We prove that Gevrey regularity is propagated by the Boltzmann equation with Maxwellian molecules, with or without angular cut-off. The proof relies on the Wild expansion of the solution to the equation and on the characterization of Gevrey regularity by the Fourier transform. This paper deals with a propagation property for the solution of the following Cauchy problem for the spatially homogeneous Boltzmann equation for Maxwellian molecules Here, f (v, t) : R3 × R+??R is the probability density of a gas which depends only on the velocity v ? R3 at the time t ? 0 and Q is the quadratic Boltzmann collision operator in the case of Maxwellian molecules: (1.1) Q(f, f )(v, t) = (f (v?, t)f (w?, t) ? f (v, t)f (w, t)) b v ? w dn dw. w?R3 n?S2 |v ? w| · n Due to the physical assumptions that the gas evolves through binary, elastic collisions which are localized both in space and time, the relations between the velocities (v?, w?) of two particles before the collision and (v, w) after it are the following: ??? v? = v +2 w + |v ?2 w| n, ?? w? = v +2 w ? |v ?2 w| n, where n is a vector in S2, the unit sphere in R3, and parametrizes all the possible pre-collisional velocities. The collision kernel b, which is supposed to be nonnegative, is the function which selects in which way the pre-collisional velocities contribute to produce particles with velocity v after the collision and is supposed (this is precisely the assumption Received by the editors November 20, 2006. 2000 Mathematics Subject Classification. Primary 76P05; Secondary 35B65.

of Maxwellian molecules) to depend only on the cosine of the deviation angle ?, namely

cos ? = |vv ?? ww| · n.

Finally, we will make the so-called non-cut-off assumption, which means that b ?/
Ll1oc([

1 b(cos ?) ? (1 ? cos ?) 45 , ? ? 0.

From a physical point of view, that means that the gas molecules repel each other with a force inversely proportional to the fifth power of their distance and a great contribution to the integral collision term is given by the grazing collisions (? ? 0). From a mathematical point of view, the more general assumption

1 (1.2) b(cos ?) ? (1 ? cos ?)? , ? ? 0, 1 < ? < 2, will be considered in this work.

The assumption that the collision kernel b is instead integrable on [

Due to the singularity of the collision kernel at the origin, the integral term (1.1) is not meaningful if f is not smooth, and so it is convenient to consider the weak form of the Boltzmann equation: for ? ? Cc?(R3), d

f (v, t)?(v) dv dt =

Q(f, f )(v, t)?(v) dv v?R3 v?R3 = f (v, t)f (w, t) (?(v?) ? ?(v)) b v ? w

v?R3 w?R3 n?S2 |v ? w| · n
or even, with another point of simplification, in the Fourier variable,
(1.3)
dn dw dv,
(1.4)
?
?tf?(?, t) = f?(?+, t)f?(??, t) ? f?(?, t)f?(0, t) b ? · n dn = Q(f, f )(?, t),
n?S2 | |
as was firstly done by Bobyl¨ev (see for instance [

?+ = 2? + |2?| n, ?? = 2? ? |2?| n.

The first results concerning the non-cut-off case for the weak equation go back
to Arkeryd [

n?S2 ?? ?l(?, 0) = f?0(?).

?l(?+, ? )?l(??, ? ) ? ?l(?, ? )?l(0, ? ) ?l A. Pulvirenti and G. Toscani proved first the existence and uniqueness of a solution ?l of the Cauchy problems (1.7). Then, letting

f?l(?, t) := ?l(?, bl?t),
they proved the convergence in a suitable setting of a subsequence of f?l to a solution
of the Cauchy problem for the non-cut-off equation. More precisely, the result is as
follows:
Theorem 1.1 (Pulvirenti, Toscani [

R3 R3

R3

R3 R3

R3 and the following Cauchy problem: (1.8) ? ?? ?tf?(?, t) =

n?S2 ?? f?(?, 0) = f?0(?),

f0(v)| log f0(v)| dv < ?, f?(?+, t)f?(??, t) ? f?(?, t)f?(0, t) b where b is a nonnegative function of Ll1oc([?1, 1[) satisfying (1.2).

Then, there exists a nonnegative solution f ? C1 [0, +?), L1(R3) to (1.8) satisfying for all t > 0 : f (v, t) dv = 1, f (v, t) vi dv = 0, i = 1, 2, 3,

R3

f (v, t)|v|2 dv = 3, f (v, t)| log f (v, t)| dv < ?.

Moreover, for all t > 0, the Fourier transform f?(·, t) of the solution is obtained as the (uniform on compact sets) limit of a subsequence of the functions ?l(·, bl?t) ? R3 C1 [0, +?), Cb(R3) , solutions of the cut-off Cauchy problems (1.7), which have the following explicit representation (called Wild?s expansion): ? k=0 ?l(k)(?)(1 ? e?? )k, where 1 k + 1 j=0 n?S2 ?l(j)(?+)?l(k?j)(??) ?l does the solution (of the cut-off or non-cut-off Boltzmann equation with Maxwellian molecules) keep on satisfying the same property? The answer is positive, provided that s ? ]0, 2], and we allow the constants K1, K2 to be different from those of the initial datum.

More precisely, the result we are going to prove is the following: Theorem 1.2. Let b be a nonnegative function of Ll1oc([?1, 1[) satisfying (1.2), and let f0 be a nonnegative function satisfying

R3 for some K1 ? 1, K2 > 0 and for some function ? : [0, +?) ? [0, +?), such that ?(r) ? r for r large enough, ?(r) ? +? for r ? +?, and satisfying the condition: (H) there exists R¯ ? 1 such that ?(?2 |?|2) ? ?2 ?(|?|2), when 0 ? ? ? 1,

¯ ? |?| ? R, then there exist R0 > 0, K > 0 such that the unique solution of the Cauchy problem (1.8) with f0 as initial datum satisfies sup |f?(?, t)|eK|?|2 ? 1, |?|<R0 sup |f?(?, t)|eK?(|?|2) ? 1, |?|?R0 t ? 0, t ? 0.

We would like to point out that condition (H) is satisfied for every concave function ? : [0, +?) ? [0, +?), such that ?(0) = 0 such as, for instance, ?(t) = |t|µ for µ ? (0, 1) and also for functions like ?(t) = |t|µ | log t|? for µ ? (0, 1), ? > 0.

Denoting by G? (R3) the space of Gevrey functions and by G0? (R3) the space of Gevrey functions with compact support (we shall recall in Section 4 their definition), we are able to deduce from the previous result the propagation along the solution of a Gevrey-type regularity satisfied by the initial datum.

Corollary 1.3. Let b be a nonnegative function of Ll1oc([?1, 1[) satisfying (1.2), and let f0 be a nonnegative function satisfying (1.9)

R3

R3

R3 i) If ? > 1 and f0 ? G0? (R3), then the solution f (·, t) of the Cauchy problem (1.8) is in G? (R3), uniformly for all t ? 0. ii) If ? ? 1, f0 ? G? (R3) ? S (R3), and moreover satisfies sup??R3 |f?0(?)| ? 1 K1e?K2|?| ? for K1 ? 1, K2 > 0, then the solution f (·, t) of the Cauchy problem (1.8) is in G? (R3), uniformly for all t ? 0.

The question whether an extra property satisfied by the initial datum f0
propagates along the solution has already been addressed concerning Sobolev or Lebesgue
regularity. In [

Note finally that many papers address the important question of the propagation of the behavior of the solution with respect to large v (that is, propagation of moments, evolution of Maxwellian tails, etc.). We do not investigate in this direction in this work.

The plan of the paper is the following: ? in Section 2, we shall present the result in a simpler form and for the socalled Kac model (which is 1-dimensional and describes radially symmetric solutions of the Boltzmann equation); ? in Section 3, we shall generalize the result both to the Boltzmann equation and to more general bounds on the initial datum; ? finally, in Section 4, we shall recall the main definitions of Gevrey functions, and we shall state the propagation result of a Gevrey-type regularity. w?R ??[? ?2 , ?2 ] = Q?(f, f )(v, t).

In this section, we present our result in the simpler case of the Kac equation. This equation, in its cut-off or non-cut-off version, is obtained when one considers radially symmetric solutions of the homogeneous Boltzmann equation for Maxwellian molecules. It reads: ?tf (v, t) =

(f (v??, t)f (w??, t) ? f (v, t)f (w, t)) b(?) d? dw Here, f (v, t) : R × R+??R is the probability density of a gas of one dimensional particles which depends only on the velocity v ? R at the time t ? 0, and which evolves through collisions which conserve energy but not momentum. The relations between the velocities (v??, w??) of two particles before the collision and (v, w) after it are the following: v?? = v cos ? + w sin ?, w?? = ?v sin ? + w cos ?.

We shall make the following assumption on the collision kernel b: the function b is even, locally integrable on ]0, ?2 ], and such that

cos ? (2.1) b(?) = O??0 , for some ? ? ]1, 3[.

| sin ?|?
Actually, this kind of assumption for the Kac equation was introduced by
Desvillettes in [

f?(? cos ?, t)f?(? sin ?, t) ? f?(?, t)f?(0, t) b(?) d? ? ?????

?tf?(?, t) = ????? f?(?, 0) = f?0(?), (2.2) (2.3) where the even initial datum f0 ? 0 satisfies the assumptions: ?? ?l(?, 0) = f?0(?), where each ?l(?) is a bounded function defined as in (1.4), (1.5) and (1.6), it is possible to prove that each Cauchy problem (2.4) has a unique solution ?l, which has the following explicit representation, called Wild?s expansion: where

?l(j)(? cos ?)?l(n?j)(? sin ?) ?l(?) d?.

f?l(?, t) := ?l(?, bl?t), it is possible to establish the (uniform on compact sets) convergence of a subsequence of f?l to a solution f? of the Cauchy problem for the (not necessarily cut-off) equation (2.2).

Let us suppose now that the initial datum f0 satisfies the extra property: |f?0(?)| ? e?K|?|s ,

? ? R, K > 0, s ? (0, 2].

Thanks to the representation of the solution of the cut-off equation (2.4) in Wild?s expansion, it is straightforward to prove that the solution itself satisfies the same upper bound. Indeed, by a direct computation, we have that |?l(1)(?)| ? e?K|?|s , since

?l(1)(?)eK|?|s =

eK|?|s?K|? cos ?|s?K|? sin ?|s f?0(? cos ?)eK|? cos ?|s f?0(? sin ?)eK|? sin ?|s ?l(?) d?, ? and 1 ? | cos ?|s ? | sin ?|s ? 0, for s ? (0, 2]. Hence by an immediate iteration argument, the same inequality holds for any ?(n)(?), and finally for the solution l of the cut-off equation for any t ? 0. Passing to the limit when l ? +? in the estimate ?l(?, bl?t) ? e?K|?|s , we see that the inequality also holds for the solution of the (not necessarily cut-off) equation (2.2).

Due to the nonlinearity of the collision operator, if we now consider the weaker assumption the same argument allows us to prove that the solution of each cut-off equation (2.4) satisfies the same upper bound, but only for a finite interval of time. In this case, by letting l go to infinity, the interval of time where the estimate is true can reduce to nothing.

In spite of this, we prove in this section that the condition |f?0(?)| ? K1e?K2|?|s propagates (though possibly with different constants K1 and K2 ) along the solution of the (not necessarily cut-off) Kac equation. 2.1. Some preliminary properties of initial data. In this section we emphasize some useful properties satisfied by any even, nonnegative function g such that R g(v) dv = 1 and R v2 g(v) dv = 1.

Lemma 2.1. Let g be a nonnegative, even function, satisfying

R

Then, there exist ? > 0 and K? > 0 such that for all |?| ? ?: Proof. We observe that under the hypotheses of the lemma, g? is of class C2 and satisfies the following property: g?(0) = 1, g? (0) = 0, and g? (0) = ?1. Using a Taylor expansion of g? at order 2, we obtain g?(?) = 1 ? 21 ?2 + o(?2) when ? ? 0. Then, the estimate of the lemma holds for any K? ? ]0, 12 [.

Then, we prove the following lemma, which is a simplified, one-dimensional
version of Lemma 3 in [

Lemma 2.2. Let g ? 0 such that R g(v) dv = 1. Then, for all r > 0, there exist Cr ? (0, 12 ) and C?r ? (0, 12 ) such that

R R g(v) sin2 v?

2 g(v) cos2 v? 2 dv ? Cr, |?| > r, dv ? C?r, |?| > r.

Proof. We only prove the first inequality, since the second one can be proven in exactly the same way.

Thanks to Lebesgue?s dominated convergence theorem and thanks to the absolute continuity of the measure ?(E) := E g(v) dv with respect to the Lebesgue measure, there exist R > 0 and ? > 0 such that for all measurable sets A ? R such that |A| ? ?, we have (2.5) Let ? ? R be fixed. For µ ? ]0, ?/2[, we define

Ac?B(0,R)

1 g(v) dv ? 2 .

v?

Kµ,R := v ? R, |v| ? R and ? k ? Z, 2 ? k? ? µ .

Kµc,R?B(0,R) We can therefore conclude that 2.2. The propagation theorem. We are now in position to state the theorem. Theorem 2.3. Let f0 be a nonnegative, even function, satisfying

R R We suppose that f0 is such that for some K1 ? 1, K2 > 0, and 0 < s ? 2. Then there exist R0 > 0, K > 0 such that the unique solution of the Cauchy problem (2.2) (under condition (2.1)) satisfies:

Let us suppose moreover that, for given s ? (0, 2], K1 > 1 and K2 > 0, g satisfies the following bound:

|g?(?)| ? K1e?K2|?|s , ? ? R.

Then, there exists ? > 0 such that for all R > ?, there exists K > 0 (depending on R) such that Then, g? is real and |g?(?)| ? 1 for all ? ? R. Moreover

R dv =

g(v) cos(?v) dv.

R

R 1 ? g?(?) =

g(v)(1 ? cos(?v)) dv = 2 g?(?) + 1 = g(v)(1 + cos(?v)) dv = 2

R R g(v) sin2 ?v

2 g(v) cos2 ?v 2 dv, dv.

1 ? g?(?) ? 2 C?, |?| > ?, g?(?) + 1 ? 2 C??, |?| > ?.

|g?(?)| ? 1 ? min(2 C?, 2 C??), |?| > ?. Now, there exists K4 > 0 such that

1 ? min(2 C?, 2 C??) ? e?K4R2 , which implies

|g?(?)| ? 1 ? min(2 C?, 2 C??) ? e?K4R2 ? e?K4|?|2 , ? < |?| ? R.

We can conclude the proof by letting K = min(K? , K3, K4).

Proof of Theorem 2.3. The cut-off case. There exists ? > 0 such that for any R0 > ?, there exists a strictly positive K such that the initial datum f?0 satisfies In order to prove the theorem for the cut-off case, it is enough to establish that any ?(n) in Wild?s sums satisfies (2.7). Let us check that this is true for ?l(1). Let us l define sup |f?0(?)|eK|?|2 ? 1, |?|<R0 sup |f?0(?)|eK|?|s ? 1.

|?|?R0

× eH(|? sin ?|)|f?0(? sin ?)|eH(|? cos ?|)|f?0(? cos ?)|?l(?)d? eH(|?|)?H(|? sin ?|)?H(|? cos ?|)?l(?)d?.

Condition (2.7) on the initial datum f0 reads therefore

eH(|?|)?l(1)(?) ? Since ? ?l(?)d? = 1, we end the estimate by proving that H(|?|) ? H(|? sin ?|) ? H(|? cos ?|) ? 0 for ? ? R and ? ? [? ?2 , ?2 ] if R0 ? 1. Thanks to the symmetries of the function H with respect to ?, we can restrict ourselves to the interval [0, ?4 ]. Now, when |?| < R0, we have

H(|?|) ? H(|? sin ?|) ? H(|? cos ?|) = K|?|2(1 ? (sin ?)2 ? (cos ?)2) = 0. If |?| ? R0 and |? sin ?| ? R0, |? cos ?| ? R0, then

H(|?|) ? H(|? sin ?|) ? H(|? cos ?|) = K|?|s(1 ? (sin ?)s ? (cos ?)s) ? 0 for 0 < s ? 2. Whenever |?| ? R0 and |? sin ?| < R0, |? cos ?| < R0 we have H(|?|) ? H(|? sin ?|) ? H(|? cos ?|) = K |?|s ? |?|2 (sin ?)2 + (cos ?)2 = K |?|s ? |?|2 .

If we choose R0 ? 1, we can conclude, since |?| ? R0, that | | ? |?|2 ? 0. ? s If now |?| ? R0 and |? sin ?| < R0, |? cos ?| ? R0, we have

H(|?|) ? H(|? sin ?|) ? H(|? cos ?|) = K | | ? |?|2(sin ?)2 ? |?|s(cos ?)s ? s ? K | | ? |?|s (sin ?)2 ? |?|s (cos ?)2 = 0.

? s Note that, since 0 ? ? ? ?/4, there is no other case to treat.

The non-cut-off case. As we have recalled in the introduction of section 2, the solution f?(?, t) of (2.2) in the non-cut-off case is obtained as the limit, uniform on compact sets, of a subsequence of ?l(?, bl?t), where ?l(?, ? ) is the solution of the Cauchy problem (2.4). Here bl? = ? min(b(?), l) d?. Since the estimate on ?l(?, ? ) holds true uniformly for ? ? 0, the same is valid for ?l(?, bl?t) and hence for f?(?, t).

Remark 2.5. In Theorem 2.3, the hypothesis that f0 is even could be replaced by the weaker hypothesis that R f0(v) v dv = 0. Since the Kac equation comes from the Boltzmann equation when one considers radially symmetric solutions, it is however natural to study only even initial data.

We would like to extend to the solution of the Boltzmann equation (1.3) the results proven in the previous section for the solution of the Kac equation. Two kinds of extensions are in order: first, we have to pass from the one-dimensional to the three-dimensional setting; second we would like to state the result considering not only functions such as e?|?|s , but also such as e??(|?|2), where ? is a suitable function. We now begin by restating the lemmas of the previous section in three dimensions. We shall only indicate the major modifications in the proofs. Lemma 3.1. Let g : R3 ? R be a nonnegative function satisfying R3

R3

Then there exist ? > 0 and K? > 0 such that for all |?| ? ?:

Proof. We observe that the result of the lemma is not changed when g is replaced by g ? R, where R is any rotation of R3. As a consequence, we can suppose that the symmetric matrix ( R3 g(v) vi vj dv)i,j?{1,2,3} is diagonal. Moreover, since g ? L1(R3) and v?R3 g(v) |v|2 dv = 3, we see that v?R3 g(v) vi2 dv > 0 for i = 1, 2, 3.

Then, g?(?) = 1 ? j3=1 ?j ?j2 + o??0(|?|2), with ?j > 0 for j = 1, 2, 3, and we conclude the proof as in Lemma 2.1.

Lemma 3.2. Let g : R3 ? R be a nonnegative function satisfying R3 g(v) dv = 1. Then, for all r > 0, there exists Cr ? (0, 12 ) such that for all ? ? R, R3 g(v) sin2

Proof. The proof follows the same lines as that of Lemma 2.2 and is a simplified
version of Lemma 3 in [

Ac?Q(0,R)

Kµ,R,? := v ? R3, v ? Q(0, R) and ? k ? Z, Thanks to the choice of the coordinate system, we have

Kµ,R,? = v ? R3, |vi| ? R, i = 1, 2, 3 and ? k ? Z, So, it is easy to see that

|Kµ,R,?| ? |??|R + 1 4|?µ| R2 = 4µR2 R? + |1?| and we can conclude the proof as in the one-dimensional case.

, Proposition 3.3. Let g be a nonnegative function satisfying where K1 ? 1, K2 > 0 and ? : [0, +?) ? [0, +?) satisfies limt?+? ?(t) = +?. Then, there exists ? > 0 such that for all R > ?, there exists K > 0 (depending on R) such that

Proof. The proof of this proposition is only a slight modification of that of Proposition 2.4. We only point out the few differences.

First, we explain how to find ? > 0: we can fix K3 ? (0, K2) and let ? = ?(K3) 1 be a positive constant such that ?(|?|) ? K2?K3 log K1 for every |?| ? ?; then of course for every R ? ?,

K1e?K2?(|?|) ? e?K3?(|?|), |?| ? R.

Second, we observe that for all ? ? R3 it is possible to find ? ? R (depending on ?) such that

|g?(?)| = g?(?)ei? =

R3

g(v) cos (? · v + ?) dv. 1 ? |g?(?)| = 2

R3 Thanks to Lemma 3.2, there exists C? such that

R3 g(v) sin2 ? · v + ? 2 dv ? 2 C?, |?| > ?.

Then, we can conclude as in the proof of Proposition 2.4.

Proof of Theorem 1.2. As we did for the Kac equation, for each of the Cauchy problems (1.7), we write the solution ?l under the form of a Wild?s expansion.

In order to prove the bound for the solution, it is enough to prove it for every term ?(n) in the sum. We define l

We denote ?? = ?2 . Thanks to the symmetries of the functions H(|?|) ? H(|? sin ??|) ? H(|? cos ??|) with respect to ??, we can restrict ourselves to the interval (0, ?4 ). Now, the case |?| < R0 is the same as in Theorem 2.3. If |?| ? R0 and |? sin ??| ? R0, |? cos ??| ? R0, then

H(|?|) ? H(|? sin ??|) ? H(|? cos ??|) = K(?(|?|2) ? ?(|? sin ??|2) ? ?(|? cos ??|2)). Thanks to the (H) property of ? we have ?(|?|2(sin ??)2) ? (sin ??)2?(|?|2) and ?(|?|2(cos ??)2) ? (cos ??)2?(|?|2). Hence ?(|?|2) ? ?(|? sin ??|2) ? ?(|? cos ??|2) ? ?(|?|2) ? (sin ??)2?(|?|2) ? (cos ??)2?(|?|2) = 0. Whenever |?| ? R0 and |? sin ??| < R0, |? cos ??| < R0 we have

H(|?|) ? H(|? sin ??|) ? H(|? cos ??|) = K(?(|?|2) ? |?|2((sin ??)2 + (cos ??)2)) If we choose R0 large enough, thanks to the assumption that ?(r) ? r for r large enough, and since |?| ? R0, we can conclude that

By using now the (H) property of ? and the fact that ?(r) ? r, if now |?| ? R0 and |? sin ??| < R0, |? cos ??| ? R0 we have

H(|?|) ? H(|? sin ??|) ? H(|? cos ??|) = K(?(|?|2) ? |?|2(sin ??)2 ? ?(|? cos ??|2)) = K(?(|?|2) ? |?|2). ? K(?(|?|2) ? ?(|?|2) (sin ??)2 ? ?(|? cos ??|2)) ? K?(|?|2)(1 ? (sin ??)2 ? (cos ??)2) = 0.

We end the proof by first noticing that a simple induction shows the estimate ?l(n)(?)e H(|?|) ? 1 when n ? 1, and so for the solution ?l(?, ? ) of the cut-off Cauchy problem, uniformly in ? ? 0. Then we may pass to the limit when l ? +? for a subsequence of ?l(?, bl?t) getting the same result for the solution f?(?, t) of the non-cut-off Cauchy problem.

In this section, we translate the propagation result obtained in the previous
sections in terms of Gevrey regularity for the solutions of the Boltzmann equation.
Let us begin by recalling the classical definitions of Gevrey functions and a useful
characterisation of these functions through their Fourier transform. For more
information, the interested reader can consult for instance the book by Rodino [

Let ? ? Rn be an open set and let ? ? 1 be a fixed real number.

Definition 4.1. The class G? (?) of Gevrey functions of order ? in ? is the set of functions f ? C?(?) satisfying the following property: for every compact subset K of ?, there exists a positive constant C = C(K) such that for all l ? Nn and all x ? K, (4.1)

Assumption (4.1) can be replaced by other equivalent assumptions, for example |?lf (x)| ? C|l|+1(l!)? .

|?lf (x)| ? RC|l|(l!)? , where R and C are two positive constants independent of l and x ? K. It is easy to recognize that G1(?) = A(?), the space of all analytic functions in ?, and that for ? ? ? , one has G? (?) ? G? (?). Moreover, it is interesting to underline the following inclusions:

A(?) ?

G? (?) , ?>1 ??1

G? (?) ? C?(?) , which are strict in both cases. We also recall that the Gevrey class G? (?) is closed under differentiation.

In what follows, we shall also need the following. Definition 4.2. Assume ? > 1. We shall denote by G0? (?) the vector space of all f ? G? (?) with compact support in ?.

The exclusion of ? = 1 in the previous definition is mandatory, because there are no analytic test functions other than the zero function. As for the other values ? > 1, one could wonder whether such compact supported functions do exist. An example in R is the following: let r > 0, ? > 1, d = 1?? and 1 ?(t) = e?td 0 t > 0, t ? 0.

f (x) = ?(x + r)?(x ? r)

is then in G0? (R) ([

The result that we are going to use in order to relate our propagation result to Gevrey regularity is the following.

Theorem 4.3 ([

i) Let ? > 1. If ? ? G0? (Rn), then there exist positive constants C and ? such that (4.2) 1 |??(?)| ? Ce??|?| ? , ? ? Rn. ii) Let ? ? 1. If the Fourier transform of ? ? S (Rn) satisfies (4.2), then ? ? G? (Rn).

We can therefore deduce Corollary 1.3 concerning the regularity of the solutions of the Boltzmann equation.

Proof of Corollary 1.3. The result is straightforward from Theorem 4.3 and Theorem 1.2. It is enough to notice that one can replace the uniform estimate sup |f?(?, t)|eK|?|2 ? 1, t ? 0, |?|<R0 sup |f?(?, t)|eK|?|s ? 1, t ? 0, |?|?R0 obtained in Theorem 1.2, by the following uniform estimate: sup |f?(?, t)|eK?2|?|s ? K?1, t ? 0, ??R3 for K?1 ? 1 and K?2 > 0 properly chosen. Now letting ? = 1s , one immediately gets the result. This ends the proof.

We end this section by comparing the regularity result we have just obtained with
the one obtained by Ukai in [

???,? = {g : g ?,?,? = l?N3 (?l!|)l|? vs?uRp3 e?(1+|v|2) 21 |?lg(v)| < ?}.

Comparing this space with the spaces in Definition 4.1, one can deduce that initial data in the Ukai setting are indeed in the Gevrey space G? (R3), but also decay very strongly at infinity together with all their derivatives. Ukai was able to prove by a fixed point argument that there exists a unique, local in time solution f belonging at every time to a functional space of the same kind as the initial datum but in which the indices change with t. More precisely, he proved that there exist T > 0, ? > 0, ? > 0 such that

Since initial data which belong to G0? (R3) also belong to the Ukai space, the question of comparing the two results is meaningful. Let us consider a nonnegative function f0 ? G0? (R3). If we suppose moreover that f0 satisfies assumptions (1.9), then by Theorem 1.1 we know that there is a solution f ? C1 [0, +?), L1(R3) which is unique by Toscani?Villani?s result. Since G0? (R3) ? ???,? for all ? ? 0, we can deduce from the Ukai result that for all ? ? 0 there is a time T = T (?) > 0 (possibily finite) such that this solution stays in the class (4.3) for t ? [0, T (?)], but this space is not uniform in time (in addition to the fact that for all ? it is difficult to compare the life times T (?)). Our result says instead that the solution stays for t ? [0, ?) in the same Gevrey class as its initial datum, without any information about the decay at infinity and moreover that all the estimates on the Gevrey seminorms are uniform in time.

The authors would like to thank G. Toscani for useful discussions about this problem.