# Well-posedness in any dimension for Hamiltonian flows with non BV force terms

Nicolas Champagnat, Pierre-Emmanuel Jabin

► To cite this version:

Nicolas Champagnat, Pierre-Emmanuel Jabin. Well-posedness in any dimension for Hamiltonian flows with non BV force terms. Communications in Partial Differential Equations, 2010, 35 (5), pp.786-816. 10.1080/03605301003646705 . inria-00373784v2

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Submitted on 10 Mar 2016

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# Well posedness in any dimension for Hamiltonian flows with non BV force terms

Nicolas Champagnat<fnref n="1" />, Pierre-Emmanuel Jabin<fnref n="1" />,<fnref n="2" />

## Abstract

We study existence and uniqueness for the classical dynamics of a particle in a force field in the phase space. Through an explicit control on the regularity of the trajectories, we show that this is well posed if the force belongs to the Sobolev space $H^{3/4}$.

MSC 2000 subject classifications: 34C11, 34A12, 34A36, 35L45, 37C10

Key words and phrases: Flows for ordinary differential equations, Kinetic equations, Stability estimates

## 1 Introduction

This paper studies existence and uniqueness of a flow for the equation

$$ \begin{cases} \partial_t X(t, x, v) = V(t, x, v), & X(0, x, v) = x, \\ \partial_t V(t, x, v) = F(X(t, x, v)), & V(0, x, v) = v, \end{cases} \qquad (1.1) $$

where $x$ and $v$ are in the whole $\mathbb{R}^d$ and $F$ is a given function from $\mathbb{R}^d$ to $\mathbb{R}^d$. Those are of course Newton's equations for a particle moving in a force field $F$. For many applications the force field is in fact a potential

$$ F(x) = -\nabla \phi(x), $$

even though we will not use the additional Hamiltonian structure that this is providing.

<footnote n="1">¹TOSCA project-team, INRIA Sophia Antipolis - Méditerranée, 2004 route des Lucioles, B.P. 93, 06902 Sophia Antipolis Cedex, France, E-mail: Nicolas.Champagnat@sophia.inria.fr</footnote>

<footnote n="2">²Laboratoire J.-A. Dieudonné, Université de Nice - Sophia Antipolis, Parc Valrose, 06108 Nice Cedex 02, France, E-mail: jabin@unice.fr</footnote>
This is a particular case of a system of differential equations

$$ \partial_t \Xi(t, \xi) = \Phi(\Xi), \qquad (1.2) $$

with $\Xi = (X, V)$, $\xi = (x, v)$, $\Phi(\xi) = (v, F(x))$. Cauchy-Lipschitz' Theorem applies to (1.1) and gives maximal solutions if $F$ is Lipschitz. Those solutions are in particular global in time if for instance $F \in L^\infty$. Moreover because of the particular structure of Eq. (1.1), this solution has the additional

**Property 1** For any $t \in \mathbb{R}$ the application

$$ (x, v) \in \mathbb{R}^d \times \mathbb{R}^d \mapsto (X(t, x, v), V(t, x, v)) \in \mathbb{R}^d \times \mathbb{R}^d \qquad (1.3) $$

is globally invertible and has Jacobian 1 at any $(x, v) \in \mathbb{R}^d \times \mathbb{R}^d$. It also defines a semi-group

$$ \begin{array}{ll} \forall s, t \in \mathbb{R}, & X(t + s, x, v) = X(s, X(t, x, v), V(t, x, v)), \\ \text{and} & V(t + s, x, v) = V(s, X(t, x, v), V(t, x, v)). \end{array} \qquad (1.4) $$

In many cases this Lipschitz regularity is too demanding and one would like to have a well posedness theory with a less stringent assumption on $F$. That is the aim of this paper. More precisely, we prove

**Theorem 1.1** Assume that $F \in H^{3/4} \cap L^\infty$. Then, there exists a solution to (1.1), satisfying Property 1. Moreover this solution is unique among all limits of solutions to any regularization of (1.1).

Many works have already studied the well posedness of Eq. (1.2) under weak conditions for $\Phi$. The first one was essentially due to DiPerna and Lions [19], using the connection between (1.2) and the transport equation

$$ \partial_t u + \Phi(\xi) \cdot \nabla_{\xi} u = 0. \qquad (1.5) $$

The notion of renormalized solutions for Eq. (1.5) provided a well posedness theory for (1.2) under the conditions $\Phi \in W^{1,1}$ and $\text{div}_\xi \Phi \in L^\infty$. This theory was generalized in [28], [27] and [24].

Using a slightly different notion of renormalization, Ambrosio [2] obtained well posedness with only $\Phi \in BV$ and $\text{div}_\xi \Phi \in L^\infty$ (see also the papers by Colombini and Lerner [12], [13] for the $BV$ case). The bounded divergence condition was then slightly relaxed by Ambrosio, De Lellis and Malý in [4] with only $\Phi \in SBV$ (see also [17]).
Of course there is certainly a limit to how singular $\Phi$ may be and still provide uniqueness, as shown by the counterexamples of Aizenman [1] and Bressan [10]. The example by De Pauw [18] even suggests that for the general setting (1.2), $BV$ is probably close to optimal.

But as (1.1) is a very special case of (1.2), it should be easier to deal with. And for instance Bouchut [6] got existence and uniqueness to (1.1) with $F \in BV$ in a simpler way than [2]. Hauray [23] handled a slightly less than $BV$ case ($BV_{loc}$).

In dimension $d = 1$ of physical space (dimension 2 in phase space), Bouchut and Desvillettes proved well posedness for Hamiltonian systems (thus including (1.1) as $F$ is always a derivative in dimension 1) without any additional derivative for $F$ (only continuity). This was extended to Hamiltonian systems in dimension 2 in phase space with only $L^p$ coefficients in [22] and even to any system (non necessarily Hamiltonian) with bounded divergence and continuous coefficient by Colombini, Crippa and Rauch [11] (see also [14] for low dimensional settings and [9] with a very different goal in mind).

Unfortunately in large dimensions (more than 1 of physical space or 2 in the phase space), the Hamiltonian or bounded divergence structure does not help so much. To our knowledge, Th. 1.1 is the first result to require less than one derivative on the force field $F$ in any dimension. Note that the comparison between $H^{3/4}$ and $BV$ is not clear as $BV \not\subset H^{3/4}$ and $H^{3/4} \not\subset BV$. Even if one considers the stronger assumption that the force field be in $L^\infty \cap BV$, that space contains by interpolation $H^s$ for $s < 1/2$ and not $H^{3/4}$. As the proof of Th. 1.1 uses orthogonality arguments, we do not know how to work in spaces non based on $L^2$ norms ($W^{3/4,1}$ for example). Therefore strictly speaking Th. 1.1 is neither stronger nor weaker than previous results.

We have no idea whether this $H^{3/4}$ is optimal or in which sense. It is striking because it already appears in a question concerning the related Vlasov equation

$$ \partial_t f + v \cdot \nabla_x f + F \cdot \nabla_v f = 0. \qquad (1.6) $$

Note that this is the transport equation corresponding to Eq. (1.1), just as Eq. (1.5) corresponds to (1.2). As a kinetic equation, it has some regularization property namely that the average

$$ \rho(t, x) = \int_{\mathbb{R}^d} f(t, x, v) \psi(v) dv, \quad \text{with } \psi \in C_c^\infty(\mathbb{R}^d), $$

is more regular than $f$. And precisely if $f \in L^2$ and $F \in L^\infty$ then $\rho \in H^{3/4}$; we refer to Golse, Lions, Perthame and Sentis [21] for this result, DiPerna,
Lions, Meyer for a more general one [20], or [26] for a survey of averaging lemmas. Of course we do not know how to use this kind of result for the uniqueness of (1.6) or even what is the connection between the $H^{3/4}$ of averaging lemmas and the one found here. It could just be a scaling property of these equations.

Note in addition that the method chosen for the proof may in fact be itself a limitation. Indeed it relies on an explicit control on the trajectories : for instance, we show that $|X(t, x, v)-X^{\delta}(t, x, v)|$ and $|V(t, x, v)-V^{\delta}(t, x, v)|$ remain approximately of order $|\delta|$ if $\delta = (\delta_1, \delta_2)$ and

$$X^{\delta}(t, x, v) = X(t, x + \delta_1, v + \delta_2), \quad V^{\delta}(t, x, v) = V(t, x + \delta_1, v + \delta_2).$$

However the example given in Section 3 demonstrates that such a control is not always possible: Even in the physical space is one-dimensional, it requires at least 1/2 derivative on the force term ($F \in W_{loc}^{1/2,1}$) whereas well posedness is known with essentially $F \in L^p$ (see the references above). Note moreover that if $X, V$ satisfies (1.3) and (1.4), then $X^{\delta}$ and $V^{\delta}$ satisfy (1.3) but not (1.4). However, the functions $X^{\delta}(t, x - \delta_1, v - \delta_2)$ and $V^{\delta}(t, x - \delta_1, v - \delta_2)$ of course satisfy (1.4).

This kind of control is obviously connected with regularity properties of the flow (differentiability for instance), which were studied in [5] (see also [3]). The idea to prove them directly and then use them for well posedness is quite recent, first by Crippa and De Lellis in [16] with the introduction and subsequent bound on the functional

$$\int_{\Omega} \sup_{r} \int_{|\delta| \le r} \log \left( 1 + \frac{|\Xi(t, \xi) - \Xi(t, \xi + \delta)|}{|\delta|} \right) d\delta dx. \quad (1.7)$$

This gave existence/uniqueness for Eq. (1.2) with $\Phi \in W_{loc}^{1,p}$ for any $p > 1$ and a weaker version of the bounded divergence condition. This was extended in [7] and [25].

We use here a modified version of (1.7) which takes the different roles of $x$ and $v$ into account. The way of bounding it is also quite different as we essentially try to integrate the oscillations of $F$ along a trajectory.

The paper is organized as follows: The next section introduces the functional that is studied, states the bounds that are to be proved and briefly explains the relation with the well posedness result Th. 1.1. The section after that presents the example in 1d and the last and longer section the proof of the bound.
## Notation

* $u \cdot v$ denotes the usual scalar product of $u \in \mathbb{R}^d$ and $v \in \mathbb{R}^d$.

* $S^{d-1}$ denotes the $(d-1)$-dimensional unit sphere in $\mathbb{R}^d$.

* $B(x, r)$ is the closed ball of $\mathbb{R}^d$ for the standard Euclidean norm with center $x \in \mathbb{R}^d$ and radius $r \ge 0$.

* $C$ denotes a positive constant that may change from line to line.

# 2 Preliminary results

## 2.1 Reduction of the problem

In the sequel, we give estimates on the flow of Eq. (1.1) for initial values $(x, v)$ in a compact subset $\Omega = \Omega_1 \times \Omega_2$ of $\mathbb{R}^{2d}$ and for time $t \in [0, T]$. Fix some $A > 0$ and consider any $F \in L^\infty$ with $\|F\|_{L^\infty} \le A$. Then for any solution to Eq. (1.1)

$$ \begin{gathered} |V(t, x, v) - v| \le \|F\|_{L^\infty} t \le At \\ \text{and } |X(t, x, v) - x| \le vt + \frac{\|F\|_{L^\infty} t^2}{2} \le \frac{vt + At^2}{2}. \end{gathered} $$

Therefore, for any $t \in [0, T]$ and for any $(x, v)$ at a distance smaller than 1 from $\Omega$, $(X(t, x, v), V(t, x, v)) \in \Omega' = \Omega'_1 \times \Omega'_2$ for some compact subset $\Omega'$ of $\mathbb{R}^{2d}$. Moreover $\Omega'$ depends only on $\Omega$ and $A$. Similarly, we introduce $\Omega''$ a compact subset of $\mathbb{R}^{2d}$ such that the couple $(X(-t, x, v), V(-t, x, v))$ belongs to $\Omega''$ for any $t \in [0, T]$ and any $(x, v)$ at a distance smaller than 1 from $\Omega'$.

For $T > 0$ and $\delta \in \mathbb{R}^2 \setminus \{0\}$, define the quantity

$$ Q_{\delta}(T) = \iint_{\Omega} \log \left( 1 + \frac{1}{|\delta|^2} \left( \begin{aligned} & \sup_{0 \le t \le T} |X(t, x, v) - X^{\delta}(t, x, v)|^2 \\ & + \int_{0}^{T} |V(t, x, v) - V^{\delta}(t, x, v)|^2 dt \end{aligned} \right) \right) dx \, dv, $$

where $X, V$ and $X^\delta, V^\delta$ are two solutions to (1.1), satisfying

$$ \left\{ \begin{array}{ll} X(0, x, v) = x, & V(0, x, v) = v, \quad (X, V) \text{ satisfies (1.3), (1.4)} \\ \text{and either } X^{\delta}(0, x, v) = x, & V^{\delta}(0, x, v) = v, \quad (X, V) \text{ satisfies (1.3), (1.4)} \\ \text{or } X^{\delta}(0, x, v) = x + \delta_1, & V^{\delta}(0, x, v) = v + \delta_2, \\ & (X, V)(t, x - \delta_1, v - \delta_2) \text{ satisfies (1.3), (1.4).} \end{array} \right. \qquad (2.1) $$
We prove the following result

**Proposition 2.1** Fix $T > 0$, any $A > 0$ and $\Omega \in \mathbb{R}^{2d}$ compact. Define $\Omega'$ and $\Omega''$ as above. There exists a constant $C > 0$ depending only of $\mathrm{diam}(\Omega')$, $|\Omega''|$, $T$ and $A$, such that, for any $a \in (0, 1/4)$, $F \in H^{3/4+a}$ with $\|F\|_{L^\infty} \le A$ and any solutions $(X, V)$ and $(X^\delta, V^\delta)$ to (1.1) such that $(X, V)$ and $(X^\delta, V^\delta)$ satisfy (2.1), one has for any $|\delta| < 1/e$,

$$Q_{\delta}(T) \le C \left( \log \frac{1}{|\delta|} \right)^{1-2a} \left( 1 + \|F\|_{H^{3/4+a}(\Omega'')} \right).$$

This result can be actually extended without difficulty to any $F \in L^\infty$ such that

$$\int_{\mathbb{R}^d} |k|^{3/2} |\alpha(k)|^2 f(k) dk < \infty, \quad (2.2)$$

for some function $f \ge 1$ such that $f(k) \to +\infty$ when $|k| \to +\infty$, where $\alpha(k)$ is the Fourier transform of $F$. By de la Vallée Poussin, such an $f$ may be found for any $F \in H^{3/4}$. The modified proposition hence holds

**Proposition 2.2** Fix $T > 0$, $A > 0$, $\Omega \in \mathbb{R}^{2d}$ compact and any $f \ge 1$ such that $f(k) \to +\infty$ when $|k| \to +\infty$. Define $\Omega'$ and $\Omega''$ as above. There exists a continuous, increasing function $\varepsilon(\delta)$ with $\varepsilon(0) = 0$ s.t. for any $F \in H^{3/4} \cap L^\infty$ satisfying (2.2), with $\|F\|_{L^\infty} \le A$, for any solutions $(X, V)$ and $(X^\delta, V^\delta)$ to (1.1) such that $(X, V)$ and $(X^\delta, V^\delta)$ satisfy (2.1), one has for any $|\delta| < 1/e$,

$$Q_{\delta}(T) \le \log \left( \frac{1}{|\delta|} \right) \varepsilon(|\delta|) \left( 1 + \int_{\mathbb{R}^d} |k|^{3/2} |\alpha(k)|^2 f(k) dk \right)^{1/2},$$

with $\alpha$ the Fourier transform of $F$.

For reasons of simplicity we essentially give the proof of Prop. 2.1 in this paper. However let us briefly indicate here how one may obtain Prop. 2.2.

The first point is that the estimates in the proof of Prop. 2.1 are in fact linear. More precisely, after differentiating in time $Q_\delta$, one bounds (see the beginning of the proof of Prop. 2.1 for details)

$$Q_{\delta}(T) \le C \int_{0}^{T} \int_{\Omega} \frac{V(t, x, v) - V^{\delta}(t, x, v)}{A_{\delta}(t, x, v)} \\ \int_{0}^{t} \int_{\mathbb{R}^{d}} \alpha(k) \left( e^{ik \cdot X(s, x, v)} - e^{ik \cdot X^{\delta}(s, x, v)} \right) dk \, ds \, dx \, dv \, dt,$$
with

$$A_{\delta} = |\delta|^2 + \sup_{0 \le s \le t} |X(s, x, v) - X^{\delta}(s, x, v)|^2 + \int_{0}^{t} |V(s, x, v) - V^{\delta}(s, x, v)|^2 ds.$$

The estimation is linear because after that point, we never use the relation between $X, V, X^{\delta}, V^{\delta}$ and the Fourier transform $\alpha$ of $F$ involved in the previous expression. The only information on $F$ that is used is the fact that $\|F\|_{L^{\infty}}$ is finite (when bounding $|V(t, x, v)|$ in the proof of Lemma 4.3). So in fact for any given $X, V, X^{\delta}, V^{\delta}$, we may define

$$Z(\tilde{F}) = \int_{0}^{T} \int_{\Omega} \frac{V(t, x, v) - V^{\delta}(t, x, v)}{A_{\delta}(t, x, v)} \int_{0}^{t} \int_{\mathbb{R}^{d}} \tilde{\alpha}(k) (e^{ik \cdot X(s,x,v)} - e^{ik \cdot X^{\delta}(s,x,v)}) dk \, ds \, dx \, dv \, dt$$

where $\tilde{\alpha}$ is the Fourier transform of $\tilde{F}$. What we prove is that if $(X, V)$ and $(X^{\delta}, V^{\delta})$ satisfy (2.1), then for any $A > 0$, there exists a constant $C$ depending only on $F$ and $A$ such that for any $\tilde{F} \in H^{3/4+a} \cap L^{\infty}$ with $\|\tilde{F}\|_{L^{\infty}} \le A$, one has

$$Z(\tilde{F}) \le C (\log 1/|\delta|)^{1-2a} \left( 1 + \int_{\mathbb{R}^d} |k|^{3/2+2a} |\tilde{\alpha}(k)|^2 dk \right)^{1/2} . \quad (2.3)$$

The key point is now to notice that the numerical constant $C$ in (2.3) may be taken independently of $a$. It is of course not obvious and has to be checked carefully in the proof of Prop 2.1. With that, one deduces the limit inequality by letting $a \to 0$

$$Z(\tilde{F}) \le C \log(1/|\delta|) \left( 1 + \int_{\mathbb{R}^d} |k|^{3/2} |\tilde{\alpha}(k)|^2 dk \right)^{1/2} . \quad (2.4)$$

In itself this inequality is useless as $Z(F)$ is trivially dominated by a constant times $\log(1/|\delta|)$. Nevertheless this allows us to interpolate with (2.3).

For any $F \in H^{3/4} \cap L^{\infty}$ satisfying (2.2), fix some $M$ to be precised later and write

$$F = F_1 + F_2, \quad \int_{\mathbb{R}^d} e^{-ik \cdot x} F_1(x) \, dx = \alpha(k) \mathbb{I}_{|k| \le M}, \\ \int_{\mathbb{R}^d} e^{-ik \cdot x} F_2(x) \, dx = \alpha(k) \mathbb{I}_{|k| > M}.$$
Now apply (2.3) to $F_1$ to find

$$ 
\begin{aligned}
Z(F_1) & \le C (\log 1/|\delta|)^{1-2a} (1 + \|F_1\|_{H^{3/4+a}}^2)^{1/2} \\
& \le C M^a (\log 1/|\delta|)^{1-2a} (1 + \|F\|_{H^{3/4}}^2)^{1/2}.
\end{aligned}
$$

Apply (2.4) to $F_2$

$$ 
\begin{aligned}
Z(F_2) & \le C \log(1/|\delta|) (1 + \|F_2\|_{H^{3/4}}^2)^{1/2} \\
& \le \frac{C}{\inf_{|k| \ge M} \sqrt{f(k)}} \log(1/|\delta|) \left( 1 + \int_{\mathbb{R}^d} |k|^{3/2} |\alpha(k)|^2 f(k) dk \right)^{1/2}.
\end{aligned}
$$

It simply remains to take for example $M = \log(1/|\delta|)^{1/4}$ to deduce

$$ Z(F) \le \log(1/|\delta|) \varepsilon(|\delta|) \left( 1 + \int_{\mathbb{R}^d} |k|^{3/2} |\alpha(k)|^2 f(k) dk \right)^{1/2}, $$

for an $\varepsilon$ which depends only on $f$, which ends the proof of Prop 2.2.

## 2.2 From Prop. 2.1 and 2.2 to Th. 1.1

It is well known how to pass from an estimate like the one provided by Prop. 2.1 to a well posedness theory (see [16] for example) and therefore we only recall the main steps. Take any $F \in H^{3/4+a} \cap L^\infty$.

We start by the existence of a solution. For that, define $F_n$ a regularizing sequence of $F$; $F_n$ is hence uniformly bounded in $H^{3/4+a} \cap L^\infty$. Denote $X_n$, $V_n$ the solution to (1.1) with $F_n$ instead of $F$ and $(X_n, V_n)(t=0) = (x, v)$. For any $\delta = (\delta_1, \delta_2)$ in $\mathbb{R}^{2d}$, put

$$ (X_n^\delta, V_n^\delta)(t, x, v) = (X_n, V_n)(t, x + \delta_1, v + \delta_2). $$

The function $F_n$ and the solutions $(X_n, V_n)$, $(X_n^\delta, V_n^\delta)$ satisfy all the assumptions of Prop. 2.1, as $F_n \in W^{1,\infty}$, using Property 1. Since $F_n$ is uniformly bounded in $L^\infty \cap H^{3/4+a}$, the proposition then shows that

$$ Q_{\delta,n}(T) = \iint_{\Omega} \log \left( 1 + \frac{1}{|\delta|^2} \left( \begin{aligned} & \sup_{0 \le t \le T} |X_n(t, x, v) - X_n^\delta(t, x, v)|^2 \\ & + \int_0^T |V_n(t, x, v) - V_n^\delta(t, x, v)|^2 dt \end{aligned} \right) \right) dx dv $$
is uniformly bounded in $n$ by

$$C \left( \log \frac{1}{|\delta|} \right)^{1-2a}.$$

This is enough to prove that the sequence $(X_n, V_n)$ is compact in $L_{loc}^1(\mathbb{R}_+ \times \mathbb{R}^{2d})$. Indeed, by Cauchy-Schwartz inequality,

$$\int_0^T \iint_{\Omega} (|X_n(t, x, v) - X_n^\delta(t, x, v)| + |V_n(t, x, v) - V_n^\delta(t, x, v)|) dx dv dt \\ \le C |\Omega|^{1/2} \left( \iint_{\Omega} A_\delta(x, v) dx dv \right)^{1/2},$$

with

$$A_\delta(x, v) = \sup_{t \le T} |X_n(t, x, v) - X_n^\delta(t, x, v)|^2 + \int_0^T |V_n(t, x, v) - V_n^\delta(t, x, v)|^2 dt.$$

Now $A_\delta(x, v)$ is uniformly bounded by a finite constant $A$. Hence

$$\int_0^T \iint_{\Omega} (|X_n(t, x, v) - X_n^\delta(t, x, v)| + |V_n(t, x, v) - V_n^\delta(t, x, v)|) dx dv dt \\ \le C |\Omega|^{1/2} \left( \iint_{\Omega} \frac{A \log (1 + A_\delta(x, v)/|\delta|^2)}{\log(1 + A|\delta|^{-2})} dx dv \right)^{1/2} \\ \le C \left( \frac{|\Omega| Q_{\delta,n}(T)}{\log(1 + |\delta|^{-1})} \right)^{1/2} \le C |\Omega|^{1/2} \left( \log \frac{1}{|\delta|} \right)^{-a}$$

if $|\delta|$ is small enough, by Prop. 2.1. This shows that the first integral converges to 0 as $|\delta|$ tends 0, uniformly in $n$, which directly implies the compactness of the sequence $(X_n, V_n)$ in $L_{loc}^1$.

Denoting by $(X, V)$ an extracted limit, one directly checks that $(X, V)$ is a solution to (1.1) by taking the limit in each term. In addition, as $(X_n, V_n)$ satisfies (1.3) uniformly in $n$ then so does $(X, V)$ and finally passing to the limit in (1.4), one concludes that $(X, V)$ verifies this last property. Thus existence is proved.

For uniqueness, consider another solution $(X^\delta, V^\delta)$ to (1.1), which is also the limit of solutions to a regularized equation (such as the one given by $F_n$ or by another regularizing sequence of $F$). Then with the same argument, $(X^\delta, V^\delta)$ also satisfies (1.3) and (1.4). Moreover

$$X(0, x, v) - X^\delta(0, x, v) = x - x = 0, \quad V(0, x, v) - V^\delta(0, x, v) = v - v = 0,$$
so that $(X, V)$ and $(X^\delta, V^\delta)$ also verify (2.1) for any $\delta \neq 0$. Applying again Prop. 2.1, one finds that

$$Q_{\delta,n}(T) = \iint_{\Omega} \log \left( 1 + \frac{1}{|\delta|^2} \left( \sup_{0 \le t \le T} |X(t, x, v) - X^\delta(t, x, v)|^2 + \int_0^T |V(t, x, v) - V^\delta(t, x, v)|^2 dt \right) \right) dx \, dv$$

is bounded by

$$C \left( \log \frac{1}{|\delta|} \right)^{1-2a}.$$

Letting $\delta$ go to 0, one concludes that $X = X^\delta$ and $V = V^\delta$, a.e. in $t, x, v$.

Note from this sketch that one has uniqueness among all solutions to (1.1) satisfying (1.3) and (1.4) and not only those which are limit of a regularized problem. However not all solutions to (1.1) necessarily satisfy those two conditions so that the uniqueness among all solutions to (1.1) is unknown. Indeed in many cases, it is not true, as there is a hidden selection principle in (1.3) (see the discussion in [4], [15] or [17]).

Finally if $F \in H^{3/4}$ only, then one first applies the De La Vallée Poussin's lemma to find a function $f$ s.t. $f(k) \to +\infty$ when $|k| \to +\infty$ and

$$\int_{\mathbb{R}^d} |k|^{3/2} f(k) |\alpha(k)|^2 dk < +\infty. \qquad (2.5)$$

One proceeds as before with a regularizing sequence $F_n$ which now has to satisfy uniformly the previous estimate. Using Prop. 2.2 instead of Prop. 2.1, the rest of the proof is identical.

### 3 The question of optimality : An example

It is hard to know whether the condition $F \in H^{3/4}$ is optimal and in which sense (see the short discussion in the introduction). Instead the purpose of this section is to give a simple example showing that $F \in W^{1/2,1}$ is a necessary condition in order to use the method followed in this paper; namely a quantitative estimate on $X - X^\delta$ and $V - V^\delta$. More precisely, for any $\alpha < 1/2$, we are going to construct a sequence of force fields $(F_N)_{N \ge 1}$ uniformly bounded in $W^{\alpha,1} \cap L^\infty$ and a sequence $(\delta_N)_{N \ge 1}$ in $\mathbb{R}^2$ converging to 0 such that functionals like $Q_\delta(T)$ cannot be uniformly bounded in $N$.

This example is one dimensional (two in phase space) where it is known that much less is required to have uniqueness of the flow (F continuous by [8]
and even $F \in L^2$ by [22]). So this indicates in a sense that the method itself is surely not optimal. Moreover what this should imply in higher dimensions is not clear...

Through all this section we use the notation $f = O(g)$ if there exists a constant $C$ s.t.

$$|f| \leq C|g| \quad \text{a.e.}$$

In dimension 1 all functions $F$ derive from a potential so take

$$\phi(x) = x + \frac{h(Nx)}{N^{\alpha+1}}, \quad F = -\phi'(x)$$

with $h$ a periodic and regular function ($C^2$ at least) with $h(0) = 0$.

As $\phi$ is regular, we know that the solution $(X, V)$ with initial condition $(x, v)$ and the shifted one $(X^\delta, V^\delta)$ corresponding to the initial condition $(x, v) + \delta$ with $\delta = (0, \delta_2)$, $\delta_2 > 0$, satisfy the conservation of energy or

$$V^2 + 2\phi(X) = v^2 + 2\phi(x), \quad |V^\delta|^2 + 2\phi(X^\delta) = |v + \delta_2|^2 + 2\phi(x).$$

As $\phi$ is defined up to a constant, we do not need to look at all the trajectories and may instead restrict ourselves to the one starting at $x$ s.t. $v^2+2\phi(x) = 0$. By symmetry, we may assume $v > 0$.

Let $t_0$ and $t_0^\delta$ be the first times when the trajectories stop increasing: $V(t_0) = 0$ and $V^\delta(t_0^\delta) = 0$. As both velocities are initially positive, they stay so until $t_0$ or $t_0^\delta$. So for instance

$$\dot{X} = V = \sqrt{-2\phi(X)}.$$

Hence $t_0$ is obtained by

$$t_0 = \int_0^{t_0} \frac{\dot{X}}{\sqrt{-2\phi(X)}} dt = \int_x^{x_0} \frac{dy}{\sqrt{-2\phi(y)}} \\ = \int_x^{x_0} \frac{dy}{\sqrt{-2y - 2h(Ny)N^{-1-\alpha}}},$$

if $x_0 = X(t_0)$. Of course by energy conservation $\phi(x_0) = 0$ and again as we are in dimension 1 we can assume for convenience that $x_0 = 0$.

We have the equivalent formula for $t_0^\delta$ with $x_0^\delta$ (which we may not assume equal to 0). Put

$$K_\delta = |v + \delta_2|^2 + 2\phi(x) = \delta_2^2 + 2v\delta_2, \quad \eta = N(x_0^\delta - x_0) = N x_0^\delta$$
<figure><img src="image_2.png" /><figcaption>Figure 1: The potential $\phi$ and the construction of $x_0$ and $x_0^\delta$</figcaption></figure>

and note that $2\phi(x_0^\delta) - 2\phi(x_0) = K_\delta$, so that $|x_0^\delta - x_0| = |x_0^\delta| \le C\delta_2$ since $\phi' \ge 1/2$ for $N$ large enough. Then

$$ t_0^\delta = \int_x^{x_0^\delta} \frac{dy}{\sqrt{K_\delta - 2\phi(y)}} = \int_{x-\eta/N}^{x_0} \frac{dy}{\sqrt{K_\delta - 2y - 2\eta/N - 2N^{-1-\alpha}h(Ny+\eta)}} \\ = O(\delta_2) + \int_x^{x_0} \frac{dy}{\sqrt{-2y - 2N^{-1-\alpha}(h(Ny+\eta) - h(\eta))}}, $$

as the integral between $x$ and $x - \eta/N$ is bounded by $O(\delta_2)$ (the integrand is bounded here) and

$$ K_\delta = 2\phi(x_0^\delta) = 2x_0^\delta + \frac{2}{N^{1+\alpha}}h(Nx_0^\delta) = 2\frac{\eta}{N} + \frac{2}{N^{1+\alpha}}h(\eta). $$

Note that as $h$ is Lipschitz regular

$$ \frac{|h(Nx + \eta) - h(\eta)|}{N^{1+\alpha}} = O\left(\frac{x}{N^\alpha}\right), $$

$$ \frac{|h(Nx)|}{N^{1+\alpha}} = \frac{|h(Nx) - h(0)|}{N^{1+\alpha}} = O\left(\frac{x}{N^\alpha}\right). $$

So subtracting the two formula and making an asymptotic expansion

$$ t_0 - t_0^\delta = O(\delta_2) + \int_x^{x_0} \frac{dy}{(-2y)^{3/2}} \left( -\frac{2}{N^{1+\alpha}}(h(Ny) - h(Ny + \eta) + h(\eta)) \right. \\ \left. +O\left(\frac{h(Ny)}{N^{1+\alpha}\sqrt{y}}\right)^2 \right). $$
Making the change of variable $Ny = z$ in the dominant term in the integral, one finds

$$t_0 - t_0^\delta = O(\delta_2) - 2 \int_{Nx}^0 N^{1/2-1-\alpha} \frac{h(z) - h(z+\eta) + h(\eta)}{(-2z)^{3/2}} dz + O(N^{-3/2-2\alpha}).$$

Consequently as long as

$$A(\eta) = \int_{-\infty}^{0} \frac{h(z) - h(z + \eta) - h(\eta)}{(-2z)^{3/2}} dz$$

is of order 1 then $t_0 - t_0^\delta$ is of order $N^{-1/2-\alpha}$. Note that $A(\eta)$ is small when $\eta$ is, but it is always possible to find functions $h$ s.t. $A(\eta)$ is of order 1 at least for some $\eta$. One way to see this is by observing that

$$A'(\eta) = - \int_{-\infty}^{0} \frac{h'(y + \eta) + h'(\eta)}{(-2y)^{3/2}} dy$$

cannot vanish for all $\eta$ and functions $h$. Taking $h$ such that $A'(\eta) \ge 1$ for $\eta$ in some non-trivial interval, we can assume that $A$ is of order 1 for $\eta \in [\eta, \bar{\eta}]$ for some $0 < \bar{\eta} < \bar{\eta}$.

Coming back to the definition of $\eta$ and $x_0^\delta$, $\eta \in [\eta, \bar{\eta}]$ is equivalent to

$$\delta_2^2 + 2v\delta_2 \in \phi([\eta/N, \bar{\eta}/N]). \qquad (3.1)$$

Using the formula for $\phi$ and the fact that $\eta$ and $\bar{\eta}$ are independent of $N$ or $\delta_2$, we find

$$\delta_2^2 + 2v\delta_2 + O(N^{-1-\alpha}) \in [\eta/N, \bar{\eta}/N].$$

Charging $\delta_2 = 1/N$, we obtain

$$v + O(N^{-\alpha}) \in [\eta/2, \bar{\eta}/2].$$

We denote by $\mathcal{V}$ the space of initial velocities $v \in \mathcal{V}_0$ s.t. (3.1) is satisfied for $N$ large enough. We may assume that $\mathcal{V} \subset \{v > \eta/4\}$.

Since the initial position $x$ is arbitrary, for all $x$ in a given compact set $\Omega_1$, there exists $c > 0$ and $N_0 \ge 1$ such that we can construct a set $\mathcal{V}(x) \subset \{v > \eta/4\}$ satisfying $\int_{\Omega_1} |\mathcal{V}(x)|dx > 0$ and

$$cN^{-1/2-\alpha} \le |t_0(x, v) - t_0^\delta(x, v)| \le c^{-1}N^{-1/2-\alpha} \qquad (3.2)$$

for $N \ge N_0$ for all $x \in \Omega_1, v \in \mathcal{V}(x)$.
We now consider the rest of the trajectories after times $t_0$ and $t_0^\delta$. Since $F \in W^{1,\infty}$ for all $N$, uniqueness holds for the flow $(X, V)$. Therefore, for all $0 \le t \le t_0$,

$$X(t_0 + t, x, v) = X(t_0 - t, x, v), \\ V(t_0 + t, x, v) = -V(t_0 - t, x, v)$$

and similarly for $(X^\delta, V^\delta)$. In particular,

$$X(2t_0, x, v) = x, \quad \text{and} \quad X^\delta(2t_0^\delta, x, v) = x.$$

If $v \in \mathcal{V}(x)$, we have $v > \eta/4$, and since $F$ is bounded, $V(t, x, v)$ stays of order 1 when $t$ is at a distance $o(1)$ from $2t_0$. Therefore, it follows from (3.2) that

$$|X(2t_0^\delta, x, v) - x| \ge CN^{-1/2-\alpha},$$

for a sufficiently small constant $C$. Since $|\delta| = 1/N$, taking $T$ large enough, one then concludes that

$$Q_{\delta,N}(T) = \int_{x,v} \log \left( 1 + \frac{1}{|\delta|^2} \left( \sup_{0 \le t \le T} |X(t,x,v) - X^\delta(t,x,v)|^2 + \int_0^T |V(t,x,v) - V^\delta(t,x,v)|^2 dt \right) \right) dx \, dv$$

is bounded from below by

$$\int_{x \in \Omega_1} \int_{v \in \mathcal{V}(x)} \log \left( 1 + \frac{|X(2t_0^\delta, x, v) - X^\delta(2t_0^\delta, x, v)|^2}{|\delta|^2} \right) dx \, dv \\ \ge \int_{x \in \Omega_1} \int_{v \in \mathcal{V}(x)} \log (1 + C|\delta|^{2\alpha-1}) dx \, dv \ge C \log (1 + C|\delta|^{2\alpha-1}).$$

If $\alpha < 1/2$, a uniform control in $N$ of $Q_{\delta,N}$ as given by Proposition 2.1 (or Proposition 2.2) is impossible. This gives the required counter-example. The condition $\alpha \ge 1/2$ exactly corresponds to imposing $F \in W^{1/2,1}$.

# 4 Control of $Q_\delta(T)$ : Proof of Prop. 2.1

Recall the notation $\alpha$ for the Fourier transform of $F$. The assumption of Prop. 2.1 corresponds to the following bound:

$$\int_{\mathbb{R}^d} |k|^{\frac{3}{2}+2\alpha} |\alpha(k)|^2 dk = \|F\|_{H^{3/4+\alpha}(\Omega'')}^2 < +\infty.$$
## 4.1 Decomposition of $Q_\delta(T)$

Let

$$A_{\delta}(t, x, v) = |\delta|^2 + \sup_{0 \le s \le t} |X(s, x, v) - X^{\delta}(s, x, v)|^2 \\ + \int_{0}^{t} |V(s, x, v) - V^{\delta}(s, x, v)|^2 ds.$$

From (1.1), we compute

$$\frac{d}{dt} \log \left( 1 + \frac{1}{|\delta|^2} \left( \sup_{0 \le s \le t} |X(s, x, v) - X^{\delta}(s, x, v)|^2 \right. \right. \\ \left. \left. + \int_0^t |V(s, x, v) - V^{\delta}(s, x, v)|^2 ds \right) \right) \\ = \frac{2}{A_{\delta}(t, x, v)} \left( \frac{d}{dt} \left( \sup_{0 \le s \le t} |X(s, x, v) - X^{\delta}(s, x, v)|^2 \right) \right. \\ \left. + (V(t, x, v) - V^{\delta}(t, x, v)) \int_0^t (F(X(s, x, v)) - F(X^{\delta}(s, x, v))) ds \right).$$

Since, for any $f \in BV$,

$$\frac{d}{dt} \left( \max_{0 \le s \le t} f(s)^2 \right) \le 2|f(t)f'(t)| \le 4|f(t)|^2 + \frac{1}{2}|f'(t)|^2,$$

we deduce from the previous computation that

$$Q_{\delta}(T) \le \iint_{\Omega} \int_{0}^{T} \frac{4|X - X^{\delta}|^2 + |V - V^{\delta}|^2}{A_{\delta}(t, x, v)} dt \, dx \, dv + \tilde{Q}_{\delta}(T) \\ \le 4|\Omega|(1 + T) + \tilde{Q}_{\delta}(T) + \frac{1}{2} \iint_{\Omega} \int_{0}^{T} \frac{|V - V^{\delta}|^2}{A_{\delta}(t, x, v)} dt \, dx \, dv$$

where,

$$\tilde{Q}_{\delta}(T) = -2 \int_{0}^{T} \iint_{\Omega} \frac{V(t, x, v) - V^{\delta}(t, x, v)}{A_{\delta}(t, x, v)} \cdot \\ \int_{0}^{t} \int_{\mathbb{R}^{d}} \alpha(k) (e^{ik \cdot X(s,x,v)} - e^{ik \cdot X^{\delta}(s,x,v)}) \, dk \, ds \, dx \, dv \, dt.$$
Remark that

$$
\begin{align*}
\iint_{\Omega} \int_{0}^{T} \frac{|V - V^{\delta}|^2}{A_{\delta}(t, x, v)} dt \, dx \, dv &\leq \iint_{\Omega} \int_{0}^{T} \frac{\partial_t A_{\delta}(t, x, v)}{A_{\delta}(t, x, v)} dt \, dx \, dv \\
&\leq \iint_{\Omega} \log \left( \frac{A_{\delta}(T, x, v)}{|\delta|^2} \right) dx \, dv \\
&\leq Q_{\delta}(T).
\end{align*}
$$

Therefore, we have

$$
Q_{\delta}(T) \leq 8|\Omega|(1+T) + 2 \tilde{Q}_{\delta}(T),
$$

and it is enough to bound $\tilde{Q}_{\delta}(T)$.

We introduce a $C_b^\infty$ function $\chi: \mathbb{R}_+ \to [0, 1]$ such that $\chi(x) = 0$ if $x \le 1$ and $\chi(x) = 1$ if $x \ge 2$. Writing $X_t$ (resp. $V_t$) for $X(t, x, v)$ (resp. $V(t, x, v)$) and $X_t^\delta$ (resp. $V_t^\delta$) for $X^\delta(t, x, v)$ (resp. $V^\delta(t, x, v)$), and introducing

$$
\tilde{\alpha}(k) = \begin{cases} \alpha(k) & \text{if } |k| \ge (\log 1/|\delta|)^2 \\ 0 & \text{otherwise,} \end{cases} \quad (4.1)
$$

we may write

$$
\tilde{Q}_{\delta}(T) = \tilde{Q}_{\delta}^{(1)}(T) + \tilde{Q}_{\delta}^{(2)}(T) + \tilde{Q}_{\delta}^{(3)}(T) + \tilde{Q}^{(4)}(T),
$$

where

$$
\tilde{Q}_{\delta}^{(1)}(T) = -2 \int_{0}^{T} \iint_{\Omega} \int_{0}^{t} \chi \left( \frac{|X_s - X_s^{\delta}|}{|\delta|^{4/3}} \right) \frac{V_t - V_t^{\delta}}{A_{\delta}(t, x, v)} \cdot \\
\qquad \qquad \int_{\mathbb{R}^d} \tilde{\alpha}(k) \left( e^{ik \cdot X_s} - e^{ik \cdot X_s^{\delta}} \right) dk \, ds \, dx \, dv \, dt,
$$

$$
\tilde{Q}_{\delta}^{(2)}(T) = -2 \int_{0}^{T} \iint_{\Omega} \int_{0}^{t} \chi \left( \frac{|X_s - X_s^{\delta}|}{|\delta|^{4/3}} \right) \frac{V_t - V_t^{\delta}}{A_{\delta}(t, x, v)} \cdot \\
\qquad \qquad \int_{\mathbb{R}^d} (\alpha(k) - \tilde{\alpha}(k)) \left( e^{ik \cdot X_s} - e^{ik \cdot X_s^{\delta}} \right) dk \, ds \, dx \, dv \, dt,
$$

$$
\tilde{Q}_{\delta}^{(3)}(T) = -2 \int_{0}^{T} \iint_{\Omega} \int_{0}^{t} \left(1 - \chi\left(\frac{|X_s - X_s^{\delta}|}{|\delta|^{4/3}}\right)\right) \frac{V_t - V_t^{\delta}}{A_{\delta}(t, x, v)} \cdot \\
\qquad \qquad \int_{\{|k| \le |\delta|^{-4/3}\}} \alpha(k) \left(e^{ik \cdot X_s} - e^{ik \cdot X_s^{\delta}}\right) dk \, ds \, dx \, dv \, dt,
$$
and

$$ \tilde{Q}_{\delta}^{(4)}(T) = -2 \int_{0}^{T} \iint_{\Omega} \int_{0}^{t} \left(1 - \chi\left(\frac{|X_s - X_s^{\delta}|}{|\delta|^{4/3}}\right)\right) \frac{V_t - V_t^{\delta}}{A_{\delta}(t, x, v)} \cdot \\ \int_{\{|k| > |\delta|^{-4/3}\}} \alpha(k) \left(e^{ik \cdot X_s} - e^{ik \cdot X_s^{\delta}}\right) dk \, ds \, dx \, dv \, dt. $$

There are two ideas in the decomposition. The first is to classically separate the low frequency in Fourier from the high. As we wish to have a control on $\tilde{Q}_{\delta}$ in terms of power of $\log 1/|\delta|$, this leads to the cut-off in $\tilde{\alpha}$. The corresponding term with only low $k$ is $\tilde{Q}_{\delta}^{(2)}$ and will be bounded simply since it would lead to a force term having one full derivative of just the right order.

However it is also necessary in the proof to make sure that $|X_s - X_s^{\delta}|$ is not too small as this can create problems in the singular integrals that we will introduce. There is therefore another cut-off with $\chi(|X_s - X_s^{\delta}|/\delta^{4/3})$, $|\delta^{4/3}|$ being exactly the critical scale (this is of course directly connected with the $H^{3/4}$ regularity).

The corresponding term with very small $|X_s - X_s^{\delta}|$ may be easily bounded if the frequency $k$ is not too high (again by doing the usual proof and taking one full derivative of the regularized force term). Therefore we are doing a third cut-off in frequency for $|k| \le |\delta|^{-4/3}$, obtaining $\tilde{Q}_{\delta}^{(3)}$. The term $\tilde{Q}_{\delta}^{(4)}$ is the last remainder from this separation and can be controlled by the decay of $\alpha$ in $|k|$.

Finally the term $\tilde{Q}_{\delta}^{(1)}$ with all the cut-off will have to be bounded in a more subtle way and its control constitutes the heart of our proof in Subsection 4.4.

## 4.2 Control of $\tilde{Q}_{\delta}^{(4)}(T)$

Let us first state and prove a result that is used repeatedly in the sequel.

**Lemma 4.1** There exists a constant $C$ such that, for $|\delta|$ small enough,

$$ \int_s^T \frac{|V_t - V_t^\delta|}{\sqrt{A_\delta(t, x, v)}} dt \le C(\log 1/|\delta|)^{1/2}. $$
**Proof** Using Cauchy-Schwartz inequality,

$$
\begin{align*}
\int_s^T \frac{|V_t - V_t^\delta|}{\sqrt{A_\delta(t, x, v)}} dt &\le \int_s^T \frac{|V_t - V_t^\delta|}{\left(|\delta|^2 + \int_0^t |V_r - V_r^\delta|^2 dr\right)^{1/2}} dt \\
&\le \sqrt{T} \left( \int_s^T \frac{|V_t - V_t^\delta|^2}{|\delta|^2 + \int_0^t |V_r - V_r^\delta|^2 dr} dt \right)^{1/2} \\
&= \sqrt{T} \left( \log \left( \frac{|\delta|^2 + \int_0^T |V_r - V_r^\delta|^2 dr}{|\delta|^2 + \int_0^s |V_r - V_r^\delta|^2 dr} \right) \right)^{1/2} \\
&\le C \sqrt{T} (\log 1/|\delta|)^{1/2}
\end{align*}
$$

for $|\delta|$ small enough.

□

Let us define the function

$$
\tilde{F}(x) = \int_{\{|k| > |\delta|^{-4/3}\}} \alpha(k) e^{ik \cdot x} dx.
$$

Since $\sqrt{A_\delta(t, x, v)} \ge |\delta|$, we have

$$
\begin{align*}
|\tilde{Q}_{\delta}^{(4)}(T)| &\le C \int_0^T \iint_{\Omega} (|\tilde{F}(X_s)| + |\tilde{F}(X_s^{\delta})|) \\
&\qquad \times \int_s^T \frac{|V_t - V_t^{\delta}|}{|\delta|\sqrt{A_{\delta}(t, x, v)}} dt \, dx \, dv \, ds. \\
&\le C(\log 1/|\delta|)^{1/2} |\delta|^{-1} \int_0^T \iint_{\Omega'} |\tilde{F}(x)| \, dx \, dv \, ds \\
&\le C(\log 1/|\delta|)^{1/2} |\delta|^{-1} \left( \int_{\Omega_1'} |\tilde{F}(x)|^2 \, dx \right)^{1/2},
\end{align*}
$$

where the second inequality follows from Lemma 4.1 and from Property 1 applied to the change of variables $(x, v) = (X_s, V_s)$ and $(x, v) = (X_s^\delta, V_s^\delta)$.
Then, it follows from Plancherel's identity that

$$
\begin{align*}
|\tilde{Q}_{\delta}^{(4)}(T)| &\le C(\log 1/|\delta|)^{1/2} |\delta|^{-1} \left( \int_{\{|k|>|\delta|^{-4/3}\}} |\alpha(k)|^2 dk \right)^{1/2} \\
&\le C(\log 1/|\delta|)^{1/2} |\delta|^{4a/3} \left( \int_{\mathbb{R}^d} |k|^{\frac{3}{2}+2a} |\alpha(k)|^2 dk \right)^{1/2}.
\end{align*}
$$
### 4.3 Control of $\tilde{Q}_{\delta}^{(2)}(T)$ and $\tilde{Q}_{\delta}^{(3)}(T)$

We recall that the maximal function $Mf$ of $f \in L^p(\mathbb{R}^d)$, $1 \le p \le +\infty$, is defined by

$$Mf(x) = \sup_{r>0} \frac{C_d}{r^d} \int_{B(x,r)} f(z) dz, \quad \forall x \in \mathbb{R}^d.$$

We are going to use the following classical results. First (see [16] in the appendix), there exists a constant $C$ such that, for all $x, y \in \mathbb{R}^d$ and $f \in L^p(\mathbb{R}^d)$,

$$|f(x) - f(y)| \le C |x - y|(M|\nabla f|(x) + M|\nabla f|(y)). \quad (4.2)$$

Second (see [29, Ch. 1, Thm. 1]), for all $1 < p < \infty$, the operator $M$ is a continuous application from $L^p(\mathbb{R}^d)$ to itself and is sublinear, i.e. $M(f+g) \le Mf + Mg$.

We begin with the control of $\tilde{Q}_{\delta}^{(3)}(T)$. Let

$$\hat{F}(x) = \int_{\{|k| \le |\delta|^{-4/3}\}} \alpha(k) e^{ik \cdot x} dx.$$

It follows from the previous inequality that

$$\left| \int_{\{|k| \le |\delta|^{-4/3}\}} \alpha(k) (e^{ik \cdot X_s} - e^{ik \cdot X_s^\delta}) dk \right| = |\hat{F}(X_s) - \hat{F}(X_s^\delta)| \\ \le |X_s - X_s^\delta|(M|\nabla \hat{F}|(X_s) + M|\nabla \hat{F}|(X_s^\delta)).$$

Therefore, since $1 - \chi(x) = 0$ if $|x| \ge 2$, following the same steps as for the control of $\tilde{Q}_{\delta}^{(4)}(T)$,

$$\begin{align*}
|\tilde{Q}_{\delta}^{(3)}(T)| & \le C \int_0^T \iint_{\Omega} \int_s^T \frac{|V_t - V_t^\delta|}{|\delta|\sqrt{A_\delta(t, x, v)}} |\delta|^{4/3} \\
& \qquad (M|\nabla \hat{F}|(X_s) + M|\nabla \hat{F}|(X_s^\delta)) \, dt \, dx \, dv \, ds. \\
& \le C(\log 1/|\delta|)^{1/2} |\delta|^{1/3} \left( \int_{\Omega_1'} (M|\nabla \hat{F}|(x))^2 \, dx \right)^{1/2} \\
& \le C(\log 1/|\delta|)^{1/2} |\delta|^{1/3} \left( \int_{\Omega_1'} |\nabla \hat{F}|^2(x) \, dx \right)^{1/2}.
\end{align*}$$
Then

$$
\begin{aligned}
|\tilde{Q}_{\delta}^{(3)}(T)| & \le C(\log 1/|\delta|)^{1/2}|\delta|^{1/3} \left( \int_{\{|k| \le |\delta|^{-4/3}\}} |k|^2 |\alpha(k)|^2 dk \right)^{1/2} \\
& \le C(\log 1/|\delta|)^{1/2}|\delta|^{4a/3} \left( \int_{\mathbb{R}^d} |k|^{\frac{3}{2}+2a} |\alpha(k)|^2 dk \right)^{1/2}.
\end{aligned}
$$

The control of $\tilde{Q}_{\delta}^{(2)}(T)$ follows from a similar computation: introducing
$F_0(x) = \int_{\{k<(\log 1/|\delta|)^2\}} \alpha(k)e^{ik\cdot x} dx$, we obtain

$$
|\tilde{Q}_{\delta}^{(2)}(T)| \le C \int_{0}^{T} \iint_{\Omega} \int_{s}^{T} \frac{|V_t - V_t^{\delta}|}{\sqrt{A_{\delta}(t, x, v)}} \frac{|X_s - X_s^{\delta}|}{\sqrt{A_{\delta}(t, x, v)}} (M|\nabla F_0|(X_s) + M|\nabla F_0|(X_s^{\delta})) dt \, dx \, dv \, dt.
$$

Since $|X_s - X_s^\delta| \le \sqrt{A_\delta(t, x, v)}$ for all $s \le t$

$$
\begin{aligned}
|\tilde{Q}_{\delta}^{(2)}(T)| & \le C(\log 1/|\delta|)^{1/2} \int_{0}^{T} \left( \iint_{\Omega'} (M|\nabla F_0|(x))^2 dx \, dv \right)^{1/2} ds \\
& \le C(\log 1/|\delta|)^{1/2} \left( \int_{\{|k| < (\log 1/|\delta|)^2\}} |k|^2 |\alpha(k)|^2 dk \right)^{1/2} \\
& \le C(\log 1/|\delta|)^{1-2a} \left( \int_{\mathbb{R}^d} |k|^{\frac{3}{2}+2a} |\alpha(k)|^2 dk \right)^{1/2}.
\end{aligned}
$$

4.4 Control of $\tilde{Q}_{\delta}^{(1)}(T)$

The inequality (4.2) is insufficient to control $\tilde{Q}_{\delta}^{(1)}(T)$. Our estimate relies
on a more precise version of this inequality,

**Proposition 4.2** For any $\tilde{\alpha}(k)$, one has

$$
\begin{align}
& \int \tilde{\alpha}(k) (e^{ik \cdot X_s} - e^{ik \cdot X_s^\delta}) \, dk \nonumber \\
&= \int_{\mathbb{R}^d} \tilde{\alpha}(k) \int_{\mathbb{R}^d} \frac{k}{|z|^{d-1}} \cdot \psi \left( \frac{z}{|z|}, \frac{X_s - X_s^\delta}{|X_s - X_s^\delta|}, \frac{|z|}{|X_s - X_s^\delta|} \right) e^{ik \cdot (X_s^\delta + z)} \, dz \, dk \nonumber \\
&\quad - \int_{\mathbb{R}^d} \tilde{\alpha}(k) \int_{\mathbb{R}^d} \frac{k}{|z|^{d-1}} \cdot \psi \left( \frac{z}{|z|}, \frac{X_s^\delta - X_s}{|X_s^\delta - X_s|}, \frac{|z|}{|X_s^\delta - X_s|} \right) e^{ik \cdot (X_s + z)} \, dz \, dk, \tag{4.3}
\end{align}
$$
<figure><img src="image_3.png" /><figcaption>Figure 2: The graph of $\theta \mapsto X^{\theta,h}$</figcaption></figure>

where the function $\psi$ belongs to $C_b^{0,\infty,\infty}(S^{d-1}, S^{d-1}, \mathbb{R} \setminus \{0\})$ and has support in

$$ \{(u, v) \in (S^{d-1})^2 : \cos(u, v) \ge 17^{-1/2}\} \times [0, 3/4], \quad (4.4) $$

where $(u, v)$ denotes the angle between the vectors $u$ and $v$.

Note that, since $|z| \le |X_s^\delta - X_s|$ when $\psi \ne 0$ in both terms of the right-hand side of (4.3), this proposition has as consequence the control by the maximal function of the derivatives (4.2). Our case requires the stronger result of Prop. 4.2 since we need to keep all the cancellations and not simply take the absolute values.

### 4.4.1 First part of the proof of Prop. 4.2.

For any $\theta \in [0, 1]$ and $h \in \mathbb{R}^d$, we define

$$ X^{\theta,h}(t, x, v) = \theta X(t, x, v) + (1-\theta)X^{\delta}(t, x, v) + (1-(2\theta-1)^2)h, $$

and we write for simplicity $X_t^{\theta,h}$ for $X^{\theta,h}(t, x, v)$.

For any fixed $h \in \mathbb{R}^d$, by differentiation in $\theta$

$$ \int_{\mathbb{R}^d} \tilde{\alpha}(k) (e^{ik \cdot X_s} - e^{ik \cdot X_s^\delta}) dk = \int_{\mathbb{R}^d} \tilde{\alpha}(k) \int_0^1 e^{ik \cdot X_s^{\theta,h}} k \cdot (X_s - X_s^\delta + 4(1-2\theta)h) d\theta dk. \quad (4.5) $$

For any $x, y \in \mathbb{R}^d$, we introduce the hyperplane orthogonal to $x - y$

$$ H(x, y) = \{h \in \mathbb{R}^d : h \cdot (x - y) = 0\}. $$

If $x = y$, we define for example $H(x, y) = H(0, e_1)$, where $e_1 = (1, 0, \dots, 0)$.

Fix a $C_b^\infty$ function $\tilde{\psi} : \mathbb{R}_+ \to \mathbb{R}_+$ such that $\tilde{\psi}(x) = 0$ for $x > 1$ and
$\int_{H(0,e_1)} \tilde{\psi}(|h|) dh = 1$. By invariance of $|h|$ with respect to rotations, we also have

$$ \int_{H(x,y)} \tilde{\psi}(|h|) dh = 1 $$

for all $x, y \in \mathbb{R}^d$.

Since the left-hand side of (4.5) does not depend on $h$, we have

$$ \int_{\mathbb{R}^d} \tilde{\alpha}(k) (e^{ik \cdot X_s} - e^{ik \cdot X_s^\delta}) dk \\ = \frac{1}{|X - X^\delta|^{d-1}} \int_{H(X_s, X_s^\delta)} \tilde{\psi} \left( \frac{|h|}{|X - X^\delta|} \right) \int_{\mathbb{R}^d} \tilde{\alpha}(k) \\ \int_0^1 e^{ik \cdot X_s^{\theta,h}} k \cdot (X_s - X_s^\delta + 4(1 - 2\theta)h) d\theta dk dh $$

in the case where $X_s \neq X_s^\delta$. If $X_s = X_s^\delta$, the previous quantity is 0.

Let $\rho: [0, 1] \to \mathbb{R}_+$ be a $C_b^\infty$ function such that $\rho(x) = 1$ for $0 \le x \le 1/4$, $\rho(x) = 0$ for $3/4 \le x \le 1$ and $\rho(x) + \rho(1-x) = 1$ for $0 \le x \le 1$. Then, one has

$$ \int_{\mathbb{R}^d} \tilde{\alpha}(k) (e^{ik \cdot X_s} - e^{ik \cdot X_s^\delta}) dk = B_\delta(s, x, v) + C_\delta(s, x, v), $$

where

$$ B_{\delta}(s, x, v) = \frac{1}{|X_s - X_s^{\delta}|^{d-1}} \int_{H(X_s, X_s^{\delta})} \tilde{\psi} \left( \frac{|h|}{|X_s - X_s^{\delta}|} \right) \int_{\mathbb{R}^d} \tilde{\alpha}(k) \\ \int_0^1 \rho(\theta) e^{ik \cdot X_s^{\theta,h}} k \cdot (X_s - X_s^{\delta} + 4(1 - 2\theta)h) d\theta dk dh \quad (4.6) $$

and

$$ C_{\delta}(s, x, v) = \frac{1}{|X_s - X_s^{\delta}|^{d-1}} \int_{H(X_s, X_s^{\delta})} \tilde{\psi} \left( \frac{|h|}{|X_s - X_s^{\delta}|} \right) \int_{\mathbb{R}^d} \tilde{\alpha}(k) \\ \int_0^1 \rho(1-\theta) e^{ik \cdot X_s^{\theta,h}} k \cdot (X_s - X_s^{\delta} + 4(1-2\theta)h) d\theta dk dh. \quad (4.7) $$

We focus on $B_\delta(s, x, v)$ as by symmetry between $X$ and $X^\delta$, $C_\delta$ is dealt with in exactly the same manner.
<figure><img src="image_4.png" /><figcaption>Figure 3: The set $K(x)$</figcaption></figure>

### 4.4.2 Second part for Prop. 4.2 : Change of variable $z = X_s^{\theta,h}$

For any $x \in \mathbb{R}^d$, we introduce

$$ K(x) = \{y \in \mathbb{R}^d : \exists \theta \in [0,1], h \in H(x,0) \text{ s.t. } |h| \le |x| \\ \text{and } y = \theta(x + 4(1-\theta)h)\}. \quad (4.8) $$

Observing that

$$ \theta = \frac{y}{|x|} \cdot \frac{x}{|x|}, $$

this set may also be defined as

$$ K(x) = \left\{ y \in \mathbb{R}^d : \frac{y}{|x|} \cdot \frac{x}{|x|} \in [0,1] \right. \\ \left. \text{and} \quad \left| \frac{y}{|x|} - \left( \frac{y}{|x|} \cdot \frac{x}{|x|} \right) \frac{x}{|x|} \right| \le 4 \frac{y}{|x|} \cdot \frac{x}{|x|} \left( 1 - \frac{y}{|x|} \cdot \frac{x}{|x|} \right) \right\}. $$

Note that, for any $y \in K(x)$, taking $\theta$ and $h$ as in (4.8), we have $|y|^2 = \theta^2(|x|^2 + 16(1-\theta^2)|h|^2) \le 17\theta^2|x|^2$. Therefore,

$$ \cos(x, y) = \frac{x}{|x|} \cdot \frac{y}{|y|} = \frac{\theta|x|}{|y|} \ge 17^{-1/2}. \quad (4.9) $$

For fixed $x, y \in \mathbb{R}^d$, we now introduce the application

$$ F_{x,y} : [0,1] \times \{h \in H(x,y) : |h| \le |y-x|\} \to K(x-y) \\ (\theta,h) \mapsto \theta(x-y+4(1-\theta)h). $$
It is elementary to check that $F_{x,y}$ is a bijection when $x \neq y$, with inverse

$$F_{x,y}^{-1}(z) = \left( \frac{z}{|x-y|} \cdot \frac{x-y}{|x-y|}, \frac{z - \left( \frac{z}{|x-y|} \cdot \frac{x-y}{|x-y|} \right) (x-y)}{4 \frac{z}{|x-y|} \cdot \frac{x-y}{|x-y|} \left( 1 - \frac{z}{|x-y|} \cdot \frac{x-y}{|x-y|} \right)} \right)$$

for $z \in K(x-y)$. Moreover, $F_{x,y}$ is differentiable and its differential, written in an orthonormal basis of $\mathbb{R}^d$ with first vector $(x-y)/|x-y|$, is

$$\nabla F_{x,y}(\theta, h) = \begin{pmatrix} |x-y| & 4(1-2\theta)h \\ 0 & 4\theta(1-\theta)\mathrm{Id} \end{pmatrix}.$$

Therefore, the Jacobian of $F_{x,y}$ at $(\theta, h)$ is $(4\theta(1-\theta))^{d-1}|x-y|$.

Making the change of variable $z = F_{X_s, X_s^\delta}(\theta, h)$ in (4.6), we can now compute

$$\begin{align}
B_{\delta}(s, x, v) &= \int_{\mathbb{R}^d} \tilde{\alpha}(k) \int_0^1 \int_{H(X_s, X_s^\delta)} \frac{\rho(\theta) \tilde{\psi}\left(\frac{|h|}{|X_s - X_s^\delta|}\right)}{|X_s - X_s^\delta|^d (4\theta(1-\theta))^{d-1}} e^{ik \cdot X_s^{\theta, h}} \nonumber \\
&\qquad k \cdot (X_s - X_s^\delta + 4(1-\theta)h - 4\theta h) (4\theta(1-\theta))^{d-1} |X_s - X_s^\delta| \, dh \, d\theta \, dk \tag{4.10} \\
&= B'_{\delta}(s, x, v) - B''_{\delta}(s, x, v), \nonumber
\end{align}$$

with

$$B'_{\delta}(s, x, v) = \int_{\mathbb{R}^d} \tilde{\alpha}(k) \int_{\mathbb{R}^d} \frac{k \cdot z}{|z|^d} \psi^{(1)} \left( \frac{z}{|z|}, \frac{X_s - X_s^\delta}{|X_s - X_s^\delta|}, \frac{|z|}{|X_s - X_s^\delta|} \right) e^{ik \cdot (X_s^\delta + z)} \, dz \, dk,$$

and

$$B_{\delta}''(s, x, v) = - \int_{\mathbb{R}^d} \tilde{\alpha}(k) \int_{\mathbb{R}^d} \frac{k}{|z|^{d-1}} \cdot \psi^{(2)} \left( \frac{z}{|z|}, \frac{X_s - X_s^\delta}{|X_s - X_s^\delta|}, \frac{|z|}{|X_s - X_s^\delta|} \right) e^{ik \cdot (X_s^\delta + z)} \, dz \, dk.$$

We defined, for $(a, b, c) \in S^{d-1} \times S^{d-1} \times (\mathbb{R} \setminus \{0\})$,

$$\psi^{(1)}(a, b, c) = \frac{\tilde{\rho}((a \cdot b)c) \tilde{\psi} \left( \frac{|a - (a \cdot b)b|}{4(a \cdot b)(1 - (a \cdot b)c)} \right)}{4^{d-1} (a \cdot b)^d (1 - (a \cdot b)c)^{d-1}}$$
and

$$ \psi^{(2)}(a, b, c) = \frac{\tilde{\rho}((a \cdot b)c)\tilde{\psi}\left(\frac{|a-(a \cdot b)b|}{4(a \cdot b)(1-(a \cdot b)c)}\right)}{4^{d-1}(a \cdot b)^{d-1}(1-(a \cdot b)c)^d} c(a - (a \cdot b)b), $$

where $\tilde{\rho}(x) = \rho(x)$ if $x \in [0, 1]$, and $\tilde{\rho}(x) = 0$ otherwise.

It follows from (4.9) and from the definition of $\rho$ that these two functions have support in (4.4). Moreover, they belong to $C_b^{0,\infty,\infty}(S^{d-1}, S^{d-1}, \mathbb{R}\setminus\{0\})$. Indeed, since $\tilde{\rho}(x) = 0$ for $x \ge 3/4$, the terms $(1-(a \cdot b)c)$ in the denominators do not cause any regularity problem. Moreover, since $\tilde{\psi}(x) = 0$ for $x > 1$ and

$$ \frac{|a - (a \cdot b)b|}{|a \cdot b|} \ge \frac{1}{|a \cdot b|} - 1 $$

for all $a, b \in S^{d-1}$, the terms $a \cdot b$ in the denominators do not cause any worry either. Finally, since $\tilde{\rho} \in C_b^\infty(\mathbb{R} \setminus \{0\})$, the discontinuity of $\tilde{\rho}$ at 0 can only cause a problem in the neighborhood of points such that $a \cdot b = 0$ ($c$ being nonzero). Therefore, the previous observation also solves this difficulty.

This concludes the proof of Prop. 4.2 by putting $\psi = \frac{z}{|z|} \psi^{(1)} - \psi^{(2)}$.

### 4.4.3 Averaging along the trajectory : integration by parts in time

With the obvious symmetries in the problem we focus on only one term coming from Prop. 4.2 and in particular, with the previous notation,

$$ B_{\delta}(s, x, v) = \int_{\mathbb{R}^d} \int_{\mathbb{R}^d} \tilde{\alpha}(k) \frac{e^{ik \cdot (X_s^{\delta} + z)}}{|z|^d} k \cdot \psi_s \, dk \, dz, $$

writing $\psi_t$ for

$$ \psi\left(\frac{z}{|z|}, \frac{X_t - X_t^\delta}{|X_t - X_t^\delta|}, \frac{|z|}{|X_t - X_t^\delta|}\right). \quad (4.11) $$

The aim now is to prove

**Lemma 4.3** One has for some numerical constant $C$

$$ \int_0^T \int_\Omega \int_{\mathbb{R}^d} \frac{V_t - V_t^\delta}{A_\delta(t, x, v)} \cdot B_\delta(s, x, v) \, dx \, dv \, dt \\ \le C (\log 1/|\delta|)^{1-2a} \left( \int_{\mathbb{R}^d} |k|^{3/2+2a} |\alpha(k)|^2 \, dk \right)^{1/2}. $$
For this it is necessary to gain some regularity by integrating along the trajectories and accordingly, we decompose $B_{\delta}(s, x, v)$

$$B_{\delta}(s, x, v) = \int_{\mathbb{R}^d} \int_{\mathbb{R}^d} \tilde{\alpha}(k) \frac{e^{ik \cdot (X_s^{\delta} + z)} k \cdot \psi_s}{|z|^{d-1}} \frac{i \frac{k}{|k|} \cdot V_s^{\delta}}{|k|^{-1/2} + i \frac{k}{|k|} \cdot V_s^{\delta}} dk dz \\ + \int_{\mathbb{R}^d} \int_{\mathbb{R}^d} \tilde{\alpha}(k) \frac{e^{ik \cdot (X_s^{\delta} + z)} k \cdot \psi_s}{|z|^{d-1}} \frac{|k|^{-1/2}}{|k|^{-1/2} + i \frac{k}{|k|} \cdot V_s^{\delta}} dk dz \\ =: B_{\delta}^{1}(s, x, v) + B_{\delta}^{2}(s, x, v).$$

Now, let us write $\chi_s$ for

$$\chi\left(\frac{|X_s - X_s^{\delta}|}{|\delta|^{4/3}}\right), \qquad (4.12)$$

and let us define similarly as in (4.11) and (4.12) the notation $\nabla_2 \psi_s$ (which is a Jacobian matrix), $\nabla_3 \psi_s$ (which is a gradient vector) and $\chi'_s$.

The whole point in this somewhat artificial decomposition is that the term $i \frac{k}{|k|} \cdot V_s^{\delta} e^{ik \cdot (X_s^{\delta} + z)}$ is exactly the time derivative of $\frac{1}{|k|} e^{ik \cdot (X_s^{\delta} + z)}$. So integrating by parts in time, we obtain

$$\int_{0}^{t} \chi_{s} B_{\delta}^{1}(s, x, v) ds = I(t, x, v) - II(t, x, v) - III(t, x, v) - IV(t, x, v) - V(t, x, v),$$

with

$$I(t, x, v) = \int_{\mathbb{R}^d} \int_{\mathbb{R}^d} \tilde{\alpha}(k) \frac{k \cdot \psi_t}{|k| |z|^{d-1}} \frac{\chi_t e^{ik \cdot (X_t^{\delta} + z)}}{|k|^{-1/2} + i \frac{k}{|k|} \cdot V_t^{\delta}} dk dz,$$

$$\mathrm{II}(t, x, v) = \int_{\mathbb{R}^d} \int_{\mathbb{R}^d} \tilde{\alpha}(k) \frac{k \cdot \psi_0}{|k| |z|^{d-1}} \frac{\chi_0 e^{ik \cdot (x+\delta_1+z)}}{|k|^{-1/2} + i \frac{k}{|k|} \cdot (v+\delta_2)} dk dz,$$

$$\mathrm{III}(t, x, v) = \int_{0}^{t} \int_{\mathbb{R}^d} \int_{\mathbb{R}^d} \tilde{\alpha}(k) \frac{k \cdot \psi_s}{|k| |z|^{d-1}} \frac{\chi'_s e^{ik \cdot (X_s^{\delta} + z)}}{|k|^{-1/2} + i \frac{k}{|k|} \cdot V_s^{\delta}} \frac{(X_s - X_s^{\delta}) \cdot (V_s - V_s^{\delta})}{|\delta|^{4/3} |X_s - X_s^{\delta}|} dk dz ds,$$

correspondingly

$$\mathrm{IV}(t, x, v) = \int_{0}^{t} \int_{\mathbb{R}^d} \int_{\mathbb{R}^d} \tilde{\alpha}(k) \frac{k}{|k| |z|^{d-1}} \frac{\chi_s e^{ik \cdot (X_s^{\delta} + z)}}{|k|^{-1/2} + i \frac{k}{|k|} \cdot V_s^{\delta}} \\ \cdot \left[ -\nabla_3 \psi_s \frac{|z|}{|X_s - X_s^{\delta}|^3} (X_s - X_s^{\delta}) \cdot (V_s - V_s^{\delta}) \right. \\ \left. + \nabla_2 \psi_s \left( \frac{V_s - V_s^{\delta}}{|X_s - X_s^{\delta}|} - \frac{X_s - X_s^{\delta}}{|X_s - X_s^{\delta}|^3} (X_s - X_s^{\delta}) \cdot (V_s - V_s^{\delta}) \right) \right] dk dz ds,$$
and

$$V(t, x, v) = \int_0^t \int_{\mathbb{R}^{2d}} \tilde{\alpha}(k) \frac{k \cdot \psi_s}{|k| |z|^{d-1}} \frac{\chi_s e^{ik \cdot (X_s^\delta + z)}}{\left(|k|^{-1/2} + i \frac{k}{|k|} \cdot V_s^\delta\right)^2} i \frac{k}{|k|} \cdot F(X_s^\delta) dk dz ds.$$

Let us define

$$I(T) = \int_0^T \iint_{\Omega} \frac{V_t - V_t^\delta}{A_\delta(t, x, v)} \cdot I(t, x, v) \, dx \, dv \, dt,$$

and $II(T)$, $III(T)$, $IV(T)$ and $V(T)$ similarly.

We are going to bound each of these terms, the last one being the more worrying as it has the larger power in the denominator ($|k|^{-1/2} + i \frac{k}{|k|} \cdot V_s^\delta$). Hence Lemma 4.3 is implied by the following two

**Lemma 4.4 One has**

$$|I(T)| + |II(T)| + |III(T)| + |IV(T)| + |V(T)| \\ \le C (\log 1/|\delta|)^{1-2a} \left( \int_{\mathbb{R}^d} |k|^{3/2+2a} |\alpha(k)|^2 dk \right)^{1/2}.$$

And a control on $B_\delta^2$ is also needed

**Lemma 4.5 One has**

$$\int_0^T \iint_{\Omega} \frac{V_t - V_t^\delta}{A_\delta(t, x, v)} \cdot \int_0^t \chi_s B_\delta^2(s, x, v) \, ds \, dx \, dv \, dt \\ \le C (\log 1/|\delta|)^{1-2a} \left( \int_{\mathbb{R}^d} |k|^{3/2+2a} |\alpha(k)|^2 dk \right)^{1/2}.$$

#### 4.4.4 Proof of Lemma 4.4: Upper bound for $|V(T)|$

First, we make the change of variables $z' = z + X_s^\delta$, followed by the change of variable $(x', v') = (X_s^\delta, V_s^\delta)$ in the integral defining $V(T)$. When $(x, v) \in \Omega$, the variable $(x', v')$ belongs to the set $\Omega_s = \{(X_s^\delta(s, x, v), V_s^\delta(s, x, v)), (x, v) \in \Omega\}$. Recall that, by assumption, either $(X_s^\delta, V_s^\delta)$ satisfy (1.4), or

$$(X'(t, x, v), V'(t, x, v)) = (X^\delta(t, x - \delta_1, v - \delta_2), V^\delta(t, x - \delta_1, v - \delta_2)) \quad (4.13)$$
satisfy (1.4). In the second case, one has $X'(-s, x', v') = x + \delta_1$ and
$V'(-s, x', v') = v + \delta_2$. So for example

$$
\begin{align*}
X_t^\delta((X_s^\delta, V_s^\delta)^{-1}(x', v')) &= X_t'((X_s^\delta, V_s^\delta)^{-1}(x', v') + \delta) \\
&= X_t'((X_s^\delta, V_s^\delta)^{-1}(x', v') - \delta + \delta) = X_{t-s}'(x', v').
\end{align*}
$$

Writing for convenience $x, v, z$ instead of $x', v', z'$, it follows from these
changes of variables and from Property 1 that

$$
V(T) = \int_0^T \int_0^T \iint_{\Omega_s} \int_{\mathbb{R}^d} \int_{\mathbb{R}^d} \tilde{\chi}_s \frac{\tilde{V}_{t,s}^\delta - V_{t-s}'}{|\delta|^2 + \sup_{0 \le r \le t} |\tilde{X}_{r,s}^\delta - X_{r-s}^\prime|^2 + \int_0^t |\tilde{V}_{r,s}^\delta - V_{r-s}^\prime|^2 dr} \cdot \tilde{\alpha}(k) \frac{k \cdot \tilde{\psi}_s}{|k| |z-x|^{d-1}} e^{ik \cdot z} \frac{i^{\frac{k}{|k|}} \cdot F(x)}{\left(|k|^{-1/2} + i^{\frac{k}{|k|}} \cdot v\right)^2} dk dz dx dv ds dt,
$$

where

$$
\begin{align}
\tilde{X}_{t,s}^{\delta} &= X(t, X'(-s, x, v) - \delta_1, V'(-s, x, v) - \delta_2), \tag{4.14} \\
\tilde{V}_{t,s}^{\delta} &= V(t, X'(-s, x, v) - \delta_1, V'(-s, x, v) - \delta_2), \nonumber \\
\tilde{\psi}_s &= \psi \left( \frac{z-x}{|z-x|}, \frac{\tilde{X}_{s,s}^{\delta} - x}{|\tilde{X}_{s,s}^{\delta} - x|}, \frac{|z-x|}{|\tilde{X}_{s,s}^{\delta} - x|} \right). \nonumber
\end{align}
$$

and

$$
\tilde{\chi}_s = \chi \left( \frac{|\tilde{X}_{s,s}^{\delta} - x|}{|\delta|^{4/3}} \right).
$$

In the case where $(X^\delta, V^\delta)$ satisfy (1.4), the same formula holds but setting
$\delta = 0$ in the equations (4.13) and (4.14).

Writing the tensor (remember that $\alpha(k) \in \mathbb{R}^d$)

$$
G_V(v, z) = \int_{\mathbb{R}^d} \frac{k \otimes k}{|k|^2} \otimes \frac{\tilde{\alpha}(k) e^{ik \cdot z}}{\left(|k|^{-1/2} + i \frac{k}{|k|} \cdot v\right)^2} dk,
$$

and reminding that $\Omega_s \subset \Omega'$ for all $s \in [0, T]$, we have

$$
|V(T)| \le C \int_0^T \iint_{\Omega'} \int_0^T \int_{\mathbb{R}^d} \frac{\tilde{\chi}_s |\tilde{\psi}_s| |F(x)| \|G_V(v, z)\|}{|z-x|^{d-1} (|\delta| + |\tilde{X}_{s,s}^\delta - x|)} \frac{|\tilde{V}_{t,s}^\delta - V_{t-s}|}{(|\delta|^2 + \int_0^t |\tilde{V}_{r,s}^\delta - V_{r-s}|^2 dr)^{1/2}} dz \,ds \,dx \,dv \,dt,
$$
where $\|a\|^2 = \sum_{i,j,k=1}^d a_{ijk}^2$ for any tensor with three entries $a = (a_{ijk})$ with $1 \le i, j, k \le d$. So

$$|V(T)| \le C \int_0^T \iint_{\Omega'} \int_{\mathbb{R}^d} \frac{|\tilde{\psi}_s| \|G_V(v, z)\|}{|z-x|^{d-1} (|\delta| + |\tilde{X}_{s,s}^\delta - x|)} \\ \int_s^T \frac{|\tilde{V}_{t,s}^\delta - V_{t-s}|}{(|\delta|^2 + \int_0^t |\tilde{V}_{r,s}^\delta - V_{r-s}|^2 dr)^{1/2}} dt dz dx dv ds.$$

Now, on the one hand, following the same computation as in Lemma 4.1, the integral with respect to $t$ can be upper bounded by $C(\log 1/|\delta|)^{1/2}$ for $|\delta|$ small enough. On the other hand,

$$\iint_{\Omega'} \int_{\mathbb{R}^d} \frac{|\tilde{\psi}_s| \|G_V(v, z)\|}{|z-x|^{d-1} (|\delta| + |\tilde{X}_{s,s}^\delta - x|)} dz \, dx \, dv \\ \le \left( \iint_{\Omega'} \int_{\mathbb{R}^d} \frac{|\tilde{\psi}_s|}{|z-x|^{d-1} (|\delta| + |\tilde{X}_{s,s}^\delta - x|)} dz \, dx \, dv \right)^{1/2} \\ \left( \int_{\Omega_2'} \int_{\mathbb{R}^d} \int_{\Omega_1} \frac{|\tilde{\psi}_s| \|G_V(v, z)\|^2}{|z-x|^{d-1} (|\delta| + |\tilde{X}_{s,s}^\delta - x|)} dx \, dz \, dv \right)^{1/2},$$

and, because of the properties of the support of $\tilde{\psi}$, this last term is bounded by

$$C \left( \iint_{\Omega'} \frac{1}{|\tilde{X}_{s,s}^\delta - x|} \int_{x+K(\tilde{X}_{s,s}^\delta - x)} \frac{dz}{|z-x|^{d-1}} \, dx \, dv \right)^{1/2} \\ \left( \int_{\Omega_2'} \int_{\mathbb{R}^d} \|G_V(v, z)\|^2 \int_{\Omega_1'} \frac{dx}{|z-x|^{d-1} (|\delta| + |z-x|)} \, dz \, dv \right)^{1/2} \\ \le C(\log 1/|\delta|)^{1/2} \left( \int_{\Omega_2'} \int_{\mathbb{R}^d} \|G_V(v, z)\|^2 \, dz \, dv \right)^{1/2}, \quad (4.15)$$

where we have used that, for any $z \in x+K(\tilde{X}_{s,s}^\delta - x)$, $|z-x| \le |\tilde{X}_{s,s}^\delta - x|$, and where the last inequality can be obtained by a spherical coordinate change of variable centered at $x$ in the variable $z$ in the first term, and centered at $z$ in the variable $x$ in the second term.
Now,

$$
\begin{aligned}
& \int_{\Omega'_2} \int_{\mathbb{R}^d} \|G_V(v, z)\|^2 dz \, dv \\
&= \int_{\Omega'_2} \int_{\mathbb{R}^d} \sum_{i,j,n=1}^d \int_{\mathbb{R}^d} \int_{\mathbb{R}^d} \frac{k_i l_i k_j l_j}{|k|^2 |l|^2} \frac{\tilde{\alpha}_n(k) \overline{\tilde{\alpha}_n(l)} e^{iz \cdot (k-l)}}{\left(|k|^{-1/2} + i \frac{k}{|k|} \cdot v\right)^2} \\
& \qquad \left(|l|^{-1/2} - i \frac{l}{|l|} \cdot v\right)^{-2} dl \, dk \, dz \, dv,
\end{aligned}
$$

and integrating first in $z$ and $l$, this is equal to

$$ \int_{\Omega'_2} \sum_{i,j}^{d} \int_{\mathbb{R}^d} \frac{k_i^2 k_j^2}{|k|^4} \frac{|\tilde{\alpha}(k)|^2}{|k|^{-1/2} + i \frac{k}{|k|} \cdot v} \, dk \, dv. $$

Therefore

$$
\begin{aligned}
\int_{\Omega'_2} \int_{\mathbb{R}^d} \|G_V(v, z)\|^2 dz \, dv &\le C \int_{\mathbb{R}^d} \int_{\Omega'_2} \frac{|\tilde{\alpha}(k)|^2}{\left(\frac{1}{|k|} + \left(\frac{k \cdot v}{|k|}\right)^2\right)^2} \, dv \, dk \\
&\le C \int_{\mathbb{R}^d} |k|^2 |\tilde{\alpha}(k)|^2 \int_{-\infty}^{+\infty} \frac{dv_1}{(1 + |k|v_1^2)^2} \, dk,
\end{aligned}
$$

where we write the vector $v$ as $(v_1, \dots, v_d)$ in an orthonormal basis of $\mathbb{R}^d$ with first vector $k/|k|$. In conclusion, using the definition (4.1) of $\tilde{\alpha}$,

$$
\begin{aligned}
\int_{\Omega'_2} \int_{\mathbb{R}^d} \|G_V(v, z)\|^2 dz \, dv &\le C \int_{\{|k|>(\log 1/|\delta|)^2\}} |k|^{3/2} |\alpha(k)|^2 dk \\
&\le C (\log 1/|\delta|)^{-4a} \int_{\mathbb{R}^d} |k|^{\frac{3}{2}+2a} |\alpha(k)|^2 dk, \quad (4.16)
\end{aligned}
$$

Combining this inequality with (4.15), we finally get

$$ |V(T)| \le C (\log 1/|\delta|)^{1-2a} \left( \int_{\mathbb{R}^d} |k|^{\frac{3}{2}+2a} |\alpha(k)|^2 dk \right)^{1/2}. $$
### 4.4.5 Proof of Lemma 4.4: Upper bound for $|IV(T)|$

Applying to $IV(T)$ the same change of variable as we did for $V(T)$, we have

$$|IV(T)| \le C \int_0^T \iint_{\Omega'} \int_0^t \int_{\mathbb{R}^d} \frac{|\tilde{V}_{s,s}^\delta - v| |\tilde{V}_{t,s}^\delta - V_{t-s}|}{|\delta|^2 + \int_0^t |\tilde{V}_{r,s}^\delta - V_{r-s}|^2 dr} \frac{\tilde{\chi}_s |\hat{\psi}_s| \|G_{IV}(v, z)\|}{|z-x|^{d-1} |\tilde{X}_{s,s}^\delta - x|} dz \, ds \, dv \, dt$$

where $\|a\|^2 = \sum_{i,j=1}^d a_{ij}^2$ for any matrix $a = (a_{ij})_{1\le i,j \le d}$,

$$G_{IV}(v, z) = \int_{\mathbb{R}^d} \frac{k}{|k|} \otimes \frac{\tilde{\alpha}(k) e^{ik \cdot z}}{|k|^{-1/2} + i \frac{k}{|k|} \cdot v} dk$$

and

$$\hat{\psi}_s = -\nabla_3 \psi \left( \frac{z-x}{|z-x|}, \frac{\tilde{X}_{s,s}^\delta - x}{|\tilde{X}_{s,s}^\delta - x|}, \frac{|z-x|}{|\tilde{X}_{s,s}^\delta - x|} \right) \otimes \frac{(\tilde{X}_{s,s}^\delta - x) |z|}{|\tilde{X}_{s,s}^\delta - x|^2} \\ - \left( \nabla_2 \psi \left( \frac{z-x}{|z-x|}, \frac{\tilde{X}_{s,s}^\delta - x}{|\tilde{X}_{s,s}^\delta - x|}, \frac{|z-x|}{|\tilde{X}_{s,s}^\delta - x|} \right) \frac{\tilde{X}_{s,s}^\delta - x}{|\tilde{X}_{s,s}^\delta - x|} \right) \otimes \frac{\tilde{X}_{s,s}^\delta - x}{|\tilde{X}_{s,s}^\delta - x|} \\ + \nabla_2 \psi \left( \frac{z-x}{|z-x|}, \frac{\tilde{X}_{s,s}^\delta - x}{|\tilde{X}_{s,s}^\delta - x|}, \frac{|z-x|}{|\tilde{X}_{s,s}^\delta - x|} \right).$$

Note that, because of the properties of $\psi$ in Prop. 4.2,

$$|\hat{\psi}_s| \le C \mathbb{I}_{\{z-x \in K(\tilde{X}_{s,s}^\delta - x)\}}$$

for some constant $C$.

Then, following a similar computation as the one leading to (4.15),

$$|IV(T)| \le C \left( \int_{\Omega_2'} \int_{\mathbb{R}^d} \|G_{IV}(v, z)\|^2 \int_{\Omega_1'} \frac{dx \, dz \, dv}{|z-x|^{d-1} (|\delta|^{4/3} + |z-x|)} \right)^{1/2} \\ \left( \iint_{\Omega'} \int_0^T \int_0^t \frac{|\tilde{V}_{s,s}^\delta - v|^2 |\tilde{V}_{t,s}^\delta - V_{t-s}|^2}{|\delta|^4 + \left( \int_0^t |\tilde{V}_{r,s}^\delta - V_{r-s}|^2 dr \right)^2} \frac{1}{|\tilde{X}_{s,s}^\delta - x|} \int_{x+K(\tilde{X}_{s,s}^\delta - x)} \frac{dz}{|z-x|^{d-1}} ds \, dt \, dx \, dv \right)^{1/2}.$$
Hence

$$ |\text{IV}(T)| \le C(\log 1/|\delta|)^{1/2} \left( \int_{\Omega'_2} \int_{\mathbb{R}^d} \|G_{\text{IV}}(v, z)\|^2 \right)^{1/2} \left( \iint_{\Omega'} \int_0^T \int_0^t \frac{|\tilde{V}_{s,s}^\delta - v|^2 |\tilde{V}_{t,s}^\delta - V_{t-s}|^2}{|\delta|^4 + \left(\int_0^t |\tilde{V}_{r,s}^\delta - V_{r-s}|^2 dr\right)^2} \right)^{1/2} \quad (4.17) $$

where we have used that $|\tilde{X}_{s,s}^\delta - x| \ge |\delta|^{4/3}$ and $|\tilde{X}_{s,s}^\delta - x| \ge |z - x|$ when $\tilde{\chi}_s |\tilde{\psi}_s| \ne 0$.

Now, making the change of variable $(x', v') = (X^\delta(-s, x.v), V^\delta(-s, x, v))$ and denoting $(x', v')$ as $(x, v)$ for convenience, we have

$$ \begin{aligned} & \iint_{\Omega'} \int_0^T \int_0^t \frac{|\tilde{V}_{s,s}^\delta - v|^2 |\tilde{V}_{t,s}^\delta - V_{t-s}|^2}{|\delta|^4 + \left(\int_0^t |\tilde{V}_{r,s}^\delta - V_{r-s}|^2 dr\right)^2} \\ & \le \iint_{\Omega''} \int_0^T \frac{|V_t - V_t^\delta|^2 \int_0^t |V_s - V_s^\delta|^2 ds}{|\delta|^4 + \left(\int_0^t |V_s - V_s^\delta|^2 ds\right)^2} dt \, dx \, dv \\ & = \frac{1}{2} \iint_{\Omega''} \log \left( \frac{|\delta|^4 + \left(\int_0^T |V_s - V_s^\delta|^2 ds\right)^2}{|\delta|^4} \right) \, dx \, dv \\ & \le C \log(1/|\delta|). \end{aligned} $$

Next, similarly as in the computation leading to (4.16), we have

$$ \begin{aligned} \int_{\Omega'_2} \int_{\mathbb{R}^d} \|G_{\text{IV}}(v, z)\|^2 dz \, dv &\le C \int_{\mathbb{R}^d} \int_{\Omega'_2} \frac{|\tilde{\alpha}(k)|^2}{\frac{1}{|k|} + \left(\frac{k \cdot v}{|k|}\right)^2} \, dv \, dk \\ &\le C \int_{\mathbb{R}^d} |k| |\tilde{\alpha}(k)|^2 \int_{-\infty}^{+\infty} \frac{dv_1}{1 + |k| v_1^2} \, dk \\ &\le C \int_{\mathbb{R}^d} |k|^{1/2} |\tilde{\alpha}(k)|^2 \, dk \\ &\le C (\log 1/|\delta|)^{-2-4a} \int_{\mathbb{R}^d} |k|^{\frac{3}{2}+2a} |\alpha(k)|^2 \, dk. \end{aligned} $$

The combination of these inequalities finally yields

$$ |\text{IV}(T)| \le C \left( \int_{\mathbb{R}^d} |k|^{\frac{3}{2}+2a} |\alpha(k)|^2 dk \right)^{1/2} $$

if $|\delta| < 1/e$.
We compute as before

$$|I(T)| \le C \int_0^T \iint_{\Omega'} \int_{\mathbb{R}^d} \frac{|\tilde{V}_{t,t}^\delta - v|}{\left(|\delta|^2 + \int_0^t |\tilde{V}_{r,t}^\delta - V_{r-t}^2|^2 dr\right)^{1/2}} \frac{\tilde{\chi}_s |\tilde{\psi}_s| \|G_{\text{IV}}(v, z)\|}{|z-x|^{d-1} (|\delta| + |\tilde{X}_{t,t}^\delta - x|)} dz \, dx \, dv \, dt.$$

Next, the computation is very similar to (4.17):

$$|I(T)| \le C \left( \int_{\Omega'_2} \int_{\mathbb{R}^d} \|G_{\text{IV}}(v, z)\|^2 \int_{\Omega'_1} \frac{dx}{|z-x|^{d-1}(|\delta| + |z-x|)} dz \, dv \right)^{1/2} \left( \iint_{\Omega'} \int_0^T \frac{|\tilde{V}_{t,t}^\delta - v|^2}{|\delta|^2 + \int_0^t |\tilde{V}_{r,t}^\delta - V_{r-t}^2|^2 dr} \frac{1}{|\tilde{X}_{t,t}^\delta - x|} \int_{x+K(\tilde{X}_{t,t}^\delta - x)} \frac{dz}{|z-x|^{d-1}} dt \, dx \, dv \right)^{1/2}.$$

Proceeding again as before

$$|I(T)| \le C (\log 1/|\delta|)^{1/2} \left( \int_{\Omega'_2} \int_{\mathbb{R}^d} \|G_{\text{IV}}(v, z)\|^2 \right)^{1/2} \left( \iint_{\Omega''} \int_0^T \frac{|V_t - V_t^\delta|^2}{|\delta|^2 + \int_0^t |V_r - V_r^\delta|^2 dr} dx \, dv \right)^{1/2},$$

so that eventually

$$|I(T)| \le C \log(1/|\delta|) \left( \int_{\Omega'_2} \int_{\mathbb{R}^d} \|G_{\text{IV}}(v, z)\|^2 \right)^{1/2} \\ \le C \left( \int_{\mathbb{R}^d} |k|^{\frac{3}{2}+2a} |\alpha(k)|^2 dk \right)^{1/2}.$$

### 4.4.8 Proof of Lemma 4.5: Upper bound for $|B_\delta^{12}(T)|$

Let us define

$$B_{\delta}^{2}(T) := \int_{0}^{T} \iint_{\Omega} \frac{V_{t} - V_{t}^{\delta}}{A_{\delta}(t, x, v)} \cdot \int_{0}^{t} \chi_{s} B_{\delta}^{2}(s, x, v) \, ds \, dx \, dv \, dt.$$
As will appear below, this term is very similar to $V(T)$.

We apply the same method as before, without integrating by parts in time:

$$|B_{\delta}^{2}(T)| \le C \int_{0}^{T} \iint_{\Omega'} \int_{0}^{t} \int_{\mathbb{R}^{d}} \frac{|\tilde{V}_{t,s}^{\delta} - V_{t-s}|}{(|\delta|^{2} + \int_{0}^{t} |\tilde{V}_{r,s}^{\delta} - V_{r-s}|^{2} dr)^{1/2}} \frac{|\tilde{\psi}_{s}| \|G_{12}(v, z)\|}{|z - x|^{d-1}(|\delta| + |\tilde{X}_{s,s}^{\delta} - x|)} dz \, ds \, dx \, dv \, dt$$

where

$$G_{12}(v, z) = \int_{\mathbb{R}^d} \frac{k}{|k|^{1/2}} \otimes \frac{\tilde{\alpha}(k) e^{ik \cdot z}}{|k|^{-1/2} + i \frac{k}{|k|} \cdot v} \, dk.$$

Again,

$$\begin{aligned} |B_{\delta}^{12}(T)| & \le C \log(1/|\delta|) \left( \int_{\Omega_2'} \int_{\mathbb{R}^d} \|G_{12}(v, z)\|^2 dz \, dv \right)^{1/2} \\ & \le C \log(1/|\delta|) \left( \int_{\Omega_2'} \int_{\mathbb{R}^d} \frac{|k| |\tilde{\alpha}(k)|^2}{\frac{1}{|k|} + \left(\frac{k \cdot v}{|k|}\right)^2} dk \, dv \right)^{1/2} \\ & \le C \log(1/|\delta|) \left( \int_{\mathbb{R}^d} |k|^{3/2} |\tilde{\alpha}(k)|^2 dk \right)^{1/2} \\ & \le C (\log 1/|\delta|)^{1-2a} \left( \int_{\mathbb{R}^d} |k|^{\frac{3}{2}+2a} |\alpha(k)|^2 dk \right)^{1/2}. \end{aligned}$$

### 4.4.9 Conclusion

It follows from Lemma 4.4 that

$$|B_{\delta}^{1}(T)| \le C (1 + (\log |1/|\delta|)^{1-2a}) \left( \int_{\mathbb{R}^d} |k|^{\frac{3}{2}+2a} |\alpha(k)|^2 dk \right)^{1/2},$$

where

$$B_{\delta}^{1}(T) := \int_{0}^{T} \iint_{\Omega} \frac{V_{t} - V_{t}^{\delta}}{A_{\delta}(t, x, v)} \cdot \int_{0}^{t} \chi_{s} B_{\delta}^{1}(s, x, v) \, ds \, dx \, dv \, dt.$$

Combining this with Lemma 4.5, we obtain that

$$|B_{\delta}(T)| \le C (1 + (\log 1/|\delta|)^{1-2a}) \left( \int_{\mathbb{R}^d} |k|^{\frac{3}{2}+2a} |\alpha(k)|^2 dk \right)^{1/2}$$
where

$$B_{\delta}(T) := \int_{0}^{T} \iint_{\Omega} \frac{V_t - V_{t}^{\delta}}{A_{\delta}(t, x, v)} \cdot \int_{0}^{t} \chi_{s} B_{\delta}(s, x, v) \, ds \, dx \, dv \, dt.$$

Finally, the term $C_{\delta}(s, x, v)$ of (4.7) can be bounded exactly as $B_{\delta}(s, x, v)$ by simply exchanging the roles of $X_s$ and $X_s^{\delta}$. This ends the proof of Proposition 2.1.

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