An $L^{p}$-Approach to the Well-Posedness of Transport Equations Associated to a Regular Field: Part II

Abstract

This work represents the continuation of our paper (Arlotti and Lods, in An $L^{p}$-approach to the well-posedness of transport equations associated to a regular field—part I. Mediterr J Math. 2018. https://doi.org/10.1007/s00009-019-1425-8). In $L^{p}$-spaces $1< p <\infty$ we investigate well-posedness of transport equations with general external Lipschitz fields and general measures associated to a large variety of boundary conditions modelled by abstract boundary operators H. In particular, a new explicit formula for the corresponding transport semigroup is given. Some applications are also presented.

Appendix A: On the Family \((U_{k}(t))_{t \geqslant 0}\)

We prove that the family of operators $(U_{k}(t))_{t \geqslant 0}$ introduced in Definition 3.6 is well defined and satisfies Theorem 3.8. We first need to establish general properties of ${\mathcal {T}}_{\mathrm {max},\,p}$.

1.1 A.1 Additional Properties of ${\mathcal {T}}_{\mathrm {max},\,p}$

We establish here several results, reminiscent of [4] about how ${\mathcal {T}}_{\mathrm {max},\,p}$ and some strongly continuous family of operators can interplay. We start with the following where we recall that ${\mathscr {D}}_{0}$ has been defined in the beginning of Section 3.

Proposition A.1

Let $(U(t))_{t\geqslant 0}$ be a strongly continuous family of ${\mathscr {B}}(X)$. For any $f \in X$, set

$$\begin{aligned} I_{t}[f]=\int _{0}^{t}U(s)f\mathrm {d}s, \quad \forall t \geqslant 0. \end{aligned}$$

Assume that

• (i) For any $f \in {\mathscr {D}}_{0},$ the mapping $t \in [0,\infty ) \mapsto U(t)f \in X$ is differentiable with

  $$\begin{aligned} \dfrac{\mathrm {d}}{\mathrm {d}t}U(t)f=U(t){\mathcal {T}}_{\mathrm {max},\,p}f, \quad t \geqslant 0. \end{aligned}$$

• (ii) For any $f \in {\mathscr {D}}_{0}$ and any $t \geqslant 0$, it holds that $U(t)f \in {\mathscr {D}}({\mathcal {T}}_{\mathrm {max},\,p})$ with ${\mathcal {T}}_{\mathrm {max},\,p}U(t)f=U(t){\mathcal {T}}_{\mathrm {max},\,p}f.$

Then, the following holds

• (1) for any $f \in X$ and $t >0$, $I_{t}[f] \in {\mathscr {D}}({\mathcal {T}}_{\mathrm {max},\,p})$ with

  $$\begin{aligned} {\mathcal {T}}_{\mathrm {max},\,p}I_{t}[f]=U(t)f-U(0)f. \end{aligned}$$

• (2) for any $f \in {\mathscr {D}}_{0}$ the mapping $t \in [0,\infty ) \mapsto {\mathsf {B}}^{\pm }U(t)f \in Y_{p}^{\pm }$ is continuous and,

  $$\begin{aligned} {\mathsf {B}}^{\pm }I_{t}[f]=\int _{0}^{t}{\mathsf {B}}^{\pm }U(s)f \mathrm {d}s \quad \forall t >0. \end{aligned}$$

Let now $f \in X$ be such that ${\mathsf {B}}^{-}I_{t}[f] \in L^p_-$ for all $t >0$ with $t \in [0,\infty ) \mapsto {\mathsf {B}}^{-}I_{t}[f] \in L^p_-\text { continuous,}$ then,

$$\begin{aligned} {\mathsf {B}}^{+}I_{t}[f] \in L^p_+\quad \text { and } \quad t \in [0,\infty ) \mapsto {\mathsf {B}}^{+}I_{t}[f] \in L^p_+\text { continuous.} \end{aligned}$$

Proof

Under assumptions $i)-ii)$, for any $f \in {\mathscr {D}}_{0}$, since both the mappings $t \mapsto U(t)f$ and $t \mapsto {\mathcal {T}}_{\mathrm {max},\,p}U(t)f$ are continuous and ${\mathcal {T}}_{\mathrm {max},\,p}$ is closed one has $I_{t}[f] \in {\mathscr {D}}({\mathcal {T}}_{\mathrm {max},\,p})$ with

$$\begin{aligned} {\mathcal {T}}_{\mathrm {max},\,p}I_{t}[f]=\int _{0}^{t}{\mathcal {T}}_{\mathrm {max},\,p}U(s)f\mathrm {d}s=\int _{0}^{t}\dfrac{\mathrm {d}}{\mathrm {d}s}U(s)f \mathrm {d}s=U(t)f-U(0)f. \end{aligned}$$

This proves that (1) holds for $f \in {\mathscr {D}}_{0}$ and, since ${\mathscr {D}}_{0}$ is dense in X, the result holds for any $f \in X$.

Let us prove (2). Pick $f \in {\mathscr {D}}_{0}$. Since the mapping $t \geqslant 0\mapsto U(t)f \in {\mathscr {D}}({\mathcal {T}}_{\mathrm {max},\,p})$ is continuous for the graph norm on ${\mathscr {D}}({\mathcal {T}}_{\mathrm {max},\,p})$ while ${\mathsf {B}}^{\pm }\,:\,{\mathscr {D}}({\mathcal {T}}_{\mathrm {max},\,p}) \mapsto Y_{p}^{\pm }$ is continuous (see [5, Remark 3.4]), the mapping $t \geqslant 0 \mapsto {\mathsf {B}}^{\pm }U(t)f \in Y_{p}^{\pm }$ is continuous. Moreover, ${\mathsf {B}}^{\pm }I_{t}[f]=\int _{0}^{t}{\mathsf {B}}^{\pm }U(s)f \mathrm {d}s$ still thanks to the continuity on ${\mathscr {D}}({\mathcal {T}}_{\mathrm {max},\,p})$ with the graph norm and point (1).

Let now $f \in X$ be given such that ${\mathsf {B}}^{-}I_{t}[f] \in L^p_-$ for all $t >0$ with $t \geqslant 0 \mapsto {\mathsf {B}}^{-}I_{t}[f] \in L^p_-$ continuous. For all $t \geqslant 0,$ $h \geqslant -t$, denote now

$$\begin{aligned} I_{t,t+h}[f]=\int _{t}^{t+h}U(s)f\mathrm {d}s=I_{t+h}[f]-I_{t}[f]. \end{aligned}$$

Since $I_{t}[f] \in {\mathscr {D}}({\mathcal {T}}_{\mathrm {max},\,p})$ and ${\mathsf {B}}^{-}I_{t}[f] \in L^p_-$, one has clearly $I_{t,t+h}[f] \in {\mathscr {D}}({\mathcal {T}}_{\mathrm {max},\,p})$ and ${\mathsf {B}}^{-}I_{t,t+h}[f] \in L^p_-$ and Green’s formula (see [5, Equation (2.12)]) yields

$$\begin{aligned} \Vert {\mathsf {B}}^{+}I_{t,t+h}[f]\Vert _{L^p_+}^{p}= & {} \Vert {\mathsf {B}}^{-}I_{t,t+h}[f]\Vert _{L^p_-}^{p}\\&-p\int _{\mathbf {\Omega }}\left| I_{t,t+h}[f]\right| ^{p-1}\mathrm {sign}(I_{t,t+h}[f]){\mathcal {T}}_{\mathrm {max},\,p}I_{t,t+h}[f]\mathrm {d}\mu \\\leqslant & {} \Vert {\mathsf {B}}^{-}I_{t,t+h}[f]\Vert _{L^p_-}^{p}+p\Vert I_{t,t+h}[f]\Vert _{p}^{p-1}\,\Vert {\mathcal {T}}_{\mathrm {max},\,p}I_{t,t+h}[f]\Vert _{p}. \end{aligned}$$

Since ${\mathcal {T}}_{\mathrm {max},\,p}I_{t,t+h}[f]=U(t+h)f-U(t)f$, one gets

$$\begin{aligned} \Vert {\mathsf {B}}^{+}\left( I_{t+h}[f]-I_{t}[f]\right) \Vert _{L^p_+}^{p}\leqslant & {} \Vert {\mathsf {B}}^{-}\left( I_{t+h}[f]-I_{t}[f]\right) \Vert _{L^p_-}^{p}\nonumber \\&+p\Vert I_{t+h}[f]-I_{t}[f]\Vert _{p}^{p-1}\,\Vert U(t+h)f-U(t)f\Vert _{p}.\nonumber \\ \end{aligned}$$

(A.1)

The continuity of $s \geqslant 0\mapsto U(s)f \in X$ together with the one of $s \geqslant 0 \mapsto {\mathsf {B}}^{-}I_{s}[f] \in L^p_-$ gives then that

$$\begin{aligned} \lim _{h \rightarrow 0}\Vert {\mathsf {B}}^{+}\left( I_{t+h}[f]-I_{t}[f]\right) \Vert _{L^p_+}=0 \end{aligned}$$

i.e. $t \geqslant 0 \mapsto {\mathsf {B}}^{+}I_{t}[f] \in L^p_+$ is continuous. $\square$

We can complement the above with the following whose proof is exactly as that of [4, Proposition 3] and is omitted here:

Proposition A.2

Let $(U(t))_{t\geqslant 0}$ be a strongly continuous family of ${\mathscr {B}}(X)$ satisfying the following, for any $f \in {\mathscr {D}}_{0}$:

• (i) For any $t \geqslant 0$,

  $$\begin{aligned}{}[U(t)f]({\mathbf {x}})=0 \quad \forall {\mathbf {x}}\in \mathbf {\Omega }\text { such that } \tau _{-}({\mathbf {x}}) \geqslant t. \end{aligned}$$

• (ii) For any ${\mathbf {y}}\in \Gamma _{-}$, $t >0$, $0< r< s < \tau _{+}({\mathbf {y}})$, it holds

  $$\begin{aligned} \left[ U(t)f\right] (\Phi ({\mathbf {y}},s))=\left[ U(t-s+r)f\right] (\Phi ({\mathbf {y}},r)). \end{aligned}$$

• (iii) the mapping $t \geqslant 0 \mapsto U(t)f \in X$ is differentiable with $\frac{\mathrm {d}}{\mathrm {d}t}U(t)f=U(t){\mathcal {T}}_{\mathrm {max},\,p}f$ for any $t \geqslant 0.$

Then, the following properties hold

• (1) For any $f \in X$ and any $t \geqslant 0$ and $\mu _{-}$-a.e. ${\mathbf {y}}\in \Gamma _{-}$, given $0< s_{1}< s_{2} < \tau _{+}({\mathbf {y}})$, there exists $0< r < s_{1}$ such that

  $$\begin{aligned} \int _{s_{1}}^{s_{2}}\left[ U(t)f\right] (\Phi ({\mathbf {y}},s))\mathrm {d}s=\int _{t-s_{2}+r}^{t-s_{1}+r}\left[ U(\tau )f\right] (\Phi ({\mathbf {y}},r))\mathrm {d}\tau . \end{aligned}$$

• (2) For any $f \in {\mathscr {D}}_{0}$ and $t \geqslant 0$, one has $U(t)f \in {\mathscr {D}}({\mathcal {T}}_{\mathrm {max},\,p})$ with ${\mathcal {T}}_{\mathrm {max},\,p}U(t)f=U(t){\mathcal {T}}_{\mathrm {max},\,p}f.$

1.2 A.2 Proof of Theorem 3.8

We now come to the proof of Theorem 3.8 which will consist of showing that the family of operators $(U_{k}(t))_{t \geqslant 0}$ introduced in Definition 3.6 is well defined and satisfies the properties listed in Theorem 3.8. The proof is made by induction and we start with a series of Lemmas (one for each of the above properties in Theorem 3.8) showing that $U_{1}(t)$ enjoys all the listed properties.

As already mentioned, the fact that the mapping

$$\begin{aligned} \Phi \,:\,\{({\mathbf {y}},s) \in \Gamma _{-}\times (0,\infty )\;;\;0< s < \tau _{+}({\mathbf {y}}) \} \rightarrow \mathbf {\Omega }_{-} \end{aligned}$$

is a measure isomorphism, for any $f \in {\mathscr {D}}_{0}$ and $t >0$, the function $U_{1}(t)f$ is well defined and measurable on $\mathbf {\Omega }$. Moreover, using [5, Proposition 2.2] and Fubini’s Theorem:

$$\begin{aligned} \begin{aligned} \Vert U_{1}(t)f\Vert _{p}^{p}&=\int _{\Gamma _{-}}\mathrm {d}\mu _{-}({\mathbf {y}})\int _{0}^{\tau _{+}({\mathbf {y}})}\left| [U_{1}(t)f](\Phi ({\mathbf {y}},s))\right| ^{p}\mathrm {d}s\\&=\int _{\Gamma _{-}}\mathrm {d}\mu _{-}({\mathbf {y}})\int _{0}^{\min (t,\tau _{+}({\mathbf {y}}))}\left| \left[ H({\mathsf {B}}^{+}U_{0}(t-s)f)\right] ({\mathbf {y}})\right| ^{p}\mathrm {d}s\\&\leqslant \int _{0}^{t} \Vert H({\mathsf {B}}^{+}U_{0}(t-s)f)\Vert _{L^p_-}^{p}\mathrm {d}s. \end{aligned} \end{aligned}$$

Therefore,

$$\begin{aligned} \Vert U_{1}(t)f\Vert _{p}^{p} \leqslant {\left| \left| \left| H \right| \right| \right| }^{p}\int _{0}^{t}\Vert {\mathsf {B}}^{+}U_{0}(t-s)f\Vert _{L^p_+}^{p}\mathrm {d}s={\left| \left| \left| H \right| \right| \right| }^{p}\left( \Vert f\Vert _{p}^{p}-\Vert U_{0}(t)f\Vert _{p}^{p}\right) \end{aligned}$$

thanks to Proposition 3.5. Therefore $\Vert U_{1}(t)f\Vert _{p} \leqslant {\left| \left| \left| H \right| \right| \right| }\,\Vert f\Vert _{p}$ for all $f \in {\mathscr {D}}_{0}$ with moreover

$$\begin{aligned} \lim _{t\rightarrow 0^{+}}\Vert U_{1}(t)f\Vert _{p}=0 \quad \forall f \in {\mathscr {D}}_{0}.\end{aligned}$$

(A.2)

Since ${\mathscr {D}}_{0}$ is dense in X, this allows to define a unique extension operator, still denoted by $U_{1}(t) \in {\mathscr {B}}(X)$ with

$$\begin{aligned} \Vert U_{1}(t)\Vert _{{\mathscr {B}}(X)} \leqslant {\left| \left| \left| H \right| \right| \right| }, \quad \forall t \geqslant 0. \end{aligned}$$

Now, one has the following

Lemma A.3

The family $(U_{1}(t))_{t\geqslant 0}$ is strongly continuous on X.

Proof

Let $t >0$ be fixed. Set $\mathbf {\Omega }_{t}=\{{\mathbf {x}}\in \mathbf {\Omega }_{-}\,;\;\tau _{-}({\mathbf {x}}) \leqslant t \}.$ One has $[U_{1}(t)f]({\mathbf {x}})=0$ for any ${\mathbf {x}}\in \mathbf {\Omega }{{\setminus }} \mathbf {\Omega }_{t}$ and any $f \in X.$ Let us fix $f \in {\mathscr {D}}_{0}$ and $h >0.$ One has

$$\begin{aligned} \Vert U_{1}(t+h)f-U_{1}(t)f\Vert _{p}^{p}= & {} \int _{\mathbf {\Omega }_{t}}\left| U_{1}(t+h)f-U_{1}(t)f\right| ^{p}\mathrm {d}\mu \nonumber \\&+ \int _{\mathbf {\Omega }_{t+h}{\setminus }\mathbf {\Omega }_{t}}|U_{1}(t+h)f|^{p}\mathrm {d}\mu . \end{aligned}$$

(A.3)

Now,

$$\begin{aligned}&\int _{\mathbf {\Omega }_{t}}\left| U_{1}(t+h)f-U_{1}(t)f\right| ^{p}\mathrm {d}\mu \\&\quad =\int _{ \Gamma _{-}}\mathrm {d}\mu _{-}({\mathbf {y}})\int _{0}^{\tau _{+}({\mathbf {y}})}\left| \left[ U_{1}(t+h)f-U_{1}(t)f\right] (\Phi ({\mathbf {y}},s))\right| ^{p}\mathrm {d}s \end{aligned}$$

and, repeating the reasoning before Lemma A.3 one gets

$$\begin{aligned} \int _{\mathbf {\Omega }_{t}}\left| U_{1}(t+h)f-U_{1}(t)f\right| ^{p}\mathrm {d}\mu \leqslant {\left| \left| \left| H \right| \right| \right| }^{p}\int _{0}^{t}\Vert {\mathsf {B}}^{+}\left( U_{0}(s+h)f-U_{0}(s)f\right) \Vert _{L^p_+}^{p}\mathrm {d}s. \end{aligned}$$

Since $U_{0}(s+h)f-U_{0}(s)f=U_{0}(s)\left( U_{0}(h)f-f\right)$ one gets from Proposition 3.5 that

$$\begin{aligned} \int _{\mathbf {\Omega }_{t}}\left| U_{1}(t+h)f-U_{1}(t)f\right| ^{p}\mathrm {d}\mu \leqslant {\left| \left| \left| H \right| \right| \right| }^{p}\left( \Vert U_{0}(h)f-f\Vert _{p}^{p}-\Vert U_{0}(t)\left( U_{0}(h)f-f\right) \Vert _{p}^{p}\right) . \end{aligned}$$

This proves that

$$\begin{aligned} \lim _{h\rightarrow 0^{+}}\int _{\mathbf {\Omega }_{t}}\left| U_{1}(t+h)f-U_{1}(t)f\right| ^{p}\mathrm {d}\mu =0. \end{aligned}$$

Let us investigate the second integral in (A.3). One first notices that, given ${\mathbf {x}}=\Phi ({\mathbf {y}},s)$ with ${\mathbf {y}}\in \Gamma _{-}$, $0< s < \min (t,\tau _{+}({\mathbf {y}}))$, it holds

$$\begin{aligned} \begin{aligned} \left[ U_{0}(t)U_{1}(h)f\right] ({\mathbf {x}})&=\chi _{\{t< \tau _{-}({\mathbf {x}})\}}\left[ U_{1}(h)f\right] (\Phi ({\mathbf {x}},-t))\\&=\chi _{(t,\infty )}(s)\,\left[ U_{1}(h)f\right] (\Phi ({\mathbf {y}},s-t))\\&=\chi _{(t,t+h]}(s)\left[ H\left( {\mathsf {B}}^{+}U_{0}(t+h-s)f\right) \right] ({\mathbf {y}})\\&=\chi _{(t,t+h]}(s)\left[ U_{1}(t+h)f\right] (\Phi ({\mathbf {y}},s))\\&=\chi _{\{t < \tau _{-}({\mathbf {x}})\}}\left[ U_{1}(t+h)f\right] ({\mathbf {x}}). \end{aligned} \end{aligned}$$

(A.4)

Therefore

$$\begin{aligned} \int _{\mathbf {\Omega }_{t+h}{\setminus } \mathbf {\Omega }_{t}}|U_{1}(t+h)f|^{p}\mathrm {d}\mu =\Vert U_{0}(t)U_{1}(h)f\Vert _{p}^{p} \end{aligned}$$

and, since $(U_{0}(t))_{t\geqslant 0}$ is a contraction semigroup, we get

$$\begin{aligned} \int _{\mathbf {\Omega }_{t+h}{\setminus } \mathbf {\Omega }_{t}}|U_{1}(t+h)f|^{p}\mathrm {d}\mu \leqslant \Vert U_{1}(h)f\Vert _{p}^{p}. \end{aligned}$$

Using (A.2), we get $\lim _{h\rightarrow 0^{+}}\int _{\mathbf {\Omega }_{t+h}{\setminus } \mathbf {\Omega }_{t}}|U_{1}(t+h)f|^{p}\mathrm {d}\mu =0$ and we obtain finally that $\lim _{h \rightarrow 0^{+}}\Vert U_{1}(t+h)f-U_{1}(t)f\Vert _{p}^{p}=0.$ One argues in a similar way for negative h and gets

$$\begin{aligned} \lim _{h \rightarrow 0}\Vert U_{1}(t+h)f-U_{1}(t)f\Vert _{p}=0, \quad \forall f \in {\mathscr {D}}_{0}. \end{aligned}$$

Since ${\mathscr {D}}_{0}$ is dense in X and $\Vert U_{1}(t)\Vert _{{\mathscr {B}}(X)} \leqslant {\left| \left| \left| H \right| \right| \right| }$ we deduce that above limit vanishes for all $f \in X$. This proves the result. $\square$

One has also the following

Lemma A.4

For all $t \geqslant 0,$ $h \geqslant 0$ and $f \in X$ it holds

$$\begin{aligned} U_{1}(t+h)f=U_{0}(t)U_{1}(h)f+U_{1}(t)U_{0}(h)f. \end{aligned}$$

Proof

It is clearly enough to consider $t >0$, $h >0$ since $U_{1}(0)f=0$ while $U_{0}(0)$ is the identity operator. Notice that, for any $f \in {\mathscr {D}}_{0}$ and any $0 \leqslant t_{1} \leqslant t_{2}$, for ${\mathbf {x}}=\Phi ({\mathbf {y}},s) \in \mathbf {\Omega }_{t_{1}}$ we have

$$\begin{aligned} \int _{t_1}^{t_2}[U_1(\tau )f]({\mathbf {x}})\mathrm {d}\tau =\left[ H{\mathsf {B}}^{+}\int _{t_{1}-s}^{t_{2}-s}U_0(\tau )f\mathrm {d}\tau \right] ({\mathbf {y}}). \end{aligned}$$

Now, given $f \in X$ and $0 \leqslant t_{1} \leqslant t_{2}$ the above formula is true for almost every ${\mathbf {x}}=\Phi ({\mathbf {y}},s) \in \mathbf {\Omega }_{t_{1}}$ by a density argument. Therefore, for almost every ${\mathbf {x}}=\Phi ({\mathbf {y}},s) \in \mathbf {\Omega }_{t}$ and any $\delta >0$ it holds

$$\begin{aligned} \int _{t}^{t+\delta }[U_1(r+h)f]({\mathbf {x}})\mathrm {d}r= & {} \left[ H{\mathsf {B}}^+\int _{t-s}^{t+\delta -s}U_0(r+h)f\mathrm {d}r\right] ({\mathbf {y}})\\= & {} \left[ H{\mathsf {B}}^+\int _{t-s}^{t+\delta -s}U_0(r)U_0(h)f\mathrm {d}r\right] ({\mathbf {y}}) \end{aligned}$$

so that, using the definition of $U_{1}(r)$ again

$$\begin{aligned} \int _{t}^{t+\delta }[U_1(r+h)f]({\mathbf {x}})\mathrm {d}r=\int _{t}^{t+\delta }[U_1(r)U_0(h)f]({\mathbf {x}})\mathrm {d}r \quad \forall \delta >0 \end{aligned}$$

from which we deduce that $U_1(t+h)f({\mathbf {x}})=U_1(t)U_0(h)f({\mathbf {x}})$ for almost any ${\mathbf {x}}\in \mathbf {\Omega }_{t}.$With the notations of the previous proof, one has from (A.4) that

$$\begin{aligned} U_{1}(t+h)f({\mathbf {x}})=\left[ U_{0}(t)U_{1}(h)f\right] ({\mathbf {x}}) \quad \text { for a.e. } {\mathbf {x}}\in \mathbf {\Omega }_{t+h}{\setminus } \mathbf {\Omega }_{t} \end{aligned}$$

This proves the result, since $U_{1}(t)f$ vanishes on $\mathbf {\Omega }_{t+h}{{\setminus }}\mathbf {\Omega }_{t}$ while $U_{0}(t)f$ vanishes on $\mathbf {\Omega }_{t}.$ $\square$

One has now the following

Lemma A.5

For any $f \in {\mathscr {D}}_{0}$, the mapping $t \geqslant 0 \mapsto U_{1}(t)f \in X$ is differentiable with $\frac{\mathrm {d}}{\mathrm {d}t}U_{1}(t)f=U_{1}(t){\mathcal {T}}_{\mathrm {max},\,p}f$ for any $t \geqslant 0.$

Proof

In virtue of the previous Lemma, it is enough to prove that $t \geqslant 0 \mapsto U_{1}(t)f \in X$ is differentiable at $t=0$ with

$$\begin{aligned} \frac{\mathrm {d}}{\mathrm {d}t}U_{1}(t)f\vert _{t=0}=U_{1}(0){\mathcal {T}}_{\mathrm {max},\,p}f=0. \end{aligned}$$

Consider $t >0$. One has

$$\begin{aligned} \Vert U_{1}(t)f\Vert _{p}^{p}\leqslant {\left| \left| \left| H \right| \right| \right| }^{p}\,\int _{0}^{t}\Vert {\mathsf {B}}^{+}U_{0}(s)f\Vert _{L^p_+}^{p}\mathrm {d}s. \end{aligned}$$

Now, since $f \in {\mathscr {D}}({\mathcal {T}}_{0,\,p})$, one has from (3.9)

$$\begin{aligned} \int _{0}^{t}\Vert {\mathsf {B}}^{+}U_{0}(s)f\Vert _{L^p_+}^{p}\mathrm {d}s \leqslant \frac{t^{p}}{p}\int _{0}^{t}\left( \int _{\Gamma _{+}}\left| [{\mathcal {T}}_{\mathrm {max},\,p}f](\Phi ({\mathbf {z}},-s))\right| ^{p} \chi _{\{s < \tau _{-}({\mathbf {z}})\}}\mathrm {d}\mu _{+}({\mathbf {z}})\right) \mathrm {d}s \end{aligned}$$

so that

$$\begin{aligned} \frac{\Vert U_{1}(t)f\Vert _{p}^{p}}{t^{p}} \leqslant \frac{ {\left| \left| \left| H \right| \right| \right| }^{p}}{p}\int _{0}^{t}\left( \int _{\Gamma _{+}}\left| [{\mathcal {T}}_{\mathrm {max},\,p}f](\Phi ({\mathbf {z}},-s))\right| ^{p} \chi _{\{s < \tau _{-}({\mathbf {z}})\}}\mathrm {d}\mu _{+}({\mathbf {z}})\right) \mathrm {d}s.\nonumber \\ \end{aligned}$$

(A.5)

Using [5, Proposition 2.2], one has

$$\begin{aligned} \int _{0}^{\infty }\left( \int _{\Gamma _{+}}\left| [{\mathcal {T}}_{\mathrm {max},\,p}f](\Phi ({\mathbf {z}},-s))\right| ^{p} \chi _{\{s< \tau _{-}({\mathbf {z}})\}}\mathrm {d}\mu _{+}({\mathbf {z}})\right) \mathrm {d}s=\Vert {\mathcal {T}}_{\mathrm {max},\,p}f\Vert _{p}^{p} < \infty \end{aligned}$$

so that (A.5) yields

$$\begin{aligned} \lim _{t \rightarrow 0^{+}}\frac{\Vert U_{1}(t)f\Vert _{p}}{t}=0. \end{aligned}$$

This proves the result.$\square$

Lemma A.6

For any $f \in {\mathscr {D}}_{0}$ and any $t >0$, one has $U_{1}(t)f \in {\mathscr {D}}({\mathcal {T}}_{\mathrm {max},\,p})$ with ${\mathcal {T}}_{\mathrm {max},\,p}U_{1}(t)f=U_{1}(t){\mathcal {T}}_{\mathrm {max},\,p}f$.

Proof

The proof follows from a simple application of Proposition A.2 where the assumptions (i)–(iii) are met thanks to the previous Lemmas. $\square$

Let us now establish the following

Lemma A.7

For any $f \in X$ and any $t >0$, one has ${\mathcal {I}}^{1}_{t}[f]:=\int _{0}^{t}U_{1}(s)f \mathrm {d}s \in {\mathscr {D}}({\mathcal {T}}_{\mathrm {max},\,p})$ with

$$\begin{aligned} {\mathcal {T}}_{\mathrm {max},\,p}{\mathcal {I}}^{1}_{t}[f]=U_{1}(t)f, \end{aligned}$$

and ${\mathsf {B}}^{\pm } {\mathcal {I}}_{t}^{1}[f] \mathrm {d}s \in L^{p}_{\pm }$,

$$\begin{aligned} {\mathsf {B}}^{-}{\mathcal {I}}_{t}^{1}[f]=H{\mathsf {B}}^{+}\int _{0}^{t}U_{0}(s)f \mathrm {d}s. \end{aligned}$$

(A.6)

Moreover the mappings $t \geqslant 0\mapsto {\mathsf {B}}^{\pm } {\mathcal {I}}_{t}^{1}[f] \mathrm {d}s \in L^{p}_{\pm }$ are continuous. Finally, for any $f \in {\mathscr {D}}_{0}$ and any $t \geqslant 0$, the traces ${\mathsf {B}}^{\pm }U_{1}(t)f \in L^{p}_{\pm }$ and the mappings $t \geqslant 0\mapsto {\mathsf {B}}^{\pm }U_{1}(t)f \in L^{p}_{\pm }$ are continuous.

Proof

Thanks to the previous Lemmas, the family $(U_{1}(t))_{t\geqslant 0}$ satisfies assumptions (i)–(ii) of Proposition A.1. One deduces then from the same Proposition (point (1)) that, for any $f \in X$ and any $t >0$, ${\mathcal {I}}^{1}_{t}[f] \in {\mathscr {D}}({\mathcal {T}}_{\mathrm {max},\,p})$ with ${\mathcal {T}}_{\mathrm {max},\,p}{\mathcal {I}}_{t}^{1}[f]=U_{1}(t)f-U_{1}(0)f=U_{1}(t)f.$

To show that ${\mathsf {B}}^{-}{\mathcal {I}}_{t}^{1}[f]$ can be expressed through formula (A.6) we first suppose $f \in {\mathscr {D}}_{0}$. For such an f both $U_{0}(t)f$ and $U_{1}(t)f$ belong to ${\mathscr {D}}({\mathcal {T}}_{\mathrm {max},\,p})$ for any $t\geqslant 0$ with ${\mathsf {B}}^{-}U_{1}(t)f=H{\mathsf {B}}^{+}U_{0}(t)f \in L^p_-$. Using this equality, the continuity of H and Proposition A.1 (point 2) applied both to $(U_{1}(t))_{t\geqslant 0}$ and $(U_{0}(t))_{t\geqslant 0}$ one gets

$$\begin{aligned} {\mathsf {B}}^{-}{\mathcal {I}}^{1}_{t}[f]= & {} \int _{0}^{t}{\mathsf {B}}^{-}U_{1}(s)f \mathrm {d}s=\int _{0}^{t}H{\mathsf {B}}^{+}U_{0}(s)f\mathrm {d}s\\= & {} H\left( \int _{0}^{t}{\mathsf {B}}^{+}U_{0}(s)f\mathrm {d}s\right) =H{\mathsf {B}}^{+}{\mathcal {I}}_{t}^{0}[f] \end{aligned}$$

i.e., (A.6) for $f \in {\mathscr {D}}_{0}.$

Consider now $f \in X$ and let $(f_{n})_{n} \in {\mathscr {D}}_{0}$ be such that $\lim _{n}\Vert f_{n}-f\Vert _{p}=0.$ According to Eq. (3.8), the sequence $({\mathsf {B}}^{+}{\mathcal {I}}_{t}^{0}[f_{n}])_{n}$ converges in $L^p_+$ towards ${\mathsf {B}}^{+}{\mathcal {I}}_{t}^{0}[f]$. Since (A.6) holds true for $f_{n}$, and H is continuous, then the sequence $({\mathsf {B}}^{-}{\mathcal {I}}_{t}^{1}[f_{n}])_{n}$ converges in $L^p_-$ to $H{\mathsf {B}}^{+}{\mathcal {I}}_{t}^{0}[f].$ One deduces from this that ${\mathsf {B}}^{-}{\mathcal {I}}_{t}^{1}[f] \in L^p_-$ with (A.6).

Moreover the mapping $t\geqslant 0 \mapsto {\mathsf {B}}^{-}I_{t}^{1}[f] \in L^p_-$ is continuous since both H and the mapping $t \geqslant 0\mapsto {\mathsf {B}}^{+}{\mathcal {I}}_{t}^{0}[f] \in L^p_+$ are continuous (see Proposition 3.3). This property and Proposition A.1 imply that ${\mathsf {B}}^{+}{\mathcal {I}}_{t}^{1}[f] \in L^p_+$ and that the mapping $t \mapsto {\mathsf {B}}^{+}{\mathcal {I}}_{t}^{1}[f] \in L^p_+$ is continuous too.

Finally observe that, if $f \in {\mathscr {D}}_{0}$, then $f \in {\mathscr {D}}({\mathcal {T}}_{\mathrm {max},\,p})$ and for any $t \geqslant 0$ one has

$$\begin{aligned} {\mathcal {I}}_{t}^{1}\left[ {\mathcal {T}}_{\mathrm {max},\,p}f\right] =U_{1}(t)f. \end{aligned}$$

Thus one can state that for any $f \in {\mathscr {D}}_{0}$ and any $t \geqslant 0$, the traces ${\mathsf {B}}^{\pm }U_{1}(t)f \in L^{p}_{\pm }$ and the mappings $t \geqslant 0\mapsto {\mathsf {B}}^{\pm }U_{1}(t)f \in L^{p}_{\pm }$ are continuous. $\square$

Let us now investigate Property (7):

Lemma A.8

One has

$$\begin{aligned} \int _{0}^{t}\Vert {\mathsf {B}}^{+}U_{1}(s)f\Vert _{L^p_+}^{p}\mathrm {d}s \leqslant {\left| \left| \left| H \right| \right| \right| }^{p}\,\int _{0}^{t}\Vert {\mathsf {B}}^{+}U_{0}(s)f\Vert _{L^p_+}^{p}\mathrm {d}s, \quad \forall t \geqslant 0, \forall f \in {\mathscr {D}}_{0}. \end{aligned}$$

Proof

Given $f \in {\mathscr {D}}_{0}$, for any $s >0$ and $\mu _{+}$-a.e. ${\mathbf {z}}\in \Gamma _{+}$:

$$\begin{aligned} \left[ {\mathsf {B}}^{+}U_{1}(s)f\right] ({\mathbf {z}})=\left[ H\left( {\mathsf {B}}^{+}U_{0}(s-\tau _{-}({\mathbf {z}}))f\right) \right] (\Phi ({\mathbf {z}},-\tau _{-}({\mathbf {z}}))\chi _{(0,s)}(\tau _{-}({\mathbf {z}})) \end{aligned}$$

Thus,

$$\begin{aligned} J:= & {} \int _{0}^{t}\Vert {\mathsf {B}}^{+}U_{1}(s)f\Vert _{L^p_+}^{p}\mathrm {d}s\\= & {} \int _{0}^{t}\left( \int _{\Gamma _{+}}\left| \left[ H\left( {\mathsf {B}}^{+}U_{0}(s-\tau _{-}({\mathbf {z}}))\right) f\right] (\Phi ({\mathbf {z}},-\tau _{-}({\mathbf {z}})))\right| ^{p}\chi _{(0,s)}(\tau _{-}({\mathbf {z}}))\mathrm {d}\mu _{+}({\mathbf {z}})\right) \mathrm {d}s. \end{aligned}$$

Now, using Fubini’s Theorem and, for a given ${\mathbf {z}}\in \Gamma _{+}$, the change of variable $s \mapsto s-\tau _{-}({\mathbf {z}})$, we get

$$\begin{aligned} J=\int _{\Gamma _{+}}\left( \int _{0}^{\max (0,t-\tau _{-}({\mathbf {z}}))}\left| \left[ H\left( {\mathsf {B}}^{+}U_{0}(s)\right) f\right] (\Phi ({\mathbf {z}},-\tau _{-}({\mathbf {z}})))\right| ^{p}\mathrm {d}s\right) \mathrm {d}\mu _{+}({\mathbf {z}}). \end{aligned}$$

Using Fubini’s Theorem again

$$\begin{aligned} \begin{aligned} J&\leqslant \int _{0}^{t}\left( \int _{\Gamma _{+}}\left| \left[ H \left( {\mathsf {B}}^{+}U_{0}(s)\right) f\right] (\Phi ({\mathbf {z}},-\tau _{-}({\mathbf {z}})))\right| ^{p}\mathrm {d}\mu _{+}({\mathbf {z}})\right) \mathrm {d}s\\&\leqslant \int _{0}^{t}\left( \int _{\Gamma _{-}}\left| \left[ H \left( {\mathsf {B}}^{+}U_{0}(s)\right) f\right] ({\mathbf {y}})\right| ^{p}\mathrm {d}\mu _{-}({\mathbf {y}})\right) \mathrm {d}s\end{aligned} \end{aligned}$$

where we used [5, Eq. (2.5) in Prop. 2.2]. Therefore, it is easy to check that

$$\begin{aligned} J \leqslant \left\| H \right\| _{{\mathscr {B}}(L^p_+,L^p_+)}^{p}\int _{0}^{t}\Vert {\mathsf {B}}^{+}U_{0}(s)f\Vert _{L^p_+}^{p}\mathrm {d}s. \end{aligned}$$

which is the desired result. $\square$

We finally have the following

Lemma A.9

Given $\lambda >0$ and $f \in X$, set $F_{1}=\int _{0}^{\infty }\exp (-\lambda t)U_{1}(t)f\mathrm {d}t.$ Then $F_{1} \in {\mathscr {D}}({\mathcal {T}}_{\mathrm {max},\,p})$ with ${\mathcal {T}}_{\mathrm {max},\,p}F_{1}=\lambda \,F_{1}$ and ${\mathsf {B}}^{\pm }F_{1} \in L^{p}_{\pm }$ with

$$\begin{aligned} {\mathsf {B}}^{-}F_{1}=H{\mathsf {B}}^{+}C_{\lambda }f = HG_{\lambda }f\quad {\mathsf {B}}^{+}F_{1}=(M_{\lambda }H)G_{\lambda }f. \end{aligned}$$

Proof

Let us first assume $f \in {\mathscr {D}}_{0}$. Then, for any ${\mathbf {y}}\in \Gamma _{-}$, $s \in (0,\tau _{+}({\mathbf {y}}))$:

$$\begin{aligned} \begin{aligned} F_{1}(\Phi ({\mathbf {y}},s))&=\int _{s}^{\infty }\exp (-\lambda t)\left[ H{\mathsf {B}}^{+}U_{0}(t-s)f\right] ({\mathbf {y}})\mathrm {d}t\\&=\exp (-\lambda s)\,\int _{0}^{\infty }\exp (-\lambda t)\left[ H{\mathsf {B}}^{+}U_{0}(t)f\right] ({\mathbf {y}})\mathrm {d}t\\&=\exp (-\lambda s)\left[ H{\mathsf {B}}^{+}\left( \int _{0}^{\infty }\exp (-\lambda t)U_{0}(t)f\mathrm {d}t\right) \right] ({\mathbf {y}}) \end{aligned} \end{aligned}$$

i.e. $F_{1}(\Phi ({\mathbf {y}},s))=\exp (-\lambda s)\left[ H{\mathsf {B}}^{+}C_{\lambda }f\right] ({\mathbf {y}}).$ This exactly means that $F_{1}=\Xi _{\lambda }HG_{\lambda }f$. By a density argument, this still holds for $f \in X$ and we get the desired result easily using the properties of $\Xi _{\lambda }$ and $G_{\lambda }.$ $\square$

The above lemmas prove that the conclusion of Theorem 3.8 is true for $k=1.$ One proves then by induction that the conclusion is true for any $k \geqslant 1$ exactly as above. Details are left to the reader.