Littlewood-Paley theory

Chapitre 6 Littlewood-Paley theory Introduction The purpose of this chapter is the introduction by this theory which is nothing but a precise way of counting derivatives using the localization in the frequency space For instance, if we look to the dispersive estimate for the wave equation, we see that this hypothesis of localization appears naturally 6 Localization in frequency space The very basic idea of this theory consists in a localization procedure in the frequency space The interest of this method is that the derivatives (or more generally the Fourier multipliers act in a very special way on distributions the Fourier transform of which is supported in a ball or a ring More precisely, we have the following lemma 6 Bernstein inequalities Lemme 6 (of localization Let C be a ring, B a ball A constant C exists so that, for any non negative integer k, any smooth homogeneous function σ of degree m, any couple of real (a, b so that b a and any function u of L a, we have Supp û λb sup α u L b C k+ λ k+d( a b u L a; α=k Supp û λc C k λ k u L a sup α u L a C k+ λ k u L a; α=k Proof of Lemma 6 Using a dilation of size λ, we can assume all along the proof that λ = Let φ be a function of D(R d the value of which is near B As û(ξ =φ(ξû(ξ, we can write, if g denotes the inverse fourier transform of φ, α u = α g u Applying Young inequalities the result follows through α g L c α g L + α g L 2 ( + 2 d α g L 2 (Id d (( α φ L C k+ 63

To prove the second assertion, let us consider a function φ which belongs to D(R d \{0} the value of which is identically near the ring C Using the algebraic Using the following algebraic identity ξ 2k = ξj 2 ξj 2 k j,,j k d = (iξ α ( iξ α, (6 α =k and stating g α = F (iξ j α ξ 2k φ(ξ, we can write, as û = φû that which implies that û = ( iξ α ĝ α û, α =k u = α =k and then the result This proves the whole lemma 62 Dyadic partition of unity g α α u (62 Now, let us ine a dyadic partition of unity We shall use it all along this text Proposition 6 Let us ine by C the ring of center 0, of small radius 3/4 and great radius 8/3 It exists two radial functions χ and ϕ the values of which are in the interval [0, ], belonging respectively to D(B(0, 4/3 and to D(C such that ξ R d,χ(ξ+ j 0 ϕ(2 j ξ =, (63 ξ R d \{0}, j Z ϕ(2 j ξ =, (64 j j 2 Supp ϕ(2 j Supp ϕ(2 j =, (65 q Supp χ Supp ϕ(2 j =, (66 If C = B(0, 2/3 + C, then C is a ring and we have j j 5 2 j C 2 j C =, (67 ξ R d, 3 χ2 (ξ+ ϕ 2 (2 j ξ, (68 j 0 ξ R d \{0}, 2 ϕ 2 (2 j ξ (69 j Z Proof of Proposition 6 Let us choose α in the interval ], 4/3[ let us denote by C the ring of small radius α and big radius 2α Let us choose a smooth function θ, radial with value in [0, ], supported in C with value in the neighbourhood of C The important point is the following For any couple of integers (p, q we have j j 2 2 j C 2 j C = (60 64

Let us suppose that 2 j C 2 j C = and that p q It turns out that 2 j 3/4 4 2 j+ /3, which implies that j j Now let us state S(ξ = j Z θ(2 j ξ Thanks to (60, this sum is locally finite on the space R d \{0} Thus the function S is smooth on this space As α is greater than, 2 j C = R d \{0} j Z As the function θ is non negative and has value near C, it comes from the above covering property that the above function is positive Then let us state ϕ = θ S (6 Let us check that ϕ fits It is obvious that ϕ D(C The function j 0 ϕ(2 j ξ is smooth thanks to (60 As the support of θ is included in C, we have ξ 4 3 j 0 ϕ(2 j ξ = (62 thus stating χ(ξ = j 0 ϕ(2 j ξ, (63 we get Identites (63and (65 Identity (66 is a obvious consequence of (60 and of (62 Now let us prove (67 which will be useful in Section 65 It is clear that the ring C is the ring of center 0, of small radius /2 and of big radius 0/3 Then it turns out that 2 j C 2 j C ( 3 4 2j 2 j 0 3 ou 2 2j 2 j 8, 3 and (67 is proved Now let us prove (68 As χ and ϕ have their values in [0, ], it is clear that χ 2 (ξ+ j 0 ϕ 2 (2 j ξ (64 Let us bound from below the sum of squares The notation a b(2 means that a b is even So we have Σ 0 (ξ = = (χ(ξ+σ 0 (ξ+σ (ξ 2 with ϕ(2 j ξ and Σ (ξ = j 0(2,q 0 j (2,q 0 ϕ(2 j ξ From this it comes that 3(χ 2 (ξ+σ 2 0 (ξ+σ2 (ξ But thanks to (65, we get Σ 2 i (ξ = ϕ 2 (2 j ξ and the proposition is proved j 0,q i(2 65

We shall consider all along this book two fixed functions χ and ϕ satisfying the assertions (63 (68 Now let us to fix the notations that will be used in all the following of this text Notations h = F ϕ and h = F χ, u = χ(du = F (χ(ξû(ξ, if j 0, j u = ϕ(2 j Du =2 jd R d h(2j yu(x ydy, S j u = j j if j 2, j u =0, j u = χ(2 j Du =2 jd R d h(2 j yu(x ydy, if j Z, j u = ϕ(2 j Du =2 R jd h(2j yu(x ydy, d if j Z, Ṡ j u = j u j j Remark Let us point that all the above operators j and S j maps L p into L p with norms which do not depend on j This fact will be used all along this book Now let us have a look of the case when we may write Id = j This is described by the following proposition j or Id = j Proposition 62 Let u be in S (R d Then, we have, in the sense of the convergence in the space S (R d, u = lim j S j u Proof of Proposition 62 Let f S(R d We have <u S j u, f >=< u, f S j f > Thus it is enough to prove that in the space S(R d, we have f = lim j S j f We shall use the family of semi norms k,s of S ined by j f k,s = sup ( + ξ k α f(ξ α k ξ R d Thanks to Leibnitz formula, we have { f S j f k,s sup ( + ξ k( χ(2 j ξ α f(ξ α k ξ R d + 0<βα As χ equals to near the origin it turns out that The proposition is proved f S j f k,s C α 2 j f k+,s 66 C β α2 q β ( β χ(2 j ξ α β f(ξ }

The following proposition tells us that the condition of convergence in S is somehow weak for series, the Fourier transform of which is supported in dyadic rings Proposition 63 Let (u j j N be a sequence of bounded functions such that the Fourier transform of u j is supported in 2 j C where C is a given ring Let us assume that Then the series (u j j N is convergent in S u j L C2 jn Proof of Proposition 63 Let us use the relation (62 After rescaling it can be written as u j =2 jk 2 jd g α (2 j α u j α =k Then for any test function φ in S, let us write that u j,φ = 2 jk u j, 2 jd g α (2 j α φ (65 α =k C2 jk α =k 2 jn α φ L Let us choose k > N Then ( u j,φ j N is a convergent series, the sum of which is less than C φ M,S for some integer M Thus the formula ines a tempered distribution u, φ = lim j j j j u, φ Définition 6 Let us denote by S h the space of tempered distribution such that lim Ṡ j u =0 in S j Examples The space S h is exactly the space of tempered distributions for which we may write u = j Z j u If a tempered distribution u is such that its Fourier transform û is locally integrable near 0, then u belongs to S h If u is a tempered distribution such that for some function θ in D(R d with value near the origin, we have θ(du in L p for some p [, + [, then u belongs to S h In particular, when p if finite, any inhomogeneous Besov space B s is included in S h A non zero constant function u does not belong to S h because Ṡju = u for any j in Z Thus, if u belongs to L p + L q with finite p and q, then u is in S h Remarks The fact that u belongs or not to S h is an information about low frequencies The space S h is not a closed subspace of S for the topology of weak convergence It is an exercice left to the reader to prove that u belongs to S h if and only if, for any θ in D(R d with value near the origin, we have lim θ(λdu = 0 in λ S 67

62 Homogeneous Besov spaces Définition 62 Let u be a tempered distribution, s a real number, and p and r two real numbers greater than Then we state u Ḃs = ( j Z 2 rjs j u r L p r For the semi-norms we have ined, we can prove the following inequalities Théorème 62 A constant C exists such that r r 2 u Ḃs 2 u Ḃs and p p 2 u Ḃs p2,r C u Ḃ s d ( p Moreover we have the following interpolation inequalities : For any θ [0, ], u θ Ḃ s For any θ in ]0, [, u Ḃθs +( θs 2 p, u Ḃθs +( θs 2 p,r u θ Ḃ s 2 p 2 C ( s 2 s θ + u θ θ Ḃ s u θ Ḃ s 2 Proof of Theorem 62 In order to prove this result, we again apply Lemma 6 which tells us that ( j u L p 2 C2 jd p p 2 j u L p Considering that l r (Z l r 2 (Z, the first part of the theorem is proved Now let us estimate in a different way low frequencies and high frequencies More precisely, let us write u θs B +( θs 2 = 2 j(θs +( θs 2 j u L p + 2 j(θs +( θs 2 j u L p p, jn By inition of the Besov norms, we have Thus we infer that u B θs +( θs 2 p, Choosing N such that implies the theorem j>n 2 j(θs +( θs 2 j u L p 2 j( θ(s 2 s u B s 2 j(θs +( θs 2 j u L p 2 qθ(s 2 s u B s 2 u B s u B s u B s 2 u B s jn 2 j( θ(s 2 s + u B s 2 2 N( θ(s 2 s 2 ( θ(s 2 s + u B s 2 2 N(s 2 s < 2 u B s 2 u B s 68 and 2 qθ(s 2 s j>n 2 Nθ(s 2 s 2 θ(s 2 s

There are two important facts to point out The first one is about the homogeneity If u is a tempered distribution, then let us consider for any integer N, the tempered distribution u N ine by u N = u(2 N Then we have Proposition 62 If u Ḃs is finite, so it is for u N and we have u N Ḃs =2 N(s d p u Ḃs To proof this, we have to go back to the inition of the operator j j u N (x = 2 jd = 2 jd h(2 j (x yu N (ydy h(2 j (x yu(2 N ydy By the change of variables z =2 N y, we get that j u N (x = 2 (q Nd h(2 j N (2 N x zu(zdz = ( j N u(2 N x So it turns out that we deduce from this that j u N L p =2 N d p j N u L p 2 js j u N L p =2 N(s d p 2 (q Ns j N u L p And the proposition follows immediately by summation The second one is that Ḃs is actually a semi-norm in the sense that, if u is a polynomial, then the support of its Fourier transform is exactly the origin Then, for any integer j, we have j u = 0, and so u Ḃs = 0 Let us state the initon of the homogeneous Besov spaces Définition 622 Let s be a real number and (p, r be in [, ] 2 The space Ḃs is the set of the tempered distribution u in S h such that u Ḃ is finite s Proposition 622 The space (Ḃs, Ḃs is a normed space It is obvious that Ḃs is a semi-norm Let us assume that for some u in S h, u Ḃ s = 0 This implies that Supp û {0} and thus that, for any j Z, Ṡ j u = u As u belongs to S h, this implies that u = 0 Let us notice that there is no monotony property with respect to s for homogeneous Besov spaces The reason why is that homogeneous Besov spaces carry on informations about both low and high frequencies 69

Proposition 623 A constant C exists which satisfies the following properties Let (s, p, r be in (R \{0} [, ] 2 and u a distribution in S h This distribution u belongs to Ḃs if and only if (2 js Ṡju L p j N l r Moreover, we have C s + u Ḃs Proof of Proposition 623 Let us write that ( (2 js l Ṡju L p j C + u Ḃs r s 2 js j u L p 2 js ( S j+ u L p + S j u L p 2 s 2 (q+s S j+ u L p +2 js S j u L p This proves the inequality on the left But now we can write that 2 js S j u L p 2 js j u L p As s is negative, we get the result j j j j 2 (q q s 2 j s j u L p Théorème 622 If s< d d p, then (Ḃs, Ḃs is a Banach space For any p, the space Ḃ p p, is also a Banach space First let us prove that those spaces are continuously embedded in S Thanks to Lemma 6, the series ( j u j Z is convergent in L when u belongs to Ḃ p p, As u belongs to S h, this implies that u belongs to L Using Proposition 62, we have that u L C u Ḃ0 d, C u Ḃ p (66 p, d Following (620 and using Propositions 62 and 63 and the fact that s < d/p, we have, for large enough N and M and non negative j, j u, φ 2 j u B N, φ M,S 2 j u B s d φ p M,S, For negative j, let us write that, for large enough M, C s 2 j u Ḃs d φ p M,S,, j u, φ 2 j( d s p u Ḃs p d φ L, As u belongs to S h, we have, for large enough M, C2 j( d p s u Ḃs φ M,S u, φ C s u Ḃs φ M,S (67 70

Let us consider a Cauchy sequence (u n n N in Ḃs for (s, p, r satisfying the hypothesis of the theorem Using (66 or (67, this implies that a tempered distribution u exists such that the sequence (u n n N converges to u in S The main point of the proof consists in proving that this distribution u belongs to S h If s < d/p, for any n, u n belongs to S h Thanks to (67, we have j Z, n N, Ṡju n,φ C s 2 j( d s p sup u n Ḃs φ n M,S As the sequence (u n n N tends to u in S, we have j Z, Ṡju, φ C s 2 j( d s p sup u n Ḃs φ n M,S Thus u belongs to S h d p The case when u belongs to Ḃp, is a little bit different As (u n n N is d a Cauchy sequence in Ḃ p p, Ḃ0,, for any positive ε, an integer n 0 exists such that j Z, n n 0, k u n L ε 2 + k u n0 L kj kj Let us choose j 0 small enough such that j j 0, kj k u n0 L ε 2 As u n belongs to S h, we have j j0, n n0, Ṡjun L ε We now that the sequence (u n n N tends to u in L This implies that j j 0, Ṡju L ε This proves that u belongs to S h By inition of the norm of Bs the sequence ( j u (n n N is a Cauchy one in L p for any j Thus an element u j of L p exists such that ( j u (n n N converges to u j in L p As the sequence (u (n n N converges to u in S we have j u = u j Let us ine a (n j =2 js j u (n L p and a j =2 js j u L p For any j, we have lim n a(n j = a j As (a (n j n N is a bounded sequence of l r (Z, a =(a j j Z belongs to l r (Z and thus u is in B s Moreover the fact that (u (n n N is a Cauchy sequence in B s implies that, for any positive ε, an integer n 0 exists such that, for any n n 0 and any m, a (n+m a (n l r (Z ε As (a (n tends weakly to a in l r (Z, we get, passing to the (weaklimit when m tends to infinity in the above inequality that The proposition is proved u (n u B s = a a (n l r (Z ε Exercice 62 Let φ a function of S(R d and ω in S d Then, if s ]0, d/p[, then (x ω lim ei ε φ =0 in the space Ḃp, s ε 0 7

63 Inhomogeneous Besov spaces 63 Definition and examples Définition 63 Let s be a real number, and p and r two reals numbers greater than The Besov spaces B s is the space of all tempered distributions so that u B s = (2 js l j u L p j Z < + r (Z The following proposition (the proof of which is straighforward and omitted describes the relations between homogeneous and inhomogeneous spaces Proposition 63 Let s be a negative number Then Ḃs is a subset of B s and a constant C (independent of s exists so that, for any u belonging to Ḃs, we have u B s C s u Ḃ s Let s be a positive number Then B s is a subset of Ḃ s when p is finite, B,r s S h is a subset of Ḃ,r s and a constant C exists (independent of s so that, for any u belonging to B, s we have u Ḃs C s u B s Lemme 63 If r is finite, then for any u in B s, we have lim S ju u B s j =0 The proof of this proposition is an easy exercise left to the reader Let us give the first example for Besov space, the Sobolev spaces H s We have the following result Théorème 63 The two spaces H s and B2,2 s are equal and the two norms satisfies C s + u B s 2,2 u H s C s + u B s 2,2 As the support of the Fourier transform of j u is included in the ring 2 j C, it is clear, as j 0, that a constant C exists such that, for any real s and any u such that û belongs to L 2 loc, C s + 2js j u L 2 j u H s C s + 2 js j u L 2 (68 Using Identity (68, we get 3 u 2 H s χ 2 (ξ( + ξ 2 s û(ξ 2 dξ + ϕ 2 (2 j ξ( + ξ 2 s û(ξ 2 dξ u 2 H s j 0 which proves the theorem Proposition 632 The space B 0 p, is continuously embedded in Lp and the space L p is continuously embedded in B 0 The proof is trivial The first inclusion comes from the fact that the series ( j u j Z is convergent in L p The second one comes from the fact that for any p, we have j u L p C u L p 72

632 Basic properties The first point to look at is the invariance with respect to the choice of the dyadic partition of unity choosen to ine the space Most of the properties of the Besov spaces are based on the following lemma Lemme 632 Let C be a ring in R d ; let s be a real number and p and r two real numbers greater than Let (u j j N be a sequence of smooth functions such that Supp û j 2 j C and (2 js l u j L p j N < + r Then we have u = j N u j B s and u B s C s (2 js u j L p j N l r This immediately implies the following corollary Corollaire 63 The space B s does not depend on the choice of the functions χ and ϕ used in the Definition 63 In order to prove the lemma, let us first observe that (u j j N is a convergente series in S Indeed using Lemma 6, we get that u j L C2 j( d s p Proposition 63 implies that (u j j N is a convergente series in S Then, let us study j u As C and C are two rings, an integer N 0 exists so that j j N 0 = 2 j C 2 j C = φ Here C is the ring ined in the Proposition 6 Now, it is clear that Now, we can write that So, we obtain that We deduce from this that j j N 0 = F( j u j = 0 = j u j =0 j u L p = j u j L p j j <N 0 C u j L p j j <N 0 2 j s j u L p C C j j j N 0 j j j N 0 2 j s u j L p 2 js u j L p 2 j s j u L p (c k k Z (d l l Z with c k = [ N0,N 0 ](k and d l = N (l2 ls u l L p The classical property of convolution between l (Z and l r (Z gives that which proves the lemma ( u B s C j N 73 2 rqs u j r L p r,

The following theorem is the equivalent of Sobolev embedding (see Theorem?? page?? Théorème 632 Let p p 2 and r( r 2 Then for any real number s the space Bp s,r is continuously embedded in B s d p p 2 p 2,r 2 In order to prove this result, we again apply Lemma 6 which tells us that S 0 u L p 2 C u L p and ( j u L p 2 C2 jd p p 2 j u L p Considering that l r (Z l r 2 (Z, the theorem is proved Proposition 633 The space B s is continuously embedded in S By inition B s is a subspace of S Thus we have only to proof of a constante C and an integer M exists such that for any test function φ in S we have u, φ C u B s φ M,S (69 Using the above Theorem 632 and the relation (65, we can write, if N is a large enough integer, j u, φ = 2 q(n+ j u, 2 jd g α (2 j α φ α =N+ 2 j u B N sup α φ, L (620 α =N+ C2 j u B s φ M,S Now Proposition 62 implies the Inequality (69 Proposition 634 The space B s equipped with the norm B s ined above is a Banach space The end of the proof done for the homogeneous case can be repeated words for words Proposition 635 A constant C exists which satisfies the following properties Let (s, p, r be in (R \{0} [, ] 2 and u a tempered distribution This distribution u belongs to B s if and only if (2 js S j u L p j N l r Moreover, we have Let us write that C s + u B s ( (2 js l S j u L p j C + u r B s s 2 js j u L p 2 js ( S j+ u L p + S j u L p 2 s 2 (q+s S j+ u L p +2 js S j u L p This proves the inequality on the left But now we can write that 2 js S j u L p 2 js j u L p As s is negative, we get the result j j j j 74 2 (q q s 2 j s j u L p

In order to conclude this paragraph, let us state interpolation inequalities Théorème 633 A constant C exists which satisfies the following properties If s and s 2 are two real numbers such that s <s 2, if θ ]0, [, if r is in [, ], then we have u B θs +( θs 2 u B θs +( θs 2 p, u θ B s u θ B s 2 and C s 2 s ( θ + θ u θ B s u θ B s 2 As very often in this book, we shall estimate in a different way low frequencies and high frequencies More precisely, let us write u B θs +( θs 2 p, = 2 j(θs +( θs 2 j u L p + 2 j(θs +( θs 2 j u L p jn By inition of the Besov norms, we have j>n 2 j(θs +( θs 2 j u L p 2 j( θ(s 2 s u B s 2 j(θs +( θs 2 j u L p 2 qθ(s 2 s u B s 2 and Thus we infer that u B θs +( θs 2 p, Choosing N such that implies the theorem u B s u B s u B s 2 u B s jn 2 j( θ(s 2 s + u B s 2 2 N( θ(s 2 s 2 ( θ(s 2 s + u B s 2 2 N(s 2 s < 2 u B s 2 u B s 2 qθ(s 2 s j>n 2 Nθ(s 2 s 2 θ(s 2 s 64 The case of Hölder type spaces Another relevant example of Besov spaces are Hölder spaces ined page?? We have the following result Proposition 64 For any k in N, a constant C k exists such that for any ρ ]0, ] and any function u belonging to C k,ρ, we have sup 2 j(k+ρ j u L C k u C k,ρ j To prove this, let us first observe that, when j =, we have that S 0 u L C u L When j is non negative, let us write the operator j of the convolution form j u(x = 2 jd h(2 j (x yu(ydy 75

The fact that the function ϕ is identically 0 near the origin implies that for any α N d, x α h(xdx =0 Thus we can write j u(x = 2 jd h(2 j (x y (u(y u(x Taylor formula at order k implies that u(y u(x 0< α k = α! α u(x(y x α 0 k( t k α =k 0< α k (y x α The fact that the functions α u belong to the space C ρ implies that u(y u(x 0< α k α! α! α u(x(y x α dy (62 ( α u(x + t(y x α u(x dt α! α u(x(y x α Ck y x k+ρ u C k,ρ Then it comes from (62 that j u(x C k u C k,ρ2 jd x y k+ρ h(2 j (x y dy and the proposition is proved Let us study the reciproque of Proposition 64 We have the following proposition Proposition 642 Let r be in R + \ N and let u be in B, r Then u is in C k,ρ with k =[r] and ρ = r [r] Moreover we have ( u C k,ρ C r u B r, with C r = C k ρ + ρ To star with, let us observe that, thanks to Lemma 6, we have, for any α the length of which is less than r, j α u L C k+ 2 q(r α u B r, Thus the series ( j α u j N is convergent in the space L and we have α u L C k+ ρ u B r, (622 Now let us study the derivative of order k We can write, for a positive integer N which will be chosen later on, that α u(x α u(y N j=0 By Taylor inequality, we have that α j u(x α j u(y + α j u(x α j u(y j N α j u(x α j u(y C x y sup β j u L β =k+ 76

Using Lemma 6, we get α j u(x α j u(y C k u B r, x y 2 q(ρ (623 The high frequency terms are estimated very roughly writing Then it comes from (623 that α j u(x α j u(y C k 2 qρ u B r, ( α u(x α N u(y C k u B r, 2 q(ρ x y + j=0 Thanks to (622, we may assume that x y Choosing N =[ log 2 x y ]+, in the above inequality, we conclude the proof of the proposition Propositions 64 and 642 together imply the following theorem j N+ 2 qρ Théorème 64 Let r be in R + \ N Then the spaces B, r and C [r],r [r] are equal and we have ( C [r] u B, r u C [r],r [r] C [r] r [r] + u B r (r [r], Proposition 642 turns out to be false when r is an integer Théorème 642 The space B, is not included in the space C 0, of bounded Lipschitz functions 65 Paradifferential calculus In this section, we are going to study the way how the product acts on Besov spaces Of course, we shall use the dyadic decomposition constructed in the Section 62 65 Bony s decomposition Let us consider two tempered distributions u and v, we write u = j j u and v = j j v Formally, the product can be written as uv = j,j j u j v Now, let us introduce Bony s decomposition 77

Définition 65 We shall designate paraproduct (respectively homogeneous paraproduct of v by u and shall denote by T u v (respectively T u v the following bilinear operator : T u v = j S j u j v respectively T u v = j Ṡ j u j v We shall designate remainder (respectively homogeneous remainder of u and v and shall denote by R(u, v (respectively Ṙ(u, v the following bilinear operator : R(u, v = j j j u j v respectively Ṙ(u, v = j j j u j v Just by looking at the inition, it is clear that uv = T u v + T v u + R(u, v = T u v + T v u + Ṙ(u, v (624 The way how paraproduct and remainder act on Besov spaces is described by the following theorem Théorème 65 A constant C exists which satisfies the following inequalities for any couple of real numbers (s, t with t negative and for any (p, r,r 2 in [, + ] 3 T L(L B s ;B s C s +, T L(L Ḃs ;Ḃs C s +, T L(B t,r B s 2 ;B s+t t 2 C s+t + T L( Ḃ,r t Ḃs ;Ḃs+t 2 2 C s+t + with t = min {, + } r r 2 r 2, To prove this theorem, we are going to use the proposition 6 about the contruction of the dyadic partition of unity From the assertion (67, the Fourier transform of S j u j v and also of Ṡ j u j v is supported in 2 j C So, the only thing that we have to do is to estimate The lemma 6 says that, for any integer j, S j u j v L p and Ṡj u j v L p S j u L C u L and Ṡj u L C u L Propositions 635 and 623 tells us that S j u L C t c j,r 2 qt u B t,r and Ṡj u L C t c j,r 2 qt u Ḃt,r (625 where (c j,r j Z denotes an element of the unit sphere of l r (Z Using Lemma 632, the estimates about paraproduct are proved 78

Now we shall study the behaviour of operators R Here we have to consider terms of the type j u j v The Fourier transform of such terms is not supported in ring but in balls of the type 2 j B Thus to prove that remainder terms belong to some Besov spaces, we need the following lemma Lemme 65 Let B be a ball of R d, s a positive real number and (p, r in [, ] 2 Let (u j j N be a sequence of smooth functions such that Then we have Supp û j 2 j B and (2 js u j L p j N l r < + u = j N u j B s and u B s C s (2 js u j L p j N l r We have for any j u j L p C2 qs As s is positive, (u j j N is a convergent series in L p We then study j u j As C is a ring (ined in the proposition 6 and B is a ball, an integer N exists so that So it is clear that q q + N = 2 j C 2 j B = φ q q + N = F( j u j = 0 = j u j =0 Now, we write that So, we get that j u L p = 2 j s j u L p j u j j q N L p j u j L p j q N u j L p j q N 2 j s u j L p j q N 2 (q qs 2 js u j L p j q N So, we deduce from this that 2 j s j u L p (c k (d l with c k = [ N,+ [(k2 ks and d l =2 ls u l L p So, the lemma is proved 79

Théorème 652 A constant C exists which satisfies the following inequalities For any (s,s 2, any (p,p 2,p and any (r,r 2 such that we have that s + s 2 > 0, p p + p 2 R L(B s p,r B s 2 p 2,r 2 ;B σ 2 C s +s 2 + s + s 2 and and r + r 2 = r, Ṙ L(Ḃs p,r Ḃs 2 p 2,r ;Ḃσ 2 2 C s +s 2 + with σ 2 = s + s 2 d s + s 2 And for any (s,s 2, any (p,p 2,p and any (r,r 2 so that we have that s + s 2 0, R L(B s p,r B s 2 p 2,r 2 ;B σ 2 C s +s 2 + Ṙ L(Ḃs p,r Ḃs 2 p 2,r 2 ;Ḃσ 2 C s +s 2 + p p + p 2 and and again with r + r 2 =, ( + p p 2 p ( σ 2 = s + s 2 d + p p 2 p Now, let us study the remainder operator in the homogeneous case (the inhomogeneous one can be treated exactly along the same lines By inition of the remainder operator, R(u, v = j R j with R j = j i u j v i= By inition of j, the support of the Fourier transform of R j is included in 2 j B(0, 24 So, by construction of the dyadic partition of unity, it exists an integer N 0 so that From this, we deduce that Let us ine p 2 in [, + ] by j j N 0 j R j =0 (626 j R(u, v = j R j j q N 0 p 2 = p + p 2 By hypothesis, p 2 is smaller than p So, using the lemma 6 of localization, we may write ( j R j L p 2 j d p 2 p R j L p 2 ( 2 j d p 2 p ( 2 j d p 2 p C2 j ( s s 2 +d j i u j v L p 2 i= j i u L p j v L p 2 i= ( p 2 p 2 (j j (s +s 2 2 (j is j i u L p 2 js 2 j v L p 2 i= 80

Let us ine (r q q Z by ( r j =2 j s s 2 +d ( p 2 p j R(u, v L p Using the assertion (626, we have that r j C(b ( b (2 j with b ( j = 2 q(s +s 2 N N0 (q and (627 b (2 j = 2 (q is j i u L p 2 js 2 j v L p 2 (628 i= Let us assume that s + s 2 is strictly positive By hypothesis, the sequence (b (2 j j Z belongs to l r (Z Moreover, the sequence (b ( j j Z belongs to l (Z and (b ( j j Z L Cs +s 2 s + s 2 So the theorem is proved in this case If we assume that then, if s + s 2 is positive, we have that r + r 2 =, (b ( j j Z l (Z C and (b (2 j j Z l (Z C u B s p,r v B s 2 p 2,r 2 So the theorem is proved Now, we are going to infer from this theorem the following corollary Corollaire 65 A constant C exists so that it satisfies the following properties For any positive real number s, we have uv B s Cs+ ( u L v B s s + u B s v L, uv Ḃs Cs+ ( u L v Ḃs + u Ḃs v s L Moreover, for any (s,s 2, any p 2 and any r 2 such that s + s 2 > 0 and any s < d p, we have that with uv B s p2 Cs +s 2 + ( u s,r 2 s + s v B p 2, B s 2 + u s p 2,r 2 B 2 v s p 2,r 2 B, p, uv Ḃs p2,r 2 Cs +s 2 + s + s 2 ( u Ḃs p, v Ḃ s 2 p 2,r 2 + u Ḃs 2 p 2,r 2 v Ḃs p, s = s + s 2 d p Moreover, for any (s,s 2, any p 2 and any (r,r 2 such that s + s 2 0,s < d p 8 and r + r 2 =,

we have that uv B s C s +s 2 + ( u B s p,r v B s 2 p 2,r 2 + u B s 2 p 2,r 2 v B s p,r, uv Ḃs C s +s 2 + ( u Ḃs p,r v Ḃs 2 p 2,r 2 + u Ḃs 2 p 2,r 2 v Ḃs p,r Moreover, for any (s,s 2, any (p,p 2,p and any (r,r 2 such that s + s 2 > 0,s j < d p j and p max{p,p 2 }, we have that uv B s Cs +s 2 + s + s 2 u B s p,r v B s 2 p 2,r 2 and uv Ḃs Cs +s 2+ s + s 2 u Ḃs p,r v Ḃs 2 p 2,r 2 with ( s = s + s 2 d + p p 2 p and r = min {, r + r 2 } Moreover, for any (s,s 2, any (p,p 2,p and any (r,r 2 such that s + s 2 0,s j < d p j,p max{p,p 2 } and r + r 2 =, we have that uv B s C u B s p,r v B s 2 p 2,r 2 and uv Ḃs C u Ḃs p,r v Ḃs 2 p 2,r 2 with ( s = s + s 2 d + p p 2 p and r = max{r,r 2 } The proof is nothing but the systematic use of Bony s decomposition and the application of Theorems 65 and 652 652 Paralinearization theorem In this paragraph we shall study the action of smooth functions on the space B s More precisely, if f is a smooth function vanishing at 0, and u a function of B s, does f u belongs to B s? The answer is given by the following theorem Théorème 653 Let f be a smooth function and s a positive real number and (p, r in [, ] 2 If u belongs to B s L, then f u belongs to B s and we have f u B s C(s, f, u L u B s Before proving this theorem, let us notice that if s > d/p or if s = d/p and r =, then the space B s is included into L Thus in those cases, the space B s is stable under the action of f by composition This is in particular the case for the Sobolev space H s with s > d/2 82

Let us prove the theorem We shall use the argument of the so called telescopic series As the sequence (S j u j N converges to u in L p and f(0 = 0, then we have f(u = j f j with f j = f(s j+ u f(s j u (629 Using Taylor formula at order, we get f j = m j j u with m j = 0 f (S j u + t j udt (630 At this point of the proof, let us point out that there is no hope for the Fourier transform of f j to be compactly supported Thus Lemma 65 is not efficient in this case We shall prove the following improvement of this lemma Lemme 652 Let s be a positive real number and (p, r in [, ] 2 A constant C s exists such that if (u j j N is a sequence of smooth functions which satisfies ( 2 j(s α α u j L p j lr, sup α [s]+ then we have u = ( u j B s and u B s C s sup 2 j(s α α u j L p j N α [s]+ j l r As s is positive, the series (u q q N is convergent in L p Let us denote its sum by u and let us write that j u = j u j + j u j j j j >q Using that j u j L p u j L p, we get that 2 js j u j 2 js u j L p j >q j >q L p j >q 2 (q qs 2 j s u j L p (63 Then using Lemma 6, we write that Then we write 2 js j u j L j p j j u j L p C2 q([s]+ sup α =[s]+ 2 (j j([s]+ s sup j j α =[s]+ This inequality together with (63 implies that This proves the lemma 2 js j u L p (a b j with a j b j α u j L p 2 j (s α α u j L p = N (j2 qs + N (j2 j([s]+ s and = 2 js u j L p + sup 2 j(s α α u j L p α =[s]+ 83

Let us go back to the proof of Theorem 653 Let us admit for a while that α N d, α m j L C α (f, u L 2 j α (632 Thus using Leibnitz formula and Lemma 6, we get that α f j L p Cβ α 2 j β C β (f, u L 2 j( α β j u L p βα Thus we get that α f j L p C α (f, u L 2 j α j u L p c j C α (f, u L 2 j(s α u B s (633 with (c j l r = We apply Lemma 652 and the theorem is proved provided we check (632 In order to do it, let us recall Faa-di-Bruno s formula α g(a = From this formula, we infer that α m j = α + +α p=α α j Using Lemma 6, we get that ( p 0 α + +α p=α α j k= α m j L C α (f This proves (632 and thus the theorem ( p k= α k a g (p (a α k (S j u + t j u f (p+ (S j u + t j udt α + +α p=α α j C α (f, u L 2 j α ( p 0 k= 2 j α k u L A slightly better estimate than estimate (632 can be obtained For α k, we have and thus instead of (632, we have α k (S j u + tu j L C2 j α k u B, α N d, α m j L C α (f, u B, 2j α (634 This leads us to the following theorem which is a small modification of the previous one Théorème 654 Let f be a function of Cb (R the space of smooth bounded functions and s a positive real number and (p, r in [, ] 2 If u belongs to B s and the first derivatives of u belongs to B,, then f u belongs to B s and we have f u B s C(s, f, u B, u B s d p Let us notice that if u belongs to the space B, then the first order derivatice of u belongs d to B, p Thus the space B is stable under the action of functions of Cb This is in particular the case for the Sobolev space H d 2 = B d 2 2,2 84 by composition

When the function u is more regular, we are able to precize the above theorem In fact, up to an error term which will be more regular than u, we will write f u as a paradroduct of u More precisely we shall prove the following theorem Théorème 655 Let s and ρ be two positive real numbers and f a smooth function Let us assume that ρ is not an integer Let p, r and r 2 be in [, ] such that r 2 r Then for any function u in B s B,r ρ 2 we have f u T f uu B s+ρ 2 C(f, u L u B ρ,r2 u B s with r 2 = r + r 2 The proof of this theorem follows the same lines of the proof of Theorem 653 We write again that f(u = f j with f j = f(s j+ u f(s j u j Using Taylor formula at order 2, we get f j = f (S j u j u + M j ( j u 2 with M j = Following exactly the same lines as for proving (632, we get 0 ( tf (S j u + t j udt α N d, α M j L C α (f, u L 2 j α (635 Using Leibnitz formula, we can write that α (M j ( j u 2 = CαC β γ β α β M j β γ j u γ j u γβα Using Lemma 6 and inequality (635, we get α β M j β γ j u γ j u L p C α (f, u L 2 j α j u L j u L p Thus by inition of the Besov spaces, we get that α M j ( j u 2 L p C α (f, u L c j,α 2 j(s+ρ α u B ρ,r2 u B s (636 with (c j l r 2 = Now let us study the term f (S j u j u It does not look exactly like a paraproduct Let us state µ j = f (S j u S j (f u Obviously we have Let us admit for a while that f j = S j (f u j u + µ j j u + M j ( j u 2 α µ j L c j 2 q(s+ρ α C α (f, u L u B ρ,r2 (637 with (c j l r 2 = Then using (636, we have α (f j S j (f u j u L C α (f, u L c j 2 j(s+ρ α u B ρ,r2 u B s with (c j l r 2 = and the theorem is proved provided of course that we prove the Inequality (637 85

Let us first investigate the case when α <ρ Let us write that µ j = µ ( j + µ (2 with µ ( j Thanks to Theorem 653, we have Thus, we can write that = f (S j u f (u and µ (2 = f (u S j (f (u α f (u B ρ α,r 2 and α f (u B ρ α,r 2 C α (f, u L u B ρ,r2 α (Id S j f (u L j q j α f (u L C α (f, u L u B ρ,r2 c j 2 j (ρ α j j c j 2 j(ρ α C α (f, u L u B ρ,r2 with (c j l r 2 = Using a gain the argument of telescopic series (see (629, we have f (u f (S j (u = j j f j with f j = f (S j +u f (S j u Applying Inequality (633, we get α f j L c j 2 j (ρ α C α (f, u L u B ρ,r2 with (c j l r 2 = (638 Then, by summation, we infer that, when α <ρ, α (f (u f (S j u L c j 2 j(ρ α C α (f, u L u B ρ,r2 with (c j l r 2 = Thus (637 is proved when α <ρ The case when α >ρ is treated in a different way As α f (u belongs to B ρ α,r 2, we have, using Proposition 635 and Theorem 653, α S j f (u L c j 2 j(ρ α C α (f, u L u B ρ,r2 with (c j l r 2 = In order to estimate α f (S j u, we use again the argument of the telescopic series Let us write that f (S j (u = j j f j with f j = f (S j +u f (S j u The using (637, we get, as α >ρ, α f (S j u L α f j L j j C α (f, u L u B ρ,r2 c j 2 j (ρ α j q C α (f, u L u B ρ,r2 c j 2 j(ρ α with (c j l r 2 = Inequality (637 and thus Theorem 655 is proved 86

653 Application of some embeddings theorems The purpose if this subsection is to give a short non linear proof to the following wellknown (and quite useful theorem Théorème 656 For any p [2, [, the space Ḃ0 p,2 is continuously included in Lp and the space L p is continuously included in Ḃ0 p,2 Proof of Theorem 656 Let us state F p (x = x p Let us notice that, as p 2, this function is a C 2 function Thus in the spirit of Formulas (629 and (630, let us write that u p L p = F p (udx = j j u, m j with m j (x = 0 F p (Ṡj u(x+t j u(x dt Using Plancherel formula, and stating j the convolution operator by the inverse Fourier transform of θ(2 j where θ is in D(R d \{0} with value near the support of ϕ, we can write that j u, m j = j u, j m j Then, by using Lemma 6, we infer that The chain rule and the Hölder inequality imply that j m j L p C2 j sup l m j L p (639 ld l m j L p l (Ṡju + t j uf p (Ṡj + t j u dt L p 0 0 l (Ṡju + t j u L p F p (Ṡju + t j u L p p 2 dt As F p (x =p(p x p 2, we get immediately that, for all t [0, ], F p (Ṡju + t j u L p p 2 Using Lemma 6, we infer that F p (S j u + t j u L p p 2 p(p Ṡju + t j u p 2 L p Cp p(p u p 2 L p (640 Now, by inition of Ṡj, we have, using Lemma 6 and the Young s inequality on series l (Ṡju + t j u L p kj l k u L p 2 j kj 2 k j k u L p Cc j 2 j u Ḃ0 p,2 with c 2 j = j With (639 and (640, we deduce that j m j L p Cp p(p c j u p 2 L p u Ḃ 0 p,2 87 with c 2 j = j

As we have u p L p = j u, m j, we infer that j u 2 L p Cp p(p u Ḃ0 c j j L p p,2 This concludes the proof of the theorem once observed that the embeddings about L p follows by duality j 88