Anisotropic Hardy Spaces and Wavelets

Marcin Bownik

Author address:

Department of Mathematics, University of Michigan, 525 East Uni- versity Ave., Ann Arbor, MI 48109 E-mail address: [email protected]

1 viii ANISOTROPIC HARDY SPACES AND WAVELETS

Abstract

In this paper, motivated in part by the role of discrete groups of dilations in wavelet theory, we introduce and investigate the anisotropic Hardy spaces associated with very general discrete groups of dilations. This formulation includes the classical isotropic Hardy space theory of Fefferman and Stein and parabolic Hardy space theory of Calderón and Torchinsky. Given a dilation A, that is an n n matrix all of whose eigenvalues λ satisfy λ > 1, define the radial maximal function× | | 0 −k −k Mϕf(x) := sup (f ϕk)(x) , where ϕk(x)= det A ϕ(A x). k∈Z | ∗ | | | Here ϕ is any test function in the Schwartz class with ϕ =0. For 0 sequence space characterizing wavelet coefficients of elements of the anisotropic Hardy space.

2000 Mathematics Subject Classification. Primary 42B30, 42C40; Secondary 42B20, 42B25. Key words and phrases. anisotropic Hardy space, radial maximal function, nontangential maximal function, grand maximal function, Calderón-Zygmund decomposition, atomic decomposition, Calderón-Zygmund operator, Campanato space, wavelets, frame, unconditional basis. The author thanks his advisor, Prof. Richard Rochberg, for his support, guidance and encouragement. ANISOTROPIC HARDY SPACES AND WAVELETS ix

Contents

Chapter 1. Anisotropic Hardy Spaces 1. Introduction 1 2. The space of homogeneous type associated with the discrete group of dilations 4 3. The grand maximal definition of anisotropic Hardy spaces 10 4. The atomic definition of anisotropic Hardy spaces 18 5. The Calderón-Zygmund decomposition for the grand maximal function 22

6. The atomic decomposition of Hp 34 7. Other maximal definitions 41 8. Duals of Hp 48 9. Calderón-Zygmund singular integrals on Hp 58 10. Classification of dilations 68

Chapter 2. Wavelets 1. Introduction 80 2. Wavelets in the Schwartz class 83 3. Limitations on orthogonal wavelets 87 4. Non-orthogonal wavelets in the Schwartz class 92 5. Regular wavelets as an unconditional basis for Hp 94 6. Characterization of Hp in terms of wavelet coeﬃcients 102 Notation Index 114 Bibliography 116 1. INTRODUCTION 1

CHAPTER 1 Anisotropic Hardy spaces

1. Introduction In the ﬁrst chapter of this monograph we develop the real variable theory of Hardy spaces Hp (0

∞ 1/p p f p := sup f(x + iy) dx < . || ||H | | ∞ 0 1. However, for p 1 these spaces diﬀer from Lp(Rn) and are better suited for the purposes of harmonic≤ analysis than Lp(Rn). Indeed, singular integral operators and multiplier operators turn out to be bounded on Hp, see Stein [St1]. The beginning of the 1970’s marked the birth of real-variable theory of Hardy spaces as we know it today. First, Burkholder, Gundy, and Silverstein in [BGS] using Brownian motion methods showed that f belongs to the classical Hardy space 2 1. ANISOTROPIC HARDY SPACES

Hp if and only if the nontangential maximal function of Re f belongs to Lp. The real breakthrough came in the work of C. Feﬀerman and Stein [FS2]. They showed that Hp in n dimensions can be deﬁned as the space of tempered distributions f n 0 on R whose radial maximal function Mϕ or nontangential maximal function Mϕ belong to Lp(Rn), where

0 Mϕf(x) := sup (f ϕt)(x) , 0

Mϕf(x) := sup sup (f ϕt)(y) , 0

−n and ϕt(x) := t ϕ(x/t). Here ϕ is any test function in the Schwartz class with ϕ = 0 or the Poisson kernel ϕ(x)=(1+ x 2)−(n+1)/2 (in this case f is restricted to bounded6 distribution) and the definition| of| Hp does not depend on this choice. RTo prove this impressive result C. Fefferman and Stein introduced a very important tool, the grand maximal function, which can also be used to define Hardy spaces Hp. The real analysis methods also played a decisive role in the well-known C. Fefferman duality theorem between H1 and BMO—the space of functions of bounded mean oscillation. Further insight into the theory of Hardy spaces came from the works of Coifman [Co] (n = 1) and Latter [La] (n 1) where the atomic decomposition of elements in Hp(Rn) (p 1) was exhibited. Atoms≥ are compactly supported functions satisfying certain boundedness≤ properties and some number of vanishing moments. The Hardy space Hp(Rn) can be thought of in terms of atoms and many important theorems can be reduced to easy verifications of statements for atoms. Other developments followed. Coifman and Weiss in [CW2] introduced Hardy spaces Hp for the general class of spaces of homogeneous type using as a definition atomic decompositions. Since there is no natural substitute for polynomials, the Hardy spaces Hp on spaces of homogeneous type can be defined only for p 1 sufficiently close to 1. Another approach started in the work of Calderón≤ and Torchinsky [CT1, CT2] who developed theory of Hardy spaces Hp (0 norm ρ on R satisfies ρ(Atx)= tρ(x). The parabolic Hardy space Hp (0

− tr P −1 Mϕf(x) := sup (f ϕt)(x) , ϕt(x)= t ϕ(At x), ρ(x−y)

ϕ is any test function with ϕ = 0. Calderón and Torchinsky also obtain equivalent formulations of parabolic Hardy6 spaces using the grand maximal functions, Luzin functions, and Littlewood-PaleyR functions. The atomic decomposition for theses spaces was done by Calderón [Ca, Ga] and Latter and Uchiyama [LU]. It is worth noting that the general setup for defining Hardy spaces on homogeneous groups developed by Folland and Stein [FoS] presupposes that the dilation group At : 0 ∈1.} We investigate the properties× of the space of homogeneous type induced| | by this group of dilations in Section 2. p n In the next section we define the anisotropic Hardy space HA(R ) as a space of tempered distributions f whose grand maximal function belongs to Lp(Rn). The most straightforward definition of these spaces is obtained using the radial maximal function. That is, for a dilation A and 0

2. The space of homogeneous type associated with the discrete group of dilations The concept of a dilation is a fundamental one in our work. Definition 2.1. A dilation is n n real matrix A, such that all eigenvalues λ of A satisfy λ > 1. × | | It is clear that A is a dilation if and only if A−j 0 as j . We could alternatively define a dilation as a matrix whose all|| eigenva|| →lues λ satisfy→ ∞ 0 = λ < 1. The inverse of this matrix becomes then a dilation in the sense of Definition6 | | 2.1. For any dilation A we consider the corresponding discrete group of linear trans- formations Aj : j Z which induces a natural structure of a space of homogeneous type{ on Rn. Unless∈ } the dilation A is very special (to be made precise later) this structure is different than the usual isotropic structure of Rn. In this section we present the underlying ideas. To start we suppose λ1,... ,λn are eigenvalues of A (taken according to the multiplicity) so that 1 < λ . . . λ . Let λ , λ be any numbers so that | 1| ≤ ≤ | n| − + 1 < λ− < λ1 λn < λ+. Then we can find a constant c > 0 so that for all x Rn we have| | ≤ | | ∈ (2.1) 1/cλj x Aj x cλj x for j 0, −| | ≤ | |≤ +| | ≥ (2.2) 1/cλj x Aj x cλj x for j 0, +| | ≤ | |≤ −| | ≤ where is a standard norm in Rn. Note that (2.2) is a consequence of (2.1) and vice| · | versa. Furthermore, if for any eigenvalue λ of A with λ = λ (or | | | 1| λ = λn ) the matrix A does not have Jordan blocks corresponding to λ, i.e., | | | | 2 ker(A λI) = ker(A λI) then we can set λ− = λ1 (λ+ = λn ). A− set ∆ Rn is said− to be an ellipsoid if | | | | ⊂ (2.3) ∆= x Rn : P x < 1 , { ∈ | | } for some nondegenerate n n matrix P , where denotes the standard norm in Rn. × | · | In general, we can not expect that the dilation A is expansive in the standard norm, i.e., Ax x for all x Rn. Nevertheless, by [Sz, Lemma 1.5.1], there exists a scalar| product| ≥ | | with the∈ induced norm and r> 1 so that | · |∗ (2.4) Ax r x for x Rn. | |∗ ≥ | |∗ ∈ We present the proof of this result for the sake of completeness. Lemma 2.2. Suppose A is a dilation. Then there exists an ellipsoid ∆ and r> 1 such that

(2.5) ∆ r∆ A∆. ⊂ ⊂

Proof. Deﬁne the inner product , by h· ·i∗ x, y = x, y + A−1x, A−1y + . . . + A−kx, A−ky , h i∗ h i h i h i 2. THE SPACE OF HOMOGENEOUS TYPE 5

where k is an integer satisfying k > 2 ln c/ ln λ−, and c, λ− are as in (2.1). We 1/2 claim that the norm x = x, x ∗ satisfies (2.4). Indeed, by (2.1) and (2.2) | |∗ h i Ax 2 = Ax 2 + x 2 + . . . + A−k+1x 2 = x 2 + Ax 2 A−kx 2 | |∗ | | | | | | | |∗ | | − | | x 2 +1/c2 x 2 c2λ−2k x 2 = x 2(1 + (c−2 c2λ−2k) x 2/ x 2) ≥ | |∗ | | − − | | | |∗ − − | | | |∗ c−2 c2λ−2k c−4 λ−2k = x 2 1+ − − x 2 x 2 1+ − − = r2 x 2, | |∗ x 2 + . . . + A−kx 2 | | ≥ | |∗ 1+ . . . + λ−2k | |∗ | | | | − where r is the square root of the last bracket. By a simple application of the Riesz Lemma there is a matrix Q so that Qx,y = x, y ∗. Clearly, Q is self-adjoint and h1/2 i h 2i positive definite. If we take P = Q then P x = Qx, x = x ∗. Define ∆ by n | −1| h i | | −1 −1 (2.3), i.e., ∆ = x R : x ∗ < 1 . Since A x ∗ x ∗/r then A ∆ r ∆ and hence (2.5){ holds.∈ | | } | | ≤ | | ⊂

By a scaling we can additionally assume that ellipsoid ∆ in Lemma 2.2 satisfies ∆ = 1. We define a family of balls around the origin as the sets | | (2.6) B := Ak∆ for k Z. k ∈ By (2.5) we have (2.7) B rB B , B = bk, where b := det A > 1. k ⊂ k ⊂ k+1 | k| | | Even though the choice of the expansive ellipsoid ∆ is not unique, from this point we fix one choice of ∆, and consequently the Bk’s, for a given dilation A. Next we introduce the natural concept of a quasi-norm which generalizes the usual norm on Rn. A quasi-norm satisfies a discrete homogeneity property with respect to A and a triangle inequality up to a constant. Definition 2.3. A homogeneous quasi-norm associated with a dilation A is a measurable mapping ρ : Rn [0, ), so that (i) ρ(x)=0 x = 0, → ∞ (ii) ρ(Ax)= bρ⇐⇒(x) for all x Rn, (iii) there is c> 0 so that ρ(∈x + y) c(ρ(x)+ ρ(y)) for all x, y Rn. ≤ ∈ It turns out that all quasi-norms associated to a fixed dilation A are equivalent.

Lemma 2.4. Any two homogeneous quasi-norms ρ1, ρ2 associated with a dilation A are equivalent, i.e., there exists a constant c> 0 so that (2.8) 1/cρ (x) ρ (x) cρ (x) for x Rn. 1 ≤ 2 ≤ 1 ∈ Proof. It suﬃces to show that for every quasi-norm ρ we have (2.9) 0 < inf ρ(x) sup ρ(x) < . x∈B1\B0 ≤ x∈B1\B0 ∞ Suppose on the contrary that sup ρ(x)= . Then we can ﬁnd a sequence x∈B1\B0 ∞ (x ) Rn such that x 0, ρ(x ) as i . Choose M > 0 so that the set i ⊂ | i| → i → ∞ → ∞ Ω= x B B : ρ(x) M , { ∈ 1 \ 0 ≤ } has measure strictly bigger than (b 1)/2. Choose N, so that ρ(xi) > 2cM for i>N. Since − ρ(x) 1/cρ(x ) ρ(x x), ≥ i − i − 6 1. ANISOTROPIC HARDY SPACES

Figure 1. The family of dilated balls B : k Z . { k ∈ } we have ρ(x) >M for x x Ω, i.e., ∈ i − (x Ω) Ω= . i − ∩ ∅ Since Ω B B and x 0 we have − ⊂ 1 \ 0 i → (x Ω) (B B ) Ω > (b 1)/2 as i . | i − ∩ 1 \ 0 | → | | − → ∞ Hence, b 1 < (x Ω) (B B ) + Ω B B = b 1, − | i − ∩ 1 \ 0 | | | ≤ | 1 \ 0| − for suﬃciently large i, which is a contradiction.

Finally, suppose on the contrary that infx∈B1\B0 ρ(x) =0. Then we can ﬁnd a sequence (xi) B1 B0 such that ρ(xi) 0 as i . By selecting a subsequence we can assume⊂ (x )\ converges to some point→ x =→ 0. ∞ Since i 6 ρ(x) c(ρ(x x )+ ρ(x )) 0 as i , ≤ − i i → → ∞ we have ρ(x) = 0 which is a contradiction. Therefore (2.9) holds.

The natural question is whether for two distinct dilations A1 and A2, the corresponding quasi-norms ρ1 and ρ2 are equivalent, i.e., (2.8) holds for some c> 0. One 2. THE SPACE OF HOMOGENEOUS TYPE 7 may wish to state the classification theorem in terms of eigenvalues and eigenspaces corresponding to Jordan blocks, which we postpone until Section 10, see Theorem 10.3. Instead, we consider the simplest situation. Example. Suppose A = d Id for some real d > 1. It is clear that ρ(x)= x n satisfies properties of the quasi-norm. Therefore| | all matrices of this form induce| | equivalent quasi-norms. Here it becomes clear why we impose ρ(Ax)= bρ(x) with b = det A . It is not hard to see that any matrix A without Jordan blocks and whose| all of| eigenvalues λ satisfy λ = d for some fixed d> 1 has the corresponding quasi-norm ρ which is equivalent| to| n. And vice versa, any other matrix whose all of eigenvalues are not equal in absolute| · | value or with Jordan block(s) has a quasi-norm which is not equivalent to n. Equivalently, a quasi-norm induced by a dilation A is equivalent to n |if · |and only if A is diagonalizable over C with all eigenvalues equal in the absolute| · | value. This provides the classification of dilations inducing the usual isotropic homogeneous structure on Rn announced in the beginning of this section. For a fixed dilation A we define the “canonical” quasi-norm used throughout this chapter. Definition 2.5. Define the step homogeneous quasi-norm ρ on Rn induced by the dilation A as j b if x Bj+1 Bj (2.10) ρ(x)= ∈ \ 0 if x =0. Here B = Ak∆, where ∆ is an ellipsoid from Lemma 2.2 and ∆ = 1. k | | Indeed, ρ clearly satisfies (i) and (ii) of Definition 2.2. Let ω be the smallest ω integer so that 2B0 A B0 = Bω. The existence of ω is guaranteed by (2.5). n ⊂ i j Suppose x, y R , ρ(x)= b , ρ(y)= b for some i, j Z. Thus, x + y Bi + Bj 2B A∈ωB . Therefore ∈ ∈ ⊂ max(i,j) ⊂ max(i,j) ρ(x + y) bωbmax(i,j) bω(bi + bj )= bω(ρ(x)+ ρ(y)), ≤ ≤ and (iii) holds with the constant c = bω. Therefore, we can summarize that for each i Z we have ∈ x Bi and y Bi = x + y Bi+ω, (2.11) ∈ ∈ ⇒ ∈ x B and y B = x + y B . 6∈ i+ω ∈ i ⇒ 6∈ i We also record two useful inequalities (2.12) max(1,ρ(x + y)) bω max(1,ρ(x)) max(1,ρ(y)) for all x, y, Rn, ≤ ∈ (2.13) max(1,ρ(Aj x)) bj max(1,ρ(x)) for j 0, x Rn. ≤ ≥ ∈ Remark. The notion of a quasi-norm put forth in Definition 2.3 seems to be missing in the literature. Lemarié-Rieusset in [LR] considered a closely related quasi-norm satisfying (i) and (ii) of Definition 2.3 which is C∞ on Rn except at the origin. He also gives a construction of such a quasi-norm for an arbitrary dilation A, see [LR]. Quasi-norms of Lemarié-Rieusset automatically satisfy the triangle inequality up to a constant. In our definition, which was motivated by 8 1. ANISOTROPIC HARDY SPACES

[LR], we give up the smoothness of quasi-norms to include the important class of step homogeneous quasi-norms used throughout this chapter. We need two covering lemmas which hold in arbitrary spaces of homogeneous type, see [CW1, CW2]. However, we present them in a form adapted to our setting. Lemma 2.6 (Wiener). Suppose Ω Rn, and r : Ω Z is an arbitrary function. Assume that either (a) Ω is bounded,⊂ or (b) Ω is→ open, Ω < , and | | ∞ x+Br(x) Ω for all x Ω. Then there exists a sequence (xj ) Ω (finite or infinite) so that the⊂ balls x + B∈ are mutually disjoint and Ω ⊂(x + B ). j r(xj ) ⊂ j j r(xj )+ω Proof. In case (a), if sup r(x) = then we canS find x Ω so that x∈Ω ∞ ∈ Ω x + Br(x), and we are done. Thus, we can assume supx∈Ω r(x) < . In case (b),⊂ since Ω < we also must have sup r(x) < . Pick x Ω∞ such that | | ∞ x∈Ω ∞ 1 ∈ r(x1) = supx∈Ω r(x). IfΩ x1 +Br(x1)+ω then we are done. Otherwise, we proceed ⊂ ′ j inductively. Assume we have picked x1,... ,xj , set Ω =Ω i=1(xi + Br(xi)+ω). ′ \ If Ω = we are done. If not, pick xj+1 Ω such that r(xj+1) = supx∈Ω′ r(x). Suppose∅ i < j and (x + B ) (x +∈B ) = thenSx x B i r(xi) ∩ j r(xj ) 6 ∅ j − i ∈ r(xi) − B 2B B , since r(x ) r(x ). Thus, x B , which is a r(xj ) ⊂ r(xi) ⊂ r(xi)+ω i ≥ j j ∈ r(xi)+ω contradiction. Therefore, balls xj + Br(xj ) are mutually disjoint. If the sequence (x ) is finite then clearly Ω (x + B ). If not we j ⊂ j j r(xj )+ω have r(x ) as j since Ω has finite measure and balls x + B j S j r(xj ) are mutually→ disjoint. −∞ Suppose→ ∞ on the contrary that there exists x Ω such that x (x + B ). For sufficiently large j we have r(x) r(x∈), which is a 6∈ j j r(xj )+ω ≥ j contradiction of the choice of xj ’s. S Lemma 2.7 (Whitney). Suppose Ω Rn is open, and Ω < . For every ⊂ | | ∞ integer d 0 there exists a sequence of points (x ) N Ω, and a sequence of ≥ j j∈ ⊂ integers (l ) N Z, so that j j∈ ⊂ (i) Ω= (xj + Blj ),

(ii) xj + Blj −ω are pairwise disjoint for j N, S c ∈ c (iii) for every j N, (xj + Blj +d) Ω = , but (xj + Blj +d+1) Ω = , (iv) if (x + B ∈ ) (x + B ∩ ) =∅ then l l ω, ∩ 6 ∅ i li+d−2ω ∩ j lj +d−2ω 6 ∅ | i − j |≤ (v) for each j N, the cardinality of i N : (xi +Bli+d−2ω) (xj +Blj +d−2ω) = is less than∈ L, where L is a constant{ ∈ depending only on ∩d. 6 ∅} Proof. For every x Ω define function ∈ r(x) := sup r Z : x + B Ω . { ∈ r+d+ω ⊂ } Since Ω is open and has finite measure the supremum is always finite. By the Wiener Lemma applied to the function r(x) we can find a sequence of (xj ) Ω so that (x + B ) are mutually disjoint, and (x + B ) Ω cover the⊂ set Ω. j r(xj ) j r(xj )+ω ⊂ Therefore, if we define lj := r(xj ) ω then (i) and (ii) hold. Clearly (iii) also holds by the choice of r(x). If d 1 then− we have ≥ Ω= (x + B )= (x + B ) (x + rB ), j lj j lj +1 ⊃ j lj j j j [ [ [

where r is the same as in Lemma 2.2, and therefore the sequence (xj ) is necessarily inﬁnite. 2. THE SPACE OF HOMOGENEOUS TYPE 9

To show (iv), suppose y (xi +Bli+d−2ω) (xj +Blj +d−2ω) and lj li +ω +1. Then ∈ ∩ ≥ x x y B y + B B . i − j ∈ − li+d−2ω − lj +d−2ω ⊂ lj +d−ω Since x + B = (x x )+ x + B x + B + B x + B , i li+d+1 i − j j li+d+1 ⊂ j lj +d−ω li+d+1 ⊂ j lj +d we have by (iii) c c = (x + B ) Ω (x + B ) Ω , ∅ 6 i li+d+1 ∩ ⊂ j lj +d ∩ which is a contradiction of (iii). Therefore lj li + ω, and by symmetry we obtain (iv). ≤ Fix j N and consider ∈ I = i N : (x + B ) (x + B ) = . { ∈ i li+d−2ω ∩ j lj +d−2ω 6 ∅} If i I then by (iv) ∈ x + B (x x )+ x + B x + B + B + B i li−ω ⊂ i − j j li−ω ⊂ j li+d−2ω lj +d−2ω li−ω x + B + B + B x + B . ⊂ j lj +d−ω lj +d−2ω lj ⊂ j lj +d+ω

Since for i I, xi + Blj −2ω xi + Bli−ω we conclude by (ii) that the cardinality of I is less than∈ ⊂ B / B = bd+3ω = L. | lj +d+ω| | lj −2ω| 10 1. ANISOTROPIC HARDY SPACES

3. The grand maximal deﬁnition of anisotropic Hardy spaces Definition 3.1. We say that a C∞ complex valued function ϕ on Rn belongs to the Schwartz class , if for every multi-index α and integer m 0 we have S ≥ m α (3.1) ϕ α,m := sup ρ(x) ∂ ϕ(x) < . || || x∈Rn | | ∞

The space endowed with pseudonorms α,m becomes a (locally convex) topological vectorS space. The dual space of bounded|| · || functionals on , i.e., the space of tempered distributions on Rn, is denoted by ′. S S If f ′ and ϕ we denote the evaluation of f on ϕ by f, ϕ . Sometimes we will pretend∈ S that∈ distributions S are functions by writing f, ϕh = i f(x)ϕ(x)dx. Convergence in ′ will always mean weak convergence, i.e.,h f i f in ′ if and S j → R S only if fj , ϕ f, ϕ for all ϕ . For fundamentals on distributions, see [Ru3]. Theh Schwartzi → h classi given in∈ S the above deﬁnition overlaps with the usual one by virtue of the lemmaS essentially due to Lemari´e-Rieusset, see [LR]. Lemma 3.2. Suppose ρ is a homogeneous quasi-norm associated with dilation A. Then

(3.2) 1/c′ρ(x)ln λ−/ ln b x c′ρ(x)ln λ+/ ln b for ρ(x) 1, ≤ | |≤ ≥ (3.3) 1/c′ρ(x)ln λ+/ ln b x c′ρ(x)ln λ−/ ln b for ρ(x) 1, ≤ | |≤ ≤ ′ where c is some constant, and λ−, λ+ satisfy (2.1) and (2.2). Proof. By Lemma 2.4, without loss of generality we can assume ρ is the step homogeneous quasi-norm. For every integer j 0 we have by (2.1) ≥ j −j j j ln λ+/ ln b sup x = sup A A x c sup x λ+ = c sup x b , x∈Bj+1\Bj | | x∈Bj+1\Bj | |≤ x∈B1\B0 | | x∈B1\B0 | | and analogously

j j ln λ−/ ln b inf x 1/c inf x λ− =1/c inf x b . x∈Bj+1\Bj | |≥ x∈B1\B0 | | x∈B1\B0 | | Therefore, (3.2) holds for x B B , where j 0, so for all x (B )c. ∈ j+1 \ j ≥ ∈ 0 Considering x Bj+1 Bj , where j 0 and using (2.2) we obtain (3.3) for x B . ∈ \ ≤ ∈ 1 An important role in our investigation is played by the unit ball with respect to a particular ﬁnite family of pseudonorms of . It is critical that we use (3.1) to alter the standard deﬁnition of the pseudonormsS in . Otherwise the grand maximal function introduced below would not behave nicelyS with respect to dilations and many results in Section 5 could not hold. A posteriori, we see that this is only a technical issue by virtue of Lemma 3.2. Definition 3.3. For an integer N 0 consider family ≥ (3.4) := ϕ : ϕ 1 for α N,m N . SN { ∈ S || ||α,m ≤ | |≤ ≤ } Equivalently

N α (3.5) ϕ N ϕ SN := sup sup max(1,ρ(x) ) ∂ ϕ(x) 1. ∈ S ⇐⇒ || || x∈Rn |α|≤N | |≤ 3. THE GRAND MAXIMAL DEFINITION OF ANISOTROPIC HARDY SPACES 11

For ϕ , k Z define the dilate of ϕ to the scale k by ∈ S ∈ −k −k (3.6) ϕk(x)= b ϕ(A x). Definition 3.4. Suppose ϕ , and f ′. The nontangential maximal function of f with respect to ϕ is defined∈ S as ∈ S (3.7) M f(x) := sup f ϕ (y) : x y B , k Z . ϕ {| ∗ k | − ∈ k ∈ } The radial maximal function of f with respect to ϕ is defined as 0 (3.8) Mϕf(x) := sup f ϕk(x) . k∈Z | ∗ | For given N N we define the nontangential grand maximal function of f as ∈ (3.9) MN f(x) := sup Mϕf(x). ϕ∈SN The radial grand maximal function of f is 0 0 (3.10) MN f(x) := sup Mϕf(x). ϕ∈SN Finally, given N˜ > 0 we define the tangential maximal function of f with respect to ϕ as ˜ T N f(x) = sup f ϕ (y) / max(1,ρ(A−k(x y)))N : y Rn, k Z . ϕ {| ∗ k | − ∈ ∈ } Remark. It is immediate that we have the following pointwise estimates between radial, nontangential, and tangential maximal functions M 0f(x) M f(x) T N f(x) for all x Rn. ϕ ≤ ϕ ≤ ϕ ∈ Moreover, radial and nontangential grand maximal functions are pointwise equivalent by virtue of Proposition 3.10. In Section 7, we will also see that for sufficiently large N (depending on N˜) and ϕ with ϕ = 0, the tangential maximal function N˜ ∈ S 6 Tϕ f(x) dominates pointwise the grand maximalR function MN f(x), see Lemma 7.5. Proposition 3.5. For f ′, let Mf denote any of the maximal functions introduced in Definition 3.4. Then∈ S Mf : Rn [0, ] is lower semicontinuous, function, i.e., for all λ> 0, x Rn : Mf(x) >→ λ is∞ open. { ∈ } Proof. If ϕ and f ′ then f ϕ is a continuous (even C∞) function on Rn. Note that ∈ S ∈ S ∗ M f(x) = sup f ϕ (y) : x y B Qn, k Z . ϕ {| ∗ k | − ∈ k ∩ ∈ } Furthermore, since the Schwartz class is separable with respect to pseudonorms S α,m, we can substitute N by a countable, dense subset in the definition of ||the · || grand maximal function.S Therefore, in each case Mf can be computed as a supremum of a countable family of continuous functions. Therefore, Mf is lower semicontinuous.

We start the investigation of maximal functions with the fundamental Maximal Theorem 3.6. This and the following results are relatively mechanical conversions of the well-known classical results which also hold for homogeneous groups, see [FoS]. 12 1. ANISOTROPIC HARDY SPACES

Figure 2. The nontangential maximal function Mϕf(x) is computed by taking suprema of f ϕk(y) over ellipses x + B at all scales k Z. | ∗ | k ∈ The biggest novelty is the use of discrete (instead of continuous) dilations of fairly general form. Theorem 3.6 (The Maximal Theorem). For a ﬁxed s> 1 consider family (3.11) := ϕ L∞(Rn) : ϕ(x) (1 + ρ(x))−s . F { ∈ | |≤ } For 1 p and f Lp(Rn) deﬁne the maximal function ≤ ≤ ∞ ∈ (3.12) Mf(x)= MF f(x) := sup Mϕf(x). ϕ∈F Then there exists a constant C = C(s) so that (3.13) x : Mf(x) > λ C f /λ for all f L1(Rn), λ> 0, |{ }| ≤ || ||1 ∈ (3.14) Mf Cp/(p 1) f for all f Lp(Rn), 1

R Proof. For λ,R > 0 consider Ωλ = x : Mf(x) > λ, x < R . For every R n { | | } x Ωλ take y = y(x) R , k = k(x) Z, and ϕ such that x y Bk and f ∈ ϕ (y) > λ. ∈ ∈ ∈ F − ∈ | ∗ k | −k −k λ< f ϕk(y) f(z) ϕk(z y) dz = b f(z) ϕ(A (z y)) dz | ∗ |≤ Rn | || − | | || − | Z Zy+Bk ∞ + b−k f(z) ϕ(A−k(z y)) dz | || − | j=1 y+Bk+j \Bk+j−1 X Z ∞ b−k sup ϕ(z) f(z) dz + sup ϕ(z) f(z) dz ≤ | | | | | | | | z∈B0 y+Bk j=1 z∈Bj+1\Bj y+Bk+j Z X Z 3. THE GRAND MAXIMAL DEFINITION OF ANISOTROPIC HARDY SPACES 13

∞ 1+ bj sup ρ(z) −s sup b−k−j f(z) dz ≤ | | | | j=1 z∈Bj+1\Bj j≥0 y+Bk+j X Z ∞ bj(1−s) sup b−k−j f(z) dz. ≤ | | j=0 j≥0 y+Bk+j X Z Thus, there is j 0 so that 0 ≥ 1 f(z) dz > λ/C, k+j0 b y+B | | Z k+j0 where C = ∞ bj(1−s). Since x y B , j=0 − ∈ k Py + B x + B + B x +2B x + B . k+j0 ⊂ k k+j0 ⊂ k+j0 ⊂ k+j0 +ω So for each x ΩR there is r = r(x)= k + j + ω, such that ∈ λ 0 λ f(z) dz > B . | | Cbω | r| Zx+Br R By the Wiener Lemma there is sequence (xj )inΩλ so that xj +Br(xj) are mutually R disjoint and xj + Br(xj )+ω cover Ωλ . Therefore, Cb2ω ΩR B bω B f . | λ |≤ | r(xj )+ω|≤ | r(xj )|≤ λ || ||1 j j X X By letting R we obtain (3.13). → ∞ −s ∞ Since C = Rn (1 + ρ(x)) dx < , we have Mf ∞ C f ∞ for f L . By the Marcinkiewicz Interpolation Theorem∞ we obtain|| || (3.14≤ ). || || ∈ R As a consequence of Theorem 3.6 we conclude that the Hardy-Littlewood maximal function M = MHL given by 1 (3.15) MHLf(x) := sup sup f(z) dz Z B | | k∈ y∈x+Bk | k| Zy+Bk satisﬁes (3.13) and (3.14). Naturally, this is also a consequence of general results for spaces of homogeneous type, see [St2, Theorem 1 in Chapter 1]. Theorem 3.7. Suppose ϕ L∞(Rn) satisﬁes ϕ(x) C(1+ρ(x))−s for some s> 1, and let c = ϕ. Then for∈ f Lp(Rn), 1 p| |≤, ∈ ≤ ≤ ∞ n (3.16)R lim f ϕk(y)= cf(x) for a.e. x R . k→−∞ ∗ ∈ y∈x+Bk

Proof. Suppose p = 1 and f L1(Rn). By the Luzin (Luzin) Theorem given ε > 0 we can ﬁnd a continuous∈ function g with compact support such that f g <ε. Clearly || − ||1 n lim g ϕk(y)= cg(x) for all x R . k→−∞ ∗ ∈ y∈x+Bk 14 1. ANISOTROPIC HARDY SPACES

Since

lim sup f ϕk(y) cf(x) = lim sup f ϕk(y) g ϕk(x)+ g ϕk(x) cf(x) k→−∞ | ∗ − | k→−∞ | ∗ − ∗ ∗ − | y∈x+Bk y∈x+Bk

sup sup f ϕk(y) g ϕk(x) + c g(x) f(x) ≤ k∈Z y∈x+Bk | ∗ − ∗ | | || − | M (f g)(x)+ c f(x) g(x) , ≤ ϕ − | || − | by Theorem 3.6 and the Chebyshev (Qebyxev) inequality we have for any λ> 0

x : lim sup f ϕk(y) cf(x) > λ |{ k→−∞ | ∗ − | }| y∈x+Bk x : M (f g)(x) > λ/2 + x : c f(x) g(x) > λ/2 ≤ |{ ϕ − }| |{ | || − | }| C f g /λ < Cε/λ. ≤ || − ||1 Since λ> 0 is arbitrary we have (3.16). Suppose p> 1 and f Lp(Rn). Given j Z let g = f1 . Since g L1(Rn), ∈ ∈ Bj ∈

lim g ϕk(y)= cg(x)= cf(x) for a.e. x Bj . k→−∞ ∗ ∈ y∈x+Bk

If x Bj we can choose K Z so that x + BK+ω Bj . If y x + Bk for some ∈ c ∈ c ⊂ c ∈ k K and z (Bj ) x + (BK+ω) then y z (BK ) by (2.11). Hence by H¨older’s≤ inequality,∈ 1/p⊂+1/q = 1, − ∈

1/q q (f g) ϕk(y) = f(z)ϕk(y z)dz f p ϕk(y z) dz | − ∗ | c − ≤ || || c | − | Z(Bj ) Z(Bj ) 1/q 1/q q −k +k/q q f p ϕk (z) dz f pb ϕ(z) dz ≤ || || c | | ≤ || || c | | Z(BK ) Z(BK−k ) 1/q −k+k/q −qs −k+k/q (K−k)(−s+1/q) b f p ρ(z) dz Cb b ≤ || || c | | ≤ Z(BK−k) = CbK(−s+1/q)bk(s−1) 0 as k . → → −∞ Therefore, (3.16) holds for a.e. x Bj and since j is arbitrary it holds for a.e. x Rn. ∈ ∈

The next lemma is a basic approximation of identity result for the space of tempered distributions ′. Since it is a more general variant of the result common in the literature we includeS its proof. Lemma 3.8. Suppose ϕ and ϕ =1. Then for any ψ and f ′ we have ∈ S ∈ S ∈ S R (3.17) ψ ϕ ψ in as k , ∗ k → S → −∞ (3.18) f ϕ f in ′ as k . ∗ k → S → −∞

In Lemma 3.8 we can relax the condition that ϕk’s are given by (3.6). Instead, we only need to assume that ϕ (x) = det A ϕ(A x) for some nondegenerate k | k| k matrices A such that A−1 0 as k . k || k || → → −∞ 3. THE GRAND MAXIMAL DEFINITION OF ANISOTROPIC HARDY SPACES 15

Proof. It is clear that ψ ϕk(x) ψ(x) pointwise as k . Since α α ∗ α → → −∞ ∂ (ψ ϕk)(x) = (∂ ψ) ϕk(x) ∂ ψ(x) pointwise as k for every multi- index∗α, it suﬃces to show∗ → → −∞ N (3.19) sup x ψ ϕk(x) ψ(x) 0 as k , x∈Rn | | | ∗ − | → → −∞ for every integer N 0. Given ε> 0 ﬁnd≥δ > 0 so that x N ψ(y x) ψ(x) <ε for all y δ. Let K be such that | | | − − | | |≤ (1 + y )N ϕ (y) dy <ε for k K. | | | k | ≤ Z|y|>δ If k K ≤ N N x ψ ϕk(x) ψ(x) = x (ψ(y x) ψ(x))ϕk (y)dy | | | ∗ − | Rn | | − − Z

x N ψ(y x) ψ(x) ϕ (y) dy + x N ψ(y x ) ψ(x) ϕ (y) dy ≤ | | | − − || k | | | | − − || k | Z|y|≤δ Z|y|>δ N N ε ϕk(y) dy + sup x ψ(x) ϕk(y) dy ≤ Rn | | Rn | | | | | | Z x∈ Z|y|>δ + C x N (1 + x y )−N (1 + y )−N (1 + y )N ϕ (y) dy | | | − | | | | | | k | Z|y|>δ C′ε + C′ (1 + y )N ϕ (y) dy 2C′ε. ≤ | | | k | ≤ Z|y|>δ This shows (3.19) and thus (3.17). (3.18) follows from (3.17) since f ϕk, ψ = f, ψ ϕ˜ , whereϕ ˜ (x)= ϕ ( x). h ∗ i h ∗ ki k k − The next theorem can be thought as a converse to the Maximal Theorem 3.6 for p> 1. Theorem 3.9. Suppose f ′, ϕ , ϕ = 0, and 1 p . If M 0f ∈ S ∈ S 6 ≤ ≤ ∞ ϕ ∈ Lp(Rn) then f Lp(Rn). ∈ R Proof. Without loss of generality we can assume that ϕ = 1. Suppose p p > 1. Since the set f ϕk : k Z is bounded in L , by the Alaoglu Theorem { ∗ ∈ } R p there is a sequence kj such that f ϕkj converges weak- in L , and hence in ′. By Lemma 3.8 this→ −∞ limit is f and thus∗ f Lp. ∗ SSuppose now p = 1. By [Wo1, Theorem III.C.12]∈ the set f ϕ : k Z is { ∗ k ∈ } relatively weakly compact in L1 and by the Eberlein-Smulian˘ (Xmul n) Theorem there is a sequence k such that f ϕ converges weakly in L1, and hence j → −∞ ∗ kj in ′. By Lemma 3.8 this limit is f and thus f L1. Alternatively, we could think S ∈ of the set f ϕk : k Z as a bounded set in the space of ﬁnite complex Borel measures on{ R∗n and use∈ Alaoglu} Theorem to ﬁnd a sequence k such that j → −∞ f ϕkj converges weak- to an absolutely continuous measure. This measure is f(∗x)dx by Lemma 3.8. ∗

Remark. Theorems 3.6 and 3.9 together assert that for p > 1 the subclass of regular Lp integrable distributions in ′ is invariant with respect to maximal functions introduced in Deﬁnition 3.4. Indeed,S it follows from Theorems 3.6 and 3.9 that for 1

f is regular and belongs to Lp, • M 0f Lp (or M f Lp) for every ϕ , • ϕ ∈ ϕ ∈ ∈ S M 0f Lp (or M f Lp) for some ϕ with ϕ = 0, • ϕ ∈ ϕ ∈ 6 M 0 f Lp for some (or every) N 2. • N ∈ ≥ R The next result asserts that radial and nontangential grand maximal functions are pointwise equivalent. Proposition 3.10. For every N 0 there is a constant C = C(N) so that for all f ′, ≥ ∈ S (3.20) M 0 f(x) M f(x) CM 0 f(x) for x Rn. N ≤ N ≤ N ∈ Proof. The ﬁrst inequality is obvious. To see the second inequality, note that M f(x) = sup (f ϕ )(x + Aky) : k Z,y B , ϕ N {| ∗ k | ∈ ∈ 0 ∈ SN } (3.21) = sup (f φ )(x) : k Z, φ(z)= ϕ(z + y) for some y B , ϕ {| ∗ k | ∈ ∈ 0 ∈ SN } = sup M 0(x) : φ(z)= ϕ(z + y) for some y B , ϕ . { φ ∈ 0 ∈ SN } By (2.12) if φ(x)= ϕ(x + y) for some y B then ∈ 0 N α φ SN = sup sup max(1,ρ(x) ) ∂ ϕ(x + y) || || x∈Rn |α|≤N | | N α (3.22) = sup sup max(1,ρ(x y) ) ∂ ϕ(x) x∈Rn |α|≤N − | | ωN N α ωN b sup sup max(1,ρ(x) ) ∂ ϕ(x) = b ϕ SN . ≤ x∈Rn |α|≤N | | || || Combining (3.21) and (3.22) we have M f(x) sup M 0f(x) : φ , φ bωN bωN M 0 f(x), N ≤ { φ ∈ S || ||SN ≤ }≤ N which shows (3.20).

We are now ready to state the definition of anisotropic Hardy spaces. Definition 3.11. For a given dilation A and 0 1. For every N Np we define the anisotropic Hardy space associated with the dilation A as ≥ (3.23) Hp(Rn)= Hp (Rn)= f ′ : M f Lp , A { ∈ S N ∈ } with the quasi-norm f p = M f . || ||H || N ||p Since the dimension n and the dilation A remain constant throughout this chapter (except Section 10) we are going to denote the anisotropic Hardy space by p p p H or H(N). Even though quasi-norms on H depend on the choice of N, it follows from Theorems 4.2 and 6.4 that the above definition of Hp does not depend on the choice of N as long as N N . To escape possible ambiguity and to fix the ≥ p attention the reader can think that N = Np in (3.23). By the above Remark we have Hp = Lp for p > 1 irrespective of the dilation A. Moreover, by Theorem 3.9 we have H1 L1. Therefore, only for 0

Proposition 3.12. The space Hp is complete. Proof. Since Hp = Lp for p> 1 we only need to consider p 1. ′ ≤ For every ϕ and every sequence (fi)i∈N in such that i fi converges in ′ ∈ S S to the tempered distribution f, the series i fi ϕ(x) converges pointwise to fS ϕ(x) for each x Rn. Thus, ∗ P ∗ ∈ P p M f(x)p M f (x) (M f (x))p for all x Rn, N ≤ N i ≤ N i ∈ i i X X p p and we have f Hp i fi Hp . || || ≤p || || To prove that H is complete, it suﬃces to show that for every sequence (fi) −Pi p such that fi Hp < 2 for i N, the series i fi converges in H . Since partial || || p∈ ′ sums i fi are Cauchy in H , hence they are Cauchy in , and i fi converges in ′ to some f, because ′ is complete. Therefore,P S S P S P j ∞ ∞ ∞ p p p −ip f f p = f p f p 2 0 as j . || − i||H || i||H ≤ || i||H ≤ → → ∞ i=1 i=j+1 i=j+1 i=j+1 X X X X This ﬁnishes the proof. 18 1. ANISOTROPIC HARDY SPACES

4. The atomic deﬁnition of anisotropic Hardy spaces Definition 4.1. We say a triplet (p,q,s) is admissible (with respect to the dilation A) if 0

Proof. Since N Np , we have MN a(x) MNp a(x), and we can assume S ⊂ S ≤ ˜ 0 that N = Np. By Proposition 3.10 it suffices to show Ma p C, where M = MN is the radial grand maximal function. It also suffices to|| consi|| der≤ only (p,q,s)-atoms a with minimal value of s = (1/p 1) ln b/ ln λ− . ⌊ − ⌋ n Suppose an atom a is associated with the ball x0 + Bj for some x0 R , j Z. c ∈ ∈ We estimate separately on x0 + Bj+ω and (x0 + Bj+ω) . Case I. On Bj+ω we use the Maximal Theorem 3.6 for q> 1 and Hölder’s inequality

p/q Ma(x)pdx Ma(x)q dx B 1−p/q ≤ | j+ω| Zx+Bj+ω Zx+Bj+ω C a p B 1−p/q C. ≤ || ||q | j | ≤ If q = 1, take λ> 0, and consider Ω = x : Ma(x) > λ . Then λ { } Ω B min( Ω , B ) min(C a /λ, B ) | λ ∩ j+ω|≤ | λ| | j+ω| ≤ || ||1 | j+ω| = min(C B 1−1/p/λ,bω B ) C min( B 1−1/p/λ, B ). | j | | j | ≤ | j | | j | −1/p We have equality between the two terms in the last minimum if λ = Bj . Therefore, | | ∞ Ma(x)pdx = pλp−1 Ω B dλ | λ ∩ j+ω| ZBj+ω Z0 −1/p |Bj | ∞ p−1 1−1/p p−2 C Bj pλ dλ + Bj pλ dλ = C. ≤ | | −1/p | | Z0 Z|Bj | c Case II. Take x (x0 + Bj+ω) , ϕ N , and k Z. Suppose P is a polynomial of degree s to be∈ speciﬁed later. Then∈ S we have ∈ ≤ −k −k (a ϕk)(x) = b a(y)ϕ(A (x y))dy | ∗ | Rn − Z

= b−k a(y)(ϕ(A−k(x y)) P (A−k(x y)))dy − − − Zx0+Bj

4. THE ATOMIC DEFINITION OF ANISOTROPIC HARDY SPACES 19

1/q′ ′ b−k a ϕ(A−k(x y)) P (A−k(x y)) q dy ≤ || ||q | − − − | Zx0+Bj 1/q′ ′ ′ b−kbj(1/q−1/p)bk/q ϕ(y) P (y) q dy (4.4) ≤ A−k(x−x )+B | − | Z 0 j−k ′ ′ b−kbj(1/q−1/p)bk/q b(j−k)/q sup ϕ(y) P (y) −k ≤ y∈A (x−x0)+Bj−k | − | = b−j/pbj−k sup ϕ(y) P (y) , −k y∈A (x−x0)+Bj−k | − | where q′ denotes the conjugate power to q, 1/q +1/q′ = 1. Even though the above calculation holds for q > 1 it can be easily adjusted for the case q = 1 to yield the same estimate. −k Suppose x x0 + Bj+ω+m+1 Bj+ω+m for some integer m 0. Then A (x ∈−k \ j−k ≥ − x0)+Bj−k A (Bj+ω+m+1 Bj+ω+m)+Bj−k = A ((Bω+m+1 Bω+m)+B0) j−k c⊂ \ \ ⊂ A (Bm) by virtue of (2.11). We consider two cases. If j k then we choose polynomial P = 0, and ≥

(4.5) sup ϕ(y) sup min(1,ρ(y)−N ) b−N(j−k+m). −k −k y∈A (x−x0)+Bj−k | |≤ y∈A (x−x0)+Bj−k ≤

If j < k then we choose polynomial P to be the Taylor expansion of ϕ at the point A−k(x x ) of order s. By the Taylor Remainder Theorem and (2.2) we have − 0 sup ϕ(y) P (y) −k y∈A (x−x0)+Bj−k | − | C sup sup ∂αϕ(A−k(x x )+ z) z s+1 ≤ | − 0 || | (4.6) z∈Bj−k |α|=s+1 (s+1)(j−k) −N Cλ− sup min(1,ρ(y) ) −k ≤ y∈A (x−x0)+Bj−k Cλ(s+1)(j−k) min(1,b−N(j−k+m)). ≤ − Combining (4.4), (4.5) and (4.6) we have for x x + B B , m 0, ∈ 0 j+ω+m+1 \ j+ω+m ≥

p p −j p(j−k) −Np(j−k+m) Ma(x) = sup sup (a ϕk)(x) b max sup b b Z | ∗ | ≤ Z ϕ∈SN k∈ k∈ ,k≤j p(j−k) p(s+1)(j−k) −Np(j−k+m) +C sup b λ− min(1,b ) . Z k∈ ,k>j Note that the supremum over k j has the largest value for k = j. Since N s+2 s+1 N ≤ ≥ we have bλ− b , and the supremum over k > j is attained when j k + m = 0. Indeed, it suﬃces≤ to check for the maximum value in the range j < k − j + m and k j + m. Therefore, ≤ ≥ Ma(x)p b−j max(b−Npm, C(bλs+1)−pm) Cb−j (bλs+1)−pm, ≤ − ≤ − 20 1. ANISOTROPIC HARDY SPACES again by bλs+1b−N 1. Since s = (1/p 1) ln b/ ln λ , we have s > (1/p − ≤ ⌊ − −⌋ − 1) ln b/ ln λ 1, and thus λs+1b1−1/p > 1. Therefore, − − − ∞ p p Ma(x) dx = Bj+ω+m+1 Bj+ω+m sup Ma(x) c | \ | Z(x0+Bj ) m=0 x∈x0+Bj+ω+m+1\Bj+ω+m ∞ X ∞ C bj+ω+m+1b−j (bλs+1)−pm = Cbω+1 (bp−1λ(s+1)p)−m = C′ < . ≤ − − ∞ m=0 m=0 X X

Combining case I and II and we see that Ma p C˜ for some constant C˜ independent of the choice of a (p,q,s)-atom a. || || ≤

The proof of this result could be simplified by virtue of Theorem 7.1, since it is 0 easier to estimate Mϕf than MN f. We are now ready to introduce the anisotropic Hardy spaces in terms of atomic decompositions. Definition 4.3. For a given dilation A and an admissible triplet (p,q,s) we p define the atomic anisotropic Hardy space Hq,s associated with dilation A as the ′ ∞ set of all tempered distributions f of the form i=1 κiai with convergence in ′ ∞ p ∈ S , where i=1 κi < , and ai is a (p,q,s)-atom for each i N. The quasi-norm Sof f Hp is defined| | as∞ P ∈ ∈ q,sP ∞ 1/p ∞ p f p = inf κ : f = κ a , a is a (p,q,s)-atom for i N . || ||Hq,s | i| i i i ∈ i=1 i=1 X X

∞ ′ It follows from Theorem 4.5 that the series i=1 κiai converges in for every ∞ p S choice of κi and (p,q,s)-atoms ai with i=1 κi < . The representation of f ′ ∞ | P| ∞ ∈ in this form κiai is referred as an atomic decomposition of f. Furthermore, S i=1 P if f p = 0 then necessarily f = 0. || ||Hq,s P Proposition 4.4. Suppose the triplet (p,q,s) is admissible. Then the space p Hq,s is complete. The proof of Proposition 4.4 is a routine, see Proposition 3.12.

p Theorem 4.5. If the triplet (p,q,s) is admissible and N Np then Hq,s p ′ p ≥ ⊂ H , where H is deﬁned using maximal function MN . Moreover, the inclusion maps⊂ are S continuous. Proof. Suppose f Hp has an atomic decomposition f = ∞ κ a . Then ∈ q,s i=1 i i ∞ ∞ P ∞ p p p p ˜ p f Hp = (MN ( κiai)(x)) dx κi MN ai(x) dx C κi , || || Rn ≤ | | Rn ≤ | | i=1 i=1 i=1 Z X X Z X ˜ ∞ p 1/p where C is the constant in Theorem 4.2. Since we can choose ( i=1 κi ) 1/p | | p p ˜ p to be arbitrarily close to f Hq,s thus f H C f Hq,s , and the inclusion p p || || || || ≤ || || P .Hq,s ֒ H is continuous Suppose→ next f Hp, and ϕ . ∈ ∈ S f, ϕ = f ϕ˜(0) M f(x) for x B , |h i| | ∗ |≤ ϕ˜ ∈ 0 4. THE ATOMIC DEFINITION OF ANISOTROPIC HARDY SPACES 21 whereϕ ˜(x)= ϕ( x). Therefore, − p p p p f, ϕ = B0 f, ϕ Mϕ˜f(x) dx Mϕ˜f(x) dx |h i| | ||h i| ≤ ≤ Rn ZB0 Z p p p p ϕ˜ M f(x) dx = ϕ˜ f p . SN N SN H ≤ || || Rn || || || || Z p If a sequence (fi) converges to f in H , then fi, ϕ f, ϕ as i , and the ∞ → inclusion Hp ֒ ′ is continuous. h i → h i → S 22 1. ANISOTROPIC HARDY SPACES

5. The Calderón-Zygmund decomposition for the grand maximal function In this section we define and investigate extensively the Calderón-Zygmund decomposition. The excellent exposition of this decomposition in the setting of homogeneous groups is in [FoS]. Even though we work only in Rn we allow much more general dilation structures than in [FoS] and therefore we need to build the machinery from scratch. We try to follow as closely as possible the skeleton of the construction in [FoS] to give the reader familiar with this book a comfortable path through numerous technical nuances. Throughout this section we consider a tempered distribution f so that x : Mf(x) > λ < for all λ> 0, |{ }| ∞ where M = MN for some fixed integer N 2. Later with regard to the Hardy p ≥ space H (0

0≤ we set Ω= x : Mf(x) > λ . The≥ Whitney⌊ Lemma− applied⌋ to Ω { } with d =4ω yields a sequence (xj )j∈N Ω, and a sequence of integers (lj )j∈N, so that ⊂

(5.1) Ω= (xj + Blj ), N j[∈

(5.2) (x + B ) (x + B )= for i = j, i li−ω ∩ j lj −ω ∅ 6 c c (5.3) (x + B ) Ω = , (x + B ) Ω = for j N, j lj +γ−1 ∩ ∅ j lj +γ ∩ 6 ∅ ∈

(5.4) (x + B ) (x + B ) = = l l ω, i li+2ω ∩ j lj +2ω 6 ∅ ⇒ | i − j |≤

(5.5) # j N : (x + B ) (x + B ) = L for i N, { ∈ i li+2ω ∩ j lj +2ω 6 ∅} ≤ ∈ where γ := d +1=4ω + 1, and L is some constant independent of Ω. ∞ n Fix θ C (R ) such that supp θ Bω, 0 θ 1, and θ 1 on B0. For every j N∈ deﬁne ⊂ ≤ ≤ ≡ ∈ (5.6) θ (x)= θ(A−lj (x x )). j − j One should think of θj as the localized version of θ to the scale lj centered at xj , and corresponding to the ball xj + Blj in the Whitney decomposition. Clearly supp θ x + B , θ 1 on x + B , and by (5.1) and (5.5) j ⊂ j lj +ω j ≡ j lj

(5.7) 1 θj (x) L for x Ω. ≤ N ≤ ∈ Xj∈ For every i N deﬁne ∈ θi(x)/ j θj (x) x Ω, (5.8) ζi(x)= ∈ 0 x Ω. P 6∈ ∞ n We have ζi C (R ), supp ζi xi + Bli+ω, 0 ζi 1, ζi 1 on xi + Bli−ω by (5.2), and ∈ ζ = 1 . Therefore,⊂ the family≤ ζ ≤forms a≡ smooth partition of i∈N i Ω { i} unity which is subordinate to the covering of Ω by the balls xi + Bl +ω . P { i } 5. THE CALDERON-ZYGMUND´ DECOMPOSITION 23

Let s denote the linear space of polynomials in n variables of degree s, where s P 0 is a fixed integer. For each i N we introduce the norm in the space≤ by setting≥ ∈ Ps 1/2 1 2 (5.9) P = P (x) ζi(x)dx for P s, || || ζ Rn | | ∈P i Z ′ which makes s a finite dimensionalR Hilbert space. The distribution f induces a linear functionalP on by ∈ S Ps 1 Q f,Qζi for Q s, 7→ ζi h i ∈P which is represented by the RieszR Lemma by the unique polynomial Pi s such that ∈ P 1 1 1 f,Qζi = Pi, Qζi = Pi(x)Q(x)ζi(x)dx for all Q s. ζ h i ζ h i ζ Rn ∈P i i i Z ForR every i N defineR distributionsR bi := (f Pi)ζi. ∈ − We will show that for a suitable choice of s and N the series i bi converges ′ in . Then we can define g := f bi. S − i P Definition 5.1. The representationP f = g + bi, where g and bi are as above is a Calderón-Zygmund decomposition of degree s and height λ associated P with MN f.

Intuitively, one should think of g as the good part of f and of bi’s as the bad parts of f. For a suitable choice of s and N, the good part g behaves very nicely. In particular, g is in L1 and, in addition, it is in L∞ whenever f is in L1, see Lemma 5.10. On the other hand, the bad parts bi’s are well-localized on the balls coming from the Whitney decomposition; they have a certain number of vanishing moments, and thus they can be nicely controlled in terms of the grand maximal function of f, see Lemma 5.7. The rest of this section consists of a series of lemmas. In Lemmas 5.2 and 5.3 we show properties of the smooth partition of unity ζ . In Lemmas 5.4, 5.6, 5.7, { i} and 5.8 we derive the estimates for the bad parts bi’s using Lemma 5.5, which is a result about Taylor polynomials of independent interest. Lemmas 5.9 and 5.10 give controls for the good part g. Finally, the culmination of this section is Corollary 5.11 showing the density of L1 Hp functions in Hp spaces. ∩ Lemma 5.2. There exists a constant A1 > 0 depending only on N, so that for all i N and l l ∈ ≤ i α l (5.10) sup sup ∂ ζ˜(x) A1, where ζ˜(x) := ζi(A x). |α|≤N x∈Rn | |≤

Proof. For a given i N consider J = j N : (xi + Bli+ω) (xj + Blj +ω) = ∈ ˜ { ∈−l ∩ 6 . Since supp ζi xi + Bli+ω, thus supp ζ A xi + Bli−l+ω. Therefore, we need ∅} ⊂ −l ⊂ l to consider only x A xi +Bli−l+ω in the supremum in (5.10). If A x xi +Bli+ω then θ (Alx)=0 for∈ j N J and we have ∈ j ∈ \ l l l−li −li l θi(A x) θi(A x) θ(A x A xi) ζ˜(x)= ζi(A x)= = = − . θ (Alx) θ (Alx) θ(Al−lj x A−lj x ) j∈N j j∈J j j∈J − j P P P 24 1. ANISOTROPIC HARDY SPACES

By (5.7) the denominator is 1 and has at most L terms by (5.5). Furthermore, ≥ for j J we have l lj li lj ω by (5.4). The estimate (5.10) follows now from the∈ iterative application− ≤ − of the≤ quotient rule combined with (5.11) sup sup sup ∂α(θ(Aj ))(x) = C < . x∈Rn j≤ω |α|≤N | · | ∞ To ﬁnish the proof we must show (5.11). For a given C∞ function Θ on Rn and an integer k 0, DkΘ(x) denotes the derivative of Θ of order k at the point x thought of as≥ a symmetric, multilinear functional, i.e., DkΘ(x) : (Rn)k = Rn . . . Rn R. The norm is given by × × → k k D Θ(x) = sup D Θ(x)(v1,... ,vk) . Rn || || vi∈ ,|vi|=1, | | i=1,... ,k n Suppose e1,... ,en is the standard basis of R , and σ is any mapping from 1,...,k { to 1,...,n} . Then { } { } α1 αn ∂ ∂ (α1,... ,αn) Θ(x)(eσ(1),... ,eσ(k))= α1 αn Θ(x)= ∂ Θ(x), ∂x1 · · · ∂xn j where αj = # i : σ(i) = j . If Θ(x) = θ(A x) then by the chain rule for any integer k 1 { } ≥ k k j j j D Θ(x)(v1,... ,vk)= D θ(A x)(A v1,... ,A vk), n for any vectors v1,... ,vk R . For an integer j ω denote∈ Θ(x)= θ(Aj x). Then ≤ sup sup ∂αΘ(x) sup sup D|α|Θ(x) sup sup D|α|θ(Aj x) Aj |α| x∈Rn |α|≤N | |≤ x∈Rn |α|≤N || || ≤ x∈Rn |α|≤N || |||| || (sup Aj )N sup sup Dkθ(x) = C < . ≤ j≤ω || || x∈Rn k=0,... ,N || || ∞ j ω Indeed, the above suprema are ﬁnite since for j ω we have A c(λ+) by (2.1) and (2.2), and θ is C∞ with compact support.≤ Thus, (5.11)|| holds|| ≤ and hence (5.10).

′ Lemma 5.3. There exists a constant A2 > 0 independent of f , i N, and λ> 0 so that ∈ S ∈

(5.12) sup Pi(y)ζi(y) A2λ. y∈Rn | |≤

Proof. Let π1,... ,πm (m = dim s) be an orthonormal basis of s with respect to the norm (5.9). We have P P m 1 (5.13) Pi = f(x)πk(x)ζi(x)dx πk, ζi Xk=1 Z where the integral is understoodR as f, π ζ . Hence h k ii 1 2 1 2 1= πk(x) ζi(x)dx πk(x) ζi(x)dx ζi | | ≥ Bli+ω x +B −ω | | (5.14) Z | | Z i li R = b−ω π˜ (x) 2dx, | k | ZB−ω 5. THE CALDERON-ZYGMUND´ DECOMPOSITION 25 whereπ ˜ (x) = π (x + Ali x). Since is ﬁnite dimensional all norms on are k k i Ps Ps equivalent, there exists C1 > 0 such that

1/2 sup sup ∂αP (z) C P (z) 2dz for all P . | |≤ 1 | | ∈Ps |α|≤s z∈Bω ZB−ω Therefore by (5.14),

α ω/2 (5.15) sup sup ∂ π˜k(z) C1b for k =1,... ,m. |α|≤s z∈Bω | |≤ For k =1,...,m deﬁne

li b li li Φk(y)= πk(z A y)ζi(z A y), ζi − − c where z is some point in (xi + BR li+γ ) Ω by (5.3). We claim there exists a constant ∩C > 0 such that 1/C Φ for all k = 2 2 k ∈ SN 1,... ,m. Indeed, if y supp Φ , then z Ali y x + B , Ali y x z + ∈ j − ∈ i li+ω − ∈ i − Bli+ω Bli+γ + Bli+ω, so y Bγ Bω. Hence supp Φk is bounded. We⊂− can write ∈ − l l i i i b li l b −li −li Φk(y)= πk(xi +˜z A y)ζi(xi +˜z A y)= π˜k(A z˜ y)ζ˜i(A z˜ y), ζi − − ζi − −

li li whereπ ˜k(Rx)= πk(xi + A x), ζ˜i(x)= ζi(xi + A x)R and z = xi +˜z. Consider

li li θi(xi + A x) θ(x) ζ˜i(x)= ζi(xi + A x)= = . θ (x + Ali x) θ(A−lj (x x )+ Ali−lj x) j j i j i − j Clearly supp ζ˜ B and byP Lemma 5.2 P i ⊂ ω α sup sup ∂ ζ˜i(z) A1. |α|≤N z∈Rn | |≤

By the product rule, (5.15) and supp ζ˜ B we can ﬁnd a constant C so that i ⊂ ω 3 α sup sup ∂ (˜πk(z)ζ˜i(z)) C3. |α|≤N z∈Rn | |≤

Since bli / ζ bli /bli−ω = bω, supp Φ B + B , by the above estimate and the i ≤ k ⊂ γ ω deﬁnition of Φk we can ﬁnd a constant C2 so that Φk SN C2. Since R || || ≤

(f (Φ ) )(z)= f(y)b−li Φ (A−li (z y))dy ∗ k li k − Z 1 = f(y)π (z Ali (A−li (z y)))ζ (z Ali (A−li (z y)))dy ζ k − − i − − i Z 1 R = f(y)π (y)ζ (y)dy, ζ k i i Z we have R 1 f(y)π (y)ζ (y)dy Mf(z) Φ C λ. ζ k i ≤ || k||SN ≤ 2 i Z

R 26 1. ANISOTROPIC HARDY SPACES

By (5.13), (5.15) and the above estimate (5.16) sup P (z) mC λC bω/2, | i |≤ 2 1 z∈xi+Bli+ω

ω/2 and therefore we have (5.12) with A2 = mb C1C2.

Lemma 5.4. There exists a constant A3 > 0 such that (5.17) Mb (x) A Mf(x) for x x + B . i ≤ 3 ∈ i li+2ω

Proof. Take ϕ N , and x xi + Bli+2ω. Case I. For l l we∈ write S ∈ ≤ i (5.18) (b ϕ )(x) = (f Φ )(x) ((P ζ ) ϕ )(x), i ∗ l ∗ l − i i ∗ l where Φ(z) := ϕ(z)ζ (x Alz). Deﬁne ζ˜ (z) := ζ (x Alz). By Lemma 5.2 we have i − i i − α sup sup ∂ ζ˜i(z) A1. |α|≤N z∈Rn | |≤ By the product rule there is a constant C depending only on N so that N α Φ SN = sup max(1,ρ(z) ) sup ∂ Φ(z) || || z∈Rn |α|≤N | | N α = sup max(1,ρ(z) ) sup ∂ (ϕζ˜i)(z) z∈Rn |α|≤N | | N α α C sup max(1,ρ(z) ) sup ∂ ϕ(z) sup ∂ ζ˜i(z) A1C. ≤ z∈Rn |α|≤N | | |α|≤N | |≤

′ ′ Note that for N 2 there is a constant C > 0 so that ϕ 1 C for all ϕ N . Therefore, by Lemma≥ 5.3 and (5.18), we have || || ≤ ∈ S b ϕ (x) Φ Mf(x)+ A λ ϕ A CMf(x)+ A C′λ | i ∗ l | ≤ || ||SN 2 || ||1 ≤ 1 2 (A C + A C′)Mf(x), ≤ 1 2 since Mf(x) > λ for x Ω. ∈ Case II. For l>li by a simple calculation we can write (b ϕ )(x)= bli−l(f Φ )(x) ((P ζ ) ϕ )(x), i ∗ l ∗ li − i i ∗ l where Φ(z) := ϕ(Ali−lz)ζ (x Ali z). Defineϕ ˜(z) := ϕ(Ali−lz) and ζ˜ (z) := ζ (x i − i i − Ali z). If z supp Φ then x Ali z x +B , so Ali z x x +B B + ∈ − ∈ i li+ω ∈ − i li+ω ⊂ li+2ω B B . Hence, supp Φ B . By Lemma 5.2 and since Ali−l c by li+ω ⊂ li+3ω ⊂ 3ω || || ≤ (2.2) we can find a constant C > 0 independent of l>li so that α α sup sup ∂ ϕ˜(z) C, sup sup ∂ ζ˜i(z) A1. |α|≤N z∈Rn | |≤ |α|≤N z∈Rn | |≤ By the product rule and the boundedness of the support of Φ we can find a constant ′′ ′′ ′ ′ C so that Φ SN C . Let C be the constant such that ϕ 1 C for ϕ N for N 2. As|| || in the≤ case I || || ≤ ∈ S ≥ (b ϕ )(x) bli−l Φ Mf(x)+ A λ ϕ (C′′ + A C′)Mf(x). | i ∗ l |≤ || ||SN 2 || ||1 ≤ 2 By combining both cases we arrive at (5.17). 5. THE CALDERON-ZYGMUND´ DECOMPOSITION 27

Lemma 5.5. Suppose Q Rn is bounded, convex, and 0 Q, and N is a positive integer. Then there is⊂ a constant C depending only on Q∈and N such that for every φ and every integer s, 0 s

∂βφ(y) φ(y + z)= zβ + R (z). β! y |βX|≤s For any multi-index α s we have | |≤ ∂βφ(y) ∂α+βφ(y) ∂αR (z)= ∂αφ(y + z) zβ−α = ∂αφ(y + z) zβ, y − (β α)! − β! |β|≤s, − |β|≤s−|α| Xβ≥α X where (α1,... ,αn)= α β = (β1,... ,βn) means αi βi for all i =1,... ,n. We used here ≤ ≤ 1 zβ−α/(β α)! α β, ∂βzα = − ≤ β! 0 otherwise. α α Therefore, ∂ Ry is the Taylor remainder of ∂ φ of order s α expanded at y and by the Taylor Remainder Theorem − | |

α α+β s−|α|+1 ∂ Ry(z) C sup sup ∂ φ(w) z , | |≤ w∈[y,y+z] |β|=s−|α|+1 | || | where [y,y + z] is the line segment connecting y and y + z. Thus,

α ′ α sup ∂ Ry(z) C sup sup ∂ φ(w) . z∈Q | |≤ w∈y+Q |α|=s+1 | |

α α If s < α N then ∂ Ry(z) = ∂ φ(y + z), and the estimate (5.19) follows immediately.| | ≤

Lemma 5.6. Suppose 0 s 0 so that for i N and all integers t ≤0 ∈ ≥ (5.20) Mb (x) A λλ−t(s+1) for x x + B B . i ≤ 4 − ∈ i t+li+2ω+1 \ t+li+2ω

c Proof. Suppose ϕ N , and l Z. Pick some w (xi + Bli+γ) Ω . Case I. For l l we have∈ S ∈ ∈ ∩ ≤ i (5.21) b ϕ (x)= f Φ (w) (P ζ ) ϕ (x), i ∗ l ∗ l − i i ∗ l where Φ(z) := ϕ(z+A−l(x w))ζ (w Alz). If z supp Φ then z supp ζ (w Al ), − i − ∈ ∈ i − · i.e., w Alz x + Ali B , so z Ali−l(B + A−li (w x )) Ali−l(B B ) − ∈ i ω ∈ ω − i ⊂ ω − γ ⊂ 28 1. ANISOTROPIC HARDY SPACES

Ali−lB . Suppose x x + B B for some integer t 0. If γ+ω ∈ i t+li+2ω+1 \ t+li+2ω ≥ z supp Φ then z Ali−lB + A−l(w x ) and by (2.11) ∈ ∈ ω − i z + A−l(x w) Ali−lB + A−l(w x )+ A−l(x w) − ∈ ω − i − li−l −l li−l (5.22) = A Bω + A (x xi) A (Bω + Bt+2ω+1 Bt+2ω) c − ⊂ c \ c Ali−l((B ) + B ) Ali−l((B ) ) = (B ) . ⊂ t+2ω ω ⊂ t+ω t+li−l+ω By Lemma 5.2 the partial derivatives of ζ (w Al ) of order up to N are bounded i − · by A1. Using supp Φ Bli−l+γ+ω, ϕ SN 1, (5.22) and the product rule we have ⊂ || || ≤ N α Φ SN = sup sup max(1,ρ(z) ) ∂ Φ(z) || || z∈supp Φ |α|≤N | | (bli−l+γ+ω)N A C sup sup ∂αϕ(z + A−l(x w)) ≤ 1 | − | (5.23) |α|≤N z∈supp Φ N(li−l+γ+ω) −l −N A1Cb sup max(1,ρ(z + A (x w))) l −l −l ≤ z∈A i Bω+A (w−xi) − A CbN(li−l+γ+ω)b−N(t+li−l+ω) = A CbNγb−Nt. ≤ 1 1 This enables us to estimate the ﬁrst term in (5.21). For the second term, note c that if x xi + (Bt+li+2ω) for some integer t 0, and y xi + Bli+ω then ∈ c c≥ ∈ A−l(x y) Ali−l((B ) B ) (B ) by (2.11). Since ϕ 1 − ∈ t+2ω − ω ⊂ t+li−l+ω || ||SN ≤ (P ζ ) ϕ (x) = (P ζ )(y)b−lϕ(A−l(x y))dy | i i ∗ l | i i − Z

−l −l b (Piζi)(y) ϕ(A (x y)) dy ≤ x +B | || − | Z i li+ω b−lbli+ωA λ(bt+li−l+ω)−N A λb−Nt. ≤ 2 ≤ 2 Combining the above, (5.21) and (5.23) we obtain b ϕ (x) Φ Mf(w)+ A λb−Nt (A CbNγ + A )λb−Nt. | i ∗ l | ≤ || ||SN 2 ≤ 1 2 li−l Case II. For ﬁxed l>li and ϕ N deﬁne φ(z) = ϕ(A z). Suppose x x + B B for some∈ integer S t 0. Consider the Taylor expansion∈ i t+li+2ω+1 \ t+li+2ω ≥ of φ of order s at the point y := A−li (x w), ∈ S − ∂αφ(y) φ(y + z)= zα + R (z), α! y |αX|≤s where Ry denotes the reminder. Since the distribution bi annihilates polynomials of degree s we have ≤ (b ϕ )(x)= b−l b (z)ϕ(A−l(x z))dz = b−l b (z)φ(A−li (x z))dz i ∗ l i − i − Z Z −l −li (5.24) = b b (z)R −l (A (w z))dz i A i (x−w) − Z li−l −l −li = b (f Φ )(w)+ b P (z)ζ (z)R −l (A (w z))dz, ∗ li i i A i (x−w) − Z where

li (5.25) Φ(z) := R −l (z)ζ (w A z). A i (x−w) i − 5. THE CALDERON-ZYGMUND´ DECOMPOSITION 29

li If z supp Φ then w A z xi + Bli+ω, so z Bγ + Bω and supp Φ Bγ+ω. ∈ − ∈ li−l ∈ −li ⊂ Apply Lemma 5.5 to φ(z)= ϕ(A z), y = A (x w) and Q = Bγ+ω. First assume t γ + ω and x x + B B . Since− ≥ ∈ i t+li+2ω+1 \ t+li+2ω −li y + Bγ+ω A ((Bt+li+2ω+1 Bt+li+2ω)+ Bli+γ )+ Bγ+ω ⊂ \ c c = (B B )+ B (B ) + B (B ) , t+2ω+1 \ t+2ω γ+2ω ⊂ t+2ω t+ω ⊂ t+ω we have α α sup sup ∂ Ry(z) C sup sup ∂ φ(z) z∈Bγ+ω |α|≤N | |≤ z∈y+Bγ+ω s+1≤|α|≤N | |

(li−l)(|α|+1) α li−l C sup sup λ− ∂ ϕ(A z) ≤ z∈y+Bγ+ω s+1≤|α|≤N | |

(li−l)(s+1) α li−l Cλ− sup sup ∂ ϕ(A z) ≤ z∈y+Bγ+ω s+1≤|α|≤N | |

(li−l)(s+1) li−l −N Cλ− sup max(1,ρ(A z)) c ≤ z∈(Bt+ω)

(li−l)(s+1) −N(li−l+t+ω) = Cλ− min(1,b ).

s+1 −N Since λ− b 1 the last expression is maximized over l>li if li l + t + ω = 0, i.e., l = l + t +≤ω. Therefore, we have for t γ + ω − i ≥ α −t(s+1) sup sup ∂ Ry(z) Cλ− . z∈Bγ+ω |α|≤N | |≤

If 0 t<γ + ω and x B B we can estimate as before ≤ ∈ t+li+2ω+1 \ t+li+2ω

α (li−l)(s+1) α li−l sup sup ∂ Ry(z) Cλ− sup sup ∂ ϕ(A z) C. z∈Bγ+ω |α|≤N | |≤ z∈y+Bγ+ω s+1≤|α|≤N | |≤ Therefore, we have for all t 0 ≥ α (γ+ω)(s+1) −t(s+1) (5.26) sup sup ∂ Ry(z) Cλ− λ− . z∈Bγ+ω |α|≤N | |≤ By (5.25), (5.26), Lemma 5.2 and the product rule we can ﬁnd a constant C′ so that Φ C′λ−t(s+1) for t 0. || ||SN ≤ − ≥ Therefore by (5.24),

li−l −l −li (b ϕ )(x) b f Φ (w) + b P (z)ζ (z)R −l (A (w z)) dz | i ∗ l |≤ | ∗ li | | i i A i (x−w) − | Z li−l −l li+ω b Mf(w) Φ SN + b b A2λ sup RA−li (x−w)(z) . ≤ || || z∈Bγ+ω | | Thus (b ϕ )(x) Cλλ−t(s+1) for x B B , t 0. | i ∗ l |≤ − ∈ t+li+2ω+1 \ t+li+ω ≥ This ends the proof of case II. Combining the two cases we arrive at the estimate

(b ϕ )(x) A λλ−t(s+1) for all l Z. | i ∗ l |≤ 4 − ∈ Hence we have (5.20). 30 1. ANISOTROPIC HARDY SPACES

The above proof also works in the case s N and yields the estimate ≥ Mb (x) A λλ−tN for x x + B B , i ≤ 4 − ∈ i t+li+2ω+1 \ t+li+2ω for all i N and all integers t 0. ∈ ≥ p Lemma 5.7. Suppose 0 s and f H , p ≤ ≥ ⌊ ⌋ ∈ where H is given by (3.23). Then there exists a constant A5 independent of f Hp, i N, and λ> 0 such that ∈ ∈ p p (5.27) Mbi(x) dx A5 Mf(x) dx. Rn ≤ x +B Z Z i li+2ω p Moreover, the series i bi converges in H , and

P p p (5.28) M bi (x) dx LA5 Mf(x) dx, Rn ≤ i Ω Z X Z where L is as in (5.5). Proof. By Lemma 5.4

p p p p Mbi(x) dx (A3) Mf(x) dx + Mbi(x) dx. Rn ≤ x +B (x +B )c Z Z i li+2ω Z i li+2ω By Lemma 5.6

∞ p p Mbi(x) dx = Mbi(x) dx (x +B )c x +B \B Z i li+2ω t=0 Z i t+li+2ω+1 t+li+2ω ∞ X ∞ −t(s+1)p −t(s+1)p bt+li+2ω+1(A )pλpλ (A )pb btλ Mf(x)pdx ≤ 4 − ≤ 4 − t=0 t=0 xi+Bli+2ω X X Z C′ Mf(x)pdx, ≤ x +B Z i li+2ω (s+1)p because Mf(x) > λ for x xi + Bli+2ω and λ− > b. Combining the last two estimates we obtain (5.27).∈ Since Hp is complete and

p p p Mbi(x) dx A5 Mf(x) dx LA5 Mf(x) dx, Rn ≤ ≤ i i xi+Bli+2ω Ω X Z X Z Z b converges in Hp. Therefore, b converges in ′ and as in the proof of i i i i S Proposition 3.12 we obtain M( bi)(x) Mbi(x) and thus (5.28) holds. P i P ≤ i P P Lemma 5.8. Suppose s 0, N 2, and f L1(Rn). Then the series b ≥ ≥ ∈ i∈N i converges in L1(Rn). Moreover, there is a constant A , independent of f, i, λ, such 6 P that

(5.29) bi(x) dx A6 f(x) dx. Rn N | | ≤ Rn | | Z Xi∈ Z 5. THE CALDERON-ZYGMUND´ DECOMPOSITION 31

Proof. By Lemma 5.3

bi(x) dx = (f(x) Pi(x))ζi(x) dx Rn | | Rn | − | Z Z li+ω f(x) dx + Pi(x)ζi(x) dx f(x) dx + A2λb . ≤ x +B | | x +B | | ≤ x +B | | Z i li+ω Z i li+ω Z i li+ω Therefore, by (5.2) and (5.5)

2ω bi(x) dx L f(x) dx + A2b λ Ω A6 f(x) dx, N Rn | | ≤ Ω | | | |≤ Rn | | Xi∈ Z Z Z by the Maximal Theorem 3.6.

′ Lemma 5.9. Suppose 0 s 0, such that ∈ S P −ti(s+1) (5.30) Mg(x) A λ λ + Mf(x)1 c (x), ≤ 7 − Ω i X where

t if x Bt+li+2ω+1 Bt+li+2ω for some t 0, (5.31) ti = ti(x) := ∈ \ ≥ 0 otherwise. Proof. If x Ω then as in the proof of Proposition 3.12 we have 6∈ −ti(s+1) Mg(x) Mf(x)+ Mbi(x) Mf(x)+ A4λλ− , ≤ N ≤ N Xi∈ Xi∈ by Lemma 5.6—which is exactly what we need. If x Ω choose k N such that x x + B . Let J := i N : (x + B ) ∈ ∈ ∈ k lk { ∈ i li+2ω ∩ (xk + Blk +2ω) = . By (5.5) the cardinality of J is bounded by L. By Lemma 5.6 we have 6 ∅}

(5.32) Mb (x) A λ λ−ti(s+1). i ≤ 4 − Xi6∈J Xi6∈J It suﬃces to estimate the grand maximal function of g + b = f b . i6∈J i − i∈J i Take ϕ N and l Z. Case I. For∈l S l we write∈ P P ≤ k (f b ) ϕ (x) = (fη) ϕ (x) + ( P ζ ) ϕ (x) − i ∗ l ∗ l i i ∗ l i∈J i∈J (5.33) X X = f Φ (w) + ( P ζ ) ϕ (x), ∗ l i i ∗ l Xj∈J where w (x + B ) Ωc, η =1 ζ , and ∈ k lk +γ ∩ − i∈J i (5.34) Φ(z) := ϕ(z + A−Pl(x w))η(w Alz). − −

By the deﬁnition of J, η 0 on xk + Blk +2ω. Thus, if z supp Φ then w Alz (x + B )c, and ≡z + A−l(x w)= A−l(Alz w + x)∈ A−l( x + − ∈ k lk+2ω − − ⊂ − k 32 1. ANISOTROPIC HARDY SPACES

c −l c c −l (Blk +2ω) + xk + Blk ) A (Blk+ω) = (Blk−l+ω) by (2.11). Since A (x w) A−l(B B )= B⊂ B B we have − ⊂ lk − lk+γ lk−l − lk−l+γ ⊂ lk −l+γ+ω (5.35) ρ(z) bω(ρ(z + A−l(x w)) + ρ( A−l(x w))) c′ρ(z + A−l(x w)), ≤ − − − ≤ − ′ γ+ω for z suppΦ, where c =2b . By (5.4) li lk ω hence l li + ω. By Lemma 5.2 applied∈ for every i J, the chain rule, and≥ (5.5)− the derivatives≤ of η(w Al ) of order s are bounded∈ by some universal constant C. For z supp Φ by− the· product rule,≤ (5.34), and (5.35) ∈

max(1,ρ(z)N ) sup ∂αΦ(z) C max(1,ρ(z)N ) sup ∂αϕ(z + A−l(x w)) |α|≤N | |≤ |α|≤N | − | C max(1,ρ(z)N )/ max(1,ρ(z + A−l(x w))N ) C(c′)N , ≤ − ≤ and therefore Φ C(c′)N . Hence || ||SN ≤ (f Φ )(w) Mf(w) Φ C(c′)N λ. | ∗ l |≤ || ||SN ≤ ′ ′ Since for N 2 there is a constant C > 0 so that ϕ 1 C for all ϕ N and Lemma 5.2 ≥ || || ≤ ∈ S P ζ ϕ (x) LA C′λ. i i ∗ l ≤ 2 i∈J X ′ N By the two estimates above and (5.33) we have (f i∈J bi) ϕl(x) (C(c ) + ′ | − ∗ |≤ LA2C )λ. −l P c Case II. For l>lk we let Φ(z) := ϕ(z +A (x w)), where w (xk +Blk+γ ) Ω . −l − ∈ γ∩+2ω Since A (x w) Blk −l Blk−l+γ B0 Bγ Bγ+ω we have Φ SN b . Therefore, − ∈ − ⊂ − ⊂ || || ≤

(f ϕ )(x) = (f Φ )(w) Mf(w) Φ bγ+2ωλ. | ∗ l | | ∗ l |≤ || ||SN ≤ By Lemma 5.6

(b ϕ )(x) = (b Φ )(w) Φ Mb (w) bγ+2ωLA λ, | i ∗ l | | i ∗ l | ≤ || ||SN i ≤ 4 Xi∈J Xi∈J Xi∈J because w xi + Bli+2ω for all i N. By the two estimates above we have (f b6∈) ϕ (x) bγ+2ω(LA +∈ 1)λ. | − i∈J i ∗ l |≤ 4 Combining cases I and II we have M(f bi)(x) Cλ, where C is the P − i∈J ≤ maximum of the two constants in these cases. Since tk = 0 the kth term in the P sum (5.30) takes care of the estimate of the maximal function of f i∈J bi and by (5.32) we obtain (5.30). − P

Lemma 5.10. (i) Suppose 0 s, and Mf p 1 ≤ ≥ ⌊ ⌋ ∈ L . Then Mg L , and there exists a constant A8 (independent of f and λ) such that ∈

1−p p (5.36) Mg(x)dx A8λ Mf(x) dx. Rn ≤ Rn Z Z (ii) Without an assumption on N (N 2), if f L1 then g L∞ and there exists a constant A (independent of f and λ≥) such that∈ g A∈λ. 9 || ||∞ ≤ 9 5. THE CALDERON-ZYGMUND´ DECOMPOSITION 33

Proof. (i) By Lemmas 5.7 and 5.9

−ti(x)(s+1) Mg(x)dx A7λ λ− dx + Mf(x)dx, c Rn ≤ N Rn Ω Z Xi∈ Z Z where t (x) is defined as in Lemma 5.9. For fixed i N i ∈ ∞ −ti(x)(s+1) −t(s+1) λ− dx = dx + λ− dx Rn x +B x +B \B Z Z i li+2ω t=0 Z i t+li+2ω+1 t+li+2ω ∞ X ∞ −t(s+1) bli+2ω + bt+li+2ω+1λ = bli+2ω 1+ b (bλ−s−1)t = C B , ≤ − − | li+2ω| t=0 t=0 X X where C represents the value of the expression in the bracket. Taking the sum over all i N we obtain ∈ 3ω Mg(x)dx A7λCb Bli−ω + Mf(x)dx c Rn ≤ N | | Ω Z Xi∈ Z 3ω A7λCb Ω + Mf(x)dx ≤ | | c ZΩ 3ω −p p 1−p p A7λCb λ Mf(x) dx + λ Mf(x) dx ≤ c ZΩ ZΩ 1−p p A8λ Mf(x) dx. ≤ Rn Z 1 1 (ii) If f L then g and bi’s are functions and i bi converges in L by Lemma 5.8 (and∈ thus in ′). Since S P g = f bi = f(1 ζi)+ Piζi = f1Ωc + Piζi, − N − N N N Xi∈ Xi∈ Xi∈ Xi∈ by Lemma 5.3 and (5.5) for x Ω we have g(x) LA2λ. Since g(x) = f(x) Mf(x) λ for a.e. x Ωc, therefore∈ g | LA|≤λ = A λ. | | | |≤ ≤ ∈ || ||∞ ≤ 2 9 Corollary 5.11. If 0 ln b/(p ln λ ) . ∩ ⌊ − ⌋ Proof. If f Hp and λ > 0, let f = gλ + bλ be the Calderón-Zygmund ∈ i i decomposition of f of degree s, ln b/(p ln λ−) s

λ p bi LA5 Mf(x) dx. p ≤ i∈N H Z{x:Mf(x)>λ} X λ p λ 1 Therefore, g f in H as λ . But by Lemma 5.10, Mg L , so by Theorem 3.9, g→λ L1. → ∞ ∈ ∈ 34 1. ANISOTROPIC HARDY SPACES

6. The atomic decomposition of Hp We will continue to closely follow the proof of atomic decomposition as pre- sented by Folland and Stein in [FoS]. The proofs of Lemmas 6.1 and 6.2 still require some technical arguments, whereas the proofs of Lemma 6.3 and Theorems 6.4 and 6.5 are copied almost verbatim from [FoS], see also [LU], since all the necessary preparatory work is already done. p Suppose f is a tempered distribution such that Mf = MN f L for some 0 s := ln b/(p ln λ ) . Thus N N . For each k∈ Z consider ≤ ⌊ − ⌋ ≥ p ∈ the Calderón-Zygmund decomposition of f of degree s and height λ =2k associated to Mf, k k f = g + bi , where N Xi∈ k k k k k k k (6.1) Ω := x : Mf(x) > 2 , bi := (f Pi )ζi , Bi := xi + Blk . { } − i k k k Recall that for fixed k Z, (xi = xi )i∈N is a sequence in Ω and (li = li )i∈N a ∈ k k sequence of integers such that (5.1)–(5.5) hold for Ω = Ω , ζi = ζi is given by (5.8) and P = P k is the projection of f onto with respect to the norm given by (5.9). i i Ps Define a polynomial P k+1 as an orthogonal projection of (f P k+1)ζk on ij − j i Ps with respect to the norm 1 (6.2) P 2 = P (x) 2ζk+1(x)dx, || || ζk+1 | | j j Z that is P k+1 is the unique elementR of such that ij Ps

k+1 k k+1 k+1 k+1 (f(x) Pj (x))ζi (x)Q(x)ζj (x)= Pij (x)Q(x)ζj (x)dx. Rn − Rn Z Z k+1 k k+1 Here the ﬁrst integral is understood as (f Pj )ζi , Qζj in the case that f is h − k i k not a regular distribution. For convenience we denote Bˆ := x + B k . i i li +ω ˆk+1 ˆk k+1 k ˆk+1 k Lemma 6.1. (i) If B Bi = then l li +ω, and B xi +Blk+4ω. j ∩ 6 ∅ j ≤ j ⊂ i (ii) For each j N the cardinality of i N : Bˆk+1 Bˆk = is bounded by 2L, ∈ { ∈ j ∩ i 6 ∅} where L is the constant in (5.5). Proof. The proof of (i) follows along the lines of the proof of (iv) in Lemma ˆk+1 ˆk k+1 k k+1 k 2.7. Suppose y Bj Bi = (xj +Blk+1+ω) (xi +Blk+ω) and lj li +ω+1. ∈ ∩ j ∩ i ≥ Then k k+1 xi xj y Blk+ω y + Blk+1+ω Blk+1+2ω. − ∈ − i − j ⊂ j Since γ =4ω + 1 and lk + γ lk+1 + γ ω, i ≤ j − k k k+1 k+1 k+1 xi + Blk+γ = (xi xj )+ xj + Blk+γ xj + Blk+1+2ω + Blk +γ i − i ⊂ j i xk+1 + B , ⊂ j lj +γ−1 by (2.11). Therefore, by (5.3) and Ωk+1 Ωk, ⊂ k k c k+1 k+1 c = (xi + Blk+γ ) (Ω ) (xj + Blk+1+γ−1) (Ω ) , ∅ 6 i ∩ ⊂ j ∩ 6. THE ATOMIC DECOMPOSITION OF Hp 35 which is a contradiction of (5.3). Hence lk+1 lk + ω. Clearly, by (2.11) j ≤ i ˆk+1 k+1 k+1 k k Bj = xj + Blk+1+ω = (xj xi )+ xi + Blk+1+ω j − j k xi + (y Blk+1+ω y + Blk +ω)+ Blk+1+ω ⊂ − j − i j k k xi + Blk+1+2ω + Blk +ω xi + Blk +4ω. ⊂ j i ⊂ i k+1 i ˆk+1 k Moreover, if l l ω then clearly B xi + Blk +2ω. j ≤ k − j ⊂ i To show (ii) ﬁx j N and consider ∈ I = i N : Bˆk+1 Bˆk = and lk lk+1 + ω . 1 { ∈ j ∩ i 6 ∅ i ≥ j } ˆk+1 k k+1 k By the above I1 i N : B xi + Blk+2ω i N : x xi + Blk+2ω ⊂{ ∈ j ⊂ i }⊂{ ∈ j ∈ i } and by (5.1) and (5.5) the cardinality of I1 is at most L. Finally, consider I = i N : Bˆk+1 Bˆk = and lk lk+1 + ω . 2 { ∈ j ∩ i 6 ∅ i ≤ j } As in the proof of (v) in Lemma 2.7, if i I then ∈ 2 k k k+1 k+1 k+1 xi + Blk −ω (xi xj )+ xj + Blk −ω xj + Blk+ω + Blk+1+ω + Blk−ω i ⊂ − i ⊂ i j i k+1 k+1 xj + Blk+1+2ω + Blk+1+ω + Blk+1 xj + Blk+1+3ω. ⊂ j j j ⊂ j k k Since for i I2, xi +Blk+1−2ω xi +Blk −ω we conclude by (5.2) that the cardinality ∈ j ⊂ i of I2 is less than 5ω Blk+1+3ω / Blk+1−2ω = b L. | j | | j | ≤

The proof of (ii) in Lemma 6.1 could be slightly shortened and simpliﬁed if we use the Whitney Lemma with d = 6ω + 1 throughout Sections 5 and 6 instead of d =4ω + 1.

Lemma 6.2. There exists a constant A10 independent of i, j N, and k Z, such that ∈ ∈

k+1 k+1 k+1 (6.3) sup Pij (y)ζj (y) A102 . y∈Rn | |≤

Proof. Let π1,... ,πm (m = dim s) be an orthonormal basis of s with respect to the norm P P 1 P 2 = P (x) 2ζk+1(x)dx. || || ζk+1 | | j j Z We have R m k+1 1 k+1 k k+1 (6.4) Pij = k+1 (f(x) Pj (x))ζi (x)πl(x)ζj (x)dx πl. ζj − Xl=1 Z By Lemma 5.3 applied forR λ =2k+1, (5.15) and (5.16)

(6.5) P k+1(y) C2k+1, π (y) C, for y Bˆk+1. | j |≤ | l |≤ ∈ j 36 1. ANISOTROPIC HARDY SPACES

Hence

1 k+1 k k+1 k+1 (6.6) P (y)πl(y)ζ (y)ζ (y)dy C2 . ζk+1 j i j ≤ j Z

Therefore, by (6.4), R (6.5), and (6.6) we need to show

1 k k+1 k+1 (6.7) f(y)πl(y)ζi (y)ζj (y)dy = f Φlk+1 (w) C2 , ζk+1 | ∗ j |≤ j Z

R k+1 k+1 c for some constant C, where w (xj + Blk+1 +γ) (Ω ) , and ∈ j ∩ lk+1 b j k+1 k k+1 lj (6.8) Φ(z) := k+1 (πlζi ζj )(w A z). ζj −

To see (6.7) it suﬃces to showR that Φ SN C. However, we need only consider ˆk||+1 || ˆk≤ k k+1 those values of i and j such that Bj Bi = , since otherwise ζi ζj vanishes everywhere. Deﬁne ∩ 6 ∅

k+1 lk+1 k+1 k+1 k+1 lk+1 k k lk+1 π˜ (z)= π (x +A j z), ζ˜ (z)= ζ (x +A j z), ζ˜ (z)= ζ (w A j z). l l j j j j i i − Since supp ζ˜k+1 B , by (5.15), Lemma 5.2, and the product rule, the partial j ⊂ ω derivatives ofπ ˜ ζ˜k+1 of order N are bounded by some universal constant. Since l j ≤ k+1 k ˜k lj li + ω by the chain rule and Lemma 5.2, the partial derivatives of ζi of order≤ N are also bounded by some universal constant. Hence, the function Φ can be≤ written as lk+1 b j k+1 ˜k+1 −lj k+1 ˜k Φ(z)= k+1 (˜πlζj )(A (w xj ) z)ζi (z). ζj − − ω Since the above fractionR is b and supp Φ Bγ +Bω we can find another universal constant independent of i,j,k,l≤ so that Φ⊂ C. This shows (6.7). || ||SN ≤ Lemma 6.3. For every k Z, ( P k+1ζk+1)=0, where the series ∈ i∈N j∈N ij j converges pointwise and in ′. S P P Proof. By (5.5) the cardinality of j N : ζk+1(x) = 0 L. Moreover, { ∈ j 6 } ≤ k+1 ˆk+1 ˆk Pij = 0 if Bj Bi = . By Lemma 6.1(ii) for fixed x N the series k+1 ∩k+1 ∅ 2 ∈ i∈N j∈N Pij (x)ζj (x) contains at most 2L nonzero terms. By Lemma 6.2 P P k+1 k+1 2 k+1 (6.9) Pij (x)ζj (x) 2L A102 . N N | |≤ Xi∈ Xj∈ k+1 k+1 Hence, by the Lebesgue Dominated Convergence Theorem i∈N j∈N Pij ζj converges unconditionally in ′. To conclude the proof it suffices to show that S P P k+1 k+1 Pij = Pij =0 forevery j N, N ∈ Xi∈ Xi∈I where I = i N : Bˆk+1 Bˆk = . Indeed, for fixed j N, P k+1 is { ∈ j ∩ i 6 ∅} ∈ i∈N ij an orthogonal projection of (f P k+1) ζk onto with respect to the norm − j i∈I i Ps P (6.2). Since ζk(x) =1 for x Bˆk+1, P k+1 is an orthogonal projection i∈I i ∈ j P i∈N ij P P 6. THE ATOMIC DECOMPOSITION OF Hp 37

k+1 of (f Pj ) with respect to the norm (6.2) which is zero by the deﬁnition of k+1 − Pj .

Theorem 6.4 (The atomic decomposition). If 0

s. Then H H∞,s, where H is given by (3.23). The inclusion map is continuous. ⊂ Proof. Suppose ﬁrst f Hp L1. Consider the Calder´on-Zygmund decom- ∈ ∩ k position of f of degree s ln b/(p ln λ−) and height 2 associated to MN f, f = gk + bk. By Lemma≥ ⌊ 5.7 (5.28), g⌋k f in Hp as k . By Lemma i∈N i → → ∞ 5.10(ii), gk 0 as k . Therefore, || P||∞ → → −∞ ∞ (6.10) f = gk+1 gk in ′. − S k=X−∞ k k+1 k+1 k+1 By Lemma 6.3 and i∈N ζi bj = 1Ωk bj = bj ,

k+1 Pk k+1 k g g = (f bj ) (f bj ) − − N − − N Xj∈ Xj∈ k k+1 k+1 k+1 = bj bj + ( Pij ζj ) N − N N N Xj∈ Xj∈ Xi∈ Xj∈ k k k+1 k+1 k+1 k = bi (ζi bj Pij ζj ) = hi , N − N − N Xi∈ Xj∈ Xi∈ where all the series converge in ′ and S k k k k+1 k k+1 k+1 hi = (f Pi )ζi ((f Pj )ζi Pij )ζj . − − N − − Xj∈ k k+1 By the choice of Pi and Pij

k (6.11) hi (x)P (x)dx = 0 for all P s. Rn ∈P Z k+1 k Moreover, since j∈N ζj = 1Ωk+1 we can write hi as

k P k k k k+1 k k+1 k+1 k+1 hi = f1(Ωk+1)c ζi Pi ζi + Pj ζi ζj + Pij ζj . − N N Xj∈ Xj∈ k+1 k+1 c By Theorem 3.7, f(x) C1MN f(x) c2 for a.e. x (Ω ) and by Lemmas 5.3, 6.1, and 6.2 | |≤ ≤ ∈

(6.12) hk c2k+1 + A 2k +2LA 2k+1 +2LA 2k+1 = C 2k. || i ||∞ ≤ 2 2 10 2 k+1 ˆk+1 ˆk k+1 ˆk+1 Recall that Pij = 0 implies Bj Bi = 0 and hence supp ζj Bj k 6 ∩ 6 ⊂ ⊂ x + B k . Therefore i li +4ω

k k (6.13) supp hi xi + Blk+4ω. ⊂ i 38 1. ANISOTROPIC HARDY SPACES

k k k k By (6.11), (6.12) and (6.13), hi is a multiple of a (p, ,s) atom ai , i.e., hi = κk,iai , k ∞ k k 1/p k (li +4ω)/p k li /p where κk,i = C22 Blk+4ω = C22 b = C32 b . By (5.2) | i | ∞ ∞ ∞ p p kp ω k p ω kp k (κk,i) = (C3) 2 b Bi−ω (C3) b 2 Ω N N | |≤ | | k=X−∞ Xi∈ k=X−∞ Xi∈ k=X−∞ ∞ (C )pbω2/p p2k(p−1) Ωk 2k−1 (6.14) ≤ 3 | | k=−∞ ∞ X p−1 C4 pλ x : MNp f(x) > λ dλ ≤ 0 |{ }| Z p p = C M f = C f p . 4|| Np ||p 4|| ||H ∞ k ∞ k Therefore, f = k=−∞ i∈N hi = k=−∞ i∈N κk,iai defines an atomic decomposition of f Hp L1. P P P P If f is a general∈ ∩ element of Hp then by Corollary 5.11 we can find a sequence p 1 p 2−m p f N H L so that f p 2 f p and f = f . For every { m}m∈ ⊂ ∩ || m||H ≤ || ||H m∈N m m N let f = κ ai be an atomic decomposition constructed above with m i∈N i,m m P the∈ summation enumerated over one index i. By (6.14) P p p p 2−m p (κi,m) C4 fm Hp C4 f Hp 2 =4C4 f Hp , N N ≤ N || || ≤ || || N || || mX∈ Xi∈ mX∈ mX∈ and f = κ ai defines an atomic decomposition of f Hp. m∈N i∈N i,m m ∈

TheP next theoremP shows that different choices of N Np in Definition 3.11 and admissible triplets (p,q,s) in Definition 4.3 yield the same≥ Hardy space Hp. Theorem 6.5 (The equivalence of norms). Suppose 0 0 so that ≥ p 1 2 p (6.15) M f C f p C f p C M f for f H . || Np ||p ≤ 1|| ||Hq,s ≤ 1|| ||H∞,s ≤ 2|| N || ∈ Proof. Note that if N M N then Hp Hp , where the subscript ≥ ≥ p (M) ⊂ (N) corresponds to the choice of the grand maximal function in Definition 3.11. By p p Definition 4.1 every (p, ,s) atom is also a (p,q,s) atom, and hence H∞,s Hq,s. Therefore, if we take N>s∞ then by Theorems 6.4 and 4.5 we have ⊂ (6.16) Hp Hp Hp Hp Hp . (N) ⊂ ∞,s ⊂ q,s ⊂ (Np) ⊂ (N) p p p Thus, H = H = H for all N Np and all admissible triplets (p,q,s). (N) (Np) q,s ≥ Furthermore, since the inclusion maps in (6.16) are continuous we have (6.15).

Theorem 6.5 encompasses fundamental properties of anisotropic Hardy spaces. As a ﬁrst application of this theorem, we prove that smooth functions with certain number of vanishing moments are dense in Hp. Lemma 6.6. Suppose ψ . For any N N, there exists C > 0 such that for all f ′ we have ∈ S ∈ ∈ S n (6.17) sup MN+2(f ψl)(x) CMN f(x) for all x R . l∈Z ∗ ≤ ∈ 6. THE ATOMIC DECOMPOSITION OF Hp 39

Proof. We claim that for any N N there exists a constant C > 0 so that ∈

(6.18) sup ϕl ψ SN C ϕ SN+2 ψ SN for any ϕ, ψ . l∈Z,l≤0 || ∗ || ≤ || || || || ∈ S Indeed, take any multi-index α, α N, l Z, l 0, and x Rn. By (2.12) | |≤ ∈ ≤ ∈ max(1,ρ(x))N ∂α(ϕ ψ)(x) = max(1,ρ(x))N (ϕ ∂αψ)(x) | l ∗ | | l ∗ | ωN N α N b max(1,ρ(x y)) ∂ ψ(x y) max(1,ρ(y)) ϕl(y) dy ≤ Rn − | − | | | Z ωN N −l −N−2 −l b ψ SN max(1,ρ(y)) max(1,ρ(A y)) b ϕ SN+2 dy ≤ || || Rn || || Z ωN −l −l −2 b ψ SN ϕ SN+2 b max(1,ρ(A y)) dy = C ϕ SN+2 ψ SN . ≤ || || || || Rn || || || || Z For any f ′, ψ , ϕ , and x Rn. By (6.18) ∈ S ∈ S ∈ SN+2 ∈ sup (f ψl) ϕk(x) sup f (ψl ϕk)(x) + sup f (ψl ϕk)(x) k,l∈Z | ∗ ∗ |≤ k,l∈Z | ∗ ∗ | k,l∈Z | ∗ ∗ | k≤l k>l

sup f (ψ ϕk−l)l(x) + sup f (ψl−k ϕ)k(x) ≤ k,l∈Z | ∗ ∗ | k,l∈Z | ∗ ∗ | k−l≤0 l−k<0 C ψ ϕ M f(x)+ C ψ ϕ M f(x) ≤ || ||SN || ||SN+2 N || ||SN+2 || ||SN N 2C ψ M f(x), ≤ || ||SN+2 N which shows (6.17).

As an immediate consequence of Lemma 6.6 we obtain Corollary 6.7. If f Hp and ϕ then f ϕ Hp. ∈ ∈ S ∗ ∈ A less immediate consequence of Lemma 6.6 is Theorem 6.8. Suppose ψ and ψ =1. For any 0 0 we can ﬁnd a continuous function g with compact support such that f g <ε. By Lemma 6.6 || − ||1 lim sup MN (f ψl f)(x) l→−∞ ∗ −

sup MN ((f g) ψl)(x) + lim sup MN (g ψl g)(x)+ MN (f g)(x) ≤ l∈Z − ∗ l→−∞ ∗ − − CM (f g)(x) ≤ Np − 40 1. ANISOTROPIC HARDY SPACES

By Theorem 3.6 for any λ> 0

x : lim sup MN (f ψl f)(x) > λ |{ l→−∞ ∗ − }| x : M (f g)(x) > λ/C C′ f g /λ 0 is arbitrary we have (6.20) for every f L1 Hp. Recall that L1 Hp is dense in Hp. This follows∈ from∩ Corollary 5.11 (0

0 we can ﬁnd 1 p p g L H such that f g p <ε. Therefore, ∈ ∩ || − ||H p lim sup f ψl f Hp l→−∞ || ∗ − || p p p sup (f g) ψl Hp + f g Hp + lim sup g ψl g Hp ≤ l∈Z || − ∗ || || − || l→−∞ || ∗ − || p C f g p Cε. ≤ || − ||H ≤ Since ε> 0 was arbitrary we obtain (6.19).

Finally, we prove that smooth functions with compact support are dense in Hp. p ∞ p ∞ ∞ Theorem 6.9. H C0 is dense in H , where C0 denotes C0 functions with compact support. ∩ ∞ Proof. Suppose f is a finite linear combination of atoms. Take ψ C0 . ∞ p ∈ Clearly, f ψl C0 for every l Z and f ψl f in H as l by Theorem 6.8. This∗ finishes∈ the proof since∈ finite linear∗ combinations→ of→ atoms −∞ are dense in Hp.

∞ p Remark. C0 is not a subset of H for p 1. It is not hard to see that ∞ p≤ ∞ ψ C0 with ψ = 0 does not belong to any H for p 1. However, if ψ C0 has∈ vanishing moments6 of order s, i.e., ψP = 0 for≤ all P , where∈ s R ∈ Ps ≥ (1/p 1) ln b/ ln λ− , then ψ is a constant multiple of a (p, ,s) atom and hence ⌊ψ H−p. ⌋ R ∞ ∈ 7. OTHER MAXIMAL DEFINITIONS 41

7. Other maximal deﬁnitions The goal of this section is to prove the characterization of Hardy spaces using the radial and nontangential maximal functions of a single test function ϕ , ϕ = 0. ∈ S 6 ′ R Theorem 7.1. Suppose ϕ , ϕ =0, and 0 λ is open → ∞ ∗ n { } for all λ > 0. Indeed, if Fl (x) > λ for some x R , i.e., there is k K and ′ ∈ ′ ≤ y x + Bk+l so that F (y, k) > λ, then y x + Bk+l for x in a suﬃciently small ∈ ∗ ′ ∈ neighborhood of x. Thus Fl (x ) > λ. Here we need that the balls Bj are open for all j Z. ∈ Lemma 7.2. There exists a constant C > 0 so that for all functions F : Rn Z [0, ), λ> 0, l l′ Z, and K Z we have × → ∞ ≥ ∈ ∈ ∪ {∞} ∗K l−l′ ∗K (7.5) x : F (x) > λ Cb x : F ′ (x) > λ . |{ l }| ≤ |{ l }| In particular,

∗K l−l′ ∗K (7.6) Fl (x)dx Cb Fl′ (x)dx. Rn ≤ Rn Z Z ∗ ∗ n Proof. Let Ω = x : F ′ (x) > λ . Suppose F (x) > λ for some x R . Then { l } l ∈ there exist k K, y x+B such that F (y, k) > λ. Clearly, y +B ′ Ω. Also ≤ ∈ k+l k+l ⊂ y + Bk+l′ x + Bk+l + Bk+l′ x + Bk+l+ω. Hence y + Bk+l′ Ω (x + Bk+l+ω) and ⊂ ⊂ ⊂ ∩ k+l′ k+l+ω l′−l−ω Ω (x + Bk+l+ω) b = b b . | ∩ l′−l−ω |≥ Therefore, MHL(1Ω)(x) b , where MHL is the centered Hardy-Littlewood ≥ −m maximal function MHLf(y) := supm∈Z b f(z) dz. By Theorem 3.6 y+Bm | | ′ ′ ′ x : F ∗(x) > λ x : M (1 )(x) bl −Rl−ω Cbl−l +ω 1 = C′bl−l Ω , |{ l }| ≤ |{ HL Ω ≥ }| ≤ || Ω||1 | | hence (7.5) holds. Integrating (7.5) on (0, ) with respect to λ yields (7.6). ∞ The following result enables us to pass from one function in to the sum of negative dilates of another function in with nonzero mean. S S 42 1. ANISOTROPIC HARDY SPACES

Lemma 7.3. Suppose ϕ with ϕ =0. For every ψ there is a sequence ∈ S 6 ∈ S of test functions (ηj )∞ , ηj such that j=0 ∈ S R ∞ (7.7) ψ = ηj ϕ , ∗ −j j=0 X where the convergence is in . Furthermore, for any integersS L,N > 0 there exist a constant C and an integer M > 0 (depending on L and N, but independent of the choice of ψ) such that (7.8) ηj Cb−jL ψ for all j 0. || ||SN ≤ || ||SM ≥ Proof. For the purpose of the proof we will work with the standard deﬁnition of N , that is S N α η SN = sup sup (1 + x ) ∂ η(x) . || || x∈Rn |α|≤N | | | |

By Lemma 3.2 this implies the corresponding result for the usual homogeneous N . We claim that for every integer N > 0 there is a constant C so that S (7.9) ηˆ C η . || ||SN ≤ || ||SN+n+1 Indeed, for any multi-indices α , β N we have | | | |≤ (2πi)|β|ξβ∂αηˆ(ξ) = ( 2πi)|α| e−2πihx,ξi∂β(xαη(x))dx. − Rn Z Hence, by multiplying and dividing the right hand side by (1 + x )n+1 we have | | ξβ∂αηˆ(ξ) (2π)|α|−|β| sup (1 + x )n+1 ∂β(xαη(x)) (1 + x )−n−1dx, | |≤ Rn | | | | Rn | | x∈ Z which implies (7.9). Let ∆ˆ be an expansive ellipsoid for the dilation B = AT , i.e., ∆ˆ r∆ˆ B∆,ˆ ⊂ ⊂ guaranteed by Lemma 2.2. By scaling of ϕ and ∆ˆ we can assume that ϕ = 1 and ϕˆ(ξ) 1/2 for ξ B∆.ˆ Consider a C∞ function ζ such that ζ 1 on ∆ˆ and | | ≥ ∈ ≡ R supp ζ B∆.ˆ Deﬁne a sequence of functions (ζ )∞ by ζ = ζ, ⊂ j j=0 0 ζ (ξ)= ζ(B−j ξ) ζ(B−j+1ξ) for j 1. j − ≥ Clearly, ∞ ζ (ξ) = 1 for all ξ Rn. j ∈ j=0 X Thus ∞ ∞ ζ (ξ) ψˆ(ξ)= ζ (ξ)ψˆ(ξ)= j ψˆ(ξ)ˆϕ(B−j ξ). j ϕˆ(B−j ξ) j=0 j=0 X X For j 0 deﬁne ηj by ≥ ζ (ξ) (7.10)η ˆj (ξ)= j ψˆ(ξ). ϕˆ(B−j ξ) Then, by the choice of ηj ’s, (7.7) holds. Moreover, the convergence in of the series (7.7) is a direct consequence of (6.18) and (7.8). Therefore, it remainsS to show (7.8). 7. OTHER MAXIMAL DEFINITIONS 43

To show (7.8), take M = N +2(n+1)+L ln b/ ln λ− , where λ− is the same as in (2.1). Our goal is to show that there exists⌈ a constant⌉C (independent of ψ ) such that ∈ S ηˆj Cb−jL ψ for j 0. || ||SN+n+1 ≤ || ||SM ≥ Let ζ˜(ξ) = (ζ(ξ) ζ(Bξ))/ϕˆ(ξ). Since supp ζ˜ B∆ˆ and ϕˆ(ξ) 1/2 for ˆ − α ˜ ⊂ | | ≥ ξ B∆ we have supξ sup|α|≤N+n+1 ∂ ζ(ξ) C for some constant C. Since ∈−j −j | | ≤ ζ˜(B ξ)= ζj (ξ)/ϕˆ(B ξ) by the chain rule we have

α −j sup sup ∂ (ζj ( )/ϕˆ(B ))(ξ) C. ξ |α|≤N+n+1 | · · |≤

By (7.10), the product rule and since ζ (ξ)=0 for ξ Bj−1∆ˆ j ∈ j N+n+1 α ˆ ηˆ SN+n+1 C sup sup (1 + ξ ) ∂ ψ(ξ) || || ≤ ξ6∈Bj−1∆ˆ |α|≤N+n+1 | | | | C sup (1 + ξ )−L⌈ln b/ ln λ−⌉ sup (1 + ξ )N+n+1+L⌈ln b/ ln λ−⌉ ∂αψˆ(ξ) ≤ ξ6∈Bj−1 ∆ˆ | | |α|≤N+n+1 | | | | Cb−jL ψˆ Cb−jL ψ , ≤ || ||SM−n−1 ≤ || ||SM because for ξ ∆,ˆ (1 + Bj−1ξ ) 1/cλj−1 inf ξ . This ﬁnishes the proof of 6∈ | | ≥ − ξ6∈∆ˆ | | (7.8). Since our choice of N is arbitrary in (7.8) the series in (7.7) converges in . S We now introduce maximal functions obtained from truncation with an additional extra decay term. Namely, for an integer K representing the truncation level and real number L 0 representing the decay level, we deﬁne radial, nontangential, tangential, radial grand,≥ and nontangential grand maximal functions, respectively as

0(K,L) −K −L −k−K −L Mϕ f(x) = sup (f ϕk)(x) max(1,ρ(A x)) (1 + b ) , k∈Z | ∗ | k≤K (K,L) −K −L −k−K −L Mϕ f(x) = sup sup (f ϕk)(y) max(1,ρ(A y)) (1 + b ) , Z k∈ y∈x+Bk | ∗ | k≤K −k−K −L N(K,L) (f ϕk)(y) (1 + b ) Tϕ f(x) = sup sup | ∗ −k | N −K L , k∈Z y∈Rn max(1,ρ(A (x y))) max(1,ρ(A y)) k≤K − 0(K,L) 0(K,L) MN f(x) = sup Mϕ f(x). ϕ∈SN (K,L) (K,L) MN f(x) = sup Mϕ f(x). ϕ∈SN

The next lemma guarantees the control of the tangential by the nontangential maximal function independent of K and L. Lemma 7.4. Suppose p> 0, N > 1/p, and ϕ . Then there exists a constant C so that for all K Z, L 0, and f ′, ∈ S ∈ ≥ ∈ S (7.11) T N(K,L)f C M (K,L)f . || ϕ ||p ≤ || ϕ ||p 44 1. ANISOTROPIC HARDY SPACES

Proof. Consider function F : Rn Z [0, ) given by × → ∞ F (y, k)= (f ϕ )(y) p max(1,ρ(A−K y))−pL(1 + b−k−K )−pL. | ∗ k | Fix x Rn. If k K and x y B then ∈ ≤ − ∈ k+1 F (y, k) max(1,ρ(A−k(x y)))−pN F ∗K (x). − ≤ 0 If x y B B for some j 1 then − ∈ k+j+1 \ k+j ≥ F (y, k) max(1,ρ(A−k(x y)))−pN F ∗K (x)b−jNp. − ≤ j By taking supremum over all y Rn, k K we have ∈ ≤ ∞ (T N(K,L)f(x))p F ∗K (x)b−jNp. ϕ ≤ j j=0 X Therefore by Lemma 7.2,

∞ N(K,L) p −jNp ∗K Tϕ f p b Fj (x)dx || || ≤ Rn j=0 X Z ∞ −jNp j ∗K ′ (K,L) p C b b F0 (x)dx C Mϕ f p, ≤ Rn ≤ || || j=0 X Z where C′ = C ∞ bj(1−Np) < . j=0 ∞ P Lemma 7.5 gives the pointwise majorization of the grand maximal function by the tangential one. Lemma 7.5. Suppose ϕ and ϕ = 0. For given N,L 0 there exists an integer M > 0 and a constant∈C S > 0 so that6 for all f ′ and≥ integers K 0 we have R ∈ S ≥

(7.12) M 0(K,L)f(x) CT N(K,L)f(x) for all x Rn. M ≤ ϕ ∈

Proof. Take any ψ . By Lemma 7.3 there exists a sequence of test functions (ηj )∞ so that (7.7)∈ holds. S For a ﬁxed integer k K and x Rn, j=0 ≤ ∈ ∞ ∞ (f ψ )(x) = f (ηj ϕ ) (x) = f (ηj ) ϕ (x) | ∗ k | ∗ ∗ −j k ∗ k ∗ −j+k j=0 j=0 X X ∞ ∞ j j (f (η )k ϕ−j+k)(x) (f ϕ−j+k)(x y) (η )k(y) dy ≤ | ∗ ∗ |≤ Rn | ∗ − || | j=0 j=0 X X Z T N(K,L)f(x) ≤ ϕ ∞ j−k N −K L j−k−K L j max(1,ρ(A y)) max(1,ρ(A (x y))) (1 + b ) (η )k(y) dy. · Rn − | | j=0 X Z 7. OTHER MAXIMAL DEFINITIONS 45

Therefore ∞ 0(K,L) N(K,L) j−k N Mψ f(x) Tϕ f(x) sup max(1,ρ(A y)) ≤ k∈Z Rn j=0 k≤K X Z max(1,ρ(A−K (x y)))L(1 + bj−k−K )L − (ηj ) (y) dy. max(1,ρ(A−K x))L(1 + b−k−K )L | k |

Using the inequalities (2.12), (2.13) and the change of variables we can estimate the last sum by

∞ max(1,ρ(Aj−ky))N bωL max(1,ρ(A−K y))L2LbjLb−k ηj (A−ky) dy Rn | | j=0 X Z ∞ ∞ L ωL j(N+L) N+L j L ωL j(N+L) j 2 b b max(1,ρ(y)) η (y) dy 2 b b η SN+L . ≤ Rn | | ≤ || || j=0 j=0 X Z X

By Lemma 7.3 there exists an integer M 0 so that ≥ ηj Cb−j(N+L+1) ψ , || ||SN+L ≤ || ||SM and hence ∞ M 0(K,L)f(x) = sup M 0(K,L)f(x) 2LbωL Cb−j T N(K,L)f(x) M ψ ≤ ϕ ψ∈SM j=0 X ′ N(K,L) = C Tϕ f(x). This shows (7.12).

Lemma 7.6. Suppose p> 0, ϕ , and f ′. Then for every M > 0 there exists L> 0 so that ∈ S ∈ S

(7.13) M (K,L)f(x) C max(1,ρ(x))−M for all x Rn, ϕ ≤ ∈ for some constant C = C(K) dependent on K 0. ≥ Proof. There exists an integer N > 0 and a constant C > 0 so that

f ϕ(x) C ϕ max(1,ρ(x))N for all x Rn, ϕ . | ∗ |≤ || ||SN ∈ ∈ S If k 0 then by the chain rule ≤ N −k α −k ϕk SN = sup sup max(1,ρ(z)) b ∂ ϕ(A )(z) || || z∈Rn |α|≤N | · | −k N α −k −k N −k −kN b sup sup max(1,ρ(z)) ∂ ϕ(A z) C A Cb λ+ ϕ SN . ≤ z∈Rn |α|≤N | | || || ≤ || ||

If 0 < k K then by the chain rule ≤ N α −k KN ϕk SN C sup sup max(1,ρ(z)) ∂ ϕ(A z) Cb ϕ SN . || || ≤ z∈Rn |α|≤N | |≤ || || 46 1. ANISOTROPIC HARDY SPACES

Therefore, if we take L = N + M, then for any k K and y x + B , ≤ ∈ k f ϕ (y) max(1,ρ(A−K y))−L(1 + b−k−K )−L | ∗ k | C max(1,ρ(y))N ϕ bKL max(1,ρ(y))−Lb(k+K)L ≤ || k||SN C max(1,ρ(y))N−Lb3KL+KN ϕ ≤ || ||SN Cb(K+ω)M b3KL+KN max(1,ρ(x))−M ϕ . ≤ || ||SN This shows (7.13) and ﬁnishes the proof of Lemma 7.6.

Note that the above argument gives the same estimate for the truncated grand 0(K,L) maximal function MN f. As a consequence of Lemma 7.6 we have that for any choice of K 0 and any f ′ we can find an appropriate L > 0 so that ≥ (K,L∈) S p n the maximal function, say Mϕ f, is bounded and belongs to L (R ). This becomes crucial in the proof of Theorem 7.1, where we work with truncated maximal functions. The complexity of the preceding argument stems from the fact that a 0 p p priori we do not know whether Mϕf L implies Mϕf L . Instead we must work with variants of maximal functions∈ for which this is sat∈ isfied. Proof of Theorem 7.1. Clearly, (7.1) = (7.2) = (7.3). By Lemma ⇒ ⇒ 7.5 applied for N > 1/p and L = 0 we have pointwise estimate M 0(K,0)f(x) M ≤ CT N(K,0)f(x) for all f ′ and integers K 0. By Lemma 7.4 we have another constant C so that ∈ S ≥ M 0(K,0)f C M (K,0)f for f ′,K 0. || M ||p ≤ || ϕ ||p ∈ S ≥ 0 As K we obtain MM f p C Mϕf p by the Monotone Convergence The- orem.→ It remains∞ to show|| (7.3)|| =≤ (7|| .2). || Suppose now M 0f Lp. By⇒ Lemma 7.6 we can find L > 0 so that (7.13) ϕ ∈ (K,L) p holds, i.e., Mϕ f L for all K 0. By Lemmas 7.4 and 7.5 we can find M > 0 0(K,L) ∈ (K,L)≥ so that MM f p C1 Mϕ f p with a constant C1 independent of K 0. For given|| K 0 let|| ≤ || || ≥ ≥ (7.14) Ω = x Rn : M 0(K,L)f(x) C M (K,L)f(x) , K { ∈ M ≤ 2 ϕ } 1/p where C2 =2 C1. We claim that

(K,L) p (K,L) p (7.15) Mϕ f(x) dx 2 Mϕ f(x) dx. Rn ≤ Z ZΩK Indeed, this follows from

(K,L) p −p (K,L) p p (K,L) p Mϕ f(x) dx C2 MM f(x) dx (C1/C2) Mϕ f(x) dx, Ωc ≤ Ωc ≤ Rn Z K Z K Z p and (C1/C2) =1/2. We also claim that for 0 0 so that for all integers K 0 ≥ (7.16) M (K,L)f(x) C (M (M 0(K,L)f( )q)(x))1/q for x Ω . ϕ ≤ 3 HL ϕ · ∈ K Indeed, let F (y, k)= (f ϕ )(y) max(1,ρ(A−K y))−L(1 + b−k−K )−L. | ∗ k | 7. OTHER MAXIMAL DEFINITIONS 47

Suppose x Ω . There is k K and y x + B such that ∈ K ≤ ∈ k F (y, k) F ∗(x)/2= M (K,L)f(x)/2. ≥ 0 ϕ Consider x′ y + B for some integer l 0 to be speciﬁed later. We have ∈ k−l ≥ f ϕ (x′) f ϕ (y)= f Φ (y), where Φ(z)= ϕ(z + A−k(x′ y)) ϕ(z). ∗ k − ∗ k ∗ k − − By the Mean Value Theorem

Φ SM sup ϕ( + h) ϕ( ) SM || || ≤ h∈B−l || · − · || = sup sup sup max(1,ρ(z)M ) ∂αϕ(z + h) ∂αϕ(z) n h∈B−l z∈R |α|≤M | − | M α −l C sup sup sup max(1,ρ(z + h) ) ∂ ϕ(z + h) sup h C4λ− , n ≤ h∈B−l z∈R |α|≤M+1 | | · h∈B−l | |≤

−K ′ ω −K where C4 does not depend on L. Since max(1,ρ(A x )) b max(1,ρ(A y)) we have ≤ bωLF (x′, k) ( f ϕ (y) f Φ (y) ) max(1,ρ(A−K y))−L(1 + b−k−K )−L ≥ | ∗ k | − | ∗ k | F (y, k) M (K,L)f(x) Φ M (K,L)f(x)/2 bωM M 0(K,L)f(x)C λ−l ≥ − M || ||SM ≥ ϕ − M 4 − M (K,L)f(x)/2 bωM C λ−lC M (K,L)f(x) M (K,L)f(x)/4, ≥ ϕ − 4 − 2 ϕ ≥ ϕ if we choose for l the smallest integer l 0 so that bωM C λ−lC 1/4. Here ≥ 4 − 2 ≤ we used the fact that x ΩK and the pointwise majorization of nontangential by radial grand maximal function,∈ M (K,L)f(x) bωM M 0(K,L)f(x), M ≤ M see Proposition 3.10. Therefore for x Ω , ∈ K 4qbωLq M (K,L)f(x)q F (z, k)qdz ϕ ≤ B | k−l| Zy+Bk−l bω+l 4qbωLq M 0(K,L)f(z)qdz C M (M 0(K,L)f( )q)(x), ≤ B ϕ ≤ 3 HL ϕ · | k+ω| Zx+Bk+ω which shows (7.16). Finally, by (7.15), (7.16), and the Maximal Theorem 3.6

(K,L) p (K,L) p Mϕ f(x) dx 2 Mϕ f(x) dx Rn ≤ (7.17) Z ZΩK p/q 0(K,L) q p/q 0(K,L) p 2C3 MHL(Mϕ f( ) )(x) dx C5 Mϕ f(x) dx, ≤ · ≤ Rn ZΩK Z where the constant C5 depends on p/q > 1 and L 0, but is independent of K 0. This inequality is crucial as it gives a bound of≥ nontangential by radial maximal≥ function in Lp. The rest of the proof is immediate. (K,L) Since Mϕ f(x) converges pointwise and monotonically to Mϕf(x) for all n p n x R as K , Mϕf L (R ) by (7.17) and the Monotone Convergence The- orem.∈ Therefore,→ ∞ we can now∈ choose L = 0 and again by (7.17) and the Monotone Convergence Theorem we have M f p C M 0f p, where now C corresponds || ϕ ||p ≤ 5|| ϕ ||p 5 to L = 0 and is independent of f ′. This concludes the proof of Theorem 7.1. ∈ S 48 1. ANISOTROPIC HARDY SPACES

8. Duals of Hp In this section we provide the description of the duals of anisotropic Hp spaces, 0 < p 1, in terms of Campanato spaces. This description is a consequence of the atomic≤ decomposition of Hp and some functional analysis and approximation arguments. The dual of Hp space of holomorphic functions on the unit disc for 0

−l (8.2) g l := sup inf B ess sup g(x) P (x) < . (q = ) C∞,s x∈B || || B∈B P ∈Ps | | | − | ∞ ∞

Here s denotes the space of all polynomials (in n variables) of degree at most s. P l We identify two elements of Cq,s if they are equal almost everywhere. l One may easily verify that: l is a seminorm, C , and g l = Cq,s s q,s Cq,s ||l · || P ⊂ || || 0 g s. Therefore, Cq,s/ s is a normed linear space. Moreover, the ⇐⇒ ∈ P P l standard arguments show that this space is also complete; hence Cq,s/ s is a Banach space. P The conditions (8.1) and (8.2) can be also written in terms of a quasi-norm ρ l associated to a dilation A. For example we can equivalently deﬁne Cq,s as the space of all locally Lq functions g on Rn so that

1/q −l −1 q (8.3) g l := sup inf r r g(x) P (x) dx < . Cq,s x ∈Rn P ∈Ps || || 0 ρ(x−x0)≤r | − | ∞ r>0 Z Indeed, this exactly the case when ρ is the step homogeneous quasi-norm; the general case follows from Lemma 2.4. The main goal of this section is to prove that the dual of anisotropic Hardy p 1/p−1 space HA is isomorphic to the Campanato space Cq,s / s, where (p,q,s) is an admissible triplet in the sense of Definition 4.1, see TheoremP 8.3. As a consequence, l this shows that Campanato spaces Cq,s depend effectively only the choice of l and not on q or s, see Corollary 8.6. Analogous results for Hardy spaces on homogeneous groups were obtained by Folland and Stein [FoS, Chapter 5]. However, careful 8. DUALS OF Hp 49 examination of the arguments in [FoS] reveals a gap in the first part of the proof of [FoS, Theorem 5.3]. The problem is with [FoS, Lemma 5.1] which does not hold unless we assume that a functional L is bounded. Hence, an additional argument is needed. This gap can be filled in the setting of anisotropic Hardy spaces by applying a rather subtle approximation argument inspired by [GR, Chapter III.5]. We start with a basic lemma. p p Lemma 8.2. Suppose L is a continuous linear functional on H = Hq,s, and (p,q,s) is admissible. Then

(8.4) L p ∗ := sup Lf : f p 1 = sup La : a is (p,q,s)-atom . || ||(Hq,s) {| | || ||Hq,s ≤ } {| | }

Proof. Since every (p,q,s)-atom a satisﬁes a p 1 it suﬃces to show || ||Hq,s ≤ (8.5) sup Lf : f p 1 sup La : a is (p,q,s)-atom . {| | || ||Hq,s ≤ }≤ {| | } p Take any f H with f p 1. For every ε> 0 there is an atomic decomposi- ∈ || ||Hq,s ≤ tion f = κ a with κ p 1+ ε. Therefore i i i i | i| ≤ P P 1/p Lf κ La sup La : a is (p,q,s)-atom κ p | |≤ | i|| i|≤ {| | } | i| i i X X (1 + ε)1/p sup La : a is (p,q,s)-atom . ≤ {| | } Since ε> 0 is arbitrary we have (8.5).

p ∗ Clearly, p ∗ is a norm on (H ) (= space of continuous functionals on || · ||(Hq,s) Hp) which makes (Hp)∗ into a Banach space. Theorem 8.3. Suppose (p,q,s) is admissible. Then

p ∗ p ∗ l ′ (8.6) (H ) = (H ) C ′ / , where 1/q +1/q =1, l =1/p 1. q,s ≡ q ,s Ps − l If g C ′ and f is a ﬁnite linear combination of (p,q,s)-atoms, let L f = gf. ∈ q ,s g Then L extends continuously to Hp, and every L (Hp)∗ is of this form. More- g R over, ∈ l l p ∗ ′ (8.7) g C = Lg (Hq,s ) for all g Cq ,s. || || q′,s || || ∈

1 Proof. For any B deﬁne πB : L (B) s the natural projection deﬁned using the Riesz Lemma ∈B →P

(8.8) (π f(x))Q(x)dx = f(x)Q(x)dx for all f L1(B),Q . B ∈ ∈Ps ZB ZB We claim that there is a universal constant (depending only on s) such that 1 (8.9) sup π f(x) C f(x) dx. | B |≤ B | | x∈B | | ZB Indeed, if B = B0 then we ﬁnd an orthonormal basis Qα : α s of s with 2 { | | ≤ } P respect to the L (B0) norm. Since

πB0 f = f(x)Qα(x)dx Qα. B0 |αX|≤s Z 50 1. ANISOTROPIC HARDY SPACES the claim follows for B = B . Since π f = (D −k π D k )f and π f = 0 Bk A ◦ B0 ◦ A y+Bk (τy πBk τ−y)f the claim (8.9) follows for arbitrary B = y + Bk. ◦ ◦ q q For 1 q and B we deﬁne the closed subspace L0(B) L (B) q ≤ ≤ ∞q ∈ B q ⊂ by L0(B) = f L (B) : πBf = 0 . We will identify L (B) with the subspace of Lq(Rn) consisting{ ∈ of functions vanishing} outside B. With this identiﬁcation, if f Lq(B) then B 1/q−1/p f −1f is a (p,q,s)-atom. ∈ 0 | | || ||q Suppose L (Hp)∗ = (Hp )∗. By Lemma 8.2 ∈ q,s 1/p−1/q q (8.10) Lf L p ∗ B f , for f L (B). | | ≤ || ||(Hq,s) | | || ||q ∈ 0 q Therefore, L provides a bounded linear functional on L0(B) which can be extended by the Hahn-Banach Theorem to the whole space Lq(B) without increasing its ′ norm. Suppose, momentarily, that q< . By the duality Lq(B)∗ = Lq (B), where ′ q′ ∞ q 1/q +1/q = 1, there exists h L (B) such that Lf = fh for all f L0(B). ∈ ′ ∈ In particular, L∞(B) Lq(B) implies there is h Lq (B) L1(B) such that ⊂∞ ∈ R ⊂ q′ Lf = fh for all f L0 (B). Therefore, also for q = , there exists h L (B) ∈ ∞ ′ ∈ such that Lf = fh for all f Lq(B). If h′ is another element of Lq (B) such that R 0 Lf = fh′ for all f Lq(B)∈ then h h′ . We can say even more. Suppose R ∈ 0 − ∈ Ps that for some h,h′ L1(B) we have Lf = fh = fh′ for all f L∞(B) then R B B 0 h h′ . Indeed,∈ for all f L∞(B) ∈ − ∈Ps ∈ R R 0= (f π f)(h h′)= f(h h′) (π f)(π (h h′)) − B − − − B B − ZB ZB ZB = f(h h′) f(π (h h′)) = f((h h′) π (h h′)), − − B − − − B − ZB ZB ZB ′ ′ hence (h h )(x)= πB (h h )(x) for a.e. x B. Therefore, after changing values −′ − ∈ ′ of h (or h ) on a set of measure zero we have h h s. Hence, the function h is unique up to a polynomial of degree at most− s ∈regardless P of the exponent q ∞ 1 q . Therefore, h q<∞ L (B) for p = 1, and h L (B) for 0

(8.12) g q ∗ = inf g P q′ . L0(B) L (B) || || P ∈Ps || − || This is, at least for q < , an immediate consequence of an elementary fact from functional analysis ∞

′ Lq(B)∗ = Lq(B)∗/Lq(B)⊥ = Lq (B)/ , 0 0 Ps q ⊥ q ∗ where L (B) = h L (B) : h Lq(B) = 0 denotes the annihilator of the sub- 0 { ∈ | 0 } space Lq(B) Lq(B). The special case of (8.12) for q = requires similar 0 ⊂ ∞ 8. DUALS OF Hp 51

1 arguments and the fact that g L (B) and g P L∞(B)∗ = g P L1(B) for any P . Combining (8.11) and∈ (8.12) we have|| − for l||=1/p 1|| − || ∈Ps − −l−1/q′ (8.13) g l = sup B g q ∗ L p ∗ , C ′ L0(B) (Hq,s) || || q ,s B∈B | | || || ≤ || || which completes the ﬁrst part of the proof of Theorem 8.3. l Conversely, suppose g Cq′,s. Our goal is to show that functional Lgf = gf deﬁned initially for f Θq∈, where ∈ s R (8.14) Θq = f Lq(Rn) : supp f is compact and f(x)xαdx = 0 for α s , s { ∈ Rn | |≤ }

p R p extends to a bounded functional on Hq,s and Lg (H )∗ g Cl . || || q,s ≤ || || q′,s Suppose a is (p,q,s)-atom associated to the dilated ball B = x0 + Bk. Since ga = (g P )a for all P we have − ∈Ps R R Lga = ga = inf (g P )a | | P ∈Ps − Z Z 1/q 1/q′ ′ (8.15) a q inf g P q ≤ | | P ∈Ps | − | ZB Z 1/q′ 1/q−1/p l+1/q′ q′ B B inf g P = g l . C ′ ≤ | | | | P ∈Ps | − | || || q ,s Z At this moment the reader might be tempted to use Lemma 8.2. That is we can try to deﬁne L f for arbitrary f Hp by using the atomic decomposition of f, g ∈ ∞ ∞ (8.16) Lgf = κiLgai, where f = κiai. i=1 i=1 X X ∞ p Since for every ε > 0 we can choose a decomposition so that i=1 κi (1 + p | | ≤ ε) f p we have || ||Hq,s P ∞ ∞ 1/p p (8.17) Lgf κi Lgai g Cl κi (1 + ε) g Cl f Hp . | |≤ | || | ≤ || || q′,s | | ≤ || || q′,s || || q,s i=1 i=1 X X This may seem to show that Lg is bounded with the appropriate constant. The problem with this argument is the issue of well-deﬁnedness of Lg. Namely, l q given g Cq′ ,s, we ought to make sure that if f Θs has an atomic decomposition ∞∈ ∈ f = i=1 κiai then necessarily P ∞ fg = κi aig. i=1 Z X Z As we will see this fact is very non-trivial and it requires a rather subtle approximation argument. For the classical isotropic case, the interested reader is directed to [GR, Section III.5], where detailed exposition of this fact can be found. Note that there is no problem if g belongs to the Schwartz class, since the atomic decomposition of f converges in the sense of distributions. The following lemma comes to the rescue. 52 1. ANISOTROPIC HARDY SPACES

q q Lemma 8.4. Suppose that (p,q,s) is admissible and f Θs, where Θs is given l ′ ∈ by (8.14). Suppose g Cq′,s, 1/q +1/q = 1, l = 1/p 1. There exists s˜ s so ∈ ∞ ∞− p ≥ that if f is decomposed into f = i=1 κiai, where i=1 κi < and ai’s are (p,q, s˜)-atoms, then | | ∞ P P ∞ ∞ (8.18) Lgf = fg = κi aig = κiLgai. i=1 i=1 Z X Z X

q Given Lemma 8.4 the rest of the proof is immediate. Suppose f Θs. By ∞ ∈ Theorem 6.5 we can ﬁnd an atomic decomposition of f = i=1 κiai, where

∞ P p 1/p ( κi ) 2 f Hp C f Hp , | | ≤ || || q,s˜ ≤ || || q,s i=1 X and ai’s are (p,q, s˜)-atoms. By (8.15) and (8.18)

∞ ∞ 1/p p l l p (8.19) Lgf κi Lgai g C κi C g C f Hq,s . | |≤ | || | ≤ || || q′,s | | ≤ || || q′,s || || i=1 i=1 X X p Therefore, Lg extends uniquely to a bounded functional on Hq,s. Furthermore, by Lemma 8.2 and (8.15) Lg (Hp )∗ g Cl . This ﬁnishes the proof of Theorem || || q,s ≤ || || q′,s 8.3.

Remark. As a consequence of Theorem 8.3 we conclude that if we write f = ∞ q ∞ p i=1 κiai Θs, where i=1 κi < and ai’s are (p,q,s)-atoms then (8.18) still holds.∈ As an indication why| | this result∞ is non-trivial we recall the following Pobservation due to Meyer,P see [GR, MTW]. p In Definition 4.3 of the atomic norm f Hq,s , it is not legitimate to take the infimum only over finite linear combinations|| of|| atoms even in the case when f admits such a finite decomposition, e.g., when f is itself an atom. Indeed, such infimum may be much larger than f p , which is evidenced by an example due to Meyer, || ||Hq,s see [GR, Chapter III.8.3]. Therefore, even though finite linear combinations of atoms are dense in Hp, in order to compute their Hp norm, it is not enough to look only at finite decompositions. To show Lemma 8.4 we need the following approximation lemma.

l ′ Lemma 8.5. Suppose g C ′ , where l 0, 1 q , s = 0, 1,... There ∈ q ,s ≥ ≤ ≤ ∞ exists s˜ s, a constant C > 0 independent of g, and a sequence of test functions ≥ (g ) N so that k k∈ ⊂ S

(8.20) gk Cl C g Cl for all k N, || || q′,s˜ ≤ || || q′,s ∈ and (1/q +1/q′ = 1),

q (8.21) lim f(x)gk(x)dx = f(x)g(x)dx for all f Θs, . k→∞ Rn Rn ∈ Z Z 8. DUALS OF Hp 53

We are going to use two simple observations about Campanato spaces. Firstly, note that if DAg(x)= g(Ax) then

1/q′ −l 1 q′ D g l = sup inf B g(Ax) P (x) dx A C ′ k q ,s x ∈Rn P ∈Ps || || 0 | | Bk x0+Bk | − | k∈Z | | Z 1/q′ (8.22) l −l 1 q′ = sup inf b Bk+1 g(x) P (x) dx x ∈Rn P ∈Ps 0 | | Bk+1 Ax0+Bk+1 | − | k∈Z | | Z l = b g Cl . || || q′,s l Secondly, we can deﬁne an equivalent norm on Cq′,s by setting

1/q′ −l 1 q′ ′ (8.23) g Cl = sup B g(x) πBg(x) dx (1 q < ), ||| ||| q′,s | | B | − | ≤ ∞ B∈B | | ZB

−l ′ (8.24) g l = sup B ess sup g(x) π g(x) (q = ). C∞,s x∈B B ||| ||| B∈B | | | − | ∞ where πB g is the natural projection on s given by (8.8). Indeed, for any B and P by the Minkowski inequalityP we have ∈ B ∈Ps 1/q′ 1 ′ g(x) π g(x) q dx B | − B | | | ZB 1/q′ 1/q′ 1 ′ 1 ′ g(x) P (x) q dx + P (x) π g(x) q dx ≤ B | − | B | − B | | | ZB | | ZB 1/q′ 1 ′ 1 g(x) P (x) q dx + C g(x) P (x) dx ≤ B | − | B | − | | | ZB | | ZB 1/q′ 1 ′ (C + 1) g(x) P (x) q dx . ≤ B | − | | | ZB using P (x) π g(x)= π (P g)(x) and (8.9). Therefore we have − B B − l (8.25) g Cl g Cl (C + 1) g Cl for all g Cq′,s. || || q′,s ≤ ||| ||| q′,s ≤ || || q′,s ∈ l Proof of Lemma 8.5. Suppose g C ′ . Take a nonnegative function ϕ ∈ q ,s ∈ C∞ with compact support and ϕ = 1. Let ϕ (x)= b−kϕ(A−kx). If q′ < then k ∞ ′ g ϕ gR in Lq (Rn) as k . ∗ k → loc → −∞ q′ Indeed, it suﬃces to apply Theorem 6.8 in L for truncations g1B(0,r) for suﬃciently large r> 0. Therefore

q (8.26) f(x)(g ϕk)(x)dx f(x)g(x)dx as k for all f Θs. Rn ∗ → Rn → ∞ ∈ Z Z ′ If q = we use Theorem 3.7 applied for truncations g1B(0,r) for suﬃciently large r> 0 to∞ obtain (g ϕ )(x) g(x) as k for a.e. x Rn. ∗ k → → −∞ ∈ 54 1. ANISOTROPIC HARDY SPACES

By the Lebesgue Dominated Convergence Theorem (8.26) holds also for q′ = . Clearly, g ϕ C∞ and moreover ∞ ∗ k ∈

(8.27) g ϕk Cl g Cl for all k N. || ∗ || q′,s ≤ ||| ||| q′,s ∈ Indeed, for every B and k N deﬁne a function P by ∈B ∈ k

Pk(x)= π−y+Bg(x y)ϕk(y)dy. Rn − Z Since we can write π g(x y)= c (y)(x y)α and the coeﬃcients c (y) −y+B − |α|≤s α − α are continuous functions of y, P is a polynomial of degree s. By the Minkowski k P inequality ≤

1/q′ 1 q′ (g ϕk)(x) Pk(x) dx B B | ∗ − | | | Z ′ ′ 1 q 1/q = (g(x y) π−y+Bg(x y))ϕk(y)dy dx B Rn − − − | | ZB Z 1/q′ 1 q′ g(x y) π−y+Bg(x y) dx ϕk(y) dy ≤ Rn B | − − − | | | Z | | ZB 1/q′ 1 q′ l = g(z) π g(z) dz ϕ (y)dy g l B . −y+B k C ′ Rn y + B | − | ≤ ||| ||| q ,s | | Z |− | Z−y+B This shows (8.27). Formulae (8.26) and (8.27) suggest that it is enough to show the lemma for l ∞ ∞ g C ′ C . That is indeed the case. Let φ C be such that supp φ B , ∈ q ,s ∩ ∈ ⊂ 0 0 φ(x) 1, and φ(x)=1 for x B−1. We claim that there is a constant C and s˜ ≤ s such≤ that ∈ ≥ l ∞ (8.28) (g πB0 g)φ Cl C g Cl for all g Cq′ ,s C . || − || q′,s˜ ≤ || || q′,s ∈ ∩

l ∞ Indeed, take any g Cq′,s C with g Cl 1. For brevity, we only consider ∈ ∩ ||| ||| q′,s ≤ ′ ′ the case q < ; the case q = uses a similar argument. Let G = g πB0 g. ∞ q′ ∞ q′ lq′ +1 − Since supp φ B0, Rn Gφ G B0 = 1. Therefore, if we take ⊂ | | ≤ B0 | | ≤ | | ball B , B = x + B and k 0 then ∈B 0 R k ≥R 1/q′ 1 ′ B −l G(x)φ(x) q dx 1. | | B | | ≤ | | ZB

Hence, to show (8.28) we can restrict to balls B = x0 + Bk with k < 0. Let P1 = πB g = πBG. By (8.9)

1/q′ 1/q′ 1 ′ 1 ′ ′ (8.29) P (x) q dx C G(x) q dx Cb−k/q . B | 1 | ≤ B | | ≤ | | ZB | | ZB

Let P2(x) be the Taylor polynomial of φ at x0 of degree r (to be speciﬁed later), i.e., P (x) = ∂αφ(x )(x x )α/α!. By the Taylor Theorem the remainder 2 |α|≤r 0 − 0 satisﬁes φ(x) P (x) C′ x x r+1 with the constant C′ independent of x . | P− 2 | ≤ | − 0| 0 8. DUALS OF Hp 55

Finally, let P (x) = P1(x)P2(x) be a polynomial of degree at mosts ˜ = s + r. By the Minkowski inequality and (8.29)

1/q′ ′ G(x)φ(x) P (x) q dx | − | ≤ ZB 1/q′ 1/q′ ′ ′ G(x)φ(x) P (x)φ(x) q dx + P (x)φ(x) P (x)P (x) q dx | − 1 | | 1 − 1 2 | ZB ZB 1/q′ 1/q′ ′ ′ φ G(x) P (x) q dx + sup φ(x) P (x) P (x) q dx ≤ || ||∞ | − 1 | | − 2 | | 1 | ZB x∈B ZB l+1/q′ −k/q′ ′ r+1 l+1/q′ −k/q′ ′ k r+1 B + b C sup x x0 B + b C (cλ−) ≤ | | x∈x0+Bk | − | ≤ | | ′ ′ ′ ′ B l+1/q + Cb−k/q bk(l+2/q ) = (C + 1) B l+1/q . ≤ | | | | r+1 l+2/q′ Here we need to choose r large enough such that λ− b , i.e., take r = ′ ≥ (l +2/q ) ln b/ ln λ− . Since B is arbitrary this shows (8.28) withs ˜ = s + r. ⌊ ⌋ ∈B l ∞ We are now ready to deﬁne a sequence (g ) N for g C ′ C . Let k k∈ ⊂ S ∈ q ,s ∩ g˜ = D k g and g = D −k ((˜g π g˜ )φ). By (8.22) and (8.28) k A k A k − B0 k kl (˜gk πB0 g˜k)φ Cl C g˜k Cl = Cb g Cl || − || q′ ,s˜ ≤ || || q′,s || || q′,s Therefore (8.20) holds, since

−kl (8.30) gk Cl = b (˜gk πB0 g˜k)φ Cl C g Cl . || || q′,s˜ || − || q′ ,s˜ ≤ || || q′,s Moreover,

(8.31) g (x)= g(x) (D −k π D k )g(x)= g(x) π g(x) for x B . k − A ◦ B0 ◦ A − Bk ∈ k−1 Thus (8.21) also holds. ∞ l To end the proof we must relax the assumption that g C . Suppose g C ′ ∈ ∈ q ,s is arbitrary. Deﬁne the sequence (g ) N by g = D −k ((˜g π g˜ )φ), where k k∈ ⊂ S k A k − B0 k g˜k = DAk (g ϕk). Combining (8.27) and (8.30) yields (8.20), whereas (8.26) and (8.31) yield (8.21),∗ completing the proof of Lemma 8.5.

To completely ﬁnish the proof of the duality of Hp spaces we must establish Lemma 8.4.

q ∞ Proof of Lemma 8.4. Suppose f Θs is decomposed into f = i=1 κiai, ∞ p ∈ where i=1 κi < and ai’s are (p,q, s˜)-atoms, wheres ˜ s is the same as | | ∞ l ′ ≥ P in Lemma 8.5. Suppose also that g C ′ , 1/q +1/q = 1, l = 1/p 1. Let P ∈ q ,s − (g ) N be a sequence guaranteed by Lemma 8.5. For every k N we have k k∈ ⊂ S ∈ ∞ ∞

(8.32) Lgk f = fgk = κi aigk = κiLgk ai, i=1 i=1 Z X Z X since convergence in Hp implies convergence in ′ by Theorem 4.5. By (8.21) S

lim ai(x)gk(x)dx = ai(x)g(x)dx for all i N. k→∞ Rn Rn ∈ Z Z 56 1. ANISOTROPIC HARDY SPACES

∞ l l By (8.15) and (8.20) we have Lgk ai gk C C g C . Since i=1 κi | | ≤ || || q′,s˜ ≤ || || q′,s | | ≤ ∞ p 1/p ( i=1 κi ) < we can take limit as k in (8.32) by the LebesgueP Dom- inated| Convergence| ∞ Theorem applied to the→ counting ∞ measure on N. This shows (8.18).P

As an immediate consequence of Theorems 6.5 and 8.3 we have Corollary 8.6. Suppose that l 0, 1 q, q′ (q, q′ < if l = 0), and ′ l ≥l ≤ ≤ ∞ ∞ s,s l ln b/ ln λ− . Then Cq,s = Cq′,s′ and the seminorms Cl , Cl ≥ ⌊ ⌋ || · || q,s || · || q′,s′ 0 0 are equivalent. If l = 0 then Cq,s = C1,0 is the space of BMO (bounded mean oscillation). The following simple proposition is very useful. Proposition 8.7. Suppose g CN (Rn) is bounded and all partial derivatives α ∈ ∂ g of order N, α = N, are bounded. Then for every 0 l N ln λ−/ ln b and s l ln b/ ln λ ,| g| Cl . ≤ ≤ ≥⌊ −⌋ ∈ ∞,s Proof. The case N = 0 is trivial. If N 1 then by Corollary 8.6 it suﬃces l ≥ to show that g C∞,N−1. We need to show that there exists a constant C such that for every dilated∈ ball B , ∈B (8.33) inf sup g(x) P (x) C B l. P ∈PN−1 x∈B | − |≤ | | Since g is bounded, (8.33) trivially holds for large balls with B 1. Take any n | | ≥ B = x0 + Bk , where x0 R and k < 0. Let P (x) N−1 be the Taylor polynomial of g∈at B the point x ∈of order N 1. By the Taylor∈ PRemainder Theorem 0 − α N Nk g(x) P (x) C sup ∂ g ∞ sup z C(λ−) for all x B. | − |≤ |α|=N || || z∈Bk | | ≤ ∈

Since (λ )N bl, we obtain (8.33). − ≥

There are redundant alternative ways of defining Campanato spaces. Let ∆h n denote the difference operator by a vector h R , i.e., ∆hg(x)= g(x + h) g(x). l ∈ − Suppose that l > 0 and s 1. Define by Cs the space of all continuous functions g on Rn such that ≥

∆ . . . ∆ g(x) c(ρ(h )+ . . . + ρ(h ))l for all x Rn, (h ,... ,h ) (Rn)s | h1 hs |≤ 1 s ∈ 1 s ∈ for some constant c. The infimum over all constants c satisfying the above de- l l l fines the seminorm of g in Cs. Some effort is needed to show that Cq,s = Cs for sufficiently large s and 1 q , see [Gr, Ja, JTW, Kr] for the isotropic case. ≤ ≤ ∞ Example. The best known example of Campanato spaces occurs when the dilation A defines the usual isotropic structure on Rn, for example if A = 2Id. In l n this case the Campanato spaces Cq,s(R ) coincide with the homogeneous Hölder γ n n spaces C˙ (R ) (sometimes also called Lipshitz spaces Λγ (R )) with γ = nl, for the proof see [GR, Section III.5]. The space C˙ γ (Rn) for γ > 0 is defined as follows. Let γ = γ + γ′ 1, ⌈ ⌉ − 0 < γ′ 1. The space C˙ γ (Rn) consists of all functions g in C⌈γ⌉−1(Rn) ( γ 1 ≤ ⌈ ⌉− 8. DUALS OF Hp 57 times continuously differentiable) with the norm

−γ′ α ′ g C˙ γ (Rn) = sup sup h ∆h∂ g ∞ if γ < 1, || || |α|=⌈γ⌉−1 h∈Rn\{0} | | || || −1 2 α ′ g C˙ γ (Rn) = sup sup h ∆h∂ g ∞ if γ =1. || || |α|=⌈γ⌉−1 h∈Rn\{0} | | || || For example, if 0 <γ< 1 then C˙ γ (Rn) is the usual H¨older space consisting of all functions g satisfying g(x + h) g(x) c h γ for all x, h Rn, | − |≤ | | ∈ and if γ = 1 then C˙ 1(Rn) is the Zygmund class [Zy] consisting of all functions g satisfying g(x +2h) 2g(x + h)+ g(x) c h for all x, h Rn. | − |≤ | | ∈ 58 1. ANISOTROPIC HARDY SPACES

9. Calderón-Zygmund singular integrals on Hp In this section we present the theory of Calderón-Zygmund singular integrals on the anisotropic Hp spaces for 0 1 this theory reduces to studying Lp spaces and therefore follows≤ from the general theory of Calderón- Zygmund singular integrals on the spaces of homogeneous type, see [St2, Chapter 1.5]. We start with some preliminaries. Let T : (Rn) ′(Rn) be a continuous linear operator. By the Schwartz Kernel TheoremS there→ exist S s S ′(Rn Rn) such that ∈ S × (9.1) T (f),g = S,g f for all f,g (Rn). h i h ⊗ i ∈ S Let Ω = (x, y) Rn Rn : x = y . We say that a distribution S is regular on Ω if there exists{ a∈ locally× integrable6 function} K(x, y) on Ω such that

(9.2) S(G)= K(x, y)G(x, y)dxdy for all G (Rn Rn), supp G Ω. ∈ S × ⊂ ZΩ Definition 9.1. Let T : (Rn) ′(Rn) be a continuous linear operator. We say that T is a Calderón-ZygmundS → operator S (with respect to a dilation A with a quasi-norm ρ) if there are constants C > 0, γ > 0 such that (i) a distribution S given by (9.1) is regular on Ω with kernel K satisfying (9.3) K(x, y) C/ρ(x y), | |≤ − (ii) If (x, y) Ω and ρ(x′ x) ρ(x y)/b2ω then ∈ − ≤ − ρ(x′ x)γ (9.4) K(x′,y) K(x, y) C − , | − |≤ ρ(x y)1+γ − (iii) If (x, y) Ω and ρ(y′ y) ρ(x y)/b2ω then ∈ − ≤ − ρ(y′ y)γ (9.5) K(x, y′) K(x, y) C − , | − |≤ ρ(x y)1+γ − (iv) T extends to a continuous linear operator on L2(Rn) with T C. || || ≤ A few remarks are needed. It can be shown that operators in the above class are bounded from L1 to weak-L1. This follows from the general theory of Calderón- Zygmund operators defined on arbitrary spaces of homogeneous type, see [St2, Chapter 1]. By the Marcinkiewicz Interpolation Theorem they are bounded from Lp into Lp for 1 < p 2. By taking duals they are also bounded for 2 p < , hence in the range 1

Since we are interested in boundedness results on Hp spaces, 0

0 and a = a1 + . . . + an. Then ρ given by ρ(x1,... ,xn) = max x a/ai is a quasi-norm associated with A. Pick any x, y Ω with 1≤i≤n | i| ∈ det A k ρ(x y) < det A k+1 for some k Z. Since | | ≤ − | | ∈ ∂α[K( , Ak )](x, A−ky)= ek(α1a1+...+αnan) ∂αK(x, y) Cρ(x y)−1, | y · · | y |≤ − where α = (α1,... ,αn), the condition (9.6) takes a more familiar form ∂αK(x, y) Cρ(x y)−1−(α1a1+...+αnan)/a for α m, | y |≤ − | |≤ see [St2, Chapter 13.5]. However, if A has some non-trivial blocks in its Jordan decomposition then (9.6) does not have a more explicit form due to the complexity of the action of A on Rn. The following basic fact provides a suﬃcient condition for a function to belong to Hp.

Lemma 9.3. Let (p,q,s) be admissible and δ > max(1/p,s ln λ+/ ln b + 1). Suppose that f is a measurable function on Rn such that for some constant C > 0 we have 1 1/q (9.7) f(x) q dx C B −1/p for some k Z, B | | ≤ | k| ∈ | k| ZBk c (9.8) f(x) C B −1/pρ(A−kx)−δ for x B , | |≤ | k| ∈ k

(9.9) f(x)xαdx =0 for α s. Rn | |≤ Z p ′ ′ Then f H with f p C , where C depends only on C. ∈ || ||Hq,s ≤ 60 1. ANISOTROPIC HARDY SPACES

We remark that (9.9) is meaningful since f(x) (1 + x s) is integrable by (9.7) and (9.8). Indeed, by Lemma 3.2 | | | |

f(x) x s Cρ(A−kx)−δc′ρ(x)s ln λ+/ ln b = Cc′bkδρ(x)s ln λ+/ ln b−δ for ρ(x) 1. | || | ≤ ≥ Proof. Given a ball B consider the natural projection π : L1(B) ∈B B →Ps given by (8.8). Deﬁne the complementary projectionπ ˜B = Id πB , i.e.,π ˜B f = f π f. By (8.9)π ˜ is bounded on Lq(B), i.e., − − B B

(9.10) π˜ f q C f q , || B ||L (B) ≤ 0|| ||L (B) with the constant C independent of B . Moreover, 0 ∈B π˜ f(x)xα =0 for α s. B | |≤ ZB We want to represent f as a combination of atoms. To do this deﬁne the ∞ sequence of functions (gj )j=k by gj =π ˜Bj (f)1Bj . Clearly, supp gj Bj for j k. 1/q−1/p ⊂ ≥ Since gk q C0 f1Bk q C0C Bk and gk has vanishing moments up to || || ≤ || || ≤ | | −1 order s, gk is at most C0C multiplicity of some (p,q,s)-atom (namely (C0C) gk). We claim that g f in L1 (and hence in ′) as j . It suﬃces to show j → S → ∞ 1 that πBj f L (Bj ) 0 as j . Indeed, let Qα : α s be an orthonormal || || → → ∞2 { | | ≤ } basis of s with respect to the L (B0) norm. By the argument used to show (8.9) we have P

π f =(D −j π D j )f = D j f(x)Q (x)dx D −j Q Bj A ◦ B0 ◦ A A α A α |α|≤s ZB0 (9.11) X −j −j = f(x)Qα(A x)dx b DA−j Qα, Bj |αX|≤s Z j j where D f(x)= f(Ax). We also have D −j Q 1 = b Q 1 b and A || A α||L (Bj ) || α||L (B0) ≤

−j −j f(x)Qα(A x)dx = f(x)Qα(A x)dx 0 as j , B − Bc → → ∞ Z j Z j −j by (9.9) and the uniform boundedness of coeﬃcients of the polynomials Qα(A x) for j 0. This shows ≥ ∞ (9.12) f = g + (g g ) with convergence in L1(Rn). k j+1 − j Xj=k p In fact, we will prove that we also have convergence in H by showing that gj+1 gj are appropriate multiples of (p, ,s)-atoms supported on B . Indeed, − ∞ j+1 g g = π˜ (f)1 π˜ (f)1 || j+1 − j||∞ || Bj+1 Bj+1 − Bj Bj ||∞ (9.13) = f1 1 π f + 1 π f || Bj+1\Bj − Bj+1 Bj+1 Bj Bj ||∞ f1 + 1 π f + 1 π f . ≤|| Bj+1\Bj ||∞ || Bj+1 Bj+1 ||∞ || Bj Bj ||∞ By (9.8) f1 Cb−k/pbδ(k−j) = Cb−j/pb(δ−1/p)(k−j). || Bj+1\Bj ||∞ ≤ 9. CALDERON-ZYGMUND´ SINGULAR INTEGRALS ON Hp 61

−j Since 1Bj DA Qα ∞ = Qα1B0 ∞ C1 for all α s, by (9.9) and (9.11) we have || || || || ≤ | | ≤ −j −j 1Bj πBj f ∞ C1b f(x)Qα(A x)dx . || || ≤ Bc |α|≤s Z j X Since Q (x) C x s for x Bc for some constant C > 0 | α |≤ 2| | ∈ 0 2

−j −j s f(x)Qα(A x)dx C2 f(x) A x dx Bc ≤ Bc | || | Z j Z j

−1/p −k −δ −j s ln λ+/ ln b C2 C Bk ρ( A x) ρ(A x) dx ≤ Bc | | Z j

−k/p δ(k−j) −j −δ+s ln λ+/ ln b = C2Cb b ρ(A x) dx Bc Z j

−k/p δ(k−j) j −δ+s ln λ+/ ln b j(1−1/p) (δ−1/p)(k−j) C2Cb b b ρ(x) dx = C3b b . ≤ Bc Z 0 Inserting the last three inequalities into (9.13) we conclude that

g g C b−(j+1)/pb(δ−1/p)(k−j) for j k, || j+1 − j ||∞ ≤ 4 ≥ for some constant C . Since g ’s have vanishing moments up to order s, g g 4 j j+1 − j is a κj multiple of a (p, ,s)-atom aj supported on Bj+1, where gj+1 gj = κj aj , (δ−1/p)(k−j) ∞ − κj = C4b . By (9.12)

∞ 1/p ∞ 1/p p p p p (pδ−1)(k−j) ′ f p (C C) + κ = (C C) + C b = C , || ||Hq,s ≤ 0 | j | 0 4 Xj=k Xj=k which ends the proof of the lemma.

Remark. Since translations are isometries in Hp we can immediately gener- n alize Lemma 9.3 to functions centered at arbitrary x0 R . Conditions (9.7) and (9.8) can be substituted then by ∈

1 1/q (9.14) f(x) qdx C B −1/p, B | | ≤ | k| k x0+Bk | | Z c (9.15) f(x) C B −1/pρ(A−k(x x ))−δ for x x + B . | |≤ | k| − 0 ∈ 0 k A function f satisfying (9.9), (9.14), and (9.15) with C = 1 is refferred to as a molecule localized around the dilated ball x0 + Bk . Hence, a molecule satisfies all the properites of an atom as in Definition∈ 4.1 B with the exception of compact support condition, which is replaced by a suitable decay condition (9.15). Therefore, Lemma 9.3 says that, for fixed (p,q,s) and decay exponent δ, every molecule f belongs to Hp with the Hp norm bounded by some constant depending only on (p,q,s) and δ. Moreover, by examining the arguments of Lemma 9.3, one can easily see that, for fixed 0

Our next goal is to show that Calderón-Zygmund operators map atoms into molecules. Generally, we can not expect this unless we also assume that our operator preserves vanishing moments. The precise meaning of this is given in Definition 9.4. Definition 9.4. We say that a Calderón-Zygmund operator T of order m sat- ∗ α 2 isfies T (x ) = 0 for all α s, where s ∗ α max((1/p 1) ln b/ ln λ−,s ln λ+/ ln λ−). Assume T is (CZ-m) and T (x )=0 for α s. Then− there exists a constant C, depending only on the Calderón-Zygmund | |≤ norm T (m) of T , such that Ta Hp C for every (p, 2,m 1)-atom a. || || || || 2,s ≤ − Proof. Suppose a (p, 2,m 1)-atom a is supported in the ball x0 + Bk. We estimate Ta separately around and− away from the support of an atom a.

2 2 2 1−2/p (9.16) Ta(x) dx Ta(x) dx C a 2 B . | | ≤ Rn | | ≤ || || ≤ | | Zx0+Bk+ω Z Suppose x x0 + Bk+l+ω+1 Bk+l+ω for some l 0 and y x0 + Bk. Then x y B ∈ but x y \ B . Hence, by (9.6)≥ and the chain∈ rule − ∈ k+l+2ω+1 − 6∈ k+l (9.17) ∂α[K( , Ak+l )](x, A−k−ly) C′/ρ(x y)= C′b−k−l for α m, | y · · |≤ − | |≤ where C′ depends only on the constant C in (9.6). Away from the support of the atom a, we estimate Ta by

(9.18) Ta(x) = K(x, y)a(y)dy = K˜ (x, A−k−ly)a(y)dy , | | Zx0+Bk Zx0+Bk

˜ k+l ˜ where K(x, y)= K(x, A y). Now we expand K(x, y) into the Taylor polynomial of degree m 1 (only in y variable) at the point (x, A−k−lx ). That is, − 0 α −k−l ∂ K˜ (x, A x0) (9.19) K˜ (x, y′)= y (y′ A−k−lx )α + R (y′). α! − 0 m |α|≤Xm−1 ′ −k−l Here we think that y = A y and that y ranges over x0 + Bk as in (9.18). Since we are going to apply (9.19) for y′ A−k−lx + B , the remainder R satisﬁes ∈ 0 −l m ′ ˜ α ˜ ′ −k−l m Rm(y ) C sup sup ∂y K(x, z) y A x0 | |≤ z∈A−k−lx +B |α|=m | || − | (9.20) 0 −l C′b−k−l sup z m, ≤ z∈B−l | | 9. CALDERON-ZYGMUND´ SINGULAR INTEGRALS ON Hp 63 because the partial derivatives of K˜ (x, y) in the variable y of order m satisfy (9.17). Combining (9.18), (9.19), and (9.20) and using the moment condition of atoms we have

Ta(x) = R (A−k−ly)a(y)dy R (A−k−ly)a(y) dy | | m ≤ | m | Zx0+Bk Zx0+Bk

C′b−k−l sup z m a (y) dy C′b−k−l(cλ−l)mbk−k/p ≤ | | | | ≤ − z∈B−l Zx0+Bk ′′ −k/p −l −lm ′′ −k/p −lδ = C b b λ− = C b b , where δ = m ln λ−/ ln b +1. Therefore, Ta satisfies (9.14) and (9.15). Furthermore, we have T ∗(xα) = 0 for α s meaning Tf has vanishing moments up to order s whenever f L2 with compact| | ≤ support has vanishing moments up order m 1, i.e., f is a multiple∈ of a (p, 2,m 1)-atom. By Lemma 9.3 this implies that there− − is a constant C independent of a so that Ta Hp C. || || 2,s ≤ Lemma 9.5 strongly suggests that any Calderón-Zygmund operator T of order m extends to a bounded operator on the Hardy space Hp, where m is sufficiently large and depends on 0

(9.21) T f(x)= K (x, y)f(y)dy for x Rn, i i ∈ Z where f L2 has compact support, and ∈ (9.22) T f Tf in L2 as i for all f L2. i → → −∞ ∈ ∗ α ∗ α Furthermore, if T (x )=0 for α s then also (Ti) (x )=0 for all α s and i N. | |≤ | |≤ ∈ One possible approximation technique for Calder´on-Zygmund operators in- volves the truncation of the kernel. Given a C∞ function ϕ such that ϕ(x)=0 c for x B−1, ϕ(x) = 1 for x (B0) we can deﬁne kernels Ki by Ki(x, y) = K(x, y∈)ϕ(A−i(x y)). By adapting∈ arguments of Meyer in [MC] it is possible to show, though it− is not automatic, that the family of the corresponding operators

(Ti) is (CZ-m) with uniform constants. Furthermore, Tij ’s (after taking a subsequence) converge weakly to Mh + T as ij , where Mh denotes the operator of multiplication by a function h L∞, see→ [MC, −∞ Chapter 7, Proposition 3]. How- ∈ ever, if K(x, y) is not a convolution kernel then in general the Ti’s might not preserve vanishing moments. Therefore, this approximation is not suitable for showing 64 1. ANISOTROPIC HARDY SPACES boundedness of T : Hp Hp. It can be used, though, to show boundedness of T : Hp Lp. → Instead→ we are going to use an approximation based on smoothing by a convolution with a compactly supported smooth function ϕ. Indeed, suppose ϕ is C∞, ′ ′ supp ϕ B0 and ϕ = 1. Forany i Z deﬁne a convolution operator Ri : , by ⊂ ∈ S → S R (9.23) R f = f ϕ . i ∗ i By Lemma 6.6 for every 0

Proof of Theorem 9.6. We define operators Ti by Ti = RiT Ri, where Ri is given by (9.23). Since T : ′ it has a kernel K ′(Rn Rn) by the Schwartz i S → S i ∈ S × Kernel Theorem. We claim that Ki is a regular distribution which is identified with the function Ki given by (9.26) K (x, y)= T (τ ϕ ), τ ϕ˜ , i h y i x ii whereϕ ˜ (z)= ϕ ( z). Indeed, recall that for any f ′, R f = f ϕ is a regular i i − ∈ S i ∗ i distribution identified with Rif(x) = f(τxϕ˜i). For any ψ , RiT Ri(ψ) is also regular distribution and ∈ S (R T R )ψ(x)= T R (ψ), τ ϕ˜ = T (ϕ ψ), τ ϕ˜ i ◦ ◦ i h i x ii h i ∗ x ii = T ψ(y)ϕ ( y)dy , τ ϕ˜ = ψ(y)T (τ ϕ )( )dy, τ ϕ˜ i ·− x i y i · x i Z Z = ψ(y) T (τ ϕ ), τ ϕ˜ dy. h y i x ii Z The next to last equality is justified by approximating the integral in L2 by finite linear combinations of functions ϕ( y) for y Rn. This shows (9.26). Moreover, if x y 2B B then supports·− of τ ϕ and∈ τ ϕ˜ are disjoint and − 6∈ i ⊂ i+ω y i x i

Ki(x, y)= T (τyϕi), τxϕ˜i = K(u, v)ϕi(v y)dvϕi(x u)du h i Rn Rn − − (9.27) Z Z = K(x u,y + v)ϕi(u)ϕi(v)dudv. Rn Rn − Z Z Fix (x ,y ) Ω and suppose that x y B B for some l i. 0 0 ∈ 0 − 0 ∈ l+2ω+1 \ l+2ω ≥ Since u and v range over Bi in (9.27) x0 u (y0 + v) = x0 y0 u v Bc + B Bc . Also x y u v− B− . Note that− for α− −m ∈ l+2ω i+ω ⊂ l+ω 0 − 0 − − ∈ l+3ω+1 | |≤ (9.28) ∂α[K ( , Al+2ω )](x, y)= ∂α[K( u, Al+2ω +v)](x, y)ϕ (u)ϕ (v)dudv, y i · · y ·− · i i Z Z by moving the diﬀerentiation inside the integral. By the chain rule and (9.6) there is a constant C′ so that (9.29) ∂α[K( , Al+2ω )](x u, A−l−2ω(y + v)) C′b−l−2ω | y · · 0 − 0 |≤ 9. CALDERON-ZYGMUND´ SINGULAR INTEGRALS ON Hp 65 for all α m and u, v B . Combining (9.28) and (9.29) we obtain | |≤ ∈ i (9.30) ∂α[K ( , Al+2ω )](x , A−l−2ωy ) C′b−l−2ω = C′ρ(x y )−1. | y i · · 0 0 |≤ 0 − 0 Suppose next that x0 y0 Bl+1 Bl for some l

Proof. For any α m˜ by the Fubini Theorem | |≤ α α x Rif(x)dx = x f(x y)ϕi(y)dydx Rn Rn Rn − Z Z Z α = (x + y) f(x)ϕi(y)dxdy =0, Rn Rn Z Z because the integrand belongs to L1(Rn Rn) since the support of ϕ is bounded. × We are now ready to prove the main result of this section. Theorem 9.8. Suppose T is a Calderón-Zygmund operator of order m. If p satisfies (ln λ )2 (9.33) 0 1/p 1 < − m, ≤ − ln b ln λ+ then T extends to the continuous linear operator T : Hp(Rn) Hp(Rn) provided T ∗(xα)=0 for α s = (1/p 1) ln b/ ln λ . → | |≤ ⌊ − −⌋ Proof. Note that (9.33) guarantees that p and m satisfy the assumptions of Lemma 9.5 and Theorem 9.6, i.e., m > max((1/p 1) ln b/ ln λ ,s ln λ / ln λ ). − − + − Let (Ti)i∈Z be the sequence of operators with kernels Ki(x, y) given by Theorem 9.6. We claim that for any (p, 2,m 1)-atom a we have − (9.34) T a Ta in Hp as i . i → → −∞ Indeed, if a is supported in the ball x0 + Bk then Ria is supported in x0 + Bk+ω for i k. Since a Ria 2 0 as i , therefore a Ria is a κi multiple of some≤ (p, 2,m 1)|| atom− for|| i→ k by Lemma→ −∞ 9.7. Furthermore,− κ 0 as i . − ≤ i → → −∞ By Lemma 9.5, T (a Ria) Hp 0 as i . By (9.24) and (9.25), || − || 2,s → → ∞ p p p Ta R T R a p Ta R Ta p + R (Ta T R a) p 0 as i , || − i i ||H ≤ || − i ||H || i − i ||H → → −∞ which shows (9.34). ∞ The kernel function Ki(x, y) of Ti, given by (9.26), is C . Furthermore, by the Cauchy-Schwarz inequality all partial derivatives of K (x, y) of order N are i ≤ uniformly bounded for any natural number N. By taking N (1/p 1) ln b/ ln λ− we conclude by Proposition 8.7 that K (x, ) belongs to the Campanato≥ − space Cl i · ∞,s with l =1/p 1 for any fixed x Rn. Furthermore, − ∈ n (9.35) Ki(x, ) 1/p−1 C(i) for all x R , || · ||C∞,s ≤ ∈ and for some constant C = C(i) depending only on i Z. Take any f L2 with compact support and vanishing∈ moments up to order ∈ 2 m 1, that is f Θm−1. Let f = j∈N κj aj be an atomic decomposition of f into − ∈ p p (p, 2,m 1) atoms with j∈N κj 2 f Hp . By the duality Theorem 8.3, − | P| ≤ || || 2,m−1 K (x, ) defines a bounded functional on Hp and by (8.18) and (9.35), i · P n (9.36) Tif(x)= κj Tiaj (x) forevery x R . N ∈ Xj∈ Furthermore, the convergence in (9.36) is uniform on Rn by (8.15) and (9.35), and hence the series in (9.36) converges in ′. By Theorem 9.6, T ’s are uniformly S i 9. CALDERON-ZYGMUND´ SINGULAR INTEGRALS ON Hp 67 bounded (CZ-m) and hence by Lemma 9.5, Tiaj Hp C for some constant C independent of i and j. Therefore, || || ≤

p (9.37) Tif = κj Tiaj convergence in H , N Xj∈ Moreover, p p p p p Tif Hp κj Tiaj Hp 2C f p . H2,m−1 || || ≤ N | | || || ≤ || || Xj∈ Combining this with (9.34), (9.37), and letting i , we obtain → −∞ ′ 2 (9.38) Tf p C f p for f Θ || ||H ≤ || ||H ∈ m−1 for some constant C′ independent of f. Moreover,

p (9.39) Tf = κj Taj convergence in H . N Xj∈ 2 p Since Θm−1 is a dense subspace of H , T extends uniquely to a bounded operator in Hp by (9.38).

As a consequence we conclude that for any f Hp with f = κ a , where ∈ j∈N j j κ p < and a ’s are (p, 2,s)-atoms, we can compute Tf by applying the j∈N | j | ∞ j P formula (9.39). P In the case when T does not necessarily satisfy T ∗(xα)=0 for α s, we still have a boundedness result which is analogous to Theorem 9.8. | |≤ Theorem 9.9. Suppose T is a Calderón-Zygmund operator of order m. If p satisfies (9.33) then T extends to the continuous linear operator T : Hp(Rn) Lp(Rn). → Proof. The proof follows along the lines of the proof of Theorem 9.8 with the exception that (9.34) may not hold, since Ta may not even belong to Hp for a (p, 2,m 1) atom a. Nevertheless, Ta(x) satisfies the same size estimates (9.14) and − (9.15) as in Lemma 9.5. Therefore, Ta Lp C for some constant C depending on || || ≤ 2 T (m), but independent of an atom a. Moreover, (9.36) still holds for f Θm−1 || || 2 ∈ and consequently Tif Lp C f Hp for all i Z and f Θm−1. By (9.22) || || ≤ || || n ∈ ∈ we have Tif(x) Tf(x) for a.e. x R as i (possibly after taking a → ∈ → −∞ 2 subsequence). Therefore, by Fatou’s Lemma Tf Lp C f Hp for all f Θm−1. Therefore, T has a unique extension from H||p into|| Lp≤, which|| || concludes the∈ proof of Theorem 9.9. 68 1. ANISOTROPIC HARDY SPACES

10. Classiﬁcation of dilations In this section we investigate the question when two diﬀerent dilations generate the same anisotropic Hardy space Hp for some 0

0 then there is a constant c3 = c3(c1,c2) > 0 independent of D such that Dz c z for all z Rn. Similarily, if Dz c z for all z Rn | | ≥ 3| | ∈ | | ≥ 1| | ∈ and det D c2 for some c1,c2 > 0 then there is a constant c3 = c3(c1,c2) > 0 independent| | ≤of D such that D c . || || ≤ 3 n Lemma 10.2. Suppose we have two dilations A1 and A2 on R . Let ρ1 and ρ2 be the quasi-norms associated to A1 and A2, respectively. Then ρ1 and ρ2 are equivalent if and only if

k ⌊ǫk⌋ (10.1) inf inf A1 z / A2 z > 0, z∈Rn\{0} k∈Z | | | | or k ⌊ǫk⌋ (10.2) sup sup A1 z / A2 z < , z∈Rn\{0} k∈Z | | | | ∞ where (10.3) ε = ε(A , A ) = ln det A / ln det A . 1 2 | 1| | 2| Proof. Note that for every k Z ∈ k −ǫk k −⌊ǫk⌋ 1= det A1 det A2 det A1 det A2 (10.4) | | | | ≤ | | | | = det(AkA−⌊ǫk⌋) det A k det A −ǫk+1 = det A . | 1 2 | ≤ | 1| | 2| | 2| k −⌊ǫk⌋ By Lemma 10.1 applied to A1 A2 we obtain (10.1) is equivalent to (10.2). Assume that (10.1) and hence (10.2) holds. Let c and d denote the values of (10.1) and (10.2), respectively. Fix r> 0 so that for every z Rn 0 , and i =1, 2, k ∈ \{ } there exists k Z such that 1 Ai z r. Clearly, r = max( A1 , A2 ) works. Denote ∈ ≤ | |≤ || || || || c = inf ρ (z):1 z r , d = sup ρ (z):1 z r , 1 { 1 ≤ | |≤ } 1 { 1 ≤ | |≤ } c = inf ρ (z):1/d z r/c , d = sup ρ (z):1/d z r/c . 2 { 2 ≤ | |≤ } 2 { 2 ≤ | |≤ } Fix x Rn 0 and choose k Z such that 1 Akx r. Clearly ∈ \{ } ∈ ≤ | 1 |≤ (10.5) det A −kc ρ (x) det A −kd . | 1| 1 ≤ 1 ≤ | 1| 1 By (10.1) and (10.2) 1/d A⌊ǫk⌋x r/c, ≤ | 2 |≤ thus by (10.4) and (10.5) ρ (x)= det A −⌊ǫk⌋ρ (A⌊ǫk⌋x) det A −⌊ǫk⌋ sup ρ (z):1/d z r/c 2 | 2| 2 2 ≤ | 2| { 2 ≤ | |≤ } det A −k det A d ρ (x)c−1d det A . ≤ | 1| | 2| 2 ≤ 1 1 2| 2| Similarily ρ (x) ρ (x)d−1 inf ρ (z):1/d z r/c ρ (x)d−1c . 2 ≥ 1 1 { 2 ≤ | |≤ }≥ 1 1 2 10. CLASSIFICATION OF DILATIONS 69

Since x Rn 0 was arbitrary the quasi-norms ρ and ρ are equivalent. ∈ \{ } 1 2 Conversely, assume ρ1 and ρ2 are equivalent, i.e., there is a constant C > 0 so that 1/Cρ (x) ρ (x) Cρ (x). For any z Rn with z = 1 we have by (10.4) 2 ≤ 1 ≤ 2 ∈ | | ρ (AkA−⌊ǫk⌋z)= det A kρ (A−⌊ǫk⌋z) C det A kρ (A−⌊ǫk⌋z) (10.6) 1 1 2 | 1| 1 2 ≤ | 1| 2 2 C det A k det A −⌊ǫk⌋ρ (z) C det A sup ρ (x) : x =1 = D. ≤ | 1| | 2| 2 ≤ | 2| { 2 | | } n Clearly, for any r > 0, x : ρ1(x) r is a bounded set in R if ρ1 is a step homogeneous quasi-norm.{ By Lemma≤ 2.4} the same is true for any quasi-norm. Hence there is a constant d> 0 so that x : ρ1(x) D x : x d . Therefore k −⌊ǫk⌋ { ≤ }⊂{ | |≤ } by (10.6), A1 A2 z d. Since z was arbitrary we obtain (10.2). This ﬁnishes the proof of| the lemma.| ≤

Our next goal is to show the following theorem.

Theorem 10.3. Let ρ1 and ρ2 be the quasi-norms associated to dilations A1 and A2, respectively. Then ρ1 and ρ2 are equivalent if and only if for all r> 1 and all m =1, 2,...

m m (10.7) span ker(A1 λId) = span ker(A2 λId) , ǫ − − |λ[|=r |λ[|=r n where ǫ is given in (10.3). In (10.7) we think of A1 and A2 as linear maps on C .

Since Ai (i =1, 2) is a real matrix then the complex subspace span(ker(A λId)m ker(A λId)m) Cn, i − ∪ i − ⊂ is in fact a complexiﬁcation of the real subspace span(ker(A λId)m ker(A λId)m) Rn, i − ∪ i − ∩ for any λ C, and m = 1, 2,... Moreover, the conditions (10.1) and (10.2) are equivalent∈ to respectively

k ⌊ǫk⌋ (10.8) inf inf A1 z / A2 z > 0, z∈Cn\{0} k∈Z| | | | k ⌊ǫk⌋ (10.9) sup sup A1 z / A2 z < . z∈Cn\{0} k∈Z| | | | ∞ Lemma 10.4. Suppose A is n n complex matrix. For any z Cn, r> 0, and m =0, 1,... × ∈ (10.10) z E(r, m + 1) E(r, m), where ∈ \ (10.11) E(r, m)= E(A; r, m) = span ker(A λId)m ker(A λId)n , − ∪ − |λ[|=r |λ[| 0 and k0 N so that | |∼ → ∞ ∈ (10.12) 1/ckmrk Akz ckmrk for all k k . ≤ | |≤ ≥ 0 70 1. ANISOTROPIC HARDY SPACES

Proof. Suppose J is a j by j Jordan block associated with an eigenvalue λ, i.e.,

λ 1 0 ...... λ 1 0 . . .  . . .  (10.13) J = ......    λ 1     λ    The kth iterate of J is equal to

k k−1 k−2 k−j+1 p0(k)λ p1(k)λ p2(k)λ ... pj−1(k)λ . k k−1 k−2 .  p0(k)λ p1(k)λ p2(k)λ .  . . . .  ......  k   (10.14) J =  .. .. .  ,  . . .   . .   .. ..     k k−1   p0(k)λ p1(k)λ   k   p0(k)λ    where pi(k) is given by the recursive formula

1 i =0, k =0, 1,..., p (k)= 0 i =1,... ,j 1, k =0,... ,i 1, i  − −  p (k 1)+ p (k 1) i =1,... ,j 1, k = i,i +1,... i−1 − i − − In particular, p(k)= k, p (k)= k(k 1)/2, and by induction 1 2 − k(k 1) . . . (k i + 1) (10.15) p (k)= − − . i i! Take any 0 = z = (z ,... ,z ) Cj . Let m =0,... ,j 1 be such that 6 1 j ∈ − z ker(J λId)m+1 and z ker(J λId)m. ∈ − 6∈ −

Equivalently, m is the unique index which satisfies zm+1 = 0 and zi+1 = 0 for all i>m. If λ = 0 then by (10.14) and (10.15) 6 6 J kz z | | | m+1| as k . km λ k → λ mm! → ∞ | | | | This shows Lemma 10.4 in the case when the matrix A is a single Jordan block. In the general case we can use the Jordan Theorem to write A as A = UBU −1, where U is a nonsingular n n matrix and B = p J , where J is a Jordan block × i=1 i i of size ji associated with the eigenvalue λi. We can assume that λ1 . . . λp L | |≤ ≤ | | and j1 + . . . + jp = n. For the convenience we define l1 = 0, li = j1 + . . . + ji−1 for i = 2,... ,p. We define the basis vl : l = 1,... ,n of Jordan decomposition by { } n vl = Uel, where el : l =1,... ,n denotes the standard basis in C . Note that E{(r, m) E(r′,m} ′) if r < r′ or if r = r′ and m m′, where E(r, m) is given by (10.11).⊂ Also if r = λ for all i = 1,... ,p, or ≤m = 0, then 6 | i| 10. CLASSIFICATION OF DILATIONS 71

E(r, m)= 0 if r< λ1 and otherwise E(r, m)= E( λi ,n), where i is the largest index so that{ } λ < r|. Since| E( λ ,n)= Cn therefore| we| can express Cn as | i| | p| p n−1 (10.16) Cn = E(0,n) E( λ ,m + 1) E( λ ,m). ∪ | i| \ | i| i=1 m=0 |λ[i|6=0 [ If z E(0,n) then (10.10) can not be fulﬁlled for any r> 0 and Akz = 0 for all k n∈. Suppose that z E( λ ,m + 1) E( λ ,m ) for some i| =1| ,... ,p, ≥ ∈ | i0 | 0 \ | i0 | 0 0 λi0 = 0, and m0 = 0,... ,n 1. Let i1 = min i : λi = λi0 , and i2 = max i : |λ |= 6 λ . Observe that E(−λ ,m) consists precisely{ | | of vectors| |} n a v , where{ | i| | i0 |} | i0 | l=1 l l al C, and al = 0 if either li2

Since z E( λi0 ,m0 +1) E( λi0 ,m0), we have al = 0 if li +1+m0 +1 l li +ji ∈ | | \ | | ′ ′ ≤ ≤ for some i, i1 i i2. Also there exists i , i1 i i2, m0 +1 ji′ such that ≤ ′ ≤ ≤ ≤ ≤ al′ = 0, where l = li′ +1+ m0. Therefore, all the coefficients of the vl’s in 6 m0 k the brackets (10.17) are dominated asymptotically by k λ0 as k and at | | → ∞m0 k least one coefficient (the coefficient of vli′ +1) behaves asymptotically as k λ0 as k by (10.15). Since the norm U is equivalent to the standard norm| | , this shows→ ∞ that there exists a constant| c>· | 0, and k N so that | · | 0 ∈ 1/ckm0 λ k Akz ckm0 λ k for all k k , | 0| ≤ | |≤ | 0| ≥ 0 i.e., (10.12) holds. This combined with (10.16) also shows the converse implication.

Proof of Theorem 10.3. Lemma 10.2 says that the quasi-norms ρ1 and ρ2 are equivalent if and only if (10.8) and/or (10.9) hold. Assume that this hap- pens. For r 0 and m = 0, 1,... , let E(A ; r, m) be the linear space given ≥ i by (10.11) corresponding to the dilation Ai, i = 1, 2. By Lemma 10.4, if z k m k ∈ E(A1; r, m+1) E(A1; r, m) then A1 z k r as k and by (10.8) and (10.9), ⌊ǫk⌋ m \k k m | k/ǫ |∼ → ∞ 1/ǫ A2 z k r , so A2 z k r as k . Therefore, z E(A2; r ,m + | | ∼ 1/ǫ | | ∼ → ∞ ∈ 1) E(A2; r ,m). Since the sets E(Ai; r, m + 1) E(Ai; r, m), r> 1, m =0, 1,... \ n \ǫ partition C 0 for i = 1, 2, we have E(A1; r ,m) = E(A2; r, m) for all r > \{ } −1 −1 0, m = 0, 1,... . Analogously, by considering matrices A1 and A2 we have E(A−1; rǫ,m) = E(A−1; r, m) for all r > 0, m = 0, 1,... Since E(A ; r, m) 1 2 i ∩ E(A−1; r−1,m) = span ker(A λId)m for all r > 0, m =0, 1,... , we have i |λ|=r i − (10.7). By reversing this argument we obtain the converse implication. S We are ready to state the classiﬁcation theorem for dilations generating the same anisotropic Hardy space Hp, 0

n Theorem 10.5. Suppose we have two dilations A1 and A2 on R . The following are equivalent: (i) the quasi-norms ρ1 and ρ2 associated to A1 and A2, respectively, are equivalent, (ii) (10.7) holds for all r> 1, m =1, 2,... , p (iii) the anisotropic Hardy spaces H associated to A1 and A2 are the same for some 0

0, x Rn, ⊂{ 0 − ≤ } 0 ∈ (10.19) a r1/q−1/p, || ||q ≤ (10.20) a(x)xαdx =0 for α s. Rn | |≤ Z Indeed, if ρ is the step homogeneous quasi-norm then the above conditions for r = bj, j Z coincide with Definition 4.1. If ρ is a general quasi-norm then by Lemma 2.4∈ there is a constant c > 0 so that for every atom a satisfying (10.18), (10.19), and (10.20), ca is an atom in the sense of Definition 4.1. And vice versa. Therefore, the definition of atoms is independent of the choice of a quasi-norm up to the equivalence of a multiplicative constant. We can also replace conditions (10.18) and (10.19) by (10.21) supp a x + Aj B(0, 1) forsome j Z, x Rn, ⊂ 0 ∈ 0 ∈ (10.22) a Aj B(0, 1) 1/q−1/p, || ||q ≤ | | where B(0, 1) = x : x < 1 . { | | } Proof. Theorem 10.3 states that (i) (ii). (i) = (iv) is a consequence of the above Remark. (iv) = (iii) is automatic.⇐⇒ It suffices⇒ to show (iii) = (ii). Assume that (iii) holds for⇒ some 0 0 so that A A1 A2 p p p p p (10.23) 1/C f H f H C f H for all f HA = HA . || || A1 ≤ || || A2 ≤ || || A1 ∈ 1 2 By Theorem 6.9 and the following Remark, (10.23) is equivalent to the same condition being satisfied for all bounded, compactly supported functions f with vanishing moments up to order s, where s is large enough so that triplet (p, ,s) is admissible ∞ for both A1 and A2. The family of such functions consists precisely of scalar multiples of (p, ,s)-atoms (associated with A1 or A2). It is not hard to see that the equivalence∞ of norms for atoms implies the corresponding statement for all elements of Hp = Hp , i.e., (10.23). A1 A2 10. CLASSIFICATION OF DILATIONS 73

p If (10.23) fails then we could ﬁnd a sequence of elements (f ) N in H so i i∈ A1 −i that fi Hp 2 and fi Hp = 1 for all i N (or the similar statement with A1 A2 || || ≤ || || ∈ n the norms Hp , Hp interchanged). For any choice of vectors ki R the A1 A2 || · || || · || p ′ p ∈ series τ f converges in H (and hence in ), since f p < , i∈N ki i A1 i∈N i H S || || A1 ∞ see PropositionP 4.5. Here τkf(x) = f(x k) denotes the translationP of f by the n − vector k R . We claim that for some choice of ki’s, i∈N τki fi does not belong p ∈ to H . Indeed, if M denotes the (grand) maximal function associated to A2 then A2 P ﬁnd the sequence of numbers (ri)i∈N so that

p −i−1 (10.24) Mfi(x) dx > 1 2 for all i N. B − ∈ Z (0,ri)

Choose ki’s so that the balls B(ki, ri) are mutually disjoint. Let f = i∈N τki fi ′ ′ ∈ . For each j N, τk fj = f τk fi with convergence in , hence S ∈ j − i6=j i S P P 1/p M(τ f )(x) Mf(x)+ M(τ f )(x) Mf(x)p + M(τ f )(x)p , kj j ≤ ki i ≤ ki i Xi6=j Xi6=j for all x Rn. In particular ∈ Mf(x)p M(τ f )(x)p M(τ f )(x)p for x B(k , r ). ≥ kj j − ki i ∈ j j Xi6=j Integrating the above and using τkM = Mτk we have

p p p Mf(x) dx M(τkj fj )(x) M(τki fi)(x) dx B(kj ,rj ) ≥ B(kj ,rj ) − Z Z Xi6=j p p = Mfj (x) dx Mfi(x) dx B(0,rj ) − B(kj −ki,rj ) Z Xi6=j Z −j−1 p −i−1 1 2 Mfi(x) dx 3/4 2 =1/4, B c ≥ − − (0,ri) ≥ − N Xi6=j Z Xi∈ by (10.24) and Mf (x)pdx = 1. Summing the above over j N we have Rn i ∈ R Mf(x)pdx Mf(x)pdx = . n B R ≥ N (kj ,rj ) ∞ Z Xj∈ Z Hence, f Hp which is a contradiction of (iii). Therefore (10.23) holds. A2 6∈ 1/p We remark that if A is a dilation then DAf(x)= det A f(Ax) is an isometry on Hp . More precisely, if f ′ and ϕ then we| deﬁne| D f by A ∈ S ∈ S A D f, ϕ = det A 1/p−1 f, ϕ(A−1 ) . h A i | | h · i Indeed, by a simple calculation we have

M 0f(Ax)= det A −1/pM 0(D f)(x) for x Rn, ϕ | | ϕ A ∈ and by the deﬁnition of the grand maximal function and a change of variables

p (10.25) f Hp = DAf Hp for all f H . || || A || || A ∈ A 74 1. ANISOTROPIC HARDY SPACES

Consider the family of functions f on Rn so that

(10.26) supp f x + Aj1 Aj2 B(0, 1) forsome j , j Z, x Rn, ⊂ 0 1 2 1 2 ∈ 0 ∈

(10.27) f det A −j1/p det A −j2/p, || ||∞ ≤ | 1| | 2|

(10.28) f(x)xαdx = 0 for all α s. Rn | |≤ Z ′ ′ We claim that there is a constant C > 0 so that f Hp C for every f satisfying || || A1 ≤ (10.26)–(10.28). Indeed, for any such f, D j2 D j1 f has support contained in A2 A1 −j2 −j1 ∞ A2 A1 x0 + B(0, 1), has L norm less than 1, and has vanishing moments up to order s, hence is a constant multiple of an atom (satisfying (10.20)–(10.22) with j = 0 and q = ). The claim now follows from (10.23) and the iterative form of ∞ (10.25). Finally, let f0 be a ﬁxed function satisfying (10.26)–(10.28) with x0 = 0, j1 = j2 = 0,

δ0 for all x B(3/4e1, 1/4), (10.29) f0(x)= ∈ 0 for all x B(0, 1/2) B(3/4e , 1/4). 6∈ ∪ 1 It is clear that if δ0 > 0 is suﬃciently small a function f0 satisfying the above constraints exists. To ﬁnish the proof, assume on the contrary that (ii) fails. By (10.2) this means that either

k −⌊ǫk⌋ k −⌊ǫk⌋ lim sup A1 A2 = or lim sup A1 A2 = . k→∞ || || ∞ k→−∞ || || ∞ It is not hard to see the lim sup can be replaced by lim using the idea in the proof of Theorem 10.3. For any k Z deﬁne d(k) as the smallest integer so that ∈ AkA−⌊ǫk⌋−d(k) 1. || 1 2 || ≤ Clearly, we have

AkA−⌊ǫk⌋−d(k) = AkA−⌊ǫk⌋−d(k) A A −1 A −1. || 1 2 || || 1 2 |||| 2|||| 2|| ≥ || 2|| We either have d(k) as k or d(k) as k . We will ﬁx our attention on the case→ when ∞ d(k) → ∞ as k →; ∞the other→ case −∞ is identical. →k ∞−⌊ǫk⌋−d→(k) ∞ n For simplicity denote Qk = A1 A2 . Let zk R , zk = 1, be such that Q z = Q =: c(k). We know that A −1 c(k∈) 1.| Let| U be a unitary | k k| || k|| || 2|| ≤ ≤ k matrix such that Uke1 = zk. Consider function

(10.30) fk = D −1 −1 f0 = D −1 D −1 f0, Uk Qk Qk Uk The function f clearly satisﬁes (10.26)–(10.28) with x = 0, j = k, j = ǫk k 0 1 2 −⌊ ⌋− d(k). Since QkUkB(0, 1/2) B(0,c(k)/2) and QkUkB(3/4e1, 1/4) = QkB(3/4zk, 1/4) then by (10.29) and (10.30)⊂ c x B(0,c(k)/2) and fk(x) =0 = (10.31) ∈ 6 ⇒ x Q B(3/4z , 1/4) and f (x)= δ , ∈ k k k k 10. CLASSIFICATION OF DILATIONS 75

c(k)/4

3/4Q z

k k

c(k)/2 0

Figure 3. The support of fk is contained in the shaded regions. The function fk is equal to δk in the darker ellipse.

−k/p ⌊ǫk⌋/p+d(k)/p where δk = δ0 det A1 det A2 . Let ϕ be a nonnegative function such that| | | | ∈ S −1 1 for x B(0, 1/8 A2 ), (10.32) ϕ(x)= ∈ || || 0 for x B(0, 3/16 A −1). 6∈ || 2|| 0 −1 We claim that Mϕfk p as k . Indeed, if z B(3/4Qkzk, 1/16 A2 ) then || || → ∞ → ∞ ∈ || ||

0 Mϕfk(z) fk ϕ(z) = fk(x)ϕ(z x)dx ≥ | ∗ | Rn − Z (10.33) = δ 1 B ( x)ϕ(z x)dx k Qk (3/4zk,1/4) Rn − Z δ B(z, 1/8 A −1) Q B(3/4z , 1/4) , ≥ k| || 2|| ∩ k k | by (10.31) and (10.32) since ϕ(z x) = 0 implies − 6 x B(0, 3/16 A −1)+ B(3/4Q z , 1/16 A −1)= B(3/4Q z , 1/4 A −1), ∈ || 2|| k k || 2|| k k || 2|| c and hence x B(0,c(k)/2) . By (10.33) and Lemma 10.6 applied to P = 1/4Qk and r = (2c(k∈) A )−1 || 2|| M 0f (z) δ B(z 3/4Q z , 1/8 A −1) P B(0, 1) δ (r/2)n P B(0, 1) ϕ k ≥ k| − k k || 2|| ∩ |≥ k | | det Q B(0, 1/4) δ (4 A )−n ≥ | k|| | k || 2|| δ det A (d(k)−1)(1/p−1) B(0, 1/4) (4 A )−n = c det A d(k)(1/p−1), ≥ 0| 2| | | || 2|| | 2| 76 1. ANISOTROPIC HARDY SPACES by (10.4). Therefore, if p< 1 then

0 p 0 p Mϕfk(z) dz Mϕfk(z) dz Rn ≥ B −1 Z Z (3/4Qk zk,1/16||A2|| ) cp B(0, 1/16 A −1) det A d(k)(1−p) as k . ≥ | || 2|| || 2| → ∞ → ∞ ′ By Theorem 7.1, fk Hp as k which contradicts fk Hp C . || || A1 → ∞ → ∞ || || A1 ≤ Therefore, (ii) must necessarily hold if p< 1. The case p = 1 requires a special argument. We have f = f and || k||1 || 0||1 supp fk QkB(0, 1) for all k N. Since is a separable space there is a subse- ⊂ ∈ ′ S quence (fki )i∈N converging to some f , and r = limi→∞ c(ki) exists. Clearly f is a regular Borel measure with compact∈ S support which is singular with respect to the Lebesgue measure since supp fk 0 as k . Furthermore, f = 0 which can be veriﬁed by testing f |with an| appropriate → → (nonnegative, ∞ radial)6 test function ϕ vanishing on B(0,r/2). Let M denotes the (grand) maximal function ∈ S associated to A1. By Fatou’s Lemma

Mf(x)dx lim inf Mfki (x)dx lim inf Mfki (x)dx Rn ≤ Rn i→∞ ≤ i→∞ Rn Z Z Z ′ 1 lim inf fki H C . ≤ i→∞ || || A1 ≤ 1 1 1 Rn 1 Rn Hence f HA1 . But f L which is a contradiction of HA1 ( ) L ( ). This ﬁnishes the∈ proof of Theorem6∈ 10.5. ⊂

Lemma 10.6. Suppose Γ= P B(0, 1) = x Rn : P −1x < 1 is an ellipsoid, where P is some nondegenerate n n matrix.{ For∈ any| 0 < r| 1/}2 we have × ≤ B(z, P r) Γ | || || ∩ | (r/2)n for all z B(0, P r/2). Γ ≥ ∈ || || | | Proof. Let c = P . For any z B(0,cr/2), B(0,cr/2) B(z,cr) hence || || ∈ ⊂ B(z,cr) Γ B(0,cr/2) P B(0, 1) P −1B(0,cr/2) B(0, 1) | ∩ | | ∩ | = | ∩ | Γ ≥ P B(0, 1) B(0, 1) | | | | | | B(0,r/2) B(0, 1) | ∩ | = (r/2)n. ≥ B(0, 1) | |

We close this section with a simple example of a distribution that diﬀerentiates most anisotropic Hp spaces. Although this example gives only a necessary condition for Hp = Hp , 0

0 Proof. Since Mϕδ0(x) 1/ρ(x), where ρ is a quasi-norm associated to A, 0 p ≈ 0 Mϕδ0 is locally in L (0

ϕ(x), w pdx > 0 forany w Rn 0 , |h∇ i| ∈ \{ } ZB1\B0 where ϕ is the gradient of ϕ. By continuity ∇ inf ϕ(x), w pdx > 0. |w|=1 |h∇ i| ZB1\B0 Since ϕ(x) ϕ(x w) ϕ(x), w / w 0 as w 0 uniformly over x B1 B0, there| is a constant− − c>−h∇0 so that i| | | → | | → ∈ \

(10.35) ϕ(x) ϕ(x w) pdx c w p for all w 1. | − − | ≥ | | | |≤ ZB1\B0 We have

0 −k −k Mϕf(x) = sup b f(y)ϕ(A (x y))dy Z − (10.36) k∈ Z −k −k −k = sup b ϕ(A x) ϕ(A (x z)) , k∈Z | − − | where b = det A . Without loss of generality we can assume that z 1. Suppose | | | |≤c x Bl+1 Bl for some l l0, where l0 is suﬃciently large so that (Bl) + B(0, 1) ∈ c \ ≥ c ⊂ (Bl−1) for all l l0. Since x z (Bl−1) the supremum in (10.36) runs eﬀectively only for k l ≥1, and − ∈ ≥ − 0 1−l −k −l −l Mϕf(x) b sup ϕ(y) sup A z Cb A z . ≤ y∈Rn |∇ | k≥l−1 | |≤ | | Therefore

M 0f(x)pdx Cpbl+1b−lp A−lz p = Cpb(1−p)l+1 A−lz p. ϕ ≤ | | | | ZBl+1\Bl On the other hand by (10.35)

Proof. Corollary follows easily from Lemmas 10.4 and 10.7.

i i Example. Suppose A1 and A2 are dilations. Let λ1,... ,λn be eigenvalues of i i Ai (taken according to multiplicity) so that λ1 . . . λn , i =1, 2. Suppose that p p | |≤ ≤ | | 1/p−1 H = H for all 0 1, 1 − 2 − 1/ε |λ[|>r |λ|>r[ where ǫ is given by (10.3). In particular, by counting dimensions we have

(10.37) λ1 1/ ln | det A1| = λ2 1/ ln | det A2| for all j =1,... ,n. | j | | j | Naturally this condition falls short of suﬃciency (in comparison with (10.7), for example). Nevertheless, it does give a quick way of checking whether two dilations might generate the same anisotropic Hp spaces. In fact, (10.37) comes very close to characterizing those Hardy spaces which are equivalent up to a linear transformation, as we will see in Theorem 10.10. Definition 10.9. We say that Hp and Hp are equivalent up to a linear A1 A2 transformation, and write Hp = Hp if there is a nonsingular n n matrix P such A1 ∼ A2 × that D is an isomorphism between Hp and Hp . We say that two quasi-norms P A1 A2 are equivalent up to a linear transformation if there is a constant c > 0 and a nonsingular n n matrix P such that × 1/cρ (x) ρ (P x) cρ (x) for all x Rn. 1 ≤ 2 ≤ 1 ∈

n Theorem 10.10. Suppose we have two dilations A1 and A2 on R . The following are equivalent: (i) the quasi-norms ρ1 and ρ2 associated to A1 and A2, respectively, are equivalent up to a linear transformation, (ii) for all r> 1 and m =1, 2,... we have

m m (10.38) dim ker(A1 λId) = dim ker(A2 λId) , ǫ − − |λX|=r |λX|=r where ǫ is given by (10.3), (iii) Hp = Hp for all 0

1, m =1, 2,... by Theorem 10.5. Hence (ii) holds. 10. CLASSIFICATION OF DILATIONS 79

Assume (ii) holds, i.e., (10.39) dim ker(A rId)m = dim ker(A rId)m, for all r> 1,m =1, 2,... 1 − 2 − If r is an eigenvalue of A1 then the number of Jordan blocks of size m corresponding to r is equal to dim ker(A rId)m dim ker(A rId)m−1.≥ If r is not 1 − − 1 − an eigenvalue of A1 this number equals 0 regardless of m. By (10.39) the number of Jordan blocks of size m corresponding to the eigenvalue r is the same for both A1 and A2 and therefore the matrices A1 and A2 have equivalent Jordan decompo- −1 sitions. So there is a nonsingular n n matrix P such that A1 = P A2P . Choose any ϕ with ϕ = 0. Let ϕ ×(x) = b−kϕ(A−kx) and ψ (x) = b−kψ(A−kx), ∈ S 6 k 1 k 2 where ψ(x)= det P −1 ϕ(P −1x). Given any f ′ we have | R | ∈ S (f ϕ )(x)= b−k f(y)ϕ(P −1A−kP (x y))dy ∗ k 2 − Z = b−k det P −1 f(P −1y)ϕ(P −1A−k(P x y))dy | | 2 − Z −1 1/p = f(P y)ψ (P x y)dy = det P (D −1 f ψ )(P x). k − | | P ∗ k Z 0 0 Therefore, Mϕf p = MψDP −1 f p, where the maximal functions are associated || || || || p p p to A1 and A2, respectively. Thus, f H DP −1 f H for any f, and HA = || || A1 ∼ || || A2 1 ∼ Hp for any 0

(10.42) f(x)xαdx = 0 for all α s. Rn | |≤ Z ′ ′ We claim that there is a constant C > 0 so that f Hp C for every f satisfying || || A1 ≤ (10.40)–(10.42). Indeed, for any such f, D j2 DP −1 D j1 f has support contained in A2 A1 −j2 −j1 ∞ −1/p A2 P A1 x0 + P B(0, 1), has L norm less than det P , and has vanishing moments up to order s, hence is a constant multiple| of an| atom. The claim now follows from the hypothesis and the iterative form of (10.25). By the proof of −1 Theorem 10.5 we conclude that the dilations A1 and P A2P have equivalent −1 quasi-norms ρ1 andρ ˜2, respectively. Deﬁne ρ2(x)=ρ ˜2(P x). Clearly ρ2 is a quasi-norm associated with the dilation A2 and ρ1 and ρ2(P ) are equivalent. This shows (i) and ends the proof. · 80 1. ANISOTROPIC HARDY SPACES

CHAPTER 2 Wavelets

1. Introduction In this chapter we present constructions of orthogonal and tight frame wavelets in the Schwartz class. We investigate the limitations on regularity of orthogonal wavelets imposed by the form of a dilation. We show that sufficiently regular wavelets form an unconditional basis for the anisotropic Hardy space associated this dilation. Historical background. The theory of wavelets is a relatively new area of mathematics. The first example of an object now called a wavelet is the Haar function introduced by Haar [Ha] in 1910. Haar showed that the appropriate translates and dilates of Haar function form an orthonormal basis of L2([0, 1]). The second example of wavelets has been introduced by Strömberg [Sö] in 1981. Strömberg wavelets can be constructed to be Cr smooth with exponential decay at infinity for any integer r 1. Strömberg has also shown that they form an unconditional basis for the Hardy≥ space Hp(Rn), 0

In Section 3 we show that if move beyond the class of dilations preserving some lattice, regular wavelets may not exists. In fact, we show that for a large class of dilations all multiwavelets must be combined minimally supported in frequency. Nevertheless, in Section 4 we show that -regular tight frame wavelets exist for any dilation matrix. ∞ In Section 5 we show that r-regular wavelets associated with a dilation A form p an unconditional basis for the anisotropic Hardy space HA associated with A. This generalizes the result of Meyer [Me] who showed this in the isotropic case. We remark that unconditionality of multiwavelets with arbitrary dilations in Lp space (1 sequence space of coeﬃcients of elements of Hp in the wavelet expansion. In the isotropic case this sequence space was studied by Meyer [Me] and Frazier and Jawerth [FJ1, FJ2] in the scale of Triebel-Lizorkin spaces. Wavelet preliminaries. Throughout Section 2, we are going to assume that we have a lattice Γ (Γ = P Zn for some nondegenerate n n matrix P ) and a dilation matrix A preserving Γ, i.e., all eigenvalues λ of A×satisfy λ > 1, and AΓ Γ. Without loss of generality, we will assume that Γ = Zn. | | ⊂ Definition 1.1. Let Ψ be a ﬁnite family of functions Ψ = ψ1,... ,ψL L2(Rn). We say that Ψ is a wavelet family (or a multiwavelet) if {ψl : j Z,} k ⊂ { j,k ∈ ∈ Zn,l =1,...,L is an orthonormal basis for L2(Rn). Here, for ψ L2(Rn) we use the convention } ∈

j/2 j n ψ (x)= D j τ ψ(x)= det A ψ(A x k) j Z, k Z , j,k A k | | − ∈ ∈ where τ f(x) = f(x y) is a translation operator by the vector y Rn, and y − ∈ D f(x)= det A f(Ax) is a dilation operator by the matrix A. A | | Definitionp 1.2. By a multiresolution analysis we mean a sequence of closed 2 n subspaces (Vi)i∈Z L (R ) satisfying: (i) V V for i⊂ Z, i ⊂ i+1 ∈ (ii) Vi = DAi V0 for i Z, 2 n ∈ (iii) i∈Z Vi = L (R ), (iv) Z Vi = 0 , Si∈ { } (v) there exists ϕ called a scaling function such that τkϕ k∈Zn is an orthonormal T { } basis of V0. We say that a wavelet family Ψ = ψ1,... ,ψL is associated with a multiresolution analysis (MRA), if the spaces { }

(1.1) V = W , where W = span ψl : k Zn,l =1,...,L , i j j { j,k ∈ } j

L ∞ (1.2) D (ξ)= ψˆl(Bj (ξ + k)) 2, Ψ | | j=1 Zn Xl=1 X kX∈ 82 2. WAVELETS where B = AT , see [Ba, BRS]. Since

L 2 (1.3) DΨ(ξ)dξ =1/(b 1) ψl , (0,1)n − || || Z Xl=1 a wavelet family Ψ = ψ1,... ,ψL can be associated with an MRA only if L = b 1, where b = det A . { } − | | Definition 1.3. We say that a function f on Rn is r-regular, if f is of class Cr, r =0, 1,... , and ∞ (1.4) ∂αf(x) c (1 + x )−k, | |≤ α,k | | for each k N, and each multi-index α, with α r. A wavelet family Ψ = ψ1,... ,ψL∈ is r-regular, if ψ1,... ,ψL are r-regular| | ≤ functions. An MRA is r- { } regular if the subspace V0 given by (1.1) has an orthonormal basis of the form τ ϕ : k Zn for some r-regular scaling function ϕ. { k ∈ } If a wavelet family Ψ is r-regular for r suﬃciently large, or more precisely ψˆl(ξ) are continuous and ψˆl(ξ) C(1 + ξ )−n/2−ε for some ε> 0, then the sum (1.2)| converges| uniformly on| compact|≤ subsets| | of Rn Zn to the continuous function n \ DΨ(ξ) having integer values. Therefore, by Z -periodicity DΨ is constantly equal d for some d N. If d = 1 then r-regular wavelet family Ψ comes from some MRA (more generally,∈ Ψ comes from an MRA with multiplicity d). This result was essentially shown by Auscher [Au2, Au3, Theorem 10.1] (under slightly stronger assumptions on ψˆl’s). In general, we can not expect that this MRA is also r-regular; for a counterexample see [MC, Chapter 8, Proposition 2]. Conversely, having an r-regular MRA we can not, in general, deduce the existence of r-regular wavelet family associated with it, see [Wo2, Theorem 5.10, Remark 5.6]. Nevertheless, we can deduce the existence of r-regular wavelet family by using the following result, see [Wo2, Corollary 5.17] which also holds for r = . ∞ Proposition 1.4 (Wojtaszczyk). Assume that we have a multiresolution analysis on Rn associated with an integral dilation A with det A = b. Assume that this MRA has an r-regular scaling function ϕ(x) such that| ϕˆ(|ξ) is real for some r = 0, 1,..., . Then there exists a wavelet family associated with this MRA consisting of a∞(b 1) r-regular function. − Starting in Section 3 we relax the assumption that a dilation A has integer entries and we only assume that A is expansive in the sense of Deﬁnition 2.1 in Chapter 1. In Section 3 we show that in this setting r-regular orthogonal wavelet bases may not exist in general. Consequently, starting in Section 4 we consider also tight frame wavelets. 2. WAVELETS IN THE SCHWARTZ CLASS 83

2. Wavelets in the Schwartz class In this section we are going to construct an MRA which has a scaling function ϕ in the Schwartz class andϕ ˆ(ξ) is real for some special class of dilations satisfying a kind of expansiveness property. By Proposition 1.4, we can then find a wavelet family in the Schwartz class associated with this MRA. Definition 2.1. We say that the integral dilation B is strictly expansive if there exists a compact set K Rn such that 0 K◦, where K◦ is the interior⊂ of K, • ∈ n K (l + K) = δl,0 for l Z , •K | ∩ BK◦. | ∈ • ⊂ Definition 2.2. Given a set Y Rn and ε > 0 we define its ε-interior Y −ε and ε-neighborhood Y +ε by ⊂ Y −ε = ξ Rn : B(ξ,ε) Y , { ∈ ⊂ } Y +ε = ξ Rn : B(ξ,ε) Y = . { ∈ ∩ 6 ∅} Note that Y −ε is closed, Y +ε is open, and the interior of Y satisfies Y ◦ = −ε ε>0 Y . If the dilation B is strictly expansive and the compact set K satisfies Definition 2.1, then there exists ε> 0 such that S (2.1) K+ε B(K−ε). ⊂ We shall prove the following existence theorem. Theorem 2.3. Suppose A is a dilation matrix with AZn Zn with b = det A . If the dilation B = AT is strictly expansive then there exists⊂ a multiresolution| | analysis with a scaling function and an associated wavelet family of (b 1) functions in the Schwartz class. − For the sake of completeness, we recall the proof of Theorem 2.3 from [BS1, Theorem 3.2]. Proof. Let K and ε > 0 satisfy Definition 2.1 and (2.1). Choose a C∞ function g : Rn [0, ) such that g = 1 and → ∞ Rn (2.2) supp g := ξ RRn : g(ξ) =0 = B(0,ε). { ∈ 6 } Define the function f by (2.3) f(ξ) = (1 g)(ξ). K ∗ Clearly f is in the class C∞, 0 f(ξ) 1, and ≤ ≤ (2.4) supp f = ξ Rn : f(ξ) =0 K+ε, { ∈ 6 }⊂ (2.5) ξ Rn : f(ξ)=1 = K−ε. { ∈ } Moreover,

n (2.6) f(ξ + k)= 1K(ξ + k η)g(η)dη = 1 for all ξ R , n Zn Zn R − ∈ kX∈ kX∈ Z n since n 1 (ξ + k) =1 for a.e. ξ R by Deﬁnition 2.1. k∈Z K ∈ P 84 2. WAVELETS

Finally, deﬁne the function m : Rn [0, 1] by → (2.7) m(ξ)= f(B(ξ + k)). Zn skX∈ Claim 2.4. The function m given by (2.7) is C∞, Zn-periodic, and (2.8) m(ξ + B−1d) 2 =1 for all ξ Rn, | | ∈ dX∈D (2.9) m(ξ) > 0 = ξ Zn + B−1(K+ε), ⇒ ∈ (2.10) m(ξ)=0 for ξ (B−1Zn Zn)+ B−1(K−ε), ∈ \ T where B = A , and = d1,... ,db is the set of representatives of diﬀerent cosets of Zn/BZn, where bD= det{ A . } | | Proof of Claim 2.4. To guarantee that m is C∞, the function f must “van- α ishes strongly”, i.e., if f(ξ0) = 0 for some ξ0 then δ f(ξ0) = 0 for any multi-index α. It is clear that if nonnegative function f in C∞ “vanishes strongly” then √f is also C∞. The condition (2.8) is a consequence of

m(ξ + B−1d) 2 = f(B(ξ + B−1d + k)) = f(ξ + d + Bk)=1, | | Zn Zn dX∈D kX∈ dX∈D kX∈ dX∈D by (2.6). To see (2.9), take ξ such that m(ξ) > 0. By (2.4) and (2.7), B(ξ + k) K+ε for some k Zn, and hence (2.9) holds. ∈ We claim∈ that (2.10) follows from (2.8) and (2.11) m(ξ)=1 for ξ Zn + B−1(K−ε). ∈ Indeed, if ξ B−1d + k + B−1(K−ε) for some d BZn and k Zn, then by (2.11) we∈ have m(ξ B−1d) = 1. Hence by (2.8)∈ D\m(ξ) = 0 and (2.10)∈ holds. Finally, (2.11) is the immediate− consequence of (iii) and (2.7). This ends the proof of the claim.

We can write m in the Fourier expansion as

1 −2πihk,ξi (2.12) m(ξ)= hke , det A Zn | | kX∈ where we include the factor detpA −1/2 outside the summation as in [Bo1]. Since ∞ | | m is C , the coefficients hk decay polynomially at infinity, that is for all N > 0 there is CN > 0 so that h C k −N for k Zn 0 . | k|≤ N | | ∈ \{ } Since m satisfies (2.8) and m(0) = 1, m is a low-pass filter which is regular in the sense of the definition following [Bo1, Theorem 1]. By [Bo1, Theorem 5] ϕ L2(Rn) defined by ∈ ∞ (2.13)ϕ ˆ(ξ)= m(B−j ξ), j=1 Y 2. WAVELETS IN THE SCHWARTZ CLASS 85 has orthogonal translates, i.e., ϕ, τ ϕ = δ for l Zn, h l i l,0 ∈ if and only if m satisfies the Cohen condition, that is there exists compact set K˜ Rn such that ⊂K˜ contains a neighborhood of zero, • K˜ (l + K˜ ) = δ for l Zn, • | ∩ | l,0 ∈ m(B−j ξ) = 0 for ξ K,˜ j 1. • 6 ∈ ≥ The first guess for K˜ to be K is in general incorrect, e.g. if K has isolated points. Instead we claim that there is 0 <δ< 1 so that (2.14) K˜ = ξ K : B(ξ,ε) K δ B(ξ,ε) { ∈ | ∩ |≥ | |} does the job. Clearly, if ξ K˜ then h(ξ) = 0, hence f(ξ) = 0 and thus m(B−1ξ) = ∈ 6 6 6 0. By (2.1) B−1K˜ B−1K+ε K−ε K˜ and thus m(B−j ξ) = 0 for all j 1. Finally, it suffices to⊂ check that⊂ ⊂ 6 ≥

n (2.15) 1K˜ (ξ + k) 1 for all ξ R . Zn ≥ ∈ kX∈ By the compactness of K there is a finite index set I Zn such that ⊂ (2.16) 1 (ξ + k) 1 for all ξ [ 1, 1]n. K ≥ ∈ − Xk∈I Take any ξ [ 1/2, 1/2]n and integrate (2.16) over B(ξ,ε) to obtain ∈ − B(ξ + k,ε) K B(ξ,ε) . | ∩ | ≥ | | Xk∈I Therefore, if we take δ = 1/#I then there is k I such that B(ξ + k,ε) K ∈ | ∩ | ≥ δ B(ξ,ε) and hence ξ + k K˜ . Thus, (2.15) holds and K˜ given by (2.14) satisfies the| Cohen| condition. Therefore,∈ ϕ is a scaling function for the multiresolution analysis (Vj )j∈Z defined by n V = span D j τ ϕ : l Z for j Z. j { A l ∈ } ∈ It remains to show that ϕ . We are going to prove that ϕ is band-limited, i.e.,ϕ ˆ is compactly supported.∈ By S (2.9) and (2.13) (2.17)ϕ ˆ(ξ) =0 = ξ BZn + K+ε. 6 ⇒ ∈ On the other hand, by (2.10) m(B−j ξ)=0 for ξ Bj−1Zn Bj Zn + Bj−1(K−ε). Since ∈ \ ∞ (Bj−1Zn Bj Zn)= BZn 0 , \ \{ } j=2 [ and K+ε B(K−ε) Bj−1(K−ε) for j 2, ⊂ ⊂ ≥ we have (2.18)ϕ ˆ(ξ)=0 for ξ BZn 0 + K+ε. ∈ \{ } Combining (2.17) and (2.18) we haveϕ ˆ(ξ) = 0 for ξ (K+ε)c. Therefore, suppϕ ˆ K+ε and ϕ is in the Schwartz class. To conclude∈ the proof it suffices to use Proposition⊂ 1.4 for r = . ∞ 86 2. WAVELETS

As a corollary of Theorem 2.3 we have the following. Corollary 2.5. Suppose A is a dilation with integer entries with b = det A . Then there exists m N and a multiresolution analysis with a scaling function| | and a wavelet family of∈ (bm 1) functions in the Schwartz class associated to the dilation Am. − Proof. It suffices to notice that Bm = (AT )m is strictly expansive for sufficiently large m N and K = [ 1/2, 1/2]n. ∈ − Even though the number of functions in the multiwavelet in the above corollary could be much bigger than b 1, Corollary 2.5 still has great significance. By results of Section 5, a multiwavelet Ψ− in the Schwartz class generates an unconditional basis p p of HA = HAm for the whole range of 0

3. Limitations on orthogonal wavelets The construction of Section 2 applies only when the dilation A has integer entries, or more generally, A preserves some lattice Γ = P Zn, where P is an n n nonsingular matrix. If the dilation A does not preserve any lattice then there× could be no well localized wavelets, even 0-regular. Chui and Shi [CS] have shown that all orthogonal wavelets associated with “almost any” irrational dilation in the dimension n = 1 must be MSF (minimally supported in frequency). We are going to show that the analogous statement holds for general multiwavelets. We say that a multiwavelet Ψ = ψ1,... ,ψL associated with A is MSF, if ˆl { } ψ = 1Wl for some measurable sets Wl. By [BRS, Theorem 2.4] these sets are |characterized| by

n ′ 1 1 ′ R Wl (ξ + k) Wl′ (ξ + k)= δl,l a.e. ξ , l,l =1,... ,L, Zn ∈ kX∈ (3.1) L j n 1Wl (B ξ)=1 a.e. ξ R , Z ∈ Xj∈ Xl=1 where B = AT . We say that a multiwavelet Ψ = ψ1,... ,ψL associated with A is combined MSF if { }

L (3.2) ψˆl(ξ) 2 = 1 (ξ) for a.e. ξ Rn, | | W ∈ Xl=1

L for some multiwavelet set W of order L, i.e., W = l=1 Wl for some W1,... ,WL satisfying (3.1). By [BRS, Theorem 2.6] a multiwavelet set W of order L is characterized by S

n (3.3) 1W (ξ + k)= L a.e. ξ R , Zn ∈ kX∈ j n (3.4) 1W (B ξ)=1 a.e. ξ R , Z ∈ Xj∈ where B = AT .

Theorem 3.1. Suppose that a dilation A is such that for every integer j 1 the rows of A−j (treated as vectors in Rn) together with the standard basis vectors≥ e1,... ,en are linearly independent over Q. Then any orthogonal multiwavelet Ψ= ψ1,... ,ψL associated with A is combined MSF, i.e., (3.2) holds for some set W satisfying{ (3.3)} and (3.4).

Proof. By the orthogonality

′ l l j/2 l j l′ ′ δl,l′ δj,0δk,k′ = ψj,k, ψ0,k′ = b ψ (A x k)ψ (x k )dx h i Rn − − Z = bj/2 ψl(Aj x + Aj k′ k)ψl′ (x)dx Rn − Z 88 2. WAVELETS for all j Z, k, k′ Zn, l,l′ =1,...,L. By Plancherel’s formula ∈ ∈

−j j ′ 0= ψˆl′ (ξ)ψˆl(B−j ξ)e2πihB ξ,A k −kidξ Rn (3.5) Z ′ −j = ψˆl′ (ξ)ψˆl(B−j ξ)e2πihξ,k −A kidξ. Rn Z By Lemma 3.2 we conclude that Zn +A−j Zn is dense in Rn for all j 1. Therefore, ′ ≥ by the Fourier Inversion Formula and (3.5), ψˆl (ξ)ψˆl(B−j ξ) = 0 for a.e. ξ Rn and ′ L ˆl j ∈ all j 1, l,l = 1,...,L. Let W = l=1 supp ψ . We have W B W = 0 for j 1,≥ and hence for all j Z 0 . Now (3.4) holds because| the∩ multiwavelet| ≥ ∈ \{ } S l ˆl j 2 Ψ satisﬁes the discrete Calder´on formula l=1 ψ (B ξ) = 1 for a.e. ξ, see [Bo2, CMW]. For j Z let | | ∈ P W = span ψl : k Zn, l =1,...,L . j { j,k ∈ } Naturally we have W f L2 : supp fˆ W . But by (3.4) and W = 0 ⊂ { ∈ ⊂ } j∈Z j L2(Rn) we must have L (3.6) W = f L2 : supp fˆ W . 0 { ∈ ⊂ }

The range function of a shift invariant space W0 is given by

l 2 n J(ξ) = span (ψˆ (ξ + k)) Zn : l =1,...,L ℓ (Z ), { k∈ }⊂ see [BDR, Bo3, Proposition 1.5]. Also by [Bo3, Proposition 1.5] and (3.6),

2 n J(ξ)= (s ) Zn ℓ (Z ) : s = 0 for ξ + k W . { k k∈ ∈ k 6∈ } l Since the vectors (ψˆ (ξ + k))k∈Zn , l = 1,...,L form an orthonormal basis of J(ξ) we have

L l l s, (ψˆ (ξ + m)) Zn (ψ (ξ + k)) Zn = s for all s J(ξ). h m∈ i k∈ ∈ Xl=1

By taking s = ek J(ξ), where ξ + k W and ek’s are the standard basis vectors 2 n ∈ L ˆl ∈ 2 of ℓ (Z ), we conclude that l=1 ψ (ξ + k) = 1 for ξ + k W . This shows (3.2). | | ˆl 2 ∈ (3.3) is a consequence of (3.2) since n ψ (ξ + k) = 1 for a.e. ξ. P k∈Z | | P It is worth noting that the above argument may be simpliﬁed if one applies the spectral function introduced by Rzeszotnik [BR, Rz]. Lemma 3.2. Suppose D is n n real matrix such that the rows of D (treated n × as vectors in R ) together with vectors e1,... ,en are linearly independent over Q. Then the set Zn + DZn is dense in Rn. Since the author does not know an elementary proof of this result we are going to show something more, namely that that a sequence obtained by a certain ordering of DZn is uniformly distributed mod 1. We are going to follow the excellent book of Kuipers and Niederreiter [KN]. 3. LIMITATIONS ON ORTHOGONAL WAVELETS 89

Definition 3.3. Suppose a = (a ,... ,a ), b = (b ,... ,b ) Rn. We say 1 n 1 n ∈ that a < b (a b) if aj

N 1 h x (3.8) lim e2πih , ki =0. N→∞ N kX=1 n Therefore, a sequence (xk)k∈N R is u. d. mod 1 if and only if for every n ⊂ h Z 0 , the sequence of real numbers ( h, x ) N is u. d. mod 1. By the ∈ \{ } h ki k∈ Weyl Criterion, see [KN, Example 6.1, Chapter 1], given θ = (θ1,... ,θm) with the property that the real numbers 1,θ1,... ,θm are linearly independent over Q, the sequence (kθ)k∈N is u. d. mod 1. We are now ready to present the proof of Lemma 3.2.

n Proof of Lemma 3.2. Given k = (k1,... ,kn) R let k ∞ = max( k1 , ... , k ). Let k , k ,... be the ordering of all elements∈ of Zn such|| || that k | <| | n| 1 2 || i||∞ kj ∞ implies i < j. It suﬃces to show that the sequence (xi)i∈N = (Dki)i∈N is ||u. d.|| mod 1. By (3.7) it suﬃces to show that

n #([a, b); (2N + 1)n) lim = (bj aj) N→∞ (2N + 1)n − j=1 Y for all intervals [a, b) In = [0, 1)n. Therefore, by the Weyl Criterion (3.8) we must show that for every⊂ h Zn 0 , ∈ \{ } 1 h Dk (3.9) lim e2πih , i =0. N→∞ (2N + 1)n k∈Zn ||k||X∞≤N

n Fix any h Z 0 . By our hypothesis there exists m0 = 1,... ,n such that n ∈ \{ } θ = h d Q, where D = (d )m=1,... ,n. Otherwise we would have j=1 j j,m0 6∈ j,m j=1,... ,n n h (d ,... ,d ) Qn, which contradicts the linear independence of the rows j=1Pj j,1 j,n ∈ of D together with the standard basis vectors of Rn over Q. Since the sequence P (kθ) N is u. d. mod 1, for every ε> 0 there is M such that for all N M k∈ ≥ N 1 (3.10) e2πiθk < ε. 2N +1 k=−N X

90 2. WAVELETS

We also have

1 h Dk e2πih , i (2N + 1)n k∈Zn ||kX||∞≤N N N 1 2πi Pn Pn h d k = . . . e j=1 m=1 j j,m m (2N + 1)n k1X=−N knX=−N N N n 1 2πi P P hj dj,mkm 1 2πiθk = e j=1 m6=m0 e m0 (2N + 1)n−1 2N +1 k ,...kˇ ... ,k =−N km =−N 1 m0X n X0 ˇ where km0 means we omit the index km0 . Therefore, by (3.10) we have

1 h Dk e2πih , i <ε for all N M. (2N + 1)n ≥ k∈Zn ||k||X∞≤N

Since ε> 0 is arbitrary this shows (3.9) and ends the proof of the lemma.

Remarks. It is not hard to see that linear independence over Q of the rows −j −j n n of A together with basis vectors e1,... ,en is equivalent to B Q Q = 0 , where B = AT . This, in turn, is equivalent to B−j Zn Zn = 0 . Therefore,∩ { the} hypothesis of Theorem 3.1 can be written in a short form∩ as { } (3.11) (AT )j Zn Zn = 0 for all j Z 0 . ∩ { } ∈ \{ } Combined MSF multiwavelets are not well localized in the direct domain. In- deed, if ψl L1 for all l =1,...,L then ψˆl is continuous and (3.2) can never hold. Hence, combined∈ MSF multiwavelets are of little use in other spaces than L2. A ˆ notable exception is the Shannon wavelet ψ(ξ)= 1[−1,−1/2]∪[1/2,1], which generates an unconditional basis for Lp spaces with 1

L a1 (ξ) ψˆl(ξ) 2 c1 (ξ) for a.e. ξ Rn, W ≤ | | ≤ W ∈ Xl=1 for some multiwavelet set W of order L and 0

Another interesting issue is to determine the class of dilation matrices that allow regular orthogonal wavelets. As we have seen in Theorems 2.6 and 3.1, this class certainly includes all dilations with integer entries and excludes all dilations satisfying (3.11) and thus far from preserving the lattice Zn. This still leaves a large class of dilations for which it is not known whether there exist regular wavelets. In one dimension, the above problem goes back to Daubechies [Da2] who asked whether there exist orthonormal wavelet bases with good time-frequency localization for irrational dilation factors a. This question was partially answered by Chui and Shi [CS]. The complete answer was given by the author who showed that all orthonormal wavelets associated with irrational dilation factors have poor time-frequency localization, see [Bo6]. Combining this with the construction of -regular orthogonal wavelets for rational dilations due to Auscher [Au1, KL], we ∞obtain the complete picture in one dimension. The higher dimensional version of this problem remains open. It is not even known whether there are regular wavelets for diagonal 2 2 dilations such that one entry is rational while the other is irrational. Therefore,×it is perfectly conceivable that there are dilations A such that all other dilations with the same quasi-norm a 0 do not allow regular wavelet bases. A good candidate is the dilation A = , 0 2 where a > 0 satisﬁes the condition that r > 0, 2r Q = ar Q. By ∀′ p ∈ p ⇒ 6∈ Theorem 10.5 in Chapter 1, all dilations A with H = H ′ , 0 0. Hence, either arj or 2rj is ±0 2r irrational for all integers j± 1. Thus, it is very likely that there are no regular wavelets associated with such≥ dilations A′. This shows the importance of having a substitute for orthonormal multiwavelets. 92 2. WAVELETS

4. Non-orthogonal wavelets in the Schwartz class The results of Section 3 show the demand for an alternative to orthogonal wavelets. In the theory of wavelets a natural substitute for an orthogonal basis is a frame. In this section we show the existence of tight frame wavelets for all dilations. This fact is probably a part of the folklore. Frames were originally introduced by Duffin and Schaeffer [DS] to study nonharmonic Fourier series. More recently, frames have found applications in many other areas such as wavelets, Weyl-Heisenberg (Gabor) systems, sampling theory, signal processing, etc. For a good introduction to frames, see [Da2, HeW, Yo]. For a more extensive treatment of frame theory, see [Cz, HL]. We start by recalling the notion of a Bessel family and a frame. Definition 4.1. A subset X of a Hilbert space is a Bessel family if there exists c> 0 so that H (4.1) f, η 2 c f 2 for f . |h i| ≤ || || ∈ H ηX∈X If in addition there exists 0 0 for all ξ =0andη ˜ is C∞ on R∈n 0\{. Define} ψ L2(Rn) j∈Z 6 \{ } ∈ ˆ by ψ(ξ)=P η(ξ)/η˜(ξ). Clearly, (4.3) p ψˆ(Bj ξ) 2 = η(Bj ξ)/η˜(Bj ξ)= η(Bj ξ)/η˜(ξ)=1. Z | | Z Z Xj∈ Xj∈ Xj∈ 4. NON-ORTHOGONAL WAVELETS IN THE SCHWARTZ CLASS 93

By [Bo2, Lemma 3.1] we have

2 j j 2 2 f, ψj,k = b fˆ(B ξ) ψˆ(ξ) dξ |h i| Rn | | | | j∈Z k∈Zn j∈Z Z (4.4) X X X + bj fˆ(Bj ξ)ψˆ(ξ) fˆ(Bj (ξ + m))ψˆ(ξ + m) dξ, Rn Z Zn Xj∈ Z m∈X\{0} for any f , where ∈ D = f L2(Rn) : fˆ L∞(Rn), supp fˆ K for some compact K Rn 0 , D { ∈ ∈ ⊂ ⊂ \{ }} is a dense subspace of L2(Rn). Since supp ψˆ B(0, 1/2) ( 1/2, 1/2)n we have ⊂ ⊂ − 2 2 f, ψj,k = f 2 for all f , Z Zn |h i| || || ∈ D Xj∈ kX∈ 2 n and hence for all f L (R ) by (4.3) and (4.4). Therefore, ψj,k j∈Z,k∈Zn forms a tight frame with constant∈ 1 in L2(Rn). Note that this frame{ is not} an orthogonal basis since ψ < 1. || ||2 94 2. WAVELETS

5. Regular wavelets as an unconditional basis for Hp Preliminaries in functional analysis. In this subsection we recall some basic facts about quasi-Banach spaces. Definition 5.1. Suppose X is a vector space over K (K is either R or C). We say that the map : X [0, )is a quasi-norm on X if (i) x =0 ||x ·= || 0, → ∞ (ii) ||αx|| = α⇐⇒ x for all α K, x X, (iii)|| there|| is c>| | ·0 || so|| that x +∈y c∈max( x , y ) for all x, y X. We say that (X, )|| is a quasi-Banach|| ≤ || space|| || ||if is a quasi-norm∈ and X equipped with is|| complete · || as a metric linear space.|| · || || · || It is well-known, see [KPR], that the (Hausdorﬀ) topological linear space which is locally bounded, i.e., it has a bounded neighborhood of the origin, has a quasi- norm. Conversely, the topology associated with any quasi-norm is locally bounded. Obviously the condition (iii) is equivalent to the more common (iii) there is c′ > 0 so that x + y c′( x + y ) for all x, y X. || || ≤ || || || || ∈ Definition 5.2. For 0

The smallest such constant K is called an unconditional basis constant of the system ∗ (em,em)m∈M . ∗ Since actually the elements (em)m∈M determine the functionals (em)m∈M it is customary to speak about (em)m∈M as being an unconditional basis. The following fact is going to be very useful. 5. REGULAR WAVELETS AS AN UNCONDITIONAL BASIS FOR Hp 95

Lemma 5.4. For a series m∈N xm in a quasi-Banach space X, the following are equivalent: P (i) the series is unconditionally convergent, i.e., m∈N βmxm converges for any choice of βm 0, 1 , ∈{ } P (ii) for each permutation σ of N the series m∈N xσ(m) is convergent, (iii) for each bounded sequence (βm)m∈N of scalars the series m∈M βmxm is convergent. Furthermore, there is a constantP C > 0 such that P

(5.3) βmxm C sup βm . ≤ N | | m∈N m∈ X

Proof. The equivalence of (i) and (ii) is a well known fact due to Orlicz, see [Ro, Theorem 3.8.2]. (iii) = (i) is trivial, whereas (iii) = (ii) follows from an argument involving binary expansions.⇒ ⇒ Indeed, let be a p-subadditive quasi-norm defining the topology on X, 0 < p 1. For|| a · ||finite subset F N denote x(F ) = x . Without loss ≤ ⊂ m∈F m of generality we may assume 0 β 1. For each m let β = ∞ β (k)2−k, ≤ m ≤ P m k=1 m βm(k) 0, 1 be a dyadic expansion of βm. Since the series m∈N xm is unconditionally∈{ Cauchy,} given ε> 0 there exists N such that x(F ) ε Pwhenever F N is finite and min F N. For any such F we have || ||P ≤ ⊂ ≥ ∞ ∞ −k −k βmxm = βm(k)2 xm = 2 x(Fk), mX∈F Xk=1 mX∈F Xk=1 where F = m F : β (k)=1 . Therefore, k { ∈ m } ∞ ∞ β x 2−kp x(F ) ε 2−kp. m m ≤ || k || ≤ m∈F k=1 k=1 X X X

Hence, the series m∈N βmxm is unconditionally Cauchy. Since the last estimate is independent of the choice 0 βm 1 we have (5.3). P ≤ ≤ Calderón-Zygmund operators associated with wavelet expansions. In this subsection we are going to show that a wide class of operators associated with wavelet expansions are Calderón-Zygmund, and hence they are bounded on Hp. The first result of this type was shown by Strömberg [Sö] for a particular class of wavelets with exponential decay at infinity. Meyer extended this result to the class of (isotropic) dyadic r-regular wavelets by Meyer [Me, Chapter 6]. Our goal is to show that Meyer’s approach works also for non-isotropic wavelet expansions for general dilations. The results of this subsection are valid if we replace the orthogonality condition of wavelets by a Bessel condition. Suppose that Ψ = ψ1,... ,ψL L2(Rn) and Φ = φ1,... ,φL are two Bessel multiwavelets, see{ Section 4. We} ⊂ assume that Ψ and Φ{ which are}r-regular for some r N. We will also require that all ψl’s and φl’s have vanishing moments up to a certain,∈ see Theorem 5.6. Define the index set Λ= (l, j, k) : l =1,... ,L,j Z, k Zn . { ∈ ∈ } We consider the operators being a composition of three simple operations of analysis, multiplication of the sequence space, and synthesis. Namely an analysis operator 96 2. WAVELETS is a mapping

(5.4) L2(Rn) f ( f, ψl ) ℓ2(Λ). ∋ 7→ h j,ki (l,j,k)∈Λ ∈ A multiplication operator by a sequence of bounded scalars ǫ = (ǫl ) j,k (l,j,k)∈Λ ∈ ℓ∞(Λ) is a mapping

(5.5) ℓ2(Λ) (sl ) (ǫl sl ) ℓ2(Λ). ∋ j,k (l,j,k)∈Λ 7→ j,k j,k (l,j,k)∈Λ ∈ Finally, a synthesis operator is a mapping

(5.6) ℓ2(Λ) (sl ) sl φl L2(Rn). ∋ j,k (l,j,k)∈Λ 7→ j,k j,k ∈ (l,j,kX)∈Λ Given two Bessel multiwavelets Ψ and Φ, and a sequence of bounded scalars ǫ = (ǫl )l=1,... ,L we deﬁne the operator T : L2(Rn) L2(Rn) by j,k (j,k)∈Z×Zn ǫ → (5.7) T (f)= ǫl f, ψl φl . ǫ j,kh j,ki j,k (l,j,kX)∈Λ

The operator Tǫ is bounded as a composition of bounded operators. Lemma 5.5. Suppose Ψ and Φ are r-regular Bessel multiwavelets. Then for any sequence of scalars ǫ = (ǫl ) with ǫl 1 the operators T given by (5.7) j,k (l,j,k)∈Λ | j,k|≤ ǫ are Calder`on-Zygmund of order r with uniformly bounded constants independent of ǫ. Moreover, the kernels Kǫ(x, y) of Tǫ satisfy the symmetric (CZ-r) condition, i.e., there exists a constant C such that for every x = y 6 (5.8) ∂α∂β[K (Al , Al )](A−lx, A−ly) C/ρ(x y)= Cb−l for α , β r, | y x ǫ · · |≤ − | | | |≤ where l Z is the unique integer such that x y Bl+1 Bl. Furthermore,∈ if Φ consists of functions with− ∈ vanishing\ moments up to order s < r ln λ / ln λ then T ∗(xα)=0 for all α s. − + ǫ | |≤ Proof. For the sake of simplicity we are going to present the calculations only when Ψ and Φ consist of a single function. The kernel Kǫ of the operator Tǫ given by (5.7) is given by

Kǫ(x, y)= ǫj,kφj,k(x)ψj,k(y) (j,k)∈Z×Zn (5.9) X = ǫ bj φ(Aj x k)ψ(Aj y k). j,k − − Z Zn (j,k)X∈ ×

The formula (5.9) makes perfect sense if all but ﬁnitely many of ǫj,k’s are zeroes. In general, the above series converges absolutely for all x = y and for any boundedly supported f L2(Rn) 6 ∈

(5.10) Tǫf(x)= Kǫ(x, y)f(y)dy for x supp f. Rn 6∈ Z Furthermore, we claim that the convergence in (5.9) is uniform on compact subsets of Ω= (x, y) Rn Rn : x = y . Suppose that ρ(x y) bl, i.e., x y (B )c { ∈ × 6 } − ≥ − ∈ l 5. REGULAR WAVELETS AS AN UNCONDITIONAL BASIS FOR Hp 97 for some l Z. By the Cauchy-Schwarz inequality ∈ j j j ǫj,kφj,k(x)ψj,k(y) = b φ(A x k)ψ(A y k) Zn | | Zn | − − | jX≤−l kX∈ jX≤−l kX∈ 1/2 1/2 j j 2 j 2 (5.11) b φ(A x k) ψ(A y k) ≤ Zn | − | Zn | − | jX≤−l kX∈ kX∈ b C bj = C b−l. ≤ b 1 jX≤−l − Here the constant C is such that ψ(z k) 2 C and φ(z k) 2 C for all z Rn. Zn | − | ≤ Zn | − | ≤ ∈ kX∈ kX∈ To estimate the sum over j > l we use the inequality − (5.12) φ(Aj x k)ψ(Aj y k) C(1 + Aj x k )−n−1(1 + Aj (x y) )−N , | − − |≤ | − | | − | −N for some N satisfying bλ− < 1. (5.12) holds since both φ and ψ decay polynomially fast and (1 + z )(1 + z′ ) 1+ z z′ for any z,z′ Rn. | | | | ≥ | − | ∈ By (5.12)

ǫj,kφj,k(x)ψj,k(y) Zn | | j>X−l kX∈ C bj (1 + Aj x k )−n−1(1 + Aj (x y) )−N ≤ | − | | − | j>−l k∈Zn (5.13) X X C bj Aj (x y) −N C bj(1/cλj+l)−N ≤ | − | ≤ − j>X−l j>X−l Cb−l (bλ−N )j Cb−l, ≤ − ≤ j>0 X since Aj (x y) (B )c. Combining (5.11) and (5.13) we see that the series in − ∈ j+l (5.9) converges absolutely and uniformly in the region ρ(x y) bl. Since l Z is arbitrary, we have uniform convergence on compact subset−s of≥ Ω. Furthermore,∈ the estimate (5.8) holds for α = β = 0. Since (5.10) holds for ǫ’s with all but ﬁnitely many nonzero coeﬃcients it holds for any ǫ by the Lebesgue Dominated Convergence Theorem. Suppose now that x y Bl+1 Bl for some l Z. Formally, for any multi- indices α , β r we have− ∈ \ ∈ | | | |≤ ∂α∂β[K(Al , Al )](A−lx, A−ly) y x · · (5.14) j β j+l −l α j+l −l = ǫj,kb ∂ [φ(A k)](A x)∂ [ψ(A k)](A y). Z Zn ·− ·− (j,k)X∈ × To justify (5.14) we will show that the above series converges absolutely to some value smaller than Cb−l. As before we estimate separately the sum over j l and j > l. By the chain rule ≤ − − ∂α[ψ(Aj+l k)](A−ly) Aj+l |α| D|α|ψ(Aj y k) C D|α|ψ(Aj y k) , | ·− | ≤ || || || − || ≤ || − || 98 2. WAVELETS for j l. Since φ and ψ are r-regular, by the Cauchy-Schwarz inequality ≤− bj ∂β[φ(Aj+l k)](A−lx) ∂α[ψ(Aj+l k)](A−ly) Zn | ·− || ·− | jX≤−l kX∈ 1/2 1/2 (5.15) C bj D|β|φ(Aj x k) 2 D|α|ψ(Aj y k) 2 ≤ Zn || − || Zn || − || jX≤−l kX∈ kX∈ Cb−l. ≤ To estimate the sum over j > l we use the inequality − ∂β[φ(Aj+l k)](A−lx) ∂α[ψ(Aj+l k)](A−ly) | ·− || ·− | C2 Aj+l 2r D|β|φ(Aj x k) D|α|ψ(Aj y k) ≤ || || || − || · || − || Cλ2r(j+l)(1 + Aj x k )−n−1(1 + Aj (x y) )−N , ≤ + | − | | − | 2r −N for some N satisfying bλ+ λ− < 1. The above holds since both φ and ψ are r-regular. Therefore

bj ∂β[φ(Aj+l k)](A−lx) ∂α[ψ(Aj+l k)](A−ly) Zn | ·− || ·− | jX≤−l kX∈ 2r(j+l) j j −n−1 j −N (5.16) Cλ+ b (1 + A x k ) A (x y) ≤ Zn | − | | − | jX≤−l kX∈ Cb−l (bλ2rλ−N )j Cb−l, ≤ + − ≤ j>0 X j since A (x y) Bj+l+1 Bj+l. Thus, the series in (5.14) converges absolutely and we have− equality∈ there.\ By (5.15) and (5.16) the estimate (5.8) holds. Finally, suppose φ has vanishing moments up to order s < r ln λ−/ ln λ+, i.e., φ(x)xαdx = 0 for α s. Take any f L2 with compact support and vanishing | |≤ ∈ moments up to order r 1. Assume that supp f Bl for some l Z. Since Tε are R(CZ-r) with uniform constants− ⊂ ∈

T f(x) 2dx C | ǫ | ≤ (5.17) ZBl+ω c T f(x) Cρ(A−l−ωx)−δ for x (B ) | ǫ |≤ ∈ l+ω by the proof of Lemma 9.5 in Chapter 1, where the constant C is independent of the choice of ǫ. Here δ = r ln λ−/ ln b + 1. To guarantee the integrability of s Tǫf(x)(1 + x ) we must have δ>s ln λ+/ ln b + 1 and thus s < r ln λ−/ ln λ+. By (5.7) we have| |

(5.18) T f(x)xαdx =0 for α s, ǫ | |≤ Z if ǫj,k = 0 for all but ﬁnitely many (j, k)’s. Given a general ǫ = (ǫj,k) with ǫj,k 1 i | |≤ we deﬁne the sequence (ǫ )i∈N by

ǫj,k if j , k i, ǫi = | | | |≤ j,k 0 otherwise. 5. REGULAR WAVELETS AS AN UNCONDITIONAL BASIS FOR Hp 99

l Since Kǫi (x, y) converges to Kǫ(x, y) as i uniformly on (x, y) : ρ(x y) b we have → ∞ { − ≥ } c T i f(x) T f(x) as i for x (B ) . ǫ → ǫ → ∞ ∈ l+ω α Since Tǫi f(x)x dx = 0 for all α s, i N we have (5.18) by the Lebesgue Dominated Convergence Theorem| and| ≤ (5.17).∈ This ends the proof of Lemma 5.5. R Theorem 5.6. Suppose Ψ and Φ are r-regular Bessel multiwavelets for some r. Suppose also that p satisﬁes

(ln λ )2 (5.19) 0 1/p 1 < − r. ≤ − ln b ln λ+ If Φ consists of functions with vanishing moments up to order s = (1/p l l⌊ − 1) ln b/ ln λ− then for any sequence of scalars ǫ = (ǫj,k)(l,j,k)∈Λ with ǫj,k 1 ⌋ | p| ≤n then the operator Tǫ given by (5.7) extends to a bounded operator from HA(R ) p n into HA(R ) with the norm independent of ǫ. Proof. The proof is an immediate consequence of Lemma 5.5 and Theorem 9.8 in Chapter 1 since s < r ln λ−/ ln λ+.

We shall need one more result. Lemma 5.7. With all the assumptions of Theorem 5.6 we also assume that i,l we have a sequence (ǫi) N in the unit ball of ℓ∞(Λ), i.e., ǫ 1 for i N, i∈ | j,k| ≤ ∈ (l, j, k) Λ such that ǫi ǫ weak- in ℓ∞ as i . In other words, ∈ → ∗ → ∞ (5.20) ǫi,l ǫl as i for all (l, j, k) Λ. j,k → j,k → ∞ ∈ Then for any f Hp (Rn) ∈ A p (5.21) T i f T f in H as i . ǫ → ǫ A → ∞

Proof. Suppose ﬁrst that f L2 is compactly supported with vanishing ∈ moments up to order s, say supp f Bl for some l Z. By (5.7), (5.17), and (5.20) ⊂ ∈

2 (T i T )f(x) dx 0 as i , | ǫ − ǫ | → → ∞ ZBl+ω (5.22) −l−ω −δ c Tǫi f(x) , Tǫf(x) Cρ(A x) for x (Bl+ω) , | | | |≤ ∈ c (T i T )f(x) 0 as i for x (B ) . | ǫ − ǫ | → → ∞ ∈ l+ω α Furthermore (T i T )f(x)x dx = 0 for α s. To show that ǫ − ǫ | |≤

(5.23) R (Tǫi Tǫ)f Hp 0 as i , || − || 2,s → → ∞ a slight modiﬁcation of the proof of Lemma 9.3 in Chapter 1 is required. The proof proceeds in the same manner, except that we are going to show that gj+1 gj are appropriate multiples of (p, 2,s)-atoms. We not only have −

g g C b(j+1)(1/2−1/p)b(δ−1/p)(l+ω−j) for j l + ω, || j+1 − j||2 ≤ 4 ≥ 100 2. WAVELETS

i where gj =π ˜Bj ((Tǫ Tǫ)f)1Bj , but also gj+1 gj 2 0 as i by (5.22) and the Lebesgue Dominated− Convergence|| Theorem.− || This→ shows→(5.23) ∞ for all (p, 2,s)-atoms f. In particular, for any (p, 2,s)-atom a we have

(5.24) T (a)= ǫl a, ψl φl , ǫ j,kh j,ki j,k (l,j,kX)∈Λ p p with the unconditional convergence in HA. Given a general element f HA we ﬁnd its atomic decomposition f = κ a with κ p < and∈a ’s are j∈N j j j∈N | j | ∞ j (p, 2,s)-atoms. By (5.23) and the uniform boundedness of T i ’s P P ǫ

Tǫi f = κj Tǫi aj κjTǫaj = Tǫf as i . N → N → ∞ Xj∈ Xj∈ Therefore (5.21) holds. This ends the proof of Lemma 5.7.

As a corollary of the above considerations we have for all f Hp ∈ A T (f)= κ T a = κ ǫl a , ψl φl ǫ i ǫ i i j,kh i j,ki j,k i∈N i∈N (l,j,k)∈Λ (5.25) X X X l l l l l l = ǫj,k κi ai, ψj,k φj,k = ǫj,k f, ψj,k φj,k, N h i h i (l,j,kX)∈Λ Xi∈ (l,j,kX)∈Λ l since the above series converges unconditionally and the ψj,k’s belong to the Cam- 1/p−1 panato space C∞,s . The formula (5.25) is of practical importance because it says that (5.7) holds also if f Hp for an appropriate range of p’s. ∈ A Wavelets as an unconditional basis. We are now ready to harvest the fruit of our labor. Theorem 5.8. Suppose Ψ is an r-regular multiwavelet consisting of functions l with vanishing moments up to the order r ln λ−/ ln λ+ . Then (ψj,k)(l,j,k)∈Λ forms p ⌊ ⌋ an unconditional basis for HA for all p satisfying (5.19). Proof. We can apply Theorem 5.6 and Lemma 5.7, since

s = (1/p 1) ln b/ ln λ r ln λ / ln λ . ⌊ − −⌋≤⌊ − +⌋ By Lemma 5.7 and (5.25)

f = f, ψl ψl , h j,ki j,k (l,j,kX)∈Λ p l and the convergence is unconditional in HA. Therefore, the system (ψj,k)(l,j,k)∈Λ p is a basis for HA, since Ψ is an orthonormal multiwavelet.

We remark that the assumption in Theorem 5.8 that Ψ consists of functions with vanishing moments is, in fact, a consequence of Ψ being an r-regular multiwavelet. This is a well-known fact for dyadic wavelets which can be found in [Bt, Da2, Me]. This result was extended to general dilations with integer entries by the author [Bo4, Theorem 4]. 5. REGULAR WAVELETS AS AN UNCONDITIONAL BASIS FOR Hp 101

Theorem 5.9. Suppose A is a dilation with integer entries and Ψ is an r- regular orthonormal multiwavelet for some r N. Then ∈ α l x ψ (x)dx =0 for all l =1,... ,L, α < r ln λ−/ ln λ+. Rn | | Z Therefore, the assumption in Theorem 5.8 that Ψ has vanishing moments is an automatic consequence of r-regularity of Ψ. Even though the proof of Theorem 5.9 in [Bo4] works only for integer dilations, it is very likely that Theorem 5.9 holds for all dilations. For example, Theorem 5.9 holds trivially for dilations A satisfying (3.11), since such dilations do not allow r-regular wavelets. Hence, we conjecture that Theorem 5.9 is valid for all dilations. By Theorem 5.8 and standard interpolation theory we can also conclude that a 1-regular multiwavelet generates an unconditional basis in Lp for 1

6. Characterization of Hp in terms of wavelet coefficients In this section we will characterize the sequence space of wavelet expansion l p n coefficients ( f, ψj,k ) for f HA, whereΛ = (l, j, k) : l =1,... ,L,j Z, k Z . l h i ∈ { ∈ ∈ } Let (ǫj,k)(l,j,k)∈Λ be the sequence of independent identically distributed (i.i.d.) l l random Bernoulli variables with P (ǫj,k = 1) = P (ǫj,k = 1) = 1/2. In other − l words, if we consider the probability space ([0, 1], dt) then we can think of ǫj,k(t)= rσ(l,j,k)(t), where σ :Λ N is a bijection and (ri(t))i∈N are the Rademacher → i functions defined by ri(t) = sign(sin(2 πt)) for t [0, 1]. The fundamental result concerning the means∈ of i.i.d. random Bernoulli variables is Khinchin’s (Hinqin) inequality.

Theorem 6.1. For any 0

1/2 1 p 1/p 1/2 (6.1) A c 2 c r (t) dt B c 2 , p | i| ≤ i i ≤ p | i| i∈N Z0 i∈N i∈N X X X for any sequence of scalars (ci).

p The inequality (6.1) means that i ciri converges unconditionally in L if and 2 2 p only if i ciri converges in L , i.e., i ci < . It diverges in the L norm to 2 P | | ∞ inﬁnity if i ci = . WeP shall also| | need∞ the Feﬀerman-SteinP vector-valued maximal inequality, for a proof see [FS1,P St2]. Here M is the Hardy-Littlewood maximal operator on a space of homogeneous type. Theorem 6.2. Suppose 1

1/q 1/q q q (6.2) Mfi Cp,q fi . | | p ≤ | | p i∈N L i∈N L X X

Our goal is to show the following lemma. Lemma 6.3. Suppose 0

l l p (6.3) cj,kψj,k converges unconditionally in HA, (l,j,kX)∈Λ 1/2 (6.4) cl 2 ψl (x) 2 Lp(Rn), | j,k| | j,k | ∈ (l,j,kX)∈Λ 1/2 (6.5) cl 2 (1 ) (x) 2 Lp(Rn), | j,k| | El j,k | ∈ (l,j,kX)∈Λ for any (or some) bounded measurable sets E Rn with E > 0, l =1,...,L. l ⊂ | l| 6. CHARACTERIZATION OF Hp IN TERMS OF WAVELET COEFFICIENTS 103

Before proceeding with the proof of Lemma 6.3 we need to introduce some terminology. Let be the family of dilated cubes, i.e., I = Aj ([0, 1]n + k) : j Z, k Zn . I { ∈ ∈ } Given I = Aj ([0, 1]n + k) we deﬁne the scale of I by scale(I)= j. Let g be the ∈ I n square function given by (6.5) for El = [0, 1] ,

1/2 L 1/2 (6.6) g(x)= cl 2 (1 ) (x) 2 = cl 2 I −11 (x) , | j,k| | El j,k | | I | | | I (l,j,kX)∈Λ Xl=1 IX∈I where we use the convention cl = cl for a cube I = Aj ([0, 1]n + k), j Z, I −j,k ∈ k Zn. ∈ Definition 6.4. Suppose ′ . We say that the cube I is stacked below I′ ⊂ I the cube J within the family , and write I 4I′ J, if there is a chain of cubes I = I , I ,... ,I = J ′ suchI that 0 1 s ∈ I scale(I ) < scale(I ) and I I = 0 for all i =0,... ,s 1. i i+1 | i ∩ i+1| 6 − ′ ′ The relation 4I′ induces a partial order in . Let max( ) be the set of maximal ′ I I elements in with respect to 4 ′ . I I ′ If a subfamily does not contain arbitrary large cubes, i.e., supI∈I′ scale(I) < I ′ ′ , then for any cube I there is always a cube J max( ) with I 4I′ J. In general,∞ a maximal cube∈ is I not unique unless, for example,∈ thIe dilation A = 2Id and we work with nicely nested dyadic cubes. We shall need a simple lemma. Lemma 6.5. There is a universal constant η N such that whenever we have ′ j n ∈ j two cubes I, J with I 4 ′ J = A ([0, 1] + k) then I A (B + k). ∈ I I ⊂ η In the above lemma a family ′ does not play any role and can be substituted by the whole family . I I Proof. Suppose ﬁrst that J = [0, 1]n + k for some k Z. For any integer the diameter of a cube with scale(I)= j is at most ∈ diam(I) = diam(Aj [0, 1]n) Aj diam([0, 1]n)=2n Aj . ≤ || || || || Whenever we have a chain of cubes I0, I1,... ,Is = J satisfying Deﬁnition 6.4 then ∈ I s s 0 diam I diam(I ) 2n Aj < . i ≤ i ≤ || || ∞ i=0 i=0 j=−∞ [ X X Therefore, there exists η N such that I Bη + k whenever I 4I J. If J = Aj ([0, 1]n +k) is arbitrary then∈ I 4 J ⊂A−j I 4 A−j J. Thus, A−j I B +k I ⇐⇒ I ⊂ η and hence I Aj (B + k). ⊂ η We also need the following elementary lemma.

Lemma 6.6. Suppose r1, r2 > 0 and P is an n n real matrix with P x r1 x n × | |≥ | | for all x R . Then there is a constant c = c(r1, r2) depending only on r1 and r2 such that∈ #(Zn P B(z, r )) c det P for any z Rn. ∩ 2 ≤ | | ∈ 104 2. WAVELETS

Proof. Note that

#(Zn P B(z, r )) = #(P −1Zn B(z, r )) ∩ 2 ∩ 2 # k Zn : P −1(k + (0, 1)n) B(0, r ) = ≤ { ∈ ∩ 2 6 ∅} # k Zn : P −1(k + (0, 1)n) B(0, r +2n/r ) ≤ { ∈ ⊂ 2 1 } B(0, r +2n/r ) | 2 1 | = B(0, r +2n/r ) det P , ≤ P −1((0, 1)n) | 2 1 | · | | | | −1 n n −1 n n since diam P ((0, 1) ) 2 /r1 and the family P (k+(0, 1) ) : k Z consists of pairwise disjoint sets.≤ { ∈ }

l Proof of Lemma 6.3. Assume that (6.3) holds. Let (ǫj,k) be a sequence i.i.d. random Bernoulli variables. By Lemma 5.4 there exists a constant C such that for any subset F Λ ⊂

l l l (6.7) ǫj,k(t)cj,kψj,k C for any t [0, 1]. p ≤ ∈ (l,j,k)∈F H X

The identity on L2(Rn) is a Calder´on-Zygmund operator (of any order) hence it extends to a bounded operator from Hp to Lp with norm 1 by Theorem 9.9 in Chapter 1. If F Λ is ﬁnite then by Khinchin’s inequality, the Fubini Theorem and (6.7) ⊂

p/2 1 p l 2 l 2 l l l Ap cj,k ψj,k(x) dx ǫj,k(t)cj,kψj,k(x) dtdx Rn | | | | ≤ Rn Z (l,j,k)∈F Z Z0 (l,j,k)∈F X X 1 p l l l p ǫj,k(t)cj,kψj,k dt C . ≤ p ≤ Z0 (l,j,k)∈F L X

Since F Λ is arbitrary, by Fatou’s Lemma we have ⊂ p/2 l 2 l 2 p cj,k ψj,k(x) dx C /Ap, Rn | | | | ≤ Z (l,j,kX)∈Λ which shows (6.4). Assume now that (6.4) holds. Let δ > 0 be suﬃciently small so that E˜l = x Rn : ψl(x) >δ has nonzero measure for every l =1,...,L. Since { ∈ | | } l 2 l 2 2 l 2 2 n c ψ (x) δ c (1 ˜ )j,k(x) for all x R , | j,k| | j,k | ≥ | j,k| | El | ∈ (l,j,kX)∈Λ (l,j,kX)∈Λ

(6.5) holds for some E˜l’s. Choose any other bounded El’s with nonzero measure. We can then ﬁnd δ′ > 0 so that M(1 )(x) δ′1 (x) for all x Rn, where M E˜l El is the Hardy-Littlewood maximal operator associated≥ to the dilation∈ A given by (3.15) in Chapter 1. Since M commutes with translations and dilations by the power of A we have (M(1 )) (x)= M((1 )r )1/r(x) for any r > 0. Hence, for E˜l j,k E˜l j,k 6. CHARACTERIZATION OF Hp IN TERMS OF WAVELET COEFFICIENTS 105

p/2 ′ p l 2 2 (δ ) cj,k (1El )j,k(x) dx Rn | | | | Z (l,j,kX)∈Λ p/2 cl 2M((1 )r )2/r(x) dx j,k E˜l j,k ≤ Rn | | Z (l,j,kX)∈Λ (r/2)(p/r) = M( cl r(1 )r )(x) 2/r dx j,k E˜l j,k Rn | | | | Z (l,j,kX)∈Λ p/2 (C )p/r cl 2 (1 ) (x) 2 dx, p/r,2/r j,k E˜l j,k ≤ Rn | | | | Z (l,j,kX)∈Λ which shows (6.5). n Finally, assume (6.5) holds with El = [0, 1] for l = 1,...,L. We shall show that after some rearrangement the series in (6.3) is an appropriate combination of molecules. n r For each r Z define Ωr = x R : g(x) > 2 , where g(x) is given by (6.6). Clearly, Ω ∈Ω and { ∈ } r+1 ⊂ r ∞ s 2pr Ω = 2pr Ω Ω = 2pr Ω Ω | r| | s+1 \ s| | s+1 \ s| Z Z s=r Z r=−∞ (6.8) Xr∈ Xr∈ X Xs∈ X 1 ps 1 p = −p 2 Ωs+1 Ωs −p g(x) dx. 1 2 Z | \ |≤ 1 2 Rn | | − Xs∈ − Z We also define a subfamily of dilated cubes = I : I Ω I /2 and I Ω < I /2 . Ir { ∈ I | ∩ r| ≥ | | | ∩ r+1| | | } Note that for any I with cl = 0 for some l =1,...,L, there is a unique r Z ∈ I I 6 ∈ such that I r. For each∈r I Z, we have ∈ = I : I 4 J , Ir { ∈ Ir Ir } J∈max([ Ir) since the subfamily r does not contain arbitrary large cubes by Ωr < . By induction on J max(I ) we can find pairwise disjoint subfamilies | | satisfying∞ ∈ Ir Ir,J I : I 4 J , = , Ir,J ⊂{ ∈ Ir Ir } Ir Ir,J (6.9) J∈max(Ir) [ ′ ′ = for J = J max( ). Ir,J ∩ Ir,J ∅ 6 ∈ Ir We are now ready to rearrange the formal series in (6.3) as a combination of l l j n molecules. Using the convention cI = c−j,k, ψI = ψ−j,k for a cube I = A ([0, 1] +k) we can write L L l l l l l l cj,kψj,k = cI ψI = cI ψI , Z (l,j,kX)∈Λ Xl=1 XI∈I Xr∈ J∈max(X Ir) Xl=1 I∈IXr,J 106 2. WAVELETS

p where r and r,j are given by (6.9). To show that this series converges in HA it I I p suﬃces to show that cl ψl is an element of H and I∈Ir,J I I A P L p l l (6.10) cI ψI < , Z Hp ∞ r∈ J∈max(Ir) l=1 I∈Ir,J X X X X p by the completeness of HA and p-subadditivity of Hp . j0 n || · || Fix r Z and J = A ([0, 1] + k0) max( r). We claim that there is a constant C∈(independent of r and J) such that∈ I

L l l r 1/p (6.11) cI ψI C2 J . Hp ≤ | | l=1 I∈Ir,J X X

By Lemma 6.5 we have L L cl 2 2 cl 2 I −1 I Ω | I | ≤ | I | | | | \ r+1| Xl=1 I∈IXr,J Xl=1 I∈IXr,J L l 2 −1 2 cI I 1I (x)dx (6.12) j ≤ (A 0 (Bη +k0))\Ωr+1 | | | | Xl=1 I∈IXr,J Z 2 g(x) 2dx 2bj0+η22(r+1) ≤ j0 | | ≤ Z(A (Bη+k0))\Ωr+1 =8bj0+η22r,

r+1 since outside of Ωr+1 we have g(x) 2 . Since ψl is a Bessel family by≤ (6.12) we have { I } L 2 1 l l cI ψI (x) dx j Bj0+η+ω Bj +η+ω+A 0 k0 Z 0 l=1 I∈Ir,J | | X X L 2 L 1 C cl ψl cl 2 C22r. j0+η+ω I I j0+η+ω I ≤b L2 ≤ b | | ≤ l=1 I∈Ir,J l=1 I∈Ir,J X X X X j n j0 For any I r,J by Lemma 6.5 we have I = A ([0, 1] + k) A (Bη + k0) j ∈ Ij0 ⊂ and thus A k A k0 + Bj0+η. Therefore, for ﬁxed j < j0 the number of such k’s is bounded by∈ Lemma 6.6 (6.13) #(Zn (Aj0 −j k + B )) cbj0−j+η . ∩ 0 j0−j+η ≤ c Also for x Aj0 k + (B ) we have ∈ 0 j0 +η+ω ρ(x Aj0 k ) bωρ(x Aj k), − 0 ≤ − since Aj0 k Aj k B . Since Ψ is 0-regular, for any δ > 0 there is C = C(δ) > 0 0 − ∈ j0+η such that ψl(x) Cρ(x)−δ for all x Rn. In particular, we can take δ satisfying | |≤ ∈ c the hypothesis of Lemma 9.3 in Chapter 1. Therefore, for x Aj0 k + (B ) , ∈ 0 j0 +η+ω ψl (x) 2 = ψl (x) 2 = b−j ψl(A−j x k) 2 Cb−j ρ(A−j x k)−2δ | I | | −j,k | | − | ≤ − Cb2δωb−j ρ(A−j (x Aj0 k ))−2δ ≤ − 0 =Cb−j b−2δ(j0−j+η)ρ(A−j0 −η−ω(x Aj0 k ))−2δ. − 0 6. CHARACTERIZATION OF Hp IN TERMS OF WAVELET COEFFICIENTS 107

Combining the above with (6.12) and (6.13), by the Cauchy-Schwarz inequality we have L L 1/2 L 1/2 cl ψl (x) cl 2 ψl (x) 2 I I ≤ | I | | I | l=1 I∈Ir,J l=1 I∈Ir,J l=1 I∈Ir,J X X X X X X 1/2

b(j0+η)/22r cbj0−j+η Cb−jb−2δ(j0−j+η)ρ(A−j0−η−ω(x Aj0 k ))−2δ ≤ − 0 j

This proves (6.11). To show (6.10) we use J 2 J Ωr for J max( r) to obtain | |≤ | ∩ | ∈ I L p l l p pr cI ψI 2C 2 J Ωr Z Hp ≤ Z | ∩ | r∈ J∈max(Ir) l=1 I∈Ir,J r∈ J∈max(Ir) X X X X X X p p pr 2C p 2C 2 Ωr −p g Lp . ≤ Z | |≤ 1 2 || || Xr∈ − l Since the above estimate depends only on a magnitude of coeﬃcients cI ’s, we obtain the unconditional convergence in (6.3). This ends the proof of Lemma 6.3.

Note that the assumption that Ψ is 0-regular in Lemma 6.3 can be slightly relaxed. It follows from the proof that it suffices to assume ψl(x) Cρ(x)−δ for | | ≤ l = 1,...,L with δ > max(1/p,s ln λ+/ ln b + 1) in order to apply Lemma 9.3 in Chapter 1. We are now ready to prove the characterization of the sequence space of wavelet p expansion coefficients of elements in HA. Theorem 6.7. Suppose p satisfies (5.19), and Ψ is an r-regular tight frame multiwavelet with vanishing moments up to order s = (1/p 1) ln b/ ln λ− . Then for any f Hp the series f, ψl ψl converges⌊ unconditionally− ⌋ to f in ∈ A (l,j,k)∈Λh j,ki j,k Hp , and A P 1/2 l 2 l 2 f p f, ψ ψ ( ) HA j,k j,k || || ∼ |h i| | · | p (l,j,k)∈Λ L (6.14) X 1/2 l 2 2 f, ψj,k (1El )j,k( ) , ∼ |h i| | · | p (l,j,k)∈Λ L X

108 2. WAVELETS

n for any (or some) bounded measurable sets El R with El > 0, l = 1,...,L. The equivalence constants in (6.14) do not depend⊂ on f. | | l If, in addition, Ψ is an orthonormal multiwavelet then for any (cj,k)(l,j,k)∈Λ satisfying

1/2 (6.4) cl 2 ψl (x) 2 Lp(Rn), | j,k| | j,k | ∈ (l,j,kX)∈Λ there is a unique f Hp such that cl = f, ψl for all (l, j, k) Λ and (6.14) ∈ A j,k h j,ki ∈ holds. Proof. By Theorem 5.10 we have (5.26). By Lemma 6.3 we have (6.14). The fact that the equivalence constants in (6.14) do not depend on the choice of f follows by analyzing the proof of Lemma 6.3. Suppose, in addition, that Ψ is an orthogonal multiwavelet. Given a sequence l l l (cj,k)(l,j,k)∈Λ satisfying (6.4) by Lemma 6.3, (l,j,k)∈Λ cj,kψj,k converges unconditionally to some element f Hp . Since ψl’s belong to the dual of Hp we conclude A P A that for any (l′, j′, k′) Λ ∈ ∈ l′ l l l′ l l l′ l′ f, ψ ′ ′ = c ψ , ψ ′ ′ = c ψ , ψ ′ ′ = c ′ ′ . h k ,j i j,k j,k j ,k j,kh j,k j ,k i j ,k (l,j,kX)∈Λ (l,j,kX)∈Λ Finally, (6.14) holds by Lemma 6.3.

Wavelet expansion coeﬃcients for Lp. It comes as no surprise that The- orems 5.8, 5.10, and 6.7 can be extended for exponents p> 1. In fact, orthogonal wavelets with very mild decay conditions already form an unconditional basis for Lp, 1

Lemma 6.8. Suppose 0

Proof. By Khinchin’s inequality and the Fubini Theorem we have

1/2 p 1 p 1/2 p A f 2 r (t)f dt B f 2 , p | i| ≤ i i ≤ p | i| i∈F p Z0 i∈F p i∈F p X X X p for any ﬁnite F N. If i∈N fi converges unconditionally in L then by Lemma 5.4, ⊂ P C = sup r (t)f < . i i ∞ t∈[0,1] i∈N p X Hence, A ( f 2)1/2 p Cp. Since F N is arbitrary this shows (6.15). p|| i∈F | i| ||p ≤ ⊂ P 6. CHARACTERIZATION OF Hp IN TERMS OF WAVELET COEFFICIENTS 109

Lemma 6.9. Suppose 0 0 and all i N. Then there is a sequence of coeﬃcients (ǫi)i∈N, ǫi = 1 such that ∈ ±

N (6.16) lim sup ǫ f = . i i ∞ N→∞ i=1 p X

Proof. We claim that it suﬃces to show p/2 2 (6.17) fi = . N | | ∞ Z Xi∈ Assume for the time being that (6.17) holds. We claim that for any j N and r> 0 there is N j and a sequence ǫ ,... ,ǫ with ǫ = 1 such that ∈ ≥ j N i ± N ǫifi > r. i=j p X

Indeed, it suﬃces to take N such that ( N f 2) > r(A )−1/p and apply || i=j | i| ||p p (6.15). By a simple induction we can now produce a sequence (ǫ ) N, ǫ = 1 P i i∈ i satisfying (6.16). ± In order to show (6.17) we need to consider two cases. Suppose ﬁrst that

(6.18) lim sup x : fi(x) > η =0 forevery η> 0. i→∞ |{ | | }|

If (6.18) holds then there is a subsequence (fij ) and a decreasing sequence of positive numbers 1 = η0 > η1 >... , such that

p (6.19) fij > δ/2 for all j =1, 2,... η <|f |<η | | Z j ij j−1 p Indeed, we are going to proceed by induction. Let i1 be such that |f |<1 fi1 > i1 | | p 3/4δ. Choose η1 such that |f |≤η fi1 < 1/4δ. Hence, (6.19) holds for j = 1. i1 1 | | R Assume (6.19) holds up to some j > 1. By (6.18) we can pick ij+1 such that p R p |f |<η fij+1 > 3/4δ. Choose ηj+1 such that |f |≤η fij+1 < 1/4δ. This ij+1 j | | ij+1 j+1 | | completesR the induction step since R

p fij+1 > δ/2. η <|f |<η | | Z j+1 ij+1 j Deﬁne sets S = x : η < f (x) < η . Again by induction we can ﬁnd a j { j+1 | ij+1 | j } subsequence (jk) such that

(6.20) f p > δ/4 for all k =1, 2,... ijk S \(S ∪...∪S ) | | Z jk j1 jk−1 Indeed, by (6.18) and (6.19) for any set S with S < , | | ∞ p lim inf fi δ/2, j→∞ | j | ≥ ZSj \S 110 2. WAVELETS since η 0 as j . Therefore by (6.20), j → → ∞ p/2 f 2 f p = . i ijk N | | ≥ N Sj \(Sj ∪...Sj ) | | ∞ Z Xi∈ kX∈ Z k 1 k−1 On the other hand, if (6.18) fails then there exist η,µ > 0 such that x : |{ fi(x) > η > µ for infinitely many i’s. After taking a subsequence we can assume |that this| holds}| for all i N. Let S = x : f (x) > η . If S < ∈ i { | i | } | i∈N i| ∞ then necessarily i∈N 1Si (x) = on a set of positive measure. Therefore, the square function in (6.17) is infinite∞ on a set of positive measure.S Otherwise, if P i∈N Si = then the square function is at least η on a set of infinite measure. |This shows| (6.17)∞ and ends the proof. S The analog of Lemma 6.3 in the range 1

l l p (6.21) cj,kψj,k converges unconditionally in L , (l,j,kX)∈Λ 1/2 (6.22) cl 2 ψl (x) 2 Lp(Rn), | j,k| | j,k | ∈ (l,j,kX)∈Λ 1/2 (6.23) cl 2 (1 ) (x) 2 Lp(Rn), | j,k| | El j,k | ∈ (l,j,kX)∈Λ for any (or some) bounded measurable sets E Rn with E > 0, l =1,...,L. l ⊂ | l| Before we present the proof we need to introduce the standardized version of the sequence space under consideration. We also need a description of its dual which can be found in [NT, Section 9]. Definition 6.11. Given 0 sequences c = (cj,k)(l,j,k)∈Λ such that

1/2 l 2 2 c ℓp = cj,k (1[0,1]n )j,k || || A | | | | (l,j,k)∈Λ p (6.24) X L 1/2 = cl 2 I −11 < . | I | | | I ∞ l=1 I∈I p X X

p If the dilation A = 2Id and L = 1 then the sequence space ℓA coincides with ˙0,2 the space fp introduced by Frazier and Jawerth in [FJ2]. Obviously we could ˙ α,q have deﬁned a whole scale of spaces fp for a general dilation A. However, the ˙ α,q investigation of anisotropic homogeneous Triebel-Lizorkin spaces Fp is beyond the scope of this work. For the isotropic theory we refer the interested reader to [FJW, Tr1, Tr2]; for the function space theory on spaces of homogeneous type, see [Hn, HS, HW]. 6. CHARACTERIZATION OF Hp IN TERMS OF WAVELET COEFFICIENTS 111

p Lemma 6.12. Suppose A is a dilation and 1

l l −1/2 (6.26) f (x)j = cI I 1I (x) for l =1,... ,L. I∈I | | scale(XI)=j p p 2 This identiﬁcation deﬁnes an isometric inclusion of ℓA into L (ℓ ). Functional analysis tells us that the dual of Lp(ℓ2) is isometrically isomorphic to Lq(ℓ2), where 1/p +1/q = 1. Furthermore, the duality is given by

L (6.27) f,g = f l(x),gl(x) dx for f Lp(ℓ2),g Lq(ℓ2). h i Rn h i ∈ ∈ Z Xl=1 p ∗ q 2 p Also (ℓA) is isomorphic with the quotient of L (ℓ ) and the annihilator of ℓA, q 2 l which consists of functions f L (ℓ ) with I f (x)j dx = 0 for all l = 1,...,L, ∈ p q I , j = scale(I). Therefore, the dual of ℓA can be identiﬁed with ℓA. Moreover, p R p the∈ dualI of the inclusion ℓ ֒ Lp(ℓ2) is a projection P : Lq(ℓ2) (ℓ )∗ given by A → → A L (Pf)(x) = I −1 f l(y) dy 1 . j | | i I l=1 I∈I ZI X scale(XI)=j The duality (6.25) follows from (6.26) and (6.27).

Proof of Lemma 6.10. (6.21) = (6.22) follows from Lemma 6.8. (6.22) (6.23) is shown in the same way as in⇒ Lemma 6.3. However, (6.23) = (6.21) ⇐⇒is shown in a different manner since we do not have an atomic decomposition⇒ at our disposal for p> 1. l By Lemma 5.5 for any sequence of bounded scalars ǫ = (ǫj,k)(l,j,k)∈Λ the operator T : L2(Rn) L2(Rn) given by ǫ → (6.28) T (f)= ǫl f, ψl ψl , ǫ j,kh j,ki j,k (l,j,kX)∈Λ is Calderón-Zygmund of order 1. Hence, Tǫ is a Calderón-Zygmund operator in the sense of Definition 9.1, Chapter 1. Therefore, Tǫ extends to a bounded operator on Lp for 1

If the series f, ψl ψl does not converge unconditionally in Lp then (l,j,k)∈Λh j,ki j,k there are pairwise disjoint finite sets F Λ and coefficients ǫl = 1, (l, j, k) F P i ⊂ j,k ± ∈ i such that f = ǫl f, ψl ψl , i j,kh j,ki j,k (l,j,kX)∈Fi satisfies f >δ for some δ > 0 and all i N. By Lemma 6.9, there are coefficients || i||p ∈ ǫ = 1 such that lim sup N ǫ f = . This clearly contradicts the i ± N→∞ || i=1 i i||p ∞ uniform boundedness of T ’s. Finally, ǫl f, ψl ψl converges to T f ǫ P (l,j,k)∈Λ j,kh j,ki j,k ǫ by considering f L2 Lp. By Lemma 6.8, ∈ ∩ P p/2 l 2 l 2 f, ψj,k ψj,k(x) dx Rn |h i| | | Z (l,j,kX)∈Λ p (A )−1 sup ǫl (t) f, ψl ψl C/A f p, ≤ p j,k h j,ki j,k ≤ p|| ||p t∈[0,1] (l,j,k)∈Λ p X p l p for any f L . Therefore, ( f, ψj,k ) p C f p. In other words, the analysis ℓA ∈ p|| h p i || ≤ || || operator is bounded from L to ℓA. By Lemma 6.12, the dual of analysis operator is a synthesis operator

ℓq (cl ) cl ψl Lq(Rn), A ∋ j,k (l,j,k)∈Λ 7→ j,k j,k ∈ (l,j,kX)∈Λ where 1/p +1/q = 1. We also have

l l q cj,kψj,k C c ℓq for all c ℓ . ≤ || || A ∈ A (l,j,k)∈Λ q X

This shows (6.23) = (6.21). ⇒ We are now ready to prove the characterization of the sequence space of wavelet expansion coeﬃcients of functions in Lp. Theorem 6.13. Suppose 1 0, l = 1,...,L. The equivalence constants in (6.29) do not depend⊂ on f. | | l If, in addition, Ψ is an orthonormal multiwavelet then for any (cj,k)(l,j,k)∈Λ satisfying

1/2 (6.4) cl 2 ψl (x) 2 Lp(Rn), | j,k| | j,k | ∈ (l,j,kX)∈Λ 6. CHARACTERIZATION OF Hp IN TERMS OF WAVELET COEFFICIENTS 113 there is a unique f Lp such that cl = f, ψl for all (l, j, k) Λ and (6.29) ∈ j,k h j,ki ∈ holds. Proof. The ﬁrst part of the theorem follows from Lemma 6.10, see the proof of (6.23) = (6.21). ⇒ l Suppose, in addition, that Ψ is an orthonormal multiwavelet. If(cj,k)(l,j,k)∈Λ l l satisﬁes (6.4) then by Lemma 6.10, (l,j,k)∈Λ cj,kψj,k converges unconditionally to some f Lp. Since ψl’s belong to the dual of Lp we must have cl = f, ψl for ∈ P j,k h j,ki all (l, j, k) Λ. This completes the proof of Theorem 6.13. ∈ 114 2. WAVELETS

Notation Index

the standard norm in Rn or the Lebesgue measure of a subset in Rn |B ·( |z, r) the Euclidean ball with center z and radius r , the scalar product in Rn, L2(Rn), or f, ϕ = f(ϕ) for f ′, ϕ h· ·in nh i ∈ S ∈ S ei i=1 the standrad orthonormal basis in R {ˆ } ˆ −2πihx,ξi f(ξ) the Fourier transform of f, f(ξ)= Rn f(x)e dx supp f the support of f, supp f = x : f(x) =0 Dkf(x) the derivative of f at the point{ x thoughtR 6 } of as a symmetric multilinear operator (Rn)k Rn → n 1E the indicator function of the set E R A the dilation, i.e., n n matrix with⊂ all eigenvalues λ, λ > 1 × | | λ− the absolute value of the smallest eigenvalue of A λ+ the absolute value of the biggest eigenvalue of A b the number equal to det A | | k Bk the dilated balls of the form Bk = A ∆, k Z, where ∆ is a special ellipsoid tailored to the dilation A with ∆∈=1 the family of dilated balls of the form x | +| B , x Rn, k Z B 0 k 0 ∈ ∈ ω the smallest integer such that 2B0 Bω ρ the quasi-norm associated to the dilation⊂ A ϕ the dilate of ϕ to the scale k Z, ϕ (x)= b−kϕ(A−kx) k ∈ k ψj,k the dilate and translate of ψ given by ψ (x)= det A j/2ψ(Aj x k) where j Z, k Zn j,k | | − ∈ ∈ Ψ the collection Ψ = ψ1,... ,ψL L2(Rn) the collection of dilated{ cubes }⊂Aj ([0, 1]n + k) : j Z, k Zn I j n { ∈ ∈ } ψI is ψ−j,k for I = A ([0, 1] + k) ; the exception to this rule is 1I Λ the index set Λ = (l, j, k) : l =1∈,... I ,L,j Z, k Zn { ∈ ∈ } D the dilation operator (usually on Hp), D f(x)= det A 1/pf(Ax) A A | | τy the translation operator, τyf(x)= f(x y) (p,q,s) an admissible triplet with respect to the− dilation A, i.e., 0

s the space of polynomials of degree s Pp ≤ p L the space of functions with Rn f(x) dx < , 0

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