Product Theory on Spaces of Homogeneous Type

Product Theory on Spaces of Homogeneous Type

Yongsheng Han, Guozhen Lu and Dachun Yang

October, 2007

Abstract. Theory of multi-parameter analysis has been a central subject in harmonicanalysis and has received substantial progress in the past decades. Motivated by the classicalHp (p ≤ 1) product theory in the Euclidean spaces and the recent development of Lp (p > 1)multi-parameter analysis on groups of stratified type, we build up the Hp (p ≤ 1) theory ofmulti-parameter analysis on spaces of homogeneous type. We first establish the Littlewood-Paley theory, Calderon reproducing formulas, and Plancherel-Polya inequalities on productspaces and then introduce and develop the product Hp theory. Atomic decomposition isgiven on such Hp spaces (see Theorem 4.3) and boundedness of singular integrals on suchHp spaces and from Hp to Lp are established (see Theorems 5.1 and 5.2). A Journe typecovering lemma is also proved in the product of two homogeneous spaces which is of itsindependent interest (see Lemma 4.2). Results in this paper grow out of the product Hp

theory of two stratified groups, such as the Heisenberg groups, developed earlier by the firsttwo authors ([HL1], [HL2]).

2000 Mathematics Subject Classification: Primary 42B35; Secondary 42B30, 42B25, 43-99.Key words and phrases: spaces of homogeneous type, product space, Calderon-type reproducing formula,

approximation to the identity, Hp space.Yongsheng Han acknowledges the support of NNSF (No. 10271015) of China, Guozhen Lu acknowledges

the support of US NSF grants DMS9622996, 9970352 and 0500853 and Dachun Yang acknowledges thesupport of both NNSF (No. 10271015) and RFDP (No. 20020027004) of China.

1

2 Yongsheng Han, Guozhen Lu and Dachun Yang

1 Introduction

Our goal of this article is to develop the product theory on spaces of homogeneoustype. The classical theory of Calderon-Zygmund was described by certain singular integraloperators which commute with the one parameter dilations on Rn, given by ρδ(x) = δx

for all δ > 0. The product theory of Fourier analysis on Rn emphasizes operators built outof product acting on each R1’s, and which commute with the action of multi-parameterscaling on Rn, given by ρδ(x) = (δ1x1, δ2x2, ..., δnxn) with δ = (δ1, δ2, ..., δn), δi > 0 for 1 ≤i ≤ n. In fact, this theory has a long history: beginning with the strong maximal functionof Jessen, Marcinkiewicz and Zygmund ([JMZ]), the original form of the Marcinkiewiczmultiplier theorem ([S1]). It was then later developed in the setting of product Lp theory ofCalderon-Zygmund operators (see R. Fefferman-Stein [FS] and Journe [J1]), and of Hardyspace Hp for p < 1 and BMO spaces (see for example the works by Chang-R. Fefferman,Gundy-Stein, Journe, Pipher in [CF1], [CF2], [CF3], [F1], [F2], [F4], [Cha], [F6], [J2],[GS], [P]). Recently, the product Lp (p > 1) theory plays a crucial role in the study ofmany questions arising in multi-parameter analysis, such as Marcinkiewicz multipliers andmulti-parameter structures on Heisenberg-type groups (see works by Muller-Ricci-Stein in[MRS1, MRS2]), operators on nilpotent Lie groups given by convolution with certain flagsingular kernels (see work by Nagel-Ricci-Stein in [NRS]), etc. More recently, to estimatefundamental solutions of ¤b on certain model domains in several complex variables, Nageland Stein developed the product theory of singular integrals with non convolution kernels,namely the Lp theory for 1 < p < ∞ ([NS1, NS2, NS3]).

The main purpose of our paper is to develop a satisfactory product theory for 0 <

p ≤ 1 on product of two spaces of homogeneous type, namely, the theory of Hardy spaces(including atomic decomposition) and boundeness of singular operators on such Hardyspaces Hp and from Hp to Lp. Results in this paper include the product Hp theory,developed in [HL1] and [HL2], of two stratified groups such as the Heisenberg group as aspecial case. Our methods are quite different from those given in [NS3] for 1 < p < ∞and also in the classical product theory in Euclidean spaces in [CF1, CF2, CF3, F1, F4,F6] because we mainly establish the Hardy space theory using the Calderon reproducingformula and Littlewood-Paley analysis which hold in test function spaces in the productof homogeneous spaces, which are particularly suitable for the Hp theory when 0 < p ≤ 1.

To see how our methods work, let us recall some basic ideas and results of the producttheory on Rn. The simplest example of a product-type singular integral on Rn is the doubleHilbert transform H1H2 on R2 defined by

H1H2(f) = f ∗ 1x1x2

.

For such tensor products the Lp-boundedness for 1 < p < ∞ is trivial consequence ofFubini’s theorem. But for operators defined by T (f) = f ∗ K where K is defined onRn × Rm and satisfies all the analogous estimates to those satisfied by 1

x1x2, but cannot

Product Hp Theory on homogeneous spaces 3

be written in the tensor product form K1(x1)K2(x2), then the arguments which dealwith H1H2 fail. Fefferman and Stein developed a new method to deal with these moregeneral product-type operators with convolutional kernels. The basic idea they used isto develop the product-type Littlewood-Paley theory for Lp(Rn × Rm), 1 < p < ∞. Thisfollows from the original vector-valued Littlewood-Paley theory on Rn and an iterationargument. Their methods work for product-type convolutional operators very well, see[FS] for more details. Furthermore, Journe considered general product-type operatorswith non-convolutional kernels. He proved the T1 theorem in the product setting. A newidea Journe used is the vector-valued T1 theorem on Rn and a basic result he proved isthe so-called Journe’s covering lemma. We also refer to the work of Pipher [P] for Journetype covering lemmas of more parameters.

The product Hp theory has an extensive history. A counterexample given by Car-leson showed that the product Hp theory cannot be obtained by a routine matter ofiterating one dimensional methods ([Car]). Gundy and Stein established in [GS] the prod-uct Hp theory by using the non-tangential maximal function and the area integral ortheir probabilistic analogues resulted by introducing two-time Brownian motion, i.e., themartingale maximal function, and the corresponding square function. Chang and Feffer-man obtained atomic decomposition of the product Hp spaces ([CF1]). The key tool theyused is Calderon reproducing formula on product Rn, which follows from using the Fouriertransform. Since the support of each atom is an open set in the product Rn, one cannotuse atomic decomposition to get the Hp−Lp boundedness of product Calderon-Zygmundoperators while this worked very well on the single space Rn. Nevertheless, R. Feffermanfound that it would suffice to check the action of operators on atoms whose supports arerectangles in the product space to show the boundedness of operators on the product Hp

spaces ([F4]). As a consequence, R. Fefferman proved the Hp − Lp boundedness for aclass of operators introduced by Journe who proved the L∞ − BMO boundedness. TheJourne’s covering lemma plays a crucial role in the proofs of these results ([J1, J2]).

In the recent paper ([NS3]), Nagel and Stein considered the product space M =M1 × M2 × · · · × Mn where each factor Mi, 1 ≤ i ≤ n, is either a compact connectedsmooth manifold, or arises as the boundary of a model polynomial domain in C2. Bothare associated with the real vector fields which are of finite-type. They proved the Lp

boundedness of certain operators on these product spaces. The key idea of the proof is touse the product-type Littlewood-Paley theory on Lp, 1 < p < ∞. A crucial role to developthis theory is a reproducing formula which is constructed by use of heat kernel on each Mi

and then product-type reproducing formula is a tensor product. Since their reproducingformula holds only on L2, so it is not sufficient to use this reproducing formula to developthe product Hp(M) theory for 0 < p ≤ 1 in this setting.

To develop the product theory on spaces of homogeneous type, we first consider thatX = X1×X2, where (Xi, ρi, µi)di,θi , i = 1, 2, are spaces of homogeneous type in the senseof Coifman and Weiss ([CW1, CW2]; see also definition below). It was proved that if


Skiki∈Z , i = 1, 2, is an approximation to the identity on Xi, and set Dki = Ski − Ski−1

for ki ∈ Z, then the following Calderon reproducing formulae hold

(1.1) f(x) =∞∑

ki=−∞DkiDki(f)(x) =

∞∑

ki=−∞DkiDki(f)(x),

where the series converge both in Lp(Xi) with 1 < p < ∞ and in some test function spacesand its dual spaces on Xi, i = 1, 2, see [HS, H1] for more details.

The main tool to develop the product theory on X is the following product-typeCalderon reproducing formula

(1.2) f(x1, x2) =∞∑

k1=−∞

∞∑

k2=−∞Dk1Dk2Dk1Dk2(f)(x1, x2)

=∞∑

k1=−∞

∞∑

k2=−∞Dk1Dk2Dk1Dk2(f)(x1, x2),

where the series converge in Lp(X), 1 < p < ∞, in product test function spaces and itsdual spaces on X.

This formula together with the vector-valued Littlewood-Paley theory and an iter-ation argument yields the product Littlewood-Paley theory for Lp(X), 1 < p < ∞. Toestablish Hp(X) space, we formally introduce the product Littlewood-Paley-Stein S func-tion and define the product Hp(X) norm for distribution space mentioned above. As inthe product Rn case, the formula in (1.2) is a key tool to obtain atomic decomposition ofHp(X) space. To see the Littlewood-Paley-Stein S function and g function are equivalenton Hp(X), one needs the discrete product-type Calderon reproducing formula. In fact,the following discrete Calderon reproducing formula on each (Xi, ρi, µi)di,θi , i = 1, 2, wasproved in [H3]:

(1.3) f(x) =∞∑

k=−∞

∑

τ∈Ik

N(k,τ)∑

ν=1

µ(Qk,ντ )Dk(x, yk,ν

τ )Dk(f)(yk,ντ )

=∞∑

k=−∞

∑

τ∈Ik

N(k,τ)∑

ν=1


τ )Dk(f)(yk,ντ ),

where

Qk,ντ

k∈Z, τ∈Ik, ν=1, ···,N(k,τ)

is the collection of dyadic cubes in the sense of Christ

([Chr1, Chr2]), Sawyer-Wheeden ([SW]) and the series converge in Lp(Xi) with 1 < p < ∞,

in some test function spaces and its dual spaces on Xi and i = 1, 2, see [H2] for moredetails.


Similarly, the discrete product-type Calderon reproducing formula is given by

(1.4) f(x1, x2) =∞∑

k1=−∞

∑

τ1∈Ik1

N(k1,τ1)∑

ν1=1

∞∑

k2=−∞

∑

τ2∈Ik2

N(k2,τ2)∑

ν2=1

µ1(Qk1,ν1τ1 )µ2(Qk2,ν2

τ2 )

×Dk1(x1, yk1,ν1τ1 )Dk2(x2, y

k2,ν2τ2 )Dk1Dk2(f)(yk1,ν1

τ1 , yk2,ν2τ2 )

=∞∑

k1=−∞

∑

τ1∈Ik1

N(k1,τ1)∑

ν1=1

∞∑

k2=−∞

∑

τ2∈Ik2

N(k2,τ2)∑

ν2=1


τ2 )

×Dk1(x1, yk1,ν1τ1 )Dk2(x2, y


τ1 , yk2,ν2τ2 ),

where the series converge in Lp(X), 1 < p < ∞, in product test function spaces and itsdual spaces on X.

The formula of (1.4) implies the so-called Plancherel-Polya inequalities. As a sim-ple consequence, we show the above Hp(X) space can be characterized by the productLittlewood-Paley-Stein g function. Using Littlewood-Paley-Stein g function, we will alsoprove the product T1 theorem of the Calderon-Zygmund operators.

To explain how our results include the product Hp theory on two stratified groupssuch as the Heisenberg group, we give some preliminary introduction here.

We begin with some preliminaries concerning stratified Lie groups (or so-calledCarnot groups). We refer the reader to the books [FS] and [VSCC] for analysis on stratifiedgroups. Let G be a finite-dimensional, stratified, nilpotent Lie algebra. Assume that

G = ⊕si=1Vi ,

with [Vi, Vj ] ⊂ Vi+j for i + j ≤ s and [Vi, Vj ] = 0 for i + j > s. Let X1, · · · , Xl be a basisfor V1 and suppose that X1, · · · , Xl generate G as a Lie algebra. Then for 2 ≤ j ≤ s, wecan choose a basis Xij, 1 ≤ i ≤ kj , for Vj consisting of commutators of length j. We setXi1 = Xi, i = 1, · · · , l and k1 = l, and we call Xi1 a commutator of length 1.

If G is the simply connected Lie group associated with G, then the exponentialmapping is a global diffeomorphism from G to G. Thus, for each g ∈ G, there is x =(xij) ∈ RN for 1 ≤ i ≤ kj , 1 ≤ j ≤ s and N =

∑sj=1 kj such that

g = exp(∑

xijXij) .

A homogeneous norm function | · | on G is defined by

|g| = (∑

|xij |2s!/j)1/2s! ,

and Q =∑s

j=1 jkj is said to be the homogeneous dimension of G. The dilation δr onG is defined by

δr(g) = exp(∑

rjxijXij) if g = exp(∑

xijXij).


We call a curve γ : [a, b] → G ”a horizontal curve” connecting two points x, y ∈ Gif γ(a) = x, γ(b) = y and γ

′(t) ∈ V1 for all t. Then the Carnot-Caratheodory distance

between x, y is defined as

dcc(x, y) = infγ

∫ b

a< γ

′(t), γ

′(t) >

12 dt,

where the infimum is taken over all horizontal curves γ connecting x and y. It is knownthat any two points x, y on G can be joined by a horizontal curve of finite length and thendcc is a left invariant metric on G. We can define the metric ball centered at x and withradius r associated with this metric by

Bcc(x, r) = y : dcc(x, y) < r.

We must notice that this metric dcc is equivalent to the pseudo-metric ρ(x, y) = |x−1y|defined by the homogeneous norm | · | in the following sense (see [FS])

Cρ(x, y) ≤ dcc(x, y) ≤ Cρ(x, y).

We denote the metric ball associated with ρ as D(x, r) = y ∈ G : ρ(x, y) < r. Animportant feature of both of these distance functions is that these distances and thus theassociated metric balls are left invariant, namely,

dcc(zx, zy) = d(x, y), Bcc(x, r) = xBcc(0, r)

andρ(zx, zy) = ρ(x, y), D(x, r) = xD(0, r).

For simplicity, we will use the left invariant metric dcc to study the product theoryof two stratified groups. An important property of the metric ball is that

µ(Bcc(x, r)) = cQrQ

for all x ∈ G and r > 0, where µ is the Lebesgue measure on G and Q is the homogeneousdimension. Therefore, the space (G, dcc, µ) is a space of homogenous type.

If we consider two stratified groups (G1, d1cc, µ) and (G2, d

2cc, µ), the product Hp theory

developed in this paper includes the case of product theory on G1×G2 as a special case. Ofparticular interests are the case Hp(G1 ×G2) when G1 or G2 is the renowned Heisenberggroup. Such product Hp theory was developed earlier by the first two authors in ([HL1],[HL2]). It is this work which motivated the generalization to the Hp product theory oftwo homogeneous spaces in the current paper.

The following final remarks are in order. First of all, as we pointed out at thebeginning of the introduction, the methods employed in this paper are different fromthose classical product Hardy space theory in several ways. We develop a discrete Calderon


reproducing formula and establish a Plancherel-Polya inequalities in product spaces. Thesetools are used for the first time in product spaces. Indeed, we adapt successfully to thecase of product spaces the methods of the Littlewood-Paley theory, Calderon reproducingformula, Plancherel-Polya inequalities in the single homogeneous spaces developed by thefirst author over the past decade (see e.g., [H1], [H2], [H3]). Second, we would like to pointout that this paper is a substantial extension and expansion of the earlier unpublishedmanuscripts by the first two authors in the case of the product theory of two stratifiedgroups about ten years ago (see [HL1], [HL2]).1 We were motivated then by the importantdevelopment of multi-parameter analysis on Rn by R. Fefferman and Pipher ([FP1] and[FP2]), Ricci and Stein ([RS]), and Nagel-Ricci-Stein ([NRS]), and Nagel and Stein ([NS1],[NS2], [NS3], [NS4]), on Heisenberg-type groups of Muller-Ricci-Stein ([MRS1], [MRS2]).Third, on the one hand for the convenience of the reader, we have made every effort tomake our presentation in this paper self-contained. On the other hand, we have strived tosimplify the exposition so that similar estimates which appear in one part of this paperwill be very brief in other parts of the paper. Therefore, we must apologize for the lengthycomputations and estimates given in some parts of this paper due to the very complicatednature of product theory itself, and also some somewhat concise exposition in other parts.We hope that an interested reader will be patient enough while reading this paper.

A brief description of the contents of this paper follows. In Section 2 we first introducea class of test functions and its dual space (distribution space), and then establish aspecial product-type Calderon reproducing formula. In Section 3, we develop the productLittlewood-Paley-Stein theory for Lp, 1 < p < ∞. The product Hp space is established inSection 4, and we prove the Journe’s covering lemma for product spaces of homogeneoustype (Lemma 4.2). Atomic decomposition for Hp spaces is proved in this section (SeeTheorem 4.3). The boundedness of Calderon-Zygmund operators on Hp space and fromHp to Lp are derived in Section 5 (See Theorems 5.1 and 5.2).

1Those works in that framework have been presented by the second author in the invited talks at theinternational conference in harmonic analysis in Kiel, Germany in 1998 and also in the AMS special sessionof harmonic analysis in Chicago in 1999.


2 Special Calderon reproducing formulae

We begin with recalling some necessary definitions and notation on spaces of homo-geneous type.

A quasi-metric ρ on a set X is a function ρ : X ×X → [0,∞) satisfying that(i) ρ(x, y) = 0 if and only if x = y;

(ii) ρ(x, y) = ρ(y, x) for all x, y ∈ X;

(iii) There exists a constant A ∈ [1,∞) such that for all x, y and z ∈ X,

(2.1) ρ(x, y) ≤ A[ρ(x, z) + ρ(z, y)].

Any quasi-metric defines a topology, for which the balls

B(x, r) = y ∈ X : ρ(y, x) < rfor all x ∈ X and all r > 0 form a basis.

In what follows, we set diamX = supρ(x, y) : x, y ∈ X and Z+ = N ∪ 0. Wealso make the following conventions. We denote by f ∼ g that there is a constant C > 0independent of the main parameters such that C−1g < f < Cg. Throughout the paper,we denote by C a positive constant which is independent of the main parameters, butit may vary from line to line. Constants with subscripts, such as C1, do not change indifferent occurrences. For any q ∈ [1,∞], we denote by q′ its conjugate index, namely,1/q + 1/q′ = 1. Let A be a set and we will denote by χA the characteristic function of A.

Definition 2.1 Let d > 0 and θ ∈ (0, 1]. A space of homogeneous type, (X, ρ, µ)d,θ, is aset X together with a quasi-metric ρ and a nonnegative Borel regular measure µ on X,and there exists a constant C0 > 0 such that for all 0 < r < diamX and all x, x′, y ∈ X,

(2.2) µ(B(x, r)) ∼ rd

and

(2.3) |ρ(x, y)− ρ(x′, y)| ≤ C0ρ(x, x′)θ[ρ(x, y) + ρ(x′, y)]1−θ.

The space of homogeneous type was first introduced by Coifman and Weiss [CW1]and its theory has developed significantly in the past three decades. For a variant of thespace of homogeneous type as given in the above definition, we refer to [MS1]. In [MS1],Macias and Segovia have proved that one can replace the quasi-metric ρ of the space ofhomogeneous type in the sense of Coifman and Weiss by another quasi-metric ρ whichyields the same topology on X as ρ such that (X, ρ, µ) is the space defined by Definition2.1 with d = 1.

Throughout this section to Section 6, we will always assume that µ(X) = ∞.Let us now recall the definition of the space of test functions on spaces of homogeneous

type.


Definition 2.2 ([H1]) Let X be a space of homogeneous type as in Definition 2.1. Fixγ > 0 and β > 0. A function f defined on X is said to be a test function of type (x0, r, β, γ)with x0 ∈ X and r > 0, if f satisfies the following conditions:

(i) |f(x)| ≤ Crγ

(r + ρ(x, x0))d+γ;

(ii) |f(x)− f(y)| ≤ C

(ρ(x, y)

r + ρ(x, x0)

)β rγ

(r + ρ(x, x0))d+γ

for ρ(x, y) ≤ 12A

[r + ρ(x, x0)];

(iii)∫X f(x) dµ(x) = 0.

If f is a test function of type (x0, r, β, γ), we write f ∈ G(x0, r, β, γ), and the norm of f

in G(x0, r, β, γ) is defined by

‖f‖G(x0,r,β,γ) = infC : (i) and (ii) hold.

Now fix x0 ∈ X and let G(β, γ) = G(x0, 1, β, γ). It is easy to see that

G(x1, r, β, γ) = G(β, γ)

with an equivalent norm for all x1 ∈ X and r > 0. Furthermore, it is easy to check thatG(β, γ) is a Banach space with respect to the norm in G(β, γ). Also, let the dual space(G(β, γ))′ be all linear functionals L from G(β, γ) to C with the property that there existsC ≥ 0 such that for all f ∈ G(β, γ),

|L(f)| ≤ C‖f‖G(β,γ).

We denote by 〈h, f〉 the natural pairing of elements h ∈ (G(β, γ))′ and f ∈ G(β, γ). Clearly,for all h ∈ (G(β, γ))′ , 〈h, f〉 is well defined for all f ∈ G(x0, r, β, γ) with x0 ∈ X and r > 0.

It is well-known that even when X = Rn, G(β1, γ) is not dense in G(β2, γ) if β1 > β2,which will bring us some inconvenience. To overcome this defect, in what follows, fora given ε ∈ (0, θ], we let G(β, γ) be the completion of the space G(ε, ε) in G(β, γ) when0 < β, γ < ε.

Definition 2.3 ([H1]) Let X be a space of homogeneous type as in Definition 2.1. Asequence Skk∈Z of linear operators is said to be an approximation to the identity oforder ε ∈ (0, θ] if there exists C1 > 0 such that for all k ∈ Z and all x, x′, y and y′ ∈ X,

Sk(x, y), the kernel of Sk is a function from X ×X into C satisfying

(1) |Sk(x, y)| ≤ C12−kε

(2−k + ρ(x, y))d+ε;


(2) |Sk(x, y)− Sk(x′, y)| ≤ C1

(ρ(x, x′)

2−k + ρ(x, y)

)ε 2−kε

(2−k + ρ(x, y))d+ε

for ρ(x, x′) ≤ 12A

(2−k + ρ(x, y));

(3) |Sk(x, y)− Sk(x, y′)| ≤ C1

(ρ(y, y′)

2−k + ρ(x, y)

)ε 2−kε

(2−k + ρ(x, y))d+ε

for ρ(y, y′) ≤ 12A

(2−k + ρ(x, y));

(4) |[Sk(x, y)− Sk(x, y′)]− [Sk(x′, y)− Sk(x′, y′)]| ≤ C1

(ρ(x, x′)

2−k + ρ(x, y)

)ε

×(

ρ(y, y′)2−k + ρ(x, y)

)ε 2−kε

(2−k + ρ(x, y))d+ε

for ρ(x, x′) ≤ 12A

(2−k + ρ(x, y)) and ρ(y, y′) ≤ 12A

(2−k + ρ(x, y));

(5)∫X Sk(x, y) dµ(y) = 1;

(6)∫X Sk(x, y) dµ(x) = 1.

Moreover, A sequence Skk∈Z of linear operators is said to be an approximation to theidentity of order ε ∈ (0, θ] having compact support if there exist constants C2, C3 > 0 suchthat for all k ∈ Z and all x, x′, y and y′ ∈ X, Sk(x, y), the kernel of Sk is a function fromX ×X into C satisfying (5), (6) and

(7) Sk(x, y) = 0 if ρ(x, y) ≥ C22−k and ‖Sk‖L∞(X×X) ≤ C32kd;

(8) |Sk(x, y)− Sk(x′, y)| ≤ C32k(d+ε)ρ(x, x′)ε;

(9) |Sk(x, y)− Sk(x, y′)| ≤ C32k(d+ε)ρ(y, y′)ε;

(10) |[Sk(x, y)− Sk(x, y′)]− [Sk(x′, y)− Sk(x′, y′)]| ≤ C32k(d+2ε)ρ(x, x′)ερ(y, y′)ε.

Remark 2.1 By Coifman’s construction in [DJS], one can construct an approximationto the identity of order θ having compact support satisfying the above Definition 2.3.

We now recall the continuous Calderon reproducing formulae on spaces of homoge-neous type in [HS, H1].

Lemma 2.1 Let X be a space of homogeneous type as in Definition 2.1, ε ∈ (0, θ], Skk∈Zbe an approximation to the identity of order ε and Dk = Sk − Sk−1 for k ∈ Z. Then there


are families of linear operators Dkk∈Z and Dkk∈Z such that for all f ∈ G(β, γ) withβ, γ ∈ (0, ε),

(2.4) f =∞∑

k=−∞DkDk(f) =

∞∑

k=−∞DkDk(f),

where the series converge in the norm of both the space G(β′, γ′) with 0 < β′ < β and0 < γ′ < γ and the space Lp(X) with p ∈ (1,∞). Moreover, Dk(x, y), the kernel of Dk

for all k ∈ Z satisfies the conditions (i) and (ii) of Definition 2.3 with ε replaced by anyε′ ∈ (0, ε), and

(2.5)∫

XDk(x, y) dµ(y) = 0 =

∫

XDk(x, y) dµ(x);

Dk(x, y), the kernel of Dk satisfies the conditions (i) and (iii) of Definition 2.3 with ε

replaced by any ε′ ∈ (0, ε) and (2.5).

By an argument of duality, Han and Sawyer in [HS, H1] also establish the following

continuous Calderon reproducing formulae on spaces of distributions,(G(β, γ)

)′with

β, γ ∈ (0, ε).

Lemma 2.2 With all the notation as in Lemma 2.1, then for all f ∈(G(β, γ)

)′with

β, γ ∈ (0, ε), (2.4) holds in(G(β′, γ′)

)′with β < β′ < ε and γ < γ′ < ε.

Let now (Xi, ρi, µi)di,θifor i = 1, 2 be two spaces of homogeneous type as in Definition

2.1 and ρi satisfies (2.1) with A replaced by Ai for i = 1, 2. We now introduce the spaceof test functions on the product space X1 ×X2 of spaces of homogeneous type.

Definition 2.4 For i = 1, 2, fix γi > 0 and βi > 0. A function f defined on X1 ×X2 issaid to be a test function of type (β1, β2, γ1, γ2) centered at (x0, y0) ∈ X1 ×X2 with widthr1, r2 > 0 if f satisfies the following conditions:

(i) |f(x, y)| ≤ Crγ11

(r1 + ρ1(x, x0))d1+γ1

rγ22

(r2 + ρ2(y, y0))d2+γ2;

(ii) |f(x, y)−f(x′, y)| ≤ C

(ρ1(x, x′)

r1 + ρ1(x, x0)

)β1 rγ11

(r1 + ρ1(x, x0))d1+γ1

rγ22

(r2 + ρ2(y, y0))d2+γ2

for ρ1(x, x′) ≤ 12A1

[r1 + ρ1(x, x0)];

(iii) |f(x, y)− f(x, y′)| ≤ Crγ11

(r1 + ρ1(x, x0))d1+γ1

(ρ2(y, y′)

r2 + ρ2(y, y0)

)β2 rγ22

(r2 + ρ2(y, y0))d2+γ2

for ρ2(y, y′) ≤ 12A2

[r2 + ρ2(y, y0)];


(iv)

|[f(x, y)− f(x′, y)]− [f(x, y′)− f(x′, y′)]|

≤ C

(ρ1(x, x′)

r1 + ρ1(x, x0)

)β1 rγ11

(r1 + ρ1(x, x0))d1+γ1

×(

ρ2(y, y′)r2 + ρ2(y, y0)

)β2 rγ22

(r2 + ρ2(y, y0))d2+γ2

for ρ1(x, x′) ≤ 12A1

[r1 + ρ1(x, x0)] and ρ2(y, y′) ≤ 12A2

[r2 + ρ2(y, y0)];

(v)∫X1

f(x, y) dµ1(x) = 0 for all y ∈ X2;

(vi)∫X2

f(x, y) dµ2(y) = 0 for all x ∈ X1.

If f is a test function of type (β1, β2, γ1, γ2) centered at (x0, y0) ∈ X1 × X2 with widthr1, r2 > 0, we write f ∈ G(x0, y0; r1, r2; β1, β2; γ1, γ2) and we define the norm of f by

‖f‖G(x0,y0;r1,r2;β1,β2;γ1,γ2) = infC : (i), (ii), (iii) and (iv) hold.

Remark 2.2 In the sequel, if β1 = β2 = β and γ1 = γ2 = γ, we will then simply write

f ∈ G(x0, y0; r1, r2; β; γ)

and similar for any other parameter.

We now denote by G(β1, β2; γ1, γ2) the class of G(x0, y0; r1, r2; β1, β2; γ1, γ2) with r1 =r2 = 1 for fixed (x0, y0) ∈ X1 ×X2. It is easy to see that

G(x1, y1; r1, r2;β1, β2; γ1, γ2) = G(β1, β2; γ1, γ2)

with an equivalent norm for all (x1, y1) ∈ X1 × X2. We can easily check that the spaceG(β1, β2; γ1, γ2) is a Banach space. Also, we denote by (G(β1, β2; γ1, γ2))′ its dual spacewhich is the set of all linear functionals L from G(β1, β2; γ1, γ2) to C with the propertythat there exists C ≥ 0 such that for all f ∈ G(β1, β2; γ1, γ2),

|L(f)| ≤ C‖f‖G(β1,β2;γ1,γ2).

We denote by 〈h, f〉 the natural pairing of elements h ∈ (G(β1, β2; γ1, γ2))′ and f ∈G(β1, β2; γ1, γ2). Clearly, for all h ∈ (G(β1, β2; γ1, γ2))′, 〈h, f〉 is well defined for allf ∈ G(x0, y0; r1, r2;β1, β2; γ1, γ2) with (x0, y0) ∈ X1 × X2, r1 > 0 and r2 > 0. By thesame reason as the case of non product spaces, we denote by G(β1, β2; γ1, γ2) the comple-tion of the space G(ε1, ε2) in G(β1, β2; γ1, γ2) when 0 < β1, γ1 < ε1 and 0 < β2, γ2 < ε2.


Lemma 2.3 Let (x1, x2) ∈ X1 ×X2, ri > 0, εi ∈ (0, θi] and 0 < βi, γi < εi for i = 1, 2.If the linear operators T1 and T2 are respectively bounded on the spaces G(x1, r1, β1, γ1)and G(x2, r2, β2, γ2) with operator norms C4,1 and C4,2, then the operator T1T2 is boundedon G(x1, x2; r1, r2; β1, β2; γ1, γ2) with an operator norm C4,1C4,2.

Proof. Let f ∈ G(x1, x2; r1, r2; β1, β2; γ1, γ2). By Definition 2.4, we see that for any fixedx ∈ X1, f(x, ·) ∈ G(x2, r2, β2, γ2) and

(2.6) ‖f(x, ·)‖G(x2,r2,β2,γ2) ≤ ‖f‖G(x1,x2;r1,r2;β1,β2;γ1,γ2)rγ11

(r1 + ρ1(x, x1))d1+γ1;

and for any x, x′ ∈ X1 with ρ(x, x′) ≤ 12A1

[r1+ρ1(x, x1)], f(x, ·)−f(x′, ·) ∈ G(x2, r2, β2, γ2)and

(2.7) ‖f(x, ·)− f(x′, ·)‖G(x2,r2,β2,γ2)≤ ‖f‖G(x1,x2;r1,r2;β1,β2;γ1,γ2)

(ρ1(x, x′)

r1 + ρ1(x, x1)

)β1

× rγ11

(r1 + ρ1(x, x1))d1+γ1.

The assumption on T2, (2.6) and (2.7) yield that for any x ∈ X1, T2f(x, ·) ∈ G(x2, r2, β2, γ2)and

(2.8) ‖T2f(x, ·)‖G(x2,r2,β2,γ2) ≤ C4,2‖f‖G(x1,x2;r1,r2;β1,β2;γ1,γ2)rγ11

(r1 + ρ1(x, x1))d1+γ1;

and for any x, x′ ∈ X1 with ρ(x, x′) ≤ 12A1

[r1 + ρ1(x, x1)], T2f(x, ·) − T2f(x′, ·) ∈G(x2, r2, β2, γ2) and

(2.9) ‖T2f(x, ·)− T2f(x′, ·)‖G(x2,r2,β2,γ2)≤ C4,2‖f‖G(x1,x2;r1,r2;β1,β2;γ1,γ2)

×(

ρ1(x, x′)r1 + ρ1(x, x1)

)β1 rγ11

(r1 + ρ1(x, x1))d1+γ1.

The estimates (2.8) and (2.9) imply that for any y ∈ X2, T2f(·, y) ∈ G(x1, r1, β1, γ1) and

(2.10) ‖T2f(·, y)‖G(x1,r1,β1,γ1) ≤ C4,2‖f‖G(x1,x2;r1,r2;β1,β2;γ1,γ2)rγ22

(r2 + ρ2(y, x2))d2+γ2;

and for any y, y′ ∈ X2 with ρ(y, y′) ≤ 12A2

[r2 + ρ2(y, x2)], T2f(·, y) − T2f(·, y′) ∈G(x1, r1, β1, γ1) and

(2.11) ‖T2f(·, y)− T2f(·, y′)‖G(x1,r1,β1,γ1)≤ C4,2‖f‖G(x1,x2;r1,r2;β1,β2;γ1,γ2)

×(

ρ2(y, y′)r2 + ρ2(y, x2)

)β2 rγ22

(r2 + ρ2(y, x2))d2+γ2.


From the assumption on T1, (2.10) and (2.11), it follows that for any y ∈ X2, T1T2f(·, y) ∈G(x1, r1, β1, γ1) and

(2.12) ‖T1T2f(·, y)‖G(x1,r1,β1,γ1)

≤ C4,1C4,2‖f‖G(x1,x2;r1,r2;β1,β2;γ1,γ2)rγ22

(r2 + ρ2(y, x2))d2+γ2;

and for any y, y′ ∈ X2 with ρ(y, y′) ≤ 12A2

[r2 + ρ2(y, x2)], T1T2f(·, y) − T1T2f(·, y′) ∈G(x1, r1, β1, γ1) and

(2.13) ‖T1T2f(·, y)− T1T2f(·, y′)‖G(x1,r1,β1,γ1)

≤ C4,1C4,2‖f‖G(x1,x2;r1,r2;β1,β2;γ1,γ2)

×(

ρ2(y, y′)r2 + ρ2(y, x2)

)β2 rγ22

(r2 + ρ2(y, x2))d2+γ2.

The estimates (2.12) and (2.13) actually tell us that T1T2f ∈ G(x1, x2; r1, r2; β1, β2; γ1, γ2)and

‖T1T2f‖G(x1,x2;r1,r2;β1,β2;γ1,γ2) ≤ C4,1C4,2‖f‖G(x1,x2;r1,r2;β1,β2;γ1,γ2),

which completes the proof of Lemma 2.3.

To establish special continuous Calderon reproducing formulae on the product spacesX1 × X2, we first need to recall some details of the proof of the same formulae for thenon-product space case in [H1], namely Lemma 2.1. One of the keys for establishing theseformulae is Coifman’s idea in [DJS]. Let X be a space of homogeneous type as in Definition2.1, Skk∈Z be an approximation to the identity of order ε ∈ (0, θ] on X as in Definition2.3 and Dk = Sk − Sk−1 for k ∈ Z. Then, it is easy to see that

(2.14) I =∞∑

k=−∞Dk in L2(X)

Let N ∈ N. Coifman’s idea is to rewrite (2.14) into

(2.15) I =

( ∞∑

k=−∞Dk

)

∞∑

j=−∞Dj

=∑

|j|>N

∞∑

k=−∞Dk+jDk +

∞∑

k=−∞

∑

|j|≤N

Dk+jDk

= RN + TN ,


where

(2.16) RN =∑

|j|>N

∞∑

k=−∞Dk+jDk

and

(2.17) TN =∞∑

k=−∞DN

k Dk

withDN

k =∑

|j|≤N

Dk+j .

It was proved in [H1] that there are constants C4 > 0 and δ > 0 independent of N ∈ Nsuch that for all f ∈ G(x1, r, β, γ) with x1 ∈ X, r > 0 and 0 < β, γ < ε,

(2.18) ‖RNf‖G(x1,r,β,γ) ≤ C42−Nδ‖f‖G(x1,r,β,γ).

Thus, if we choose N ∈ N such that

(2.19) C42−Nδ < 1,

then TN in (2.17) is invertible in the space G(x1, r, β, γ), namely, T−1N exists in the space

G(x1, r, β, γ) and there is a constant C > 0 such that for all f ∈ G(x1, r, β, γ),

‖T−1N f‖G(x1,r,β,γ) ≤ C‖f‖G(x1,r,β,γ).

For such chosen N ∈ N, letting

(2.20) Dk = T−1N DN

k ,

we then obtain the first formula in (2.4) by (2.17). The proof of the second formula in(2.4) is similar.

Using this idea, we can obtain the following continuous Calderon reproducing formulaof separable variable type on product spaces of homogeneous-type spaces, which is alsothe main theorem of this section.

Theorem 2.1 Let i = 1, 2, εi ∈ (0, θi], Skiki∈Z be an approximation to the identity oforder εi on space of homogeneous type, Xi, and Dki = Ski − Ski−1 for all ki ∈ Z. Thenthere are families of linear operators Dkiki∈Z on Xi such that for all f ∈ G(β1, β2; γ1, γ2)with βi, γi ∈ (0, εi) for i = 1, 2,

(2.21) f =∞∑

k1=−∞

∞∑

k2=−∞Dk1Dk2Dk1Dk2(f),


where the series converge in the norm of both the space G(β′1, β′2; γ

′1, γ

′2) with β′i ∈ (0, βi)

and γ′i ∈ (0, γi) for i = 1, 2, and Lp(X1 × X2) with p ∈ (1,∞). Moreover, Dki(xi, yi),the kernel of Dki for xi, yi ∈ Xi and all ki ∈ Z satisfies the conditions (1) and (2) ofDefinition 2.3 with εi replaced by any ε′i ∈ (0, εi), and

∫

Xi

Dki(xi, yi) dµi(yi) = 0 =∫

Xi

Dki(xi, yi) dµi(xi),

where i = 1, 2.

Proof. We prove (2.21) by taking advantage of its nature of separation of variables. Fori = 1, 2, let Ii be the identity operator on L2(Xi). We rewrite (2.15) into

(2.22) Ii = RNi + TNi ,

where RNi and TNi are defined by (2.16) and (2.17) instead of Dk and N there respectivelyby Dki

and Ni ∈ N. By (2.18), we know that there are constants C5,i > 0 and δi > 0independent of Ni ∈ N such that for all f ∈ G(xi, ri, βi, γi) with xi ∈ Xi, ri > 0, β′i ∈ (0, βi]and γ′i ∈ (0, γi],

(2.23) ‖RNif‖G(xi,ri,β′i,γ′i)≤ C5,i2−Niδi‖f‖G(xi,ri,β′i,γ

′i).

We now choose Ni ∈ N large enough such that

(2.24) C5,i2−Niδi < 1,

and thus TNi is invertible in the space G(xi, ri, β′i, γ

′i), namely, T−1

Niexists in G(xi, ri, β

′i, γ

′i)

and there is a constant C > 0 such that for all f ∈ G(xi, ri, β′i, γ

′i),

(2.25) ‖T−1Ni

f‖G(xi,ri,β′i,γ′i)≤ C‖f‖G(xi,ri,β′i,γ

′i).

Similarly to (2.20), we now define

(2.26) Dki = T−1Ni

DNiki

for ki ∈ Z. With these Dki as in (2.26), we now verify (2.21) holds in G(β′1, β′2; γ

′1, γ

′2). Let

f ∈ G(β1, β2; γ1, γ2) and Li ∈ N. We wish to show that

(2.27) limL1, L2→∞

∥∥∥∥∥∥f −

∑

|k1|<L1

∑

|k2|<L2

Dk1Dk2Dk1Dk2(f)

∥∥∥∥∥∥G(β′1,β′2;γ′1,γ′2)

= 0.


By (2.26) and Lemma 2.3, we can write

f −∑

|k1|<L1

∑

|k2|<L2

Dk1Dk2Dk1Dk2(f)

= f −∑

|k1|<L1

∑

|k2|<L2

T−1N1

DN1k1

T−1N2

DN2k2

Dk1Dk2(f)

= f − T−1N1

T−1N2

∑

|k1|<L1

∑

|k2|<L2

DN1k1

DN2k2

Dk1Dk2(f)

= f − T−1N1

T−1N2

TN1 −

∑

|k1|≥L1

DN1k1

Dk1

TN2 −

∑

|k2|≥L2

DN2k2

Dk2

(f)

=[f − T−1

N1T−1

N2TN1TN2(f)

]+ T−1

N1T−1

N2TN1

∑

|k2|≥L2

DN2k2

Dk2(f)

+T−1N1

T−1N2

∑

|k1|≥L1

DN1k1

Dk1TN2,2(f)

−T−1N1

T−1N2

∑

|k1|≥L1

DN1k1

Dk1

∑

|k2|≥L2

DN2k2

Dk2(f)

= F1 + F2 + F3 − F4.

We will estimate each term separately. We first estimate F1. By (2.22), we can write

F1 = f − T−1N1

T−1N2

TN1TN2(f)

= f − T−1N1

(I1 −RN1)T−1N2

(I2 −RN2)(f)

= f − lim

m1→∞

m1−1∑

l1=0

Rl1N1

(I1 −RN1)

lim

m2→∞

m2−1∑

l2=0

Rl2N2

(I2 −RN2)

(f)

= f −(

I1 − limm1→∞

Rm1N1

)(I2 − lim

m2→∞Rm2

N2

)(f)

= limm1→∞

Rm1N1

I2(f) + limm2→∞

I1Rm2N2

(f)− limm1, m2→∞

Rm1N1

Rm2N2

(f).

The estimate (2.23) and Lemma 2.3 tell us that∥∥∥Rm1

N1I2(f)

∥∥∥G(β′1,β′2;γ′1,γ′2)

≤(C5,12−N1δ1

)m1 ‖f‖G(β′1,β′2;γ′1,γ′2),

∥∥∥I1Rm2N2

(f)∥∥∥G(β′1,β′2;γ′1,γ′2)

≤(C5,22−N2δ2

)m2 ‖f‖G(β′1,β′2;γ′1,γ′2)


and∥∥∥Rm1

N1Rm2

N2(f)

∥∥∥G(β′1,β′2;γ′1,γ′2)

≤(C5,12−N1δ1

)m1(C5,22−N2δ2

)m2 ‖f‖G(β′1,β′2;γ′1,γ′2).

Thus, the assumption (2.24) leads us that

(2.28) ‖F1‖G(β′1,β′2;γ′1,γ′2)≤ limm1→∞

∥∥∥Rm1N1

I2(f)∥∥∥G(β′1,β′2;γ′1,γ′2)

+ limm2→∞

∥∥∥I1Rm2N2

(f)∥∥∥G(β′1,β′2;γ

′1,γ′2)

+ limm1, m2→∞

∥∥∥Rm1N1

Rm2N2

(f)∥∥∥G(β′1,β′2;γ′1,γ′2)

= 0.

We now assume β′i ∈ (0, βi) and γ′i ∈ (0, γi) for i = 1, 2 to be the same as in thetheorem. To estimate F2, F3 and F4, we first recall that there exist constants σi > 0 andC6,i > 0 independent of fi and Li such that for all fi ∈ G(βi, γi),

(2.29)

∥∥∥∥∥∥∑

|ki|≥Li

DNiki

Dki(f)

∥∥∥∥∥∥G(β′i,γ

′i)

≤ C6,i2−Liσi‖f‖G(βi,γi),

where i = 1, 2; see [H1, pp. 72-76] for a proof of this fact.We now estimate F2. By (2.22), we can write

F2 = T−1N1

(I1 −RN1)T−1N2

∑

|k2|≥L2

DN2k2

Dk2(f)

= limm1→∞

m1−1∑

j=0

RjN1

(I1 −RN1)T−1N2

∑

|k2|≥L2

DN2k2

Dk2(f)

=(

I1 − limm1→∞

Rm1N1

)T−1

N2

∑

|k2|≥L2

DN2k2

Dk2(f).

The estimates (2.29), (2.23) and (2.25), and Lemma 2.3 yield

‖F2‖G(β′1,β′2;γ′1,γ′2) ≤ CC5,22−L2σ2

1 + lim

m1→∞

(C5,12−N1δ1

)m1‖f‖G(β′1,β′2;γ′1,γ′2).

Then the assumption (2.24) further implies that

(2.30) limL2→∞

‖F2‖G(β′1,β′2;γ′1,γ′2) = 0.

The estimate for F3 is similar to that for F2 by symmetry.


Finally, the estimates (2.29) and (2.25), and Lemma 2.3 lead us that

(2.31) limL1, L2→∞

‖F4‖G(β′1,β′2;γ′1,γ′2)

≤ limL1, L2→∞

C2C6,1C6,22−L1σ12−L2σ2‖f‖G(β′1,β′2;γ′1,γ′2)

= 0.

The estimates (2.28), (2.30) and (2.31) yield (2.27) and we have verified that (2.21)holds in G(β′1, β

′2; γ

′1, γ

′2) with β′i ∈ (0, βi) and γ′i ∈ (0, γi) for i = 1, 2.

We now verify (2.21) also holds in Lp(X1 ×X2) for p ∈ (1,∞). Instead of (2.27), weneed to show that

(2.32) limL1, L2→∞

∥∥∥∥∥∥f −

∑

|k1|<L1

∑

|k2|<L2

Dk1Dk2Dk1Dk2(f)

∥∥∥∥∥∥Lp(X1×X2)

= 0.

To see (2.32) is true, we only need to note that the following facts are true:

(i) If Ti is bounded in Lp(Xi) for p ∈ (1,∞) with an operator norm C7,i for i = 1, 2,then T1T2 is also bounded in Lp(X1 ×X2) with an operator norm C7,1C7,2.

(ii) Let i = 1, 2. The operator RNi in (2.22) is bounded in Lp(Xi) with an operatornorm C8,i2−Niδi , where δi is the same as in (2.23). This fact was proved in [H1, p.76]. Therefore, if we choose Ni ∈ N such that

C8,i2−Niδi < 1,

then T−1Ni

exists and is also bounded in Lp(Xi) for p ∈ (1,∞) with an operator norm∑∞j=0

(C8,i2−Niδi

)j .

(iii)

limLi→∞

∥∥∥∥∥∥∑

|ki|≥Li

DNiki

Dki(f)

∥∥∥∥∥∥Lp(Xi)

= 0,

which was proved in [H1, p. 77] by a result in [DJS].

Using these facts and repeating the procedure of the proof of (2.27), we can prove(2.32) holds.

This completes the proof of Theorem 2.1.

By a procedure similar to the proof of Theorem 2.1, we can establish another con-tinuous Calderon reproducing formulae. We leave the details to the reader.


Theorem 2.2 Let i = 1, 2 and Dkiki∈Z be the same as in Theorem 2.1. Then thereare families of linear operators Dkiki∈Z on Xi such that for all f ∈ G(β1, β2; γ1, γ2) withβi, γi ∈ (0, εi) for i = 1, 2,

f =∞∑

k1=−∞

∞∑

k2=−∞Dk1Dk2Dk1Dk2(f),


′1, γ

′2) with β′i ∈ (0, βi)

and γ′i ∈ (0, γi) for i = 1, 2, and Lp(X1 × X2) with p ∈ (1,∞). Moreover, Dki(xi, yi),the kernel of Dki for xi, yi ∈ Xi and all ki ∈ Z satisfies the conditions (1) and (3) ofDefinition 2.3 with εi replaced by any ε′i ∈ (0, εi), and

∫

Xi

Dki(xi, yi) dµi(yi) = 0 =

∫

Xi

Dki(xi, yi) dµi(xi),

where i = 1, 2.

To establish the following continuous Calderon reproducing formulae in spaces ofdistributions, we need to use the theory of Calderon-Zygmund operators on these spacesdeveloped in [H1]. We first recall some definitions.

Let X be a space of homogeneous type as in Definition 2.1. For η ∈ (0, θ], we defineCη

0 (X) to be the set of all functions having compact support such that

‖f‖Cη0 (X) = sup

x6=y

|f(x)− f(y)|ρ(x, y)η

< ∞.

Endow Cη0 (X) with the natural topology and let (Cη

0 (X))′ be its dual space.

Definition 2.5 Let ε ∈ (0, θ] and X be a space of homogeneous type as in Definition 2.1.A continuous complex-valued function K(x, y) on

Ω = (x, y) ∈ X ×X : x 6= yis called a Calderon-Zygmund kernel of type ε if there exist a constant C9,1 > 0 such that

(i) |K(x, y)| ≤ C9,1ρ(x, y)−d,

(ii) |K(x, y)−K(x′, y)| ≤ C9,1ρ(x, x′)ερ(x, y)−d−ε for ρ(x, x′) ≤ ρ(x, y)2A

,

(iii) |K(x, y)−K(x, y′)| ≤ C9,1ρ(y, y′)ερ(x, y)−d−ε for ρ(y, y′) ≤ ρ(x, y)2A

.

A continuous linear operator T : Cη0 (X) → (Cη

0 (X))′ for all η ∈ (0, θ] is a Calderon-Zygmund singular integral operator of type ε if there is a Calderon-Zygmund kernel K(x, y)of the type ε as above such that

〈Tf, g〉 =∫

X

∫

XK(x, y)f(y)g(x) dµ(x) dµ(y)

for all f, g ∈ Cη0 (X) with disjoint supports. In this case, we write T ∈ CZO(ε).


We also need the following notion of the strong weak boundedness property in [HS].

Definition 2.6 Let X be a space of homogeneous type as in Definition 2.1. A Calderon-Zygmund singular integral operator T of the kernel K is said to have the strong weakboundedness property, if there exist η ∈ (0, θ] and constant C9,2 > 0 such that

|〈K, f〉| ≤ C9,2rd

for all r > 0 and all continuous f on X×X with supp f ⊆ B(x1, r)×B(y1, r), where x1 andy1 ∈ X, ‖f‖L∞(X×X) ≤ 1, ‖f(·, y)‖Cη

0 (X) ≤ r−η for all y ∈ X and ‖f(x, ·)‖Cη0 (X) ≤ r−η

for all x ∈ X. We denote this by T ∈ SWBP.

The following theorem is the variant on space of homogeneous type of Theorem 1.19in [H1].

Lemma 2.4 Let ε ∈ (0, θ] and X be a space of homogeneous type as in Definition 2.1.Let T ∈ CZO(ε), T (1) = T ∗(1) = 0, and T ∈ SWBP . Furthermore, K(x, y), the kernelof T , satisfies the following smoothness condition

(2.33) |[K(x, y)−K(x′, y)]− [K(x, y′)−K(x′, y′)]|

≤ C9,3ρ(x, x′)ερ(y, y′)ερ(x, y)−d−2ε

for all x, x′, y, y′ ∈ X such that ρ(x, x′), ρ(y, y′) ≤ ρ(x,y)3A2 . Then for any x0 ∈ X,

r > 0 and 0 < β, γ < ε, T maps G(x0, r, β, γ) into itself. Moreover, if we let ‖T‖ =maxC9,1, C9,2, C9,3, then there exists a constant C9,4 > 0 such that

‖Tf‖G(x0,r,β,γ) ≤ C9,4‖T‖‖f‖G(x0,r,β,γ).

We also need the following construction given by Christ in [Chr2], which provides ananalogue of the grid of Euclidean dyadic cubes on spaces of homogeneous type. A similarconstruction was independently given by Sawyer and Wheeden [SW].

Lemma 2.5 Let X be a space of homogeneous type as in Definition 2.1. Then there exista collection

Qkα ⊂ X : k ∈ Z, α ∈ Ik

of open subsets, where Ik is some index set, and constants δ ∈ (0, 1) and C10,1, C10,2 > 0such that

(i) µ(X \ ∪αQkα) = 0 for each fixed k and Qk

α ∩Qkβ = ∅ if α 6= β;

(ii) for any α, β, k, l with l ≥ k, either Qlβ ⊂ Qk

α or Qlβ ∩Qk

α = ∅;


(iii) for each (k, α) and each l < k there is a unique β such that Qkα ⊂ Ql

β;

(iv) diam (Qkα) ≤ C10,1δ

k;

(v) each Qkα contains some ball B(zk

α, C10,2δk), where zk

α ∈ X.

In fact, we can think of Qkα as being a dyadic cube with diameter roughly δk and

centered at zkα. In what follows, we always suppose δ = 1/2. See [HS] for how to remove

this restriction. Also, in the following, for k ∈ Z+ and τ ∈ Ik, we will denote by Qk,ντ ,

ν = 1, 2, · · · , N(k, τ), the set of all cubes Qk+jτ ′ ⊂ Qk

τ , where j is a fixed large positiveinteger. Denote by yk,ν

τ a point in Qk,ντ . For any dyadic cube Q and any f ∈ L1

loc (X), weset

mQ(f) =1

µ(Q)

∫

Qf(x) dµ(x).

Using Theorem 2.1, we now try to establish the following continuous Calderon re-producing formulae in spaces of distributions.

Theorem 2.3 Let all the notation be the same as in Theorem 2.1. Then for all f ∈(G(β1, β2; γ1, γ2)

)′,

f =∞∑

k1=−∞

∞∑

k2=−∞D∗

k1D∗

k2D∗

k1D∗

k2(f)

holds in(G(β′1, β

′2; γ

′1, γ

′2)

)′with β′i ∈ (βi, εi) and γ′i ∈ (γi, εi) for i = 1, 2, where

D∗ki

(x, y) = Dki(y, x) and D∗

ki(x, y) = Dki

(y, x).

Proof. Let f ∈(G(β1, β2; γ1, γ2)

)′and g ∈ G(β′1, β

′2; γ

′1, γ

′2) with the same notation as in

the theorem. By Theorem 2.1, we have

g =∞∑

k1=−∞

∞∑

k2=−∞Dk1Dk2Dk1Dk2(g)

holds in G(β1, β2; γ1, γ2). From this, it follows that

〈f, g〉=⟨

f,

∞∑

k1=−∞

∞∑

k2=−∞Dk1Dk2Dk1Dk2(g)

⟩

=∞∑

k1=−∞

∞∑

k2=−∞

⟨f, Dk1Dk2Dk1Dk2(g)

⟩.

To prove the theorem, we still need to show that

(2.34)⟨f, Dk1Dk2Dk1Dk2(g)

⟩=

⟨D∗

k1D∗

k2D∗

k1D∗

k2(f), g

⟩.


Let M1, M2 ∈ N be large enough, B1(x0,M1) = x1 ∈ X1 : ρ1(x1, x0) < M1 andB2(y0,M2) = x2 ∈ X2 : ρ2(x2, y0) < M2. For any fixed k1, k2 ∈ Z, we then define

gM1,M2(x1, x2)

=∫

B1(x0,M1)

∫

B2(y0,M2)Dk1(x1, y1)Dk2(x2, y2)Dk1Dk2(g)(y1, y2) dµ1(y1) dµ2(y2).

We first claim that

(2.35) limM1→∞, M2→∞

∥∥∥Dk1Dk2Dk1Dk2(g)− gM1,M2

∥∥∥G(β1,β2;γ1,γ2)

= 0.

To verify this, we use Lemma 2.4. To this end, we write

Dk1Dk2Dk1Dk2(g)(x1, x2)− gM1,M2(x1, x2)

=∫

X1

∫

X2\B2(y0,M2)Dk1(x1, y1)Dk2(x2, y2)Dk1Dk2(g)(y1, y2) dµ1(y1) dµ2(y2)

+∫

X1\B1(x0,M1)

∫

B2(y0,M2)· · ·

= T1(g)(x1, x2) + T2(g)(x1, x2).

We first consider T1(g)(x1, x2), whose kernel is

K1(x1, x2; z1, z2)

=∫

X1

∫

X2\B2(y0,M2)Dk1(x1, y1)Dk2(x2, y2)Dk1(y1, z1)Dk2(y2, z2) dµ1(y1) dµ2(y2)

=∫

X1

Dk1(x1, y1)Dk1(y1, z1) dµ(y1)

×∫

X2\B2(y0,M2)Dk2(x2, y2)Dk2(y2, z2) dµ2(y2)

= K11(x1, z1)K12(x2, z2).

It is easy to verify that the operator T11 with the kernel K11 satisfies all the conditionof Lemma 2.4. Thus, by Lemma 2.4, we know that there is a constant C > 0 independentof M1 and M2 such that for all f ∈ G(β1, γ1),

(2.36) ‖T11(f)‖G(β1,γ1) ≤ C‖f‖G(β1,γ1).

We now verify that K12 is a Calderon-Zygmund kernel of type ε′2 > 0 on X2 asin Definition 2.5 with a constant C9,1 independent of M2, where ε′2 can be any positive


number in (0, ε2). We first have

(2.37) |K12(x2, z2)|=∣∣∣∣∣∫

X2\B2(y0,M2)Dk2(x2, y2)Dk2(y2, z2) dµ2(y2)

∣∣∣∣∣

≤ Ck2

∫

X2

∣∣∣Dk2(x2, y2)Dk2(y2, z2)∣∣∣ dµ2(y2)

≤ Ck2

1(1 + ρ(x2, z2))d2+ε′2

≤ Ck2

1ρ(x2, z2)d2

,

where Ck2 is independent of M2.To verify K12(x2, z2) satisfies Definition 2.5 (ii), assuming that ρ2(x2, x

′2) ≤ ρ2(x2,z2)

2A2,

we have

(2.38) |K12(x2, z2)−K12(x′2, z2)|

=

∣∣∣∣∣∫

X2\B2(y0,M2)

[Dk2(x2, y2)− Dk2(x

′2, y2)

]Dk2(y2, z2) dµ2(y2)

∣∣∣∣∣

≤∫

ρ(x2,x′2)≤ 12A

(1+ρ(x2,y2))

∣∣∣Dk2(x2, y2)− Dk2(x′2, y2)

∣∣∣ |Dk2(y2, z2)| dµ2(y2)

+∫

ρ(x2,x′2)> 12A

(1+ρ(x2,y2))

[∣∣∣Dk2(x2, y2)∣∣∣ +

∣∣∣Dk2(x′2, y2)

∣∣∣]|Dk2(y2, z2)| dµ2(y2)

≤ Ck2

∫

X2

ρ(x2, x′2)

ε′2

(1 + ρ2(x2, y2))d2+2ε′2

1(1 + ρ2(y2, z2))d2+ε2

dµ2(y2)

+∫

ρ(x2,x′2)> 12A

(1+ρ(x2,y2))

[1

(1 + ρ2(x2, y2))d2+ε′2+

1(1 + ρ2(x′2, y2))d2+ε′2

]

× 1(1 + ρ2(y2, z2))d2+ε2

dµ2(y2)

≤ Ck2

ρ(x2, x′2)

ε′2

(1 + ρ2(x2, z2))d2+ε′2

≤ Ck2

ρ(x2, x′2)

ε′2

ρ2(x2, z2)d2+ε′2,

where Ck2 is independent of M2.By symmetry, similarly to the proof of (2.38), we also have that for ρ2(z2, z

′2) ≤

ρ2(x2,z2)2A2

,

(2.39)∣∣K12(x2, z2)−K12(x2, z

′2)

∣∣ ≤ Ck2

ρ(z2, z′2)

ε′2

ρ2(x2, z2)d2+ε′2,

where Ck2 is independent of M2.


We now verify that K12 satisfies (2.33). First assuming that ρ2(x2, x′2) ≤ ρ2(x2,z2)

4A22

and ρ2(z2, z′2) ≤ ρ2(x2,z2)

4A22

, we write

[K12(x2, z2)−K12(x′2, z2)]− [K12(x2, z′2)−K12(x′2, z

′2)]

=∫

X2\B2(y0,M2)

[Dk2(x2, y2)− Dk2(x

′2, y2)

] [Dk2(y2, z2)−Dk2(y2, z

′2)

]dµ2(y2).

By our assumption, we now have three cases. Case 1. ρ2(x2, x′2) ≤ 1

2A2(1+ρ2(x2, y2)) and

ρ2(z2, z′2) ≤ 1

2A2(1 + ρ2(y2, z2)). In this case, we have

(2.40) |[K12(x2, z2)−K12(x′2, z2)]− [K12(x2, z′2)−K12(x′2, z

′2)]|

≤∫

X2

∣∣∣Dk2(x2, y2)− Dk2(x′2, y2)

∣∣∣∣∣Dk2(y2, z2)−Dk2(y2, z

′2)

∣∣ dµ2(y2)

≤ Ck2

∫

X

ρ2(x2, x′2)

ε′2

(1 + ρ2(x2, y2))d2+2ε′2

ρ2(z2, z′2)

ε2

(1 + ρ2(y2, z2))d2+2ε2dµ2(y2)

≤ Ck2

ρ2(x2, x′2)

ε′2ρ2(z2, z′2)

ε′2

(1 + ρ2(x2, z2))d2+2ε′2

≤ Ck2

ρ2(x2, x′2)

ε′2ρ2(z2, z′2)

ε′2

ρ2(x2, z2)d2+2ε′2,

where Ck2 is independent of M2.Case 2. ρ2(x2, x

′2) ≤ 1

2A2(1 + ρ2(x2, y2)) and ρ2(z2, z

′2) > 1

2A2(1 + ρ2(y2, z2)). In this

case, we in fact have ρ2(y2, z2) < ρ2(x2,z2)2A2

, which implies that ρ2(x2, y2) ≥ ρ2(x2,z2)2A2

. Thelast fact and the fact that ρ2(z2, z

′2) > 1

2A2yield that

(2.41) |[K12(x2, z2)−K12(x′2, z2)]− [K12(x2, z′2)−K12(x′2, z

′2)]|

≤ Ck2

∫

X

ρ2(x2, x′2)

ε′2

(1 + ρ2(x2, y2))d2+2ε′2

[|Dk2(y2, z2)|+∣∣Dk2(y2, z

′2)

∣∣] dµ2(y2)

≤ Ck2

ρ2(x2, x′2)

ε′2

(1 + ρ2(x2, z2))d2+2ε′2

≤ Ck2

ρ2(x2, x′2)

ε′2ρ2(z2, z′2)

ε′2

ρ2(x2, z2)d2+2ε′2,

where Ck2 is independent of M2.Case 2. ρ2(x2, x

′2) > 1

2A2(1 + ρ2(x2, y2)) and ρ2(z2, z

′2) ≤ 1

2A2(1 + ρ2(y2, z2)). The

proof of this case is similar to the case 2 by the symmetry.If ρ2(x2,z2)

4A22

< ρ2(x2, x′2) ≤ ρ2(x2,z2)

3A22

or ρ2(x2,z2)4A2

2< ρ2(z2, z

′2) ≤ ρ2(x2,z2)

3A22

, we then candeduce that K12 satisfies (2.33) from (2.38) or (2.39), which together with (2.40) and(2.41) verifies that K12 satisfies (2.33). We omit the details.


Finally we verify that K12 has the strong weak boundedness property as in Definition2.6. Let r > 0 and f be a continuous function on X2 × X2 with supp f ⊂ B2(x21, r) ×B2(x22, r), where x21, x22 ∈ X2, ‖f‖L∞(X2×X2) ≤ 1, ‖f(·, z2)‖Cη

0 (X2) ≤ r−η for all z2 ∈ X2

and ‖f(x2, ·)‖Cη0 (X2) ≤ r−η for all x2 ∈ X2. From (2.37), it follows that

|〈K12, f〉|

=∣∣∣∣∫

X2

∫

X2

K12(x2, z2)f(x2, z2) dµ2(x2) dµ2(z2)∣∣∣∣

≤ Ck2‖f‖L∞(X2×X2)

∫

B2(x22,r)

∫

X2

1(1 + ρ(x2, z2))d2+ε′2

dµ2(x2)

dµ2(z2)

≤ Ck2rd2 ,

where Ck2 is independent of M2.Let T12 be the Calderon-Zygmund operator with the kernel K12. It is also obvious

that T12(1) = 0. Thus, T12 satisfies all the conditions of Lemma 2.4 with ‖T12‖ = Ck2

independent of M2. By Lemma 2.4, we know that there is a constant C > 0 independentof M2 such that for all f ∈ G(β2, γ2),

(2.42) ‖T12(f)‖G(β2,γ2) ≤ C‖f‖G(β2,γ2).

The estimates (2.36) and (2.42), and Lemma 2.3 tell us that T1 is bounded onG(β′1, β

′2; γ

′1, γ

′2) with an operator norm independent of M1 and M2. Similarly, we can

show that T2 has the same property. Let

g(x1, x2) = Dk1Dk2Dk1Dk2(g)(x1, x2)− gM1,M2(x1, x2).

Then g ∈ G(β′1, β′2; γ

′1, γ

′2) with a norm independent of M1 and M2. Namely, there is a

constant C > 0 independent of M1 and M2 such that

(2.43) |g(x1, x2)− g(x′1, x2)|

≤ C

(ρ1(x1, x

′1)

1 + ρ1(x1, x0)

)β′1 1(1 + ρ1(x1, x0))d1+γ′1

1(1 + ρ2(x2, y0))d2+γ′2

for ρ1(x1, x′1) ≤ 1

2A1[1 + ρ1(x1, x0)];

(2.44) |g(x1, x2)− g(x1, x′2)|

≤ C1

(1 + ρ1(x1, x0))d1+γ′1

(ρ2(x2, x

′2)

1 + ρ2(x2, y0)

)β′2 1(1 + ρ2(x2, y0))d2+γ′2


for ρ2(x2, x′2) ≤ 1

2A2[1 + ρ2(x2, y0)];

(2.45) |[g(x1, x2)− g(x′1, x2)]− [g(x1, x′2)− g(x′1, x

′2)]|

≤ C

(ρ1(x1, x

′1)

1 + ρ1(x1, x0)

)β′1 1(1 + ρ1(x1, x0))d1+γ′1

×(

ρ2(x2, x′2)

1 + ρ2(x2, y0)

)β′2 1(1 + ρ2(x2, y0))d2+γ′2

for ρ1(x1, x′1) ≤ 1

2A1[1 + ρ1(x1, x0)] and ρ2(x2, x

′2) ≤ 1

2A2[1 + ρ2(x2, y0)];

(2.46)∫

X1

g(x1, x2) dµ1(x1) = 0

for all x2 ∈ X2;

(2.47)∫

X2

g(x1, x2) dµ2(x2) = 0

for all x1 ∈ X1.Moreover, we can directly compute that

(2.48) |g(x1, x2)| ≤ Ck1,k2‖g‖G(β′1,β′2;γ′1,γ′2)

×∫

X1

∫

X2\B2(y0,M2)

∫

X1

∫

X2

1(1 + ρ1(x1, y1))ε′1

1(1 + ρ2(x2, y2))ε′2

× 1(1 + ρ1(y1, z1))ε1

1(1 + ρ2(y2, z2))ε′2

1(1 + ρ1(z1, x0))γ′1

× 1(1 + ρ1(z2, y0))γ′2

dµ1(z1) dµ2(z2) dµ1(y1) dµ2(y2)

+∫

X1\B1(x0,M1)

∫

B2(y0,M2)

∫

X1

∫

X2

· · ·

≤ Ck1,k2‖g‖G(β′1,β′2;γ′1,γ′2)

×∫

X1

∫

X2\B2(y0,M2)

1(1 + ρ1(x1, y1))ε′1

1(1 + ρ2(x2, y2))ε′2

× 1(1 + ρ1(y1, x0))γ′1

1(1 + ρ2(y2, y0))γ′2

dµ1(y1) dµ2(y2)

+∫

X1\B1(x0,M1)

∫

B2(y0,M2)· · ·

≤ Ck1,k2

1

Mγ′1−γ1

1

+1

Mγ′2−γ2

2

‖g‖G(β′1,β′2;γ′1,γ′2)

× 1(1 + ρ1(x1, x0))γ1

1(1 + ρ2(x2, y0))γ2

,


where Ck1,k2 is independent of M1 and M2.If ρ1(x1, x

′1) ≤ 1

2A1[1 + ρ1(x1, x0)], from (2.48), it follows that

(2.49) |g(x1, x2)− g(x′1, x2)|

≤ Ck1,k2

1

Mγ′1−γ1

1

+1

Mγ′2−γ2

2

‖g‖G(β′1,β′2;γ′1,γ′2)

× 1(1 + ρ1(x1, x0))d1+γ1

1(1 + ρ2(x2, y0))d2+γ2

.

Let α1 ∈ (0, 1). The geometric means between (2.43) and (2.49) then gives that

(2.50) |g(x1, x2)− g(x′1, x2)|

≤ Ck1,k2

(1

Mγ′1−γ1

1

+1

Mγ′2−γ2

2

)1−α1

‖g‖1−α1

G(β′1,β′2;γ′1,γ′2)

×(

ρ1(x1, x′1)

1 + ρ1(x1, x0)

)α1β′1 1(1 + ρ1(x1, x0))d1+γ1

1(1 + ρ2(x2, y0))d2+γ2

.

Let α2 ∈ (0, 1). Similarly, from (2.48), (2.44) and the geometric means, we candeduce that if

ρ2(x2, x′2) ≤

12A2

[1 + ρ2(x2, y0)],

then

(2.51) |g(x1, x2)− g(x1, x′2)|

≤ Ck1,k2

(1

Mγ′1−γ2

1

+1

Mγ′2−γ2

2

)1−α2

‖g‖1−α2

G(β′1,β′2;γ′1,γ′2)

× 1(1 + ρ1(x1, x0))d1+γ1

(ρ1(x2, x

′2)

1 + ρ1(x2, y0)

)α2β′2 1(1 + ρ2(x2, y0))d2+γ2

.

Let α3, α4 ∈ (0, 1). The estimates (2.50), (2.51) and (2.45) and the geometric meansimply that if ρ1(x1, x

′1) ≤ 1

2A1[1 + ρ1(x1, x0)] and ρ2(x2, x

′2) ≤ 1

2A2[1 + ρ2(x2, y0)], then

(2.52) |[g(x1, x2)− g(x′1, x2)]− [g(x1, x′2)− g(x′1, x

′2)]|

≤ Ck1,k2

(1

Mγ′1−γ1

1

+1

Mγ′2−γ2

2

)(1−α1)(1−α2)α4+(1−α3)(1−α4)

×‖g‖(1−α1)(1−α2)α4+(1−α3)(1−α4)G(β′1,β′2;γ′1,γ′2)

×(

ρ1(x1, x′1)

1 + ρ1(x1, x0)

)[α1(1−α2)+α2]α4β′1 1(1 + ρ1(x1, x0))d1+γ1

×(

ρ2(x2, x′2)

1 + ρ2(x2, y0)

)[α2α4+α3(1−α4)]β′2 1(1 + ρ2(x2, y0))d2+γ2

.


From (2.48), (2.50), (2.51), (2.52), (2.46) and (2.47), it follows that g(x1, x2) ∈G(β1, β2; γ1, γ2) if we suitably chose αi for i = 1, 2, 3, 4, and

limM1, M2→∞

‖g‖G(β1,β2;γ1,γ2) = 0,

namely (2.35) holds, which yields that


⟩= lim

M1, M2→∞〈f, gM1,M2〉 .

For J1, J2 ∈ N and any fixed M1, M2 ∈ N large enough, we define

NJ1 =

i1 ∈ IJ1 : QJ1i1∩B1(x0,M1) 6= ∅

andNJ2 =

i2 ∈ IJ2 : QJ2

i2∩B2(y0,M2) 6= ∅

,

where QJ1i1J1∈Z, i1∈IJ1

and QJ2i2J2∈Z, i2∈IJ2

are respectively the dyadic cubes of X1 andX2 as in Lemma 2.5. Then the cardinal number of NJ1 ∼ Md1

1 2J1d1 and the cardinalnumber of NJ2 ∼ Md2

2 2J2d2 . Write

gM1,M2(x1, x2)

=∑

i1∈NJ1

∫

QJ1i1∩B1(x0,M1)

∫

B2(y0,M2)

[Dk1(x1, y1)− Dk1

(x1, xQ

J1i1

)]

×Dk2(x2, y2)Dk1Dk2(g)(y1, y2) dµ1(y1) dµ2(y2)

+∑

i2∈NJ2

Dk1

(x1, xQ

J1i1

)∫

B1(x0,M1)

∫

QJ2i2∩B2(y0,M2)

[Dk2(x2, y2)− Dk2

(x2, xQ

J2i2

)]

×Dk1Dk2(g)(y1, y2) dµ1(y1) dµ2(y2)

+∑

i1∈NJ1

∑

i2∈NJ2

Dk1

(x1, xQ

J1i1

)Dk2

(x2, xQ

J2i2

)

×∫

QJ1i1∩B1(x0,M1)

∫

QJ2i2∩B2(y0,M2)

Dk1Dk2(g)(y1, y2) dµ1(y1) dµ2(y2)

= g1M1,M2

(x1, x2) + g2M1,M2

(x1, x2) + g3M1,M2

(x1, x2),

where xQ

J1i1

and xQ

J2i2

are respectively any point in QJ1i1∩B1(x0,M1) and QJ2

i2∩B2(y0, M2).

Our task now is to verify that

(2.54) limJ1, J2→∞

‖giM1,M2

‖G(β1,β2;γ1,γ2) = 0,


where i = 1, 2. The proof of (2.54) for i = 2 is similar to that for i = 1. We only verify(2.54) for i = 1, which can be deduced from Lemma 2.4 and Lemma 2.3 by a proceduresimilar to the proof of (2.35). To this end, we regard g1

M1,M2(x1, x2) as an operator T3 acts

on the functions g. The kernel K3 of T3 can be written into

K3(x1, x2; z1, z2)

=∑

i1∈NJ1

∫

QJ1i1∩B1(x0,M1)

∫

B2(y0,M2)

[Dk1(x1, y1)− Dk1

(x1, xQ

J1i1

)]

×Dk2(x2, y2)Dk1(y1, z1)Dk2(y2, z2) dµ1(y1) dµ2(y2)

=

∑

i1∈NJ1

∫

QJ1i1∩B1(x0,M1)

[Dk1(x1, y1)− Dk1

(x1, xQ

J1i1

)]

×Dk1(y1, z1) dµ1(y1)

×∫

B2(y0,M2)Dk2(x2, y2)Dk2(y2, z2) dµ2(y2)

= K31(x1, z1)K32(x2, z2).

Let T31 and T32 be respectively the operator corresponding to the kernel K31 and thekernel K32. Then complete similarly to the proof of (2.42), we can find a constant C > 0independent of M1, M2, J1 and J2 such that for all f ∈ G(β2, γ2),

(2.55) ‖T32(f)‖G(β2,γ2) ≤ C‖f‖G(β2,γ2).

We now verify the operator T31 satisfies all the conditions of Lemma 2.4. In whatfollows, let Ck1 > 0 be a constant independent of M1, M2, J1 and J2 and let ε′1 > 0 be anypositive number in (0, ε1). Noting that y1, x

QJ1i1

∈ QJ1i1

and J1 is large enough, by Lemma

2.5, we first have

(2.56) |K31(x1, z1)|

=

∣∣∣∣∣∑

i1∈NJ1

∫

QJ1i1∩B1(x0,M1)

[Dk1(x1, y1)− Dk1

(x1, xQ

J1i1

)]

×Dk1(y1, z1) dµ1(y1)

∣∣∣∣∣

≤ Ck1

∫

B1(x0,M1)

2−J1ε′1

(1 + ρ1(x1, y1))d1+2ε′1|Dk1(y1, z1)| dµ1(y1)


≤ Ck1

2−J1ε′1

(1 + ρ1(x1, z1))d1+ε′1

≤ Ck12−J1ε′1

1ρ1(x1, z1)d1

.

Assuming that ρ1(x1, x′1) ≤ ρ1(x1,z1)

2A1, we estimate

(2.57) |K31(x1, z1)−K31(x′1, z1)|

=

∣∣∣∣∣∑

i1∈NJ1

∫

QJ1i1∩B1(x0,M1)

[Dk1(x1, y1)− Dk1(x

′1, y1)

]

−[Dk1

(x1, xQ

J1i1

)− Dk1

(x′1, xQ

J1i1

)]Dk1(y1, z1) dµ1(y1)

∣∣∣∣∣

≤∫

B1(x0,M1)

∣∣∣Dk1(x1, y1)− Dk1(x′1, y1)

∣∣∣ |Dk1(y1, z1)| dµ1(y1)

+∑

i1∈NJ1

∫

QJ1i1∩B1(x0,M1)

∣∣∣∣Dk1

(x1, xQ

J1i1

)− Dk1

(x′1, xQ

J1i1

)∣∣∣∣

× |Dk1(y1, z1)| dµ1(y1)

≤∫

ρ1(x1,x′1)≤ 12A1

(1+ρ1(x1,y1)

∣∣∣Dk1(x1, y1)− Dk1(x′1, y1)

∣∣∣ |Dk1(y1, z1)| dµ1(y1)

+∫

ρ1(x1,x′1)> 12A1

(1+ρ1(x1,y1)

[∣∣∣Dk1(x1, y1)∣∣∣ +

∣∣∣Dk1(x′1, y1)

∣∣∣]

× |Dk1(y1, z1)| dµ1(y1)

+∑

i1∈NJ1

∫Q

J1i1∩B1(x0,M1)

ρ1(x1,x′1)≤ 12A1

(1+ρ1(x1,xQ

J1i1

)

∣∣∣∣Dk1

(x1, xQ

J1i1

)− Dk1

(x′1, xQ

J1i1

)∣∣∣∣

× |Dk1(y1, z1)| dµ1(y1)

+∑

i1∈NJ1

∫Q

J1i1∩B1(x0,M1)

ρ1(x1,x′1)> 12A1

(1+ρ1(x1,xQ

J1i1

)

[∣∣∣∣Dk1

(x1, xQ

J1i1

)∣∣∣∣ +∣∣∣∣Dk1

(x′1, xQ

J1i1

)∣∣∣∣]

× |Dk1(y1, z1)| dµ1(y1)

≤ Ck1

∫

X1

ρ1(x1, x′1)

ε′1

(1 + ρ1(x1, y1))d1+2ε′1

1(1 + ρ1(y1, z1))d1+ε1

dµ1(y1)

+Ck1ρ1(x1, x′1)

ε′1

∫

X1

[1

(1 + ρ1(x1, y1))d1+ε′1+

1(1 + ρ1(x′1, y1))d1+ε′1

]


× 1(1 + ρ1(y1, z1))d1+ε1

dµ1(y1)

≤ Ck1

ρ1(x1, x′1)

ε′1

(1 + ρ1(x1, z1))d1+ε′1

≤ Ck1

ρ1(x1, x′1)

ε′1

ρ1(x1, z1)d1+ε′1,

where in third-to-last inequality, we used the following facts that for any y1 ∈ QJ1i1

, andall x1, x′1 ∈ X1,

(2.58) 1 + ρ1(x1, y1) ≤ A1

(1 + ρ1(x1, xQ

J1i1

))

and

(2.59) 1 + ρ1(x′1, y1) ≤ A1

(1 + ρ1(x′1, xQ

J1i1

))

by Lemma 2.5 and the large choice on J1 ∈ N.The estimates (2.56) and (2.57) and the geometric means then tell us that for any

α5 ∈ (0, 1),

(2.60)∣∣K31(x1, z1)−K31(x′1, z1)

∣∣ ≤ Ck12−J1ε′1(1−α5) ρ1(x1, x

′1)

ε′1α5

ρ1(x1, z1)d1+ε′1α5.

Assuming that ρ1(z1, z′1) ≤ ρ1(x1,z1)

2A1, we now estimate

(2.61) |K31(x1, z1)−K31(x1, z′1)|

=

∣∣∣∣∣∣∑

i1∈NJ1

∫

QJ1i1∩B1(x0,M1)

[Dk1(x1, y1)− Dk1

(x1, xQ

J1i1

)]

× [Dk1(y1, z1)−Dk1(y1, z′1)] dµ1(y1)|

≤ Ck1

∫

ρ1(z1,z′1)≤ 12A1

(1+ρ1(x1,z1)

2−J1ε′1

(1 + ρ1(x1, y1))d1+2ε′1

× |Dk1(y1, z1)−Dk1(y1, z′1)| dµ1(y1)

+Ck1

∫

ρ1(z1,z′1)> 12A1

(1+ρ1(x1,z1)

2−J1ε′1

(1 + ρ1(x1, y1))d1+2ε′1

× [|Dk1(y1, z1)|+ |Dk1(y1, z′1)|] dµ1(y1)

≤ Ck12−J1ε′1

ρ1(z1, z′1)

ε′1

(1 + ρ1(x1, z1))d1+ε′1

≤ Ck12−J1ε′1

ρ1(z1, z′1)

ε′1

ρ1(x1, z1)d1+ε′1.


To verify that K31 satisfies (2.33), similarly to the proof of K12, we may assume thatρ1(x1, x

′1) ≤ ρ1(x1,z1)

4A31

and ρ1(z1, z′1) ≤ ρ1(x1,z1)

4A31

. Under these assumptions, we then write

|[K31(x1, z1)−K31(x′1, z1)]− [K31(x1, z′1)−K31(x′1, z

′1)]|

=

∣∣∣∣∣∑

i1∈NJ1

∫

QJ1i1∩B1(x0,M1)

[Dk1(x1, y1)− Dk1(x

′1, y1)

]

−[Dk1

(x1, xQ

J1i1

)− Dk1

(x′1, xQ

J1i1

)]

× [Dk1(y1, z1)−Dk1(y1, z′1)] dµ1(y1)

∣∣∣∣∣

≤∫

B1(x0,M1)

∣∣∣Dk1(x1, y1)− Dk1(x′1, y1)

∣∣∣∣∣Dk1(y1, z1)−Dk1(y1, z

′1)

∣∣ dµ1(y1)

+∑

i1∈NJ1

∫

QJ1i1∩B1(x0,M1)

∣∣∣∣Dk1

(x1, xQ

J1i1

)− Dk1

(x′1, xQ

J1i1

)∣∣∣∣

× |Dk1(y1, z1)−Dk1(y1, z′1)| dµ1(y1)

= O1 + O2.

For O1, we only have the following three cases:

(i) ρ1(x1, x′1) ≤ 1

2A1(1 + ρ1(x1, y1)) and ρ1(z1, z

′1) ≤ 1

2A1(1 + ρ1(y1, z

′1));

(ii) ρ1(x1, x′1) > 1

2A1(1 + ρ1(x1, y1)) and ρ1(z1, z

′1) ≤ 1

2A1(1 + ρ1(y1, z

′1));

(iii) ρ1(x1, x′1) ≤ 1

2A1(1 + ρ1(x1, y1)) and ρ1(z1, z

′1) > 1

2A1(1 + ρ1(y1, z

′1)).

For O2, by (2.58), we also only have the following three cases:

(i) ρ1(x1, x′1) ≤ 1

2A1(1 + ρ1(x1, xQ

J1i1

)) and ρ1(z1, z′1) ≤ 1

2A1(1 + ρ1(y1, z

′1));

(ii) ρ1(x1, x′1) > 1

2A1(1 + ρ1(x1, xQ

J1i1

)) and ρ1(z1, z′1) ≤ 1

2A1(1 + ρ1(y1, z

′1));

(iii) ρ1(x1, x′1) ≤ 1

2A1(1 + ρ1(x1, xQ

J1i1

)) and ρ1(z1, z′1) > 1

2A1(1 + ρ1(y1, z

′1)).

Then a procedure similar to that for K12 tells us that

(2.62) |[K31(x1, z1)−K31(x′1, z1)]− [K31(x1, z′1)−K31(x′1, z

′1)]|

≤ Ck1

ρ1(x1, x′1)

ε′1ρ1(z1, z′1)

ε′1

(1 + ρ1(x1, z1))d1+2ε′1

≤ Ck1

ρ1(x1, x′1)

ε′1ρ1(z1, z′1)

ε′1

ρ1(x1, z1)d1+2ε′1.


The geometric means between (2.61) and (2.62) yields that

(2.63) |[K31(x1, z1)−K31(x′1, z1)]− [K31(x1, z′1)−K31(x′1, z

′1)]|

≤ Ck12−J1ε′1(1−α5) ρ1(x1, x

′1)

ε′1α5ρ1(z1, z′1)

ε′1

ρ1(x1, z1)d1+ε′1α5+ε′1.

Finally we verify that K31 has the strong weak boundedness property as in Definition2.6. Let r > 0 and f be a continuous function on X1 × X1 with supp f ⊂ B1(x11, r) ×B1(x12, r), where x11, x12 ∈ X1, ‖f‖L∞(X1×X1) ≤ 1, ‖f(·, z1)‖Cη

0 (X1) ≤ r−η for all z1 ∈ X1

and ‖f(x1, ·)‖Cη0 (X1) ≤ r−η for all x1 ∈ X1. From (2.56), it follows that

(2.64) |〈K31, f〉|

=∣∣∣∣∫

X1

∫

X1

K31(x1, z1)f(x1, z1) dµ1(x1) dµ1(z1)∣∣∣∣

≤ Ck12−J1ε′1‖f‖L∞(X1×X1)

×∫

B1(x12,r)

∫

X1

1(1 + ρ(x1, z1))d1+ε′1

dµ1(x1)

dµ1(z1)

≤ Ck12−J1ε′1rd1 .

Obviously T31(1) = 0, which together with the estimates (2.56), (2.60), (2.61), (2.63)and (2.64) tells us that T31 satisfies all the conditions of Lemma 2.4 with

‖T31‖ ≤ Ck12−J1ε′1(1−α5).

Thus, for all f ∈ G(β1, γ1),

(2.65) ‖T31(f)‖G(β1,γ1) ≤ Ck12−J1ε′1(1−α5)‖f‖G(β1,γ1).

From Lemma 2.3, (2.65) and (2.55), it follows that T3 is bounded on G(β1, β2; γ1, γ2)and for all g ∈ G(β1, β2; γ1, γ3),

‖T3(g)‖G(β1,β2;γ1,γ3) ≤ Ck12−J1ε′1(1−α5)‖g‖G(β1,β2;γ1,γ3),

which just means (2.54) is true for i = 1, and therefore, it is also true for i = 2.By (2.53), (2.54) and Lemma 2.5, we obtain


⟩


= limM1, M2→∞

limJ1, J2→∞

⟨f, g3

M1,M2

⟩

= limM1, M2→∞

limJ1, J2→∞

∑

i1∈NJ1

∑

i2∈NJ2

D∗k1

D∗k2

(f)(

xQ

J1i1

, xQ

J2i2

)

×∫

QJ1i1∩B1(x0,M1)

∫

QJ2i2∩B2(y0,M2)

Dk1Dk2(g)(y1, y2) dµ1(y1) dµ2(y2)

=∫

X1

∫

X2

D∗k1

D∗k2

(f) (y1, y2) Dk1Dk2(g)(y1, y2) dµ1(y1) dµ2(y2)

+ limM1, M2→∞

limJ1, J2→∞

∑

i1∈NJ1

∑

i2∈NJ2

∫

QJ1i1∩B1(x0,M1)

×∫

QJ2i2∩B2(y0,M2)

[D∗

k1D∗

k2(f)

(x

QJ1i1

, xQ

J2i2

)− D∗

k1D∗

k2(f)

(x

QJ1i1

, y2

)]

+[D∗

k1D∗

k2(f)

(x

QJ1i1

, y2

)− D∗

k1D∗

k2(f) (y1, y2)

]

×Dk1Dk2(g)(y1, y2) dµ1(y1) dµ2(y2).

Using the size and smooth conditions of Dk1 and Dk2 , by the geometric means, we canverify that the functions on x1 and x2,

∑

i1∈NJ1

∑

i2∈NJ2

Dk1

(x1, xQ

J1i1

)[Dk2

(x2, xQ

J2i2

)− Dk2(x2, y2)

]

×χQ

J1i1∩B1(x0,M1)

(y1)χQJ2i2∩B2(y0,M2)

(y2)

and∑

i1∈NJ1

∑

i2∈NJ2

[Dk1

(x1, xQ

J1i1

)− Dk1 (x1, y1)

]Dk2 (x2, y2)

×χQ

J1i1∩B1(x0,M1)


(y2)

belong to the space G(β1, β2; γ1, γ2) uniformly in y1 and y2, and there exist some constantsα6 > 0 and C > 0 independent of Mi, Ji and yi for i = 1, 2, such that

∥∥∥∥∥∥∑

i1∈NJ1

∑

i2∈NJ2

Dk1

(·1, xQ

J1i1

)[Dk2

(·2, xQ

J2i2

)− Dk2(·2, y2)

]

×χQ

J1i1∩B1(x0,M1)


(y2)∥∥∥∥G(β1,β2;γ1,γ2)

≤ C2−α6J2


and ∥∥∥∥∥∥∑

i1∈NJ1

∑

i2∈NJ2

[Dk1

(·1, xQ

J1i1

)− Dk1 (·1, y1)

]Dk2 (·2, y2)

×χQ

J1i1∩B1(x0,M1)


(y2)∥∥∥∥G(β1,β2;γ1,γ2)

≤ C2−α6J1 ,

which imply that

(2.67)∣∣∣∣D∗

k1D∗

k2(f)

(x

QJ1i1

, xQ

J2i2

)− D∗

k1D∗

k2(f)

(x

QJ1i1

, y2

)∣∣∣∣ ≤ C2−α6J2

and

(2.68)∣∣∣∣D∗

k1D∗

k2(f)

(x

QJ1i1

, y2

)− D∗

k1D∗

k2(f) (y1, y2)

∣∣∣∣ ≤ C2−α6J1 ,

where C > 0 is independent of Mi, Ji and yi for i = 1, 2. Moreover, it is easy to check thatDk1Dk2(g) ∈ L1(X1 × X2), which together with (2.67), (2.68), (2.66) and the Lebesguedominated convergence theorem yields that

⟨f, Dk1Dk2Dk1Dk2(g)

⟩

=∫

X1

∫

X2

D∗k1

D∗k2

(f) (y1, y2) Dk1Dk2(g)(y1, y2) dµ1(y1) dµ2(y2)

=⟨D∗

k1D∗

k2D∗

k1D∗

k2(f), g

⟩,

where the last equality can be obtained by repeating the above procedure on Dk1 and Dk2 .This proves (2.34) and we complete the proof of Theorem 2.3.

Similarly, from Theorem 2.2, we can deduce the following continuous Calderon re-producing formulae in spaces of distributions.


)′,

f =∞∑

k1=−∞

∞∑

k2=−∞D∗k1

D∗k2

D∗k1

D∗k2

(f)


′2; γ

′1, γ

′2)

)′with β′i ∈ (βi, εi) and γ′i ∈ (γi, εi) for i = 1, 2, where

D∗ki

(x, y) = Dki(y, x) and D∗ki

(x, y) = Dki(y, x).


Let i = 1, 2. Note that D∗ki

, D∗ki

and D∗ki

respectively have the same propertiesas Dki , Dki and Dki . From this, it is easy to see that we can re-state Theorem 2.3 andTheorem 2.4 as the following theorem, which will simplify the notation in the followingapplications of these formulae.

Theorem 2.5 Let all the notation be the same as in Theorem 2.1 and Theorem 2.2. Thenfor all f ∈

(G(β1, β2; γ1, γ2)

)′,

f =∞∑

k1=−∞

∞∑

k2=−∞Dk1Dk2Dk1Dk2(f) =

∞∑

k1=−∞

∞∑

k2=−∞Dk1Dk2Dk1Dk2(f)


′2; γ

′1, γ

′2)

)′with β′i ∈ (βi, εi) and γ′i ∈ (γi, εi) for i = 1, 2.

We now recall the discrete Calderon reproducing formulae on spaces of homogeneoustype in [H3].

Lemma 2.6 With all the notation as in Lemma 2.1, then for all f ∈ G(β, γ) with β, γ ∈(0, ε) and any yk,ν

τ ∈ Qk,ντ ,

(2.69) f(x) =∞∑

k=−∞

∑

τ∈Ik

N(k,τ)∑

ν=1


τ )Dk(f)(yk,ντ )

=∞∑

k=−∞

∑

τ∈Ik

N(k,τ)∑

ν=1


τ )Dk(f)(yk,ντ ),

where the series converge in the norm of both the space G(β′, γ′) with 0 < β′ < β and0 < γ′ < γ and the space Lp(X) with p ∈ (1,∞).

By an argument of duality, Han in [H3] also established the following discrete Calderon

reproducing formulae on spaces of distributions,(G(β, γ)

)′with β, γ ∈ (0, ε).

Lemma 2.7 With all the notation as in Lemma 2.6, then for all f ∈(G(β, γ)

)′with

β, γ ∈ (0, ε), (2.69) holds in(G(β′, γ′)

)′with β < β′ < ε and γ < γ′ < ε.

By a procedure similar to the proofs of Theorems 2.1, 2.2, 2.3 and 2.4, using Lemma2.6 and Lemma 2.7, we can also establish the following discrete Calderon reproducingformulae on product spaces of homogeneous-type spaces. We only state the results andleave the details to the reader; see also [HY].

Theorem 2.6 Let all the notation as in Theorems 2.1 and 2.2, and

Qk1,ν1τ1 : k1 ∈ Z, τ1 ∈ Ik1 , ν1 = 1, · · · , N(k1, τ1)


and Qk2,ν2τ2 : k2 ∈ Z, τ2 ∈ Ik2 , ν2 = 1, · · · , N(k2, τ2) respectively be the dyadic cubes of

X1 and X2 defined above with j1, j2 ∈ N large enough. Then for all f ∈ G(β1, β2; γ1, γ2)with βi, γi ∈ (0, εi) for i = 1, 2 and any yk1,ν1

τ1 ∈ Qk1,ν1τ1 and yk2,ν2

τ2 ∈ Qk2,ν2τ2 ,

(2.70) f(x1, x2) =∞∑

k1=−∞

∑

τ1∈Ik1

N(k1,τ1)∑

ν1=1

∞∑

k2=−∞

∑

τ2∈Ik2

N(k2,τ2)∑

ν2=1


τ2 )

×Dk1(x1, yk1,ν1τ1 )Dk2(x2, y


τ1 , yk2,ν2τ2 )

=∞∑

k1=−∞

∑

τ1∈Ik1

N(k1,τ1)∑

ν1=1

∞∑

k2=−∞

∑

τ2∈Ik2

N(k2,τ2)∑

ν2=1


τ2 )

×Dk1(x1, yk1,ν1τ1 )Dk2(x2, y


τ1 , yk2,ν2τ2 ),


′1, γ

′2) with β′i ∈ (0, βi)

and γ′i ∈ (0, γi) for i = 1, 2, and Lp(X1 ×X2) with p ∈ (1,∞).


)′, (2.70) holds in

(G(β′1, β

′2; γ

′1, γ

′2)

)′with β′i ∈ (βi, εi) and γ′i ∈ (γi, εi)

for i = 1, 2.


3 Littlewood-Paley theory

We first establish the Littlewood-Paley theorem on product spaces of spaces of ho-mogeneous type. To this end, we recall the Littlewood-Paley theorem on spaces of homo-geneous type in [DJS].Lemma 3.1 Let X be a space of homogeneous type as in Definition 2.1, ε ∈ (0, θ), Skk∈Zbe an approximation to the identity of order ε as in Definition 2.3 and Dk = Sk − Sk−1

for k ∈ Z. If 1 < p < ∞, then there is a constant Cp > 0 such that for all f ∈ Lp(X),

(3.1) C−1p ‖f‖Lp(X) ≤

∥∥∥∥∥∥

∞∑

k=−∞|Dk(f)|2

1/2∥∥∥∥∥∥

Lp(X)

≤ Cp‖f‖Lp(X).

The Littlewood-Paley theorem on product spaces of homogeneous-type spaces can bestated as follows, whose proof can be deduced from the well-known discrete vector-valuedLittlewood-Paley theorem on spaces of homogenous type, Lemma 3.1 and the Calderonreproducing formulae, Theorem 2.1; see also the proof of Theorem 2 in [FS].

Theorem 3.1 Let i = 1, 2, Xi be a space of homogeneous type as in Definition 2.1, εi ∈(0, θi], Skiki∈Z be an approximation to the identity of order εi on space of homogeneoustype, Xi, and Dki = Ski − Ski−1 for all ki ∈ Z. If 1 < p < ∞, then there is a constantCp > 0 such that for all f ∈ Lp(X1 ×X2),

(3.2) C−1p ‖f‖Lp(X1×X2) ≤ ‖g2(f)‖Lp(X1×X2) ≤ Cp‖f‖Lp(X1×X2),

where gq(f) for q ∈ (0,∞) is called the discrete Littlewood-Paley g-function defined by

gq(f)(x1, x2) =

∞∑

k1=−∞

∞∑

k2=−∞|Dk1Dk2(f)(x1, x2)|q

1/q

for x1 ∈ X1 and x2 ∈ X2.

Proof. We first prove the second inequality in (3.2). To do so, we use the known vector-valued Littlewood-Paley theorem on X2 and Lemma 3.1 on X1. Let f ∈ Lp(X1 ×X2),F (x1, x2) = Fk1(x1, x2)k1∈Z and Fk1(x1, x2) = Dk1 [f(·, x2)] (x1) for k1 ∈ Z. Set

‖F (x1, x2)‖l2(Z) =

∞∑

k1=−∞|Fk1(x1, x2)|2

1/2

.

For k2 ∈ Z, define Dk2F (x1, x2) = Dk2 [Fk1(x1, ·)] (x2)k1∈Z and the discrete l2(Z)-valuedLittlewood-Paley g2,2 function on X2 by

(3.3) g2,2[F (x1, ·)](x2) =

∞∑

k2=−∞‖Dk2F (x1, x2)‖2

l2(Z)

1/2

.


Obviously,

(3.4) g2,2[F (x1, ·)](x2) = g2(f)(x1, x2).

Taking the Lp(X2)-norm on both sides of (3.3) and by the l2(Z)-valued Littlewood-Paleytheorem on Lp(X2), we obtain

(3.5)∫

X2

|g2[F (x1, ·)](x2)|p dµ2(x2) ≤ Cpp

∫

X2

‖F (x1, x2)‖pl2(Z) dµ2(x2).

Taking the Lp(X1)-norm on both sides of (3.5), using (3.4) and Lemma 3.1 on Lp(X1),and exchanging the order of the integrals on X1 and X2 yield the desired second inequalityof (3.2).

We now prove the first inequality in (3.2) by using Theorem 2.1 and the secondinequality of (3.2). Let f ∈ Lp(X1 ×X2) and all the notation be the same as in Theorem2.1. By Theorem 2.1, we have

(3.6) f =∞∑

k1=−∞

∞∑

k2=−∞Dk1Dk2Dk1Dk2(f).

Let 1/p + 1/p′ = 1 and g ∈ Lp′(X1 ×X2). The Holder inequality tells us that

(3.7) |〈f, g〉|=∣∣∣∣∣∣

⟨ ∞∑

k1=−∞

∞∑

k2=−∞Dk1Dk2Dk1Dk2(f), g

⟩∣∣∣∣∣∣

=

∣∣∣∣∣∣

∞∑

k1=−∞

∞∑

k2=−∞

⟨Dk1Dk2(f), D∗

k1D∗

k2(g)

⟩∣∣∣∣∣∣

≤ ‖g2(f)‖Lp(X1×X2)

∥∥∥∥∥∥∥

∞∑

k1=−∞

∞∑

k2=−∞

∣∣∣D∗k1

D∗k2

(g)∣∣∣2

1/2∥∥∥∥∥∥∥

Lp′ (X1×X2)

,

where D∗ki

(xi, yi) = Dki(yi, xi) for i = 1, 2. It is well-known that for ki, k′i ∈ Z,

(3.8)∣∣∣D∗

kiDk′i(xi, zi)

∣∣∣ ≤ C2−|ki−k′i|ε′i 2−(ki∧k′i)ε′i

(2−(ki∧k′i) + ρi(xi, zi))di+ε′i,

where i = 1, 2, εi ∈ (0, εi) and a ∧ b = min(a, b) for any a, b ∈ R; see [H1] for a proof.Let Mi be the Hardy-Littlewood maximal function on Xi. Applying (3.6) to the functiong and using (3.8) and the Holder inequality lead us that

(3.9)

∞∑

k1=−∞

∞∑

k2=−∞

∣∣∣D∗k1

D∗k2

(g)∣∣∣2

1/2


=

∞∑

k1=−∞

∞∑

k2=−∞

∣∣∣∣∣∣

∞∑

k′1=−∞

∞∑

k′2=−∞D∗

k1D∗

k2Dk′1Dk′2Dk′1Dk′2(g)

∣∣∣∣∣∣

2

1/2

≤ C

∞∑

k1=−∞

∞∑

k2=−∞

∞∑

k′1=−∞

∞∑

k′2=−∞2−|k1−k′1|ε′12−|k2−k′2|ε′2

× M1M2

[Dk′1Dk′2(g)

])21/2

≤ C

∞∑

k′1=−∞

∞∑

k′2=−∞

(M1M2

[Dk′1Dk′2(g)

])2

1/2

.

By taking the Lp′(X1 × X2)-norm on both sides of (3.9) and an iterative application ofthe Fefferman-Stein vector-valued inequality in [FeS] on Lp′(X1) and Lp′(X2), we obtain

(3.10)

∥∥∥∥∥∥∥

∞∑

k1=−∞

∞∑

k2=−∞

∣∣∣D∗k1

D∗k2

(g)∣∣∣2

1/2∥∥∥∥∥∥∥

Lp′ (X1×X2)

≤ C

∥∥∥∥∥∥∥

∞∑

k′1=−∞

∞∑

k′2=−∞

(M1M2

[Dk′1Dk′2(g)

])2

1/2∥∥∥∥∥∥∥

Lp′ (X1×X2)

≤ Cp′

∥∥∥∥∥∥∥

∞∑

k′1=−∞

∞∑

k′2=−∞

(M2

[Dk′1Dk′2(g)

])2

1/2∥∥∥∥∥∥∥

Lp′ (X1×X2)

≤ Cp′

∥∥∥∥∥∥∥

∞∑

k′1=−∞

∞∑

k′2=−∞

∣∣∣Dk′1Dk′2(g)∣∣∣2

1/2∥∥∥∥∥∥∥

Lp′ (X1×X2)

≤ Cp′‖g‖Lp′ (X1×X2),

where in the last step, we used the second inequality in (3.2).Taking the supremum on g ∈ Lp′(X1 ×X2) with ‖g‖Lp′ (X1×X2) ≤ 1 on both sides of

(3.7) and applying (3.10) then imply the first inequality in (3.2). This finishes the proofof Theorem 3.1.

Remark 3.1 Let Dki for i = 1, 2 be the same as in Theorem 2.2. In the proof of Theorem3.1, we actually prove that the second inequality in (3.2) still holds if we replace Dki thereby those Dki for i = 1, 2; see (3.10). This is well known fact for the second inequality in(3.1); see also [DJS, H2].

Let all the notation be the same as in Theorem 3.1. We now define the Littlewood-


Paley S-function Sq on the product space X1 ×X2 by

(3.11) Sq(f)(x1, x2)

=

∞∑

k1=−∞

∞∑

k2=−∞

∫

ρ1(x1,y1)≤C11,12−k1

∫

ρ2(x2,y2)≤C11,22−k2

2k1d1+k2d2

× |Dk1Dk2(f)(y1, y2)|q dµ1(y1) dµ2(y2)

1/q

for x1 ∈ X1 and x2 ∈ X2.We have the following relation theorem on the Littlewood-Paley S-function Sq and

the Littlewood-Paley g-function gq.

Lemma 3.2 Let 1 < p, q < ∞. Then there exists a constant Cp,q > 0 such that for allf ∈ Lp(X1 ×X2),

‖Sq(f)‖Lp(X1×X2) ≤ Cp,q ‖gq(f)‖Lp(X1×X2) .

Proof. Let f ∈ Lp(X1 ×X2) and all the notation be the same as in Theorem 2.1. ByTheorem 2.1, we write f as in (3.6). Let ρ1(x1, y1) ≤ C11,12−k1 and ρ2(x2, y2) ≤ C11,22−k2 ,which imply that

2−(ki∧k′i) + ρi(xi, zi) ≤ C(2−(ki∧k′i) + ρi(yi, zi)

)

for i = 1, 2. From this and an estimate similar to (3.8) with D∗ki

there replaced by Dki ,it follows that

(3.12) |Dk1Dk2(f)(y1, y2)|

=

∣∣∣∣∣∣

∞∑

k′1=−∞

∞∑

k′2=−∞Dk1Dk2Dk′1Dk′2Dk′1Dk′2(f)(y1, y2)

∣∣∣∣∣∣

≤ C

∞∑

k′1=−∞

∞∑

k′2=−∞2−|k1−k′1|ε′12−|k2−k′2|ε′2

×∫

X1

∫

X2

2−(k1∧k′1)ε′1

(2−(k1∧k′1) + ρ1(y1, z1))d1+ε′1

2−(k2∧k′2)ε′2

(2−(k2∧k′2) + ρ2(y2, z2))d2+ε′2

× |Dk1Dk2(f)(z1, z2)| dµ1(z1) dµ2(z2)

≤ C∞∑

k′1=−∞

∞∑

k′2=−∞2−|k1−k′1|ε′12−|k2−k′2|ε′2

×∫

X1

∫

X2

2−(k1∧k′1)ε′1

(2−(k1∧k′1) + ρ1(x1, z1))d1+ε′1

2−(k2∧k′2)ε′2

(2−(k2∧k′2) + ρ2(x2, z2))d2+ε′2

× |Dk1Dk2(f)(z1, z2)| dµ1(z1) dµ2(z2)


≤ C∞∑

k′1=−∞

∞∑

k′2=−∞2−|k1−k′1|ε′12−|k2−k′2|ε′2M1M2 [Dk1Dk2(f)] (x1, x2).

Using (3.12) and an iterative application of the Fefferman-Stein vector-valued in-equality in [FeS] on Lp′(X1) and Lp′(X2) yield that

‖Sq(f)‖Lp(X1×X2)

≤ C

∥∥∥∥∥∥

∞∑

k1=−∞

∞∑

k2=−∞

∞∑

k′1=−∞

∞∑

k′2=−∞2−|k1−k′1|ε′12−|k2−k′2|ε′2

×M1M2 [Dk1Dk2(f)])q

1/q∥∥∥∥∥Lp(X1×X2)

≤ C

∥∥∥∥∥∥∥

∞∑

k′1=−∞

∞∑

k′2=−∞(M1M2 [Dk1Dk2(f)])q

1/q∥∥∥∥∥∥∥

Lp(X1×X2)

≤ Cp,q ‖gq(f)‖Lp(X1×X2) ,

which completes the proof of Lemma 3.2.

Remark 3.2 It is easy to see from the proof of Lemma 3.2 that if we replace Dki in thedefinition of Sq(f)(x1, x2), (6.32), by Dki as in Theorem 2.2 for i = 1, 2, then Lemma 3.2still holds. This is because the contribution of Dki to the inequality in the lemma comesfrom an estimate similar to (3.8), which still holds if we replace Dki by Dki for i = 1, 2.

Lemma 3.3 Let 1 < p, q < ∞. Then there exists a constant Cp > 0 such that for allf ∈ Lp(X1 ×X2),

‖f‖Lp(X1×X2) ≤ Cp ‖S2(f)‖Lp(X1×X2).

Proof. Let f ∈ Lp(X1 ×X2) and all the notation be the same as in Theorem 2.1. ByTheorem 2.1, we write f as in (3.6). Let g ∈ Lp′(X1 × X2). The Holder inequality,changing the order of the integrals, Remark 3.2 and Theorem 3.1 tell us that

(3.13) |〈f, g〉|=∣∣∣∣∣∣

⟨ ∞∑

k1=−∞

∞∑

k2=−∞Dk1Dk2Dk1Dk2(f), g

⟩∣∣∣∣∣∣

=

∣∣∣∣∣∣

∞∑

k1=−∞

∞∑

k2=−∞

⟨Dk1Dk2(f), D∗

k1D∗

k2(g)

⟩∣∣∣∣∣∣

≤ C∞∑

k1=−∞

∞∑

k2=−∞

∫

X1×X2

∣∣∣Dk1Dk2(f)(y1, y2)D∗k1

D∗k2

(g)(y1, y2)∣∣∣

×

2k1d1

∫

X1

χB1(y1,C11,12−k1 )(x1) dµ1(x1)


×

2k2d2

∫

X2

χB2(y2,C11,22−k2)(x2) dµ2(x2)

dµ1(y1) dµ2(y2)

≤ C∞∑

k1=−∞

∞∑

k2=−∞

∫

X1×X2

∫

ρ1(x1,y1)≤C11,12−k1

∫

ρ2(x2,y2)≤C11,22−k2

2k1d1+k2d2

×∣∣∣Dk1Dk2(f)(y1, y2)D∗

k1D∗

k2(g)(y1, y2)

∣∣∣ dµ1(y1) dµ2(y2) dµ1(x1) dµ2(x2)

≤ C

∫

X1×X2

Sq(f)(x1, x2)

×

∞∑

k1=−∞

∞∑

k2=−∞

∫

ρ1(x1,y1)≤C11,12−k1

∫

ρ2(x2,y2)≤C11,22−k2

2k1d1+k2d2

×∣∣∣D∗

k1D∗

k2(g)(y1, y2)

∣∣∣q′

dµ1(y1) dµ2(y2)

1/q′

dµ1(x1) dµ2(x2)

≤ C ‖S2(f)‖Lp(X1×X2)

×

∞∑

k1=−∞

∞∑

k2=−∞

∫

ρ1(x1,y1)≤C11,12−k1

∫

ρ2(x2,y2)≤C11,22−k2

2k1d1+k2d2

×∣∣∣D∗

k1D∗

k2(g)(y1, y2)

∣∣∣2

dµ1(y1) dµ2(y2)

]p′/2

dµ1(x1) dµ2(x2)

1/p′

≤ C ‖S2(f)‖Lp(X1×X2) ‖g2(g)‖Lp′ (X1×X2)

≤ C ‖g‖Lp′ (X1×X2) ‖S2(f)‖Lp(X1×X2) .

Taking the supremum on g ∈ Lp′(X1×X2) with ‖g‖Lp′ (X1×X2) ≤ 1 on both sides of (3.13)yields the conclusion of the lemma, which completes the proof of Lemma 3.3.

Lemma 3.2, Lemma 3.3 and Theorem 3.1 imply the following equivalence of theLittlewood-Paley S-function and g-function in Lp(X1 ×X2)-norm.

Theorem 3.2 Let all the notation be the same as in Theorem 3.1, g2 and S2 be definedrespectively as in Theorem 3.1 and (6.32). If 1 < p < ∞, then there is a constant Cp > 0such that for all f ∈ Lp(X1 ×X2),

C−1p ‖S2(f)‖Lp(X1×X2) ≤ ‖g2(f)‖Lp(X1×X2) ≤ Cp ‖S2(f)‖Lp(X1×X2) .


4 Hp spaces

In this section, we first apply the discrete Calderon reproducing formulae, Theorem2.7, to establish the equivalence between the Littlewood-Paley S-function and g-functionin Lp(X1 ×X2)-norm with p ≤ 1, which generalizes Theorem 3.2. Such a result for non-product spaces was already obtained in [H2] via a Plancherel-Polya inequality. We usethe same ideas as in [H2] here. Thus, we first establish a product-type Plancherel-Polyainequality. To this end, we need the following lemma which can be found in [FJ, pp.147-148] for Rn and [HS, p. 93] for spaces of homogeneous type.Lemma 4.1. Let X be a space of homogeneous type as in Definition 2.1, 0 < r ≤ 1, k,

η ∈ Z+ with η ≤ k and for any dyadic cube Qk,ντ ,

|fQk,ν

τ(x)| ≤ (1 + 2ηρ(x, yk,ν

τ ))−d−γ ,

where x ∈ X, yk,ντ is any point in Qk,ν

τ and γ > d(1/r − 1). Then

∑

τ∈Ik

N(k,τ)∑

ν=1

|λQk,ν

τ||f

Qk,ντ

(x)| ≤ C2(k−η)d/r

M

∑

τ∈Ik

N(k,τ)∑

ν=1

|λQk,ν

τ|rχ

Qk,ντ

(x)

1/r

,

where C is independent of x, k and η, and M is the Hardy-Littlewood maximal operatoron X.

Theorem 4.1 Let the notation be the same as in Theorem 2.6. Moreover, let

Qk′1,ν′1τ ′1

: k′1 ∈ Z, τ ′1 ∈ Ik′1 , ν ′1 = 1, · · · , N(k′1, τ′1)

and Qk′2,ν′2τ ′2

: k′2 ∈ Z, τ ′2 ∈ Ik′2 , ν′2 = 1, · · · , N(k′2, τ′2) respectively be another set of

dyadic cubes of X1 and X2 defined above with j′1, j′2 ∈ N large enough, let Pkiki∈Z

be another approximation to the identity of order εi on homogeneous-type space Xi andEki

= Pk1 − Pki−1 for ki ∈ Z and i = 1, 2. If max

d1d1+ε1

, d2d2+ε2

< p, q ≤ ∞, then

there is a constant C > 0 such that for all f ∈(G(β1, β2; γ1, γ2)

)′with βi, γi ∈ (0, εi) for

i = 1, 2,

(4.1)

∥∥∥∥∥∥

∞∑

k1=−∞

∑

τ1∈Ik1

N(k1,τ1)∑

ν1=1

∞∑

k2=−∞

∑

τ2∈Ik2

N(k2,τ2)∑

ν2=1

supz1∈Q

k1,ν1τ1

,z2∈Qk2,ν2τ2

|Dk1Dk2(f)(z1, z2)|q

×χQ

k1,ν1τ1

(·)χQ

k2,ν2τ2

(·)1/q∥∥∥∥∥

Lp(X1×X2)

≤ C

∥∥∥∥∥∥∥

∞∑

k′1=−∞

∑

τ ′1∈Ik′1

N(k′1,τ ′1)∑

ν′1=1

∞∑

k′2=−∞

∑

τ ′2∈Ik′2

N(k′2,τ ′2)∑

ν′2=1

× infz1∈Q

k′1,ν′1τ ′1

,z2∈Qk′2,ν′2τ ′2

∣∣∣Ek′1Ek′2(f)(z1, z2)∣∣∣qχ

Qk′1,ν′1τ ′1

(·)χQ

k′2,ν′2τ ′2

(·)1/q∥∥∥∥∥

Lp(X1×X2)

.


Proof. We first choose ε′i ∈ (0, εi) for i = 1, 2 and r ∈ (0, 1] such that

(4.2) max

d1

d1 + ε′1,

d2

d2 + ε′2

< r < min(p, q).

Let f ∈(G(β1, β2; γ1, γ2)

)′with βi, γi ∈ (0, εi) for i = 1, 2. By Theorem 2.7, we have

(4.3) f(x1, x2) =∞∑

k′1=−∞

∑

τ ′1∈Ik′1

N(k′1,τ ′1)∑

ν′1=1

∞∑

k′2=−∞

∑

τ ′2∈Ik′2

N(k′2,τ ′2)∑

ν′2=1

µ1(Qk′1,ν′1τ ′1

)µ2(Qk′2,ν′2τ ′2

)

×Ek′1(x1, yk′1,ν′1τ ′1

)Ek′2(x2, yk′2,ν′2τ ′2

)Ek′1Ek′2(f)(yk′1,ν′1τ ′1

, yk′2,ν′2τ ′2

)


′2; γ

′1, γ

′2)

)′with β′i ∈ (βi, εi) and γ′i ∈ (γi, εi), where Ek′i satisfies the same

properties as Dki in Theorem 2.7 and i = 1, 2. From (4.3) and an estimate similar to (3.8)with D∗

kiand Dk′i respectively replaced by Dki and Ek′i , it follows that for any k1, k2 ∈ Z,

(4.4) |Dk1Dk2(f)(z1, z2)|

=

∣∣∣∣∣∞∑

k′1=−∞

∑

τ ′1∈Ik′1

N(k′1,τ ′1)∑

ν′1=1

∞∑

k′2=−∞

∑

τ ′2∈Ik′2

N(k′2,τ ′2)∑

ν′2=1

µ1(Qk′1,ν′1τ ′1

)µ2(Qk′2,ν′2τ ′2

)

×Dk1Ek′1(z1, yk′1,ν′1τ ′1

)Dk2Ek′2(z2, yk′2,ν′2τ ′2

)Ek′1Ek′2(f)(yk′1,ν′1τ ′1

, yk′2,ν′2τ ′2

)

∣∣∣∣∣

≤ C∞∑

k′1=−∞

∑

τ ′1∈Ik′1

N(k′1,τ ′1)∑

ν′1=1

∞∑

k′2=−∞

∑

τ ′2∈Ik′2

N(k′2,τ ′2)∑

ν′2=1

2−k′1d1−k′2d22−|k1−k′1|ε′1−|k2−k′2|ε′2

× 2−(k1∧k′1)ε′1

(2−(k1∧k′1) + ρ1(z1, yk′1,ν′1τ ′1

))d1+ε′1

2−(k2∧k′2)ε′2

(2−(k2∧k′2) + ρ2(z2, yk′2,ν′2τ ′2

))d2+ε′2

×∣∣∣Ek′1Ek′2(f)(yk′1,ν′1

τ ′1, y

k′2,ν′2τ ′2

)∣∣∣ .

Lemma 4.1 with (4.2) and the estimate (4.4) tell us that

(4.5)

∞∑

k1=−∞

∑

τ1∈Ik1

N(k1,τ1)∑

ν1=1

∞∑

k2=−∞

∑

τ2∈Ik2

N(k2,τ2)∑

ν2=1

supz1∈Q

k1,ν1τ1

,z2∈Qk2,ν2τ2


×χQ

k1,ν1τ1

(x1)χQk2,ν2τ2

(x2)

1/q

≤ C

∞∑

k1=−∞

∞∑

k2=−∞

∞∑

k′1=−∞

∑

τ ′1∈Ik′1

N(k′1,τ ′1)∑

ν′1=1


×∞∑

k′2=−∞

∑

τ ′2∈Ik′2

N(k′2,τ ′2)∑

ν′2=1

2−k′1d1−k′2d22−|k1−k′1|ε′1−|k2−k′2|ε′2

× 2−(k1∧k′1)ε′1

(2−(k1∧k′1) + ρ1(x1, yk′1,ν′1τ ′1

))d1+ε′1

2−(k2∧k′2)ε′2

(2−(k2∧k′2) + ρ2(x2, yk′2,ν′2τ ′2

))d2+ε′2

×∣∣∣Ek′1Ek′2(f)(yk′1,ν′1

τ ′1, y

k′2,ν′2τ ′2

)∣∣∣]q1/q

≤ C

∞∑

k1=−∞

∞∑

k2=−∞

∞∑

k′1=−∞

∞∑

k′2=−∞2−k′1d1−k′2d22−|k1−k′1|ε′1−|k2−k′2|ε′2

×2(k1∧k′1)d1+[k′1−(k1∧k′1)]d1/r2(k2∧k′2)d2+[k′2−(k2∧k′2)]d2/r

×

M1

∑

τ ′1∈Ik′1

N(k′1,τ ′1)∑

ν′1=1

M2

∑

τ ′2∈Ik′2

N(k′2,τ ′2)∑

ν′2=1

∣∣∣Ek′1Ek′2(f)(yk′1,ν′1τ ′1

, yk′2,ν′2τ ′2

)∣∣∣r

×χQ

k′2,ν′2τ ′2

)(x2)χ

Qk′1,ν′1τ ′1

](x1)

)1/r]q1/q

≤ C

∞∑

k′1=−∞

∞∑

k′2=−∞

M1

∑

τ ′1∈Ik′1

N(k′1,τ ′1)∑

ν′1=1

M2

∑

τ ′2∈Ik′2

×N(k′2,τ ′2)∑

ν′2=1

∣∣∣Ek′1Ek′2(f)(yk′1,ν′1τ ′1

, yk′2,ν′2τ ′2

)∣∣∣rχ

Qk′2,ν′2τ ′2

)(x2)χ

Qk′1,ν′1τ ′1

](x1)

)q/r1/q

,

where if q ≤ 1, we used the following facts that for all ai ∈ C,

(4.6)

(∑

i

|ai|)q

≤∑

i

|ai|q,

and for i = 1, 2,∞∑

ki=−∞2−k′idiq−|ki−k′i|ε′iq+(ki∧k′i)diq+[k′i−(ki∧k′i)]qdi/r ≤ C;

if q > 1, we used the Holder inequality and the facts that for i = 1, 2,

∞∑

ki=−∞+

∞∑

k′i=−∞

2−k′idi−|ki−k′i|ε′i+(ki∧k′i)di+[k′i−(ki∧k′i)]di/r ≤ C.

Taking the Lp(X1 ×X2)-norm on both sides of (4.5) and an iterative application ofthe Fefferman-Stein vector-valued inequality in [FeS] on Lp/r(X1) and Lp/r(X2) togetherwith the arbitrariness of y

k′1,ν′1τ ′1

and yk′2,ν′2τ ′2

give us the desired (4.1), which completes theproof of Theorem 4.1.


Remark 4.1 If we replace Dki in (4.1) by Dki for i = 1, 2 as in Theorem 2.2, then(4.1) still holds. The reason for this is that the contribution to (4.1) of Dki is given by anestimate similar to (3.8), which still holds if we replace Dki by Dki for i = 1, 2.

We now can generalize Theorem 3.2 to the case p, q ≤ 1.

Theorem 4.2 Let all the notation be the same as in Theorem 3.2. If

max

d1

d1 + ε1,

d2

d2 + ε2

< p, q ≤ ∞,

then there is a constant Cp,q > 0 such that for all f ∈(G(β1, β2; γ1, γ2)

)′with βi, γi ∈

(0, εi) for i = 1, 2,

(4.7) C−1p,q ‖Sq(f)‖Lp(X1×X2) ≤ ‖gq(f)‖Lp(X1×X2) ≤ Cp,q ‖Sq(f)‖Lp(X1×X2) .

Proof. We begin with proving the first inequality in (4.7). Let f ∈(G(β1, β2; γ1, γ2)

)′

with βi, γi ∈ (0, εi) for i = 1, 2. By Theorem 2.7, we have

(4.8) f(x1, x2) =∞∑

k′1=−∞

∑

τ ′1∈Ik′1

N(k′1,τ ′1)∑

ν′1=1

∞∑

k′2=−∞

∑

τ ′2∈Ik′2

N(k′2,τ ′2)∑

ν′2=1

µ1(Qk′1,ν′1τ ′1

)µ2(Qk′2,ν′2τ ′2

)

×Dk′1(x1, yk′1,ν′1τ ′1

)Dk′2(x2, yk′2,ν′2τ ′2

)Dk′1Dk′2(f)(yk′1,ν′1τ ′1

, yk′2,ν′2τ ′2

)


′2; γ

′1, γ

′2)

)′with β′i ∈ (βi, εi) and γ′i ∈ (γi, εi), where Dk′i satisfies the same

properties as Dki in Theorem 2.7 and i = 1, 2.In what follows, if Q is a dyadic cube and C > 0 is a constant, let CQ be the dyadic

cube with the same center as Q and diameter C diam (Q).From (4.8) and an estimate similar to (3.8) with D∗

kiand Dk′i respectively replaced

by Dki and Dk′i , it follows that for some given constant C12 > 0 and any k1, k2 ∈ Z,

(4.9) supz1∈C12Q

k1,ν1τ1

,z2∈C12Qk2,ν2τ2

|Dk1Dk2(f)(z1, z2)|χQk1,ν1τ1

(x1)χQk2,ν2τ2

(x2)

=

∣∣∣∣∣∞∑

k′1=−∞

∑

τ ′1∈Ik′1

N(k′1,τ ′1)∑

ν′1=1

∞∑

k′2=−∞

∑

τ ′2∈Ik′2

N(k′2,τ ′2)∑

ν′2=1

µ1(Qk′1,ν′1τ ′1

)µ2(Qk′2,ν′2τ ′2

)

×Dk1Dk′1(z1, yk′1,ν′1τ ′1

)Dk2Dk′2(z2, yk′2,ν′2τ ′2

)Dk′1Dk′2(f)(yk′1,ν′1τ ′1

, yk′2,ν′2τ ′2

)

∣∣∣∣∣

×χQ

k1,ν1τ1

(x1)χQk2,ν2τ2

(x2)

≤ C

∞∑

k′1=−∞

∑

τ ′1∈Ik′1

N(k′1,τ ′1)∑

ν′1=1

∞∑

k′2=−∞

∑

τ ′2∈Ik′2

N(k′2,τ ′2)∑

ν′2=1

2−k′1d1−k′2d22−|k1−k′1|ε′1−|k2−k′2|ε′2


× 2−(k1∧k′1)ε′1

(2−(k1∧k′1) + ρ1(x1, yk′1,ν′1τ ′1

))d1+ε′1

2−(k2∧k′2)ε′2

(2−(k2∧k′2) + ρ2(x2, yk′2,ν′2τ ′2

))d2+ε′2

×∣∣∣Dk′1Dk′2(f)(yk′1,ν′1

τ ′1, y

k′2,ν′2τ ′2

)∣∣∣ ;

see also the proof of (4.4).Instead of (4.4) by (4.9) and repeating the proof of (4.1) yield a variant of (4.1),

namely, there is a constant C > 0 such that for all f ∈(G(β1, β2; γ1, γ2)

)′with βi, γi ∈

(0, εi) for i = 1, 2,

(4.10)

∥∥∥∥∥∥

∞∑

k1=−∞

∑

τ1∈Ik1

N(k1,τ1)∑

ν1=1

∞∑

k2=−∞

∑

τ2∈Ik2

N(k2,τ2)∑

ν2=1

× supz1∈C12Q

k1,ν1τ1

,z2∈C12Qk2,ν2τ2


×χQ

k1,ν1τ1

χQ

k2,ν2τ2

1/q∥∥∥∥∥Lp(X1×X2)

≤ C

∥∥∥∥∥∥

∞∑

k1=−∞

∑

τ1∈Ik1

N(k1,τ1)∑

ν1=1

∞∑

k2=−∞

∑

τ2∈Ik2

N(k2,τ2)∑

ν2=1

× infz1∈Q

k1,ν1τ1

,z2∈Qk2,ν2τ2

|Dk1Dk2(f)(z1, z2)|q χQ

k1,ν1τ1

χQ

k2,ν2τ2

1/q∥∥∥∥∥Lp(X1×X2)

,

where p, q are the same as in the theorem.From (4.10) with suitably chosen C12 and the definition of the Littlewood-Paley

S-function, (3.13), it follows that


=

∥∥∥∥∥∥

∞∑

k1=−∞

∑

τ1∈Ik1

N(k1,τ1)∑

ν1=1

∞∑

k2=−∞

∑

τ2∈Ik2

N(k2,τ2)∑

ν2=1

∫

ρ1(·1,y1)≤C11,12−k1

×∫

ρ2(·2,y2)≤C11,22−k2

2k1d1+k2d2 |Dk1Dk2(f)(y1, y2)|q

×χQ

k1,ν1τ1

(·1)χQk2,ν2τ2

(·2) dµ1(y1) dµ2(y2)

1/q∥∥∥∥∥Lp(X1×X2)

≤ C

∥∥∥∥∥∥

∞∑

k1=−∞

∑

τ1∈Ik1

N(k1,τ1)∑

ν1=1

∞∑

k2=−∞

∑

τ2∈Ik2

N(k2,τ2)∑

ν2=1

× supz1∈C12Q

k1,ν1τ1

,z2∈C12Qk2,ν2τ2


k1,ν1τ1

χQ

k2,ν2τ2

1/q∥∥∥∥∥Lp(X1×X2)


≤ C

∥∥∥∥∥∥

∞∑

k1=−∞

∑

τ1∈Ik1

N(k1,τ1)∑

ν1=1

∞∑

k2=−∞

∑

τ2∈Ik2

N(k2,τ2)∑

ν2=1

× infz1∈Q

k1,ν1τ1

,z2∈Qk2,ν2τ2


k1,ν1τ1

χQ

k2,ν2τ2

1/q∥∥∥∥∥Lp(X1×X2)

≤ C ‖gq(f)‖Lp(X1×X2) ,

which proves the first inequality in (4.7).We now turn to the proof of the second inequality in (4.7). By Theorem 4.1, we have


=

∥∥∥∥∥∥

∞∑

k1=−∞

∑

τ1∈Ik1

N(k1,τ1)∑

ν1=1

∞∑

k2=−∞

∑

τ2∈Ik2

N(k2,τ2)∑

ν2=1

∫

ρ1(·1,y1)≤C11,12−k1

×∫

ρ2(·2,y2)≤C11,22−k2

2k1d1+k2d2 |Dk1Dk2(f)(y1, y2)|q

×χQ

k1,ν1τ1

(·1)χQk2,ν2τ2

(·2) dµ1(y1) dµ2(y2)

1/q∥∥∥∥∥Lp(X1×X2)

≥∥∥∥∥∥∥

∞∑

k1=−∞

∑

τ1∈Ik1

N(k1,τ1)∑

ν1=1

∞∑

k2=−∞

∑

τ2∈Ik2

N(k2,τ2)∑

ν2=1

∫

ρ1(·1,y1)≤C11,12−k1

×∫

ρ2(·2,y2)≤C11,22−k2

2k1d1+k2d2 infz1∈Q

k1,ν1τ1

,z2∈Qk2,ν2τ2


×χQ

k1,ν1τ1

(·1)χQk2,ν2τ2

(·2) dµ1(y1) dµ2(y2)

1/q∥∥∥∥∥Lp(X1×X2)

≥ C

∥∥∥∥∥∥

∞∑

k1=−∞

∑

τ1∈Ik1

N(k1,τ1)∑

ν1=1

∞∑

k2=−∞

∑

τ2∈Ik2

N(k2,τ2)∑

ν2=1

× infz1∈Q

k1,ν1τ1

,z2∈Qk2,ν2τ2


k1,ν1τ1

(·1)χQk2,ν2τ2

(·2)1/q∥∥∥∥∥

Lp(X1×X2)

≥ C

∥∥∥∥∥∥

∞∑

k1=−∞

∑

τ1∈Ik1

N(k1,τ1)∑

ν1=1

∞∑

k2=−∞

∑

τ2∈Ik2

N(k2,τ2)∑

ν2=1

× supz1∈Q

k1,ν1τ1

,z2∈Qk2,ν2τ2


k1,ν1τ1

(·1)χQk2,ν2τ2

(·2)1/q∥∥∥∥∥

Lp(X1×X2)

≥ C ‖gq(f)‖Lp(X1×X2) ,

where in the second inequality, we used the following fact that if xi, yi ∈ Qki,νiτi , then


ρi(xi, yi) ≤ C10,12−ki−ji , and therefore, if ji ∈ N is large enough, then ρi(xi, yi) ≤ C11,i2−ki

for i = 1, 2. This verifies the second inequality in (4.7) and completes the proof of Theorem4.2.

Remark 4.2 By Remark 4.1, it is easy to see that the first inequality in (4.7) still holdsif we replace Dki in the definition of Sq(f) by Dki for i = 1, 2 as in Theorem 2.2, whichis useful in applications.

We can now introduce the Hardy spaces Hp(X1 ×X2) for some p ≤ 1 and establishtheir atomic decomposition characterization.

Definition 4.1 Let Xi be a homogeneous-type space as in Definition 2.1, εi ∈ (0, θi] andDkiki∈Z be the same as in Theorem 3.1 for i = 1, 2. Let

max

d1

d1 + ε1,

d2

d2 + ε2

< p < ∞

and for i = 1, 2,

(4.11) di(1/p− 1)+ < βi, γi < εi.

The Hardy space Hp(X1 ×X2) is defined to be the set of all f ∈(G(β1, β2; γ1, γ2)

)′such

that ‖g2(f)‖Lp(X1×X2) < ∞, and we define

‖f‖Hp(X1×X2) = ‖g2(f)‖Lp(X1×X2) ,

where g2(f) is defined as in Theorem 3.1.

We first consider the reasonability of the definition of the Hardy space Hp(X1 ×X2).

Proposition 4.1 Let all the notation be the same as in Definition 4.1. Then the definitionof the Hardy space Hp(X1 ×X2) is independent of the choice of the approximations to theidentity and the spaces of distributions with βi and γi satisfying (4.11), where i = 1, 2.

Proof. We first verify that the definition of the Hardy space Hp(X1 ×X2) is independentof the choice of approximations to the identity, which is a corollary of Theorem 2.1. Infact, let all the notation be the same as in Theorem 4.1. Then, Theorem 4.1 tells us that

∥∥∥∥∥∥∥

∞∑

k1=−∞

∞∑

k2=−∞|Dk1Dk2(f)|2

1/2∥∥∥∥∥∥∥

Lp(X1×X2)

≤∥∥∥∥∥∥

∞∑

k1=−∞

∑

τ1∈Ik1

N(k1,τ1)∑

ν1=1

∞∑

k2=−∞

∑

τ2∈Ik2

N(k2,τ2)∑

ν2=1

supz1∈Q

k1,ν1τ1

,z2∈Qk2,ν2τ2


×χQ

k1,ν1τ1

(·)χQ

k2,ν2τ2

(·)1/q∥∥∥∥∥

Lp(X1×X2)


≤ C

∥∥∥∥∥∥

∞∑

k1=−∞

∑

τ1∈Ik1

N(k1,τ1)∑

ν1=1

∞∑

k2=−∞

∑

τ2∈Ik2

N(k2,τ2)∑

ν2=1

× infz1∈Q

k1,ν1τ1

,z2∈Qk2,ν2τ2

|Ek1Ek2(f)(z1, z2)|q χQ

k1,ν1τ1

(·)χQ

k2,ν2τ2

(·)1/q∥∥∥∥∥

Lp(X1×X2)

≤

∥∥∥∥∥∥∥

∞∑

k1=−∞

∞∑

k2=−∞|Ek1Ek2(f)|2

1/2∥∥∥∥∥∥∥

Lp(X1×X2)

.

By the symmetry, we further obtain∥∥∥∥∥∥∥

∞∑

k1=−∞

∞∑

k2=−∞|Dk1Dk2(f)|2

1/2∥∥∥∥∥∥∥

Lp(X1×X2)

∼

∥∥∥∥∥∥∥

∞∑

k1=−∞

∞∑

k2=−∞|Ek1Ek2(f)|2

1/2∥∥∥∥∥∥∥

Lp(X1×X2)

.

Thus, the definition of the Hardy space Hp(X1 ×X2) is independent of the choice ofapproximations to the identity.

Let βi, γi and β′i, γ′i for i = 1, 2 both satisfy (4.11) and f ∈(G(β1, β2; γ1, γ2)

)′

with ‖g2(f)‖Lp(X1×X2) < ∞. We now verify that f ∈(G(β1, β2; γ1, γ2)

)′. To this end, let

ψ ∈ G(ε1, ε2) and the notation be the same as in Theorem 2.7. Let γ′1 ∈ (0, γ1, γ′2 ∈ (0, γ2)and y1 ∈ X1, y2 ∈ X2. We claim that for any k1, k2 ∈ Z+,

(4.12)∣∣∣⟨Dk1(·1, y1)Dk2(·2, y2), ψ

⟩∣∣∣

≤ C2−k1β1−k2β2‖ψ‖G(β1,β2;γ1,γ2)1

(1 + ρ1(y1, x0))d1+γ1

1(1 + ρ2(y2, y0))d2+γ2

;

for any k1 ∈ Z+ and any k2 ∈ Z \ Z+,

(4.13)∣∣∣⟨Dk1(·1, y1)Dk2(·2, y2), ψ

⟩∣∣∣

≤ C2k2γ′2−k1β1‖ψ‖G(β1,β2;γ1,γ2)1

(1 + ρ1(y1, x0))d1+γ1

2−k2γ2

(2−k2 + ρ2(y2, y0))d2+γ2;

for any k1 ∈ Z \ Z+ and any k2 ∈ Z+,

(4.14)∣∣∣⟨Dk1(·1, y1)Dk2(·2, y2), ψ

⟩∣∣∣

≤ C2k1γ′1−k2β2‖ψ‖G(β1,β2;γ1,γ2)2−k1γ1

(2−k1 + ρ1(y1, x0))d1+γ1

1(1 + ρ2(y2, y0))d2+γ2

;


and for any k1, k2 ∈ Z \ Z+,

(4.15)∣∣∣⟨Dk1(·1, y1)Dk2(·2, y2), ψ

⟩∣∣∣

≤ C2k1γ′1+k2γ′2‖ψ‖G(β1,β2;γ1,γ2)2−k1γ1

(2−k1 + ρ1(y1, x0))d1+γ1

× 2−k2γ2

(2−k2 + ρ2(y2, y0))d2+γ2.

We first verify (4.12). In this case, we have

∣∣∣⟨Dk1(·1, y1)Dk2(·2, y2), ψ

⟩∣∣∣

=∣∣∣∣∫

X1×X2

Dk1(z1, y1)Dk2(z2, y2)ψ(z1, z2) dµ1(z1) dµ2(z2)∣∣∣∣

=∣∣∣∣∫

X1×X2

Dk1(z1, y1)Dk2(z2, y2)

×[ψ(z1, z2)− ψ(y1, z2)]− [ψ(z1, y2)− ψ(y1, y2)] dµ1(z1) dµ2(z2)

∣∣∣∣∣

≤∫

ρ1(z1,y1)≤ 12A1

(1+ρ1(y1,x0))

ρ2(z2,y2)≤ 12A2

(1+ρ2(y2,y0))

∣∣∣Dk1(z1, y1)Dk2(z2, y2)∣∣∣

× |[ψ(z1, z2)− ψ(y1, z2)]− [ψ(z1, y2)− ψ(y1, y2)]| dµ1(z1) dµ2(z2)

+∫

ρ1(z1,y1)≤ 12A1

(1+ρ1(y1,x0))

ρ2(z2,y2)> 12A2

(1+ρ2(y2,y0))

· · ·+∫

ρ1(z1,y1)> 12A1

(1+ρ1(y1,x0))

ρ2(z2,y2)≤ 12A2

(1+ρ2(y2,y0))

· · ·

+∫

ρ1(z1,y1)> 12A1

(1+ρ1(y1,x0))

ρ2(z2,y2)> 12A2

(1+ρ2(y2,y0))

· · ·

= Q11 + Q12 + Q13 + Q14.

For Q11, by the second difference condition (iv) satisfied by ψ as in Definition 2.4,we have

Q11≤ C‖ψ‖G(β1,β2;γ1,γ2)

∫ρ1(z1,y1)≤ 1

2A1(1+ρ1(y1,x0))

ρ2(z2,y2)≤ 12A2

(1+ρ2(y2,y0))

∣∣∣Dk1(z1, y1)Dk2(z2, y2)∣∣∣

×(

ρ1(z1, y1)1 + ρ1(y1, x0)

)β1 1(1 + ρ1(y1, x0))d1+γ1

×(

ρ1(z2, y2)1 + ρ2(y2, y0)

)β2 1(1 + ρ2(y2, y0))d2+γ2

dµ1(z1) dµ2(z2)

≤ C2−k1β1−k2β2‖ψ‖G(β1,β2;γ1,γ2)1

(1 + ρ1(y1, x0))d1+γ1

1(1 + ρ2(y2, y0))d2+γ2

,


which is a desired estimate.Definition 2.4 (ii) tells us that

Q12≤ C‖ψ‖G(β1,β2;γ1,γ2)

∫ρ1(z1,y1)≤ 1

2A1(1+ρ1(y1,x0))

ρ2(z2,y2)> 12A2

(1+ρ2(y2,y0))

∣∣∣Dk1(z1, y1)Dk2(z2, y2)∣∣∣

×(

ρ1(z1, y1)1 + ρ1(y1, x0)

)β1 1(1 + ρ1(y1, x0))d1+γ1

×

1(1 + ρ2(z2, y0))d2+γ2

+1

(1 + ρ2(y2, y0))d2+γ2

dµ1(z1) dµ2(z2)

≤ C2−k1β1−k2β2‖ψ‖G(β1,β2;γ1,γ2)1

(1 + ρ1(y1, x0))d1+γ1

1(1 + ρ2(y2, y0))d2+γ2

,

which is also a desired estimate.The estimate for Q13 is similar to that for Q12. We omit the details.For Q14, the size condition satisfied by ψ tells us that

Q14≤ C‖ψ‖G(β1,β2;γ1,γ2)

∫ρ1(z1,y1)> 1

2A1(1+ρ1(y1,x0))

ρ2(z2,y2)> 12A2

(1+ρ2(y2,y0))

∣∣∣Dk1(z1, y1)Dk2(z2, y2)∣∣∣

×

1(1 + ρ2(z2, y0))d2+γ2

[1

(1 + ρ1(z1, x0))d1+γ1+

1(1 + ρ1(y1, x0))d1+γ1

]

+1

(1 + ρ2(y2, y0))d2+γ2

×[

1(1 + ρ1(z1, x0))d1+γ1

+1

(1 + ρ1(y1, x0))d1+γ1

]dµ1(z1) dµ2(z2)

≤ C2−k1β1−k2β2‖ψ‖G(β1,β2;γ1,γ2)1

(1 + ρ1(y1, x0))d1+γ1

1(1 + ρ2(y2, y0))d2+γ2

,

which completes the proof of (4.12).We now prove (4.13). In this case, we write

∣∣∣⟨Dk1(·1, y1)Dk2(·2, y2), ψ

⟩∣∣∣

=∣∣∣∣∫

X1×X2


=∣∣∣∣∫

X1×X2

Dk1(z1, y1)[Dk2(z2, y2)− Dk2(y0, y2)

]

× [ψ(z1, z2)− ψ(y1, z2)] dµ1(z1) dµ2(z2)

∣∣∣∣∣

≤∫

ρ1(z1,y1)≤ 12A1

(1+ρ1(y1,x0))

ρ2(z2,y0)≤ 12A2

(2−k2+ρ2(y2,y0))

∣∣∣Dk1(z1, y1)∣∣∣∣∣∣Dk2(z2, y2)− Dk2(y0, y2)

∣∣∣


× |ψ(z1, z2)− ψ(y1, z2)| dµ1(z1) dµ2(z2)

+∫

ρ1(z1,y1)≤ 12A1

(1+ρ1(y1,x0))

ρ2(z2,y0)> 12A2

(2−k2+ρ2(y2,y0))

· · ·+∫

ρ1(z1,y1)> 12A1

(1+ρ1(y1,x0))

ρ2(z2,y0)≤ 12A2

(2−k2+ρ2(y2,y0))

· · ·

+∫

ρ1(z1,y1)> 12A1

(1+ρ1(y1,x0))

ρ2(z2,y0)> 12A2

(2−k2+ρ2(y2,y0))

· · ·

= Q21 + Q22 + Q23 + Q24.

The regularity of Dk2 and ψ and the size condition of Dk1 yield that

Q21≤ C‖ψ‖G(β1,β2;γ1,γ2)

∫ρ1(z1,y1)≤ 1

2A1(1+ρ1(y1,x0))

ρ2(z2,y0)≤ 12A2

(2−k2+ρ2(y2,y0))

∣∣∣Dk1(z1, y1)∣∣∣

×(

ρ2(z2, y0)2−k2 + ρ2(y0, y2)

)γ′2 2−k2ε′2

(2−k2 + ρ2(y0, y2))d2+ε′2

×(

ρ1(z1, y1)1 + ρ1(y1, x0)

)β1 1(1 + ρ1(y1, x0))d1+γ1

× 1(1 + ρ2(z2, y0))d2+γ2

dµ1(z1) dµ2(z2)

≤ C2k2γ′2−k1β1‖ψ‖G(β1,β2;γ1,γ2)1

(1 + ρ1(y1, x0))d1+γ1

2−k2γ2

(2−k2 + ρ2(y2, y0))d2+γ2,

which is a desired estimate.Similarly we have

Q22≤ C‖ψ‖G(β1,β2;γ1,γ2)

∫ρ1(z1,y1)≤ 1

2A1(1+ρ1(y1,x0))

ρ2(z2,y0)> 12A2

(2−k2+ρ2(y2,y0))

∣∣∣Dk1(z1, y1)∣∣∣

×

2−k2ε′2

(2−k2 + ρ2(y2, z2))d2+ε′2+

2−k2ε′2

(2−k2 + ρ2(y0, y2))d2+ε′2

×(

ρ1(z1, y1)1 + ρ1(y1, x0)

)β1 1(1 + ρ1(y1, x0))d1+γ1

× 1(1 + ρ2(z2, y0))d2+γ2

dµ1(z1) dµ2(z2)

≤ C2k2γ′2−k1β1‖ψ‖G(β1,β2;γ1,γ2)1

(1 + ρ1(y1, x0))d1+γ1

2−k2γ2

(2−k2 + ρ2(y2, y0))d2+γ2,

which is also a desired estimate.The estimate for Q23 is similar to that for Q22 by symmetry.


Finally the size conditions of Dk1 , Dk2 and ψ imply that

Q24≤ C‖ψ‖G(β1,β2;γ1,γ2)

∫ρ1(z1,y1)> 1

2A1(1+ρ1(y1,x0))

ρ2(z2,y0)> 12A2

(2−k2+ρ2(y2,y0))

∣∣∣Dk1(z1, y1)∣∣∣

×

2−k2ε′2

(2−k2 + ρ2(y2, z2))d2+ε′2+

2−k2ε′2

(2−k2 + ρ2(y0, y2))d2+ε′2

×

1(1 + ρ1(z1, x0))d1+γ1

+1

(1 + ρ1(y1, x0))d1+γ1

× 1(1 + ρ2(z2, y0))d2+γ2

dµ1(z1) dµ2(z2)

≤ C2k2γ′2−k1β1‖ψ‖G(β1,β2;γ1,γ2)1

(1 + ρ1(y1, x0))d1+γ1

2−k2γ2

(2−k2 + ρ2(y2, y0))d2+γ2,

which completes the proof of (4.13).The verification of (4.14) is similar to that for (4.13) by symmetry.We now prove (4.15). Write

∣∣∣⟨Dk1(·1, y1)Dk2(·2, y2), ψ

⟩∣∣∣

=∣∣∣∣∫

X1×X2


=∣∣∣∣∫

X1×X2

[Dk1(z1, y1)− Dk1(x0, y1)

] [Dk2(z2, y2)− Dk2(y0, y2)

]

×ψ(z1, z2) dµ1(z1) dµ2(z2)

∣∣∣∣∣

≤∫

ρ1(z1,x0)≤ 12A1

(2−k1+ρ1(y1,x0))

ρ2(z2,y0)≤ 12A2

(2−k2+ρ2(y2,y0))

∣∣∣Dk1(z1, y1)− Dk1(x0, y1)∣∣∣∣∣∣Dk2(z2, y2)− Dk2(y0, y2)

∣∣∣

× |ψ(z1, z2)| dµ1(z1) dµ2(z2) +∫

ρ1(z1,x0)≤ 12A1

(2−k1+ρ1(y1,x0))

ρ2(z2,y0)> 12A2

(2−k2+ρ2(y2,y0))

· · ·

+∫

ρ1(z1,x0)> 12A1

(2−k1+ρ1(y1,x0))

ρ2(z2,y0)≤ 12A2

(2−k2+ρ2(y2,y0))

· · ·+∫

ρ1(z1,x0)> 12A1

(2−k1+ρ1(y1,x0))

ρ2(z2,y0)> 12A2

(2−k2+ρ2(y2,y0))

· · ·

= Q31 + Q32 + Q33 + Q34.

The regularity satisfied by Dk1 and Dk2 and the size condition satisfied by ψ tell usthat

Q31≤ C‖ψ‖G(β1,β2;γ1,γ2)

∫ρ1(z1,x0)≤ 1

2A1(2−k1+ρ1(y1,x0))

ρ2(z2,y0)≤ 12A2

(2−k2+ρ2(y2,y0))

(ρ1(z1, x0)

2−k1 + ρ1(x0, y1)

)γ′1


× 2−k1ε′1

(2−k1 + ρ1(x0, y1))d1+ε′1

(ρ2(z2, y0)

2−k2 + ρ2(y0, y2)

)γ′2 2−k2ε′2

(2−k2 + ρ2(y0, y2))d2+ε′2

× 1(1 + ρ1(z1, x0))d1+γ1

1(1 + ρ2(z2, y0))d2+γ2

dµ1(z1) dµ2(z2)

≤ C2k1γ′1+k2γ′2‖ψ‖G(β1,β2;γ1,γ2)2−k1γ1

(2−k1 + ρ1(y1, x0))d1+γ1

× 2−k2γ2

(2−k2 + ρ2(y2, y0))d2+γ2,

which is a desired estimate.Similarly, for Q32, we have

Q32≤ C‖ψ‖G(β1,β2;γ1,γ2)

∫ρ1(z1,x0)≤ 1

2A1(2−k1+ρ1(y1,x0))

ρ2(z2,y0)> 12A2

(2−k2+ρ2(y2,y0))

(ρ1(z1, x0)

2−k1 + ρ1(x0, y1)

)γ′1

× 2−k1ε′1

(2−k1 + ρ1(x0, y1))d1+ε′1

2−k2ε′2

(2−k2 + ρ2(z2, y2))d2+ε′2+

2−k2ε′2

(2−k2 + ρ2(y0, y2))d2+ε′2

× 1(1 + ρ1(z1, x0))d1+γ1

1(1 + ρ2(z2, y0))d2+γ2

dµ1(z1) dµ2(z2)

≤ C2k1γ′1+k2γ′2‖ψ‖G(β1,β2;γ1,γ2)2−k1γ1

(2−k1 + ρ1(y1, x0))d1+γ1

× 2−k2γ2

(2−k2 + ρ2(y2, y0))d2+γ2,

which is also a desired estimate.The symmetry of Q33 with Q32 implies a desired estimate for Q33.We now estimate Q34 by

Q34≤ C‖ψ‖G(β1,β2;γ1,γ2)

∫ρ1(z1,x0)> 1

2A1(2−k1+ρ1(y1,x0))

ρ2(z2,y0)> 12A2

(2−k2+ρ2(y2,y0))

2−k1ε′1

(2−k1 + ρ1(z1, y1))d1+ε′1

+2−k1ε′1

(2−k1 + ρ1(x0, y1))d1+ε′1

×

2−k2ε′2

(2−k2 + ρ2(z2, y2))d2+ε′2+

2−k2ε′2

(2−k2 + ρ2(y0, y2))d2+ε′2

× 1(1 + ρ1(z1, x0))d1+γ1

1(1 + ρ2(z2, y0))d2+γ2

dµ1(z1) dµ2(z2)

≤ C2k1γ′1+k2γ′2‖ψ‖G(β1,β2;γ1,γ2)2−k1γ1

(2−k1 + ρ1(y1, x0))d1+γ1

× 2−k2γ2

(2−k2 + ρ2(y2, y0))d2+γ2,

which verified (4.15).


Theorem 2.7 now tells us that

|〈f, ψ〉|=∣∣∣∣∣∣

∞∑

k1=−∞

∑

τ1∈Ik1

N(k1,τ1)∑

ν1=1

∞∑

k2=−∞

∑

τ2∈Ik2

N(k2,τ2)∑

ν2=1


τ2 )

×Dk1Dk2(f)(yk1,ν1τ1 , yk2,ν2

τ2 )⟨Dk1(·1, yk1,ν1

τ1 )Dk2(·2, yk2,ν2τ2 ), ψ

⟩ ∣∣∣∣∣

≤∞∑

k1=0

∑

τ1∈Ik1

N(k1,τ1)∑

ν1=1

∞∑

k2=0

∑

τ2∈Ik2

N(k2,τ2)∑

ν2=1


τ2 )

×∣∣∣Dk1Dk2(f)(yk1,ν1

τ1 , yk2,ν2τ2 )

∣∣∣∣∣∣⟨Dk1(·1, yk1,ν1

τ1 )Dk2(·2, yk2,ν2τ2 ), ψ

⟩∣∣∣

+∞∑

k1=0

∑

τ1∈Ik1

N(k1,τ1)∑

ν1=1

−1∑

k2=−∞

∑

τ2∈Ik2

N(k2,τ2)∑

ν2=1

· · ·

+−1∑

k1=−∞

∑

τ1∈Ik1

N(k1,τ1)∑

ν1=1

∞∑

k2=0

∑

τ2∈Ik2

N(k2,τ2)∑

ν2=1

· · ·

+−1∑

k1=−∞

∑

τ1∈Ik1

N(k1,τ1)∑

ν1=1

−1∑

k2=−∞

∑

τ2∈Ik2

N(k2,τ2)∑

ν2=1

· · ·

= Q41 + Q42 + Q43 + Q44.

If p ≤ 1, the estimate (4.12), the inequality (4.6), the arbitrariness of yk1,ν1τ1 and

yk2,ν2τ2 , the assumption on β1 and β2 and the Holder inequality imply that

Q41≤ C‖ψ‖G(β1,β2;γ1,γ2)

∞∑

k1=0

∑

τ1∈Ik1

N(k1,τ1)∑

ν1=1

∞∑

k2=0

∑

τ2∈Ik2

N(k2,τ2)∑

ν2=1

2−k1β1−k2β2

×µ1(Qk1,ν1τ1 )µ2(Q

k2,ν2τ2 )

∣∣∣Dk1Dk2(f)(yk1,ν1τ1 , yk2,ν2

τ2 )∣∣∣

× 1

(1 + ρ1(yk1,ν1τ1 , x0))d1+γ1

1

(1 + ρ2(yk2,ν2τ2 , y0))d2+γ2

≤ C‖ψ‖G(β1,β2;γ1,γ2)

∞∑

k1=0

∞∑

k2=0

2−k1β1−k1d1(1−1/p)−k2β2−k2d2(1−1/p)

× ∑

τ1∈Ik1

N(k1,τ1)∑

ν1=1

∑

τ2∈Ik2

N(k2,τ2)∑

ν2=1


τ2 )

×∣∣∣Dk1Dk2(f)(yk1,ν1

τ1 , yk2,ν2τ2 )

∣∣∣p]1/p

≤ C‖ψ‖G(β1,β2;γ1,γ2)

∞∑

k1=0

∞∑

k2=0

‖Dk1Dk2(f)‖2Lp(X1×X2)

1/2

;


while if p > 1, similarly we have

Q41≤ C‖ψ‖G(β1,β2;γ1,γ2)

∞∑

k1=0

∑

τ1∈Ik1

N(k1,τ1)∑

ν1=1

∞∑

k2=0

∑

τ2∈Ik2

N(k2,τ2)∑

ν2=1

2−k1β1−k2β2

×µ1(Qk1,ν1τ1 )µ2(Q

k2,ν2τ2 )


τ2 )∣∣∣p

× 1

(1 + ρ1(yk1,ν1τ1 , x0))d1+γ1

1

(1 + ρ2(yk2,ν2τ2 , y0))d2+γ2

1/p

×

∞∑

k1=0

∞∑

k2=0

2−k1β1−k2β2

∫

X1×X2

1(1 + ρ1(y1, x0))d1+γ1

× 1(1 + ρ2(y2, y0))d2+γ2

dµ1(y1) dµ2(y2)1/p′

≤ C‖ψ‖G(β1,β2;γ1,γ2)

∞∑

k1=0

∞∑

k2=0

2−k1β1−k2β2 ‖Dk1Dk2(f)‖pLp(X1×X2)

1/p

≤ C‖ψ‖G(β1,β2;γ1,γ2)

∞∑

k1=0

∞∑

k2=0

‖Dk1Dk2(f)‖max(p,2)Lp(X1×X2)

1/ max(p,2)

.

Thus, we always have

(4.16) Q41 ≤ C‖ψ‖G(β1,β2;γ1,γ2)

∞∑

k1=0

∞∑

k2=0


1/ max(p,2)

.

If p ≤ 1, instead of (4.12) by (4.13), similarly to the estimate for Q41, we have

Q42≤ C‖ψ‖G(β1,β2;γ1,γ2)

∞∑

k1=0

∑

τ1∈Ik1

N(k1,τ1)∑

ν1=1

−1∑

k2=−∞

∑

τ2∈Ik2

N(k2,τ2)∑

ν2=1

2k2γ′2−k1β1

×µ1(Qk1,ν1τ1 )µ2(Q

k2,ν2τ2 )


τ2 )∣∣∣

× 1

(1 + ρ1(yk1,ν1τ1 , x0))d1+γ1

2−k2γ2

(2−k2 + ρ2(yk2,ν2τ2 , y0))d2+γ2

≤ C‖ψ‖G(β1,β2;γ1,γ2)

∞∑

k1=0

−1∑

k2=−∞2k2γ′2−k1β1−k1d1(1−1/p)+k2d2/p

× ∑

τ1∈Ik1

N(k1,τ1)∑

ν1=1

∑

τ2∈Ik2

N(k2,τ2)∑

ν2=1


τ2 )∣∣∣Dk1Dk2(f)(yk1,ν1

τ1 , yk2,ν2τ2 )

∣∣∣p

1/p

≤ C‖ψ‖G(β1,β2;γ1,γ2)

∞∑

k1=0

−1∑

k2=−∞‖Dk1Dk2(f)‖2

Lp(X1×X2)

1/2

;



Q42≤ C‖ψ‖G(β1,β2;γ1,γ2)

∞∑

k1=0

∑

τ1∈Ik1

N(k1,τ1)∑

ν1=1

−1∑

k2=−∞

∑

τ2∈Ik2

N(k2,τ2)∑

ν2=1

2k2γ′2−k1β1

×µ1(Qk1,ν1τ1 )µ2(Q

k2,ν2τ2 )


τ2 )∣∣∣p

× 1

(1 + ρ1(yk1,ν1τ1 , x0))d1+γ1

2−k2γ2

(2−k2 + ρ2(yk2,ν2τ2 , y0))d2+γ2

1/p

×

∞∑

k1=0

−1∑

k2=−∞2k2γ′2−k1β1

∫

X1×X2

1(1 + ρ1(y1, x0))d1+γ1

× 2−k2γ2

(2−k2 + ρ2(y2, y0))d2+γ2dµ1(y1) dµ2(y2)

1/p′

≤ C‖ψ‖G(β1,β2;γ1,γ2)

∞∑

k1=0

−1∑

k2=−∞2k2γ′2−k1β1‖Dk1Dk2(f)‖p

Lp(X1×X2)

1/p

≤ C‖ψ‖G(β1,β2;γ1,γ2)

∞∑

k1=0

−1∑

k2=−∞‖Dk1Dk2(f)‖max(p,2)

Lp(X1×X2)

1/ max(p,2)

.


(4.17) Q42 ≤ C‖ψ‖G(β1,β2;γ1,γ2)

∞∑

k1=0

−1∑


Lp(X1×X2)

1/ max(p,2)

.

By instead of (4.13) by (4.14) and the symmetry with Q42, we can verify that

(4.18) Q43 ≤ C‖ψ‖G(β1,β2;γ1,γ2)

−1∑

k1=−∞

∞∑

k2=0


1/ max(p,2)

.

If p ≤ 1, the estimate (4.15) and some similar computation to the above yield that

Q44≤ C‖ψ‖G(β1,β2;γ1,γ2)

−1∑

k1=−∞

∑

τ1∈Ik1

N(k1,τ1)∑

ν1=1

−1∑

k2=−∞

∑

τ2∈Ik2

N(k2,τ2)∑

ν2=1

2k1γ′1+k2γ′2

×µ1(Qk1,ν1τ1 )µ2(Q

k2,ν2τ2 )


τ2 )∣∣∣

× 2−k1γ1

(2−k1 + ρ1(yk1,ν1τ1 , x0))d1+γ1

2−k2γ2

(2−k2 + ρ2(yk2,ν2τ2 , y0))d2+γ2


≤ C‖ψ‖G(β1,β2;γ1,γ2)

−1∑

k1=−∞

−1∑

k2=−∞2k1γ′1+k1d1/p+k2γ′2+k2d2/p

× ∑

τ1∈Ik1

N(k1,τ1)∑

ν1=1

∑

τ2∈Ik2

N(k2,τ2)∑

ν2=1


τ2 )∣∣∣Dk1Dk2(f)(yk1,ν1

τ1 , yk2,ν2τ2 )

∣∣∣p

1/p

≤ C‖ψ‖G(β1,β2;γ1,γ2)

−1∑

k1=−∞

−1∑

k2=−∞‖Dk1Dk2(f)‖2

Lp(X1×X2)

1/2

;


Q44≤ C‖ψ‖G(β1,β2;γ1,γ2)

−1∑

k1=−∞

∑

τ1∈Ik1

N(k1,τ1)∑

ν1=1

−1∑

k2=−∞

∑

τ2∈Ik2

N(k2,τ2)∑

ν2=1

2k1γ′1+k2γ′2

×µ1(Qk1,ν1τ1 )µ2(Q

k2,ν2τ2 )


τ2 )∣∣∣p

× 2−k1γ1

(2−k1 + ρ1(yk1,ν1τ1 , x0))d1+γ1

2−k2γ2

(2−k2 + ρ2(yk2,ν2τ2 , y0))d2+γ2

1/p

×

−1∑

k1=−∞

−1∑

k2=−∞2k1γ′1+k2γ′2

∫

X1×X2

2−k1γ1

(2−k1 + ρ1(y1, x0))d1+γ1

× 2−k2γ2

(2−k2 + ρ2(y2, y0))d2+γ2dµ1(y1) dµ2(y2)

1/p

≤ C‖ψ‖G(β1,β2;γ1,γ2)

−1∑

k1=−∞

−1∑

k2=−∞2k1γ′1+k2γ′2 ‖Dk1Dk2(f)‖p

Lp(X1×X2)

1/p

≤ C‖ψ‖G(β1,β2;γ1,γ2)

−1∑

k1=−∞

−1∑


Lp(X1×X2)

1/ max(p,2)

.


(4.19) Q44 ≤ C‖ψ‖G(β1,β2;γ1,γ2)

−1∑

k1=−∞

−1∑


Lp(X1×X2)

1/ max(p,2)

.

Combining (4.16), (4.17), (4.18) and (4.19) tells us that

(4.20) |〈f, ψ〉| ≤ C‖ψ‖G(β1,β2;γ1,γ2)

∞∑

k1=−∞

∞∑


Lp(X1×X2)

1/ max(p,2)

≤ C‖ψ‖G(β1,β2;γ1,γ2)

∥∥∥∥∥∥∥

∞∑

k1=−∞

∞∑

k2=−∞|Dk1Dk2(f)|2

1/2∥∥∥∥∥∥∥

Lp(X1×X2)

≤ C‖ψ‖G(β1,β2;γ1,γ2) ‖g2(f)‖Lp(X1×X2) ,


where in the second equality we used the Minkowski inequality on the series and integralif p ≤ 2 and we used (4.6) if p ≥ 2.

Suppose now ψ ∈ G(β1, β2; γ1, γ2). Then there is a sequence of test functions,ψnn∈N, hn ∈ G(ε1, ε2) such that

‖ψn − ψ‖G(β1,β2;γ1,γ2) → 0

as n →∞. The estimate (4.20) then implies that for any n, m ∈ N,

|〈f, ψn − ψm〉| ≤ C ‖g2(f)‖Lp(X1×X2) ‖ψn − ψm‖G(β1,β2;γ1,γ2),

which shows limn→∞〈f, ψn〉 exists and the limit is independent of the choice of ψnn∈N.Therefore, we define

〈f, ψ〉 = limn→∞〈f, ψn〉.

Then, the estimate (4.20) again tells us that for all ψ ∈ G(β1, β2; γ1, γ2),

|〈f, ψ〉| ≤ C ‖g2(f)‖Lp(X1×X2) ‖ψ‖G(β1,β2;γ1,γ2),

which indicates that f ∈(G(β1, β2; γ1, γ2)

)′and we complete the proof of Proposition 4.1.

Thus, Definition 4.1 is reasonable by Proposition 4.1. We remark that in the proofof Proposition 4.1, we actually only require that 0 < γi < εi for i = 1, 2. However, if γi

and βi for i = 1, 2 are as in (4.11), we then can verify that the space of test functions,G(β1, β2; γ1, γ2), is contained in the Hardy space Hp(X1 ×X2). To be precise, we have thefollowing proposition.

Proposition 4.2 Let p and the space Hp(X1 ×X2) be the same as in Definition 4.1. If0 < βi < εi and di(1/p− 1)+ < γi < εi for i = 1, 2, then

G(β1, β2; γ1, γ2) ⊂ Hp(X1 ×X2).

Proof. Let ψ ∈ G(β1, β2; γ1, γ2) and γ′i ∈ (0, γi) such that di(1/p − 1)+ < γ′i < εi fori = 1, 2. It is easy to see that the estimates (4.12), (4.13), (4.14) and (4.15) with Dki

replaced by Dki for i = 1, 2 still holds. By these estimates and a proof similar to that for(4.20), we obtain

‖ψ‖Hp(X1×X2) =

∥∥∥∥∥∥∥

∞∑

k1=−∞

∞∑

k2=−∞|Dk1Dk2(f)|2

1/2∥∥∥∥∥∥∥

Lp(X1×X2)

≤ C

∞∑

k1=−∞

∞∑

k2=−∞‖Dk1Dk2(f)‖min(p,2)

Lp(X1×X2)

1/ min(p,2)


≤ C‖ψ‖G(β1,β2;γ1,γ2)

∞∑

k1=0

∞∑

k2=0

2−(k1β1+k2β2)min(p,2)

+∞∑

k1=0

−1∑

k2=−∞2[k2γ′2+k2d2(1−1/p)−k1β1]min(p,2)

+−1∑

k1=−∞

∞∑

k2=0

2[k1γ′1+k1d1(1−1/p)−k2β2]min(p,2)

+−1∑

k1=−∞

−1∑

k2=−∞2[k1γ′1+k1d1(1−1/p)+k2γ′2+k2d2(1−1/p)]min(p,2)

1/ min(p,2)

≤ C‖ψ‖G(β1,β2;γ1,γ2),

which completes the proof of Proposition 4.2.

From Theorem 3.1, it is easy to deduce the following result.

Proposition 4.3 If 1 < p < ∞, then the space Hp(X1 ×X2) is the same space as thespace Lp(X1 ×X2) with an equivalent norm.

Theorem 4.2 tells the following fact.

Proposition 4.4 Let p and the space Hp(X1 ×X2) be the same as in Definition 4.1,and S2 be defined as in (6.32) with q = 2. If βi and γi with i = 1, 2 are as in (4.11),

then f ∈ Hp(X1 ×X2) if and only if f ∈(G(β1, β2; γ1, γ2)

)′and S2(f) ∈ Lp(X1 ×X2).

Moreover,

‖f‖Hp(X1×X2) ∼ ‖S2(f)‖Lp(X1×X2) .

We now use Proposition 4.4 to obtain the atomic decomposition of the Hardy spaceHp(X1 ×X2). First, we need to establish Journe’s covering lemma in the setting ofhomogeneous-type spaces.

We recall some notation. Let Qkiαi⊂ Xi : ki ∈ Z, αi ∈ Iki for i = 1, 2 be the same

as in Lemma 2.5. Then the open set Qk1α1×Qk2

α2for k1, k2 ∈ Z, α1 ∈ Ik1 and α2 ∈ Ik2 is

called a dyadic rectangle of X1 ×X2. Let Ω ⊂ X1 ×X2 be an open set of finite measureand Mi(Ω) denote the family of dyadic rectangles R ⊂ Ω which are maximal in the xi

“direction”, where i = 1, 2. In what follows, we denote by R = B1 × B2 any dyadicrectangle of X1 ×X2. Given R = B1 × B2 ∈ M1(Ω), let B2 = B2(B1) be the “longest”dyadic cube containing B2 such that

(4.21) (µ1 × µ2) (B1 × B2 ∩ Ω) >12

(µ1 × µ2) (B1 × B2);


and given R = B1×B2 ∈M2(Ω), let B1 = B1(B2) be the “longest” dyadic cube containingB1 such that

(4.22) (µ1 × µ2) (B1 ×B2 ∩ Ω) >12

(µ1 × µ2) (B1 ×B2).

If Bi = Qkiαi⊂ Xi for some ki ∈ Z and some αi ∈ Iki , (Bi)k for k ∈ N is used to denote

any dyadic cube Qki−kαi

containing Qkiαi

and (Bi)0 = Bi, where i = 1, 2. Also, let w(x)be any increasing function such that

∑∞j=0 jw(C132−j) < ∞, where C13 > 0 is any given

constant. In particular, we may take w(x) = xδ for any δ > 0.The main idea of the following variant of Journe’s covering lemma in the setting of

homogeneous type comes from Pipher [P].

Lemma 4.2 Assume that Ω ⊂ X1 × X2 is an open set with finite measure. Let all thenotation be the same as above and µ = µ1 × µ2. Then

(4.23)∑

R=B1×B2∈M1(Ω)

µ(R)w

(µ2(B2)

µ2(B2)

)≤ Cµ(Ω)

and

(4.24)∑

R=B1×B2∈M2(Ω)

µ(R)w

(µ1(B1)

µ1(B1)

)≤ Cµ(Ω).

Proof. We only verify (4.24) and the proof of (4.23) is similar. Let R = B1×B2 ∈M2(Ω)and for k ∈ N, let

(4.25) AB1,k =⋃

B2 : B1 ×B2 ∈M2(Ω) and B1 = (B1)k−1

.

Then

(4.26)∑

R=B1×B2∈M2(Ω)

µ(R)w

(µ1(B1)

µ1(B1)

)

=∑

R=B1×B2∈M2(Ω)

µ1(B1)µ2(B2)w

(µ1(B1)

µ1(B1)

)

=∑

B1: B1×B2∈M2(Ω)µ1(B1)

∞∑

k=1

∑

B2: B2∈AB1,kµ2(B2)w

(µ1(B1)

µ1(B1)

)

≤∑

B1: B1×B2∈M2(Ω)µ1(B1)

∞∑

k=1

w(C132−k

) ∑

B2: B2∈AB1,kµ2(B2)

=∑

B1: B1×B2∈M2(Ω)µ1(B1)

∞∑

k=1

w(C132−k

)µ2 (AB1,k) ,


since B2 : B2 ∈ AB1,k are disjoint by their “maximality”, where C13 > 0 depends onlyon the constants of µ1 appearing in (2.2) and the constants C10,1 and C10,2 in Lemma 2.5for X1.

Set

EB1(Ω) =⋃B2 : B1 ×B2 ⊂ Ω .

If x2 ∈ AB1,k, then there is some dyadic cube B1 × B2 ∈ M2(Ω) and some k ∈ N suchthat x2 ∈ B2 and B1 = (B1)k−1 by (4.25). By (4.22) and the maximality of B1, we have

µ ((B1)k−1 ×B2 ∩ Ω) >12µ ((B1)k−1 ×B2)

and

µ ((B1)k ×B2 ∩ Ω) ≤ 12µ ((B1)k ×B2) ,

which implies that

µ((B1)k ×B2 ∩

((B1)k × E(B1)k

)) ≤ 12µ ((B1)k ×B2)

and further

µ((B1)k ×

(B2 ∩ E(B1)k

)) ≤ 12µ ((B1)k ×B2) .

Therefore,

µ2

(B2 ∩ E(B1)k

) ≤ 12µ2(B2),

which in turn tells us that

(4.27) µ2

(B2 ∩

(E(B1)k

)c)>

12µ2(B2),

where(E(B1)k

)c = X2 \E(B1)k. From (4.27), it follows that

M2

(χEB1

\E(B1)k

)(x2) >

12

and therefore

AB1,k ⊂

x2 ∈ X2 : M2

(χEB1

\E(B1)k

)(x2) >

12

,

which implies that

(4.28) µ2 (AB1,k)≤ µ2

(x2 ∈ X2 : M2

(χEB1

\E(B1)k

)(x2) >

12

)

≤ Cµ2

(EB1 \ E(B1)k

).


Combining (4.26) with (4.28) yields that

∑

R=B1×B2∈M2(Ω)

µ(R)w

(µ1(B1)

µ1(B1)

)

≤ C∑

B1: B1×B2∈M2(Ω)µ1(B1)

∞∑

k=1

w(C132−k

)µ2

(EB1 \ E(B1)k

)

≤ C∑

B1: B1×B2∈M2(Ω)µ1(B1)

∞∑

k=1

w(C132−k

)

×µ2

(EB1 \E(B1)1

)+ · · ·+ µ2

(E(B1)k−1

\ E(B1)k

)

≤ C∑

B1: B1×B2∈M2(Ω)µ1(B1)

∞∑

k=1

w(C132−k

)

×∑

B0 dyadic cube: B1⊆B0((B1)kB0×(EB0

\E(B0)1)⊂Ω

µ2

(EB0 \E(B0)1

)

≤ C

∞∑

k=1

w(C132−k

) ∑B0 dyadic cube

B0×(EB0\E(B0)1

)⊂Ω

µ1(B0)µ2

(EB0 \E(B0)1

)

×∑

B1 dyadic cube: B1⊂B0⊂(B1)k

µ1(B1)µ1(B0)

≤ C∞∑

k=1

w(C132−k

) ∑B0 dyadic cube

B0×(EB0\E(B0)1

)⊂Ω

µ1(B0)µ2

(EB0 \E(B0)1

)

×k∑

j=1

∑

B1 dyadic cube: µ1(B1)∼2−jdµ1(B0)

µ1(B1)µ1(B0)

≤ C∞∑

k=1

kw(C132−k

) ∑B0 dyadic cube

B0×(EB0\E(B0)1

)⊂Ω

µ1(B0)µ2

(EB0 \E(B0)1

)

≤ C∞∑

k=1

kw(C132−k

)µ(Ω),

since ∑B0 dyadic cube

B0×(EB0\E(B0)1

)⊂Ω

µ1(B0)µ2

(EB0 \ E(B0)1

) ≤ Cµ(Ω)

by noting that the setsB0 ×

(EB0 \ E(B0)1

) ⊂ Ω : B0 is any dyadic cube

are disjoint,which finishes the proof of Lemma 4.2.

We now introduce the Hp(X1 ×X2)-atom. In what follows, for any open set Ω, wedenote by M(Ω) the set of all maximal dyadic rectangles contained in Ω.


Definition 4.2 Let all the notation be the same as in Definition 4.1 and µ = µ1 × µ2. Afunction a(x1, x2) on X1 ×X2 is called a (p, 2)-atom of Hp(X1 ×X2), if it satisfies

(1) supp a ⊂ Ω, where Ω is an open set of X1 ×X2 with finite measure;

(2) a can be further decomposed into

a =∑

R∈M(Ω)

aR,

where

(i) supposing R = Q1 ×Q2 with diamQ1 ∼ 2−k1 and diamQ2 ∼ 2−k2, then

supp aR ⊂ B1(z1, A1

(C2,1 + C1

10,1

)2−k1)×B2(z2, A2

(C2,2 + C2

10,2

)2−k2),

where zi is the center of Qi for i = 1, 2, C110,1 and C2

10,1 mean the constantC10,1 in Lemma 2.5, respectively, for X1 and X2, and C2,1 and C2,2 means theconstant C2 in Definition 2.3, respectively, for X1 and X2.

(ii) for all x1 ∈ X1, ∫

X2

aR(x1, x2) dµ2(x2) = 0

and for all x2 ∈ X2, ∫

X1

aR(x1, x2) dµ1(x1) = 0;

(iii) ‖a‖L2(X1×X2) ≤ µ(Ω)1/2−1/p and

∑

R∈M(Ω)

‖aR‖2L2(X1×X2)

1/2

≤ µ(Ω)1/2−1/p.

Moreover, aR is called an Hp(X1 ×X2) (p, 2)-rectangle atom, if aR satisfies (i), (ii) and

(iv) ‖aR‖L2(X1×X2) ≤ µ(R)1/2−1/p.

The atomic decomposition of the Hardy space Hp(X1 ×X2) is stated in the followingtheorem.

Theorem 4.3 Let i = 1, 2, Xi be a homogeneous-type space as in Definition 2.1, εi ∈(0, θi] and

max

d1

d1 + ε1,

d2

d2 + ε2

< p ≤ 1.


Then f ∈ Hp(X1 ×X2) if and only if f ∈(G(β1, β2; γ1, γ2)

)′for some βi, γi satisfying

(4.11), where i = 1, 2, and there is a sequence of numbers, λkk∈Z, and a sequence of(p, 2)-atoms of Hp(X1 ×X2), akk∈Z, such that

∑∞k=−∞ |λk|p < ∞ and

f =∞∑

k=−∞λkak

in(G(β1, β2; γ1, γ2)

)′. Moreover, in this case,

‖f‖Hp(X1×X2) ∼ inf

[ ∞∑

k=−∞|λk|p

]1/p ,

where the infimum is taken over all the decompositions as above.

Proof. Let f ∈ Hp(X1 ×X2). By Definition 4.1, f ∈(G(β1, β2; γ1, γ2)

)′for some βi, γi

satisfying (4.11), where i = 1, 2. We use Theorem 2.5 and Proposition 4.4 to get the atomicdecomposition of f . To this end, for i = 1, 2, let Skiki∈Z be an approximation to theidentity of order εi having compact support as in Definition 2.3 on space of homogeneoustype, Xi, and Dki = Ski − Ski−1 for all ki ∈ Z. Then, by Theorem 2.5, there exist twofamilies of linear operators Dkiki∈Z on Xi as in Theorem 2.2 such that

(4.29) f =∞∑

k1=−∞

∞∑


in(G(β′1, β

′2; γ

′1, γ

′2)

)′with β′i ∈ (βi, εi) and γi ∈ (γi, εi) for i = 1, 2. By Definition 4.1,

Theorem 4.2 and Remark 4.1, we have

‖S2(f)‖Lp(X1×X2) ≤ C‖f‖Hp(X1×X2),

where S2(f) is defined by (6.32) with Dkireplaced by by Dki

for i = 1, 2 as in Theorem2.2, q = 2, C11,1 ≥ C1

10,1 and C11,2 ≥ C210,1. For any k ∈ Z, let

Ωk =

(x1, x2) ∈ X1 ×X2 : S2(f)(x1, x2) > 2k

.

Let µ = µ1 × µ2,

R = R = Q1 ×Q2 : Q1 and Q2 are dyadic cubes, respectively, of X1 and X2 ,

and for k ∈ Z,

Rk =

R ∈ R : µ(R ∩ Ωk) >12µ(R) and µ(R ∩ Ωk+1) ≤ 1

2µ(R)

.


Obviously, for any R ∈ R, there is a unique k ∈ Z such that R ∈ R. Thus, we canreclassify the set of all dyadic cubes in X1 ×X2 by

(4.30)⋃

R∈RR =

⋃

k∈Z

⋃

R∈Rk

R.

In what follows, for i = 1, 2, if Qki is a dyadic cube and diamQki ∼ 2−ki , we rewriteDki and Dki , respectively, by DQki

and DQki. Then, from (4.29) and (4.30), it follows

that

(4.31) f =∞∑

k1=−∞

∞∑


=∞∑

k1=−∞

∑

diam Qk1∼2−k1

∞∑

k2=−∞

∑

diam Qk2∼2−k2

∫

Qk1

∫

Qk2

DQk1(x1, y1)DQk2

(x2, y2)

×DQk1DQk2

(f)(y1, y2) dµ1(y1) dµ2(y2)

=∑

R=Qk1×Qk2

∈R

∫

RDQk1

(x1, y1)DQk2(x2, y2)

×DQk1DQk2

(f)(y1, y2) dµ1(y1) dµ2(y2)

=∞∑

k=−∞

∑

R=Qk1×Qk2

∈Rk

∫

RDQk1

(x1, y1)DQk2(x2, y2)

×DQk1DQk2

(f)(y1, y2) dµ1(y1) dµ2(y2)

=∞∑

k=−∞λkak(x1, x2),

where

(4.32) λk =2kµ(Ωk)1/p

C14,1

and

(4.33) ak(x1, x2) =C14,1

2kµ(Ωk)1/p

∑

R=Qk1×Qk2

∈Rk

∫

RDQk1

(x1, y1)DQk2(x2, y2)

×DQk1DQk2

(f)(y1, y2) dµ1(y1) dµ2(y2),

where C14,1 > 0 is a constant which will be determined later.


We now verify that λkk∈Z and akk∈Z satisfy the requirement of the theorem.First, some trivial computation tells us that

(4.34)∞∑

k=−∞|λk|p =

1Cp

14,1

∞∑

k=−∞2kpµ(Ωk)

≤ C∞∑

k=−∞2kpµ(Ωk \ Ωk+1)

≤ C ‖S2(f)‖pLp(X1×X2) ,

which is a desired estimate.

Let C14,2 ∈ (0, 1/2) be a small constant which will be determined later and for k ∈ Z,

Ωk = (x1, x2) ∈ X1 ×X2 : MsχΩk(x1, x2) > C14,2 .

Then, the Lq(X1 ×X2)-boundedness of Ms with q ∈ (1,∞) implies that

(4.35) µ(Ωk) ≤ Cµ(Ωk).

Moreover, if C14,2 is chosen to be small enough which depends on A1, A2, C2 in Definition2.3, the constants concealed in (2.2), C1

10,1 and C210,1, then it is easy to check that

(4.36) supp ak ⊂ Ωk.

Let now h ∈ L2(X1 ×X2) with ‖h‖L2(X1×X2) ≤ 1. The Holder inequality, (4.30) andTheorem 3.1 tell us that

(4.37) |〈ak, h〉|

=

∣∣∣∣∣∣C14,1

2kµ(Ωk)1/p

∑

R=Qk1×Qk2

∈Rk

∫

X1×X2

∫

RDQk1

(x1, y1)DQk2(x2, y2)

×DQk1DQk2

(f)(y1, y2)h(x1, x2) dµ1(y1) dµ2(y2) dµ1(x1) dµ2(x2)

∣∣∣∣∣

=C14,1

2kµ(Ωk)1/p

∣∣∣∣∣∣∑

R=Qk1×Qk2

∈Rk

∫

RDQk1

DQk2(f)(y1, y2)

×D∗Qk1

D∗Qk2

(h)(y1, y2) dµ1(y1) dµ2(y2)

∣∣∣∣∣


≤ C14,1

2kµ(Ωk)1/p

×

∑

R=Qk1×Qk2

∈Rk

∫

R

∣∣∣DQk1DQk2

(f)(y1, y2)∣∣∣2

dµ1(y1) dµ2(y2)

1/2

×

∑

R=Qk1×Qk2

∈Rk

∫

R

∣∣∣D∗Qk1

D∗Qk2

(h)(y1, y2)∣∣∣2

dµ1(y1) dµ2(y2)

1/2

≤ C14,1

2kµ(Ωk)1/p

×

∑

R=Qk1×Qk2

∈Rk

∫

R

∣∣∣DQk1DQk2

(f)(y1, y2)∣∣∣2

dµ1(y1) dµ2(y2)

1/2

×

∞∑

k1=−∞

∞∑

k2=−∞

∫

X1×X2

∣∣∣D∗Qk1

D∗Qk2

(h)(y1, y2)∣∣∣2

dµ1(y1) dµ2(y2)

1/2

≤ C14,1‖h‖L2(X1×X2)

2kµ(Ωk)1/p

×

∑

R=Qk1×Qk2

∈Rk

∫

R

∣∣∣DQk1DQk2

(f)(y1, y2)∣∣∣2

dµ1(y1) dµ2(y2)

1/2

≤ C14,1

2kµ(Ωk)1/p

×

∑

R=Qk1×Qk2

∈Rk

∫

R

∣∣∣DQk1DQk2

(f)(y1, y2)∣∣∣2

dµ1(y1) dµ2(y2)

1/2

.

We now claim that

(4.38)

∑

R=Qk1×Qk2

∈Rk

∫

R

∣∣∣DQk1DQk2

(f)(y1, y2)∣∣∣2

dµ1(y1) dµ2(y2)

1/2

≤ C2kµ(Ωk)1/2.

To see this, we note that the estimate (4.35) indicates that

(4.39)∫

Ωk\Ωk+1

[S2(f)(x1, x2)]2 dµ1(x1) dµ2(x2) ≤ 22(k+1)µ(Ωk \ Ωk+1) ≤ C22kµ(Ωk).

On the other hand, we have

(4.40)∫

Ωk\Ωk+1

[S2(f)(x1, x2)]2 dµ1(x1) dµ2(x2)

=∫

Ωk\Ωk+1

∞∑

k1=−∞

∞∑

k2=−∞

∫

ρ1(x1,y1)≤C11,12−k1


×∫

ρ2(x2,y2)≤C11,22−k2

2k1d1+k2d2∣∣Dk1Dk2(f)(y1, y2)

∣∣2 dµ1(y1) dµ2(y2)

× dµ1(x1) dµ2(x2)

=∞∑

k1=−∞

∞∑

k2=−∞

∫

X1×X2

2k1d1+k2d2µ(

(x1, x2) ∈ Ωk \ Ωk+1 :

×ρ1(x1, y1) ≤ C11,12−k1 , ρ2(x2, y2) ≤ C11,22−k2

)

× ∣∣Dk1Dk2(f)(y1, y2)∣∣2 dµ1(y1) dµ2(y2)

≥∑

R=Qk1×Qk2

∈Rk

∫

R2k1d1+k2d2µ

((x1, x2) ∈ Ωk \ Ωk+1 :

×ρ1(x1, y1) ≤ C11,12−k1 , ρ2(x2, y2) ≤ C11,22−k2

)

× ∣∣Dk1Dk2(f)(y1, y2)∣∣2 dµ1(y1) dµ2(y2).

If (y1, y2), (x1, x2) ∈ R = Qk1 × Qk2 ∈ Rk, by Lemma 2.5 and our choice that C11,1 ≥C1

10,1 and C11,2 ≥ C210,1, we have ρ1(x1, y1) ≤ C11,12−k1 and ρ2(x2, y2) ≤ C11,22−k2 , and

moreover, R ∈ Rk implies that R ⊂ Ωk, since C14,2 < 1/2. These facts lead us that

(4.41) µ(

(x1, x2) ∈ Ωk \ Ωk+1 :

×ρ1(x1, y1) ≤ C11,12−k1 , ρ2(x2, y2) ≤ C11,22−k2

)

≥ µ[R ∩

(Ωk \ Ωk+1

)]

= µ [R \ (R ∩ Ωk+1)]

= µ(R)− µ (R ∩ Ωk+1)

≥ 12µ(R)

= C2−k1d1−k2d2 .

Combining (4.40), (4.41) with (4.39) yield our claim (4.38). Using (4.38) and (4.35) andtaking the supremum in both sides of (4.37) on h ∈ L2(X1 ×X2) with ‖h‖L2(X1×X2) ≤ 1tells us that

(4.42) ‖ak‖L2(X1×X2) ≤ CC14,1µ(Ωk)1/2−1/p ≤ µ(Ωk

)1/2−1/p,

if we choose C14,1 > 0 such that CC14,1 < 1, which is a desired estimate.


Obviously if R ∈ Rk, then R ⊂ Ωk. From this, it is easy to see that we can furtherdecompose ak(x1, x2) into

(4.43) ak(x1, x2) =C14,1

2kµ(Ωk)1/p

∑

R=Qk1×Qk2

∈Rk

∫

RDQk1

(x1, y1)DQk2(x2, y2)

×DQk1DQk2

(f)(y1, y2) dµ1(y1) dµ2(y2)

=C14,1

2kµ(Ωk)1/p

∑

R∈M(Ωk)

∑R=Qk1

×Qk2∈Rk

R⊂R

∫

RDQk1

(x1, y1)DQk2(x2, y2)

×DQk1DQk2

(f)(y1, y2) dµ1(y1) dµ2(y2)

=∑

R∈M(Ωk)

C14,1

2kµ(Ωk)1/p

∑R=Qk1

×Qk2∈Rk

R⊂R

∫

RDQk1

(x1, y1)DQk2(x2, y2)

×DQk1DQk2

(f)(y1, y2) dµ1(y1) dµ2(y2)

=∑

R∈M(Ωk)

αR(x1, x2).

Let R = Q1 ×Q2 with diamQ1 ∼ 2−k′1 and diamQ2 ∼ 2−k′2 and zi be the center ofQi with i = 1, 2. Then k′i ≤ ki for i = 1, 2. From this, it is easy to verify that

(4.44) supp aR⊂ B1(z1, A1 (C2,1 + C10,1) 2−k′1)×B2(z2, A2 (C2,2 + C10,2) 2−k′2).

Obviously, we have that for all x2 ∈ X2,

(4.45)∫

X1

αR(x1, x2) dµ1(x1) = 0,

and for all x1 ∈ X1,

(4.46)∫

X2

αR(x1, x2) dµ2(x2) = 0.

Let h be the same as in (4.37). Similarly to the estimate for (4.37), we have

∣∣⟨αR, h

⟩∣∣

=C14,1

2kµ(Ωk)1/p

∣∣∣∣∣∣∣∣

∫

X1×X2

∑R=Qk1

×Qk2∈Rk

R⊂R

∫

RDQk1

(x1, y1)DQk2(x2, y2)

×DQk1DQk2

(f)(y1, y2)h(x1, x2) dµ1(y1) dµ2(y2) dµ1(x1) dµ2(x2)

∣∣∣∣∣


≤ C14,1

2kµ(Ωk)1/p

∑R=Qk1

×Qk2∈Rk

R⊂R

∫

R

∣∣∣DQk1DQk2

(f)(y1, y2)∣∣∣2

dµ1(y1) dµ2(y2)

1/2

×

∑R=Qk1

×Qk2∈Rk

R⊂R

∫

R

∣∣∣D∗Qk1

)D∗Qk2

(h)(y1, y2)∣∣∣2

dµ1(y1) dµ2(y2)

1/2

≤ CC14,1‖h‖L2(X1×X2)

2kµ(Ωk)1/p

×

∑R=Qk1

×Qk2∈Rk

R⊂R

∫

R

∣∣∣DQk1DQk2

(f)(y1, y2)∣∣∣2

dµ1(y1) dµ2(y2)

1/2

≤ CC14,1

2kµ(Ωk)1/p

∑R=Qk1

×Qk2∈Rk

R⊂R

∫

R

∣∣∣DQk1DQk2

(f)(y1, y2)∣∣∣2

dµ1(y1) dµ2(y2)

1/2

.

From this, it follows that

∥∥αR

∥∥L2(X1×X2)

≤ CC14,1

2kµ(Ωk)1/p

∑R=Qk1

×Qk2∈Rk

R⊂R

∫

R

∣∣∣DQk1DQk2

(f)(y1, y2)∣∣∣2

dµ1(y1) dµ2(y2)

1/2

.

Thus, (4.38) and (4.35) tell us that

∑

R∈M(Ωk)

∥∥αR

∥∥2

L2(X1×X2)

1/2

≤ CC14,1

2kµ(Ωk)1/p

×

∑

R∈M(Ωk)

∑R=Qk1

×Qk2∈Rk

R⊂R

∫

R

∣∣∣DQk1DQk2

(f)(y1, y2)∣∣∣2

dµ1(y1) dµ2(y2)

1/2

≤ CC14,1

2kµ(Ωk)1/p

∑

R=Qk1×Qk2

∈Rk

∫

R

∣∣∣DQk1DQk2

(f)(y1, y2)∣∣∣2

dµ1(y1) dµ2(y2)

1/2

≤ CC14,1µ(Ωk)1/2−1/p


≤ µ(Ωk

)1/2−1/p,

if we chose C14,1 small enough such that CC14,1 < 1. This, together with (4.34), (4.36),(4.42), (4.44), (4.45) and (4.46), tells us that we have obtained a desired atomic decom-position for f .

We now consider the converse. To this end, by Definition 4.1 and (4.6), we easily seethat it suffices to verify that there is a constant C > 0 such that for any (p, 2)-atom ofHp(X1 ×X2), a,

(4.47) ‖g2(a)‖Lp(X1×X2) ≤ C,

where

g2(a)(x1, x2) =

∞∑

k1=−∞

∞∑

k2=−∞|Dk1Dk2(a)(x1, x2)|2

1/2

and we choose Dkiki∈Z as in (4.29).We suppose supp a ⊂ Ω and define

Ω = (x1, x2) ∈ X1 ×X2 : MsχΩ(x1, x2) > 1/2and

Ω =(x1, x2) ∈ X1 ×X2 : MsχΩ

(x1, x2) > 1/2

.

Moreover, suppose a =∑

R∈M(Ω) aR. For any R = Q1 × Q2 ∈ M(Ω), we define R =

Q1 ×Q2 ∈M1

(Ω

)such that

µ(R ∩ Ω

)>

12µ

(R

)

and R = Q1 × Q2 ∈M2

(Ω

)such that

µ(R ∩ Ω

)>

12µ

(R

).

Let C15,1 ≥ 1 and C15,2 ≥ 1 be two constants which will be determined later and we set

100−→C R = 100C15,1Q1 × 100C15,2Q2,

where 100C15,iQi means the “cube” with the same center as Qi but with diameter 100C15,i

times the diameter of Qi. We also denote by zi the center of Qi for i = 1, 2.We now control ‖g2(a)‖Lp(X1×X2) by

‖g2(a)‖pLp(X1×X2)

=∫

∪R′∈M(Ω)100−→C R

′g2(a)(x1, x2)p dµ1(x1) dµ2(x2) +

∫

(∪R′∈M(Ω)100−→C R

′)c

· · ·

= U11 + U12,


where(∪R′∈M(Ω)100−→C R

′)c= (X1 ×X2) \

(∪R′∈M(Ω)100−→C R

′). The Holder inequality,Lemma 2.5 and Theorem 3.1 imply that

(4.48) U11≤ µ(∪R′∈M(Ω)100−→C R

′)1−p/2∫

X1×X2

g2(a)(x1, x2)2 dµ1(x1) dµ2(x2)p/2

≤ Cµ(Ω)1−p/2‖a‖pL2(X1×X2)

≤ C,

which is a desired estimate.We further control U12 by

(4.49) U12 =∫

(∪R′∈M(Ω)100−→C R

′)c

g2(a)(x1, x2)p dµ1(x1) dµ2(x2)

≤∑

R∈M(Ω)

∫

(∪R′∈M(Ω)100−→C R

′)c

g2(aR)(x1, x2)p dµ1(x1) dµ2(x2)

≤∑

R∈M(Ω)

∫

x1 /∈100C15,1Q1

∫

X2

g2(aR)(x1, x2)p dµ1(x1) dµ2(x2)

+∫

X1

∫

x2 /∈100C15,2Q2

· · ·

=∑

R∈M(Ω)

(U1R + U2R) .

The estimate for U2R is similar to the estimate for U1R by symmetry. Thus, we onlyestimate U1R and leave the details for the estimate of U2R to the reader. To estimate U1R,we further decompose it into

U1R =∫

x1 /∈100C15,1Q1

∫

X2

g2(aR)(x1, x2)p dµ1(x1) dµ2(x2)

=∫

x1 /∈100C15,1Q1

∫

x2∈100C15,2Q2

g2(aR)(x1, x2)p dµ1(x1) dµ2(x2)

+∫

x1 /∈100C15,1Q1

∫

x2 /∈100C15,2Q2

· · ·

= U1R1 + U1R2.

The Holder inequality implies that

(4.50) U1R1≤ µ2 (100C15,2Q2)1−p/2

×∫

x1 /∈100C15,1Q1

[∫

x2∈100C15,2Q2

g2(aR)(x1, x2)2 dµ2(x2)

]p/2

dµ1(x1)


≤ Cµ2(Q2)1−p/2

∫

x1 /∈100C15,1Q1

[∫

X2

g2(aR)(x1, x2)2 dµ2(x2)]p/2

dµ1(x1).

By Lemma 3.1 for X2, we obtain

(4.51)∫

X2

g2(aR)(x1, x2)2 dµ2(x2)

=∫

X2

∞∑

k1=−∞

∞∑

k2=−∞|Dk1Dk2(aR)(x1, x2)|2 dµ2(x2)

≤ C∞∑

k1=−∞

∫

X2

|Dk1 [aR(·, x2)] (x1)|2 dµ2(x2)

= C∞∑

k1=−∞

∫

X2

∣∣∣∣∫

X1

Dk1(x1, y1)aR(y1, x2) dµ1(y1)∣∣∣∣2

dµ2(x2).

Suppose diamQi ∼ 2−ki,0 and diam Qi ∼ 2−ki,0 for some ki,0, ki,0 ∈ Z and i = 1, 2. Thenki,0 ≤ ki,0. From

supp aR ⊂ B1(z1, A1(C2,1 + C110,1)2

−k1,0)×B2(z2, A2(C2,2 + C210,2)2

−k2,0),

where zi is the center of Qi, it follows that

ρ1(y1, z1) ≤ A1(C2,1 + C110,1)2

−k1,0 ,

which combines the fact that ρ1(x1, y1) ≤ C2,12−k1 tells us that

(4.52) ρ1(x1, z1)≤ A1C2,12−k1 + A1(C2,1 + C110,1)2

−k1,0

≤ A1C2,12−k1 + A1(C2,1 + C110,1)2

−k1,0 .

On another hand, since z1 ∈ Q1 ⊂ Q1 and x1 /∈ 100C15,1Q1, we then have

(4.53) ρ1(x1, z1) ≥(

1− C110,1

100C110,2C15,1

)100C1

10,2C15,12−k1,0 .

If we choose C15,1 large enough, then (4.52) and (4.53) tell us that k1 ≤ k1,0 in (4.51).Thus, by the Holder inequality, we further have

(4.54)∫

X2

g2(aR)(x1, x2)2 dµ2(x2)


≤ C

k1,0∑

k1=−∞

∫

X2

∣∣∣∣∫

X1

Dk1(x1, y1)aR(y1, x2) dµ1(y1)∣∣∣∣2

dµ2(x2)

= C

k1,0∑

k1=−∞

∫

X2

∣∣∣∣∫

X1

[Dk1(x1, y1)−Dk1(x1, z1)] aR(y1, x2) dµ1(y1)∣∣∣∣2

dµ2(x2)

≤ C

k1,0∑

k1=−∞

2−2k1,0ε122k1ε′1

ρ1(x1, z1)2d1+2(ε1−ε′1)

∫

X2

[∫

X1

|aR(y1, x2)| dµ1(y1)]2

dµ2(x2)

≤ C2−2k1,0ε122k1,0ε′1

ρ1(x1, z1)2d1+2(ε1−ε′1)µ1(Q1)‖aR‖2

L2(X1×X2),

where we chose ε′1 ∈ (0, ε1) such that d1p + (ε1 − ε′1)p > d1. Noting that if C15,1 is largeenough, then x1 /∈ 100C15,1Q1 implies that ρ1(x1, z1) ≥ ρ1(x1, z1), which, together (4.54)with (4.50) indicates that

U1R1≤ Cµ2(Q2)1−p/2µ1(Q1)p/2‖aR‖pL2(X1×X2)

×∫

x1 /∈100C15,1Q1

2−pk1,0ε12pk1,0ε′1

ρ1(x1, z1)pd1+p(ε1−ε′1)dµ1(x1)

≤ Cµ2(Q2)1−p/2µ1(Q1)p/2‖aR‖pL2(X1×X2)

×∫

x1 /∈100C15,1Q1

2−pk1,0ε12pk1,0ε′1

ρ1(x1, z1)pd1+p(ε1−ε′1)dµ1(x1)

≤ Cµ(R)1−p/2‖aR‖pL2(X1×X2)

(µ1(Q1)

µ1(Q1)

)p(d1+ε1)/d1−1

.

From this, the Holder inequality and Lemma 4.2, it follows that

(4.55)∑

R∈M(Ω)

U1R1≤ C

∑

R∈M(Ω)

‖aR‖2L2(X1×X2)

p/2

×

∑

R∈M(Ω)

µ(R)

(µ1(Q1)

µ1(Q1)

)[p(d1+ε1)/d1−1]2/2−p

1−p/2

≤ Cµ(Ω)p/2−1µ(Ω)1−p/2

≤ C,

which is a desired estimate.We now estimate U1R2. For x1 /∈ 100C15,1Q1 and x2 /∈ 100C15,2Q2, similarly to the


estimate for U1R1, if we choose C15,1 and C15,2 large enough, we then have

g2(aR)(x1, x2)

=

∞∑

k1=−∞

∞∑

k2=−∞

∣∣∣∣∫

X1×X2

Dk1(x1, y1)Dk2(x2, y2)aR(y1, y2) dµ1(y1) dµ2(y2)∣∣∣∣2

1/2

≤

k1,0∑

k1=−∞

k2,0∑

k2=−∞

∣∣∣∣∫

X1×X2

Dk1(x1, y1)Dk2(x2, y2)aR(y1, y2) dµ1(y1) dµ2(y2)∣∣∣∣2

1/2

=

k1,0∑

k1=−∞

k2,0∑

k2=−∞

∣∣∣∣∫

X1×X2

[Dk1(x1, y1)−Dk1(x1, z1)]

× [Dk2(x2, y2)−Dk2(x2, z2)] aR(y1, y2) dµ1(y1) dµ2(y2)∣∣∣∣21/2

≤ C

k1,0∑

k1=−∞

k2,0∑

k2=−∞

22k1ε′1−2k1,0ε1

ρ1(x1, z1)2d1+2(ε1−ε′1)

22k2ε′2−2k2,0ε2

ρ2(x2, z2)2d2+2(ε2−ε′2)

×[∫

X1×X2

|aR(y1, y2)| dµ1(y1) dµ2(y2)]2

1/2

≤ C ‖aR‖L2(X1×X2) µ(R)1/2 2k1,0ε′1−k1,0ε1

ρ1(x1, z1)d1+ε1−ε′1

2k2,0(ε′2−ε2)

ρ2(x2, z2)d2+ε2−ε′2,

where we choseε′1 ∈ (0, ε1) and ε′2 ∈ (0, ε2) such that p(d1+ε1−ε′1) > d1 and p(d2+ε2−ε′2) >

d2. From this and the fact ρ1(x1, z1) ≥ Cρ1(x1, z1), it follows that

(4.56) U1R2≤ C ‖aR‖pL2(X1×X2)

µ(R)p/2

∫

x1 /∈100C15,1Q1

2k1,0pε′1−k1,0pε1

ρ1(x1, z1)(d1+ε1−ε′1)pdµ1(x1)

×∫

x2 /∈100C15,2Q2

2k2,0(ε′2−ε2)p

ρ2(x2, z2)(d2+ε2−ε′2)pdµ2(x2)

≤ Cµ(R)1−p/2‖aR‖pL2(X1×X2)

(µ1(Q1)

µ1(Q1)

)p(d1+ε1)/d1−1

.

Thus, similarly to the estimate for (4.55), the estimate (4.56), the Holder inequality, andLemma 4.2 tell us that

(4.57)∑

R∈M(Ω)

U1R2 ≤ C.

Combining the estimate (4.55) with (4.57), we obtain∑

R∈M(Ω)

U1R ≤ C,

which completes the proof of Theorem 4.3.


5 Singular integrals

We first recall some notation. Let Ω be an open set in X1 ×X2. As in the proof ofTheorem 4.3, we define

Ω = (x1, x2) ∈ X1 ×X2 : MsχΩ(x1, x2) > 1/2and

Ω =(x1, x2) ∈ X1 ×X2 : MsχΩ

(x1, x2) > 1/2

.

For any R = Q1 ×Q2 ∈M(Ω), we define R = Q1 ×Q2 ∈M1

(Ω

)such that

(5.1) µ(R ∩ Ω

)>

12µ

(R

)

and R = Q1 × Q2 ∈M2

(Ω

)such that

(5.2) µ(R ∩ Ω

)>

12µ

(R

).

Let C16,1 ≥ 1 and C16,2 ≥ 1 be two constants which are large enough and we set

(5.3) −→C R = C16,1Q1 × C16,2Q2,

where C16,iQi means the “cube” with the same center as Qi but with diameter C16,i timesthe diameter of Qi. We also denote by zi the center of Qi for i = 1, 2.

We first have the following general theorem on the boundedness of linear operatorsfrom Hp(X1 ×X2) to Lp(X1 ×X2) with p ∈ (p0, 1], when the linear operators are assumedto be bounded on L2(X1×X2). This is a generalization of R. Fefferman’s theorem in pureproduct setting in Euclidean spaces, see Theorem 1 in [F4]. Here p0 is some positivenumber less than 1.Theorem 5.1 Suppose that T is a bounded linear operator on L2(X1×X2). Let εi ∈ (0, θi]and

max

d1

d1 + ε1,

d2

d2 + ε2

< p ≤ 1.

Suppose further that if aR is an Hp(X1 ×X2) (p, 2)-rectangle atom as in Definition 4.2and R = Q1 ×Q2. Let Q1 and Q2 be the same as in (5.1) and (5.2). If there exist fixedconstant δ > 0 and some fixed large enough constants C16,1 ≥ 1 and C16,2 ≥ 1 such thatfor all R = Q1 ×Q2,

(5.4)∫

X2

∫

(C16,1Q1)|T (aR)(x1, x2)|p dµ1(x1) dµ2(x2) ≤ C

(µ1(Q1)

µ1(Q1)

)δ

and

(5.5)∫

(C16,2Q2)

∫

X1

|T (aR)(x1, x2)|p dµ1(x1) dµ2(x2) ≤ C

(µ2(Q2)

µ2(Q2)

)δ

,


then T is a bounded operator from Hp(X1 ×X2) to Lp(X1 ×X2), where

(C16,iQi

)= Xi \ C16,iQi, i = 1, 2.

Proof. It suffices to prove that there is a constant C > 0 such that for all (p, 2)-atomsof Hp(X1 ×X2),

(5.6) ‖T (a)‖Lp(X1×X2) ≤ C.

Use all the notation the same as in Definition 4.2, in particular, suppose suppa ⊂ Ω and

a =∑

R∈M(Ω)

aR.

For R = Q1 × Q2, let Q1, Q2 and −→C R be the same as in (5.1), (5.2) and (5.3). By theHolder inequality and L2(X1 ×X2)-boundedness of T , we can estimate

(5.7)∫

∪R′∈M(Ω)

−→C R

′|T (a)(x1, x2)|p dµ1(x1) dµ2(x2)

≤ µ

⋃

R′∈M(Ω)

−→C R

′

1−p/2 ∫

X1×X2

|T (a)(x1, x2)|p dµ1(x1) dµ2(x2)p/2

≤ Cµ(Ω

)1−p/2 ‖a‖pL2(X1×X2

≤ Cµ(Ω)1−p/2µ(Ω)(1/2−1/p)p

≤ C,

which is a desired estimate.We now write

∫

(∪R′∈M(Ω)

−→C R

′)|T (a)(x1, x2)|p dµ1(x1) dµ2(x2)

≤∑

R∈M(Ω)

∫

(∪R′∈M(Ω)

−→C R

′)|T (aR)(x1, x2)|p dµ1(x1) dµ2(x2)

≤∑

R∈M(Ω)

∫

X2

∫

(C16,1Q1)|T (aR)(x1, x2)|p dµ1(x1) dµ2(x2)

+∑

R∈M(Ω)

∫

(C16,2Q2)

∫

X1

|T (aR)(x1, x2)|p dµ1(x1) dµ2(x2)

= J1 + J2.


Note that aRµ(R)1/2−1/p‖aR‖−1L2(X1×X2)

is an Hp(X1 ×X2) (p, 2)-rectangle atom.The assumption (5.1), the Holder inequality and Lemma 4.2 tell us that

(5.8) J1≤ C∑

R∈M(Ω)

‖aR‖pL2(X1×X2)

µ(R)(1/p−1/2)p

(µ1(Q1)

µ1(Q1)

)δ

≤ C

∑

R∈M(Ω)

‖aR‖2L2(X1×X2)

p/2 ∑

R∈M(Ω)

µ(R)

(µ1(Q1)

µ1(Q1)

)2δ/(2−p)

1−p/2

≤ Cµ(Ω)(1/2−1/p)p

∑

R∈M2(Ω)

µ(R)

(µ1(Q1)

µ1(Q1)

)2δ/(2−p)

1−p/2

≤ Cµ(Ω)p/2−1µ(Ω)1−p/2

≤ C,

which is a desired estimate.Finally, we note that if R′ = Q′

1 ×Q′2 ∈M(Ω) and R = Q1 ×Q2 ∈M(Ω) such that

R′ = R ∈M1

(Ω

), then R′ = R or R′ ∩R = ∅. From this fact, the assumption (5.2), the

Holder inequality and Lemma 4.2, it follows that

(5.9) J2≤ C∑

R∈M(Ω)

‖aR‖pL2(X1×X2)

µ(R)(1/p−1/2)p

(µ2(Q2)

µ2(Q2)

)δ

≤ C

∑

R∈M(Ω)

‖aR‖2L2(X1×X2)

p/2 ∑

R∈M(Ω)

µ(R)

(µ2(Q2)

µ2(Q2)

)2δ/(2−p)

1−p/2

≤ Cµ(Ω)p/2−1

∑

S∈M1(Ω)

∑

R=SR∈M(Ω)

µ(R)

[µ2(Q2)

µ2(Q2)

]2δ/(2−p)

1−p/2

≤ Cµ(Ω)p/2−1

∑

S∈M1(Ω)

µ(S)

[µ2(Q2)

µ2(Q2)

]2δ/(2−p)

1−p/2

≤ Cµ(Ω)p/2−1µ(Ω

)1−p/2

≤ Cµ(Ω)p/2−1µ(Ω)1−p/2

≤ C,

which is a desired estimate.Combining (5.7), (5.8) and (5.9) gives us (5.6) which completes the proof of Theorem

5.1.


Remark 5.1 We mention that the examples where Theorem 5.1 applies, if X1 and X2

are Euclidean spaces, are the double Hilbert transform, product versions of commutatorsas in [F6], and the class introduced by Fefferman and Stein in [FS]; see also [F4].

We now consider the boundedness on Hp space for a certain range of p ∈ (p0, 1] fora class of singular integrals similar to [NS3].

Let ηi ∈ (0, θi], i = 1, 2. We define Cη1,η20 (X1×X2) = Cη1

0 (X1)⊗Cη20 (X2). Also, for

i = 1, 2, we say ϕ is a bump function on Xi associated to a ball B(xi, δi), if it is supportedin that ball, and satisfies ‖ϕ‖L∞(Xi) ≤ 1 and ‖ϕ‖Cη

0 (Xi) ≤ Cδηi for all η ∈ (0, θi], where

C ≥ 0 is independent of δi and xi. In what follows, for its convenience, if f ∈ L∞(Xi), wewrite f ∈ C0(Xi) and define

‖f‖C0(Xi) = ‖f‖L∞(Xi),

and for ηi ∈ (0, θi],

‖f‖Cηi (Xi) = supxi,yi∈Xi

|f(xi)− f(yi)|ρi(xi, yi)ηi

, i = 1, 2.

Definition 5.1 Let ηi ∈ (0, θi], i = 1, 2. A linear operator T initially defined fromCη1,η2

0 (X1 × X2) = Cη10 (X1) ⊗ Cη2

0 (X2) to its dual is called a singular integral if T hasan associated distribution kernel K(x1, x2; y1, y2) which is locally integrable away from the“cross”

(x1, x2; y1, y2) : x1 = y1, or x2 = y2satisfying the following additional properties

(i)

〈T (ϕ1 ⊗ ϕ2), ψ1 ⊗ ψ2〉

=∫

X1×X2×X1×X2

K(x1, x2; y1, y2)ϕ1(y1)ϕ2(y2)

×ψ1(x1)ψ2(x2) dµ1(y1) dµ2(y2) dµ1(x1) dµ2(x2)

whenever ϕ1, ψ1 ∈ Cη10 (X1) and have disjoint supports, and ϕ2, ψ2 ∈ Cη2

0 (X2) andhave disjoint supports;

(ii) For each bump function ϕ2 on X2 and each x2 ∈ X2, there exists a singular integralTϕ2,x2 (of the one-factor type) on X1, so that x2 → Tϕ2,x2 is smooth in the sensemake precise below, and so that

〈T (ϕ1 ⊗ ϕ2), ψ1 ⊗ ψ2〉 =∫

X2

〈Tϕ2,x2ϕ1, ψ1〉ψ2(x2) dµ2(x2).

Moreover, we require that Tϕ2,x2 uniformly satisfies the following conditions thatTϕ2,x2 has a distribution kernel Kϕ2,x2(x1, y1) having the following properties:


(ii)1 If ϕ1, ψ1 ∈ Cη10 (X1) have disjoint supports, then

〈Tϕ2,x2ϕ1, ψ1〉 =∫

X1×X1

Kϕ2,x2(x1, y1)ϕ1(x1)ψ1(y1) dµ1(x1) dµ1(y1);

(ii)2 If ϕ1 is a bump function associated to the ball B(x1, r1), then

‖Tϕ2,x2ϕ1‖Ca1 (X1) ≤ Cr−a11

for all a1 ∈ [0, θ1], where C ≥ 0 is independent of ϕ2, x2, and r1. Precisely, thismeans that for each a1 ≥ 0, there is a b1 ≥ 0 and a constant Ca1,b1, independentof ϕ2, x2 and r1, so that whenever ϕ ∈ Cθ1

0 (X1) supported in a ball B(x1, r1),then

ra11 ‖Tϕ2,x2ϕ1‖Ca1 (X1) ≤ Ca1,b1 sup

c1≤b1

rc11 ‖Tϕ2,x2ϕ1‖Cc1 (X1);

(ii)3 There is a constant C > 0 independent of ϕ2, x2, and r1 such that

(ii)31 |Kϕ2,x2(x1, y1)| ≤ Cρ1(x1, y1)−d1,(ii)32 |Kϕ2,x2(x1, y1)−Kϕ2,x2(x′1, y1)| ≤ Cρ1(x1, x

′1)

η1ρ1(x1, y1)−d1−η1 for

ρ1(x1, x′1) ≤

ρ1(x1, y1)2A1

,

(ii)33 |Kϕ2,x2(x1, y1)−Kϕ2,x2(x1, y′1)| ≤ Cρ1(y1, y

′1)

η1ρ1(x1, y1)−d1−η1 for

ρ1(y1, y′1) ≤

ρ1(x1, y1)2A1

;

(ii)4 If ϕ2 is a bump function associated to B(x2, r2), then for a2 ∈ (0, θ2],

ra22 ρ2(x2, u2)−a2 [Tϕ2,x2 − Tϕ2,u2 ]

also uniformly satisfies properties (ii)1 through (ii)3;

(ii)5 Properties (ii)1 through (ii)4 also hold with x1 and y1 interchanged. That is,there properties also hold for the adjoint operator (Tϕ2,x2)t defined by

⟨(Tϕ2,x2)tϕ, ψ

⟩= 〈Tψ, ϕ〉;

(iii) The property (ii) hold when the index 1 and 2 are interchanged, namely, if the rolesof X1 and X2 are interchanged;

(iv) There is a constant C > 0 such that for all bump functions ϕ1 and ϕ2, respectively,associated to B(x1, r1) and B(x2, r2),

|[T (ϕ1 ⊗ ϕ2)(x1, x2)− T (ϕ1 ⊗ ϕ2)(u1, x2)]

− [T (ϕ1 ⊗ ϕ2)(x1, u2)− T (ϕ1 ⊗ ϕ2)(u1, u2)]|

≤ Cr−a11 r−a2

2 ρ1(x1, u1)a1ρ2(x2, u2)a2

for all a1 ∈ (0, θ1] and all a2 ∈ (0, θ2];


(v) The kernel K(x1, x2; y1, y2) satisfies the following conditions:

(v)1 |K(x1, x2; y1, y2)| ≤ Cρ1(x1, y1)−d1ρ2(x2, y2)−d2,

(v)2 |K(x1, x2; y1, y2)−K(x1, x′2; y1, y2)| ≤ C

1ρ1(x1, y1)d1

ρ2(x2, x′2)

η2

ρ2(x2, y2)d2+η2for

ρ2(x2, x′2) ≤

ρ2(x2, y2)2A2

,

(v)3 |K(x1, x2; y1, y2)−K(x1, x2; y1, y′2)| ≤ C

1ρ1(x1, y1)d1

ρ2(y2, y′2)

η2

ρ2(x2, y2)d2+η2for

ρ2(y2, y′2) ≤

ρ2(x2, y2)2A2

,

(v)4

|[K(x1, x2; y1, y2)−K(x′1, x2; y1, y2)]

−[K(x1, x′2; y1, y2)−K(x′1, x

′2; y1, y2)]|

≤ Cρ1(x1, x

′1)

η1

ρ1(x1, y1)d1+η1

ρ2(x2, x′2)

η2

ρ2(x2, y2)d2+η2

for ρ1(x1, x′1) ≤

ρ1(x1, y1)2A1

and ρ2(x2, x′2) ≤

ρ2(x2, y2)2A2

,

(v)5

|[K(x1, x2; y1, y2)−K(x′1, x2; y1, y2)]

−[K(x1, x2; y1, y′2)−K(x′1, x2; y1, y

′2)]|

≤ Cρ1(x1, x

′1)

η1

ρ1(x1, y1)d1+η1

ρ2(y2, y′2)

η2

ρ2(x2, y2)d2+η2

for ρ1(x1, x′1) ≤

ρ1(x1, y1)2A1

and ρ2(y2, y′2) ≤

ρ2(x2, y2)2A2

,

(v)6

|[K(x1, x2; y1, y2)−K(x1, x2; y′1, y2)]

−[K(x1, x2; y1, y′2)−K(x1, x2; y′1, y

′2)]|

≤ Cρ1(y1, y

′1)

η1

ρ1(x1, y1)d1+η1

ρ2(y2, y′2)

η2

ρ2(x2, y2)d2+η2

for ρ1(y1, y′1) ≤

ρ1(x1, y1)2A1

and ρ2(y2, y′2) ≤

ρ2(x2, y2)2A2

,


(v)7 The properties (iii)2 to (iii)6 hold when the index 1 and 2 are interchanged, thatis, if the roles of X1 and X2 are interchanged.

(vi) The same properties are assumed to hold for the 3 “transposes” of T , i. e. thoseoperators which arise by interchanging x1 and y1, or interchanging x2 and y2, ordoing both interchanges.

We can now establish the Hp-boundedness of these singular operators as defined inDefinition 5.1 as follows.

Theorem 5.2 Let 0 < εi, ηi ≤ θi, i = 1, 2, and

max

d1

d1 + ε1,

d2

d2 + ε2,

d1

d1 + η1,

d2

d2 + η2

< p < ∞.

Each product singular integral as in Definition 5.1 extends to a bounded operator onHp(X1 ×X2) to itself.

Proof. Let all the notation be the same as in Theorem 3.1 and Theorem 2.2. Forf ∈ Hp(X1 ×X2), by Theorem 2.7, for k1, k2 ∈ Z, we have

(5.10) Dk1Dk2Tf =∞∑

k′1=−∞

∑

τ ′1∈Ik′1

N(k′1,τ ′1)∑

ν′1=1

∞∑

k′2=−∞

∑

τ ′2∈Ik′2

N(k′2,τ ′2)∑

ν′2=1

µ1(Qk′1,ν′1τ ′1

)µ2(Qk′2,ν′2τ ′2

)

×Dk1Dk2TDk′1Dk′2(x1, x2; yk′1,ν′1τ ′1

, yk′2,ν′2τ ′2

)Dk′1Dk′2(f)(yk′1,ν′1τ ′1

, yk′2,ν′2τ ′2

).

We now prove that there is constants C > 0, δ1 > 0 and δ2 > 0 such that for allk1, k2, k′1, k′2 ∈ Z and all xi, yi ∈ Xi, i = 1, 2,

(5.11)∣∣∣Dk1Dk2TDk′1Dk′2(x1, x2; y1, y2)

∣∣∣

≤ C2−|k1−k′1|δ12−|k2−k′2|δ2 2−(k1∧k′1)η1

(2−(k1∧k′1) + ρ1(x1, y1))d1+η1

× 2−(k2∧k′2)η2

(2−(k2∧k′2) + ρ2(x2, y2))d2+η2.

To verify (5.11), we need to consider several cases. We only prove the case that k1 ≥ k′1and k2 ≥ k′2 and leave the other cases to the reader. Under this assumption, we againneed to consider several cases. Let l1, l2 ∈ N be large enough which will be decided later.

Case 1. ρ1(x1, y1) ≥ 2l1−k′1 and ρ2(x2, y2) ≥ 2l2−k′2 . In this case, by

(5.12)∫

Xi

Dki(xi, ui) dµi(ui) = 0, i = 1, 2,


we can write

Dk1Dk2TDk′1Dk′2(x1, x2; y1, y2)

=∫

X1×X2

∫

X1×X2

Dk1(x1, u1)Dk2(x2, u2)K(u1, u2; z1, z2)

×Dk′1(z1, y1)Dk′2(z2, y2) dµ1(u1) dµ2(u2) dµ1(z1) dµ2(z2)

=∫

X1×X2

∫

X1×X2

Dk1(x1, u1)Dk2(x2, u2) [K(u1, u2; z1, z2)−K(x1, u2; z1, z2)]

− [K(u1, x2; z1, z2)−K(x1, x2; z1, z2)]

×Dk′1(z1, y1)Dk′2(z2, y2) dµ1(u1) dµ2(u2) dµ1(z1) dµ2(z2).

We choose l1, l2 ∈ N large enough, depending on A1 and A2, such that in this case, wehave

(5.13) ρi(ui, zi) ≥ Cρi(xi, yi),

where i = 1, 2. A property similar to (v)6 in Definition 5.1 tells us that∣∣∣Dk1Dk2TDk′1Dk′2(x1, x2; y1, y2)

∣∣∣

≤ C1

ρ1(x1, y1)d1+η1

1ρ2(x2, y2)d2+η2

×∫

X1×X2

∫

X1×X2

∣∣∣Dk1(x1, u1)Dk2(x2, u2)Dk′1(z1, y1)Dk′2(z2, y2)∣∣∣

×ρ1(u1, x1)η1ρ2(u2, x2)η2 dµ1(u1) dµ2(u2) dµ1(z1) dµ2(z2)

≤ C2−k1η1

ρ1(x1, y1)d1+η1

2−k2η2

ρ2(x2, y2)d2+η2

≤ C2−(k1−k′1)η12−(k2−k′2)η22−k′1η1

ρ1(x1, y1)d1+η1

2−k′2η2

ρ2(x2, y2)d2+η2,

which is what expect to derive.Case 2. ρ1(x1, y1) < 2l1−k′1 and ρ2(x2, y2) < 2l2−k′2 . In this case, by (5.12), we can

write


=∫

X1×X2

∫

X1×X2

Dk1(x1, u1)Dk2(x2, u2)

×[

TDk′1Dk′2(u1, u2; y1, y2)− TDk′1Dk′2(x1, u2; y1, y2)]

−[TDk′1Dk′2(u1, x2; y1, y2)− TDk′1Dk′2(x1, x2; y1, y2)

]dµ1(u1) dµ2(u2).


Noting that 2−k′1d1Dk′1 and 2−k′2d2Dk′2 are bump functions, respectively, associated toB(y1, C2−k′1) and B(y2, C2−k′2) with an absolute constant, by the property (iv) of Defini-tion 5.1, we obtain

∣∣∣Dk1Dk2TDk′1Dk′2(x1, x2; y1, y2)∣∣∣

≤ C2k′1(d1+η1)2k′2(d2+η2)

∫

X1×X2

|Dk1(x1, u1)Dk2(x2, u2)|

×ρ1(u1, x1)η1ρ2(u2, x2)η2 dµ1(u1) dµ2(u2)

≤ C2−(k1−k′1)η12−(k2−k′2)η22k′1d12k′2d2 ,

which is a desired estimate.Case 3. ρ1(x1, y1) < 2l1−k′1 and ρ2(x2, y2) ≥ 2l2−k′2 . In this case, by (5.12) and

property (ii) of Definition 5.1, we can write


= 2k′1d1

∫

X1×X2

∫

X1×X2

Dk1(x1, u1)Dk2(x2, u2)

×[

K2−k′1d1Dk′1

,u1(u2, z2)−K2−k′1d1Dk′1

,x1(u2, z2)]

−[K

2−k′1d1Dk′1,u1(x2, z2)−K

2−k′1d1Dk′1,x1(x2, z2)

]

×Dk′1(z1, y1)Dk′2(z2, y2) dµ1(u1) dµ2(u2) dµ1(z1) dµ2(z2).

Choose l2 ∈ N large enough such that (5.13) holds. Then this choice and the property(ii)4 yield that

∣∣∣Dk1Dk2TDk′1Dk′2(x1, x2; y1, y2)∣∣∣

≤ C2k′1(d1+η1)

∫

X1×X2

∫

X1×X2

∣∣∣Dk1(x1, u1)Dk2(x2, u2)Dk′1(z1, y1)Dk′2(z2, y2)∣∣∣

×ρ1(u1, x1)η1ρ2(u2, x2)η2

ρ2(u2, z2)d2+η2dµ1(u1) dµ2(u2) dµ1(z1) dµ2(z2)

≤ C2−(k1−k′1)η12−(k2−k′2)η22k′1d12−k′2η2

ρ2(x2, y2)d2+η2,

which is also a desired estimate.Case 4. ρ1(x1, y1) ≥ 2l1−k′1 and ρ2(x2, y2) < 2l2−k′2 . The proof for this case is similar

to Case 3. We omit the details.


Using (5.11), Lemma 4.1, Remark 4.1 and the Fefferman-Stein vector-valued inequal-ity and some computation similar to the proof of Theorem 4.1, we can verify

‖Tf‖Hp(X1×X2) ≤ C‖f‖Hp(X1×X2).

This completes the proof of Theorem 5.2.


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Yongsheng Han:Department of Mathematics, Auburn University, Auburn, AL 36849-5310, USA

E-mail address: [email protected]

Guozhen Lu:Department of Mathematics, Wayne State University, Detroit, MI 48202, USA


Dachun Yang:School of Mathematical Sciences, Beijing Normal University, Beijing 100875, People’sRepublic of China


Product Theory on Spaces of Homogeneous Type

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