On Exponential Moments of the Homogeneous Boltzmann Equation for Hard Potentials Without Cutoff

Abstract

We consider the spatially homogeneous Boltzmann equation for hard potentials without cutoff. We prove that an exponential moment of order $\rho =\min \{2\gamma /(2-\nu ),2\}$, with the usual notation, is immediately created. This is stronger than what happens in the case with cutoff. We also show that exponential moments of order $\rho \in (0,2]$ are propagated.

1 Introduction and Results

1.1 The Boltzmann equation

We consider a spatially homogeneous gas modeled by the Boltzmann equation: the density $f_t(v)$ of particles with velocity $v\in {\mathbb {R}}^3$ at time $t\ge 0$ solves

$$\begin{aligned} \partial _t f_t(v) = \int _{{\mathbb {R}}^3}\mathrm{d}{v_*}\int _{{\mathbb {S}^2}} \mathrm{d}\sigma B(|v-{v_*}|,\cos \theta ) \big [f_t(v')f_t(v'_*) -f_t(v)f_t({v_*})\big ], \end{aligned}$$

(1)

where

$$\begin{aligned} v'=\frac{v+{v_*}}{2} + \frac{|v-{v_*}|}{2}\sigma , \quad v'_*=\frac{v+{v_*}}{2} -\frac{|v-{v_*}|}{2}\sigma \quad \hbox {and} \quad \cos \theta = \frac{v-{v_*}}{|v-{v_*}|} \cdot \sigma .\nonumber \\ \end{aligned}$$

(2)

We refer to the book of Cercignani [7] and to the long review papers of Villani [19] and Alexandre [1] for some detailed and complete accounts of what is known, both from the physical and mathematical points of view, about this equation. One may assume without loss of generality that the initial condition satisfies

$$\begin{aligned} \int _{{\mathbb {R}}^3}f_0(v)\mathrm{d}v = 1, \quad \int _{{\mathbb {R}}^3}vf_0(v)\mathrm{d}v = 0 \quad \hbox {and}\quad \int _{{\mathbb {R}}^3}|v|^2 f_0(v)\mathrm{d}v = 1, \end{aligned}$$

and these quantities, namely the mass, momentum and kinetic energy, are constant, at least informally, as time evolves.

1.2 Assumptions

We will suppose that for some $\gamma \in (0,1]$ and some $\nu \in (0,2)$,

$$\begin{aligned}&(H_1(\gamma )) \qquad B(|v-{v_*}|,\cos \theta )\sin \theta = |v-{v_*}|^\gamma \beta (\theta )\\&\qquad \quad \quad \quad \quad \,\, \hbox {for some measurable } \beta :(0,\pi ]\rightarrow {\mathbb {R}}_+,\\&(H_2(\nu )) \qquad \exists \; \kappa _1,\kappa _2\in (0,\infty ),\; \forall \; \theta \in (0,\pi ], \;\; \kappa _1 \theta ^{-\nu -1} \le \beta (\theta ) \le \kappa _2 \theta ^{-\nu -1} . \end{aligned}$$

We consider any possible values of $\gamma \in (0,1]$ and $\nu \in (0,2)$, but the only physically relevant situations are the following. As explained in [1, 7, 19], when particles interact through a repulsive force in $1/r^s$ for some $s> 2$, we have $(H_1(\gamma ))$ and $(H_2(\nu ))$ with $\gamma =(s-5)/(s-1)$ and $\nu =2/(s-1)$. When $\gamma \in (0,1)$ (i.e. $s>5$), one speaks of hard potentials.

One speaks of hard potentials with cutoff when we have $(H_1(\gamma ))$ for some $\gamma \in (0,1]$ and when $(H_2(\nu ))$ is replaced by the condition $\int _0^\pi \beta (\theta )\mathrm{d}\theta \in (0,\infty )$.

1.3 Weak solutions

First, we parameterize (2) as in [9]. For every $X\in {\mathbb {R}}^3\setminus \{0\}$, we introduce $I(X),J(X)\in {\mathbb {R}}^3$ such that $(\frac{X}{|X|},\frac{I(X)}{|X|},\frac{J(X)}{|X|})$ is an orthonormal basis of ${\mathbb {R}}^3$. We also put $I(0)=J(0)=0$. For $X,v,{v_*}\in {\mathbb {R}}^3$, $\theta \in (0,\pi ]$ and $\varphi \in [0,2\pi )$, we set

$$\begin{aligned} \left\{ \begin{array}{l} \Gamma (X,\varphi )=(\cos \varphi ) I(X) + (\sin \varphi )J(X), \\ \displaystyle v'= v - \frac{1-\cos \theta }{2} (v-{v_*}) + \frac{\sin \theta }{2}\Gamma (v-{v_*},\varphi ),\\ \displaystyle v'_*= {v_*}+ \frac{1-\cos \theta }{2} (v-{v_*}) - \frac{\sin \theta }{2}\Gamma (v-{v_*},\varphi ). \end{array} \right. \end{aligned}$$

(3)

We denote by ${\mathcal {P}}({{\mathbb {R}}^3})$ the set of probability measures on ${{\mathbb {R}}^3}$. For $p\in {\mathbb {R}}_+$ and $f\in {\mathcal {P}}({{\mathbb {R}}^3})$, we introduce the moment of order p of f:

$$\begin{aligned} m_p(f)=\int _{{\mathbb {R}}^3}|v|^p f(\mathrm{d}v). \end{aligned}$$

We use the following classical notion of weak solutions.

Definition 1

Assume $(H_1(\gamma ))$ and $(H_2(\nu ))$ for some $\gamma \in (0,1]$ and some $\nu \in (0,2)$. A weakly continuous family $(f_t)_{t\ge 0}$ of probability measures on ${{\mathbb {R}}^3}$ is a weak solution to (1) if for all $t\ge 0$,

$$\begin{aligned} \int _{{\mathbb {R}}^3}vf_t(\mathrm{d}v)=0 \quad \hbox {and} \quad m_2(f_t)=1 \end{aligned}$$

and if for any $\phi \in C^2_b({{\mathbb {R}}^3})$ and any $t\ge 0$, using the parameterization (3),

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\!\int _{{\mathbb {R}}^3}\!\!\phi (v)f_t(\mathrm{d}v)\! =\!\! \int _{{\mathbb {R}}^3}\!\int _{{\mathbb {R}}^3}\! \int _0^\pi \!\!\!\int _0^{2\pi } \![\phi (v')\!+\phi ({v_*}')\!-\phi (v)\!-\phi ({v_*})]|v\!-\!{v_*}|^\gamma \mathrm{d}\varphi \beta (\theta )\mathrm{d}\theta f_t(\mathrm{d}{v_*}) f_t(\mathrm{d}v). \end{aligned}$$

(4)

As shown by Lu–Mouhot in [13], see also Villani [19], weak solutions exist starting from any given initial condition $f_0\in {\mathcal {P}}({{\mathbb {R}}^3})$ such that $\int _{{\mathbb {R}}^3}v f_0(\mathrm{d}v)=0$ and $m_2(f_0)=1$, and they satisfy

$$\begin{aligned} \textstyle \hbox {for all } p\ge 0,\hbox { all } t_0>0, \quad \sup _{t\ge t_0} m_p(f_t)<\infty . \end{aligned}$$

(5)

Let us mention the recent uniqueness result of Heydecker [12], which concerns the case where $\nu \in (0,1)$, assuming only that $m_p(f_0)<\infty$ for some sufficiently large p.

1.4 Main result

Here is our main result.

Theorem 2

Assume $(H_1(\gamma ))$ and $(H_2(\nu ))$ for some $\gamma \in (0,1]$ and some $\nu \in (0,2)$. Consider any weak solution $(f_t)_{t\ge 0}$ to (1).

(i) Put $\rho =\min \{2\gamma /(2-\nu ),2\}$. There are some constants $T>0$ and $\sigma >0$, depending only on $\gamma ,\nu ,\kappa _1,\kappa _2$, such that

$$\begin{aligned} \sup _{t\in [0,T]} \int _{{\mathbb {R}}^3}\exp [\sigma t^{\rho /\gamma } |v|^{\rho }] f_t(\mathrm{d}v) \le 4. \end{aligned}$$

(ii) For any $A>0$, any $\sigma _0>0$, any $\rho \in (0,2]$, there is a constant $\sigma >0$, depending only on $\gamma ,\nu ,\kappa _1,\kappa _2,\rho ,\sigma _0,A$, such that

$$\begin{aligned} \int _{{\mathbb {R}}^3}\exp [\sigma _0 |v|^{\rho }] f_0(\mathrm{d}v)\le A \quad \Longrightarrow \quad \sup _{t\ge 0} \int _{{\mathbb {R}}^3}\exp [\sigma |v|^{\rho }] f_t(\mathrm{d}v) \le 6. \end{aligned}$$

Since $\min \{2\gamma /(2-\nu ),2\}>\gamma$, point (i) is stronger than in the cutoff case where, as we will see in the next subsection, only exponential moments of order $\rho =\gamma$ are created. In (ii), we have a possible deterioration of the constant $\sigma$, as in all the references below.

By (ii), (i) can be extended to: there is $\sigma >0$, depending only on $\gamma ,\nu ,\kappa _1,\kappa _2$, such that

$$\begin{aligned} \sup _{t\ge 0} \int _{{\mathbb {R}}^3}\exp [\sigma (t\wedge 1)^{\rho /\gamma } |v|^{\rho }] f_t(\mathrm{d}v) \le 6 \quad \hbox {with}\quad \rho =\min \left\{ \frac{2\gamma }{2-\nu },2\right\} . \end{aligned}$$

Remark 3

In the physically relevant case where particles interact through a repulsive force in $1/r^s$ for some $s>5$, we have $(H_1(\gamma ))$ and $(H_2(\nu ))$ with $\gamma =(s-5)/(s-1) \in (0,1)$ and $\nu =2/(s-1) \in (0,1/2)$. We thus can apply Theorem 2: there is $\sigma >0$ depending only on $s,\kappa _1,\kappa _2$ such that for any weak solution $(f_t)_{t\ge 0}$ to (1),

$$\begin{aligned} \sup _{t\ge 0} \int _{{\mathbb {R}}^3}\exp [\sigma (t\wedge 1)^{(s-1)/(s-2)} |v|^{(s-5)/(s-2)}] f_t(\mathrm{d}v) \le 6. \end{aligned}$$

Observe that $\frac{s-1}{s-2}=\frac{4}{3+\gamma }$ and $\frac{s-5}{s-2}=\frac{4\gamma }{3+\gamma }$.

By Bobylev–Gamba [4], point (ii) of Theorem 2 also holds true for Maxwell molecules, i.e. when $\gamma =0$ and $\nu =1/2$. Actually, Maxwell molecules with cutoff are studied in [4], but the constants do not depend on the cutoff parameter, as noted in [4, Remark 3.2]. Point (i) is irrelevant in this case, since $\rho =0$.

1.5 References

There is a large literature on the subject, because exponential moments can be used for different purposes, such as estimating the rate of convergence to equilibrium, see Mouhot [15], or uniqueness, see [10].

Using the famous Povzner inequality [17], Wennberg [20] discovered that polynomial moments are immediately created by the homogeneous Boltzmann equation for cutoff hard potentials (CHP in short), i.e. when the angular cross section $\beta$ is assumed to be integrable on $[0,\pi ]$. This really requires that $\gamma >0$ and the main intuition is that particles with large velocities are quickly slowed down, because they collide at large rate (since $\gamma >0$) with slow particles. Studying a linearized Boltzmann equation, Mouhot–Strain [16] showed that this property is stronger for non cutoff hard potentials (NCHP in short), in the sense that the rate of decay of polynomial moments is higher for NCHP than for CHP.

In his seminal paper [5], Bobylev proved that Gaussian moments ($\rho =2$) are propagated in the case of hard spheres, i.e. when $\gamma =1$ and $\beta (\theta )=\sin \theta$. This relies on very tight computations involving a recursive ODE argument to estimate the polynomial moments, which are then summed to estimate Gaussian moments. This was generalized to all CHP by Gamba–Panferov–Villani [11], who also proved some local estimates that we will mention later. Let us also cite Bobylev–Gamba–Panferov [6] who studied inelastic collisions. Following the ideas of [5], Mischler, Mouhot, and Rodriguez Ricard [14] and Mouhot [15] managed to create exponential moments of order $\rho =\gamma /2$, still for CHP. Following the same approach, Lu–Mouhot [13] were able to create exponential moments of order $\rho =\gamma$ for CHP and NCHP. Alonso–Cañizo–Gamba–Mouhot [2] found a much simpler method to create exponential moments of order $\rho =\gamma$ and propagate exponential moments of order $\rho \in (0,2]$, for CHP. Let us finally quote Alonso–Gamba–Tasković [3], who studied some much stronger Lebesgue and Sobolev norms of $f_t$ with exponential weights for CHP.

Concerning NCHP, there is the work of Lu–Mouhot [13] already mentioned. The proof of Lemma 4.1 in Fournier–Mouhot [10], which concerns exponential moments for NCHP, is unfortunately false, there is a major gap (the function $\delta (\eta )$ in (4.6) actually depends on p). What was required there for the uniqueness criterion for NCHP to imply a well-posedness result, was the propagation of exponential moments of order $\rho =\gamma$. Tasković–Alonso–Gamba–Pavlović [18] have shown, for NCHP, creation of exponential moments of order $\rho =\gamma$ and propagation of exponential moments of order $\rho \in (0,4/(2+\nu )]$ (which contains $\rho =\gamma \in (0,1]$ and thus fixes the issue in [10]).

It thus seems that concerning NCHP, [18] contains the best available results, and Theorem 2 is stronger both for creation and propagation. In particular, we show that NCHP create more exponential moments than CHP.

The homogeneous Landau equation for hard potentials, which often behaves in a similar way as the Boltzmann equation, but which is considerably simpler in many points, immediately creates Gaussian moments ($\rho =2$) for any value of $\gamma \in (0,1]$, see [8]. Once this is observed, it is natural to wonder if NCHP create more exponential moments than CHP. The answer is not intuitively clear, because the (considerably many) additional collisions caused by the singularity of $\beta$ near 0 involve some (considerably) small values of $\theta$ and do not much slow down particles with high velocity. Actually, the effect is strong enough to modify the behavior of the solutions: exponential moments of order $\rho =\min \{2\gamma /(2-\nu ),2\}>\gamma$ are created by NCHP, while only exponential moments of order $\rho =\gamma$ are created by CHP.

By the way, Theorem 2-(i) implies that for any $\gamma \in (0,1]$, Gaussian moments ($\rho =2$) are created by NCHP if the angular cross section is singular enough, namely if $\nu \ge 2-\gamma$, but this does never happen in the physical situations described in Remark 3.

Let us finally mention some closely related works: Gamba–Panferov–Villani [11] showed that for CHP, if there are $A,a>0$ such that $f_0(v)\le A e^{-a|v|^2}$ for all $v \in {\mathbb {R}}^3$, then there are $B,b>0$ such that $f(t,v)\le B e^{-b|v|^2}$ for all $t\ge 0$ and all $v\in {\mathbb {R}}^3$. Bobylev–Gamba [4] established the same result for Maxwell molecules with cutoff. To our knowledge, such a local Gaussian estimate is open in all the other situations.

1.6 Strategy

Our strategy is the same as that of [18]: we adapt the ideas of [2] to NCHP, taking advantage of the simplicity of the method. The present paper resembles [18] in several points. In particular, some Mittag–Leffler moments of the form

$$\begin{aligned} \sum _{n\ge 0}\frac{a^n m_{2n}(f)}{\Gamma (\alpha n+1)}, \end{aligned}$$

with $a>0$ and $\alpha \ge 1$, are used in [18], while we are led to use some series of the form

$$\begin{aligned} \sum _{n\ge 0} \frac{a^n m_{2n}(f)}{(n!)^\alpha }, \end{aligned}$$

with $a>0$ and $\alpha \ge 1$. This is almost the same thing and in both cases, this is enough to control some exponential moments of the form $\int _{{\mathbb {R}}^3}\exp (b |v|^\rho ) f(\mathrm{d}v)$, with $\rho =2/\alpha$, see Lemma 11. The main advantage of using such series is that it then suffices to study integer moments $m_{2n}(f)$, which leads to much more explicit computations than if using non-integer moments, as is done e.g. in [2], where $\int _{{\mathbb {R}}^3}\exp (b |v|^\rho ) f(\mathrm{d}v)$ is more naturally studied through $\sum _{n\ge 0} (n!)^{-1}b^n m_{\rho n}(f)$.

However, we try to really take advantage of the singularity of the cross section to establish a stronger Povzner inequality than in the cutoff case, see Lemma 4 and the paragraph below. We then have to adapt suitably the proof of [2], on the one hand because we can only deal with integer moments, and on the other hand because we have to exploit the new Povzner inequality.

1.7 About optimality

Since the solutions to (1) converge to some Maxwell (Gaussian) distributions, which are stationary solutions, we cannot expect to create or propagate exponential moments of order $\rho >2$. The propagation result thus seems optimal. Concerning creation, one may get convinced, following the proofs of Lemmas 4 and 7, that for some constant $c>0$, setting $m_{2n}'(f_t)=\frac{\mathrm{d}}{\mathrm{d}t}m_{2n}(f_t)$,

$$\begin{aligned} \forall \, n\ge 2,\quad m_{2n}'(f_t) \ge -cn^{\nu /2}m_{2n+\gamma }(f_t). \end{aligned}$$

Admitting, and this is not so clear, that the Hölder inequality is sharp enough so that we have $m_{2n+\gamma }(f_t)\simeq [m_{2n}(f_t)]^{1+\gamma /2n}$, we end with $m_{2n}'(f_t) \gtrsim - n^{\nu /2} [m_{2n}(f_t)]^{1+\gamma /2n}$, from which we easily conclude, if $m_{2n}(0)=\infty$, that $m_{2n}(f_t) \gtrsim [n^{1-\nu /2}/ t]^{2n/\gamma }$. Still informally, this should tell us that $m_{\rho n}(f_t) \gtrsim [n^{1-\nu /2}/ t]^{\rho n/\gamma }$, so that

$$\begin{aligned} \int _{{\mathbb {R}}^3}\exp [\sigma |v|^\rho ]f_t(\mathrm{d}v) = \sum _{n\ge 0} \frac{\sigma ^n m_{\rho n}(f_t) }{n!} \gtrsim \sum _{n\ge 0} \frac{\sigma ^n n^{\rho (1-\nu /2)n/\gamma }}{t^{\rho n/\gamma } n!}. \end{aligned}$$

By Stirling’s formula $n!\sim \sqrt{2\pi n}(n/e)^n$, this series is divergent, for any value of $\sigma >0$, when $\rho (1-\nu /2)/\gamma >1$, i.e. when $\rho > 2\gamma /(2-\nu )$, which is coherent with Theorem 2-(i).

1.8 About uniqueness

Assume $(H_1(\gamma ))$ and $(H_2(\nu ))$ for some $\gamma \in (0,1]$ and some $\nu \in (0,1)$. Using Theorem 2 and [10, Theorem 2.2], it seems possible to prove, in a few pages, the well-posedness of (1) assuming that the initial condition satisfies $\int _{{\mathbb {R}}^3}\exp {(|v|^\delta )} f_0(\mathrm{d}v)<\infty$ for some $\delta >0$. This is stronger than [10], where we assumed that $\int _{{\mathbb {R}}^3}\exp {(|v|^\gamma )} f_0(\mathrm{d}v)<\infty$, but weaker than the recent result of Heydecker [12], who only assumes that $m_p(f_0)<\infty$ for some large explicitable p.

1.9 Plan

The paper is technical and we are guided by computations rather than intuition.

In Sect. 2, we establish a Povzner lemma, which is stronger than what is known in the cutoff case. We handle the whole computation as explicitly as possible, not relying on any previous Povzner estimate, because this is required if we really want to show that the singular part of the cross section accelerates the slowing down of particles.

In Sect. 3, we derive some differential inequalities for the even integer moment from the Povzner inequality, and we prove some first rough estimates about these moments.

In Sect. 4, we quickly study how to control exponential moments by even integer moments and vice-versa.

Finally, we adapt the proofs of [2] to show Theorem 2-(i) in Sect. 5 and Theorem 2-(ii) in Section 6. This requires some work, because we can only use integer moments, and because we start from a different Povzner estimate.

1.10 Notation

We use the convention that ${\mathbb {N}}=\{0,1,...\}$. For $a,b\in {\mathbb {N}}$ with $a\le b$, we set $[\![a,b ]\!]=\{a,a+1,\dots ,b\}$. In the whole paper, $(f_t)_{t\ge 0}$ is a given weak solution satisfying $m_0(f_t)=m_2(f_t)=1$ and $\int _{{\mathbb {R}}^3}v f_t(\mathrm{d}v)=0$ for all $t\ge 0$. For $p\in {\mathbb {R}}_+$ and $t\ge 0$, we set $m_p(t)=m_p(f_t)$.

2 A Non-cutoff Povzner Lemma

The goal of this section is to establish the following Povzner inequality.

Lemma 4

Assume $(H_2(\nu ))$ for some $\nu \in (0,2)$. There are some constants $\lambda _1,\lambda _2\in (0,\infty )$, depending only on $\nu ,\kappa _1,\kappa _2$, such that for all integer $n\ge 2$, all $v,{v_*}\in {{\mathbb {R}}^3}$,

$$\begin{aligned} D_n(v,{v_*}):=&\int _0^\pi \int _0^{2\pi } [|v'|^{2n}+|v'_*|^{2n}-|v|^{2n}-|{v_*}|^{2n} ] \mathrm{d}\varphi \beta (\theta )\mathrm{d}\theta \\ \le&-\lambda _1 n^{\nu /2} (|v|^{2n}+|{v_*}|^{2n}) \\&+ \lambda _2 \sum _{a=1}^{n-1} \left( {\begin{array}{c}n\\ a\end{array}}\right) \Big (\frac{n^{\nu /2}}{(n-a)^{\nu /2+1}}+\frac{1}{a}\Big ) [|v|^{2a}|{v_*}|^{2(n-a)}+|v|^{2(n-a)}|{v_*}|^{2a}]. \end{aligned}$$

In the case with cutoff, see e.g. [2], one gets (roughly) something like

$$\begin{aligned} D_n(v,{v_*}) \le - (|v|^{2n}+|{v_*}|^{2n}) + \epsilon _n \sum _{a=1}^{n-1} \left( {\begin{array}{c}n\\ a\end{array}}\right) [|v|^{2a}|{v_*}|^{2(n-a)}+|v|^{2(n-a)}|{v_*}|^{2a}], \end{aligned}$$

with $\epsilon _n\rightarrow 0$ as $n\rightarrow \infty$. Here the negative term is reinforced by the factor $n^{\nu /2}$, and this is the main fact we will have to exploit. We will also have to play tightly with the positive term, showing that despite the fact it is not clearly multiplied by a small factor, it can be absorbed, in some sense, by the negative term.

We start with an explicit computation of the $\varphi$-average.

Lemma 5

For any integer $n\ge 2$, any $v,{v_*}\in {{\mathbb {R}}^3}$, any $\theta \in (0,\pi ]$, we have

$$\begin{aligned} \Theta _n(v,{v_*},\theta )&:=\frac{1}{2\pi }\int _0^{2\pi } |v'|^{2n}\mathrm{d}\varphi = \left( \frac{1+\cos \theta }{2}\right) ^n |v|^{2n}\\&\quad +\Bigg (\frac{1-\cos \theta }{2}\Bigg )^n |{v_*}|^{2n}+\Lambda _n(v,{v_*},\theta ), \end{aligned}$$

where, setting ${\mathcal {A}}_n=\{(i,j,k)\in {\mathbb {N}}^2 : i+j+k=n, \; i\le n-1, \; j\le n-1,\;k \in 2{\mathbb {N}}\}$,

$$\begin{aligned} \Lambda _n(v,{v_*},\theta )= & {} \!\!\!\!\!\! \sum _{(i,j,k)\in {\mathcal {A}}_n} \!\!\! \frac{n!}{i!j! [(k/2)!]^2}\Bigg (\frac{1+\cos \theta }{2}\Bigg )^i\\&\Bigg (\frac{1-\cos \theta }{2}\Big )^j\Big (\frac{\sin \theta }{2}\Bigg )^k |v|^{2i}|{v_*}|^{2j} \Bigg (|v|^2|{v_*}|^2-\big (v\cdot {v_*}\big )^2\Bigg )^{k/2}. \end{aligned}$$

Proof

We fix $n\ge 2$ and divide the proof into 3 steps.

Step 1. Recalling from (3) that $v'=v-\frac{1-\cos \theta }{2}(v-{v_*})+\frac{\sin \theta }{2} \Gamma (v-{v_*},\varphi )$, that $| \Gamma (v-{v_*},\varphi )|=|v-{v_*}|$ and that $(v-{v_*})\cdot \Gamma (v-{v_*},\varphi )=0$, we find

$$\begin{aligned} |v'|^2=&|v|^2\!+\!\Bigg (\frac{1\!-\!\cos \theta }{2}\Bigg )^2|v-{v_*}|^2\!+\!\Bigg (\frac{\sin \theta }{2}\Bigg )^2|v-{v_*}|^2 \!-\!(1-\cos \theta )v\cdot (v-{v_*})\\&+(\sin \theta )v\cdot \Gamma (v-{v_*},\varphi )\\ =&\frac{1+\cos \theta }{2} |v|^2 + \frac{1-\cos \theta }{2}|{v_*}|^2 + (\sin \theta ) v\cdot \Gamma (v-{v_*},\varphi ). \end{aligned}$$

Hence, by Newton’s trinomial expansion, setting ${\mathcal {B}}_n=\{(i,j,k)\in {\mathbb {N}}^3 : i+j+k=n\}$,

$$\begin{aligned} |v'|^{2n}=&\sum _{(i,j,k)\in {\mathcal {B}}_n} \frac{n!}{i!j!k!} \Bigg (\frac{1+\cos \theta }{2}\Bigg )^i\Bigg (\frac{1-\cos \theta }{2}\Bigg )^j\\&(\sin \theta )^k |v|^{2i}|{v_*}|^{2j} (v\cdot \Gamma (v-{v_*},\varphi ))^k. \end{aligned}$$

Step 2. We now prove that for $k\in {\mathbb {N}}$,

$$\begin{aligned} \frac{1}{2\pi }\int _0^{2\pi } (v\cdot \Gamma (v-{v_*},\varphi ))^k \mathrm{d}\varphi = \hbox \mathrm{1}{}\hbox \mathrm{I}_{\{k \in 2{\mathbb {N}}\}} \frac{k!}{2^k [(k/2)!]^2}(|v|^2|{v_*}|^2-(v\cdot {v_*})^2)^{k/2}. \end{aligned}$$

We have $v\cdot \Gamma (v-{v_*},\varphi )=a \cos \varphi + b \sin \varphi$, where $a=v\cdot I(v-{v_*})$ and $b=v\cdot J(v-{v_*})$, whence

$$\begin{aligned} v\cdot \Gamma (v-{v_*},\varphi )=\sqrt{a^2+b^2} \sin (\varphi +\varphi _0), \end{aligned}$$

for $\varphi _0$ such that $\frac{a}{\sqrt{a^2+b^2}}=\sin \varphi _0$ and $\frac{b}{\sqrt{a^2+b^2}}=\cos \varphi _0$. We thus recognize a Wallis integral:

$$\begin{aligned} \frac{1}{2\pi }\int _0^{2\pi } (v\cdot \Gamma (v-{v_*},\varphi ))^k \mathrm{d}\varphi= & {} \frac{2}{\pi }(a^2+b^2)^{k/2} \hbox \mathrm{1}{}\hbox \mathrm{I}_{\{k \in 2{\mathbb {N}}\}} \int _0^{\pi /2} \sin ^k \varphi \mathrm{d}\varphi \\= & {} \hbox \mathrm{1}{}\hbox \mathrm{I}_{\{k \in 2{\mathbb {N}}\}} \frac{k!}{2^k [(k/2)!]^2}(a^2+b^2)^{k/2}. \end{aligned}$$

To complete the step, we recall that $\left( \frac{v-{v_*}}{|v-{v_*}|},\frac{I(v-{v_*})}{|v-{v_*}|}, \frac{J(v-{v_*})}{|v-{v_*}|}\right)$ is an orthonormal basis, whence

$$\begin{aligned} |v|^2= & {} \frac{[v\cdot (v-{v_*})]^2+[v\cdot I(v-{v_*})]^2+[v\cdot J(v-{v_*})]^2 }{|v-{v_*}|^2}\\= & {} \frac{[v\cdot (v-{v_*})]^2+a^2+b^2}{|v-{v_*}|^2} \end{aligned}$$

and thus $a^2+b^2=|v|^2|v-{v_*}|^2-[v\cdot (v-{v_*})]^2=|v|^2|{v_*}|^2-(v\cdot {v_*})^2$.

Step 3. Gathering Steps 1 and 2 and setting ${\mathcal {C}}_n=\{(i,j,k)\in {\mathbb {N}}^3 : i+j+k=n,\; k\in 2{\mathbb {N}}\}$,

$$\begin{aligned} \Theta _n(v,{v_*},\theta )\!=&\!\!\!\!\!\! \sum _{(i,j,k)\in {\mathcal {C}}_n} \!\!\! \frac{n!}{i!j! [(k/2)!]^2}\Big (\frac{1\!+\!\cos \theta }{2}\Big )^i\\&\Big (\frac{1\!-\!\cos \theta }{2}\Big )^j\Big (\frac{\sin \theta }{2}\Big )^k |v|^{2i}|{v_*}|^{2j} \Big (|v|^2|{v_*}|^2\!-(v\cdot {v_*})^2\Big )^{k/2}. \end{aligned}$$

It then suffices to isolate the two extreme terms $(i,j,k)=(n,0,0)$ and $(i,j,k)=(0,n,0)$. $\square$

We next estimate, sharply, some integrals in $\theta$.

Lemma 6

Assume $(H_2(\nu ))$. There are $\zeta _1,\zeta _2 \in (0,\infty )$, depending only on $\nu ,\kappa _1,\kappa _2$, such that:

(i) for all integer $n\ge 2$,

$$\begin{aligned} a_n:=\int _0^\pi \Big (1-\Bigg [\frac{1+\cos \theta }{2}\Bigg ]^n-\Bigg [\frac{1-\cos \theta }{2}\Bigg ]^n\Bigg ) \beta (\theta ) \mathrm{d}\theta \ge \zeta _1 n^{\nu /2}; \end{aligned}$$

(ii) for all integers a, n such that $1\le a \le n-1$, setting

$$\begin{aligned}&J_{n,a}:=\int _0^\pi \Bigg (\frac{1+\cos \theta }{2}\Bigg )^a\Bigg (\frac{1-\cos \theta }{2}\Bigg )^{n-a} \beta (\theta ) \mathrm{d}\theta ,\\&\quad \hbox {we have}\quad \left( {\begin{array}{c}n\\ a\end{array}}\right) J_{n,a} \le \zeta _2 \Bigg [\frac{n^{\nu /2}}{(n-a)^{\nu /2+1}} + \frac{1}{a} \Bigg ]. \end{aligned}$$

Proof

We start with (i): the integrand in $a_n$ is nonnegative, because for $x=(1+\cos \theta )/2\in [0,1]$, we have $x^n+(1-x)^n\le 1$. Hence recalling $(H_2(\nu ))$,

$$\begin{aligned} a_n \ge \kappa _1 \int _0^{n^{-1/2}}\Bigg (1-\Bigg [\frac{1+\cos \theta }{2}\Bigg ]^n-\Bigg [\frac{1-\cos \theta }{2}\Bigg ]^n\Bigg ) \theta ^{-\nu -1} \mathrm{d}\theta . \end{aligned}$$

For all $\theta \in [0,n^{-1/2}]$, we have $(1+\cos \theta )/2\le 1-\theta ^2/5$, whence $[(1+\cos \theta )/2]^n \le (1-\theta ^2/5)^n\le \exp (-n \theta ^2/5)$ and thus $1-[(1+\cos \theta )/2]^n \ge n \theta ^2/10$. Next, still for $\theta \in [0,n^{-1/2}]$, we have $(1-\cos \theta )/2\le \theta ^2/4$, whence, since $n\ge 2$,

$$\begin{aligned}&1-\Bigg [\frac{1+\cos \theta }{2}\Bigg ]^n-\Bigg [\frac{1-\cos \theta }{2}\Bigg ]^n \ge \frac{n \theta ^2}{10} -\frac{\theta ^{2n}}{4^n}\\&\ge \Bigg (\frac{n}{10}-\frac{1}{16} \Bigg )\theta ^2 \ge \Bigg (\frac{n}{10}-\frac{n}{32} \Bigg ) \theta ^2 \ge \frac{n}{20} \theta ^2. \end{aligned}$$

Consequently,

$$\begin{aligned} a_n \ge \kappa _1 \frac{n}{20} \int _0^{n^{-1/2}} \theta ^{1-\nu } \mathrm{d}\theta = \frac{\kappa _1}{20(2-\nu )} n^{\nu /2}. \end{aligned}$$

For (ii), we first use $(H_2(\nu ))$ to write $J_{n,a}\le \kappa _2 K_{\nu ,n,a}+\kappa _2 L_{\nu ,n,a}$, where

$$\begin{aligned} K_{\nu ,n,a}=&\int _0^{\pi /2} \Bigg (\frac{1+\cos \theta }{2}\Bigg )^a\Bigg (\frac{1-\cos \theta }{2}\Bigg )^{n-a} \theta ^{-\nu -1} \mathrm{d}\theta ,\\ L_{\nu ,n,a} =&\Bigg (\frac{2}{\pi }\Bigg )^{\nu +1} \int _{\pi /2}^\pi \Bigg (\frac{1+\cos \theta }{2}\Bigg )^a\Bigg (\frac{1-\cos \theta }{2}\Bigg )^{n-a}\mathrm{d}\theta . \end{aligned}$$

Using the substitution $\theta \rightarrow \pi -\theta$, we see that

$$\begin{aligned} L_{\nu ,n,a}=&\Bigg (\frac{2}{\pi }\Bigg )^{\nu +1} \int _{0}^{\pi /2} \Bigg (\frac{1+\cos \theta }{2}\Bigg )^{n-a}\Bigg (\frac{1-\cos \theta }{2}\Bigg )^{a}\mathrm{d}\theta \\ \le&\Bigg (\frac{2}{\pi }\Bigg )^{\nu } \int _{0}^{\pi /2} \Bigg (\frac{1+\cos \theta }{2}\Bigg )^{n-a}\Bigg (\frac{1-\cos \theta }{2}\Bigg )^{a}\theta ^{-1}\mathrm{d}\theta =\Bigg (\frac{2}{\pi }\Bigg )^{\nu } K_{0,n,n-a}. \end{aligned}$$

We will prove that for $\nu \in [0,2)$, there is a constant $A_\nu \in (0,\infty )$ such that for all $1\le a \le n-1$,

$$\begin{aligned} \left( {\begin{array}{c}n\\ a\end{array}}\right) K_{\nu ,n,a} \le A_\nu \frac{n^{\nu /2}}{(n-a)^{1+\nu /2}}. \end{aligned}$$

(6)

We will deduce that $\left( {\begin{array}{c}n\\ a\end{array}}\right) L_{\nu ,n,a}\le [2/\pi ]^{\nu } \left( {\begin{array}{c}n\\ a\end{array}}\right) K_{0,n,n-a}\le [2/\pi ]^{\nu }A_0 a^{-1}$ and this will end the proof. For $\theta \in (0,\pi /2]$, we have $\theta \le 2\sin \theta$ and $\theta ^{-1}\le [(1-\cos \theta )/2]^{-1/2}$, so that

$$\begin{aligned} \theta ^{-\nu -1} \le 2 \theta ^{-\nu -2} \sin \theta \le 2 \Bigg ( \frac{1-\cos \theta }{2}\Bigg )^{-\nu /2-1} \sin \theta \end{aligned}$$

and thus

$$\begin{aligned}&K_{\nu ,n,a}\le 2 \int _0^{\pi /2} \Bigg (\frac{1+\cos \theta }{2}\Bigg )^a\Bigg (\frac{1-\cos \theta }{2}\Bigg )^{n-a-\nu /2-1}\\&\sin \theta \mathrm{d}\theta =4 \int _{1/2}^1 x^a (1-x)^{n-a-\nu /2-1}\mathrm{d}x, \end{aligned}$$

using the change of variables $x=(1+\cos \theta )/2$. Hence

$$\begin{aligned} K_{\nu ,n,a}\le 4 \int _0^{1} x^a (1-x)^{n-a-\nu /2-1}\mathrm{d}x = 4 \frac{\Gamma (a+1)\Gamma (n-a-\nu /2)}{\Gamma (n+1-\nu /2)}, \end{aligned}$$

where $\Gamma$ is Euler’s Gamma function. Using that $\Gamma (k+1)=k!$, that $(n-a-\nu /2)\Gamma (n-a-\nu /2)=\Gamma (n-a+1-\nu /2)$ and setting $u_{\nu ,k}= \Gamma (k+1)/\Gamma (k+1-\nu /2)$, we realize that

$$\begin{aligned} \left( {\begin{array}{c}n\\ a\end{array}}\right) K_{\nu ,n,a} \le 4 \frac{n!\Gamma (n-a-\nu /2)}{\Gamma (n+1-\nu /2)(n-a)!} = 4 \frac{ u_{\nu ,n}}{(n-a-\nu /2)u_{\nu ,n-a}}. \end{aligned}$$

But using Stirling’s formula $\Gamma (x+1)\sim \sqrt{2\pi x} (x/e)^x$ as $x\rightarrow \infty$, one can verify that $u_{\nu ,k}\sim k^{\nu /2}$ as $k\rightarrow \infty$ so that there is a constant $A_\nu \in (1,\infty )$ such that for all $k\ge 1$,

$$\begin{aligned} A_\nu ^{-1} k^{\nu /2}\le u_{\nu ,k} \le A_\nu k^{\nu /2}. \end{aligned}$$

We conclude that for all $1\le a \le n-1$,

$$\begin{aligned} \left( {\begin{array}{c}n\\ a\end{array}}\right) K_{\nu ,n,a}\le 4A_\nu ^2 \frac{n^{\nu /2}}{(n-a)^{\nu /2}(n-a-\nu /2)} \le \frac{4A_\nu ^2}{1-\nu /2} \times \frac{n^{\nu /2}}{(n-a)^{\nu /2+1}} \end{aligned}$$

because $n-a\ge 1$ implies that $n-a-\nu /2 \ge (n-a)(1-\nu /2)$. We have checked (6) and the proof is complete. $\square$

We can now handle the

Proof of Lemma 4

We fix $n\ge 2$. Using (3), we realize that, with the notation of Lemma 5,

$$\begin{aligned} \frac{1}{2\pi }\int _0^{2\pi } [|v'|^{2n}\!+\!|v'_*|^{2n}\!-\!|v|^{2n}\!-\!|{v_*}|^{2n} ] \mathrm{d}\varphi \!=\! \Theta _n(v,{v_*},\theta )\!+\! \Theta _n({v_*},v,\!-\!\theta )\!-\!|v|^{2n}\!-\!|{v_*}|^{2n}. \end{aligned}$$

We deduce from Lemma 5 that

$$\begin{aligned}&\frac{1}{2\pi }\int _0^{2\pi } [|v'|^{2n}+|v'_*|^{2n} -|v|^{2n} -|{v_*}|^{2n} ] \mathrm{d}\varphi \\&\quad \!=\! \!-\!\Big (1\!-\!\Big [\frac{1\!+\!\cos \theta }{2}\Big ]^n \!-\!\Big [\frac{1\!-\!\cos \theta }{2}\Big ]^n\Big )\times (|v|^{2n}\!+\!|{v_*}|^{2n}) \!+\! \Lambda _n(v,{v_*},\theta )\!+\!\Lambda _n({v_*},v,\!-\!\theta ). \end{aligned}$$

Hence $D_n(v,{v_*})= -D_{n,1}(v,{v_*})+D_{n,2}(v,{v_*})$, where

$$\begin{aligned} D_{n,1}(v,{v_*})=&2\pi \int _0^\pi \Big (1-\Big [\frac{1+\cos \theta }{2}\Big ]^n-\Big [\frac{1-\cos \theta }{2}\Big ]^n\Big ) \beta (\theta ) \mathrm{d}\theta \times (|v|^{2n}+|{v_*}|^{2n}),\\ D_{n,2}(v,{v_*})=&2\pi \int _0^\pi [\Lambda _n(v,{v_*},\theta )+\Lambda _n({v_*},v,-\theta )] \beta (\theta )\mathrm{d}\theta . \end{aligned}$$

We now divide the proof into 5 steps.

Step 1. By Lemma 6-(i), we have $D_{n,1}(v,{v_*})\ge 2\pi \zeta _1 n^{\nu /2}(|v|^{2n}+|{v_*}|^{2n})$.

Step 2. We next roughly bound $|v|^2|{v_*}|^2-(v\cdot {v_*})^2$ by $|v|^2|{v_*}|^2$ in the expression of $\Lambda _n$ to find

$$\begin{aligned} D_{n,2}(v,{v_*})\le 2\pi \sum _{(i,j,k)\in {\mathcal {A}}_n} \!\!\! \frac{n!}{i!j! [(k/2)!]^2}I_{i,j,k} [|v|^{2i+k}|{v_*}|^{2j+k}+|v|^{2j+k}|{v_*}|^{2i+k}], \end{aligned}$$

where

$$\begin{aligned} I_{i,j,k}=\int _0^{\pi } \Bigg (\frac{1+\cos \theta }{2}\Bigg )^i \Bigg (\frac{1-\cos \theta }{2}\Bigg )^j\Bigg (\frac{\sin \theta }{2}\Bigg )^k \beta (\theta )\mathrm{d}\theta . \end{aligned}$$

This can be rewritten as

$$\begin{aligned} D_{n,2}(v,{v_*})\le 2\pi \sum _{a=0}^n K_{n,a} [|v|^{2a}|{v_*}|^{2(n-a)}+|v|^{2(n-a)}|{v_*}|^{2a}] \end{aligned}$$

where, setting ${\mathcal {A}}_{n,a}=\{(i,j,k)\in {\mathcal {A}}_n : i+k/2=a$ (whence $j+k/2=n-a)\}$,

$$\begin{aligned} K_{n,a}= \sum _{(i,j,k)\in {\mathcal {A}}_{n,a}} \frac{n!}{i!j! [(k/2)!]^2} I_{i,j,k}. \end{aligned}$$

Step 3. We have $K_{n,n}=K_{n,0}=0$. Indeed, we e.g. have ${\mathcal {A}}_{n,n}=\emptyset$ because for $(i,j,k)\in {\mathcal {A}}_{n,n}$, we have $i+j+k=n$ and $j+k/2=0$, whence $i=n$, which is forbidden since $(i,j,k)\in {\mathcal {A}}_n$.

Step 4. We now fix $a\in [\![1,n-1 ]\!]$. Then $(i,j,k) \in {\mathcal {A}}_{n,a}$ if and only if there is $\ell \in {\mathbb {N}}$ such that $k=2\ell \in [\![0,n ]\!]$, $i=a-\ell \in [\![0,n-1]\!]$ and $j=(n-a)-\ell \in [\![0,n-1 ]\!]$, so that

$$\begin{aligned} K_{n,a}=\sum _{\ell =0}^{a\wedge (n-a)} \frac{n!}{(a-\ell )!(n-a-\ell )!(\ell !)^2} I_{a-\ell ,n-a-\ell ,2\ell }. \end{aligned}$$

But for any $\ell \in [\![0,a\wedge (n-a)]\!]$, it holds that

$$\begin{aligned} I_{a-\ell ,n-a-\ell ,2\ell }=\int _0^{\pi } \Bigg (\frac{1+\cos \theta }{2}\Bigg )^{a-\ell } \Bigg (\frac{1-\cos \theta }{2}\Bigg )^{n-a-\ell }\Bigg (\frac{\sin \theta }{2}\Bigg )^{2\ell } \beta (\theta )\mathrm{d}\theta =J_{n,a}, \end{aligned}$$

where

$$\begin{aligned} J_{n,a} =\int _0^{\pi } \Bigg (\frac{1+\cos \theta }{2}\Bigg )^{a}\Bigg (\frac{1-\cos \theta }{2}\Bigg )^{n-a}\beta (\theta )\mathrm{d}\theta , \end{aligned}$$

because $(1+\cos \theta )(1-\cos \theta )=\sin ^2\theta$. Thus

$$\begin{aligned} K_{n,a}= & {} J_{n,a}\sum _{\ell =0}^{a\wedge (n-a)} \frac{n!}{(a-\ell )!(n-a-\ell )!(\ell !)^2} =J_{n,a} \left( {\begin{array}{c}n\\ a\end{array}}\right) \sum _{\ell =0}^{a\wedge (n-a)} \left( {\begin{array}{c}a\\ \ell \end{array}}\right) \left( {\begin{array}{c}n-a\\ \ell \end{array}}\right) \\= & {} J_{n,a}\Big [\left( {\begin{array}{c}n\\ a\end{array}}\right) \Big ]^2. \end{aligned}$$

We finally used the well-known identity $\sum _{\ell =0}^{a\wedge b}\left( {\begin{array}{c}a\\ \ell \end{array}}\right) \left( {\begin{array}{c}b\\ \ell \end{array}}\right) = \left( {\begin{array}{c}a+b\\ a\end{array}}\right)$, which can be shown noting that $\left( {\begin{array}{c}a+b\\ a\end{array}}\right)$ is the coefficient in front of $X^a$ of $(1+X)^{a+b}$, while $\sum _{\ell =0}^{a\wedge b}\left( {\begin{array}{c}a\\ \ell \end{array}}\right) \left( {\begin{array}{c}b\\ \ell \end{array}}\right) =\sum _{\ell =0}^{a\wedge b}\left( {\begin{array}{c}a\\ \ell \end{array}}\right) \left( {\begin{array}{c}b\\ b-\ell \end{array}}\right)$ is the coefficient in front of $X^a$ of $(1+X)^{a}(1+X)^{b}$.

Step 5. Gathering Steps 2–3–4, we have checked that

$$\begin{aligned} D_{n,2}(v,{v_*})\le&2\pi \sum _{a=1}^{n-1} J_{n,a} \Bigg [ \left( {\begin{array}{c}n\\ a\end{array}}\right) \Bigg ]^2 [|v|^{2a}|{v_*}|^{2(n-a)}+|v|^{2(n-a)}|{v_*}|^{2a}]. \end{aligned}$$

The conclusion then follows from Lemma 6-(ii), from which $\left( {\begin{array}{c}n\\ a\end{array}}\right) J_{n,a}\le \zeta _2 \Big [\frac{n^{\nu /2}}{(n-a)^{\nu /2+1}} + \frac{1}{a}\Big ]$. $\square$

3 Even Integer Moments

Using the previous Povzner inequality, we can derive the even integer moments.

Lemma 7

Assume $(H_1(\gamma ))$ and $(H_2(\nu ))$ for some $\gamma \in (0,1]$ and some $\nu \in (0,2)$. For any integer $n\ge 2$, any $t>0$,

$$\begin{aligned} m_{2n}'(t) \le - c_1 n^{\nu /2} m_{2n+\gamma }(t)+ c_2 S_n(t) + 2c_1 n^{\nu /2}2^{2n/\gamma }, \end{aligned}$$

where $c_1=\lambda _1$ and $c_2=16\lambda _2$ (see Lemma 4) depend only on $\nu ,\kappa _1,\kappa _2$ and where

$$\begin{aligned} S_n(t) = \sum _{a=1}^{\lfloor n/2\rfloor } \left( {\begin{array}{c}n\\ a\end{array}}\right) \frac{n^{\nu /2}}{a^{\nu /2+1}} m_{2a}(t)m_{2(n-a)+\gamma }(t). \end{aligned}$$

Proof

We fix $n\ge 2$ and use the weak formulation (4) with $\phi (v)=|v|^{2n}$, which is licit thanks to (5), to find, for any $t>0$ and with the notation of Lemma 4,

$$\begin{aligned} m_{2n}'(t)= \int _{{\mathbb {R}}^3}\int _{{\mathbb {R}}^3}D_n(v,{v_*}) |v-{v_*}|^\gamma f_t(\mathrm{d}{v_*}) f_t(\mathrm{d}v). \end{aligned}$$

Hence using Lemma 4, $m_{2n}'(t)\le -A_n(t)+B_n(t)$, where

$$\begin{aligned} A_n(t)=&\lambda _1 n^{\nu /2} \int _{{\mathbb {R}}^3}\int _{{\mathbb {R}}^3}(|v|^{2n}+|{v_*}|^{2n}) |v-{v_*}|^\gamma f_t(\mathrm{d}{v_*}) f_t(\mathrm{d}v),\\ B_n(t)=&\lambda _2 \sum _{a=1}^{n-1} \left( {\begin{array}{c}n\\ a\end{array}}\right) \Big (\frac{n^{\nu /2}}{(n-a)^{\nu /2+1}}+\frac{1}{a}\Big ) \int _{{\mathbb {R}}^3}\int _{{\mathbb {R}}^3}[|v|^{2a}|{v_*}|^{2(n-a)}+|v|^{2(n-a)}|{v_*}|^{2a}] \\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad |v-{v_*}|^\gamma f_t(\mathrm{d}{v_*}) f_t(\mathrm{d}v). \end{aligned}$$

We now divide the proof into 2 steps.

Step 1. Using a symmetry argument and that $|v-{v_*}|^\gamma \ge |v|^\gamma -|{v_*}|^\gamma$, we see that

$$\begin{aligned} A_n(t)&=2\lambda _1 n^{\nu /2} \int _{{\mathbb {R}}^3}\int _{{\mathbb {R}}^3}|v|^{2n} |v-{v_*}|^\gamma f_t(\mathrm{d}{v_*}) f_t(\mathrm{d}v)\\&\ge 2\lambda _1 n^{\nu /2} (m_{2n+\gamma }(t)- m_{2n}(t)m_{\gamma }(t)). \end{aligned}$$

By Hölder’s inequality, $m_\gamma (t)\le [m_2(t)]^{\gamma /2}=1$. Using moreover that $x^{2n} \le \frac{1}{2} x^{2n+\gamma } + 2^{2n/\gamma }$ for all $x\ge 0$ (separate the cases $x \le 2^{1/\gamma }$ and $x\ge 2^{1/\gamma }$),

$$\begin{aligned} m_\gamma (t)m_{2n}(t)\le & {} m_{2n}(t)=\int _{{\mathbb {R}}^3}|v|^{2n}f_t(\mathrm{d}v) \le \int _{{\mathbb {R}}^3}\Bigg (\frac{1}{2}|v|^{2n+\gamma }+2^{2n/\gamma }\Bigg )f_t(\mathrm{d}v)\\= & {} \frac{1}{2}m_{2n+\gamma }(t) + 2^{2n/\gamma }. \end{aligned}$$

All in all,

$$\begin{aligned} A_n(t)\ge \lambda _1 n^{\nu /2} m_{2n+\gamma }(t) - 2\lambda _1 n^{\nu /2}2^{2n/\gamma }. \end{aligned}$$

Step 2. Using a symmetry argument and that $|v-{v_*}|^\gamma \le |v|^\gamma +|{v_*}|^\gamma$, we find

$$\begin{aligned} B_n(t) \le&2\lambda _2\sum _{a=1}^{n-1} \left( {\begin{array}{c}n\\ a\end{array}}\right) \Bigg (\frac{n^{\nu /2}}{(n-a)^{\nu /2+1}}+\frac{1}{a}\Bigg ) [m_{2a+\gamma }(t)m_{2(n-a)}(t)+m_{2a}(t)m_{2(n-a)+\gamma }(t)]\\ =&2\lambda _2\sum _{a=1}^{n-1} \left( {\begin{array}{c}n\\ a\end{array}}\right) \Bigg (\frac{n^{\nu /2}}{a^{\nu /2+1}}+\frac{1}{a}\Bigg ) [m_{2a+\gamma }(t)m_{2(n-a)}(t)+m_{2a}(t)m_{2(n-a)+\gamma }(t)] \end{aligned}$$

by symmetry again. Since now $1/a \le n^{\nu /2}/a^{\nu /2+1}$ and since $a\rightarrow n^{\nu /2}/a^{\nu /2+1}$ is decreasing,

$$\begin{aligned} B_n(t) \le 8\lambda _2 \sum _{a=1}^{\lfloor n/2\rfloor }\left( {\begin{array}{c}n\\ a\end{array}}\right) \frac{n^{\nu /2}}{a^{\nu /2+1}} [m_{2a+\gamma }(t)m_{2(n-a)}(t)+m_{2a}(t)m_{2(n-a)+\gamma }(t)]. \end{aligned}$$

But for $a\in [\![1, \lfloor n/2\rfloor ]\!]$, we have $a\le n-a$ whence $m_{2a+\gamma }(t)m_{2(n-a)}(t) \le m_{2a}(t)m_{2(n-a)+\gamma }(t)$ by Lemma 8 below, so that

$$\begin{aligned} B_n(t) \le 16\lambda _2 \sum _{a=1}^{\lfloor n/2\rfloor }\left( {\begin{array}{c}n\\ a\end{array}}\right) \frac{n^{\nu /2}}{a^{\nu /2+1}} m_{2a}(t)m_{2(n-a)+\gamma }(t)= 16\lambda _2 S_n(t). \end{aligned}$$

Recalling that $m_{2n}'(t)=-A_n(t)+B_n(t)$ and Step 1, we have reached our goal.$\square$

We now prove a classical lemma that we used a few lines above.

Lemma 8

For any $b\ge a \ge 0$, any $\alpha >0$, any $f\in {\mathcal {P}}({{\mathbb {R}}^3})$,

$$\begin{aligned} m_{a+\alpha }(f)m_b(f)\le m_{b+\alpha }(f)m_a(f). \end{aligned}$$

Proof

We fix $\alpha >0$ and $f\in {\mathcal {P}}({{\mathbb {R}}^3})$ and have to prove that the function $u(a)=m_{a+\alpha }(f)/m_a(f)$ is nondecreasing on ${\mathbb {R}}_+$. Observing that $\frac{\mathrm{d}}{\mathrm{d}a}m_a(f)=\int _{{\mathbb {R}}^3}(\log |v|) |v|^a f(\mathrm{d}v)$, we find

$$\begin{aligned} u'(a)&=\frac{1}{m_a(f)}\int _{{\mathbb {R}}^3}(\log |v|) |v|^{a+\alpha } f(\mathrm{d}v)\\&\,\,\quad - \frac{1}{m_a^2(f)} \Bigg (\int _{{\mathbb {R}}^3}(\log |v|) |v|^{a} f(\mathrm{d}v)\Bigg )\Bigg (\int _{{\mathbb {R}}^3}|v|^{a+\alpha } f(\mathrm{d}v)\Bigg ). \end{aligned}$$

Setting $g_a(\mathrm{d}v)= |v|^{a} f(\mathrm{d}v)/m_a(f)$, which is a probability measure,

$$\begin{aligned} u'(a)=\frac{1}{\alpha }\int _{{\mathbb {R}}^3}(\log |v|^\alpha ) |v|^{\alpha } g_a(\mathrm{d}v)- \frac{1}{\alpha }\Bigg (\int _{{\mathbb {R}}^3}(\log |v|^\alpha ) g_a(\mathrm{d}v)\Bigg )\Bigg (\int _{{\mathbb {R}}^3}|v|^{\alpha } g_a(\mathrm{d}v)\Bigg ). \end{aligned}$$

Hence $u'(a)\ge 0$ by the Jensen inequality.$\square$

We will also use the following remark.

Remark 9

For $f \in {\mathcal {P}}({{\mathbb {R}}^3})$ such that $m_2(f)=1$, for all $r\ge s \ge 2$, by Hölder’s inequality,

$$\begin{aligned} m_{s}(f)&=\int _{{\mathbb {R}}^3}|v|^{s-2} |v|^2 f(\mathrm{d}v) \le \Bigg (\int _{{\mathbb {R}}^3}|v|^{r-2} |v|^2 f(\mathrm{d}v)\Bigg )^{(s-2)/(r-2)}\\&=[m_r(f)]^{(s-2)/(r-2)}. \end{aligned}$$

We finally quickly prove some moment estimates that are more or less well-known.

Lemma 10

Assume $(H_1(\gamma ))$ and $(H_2(\nu ))$ for some $\gamma \in (0,1]$ and some $\nu \in (0,2)$.

(i) For all $r>0$, there is $K_r\in (0,\infty )$, depending only on $\gamma ,\nu ,\kappa _1,\kappa _2,r$, such that for all $t>0$,

$$\begin{aligned} m_r(t)\le K_r(1+t^{-(r-2)/\gamma }). \end{aligned}$$

(ii) For all integer $n\ge 2$, all $A\ge 1$, there is $K_{2n,A}\in (0,\infty )$, depending only on $\gamma ,\nu ,\kappa _1,\kappa _2,n,A$, such that

$$\begin{aligned} m_{2n}(0)\le A \quad \Longrightarrow \quad \sup _{t\ge 0}m_{2n}(t)\le K_{2n,A}. \end{aligned}$$

Proof

We first prove that for any fixed integer $n\ge 2$, there is a constant $A_n$, depending only on $\gamma ,\nu ,\kappa _1,\kappa _2,n$, allowed to vary from line to line, such that for all $t>0$,

$$\begin{aligned} m_{2n}'(t) \le -\frac{c_1}{2} [m_{2n}(t)]^{1+\gamma /(2n-2)}+A_n. \end{aligned}$$

(7)

Using Lemma 7, a rough upperbound, and then Lemma 8, we get

$$\begin{aligned} m_{2n}'(t) \!\le & {} \! -\,c_1 m_{2n+\gamma }(t) + A_n \! \sum _{a=1}^{\lfloor n/2\rfloor } \!m_{2a}(t)m_{2(n-a)+\gamma }(t) +A_n \\\le & {} -c_1 m_{2n+\gamma }(t) +A_nm_{2(n-1)+\gamma }m_2(t) +A_n. \end{aligned}$$

Using now Remark 9 and that $m_2(t)=1$, we find

$$\begin{aligned} m_{2n}'(t)\le - c_1[m_{2n}(t)]^{1+\gamma /(2n-2)} + A_n [m_{2n}(t)]^{1-(2-\gamma )/(2n-2)} +A_n, \end{aligned}$$

from which (7) follows. Point (ii) clearly follows from 7. When $r=2n\ge 4$ is an even integer, point (i) also follows from 7. For a general $r>2$, we consider an integer $n_r\ge 2$ such that $2n_r\ge r$ and we conclude from Remark 9 that

$$\begin{aligned} m_r(t)\le [m_{2n_r}(t)]^{(r-2)/(2n_r-2)} \le (K_{2n_r}(1+t^{-(2n_r-2)/\gamma }))^{(r-2)/(2n_r-2)}, \end{aligned}$$

whence the result. Finally, for $r\in (0,2]$, we obviously have $m_r(t)\le [m_2(t)]^{r/2}\le 1$. $\square$

4 Control of Exponential Moments by Even Integer Moments

We explain how to control exponential moments from even integer moments and vice versa.

Lemma 11

• (i) For $f \in {\mathcal {P}}({{\mathbb {R}}^3})$, for $\sigma _0\in (0,\infty )$, $\alpha \in [1,\infty )$ and $K \in [1,\infty )$,

  $$\begin{aligned} \sup _{n\ge 0} \frac{\sigma _0 ^nm_{2n}(f)}{ (n!)^\alpha } \le K \quad \Longrightarrow \quad \int _{{\mathbb {R}}^3}\exp (\sigma _0^{1/\alpha }|v|^{2/\alpha }/2) f(\mathrm{d}v) \le 2K^{1/\alpha }. \end{aligned}$$

• (ii) For $\rho \in (0,2]$, $\sigma _0\in (0,1]$ and $K\in (1,\infty )$, there is $\sigma _1\in (0,\infty )$, depending only on $\rho ,\sigma _0,K$, such that for all $f \in {\mathcal {P}}({{\mathbb {R}}^3})$,

  $$\begin{aligned} \int _{{\mathbb {R}}^3}\exp (\sigma _0|v|^{\rho }) f(\mathrm{d}v) \le K \quad \Longrightarrow \quad \sup _{n\ge 0} \frac{\sigma _1^nm_{2n}(f)}{ (n!)^{2/\rho }} \le 1. \end{aligned}$$

Proof

We start with (i). By Hölder’s inequality, since $\alpha \ge 1$,

$$\begin{aligned} m_{2n/\alpha }(f) \le [m_{2n}(f)]^{1/\alpha }\le K^{1/\alpha } \sigma _0^{-n/\alpha } n!, \end{aligned}$$

whence

$$\begin{aligned} \int _{{\mathbb {R}}^3}\exp [\sigma _0^{1/\alpha }|v|^{2/\alpha }/2] f(\mathrm{d}v)=\sum _{n\ge 0} \frac{\sigma _0^{n/\alpha } m_{2n/\alpha }(f) }{2^nn!} \le K^{1/\alpha } \sum _{n\ge 0} 2^{-n}=2K^{1/\alpha }. \end{aligned}$$

We now check (ii). We know that

$$\begin{aligned} \sup _{n\ge 0} \frac{\sigma _0^n m_{\rho n}(f)}{n!}\le \sum _{n\ge 0} \frac{\sigma _0^n m_{\rho n}(f)}{n!}=\int _{{\mathbb {R}}^3}\exp (\sigma _0|v|^{\rho }) f(\mathrm{d}v)\le K. \end{aligned}$$

For $n\ge 1$, we set $k_n=\lceil 2n/\rho \rceil \in [2n/\rho ,2n/\rho +1)$ and write, using that $\rho k_n \ge 2n$,

$$\begin{aligned} m_{2n}(f) \le 1+ m_{\rho k_n}(f) \le 1+\frac{K k_n!}{\sigma _0^{k_n}}, \quad \hbox {so that}\quad \frac{\sigma _1^nm_{2n}(f)}{(n!)^{2/\rho }} \le \frac{\sigma _1^n}{(n!)^{2/\rho }}+ \frac{\sigma _1^n K k_n!}{\sigma _0^{k_n} (n!)^{2/\rho }}. \end{aligned}$$

By Stirling’s formula $n!\sim \sqrt{2\pi n} (n/e)^n$ and since $k_n=\lceil 2n/\rho \rceil$, we find that, for some constant $A\in (0,\infty )$, depending only on $\rho$ and allowed to vary, for all $n\ge 1$,

$$\begin{aligned} \frac{k_n!}{(n!)^{2/\rho }} \!\le & {} \! A \frac{n^{1/2} [(2n/\rho +1)/e]^{2n/\rho +1}}{n^{1/\rho }[n/e]^{2n/\rho }} \!\le \! A n^{3/2-1/\rho } \frac{[(3n)/(e\rho )]^{2n/\rho }}{[n/e]^{2n/\rho }}\!\le \!A n^{3/2}\Bigg (\frac{3}{\rho }\Bigg )^{2n/\rho } \!\!\! \\\le & {} A \Bigg (\frac{4}{\rho }\Bigg )^{2n/\rho }. \end{aligned}$$

Observing next that $\sigma _0^{k_n} \ge \sigma _0^{2n/\rho +1}$ (since $\sigma _0\in (0,1]$), we end with

$$\begin{aligned} \hbox {for all }n\ge 1,\quad \frac{\sigma _1^n m_{2n}(f)}{(n!)^{2/\rho }} \le \frac{\sigma _1^n}{(n!)^{2/\rho }}+\frac{KA}{\sigma _0} \frac{\sigma _1^n}{\sigma _0^{2n/\rho }} \Bigg (\frac{4}{\rho }\Bigg )^{2n/\rho }. \end{aligned}$$

This last quantity is bounded by 1 if $\sigma _1>0$ is small enough (depending on $\rho$, K, and $\sigma _0$). $\square$

5 Creation of Exponential Moments

The following estimate will allow us to prove the creation result by Lemma 11.

Lemma 12

Assume $(H_1(\gamma ))$ and $(H_2(\nu ))$ for some $\gamma \in (0,1]$ and some $\nu \in (0,2)$ and set $\alpha =\max \{1,(2-\nu )/\gamma \}$. There are $\sigma \in (0,1]$ and $T>0$, depending only on $\gamma ,\nu ,\kappa _1,\kappa _2$, such that

$$\begin{aligned} \sup _{t\in [0,T]} \sum _{n=0}^\infty \frac{(\sigma t)^{2n/\gamma }m_{2n}(t) }{ (n!)^\alpha } \le 2. \end{aligned}$$

Proof

We recall that by Lemma 7, for any integer $n\ge 2$, any $t>0$,

$$\begin{aligned} m_{2n}'(t) \le - c_1 n^{\nu /2} m_{2n+\gamma }(t)+ c_2 S_n(t) + 2c_1 n^{\nu /2}2^{2n/\gamma }, \end{aligned}$$

(8)

where $S_n(t) = \sum _{a=1}^{\lfloor n/2\rfloor } \left( {\begin{array}{c}n\\ a\end{array}}\right) \frac{n^{\nu /2}}{a^{\nu /2+1}} m_{2a}(t)m_{2(n-a)+\gamma }(t)$.

Step 1. We introduce, for $\sigma \in (0,1]$ to be chosen later, for $p\ge 2$ and $t\ge 0$,

$$\begin{aligned} E_p(t)=\sum _{n=0}^p \frac{(\sigma t)^{2n/\gamma }m_{2n}(t)}{(n!)^\alpha }. \end{aligned}$$

By Lemma 10-(i) and since $m_0(t)=1$, it holds that for some constant $C_p\in (0,\infty )$,

$$\begin{aligned} 1\le & {} E_p(t) \le 1 + C_p \sum _{n=1}^p t^{2n/\gamma }m_{2n}(t)\le 1+C_p\sum _{n=1}^p t^{2n/\gamma }[1+t^{-(2n-2)/\gamma }]\\= & {} 1+C_p\sum _{n=1}^p [t^{2n/\gamma } +t^{2/\gamma }], \end{aligned}$$

whence $\lim _{t\rightarrow 0} E_p(t)=1$.

Step 2. Since $m_0'(t)=m_{2}'(t)=0$, we deduce from (8) that for all $p\ge 2$, all $t\in [0,1]$,

$$\begin{aligned} E_p'(t) \le -c_1 F_p(t) + c_2 G_p(t)+\frac{2\sigma }{\gamma }H_p(t) + C, \end{aligned}$$

where

$$\begin{aligned} F_p(t)= & {} \!\sum _{n=2}^p n^{\nu /2} \frac{(\sigma t)^{2n/\gamma }m_{2n+\gamma }(t) }{(n!)^\alpha },\quad G_p(t)=\!\sum _{n=2}^p \frac{(\sigma t)^{2n/\gamma }S_n(t)}{(n!)^\alpha },\quad \\ H_p(t)= & {} \!\sum _{n=1}^p \frac{n (\sigma t)^{2n/\gamma -1} m_{2n}(t)}{(n!)^\alpha } \end{aligned}$$

and where, since $\sigma t\le 1$,

$$\begin{aligned} C= 2 c_1 \sum _{n=2}^\infty \frac{n^{\nu /2}2^{2n/\gamma }}{(n!)^\alpha }<\infty . \end{aligned}$$

Step 3. We first prove that for all $\epsilon \in (0,\infty )$, there is $A_\epsilon \in (0,\infty )$, depending only on $\gamma ,\nu ,\kappa _1,\kappa _2,\epsilon$, such that for any choice of $\sigma \in (0,1]$, for all $p\ge 2$, all $t\in [0,1]$,

$$\begin{aligned} G_p(t) \le \epsilon E_p(t)F_p(t) + \sigma ^{2/\gamma } A_\epsilon (F_p(t)+1). \end{aligned}$$

We start from

$$\begin{aligned} G_p(t)=&\sum _{n=2}^p \sum _{a=1}^{\lfloor n/2\rfloor } \frac{(\sigma t)^{2n/\gamma }}{(n!)^\alpha }\frac{n^{\nu /2}}{a^{\nu /2+1}} \left( {\begin{array}{c}n\\ a\end{array}}\right) m_{2a}(t)m_{2(n-a)+\gamma }(t). \end{aligned}$$

Since $\alpha \ge 1$ and since $a \le \lfloor n/2\rfloor$ implies $n \le 2(n-a)$ and thus $n^{\nu /2} \le 2^{\nu /2}(n-a)^{\nu /2}$,

$$\begin{aligned} G_p(t) \le&2^{\nu /2} \sum _{n=2}^p \sum _{a=1}^{\lfloor n/2\rfloor } \frac{(\sigma t)^{2n/\gamma }}{(n!)^\alpha }\frac{(n-a)^{\nu /2}}{a^{\nu /2+1}} \Bigg [\left( {\begin{array}{c}n\\ a\end{array}}\right) \Bigg ]^\alpha m_{2a}(t)m_{2(n-a)+\gamma }(t)\\ =&2^{\nu /2} \sum _{n=2}^p \sum _{a=1}^{\lfloor n/2\rfloor } (n-a)^{\nu /2} \frac{(\sigma t)^{2(n-a)/\gamma }m_{2(n-a)+\gamma }(t)}{((n-a)!)^\alpha } \frac{(\sigma t)^{2a/\gamma }m_{2a}(t)}{(a!)^\alpha a^{\nu /2+1}}\\ =&2^{\nu /2} \sum _{a=1}^{\lfloor p/2\rfloor } \sum _{n=2a}^{p} (n-a)^{\nu /2} \frac{(\sigma t)^{2(n-a)/\gamma }m_{2(n-a)+\gamma }(t)}{((n-a)!)^\alpha } \frac{(\sigma t)^{2a/\gamma }m_{2a}(t)}{(a!)^\alpha a^{\nu /2+1}}. \end{aligned}$$

Using the change of indices $n \rightarrow \ell =n-a$

$$\begin{aligned} G_p(t) \le&2^{\nu /2} \sum _{a=1}^{\lfloor p/2\rfloor } \sum _{\ell =a}^{p-a} \ell ^{\nu /2} \frac{(\sigma t)^{2\ell /\gamma }m_{2\ell +\gamma }(t)}{(\ell !)^\alpha } \frac{(\sigma t)^{2a/\gamma }m_{2a}(t)}{(a!)^\alpha a^{\nu /2+1}}\\ \le&\Bigg (\sum _{\ell =1}^{p} \ell ^{\nu /2} \frac{(\sigma t)^{2\ell /\gamma }m_{2\ell +\gamma }(t)}{(\ell !)^\alpha }\Bigg ) \times 2^{\nu /2} \Bigg (\sum _{a=1}^p \frac{(\sigma t)^{2a/\gamma }m_{2a}(t)}{(a!)^\alpha a^{\nu /2+1}} \Bigg )\\ =&\Big (F_p(t)+ (\sigma t)^{2/\gamma }m_{2+\gamma }(t) \Big ) I_p(t), \end{aligned}$$

where

$$\begin{aligned} I_p(t)= 2^{\nu /2} \sum _{a=1}^p \frac{(\sigma t)^{2a/\gamma }m_{2a}(t)}{(a!)^\alpha a^{\nu /2+1}} . \end{aligned}$$

But setting $N_\epsilon = \lceil 2^{\nu /(\nu +2)} \epsilon ^{-2/(\nu +2)}\rceil$, it holds that $a\ge N_\epsilon$ implies $2^{\nu /2}a^{-\nu /2-1} \le \epsilon$, whence

$$\begin{aligned} I_p(t)\le \epsilon E_p(t)+ J_{\epsilon }(t), \quad \hbox {where}\quad J_{\epsilon }(t)=2^{\nu /2} \sum _{a=1}^{N_\epsilon } \frac{(\sigma t)^{2a/\gamma }m_{2a}(t)}{(a!)^\alpha }. \end{aligned}$$

By Lemma 10-(i), we see that for some constants $A,A_\epsilon \in (0,\infty )$, depending only on $\gamma ,\nu ,\kappa _1,\kappa _2,\epsilon$ and allowed to vary, for any choice of $\sigma \in (0,1]$, for all $t\in [0,1]$,

$$\begin{aligned} (\sigma t)^{2/\gamma }m_{2+\gamma }(t)\le & {} A (\sigma t)^{2/\gamma }(1+t^{-1})\le A \sigma ^{2/\gamma } \;\; \hbox {and}\;\; J_\epsilon (t) \\\le & {} A_\epsilon \sum _{a=1}^{N_\epsilon } (\sigma t)^{2a/\gamma } [1+t^{-(2a-2)/\gamma }] \le A_\epsilon \sigma ^{2/\gamma }. \end{aligned}$$

All in all, we have proved that for any choice of $\sigma \in (0,1]$, for all $t\in [0,1]$,

$$\begin{aligned} G_p(t)\le & {} \Bigg (F_p(t)+A\sigma ^{2/\gamma }\Bigg )\Bigg (\epsilon E_p(t) + A_\epsilon \sigma ^{2/\gamma } \Bigg ) \\\le & {} \epsilon E_p(t)F_p(t)+ ~\sigma ^{2/\gamma }[\epsilon A E_p(t)+ A_\epsilon F_p(t)+AA_\epsilon ]. \end{aligned}$$

Since $E_p(t)\le m_0(t)+ \sigma t m_{2}(t)+F_p(t)\le 2+F_p(t)$, we conclude that, as desired,

$$\begin{aligned} G_p(t) \le \epsilon E_p(t)F_p(t)+ \sigma ^{2/\gamma }[(\epsilon A+A_\epsilon )F_p(t)+AA_\epsilon +2\epsilon A]. \end{aligned}$$

Step 4. We now verify that there are some constants $\kappa , B\in (0,\infty )$, depending only on $\gamma ,\nu ,\kappa _1,\kappa _2$, such that for any choice of $\sigma \in (0,1]$, for all $p\ge 2$, all $t\in [0,1]$,

$$\begin{aligned} H_p(t) \le \kappa F_p(t)+B. \end{aligned}$$

We first observe that, for $\kappa >0$ to be chosen later,

$$\begin{aligned} \frac{n m_{2n}(t)}{\sigma t} \le \kappa n^{\nu /2} [m_{2n}(t)]^{1+\gamma /(2n-2)} + \frac{n}{\sigma t}\Bigg ( \frac{n^{1-{\nu /2}}}{\kappa \sigma t}\Bigg )^{(2n-2)/\gamma }. \end{aligned}$$

Indeed, $nm_{2n}(t)/(\sigma t)$ is bounded by the second term if $m_{2n}(t) \le [n^{1-{\nu /2}}/(\kappa \sigma t)]^{(2n-2)/\gamma }$ and by the first term else. Since $[m_{2n}(t)]^{1+\gamma /(2n-2)}\le m_{2n+\gamma }(t)$ by Remark 9, we conclude that

$$\begin{aligned}{} H_p(t) =&\sum _{n=1}^p \frac{n m_{2n}(t)}{\sigma t} \frac{(\sigma t)^{2n/\gamma }}{(n!)^\alpha }\\ \le&\kappa \sum _{n=1}^p n^{\nu /2} \frac{(\sigma t)^{2n/\gamma }m_{2n+\gamma }(t) }{(n!)^\alpha } + \sum _{n=1}^p \frac{n}{\sigma t} \Bigg ( \frac{n^{1-{\nu /2}}}{\kappa \sigma t}\Bigg )^{(2n-2)/\gamma } \frac{(\sigma t)^{2n/\gamma }}{(n!)^\alpha }\\ =&\,\kappa F_p(t)+ \kappa (\sigma t)^{2/\gamma }m_{2+\gamma }(t)+(\sigma t)^{2/\gamma -1}\sum _{n=1}^p \frac{n^{(2-\nu )n / \gamma - (2-\nu )/ \gamma +1} }{(n!)^\alpha \kappa ^{(2n-2)/\gamma }}. \end{aligned}$$

But by Lemma 10-(i), there is $A\in (0,\infty )$, depending only on $\gamma ,\nu ,\kappa _1,\kappa _2$, such that for all $t \in [0,1]$,

$$\begin{aligned} t^{2/\gamma }m_{2+\gamma }(t) \le A t^{2/\gamma }[1+t^{-1}] \le 2A. \end{aligned}$$

Moreover, since $\alpha =\max \{1,(2-\nu )/\gamma \}\ge (2-\nu )/\gamma$, the series

$$\begin{aligned} S:=\sum _{n=1}^\infty \frac{n^{(2-\nu )n / \gamma - (2-\nu )/ \gamma +1} }{(n!)^\alpha \kappa ^{(2n-2)/\gamma }} \le \kappa ^{2/\gamma } \sum _{n=1}^\infty \frac{n^{\alpha n +1} }{(n!)^{\alpha }\kappa ^{2n/\gamma }} \end{aligned}$$

is convergent if $\kappa =2e^{\alpha \gamma /2}$, by Stirling’s formula $n!\sim \sqrt{2\pi n}(n/e)^n$. Hence for any choice of $\sigma \in (0,1]$, for all $t\in [0,1]$,

$$\begin{aligned} H_p(t) \le \kappa F_p(t) + 2\kappa A + S. \end{aligned}$$

Step 5. By Steps 2–3–4, for any choice of $\epsilon \in (0,\infty )$ and $\sigma \in (0,1]$, for all $p\ge 2$, all $t\in (0,1]$,

$$\begin{aligned} E_p'(t) \le&-c_1 F_p(t)+ c_2 G_p(t)+\frac{2\sigma }{\gamma }H_p(t)+C\\ \le&-c_1F_p(t)+c_2 [\epsilon E_p(t)F_p(t) + \sigma ^{2/\gamma } A_\epsilon (F_p(t)+1)] + \frac{2\sigma }{\gamma }[\kappa F_p(t)+B] + C. \end{aligned}$$

Choosing $\epsilon =c_1/(4c_2)$ and $\sigma \in (0,1]$ small enough so that $c_2 \sigma ^{2/\gamma } A_\epsilon + 2\kappa \sigma /\gamma \le c_1/2$, we find

$$\begin{aligned} E_p'(t) \le -\frac{c_1}{2} F_p(t) + \frac{c_1}{4} E_p(t)F_p(t) + D, \end{aligned}$$

for some constant $D\in (0,\infty )$. By Step 1, we know that $E_p(0)=\lim _{t\rightarrow 0}E_p(t)=1$, whence

$$\begin{aligned} T_p = \sup \{ t>0 : E_p(t)\le 2\} >0. \end{aligned}$$

For $p\ge 2$ and $t \in [0,T_p\wedge 1)$, we have $E_p'(t) \le D$. Recalling that $E_p(0)=1$, we deduce that $T_p\ge \min \{1,1/D\}=:T.$ In other words, with our choice of $\sigma$, all $t\in [0,T]$, $\sup _{p\ge 2}E_p(t) \le 2$. $\square$

We can finally give the

Proof of Theorem 2-(i)

We assume $(H_1(\gamma ))$ and $(H_2(\nu ))$ for some $\gamma \in (0,1]$ and some $\nu \in (0,2)$. We fix $\rho =\min \{2\gamma /(2-\nu ),2\}$ and observe that $\alpha :=2/\rho = \max \{1,(2-\nu )/\gamma \}$. Thus by Lemma 12, there are $\sigma \in (0,1]$ and $T>0$, depending only on $\gamma ,\nu ,\kappa _1,\kappa _2$, such that

$$\begin{aligned} \sup _{[0,T]} \sum _{n=0}^\infty \frac{(\sigma t)^{2n/\gamma }m_{2n}(t) }{ (n!)^\alpha } \le 2. \end{aligned}$$

We deduce from Lemma 11-(i) (with $\sigma _0=(\sigma t)^{2/\gamma }$) that

$$\begin{aligned}&\sup _{[0,T]} \int _{{\mathbb {R}}^3}\exp [(\sigma t)^{\rho /\gamma } |v|^{\rho }/2] f_t(\mathrm{d}v)\\&= \sup _{[0,T]} \int _{{\mathbb {R}}^3}\exp [(\sigma t)^{2/(\gamma \alpha )} |v|^{2/\alpha }/2] f_t(\mathrm{d}v ) \le 2^{1+1/\alpha }\le 4. \end{aligned}$$

as desired. $\square$

6 Propagation of Exponential Moments

We proceed as in the previous section.

Lemma 13

Assume $(H_1(\gamma ))$ and $(H_2(\gamma ))$ for some $\gamma \in (0,1]$ and some $\nu \in (0,2)$. For any $\sigma _0 \in (0,\infty )$ and any $\alpha \ge 1$, there is $\sigma \in (0,\sigma _0]$, depending only on $\gamma ,\nu ,\kappa _1,\kappa _2,\sigma _0$, such that

$$\begin{aligned} \sup _{n\ge 0} \frac{\sigma _0^nm_{2n}(0)}{(n!)^\alpha }\le 1 \quad \Longrightarrow \quad \sup _{t\ge 0} \sum _{n=0}^\infty \frac{\sigma ^n m_{2n}(t)}{ (n!)^\alpha }\le 3. \end{aligned}$$

Proof

We fix $\alpha \ge 1$ and $\sigma _0\in (0,\infty )$ and assume that $\sup _{n\ge 0} (n!)^{-\alpha } \sigma _0^nm_{2n}(0)\le 1$.

Step 1. We introduce, for $\sigma \in (0,1]$ to be chosen later, for $p\ge 2$ and $t\ge 0$,

$$\begin{aligned} E_p(t)=\sum _{n=0}^p \frac{\sigma ^n m_{2n}(t)}{(n!)^\alpha }. \end{aligned}$$

If $\sigma \in (0,\sigma _0/2]$, $E_p(0) \le 2$ for all $p\ge 2$, because

$$\begin{aligned} E_p(0)\le 1+\sum _{n\ge 1} \frac{(\sigma _0/2)^n m_{2n}(0)}{(n!)^\alpha } \le 1+\sum _{n\ge 1} 2^{-n}=2. \end{aligned}$$

Step 2. By Lemma 7, since $m_0'(t)=m_2'(t)=0$ and since $\sigma \in (0,1]$,

$$\begin{aligned} E_p'(t) \le -c_1 F_p(t) + c_2 G_p(t)+ C, \end{aligned}$$

where

$$\begin{aligned} F_p(t)=\sum _{n=2}^p n^{\nu /2} \frac{\sigma ^n m_{2n+\gamma }(t) }{(n!)^\alpha },\qquad G_p(t)=\sum _{n=2}^p \frac{\sigma ^nS_n(t)}{(n!)^\alpha }, \qquad C= 2c_1 \sum _{n=2}^\infty \frac{n^{\nu /2}2^{2n/\gamma }}{(n!)^\alpha }. \end{aligned}$$

Step 3. We prove that for all $\epsilon \in (0,\infty )$, there is a constant $A_\epsilon \in (0,\infty )$, depending only on $\epsilon ,\gamma ,\nu ,\kappa _1,\kappa _2,\alpha ,\sigma _0$, such that for any choice of $\sigma \in (0,1]$, for all $p\ge 2$, all $t\ge 0$,

$$\begin{aligned} G_p(t) \le \epsilon E_p(t)F_p(t) + \sigma A_\epsilon (F_p(t)+1). \end{aligned}$$

Exactly as in the previous section, Step 3, we have

$$\begin{aligned} G_p(t) \le&\Big (F_p(t)+ \sigma m_{2+\gamma }(t) \Big ) \Big (\epsilon E_p(t)+ J_{\epsilon }(t) \Big ), \end{aligned}$$

where, setting $N_\epsilon = \lceil 2^{\nu /(\nu +2)} \epsilon ^{-2/(\nu +2)}\rceil$,

$$\begin{aligned} J_{\epsilon }(t)=2^{\nu /2} \sum _{a=1}^{N_\epsilon } \frac{\sigma ^{a}m_{2a}(t)}{(a!)^\alpha }, \end{aligned}$$

For all $n\ge 2$, $m_{2n}(0)\le \sigma _0^{-n}(n!)^\alpha$, so that by Lemma 10-(ii), there are some constants $A,A_\epsilon \in (0,\infty )$, depending only on $\epsilon ,\gamma ,\nu ,\kappa _1,\kappa _2,\alpha ,\sigma _0$ and allowed to vary, such that for all $t\ge 0$,

$$\begin{aligned} m_{2+\gamma }(t)\le [m_4(t)]^{(2+\gamma )/4}\le A \quad \hbox {and}\quad J_\epsilon (t) \le A_\epsilon \sum _{a=1}^{N_\epsilon } \sigma ^{a} \le A_\epsilon \sigma . \end{aligned}$$

All in all, we have proved that for any choice of $\sigma \in (0,1]$, for all $t\ge 0$,

$$\begin{aligned} G_p(t)&\le (F_p(t)+A \sigma )(\epsilon E_p(t) + A_\epsilon \sigma )\\&\le \epsilon E_p(t)F_p(t)+ \sigma [\epsilon A E_p(t)+ A_\epsilon F_p(t)+AA_\epsilon ]. \end{aligned}$$

The conclusion follows since $E_p(t)\le m_0(t)+ \sigma m_{2}(t)+F_p(t)\le 2+F_p(t)$.

Step 4. We now prove that for all $p\ge 2$, all $t\ge 0$,

$$\begin{aligned} F_p(t)\ge \sigma ^{-\gamma /2}(E_p(t)-e). \end{aligned}$$

We write

$$\begin{aligned} F_p(t)=&\sum _{n=2}^p n^{\nu /2} \frac{\sigma ^nm_{2n+\gamma }(t)}{(n!)^\alpha }\ge \frac{1}{\sigma ^{\gamma /2}} \sum _{n=2}^p \frac{\sigma ^{n+\gamma /2}m_{2n+\gamma }(t)}{(n!)^\alpha }\ge \frac{1}{\sigma ^{\gamma /2}} \sum _{n=2}^p \frac{\sigma ^{n}m_{2n}(t) -1}{(n!)^\alpha }, \end{aligned}$$

because

$$\begin{aligned} \sigma ^{n+\gamma /2}m_{2n+\gamma }(t)=\int _{{\mathbb {R}}^3}(\sigma |v|^2)^{n+\gamma /2}f_t(\mathrm{d}v) \ge \int _{{\mathbb {R}}^3}[(\sigma |v|^2)^{n}-1]f_t(\mathrm{d}v) = \sigma ^{n}m_{2n}(t) -1. \end{aligned}$$

Hence, since $m_0(t)=m_2(t)=1$, since $\sigma \in (0,1]$ and since $\alpha \ge 1$,

$$\begin{aligned} F_p(t)\ge & {} \frac{1}{\sigma ^{\gamma /2}}\Bigg [E_p(t)-m_0(t)-\sigma m_{2}(t) -\sum _{n=2}^p \frac{1}{(n!)^\alpha }\Bigg ]\\\ge & {} \frac{1}{\sigma ^{\gamma /2}}\Bigg [E_p(t)-\sum _{n=0}^p \frac{1}{(n!)^\alpha }\Bigg ]\ge \frac{1}{\sigma ^{\gamma /2}}[E_p(t)-e]. \end{aligned}$$

Step 5. By Steps 2–3–4, for any choice of $\epsilon \in (0,\infty )$ and $\sigma \in (0,1]$, for all $p\ge 2$, all $t\ge 0$,

$$\begin{aligned} E_p'(t) \le&-c_1 F_p(t) + c_2 G_p(t)+C \le -c_1 F_p(t) + c_2[\epsilon E_p(t)F_p(t)+ \sigma A_\epsilon (F_p(t)+1)]+C. \end{aligned}$$

Choosing $\epsilon =c_1/(16c_2)$ and then $\sigma _1 \in (0,1]$ small enough so that $c_2 \sigma _1 A_\epsilon \le c_1/2$, we conclude that for some constant $D\in (0,\infty )$, for any choice of $\sigma \in (0,\sigma _1]$, any $t\ge 0$,

$$\begin{aligned} E_p'(t) \le -\frac{c_1}{2} F_p(t) + \frac{c_1}{16} E_p(t)F_p(t) +D. \end{aligned}$$

We now recall from Step 1 that for $\sigma \in (0,\sigma _0/2]$, we have $E_p(0)\le 2$ and by continuity,

$$\begin{aligned} T_p= \sup \{t\ge 0 : E_p(t)\le 4\}>0. \end{aligned}$$

But for all $t \in [0,T_p)$, if $\sigma \in (0,(\sigma _0/2)\wedge \sigma _1]$,

$$\begin{aligned} E_p'(t) \le -\frac{c_1}{4}F_p(t) + D \le -\frac{c_1}{4\sigma ^{\gamma /2}}[E_p(t)-e]+D \end{aligned}$$

by Step 4. Since $E_p(0)\le 2$, this implies that if $\sigma \in (0,(\sigma _0/2)\wedge \sigma _1]$, for all $p\ge 2$ and all $t\in [0,T_p)$,

$$\begin{aligned} E_p(t) \le e+ \frac{4\sigma ^{\gamma /2}D}{c_1}. \end{aligned}$$

Choosing $\sigma \in (0,(\sigma _0/2)\wedge \sigma _1]$ small enough so that ${4\sigma ^{\gamma /2}D}/{c_1}\le 3-e$, we conclude that for all $p\ge 2$, $E_p(t)\le 3$ for all $t\in [0,T_p)$, whence $T_p=\infty$ by continuity. In other words, for all $p\ge 2$, all $t\ge 0$, $E_p(t)\le 3$, which was our goal. $\square$

We finally can give the

Proof of Theorem 2-(ii)

We assume $(H_1(\gamma ))$ and $(H_2(\nu ))$ for some $\gamma \in (0,1]$ and some $\nu \in (0,2)$. We fix $A>1$, $\rho \in (0,2]$ and $\sigma _0>0$ and assume that $\int _{{\mathbb {R}}^3}\exp (\sigma _0 |v|^\rho )f_0(\mathrm{d}v)\le A$.

We set $\alpha =2/\rho \ge 1$. By Lemma 11-(ii), for some $\sigma _1\in (0,\infty )$, depending only on $\rho ,\sigma _0,A$,

$$\begin{aligned} \sup _{n\ge 0} \frac{\sigma _1^nm_{2n}(0)}{(n!)^\alpha } \le 1. \end{aligned}$$

We thus may apply Lemma 13: there is $\sigma _2\in (0,\sigma _1]$, depending only on $\gamma ,\nu ,\kappa _1,\kappa _2,\sigma _1$, such that

$$\begin{aligned} \sup _{t\ge 0} \sum _{n=0}^\infty \frac{\sigma _2^nm_{2n}(t) }{ (n!)^\alpha } \le 3. \end{aligned}$$

We deduce from Lemma 11-(i) that, setting $\sigma _3=\sigma _2^{1/\alpha }/2$,

$$\begin{aligned} \sup _{t\ge 0} \int _{{\mathbb {R}}^3}\exp [\sigma _3 |v|^{\rho }] f_t(\mathrm{d}v) = \sup _{t\ge 0} \int _{{\mathbb {R}}^3}\exp [\sigma _2^{1/\alpha } |v|^{2/\alpha }/2] f_t(\mathrm{d}v) \le 2.3^{1/\alpha }\le 6 \end{aligned}$$

as desired. $\square$