Littlewood-Paley theory - IMJ-PRGisabelle.gallagher/littlewoodpaley.pdfLittlewood-Paley theory Introduction The purpose of this chapter is the introduction by this theory which is

Chapitre 6

Littlewood-Paley theory

Introduction

The purpose of this chapter is the introduction by this theory which is nothing but aprecise 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 oflocalization appears naturally.

6.1 Localization in frequency space

The very basic idea of this theory consists in a localization procedure in the frequencyspace. The interest of this method is that the derivatives (or more generally the Fourier mul-tipliers) act in a very special way on distributions the Fourier transform of which is supportedin a ball or a ring. More precisely, we have the following lemma.

6.1.1 Bernstein inequalities

Lemme 6.1.1 (of localization) Let C be a ring, B a ball. A constant C exists so that, forany non negative integer k, any smooth homogeneous function ! of degree m, any couple ofreal (a, b) so that b ! a ! 1 and any function u of La, we have

Supp !u " "B # sup!=k

$#!u$Lb % Ck+1"k+d( 1a!

1b )$u$La ;

Supp !u " "C # C!k!1"k$u$La % sup!=k

$#!u$La % Ck+1"k$u$La ;

Proof of Lemma 6.1.1 Using a dilation of size ", we can assume all along the proofthat " = 1. Let $ be a function of D(Rd) the value of which is 1 near B. As !u(%) = $(%)!u(%),we can write, if g denotes the inverse fourier transform of $,

#!u = #!g & u.

Applying Young inequalities the result follows through

$#!g$Lc % $#!g$L! + $#!g$L1

% 2$(1 + | · |2)d#!g$L!

% 2$(Id&!)d((·)!$)$L1

% Ck+1.

63

To prove the second assertion, let us consider a function "$ which belongs to D(Rd \{0}) thevalue of which is identically 1 near the ring C. Using the algebraic Using the following algebraicidentity

|%|2k =#

1"j1,···,jk"d

%2j1 · · · %

2jk

=#

|!|=k

(i%)!(&i%)!, (6.1)

and stating g!def= F!1(i%j)!|%|!2k "$(%), we can write, as !u = "$!u that

!u =#

|!|=k

(&i%)!!g!!u,

which implies thatu =

#

|!|=k

g! & #!u (6.2)

and then the result. This proves the whole lemma.

6.1.2 Dyadic partition of unity

Now, let us define a dyadic partition of unity. We shall use it all along this text.

Proposition 6.1.1 Let us define by C the ring of center 0, of small radius 3/4 and greatradius 8/3. It exists two radial functions ' and ( the values of which are in the interval [0, 1],belonging respectively to D(B(0, 4/3)) and to D(C) such that

'% ( Rd , '(%) +#

j#0

((2!j%) = 1, (6.3)

'% ( Rd \{0} ,#

j$Z

((2!j%) = 1, (6.4)

|j & j%| ! 2 # Supp ((2!j ·) ) Supp ((2!j" ·) = *, (6.5)

q ! 1 # Supp ' ) Supp ((2!j ·) = *, (6.6)

If "C = B(0, 2/3) + C, then "C is a ring and we have

|j & j%| ! 5 # 2j" "C ) 2jC = *, (6.7)

'% ( Rd ,13% '2(%) +

#

j#0

(2(2!j%) % 1, (6.8)

'% ( Rd \{0} ,12%

#

j$Z

(2(2!j%) % 1. (6.9)

Proof of Proposition 6.1.1 Let us choose ) in the interval ]1, 4/3[ let us denote by C% thering of small radius )!1 and big radius 2). Let us choose a smooth function *, radial withvalue in [0, 1], supported in C with value 1 in the neighbourhood of C%. The important pointis the following. For any couple of integers (p, q) we have

|j & j%| ! 2 # 2jC ) 2j"C = *. (6.10)

64

Let us suppose that 2j"C ) 2jC += * and that p ! q. It turns out that 2j" , 3/4 % 4, 2j+1/3,which implies that j% & j % 1. Now let us state

S(%) =#

j$Z

*(2!j%).

Thanks to (6.10), this sum is locally finite on the space Rd \{0}. Thus the function S issmooth on this space. As ) is greater than 1,

$

j$Z

2jC% = Rd \{0}.

As the function * is non negative and has value 1 near C%, it comes from the above coveringproperty that the above function is positive. Then let us state

( =*

S· (6.11)

Let us check that ( fits. It is obvious that ( ( D(C). The function 1&#

j#0

((2!j%) is smooth

thanks to (6.10). As the support of * is included in C, we have

|%| ! 43#

#

j#0

((2!j%) = 1. (6.12)

thus stating'(%) = 1&

#

j#0

((2!j%), (6.13)

we get Identites (6.3)and (6.5). Identity (6.6) is a obvious consequence of (6.10) and of (6.12).Now let us prove (6.7) which will be useful in Section 6.5. It is clear that the ring "C is thering of center 0, of small radius 1/12 and of big radius 10/3. Then it turns out that

2j" "C ) 2jC += * #%34, 2j % 2j" , 10

3ou

112, 2j" % 2j 8

3

&,

and (6.7) is proved. Now let us prove (6.8). As ' and ( have their values in [0, 1], it is clearthat

'2(%) +#

j#0

(2(2!j%) % 1. (6.14)

Let us bound from below the sum of squares. The notation a - b(2) means that a& b is even.So we have

1 = ('(%) + "0(%) + "1(%))2 with"0(%) =

#

j&0(2),q#0

((2!j%) and "1(%) =#

j&1(2),q#0

((2!j%).

From this it comes that 1 % 3('2(%) + "20(%) + "2

1(%)). But thanks to (6.5), we get

"2i (%) =

#

j#0,q&i(2)

(2(2!j%)

and the proposition is proved.

65

We shall consider all along this book two fixed functions ' and ( satisfying the asser-tions (6.3)–(6.8). Now let us to fix the notations that will be used in all the following of thistext.

Notations

h = F!1( and "h = F!1',

!!1u = '(D)u = F!1('(%)!u(%)),

if j ! 0 , !ju = ((2!jD)u = 2jd'

Rdh(2jy)u(x& y)dy,

if j % &2 , !ju = 0,

Sju =#

j""j!1

!j"u = '(2!jD)u = 2jd'

Rd

"h(2jy)u(x& y)dy,

if j ( Z , !ju = ((2!jD)u = 2jd'

Rdh(2jy)u(x& y)dy,

if j ( Z , Sju =#

j""j!1

!j"u.

Remark Let us point that all the above operators !j and Sj maps Lp into Lp with normswhich 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

!j or Id =#

j

!j .

This is described by the following proposition.

Proposition 6.1.2 Let u be in S %(Rd). Then, we have, in the sense of the convergence inthe space S %(Rd),

u = limj'(

Sju.

Proof of Proposition 6.1.2 Let f ( S(Rd). We have < u & Sju, f >=< u, f & Sjf >.Thus it is enough to prove that in the space S(Rd), we have

f = limj'(

Sjf.

We shall use the family of semi norms $ · !$k,S of S defined by

$f!$k,Sdef= sup

|!|"k"$Rd

(1 + |%|)k|#! !f(%)|.

Thanks to Leibnitz formula, we have

$f & Sjf!$k,S % sup|!|"k"$Rd

((1 + |%|)k

%|1& '(2!j%)|,| #! !f(%)|)

+#

0<#"!

C#!2!q|#||(##')(2!j%)|,| #!!# !f(%)|

&).

As ' equals to 1 near the origin it turns out that

$f & Sjf!$k,S % C!2!j$f!$k+1,S .

The proposition is proved.

66

The following proposition tells us that the condition of convergence in S % is somehow weakfor series, the Fourier transform of which is supported in dyadic rings.

Proposition 6.1.3 Let (uj)j$N be a sequence of bounded functions such that the Fourier

transform of uj is supported in 2j "C where "C is a given ring. Let us assume that

$uj$L! % C2jN .

Then the series (uj)j$N is convergent in S %.

Proof of Proposition 6.1.3 Let us use the relation (6.2). After rescaling it can be writtenas

uj = 2!jk#

|!|=k

2jdg!(2j ·) & #!uj .

Then for any test function $ in S, let us write that

.uj ,$/ = &2!jk#

|!|=k

.uj , 2jdg!(2j ·) & #!$/ (6.15)

% C2!jk#

|!|=k

2jN$#!$$L1 .

Let us choose k > N . Then (.uj ,$/)j$N is a convergent series, the sum of which is lessthan C$$$M,S for some integer M . Thus the formula

.u, $/ def= limj'(

#

j""j

.!j"u, $/

defines a tempered distribution.

Definition 6.1.1 Let us denote by S %h the space of tempered distribution such that

limj'!(

Sju = 0 in S %.

Examples– The space S %h is exactly the space of tempered distributions for which we may write

u =#

j$Z

!ju.

– If a tempered distribution u is such that its Fourier transform !u is locally integrablenear 0, then u belongs to S %h.

– If u is a tempered distribution such that for some function * in D(Rd) with value 1near the origin, we have *(D)u in Lp for some p ( [1,+0[, then u belongs to S %h. Inparticular, when p if finite, any inhomogeneous Besov space Bs

p,r is included in S %h.– A non zero constant function u does not belong to S %h because Sju = u for any j in Z.

Thus, if u belongs to Lp + Lq with finite p and q, then u is in Sh.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 *

in D(Rd) with value 1 near the origin, we have lim$'(

*("D)u = 0 in S %.

67

6.2 Homogeneous Besov spaces

Definition 6.2.1 Let u be a tempered distribution, s a real number, and p and r two realnumbers greater than 1. Then we state

$u$Bsp,r

def=* #

j$Z

2rjs$!ju$rLp

+ 1r

.

For the semi-norms we have defined, we can prove the following inequalities

Theoreme 6.2.1 A constant C exists such that

r1 % r2 # $u$Bsp,r2

% $u$Bsp,r1

and

p1 % p2 # $u$Bsp2,r

% C$u$B

s#d( 1p1# 1

p2 )p1,r

.

Moreover we have the following interpolation inequalities :– For any * ( [0, 1],

$u$B

!s1+(1#!)s2p,r

% $u$%B

s1p,r$u$1!%

Bs2p,r

.

– For any * in ]0, 1[,

$u$B

!s1+(1#!)s2p,1

% C

s2 & s1

%1*

+1

1& *

&$u$%

Bs1p,!$u$1!%

Bs2p,!

.

Proof of Theorem 6.2.1 In order to prove this result, we again apply Lemma 6.1.1 whichtells us that

$!ju$Lp2 % C2jd

%1

p1! 1

p2

&

$!ju$Lp1 .

Considering that +r1(Z) " +r2(Z), the first part of the theorem is proved.Now let us estimate in a di#erent way low frequencies and high frequencies. More precisely,

let us write

$u$B

!s1+(1#!)s2p,1

=#

j"N

2j(%s1+(1!%)s2)$!ju$Lp +#

j>N

2j(%s1+(1!%)s2)$!ju$Lp .

By definition of the Besov norms, we have

2j(%s1+(1!%)s2)$!ju$Lp % 2j(1!%)(s2!s1)$u$Bs1p,!

and

2j(%s1+(1!%)s2)$!ju$Lp % 2!q%(s2!s1)$u$Bs2p,!

.

Thus we infer that

$u$B

!s1+(1#!)s2p,1

% $u$Bs1p,!

#

j"N

2j(1!%)(s2!s1) + $u$Bs2p,!

#

j>N

2!q%(s2!s1)

% $u$Bs1p,!

2N(1!%)(s2!s1)

2(1!%)(s2!s1) & 1+ $u$B

s2p,!

2!N%(s2!s1)

1& 2!%(s2!s1)·

Choosing N such that$u$B

s2p,!

$u$Bs1p,!

% 2N(s2!s1) < 2$u$B

s2p,!

$u$Bs1p,!

implies the theorem.

68

There are two important facts to point out. The first one is about the homogeneity. If uis a tempered distribution, then let us consider for any integer N , the tempered distributionuN define by

uNdef= u(2N ·).

Then we have

Proposition 6.2.1 If $u$Bsp,r

is finite, so it is for uN and we have

$uN$Bsp,r

= 2N(s! dp )$u$Bs

p,r.

To proof this, we have to go back to the definition of the operator !j

!juN (x) = 2jd'

h(2j(x& y)uN (y)dy

= 2jd'

h(2j(x& y))u(2Ny)dy.

By the change of variables z = 2Ny, we get that

!juN (x) = 2(q!N)d'

h(2j!N (2Nx& z))u(z)dz

= (!j!Nu)(2Nx).

So it turns out that$!juN$Lp = 2!N d

p $!j!Nu$Lp .

we deduce from this that

$2js!juN$Lp = 2N(s! dp )2(q!N)s$!j!Nu$Lp .

And the proposition follows immediately by summation. The second one is that $ ·$ Bsp,r

isactually a semi-norm in the sense that, if u is a polynomial, then the support of its Fouriertransform is exactly the origin. Then, for any integer j, we have !ju = 0, and so $u$Bs

p,r= 0.

Let us state the definiton of the homogeneous Besov spaces.

Definition 6.2.2 Let s be a real number and (p, r) be in [1,0]2. The space Bsp,ris the set of

the tempered distribution u in S %h such that $u$Bsp,r

is finite.

Proposition 6.2.2 The space (Bsp,r, $ ·$ Bs

p,r) is a normed space.

It is obvious that $ ·$ Bsp,r

is a semi-norm. Let us assume that for some u in S %h, $u$Bsp,r

= 0.

This implies that Supp !u " {0} and thus that, for any j ( Z, Sju = 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 Besovspaces. The reason why is that homogeneous Besov spaces carry on informations about bothlow and high frequencies.

69

Proposition 6.2.3 A constant C exists which satisfies the following properties. Let (s, p, r)be in (R! \{0}) , [1,0]2 and u a distribution in S %h. This distribution u belongs to Bs

p,r ifand only if

(2js$Sju$Lp)j$N ( +r.

Moreover, we have

C!|s|+1$u$Bsp,r%

,,,(2js$Sju$Lp)j

,,,&r% C

%1 +

1|s|

&$u$Bs

p,r.

Proof of Proposition 6.2.3 Let us write that

2js$!ju$Lp % 2js($Sj+1u$Lp + $Sju$Lp)

% 2!s2(q+1)s$Sj+1u$Lp + 2js$Sju$Lp .

This proves the inequality on the left. But now we can write that

2js$Sju$Lp % 2js#

j""j!1

$!j"u$Lp

%#

j""j!1

2(q!q")s2j"s$!j"u$Lp

As s is negative, we get the result.

Theoreme 6.2.2 If s <d

p, then (Bs

p,r, $ ·$ Bsp,r

) is a Banach space. For any p, the space Bdpp,1

is also a Banach space.

First let us prove that those spaces are continuously embedded in S %. Thanks to Lemma 6.1.1,

the series (!ju)j$Z is convergent in L( when u belongs to Bdpp,1. As u belongs to S %h, this

implies that u belongs to L(. Using Proposition 6.2.1, we have that

$u$L! % C$u$B0!,1

% C$u$B

dpp,1

. (6.16)

Following (6.20) and using Propositions 6.2.1 and 6.3.1 and the fact that s < d/p, we have,for large enough N and M and non negative j,

|.!ju, $/| % 2!j$u$B#N!,!

$$$M,S

% 2!j$u$B

s# dp

!,!

$$$M,S

% Cs2!j$u$B

s# dp

!,!

$$$M,S ,

For negative j, let us write that, for large enough M ,

|.!ju, $/| % 2j-

dp!s

.$u$

Bs# d

p!,!

$$$L1

% C2j-

dp!s

.$u$Bs

p,r$$$M,S .

As u belongs to S %h, we have, for large enough M ,

|.u, $/| % Cs$u$Bsp,r$$$M,S . (6.17)

70

Let us consider a Cauchy sequence (un)n$N in Bsp,r for (s, p, r) satisfying the hypothesis of

the theorem. Using (6.16) or (6.17), this implies that a tempered distribution u exists suchthat the sequence (un)n$N converges to u in S %. The main point of the proof consists inproving that this distribution u belongs to S %h. If s < d/p, for any n, un belongs to S %h. Thanksto (6.17), we have

'j ( Z , 'n ( N , |.Sjun,$/| % Cs2j-

dp!s

.sup

n$un$Bs

p,r$$$M,S .

As the sequence (un)n$N tends to u in S %, we have

'j ( Z , |.Sju, $/| % Cs2j-

dp!s

.sup

n$un$Bs

p,r$$$M,S .

Thus u belongs to S %h. The case when u belongs to Bdpp,1 is a little bit di#erent. As (un)n$N is

a Cauchy sequence in Bdpp,1 " B0

(,1, for any positive ,, an integer n0 exists such that

'j ( Z , 'n ! n0 ,#

k"j

$!kun$L! % ,

2+

#

k"j

$!kun0$L! .

Let us choose j0 small enough such that

'j % j0 ,#

k"j

$!kun0$L! % ,

2·

As un belongs to S %h, we have

'j % j0 , 'n ! n0 , $Sjun$L! % ,.

We now that the sequence (un)n$N tends to u in L(. This implies that

'j % j0 , $Sju$L! % ,.

This proves that u belongs to S %h. By definition of the norm of Bsp,r the sequence (!ju(n))n$N

is a Cauchy one in Lp for any j. Thus an element uj of Lp exists such that (!ju(n))n$N

converges to uj in Lp. As the sequence (u(n))n$N converges to u in S % we have !ju = uj . Letus define

a(n)j = 2js$!ju

(n)$Lp and aj = 2js$!ju$Lp .

For any j, we have limn'(

a(n)j = aj . As (a(n)

j )n$N is a bounded sequence of +r(Z), a = (aj)j$Z

belongs to +r(Z) and thus u is in Bsp,r. Moreover the fact that (u(n))n$N is a Cauchy sequence

in Bsp,r implies that, for any positive ,, an integer n0 exists such that, for any n ! n0 and

any m,$a(n+m) & a(n)$&r(Z) % ,.

As (a(n)) tends weakly to a in +r(Z), we get, passing to the (weak)limit when m tends toinfinity in the above inequality that

$u(n) & u$Bsp,r

= $a& a(n)$&r(Z) % ,.

The proposition is proved.

Exercice 6.2.1 Let $ a function of S(Rd) and - in Sd!1. Then, if s (]0, d/p[, then

lim''0

ei (x|")# $ = 0 in the space Bs

p,1.

71

6.3 Inhomogeneous Besov spaces

6.3.1 Definition and examples

Definition 6.3.1 Let s be a real number, and p and r two reals numbers greater than 1. TheBesov spaces Bs

p,r is the space of all tempered distributions so that

$u$Bsp,r

def=,,,(2js$!ju$Lp)j$Z

,,,&r(Z)

< +0.

The following proposition (the proof of which is straighforward and omitted) describes therelations between homogeneous and inhomogeneous spaces.

Proposition 6.3.1 Let s be a negative number. Then Bsp,r is a subset of Bs

p,r and a constant C

(independent of s) exists so that, for any u belonging to Bsp,r, we have

$u$Bsp,r% C

&s$u$Bs

p,r.

Let s be a positive number. Then Bsp,r is a subset of Bs

p,r when p is finite, Bs(,r)S %h is a subset

of Bs(,r and a constant C exists (independent of s) so that, for any u belonging to Bs

p,r, wehave

$u$Bsp,r% C

s$u$Bs

p,r.

Lemme 6.3.1 If r is finite, then for any u in Bsp,r, we have

limj'(

$Sju& u$Bsp,r

= 0

The proof of this proposition is an easy exercise left to the reader. Let us give the first examplefor Besov space, the Sobolev spaces Hs. We have the following result.

Theoreme 6.3.1 The two spaces Hs and Bs2,2 are equal and the two norms satisfies

1C |s|+1

$u$Bs2,2% $u$Hs % C |s|+1$u$Bs

2,2.

As the support of the Fourier transform of !ju is included in the ring 2jC, it is clear, as j ! 0,that a constant C exists such that, for any real s and any u such that !u belongs to L2

loc,

1C |s|+1

2js$!ju$L2 % $!ju$Hs % C |s|+12js$!ju$L2 . (6.18)

Using Identity (6.8), we get

13$u$2

Hs %'

'2(%)(1 + |%|2)s|!u(%)|2d% +#

j#0

'(2(2!j%)(1 + |%|2)s|!u(%)|2d% % $u$2

Hs

which proves the theorem.

Proposition 6.3.2 The space B0p,1 is continuously embedded in Lp and the space Lp is

continuously embedded in B0p,(.

The proof is trivial. The first inclusion comes from the fact that the series (!ju)j$Z isconvergent in Lp. The second one comes from the fact that for any p, we have $!ju$Lp %C$u$Lp .

72

6.3.2 Basic properties

The first point to look at is the invariance with respect to the choice of the dyadic partitionof unity choosen to define the space. Most of the properties of the Besov spaces are based onthe following lemma.

Lemme 6.3.2 Let C% be a ring in Rd ; let s be a real number and p and r two real numbersgreater than 1. Let (uj)j$N be a sequence of smooth functions such that

Supp !uj " 2jC% and,,,(2js$uj$Lp)j$N

,,,&r

< +0.

Then we have

u =#

j$N

uj ( Bsp,r and $u$Bs

p,r% Cs

,,,(2js$uj$Lp)j$N

,,,&r

.

This immediately implies the following corollary.

Corollaire 6.3.1 The space Bsp,r does not depend on the choice of the functions ' and (

used in the Definition 6.3.1.

In order to prove the lemma, let us first observe that (uj)j$N is a convergente series in S %.Indeed using Lemma 6.1.1, we get that $uj$L! % C2j

-dp!s

.. Proposition 6.1.3 implies that

(uj)j$N is a convergente series in S %. Then, let us study !j"u. As C and C% are two rings, aninteger N0 exists so that

|j% & j| ! N0 =# 2jC ) 2j"C% = $.

Here C is the ring defined in the Proposition 6.1.1. Now, it is clear that

|j% & j| ! N0 =# F(!j"uj) = 0=# !j"uj = 0.

Now, we can write that

$!j"u$Lp = $#

|j!j"|<N0

!j"uj$Lp

% C#

|j!j"|<N0

$uj$Lp .

So, we obtain that

2j"s$!j"u$Lp % C#

j"#!1|j!j"|"N0

2j"s$uj$Lp

% C#

j"#!1|j!j"|"N0

2js$uj$Lp .

We deduce from this that

2j"s$!j"u$Lp % (ck)k$Z & (d&)&$Z with ck = 1[!N0,N0](k) and d& = 1N(+)2&s$u&$Lp .

The classical property of convolution between +1(Z) and +r(Z) gives that

$u$Bsp,r% C

* #

j$N

2rqs$uj$rLp

+ 1r

,

which proves the lemma.

73

The following theorem is the equivalent of Sobolev embedding (see Theorem ?? page ??).

Theoreme 6.3.2 Let 1 % p1 % p2 % 0 and 1 % r1 % r2 % 0. Then for any real number s

the space Bsp1,r1

is continuously embedded in Bs!d

%1

p1! 1

p2

&

p2,r2 .

In order to prove this result, we again apply Lemma 6.1.1 which tells us that

$S0u$Lp2 % C$u$Lp1 and

$!ju$Lp2 % C2jd

%1

p1! 1

p2

&

$!ju$Lp1 .

Considering that +r1(Z) " +r2(Z), the theorem is proved.

Proposition 6.3.3 The space Bsp,r is continuously embedded in S %.

By definition Bsp,r 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 $ in S we have

.u, $/ % C$u$Bsp,r$$$M,S . (6.19)

Using the above Theorem 6.3.2 and the relation (6.15), we can write, if N is a large enoughinteger,

.!ju, $/ = &2!q(N+1)#

|!|=N+1

.!ju, 2jdg!(2j ·) & #!$/

% 2!j$u$B#N!,!

sup|!|=N+1

$#!$$L1 (6.20)

% C2!j$u$Bsp,r$$$M,S .

Now Proposition 6.1.2 implies the Inequality (6.19).

Proposition 6.3.4 The space Bsp,r equipped with the norm $·$Bs

p,rdefined above is a Banach

space.

The end of the proof done for the homogeneous case can be repeated words for words.

Proposition 6.3.5 A constant C exists which satisfies the following properties. Let (s, p, r)be in (R! \{0}), [1,0]2 and u a tempered distribution. This distribution u belongs to Bs

p,r

if and only if(2js$Sju$Lp)j$N ( +r.

Moreover, we have

C!|s|+1$u$Bsp,r%

,,,(2js$Sju$Lp)j

,,,&r% C

%1 +

1|s|

&$u$Bs

p,r.

Let us write that

2js$!ju$Lp % 2js($Sj+1u$Lp + $Sju$Lp)

% 2!s2(q+1)s$Sj+1u$Lp + 2js$Sju$Lp .

This proves the inequality on the left. But now we can write that

2js$Sju$Lp % 2js#

j""j!1

$!j"u$Lp

%#

j""j!1

2(q!q")s2j"s$!j"u$Lp

As s is negative, we get the result.

74

In order to conclude this paragraph, let us state interpolation inequalities.

Theoreme 6.3.3 A constant C exists which satisfies the following properties. If s1 and s2

are two real numbers such that s1 < s2, if * (]0, 1[, if r is in [1,0], then we have

$u$B

!s1+(1#!)s2p,r

% $u$%B

s1p,r$u$1!%

Bs2p,r

and

$u$B

!s1+(1#!)s2p,1

% C

s2 & s1

%1*

+1

1& *

&$u$%

Bs1p,!$u$1!%

Bs2p,!

.

As very often in this book, we shall estimate in a di#erent way low frequencies and highfrequencies. More precisely, let us write

$u$B

!s1+(1#!)s2p,1

=#

j"N

2j(%s1+(1!%)s2)$!ju$Lp +#

j>N

2j(%s1+(1!%)s2)$!ju$Lp .

By definition of the Besov norms, we have

2j(%s1+(1!%)s2)$!ju$Lp % 2j(1!%)(s2!s1)$u$Bs1p,!

and

2j(%s1+(1!%)s2)$!ju$Lp % 2!q%(s2!s1)$u$Bs2p,!

.

Thus we infer that

$u$B

!s1+(1#!)s2p,1

% $u$Bs1p,!

#

j"N

2j(1!%)(s2!s1) + $u$Bs2p,!

#

j>N

2!q%(s2!s1)

% $u$Bs1p,!

2N(1!%)(s2!s1)

2(1!%)(s2!s1) & 1+ $u$B

s2p,!

2!N%(s2!s1)

1& 2!%(s2!s1)·

Choosing N such that$u$B

s2p,!

$u$Bs1p,!

% 2N(s2!s1) < 2$u$B

s2p,!

$u$Bs1p,!

implies the theorem.

6.4 The case of Holder type spaces

Another relevant example of Besov spaces are Holder spaces defined page ??. We have thefollowing result.

Proposition 6.4.1 For any k in N, a constant Ck exists such that for any . (]0, 1] and anyfunction u belonging to Ck,(, we have

supj

2j(k+()$!ju$L! % Ck$u$Ck,$ .

To prove this, let us first observe that, when j = &1, we have that $S0u$L! % C$u$L! .When j is non negative, let us write the operator !j of the convolution form

!ju(x) = 2jd'

h(2j(x& y))u(y)dy.

75

The fact that the function ( is identically 0 near the origin implies that for any ) ( Nd,'

x!h(x)dx = 0.

Thus we can write

!ju(x) = 2jd'

h(2j(x& y))%u(y)& u(x)&

#

0<|!|"k

1)!

#!u(x)(y & x)!&dy. (6.21)

Taylor formula at order k & 1 implies that

u(y)& u(x)&#

0<|!|"k

1)!

#!u(x)(y & x)!

=' 1

0k(1& t)k!1

#

|!|=k

(y & x)!

)!

%#!u(x + t(y & x))& #!u(x)

&dt.

The fact that the functions #!u belong to the space C( implies that///u(y)& u(x)&

#

0<|!|"k

1)!

#!u(x)(y & x)!/// % Ck|y & x|k+($u$Ck,$ .

Then it comes from (6.21) that

|!ju(x)| % Ck$u$Ck,$2jd'

|x& y|k+(|h(2j(x& y))|dy

and the proposition is proved.

Let us study the reciproque of Proposition 6.4.1. We have the following proposition.

Proposition 6.4.2 Let r be in R+ \N and let u be in Br(,(. Then u is in Ck,( with k = [r]

and . = r & [r]. Moreover we have

$u$Ck,$ % Cr$u$Br!,! with Cr = Ck

*1.

+1

1& .

+·

To star with, let us observe that, thanks to Lemma 6.1.1, we have, for any ) the length ofwhich is less than r,

$!j#!u$L! % Ck+12!q(r!|!|)$u$Br

!,! .

Thus the series (!j#!u)j$N is convergent in the space L( and we have

$#!u$L! % Ck+1 1.$u$Br

!,! . (6.22)

Now let us study the derivative of order k. We can write, for a positive integer N which willbe chosen later on, that

|#!u(x)& #!u(y)| %N!1#

j=0

|#!!ju(x)& #!!ju(y)| +#

j#N

|#!!ju(x)& #!!ju(y)|.

By Taylor inequality, we have that

|#!!ju(x)& #!!ju(y)| % C|x& y| sup|#|=k+1

$##!ju$L! .

76

Using Lemma 6.1.1, we get

|#!!ju(x)& #!!ju(y)| % Ck$u$Br!,! |x& y|2!q((!1). (6.23)

The high frequency terms are estimated very roughly writing

|#!!ju(x)& #!!ju(y)| % Ck2!q($u$Br!,! .

Then it comes from (6.23) that

|#!u(x)& #!u(y)| % Ck$u$Br!,!

% N#

j=0

2!q((!1)|x& y| +#

j#N+1

2!q(&.

Thanks to (6.22), we may assume that |x& y| % 1. Choosing

N = [& log2 |x& y|] + 1,

in the above inequality, we conclude the proof of the proposition.

Propositions 6.4.1 and 6.4.2 together imply the following theorem.

Theoreme 6.4.1 Let r be in R+ \N. Then the spaces Br(,( and C [r],r![r] are equal and we

have

C!1[r] $u$Br

!,! % $u$C[r],r#[r] % C[r]

* 1r & [r]

+1

1& (r & [r])

+$u$Br

!,! .

Proposition 6.4.2 turns out to be false when r is an integer.

Theoreme 6.4.2 The space B1(,( is not included in the space C0,1 of bounded Lipschitz

functions.

6.5 Paradi!erential calculus

In this section, we are going to study the way how the product acts on Besov spaces. Ofcourse, we shall use the dyadic decomposition constructed in the Section 6.1.2.

6.5.1 Bony’s decomposition

Let us consider two tempered distributions u and v, we write

u =#

j"!j"u and v =

#

j

!jv.

Formally, the product can be written as

uv =#

j,j"!j"u!jv.

Now, let us introduce Bony’s decomposition.

77

Definition 6.5.1 We shall designate paraproduct (respectively homogeneous paraproduct)of v by u and shall denote by Tuv (respectively Tuv) the following bilinear operator :

Tuvdef=

#

j

Sj!1u!jv respectively Tuvdef=

#

j

Sj!1u!jv.

We shall designate remainder (respectively homogeneous remainder) of u and v and shalldenote by R(u, v) (respectively R(u, v)) the following bilinear operator :

R(u, v) =#

|j!j"|"1

!j"u!jv respectively R(u, v) =#

|j!j"|"1

!j"u!jv.

Just by looking at the definition, it is clear that

uv = Tuv + Tvu + R(u, v) = Tuv + Tvu + R(u, v). (6.24)

The way how paraproduct and remainder act on Besov spaces is described by the followingtheorem.

Theoreme 6.5.1 A constant C exists which satisfies the following inequalities for any coupleof real numbers (s, t) with t negative and for any (p, r1, r2) in [1,+0]3.

$T$L(L!)Bsp,r;Bs

p,r) % C |s|+1,

$T$L(L!)Bsp,r;Bs

p,r) % C |s|+1,

$T$L(Bt!,r1

)Bsp,r2

;Bs+tp,r12

) % C |s+t|+1

&t,

$T$L(Bt!,r1

)Bsp,r2

;Bs+tp,r12

) % C |s+t|+1

&twith

1r12

def= min01,

1r1

+1r2

1·

To prove this theorem, we are going to use the proposition 6.1.1 about the contruction of thedyadic partition of unity. From the assertion (6.7), the Fourier transform of Sj!1u!jv andalso of Sj!1u!jv is supported in 2j "C. So, the only thing that we have to do is to estimate

$Sj!1u!jv$Lp and $Sj!1u!jv$Lp .

The lemma 6.1.1 says that, for any integer j,

$Sj!1u$L! % C$u$L! and $Sj!1u$L! % C$u$L!

Propositions 6.3.5 and 6.2.3 tells us that

$Sj!1u$L! % C

&tcj,r12

!qt$u$Bt!,r1

and $Sj!1u$L! % C

&tcj,r12

!qt$u$Bt!,r1

(6.25)

where (cj,r)j$Z denotes an element of the unit sphere of +r(Z). Using Lemma 6.3.2, the esti-mates about paraproduct are proved.

78

Now we shall study the behaviour of operators R. Here we have to consider terms of thetype !ju!jv. The Fourier transform of such terms is not supported in ring but in balls ofthe type 2jB. Thus to prove that remainder terms belong to some Besov spaces, we need thefollowing lemma.

Lemme 6.5.1 Let B be a ball of Rd, s a positive real number and (p, r) in [1,0]2. Let(uj)j$N be a sequence of smooth functions such that

Supp !uj " 2jB and,,,(2js$uj$Lp)j$N

,,,&r

< +0.

Then we have

u =#

j$N

uj ( Bsp,r and $u$Bs

p,r% Cs

,,,(2js$uj$Lp)j$N

,,,&r

.

We have for any j

$uj$Lp % C2!qs.

As s is positive, (uj)j$N is a convergent series in Lp. We then study !j"uj . As C is a ring(defined in the proposition 6.1.1) and B is a ball, an integer N1 exists so that

q% ! q + N1 =# 2j"C ) 2jB = $.

So it is clear that

q% ! q + N1 =# F(!j"uj) = 0=# !j"uj = 0.

Now, we write that

$!j"u$Lp =,,,,

#

j#q"!N1

!j"uj

,,,,Lp

%#

j#q"!N1

$!j"uj$Lp

%#

j#q"!N1

$uj$Lp .

So, we get that

2j"s$!j"u$Lp %#

j#q"!N1

2j"s$uj$Lp

%#

j#q"!N1

2(q"!q)s2js$uj$Lp .

So, we deduce from this that

2j"s$!j"u$Lp % (ck) & (d&) with ck = 1[!N1,+([(k)2!ks and d& = 2&s$u&$Lp .

So, the lemma is proved.

79

Theoreme 6.5.2 A constant C exists which satisfies the following inequalities. For any (s1, s2),any (p1, p2, p) and any (r1, r2) such that

s1 + s2 > 0 ,1p% 1

p1+

1p2

and1r1

+1r2

def=1r% 1,

we have that

$R$L(Bs1p1,r1

)Bs2p2,r2

;B%12p,r ) % C |s1+s2|+1

s1 + s2and

$R$L(Bs1p1,r1

)Bs2p2,r2

;B%12p,r ) % C |s1+s2|+1

s1 + s2with !12

def= s1 + s2 & d* 1

p1+

1p2& 1

p

+·

And for any (s1, s2), any (p1, p2, p) and any (r1, r2) so that

s1 + s2 ! 0 ,1p% 1

p1+

1p2

and1r1

+1r2

= 1,

we have that

$R$L(Bs1p1,r1

)Bs2p2,r2

;B%12p,!) % C |s1+s2|+1 and

$R$L(Bs1p1,r1

)Bs2p2,r2

;B%12p,!) % C |s1+s2|+1 again with !12

def= s1 + s2 & d* 1

p1+

1p2& 1

p

+·

Now, let us study the remainder operator in the homogeneous case (the inhomogeneousone can be treated exactly along the same lines). By definition of the remainder operator,

R(u, v) =#

j

Rj with Rj =1#

i=!1

!j!iu!jv.

By definition of !j , the support of the Fourier transform of Rj is included in 2jB(0, 24). So,by construction of the dyadic partition of unity, it exists an integer N0 so that

j% ! j &N0 # !j"Rj = 0. (6.26)

From this, we deduce that!j"R(u, v) =

#

j#q"!N0

!j"Rj .

Let us define p12 in [1,+0] by1

p12

def=1p1

+1p2

·

By hypothesis, p12 is smaller than p. So, using the lemma 6.1.1 of localization, we may write

$!j"Rj$Lp % 2j"d

%1

p12! 1

p

&

$Rj$Lp12

% 2j"d

%1

p12! 1

p

&1#

i=!1

$!j!iu!jv$Lp12

% 2j"d

%1

p12! 1

p

&1#

i=!1

$!j!iu$Lp1$!jv$Lp2

% C2j"%!s1!s2+d

%1

p12! 1

p

&&1#

i=!1

2!(j!j")(s1+s2)2(j!i)s1$!j!iu$Lp12js2$!jv$Lp2 .

80

Let us define (rq)q$Z by

rjdef= 2

j%!s1!s2+d

%1

p12! 1

p

&&

$!jR(u, v)$Lp .

Using the assertion (6.26), we have that

rj % C(b(1) & b(2))j with

b(1)j = 2!q(s1+s2)1N!N0(q) and (6.27)

b(2)j =

1#

i=!1

2(q!i)s1$!j!iu$Lp12js2$!jv$Lp2 . (6.28)

Let us assume that s1 + s2 is strictly positive. By hypothesis, the sequence (b(2)j )j$Z belongs

to +r(Z). Moreover, the sequence (b(1)j )j$Z belongs to +1(Z) and

$(b(1)j )j$Z$L1 % Cs1+s2

s1 + s2·

So the theorem is proved in this case. If we assume that

1r1

+1r2

= 1,

then, if s1 + s2 is positive, we have that

$(b(1)j )j$Z$&!(Z) % C and $(b(2)

j )j$Z$&1(Z) % C$u$Bs1p1,r1

$v$Bs2p2,r2

.

So the theorem is proved.Now, we are going to infer from this theorem the following corollary.

Corollaire 6.5.1 A constant C exists so that it satisfies the following properties. For anypositive real number s, we have

$uv$Bsp,r

% Cs+1

s($u$L!$v$Bs

p,r+ $u$Bs

p,r$v$L!),

$uv$Bsp,r

% Cs+1

s($u$L!$v$Bs

p,r+ $u$Bs

p,r$v$L!).

Moreover, for any (s1, s2), any p2 and any r2 such that s1 + s2 > 0 and any s1 <d

p1

, we have

that

$uv$Bsp2,r2

% Cs1+s2+1

s1 + s2

%$u$B

s1p1,!

$v$Bs2p2,r2

+ $u$Bs2p2,r2

$v$Bs1p1,!

&,

$uv$Bsp2,r2

% Cs1+s2+1

s1 + s2

%$u$B

s1p1,!

$v$Bs2p2,r2

+ $u$Bs2p2,r2

$v$Bs1p1,!

&

with

s = s1 + s2 &d

p1·

Moreover, for any (s1, s2), any p2 and any (r1, r2) such that

s1 + s2 ! 0 , s1 <d

p1and

1r1

+1r2

= 1,

81

we have that

$uv$Bsp,! % Cs1+s2+1($u$B

s1p1,r1

$v$Bs2p2,r2

+ $u$Bs2p2,r2

$v$Bs1p1,r1

),

$uv$Bsp,!

% Cs1+s2+1($u$Bs1p1,r1

$v$Bs2p2,r2

+ $u$Bs2p2,r2

$v$Bs1p1,r1

)

Moreover, for any (s1, s2), any (p1, p2, p) and any (r1, r2) such that

s1 + s2 > 0 , sj <d

pjand p ! max{p1, p2},

we have that

$uv$Bsp,r% Cs1+s2+1

s1 + s2$u$B

s1p1,r1

$v$Bs2p2,r2

and $uv$Bsp,r% Cs1+s2+1

s1 + s2$u$B

s1p1,r1

$v$Bs2p2,r2

.

with

s = s1 + s2 & d* 1

p1+

1p2& 1

p

+and

1r

= min(

1,1r1

+1r2

)·

Moreover, for any (s1, s2), any (p1, p2, p) and any (r1, r2) such that

s1 + s2 ! 0 , sj <d

pj, p ! max{p1, p2} and

1r1

+1r2

= 1,

we have that

$uv$Bsp,! % C$u$B

s1p1,r1

$v$Bs2p2,r2

and $uv$Bsp,!

% C$u$Bs1p1,r1

$v$Bs2p2,r2

with

s = s1 + s2 & d* 1

p1+

1p2& 1

p

+and r = max{r1, r2}.

The proof is nothing but the systematic use of Bony’s decomposition and the application ofTheorems 6.5.1 and 6.5.2.

6.5.2 Paralinearization theorem

In this paragraph we shall study the action of smooth functions on the space Bsp,r. More

precisely, if f is a smooth function vanishing at 0, and u a function of Bsp,r, does f 1u belongs

to Bsp,r ? The answer is given by the following theorem.

Theoreme 6.5.3 Let f be a smooth function and s a positive real number and (p, r) in [1,0]2.If u belongs to Bs

p,r ) L(, then f 1 u belongs to Bsp,r and we have

$f 1 u$Bsp,r% C(s, f, $u$L!)$u$Bs

p,r.

Before proving this theorem, let us notice that if s > d/p or if s = d/p and r = 1, then thespace Bs

p,r is included into L(. Thus in those cases, the space Bsp,r is stable under the action

of f by composition. This is in particular the case for the Sobolev space Hs with s > d/2.

82

Let us prove the theorem. We shall use the argument of the so called ”telescopic series”.As the sequence (Sju)j$N converges to u in Lp and f(0) = 0, then we have

f(u) =#

j

fj with fjdef= f(Sj+1u)& f(Sju). (6.29)

Using Taylor formula at order 1, we get

fj = mj!ju with mjdef=

' 1

0f %(Sju + t!ju)dt. (6.30)

At this point of the proof, let us point out that there is no hope for the Fourier transform of fj

to be compactly supported. Thus Lemma 6.5.1 is not e$cient in this case. We shall prove thefollowing improvement of this lemma.

Lemme 6.5.2 Let s be a positive real number and (p, r) in [1,0]2. A constant Cs exists suchthat if (uj)j$N is a sequence of smooth functions which satisfies

%sup

|!|"[s]+12j(s!|!|)$#!uj$Lp

&

j( +r,

then we have

u =#

j$N

uj ( Bsp,r and $u$Bs

p,r% Cs

,,,%

sup|!|"[s]+1

2j(s!|!|)$#!uj$Lp

&

j

,,,&r

.

As s is positive, the series (uq)q$N is convergent in Lp. Let us denote its sum by u and let uswrite that

!ju =#

j""j

!juj" +#

j">q

!juj" .

Using that $!juj"$Lp % $uj"$Lp , we get that

2js,,,

#

j">q

!juj"

,,,Lp

% 2js#

j">q

$uj"$Lp

%#

j">q

2!(q"!q)s2j"s$uj"$Lp . (6.31)

Then using Lemma 6.1.1, we write that

$!juj"$Lp % C2!q([s]+1) sup|!|=[s]+1

$#!uj"$Lp .

Then we write

2js,,,

#

j""j

!juj"

,,,Lp

%#

j""j

2(j"!j)([s]+1!s) sup|!|=[s]+1

2j"(s!|!|)$#!uj"$Lp .

This inequality together with (6.31) implies that

2js$!ju$Lp % (a &b )j with

ajdef= 1N(j)2!qs + 1N(j)2!j([s]+1!s) and

bjdef= 2js$uj$Lp + sup

|!|=[s]+12j(s!|!|)$#!uj$Lp .

This proves the lemma.

83

Let us go back to the proof of Theorem 6.5.3. Let us admit for a while that

') ( Nd , $#!mj$L! % C!(f, $u$L!)2j|!|. (6.32)

Thus using Leibnitz formula and Lemma 6.1.1, we get that

$#!fj$Lp %#

#"!

C!# 2j|#|C#(f, $u$L!)2j(|!|!|#|)$!ju$Lp

Thus we get that

$#!fj$Lp % C!(f, $u$L!)2j|!|$!ju$Lp

% cjC!(f, $u$L!)2!j(s!|!|)$u$Bsp,r

(6.33)

with $(cj)$&r = 1. We apply Lemma 6.5.2 and the theorem is proved provided we check (6.32).In order to do it, let us recall Faa-di-Bruno’s formula.

#!g(a) =#

!1+···+!p=!|!j |#1

% p2

k=1

#!ka&g(p)(a).

From this formula, we infer that

#!mj =#

!1+···+!p=!|!j |#1

' 1

0

% p2

k=1

#!k(Sju + t!ju)&f (p+1)(Sju + t!ju)dt.

Using Lemma 6.1.1, we get that

$#!mj$L! % C!(f)#

!1+···+!p=!|!j |#1

' 1

0

% p2

k=1

2j|!k|$u$L!

&

% C!(f, $u$L!)2j|!|.

This proves (6.32) and thus the theorem.

A slightly better estimate than estimate (6.32) can be obtained. For |)k| ! 1, we have

$#!k(Sju + tuj)$L! % C2j|!k|$2u$B#1!,!

and thus instead of (6.32), we have

') ( Nd , $#!mj$L! % C!(f, $2u$B#1!,!

)2j|!|. (6.34)

This leads us to the following theorem which is a small modification of the previous one.

Theoreme 6.5.4 Let f be a function of C(b (R) the space of smooth bounded functions and sa positive real number and (p, r) in [1,0]2. If u belongs to Bs

p,r and the first derivatives of ubelongs to B!1

(,(, then f 1 u belongs to Bsp,r and we have

$f 1 u$Bsp,r% C(s, f, $2u$B#1

!,!)$u$Bs

p,r.

Let us notice that if u belongs to the space Bdpp,r, then the first order derivatice of u belongs

to B!1(,(. Thus the space B

dpp,r is stable under the action of functions of C(b by composition.

This is in particular the case for the Sobolev space Hd2 = B

d22,2.

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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 1 u as a paradroductof u. More precisely we shall prove the following theorem.

Theoreme 6.5.5 Let s and . be two positive real numbers and f a smooth function. Let usassume that . is not an integer. Let p, r1 and r2 be in [1,0] such that r2 ! r1. Then for anyfunction u in Bs

p,r1)B(

(,r2we have

$f 1 u& Tf "*uu$Bs+$p,r12

% C(f, $u$L!)$u$B$!,r2

$u$Bsp,r1

with1

r12

def=1r1

+1r2·

The proof of this theorem follows the same lines of the proof of Theorem 6.5.3. We write againthat

f(u) =#

j

fj with fjdef= f(Sj+1u)& f(Sju).

Using Taylor formula at order 2, we get

fj = f %(Sju)!ju + Mj(!ju)2 with Mjdef=

' 1

0(1& t)f %%(Sju + t!ju)dt.

Following exactly the same lines as for proving (6.32), we get

') ( Nd , $#!Mj$L! % C!(f, $u$L!)2j|!|. (6.35)

Using Leibnitz formula, we can write that

#!(Mj(!ju)2) =#

)"#"!

C#!C)

##!!#Mj##!)!ju#)!ju.

Using Lemma 6.1.1 and inequality (6.35), we get

$#!!#Mj##!)!ju#)!ju$Lp % C!(f, $u$L!)2j|!|$!ju$L!$!ju$Lp .

Thus by definition of the Besov spaces, we get that

$#!Mj(!ju)2$Lp % C!(f, $u$L!)cj,!2!j(s+(!|!|)$u$B$!,r2

$u$Bsp,r1

(6.36)

with $(cj)$&r12 = 1. Now let us study the term f %(Sju)!ju. It does not look exactly like aparaproduct. Let us state

µjdef= f %(Sju)& Sj!1(f % 1 u)

Obviously we havefj = Sj!1(f % 1 u)!ju + µj!ju + Mj(!ju)2.

Let us admit for a while that

$#!µj$L! % cj2!q(s+(!|!|)C!(f, $u$L!)$u$B$!,r2

(6.37)

with $(cj)$&r2 = 1. Then using (6.36), we have

$#!(fj&Sj!1(f %1u)!ju)$L! % C!(f, $u$L!)cj2!j(s+(!|!|)$u$B$!,r2

$u$Bsp,r1

with $(cj)$&r12 = 1

and the theorem is proved provided of course that we prove the Inequality (6.37).

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Let us first investigate the case when |)| < .. Let us write that

µj = µ(1)j + µ(2) with

µ(1)j

def= f %(Sju)& f %(u) and

µ(2) def= f %(u)& Sj!1(f %(u)).

Thanks to Theorem 6.5.3, we have

#!f %(u) ( B(!|!|(,r2

and $#!f %(u)$B

$#|&|!,r2

% C!(f, $u$L!)$u$B$!,r2

.

Thus, we can write that

$#!(Id&Sj)f %(u)$L! %#

j"#q

$!j"#!f %(u)$L!

% C!(f, $u$L!)$u$B$!,r2

#

j"#j

cj"2!j"((!|!|)

% cj2!j((!|!|)C!(f, $u$L!)$u$B$!,r2

with $(cj)$&r2 = 1. Using a gain the argument of ”telescopic series” (see (6.29)), we have

f %(u)& f %(Sj(u)) =#

j"#j

"fj" with

"fj"def= f %(Sj"+1u)& f %(Sj"u).

Applying Inequality (6.33), we get

$#! "fj"$L! % cj"2!j"((!|!|)C!(f, $u$L!)$u$B$!,r2

with $(cj)$&r2 = 1. (6.38)

Then, by summation, we infer that, when |)| < .,

$#!(f %(u)& f %(Sju))$L! % cj2!j((!|!|)C!(f, $u$L!)$u$B$!,r2

with $(cj)$&r2 = 1.

Thus (6.37) is proved when |)| < .. The case when |)| > . is treated in a di#erent way.As #!f %(u) belongs to B(!|!|

(,r2 , we have, using Proposition 6.3.5 and Theorem 6.5.3,

$#!Sj!1f%(u)$L! % cj2!j((!|!|)C!(f, $u$L!)$u$B$

!,r2with $(cj)$&r2 = 1.

In order to estimate #!f %(Sju), we use again the argument of the telescopic series. Let uswrite that

f %(Sj(u)) =#

j""j

"fj" with

"fj"def= f %(Sj"+1u)& f %(Sj"u).

The using (6.37), we get, as |)| > .,

$#!f %(Sju)$L! %#

j""j

$#! "fj"$L!

% C!(f, $u$L!)$u$B$!,r2

#

j""q

cj"2!j"((!|!|)

% C!(f, $u$L!)$u$B$!,r2

cj2!j((!|!|)

with $(cj)$&r2 = 1. Inequality (6.37) and thus Theorem 6.5.5 is proved.

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6.5.3 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.

Theoreme 6.5.6 For any p ( [2,0[, the space B0p,2 is continuously included in Lp and the

space Lp" is continuously included in B0p",2.

Proof of Theorem 6.5.6 Let us state Fp(x) = |x|p. Let us notice that, as p ! 2, thisfunction is a C2 function. Thus in the spirit of Formulas (6.29) and (6.30), let us write that

$u$pLp =

'Fp(u)dx =

#

j

.!ju, mj/ with mj(x) def=' 1

0F %p

%Sju(x) + t!ju(x)

&dt.

Using Plancherel formula, and stating "!j the convolution operator by the inverse Fouriertransform of *(2!j ·) where * is in D(Rd \{0}) with value 1 near the support of (, we canwrite that

.!ju, mj/ = .!ju, "!jmj/.

Then, by using Lemma 6.1.1, we infer that

$ "!jmj$Lp" % C2!j sup1"&"d

$#&mj$Lp" . (6.39)

The chain rule and the Holder inequality imply that

$#&mj$Lp" %' 1

0

,,,#&(Sju + t!ju)F %%p (Sj + t!ju),,,

Lp"dt

%' 1

0$#&(Sju + t!ju)$Lp$F %%p (Sju + t!ju)$

Lp

p#2dt.

As F %%p (x) = p(p& 1)xp!2, we get immediately that, for all t ( [0, 1],

$F %%p (Sju + t!ju)$L

pp#2

% p(p& 1)$Sju + t!ju$p!2Lp .

Using Lemma 6.1.1, we infer that

$F %%p (Sju + t!ju)$L

pp#2

% Cpp(p& 1)$u$p!2Lp . (6.40)

Now, by definition of Sj , we have, using Lemma 6.1.1 and the Young’s inequality on series

$#&(Sju + t!ju)$Lp %#

k"j

$#&!ku$Lp

% 2j#

k"j

2k!j$!ku$Lp

% Ccj2j$u$B0p,2

with#

j

c2j = 1.

With (6.39) and (6.40), we deduce that

$ "!jmj$Lp" % Cpp(p& 1)cj$u$p!2Lp $u$B0

p,2with

#

j

c2j = 1.

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As we have $u$pLp =

#

j

.!ju, mj/, we infer that

$u$2Lp % Cpp(p& 1)$u$B0

p,2

#

j

cj$!j$Lp .

This concludes the proof of the theorem once observed that the embeddings about Lp" followsby duality.

88