From an initial data to a global solution of the nonlinear Schrödinger equation: a building process

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1 From an initial data to a global solution of the nonlinear Schrödinger equation: a building process Jean-Yves Chemin, Claire David November 7, 21 Université Pierre et Marie Curie-Paris 6 Laboratoire Jacques Louis Lions - UMR 7598 Boîte courrier 187, place Jussieu, F Paris cedex 5, France Abstract The purpose of this work is to construct a continuous map from the homogeneous Besov space Ḃ 2,R 2 in the set G of initial data in Ḃ 2,R 2 which gives birth to global solution of the mass critical non linear Schrödinger equation in the space L R 1+2 We use the fact that solutions of scale which are different enough almost do not interact; the main point is that we determine a condition about the size of the scale which depends continuously on the data 1 Introduction We consider in the space R 2 the non linear Schrödinger equation of the type GNLS m { i t u + u = P 3 u, u u t= = u x 1 where P 3 is a homogeneous polynomial of order 3 in z and z Let us observe that if we ine u λ t, x = λuλ 2 t, λx for a solution of GNLS m then u λ is also a solution As we work in R 2, the L 2 norm, which represents the total mass, is invariant under this scaling transformation This family of equations is thus called "mass critical" As done for instance in [], Strichartz estimate for the Schrödinger equation claims that for any couples p j, q j such that 2 p j + 2 q j = 1 and p j, q j 2,, there exists a constant C such that for any interval I = [I, I + ] of R, one has u L p 1 I;L q 1 R 2 ui L 2 R 2 + i t + u L p 2 I;L q 2 R 2 2 It is now classical to prove, with a fixed point argument, that Equation GNLS m is locally well posed for initial data in L 2 R 2, and globally well posed for small initial data in L 2 R 2 More precisely, we have the following theorem see for instance [] Theorem 11 Let u be an initial data in L 2 R 2 Then a positive time T exists such that there is a unique solution to GNLS m in L [, T ] R 2 Moreover, a positive constant c exists such that, if u Ḃ 2, R 2 is less than or equal to c, the solution is global and belongs to L R 1+2 and R 1+2 ut, x dxdt u L 2 R 2 3 1

2 Let us mention that some important particular cases of GNLS m have been intensively studied: the case when P 3 u, u = u 2 u, called ocusing case and the case when P 3 u, u = u 2 u called the focusing case In the ocusing case, B Dodson proves in [7] that for any initial data in L 2 R 2, a unique global solution exists in L R 1+2 ; he proved in [8] that it was also the case in the ocusing case when the mass of the initial data was less than the mass of the ground state In the works [11] and [1], refined blow phenomena are described for initial data closed to the ground state In the present work, we proved general results far away from using the refined structure of the focusing or ocusing case As observed by P Gérard in an published work see the text [13] by F Planchon and references therein, it is possible to improve a little this result using Besov spaces For technical reasons, we prefer to use a continuous version of the Littlewood-Paley theory Let us consider a radial function φ in DR d \ {} such that φξ = f ξ with fr 1 dr r = 1 The fact that the measure µ 1 dµ is invariant under dilatation gives, for any ξ in R d \ {}, φµ 1 ξ dµ µ = 1 Let us ine µ a = F 1 φµ 1 ξâξ Now let us recall the inition of Besov spaces Definition 11 We ine the Besov space as the space of distributions such that u Ḃs p,r R d = µ s µ u L p L r R + ; dµ µ The theory is strictly analogous to the one of the discrete case In this context, the norm Ḃ 2, R2 is ined by u = Ḃ2, R2 µ u dµ L 2 R 2 µ Let us observe that the space Ḃ 2, R2 is slightly bigger than L 2 R 2 = Ḃ 2,2 R2, but has the same scaling invariant We have the following result see [13] Theorem 12 Let u be an initial data in Ḃ 2, R2 Then a positive time T exists such that there is a unique solution to GNLS m in L [, T ] R 2 Moreover, a positive constant c exists such that, if u Ḃ 2, R 2 is less than or equal to c, the solution is global and belongs to L R 1+2 and ut, x dxdt u R 1+2 Ḃ2, Let us give some sketchy indications about the proof of this theorem Using an improvement of Strichartz inequality 2 proved by J Bourgain in [2], we get e it u L R 1+2 u Ḃ 2, R 2 5 Then it is possible to make a fixed point theorem in the space L [, T ] R 2 Let us point out that if u is a solution to GNLS m in the space L [, T ] R 2 then w = u e it u satisfies i t w + w = P 3 u, u L 3 [, T ] R 2 and w t= = Strichartz estimate 2 with p 2, q 2 =, and p 1, q 1 =, 2 implies that u e it u belongs to the space L [, T ]; L 2 R 2 Definition 12 Let us denote by G the set of all initial data in Ḃ 2, R2 such that a global solution of GNLS m exists in L R 1+2 2

3 It is classical that the set G is an open subset of Ḃ 2, R2 Indeed, if u belongs to G, and v to Ḃ 2, R2, one can search the solution associated with u + v under the form u + v, where u is the solution associated with u Elementary computations lead to i t v + v = Eu, v with Eu, vt, x vt, x ut, x 2 + vt, x 2 Inequalities 5 and 2 allow to prove that if v is small enough in the norm Ḃ 2, R2, then fixed point argument works in the space L R 1+2 for v The purpose of this work is to prove the following theorem Theorem 13 A continuous function F from Ḃ 2, R2 into G, and a constant C exist, such that C 1 v Ḃ 2, R2 F v Ḃ 2, R2 C v Ḃ 2, R 2 As far as we know, there is no results of this type in the literature The motivation of this result is to catch some informations about the set G in the spirit of the works [5] by J-Y Chemin and I Gallagher or [6] by J-Y Chemin, I Gallagher and P Zhang is the context of homogeneous incompressible Navier-Stokes equation 2 Structure of the proof The idea is to make the function v small by multiplication, and then to copy, as many times as required, a dilatation of the now small function Let us introduce, for p in [1, [, the following notation: v L 2 R d, π p S v = v v L p and π B S v = v v Ḃ 2, The first step consists in the inition of a function F on Ḃ 2, R2 R + R +, with values in Ḃ 2, R2, which make explicit the idea explained above Definition 21 Let us consider a smooth function χ ined on R + such that χ 1 [, ] = and χ [ 3 = 1,+ [ Let us ine F from Ḃ 2, R2 R + R + into Ḃ 2, R2 as follows [λ] Fv,, λ = j v j + χ λ [λ] [λ] 1 v [λ] 1 j= The main properties of F are described hereafter Proposition 21 The map F is continuous from Ḃ 2, R2 R + R + into Ḃ 2, R2 Moreover, there exists a positive constant C and a continuous function from ]1, + [ SḂ 2, R2 into R + such that, for any v in Ḃ 2, R2 different from, we have λ + 1, π B S v = 1 C λ + 1 v Ḃ 2, R2 Fv,, λ Ḃ 2, R2 Cλ + 1 v Ḃ 2, R2 6 This proposition means that functions, the scales of which are different enough, are, in some sense, "orthogonal" in the Besov space Ḃ 2, R2 The main point of the proposition is that the choice of the scales size can be made continuously with respect to the function v Proposition 22 Let B c be the open centered ball, the radius c of which is given by Theorem 12 There exists a continuous function 1 from ]1, [ B ρ into ], + [ such that for any v in B c, 1 λ + 1, v = Fv,, λ G 3

4 The proof of this proposition relies on the idea that two solutions of an evolution equations with scales that are different enough almost do not interact The idea which is now currently used goes back to the works of H Bahouri and P Gérard see [1] and F Merle and L Vega see [12] Again, it is related to the fact that the choice of the size can be made continuously with respect to the function v These two propositions imply Theorem 13 Indeed, if C is the constant given by Theorem 12, we ine F v = 1 χ ρ 1 v Ḃ 2, R v + χ ρ 1 2 v Ḃ 2, R2 F ρ π B S v, v, v Ḃ 2, R2 with v = v Ḃ 2, R 2 + 1, πb S v + 1 v Ḃ 2, R 2 + 1, ρ πb S v and then apply the above two propositions The idea of the proof of Proposition 21 is elementary: we develop the quantity Fv,, λ Ḃ 2, R2 and we prove that the cross terms are small enough thanks to the following lemma Lemma 23 Let us denote by S L p R d the unit sphere of the Lebesgue space L p R d For any p in ]1, + [, there exists a continuous function p from ], 1[ SL p R d SL p R d such that for any couple f, g of L p R d \ {} L p R d \ {}, one has p ε, π p S f, πp S g = fx d p g 1 x dx ε f L p R d g L p R d R d If one does not take continuity into account, the property is a very classical one: it is enough to approach, up to ε, the functions π p S f and πp S g respectively in L p and L p, by ring-supported functions Then, one chooses large enough such that the integral of the truncated functions is The question is: can the choice of the cut off be made continuously with respect to the functions? The answer is no, as shown by the following picture: y f x > t t Μ f t x Figure 1: The graph of a non-negative function and measurable function f If the function is radial and non increasing, it becomes possible The use of the non-increasing reordering enables one to prove Lemma 23 Section 3 is, first, devoted to brief recalls on the inition and main properties of the symmetric nonincreasing rearrangement One proves, then, Lemma 23, which enables one to continuously truncate the rearrangements Once it is done, one can prove Proposition 21 Section concentrates on the proof of Proposition 22, where Lemma 23 plays a major role regarding continuity Section 5 is devoted to the proof of basic results on the non-increasing reordering which are collected here for the reader s convenience

5 y f x > t t x Figure 2: The graph of the rearrangement f 3 Non-increasing reordering and almost-orthogonality To begin with, let us work in the space R d Let us, first, recall basic notations, that we will require later on For any measurable function f, we set f > t = { x R d / fx > t } and f t = { x R d / fx t } Let us consider a non-negative function and measurable function f: R d R, which vanishes at infinity, in the sense that, for any strictly positive real number t, the set f > t is of finite measure Definition 31 Let A be a measurable set of R d, the Lebesgue measure of which is finite denotes the open centered ball whose measure is the same of the one of A Let us consider a realvalued, measurable function f, ined on R d, which vanishes at infinity, in the sense that, for any strictly positive real number t, the set f > t is of finite measure The symmetric non-increasing rearrangement of f is the measurable and non negative function f, ined on [, + [ by f x = + 1 f >t x dt More detailed information can be found, for instance, in [9] and [16] One can already notice that f = f, and that the function f is a radially symmetric and non-increasing one, in the sense that f x = f y if x = y, and f x f y if x y We will require the non-increasing reordering properties described hereafter Proposition 31 Let us consider two real-valued, measurable functions f and g, ined on R d, which vanishes at infinity For any strictly positive number s, fs = f s, 7 fxgxdx f xg xdx and 8 R d R d f x1 g sxdx fx1 g s xdx 9 R d R d If, moreover, f and g belong to the space L p, it is also the case of f and g ; in addition, f L p = f L p and f g L p f g L p For the convenience of the reader, this proposition is proved in Section 5 Proof of Lemma 23 Thanks to inequalities 7 and 8, one is led to find an upper bound to fx d p g 1 xdx I = f x d p g 1 xdx R d R d As explained in the introduction, this means truncating the functions f and g in rings, which can be done using the following lemma 5 A

6 Lemma 32 For any p in [1, [, there exists a continuous function R from ], 1[ SL p to R + such that, for any f in SL p, ϕ R ε,ff L p = ε with ϕ R x = 1 1 {R 1 x R}x Proof Let us first prove that, for any positive valued function h, radially symmetric, non-increasing, the L p norm of which is equal to 1, and for any real number ε in ], 1[, there exists a unique Rε, h such that ϕ Rε,h xh p xdx = ε R d From the dominated convergence theorem, the function ρ h R = ϕ R xh p xdx R d is continuous and non-increasing on [1, [ Moreover : ρ h 1 = h p L = 1, and ρ p h + = Let us denote by R h the maximum of numbers R satisfying ρ h R = 1 The function ρ h is decreasing on the interval [R h, [ Thus, there exists a unique R h such that ρ h R h = ε Applying this property to the function h = f enables one to ine the function R In order to prove the continuity of R, let us consider a sequence ε n, f n of ], 1[ SL p, which converges towards ε, f in ], 1[ SL p It is enough to prove that any convergent subsequence converges towards R ε, f Let us denote by R the limit of a convergent subsequence of R ε n, f n For sake of simplicity, extraction will not be distinguished The triangle inequality leads to ϕ R ε n,f n f n L p f f n L p ϕ R ε n,f n f L p ϕ R ε n,f n f n L p + f f n L p By inition of R, ε n f f n L p ϕ R ε n,f n f L p ε n + f f n L p Since the symmetric non-increasing rearrangement is 1 lipschtizian on the Lebesgue spaces L p, one has ε n f f n L p ϕ R ε n,f n f L p ε n + f f n L p 1 The Lebesgue theorem implies then that: lim ϕr ε n,f n f L p = ϕ R f L p n + By passing through the limit in 1, we obtain ϕ R f L p R d = ε, which, by inition of R, leads to R = R ε, f Then, the function R is a continuous one The lemma is proved Conclusion of the proof of Lemma 23 One can estimate I Thanks to the identities f = ϕ R ε,ff R ε,f x R ε,f f and g = ϕ R ε,g g R ε,g x R ε,g g, we have I ϕr ε,ff x d p g x dx R d R d R ε,f x R ε,f f x d p ϕ R ε,gg x dx R d R ε,f x R ε,f f x d p 1 1 R ε,g x R ε,g g x dx ϕ R ε,ff L p g L p + f L p ϕ R ε,gg L p R d R ε,f x R ε,f f x d p 1 1 R ε,g x R ε,g g x dx 6

7 As f L p = f L p = g L p = g L p = 1, this leads to I 2ε ε,f x R R d R ε,f f x d p 1 1 R ε,g x R ε,g g x dx 2ε + R ε,f x R ε,f x 1 R ε,g x R ε,g x f x d p g x dx then Let us notice that if { x R d, R d } { R ε, f x R ε, f x R d, 2R ε, f R ε, g 11 } R ε, g x R ε, g = Thus, provided the condition 11 is satisfied, one has I f, g 2ε For f, g in SL p SL p, let us ine p ε, f, g ε ε = 2R 2, f R 2, g This function is suitable and Lemma 23 is thus proved Proof of Proposition 21 A basic expansion and the use of the scaling invariance lead to Fv,, λ [λ] χλ [λ] v E Ḃ2, Ḃ2, R2 v,,λ where = µ j 1 v j1 L 2 µ j 2 v j2 L 2 As we have E v,,λ J j 1,j 2,j 3,j J µ j 3 v j3 L 2 µ j v j L 2 = {,, [λ] + 1 } \ { j, j, j, j, j {,, [λ] + 1} } one gets, due to the scaling of the L 2 norm, that Ev,, λ = µ j v j = j µ jv j, j 1,j 2,j 3,j J dµ µ µ j 1 v L 2 µ j 2 v L 2 µ j 3 v L 2 µ j v L 2 Without any loss of generality, we can assume that j 2 is smaller than j 1 in the above integral Hölder inequality, for the measure dµ, and the scaling invariance of this same measure, imply that µ µ j 1 v L 2 µ j 2 v L 2 µ j 3 v L 2 µ j v dµ L 2 µ v 2 Ḃ2, R2 One deduces that: Ev,, λ [λ] v 2 Ḃ 2, R2 [λ]+1 The problem is reduced to the study of an integral of the form j= fµg 1 µ dµ µ 7 with dµ µ µ v 2 L 2 µ j 1 j 2 v 2 L 2 dµ µ µ v 2 L 2 µ jv 2 L 2 dµ µ

8 with f and g in L 2 ], [; µ 1 dµ We ine Let us now write that fµ = 1 [, [ µ fµ µ 1 2 fµg 1 µ dµ µ = R fµ 1 2 g 1 µ dµ Let us apply Lemma 23 with d = 1, p = 2 and let us choose ε = C[λ] +1 ; this gives Inequality 6 In order to prove the entire proposition, let us concentrate on the continuity of F which comes mostly from the following lemma Lemma 33 The map is a continuous one D { Ḃ 2, R 2 R + Ḃ 2, R2 v, δ δ 1 v δ 1 Proof It is worth noting that this map is an isometry in the sense that Dv, δ Ḃ 2, = v Ḃ 2, For any v, δ and v, δ belonging to Ḃ 2, R2 R +, Dv, δ Dv, δ Ḃ 2, R 2 δ 1 v v δ 1 Ḃ 2, R 2 + δ 1 v δ 1 δ 1 v δ 1 Ḃ 2, R 2 v v Ḃ 2, R 2 + v δ δ 1 v δ δ 1 Ḃ 2, R 2 Let us then consider a strictly positive real number ε There exists a function v,ε of DR 2 such that v v,ε Ḃ 2, R 2 ε This yields then, for any strictly positive real number δ, v δ 1 v δ 1 Ḃ 2, R 2 v v,ε Ḃ 2, R 2 + v,ε δ 1 v,ε δ 1 Ḃ 2, R 2 + δ 1 v,ε δ 1 δ 1 v δ 1 Ḃ 2, R 2 2 v v,ε Ḃ 2, R 2 + v,ε δ 1 v,ε δ 1 Ḃ 2, R 2 2 ε + C v,ε δ 1 v,ε δ 1 L 2 R 2 As lim v,ε δ 1 v,ε δ 1 δ 1 L =, one can infer that the mapping D is a continuous one 2 R 2 Conclusion of the proof of Proposition 21 The above lemma assures that the function F is continuous on L 2 R 2 R + R + \ N In order to conclude the proof, let us observe that, for any positive integer k, ] λ 1 + k, k + 1 [ k, Fv,, λ = j v j 12 which, thanks to Lemma 33, ensures the continuity of the function F on Ḃ 2, R2 R + R + If λ belongs to the interval ] 1/ + k, k[, then [λ] = k 1, χλ [λ] = 1, and Identity 12 is satisfied If, now, λ belongs to the interval [k, k + 1/[, then [λ] = k, χλ [λ] =, and the identity 12 is, again, satisfied Proposition 21 is then proved j= 8

9 Almost-orthogonality and making of global solutions We hereafter aim at proving Proposition 22 Classically, one searches the solution related to Fv,, λ under the form u app + R, where the approached solution u app t, x is given by u app t, x V j, t, x V [λ]+1, t, x = [λ]+1 j= V j, t, x with = j N LSv 2 j t, j x for j in {1,, [λ]} and = [λ]+1 N LS χλ [λ]v 2[λ]+1 t, [λ]+1 x and where the term R has to be understood as an error one The idea is that if we choose the scaling parameter large enough, then the error term R is small, due to the fact that solutions of very different scale almost do not interact The point here is thus to prove that the choice of the size of the parameter can be made continuously when the function v varies In order to do so, let us write the equation satisfied by the error term R One has { i t R + R = Ẽ + ΦR with R t= = Ẽt, x E t, x = V j, t, x V k, t, x V l, t, x and j,k,l [λ]+1 j,k,l j,j,j ΦRt, x Rt, x 3 + u app t, x 2 Rt, x Here Φ is a generic notation used to denote a function of R that will not be explicitly given, for the sake of simplicity Let us estimate E L 3 R 1+2 By inition of the functions V j,, we get using the Hölder inequality in the space variable E j,k,l = V j, t, x 3 Vk, t, x 3 Vl, t, x 3 dxdt R j+k+l 3 N LSv 2 j t, 3 L R 2 R N LSv 2 k t, 3 L R 2 N LSv 2 l t, 3 L R 2 dt As j, k, l is not of the type j, j, j, we can assume up to a perturbation of indices that k is smaller than l Hölder inequality and then a change of variable imply that E j,k,l R N LSv 2 j t, 1 3 L R 2 2j dt N LSv 2 k t, 2 L R R 2 l k N LSv 2 l t, L R 2 2k dt N LSv 3 L R 1+2 N LSvt, 2 L R 2 l k N LSv 2l k t, 2 2 L R 2 dt R Applying Lemma 23, with d = 1, p = 2, f = g = N LSvt, 2 L R 2, one gets F ε, v 2 = 2 ε, π S N LSvt, 2 L R 2, π2 S N LSvt, 2 L R 2 9 = E j,k,l ε 2 3 N LSv L R 1+2 3

10 As the function v is assumed to be in the ball B ρ, we have N LSv L R 1+2 v Ḃ 2, R 2 Moreover the map { Bρ L 2 R v N LSvt, 2 L R 2 is continuous Thus the map ε, v F ε, v is continuous from ], [ B ρ into ], [, and we have By inition of E and E j,k,l λ, we have With 13 this leads to F ε, v = E j,k,l ε 2 3 v Ḃ 2, R 2 13 E 3 L 3 R 1+2 j,k,l [λ]+1 j,k,l j,j,j E j,k,l λ F ε, v = E L 3 R 1+2 Cε 1 2 [λ] v 3Ḃ 2, R 2 1 The estimate of u app L R 1+2 follows the same lines Indeed, let us write that [λ] u app L R 1+2 V j, L R 1+2 j= j 1,j 2,j 3,j [λ]+1 j 1,j 2,j 3,j j,j,j,j R 1+2 V j1,t, x V j2,t, x V j3,t, x V j,t, x dxdt Up to a permutation of indices, we can assume that j 1 j 2 Hölder inequalities and scaling invariance lead to [λ]+1 u app L R 1+2 V j, L R 1+2 j= [λ] v 2 Ḃ 2, R2 [λ]+1 j=1 R N LSvt, 2 L R 2 j N LSv 2j t, 2 L R 2 dt 1 2 Applying Lemma 23, with d = 1, p = 2 and f = g = N LSvt, 2 L R 2 gives F ε, v = u app L R 1+2 [λ] + 1 N LSv L R 1+2 C[λ] + 1 ε 1 2 v Ḃ 2, R 2 One deduces, then ε 1 [λ] and F ε, v = u app L R 1+2 2C[λ] + 1 v Ḃ2, R2 15 Let us admit for a while that, for any time T smaller than the maximal time of existence T, R L [,T ] R 2 C E L 3 R R 2 L [,T ] R 2 exp uapp L R A positive real number η being given, let us introduce the time T η as T η = sup { T < T / T Rt, L R 2 dt η} 1

11 If we prove that T = T then the maximal time of existence T will be equal to + Inequality 16 we get that for any T smaller than T η, Using R L [,T ] R 2 C E L 3 R η2 exp C u app L R 1+2 Let us choose η and ε such that η = 1 C exp C u app L R 1+2 ε = η 2 C [λ] v 6 Ḃ 2, R2 and exp C[λ] + 1 v Ḃ 2, R2 Assertions 1, 15 and 16 lead to R L [,T ] R 2 3/ Thus, T η = T which leads to T = + Our theorem is proved provided it is the case of inequality 16 Let us introduce the increasing sequence T m m M+1 such that T =, T M+1 = +, and m < M, Tm+1 T m u app t L R 2 dt = c and TM+1 T M u app t L R 2 dt c 17 for some given positive real number c, which will be chosen small enough later on Obviously, we have Mc TM + u app t L R 2 dt u app t L R 2 dt M + 1c 18 Thus the number M of T m such that T m is finite is less than c u app L R 1+2 Let us set RT = R L [,T ];L 2 R 2 + R L [,T ] R 2 In order to prove, by induction, that for any T smaller than min{t m, T }, one has RT C m+1 E L /3 R 1+2 Let us now us consider a time T smaller than min{t m+1, T } Strichartz estimate to the interval [T m, T ] gives R L [T m,t ];L 2 R 2 + R L [T m,t ] R 2 C RT m L 2 R 2 + E L 3 R 1+2 By inition of c, we get + R 3 L [T m,t ] R 2 + R L [T m,t ] R 2 u app 2 L [T m,t m+1 ] R 2 R L [T m,t ];L 2 + R L [T m,t ] R 2 C RT m L 2 R 2 + E L 3 R 1+2 If C η 2 + c 2 1/2, one obtains +η 2 + c 2 R L [T m,t ] R 2 R L [T m,t ];L 2 + R L [T m,t R 2 2C RT m L 2 R 2 + E L 3 R 1+2 With the choice C = 2C + 1, the induction hypothesis immediately yields RT 2C + 1C m+1 E L 3 R 1+2 C m+2 E L 3 R 1+2 By induction, we deduce that, for any T less than T,we have RT C M+2 E Using L 3 R 1+2 Inequality 18, it turns out that RT C E exp L 3 C u app t R 1+2 L R 2 dt which imply Inequality 16 Proposition 22 is proved 11

12 5 Some non-increasing reordering properties We hereafter aim at proving proposition 31 More information can be found in [9], [16] We nonetheless recall the main results useful to our study Various proofs of those results can be found in the above references In the following, we concentrate on positive-valued functions We will frequently refer to the fact that, for any function h, hx = 1 h>t xdt 19 To prove 7, let us note that fs > t = s 1 f > t Let R t denote the radius such that the volume of f > t is the same as the one of the centered ball of radius R t Thus we have 1 fs >t x = 1 B,s 1 R t x = 1 f>t sx One obtains inequality 7 by integration in t The proof of Inequality 8 relies on the following observation Let A and B denote two sets of R d of finite measure If A and B are two centered balls, one has measa B = min { meas A, meas B } = min { meas A, meas B } meas A B 2 This can be illustrated by the following drawing A A B B Figure 3: The sets A et B, with or without an intersection A B Figure : The centered balls A and B Thanks to Fubini s theorem, and to inequality 2, one has, from the inition of f and g, f xg xdx = 1 f>t xdt 1 g>s xds dx R d R d = meas f > t g > s dtds 12

13 By applying 2 in the above relation, one gets: f xg xdx R d If we apply then 19, we obtain: R d f xg xdx which ensures the required inequality R d meas f > t g > s dtds 1 f>t x1 g>s xdtdsdx R d 1 f>t xdt 1 g>s xds dx R d fxgx dx The proof of Inequality 9 is a very close one According to 19 and Fubini s theorem, we get fx1 g s xdx = 1 f>t x1 g s xdxdt R d R d = As 1 f>t 1 g s = 1 f>t 1 f>t 1 g>s, an integration leads to: meas f > t g s dt 21 meas f > t g s = measf > t meas f > t g > s By applying 2 to the sets f > t and g > s, one deduces meas f > t g s meas f > t meas f > t g > s meas f > t meas f > t g > s meas f > t g s Using it in inequality 21 enables one to write, thanks to Fubini s theorem, that fx1 g s xdx 1 f>t x1 g sxdxdt R d R d f x1 g sxdx R d which is Inequality 9 To prove the equality of the L p norms, let us note that, according to Fubini s theorem, for any positive valued function h, and any p in [1, [, one has h p L = p 1 p h>t xt p 1 dtdx = p R d meash > tt p 1 dt By inition of operation on sets, f > t and f > t are of same measure The above formula ensures immediately the equality of the L p norms The fact that the process is 1-lipschitzian on L p R d comes from the more general following result Lemma 51 Let J denote a convex function from R to R + such that J = If f and g are two positive-valued functions, then J f x g x dx J fx gx dx R d R d 13

14 Proof Let us write J = J + + J with J + = 1 R +J and J = 1 R \{}J The two functions J + and J are convex, and positive valued One proves the inequality for J +, the proof for J being strictly analogous Let h 1 and h 2 denote two non negative functions One has J + h 1 x h 2 x = = h1 x h 2 x + h 2 x J + h 1 x s ds J + h 1 x s 1 h2 sx ds 22 If H is an increasing function on R +, its derivative in the sense of distributions is a non negative measure which we denote by dh and which satisfies Hy s = 1 ]s+t, [ ydht Thanks to Fubini s theorem, we deduce that, for any set of functions h 1, h 2, Hh 1 x s1 h2 sxdx = 1 h1 >s+tx1 h2 sxdx dht 23 R d R d By applying this formula with h 1, h 2 = f, g, in conjunction with Inequality 9, one obtains, thanks to the positivity of the measure dht, Hf x s1 g sxdx = 1 f >s+tx1 g sxdxdht R d R d [, [ 1 f>s+t x1 g s xdx dht R d If, now, we apply Formula 23 with the set of functions f, g, we get Hf x s1 g sxdx Hfx s1 g s xdx R d R d By applying this inequality with H = J +, it ensures the required result, thanks to Formula 22 References [1] H Bahouri and P Gérard, High frequency approximation of solutions to critical nonlinear wave equations, American Journal of Math, 121, 1999, pages [2] J Bourgain, Refinements of Strichartz inequality and applications to 2D-NLS with critical nonlinearity, International Mathematical Research Notices, 5, 1998, pages [3] N Burq, Explosion pour l équation de Schrödinger au régime du "log-log", d après Merle- Raphael, Séminaire Bourbaki 25/6, Astérisque 311, 27, pages [] T Cazenave and F Weissler, Some remarks on the nonlinear Schrödinger equation in the critical case, Nonlinear Semigroups, Partial Differential Equations and Attractors Lecture Notes in Mathematics, 139, Berlin, 1989, pages [5] avec I Gallagher Wellposedness and stability results for the Navier-Stokes equations in R 3, Annales de l Institut Henri Poincaré, Analyse Non Linéaire, 26, 29, pages

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