# Littlewood–Paley Characterization for Musielak–Orlicz–Hardy Spaces Associated with Self-Adjoint Operators

Authors: Jiawei Shen; Shunchao Long; Yu-long Deng
Journal: Journal of Function Spaces
 (2022)

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## Abstract

LetX,d,μ be a metric measure space endowed with a metric d and a non-negative Borel doubling measure μ. Let L be a non-negative self-adjoint operator on L2X. Assume that the (heat) kernel associated to the semigroup e−tL satisfies a Gaussian upper bound. In this paper, we prove that the Musielak–Orlicz–Hardy space Hφ,LX associated with L in terms of the Lusin-area function and the Musielak–Orlicz–Hardy space HL,G,φX associated with L in terms of the Littlewood–Paley function coincide and their norms are equivalent. To do this, we first establish the discrete characterization of these two spaces. It improves the known results in the literature.


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## Body

## 1. Introduction

The metric measure spaceX,d,μ is a set X equipped with a metric d and a non-negative Borel doubling measure μ on X. Let f∈L2X and L be a densely defined operator on L2X which satisfies the following two conditions:(i)
(H1)L is a non-negative self-adjoint operator on L2X.(ii)
(H2) The kernel of e−tL, denoted by ptx,y, is a measurable function on X×X satisfying the Gaussian estimates, i.e., there exist C1,C2>0 such that(1)ptx,y≤C1Vx,te−dx,y2/C2t,holds for all t>0 and x,y∈X, where Vx,t=μBx,t.The Littlewood–Paley functionGLf and Lusin-area function SLf associated with the heat semigroup generated by L are given by(2)GLfx=∫0∞t2Le−t2Lfx2dtt1/2,SLfx=∫0∞∫dx,y<tt2Le−t2Lfx2dμyμBx,tdtt1/2.In this paper, we focus on the characterization of the Musielak–Orlicz–Hardy spacesHφ,L and HL,G,φ, where the operator L satisfies (H1) and (H2) and φ is a growth function (cf. Definition 6 below).Definition 1.
Suppose that the operatorL satisfies (H1) and (H2) and φ is a growth function. A function f∈H2X is said to be in H˜φ,LX if SLf∈LφX (cf. Definition 7 below). Moreover, we define(3)fH˜φ,LX=SLfLφ=infλ∈0,∞:∫Xφx,SLfxλdμx≤1.
The Musielak–Orlicz–Hardy spaceHφ,LX is defined to be the complement space of H˜φ,LX.Definition 2.
Suppose that the operatorL satisfies (H1) and (H2) and φ is a growth function. A function f∈H2X is said to be in H˜L,G,φX if GLf∈LφX. Moreover, we define(4)fH˜L,G,φX=GLfLφ=infλ∈0,∞:∫Xφx,GLfxλdμx≤1.
The Musielak–Orlicz–Hardy spaceHL,G,φX is defined to be the complement space of H˜L,G,φX.Recently, the study of Hardy spaces associated with operators has been attracting great interest. It was initiated by Auscher et al. who studied the Hardy spaceHL1ℝn with operators L in [1], where the heat kernel of L satisfies the pointwise Poisson upper bounded condition. Later on, Duong and Yan [2, 3] presented the adapted BMO theory on condition that the heat kernel of L satisfies the pointwise Gaussian estimate. In [4], Yan established the theory of Hardy space HLpℝn for 0<p<1 associated with the operator L satisfying Davies–Gaffney estimates.It is a natural question to ask the behavior of weighted Hardy spaceHL,ωpℝn associated with an operator L and an appropriate weight ω. A pioneering investigation work of the weighted Hardy space HL,ω1ℝn associated with the Schrödinger operator L was the paper by Song and Yan [5]. In 2016, Duong et al. [6] considered the weighted Hardy spaces HL,S,ωpℝn and HL,G,ωpℝn on homogeneous space X for 0<p≤1 and obtained the equivalence of these two kinds by adding Moser-type conditions, where the operator L has the kernel satisfying Gaussian upper bound. Shortly after that, the equivalence of these two kinds spaces was characterized by Hu [7] without assuming the Moser-type boundedness condition.In 2014, Ky [8] introduced the Musielak–Orlicz–Hardy space Hφℝn by using growth function φ. Naturally, the Musielak–Orlicz–Hardy space Hφ,L which is defined by means of the Lusin-area function associated with an operator L was introduced and studied in [9], where L satisfies Davies–Gaffney estimates. Unfortunately, the characterization of Hφ,L required an extra assumption that φ satisfies the uniformly reverse Hölder condition (cf. [9]).Motivated by the above, we are concerned with the Musielak–Orlicz spacesHφ,LX and HL,G,φX which we define by means of the Lusin-area function and the Littlewood–Paley function on homogeneous space X. Our aim in the present paper is to prove that the two kinds of Musielak–Orlicz spaces are equivalent. Our main result is stated as follows.Theorem 1.
Suppose that the operatorL satisfies (H1) and (H2) and φ is a growth function of uniformly lower type p1. Then, the spaces Hφ,LX and HL,G,φX coincide and their norms are equivalent.Theorem1 obtains the behavior of Littlewood–Paley g-function GL on Hφ,L and partly improves the result in [9]. To make it clear, we first establish the discrete characterization of the Musielak–Orlicz spaces Hφ,LX and HL,G,φX and state these results as follows.Theorem 2.
Suppose that the operatorL satisfies (H1) and (H2) and φ is a growth function of uniformly lower type p1. Let f∈Hφ,LX∩L2X. Then, for all M∈ℕ with M>nqφ/2p1, f has an ATℒ,M-expansion such that(5)fHφ,LX≅WfLφX.Theorem 3.
Suppose that the operatorL satisfies (H1) and (H2) and φ is a growth function of uniformly lower type p1. Let f∈HL,G,φX∩L2X. Then, for all M∈ℕ with M>nqφ/2p1, f has an ATℒ,M-expansion such that(6)fHL,G,φX≅WfLφX.Theorems2 and 3 extend the results in [6, 7], respectively. Also, we extend the results in [9] by removing the assumption of uniformly reverse Hölder condition. As a consequence of Theorems 2 and 3, we immediately get Theorem 1.The paper is organized as follows. Section2 contains some basic definitions and lemmas concerning metric measure spaces, growth functions, Musielak–Orlicz space, and ATℒ,M-family. The aim of Section 3 is to prove Theorem 2 and establish the characterization of Musielak–Orlicz–Hardy space Hφ,L. We develop a method to unify the different control terms of inner integral. The aim of Section 4 is to prove Theorem 3 and set up the characterization of Musielak–Orlicz–Hardy space HL,G,φ. We borrow the ideas from [6,10]. Consequently, we get that the characterization of Musielak–Orlicz–Hardy space by means of Hφ,L and HL,G,φ is equivalent.Most of the notations we use are standard.C denotes a positive constant that may change from line to line and we use the subscript for the sake of eliminating confusion. We write A≅B if there exist constants C1 and C2 which are independent of A and B such that C1B≤A≤C2B. For a measurable set A, A denotes the Lebesgue measure of A and χA is the characteristic function.


## 2. Basic Concepts and Lemmas

### 2.1. Metric Measure Spaces

A metric measure spaceX,d,μ is a set X equipped with a metric d and a non-negative Borel doubling measure μ on X. Fix x∈X and let r∈0,∞, and we denote the open ball centered at x with radius r by(7)Bx,r=y∈X:dx,y<r,and set Vx,r=μBx,r.Definition 3.
A space of homogeneous typeX,d,μ is a set X with a metric d and a non-negative measure μ on X, so that there exists a constant CD∈1,∞ such that for all x∈X and r>0,(8)Vx,2r≤CDVx,r<∞.Definition3 was introduced by Coifman and Weiss [11]. The property of μ in (8) is the doubling condition and it implies the strong n homogeneity property, i.e., for some constant C>0 and homogeneity n,(9)Vx,λr≤CλnVx,rholds uniformly for all λ∈1,∞, x∈X, and r>0.LetCD be as in (8) and set m=log2CD, and Grigor’yan et al. have shown that (see [12])(10)Vx,R≤CDR+dx,yrmVy,rholds for all x,y∈X and 0<r≤R<∞. It is easy to verify, by doubling condition (8), that for any N>n, there exists a constant CN such that for all x∈X and t>0,(11)∫X1+t−1dx,y−Ndμy≤CNVx,t.The dyadic cube decomposition on spaces of homogeneous type comes from Christ [13] as follows:Lemma 1.
LetX,d,μ be a space of homogeneous type. Then, there exists a collection of open subsets Qαk⊂X:k∈ℤ,α∈Ik and constants δ∈0,1 and 0<C1,C2<∞ such that(i)
μX∖∪αQαk=0,∀k; for each fixed k, if α≠β, then Qαk∩Qβk=∅.(ii)
For anyα,β,k,l, if k≤l, then either Qβl⊂Qαk or Qβl∩Qαk=∅.(iii)
For eachk,α and each l<k, there is a unique β∈Il such that Qαk⊂Qβl.(iv)
DiameterQαk≤C1δk.(v)
EachQαk contains some ball Bzαk,C2δk, where zαk∈X.The setsQαk are analogues of the Euclidean dyadic cubes; it may help to think of Qαk as being essentially a cube of ball of diameter roughly δk with center zαk. We then set ℓQαk=C1δk. It is worthy pointing out that the precise value of C1 is non-essential (cf. Christ [13]). Here and in what follows, we assume C1=δ−1.


### 2.2. Growth Functions

We first recall the Orlicz function. A non-decreasing functionΦ:0,∞⟶0,∞ is called an Orlicz function if Φ0=0, Φt>0 for all t∈0,∞ and limt⟶∞Φt=∞ (cf. Yang [9]).The functionΦ is said to be of upper type p (resp., lower type p) for some p∈0,∞, if for all t∈1,∞ (resp., t∈0,1) and s∈0,∞, there is a constant C>0 such that Φst≤CtpΦs. Φ is said to be of type p1,p2 if it is of both upper type p1 and lower type p2.Given a functionφ:X×0,∞⟶0,∞, for any x∈X, φx,⋅ is an Orlicz function. If there exists a constant C>0 such that for all x∈X, t∈1,∞ (resp., t∈0,1)) and s∈0,∞,(12)φx,st≤Ctpφx,s,then φ is said to be of uniformly upper type p (resp., uniformly lower type p). Moreover, φ is said to be of positive uniformly upper type (resp., uniformly lower type) if it is of uniformly upper type (resp., uniformly lower type) p for some p∈0,∞.Letφ:X×0,∞⟶0,∞. If for all t∈0,∞, x↦φx,t is measurable and for all bounded subsets K of X,(13)∫Ksupt∈0,∞φx,t∫Kφy,tdμy−1dμx<∞.Then,φ⋅,t is said to be uniformly locally integrable (cf. [8]).We next recall the uniformly Muckenhoupt condition in [9, 14].Definition 4.
Letφ:X×0,∞⟶0,∞ be uniformly locally integrable. The function φ⋅,t is said to satisfy the uniformly Muckenhoupt condition for some q∈1,∞, denoted by φ∈AqX, if(14)Aqφ=supt∈0,∞supB⊂X1μB∫Bφx,tdμx1μB∫Bφy,t−q′/qdμxq/q′<∞,when q∈1,∞ and 1/q+1/q′=1, or(15)A1φ=supt∈0,∞supB⊂X1μB∫Bφx,tdμxesssupy∈Bφy,t−1<∞.
Here the first supremum is taken over allt∈0,∞ and the second one is taken over all balls B⊂X.We defineA∞X=∪q∈1,∞AqX and let(16)qφ=infq∈1,∞:φ∈AqXbe the critical indices of φ. Moreover, we denote(17)φE,t=∫Eφx,tdμx,for any measurable subset E of X and t∈0,∞. Let ℳ be the Hardy–Littlewood maximal function on X, namely, for all x∈X,(18)ℳfx=supB∋x1μB∫Bfydμy,where the supremum is taken over all balls B containing x. The following lemma on the properties of A∞X is Lemma 2.8 in [9].Lemma 2.
(i)
A1X⊂ApX⊂AqX for 1≤p≤q<∞.(ii)
Ifφ∈ApX with p∈1,∞, then there exist some q∈1,p such that φ∈AqX.(iii)
Ifφ∈ApX with p∈1,∞, then there exists a constant C>0 such that for all measurable functions f on X and t∈0,∞,(19)∫Xℳfxpφx,tdμx≤C∫Xfxpφx,tdμx.(iv)
Ifφ∈ApX with p∈1,∞, then there exists a constant C>0 such that for all balls B⊂X and measurable set E⊂B and t∈0,∞,(20)φB,tφE,t≤CμBμEp.We now introduce the notion of growth functions (cf. [8, 9]).Definition 5.
Letφ:X×0,∞⟶0,∞. Then, φx,t is a Musielak–Orlicz function, if(i)
φx,⋅:0,∞⟶0,∞ is an Orlicz function for all x∈X.(ii)
φ⋅,t is a measurable function for all t∈0,∞.Definition 6.
Letφ:X×0,∞⟶0,∞. Then, φ is called a growth function, if the following hold.(i)
φ is a Musielak–Orlicz function.(ii)
φ∈A∞X.(iii)
φ is of positive uniformly upper type p1 for some p1∈0,1 and of uniformly lower type p2 for some p2∈0,1.Lemma 3.
Letφ be a growth function and set φ˜x,t=∫0tφx,s/sds for all x,t∈X×0,∞. Then, φ˜ is a growth function, which is equivalent to φ, and φ˜x,⋅ is continuous and strictly increasing.


### 2.3. Musielak–Orlicz Spaces

In this section, we study the Musielak–Orlicz spaces associated with the growth functionφ.Definition 7.
The Musielak–Orlicz spaceLφX denotes the set of all measurable function f on X with ∫Xφx,fxdμx<∞ and the Luxembourg norm(21)fLφX=infλ∈0,∞:∫Xφx,fxλdμ≤1.The spaceLφℓp,X is defined to be the set of all fjj∈ℤ satisfying ∑jfjp1/p∈LφX and let(22)fjjLφℓp,X=∑jfjp1/pLφX.We have the following Fefferman–Stein vector-valued inequality of Musielak–Orlicz type (cf. [15]).Lemma 4.
Letp∈1,∞, φ be a Musielak–Orlicz function with uniformly lower type p1 and upper type p2, q∈1,∞, and φ∈AqX. If qφ<p1≤p2<∞, then there exists a constant C>0 such that, for all fjj∈ℤ∈Lφℓp,X,(23)∫Xφx,∑jℳfjxp1/pdμx≤C∫Xφx,∑jfjxp1/pdμx.Corollary 1.
Letp,φ be as in Lemma 4. Then, for all r∈0,p1/qφ and fjj∈ℤ∈Lφℓp,X, there exists a constant C>0 such that(24)∫Xφx,∑jℳfjxp1/rpdμx≤C∫Xφx,∑jfjxp1/rpdμx.Proof.
Fixr∈0,p1/qφ and let φ˜x,t=φx,t1/r. We claim that φ˜ is of uniformly lower type p1/r and upper type p2/r. In fact, there exist constants C1,C2>0 such that(25)φ˜x,st=φx,s1/rt1/r≤C1tp1/rφx,s1/r=C1tp1/rφ˜x,s,φ˜x,st=φx,s1/rt1/r≤C2tp2/rφx,s1/r=C2tp2/rφ˜x,s,for all t∈1,∞,x∈X and s∈0,∞. Meanwhile, one has(26)qφ˜<p1r≤p2r<∞,since qφ=qφ˜. Therefore, Lemma 4 yields(27)∫Xφx,∑jℳfjxp1/rpdμx=∫Xφ˜x,∑jℳfjxp1/pdμx≤C∫Xφ˜x,∑jfjxp1/pdμx=C∫Xφx,∑jfjxp1/rpdμx.
It finishes the proof of Corollary1.


### 2.4.ATℒ,M-Family and Decomposition Theorem

In this section, we assume that the spaceX satisfies the strong homogeneity property (9) with homogeneous dimension n. In view of Lemma 1, the space X possesses a dyadic decomposition analogous to the Euclidean dyadic cubes, i.e., there exists a collection of open subsets Qαk⊂X:k∈ℤ,α∈Ik such that for every k∈ℤ,(28)X=∪α∈IkQαk,where Ik is some index set and Qαk has the properties as in Lemma 1. Such open subsets Qαk⊂X:k∈ℤ,α∈Ik are said to be a family of dyadic cubes of X (cf. [6]).Definition 8.
Suppose that the operatorL satisfies (H1) and (H2) and M∈ℕ. Then, a collection of functions aQQ:Dyadic in L2X is said to be an ATℒ,M-family associated with an operator L, if for every dyadic Q, there exists a function DL2M such that(i)
aQ=LMbQx.(ii)
suppLkbQ⊂3Q,k=0,1,…,2M.(iii)
ℓQ2LkbQx≤ℓQ2MVQ−1/2,k=0,1,…,2M.
Here,DT denotes the domain of an unbounded operator T and Tk, which is the k-fold composition of T with itself, in the sense of unbounded operators.For a functionf in L2X, if there exists sequence s=sQQ:dyadic,0≤sQ<∞, and an ATℒ,M-family aQQ:dyadic in L2X such that(29)f=∑Q:dyadicsQaQ,we say that f has an ATℒ,M-expansion. Then, we denote the function related to the sequence s=sQQ:dyadic by Wfx and(30)Wfx=∑Q:dyadicVQ−1/2sQχQx21/2.With the notation above, we have the following characterization ofL2X.Proposition 1.
Suppose that the operatorL satisfies (H1) and (H2). Let f∈L2X. Then, for M∈ℕ, f has an ATℒ,M-expansion(31)f=∑Q:dyadicsQaQ.
Moreover, letQαk and δ be as in Lemma 1. Then,(32)sQαk=∫δk+1δk∫Qαkt2Le−t2Lfy2dμydtt2.Proof.
The proof of Proposition1 can be found in [6, Theorem 3.2].


## 2.1. Metric Measure Spaces

A metric measure spaceX,d,μ is a set X equipped with a metric d and a non-negative Borel doubling measure μ on X. Fix x∈X and let r∈0,∞, and we denote the open ball centered at x with radius r by(7)Bx,r=y∈X:dx,y<r,and set Vx,r=μBx,r.Definition 3.
A space of homogeneous typeX,d,μ is a set X with a metric d and a non-negative measure μ on X, so that there exists a constant CD∈1,∞ such that for all x∈X and r>0,(8)Vx,2r≤CDVx,r<∞.Definition3 was introduced by Coifman and Weiss [11]. The property of μ in (8) is the doubling condition and it implies the strong n homogeneity property, i.e., for some constant C>0 and homogeneity n,(9)Vx,λr≤CλnVx,rholds uniformly for all λ∈1,∞, x∈X, and r>0.LetCD be as in (8) and set m=log2CD, and Grigor’yan et al. have shown that (see [12])(10)Vx,R≤CDR+dx,yrmVy,rholds for all x,y∈X and 0<r≤R<∞. It is easy to verify, by doubling condition (8), that for any N>n, there exists a constant CN such that for all x∈X and t>0,(11)∫X1+t−1dx,y−Ndμy≤CNVx,t.The dyadic cube decomposition on spaces of homogeneous type comes from Christ [13] as follows:Lemma 1.
LetX,d,μ be a space of homogeneous type. Then, there exists a collection of open subsets Qαk⊂X:k∈ℤ,α∈Ik and constants δ∈0,1 and 0<C1,C2<∞ such that(i)
μX∖∪αQαk=0,∀k; for each fixed k, if α≠β, then Qαk∩Qβk=∅.(ii)
For anyα,β,k,l, if k≤l, then either Qβl⊂Qαk or Qβl∩Qαk=∅.(iii)
For eachk,α and each l<k, there is a unique β∈Il such that Qαk⊂Qβl.(iv)
DiameterQαk≤C1δk.(v)
EachQαk contains some ball Bzαk,C2δk, where zαk∈X.The setsQαk are analogues of the Euclidean dyadic cubes; it may help to think of Qαk as being essentially a cube of ball of diameter roughly δk with center zαk. We then set ℓQαk=C1δk. It is worthy pointing out that the precise value of C1 is non-essential (cf. Christ [13]). Here and in what follows, we assume C1=δ−1.


## 2.2. Growth Functions

We first recall the Orlicz function. A non-decreasing functionΦ:0,∞⟶0,∞ is called an Orlicz function if Φ0=0, Φt>0 for all t∈0,∞ and limt⟶∞Φt=∞ (cf. Yang [9]).The functionΦ is said to be of upper type p (resp., lower type p) for some p∈0,∞, if for all t∈1,∞ (resp., t∈0,1) and s∈0,∞, there is a constant C>0 such that Φst≤CtpΦs. Φ is said to be of type p1,p2 if it is of both upper type p1 and lower type p2.Given a functionφ:X×0,∞⟶0,∞, for any x∈X, φx,⋅ is an Orlicz function. If there exists a constant C>0 such that for all x∈X, t∈1,∞ (resp., t∈0,1)) and s∈0,∞,(12)φx,st≤Ctpφx,s,then φ is said to be of uniformly upper type p (resp., uniformly lower type p). Moreover, φ is said to be of positive uniformly upper type (resp., uniformly lower type) if it is of uniformly upper type (resp., uniformly lower type) p for some p∈0,∞.Letφ:X×0,∞⟶0,∞. If for all t∈0,∞, x↦φx,t is measurable and for all bounded subsets K of X,(13)∫Ksupt∈0,∞φx,t∫Kφy,tdμy−1dμx<∞.Then,φ⋅,t is said to be uniformly locally integrable (cf. [8]).We next recall the uniformly Muckenhoupt condition in [9, 14].Definition 4.
Letφ:X×0,∞⟶0,∞ be uniformly locally integrable. The function φ⋅,t is said to satisfy the uniformly Muckenhoupt condition for some q∈1,∞, denoted by φ∈AqX, if(14)Aqφ=supt∈0,∞supB⊂X1μB∫Bφx,tdμx1μB∫Bφy,t−q′/qdμxq/q′<∞,when q∈1,∞ and 1/q+1/q′=1, or(15)A1φ=supt∈0,∞supB⊂X1μB∫Bφx,tdμxesssupy∈Bφy,t−1<∞.
Here the first supremum is taken over allt∈0,∞ and the second one is taken over all balls B⊂X.We defineA∞X=∪q∈1,∞AqX and let(16)qφ=infq∈1,∞:φ∈AqXbe the critical indices of φ. Moreover, we denote(17)φE,t=∫Eφx,tdμx,for any measurable subset E of X and t∈0,∞. Let ℳ be the Hardy–Littlewood maximal function on X, namely, for all x∈X,(18)ℳfx=supB∋x1μB∫Bfydμy,where the supremum is taken over all balls B containing x. The following lemma on the properties of A∞X is Lemma 2.8 in [9].Lemma 2.
(i)
A1X⊂ApX⊂AqX for 1≤p≤q<∞.(ii)
Ifφ∈ApX with p∈1,∞, then there exist some q∈1,p such that φ∈AqX.(iii)
Ifφ∈ApX with p∈1,∞, then there exists a constant C>0 such that for all measurable functions f on X and t∈0,∞,(19)∫Xℳfxpφx,tdμx≤C∫Xfxpφx,tdμx.(iv)
Ifφ∈ApX with p∈1,∞, then there exists a constant C>0 such that for all balls B⊂X and measurable set E⊂B and t∈0,∞,(20)φB,tφE,t≤CμBμEp.We now introduce the notion of growth functions (cf. [8, 9]).Definition 5.
Letφ:X×0,∞⟶0,∞. Then, φx,t is a Musielak–Orlicz function, if(i)
φx,⋅:0,∞⟶0,∞ is an Orlicz function for all x∈X.(ii)
φ⋅,t is a measurable function for all t∈0,∞.Definition 6.
Letφ:X×0,∞⟶0,∞. Then, φ is called a growth function, if the following hold.(i)
φ is a Musielak–Orlicz function.(ii)
φ∈A∞X.(iii)
φ is of positive uniformly upper type p1 for some p1∈0,1 and of uniformly lower type p2 for some p2∈0,1.Lemma 3.
Letφ be a growth function and set φ˜x,t=∫0tφx,s/sds for all x,t∈X×0,∞. Then, φ˜ is a growth function, which is equivalent to φ, and φ˜x,⋅ is continuous and strictly increasing.


## 2.3. Musielak–Orlicz Spaces

In this section, we study the Musielak–Orlicz spaces associated with the growth functionφ.Definition 7.
The Musielak–Orlicz spaceLφX denotes the set of all measurable function f on X with ∫Xφx,fxdμx<∞ and the Luxembourg norm(21)fLφX=infλ∈0,∞:∫Xφx,fxλdμ≤1.The spaceLφℓp,X is defined to be the set of all fjj∈ℤ satisfying ∑jfjp1/p∈LφX and let(22)fjjLφℓp,X=∑jfjp1/pLφX.We have the following Fefferman–Stein vector-valued inequality of Musielak–Orlicz type (cf. [15]).Lemma 4.
Letp∈1,∞, φ be a Musielak–Orlicz function with uniformly lower type p1 and upper type p2, q∈1,∞, and φ∈AqX. If qφ<p1≤p2<∞, then there exists a constant C>0 such that, for all fjj∈ℤ∈Lφℓp,X,(23)∫Xφx,∑jℳfjxp1/pdμx≤C∫Xφx,∑jfjxp1/pdμx.Corollary 1.
Letp,φ be as in Lemma 4. Then, for all r∈0,p1/qφ and fjj∈ℤ∈Lφℓp,X, there exists a constant C>0 such that(24)∫Xφx,∑jℳfjxp1/rpdμx≤C∫Xφx,∑jfjxp1/rpdμx.Proof.
Fixr∈0,p1/qφ and let φ˜x,t=φx,t1/r. We claim that φ˜ is of uniformly lower type p1/r and upper type p2/r. In fact, there exist constants C1,C2>0 such that(25)φ˜x,st=φx,s1/rt1/r≤C1tp1/rφx,s1/r=C1tp1/rφ˜x,s,φ˜x,st=φx,s1/rt1/r≤C2tp2/rφx,s1/r=C2tp2/rφ˜x,s,for all t∈1,∞,x∈X and s∈0,∞. Meanwhile, one has(26)qφ˜<p1r≤p2r<∞,since qφ=qφ˜. Therefore, Lemma 4 yields(27)∫Xφx,∑jℳfjxp1/rpdμx=∫Xφ˜x,∑jℳfjxp1/pdμx≤C∫Xφ˜x,∑jfjxp1/pdμx=C∫Xφx,∑jfjxp1/rpdμx.
It finishes the proof of Corollary1.


## 2.4.ATℒ,M-Family and Decomposition Theorem

In this section, we assume that the spaceX satisfies the strong homogeneity property (9) with homogeneous dimension n. In view of Lemma 1, the space X possesses a dyadic decomposition analogous to the Euclidean dyadic cubes, i.e., there exists a collection of open subsets Qαk⊂X:k∈ℤ,α∈Ik such that for every k∈ℤ,(28)X=∪α∈IkQαk,where Ik is some index set and Qαk has the properties as in Lemma 1. Such open subsets Qαk⊂X:k∈ℤ,α∈Ik are said to be a family of dyadic cubes of X (cf. [6]).Definition 8.
Suppose that the operatorL satisfies (H1) and (H2) and M∈ℕ. Then, a collection of functions aQQ:Dyadic in L2X is said to be an ATℒ,M-family associated with an operator L, if for every dyadic Q, there exists a function DL2M such that(i)
aQ=LMbQx.(ii)
suppLkbQ⊂3Q,k=0,1,…,2M.(iii)
ℓQ2LkbQx≤ℓQ2MVQ−1/2,k=0,1,…,2M.
Here,DT denotes the domain of an unbounded operator T and Tk, which is the k-fold composition of T with itself, in the sense of unbounded operators.For a functionf in L2X, if there exists sequence s=sQQ:dyadic,0≤sQ<∞, and an ATℒ,M-family aQQ:dyadic in L2X such that(29)f=∑Q:dyadicsQaQ,we say that f has an ATℒ,M-expansion. Then, we denote the function related to the sequence s=sQQ:dyadic by Wfx and(30)Wfx=∑Q:dyadicVQ−1/2sQχQx21/2.With the notation above, we have the following characterization ofL2X.Proposition 1.
Suppose that the operatorL satisfies (H1) and (H2). Let f∈L2X. Then, for M∈ℕ, f has an ATℒ,M-expansion(31)f=∑Q:dyadicsQaQ.
Moreover, letQαk and δ be as in Lemma 1. Then,(32)sQαk=∫δk+1δk∫Qαkt2Le−t2Lfy2dμydtt2.Proof.
The proof of Proposition1 can be found in [6, Theorem 3.2].


## 3. The Proof of Theorem2

In this section, we establish a characterization of the Musielak–Orlicz–Hardy spaceHφ,L, where the operator L satisfies (H1) and (H2) and φ is a growth function.For everyv∈0,∞ and x∈X, let Γvx=y,t∈X×0,∞:dx,y<vt be the cone of aperture v and vertex x∈X. For any closed subset F of X, we denote the union of all cones with vertices in F by(33)ℛvF=∪x∈FΓvx.Whenv=1, Γx and ℛF stand for Γ1x and ℛ1x, respectively. Given an open subset O of X, we establish Lemma 5 of ℛO∁ on the geometric properties. We also remark that Aguilera and Segovia [16] obtained the same result in the case of Euclidean space.Lemma 5.
Suppose thatX,d,μ is a space of homogeneous type and there exists a constant CD>1 such that (8) holds. Let O be an open subset of X and F=O∁. For v>1, we denote O∗ by(34)O∗=x∈X:ℳχOx>4v−2log2CD,and write F∗=O∗∁. Then,(i)
ℛvF∗⊂ℛF.(ii)
There exists a constantCv such that(35)Vz,t<CvμBz,t∩Fhods for z,t∈ℛvF∗.Proof.
It suffices to show that the lemma holds whenℛvF∗≠∅ since it is trivial if ℛvF∗=∅. We first prove (i) on the condition that ℛvF∗≠∅, which implies O≠X.
Letz,t∈ℛvF∗. We thus have z∈F or z∈O. It is easy to see that z,t∈ℛF since dz,z=0<t in the case z∈F and then ℛvF∗⊂ℛF holds. The proof of (i) is reduced to the verification in the case z∈O.
Supposez∈O and let δ=distz,F. Then, 0<δ<∞ and Bz,δ⊂O since F is closed and non-empty. For every z,t∈ℛvF∗, we have y∈F∗ such that dz,y<vt. Thus, writing r=δ+dz,y, we get Bz,δ⊂By,r and(36)Bz,δ⊂Bz,δ∩O⊂By,r∩O.
Hence,(37)Vz,δ≤μBy,r∩O≤4v−2log2CDVy,r.
By using (10) twice, we have(38)Vy,r≤CDrδ−1log2CDVy,δ≤CD2rδ−1log2CD1+δ−1dy,zlog2CDVz,δ=2rδ−12log2CDVz,δ,and then(39)δ≤r2v=δ+dz,y2v<δ+vt2v.
It follows thatδ<t since v>1. Recalling the definition of δ, we get x∈F such that dx,z<t, which implies z,t∈ℛF. It completes the proof of (i).
Next, we prove (ii). Givenz,t∈ℛvF∗, we get y∈F∗ such that dz,y<vt. Thus, Bz,t⊂By,1+vt and(40)μBz,t∩O≤μBy,1+vt∩O≤4v−2log2CDVy,1+vt.
Therefore,(41)μBz,t∩O≤4v−2log2CDCDVy,t≤CD24v−2log2CD1+vlog2CD1+t−1dy,zlog2CDVz,t<1+v2v2log2CDVz,t.
We obtain(42)1−1+v2v2log2CDVz,t<μBz,t∩F,since Vz,t=μBz,t∩O+μBz,t∩F, and complete the proof of (ii). It finishes the proof of Lemma 5.For allv∈0,∞,f∈L2X and x∈X, the variant Lusin-area function associated with L is given by(43)SL,vfx=∫0∞∫dx,y<vtt2Le−t2Lfy2dμyVx,tdtt1/2.Lemmas6 and 7 extend the results in [14, 16] for the operator SL,v.Lemma 6.
Suppose that the operatorL satisfies (H1) and (H2). Let φ∈ApX for 1≤p<∞ and O,O∗,F,F∗ be as in Lemma 5. Then, there exists a finite constant C, which is independent of O, such that for all λ∈0,∞ and f∈L2X,(44)∫F∗SL,vfx2φx,λdμx≤C∫FSLfx2φx,λdμx,where SL is the short hand of SL,1.Proof.
Givenx∈F∗ and y,t∈Γvx, we observe that dx,y<vt and hence by (10),(45)Vx,t−1≤CD1+t−1dx,ylog2CDVy,t−1<CD1+vlog2CDVy,t−1.
It follows that(46)∫F∗SL,vfx2φx,λdμx≤CD1+vlog2CD∫F∗∫Γvxt2Le−t2Lfy2dμydtVy,ttφx,λdμx=Cv,D∫ℛvF∗t2Le−t2Lfy2Vy,t−1φBy,vt∩F∗,λdμydtt.
Then, applying Lemma2 to the sets By,t and By,vt, By,t∩F and By,t, respectively, we get(47)φBy,vt,λ≤C2vlog2CDφBy,t,λ,(48)φBy,t,λ≤CVy,tμBy,t∩FpφBy,t∩F,λ.
Therefore, (47), (48), and Lemma 5 yield(49)φBy,vt,λ≤CφBy,t∩F,λ.
Thus,∫F∗SL,vfx2φx,λdμx is bounded by(50)C∫ℛvF∗t2Le−t2Lfy2Vy,t−1φBy,t∩F,λdμydtt.
Finally, in view ofℛvF∗⊂ℛF (see Lemma 5), it follows immediately that (50) is bounded by(51)C∫ℛFt2Le−t2Lfy2Vy,t−1φBy,t∩F,λdμydtt=C∫F∫Γvxt2Le−t2Lfy2dμydtVy,ttφx,λdμx≤C∫FSLfx2φx,λdμx,where we use the fact that(52)Vy,t−1≤CD1+t−1dx,ylog2CDVx,t−1<CD2Vx,t−1,for y,t∈Γvx in the last line. It finishes the proof of Lemma 6.Lemma 7.
Suppose that the operatorL satisfies (H1) and (H2). Let φ be a growth function and φ∈AqX for 1<q<∞. Then, there exists a constant Cv>0 such that(53)∫Xφx,SL,vfxdμx≤Cv∫Xφx,SLfxdμxholds for all v∈0,∞ and all measurable functions f.Proof.
It suffices to show that Lemma7 holds in the case v∈1,∞ since the conclusion is trivial if v∈0,1. Given λ∈0,∞, we introduce the notations(54)Oλ=x∈X:SLfx>λ,Oλ∗=x∈X:ℳχOλx>4v−log2CD,where ℳ is the Hardy–Littlewood maximal function. Noting that φ∈AqX, Lemma 2 yields(55)φOλ∗,λ=φx∈X:ℳχOλx>4v−log2CD,λ≤∫X4vqlog2CDℳχOλxqφx,λdμx≤CφOλ,λ.
WritingFλ=Oλ∁,Fλ∗=Oλ∗∁ and applying Lemma 6, we obtain(56)∫Fλ∗SL,vfx2φx,λdμx≤∫FλSLfx2φx,λdμx.
Thus, by using (55) and (56), we have(57)φx∈X:SL,vfx>λ,λ≤φOλ∗,λ+φx∈Fλ∗:SL,vfx>λ,λ≤CφOλ,λ+1λ2∫Fλ∗SL,vfx2φx,λdμx≤CφOλ,λ+1λ2∫FλSLfx2φx,λdμx≤CφOλ,λ+1λ2∫0ttφx∈X:SLfx>t,λdt.
Therefore, we employ (57) together with the assumption v∈1,∞, Lemma 3, and the uniformly upper type 1 of φ to get(58)∫Xφx,SL,vfxdμx≤C∫0∞1λφx∈X:SL,vfx>λ,λdλ≤C∫0∞1λφOλ,λdλ+C∫0∞1λ3∫0ttφx∈X:SLfx>t,λdtdλ≤C∫0∞1λφx∈X:SLfx>λ,λdλ+C∫0∞1λ3∫0ttφx∈X:SLfx>t,λdtdλ≤C∫Xφx,SLfxdμx+C∫0∞φx∈X:SLfx>t,t∫t∞dλλ2dt≤C∫Xφx,SLfxdμx.
It finishes the proof of Lemma7.Lemma8 says that the sequence sQαkα∈Ik can be majorized by the Hardy–Littlewood maximal operator ℳ on X,d,μ (cf. [17], pp.147, where we take r=1).Lemma 8.
Suppose0<q≤1 and N>n/q. Fix k∈ℤ and let sQαkα∈Ik be as in Proposition 1. Then, for any subsequence Ik′⊂Ik and for every x∈X,(59)∑α∈Ik′sQαk1+ℓQαk−1dx,yαk≤Cℳ∑α∈Ik′sQαkqχQαk1/q,where yαk is the center of Qαk and C depends only on n and N−n/q.Proof.
of Theorem2. Fix f∈Hφ,LX∩L2X, and we let λ0=fHφ,LX and λ1=WfLφX. It suffices to show that for all λ∈0,∞, we have(60)∫Xφx,SLfxλdμx≅∫Xφx,Wfxλdμx.
In fact, since (60) holds for all λ∈0,∞, there exists a constant C0 such that(61)∫Xφx,SLfxλ1dμx≤C0∫Xφx,Wfxλ1dμx≤C0.
Using (12) and (61), we have(62)∫Xφx,GLfxC1λ1dμx≤1,for some constant C1, which implies λ0≤C1λ1. Analogously, there exists a constant C2 such that λ1≤C2λ0 and we get the desired result.
We now turn to prove (60). Given x,k∈X×ℤ, by Lemma 1, there exists a unique α∈Ik such that x∈Qαk. We denote such Qαk by Qxk and write(63)Wfx=∑k∈ℤ∑α∈IkμQαk−1/2sQαkχQαkx21/2=∑k∈ℤμQxk−1sQxk21/2=∑k∈ℤ∫δk+1δkμQxk−1∫Qxkt2Le−t2Lfy2dμydtt1/2,where constants δ∈0,1 satisfy Lemma 1 and the last line is obtained by using Proposition 1.
Moreover, for any fixedx,k∈X×ℤ, Lemma 1 also tells us that there are zxk∈Qxk and constants C1∈0,1,C2=δ−1 such that(64)Bzxk,C1δk⊂Qxk⊂Bx,C2δk⊂Bx,C2δ−1t,for all t∈δk+1,δk. Consequently,(65)μQαk−1≤Vzxk,C1δk−1≤C1+dx,zxkC1δkmVx,C1δk−1≤CVx,C1δk−1≤CVx,δk−1≤CVx,t−1,where we use (10) and the fact that Vx,δk≤CDC1−mVx,C1δk with C1∈0,1. Hence, (63) and (65) yield(66)Wfx≤C∑k∈ℤ∫δk+1δkVx,t−1∫Bx,C2δ−1tt2Le−t2Lfy2dμydtt1/2=C∫0∞∫dx,y<δ−2tt2Le−t2Lfy2dμyVx,tdtt1/2=CSL,δ−2fx.
Thus, by using Lemma7, we deduce that(67)∫Xφx,SLfxλdμx≥C∫Xφx,Wfxλdμx.
It remains to establish the reverse inequality of (67). Let δ be as in Lemma 1. In view of Proposition 1, we write(68)f=∑k∈ℤ∑α∈IksQαkaQαk,and get(69)SLfx=∫0∞∫dx,y<tt2Le−t2Lfy2dμyVx,tdtt1/2=∫0∞∫dx,y<tt2Le−t2L∑k∈ℤ∑α∈IksQαkaQαky2dμyVx,tdtt1/2=∑j∈ℤ∫δjδj−1∫dx,y<tt2Le−t2L∑k∈ℤ∑α∈IksQαkaQαky2dμyVx,tdtt1/2≤∑j∈ℤ∫δjδj−1∫dx,y<tt2Le−t2L∑k>j∑α∈IksQαkaQαky21/2dμyVx,tdtt+∑j∈ℤ∫δjδj−1∫dx,y<tt2Le−t2L∑k≤j∑α∈IksQαkaQαky2dμyVx,tdtt1/2=I1+I2.
We firstly estimate the inner integral ofI1. For any k>j and α∈Ik, noting aQαk=LMbQαk, we have(70)t2Le−t2LaQαky=t2LM+1e−t2LbQαky=t−2Mt2LM+1e−t2LbQαky.
SinceM>nqφ/2p1 with n given as in (9), we can choose some q satisfying Corollary 1 such that 2M>n/q. Thus, there is some N>0 such that 2M>N>n/q. Then, applying Definition 8, the upper bound of the kernel t2LM+1e−t2L (cf. [18], Proposition 3.1), and (11), we get(71)t2Le−t2LaQαky≤C5Vy,tt−2MℓQαk2MμQαk−1/2∫3Qαke−dy,z2/C6t2dμz≤Ct−2MℓQαk2MμQαk−1/2tt+dy,zαkN,where zαk is the center of Qαk. Since dx,y<t, we further have(72)∫dx,y<tt2Le−t2LaQαky2dμyVx,t1/2≤Ct−2MℓQαk2MμQαk−1/2∫dx,y<tt+dx,yt+dx,zαk2NdμyVx,t1/2≤Ct−2MℓQαk2MμQαk−1/21+t−1dx,zαk−N,
Hence, Lemma8 yields the inner integral of I1 which is bounded by(73)∫dx,y<tt2Le−t2L∑k>j∑α∈IksQαkaQαky2dμyVx,t1/2≤C∑k>j∑α∈Ikt−2MℓQαk2MμQαk−1/2sQαk1+t−1dx,zαkN≤C∑k>jδ2M−Nk−j∑α∈IkμQαk−1/2sQαk1+ℓQαk−1dx,zαkN≤C∑k>jδ2M−Nk−jℳ∑α∈IksQαkqμQαk−q/2χQαkx1/q.
Secondly, we estimate the inner integral ofI2. For any k≤j and α∈Ik, we write(74)t2Le−t2LaQαky=t2e−t2LLaQαky.
Then, using Definition8, Gaussian estimate (1), and inequality (11), we obtain(75)t2Le−t2LaQαky≤C5Vy,tt2ℓQαk−2μQαk−1/2∫3Qαke−dy,z2/C6t2dμz≤Ct2ℓQαk−2μQαk−1/21+ℓQαk−1dy,zαk−N.
Sincedx,y<t≤ℓQαk, we further have(76)∫dx,y<tt2Le−t2LaQαky2dμyVx,t1/2≤Ct2ℓQαk−2μQαk−1/2∫dx,y<tℓQαk+dx,yℓQαk+dx,zαk2NdμyVx,t1/2≤Ct2ℓQαk−2μQαk−1/21+ℓQαkdx,zαk−N,
Hence, Lemma8 yields the inner integral of I2 which is bounded by(77)∫dx,y<tt2Le−t2L∑k≤j∑α∈IksQαkaQαky2dμyVx,t1/2≤C∑k≤j∑α∈Ikt2ℓQαk−2μQαk−1/2sQαk1+ℓQαk−1dx,zαkN≤C∑k≤jδ2j−k∑α∈IkμQαk−1/2sQαk1+ℓQαk−1dx,zαkN≤C∑k≤jδ2j−kℳ∑α∈IksQαkqμQαk−q/2χQαkx1/q.
Fixj∈ℤ, and we let β>0 and(78)τ=1,k>j;−1,k≤j.
WritingFkx=ℳ∑α∈IksQαkqμQαk−q/2χQαkx, we now turn to estimate(79)J=∑kδβτk−jFkx1/q2.
Since(80)δβτk−j=βδβ1−δβ∫δτk−jδτk−j−1sβ−1ds,we have(81)J=Cβ∑k∫δτk−jδτk−j−1Fkx1/qsβ−1ds2=Cβ∫01∑kχEksFkx1/qsβ−1ds2≤Cβ∫01sβ−1ds∫01∑kχEksFkx1/q2sβ−1ds≤Cβ∫01∑kχEksFkx1/q2sβ−1ds=Cβ∑k∫δτk−jδτk−j−1Fkx2/qsβ−1ds=Cβ∑kδβτk−jFkx2/q,where Ek=δτk−j,δτk−j−1. In view of inequalities (73)–(81), taking β=2M−N,τ=1, and β=2,τ=−1 respectively, we get(82)SLfx≤C∑j∈ℤ∫δjδj−1∑k>jδ2M−Nk−jFkx1/q2dtt+∑j∈ℤ∫δjδj−1∑k≤jδ2j−kFkx1/q2dtt1/2≤C∑j∈ℤ∫δjδj−1∑k>jδ2M−Nk−jFkx2/qdtt+∑j∈ℤ∫δjδj−1∑k≤jδ2j−kFkx2/qdtt1/2=C∑k∈ℤFkx2/q∑k>jδ2M−Nk−j+∑k≤jδ2j−k1/2≤C∑k∈ℤFkx2/q1/2.
Therefore, (82) and Corollary 1 yield(83)∫Xφx,SLfxλdμx≤C∫Xφx,λ−1∑k∈ℤFkx2/q1/2dμx≤C∫Xφx,λ−1∑k∈ℤ∑α∈IksQαkqμQαk−q/2χQαkx2/q1/2dμx=C∫Xφx,λ−1∑k∈ℤ∑α∈IksQαk2μQαk−1χQαkx1/2dμx=C∫Xφx,Wfxλdμx,which gives the reverse inequality of (67). It finishes the proof of Theorem 2.


## 4. The Proof of Theorem3

In this section, we establish a characterization of the Musielak–Orlicz–Hardy spaceHL,G,φ, where the operator L satisfies (H1) and (H2) and φ is a growth function. Our proof will borrow some ideals from Duong et al. [6].We first recall some basic definitions and facts about Fefferman–Stein type maximal function, referring to [7] for a complete account.Givenf∈L2X, a>0, and x,t∈X×0,∞, the Fefferman–Stein type maximal function is defined as(84)ℳa,L∗fx,t=esssupy∈Xt2Le−t2Lfy1+t−1dx,ya.Lemma9 is useful (cf. [7]).Lemma 9.
Suppose the operatorL satisfies (H1) and (H2). Let m be as in (10). Then, for any β,r>0 and a>m/2, there exists a constant C>0 such that(85)ℳa,L∗fx,2−ltr≤C∑j=l∞2−j−lβr∫X2−jt2Le−2−jt2LfxrVz,2−l1+2ldx,zardμzholds for all f∈L2X,l∈ℤ,x∈X, and t∈1,2.We also need Lemma10, and its proof is standard, which we omit here.Lemma 10.
Letn and m be as in (9) and (10), and N>n+m. Then, there exists a constant C>0 such that(86)∫XfxVx,t1+t−1dx,yNdμx≤Cℳfyholds for all measurable functions f on X,d,μ, t>0, and each y∈X.Proof.
of Theorem3. Fix f∈HL,G,φX∩L2X, and we let λ0=fHL,G,φX and λ1=WfLφX. It suffices to show that for all λ∈0,∞, we have(87)∫Xφx,GLfxλdμx≅∫Xφx,Wfxλdμx.
In fact, since (87) holds for all λ∈0,∞, there exists a constant C0 such that(88)∫Xφx,GLfxλ1dμx≤C0∫Xφx,Wfxλ1dμx≤C0.
Using (12) and (88), we have(89)∫Xφx,GLfxC1λ1dμx≤1,for some constant C1, which implies λ0≤C1λ1. Analogously, there exists a constant C2 such that λ1≤C2λ0 and we get the desired result.
We now fix arbitraryλ∈0,∞ and turn to prove (88). Given x,k∈X×ℤ, by Lemma 1, there exists a unique α∈Ik such that x∈Qαk. We denote such Qαk by Qxk and write(90)Wfx=∑k∈ℤ∑α∈IkμQαk−1/2sQαkχQαkx21/2=∑k∈ℤμQxk−1sQxk21/2=∑k∈ℤ∫δk+1δkμQxk−1∫Qxkt2Le−t2Lfy2dμydtt1/2,where constants δ∈0,1 satisfy Lemma 1 and the last line is obtained by using Proposition 1.
Moreover, for any fixedx,k∈X×ℤ, Lemma 1 also tells us that there are zxk∈Qxk and constants C3∈0,1,C4>0 such that(91)Bzxk,C3δk⊂Qxk⊂Bx,C4δk⊂Bx,C4δ−1t=Bx,for all t∈δk+1,δk. Consequently, by inequalities (9) and (10), we have(92)∫δk+1δkμQxk−1∫Qxkt2Le−t2Lfy2dμydtt≤∫δk+1δkμBzxk,C3δk−1∫Bx,C4δkt2Le−t2Lfy2dμydtt≤C∫δk+1δkμBx,C4δkμBzxk,C3δkesssupy∈Bx,C4δkt2Le−t2Lfy2dtt≤C∫δk+1δkesssupy∈Bxt2Le−t2Lfy2dtt≤C∫δk+1δkℳa,L∗fx,t2dtt,for some appropriate constant C, where ℳa,L∗fx,t is the Fefferman–Stein type maximal function with some large enough constant a to be chosen, and the last line follows from(93)esssupy∈Bxt2Le−t2Lfy2=esssupy∈Bxt2Le−t2Lfy21+t−1dx,y2a1+t−1dx,y2a≤1+C1δ−1ℳa,L∗fx,t2.
Hence, (90) and (92) yield(94)Wfx≤C∫0∞ℳa,L∗fx,t2dtt1/2≤C∑k∈ℤ∫2−k2−k+1ℳa,L∗fx,t2dtt1/2=C∑k∈ℤ∫12ℳa,L∗fx,2−kt2dtt1/2.
Thus, by using Lemma9, we deduce that for any β,r>0 and a>m/2, there exists a constant C such that(95)ℳa,L∗fx,2−ktr≤C∑j=k∞2−j−kβr∫X2−jt2Le−2−jt2LfzrVz,2−k1+2kdx,zardμz.
As in Corollary1, let r∈0,1 with p=2/r>1. Fix β>0 and choose a>m/2 such that ar>m+n. Then, we integrate on both sides of (95) and employ Minkowski inequality and Lemma 10 to get(96)∫12ℳa,L∗fx,2−ky2dttr/2≤C∫12∑j=k∞2−j−kβr∫X2−jt2Le−2−jt2LfzrVz,2−k1+2kdx,zardμz2/rdttr/2=C∑j=k∞2−j−kβr∫12∫X2−jt2Le−2−jt2LfzrVz,2−k1+2kdx,zardμz2/rdttr/2≤C∑j=k∞2−j−kβr∫X∫122−jt2Le−2−jt2Lfz2dt/tr/2Vz,2−k1+2kdx,zardμz=C∫X∑j=k∞2−j−kβr∫122−jt2Le−2−jt2Lfz2dt/tr/2Vz,2−k1+2kdx,zardμz≤Cℳ∑j=k∞2−j−kβr∫122−jt2Le−2−jt2Lf⋅2dttr/2x=CℳGkx.
Thus, by using Corollary1, we have(97)∫Xφx,Wfxλdμx≤∫Xφx,C∑k∈ℤ∫12ℳa,L∗fx,2−kt2dt/t1/2λdμx=∫Xφx,C∑k∈ℤ∫12ℳa,L∗fx,2−kt2dt/tr/22/r1/2λdμx≤∫Xφx,C∑k∈ℤℳGkx2/r1/2λdμx≤C∫Xφx,∑k∈ℤGkx2/r1/2λdμx.
We now turn to estimateGk2/rx. Since(98)2−j−kβr=βr1−2−βr∫2j−k2j−k+1s−βr−1ds,for any k∈ℤ, we have(99)Gkx=C∑j=k∞∫2j−k2j−k+1∫122−jt2Le−2−jt2Lfx2dttr/2dssβr+1=C∑j=k∞∫1∞∫122−jt2Le−2−jt2Lfx2dttr/2χEjsdssβr+1=C∫1∞∑j=k∞∫122−jt2Le−2−jt2Lfx2dttr/2χEjsdssβr+1,where Ej=2j−k,2j−k+1. Using Hölder’s inequality, we obtain(100)Gkx2/r=C∫1∞∑j=k∞∫122−jt2Le−2−jt2Lfx2dttr/2χEjsdssβr+12/r≤C∫1∞s−βr−1ds4/r2−r×∫1∞∑j=k∞∫122−jt2Le−2−jt2Lfx2dttr/2χEjs2/rdssβr+1=C∫1∞∑j=k∞∫122−jt2Le−2−jt2Lfx2dttχEjsdssβr+1=C∑j=k∞∫2j−k2j−k+1dssβr+1∫122−jt2Le−2−jt2Lfx2dtt=C∑j=k∞2−j−kβr∫122−jt2Le−2−jt2Lfx2dtt.
Summarizing allk∈ℤ, we have(101)∑k∈ℤGkx2/r≤C∑k∈ℤ∑j≥k2−j−kβr∫122−jt2Le−2−jt2Lfx2dtt=C∑j∈ℤ∑k≤j2−j−kβr∫122−jt2Le−2−jt2Lfx2dtt=C1−2−βr−1∑j∈ℤ∫122−jt2Le−2−jt2Lfx2dtt=C∑j∈ℤ∫2−j2−j+1t2Le−t2Lfx2dtt=C∫0∞t2Le−t2Lfx2dtt=CGLfx2.
Therefore, (11), (97), and (101) yield(102)∫Xφx,GLfxλdμx≥C∫Xφx,Wfxλdμx.
It reduces to show the reverse inequality of (102). Let δ be as in Lemma 1. In view of Proposition 1, we write(103)f=∑k∈ℤ∑α∈IksQαkaQαk,and get(104)GLfx=∫0∞t2Le−t2Lfx2dtt1/2=∫0∞t2Le−t2L∑k∈ℤ∑α∈IksQαkaQαkx2dtt1/2=∑j∈ℤ∫δjδj−1t2Le−t2L∑k∈ℤ∑α∈IksQαkaQαkx2dtt1/2≤∑j∈ℤ∫δjδj−1t2Le−t2L∑k>j∑α∈IksQαkaQαkx2dtt1/2+∑j∈ℤ∫δjδj−1t2Le−t2L∑k≤j∑α∈IksQαkaQαkx2dtt1/2=I3+I4.
We now estimate the integrand function ofI3. For any k>j and α∈Ik, noting aQαk=LMbQαk, we have(105)t2Le−t2LaQαkx=t2LM+1e−t2LbQαkx=t−2Mt2LM+1e−t2LbQαkx.
SinceM>nqφ/2p1 with n given as in (9), we can choose some q satisfying Corollary 1 such that 2M>n/q. Thus, there is some N>0 such that 2M>N>n/q. Then, applying Definition 8, the upper bound of the kernel t2LM+1e−t2L (cf. [18], Proposition 3.1), and (11), we get(106)t2Le−t2LaQαkx≤C5Vx,tt−2MℓQαk2MμQαk−1/2∫3Qαke−dx,y2/C6t2dμy≤Ct−2MℓQαk2MμQαk−1/2tt+dx,yαkN,where yαk is the center of Qαk. Hence, by Lemma 8, we have(107)t2Le−t2L∑k>j∑α∈IksQαkaQαkx≤C∑k>j∑α∈Ikt−2MℓQαk2MμQαk−1/2sQαk1+t−1dx,yαkN≤C∑k>jδ2M−Nk−j∑α∈IkμQαk−1/2sQαk1+ℓQαk−1dx,yαkN≤C∑k>jδ2M−Nk−jℳ∑α∈IksQαkqμQαk−q/2χQαkx1/q.
For the integrand function ofI4 in the case k≤j and α∈Ik, we write(108)t2Le−t2LaQαkx=t2e−t2LLaQαkx.
Then, using Definition8, Gaussian estimate (1), and inequality (11), we obtain(109)t2Le−t2LaQαkx≤C5Vx,tt2ℓQαk−2μQαk−1/2∫3Qαke−dx,y2/C6t2dμy≤Ct2ℓQαk−2μQαk−1/21+ℓQαk−1dx,yαk−N.
Consequently,(110)t2Le−t2L∑k≤j∑α∈IksQαkaQαkx≤C∑k≤j∑α∈Ikt2ℓQαk−2μQαk−1/2sQαk1+ℓQαk−1dx,yαkN≤C∑k≤jδ2j−k∑α∈IkμQαk−1/2sQαk1+ℓQαk−1dx,yαkN≤C∑k≤jδ2j−kℳ∑α∈IksQαkqμQαk−q/2χQαkx1/q.
Similar to the discussion in (82) and combining (104)–(110), we get(111)GLfx≤C∑j∈ℤ∫δjδj−1∑k>jδ2M−Nk−jGkx1/q2dtt+∑j∈ℤ∫δjδj−1∑k≤jδ2j−kGkx1/q2dtt1/2≤C∑j∈ℤ∫δjδj−1∑k>jδ2M−Nk−jGkx2/qdtt+∑j∈ℤ∫δjδj−1∑k≤jδ2j−kGkx2/qdtt1/2=C∑k∈ℤGkx2/q∑k>jδ2M−Nk−j+∑k≤jδ2j−k1/2≤C∑k∈ℤGkx2/q1/2.
Therefore, (111) and Corollary 1 yield(112)∫Xφx,GLfxλdμx≤C∫Xφx,λ−1∑k∈ℤGkx2/q1/2dμx≤C∫Xφx,λ−1∑k∈ℤ∑α∈IksQαkqμQαk−q/2χQαkx2/q1/2dμx=C∫Xφx,λ−1∑k∈ℤ∑α∈IksQαk2μQαk−1χQαkx1/2dμx=C∫Xφx,Wfxλdμx,which gives the reverse inequality of (102). It finishes the proof of Theorem 2.



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*Source: 5112954-2022-03-17.xml*