Xuguang Lu
email: xglu@math.tsinghua.edu.cn

C L Ément Mouhot

Clément Mouhot
email: c.mouhot@dpmms.cam.ac.uk

ON MEASURE SOLUTIONS OF THE BOLTZMANN EQUATION, PART I: MOMENT PRODUCTION AND STABILITY ESTIMATES

Keywords: Mathematics Subject Classification (2000): 35Q Equations of mathematical physics and other areas of application [See also 35J05, 35J10, 35K05, 35L05], 76P05 Rarefied gas flows, Boltzmann equation [See also 82B40, 82C40, 82D05] Boltzmann equation, spatially homogeneous, hard potentials, hard spheres, long-range interactions, measure solution, moment estimate, moment production, exponential moment, stability estimate, Mehler transform

The spatially homogeneous Boltzmann equation with hard potentials is considered for measure valued initial data having finite mass and energy. We prove the existence of weak measure solutions, with and without angular cutoff on the collision kernel; the proof in particular makes use of an approximation argument based on the Mehler transform. Moment production estimates in the usual form and in the exponential form are obtained for these solutions. Finally for the Grad angular cutoff, we also establish uniqueness and strong stability estimate on these solutions.

In this paper we study the spatially homogeneous Boltzmann equation for hard interaction potentials with or without angular cutoff. The initial data are assumed to be positive Borel measures having finite moments up to order 2. Our main results are the existence and stability of measure solutions that have polynomial and exponential moment production properties.

1.1. The spatially homogeneous Boltzmann equation.

∂ ∂t f t (v) = Q(f t , f t )(v) , (v, t) ∈ R N × (0, ∞) , N ≥ 2
with some given initial data f t (v)| t=0 = f 0 (v) and Q is the collision integral defined by

(1.2) Q(f, f )(v) = R N ×S N-1 B(v -v * , σ) f (v ′ )f (v ′ * ) -f (v)f (v * ) dσdv * ,
where v, v * and v ′ , v ′ * stand for velocities of two particles respectively after and before their collision,

(1.3) v ′ = v + v * 2 + |v -v * | 2 σ , v ′ * = v + v * 2 - |v -v * | 2 σ , σ ∈ S N -1 .
The above relation between v, v * and v ′ , v ′ * shows that the collision is elastic:

v ′ + v ′ * = v + v * , |v ′ | 2 + |v ′ * | 2 = |v| 2 + |v * | 2 . 1.1.2.
The collision kernel. The collision kernel B(z, σ) under consideration is assumed to be a function of (|z|, z |z| • σ), i.e.

(1.4)

B(z, σ) = B(|z|, cos θ), cos θ = z |z| • σ, θ ∈ [0, π]
where (r, t) → B(r, t) is a non-negative Borel function on [0, ∞) × [-1, 1] satisfying (1.5) ∀ t ∈ (-1, 1), r → B(r, t) is continuous on [0, ∞),

(1.6) B(r, t) ≤ (1 + r 2 ) γ/2 b(t), 0 < γ ≤ 2.

In this paper most of the results are concerned with the case (1.7) B(z, σ) = |z| γ b(cos θ), 0 < γ ≤ 2 which corresponds to the so-called hard potential molecular interactions. The function t → b(t) in (1.6)-(1.7) has some weighted integrability. We shall consider several options for the assumptions on b(•). Our strongest assumption is that b(•) as a function of σ is integrable on the sphere S N -1 , which means π 0 b(cos θ) sin N -2 θ dθ < ∞ which is the Grad's angular cutoff. However more singular situations can be considered. The minimal assumption is that b(cos θ) sin 2 θ is integrable on the sphere as a function of σ (this corresponds physically to an angular momentum), i.e. In dimension N = 3, it is well known that for the hard spheres model the function b(•) is constant, whereas for hard potential models (without angular cutoff), there is only weighted integrability: More precisely, given an interaction potentialφ(r) = C r 1-s for C > 0 and s > 3, we obtain the following formula from the physics literature [START_REF] Cercignani | The Boltzmann equation and its applications[END_REF] in dimension N = 3:

B(z, σ) = |z| γ b(cos θ), γ = s -5 s -1 ; b(cos θ) sin θ ∼ C ′ θ -1-2 s-1 (θ → 0 + )
for some constant C ′ > 0, and hard potential interactions correspond to s > 5.

In this paper we consider the following different assumptions: Then for any g ∈ L 1 (S N -1 ) or g ≥ 0 (measurable) on S N -1 we have

(H0) 0 < γ ≤ 2 , A 2 := S N -2 π 0 b(cos θ) sin N θ dθ < ∞, (
S N-1 g(σ)dσ = π 0 sin N -2 θ S N-2 (n)
g(cos θn + sin θ ω)dω dθ where dω is the Lebesgue spherical measure on S N -2 (n) and in case N = 2 we define S 0 (n) g(ω)dω = g(-n ⊥ ) + g(n ⊥ ) .

Let |S N -2 (n)| = S N-2 (n) dω, etc. Then |S N -2 (n)| = |S N -2 | for N ≥ 3, |S 0 (n)| = |S 0 | = 2 for N = 2.
By classical calculation one has Observe that when assuming one of the assumptions (H0), (H1), (H2) (non-cutoff cases), the collision operator in the dual form (1.8) above is well-defined thanks to the cancellations in the symmetric difference ∆ϕ of ϕ ∈ C 2 (R N ). Basic estimates on ∆ϕ are as follows (see for instance [START_REF] Carlen | On strong convergence to equilibrium for the Boltzmann equation with soft potentials[END_REF]Lemma 3.2]): For all (v, v * , σ) ∈ R N × R N × S N -1 one has (1.11)

(1.8) Q(f, g), ϕ := R N Q(f, g)(v)ϕ(v)dv = 1 2 R N ×R N L B [∆ϕ](v, v * )f (v)g(v * )dvdv * where ∆ϕ := ∆ϕ(v, v * , v ′ , v ′ * ) = ϕ(v ′ ) + ϕ(v ′ * ) -ϕ(v) -ϕ(v * ) , (1.9
S N-2 (n) ∆ϕ dω ≤ |S N -2 | max |ξ|≤ √ |v| 2 +|v * | 2 |H ϕ (ξ)| |v -v * | 2 sin 2 θ ,
where ∇ϕ, H ϕ are gradient and Hessian matrix of ϕ. Consequently the Boltzmann equation (1.1) in a weak form can be written (1.12)

R N ϕ(v)f t (v)dv = R N ϕ(v)f 0 (v)dv + t 0 Q(f τ , f τ ), ϕ dτ .
From the estimate (1.11) it is easily seen that if A 2 < ∞ (minimal assumption) then L B [∆ϕ] is well-defined for all ϕ ∈ C 2 (R N ).

In fact we shall prove in Proposition 2.1 (see Section 2) that (v, v * ) → L B [∆ϕ](v, v * ) is also continuous on R N × R N . Furthermore if π 0 b(cos θ) sin N -1 θ dθ < ∞ then from the estimate (1.10) one sees that

L B [|∆ϕ|](v, v * ) = S N-1 B(v -v * , σ)|∆ϕ| dσ < ∞ so that L B coincides with the simpler formula (1.13) L B [∆ϕ](v, v * ) = S N-1 B(v -v * , σ)∆ϕ dσ .
The collision integral (1.8) and the equation (1.12) for L 1 -functions are naturally extended to finite Borel measures. For every 0 ≤ s < ∞, let B s (R N ) = (B s (R N ), • s ) be the Banach space of real Borel measures on R N having finite total variations up to order s, i.e.

µ s := R N v s d|µ|(v) < ∞, v := (1 + |v| 2 ) 1/2
where the positive Borel measure |µ| is the total variation of µ. In particular µ = µ 0 = |µ|(R N ) is simply the total variation of µ. Let

B + s (R N ) = µ ∈ B s (R N ) | µ ≥ 0 .
In accordance with (1.8) we now define for every µ, ν ∈ B s (R N ) and every suitable smooth function ϕ

(1.14) Q(µ, ν), ϕ := 1 2 R N ×R N L B [∆ϕ](v, v * )dµ(v)dν(v * ) .
Our test function space for defining measure weak solutions is chosen C 2 b (R N ), where

C k b (R N ) =    ϕ ∈ C k (R N ) |α|≤k sup v∈R N |∂ α ϕ(v)| < ∞    .
Finally by analogy with B s (R N ) we introduce the class L ∞ -s (R N ) of locally bounded Borel functions such that

ψ ∈ L ∞ -s (R N ) ⇐⇒ ψ L ∞ -s := sup v∈R N |ψ(v)| v -s < ∞
and we define

L ∞ -s ∩ C k (R N ) =    ϕ ∈ C k (R N ) |α|≤k ∂ α ϕ L ∞ -s < ∞    , s ≥ 0, k ∈ N .
1.2. Previous results and references. Let us give a short (and non-exhaustive) overview of the main previous results and references related to the subject of this paper.

1.2.1. Cauchy theory for the spatially homogeneous Boltzmann equation for hard potentials with cutoff. The first rigorous mathematical result is due to Carleman [START_REF] Carleman | Sur la théorie de l'équation intégrodifférentielle de Boltzmann[END_REF][START_REF] Carleman | Problèmes mathématiques dans la théorie cinétique des gaz[END_REF] who proved existence and uniqueness of solutions in L 1 ∩ L ∞ with pointwise moment bounds, for hard spheres interactions. A general Cauchy theory was later developed by Arkeryd [START_REF] Arkeryd | On the Boltzmann equation. I. Existence[END_REF][START_REF] Arkeryd | On the Boltzmann equation. II. The full initial value problem[END_REF] who proved existence and uniqueness of solutions in L 1 ∩ L log L with L 1 moment bounds.

More recently optimal results were obtained by Mischler and Wennberg [START_REF] Mischler | On the spatially homogeneous Boltzmann equation[END_REF] (see also Lu [START_REF] Lu | Conservation of energy, entropy identity, and local stability for the spatially homogeneous Boltzmann equation[END_REF]), and we refer to the references therein for a more extensive bibliography.

1.2.2. Cauchy theory for the spatially homogeneous Boltzmann equation for hard potentials without cutoff. This theory is much more recent, and not complete at now. As far as existence of solutions is concerned let us mention the seminal works of Villani [START_REF] Villani | On a new class of weak solutions to the spatially homogeneous Boltzmann and Landau equations[END_REF] and then Alexandre and Villani [START_REF] Alexandre | On the Boltzmann equation for long-range interactions[END_REF]. As far as uniqueness of solutions is concerned (in the general far from equilibrium regime), let us mention the works [START_REF] Toscani | Probability metrics and uniqueness of the solution to the Boltzmann equation for a Maxwell gas[END_REF][START_REF] Fournier | Uniqueness for a class of spatially homogeneous Boltzmann equations without angular cutoff[END_REF][START_REF] Fournier | On the well-posedness of the spatially homogeneous Boltzmann equation with a moderate angular singularity[END_REF][START_REF] Fournier | On the uniqueness for the spatially homogeneous Boltzmann equation with a strong angular singularity[END_REF] based on Wasserstein metrics and probabilistic tools, and the work [START_REF] Desvillettes | Stability and uniqueness for the spatially homogeneous Boltzmann equation with long-range interactions[END_REF] based on a priori estimates. Finally let us mention the related recent works in the perturbative close-to-equilibrium regime (but without assuming spatial homogeneity) of Gressman and Strain [START_REF] Gressman | Global classical solutions of the Boltzmann equation with long-range interactions[END_REF] on the one hand, and Alexandre, Morimoto, Ukai, Xu, Yang [START_REF] Alexandre | Global well-posedness theory for the spatially inhomogeneous Boltzmann equation without angular cutoff[END_REF] on other hand.

1.2.3. Polynomial moment bounds. The first seminal result of the propagation of polynomial moments that exists initially for "variable hard spheres" (hard potentials with angular cutoff) is due to Elmroth [START_REF] Elmroth | Global boundedness of moments of solutions of the Boltzmann equation for forces of infinite range[END_REF] and makes use of so-called "Povzner's inequalities" [START_REF] Povzner | On the Boltzmann equation in the kinetic theory of gases[END_REF]. Then Desvillettes [START_REF] Desvillettes | Some applications of the method of moments for the homogeneous Boltzmann and Kac equations[END_REF] proved, for the same model, the appearance of any polynomial as soon as a moment of order strictly higher than 2 exists initially (see also [START_REF] Wennberg | On moments and uniqueness for solutions to the space homogeneous Boltzmann equation[END_REF]). Finally optimal results were obtained in [START_REF] Mischler | On the spatially homogeneous Boltzmann equation[END_REF] again.

1.2.4. Exponential moment bounds. The first seminal result of propagation of moments of exponential form is due to Bobylev [START_REF] Bobylev | Moment inequalities for the Boltzmann equation and applications to spatially homogeneous problems[END_REF], still in the case of short-ranged interactions. Significant improvements of these results were later obtained in [START_REF] Bobylev | Moment inequalities and high-energy tails for Boltzmann equations with inelastic interactions[END_REF]. Let us also mention the related result of propagation of pointwise Maxwellian bound in [START_REF] Gamba | Upper Maxwellian bounds for the spatially homogeneous Boltzmann equation[END_REF]. Inspired by the same techniques, the appearance of exponential moments was first obtained by the second author together with Mischler in [START_REF] Mischler | Cooling process for inelastic Boltzmann equations for hard spheres. II. Self-similar solutions and tail behavior[END_REF][START_REF] Mouhot | Rate of convergence to equilibrium for the spatially homogeneous Boltzmann equation with hard potentials[END_REF], see also the recent work [START_REF] Alonso | A simplified approach to the creation and propagation of exponential moments in the Boltzmann equation[END_REF].

Let F 0 ∈ B + 2 (R N ) and {F t } t≥0 ⊂ B + 2 (R N ).
We say that {F t } t≥0 , or simply F t , is a measure weak solution of Eq. (1.1) associated with the initial datum F 0 , if it satisfies the following (i)-(ii):

(i) sup

t≥0 F t 2 < ∞ . (ii) For every ϕ ∈ C 2 b (R N ),                R N ×R N |L B [∆ϕ](v, v * )|dF t (v)dF t (v * ) < ∞ , ∀ t > 0 t → Q(F t , F t ), ϕ belongs to C((0, ∞)) ∩ L 1 loc ([0, ∞)) R N ϕ(v)dF t (v) = R N ϕ(v)dF 0 (v) + t 0 Q(F τ , F τ ), ϕ dτ ∀ t ≥ 0 .
Moreover a measure weak solution F t is called a conservative solution if it conserves the mass, momentum and energy, i.e.

R N   1 v |v| 2   dF t (v) = R N   1 v |v| 2   dF 0 (v) ∀ t ≥ 0 .
Note that every measure weak solution conserves the mass because the constant ϕ = 1 belongs to C 2 b (R N ) and ∆ϕ = 0. The conservations of the momentum and energy are formally true since one also has ∆ϕ = 0 for ϕ = v j , j = 1, 2, . . . , N and ϕ = |v| 2 , but these ϕ do not belong to C 2 b (R N ). In fact under the assumption (H1), one can follow the same argument in [START_REF] Lu | Solutions with increasing energy for the spatially homogeneous Boltzmann equation[END_REF] to construct a weak solution of Eq. (1.1) such that the energy is increasing. Now let us consider a stronger notion of measure strong solutions under the angular cutoff assumption (H4). Let B(z, σ) be given by (1.4)-(1.5)-(1.6) with b(•) satisfying A 0 < ∞. Then we can define bilinear operators (see Proposition 2.3 below)

Q ± : B s+γ (R N ) × B s+γ (R N ) → B s (R N ) (s ≥ 0) and (1.15) Q(µ, ν) := Q + (µ, ν) -Q -(µ, ν) through Riesz's representation theorem by (1.16) R N ψ(v)dQ + (µ, ν)(v) = R N ×R N L B [ψ](v, v * )dµ(v)dν(v * ) , (1.17)
R N ψ(v)dQ -(µ, ν)(v) = R N ×R N A(v -v * )ψ(v)dµ(v)dν(v * ) for all ψ ∈ L ∞ -s ∩ C(R N ), where (1.18) L B [ψ](v, v * ) = S N-1 B(v -v * , σ)ψ(v ′ ) dσ, A(z) = S N-1 B(z, σ)dσ and recall that n = (v -v * )/|v -v * | in b(n • σ) is replaced by a fixed unit vector e 1 for v = v * .
Recall that the norm µ s of µ ∈ B s (R N ) (s ≥ 0) can be estimated in terms of compactly smooth test functions: For all k ≥ 0 (1. [START_REF] Gressman | Global classical solutions of the Boltzmann equation with long-range interactions[END_REF])

µ s = sup ϕ∈C k c (R N ), ϕ L ∞ -s ≤1 R N ϕdµ .
We are now ready for stating the definition of measure strong solutions, for which some time-differentiability is assumed in total variation topology.

Let F 0 ∈ B + 2 (R N ) and {F t } t≥0 ⊂ B + 2 (R N ).
We say that F t is a measure strong solution of Eq.(1.1) associated with the initial datum

F t | t=0 = F 0 , if it satisfies the following (i)-(ii): (i) sup t≥0 F t 2 < ∞ . (ii) t → F t ∈ C([0, ∞); B 2 (R N )) ∩ C 1 ([0, ∞); B 0 (R N )) and (1.20) d dt F t = Q(F t , F t ) , t ∈ [0, ∞) . Note that from (2.18)-(2.19)-(2.20) in Proposition 2.3 the strong continuity of t → F t ∈ C([0, ∞); B 2 (R N )) implies the strong continuity t → Q(F t , F t ) ∈ C([0, ∞); B 0 (R N )),
(E) = d dt µ t (E) , b a ν t dt (E) = b a ν t (E)dt for all Borel sets E ⊂ R N .
Note also that if a strong measure solution F t is absolutely continuous with respect to the Lebesgue measure for all t ≥ 0, i.e. dF t (v) = f t (v)dv, then it is easily seen that f t (after modification on a v-null set) is a mild solution of Eq. ( 1

.1). That is, (t, v) → f t (v) is nonnegative and Lebesgue measurable on [0, ∞) × R N and for every t ≥ 0, v → f t (v) belongs to L 1 2 (R N ), sup t≥0 f t L 1 2 < ∞, and there is a Lebesgue null set Z 0 ⊂ R N (which is independent of t) such that          t 0 Q ± (f τ , f τ )(v)dτ < ∞ ∀ t ∈ [0, ∞) , ∀ v ∈ R N \ Z 0 f t (v) = f 0 (v) + t 0 Q(f τ , f τ )(v)dτ , ∀ t ∈ [0, ∞) , ∀ v ∈ R N \ Z 0 .
Here

L 1 s (R N ) = f ∈ L 1 (R N ) | f L 1 s := R N |f (v)| v s dv < ∞ , s ≥ 0 .
From classical measure theory [START_REF] Rudin | Real and complex analysis[END_REF]Theorem 6.13,page 149

]: if dµ(v) = f (v)dv for f ∈ L 1 s (R N ), then d|µ|(v) = |f (v)|dv and hence µ s = f L 1 s .
For any positive measure µ ∈ B + 2 (R N ) we finally introduce the following continuous function r → Ψ µ (r) on [0, ∞):

(1.22) Ψ µ (r) = r + r 1/3 + |v|>r -1/3 |v| 2 dµ(v) , r > 0 with Ψ µ (0) = 0
which quantifies the localization of the energy of µ.

F t 2 ≤ F 0 2 ∀ t > 0; sup t≥t 0 F t s < ∞ ∀ t 0 > 0, ∀ s > 2.
Then F t is conservative, i.e. F t conserves the mass, momentum, and energy. (c) The Eq. (1.1) always has a conservative measure weak solution F t with F t | t=0 = F 0 which satisfies the following moment production estimate:

(1.24)

F t s ≤ K s (F 0 ) 1 + 1 t s-2 γ ∀ t > 0 , ∀ s ≥ 2
where

(1.25) K s (F 0 ) = F 0 2 2 s+7 F 0 2 F 0 0 1 + 1 16 F 0 2 A 2 γ s-2 γ
.

(d) If in addition either 0 < γ ≤ 1 or one of the assumptions (H2), (H3) is satisfied, then every solution F t in part (c) (or generally in part (b)) satisfies a moment production estimate of exponential form:

(1.26) R N e α(t) v γ dF t (v) ≤ 2 F 0 0 ∀ t > 0 ,
where

α(t) = 2 -s 0 F 0 0 F 0 2 (1 -e -βt ) , β = 16 F 0 2 A 2 γ > 0 and 1 < s 0 < ∞ depends only on b(•) and γ.
It is possible to deduce from the previous theorem some more conventional moment estimates in exponential form where the constant in the argument of the exponential moment remains time-dependent: Corollary 1.4. Under the same assumptions on B(z, σ) and the initial datum F 0 in Theorem 1.3, there exists a conservative measure weak solution F t of Eq. (1.1) such that for any 0 < s < γ and any c > 0

R N e c v s dF t (v) ≤ e αs(t) + 2 F 0 0 ∀ t > 0 where α s (t) = c c α(t) s γ-s .
Proof of Corollary 1.4. The proof of this Corollary is quite short and we can present it here. As a consequence of Theorem 1.3 there exists a conservative measure weak solution F t of Eq. (1.1) such that F t satisfies (1.26). For any t > 0, by definition of α s (t) and 0 < s < γ we have

c v s > α s (t) =⇒ c v s = α s (t) c v s γ-s s α(t) v γ < α(t) v γ . Thus R N e c v s dF t (v) = {c v s ≤αs(t)} e c v s dF t (v) + {c v s >αs(t)} e c v s dF t (v) ≤ e αs(t) F 0 0 + {c v s >αs(t)} e α(t) v γ dF t (v) ≤ e αs(t) F 0 0 + 2 F 0 0 .
Our second main result of this paper is Theorem 1.5 (Uniqueness and stability estimates for locally integrable b(•)). Let B(z, σ) = |z| γ b(cos θ) satisfy (H4). Given any initial datum F 0 ∈ B + 2 (R N ) with F 0 0 = 0, we have (a) Every conservative measure weak solution of Eq. (1.1) is a strong solution, while every measure strong solution of Eq. (1.1) is a measure weak solution. (b) Let F t be a measure strong solution of Eq. (1.1) with the initial datum F 0 satisfying F t 2 ≤ F 0 2 for all t ≥ 0. Then F t in fact conserves the mass, momentum and energy. (c) There exists a unique conservative measure strong solution F t of Eq. (1.1) such that F t | t=0 = F 0 . Therefore F t satisfies the moment production estimates in Theorem 1.3 . (d) Let F t be the unique conservative measure strong solutions of Eq. (1.1) with the initial datum F 0 and let G t be a conservative measure strong solutions of Eq. (1.1)

on the time interval [τ, ∞) with an initial datum G t | t=τ = G τ ∈ B + 2 (R N )
for some τ ≥ 0. Then:

-If τ = 0, then

(1.27) F t -G t 2 ≤ Ψ F 0 ( F 0 -G 0 2 )e C(1+t) , t ≥ 0
where Ψ F 0 is given by (1.22), C = R(γ, A 0 , A 2 F 0 0 , F 0 2 ) is an explicit positive continuous function on (R >0 ) 5 .

-If τ > 0, then

(1.28) F t -G t 2 ≤ F τ -G τ 2 e cτ (t-τ ) , t ∈ [τ, ∞)
where

c τ = 4A 0 (K 2+γ (F 0 ) + F 0 2 )(1 + 1 τ ), K 2+γ (F 0 ) is given in (1.25) with s = 2 + γ.
(e) If F 0 is absolutely continuous with respect to the Lebesgue measure, i.e. dF 0 (v) = f 0 (v)dv with 0 ≤ f 0 ∈ L 1 2 (R N ), then the unique conservative measure strong solution F t with the initial datum F 0 is also absolutely continuous with respect to the Lebesgue measure: dF t (v) = f t (v)dv for all t ≥ 0, and f t is the unique conservative mild solution of Eq. (1.1) with the initial datum f 0 . (f) If F 0 is not a Dirac mass and let F t be the unique measure strong solution of Eq. (1.1) with the initial datum F 0 , then there is a sequence

{f n t } of conservative L 1 -solutions of Eq. (1.1) with initial data 0 ≤ f n 0 ∈ L 1 2 (R N ) satisfying (1.29) R N   1 v |v| 2   f n 0 (v)dv = R N   1 v |v| 2   dF 0 (v) , n = 1, 2, . . . such that (1.30) lim n→∞ R N ϕ(v)f n t (v)dv = R N ϕ(v)dF t (v) ∀ ϕ ∈ C b (R N ), ∀ t ≥ 0 .
Remark 1.6. The trivial case F 0 0 = 0, i.e. F 0 = 0, is excluded from the above theorems since F 0 = 0 implies that F t ≡ 0 is the unique conservative measure solution of Eq. (1.1).

Remark 1.7. An application of the estimate (1.28) for solutions with different initial times will be seen in our next paper concerning the rate of convergence to equilibrium.

Remark 1.8. In the second part of this work we shall prove the exponential convergence to equilibrium (for bounded angular function b(•)): F t -M 0 ≤ Ce -ct where M is the Maxwellian (Gaussian) with the same mass, momentum and energy as F 0 (assuming that F 0 is not a single Dirac mass and F 0 0 = 0), C, c > 0 are constants depending only on N, b(•), γ and the mass, momentum and energy of F 0 . This result will allow us to improve the stability estimate (1.27) to be uniform in time:

sup t≥0 F t -G t 2 ≤ Ψ F 0 ( F 0 -G 0 2 )
for some explicit continuous function Ψ F 0 (r) on [0, ∞) satisfying Ψ F 0 (0) = 0.

1.5. Strategy and plan of the paper. We shall first in Section 2 prove some continuity and Lipschitz estimates on the collision operator Q in (weighted) total variation topology.

In Section 3 we shall prove moment estimates, first on the kernel L B and then on the collision operator Q, plus several technical lemmas on fractional binomial expansions, on the beta function and on some ODE estimates. After these two sections which remain purely at the level of functional inequalities, we shall start considering the time evolution problem and tackle the proof of the first main Theorem 1.3 in Section 4: the main step in the construction of weak measure solutions is based on an approximation argument with the help of the Mehler transform, and the moment estimates on the solutions will be proved with the help of the functional results in the previous section. Finally in Section 5 we shall prove the second main Theorem 1.5 by carefully revisiting the uniqueness estimates known for functions in the case of measures.

Regularity estimates on the collision operator

We shall prove in this section some continuity and Lipschitz estimates on the collision operator in the (weighted) total variation topology. It will be useful for defining measure weak solutions of Eq. (1.1) as we mentioned in Section 1, but also for proving weak convergence of approximate solutions, which leads to the existence of measure weak solutions. We start with a preliminary useful representation of the collision velocities.

2.1. Representations of v ′ 2 , v ′ * 2 . We first begin this section with a preliminary technical computation.

For any v, v * ∈ R N , let us define

h = v + v * |v + v * | for v + v * = 0 ; h = e 1 = (
       v ′ 2 := 1 + |v ′ | 2 = v 2 + v * 2 2 + |v + v * ||v -v * | 2 (h • σ) v ′ * 2 := 1 + |v ′ * | 2 = v 2 + v * 2 2 - |v + v * ||v -v * | 2 (h • σ) .
Let us also define the unit vector

j = h -(h • n)n 1 -(h • n) 2 for |h • n| < 1 and j = e 1 for |h • n| = 1 .
Then with the change of variables σ = cos θn + sin θ ω , ω ∈ S N -2 (n), we have

h • σ = (h • n) cos θ + 1 -(h • n) 2 sin θ (j • ω) , ω ∈ S N -2 (n)
so that we get another representation:

(2.2)

   v ′ 2 = v 2 cos 2 θ/2 + v * 2 sin 2 θ/2 + |v| 2 |v * | 2 -(v • v * ) 2 sin θ (j • ω) v ′ * 2 = v 2 sin 2 θ/2 + v * 2 cos 2 θ/2 -|v| 2 |v * | 2 -(v • v * ) 2 sin θ (j • ω) .

Continuity estimate on the collision operator.

Proposition 2.1 (Continuity of the collision operator). Let B(z, σ) be given by

(1.4)- (1.5)-(1.6) with b(•) satisfying (H0). Then (I) The function (v, v * ) → L B [∆ϕ](v, v * ) is continuous on R N × R N for all ϕ ∈ C 2 (R N ). (II) Let B n (z, σ) = Bn (|z|, cos θ) satisfy (1.5) and (2.3) Bn (r, t) ր B(r, t) (n → ∞) ∀ (r, t) ∈ [0, ∞) × (-1, 1). Then for any ϕ ∈ C 2 (R N ) and any 0 < R < ∞ (2.4) sup |v|+|v * |≤R |L Bn [∆ϕ](v, v * ) -L B [∆ϕ](v, v * )| → 0 (n → ∞). Moreover let ϕ n ∈ C 2 (R N ) satisfy (2.5) lim n→∞ ϕ n (v) = ϕ(v) ∀ v ∈ R N ; sup n≥1 sup |v|≤R |α|≤2 |∂ α ϕ n (v)| < ∞ ∀ R < ∞ . Then (2.6) L Bn [∆ϕ n ](v, v * ) → L B [∆ϕ](v, v * ) (n → ∞) ∀ (v, v * ) ∈ R N × R N . Proof of Proposition 2.1. Let us write (2.7) L B [∆ϕ](v, v * ) = π 0 B(|v -v * |, cos θ) sin N θ L[∆ϕ](v, v * , θ)dθ
where

L[∆ϕ](v, v * , θ) = 1 sin 2 θ S N-2 (n) ∆ϕ dω, 0 < θ < π.
Recalling (1.11) we have

(2.8) sup 0<θ<π |L[∆ϕ](v, v * , θ)| ≤ |S N -2 | max |ξ|≤ √ |v| 2 +|v * | 2 |H ϕ (ξ)| |v -v * | 2 .
Part (I). For any 0 < R < ∞, consider decomposition

B(z, σ) = B(z, σ) ∧ R + (B(z, σ) -R) +
where x ∧ y = min{x, y}, (x -y) + = max{x -y, 0}. We have

L B [∆ϕ](v, v * ) = L B∧R [∆ϕ](v, v * ) + L (B-R) + [∆ϕ](v, v * ), L B∧R [∆ϕ](v, v * ) = S N-1 [B(v -v * , σ) ∧ R]∆ϕ dσ. Fix any (v 0 , v * 0 ) ∈ R N × R N . Applying (2.7)-(2.8) to L (B-R) + [∆ϕ]
and recalling the assumption (1.6) we have

sup |v-v 0 | 2 +|v * -v * 0 | 2 ≤1 |L (B-R) + [∆ϕ](v, v * )| ≤ C ϕ π 0 C γ b(cos θ) -R + sin N θ dθ =: I ϕ,γ (R)
where C ϕ , C γ are finite constants depending only on ϕ, γ, v 0 , v * 0 . Therefore

(2.9) |L B [∆ϕ](v, v * ) -L B [∆ϕ](v 0 , v * 0 )| ≤ |L B∧R [∆ϕ](v, v * ) -L B∧R [∆ϕ](v 0 , v * 0 )| + I ϕ,γ (R) ∀ |v -v 0 | 2 + |v * -v * 0 | 2 ≤ 1. Let (∆ϕ) 0 = ϕ(v 0 ′ ) + ϕ(v * ′ 0 ) -ϕ(v 0 ) -ϕ(v * 0 ). Applying (2.7) to L B∧R [∆ϕ]
and using the assumption (1.5) we have

|L B∧R [∆ϕ](v, v * ) -L B∧R [∆ϕ](v 0 , v * 0 )| ≤ C ϕ |S N -2 | π 0 B(|v -v * |, cos θ) ∧ R -B(|v 0 -v * 0 |, cos θ) ∧ R sin N -2 θ dθ +R S N-1 ∆ϕ -(∆ϕ) 0 dσ → 0 as (v, v * ) → (v 0 , v * 0 ).
Also by assumption π 0 b(cos θ) sin N θ dθ < ∞ we have I ϕ,γ (R) → 0 as R → +∞. Thus from (2.9), by first letting (v, v * ) → (v 0 , v * 0 ) and then letting R → +∞, we obtain lim sup 3) and dominated convergence and then using Dini's theorem we conclude that for any 0

(v,v * )→(v 0 ,v * 0 ) |L B [∆ϕ](v, v * ) -L B [∆ϕ](v 0 , v * 0 )| = 0 .
< R < ∞ π 0 B(r, cos θ) -Bn (r, cos θ) sin N θ dθ → 0 (n → 0) uniformly in r ∈ [0, R]. Therefore applying (2.7)-(2.8) to L B-Bn [∆ϕ] we have, for any 0 < R < ∞, sup |v|+|v * |≤R |L B [∆ϕ](v, v * ) -L Bn [∆ϕ](v, v * )| = sup |v|+|v * |≤R |L B-Bn [∆ϕ](v, v * )| ≤ C ϕ,R sup r∈[0,R] π 0 B(r, cos θ) -Bn (r, cos θ) sin N θ dθ → 0 (n → ∞)
where

C ϕ,R = sup |ξ|≤R |H ϕ (ξ)|R 2 .
Finally for any (v, v * ) ∈ R N × R N , using (2.5) and denoting r = |v -v * | we have by dominated convergence that

|L B [∆ϕ](v, v * ) -L Bn [∆ϕ n ](v, v * )| ≤ |L B [∆(ϕ -ϕ n )](v, v * )| + |L B-Bn [∆ϕ n ](v, v * )| ≤ π 0 B(r, cos θ) sin N θ |L[∆(ϕ -ϕ n )](v, v * , θ)| dθ + C π 0 B(r, cos θ) -Bn (r, cos θ) sin N θ dθ -→ 0 (n → ∞)
which concludes the proof.

2.3.

A continuity estimate for product measures. We shall now prove a continuity property for product measures which will prove useful for the construction of weak measure solutions.

Proposition 2.2 (A continuity property of product measures

). Let 0 ≤ s j < ∞, {µ n j } ∞ n=1 ⊂ B + s j (R N j ) , µ j ∈ B + 0 (R N j ) satisfy (2.10) sup n≥1 µ n j s j < ∞, j = 1, 2, . . . , k ; (2.11) lim n→∞ R N j ϕ j dµ n j = R N j ϕ j dµ j , ∀ ϕ j ∈ C ∞ c (R N j ), j = 1, 2, . . . , k .

Then

(2.12)

µ j ∈ B + s j (R N j ), µ j s j ≤ lim inf n→∞ µ n j s j , j = 1, 2, . . . , k . Moreover if Ψ n , Ψ ∈ C(R N 1 × R N 2 × • • • × R N k ) satisfy (2.13) lim |x|→∞ sup n≥1 |Ψ n (x)| k j=1 x j s j = 0 , lim n→∞ sup |x|≤R |Ψ n (x) -Ψ(x)| = 0 for all 0 < R < ∞, where x = (x 1 , x 2 , . . . , x k ) ∈ k j=1 R N j , then (2.14) lim n→∞ k j=1 R N j Ψ n d(µ n 1 ⊗ µ n 2 ⊗ • • • ⊗ µ n k ) = k j=1 R N j Ψd(µ 1 ⊗ µ 2 ⊗ • • • ⊗ µ k ) .
Proof of Proposition 2.2. First (2.12) easily follows from Fatou's Lemma. Let us prove (2.14). Let

M = sup n≥1 { µ n 1 s 1 , µ n 2 s 2 , . . . , µ n k s k } , ν n = µ n 1 ⊗ µ n 2 ⊗ • • • ⊗ µ n k , ν = µ 1 ⊗ µ 2 ⊗ • • • ⊗ µ k . By assumption on Ψ n , Ψ, for any ε > 0 there exist R ≥ 1, n ε ≥ 1 such that (2.15) |Ψ n (x)| , |Ψ(x)| < ε k j=1 x j s j , ∀ |x| > R, ∀ n ≥ n ε ; (2.16) |Ψ n (x) -Ψ(x)| < ε, ∀ |x| ≤ 2kR , ∀ n ≥ n ε .
On the other hand, by polynomial approximation, there exists a polynomial P (x) such that (2.17)

|Ψ(x) -P (x)| < ε ∀ |x| ≤ 2kR . Choose χ R j ∈ C ∞ c (R N j ) satisfying 0 ≤ χ R j (x j ) ≤ 1 on R N j and χ R j (x j ) = 1 for |x j | ≤ R and χ R j (x j ) = 0 for |x j | ≥ 2R. If we write P (x) = m i=1 k j=1 P i,j (x j
) where m ∈ N and P i,j (x j ) are polynomials in x j , then

P (x) k j=1 χ R j (x j ) = m i=1 k j=1 ϕ i,j (x j )
where ϕ i,j (x j ) = P i,j (x j )χ R j (x j ). Then consider the decomposition:

k j=1 R N j Ψ n dν n - k j=1 R N j Ψdν = k j=1 R N j Ψ n 1 - k j=1 χ R j dν n + k j=1 R N j (Ψ n -Ψ) k j=1 χ R j dν n + k j=1 R N j (Ψ -P ) k j=1 χ R j dν n + m i=1 k j=1 k j=1 R N j ϕ i,j dµ n j - m i=1 k j=1 k j=1 R N j ϕ i,j dµ j + k j=1 R N j (P -Ψ) k j=1 χ R j dν + k j=1 R N j Ψ k j=1 χ R j -1 dν := I n,1 + I n,2 + I n,3 + I n,4 + I 5 + I 6 . Since 1 -k j=1 χ R j (x j ) = 0 for all |x| ≤ R, and k j=1 χ R j (x j ) = 0 for all |x| > 2kR, it follows from (2.15)-(2.16)-(2.17) that for all n ≥ n ε |I n,1 | + |I 6 | ≤ 2ε |x|>R k j=1 x j s j dν n ≤ 2εkM k , |I n,2 | + |I n,3 | + |I 5 | ≤ 2ε |x|≤2kR dν n + ε |x|≤2kR dν ≤ 3εM k . For I n,4 , since ϕ i,j ∈ C ∞ c (R N j )
, it follows from the assumption of the lemma that

I n,4 = m i=1 k j=1 R N j ϕ i,j dµ n j - k j=1 R N j ϕ i,j dµ j → 0 (n → ∞). Therefore lim sup n→∞ k j=1 R N j Ψ n dν n - k j=1 R N j Ψdν ≤ 5kM k ε .
This proves (2.14) by letting ε → 0 + .

2.4.

Weighted Lipschitz regularity of the collision operator. Let us prove some (weighted) Lipschitz properties on the collision operator acting on Borel measures, in the (weighted) total variation topology. Then

Q ± : B s+γ (R N ) × B s+γ (R N ) → B s (R N ) (s ≥ 0)
are bounded and

(2.18) Q ± (µ, ν) s ≤ 2 (s+γ)/2 A 0 ( µ s+γ ν 0 + µ 0 ν s+γ ) , (2.19)
Q ± (µ, µ) -Q ± (ν, ν) s ≤ 2 (s+γ)/2 A 0 ( µ + ν s+γ µ -ν 0 + µ + ν 0 µ -ν s+γ )
and hence

(2.20) Q(µ, µ) -Q(ν, ν) 0 ≤ 2 1+(s+γ)/2 A 0 ( µ + ν γ µ -ν 0 + µ + ν 0 µ -ν γ ) .
Finally for all µ ∈ B γ (R N ) and all ϕ ∈ C 2 b (R N ), there holds

(2.21) Q(µ, µ), ϕ = R N ϕdQ(µ, µ)
where the left-hand side of (2.21) is defined in (1.14).

Proof of Proposition 2.3. By elementary inequalities

v ′ s ≤ ( v 2 + v * 2 ) s/2 , 1 + |v -v * | 2 γ/2 ≤ 2 γ/2 v 2 + v * 2 γ/2
and the assumption on B we have for any

ϕ ∈ C c (R N ) with ϕ L ∞ -s ≤ 1 |ϕ(v ′ )|B(v -v * , σ) ≤ v ′ s 1 + |v -v * | 2 γ/2 b(cos θ) ≤ 2 (s+γ)/2 ( v s+γ + v * s+γ )b(cos θ)
and hence

R N ×R N L B [|ϕ|] (v, v * )d(|µ| ⊗ |ν|) ≤ A 0 2 (s+γ)/2 ( µ s+γ ν 0 + µ 0 ν s+γ ) , R N ×R N A(v -v * )|ϕ(v)|d(|µ| ⊗ |ν|) ≤ A 0 2 (s+γ)/2 ( µ s+γ ν 0 + µ 0 ν s+γ ) .
These imply (2.18). The inequality (2.19) follows from (2.18) and the following identities:

Q ± (µ, µ) -Q ± (ν, ν) = 1 2 Q ± (µ + ν, µ -ν) + 1 2 Q ± (µ -ν, µ + ν) . Next recall B(v -v * , σ) = B(|v -v * |, v-v * |v-v * | • σ)
. By changing variables σ → -σ, v ↔ v * and using Fubini's theorem we have

R N ϕdQ + (µ, µ) = 1 2 R N ×R N S N-1 B(v -v * , σ) ϕ(v ′ ) + ϕ(v ′ * ) dσ dµ(v)dµ(v * ).
A similar symmetry for R N ϕdQ -(µ, µ) is obvious. The difference of the two is equal to Q(µ, µ), ϕ . This proves (2.21).

Moment estimates on the collision operator

In this section we shall prove several inequalities on the moments of the collision operator which will be useful for the moment estimates of the weak measure solutions we shall construct.

3.1. Analytical toolbox. Let us first collect and prove some useful analytical results. Lemma 3.1 (Fractional binomial expansion). Let p ≥ 1 and k p = [(p + 1)/2] the integer part of (p + 1)/2. Then for all x, y ≥ 0

kp-1 k=0 p k x k y p-k + x p-k y k ≤ (x + y) p ≤ kp k=0 p k x k y p-k + x p-k y k where p k = p(p -1) • • • (p -k + 1) k! , k ≥ 1 ; p 0 = 1 .
Proof of Lemma 3.1. We refer to [7, Lemma 2] for the proof.

Let p ≥ 1 and n ∈ {1, 2, . . . , [p]}. Then using Taylor's formula for the function x → (1 + x) p one has

n k=0 p k x k ≤ (1 + x) p ∀ x ≥ 0 .
In particular

(3.1) n k=0 p k ≤ 2 p , 1 ≤ n ≤ p .
Let Γ(x), B(x, y) be the gamma and beta functions:

Γ(x) = ∞ 0 t x-1 e -t dt , x > 0 ; B(x, y) = 1 0 t x-1 (1 -t) y-1 dt , x, y > 0 .
It is well known that

(3.2) Γ(x)Γ(y) = Γ(x + y)B(x, y) , ∀ x, y > 0 .
Other relations that we shall also use are: For any integer k ≥ 1 and for any real number p ≥ k we have

(3.3) p k = Γ(p + 1) Γ(p -k + 1)Γ(k + 1)
.

And (3.4) B(x + 1, y) + B(x, y + 1) = B(x, y) , x, y > 0 . Lemma 3.2 (A stationary phase result). Let 0 < α, R < ∞, g ∈ C([0, R]) and S ∈ C 1 ([0, R]) such that S(0) = 0, S ′ (x) < 0 ∀ x ∈ [0, R) . Then for any λ ≥ 1 we have R 0 x α-1 g(x)e λS(x) dx = Γ(α) 1 -λS ′ (0) α g(0) + o(1)
where o(1) → 0 as λ → ∞.

Proof of Lemma 3.2. This is classical stationary phase type of analysis, we omit the proof for the sake of conciseness of this paper. More generally for any a > 1 we have

(3.6) kp k=1 p k B(ak, a(p -k)) ≤ C a (ap) 1-a , (3.7
)

kp-1 k=0 p -2 k B(a(k + 1), a(p -k -1)) ≤ C a (ap) -a
where 0 < C a < ∞ only depends on a.

Proof of Lemma 3.3. Since p ≥ 3 we have

kp k=1 p k B(k, p -k) = kp k=1 p k(p -k) = kp k=1 1 k + 1 p -k ≤ 2 kp k=1 1 k ≤ 4 log p . Now suppose a > 1. Let kp k=1 p k B(ak, a(p -k)) = I a (p) + I a (p, k p )
where

I a (p) = kp-1 k=1 p k B(ak, a(p -k)), I a (p, k p ) = p k p B(ak p , a(p -k p )).
For the first term I a (p) we use the symmetry (w.r.t x = 1/2) and Lemma 3.1 to get

I a (p) = 1 2 1 0 1 x(1 -x) kp-1 k=1 p k x ak (1 -x) a(p-k) + x a(p-k) (1 -x) ak dx ≤ 1 2 1 0 1 x(1 -x) x a + (1 -x) a p -x ap -(1 -x) ap dx = 1/2 0 1 x(1 -x) x a + (1 -x) a p -x ap -(1 -x) ap dx .
Omitting the negative term -x ap we have

(x a + (1 -x) a ) p -x ap -(1 -x) ap ≤ p(x a + (1 -x) a ) p-1 x a so that I a (p) ≤ p 1/2 0 x a-1 g 1 (x)e pS(x) dx where g 1 (x) = (1 -x) -1 (x a + (1 -x) a ) -1 and S(x) = log(x a + (1 -x) a ), x ∈ [0, 1/2]. Since g 1 (0) = 1, S(0) = 0 and S ′ (0) = -a , S ′ (x) = a x a-1 -(1 -x) a-1 x a + (1 -x) a < 0 ∀ x ∈ [0, 1/2)
(because a > 1) it follows from Lemma 3.2 that for all p ≥ 3

I a (p) ≤ C a pΓ(a) 1 pa a = C a (ap) 1-a .
For the second term I a (p, k p ) we use Stirling's formula

Γ(x) = x e x 2π x e θx 12x , Γ(x + 1) = xΓ(x) = x e x √ 2πx e θx 12x , x ≥ 1 (0 < θ x < 1) to compute I a (p, k p ) = Γ(p + 1) Γ(k p + 1)Γ(p -k p + 1) • Γ(ak p )Γ(a(p -k p )) Γ(ap) (3.8) ≤ e 1/4 √ a ap k p p (a-1)kp p -k p p (a-1)(p-kp) p k p p p -k p ≤ C a 1 ap 1 2
(a-1)p

.

Here in the last inequality we used the simple estimates

p -1 2 ≤ p -k p ≤ p + 1 2 for p ≥ 3. This proves (3.6) because a > 1.
In order to prove (3.7) we consider again a decomposition

kp-1 k=0 p -2 k B(a(k + 1), a(p -k -1)) = J a (p) + J a (p, k p )
where for the first term J a (p) we use that

k p -2 = [(p -1)/2] -1 = k p-2 -1 and Lemma 3.1 to get J a (p) := kp-2 k=0 p -2 k B(a(k + 1), a(p -k -1)) = 1 2 1 0 x a-1 (1 -x) a-1 k p-2 -1 k=0 p -2 k x ak (1 -x) a(p-2-k) + x a(p-2-k) (1 -x) ak dx ≤ 1 2 1 0 x a-1 (1 -x) a-1 x a + (1 -x) a p-2 dx = 1/2 0 x a-1 g 2 (x)e pS(x) dx with g 2 (x) = (1 -x) a-1 (x a + (1 -x) a ) -2 . Since a > 1, it follows from Lemma 3.2 that J a (p) ≤ C a 1 ap a .
For the second term J a (p, k p ) we use (3.8) to get

J a (p, k p ) := p -2 k p -1 B(ak p , a(p -k p )) = (p -k p )k p p(p -1) I a (p, k p ) ≤ C a 1 ap 1 2
(a-1)p

.

Since a > 1, this proves the lemma.

3.2. An estimate of the angular cutoff reminder.

Lemma 3.4. Suppose b(•) satisfies the assumption (H0). For all p ≥ 3 we define

(3.9) ε p := 2 A 2 S N -2 π 0 1 0 t 1 - sin 2 θ 2 t p-2 dt b(cos θ) sin N θ dθ (≤ 1) . Then ε p → 0 (p → ∞). Furthermore, if either 0 < γ ≤ 1 or (H2) is satisfied, then (3.10) p 2-2/γ ε p → 0 (p → ∞) .
Proof of Lemma 3.4. Under the assumption (H0), the convergence ε p → 0 (p → ∞) is obvious and hence (3.10) holds for 0 < γ ≤ 1. Suppose (H2) is satisfied, which means that ν = 2 -2/γ ∈ (0, 1) and θ → b(cos θ) sin N -2ν θ is integrable on [0, π]. For all p ≥ 3 we have

(3.11) p ν ε p ≤ C π 0 1 0 (p -2) sin 2 θ 2 t ν 1 - sin 2 θ 2 t p-2 dt b(cos θ) sin N -2ν θ dθ
where C depends only on N, A 2 and ν. Applying elementary estimates

0 ≤ (λx) ν (1 -x) λ < 1 , (λx) ν (1 -x) λ → 0 (λ → ∞) ∀ x ∈ [0, 1]
to λ = p -2 and x = sin 2 θ 2 t we conclude from (3.11) and the dominated convergence theorem that p ν ε p → 0 (p → ∞).

Remark 3.5. It is easily calculated that if the assumption (H4) is satisfied, i.e. if A 0 < ∞, then ε p ≤ 16A 0 A 2 1 p for all p ≥ 3, so that in case 0 < γ < 2 we have p 2-2/γ ε p ≤ 16A 0 A 2 p 1-2/γ .

3.3.

Moment estimates on the kernel L B . In this subsection we shall prove moment estimates on the kernel L B as defined in (1.9).

Lemma 3.6. Let B(z, σ) = |z| γ b(cos θ).

(I) Under the assumption (H0) we have for all p ≥ 3

L B ∆ • 2p (v, v * ) (3.12) ≤ - A 2 4 v 2p+γ + v * 2p+γ + A 2 2 v 2p v * γ + v * 2p v γ +A 2 kp k=1 p k v 2k+γ v * 2(p-k) + v 2(p-k)+γ v * 2k +A 2 kp k=1 p k v 2k v * 2(p-k)+γ + v 2(p-k) v * 2k+γ +2p(p -1)A 2 ε p kp-1 k=0 p -2 k v 2(k+1)+γ v * 2(p-1-k) + v 2(p-1-k)+γ v * 2(k+1) +2p(p -1)A 2 ε p kp-1 k=0 p -2 k v 2(k+1) v * 2(p-1-k)+γ + v 2(p-1-k) v * 2(k+1)+γ .
(II) Under the assumption (H3) which is rewritten in the form

(3.13) γ = 2, 1 < p 1 < ∞ , A * p 1 := S N -2 π 0 [b(cos θ)] p 1 sin N -2 θ dθ 1/p 1 < ∞
and let

(3.14) q 1 = p 1 p 1 -1 , η = 1 2q 1 .
Then

L B ∆ • 2p (v, v * ) (3.15) ≤ 12A * p 1 p η kp k=1 p k v 2(k+1) v * 2(p-k) + v 2(p-k+1) v * 2k + 12A * p 1 p η kp k=1 p k v 2k v * 2(p-k+1) + v 2(p-k) v * 2(k+1) + A 0 2 v 2p v * 2 + A 0 2 v * 2p v 2 - A 0 4 v 2(p+1) - A 0 4 v * 2(p+1) for all p ≥ 12A * p 1 /A 0 2q 1 .
Proof of Lemma 3.6.

Part (I) Let us write

L B ∆ • 2p (v, v * ) = |v -v * | γ S N -2 π 0 b(cos θ) sin N θ L p (v, v * , θ) dθ with L p (v, v * , θ) := 1 sin 2 θ |S N -2 | S N-2 (k) v ′ 2p + v ′ * 2p -v 2p -v * 2p dω.
We first prove that

L p (v, v * , θ) ≤ - 1 2 v 2p + v * 2p (3.16) + 1 2 kp k=1 p k v 2k v * 2p-2k + v 2p-2k v * 2k +2p(p -1) 1 0 t 1 - sin 2 θ 2 t p-2 dt × kp-1 k=0 p -2 k v 2(k+1) v * 2(p-1-k) + v 2(p-1-k) v * 2(k+1) .
To do this we denote the shorthand

E(θ) = v 2 cos 2 θ/2 + v * 2 sin 2 θ/2 , h = |v| 2 |v * | 2 -v, v * 2 .
Then by (2.2)

v ′ 2 = E(θ) + h sin θ (j • ω) , v ′ * 2 = E(π -θ) -h sin θ (j • ω) .
By Taylor's formula we have

E(θ) ± h sin θ (j • ω) p = E(θ) p ± q E(θ) p-1 h sin θ (j • ω) +p(p -1) 1 0 (1 -t) E(θ) ± th sin θ j, ω p-2 dt(h sin θ j, ω ) 2 .
Look at the last term: We have for all θ ∈ (0, π), t ∈ [0, 1]

E(θ) + th sin θ |(j • ω)| ≤ E(θ) + E(π -θ) t = v 2 + v * 2 -E(π -θ) (1 -t) ≤ v 2 + v * 2 1 - 1 -t 2 sin 2 θ
where we used

E(π -θ) ≥ ( v 2 + v * 2 ) min{cos 2 θ/2 , sin 2 θ/2} ≥ ( v 2 + v * 2 ) sin 2 θ 2 . Since S N-2 (n) (j • ω)dω = 0 it follows that L p (v, v * , θ) ≤ 1 sin 2 θ (E(θ)) p + (E(π -θ)) p -v 2p -v * 2p (3.17) +2p(p -1) v 2 + v * 2 p-2 h 2 1 0 t 1 - sin 2 θ 2 t p-2
dt .

We need to prove that for p ≥ 3 and

k p = [(p + 1)/2] 1 sin 2 θ (E(θ)) p + (E(π -θ)) p -v 2p -v * 2p (3.18) ≤ - 1 2 v 2p + v * 2p + 1 2 kp k=1 p k v 2k v * 2p-2k + v 2p-2k v * 2k .
In fact using Lemma 3.1 we have

(E(θ)) p + (E(π -θ)) p ≤ kp k=0 p k v 2 cos 2 (θ/2) k v * 2 sin 2 (θ/2) p-k + v 2 cos 2 (θ/2) p-k v * 2 sin 2 (θ/2) k + kp k=0 p k v 2 sin 2 (θ/2) k v * 2 cos 2 (θ/2) p-k + v 2 sin 2 (θ/2) p-k v * 2 cos 2 (θ/2) k ≤ sin 2 θ 2 kp k=1 p k v 2k v * 2p-2k + v 2p-2k v * 2k + v 2p + v * 2p cos 2p (θ/2) + sin 2p (θ/2)
where we used the fact that p ≥ 3 =⇒ p -k p ≥ 1 so that

cos 2k (θ/2) sin 2p-2k (θ/2), sin 2k (θ/2) cos 2p-2k (θ/2) ≤ 1 4 sin 2 θ for all k ∈ {1, 2, ..., k p }. Since p ≥ 3 implies cos 2p (θ/2) + sin 2p (θ/2) ≤ cos 4 (θ/2) + sin 4 (θ/2) = 1 - 1 2 sin 2 (θ)
this gives (3.18). Note that h 2 ≤ v 2 v * 2 . Then using Lemma 3.1 again and recalling 2(k+1) .

k p -1 = k p-2 = [(p -1)/2] we have v 2 + v * 2 p-2 h 2 ≤ kp-1 k=0 p -2 k v 2(k+1) v * 2(p-1-k) + v 2(p-1-k) v *
This together with (3.17)-(3.18) concludes the proof of (3.16). Now using (3.16) and the definitions of L B [∆ϕ], A 2 and ε p we obtain

L B ∆ • 2p (v, v * ) ≤ - A 2 2 v 2p + v * 2p |v -v * | γ (3.19) + A 2 2 kp k=1 p k v 2k v * 2p-2k + v 2p-2k v * 2k |v -v * | γ +p(p -1)A 2 ε p kp-1 k=0 p -2 k v 2(k+1) v * 2(p-1-k) + v 2(p-1-k) v * 2(k+1) |v -v * | γ .
Next by 0 < γ ≤ 2 we have

(3.20) |v -v * | γ ≥ 1 2 v γ -v * γ , |v -v * | γ ≥ 1 2 v * γ -v γ .
Thus

v 2p + v * 2p |v -v * | γ = v 2p |v -v * | γ + v * 2p |v -v * | γ ≥ v 2p 1 2 v γ -v * γ + v * 2p 1 2 v * γ -v γ = 1 2 v 2p+γ + 1 2 v * 2p+γ -v 2p v * γ -v * 2p v γ . Since (3.21) |v -v * | γ ≤ 2( v γ + v * γ )
it follows that

v 2k v * 2(p-k) + v 2(p-k) v * 2k |v -v * | γ ≤ 2 v 2k v * 2(p-k) + v 2(p-k) v * 2k ( v γ + v * γ ) = 2 v 2k+γ v * 2(p-k) + v 2(p-k)+γ v * 2k +2 v 2k v * 2(p-k)+γ + v 2(p-k) v * 2k+γ .
And similarly

v 2(k+1) v * 2(p-1-k) + v 2(p-1-k) v * 2(k+1) |v -v * | γ ≤ 2 v 2(k+1) v * 2(p-1-k) + v 2(p-1-k) v * 2(k+1) ( v γ + v * γ ) = 2 v 2(k+1)+γ v * 2(p-1-k) + v 2(p-1-k)+γ v * 2(k+1) +2 v 2(k+1) v * 2(p-1-k)+γ + v 2(p-1-k)+γ v * 2(k+1)+γ .
These together with (3.19) yield the estimate (3.12).

Part (II)

For any p ≥ 1 we have

|v -v * | -2 L B [∆ • 2p ](v, v * ) = 2 S N-1 b(cos θ) v ′ 2p dσ -A 0 v 2p + v * 2p ≤ 2A * p 1 1 |S N -2 | S N-1 v ′ 2pq 1 dσ 1/q 1 -A 0 v 2p + v * 2p
where we used Hölder's inequality. We have to prove that

(3.22) 1 |S N -2 | S N-1 v ′ 2pq 1 dσ 1/q 1 ≤ 3 p η v 2 + v * 2 p
.

To do this we denote λ = pq 1 (> 1). Then using elementary inequalities

1 + x 2 λ + 1 -x 2 λ ≤ 1 + y 2 λ + 1 -y 2 λ , x, y ∈ [-1, 1], |x| ≤ |y| ; |v + v * ||v -v * | ≤ v 2 + v * 2 ,
and the formula (2.1) we compute (recall that N ≥ 2)

1 |S N -2 | S N-1 v ′ 2λ dσ = v 2 + v * 2 λ π 0 sin N -2 θ 1 2 + |v + v * ||v -v * | 2( v 2 + v * 2 ) cos θ λ dθ ≤ v 2 + v * 2 λ π 0 1 -cos θ 2 λ dθ ≤ v 2 + v * 2 λ 2π λ
where we used the well-known inequality

π/2 0 sin n θ dθ < π 2n
with n = 2[λ]. This yields (3.22). From this and using Lemma 3.1 we obtain that for all p ≥ 3

|v -v * | -2 L B ∆ • 2p (v, v * ) ≤ 6A * p 1 p η v 2 + v * 2 p -A 0 v 2p + v * 2p ≤ 6A * p 1 p η kp k=1 p k v 2k v * 2(p-k) + v 2(p-k) v * 2k -A 0 - 6A * p 1 p η v 2p + v * 2p . Since p ≥ 12A * p 1 /A 0 2q 1 ⇐⇒ 6A * p 1 /p η ≤ A 0 /2, it follows that L B ∆ • 2p (v, v * ) ≤ 6A * p 1 p η kp k=1 p k v 2k v * 2(p-k) + v 2(p-k) v * 2k |v -v * | 2 - A 0 2 v 2p + v * 2p |v -v * | 2 ∀ p ≥ 12A * p 1 /A 0 2q 1 .
Therefore as shown in the above using (3.20)-(3.21) with γ = 2 we obtain (3.15).

3.4.

Moment estimates on the collision operator. We shall now deduce from the moment estimates on L B in the previous Lemma 3.6 some moment estimates on the collision operator.

Q(µ, µ), • 2p ≤ 2 2p+1 A 2 µ 2 µ 2p - 1 4 A 2 µ 0 µ 2p+γ . (3.23) Furthermore if 0 < γ < 2, then Q(µ, µ), • 2p Γ(q) µ 0 (3.24) ≤ C a q 2-a + C a q 3-a ε p A 2 µ 0 Z * p + 1 2 µ 2 A 2 Z q - q 16 A 2 µ 0 Z 1+ 1 q q
where q = ap , a = 2/γ, (3.25)

Z q = µ γq Γ(q) µ 0 , Z * p = max k∈{1,2,...,kp} {Z ak+1 Z a(p-k) , Z ak Z a(p-k)+1 }
and the constant 0 < C a < ∞ only depends on a.

(II) If γ = 2 and b(cos θ) satisfies (H3) which is rewritten as in (3.13), and let p 1 , q 1 , η be given in (3.13)-(3.14), then

(3.26) Q(µ, µ), • 2p Γ(p) µ 0 ≤ 48A * p 1 p 1-η (log p) µ 0 Z * p + 12A * p 1 p 1-η + A 0 4 µ 2 Z p - p 16 A 0 µ 0 Z 1+ 1 p p
for all p ≥ 12A * p 1 /A 0 2q 1 , where

(3.27) Z p = µ 2p Γ(p) µ 0 , Z * p = max k∈{1, 2,..., kp} Z k+1 Z p-k .
Proof of Lemma 3.7. By replacing µ with µ/ µ 0 we can assume that µ 0 = 1. Part (I). By part (I) of Lemma 3.6 we have

Q(µ, µ), • 2p = 1 2 R N ×R N L B ∆ • 2p (v, v * )dµ(v)dµ(v * ) (3.28) ≤ A 2 kp k=1 p k µ 2k+γ µ 2(p-k) + µ 2k µ 2(p-k)+γ +2p(p -1)A 2 ε p kp-1 k=0 p -2 k µ 2(k+1)+γ µ 2(p-1-k) + µ 2(k+1) µ 2(p-1-k)+γ + A 2 2 µ 2p µ γ - A 2 4 µ 2p+γ .
Using Hölder's inequality we have (for s > 2)

(3.29) µ r ≤ µ s-r s-2 2 µ r-2 s-2 s
, 2 ≤ r ≤ s from which we obtain for all s 1 , s 2 ≥ 2 satisfying

s 1 + s 2 ≤ 2p + 2 µ s 1 µ s 2 ≤ µ 2p-s 1 +2p-s 2 2p-2 2 µ s 1 +s 2 -4 2p-2 2p ≤ µ 2 µ 2p
where we used µ 2 ≤ µ 2p . Thus

Q(µ, µ), • 2p ≤ 4A 2 kp k=1 p k + 2p(p -1) kp-1 k=0 p -2 k µ 2 µ 2p + A 2 2 µ 2 µ 2p - A 2 4 µ 2p+γ ≤ 4A 2 2 p -1 + p(p -1)2 p-1 µ 2 µ 2p + A 2 2 µ 2 µ 2p - A 2 4 µ 2p+γ ≤ 2 2p+1 A 2 µ 2 µ 2p - A 2 4 µ 2p+γ
which proves (3.23) for µ 0 = 1, where we used the inequality (3.1) and

2 p + p(p -1)2 p-1 ≤ 2 2p-1 , p ≥ 3 .
Now suppose that 0 < γ < 2. This implies a = 2/γ > 1. Recall definitions of Z q and Z * p in (3.25). Then applying (3.2) and (3.4) we compute for all k ∈ {1, 2, ...,

k p } µ 2k+γ µ 2(p-k) + µ 2k µ 2(p-k)+γ = µ γ(ak+1) µ γa(p-k) + µ γak µ γ(a(p-k)+1) = Z ak+1 Z a(p-k) Γ(ak + 1)Γ(a(p -k)) + Z ak Z a(p-k)+1 Γ(ak)Γ(a(p -k) + 1) ≤ Z * p Γ(ap + 1) B(ak + 1, a(p -k)) + B(ak, a(p -k) + 1) = Z *
p Γ(q + 1)B(ak, a(p -k)) , and for all k ∈ {0, 1, ..., k p -1}

µ 2(k+1)+γ µ 2(p-1-k) + µ 2(k+1) µ 2(p-1-k)+γ = Z a(k+1)+1 Z a(p-k-1) Γ(a(k + 1) + 1)Γ(a(p -1 -k)) +Z a(k+1) Z a(p-1-k)+1 Γ(a(k + 1))Γ(a(p -1 -k) + 1) ≤ Z * p Γ(q + 1)B(a(k + 1), a(p -1 -k))
. This together with Γ(q + 1)/Γ(q) = q and Lemma 3.3 gives from (3.28) that

Q(µ, µ), • 2p Γ(q) ≤ Z * p qA 2 kp k=1 p k B(ak, a(p -k)) (3.30) +Z * p 2qp(p -1)A 2 ε p kp-1 k=0 p -2 k B(a(k + 1), a(p -1 -k)) + A 2 µ 2 2 Z q - A 2 4 µ 2p+γ Γ(q) ≤ Z * p A 2 C a q 2-a + Z * p A 2 C a q 3-a ε p + A 2 µ 2 2 Z q - A 2 4 µ 2p+γ Γ(q) .
For the negative term we use Hölder's inequality, µ 0 = 1, and q = ap = 2p γ to get

µ 2p+γ ≥ µ 1+ γ 2p 2p = µ 1+ 1 q γq
and so

µ 2p+γ Γ(q) ≥ Γ(q) 1 q µ γq Γ(q) 1+ 1 q = Γ(q) 1 q Z 1+ 1 q q ≥ q 4 Z 1+ 1 q q (3.31)
where we have used the inequality Γ(q) 1 q ≥ q/4. Thus (3.25) (with µ 0 = 1) follows from (3.30). Part (II). In this case we have γ = 2, i.e. a = 1 so that q = p and hence (3.31) becomes

µ 2(p+1) Γ(p) ≥ p 4 Z 1+ 1 p p .
By part (II) of Lemma 3.6 we have, as shown above, that (the special term µ 2k µ 2(p-k+1) for k = 1 in the sum should be treated separately)

Q(µ, µ), • 2p Γ(p) = 1 2Γ(p) R N ×R N L B ∆ • 2p (v, v * )dµ(v)dµ(v * ) ≤ 1 Γ(p) • 12A * p 1 p η kp k=1 p k µ 2(k+1) µ 2(p-k) + µ 2k µ 2(p-k+1) + 1 4Γ(p) A 0 µ 2 µ 2p - A 0 4 µ 2(p+1) Γ(p) = 1 Γ(p) • 12A * p 1 p η kp k=2 p k µ 2(k+1) µ 2(p-k) + µ 2k µ 2(p-k+1) + 1 Γ(p) • 12A * p 1 p η p 1 µ 4 µ 2(p-1) + 1 Γ(p) • 12A * p 1 p η p 1 µ 2 µ 2p + A 0 4 µ 2 µ 2p Γ(p) - A 0 4 µ 2(p+1) Γ(p) ≤ Z * p 12A * p 1 p η • p kp k=2 p k B(k, p -k) + Z * p • 12A * p 1 p η p p 1 B(2, p -1) + 12A * p 1 p 1-η + A 0 4 µ 2 Z p - p 16 A 0 Z 1+ 1 p p ≤ Z * p 12A * p 1 p η • p kp k=1 p k B(k, p -k) + 12A * p 1 p 1-η + A 0 4 µ 2 Z p - p 16 A 0 Z 1+ 1 p p ≤ 48A * p 1 p 1-η (log p) Z * p + 12A * p 1 p 1-η + A 0 4 µ 2 Z p - p 16 A 0 Z 1+ 1 p p
where in the last inequality we used Lemma 3.3. This proves (3.26) for µ 0 = 1.

An ODE comparison inequality.

Finally we shall conclude this section by proving an ODE comparison inequality which will be useful for proving moment production estimates.

Lemma 3.8. Given any A > 0, B > 0, ε > 0, we have:

(I) The function Y (t) = A B(1 -e -εAt ) 1/ε , t > 0 is the unique positive C 1 -solution of the equation d dt Y (t) = AY (t) -BY (t) 1+ε , t > 0 ; Y (0+) = ∞ .
(II) Let u(t) be a non-negative function in (0, ∞) with the properties that u is absolutely continuous on every bounded closed subinterval of (0, ∞) and

d dt u(t) 1 {u(t)>Y (t)} ≤ Au(t) -Bu(t) 1+ε 1 {u(t)>Y (t)} a.e. t ∈ (0, ∞) .
Then u(t) ≤ Y (t) for all t ∈ (0, ∞).

Proof of Lemma 3.8. Part (I) is obvious. To prove part (II) we use the assumption on u and notice that the function x → Bx 1+ε -Ax is increasing in ((A/B) 1/ε , ∞) and Y (t) > (A/B) 1/ε . Then it follows from the assumption of the lemma that

d dt u(t) - d dt Y (t) 1 {u(t)>Y (t)} ≤ BY (t) 1+ε -AY (t) -Bu(t) 1+ε + Au(t) 1 {u(t)>Y (t)} ≤ 0 a.e. t ∈ (0, ∞).
Thus by the absolute continuity of u we have for any t > t * > 0

(u(t) -Y (t)) + = (u(t * ) -Y (t * )) + + t t * d dτ u(τ ) - d dτ Y (τ ) 1 {u(τ )>Y (τ )} dτ ≤ (u(t * ) -Y (t * )) + .
From this we see it is enough to prove that for any t > 0 there is t * ∈ (0, t) such that u(t * ) ≤ Y (t * ). Otherwise there were t 0 > 0 such that u(t) > Y (t) for all t ∈ (0, t 0 ). By assumption on u, this implies d dt u(t) ≤ Au(t) -Bu(t) 1+ε a.e. t ∈ (0, t 0 ) .

On the other hand, from the lower bound Y (t) > (A/B) 1/ε we see that the function t → u -ε (t) is absolutely continuous on every closed subinterval of (0, t 0 ]. We then compute for a.e. t ∈ (0, t 0 )

d dt (u -ε (t)) ≥ -εAu -ε (t) + εB
and hence for any 0 < τ < t 0 we have by the absolute continuity of t → u -ε (t)e εAt on [τ, t 0 ] that

u -ε (t)e εAt ≥ u -ε (τ )e εAτ + B(e εAt -e εAτ ) A , ∀ t ∈ [τ, t 0 ] .
Omitting the positive term u -ε (τ )e εAτ and letting τ → 0+ leads to

u -ε (t)e εAt ≥ B(e εAt -1) A , ∀ t ∈ (0, t 0 ] i.e. u(t) ≤ A B(1 -e -εAt ) 1/ε = Y (t) ∀ t ∈ (0, t 0 ]
which contradicts the assertion "u(t) > Y (t) for all t ∈ (0, t 0 )". This prove the existence of t * ∈ (0, t) for all t > 0 and therefore concludes the proof of the lemma. For notation convenience we denote

R N ϕdF t = R N ϕ(v)dF t (v), etc.
And note that if F t is a measure weak solution of Eq. (1.1), then for any ϕ ∈ C 2 b (R N ) we have (4.1)

R N ϕdF t = R N ϕdF t 0 + t t 0 Q(F τ , F τ ), ϕ dτ ∀ t > t 0 > 0 .
Our proofs of the parts (a)-(b)-(c)-(d) of Theorem 1.3 are contained in the following three steps.

Step 1. A priori estimates for measure weak solutions. We first prove part (b) and moreover we prove that the solution F t in part (b) satisfies that for any s ≥ 0 and

any ϕ ∈ L ∞ -s ∩ C 2 (R N ), (4.2) t → Q(F t , F t ), ϕ is continuous in (0, ∞) and (4.3) d dt R N ϕdF t = Q(F t , F t ), ϕ ∀ t > 0 .
And these integrals are absolutely convergent for any t > 0. Then we prove that F t satisfies the moment production estimates in parts (c) and (d) of Theorem 1.3. Now let F t satisfy the assumptions in part (b). Recall that F t already conserves the mass as mentioned in Definition 1.1. Therefore the assumption F t 2 ≤ F 0 2 (∀ t > 0) is equivalent to the energy inequality (4.4)

R N |v| 2 dF t (v) ≤ R N |v| 2 dF 0 (v) ∀ t > 0.
Since our test function space for defining measure weak solutions is only

C 2 b (R N ), we need a truncation-mollification approximation. Let χ ∈ C ∞ c (R N ) satisfy 0 ≤ χ ≤ 1 on R N and χ(v) = 1 for |v| ≤ 1, χ(v) = 0 for |v| ≥ 2. Given any s ≥ 0 and any ϕ ∈ L ∞ -s ∩C 2 (R N ), let ϕ n (v) := ϕ(v)χ(v/n). It is easily seen that ϕ n ∈ C 2 c (R N ) ⊂ C 2 b (R N )
and their Hessian matrices satisfy sup

n≥1 |H ϕn (v)| ≤ C ϕ v s .
Thus by (1.11) we have for any

s 1 > s + 2 + γ sup n≥1 |L B [∆ϕ n ] (v, v * )| v s 1 + v * s 1 ≤ C ϕ A 2 ( v s + v * s )|v -v * | 2+γ v s 1 + v * s 1 → 0 as |v| 2 + |v * | 2 → ∞,
L B [∆ϕ n ] (v, v * ) = L B [∆ϕ] (v, v * ) ∀ (v, v * ) ∈ R N × R N .
Thus by (4.1), the assumption (1.23) and the dominated convergence theorem we obtain

lim n→∞ t t 0 Q(F τ , F τ ), ϕ n dτ = t t 0 Q(F τ , F τ ), ϕ dτ ∀ t > t 0 > 0 and thus (4.1) holds for all ϕ ∈ s≥0 L ∞ -s ∩ C 2 (R N ). Since ψ j (v) = v j , j = 1, . . . , N, and ψ(v) = |v| 2 belong to L ∞ -2 ∩ C 2 (R N
) and ∆ψ j = ∆ψ = 0, it follows from (4.1) that F t conserves the momentum and energy in the open interval (0, ∞). Therefore in order to prove the conservation of momentum and energy in the closed interval [0, ∞), we only have to prove that (4.5) lim

t→0 + R N v j dF t (v) = R N v j dF 0 (v), lim t→0 + R N |v| 2 dF t (v) = R N |v| 2 dF 0 (v)
for j = 1, 2, . . . , N.

Let χ(v) be given above and let

ε > 0. Then v → v j χ(εv), v → |v| 2 χ(εv) belong to C 2 c (R N ) ⊂ C 2 b (R N
) so that, by definition of measure weak solutions, the functions

t → R N v j χ(εv)dF t (v) and t → R N |v| 2 χ(εv)dF t (v)
are all continuous on [0, ∞). Since

|v j -v j χ(εv)| ≤ |v|1 {|v|≥1/ε} ≤ ε|v| 2 and C := sup t≥0 R N |v| 2 dF t (v) ≤ R N |v| 2 dF 0 (v) < ∞, it follows that R N v j dF t (v) = R N v j χ(εv)dF t (v) + O(ε) ∀ t ≥ 0 where |O(ε)| ≤ Cε. Thus letting t → 0 + gives lim t→0 + R N v j dF t (v) = R N v j χ(εv)dF 0 (v) + O(ε).
Then letting ε → 0 + leads to the first equality in (4.5) for j = 1, 2, . . . , N. Next using |v| 2 ≥ |v| 2 χ(εv) and the inequality (4.4) we have

R N |v| 2 dF 0 ≥ lim t→0 + R N |v| 2 dF t ≥ lim t→0 + R N |v| 2 χ(εv)dF t (v) = R N |v| 2 χ(εv)dF 0 (v)
which leads to the second equality in (4.5) by letting ε → 0 + . Next let's prove (4.2) and (4.3). Given any s ≥ 0 and ϕ ∈ L ∞ -s ∩ C 2 (R N ). For any 0 < δ < T < ∞, by denoting

C δ,T,s = sup δ≤t≤T F t 2 s < ∞
and using (1.11) we have

R N ϕdF t 1 - R N ϕdF t 2 ≤ C ϕ A 2 C δ,T,s |t 1 -t 2 | ∀ t 1 , t 2 ∈ [δ, T ] . So (4.6) t → R N ϕdF t is continuous in t ∈ (0, ∞) .
In order to prove (4.2), we need only to show that for any fixed t > 0 and any sequence

{t n } ⊂ [t/2, 3t/2] satisfying t n → t (n → ∞) we have (4.7) lim n→∞ Q(F tn , F tn ), ϕ = Q(F t , F t ), ϕ .
This is an application of Proposition 2.2. In fact by Proposition 2.

1 we know that (v, v * ) → L B [∆ϕ](v, v * ) is continuous on R N × R N
, and as shown above

|L B [∆ϕ] (v, v * )| v s 1 + v * s 1 ≤ C ϕ A 2 ( v s + v * s )|v -v * | 2+γ v s 1 + v * s 1 → 0 for all s 1 > s + 2 + γ as |v| 2 + |v * | 2 → ∞. Since sup t/2≤τ ≤3t/2 F τ s 1 < ∞,
it follows from Proposition 2.2 and the weak-star convergence

F tn ⇀ F t (n → ∞) (see (4. 6
)) that (4.7) and therefore (4.2) hold true. The differential equation ( 4.3) follows from the continuity property (4.2) and from the equation (4.1) which has been proven to hold for all ϕ ∈ L ∞ -s ∩ C 2 (R N ). Now for any s ≥ 6, applying (4.3) to ϕ(v) = v s , which belongs to L ∞ -s ∩ C 2 (R N ), and applying Lemma 3.7 with p = s/2 we have for any t > 0

d dt F t s = Q(F t , F t ), • s ≤ 2 s+1 A 2 F 0 2 F t s - 1 4 A 2 F 0 0 F t s+γ .
Since, by using the inequality (3.29),

F t s+γ ≥ ( F 0 2 ) -γ s-2 ( F t s ) 1+ γ s-2 it follows that d dt F t s ≤ 2 s+1 A 2 F 0 2 F t s - 1 4 A 2 F 0 0 ( F 0 2 ) -γ s-2 ( F t s ) 1+ γ s-2 ∀ t > 0 .
Thus using Lemma 3.8 we obtain

F t s ≤   2 s+1 A 2 F 0 2 1 4 A 2 F 0 0 ( F 0 2 ) -γ s-2 1 -exp(-γ s-2 2 s+1 A 2 F 0 2 t)   s-2 γ ∀ t > 0 . Since s ≥ 6 implies 2 s ≥ 8(s -2), this gives γ s -2 2 s+1 A 2 F 0 2 ≥ 16A 2 F 0 2 γ =: β
and hence

F t s ≤ F 0 2 F 0 2 F 0 0 • 2 s+3 1 -e -βt s-2 γ
, t > 0 , s ≥ 6 .

Applying this estimate to s = 6 we also obtain that for any 2 ≤ s < 6

F t s ≤ ( F 0 2 ) 6-s 4 ( F t 6 ) s-2 4 ≤ ( F 0 2 ) 6-s 4 ( F 0 2 ) s-2 4 F 0 2 F 0 0 • 2 9 1 -e -βt 4 γ × s-2 4 = F 0 2 F 0 2 F 0 0 • 2 9 1 -e -βt s-2 γ
.

Maximizing the two cases gives max{2 s+3 , 2 9 } ≤ 2 s+7 for all s ≥ 2 and thus (4.8)

F t s ≤ F 0 2 F 0 2 F 0 0 • 2 s+7 1 -e -βt s-2 γ ∀ t > 0 , ∀ s ≥ 2.
The estimate (1.24) now follows from (4.8) since by using the inequality

1 1 -e -βt ≤ 1 + 1 β 1 + 1 t
we have

F t s ≤ F 0 2 2 s+7 F 0 2 F 0 0 1 + 1 β s-2 γ 1 + 1 t s-2 γ = K s (F 0 ) 1 + 1 t s-2 γ .
Note that from (4.8) and 0 < γ ≤ 2 we also have (4.9)

F t s ≤ F 0 0 2 s+7 F 0 2 F 0 0 • 2 s+7 (1 -e -βt ) s γ ∀ t > 0 , ∀ s ≥ 2
which will be used below. Now we are going to prove the exponential moment production estimate (1.26). Let p, q be defined through the following relation (as used in Lemma 3.7) q = ap with a = 2 γ .

Also recall that F t conserves the mass and energy, i.e. F t 0 = F 0 0 , F t 2 = F 0 2 . We consider two cases: Case 1. 0 < γ < 2. In this case we have a > 1. By Lemma 3.7 we have for all t > 0 and q ≥ 3a (i.e. for all p ≥ 3)

d dt Z q (t) = Q(F t , F t ), • 2p Γ(q) F 0 0 ≤ C a q 2-a + C a q 3-a ε p A 2 F 0 0 Z * p (t) + 1 2 A 2 F 0 2 Z q (t) - q 16 A 2 F 0 0 Z q (t) 1+ 1 q , where Z q (t) = F t γq Γ(q) F 0 0 , Z * p (t) = max k∈{1,2,...,kp} {Z ak+1 (t)Z a(p-k) (t) , Z ak (t)Z a(p-k)+1 (t)} .
Using a = 2/γ > 1 and Lemma 3.4 we have

C a q 2-a + C a q 3-a ε p = o(1)q (q → ∞)
so that there is a positive integer n 0 , depending only on b(•) and γ, such that n 0 δ ≥ 3a and C a q 2-a + C a q 3-a ε p ≤ q 32 ∀ q ≥ n 0 δ, where δ = a -1 .

Since

q ≥ n 0 δ =⇒ 1 2 A 2 F 0 2 < 16A 2 F 0 2 γq = βq, it follows that (4.10) d dt Z q (t) ≤ A 2 F 0 0 q 32 Z * p (t) + βqZ q (t) - q 16 A 2 F 0 0 Z q (t) 1+ 1 q ∀ q ≥ n 0 δ . Let Θ = 2 γn 0 δ+7 F 0 2 F 0 0 , Y q (t) = Θ 1 -e -βt q , t > 0 .
Then Y q satisfies the equation

d dt Y q (t) = βqY q (t) - βq Θ (Y q (t)) 1+ 1 q , t > 0 ; Y q (0+) = ∞ .
We now prove that

(4.11) Z q (t) ≤ Y q (t) ∀ t > 0 , ∀ q ≥ 1 .
To do this, it suffices to show that (4.12)

Z q (t) ≤ Y q (t) ∀ t > 0 , ∀ q ∈ [1, nδ] , n = n 0 , n 0 + 1, n 0 + 2, . . . .
First of all it is easily seen that (4.12) holds for n = n 0 . In fact by definitions of Z q (t) and Y q (t) and using the inequality Γ(q) > 1/2 (∀ q ≥ 1) and (4.9) we have for all 1 ≤ q ≤ n 0 δ

Z q (t) ≤ 2 F t γq F 0 0 ≤ F 0 2 F 0 0 • 2 γq+7 1 -e -βt q ≤ Y q (t) ∀ t > 0.
Suppose that (4.12) holds for an integer n ≥ n 0 . Take any q ∈ [nδ , (n + 1)δ] . Then q ≥ nδ ≥ n 0 δ and so (4.10) holds for such q. Recall that ap = q. Since for all integer 1 ≤ k ≤ k p = [(p + 1)/2] there hold

1 < ak < ak + 1 ≤ (n+1)δ+a 2 + 1 < nδ, 1 < a(p -k) < a(p -k) + 1 ≤ q -δ ≤ nδ it follows from the inductive hypothesis that Z ak+1 (t)Z a(p-k) (t) ≤ Y ak+1 (t)Y a(p-k) (t) = Y q+1 (t) , Z ak (t)Z a(p-k)+1 (t) ≤ Y ak (t)Y a(p-k)+1 (t) = Y q+1 (t) .
Therefore by definitions of Z * p (t), Y q (t) we obtain

Z * p (t) ≤ Y q+1 (t) = Y q (t) 1+ 1 q , ∀ t > 0 , ∀ q ∈ [nδ, (n + 1
)δ] and hence by (4.10)

d dt Z q (t) ≤ βqZ q (t) + A 2 F 0 0 32 qY q (t) 1+ 1 q - A 2 F 0 0 16 qZ q (t) 1+ 1 q ∀ t > 0
for all q ∈ [nδ, (n + 1)δ]. From this we obtain the following inequality:

d dt Z q (t) 1 {Zq(t)>Yq(t)} ≤ βqZ q (t) - βq Θ Z q (t) 1+ 1 q 1 {Zq(t)>Yq (t)} ∀ t > 0
where we used the obvious fact that

A 2 F 0 0 32 > β Θ .
Thus applying Lemma 3.8 we conclude Z q (t) ≤ Y q (t) for all t > 0. This together with the inductive hypotheses implies that Z q (t) ≤ Y q (t) for all t > 0 and all q ∈ [1 , (n + 1)δ] . This proves (4.12) and thus (4.11) holds true. Now let (4.13)

α(t) = 1 -e -βt 2Θ , t > 0 .
Then by definitions of Z q (t), Y q (t) and Z q (t) ≤ Y q (t) we have for all t > 0 (α(t)) q F t γq q! F 0 0 ≤ (α(t)) q Z q (t) ≤ (α(t)) q Y q (t) = 1 2 q , q = 1, 2, . . . and thus

R N e α(t) v γ dF t (v) = F 0 0 + ∞ q=1 (α(t)) q q! F t γq ≤ 2 F 0 0 . Case 2. γ = 2.
In this case we have a = 1 hence q = p. From part (II) of Lemma 3.7 with p 1 , q 1 and η given in (3.13)-(3.14), we have for all p ≥ (12A * p 1 /A 0 ) 2q 1 (which is larger than 5)

d dt Z p (t) ≤ 48A * p 1 p 1-η (log p) F 0 0 Z * p (t) + 12A * p 1 p 1-η + A 0 4 F 0 2 Z p (t) - A 0 F 0 0 16 pZ p (t) 1+ 1 p where Z p (t) = F t 2p Γ(p) F 0 0 , Z * p (t) = max k∈{1,2,..., kp} Z k+1 (t)Z p-k (t) , t > 0 .
Let us fix an integer n 0 ≥ (12A * p 1 /A 0 ) 2q 1 such that

48A * p 1 p 1-η log p ≤ A 2 32 p , 12A * p 1 p 1-η + A 0 4 ≤ 32A 2 p ∀ p ≥ n 0 . Recalling β = 32A 2 F 0 2 for γ = 2, this gives (4.14) d dt Z p (t) ≤ A 2 F 0 0 32 p Z * p (t) + βpZ p (t) - A 2 F 0 0 16 pZ p (t) 1+ 1 p ∀ p ≥ n 0 .
It will be clear that in the present case all p can be chosen integers. Let

Θ = 2 2n 0 +7 F 0 2 F 0 0 , Y p (t) = Θ 1 -e -βt p , t > 0 ; p ≥ 1 . Then Y p satisfies the equation d dt Y p (t) = βpY p (t) - βp Θ Y p (t) 1+ 1 p , t > 0 ; Y p (0+) = ∞ .
We now prove that

(4.15) Z p (t) ≤ Y p (t) ∀ t > 0 , p = 1, 2, 3, . . .
As shown in the Case 1 one sees that (4.15) holds for all integer 1 ≤ p ≤ n 0 . Suppose that (4.15) holds true for some integer p -1 ≥ n 0 . Let us check the case p. By p -

1 ≥ n 0 > 5 we have k p +1 ≤ (p+1)/2+1 ≤ p-1 and so Z k+1 (t)Z p-k (t) ≤ Y k+1 (t)Y p-k (t) = (Y p (t)) 1+ 1 p hold for all k ∈ {1, 2, ..., k p }. So Z * p (t) = max k∈{1, 2,..., kp} Z k+1 (t)Z p-k (t) ≤ Y p (t) 1+ 1 p hence from (4.14) we obtain d dt Z p (t) ≤ βpZ p (t) + A 2 F 0 0 32 pY p (t) 1+ 1 p - A 2 F 0 0 16 pZ p (t) 1+ 1 p ∀ t > 0 which together with A 2 F 0 0 32 > β Θ implies the inequality d dt Z p (t) 1 {Zp(t)>Yp(t)} ≤ βpZ p (t) - βp Θ Z p (t) 1+ 1 p 1 {Zp(t)>Yp(t)} ∀ t > 0.
Applying Lemma 3.8 we then conclude that Z p (t) ≤ Y p (t) ∀ t > 0 . This proves (4.15).

As shown above we obtain with the function α(t) defined in (4.13) that

R N e α(t) v 2 dF t (v) ≤ 2 F 0 0 ∀ t > 0 .
This completes Step 1.

Step 2. Construction of solutions for absolutely continuous measures. Suppose that F 0 is absolutely continuous with respect to the Lebesgue measure, i.e. dF 0 (v) = f 0 (v)dv, and suppose that (moment bounds and finite entropy)

0 ≤ f 0 ∈ s≥0 L 1 s (R N ) and 0 < R N f 0 (v)| log f 0 (v)|dv < ∞ .
In this case we prove that there exists {f t } t≥0 ⊂ s≥0 L 1 s (R N ) such that the measure F t defined by dF t (v) = f t (v)dv is a conservative measure weak solution of Eq. (1.1) associated with the initial datum F 0 and F t satisfies the moment production estimates (1.24) and (1.26).

To do this we consider some bounded truncations B n of the kernel B:

B n (z, σ) = min{|z| γ , n} min{b(cos θ), n} , n = 1, 2, . . .
It is well known that for every n ≥ 1 the Eq. (1.1) with the bounded kernel B n has a unique conservative solution

f n t (v) satisfying f n 0 (v) = f 0 (v) and f n ∈ C 1 ([0, ∞); L 1 s (R N ))∩ L ∞ loc ([0, ∞); L 1 s (R N
)) for all s ≥ 0, and

(4.16) sup n≥1, t≥0 R N f n t (v) 1 + |v| 2 + | log f n t (v)| dv < ∞ .
Let Q Bn (•, •) (collision operator) and A n,2 (angular momentum defined in (H0)) correspond to the kernel B n , and define dF n t (v) = f n t (v)dv. Then F n t 2 = F n 0 2 = F 0 2 and from the proof of Lemmas 3.6-3.7 we see that by omitting the negative term in the proofs of the two lemmas and noting that A n,2 ≤ A 2 we have for all p ≥ 3 d dt

F n t 2p = Q Bn (F n t , F n t ), • 2p ≤ 2 2p+1 A 2 F 0 2 F n t 2p
. Thus for all s ≥ 6, letting p = s/2 and recalling f n t L 1 s = F n t s we obtain sup

n≥1 f n t L 1 s ≤ f 0 L 1 s exp 2 s+1 A 2 F 0 2 t ∀ t ≥ 0 .
From this and the basic estimate (1.11) we get for any ϕ ∈ C 2 b (R N ) and any T ∈ (0, ∞)

R N ϕ(v)f n t 1 (v)dv - R N ϕ(v)f n t 2 (v)dv ≤ C ϕ,T |t 1 -t 2 | ∀ t 1 , t 2 ∈ [0, T ] .
This together with (4.16) implies for any ψ ∈ L ∞ (R N ) and any T ∈ (0, ∞) (4.17) sup

t 1 ,t 2 ∈[0,T ], |t 1 -t 2 |≤δ; n≥1 R N ψf n t 1 dv - R N ψf n t 2 dv → 0 as δ → 0 + .
Since (4.16) implies that for every t ≥ 0, {f n t } ∞ n=1 is L 1 -weakly relatively compact, it follows from diagonal argument and (4.17) that there is a subsequence of {n} (independent of t), still denoted as {n}, and a nonnegative measurable function

(t, v) → f t (v) on [0, ∞) × R N satisfying f t ∈ L 1 (R N ) (∀ t ≥ 0) such that for all ψ ∈ L ∞ (R N ) (4.18) lim n→∞ R N ψf n t dv = R N ψf t dv ∀ t ≥ 0 .
And consequently

f t ∈ s≥0 L 1 s (R N ) ∀ t ≥ 0 , and (4.19) sup t≥0
f t L 1 2 ≤ f 0 L 1 2 , sup 0≤t≤T f t L 1 s < ∞ ∀ 0 < T < ∞ , ∀ s ≥ 0 ,
and for any s > 0 and any

ψ ∈ L ∞ (R N ) (4.20) t → R N ψf t dv is continuous on [0, ∞) .
Now we are going to show that f t (or equivalently the measure F t defined by dF t (v) = f t (v)dv) is a conservative weak solution of Eq. (1.1) with the kernel B. Given any ϕ ∈ C 2 b (R N ), we have by (1.11) and

B n ≤ B sup n≥1 |L Bn [∆ϕ] (v, v * )| v s + v * s ≤ A 2 C ϕ |v -v * | 2+γ v s + v * s → 0 (|v| 2 + |v * | 2 → ∞) for s > 2 + γ. Moreover by Proposition 2.1, L Bn [∆ϕ](v, v * ), L B [∆ϕ](v, v * ) are all contin- uous on (v, v * ) ∈ R N × R N , and lim n→∞ sup |v|+|v * |≤R |L Bn [∆ϕ] (v, v * ) -L B [∆ϕ] (v, v * )| = 0 ∀ 0 < R < ∞ .
It follows from (4.18) and Proposition 2.2 that

sup 0≤t≤T R N ×R N |L B [∆ϕ] (v, v * )| f t (v)f t (v * )dvdv * < ∞ ∀ 0 < T < ∞ , Q Bn (f n t , f n t ), ϕ → Q B (f t , f t ), ϕ (n → ∞) ∀ t ≥ 0 . Again using Proposition 2.2 and (4.20) we conclude that t → Q B (f t , f t ), ϕ is continuous on [0, ∞) .
Finally using the dominated convergence theorem (in the t variable) we conclude that

R N ϕf t dv = R N ϕf 0 dv + t 0 Q B (f τ , f τ ), ϕ dτ ∀ t ≥ 0 .
Thus f t is a weak solution of Eq. (1.1). Let F t be defined by dF t (v) = f t (v)dv. Then from F t s = f t L 1 s , (4.19), and Step 1 we conclude that F t is a conservative measure weak solution of Eq. (1.1) associated with the initial datum F 0 and satisfies the moment production estimates (1.24) and (1.26).

Step 3. The approximation argument and conclusion. Let F 0 be the given measure in B + 2 (R N ) with F 0 0 = 0. We shall prove the existence of a measure weak solution F t that has all properties listed in the theorem.

First if F 0 = cδ v=v 0 (c > 0) is a Dirac mass, then it is easily checked that the measure F t ≡ cδ v=v 0 is a measure weak solution of Eq.(1.1) and apparently it conserves the mass, momentum and energy and has finite moments of all orders. By Step 1 we conclude that F t satisfies the moment production estimates (1.24)- (1.26).

Suppose F 0 is not a Dirac mass. We shall use Mehler transform:

Let (4.21) ρ = F 0 0 , v 0 = 1 ρ R N vdF 0 (v) , T = 1 Nρ R N |v -v 0 | 2 dF 0 (v) .
Then T > 0 so that the Maxwellian used in the Mehler transform can be defined:

(4.22) M(v) = e -|v| 2 /2T (2πT ) N/2 , v ∈ R N .
The Mehler transform of F 0 is defined by (4.23)

f n 0 (v) = e N n R N M e n v -v 0 - √ 1 -e -2n (v * -v 0 ) dF 0 (v * ) , n ≥ 1 .
It is well known that

R N   1 v |v| 2   f n 0 (v)dv = R N   1 v |v| 2   dF 0 (v) and for all ψ ∈ L ∞ -2 ∩ C(R N ) lim n→∞ R N ψ(v)f n 0 (v)dv = R N ψ(v)dF 0 (v) .
For every n, choose K n > n such that (4.24)

R N f n 0 (v) -min{f n 0 (v), K n }e -|v| 2 Kn v 2 dv ≤ F 0 0 2n . Then let f n 0 (v) = min{f n 0 (v), K n }e -|v| 2 /n , dF n 0 (v) = f n 0 (v)dv . We need to prove that (4.25) lim n→∞ R N ψdF n 0 = R N ψdF 0 ∀ ψ ∈ L ∞ -2 C(R N ) .
Indeed we have

R N ψdF n 0 - R N ψdF 0 ≤ R N ψ( f n 0 -f n 0 )dv + R N ψf n 0 dv - R N ψdF 0 .
The second term converges to zero (n → ∞). The first term also goes to zero: By (4.24) we have

R N ψ( f n 0 -f n 0 )dv ≤ C R N v 2 | f n 0 -f n 0 |dv ≤ C 2n .
Since for every n, f n 0 satisfies the condition in the Step 2, there is a conservative measure weak solution F n t of Eq. (1.1) with the kernel B and the initial data F n 0 , such that F n t satisfies the moment estimates

F n t s ≤ K s (F n 0 )(1 + 1/t) s-2 γ ∀ t > 0 , ∀ s ≥ 2.
Here recall that K s (•) is defined in (1.25). By the convergence (4.25) we have

lim n→∞ K s (F n 0 ) = K s (F 0 ) ∀ s ≥ 2 .
Thus for any s ≥ 2,

C * s := sup n≥1 K s (F n 0 ) < ∞ and hence (4.26) sup n≥1 F n t s ≤ C * s (1 + 1/t) s-2 γ ∀ t > 0 , ∀ s ≥ 2 .
Next we prove the equi-continuity of

{F n t } ∞ n=1 in t ∈ [0, ∞) (in
(R N ) S N-2 (n) ∆ϕ dω ≤ C ϕ S N-2 (n) ∆ϕ dω 2-γλ(θ) 2 ≤ C ϕ |v -v * | 2-γλ(θ) (sin θ) 2-γλ(θ)
where here and below C ϕ only depends on ϕ and N. Then by using

|v -v * | γ+2-γλ(θ) ≤ 8 v γ+2-γλ(θ) + v * γ+2-γλ(θ)
and (sin θ) -γλ(θ) = e γ(1-λ(θ)) ≤ e 2 and recalling (1.9) we obtain

|L B [∆ϕ] (v, v * )| ≤ C ϕ π 0 b(cos θ) sin N θ v γ+2-γλ(θ) + v * γ+2-γλ(θ) dθ .
So for all t > 0 (using Fubini's theorem and (4.26))

R N ×R N |L B [∆ϕ] (v, v * )| dF n t (v)dF n t (v * ) (4.27) ≤ C ϕ F 0 0 π 0 b(cos θ) sin N θ F n t γ+2-γλ(θ) dθ ≤ C ϕ,F 0 π 0 b(cos θ) sin N θ 1 + 1 t 1-λ(θ)
dθ .

Thus for all t 1 , t 2 ∈ [0, ∞) we compute (assuming t 1 < t 2 )

t 2 t 1 dt R N ×R N |L B [∆ϕ] (v, v * )| dF n t (v)dF n t (v * ) (4.28) ≤ C ϕ,F 0 π 0 b(cos θ) sin N θ dθ (1 + t 2 -t 1 ) 1-λ(θ) t 2 -t 1 0 t λ(θ)-1 dt = C ϕ,F 0 π 0 b(cos θ) sin N θ (1 + | log(sin θ)|) (1 + t 2 -t 1 ) 1-λ(θ) (t 2 -t 1 ) λ(θ) dθ =: C ϕ,F 0 Ω(t 2 -t 1 ). Since | Q(F n t , F n t ), ϕ | ≤ R N ×R N |L B [∆ϕ] (v, v * )| dF n t (v)dF n t (v * ) , it follows that sup n≥1 R N ϕdF n t 2 - R N ϕdF n t 1 ≤ sup n≥1 t 2 t 1 | Q(F n t , F n t ), ϕ |dt ≤ C ϕ,F 0 Ω(|t 2 -t 1 |) → 0 as |t 1 -t 2 | → 0. We then deduce for any ψ ∈ C c (R N ) that (4.29) Λ ψ (δ) := sup |t 1 -t 2 |≤δ; n≥1 R N ψdF n t 1 - R N ψdF n t 2 → 0 as δ → 0 + . Since C c (R N
) is separated, it follows from a diagonal argument that there is a subsequence of {n} (independent of t), still denoted by {n}, and a family {F t } t≥0 ⊂ B + 2 (R N ), such that (4.30)

lim n→∞ R N ψdF n t = R N ψdF t ∀ t ≥ 0 , ∀ ψ ∈ C c (R N ) .
Using (4.26) and the fact that F n t are conservative solutions we have (4.31)

F t 2 ≤ F 0 2 , F t s ≤ C * s (1 + 1/t) s-2 γ ∀ t > 0 , ∀ s ≥ 2 .
Also by (4.30) and (4.29) we have

R N ψdF t 1 - R N ψdF t 2 ≤ Λ ψ (|t 1 -t 2 |) . Hence (4.32) t → R N ψdF t is continuous on [0, ∞) ∀ ψ ∈ C c (R N ) .
We now prove that F t is a measure weak solution of Eq. (1.1). Given any ϕ ∈ C 2 b (R N ), by (4.31) we see that the derivation of (4.27) holds also for F t and so

R N ×R N |L B [∆ϕ] (v, v * )| dF t (v)dF t (v * ) < ∞ ∀ t > 0 . Next by Proposition 2.1 the function (v, v * ) → L B [∆ϕ](v, v * ) is continuous on R N × R N and (4.33) |L B [∆ϕ] (v, v * )| v s + v * s ≤ C ϕ A 2 |v -v * | 2+γ v s + v * s → 0 (|v| 2 + |v * | 2 → ∞)
for all s > 2 + γ.

t → Q(F t , F t ), ϕ is continuous in (0, ∞) .
Note that the derivation of (4.28) also holds for F t and hence we have for all T ∈ (0, ∞)

(4.36) T 0 dτ R N ×R N |L B [∆ϕ] (v, v * )| dF t (v)dF t (v * ) ≤ C ϕ,F 0 Ω(T ) < ∞ . Thus t → Q(F t , F t ), ϕ belongs to C((0, ∞)) ∩ L 1 loc ([0, ∞))
. And it also follows from (4.28)-(4.34) and the dominated convergence theorem that for all t > 0 we have

t 0 Q(F n τ , F n τ ), ϕ dτ → t 0 Q(F τ , F τ ), ϕ dτ (n → ∞) .
Thus in the integral equation of measures solutions F n t , letting n → ∞ gives

R N ϕdF t = R N ϕdF 0 + t 0 Q(F τ , F τ ), ϕ dτ ∀ t > 0 .
We have proved that F t satisfies the conditions (i)-(ii) in the Definition 1.1 of measure weak solutions. So F t is a measure weak solution of Eq. (1.1) associated with the initial datum F 0 . Finally from the moment estimates in (4.31) and Step 1 we conclude that the solution F t conserves mass, momentum and energy, and satisfies the moment production estimates (1.24)- (1.26). This completes the proof of Theorem 1.3.

5.

Uniqueness and stability for angular cutoff: Proof of Theorem 1.5

This section is devoted to the proof of Theorem 1.5. We shall first prove some lemmas on how the sign decomposition of measures behaves with time integration and with the action of the collision operator.

Sign decomposition of measures. As usual we denote

B(R N ) = B 0 (R N ), µ = µ 0 = |µ|(R N ).
For any µ ∈ B(R N ), let µ + , µ -be the positive and negative parts of µ, i.e. µ ± = 1 2 (|µ|±µ). Let h : R N → R be the Borel function satisfying |h(v)| ≡ 1 such that dµ = hd|µ|. We may call h the sign function of µ. Then dµ + = 1 2 (1 + h)dµ. So for any µ, ν ∈ B(R N ), we have (5.1) |µ -ν| = ν -µ + 2(µ -ν) + .

Let us now prove that this sign decomposition behaves well with the time integration.

R N ψdν t = R N ψdν a + t a ds R N ψdµ s , (5.3)
R N ψd|ν t | = R N ψd|ν a | + t a ds R N ψh s dµ s , (5.4)
R N ψdν + t = R N ψdν + a + t a ds R N ψκ s dµ s .
Proof of Lemma 5.1. Since the half-sum of (5.2) and (5.3) is equal to (5.4), we only have to prove (5.2) and (5.3). The proof of (5.2) is easy and similar to that of (5.3). By simple function approximation, the proof of (5.3) can be reduced to the proof of that for any Borel set E ⊂ R N , t → E h t dµ t belongs to L 1 loc ([a, ∞)) (and so does t → R N ψh t dµ t for any bounded Borel function ψ on R N ) and

(5.5)

|ν t |(E) = |ν a |(E) + t a ds E h s dµ s , t ∈ [a, ∞) .
By assumption on µ t , the strong derivative d dt ν t = µ t exists, and

ν t 1 -ν t 2 ≤ t 2 t 1 µ s ds ∀ a ≤ t 1 ≤ t 2 < ∞ .
This implies that for any Borel set

E ⊂ R N , t → |ν t |(E) is Lipschitz on every bounded interval [a, T ] ⊂ [a, ∞): For all a ≤ t 1 ≤ t 2 ≤ T ||ν t 1 | (E) -|ν t 2 | (E)| ≤ |ν t 1 -ν t 2 |(E) ≤ t 2 t 1 µ s ds ≤ C T |t 1 -t 2 |
and so t → |ν t |(E) is differentiable for almost every t ∈ [a, ∞) and satisfies

|ν t |(E) = |ν a |(E) + t a d ds |ν s |(E)ds ∀ t ∈ [a, ∞) .
Therefore in order to prove (5.

s > t =⇒ |ν s |(E) -|ν t |(E) s -t ≥ E h t d ν s -ν t s -t , s < t =⇒ |ν s |(E) -|ν t |(E) s -t ≤ E h t d ν s -ν t s -t . Since (ν s -ν t )/(s -t) → µ t (s → t) in norm • , it follows that d dt |ν t |(E) = lim s→t |ν s |(E) -|ν t |(E) s -t = E h t dµ t .
This proves (5.6) and completes the proof.

Let us now prove that the sign decomposition on differences of product measures preserves the invariance by exchanging v and v * .

∈ L ∞ -s (R N × R N ) we have (5.8) R N ×R N ψ(v, v * )d(µ ⊗ µ -ν ⊗ ν) = R N ×R N ψ(v * , v)d(µ ⊗ µ -ν ⊗ ν) , (5.9)
R N ×R N ψ(v, v * )d|µ ⊗ µ -ν ⊗ ν| = R N ×R N ψ(v * , v)d|µ ⊗ µ -ν ⊗ ν| , (5.10)
R N ×R N ψ(v, v * )d(µ ⊗ µ -ν ⊗ ν) + = R N ×R N ψ(v * , v)d(µ ⊗ µ -ν ⊗ ν) + .
Proof of Lemma 5.2. Equality (5.8) easily follows from Fubini's theorem. Equality (5.10) follows from (5.9) and the relation

d(µ ⊗ µ -ν ⊗ ν) + = 1 2 d|µ ⊗ µ -ν ⊗ ν| + d(µ ⊗ µ -ν ⊗ ν) .
So we only have to prove (5.9). To do this we split ψ as ψ = ψ + -(-ψ) + so that we can assume that ψ ≥ 0. Let h(v, v * ) be the sign function of the measure µ ⊗ µ -ν ⊗ ν. Then applying (5.8) to ψ(v, v * )h(v, v * ) we have

R N ×R N ψ(v, v * )d|µ ⊗ µ -ν ⊗ ν| = R N ×R N ψ(v, v * )h(v, v * )d(µ ⊗ µ -ν ⊗ ν) = R N ×R N ψ(v * , v)h(v * , v)d(µ ⊗ µ -ν ⊗ ν) ≤ R N ×R N ψ(v * , v)d|µ ⊗ µ -ν ⊗ ν| .
Replacing ψ(v, v * ) with ψ(v * , v) we also obtain the reversed inequality. This proves (5.9).

Finally let us prove a signed estimate on the collision operator.

so that κd(µ -ν) = d(µ -ν) + . Then for any ϕ ∈ C b (R N ) satisfying 0 ≤ ϕ(v) ≤ v 2 we have R N ϕ(v)κ(v)d(Q(µ, µ) -Q(ν, ν))(v) (5.11) ≤ E ϕ + 2 γ/2 A 0 µ 2+γ µ -ν 0 + µ 2 µ -ν γ where E ϕ = A 0 2 γ µ γ R N ( v 2 -ϕ(v)) v γ dµ(v).
Proof of Lemma 5.3. Since ϕ is bounded, there is no problem of integrability in the following derivation. For instance we can write

R N ϕ(v)κ(v)d Q(µ, µ) -Q(ν, ν) (v) = I (+) -I (-) (5.12)
where

I (+) = R N ×R N L B [ϕκ](v, v * )d(µ ⊗ µ -ν ⊗ ν), I (-) = R N ×R N A(v -v * )ϕ(v)κ(v)d(µ ⊗ µ -ν ⊗ ν). By definition of B(v -v * , σ) and ϕ(v)κ(v) ≤ v 2 we have L B [ϕκ](v, v * ) + L B [ϕκ](v * , v) ≤ S N-1 B(v -v * , σ)( v ′ 2 + v ′ * 2 )dσ = A(v -v * )( v 2 + v * 2 ).
Then using d(µ ⊗ µ -ν ⊗ ν) ≤ d(µ ⊗ µ -ν ⊗ ν) + and Lemma 5.2 we compute

I (+) ≤ 1 2 R N ×R N (L B [ϕκ](v, v * ) + L B [ϕκ](v * , v))d(µ ⊗ µ -ν ⊗ ν) + ≤ R N ×R N A(v -v * ) v 2 d(µ ⊗ µ -ν ⊗ ν) + . Since A(v -v * ) ≤ A 0 2 γ v γ v * γ , v 2 -ϕ(v) ≥ 0, and (µ ⊗ µ -ν ⊗ ν) + ≤ µ ⊗ µ, it follows that R N ×R N A(v -v * )( v 2 -ϕ(v))d(µ ⊗ µ -ν ⊗ ν) + ≤ A 0 2 γ R N ×R N v γ v * γ ( v 2 -ϕ(v))d(µ ⊗ µ) = A 0 2 γ µ γ R N v γ ( v 2 -ϕ(v))dµ(v) = E ϕ . Therefore using d(µ ⊗ µ -ν ⊗ ν) + (v, v * ) ≤ dµ(v)d(µ -ν) + (v * ) + d(µ -ν) + (v)dν(v * )
we have

I (+) ≤ E ϕ + R N ×R N A(v -v * )ϕ(v)dµ(v)d(µ -ν) + (v * ) (5.13) + R N ×R N A(v -v * )ϕ(v)d(µ -ν) + (v)dν(v * ) . Similarly using d(µ ⊗ µ -ν ⊗ ν)(v, v * ) = dµ(v)d(µ -ν)(v * ) + d(µ -ν)(v)dν(v * ) and κ(v)d(µ -ν)(v) = d(µ -ν) + (v) we have I (-) = R N ×R N A(v -v * )ϕ(v)κ(v)dµ(v)d(µ -ν)(v * ) (5.14) + R N ×R N A(v -v * )ϕ(v)d(µ -ν) + (v)dν(v * ).
Canceling the common term in (5.13) and (5.14) and noticing that

d(µ -ν) + (v * ) ≤ d(µ -ν)(v * ) + d|µ -ν|(v * )
we obtain from (5.12),(5.13),(5.14) that

R N ϕ(v)κ(v)d(Q(µ, µ) -Q(ν, ν)) (5.15) ≤ E ϕ + R N ×R N A(v -v * )ϕ(v)dµ(v)d|µ -ν|(v * ) . Since A(v -v * )ϕ(v) ≤ A 0 2 γ/2 ( v γ + v * γ ) v 2 , it follows that R N ×R N A(v -v * )ϕ(v)dµ(v)d|µ -ν|(v * ) ≤ A 0 2 γ/2 ( µ 2+γ µ -ν 0 + µ 2 µ -ν γ )
which together with (5.15) proves (5.11).

5.2.

Proof of Theorem 1.5. We shall consider each part step by step.

Proof of part (a). Recall that B(z, σ) = |z| γ b(cos θ) satisfies A 0 < ∞ and 0 < γ ≤ 2. Let F t be a conservative measure weak solution of Eq. (1.1) with

F t | t=0 = F 0 ∈ B + 2 (R N ). We prove that F t is a measure strong solution.
First of all by F t 0 , F t γ ≤ F 0 2 and Proposition 2.3 we have

Q ± (F t , F t ) 0 ≤ 4A 0 F 0 2 2 , ∀ t ≥ 0, Q(F t , F t ), ϕ = R N ϕdQ(F t , F t ) ∀ ϕ ∈ C 2 b (R N ), ∀ t ≥ 0. Since t → R N ϕdQ(F t , F t ) = Q(F t , F t ), ϕ belongs to C((0, ∞)) ∩ L 1 loc ([0, ∞))
there is no problem of integrability and the integral equation for a measure weak solutions becomes (5.16)

R N ϕdF t = R N ϕdF 0 + t 0 ds R N ϕdQ(F s , F s ) . Now take any ϕ ∈ C 2 c (R N ) satisfying ϕ L ∞ ≤ 1. We have R N ϕdQ(F t , F t ) ≤ Q(F t , F t ) 0 ≤ 8A 0 F 0 2 2 , ∀ t ≥ 0 .
and thus using (5.16), for all 0 ≤ t

1 < t 2 < ∞ R N ϕd(F t 2 -F t 1 ) ≤ t 2 t 1 R N ϕdQ(F s , F s ) ds ≤ 8A 0 F 0 2 2 |t 1 -t 2 | .
Applying (1.19) this gives (5.17)

F t 1 -F t 2 0 ≤ 8A 0 F 0 2 2 |t 1 -t 2 | , ∀ t 1 , t 2 ∈ [0, ∞)
which enables us to prove the strong continuity:

(5.18) t → F t ∈ C([0, ∞); B 2 (R N )), t → Q ± (F t , F t ) ∈ C([0, ∞); B 0 (R N )) .
In fact applying the inequality (2.19) in Proposition 2.3 with s = 0 (recall that 0 < γ ≤ 2) we have

Q ± (F t , F t ) -Q ± (F t 0 , F t 0 ) 0 ≤ 8A 0 F 0 2 F t -F t 0 2 , t, t 0 ≥ 0. (5.19)
Fix t 0 ∈ [0, ∞). Using (5.1), the conservation of mass and energy, d(F t 0 -F t ) + ≤ dF t 0 , and (5.17) we have for any R ≥ 1

F t -F t 0 2 = 2 R N v 2 d(F t 0 -F t ) + (v) ≤ 2R 2 v ≤R d(F t 0 -F t ) + (v) + 2 v >R v 2 dF t 0 (v) ≤ 2 4 A 0 R 2 |t -t 0 | + 2 v >R v 2 dF t 0 (v) .
Thus first letting t → t 0 and then letting R → ∞ leads to lim sup t→t 0 F t -F t 0 2 = 0 . This together with (5.19) proves (5.18).

From the strong continuity in (5.18) we have for all

ϕ ∈ C 2 b (R N ) t 0 ds R N ϕdQ(F s , F s ) = R N ϕd t 0 Q(F s , F s )ds
which together with (5.16) yields

R N ϕdF t = R N ϕdF 0 + R N ϕd t 0 Q(F s , F s )ds .
Therefore applying (1.19) we obtain

F t = F 0 + t 0 Q(F s , F s )ds , t ≥ 0 . Since t → Q ± (F t , F t ) ∈ C([0, ∞); B 0 (R N )), it follows that t → F t ∈ C 1 ([0, ∞); B 0 (R N )) and d dt F t = Q(F t , F t ), t ≥ 0 .
So F t is a measure strong solution.

The converse is obvious because of (1.21) and (2.18) with s = 0: Every measure strong solution is a measure weak solution.

Proof of parts (b)-(c)-(d).

The proof of these three parts can be reduced to the proof of the following lemma: Lemma 5.4. Let F 0 ∈ B + 2 (R N ) with F 0 0 = 0 and let F t be a conservative measure strong solution of Eq. (1.1) with the initial datum F 0 and satisfy the moment production estimate (1.24)-(1.25) in Theorem 1.3. Let G t be any measure strong solution of Eq. (1.1) on the time interval [τ, ∞) with initial data

G t | t=τ = G τ ∈ B +
2 (R N ) for some τ ≥ 0, and satisfying G t 2 ≤ G τ 2 for all t ∈ [τ, ∞).

Then the stability estimates (1.27) (for τ = 0) and (1.28) (for τ > 0) hold true.

Note that the existence of such a solution F t as in the statement has been proven by Theorem 1.3 and part (a) of the present theorem. Therefore if Lemma 5.4 holds true, then by taking G 0 = F 0 (for the case τ = 0) we get G t ≡ F t on [0, ∞) and hence this proves parts (b), (c) and (d).

Proof of Lemma 5.4. Our proof is divided into several steps. First of all for notation convenience we denote H t = F t -G t .

Step 1. Given any 0 < r ∈ [τ, ∞). We prove that Here K 2+γ (F 0 ) is the constant in (1.25) with s = 2 + γ. To prove (5.20), we consider approximation: By d|H t | = dG t -dF t + 2d(H t ) + we have

H t 2 ≤ G τ 2 -F τ 2 + 2 (H r ) +
H t 2 = G t 2 -H t 2 + 2 lim n→∞ R N v 2 n d(H t ) + with v 2 n = min{ v 2 , n}.
Let v → h t (v) be the sign function of H t and κ t (v) = 1 2 (1+h t (v)) so that κ t dH t = d(H t ) + . Then applying Lemma 5.1 to the measure

H t = H r + t r (Q(F s , F s ) -Q(G s , G s )
)ds for t ≥ r and then using Lemma 5.3 we have with c τ = 4A 0 (K 2+γ (F 0 ) + F 0 2 )(1 + 1 τ ). This gives (1.28) by Gronwall's Lemma. The remaining steps deal with the case τ = 0 and prove (1.27).

R N v 2 n d(H t ) + = R N v 2 n d(H r ) + + t r ds R N v 2 n κ s (v)d Q(F s , F s ) -Q(G s , G
Step 3. If H 0 2 ≥ 1, then using F t 2 = F 0 2 , G t 2 ≤ G 0 2 we have

H t 2 ≤ (1 + 2 F 0 2 ) H 0 2 ∀ t ∈ [0, ∞) .
So in the following we assume that H 0 2 < 1 . Note that in this case we have (5.21)

F t ± G t 2 ≤ 1 + 2 F 0 2 =: C 0 ∀ t ≥
+ 8A 0 C 2 0 )R 2 r ∀ t ∈ [0, r].
Using the conservation of mass and energy we compute Step 5. In the following we denote C i = R i (γ, A 0 , A 2 , F 0 0 , F 0 2 ) for (i = 1, 2, . . . , 6), where R i (x 1 , x 2 , . . . , x 5 ) are some explicit positive continuous functions in (R >0 ) 5 . In (5.20) setting τ = 0, r = 1 we have Finally if H 0 2 = 0, then in (5.34) letting r → 0+ leads to H t 2 ≡ 0; if H 0 2 > 0, we take r = H 0 2 . This proves (1.27) and completes the proof of the lemma.

|v|>R v 2 dF t (v) = R N v 2 dF t (v) - |v|≤R v 2 dF t (v) = R N v 2 dF 0 (v) -
H t
Proof of part (e). Let dF 0 (v) = f 0 (v)dv with 0 ≤ f 0 ∈ L 1 2 (R N ), and let F t be the unique conservative measure strong solution of Eq. (1.1) with the initial datum F 0 . By the Lebesgue-Radon-Nikodym theorem, for every t ≥ 0 we have a decomposition dF t (v) = f t (v)dv + dµ t (v) where 0 ≤ f t ∈ L 1 2 (R N ), µ t ∈ B + 2 (R N ) and µ t concentrates on a Lebesgue null set. By the uniqueness of F t we can assume that f 0 L 1 = 0. Let f n 0 (v) = min{f 0 (v), n}e -|v| 2 /n , and dF n 0 (v) = f n 0 (v)dv .

By

Step 2 of the proof of Theorem 1.3, for every n there is a conservative measure weak solution F n t with the initial datum F n 0 and dF n t (v) = f n t (v)dv, 0 ≤ f n t ∈ L 1 2 (R N ) for all t ≥ 0. By part (a), F n t is also a measure strong solution. Since d(F t -F n t ) = (f t -f n t )dv + dµ t we have

F t -F n t 2 = f t -f n t L 1 2 + µ t 2 . Since F 0 -F n 0 2 = f 0 -f n 0 L 1 2 ≤ f 0 (v)>n f 0 (v) v 2 dv + R N f 0 (v)(1 -e -|v| 2 /n ) v 2 dv → 0 (n → ∞)
it follows from the stability estimate that for every fixed t ≥ 0 we have

f t -f n t L 1 2 + µ t 2 = F t -F n t 2 ≤ e C(1+t) Ψ F 0 ( F 0 -F n 0 2 ) --→ n→0 0
and therefore µ t ≡ 0. Thus dF t (v) = f t (v)dv for all t ≥ 0 and hence f t is the unique conservative mild solution of Eq. (1.1) associated with the initial datum f 0 . This proves part (e).

5 .

 5 estimates on the collision operator 11 3. Moment estimates on the collision operator 16 4. Construction of weak measure solutions: Proof of Theorem 1.3 29 Uniqueness and stability for angular cutoff: Proof of Theorem 1

1. 1 . 1 .

 11 The equation. Before introducing the main results, let us recall the Boltzmann equation for L 1 -solutions and basic notations. The equation for the space homogeneous solution takes the form(1.1)

  θ) sin N θ dθ < ∞ .

π 0 b

 0 (cos θ) sin θ dθ = ∞, π 0 b(cos θ) sin 2 θ dθ < ∞ .

1 . 1 . 3 .

 113 θ) sin N θ (1 + | log(sin θ)|)dθ < ∞, (H2) 1 < γ < 2 , π 0 b(cos θ) sin N -2ν θ dθ < ∞ , ν = 2 -2/γ ∈ (0, 1), (H3) γ = 2 , ∃ p ∈ (1, ∞) s.t. π 0 [b(cos θ)] p sin N -2 θ dθ < ∞, (H4) 0 < γ ≤ 2 , A 0 := S N -2 π 0 b(cos θ) sin N -2 θ dθ < ∞. Observe that (H3)| b ⇒ (H4)| b ⇒ (H2)| b ⇒ (H1)| b ⇒ (H0)| b ,where for instance (H3)| b denotes the assumption with respect to b(•) in (H3). Note also that (H3)| b and (H4)| b corresponds to the angular cutoff case (short-range interactions), whereas (H0)| b , (H1)| b and (H2)| b allow for non-locally integrable functions b(•) on the sphere, i.e. non-cutoff cases (long-range interactions). Dual form of the collision operator. For any n ∈ S N -1 , letS N -2 (n) = {ω ∈ S N -1 | ω • n = 0 } (N ≥ 3)and in dimension N = 2 let S 0 (n) = {-n ⊥ , n ⊥ } where n ⊥ ∈ S 1 satisfies n ⊥ • n = 0 .

0 B

 0 ) L B [∆ϕ](v, v * ) := π (|v -v * |, cos θ) sin N -2 θ S N-2 (n) ∆ϕ dω dθ and σ = cos θ n + sin θ ω , n = (v -v * )/|v -v * | for v = v * ; n = e 1 = (1, 0, . . . , 0) for v = v * .

|v| 2

 2 +|v * | 2 |∇ϕ(ξ)| |v -v * | sin θ ;

1. 3 .Definition 1 . 1 (

 311 Definitions of measure solutions. Let us start with a notion of measure weak solutions, where the time evolution is defined in the integral sense. Measure weak solutions). Let B(z, σ) be given by (1.4)-(1.5)-(1.6) with γ and b(•) satisfying (H0).

Definition 1 . 2 (

 12 Measure strong solutions). Let B(z, σ) be given by (1.4)-(1.5)-(1.6) with γ and b(•) satisfying (H4).

0 0 B

 0 Part (II). By assumption (2.3) and (1.6) we have Bn (r, cos θ) ≤ Bn+1 (r, cos θ) ≤ B(r, cos θ) ≤ (1 + r 2 ) γ/2 b(cos θ) which together with (1.5) implies that the functions r → π Bn (r, cos θ) sin N θ dθ, r → π (r, cos θ) sin N θ dθ are all continuous on [0, ∞). Thus by first using (2.

Proposition 2 . 3 (

 23 A weighted Lipschitz property on the collision operator). Let B(z, σ) be given by (1.4)-(1.5)-(1.6) with b(•) satisfying (H4).

Lemma 3 . 3 (

 33 An estimate on the beta function). Let p ≥ 3 and k p = [(p + 1)/2]. Then (3.5) kp k=1 p k B(k, p -k) ≤ 4 log p .

Lemma 3 . 7 .

 37 Let B(z, σ) = |z| γ b(cos θ), µ ∈ B + s (R N ) with µ 0 = 0, s ≥ γ + 2p, 0 < γ ≤ 2,and p ≥ 3. (I) If b(cos θ) satisfies the assumption (H0), then

4 .

 4 Construction of weak measure solutions: Proof of Theorem 1.3

  particular in the neighborhood of t = 0). It is only in this part that the logarithm | log(sin θ)| comes into play. Let λ(θ) := 1 1 + | log(sin θ)| , 0 < θ < π . By (1.11) and 0 < γλ(θ) ≤ γ ≤ 2 we have for any ϕ ∈ C 2 b

Lemma 5 . 2 (

 52 Sign decomposition and exchange of particles). For any µ, ν ∈ B + s (R N ) (s ≥ 0) and any locally bounded Borel function ψ

Lemma 5 . 3 .

 53 Let B(z, σ) be given by (1.4)-(1.5)-(1.6) with b(•) satisfying (H4). Let µ ∈ B + 2+γ (R N ) , ν ∈ B + 2γ (R N ), and let h(v) be the sign function of µ-ν and let κ = 1 2 (1+h)

2 ( 5

 25 1/s) H s 0 ds + F 0 2 t r H s γ ds , t ≥ r .

|v|≤R v 2 2 2 2 2 2 4 |v|>R |v| 2 4 |v|>r - 1 / 3 |v| 2

 22222424132 dF 0 (v) -dQ -(F s , F s ) .For the last term we use |v -v * | γ ≤ v γ v * γ ≤ v 2 v * 2 to get for all t ∈ [0, r] dQ -(F s , F s ) ≤ 2R 2 t 0 ds R N dQ -(F s , F s ) ≤ 2A 0 F dF t (v) ≤ |v|>R v 2 dF 0 (v) + 2A 0 F 0 R 2 r ∀ t ∈ [0, r] .Combining (5.25)-(5.26)-(5.27) gives(5.28) 2 (H t ) + 2 ≤ 4(1 + 9A 0 C 2 0 )R 2 r + dF 0 (v) , t ∈ [0, r] .Now choose R = r -1/3 . Then from (5.24), (5.28) we obtainH t 2 ≤ r + 4(1 + 9A 0 C 2 0 )r 1/3 + dF 0 (v) , t ∈ [0, r] .This gives (5.23) by definition of Ψ F 0 (r) in(1.22).

  1.4. Main results. Our first main result is the following Theorem 1.3 (Existence of solutions and moment production estimates without cutoff).

	2 (R N ) Suppose that B(z, σ) = |z| γ b(cos θ) satisfies (H1). Given any initial datum F 0 ∈ B + with F 0 0 = 0, we have
	(a) The Eq. (1.1) always has a conservative measure weak solution F t satisfying F t | t=0 = F 0 .
	(b) Let F t be a measure weak solution of Eq. (1.1) associated with the initial datum
	F 0 satisfying
	(1.23)

  and by part (II) of Proposition 2.1, we deduce lim n→∞

  and let v → h t (v) be the sign function of the measure ν t and let κ t = (1 + h t )/2 so that dν + t = κ t dν t .Then for any bounded Borel function ψ on R N , the functions

	t →	R N	ψdµ t , t →	R N	ψd|µ t | and t →	R N	ψdµ + t
	all belong to L 1 loc ([a, ∞)) and for any t ∈ [a, ∞) we have
	(5.2)
	Lemma 5.1 (Sign decomposition and time integration). Let µ t ∈ C([a, ∞); B(R N )), ν a ∈ B(R N ), and
			ν t = ν a +

t a µ s ds , t ≥ a,

  5) we only have to show that for every Borel set E ⊂ R N Now take any t ∈ (a, ∞) such that the derivative d dt |ν t |(E) exists. By (5.7) we have

	(5.6)	d dt	|ν t |(E) =	E	h t dµ t , a.e. t ∈ [a, ∞)
	which also implies that t → E h t dµ t belongs to L 1 loc ([a, ∞)). For any t, s ∈ [a, ∞), using
			|ν s |(E) =
	we have
	(5.7)	|ν

E d|ν s | ≥ E h t dν s s |(E) -|ν t |(E) ≥ E h t d(ν s -ν t ) .

  (t) = 0 and thusH t 2 ≤ G t 2 -F t 2 + 2 (H r ) + H s γ ds , ∀ t ∈ [r, ∞) .By assumption on F t and G t we have G t 2 -F t 2 ≤ G τ 2 -F τ 2 and F s 2+γ ≤ K 2+γ (F 0 )(1 + 1/s). This proves(5.20).Step 2. Suppose τ > 0. Then taking r = τ in (5.20) and usingG τ 2 -F τ 2 + 2 (H τ ) + 2 = H τ 2 we obtain H t 2 ≤ H τ 2 + c τ

	≤ (H r ) + where	2 + E n (t) + 2A 0 E n (t) = 4A 0 + 4A 0 t r F s 2+γ H s 0 ds + F 0 2 t r F s 2+γ H s 0 ds + F 0 2 t r F s γ t r t R 2 τ H s 2 ds ∀ t ∈ [τ, ∞) t r H s γ ds , t ∈ [r, ∞)

s ) N ( v 2 -v 2 n ) v γ dF s ds.

Since, by moment estimate (1.24), t r

F s γ R N v 2+γ dF s (v) ds ≤ F 0 2 t r F s 2+γ ds < ∞ , t ∈ [r, ∞) ,

it follows from dominated convergence that lim n→∞ E n

  0 . Step 4. Let r > 0 satisfy H 0 2 ≤ r ≤ 1. We prove that ≤ 4(1 + 9A 0 C 2 0 )Ψ F 0 (r) .First of all using (5.1) andG t 2 -F t 2 ≤ G 0 2 -F 0 2 ≤ r we have (5.24) H t 2 = G t 2 -F t 2 + 2 (H t ) + 2 ≤ r + 2 (H t ) +

	Using Proposition 2.3 we have
	H t 0 ≤ H 0 0 +
	(5.23)	U(r) := sup
		2
	and for any R ≥ 1
	(5.25)	2 (H

t 0 Q(F s , F s ) -Q(G s , G s ) 0 ds ≤ H 0 0 + 4A 0 t 0 F s + G s γ H s 0 + F s + G s 0 H s γ ds

and thus by 0 < γ ≤ 2 and (5.21) we obtain

(5.22

)

H t 0 ≤ H 0 0 + 8A 0 C 0 t 0 H s 2 ds , ∀ t ≥ 0. 0≤t≤r H t 2 t ) + 2 ≤ 4R 2 H t 0 + 2 |v|>R v 2 dF t (v) .

Next by H 0 2 ≤ r and (5.22) we have (5.26)

4R 2 H t 0 ≤ 4(1

  2 ≤ H 0 2 + 2 H 1 2 + C 1 H t 2 ≤ H 0 2 + 2 H 1 2 exp(C 1 (t -1)) , t ≥ 1 .Now we concentrate our estimate fort ∈ [0, 1] .In what follows we assume r satisfy (5.30) r > 0 , H 0 2 ≤ r < 1 .Using (5.20) (with τ = 0), G 0 2 -F 0 2 ≤ H 0 2 ≤ r, and H r 2 ≤ U(r) we haveH t 2 ≤ r + 2U(r) + C 2Thus for all t ∈ [r, 1](5.31)H t 2 ≤ r + 2U(r) + C 2 r| log r| + C 3 t 0 H s 2 (1 + | log s|)ds .Since H t 2 ≤ U(r) for all t ∈ [0, r], the inequality (5.31) holds for all t ∈ [0, 1]. Therefore by Gronwall's Lemma we conclude (5.32)H t 2 ≤ C 4 (r + U(r) + r| log r| ) ∀ t ∈ [0, 1] .In particular taking t = 1 yields the estimate for H 1 2 and thus from (5.29)-(5.30) we obtain(5.33)H t 2 ≤ C 5 (r + U(r) + r| log r| ) exp(C 1 (t -1)) , ∀ t ∈ [1, ∞) .Combining (5.32)-(5.33) and the inequality r| log r| ≤ r 1/3 we conclude (5.34) H t 2 ≤ Ψ F 0 (r) exp(C 6 (1 + t)) ∀ t ≥ 0 .

				t
				1	H s 2 ds , t ≥ 1
	so that Gronwall's Lemma applies to get
	(5.29)
	r	t	1 s	H s 0 ds +

t r H s γ ds , t ∈ [r, 1] . Further, using (5.22) we compute for all t ∈ [r, 1] t r 1 s H s 0 ds ≤ r log(t/r) + 8A 0 C 0 t r 1 s s 0 H τ 2 dτ ds ≤ r| log r| + 8A 0 C 0 t 0 H τ 2 | log τ |dτ .

c 2011 by the authors. This paper may be reproduced, in its entirety, for non-commercial purposes.

Acknowledgements. This work was started while the first author was visiting the University Paris-Dauphine as an invited professor during the autumn 2006, and the support of this university is acknowledged. The first author also acknowledges support of National Natural Science Foundation of China, Grant No. 10571101.

Proof of part (f ). Suppose F 0 ∈ B + (R N ) is not a Dirac mass. We can assume that F 0 0 = 0. Let f n 0 (v) be defined by (4.21)-(4.23) (the Mehler transform of F 0 ). By part (e), for every n ≥ 1 there exists a unique conservative L 1 -solution f n t of Eq.(1.1) associated with the initial datum f n t | t=0 = f n 0 . If we define F n 0 , F n t by dF n 0 (v) = f n 0 (v)dv and dF n t (v) = f n t (v)dv, then by uniqueness and Theorem 1.3 we see that F n t satisfies the moment production estimates. Thus it is easily checked that the Step 3 (where there is no need of introducing f n 0 for the present case) in the proof of Theorem 1.3 is totally valid here. Therefore there is a subsequence, which we still denote as {f n t } ∞ n=1 , such that for the unique measure solution F t of Eq. (1.1) with F t | t=0 = F 0 , the weak convergence (1.30) holds true. This completes the proof of Theorem 1.5.