Appendix
Proof of Lemma 4.3
Let put
$$\begin{aligned} \Delta _{\sigma (\cdot )}(\lambda ,v,v')={1\over (v-v')\lambda +[v\sigma (v')-v'\sigma (v)]},\ \lambda \in {{F_{-\sigma _0}}}\ \text { and }\ v,v'\in (0,+\infty ). \end{aligned}$$
Naturally, the proof is divided into the following four steps:
Step 1. For every \(\lambda \in {{F_{-\sigma _0}}}\) and \(\varphi \in \Lambda _{\sigma (\cdot ),p}\), we have
$$\begin{aligned}{} & {} e^{-{\lambda +\sigma (v)\over v}\,a}\, \int _0^a e^{-\big ({\lambda +\sigma (v')\over v'}-{\lambda +\sigma (v)\over v}\big )\,\mu '}\, d\mu '\\{} & {} \quad = vv'\Delta _{\sigma (\cdot )}(\lambda ,v,v')\,\big [e^{-{\lambda +\sigma (v)\over v}\,a}-e^{-{\lambda +\sigma (v')\over v'}\,a}\big ].\end{aligned}$$
Further,
$$\begin{aligned}{} & {} \pi _\lambda {\mathbb {B}}_\varepsilon Q_\lambda \varphi (a,v)=\displaystyle {f(v)\over v}\displaystyle \int _0^a d\mu '\, e^{-{\lambda +\sigma (v)\over v}\,(a-\mu ')}\\{} & {} \quad \int _0^{+\infty } dv'\, g(v')\,\Lambda _\varepsilon (v,v')\,e^{-{\lambda +\sigma (v')\over v'}\,\mu '}\,\varphi (0,v').\end{aligned}$$
Then, by applying Fubini’s theorem we get:
$$\begin{aligned}{} & {} \pi _\lambda {\mathbb {B}}_\varepsilon Q_\lambda \varphi (a,v)\\{} & {} \quad =\displaystyle {f(v)\over v}\displaystyle \int _0^{+\infty } dv'\, g(v')\,\Lambda _\varepsilon (v,v')\,\varphi (0,v')\,e^{-{\lambda +\sigma (v)\over v}\,a}\int _0^a d\mu '\, e^{-\big ({\lambda +\sigma (v')\over v'}-{\lambda +\sigma (v)\over v}\big )\,\mu '}\\{} & {} \quad = \displaystyle {f(v)}\displaystyle \int _0^{+\infty } dv'\, g(v')\,\Lambda _\varepsilon (v,v')\,\varphi (0,v')\,{v'\Delta _{\sigma (\cdot )}(\lambda ,v,v')}\big [e^{-{\lambda +\sigma (v)\over v}\,a}-e^{-{\lambda +\sigma (v')\over v'}\,a}\big ]. \end{aligned}$$
Let put \(\pi _\lambda =\pi _{(\lambda ,\sigma (\cdot ))}\), \(Q_\lambda =Q_{(\lambda ,\sigma (\cdot ))}\) and notice that
$$\begin{aligned}{} & {} |\pi _\lambda {\mathbb {B}}_\varepsilon Q_\lambda \varphi (a,v)|\\{} & {} \quad \le \pi _{(Re\lambda ,\sigma _{0})} {\mathbb {B}}_\varepsilon Q_{(Re\lambda ,\sigma _{0})} |\varphi |(a,v) \\{} & {} \quad \le 2\displaystyle \displaystyle {f(v)}\displaystyle \int _0^{+\infty } dv'\, g(v')\,\Lambda _\varepsilon (v,v')\,|\varphi (0,v')|\,{v'\Delta _{\sigma _0}(Re\lambda ,v,v')} \\{} & {} \quad \le 2 d^{1/q}\displaystyle \displaystyle {f(v)}\displaystyle \int _0^{+\infty } dv'\, {g(v')\over \sigma (v')^{1/p}}\,\Lambda _\varepsilon (v,v')\,\sigma (v')^{1/p}|\varphi (0,v')|\,{v'}^{1/p}\Delta _{\sigma _0}(Re\lambda ,v,v'). \end{aligned}$$
This, by using Hölder’s inequality, implies that
$$\begin{aligned}{} & {} \Vert \pi _\lambda {\mathbb {B}}_\varepsilon Q_\lambda \varphi \Vert _{\Lambda _{\sigma (\cdot ),p}}\\{} & {} \quad \le {2^{1/q}\,d^{(p+1)/p}\,\varepsilon ^{-1}\over Re\lambda +\sigma _0 }\,\Vert f(\cdot )\Vert _{L^\infty (0,+\infty )}\,\Vert \sigma (\cdot )_{|supp(f(\cdot ))}\Vert ^{1/p}_{L^\infty (0,+\infty )}\\{} & {} \quad \Vert {g(\cdot )\over \sigma (\cdot )^{1/p}}\Vert _{L^q(0,+\infty )} \,\Vert \varphi \Vert _{\Lambda _{\sigma (\cdot ),p}}, \end{aligned}$$
which ends the first step.
Step 2. For every \(\lambda \in {{F_{-\sigma _0}}}\) and \(\varphi \in {\mathbb {X}}_{\sigma (\cdot ),p}\) we have
$$\begin{aligned}{} & {} \displaystyle e^{{\lambda +\sigma (v')\over v'}\,\mu "}\,e^{-{\lambda +\sigma (v)\over v}\,a}\, \int _{\mu "}^a e^{-\big ({\lambda +\sigma (v')\over v'}-{\lambda +\sigma (v)\over v}\big )\,\mu '}\, d\mu '\\{} & {} \quad \displaystyle = vv'\Delta _{\sigma (\cdot )}(\lambda ,v,v')\,\big [e^{-{\lambda +\sigma (v')\over v'}\,(a-\mu ")}-e^{-{\lambda +\sigma (v)\over v}\,(a-\mu ")}\big ]. \end{aligned}$$
Moreover,
$$\begin{aligned} \pi _\lambda {\mathbb {B}}_\varepsilon \Xi _\lambda \varphi (a,v)= & {} \displaystyle {f(v)\over v}\displaystyle \!\int _0^a d\mu '\, e^{-{\lambda +\sigma (v)\over v}\,(a-\mu ')}\!\int _0^{+\infty } dv'\, g(v')\,\Lambda _\varepsilon (v,v')\\{} & {} \times \displaystyle {1\over v'}\!\int _0^{\mu '}d\mu "\,e^{-{\lambda +\sigma (v')\over v'}\,(\mu '-\mu ")}\,\varphi (\mu ",v'). \end{aligned}$$
This, by applying Fubini’s theorem, yields to
$$\begin{aligned}{} & {} \pi _\lambda {\mathbb {B}}_\varepsilon \Xi _\lambda \varphi (a,v)=\displaystyle {f(v)\over v}\displaystyle \int _0^{+\infty } dv'\, g(v')\,\Lambda _\varepsilon (v,v'){1\over v'}\\{} & {} \qquad \times \displaystyle \int _0^{a}d\mu "\int _{\mu "}^a d\mu '\, e^{-{\lambda +\sigma (v)\over v}\,(a-\mu ')} \,e^{-{\lambda +\sigma (v')\over v'}\,(\mu '-\mu ")}\,\varphi (\mu ",v')\\{} & {} \quad =\displaystyle {f(v)\over v}\displaystyle \int _0^{+\infty } dv'\, g(v')\,\Lambda _\varepsilon (v,v'){1\over v'}\int _0^{a}d\mu " \,\varphi (\mu ",v')\\{} & {} \qquad \times \displaystyle e^{{\lambda +\sigma (v')\over v'}\,\mu "}\,e^{-{\lambda +\sigma (v)\over v}\,a}\int _{\mu "}^a d\mu '\, e^{-\big ({\lambda +\sigma (v')\over v'}-{\lambda +\sigma (v)\over v}\big )\,\mu '}\\{} & {} \quad =\displaystyle {f(v)}\displaystyle \int _0^{+\infty } dv'\, g(v')\,\Lambda _\varepsilon (v,v')\,\Delta _{\sigma (\cdot )}(\lambda ,v,v')\\{} & {} \qquad \times \displaystyle \int _0^{a}d\mu " \,\varphi (\mu ",v')\displaystyle \,\big [e^{-{\lambda +\sigma (v')\over v'}\,(a-\mu ")}-e^{-{\lambda +\sigma (v)\over v}\,(a-\mu ")}\big ]. \end{aligned}$$
Let put \(\Xi _\lambda =\Xi _{(\lambda ,\sigma (\cdot ))}\) and notice that
$$\begin{aligned}{} & {} |\pi _\lambda {\mathbb {B}}_\varepsilon \Xi _\lambda \varphi (a,v)|\le \pi _{(Re\lambda ,\sigma _{0})} {\mathbb {B}}_\varepsilon \Xi _{(Re\lambda ,\sigma _{0})}|\varphi | (a,v) \\{} & {} \quad \le 2\displaystyle \displaystyle {f(v)}\displaystyle \int _0^{+\infty } dv'\, g(v')\,\Lambda _\varepsilon (v,v')\,\Delta _{\sigma _0}(Re\lambda ,v,v')\int _0^{a}d\mu " \,|\varphi (\mu ",v')|\displaystyle . \end{aligned}$$
This, by using Hölder’s inequality, implies that
$$\begin{aligned}{} & {} \Vert \pi _\lambda {\mathbb {B}}_\varepsilon \Xi _\lambda \varphi \Vert _{\Lambda _{\sigma (\cdot ),p}}\\{} & {} \quad \le {d^2\,\varepsilon ^{-1}\over Re\lambda +\sigma _0 }\,\Vert f(\cdot )\Vert _{L^\infty (0,+\infty )}\,\Vert \sigma (\cdot )_{|supp(f(\cdot ))}\Vert _{L^\infty (0,+\infty )}\, \Vert {g(\cdot )\over \sigma (\cdot )^{1/p}}\Vert _{L^q(0,+\infty )} \,\Vert \varphi \Vert _{{\mathbb {X}}_{\sigma (\cdot ),p}}, \end{aligned}$$
which ends the second step.
Step 3. For every \(\lambda \in {{F_{-\sigma _0}}}\) and \(\varphi \in \Lambda _{\sigma (\cdot ),p}\) we have
$$\begin{aligned}{} & {} e^{-{\lambda +\sigma (v)\over v}\,\mu }\, \int _0^\mu e^{-\big ({\lambda +\sigma (v')\over v'}-{\lambda +\sigma (v)\over v}\big )\,\mu '}\, d\mu '\\{} & {} \quad = vv'\Delta _{\sigma (\cdot )}(\lambda ,v,v')\,\big [e^{-{\lambda +\sigma (v)\over v}\,\mu }-e^{-{\lambda +\sigma (v')\over v'}\,\mu }\big ].\end{aligned}$$
Further,
$$\begin{aligned}{} & {} \Xi _\lambda {\mathbb {B}}_\varepsilon Q_\lambda \varphi (\mu ,v)=\displaystyle {f(v)\over v}\displaystyle \int _0^\mu d\mu '\, e^{-{\lambda +\sigma (v)\over v}\,(\mu -\mu ')}\\{} & {} \quad \int _0^{+\infty } dv'\, g(v')\,\Lambda _\varepsilon (v,v')\,e^{-{\lambda +\sigma (v')\over v'}\,\mu '}\,\varphi (v'). \end{aligned}$$
Then, by applying Fubini’s theorem, we get
$$\begin{aligned}{} & {} \Xi _\lambda {\mathbb {B}}_\varepsilon Q_\lambda \varphi (\mu ,v)\\{} & {} \quad =\displaystyle {f(v)\over v}\displaystyle \int _0^{+\infty } dv'\, g(v')\,\Lambda _\varepsilon (v,v')\,\varphi (v')\,e^{-{\lambda +\sigma (v)\over v}\,\mu }\int _0^\mu d\mu '\, e^{-\big ({\lambda +\sigma (v')\over v'}-{\lambda +\sigma (v)\over v}\big )\,\mu '}\\{} & {} \quad = \displaystyle {f(v)}\displaystyle \int _0^{+\infty } dv'\, g(v')\,\Lambda _\varepsilon (v,v')\,\varphi (v')\,{v'\Delta _{\sigma (\cdot )}(\lambda ,v,v')}\big [e^{-{\lambda +\sigma (v)\over v}\,\mu }-e^{-{\lambda +\sigma (v')\over v'}\,\mu }\big ]. \end{aligned}$$
Notice that
$$\begin{aligned}{} & {} |\Xi _\lambda {\mathbb {B}}_\varepsilon Q_\lambda \varphi (\mu ,v)|\\{} & {} \quad \le \Xi _{(Re\lambda ,\sigma _{0})} {\mathbb {B}}_\varepsilon Q_{(Re\lambda ,\sigma _{0})} |\varphi |(\mu ,v) \\{} & {} \quad \le 2\displaystyle \displaystyle {f(v)}\displaystyle \int _0^{+\infty } dv'\, g(v')\,\Lambda _\varepsilon (v,v')\,|\varphi (v')|\,{v'\Delta _{\sigma _0}(Re\lambda ,v,v')} \\{} & {} \quad \le 2d^{1/q}\displaystyle \displaystyle {f(v)}\displaystyle \int _0^{+\infty } dv'\, {g(v')\over \sigma (v')^{1/p}}\,\Lambda _\varepsilon (v,v')\, \sigma (v')^{1/p}|\varphi (v')|\,{v'}^{1/p}\Delta _{\sigma _0}(Re\lambda ,v,v'). \end{aligned}$$
This, by using Hölder’s inequality, implies that
$$\begin{aligned}{} & {} \Vert \Xi _\lambda {\mathbb {B}}_\varepsilon Q_\lambda \varphi \Vert _{{\mathbb {X}}_{\sigma (\cdot ),p}}\\{} & {} \quad \le {2\,a^{1/p}\,d\,\varepsilon ^{-1}\over Re\lambda +\sigma _0 }\,\Vert f(\cdot )\Vert _{L^\infty (0,+\infty )}\,\Vert \sigma (\cdot )_{|supp(f(\cdot ))}\Vert _{L^\infty (0,+\infty )}\,\Vert {g(\cdot )\over \sigma (\cdot )^{1/p}}\Vert _{L^\infty (0,+\infty )} \,\Vert \varphi \Vert _{\Lambda _{\sigma (\cdot ),p}}, \end{aligned}$$
which ends the third step.
Step 4. For every \(\lambda \in {{F_{-\sigma _0}}}\) and \(\varphi \in {\mathbb {X}}_{\sigma (\cdot ),p}\) we have
$$\begin{aligned}{} & {} e^{-{\lambda +\sigma (v)\over v}\,\mu }\,e^{{\lambda +\sigma (v')\over v'}\,\mu "} \int _{\mu "}^\mu e^{-\big ({\lambda +\sigma (v')\over v'}-{\lambda +\sigma (v)\over v}\big )\,\mu '}\, d\mu '\\{} & {} \quad = vv'\Delta _{\sigma (\cdot )}(\lambda ,v,v')\,\big [e^{-{\lambda +\sigma (v')\over v'}\,(\mu -\mu ")}-e^{-{\lambda +\sigma (v)\over v}\,(\mu -\mu ")}\big ].\end{aligned}$$
Further,
$$\begin{aligned}{} & {} \Xi _\lambda {\mathbb {B}}_\varepsilon \Xi _\lambda \varphi (\mu ,v)\\{} & {} \quad =\displaystyle {f(v)\over v}\displaystyle \int _0^\mu d\mu '\, e^{-{\lambda +\sigma (v)\over v}\,(\mu -\mu ')}\int _0^{+\infty } {dv'\over v'}\, g(v')\,\Lambda _\varepsilon (v,v')\\{} & {} \quad \int _0^{\mu '}d\mu "\,e^{-{\lambda +\sigma (v')\over v'}\,(\mu '-\mu ")}\,\varphi (\mu ",v'). \end{aligned}$$
Then, by applying Fubini’s theorem, we deduce that
$$\begin{aligned}{} & {} \Xi _\lambda {\mathbb {B}}_\varepsilon \Xi _\lambda \varphi (\mu ,v)=\displaystyle {f(v)\over v}\displaystyle \int _0^{+\infty } {dv'\over v'}\, g(v')\,\Lambda _\varepsilon (v,v')\,e^{-{\lambda +\sigma (v)\over v}\,\mu }\\{} & {} \qquad \int _0^\mu d\mu "\, e^{{\lambda +\sigma (v')\over v'}\,\mu "}\varphi (\mu ",v')\\{} & {} \qquad \times \,\displaystyle \int _{\mu "}^\mu d\mu '\, e^{-\big ({\lambda +\sigma (v')\over v'}-{\lambda +\sigma (v)\over v}\big )\,\mu '}\\{} & {} \quad = \displaystyle {f(v)}\displaystyle \int _0^{+\infty } dv'\, g(v')\,\Lambda _\varepsilon (v,v')\int _0^\mu d\mu "\,\varphi (\mu ",v')\,{\Delta _{\sigma (\cdot )}(\lambda ,v,v')}\\{} & {} \qquad \times \,\big [e^{-{\lambda +\sigma (v')\over v'}\,(\mu -\mu ")}-e^{-{\lambda +\sigma (v)\over v}\,(\mu -\mu ")}\big ]. \end{aligned}$$
Notice that
$$\begin{aligned}&|\Xi _\lambda {\mathbb {B}}_\varepsilon \Xi _\lambda \varphi (\mu ,v)|\\&\quad \le \Xi _{(Re\lambda ,\sigma _{0})} {\mathbb {B}}_\varepsilon \Xi _{(Re\lambda ,\sigma _{0})} |\varphi |(\mu ,v) \\&\quad \le 2\displaystyle \displaystyle {f(v)}\displaystyle \int _0^{+\infty } dv'\, g(v')\,\Lambda _\varepsilon (v,v')\int _0^\mu d\mu "\,|\varphi (\mu ",v')|\,{\Delta _{\sigma _0}(Re\lambda ,v,v')} \\&\quad = 2\displaystyle \displaystyle {f(v)}\int _0^\mu d\mu "\,\displaystyle \int _0^{+\infty } dv'\, {g(v')\over \sigma (v')^{1/p}}\,\Lambda _\varepsilon (v,v')\,\sigma (v')^{1/p}|\varphi (\mu ",v')|\,{\Delta _{\sigma _0}(Re\lambda ,v,v')} \end{aligned}$$
by using Fubini’s theorem, again. This, by using Hölder’s inequality, amounts to
$$\begin{aligned}\begin{array}{c} \Vert \Xi _\lambda {\mathbb {B}}_\varepsilon \Xi _\lambda \varphi \Vert _{{\mathbb {X}}_{\sigma (\cdot ),p}}\\ \le {2\,a^{1/p}\,d^{1/p}\,\varepsilon ^{-1}\over Re\lambda +\sigma _0 }\,\Vert f(\cdot )\Vert _{L^\infty (0,+\infty )}\,\Vert \sigma (\cdot )_{|supp(f(\cdot ))}\Vert ^{1/p}_{L^\infty (0,+\infty )}\,\Vert {g(\cdot )\over \sigma (\cdot )^{1/p}}\Vert _{L^\infty (0,+\infty )} \,\Vert \varphi \Vert _{{\mathbb {X}}_{\sigma (\cdot ),p}},\end{array}\end{aligned}$$
which proves the forth step and finishes the proof of Lemma 4.3. \(\square \)
