, and every maximiser has $t,s<T$. Evaluating ${\mathrm{\Phi }}_{\alpha \mathrm{/}2}$ at a maximiser of ${\mathrm{\Phi }}_{\alpha }$ again gives $\frac{\alpha }{4}{d}^2\le M_{\alpha \mathrm{/}2}-M_{\alpha }$, hence $d\le {\delta }_{\alpha }\to 0$ uniformly over maximisers, and with $p=q=\alpha (x-y)$ in case (b) one has $(1+\mathrm{\mid }p\mathrm{\mid })d\le d+\alpha d^2\le {\delta }_{\alpha }+4(M_{\alpha \mathrm{/}2}-M_{\alpha })\to 0$. Since $\mathrm{\mid }x-y\mathrm{\mid }+\mathrm{\mid }t-s\mathrm{\mid }\le 2d$, it follows that $(1+\mathrm{\mid }p\mathrm{\mid })(\mathrm{\mid }x-y\mathrm{\mid }+\mathrm{\mid }t-s\mathrm{\mid })\le 2{\delta }_{\alpha }+8(M_{\alpha \mathrm{/}2}-M_{\alpha })\to 0$.

Case (b): exclusion of the parabolic boundary and the contradiction. Let $\alpha$ be large enough that $\varpi (2{\delta }_{\alpha })<B\mathrm{/}T$. If a maximiser had $t=0$, then $(x,0)\in \mathrm{\Gamma }$ gives $u(x,0)\le v(x,0)$ and $\mathrm{\mid }x-y\mathrm{\mid }+s\le 2d\le 2{\delta }_{\alpha }$, so ${\mathrm{\Phi }}_{\alpha }\le u(x,0)-v(y,s)-\frac{B}{T}\le \varpi (2{\delta }_{\alpha })-\frac{B}{T}<0,$ contradicting $M_{\alpha }\ge \sigma \mathrm{/}2$. If $x\in \mathrm{\partial }O$ with $t,s>0$, then $(x,t)\in \mathrm{\Gamma }$ gives $u(x,t)\le v(x,t)$ and $\mathrm{\mid }x-y\mathrm{\mid }+\mathrm{\mid }t-s\mathrm{\mid }\le 2d\le 2{\delta }_{\alpha }$, so ${\mathrm{\Phi }}_{\alpha }\le \varpi (2{\delta }_{\alpha })-B\mathrm{/}T<0$; the case $y\in \mathrm{\partial }O$ is symmetric, as is $s=0$ using the initial inequality at $(y,0)$ and the modulus of $u$. Thus all maximisers for large $\alpha$ have $0<t,s<T$ and $x,y\in O$. At such a maximiser the penalty inequalities of [F1] hold at the jets $p=q$, $p_0$ of case (b), and subtracting them gives $\frac{B}{T^2}\le H(y,s,p)-H(x,t,p)\le \omega ((1+\mathrm{\mid }p\mathrm{\mid })(\mathrm{\mid }x-y\mathrm{\mid }+\mathrm{\mid }t-s\mathrm{\mid }))\le \omega (2{\delta }_{\alpha }+8(M_{\alpha \mathrm{/}2}-M_{\alpha })),$ which tends to $0$ as $\alpha \to \mathrm{\infty }$ by step 1.2 and $\omega (0+)=0$; this contradicts $B\mathrm{/}{T}^2>0$. Hence $\sigma \le 0$, that is $u\le v$ on $\bar{Z}$.

Case (a): exclusion of the initial faces and the contradiction. Fix the $\rho$ of step 1.1 and let $K$ be the closed ball containing every spatial coordinate of every maximiser. On the compact set $K\times K\times [0,T\mathrm{/}2]$, the functions $f_0(x,y,s):=u(x,0)-v(y,s)$ and $f_1(x,y,t):=u(x,t)-v(y,0)$ are upper semicontinuous by [F3, F5], and both are nonpositive on the diagonal sets $(z,z,0)$ by the initial inequality. There is ${\delta }_0>0$ such that $f_0(x,y,s)<3\sigma \mathrm{/}8$ whenever $\mathrm{\mid }x-y\mathrm{\mid }+s<{\delta }_0$, and likewise ${\delta }_1>0$ for $f_1$ whenever $\mathrm{\mid }x-y\mathrm{\mid }+t<{\delta }_1$: otherwise the closed superlevel sets $\{f_i\ge 3\sigma \mathrm{/}8\}$ intersected with the nested closed sets where the corresponding distance is at most $1\mathrm{/}m$ would be nonempty compact sets with the finite-intersection property, so [F4] would give a point $(z,z,0)$ in the superlevel set, a contradiction. For $\alpha$ large enough that $2{\delta }_{\alpha }<\min \{{\delta }_0,{\delta }_1,T\mathrm{/}2\}$, if a maximiser had $t=0$, then ${\mathrm{\Phi }}_{\alpha }\le f_0(x,y,s)$ because all remaining penalties are nonpositive, while $\mathrm{\mid }x-y\mathrm{\mid }+s\le 2d\le 2{\delta }_{\alpha }$; this contradicts ${\mathrm{\Phi }}_{\alpha }=M_{\alpha }\ge 3\sigma \mathrm{/}8$. If $s=0$, similarly ${\mathrm{\Phi }}_{\alpha }\le f_1(x,y,t)$ and $\mathrm{\mid }x-y\mathrm{\mid }+t\le 2{\delta }_{\alpha }$, again a contradiction. Hence for all sufficiently large $\alpha$ every maximiser has $0<t,s<T$. At such a maximiser the contact inequalities of [F1] apply at the jets $p,q,p_0$ of [F2]: $p_0+H(x,t,p)\le -\eta \mathrm{/}{T}^2$ and $p_0+H(y,s,q)\ge {\eta }^{\mathrm{\prime }}\mathrm{/}{T}^2$. Subtracting and using the two Lipschitz conditions of case (a) gives $B\mathrm{/}{T}^2\le H(y,s,q)-H(x,t,p)\le C\mathrm{\mid }q-p\mathrm{\mid }+C(1+\mathrm{\mid }p\mathrm{\mid })(\mathrm{\mid }x-y\mathrm{\mid }+\mathrm{\mid }t-s\mathrm{\mid })$, and by step 1.1 the right-hand side is at most $4\rho C{C}_{\rho }+2C({\delta }_{\alpha }+4(M_{\alpha \mathrm{/}2}-M_{\alpha })+2\rho C_{\rho }{\delta }_{\alpha })$, using $\mathrm{\mid }x-y\mathrm{\mid }+\mathrm{\mid }t-s\mathrm{\mid }\le 2d$ for every maximiser with $\alpha$ large. Letting $\alpha \to \mathrm{\infty }$ gives $B\mathrm{/}{T}^2\le 4\rho C{C}_{\rho }=4C(\rho (\sup u-\inf v-3\sigma \mathrm{/}8){)}^{1\mathrm{/}2}$, and then letting $\rho \downarrow 0$ gives $B\mathrm{/}{T}^2\le 0$, a contradiction. Thus $\sigma \le 0$ and $u\le v$ on $Z$.

Conclusion. Case (a) is step 2.2 and case (b) is step 2.1; in both cases the contradiction is obtained by uniform estimates over the maximiser sets, so no maximiser, subsequence or index is selected and no choice principle is used.