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VII Calder6n, or the Cauchy integral operator on a Lipschitz graph. We shall concentrate, in this part, on an L2boundedness criterion (the T(b)theorem), and then give a few of the classical applications (essentially, those that are related to our part 3). It is interesting to note that we shall use, for this part, the same sort of techniques as in Part 1. Our proof of the T(b)theorem, for instance, is a very nice illustration of how ideas coming from wavelets can be used with success on Calder6nZygmund operators. This proof is due to R. Coifman and S. Semmes, and its idea is the following. When T is such that both T and its transpose send the function b to 0 , one finds a suitable Riesz basis (the Haar system works when b = 1), and then one proves that the coefficients of the matrix of T in that basis decay, sufficiently fast, away froLn the diagonal. The choice of a basis made of step functions simplifies the computations a great deal. After a first section of introduction, we shall construct, in Section 2, the Riesz basis associated to a given paraaccretive function b. This basis will then be used to prove, with the help of Schur's lemma, that T is bounded in the special case when Tb = Ttb = 0 (Section 3). The general case will follow from this and the construction of a paraproduct (Section 4). The proof extends very nicely to spaces of homogeneous type (see the remarks of Section 5). Most of the few applications mentioned in the final section will be variants of the theorem of Coifman, McIntosh and Meyer that says that the Cauchy integral defines a bounded operator on L 2 for any Lipschitz graph. For a broader point of view, we refer to [Chl] or [My4]. Our third part is, in a way, a continuation of the second one. Can one extend Coifman, McIntosh and Meyer's theorem to curves other than Lipschitz graphs, and then, even worse, to higherdimensional nonsmooth surfaces? There are a few approaches to try to answer the question, but by lack of time we shall give most of our attention to only one of them (the author's favorite). The idea is the following. If eg is a kdimensional Lipschitz graph in IRn , then the singular kernels we want to consider always define bounded operators on L2(,~). This is easily deduced from Coffman, McIntosh and Meyer's result on the Cauchy kernel in one dimension. One can then try to use the standard estimates on the kernel and "good X inequalities" to extend the L2boundedness results from Lipschitz graphs to more and more general sets (this is pretty much in the spirit of the pioneers of CalderdnZygmund theory). The systematic part of this approach will be described in Sections 2 and 3. Once this is done, one is left with a problem on the geometry of the nonsmooth surface e6 (something like this : does eg contain big pieces of Lipschitz graphs?). This problem is simple enough when eg is a curve, and one can then give a necessary and sufficient co~adition, on a rectifiable curve eg, for the Cauchy integral to define a bounded operator on L2(,~) : eg must be Ahlforsregular (see Definition 41). Also, a minor modification of the argument for regular curves can be used to
VIII deduce the boundedness of the Cauchy integral on any Lipschitz graph from the case when the Lipschitz norm is small enough (i.e., from Calder6n's theorem). This is done in Section 4. In dimensions higher than one, or even for onedimensional nonconnected sets, things are not so simple. In Section 5, we shall say a few words about Garnett's counterexample to convince the reader that the geometry of the situation is much more complicated (and interesting) than for curves. We shall then give, in Section 6, three classes of surfaces on which, for various reasons, singular integrals define bounded operators. We shall mention in Section 7 a few cases when the "good ~. approach" described above is successful. 3ection 8 contains the proof of a nice result of P. Jones concerning the existence of sets on wh~ch a given Lipschitz function has a bilipschitz restriction. About one year after the lectures, a new Section 9 was added to the others. It concerns more recent results, and more particularly another way to deal with our problem. In [Jn2], in order to study the L2boundedness of the Cauchy integral operator on curves, P. Jones showed that Lipschitz graphs satisfy what he calls a geometric lemma. It was discovered later that, even in higher dimensions, the geometric lemma was, as P. Jones put it, lurking in the background. A second tool used by P. Jones in [Jn2] is a variant of the stoppingtime argument often referred to as the corona construction. Most of Section 9 is concerned with recent results using those ideas. Part 3 ends with a Section 10 on further questions. The official author of these notes would like to thank Peng Lin and JiuPing Zhong, who helped preparing this manuscript. He is also very thankful to Paule Truc, from Ecole Polytechnique, and Josette Dumas, From Paris 11 (Orsay), Wao kindly accepted to type it. The author is very grateful to Y. Meyer for allo~ving him to use, quite extensively, a preliminary version of his book [My4] for the preparation of the first part of these notes ; he is even more grateful for all the time Y. Meyer spent explaining him the little he knows about wavelets. It is also a pleasure to thank R. Coifman, P. Jones and S. Semmes for many enriching conversations, and for providing the author with the details of the various proofs that can be found in these notes. Finally, the author wishes to thank Professor S. S. Chern, Professor M. T. Cheng and the Nankai Institute for inviting him, and his patient audience for their friendly welcome. He hopes to be able to return to China, in such a pleasant atmosphere, very soon. The kind reader is asked not to pay too much attention to the quality of english used in these notes.
TABLE OF CONTENTS
P a r t I : Wavelets 1. 2. 3. 4. 5. 6. 7. 8.
Introduction Multiscale analysis The function Construction of ¢ in dimension 1 Wavelets in higher dimensions : tensor product Other wavelets in higher dimensions Compactly supported wavelets Properties of the wavelet transform, extensions
1 3 6 8 11 12 15 23
Part II : Singular integral operators 1. 2. 3. 4. 5. 6.
Introduction and generalities First step of CoifmanSemmes' proof : a Riesz basis Proof of the T(b)theorem when T b l = Ttb2 = 0 End of the proof : paraproducts Comments on T ( b ) , spaces of homogeneous type Applications A.More wavelets B. The Cauchy integral and related operator
26 32 38 44 47 48 49 50
P a r t I I I : S i n g u l a r i n t e g r a l s on c u r v e s a n d s u r f a c e s 1. 2. 3. 4.
5. 6.
7. 8.
Introduction and notations CalderSnZygmund techniques The "good ~" method Regular curves and Lipschitz graphs A. Regular curves B. The Cauchy integral on Lipschitz graphs again Garnett's example Three classes of surfaces A. Chordarc surfaces with small constant B. wregular surfaces C. Stephen Semmes surfaces Finding big pieces : clouds and shadows Bilipschitz mappings inside Lipschitz functions
55 58 60 64 64 66 67 69 69 71 72 76 79
x 9. Square functions, geometric lemma and the corona construction A. Square function estimates B. P. Jones' geometric lemma C. The corona construction D. A characterization 10. A few questions
Appendix Appendix References
1 : Construction of dyadic cubes on a regular set 2 : Two recapitulatory diagrams
83 83 84 86 88 89
93 97
100
Part
1.
I : WAVELETS
Introduction
As was mentioned earlier, this part is nothing more than extracts from Y. Meyer's book [My4]. Our approach here, however, will be strongly biased : I will try to take the point of view of someone interested mostly in Calder6nZygmund operators, rather than numerical analysis. For a more natural point of view, it is strongly recommended to anyone that can read some french to report directly to "the book" [My4]. Let us say a few words about motivations. The term "wavelets " is meant to suggest "small waves", and the idea is to oppose them to the "long ones", like the sine and cosine. We wish to find ways of decomposing a function f into a series (or an integral) involving functions (say, ¢ I for I C ~) given in advance. For instance, we could take the eiz'¢, ~ C R n, and use the Fourier transform to get
f(x) = f T(~)elZ'¢d~ (note that the constant (2~r)n/2 was omitted ; we shall go on omitting constants like this, and we'll try to compensate by always taking an even number of Fourier transforms). Another example, for periodic functions, is the Fourier expansion f(x) =
~
kEZ n
cke ik'z
where, this time, the sum is discrete. We will try to find expansions like (1)
f  ~ cI ¢1 IEA
(where
.4
is some set of indices)
It will definitely be a good thing if el can be computed, for instance by el = f f~I (it will not be too bad if cz = f f ¢ I , for some other set of ¢i's). Of course, it will be even better if the ~b1's are an orthonormal basis, because then the the c1's will be unique, and also, perhaps, because in this case the ci's are a minimal amount of information that allows to recover f . We also want to ask the functions Cz to be obtained from a single one (or may be a finite family) by translations and dilations. Decompositions like (1) have been known for quite some time. For instance, recall Calder6n's reproducing formula. Let ¢ E L I ( R n) be real, radial and such that f ¢ = 0 and f0°° I¢(t~)12~ = 1. Set Ct(x) = t'~¢(x/t) for t > 0. The formula is
(2)
f:
fo ° ¢ ~ * ¢ ~ * f
dt T
'f°r
fEL2(R~).
Now, we can see (2) as a way to associate coefficients to f , and then reconstruct f from the coefficients : for a > 0 and b E R '~, call ~ , ~ = a"/2~(z~b) (the "wavelet" with "scale" a and "centered" at b). The "wavelet transform" can be defined by
(3)
F(a, b) = / f(x) ~a,b(x)dx,
and (2) can be seen as the "inversion formula"
(4)
f(z) =
/
F(a, b)~a,b(~)
dadb
a>O bER
'~
(we leave the details as an exercise). This formula is far from being recent. The reader should be warned that, for most of the applications of wavelets , the formula (2) would be quite sufficient. However, introducing orthogonal wavelets will, in most cases, make computations much simpler and, in some cases, the fact that one has a basis will be really necessary. Note that, with a more clever choice of ~b, one can simplify (2) a little, and in particular replace the integral by a discrete sum (see [FJ1,2,3]). Also, reproducing formulae such as (2) were introduced later (and independently) by A. Grossman (for theoretical physics reasons) and J. Morlet (for oil prospection reasons). We shall look more specifically for functions ~b, defined in R , such that the functions ~bj,k(x) = 2H2~b(2Jz k), j,k C Z are an orthonormal basis of L2(R). Furthermore, we shall ask ~b to be rapidly decreasing, and reasonably smooth. Let us rapidly explain why. An example of function ~b would be 1~o,1/2]  111/2,1], which would give the Haar basis. When we take a function f , and write it as a series E cj,k~bj,k, the partial sums are never continuous (even if f is very regular), and will, in general, converge to f only in L~norm. If the function ~b is smooth, then the partial sums are smooth too, and it is possible to hope that if f is regular, then the partial sums will converge to f in a much smaller space than L 2. Without getting ahead of ourselves, let us mention that it is indeed the case : if ~b is smooth enough (and rapidly decreasing) and f is in the Sobolev space H s, then partial sums will indeed converge to f for the topology of H m ! A connected fact is that many function spaces will be characterized (again if ~b is smooth enough, and rapidly decreasing) by the fact that the coefficients ej,k = f ~j,k vanish at a certain rate when j tends to +co. We shall be a little more precise later. Before we start constructing wavelets , let us mention a little bit of history. The first occurence of "smooth, orthogonal wavelets ", is due to O. Stromberg, in 1981 : he found, for each integer r, a function ~b of class C r that decays exponentially at co, and such that the ~bj,k's are an orthonormal basis of L2(R) [Sg]. A function ~b E S(R) with the same property was discovered (again independently) by Y. Meyer (in 1985), and rapidly generalized to r~ dimensions by P.G. Lemari6 [LM]. The approach that will be described here is quite different from the initial ones. It relies on the notion, introduced by S. Mallat, of "multiscale analysis".
2. M u l t i s c a l e A n a l y s i s .
The following notion will help us understand the construction of the wavelets. It will also make the proofs a little longer : in some cases, we could just pull out a formula for ¢, and ask the reader to check that it works (actually, the first proofs looked a little like this to the unexperienced reader). Let us mention, before we start, that the construction below is due to S. Mallat and Y. Meyer. D e f i n i t i o n 2 . 1 .   A multiscale analysis of L 2 ( R n) is an increasing sequence Pry, j E Z, of closed subspaces of L 2 (R'~), with the following properties : (5) j ~ . V,. = {0} and Je~UV,. is dense
(6)
e vi if and only if l(2 ) e 5 + , ;
(7) f(x) e Vo if and only if f ( x  k) E Vo for each k • Z "~ ; (8) There is a function g • Vo such that the functions g(x  k), k • Z n, form a Riesz basis of Vo. Recall that (8) means that the g(x  k) generate a dense subspace of Vo, and that there is a C > 0 such that
for all £2sequences ak (or all finite sequences ak). Equivalently, it means that (ak) ~* ~ Ctkg(x  k) defines an isomorphism from £2(Z'~) kEZ" onto V0. Comments :  If we k n o w Vo, (6) tells us h o w to define the Vj's, and so we only need to check (5),
(7), (8) and the monotonicity of the Vi's.  The function g is not unique. analysis.
Many different g's could give the same multiscale
 A trivial consequence of (6) and (7) is that f ( x ) E Vj if and only if f ( x  k2  i ) E Vi for all k E Z'*.  Here is a vague interpretation of the role of the Vi's. For each j , call P1 the orthogonal projection on Vi ; (5) says that P i f * 0 when j ~  o o , and P i f ~ f when j * +oo (in both case, with strong convergence [Exercise[). Each j corresponds roughly to the size 2  i , and the projection P i f we'll give details about f , up to the size  2  i . So, PYf should give more and more precise details (we'll see this later, in examples). Example
1 :
splines of order r .
For this example, the dimension is n = I. W e define V0 = {f E L2(R) : / is of class C r1 and the restriction of f to each interval ]k, k + 11 is a polynomial of degree _< r). For instance, if r  0 we get step functions, whereas r = 1 gives piecewise affine functions that are continuous. The spaces Vj are defined using (6), and the properties (5) and (7) are clear. We'll see soon that (8) is true, and even that we can pick g = X * X ... * X (r + 1 times), where X is the characteristic function of [0,1].
D e f i n i t i o n 2 . 2 .   Let r ~ N . W e T say that the multiscale analysis (Vj) is rregular if one can choose the function g in (8) so that
(9)
la"g(~)l <_ CM,~(1 + Ixl) M for all M > 0 and all multiindices a such that [a[ < r.
Note that we only ask Oag to be in L c~ but not necessarily continuous. If we believe that g = 27 * ... * Z gives a Riesz basis for V0, then splines of order r give a M.S.A. (multiscale analysis) which is rregular. Example 2 : Let us try Vo = { f E Lz(R) : f i s supported on [Tr, lr]} in dimension 1. It is easy to check that Vo is stable by translations by integers and that (5) is true (with the obvious definition of Vh). Also, the Vj's are an increasing sequence. The problem with this example will be to find a suitable function g. If we look for an orthonormal basis of 110, we might as well use the Fourier transform, because the l[_r,rl e ik~, k E Z, are an orthogonal basis of V0. Applying ~r1 gives a function g ( x )  sin,vx  7 ~ " But g does not decay rapidly at oo. Furthermore, no other choice of g E Vo would both give a Riesz basis, and enough decay at oo (we'll leave this as an exercise). So we'll have to exclude this example from the authorized MSA's. Note, by the way, that since g(0) = 1 and g(k) = 0 f o r Akg(x  k) is quite easy: ~1¢ = f(k).
k¢
0, writing f E Vo as
k
Example 3 : This one is a smoothedup version of Example 2 ; this is the example that will give Y. Meyer's basis. Let 0(~) be a function in D(R) = C ~ ( R ) , with the properties (10) 0 _< 0(~) _< 1 everywhere ; (11)
0 is even ;
(12)
O(~) = 1 for ~ E r2~ ~1. t 3 ' 31'
(13) 0 ( ~ ) = 0 out of t[  48 r ' 34_~x 1 and 1
(14) o2(~) + o;(2~  ~)  1 for o <_ ~ _< 2~ (see the picture).
J
 4n/'3
of 2~/3
N~,A~
I0
21 / 3
Local point of 2 symmetry for 0
~ /3
Note t h a t if g is defined by ~ = 0, then the g(z  k), k • Z, are an orthogonal s y s t e m (we leave this as an exercise, with the following hint : t r y on the Fourier t r a n s f o r m side !). Define V0 to be the closed space spanned by the g(x  k), and Vj, j • Z, by the rule (6). To see t h a t we have a MSA, we still need to check (5) and the fact t h a t the Vj's are increasing. Since V0 = { f • L 2 : f(m) = ~  ~ k g C m   k)
for some sequence
(a/c) • 12(Z)},
we see t h a t V0 = ( ] ' • L 2: f =
=(]'•L2:f'=rnO,
~
ak e~k~0(~)
where
rn
is
for some sequence 27rperiodic
(ak) • 12(Z)}
andin
LZ(lr,r])}.
Since the g(x  k) are (a constant times) an o r t h o n o r m a l basis, we even get t h a t IlfllL2 = cl[rn[IL2([~,,~]), where m is the 2rperiodic function such t h a t f ' = rn0. Now (5) follows because Vj. = ( f ' • L 2 : f(~) = rn(2J~)O(2J~) for some 2rperiodic rn • n~o¢}. The fact t h a t Vj C Vi+1 also follows from that description of Vj and the fact t h a t 0(~) = rn(~/2)O(~/2) for a suitable rn. So the Vi's are a MSA, which is as regular as we want because g • S (R). Example
4.
(tensor product)
:
Suppose the Vi's are a MSA on LZ(R), and let us build a MSA on L2(R2). Let g be the function defined in (8), and consider the function g(x)g(y), defined on R 2. One easily checks t h a t the functions g(x  k)g(y  l), (k,£) • Z 2, are a Riesz system. Denote by V0 ® V0 the closed subspace of L 2 ( R 2) spanned by this Riesz s y s t e m : V0 @ Vo = { f • L 2 ( R 2) : f ( x , y ) = ~ ~
ak,tg(x


k)g(y  g) for a sequence
ak,~ • ~2(Z × Z)}. Calling Vj @ Vj the subspace of L 2 ( R 2) defined from Vo ® Vo by the rule (6), we see that the V1 @ Vj's are an increasing sequence of closed subspaces, t h a t satisfy (5)(8), and which defines a MSA of L 2 ( R 2) of the same regularity as the Vi's. Exercise :
define a MSA for L2(Rn).
Let us now go back to onedimensional spline functions of order r, and prove w h a t we said earlier a b o u t the function g = X * . . . * X ( r + l times). R e m e m b e r t h a t Vo is composed of the L2functions t h a t are of class C r  1, and whose restriction to any interval d ~r+l [k, k + 1[ is a polynomial of degree < r. If f E Vo, then {~T~z J ]. ~ c k/~k, where //k is k
the Dirac mass at k • Z. Since f , restricted to [k  1,k[ and [k,k + 1[, is a polynomial 2 1 k+l 2 of degree < r, and f is of class C r  l , one easily check t h a t c k < ~ f ~ _ l Ill and so ICkl2 < ÷oo. Then, the function q(~) = (i~)r+l~'(~) is a 2rperiodic function, with k
Ilql]L2([g,~]) _< Cllfll2. Next, define rn(~) = q(~)(1  ei~)  r  1 : this is also a 2~periodic function, and it is in L 2 because, near 0, m(~) ~ T(~). As a consequence, we can write T(~)  m(~)
for some 2~rperiodic m • n2([0,2r[).
Conve
ely, if
l_ei¢ ,~ r + l
is of the form
/
for some 2
periodic
e L OC'
then f • L 2 and ( ~ ) r + l f
= ~] ck6k for a sequence (ck) • ~2(Z), and it is easily checked
that f • Vo (exercise). So
V0 = {rn(~) \
i~ ]
: m is 2~rperiodic and in L2([0,27r])}.
Note t h a t this looks a lot like our description of V0 in the Examples 2 and 3, except that this time 0 does not decrease rapidly at infinity. In fact, the rapidity of the decay depends on r, and this will account for the regularity of g and of the MSA. From our description of V0, we see t h a t we can choose g = X * . . . * X (r + I times) to define a Riesz basis of V0 (g is indeed the inverse Fourier transform of  ( ~  ) r + l ) . Remark. W h e n r is odd, one can find g • V0 such t h a t g(0)  1 and g(k)  0 for k ¢ 0 ; when r is even, it is impossible to find such a g (and so the c o m p u t a t i o n of the coefficients of f • V0 in the basis defined by 9 is not so easy). For more details, see the book [My4]. Example : Whenr2xfor 1 < x<2.
xfor0
1, and
C o m p u t e g(x) for r = 3 (the solution is in [My4]).
Exercise. 3.
1, g(x) = ~ ( * X ( x ) i s 0 f o r x < 0 o r x > 2 ,
T h e f u n c t i o n iO •
We wish to replace the function g of (8) by a function ~o • V0 such t h a t the ~0(x  k), k • Z ", will be an o r t h o n o r m a l basis of V0. One systematic way of doing this would be to use G r a m matrices (let us call t h a t G r a m ' s orthogonalization process). The idea is to apply to the basis (g(x  k)) the inverse of the square root of the positive matrix of scalar products, so t h a t the result is an orthonormal basis. T h e advantage of this process is that, countrary to the G r a m  S c h m i d t orthogonalization process, this one does not put any order on the elements of the basis. As a consequence, if one starts with a basis which is invariant under the action of a group (like here, with the translations of Z~), the result is also invariant under the action of the group. In the case t h a t we want to study, one can use the Fourier transform to get precise formulae, and this gives a slightly faster construction (one can check t h a t the construction given below, however, coincides with the one obtained from G r a m ' s process). In more general cases, the Fourier transform cannot be used any longer, and G r a m ' s process becomes the providential tool. To find our function ~o • V0, let us write ~(x) = Y]~ck g(x + k) for some £2sequence k (ck). T h e n ~(~) = ~ c k e i k ~ ( ~ ) = m(~)~(~) for some 2~rperiodic (in all directions) function rn(~). One even has l[to[]2 ~ [[rn[[L2([o,2~]). Now let us c o m p u t e scalar products. If h,£ are in Z 't,
J
J
=
Calling G(O =
(15)
[
E
jEZ ~
I~(~ + 2~i)1 ~,
+
+ h+
J
we get
:/" J{o
Taking h = l = 0, w e g e t
1199111 f[0,2.{
G ( ( ) [ m ( ( ) { 2d(" Since this formula is valid
for any function 99 E Vo, and since 119911~ ___ IImliL we see t h a t there is a constant c > 0 such G(~) _> c almost everywhere. Let us pause in our search for 99 to study G a little more. Lemma
If the M S A (Vy) is 0regular, G(~) is of class C °°.
3.1.
To prove this, we'll only use the fact that G(~) =
~ {~(~ + 27rj)] 2, where g is yez a hounded function t h a t decreases rapidly at infinity. More precisely, for any M > 0, f lg(x)2l(1 + ( x ) ) 2 ~ d x < + c o , and so ~ is in the Soholev space H M ( R " ) . Let 0 E C ~ ( R n) be such t h a t O(x)  1 on B ( 0 , 2 0 n ) . For each M > 0, the sequence k ~ II~(00(~  k ~ ) l l ~ ~ in e2(z ") [exercise ; use Leihnitz'rule and the fact that if functions have disjoint s u p p o r t s and their s u m is in L 2, then their L2norms form a l 2sequence]. By Soholev's embedding's theorem, we see t h a t for all M ~ > 0, the sequence k + II~(00(~  k~)llo~, is also in e2(Z'~). This is enough to conclude t h a t the series
~ Ig(( + 27rj)[2 converges uniformly on jEZ ~ each compact. T h e same thing would be true if ~ were replaced by any of its derivatives. Therefore, G is in COO(Rn), and we proved L e m m a 3.1. Since we know t h a t G > c > 0 a.e., it follows t h a t G(~) 1/2 is a welldefined, positive and Coo function ]the expert reader will note the similarity with G r a m ' s process]. As (15) would suggest, let us try to define 99 by (16)
~(~) = m ( ~ ) ~ ( ~ ) ,
where
m(~) : G(~)  1 / 2 .
Lemma 3.2.The function 99 defined by (16) satisfies the regularity conditions (9) with the same r as g, and 99(x  k), k E Z '~, is an orthonormal basis of Vo. ~ akc ik~, kE ~.,* where the ak's are decaying as fast as we wish at infinity. It is then an easy exercise to check t h a t 99{x) : ~ a k g ( x + k), and its derivatives of order _< r, have the same decay as the derivatives of order <_ r of g. Let us first prove (9) for 99. Since rn(~) is smooth, it can be written as
The fact t h a t the 99(x  k) are an orthonormal s y s t e m is not a surprise, hecause J
J[o ,2W] n
by (15) and (16).
If we didn't normalize the wrong way, this means t h a t the ~ ( x  k] axe an o r t h o n o r m a l basis of some suhspace Vo of V0. To check t h a t Vo = Vo, note t h a t V0 is c o m p o s e d of all products n ( f ) ~ ( f ) , where n is 2~rperiodic in all directions and locally in L 2. T h e same c o m p u t a t i o n shows t h a t (Vo) ^ is composed of all products n ' ( f ) ~ ( f ) = n ~ ( f ) m ( f ) ~ ( f ) , where n ~ is also 2~rperiodic and locally in L 2. Since m is C °°, 2~rperiodic and invertihle, Vo : Vo, and we proved the lemma.
Remark about uniqueness. If we wanted ~o to define an orthonormal system, then we had to choose G(~)irn(~)l 2 so that all its Fourier coefficients (except the 0 th) would be zero. So Im(~)[ 2  G(~) 1 was necessary. On the other hand, multiplying rn(~) by any 27rperiodic, C °o function of modulus 1 would have given a function ~ with exactly the same properties. Finally, one could always try to multiply m(~) by unimodular functions that are not smooth, which could give the stupid result of finding a ~ which does not have enough decay ! Taking for m(~) the positive square root of G(~) ~ has the advantage of being natural, and of giving the same function ~ as G r a m ' s orthogonalization process. 4.
C o n s t r u c t i o n o f ¢ in d i m e n s i o n 1.
We are now ready to construct the wavelets in dimension 1. Remember that we want to find a function ¢ such that the ¢i,k(x) = 2J/2¢(2Jx  k), for j G Z and k G Z, form an orthonormal basis of L2(R). The idea is to find a ¢ such that, for each j , the 2J/2¢(2Jx  k), k E Z, form an orthonormal basis of the orthogonal complement of Vo in
v:. Call Wo the orthogonal complement of V0 in V1, and let us state what we want as a theorem. T h e o r e m 4 . 1 .   Let (Vy) be a rregular multiscale analysis of L2(R). There is a function ¢ E V:, satisfying (9), and such that the ~b(x  k), k E Z, form an orthonormal basis of
Wo. For this function ¢, the ¢i,k(x) = 2i/2~b(2ix  k), (j,k) E Z 2, form an orthonormal basis of L2(R). Let us first check that the last statement follows from the first one. Using the rule (6), we see that, for each j , the ¢z',k, k E Z, form an orthonormal basis of Wi, the orthogonal complement of Vi in Vy+I. We now use condition (5), which implies that L2(R) is the orthogonal sum of all the Wi's , j E Z, to see that the ~bi,k's, (j, k) E Z 2, are a basis of L 2. Let us now look for the function ¢. Since we prefer working with functions of V0, we will rather look for a function ¢1 such that the ¢ ' ( x  2k), k E Z, form an orthonormal basis of W1. First, note that ~ ( ~ ) E V1 C Vo, and thus one can find a sequence ak E ~2(Z) such that ~o(~) 1 x = E ak~(X + k). Moreover, since ak  ½f ~(~)~(x + k) , the sequence •k is rapidly decreasing. Applying a Fourier transform, we get (17)
~(2~) = mo(~) ~3(~),
where mo(~) = ~ ak e ikq is 27rperiodic and C °°. Lemma 4.2.
Ira0(~)[ 2 + lmo(~ + ~r)12 : 1.
Remember that, by definition of ~, +
k
2k )l
I (e)l l Ce
=
k
+
2k )l
=
Im(e)l C(, ')
=
1,
and therefore, ~ 1~(2~ + 2k~)l 2 = 1, too. Using (17), k
Imo(~ + k~)121~(~ + k~)l 2 = 1 k
We now split the sum and use the fact that m0 is 2rperiodic, to obtain
Im0(~)l z ~
I~(~ + 2l~)1 ~ + Im0(~ + ~)P ~ l
I~(~ + ~ + 2l~)1 ~ = 1,
l
and so, [rn0(~)l 2 + Im0(~ ÷ 7r)l 2 = 1, as promised. From the fact that the 2   !2to(yZ  k) form an orthonormal basis of V  l , we see t h a t
V1 ={rn(2~)~(2~) :rn
is
={m(2~)rno(~)~(~) :rn
2~periodic is
and locally in
2~periodic
L 2}
and locailyin
L2}.
We can use this description of I2_1, together with the fact t h a t for each f E V0, T c a n be written as ]'(~) = m(~)~(~) for a 27rperiodic m(~) such t h a t [[fll = IlmllL2([0,2~[) (SO that the map f ~ rn is unitary), to find W1. Let us determine which 27rperiodic functions £(~) are orthogonai to all the m (2 ~) mo (~), where m is 27rperiodic on [0, 2~r]. The condition is
L
" ,,,(2¢){~o(¢)l(¢) + ~o(¢ + ~)~(¢ + ~)}d,~ = o,
and so m0(¢)l~ + mo(¢ + ~ r ) ~ = 0 a.e. on [0, ~r]. By L e m m a 4.2, the vector with coordinates rn0(~) and mo(~ + ~r) is always of length 1, and so the orthogonaiity condition is equivalent to (18)
l(¢ + ~r)
= )~(¢)ei¢ ~, m6(~)
a.e. on
,
for some function A(~) [we added the e i¢ for our later convenience]. Note t h a t Ill(~)11r.2([o,2,]) = IIA(~)IIL2([o,,]) by L e m m a 4.2. Also, the first line of (18) is l(¢) = A ( ¢ ) e  i ~ o ( ¢ + 7r), and the second line is l(¢ + ~r) = A(¢)eiC~+~)~o(¢ + 2~r), so that (18) is equivalent to
(19)
e(¢) = )~(¢)ei~'~o(~¢+ 7r) a.e,
for some 7rperio.dic function A . The space W1 is thus composed of all functions £(~)~(~), where £ is as in (19). T h e equality between norms even allows us to get an orthonormal basis for W_~ from a basis of L2([0,~T]) : we can take the functions A(~)  vz2c ~ , k E Z, get the functions
~(¢)~(¢) = v %  ' ~ o ( ¢
+ ~)~(¢)a '~ , k e Z,
in W_~, and then an orthonormal basis of W_, is given by the ¢ ' ( x  2k), where ¢'(~) = v ~ e  i ¢ ~ 0 ( ~ + ~r)~(~). By a dilation, we see that a basis of Wo is the ¢ ( x  k), k E Z, where
(20)
~(,~) = e~U"mo,,~ '¢ + ~),Z(¢).
As before, since mo is C °° and 2~rperiodic, we see that ¢ satisfies (9) because tP does and ¢ is a linear combination of translates of/o with coefficients t h a t decay rapidly at oo. This finishes the proof of Theorem 4.1. Let us end this section with a few examples.
10 E x a m p l e 1 : ( S p l i n e f u n c t i o n s o f o r d e r 0) In this case, we can take ~o = g = l[o,zl. To c o m p u t e m0, note t h a t 1 x 1 1 5~,(~) = 5 ~ ( x ) + ~ , ( x  1), so t h a t 1 + ei~)~((), ~ ( 2 ( ) = {(1
and
m 0 ( ( ) = {1 (1 + e i~)
Next,
¢C,') = e'~/~ 1(1  ei~/~)~(~12) 2
and so ¢(x) = @(2z  1)  99(2x)  l[z/2,z]  l[o,z/~ I • We miraculously recovered the H a a r system ! Exercise : Example
C o m p u t e w h a t h a p p e n s for splines of order 2, or look in [My4].
2 :
(Y. Meyer's wavelets).
Let us consider the multiscale analysis of E x a m p l e 3, Section 2. In this case, we can take ~o = g = ~[, and so
2x 1 where m0 is such t h a t 0(2~) = rno(~)O(~). Note t h a t 0(2~) is s u p p o r t e d on ft   g2~" , Y , , at the place where 0(() = 1. So rn0(~) = 0(2() for ~r < ~ < ~r (and is defined everywhere else by 2~rperiodicity), and so
=
+
+
 2
)1.
Note t h a t when one of the two functions 0 ( ( / 2 ) and [0(( + 2 r ) + 0 ( ~  2~r)] vary, then the other one is constant and equal to 0 or 1 (see fig.l). Thus, ¢(~) looks like the p r o d u c t by e  i ~ / 2 of the function 01 of fig.2.
e(,~+2~)
0({/2)
o(~2rO
..
%
O,¢
z ,,,
,,,
..," 
8&/3
2&
4k/3
2k/3
'0
2~/3
4~/3
Figure 1: 0(~2) and [0(~+2n) + 0(~2~)] .
{n
%
k
8ff/3 "
11 local points of ~ I , s y m m e t r y f°r 02 ~
8~/3'" 2[
4~3
~:~/3 Figure
I0 2:
4
2rt
8~nn/3"
=
The graph is also (locally) invariant by (x, y)
(
 x, y).
R e m a r k : Calling A1 the orthogonal projection on Wy, note that A i is a projection of LittlewoodPaley type. This is certainly a justification of our earlier statement that "Vj contains the part of functions with frequencies lower than 2i" . 5.
Wavelets in higher dimensions : tensor p r o d u c t . Remember that if (Vj) is a MSA on L2(R), we have a MSA of L2(R2), denoted by Vi ®Vi, which is obtained as in Example 4 of §2. Let us build a basis of L2(R2), using the Cj,k's given by Theorem 4.1 (applied to (Vi)). The following construction is due to P.G. Lemarid. From V1 = V0 _l_Wo (the symbol means "orthogonal sum") we get V1 ®V1  (rio ®V0) _l_ (Vo ®W0) ± (Wo ® V0) _l_~V0 ®W0) (exercise). Now, the ~ ( x  k ) ¢ ( y  l ) , (k, ~) E Z 2, are an orthonormal basis of V0®W0, and the ¢ ( x  k ) ~ ( y  l ) (respectively, ¢ ( x  k)¢(y  £) are an orthonormal basis of W0 ® V0 (respectively, Wo x W0). Thus, calling ¢0,1(x,y) = ~(x)~l,(y), ¢l,0(x,y)  ¢(x)~(y) and C LI(x,Y) = ~b(x)¢(y), we see that an orthonormal basis of L2(R 2) is given by the 21¢c(2~x  k l , 2 i y  k2) where E {(0,1), (1,0)(1,1)} and j, k l , k 2 E Z. The case of L 2(R n) is similar, so we'll just give the formula, and leave the verifications as an easy exercise. Call E the set of sequences ( E l , . . . , e , ) E {0,1} '~ that are different from (0,0...0). For e E E, set Ce(x) = ¢ ~ ( x l , . . . , X n )
= ¢~(xl)¢
e~ ( x 2 ) . . . ¢ ~ " ( x n )
where ¢o = ~o and ¢1 = ¢. An orthonormal basis of L2(R n) is obtained by taking all the functions 2~J/2¢~(2Yxk)
for e E E
,
jEZ,
and
k E Z ~.
12 Example. W h e n (Vi) corresponds to spline functions of order 0, one gets the usual twodimensional Haar system on L 2 (R2), defined with the functions ¢0,1
=
=
l 
Xlo,ll×lo,
j ,
¢~,o = ~o(x)¢(y) = l[½,x]xio,x ]  l[o,½}x[o,~ ] , and ¢1,1 = 1Q  1R, with Q = [~,1] 2 u [0, ~]2 and R = [0,1] 2 \ Q. Exercise : MSA's ? 6.
Can one take the tensor p r o d u c t of bases L 2 ( R ) coming f r o m different
O t h e r w a v e l e t s in h i g h e r d i m e n s i o n s .
Suppose we have a MSA on L 2 ( R ") which is not obtained by taking the tensor p r o d u c t of onedimensional MSA's. We still want to find an o r t h o n o r m a l basis of L 2 ( R n) a d a p t e d to our MSA. As in the tensor p r o d u c t case, we'll need 2 n  1 different functions ¢, from which we'll build a basis of Wy (we still call (Vj) our MSA, and W i the orthogonal complement of Vy in Vy+I) and then a basis of L 2 ( R n) by the usual scheme. As far as I know, the following construction is due to K. Grbchenig and Y. Meyer. Theorem 6.1.Let (1Ii) be a rregular multiscale analysis of L 2 ( R " ) . One can find 2 n  1 functions Ce,~ E E = {0, 1 ) " \ {0}, such that each Ce satist~es (g), and the Ce ( x  k ) , e E E and k E Z ", are an orthonormal basis of Wo. Then, the 2'~Y/2¢e(2ix  k ) , ¢ E E, j C Z and k E Z ~ are an orthonormal basis of L 2 ( R " ) . The last statement follows from the first one exactly like in dimension 1. Finding the ~be's is a little more complicated than in dimension 1, but the spirit is the same. R e m a r k : Our a r g u m e n t will use the action of a group of translations, and the existence of a subgroup of index 2 n ; it could be generalized to other situations with other groups (see [My4]). Although we shall not use t h a t generality, it is a good thing to have in mind that we are using group actions. As before, we can write ~(2~) = mo(~)~(~) for some 2~rperiodic function mo and, since m0(~) = ~ a k e ik~ with a sequence ak = ½f~o(~)~(x + k)dx which is rapidly decreasing, rno is C °°. Since the functions ~be, ~ E E, t h a t .we are looking for are in V1, we can write (21)
=
for some function meC~) which is 2~rperiodic in all directions. To simplify the notations, we shall add ¢(o,o ..... o) = ~o to our set of functions, and rn(0,0 ..... 0) = mo, so t h a t (21) holds for all e C E ' = {0, 1} n. Let us cut the series giving each me into 2 n pieces, and write (22)
me=
Z
eiV~me,"(2~)'
flEE'
where, for each class z/ in Zn//(2Z) n, we group all terms of the Fourier series of me in the class defined by r/. The functions me,, are 2~periodic.
13 Our problem is to find functions m e , , , ~ ~ E and ~7 ~ E ~, in such a way t h a t the Ce(x  k), ~ ~ E ~ and k ~ Z n, form an o r t h o n o r m a l basis of V1. Let us denote by U(4) the 2 '~ x 2 ~ matrix with rows and columns indexed by e ~ E ~ and 77 ~ E ~, and general t e r m m~,,(4).
The functions ¢~(x  k), e ~ E ' and k ~ Z n, form an orthonorma! of V 1 if and only if the matrix 2n/2U(4) /s unitary fort almost every 4.
Lemma 6.2.
basis
To prove the lemma, let us first c o m p u t e fCe,(x÷k')¢e(x ÷ k)dx for 6 , ~ E E ' and k, k ~ ~ Z ~. As before, it is convenient to c o m p u t e scalar products on the Fourier transform side. Using ]I{rn~}Vll = llmlI~([o,2,r].), then replacing 4 by 24, and then using the f a c t that rn is 2~periodic, we get
/ Ce'(24)(¢e)(24)a Ck'k) d4
=
by
=
(21)
J [o,2~}s [o,2~t
= c ~[0
rne, C4/2)rne(4/2)e"i(k'k)~ d4
)4"~'] n
(4)d4 (by 22).
= c[ ~[0 ,4~'] n
.i
.
Notice t h a t all the me,, and me,,,, are 2~periodic in all variables. If we perform the integral with respect to 4,~ first, and split [0,4r] in two equal parts, we will get opposite results unless rl,~ = y~. The same thing would be true if we integrated with respect to any other variable, and so the only terms left correspond to r / = ~/s :
/~be'(x+k')¢e(x+k)dx=2'~e f[ o )27r] n
eiCk'k)~Z .EE
me',"(4)rn~,"(4)d4 ~
= C~[o,2,r]"ei(k'k)¢(v'(4)'v(4)}d4' where v(4) is the vector with coordinates 2=/2me,~(4), ~/ C E ' , v'(4) is the vector with coordinates 2'~/2me, ~(4), and ( ) is the usual scalar p r o d u c t in C 2". If the ¢ e ( x  k) are an o r t h o n o r m a t basis , our c o m p u t a t i o n shows t h a t the Fourier series of (v'(4),v(4)) must be zero when e' # e and so (v'(4),v(4)) = 0 for almost every 4. Also, for e' = ~, the Fourier series of (v'(4), v(4)) is the same as the Fourier series of 1, so [Iv(4)][ = 1 for almost every 4. So the matrix 2n/2U(4) is u n i t a r y almost everywhere, and the condition of the lemma is necessary.
14 Conversely, if 2"*/zU(~) is unitary a.e., the c o m p u t a t i o n above shows t h a t the C t ( x  k ) are an o r t h o n o r m a l basis of some subspace of V1, and we only need to check t h a t it is the whole V1. Let us proceed by contradiction, and suppose t h a t some f~.E VI is orthogonal to all the C o ( x  k). Let rn(~) be the 2~rperiodic function such t h a t f(2~) = rn(~)~(~). Let us decompose rn(~) like in (22), to obtain rn(~) = ~ einqrn~(2~) for 2"* functions rnn t h a t ~EE'
are 2~rperiodic. Following the same c o m p u t a t i o n as above, we get
f f(x)¢~(x + k)dx = flo,2,~] eik¢<w(¢)' v(~)>d~, where w(~) is the vector with coordinates m~(~), and v(~) is as before. Since this is zero for all e E E ' and k E Z n, we get (w(~),v(~)) = 0 for all e E E ' and a.e. ~. Since 2'*/2U(~) is unitary, the v(~) span the whole space, and w(~) = 0 a.e. This of course implies t h a t f = 0, and proves L e m m a 6.2. Now we see t h a t finding the ¢~'s (for e E E) reduces to the following problem : we are given the first row of a 2 n × 2 ~ matrix, and we want to find 2 n  1 other rows, so t h a t the resulting matrix be unitary. Also, we want to do this for each ~, but in a continuous way.
First note that our first row is of length 1 for almost all ~, because the ~ ( x  k), k E Z n, are an o r t h o n o r m a l basis of V0 (just apply the c o m p u t a t i o n of L e m m a 6.2 with = ~ = 0). A natural a t t e m p t to complete the matrix would be to associate, to each vector of the unit sphere of R 2" (or R 2~+~ in the complex case), a unitary matrix which has this vector as its first row. T h e trouble is t h a t such a m a p p i n g from vectors to matrices, if we also require it to be continuous, does not always exist ! To avoid this difficulty, we shall use a trick, due to K. Gr6chenig. T h e idea is t h a t our first row will not take all possible values in the unit sphere, and so we only need to define the m a p p i n g from vectors to matrices in a proper subset of the unit sphere (and then the topological obstructions disappear). Note t h a t ~ ~* (2"/2rn0, ~ (~))~eE' is a C °o mapping from R n to a sphere of dimension 2"*  1 > n (because n > 2) in the real valuedcase, and 2 "*+1  1 > n in the complex case. Consequently, the image by (2"*/2rno,n)neE, of [0, 2r]'* has zero surface measure and, since it is compact, there is a small ball, centered on the sphere, which does not meet the image [2"*/2(mo,n),eE,}([O, 27r]'*). We shall use the following lemma. 6.3.Let S be the unit ball of R 2" (respectively C2"), and B a small ball centered on S. There is a mapping F, defined and C °o on S \ B, with values in the set of unitary, 2"* × 2"* matrices with real (respectively complex) coefficients, and such that for all x C S \ B, x is the first row of f ( x ) .
Lemma
Let us first see how to deduce T h e o r e m 6.1 from this lemma. Let us take, for our matrix ((2'~/2me,,)), the function F((2~'/2rno,,)) given by the l e m m a [where B is a ball that does meet (2"*/2rno,,([O, 2r]'*)) (2"*/2mo,tl(R'*))]. T h e n all the me,n are Coo, 2~rperiodic in all directions, and the matrix 2"*/2U(~) of L e m m a 6.2 is unitary for all ~. It follows t h a t the functions ¢c, ¢ E E ' defined by (21) and (22) give an o r t h o n o r m a l basis =
15 of V1, and so the ¢ ~ ( x  k), e E E and k E Z n, are an o r t h o n o r m a l basis of W0. Also, each rn~(~) is C °°, and so the functions Cr defined by (21) satisfy (9) because ~o does (we have seen the a r g u m e n t a few times already). Thus T h e o r e m 6.1 follows from L e m m a 6.3. Let us prove the lemma in the complexvalued case (in the real case, one would just remove the bars). Call q = 2 n  1 and let ( Z l , . . . ,zq,zq+l) denote the coordinates of a vector z E S c C q+l. We can safely assume that B is centered at ( 0 , . . . ,0, 1). Our first step will be to find q vectors w l , . . . ,wq, so t h a t Z, W l , . . . ,Wq form a basis of C q+l. Let us take for w I the (3" + 1) at column vector of the matrix
z2
a 0
0 a
... ...
zq Zq+l
0 ~l
... ...
a Y~q
zl
(23)
A(z) =
where a > 0 is chosen as a function of the size of B. We just have to show t h a t det (A(z)) 0 for all z ~ B , if a is small enough. Exercise:
Prove t h a t det (A(z)) = (l)qql{•ql([zl] 2 J r ' ' ' ~[Zq] 2)  olqzq+l}.
Let us check t h a t if z ~ B and a is small enough, det A(z) ~ O. If this was not true, one would have aZq+l = [Zl[ 2 + .  . + IZq[2, and so Zq+l :> 0 and also, since [Zq+l[ _< 1, [1  Zq+l[ 2 = [zll 2 +  . . + [zq[2 <_ a. This is impossible since z is not too close to ( 0 , 0 , . . . , 1). So, if a is small enough, the vectors Z, W l , . . . ,Wq form a basis of C q+l. Let us now apply the GramSchmidt orthonormalization process, starting with z so as not to change it. We get a matrix F(z),which has z as its first row, is unitary, and also depends in a C °O m a n n e r on z, because A(z) does (in the orthonormalization process, one never divides by 0, or takes the square root of 0). This proves the lemma. 7.
Compactly
supported
wavelets .
We still p o s t p o n e the description of the various properties of the various wavelets that we constructed (the interested reader can go directly to the relevant section), and construct some more. Let us come back to one dimension, and try to construct compactly s u p p o r t e d wavelets. The first construction t h a t was similar to the one we are going to present was due to Tchamitchian [Tcl,2]. He constructed two functions 0 and ~b, with c o m p a c t s u p p o r t and prescribed regularity, such t h a t the 0j,k and the ¢i,k (with the notations of Sections 1 and 4) are two (dual) Riesz bases of L 2 ( R ) , and such t h a t every f E L 2 ( R ) can be written (24)
f = E
cY,k 01,~ with
Cj,k = / f(x)¢y,k(x)dx.
The following, more precise theorem is due to I. Daubechies [Dbl]. T h e o r e m 7 . 1 .   There is a constant C >>0 such that, for each integer r > 0, there is a multiscale analysis of L 2 ( R ) , of regularity r, such that the functions ~o and ¢ of Section 3 and Section 4 are compactly supported, with supports in [Cr, Cr].
16 Note t h a t finding a ¢ E C ~ is impossible (exercise : it would follow f r o m the fact t h a t the ¢),k form an orthonormal basis t h a t all moments of ¢ are zero, which is impossible if ¢ is supported in [  M , M ] by density of the polynomials). The fact t h a t ¢ has c o m p a c t support is a nice property, because it makes the "wavelet transform" completely local (instead of almost local). One can hope t h a t this p r o p e r t y will be useful for practical problems (for instance involving time !). The proof we are going to see is not the original one. As usual, it was c o m m u n i c a t e d to me by Y. Meyer, and it involves ideas of S. Mallat, Y. Meyer, JP. K a h a n e and Y. Katznelson. The construction is also similar to Tchamitchian's work mentioned above. The idea is the following. If (Vi) is a MSA of L 2 ( R ) , we have seen in Section 4 t h a t there is a 2~rperiodic, C ~ function mo(~), such t h a t ~(2~)  mo(~)~(~) (this is (17)). We shall start from a function m0, and try to build a MSA with it. T h e point is t h a t the size of the supports of ¢ and to is easier to control from m0. So, let us give ourselves a function m0 with the following properties :
(25)
{
m0(~) is C ~ and 2 ~ r  periodic, Im0(e)l 2 + Imo(e + )12 = 1 for all too(0) = 1.
We have seen in L e m m a 4.2 that the first conditions are necessary if we want m0 to arise from (17) for a MSA. It would not be too hard to check t h a t the third condition is necessary too (because of (17), one only needs to check t h a t ~(0)  1). Since we are only interested in the converse, we will leave the verification to the reader. If we want to build a MSA from too, we will need a function to, and here is a good candidate : (26)
=
m0C2J
).
j=l Note t h a t the infinite p r o d u c t converges because m0(0)  1 ; indeed, m 0 ( 2  i { ) = 1 + O ( 2  J ) , so that ~ L o g m 0 ( 2  i { ) converges uniformly on c o m p a c t a , and 3' so ~({) is welldefined. We still want to check t h a t ~ ( x  k), k E Z, is an o r t h o n o r m a l system, and t h a t its closed span V0 is part of a MSA. Finally, we shall have to choose m0 so t h a t the resulting to is regular enough, and compactly supported. If we want ~p to be compactly supported, we'Ll have to choose for m0 a finite trigonometric sum, because
mo(~) = ~ O~keik~, where On the other hand, is compactly supported. which are s u p p o r t e d in the supports, i.e. [  2 c ,
ak = ~1 X to(])~o(x x   + k)dx
is also c o m p a c t l y supported.
suppose m0 is a (finite) trigonometric sum, and let us show t h a t to By (26), to is the (infinite) convolution p r o d u c t of the [2Jrh0(2Jx)], [  c 2  J , c 2  J ] , so that the support of ~a is contained in the s u m of 2c]. [Exercise : check the argument, and in particular make sure
17 it has a sense distributionwise, l Finally, if ~o is c o m p a c t l y s u p p o r t e d and m0 is a finite trigonometric sum, ¢ is also c o m p a c t l y s u p p o r t e d because of (20). So we'll choose for m0(~) a trigonometric polynomial. Let us come back to the p r o b l e m of defining a M S A from a function m0(~) defining
(25). Lemma
7.2.
If ~ is defined by (26) for a function m o satisfying (25), then [[~o][2 <_ 1.
We intend to prove later t h a t the n o r m is exactly 1, but let us s t a r t with this. Call 2N~r
N
IN = [
Irb,(~)12d~,
where
HN(~)=
lI m°(2i~)" 3'=1
J 2Nlr
Let us compute IN by induction. Since fIN is 2N+17rperiodic, ~0 2N+ J"~"
IN =
/ 2N~"
InNCe)12d,' . . . .
/ 2 N+l 
+ ]=,,,~
"".
J0
In the second integral, we write ~ = 2N~r + u and get
IN : fo ~'~ IYINX(¢)I 2 (Imo(~N)12 + Imo(~N +Tr)l 2) de / , 2N~r
I H N  I ( ~ ) I 2 d ~ I N  I . . . . .
 I
Ii
J0
=
[too
12d~ : 2
21r
= 2 Since
Imo(~)l <
/;
[mo(~)12d~ 7r
(Imo(~)l 2 + Imo(~ + 7r)12)d~ = 2~.
~, I~(~)I <
IY[N(~)I, a~d so
I" 2N~
[
1~(¢)[2d¢_< IN =
27r;
J 2N1r
letting N ~ + c o , we get
f~_~ I~(x)12dx< 2~, and so f ~ I~(x)12dx< 1.
At this stage, we rteed to be a little more careful, because of the following two examples : 1)
if rn0(¢) 1+ei¢2, then ~(~) = e IU2 ¢/2 , and ~ = l[x,ol (we leave the c o m p u t a tions as an exercise, but you can use E x a m p l e 1 of Section 4 if you axe lazy). In this case, one obtains H a a r ' s basis, and everything works smoothly.
2)
1 (change variables in if mo(~) = l+e31t (25) is still satisfied, but now ~o = ~1[3,o] 2 ' the c o m p u t a t i o n s of case 1)). T h e / o ( x  k) are no longer an o r t h o n o r m a l b a s i s , so something went wrong !
=
Fortunately, the m i s h a p p e n i n g of case 2) is rather easy to prevent : we shall see t h a t it is enough to ask mo not to take the value 0 between =~ and ~.
18
L e m m a 7 . 3 .   If mo is a trigonometric polynomial satisfying (25), and if mo( ~) # 0 for ~ E [=~, ~], then the ~o(x  k), k E Z, are an orthonormal sequence. Note t h a t ~o E L 2 (R) because of L e m m a 7.2, and is compactly s u p p o r t e d because n o is oo
a trigonometric polynomial, so ~ is in all the Sobolev spaces. Let G(~) = ~
[~(~+2klr)I s.
OO
This is the same function as in Section 3, except that now the dimension is 1 and ~o = g. The proof of L e m m a 3.1 still shows t h a t G is C °°, and the c o m p u t a t i o n giving (15) (with m = 1) shows t h a t the ~ ( x  k ) , k E Z, are an orthonormal system if and only if G(~) ~ 1. Let us first check t h a t G(0) = 1. Note t h a t ~ ( 2 £ r ) = 0 if £ is a nonzero integer because if ~ = 2Pq for some odd integer q, then mo(qr) shows up in the p r o d u c t giving ~(2£~r), b u t mo(qTr) = m0(~r) = 0 since too(0) = 1 and [m0(0)[ 2 + [ m o ( ~ ) [ 2 = 1. It follows that G(0) = 1, as promised. To prove L e m m a 7.3, we only need to show t h a t G is a constant. To do so, we shall use an a r g u m e n t of Tchamitchian. Call a(~) : [m0(~)[ 2, so t h a t 0 < a(~) < 1, a(~) + a(~ + ~r)  1, and
I~Ce)l ~ = IIj%~ ~(2ie). Let us first prove t h a t (27)
c(2e) = ~(~) VCe) + ace + ~) e(~ + ~)
(so G(2~) is a barycentre of G(~) and G(~ + 7r)). Write
GC2~) = ~
!~(2~ + 2k~r)l2 = ~
k
=Z
Imo(~ + k~)l~l~(~ + k~)l ~
k
+Z
k even
= I",o(~)1 :~Z
k odd
I~(~ + 2k~)l ~ + I~oC~ + ~)1 ~ ~
= a(~)G(~) + a(~ + ~r)G(~ + ~r),
I~C¢ + ~ + 2k")l ~
as promised.
Call m the m i n i m u m and M the m a x i m u m of G on 1t (i.e., on [~r, ~r])). Let ~o E [zr, 7r], such t h a t G((o) = m, and apply (27) to ~ = ~0/2. Since a ( ( ) > 0 by hypothesis (~ E [ : ~ , ~ ] ) , we get G(~) = rn, too. Applying this a r g u m e n t to ~o/2, and then ~o/4, and so on, we get G ( ~ 0 / 2 " ) = rn for all n and, since G is continuous, G(0) = m. We can also use the argument with a ~o such t h a t G(~0)  M , and obtain G(1) : M . Consequently, G is a constant, and L e m m a 7.3 is established. We'll need the following lemma. 7.4.Given r > O, one can find a finite trigonometric sum mo( ~), satisfying the hypotheses of Lemma 7.3, and such that the function ~o has r bounded derivatives.
Lemma
Of course, since ~o is compactly supported, it will also satisfy (9). Let us momentarily admit the lemma, and see how we get a multiscale analysis from mo.
19
Call V0 the closed span of the ~o(x  k), k E Z. Because of L e m m a 7.3, the ~o(x  k) are an o r t h o n o r m a l system, so t h a t (8) is automatically satisfied. Define Vj, j E Z, by the rule (6). To prove t h a t the Vj's are an increasing sequence, we only need to check t h a t V1 c Vo, which is true because l~ofz~ ,~j C Vo since ~(2~) = m o ( ~ ) ~ ( ~ ) by definition of p . We still need to check (5). Call E j
the orthogonal projection from L 2 onto Vi. We need to show t h a t for each
f E L 2, E j f , at as j , +co, and E i at * 0 as j * oo. We can c o m p u t e the kernel function of E j , because
E y I = ~'(at, ~oy,k)~oy,k k with the usual notations. The kernel of E 1 is S i ( x , y ) = 21Eo(21x,21y),
Eo(x,y) = Z ~ C x k)~oCy k). k Since ~o is b o u n d e d and compactly supported, lEo(x, Y)I ~ C, and also Eo(x,y) = 0 where
if Ix  y] > 2 diam (supp ~o). Next,
k because ~(0) = 0. Since ~ ~ ( x  k) is periodic, and k  k)
}
=
k
J/
=
= 6 =0,
we get 1. k With these properties of Eo, it is easy to show t h a t E j f ~ f when j 4 + c ~ and E j f ~ 0 as j ~  o o whenever at E C ~ , say. Since the Ei's are uniformly b o u n d e d (again by the properties of E0, or simply because they are projections), one gets /EoCx,
E j f ~
y)dy =
{ fo as j_~{+co_co forall
f G L 2,
and so the Vj's are a multiscale analysis. Since the ~ ( x  k), k E Z, are an orthonormal basis of Vo, ~ is really the function of Section 3, and so the MSA (Vi) is rregular if m0 was chosen with the help of L e m m a T 7.4. Furthermore, if m0 can be written ~ ake ~k~ for some T, then the supports of the T functions ~o and ¢ are contained in [  C T , CT] for some C _~ 0. So we see t h a t we'll have established the theorem if we can prove L e m m a 7.4, with a trigonometric polynomial m0 of degree _< Cr. We shall deduce the fact t h a t ~ has r b o u n d e d derivatives from the simpler inequality (28)
I~(~)l < c ( 1 + I~]) 8
with, say,
s = r + 1,
and Sobolev's embedding theorem. The following L e m m a will help us. It is due to F. Riesz, and its proof is left to the reader as an exercise.
20 T g( ~) = y~ ,,liceik~ be a trigonometric polynomial which is nonT negative on R . T h e n there is a trigonometric polynomial o f the form T mo(~) = y ~ c t k e ike such that Im0(~)l 2 = g(~). T Taking this l e m m a into account, we only have to find a nonnegative trigonometric T polynomial g(~) = ~ "/ke ~k~, with T <_ Cr, with the following properties : T Lemma
7.5.
(29)
Let
g(~) + g(~ + ~r) = 1
(30)
for all
g(~) > 0 on
~,
Tr
and
[~,~1,
g(o) = o;
and
oo
(31)
H g(2J() < C(1 +
I~1) 2s
y=l (condition (29) and (30) will give the hypotheses of L e m m a 7.3, and (31) ensures (28)). By the way, we shall obtain a g with real coefficients and, looking more closely at the proof of Riesz's lemma, one sees t h a t one can get a m0 with real coefficients ak (we shall not need t h a t precision). Let us t r y to m o t i v a t e the definition of g t h a t follows : if we did not have to select a trigonometric polynomial, and were just interested in (31), we could chose k T h e infinite product of (31) would then be s u p p o r t e d in [~r,~r I because if I¢1 > ~, then one of the ~/21 falls in +~  ~   ,  ~ [ and the product is zero. One can see the definition of g below as an a t t e m p t to imitate this example, and choose g as large as possible near 1 and as small as possible near 7r, so t h a t the infinite p r o d u c t will have more chances to drop sharply. To define g, select a large k E N , and call Ck =
{/0"(sin t) 2k+l dt }1
We shall define g by (32)
g(~) = 1  ck r]n~(sin t) 2k+1 dt.
Note t h a t g is a trigonometric polynomial of degree < 2k + 1, t h a t 0 < g(~) < 1 everywhere and g(~) > 0 on ]  ~r, ~r[ by our choice of ck, and t h a t g(~) + g(~ + It) = 1. So we'll just have to check t h a t oo
G(~)= Lemma
7.6.
1I 9 ( 2  j ~ ) j=l
We have ck < Cx/k.
satisfies
(31)
for some
s>k/C.
21 Choose a small a > O, and write
f,_~÷a/V~
1 _> (sin t)2k+'dt > 2a ~ [sin ck ~a/v'~
(2
~k)] 2k+1
+
V~ COS Since (2k+l)
Log
(
cos
~(2k+l)~
~ 2 a 2,
we obtain e~1 > ( C v ~ ) 1, and so the lemma is true. Lemma
7.7.
If k is large enough,
(33)
[4~ k g(t) < \ 5 ]
(34)
g(t)
2.
2k÷2
<4. ~,
and
for 0 < t < 2  .
To prove (34), we write g(5) = ek f [ ( s i n t ) 2 k + l d t for 5 ~ [0, 2~r], and, since I sin u t _< lTr  u I, we get
g(~) <__ckl~  ~12~+2/2k + 2 ___ I~  "12~+2 if k is large enough (because of Lemma 7.6). To prove (33), we write
g(() < Ck
(sin
)2k+ldu ~ cV/~(
)2k+l N ( )k
if k is large enough. Let us summarize in a lemma a few computations that we shall do later. L e m m a 7 . 8 .   Given 0 O, there exists a > 0 such that the following is true. Let f(t) be continuous, 1periodic, such that 0 < f(t) < 1 and 1 <_ t < _ 2 ~
(35)
o < f(t) < 8
(36)
0 < f(t) < C I t  ~l
1
Then Fj(t) : f(t) f ( 2 t ) . . , f ( 2 J  ' t )
(37)
for
1sup ! Fj(t)
z
for
and
0 < t < 1.
satisfies <_ 2 "j
for j >_ 1.
22 Before proving this lemma, let us show how it implies that G(~) satisfies (31). A change of variables shows that sup
f(~)...f(~f)
sup
Fi(t) =
i1 < t < ~ 1
2~+2
and so (37) implies that co
(38)
H
f ( 2  t t ) < 2=Y
for
t C [2i2,211]
and
j _> 1.
t:l
Let us take fCt) = [gC2~rt)]I/(2k+2), so that (33) and (34) imply (35) and (36) with suitable values of 5 and C. Given ~ E [2Jl~,2JTr], we can apply (38) to t = ~/2~ and obtain G(~) : H [ f ( 2  l t ) ] 2k+2 < 2 ~(2k+2)j ~_ C(1 + I~[) 2~(k+1) . l=l A similar estimate would be true for ~ 6 [  2 i  1 , 212], and so we proved (31) with s = a(k + 1). As we said before, this proves that ~o has a(k + 1)  1 bounded derivations, and so, if we choose k > a  l r + a 1 we get the function m0(~) required for Lemma 7.4, therefore proving the theorem. Thus we only need to prove Lemma 7.8. To do so, it is more convenient to consider the function h(t) = f(t) f(2t). If we can prove that h(t) h ( 2 t ) . . , h(211t) <_ 2  a j , this will imply that f(t)f(2t)2.., f(211t) 2 f(2 jr) < 2 ~1, and then f(t).., f(21t) < 2 ~j/2, which will imply (37) with a different value of a. The function h is a little better than f , because it satisfies (39)
O
(40)
0~h(t) ~ Clt~l
1
We want to show that (41)
for
~l < t < 5
~
for
0
and
1.
H(t) = h(t) h ( 2 t ) . . , h(21lt) satisfies H(t) < 2 ~3• for ~l <_t < _
1 2
(the case of t = ½ is not a problem because h(½) : 0). W r i t e t in base 2 : t = 0 . ~ +11  ~ + 1 that if q > 0, h(2qt) : h(aq+l, aq+2,'")
a 3 ~i + ' " o r , i n s h o r t , t : (0, I, a s  ). Note by periodicity.
Given t, call F C {0, 1 , . . . (j  1)} the set of q's such that aq+l ~ aq+2. Let us see why h(2qt) should be small when q e F. If q C F, 2qt is of the form ( 1 , 0 , . . . ) or (0, 1,...) modulo 1, and so it is between ½ and ½ + ¼ < s in the first case, and between ¼ and ½ in the second case. In both cases, (39) yields (42)
h(2qt) < ~.
23 If there are enough q E F, this estimate will be good enough. Otherwise, we will have to use (40) a few times. Let q E F. If 2qt is of the form ( 1 , 0 . . . ) modulo 1, call lq the number of consecutive zeros that are just after the 1 ; if 2qt is of the form Co, 1 . . . ) modulo 1, let £q be the number of consecutive l's after the initial zero. Let us give ourselves a large constant A > 0 (we shall decide on its value in a moment). If £q > A, we shall use (40). Indeed, 2qt looks like ( 1 , 0 , . . . 0 . . . . ) (with lq zeros) or ( 0 , 1 , 1 . . . 1 , . . . ) (with lq l's) modulo 1, and so it is at distance < 2 l~ from ½ (rood 1). Therefore,
(43)
h(2qt) < C2 tq .
If A is large enough, (43) implies that h(2qt) < 2 lq/2 for lq > A. We can then choose ct so that (42) implies t h a t h(2qt) <_ 2 al~ also when lq <_ A. Taking the product, we see that
H(t) < 1I
h(2qt) < 2
~,
,
and
qEF
so (41) will follow at once if we prove that ~
£q > j.
qEF
The proof is easy : split the interval {2, 3 , . . . j + 1} into intervals I t which are maximal and such that the sequence aq is constant on It. Each I t starts at q + 2 for a q E F and its length is less than lq. Also, j = ~ Iltl, and so ~ £q >_ j , as we wanted. F
This proves (41), which implies Lemma 7.8, which was our last missing piece for the proof of Theorem 7.1, which is thus established. 8.
Properties of the wavelet transform, extensions.
It is probably a good idea to justify all the work done in the previous sections by a few comments on the advantages and applications of the wavelet transforms. However, for reasons of relative conciseness, we do not wish to give anything close to a complete description of all applications of wavelets . Such a description can be found in [My4], or in the references therein. What we'll do instead is a short impressionistic picture of some of the advantages of wavelets . For convenience, we shall use the notations of dimension 1, and call
¢i,k = ~Sl~,/:(2sz  k}, j,k ~ Z an orthonormal basis of L2(R). With these notations, every f E L2(R) can be written (44)
f=y~Ci,kCy,k where
(45)
Ci,k = /f(X) Cj,k(x)dx. For practical purposes, it is often easier to use the formula
(46)
.,f = ~ ~ Cj,k~.,/,kI ~ "'/c~(Z k), j>0 ~ z
~Ez
24 where the cj, k are given by (45) and
(47)
= f f(z)
(x  k ) d x .
(Note that, if ~b and ~ are compactly supported, the cj,k and the ak can all be computed with a local knowledge of f !) Suppose that the MSA is very regular (for instance, suppose that ¢ is Y. Meyer's wavelet of Section 4, Example 2). The first very nice feature of (44) (or (46)) is that the coefficients ci,k are very small, as j becomes large, when f is smooth (we shall give one or two examples later). To see this, one first proves that ¢ has almost as m a n y vanishing moments as derivatives, and then one simply integrates by parts in (45). For instance, if IIf'll < 1, then P
ej,k = 2i/2 /
¢(2J
= 2j / 2 / I f ( x )
 k)d
 f(k/2J)] ¢ ( 2 1 x  k)dx
< 2  i / 2 / 1 2 1 x  k I ] ¢ ( 2 i x  k)ldx <_ C 2 3j/2 . As a general rule, each derivative of f allows one to win a factor of 2  i on the size of c1, k. A slightly more surprising fact is that many of the currently used function spaces can be characterized by the behaviour of the cj,k's (as j ~ + ~ ) . This is surprising if one compares the "wavelet transform" to the Fourier transform, for which many such characterizations are false. On the other hand, the reader who has been exposed to some LittlewoodPaley theory will be less surprised. In some sense, one can say that the "wavelet transform", which uses decompositions of frequencies in dyadic blocs, has all the advantages of LittlewoodPaley theory, with the simplicity of the Fourier transform (due to the fact that the coefficients are unique, for instance). Let us give three examples of characterizations of spaces by the size of the Cj,k'S ; we'll omit the proofs (they are very close to the usual LittlewoodPaley manipulations). E x a m p l e 1.
A function f is in the Sobolev space H s(R) if and only if
i
~(1 k
+ 2J)281cj,k[2 < + c o .
The condition would be the same in R n. The condition also holds, by duality, for s < 0. E x a m p l e 2. The function f is in LP(R), 1 < p < +co (or in H 1, the Hardy space, when p = 1) if and only if the following square function g(x) is in LP(R) :
=
2Jlcj, l 2 (,elk2',(k+l)~il
25 E x a m p l e 3. T h e function f is in B M O ( R ) if a n d only if t h e cy,k's satisfy t h e following Carleson m e a s u r e c o n d i t i o n : t h e r e is a c o n s t a n t C > 0 such t h a t , for each d y a d i c interval I,
~
Icj,k[2 <_ C[I[.
j" , k [~2J,(k+l)2~lct Of course, these c h a r a c t e r i z a t i o n s are also true, w i t h m i n o r modifications, in I t n. Also, m a n y m o r e function spaces have a c h a r a c t e r i z a t i o n like this (in p a r t i c u l a r , t h e various Besov spaces). T h e only ones to be e x c l u d e d are, m o r e or less, the spaces w i t h a L I or a LC~norm in their definition. We refer to [LM], or [My4], for a list. Also see [FJ3] a n d [FJW] for m o r e extensive results in t h e f r a m e w o r k of t h e ~otransform. A n easy corollary of these c h a r a c t e r i z a t i o n s is t h a t , for m o s t function spaces E , w i t h t h e n o t a b l e exception of B M O , if f E E , t h e n t h e p a r t i a l sums ~ Cj,k¢j,k converge to y


Periodic wavelets . It is n o t h a r d to find an o r t h o n o r m a l basis of L 2 ( T 1) (the 1d i m e n s i o n a l torus), say, w i t h t h e s a m e sort of s t r u c t u r e as a wavelet basis. For instance, one can t a k e t h e " s t a n d a r d " wavelets and p e r i o d i z e t h e m b r u t a l l y . See [My4], for instance. One can define M S A ' s on LZ(R2), say, t h a t are not given by a t e n s o r p r o d u c t , or even are b a s e d on a different group s t r u c t u r e (like t h e one a s s o c i a t e d w i t h a p a v i n g of R 2 by hexagons or triangles). In cases t h a t are not t o o simple, one has to replace t h e F o u r i e r t r a n s f o r m calculations of Sections 26 by a s u i t a b l e use of G r a m ' s o r t h o g o n a l i z a t i o n process. Still see [My4], for m o r e information. One c a n still go further, a n d b u i l d wavelets on various d o m a i n s of R n. T h e n t h e group s t r u c t u r e d i s a p p e a r s (and so does t h e Fourier t r a n s f o r m ) , b u t G r a m ' s process still allows one to b u i l d interesting bases. For m o r e details, see [JAM1].

In some cases, t h e fact t h a t t h e ~by,k's form a basis is not quite needed, a n d t h e n o t i o n of a "frame" is m o r e relevant. Let us not even say w h a t a frame is, a n d refer directly to [DbGM], or [My4].

T h e list above is not exhaustive, b u t since we d o n ' t w a n t P a r t I to be much longer t h a n t h e two others, we'll have to leave the wavelets here (we'll use a g a i n some of the ideas in p a r t II, however).
P A R T II SINGULAR
INTEGRAL
OPERATORS
ooOoo'
1. I n t r o d u c t i o n a n d g e n e r a l i t i e s As with Part I, we do not wish to give a complete description of the subject, but to try to explain in detail a small number of techniques that show up in the study of singular integral operators of"Calder6nZygmund type". In particular, we'll spend some time discussing a new proof of the Tbtheorem by Coifman and Semmes. Let us start this part with a short review on definitions and basic properties of Calder6nZygmund operators. D e f i n i t i o n 1.1. A singular integral operator will be, in these notes, a bounded linear operator from the space P  C ~ (]R•) of testfunctions to its dual P ' (the space of distributions), such that there is a "standard kernel" (the definition is given below) K(x,y) for which (1)
< r:,: > :
//
whenever f, g E P have disjoint supports. Here, < T f, g > denotes the effect of the distribution T f on the function g. Note that we could also have defined T as a bilinear, continuous form on P × P. Finally, (1) is the same thing as asking that Tf(x) is defined, for x ~ suppf, by Tf(x) = f g(x,y)f(y)dy. D e f i n i t i o n 1.2. A "standard kernel" is a function K(x,y), defined on ~ n x ~{'~\((x, y) : x = y}, such that, for some constants 0 < 6 < 1 and Co _> 0,
(2)
[g(.,y) ]~ Co I z  y [~
and
Xt_X [6 (3) for I x l  x
I KC*,y) 
KC.',y) I+ lK(y,*) K(y,g)I_< Co l._~ r.~6 
I< 1 i x _ y l .
Remarks. • One could be a little more general than this when defining "standard kernels", but there are not so many examples where that generality would help.
27 • If T is a singular integral operator (we'll also say a SIO), there is only one kernel associated to T. However, the same kernel can be associated to m a n y different operators. [EXERCISE : show that the operator of multiplication by a given function, or any differential operator are SIO's associated with K = 0]. • For the Tbtheorem, a slightly different definition is more natural : given two bounded functions bl and b2, we shall ask that T be defined from blP to (b2P) ~ (or be a bilinear form defined on blP x b2P), rather than from D to P~. The rest of the definition is the same. • Note the nice invariance properties of the class of SIO's by the action of translations and dilations. E x a m p l e 1.3. (Principal value operator defined by an antisymmetric kernel). Suppose that K(x, y) satisfies (2), and also K(x, y) =  K ( y , x) for all x ¢ y. Then one can define an operator T (the "principal value operator defined by K " ) by the formula (4)
<
Tf, g > =
1f f K(x,y)[f(y)gCz)fCx)gCy)]dzdy.
E x e r c i s e : Check that the integral converges because the singularity of K is killed by the fact that I f(y)g(x)  fCx)g(y) ]< C(f,g) [ x  y I . Of course, if K also satisfies (3), then T is a singular integral operator with kernel K. We shall sometimes abbreviate and write Tf(x) = p. v. f g(x,y)f(y)dy. Warning : the terminology above is quite convenient for us, but is not used with the same meaning by all mathematicians : in many sources, "singular integrals", "principal values" means something different. D e f i n i t i o n 1.4. (what is meant by T1). Let T be a SIO, and b be a bounded, C °o function, but not necessarily compactly supported. Let us define Tb : it will be a distribution known modulo an additive constant, i. e. a continuous linear form defined on the space of functions g G P such t h a t f g = 0. Given g E P such that f g = 0, write b = bl + b2, where bl E P and b2 is zero on a neighborhood of suppg. We define < Tb, g > by
< Tb, g > =< Tbl,g > + f f K(z,y)b2(y)g(z)dydz
(s) =<
Tb,,g > + / / {K(x,y)  K(xo,y)}b2(y)g(x)dydx,
where xo is any point in the support of g (the first integral is not actually defined, except in terms of the second one). Exercise : Check that the second integral in (5) converges because of (3), that the definition does not depend on z0 or the decomposition b = bl + b2, that Tb is a linear
28 function of b, and t h a t it coincides with the definition of T b when b E P or when T is given by an integrable kernel. Let us now quote, without proofs, a few classical results. D e f i n i t i o n 1.5. If the SIO T can be extended into a b o u n d e d o p e r a t o r of L 2 ( R '~) (or, in short, if T is b o u n d e d on L2(IR)), we'll say t h a t T is a C a l d e r 6 n  Z y g m u n d o p e r a t o r (or, more briefly, a " C Z O " ) . Let us say again t h a t this definition is not universally used. 1.6. f i t is a CZO, then T has bounded extensions from LP(IRn) to LP(N n) for 1 < p < +oo, from the Hardy space HI(IR n) to L I ( I R " ) , and from L°°(lR ") to B M O ( R '~) (defined below).
Theorem
This t h e o r e m is due to Calder6n and Zygmund for the LPboundedness, and to Peetre, Spanne and Stein for the LC°BMO boundedness.
lyQ I D e f i n i t i o n 1.7. A locally integrable function f is in BMO(IR n) if I[ f [[BMO= sup ~[ Q f(x)  fQ ] dx < +oo, where the s u p r e m u m is taken over all cubes Q c IR n, and fQ =
Remarks 1. A little more is true. Define truncated operators by (6)
T J ( x ) = f l z  u l > , K(x,y)f(y)dy for e > 0 and f e L 2,
and then the m a x i m a l operator (7)
=
sup E>0
ITd(
)I
If T is a CZO, then T. is b o u n d e d o n / 2 , 1 < p < +oo, and in particular the Te's are uniformly bounded. This result is obtained by means of " C o t l a r ' s inequality" (we shall actually prove this inequality, in a more general context, in P a r t III). 2. If the Te's are uniformly bounded o n L 2, say, then one can extract a sequence t h a t converges weakly to some bounded o p e r a t o r T. It could be t h a t T is not bounded (for instance, if T is a differential operator). If T is bounded, then T  :F is bounded, too, has kernel 0, and so (exercise !) T  T is the o p e r a t o r of pointwise multiplication by a b o u n d e d function [Hint for the exercise : show t h a t the function is, locally, ( T  ~%)(~[_N,N],~)]. Note t h a t the Te's do not necessarily converge when e * 0 dimensional kernel g ( x , y) =[ x  y [I+,T .
: consider the one
29 3. If T is a CZO, and T1 = 0 (modulo constants), then Theorem 1.6 can be slightly improved : T is also bounded from BMO to BMO. Also, if T is a CZO such that T t l = O, then T is bounded on H 1 (the transposed operator T ~ is defined by < T t f , g > = < Tg, f >; it is of course a SIO with kernel K ( y , x)). We have seen already that our lack of knowledge on the restriction of the "distributionkernel" of T to the diagonal is a potential source of trouble. In particular, it authorizes behaviors that do not have the correct invariance under the action of translations and dilations (think about the pointwise multiplication by a wild function, or T f = f'). The following notion of "weak boundedness" was introduced by Y. Meyer to try and compensate this lack of knowledge. D e f i n i t i o n 1.8. For t > 0 and x C ~ n , call ~It,z the operator of translation by x and dilation by t defined by ~qt,zf(u) = t'~12f(tu  x). We shall say that the operator T
: P ~
P~ is weakly bounded if the operators
~ t,1z T ~ t , z are uniformly bounded from P to P ' for t > 0 and x C IRn. This means that, for each compact set K, there is a suitable seminorm NI" Ill such that
I<
iT
'%J',g >l
c
II1 .r III III g III
for all f and g supported in K. Let us give an equivalent definition. For each B C ]Rn and each integer q > 0, define, for f supported in B,
(8)
N~(f) = R n/2 ~
R I'~IIIO'~f Iloo, l,~l_
where R is the radius of B. Note that if B = B ( x o , R ) and f is supported in B, then N B f = .• 1.q/ 2 B y, where g(x) = 2 " / z f ( 2 x  xo). We can use this invariance, and invariance by translation, to assert that T is weakly bounded if and only if there exist q > 0 and C > 0 such that, for all balls B and all functions f , g E P supported in B,
[< T f , g >[< C N B ( f ) N B ( g ) .
(9)
Exercises 1. Check what was just said. 2. Show that if T is bounded on L 2 (or LP), then T is weakly bounded. 3. Check that the operator defined by T f = f ' in dimension 1 is not weakly bounded. 4. Show that T f = m f r n E L °°.
(for some m E L~oc(IRn)) is weakly bounded if and only if
30 5. Show t h a t if the kernel K(x, y) satisfies (2) and is a n t i s y m m e t r i c (i.e., K(y, x) =  g ( z , y)), then the principal value o p e r a t o r defined by (4) is weakly b o u n d e d (we shall use this exercise later). R e m a r k . If T is associated by (1) to a kernel K(x, y) verifying (2), and if (9) is satisfied for some q > 0, one can show t h a t it is true for all q > 0 (with suitable modifications when q is not an integer). T h e proof is not too hard, but takes some space. This means t h a t T is also defined as a bilinear form on any space C ' = { f E L °° : [ f(x)  f(y) [<_ C [ x  y I" for all x, y}. We could have changed the definitions above to take this into account (this is actually w h a t happens on spaces of homogeneous type where C °O does not make sense), but we'll often find it more convenient to define T on testfunctions. We now have enough definitions to state the " T l  t h e o r e m " . 1.9 ([DJ~]). Let T be a singular integral operator (like in Definition 1.1). Then bounded extension to L2(]R n) if and only if T is weakly bounded (see Definition 1.8) ; T1 C B M O (see Definition 1.4) ; Ttl E BMO.
Theorem
T has a i) ii) iii)
Recall t h a t T t is the SIO defined by < T t f , g > = < TO, f > • T h e direct implication follows from Peetre, Spanne and Stein's result ( T h e o r e m 1.6) and the exercise 2 above. We shall prove this t h e o r e m later, and we'll discuss applications later still. Before, let us state a second result (the " T b  t h e o r e m ' ) . D e f i n i t i o n 1.10 (Paraaccretivity). Let b be a b o u n d e d function from IR n to C. We'll say t h a t b is accretive if there is a 6 > 0 such t h a t Re b(x) _> 5 a. e. We'll say t h a t b is "paraaccretive" if there are constants C _> 0 and 6 > 0 such that, for all x E ]R n and r > 0, there exists a c u b e Q such t h a t dist(x, Q) < Cr, t r < d i a m Q _< Cr, and such t h a t (10) We'll even say t h a t b is "special paraaccretive" if (10) is true for all dyadic cubes. E x e r c i s e . Check t h a t  if b is paraaccretive, then I b I_> 5 a. e. ;  in dimension 1, e iz is not paraaccretive ;  cubes can be replaced by balls in the above definition, without changing the notion. T h e next t h e o r e m says that, in T h e o r e m 1.9, the function 1 can be replaced by any paraaccretive function b. T h e definitions are slightly more complicated, because it is more natural to define T from bD to (bP) I in this case.
31
Notation. If b E L ~ , we shall denote by Mb the operator of pointwise multiplication byb.
Theorem 1.11 ([DJS]). Let bl and b2 be two paraaccretive functions. Let T be a (bounded) operator from bid to (b2P)~. Suppose that T is associated to a standard kernel K(x,y) in the sense that < Tf, g > = f f K(x,y)f(y)g(x)dydx whenever f ~ btP and g E b2P have disjoint supports. Also suppose that Mb2TMba is weakly bounded, that Tbl E B M O and Ttb2 E BMO. Then T extends to a bounded operator on L2(IRn). This t h e o r e m is actually a generalization of a result by A. McIntosh and Y. Meyer [McM] : the i m p o r t a n t special case when Tbl = Ttb2 = 0 for accretive functions bl and b 2. Their proof uses a formulation in terms of interpolation of the L2boundedness of the Cauchy integral on Lipschitz graphs [CMM]. Remarks 1. We didn't define Tbl (or Ttb2) in this case. This time, it will be a linear form on b2 P, defined modulo an additive constant. This means t h a t < Tbl, b2~o > will be defined if is a test function such t h a t f ~o(x)b2(x)dx = O. The definition of < Tbl,b2ta > is similar to w h a t was done in Definition 1.4. One picks a function X E P such t h a t X = 1 on a neighborhood of supp ~a, and one writes
< Tbl,b2~o > = < T(blX),b2~o > + f ] KCx, y)[b,(1 x)]Cy)b2(x)~o(x)dy dx = < T(blX),b2~ > + f / { K ( x , y )  K(xo,y)i{bl(1  X ) } (y)b2(z)~(x)dy dx. The last integral converges because of (3), as before. 2. For b o t h theorems, one actually gets a control on T, like I[T I1L2,L2~_k I] Tbl IIBMO +k II Ttb2 IIBMO +kCo + kC, where Co is the constant of (2) and (3), and C comes from the weak boundedness (C is any constant t h a t shows up in an inequality (9)). This easily follows from the proof, or from Baire's Category theorem. 3. T h e converse of T h e o r e m 1.11 is, as for the T l  t h e o r e m , a direct consequence of T h e o r e m 1.6. 4. Paraaccretivity is, in some sense, an optimal notion : one can show t h a t if b is such t h a t the t h e o r e m is true for bl = b2 = b (and all operators !), then b is paraaccretive (see [DJS]). The next sections are devoted to giving a short, new proof of the T b  t h e o r e m which is due to R. Coifman and S. Semmes [CJS]. It is also a nice coincidence t h a t it uses ideas related to wavelets.
32 2. F i r s t s t e p o f C o i f m a n  S e m m e s ' a Riesz basis
proof :
This section is needed only for the Tbtheorem. If bl  b2  1, then we shall construct the Haar system in this section ! Given a paraaccretive function b, we intend to build a Riesz basis of L2(~'~). We shall first consider the case of a "special paraaccretive" function b, i.e. a b o u n d e d function such that, for some b > 0, (10) is true for all dyadic cubes. Let us first define projection operators : for k E ~ , let E k be such t h a t E k f ( x )  1
(11)
/q f(t)dt,
where Q is the dyadic cube of side length 2  k t h a t contains x. (This is the projection operator associated to the Haar s y s t e m ; note t h a t it is a typical martingale projection.) Also define the difference D k = E k + l  E k . Now define corresponding projection operators "relative to b" : 1
(12)


F k f ( x )  EkbCx)
where again Q is the dyadic cube of size 2  k containing x ; note t h a t the first denominator is b o u n d e d below because of (10). Finally, call Ak Fk+l  Fk. Let us check a few easy facts. First, note t h a t f o f ( t ) b ( t ) d t = fQ F k f ( t ) b ( t ) d t for all dyadic cubes Q of size 2  k (the "martingale property;'~. Next, (13)
F i F k = Fi,.,k.
If k < j, then F k f is constant on each cube of size 2  k , and so it is constant on cubes of size 2  i , and F i does not change it. If k > j, we use the martingale property above for Fk : the integral of Fk on cubes of size 2  k is the same as the integral of f ; this remains true for cubes of size 2  i and so F j F k f = F i f . (14)
A j A k = bj,kA i .
Just expanding gives A i A k = (Fi+ x  F j ) ( F k + 1  F k ) = F j + I F k + 1  1 ? j F k + l  F i + I F k ÷FjFk. If j > k, w h a t remains is Fk+l  Fk+l  Fk + Fk = 0 ; if j < k, one also gets 0 (by s y m m e t r y ) and if j = k, one is left with F j + I  Fj  Fj ÷ F i = A j . If u and v are c o m p a c t l y supported and k ~ £,
(15)
/
 O.
33 We can assume k > l. T h e n A t v is constant on each cube of size 2  k . On such a cube Q, preserve the integral against b, and so fo(Aku)b 0. We just have to sum
Fk+l and Fk orI Q.
Let us now prove a square function estimate. Lemma
2.1. There is a
(16)
0 1
constant C > 0 such that,
for
all f E L 2,
II/II~<f~. ~ IAkf(~)12d~ <c IIf II~. kEZ
Let us first prove the second inequality. Write
Akf = (Ek+lb)l Ek+l(b f)  (Ekb)I Ek(bf) = [ ( E k + l b )  I _ (Ekb)llEk+l (bf) 4 CEkb)I [Ek+l (bf)  EkCbF)]. Since
[Ekb Ek+lb] 1
is b o u n d e d because of (10), we get
I Akf 12<  C[Dkb 121E}+:(bf) 12 +OIDk(bf ) [2 and
< I+II. For the second part, w e use the usual square function estimate (even simplified) : note t h a t bf = ~ k Dk(bf), and t h a t the pieces are orthonormal to each other because Dk is selfadjoint and
n k n t = ,Sk.tnk.
So [[ bf
[122=~ II Dk(bf) Ill = f k
Z
I Dk(b/) 12 .
For the first part, we use the fact t h a t I Dkb 12 dx, times the counting measure on {2  k : k E 7Z,}, is a Carleson measure. To check this easy fact, one must prove that, if Q is a cube and R is its sidelength,
q
~
IDkb(~)12d~<CIQl
•
k:2k<_R
But the left h a n d side is equal to
/Q ~k< [ Dk{b(t)~cQ(t)}2dx < /Q ~k ] Dk(bllcQ)12 dx 2
R
< C II b~cq
1122 ( b y the s t a n d a r d square function estimate)
< C I[ b IIc¢[ Q [, as needed.
34 The fact that, if ~ k [ Dkb
{2 dxd~2J,
is a Carleson measure, then
<_ II bf
Ek+l(bf)
II f
is a standard fact about Carleson measures (see Wh. 1.5.6 in [Ga3]). So we proved the second inequality. The first one will follow by duality. Note that b1 is bounded because b is paraaccretive, and
k
l
= / k(Sb') k(Ilbex (hy(15)). If you are worried about the convergence, note that the formula is true w h e n f is a finite
linear combination of characteristic functions of dyadic cubes ; we'll be able to pass to the limit easily, since both sides of the inequality are continuous in L 2  n o r m . Apply Schwarz' inequality :
1{f {{2 2~ C
(/
}x.(j
Z {hk(fbl)12 k
~__4{Akf {2 k
Dividing by II f 112 if it is not zero, one gets the first inequality of L e m m a 2.1. So this lemma is true. Remember that we want to find a Riesz basis that looks like the ttaar system, but would be "orthogonal" with respect to the b  s c a l a r product. One should be careful : if b is not real, this is just an analogy, and I heard from a few short proofs that were a little too short precisely because the analogy was pushed a little too far. For each cube Q of the k th generation (i.e., each dyadic cube of side length 2k), call Qn, ,7 c I , the cubes of the (k + 1) at generation that are contained in Q (here, I has 2 n elements). For each ,7, call/~n = fQ,~ b(x)dx. By (10), {fin I 2  n ~ { Q {" Let us call 0 one of the elements of I, and I  I\{0}. We want to define, for each e E I, a function h~ which is constant on each Qn, supported on Q, and has the properties (17)
(18)
[ % ( x ) b ( x ) d x = 0, Jq
/ h~(~)h~(~)b(~)d~ = 0 for all ~' ~
in i,
35 and the normalization
Exactly like in P a r t I, Section 6, it is more convenient to formulate the question in terms of matrices. For e E I and r/ E I , call ae,n the value of h~ on Qn. Let us complete the matrix by taking ao,,1 =
,~ b(x)dx
for all r/. Also call v~, e E I, the vector with
coordinates (ae,,)ne~. T h e conditions (17) and (18) say t h a t the v~, e E I , are mutually "orthogonal" for the bilinear form < (an) , (bn) > Z = ~ n a'~b'Tfln" We are given a first vector v0, and have to "complete the basis". First consider v~  {w = (wn) : ~ Wnao,,~ n  0}. This is a space of dimension 2 '~  1 ; the restriction to v~ of the bilinear form < , > 8 cannot be zero, because it w o u l d imply t h a t each element of v0a is "orthogonal" to the whole space (but the linear form < w,. >~ does not identically vanish unless w = 0). So one can find two elements wl and w2 of v~, such t h a t < w l , w 2 > ~ 0. If b o t h < wl~wl >~ and < w2,w2 >~ vanish, then < wl + w 2 , w l + w2 > Z ¢ 0. In any case, we can find a vector vl C v~ such that < v l , v l > ~ ¢ 0 and, after normalization, one can even assume t h a t < v l , v l > ~   1 (we are implicitly assuming here t h a t I  {0, 1 , . . . } ; otherwise, we would give Vl is a more complicated name). Let us do the same thing a g a i n : v~ Mv~ is a (2 '~  2 )  d i m e n s i o n a l space, and (unless n = 1) the bilinear f o r m < , > p cannot vanish on this whole space. So we can find Wl and w2 E v~ M v~ such that < w l , w 2 > ~ 0. T h e n there is a v2 C v~ n v~, with < v2, v2 >Z 1. Applying this argument enough times, one gets vectors re, e E I, such t h a t < re,re, > ~   6c,~,. Translating things back in terms in terms of functions, we just proved the existence of the h~'s. Note t h a t the ve's are independent : if ~ )~v~ = 0, t h e n taking the twisted scalar p r o d u c t with re, gives ~e,  0. Hence, the h ~ ' s are independent from each other, and are also independent from the constant function (coming from v0). L e r n m a 2.2. Call VQ the space of a11 functions f that are supported on Q, constant on each Q , , and such that f fb = O. Then every f E VQ can be written
(20)
S =
<
>b
eel. where < f , g > b = correctly chosen, (21)
S fgb
is a notation for the twisted scalar product. Also, ff the h~'s are
c  ' II S I1_<
I< f, ha >bl < C II f
for all f E VQ, and a constant C that does not depend on Q or e.
36 To prove (20), just note that, since the h~'s are independent, they form a basis of VQ. So every f E VQ can be written f = ~ Ceh~, and then
/,
hQhQb = C~,
f hqb = E
(by (18) and (19)), which gives (20). A short glance at the construction of the h~'s is enough to convince oneself that one can choose the ve's so that their coordinates (the ae,,'s) are less than C
/
min I ~ , l t r/
}"
<cIQ[ 1/~.
Consequently, one can choose that hb's so that II hb IIo~< c I Q 11/2 • The second inequality of (21) is then trivial, and the first one follows from (20). This proves Lemma 2.2. L e m m a 2.3. The h~ 's, where Q is a dyadic cube and e E ], are a Riesz basis of L2(]Rn). More precisely, each function f E L 2 can be written
¢22)
: = ~ ~ < :,h~ >, h~, Q
•
and
c' II: II~<~
(23)
I
Q
By a simple density argument, we only have to prove (22) and (23) for a dense set, like the finite linear combinations of characteristic functions of dyadic cubes. For such a function f, f  ~ k A k f , with H f I1~ ~ )~k ]1 A k f []~ by Lemma 2.1. Let us now fix k, and decompose gk = Akf a little further. By definition of Ak, gk is constant on each cube of generation k + 1, and its integral on each cube of generation k is zero. So we can write
gk ~ E llQgk' where one sums on all dyadic cubes of sidelength 2 k, and then IQgk E VQ. Q
S°gk = E { ~ < lQgk,h~2 >bh~ } by Lemma2.2. But < lQgk,heQ >b=< gk,h~ >b= Q < Fk+lf, h~ >b  < Fkf, h~ >b< Fk+lf, h~Q >b (since Fkf is constant on Q and h~ satisfies (17)). This is equal to < f, h~ >b (because h~ is constant on each Qn). Hence (22) is true. Furthermore, [[ gk [[2=
E
:
E Q of g e n e r a t i o n k
E e
l<
[< lLQgk,h~ >12 (by (21))
I[ dQgk []22~ E E
Q of generation k
f,h~ >l 2, from which
Q
(23) follows.
e
37 Up to now, we associated to each specialparaaccretive function b a Riesz basis (h~), with the following nice properties : for each Q and each e, h~ is supported on Q, and is constant on each cube of the next generation ; also, the coefficient of f relative to h~ is the twisted scalar product f f (x) h~ (x) b(x)dx. To conclude this section, let us indicate how the construction above can be modified in the general case (when we only know that b is paraaccretive). The idea will be to keep the same formalism, and modify the dyadic cubes. The following lemma will help us take care of the geometry. L e m m a 2.4. Let b be a paraaccretive function. There exist constants A E ~q a n d / ~ > 0 such that, if z C IR~ and k E 2~, there is a dyadic cube Q, of sidelength 2k, at distance < 2A+k from x, and such that
1 /Q b > ~l.
(24)
This is a very easy consequence of Definition 1.10, and is left as an exercise. Let £ denote the set of 2A+2adic cubes, and ~0 C R the subset composed of the "good" cubes, i.e. cubes Qo such that ] ~
fQo b > ½~'2 (A+2)~, where g' is as in Lemma
2.4. If Qo E ~ \ ~ o is not good, we can use Lemma 2.4 to find a cube Q E ~ contained in Q0, with length(Q) = 2A21ength(Q0), and satisfying (24). Call A(Qo) = Q, and B(Qo) = Qo\Q ; since Q0 was not a good cube, we get 1
[
b > 1~2(A+2)n
"
Our new notion of "cubes" is defined as follows : we shall call "cubes of generation k" all cubes of Qo c Ro with sidelength 2 k(A+2), and all sets of the form A(Qo) or B(Qo) for a . , then every Q0 of sidelength 2 k(A+2) which is in ~ \ R 0 . Note that, if ~'~ = ~v 1,~(A+2),~ "cube of generation k" R satisfies
(25)
~~R~]/R b > i~'.
Also, all the cubes of a given generation have roughly the same size, and they form a partition of ]R n. Finally, if R is a cube of generation k, all the cubes of generation k + 1 that meet R are contained in R, and the number of these cubes is < C. We can now apply the same argument as in the beginning of this section, and get a Riesz basis h~, where this time Q runs in the new set of cubes, and for each Q, e is in a set [(Q) (which this time does not always have the same number of elements). The formulas (22) and (23), in particular, are still true in this context.
38 3. P r o o f o f T h e o r e m
1.11 when
Tbl = Ttb2 : 0
To simplify notations a little, we shall do the proof when b 1 : b2 = b (we'll say a few words later on how to modify the argument in the general case). We'll also suppose that b is "special paraaccretive', the modifications to obtain the general case would be trivial. The idea of the proof is, simply, to estimate the coefficients of the matrix of TMb in the basis (h~) of the previous section, and to show that they decrease rapidly enough away from the diagonal. Let us isolate in a first lemma the only part of the proof where we'll need the regularity estimate (3) on the kernel of T. L e m m a 3.1. Let Q be a dyadic cube, and let xQ be its center. Let h be supported on Q,
and such that (26)
fQ h b = O. Then, 17"h(x)
for x ~[ 2Q,
I_< C lO la+ l
xQ I" li h It ,
where T = MbTMb. In particular, if h is one of the h~ 's, we get (27)
I :Fh~(z) l< C I Q 1½+~l x  xq I "  8
for x • 2Q.
To prove the lemma, write
Th(x) = b(x) f o K ( x , y)b(y)h(y)dy = b(x) / { K ( x , Jq
y)  K ( x , xQ)]b(y)h(y)dy,
and (26) follows from (3) ; (27) follows from (26) and the fact that [I h~ lloo< C [ Q ]1/2. R e m a r k 3.2. This is the only place where (3) is used (we'll use this remark in Section 4). Call CQ, R = < T h Q , h n > . Because h~ and h n a r e not smooth, we did not say yet how to define this number. When the closures of Q and R are disjoint, we can of course use the kernel of T ; when T is the principal value operator associated with an antisymmetric kernel, it is easy to check that the formula (4) can still be used. For the general case, we shall have to use a small limiting argument and the weak boundedness of T. We postpone this definition to the proof of Lemma 3.4. Since the indices e and e~ do not play any role in the estimates, we shall simplify and write CQ,R = < 7"ho,hn > instead (the e's will reappear only when we need them). Let us summarize the estimates on the CQ,R that we want. L e m m a 3.3. Let Q and R be dyadic cubes, and let the R , ' s be the cubes of the next generation that are contained in R. Then
(28)
if I Q I<1 R I, then I C ,R I< C lQ I'/'i R 11/5
t Q I'/"
I Q 16/" + d i s t ( Q , U , ORn) 6 '
39
(29) it" Q I<_I R I and dist(Q,R) >i R I v', the~
I CQ R I<_ C lQ ll/~l R I '/2 I Q I~/" ' dist(Q, R)'~+6 ; (30) it" Q I>_I R I, th~n I CQ,R I<_ C IR IV21 Q I~/~ (31) it" Q II R I and
I R I~I " IR
151'~ +dist(R,U, OQ,) ~ ;
dist(Q,R) >I Q Iv~, *he~ I c<~,R I<_ c I Q lW~l R lln
I R ]8/"
dist (Q, R)'+5"
It will be clear from the proof of (28) and (29) that (30) and (31) follow, just by exchanging Q and R. Let us start with the case when [ Q I<[ R I, Q is not contained in R, and even dist(Q,R) >I Q 11/n. Integrating (27) gives
I CQ,R I 
1½+51 ~  ~ Q I"~I hR(~) lax
_< c I Q I~+51R IW ~/o I x  :~Q I  "  ' d~ < C lQ
I~+~I
R
dist(Q,R) ~,
as wanted for (28). If we even know that dist(Q,R) >I R I ~, then we simply can replace the last line by C I Q 1½+~l R I1/~l R I dist(Q,R) '~+~, which gives (29). Next, suppose that I Q I<1R I, Q is not contained in R, but dist(Q,R) ~1 Q I~ • In this case, we write h R : hi + h2, where hi is supported in 3Q and h2 lives outside 2Q. The estimate above still gives I< Thq,h2 > l < C I Q ll/21R 11/2 . For < Thq, hi >, one can use a small limiting argument, which is very similar to the argument in Lemma 3.4 below (but simpler), and thus left as an exercise, to show that
~_ C / / [ x  y [nl Q l1/21R [i/2 1Q(y)13Q\Q(x)dxdy <_ C IQ I~I~IR l~n_< c I Q l~nlR I~n
I Q I~I" I Q 161" +di~t(Q, U. R.)~
and complete the proof of (28) in this second case. We are now left with the case when Q is contained in R. Let us start with the case when Q is strictly contained in R, and call R0 the R , which contains Q. Note that hR
40
is constant on R , . We can use the fact t h a t T t l : 0 (or, equivalently, Ttb = 0), so that < ThQ,, ha > = < ThQ, hRc >, where c is the value of h R on R0. Note t h a t this is not quite conform to our definition of Ttb (the first remark after the s t a t e m e n t of T h e o r e m 1.11) : we said t h a t < T(bhQ),b > is defined when hQ is a test function such t h a t f hQb = O. Here, we still have f hQb = 0, but hQ is not smooth. A small covering a r g u m e n t allows one to extend the definition to the case when hQ is a finite s u m of characteristic functions of cubes (the idea is t h a t in this case, the singularity of hq is not too bad). Once again, we leave the details to the reader, because we shall give a slightly m o r e complete a r g u m e n t in L e m m a 3.4 below. Note t h a t h R  c vanishes on R0 D Q, and so we'll be able to use the kernel again :
IC~,R I< C [
IThQ(x) IIR 11/~dx+ [
J2 QAR~
IQ
IThQ(x) IIR 11/~dx
J(2Q)'nRg
[1/2[
R ]1/21
x  y I'~ 1Q(y)12QnRg(x)dydx
+C[QI}+5IRI1/2f(
2Q)cnR~)
]xxo]"6
dx
(we used (2) and (27)). Therefore,
[CQ,R I~ C [ Q I1/~[ R
11/ +CIQ
R [1/2 { d i s t ( Q , R ~ ) + i Q 1~}6,
and so (28) is also true in this case (the first t e r m disappears when 2Q f3 R~) is empty). T h e inequality (28) will be completely established as soon as we prove it when Q = R. Since (30) and (31) are a trivial corollary of the proof, L e m m a 3.3 will be a direct consequence of the following lemma. Lemma
3.4. l<
~'hQ,hQ >]< C.
Note t h a t hQ can be written ho. = ~ AilR~, where the s u m has less t h a n C terms, t h e R i ' s are disjoint cubes of roughly the same size as Q, and I Ai [< C I Q ]1/2 . Since 1< T1R,, 1Rj >1_< C I Q [ when i # j by a brutal majoration of f fR,×Rj I x  y I'~ dxdy, L e m m a 3.4 will follow if we prove that I< T1R,1R > l < C [ R I for all cubes. Also, replacing the second 1R by a s m o o t h function (I) which is 1 on R and is s u p p o r t e d on 2R only brings errors < C ] R ] (the difference is < CfR f2R\R [ X y I"< C I R I)" So we only need to prove t h a t
(32)
I< T1R, ~ >1< C IR I
for such a ~. The idea is the following : if IR were smooth, then (32) would be nothing more than the weak boundedness of T. We'll show t h a t the singularity of 1R is not bad enough to destroy t h a t estimate.
41 More precisely, we'll use the fact t h a t the singularity lives on a small set (namely
OR). We'll write 1R as a sum of s m o o t h functions, and show t h a t the series giving (32) converges. Let ~o be a s m o o t h function, supported on B(0, r), where r =1 R [1/2, and with integral 1. We can choose !o so t h a t
10a79 I< _ Ca r n[al
(33)
for any multiindex a. Write 1R = ~m>_Ofm, where f0  1R {2m~99(2mX)  2 ( m  a ) " ~ ( 2 m  l X ) } = 1R * era, in B(O,2m+lr). Next, cut each fm into m a n y pieces, using supported in balls of radius 2  m r . We obtain supported in a ball of radius 2mr. We can do
* p , and then, for m > 1, f m = 1R * where Cm has integral 0 and is supported a partition of unity composed of functions f m = ~keI(m)gin,k, where each 9m,k is this with a set of indices I(m) such t h a t
FI(m) I c2m("x),
(34)
because fm is supported on {x : dist(x, OR) < 2  m + l r } . Also, we can m a n a g e so that each gm,k satisfy
(3s)
II 0 gm,k
C (2mr) I l for all
a E
~l".
Now, let us estimate each < 7"gm,k,¢ > • We choose a point x0 in the s u p p o r t of gin,k, and write ¢ = ¢1 + ¢2, where ¢1 is supported in B(xo, lOr2m), and ¢2 is zero on B(xo,br2m). We can easily manage to have [ ¢2 [< C, and [[ Oa¢l [[oo< C(2mr) [al for all a. Applying the weak boundedness of T, we get [< 7"gm,k,¢l > [ < C2 mnrn. On the other hand, l< Tgm,k,¢2 >[_< C f f D I x  y I'~ d y d x , w h e r e D = { ( x , y ) : I Y  Xo I < 2  m + l r , I x  Xo I~ 5r2m and x E 2R}. This is less t h a n c2mnLog(m+ 1)r ~, and so I< Tgm,k, ¢ >[_< C 2mnLog(m + 1)r ". Summing on k (using (34)) and then on m gives l< T 1 R , ¢ >1< Cr" = C I R I . This proves L e m m a 3.4 and, by the same token, L e m m a 3.3. R e m a r k . L e m m a 3.4 seems a little long to prove (although the proof does not require too much thought), but it seems t h a t one very often has to prove something like this. For instance, Y. Meyer's " c o m m u t a t i o n lemma" (see [Myl]) is proved very much like this, and its use seems unavoidable in some instances. The usual way to solve the problem is of course to leave the easy proof as an exercise. The estimates of L e m m a 3.3 will be enough to imply t h a t T is bounded. We want to apply Shur's lemma, but we cannot s u m our estimates on I CQ,R I directly, because there ~S are m a n y more small cubes t h a n large ones. So we'll sum the [< :Fh~,,ho, >[, for each Q and e, against an appropriate weight that depends on the size of Q~. Let us first take sums
42 on sets of cubes of the same size. Let Q be a given cube, with sidelength r. Call ~j(Q), j ~ 2~, the set of all dyadic cubes with sidelength equal to 2it. L e m m a 3.5.
(36)
C2 sÈn if j > 0 j < 0 and ~ # 1 C2i~/2 I J { 2J if j < 0 but ~ = 1.
X: (ICQ,RI+ICR,~I)
n e ~ (q]
Let us just prove the estimate for I CQ,R ] ; the other one will follow by symmetry. First suppose that j > 0 (so that [ R ]>l Q ])There are only a finite number of R's such that dist(Q,R) <1 R 11/n and for these R's, (28) gives I CQ,R I<_ C I Q 11/21 R I1/2<_ C2 i'~/2. For the other cubes R (those which are at distance _> 2Jr from Q), note that for a given k _> 0, the number of cubes R such that dist(Q,R) ~ 2k2/r is < C2 '~k. For each of these cubes,
I CQ,R [<~C I Q 11/21 R 11/2 r6(2k2Jr) n6 ~ C2k(n+~)2 1(~+6) ~_ C2k(n+6)2 in~2.
We now sum on all these cubes, and then on k, and get less than C 2 i'~/2. So (36) is valid in this case. Let us now deal with the case when j < 0 (so that I Q I>1 R [). First consider the cubes R that are at distance > r from Q. For a given k > 0, the number of R's such that 2kr _< dist(Q,R) < 2k+lr is less than C 2 n k ~ = C2~k2 '~i. For each of these cubes, [CQ,R t<_ C I Q 11/21R 11/21R 16/" [2kr] '~8 _< C2k("+6)2 i(t+~). Summing on allthese cubes, and then on k gives less than 2J(~ ~), which is compatible with (36). We are left with the cubes R such that dist(Q, R) < r. For each 0 < k <  j , consider the set ~j,k composed of all the cubes R such that 2k2Jr ~ dist(R, U~ t)Q,) < 2k+12/r (for k = 0, let us even include the cubes that are at distance _< 2Jr from U~ OQn). To estimate the number of elements of Nj,k, note that the volume they cover is < Cr"l(2k+12Jr) (the surface of [.J, 9Q, is < Cr '~1, and we have to multiply by the approximate width). Consequently, #~j,k <_ C 2k2J2  h i . For each cube R E ~j,k,
R 161'~(2k2ir)~
icq, I< c R I'/I Q _ C 2k62j"/2.
Summing on Aj,k, we obtain less than C2k(x8)212 jn/2. We can now sum on k. If ~ = 1, we have  j  1 equal terms of 2] 2 in/2, which gives (36) ; if ~ < 1, we get less than C 2J(16)2J2 in/2, which again gives (36). This completes the proof of Lemma 3.5. To apply Shur's lemma, let us compute the matrix of the operator TMbin the basis (h~) of the previous section. By (22), the coefficient of coordinates (Q, e) and (Q', e') of this matrix is < TMbhQ,ho, >b=<
on L 2 if and only if the matrix ((..')) Cq,q,
>=
Also, by (23}, TMs is bounded
defines a bounded operator on £2(i), where I
is the set of couples (Q, e). As we said earlier, rather than compute ~ x [ Cq,Q, [, we want to sum against an appropriate weight.
43 To each (Q,e) E I, let us associate the positive number w(Q,e) =1Q 1½ ~~ L e m m a 3.6. There is a constant C > 0 such that ~j~e
(37)
1%,q, I
each (Q,,)
<
(Q',e')EI
and
~
(38)
~6 s
I CQ,Q, I~(Q,e) <_ Cw(Q',e') for each (Q',~').
(Q ,e)El This is easily obtained from L e m m a 3.5. To prove (37), let us first choose j > 0 and sum on all cubes Q' E ~j(Q) and all e' (remember that, for each Q', there are less than C ~ ~, 2 i~/2. Summing on all indices d). We get less than c{2nYrn}½~2i'~/2 _< C I Q ]~~. j > 0 gives less than Cw(Q, e). Next, pick j < 0, and sum on ( Q ' , d ) , where Q' E "~J(Q)" If 6 ~ 1, we get < C{2nJrn}½~2Jn/228i < C I Q I~  ~ 2Y8/2, and summing on j still gives less than Cw(Q,e). This remains true with trivial modifications when 6 = 1, and so (37) is true. L e m m a 3.6 follows because our estin~tes on the CQ,Q, s are symmetric. It immediatly follows from Shur's lemma and L e m m a 3.6 that TMb (and therefore T) is bounded on L2(]R). Let us even give a proof of Shur's criterion for the convenience of the reader. 3.7. (Shur's criterion). We are given a set of indices I and, for each i E I, a number wi > O. Suppose that, for some C >_ O, the matrix ((Ci,j)) satisfies
Lemma
(39)
~ I Cid
[ wi <_ Cwl for each i
J and (40)
~
[ Ci,i I wi < Cwj for each j. i
Then the matrix ((Ci,j)) defines a bounded operator on l~(I). Proof. Given a vector x = (xy) , we want to estimate II y H, where y is the vector with coordinates Yi = Z CiJxJ" We write Yi = ~ i i Schwarz :
(el,i112~1.112~) (~i'1/2r~1/2~'i,1 ~Y]~ and apply
44 by (39). Then
11y I? < c ~ , Y
I c~,j 1~71 l~y i2< c)~J~71 I~y 12= c II • II2, which ~
j proves the lemma ! The proof of Theorem 1.11 is now complete in the special case when Tb = Ttb = 0 for some paraaccretive function b. Let us say how the argument above should be modified when Tbl  Ttb2 ~ 0 for two different paraaccretive functions. One considers two different bases (h~l), (Q,E) E 11 and (h~2), (Q,e) E 12 (the two sets of indices could be different if bl and b2 are not "special paraaccretive"), the first one adapted to bl, and the second one adapted to b2. Then, one computes the coefficients of TMb~ in these two bases. One l . e , l ~ l. es, 2 l ~ e , 1 I~ ~, 2 gets the numbers < "P/l, ~I,~Q ],,~Q, > b 2 : < ~~..q ,..Q >, where T : Mb~TMb~. The computations are then done exactly as above.
4. E n d o f t h e p r o o f : p a r a p r o d u c t s To complete the proof of the Tbtheorem, we shall build some sort of a paraproduct. Here, too, ideas related to wavelets will help us : we shall use a simple modification of the paraproduct using wavelets defined by Y. Meyer. To simplify notations, we shall still consider the case when bl = b2 = b ; the general case would only require minor modifications. Let us first define the equivalent of the function ~ of Part I. For each Q, call
OQ(x) = ( / Q b ( t ) d t l  l lQ(x).
(41)
Note that ]10Q ][oo< C [ Q [1, and that f #Qb ~ 1 ; if b is not "specialparaaccretive', Q runs through the modified cubes of the end of Section 2. Next, let us recall how one proves that the "wavelet coefficients" of a function of BMO satisfy a Carleson measure condition. L e m m a 4.1. If/3 E B M O , and if the C~ = in the basis (h~Q), then, for each cube R,
(4~)
f
~(x)h~Q(x)b(x)dx are the coefficients of fl
~ ~ I c~ Is<_ c I R I, Q c
where the sum is taken over all (Q, e) such that Q c R. The (very classical) proof is similar to our earlier proof of the square function estimate. Note that f fl(x)h~(x)b(x)dx, fortunately, does not change when a constant is added to ft. If Q c R, then C~  / ~(x)1R (x)h~ (x)b(x)dx
=  / [ f l ( x )  mR~]lR(x)hb(x)b(x)dx < ( 8  mR~)lR,h b >b . J
45 Consequently, it foIlows from (23) that
~ QCR
I C~ 12< C II ( ~  mRI3)IR II~<_C I R I,
e
and so (42) is true. Next, for each sequence C~ satisfying (42), we shall define an operator P ((C~))  P by its kernel
P(x,y) = ~ ~ C~h~Q(x)OQ(y). Q
(43)
L e m m a 4.2. I P(x,y) I< C I x  y In .
Proof. Let x ¢ y be given. Note that h~Q(x)OQ(y) • 0 implies that x and y are in Q. Then the size of Q is larger than ] x  y I/C and also, for each size 2 k >I x  y I/C, there is at most one cube Q of size 2 k for which h~(x)Oq(y) ~ O. Summing the absolute values gives less than ~
~ C I C~ I 2kn/22k~ < C ~ 2k'~/223k'~/2 (because (42) implies
k
e
k
that I C~ I< C IQ ['/2) < C ~~ 2kn < C I x  y
In, as needed.
k
Next, we wish to show that (43) defines a bounded operator from bP to its dual. This is because, if f and g are testfunctions, one can show that the series
Q
i
converges. Here is why : if Q is a large cube, the corresponding terms are quite small because f and g have a fixed compact support, and the L ° °  n o r m s of 0Q and h~ tend to zero fast enough ; when Q is small, one uses the smoothness of g and the fact that f g(xo)b(x)heQ(x) = 0 if x0 is any point of Q. So P is welldefined. Now, we want some weak smoothness property for the kernel (P(x, y) does not satisfy (3) because of the jumps of h~ and 0q). L e m m a 4.3. Let Q be a dyadic cube and h a bounded function, supported on Q and
such that f h(x)b(x)dx = O. Then, for all x ~ 2Q, P(bh)(x) = P~(bh)(x) = O. Note that this estimate is much better than the conclusion of Lemma 3.1. To prove Lemma 4.3, note that P(bh)(x) is a sum of terms of the form C~, h~, (x) f OQ,(y)h(y)b(y)dy, and all these terms are zero, unless QI contains x, and also some point y E Q. Since x ~ Q, Q' contains Q strictly. Then 0Q, is constant on Q, and the integral is zero, so P(bh)(x) = 0 for all x ¢ Q.
46 For the transpose, we write pt(bh)(x) as a sum of terms of the f o r m
as before, Q ' must contain Q strictly, and then h~,(y) is constant on Q. Altogether,
p t (bh) (x) = O. Lemma
4.4. If we choose C~ =
f
13(x)h~(x)b(x)dx, then Pb = 19 and ptb = O.
Let us say w h a t we mean by this. We want to show that, if R is a large cube tending to o0 (for instance, take for R the union of the 2 '~ dyadic cubes of size 2 M touching 0 and let M ~ + o o ) , then < P(blR),bg > tends to f l~(x)b(x)g(x) for each g C P such t h a t f bg = 0. A similar meaning is given to Ptb. Note t h a t if T is a SIO, and Tb = fl, then < T(blR), bg > tends to f l?(x)b(x)g(x), too, for each g E D such t h a t f bg = O. This can easily be checked from the definition of Tb (Remark 1 following T h e o r e m 1.11). To prove L e m m a 4.4, we shall do the c o m p u t a t i o n formally, and leave the details of the limiting a r g u m e n t (involving the large cube R) to the reader. First,
e'b(y) = f because each
f
: o
Q
h~Q(x)b(x)dx is zero. Also,
Q
~
Q
(by definition of OQ). This is B(x) by definition of the C~'s, and (22). We also need to show t h a t P is bounded. Lemma
4.5. P is bounded on L 2, with a norm <_ C l[ fl [[BMO .
Proof. If f 6 bP, Pf = ~ ~ q and so, by (23),
Q
O~2heQ/ 6Q(y)f(y)dy,
47 by Carleson's theorem, because the I C~ 12 satisfy the Carleson measure condition (42) (again see [Ga3], Th. 1.5.6). Let us now put everything together. Let T be like in the theorem, and suppose that
Tb = fix E B M O and Ttb = f12 E BMO. Construct operators PI and P2 with the functions /31 and f12, and consider the difference T I = T  PI  P~Since each piece does, T ~ is associated to a kernel that satisfies (2), and it also satisfies the conclusion of Lemma 3.1 (use L e m m a 4.3). Next, MbTtMb is weakly bounded, because Mb(P1 + P~)Mb is bounded and MbTMb is weakly bounded. Finally, L e m m a 4.4 implies t h a t if R is a large cube tending to oo (like in Lemma 4.4), then < T ' ( b l R ) , bg > tends to 0 when g is a test function such t h a t f g(x)b(x) = O. A similar estimate is true for T *t, and so we can apply the proof of Section 3, and obtain that T t is bounded. Since P1 and/)2 are bounded, we get the boundedness of T by substraction. If the functions bl and b2 were different, one would have to define the paraproducts a little differently. For instance, the kernel of PI would be
Pl(x,y) = E
E OQhQ (z)OQ(y),
(Q,~) e l where the set of cubes Q (replacing the dyadic cubes when bx or b2 is not "special paraaccretive') is adapted both to bl and b2 [the construction of such a set is just an iteration of the end of Section 2]. The rest of the proof is the same. We finally completed the proof of Theorem 1.11.
5. C o m m e n t s o n
Tb, s p a c e s o f h o m o g e n e o u s t y p e .
Let us start with a comment on the proof. A first proof of (part of) the Tbtheorem using wavelets was given by Tchamitchian [Tc3]. However, this proof was not quite as simple. The proof given above is quite stricking, because it gives directly the "full" Tbtheorem. A first glance at the number of pages could make the reader mistakenly think that it is not so simple, after all. Let us suggest to the reader that would be tempted by such a thought to go and have a look at [DJS] ! Also, we did our best to give all the relevant details about the proof : it would be much shorter ff we had restricted ourselves to the case when bl  b2  b is accretive and defined on the real line. For a compact proof, see [CJS]. Another nice feature about this proof is that it extends rather easily to spaces of homogeneous type. We do not wish to give a detailed explanation, and so we'll not even say what a space of homogeneous type is. We refer to [CW], [McS1] and [McS2] for this. The point is that the notion of a singular integral operator extends nicely to such a space (see [McSI& 2]) and there is even an extension of the Tbtheorem to the case of spaces of homogeneous type. If we look at the proof given in Sections 2, 3 and 4, we see that we only used the existence on ]R n of something like dyadic cubes. If the space of homogeneous
48 type E has a family Rk, k E 2Z, of partitions of E with the properties below, the proof given in the previous sections will extend to give a T b  t h e o r e m on L2(E) : we used the fact t h a t  if Q 6 )~k and Q ' E £k, for a k < k', then Q contains Q ' as soon as Q (1 Q ' ¢ 0 ;  all the cubes in R~ have comparable mass : Q, Q ' c ~k implies t h a t I Q I< C I Q ' I ;  each cube of Rk contains < C cubes of Rk+l ;  each cube Q E Rk has a "sufficiently small b o u n d a r y " . We used the last condition in L e m m a 3.4, to make sure t h a t 1Q would be regular enough to allow the use of the weak boundedness, and in L e m m a 3.6 to control the n u m b e r of small cubes R t h a t are close to Un OQn. A condition that would be enough, for instance, would be the existence of an a > 0 such that, if Q is one of the cube and r is its diameter, then [ { x : dist(x, OQ) < 2kr} [< C2 k~ [ Q t .
The existence of such cubes is not hard to establish in the special case of an Ahlforsregular subset of some ]R n, i.e. a closed subset E of ]R '~ with a measure # (generally the restriction to E of the k  d i m e n s i o n a l Hausdorff measure) such that, for some positive real n u m b e r k > 0 and some C > O, C  l r k < # ( E M B ( x , r ) ) < Cr ~ for all x E E and r > 0. A construction is given in [Dv6], but the a u t h o r is not very proud of it (it is more complicated t h a n necessary), and since it is used in a few different places in this book, a slightly simplified version will be added as an appendix. W h a t about all the other spaces of homogeneous type, then ? It turns out t h a t another construction can be given, and t h a t this construction extends to all spaces of homogeneous type (see [Ch.2]) ! Moreover, M. Christ's construction is not much more complicated. He uses it to prove a variant of the Tbtheorem, b u t with variable b (see [Ch.2], but also [Ch. 1] for a statement of the result). A final c o m m e n t on this proof : the fact t h a t we do not have to deal with complicated approximations of the identity makes it easier to extend the t h e o r e m to the case of matrixvalued kernels. You will see an example of this in P a r t III, T h e o r e m 6.7. Let us not insist more.
6. A p p l i c a t i o n s We'll make this chapter a little shorter t h a n it should he. T h e kind reader is referred to [My4] and [Ch.1] for most applications.
49
A. More wavelets We start with two results at the interface between singular integrals and wavelets. T h e o r e m 6.1. Let ¢ be a C 1  f u n c t i O n on IR, with rapid decay at oo, and such that the 21/2¢(2ix  k), (j, k) E ~2, form an orthonormal basis of L2(IR) (for instance, take Y. Meyer's wavelet). Then the 2 i ¢ ( 2 1 x  k) form an unconditional basis of the atomic space HI(1R).
Proof. One needs to show that if f E H 1 is a finite s u m of the form cj,k[2J¢(2J
j,k
 ,)1 =
c,¢,
I
(with a selfexplanatory change of notations), and if (el) is of modulus < 1, then £ = E I £ICI•I satisfies I[ f~ IIHI ~ not depend on f or (er) !]. Call T~ the operator which sends ¢ I to e~¢l. Te is n o r m < 1 because the 2J/2¢(2Jx  k) are an orthonormal
a sequence of complex numbers C II f IIHI [with a C t h a t does well defined on L 2, and has a basis of L 2. The kernel of T~ is
~~Y,k 2JeJ,k¢(21x  k) ¢(21y  k), and so one checks easily t h a t Te is a singular integral operator. Finally, T~I = 0, and it follows from the R e m a r k 3 after T h e o r e m 1.6 that the Te's are uniformly b o u n d e d on H I. (Hence we proved the theorem.) We leave the estimates on the kernel as an easy exercise. Remark. We stated this theorem because it is rather easy ; the same sort of proof would have shown t h a t the ¢ i ' s are an unconditional basis for m a n y other spaces (for H I, however, the existence of such a basis is not so recent). T h e o r e m 6.1 also generalizes to H 1 of the bidisk (the idea is to use the theory of Calder6nZygmund operators on p r o d u c t spaces). See [Le]. Our second result is a counterexample due to Lemari~ (a first example was found, shortly before, by P. Tchamitchian [Tcl]). The question was the following : if T is a b o u n d e d singular integral operator, and is invertible on L 2, is it true t h a t it is also invertible on L p, 1 < p < + c ~ ? The answer is no : for each p, there is an operator T like t h a t which is not invertible on L p. Note t h a t one cannot hope a better counterexample : an easy use of complex interpolation shows t h a t if T is b o u n d e d on L p for P0 _< P < 4Po 1 for some P0 < 2, and is invertible on L 2, then it is also invertible on L p for p in a small neighborhood of 2 (which of course depends on the norms of T on the LP's). This was observed by Calder6n [Ca2]. Here is the counterexample. Consider Y. Meyer's basis of L2(]R) (for instance) 2i/2¢(2Jx  k), and index it by dyadic intervals I = [k2  j , (k + 1)2J], so as to get a basis ( ¢ I ) i dyadic. If I is a dyadic interval, call ¢ ( I ) = [0,2 i+1] if I = [0,2J] and ~ ( I ) = 0 otherwise. Now let To be the only operator such t h a t To¢l = 0 if ~ ( I ) = 0, and To~bl = ¢¢(x) otherwise. Clearly, To is a b o u n d e d operator on L 2, with n o r m 1. Also, the kernel of To is
50 ~]j. ~b[0,2j+,](x)d]10,2j](y) ; i t satisfies (2) and (3) by the usual computation, and so To is a SIO. Thus, if 0 < r < 1, the operator T  1  rT0 is a SIO, and is invertible on L~(lR). The inverse is T 1 = ~ k > o rkTok, and in particular T1(¢I°,1} ) = Z rkOek([o,']) = Z rk2~/2¢(2~x)" k>o k>o We leave the fact that, given a p < 2, one can find r ,< 1 such that this function is not in L p as an exercise (one can do it directly, or use the characterization of L p by wavelet coefficients given in Part I, § 8, Example 2). If one wishes a counterexample for p > 2, one can use the transposed operator (which corresponds to taking ¢([0, 2k]) :: [0, 2~1]). In spite of this example, it is possible to prove positive results concerning the inversion of singular integral operators. See [Tc4], for instance. B. T h e C a u c h y i n t e g r a l a n d r e l a t e d o p e r a t o r s . In most of the following examples, the operators will be principal value operators defined by an antisymmetric kernel (Example 1.3). For these operators, we said in Section 1 that the weak boundedness property is always satisfied. Since T t l =  T 1 , one will only have to check that T1 C B M O to apply the T l  t h e o r e m (a similar remark can be formulated concerning the Tbtheorem). E x a m p l e 6.2. (Calder6n's commutators). Let A be a Lipschitz function on the real line (i.e., a function such that I A(x)  A(y) I~ C I x  y I). Set g n ( x , y ) = (A(~)A(Y))" (~_~).+l , and let Tn be the principal valeur operator defined by Kn. E x e r c i s e : prove that Tn is bounded on L 2 ( ~ n ) , with a norm < C n+l II A' I]~o for some C_>0. Hint : do this by induction. After checking that K satisfies (2) and (3) with Co _~
C(n + 1) I] A' lifo, show that T,,(1)  T~I(A') and conclude. C o m m e n t s . The first proof of this estimate dates from Calder6n [Call, in 1977. In their famous paper [CMM], Coifman, McIntosh and Meyer proved a polynomial estimate, namely II Tn ]1~_ C(1 ÷ n 4) I] A' li~, ; this estimate was then improved by Mural [Mul], by a completely different technique. The best estimate, up to now, is due to Christ and Journ6 : for each e > 0, (44)
II Tn II_ ~ C¢(I+ II A' Hoo)T M
(see [ChJ]).
E x a m p l e 6.3. (The Cauchy integral on a Lipschitz graph).
51 Let A
:
IR * IR be Lipschitz.
T h e antisymmetric s t a n d a r d kernel K(x,y) =
[x + iA(x)  y  iA(y)]  l defines a principal value o p e r a t o r CA. Exercise • Write
I [I + i A ( ~  A(Y) ]  I K(x,Y)  x _ y  y
n
~

'
n
'
deduce Calderdn's result from the estimate of E x a m p l e 6.2 : CA is b o u n d e d as soon as II A' IIoo is small enough. • Prove Coifman, Mclntosh and Meyer's result directly : CA is b o u n d e d for all Lipschitz A's ( H i n t : c o m p u t e CA(1 + iA')). Comments. T h e first proof of the boundedness of CA when II A' Iloo is small did not use c o m m u t a t o r s . Calderdn [Call showed t h a t the II CtA II, where A is given and 0 _< t < 1, satisfy a differential relation which allowed h i m to estimate CA ; the estimates on c o m m u t a t o r s are a consequence of the boundedness of CA. Since the first proof of the boundedness of CA in the general case [CMM], it has become fashionable to give new proofs of the boundedness of Ca. T h e reader is invited to produce one or two of his own. For a nice collection, see T. Mural's book [Mu2] ; for the "shortest" proof, see [CJS] (it is also in [Mu2]). T h e best estimate on the n o r m of CA is (45)
II CA II<_C ( I + II A' 11oo)1/2.
It is due to T. Murai [Mul] (also see [Mu2]) ; the proof uses a p e r t u r b a t i o n a r g u m e n t similar to the one presented in the first sections of Part III (but sharper !). T h e estimate (45) is quite good, because one can find functions ,4, with II A' I1~~ +oo, such that II CA It> a0 4 II A' II~ 2 . The idea is to approximate G a r n e t t ' s example (see Part III) by a lipschitz graph [Dv3]. Note t h a t it is not a bad idea to have a very good estimate for CA, because m a n y other operators can be constructed from CA. See E x a m p l e 6.5 below, for instance. E x a m p l e 6.4. (Chordarc curves). A rectifiable, connected curve r C C is called chordarc if, for some constant C > 0 and any two points A , B E r , the length of the piece of r connecting A to B is less than C times I B  A I • (This is for an open curve r ; a small modification, left to the reader, has to be done for closed curves.) Let I" be a chordarc curve, and denote by ds the arclength measure on r . One can try to define a Cauchy integral on r by Crf(x) = p . v . f r z@~f(w)ds(w) for f E L~(F,ds}. E x e r c i s e . Show t h a t Cr is a bounded o p e r a t o r on L 2 ( r , d s ) . (Do not pay too much attention to the definition of the principal value.) Hint : call t ~ z(t) a p a r a m e t r i z a t i o n
52 of I" by arclength. The problem is equivalent to showing t h a t K(x,V) = [z(x)  z ( v ) ]  1 defines a b o u n d e d operator T on L 2 (JR), and this is easily done because Tz' = 0 and z' is "special p a r a a c c r e t i v e ' . Example
6.5. Let A : IR ~ IR N be a Lipschitz function, and let F be a C °o function
from ]R N to C. T h e k e r n e l
1LF xy \(A(z):A(y)] ~ u /
defines a b o u n d e d singular i n t e g r a l o p e r a t o r
on L 2 (JR). One can find a first proof of this fact in [CDM]. The idea is to write F as a Fourier series, and then to s t u d y the case of F(z) = ei(~1¢~+'''+xN~N). One is trivialy reduced to the case of one dimension, and then one writes the exponential e i~ in terms of an integral of objects like (1 + ix)I (using the C auchy formula). One gets an estimate on the operator with kernel 12exn zy 
(iB(~)B(u)] \ zy ]
which grows polynomially in ]] B ' ][oo, and this estimate
is enough to sum the operators coming from the Fourier series of F, provided t h a t the Fourier coefficients have enough decay. Of course, the condition " F E Coo " is not needed ; the better one can estimate the n o r m of CA, the less derivatives will be needed on F. So it is advisable to use Mural's result (45). For a reasonably sharp estimate, see [DS1]. Using this example, and the "rotation method" to get operators on L 2 (IRn), one can get quite a collection of Calder6nZygmund operators. E x a m p l e 6.6 ("Stable kernels"). Let K(x,y) be a s t a n d a r d kernel on the line ; we'll say t h a t K is a Calder6nZygmund kernel if there is a b o u n d e d s i n such t h a t K is its kernel. We'll say that K is "stable" if, for all Lipschitz functions A : ]R * IR, the kernel K(x,y) A(z)A(u) is a Calder6nZygmund kernel (so, taking A(x) x, K is Calder6nXy Zygmund). It was first observed by Journ6 that if K is an antisymetric Calder6nZygmund kernel, then it is stable. E x e r c i s e : use the T l  t h e o r e m to prove it. One can easily characterize the Calder6nZygmund kernels t h a t are stable. First, the notion only depends on the kernel, and can be studied on truncated kernels K(x, y)l{e<_[z_yl<M}. Thus one can restrict to compactly supported kernels (and prove uniform estimates). Then, one applies the T l  t h e o r e m and does an integration by parts, and one finds t h a t the Calder6nZygmund kernel K is stable if and only if the function/3 defined by (46)
/3(u) =
f fz
K(x,y)dxdy
is in
Z
/ fy
y
K(x,y)dxdy
x
y
BMO(IR).
Note that, if K is a convolution kernel, or if K is antisymmetric, the condition is automatically satisfied. The amusing thing is that, if K is stable, then K(x, y) A(z)A(y) is also stable when xy A is Lipschitz, so that one can prove t h a t the kernel
K(x, y) (A(~)A(u)] zy / k
N
also defines a
53 b o u n d e d singular integral operator, with n o r m < C(K) N+I bl A' I1~ One can then build other bounded operators, like when K(x, y) = ~xZ'~'l by s u m m i n g series. This is a joint work with Coifman, Journfi and Semmes, but we can only refer to [Dv2] for details. There was a partial result by Mural. E x a m p l e 6.7 (singular integrals on Lipschitz graphs). Let 0 < d < n be integers, and let k(z) be a C ~ function, defined on JR"\{0}, and such t h a t I V J k ( x ) ]
(47)
for a l l j _ > O
and f
(48)
/
sup
0 < e< M Let A
]Rd ""+
:
JR, n  d
k(tO) It Id1 dt <_C for all 0 E S n1.
Je
be a Lipschitz function. T h e n the kernel
K(u,v) = k(u  v , A(u)  A(v)) defines a b o u n d e d singular integral o p e r a t o r on L 2 (IRd). T h e proof is a nice application of the techniques we have just seen (and, unfortunately, of techniques t h a t will be seen in P a r t III). We shall write it as a series of exercises. E x e r c i s e s . Let k : JR\{0} ~ C be such that
i k(x) I~ Co Ix 11 and I k'(~) I~ Co I~ Is,
(49) and also
(50)
sup 0<e<M
f
[
k(x)dx < Co.
Je
1. Show t h a t K(x,y) = k(x  y) is a Calder6nZygmund kernel. You will not even need to apply the T l  t h e o r e m , because the Fourier t r a n s f o r m of t r u n c a t e d kernels of the form X(X)k(x) can be estimated directly. 2. Show t h a t k(xy) is even a stable kernel, and t h a t K(x, y) = k(xy) \( A(,:)A(u)z_y) N defines a b o u n d e d o p e r a t o r on L2(]R), with n o r m _< cN+lCo ]] A' I]~ for some C. 3.
show that,
A is Lip ch tz, the kernel k/x
y)expi
ao ne
bounded
operator, with n o r m < C(ll A' II~)C0. T h e e s t i m a t e above will not give a constant C(I ] A' II~) which is only polynomial in t e r m s of II A~ Hoo • One now has to apply the technique of P a r t III, § 4.B, which shows
54 t h a t one can take C(]] A ' Iloo) < C ( l + II A' lloo) M for some M . We therefore postpone this part of the exercise, and suppose we get a polynomial estimate in II A ' IIoo • Now consider the kernel of Example 6.7, in the special case when d = 1. After a change of coordinates in IRn of the form (xl,.. ", xr,) ~ (Xl, Cx2,'" "Cxn), we are reduced to the case when [I A ' ][oo< 1. The values of k(x) = k(u,x') (we shall systematically write x = (u,x') with u • IR and x' • ]R n  l ) when I x' I>l u I do not matter, and so we can modify t h e m at will. Write k(x) k(u,O), where 0 = d has its values in IR "  1 . We can assume t h a t this function is 27rperiodic (in all variables) in 0. 4. Check that /c(., 0) is a C °O function of O, valued in the Banach space of kernels satisfying (49) and (50). Thus, one can write
(51)
~(u,O) = ~ vE~
k.(u)~ ~''0.', ~
where the kernels kv satisfy (49) and (50) with constants Co() that decrease rapidly as l " I' +oo. 5. E , kv(~
Coming back to K(u,v)
= k(u  v,A(u)  A(v)), use the fact t h a t K(u,v) = 2i7r,. (a(u2A(~,)~ to deduce the c a s e d = 1 of Example 6.7. v)exp
B. Finally, deduce the general case from the case when d = 1 by the " m e t h o d of rotations" : write truncated operators
os(u) =/o_o>
=
r,:s(,,)do,
where
To:f (u) = fit I>, K ( u ' u + tO)f(u + tO) l t Id1 dt can be controlled by the 1dimensional result. The reader who is not too familiar with the m e t h o d of rotations will find descriptions in m a n y sources, and a proof of the exercise can be found in [Dv5], p. 245. R e m a r k . W h e n k(x) is odd and homogeneous of degree  d , most of the exercise can be avoided : one can prove the result by a direct application of Example 6.5 and the m e t h o d of rotations.
PART III SINGULAR
INTEGRALS
ON CURVES AND SURFACES
oo0oo
1. I n t r o d u c t i o n a n d n o t a t i o n s We would like to extend Coifman, McIntosh and Meyer's nice result to as many higherdimensional objects as possible. In the following, we'll be given 0 < k < n, and a kdimensional object $ c ]R n. We'll always call $ a "surface", but we do not assume any smoothness on $ : $ is a set. Also, S will come with a non negative Radon measure on lR'~, such that supp ~u  $. For instance, we could take for S a smooth kdimensional surface, with for ~ the kdimensional surface measure on $. The singular integrals we wish to consider are generalizations of the Cauchy integral (defined on curves of the complex planes). D e f i n i t i o n 1.1. The function K(z), defined on IR"\{O}, will be called a "good kernel" if g (  z ) = K(z) for all z # 0, g is Coo and (1)
I V J K ( x ) I< C ( j )
[xl kj
for all j_>O.
R e m a r k 1.2. The choice of the class of "good kernels" is not extremely important: we only have to take it small enough, so that defining singular integrals on lipschitz graphs will not be a problem. We could also have taken the slightly larger class where the antisymmetry condition is replaced by (2)
sup
0<e<M
[
g(to) [tl k  l d t
Je
The cancellation still has to occur on each line, because we want cancellation on all kplanes, not matter how oriented. We could also have asked for the more restrictive condition that K be C °°, odd, and such that K(.~x) = .~kK(x) for all x ~ 0 and ,~ > 0. This class would not be enough, however, for the current proof of the converse result mentioned in Section 9 (Theorem 9.5) to work. Given a good kernel and a Radon measure # >_ 0 on lR n, we would like to define an operator T by a formula like Tf(x) = f g ( x  y)f(y)d#(y). We'll always assume that # is in the following class A. D e f i n i t i o n 1.3. The letter A will denote the class of non negative Radon measures # on ]R'~ such that, for some C _> 0, (3)
<__ c ,
for all •
and
," > O.
56 We'll see soon t h a t /z E A is necessary if we want all g o o d kernels to define b o u n d e d operators on L2(lRn,dtz). Let us first define a truncated and a maximal operator. For e > 0 and f E Cc(IR'~), let
(4)
= fix
K(x  y)f(y)dl (y)
and
(5)
sup
I
e>O
If the maximal operator T~ is b o u n d e d on L2(IR n, d#), the T~'s are uniformly bounded, and we'll be able, if we want, to define an operator T by extracting a weakly convergent sequence. On the other hand, studying T~ will not be harder t h a n just showing the boundedness of some limit of T~'s ; in fact, we'll prove later that, with a small additional hypothesis on #, the L 2  b o u n d e d n e s s of such a limit implies the boundedness of T*/ z " Our main question, in its general form, is the following : for which measures # is T~ b o u n d e d on L2(]Rn,d#) ? W h e n k = 1 and n = 2, we could consider the special case of K ( z ) = ~, ' and ask for which measures # the Cauchy kernel defines a b o u n d e d operator on L2(dtz). By the way, we are just talking about L 2 for convenience, but L p, 1 < p < + c o , would do as well (and give the same measures #), assuming/z is in the class ~ of Definition 2.3 below. We'll see on the way t h a t the question is not so simple, even when k  1 and n = 2. Let us at least justify our decision to restrict to measures # E A. P r o p o s i t i o n 1.4. Let Ki, i E I, be a finite family of good kernels such that, for some 6 > 0 and all x ~ O, [ Ki(x) [> i5 [ x [k for some i E I. I f # is a nonatomic measure such that the T~'s corresponding to the various K i ' s are uniformy bounded (in i and e), then /zEA. A minor variant of this result is proved in [Se3] allegedly using a wellknown method; let us give here a slightly different proof. Let us suppose t h a t the 7~'s are uniformly bounded, and t h a t / , ~/A, and let us find an atom. L e m m a 1.5. There is a constant Co such that if Qo c N n is a cube, r0 /s its sidelength, mo ~ #(Qo) and Ao moro k, then one can find a cube Q1 c Qo, of radius rl = ro/lO0, =
and such that m I = b t ( Q 1 ) > (1  ~  ) \
rn 0 .
'o/
Of course, this will be useful only when A0 is large enough ! Let us prove the lemma by contradiction. Split Q0 into C~ cubes of size Ci~r0 ; one of the cubes (call it R) is such t h a t # ( R ) > C T n m o . Also, if the lemma is not true for Q0, one can find another cube S C Q0, of size C ~ ' r o , and at distance >_ ~
from R, with a measure/z(S) _> [ ~ m 0 ] t"o
J
C 7 ~.
57 Take the testfunction f = l s , and look at the T~f(x), for e < < r, on the cube R. If C1 is large enough, then there is an i E I such t h a t the corresponding T~f satisfy ok on R [this is because, the Ki(x, y) being good kernels, each of t h e m [ T~f(x) 1>_ U/z(S)r ~ is almost constant when y varies in S]. So
II TLt
uCR)I/ uCS)ro
>_
ClZCR)l/2 (S)l/2ro k II .," llz, ~(R)'/zU(S)X/Zro k <_ C. Using our < Crko and, if Co is chosen large enough,
and since the 7~'s are uniformly bounded, we get
estimates on u(R) and U(S), we get r n o C U ' ~ ~_zxo< ~ro,1 k which contradicts the definition of ,%. So L e m m a 1.5 is true. Let us now use the lemma to show that, if Ao is large enough, there is a sequence Q o ::) Q 1 D .  ::) Q i z:) . . . o f c u b e s , w i t h sidelengths r~ = (lO0)~ro a n d m e a s u r e s ml = #(Qi) such t h a t )~i = miri k > 10i)~0 • We certainly have a cube Q0 because # ¢ A. Suppose we constructed Qi, and apply the l e m m a to Qi. We find a cube Qi+l c Qi, with sidelength ri+l = ri/10o and mass /
%
mi+l >_ _(1  c._¢_~? .) mi > 1m'2, if A0 is large enough. T h e n ~i+1 = (100)km~+~ ~ i r n , > 10AI, as promised. We get a b e t t e r e s t i m a t e for rni t h a n what was just written :
ml > (1  A I C ° I ) ( 1  A ~ )
too> 
1~(100)
..
1
too:>
2
m
if we choose A02 > 10Co, for instance. T h e n Ni Qi, which is reduced to one point, has a m e a s u r e >_ ~0_ > 0, and the proposition is true. R e m a r k 1.6. If # E A and K is a good kernel, one can define T~f(x) everywhere for any f e L2(d/z) (use Schwarz' inequality, and the estimate
ut>,
Ix  y I
du(y)
=
oo
I x  y 12k d#(y) < ~ C(2Q)2k(2Q) k < +oo). £=0
E x a m p l e 1.7. Let z : ]Rk * ]R n be a continuous function. We think of z as being the p a r a m e t r i z a t i o n of some surface. We can define a measure # on ]R n by # ( f )  f ~ . f d# f~tk f(z(x))dx (for f e ¢c(Nn)). We'll often call # the " m e a s u r e associated with z". Saying t h a t / z E A amounts to saying t h a t
58 for all w E JR" and all r > 0.
2. C a l d e r 6 n  Z y g m u n d
techniques
We shall need extensions to our context, of some of the classical results of "Calder6nZ y g m u n d theory". By lack of time, we'll only recall the m a i n steps of the proof ; if the reader has problems filling the gaps, he will find all the details in [Dvl], [Dv2], or [Dv5]. One of the useful tricks t h a t will be used here is distinguishing between the measure # t h a t comes into the definition of T~, and the measure (call it a) on which we want to integrate T~ (x). This is a little like studying a bilinear form instead of a quadratic form. The advantage is t h a t we'll be able to change # (or o) without affecting the other measure. We need an easy extension of the usual maximal theorem concerning the HardyLittlewood maximal function. D e f i n i t i o n 2.1. If # is a nonnegative R a d o n measure and f is measurable, define a maximal function by f
M~.f(z) =
sup r k ] r > 0 Slzwl
I f ( w ) I dtt(w).
Note t h a t M ~ f is allowed to be + o o at some points ; w h e n k = n, M , f HardyLittlewood maximal function of f d/z. Lemma
2.2.
is the usual
If # and a are two measures in A , 1 < p < +co and f E LP(d#), then
M , f E L p (da), with
(7)
I1M~,f IILp(da)<_ C(p, lz, a) [] f IIL,(d•) .
The proof is the same as usual : one uses interpolation to reduce to showing t h a t M~, sends n°°(dlz) into L°°(da) (which is trivial), and also L'(dtz) into weakLl(da), which can be proved easily with the help of the Besicovitch covering lemma. D e f i n i t i o n 2.3. We shall denote by y~ the subset of A composed of the measures a t h a t also satisfy, for some "~ > 0,
(s)
a ( B ( w , r ) ) > "~rk for all w E s u p p e and r > O.
E x a m p l e 2.4. If z : ~ k ~ ]R n is such t h a t the m e a s u r e / z associated to z is in A, and if z is Lipschitz (i.e., [ z(x)  z(y) [<_ C [ x  y I), then # E ~ .
59
Proof. If w e supp/~ and x is such that z(x) : w, then ~ ( B ( w , r ) ) : l ~  l ( B ( w , r ) ) ]
>1
l> c'r k
In particular, if/z is the arclength measure on a connected, rectifiable curve of infinite length, the condition (8) is always satisfied. The following lemma can be seen as an analogue of the result on quadratic forms that says that a bilinear form is dominated by the corresponding quadratic form. L e m m a 2.5. Let K be a good kernel, tz E A and a C ~ . Suppose that, for MI 1 < p < +oo, T* is bounded on LP(da). Then, for 1 < p < +oo, T~ is bounded from LP(da) to LP(d#), and T~ is bounded from LP(d~) to LP(da). The proof of L e m m a 2.5 relies on pointwise inequalities. For instance, the boundedness of T~ from LP(da) to LP(d#) follows from the inequality
(9)
T~ f(z) ~ C [Ma(T* f)] (z) + C M a f ( z )
for all z C IR'~, and from Lemma 2.2. To prove that T~ sends LP(d#) into LP(do), one first proves that there is an operator T~, which is obtained as a weak limit of T~'s, and which is bounded from LP(d~) to LP(da). The existence of T~ follows from the fact that the Tae's are uniformly bounded from LP(da) to LP(d#) (which follows from the boundedness of T* on the same spaces), and duality. To deduce the boundedness of T~ (from LP(d#) to LP(da)) from the existence of T~, one uses the following generalization of Cotlar's inequality :
(lo)
r;I(z) <
+ C [M. I I
The boundedness of T~ easily follows from (10) and L e m m a 2.2. The proof of the pointwise inequalities (9) and (10) is relatively straightforward. The only things that are used are the estimates on K and its first derivative. The proofs are given in full details in [Dvl] in dimension 1 (but the argument is the same in all dimensions). One can also look at [Dv21 or [Dvh]. R e m a r k . In the statement of L e m m a 2.5, we assumed that T~ is bounded on all LV(dcr)'s. This is only for convenience, because one can prove that the LPboundedness for one p implies the LPboundedness for all other p, 1 < p < +eo. The proof is a routine Calder6nZygmund argument, using the fact that if a E ~ , the support of a is a space of homogeneous type. The machine which will be described in the next section will need an initial result to get started. We shall feed it with the following special case. L e m m a 2.6. If K is a good kernel and a is the surface measure on the graph of a Lipschitz function A : IR k ~ IR,~k, then the operator T* is bounded on LP(da) for I < p < +oo.
60 (Of course, the image by a (linear) isometry of such a graph would work as well !) W h e n the kernel K is antisymmetric and homogeneous of degree k, this l e m m a is a rather easy consequence of Coifman, McIntosh and Meyer's t h e o r e m (more precisely, of E x a m p l e 8.5 of P a r t II) and the rotation method. W h e n K is a general " g o o d kernel" (as in Definition 1.1), L e m m a 2.6 follows easily from E x a m p l e 6.7 of P a r t II : the only difference is t h a t we replaced the Lebesgue measure on ]Rk by the surface m e a s u r e on the graph, and also t h a t we used a slightly different truncation from the usual one (the difference is easily estimated by a m a x i m a l function).
3. T h e " g o o d A" m e t h o d . Good A inequalities are an impressive tool, invented some time ago by D. Burkholder and R. Gundy. Here is a version of their lemma. Lemma
3.1 Let X be a set and # a measure on X . L e t u : X ~ [0, +co] be measurable. Suppose u is equal, e x c e p t p e r h a p s on a set o f finite measure, to a function o f L P ( X , d#). Let v E L P ( X , d # ) . S u p p o s e t h a t there is a 0 < r / < 1 and, for each e > O, a constant ff > 0 such that, for all ~ > O, (11)
kt({x e X : u ( x ) > )~ + eA and v(x) <_ 7A}) < (1  ~/)#({x e X : u ( x ) > A}).
Then u C LP(X,d,)
and [] u
llp<_C ( p , e , ~ , 7 ) IIv llp.
We'll leave the proof as an exercise (one can also go and look at [Dvl] or [DvZ]). We do not need every e, but the ¢'s t h a t work depend on r/ and p ; it will be just as easy to find a "7 for each e :> 0. The hypothesis of the l e m m a is some reasonably subtle way of saying t h a t u is dominated by v. T h e surprising fact is how well L e m m a 3.1 works. T h e following result might seem quite technical. We shall see later how to use it (see Definition 3.4 and Corollary 3.6 below). P r o p o s i t i o n 3.2. L e t tz E ~ ; s u p p o s e t h a t there exist constants 0 < 0 < for each 1 < p < + c o , A v ~_ 0 such that the following is true. For each ball the s u p p o r t o f # , one can find a measure a E ~ and a c o m p a c t set E c B that (12) a satisfies (3) with the constant C and (8) with the constant 3 = ~1 (13)
# ( E ) >_ O#(B) ;
(14)
1Elz ~_ a (i.e.: a is larger than the restriction o f # to E ) , and
(15)
IIT~f llL'(d~,)~ Ap IIf llL,(d<~)for f E Ce(~n). Then, there are constants Cp, 1 < p < +oo, such t h a t
1, C >_ 0 and, B centered on N s u p p # such .
,
61
II T~f IlL,(d~,)~ Cp II f IlL~(d~,) for f C C~(IR").
(16)
T h e proof is quite standard, but we give it to convince the unexperienced reader t h a t good ), inequalities are a really powerful tool. Note t h a t the proposition says, in some sense, t h a t local information on a small piece of the s u p p o r t of tt, at all scales, is enough to obtain a global result of boundedness on L2(d#). This is a little similar to John and Nirenberg's t h e o r e m on BMO.
Proof. Given f E Ce(]Rn), we want to apply L e m m a 3.1 with u(x) = T~f(x). Let us start by checking the qualitative assumption. Lemma
3.3. The function u is equal, outside of a compact, to a function of LP(d#).
Indeed, if I z I is large enough, I T~,f(z) IG fsupp/ I K ( z < C M v f ( z ) , which belongs to LP(d#) by L e m m a 2.2.
w) I[ f(w) I d#(w)
Let us now choose the function v : (17)
v(x) = M j ( x )
+ {M~( I f I~)}l/'(x),
where r = V~, for instance. We want to prove (11). Let A > 0, and call fl = {x • IR n : u(x) > ).} ; fl is open because T~ is lower semicontinuous. Cover I2 n s u p p # by the balls B(x) = B (x, ½dist(x, fie)) , where x • s u p p # . A covering l e m m a of Vitali type (see [St], p. 910) gives a sequence of points xi such t h a t the B(xi) are disjoint, but fl n supptz C U i ~ 1 1 0 B ( x i ) . If we prove t h a t , for each i, (18)
#({x • B(xi)
0
: u(x) < )t + eA or v(x) > "~A}) _> ~#(B(xl)),
then (11) will follow because s u m m i n g on i gives 0 # ( ( x • fl : u(x) < )~ + e,k or vCx) > qA}) > ~ E # ( B C x , ) ) i (because B(xi) is centered on s u p p # and tt • E ) Thus, we only need to find, for each e > 0, a i. If v,(x) > qA for all x • B(xi), there is nothing is a ~ • Y(xl) such t h a t v(~) < /A. Write f = f l + f2, with fx = 1B(~,Rlf, and constant C1. For f2, one uses the regularity of T~f2(:c) < T~f(a) + CM~f(~) for all x • S(xi) as an exercise.) Let us choose a point a • 3B(xi)
(19)
0
> ~ E t t C 1 0 B C x / ) ) d
> ~tt(fl). This is (11), with r / = ~. ff > 0 such t h a t (18) is satisfied for each to prove. So we can suppose that there where R = Cldist(~,12 c) for some large K to show t h a t , if C1 is large enough, and a • 3 S ( x i ) . (The verification is left n fl c, so t h a t we get
eA T ; A ( x ) _< A + ~  for all x ~ B ( x , ) ,
62 provided we choose ? small enough. Therefore, we only need to prove t h a t
(20)
•
:
<
>
Let us use the c o m p a c t set E and the measure a of the hypothesis. Since/~(E) > 8#(B(xi)), . e)~ we only have to show t h a t #({x E E : T~fl(x) > ~}) < ~#(B(xl)). By (14), the left hand side is less t h a n a({x • E : T~fl(x) > ~ } ) < 2rerA ~ [[ T~f, [[~'(d~) • By (16), 7'* is b o u n d e d on LP(da) for 1 < p < +c~. By L e m m a 2.5, T~ sends r < Lr(dtz) to n r ( d a ) , and so /~({x • E : T~fl(x) > ~}) _< Cer)~ r ]l fl ][Lr(dp)Cer)~'Rk{M~,([ f ]r)}(~) < C,r)~~#(B(x,))q~)V < ~#(B(x~)) if we choose 7 small enough. This proves (20), and so (11) and the proposition follow. Let us say how we intend to use Proposition 3.2. Let $ C IRa be given with a measure # • ~ such t h a t $ = supp #. D e f i n i t i o n 3.4. We shall say t h a t S "contains big pieces of Lipschitz graphs" (and often write $ C B P L G ) if there are constants 0 < 8 < 1 and M > 0 such t h a t , for each x E S and r > 0, there is a compact set E C S N B ( x , r ) , with # ( E ) > 8r k, and which is contained in the image by a linear isometry of IRa of the g r a p h of some Mlipschitz function A : IR k ~ IRnk. R e m a r k 3.5. T h e notion is really a notion concerning S, and not #, because if S C IRa is the s u p p o r t of some measure # E ~~.,then/~ is equivalent to the kdimensional Hausdorff measure on S (the verification is a s t a n d a r d covering argument, which is left as an exercise). Of course, the p r o p e r t y of Definition 3.4 does not change when/~ is replaced by any measure a such t h a t C  1 # < a < C/z ! C o r o l l a r y 3.6. If S CBPLG, and I~ 6 ~ is such that supp lz = $, then T~ is bounded on LP(S,dlz) for 1 < p < +c~. To prove the corollary, one applies Proposition 3.2, with for a the kdimensional surface measure on the image by an isometry of the Lipschitz graph of the definition. T h e estimate (15) then comes from L e m m a 2.6, and (14) follows because b o t h 1E/z and l e a are equivalent to the kdimensional Hausdorff measure on E (see the r e m a r k above, or the a r g u m e n t in [Dv5], p. 252). E x e r c i s e . Show that i f z : IR k * IRa is a bilipschitz m a p p i n g (i.e. satisfies C 1 I u  v l< I z(u)  z ( v ) t< C l u  v I for some C > 0), then S = z(iR k) C B P L G (and so Corollary 3.6 applies to S). [Warning : this is more challenging t h a n most of the previous exercises. If you don't find, just go on reading !]. This is not the most striking application of Corollary 3.6, because the boundedness of the singular integral operators on S can be proved directly, using the p a r a m e t r i z a t i o n of S and the rotation m e t h o d , as for L e m m a 2.6 or E x a m p l e 8.5 of Part II.
63 R e m a r k 3.7. One can imagine a more complicated corollary. Let us say t h a t " $ (CBP)2LG" if $ "contains big pieces of surfaces that C B P L G uniformly" (with a definition similar to Definition 3.4). A second use of Proposition 3.2 gives, exactly like in the proof of Corollary 3.6, that T~ is bounded on all LP($). One can of course go further, and prove the boundedness of T~ whenever $ (CBP)'~LG for some m (with obvious definitions). We shall come back to this in later sections. We conclude this section by a remark on the existence of principal values. Up to now, we tried to avoid talking about "the operator defined by the kernel K ' , because the integral f g ( x  y ) f ( y ) d ~ ( y ) does not converge in general. It turns out that, once we know that T~ is bounded, very little is needed to conclude that T~ converges to an operator T~ as e tends to 0. P r o p o s i t i o n 8.8. Suppose k is an integer, K is a good kernel and t~ E ~ is such that T~ is bounded on L2(dl~). Also suppose that the support S of# is a rectifiable set. Then, for all f C L 2 ( d , ) , T~,f(x) = lim T~f(x) exists for t~almost every x. ¢+0
t,.
There are various (nontrivially) equivalent definitions of a rectifiable set, and we'll choose the one that makes the proposition easy. We'll say that S is rectifiable if it is contained in the union of a set of measure 0 and a countable number of (rotated) Lipschitz graphs. For more information about rectifiable sets, see [Fe], [Mt] or [Fa]. E x e r c i s e . Show that if $ CBPLG, or even if S ( C B p ) m L G (see Remark 3.7), then $ is rectifiable. We stated an L2result here, but the corresponding LPresult also holds, even up to p = 1 included. The proof of Proposition 3.8 (or of its LPvariant) only requires fairly standard techniques, so we'll leave it as an exercise, with the following instructions. 1. First reduce to the ease when # is the restriction to S of the kdimensional Hausdorff measure (remember Remark 3.5). 2. Next, note that it is enough to prove that lim T,~f(x) a. e. for a dense class of ~*0
r
functions f. Indeed, given f E L2(d#), it is enough to show that
=1 lim ~:~Ct ,*0

T~" f(2:) I
has an L~norm which is equal to 0. This is shown by writing O f _ Og + 2T~.(f  g), and by choosing g in the dense class and close enough to f. This argument is a fairly standard way to use maximal estimates (see [St], p. 8 or 45, for instance). 3. Now suppose the result was established when S is a Lipschitz graph. Take for your dense class the linear combinations of functions supported on the intersection of $ with Lipschitz graphs (here we use the rectifiability of S). Since elira T~f(x) always exists when *0 rx ~ supp f , the general case will follow.
64 4. Prove the proposition when k  1 and S is a straight line. For this case, compactly supported s m o o t h functions are a nice enough dense class. 5. Show that, when k is odd, the operator of Example 6.7 of P a r t II is a principal value operator in the classical sense : for f E n 2, lim fl~ vl>~ g ( u , v ) f ( v ) d v exists for ~:* 0
a . e . u . To do this, first note t h a t each time the boundedness of an o p e r a t o r is established, the corresponding maximal operator is also bounded, because of Cotlar's inequality (or (III.10)). Therefore, one only has to follow the proof and apply the d o m i n a t e d convergence theorem as often as needed. 6. Conclude. You m a y have to use the fact t h a t the g r a p h of a Lipschitz function has tangent planes almost everywhere to go from the operator of Example II.6.7 to the operator of L e m m a 2.6.
4. R e g u l a r c u r v e s a n d L i p s c h i t z g r a p h s Let us pause in a m o m e n t to see how the " g o o d A m e t h o d " works when k = 1. Let us first consider the case of a connected, rectifiable curve F ; for simplicity, we shall restrict to the case when F has infinite length. A  Regular curves. D e f i n i t i o n 4.1. A (connected, rectifiable) curve F is regular if there is a constant C > 0 such that, for all x E ]R '~ and r > 0, the total length of F M B(x, r) is less t h a n Cr. The notion is due to Ahlfors [Ah]. If # is the arclength measure on F, we see t h a t F is regular if and only if # E A. Because F is connected,/~ also satisfies (8) automatically, and so # E ~ . Example.
Chordarc curves are regular ; a parabola in ]R 2 is regular (but not chordarc).
T h e o r e m 4.2. Let F C ~ n be a regular curve, and ~ the arclength measure on F. Then T~ (as defined by (4) and (5)) is bounded on LP(F,d/~) for 1 < p < +oo, when K is any "good kernel" (see Definition 1.1, with k  1). Note t h a t Proposition 1.4 says t h a t the condition is also necessary. To prove the theorem, we shall appeal to Corollary 3.6, and show t h a t F C B P L G . Let x E F and r > 0. Call z(s), 0 < s < so a parametrization by arclength of the piece of r between x and the first time r gets out of B(x, r). Thus, z(0) = x and [ Z(So)  x I= r. Since F is regular, we get so _~ C0r for some Co > 0. We shall find our big piece of Lipsehitz graph in z([0, so]). After a possible change of coordinates, we can assume t h a t x = 0 and z(so) = (r, 0 , . . . 0 ) . Let 6 = 2 CIo "
65 Lemma
4.3. There are a s e r E C [0, so] and a Lipschitz function h : [0, so] ~ ~t, such
that h(s) = zl(s) on E (zl(s) is the/irst coordinate of z(s)),
(21)
IE
I>
r
and (22)
g < h' <_ 1.
If we can prove this lemma, we will be finished for the following reason : s will be a Lipschitz function of h(s) on [0, h(s0)], and so all the coordinates of ~(s) = (h(s), z~(s),...zn(s)) will be Lipschitz functions of the first coordinate, so that ,~ will be the p a r a m e t r i z a t i o n of a Lipschitz graph. T h e n z(E) = ~.(E) will be the big piece we are looking for, because
I z(E) I=l ~.(E)]> if[E I> ~. To prove L e m m a 4.3, let us try the function h(s) =
sup [zl(t)  ~ft] + ~s. O
Certainly, h is Lipschitz and h'(s) > ~. We have to check t h a t if E  {s E [0, so] :
zl(s) = h(s)}, then I E I> ~. Note t h a t
E:
tO,:oj : o<,<.sup
As the reader might have guessed, we are a b o u t to apply the rising sun lemma. Let f(s) = zl(s)  ~s and F(s) = sup f(t) (see ~gure 1). O
F
..... /
SO
Figure 1
0
Figure 2
T h e set 12 = [0, so]kE is open, and so 12 = Uk Ik, where the Ik's are disjoint open intervals. Calling Ik =]ak, bk[, we have f(ak) > f(bk) (with equality, except p e r h a p s on the last
66 component). Then f ( s o )  f(O) = r  6so >_ r/2, and so ~ < f o ° f ' ( t ) d t = f E f ' ( t ) d t + ~k
f~: f ' ( t ) d t
<1 E ] + )  ~ ( f ( b k )


f ( a k ) ) <[ E [, which is (21).
This proves Theorem 4.2. B  The Cauchy integral on Lipschitz graphs again Proposition 3.2 can also be used to derive Coifman, McIntosh and Meyer's theorem from Calder6n's. The idea will be to show that MLipschitz graphs (i.e., graphs of MLipschitz functions) "contain big pieces of 9~0 Lipschitz graphs" (with a definition similar to Definition 3.4). L e m m a 4.4. L e t A : ]R ~ ~t be a L i p s c h i t z f u n c t i o n , and let M =11 A' IIoo • For each c o m p a c t interval I C ~ , there is a L i p s c h i t z f u n c t i o n ~t and a c o m p a c t E c I such t h a t A ( x )  A ( x ) for x G E,
(23)
I E I~1/I/10,
and
(24)
 M < .~'(x) < 4M 

a. e., or else
5
   4M < A l ( x ) < M 5

a. e.

To prove the lemma, one can easily reduce to the case when M = 1, and also I = [0,1]. First suppose that A(1) _> A(0). Pick A(x) = sup [A(t) + 4 t ]  gx4 (see Figure 2) ; then O
_ 45 < .4'(x) < 1, and the same argument as in Lemma 4.3 shows that E = {x E [0,1] : A(x) = A(x)} satisfies [ E [> 1 . If A(1) < A(0), we consider  A , and get a function vi such that  1 _< ~, _< g,4 and = A on a set of measure > 1 . Lemma 4.4 follows. Note that, in all cases, the graph F of the function A is the image by a rotation of the graph of some ~ 0 Lipschitz function. Thus, Lemma 4.4 really says that the graph of A "contains big pieces of ~ 0 Lipschitz graphs". Exactly as for Corollary 3.6, the boundedness of T~ on Lg(d#), where d# is the arclength measure on any MLipschitz graph, follows from the same result with 9_1_0MM_ Lipschitz graphs. Applying this enough times, we get the boundedness of T~, for any Lipschitz graph, as a consequence of the case when the Lipschitz constant is < e (where e is as small as we wish). In the special case when K ( z ) = z1 (with obvious complex notations) the boundedness of T~ on L 2 (d/z), where # is the arclength measure on the graph of the Lipschitz function A, is equivalent to the boundedness of the operator CA of Example II.6.3. Thus we deduced CoifmanMcIntoshMeyer's result from Calder6n's (CA is bounded if [[ A' []co is small enough).
67 E x e r c i s e . Check that if we know that II T ; I1~ ~ for 9i~0Lipschitz graph, the proof gives [[ T ; [[< C(A + M + 1) for MLipschitz graphs. Thus, we get the estimate [[ T~ I[< C(1 t M) N for some N. You can now complete the proof of Example 6.7 of Part II. Of course, we did not use Lemma 2.6 in part B of this section, and so the reader will agree that we did not introduce any vicious circle. R e m a r k . The argument above seems quite crude, and also quite complicated, because we used the machinery of Sections 2 and 3. However, in the special case of Lipschitz curves, the argument can be simplified a great deal. What is more surprising is that one can get very good estimates with an argument like that ! After partial results by Mural and Tchamitchian, Mural proved the best estimate for CA : 1] CA [1_
5. G a r n e t t ' s e x a m p l e The following example was introduced by Ivanov (in his thesis), and J. Garnett [Gal] to study (related) problems of analytic capacity. It is the Cantor set H = k x k C ]R2 ~ C, where k is the "middle half' Cantor set on the line. More precisely, k  Nn kn, where k0 = [0,1], kl = [0, ¼]U [3, 1], kn is composed of 2 n intervals of length 4  n , and kn+l is obtained from kn by removing, from each interval of kn, the open interval of length ½4 n situated in its center (see a picture of k 2 × k2 below). The constant ¼ is chosen so that H is a onedimensional set. There is a natural measure on H : it is the measure #, supported on H, such that, for each n, the measure of each of the 4 n squares of sidelength 4  n composing k,~ × kn is 4  n (/~ is proportional to the 1dimensional Hausdorff measure on H). Note that # E ~ (only locally, since (8) is not satisfied for large r's ; one could easily modify the example to get a # E ~ ) . The main interest for us of the set H is that, although tz E ~ , the Cauchy integral does not define a bounded operator on L 2 (d/~). One can prove this by showing that H has zero analytic capacity (see [Gal]) and then using duality. Let us very rapidly sketch the idea of a direct proof (the computations can be found in [Dv3]). Let Tr, be defined by
(25)
Tnf(z) " f~4_~,
(if z is in H, this consists in integrating on the squares of kn × k,~ that do not contain z). The idea is that the functions fn = Tnl have the same L2norm, and behave a little as if they were independent (because they vary at different scales and have integral zero). The proof then consists in estimating the scalar products < frn, fn >, and showing that
68
they vanish fast enough, so that one gets
(26)
II ~
no
Tnl 1122>~.
O
There is also a much less computational proof, by P. Jones, of the fact that the Cauchy kernel does not define a bounded operator on L2(d#), which has the advantage of being much more flexible (see [Jnl]). Unfortunately, I don't think it gives the estimate (26). R e m a r k 5.1. It is not too hard to transform (26) into an estimate for the Cauchy integral on Lipschitz graphs. Notice that, if we change coordinates (or rotate H), the set H (minus a countable set of corners) becomes the graph of a function (see the picture).
"10
on
30
O
3
O
3 [] 0
O
[] [] i
The set N x K2
L:o
"¢°
DO v
0
~0
1
Also, # is a constant times the pullback by the projection of the Lebesgue measure on the line. This means that there is a function A such that the Cauchy integral on the graph of A (the operator CA of Part II, 6.3) is not bounded. One can even approximate A by a sequence An of nLipschitz functions, and deduce from (26) that II CA., 11>_cn 1/2, therefore showing that Murai's estimate (inequality (45) of Part II) is sharp (again see [Dv3] for more details). R e m a r k 5.2. The fact that the Cauchy integral is not bounded on LZ(d/z) shows that the connectedness assumption of Definition 4.1 and Theorem 4.2 was not superfluous. If r n is the boundary of kn x kn, and #n is the arclength measure on r,~, we see that #n E uniformly in n ; however, since #n ~ / ~ , the T~.'s are not uniformly bounded ! One can go a little further, and build a counterexamaple in higher dimensions. For instance, let S~ be the 2dimensional surface of ]R3 defined by Sn = kn x k,t x {1} u
69 Fn x [0, 1] U
[lR2\(kn x
k,)] x {0} (see the picture).
A sketch of the surface S2 Note that the surface measure on S,~ is a measure/zn which is in ~ , uniformly in n, but it is not hard to find a kernel K such that the corresponding T~. 's are not uniformly bounded on L2(/zn). Thus, in higher dimensions, something stronger than conneetedness is required (see Sections 9 and 10 for further comments).
6. Three classes of surfaces We wish to describe, rather rapidly, three classes of Kdimensional objects in ~ n where something positive can be said about the operators T~'s. A  Chordarc
surfaces
with
small constant
This first example was introduced by S. Semmes ([Se3] and [Se4]). For this example, k = n  1, and also one is interested in more refined properties than the boundedness of the T~'s. We shall try to say, as fast as possible, what the point is (and leave all the definitions, precise statements, and proofs). Let $ be a hypersurface in IRk+l. We shall make the a priori assumption that S is smooth (including at c¢), oriented, and separates IR k+l into two connected components (call them 12+ and f l  ) . Let d/z denote the surface measure on S. One can define on S the analogue of the Cauchy operator. This time, the kernel takes values in the Clifford algebra (instead of C when k = 1). Let us call ¢s the resulting operator (see Semmes' papers for definitions).
70 One of the nice things about the Clifford algebra setting is that there is a notion (we'll call it "Cliffordanalyticity') that generalizes analyticity. One can then define Hardy spaces H2(12 +) and (H2(fl  ) of traces on S of Cliffordanalytic functions in fl + and f l that have boundary values in L2(S,d#), and one can show that L2($,d~) is the direct sum of H2(fl +) and H2(fl  ) if and only if Cs is bounded on L2(S,d#). The kernel of Cs is K ( x  y ) , where K is Clifford valued, C °°, odd, and homogeneous of degree  k , and so Cs is one of the operators of the previous sections. S. Semmes asks a more precise question : when is L2($, d~) the "almostorthogonal" sum of H2(n+) and H2(f] +) ? By "almost orthogonal', we mean that
(27)
[< f , g >[___e IIf IIIIg II
for all f G H2(fl +) and g E H 2 ( n  ) , and where e is small enough. In terms of operators, this means that a slightly different version Cs of the CliffordCauchy operator is "almost antiselfadjoint" (i.e. I[ Cs + C; [[< e' for an e' ~ e). T h e o r e m 6.1. ([Se3], [Se4] and [Se7]). Let 5 be as above. If (27) is true with a small enough e, then there is an Q ~ Ce such that (28) the unit normal n(y) to $, pointing towards fl +, is in B M O ( S, d#), with norm ~_ el. Conversely, if (28) is true for a small enough el, then there is an E ~_ Cel such that (27) is true. The definition of B M O ( $ ) is not a surprise : we ask for (29)
f
.(BCx, r))' !
JB (~,r)
where nx,r :
IN(y) 
fBc ,r nCy)d Cy), • e S and
I d.Cy) _< El, > 0
R e m a r k 6.2. When k  1, the result says that the Cauchy operator on a rectifiable Jordan curve F is almost antiself adjoint (or the Hardy spaces are almost orthogonal) if and only if F is a "chordarc curve with small constant" (with the definition of II, 8.4, we should say "constant close to 1"). This was essentially proved by Coifman and Meyer [CM2] (also see [Dv01 for the "only if" part). For this reason, S. Semmes decided to call the surfaces that satisfy (27) (or (28)) "chordarc surfaces with small constant" (we'll write CASSC's).
R e m a r k 6.3. There are many other equivalent geometric definitions of CASSC's. For instance, (27) implies (30) for all x C S and r > 0, there is a hyperplane H such that each y ~ $ N B(x, r) is at distance < e2r from H ;
71 (31) for all x E $ and r > 0, I P(S M B ( x , r ) )  akr k I_< ~3r k, where akr k is the volume of a ball of radius r in IRk . (32) if D(x, y) denotes the geodesic distance between x and y E S, then I x  y (1 q  E 4 ) I ~  y l ;
I<  D(x, y) <
(33) for all x E $ and r > 0, there is a surface S, which is the image of an esLipschitz graph by an isometry of IRk+l, and such t h a t # ( S N B ( x , r ) N go) < esrk. All these properties are various ways of saying that S looks very much like a hyperplane at all scales. N6te t h a t (33) makes it easy to show t h a t Cs is b o u n d e d (once (33) is deduced from (28)!). Of course, they are all true for chordarc curves. One thing we are missing here, however is a very good p a r a m e t r i z a t i o n (we'll talk a b o u t this in Section 10). Finally, (31) and (32) (with e3 b e4 small enough) imply (27), too. B  wregular surfaces Let us mention now a generalization of T h e o r e m 4.2 above. T h e main difficulty of studying surfaces is avoided by assuming we have a very nice p a r a m e t r i z a t i o n of S. D e f i n i t i o n 6.4. Let w C Aoo(IR k) be a weight of the M u c k e n h o u p t class. T h e function z : IRk __~ IRn will be called "wregular" if there is a constant C _> 0 such t h a t
I z'Cx)I~ cw(x) ~/~
(34) and
(35)
I {~ e IRk : z(~) ~ B(w, r)} I~d~< c r k
for all w E IRn and r > 0. C o m m e n t s . T h e condition (34) is m e a n t "distributionwise'. Since w is a weight in A~o, it is equivalent to
(36)
I zCx)  z(y) I< c
wC~)d~
,
where B = B(x, [ x  y [), for instance. The reader will find all needed knowledge on Aooweights in [Jg] or [GR]. He can also content himself with the case when w = 1 (in this case, z is Lipsehitz). If we define a measure # on IRn by p ( f ) = f ~ f ( z ( x ) ) w ( x ) d x , condition (35) means t h a t tz e A. Furthermore, (34) (or (36)) implies t h a t p E ~ . T h e o r e m 6.5 [Dv5]. Let z : IR k ~ IR" be wregular, and define # by fR~ f d # = fRk f o Z W dx. Then the operators T~ defined in Section 1 are bounded on nP(d#) for l
72 The initial proof used Proposition 3.2, but not Corollary 3.6 directly, because the author was not able to find directly ~big pieces of Lipschitz graphs" in $ = z(IRk). Instead, one showed that $ contains big pieces of something that contains big pieces of .... something that contains big pieces of Lipschitz graphs, and applied a few times an analogue of Corollary 3.6. We'll see a more direct proof in Sections 7 and 8. R e m a r k . We did not introduce weights in Definition 6.4 merely by thirst of generalizations. First, a converse which is stated in Section 9 (Theorem 9.5) apparently needs weights w other than 1. Similarly, for some sets $, we know how to find an wregular parametrization of $, but we do not know if they admit 1regular parametrizations (see Sections 9 and 10). Also, the wregular mappings have a few interesting properties of their own, in particular in relation with quasiconformal mappings. One easily checks that if z is wregular, then its differential Dz exists almost everywhere, and is quasisymmetric : for each x such that Dz(x) exists, there is a $ > 0 such that $ I v I_~] D z ( x ) . v I< C$ I v I for all v e ]Rk (C does not depend on x). Also, if z is wregular and h is a quasiconformal mapping of lR k, then zoh is ¢5regular, where ~(x) = w(h(x)).Jacobian(h)(x). (Cz is an Aooweight because the Jacobian of h is in Aoo, by a result of F. Gehring [Ge].) We do not know at this time the precise class of weights w such that there is an wregular mapping z. It is proved in [DS2] that w must belong to the socalled "strongAoo class", and Semmes recently proved that some condition between A1 and strongAoo is sufficient [Sell]. See Section 10 for other questions related to this one. C  Stephen Semmes surfaces D e f i n i t i o n 6.6. Let us denote by S(k) the class of hypersurfaces $ C lR k+l such that there is a # C ~ with supp # = $, and a constant C > 0 such that
(37) for all x E $ and r :> 0, there exist points Xl and x2, that lie in different connected components of l R k + l \ $ , and such that I x i  x [~_ r, but dist(xi, $) > C  l r for i  1,2. Examples. * In dimension k = 1, Lipschitz graphs, chordarc curves are in S(1). The example of Figure 1 is in S(1), but not the example of Figure 2. • When k > 2, it is no longer true that all $ E S(k) are of the form z(IR k) for an wregular z. For instance, the topology of S can be quite complicated. A simple example
73 with a few handles is suggested in Figure 3.
iiiii Figure 1
Figure 2
Figure 3
• Also, note that Definition 6.6 does not put any restriction on the number of components of S or its complement. • A compactness argument shows that the image of iR k by a bilipschitz mapping from lRk to lR n is in S(k) ([V~I] p. 111). T h e o r e m 6.7 [Seb]. If S E S(k) and K is a good kernel, the operator T~ of Section i is bounded on LP($,dp) for 1 < p < +co. Actually, Semmes' proof gives a slightly less general result. To describe that proof, we shall make the a priori assumption that $ be a smooth, compact submanifold of ]R k+l, and also the topological assumptions that $ be connected, orientable, and that its complement in IRk+l have exactly two connected components. It is easy to get rid of the smoothness assumptions by a limiting argument, but the topological assumptions cannot be dealt with so easily. Also, Semmes'proof does not work for all kernels, but only for a (large enough) class of realanalytic kernels. Theorem 6.7 is still true as we stated it, though, because it can be shown that 5 C B P L G as soon as 5 E S(k) (see Section 7). Let us now describe Semmes'proof. The idea is quite amusing : apply the Tbtheorem just like when k = 1. To do so, one first has to find a substitute for the Cauchy formula. Let us view ]R k+l as a subset of ¢k(]R), the Clifford algebra with k generators over ]R (for a precise definition of Ck(~) and its basic properties used below, see [BDS]). Call e0 the unit, and el," " e k the generators (recall that ei2 =  e 0 and ele I : eyei for i , j y~ 0 and k i ~ j). Every point of ]Rk is identified with the element xoeo + ~ = 1 xiei of Ck(lR), and the Cliffordvalued analogue of the Cauchy kernel is defined by =ll
x fl  k  1
,0e0 
.
Our assumptions on S allow us to integrate by parts, and in particula r to use the following consequence of the "CauchyClifford formula" : 0 if X E l~(38) [ K ( x  y)n(y)dy = [ n ( y ) K ( x  y)dy = a if x C fl +, Js Js
74 where n(y) is the unit normal to S at y, dy is the surface measure on S, 12 (respectively 12+) is the unbounded (resp. bounded) component of the complement of S, and a ~= 0 is a constant. The products between vectors are taken in the Clifford algebra. Next, we want to check that the T b  t h e o r e m can be applied on $. Because $ C S(k), the surface measure on $ is equivalent to the measure # of Definition 6.6 and thus is in the class ~ . To be precise, this is not quite true because we decided t h a t $ should be compact, and so we can only suppose that (3), (8) and (37) are true for • _< diam S. However, this does not change any of the proofs significantly. Thus $, with the surface measure and the Euclidian distance, is a space of homogeneous type, and the construction given in Appendix 1 even provides us with dyadic cubes on $. As a consequence, the Tbtheorem is valid on S, and we can even use the simpler proof given in P a r t II, Sections 2 and 3 (see the remarks of Section 5). The next difficulty is that we want to apply the T b  t h e o r e m with a function b (the unit normal to S) which is no longer complex, but takes Clifordvector values. Since K is also vectorvalued, let us precise some of the notations. Denote by Cs the principal value operator defined by the antisymmetric kernel K ( x  y). The definition of Cs is not a problem : we can use a formula like (4) or the smoothness of S. The o p e r a t o r Cs maps realvalued functions on vectorvalued functions, but we can also let it act (on the left or on the right) on vectorvalued functions (such as the unit normal to S) : we just take the p r o d u c t s of vectors in the Clifford algebra. T h e result is then a function with values in Ck (R). With these conventions, it follows from (38), and a small limiting argument made easier by our smoothness assumption, t h a t Csn = 0 in BMO, whether Cs acts on the left or on the right. Let us now check that the unit normal is paraaccretive. T h a t is, let us find constants C > 0 and ~ > 0 such that, for each x E S and 0 < • < diam S, there is a "dyadic cube" Q c S such t h a t dist(x, Q) _< Cr, ~rl < diam Q <<_Cr, and such t h a t
V~ fQ n(y)dy
> ~f.
Of course, we do not want the constants C and ~ to depend on the constants in the smoothness assumption, or on the diameter of S. This is similar to the definition of paraaccretivity given in page 35, but we now use a different set of cubes, and modules of complex numbers have been replaced by lengths of vectors. To prove t h a t n is paraaccretive, pick a point x C S and a radius • > 0, and apply (38) to the points xl and x2 of Definition 6.6. Taking the difference yields
fs [K( I  y) 
 y)] n(y)dy = +a.
Note t h a t the kernel p(y) ~ K(xl  y )  K ( x 2  y) satisfies the same estimates as a Poisson k e r n e l : ]l P(Y) II< Cr(r+ I x  y I)  k  1 and II V p ( y ) I 1 ~ Cr2( r+ I x  y I)  k  2 . Fix a large constant ), :> 0, and cover S by disjoint cubes Qi of diameters comparable to ) ,  l r , and then replace p(y) by ~5(y), where ~(y) is obtained as follows. Let Qi(y) be the cube Qi t h a t contains y, and let ~5(y) be the mean value of p on Qi(y) if dist(x, Qi(y)) <_ )~r, and 0 otherwise. If )~ is large enough, fi is so close to p t h a t one still has Ills P(Y)n(y)dyll > [~l" T h e paraaccretivity of n then follows by replacing i~ by one of its coordinates, and then choosing one of the Qi's on which the integral is largest.
75 Our last task is to modify the proof of Part II, Sections 2 and 3, to make it work in a Cliffordvalued context. We still want to use the same computations, but we shall have to be careful because Clifford vectors do not commute. The delicate part of the proof is the construction of the Riesz basis : the estimates for CQQ, will essentially be the same. As in the general paraaccretive case, we first modify our set of cubes in such a way that
fq b(y)dy
> ~ for each dyadic cube Q. Note that a nonzero vector of IRk+l is always
invertible in Ck (JR) : its inverse is its ~conjugate', divided by the square of its norm. Thus, b(y)dy is invertible, and its inverse is a Clifford vector with length ~ C6 x [ Q I1 . shall replace the projection operators of Part II by two operators Fk and F~ defined
~,~e by
Fkf(x) =
[fQf(t)b(t)dt] [fQb(t)dt]I
and
F~f(x) =
[fQb(t)dt]I
[fQ b(t)f(t)dt],
where Q is, as before, the dyadic cube of size 2  k that contains x. If we set Ak = Fk+l  Fk and A~ = F~+ 1  F~, the same computations as before yield (13) and (14) for Fk and Ak, and F~ and A~. The identity (15) has to be modified a little: one writes f Aku(x)b(x)A*~v(x)dx = 0 instead. L e m m a 2.1 also stays true for both Ak and A~ : the first inequality is proved as before and, for the second inequality, one modifies the duality argument and writes
k
g
The rest of the estimate is as before. Note that the argument also works when f has values in IRk+l. The conclusion of these computations is that every realvalued function can be decomposed as f = ~ k A k f , with 1t f 112~ ~ II A k f II2 • We can decompose f even further wy writing Alcf = ~ Q a q f Q , where the sum is over all dyadic cubes of size 2  k , aQ =1[ l l Q A k f ][2, and fQ = a ~ l l Q A k f . This yields the decomposition f = ~ aqfQ, where this time Q runs over all dyadic cubes, the coefficients aQ are such that [[ f [[2~ [ aQ ]2, and the functions fQ have the following properties. First, fQ is supported on Q, and is constant on each of the cubes of the next generation. As a consequence, 1[ fQ 11oo< C [ Q I'/2[[ fQ 112< C [ Q [1/2 . Finally, f q satisfies the cancellation condition f fq(x)b(x)dx = 0 because fo. Akf(x)b(x)dx = 0 for every cube Q of size 2  k . Note the we could have used the A~ and obtained a similar decomposition, but with the cancellation condition f b(x)fc2(x)dx = O. At this stage, it would probably be possible to go on as in Part ! I, and express the functions fQ in terms of some Riesz basis. The author of these notes is far too afraid of things that do not commute to proceed in this direction, and so we shall keep our functions fQ. This is not too bad, because they satisfy exactly the same estimates as the h ~ ' s of Part II, Section 2. Let us come back to our operator ¢s, and let us show that the operator T = MbCsMb is bounded on L z. The boundedness of ¢s will of course follow, since b is invertible. Because we want to consider testfunctions which are vectorvalued, we shall use the notation < fT, g > to denote the effect on the effect on g (by an action on the left) of the image of f by T (acting on the right). In other words, if K were integrable, we would have < fT, g > = f f f(y)b(y)g(x  y)b(x)g(x)dx dy.
76 We now use the Ak's to write f = ~
aQfq,
and the A~¢'s to write g = ~
13Qgq, so
that < fT, 9
>= E Z c~Q/~Q,< fQT, gQ, Q
>.
Q'
The coefficients CqQ, = < fQT, gQ, > can be estimated like in P a r t II, Section 3, using the s t a n d a r d estimates on K, the properties of dyadic cubes, and the facts t h a t T1 = T t l = 0 and f fQb = f bgQ = 0. The result is t h a t the matrix with coefficients H CQQ, II defines a b o u n d e d o p e r a t o r on £2, and t h a t
I1< f T , g >][< (7
I~Q I~
I / ~ 12
_< c II f I111g II.
This proves the boundedness of Cs. As we said earlier, the proof of S. Semmes also works for other kernels t h a n the CauchyClifford kernel K. The idea is to use the T b  t h e o r e m again, and to c o m p u t e the image of the unit normal by an integration by parts. If the result is the image of a b o u n d e d function by an operator we already know how to treat, we are in business. The details can be found in [Se5]. E x e r c i s e . Complete the proof of the T b  t h e o r e m in the case of a function b with values in IR k+l C Ck(lR). Use a p a r a p r o d u c t with a kernel of the form
P(x, y) = E c~QfQ(x)Oq(y) Q (where the f Q ' s are obtained by decomposing a function of B M O with the A~'s), or of the form
P(x, y) = E aQOQ(x)fq(y) q (where the
fQ's
are obtained with the Ak'S).
7. F i n d i n g b i g p i e c e s : c l o u d s a n d s h a d o w s Let us come back to the good~ method. We are given a "surface" $ C ]R n, and a measure # C Y]~ s u p p o r t e d on $, and we wish to apply Corollary 3.6, so we wish to show that $ C B P L G . A first step in t h a t direction is to show t h a t $ "has big projections" in the following sense : there is a 0 > 0 such t h a t for each x E $ and r > 0, there is a kdimensional vector space V such that, if ~r is the orthogonal projection on V, then (39)
I r(S f1B(x,r))]>
Ork
77 (where I ] denotes the Lebesgue measure on V). It is clear that, if S CBPLG, then S "has big projections" : if E C $ n B ( x , r ) is contained in the graph of an MLipschitz function from V to V ±, then ] ~r(E) ]> CI(M + 1)lg(E). In most known cases, showing that S has big projections is the easy part. For instance, if S is in S(k) (see Definition 6.6), we can take for V the hyperplane orthogonal to the line L that contains the points xl and x2 of Definition 6.6. Indeed, if L' is any line that meets B ( x l , C  l r ) and B ( x 2 , C  l r ) , then n' N $ N B(x,2r) is not empty. From this, we deduce that r ( S N B(x,2r)) contains a ball of radius C  l r centered at ~r(xl), which gives the "big projection" for 2r (see the picture).
i
Let us assume now that we proved that S "has big projections". We still have to find a piece of Lipschitz graph in S A B(x, r) or, equivalently, a subset E c S A B(x, r) such that ~rl~ is bilipschitz (the "Clouds and shadows problem"). A condition is given in [Dv6], that allows in certain cases to solve the "Clouds and shadows problem". The condition is not very pleasant, and the proof by a stopping time argument is not too simple, so we'll just mention a corollary : T h e o r e m 7.1. If $ = z(iR k) for some wregular function z : IRk + IR'~, or if $ E S(k), then $ CBPLG (see Definitions 6.4, 6.6 and 3.4 for the jargon). This theorem gives a new proof of Theorems 6.5 and 6.7. One can also define higher codimension analogues of S(k) for which Theorem 7.1 also holds (see [Dv6]). We shall see in Section 8 a faster proof when $ = z(iR k) for an wregular z. In the case of S(k), there is another proof, too ([DJe]), which also gives a little more : T h e o r e m 7.2. Let $ ~ S(k). Assume that 0 E S, and that the bail B with center ( 0 , . .  , 0 , 1 ) and radius C 1 is entirely contained in one of the connected components of IRk+I\s (call this component 0). Then there is a constant C1 > O, that depends only on C and the constants of Definition 6.6, and a C1Lipschitz function A, defined on B(0, (2C)1) C IRk, and such that the graph F of A satisfies :
(40)
r c 0
78
and (41)
u(r
n
s) > c i 1
(see the picture).
oC The proof of T h e o r e m 7.2 is reasonably simple, but we shall not have enough time to give it. T h e new piece of information is t h a t F is "above" $, and this allows one to obtain the following harmonic measure estimates. C o r o l l a r y 7.3. Let $ E S(k), and 0 be one of the components of IRk+lk$. If 0 is a NonTangential Access domain, then the harmonic measure on O0 (relative to O) is in the Muckenhoupt class Aoo with respect to surface measure on $. We refer to [DJe] or [JK] for the definition of NTA domains. Let us mention t h a t S. Semmes has another proof of Corollary 7.3, obtained independently by a m e t h o d related to the "corona construction" [Seg]. This m e t h o d had been used by P. Jones and S. Semmes to treat the case of the image of a bilipschitz m a p p i n g from IR k into IRk+l. Semmes obtained the special case of Corollary 7.3 when b o t h sides of S are N T A at the same time as [DJe], and the general case a little later. The condition $ E S(k) is not necessary for the proofs of T h e o r e m 7.2 and Corollary 7.3 ; in particular, the conclusion of the corollary is still true if the kdimensional Hausdorff measure on S is in ~ , if for each x E S and r > 0, there are two kdimensional disks D1 and D2, with radius ~,r at distance _< r from x, and t h a t are contained in different components of I R k + l \ $ (of course, the NTA condition on Harnack chains is still needed, too). The idea of the proof is to compare harmonic measure on 0 with harmonic measure on a Lipschitz domain fl C 0 such t h a t I 0fl N O0 I is relatively large. One uses the t h e o r e m to find fl, and techniques of [JK] to conclude. More details can be found in [DJe].
79 8. B i l i p s c h i t z m a p p i n g s inside L i p s c h i t z f u n c t i o n s The following theorem of P. Jones will allow us to solve the "clouds and shadovs problem" for regular mappings. T h e o r e m 8.1 [Jn2]. Let I be the unit cube ofIR k and f : I ~ IRk be a Lipschitz function. For each e > O, there is an integer M = M(I I V f IIoo,e), a constant C(I I V f IIoo,e) and sets E l , " "EM C I such that
(42)
I f(x)
 f ( y ) I> C(ll vf
Iloo, ~)' I •  v I
whenever z and y are in the same Ei, and (43)
I f(I
n E~ n . . . n E ~ ) I < ,.
We shall give a proof of this theorem, but let us see how it implies that the image of a 1regular mapping always contains big pieces of Lipschitz graphs. C o r o l l a r y 8.2. If f : I ~ i r k is Lipschitz, and i f I f(Z) f > ~, then there is a compact set E C I such that [ E [> 0(11 v f Iloo,~) > 0 and I f(~)  fCy) I > C(ll v/11~,6)
~ I~  v I for x,y E E.
The corollary clearly follows from the Theorem (choose e = ~). A first proof of the Corollary was given in [Dv6], but it is a little less engaging. Let us now use the corollary to prove that if z : IR k * IR'~ is 1regular (i.e., wregular with the weight w = 1), then $ = z(IR k) CBPLG. The general result (when w is any Aooweight) would require slightly different versions of Theorem 8.1 and Corollary 8.2, which are both true (and can be deduced easily from these results), but we decided to simplify the statements and take w = 1. Let w E $ and r > 0 be given. Choose a point x0 E IRk such that z(zo) = w. If C is large enough, the cube Q centered at z0 and with sidelength ~ is such that z(Q) c B ( w , r). L e m m a 8.3. There is a kdimensional subspaee V of iR n such that, if ~r is the orthogonal projection onto V, [ ~(z(Q)) 1> ~ I Q I. Of course, we do not want ~ to depend on Q. If ~ were allowed to depend on Q, the lemma would be completely trivial : one can find a point x in the interior of Q, such that z is differentiable at x ; then the derivative of z at z is of rank k (because z is 1regular), and if V is chosen well, ~r(z(Q)) will contain a small ball around ~r(z(x)) (by elementary degree theory). The fact that we can choose ~ independent of Q and z (provided the regularity constants of z stay bounded) now follows from a simple compactness argument, which is left as an exercise (take a sequence (z,t, Qn) for which the best ~n tends to O, extract a
80 convergent subsequence after normalization, and use the argument above and some more elementary degree theory). More details can be found in [Dv6], lemme 10, p. 96. Apply Corollary 8.2 to the function f ( x ) = 7r(z(x)) (possibly after an affine change of variables if Q is not the unit cube). We get a s e t E c Q, with [ E I> 0 ] Q [, and such that r o z is bilipschitz on E. The set /~ = z(E) satisfies # ( 2 ) _> Ork/C, and rl~ is also bilipschitz, so/~ is the piece of Lipschitz graph we were looking for.
Let us now prove Theorem 8.1. We shall need a way to control the situations where ] f(x)f(y) I< 6 ] x  y [ (6 > 0 i s a small constant, to be chosen later). The general idea is that, in such a case, either ] f ( B ) ] is very small for some ball B containing x and y, or else f is not close to being affine on B, and so f " must be large somewhere in B. We intend to prove that the second possibility cannot occur very often by associating to B a large "wavelet coefficient" for f~. To simplify notations, let us assume that f is 1Lipschitz, and also is defined on the whole IR k. We need a few notations before we start the stopping time argument. Let us fix a constant C > 1 ; we'll say that two dyadic cubes Q1 and Q2 c I are "semiadjacent" if[ Q1 [=[ Q2 ], and if ] Q1 ]l/k<_ dist(Q1,Q2) < C
I Q~ I~/k.
Also, if Q is a dyadic cube, the father of Q (i.e., the only dyadic cube of sidelength 2 t Q t 1/~ which contains Q) will be denoted by Q*. Finally, let us choose a wavelet basis (like in Part I), and let us say for convenience that we use compactly supported wavelets of class C 4 (many other things would work just as well). Let us write V f = ~~q ~~ea ~ ¢ ~ the wavelet expansion of V f in the basis (¢~) (note that the a~'s are vectorvalued). The following lemma will be used to control the number of places where f is clearly far from being affine. L e m m a 8.4. If the constant Co >_ 0 is large enough, then the following result is true. Let QI and Q2 be semiadjacent dyadic cubes, and suppose that, for some a > O, [ f(Q1) [> _ a [ Q1 [ • Also suppose that I f ( x l )  f(x2) [_ ~ Color I Xl x2 I for some xleQ1 and x2 E Q2. Then, there is a dyadic cube Q c C ( a ) Q , , with [ Q [>_ C(a) 1 [ Q1 [, and such that [ a~Q [>_ C(a) 1 [ Q1 [1/2 for some e. 

81
This lemma can be proved directly, but it is just as simple for us to use a compactness argument. If Lemma 8.4 is false, there is a sequence fn of 1Lipschitz functions, such that the hypotheses are satisfied for the unit cube Q1 and some semiadjacent Q~ (and the same a > 0), but such that [ a ~ [< ~1 for all Q of size > 2  n contained in 2~Q1. Extract a subsequence fn(i) that converges uniformly to a function foo, and such that the Vfn(i) also converge (locally) in all L p, p < boo. The wavelets coefficients of Vfoo are all 0, and so foo is affine. Since foo is 1Lipschitz, and there are points xl E Q1 and x2 E Q2 such that ] f~o(xl)  foo(x2) [_< 2C~1a ]Xl  x2 [, one gets [ foo(Q1) [~ CCola < ~ [ Q, [ (if Co is chosen correctly). On the other hand, we know that ] f~(Q,) [>_ a [ Q1 [ for all n, and this implies that [ foo(Q,) ]_> a [ Q, [ ; the ensuing contradiction proves the lemma. E x e r c i s e . Prove what was just said. Hint : if I foo(Q1) I is small, then f~(QI) is contained in an open set with small measure, and the same must be true of f,~(Q1) for n large enough. R e m a r k . We decided to use wavelets, but there are many other options : about any result giving some control on how the function f is approximated by affine functions at all scales would probably do as well. Next, let us introduce two categories of cubes where something unpleasant happens. Call 81 the set of dyadic cubes Q1 E I such that there is a semiadjacent cube Q2, for which I f(Q1) I_ > e/2 I Q1 I, but such that there are points xl E Q1 and x2 E Q2 with [ f ( x l )  f(x2) I_( C o l e [.T1   X2 [ /2 (the number c is the same as in the theorem). Call 82 the set of dyadic cubes Q c I such that [ f(Q) [< ~ [ Q [. We can throw away all the cubes of 82, because [ f ( U Q) I~ .e We also intend to 2 E~ throw away the points of I that belong to too many cubes of ~'1. We know from Lemma 8.4 that if Q1 E £1, there is a wavelet coefficient aS, where Q is not too far from Q1 for the hyperbolic distance, which is >[ Q1 [1/2/C. Therefore, [ Q1 ]_ C ~ QI EC1
[ a S ]2( C (since ][ V f [[oo< 1). If we choose N large enough, we'll
QCCI
get [ B [< ~ , w h e r e B  {x e I 1Lipschitz, I f ( B ) I < ~.
: x b e l o n g s to more than N cubes of ~1}. Since f is
Call G the complement in I of B U (~~q~82 Q ) " We just have to split G into M sets
El,'" "EM on which f will be bilipschitz. We want to associate, to each cube Q ~ ~q'e~'~ Q~ a certain sequence a(Q) of O's and l's ; the length of the sequence will be called £(Q). We want to define a(Q) by induction, starting from I, and then defining a(Q), for each Q, in terms of a(Q*) (Q* is the father of Q). First, let us introduce more notations. For each n _> 0, call J[n the set of all dyadic cubes Q E ~¢1 of sidelength 2  n that are not contained in any cube of ~2. Call Pn the set of all (unordered) paris (Q1, Q2), where QI and Q2 are semiadjacent cubes of .~, ; choose any order on Pn, and call P,~,I, P,,,2, "" "Pn,e""
82 the elements of P,~. We are now ready to define a(Q) for Q of sidelength 2'*, assuming this was done for cubes of sidelength 2  ( n  l ) (we take for a(I) the e m p t y sequence). C a s e 1. If Q ~ ~'1, and is not contained in any cube of ~2, we simply set a(Q) = Otherwise, we will define (or change) a(Q) as follows. We take the set P , in order, and define (or modify) a(Q1) and a(Q2) (where pn,t  (Q1, Q2)) successively for £  1,.. Suppose we already went through Pn,1,'"P,,,t1, and let Pn,~  (Qx,Q2), where Q1 and Q2 are semiadjacent cubes of ~1. The cubes Q~ and Q~ are not contained in any cube of ~2, and so a(Q~) and a(Q~) are well defined. C a s e 2. First suppose that £(Q~) = £(Q~). A  If ~(Q~) ¢ a(Q~), we leave a(Q,) = c~(Q*~) and a(Q2) = a(Q[). B  If a(Q~) = a(Q~) = ( e l , " "e~Cq;)), we set a(Qx) = (cx,"ElCq;),0) and a(Q2) =
(E~, .E~(Q;), 1). C a s e 3. Now suppose that £(Q~) ¢ l(Q~). If l(Q~) > l(Q~), a(Q~) = and (Q2) ' * •• * = Q(Q~)), we keep a(Q1) = a(Q1), and set a(Q2) = (el,. • . e ~' ( Q ~ ) , e'),where e' ~ et(Q~)+l. If l(Q1) * < t(Q2) * , we exchange Q1 and Q2, and do as above. If the same cube Q appears in more than one pair Pn,t, we do not exactly do as above, but replace a(Q*) by the last defined value of a(Q) (so, most of the time, a(Q) will change a few times during the definition process). Note that, if Q c Q', then the sequence a(Q) starts with the sequence a(Q'). Also, if x is in the good set G, and I D Q(D . .. D Q(n) D • • • is the sequence of dyadic cubes containing x, then the sequence or(Q('*)) is well defined because x ~ [Jc2 Q, and changes less than CN times during the whole definition process (it only changes for those n's such that Q(n) E ~1, and even for those n's, it changes less than C times because there is _< C pairs of semiadjacent cubes containing Q(n)). Therefore, for each x E G, the limit of the a(Q('~))'s exists, and has length _ CN. Call a(x) this limit. Let us call E the set of all sequences of length < CN, composed of O's and l's. For each i C E, let Fi = {x C G : a(x) = i}, and let us check that, for each i, (44) Let yE was not
If(x)f(Y)
I>Co 1EIxyl 2
for x, y e F i .
x,y E Fi for some i, and call Q1, Q2 two semiadjacent cubes such that x E Qx and
Q2 [such cubes exist, provided the constant C in the definition of semiadjacent cubes large enough]. Since a(x) = or(y), the sequence a(Q1) is the beginning of a(Q2) if it is longer, and otherwise a(Q2) is the beginning of a(Q1). We made sure, when defining a(Q1) and a(Q2), that this would never happen if Q1 and Q2 axe in £1. So Q1 (or Q2) is
83 not in ~'1, and this implies (44) by definition of ~'1. This concludes the proof of Theorem 8,1; because the number of sets Fi is less than CN2 CN. E x e r c i s e . State and prove an analogue of Theorem 8.1, where Lipschitz functions are replaced by functions satisfying (34) (or (36)) for some weight weAoo. Deduce a proof of the fact that z(]R k) C B P L G when z is wregular. R e m a r k . The proof above has the advantage of using very little of the structure of ]Rk. This is used in [DS4] to extend Theorem 8.1 to some functions defined on a regular set, and to give one more proof of the fact that every $ C S(d) contains big pieces of Lipschitz graphs.
9. S q u a r e f u n c t i o n s , g e o m e t r i c l e m m a a n d t h e c o r o n a c o n s t r u c t i o n In this section, we wish to take advantage of a small delay in the preparation of the manuscript to present a few more results (some of them are actually posterior to the lectures). We shall not have time to see any proof, but we'll try to introduce a few notions that are relevant to our study of singular integrals on subsets of ]R n. A  Square function estimates Let # E ~ , and $ C ]R n be the support of #. An easy argument using Rademacher functions shows that, if all the good kernels K of Definition 1.1 give operators T~ that are bounded on L2(d#), then the following square function estimates hold. Let (~ be the class of all odd functions ¢ E ¢~(]R n) ; then for each ¢ E (~, there is a constant C = C¢ such that
P
< C Js If(Y) i2 dlz(y)
for all f E L2(d#).
If, instead of asking for oddness, we had defined the good kernels by the condition (2), we would have the square function estimate (45) with the slightly larger class & = {¢ E C ~ ( ] R n) : f ~ ¢(t0) I t Id1 at = 0 for all 0 e S n  l } instead. See [DS3] for more details. We shall see later that (45) implies the boundedness of all T~'s in return. The interest of (45) is that it is slightly easier to use, because it can be stated in terms of Carleson measures, like many of the properties that will be mentioned in this section. Also, it is closer to a concept dear to S. Semmes : "LittlewoodPaley theory on the set S ' . Before we switch to a different notion, let us mention that Semmes proved slightly different square function estimates, for surfaces S E S(k) (see Definition 6.6).
Theorem 9.1 [Se5]. Let $ C S(k), K ( x ) = x* tl x. I1 k  1 be the CauchyoClifford kernel, ~ d C/(~) = fs K ( x  y)n(y)/(y)dy be the CauchyCli~rord integral o f f Ot is denned ~or
84
/ E L2($) and x f[ $). Then
(46)
fm~+, I vCCf)(*) 12distCx, S)ds
< C
fs Ill
The proof of (46) uses the same ideas as the proof of the boundedness of the CauchyClifford operator on $. A simpler proof can be found in [Se6], and variants and applications can be found in [SeS]. B  P. J o n e s ' g e o m e t r i c l e m m a For x E S c IR'~ and r > 0, define/3(x, r) by (47)
~(x, r) = inf
sup
r  l d i s t ( y , P),
P yGSABCx,r )
where the infimum is taken over all kdimensional affine subspaces P of ]R n. The number $(x, r) measures how well S can be approximated, near x and at the scale r, by kplanes. If S is a Lipschitz graph, for instance, a very brutal estimate only gives fl(x,r) < C < 1, but the point is that, for most couples (x,r), ~ ( x , r ) is much smaller than that. The $ ( x , r ) ' s were introduced in [Jnl], to obtain a quite interesting new proof of the boundedness of the Cauchy integral operator on Lipschitz graphs and Ahlforsregular curves. In particular, he used the following estimate (now known as the "geometric lemma") to compare S to straight lines. T h e o r e m 9.2 [Do], [Jnl]. If S is a Lipschitz curve in IR 2, then ~(~" , . ~ 2 ~ measure on $ x IR+, i.e.
ESnB(X,R)
is a Carleson
r
for all X E $ and R > O. This theorem is actually an easy consequence of a result of Dorronsoro on affine approximations of functions. P. Jones' main contribution is not the fact that he rediscovered it, but that he had the idea to use it in the context of singular integrals on curves. His idea of measuring the regularity of S with the function fl(x, r) (again, some sort of "geometric LittlewoodPaley theory") has been the source of a lot of recent work (see Theorem 9.4 and 9.5 below, for instance). In higher dimensions, it was noted by X. Fang [Fn] that the analogue of (48) does not always hold for Lipschitz graphs. Fortunately, only minor modifications are needed. For l
(49)
= i¢{"k
P)IPd @}'/P
85 be the LVanalogue of fl(x, r) = ~oo(x, r), where the inf is still taken over all kplanes. The geometric lemma generalizes as follows. 9.3 [Do], [Fn]. If S is a kdimensional Lipschitz g r a p h in ~ n , and if 1 < p < ~_k2, then t3p(X, r) 2 d/~(rz)ar is a Carleson measure on S x ~ + , which means that
Theorem
(50)
~zeSnB(g,R) fO
for alI X E $ an d R > 0 . We shall state, in a later subsection, a converse to T h e o r e m 9.3. In the mean time, let us say a few words about P. Jones' traveling salesman theorem. T h e question is to determine for which subsets S of ~ 2 it is possible to find a curve with finite length, t h a t passes t h r o u g h all points of S. The answer is as follows. T h e o r e m 9.4 [Jn3]. Let S be a compact subset oflR 2, with finite 1dimensional Hausdorff measure. There exists a curve F C IR 2 with finite length, and such that S C F, if and only if
(511
< +oo,
where fl(x,r) is defined as in (47) even when x ~ $, and is set equal to 0 when $ A B ( x , r ) is empty. P. Jones also characterizes the sets S c IR 2 t h a t are contained in an Ahlforsregular curve : the condition (51) is then replaced by an appropriately scaleinvariant Carleson measure condition. See [Jn3] for more details, and [B J], say, for a nice application to harmonic measure. W h e n S = supp # for some # C ~ (or, equivalently, if the restriction to ,q of the onedimensional n a u s d o r f f measure is in ~ ) , then (51) simply means t h a t
(52)
fs fo
r) 2 
<
+¢¢.
r
Also, the condition for $ to be contained in an Ahlforsregular curve is then (48) (again under the condition t h a t $ = supp # for a # E ~~). In our study of singular integrals on a set .q, the following very weak version of (48), or (50), is sometimes already useful. We shall say t h a t " $ satisfies a weak geometric lemma" if, for each c > 0, one can find a constant C(e) such that, if • = {(x,t) C 5 × IR+ : ~(x,r) > e}, then the measure l s ( x , r ) d # ( x ) ~  is a Carleson measure with n o r m < C(e). In other words,
SnB(X,R)l×lo,Rlna
r
86 for a l l X E S a n d R > 0 . It is easily proved that Lipschitz graphs always satisfy (53). Also, sets in S(k) satisfy a little more than (53). This is proved by Semmes in [Se8] ; amusingly, the proof comes from a variant of the square function estimate (46), applied to the unit normal n(y), and so uses much more "operator theory" than geometrical arguments. We shall also see in a later subsection that if $ = supp # for a ~z E ~ , and if the square function estimates (45) hold, then $ satisfies a weak geometric lemma. Conversely, a weak geometric lemma is not enough by itself to imply that $ is rectifiable. However, if $  s u p p # for some ~ E ~ , if $ satisfies a weak geometric lemma, and furthermore $ "has big projections" (as defined in the beginning of Section 7), then $ C B P L G (see Definition 3.4). The proof is a minor modification of P. Jones' proof of Theorem 8.1 given above (only Lemma 8.4 needs to be modified to work when f is a projection defined on S). Proving directly that all surfaces S E S(k) satisfy a weak geometric lemma is not too hard, and this can be used to give yet another proof of the fact that $ E S(k) always CBPLG. See [DS4] for details. C  The corona construction If S satisfies the weak geometric lemma, then the methods of Carleson's celebrated construction provide a very powerful tool for analyzing the geometry of $ and relating it to analytical issues. The idea for this comes from work of Garnett and Jones [GT] and Jones [Jnl], in which they use the corona construction to decompose complicated domains into simpler pieces, with an estimate on how the pieces fit together. Here, we intend to do the stopping time arguments directly on the set $ (rather than through a conformal mapping, for instance). Thus it is a good idea to use an analogue of dyadic cubes on $. Let us recall what we mean by this. We shall use a collection ~ of subsets Q c $, with the following properties. First, ,1~ = Uic~z ~ i , where £ i will be thought of as the set of dyadic cubes of size 2  j . We want that
(54)
for e a c h j E ~ ,
S = U Qisapartiti°n°f$ ;
(55)
if j _< j ' , Q E ~ i and Q' E ~ i ' , then either Q contains Q', or else Q n Q' = 0 ;
(56)
if 3" E ~ and Q E )~i, then C  1 2  j < d i a m Q _< C2 i, and C12 ki < #(Q)
(57)
< C2ki ; if j E ~, Q E ~i, and 0 < r < 1, then #{x E Q : dist(x, $\Q) <_r2 1) < Crl/c2 ki.
The condition (57) is a precise way of saying that the cubes of ~ have small boundaries. It is useful in some applications, but it is not really needed in what follows. The existence of a collection of cubes ~ such that (54)(57) hold was proved in [Dv6]. However, since the proof was unduely complicated, we shall give a simpler variant in the
87 appendix. The existence of ~ was recently generalized by M. Christ to the case of any space of homogeneous type. Moreover, his proof is not significantly more complicated, so the reader can also consult [Ch2] directly. If S satisfies a weak geometric lemma, it is possible, for each e > 0, to split E into  B u ~, where B satisfies the Carleson measure condition
(58)
,(Q) <
for each cube R E
QEB
QcR and (59)
for each Q E .G, there is a kplane PQ such that dist(y, Pq) _< e d i a m Q for all y E $ such that dist(y, Q) < d i a m Q .
Given a small ~f > e, it is also possible to split further the set ~ into disjoint "stopping time regions" $1, j E J, in such a way that ~ = U i e J s j , but also
Q(Si)
(60)
each Sj has a maximal element, for inclusion, denoted by Q' E ~ is such that Q c Q' c Q(Sj), then Q' E S 1, too ;
and, if Q E Sj and
(61)
if Q E Sj, then
(62)
if Q is one of the minimal cubes of Sj (again for inclusion), then one of the sons of Q is in the bad set B, or else Angle(Pq, PQ(sj)) ~_ 6
Angle(PQ,Pq(sj)) <_~ ;
(if Q E £ j is a cube of size 2  i , the sons of Q are the cubes Q~ E £ i + 1 such that Q' c Q ; note that in our case, Q might have only one son, but it always has < C sons). Up to now, if # E ~ , $ = supp # and 5 satisfies a weak geometric lemma, it is not hard to construct, for all 0 < e < 6 small enough, the sets B, ~, and $i, j E J, in such a way that (58)  (62) be satisfied. We shall say that $ "admits a corona decomposition" if, furthermore, it is possible to choose B, ~, and the Sj's in such a way that the Q(Si)'s satisfy the following Carleson measure packing condition :
,(Q(Sj)) <
(63)
iEJ:Q(Sj)cR for each cube R E £. This new condition looks a little technical, but it is extremely useful. We shall try to say a few words to that extent in the next section, but let us mention now that a very similar construction was already used in [Seg] to give a new proof of the boundedness of the operator T~ when g is any "good kernel" and S E S(k). The boundedness of T~ is obtained from the "corona decomposition" above, which is itself obtained using square functions estimates such as (46). The fact that S E S(k) admits a corona decomposition
88 also allows Semmes to prove the estimate on harmonic measure mentioned in Section 7 (see Corollary 7.3). Let us finally mention that the first use of the corona construction in this context is due to P. Jones, in his study of Ahlforsregular curves (see [Jnl]) ; however, the argument still used a conformal mapping, and the stopping time was performed on the unit disk. D  A characterization
T h e o r e m 9.5 [DS3]. For a kdimensional set S C ]Ft'~ such that S = supp # for a lZ E ~ , the following properties are equivalent : a) for each good kernel (see Definition 1.1), the operators T~, e > 0, defined by (4) are uniformly bounded on L2($, d/~) ; b) for each good kernel K , the operator T~ defined by (5) is bounded on L P ( S , d # ) for l
e) the measure 131(x, r) 2 rd_~a of~l) ;
is a Carleson measure on S × ]R + (see (49) for a definition
f ) the measure flp(x,r) 2dMz)d~ is a Carleson measure for 1 < p < ~ r for 1 <_ p <_ +oo when k = 1 (in other words, (50) is satisfied) ;
when k > 1, and
g) letting nl = M a x ( n , 2k + 1), for each e > 0 there exists a constant M >_ 1 and, for all x E $ and r > 0, an Mbilipschitz mapping z : ]Rk ~ ~ n l such that # ( $ n B ( x , r ) \ z ( l R k ) ) < er k ( i f n l > n, IR n is e m b e d d e d in IR m in the natural way) ; h) letting n2 = n if n >_ 2k and n2 = n + 1 otherwise, there is an wregular function z : IRk __. jR,,2 (see Definition 6.4) such that $ C z ( ~ k ) . Let us first comment rapidly on the proofs. The equivalence of a) and b) is standard Calder6nZygmund theory. The fact that a) implies c) is easy, and the fact that g), or h) implies b) is easily deduced from the results of previous sections (see Proposition 3.2, Corollary 3.6 or Theorem 6.5, for example). What really helps proving the theorem, however, is the condition d). Indeed, if S i is one of the stoppingtime regions with the properties (59), (60) and (61), then one can construct a Lipschitz graph I'i, which is a very good approximation of $, at the scale of all the cubes of S i. To prove that d) implies f), g), or h), one uses the fact that f), g) and h) are satisfied by the Lipschitz graphs r i , and one tries to glue together the various pieces coming from the various S i's. The inequalities
89 (58) and (63) precisely control the a m o u n t of gluing t h a t has to be done, and altogether deducing f), g) and h) from d) is relatively easy. The hard p a r t of the t h e o r e m is deducing d) from c) and, to a lesser extent, from e). W h a t saves us in this case is t h a t it is not necessary to have d) to construct the Sis, and then the lipschitz graphs Fj : only a weak geometric l e m m a is needed there. T h e technical p a r t of the proof is moving square function estimates from $ onto the lipschitz graphs F i ; the idea of the proof is that, because of the second p a r t of (62), the square function estimate on each F i cannot be too good (one does not expect Fj to be much better t h a n a 51ipschitz graph). One then controls the n u m b e r of stoppingtime regions S i by associating, to each of them, a piece of square function which is not too small ; (63) then follow from (45). A few more details can be found in [DS3]. Let us conclude this section with a few c o m m e n t s on T h e o r e m 9.5. In condition h), one can take w E Al(lRk). We do not know, however, if one can take w=l. The conditions e) and f) are not t h a t much different from each other ; they can be seen as quantified ways of saying t h a t .q is "rectifiable", or "regular" in the sense of Besicovitch (see [Fel, [Mt], or [Fa] for definitions). Actually, conditions like " $ C B P L G ' , or g), or even h), can be seen as stronger rectifiability properties, too (compare t h e m to the p r o p e r t y of being contained in the union of a countable family of Clsurfaces, which is one of the definitions of rectifiability). Also, the fact t h a t e) implies h) can be seen as an extension of P. Jones traveling salesman t h e o r e m ( T h e o r e m 9.4 below) : we characterize the Ahlforsregular sets .q t h a t are contained in the image of an wregular function. Note that, when k = 1 and n = 2, our result is weaker t h a n Jones', because we restrict ourselves to Ahlforsregular sets. On the other hand, we can use the slightly different functions/~n, 1 < p < + o o , instead of just /~oo. T h e numbers nl of g) and n2 of h) were introduced for technical reasons ; we do not know if they are sharp, or even if it is impossible to take nl = n2 rt (provided t h a t =
k
10. A f e w q u e s t i o n s Let us start with questions relative to parametrizations. W h e n k = 2 (and n = 3), S. Semmes proved t h a t every CASSC (see Section 6.A) is the image of ][{2 by a quasisymmetric m a p p i n g and so, in particular, can be parametrized in an wregular way (see Section 6B).
90 His proof relies on the existence of a conformal mapping from IR2 to $, and does not extend to higher dimensions (see [Se4]). Is it possible to parametrize all CASSC's of dimension k = 2 by 1regular mappings? Stated differently, this is a question on the class I0 of weights w E Aoo t h a t show up as the Jacobian of a quasiconformal mapping of the plane : is it true t h a t if w is the Jacobian of a conformal m a p p i n g from IR2 onto a CASSC $, then w = bwo for some w0 C To and some b such t h a t log b E Loo ? If so, a quasiconformal change of variable in IR2 gives a 1regular parametrization of $ from any conformal parametrization. One can also try to characterize the weights w E I0, or, less ambitiously, their classes modulo multiplication by the exponential of a b o u n d e d function. Obviously, not all A o o  w e i g h t s are good (exercise : if w is continuous, and its restriction to a C °o curve is zero, then w cannot be in i0). In fact, if w is the Jacobian of a quasiconformal mapping of ]R k, k _> 2, or even if w C Aoo(]R k) is such that there exists an wregular mapping, then w has to satisfy the following "strongA~o condition" : there is a constant C > 0 such that, if B C IR k is a ball and q is a p a t h t h a t joins the center of B to its boundary, then
f..,lwl/k _:>_cl { fBW} 1/k .
(64)
We do not know, however, if every strongAoo weight(or even every strongAoo weight with II log w IIBMO small enough) is such t h a t there exists an wregular mapping. For more details on strongAoo weights, see [DS2], and for a partial result in this direction, see [Sell]. It seems reasonable to conjecture that every CASSC of dimension k > 2 admits an wregular parametrization (or even a quasisymmetric one). Reasonable parametrizations were found by Semmes [Se4], but they do not quite have the right scale invarianee ; also, note that part h) of T h e o r e m 9.5 only gives an wregular parametrization of a set which is larger t h a n $. One can try to relate Sobolev estimates on $ to the boundedness of singular integrals on L2($). Let us give an example of Sobolev estimates. Suppose t h a t $ E S(k), and make all the necessary a priori assumptions. Define
gI(x)
= sup ~~ r)>O
~k
NB(z,r)
I I(Y)  f(x) 12 a.(y)
and
o~I(x,~)
= i~f
~~
nB(x,r)
I I(y)  A(yl I~ e . ( y )
,
where f is, say, of class C 1 on $ and the infimum is taken over all Cliffordanalytic, affine functions A. Semmes proves t h a t
91 Furthermore, if each of the two connected components of ]Rk+l\$ is a NTA domain, then the two quantities above are equivalent to fs ] df ]2, where df denotes the differential of f on S. In the general case, $ might have points where it chokes, and then I f(x)  f(y) I could be large even if x and y are close and ] df I is small (think about a curve which is not chordarc), and we cannot expect as much. It is still possible to give an estimate, by controlling the number of choking points of S. We refer to [SeS] for more details. It is not clear whether we should expect the right Sobolev estimates to imply that singular integrals are bounded. This looks like a reasonable way to replace the assumption of connectedness (or local connectedness) which implies the boundedness of singular integrals when S is Ahlforsregular and onedimensional. We know that, when k >_ 2, local connectedness is not enough : see [Dv6, p. 113] for the picture of an example. It would probably be interesting to study a little more the theory of differentiability of functions on one of our preferred surfaces S (for instance, is there anything like LittlewoodPaley theory on S ?). Note that the corona decomposition already gives some information. Theorem 9.5 does not tell us everything we wanted to know about singular integrals on surfaces. The most important problem is that conditions a) and c) use much more kernels and functions ¢ than we would like. For instance, if k  1 and n  2, we would really like to characterize the Ahlforsregular sets S such that the Cauchy kernel (alone !) defines a bounded operator on L2(S). We do not even know if it is enough to consider, in a), all the odd kernels K that are homogeneous of degree  k . When k = 1 and n  2, knowing whether the boundedness of T~ when K is the Cauchy kernel implies the boundedness of all other T~'s would have nice applications to the theory of analytic capacity. Even if we decide to consider all good kernels, the equivalent conditions of Theorem 9.5 are not necessarily easy to check. In fact, it seems that in all known examples, it is just as simple to prove directly that S C B P L G ! Here is an example of a class of surfaces, for which we do not know whether Theorem 9.5 applies. Suppose $ is a kdimensional surface, homeomorphic to ]R k, contained in lR k+l, and such that $  supp ~ for some # E ~~. Also suppose that for each x C S and r > 0, there is a topological ball of dimension k, contained in $ NB(x, r), but which contains S MB(x, r). What can be said about S ? When k  2, Semmes used a conformal mapping and modulus estimates to show that $ is the image of lR 2 by a quasisymmetric mapping (and so, S CBPLG), but in higher dimensions nothing is very clear. Some consequences of Theorem 9.5 have, at this moment, surprisingly indirect proofs. Let us give a few examples to amuse the reader. To prove that a), or b), is invariant when S is replaced by its image under a bilipsehitz mapping, it seems that we need the equivalence with one of the last five conditions of the theorem. Similarly, if we want to know that, if S "contains big pieces of sets that contain big pieces .... of Lipschitz graphs", then S contains big pieces of sets that contain big pieces of Lipschitz graphs, or even d), e), f), g) or h), we seem to have to go through singular integral operators ! Finally, Proposition 3.8 tells us that, if the equivalent conditions of Theorem 9.5 are satisfied, then e*0 lim T,~f(x) cexists almost everywhere for every good kernel K. It is not clear how to prove this directly (and in particular under the weaker assumption that T~ is bounded for a single K). We still do not know whether the equivalent properties of Theorem 9.5 also imply that S CBPLG. Since $ satisfies a weak geometric lemma, this is equivalent to asking whether
92 S has big projections. Using condition g), we see that it would be enough to prove the following. If S is the image of IRk by a bilipschitz mapping, and if 0 > 0 is given, there is a constant t / > 0 (that depends only on k, n, the bilipschitz constant and 0) such that, whenever x E S, r :> 0, and E C S M B(x, r) is a compact set satisfying ~(E) > Ork, there is a kplane Y such that I Try(E) [> vr k (~r~ denotes the orthogonal projection on Y). Amusingly enough, this does not even seem to be known (or known to be wrong) when
k=landn=2! Let us finally quote a variant of Vituschkin's conjecture on sets of vanishing analytic capacity. Let us restrict to k = 1 and n = 2 (the question is probably hard enough in that case !). Suppose/~ E ~ , S = supp #, and 0 e S, and consider So  S M B(0,1). The Favard length of S0 is the number FL(So) = f o I ~ro(So) [ dO, where, for each 0 E [0,~r], 7r0 is the orthogonal projection on a line that makes the angle 0 with the real axis. Is it true that if FL($o) > a > 0, then there is an image by a rotation of a Lipschitz graph, F, such that #(F N So) _> l / > 0 ? We would like v, and the Lipschitz constant of F, to depend only on a and the Ahlforsregularity constant of #. Would it help if we supposed that the Favard length is large at all scales ? What about supposing that I ~ro(So) I> _ a >0 for all 0 in a small interval ? Of course, we are interested in this question because the existence of F would imply that the analytic capacity of $ is > 0. If these questions are not hard enough, one can always try to say something sensible in the case when # E A, but # is not necessarily in ~ . For instance, does a variant of the implication e) =~ h) (the traveling salesman theorem in higher dimensions) hold when one does not assume that S = supp ~z for a # E ~ ? I hope, too, that at this stage, the reader will be tempted to add a few arrows of his own (or, better, a few boxes) to the diagrams of Appendix II.
APPENDIX I C o n s t r u c t i o n o f d y a d i c c u b e s o n a r e g u l a r set
Let/~ E ~ (see Definitions III.2.3 and III.1.3), and let ,~ be the support of )~. We want to construct a family ~ of"dyadic dubes" with the properties (54)(57) described in Section III.9. The proof that follows is a minor modification of the argument in [Dv6] ; for an extension to spaces of homogeneous type (or a different proof), see [Ch 2]. Let us start by constructing balls with reasonably small boundaries. Let r} > 0 be a small constant (to be chosen later). L e m m a A . 1 . For each x E $ and j E 2~, there is a radius r ~ ( 2  J , (1 + ~ ) 2  i ) , and such that (A1)
ball By(x) centered at x and with a
#({V E Bj(x) : dist(y, S\Bi(x)) < T122i}) +/z({y E $\Bi(z) : dist(y, Bi(x)) < 17~2i}) <
C~12ki.
This is fairly easy : one can find more than ~ radii r such that the sets that arise in (A1) are a~ll disjoint. Since all these sets are contained in B(x, 2Y+l), their total mass is ~ C2 kj, and (A1) follows by the piegonhole principle. Let us now construct coverings of $. For each j E 2Z, let Ao(j) be a maximal set of points x E S with the property that I x'  x I_~ 2 1 for x ~ x' E Ao(j). By the maximality of A0(j), the balls Bi(x), x E A0(j), are a covering of $. Also, since the balls B(x, 2  1  1 ) are disjoint and have a mass _~ "~2k12  k , there are never more than C points y E Ao(j)
such that BI(Y) meets B(x,2;+l). Our coverings are not partitions yet. To fix this, we put an order (denoted by <) on each A0 (j), and we replace B i (x) by
(A2)
= BI( ) n
BI(y)
(by what we just said, the union is locally finite). The 5 n Bj(x), x E Ao(j), are now a partition of S, but we could have removed so much from B i (x) that B~.(x) will no longer satisfy (56). So we'll now glue some of the B~.(x)'s together. To do so, let A(j) be the set of x E Ao(j) such that Iz(BJ(x)) > Co12 ki, where t h e c o n s t a n t Co will be chosen soon. To each z E Ao(J)XA(J), let us associate a point h(x) E A(j) as follows. Note that there are less than C points y E Ao(j) such that Bj.(y) meets B(x, 21), and so one of them must be such that #(B(x, 2 1) n Bj(y)) >_ C1"12 ik (this is because the Bj.(y) M 5 are a partition of .q). We let h(x) be such a y, and we choose C0 > C~/1, so that h(x) E A(j). Note that I h(x)  x [< 2  i + 2 . For each y e A(j), we set (A3)
Di(Y) = S N { BJ'(y) U [ Ua(~)=~B~'(x)]} "
94 The
Di(y), y E A(j), are a new partition of $, and they satisfy D 1(y) < 2  i + 3 ,
(A4)
diam
(A5)
I~(Dj(y)) >
Co'2 ik,
and (A6)
#({u E Di(y ) : dist(u, $\Di(y)) < r/221}) +/~({u E S\Di(y ) : dist(u, Dj(y)) < r/22i}) <
Crl2ki.
Let us try now to make the various generations of cubes interact nicely with each other. To do so, it will be easier to skip some of the generations. Let N be a large integer (it will be chosen soon, so t h a t 2  N is smaller t h a n r/2) ; we shall restrict ourselves to the j ' s that are multiples of N. If j e NTZ and x E A ( j ) , let 99(x) be the point y E A(j  N) such t h a t x E D]N(y ). Next, for j E NT], and d E IN, let Ed(x)  {y E A ( j + N d ) : 99d(y) = x}, where 99d = 99 o •  o 99. We now replace D 1 (x) by unions of smaller cubes :
(AT)
U yEEd (z)
Note t h a t the Hausdorff distance between the sets D 1 (x) and D],I(X) is trivially < C2 y. Then, since Di,d+x(x) : UyeE~(x)[Dj+Nd, I(Y)], the Hausdorff distance between the sets Di,d(X ) and Di,d+l(X ) is less than C2 jNd. Consequently, for j E NTZ and x E A(j) fixed, the sequence of the closures Dj,d(X ) converges, for the Hausdorff topology, to a c o m p a c t set R](x). First note that the Rj(x), x E A(j), are still a covering of $. Indeed, fix u E S and j ; for each d _> 0, there is an x E A(j) such t h a t u E Di,d(X ). Since there is only a finite set of points x t h a t can arise this way, u is in infinitely m a n y Dj,d(X) for some x, and so u E Ri(x ). However, because we took closures, the R j ( x ) ' s are no longer a partition of S. We could content ourselves with t h e m and show the intersections have measure 0, b u t let us go ahead and modify the R i (x)'s again. For each j > 0, put an order (denoted by <) on A(j), and do so in such a way t h a t x < y as s o o n a s 99(X) < 99(y). For all j > 0 and x E A ( j ) , set
(AS)
Qi(x)  Rj(x) M
U
Ri(Y) I
y E A(j)
y<x
j
)
Note that, as before, the union is locally finite. Obviously, for each j > 0, the Qj(x), x E A(j), are a partition of S. Let us check t h a t they also satisfy the conditions (55), (56) and (57) of P a r t II1.9.
95 To do so, we first check how close Qy(z) is to Dj(x ). Let j • NTZ, x • A(j) and u • ny, l(x ). If y is the point of A(j + N) such that u • Dy+N(y), then y • ny(x ) and so dist(u, Di(x)) <__2  i  N + I . We have seen before that the Hausdorff distance between Di, ~ and any Dy,d, d > 1, is _< C21~r, and so dist(u, Dj(x)) < C2  y  N for every u • Ry(x ). Next, if u • n i ( x ) n Qi(x) ¢, then u • Ri(x' ) for some x' ¢ x and consequently dist(u, $\Di(x)) <_ C 2  i  N . We choose N so that 2  i  N < ~ 2 2  i  2 . It follows from (A6) that the symmetric difference between Di(x ) and Q~(x) has measure < C~2 ki. Using this and (Ah), we get that ~(Qi(x)) > C~'2 kJ1 (provided ~ is small enough). Since diamQi(x ) < 2  i + 4 and ~ • ~ , we see that Qi(x) satisfies (III.56). Moreover, let us check that #({u • Qy(x) : dist(u, $\Qy(x)) <_y 2 2  i  1 } ) (A9) +jz({u • $\Qi(x) : dist(u, Qi(x)) < y22Y1}) < C~2 ki. Indeed, for the first set, note that if u • Di(x ) and dist(u, $\Qi(x)) < rl22i1, then dist(u, Di(x')) < ~22i for some x' ¢ x. The corresponding set has small measure by (A6). Since Qi(x) N Di(x) c is also small, we get the desired estimate for the first set. The second set is a finite union of sets like the first one ; (A9) follows. We did not define Qi(x) when j < 0 yet. Let us do it now : if j =  N d for some d > 0 and x • A(j), let (A10)
Qj(x) =
[,J Qo(y). • A(O)
For the same reasons as above, the Qi(x), x • Aft), are partitions of $, satisfy (IiI.55), (III.56) and (A9). We shall now check that the more precise property (III.57) follows from this. Fix j • N ~ , x • A(j) and r > 0, and let
F = {u • Qi(x) : dist(u, $\Qi(x)) <_v2 ~i} U{u • $\Qi(x) : dist(u, Qj(x)) < r21}. For every d >_ 0, let B(d) be the set of all y • A(i+ Nd) such t h a t Qi+Nd(Y) meets F, and let Fd = (JB(d) Qi+Nd(Y). Obviously, Fo ~ FI " " ~ Fd ~ F. Now suppose that r < 2 N(d+1) and let y • B(d). If u • Fd+l n Qi+Nd(Y), then d i s t ( u , f ) < 24]2 N(d+l) and so gist(u, S\Q]+Nd(Y)) iS less t h a n gist(u, S\Qi(x)) if u • Qi(x) and than dist(u, Qi(x)) otherwise. Consequently, dist(u,S\Qi+Nd(y)) < 2412 N(d+l) + r2 1 < ~22Nd21 by our choice of N. By (A9), #(Fd+I N Qi+Nd(Y)) < 1A6D(Qi+Nd(Y)), provided we chose ~ much smaller than C~1. Summing on Y0 • S(d) gives #(Fd+l) <_ 1A61z(Fd), and repeating the argument gives #(Fd+l) _< 1 0  d  l ~ ( F 0 ) _< ClOd2 kj. ThiS is true as long as r < 2  N ( a + l ) , and so we get ~ ( F ) < crl/C2 ki, as needed.
96
We now take for ~j. the set of Qy (x), x E A(j) when j E N2Z, and ~aNbb ~   ~ctN whenever 0 ~ b < N ; we just proved that the R j ' s satisfy the conditions (54)  (57) of III.9. R e m a r k . In our construction, many of the cubes have only one child (themselves). This does not create any problem in the arguments, and cannot be avoided if you want to use the dyadic scheme and the dimension k is less than 1, say.
APPENDIX II Two recapitulatory diagrams
In the following two pictures, we used the following abbreviations : Bilipschitz images are the images of ]Rk under a bilipschitz mapping z : IRk ~ ]Rn ;

 CASSC's are Chordarc surfaces with small constant, as in Section 6.A ; Sobolev inequalities are alluded to just ~fter (65) ;

 Cs is the CauchyClifford operator, as in Sections 6.A and 6.C ; "T~ is bounded" means that for every good kernel K (see Definition 1.1), the operator

defined by (4) and (S) is bounded on L2(S,d.) ; Semmes surfaces are the sets $ E S(k) introduced in Section 6.C ;

W.G.L. stands for "weak geometric lemma". It is the weak version of P. Jones' geometric lemma introduced in Section 9.B, (53) ; 
"Projections" refers to the property "S has big projections", as in the beginning of Section 7 ; 
"S C B P L G " means "S contains big pieces of Lipschitz graphs", as in Definition 3.4;

Harmonic measure estimates are estimates that say that, if (9 is one of the components of ]Rk+I\s, and if (9 is a nontangential access domain, then harmonic measure relative to (9 is Aooequivalent to surface measure on a(9 (see Corollary 7.3) ; 
Big disks surfaces are the following slight modification of the class S(/c). In the condition (37), one replaces "dist(x~, S) > ~ " by "xi is the center of a kdimensional disk of radius ~ (or even the image of such a disk by a bilipschitz mapping) that does not meet S" ; 
"Higher codimension variant" refers to a generalization of the class S(k) to higher codimension ; we refer to [Dr6] for more details ; 
 S (CBP)mLG is, as in Remark 3.7, a generalization of the condition 5 CBPLG ;  S CVBPBI (5 contains very big pieces of bilipschitz images of ]R k) is the condition g) of Theorem 9.5 ;
98  "contained in wregular" is the condition h) of Theorem 9.5 : $ is contained in the image of ~k by some wregular mapping (with perhaps values in lq:n+l) ; 
"square function estimates" are the estimates (45) (condition c) of Theorem 9.5) ;
the corona decomposition is the condition mentionned in Section 9.B (condition d) of Theorem 9.5) ; 
the Geometric Lemma is the variant of P. Jones' condition mentioned in conditions e) and f) of Theorem 9.5 ; 

wregular surfaces are the images of ~k by wregular mappings, as in Definition 6.4.
~ilipschitz)~ ??
Sobolev inequalities + NTA on t both sides [Se8]
t
( CASSC"~ .
_....I T~ is
,.~e9]
[Dv1,2,51[
bounded
[\
/
I
Semmes ~ .. Q~~surfaces
[DS4]
~I Projections I
CBPLG
[Se9] ~
Harmonic measure estimates
[DJel ~
disks
( Higvhear~t "
99
CS
is [ [Dvl,2,5~
I I_ [CMM] + [COMI
~
I NI
Seqs~.am~rete ~.
99 ""
I (CBp)mLG
Icv 'l
~k~[DS3]
I
S
/[
bounded[ 99 " '~ I
[DS3]
~
r
IIDS3] l ~ / I ~¢"
W.G.L.I
ix,. //Lioschitz"~N
Lg'raphs ~ ~
[Geometric[ Lemma I I I
i
 deCOmrpnaition
[
S
~.} ~[Jn2I+[DS4] ,] CBPLG
,~~o3~
~ov6,1 i~v,~
When[i(=1. W~e4~=2~ 0~:regular "~ ?? ~°therwise" L.... ~,,,surrac~ff
These two diagramsshow someof the implicationsbetweenproperties of a set ,~ which is the support of a measure ~t in E.
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[Ahl [Au] [AT]
[BJ] [BCJI [BDS] [Call [Ca2]
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[Gu]
M. De Guzman, Differentiation of integrals in ~ n , Lecture Notes in Mathematics 481, SpringerVerlag 1975. S. Jaffard, Construction et propri6t6s des bases d'ondelettes ; remarques sur la contr6labilit6 exacte, Doctorat, Ecole Polytechnique.
[Ja]
S. Jaffard and Y. Meyer, Bases d'ondelettes dans les ouverts de IR n , Journ. Math. Pures et Appl. 68 (1989), 95108. [Ja_N,12] S. Jaffard and Y. Meyer, Les ondelettes, Proc. Conf. on Harmonic Analysis and Partial Differential Equations, E1 Escorial 1987 (ed. J. GarcfaCuerva), p. 182192, Lecture Notes in Math. 1384, SpringerVerlag 1989. D. Jerison and C. Kenig, Boundary behaviour of harmonic functions in nontangential [JK] access domains, Adv. in Math. 46 (1982), 8014". P. Jones, Square functions, Cauchy integrals, ana!ytic capacity, and harmonic measure, [Jnl] Proc. Conf. on Harmonic Analysis and Partial Differential Equations, E1 Escorial 1987 (ed. J. GarcfaCuerva), p. 2468, Lecture Notes in Math. 1384, SpringerVerlag 1989. P. Jones, Lipschitz and bilipschitz functions, Revista Matematica Iberoamericana, [Jn2] vol.4, 1 (1988), 115122. P. Jones, Rectifiable sets and the traveling saleseman problem, Inventiones [Jn3] Mathematicae 102, 1 (1990), 116. [JnM] P. Jones and T. Mural, Positive analytic capacity but zero Buffon needle probability, Pacific J. Math. 133 (1988), 99114. JL. Joum6, Calder6nZygmund operators, pseudodifferential operators and the Cauchv [J6] integral of Calder6n, Lecture Notes in Math. 994, SpringerVerlag 1983. [Lel] PG. Lemari6, Continuit6 sur les espaces de Pesov des op6rateurs d6finis par des int6grales singuli~res, Ann. Inst. Fourier Grenoble 35 (1985), 175187.
[JAM1]
[Le2]
PG. Lemari6, Une nouvelle base inconditionnell.; de H 1(IRxlR), preprint.
104
[LM] [Ma]
[McM] [McS 1] [McS2]
[MRV] [Mt] [Mr2] [Mt3] [Mt4] [My1] [My2] [My3] [My4] [Mul] [Mu2] [Ondl] [Ond2]
[Ru] [Sel]
[Se2]
PG. Lemari6 and Y. Meyer, Ondelettes et bases Hilbertiennes, Revista Matematica Iberoamericana, Vol 2, 1(1986), 118. S. MaUat, Multiresolution approximation and wavelets, to appear, Transactions A. M. S. A. Mclntosh and Y. Meyer, Alg~bre d'op6rateurs d6finis par des int6grales singuli~res, C.R. Acad. Sci. Paris, t. 301, S6rie I, 8 (1985), 395397. R. A. Macias and C. Segovia, Lipschitz functions on spaces of homogeneous type, Adv. in Math. 33 (1979), 257270. R. A. Macias and C. Segovia, A decomposition il}to atoms of distributions on spaces of homogeneous type, Adv. in Math. 33 (1979), 271309. O. Martio, S. Rickman et J. V~is~il~i, Definitions for quasiregular mappings, Ann. Acad. Sci. Fenn., Ser. AI, 448 (1969), 140. P. Mattila, Lecture notes on geometric measure theory, Universidad de Extramadura, Spain (1986). P. Mattila, A class of sets with positive length and zero analytic capacity, Ann. Acad. Sci. Fen., 10 (1985), 387395. P. Mattila, Smooth maps, nullsets for integralgeometric measure and analytic capacity, Ann. of Math. 123 (1986), 303309. P. Mattila, Cauchy singular integrals and rectifiability of measures in the plane, preprint. Y. Meyer, Les nouveaux op6rateurs de Calder6nZygmund, Colloque en l'honneur de Laurent Schwartz, Vol. 1, Ast6risque 131 (1985), 237254. Y. Meyer, Principe d'incertitude, bases hilbertiemaes et alg~bres d'op6rateurs, S6minaire Bourbaki, 198586, 662, Ast6risque (Soci6t6 Math6matique de France). Y. Meyer, Constructions de bases orthonorm6es d'ondelettes, Revista Matematica Iberoamericana, Vol. 4 (1988), 3140. Y. Meyer, Ondelettes et Op6rateurs (3 volumes), Herman 1990. T. Murai, Boundedness of singular integral operators of Calder6n type VI, Nagoya Math. Journal 102 (1986), 127133. T. Mural, A re~l variable method for the Cauchv transform and analytic capacity, Lecture Notes in Math. 1307, SpringerVerlag 1988. Ondelettes, tempsfr6quence et espaces de phases, Proceedings Conf. Marseille 1987, to be published, L.N. in Math, SpringerVerlag. P.G. Lemari6 ed., Les Ondelettes en 1989, harmonic analysis seminar at Orsay, L.N. in Math 1438, SpringerVerlag 1990. JL. Rubio de Francia, Factorization theory and ~p weights, Am. J. Math. 106 (1984), 533547. S. Semmes, The Cauchy integral, chordarc curves, and quasiconformal mappings, proceedings of the symposium on the occasion of the proof of the Bieberbach conjecture, A.M.S. Math. Surveys 21, 1986,167184. S. Semmes, Nonlinear Fourier analysis, Bulletin A. M. S. 20 (1989), 1, 118.
105
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[Se7]
S. Semmes, Hypersurfaces in IRn whose unit normal has small BMO norm, preprint.
[Se8]
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[Se9]
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[Sel0] [St]
[St2] [Sg]
[Tcl]
[Tc2] [Tc3l [Tc4]
[Tr] [Val]
[va2] [w] [z]
O. Stromberg, A modified Franklin system and higherorder systems of IR n as unconditional bases for Hardy spaces, Conference on harmonic analysis in honor of Antoni Zygmund, Vol. II, 475493, ed. W.Beckner and al., Wadworth math. series 1983. P. Tchamitchian, Calcul symbolique sur les op6rateurs de Calder6nZygmund et bases inconditionelles de L2(~), C. R. Acad. Sci. Paris, t. 303, S6rie I, 6 (1986), 215217. P. Tchamitchian, Biorthogonalit6 et th6orie des opdrateurs, Revista Matematica Iberoamericana vol.3, 2 (1987), 163190. P. Tchamitchian, Ondelettes et int6grale de Cauct;y sur les courbes lipschitziennes, Ann. of Math. 129 (1989), 641649. P. Tchamitchian, Inversion des op6rateurs de Calder6nZygmund, preprint. H. Triebel, Theory of function spaces, Monographs in Mathematics, Vol. 78, Birkauser, Basel, 1983. J. V~is~il/i, Uniform domains, T6hoku Math. J. 40 (1988), 101118. J. V/iis/il~i, Invariants for quasisymmetric, quasim6bius and bilipschitz maps, J. d'Analyse Math. 50 (1988), 201233. M. V. Wickerhauser, Acoustic signal compression with Walshtype wavelets, preprinc A. Zygmund, Trigonometric series, Cambridge University Press 1968.
NOTATIONS
vj
3
q~
6
T e , T,
28
Mb
31
X
58
TI.te , TI.t*
56
8
mo(~)
8
hQe
34
A
55
Mg
58
S(k)
72
INDEX Ahlforsregular curve Besicovitchregular(rectifiable) Big pieces Big disks surfaces Big projections Bilipschitz mappings BMO Calder6n commutators Calder6n reproducing formula Calder6n's theorem Calder6nZygmund operator Carleson measure Coifman, McIntosh and Meyer's theorem Cauchy integral Chordarc curve Chordarc surface Clifford algebra Corona construction Cotlar inequality Daubechies' compactly supported wavelets Dyadic cubes Garnett's counterexample Geometric lemma Good kernel Good ~ inequalities Gram orthogonatisation GrSchenigMeyer construction
64,88 63,89 62,63,64,72,76,86, 88,91,97 78,97 77,79,86,91,92,97 62,79,88,97 25,28,30,31,44,70 50 1 50 28 33,44,84,88 50,51,60 50,66,67,91,99 51 69,89,97 73,83,90 86,88 59 15 48,74,86,93 51,67 84,88,99 55 60 6 12
107
Haar system H 1 (Hardy space) Harmonic measure Hausdofffmeasure Homogeneous type (space of) Jones' traveling salesman theorem Lipschitz graphs
2,10,12 24,49 78,97 62 47 85,89 50,53,59,66,84,88,97
LPboundedness Maximal function Maximal operator Meyer's wavelet Multiscale analysis Paraaccretive function Parametrizations Paraproduct Principal values Quasiconformal, quasisymmetric Regular : see Ahlfors, Besicovitch, o~Regularity (of a MSA) Rectifiable set Rising sun lemma Rotation method Semmes surface Shur's lemma Singular integral operator Spline functions Square function estimates Sobolev inequalities Space of homogeneous type Stable kernel Standard kernel Strong A,,. weight T(1) T(b) Tensor product (of MSA's) Traveling salesman Weak boundedness Weak geometric lemma orregular surface
28,49 58 28,56 10 2 3O 89 44 27,30,63 72,89,90 4 63,89 65 54 72,77,78,97 43 26 3,5,10 33,83,88,98 90,97 47 52 26 9O 27,30 31,73 5,11 85,89 29,40 85,97 71,77,88,98