The entropy solution semigroup on $L^1\cap L^{\mathrm{\infty }}$

Semigroup law. Fix $s,t\ge 0$ and a horizon $T>s+t$. The solution $S_{\tau }{u}_0$ is in $L^1$ for every $\tau \in [0,T]$: compare it with the zero solution in [F2] to get $\mathrm{\parallel }{S}_{\tau }{u}_0{\mathrm{\parallel }}_1\le \mathrm{\parallel }{u}_0{\mathrm{\parallel }}_1$; its $L^{\mathrm{\infty }}$ bound follows from [F3]. Thus $S_s{u}_0\in L^1\cap L^{\mathrm{\infty }}$ and [F1] supplies the entropy solution $z(\tau )=S_{\tau }(S_s{u}_0)$. Define $w(\tau ,x)=S_{s+\tau }{u}_0(x)$ on ${\mathrm{\Pi }}_{T-s}$: its entropy inequalities are those of the original solution with time shifted, and its strong local $L^1$ trace at $\tau =0$ is $S_s{u}_0$ by the representative's continuity in $L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1$. Both $w$ and $z$ are bounded entropy solutions with this same datum, so uniqueness [F1] gives $w=z$ almost everywhere on ${\mathrm{\Pi }}_{T-s}$. Their time-continuous representatives then agree at every time in $L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1$, so evaluating at $\tau =t<T-s$ gives $S_{s+t}{u}_0=S_t(S_s{u}_0)$; as $T$ is arbitrary, this holds for all $s,t\ge 0$.

Order, contraction and the maximum bound. Let $u_0\le v_0$ almost everywhere in $L^1\cap L^{\mathrm{\infty }}$; by [F2] and [F3], $S_t{u}_0\le S_t{v}_0$ almost everywhere, $\mathrm{\parallel }{S}_t{u}_0-S_t{v}_0{\mathrm{\parallel }}_1\le \mathrm{\parallel }{u}_0-v_0{\mathrm{\parallel }}_1$, and $\mathrm{\parallel }{S}_t{u}_0{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }{u}_0{\mathrm{\parallel }}_{\mathrm{\infty }}$ for every $t\ge 0$. This proves (ii) and (iii).

Extension to $L^1$: construction. Assume $f(0)=0$ and $f$ globally Lipschitz, and let $u_0\in L^1$. Put $u_0^m=(-m)\vee (u_0\wedge m)$ as in [F4]. For $m,\mathrm{\ell }\ge 1$ and every $t\ge 0$, step 1.2 gives $\mathrm{\parallel }{S}_t{u}_0^m-S_t{u}_0^{\mathrm{\ell }}{\mathrm{\parallel }}_1\le \mathrm{\parallel }{u}_0^m-u_0^{\mathrm{\ell }}{\mathrm{\parallel }}_1$, so $(S_t{u}_0^m{)}_m$ is Cauchy in $L^1$, uniformly in $t$; define $S_t{u}_0={\lim }_m{S}_t{u}_0^m$ in $L^1$. The definition is independent of the approximating sequence: if $w_m\in L^1\cap L^{\mathrm{\infty }}$ with $w_m\to u_0$ in $L^1$, then $\mathrm{\parallel }{S}_t{w}_m-S_t{u}_0^m{\mathrm{\parallel }}_1\le \mathrm{\parallel }{w}_m-u_0^m{\mathrm{\parallel }}_1\to 0$, so both sequences have the same limit.

Time continuity of the $L^1$ flow. First fix $w_0\in L^1\cap L^{\mathrm{\infty }}$ and $T>0$, and choose $R>0$ with $w_0^R:=w_0{\mathbf{1}}_{B(0,R)}$. Let $M=\mathrm{\parallel }{w}_0{\mathrm{\parallel }}_{\mathrm{\infty }}$ and let $L$ be a Lipschitz constant of $f$ on $[-M,M]$. By [F5], for almost every $t\in (0,T)$ the orbit $S_t{w}_0^R$ is supported in $B(0,R+Lt)$. Fix $t_0\in [0,T]$ and a compact set $K\subseteq \{x:\mathrm{\mid }x\mathrm{\mid }>R+L{t}_0\}$. Its positive distance from $\bar{B}(0,R+L{t}_0)$ lets us choose times $t_j$ from that full-measure set tending to $t_0$ with $K\subseteq \{x:\mathrm{\mid }x\mathrm{\mid }>R+L{t}_j\}$. Then $S_{t_j}{w}_0^R=0$ in $L^1(K)$, and the $L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1$ continuity [F1] gives $S_{t_0}{w}_0^R=0$ in $L^1(K)$. A countable exhaustion of the strict exterior by compact sets shows that every slice is supported in $\bar{B}(0,R+L{t}_0)$; hence all slices on $[0,T]$ are supported in the fixed ball $\bar{B}(0,R+LT)$. Local $L^1$ continuity is therefore global $L^1$ continuity for this truncated orbit. By [F2], ${\sup }_{t\in [0,T]}\mathrm{\parallel }{S}_t{w}_0-S_t{w}_0^R{\mathrm{\parallel }}_1\le \mathrm{\parallel }{w}_0-w_0^R{\mathrm{\parallel }}_1$, which tends to $0$ as $R\to \mathrm{\infty }$. Thus the $L^1$-continuous truncated orbits converge uniformly on $[0,T]$ to $t\mapsto S_t{w}_0$, proving continuity for every datum in $L^1\cap L^{\mathrm{\infty }}$. For $u_0\in L^1$ in step 2.1, the extension orbit is the uniform limit of the continuous orbits $t\mapsto S_t{u}_0^m$, since ${\sup }_{t\ge 0}\mathrm{\parallel }{S}_t{u}_0-S_t{u}_0^m{\mathrm{\parallel }}_1\le \mathrm{\parallel }{u}_0-u_0^m{\mathrm{\parallel }}_1\to 0$; hence the extension is continuous as well.

Extension: properties. The extended maps preserve order: if $u_0\le v_0$ in $L^1$, then the truncated sequences satisfy $u_0^m\le v_0^m$ and hence $S_t{u}_0^m\le S_t{v}_0^m$ almost everywhere; passing to the $L^1$ limit gives $S_t{u}_0\le S_t{v}_0$ almost everywhere. They are contractions: $\mathrm{\parallel }{S}_t{u}_0-S_t{v}_0{\mathrm{\parallel }}_1\le {\mathrm{lim\hspace{0.17em}inf}}_m\mathrm{\parallel }{S}_t{u}_0^m-S_t{v}_0^m{\mathrm{\parallel }}_1\le {\mathrm{lim\hspace{0.17em}inf}}_m\mathrm{\parallel }{u}_0^m-v_0^m{\mathrm{\parallel }}_1=\mathrm{\parallel }{u}_0-v_0{\mathrm{\parallel }}_1$, using that truncation is a contraction in $L^1$. The semigroup law passes to the limit: $S_{t+s}{u}_0={\lim }_m{S}_{t+s}{u}_0^m={\lim }_m{S}_t(S_s{u}_0^m)=S_t(S_s{u}_0)$, the last step by the contraction property just proved applied to $S_s{u}_0^m\to S_s{u}_0$. Finally, the extension agrees with the original flow on $L^1\cap L^{\mathrm{\infty }}$, because for such $u_0$ the estimate of step 2.1 with $w_m=u_0$ gives $S_t{u}_0={\lim }_m{S}_t{u}_0^m$ in the original sense as well. An order-preserving $L^1$ contraction agreeing on the dense subset $L^1\cap L^{\mathrm{\infty }}$ is unique, so the extension is unique.

Conclusion. Steps 1.1–1.2 prove (i)–(iii) for data in $L^1\cap L^{\mathrm{\infty }}$, and steps 2.1 and 3.2 construct and characterise the unique order-preserving $L^1$ contraction extension to $L^1$ when $f(0)=0$ and $f$ is globally Lipschitz, agreeing with the classical flow on $L^1\cap L^{\mathrm{\infty }}$ and satisfying the semigroup law. Step 3.1 proves strong $L^1$ continuity of these orbits. This completes the proof.