Comparison for autonomous convex superlinear Hamiltonians

Penalisation. Put ${\tilde{u}}_{\eta }(x,t)=u(x,t)-\eta \mathrm{/}(T-t)$ and ${\tilde{v}}_{{\eta }^{\mathrm{\prime }}}(y,s)=v(y,s)+{\eta }^{\mathrm{\prime }}\mathrm{/}(T-s)$ for $t,s<T$. If $\varphi$ is a $C^1$ upper test for ${\tilde{u}}_{\eta }$ at an interior point $z_0$, then $\varphi +\eta \mathrm{/}(T-t)$ is a $C^1$ upper test for $u$, so $u_t$ evaluation and [F1] give ${\varphi }_t(z_0)+\eta \mathrm{/}(T-t_0{)}^2+H(D\varphi (z_0))\le 0$, that is ${\varphi }_t+H(D\varphi )\le -\eta \mathrm{/}(T-t_0{)}^2\le -\eta \mathrm{/}{T}^2$; dually every $C^1$ lower test $\varphi$ for ${\tilde{v}}_{{\eta }^{\mathrm{\prime }}}$ satisfies ${\varphi }_t+H(D\varphi )\ge {\eta }^{\mathrm{\prime }}\mathrm{/}(T-s_0{)}^2\ge {\eta }^{\mathrm{\prime }}\mathrm{/}{T}^2$. Moreover ${\tilde{u}}_{\eta }(x,t)\le \sup u-\eta \mathrm{/}(T-t)\to -\mathrm{\infty }$ and ${\tilde{v}}_{{\eta }^{\mathrm{\prime }}}(y,s)\ge \inf v+{\eta }^{\mathrm{\prime }}\mathrm{/}(T-s)\to +\mathrm{\infty }$ as $t\uparrow T$, respectively $s\uparrow T$, uniformly in the space variable.

The doubling function and the initial-face bound. Assume for contradiction that $u(x_0,t_0)>v(x_0,t_0)$ for some $0<t_0<T$ and put $\delta :=\frac{1}{4}(u(x_0,t_0)-v(x_0,t_0))>0$. Fix $\eta ,{\eta }^{\mathrm{\prime }}>0$ with ${\tilde{u}}_{\eta }(x_0,t_0)-{\tilde{v}}_{{\eta }^{\mathrm{\prime }}}(x_0,t_0)>3\delta$ and put $B:=\eta +{\eta }^{\mathrm{\prime }}>0$. Choose $a>0$ so small that ${\omega }_0(a)<B\mathrm{/}(8T)$, and then $P>0$ with $Pa>{\sup }_{r\ge 0}{\omega }_0(r)$; this gives ${\sup }_{r\ge 0}({\omega }_0(r)-Pr)<B\mathrm{/}(8T)$ by splitting at $a$. then $\epsilon >0$ with $P\epsilon <B\mathrm{/}(8T)$, and $\psi (r):=P(\sqrt{r^2+{\epsilon }^2}-\epsilon )$, so that $\mathrm{\mid }P(x-y)\mathrm{/}\sqrt{\mathrm{\mid }x-y{\mathrm{\mid }}^2+{\epsilon }^2}\mathrm{\mid }\le P$, $\psi (r)\ge Pr-P\epsilon$ and $\psi (0)=0$; put $\chi (x):=\sqrt{1+\mathrm{\mid }x{\mathrm{\mid }}^2}$. Choose $\rho >0$ with $\rho \le 1$, $2\rho \chi (x_0)<\delta$, and such that $\mathrm{\mid }H(p)-H(q)\mathrm{\mid }<B\mathrm{/}{T}^2$ whenever $\mathrm{\mid }p-q\mathrm{\mid }\le 2\rho$ and $\mathrm{\mid }p\mathrm{\mid },\mathrm{\mid }q\mathrm{\mid }\le P+1$: the last requirement is possible because $H$ is uniformly continuous on the compact set $\{\mathrm{\mid }p\mathrm{\mid }\le P+1\}$ by [F3] and [F4]. For $\alpha >0$ define ${\mathrm{\Phi }}_{\alpha }(x,y,t,s):={\tilde{u}}_{\eta }(x,t)-{\tilde{v}}_{{\eta }^{\mathrm{\prime }}}(y,s)-\psi (\mathrm{\mid }x-y\mathrm{\mid })-\frac{\alpha }{2}\mathrm{\mid }t-s{\mathrm{\mid }}^2-\rho (\chi (x)+\chi (y))$ for $0\le t,s<T$, and set ${\mathrm{\Phi }}_{\alpha }:=-\mathrm{\infty }$ if $t=T$ or $s=T$. At the diagonal point $(x_0,x_0,t_0,t_0)$ we have ${\mathrm{\Phi }}_{\alpha }={\tilde{u}}_{\eta }(x_0,t_0)-{\tilde{v}}_{{\eta }^{\mathrm{\prime }}}(x_0,t_0)-2\rho \chi (x_0)>3\delta -\delta =2\delta$ for every $\alpha$, while ${\mathrm{\Phi }}_{\alpha }\le \sup u-\inf v<\mathrm{\infty }$. The penalty $-\rho (\chi (x)+\chi (y))$ makes the superlevel set $S_{\alpha }:=\{{\mathrm{\Phi }}_{\alpha }\ge 2\delta \}$ bounded, and $S_{\alpha }$ is closed because ${\mathrm{\Phi }}_{\alpha }$ is upper semicontinuous (it is continuous where $t,s<T$, and tends to $-\mathrm{\infty }$ at the terminal faces, where it is $-\mathrm{\infty }$) and $2\delta >-\mathrm{\infty }$; by [F3] $S_{\alpha }$ is compact and it is nonempty by the diagonal estimate. On $S_{\alpha }$ the function ${\mathrm{\Phi }}_{\alpha }$ is real-valued, and it is upper semicontinuous as a restriction of an upper semicontinuous function, so it attains on $S_{\alpha }$ a maximum $M_{\alpha }\ge 2\delta$ by [F5], and by [F3] the value $M_{\alpha }$ is finite; a maximum on $S_{\alpha }$ is a global maximum of ${\mathrm{\Phi }}_{\alpha }$ because every point outside $S_{\alpha }$ has value $<2\delta \le M_{\alpha }$, and every maximiser lies in $S_{\alpha }$, hence has $t_{\alpha },s_{\alpha }<T$. So for every $\alpha$ there is a maximiser $(x_{\alpha },y_{\alpha },t_{\alpha },s_{\alpha })$ with ${\mathrm{\Phi }}_{\alpha }(x_{\alpha },y_{\alpha },t_{\alpha },s_{\alpha })\ge 2\delta$ and $t_{\alpha },s_{\alpha }<T$.

Initial faces are excluded for large $\alpha$. Since ${\tilde{u}}_{\eta }-{\tilde{v}}_{{\eta }^{\mathrm{\prime }}}\le \sup u-\inf v$, the inequality ${\mathrm{\Phi }}_{\alpha }\ge 2\delta$ gives $\frac{\alpha }{2}\mathrm{\mid }{t}_{\alpha }-s_{\alpha }{\mathrm{\mid }}^2\le \sup u-\inf v-2\delta$, so $\mathrm{\mid }{t}_{\alpha }-s_{\alpha }\mathrm{\mid }\to 0$ as $\alpha \to \mathrm{\infty }$. If $t_{\alpha }=0$, then using $u(x_{\alpha },0)\le v(x_{\alpha },0)$, the initial modulus ${\omega }_0$, the time modulus ${\omega }_v$ and $\psi (r)\ge Pr-P\epsilon$ we get ${\mathrm{\Phi }}_{\alpha }\le {\omega }_0(\mathrm{\mid }{x}_{\alpha }-y_{\alpha }\mathrm{\mid })-P\mathrm{\mid }{x}_{\alpha }-y_{\alpha }\mathrm{\mid }+P\epsilon +{\omega }_v(s_{\alpha })-\frac{B}{T}\le \frac{B}{8T}+\frac{B}{8T}-\frac{B}{T}+{\omega }_v(s_{\alpha })<0$ for all large $\alpha$, because $s_{\alpha }\to 0$ by $\mathrm{\mid }{t}_{\alpha }-s_{\alpha }\mathrm{\mid }\to 0$; this contradicts ${\mathrm{\Phi }}_{\alpha }\ge 2\delta$. The case $s_{\alpha }=0$ is identical with ${\omega }_u$ in place of ${\omega }_v$. Hence for all sufficiently large $\alpha$ every maximiser has $0<t_{\alpha },s_{\alpha }<T$.

Contact inequalities and the contradiction. Fix $\alpha$ large enough that step 2.1 applies and $\mathrm{\mid }{t}_{\alpha }-s_{\alpha }\mathrm{\mid }<T$. At the maximiser, fixing $(y_{\alpha },s_{\alpha })$ shows that ${\varphi }^U(x,t):=\psi (\mathrm{\mid }x-y_{\alpha }\mathrm{\mid })+\rho \chi (x)+\frac{\alpha }{2}\mathrm{\mid }t-s_{\alpha }{\mathrm{\mid }}^2$ is a $C^1$ upper test for ${\tilde{u}}_{\eta }$ at $(x_{\alpha },t_{\alpha })$, and fixing $(x_{\alpha },t_{\alpha })$ shows that ${\varphi }^V(y,s):=-\psi (\mathrm{\mid }{x}_{\alpha }-y\mathrm{\mid })-\rho \chi (y)-\frac{\alpha }{2}\mathrm{\mid }{t}_{\alpha }-s{\mathrm{\mid }}^2$ is a $C^1$ lower test for ${\tilde{v}}_{{\eta }^{\mathrm{\prime }}}$ at $(y_{\alpha },s_{\alpha })$. Their derivatives are $p:=\frac{P(x_{\alpha }-y_{\alpha })}{\sqrt{\mathrm{\mid }{x}_{\alpha }-y_{\alpha }{\mathrm{\mid }}^2+{\epsilon }^2}}+\frac{\rho x_{\alpha }}{\chi (x_{\alpha })}$, $q:=\frac{P(x_{\alpha }-y_{\alpha })}{\sqrt{\mathrm{\mid }{x}_{\alpha }-y_{\alpha }{\mathrm{\mid }}^2+{\epsilon }^2}}-\frac{\rho y_{\alpha }}{\chi (y_{\alpha })}$ and $a:=\alpha (t_{\alpha }-s_{\alpha })$, with $\mathrm{\mid }p\mathrm{\mid },\mathrm{\mid }q\mathrm{\mid }\le P+\rho \le P+1$ and $\mathrm{\mid }p-q\mathrm{\mid }=\rho \mathrm{\mid }\frac{x_{\alpha }}{\chi (x_{\alpha })}+\frac{y_{\alpha }}{\chi (y_{\alpha })}\mathrm{\mid }\le 2\rho$ because $\mathrm{\mid }z\mathrm{\mid }\mathrm{/}\chi (z)\le 1$ for every $z$. Step 1.1 applied to the two tests gives $a+H(p)\le -\eta \mathrm{/}{T}^2$ and $a+H(q)\ge {\eta }^{\mathrm{\prime }}\mathrm{/}{T}^2$, hence $B\mathrm{/}{T}^2=\frac{\eta +{\eta }^{\mathrm{\prime }}}{T^2}\le H(q)-H(p)$. But $\mathrm{\mid }p-q\mathrm{\mid }\le 2\rho$ and $\mathrm{\mid }p\mathrm{\mid },\mathrm{\mid }q\mathrm{\mid }\le P+1$, so the choice of $\rho$ in step 1.2 gives $\mathrm{\mid }H(q)-H(p)\mathrm{\mid }<B\mathrm{/}{T}^2$, a contradiction. Therefore no point with $u>v$ exists in $Z$, that is $u\le v$ on $Z$.