The Hopf--Lax formula solves the Hamilton--Jacobi Cauchy problem

Boundedness and uniform continuity. For every $x$ and $t\ge 0$ we have $\inf u_0-tH(0)\le u(x,t)\le \mathrm{\parallel }{u}_0{\mathrm{\parallel }}_{\mathrm{\infty }}+tL(0)$: the upper bound is the competitor $y=x$ in [F1], and the lower bound follows from $L\ge -H(0)$. By [F4] the map $x\mapsto u(x,t)$ is uniformly continuous with modulus $\omega$ uniformly in $t$. For $t\ge s\ge 0$, [F2] and [F4] give $\mathrm{\mid }u(x,t)-u(x,s)\mathrm{\mid }\le \mathrm{\parallel }{Q}_{t-s}{u}_0-u_0{\mathrm{\parallel }}_{\mathrm{\infty }}$; and $\mathrm{\parallel }{Q}_{\tau }{u}_0-u_0{\mathrm{\parallel }}_{\mathrm{\infty }}\to 0$ as $\tau \downarrow 0$, since $Q_{\tau }{u}_0\le u_0+\tau L(0)$ and $Q_{\tau }{u}_0(x)\ge u_0(x)-{\sup }_v\{\omega (\tau \mathrm{\mid }v\mathrm{\mid })-\tau L(v)\}$, the supremum tending to $0$ as follows. For $a>0$, set $M=\mathrm{\parallel }{u}_0{\mathrm{\parallel }}_{\mathrm{\infty }}$ and choose $A=(2M+1)\mathrm{/}a$. Superlinearity and continuity of $L$ give $b\ge 0$ with $L(v)\ge A\mathrm{\mid }v\mathrm{\mid }-b$ everywhere. If $\tau \mathrm{\mid }v\mathrm{\mid }\ge a$, then $\omega (\tau \mathrm{\mid }v\mathrm{\mid })-\tau L(v)\le -1+b\tau$; if $\tau \mathrm{\mid }v\mathrm{\mid }<a$, then $L(v)\ge -H(0)$ gives the bound $\omega (a)+\tau H(0)$. Thus the limsup of the supremum is at most $\omega (a)$, which tends to zero as $a\downarrow 0$; its liminf is at least zero by the competitor $v=0$ and $\omega (0)=0$. Hence $u$ is bounded and uniformly continuous on each strip ${\mathbb{R}}^n\times [0,T]$.

The subsolution inequality. Let $\varphi \in C^1$ and let $u-\varphi$ have a strict local maximum at $(x_0,t_0)$ with $t_0>0$. Fix $v\in {\mathbb{R}}^n$ and small $h>0$, put $y:=x_0-hv$, and use the dynamic-programming identity $u(x_0,t_0)=Q_h(Q_{t_0-h}{u}_0)(x_0)\le u(y,t_0-h)+hL(v)$, the inequality coming from the competitor $y$ in the infimum defining $Q_h$. The contact inequality at $(x_0,t_0)$ gives $u(y,t_0-h)\le u(x_0,t_0)-\varphi (x_0,t_0)+\varphi (y,t_0-h)$, and combining the two gives $\varphi (x_0,t_0)-\varphi (x_0-hv,t_0-h)\le hL(v)$. Dividing by $h$ and letting $h\downarrow 0$ yields ${\varphi }_t(x_0,t_0)+⟨D\varphi (x_0,t_0),v⟩\le L(v)$ for every $v$; taking the supremum over $v$ and using $H=L^{\ast }$ of [F3] gives ${\varphi }_t(x_0,t_0)+H(D\varphi (x_0,t_0))\le 0$. Non-strict maxima are handled by strictification, so $u$ is a viscosity subsolution.

The supersolution inequality. Let $u-\varphi$ have a strict local minimum at $(x_0,t_0)$ with $t_0>0$. By [F1] there is a minimiser $y$ of $u_0(y)+t_{0}L((x_0-y)\mathrm{/}{t}_0)$; put $v:=(x_0-y)\mathrm{/}{t}_0$, so that $u(x_0,t_0)=u_0(y)+t_{0}L(v)$. For $0<h<t_0$ put $z_h:=y+\frac{t_0-h}{t_0}(x_0-y)$, so that $(z_h-y)\mathrm{/}(t_0-h)=v$ and $z_h\to x_0$. The dynamic-programming identity at $(z_h,t_0-h)$ with the competitor $y$ gives $u(z_h,t_0-h)\le u_0(y)+(t_0-h)L(v)=u(x_0,t_0)-hL(v)$, hence $u(x_0,t_0)-u(z_h,t_0-h)\ge hL(v)$. The contact inequality at the local minimum gives $u(z_h,t_0-h)\ge u(x_0,t_0)-\varphi (x_0,t_0)+\varphi (z_h,t_0-h)$; combining, $\varphi (x_0,t_0)-\varphi (z_h,t_0-h)\ge hL(v)$. Dividing by $h$ and letting $h\downarrow 0$ along $z_h\to x_0$ gives ${\varphi }_t(x_0,t_0)+⟨D\varphi (x_0,t_0),v⟩\ge L(v)$, and since $H(D\varphi )={\sup }_w(⟨D\varphi ,w⟩-L(w))\ge ⟨D\varphi ,v⟩-L(v)$ by [F3], we get ${\varphi }_t+H(D\varphi )\ge 0$. Hence $u$ is a viscosity supersolution, and with step 1.2 it is a viscosity solution of $u_t+H(Du)=0$.

The initial trace. For $\tau >0$ and every $x$, $Q_{\tau }{u}_0(x)\le u_0(x)+\tau L(0)$ and $Q_{\tau }{u}_0(x)\ge u_0(x)-{\sup }_v\{\omega (\tau \mathrm{\mid }v\mathrm{\mid })-\tau L(v)\}$, and the latter supremum tends to $0$ by the bounded-modulus estimate in step 1.1; hence ${\sup }_{{\mathbb{R}}^n}\mathrm{\mid }{Q}_{\tau }{u}_0-u_0\mathrm{\mid }\to 0$, which is the stated locally uniform (indeed uniform) attainment of the initial datum.

Uniqueness and the semigroup. Any bounded uniformly continuous viscosity solution of the Cauchy problem with datum $u_0$ is comparable with $u$ by [F5], in both orders, because both are bounded uniformly continuous and have the same continuous initial trace; hence $u$ is the unique such solution. Property (4) is [F2].