and $\frac{Te^T}{e^T-1}$ is bounded in a neighborhood of $0$, we obtain an estimate on $\Vert \partial_s \widetilde{\mathrm{X}}^{s;t}(x,v) \Vert_{\Ld^{\infty}_{x,v}}$ and then on $\Vert \partial_s \psi_{s,t}^x(v)\Vert_{\Ld^{\infty}_{x,v}}$ as before. Using the same induction procedure as for (<ref>), we finally obtain (<ref>).

From the $\W^{k, \infty}_{x,v}$-bounds we have obtained on $\mathrm{Y}^{s,t}$ along the proof, and because

\begin{align*}
\mathrm{Y}^{s;t}(x,v)=\mathrm{X}^{s;t}(x,v)-x-(1-e^{t-s})v,
\end{align*}

we can infer that

\begin{align}\label{bound:Wkinfty-X}
\underset{s,t \in [0,T]}{\sup} \, \left\Vert \partial_{x,v}^{\beta} \left(\X_{s,t}(x,v)-x-(1-e^{t-s})v\right) \right\Vert_{\Ld^{\infty}(\T^d_x \times \R^d_v)} &\leq \varphi(T) \Lambda(T,R), \ \ \vert \beta \vert \leq k.
\end{align}

Likewise, by considering

\begin{align*}
\mathrm{W}^{s,t}(x,v)=\V^{s,t}(x,v)-e^{t-s}v,
\end{align*}

one can obtain the estimate

\begin{align}\label{bound:Wkinfty-V}
\underset{s,t \in [0,T]}{\sup} \, \left\Vert \partial_{x,v}^{\beta} \left(\V_{s,t}(x,v)-e^{t-s}v \right) \right\Vert_{\Ld^{\infty}(\T^d_x \times \R^d_v)} &\leq \varphi(T) \Lambda(T,R), \ \ \vert \beta \vert \leq k.
\end{align}

Let us conclude this section by showing that in Lemma <ref>, one can consider

\begin{align*}
(T, \mathrm{F})=(T_{\eps},E_{\reg,\eps}^{u_\eps,\varrho_\eps}),
\end{align*}

for a given $\eps>0$, where $(T_{\eps},E_{\reg,\eps}^{u_\eps,\varrho_\eps})$ have been defined in the set-up of the bootstrap argument of Section <ref>. Let us prove that the assumptions of Lemma <ref> indeed hold with $k=m-2-d/2$. By the estimate (<ref>) from Lemma <ref>, we have for all $t \in (0,T_{\eps})$ and $\ell <m-1-d/2$

\begin{align*}
\LRVert{E_{\reg,\eps}^{u_\eps,\varrho_\eps}(t)}_{\W^{\ell, \infty}_{x}} &\lesssim \LRVert{E_{\reg,\eps}^{u_\eps,\varrho_\eps}(t)}_{H^{m-1}} \\
&\lesssim \Vert u_\eps(t) \Vert_{\H^{m-1}}
+ \Lambda\left(\Vert \varrho_\eps(t) \Vert_{\H^{m-2}}  \right)   \Vert \varrho_\eps (t) \Vert_{\H^{m}}.
\end{align*}

Appealing to Lemma <ref>, and using $\mathcal{N}_{m,r}(f_\eps,\varrho_\eps,u_\eps,T_\eps)\leq R$ for all $T \leq T_{\eps}$, we deduce

\begin{align*}
\Vert E_{\reg,\eps}^{u_\eps,\varrho_\eps} \Vert_{\Ld^2(0,T; \W^{k, \infty}(\T^d))} \leq \Lambda(T,R),
\end{align*}

for $k=m-2-d/2$.

§ AVERAGING OPERATORS RELATED TO THE DYNAMICS WITH FRICTION

For any smooth vector field $G(t,s,x,v) \in \R^d$, we consider the following integral operator $\mathrm{K}^{\mathrm{free}}_G$ acting on functions $H(t,x)$:

\begin{align*}
\mathrm{K}_G^{\mathrm{free}}[H](t,x)&:=\int_0^t \int_{\R^d}   [\nabla_x H](s,x-(t-s)v)\cdot G(t,s,x,v) \, \mathrm{d}v \, \mathrm{d}s.
\end{align*}

This operator, featuring an apparent loss of derivative in space, was systematically studied  in [93]. It was proven in <cit.> that this loss is only apparent, provided that the kernel is sufficiently smooth and decaying in velocity. The statement goes as follows.

Let $T>0$. If $p>1+d$ and $\sigma >d/2$ then for all $H  \in \Ld^2(0,T;\Ld^2(\T^d))$

\begin{align*}
\Big\Vert \mathrm{K}_G^{\mathrm{free}}[H]\Big\Vert_{\Ld^2(0,T;\Ld^2(\T^d))} \lesssim \underset{0 \leq s,t \leq T}{\sup} \Vert G(t,s) \Vert_{\mathcal{H}^p_{\sigma}} \LRVert{H}_{\Ld^2(0,T;\Ld^2(\T^d))} .
\end{align*}

As already noted in [93], this smoothing estimate is reminiscent of (but different from) the so-called kinetic averaging lemmas. Namely, Proposition <ref> provides the gain of one full derivative.

Averaging lemmas are well-known to provide powerful regularity and compactness results in the study of kinetic equations. Loosely speaking, moments in velocity of the solutions appear to gain some regularity compared to the solutions themselves, which are just transported along the flow of the equation. We refer to [71, 3, 69] for the introduction of the averaging lemmas, and to [60, 124, 74, 103, 9, 6, 101, 4] for several extensions of such results.

A thorough comparison between standard kinetic averaging lemmas and the estimate from Proposition <ref> can be found in [82]. We finally refer to [91] for the use of Proposition <ref> for a sligthly different purpose, as well as to [42] for an extension of this proposition.

In this section, we prove crucial smoothing estimates adaptated to kinetic equations with friction, in the spirit of Proposition <ref>. First, we define the corresponding integral operator.

For any smooth vector field $G(t,s,x,v)$, we define the following integral operators acting on functions $H(t,x)$ by

\begin{align*}
%\mathrm{K}_G^{\mathrm{free}}[F](t,x)&:=\int_0^t \int_{\R^d}   [\nabla_x F](s,x-(t-s)v)\cdot G(t,s,x,v) \, \mathrm{d}v \, \mathrm{d}s, \\
%%K_G^1[F](t,x)&:=\int_0^t \int_{\R^d}  v [\nabla_x F](s,x-(t-s)v)\cdot G(t,s,x,v) \, \mathrm{d}v \, \mathrm{d}s,
\mathrm{K}_G^{\mathrm{fric}}[H](t,x)&:=\int_0^t \int_{\R^d}   [\nabla_x H](s,x+(1-e^{t-s})v)\cdot G(t,s,x,v) \, \mathrm{d}v \, \mathrm{d}s, %\\
%\mathscr{K}_G^1[F](t,x)&:=\int_0^t \int_{\R^d}   v[\nabla_x F](s,x+(1-e^{t-s})v)\cdot G(t,s,x,v) \, \mathrm{d}v \, \mathrm{d}s, \\
\end{align*}

where $(t,x) \in \R^+ \times \T^d$.

Assuming that the kernel $G$ is sufficiently smooth and decaying in velocity, we will prove several continuity and regularization estimates for $\mathrm{K}_G^{\mathrm{free}}$ and $\mathrm{K}_G^{\mathrm{fric}}$ (see Propositions <ref>–<ref>–<ref> below).

In what follows, $\mathcal{F}_{x,v}$ will refer to the Fourier transform on $\T^d \times \R^d$ defined as

\begin{align*}
\mathcal{F}_{x,v}h(k,\xi)&=\int_{\T^d \times \R^d } e^{-i (k  \cdot x+v \cdot \xi )} h(x,v)   \, \mathrm{d}x \, \mathrm{d}v, \ \ (k, \xi) \in \Z^d \times \R^d.
\end{align*}

Our first result is the following.

There exists $C>0$ such that the followings holds. Suppose that $G_{[q]}(t,s,x,v)$ is a kernel of the form

\begin{align*}
G_{[q]}(t,s,x,v)=(t-s)^q \mathcal{G}(t,s,x,v),
\end{align*}

with $q \in \N$.
For every $T>0$ satisfying

\begin{align*}
\Vert \mathcal{G} \Vert_{T,s_1,s_2}:=\underset{0\leq t \leq T}{\sup} \left(  \sum_{m \in \Z^d}\underset{0\leq s \leq t}{\sup} \, \underset{\xi \in \R^d }{\sup} \,  \big\lbrace (1+\vert m \vert)^{s_2}(1+\vert \xi \vert)^{s_1} \vert (\mathcal{F}_{x,v} \mathcal{G})(t,s,m,\xi) \vert \big\rbrace^2 \right)^{\frac{1}{2}} < + \infty,
\end{align*}

for $s_1>1+2q$ and $s_2>d/2+2q$,

\begin{align*}
\Big\Vert \nabla^q_x \mathrm{K}_{G_{[q]}}^{\mathrm{free}}[H] \Big\Vert_{\Ld^2(0,T; \Ld^2(\T^d))} +\Big\Vert \nabla^q_x \mathrm{K}_{G_{[q]}}^{\mathrm{fric}}[H] \Big\Vert_{\Ld^2(0,T; \Ld^2(\T^d))} &\leq C \Vert \mathcal{G} \Vert_{T,s_1,s_2} \Vert H \Vert_{\Ld^2(0,T; \Ld^2(\T^d))}.%, \\[2mm]
%\Vert K_G^1[F] \Vert_{\Ld^2(0,T; \Ld^2(\T^d))} +\Vert \mathrm{K}_G^1[F] \Vert_{\Ld^2(0,T; \Ld^2(\T^d))} &\leq C \Vert v \otimes G \Vert_{T,s_1,s_2} \Vert F \Vert_{\Ld^2(0,T; \Ld^2(\T^d))}
\end{align*}

We only perform the proof in the case of $ \mathrm{K}_{G_{[q]}}^{\mathrm{fric}}$ (the proof is similar for $ \mathrm{K}_{G_{[q]}}^{\mathrm{free}}$). Writing for all $t \geq 0$

\begin{align*}
H(t,x):=\sum_{k \in \Z^d} \widehat{H}_k(t) e^{i k \cdot x} \ \ \text{in} \ \Ld^2(\T^d),
\end{align*}

we have

\begin{align*}
\mathrm{K}_{G_{[q]}}^{\mathrm{fric}}[H](t,x)&=\int_0^t  \sum_{k \in \Z^d} \widehat{H}_k(s) e^{i k \cdot x} (ik) \cdot  \int_{\R^d} e^{-ik \cdot (e^{t-s}-1)v}  G_{[q]}(t,s,x,v) \, \mathrm{d}v \, \mathrm{d}s \\
&=\int_0^t  \sum_{k \in \Z^d} \widehat{H}_k(s) e^{i k \cdot x} (ik) \cdot (\mathcal{F}_v G_{[q]})(t,s,x,k(e^{t-s}-1)) \, \mathrm{d}s.
\end{align*}

We now expand $G_{[q]}$ in Fourier series along the $x$ variable so that

\begin{align*}
\mathrm{K}_{G_{[q]}}^{\mathrm{fric}}[H](t,x)&= \sum_{k \in \Z^d}  \sum_{l \in \Z^d}e^{i (k+\ell) \cdot x}   \int_0^t \widehat{H}_k(s)  (ik) \cdot (\mathcal{F}_{x,v} G_{[q]})(t,s,\ell,k(e^{t-s}-1)) \, \mathrm{d}s \\
&=\sum_{\ell' \in \Z^d}e^{i \ell' \cdot x}  \left\lbrace \sum_{k \in \Z^d}  \int_0^t \widehat{H}_k(s)  (ik) \cdot (\mathcal{F}_{x,v} G_{[q]})(t,s,\ell'-k,k(e^{t-s}-1)) \, \mathrm{d}s \right\rbrace,
\end{align*}

and then

\begin{align*}
\nabla_x^q \mathrm{K}^{\mathrm{fric}}_{(t-s)^q \mathcal{G}}[H](t,x)=\sum_{\ell \in \Z^d}e^{i \ell \cdot x}  \left\lbrace (i \ell)^q \sum_{k \in \Z^d}  \int_0^t (t-s)^q \widehat{H}_k(s)  (ik) \cdot (\mathcal{F}_{x,v} \mathcal{G})(t,s,\ell-k,k(e^{t-s}-1)) \, \mathrm{d}s \right\rbrace.
\end{align*}

By the Parseval equality and the Cauchy-Schwarz inequality (in frequency and time), we get

\begin{align*}
\Vert \nabla_x^q \mathrm{K}^{\mathrm{fric}}_{(t-s)\mathcal{G}}[F](t) \Vert_{\Ld^2(\T^d)}^2&= \sum_{\ell \in \Z^d} \vert \ell \vert^{2q} \left\vert \sum_{k \in \Z^d}  \int_0^t (t-s)^q \widehat{H}_k(s)  (ik) \cdot (\mathcal{F}_{x,v} \mathcal{G})(t,s,\ell-k,k(e^{t-s}-1)) \, \mathrm{d}s \right\vert^2 \\
& \leq \sum_{\ell \in \Z^d}  \vert \ell \vert^{2q} \left(\sum_{k \in \Z^d}  \int_0^t (t-s)^{2q}  \vert \widehat{H}_k(s) \vert^2 \vert k \cdot (\mathcal{F}_{x,v} \mathcal{G})(t,s,\ell-k,k(e^{t-s}-1)) \vert    \, \dd s \right) \\
& \qquad     \times \left( \sum_{k \in \Z^d}  \int_0^t \vert k \cdot (\mathcal{F}_{x,v} \mathcal{G})(t,s,\ell-k,k(e^{t-s}-1)) \vert  \dd s \right).
\end{align*}

Integrating in time yields

\begin{align*}
&\Vert \nabla_x^q \mathrm{K}^{\mathrm{fric}}_{(t-s)\mathcal{G}}[F] \Vert_{\Ld^2(0,T ; \Ld^2( \times \T^d))}^2 \\
&\leq \sum_{\ell \in \Z^d} \vert \ell \vert^{2q}  \int_0^T  \int_0^t  \sum_{k \in \Z^d} (t-s)^{2q}  \vert \widehat{H}_k(s) \vert^2 \vert k \cdot (\mathcal{F}_{x,v} \mathcal{G})(t,s,\ell-k,k(e^{t-s}-1)) \vert    \, \dd s \dd t \\
& \quad \times \underset{\ell \in \Z^d}{\sup} \underset{t \in (0,T)}{\sup}  \sum_{k \in \Z^d}  \int_0^t \vert k \cdot (\mathcal{F}_{x,v} \mathcal{G})(t,s,\ell-k,k(e^{t-s}-1)) \vert  \dd s \\
&=(\mathrm{I}) \times (\mathrm{II}).
\end{align*}

A first step is to note that

\begin{align*}
\sum_{k \in \Z^d}  \int_0^t &\vert k \cdot (\mathcal{F}_{x,v} \mathcal{G})(t,s,\ell-k,k(e^{t-s}-1)) \vert  \dd s  \\
&\leq \sum_{k \in \Z^d} \underset{\substack{s \in (0,t)\\ \xi \in \R^d}}{\sup} (1+\vert \xi \vert)^{\alpha} \vert \mathcal{F}_{x,v}\mathcal{G}(t,s,\ell-k,\xi)) \vert \int_0^t \frac{\vert k \vert}{(1+\vert k \vert(e^{t-s}-1))^{\alpha}} \dd s \\
&\leq \sum_{k \in \Z^d} \underset{\substack{s \in (0,t)\\ \xi \in \R^d}}{\sup} (1+\vert \xi \vert)^{\alpha} \vert \mathcal{F}_{x,v}\mathcal{G}(t,s,\ell-k,\xi)) \vert \int_0^t \frac{\vert k \vert}{(1+\vert k \vert(t-s))^{\alpha}} \dd s,
\end{align*}

and that the change of variable $\tau=\vert k \vert (t-s)$ in the last integral yields

\begin{align*}
\int_0^t \frac{\vert k \vert}{(1+\vert k \vert(t-s))^{\alpha}} \dd s \leq \int_0^{+\infty} \frac{\dd \tau}{(1+\tau)^{\alpha}}<+\infty,
\end{align*}

provided that $\alpha >1$. With this observation, we can treat the term $(\mathrm{II})$ and obtain by the Cauchy-Schwarz inequality in $k$

\begin{align*}
(\mathrm{II}) \leq \underset{\ell \in \Z^d}{\sup} \underset{t \in (0,T)}{\sup} \sum_{k \in \Z^d} \underset{\substack{s \in (0,t)\\ \xi \in \R^d}}{\sup} (1+\vert \xi \vert)^{s_1} \vert \mathcal{F}_{x,v} \mathcal{G}(t,s,\ell-k,\xi)) \vert \leq \Vert \mathcal{G} \Vert_{T,s_1,s_2},
\end{align*}

for $s_1>1$ and $s_2>d/2$. For the term $(\mathrm{I})$, we use Fubini-Tonelli theorem and get

\begin{align*}
(\mathrm{I})&=\sum_{\ell \in \Z^d} \vert \ell \vert^{2q} \int_0^T  \int_0^t  \sum_{k \in \Z^d} (t-s)^{2q}  \vert \widehat{H}_k(s) \vert^2 \vert k \vert \vert (\mathcal{F}_{x,v} \mathcal{G})(t,s,\ell-k,k(e^{t-s}-1))\vert       \, \dd s \, \dd t \\
%&= \int_0^T \sum_{k \in \Z^d} \vert \widehat{H}_k(s) \vert^{2}   \int_s^T \sum_{\ell \in \Z^d} (t-s)^{2q}  \vert \ell \vert^{2q} \vert k \vert \vert (\mathcal{F}_{x,v} \mathcal{G})(t,s,\ell-k,k(e^{t-s}-1)) \vert        \, \dd t \, \dd s \\
&= \int_0^T \sum_{k \in \Z^d}   \vert \widehat{H}_k(s) \vert^2 \int_s^T \sum_{\ell \in \Z^d} (t-s)^{2q} \vert \ell \vert^{2q} \vert k \vert  \vert (\mathcal{F}_{x,v} \mathcal{G})(t,s,\ell-k,k(e^{t-s}-1))\vert      \, \dd t \, \dd s \\
&\leq \Vert H \Vert_{\Ld^2(0,T ; \Ld^2(\T^d))}^2  \underset{k \in \Z^d}{\sup} \underset{0\leq s \leq T}{\sup}\int_s^T \sum_{\ell \in \Z^d} (t-s)^{2q}  \vert \ell \vert^{2q} \vert k \vert  \vert (\mathcal{F}_{x,v} \mathcal{G})(t,s,\ell-k,k(e^{t-s}-1)) \vert     \, \dd t.
\end{align*}

Note that the last expression can be taken into account for $k \in \dot \Z^d$ only (indeed, the term corresponding to $k=0$ vanishes in $(\mathrm{I})$). We then have

\begin{align*}
&\underset{k \in \dot \Z^d}{\sup} \underset{0 \leq s \leq T}{\sup}\int_s^T (t-s)^{2q}  \vert k \vert  \sum_{\ell \in \Z^d}\vert \ell \vert^{2q} \vert (\mathcal{F}_{x,v} \mathcal{G})(t,s,\ell-k,k(e^{t-s}-1)) \vert         \, \dd t \\
& \leq \underset{k \in \dot \Z^d}{\sup} \underset{0\leq s \leq T}{\sup}\int_s^T \frac{(t-s)^{2q}   \vert k \vert }{(1+\vert k \vert (e^{t-s}-1))^{\alpha_1}} \sum_{\ell \in \Z^d}\vert \ell \vert^{2q}  (1+\vert k \vert (e^{t-s}-1) )^{\alpha_1})\vert (\mathcal{F}_{x,v} \mathcal{G})(t,s,\ell-k,k(e^{t-s}-1) )\vert    \, \dd t \\
&\leq  \underset{k \in \dot \Z^d}{\sup} \underset{0\leq s \leq T}{\sup}\int_s^T \frac{(t-s)^{2q}   \vert k \vert}{(1+\vert k \vert(e^{t-s}-1))^{\alpha_1}} \, \dd t  \\
& \qquad \qquad \qquad \qquad  \qquad \times  \underset{0\leq s \leq T} {\sup} \underset{s\leq t \leq T}{\sup}  \sum_{\ell \in \Z^d} \vert \ell \vert^{2q} \, \underset{ \xi }{\sup} \,  (1+\vert \xi \vert^{\alpha_1}) \vert (\mathcal{F}_{x,v} \mathcal{G})(t,s,\ell-k,\xi) \vert  \\
&\leq  \underset{k \in \dot \Z^d}{\sup} \, \underset{0\leq s \leq T}{\sup}\int_s^T \frac{(t-s)^{2q}   \vert k \vert}{(1+\vert k \vert(t-s))^{\alpha_1}} \, \dd t  \\
& \qquad \qquad \qquad \qquad  \qquad \times  \underset{0\leq s \leq T} {\sup} \underset{s\leq t \leq T}{\sup}  \sum_{\ell \in \Z^d} \vert \ell+k \vert^{2q} \, \underset{ \xi }{\sup} \,  (1+\vert \xi \vert^{\alpha_1}) \vert (\mathcal{F}_{x,v} \mathcal{G})(t,s,\ell,\xi) \vert,
\end{align*}

\begin{align*}
\underset{k \in \dot \Z^d}{\sup}& \underset{0 \leq s \leq T}{\sup}\int_s^T (t-s)^{2q}  \vert k \vert  \sum_{\ell \in \Z^d}\vert \ell \vert^{2q} \vert (\mathcal{F}_{x,v} \mathcal{G})(t,s,\ell-k,k(e^{t-s}-1)) \vert         \, \dd t\\
&\leq  \underset{k \in \dot \Z^d}{\sup} \, \underset{0\leq s \leq T}{\sup}\int_s^T \frac{(t-s)^{2q}  \vert k \vert}{(1+\vert k \vert(t-s))^{\alpha_1}} \, \dd t  \\
& \qquad \qquad \qquad \qquad  \qquad \times  \underset{0\leq s \leq T} {\sup} \underset{s\leq t \leq T}{\sup}  \sum_{\ell \in \Z^d} \vert \ell \vert^{2q} \, \underset{ \xi }{\sup} \,  (1+\vert \xi \vert^{\alpha_1}) \vert (\mathcal{F}_{x,v} \mathcal{G})(t,s,\ell,\xi) \vert  \\
& \quad + \underset{k \in \dot \Z^d}{\sup} \, \underset{0\leq s \leq T}{\sup}\int_s^T \frac{(t-s)^{2q}   \vert k \vert^{1+2q}}{(1+\vert k \vert(t-s))^{\alpha_1}} \, \dd t  \\
& \qquad \qquad \qquad \qquad  \qquad \times  \underset{0\leq s \leq T} {\sup} \underset{s\leq t \leq T}{\sup}  \sum_{\ell \in \Z^d} \, \underset{ \xi }{\sup} \,  (1+\vert \xi \vert^{\alpha_1}) \vert (\mathcal{F}_{x,v} \mathcal{G})(t,s,\ell,\xi) \vert  \\
%&= \underset{k \in \dot \Z^d}{\sup} \, \underset{0\leq s \leq T}{\sup}\int_s^T \frac{(1+\vert k \vert^2 )\vert k \vert^2 (t-s)^3 \varphi(t-s)^2}{(1+\vert k \vert(t-s))^{\alpha_1}} \, \dd t  \\
%& \qquad \qquad \qquad \qquad \qquad \times  \underset{0\leq s \leq T} {\sup} \underset{s\leq t \leq T}{\sup} \left(  \sum_{m \in \Z^d} \underset{\xi }{\sup} \,  \left\lbrace (1+\vert m \vert^{\alpha_2})(1+\vert \xi \vert^{\alpha_1}) \vert (\mathcal{F}_{x,v} \widetilde{G})(t,s,\ell,\xi) \vert\right\rbrace^2 \right)^{\frac{1}{2}}
&=\mathrm{S}_1 + \mathrm{S}_2.
\end{align*}

Let us treat these two terms separately.

$\bullet$ For $\mathrm{S}_1$, we have by the Cauchy-Schwarz inequality

\begin{align*}
\mathrm{S}_1 &\leq \underset{k \in \dot \Z^d}{\sup} \underset{0\leq s \leq T}{\sup}\int_s^T \frac{(t-s)^{2q}   \vert k \vert}{(1+\vert k \vert(t-s))^{\alpha_1}} \, \dd t  \\
& \qquad \qquad \qquad \qquad \qquad \times  \underset{0\leq s \leq T} {\sup} \underset{s\leq t \leq T}{\sup} \left(  \sum_{\ell \in \Z^d} \underset{\xi }{\sup} \,  \left\lbrace (1+\vert \ell \vert^{2q+\alpha_2})(1+\vert \xi \vert^{\alpha_1}) \vert (\mathcal{F}_{x,v} \mathcal{G})(t,s,\ell,\xi) \vert \right\rbrace^2 \right)^{\frac{1}{2}},
\end{align*}

if $\alpha_2 >d/2$. For the integral term, we write for $\alpha_1>3$.

\begin{align*}
\underset{k \in \dot \Z^d}{\sup} \underset{0\leq s \leq T}{\sup}\int_s^T \frac{(t-s)^{2q}   \vert k \vert}{(1+\vert k \vert(t-s))^{\alpha_1}} \, \dd t &=\underset{k \in \dot \Z^d}{\sup}  \frac{1}{\vert k \vert^{2q}}\underset{0\leq s \leq T}{\sup}\int_0^{\vert k \vert(T-s)} \frac{\tau^{2q} }{(1+\tau)^{\alpha_1}} \, \dd \tau \\
% &\leq \underset{k \in  \dot \Z^d}{\sup}  \frac{1}{\vert k \vert^{2q}} \underset{0\leq s \leq T}{\sup}   \int_{0}^{+\infty} \frac{\tau^{2q}}{(1+\tau)^{\alpha_1}} \, \dd \tau. \\
& \leq  \underset{k \in  \dot \Z^d}{\sup}  \frac{1}{\vert k \vert^{2q}} \int_{0}^{+\infty} \frac{\tau^{2q}}{(1+\tau)^{\alpha_1}} \, \dd \tau, \\
\end{align*}

which is a finite constant independent of $k$ and $T$ (since $\alpha_1>1+2q$) therefore

\begin{align*}
\mathrm{S}_1 \lesssim \underset{0\leq t \leq T}{\sup} \left(  \sum_{\ell \in \Z^d}\underset{0\leq s \leq t}{\sup} \ \underset{\xi \in \R^d }{\sup} \,  \left\lbrace (1+\vert \ell \vert^{2q+\alpha_2})(1+\vert \xi \vert^{\alpha_1}) \vert (\mathcal{F}_{x,v} \mathcal{G})(t,s,\ell,\xi) \vert \right\rbrace^2 \right)^{\frac{1}{2}}.
\end{align*}

$\bullet$ For $\mathrm{S}_2$ , we have for $\alpha_2 >d/2$

\begin{align*}
\mathrm{S}_2 &\leq \underset{k \in \dot \Z^d}{\sup} \underset{0\leq s \leq T}{\sup}\int_s^T \frac{(t-s)^{2q}  \vert k \vert^{1+2q} }{(1+\vert k \vert(t-s))^{\alpha_1}} \, \dd t  \\
& \qquad \qquad \qquad \qquad \qquad \times  \underset{0\leq s \leq T} {\sup} \underset{s\leq t \leq T}{\sup} \left(  \sum_{\ell \in \Z^d} \underset{\xi }{\sup} \,  \left\lbrace (1+\vert m \vert^{\alpha_2})(1+\vert \xi \vert^{\alpha_1}) \vert (\mathcal{F}_{x,v} \mathcal{G})(t,s,\ell,\xi) \vert \right\rbrace^2 \right)^{\frac{1}{2}}.
\end{align*}

The integral term now reads for $\alpha_1>3$

\begin{align*}
\underset{k \in \dot \Z^d}{\sup} \underset{0\leq s \leq T}{\sup}\int_s^T \frac{(t-s)^{2q}   \vert k \vert^{1+2q} }{(1+\vert k \vert(t-s))^{\alpha_1}} \, \dd t &=\underset{k \in \dot \Z^d}{\sup} \underset{0\leq s \leq T}{\sup}\int_0^{\vert k \vert(T-s)} \frac{\tau^{2q} }{(1+\tau)^{\alpha_1}} \, \dd \tau
%&\leq  \underset{0\leq s \leq T}{\sup}   \int_{0}^{+\infty} \frac{\tau^{2q}}{(1+\tau)^{\alpha_1}} \, \dd \tau. \\
\leq   \int_{0}^{+\infty} \frac{\tau^{2q}}{(1+\tau)^{\alpha_1}} \, \dd \tau,
\end{align*}

\begin{align*}
\mathrm{S}_1 \lesssim \underset{0\leq t \leq T}{\sup} \left(  \sum_{\ell \in \Z^d}\underset{0\leq s \leq t}{\sup} \ \underset{\xi \in \R^d }{\sup} \,  \left\lbrace (1+\vert \ell \vert^{\alpha_2})(1+\vert \xi \vert^{\alpha_1}) \vert (\mathcal{F}_{x,v} \mathcal{G})(t,s,\ell,\xi) \vert \right\rbrace^2 \right)^{\frac{1}{2}}.
\end{align*}

We have thus proven that

\begin{align*}
(\mathrm{I}) \lesssim  \Vert H \Vert_{\Ld^2(0,T ; \Ld^2(\T^d))}^2 \left( \Vert \mathcal{G} \Vert_{T,s_1,s_2}+\Vert \mathcal{G} \Vert_{T,s_1,s_2+2q} \right),
\end{align*}

for $s_1>1+2q$ and $s_2>d/2+2q$.All in all, we get

\begin{align*}
\Vert \nabla_x^q \mathrm{K}_{G_{[q]}}^{\mathrm{fric}}[H] \Vert_{\Ld^2(0,T; \Ld^2(\T^d))}\leq C \Vert \mathcal{G} \Vert_{T,s_1,s_2} \Vert H \Vert_{\Ld^2(0,T; \Ld^2(\T^d))},
\end{align*}

which ends the proof.

We then deduce the two following propositions. The first one is a direct consequence of Proposition <ref> with $q=0$, and states the continuity of
$ \mathrm{K}_G^{\mathrm{free}}$ and $ \mathrm{K}_G^{\mathrm{fric}}$ on $\Ld^2_T \Ld^2_x$.

There exists $C>0$ such that for every $T>0$, if $p>1+d$ and $\sigma >d/2$ then for all $H  \in \Ld^2(0,T;\Ld^2(\T^d))$

\begin{align*}
\Big\Vert \mathrm{K}_G^{\mathrm{free}}[H] \Big\Vert_{\Ld^2(0,T; \Ld^2(\T^d))} +\Big\Vert \mathrm{K}_G^{\mathrm{fric}}[H] \Big\Vert_{\Ld^2(0,T; \Ld^2(\T^d))} &\leq C \underset{0 \leq s,t \leq T}{\sup} \Vert G(t,s) \Vert_{\mathcal{H}^p_{\sigma}} \Vert H \Vert_{\Ld^2(0,T; \Ld^2(\T^d))}.%, \\[2mm]
%\Vert K_G^1[F] \Vert_{\Ld^2(0,T; \Ld^2(\T^d))} +\Vert \mathscr{K}_G^1[F] \Vert_{\Ld^2(0,T; \Ld^2(\T^d))} &\leq C \Vert v \otimes G \Vert_{T,s_1,s_2} \Vert F \Vert_{\Ld^2(0,T; \Ld^2(\T^d))}
\end{align*}

By Proposition <ref> with $q=0$, we have

\begin{align*}
\Big\Vert \mathrm{K}_{G}^{\mathrm{free}}[H] \Big\Vert_{\Ld^2(0,T; \Ld^2(\T^d))} +\Big\Vert  \mathrm{K}_{G}^{\mathrm{fric}}[H] \Big\Vert_{\Ld^2(0,T; \Ld^2(\T^d))} &\leq C \Vert G \Vert_{T,s_1,s_2} \Vert H \Vert_{\Ld^2(0,T; \Ld^2(\T^d))}.%, \\[2mm]
%\Vert K_G^1[F] \Vert_{\Ld^2(0,T; \Ld^2(\T^d))} +\Vert \mathrm{K}_G^1[F] \Vert_{\Ld^2(0,T; \Ld^2(\T^d))} &\leq C \Vert v \otimes G \Vert_{T,s_1,s_2} \Vert H \Vert_{\Ld^2(0,T; \Ld^2(\T^d))},
\end{align*}

for any $s_1>1$ and $s_2>d/2$. Now appealing to <cit.>, one can prove that for all $p> 1+d$ and $\sigma>d/2$, there exist $s_2>d/2$ and $s_1> 1$ such that

\begin{align*}
\Vert G \Vert_{T,s_1,s_2} &\lesssim \underset{0 \leq s,t \leq T}{\sup} \Vert G(t,s) \Vert_{\mathcal{H}^p_{\sigma}}, %\\
%\Vert v \otimes G \Vert_{T,s_1,s_2} &\lesssim \underset{0 \leq s,t \leq T}{\sup} \Vert G(t,s) \Vert_{\mathcal{H}^p_{\sigma+1}},
\end{align*}

hence the result.

When the kernel $G$ vanishes along the diagonal in time $\lbrace t=s \rbrace$, Proposition <ref> with $q=1$ leads to the following additional regularizing effect of the operators $\mathrm{K}_G$ (as already observed in [94]). Loosely speaking, the operators $\mathrm{K}_G^{\mathrm{free}}$ and $\mathrm{K}_G^{\mathrm{fric}}$ are bounded from $\Ld^2_T \Ld^2_x$ to $\Ld^2_T \dot{ \mathrm{H}}^1_x$ in this case.

There exists $C>0$ such that if the kernel $G$ satisfies

\begin{align*}
\end{align*}

the following holds. Let $T>0$. If $p>7+d$ and $\sigma >d/2$ then for all $H \in \Ld^2(0,T;\Ld^2(\T^d))$

\begin{align*}
\Big\Vert  \mathrm{K}_G^{\mathrm{free}}[H] \Big\Vert_{\Ld^2(0,T; \H^1(\T^d))}+\Big\Vert   \mathrm{K}_G^{\mathrm{fric}}[H] \Big\Vert_{\Ld^2(0,T; \H^1(\T^d))} \leq C (1+T)\underset{0 \leq s,t \leq T}{\sup} \Vert \partial_s G(t,s) \Vert_{\mathcal{H}^p_{\sigma}} \Vert H \Vert_{\Ld^2(0,T; \Ld^2(\T^d))}.
\end{align*}

Since $G(t,t,x,v)=0$, the Taylor's formula shows that

\begin{align*}
G(t,s,x,v)=(t-s)\widetilde{G}(t,s,x,v), \ \ \widetilde{G}(t,s,x,v):=-\int_0^1\partial_s G(t,t+\tau(s-t),x,v) \, \mathrm{d}\tau.
\end{align*}

By Proposition <ref> with $q=1$, we get for $s_1>3$ and $s_2>d/2+2$

\begin{align*}
\Vert \nabla_x \mathrm{K}_G^{\mathrm{free}}[H] \Vert_{\Ld^2(0,T; \Ld^2(\T^d))}+\Vert \nabla_x  \mathrm{K}_G^{\mathrm{fric}}[H] \Vert_{\Ld^2(0,T; \Ld^2(\T^d))} &\leq C \Vert \widetilde{G} \Vert_{T,s_1,s_2} \Vert H \Vert_{\Ld^2(0,T; \Ld^2(\T^d))}.
\end{align*}

To conclude, we observe that

\begin{align*}
\big\lbrace (1+\vert m \vert)^{s_2}(1+\vert \xi \vert)^{s_1} & \vert (\mathcal{F}_{x,v} \widetilde{G})(t,s,m,\xi) \vert \big\rbrace^2 \\
& \qquad  \qquad \leq \left( (1+\vert m \vert)^{s_2}(1+\vert \xi \vert)^{s_1} \int_0^1 \vert \mathcal{F}_{x,v}(\partial_s G)(t,t+\tau(s-t),m, \xi) \vert \, \mathrm{d}\tau \right)^2 \\
& \qquad  \qquad  \leq \int_0^1 \big\lbrace (1+\vert m \vert)^{s_2}(1+\vert \xi \vert)^{s_1} \vert \mathcal{F}_{x,v}(\partial_s G)(t,t+\tau(s-t),m, \xi) \vert \big\rbrace^2 \, \mathrm{d}\tau,
\end{align*}

\begin{align*}
\Vert \nabla_x \mathrm{K}_G^{\mathrm{free}}[H] \Vert_{\Ld^2(0,T; \Ld^2(\T^d))}+\Vert \nabla_x  \mathrm{K}_G^{\mathrm{fric}}[H] \Vert_{\Ld^2(0,T; \Ld^2(\T^d))} \leq C \Vert \partial_s G \Vert_{T,s_1,s_2} \Vert H \Vert_{\Ld^2(0,T; \Ld^2(\T^d))}.
\end{align*}

Now appealing to <cit.>, one can prove that for all $p>2\ell+s + 1+d$ and $\sigma>d/2$ (with $\ell,s \in \R^+$), there exist $s_2>\ell+ d/2$ and $s_1>s + 1$ such that

\begin{align*}
\Vert G \Vert_{T,s_1,s_2} &\lesssim \underset{0 \leq s,t \leq T}{\sup} \Vert G(t,s) \Vert_{\mathcal{H}^p_{\sigma}}. %\\
%\Vert v \otimes G \Vert_{T,s_1,s_2} &\lesssim \underset{0 \leq s,t \leq T}{\sup} \Vert G(t,s) \Vert_{\mathcal{H}^p_{\sigma+1}}.
\end{align*}

Since $G(t,t,x,v)=0$, we also have

\begin{align*}
\Vert G \Vert_{T,s_1,s_2} &\lesssim T \underset{0 \leq s,t \leq T}{\sup} \Vert \partial_s G(t,s) \Vert_{\mathcal{H}^p_{\sigma}}.
\end{align*}

By Proposition <ref>, and taking $\ell=s=2$, we end up with the desired conclusion.

A variant of Proposition <ref> holds in the following form: there exists $C>0$ such that for $p>7+d$ and $\sigma>d/2$, and

\begin{align*}
\end{align*}

we have for all $H \in \Ld^2(0,T; \dot \H^{-1}(\T^d))$

\begin{align*}
\Big\Vert \mathrm{K}_G^{\mathrm{free}}[H] \Big\Vert_{\Ld^2(0,T; \Ld^2(\T^d))}+\Big\Vert  \mathrm{K}_G^{\mathrm{fric}}[H] \Big\Vert_{\Ld^2(0,T; \Ld^2(\T^d))} &\leq C (1+T) \underset{0 \leq s,t \leq T}{\sup} \Vert \partial_s G(t,s) \Vert_{\mathcal{H}^p_{\sigma}} \Vert H \Vert_{\Ld^2(0,T; \dot \H^{-1}(\T^d))}.
\end{align*}

We don't detail the proof, which follows the same lines as the one of Proposition <ref>.

We finally investigate the smoothing properties of the difference operator $\mathrm{K}_G^{\mathrm{free}}-\mathrm{K}_G^{\mathrm{fric}}$. A somewhat surprising result is the fact that this operator gains one additional derivative. This is the content of the following proposition.

There exists $C>0$ such that for every $T>0$, if $p>8+d$ and $\sigma>1+d/2$ then for all $H \in \Ld^2(0,T;\Ld^2(\T^d))$,

\begin{align*}
\Big\Vert \mathrm{K}_G^{\mathrm{free}}[H]-\mathrm{K}_G^{\mathrm{fric}}[H]  \Big\Vert_{\Ld^2(0,T;\H^1(\T^d))}  &\leq  C \varphi(T)\underset{0 \leq s,t \leq T}{\sup} \Vert G(t,s) \Vert_{\mathcal{H}^p_{\sigma}} \Vert H \Vert_{\Ld^2(0,T;\Ld^2(\T^d))},
\end{align*}

where $\varphi : \R^+ \rightarrow \R^+$ is a continuous nondecreasing function.

Following the computations performed in the proof of Proposition <ref>, we have

\begin{multline*}
\mathrm{K}_G^{\mathrm{free}}[F](t,x)-\mathrm{K}_G^{\mathrm{fric}}[F](t,x)=\sum_{\ell \in \Z^d}e^{i \ell \cdot x}  \Bigg\lbrace \sum_{k \in \Z^d}  \int_0^t \widehat{F}_k(s)  (ik) \cdot \left[(\mathcal{F}_{x,v} G)(t,s,\ell-k,k(t-s)) \right. \\ \left. - (\mathcal{F}_{x,v} G)(t,s,\ell-k,k(e^{t-s}-1)) \right] \, \mathrm{d}s  \Bigg\rbrace,
\end{multline*}

therefore, if we set

\begin{align*}
\Theta(t,s,\ell,k):=(\mathcal{F}_{x,v} G)(t,s,\ell-k,k(t-s))- (\mathcal{F}_{x,v} G)(t,s,\ell-k,k(e^{t-s}-1)),
\end{align*}

we get

\begin{align*}
\Vert \nabla_x \pr{\mathrm{K}_G^{\mathrm{free}}[H]-\mathrm{K}_G^{\mathrm{fric}}[H]}(t) \Vert_{\Ld^2(\T^d)}^2
%&= \sum_{\ell \in \Z^d}  \vert \ell \vert^2 \left\vert\sum_{k \in \Z^d}  \int_0^t \widehat{H}_k(s)  (ik) \cdot \Theta(t,s,\ell,k) \, \mathrm{d}s \right\vert^2 \\
&\leq  \sum_{\ell \in \Z^d} \vert \ell \vert^2 \left(\sum_{k \in \Z^d}  \int_0^t \vert \widehat{H}_k(s) \vert  \vert k \vert  \vert \Theta(t,s,\ell,k)\vert  \, \mathrm{d}s \right)^2.
%& \leq \sum_{\ell \in \Z^d}  \left(\sum_{k \in \Z^d}  \int_0^t \vert \widehat{F}_k(s) \vert^2 \vert k \cdot \Theta(t,s,\ell,k) \vert    \, \dd s \right) \\
%& \qquad     \times \left( \sum_{k \in \Z^d}  \int_0^t \vert k \cdot \Theta(t,s,\ell,k) \vert  \dd s \right).
\end{align*}

We also have, by setting $\mathcal{G}^{t,s}_{\ell-k}=(\mathcal{F}_{x,v} G)(t,s,\ell-k,\bullet)$

\begin{align*}
\vert \Theta(t,s,\ell,k)\vert &=\vert \mathcal{G}^{t,s}_{\ell-k}(k(e^{t-s}-1))-\mathcal{G}^{t,s}_{\ell-k}(k(t-s))\vert \\
&=\vert \mathcal{G}^{t,s}_{\ell-k}(k(t-s)+k(t-s)^2\varphi(t-s))-\mathcal{G}^{t,s}_{\ell-k}(k(t-s)) \vert \\
& \leq \underset{\theta \in [0,1]}{\sup} \vert \nabla_\xi \mathcal{G}^{t,s}_{\ell-k}\left(\xi_{\theta}^{t,s}(k(t-s)) \right) \vert   \vert  k \vert (t-s)^2 \varphi(t-s),
%&=  \underset{\theta \in [0,1]}{\sup} \vert \nabla_{\xi} (\mathcal{F}_{x,v} G)(t,s,\ell-k,k (t-s)+ \theta k(t-s)^2\varphi(t-s)) \vert,
%& \qquad \qquad \qquad \qquad \times \vert  k \vert (t-s)^2 \varphi(t-s),
%& \leq \underset{\xi \in \R^d}{\sup} \vert \D \mathcal{G}(\xi) \vert \vert k(t-s)^2e^{t-s} \vert \\
%& \leq  \vert k \vert (t-s)^2e^{t-s} \underset{\xi \in \R^d}{\sup} \vert \D_{\xi}(\mathcal{F}_{x,v} G)(t,s,\ell-k,\xi) \vert
\end{align*}

where $\varphi(z)=\sum_{i \geq 0} \frac{z^i}{(i+2)!}$ and $\xi_{\theta}^{t,s}(z)=z+ \theta z(t-s)\varphi(t-s)$. By continuity, there exists $\theta^{\star} \in [0,1]$ (which may depend on all the other variables) such that

\begin{align*}
\vert \Theta(t,s,\ell,k)\vert \leq \vert \nabla_\xi \mathcal{G}^{t,s}_{\ell-k}\left(\xi_{\theta^{\star}}^{t,s}(k(t-s)) \right) \vert   \vert  k \vert (t-s)^2 \varphi(t-s).
\end{align*}

This yields

\begin{align*}
\Vert \nabla_x &\pr{\mathrm{K}_G^{\mathrm{free}}[H]-\mathrm{K}_G^{\mathrm{fric}}[H]}(t) \Vert_{\Ld^2(\T^d)}^2 \\
&\leq   \sum_{\ell \in \Z^d} \vert \ell \vert^2 \left(\sum_{k \in \Z^d}  \int_0^t \vert \widehat{H}_k(s) \vert \vert  k \vert^2 (t-s)^2 \varphi(t-s) \vert \nabla_\xi \mathcal{G}^{t,s}_{\ell-k}\left(\xi_{\theta^{\star}}^{t,s}(k(t-s)) \right) \vert    \, \mathrm{d}s \right)^2 \\
& \leq \sum_{\ell \in \Z^d} \vert \ell \vert^2  \left(\sum_{k \in \Z^d}  \int_0^t \vert \widehat{H}_k(s) \vert^2 \vert k \vert^2 (t-s)^3 \varphi(t-s)^2 \vert \nabla_\xi \mathcal{G}^{t,s}_{\ell-k}\left((\xi_{\theta^{\star}}^{t,s}(k(t-s)) \right) \vert      \, \dd s \right) \\
& \qquad     \times \left( \sum_{k \in \Z^d}  \int_0^t \vert k \vert ^2 (t-s) \vert \nabla_\xi \mathcal{G}^{t,s}_{\ell-k}\left(\xi_{\theta^{\star}}^{t,s}(k(t-s)) \right)\vert   \,  \dd s \right),
\end{align*}

thanks to the Cauchy-Schwarz inequality.

As in the proof of Proposition <ref>, we obtain by integrating in time that

\begin{align*}
\Vert \nabla_x &\pr{\mathrm{K}_G^{\mathrm{free}}[H]-\mathrm{K}_G^{\mathrm{fric}}[H]}\Vert_{\Ld^2((0,T) \times \T^d)}^2 \\
&\leq \sum_{\ell \in \Z^d} \vert \ell \vert^2 \int_0^T  \int_0^t  \sum_{k \in \Z^d} \vert \widehat{H}_k(s) \vert^2 \vert k \vert^2 (t-s)^3 \varphi(t-s)^2 \vert \nabla_\xi \mathcal{G}^{t,s}_{\ell-k}\left(\xi_{\theta^{\star}}^{t,s}(k(t-s)) \right) \vert     \, \dd s \, \dd t \\
& \quad \quad  \quad  \times \underset{\ell \in \Z^d}{\sup} \underset{t \in (0,T)}{\sup}  \sum_{k \in \Z^d}  \int_0^t \vert k \vert ^2 (t-s) \vert \nabla_\xi \mathcal{G}^{t,s}_{\ell-k}\left(\xi_{\theta^{\star}}^{t,s}(k(t-s)) \right) \vert    \dd s \\
&=(\mathrm{A}) \times (\mathrm{B}).
\end{align*}

First, we note that for all $\theta \in [0,1]$, $k \in \Z^d$ and $0 \leq s \leq t$

\begin{align*}
\vert \xi_{\theta}^{t,s}(k(t-s)) \vert & =\vert k(t-s)+ \theta k(t-s)^2\varphi(t-s) \vert \\
& = \vert k \vert (t-s) \left[ 1+ \theta (t-s) \varphi(t-s) \right] \\
& \geq \vert k \vert (t-s).
\end{align*}

For $(\mathrm{B})$, we thus have

\begin{align*}
(\mathrm{B}) &\leq \underset{\ell \in \Z^d}{\sup} \underset{t \in (0,T)}{\sup}  \sum_{k \in \Z^d}  \int_0^t  (1+ \vert \xi_{\theta^{\star}}^{t,s}(k(t-s)) \vert )^{\beta_1} \vert \nabla_\xi \mathcal{G}^{t,s}_{\ell-k}\left(\xi_{\theta^{\star}}^{t,s}(k(t-s)) \right) \vert  \frac{\vert k \vert^2(t-s) }{(1+\vert \xi_{\theta^{\star}}^{t,s}(k(t-s)) \vert )^{\beta_1}}  \dd s \\
& \leq  \underset{\ell \in \Z^d}{\sup} \underset{t \in (0,T)}{\sup}  \sum_{k \in \Z^d}   \underset{s, \xi }{\sup} \big\lbrace (1+\vert \xi \vert )^{\beta_1}\vert \nabla_\xi \mathcal{G}^{t,s}_{\ell-k}(\xi) \vert  \big\rbrace \int_0^t \frac{\vert k \vert^2 (t-s)}{(1+\vert k \vert (t-s))^{\beta_1}}  \dd s.
%& \leq  \underset{\ell \in \Z^d}{\sup} \underset{t \in (0,T)}{\sup}  \sum_{k \in \Z^d}  \int_0^t  \underset{\ \xi \in \R^d}{\sup} \big\lbrace (1+\vert \xi \vert )^{\beta_1} \vert \xi \vert \Vert \nabla_{\xi} (\mathcal{F}_{x,v} G)(t,s,\ell-k,\cdot) \Vert_{\infty} \big\rbrace \frac{\vert k \vert (t-s) \varphi(t-s) }{(1+\vert k \vert (t-s))^{\beta_1}}  \dd s \\
%&\leq  \underset{\ell \in \Z^d}{\sup} \underset{t \in (0,T)}{\sup}  \sum_{k \in \Z^d}  \underset{\substack{s \in (0,t)\\ \xi \in \R^d}}{\sup} \big\lbrace (1+\vert \xi \vert )^{\alpha_1} \vert \xi \vert \Vert \nabla_{\xi} (\mathcal{F}_{x,v} G)(t,s,\ell-k,\cdot) \Vert_{\infty} \big\rbrace \int_0^t \frac{\vert k \vert^2 (t-s)}{(1+\vert k \vert (t-s))^{\beta_1}}  \dd s.
\end{align*}

\begin{align*}
\int_0^t \frac{\vert k \vert^2 (t-s)}{(1+\vert k \vert (t-s))^{\beta_1}}  \dd s=\int_0^{\vert k \vert t} \frac{\tau}{(1+\tau)^{\beta_1}} \dd \tau \leq  \int_0^{+\infty} \frac{\tau}{(1+\tau)^{\beta_1}} \dd \tau < + \infty,
\end{align*}

if $\beta_1>2$, we get

\begin{align*}
(\mathrm{B}) \lesssim  \underset{\ell \in \Z^d}{\sup} \underset{t \in (0,T)}{\sup}  \sum_{k \in \Z^d}   \underset{s, \xi }{\sup} \big\lbrace (1+\vert \xi \vert )^{\beta_1}\vert \nabla_\xi \mathcal{G}^{t,s}_{\ell-k}(\xi) \vert  \big\rbrace.
\end{align*}

By choosing $\beta_2 >d/2$ and by the Cauchy-Schwarz inequality, this yields

\begin{align*}
(\mathrm{B}) \lesssim  \underset{t \in (0,T)}{\sup} \left( \sum_{k \in \Z^d}   \underset{s, \xi }{\sup} \big\lbrace  (1+\vert k \vert )^{\beta_2}(1+\vert \xi \vert )^{\beta_1}\vert \nabla_\xi \mathcal{G}^{t,s}_{k}(\xi) \vert  \big\rbrace^2 \right)^{\frac{1}{2}}.
\end{align*}

Let us estimate the other term $(\mathrm{A})$. By the Fubini-Tonelli theorem, we have

\begin{align*}
(\mathrm{A})&=\sum_{\ell \in \Z^d} \vert \ell \vert^2 \int_0^T  \int_0^t  \sum_{k \in \Z^d} \vert \widehat{H}_k(s) \vert^2 \vert k \vert^2 (t-s)^3 \varphi(t-s)^2 \vert \nabla_\xi \mathcal{G}^{t,s}_{\ell-k}\left(\xi_{\theta^{\star}}^{t,s}(k(t-s)) \right) \vert       \, \dd s \, \dd t \\
&= \int_0^T \sum_{k \in \Z^d} \vert \widehat{H}_k(s) \vert^2   \int_s^T \sum_{\ell \in \Z^d}  \vert \ell \vert^2 \vert k \vert^2 (t-s)^3 \varphi(t-s)^2 \vert \nabla_\xi \mathcal{G}^{t,s}_{\ell-k}\left(\xi_{\theta^{\star}}^{t,s}(k(t-s)) \right) \vert        \, \dd t \, \dd s \\
%&= \int_0^T \sum_{k \in \Z^d} \vert \widehat{H}_k(s) \vert^2   \int_s^T \sum_{\ell \in \Z^d} (1+\vert k \vert)^2 \vert \ell \vert^2 \vert k \vert^2 (t-s)^3 \varphi(t-s)^2 \vert \nabla_\xi \mathcal{G}^{t,s}_{\ell-k}(\left(\xi_{\theta^{\star}}^{t,s}(k(t-s)) \right) \vert      \, \dd t \, \dd s \\
&\leq \Vert H \Vert_{\Ld^2(0,T ; \Ld^2(\T^d))}^2  \underset{k \in \Z^d}{\sup} \underset{0\leq s \leq T}{\sup}\int_s^T \sum_{\ell \in \Z^d} \vert \ell \vert^2 \vert k \vert^2 (t-s)^3 \varphi(t-s)^2 \vert \nabla_\xi \mathcal{G}^{t,s}_{\ell-k}\left(\xi_{\theta^{\star}}^{t,s}(k(t-s)) \right) \vert     \, \dd t.
%&= \int_0^T \sum_{k \in \Z^d}  \frac{ \vert \widehat{H}_k(s) \vert^2}{(1+\vert k \vert )^2}    \int_s^T \sum_{\ell \in \Z^d} (1+\vert k \vert)^2 \vert k \vert^2 (t-s)^3 \varphi(t-s)^2 \vert \nabla_\xi \mathcal{G}^{t,s}_{\ell-k}(\left(\xi_{\theta^{\star}}^{t,s}(k(t-s)) \right) \vert      \, \dd t \, \dd s \\
%&\leq \Vert H \Vert_{\Ld^2(0,T ; \H^{-1}(\T^d))}^2  \underset{k \in \Z^d}{\sup} \underset{0\leq s \leq T}{\sup}\int_s^T \sum_{\ell \in \Z^d} (1+\vert k \vert^2 )\vert k \vert^2 (t-s)^3 \varphi(t-s)^2 \vert \nabla_\xi \mathcal{G}^{t,s}_{\ell-k}\left(\xi_{\theta^{\star}}^{t,s}(k(t-s)) \right) \vert     \, \dd t.
\end{align*}

As in the proof of Proposition <ref>, we have

\begin{align*}
\underset{k \in \Z^d}{\sup}& \underset{0 \leq s \leq T}{\sup}\int_s^T \vert k \vert^2 (t-s)^3 \varphi(t-s)^2 \sum_{\ell \in \Z^d} \vert \ell \vert^2 \vert \nabla_\xi \mathcal{G}^{t,s}_{\ell-k}\left(\xi_{\theta^{\star}}^{t,s}(k(t-s)) \right) \vert       \, \dd t \\
&\leq  \underset{k \in \dot \Z^d}{\sup} \, \underset{0\leq s \leq T}{\sup}\int_s^T \frac{\vert k \vert^2 (t-s)^3 \varphi(t-s)^2}{(1+\vert k \vert(t-s))^{\alpha_1}} \, \dd t  \\
& \qquad \qquad \qquad \qquad  \qquad \times  \underset{0\leq s \leq T} {\sup} \underset{s\leq t \leq T}{\sup}  \sum_{\ell \in \Z^d} \vert \ell \vert^2 \, \underset{ \xi }{\sup} \,  (1+\vert \xi \vert^{\alpha_1}) \vert \nabla_\xi \mathcal{G}^{t,s}_{\ell}(\xi) \vert  \\
& \quad + \underset{k \in \dot \Z^d}{\sup} \, \underset{0\leq s \leq T}{\sup}\int_s^T \frac{\vert k \vert^4 (t-s)^3 \varphi(t-s)^2}{(1+\vert k \vert(t-s))^{\alpha_1}} \, \dd t  \\
& \qquad \qquad \qquad \qquad  \qquad \times  \underset{0\leq s \leq T} {\sup} \underset{s\leq t \leq T}{\sup}  \sum_{\ell \in \Z^d} \, \underset{ \xi }{\sup} \,  (1+\vert \xi \vert^{\alpha_1}) \vert \nabla_\xi \mathcal{G}^{t,s}_{\ell}(\xi) \vert  \\
&=\mathrm{T}_1 + \mathrm{T}_2.
\end{align*}

We treat these two terms in a  separate way.

$\bullet$ In $\mathrm{T}_1$, the integral term can be bounded via

\begin{align*}
\underset{k \in \dot \Z^d }{\sup}  \underset{0\leq s \leq T}{\sup}\int_s^T \frac{\vert k \vert^2 (t-s)^3 \varphi(t-s)^2}{(1+\vert k \vert(t-s))^{\alpha_1}} \, \dd t
%&=\underset{k \in \dot \Z^d}{\sup}  \frac{1}{\vert k \vert^2}\underset{0\leq s \leq T}{\sup}\int_0^{\vert k \vert(T-s)} \frac{\tau^3 \varphi(\frac{\tau}{\vert k \vert})^2}{(1+\tau)^{\alpha_1}} \, \dd \tau \\
&\leq \underset{k \in  \dot \Z^d}{\sup}  \frac{1}{\vert k \vert^2} \underset{0\leq s \leq T}{\sup} \varphi(T-s)^2  \int_{0}^{+\infty} \frac{\tau^3}{(1+\tau)^{\alpha_1}} \, \dd \tau. \\
& \leq \varphi(T)^2 \underset{k \in  \dot \Z^d}{\sup}  \frac{1}{\vert k \vert^2} \int_{0}^{+\infty} \frac{\tau^3}{(1+\tau)^{\alpha_1}} \, \dd \tau \\
& \lesssim \varphi(T)^2,
\end{align*}

provided that $\alpha_1 >4$. This implies that for any $\alpha_2>d/2$

\begin{align*}
\mathrm{T}_1 \lesssim\varphi(T)^2 \underset{0\leq s \leq T} {\sup} \underset{s\leq t \leq T}{\sup} \left(  \sum_{m \in \Z^d} \underset{\xi }{\sup} \,  \left\lbrace (1+\vert m \vert^{2+\alpha_2})(1+\vert \xi \vert^{\alpha_1}) \vert \nabla_\xi \mathcal{G}^{t,s}_{m}(\xi) \vert \right\rbrace^2 \right)^{\frac{1}{2}}.
\end{align*}

$\bullet$ In $\mathrm{T}_2$, the integral term can be bounded in a similar way via

\begin{align*}
\underset{k \in \dot \Z^d}{\sup} \, \underset{0\leq s \leq T}{\sup}\int_s^T \frac{\vert k \vert^4 (t-s)^3 \varphi(t-s)^2}{(1+\vert k \vert(t-s))^{\alpha_1}} \, \dd t
% &=\underset{k \in \dot \Z^d}{\sup}  \underset{0\leq s \leq T}{\sup}\int_0^{\vert k \vert(T-s)} \frac{\tau^3 \varphi(\frac{\tau}{\vert k \vert})^2}{(1+\tau)^{\alpha_1}} \, \dd \tau \\
&\leq  \underset{0\leq s \leq T}{\sup} \varphi(T-s)^2  \int_{0}^{+\infty} \frac{\tau^3}{(1+\tau)^{\alpha_1}} \, \dd \tau. \\
& \leq \varphi(T)^2  \int_{0}^{+\infty} \frac{\tau^3}{(1+\tau)^{\alpha_1}} \, \dd \tau \\
& \lesssim \varphi(T)^2,
\end{align*}

provided that $\alpha_1 >4$. This implies that for any $\alpha_2>d/2$

\begin{align*}
\mathrm{T}_2 \lesssim\varphi(T)\underset{0\leq s \leq T} {\sup} \underset{s\leq t \leq T}{\sup} \left(  \sum_{m \in \Z^d} \underset{\xi }{\sup} \,  \left\lbrace (1+\vert m \vert^{\alpha_2})(1+\vert \xi \vert^{\alpha_1}) \vert \nabla_\xi \mathcal{G}^{t,s}_{m}(\xi) \vert \right\rbrace^2 \right)^{\frac{1}{2}}.
\end{align*}

All in all, we get for $\alpha_1 >4$ and $\alpha_2>2+d/2$

\begin{align*}
(\mathrm{A}) &\lesssim \varphi(T)^2 \Vert H \Vert_{\Ld^2(0,T ; \Ld^2(\T^d))}^2 \\
& \qquad \qquad  \times  \underset{0\leq s \leq T} {\sup} \underset{s\leq t \leq T}{\sup} \left(  \sum_{m \in \Z^d} \underset{\xi }{\sup} \,  \left\lbrace (1+\vert m \vert^{\alpha_2})(1+\vert \xi \vert^{\alpha_1}) \vert \nabla_\xi \mathcal{G}^{t,s}_{m}(\xi) \vert \right\rbrace^2 \right)^{\frac{1}{2}} \\
&\lesssim \varphi(T)^2 \Vert H \Vert_{\Ld^2(0,T ; \Ld^2(\T^d))}^2 \\
& \qquad \qquad  \times  \underset{0\leq t \leq T}{\sup} \left(  \sum_{m \in \Z^d}\underset{0\leq s \leq t}{\sup} \underset{\xi }{\sup} \,  \left\lbrace (1+\vert m \vert^{\alpha_2})(1+\vert \xi \vert^{\alpha_1}) \vert \nabla_\xi \mathcal{G}^{t,s}_{m}(\xi) \vert \right\rbrace^2 \right)^{\frac{1}{2}},
\end{align*}

We have thus proven that for $s_1>4$ and $s_2>2+d/2$

\begin{align*}
\LRVert{\nabla_x \left(\mathrm{K}_G^{\mathrm{free}}[H]-\mathrm{K}_G^{\mathrm{fric}}[H] \right)}_{\Ld^2(0,T ; \Ld^2(\T^d))} \lesssim \varphi(T)\LRVert{v \otimes G}_{T,s_1,s_2} \Vert H \Vert_{\Ld^2(0,T ; \Ld^2(\T^d))},
\end{align*}

where we have used the semi-norm $\Vert \cdot \Vert_{T,s_1,s_2}$ from Proposition <ref>. We can now conclude as in the proof of Proposition <ref>. We observe that for all $p>2\ell+s+1+d$ and $\sigma>1+d/2$ (with $\ell, s \in \R^+$), there exist $s_1>s+1$ and $s_2>\ell+d/2$ such that

\begin{align*}
\Vert v \otimes G \Vert_{T,s_1,s_2} &\lesssim \underset{0 \leq s,t \leq T}{\sup} \Vert G(t,s) \Vert_{\mathcal{H}^p_{\sigma}}. %\\
%\mathcal{N}_{T}^{\alpha_1,\alpha_2}(v \otimes G) &\lesssim \underset{0 \leq s,t \leq T}{\sup} \Vert G(t,s) \Vert_{\mathcal{H}^p_{\sigma+1}}.
\end{align*}

By taking $\ell=2$ ans $s=3$, and by finally using Proposition <ref>, we reach the desired conclusion.

§ ANALYSIS OF THE KINETIC MOMENTS

Following the bootstrap procedure initiated in Section <ref>, we want to control $\LRVert{\varrho_\eps}_{\Ld^{2}(0,T; \H^m)}$ uniformly in $\eps$ and for $T<T_\eps$. In view of the transport equation bearing on $\varrho_\eps$ (see Lemma <ref>), we will relate the kinetic moments $\rho_{f_\eps}$ and $j_{f_\eps}$ to the fluid density $\varrho_\eps$  itself.

In this section, to ease readability, we drop out the subscripts $\eps$ when we refer to the solution. Recall that $\Lambda$ will always stand for a nonnegative continuous function which is independent of $\eps$, nondecreasing with respect to each of its argument, that may depend implicitly on the initial data and that may change from line to line.

For all $T \in [0,T_\eps)$ small enough, the goal of this section is thus to prove the following result.

Let $T \in (0, \min \left(T_\eps(R),\overline{T}(R)\right)$. For all $|I|\leq m$, we have for any $t \in (0,T)$,

\begin{align*}
\partial_x^I \rho_{f}(t,x)&=p'(\varrho(t,x))\int_0^t  \int_{\R^d}   \nabla_x [\mathrm{J}_\eps \partial_x^I\varrho](s, x-(t-s)v) \cdot \nabla_v f(t,x,v) \, \mathrm{d}v \, \mathrm{d}s + R^I[\rho_f](t,x), \\
\partial_x^I  j_{f}(t,x)&=p'(\varrho(t,x))\int_0^t  \int_{\R^d}  v\nabla_x  [\mathrm{J}_\eps \partial_x^I \varrho](s, x-(t-s)v) \cdot \nabla_v f(t,x,v) \, \mathrm{d}v \, \mathrm{d}s+R^I[j_f](t,x),
\end{align*}

where the remainders $R^I[\rho_f]$ and $R^I[j_f]$ satisfy

\begin{align*}
\left\Vert R^I[\rho_f] \right\Vert_{\Ld^2(0,T, \H^1_x)} \leq \Lambda(T,R), \ \ \left\Vert R^I[j_f] \right\Vert_{\Ld^2(0,T, \H^1_x)} \leq \Lambda(T,R).
\end{align*}

We recall the definition of the time $T_\eps(R)$ from our bootstrap procedure settled in Section <ref>, as well as the definition of the time $\overline{T}(R)$ from Lemma <ref> in Section <ref>. Note that it is independent of $\eps$. In the rest of this section, we will always implicitly consider times $T>0$ such that

\begin{align*}
T< \min \left(T_\eps(R),\overline{T}(R) \right).
\end{align*}

From Proposition <ref>, we can immediately infer the following corollary.

For $m>2+d$, $\sigma >1+d/2$ and $|I|\leq m$, we have

\begin{align*}
\Vert \partial_x^I \rho_f \Vert_{\Ld^2(0,T;\Ld^2)}  &\leq \Lambda(T,R), \\
\Vert \partial_x^I j_f \Vert_{\Ld^2(0,T;\Ld^2)}  &\leq \Lambda(T,R).
\end{align*}

By Proposition <ref>, we can write

\begin{align*}
\partial_x^I \rho_{f}= p'(\varrho)\mathrm{K}_{G}^{\mathrm{free}}[\mathrm{J}_\eps \partial_x^I\varrho]+R^I[\rho_f],
\end{align*}

with $G(t,x,v)= \nabla_v f(t,x,v)$ and $\left\Vert R^I[\rho_f] \right\Vert_{\Ld^2(0,T, \H^1_x)} \leq \Lambda(T,R)$. Since the kernel $G$ satisfies for $p>1+d$

\begin{align*}
\underset{0 \leq t \leq T}{\sup} \Vert G\Vert_{\mathcal{H}^p_{\sigma}} \leq \Vert  f\Vert_{\Ld^{\infty}(0,T;\mathcal{H}^{m-1}_{\sigma})} \leq \Lambda(T,R),
\end{align*}

we can use the estimate from Proposition <ref> to get

\begin{align*}
\Vert \partial_x^I \rho_f \Vert_{\Ld^2(0,T;\Ld^2)} &\lesssim \Vert p'(\varrho) \Vert_{\Ld^{2}(0,T; \Ld^{\infty})} \Lambda(T,R)\Vert \partial_x^I \varrho \Vert_{\Ld^2(0,T;\Ld^2)}  + \Vert R^I[\rho_f] \Vert_{\Ld^2(0,T;\Ld^2)}  \\
& \lesssim \mathrm{C}(\Vert \varrho \Vert_{\Ld^{\infty}(0,T; \H^{m-2})}) \Lambda(T,R)  +\Lambda(T,R) \\
& \leq \Lambda(T,R),
\end{align*}

by Sobolev embedding, Proposition <ref> and Lemma <ref>. The same argument applies for $\Vert \partial_x^I j_f \Vert_{\Ld^2(0,T;\Ld^2)}$.

Our strategy to prove Proposition <ref> goes as follows:

$-$ first, we take derivatives in the Vlasov equation to obtain a system of coupled kinetic equations satisfied by the augmented unknown $(\partial^I_{x} \partial^J_{v} f_\eps)_{|I|+|J|=m-1, m}$;

$-$ next, we study the average in velocity of $\mathcal{F}$ by relying on Duhamel formula and the Lagrangian point of view of Section <ref>. We isolate the leading terms and prove estimates for the remainders, using crucially the techniques developed in Sections <ref> and <ref>.

§.§ The integro-differential system for derivatives of moments

§.§.§ Applying derivatives

We start with the following algebraic lemma, where we apply $\partial_x^{I} \partial_v^{J}$ to the Vlasov equation.

Let us recall the notation $\widehat{\alpha}^k$ and $\overline{\alpha}^k$ for shifted indices (see Definition <ref>).

For any $I=(i_1, \cdots, i_d), J=(j_1, \cdots, ,j_d) \in \N^d$ such that $ \vert I \vert+\vert J \vert \in \lbrace m-1,m \rbrace $ and for any smooth function $f(t,x,v)$, we have

\begin{align*}
\left[\partial_x^{I} \partial_v^{J},\mathcal{T}_{\reg,\eps}^{u,\varrho} \right]f= \partial_x^I \partial_v^J E_{\reg,\eps}^{u,\varrho} \cdot \nabla_v f + \mathcal{M}^{I,J} \mathcal{F} +\mathcal{R}^{I,J}_1+ \mathcal{R}^{I,J}_0,
\end{align*}

\begin{align*}
\mathcal{F}&:=\left(\partial_x^{I} \partial_v^{J}f \right)_{\substack{I,J \in \N^d, \\  \vert I \vert+\vert J \vert \in \lbrace m-1,m \rbrace} }, \\
\mathcal{M}^{I,J} \mathcal{F}&:=\sum_{p=1}^d \mathbf{1}_{j_p \neq 0}  \left( \partial_x^{\widehat{I}^p} \partial_v^{\overline{J}^p} f-\partial_x^{I} \partial_v^{J}f \right)+\mathbf{1}_{\vert  I \vert > 2} \sum_{\substack{0 <\alpha < I \\ \vert \alpha  \vert \in \lbrace 1,2 \rbrace}} \binom{I}{\alpha}\partial_x^{\alpha} E_{\reg,\eps}^{u,\varrho} \cdot \nabla_v \partial_x^{I-\alpha} \partial_v^{J}  f \\
& \quad +\mathbf{1}_{\vert I \vert =1,2}\partial_x^I E_{\reg,\eps}^{u,\varrho} \cdot \partial_v^J \nabla_v f, \\[2mm]
%&+\sum_{k=1}^d \sum_{p=1}^d I! \mathbf{1}_{i_p \neq 0} \partial_{x_p} E_k \,  \partial_x^{\overline{I}^p} \partial_v^{\widehat{J}^k}f
%+\sum_{k=1}^d \sum_{p=1}^d J! \mathbf{1}_{j_p \neq 0}  E_k \,  \partial_x^{I} \partial_v^{\overline{J}^p} \partial_{v_k} f \\
%& \quad +\sum_{k=1}^d \sum_{\substack{0 \leq \alpha \leq I \\ \vert \alpha  \vert=2}} C_{\alpha}^I\textbf{1}_{\vert I \vert + \vert J \vert=m}\partial_x^{\alpha} E_k \,  \partial_x^{I-\alpha} \partial_v^{J} \partial_{v_k} f +\sum_{k=1}^d \sum_{\substack{0 \leq \beta \leq J \\ \vert \beta \vert =2}} C_{\beta}^J \textbf{1}_{\vert I \vert + \vert J \vert=m}  E_k \,  \partial_x^{I} \partial_v^{J-\beta} \partial_{v_k} f, \\
%& \quad +\sum_{k=1}^d \sum_{\substack{0\leq\alpha \leq I \\ 0\leq \beta \leq J \\ \vert \alpha \vert =\vert \beta  \vert =1 }} C^{I,J}_{\alpha, \beta} \textbf{1}_{\vert I \vert + \vert J \vert=m} \partial_x^{\alpha} E_k \,  \partial_x^{I-\alpha} \partial_v^{J-\beta}  \partial_{v_k} f, \\[2mm]
\mathcal{R}^{I,J}_1&:=\mathbf{1}_{\substack{\vert I \vert > 2 \\ \vert J \vert \neq 0}} \partial_x^I E_{\reg,\eps}^{u,\varrho} \cdot  \nabla_v \partial_v^Jf+\mathbf{1}_{\vert I \vert>1} \sum_{\substack{0 <\alpha < I \\ \vert \alpha  \vert = m-1}} \binom{I}{\alpha}\partial_x^{\alpha} E_{\reg,\eps}^{u,\varrho} \cdot \nabla_v \partial_x^{I-\alpha} \partial_v^{J}  f,
\end{align*}

\begin{align}\label{estim1L2:R_1}
\left\Vert \mathcal{R}^{I,J}_1 \right\Vert_{\Ld^2(0,T;\mathcal{H}^0_{r})} \leq  \Lambda(T,R),
\end{align}

and where $\mathcal{R}^{I,J}_0$ is a remainder satisfying,

\begin{align}\label{estim1:R_0}
%\left\Vert \mathcal{R}^{I,J}_0 \right\Vert_{\Ld^2(0,T;\mathcal{H}^1_{r})} \lesssim  \Lambda\left(T,R,\Vert \varrho^{\mathrm{in}} \Vert_{\H^{m-2}},\frac{1}{1- \Vert \rho_{f^0} \Vert_{\Ld^{\infty}} - T R}\right).
\left\Vert \mathcal{R}^{I,J}_0 \right\Vert_{\Ld^2(0,T;\mathcal{H}^1_{r})} \leq  \Lambda(T,R).
\end{align}

Using Lemma <ref>, we have

\begin{align*}
\partial_x^{I} \partial_v^{J}(\mathcal{T}_{\reg,\eps}^{u,\varrho} f)&=\mathcal{T}_{\reg,\eps}^{u,\varrho}(\partial_x^{I} \partial_v^{J}f)+ \sum_{p=1}^d  \mathbf{1}_{j_p \neq 0} \left( \partial_x^{\widehat{I}^p} \partial_v^{\overline{J}^p} f-\partial_x^{I} \partial_v^{J}f \right)+ \left[\partial_x^{I} \partial_v^{J}, E_{\reg,\eps}^{u,\varrho}\cdot \nabla_v\right]f.
\end{align*}

Since the force $E_{\reg,\eps}^{u,\varrho}(t,x)$ does not depend on $v$, we expand the commutator as

\begin{align*}
\left[\partial_x^{I} \partial_v^{J}, E(t,x)\cdot \nabla_v\right]f
%&=\mathbf{1}_{\substack{I \neq 0 \\ J=0}}\sum_{k=1}^d \sum_{0<\alpha \leq I } \binom{I}{\alpha} \partial_x^{\alpha} (E_{\reg,\eps}^{u,\varrho})_k \,  \partial_x^{I-\alpha} \partial_{v_k} f \\
%& \qquad \qquad  \qquad \qquad  \qquad  \qquad \qquad  +\mathbf{1}_{\substack{I \neq 0 \\ J \neq 0}} \sum_{k=1}^d \sum_{0<\alpha \leq I } \binom{I}{\alpha}\partial_x^{\alpha} (E_{\reg,\eps}^{u,\varrho})_k \,  \partial_x^{I-\alpha} \partial_v^{J} \partial_{v_k} f \\
= \mathbf{1}_{I \neq 0}\partial_x^I E_{\reg,\eps}^{u,\varrho} \cdot \partial_v^J \nabla_v f   +\mathbf{1}_{\vert I \vert> 1} \sum_{0<\alpha < I } \binom{I}{\alpha}\partial_x^{\alpha} E_{\reg,\eps}^{u,\varrho} \cdot  \partial_x^{I-\alpha} \partial_v^{J} \nabla_v f.
%&= \mathrm{I}+ \mathrm{II},
\end{align*}

Note that if $J \neq 0$ then $\vert I \vert \leq m-1$, and if $\vert I \vert =1,2$ then $J \neq 0$. The terms that cannot be considered as remainders in the last sum are those such that $\vert I \vert-\vert \alpha \vert + \vert J \vert+1 \in \lbrace m-1,m \rbrace$, that is for $\vert \alpha \vert  \in \lbrace 1,2 \rbrace$. We thus have

\begin{align*}
\left[\partial_x^{I} \partial_v^{J}, E_{\reg,\eps}^{u,\varrho}\cdot \nabla_v\right]f&=\mathbf{1}_{\substack{\vert I \vert \neq 0 \\ J=0}}\partial_x^I \partial_v^J E_{\reg,\eps}^{u,\varrho} \cdot \nabla_v f+\mathbf{1}_{\vert I \vert =1,2}\partial_x^I E_{\reg,\eps}^{u,\varrho} \cdot \partial_v^J \nabla_v f   \\
& \quad +\mathbf{1}_{\vert I \vert> 2}\sum_{\substack{0 <\alpha < I \\ \vert \alpha  \vert \in \lbrace 1,2 \rbrace}} \binom{I}{\alpha}\partial_x^{\alpha} E_{\reg,\eps}^{u,\varrho} \cdot \nabla_v \partial_x^{I-\alpha} \partial_v^{J}  f \\
& \quad +  \mathcal{R}^{I,J}_1+\mathcal{R}^{I,J}_0,
\end{align*}

\begin{align*}
\mathcal{R}^{I,J}_1&:=\mathbf{1}_{\substack{\vert I \vert > 2 \\ \vert J \vert \neq 0}}\partial_x^I E_{\reg,\eps}^{u,\varrho} \cdot \partial_v^J \nabla_v f
+\mathbf{1}_{\vert I \vert> 1}\sum_{\substack{0 <\alpha < I \\ \vert \alpha  \vert = m-1}} \binom{I}{\alpha}\partial_x^{\alpha} E_{\reg,\eps}^{u,\varrho} \cdot \nabla_v \partial_x^{I-\alpha} \partial_v^{J}  f, \\
\mathcal{R}^{I,J}_0&:=\mathbf{1}_{\vert I \vert> 1} \sum_{\substack{0 <\alpha < I \\ 3 \leq \vert \alpha  \vert \leq m-2}} \binom{I}{\alpha}\partial_x^{\alpha} E_{\reg,\eps}^{u,\varrho} \cdot \nabla_v \partial_x^{I-\alpha} \partial_v^{J}  f.
\end{align*}

Let us estimate the remainder $\mathcal{R}^{I,J}_0$ in $\mathcal{H}_r^1$: setting $\chi(v)=(1+\vert v \vert^2)^{r/2}$, we have

\begin{align*}
\Vert \mathcal{R}^{I,J}_0 \Vert_{\mathcal{H}_r^1} &\lesssim \sum_{\substack{0 <\alpha < I \\ 3 \leq \vert \alpha  \vert \leq m-2}} \Vert \chi \partial_x^{\alpha} E_{\reg,\eps}^{u,\varrho} \cdot \nabla_v \partial_x^{I-\alpha} \partial_v^{J}  f \Vert_{\Ld^2_{x,v}} \\
& \quad +\sum_{\substack{0 <\alpha < I \\ 3 \leq \vert \alpha  \vert \leq m-2}}  \sum_{k=1}^d\Big( \Vert \chi \partial_x^{\widehat{\alpha}^k} E_{\reg,\eps}^{u,\varrho} \cdot \nabla_v \partial_x^{I-\alpha} \partial_v^{J}  f \Vert_{\Ld^2_{x,v}}
+\Vert \chi \partial_x^{\alpha} E_{\reg,\eps}^{u,\varrho} \cdot \partial_{x_k} \nabla_v \partial_x^{I-\alpha} \partial_v^{J}  f \Vert_{\Ld^2_{x,v}} \\
& \quad \qquad  \qquad \qquad \qquad \qquad  \qquad \qquad  \qquad \qquad \qquad \qquad  +\Vert \chi \partial_x^{\alpha} E_{\reg,\eps}^{u,\varrho} \cdot \partial_{v_k} \nabla_v \partial_x^{I-\alpha} \partial_v^{J}  f \Vert_{\Ld^2_{x,v}} \Big).
\end{align*}

$\bullet$ If $\frac{m+1}{2}\leq\vert \alpha \vert \leq m-2$, then for all $k$

\begin{align*}
\Vert \chi \partial_x^{\alpha} E_{\reg,\eps}^{u,\varrho} \cdot \nabla_v \partial_x^{I-\alpha} \partial_v^{J}  f \Vert_{\Ld^2_{x,v}}^2&+\Vert \chi \partial_x^{\widehat{\alpha}^k} E_{\reg,\eps}^{u,\varrho} \cdot \nabla_v \partial_x^{I-\alpha} \partial_v^{J}  f \Vert_{\Ld^2_{x,v}}^2 \\
&\leq \left(\Vert \partial_x^{\alpha} E_{\reg,\eps}^{u,\varrho} \Vert_{\Ld^2}^2+ \Vert \partial_x^{\widehat{\alpha}^k} E_{\reg,\eps}^{u,\varrho} \Vert_{\Ld^2}^2 \right)\int_{\R^d} \chi(v)^2 \Vert \nabla_v \partial_x^{I-\alpha} \partial_v^{J}  f \Vert_{\Ld^{\infty}}^2 \, \mathrm{d}v \\
& \lesssim \Vert  E_{\reg,\eps}^{u,\varrho} \Vert_{\H^{m-1}}^2 \int_{\R^d} \chi(v)^2 \Vert \nabla_v \partial_v^{J}  f \Vert_{\H^{\sigma}}^2 \, \mathrm{d}v,
\end{align*}

provided that $\sigma>\vert I-\alpha \vert+d/2$. Since $\vert J \vert +1 + \vert I \vert -\vert \alpha \vert+d/2 \leq m+1-\vert \alpha \vert +d/2\leq \frac{m+1+d}{2}$ and since $m>4+d$, we can find such a $\sigma$ so that

\begin{align*}
\Vert \chi \partial_x^{\alpha} E_{\reg,\eps}^{u,\varrho} \cdot \nabla_v \partial_x^{I-\alpha} \partial_v^{J}  f \Vert_{\Ld^2_{x,v}}+ \Vert \chi \partial_x^{\widehat{\alpha}^k} E_{\reg,\eps}^{u,\varrho} \cdot \nabla_v \partial_x^{I-\alpha} \partial_v^{J}  f \Vert_{\Ld^2_{x,v}} \lesssim \Vert  E_{\reg,\eps}^{u,\varrho} \Vert_{\H^{m-1}} \Vert f \Vert_{\mathcal{H}_r^{m-1}}.
\end{align*}

Likewise, we have for all $k$

\begin{align*}
\Vert \chi \partial_x^{\alpha} E_{\reg,\eps}^{u,\varrho} \cdot \partial_{x_k} \nabla_v \partial_x^{I-\alpha} \partial_v^{J}  f \Vert_{\Ld^2_{x,v}} ^2
\leq \Vert  E_{\reg,\eps}^{u,\varrho} \Vert_{\H^{m-1}}^2 \int_{\R^d} \chi(v)^2 \Vert \nabla_v \partial_v^{J}  f \Vert_{\H^{\sigma}}^2 \, \mathrm{d}v,
\end{align*}

provided that $\sigma>1+\vert I-\alpha \vert+d/2$. Since $\vert J \vert +1 + 1+\vert I \vert -\vert \alpha \vert+d/2 \leq m+2-\vert \alpha \vert +d/2\leq \frac{m+3+d}{2}$ and since $m>6+d$, there exists such a $\sigma$ so that

\begin{align*}
\Vert \chi \partial_x^{\alpha} E \cdot \partial_{x_k} \nabla_v \partial_x^{I-\alpha} \partial_v^{J}  f \Vert_{\Ld^2_{x,v}} \lesssim \Vert  E \Vert_{\H^{m-1}} \Vert f \Vert_{\mathcal{H}_r^{m-1}}.
\end{align*}

The same procedure can be applied for the terms $\Vert \chi \partial_x^{\alpha} E_{\reg,\eps}^{u,\varrho} \cdot \partial_{v_k} \nabla_v \partial_x^{I-\alpha} \partial_v^{J}  f \Vert_{\Ld^2_{x,v}}$.

$\bullet$ If $3 \leq \vert \alpha \vert < \frac{m+1}{2}$, then

\begin{align*}
\Vert \chi \partial_x^{\alpha} E_{\reg,\eps}^{u,\varrho} \cdot \nabla_v \partial_x^{I-\alpha} \partial_v^{J}  f \Vert_{\Ld^2_{x,v}}&+ \Vert \chi \partial_x^{\widehat{\alpha}^k} E_{\reg,\eps}^{u,\varrho} \cdot \nabla_v \partial_x^{I-\alpha} \partial_v^{J}  f \Vert_{\Ld^2_{x,v}} \\
&\leq \left(  \Vert \partial_x^{\alpha} E_{\reg,\eps}^{u,\varrho} \Vert_{\Ld^{\infty}} + \Vert \partial_x^{\widehat{\alpha}^k} E_{\reg,\eps}^{u,\varrho} \Vert_{\Ld^{\infty}} \right) \Vert \chi   \nabla_v \partial_x^{I-\alpha} \partial_v^{J}  f \Vert_{\Ld^2_{x,v}} \\
&\lesssim \Vert  E_{\reg,\eps}^{u,\varrho} \Vert_{\H^{\sigma}} \Vert f \Vert_{\mathcal{H}_r^{m-1}},
\end{align*}

provided that $\sigma> 1+ \vert \alpha \vert+d/2$. Since $1+ \vert \alpha \vert+d/2 \leq \frac{m+3+d}{2}$ and $m>5+d$, we can find such a $\sigma$ so that

\begin{align*}
\Vert \chi \partial_x^{\widehat{\alpha}^k} E_{\reg,\eps}^{u,\varrho} \cdot \nabla_v \partial_x^{I-\alpha} \partial_v^{J}  f \Vert_{\Ld^2_{x,v}}  \lesssim \Vert  E_{\reg,\eps}^{u,\varrho} \Vert_{\H^{m-1}} \Vert f \Vert_{\mathcal{H}_r^{m-1}}.
\end{align*}

Likewise, we have for all $k$

\begin{align*}
\Vert \chi \partial_x^{\alpha} E_{\reg,\eps}^{u,\varrho} \cdot \partial_{x_k} \nabla_v \partial_x^{I-\alpha} \partial_v^{J}  f \Vert_{\Ld^2_{x,v}} ^2
\leq \Vert \partial_x^{\alpha} E_{\reg,\eps}^{u,\varrho} \Vert_{\Ld^{\infty}} \Vert \chi \partial_x  \nabla_v \partial_x^{I-\alpha} \partial_v^{J}  f \Vert_{\Ld^2_{x,v}} \lesssim \Vert  E_{\reg,\eps}^{u,\varrho} \Vert_{\H^{\sigma}} \Vert f \Vert_{\mathcal{H}_r^{m-1}},
\end{align*}

provided that $\sigma>\vert I-\alpha \vert+d/2$. Since $\vert \alpha \vert+d/2 \leq \frac{m+2+d}{2}$ and $m>4+d$, there exists such a $\sigma$ so that

\begin{align*}
\Vert \chi \partial_x^{\alpha} E_{\reg,\eps}^{u,\varrho} \cdot \nabla_v \partial_x \partial_x^{I-\alpha} \partial_v^{J}  f \Vert_{\Ld^2_{x,v}}  \lesssim \Vert  E_{\reg,\eps}^{u,\varrho} \Vert_{\H^{m-1}} \Vert f \Vert_{\mathcal{H}_r^{m-1}}.
\end{align*}

The same procedure can be applied for the terms $\Vert \chi \partial_x^{\alpha} E_{\reg,\eps}^{u,\varrho} \cdot \partial_{v_k} \nabla_v \partial_x^{I-\alpha} \partial_v^{J}  f \Vert_{\Ld^2_{x,v}}$.

All in all, we have proven that for all $t \in [0,T]$

\begin{align*}
\Vert \mathcal{R}^{I,J}_0 (t)\Vert_{\mathcal{H}_r^1} \lesssim \Vert  E_{\reg,\eps}^{u,\varrho}(t) \Vert_{\H^{m-1}} \Vert f(t) \Vert_{\mathcal{H}_r^{m-1}} \leq \Vert f \Vert_{\Ld^{\infty}(0,T;\mathcal{H}_r^{m-1})} \Vert  E_{\reg,\eps}^{u,\varrho}(t) \Vert_{\H^{m-1}} \leq R \Vert  E(t) \Vert_{\H^{m-1}}.
\end{align*}

By the estimate (<ref>) from Lemma <ref> and by Lemma <ref>, we finally have

\begin{align*}
\left\Vert \mathcal{R}^{I,J}_0 \right\Vert_{\Ld^2(0,T;\mathcal{H}^1_{r})} \leq  \Lambda(T,R).
\end{align*}

With the same exact arguments, we easily obtain the fact that

\begin{align*}
\left\Vert \mathcal{R}^{I,J}_1 \right\Vert_{\Ld^2(0,T;\mathcal{H}^0_{r})} \leq  \Lambda(T,R),
\end{align*}

and this concludes the proof.

We will actually obtain an improved $\Ld^2(0,T;\mathcal{H}^1_{r})$ estimate for the term $\mathcal{R}^{I,J}_1$ (or, more preciely, related terms) in the end of the current section.

We can see $\mathcal{M}^{I,J} \mathcal{F}$ appearing in Lemma <ref> as a linear combination of $\mathcal{F}^{K,L}= \partial_x^{K} \partial_v^{L}f$. More precisely, we can write for all $(I,J)$,

\begin{align*}
\mathcal{M}^{I,J}\mathcal{F} &=\sum_{K,L} \mathcal{M}_{(I,J),(K,L)}\mathcal{F}^{K,L}, \\
\mathcal{M}_{(I,J),(K,L)}&:=\sum_{p=1}^d \mathbf{1}_{j_p \neq 0} (\mathbf{1}_{(K,L)=(\widehat{I}^p,\overline{J}^p)}-\mathbf{1}_{(K,L)=(I,J)} )
+\mathbf{1}_{ \vert I \vert=1,2} \sum_{p=1}^d  \mathbf{1}_{(K,L)=(0,\widehat{J}^p)} \partial_x^{I} (E_{\reg,\eps}^{u,\varrho})_p  \\
& \quad +\sum_{p=1}^d \sum_{\substack{0 <\alpha < I \\ \vert \alpha  \vert \in \lbrace 1,2 \rbrace}}  \binom{I}{\alpha} \mathbf{1}_{(K,L)=(I-\alpha,\widehat{J}^p)} \partial_x^{\alpha} (E_{\reg,\eps}^{u,\varrho})_p .
\end{align*}

Let us observe that the coefficient involved in the operator $\mathcal{M}$ involve only $0$, $1$ or $2$ derivatives of the force field $E_{\reg,\eps}^{u,\varrho}(t,x)$, but nothing coming from $f$.

We finally introduce the following additional notations which will allow us to reformulate Lemma <ref> in a compact way.

We consider the following quantities:

* $\mathcal{R}_0$ and $\mathcal{R}_1$ are the vectors defined by

\begin{align*}
\mathcal{R}_0=\left(\mathcal{R}^{I,J}_0 \right)_{\substack{I,J \in \N^d, \\  \vert I \vert+\vert J \vert \in \lbrace m-1,m \rbrace }}, \ \ \mathcal{R}_1=\left(\mathcal{R}^{I,J}_0 \right)_{\substack{I,J \in \N^d, \\  \vert I \vert+\vert J \vert \in \lbrace m-1,m \rbrace }};
% \ \ \mathcal{R}=\left(\mathcal{R}^{I,J}+\mathbf{1}_{\vert I \vert \neq m} \partial_x^I E_{\reg,\eps}^{u,\varrho} \cdot \nabla_v f \right)_{I,J \in \N^d,  \vert I \vert+\vert J \vert \in \lbrace m-1,m \rbrace };
\end{align*}

* $\mathcal{M}$ is the linear map defined by

\begin{align*}
\mathcal{M}=\left(\mathcal{M}_{(I,J),(K,L)} \right)_{\substack{ (I,J),(K,L)\\  \vert I \vert+\vert J \vert, \vert K \vert+\vert L \vert \in \lbrace m-1,m \rbrace }};
\end{align*}

* $\mathcal{L}$ is the vector defined by

\begin{align*}
%\mathcal{L}=\left(\mathbf{1}_{\vert I \vert =m} \partial_x^I E_{\reg,\eps}^{u,\varrho} \cdot \nabla_v f \right)_{I,J \in \N^d,  \vert I \vert+\vert J \vert \in \lbrace m-1,m \rbrace }.
\mathcal{L}=\left( \partial_x^I \partial_v^J E_{\reg,\eps}^{u,\varrho} \cdot \nabla_v f \right)_{\substack{I,J \in \N^d, \\  \vert I \vert+\vert J \vert \in \lbrace m-1,m \rbrace }}.
\end{align*}

§.§.§ The semi-Lagrangian approach

If $f$ satisfies the Vlasov equation (in a strong sense), we have $\partial_x^{I} \partial_v^{J}(\mathcal{T}_{\reg,\eps}^{u,\varrho} f)=0$ for any $I,J$. We can use Lemma <ref> to obtain the following coupled system of equations satisfied by the family $\mathcal{F}=(\partial_x^{I} \partial_v^{J}f)_{I,J}$:

\begin{align}\label{eq:augmentedVarF}
\mathcal{T}_{\reg,\eps}^{u,\varrho} \mathcal{F}+\mathcal{M} \mathcal{F}+\mathcal{L}=-\mathcal{R}_0-\mathcal{R}_1.
\end{align}

For any function $g(t,x,v)$, we set

\begin{align*}
\widetilde{g}(t,x,v)=g(t,\X^{t;0}(x,v),\V^{t;0}(x,v)),
\end{align*}

$$s \mapsto \mathrm{Z}^{s;t}(x,v)=\pr{\X^{s;t}(x,v),\V^{s;t}(x,v)}$$
is the solution to

\begin{equation}\label{EDO-charac}
\left\{
\begin{aligned}
\frac{\mathrm{d}}{\mathrm{d}s} \mathrm{X}^{s;t}(x,v) &=\mathrm{V}^{s;t}(x,v), \ \ \mathrm{X}^{t;t}(x,v)=x \in \T^d, \\[2mm]
\frac{\mathrm{d}}{\mathrm{d}s}\mathrm{V}^{s;t}(x,v)&=-\mathrm{V}^{s;t}(x,v)+E_{\reg,\eps}^{u,\varrho}(s,\mathrm{X}^{s;t}(x,v)), \ \ \mathrm{V}^{t;t}(x,v)=v \in \R^d.
\end{aligned}
\right.
\end{equation}

After the composition by $ (t,x,v) \mapsto (t,\mathrm{X}^{t;0}(x,v),\mathrm{V}^{t;0}(x,v))$, we thus obtain by the method of characteristics

\begin{align}\label{eq:Ftilde}
\partial_t \widetilde{\mathcal{F}}+\widetilde{\mathcal{M}} \widetilde{\mathcal{F}}+\widetilde{\mathcal{L}}=d\widetilde{\mathcal{F}} -\widetilde{\mathcal{R}}_0-\widetilde{\mathcal{R}}_1.
\end{align}

To deal with the coupling matrix $\mathcal{M}$, we introduce the following object.

For all $(x,v)$ and $s,t \geq 0$, we define the resolvant operator $\mathfrak{N}^{s,t}(x,v)$ as the solution $s \mapsto \mathfrak{N}^{s,t}(x,v)$ of

\begin{equation}\label{eq:resolvant}
\left\{
\begin{aligned}
\partial_s \mathfrak{N}^{s;t}+\left[\mathcal{M} \circ \mathrm{Z}^{s;0}-d \mathrm{Id} \right] \mathfrak{N}^{s;t}&=0,\\
\mathfrak{N}^{t;t}&= \mathrm{Id}.
\end{aligned}
\right.
\end{equation}

The resolvant is well-defined thanks to the Cauchy-Lipschitz theorem. We also have

\begin{align*}
\mathfrak{N}^{s;t}(x,v)=e^{d(s-t)} \mathrm{N}^{s;t}(x,v),
\end{align*}

\begin{equation}\label{eq:resolvant2}
\left\{
\begin{aligned}
\partial_s \mathrm{N}^{s;t}+\mathcal{M} \circ \mathrm{Z}^{s;0} \mathrm{N}^{s;t}&=0,\\
\mathrm{N}^{t;t}&= \mathrm{Id}.
\end{aligned}
\right.
\end{equation}

For the upcoming analysis, we need the following bounds on the resolvant.

For all $0 \leq k<m-3-d/2$, we have

\begin{align*}
\underset{0 \leq s,t \leq T}{\sup} \Vert \mathrm{N}^{s;t} \Vert_{\W^{k, \infty}_{x,v}} + \underset{0 \leq s,t \leq T}{\sup} \Vert \partial_s \mathrm{N}^{s;t} \Vert_{\W^{k, \infty}_{x,v}}+ \underset{0 \leq s,t \leq T}{\sup} \Vert \partial_t \mathrm{N}^{s;t} \Vert_{\W^{k, \infty}_{x,v}} \leq \Lambda(T,R).
\end{align*}

We have

$$\mathrm{N}^{s;t}= \mathrm{Id}- \int_t^s [\mathcal{M} \circ \mathrm{Z}^{\tau,0}] \mathrm{N}^{\tau;t} \, \mathrm{d}\tau.$$

By the definition of coefficients of the matrix $\mathcal{M}$, we also have

\begin{align*}
\Vert  \mathcal{M} \Vert_{\Ld^{2}(0,T;\Ld^{\infty})} \lesssim \underset{ \vert \alpha \vert \leq 2}{\sup} \Vert \partial_x^{\alpha} E \Vert_{\Ld^{2}(0,T;\Ld^{\infty})} \leq \Lambda(T,R),
\end{align*}

thanks to the estimate (<ref>) from Lemma <ref> and by Sobolev embedding (with $m-1>2+d/2$). Grönwall's lemma leads to the conclusion.

We first obtain the following decomposition for the kinetic moments of the vector $\mathcal{F}$.

We have

\begin{multline*}
\int_{\R^d} \mathcal{F}(t,x,v) \, \mathrm{d}v=\mathcal{I}_{in}^0(t,x)+\mathcal{I}_{\mathcal{R}_0}^0(t,x)+\mathcal{I}_{\mathcal{R}_1}^0(t,x)\\
-\int_0^t  e^{d(t-s)} \int_{\R^d} \mathrm{N}^{t;s}(\mathrm{Z}^{0;t}(x,v)) \mathcal{L}(s, \mathrm{Z}^{s;t}(x,v)) \, \mathrm{d}v \, \mathrm{d}s,
\end{multline*}

\begin{multline*}
\int_{\R^d} v \otimes \mathcal{F}(t,x,v) \, \mathrm{d}v=\mathcal{I}_{in}^{1}(t,x)+\mathcal{I}_{\mathcal{R}_0}^{1}(t,x)+\mathcal{I}_{\mathcal{R}_1}^{1}(t,x)\\
-\int_0^t  e^{d(t-s)} \int_{\R^d} v\otimes \mathrm{N}^{t;s}(\mathrm{Z}^{0;t}(x,v)) \mathcal{L}(s, \mathrm{Z}^{s;t}(x,v)) \, \mathrm{d}v \, \mathrm{d}s,
\end{multline*}

\begin{align*}
\mathcal{I}_{in}^0(t,x)&:= e^{dt}\int_{\R^d}  \mathrm{N}^{t;0}(\mathrm{Z}^{0;t}(x,v))\mathcal{F}_{\mid t=0}(\mathrm{Z}^{0;t}(x,v)) \, \mathrm{d}v, \\
\mathcal{I}_{\mathcal{R}_j}^0(t,x)&:=-\int_0^t e^{d(t-s)}  \int_{\R^d} \mathrm{N}^{t;s}(\mathrm{Z}^{0;t}(x,v))  \mathcal{R}_j(s, \mathrm{Z}^{s;t}(x,v)) \, \mathrm{d}v \, \mathrm{d}s, \ \ j=0,1, \\
\mathcal{I}_{in}^{1}(t,x)&:= e^{dt} \int_{\R^d} v \otimes \mathrm{N}^{t;0}(\mathrm{Z}^{0;t}(x,v))\mathcal{F}_{\mid t=0}(\mathrm{Z}^{0;t}(x,v)) \, \mathrm{d}v, \\
\mathcal{I}_{\mathcal{R}_j}^{1}(t,x)&:=-\int_0^t e^{d(t-s)} \int_{\R^d}   v \otimes \mathrm{N}^{t;s}(\mathrm{Z}^{0;t}(x,v))  \mathcal{R}_j(s, \mathrm{Z}^{s;t}(x,v)) \, \mathrm{d}v \, \mathrm{d}s; \ \ j=0,1.
\end{align*}

We only explain the case of the moment of order $0$, the other being similar. Starting from the equation (<ref>) and using the resolvant operators $\mathfrak{N}$ and $\mathrm{N}$ defined in (<ref>) and (<ref>), we have

\begin{align*}
\widetilde{\mathcal{F}}(t)&=\mathfrak{N}^{t,0} \widetilde{\mathcal{F}}_{\mid t=0}-\int_0^t \mathfrak{N}^{t,s}\left[\widetilde{\mathcal{L}}(s)+\widetilde{\mathcal{R}}_0(s)+\widetilde{\mathcal{R}}_1(s) \right] \, \mathrm{d}s \\
&=e^{dt}\mathrm{N}^{t;0} \widetilde{\mathcal{F}}_{\mid t=0}-\int_0^t e^{d(t-s)} \mathrm{N}^{t;s}\left[\widetilde{\mathcal{L}}(s)+\widetilde{\mathcal{R}}_0(s)+\widetilde{\mathcal{R}}_1(s) \right] \, \mathrm{d}s.
\end{align*}

After a composition by the map $ (t,x,v) \mapsto (t,\mathrm{X}^{0;t}(x,v),\mathrm{V}^{0;t}(x,v))$, we obtain

\begin{multline*}
\mathcal{F}(t)=e^{dt}\left[\mathrm{N}^{t;0} \circ \mathrm{Z}^{0;t}\right]\mathcal{F}_{\mid t=0} \circ \mathrm{Z}^{0;t}-\int_0^t e^{d(t-s)} \left[\mathrm{N}^{t;s} \circ \mathrm{Z}^{0;t} \right]\left(\mathcal{R}_0(s,\mathrm{Z}^{s;t}) +\mathcal{R}_1(s,\mathrm{Z}^{s;t}) \right) \, \mathrm{d}s \\
-\int_0^t  e^{d(t-s)} \left[\mathrm{N}^{t;s} \circ \mathrm{Z}^{0;t} \right] \mathcal{L}(s, \mathrm{Z}^{s;t}) \, \mathrm{d}s.
\end{multline*}

We reach the desired conclusion by integrating in velocity.

§.§ First remainders

Let us show straightaway that some of the previous terms can be considered as remainders.

We have

\begin{align}
\label{estimate:I_0-H1}\Vert \mathcal{I}_{in}^0 \Vert_{\Ld^2(0,T;\H^1)} \Vert +\Vert \mathcal{I}_{in}^{1} \Vert_{\Ld^2(0,T;\H^1)}   &\leq \Lambda(T,R) \Vert f^{\mathrm{in}} \Vert_{\mathcal{H}^{m+1}_r}, \\
\label{estimate:I_R0-H1}\Vert \mathcal{I}_{\mathcal{R}_0}^0\Vert_{\Ld^2(0,T;\H^1)} + \Vert \mathcal{I}_{\mathcal{R}_0}^{1} \Vert_{\Ld^2(0,T;\H^1)}  &\leq  \Lambda(T,R), \\
\label{estimate:I_R1-L2}\Vert \mathcal{I}_{\mathcal{R}_1}^0\Vert_{\Ld^2(0,T;\Ld^2)} + \Vert \mathcal{I}_{\mathcal{R}_1}^{1} \Vert_{\Ld^2(0,T;\Ld^2)}  &\leq  \Lambda(T,R).
\end{align}

We shall first estimate the term $\mathcal{I}_{in}^0$. We have

\begin{align*}
\vert \mathcal{I}_{in}^0(t,x) \vert \lesssim e^{dT} \underset{0 \leq t \leq T}{\sup} \Vert \mathrm{N}^{t;s} \Vert_{\Ld^{\infty}_{x,v}} \int_{\R^d} \vert \mathcal{F}_{0}(\mathrm{Z}^{0;t}(x,v)) \vert \, \mathrm{d}v \lesssim \Lambda(T,R) \sum_{I,J} \int_{\R^d} \vert \partial_x^{I} \partial_v^{J} f^{\mathrm{in}} (\mathrm{Z}^{0;t}(x,v)) \vert \, \mathrm{d}v,
\end{align*}

thanks to Lemma <ref> with $k=0$. Using the generalized Minkowski inequality and the Cauchy-Schwarz inequality, we get

\begin{align*}
\Vert \mathcal{I}_{in}^0 \Vert_{\Ld^2(0,T;\Ld^2)} &\lesssim \Lambda(T,R)  \sum_{I,J} \left\Vert \int_{\R^d} \Vert \partial_x^{I} \partial_v^{J} f^{\mathrm{in}} (\mathrm{Z}^{0;t}(\cdot,v)) \Vert_{\Ld^2} \, \mathrm{d}v \right\Vert_{\Ld^2(0,T)} \\
&\lesssim\Lambda(T,R)  \sum_{I,J} \left\Vert \left( \int_{\T^d \times \R^d} (1+ \vert v \vert^2)^r \vert \partial_x^{I} \partial_v^{J} f^{\mathrm{in}} (\mathrm{Z}^{0;t}(x,v)) \vert^2 \, \mathrm{d}x \, \mathrm{d} v \right)^{1/2} \right\Vert_{\Ld^2(0,T)},
\end{align*}

since $2r>d$. We then perform the change of variable $(x,v) \mapsto \mathrm{Z}^{0;t}(x,v)$ which yields

\begin{align*}
\int_{\T^d \times \R^d} (1+ \vert v \vert^2)^r \vert \partial_x^{I} \partial_v^{J} f^{\mathrm{in}} (\mathrm{Z}^{0;t}(x,v)) \vert^2 \, \mathrm{d}x \, \mathrm{d} v  \lesssim \Lambda(T,R) \int_{\T^d \times \R^d} (1+ \vert \mathrm{V}^{t;0}(x,v) \vert^2)^r \vert \partial_x^{I} \partial_v^{J} f^{\mathrm{in}} (x,v) \vert^2 \, \mathrm{d}x \, \mathrm{d} v .
\end{align*}

\begin{align*}
\mathrm{V}^{t;0}(x,v)=e^{-t}v + \int_0^t e^{\tau-t} E_{\reg,\eps}^{u,\varrho}(\tau, \X^{\tau;t}(x,v)) \, \mathrm{d}\tau,
\end{align*}

we have by Sobolev embedding

\begin{align*}
\vert \mathrm{V}^{t;0}(x,v) \vert^2  \leq \vert v \vert^2 + \left\vert \int_0^t \Vert E_{\reg,\eps}^{u,\varrho}(\tau) \Vert_{\Ld^{\infty}} \, \mathrm{d}\tau \right\vert^2 \lesssim \vert v \vert^2+ T \Vert E_{\reg,\eps}^{u,\varrho} \Vert_{\Ld^{2}(0,T; \H^{m-1})}^2.
\end{align*}

By the estimate (<ref>) from Lemma <ref>, we get

\begin{align*}
\vert \mathrm{V}^{t;0}(x,v) \vert^2   \leq\vert v \vert^2 \Lambda(T,R),
\end{align*}

and then

\begin{align*}
\int_{\T^d \times \R^d} (1+ \vert \mathrm{V}^{t;0}(x,v) \vert^2)^r \vert \partial_x^{I} \partial_v^{J} f^{\mathrm{in}} (x,v) \vert^2 \, \mathrm{d}x \, \mathrm{d} v &\lesssim \Lambda(T,R)\int_{\T^d \times \R^d} (1+ \vert v\vert^2)^r \vert \partial_x^{I} \partial_v^{J} f^{\mathrm{in}} (x,v) \vert^2 \, \mathrm{d}x \, \mathrm{d} v \\
& \lesssim \Lambda(T,R) \Vert f^{\mathrm{in}} \Vert_{\mathcal{H}^m_r}^2.
\end{align*}

This implies

\begin{align*}
%\Vert \mathcal{I}_{in}^0 \Vert_{\Ld^2(0,T;\Ld^2)}  \leq T^{\frac{1}{2}} \Lambda(T,R) \Vert f^{\mathrm{in}} \Vert_{\mathcal{H}^m_r}.
\Vert \mathcal{I}_{in}^0 \Vert_{\Ld^2(0,T;\Ld^2)}  \leq \Lambda(T,R) \Vert f^{\mathrm{in}} \Vert_{\mathcal{H}^m_r}.
\end{align*}

Likewise, using $2r>d+1$, we have

\begin{align*}
\Vert \mathcal{I}_{in}^{1} \Vert_{\Ld^2(0,T;\Ld^2)} \lesssim\Lambda(T,R) \Vert f^{\mathrm{in}} \Vert_{\mathcal{H}^m_r}.
\end{align*}

We also have

\begin{align*}
\left[\mathcal{I}_{in}^0 \right]_{(I,J)}(t,x)=e^{dt}\int_{\R^d} \sum_{(K,L)} \mathrm{N}^{t;0}_{(I,J),(K,L)}(\mathrm{Z}^{0;t}(x,v))\left[\mathcal{F}_{\mid t=0} \right]_{(K,L)}(\mathrm{Z}^{0;t}(x,v)) \, \mathrm{d}v,
\end{align*}

\begin{align*}
\nabla_x\left[\mathcal{I}_{in}^0 \right]_{(I,J)}(t,x)&=e^{dt}\int_{\R^d} \sum_{(K,L)} \nabla_x \mathrm{Z}^{0;t}(x,v) \nabla_x\mathrm{N}^{t;0}_{(I,J),(K,L)}(\mathrm{Z}^{0;t}(x,v))\left[\mathcal{F}_{\mid t=0} \right]_{(K,L)}(\mathrm{Z}^{0;t}(x,v)) \, \mathrm{d}v \\
& \quad + e^{dt}\int_{\R^d} \sum_{(K,L)} \mathrm{N}^{t;0}_{(I,J),(K,L)}(\mathrm{Z}^{0;t}(x,v)) \nabla_x \mathrm{Z}^{0;t}(x,v) \nabla_x \left[\mathcal{F}_{\mid t=0} \right]_{(K,L)}(\mathrm{Z}^{0;t}(x,v)) \, \mathrm{d}v.
\end{align*}

The same procedure as before, using Lemma <ref> with $k=1$ and the pointwise bounds (<ref>)-(<ref>) from Remark <ref>, gives

\begin{align*}
\Vert \nabla_x \mathcal{I}_{in}^0 \Vert_{\Ld^2(0,T;\Ld^2)}+\Vert \nabla_x\mathcal{I}_{in}^{1} \Vert_{\Ld^2(0,T;\Ld^2)} \lesssim  \Lambda(T,R) \Vert f^{\mathrm{in}} \Vert_{\mathcal{H}^{m+1}_r}.
\end{align*}

We now estimate the terms $ \mathcal{I}_{\mathcal{R}_0}^0$ and $\mathcal{I}_{\mathcal{R}_0}^{1}$. We use the same arguments as before, with an additional Cauchy-Schwarz inequality in time leading to

\begin{align*}
\Vert  \mathcal{I}_{\mathcal{R}_0}^0  \Vert_{\Ld^2(0,T;\Ld^2)} &\leq \Lambda(T,R) \sum_{I,J} \left\Vert \int_0^t \int_{\R^d} \left\Vert  \mathcal{R}_0^{I,J}(s,\mathrm{Z}^{s;t}(\cdot,v)  \right\Vert_{\Ld^2} \, \mathrm{d}s \, \mathrm{d}v  \right \Vert_{\Ld^2(0,T)} \\
& \leq  \Lambda(T,R) \left\Vert \int_0^t  \left\Vert  \mathcal{R}_0(s)  \right\Vert_{\mathcal{H}^0_r} \, \mathrm{d}s   \right \Vert_{\Ld^2(0,T)} \\
& \leq   \Lambda(T,R) T \left\Vert \mathcal{R}_0 \right\Vert_{\Ld^2(0,T; \mathcal{H}^0_r)} \\
& \leq   \Lambda(T,R),
\end{align*}

thanks to (<ref>) in Lemma <ref>. We obtain the same result for $\mathcal{I}_{\mathcal{R}_0}^{1}$. Using again (<ref>) for the first order derivative, we also have

\begin{align*}
\Vert \mathcal{I}_{\mathcal{R}_0} \Vert_{\Ld^2(0,T;\H^1)}+\Vert \mathcal{I}_{\mathcal{R}_0}^{1} \Vert_{\Ld^2(0,T;\H^1)} \lesssim  \Lambda(T,R).
\end{align*}

For the estimate in $\Ld^2(0,T; \Ld^2)$ of the last term $\mathcal{I}_{\mathcal{R}_1}^0$, we end up with the conclusion thanks to the inequality (<ref>) in Lemma <ref>, combined with the same arguments as before.

Note that the previous lemma does not give any control on $\Vert  \nabla_x\mathcal{I}_{\mathcal{R}_1}^0\Vert_{\Ld^2(0,T;\Ld^2)} + \Vert \nabla_x \mathcal{I}_{\mathcal{R}_1}^{1} \Vert_{\Ld^2(0,T;\Ld^2)} $.
The treatment of these terms requires additional arguments that we will develop in the next subsections. For now, we merely state the result and postpone the proof to the end of Subsection <ref>.

We have

\begin{align}
\label{estimate:I_R1-H1}\Vert \mathcal{I}_{\mathcal{R}_1}^0\Vert_{\Ld^2(0,T;\H^1)} + \Vert \mathcal{I}_{\mathcal{R}_1}^{1} \Vert_{\Ld^2(0,T;\H^1)}  \leq \Lambda(T,R).
\end{align}

§.§ The leading terms and the conclusion

In this section, we focus on the following two terms:

\begin{align*}
\mathfrak{L}^0(t,x):=-\int_0^t e^{d(t-s)} \int_{\R^d}  \mathrm{N}^{t;s}(\mathrm{Z}^{0;t}(x,v))  \mathcal{L}(s, \mathrm{Z}^{s;t}(x,v)) \, \mathrm{d}v \, \mathrm{d}s,
\end{align*}

\begin{align*}
\mathfrak{L}^{1}(t,x):=-\int_0^t e^{d(t-s)} \int_{\R^d}   v \otimes \mathrm{N}^{t;s}(\mathrm{Z}^{0;t}(x,v))  \mathcal{L}(s, \mathrm{Z}^{s;t}(x,v)) \, \mathrm{d}v \, \mathrm{d}s.
\end{align*}

The goal is to prove that $\mathfrak{L}^0$ and $\mathfrak{L}^1$ can be decomposed as a sum of a leading term and a remainder in $\Ld^2_T \H^1$. This will imply the result stated in Proposition <ref>. Since the treament of $\mathfrak{L}^1$ is similar, we focus on $\mathfrak{L}^0$.

First, we have the following decomposition, which introduces several remainder terms that we shall estimate later on. Recall the definition of the straightening diffeomorphism $\psi_{s,t}(x,v)$ from Lemma <ref>.

For $\vert I \vert \leq m$, we have

\begin{multline*}
[\mathfrak{L}^0]_{(I,0)}(t,x)=-\int_0^t  \int_{\R^d}   \partial_x^I E_{\reg,\eps}^{u,\varrho}(s, x-(t-s)v) \cdot \nabla_v f(t,x,v)  \, \mathrm{d}v \, \mathrm{d}s \\
+ \mathscr{R}_{I}^{\mathrm{Diff}}(t,x)+\mathscr{R}_{I,1}^{\mathrm{Duha}}(t,x)+\mathscr{R}_{I,2}^{\mathrm{Duha}}(t,x),
\end{multline*}

\begin{align*}
\mathscr{R}_{I}^{\mathrm{Diff}}(t,x)&:=-\int_0^t  \int_{\R^d}  \left[ \partial_x^I E_{\reg,\eps}^{u,\varrho}(s, x+(1-e^{t-s})v)-\partial_x^I E_{\reg,\eps}^{u,\varrho}(s, x-(t-s)v) \right] \cdot \nabla_v f(t,x,v) \, \mathrm{d}v \, \mathrm{d}s, \\
\mathscr{R}_{I,1}^{\mathrm{Duha}}(t,x)&:=-\sum_{K}\int_0^t  \int_{\R^d}   \partial_x^K E_{\reg,\eps}^{u,\varrho}(s, x+(1-e^{t-s})v) \cdot\mathrm{H}^{K,I}(s,t,x,v) \, \mathrm{d}v \, \mathrm{d}s, \\
\mathscr{R}_{I,2}^{\mathrm{Duha}}(t,x)&:=-\sum_{K}\int_0^t \int_{\R^d} \partial_x^K E_{\reg,\eps}^{u,\varrho}(s, x+(1-e^{t-s}v)) \cdot \mathfrak{H}^{K,I}(t,s,x,v)\, \mathrm{d}v \, \mathrm{d}s,
\end{align*}

where $\mathrm{H}^{K,I}$ and $\mathfrak{H}^{K,I}$ are vector fields defined by

\begin{align}\label{def:KernelH1}
\mathrm{H}^{K,I}(t,s,x,v):=\mathrm{N}^{t;s}(\mathrm{Z}^{0;t}(x,\psi_{s,t}(x,v)))_{(I,0),(K,0)}\mathrm{J}^{s,t}(x,v)   \nabla_v f(t,x,\psi_{s,t}(x,v)) -     \nabla_v f(t,x,v),
\end{align}

\begin{multline}\label{def:KernelH2}
\mathfrak{H}^{K,I}(t,s,x,v) :=\int_s^t e^{d(t-\tau)}   \mathrm{N}^{t;s}(\mathrm{Z}^{0;t}(x,\psi_{s,t}(x,v)))_{(I,0),(K,0)}   \\
\big( \nabla_x f(\tau, \mathrm{Z}^{\tau;t}(x,\psi_{s,t}(x,v))) -\nabla_v f(\tau, \mathrm{Z}^{\tau;t}(x,\psi_{s,t}(x,v))) \big) \, \mathrm{J}^{s,t}(x,v) \, \mathrm{d}\tau,
\end{multline}

\begin{align*}
\mathrm{J}^{s,t}(x,w)=\vert \det \left(\mathrm{D}_w  \psi_{s,t}(x,w) \right) \vert.
\end{align*}

Let $T \in (0, \min \left(T_\eps(R),\overline{T}(R)\right)$.
We have for $\vert I \vert \leq m$

\begin{align}\label{exp:L^0-beforeCHARACchange}
\begin{split}
&\qquad    =-\sum_{(K,L)}\int_0^t e^{d(t-s)} \int_{\R^d}  \mathrm{N}^{t;s}(\mathrm{Z}^{0;t}(x,v))_{(I,0),(K,L)} \partial^{K}_x \partial^{L}_v E_{\reg,\eps}^{u,\varrho}(s, \mathrm{X}^{s;t}(x,v)) \cdot \nabla_v f(s,\mathrm{Z}^{s;t}(x,v)) \, \mathrm{d}v \, \mathrm{d}s \\
&\qquad    =-\sum_{K}\int_0^t e^{d(t-s)} \int_{\R^d}  \mathrm{N}^{t;s}(\mathrm{Z}^{0;t}(x,v))_{(I,0),(K,0)} \partial^{K}_x E_{\reg,\eps}^{u,\varrho}(s, \mathrm{X}^{s;t}(x,v)) \cdot \nabla_v f(s,\mathrm{Z}^{s;t}(x,v)) \, \mathrm{d}v \, \mathrm{d}s,
\end{split}
\end{align}

since $E_{\reg,\eps}^{u,\varrho}$ does not depend on the $v$ variable.

By Lemma <ref>, we know that

\begin{align*}
\mathcal{T}_{\reg,\eps}^{u,\varrho}(\nabla_v f)=\nabla_v f -\nabla_x f.
\end{align*}

Invoking Duhamel formula, we get

\begin{align*}
\nabla_v f(t,x,v)=e^{d(t-s)} \nabla_v f(s, \mathrm{Z}^{s;t}(x,v)) + \int_s^t e^{d(t-\tau)} \left( \nabla_v f(\tau, \mathrm{Z}^{\tau;t}(x,v))-\nabla_x f(\tau, \mathrm{Z}^{\tau;t}(x,v)) \right) \, \mathrm{d}\tau,
\end{align*}

and therefore

\begin{align*}
e^{d(t-s)} \nabla_v f(s, \mathrm{Z}^{s;t}(x,v)) =\nabla_v f(t,x,v)+\int_s^t e^{d(t-\tau)} \left( \nabla_x f(\tau, \mathrm{Z}^{\tau;t}(x,v))-\nabla_v f(\tau, \mathrm{Z}^{\tau;t}(x,v)) \right) \, \mathrm{d}\tau.
\end{align*}

Inserting this expression in (<ref>) yields

\begin{align}\label{decompo:L}
\begin{split}
[\mathfrak{L}^0]_{(I,0)}(t,x)&=\mathscr{L}_I^1(t,x)+\mathscr{L}_I^2(t,x), \\
\end{split}
\end{align}

\begin{align*}
\mathscr{L}_I^1(t,x)&:=-\sum_{K}\int_0^t  \int_{\R^d}  \mathrm{N}^{t;s}(\mathrm{Z}^{0;t}(x,v))_{(I,0),(K,0)} \partial^{K}_x E_{\reg,\eps}^{u,\varrho}(s, \mathrm{X}^{s;t}(x,v)) \cdot\nabla_v f(t,x,v) \, \mathrm{d}v \, \mathrm{d}s, \\
\mathscr{L}_I^2(t,x)&:=-\sum_{K}\int_0^t  \int_{\R^d} \int_s^t e^{d(t-\tau)} \mathrm{N}^{t;s}(\mathrm{Z}^{0;t}(x,v))_{(I,0),(K,0)} \partial^{K}_x E_{\reg,\eps}^{u,\varrho}(s, \mathrm{X}^{s;t}(x,v)) \\
& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad  {\cdot}   \left( \nabla_x f(\tau, \mathrm{Z}^{\tau;t}(x,v))-\nabla_v f(\tau, \mathrm{Z}^{\tau;t}(x,v)) \right) \, \mathrm{d}\tau \, \mathrm{d}v \, \mathrm{d}s.
\end{align*}

Let us transform these two expressions in order to make the terms $\mathscr{R}_I^1(t,x)$, $\mathscr{R}_I^2(t,x)$ and $\mathscr{R}_I^3(t,x)$ appear.

$\bullet$ We first focus on $\mathscr{L}_I^1$, which will produce the leading term in the result. Using the change of variable $v=\psi_{s,t}(x,w)$ coming from Lemma <ref>, we have (since $t \leq \overline{T}(R)$)

\begin{align*}
\mathscr{L}_I^1(t,x)
%&=-\sum_{K}\int_0^t  \int_{\R^d}  \mathrm{N}^{t;s}(\mathrm{Z}^{0;t}(x,v))_{(I,0),(K,0)} \partial_x^K E_{\reg,\eps}^{u,\varrho}(s, \mathrm{X}^{s;t}(x,v)) \cdot \nabla_v f(t,x,v) \, \mathrm{d}v \, \mathrm{d}s \\
%&=-\sum_{K}\int_0^t  \int_{\R^d}  \mathrm{N}^{t;s}(\mathrm{Z}^{0;t}(x,\psi_{s,t}(x,w)))_{(I,0),(K,0)} \\
%& \quad \qquad  \qquad \qquad \qquad \qquad \partial_x^K E_{\reg,\eps}^{u,\varrho}(s, \mathrm{X}^{s;t}(x,\psi_{s,t}(x,w))) \cdot \nabla_v f(t,x,\psi_{s,t}(x,w))  \, \mathrm{J}^{s,t}(x,w) \, \mathrm{d}w \, \mathrm{d}s \\
&=-\sum_{K}\int_0^t  \int_{\R^d}  \mathrm{N}^{t;s}(\mathrm{Z}^{0;t}(x,\psi_{s,t}(x,v)))_{(I,0),(K,0)} \\
& \quad \qquad \qquad \qquad \qquad \qquad \qquad\partial_x^K E_{\reg,\eps}^{u,\varrho}(s, x+(1-e^{t-s})v) \cdot \nabla_v f(t,x,\psi_{s,t}(x,v))  \mathrm{J}^{s,t}(x,v) \, \mathrm{d}v \, \mathrm{d}s,
%&=-\sum_{K}\int_0^t  \int_{\R^d}   \partial_x^K E_{\reg,\eps}^{u,\varrho}(s, x+(1-e^{t-s})v) \cdot \nabla_v f(t,x,v) \, \mathrm{d}v \, \mathrm{d}s  \\
%& \quad \qquad \qquad \qquad \qquad \qquad \qquad  \qquad \qquad \qquad  -\sum_{K} \int_0^t  \int_{\R^d}  \partial_x^K E_{\reg,\eps}^{u,\varrho}(s, x+(1-e^{t-s})v) \cdot \mathrm{H}^{K,I}(s,t,x,v) \, \mathrm{d}v \, \mathrm{d}s,
\end{align*}

where $\mathrm{J}^{s,t}(x,w)=\vert \det \left(\mathrm{D}_w  \psi_{s,t}(x,w) \right) \vert$. We obtain

\begin{align*}
\mathscr{L}_I^1(t,x)&=-\sum_{K}\int_0^t  \int_{\R^d}   \partial_x^K E_{\reg,\eps}^{u,\varrho}(s, x+(1-e^{t-s})v) \cdot \nabla_v f(t,x,\psi_{t,t}(x,v)) \\
& \quad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \mathrm{N}^{t;t}(\mathrm{Z}^{0;t}(x,\psi_{t,t}(x,v)))_{(I,0),(K,0)} \mathrm{J}^{t,t}(x,v) \, \mathrm{d}v \, \mathrm{d}s \\
& \quad -\sum_{K} \int_0^t  \int_{\R^d} \partial_x^K E_{\reg,\eps}^{u,\varrho}(s, x+(1-e^{t-s})v) \cdot \mathrm{H}^{K,I}(t,s,x,v) \, \mathrm{d}v \, \mathrm{d}s,
\end{align*}

\begin{multline*}
\mathrm{H}^{K,I}(t,s,x,v):=\mathrm{N}^{t;s}(\mathrm{Z}^{0;t}(x,\psi_{s,t}(x,v)))_{(I,0),(K,0)}\mathrm{J}^{s,t}(x,v)   \nabla_v f(t,x,\psi_{s,t}(x,v)) \\[2mm]
\quad   - \mathrm{N}^{t;t}(\mathrm{Z}^{0;t}(x,\psi_{t,t}(x,v)))_{(I,0),(K,0)} \mathrm{J}^{t,t}(x,v)       \nabla_v f(t,x,v).
\end{multline*}

Now observe that since $\mathrm{N}^{t;t}= \mathrm{Id}$ and $\psi_{t,t}=\mathrm{Id}$, we have

\begin{align*}
\nabla_v f(t,x,\psi_{t,t}(x,v))\mathrm{N}^{t;t}(\mathrm{Z}^{0;t}(x,\psi_{t,t}(x,v)))_{(I,0),(K,0)} \mathrm{J}^{t,t}(x,v)=\mathbf{1}_{I=K} \nabla_v f(t,x,v),
\end{align*}

\begin{align*}
\mathscr{L}_I^1(t,x)&=-\int_0^t  \int_{\R^d}   \partial_x^I E_{\reg,\eps}^{u,\varrho}(s, x+(1-e^{t-s})v) \cdot \nabla_v f(t,x,v)  \, \mathrm{d}v \, \mathrm{d}s \\
& \quad-\sum_{K}\int_0^t  \int_{\R^d}   \partial_x^K E_{\reg,\eps}^{u,\varrho}(s, x+(1-e^{t-s})v) \cdot\mathrm{H}^{K,I}(s,t,x,v) \, \mathrm{d}v \, \mathrm{d}s \\
&=-\int_0^t  \int_{\R^d}   \partial_x^I E_{\reg,\eps}^{u,\varrho}(s, x-(t-s)v) \cdot \nabla_v f(t,x,v)  \, \mathrm{d}v \, \mathrm{d}s + \mathscr{R}_{I}^{\mathrm{Diff}}(t,x)+\mathscr{R}_{I,1}^{\mathrm{Duha}}(t,x).
\end{align*}

$\bullet$ To deal with the term $\mathscr{L}_I^2$, we apply again the change of variable $v=\psi_{s,t}(x,w)$ from Lemma <ref> and get (since $T \leq \overline{T}(R)$)

\begin{align*}
\mathscr{L}_I^2(t,x)
&=-\sum_{K}\int_0^t \int_{\R^d} \partial_x^K E_{\reg,\eps}^{u,\varrho}(s, x+(1-e^{t-s}v)) \cdot \mathfrak{H}^{K,I}(t,s,x,v)\, \mathrm{d}v \, \mathrm{d}s,
\end{align*}

\begin{multline*}
\mathfrak{H}^{K,I}(t,s,x,v) :=\int_s^t e^{d(t-\tau)}   \mathrm{N}^{t;s}(\mathrm{Z}^{0;t}(x,\psi_{s,t}(x,v)))_{(I,0),(K,0)}   \\
\big( \nabla_x f(\tau, \mathrm{Z}^{\tau;t}(x,\psi_{s,t}(x,v))) -\nabla_v f(\tau, \mathrm{Z}^{\tau;t}(x,\psi_{s,t}(x,v))) \big) \, \mathrm{J}^{s,t}(x,v) \, \mathrm{d}\tau,
\end{multline*}

which means that $\mathscr{L}_I^2(t,x)=\mathscr{R}_{I,2}^{\mathrm{Duha}}(t,x)$. Combining the previous decompositions eventually yields the conclusion.

We now have the following lemma, which is the continuation of Lemma <ref>, in which we express $[\mathfrak{L}^0]_{(I,0)}$ as a sum of a leading term and a remainder which is controlled in $\Ld^2_T \H^1$. The proof, which is rather lengthy and technical, is based on the smoothing estimates of Section <ref>. We refer to Subsection <ref> where the proof is postponed.

We have for all $|I|=m$,

\begin{align*}
[\mathfrak{L}^0]_{(I,0)}(t,x)=p'(\varrho(t,x)) \int_0^t  \int_{\R^d}   \nabla_x [\mathrm{J}_\eps \partial_x^I\varrho](s, x-(t-s)v) \cdot \nabla_v f(t,x,v) \, \mathrm{d}v \, \mathrm{d}s + \mathscr{R}_I(t,x),
\end{align*}

\begin{align*}
\Vert \mathscr{R}_I \Vert_{\Ld^2(0,T, \H^1)} \leq \Lambda(T,R).
\end{align*}

We can finally proceed with the proof of Proposition <ref>.

We only treat the case of $ \partial_x^I \rho_{f}$, the one of $ \partial_x^I j_{f}$ being similar. First invoking Lemma <ref>, we have

\begin{align*}
\partial_x^I \rho_{f}&=[\mathcal{I}_{in}^{0}]_{(I,0)}+[\mathcal{I}_{\mathcal{R}_0}^0]_{(I,0)}+[\mathcal{I}_{\mathcal{R}_1}^0]_{(I,0)}+[\mathfrak{L}^0]_{(I,0)}.
% &=[\mathcal{I}_{in}^{0}]_{(I,0)}+[\mathcal{I}_{\mathcal{R}_0}^0]_{(I,0)}+[\mathcal{I}_{\mathcal{R}_1}^0]_{(I,0)}+\mathrm{I}^I+\mathrm{II}^I
\end{align*}

Thanks to Lemmas <ref>–<ref>–<ref>, we infer

\begin{align*}
\partial_x^I \rho_{f}(t,x)&=p'(\varrho(t,x)) \int_0^t  \int_{\R^d}   \nabla_x [\mathrm{J}_\eps \partial_x^I\varrho](s, x-(t-s)v) \cdot \nabla_v f(t,x,v) \, \mathrm{d}v \, \mathrm{d}s \\
& \quad +[\mathcal{I}_0^{0}]_{(I,0)}(t,x)+[\mathcal{I}_{\mathcal{R}_0}^0]_{(I,0)}(t,x)+[\mathcal{I}_{\mathcal{R}_1}^0]_{(I,0)}(t,x)+ \mathscr{R}_I(t,x),
\end{align*}

where the previous remainders are estimated as

\begin{align*}
\Vert \mathcal{I}_{in}^0 \Vert_{\Ld^2(0,T;\H^1)} \Vert  &\leq \Lambda(T,R) \Vert f^{\mathrm{in}} \Vert_{\mathcal{H}^{m+1}_r}, \\
\Vert \mathcal{I}_{\mathcal{R}_0}^0\Vert_{\Ld^2(0,T;\H^1)}+\Vert \mathcal{I}_{\mathcal{R}_1}^0\Vert_{\Ld^2(0,T;\H^1)}   &\leq  \Lambda(T,R), \\
%\Vert \mathcal{I}_{\mathcal{R}_1}^0\Vert_{\Ld^2(0,T;\H^1)}  &\leq \Lambda(T,R), \\
\Vert \mathscr{R}_I \Vert_{\Ld^2(0,T, \H^1)} &\leq \Lambda(T,R).
\end{align*}

This concludes the proof.

§.§ Estimates of the last remainders

In this subsection, we mainly aim at giving a proof for Lemma <ref> and Lemma <ref>, that we have previously stated. We shall rely on the crucial smoothing estimates derived in Section <ref> to treat the different remainders. Broadly speaking, there are three types of terms requiring a gain of regularity:

$-$ Type I: terms that will be treated thanks to the continuity estimate of Proposition <ref>, as in [93], and of Proposition <ref>. It will be used for the terms $ \nabla_x \mathcal{I}_{\mathcal{R}_1}^0$ and $ \nabla_x \mathcal{I}_{\mathcal{R}_1}^1$ in the proof of Lemma <ref>.

$-$ Type II: terms involving a kernel vanishing on the diagonal in time. They will be treated by the regularization estimate of Proposition <ref>. Such terms will appear in the proof of Lemma <ref>, as well as for the remainders $\mathscr{R}_{I}^{\mathrm{Diff}}, \mathscr{R}_{I,1}^{\mathrm{Duha}}$ and $\mathscr{R}_{I,2}^{\mathrm{Duha}}$.

$-$ Type III: terms involving the difference between the integral operators $\mathrm{K}^{\mathrm{free}}$ and $\mathrm{K}^{\mathrm{fric}}$ (see Section <ref>. They will be handled thanks to the regularization estimate of Proposition <ref>. They will also appear in the treatment of the different remainders.

Let us mention that the previous gains require to control a fixed number of derivatives of the kernels that are involved (see Section <ref>).

Recall also the expression

\begin{align*}
E_{\reg,\eps}^{u,\varrho}(t,x)=u(t,x)-p'(\varrho)\nabla_x \left[\mathrm{J}_{\eps}\varrho \right](t,x),
\end{align*}

as well as Definition <ref> for shifted indices. In order to check that the assumptions of the smoothing estimates of Section <ref> are satisfied, it is convenient to have the following result.

For any $K \in \N^d$ such that $\vert K \vert>0$, we have

\begin{multline*}
\partial_x^K E_{\reg,\eps}^{u,\varrho}=-p'(\varrho) \nabla_x [\partial_x^K\mathrm{J}_{\eps}\varrho]+\partial_x^K u
-\sum_{\ell=0}^{\lfloor \frac{\vert K \vert-1}{2} \rfloor} \sum_{\substack{i=1,{\cdots},d \\ \beta \in \mathbb{B}_K(i ,\ell)}} \binom{K}{\widehat{\beta}^i} \nabla_x(\partial_x^{\overline{K}^i-\beta} \mathrm{J}_{\eps}\varrho)\nabla_x (\partial_x^{\beta}p'(\varrho) ) \cdot e_i  \\
-\sum_{\ell=\lfloor \frac{\vert K \vert-1}{2} \rfloor+1}^{\vert K \vert-1} \sum_{\substack{i=1,{\cdots},d \\ \beta \in \mathbb{B}_K(i ,\ell)}} \binom{K}{\widehat{\beta}^i} \nabla_x (\partial_x^{\beta}p'(\varrho) ) \cdot e_i \nabla_x(\partial_x^{\overline{K}^i-\beta}\mathrm{J}_{\eps} \varrho),
\end{multline*}

\begin{align*}
\mathbb{B}_K(i ,\ell):=\left\lbrace \beta \in \N^d \mid \vert \beta \vert=\ell, \ \ 0<\widehat{\beta}^i \leq K \right\rbrace, \ \ i=1,{\cdots},d, \ \ \ell =0,{\cdots}, \vert K \vert-1 .
\end{align*}

The proof directly follows from Leibniz formula, which provides

\begin{align*}
\partial_x^K E_{\reg,\eps}^{u,\varrho}=-p'(\varrho) \nabla_x \partial_x^K\varrho+\partial_x^K u-\sum_{\ell=0}^{\vert K \vert-1} \sum_{\substack{i=1,{\cdots},d \\ \beta \in \mathbb{B}_K(i ,\ell)}}\binom{K}{\widehat{\beta}^i} \nabla_x (\partial_x^{\beta}p'(\varrho) ) \cdot e_i \nabla_x(\partial_x^{\overline{K}^i-\beta} \mathrm{J}_\eps \varrho),
\end{align*}

and from which we infer the conclusion.

In the next lemma, we show how to obtain the leading term of Lemma <ref> (up to some good remainder) from the integral term of Lemma <ref>.

We have

\begin{multline*}
-\int_0^t  \int_{\R^d}   \partial_x^I E_{\reg,\eps}^{u,\varrho}(s, x-(t-s)v) \cdot \nabla_v f(t,x,v)  \, \mathrm{d}v \, \mathrm{d}s \\=p'(\varrho(t,x)) \int_0^t  \int_{\R^d}   \nabla_x [ \mathrm{J}_\eps\partial_x^I\varrho](s, x-(t-s)v) \cdot \nabla_v f(t,x,v) \, \mathrm{d}v \, \mathrm{d}s +\widetilde{\mathscr{R}}_{I}(t,x),
\end{multline*}

where the remainder $\widetilde{\mathscr{R}}_{I}$ satisfies

\begin{align*}
\LRVert{\widetilde{\mathscr{R}}_{I}}_{\Ld^2(0,T; \H^1)} \leq \Lambda(T,R).
\end{align*}

Let us introduce the following vector fields

\begin{align*}
G_1(s,t,x,v)=[p'(\varrho(s,x-(t-s)v)-p'(\varrho(t,x))] \nabla_v f(t,x,v),
\end{align*}

\begin{align*}
G_{3,i}^{\beta}(s,t,x,v)&:=\Big(\nabla_x (\partial_x^{\beta}p'(\varrho) )(s, x-(t-s)v) \cdot e_i \Big) \nabla_v f(t,x,v), \\[2mm]
G_{4,i }^{K, \beta}(s,t,x,v) &:=\Big( \nabla_x(\partial_x^{\overline{K}^i-\beta} \mathrm{J}_\eps \varrho)(s, x-(t-s)v)\cdot \nabla_v f(t,x,v)  \Big)e_i,
\end{align*}

\begin{align*}
\beta \in \mathbb{B}_K(i ,\ell)=\left\lbrace \beta \in \N^d \mid \vert \beta \vert=\ell, \ \ 0<\widehat{\beta}^i \leq K \right\rbrace, \ \ i=1,{\cdots},d, \ \ \ell =0,{\cdots}, \vert K \vert-1 .
\end{align*}

Thanks to Lemma <ref>, we can write

\begin{multline*}
-\int_0^t  \int_{\R^d}   \partial_x^I E_{\reg,\eps}^{u,\varrho}(s, x-(t-s)v) \cdot \nabla_v f(t,x,v)  \, \mathrm{d}v \, \mathrm{d}s \\
=p'(\varrho(t,x))\int_0^t  \int_{\R^d} \nabla_x [\mathrm{J}_\eps \partial_x^I \varrho](s, x-(t-s)v) \cdot \nabla_v f(t,x,v) \, \mathrm{d}v \, \mathrm{d}s +\mathbf{S_1}(t,x)+\mathbf{S_2}(t,x) + \mathbf{S_3}(t,x)+\mathbf{S_4}(t,x),
\end{multline*}

\begin{align*}
\mathbf{S_1}(t,x)
%&:=\int_0^t  \int_{\R^d}  \nabla_x [\partial_x^I \mathrm{J}_\eps \varrho](s, x-(t-s)v) \cdot \Big([p'(\varrho(s,x-(t-s)v)-p'(\varrho(t,x))] \nabla_v f(t,x,v) \Big) \, \mathrm{d}v \, \mathrm{d}s,
&:=\mathrm{K}_{G_1}^{\mathrm{free}}[\partial_x^I \mathrm{J}_\eps \varrho],\\
\mathbf{S_2}(t,x)&:= -\int_0^t  \int_{\R^d}  \partial_x^I u(s, x-(t-s)v) \cdot \nabla_v f(t,x,v) \, \mathrm{d}v \, \mathrm{d}s , \\
\mathbf{S_3}(t,x)&:=\sum_{\ell=0}^{\lfloor \frac{\vert I \vert-1}{2} \rfloor} \sum_{\substack{i=1,{\cdots},d \\ \beta \in \mathbb{B}_I(i ,\ell)}} \binom{I}{\widehat{\beta}^i} \mathrm{K}_{G_{3,i}^{\beta}}^{\mathrm{free}}\left[\partial_x^{\overline{I}^i-\beta} \mathrm{J}_\eps \varrho \right], \\
\mathbf{S_4}(t,x)&:=\sum_{\ell=\lfloor \frac{\vert I \vert-1}{2} \rfloor+1}^{\vert I \vert-1} \sum_{\substack{i=1,{\cdots},d \\ \beta \in \mathbb{B}_I(i ,\ell)}} \binom{I}{\widehat{\beta}^i} \mathrm{K}_{G_{4,i}^{\overline{I}^i,\beta}}^{\mathrm{free}}\Big[\partial_x^{\beta}p'(\varrho)\Big].
\end{align*}

The treatment of the term $\mathbf{S_2}$ will follow from a straightforward estimate. The terms $\mathbf{S_3}$ and $\mathbf{S_4}$ are terms of Type I and we will use the continuity estimates provided by Proposition <ref>. The term $\mathbf{S_1}$, which already contains $I$ derivatives of $\varrho$, is a term of Type II (since $G_1(t,t,x,v)=0$) and we will rely on the regularization estimate of Proposition <ref>.

$\bullet$ Estimate of $\mathbf{S_1}$: Since $G_1(t,t,x,v)=0$, we use Proposition <ref> to get

\begin{align*}
\LRVert{\mathbf{S_1}}_{\Ld^2(0,T; \H^1)} &\lesssim (1+T)\underset{0\leq s,t \leq T}{\sup} \,  \Vert \partial_s G_1(s,t) \Vert_{\mathcal{H}^{\ell}_{\sigma}}\Vert \partial_x^I\varrho \Vert_{\Ld^2(0,T; \Ld^2)} \\
& \lesssim (1+T)\underset{0\leq s,t \leq T}{\sup} \,  \Vert \partial_s G_1(s,t) \Vert_{\mathcal{H}^{\ell}_{\sigma}}\Vert \varrho \Vert_{\Ld^2(0,T; \H^m)} ,
\end{align*}

for $\ell >7+d$ and $\sigma >d/2$. A direct computation gives

\begin{align*}
\partial_s G_1(s,t,x,v)=p''(\varrho)\left[ \partial_s \varrho + v \cdot\nabla_x \varrho\right](s,x-(t-s)v) \nabla_v f(t,x,v).
\end{align*}

We thus have for all $0 \leq s,t \leq T$

\begin{align*}
\Vert \partial_s G_1(s,t,x,v) \Vert_{\mathcal{H}^{\ell}_{\sigma}}^2 &\lesssim C_T\sum_{\vert \mu \vert + \vert \nu \vert \leq \ell}\sum_{\gamma=0}^{\mu + \nu}\left( \Vert  \partial_x^{\gamma}(p''(\varrho)\partial_s \varrho (s))\Vert_{\Ld^{\infty}}^2+\Vert  \partial_x^{\gamma} (p''(\varrho)\nabla_x \varrho (s))\Vert_{\Ld^{\infty}}^2 \right)\\
& \qquad \qquad \qquad \qquad \qquad \qquad  \times  \int_{\T^d \times \R^d} \langle v \rangle^{2\sigma+2}  \vert\partial^{\mu+ \nu-\gamma}_{x,v} \nabla_v f(t,x,v) \vert^2 \, \mathrm{d}x \, \mathrm{d}v \\
& \lesssim C_T \left( \LRVert{p''(\varrho) \partial_s \varrho(s)}_{\H^k}^2 +\LRVert{p''(\varrho) \nabla_x \varrho(s)}_{\H^k}^2\right)  \Vert f(t) \Vert_{\mathcal{H}^{m-1}_{\sigma+1}}^2 \\
& \lesssim C_T \LRVert{p''(\varrho(s))}_{\H^k}^2\left( \LRVert{\partial_s \varrho(s)}_{\H^k}^2 +\LRVert{ \nabla_x \varrho(s)}_{\H^k}^2\right)  \Vert f(t) \Vert_{\mathcal{H}^{m-1}_{\sigma+1}}^2,
\end{align*}

for $k>\frac{d}{2}+\ell>\frac{d}{2}+1+d$ and $m-1>\ell>1+d$. Using the equation satisfied by $\varrho$, we get

\begin{align*}
\LRVert{p''(\varrho(s))}_{\H^k} &\leq \Lambda \left( \LRVert{\varrho(s)}_{\H^k} \right) \LRVert{\varrho(s)}_{\H^k}, \\
\LRVert{\partial_s \varrho(s)}_{\H^k} +\LRVert{ \nabla_x \varrho(s)}_{\H^k} &\lesssim \LRVert{u}_{\H^k}\LRVert{ \nabla_x \varrho}_{\H^{k}} +\left\Vert \frac{\varrho}{1-\rho_f} \right\Vert_{\H^k}  \LRVert{ \mathrm{div}_x \left(j_f-\rho_fu+u \right)}_{\H^{k}} + \LRVert{\varrho}_{\H^{k+1}},
\end{align*}

thanks to the algebra property of $\H^k$. Taking $k+1<m-3$ and using the bootstrap assumption combined with (the proof) of Lemma <ref>, we obtain

\begin{align*}
%\underset{0\leq s,t \leq T}{\sup} \,  \Vert \partial_s G(s,t) \Vert_{\mathcal{H}^{\ell}_{\sigma}} \lesssim \Lambda(T,R)  \Vert f \Vert_{\Ld^{\infty}(0,T;\mathcal{H}^{m-1}_{\sigma+1})}
\underset{0\leq s,t \leq T}{\sup} \,  \Vert \partial_s G_1(s,t) \Vert_{\mathcal{H}^{\ell}_{\sigma}} \lesssim \Lambda(T,R).
\end{align*}

$\bullet$ Estimate of $\mathbf{S_2}$:
Using the generalized Minkowski inequality followed by the Cauchy-Schwarz inequality, we have for $r>d/2$

\begin{align*}
\LRVert{\mathbf{S_2}}_{\Ld^2(0,T;\Ld^2)} &\leq \left\Vert \int_0^t  \left(\int_{\T^d \times \R^d}  (1+ \vert v \vert^2)^r \vert \partial_x^I u(s, x-(t-s)v) \vert^2 \vert \nabla_v f(t,x,v) \vert^2 \, \mathrm{d}x \, \mathrm{d}v  \right)^{1/2}\, \mathrm{d}s  \right\Vert_{\Ld^2(0,T)} \\
& \leq  \left\Vert \int_0^t  \left(\int_{\T^d \times \R^d}  (1+ \vert v \vert^2)^r \vert \partial_x^I u(s, y) \vert^2 \vert \nabla_v f(t,y+(t-s)v,v) \vert^2 \, \mathrm{d}y \, \mathrm{d}v  \right)^{1/2}\, \mathrm{d}s  \right\Vert_{\Ld^2(0,T)} \\
& \leq \underset{t \in [0,T]}{\sup} \left( \int_{\R^d} (1+ \vert v \vert^2)^r \Vert \nabla_v f(t,\cdot,v) \Vert^2_{\Ld^{\infty}} \, \mathrm{d}v \right)^{1/2} \left\Vert \int_0^t  \Vert \partial_x^I u(s) \Vert_{\Ld^2}  \mathrm{d}s  \right\Vert_{\Ld^2(0,T)} \\
& \lesssim \Vert f \Vert_{\Ld^{\infty}(0,T; \mathcal{H}^{m-1}_{r})} \Vert u \Vert_{\Ld^{2}(0,T;\H^m)},
\end{align*}

by Sobolev embedding in the last line, since $m-1>1+d/2$. This yields

\begin{align*}
\LRVert{\mathbf{S_2}}_{\Ld^2(0,T;\Ld^2)} \lesssim \Lambda(T,R).
\end{align*}

Likewise, we have since $m-1>2+d/2$

\begin{align*}
\LRVert{ \nabla_x \mathbf{S_2}}_{\Ld^2(0,T;\Ld^2)} &\lesssim \Vert f \Vert_{\Ld^{\infty}(0,T; \mathcal{H}^{m-1}_{r})} \Vert u \Vert_{\Ld^{2}(0,T;\H^m)}+ \Vert f \Vert_{\Ld^{\infty}(0,T; \mathcal{H}^{m-1}_{r})} \Vert u \Vert_{\Ld^{2}(0,T;\H^{m+1})} \\
& \leq \Lambda(T,R),
\end{align*}

which gives the conclusion.

$\bullet$ Estimate of $\mathbf{S_3}$ and $\mathbf{S_4}$: for all $0 \leq \vert \beta \vert \leq \lfloor \frac{\vert I \vert-1}{2} \rfloor$ and $i=1, \cdots, d$, we use Proposition <ref> (after taking one derivative in space) to get

\begin{align*}
%\left\Vert \mathrm{K}_{G_{3,i}^{\beta}}^{\mathrm{free}}\left[\partial_x^{\overline{I}^i-\beta} \mathrm{J}_\eps\varrho \right] \right\Vert_{\Ld^2(0,T; \Ld^2)} \lesssim \underset{0 \leq s,t \leq T}{\sup} \Vert \omega_{1}^{\beta}(t,s) \Vert_{\mathcal{H}^p_{\sigma}} \left\Vert  \partial_x^{\overline{I}^i-\beta} \varrho \right\Vert_{\Ld^2(0,T; \Ld^2)}
\left\Vert \mathrm{K}^{\mathrm{free}}_{G_{3,i}^{\beta}}\left[\partial_x^{\overline{I}^i-\beta} \mathrm{J}_\eps \varrho \right] \right\Vert_{\Ld^2(0,T; \H^1)} \lesssim \underset{0 \leq s,t \leq T}{\sup} \Vert G_{3,i}^{\beta}(t,s) \Vert_{\mathcal{H}^{\ell+1}_{\sigma}} \left\Vert  \partial_x^{\overline{I}^i-\beta} \varrho \right\Vert_{\Ld^2(0,T; \H^1)},
\end{align*}

for $\ell>1+d$ and $\sigma>d/2$ therefore

\begin{align*}
\left\Vert \mathrm{K}^{\mathrm{free}}_{G_{3,i}^{\beta}}\left[\partial_x^{\overline{I}^i-\beta}\mathrm{J}_\eps \varrho \right] \right\Vert_{\Ld^2(0,T; \H^1)} \lesssim  \underset{0 \leq s,t \leq T}{\sup} \Vert G_{3,i}^{\beta}(t,s) \Vert_{\mathcal{H}^{\ell+1}_{\sigma}} \left\Vert   \varrho \right\Vert_{\Ld^2(0,T; \H^m)}.
\end{align*}

We have for all $0 \leq s,t \leq T$

\begin{align*}
& \Vert G_{3,i}^{\beta}(t,s) \Vert_{\mathcal{H}^{\ell+1}_{\sigma}}^2 \\
%&\lesssim C_T \sum_{\vert \mu \vert + \vert \nu \vert \leq \ell+1} \sum_{\gamma=0}^{\mu + \nu} \int_{\T^d \times \R^d} \langle v \rangle^{2\sigma} \vert \partial_x^{\gamma}\nabla_x (\partial_x^{\beta}p'(\varrho) )(s, x-(t-s)v)\vert^2 \vert\partial^{\mu+ \nu-\gamma}_{x,v} \nabla_v f(t,x,v) \vert^2 \, \mathrm{d}x \, \mathrm{d}v \\
& \leq C_T\sum_{\vert \mu \vert + \vert \nu \vert \leq \ell+1}\sum_{\gamma=0}^{\mu + \nu} \Vert  \partial_x^{\gamma}\nabla_x (\partial_x^{\beta}p'(\varrho) )(s)\Vert_{\Ld^{\infty}}^2 \int_{\T^d \times \R^d} \langle v \rangle^{2\sigma}  \vert\partial^{\mu+ \nu-\gamma}_{x,v} \nabla_v f(t,x,v) \vert^2 \, \mathrm{d}x \, \mathrm{d}v \\
& \lesssim C_T  \Vert p'(\varrho) \Vert_{\Ld^{\infty}(0,T;\H^k)}^2 \Vert f(t) \Vert_{\mathcal{H}^{m-1}_{\sigma}}^2,
\end{align*}

provided that $m-1 \geq \ell +2$ and $k>\frac{d}{2	}+ \vert \gamma \vert+1 + \vert \beta\vert$. Since $\ell>1+d$ and $\frac{d}{2}+ \vert \gamma \vert+1 + \vert \beta\vert \leq \frac{d}{2}+ \ell +2 + \frac{m-1}{2}$, a condition such as $3d+9<m$ ensures that

\begin{align*}
\Vert G_{3,i}^{\beta}(t,s) \Vert_{\mathcal{H}^{\ell+1}_{\sigma}} &\lesssim C_T \Vert p'(\varrho) \Vert_{\Ld^{\infty}(0,T;\H^{m-2})} \Vert f \Vert_{\Ld^{\infty}(0,T;\mathcal{H}^{m-1}_{\sigma})} \leq \Lambda(T,R).
\end{align*}

thanks to (<ref>) from Lemma <ref>, Sobolev embedding, and  Lemma <ref>. We thus obtain

\begin{align*}
\left\Vert \mathrm{K}^{\mathrm{free}}_{G_{3,i}^{\beta}}\left[\partial_x^{\overline{I}^i-\beta} \mathrm{J}_\eps \varrho \right] \right\Vert_{\Ld^2(0,T; \H^1)} \leq  \Lambda(T,R).
\end{align*}

We proceed in the same way for $\mathbf{S_4}$: for all $\lfloor \frac{\vert I \vert-1}{2} \rfloor+1 \leq \vert \beta \vert \leq \vert I \vert-1 $ and $i=1, \cdots, d$, we take one derivative in space and use Proposition <ref> to write for $\ell>1+d$ and $\sigma>d/2$

\begin{align*}
%\left\Vert \mathrm{K}^{\mathrm{free}}_{\omega_{2}^{\overline{I}^i,\beta}}\left[\partial_x^{\beta}p'(\varrho)\right] \right\Vert_{\Ld^2(0,T; \Ld^2)} \lesssim \underset{0 \leq s,t \leq T}{\sup} \Vert \omega_{2}^{\overline{I}^i,\beta}(t,s) \Vert_{\mathcal{H}^p_{\sigma}} \left\Vert  \partial_x^{\beta}p'(\varrho) \right\Vert_{\Ld^2(0,T; \Ld^2)}
\left\Vert \mathrm{K}^{\mathrm{free}}_{G_{4,i}^{\overline{I}^i,\beta}}\left[\partial_x^{\beta}p'(\varrho)\right] \right\Vert_{\Ld^2(0,T; \H^1)}
%&\lesssim \underset{0 \leq s,t \leq T}{\sup} \Vert G_{4,i}^{\overline{I}^i,\beta}(t,s) \Vert_{\mathcal{H}^{\ell+1}_{\sigma}} \left\Vert  \partial_x^{\beta}p'(\varrho) \right\Vert_{\Ld^2(0,T; \H^1)}\\
&\leq \underset{0 \leq s,t \leq T}{\sup} \Vert G_{4,i}^{\overline{I}^i,\beta}(t,s) \Vert_{\mathcal{H}^{\ell+1}_{\sigma}} \left\Vert  p'(\varrho) \right\Vert_{\Ld^2(0,T; \H^m)}.
\end{align*}

The kernel $G_{4,i}^{\overline{I}^i,\beta}$ is estimated as before: using Leibniz rule, we have for all $0 \leq s,t \leq T$

\begin{align*}
\Vert G_{4,i}^{\overline{I}^i,\beta}(t,s) \Vert_{\mathcal{H}^{\ell+1}_{\sigma}}^2
%&\lesssim C_T \sum_{\vert \mu \vert + \vert \nu \vert \leq \ell+1} \sum_{\gamma=0}^{\mu + \nu} \int_{\T^d \times \R^d} \langle v \rangle^{2\sigma} \vert \partial_x^{\gamma}\partial_x^{\overline{I}^i-\beta} \mathrm{J}_\eps\varrho(s, x-(t-s)v)\vert^2 \vert\partial^{\mu+ \nu-\gamma}_{x,v} \nabla_v f(t,x,v) \vert^2 \, \mathrm{d}x \, \mathrm{d}v \\
%& \leq C_T \sum_{\vert \mu \vert + \vert \nu \vert \leq \ell+1}\sum_{\gamma=0}^{\mu + \nu} \Vert \vert \partial_x^{\gamma}\nabla_x (\partial_x^{\overline{I}^i-\beta} \mathrm{J}_\eps\varrho(s)\Vert_{\Ld^{\infty}}^2 \int_{\T^d \times \R^d} \langle v \rangle^{2\sigma}  \vert\partial^{\mu+ \nu-\gamma}_{x,v} \nabla_v f(t,x,v) \vert^2 \, \mathrm{d}x \, \mathrm{d}v \\
& \lesssim C_T \Vert \varrho\Vert_{\Ld^{\infty}(0,T;\H^k)}^2 \Vert f(t) \Vert_{\mathcal{H}^{m-1}_{\sigma}}^2,
\end{align*}

provided that $m-1>2+\ell$ and $k>\frac{d}{2}+\vert \gamma \vert +1 +\vert \overline{I}^i \vert- \vert \beta \vert$. Since $\ell>1+d$ and

$$\frac{d}{2}+\vert \gamma \vert +1 +\vert \overline{I}^i \vert- \vert \beta \vert \leq \frac{d+2 \ell+m+1}{2},$$

a condition such as $3d+9<m$ ensures that

\begin{align*}
\Vert G_{4,i}^{\overline{I}^i,\beta}(t,s) \Vert_{\mathcal{H}^{\ell+1}_{\sigma}} &\lesssim C_T \Vert \varrho \Vert_{\Ld^{\infty}(0,T;\H^{m-2})} \Vert f \Vert_{\Ld^{\infty}(0,T;\mathcal{H}^{m-1}_{\sigma})}.
\end{align*}

We thus obtain by Sobolev embedding, the bound (<ref>) in Lemma <ref> and Lemma <ref>.

\begin{align*}
\left\Vert \mathrm{K}^{\mathrm{free}}_{G_{4,i}^{\overline{I}^i,\beta}}\left[\partial_x^{\beta}p'(\varrho)\right] \right\Vert_{\Ld^2(0,T; \H^1)} \leq \Lambda(T,R).
\end{align*}

The remaining task is to show that the remainder terms $\mathscr{R}_{I}^{\mathrm{Diff}}, \mathscr{R}_{I,1}^{\mathrm{Duha}}$ and $\mathscr{R}_{I,2}^{\mathrm{Duha}}$ introduced in Lemma <ref> are well-controlled in $\Ld^2(0,T; \H^1)$. For the first one, we have the following lemma.

We have

\begin{align*}
\LRVert{\mathscr{R}_{I}^{\mathrm{Diff}}}_{\Ld^2(0,T; \H^1)} \leq \Lambda(T,R).
\end{align*}

In view of Lemma <ref>, let us write for $0\leq \vert \mu \vert \leq \lfloor \frac{\vert K \vert-1}{2} \rfloor $ and $ \lfloor \frac{\vert K \vert-1}{2} \rfloor+1 \leq \vert \nu \vert \leq \vert K \vert-1 $

\begin{align*}
& \big(\nabla_x [\partial_x^{\overline{K}^i-\mu} \mathrm{J}_\eps \varrho]\nabla_x[ \partial_x^{\mu}p'(\varrho)] {\cdot} e_i+ \nabla_x [\partial_x^{\mu}p'(\varrho)]{\cdot} e_i \nabla_x [\partial_x^{\overline{K}^i-\nu} \mathrm{J}_\eps \varrho] \big)(s,x-(t-s)v) \cdot \nabla_v f(t,x,v)\\[2mm]
& \qquad  -\big(\nabla_x [\partial_x^{\overline{K}^i-\mu} \mathrm{J}_\eps \varrho]\nabla_x[ \partial_x^{\mu}p'(\varrho)] {\cdot} e_i+ \nabla_x [\partial_x^{\mu}p'(\varrho)]{\cdot} e_i \nabla_x [\partial_x^{\overline{K}^i-\nu} \mathrm{J}_\eps \varrho]  \big)(s,x+(1-e^{t-s})v) \cdot \nabla_v f(t,x,v) \\[2mm]
& =\Big(\nabla_x [\partial_x^{\overline{K}^i-\mu} \mathrm{J}_\eps \varrho] (s,x-(t-s)v)  -\nabla_x [\partial_x^{\overline{K}^i-\mu} \mathrm{J}_\eps \varrho](s,x+(1-e^{t-s})v) \Big) \cdot G_{8,\mu,i}(s,t,x,v) \\
& \quad + \nabla_x [\partial_x^{\overline{K}^i-\mu} \mathrm{J}_\eps \varrho](s,x+(1-e^{t-s})v) \cdot G_{9,\mu,i}(s,t,x,v)\\
& \quad +\nabla_x [\partial_x^{\nu} p'(\varrho)](s,x-(t-s)v) \cdot G_{10,\nu,i}^K(s,t,x,v) \\
& \quad +  \Big( \nabla_x [\partial_x^{\nu}p'(\varrho)] (s,x-(t-s)v)-\nabla_x[\partial_x^{\nu}p'(\varrho)] (s,x+(1-e^{t-s})v) \Big) \cdot G_{11,\nu,i}^K(s,t,x,v).
\end{align*}

\begin{align*}
G_{8,\mu,i}(s,t,x,v)&:=\Big(\nabla_x [\partial_x^{\mu}p'(\varrho)] (s,x-(t-s)v) {\cdot} e_i \nabla_v f(t,x,v) \Big), \\
G_{9,\mu,i}(s,t,x,v)&:=\Big((\nabla_x [\partial_x^{\mu}p'(\varrho)] {\cdot} e_i-\nabla_x[\partial_x^{\mu}p'(\varrho)](s,x+(1-e^{t-s})v) {\cdot} e_i ) \nabla_v f(t,x,v) \Big), \\
G_{10,\nu,i}^K(s,t,x,v)&:=\Big((\nabla_x [\partial_x^{\overline{K}^i-\nu} \mathrm{J}_\eps \varrho](s,x-(t-s)v)-\nabla_x [\partial_x^{\overline{K}^i-\nu} \mathrm{J}_\eps \varrho](s,x+(1-e^{t-s})v))  {\cdot} \nabla_v f(t,x,v) e_i\Big), \\
G_{11,\nu,i}^K(s,t,x,v)&:=\Big(\nabla_x [\partial_x^{\overline{K}^i-\nu} \mathrm{J}_\eps \varrho](s,x+(1-e^{t-s})v) {\cdot} \nabla_x f(t,x,v) e_i \Big).
\end{align*}

By Lemma <ref>, we can thus rewrite

\begin{align*}
\mathscr{R}_{I}^{\mathrm{Diff}}=\mathbf{S_5}+\mathbf{S_6}+\mathbf{S_7}+\mathbf{S_8}+\mathbf{S_9}+\mathbf{S_{10}}+\mathbf{S_{11}},
\end{align*}

\begin{align*}
\mathbf{S_5}(t,x)&:=-\int_0^t  \int_{\R^d}  \left[ \partial_x^I u(s, x+(1-e^{t-s})v)-\partial_x^I u(s, x-(t-s)v) \right] \cdot \nabla_v f(t,x,v) \, \mathrm{d}v \, \mathrm{d}s,
\end{align*}

\begin{align*}
\mathbf{S_6}&
%:= \int_0^t  \int_{\R^d}  \left[ \nabla_x(\partial_x^I \mathrm{J}_\eps \varrho)(s, x-(t-s)v)- \nabla_x(\partial_x^I \mathrm{J}_\eps \varrho)(s, x+(1-e^{t-s})v) \right] \cdot \left( p'(\varrho)(s, x-(t-s)v)\nabla_v f(t,x,v) \right) \, \mathrm{d}v \, \mathrm{d}s, \\
:= \mathrm{K}_{G_6}^{\mathrm{free}}[\partial_x^I \mathrm{J}_\eps \varrho]-\mathrm{K}_{G_6}^{\mathrm{fric}}[\partial_x^I \mathrm{J}_\eps \varrho], \ \ \ ÷  G_6(t,s,x,v):=p'(\varrho(s,x-(t-s)v))\nabla_v f(t,x,v), \\[2mm]
\mathbf{S_7}
%&:=  \int_0^t  \int_{\R^d}  \nabla_x(\partial_x^I \mathrm{J}_\eps \varrho)(s, x+(1-e^{t-s})v)  \cdot \left[ (p'(\varrho)(s, x-(t-s)v)- p'(\varrho)(s, x+(1-e^{t-s})v))\nabla_v f(t,x,v) \right] \, \mathrm{d}v \, \mathrm{d}s \\
&:= \mathrm{K}_{G_7}^{\mathrm{fric}}[\partial_x^I \mathrm{J}_\eps \varrho],  \ \ G_7(t,s,x,v):=\left[ (p'(\varrho)(s, x-(t-s)v)- p'(\varrho)(s, x+(1-e^{t-s})v))\nabla_v f(t,x,v) \right], \\
\mathbf{S_8}&:=  \sum_{\ell=0}^{\lfloor \frac{\vert I \vert-1}{2} \rfloor} \sum_{\substack{i=1,{\cdots},d \\ \beta \in \mathbb{B}_I(i ,\ell)}} \binom{I}{\widehat{\beta}^i} \left( \mathrm{K}_{G_{8,\beta,i}}^{\mathrm{free}}[\partial_x^{\overline{I}^i-\beta} \mathrm{J}_\eps \varrho] -\mathrm{K}_{G_{8,\beta,i}}^{\mathrm{fric}}[\partial_x^{\overline{I}^i-\beta} \mathrm{J}_\eps \varrho] \right), \\
\mathbf{S_9}&:= \sum_{\ell=0}^{\lfloor \frac{\vert I \vert-1}{2} \rfloor} \sum_{\substack{i=1,{\cdots},d \\ \beta \in \mathbb{B}_I(i ,\ell)}} \binom{I}{\widehat{\beta}^i} \mathrm{K}_{G_{9,\beta,i}}^{\mathrm{fric}}\left[\partial_x^{\overline{I}^i-\beta} \mathrm{J}_\eps \varrho \right], \\
\mathbf{S_{10}}&:=\sum_{\ell=\lfloor \frac{\vert I \vert-1}{2} \rfloor+1}^{\vert I \vert-1} \sum_{\substack{i=1,{\cdots},d \\ \beta \in \mathbb{B}_I(i ,\ell)}} \binom{I}{\widehat{\beta}^i} \mathrm{K}_{G_{10,\beta,i}^I }^{\mathrm{free}}\left[\partial_x^{\beta}p'(\varrho)\right], \\
\mathbf{S_{11}}&:=\sum_{\ell=\lfloor \frac{\vert I \vert-1}{2} \rfloor+1}^{\vert I \vert-1} \sum_{\substack{i=1,{\cdots},d \\ \beta \in \mathbb{B}_I(i ,\ell)}} \binom{I}{\widehat{\beta}^i} \left( \mathrm{K}_{G_{11,\beta,i}^I}^{\mathrm{free}}[\partial_x^{\beta}p'(\varrho)] -\mathrm{K}^{\mathrm{fric}}_{G_{11,\beta,i}^I}[\partial_x^{\beta}p'(\varrho)] \right).
\end{align*}

Let us explain how to estimate each of these terms. The term $\mathbf{S_5}$ will be estimated by a direct proof. All the other terms actually require the use of the smoothing estimates coming either from Proposition <ref> or Proposition <ref>:

$-$ for $\mathbf{S_6}, \mathbf{S_8}$ and $\mathbf{S_{11}}$, we have the difference of the operators $\mathrm{K}^{\mathrm{free}}$ and $\mathrm{K}^{\mathrm{fric}}$ which appears: these terms are therefore of Type III and we will use Proposition <ref>;

$-$ since the kernels $G_7$, $G_9$ and $G_{10}$ vanish on the diagonal $\lbrace s=t \rbrace$, the terms $\mathbf{S_7}, \mathbf{S_9}$ and $\mathbf{S_{10}}$ are of Type II and we will appeal to Proposition <ref>.

Let us now turn to the estimates.

$\bullet$ Estimate of $\mathbf{S_5}$: the argument is the same as for $\mathbf{S}_2 $ above. We obtain

$$\LRVert{\mathbf{S_5}}_{\Ld^2(0,T;\Ld^2)} \lesssim \Lambda(T,R).$$

$\bullet$ Estimate of $\mathbf{S_6}, \mathbf{S_8}$ and $\mathbf{S_{11}}$: by Proposition <ref>, we have for all $\ell>8+d$ and $\sigma>1+d/2$

\begin{align*}
\LRVert{\mathbf{S_6}}_{\Ld^2(0,T; \H^1(\T^d))} &\lesssim \underset{0 \leq s,t \leq T}{\sup} \Vert G_6(t,s) \Vert_{\mathcal{H}^{\ell}_{\sigma}} \Vert \partial_x^I \varrho \Vert_{\Ld^2(0,T; \Ld^2(\T^d))} \\
& \leq \Lambda(T,R),
\end{align*}

thanks to the bound (<ref>) in Lemma <ref>, Lemma <ref> and provided that we can take $m -2\geq \frac{d}{2}+8+d$.

Likewise, for $\mathbf{S_8}$, we use Proposition <ref> to get

\begin{align*}
\LRVert{\mathbf{S_8}}_{\Ld^2(0,T;\H^1)}  \lesssim \underset{\substack{0\leq s,t \leq T\\ 0 \leq \vert \beta \vert \leq \lfloor \frac{\vert I \vert-1}{2} \rfloor \\i=1, \cdots, d}}{\sup} \,  \Vert  G_{8, \beta, i} (s,t) \Vert_{\mathcal{H}^{\ell}_{\sigma}} \Vert  \varrho \Vert_{\Ld^2(0,T; \H^{m-1})},
\end{align*}

for all $\ell >8+d$ and $\sigma >1+d/2$. As in the treatment of $\mathbf{S_3}$ above, we deduce

\begin{align*}
\LRVert{\mathbf{S_8}}_{\Ld^2(0,T;\H^1)}  \lesssim C_T \Vert p'(\varrho) \Vert_{\Ld^{\infty}(0,T;\H^{m-2})} \Vert f \Vert_{\Ld^{\infty}(0,T;\mathcal{H}^{m-1}_{\sigma})} \Vert \varrho \Vert_{\Ld^2(0,T; \H^{m-1})} \leq \Lambda(T,R),
\end{align*}

since $m>3d+21$. We argue in the same way for $\mathbf{S_{11}}$ (see the treatment of $\mathbf{S_4}$ above) and obtain

\begin{align*}
\LRVert{\mathbf{S_{11}}}_{\Ld^2(0,T;\H^1)}  \leq  \Lambda(T,R).
\end{align*}

$\bullet$ Estimate of $\mathbf{S_7},\mathbf{S_9}$ and $\mathbf{S_{10}}$: we proceed exactly as in the estimate of $\textbf{S}_1$ above, since $G_7(t,t,x,v)=0$. Here, we have

\begin{multline*}
\partial_s G_7(s,t,x,v)=\Big[ p''(\varrho)\left[ \partial_s \varrho + v \cdot\nabla_x \varrho\right](s,x-(t-s)v)\\
-p''(\varrho)\left[ \partial_s \varrho + e^{t-s}v \cdot\nabla_x \varrho\right](s,x+(1-e^{t-s})v) \Big]\nabla_v f(t,x,v).
\end{multline*}

Using triangle inequality with Proposition <ref>, we end up with

\begin{align*}
\LRVert{\mathbf{S_7}}_{\Ld^2(0,T; \H^1)} \leq \Lambda(T,R).
\end{align*}

For $\mathbf{S_9}$, using $G_{9, \beta, i}(t,t,x,v)=0$ for $0 \leq \vert \beta \vert \leq \lfloor \frac{\vert I \vert-1}{2} \rfloor$, we also have by  Proposition <ref>

\begin{align*}
\LRVert{\mathrm{K}^{\mathrm{fric}}_{G_{9,\beta,i}}\left[\partial_x^{\overline{I}^i-\beta} \mathrm{J}_\eps \varrho \right]}_{\Ld^2(0,T;\H^1)}
&\lesssim (1+T)\underset{0\leq s,t \leq T}{\sup} \,  \Vert \partial_s G_{9,\beta,i}(s,t) \Vert_{\mathcal{H}^{\ell}_{\sigma}}\Vert  \varrho \Vert_{\Ld^2(0,T; \H^{m-1})},
\end{align*}

for all $\ell >7+d$ and $\sigma >d/2$. We then proceed as in the estimate of $\mathbf{S}_1$ to deal with the time derivative, combined  with what we have done for the estimate of $\mathbf{S}_3$ (since $m>3d+19$ for instance) and get

\begin{align*}
\LRVert{\mathbf{S_9}}_{\Ld^2(0,T;\H^1)} \leq \Lambda(T,R) \Vert f \Vert_{\Ld^{\infty}(0,T;\mathcal{H}^{m-1}_{\sigma})}\Vert  \varrho \Vert_{\Ld^2(0,T; \H^{m-1})} \leq \Lambda(T,R).
\end{align*}

Finally, we use the same exact arguments as before for $\mathbf{S_{10}}$ (see the estimate of $\mathbf{S}_4$ above for instance) to get

\begin{align*}
\LRVert{\mathbf{S_{10}} }_{\Ld^2(0,T;\H^1)} \leq \Lambda(T,R) \Vert \varrho \Vert_{\Ld^2(0,T; \H^{m-1})} \leq \Lambda(T,R).
\end{align*}

The second term $\mathscr{R}_{I,1}^{\mathrm{Duha}}$ from Lemma <ref> is estimated thanks to the following lemma.

We have

\begin{align*}
\LRVert{\mathscr{R}_{I,1}^{\mathrm{Duha}}}_{\Ld^2(0,T; \H^1)} \leq \Lambda(T,R).
\end{align*}

Let us introduce the following vector fields

\begin{align*}
G_{14, i, \beta}^{K,I}(s,t,x,v)&:= \Big(\nabla_x (\partial_x^{\beta}p'(\varrho) )(s, x+(1-e^{t-s}v)) \cdot e_i \Big)\mathrm{H}^{K,I}(s,t,x,v), \\[2mm]
G_{15,i, \beta}^{K,I}(s,t,x,v)&:= \Big(\nabla_x(\partial_x^{\overline{K}^i-\beta} \mathrm{J}_\eps \varrho)(s,x+(1-e^{t-s}v)) \cdot \mathrm{H}^{K,I}(s,t,x,v) \Big) e_i,
\end{align*}

\begin{align*}
\beta \in \mathbb{B}_K(i ,\ell)=\left\lbrace \beta \in \N^d \mid \vert \beta \vert=\ell, \ \ 0<\widehat{\beta}^i \leq K \right\rbrace, \ \ i=1,{\cdots},d, \ \ \ell =0,{\cdots}, \vert K \vert-1,
\end{align*}

and where we recall the expression of the kernel $H$ defined in (<ref>) by

\begin{align*}
\mathrm{H}^{K,I}(t,s,x,v):=\mathrm{N}^{t;s}(\mathrm{Z}^{0;t}(x,\psi_{s,t}(x,v)))_{(I,0),(K,0)}\mathrm{J}^{s,t}(x,v)   \nabla_v f(t,x,\psi_{s,t}(x,v)) -    \nabla_v f(t,x,v).
\end{align*}

By Lemma <ref>, we now decompose $\mathscr{R}_{I,1}^{\mathrm{Duha}}$ as

\begin{align*}
\mathscr{R}_{I,1}^{\mathrm{Duha}}=\mathbf{S_{12}}+\mathbf{S_{13}}+\mathbf{S_{14}}+\mathbf{S_{15}},
\end{align*}

\begin{align*}
\mathbf{S_{12}}(t,x)&:=- \sum_{K} \int_0^t  \int_{\R^d}  \partial_x^K u(s, x+(1-e^{t-s})v) \cdot \mathrm{H}^{K,I}(t,s,x,v) \, \mathrm{d}v \, \mathrm{d}s,
\end{align*}

\begin{align*}
\mathbf{S_{13}}&:= \sum_{K} \mathrm{K}^{\mathrm{fric}}_{G_{13}} \left[\partial_x^K  \mathrm{J}_\eps\varrho \right], \ \ \ \  G_{13}(t,s,x,v):=p'(\varrho)(s,x+(1-e^{t-s})v)\mathrm{H}^{K,I}(t,s,x,v),\\
\mathbf{S_{14}}&:=\sum_{K}  \sum_{\ell=0}^{\lfloor \frac{\vert K \vert-1}{2} \rfloor} \sum_{\substack{i=1,{\cdots},d \\ \beta \in \mathbb{B}_K(i ,\ell)}} \binom{I}{\widehat{\beta}^i}\mathrm{K}^{\mathrm{free}}_{G_{14, i, \beta}^{K,I}}\left[\partial_x^{\overline{K}^i-\beta}\mathrm{J}_\eps \varrho \right], \\
\mathbf{S_{15}}&:= \sum_{K} \sum_{\ell=\lfloor \frac{\vert K \vert-1}{2} \rfloor+1}^{\vert K \vert-1} \sum_{\substack{i=1,{\cdots},d \\ \beta \in \mathbb{B}_K(i ,\ell)}} \binom{K}{\widehat{\beta}^i} \mathrm{K}^{\mathrm{free}}_{G_{15,i, \beta}^{K,I}} \left[\partial_x^{\beta} p'(\varrho) \right].
\end{align*}

Let us estimate each of theses terms. Note that $\mathrm{H}^{K,I}(t,t,x,v)=0$ so that the kernels appearing in $\mathbf{S_{13}}, \mathbf{S_{14}}$ and $\mathbf{S_{15}}$ vanish in the diagonal in time: these terms are therefore of Type II and we will use the regularization property from Proposition <ref> to handle them.

$\bullet$ Estimate of $\mathbf{S_{12}}$: We proceed as in the estimate for $\mathbf{S}_2$ above and first get for $k>1+\frac{d}{2}$

\begin{align*}
\LRVert{J_8}_{\Ld^2(0,T;\H^1)} \lesssim  \sum_{K} \left\Vert \mathrm{H}^{K,I} \right\Vert_{\Ld^{\infty}(0,T; \mathcal{H}^{k}_{r})} \Vert u \Vert_{\Ld^{2}(0,T;\H^{m+1})}.
\end{align*}

Now observe that for $s,t$, we have

\begin{align*}
&\left\Vert \mathrm{H}^{K,I}(s,t) \right\Vert_{\mathcal{H}^{k}_{r}} \\
&\lesssim  \LRVert{J^{s,t}}_{\mathrm{W}^{k, \infty}_{x,v}} \LRVert{\mathrm{N}^{s;t}(\mathrm{Z}^{0;t}(\cdot, \psi_{s,t})) \nabla_vf(t,\cdot, \psi_{s,t})}_{\mathcal{H}^{k}_{r}} + \left\Vert f(t)\right\Vert_{\mathcal{H}^{k+1}_{r}} \\
& \lesssim \LRVert{J^{s,t}}_{\mathrm{W}^{k, \infty}_{x,v}}\left(1+\LRVert{\mathrm{Z}^{0;t}}_{\mathrm{W}^{k, \infty}_{x,v}} \right)\left( 1+\LRVert{\psi_{s,t}}_{ \dot{\mathrm{W}}^{k, \infty}_{x,v}} \right) \LRVert{\mathrm{N}^{t;s}}_{\mathrm{W}^{k, \infty}_{x,v}} \sum_{\vert \gamma \vert \leq k} \LRVert{ \partial_{x,v}^{\gamma}(\nabla_v f)(t,\cdot, \psi_{s,t})}_{\mathcal{H}^{0}_{r}} \\
& \quad + \left\Vert f(t)\right\Vert_{\mathcal{H}^{k+1}_{r}} \\
&\lesssim \Lambda(T,R) \sum_{\vert \gamma \vert \leq k} \LRVert{ \partial_{x,v}^{\gamma}(\nabla_v f)(t,\cdot, \psi_{s,t})}_{\mathcal{H}^{0}_{r}} +\left\Vert f(t)\right\Vert_{\mathcal{H}^{k+1}_{r}},
\end{align*}

thanks to the estimate (<ref>) of Lemma <ref>, Remark <ref> and Lemma <ref>, and since $m>4+d$. To handle the last sum, we write

\begin{multline*}
\int_{\T^d \times \R^d} \vert \partial_{x,v}^{\gamma} \nabla_v f(t,x, \psi_{s,t}(x,v)) \vert^2 (1+\vert v \vert^2)^r \, \mathrm{d}x \, \mathrm{d}v \\
\leq  \int_{\T^d \times \R^d} \vert \partial_{x,v}^{\gamma} \nabla_v f(t,x, \psi_{s,t}(x,v)) \vert^2 (1+\vert v-\psi_{s,t}(x,v) \vert^2+\vert \psi_{s,t}(x,v) \vert^2)^r \, \mathrm{d}x \, \mathrm{d}v,
\end{multline*}

and use the change of variable $v\mapsto w= \psi_{s,t}(x,v)$ from Lemma <ref>, combined with the bounds (<ref>) and (<ref>) to get (choosing $k$ such that $k+1 \leq m-1$)

\begin{align*}
\left\Vert \mathrm{H}^{K,I}(s,t) \right\Vert_{\mathcal{H}^{k}_{r}} \leq \Lambda(T,R),
%&\lesssim \Lambda(T,R) \left\Vert f(t)\right\Vert_{\mathcal{H}^{k+1}_{r}} +\left\Vert f(t)\right\Vert_{\mathcal{H}^{k+1}_{r}}
\end{align*}

Taking a supremum in time we obtain

\begin{align*}
\LRVert{\mathbf{S_{12}}}_{\Ld^2(0,T;\H^1)} \leq \Lambda(T,R).
%\lesssim \Lambda(T,R) \left\Vert f(t)\right\Vert_{\Ld^{\infty}(0,T;\mathcal{H}^{m-1}_{r})}\Vert u \Vert_{\Ld^{2}(0,T;\H^{m+1})},
\end{align*}

which yields the result.

$\bullet$ Estimate of $\mathbf{S_{13}}, \mathbf{S_{14}}$ and $\mathbf{S_{15}}$: Since $\mathrm{H}^{K,I}(t,t,x,v)=0$, we observe that $G_{13}(t,t,x,v)=0$. By Proposition <ref>, we therefore have for $\ell>7+d$ and $\sigma >d/2$

\begin{align*}
\LRVert{\mathbf{S_{13}}}_{\Ld^2(0,T;\H^1)} &\lesssim (1+T) \sum_{K} \underset{0\leq s,t \leq T}{\sup} \,  \Vert \partial_s\left[ p'(\varrho(s,x+(1-e^{t-s})v))\mathrm{H}^{K,I}(s,t) \right] \Vert_{\mathcal{H}^{\ell}_{\sigma}} \Vert \partial_x^{K} \varrho \Vert_{\Ld^2(0,T; \Ld^2)}\\
& \lesssim C_T \sum_{K} \underset{0\leq s,t \leq T}{\sup} \,  \Vert  p''(\varrho)\partial_s \varrho (s,x+(1-e^{t-s})v))\mathrm{H}^{K,I}(s,t)\Vert_{\mathcal{H}^{\ell}_{\sigma}} \Vert \partial_x^{K} \varrho \Vert_{\Ld^2(0,T; \Ld^2)} \\
& \quad + C_T \sum_{K} \underset{0\leq s,t \leq T}{\sup} \, \Vert p''(\varrho)\nabla_x \varrho (s,x+(1-e^{t-s})v))\mathrm{H}^{K,I}(s,t) \Vert_{\mathcal{H}^{\ell}_{\sigma+1}} \Vert \partial_x^{K} \varrho \Vert_{\Ld^2(0,T; \Ld^2)} \\
& \quad +C_T \sum_{K} \underset{0\leq s,t \leq T}{\sup} \,  \Vert  p'(\varrho(s,x+(1-e^{t-s})v))  \partial_s\mathrm{H}^{K,I}(s,t)  \Vert_{\mathcal{H}^{\ell}_{\sigma}} \Vert \partial_x^{K} \varrho \Vert_{\Ld^2(0,T; \Ld^2)}.
\end{align*}

The two first terms can be handled by similar arguments to the ones used for $\mathbf{S}_1$ and $\mathbf{S_{12}}$, a fixed number of derivatives being involved. For the last one, we proceed as for the other terms, combined with the arguments used for $\mathbf{S}_{12}$, and write for all $t,s$

\begin{align*}
&\Vert  p'(\varrho(s,x+(1-e^{t-s})v))  \partial_s\mathrm{H}^{K,I}(s,t)  \Vert_{\mathcal{H}^{\ell}_{\sigma}} \\
&\leq C_T \Vert p'(\varrho) \Vert_{\Ld^{\infty}(0,T;\H^{m-2})} \Vert  \partial_s\mathrm{H}^{K,I}(s,t)  \Vert_{\mathcal{H}^{\ell}_{\sigma}}\\
&\leq \Lambda(T,R)  \Big( \Vert J^{s,t} \Vert_{\W^{\ell,\infty}_{x,v}} \Vert \nabla_v f(t) \Vert_{\mathcal{H}^{\ell}_{\sigma}} +\Vert \partial_s J^{s,t} \Vert_{\W^{\ell, \infty}_{x,v}} \Vert \nabla_v f(t,\cdot, \psi_{s,t}) \Vert_{\mathcal{H}^{\ell}_{\sigma}} +\Vert \nabla_v f(t) \Vert_{\mathcal{H}^{\ell+1}_{\sigma}}\Big)\\
& \leq \Lambda(T,R),
\end{align*}

since $m>3d/2+11$. This yields

\begin{align*}
\LRVert{\mathbf{S_{13}}}_{\Ld^2(0,T;\H^1)}  \leq \Lambda(T,R).
\end{align*}

Next, for $\mathbf{S_{14}}$, we use the fact that $G_{14, i, \beta}^{K,I}(t,t,x,v)=0$ for $0 \leq \vert \beta \vert \leq \lfloor \frac{\vert K \vert-1}{2} \rfloor$ and $i=1, \cdots, d$ so that by Proposition <ref>, we have

\begin{align*}
\left\Vert \mathrm{K}_{G_{14, i, \beta}^{K,I}}\left[\partial_x^{\overline{K}^i-\beta} \mathrm{J}_\eps \varrho \right] \right\Vert_{\Ld^2(0,T;\H^1)}
%&\lesssim (1+T) \underset{0 \leq t,s \leq T}{\sup} \, \left\Vert  \partial_s G_{14, i, \beta}^{K,I}(s,t) \right\Vert_{\mathcal{H}^{\ell}_{\sigma}} \left\Vert \partial_x^{\overline{K}^i-\beta} \varrho \right\Vert_{\Ld^2(0,T;\Ld^2)} \\
& \lesssim (1+T) \underset{0 \leq t,s \leq T}{\sup} \, \left\Vert  \partial_s G_{14, i, \beta}^{K,I}(s,t) \right\Vert_{\mathcal{H}^{\ell}_{\sigma}} \left\Vert  \varrho \right\Vert_{\Ld^2(0,T;\H^m)} ,
\end{align*}

for $\ell >7+d$ and $\sigma>d/2$. Next, we have for all $t,s$

\begin{multline*}
\left\Vert  \partial_s G_{14, i, \beta}^{K,I}(s,t) \right\Vert_{\mathcal{H}^{\ell}_{\sigma}} \lesssim \left\Vert \partial_s[\nabla_x(\partial_x^{\beta} p'(\varrho))(s, x+(1-e^{t-s})v)]\mathrm{H}^{K,I}(s,t) \right\Vert_{\mathcal{H}^{\ell}_{\sigma}} \\ + \left\Vert \nabla_x(\partial_x^{\beta} p'(\varrho))(s, x+(1-e^{t-s})v) \partial_s \mathrm{H}^{K,I}(s,t) \right\Vert_{\mathcal{H}^{\ell}_{\sigma}}.
\end{multline*}

The first term can be handled by arguments similar to the ones used for $\mathbf{S_9}$, $\mathbf{S}_1$ and for $\mathbf{S_{12}}$. The second can be adressed by the same procedure where one relies on the arguments used for $\mathbf{S_{13}}$. Likewise, we obtain

\begin{align*}
\LRVert{\mathbf{S_{14}}}_{\Ld^2(0,T;\H^1)}  \leq \Lambda(T,R).
\end{align*}

We eventually estimate the third term $\mathscr{R}_{I,2}^{\mathrm{Duha}}$ from Lemma <ref>.

We have

\begin{align*}
\LRVert{\mathscr{R}_{I,2}^{\mathrm{Duha}}}_{\Ld^2(0,T; \H^1)} \leq \Lambda(T,R).
\end{align*}

We proceed as before by introducing the vector fields

\begin{align*}
G_{18, i, \beta}^{K,I}(t,s,x,v)&:= \left(\nabla_x(\partial_x^{\beta} p'(\varrho))(s,x+(1-e^{t-s}v)) \cdot e_i \right)\mathfrak{H}^{K,I}(s,t,x,v), \\[2mm]
G_{19,i, \beta}^{K,I}(t,s,x,v)&:= \Big(\nabla_x(\partial_x^{\overline{K}^i-\beta} \mathrm{J}_\eps \varrho)(s,x+(1-e^{t-s}v)) \cdot \mathfrak{H}^{K,I}(s,t,x,v) \Big) e_i,
\end{align*}

\begin{align*}
\beta \in \mathbb{B}_K(i ,\ell)=\left\lbrace \beta \in \N^d \mid \vert \beta \vert=\ell, \ \ 0<\widehat{\beta}^i \leq K \right\rbrace, \ \ i=1,{\cdots},d, \ \ \ell =0,{\cdots}, \vert K \vert-1 .
\end{align*}

Let us also recall the expression of the kernel $\mathfrak{H}$ defined in (<ref>) by

\begin{multline*}
\mathfrak{H}^{K,I}(t,s,x,v):=\int_s^t e^{d(t-\tau)}   \mathrm{N}^{t;s}(\mathrm{Z}^{0;t}(x,\psi_{s,t}(x,v)))_{(I,0),(K,0)}   \\
\big( \nabla_x f(\tau, \mathrm{Z}^{\tau;t}(x,\psi_{s,t}(x,v))) -\nabla_v f(\tau, \mathrm{Z}^{\tau;t}(x,\psi_{s,t}(x,v))) \big) \, \mathrm{J}^{s,t}(x,v) \, \mathrm{d}\tau.
\end{multline*}

Thanks to Lemma <ref>, we can write

\begin{align*}
\mathscr{R}_{I,2}^{\mathrm{Duha}}=\mathbf{S}_{16}+\mathbf{S}_{17}+\mathbf{S}_{18}+\mathbf{S}_{19},
\end{align*}

\begin{align*}
\mathbf{S}_{16}(t,x) &:=-\sum_{K}\int_0^t \int_{\R^d} \partial_x^K u(s,x+(1-e^{t-s}v))\cdot \mathfrak{H}^{K,I}(t,s,x,v)\, \mathrm{d}v \, \mathrm{d}s,
\end{align*}

\begin{align*}
\mathbf{S}_{17} &:=\mathrm{K}^{\mathrm{fric}}[\partial_x^K \mathrm{J}_\eps \varrho], \ \ \ G_{17}(t,s,x,v):=p'(\varrho(s, x+(1-e^{t-s})v))\mathfrak{H}^{K,I}(t,s,x,v), \\
%&:=-\sum_{K}\int_0^t \int_{\R^d}  \nabla_x \left[\partial_x^K \mathrm{J}_\eps \varrho \right](s,\mathcal{Y}^{s,t}_z))\cdot \left(p'(\varrho(s, \mathcal{Y}^{s,t}_z))\mathfrak{H}^{K,I}(t,s,x,v) \right)\, \mathrm{d}v \, \mathrm{d}s, \\
\mathbf{S}_{18} &:= \sum_{K}\sum_{\ell=0}^{\lfloor \frac{\vert K \vert-1}{2} \rfloor} \sum_{\substack{i=1,{\cdots},d \\ \beta \in \mathbb{B}_K(i ,\ell)}} \binom{K}{\widehat{\beta}^i} \mathrm{K}_{G_{18, i, \beta}^{K,I}}\left[ \partial_x^{\overline{K}^i-\beta} \mathrm{J}_\eps \varrho \right], \\
\mathbf{S}_{19} &:= \sum_{K} \sum_{\ell=\lfloor \frac{\vert K \vert-1}{2} \rfloor+1}^{\vert K \vert-1} \sum_{\substack{i=1,{\cdots},d \\ \beta \in \mathbb{B}_K(i ,\ell)}} \binom{K}{\widehat{\beta}^i}\mathrm{K}_{G_{19, i, \beta}^{K,I}} \left[ \partial_x^{\beta}p'(\varrho) \right].
\end{align*}

Let us estimate these terms, as we have done previously. Let us observe that $ \mathfrak{H}^{K,I}(t,t,x,v)=0$ therefore the terms $\mathbf{S}_{17}, \mathbf{S}_{18}$ and $\mathbf{S}_{19}$ are of Type II and we can rely on Proposition <ref>.

$\bullet$ Estimate of $\mathbf{S}_{16}$: We mainly proceed as for $\mathbf{S}_{12}$, the kernel being changed from $\mathrm{H}$ to $\mathfrak{H}$. Hence, we only have to give an estimate for $\left\Vert \mathfrak{H}^{K,I}(s,t) \right\Vert_{\mathcal{H}^{k}_{r}}$ (for $k$ and $r$ large enough).

As in the estimate of $\mathbf{S}_{12}$, we use $\W^{k,\infty}_{x,v}$ bounds on $\psi_{s,t},  \mathrm{Z}^{s;t}$ and $ \mathrm{N}^{s;t}$ from Lemmas <ref>, Remark <ref> and Lemma <ref> to write

\begin{align*}
&\left\Vert \mathfrak{H}^{K,I}(s,t) \right\Vert_{\mathcal{H}^{k}_{r}}^2 \\
&\lesssim \Lambda(T,R) \LRVert{J^{s,t}}_{\W^{k, \infty}_{x,v}}^2 \left( 1+\LRVert{\mathrm{N}^{s;t}}_{\W^{k, \infty}_{x,v}}^2 \right)   \left\Vert \int_s^t \left[\nabla_x f(\tau, \mathrm{Z}^{\tau;t}(\cdot, \psi_{s,t}))-\nabla_v f(\tau, \mathrm{Z}^{\tau;t}(\cdot, \psi_{s,t})) \right] \, \mathrm{d}\tau \right\Vert_{\mathcal{H}^{k}_r}^2\\
& \lesssim \Lambda(T,R) \sum_{\vert \gamma \vert \leq k} \int_{\T^d \times \R^d}  \langle v \rangle^{2r} \left\vert \int_s^t \left[ \partial_{x,v}^{\gamma}(\nabla_x f)(\tau, \mathrm{Z}^{\tau;t}(\cdot, \psi_{s,t}))-\partial_{x,v}^{\gamma}(\nabla_v f)(\tau,\mathrm{Z}^{\tau;t}(\cdot, \psi_{s,t})) \right] \mathrm{d}\tau \right\vert^2 \, \mathrm{d}x \, \mathrm{d}v.
\end{align*}

By the the generalized Minkowski inequality, the last expression is bounded by

\begin{align*}
& \sum_{\vert \gamma \vert \leq k} \left( \int_s^t \left( \int_{\T^d \times \R^d} \langle v \rangle^{2r} \vert \partial_{x,v}^{\gamma}(\nabla_x f)(\tau,\mathrm{Z}^{\tau;t}(\cdot, \psi_{s,t}))-\partial_{x,v}^{\gamma}(\nabla_v f)(\tau, \mathrm{Z}^{\tau;t}(\cdot, \psi_{s,t})) \vert^2 \, \mathrm{d}x \, \mathrm{d}v \right)^{1/2} \, \mathrm{d}\tau \right)^2 \\
& \leq \Lambda(T,R)\sum_{\vert \gamma \vert \leq k} \left( \int_s^t \left( \int_{\T^d \times \R^d} \langle v \rangle^{2r} \vert \partial_{x,v}^{\gamma}(\nabla_x f)(\tau,x,v)-\partial_{x,v}^{\gamma}(\nabla_v f)(\tau,x,v) \vert^2 \, \mathrm{d}x \, \mathrm{d}v \right)^{1/2} \, \mathrm{d}\tau \right)^2,
\end{align*}

where we have performed the change of variable $v \mapsto w=\psi_{s,t}(x,v)$ from Lemma <ref> followed by $(x,w) \mapsto \mathrm{Z}^{0;t}(x,w)$. Here, we have used the bounds on the Jacobian from (<ref>), as well as the one on $\vert v-\psi_{s,t}(x,v) \vert$ via (<ref>). It follows that

\begin{align*}
\left\Vert \mathfrak{H}^{K,I}(s,t) \right\Vert_{\mathcal{H}^{k}_{r}}^2 &\lesssim\Lambda(T,R) \vert t-s \vert^2  \sup_{0 \leq \tau \leq T} \, \Big\lbrace \Vert \nabla_x f(\tau) \Vert_{\mathcal{H}^k_r}^2+\Vert \nabla_v f(\tau) \Vert_{\mathcal{H}^k_r}^2 \Big \rbrace,
\end{align*}

and therefore

\begin{align*}
\LRVert{\mathbf{S_{16}}}_{\Ld^2(0,T;\H^1)} \leq \Lambda(T,R).
%\lesssim \Lambda(T,R) \Vert  f \Vert_{\Ld^{\infty}(0,T;\mathcal{H}^{k+1}_r)} \Vert u \Vert_{\Ld^{2}(0,T;\H^{m+1})}.
\end{align*}

$\bullet$ Estimate of $\mathbf{S}_{17}, \mathbf{S}_{18}$ and $\mathbf{S}_{19}$: We mainly proceed as for $\mathbf{S}_{13}, \mathbf{S}_{14}$ and $\mathbf{S}_{15}$, using Proposition <ref>. As before, the kernel has just been changed from $\mathrm{H}$ to $\mathfrak{H}$. Hence, we only have to provide an estimate for $\Vert  \partial_s\mathfrak{H}^{K,I}(s,t)  \Vert_{\mathcal{H}^{k}_{r}} $ (for $k$ and $r$ large enough).

We have

\begin{align*}
\partial_s\mathfrak{H}^{K,I}(s,t,x,v)&=\partial_s J^{s,t} \mathrm{N}^{t;s}(\mathrm{Z}^{0;t}(x,\psi_{s,t}(x,v)))_{(I,0),(K,0)} \\
&   \qquad  \qquad \qquad     \int_s^t e^{d(t-\tau)}  \big[ \nabla_x f(\tau, \mathrm{Z}^{\tau;t}(x,\psi_{s,t}(x,v))) -\nabla_v f(\tau, \mathrm{Z}^{\tau;t}(x,\psi_{s,t}(x,v))) \big] \, \mathrm{d}\tau \\
& \quad + J^{s,t}    \partial_s \left\lbrace \mathrm{N}^{t;s}(\mathrm{Z}^{0;t}(x,\psi_{s,t}(x,v)))_{(I,0),(K,0)} \right\rbrace   \\
&   \qquad  \qquad \qquad     \int_s^t e^{d(t-\tau)} \big[ \nabla_x f(\tau, \mathrm{Z}^{\tau;t}(x,\psi_{s,t}(x,v))) -\nabla_v f(\tau, \mathrm{Z}^{\tau;t}(x,\psi_{s,t}(x,v))) \big] \, \mathrm{d}\tau \\
& \quad +J^{s,t} \mathrm{N}^{t;s}(\mathrm{Z}^{0;t}(x,\psi_{s,t}(x,v)))_{(I,0),(K,0)}   \\
&   \qquad  \qquad \qquad     \int_s^t e^{d(t-\tau)} \partial_s \big[ \nabla_x f(\tau, \mathrm{Z}^{\tau;t}(x,\psi_{s,t}(x,v))) -\nabla_v f(\tau, \mathrm{Z}^{\tau;t}(x,\psi_{s,t}(x,v))) \big] \, \mathrm{d}\tau \\
& \quad -e^{d(t-s)} \J^{s,t} \mathrm{N}^{t;s}(\mathrm{Z}^{0;t}(x,\psi_{s,t}(x,v)))_{(I,0),(K,0)}  \\
&  \qquad  \qquad \qquad  \qquad \qquad\big[ \nabla_x f(s, \mathrm{Z}^{s;t}(x,\psi_{s,t}(x,v))) -\nabla_v f(s, \mathrm{Z}^{s;t}(x,\psi_{s,t}(x,v))) \big].
\end{align*}

Using the same $\W^{k,\infty}_{x,v}$ bounds on $J^{s,t}$, $\mathrm{N}^{s;t}$ and $\dot{\W}^{k,\infty}_{x,v}$ bounds on $\psi_{s,t}$ as before, as well as on their time derivatives, we obtain

\begin{align*}
\Vert  \partial_s\mathfrak{H}^{K,I}(s,t)  \Vert_{\mathcal{H}^{k}_{r}} \leq  \Lambda(T,R).
\end{align*}

This eventually yields

\begin{align*}
\LRVert{\mathbf{S_{17}}}_{\Ld^2(0,T;\H^1)}+\LRVert{\mathbf{S_{18}}}_{\Ld^2(0,T;\H^1)} +\LRVert{\mathbf{S_{19}}}_{\Ld^2(0,T;\H^1)}  \leq \Lambda(T,R).
\end{align*}