# On the $L^{2}$ Rate of Convergence in the Limit from the Hartree to the
Vlasov–Poisson Equation

Jacky J. Chong  ,  Laurent Lafleche  Department of Mathematics, The University
of Texas at Austin, Austin, TX 78712, USA<EMAIL_ADDRESS><EMAIL_ADDRESS>and  Chiara Saffirio  Department of Mathematics and
Computer Science, University of Basel, 4051 Basel, Switzerland,
<EMAIL_ADDRESS>
###### Abstract.

Using a new stability estimate for the difference of the square roots of two
solutions of the Vlasov–Poisson equation, we obtain the convergence in the
$L^{2}$ norm of the Wigner transform of a solution of the Hartree equation
with Coulomb potential to a solution of the Vlasov–Poisson equation, with a
rate of convergence proportional to $\hbar$. This improves the
$\hbar^{3/4-\varepsilon}$ rate of convergence in $L^{2}$ obtained in [L.
Lafleche, C. Saffirio: Analysis & PDE, to appear]. Another reason of interest
of this paper is the new method, reminiscent of the ones used to prove the
mean-field limit from the many-body Schrödinger equation towards the
Hartree–Fock equation for mixed states.

###### Key words and phrases:

semiclassical limit, Hartree equation, Vlasov equation, Coulomb potential,
gravitational potential.

###### 2010 Mathematics Subject Classification:

81Q20, 35Q55, 35Q83 (82C10, 82C05).

######  Table of Contents

  1. 1 Introduction and main result
    1. 1.1 Notations
    2. 1.2 Main result
  2. 2 Stability estimates for the Vlasov–Poisson and Hartree equations
    1. 2.1 Classical case
    2. 2.2 Quantum case
  3. 3 Semiclassical limit in $\mathcal{L}^{2}$ Schatten norm
    1. 3.1 An estimate on the lack of positivity
    2. 3.2 Bound on $\left\lVert\boldsymbol{\rho}-\widetilde{\boldsymbol{\rho}}\right\rVert_{\mathcal{L}^{2}}$
    3. 3.3 Proof of the main theorems
    4. 3.4 Propagation of regularity

##  1\. Introduction and main result

In this paper, we study the limit as $\hbar\to 0$ of the Hartree equation
towards the Vlasov–Poisson equation. We focus on the cases of the three
dimensional Coulomb and gravitational interaction potentials

(1) $K(x)=\frac{\pm 1}{4\pi\left\lvert x\right\rvert},$

which are the most relevant singular potentials from a physical viewpoint and
the most challenging from a mathematical viewpoint.

We consider the Hartree equation

(2)
$i\hbar\,\partial_{t}\boldsymbol{\rho}=\left[{H_{\boldsymbol{\rho}},\boldsymbol{\rho}}\right]\quad\text{
with }\quad\boldsymbol{\rho}(0)=\boldsymbol{\rho}^{\mathrm{in}},$

where $\hbar=\frac{h}{2\pi}$ is the reduced Planck constant,
$\boldsymbol{\rho}$ is a time-dependent positive self-adjoint trace class
operator acting on $L^{2}(\mathbb{R}^{3})$, and
$H_{\boldsymbol{\rho}}=-\frac{\hbar^{2}\Delta}{2}+V_{\boldsymbol{\rho}}$ is
the Hartree Hamiltonian. The operator $V_{\boldsymbol{\rho}}$ is the operator
of multiplication by the mean-field potential
$V_{\boldsymbol{\rho}}(x)=(K*\rho)(x)$ associated to the two-body interaction
potential $K$ defined in our case by Equation (1), and $\rho(x)$ is the
quantum spatial density defined as the scaled restriction to the diagonal of
the integral kernel of $\boldsymbol{\rho}$,

$\rho(x)=\operatorname{diag}(\boldsymbol{\rho})(x):=h^{3}\boldsymbol{\rho}(x,x),$

where we adopt the same notation to denote both the operator
$\boldsymbol{\rho}$ and its integral kernel $\boldsymbol{\rho}(x,y)$. The
classical analogue of Equation (2) is the Vlasov–Poisson equation given by

(3) $\partial_{t}f+\xi\cdot\nabla_{x}f+E_{f}\cdot\nabla_{\xi}f=0\quad\text{
with }\quad f(0,x,\xi)=f^{\mathrm{in}}(x,\xi)\geq 0,$

where $E_{f}=-\nabla V_{f}$ is the self-consistent force field associated to
the mean-field potential $V_{f}(x)=(K*\rho_{f})(x)$ with $\rho_{f}$ the
spatial density given by

$\rho_{f}(x)=\int_{\mathbb{R}^{3}}f(t,x,\xi)\,\mathrm{d}\xi.$

For the well-posedness of the Hartree and the Vlasov–Poisson equations we
refer to [6] and [23, 20] respectively and references therein.

Equation (3) can be seen as the semiclassical approximation of a system of
many interacting quantum particles, as pointed out in the pioneering works by
Narnhofer and Sewell [22] and by Spohn [27] where the Vlasov equation was
obtained directly from the many-body Schrödinger equation with smooth
interaction in the combined mean-field and semiclassical regime. This has been
reconsidered in [13] and more recently in [8, 7], where the case of the
Coulomb potential with a $N$ dependent cut-off has been addressed. Moreover, a
combined mean-field and semiclassical limit for particles interacting via the
Coulomb potential has been treated in [12] for factorized initial data whose
first marginal is given by a monokinetic Wigner measure (that can be seen as
the Klimontovich solutions to the Vlasov equation), which leads to the
pressureless Euler–Poisson system.

Most of the above mentioned works rely on compactness methods that do not
allow for an explicit bound on the rate of convergence, which is essential for
applications. For this reason the Hartree equation (2) has been considered as
an intermediate step to decouple the problem into two separate parts, namely
to prove the convergence of the mean-field limit from the many-body
Schrödinger equation towards the Hartree equation, and then the semiclassical
limit from the Hartree equation to the Vlasov equation. In this paper, we are
interested in the latter problem, that has been largely studied in different
settings. It was first proven by Lions and Paul in [19], and later in [21,
10], that the Wigner transforms of the solutions of the Hartree equation (2)
converge in some weak sense to solutions of the Vlasov–Poisson equation.
Quantitative rates of convergence were then obtained, first in the case when
the Coulomb potential is replaced by a smoother potential, in Lebesgue-type
norms [2, 1, 4] and in a quantum analogue of the Wasserstein distances [11].
The case of singular interactions was then treated in [14, 15] with the same
quantum Wasserstein distances, and in [25, 26, 16] in Lebesgue-type norms. In
particular, for $K$ as in Equation (1), the explicit rate has been established
in [14] for the weak topology and in [26, 16] for the Schatten norms.

In a different setting, the semiclassical limit has also been studied for
local perturbations of stationary states in the case of infinite gases in
[18].

Our goals here are twofold. On the one hand, we want to improve the rate of
convergence in the $L^{2}$ norm in terms of $\hbar$ for the convergence of the
Wigner transform of the solution of the Hartree equation to the Vlasov–Poisson
equation. The rate $\hbar^{3/4-\varepsilon}$, for $\varepsilon>0$ small
enough, was obtained in [16], the reason being that the method used in that
previous paper was relying on a $L^{1}$ weak-strong uniqueness principle, thus
leading to the expected correct rate of order $\hbar$ in trace norm, but lower
order rates for higher Schatten norms. In this paper we recover a rate of
order $\hbar$ for the $L^{2}$ convergence, that is expected to be optimal.

On the other hand, the method used in this paper is closer to the method used
in [3, 8] to prove the semiclassical mean-field limit from the many-body
Schrödinger equation towards the Hartree–Fock equation for mixed states, in
the sense that it considers the square roots of the phase space densities in
the $L^{2}$ setting. It might give additional insights on the problem of the
derivation of the Vlasov–Poisson equation as a mean-field limit of a many-body
problem. In fact, it sheds light on the difference of requirements between the
$L^{1}$ and the $L^{2}$ method. In particular, the latter, still being a weak-
strong estimate, seems however to require the uniform boundedness in the phase
space of both the solutions of the Hartree and the Vlasov–Poisson equation, as
it is clear from our main result.

###  1.1. Notations

####  1.1.1. Functional spaces

Let us fix some notations before stating our main result. A point in the one-
particle phase space $\mathbb{R}^{3}_{x}\times\mathbb{R}^{3}_{\xi}$ is denoted
by $z=(x,\xi)$. We denote the phase space Lebesgue norm and its mixed norm
variant by

$\displaystyle\left\lVert
f\right\rVert_{L^{r}}=\Big{(}\int_{\mathbb{R}^{6}}\left\lvert
f(z)\right\rvert^{r}\mathrm{d}z\Big{)}^{\\!\frac{1}{r}},\quad\left\lVert
f\right\rVert_{L^{p}_{x}L^{q}_{\xi}}=\big{\lVert}\\!\left\lVert
f(x,\cdot)\right\rVert_{L^{q}_{\xi}(\mathbb{R}^{3})}\\!\big{\rVert}_{L^{p}_{x}(\mathbb{R}^{3})}$

for $1\leq r<\infty$ and with the usual modification when $r=\infty$. The
corresponding weighted Sobolev spaces $W^{k,p}_{n}$ are defined by the norm

$\left\lVert
f\right\rVert_{W^{k,p}_{n}}=\Big{(}\sum_{\begin{subarray}{c}\left\lvert\alpha\right\rvert\leq
k\end{subarray}}\left\lVert\left\langle\xi\right\rangle^{n}\partial^{\alpha}f\right\rVert_{L^{p}}^{2}\Big{)}^{\\!\frac{1}{2}}\quad\text{
with }\quad\begin{array}[]{l}\alpha\in\mathbb{Z}_{\geq
0}^{6},\,\left\lvert\alpha\right\rvert=\sum^{6}_{\mathrm{i}=1}\alpha_{\mathrm{i}},\\\
\text{and
}\partial^{\alpha}=\partial_{x_{1}}^{\alpha_{1}}\cdots\partial_{\xi_{3}}^{\alpha_{6}}\end{array}$

where $\left\langle z\right\rangle:=\sqrt{1+|z|^{2}}$ is the standard bracket
notation. We also use the standard notation $H^{k}_{n}=W^{k,2}_{n}$.

For any bounded linear operator $\boldsymbol{\rho}$ acting on
$L^{2}(\mathbb{R}^{3})$, we denote its operator norm by
$\left\lVert\boldsymbol{\rho}\right\rVert_{\infty}$, and for $1\leq p<\infty$,
we define its Schatten-$p$ norm by

(4)
$\displaystyle\left\lVert\boldsymbol{\rho}\right\rVert_{p}=\operatorname{Tr}\\!\left(\left\lvert\boldsymbol{\rho}\right\rvert^{p}\right)^{\frac{1}{p}}$

where
$\left\lvert\boldsymbol{\rho}\right\rvert=\sqrt{\boldsymbol{\rho}^{\ast}\boldsymbol{\rho}}$.
To obtain meaningful quantities in the semiclassical limit $\hbar\rightarrow
0$, we define the semiclassical analogue of the Schatten spaces via the
rescaled Schatten norm

(5)
$\displaystyle\left\lVert\boldsymbol{\rho}\right\rVert_{\mathcal{L}^{p}}=h^{\frac{3}{p}}\left\lVert\boldsymbol{\rho}\right\rVert_{p}=h^{\frac{3}{p}}\operatorname{Tr}\\!\left(\left\lvert\boldsymbol{\rho}\right\rvert^{p}\right)^{\frac{1}{p}}.$

These semiclassical spaces play the roles of the Lebesgue spaces on the phase
space. We will denote by $C$ a $\hbar$-independent constant that can change
from line to line.

####  1.1.2. Weyl quantization and Wigner transform

The Fourier transform is defined with the convention that

(6) $\displaystyle\widehat{g}(\xi)=\int_{\mathbb{R}^{3}}e^{-2\pi
i\,x\cdot\xi}\,g(x)\,\mathrm{d}x.$

Being consistent with our definition of the Fourier transform, we associate to
every function $f\in L^{1}(\mathbb{R}^{3}_{x}\times\mathbb{R}^{3}_{\xi})$ the
operator $\boldsymbol{\rho}_{f}$ with integral kernel

(7) $\displaystyle\boldsymbol{\rho}_{f}(x,y)=\int_{\mathbb{R}^{3}}e^{-2\pi
i\,(y-x)\cdot\xi}f(\tfrac{x+y}{2},h\xi)\,\mathrm{d}\xi.$

The operator $\boldsymbol{\rho}_{f}$ is called the Weyl quantization
associated to $f(z)$. By the Fourier inversion theorem, we could also define
the inverse mapping called the Wigner transform. More precisely, for any
operator $\boldsymbol{\rho}$ with sufficiently regular kernel
$\boldsymbol{\rho}(x,y)$, we define its Wigner transform by

(8) $\displaystyle f_{\boldsymbol{\rho}}(x,\xi)=\int_{\mathbb{R}^{3}}e^{-2\pi
i\,\xi\cdot
y/h}\boldsymbol{\rho}(x+\tfrac{y}{2},x-\tfrac{y}{2})\,\mathrm{d}y.$

###  1.2. Main result

Our main result reads as follows.

######  Theorem 1.1.

Let $\boldsymbol{\rho}\geq 0$ be a solution to the Hartree equation (2) with
initial condition
$\boldsymbol{\rho}^{\mathrm{in}}\in\mathcal{L}^{1}\cap\mathcal{L}^{\infty}$
and $f\geq 0$ be a solution to the Vlasov–Poisson equation (3) with initial
condition $f^{\mathrm{in}}$ verifying

$f^{\mathrm{in}},\,\sqrt{f^{\mathrm{in}}}\in W^{4,\infty}_{4}\cap
H^{4}_{4}\quad\text{ and
}\quad\iint_{\mathbb{R}^{6}}f^{\mathrm{in}}\left\lvert\xi\right\rvert^{n_{1}}\mathrm{d}x\,\mathrm{d}\xi<\infty$

for $n_{1}>6$. Then there exist $\Lambda,C_{1},C_{2}\in
C^{0}(\mathbb{R}_{+},\mathbb{R}_{+})$, independent of $\hbar$ and depending
only on the initial conditions of equations (2) and (3) such that

(9) $\left\lVert
f_{\boldsymbol{\rho}}-f\right\rVert_{L^{2}}\leq\mathcal{C}_{\infty}^{\frac{1}{2}}\left(\big{\lVert}f_{\sqrt{\boldsymbol{\rho}^{\mathrm{in}}}}-\sqrt{f^{\mathrm{in}}}\big{\rVert}_{L^{2}}+C(t)\,\hbar\right)e^{\Lambda(t)}+C_{2}(t)\,\hbar.$

######  Remark 1.1.

Our method also allows us to treat the Hartree–Fock equation. In that case,
the exchange term vanishes in the semiclassical limit and can be treated as an
error term as in [16, Proposition 5.1].

####  1.2.1. Strategy and explicit constants

Before giving more precise upper bounds on the functions $\Lambda$, $C_{1}$
and $C_{2}$, let us explain our strategy. The rationale of our result is a
stability estimate for the square root of the solutions of the Vlasov equation
in $L^{2}$, as explained in Section 2. Rephrasing this estimate in the quantum
context to estimate the difference between the solutions to the Hartree
equation and the Vlasov equation, we have to deal with the fact that the Weyl
quantization of a non-negative function is not always non-negative. Thus, we
have to consider an alternative quantization of $f$, sometimes called the Wick
quantization, the anti-Wick quantization, the Töplitz quantization, the Husimi
quantization or the coherent state quantization. We introduce the coherent
state at the point $z$ and its corresponding density operator

(10) $\displaystyle\psi_{z}(y):=\left(\pi\hbar\right)^{-3/4}e^{-\left\lvert
y-x\right\rvert^{2}/(2\hbar)}\,e^{iy\cdot\xi/\hbar}\quad\text{ and
}\quad\boldsymbol{\rho}_{z}:=h^{-3}\ket{\psi_{z}}\bra{\psi_{z}}.$

Then, for every function $f$ on the phase space, we associate the operator
$\widetilde{\boldsymbol{\rho}}_{f}$ defined by taking averages of the coherent
states $\boldsymbol{\rho}_{z}$ against $f(z)$, that is

(11)
$\displaystyle\widetilde{\boldsymbol{\rho}}_{f}=\int_{\mathbb{R}^{6}}f(z)\,\boldsymbol{\rho}_{z}\,\mathrm{d}z=\frac{1}{h^{3}}\int_{\mathbb{R}^{6}}f(z)\ket{\psi_{z}}\bra{\psi_{z}}\mathrm{d}z.$

It follows from this formula that $\widetilde{\boldsymbol{\rho}}_{f}$ is a
positive operator whenever $f\geq 0$. We summarize some of its other
properties which we will need in this work in the following lemma. We refer
the reader to [19, 17] for additional properties of this quantization and the
proof of the following lemma.

######  Lemma 1.1.

Let $f$ be a function of the phase space. Then the quantization (11) is the
combination of the Weyl quantization and a convolution with a Gaussian

(12)
$\widetilde{\boldsymbol{\rho}}_{f}=\boldsymbol{\rho}_{\tilde{f}}\quad\text{
where }\quad\widetilde{f}=g_{h}\ast f\text{ with
}g_{h}(z):=(\pi\hbar)^{-3}\,e^{-\left\lvert z\right\rvert^{2}/\hbar}.$

By (12), we have the estimate

(13)
$\left\lVert\boldsymbol{\rho}_{f}-\widetilde{\boldsymbol{\rho}}_{f}\right\rVert_{\mathcal{L}^{2}}=\left\lVert
f-g_{h}\ast
f\right\rVert_{L^{2}}\leq\tfrac{3\,\hbar}{2}\left\lVert\nabla^{2}f\right\rVert_{L^{2}}.$

For $1\leq p\leq\infty$, we have that

(14)
$\left\lVert\widetilde{\boldsymbol{\rho}}_{f}\right\rVert_{\mathcal{L}^{p}}\leq\left\lVert
f\right\rVert_{L^{p}}.$

This quantization will allow us to consider an intermediate equation between
the Hartree and the Vlasov–Poisson equations, depending on the mean-field
force of the Vlasov equation, but whose solution is a positive operator. More
precisely, we consider $\widetilde{\boldsymbol{\rho}}$ a solution to the
equation

$i\hbar\,\partial_{t}\widetilde{\boldsymbol{\rho}}=\left[{H_{f},\widetilde{\boldsymbol{\rho}}}\right]$

with Hamiltonian $H_{f}=-\frac{\hbar^{2}}{2}\,\Delta+V_{f}$ and initial data
$\widetilde{\boldsymbol{\rho}}^{\mathrm{in}}:=\widetilde{\boldsymbol{\rho}}_{\sqrt{f^{\mathrm{in}}}}^{2}$,
and we define

$\tilde{v}:=\sqrt{\widetilde{\boldsymbol{\rho}}},$

which satisfies the same equation. Our estimates rely on the semiclassical
regularity of this equation.

####  1.2.2. Quantum Sobolev spaces

To define the semiclassical version of Sobolev spaces on the phase space, we
introduce the following operators

(15)
$\boldsymbol{\nabla}_{\\!x}\boldsymbol{\rho}:=\left[{\nabla,\boldsymbol{\rho}}\right]\quad\text{
and
}\quad\boldsymbol{\nabla}_{\\!\xi}\boldsymbol{\rho}:=\left[{\frac{x}{i\hbar},\boldsymbol{\rho}}\right],$

which can be seen as an application of the correspondence principle of quantum
mechanics. More precisely, one observes that these operators correspond to the
gradients of the Wigner transform, that is

(16)
$f_{\boldsymbol{\nabla}_{\\!x}\boldsymbol{\rho}}=\nabla_{x}f_{\boldsymbol{\rho}}\quad\text{
and }\quad
f_{\boldsymbol{\nabla}_{\\!\xi}\boldsymbol{\rho}}=\nabla_{\xi}f_{\boldsymbol{\rho}}.$

We will refer to $\boldsymbol{\nabla}_{\\!x}\boldsymbol{\rho}$ and
$\boldsymbol{\nabla}_{\\!\xi}\boldsymbol{\rho}$ as the quantum gradients. The
semiclassical analogues of the weighted kinetic homogeneous Sobolev norms are
defined by

(17a)
$\displaystyle\left\lVert\boldsymbol{\rho}\right\rVert_{\dot{\mathcal{W}}^{k,p}(\mathsf{m})}^{p}$
$\displaystyle:=\sum_{\left\lvert\alpha\right\rvert\leq
k}\left\lVert\boldsymbol{\nabla}_{\\!z}^{\alpha}\boldsymbol{\rho}\right\rVert_{\mathcal{L}^{p}(\mathsf{m})}^{p}\quad\text{
with
}\quad\boldsymbol{\nabla}_{\\!z}^{\alpha}=\boldsymbol{\nabla}_{\\!x_{1}}^{\alpha_{1}}\cdots\boldsymbol{\nabla}_{\\!\xi_{3}}^{\alpha_{6}}$
(17b)
$\displaystyle\left\lVert\boldsymbol{\rho}\right\rVert_{\dot{\mathcal{W}}^{k,\infty}(\mathsf{m})}$
$\displaystyle:=\sup_{\left\lvert\alpha\right\rvert\leq
k}\left\lVert\boldsymbol{\nabla}_{\\!z}^{\alpha}\boldsymbol{\rho}\right\rVert_{\mathcal{L}^{\infty}(\mathsf{m})},$

and we consider the particular case of the weights defined for
$n\in\mathbb{N}$ by

(18)
$\displaystyle\mathsf{m}:=\left\langle\boldsymbol{p}\right\rangle^{n}=\big{(}1+\left\lvert\boldsymbol{p}\right\rvert^{2}\big{)}^{\\!\frac{n}{2}},$

where $\boldsymbol{p}=-i\hbar\nabla$ so that
$\left\lvert\boldsymbol{p}\right\rvert^{2}=-\hbar^{2}\Delta$. We also define
the inhomogeneous version by

(19)
$\displaystyle\left\lVert\boldsymbol{\rho}\right\rVert_{\mathcal{W}^{k,p}(\mathsf{m})}^{p}$
$\displaystyle:=\left\lVert\boldsymbol{\rho}\right\rVert_{\mathcal{L}^{p}(\mathsf{m})}^{p}+\left\lVert\boldsymbol{\rho}\right\rVert_{\dot{\mathcal{W}}^{k,p}(\mathsf{m})}^{p},$

with the usual modification when $p=\infty$. In particular, for $p=2$,
$\left\lVert\boldsymbol{\rho}\right\rVert_{\mathcal{W}^{k,2}}=\left\lVert
f_{\boldsymbol{\rho}}\right\rVert_{H^{k}}$.

####  1.2.3. A precised version of the main theorem

With the above definitions, we now give another version of the main theorem
with more explicit upper bounds on the constants appearing in Inequality (9).

######  Theorem 1.2.

Under the assumptions of Theorem 1.1, the following estimate holds

(20)
$\left\lVert\boldsymbol{\rho}-\boldsymbol{\rho}_{f}\right\rVert_{\mathcal{L}^{2}}\leq\mathcal{C}_{\infty}^{\frac{1}{2}}\left(\left\lVert\widetilde{\boldsymbol{\rho}}_{\sqrt{f^{\mathrm{in}}}}-\sqrt{\boldsymbol{\rho}^{\mathrm{in}}}\right\rVert_{\mathcal{L}^{2}}+C(t)\,\hbar\right)e^{\Lambda(t)}+C_{2}(t)\,\hbar.$

An upper bound for $\Lambda$ is given by $\Lambda(t)\leq
C\int_{0}^{t}\lambda(s)\,\mathrm{d}s$ where $\lambda$ is defined for some
$\varepsilon\in(0,1)$ by

(21)
$\lambda(t)=\left\lVert\boldsymbol{\nabla}_{\\!\xi}\tilde{v}\right\rVert_{\mathcal{W}^{1,2}}\left\lVert\rho_{f}\right\rVert_{L^{\infty}_{x}}^{\frac{1}{2}}+\mathcal{C}_{\infty}^{\frac{1}{2}}\left\lVert\boldsymbol{\nabla}_{\\!\xi}\tilde{v}\right\rVert_{\mathcal{L}^{3\pm\varepsilon}(\left\langle\boldsymbol{p}\right\rangle^{n})}.$

with $n>2$. The functions $C(t)$ and $C_{2}(t)$ are bounded above by

(22) $\displaystyle C(t)^{2}$ $\displaystyle\leq
C\int_{0}^{t}\left\lVert\boldsymbol{\nabla}_{\\!\xi}\tilde{v}\right\rVert_{\mathcal{W}^{1,2}}^{2}\left(C^{\mathrm{in}}+\left\lVert\rho_{f}\right\rVert_{W^{1,\infty}_{x}}\left\lVert\boldsymbol{\rho}_{f}\right\rVert_{\mathcal{W}^{2,2}(\left\langle\boldsymbol{p}\right\rangle^{2})}\right)^{2}$
(23) $\displaystyle C_{2}(t)$ $\displaystyle\leq
C\left(C^{\mathrm{in}}+\int_{0}^{t}\left\lVert\rho_{f}\right\rVert_{W^{1,\infty}_{x}}\left\lVert\nabla_{\xi}^{2}f\right\rVert_{L^{2}}\right)$

which remain bounded at any time $t\geq 0$, and where

(24)
$C^{\mathrm{in}}=\big{\lVert}\sqrt{f^{\mathrm{in}}}\big{\rVert}_{W^{1,4}_{1}\cap
H^{3}}^{2}+\left\lVert f^{\mathrm{in}}\right\rVert_{H^{1}_{1}\cap H^{2}}.$

The fact that these quantities (21), (22) and (23) remain bounded uniformly in
$\hbar$ at any time is proved in the last section in Proposition 3.2.

######  Remark 1.2.

Notice that the quantity

$\left\lVert\widetilde{\boldsymbol{\rho}}_{\sqrt{f^{\mathrm{in}}}}-\sqrt{\boldsymbol{\rho}^{\mathrm{in}}}\right\rVert_{\mathcal{L}^{2}}=\left\lVert\boldsymbol{\rho}_{g_{h}*\sqrt{f^{\mathrm{in}}}}-\sqrt{\boldsymbol{\rho}^{\mathrm{in}}}\right\rVert_{\mathcal{L}^{2}}=\left\lVert
g_{h}*\sqrt{f^{\mathrm{in}}}-f_{\sqrt{\boldsymbol{\rho}^{\mathrm{in}}}}\right\rVert_{L^{2}}.$

that appears in the right-hand side of Equation (20) is $0$ in the case when
$\boldsymbol{\rho}^{\mathrm{in}}=\widetilde{\boldsymbol{\rho}}_{\sqrt{f^{\mathrm{in}}}}^{2}$,
that is when $\boldsymbol{\rho}^{\mathrm{in}}$ is the square of a Wick
quantization.
Due to the strategy of the proof, that mimics the proof of the stability
estimates presented in Section 2, the initial datum for the auxiliary problem
(31) has to be chosen positive, close to $\boldsymbol{\rho}_{f}^{\mathrm{in}}$
in $\mathcal{L}^{2}$, its square root is close to
$\sqrt{\boldsymbol{\rho}^{\mathrm{in}}}$ in $\mathcal{L}^{2}$ and regular in
the sense that the quantity
$\big{\lVert}\boldsymbol{\nabla}_{\\!\xi}\sqrt{\smash[t]{\widetilde{\boldsymbol{\rho}}}^{\mathrm{in}}}\,\mathsf{m}\big{\rVert}_{\mathcal{L}^{p}}+\big{\lVert}\boldsymbol{\nabla}_{\\!x}\sqrt{\smash[t]{\widetilde{\boldsymbol{\rho}}}^{\mathrm{in}}}\,\mathsf{m}\big{\rVert}_{\mathcal{L}^{p}}$
is uniformly bounded with respect to $\hbar$. The choice
$\widetilde{\boldsymbol{\rho}}^{\mathrm{in}}=\widetilde{\boldsymbol{\rho}}_{\sqrt{f^{\mathrm{in}}}}^{2}$
guarantees these properties.

##  2\. Stability estimates for the Vlasov–Poisson and Hartree equations

###  2.1. Classical case

We start by explaining our method through examining the classical case and
obtaining a stability estimate for the Vlasov–Poisson equation.

######  Theorem 2.1.

Let $f_{1}$ and $f_{2}$ be two solutions of the Vlasov–Poisson equation
verifying $\left\lVert f_{1}\right\rVert_{L^{\infty}}=\left\lVert
f_{2}\right\rVert_{L^{\infty}}\leq\mathcal{C}_{\infty}$. Then we have the
following bound

$\Big{\lVert}\sqrt{f_{1}}-\sqrt{f_{2}}\Big{\rVert}_{L^{2}}\leq\Big{\lVert}\sqrt{f_{1}^{\mathrm{in}}}-\sqrt{f_{2}^{\mathrm{in}}}\Big{\rVert}_{L^{2}}e^{\Lambda(t)}$

where $\Lambda(t)=C\,\int_{0}^{t}\lambda(s)\,\mathrm{d}s$ with

$\lambda(t)=\left\lVert\rho_{2}\right\rVert_{L^{\infty}_{x}}^{\frac{1}{2}}\left\lVert\nabla_{\xi}\sqrt{f_{2}}\right\rVert_{L^{3}_{x}L^{2}_{\xi}}+\mathcal{C}_{\infty}^{\frac{1}{2}}\left\lVert\nabla_{\xi}\sqrt{f_{2}}\right\rVert_{L^{3,1}_{x}L^{1}_{\xi}}.$

In particular, we have the following estimate

$\left\lVert f_{1}-f_{2}\right\rVert_{L^{2}}\leq
2\,\mathcal{C}_{\infty}^{\frac{1}{2}}\left\lVert
f_{1}-f_{2}\right\rVert_{L^{1}}^{\frac{1}{2}}e^{\Lambda(t)}.$

###### Proof.

Let $v_{1}=\sqrt{f_{1}}$, $v_{2}=\sqrt{f_{2}}$ and $v:=v_{1}-v_{2}$. Then $v$
satisfies the equation

$\big{(}\partial_{t}+\xi\cdot\nabla_{x}+E_{v_{1}^{2}}\cdot\nabla_{\xi}\big{)}v=\big{(}E_{v_{2}^{2}}-E_{v_{1}^{2}}\big{)}\cdot\nabla_{\xi}v_{2}=\left(\nabla
K*\rho_{\left(v_{2}+v_{1}\right)v}\right)\cdot\nabla_{\xi}v_{2}.$

Then, by direct computation, we have that

$\displaystyle\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}\left\lVert
v\right\rVert_{L^{2}}^{2}$ $\displaystyle=-\int_{\mathbb{R}^{6}}v\left(\nabla
K*\rho_{\left(v_{2}+v_{1}\right)v}\right)\cdot\nabla_{\xi}v_{2}\,\mathrm{d}x\,\mathrm{d}\xi$
$\displaystyle=-\int_{\mathbb{R}^{12}}\big{(}\left\lvert
v(z^{\prime})\right\rvert^{2}+2\,v_{2}(z^{\prime})\,v(z^{\prime})\big{)}\nabla
K(x-x^{\prime})\cdot\left(v(z)\,\nabla_{\xi}v_{2}(z)\right)\,\mathrm{d}z\,\mathrm{d}z^{\prime}$
$\displaystyle=:I_{1}+2\,I_{2}.$

The first term is bounded by writing first

$\displaystyle I_{1}$ $\displaystyle\leq\Big{\lVert}\nabla
K*\int_{\mathbb{R}^{3}}v\,\nabla_{\xi}v_{2}\,\mathrm{d}\xi\Big{\rVert}_{L^{\infty}_{x}}\left\lVert
v\right\rVert_{L^{2}}^{2}\leq\Big{\lVert}\left\lvert\nabla
K\right\rvert*\int_{\mathbb{R}^{3}}\left\lvert\nabla_{\xi}v_{2}\right\rvert\mathrm{d}\xi\Big{\rVert}_{L^{\infty}_{x}}\left\lVert
v\right\rVert_{L^{\infty}}\left\lVert v\right\rVert_{L^{2}}^{2}.$

and then by applying Hölder’s inequality for the Lorentz spaces to get

$I_{1}\leq\left\lVert\nabla
K\right\rVert_{L^{3/2,\infty}_{x}}\left\lVert\nabla_{\xi}v_{2}\right\rVert_{L^{3,1}_{x}L^{1}_{\xi}}\left\lVert
v\right\rVert_{L^{\infty}}\left\lVert v\right\rVert_{L^{2}}^{2}.$

The second term is bounded by the Hardy–Littlewood–Sobolev inequality and then
by the Cauchy–Schwarz inequality, leading to

$\displaystyle I_{2}$ $\displaystyle\leq
C\,\Big{\lVert}\int_{\mathbb{R}^{3}}v\,v_{2}\,\mathrm{d}\xi\,\Big{\rVert}_{L^{2}_{x}}\,\Big{\lVert}\int_{\mathbb{R}^{3}}v\,\nabla_{\xi}v_{2}\,\mathrm{d}\xi\,\Big{\rVert}_{L^{6/5}_{x}}$
$\displaystyle\leq C\left\lVert\left\lVert
v\right\rVert_{L^{2}_{\xi}}\left\lVert
v_{2}\right\rVert_{L^{2}_{\xi}}\right\rVert_{L^{2}_{x}}\left\lVert\left\lVert
v\right\rVert_{L^{2}_{\xi}}\left\lVert\nabla_{\xi}v_{2}\right\rVert_{L^{2}_{\xi}}\right\rVert_{L^{6/5}_{x}}.$

Finally, by Hölder’s inequality, we obtain

$\displaystyle I_{2}$ $\displaystyle\leq C\left\lVert
v_{2}\right\rVert_{L^{\infty}_{x}L^{2}_{\xi}}\left\lVert\nabla_{\xi}v_{2}\right\rVert_{L^{3}_{x}L^{2}_{\xi}}\left\lVert
v\right\rVert_{L^{2}}^{2}=C\left\lVert\rho_{2}\right\rVert_{L^{\infty}_{x}}^{\frac{1}{2}}\left\lVert\nabla_{\xi}v_{2}\right\rVert_{L^{3}_{x}L^{2}_{\xi}}\left\lVert
v\right\rVert_{L^{2}}^{2}.$

Combining the bounds for $I_{1}$ and $I_{2}$ leads to

$\frac{\mathrm{d}}{\mathrm{d}t}\left\lVert v\right\rVert_{L^{2}}^{2}\leq
C\,\big{(}\left\lVert\rho_{2}\right\rVert_{L^{\infty}_{x}}\left\lVert\nabla_{\xi}v_{2}\right\rVert_{L^{3}_{x}L^{2}_{\xi}}+\left\lVert
v\right\rVert_{L^{\infty}}\left\lVert\nabla_{\xi}v_{2}\right\rVert_{L^{3,1}_{x}L^{1}_{\xi}}\big{)}\left\lVert
v\right\rVert_{L^{2}}^{2}.$

We conclude by using the fact that $\left\lVert
v\right\rVert_{L^{\infty}}\leq\left\lVert
f_{1}\right\rVert_{L^{\infty}}^{\frac{1}{2}}+\left\lVert
f_{2}\right\rVert_{L^{\infty}}^{\frac{1}{2}}=2\,\mathcal{C}_{\infty}^{\frac{1}{2}}$
and Grönwall’s lemma. ∎

###  2.2. Quantum case

Using the quantum analogue of gradients and Lebesgue norms of the phase space,
we deduce a similar estimate for the Hartree equation.

######  Theorem 2.2.

Let $\boldsymbol{\rho}_{1}$ and $\boldsymbol{\rho}_{2}$ be two solutions of
the Hartree equation (2) such that
$\left\lVert\boldsymbol{\rho}_{1}\right\rVert_{\mathcal{L}^{\infty}}\leq\mathcal{C}_{\infty}$
and
$\left\lVert\boldsymbol{\rho}_{2}\right\rVert_{\mathcal{L}^{\infty}}\leq\mathcal{C}_{\infty}$
and let $v_{1}=\sqrt{\boldsymbol{\rho}_{1}}$ and
$v_{2}=\sqrt{\boldsymbol{\rho}_{2}}$. Then there exists a universal constant
$C>0$ and $T>0$ such that for any $t\in[0,T]$,

(25) $\left\lVert v_{1}-v_{2}\right\rVert_{\mathcal{L}^{2}}\leq\left\lVert
v_{1}^{\mathrm{in}}-v_{2}^{\mathrm{in}}\right\rVert_{\mathcal{L}^{2}}e^{\lambda(t)}$

where $\Lambda(t)=C\int_{0}^{t}\lambda(s)\,\mathrm{d}s$ with $\lambda$ given
for some $\varepsilon\in(0,1)$ by

(26)
$\lambda(t)=\left\lVert\boldsymbol{\nabla}_{\\!\xi}v_{2}\right\rVert_{\mathcal{W}^{1,2}}\left\lVert\rho_{f}\right\rVert_{L^{\infty}_{x}}^{\frac{1}{2}}+\mathcal{C}_{\infty}^{\frac{1}{2}}\left\lVert\boldsymbol{\nabla}_{\\!\xi}v_{2}\,\mathsf{m}_{n}\right\rVert_{\mathcal{L}^{3\pm\varepsilon}},$

where $\mathsf{m}_{n}=\left\langle\boldsymbol{p}\right\rangle^{n}$ with $n>2$.
This implies the following estimate

(27)
$\left\lVert\boldsymbol{\rho}_{1}-\boldsymbol{\rho}_{2}\right\rVert_{\mathcal{L}^{2}}\leq
2\,\mathcal{C}_{\infty}^{\frac{1}{2}}\left\lVert\boldsymbol{\rho}_{1}^{\mathrm{in}}-\boldsymbol{\rho}_{2}^{\mathrm{in}}\right\rVert_{\mathcal{L}^{1}}^{\frac{1}{2}}e^{\Lambda(t)}.$

######  Remark 2.1.

It is actually not difficult to see from the interpolation argument in the
proof of Lemma 2.1 that
$\left\lVert\boldsymbol{\nabla}_{\\!\xi}v_{2}\right\rVert_{\mathcal{W}^{1,2}}=\left\lVert\nabla_{\xi}f_{v_{2}}\right\rVert_{H^{1}}$
can actually be replaced by
$\left\lVert\nabla_{\xi}f_{v_{2}}\right\rVert_{H^{1/2}}$.

######  Remark 2.2.

Contrarily to our main theorem, the result here is only local-in-time since it
is not yet known whether the global-in-time uniform in $\hbar$ propagation of
regularity holds for the Hartree equation in the case of the Coulomb
potential. For slightly less singular potentials, we have proved it in our
recent paper [9].

###### Proof.

Let $v=v_{1}-v_{2}$. Then $v$ satisfies the equation

$i\hbar\,\partial_{t}v=\left[{H_{\boldsymbol{\rho}_{1}},v}\right]+\left[{V_{\boldsymbol{\rho}_{1}}-V_{\boldsymbol{\rho}_{2}},v_{2}}\right].$

Differentiating its Hilbert–Schmidt norm with respect to time, we obtain

$\displaystyle\frac{1}{4\pi}\,\frac{\mathrm{d}}{\mathrm{d}t}\left\lVert
v\right\rVert_{\mathcal{L}^{2}}^{2}$
$\displaystyle=h^{2}\operatorname{Tr}\\!\left(\left[{V_{\boldsymbol{\rho}_{1}}-V_{\boldsymbol{\rho}_{2}},v_{2}}\right]v\right)=h^{2}\operatorname{Tr}\\!\left(\left[{V_{v_{1}^{2}-v_{2}^{2}},v_{2}}\right]v\right).$

Since
$v_{1}^{2}-v_{2}^{2}=\left(v+v_{2}\right)^{2}-v_{2}^{2}=v^{2}+v\,v_{2}+v_{2}\,v$
and
$\operatorname{diag}\\!\left(v\,v_{2}\right)=\operatorname{diag}\\!\left(v_{2}\,v\right)$,
we get

$\displaystyle\frac{1}{4\pi}\,\frac{\mathrm{d}}{\mathrm{d}t}\left\lVert
v\right\rVert_{\mathcal{L}^{2}}^{2}=h^{2}\operatorname{Tr}\\!\left(\left[{V_{v^{2}},v_{2}}\right]v\right)+2\,h^{2}\operatorname{Tr}\\!\left(\left[{V_{v_{2}\,v},v_{2}}\right]v\right)=:I_{1}+2\,I_{2}.$

To bound the first term, we introduce the notation $K_{x}:=K(x-y)$ and write

$\displaystyle
I_{1}=h^{2}\int_{\mathbb{R}^{3}}\operatorname{Tr}\\!\left(\left[{K_{x},v_{2}}\right]v\right)\rho_{v^{2}}(x)\,\mathrm{d}x$
$\displaystyle\leq\sup_{x\in\mathbb{R}^{3}}\left(h^{-1}\left\lVert\left[{K_{x},v_{2}}\right]\right\rVert_{\mathcal{L}^{1}}\left\lVert
v\right\rVert_{\mathcal{L}^{\infty}}\right)\int_{\mathbb{R}^{3}}\left\lvert\rho_{v^{2}}\right\rvert\mathrm{d}x$
$\displaystyle\leq
C\left\lVert\boldsymbol{\nabla}_{\\!\xi}v_{2}\,\mathsf{m}_{n}\right\rVert_{\mathcal{L}^{3\pm\varepsilon}}\left\lVert
v\right\rVert_{\mathcal{L}^{\infty}}\left\lVert
v\right\rVert_{\mathcal{L}^{2}}^{2}$

where we have used [16, Theorem 4.1] to bound the commutator in trace norm. To
bound the second term, we use Hölder’s inequality to get

$I_{2}\leq\left\lVert\left[{V_{v\,v_{2}},v_{2}}\right]\right\rVert_{\mathcal{L}^{2}}\left\lVert
v\right\rVert_{2}$

and then we apply the following inequality which we will prove in the
subsequent section (see Lemma 2.1)

$\left\lVert\left[{V_{v\,v_{2}},v_{2}}\right]\right\rVert_{\mathcal{L}^{2}}\leq
C\left\lVert\boldsymbol{\nabla}_{\\!\xi}v_{2}\right\rVert_{\mathcal{W}^{1,2}}\left\lVert\rho_{v\,v_{2}}\right\rVert_{L^{2}_{x}}.$

Also, notice that

$\displaystyle\left\lVert\rho_{v\,v_{2}}\right\rVert_{L^{2}_{x}}$
$\displaystyle=h^{3}\left(\int_{\mathbb{R}^{3}}\Big{|}\int_{\mathbb{R}^{3}}v(x,y)\,v_{2}(y,x)\,\mathrm{d}y\Big{|}^{2}\mathrm{d}x\right)^{\frac{1}{2}}\leq
h^{\frac{3}{2}}\left\lVert
v_{2}\right\rVert_{L^{\infty}_{x}L^{2}_{y}}\left\lVert
v\right\rVert_{\mathcal{L}^{2}}.$

Hence, since $h^{3}\left\lVert
v_{2}\right\rVert_{L^{2}_{y}}^{2}=\operatorname{diag}\\!\left(\boldsymbol{\rho}_{2}\right)=\rho_{2}$,
we deduce that

$I_{2}\leq\left\lVert\boldsymbol{\nabla}_{\\!\xi}v_{2}\right\rVert_{\mathcal{W}^{1,2}}\left\lVert\rho_{2}\right\rVert_{L^{\infty}_{x}}^{\frac{1}{2}}\left\lVert
v\right\rVert_{\mathcal{L}^{2}}^{2},$

and, as in previous proposition, applying Grönwall’s lemma proves (25).
Inequality (27) follows from the identity
$\boldsymbol{\rho}_{1}-\boldsymbol{\rho}_{2}=\frac{1}{2}(v_{1}-v_{2})(v_{1}+v_{2})+\frac{1}{2}(v_{1}+v_{2})(v_{1}-v_{2})$
and the Powers–Størmer inequality (see [24]). ∎

####  2.2.1. A Semiclassical inequality for commutators

######  Lemma 2.1.

There exists a constant $C>0$ such that for any self-adjoint trace class
operators $\boldsymbol{\rho}$ and $\boldsymbol{\mu}$, we have the following
estimate

$\frac{1}{\hbar}\left\lVert\left[{V_{\boldsymbol{\rho}},\boldsymbol{\mu}}\right]\right\rVert_{\mathcal{L}^{2}}\leq
C\left\lVert\rho\right\rVert_{L^{2}_{x}}\left\lVert\boldsymbol{\nabla}_{\\!\xi}\boldsymbol{\mu}\right\rVert_{\mathcal{W}^{1,2}}$

where $\rho=\operatorname{diag}\\!\left(\boldsymbol{\rho}\right)$.

###### Proof of Lemma 2.1.

We use a first order Taylor expansion of $V_{\boldsymbol{\rho}}$ of the form

$V_{\boldsymbol{\rho}}(x)=V_{\boldsymbol{\rho}}(y)+\left(x-y\right)\cdot\int_{0}^{1}\nabla
V_{\boldsymbol{\rho}}(z_{\theta})\,\mathrm{d}\theta$

where $z_{\theta}=\left(1-\theta\right)x+\theta\,y$. This implies that
$\frac{1}{i\hbar}\left[{V_{\boldsymbol{\rho}},\boldsymbol{\mu}}\right]$ is the
operator with kernel

$\frac{1}{i\hbar}\left[{V_{\boldsymbol{\rho}},\boldsymbol{\mu}}\right](x,y)=\int_{0}^{1}\nabla
V_{\boldsymbol{\rho}}(z_{\theta})\cdot\boldsymbol{\nabla}_{\\!\xi}\boldsymbol{\mu}(x,y)\,\mathrm{d}\theta.$

Its Hilbert–Schmidt norm being given by the $L^{2}$ norm of its kernel, and
doing the change of variables $(x,y)\mapsto(x+\theta y,x-(1-\theta)y)$ with
Jacobian equal to $1$, we obtain

$\displaystyle\left\lVert\tfrac{1}{i\hbar}\left[{V_{\boldsymbol{\rho}},\boldsymbol{\mu}}\right]\right\rVert_{\mathcal{L}^{2}}^{2}$
$\displaystyle\leq
h^{d}\int_{0}^{1}\\!\\!\\!\iint_{\mathbb{R}^{6}}\left\lvert\nabla
V_{\boldsymbol{\rho}}(x)\cdot\boldsymbol{\nabla}_{\\!\xi}\boldsymbol{\mu}(x+\theta
y,x-(1-\theta)y)\right\rvert^{2}\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}\theta$
$\displaystyle\leq h^{d}\left\lVert\nabla
V_{\boldsymbol{\rho}}\right\rVert_{L_{x}^{6}}^{2}\int_{0}^{1}\\!\\!\\!\int_{\mathbb{R}^{3}}\left\lVert\boldsymbol{\nabla}_{\\!\xi}\boldsymbol{\mu}(x+\theta
y,x-(1-\theta)y)\right\rVert_{L^{3}_{x}}^{2}\mathrm{d}y\,\mathrm{d}\theta,$

where the second inequality follows by Hölder’s inequality. The first factor
on the right-hand side is controlled using the Hardy–Littlewood–Sobolev
inequality by $\left\lVert\rho\right\rVert_{L^{2}_{x}}$. To control the second
factor, we perform the change of variable $x\mapsto x-\theta y$ to get

$\int_{0}^{1}\\!\\!\\!\int_{\mathbb{R}^{3}}\left\lVert\boldsymbol{\nabla}_{\\!\xi}\boldsymbol{\mu}(x+\theta
y,x-(1-\theta)y)\right\rVert_{L^{3}_{x}}^{2}\mathrm{d}y\,\mathrm{d}\theta=\int_{\mathbb{R}^{3}}\left\lVert\boldsymbol{\nabla}_{\\!\xi}\boldsymbol{\mu}(x,x-y)\right\rVert_{L^{3}_{x}}^{2}\mathrm{d}y.$

The $L^{3}$ norm in this term is now bounded using the
Gagliardo–Nirenberg–Sobolev inequality associated to the embedding
$H^{1}\subset L^{3}$ which yields

$\left\lVert\boldsymbol{\nabla}_{\\!\xi}\boldsymbol{\mu}(x,x-y)\right\rVert_{L^{3}_{x}}^{2}\leq
C\left(\left\lVert\boldsymbol{\nabla}_{\\!\xi}\boldsymbol{\mu}(x,x-y)\right\rVert_{L^{2}_{x}}^{2}+\left\lVert\nabla_{x}(\boldsymbol{\nabla}_{\\!\xi}\boldsymbol{\mu}(x,x-y))\right\rVert_{L^{2}_{x}}^{2}\right).$

However, notice now that the gradient with respect to $x$ appearing in the
above formula is actually also a quantum gradient of the operator since

$\nabla_{x}(\boldsymbol{\nabla}_{\\!\xi}\boldsymbol{\mu}(x,x-y))=((\nabla_{1}+\nabla_{2})\boldsymbol{\nabla}_{\\!\xi}\boldsymbol{\mu})(x,x-y)=\left[{\nabla,\boldsymbol{\nabla}_{\\!\xi}\boldsymbol{\mu}}\right](x,x-y)$

and this is exactly
$\boldsymbol{\nabla}_{\\!x}\boldsymbol{\nabla}_{\\!\xi}\boldsymbol{\mu}(x,x-y)$.
Using the fact that for both
$\boldsymbol{\nu}=\boldsymbol{\nabla}_{\\!\xi}\boldsymbol{\mu}$ and for
$\boldsymbol{\nu}=\boldsymbol{\nabla}_{\\!x}\boldsymbol{\nabla}_{\\!\xi}\boldsymbol{\mu}$,
by the Fubini theorem and the change of variables $y\mapsto x-y$ it holds

$\displaystyle
h^{d}\int_{\mathbb{R}^{3}}\left\lVert\boldsymbol{\nu}(x,x-y)\right\rVert_{L^{2}_{x}}^{2}\mathrm{d}y=h^{d}\iint_{\mathbb{R}^{6}}\left\lvert\boldsymbol{\nu}(x,x-y)\right\rvert^{2}\mathrm{d}x\,\mathrm{d}y=\left\lVert\boldsymbol{\nu}\right\rVert_{\mathcal{L}^{2}}^{2},$

we therefore arrive at the inequality

$\int_{0}^{1}\\!\\!\\!\int_{\mathbb{R}^{3}}\left\lVert\boldsymbol{\nabla}_{\\!\xi}\boldsymbol{\mu}(x+\theta
y,x-(1-\theta)y)\right\rVert_{L^{3}_{x}}^{2}\mathrm{d}y\,\mathrm{d}\theta\leq
C\left\lVert\boldsymbol{\nabla}_{\\!\xi}\boldsymbol{\mu}\right\rVert_{\mathcal{W}^{1,2}},$

which concludes the proof. ∎

##  3\. Semiclassical limit in $\mathcal{L}^{2}$ Schatten norm

We consider $\boldsymbol{\rho}$ a solution of the Hartree equation (2) and $f$
a solution of the Vlasov–Poisson equation (3). Then, in the same spirit as
[16], we will use the fact that the Weyl transform $\boldsymbol{\rho}_{f}$ of
$f$ solves the equation

(28)
$i\hbar\,\partial_{t}\boldsymbol{\rho}_{f}=\left[{H_{f},\boldsymbol{\rho}_{f}}\right]-B_{f}(\boldsymbol{\rho}_{f})\quad\text{
with
}\quad\boldsymbol{\rho}_{f}(0)=\boldsymbol{\rho}^{\mathrm{in}}_{f}:=\boldsymbol{\rho}_{f^{\mathrm{in}}}$

where $B_{f}(\boldsymbol{\rho}_{f})$ is the operator with kernel

(29)
$B_{f}(\boldsymbol{\rho}_{f})(x,y)=\left(V_{f}(x)-V_{f}(y)-\left(x-y\right)\cdot\nabla
V_{f}\Big{(}\frac{x+y}{2}\Big{)}\\!\right)\boldsymbol{\rho}_{f}(x,y).$

By the proof of [16, Proposition 4.4], we see that
$B_{f}(\boldsymbol{\rho}_{f})$ verifies the estimate

(30) $\frac{1}{\hbar}\left\lVert
B_{f}(\boldsymbol{\rho}_{f})\right\rVert_{\mathcal{L}^{2}}\leq
C\,\hbar\left\lVert\nabla
E_{f}\right\rVert_{L^{\infty}_{x}}\left\lVert\nabla_{\xi}^{2}f\right\rVert_{L^{2}},$

where $\left\lVert\nabla E_{f}\right\rVert_{L^{\infty}_{x}}$ is controlled by
$\left\lVert\rho_{f}\right\rVert_{W^{1,\infty}}$. Let $\boldsymbol{\rho}$ be a
solution to the Cauchy problem (2) and $\widetilde{\boldsymbol{\rho}}$ be a
solution to the following Cauchy problem for the linear Hartree equation

(31)
$i\hbar\,\partial_{t}\widetilde{\boldsymbol{\rho}}=\left[{H_{f},\widetilde{\boldsymbol{\rho}}}\right]\quad\text{
with
}\quad\widetilde{\boldsymbol{\rho}}(0)=\widetilde{\boldsymbol{\rho}}^{\mathrm{in}}:=\widetilde{\boldsymbol{\rho}}_{\sqrt{f^{\mathrm{in}}}}^{2}$

with Hamiltonian $H_{f}=-\frac{\hbar^{2}}{2}\,\Delta+V_{f}$.

######  Remark 3.1.

The choice
$\widetilde{\boldsymbol{\rho}}^{\mathrm{in}}=\widetilde{\boldsymbol{\rho}}_{\sqrt{f^{\mathrm{in}}}}^{2}$
guarantees the following property. Defining
$\tilde{v}^{\mathrm{in}}=\widetilde{\boldsymbol{\rho}}_{\sqrt{f^{\mathrm{in}}}}$,
then it holds

$\left\lVert\boldsymbol{\nabla}_{\\!\eta}\tilde{v}^{\mathrm{in}}\,\mathsf{m}\right\rVert_{\mathcal{L}^{p}}=\big{\lVert}\boldsymbol{\rho}_{g_{h}*\nabla_{\eta}\sqrt{f^{\mathrm{in}}}}\,\mathsf{m}\big{\rVert}_{\mathcal{L}^{p}}$

which, by [16, Proposition 3.2], is bounded by the regularity of
$\sqrt{f^{\mathrm{in}}}$.

Our aim is to show that the bound on the difference between
$\boldsymbol{\rho}$ and $\boldsymbol{\rho}_{f}$ in $\mathcal{L}^{2}$ is of
order $\hbar$. To this end, we consider $\widetilde{\boldsymbol{\rho}}$ the
solution to the auxiliary problem (31) with initial datum
$\widetilde{\boldsymbol{\rho}}^{\mathrm{in}}=\widetilde{\boldsymbol{\rho}}_{\sqrt{f^{\mathrm{in}}}}^{2}$.
Then, applying Minkowski’s inequality yields

(32)
$\left\lVert\boldsymbol{\rho}-\boldsymbol{\rho}_{f}\right\rVert_{\mathcal{L}^{2}}\leq\left\lVert\boldsymbol{\rho}-\widetilde{\boldsymbol{\rho}}\right\rVert_{\mathcal{L}^{2}}+\left\lVert\widetilde{\boldsymbol{\rho}}-\boldsymbol{\rho}_{f}\right\rVert_{\mathcal{L}^{2}}.$

The second term on the right-hand side corresponds to a measure of the defect
of positivity of the Weyl quantization of $f$. From the equations verified by
$\widetilde{\boldsymbol{\rho}}$ and $\boldsymbol{\rho}_{f}$ and Inequality
(30), it is bounded by Grönwall’s lemma by

(33)
$\left\lVert\widetilde{\boldsymbol{\rho}}-\boldsymbol{\rho}_{f}\right\rVert_{\mathcal{L}^{2}}\leq\left\lVert\widetilde{\boldsymbol{\rho}}^{\mathrm{in}}-\boldsymbol{\rho}_{f^{\mathrm{in}}}\right\rVert_{\mathcal{L}^{2}}+C\,\hbar\int_{0}^{t}\left\lVert\nabla
E_{f}(s,\cdot)\right\rVert_{L^{\infty}_{x}}\left\lVert\nabla_{\xi}^{2}f(s,\cdot)\right\rVert_{L^{2}}\mathrm{d}s,$

The first term on the right-hand side of Inequality (32) is the true nonlinear
error that corresponds to the stability estimate for the Hartree equation
proved in the previous section. It can be readily estimated in a similar
manner as in Section 2.2 by

(34)
$\left\lVert\boldsymbol{\rho}-\widetilde{\boldsymbol{\rho}}\right\rVert_{\mathcal{L}^{2}}\leq\big{\lVert}\sqrt{\boldsymbol{\rho}}-\sqrt{\widetilde{\boldsymbol{\rho}}}\,\big{\rVert}_{\mathcal{L}^{2}}\left(\left\lVert\sqrt{\boldsymbol{\rho}}\right\rVert_{\mathcal{L}^{\infty}}+\big{\lVert}\sqrt{\widetilde{\boldsymbol{\rho}}}\,\big{\rVert}_{\mathcal{L}^{\infty}}\right).$

We now estimate each term separately.

###  3.1. An estimate on the lack of positivity

For the second term on the right-hand side of Equation (33),
$\left\lVert\nabla E_{f}\right\rVert_{L^{\infty}_{x}}$ and
$\left\lVert\nabla^{2}_{\xi}f\right\rVert_{L^{2}}$ are bounded by [16,
Proposition A.1]. As for the first term, the difference between the initial
data $\widetilde{\boldsymbol{\rho}}^{\mathrm{in}}$ and
$\boldsymbol{\rho}_{f}^{\mathrm{in}}$ can be written as follows

$\widetilde{\boldsymbol{\rho}}^{\mathrm{in}}-\boldsymbol{\rho}_{f}^{\mathrm{in}}=\widetilde{\boldsymbol{\rho}}_{\sqrt{f^{\mathrm{in}}}}^{2}-\widetilde{\boldsymbol{\rho}}_{f}^{\mathrm{in}}+\widetilde{\boldsymbol{\rho}}_{f}^{\mathrm{in}}-\boldsymbol{\rho}_{f}^{\mathrm{in}}.$

Since, by Inequality (13), we have that

$\big{\lVert}\widetilde{\boldsymbol{\rho}}_{f}^{\mathrm{in}}-\boldsymbol{\rho}_{f}^{\mathrm{in}}\big{\rVert}_{\mathcal{L}^{2}}\leq\frac{3\,\hbar}{2}\left\lVert\nabla
f^{\mathrm{in}}\right\rVert_{L^{2}},$

it remains to estimate
$\widetilde{\boldsymbol{\rho}}_{\sqrt{f^{\mathrm{in}}}}^{2}-\widetilde{\boldsymbol{\rho}}_{f}^{\mathrm{in}}$
in $\mathcal{L}^{2}$. This is done via the following lemma.

######  Lemma 3.1.

Let $g\in W^{1,2p}$ with $p\in[1,\infty]$. Then

$\displaystyle\left\lVert\widetilde{\boldsymbol{\rho}}_{g^{2}}-\widetilde{\boldsymbol{\rho}}_{g}^{2}\right\rVert_{\mathcal{L}^{p}}\leq
48\,\hbar\left\lVert\nabla g\right\rVert_{L^{2p}}^{2}.$

###### Proof.

Notice that for any $(f,g)\in(L^{1}+L^{\infty})^{2}$,

$\widetilde{\boldsymbol{\rho}}_{f}\,\widetilde{\boldsymbol{\rho}}_{g}=h^{-6}\iint_{\mathbb{R}^{6}}f\,g^{\prime}\,G(z,z^{\prime})\,\ket{\psi_{z}}\bra{\psi_{z^{\prime}}}\,\mathrm{d}z\,\mathrm{d}z^{\prime}$

where $f=f(z),g^{\prime}=g(z^{\prime})$ and

$G(z,z^{\prime})=\Braket{\psi_{z}}{\psi_{z^{\prime}}}=\left(\tfrac{2}{h}\right)^{\frac{3}{2}}\int_{\mathbb{R}^{3}}e^{-\left(\left\lvert
y-x\right\rvert^{2}+\left\lvert
y-x^{\prime}\right\rvert^{2}\right)/(2\hbar)}\,e^{i\,y\cdot\left(\xi^{\prime}-\xi\right)/\hbar}\,\mathrm{d}y.$

Using the parallelogram identity, we can evaluate the integral in $y$ of $G$
as follows

$\displaystyle G(z,z^{\prime})$
$\displaystyle=\left(\tfrac{2}{h}\right)^{\frac{3}{2}}e^{-\left\lvert\frac{x-x^{\prime}}{2}\right\rvert^{2}/\hbar}\int_{\mathbb{R}^{3}}e^{-\left\lvert
y-\frac{x+x^{\prime}}{2}\right\rvert^{2}/\hbar}\,e^{i\,y\cdot\left(\xi^{\prime}-\xi\right)/\hbar}\,\mathrm{d}y$
$\displaystyle=\left(\tfrac{2}{h}\right)^{\frac{3}{2}}e^{-\left\lvert\frac{x-x^{\prime}}{2}\right\rvert^{2}/\hbar}e^{i\,\left(x+x^{\prime}\right)\cdot\left(\xi^{\prime}-\xi\right)/(2\hbar)}\operatorname{\mathcal{F}}\\!\left(e^{-2\pi\left\lvert
y\right\rvert^{2}/h}\right)(\tfrac{\xi-\xi^{\prime}}{h})$
$\displaystyle=e^{-\left\lvert\frac{z-z^{\prime}}{2}\right\rvert^{2}/\hbar}e^{i\,\left(x+x^{\prime}\right)\cdot\left(\xi^{\prime}-\xi\right)/(2\hbar)}.$

Notice also that $\widetilde{\boldsymbol{\rho}}_{1}=\mathds{1}$ is the
identity operator. Hence, we write

$\displaystyle\widetilde{\boldsymbol{\rho}}_{fg}$
$\displaystyle=\frac{\widetilde{\boldsymbol{\rho}}_{fg}\,\widetilde{\boldsymbol{\rho}}_{1}+\widetilde{\boldsymbol{\rho}}_{1}\,\widetilde{\boldsymbol{\rho}}_{fg}}{2}=h^{-6}\iint_{\mathbb{R}^{6}}\frac{f^{\prime}g^{\prime}+fg}{2}\,G(z,z^{\prime})\ket{\psi_{z}}\bra{\psi_{z^{\prime}}}\mathrm{d}z\,\mathrm{d}z^{\prime}.$

Therefore, since
$f\,g+f^{\prime}\,g^{\prime}-f\,g^{\prime}-f^{\prime}\,g=\left(f-f^{\prime}\right)\left(g-g^{\prime}\right)$,
we obtain

$\displaystyle\Lambda(f,g)$
$\displaystyle:=\widetilde{\boldsymbol{\rho}}_{fg}-\frac{\widetilde{\boldsymbol{\rho}}_{f}\,\widetilde{\boldsymbol{\rho}}_{g}+\widetilde{\boldsymbol{\rho}}_{g}\,\widetilde{\boldsymbol{\rho}}_{f}}{2}$
$\displaystyle=h^{-6}\iint_{\mathbb{R}^{6}}\frac{\left(f-f^{\prime}\right)\left(g-g^{\prime}\right)}{2}\,G(z,z^{\prime})\ket{\psi_{z}}\bra{\psi_{z^{\prime}}}\mathrm{d}z\,\mathrm{d}z^{\prime}.$

To obtain bounds on
$\Lambda(g,g)=\widetilde{\boldsymbol{\rho}}_{g^{2}}-\widetilde{\boldsymbol{\rho}}_{g}^{2}$,
we will now bound $\Lambda(f,g)$ by bilinear interpolation.

##### Trace norm estimate.

Using Minkowski’s inequality and the fact that the trace norm of
$\ket{\psi_{z}}\bra{\psi_{z^{\prime}}}$ is 1, we obtain the following trace
norm estimate

$\left\lVert\Lambda(f,g)\right\rVert_{\mathcal{L}^{1}}\leq\frac{h^{-3}}{2}\iint_{[0,1]^{2}}\iint_{\mathbb{R}^{6}\times\mathbb{R}^{6}}\left\lvert\nabla
f(z_{\theta})\right\rvert\left\lvert\nabla g(z_{\theta^{\prime}})\right\rvert
G_{2}(z-z^{\prime})\,\mathrm{d}z\,\mathrm{d}z^{\prime}\,\mathrm{d}\theta\,\mathrm{d}\theta^{\prime}\\\
\leq\frac{h^{-3}}{2}\left(\iint_{\mathbb{R}^{6}\times\mathbb{R}^{6}}\left\lvert\nabla
f(u)\right\rvert^{2}G_{2}(v)\,\mathrm{d}u\,\mathrm{d}v\right)^{\frac{1}{2}}\left(\iint_{\mathbb{R}^{6}\times\mathbb{R}^{6}}\left\lvert\nabla
g(u)\right\rvert^{2}G_{2}(v)\,\mathrm{d}u\,\mathrm{d}v\right)^{\frac{1}{2}}\\\
\leq 48\,\hbar\left\lVert\nabla f\right\rVert_{L^{2}}\left\lVert\nabla
g\right\rVert_{L^{2}}$

where $G_{2}(z)=\left\lvert
z\right\rvert^{2}e^{-\left\lvert\frac{z}{2}\right\rvert^{2}/\hbar}$ and
$z_{\theta}=\left(1-\theta\right)z+\theta\,z^{\prime}$.

##### Operator norm estimate.

Let $(\varphi,\phi)\in(L^{2})^{2}$ and define
$\psi_{\varphi}(z):=\Braket{\psi_{z}}{\varphi}$. Then

$\displaystyle\Braket{\phi}{\Lambda(f,g)\,\varphi}$
$\displaystyle=h^{-6}\iint_{\mathbb{R}^{6}\times\mathbb{R}^{6}}\frac{\left(f-f^{\prime}\right)\left(g-g^{\prime}\right)}{2}\,G(z,z^{\prime})\,\psi_{\phi}(z)\,\psi_{\varphi}(z^{\prime})\,\mathrm{d}z\,\mathrm{d}z^{\prime}$
$\displaystyle\leq h^{-6}\left\lVert\nabla
f\right\rVert_{L^{\infty}}\left\lVert\nabla
g\right\rVert_{L^{\infty}}\iint_{\mathbb{R}^{6}\times\mathbb{R}^{6}}G_{2}(z-z^{\prime})\left\lvert\psi_{\phi}(z)\right\rvert\left\lvert\psi_{\varphi}(z^{\prime})\right\rvert\mathrm{d}z\,\mathrm{d}z^{\prime}$
$\displaystyle\leq h^{-6}\left\lVert\nabla
f\right\rVert_{L^{\infty}}\left\lVert\nabla
g\right\rVert_{L^{\infty}}\left\lVert
G_{2}\right\rVert_{L^{1}}\left\lVert\psi_{\phi}\right\rVert_{L^{2}}\left\lVert\psi_{\varphi}\right\rVert_{L^{2}}$

where the last inequality follows from Young’s inequality. Observing that
$\psi_{\varphi}(z)=\operatorname{\mathcal{F}}\\!\left(\psi_{0}(x-\cdot)\,\varphi\right)(\xi/h)$,
we deduce that
$\left\lVert\psi_{\varphi}\right\rVert_{L^{2}}=h^{\frac{3}{2}}\left\lVert\varphi\right\rVert_{L^{2}}$
and so

$\left\lVert\Lambda(f,g)\right\rVert_{\mathcal{L}^{\infty}}\leq
48\,\hbar\left\lVert\nabla f\right\rVert_{L^{\infty}}\left\lVert\nabla
g\right\rVert_{L^{\infty}}.$

##### General Schatten norms.

By the complex bilinear interpolation (see e.g. [5, Section 4.4]), we deduce
that for any $p\in[1,\infty]$,

$\left\lVert\Lambda(f,g)\right\rVert_{\mathcal{L}^{p}}\leq
48\,\hbar\left\lVert\nabla f\right\rVert_{L^{2p}}\left\lVert\nabla
g\right\rVert_{L^{2p}}$

which leads to the desired result by taking $f=g$. ∎

###  3.2. Bound on
$\left\lVert\boldsymbol{\rho}-\widetilde{\boldsymbol{\rho}}\right\rVert_{\mathcal{L}^{2}}$

In Equation (34),
$\left\lVert\sqrt{\boldsymbol{\rho}}\right\rVert_{\mathcal{L}^{\infty}}$ and
$\big{\lVert}\sqrt{\widetilde{\boldsymbol{\rho}}}\big{\rVert}_{\mathcal{L}^{\infty}}$
are bounded because $\boldsymbol{\rho}\in\mathcal{L}^{\infty}$. The main
result of this section is Proposition 3.1 which gives an estimate for
$\big{\lVert}\sqrt{\boldsymbol{\rho}}-\sqrt{\widetilde{\boldsymbol{\rho}}}\big{\rVert}_{\mathcal{L}^{2}}$
in terms of $\hbar$.

To prove the proposition, we need to first establish some preliminary results.
It will be useful to exchange the role of $x$ and $\xi$ to manipulate weights
of the form $\left\langle x\right\rangle$ instead of weights of the form
$\left\langle\boldsymbol{p}\right\rangle$. The quantum analogue of exchanging
the $x$ and $\xi$ variables is obtained by conjugation with the semiclassical
Fourier transform
$\mathscr{F}_{h}\varphi(\xi):=h^{\frac{3}{2}}\,\widehat{\varphi}(h\xi)$. More
precisely, defining
$\boldsymbol{\rho}^{\star}:=\mathscr{F}^{-1}_{h}\boldsymbol{\rho}\,\mathscr{F}_{h}$,
then it holds

(35) $\boldsymbol{\rho}_{f}^{\star}=\boldsymbol{\rho}_{f^{\star}}\quad\text{
with }\quad f^{\star}(x,\xi)=f(\xi,x).$

This exchange operation is a linear automorphism which preserves the
$\mathcal{L}^{p}$ norms. Moreover, from the definition of a Fourier
multiplier,
$\left\langle\boldsymbol{p}\right\rangle\mathscr{F}_{h}=\mathscr{F}_{h}\left\langle
x\right\rangle$ and so $\left\langle
x\right\rangle^{\star}=\left\langle\boldsymbol{p}\right\rangle$ and
$\left\langle\boldsymbol{p}\right\rangle^{\star}=\left\langle x\right\rangle$.

###### Proof of (35).

Computing the kernel of $\boldsymbol{\rho}_{f}^{\star}$ yields

$\boldsymbol{\rho}_{f}^{\star}(x,y)=\frac{1}{h^{3}}\int_{\mathbb{R}^{9}}e^{2\pi
i\,\left(x\cdot x^{\prime}-y^{\prime}\cdot y\right)/h}\,e^{-2\pi
i\,\left(y^{\prime}-x^{\prime}\right)\cdot\xi}\,f\\!\left(\tfrac{x^{\prime}+y^{\prime}}{2},h\,\xi\right)\mathrm{d}x^{\prime}\,\mathrm{d}y^{\prime}\,\mathrm{d}\xi$

Therefore, by the change of variables
$\zeta=\frac{x^{\prime}+y^{\prime}}{2\,h}$ and $\eta=x^{\prime}-y^{\prime}$

$\boldsymbol{\rho}_{f}^{\star}(x,y)=\int_{\mathbb{R}^{6}}e^{-2\pi
i\,\left(y-x\right)\cdot\zeta/h}\,e^{2i\pi\eta\cdot\frac{x+y}{2\,h}}\,\mathscr{F}\\!\left(f(h\,\zeta,h\,\cdot)\right)\\!(\eta)\,\mathrm{d}\eta\,\mathrm{d}\zeta$

and we deduce Equation (35) by the Fourier inversion theorem. ∎

We can now use the above defined operation to prove the following lemma.

######  Lemma 3.2.

Let $f\in H^{1}$ be a function on the phase space. Then the following
estimates hold

(36a) $\displaystyle\left\lVert\boldsymbol{\rho}_{\left\langle\xi\right\rangle
f}-\boldsymbol{\rho}_{f}\left\langle\boldsymbol{p}\right\rangle\right\rVert_{\mathcal{L}^{2}}$
$\displaystyle\leq\tfrac{\hbar}{2}\left\lVert\nabla_{x}f\right\rVert_{L^{2}},$
(36b)
$\displaystyle\left\lVert\left\langle\boldsymbol{p}\right\rangle\boldsymbol{\rho}_{f}\left\langle\boldsymbol{p}\right\rangle-\boldsymbol{\rho}_{\left\langle\xi\right\rangle^{2}f}\right\rVert_{\mathcal{L}^{2}}$
$\displaystyle\leq\tfrac{\hbar^{2}}{4}\left\lVert\Delta_{x}f\right\rVert_{L^{2}}.$
Notice that the same inequality holds when we replace
$\boldsymbol{\rho}_{f}\left\langle\boldsymbol{p}\right\rangle$ by
$\left\langle\boldsymbol{p}\right\rangle\boldsymbol{\rho}_{f}$, which follows
by taking the adjoint.

###### Proof.

Notice the kernel of the difference $\boldsymbol{\rho}_{\left\langle
x\right\rangle f}-\boldsymbol{\rho}_{f}\left\langle x\right\rangle$ is given
by

$\int_{\mathbb{R}^{3}}e^{-2\pi
i\left(y-x\right)\cdot\xi}\left(\left\langle\tfrac{x+y}{2}\right\rangle-\left\langle
y\right\rangle\right)f\left(\tfrac{x+y}{2},h\xi\right)\mathrm{d}\xi\\\
=\frac{i\hbar}{2}\frac{(\frac{x+y}{2})+y}{\left\langle\tfrac{x+y}{2}\right\rangle+\left\langle
y\right\rangle}\cdot\int_{\mathbb{R}^{3}}e^{-2\pi
i\left(y-x\right)\cdot\xi}\,\nabla_{\xi}f\left(\tfrac{x+y}{2},h\xi\right)\,\mathrm{d}\xi=\boldsymbol{\rho}_{r}$

where $r\in L^{2}$ is defined by
$r(z)=\tfrac{i\hbar}{2}\,(a_{x}(i\hbar\,\nabla_{\xi})\cdot\nabla_{\xi}f)(z)$
where $a_{x}(i\hbar\,\nabla_{\xi})$ is a bounded, $x$-dependent Fourier
multiplier of the $\xi$ variable associated to the function
$a_{x}(\eta)=\frac{2\,x-\eta/2}{\left\langle x\right\rangle+\left\langle
x-\eta/2\right\rangle}$. More explicitly

$r(z)=\frac{i\hbar}{2}\int_{\mathbb{R}^{6}}e^{2\pi
i\left(\xi-\xi^{\prime}\right)\cdot\eta}\,a_{x}(h\eta)\cdot\nabla_{\xi}f(x,\xi^{\prime})\,\mathrm{d}\xi^{\prime}\,\mathrm{d}\eta.$

In particular, it follows that

(37) $\left\lVert\boldsymbol{\rho}_{\left\langle x\right\rangle
f}-\boldsymbol{\rho}_{f}\left\langle
x\right\rangle\right\rVert_{\mathcal{L}^{2}}=\left\lVert\boldsymbol{\rho}_{r}\right\rVert_{\mathcal{L}^{2}}=\left\lVert
r\right\rVert_{L^{2}}\leq\tfrac{\hbar}{2}\left\lVert\nabla_{\xi}f\right\rVert_{L^{2}}.$

We now use Inequality (37) together with the properties of the exchange
operation (35) to get

$\displaystyle\left\lVert\boldsymbol{\rho}_{\left\langle\xi\right\rangle
f}-\boldsymbol{\rho}_{f}\left\langle\boldsymbol{p}\right\rangle\right\rVert_{\mathcal{L}^{2}}$
$\displaystyle=\left\lVert\boldsymbol{\rho}_{\left\langle\xi\right\rangle
f}^{\star}-\boldsymbol{\rho}_{f}^{\star}\left\langle
x\right\rangle\right\rVert_{\mathcal{L}^{2}}=\left\lVert\boldsymbol{\rho}_{\left\langle
x\right\rangle f^{\star}}-\boldsymbol{\rho}_{f^{\star}}\left\langle
x\right\rangle\right\rVert_{\mathcal{L}^{2}}$
$\displaystyle\leq\hbar\left\lVert\nabla_{\xi}f^{\star}\right\rVert_{L^{2}}=\hbar\left\lVert(\nabla_{x}f)^{\star}\right\rVert_{L^{2}}=\hbar\left\lVert\nabla_{x}f\right\rVert_{L^{2}},$

which completes the proof of Inequality (36a). To get the second inequality, a
similar computation gives that

$\left\langle x\right\rangle\boldsymbol{\rho}_{f}\left\langle
x\right\rangle-\boldsymbol{\rho}_{\left\langle
x\right\rangle^{2}f}=\boldsymbol{\rho}_{r_{2}}$

with
$r_{2}(z)=\tfrac{-\hbar^{2}}{4}\,(b_{x}(i\hbar\,\nabla_{\xi})\cdot\Delta_{\xi}f)(z)$,
where $b_{x}$ is the bounded function given by

$b_{x}(2\eta)=\tfrac{2\left\lvert
x\right\rvert^{2}-2-\left\lvert\eta\right\rvert^{2}}{\left\langle
x\right\rangle^{2}+\left\langle x+\eta\right\rangle\left\langle
x-\eta\right\rangle}=\tfrac{\left\lvert
x\right\rvert^{2}-2+\left(x+\eta\right)\left(x-\eta\right)}{\left\langle
x\right\rangle^{2}+\left\langle x+\eta\right\rangle\left\langle
x-\eta\right\rangle},$

and the same reasoning leads to Inequality (36b). ∎

Another simple observation is that the commutator between convolution with the
Gaussian function $g_{h}(z)=\left(\pi\hbar\right)^{-3}e^{-\left\lvert
z\right\rvert^{2}/\hbar}$ and multiplication by a weight is also of order
$\hbar$.

######  Lemma 3.3.

Let $h\in(0,1)$. Then there exists $C>0$ such that for any $p\in[1,\infty]$
and any function of the phase space $f\in W^{1,p}$ it holds

(38a)
$\displaystyle\left\lVert\left[{\left\langle\xi\right\rangle,\sim}\right]f\right\rVert_{W^{2,p}}\leq
C\,\hbar\left\lVert f\right\rVert_{W^{3,p}}$ (38b)
$\displaystyle\left\lVert\left[{\langle\xi\rangle^{2},\sim}\right]f\right\rVert_{L^{p}}\leq
C\,\hbar\left\lVert f\right\rVert_{W^{1,p}_{1}}$

where $\left[{m,\sim}\right]f:=m\left(g_{h}*f\right)-\left(g_{h}*(mf)\right)$.

###### Proof.

For any $z=(x,\xi)\in\mathbb{R}^{6}$ and any general $m(\xi)$, it holds

$\left[{m,\sim}\right]f(z)=\int_{\mathbb{R}^{3}}g_{h}(z^{\prime})\left(m(\xi)-m(\xi-\xi^{\prime})\right)f(z-z^{\prime})\,\mathrm{d}z^{\prime}.$

By a second order Taylor expansion, this can be written as

$\left[{m,\sim}\right]f=\int_{\mathbb{R}^{3}}g_{h}(z^{\prime})\,\xi^{\prime}\cdot\left(\nabla
m(\xi)-\int_{0}^{1}\theta\,\xi^{\prime}\cdot\nabla^{2}m(\xi-\theta\xi^{\prime})\,\mathrm{d}\theta\right)f(z-z^{\prime})\,\mathrm{d}z^{\prime}=I_{1}-I_{2}.$

Next, using the fact that that $g_{h}$ is even, one can replace
$f(z-z^{\prime})$ by $f(z-z^{\prime})-f(z)$ in $I_{1}$, leading to

$I_{1}=\nabla
m(\xi)\cdot\int_{0}^{1}\\!\\!\\!\int_{\mathbb{R}^{3}}\xi^{\prime}\,g_{h}(z^{\prime})\,\xi^{\prime}\cdot\,\nabla_{\xi}f(z-\theta\xi^{\prime})\,\mathrm{d}z^{\prime}\,\mathrm{d}\theta.$

By Minkowski’s inequality, we obtain

$\displaystyle\left\lVert I_{1}\right\rVert_{L^{p}}\leq\left\lVert\nabla
m\right\rVert_{L^{\infty}}\,\big{\lVert}\\!\left\lvert\xi\right\rvert^{2}g_{h}\big{\rVert}_{L^{1}}\left\lVert\nabla_{\xi}f\right\rVert_{L^{p}}=\tfrac{3}{2}\,\hbar\,\left\lVert\nabla
m\right\rVert_{L^{\infty}}\left\lVert\nabla_{\xi}f\right\rVert_{L^{p}}.$

Similarly, $I_{2}$ is bounded by

$\displaystyle\left\lVert
I_{2}\right\rVert_{L^{p}}\leq\frac{1}{2}\left\lVert\nabla^{2}m\right\rVert_{L^{\infty}}\,\big{\lVert}\\!\left\lvert\xi\right\rvert^{2}g_{h}\big{\rVert}_{L^{1}}\left\lVert
f\right\rVert_{L^{p}}=\tfrac{3}{4}\,\hbar\,\left\lVert\nabla^{2}m\right\rVert_{L^{\infty}}\left\lVert
f\right\rVert_{L^{p}}.$

In particular, if $m=\left\langle\xi\right\rangle$, then it yields
$\left\lVert\left[{m,\sim}\right]f\right\rVert_{L^{p}}\leq 3\,\hbar\left\lVert
f\right\rVert_{W^{1,p}}$. Noticing that
$\nabla_{x}^{2}\left[{m,\sim}\right]f=\left[{m,\sim}\right]\nabla_{x}^{2}f$
also yields
$\left\lVert\nabla_{x}^{2}\left[{m,\sim}\right]f\right\rVert_{L^{p}}\leq
3\,\hbar\left\lVert\nabla_{x}^{2}f\right\rVert_{W^{1,p}}$. Finally, noticing
that

$\nabla_{\xi}^{2}\left[{m,\sim}\right]f=\left[{m,\sim}\right]\nabla_{\xi}^{2}f+2\left[{\nabla
m,\sim}\right]\nabla_{\xi}f+\left[{\nabla^{2}m,\sim}\right]f$

and using the above estimates with $m$ replaced by $\nabla m$ and
$\nabla^{2}m$ leads to

$\left\lVert\nabla_{\xi}^{2}\left[{m,\sim}\right]f\right\rVert_{L^{p}}\leq
36\,\hbar\left\lVert f\right\rVert_{H^{3}}$

and so to Inequality (38a). To get Inequality (38b), notice that
$\left[{\left\langle\xi\right\rangle^{2},\sim}\right]f=\left[{\left\lvert\xi\right\rvert^{2},\sim}\right]f$
and write

$\left[{\left\lvert\xi\right\rvert^{2},\sim}\right]f=\int_{\mathbb{R}^{6}}g_{h}(z^{\prime})\left(\xi^{\prime}\right)^{\otimes
2}:\left(f(z-z^{\prime})\,\mathrm{Id}+2\,\xi\otimes\int_{0}^{1}\nabla_{\xi}f(z-\theta\,\xi^{\prime})\,\mathrm{d}\theta\right)\mathrm{d}z^{\prime}$

so that

$\displaystyle\left\lVert\left[{\left\lvert\xi\right\rvert^{2},\sim}\right]f\right\rVert_{L^{p}}\leq\tfrac{3\,\hbar}{2}\left(\left\lVert
f\right\rVert_{L^{p}}+2\left\lVert\left\lvert\xi\right\rvert\nabla_{\xi}f\right\rVert_{L^{p}}\right)+\tfrac{16\,\hbar^{\frac{3}{2}}}{\sqrt{\pi}}\left\lVert\nabla_{\xi}f\right\rVert_{L^{p}},$

and this concludes the proof. ∎

######  Lemma 3.4.

Let $h\in(0,1)$. Then there exists $C>0$ such that for any function $g$ on the
phase space

$\left\lVert\left\langle\boldsymbol{p}\right\rangle\left(\widetilde{\boldsymbol{\rho}}_{g}^{2}-\widetilde{\boldsymbol{\rho}}_{g^{2}}\right)\left\langle\boldsymbol{p}\right\rangle\right\rVert_{\mathcal{L}^{2}}\leq
C\,\hbar\left(\left\lVert g\right\rVert_{W^{1,4}_{1}\cap
H^{3}}^{2}+\left\lVert g^{2}\right\rVert_{H^{1}_{1}\cap H^{2}}\right).$

###### Proof.

Let $\mathsf{m}:=\left\langle\boldsymbol{p}\right\rangle$ and
$m(\xi)=\left\langle\xi\right\rangle$. By the definition of the absolute value
for operators, by the proof of Lemma 3.2 and by Lemma 3.3, we can write

$\mathsf{m}\,\widetilde{\boldsymbol{\rho}}_{g}^{2}\,\mathsf{m}=\left\lvert\widetilde{\boldsymbol{\rho}}_{g}\mathsf{m}\right\rvert^{2}=\left\lvert\boldsymbol{\rho}_{\tilde{g}}\mathsf{m}\right\rvert^{2}=\left\lvert\boldsymbol{\rho}_{m\,\tilde{g}}+\boldsymbol{\rho}_{r}\right\rvert^{2}=\left\lvert\widetilde{\boldsymbol{\rho}}_{m\,g}+\boldsymbol{\rho}_{r_{1}}\right\rvert^{2}$

where $r_{1}=\left[{m,\sim}\right]g+r$, and

$\mathsf{m}\,\widetilde{\boldsymbol{\rho}}_{g^{2}}\,\mathsf{m}=\boldsymbol{\rho}_{m^{2}\widetilde{g^{2}}}+\boldsymbol{\rho}_{r_{2}}=\widetilde{\boldsymbol{\rho}}_{m^{2}\,g^{2}}+\boldsymbol{\rho}_{r_{0}}$

where $r_{0}=\left[{m^{2},\sim}\right]g^{2}+r_{2}$. It follows that

(39)
$\left\lVert\mathsf{m}\Big{(}\widetilde{\boldsymbol{\rho}}_{g}^{2}-\widetilde{\boldsymbol{\rho}}_{g^{2}}\Big{)}\mathsf{m}\right\rVert_{\mathcal{L}^{2}}=\left\lVert\left\lvert\widetilde{\boldsymbol{\rho}}_{m\,g}+\boldsymbol{\rho}_{r_{1}}\right\rvert^{2}-\widetilde{\boldsymbol{\rho}}_{m^{2}\,g^{2}}-\boldsymbol{\rho}_{r_{0}}\right\rVert_{\mathcal{L}^{2}}\\\
\leq\left\lVert\widetilde{\boldsymbol{\rho}}_{m\,g}^{2}-\widetilde{\boldsymbol{\rho}}_{m^{2}\,g^{2}}\right\rVert_{\mathcal{L}^{2}}+2\left\lVert\widetilde{\boldsymbol{\rho}}_{m\,g}\,\boldsymbol{\rho}_{r_{1}}\right\rVert_{\mathcal{L}^{2}}+\left\lVert\boldsymbol{\rho}_{r_{1}}\right\rVert_{\mathcal{L}^{4}}^{2}+\left\lVert\boldsymbol{\rho}_{r_{0}}\right\rVert_{\mathcal{L}^{2}}.$

The first term on the right-hand side of Inequality (39) is bounded by Lemma
3.1 with $p=2$, leading to

$\left\lVert\widetilde{\boldsymbol{\rho}}_{m\,g}^{2}-\widetilde{\boldsymbol{\rho}}_{m^{2}\,g^{2}}\right\rVert_{\mathcal{L}^{2}}\leq
48\,\hbar\left\lVert\nabla(m\,g)\right\rVert_{L^{4}}^{2}.$

The second term can be bounded by Hölder’s inequality and Inequality (14) as
follows

$\left\lVert\widetilde{\boldsymbol{\rho}}_{m\,g}\,\boldsymbol{\rho}_{r_{1}}\right\rVert_{\mathcal{L}^{2}}\leq\left\lVert\widetilde{\boldsymbol{\rho}}_{m\,g}\right\rVert_{\mathcal{L}^{4}}\left\lVert\boldsymbol{\rho}_{r_{1}}\right\rVert_{\mathcal{L}^{4}}\leq
C\,\hbar\left\lVert m\,g\right\rVert_{L^{4}}\left\lVert
g\right\rVert_{H^{3}}.$

The third term can be bounded by

$\left\lVert\boldsymbol{\rho}_{r_{1}}\right\rVert_{\mathcal{L}^{4}}\leq\left\lVert\boldsymbol{\rho}_{r_{1}}-\widetilde{\boldsymbol{\rho}}_{r_{1}}\right\rVert_{\mathcal{L}^{4}}+\left\lVert\widetilde{\boldsymbol{\rho}}_{r_{1}}\right\rVert_{\mathcal{L}^{4}}$

and then noticing that the Schatten-$4$ norm is controlled by the
Hilbert–Schmidt norm, one obtains
$\left\lVert\boldsymbol{\rho}_{r_{1}}-\widetilde{\boldsymbol{\rho}}_{r_{1}}\right\rVert_{\mathcal{L}^{4}}\leq
h^{-\frac{3}{4}}\left\lVert\boldsymbol{\rho}_{r_{1}}-\widetilde{\boldsymbol{\rho}}_{r_{1}}\right\rVert_{\mathcal{L}^{2}}\leq
h^{\frac{1}{4}}\left\lVert\nabla^{2}r_{1}\right\rVert_{L^{2}}$, where we used
Inequality (13) in the last inequality. Hence we deduce that

$\left\lVert\boldsymbol{\rho}_{r_{1}}\right\rVert_{\mathcal{L}^{4}}\leq\left\lVert
r_{1}\right\rVert_{L^{4}}+h^{\frac{1}{4}}\left\lVert\nabla^{2}r_{1}\right\rVert_{L^{2}}\leq
C\,\hbar\left\lVert g\right\rVert_{H^{3}}.$

Finally, the last term can be bounded by

$\displaystyle\left\lVert\boldsymbol{\rho}_{r_{0}}\right\rVert_{\mathcal{L}^{2}}\leq
C\,\hbar\left(\left\lVert
g^{2}\right\rVert_{H^{1}_{1}}+\hbar\left\lVert\Delta_{x}g^{2}\right\rVert_{L^{2}}\right).$

Combining all the inequalities leads to the desired result. ∎

######  Lemma 3.5.

Let $\boldsymbol{\rho}_{f}$ be a solution to Equation (28) and
$\widetilde{\boldsymbol{\rho}}$ be a solution to Equation (31). Then

$\left\lVert\operatorname{diag}(\widetilde{\boldsymbol{\rho}}-\boldsymbol{\rho}_{f})\right\rVert_{L^{2}_{x}}\leq
C\,\hbar\left(C^{\mathrm{in}}+\left\lVert\rho_{f}\right\rVert_{W^{1,\infty}_{x}}\left\lVert\boldsymbol{\rho}_{f}\right\rVert_{\mathcal{W}^{2,2}(\left\langle\boldsymbol{p}\right\rangle^{2})}\right),$

where $C^{\mathrm{in}}$ is given by Equation (24).

###### Proof.

Let $\mathsf{m}=\left\langle\boldsymbol{p}\right\rangle$. By the proof of [8,
Proposition 6.4] and the equation verified by
$\widetilde{\boldsymbol{\rho}}-\boldsymbol{\rho}_{f}$, we obtain

$\displaystyle\left\lVert\operatorname{diag}(\widetilde{\boldsymbol{\rho}}-\boldsymbol{\rho}_{f})\right\rVert_{L^{2}_{x}}$
$\displaystyle\leq
C\left\lVert\mathsf{m}\,(\widetilde{\boldsymbol{\rho}}-\boldsymbol{\rho}_{f})\,\mathsf{m}\right\rVert_{\mathcal{L}^{2}}$
$\displaystyle\leq
C\left\lVert\mathsf{m}\,(\widetilde{\boldsymbol{\rho}}^{\mathrm{in}}-\boldsymbol{\rho}_{f}^{\mathrm{in}})\,\mathsf{m}\right\rVert_{\mathcal{L}^{2}}+\frac{C}{\hbar}\int_{0}^{t}\left\lVert
B_{f}(\boldsymbol{\rho}_{f})\,\mathsf{m}^{2}\right\rVert_{\mathcal{L}^{2}}\mathrm{d}s.$

With our choice of the initial datum
$\widetilde{\boldsymbol{\rho}}^{\mathrm{in}}=\widetilde{\boldsymbol{\rho}}^{2}_{\sqrt{f^{\mathrm{in}}}}$,
the first term on the right-hand side is bounded in Lemma 3.4. As for the
second term, we compute explicitly the $L^{2}$ norm of the kernel of the
operator $B_{f}(\boldsymbol{\rho}_{f})\,\mathsf{m}^{2}$. Recall the definition
of $B_{f}(\boldsymbol{\rho})$ given by Formula (29) and define

$\delta^{2}V(x,y):=V_{f}(x)-V_{f}(y)-\left(x-y\right)\cdot\nabla
V_{f}\Big{(}\frac{x+y}{2}\Big{)}$

so that the integral kernel of $B$ can be written
$B_{f}(\boldsymbol{\rho})(x,y)=\delta^{2}V(x,y)\,\boldsymbol{\rho}(x,y)$. Then
by the chain rule, the integral kernel of the operator
$B_{f}(\boldsymbol{\rho}_{f})\left\lvert\boldsymbol{p}\right\rvert^{2}$ is
given by

$\displaystyle\big{(}B_{f}(\boldsymbol{\rho}_{f})\left\lvert\boldsymbol{p}\right\rvert^{2}\big{)}(x,y)$
$\displaystyle=-\hbar^{2}\,\delta^{2}V(x,y)\,\Delta_{y}\boldsymbol{\rho}_{f}(x,y)-\hbar^{2}\Delta_{y}\\!\left(\delta^{2}V(x,y)\right)\boldsymbol{\rho}_{f}(x,y)$
$\displaystyle\qquad-2\,\hbar^{2}\,\nabla_{y}\delta^{2}V(x,y)\cdot\nabla_{y}\boldsymbol{\rho}_{f}(x,y)=:J_{1}+J_{2}+J_{3}.$

The $J_{1}$ term is simply
$B_{f}(\boldsymbol{\rho}_{f}\left\lvert\boldsymbol{p}\right\rvert^{2})$ which
can be bounded in $\mathcal{L}^{2}$ by Inequality (30), leading to

$\displaystyle\left\lVert
B_{f}(\boldsymbol{\rho}_{f}\left\lvert\boldsymbol{p}\right\rvert^{2})\right\rVert_{\mathcal{L}^{2}}\leq
C\,\hbar^{2}\left\lVert\rho_{f}\right\rVert_{W^{1,\infty}_{x}}\left\lVert\boldsymbol{\nabla}_{\\!\xi}^{2}\\!\left(\boldsymbol{\rho}_{f}\left\lvert\boldsymbol{p}\right\rvert^{2}\right)\right\rVert_{\mathcal{L}^{2}}.$

For the $J_{2}$ term, since $V$ is the solution of the Poisson equation, we
have that

$\displaystyle-\hbar^{2}\,\Delta_{y}\delta^{2}V(x,y)=\pm\hbar^{2}\Big{(}\rho_{f}\Big{(}\frac{x+y}{2}\Big{)}-\rho_{f}(y)\Big{)}-\tfrac{\pm\hbar^{2}}{4}\,\nabla\rho_{f}\Big{(}\frac{x+y}{2}\Big{)}\cdot(x-y)$

which means

$\displaystyle
J_{2}(x,y)=\pm\hbar^{2}\Big{(}\rho_{f}\Big{(}\frac{x+y}{2}\Big{)}-\rho_{f}(y)\Big{)}\boldsymbol{\rho}_{f}(x,y)\pm\frac{\hbar^{3}}{4i}\nabla\rho_{f}\Big{(}\frac{x+y}{2}\Big{)}\cdot\boldsymbol{\nabla}_{\\!\xi}\boldsymbol{\rho}_{f}(x,y).$

Then we deduce that

$\left\lVert J_{2}\right\rVert_{\mathcal{L}^{2}}\leq
2\,\hbar^{2}\left\lVert\rho_{f}\right\rVert_{L^{\infty}_{x}}\left\lVert\boldsymbol{\rho}_{f}\right\rVert_{\mathcal{L}^{2}}+\tfrac{\hbar^{3}}{4}\left\lVert\nabla\rho_{f}\right\rVert_{L^{\infty}_{x}}\left\lVert\boldsymbol{\nabla}_{\\!\xi}\boldsymbol{\rho}_{f}\right\rVert_{\mathcal{L}^{2}}.$

Lastly, for the $J_{3}$ term, observe that

$\displaystyle-2\,\hbar^{2}\,\nabla_{y}\delta^{2}V(x,y)=2\,\hbar^{2}\left(E_{f}(\tfrac{x+y}{2})-E_{f}(y)\right)+\hbar^{2}\left(x-y\right)\cdot\nabla
E_{f}\left(\tfrac{x+y}{2}\right)$

which means

$J_{3}(x,y)=i\hbar^{3}\int^{1}_{0}\nabla
E_{f}(y+t\,\tfrac{x-y}{2}):\nabla_{y}\boldsymbol{\nabla}_{\\!\xi}\boldsymbol{\rho}_{f}(x,y)\,\mathrm{d}t\\\
+i\hbar^{3}\nabla
E_{f}\left(\tfrac{x+y}{2}\right):\nabla_{y}\boldsymbol{\nabla}_{\\!\xi}\boldsymbol{\rho}_{f}(x,y).$

Hence it follows that

$\displaystyle\left\lVert J_{3}\right\rVert_{\mathcal{L}^{2}}\leq
C\,\hbar^{2}\left\lVert\nabla
E_{f}\right\rVert_{L^{\infty}_{x}}\left\lVert\boldsymbol{\nabla}_{\\!\xi}\boldsymbol{\rho}_{f}\,\boldsymbol{p}\right\rVert_{\mathcal{L}^{2}}.$

This completes our proof of the lemma. ∎

######  Proposition 3.1.

Let $\boldsymbol{\rho}$ be a solution to the Hartree equation (2) with initial
datum $\boldsymbol{\rho}^{\mathrm{in}}$ and $\widetilde{\boldsymbol{\rho}}$ be
a solution to (31) with initial datum
$\widetilde{\boldsymbol{\rho}}^{\mathrm{in}}$. We denote by
$v_{1}:=\sqrt{\boldsymbol{\rho}}$,
$\tilde{v}:=\sqrt{\widetilde{\boldsymbol{\rho}}}$. Then there exists
$\lambda(t)\in C^{0}(\mathbb{R}_{+},\mathbb{R}_{+})$ such that

$\left\lVert v_{1}-\tilde{v}\right\rVert_{\mathcal{L}^{2}}^{2}\leq\left\lVert
v_{1}^{\mathrm{in}}-\tilde{v}^{\mathrm{in}}\right\rVert_{\mathcal{L}^{2}}^{2}e^{2\,\Lambda(t)}+\hbar^{2}\int_{0}^{t}c(s)^{2}\,e^{2\left(\Lambda(t)-\Lambda(s)\right)}\,\mathrm{d}s$

where $\Lambda(t)=C\int_{0}^{t}\lambda(s)\,\mathrm{d}s$ with $\lambda$ given
by (21) and $c(t)$ is given by

$c(t)=C\,\left\lVert\boldsymbol{\nabla}_{\\!\xi}\tilde{v}\right\rVert_{\mathcal{W}^{1,2}}\left(C^{\mathrm{in}}+\left\lVert\rho_{f}\right\rVert_{W^{1,\infty}_{x}}\left\lVert\boldsymbol{\rho}_{f}\right\rVert_{\mathcal{W}^{2,2}(\left\langle\boldsymbol{p}\right\rangle^{2})}\right)$

with $C^{\mathrm{in}}$ given in Equation (24).

###### Proof.

Define $v:=v_{1}-\tilde{v}$. Observe that $v_{1}$ solves

$i\hbar\,\partial_{t}v_{1}=\left[{H_{1},v_{1}}\right],\quad\quad
H_{1}:=H_{v_{1}^{2}}$

and $\tilde{v}$ solves

$i\hbar\,\partial_{t}\tilde{v}=\left[{H_{f},\tilde{v}}\right].$

Now we proceed by mimicking the proof of Theorem 2.2. By direct computation,
we have that

$i\hbar\,\partial_{t}v=\left[{H_{1},v}\right]+\big{[}{V_{\boldsymbol{\rho}}-V_{\boldsymbol{\rho}_{f}},\tilde{v}}\big{]}$

which implies

(40) $\frac{1}{4\pi}\,\frac{\mathrm{d}}{\mathrm{d}t}\left\lVert
v\right\rVert_{\mathcal{L}^{2}}^{2}=h^{2}\operatorname{Tr}\\!\left(\big{[}{V_{v_{1}^{2}}-V_{\boldsymbol{\rho}_{f}},\tilde{v}}\big{]}v\right).$

Now adding and subtracting
$h^{2}\operatorname{Tr}\\!\left(\left[{V_{\tilde{\boldsymbol{\rho}}},\tilde{v}}\right]v\right)$
to the right-hand side of Equation (40) yields

(41) $\begin{split}\frac{1}{4\pi}\,\frac{\mathrm{d}}{\mathrm{d}t}\left\lVert
v\right\rVert_{\mathcal{L}^{2}}^{2}&=h^{2}\operatorname{Tr}\\!\left(\left[{V_{\boldsymbol{\rho}}-V_{\tilde{\boldsymbol{\rho}}},\tilde{v}}\right]v\right)+h^{2}\operatorname{Tr}\\!\left(\big{[}{V_{\tilde{\boldsymbol{\rho}}}-V_{\boldsymbol{\rho}_{f}},\tilde{v}}\big{]}v\right)\\\
&=h^{2}\operatorname{Tr}\\!\left(\big{[}{V_{v_{1}^{2}-\tilde{v}^{2}},\tilde{v}}\big{]}v\right)+h^{2}\operatorname{Tr}\\!\left(\big{[}{V_{\tilde{\boldsymbol{\rho}}-\boldsymbol{\rho}_{f}},\tilde{v}}\big{]}v\right)=:J_{1}+J_{2}.\end{split}$

We bound $J_{1}$ in the same manner as in Theorem 2.2 and get that

$J_{1}\leq
h^{2}\operatorname{Tr}\\!\left(\left[{V_{v^{2}},\tilde{v}}\right]v\right)+2\,h^{2}\operatorname{Tr}\\!\left(\left[{V_{\tilde{v}v},\tilde{v}}\right]v\right)=:J_{1,1}+2\,J_{1,2},$

where the two terms are readily bounded as follows

$\displaystyle J_{1,1}$ $\displaystyle\leq
C\left\lVert\boldsymbol{\nabla}_{\\!\xi}\tilde{v}\,\mathsf{m}\right\rVert_{\mathcal{L}^{3\pm\varepsilon}}\left\lVert
v\right\rVert_{\mathcal{L}^{\infty}}\left\lVert
v\right\rVert_{\mathcal{L}^{2}}^{2},$ $\displaystyle J_{1,2}$
$\displaystyle\leq
C\left\lVert\left[{V_{v\tilde{v}},\tilde{v}}\right]\right\rVert_{\mathcal{L}^{2}}\left\lVert
v\right\rVert_{\mathcal{L}^{2}}.$

The right-hand side of $J_{1,1}$ is controlled by $\left\lVert
v\right\rVert_{\mathcal{L}^{2}}^{2}$ since
$\left\lVert\boldsymbol{\nabla}_{\\!\xi}\tilde{v}\,\mathsf{m}\right\rVert_{\mathcal{L}^{3\pm\varepsilon}}$
is bounded thanks to Proposition 3.2 and $\left\lVert
v\right\rVert_{\mathcal{L}^{\infty}}\leq\left\lVert
v_{1}\right\rVert_{\mathcal{L}^{\infty}}+\left\lVert\tilde{v}\right\rVert_{\mathcal{L}^{\infty}}$,
which are both bounded. By Lemma 2.1, similarly as in the proof of Theorem
2.2, we get

$J_{1,2}\leq
C\left\lVert\boldsymbol{\nabla}_{\\!\xi}\tilde{v}\right\rVert_{\mathcal{W}^{1,2}}\left\lVert\rho_{f}\right\rVert_{L^{\infty}_{x}}^{\frac{1}{2}}\left\lVert
v\right\rVert_{\mathcal{L}^{2}}^{2}$

where $\rho_{f}\in L^{\infty}(\mathbb{R}^{3})$ and
$\left\lVert\boldsymbol{\nabla}_{\\!\xi}\tilde{v}\right\rVert_{\mathcal{W}^{1,2}}$
is bounded in time thanks to Proposition 3.2. We are now left with the bound
on $J_{2}$. By Hölder’s inequality for Schatten norms and Young’s inequality
for products we obtain

$J_{2}=h^{2}\operatorname{Tr}\\!\left(\big{[}{V_{\tilde{\boldsymbol{\rho}}-\boldsymbol{\rho}_{f}},\tilde{v}}\big{]}v\right)\leq\big{\lVert}\tfrac{1}{h}\big{[}{V_{\tilde{\boldsymbol{\rho}}-\boldsymbol{\rho}_{f}},\tilde{v}}\big{]}\big{\rVert}_{\mathcal{L}^{2}}^{2}+\left\lVert
v\right\rVert_{\mathcal{L}^{2}}^{2}.$

We aim at showing that
$\tfrac{1}{h}\big{\lVert}\big{[}{V_{\tilde{\boldsymbol{\rho}}-\boldsymbol{\rho}_{f}},\tilde{v}}\big{]}\big{\rVert}_{\mathcal{L}^{2}}$
is small. To this end, apply Lemma 2.1 to get

(42)
$\tfrac{1}{h}\big{\lVert}\big{[}{V_{\tilde{\boldsymbol{\rho}}-\boldsymbol{\rho}_{f}},\tilde{v}}\big{]}\big{\rVert}_{\mathcal{L}^{2}}\leq
C\left\lVert\boldsymbol{\nabla}_{\\!\xi}\tilde{v}\right\rVert_{\mathcal{W}^{1,2}}\left\lVert\operatorname{diag}(\widetilde{\boldsymbol{\rho}}-\boldsymbol{\rho}_{f})\right\rVert_{L^{2}_{x}}.$

The term
$\left\lVert\operatorname{diag}(\widetilde{\boldsymbol{\rho}}-\boldsymbol{\rho}_{f})\right\rVert_{L^{2}_{x}}$
in Equation (42) is controlled in Lemma 3.5, leading to the statement of the
proposition. ∎

###  3.3. Proof of the main theorems

Now that we have a bound on
$\sqrt{\boldsymbol{\rho}}-\sqrt{\widetilde{\boldsymbol{\rho}}}$ in
$\mathcal{L}^{2}$ by the above proposition, we are ready to finish the proof
of the semiclassical limit.

###### Proof of Theorem 1.1 and Theorem 1.2.

As explained in the beginning of the section, by Lemma 3.1 with $p=2$, we get
that

(43)
$\displaystyle\left\lVert\widetilde{\boldsymbol{\rho}}_{f^{\mathrm{in}}}-\widetilde{\boldsymbol{\rho}}^{\mathrm{in}}\right\rVert_{\mathcal{L}^{2}}\leq
48\,\hbar\left\lVert\nabla\sqrt{f^{\mathrm{in}}}\right\rVert_{L^{4}}^{2},$

and so, thanks to Inequality (33) and the fact that $\left\lVert\nabla
E_{f}\right\rVert_{L^{\infty}_{x}}$ can be controlled by
$\left\lVert\rho_{f}\right\rVert_{W^{1,\infty}_{x}}$, the error of positivity
is given by

(44)
$\left\lVert\widetilde{\boldsymbol{\rho}}-\boldsymbol{\rho}_{f}\right\rVert_{\mathcal{L}^{2}}\leq
C\,\hbar\left(\left\lVert\nabla\sqrt{f^{\mathrm{in}}}\right\rVert_{L^{4}}^{2}+\int_{0}^{t}\left\lVert\rho_{f}\right\rVert_{W^{1,\infty}_{x}}\left\lVert\nabla_{\xi}^{2}f\right\rVert_{L^{2}}\right).$

On the other side, the nonlinear stability error (34) is controlled, using
Proposition 3.1 and recalling that
$\tilde{\boldsymbol{\rho}}^{\mathrm{in}}=\widetilde{\boldsymbol{\rho}}_{\sqrt{f^{\mathrm{in}}}}^{2}$,
by

(45)
$\left\lVert\boldsymbol{\rho}-\widetilde{\boldsymbol{\rho}}\right\rVert_{\mathcal{L}^{2}}\leq\mathcal{C}_{\infty}^{\frac{1}{2}}\left(\left\lVert\sqrt{\boldsymbol{\rho}^{\mathrm{in}}}-\widetilde{\boldsymbol{\rho}}_{\sqrt{f^{\mathrm{in}}}}\right\rVert_{\mathcal{L}^{2}}+\hbar\left\lVert
c\right\rVert_{L^{2}([0,t])}\right)e^{\Lambda(t)}.$

Summing up (44) and (45) and using the estimate on the initial state (43)
prove Theorem 1.2.

Moreover, we can control the difference of the square roots using Inequality
(13) which tells us that the Wick and Weyl quantization are close since
$\sqrt{f^{\mathrm{in}}}$ is sufficiently regular, and then the fact that the
Wigner transform is an isometry from $\mathcal{L}^{2}$ to $L^{2}$ to get

$\left\lVert\sqrt{\boldsymbol{\rho}^{\mathrm{in}}}-\widetilde{\boldsymbol{\rho}}_{\sqrt{f^{\mathrm{in}}}}\right\rVert_{\mathcal{L}^{2}}\leq\left\lVert\sqrt{f^{\mathrm{in}}}-f_{\sqrt{\boldsymbol{\rho}^{\mathrm{in}}}}\right\rVert_{L^{2}}+\tfrac{3\,\hbar}{2}\left\lVert\nabla^{2}\sqrt{f^{\mathrm{in}}}\right\rVert_{L^{2}}$

which leads to our main inequality (9). ∎

###  3.4. Propagation of regularity

In this section, we prove the following proposition about the semiclassical
propagation of regularity for $\tilde{v}$ assuming sufficient regularity on
the solution to the Vlasov–Poisson equation.

######  Proposition 3.2.

Let $h\in(0,1)$, $q\in[1,\infty)$, $n\in\mathbb{N}$, $f$ be a smooth solution
of the Vlasov–Poisson equation (3) and $\tilde{v}$ be the solution to the
linear equation

$i\hbar\,\partial_{t}\tilde{v}=\left[{H_{f},\tilde{v}}\right],$

where $H_{f}=-\tfrac{\hbar^{2}}{2}\Delta+V_{f}$. Define
$\mathsf{m}=\left\langle\boldsymbol{p}\right\rangle^{2n}$. Then for any
$\varepsilon\in(0,1)$, there exists $C_{q,n}$ depending only on $q$, $n$ and
$\varepsilon$, such that for every $t>0$ it holds

$\left\lVert\tilde{v}(t)\right\rVert_{\mathcal{W}^{k}(\mathsf{m})}\leq\left\lVert\tilde{v}^{\mathrm{in}}\right\rVert_{\mathcal{W}^{k}(\mathsf{m})}\exp\\!\left(C_{q,n}\int^{t}_{0}\left\lVert\rho_{f}(s,\cdot)\right\rVert_{W_{x}^{2n,3\pm\varepsilon}}\mathrm{d}s\right).$

######  Remark 3.2.

To obtain the result of Theorem 1.1, we need bounds on the norm of $\tilde{v}$
in
$\mathcal{W}^{2,2}\cap\mathcal{W}^{1,3\pm\varepsilon}(\left\langle\boldsymbol{p}\right\rangle^{2n})$
with $2n>2$. In particular, by the inequalities in [16], these norms are
bounded for the initial data
$\tilde{v}^{\mathrm{in}}=\tilde{\boldsymbol{\rho}}_{\sqrt{f^{\mathrm{in}}}}$
if

$\left\lVert\sqrt{f^{\mathrm{in}}}\right\rVert_{H^{2}\cap
W^{1,3\pm\varepsilon}_{2n}}+\hbar^{2}\left\lVert\sqrt{f^{\mathrm{in}}}\right\rVert_{W^{4,3\pm\varepsilon}}$

is bounded uniformly in $\hbar$.

###### Proof.

By the Jacobi identity, we arrive at the equation

(46)
$\partial_{t}\boldsymbol{\nabla}_{\\!x}^{k}\tilde{v}=\tfrac{1}{i\hbar}\left[{H_{f},\boldsymbol{\nabla}_{\\!x}^{k}\tilde{v}}\right]-\sum^{k}_{j=1}\binom{k}{j}\tfrac{1}{i\hbar}\left[{\boldsymbol{\nabla}_{\\!x}^{j-1}E_{f},\boldsymbol{\nabla}_{\\!x}^{k-j}\tilde{v}}\right].$

Likewise, using the fact that $\boldsymbol{\nabla}_{\\!\xi}^{j}H_{f}=0$ for
$j\geq 3$, we have that

$\partial_{t}\boldsymbol{\nabla}_{\\!\xi}^{k}\tilde{v}=\tfrac{1}{i\hbar}\left[{H_{f},\boldsymbol{\nabla}_{\\!\xi}^{k}\tilde{v}}\right]-k\,\boldsymbol{\nabla}_{\\!\xi}^{k-1}\boldsymbol{\nabla}_{\\!x}\tilde{v}$

and

$\partial_{t}\boldsymbol{\nabla}_{\\!x}^{\ell}\boldsymbol{\nabla}_{\\!\xi}^{k-\ell}\tilde{v}=\tfrac{1}{i\hbar}\left[{H_{f},\boldsymbol{\nabla}_{\\!x}^{\ell}\boldsymbol{\nabla}_{\\!\xi}^{k-\ell}\tilde{v}}\right]-\left(k-\ell\right)\boldsymbol{\nabla}_{\\!\xi}^{k-\ell-1}\boldsymbol{\nabla}_{\\!x}^{\ell+1}\tilde{v}\\\
-\sum^{\ell}_{j=1}\binom{\ell}{j}\tfrac{1}{i\hbar}\left[{\boldsymbol{\nabla}_{\\!x}^{j-1}E_{f},\boldsymbol{\nabla}_{\\!x}^{\ell-j}\boldsymbol{\nabla}_{\\!\xi}^{k-\ell}\tilde{v}}\right]$

where $\ell<k$. Let us focus on propagating moments for
$\boldsymbol{\nabla}_{\\!x}^{k}\tilde{v}$ since Equation (46) only depends on
$\boldsymbol{\nabla}_{\\!x}^{\ell}\tilde{v}$ for $\ell\leq k$. Define a weight
equivalent to $\mathsf{m}$ by
$\widetilde{\mathsf{m}}=1+\sum^{3}_{j=1}\boldsymbol{p}^{2n}_{\mathrm{i}}$ as
in [8, Lemma 6.3]. Then, by [8, Lemma 6.2], we arrive at the estimate

$\frac{\mathrm{d}}{\mathrm{d}t}\left\lVert\boldsymbol{\nabla}_{\\!x}^{k}\tilde{v}\,\widetilde{\mathsf{m}}\right\rVert_{q}\leq\frac{1}{\hbar}\left\lVert\left[{V_{f},\widetilde{\mathsf{m}}}\right]\boldsymbol{\nabla}_{\\!x}^{k}\tilde{v}\right\rVert_{q}+\frac{1}{\hbar}\sum^{k}_{j=1}\binom{k}{j}\left\lVert\left[{\boldsymbol{\nabla}_{\\!x}^{j-1}E_{f},\boldsymbol{\nabla}_{\\!x}^{k-j}\tilde{v}}\right]\widetilde{\mathsf{m}}\right\rVert_{q}$

for $q\geq 2$. Let us estimate the first term. To simplify the notation, we
write $V=V_{f}$, $E=E_{f}$, and
$\boldsymbol{\mu}=\boldsymbol{\nabla}_{\\!x}^{k}\tilde{v}$. Recall the
identity $\left[{V,\boldsymbol{p}}\right]=i\hbar\,E$. Then it follows

$\frac{1}{i\hbar}\left[{V,\boldsymbol{p}^{2n}_{\mathrm{i}}}\right]=\sum^{2n-1}_{k=0}\boldsymbol{p}^{k}_{\mathrm{i}}\,E_{\mathrm{i}}\,\boldsymbol{p}_{\mathrm{i}}^{2n-1-k}=\sum^{2n-1}_{k=0}\sum^{k}_{\ell=0}\binom{k}{\ell}g_{\ell}\,\boldsymbol{p}_{\mathrm{i}}^{2n-1-\ell}$

where
$g_{\ell}=\left(-i\hbar\right)^{\ell}\partial_{\mathrm{i}}^{\ell}E_{\mathrm{i}}$.
Using the above identity yields the estimate

$\displaystyle\frac{1}{\hbar}\left\lVert\left[{V,\boldsymbol{p}_{\mathrm{i}}^{2n}}\right]\boldsymbol{\mu}\right\rVert_{\mathcal{L}^{q}}$
$\displaystyle\leq\sum^{2n-1}_{\ell=0}\binom{2n}{\ell+1}\left\lVert
g_{\ell}\right\rVert_{L^{\infty}}\left\lVert\boldsymbol{\mu}\,\mathsf{m}\right\rVert_{\mathcal{L}^{q}}.$

Notice that

$\left\lVert g_{\ell}\right\rVert_{L^{\infty}}=\hbar^{\ell}\left\lVert\nabla
K*\partial_{\mathrm{i}}^{\ell}\rho_{f}\right\rVert_{L^{\infty}}\leq
C\,\hbar^{\ell}\left\lVert\partial_{\mathrm{i}}^{\ell}\rho_{f}\right\rVert_{L^{3\pm\varepsilon}}$

for $\varepsilon\in(0,1)$, then it follows that

$\frac{1}{\hbar}\left\lVert\left[{V,\boldsymbol{p}_{\mathrm{i}}^{2n}}\right]\boldsymbol{\mu}\right\rVert_{\mathcal{L}^{q}}\leq
C_{f}(t)\left(\tfrac{(1+\hbar)^{2n}-1}{\hbar}\right)\left\lVert\boldsymbol{\mu}\,\mathsf{m}\right\rVert_{\mathcal{L}^{q}}.$

Let us now estimate the second term. Write
$\boldsymbol{\mu}_{k-j}=\boldsymbol{\nabla}_{\\!x}^{k-j}\tilde{v}$. Using the
fact that

$\displaystyle\left[{\boldsymbol{\nabla}_{\\!x}^{j-1}E,\boldsymbol{\mu}_{k-j}}\right]\boldsymbol{p}_{\mathrm{i}}^{2n}=\left[{\boldsymbol{\nabla}_{\\!x}^{j-1}E,\boldsymbol{\mu}_{k-j}\,\boldsymbol{p}_{\mathrm{i}}^{2n}}\right]-\boldsymbol{\mu}_{k-j}\left[{\boldsymbol{\nabla}_{\\!x}^{j-1}E,\boldsymbol{p}_{\mathrm{i}}^{2n}}\right]$

then it follows

$\frac{1}{\hbar}\left\lVert\left[{\boldsymbol{\nabla}_{\\!x}^{j-1}E,\boldsymbol{\mu}_{k-j}}\right]\boldsymbol{p}_{\mathrm{i}}^{2n}\right\rVert_{\mathcal{L}^{q}}\\\
\leq\frac{1}{\hbar}\left\lVert\left[{\boldsymbol{\nabla}_{\\!x}^{j-1}E,\boldsymbol{\mu}_{k-j}\,\boldsymbol{p}_{\mathrm{i}}^{2n}}\right]\right\rVert_{\mathcal{L}^{q}}+\frac{1}{\hbar}\left\lVert\boldsymbol{\mu}_{k-j}\left[{\boldsymbol{\nabla}_{\\!x}^{j-1}E,\boldsymbol{p}_{\mathrm{i}}^{2n}}\right]\right\rVert_{\mathcal{L}^{q}}=:I_{1}+I_{2}.$

By [8, Proposition 6.5], we have that

$I_{1}\leq
C\left\lVert\nabla^{j-1}\rho_{f}\right\rVert_{W^{1,r}}\left\lVert\boldsymbol{\nabla}_{\\!\xi}\boldsymbol{\mu}_{k-j}\,\boldsymbol{p}_{\mathrm{i}}^{2n}\right\rVert_{\mathcal{L}^{q}}\quad\text{
with }\quad\frac{1}{r}=\frac{1}{2}-\frac{1}{q}<\frac{1}{3}$

and $k>j$. Term $I_{2}$ is handled in the same manner as in the case of
$\tfrac{1}{i\hbar}\left[{V,\boldsymbol{p}_{\mathrm{i}}^{2n}}\right]\boldsymbol{\mu}$,
which yields

$I_{2}\leq C_{q}\sup_{1\leq\ell\leq
2n}\left\lVert\partial_{\mathrm{i}}^{\ell}\rho_{f}\right\rVert_{L^{3\pm\varepsilon}}\left(\tfrac{(1+\hbar)^{2n}-1}{\hbar}\right)\left\lVert\boldsymbol{\mu}\,\mathsf{m}\right\rVert_{\mathcal{L}^{q}}.$

Hence it follows

(47a)
$\frac{\mathrm{d}}{\mathrm{d}t}\left\lVert\boldsymbol{\nabla}_{\\!x}^{k}\tilde{v}\,\widetilde{\mathsf{m}}\right\rVert_{q}\leq
C_{q,f,n}(t)\left(\left\lVert\boldsymbol{\nabla}_{\\!x}^{k}\tilde{v}\,\widetilde{\mathsf{m}}\right\rVert_{q}+\sum^{k}_{j=1}\binom{k}{j}\left\lVert\boldsymbol{\nabla}_{\\!\xi}\boldsymbol{\nabla}_{\\!x}^{k-j}\tilde{v}\,\widetilde{\mathsf{m}}\right\rVert_{q}\right).$
In fact, recycling the above argument yields the estimates (47b)
$\frac{\mathrm{d}}{\mathrm{d}t}\left\lVert\boldsymbol{\nabla}_{\\!\xi}^{k}\tilde{v}\,\widetilde{\mathsf{m}}\right\rVert_{q}\leq\,C_{q,f,n}(t)\left(\left\lVert\boldsymbol{\nabla}_{\\!\xi}^{k}\tilde{v}\,\widetilde{\mathsf{m}}\right\rVert_{q}+\left\lVert\boldsymbol{\nabla}_{\\!\xi}^{k-1}\boldsymbol{\nabla}_{\\!x}\tilde{v}\,\widetilde{\mathsf{m}}\right\rVert_{q}\right)$
and (47c)
$\frac{\mathrm{d}}{\mathrm{d}t}\left\lVert\boldsymbol{\nabla}_{\\!x}^{\ell}\boldsymbol{\nabla}_{\\!\xi}^{k-\ell}\tilde{v}\,\widetilde{\mathsf{m}}\right\rVert_{q}\leq
C_{q,f,n}(t)\bigg{(}\left\lVert\boldsymbol{\nabla}_{\\!x}^{\ell}\boldsymbol{\nabla}_{\\!\xi}^{k-\ell}\tilde{v}\,\widetilde{\mathsf{m}}\right\rVert_{q}\\\
+(k-\ell)\left\lVert\boldsymbol{\nabla}_{\\!\xi}^{k-1-\ell}\boldsymbol{\nabla}_{\\!x}^{\ell+1}\tilde{v}\,\widetilde{\mathsf{m}}\right\rVert_{q}+\sum^{\ell}_{j=1}\binom{\ell}{j}\left\lVert\boldsymbol{\nabla}_{\\!x}^{\ell-j}\boldsymbol{\nabla}_{\\!\xi}^{k-\ell}\tilde{v}\,\widetilde{\mathsf{m}}\right\rVert_{q}\bigg{)}.$

Combining the above estimates (47) gives us a bound of the form

$\frac{\mathrm{d}}{\mathrm{d}t}\left\lVert\tilde{v}(t)\right\rVert_{\mathcal{W}^{k}(\mathsf{m})}\leq
C_{q,f,n}^{\prime}(t)\left\lVert\tilde{v}(t)\right\rVert_{\mathcal{W}^{k}(\mathsf{m})}$

which then by Grönwall’s lemma leads to the result. ∎

Acknowledgments. J.C. was supported by the NSF through the RTG grant DMS- RTG
184031. C.S. acknowledges the NCCR SwissMAP and the support of the SNSF
through the Eccellenza project PCEFP2_181153.

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