Finite speed of dependence for Hamiltonians Lipschitz in momentum

Reduction. Put $w:=u-v$, which is bounded and upper semicontinuous. Let $\varphi \in C^1$ and suppose $w-\varphi$ has a strict local maximum at $z_{\ast }=(x_{\ast },t_{\ast })\in Z$. Choose $T_{\ast }$ with $t_{\ast }<T_{\ast }<T$ and a compact cylinder $Q⋐{\mathbb{R}}^n\times (0,T_{\ast })$ around $z_{\ast }$ on which the maximum is strict. Write $z=(x,t)$ and $z^{\mathrm{\prime }}=(y,s)$. For fixed $\rho >0$ set $G_{\rho }(z,z^{\mathrm{\prime }}):=u(x,t)-v(y,s)-\varphi (x,t)-\rho (\mathrm{\mid }x-x_{\ast }{\mathrm{\mid }}^2+\mathrm{\mid }y-x_{\ast }{\mathrm{\mid }}^2)$ and $F_{\alpha }(z,z^{\mathrm{\prime }}):=G_{\rho }(z,z^{\mathrm{\prime }})-\frac{\alpha }{2}\mathrm{\mid }z-z^{\mathrm{\prime }}{\mathrm{\mid }}^2$. By [F2]--[F3], $F_{\alpha }$ attains a finite maximum $M_{\alpha }$ on $Q\times Q$, and $M_{\alpha }\ge m_{\ast }:=(w-\varphi )(z_{\ast })$ by evaluation at $(z_{\ast },z_{\ast })$. The values $M_{\alpha }$ decrease with $\alpha$ and are bounded below by $m_{\ast }$, so they converge. For every maximiser $(z,z^{\mathrm{\prime }})$, comparison with $F_{\alpha \mathrm{/}2}$ at that same point gives $\alpha \mathrm{\mid }z-z^{\mathrm{\prime }}{\mathrm{\mid }}^2\le 4(M_{\alpha \mathrm{/}2}-M_{\alpha })\to 0$, uniformly over the maximiser sets; in particular $\mathrm{\mid }z-z^{\mathrm{\prime }}\mathrm{\mid }\to 0$ uniformly. Fix any sufficiently small open neighbourhood $V$ of $z_{\ast }$ with closure in the interior of $Q$, and put $A:=Q\setminus V$. Strictness and [F2]--[F3] give $m_A:={\max }_{z\in A}(w-\varphi )(z)<m_{\ast }$. Let $\gamma :=(m_{\ast }-m_A)\mathrm{/}2>0$ and ${\mathrm{\Delta }}_A:=\{(z,z):z\in A\}$. On ${\mathrm{\Delta }}_A$, $G_{\rho }(z,z)\le m_A=m_{\ast }-2\gamma$. The compact superlevel set $C:=\{(z,z^{\mathrm{\prime }})\in A\times Q:G_{\rho }(z,z^{\mathrm{\prime }})\ge m_{\ast }-\gamma \}$ is disjoint from ${\mathrm{\Delta }}_A$. If $C$ is nonempty, [F3] gives a positive distance $d_{\ast }>0$ between these compact sets; if it is empty, choose any $d_{\ast }>0$. Thus $G_{\rho }(z,z^{\mathrm{\prime }})<m_{\ast }-\gamma$ whenever $z\in A$ and $\mathrm{\mid }z-z^{\mathrm{\prime }}\mathrm{\mid }<d_{\ast }$. For all sufficiently large $\alpha$, every maximiser has $\mathrm{\mid }z-z^{\mathrm{\prime }}\mathrm{\mid }<d_{\ast }$, so its first slot cannot lie in $A$, since its value is at least $m_{\ast }$. As $V$ was arbitrary, all first slots converge uniformly to $z_{\ast }$; the second slots do also by the diagonal estimate. At each maximiser, fixing one slot gives $C^1$ upper and lower contacts for $u$ and $v$ with spatial gradients $p_{\alpha }:=D\varphi (z)+\alpha (x-y)+2\rho (x-x_{\ast })$ and $q_{\alpha }:=\alpha (x-y)-2\rho (y-x_{\ast })$, and time derivatives ${\varphi }_t(z)+\alpha (t-s)$ and $\alpha (t-s)$. By [F1] and the two Lipschitz bounds, ${\varphi }_t(z)\le H(y,s,q_{\alpha })-H(x,t,p_{\alpha })\le L\mathrm{\mid }{q}_{\alpha }-p_{\alpha }\mathrm{\mid }+2C(1+\mathrm{\mid }{p}_{\alpha }\mathrm{\mid })\mathrm{\mid }z-z^{\mathrm{\prime }}\mathrm{\mid }$. Since $\mathrm{\mid }z-z^{\mathrm{\prime }}\mathrm{\mid }\to 0$, $\alpha \mathrm{\mid }z-z^{\mathrm{\prime }}{\mathrm{\mid }}^2\to 0$, and $D\varphi$ is bounded on $Q$, the last error tends to zero uniformly over maximisers. Also $q_{\alpha }-p_{\alpha }=-D\varphi (z)-2\rho ((x-x_{\ast })+(y-x_{\ast }))\to -D\varphi (z_{\ast })$ uniformly for fixed $\rho$. Passing to these uniform limits gives ${\varphi }_t(z_{\ast })-L\mathrm{\mid }D\varphi (z_{\ast })\mathrm{\mid }\le 0$. The non-strict case follows by Strictification of a viscosity test function by a quartic perturbation; hence $w$ is a viscosity subsolution of $w_t-L\mathrm{\mid }Dw\mathrm{\mid }=0$ in $Z$.

The cone barrier. Let $M:=\max \{0,{\sup }_{{\mathbb{R}}^n\times [0,T)}w\}$, and let $h:\mathbb{R}\to \mathbb{R}$ be the explicit nondecreasing $C^1$ cutoff with $h=0$ on $(-\mathrm{\infty },0]$, $h(s)=3s^2-2s^3$ for $s\in [0,1]$ and $h=1$ on $[1,\mathrm{\infty })$. For $0<\epsilon <R$ and $\delta >0$ put ${\xi }_{\epsilon }(r):=Mh((r-(R-\epsilon ))\mathrm{/}\epsilon )$ and ${\psi }_{\epsilon ,\delta }(x,t):={\xi }_{\epsilon }(\sqrt{\mathrm{\mid }x-x_0{\mathrm{\mid }}^2+{\delta }^2}+Lt)$. Then ${\psi }_{\epsilon ,\delta }$ is $C^1$, bounded and nonnegative, and it is a classical supersolution of ${\psi }_t-L\mathrm{\mid }D\psi \mathrm{\mid }=0$ on ${\mathbb{R}}^n\times (0,T)$: indeed $\mathrm{\mid }D\psi \mathrm{\mid }={\xi }_{\epsilon }^{\mathrm{\prime }}\cdot \mathrm{\mid }x-x_0\mathrm{\mid }\mathrm{/}\sqrt{\mathrm{\mid }x-x_0{\mathrm{\mid }}^2+{\delta }^2}$ and ${\psi }_t=L{\xi }_{\epsilon }^{\mathrm{\prime }}$, so ${\psi }_t-L\mathrm{\mid }D\psi \mathrm{\mid }=L{\xi }_{\epsilon }^{\mathrm{\prime }}(1-\mathrm{\mid }x-x_0\mathrm{\mid }\mathrm{/}\sqrt{\mathrm{\mid }x-x_0{\mathrm{\mid }}^2+{\delta }^2})\ge 0$ because ${\xi }_{\epsilon }^{\mathrm{\prime }}\ge 0$. At $t=0$ we have $w(x,0)\le {\psi }_{\epsilon ,\delta }(x,0)$ for every $x$: for $\mathrm{\mid }x-x_0\mathrm{\mid }<R$ this uses $w(x,0)\le 0\le {\psi }_{\epsilon ,\delta }(x,0)$, and for $\mathrm{\mid }x-x_0\mathrm{\mid }\ge R$ it uses ${\psi }_{\epsilon ,\delta }(x,0)=M\ge w(x,0)$ (the cutoff argument at $t=0$ is $(\sqrt{\mathrm{\mid }x-x_0{\mathrm{\mid }}^2+{\delta }^2}-(R-\epsilon ))\mathrm{/}\epsilon \ge 1$).

Comparison with the barrier and conclusion. By step 1.1 the difference $w$ is a bounded upper semicontinuous subsolution of $w_t+G(x,t,Dw)=0$ and by step 1.2 the barrier is a bounded continuous supersolution of the same equation with ordered initial traces; comparison [F4] gives $w\le {\psi }_{\epsilon ,\delta }$ on ${\mathbb{R}}^n\times (0,T)$. At $t=0$ the desired inequality is the assumed initial order. Now fix $0<t<\min \{T,R\mathrm{/}L\}$ and $\mathrm{\mid }x-x_0\mathrm{\mid }<R-Lt$. Choose $\epsilon >0$ with $\mathrm{\mid }x-x_0\mathrm{\mid }+Lt<R-\epsilon$ and then $\delta >0$ with $\sqrt{\mathrm{\mid }x-x_0{\mathrm{\mid }}^2+{\delta }^2}+Lt\le R-\epsilon$; for these parameters the cutoff argument is at most $0$, so ${\psi }_{\epsilon ,\delta }(x,t)=0$ and comparison gives $w(x,t)\le 0$, that is $u(x,t)\le v(x,t)$. Under the additional semicontinuity assumptions in the equality clause, $v$ is an upper semicontinuous subsolution and $u$ a lower semicontinuous supersolution on the same half-closed slab. Applying the same conclusion to $(v,u)$ with the initial agreement then gives the reverse inequality and hence equality on the cone.