# Boundedness and Continuity of Several Integral Operators with Rough Kernels inWFβSn-1 on Triebel-Lizorkin Spaces

Authors: Feng Liu
Journal: Journal of Function Spaces
 (2018)

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## Abstract

A systematic treatment is given of singular integrals and Marcinkiewicz integrals associated with surfaces generated by polynomial compound mappings as well as related maximal functions with rough kernels inWFβ(Sn-1), which relates to the Grafakos-Stefanov function class. Certain boundedness and continuity for these operators on Triebel-Lizorkin spaces and Besov spaces are proved by applying some criterions of bounds and continuity for several operators on the above function spaces.


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## Body

## 1. Introduction

LetRn(n≥2) be the n-dimensional Euclidean space and Sn-1 denote the unit sphere in Rn equipped with the induced Lebesgue measure dσ. Assume that Ω∈L1(Sn-1) is a homogeneous function of degree zero and satisfies (1)∫Sn-1Ωudσu=0.For a suitable function h defined on R+≔(0,∞), a complex number ρ=ς+iτ(ς,τ∈R with ς>0), and a suitable mapping Γ:Rn→Rn, we consider the singular integral operators Th,Ω,Γ and parametric Marcinkiewicz integral operators Mh,Ω,Γ,ρ in Rn by (2)Th,Ω,Γfx=p.v.∫Rnfx-ΓyhyΩyyndy;(3)Mh,Ω,Γ,ρfx=∫0∞1tρ∫y≤tfx-ΓyhyΩyyn-ρdy2dtt1/2.Define the related maximal operators SΩ,Γ and MΩ,Γ,ρ by (4)SΩ,Γfx=suph∈K2⁡Th,Ω,Γfx,(5)MΩ,Γ,ρfx=suph∈K2⁡Mh,Ω,Γ,ρfx,where K2 is the set of all measurable functions h:R+→R with hL2(R+,r-1dr)≤1.The primary purpose of this paper is to study the bounds and continuity of the singular integral operators and Marcinkiewicz integral operators associated with surfaces generated by polynomial compound mappings as well as related maximal functions with rough kernels inWFβ(Sn-1) on the Triebel-Lizorkin spaces and Besov spaces. Before stating our main results, let us recall some pertinent definitions, notations, and backgrounds.Definition 1 (function classWFβ(Sn-1)).
Forβ>0, the function class WFβ(Sn-1) is the set of all L1(Sn-1) functions Ω which satisfy (6)supξ∈Sn-1⁡∬Sn-1×Sn-1ΩθΩu′logβ⁡2eθ-u′·ξdσθdσu′<∞.
The function classWFβ(Sn-1) was originally introduced by Fan and Sato [1]. It is closely related to the Grafakos-Stefanov function class Fβ(Sn-1), which was first introduced in [2] and is given by (7)FβSn-1≔Ω∈L1Sn-1:supξ∈Sn-1⁡∫Sn-1Ωy′logβ⁡2ξ·y′dσy′<∞,β>0.It was shown in [1, 3] that (8)FβS1⊂WFβS1,WF2βSn-1∖FβSn-1≠∅,β>0;⋃r>1LrSn-1⊂Fβ2Sn-1⊂Fβ1Sn-1,0<β1<β2<∞;⋃r>1LrSn-1⊂WFβ2Sn-1⊂WFβ1Sn-1,0<β1<β2<∞.To introduce some known results, we need to recall one more function spaceΔγ(R+).Definition 2 (function classΔγ(R+)).
For1≤γ≤∞, the function class Δγ(R+) is the set of all measurable functions h:R+→R satisfying (9)hΔγR+≔supR>0⁡R-1∫0Rhtγdt1/γ<∞.It is clear that Δγ2(R+)⊊Δγ1(R+) for 1≤γ1<γ2≤∞ and Δ∞(R+)=L∞(R+).
WhenΓ(y)=y, the operators defined in (2) reduce to the classical Calderón-Zygmund operator (10)Th,Ωfx=p.v.∫Rnfx-yhyΩyyndy,which was originally studied by Calderón and Zygmund [4] and later investigated by many authors (see [1, 2, 5, 6], etc.). In 2009, Fan and Sato [1] first studied the Lp bounds for Th,Ω with Ω which belongs to WFβ(Sn-1). More precisely, the above authors established the Lp bounds for Th,Ω with 1/p-1/2<1/max⁡{2,γ′}-1/β if h∈Δγ(R+) for γ>1 and Ω∈WFβ(Sn-1) for some β>max⁡{γ′,2}. Recently, Liu and Wu [7] extended the result of [1] to the singular integrals along polynomial compound curves in the mixed homogeneity setting.Let us recall the definitions of Triebel-Lizorkin spaces and Besov spaces.Definition 3 (Triebel-Lizorkin spaces and Besov spaces).
LetS′(Rn) be the tempered distribution class on Rn. For α∈R and 0<p,q≤∞(p≠∞), the homogeneous Triebel-Lizorkin spaces F˙αp,q(Rn) and Besov spaces B˙αp,q(Rn) are defined by (11)F˙αp,qRn≔f∈S′Rn:fF˙αp,qRn=∑i∈Z2-iαqΨi∗fq1/qLpRn<∞;(12)B˙αp,qRn≔f∈S′Rn:fB˙αp,qRn=∑i∈Z2-iαqΨi∗fLpRnq1/q<∞,where Ψi^(ξ)=ϕ(2iξ) for i∈Z and ϕ∈Cc∞(Rn) satisfies the conditions 0≤ϕ(x)≤1; supp(ϕ)⊂{x:1/2≤x≤2}; ϕ(x)>c>0 if 3/5≤x≤5/3. The inhomogeneous versions of Triebel-Lizorkin spaces and Besov spaces, which are denoted by Fαp,q(Rn) and Bαp,q(Rn), respectively, are obtained by adding the term Θ∗fLp(Rn) to the right hand side of (11) or (12) with ∑i∈Z replaced by ∑i≥1, where Θ∈S(Rn) (the Schwartz class), supp(Θ^)⊂{ξ:ξ≤2},Θ^(x)>c>0 if x≤5/3.The following properties are well-known (see [8, 9] for more details): (13)F˙0p,2Rn=LpRn,1<p<∞;(14)F˙αp,pRn=B˙αp,pRn,α∈R,1<p<∞;(15)Fαp,qRn~F˙αp,qRn∩LpRn,fFαp,qRn~fF˙αp,qRn+fLpRn,α>0;(16)Bαp,qRn~B˙αp,qRn∩LpRn,fBαp,qRn~fB˙αp,qRn+fLpRn,α>0.Recently, the investigation of the bounds for singular integrals with rough kernels inWFβ(Sn-1) on Triebel-Lizorkin spaces and Besov spaces has received some attention of many authors (see [3, 10, 11]). Particularly, Liu et al. [10] obtained the following result.Theorem A (see [10]).
LetΓ(y)=PNyy′, where PN is a real-valued polynomial with PN(0)=0 and deg(PN)=N. Suppose that h∈Δγ(R+) for some γ>1 and Ω∈WFβ(Sn-1) for some β>max⁡{2,γ′} satisfying (1). Then
( i )   T h , Ω , Γ is bounded on F˙αp,q(Rn) for α∈R and max⁡{1/p-1/2,1/q-1/2}<1/max⁡{2,γ′}-1/β;
( i i )   T h , Ω , Γ is bounded on B˙αp,q(Rn) for α∈R, q∈(1,∞), and 1/p-1/2<1/max⁡{2,γ′}-1/β.It should be pointed out that there is a gap in the proof of part (i) in Theorem A. To the best of my knowledge, it is unknown whether Theorem A(i) holds. However, we can obtain the following result.Theorem 4.
LetΓ(y)=(P1(y)y1′,…,Pn(y)yn′) with each Pj being a real-valued polynomial on R satisfying Pj(0)=0. Suppose that h∈Δγ(R+) for some γ>1 and Ω∈WFβ(Sn-1) for some β>max⁡{2,γ′} satisfying (1).
( i ) Then, for α∈R and (1/p,1/q)∈Rγ,β, there exists a constant C>0 such that (17)Th,Ω,ΓfF˙αp,qRn≤CfF˙αp,qRnfor all f∈F˙αp,q(Rn), where C=Cn,α,p,q,γ,β is independent of the coefficients of {Pj}j=1n. Here Rγ,β is the set of all interiors of the convex hull of three squares (1/2,1/2+1/max⁡{2,γ′}-1/β)2, (1/2-1/max⁡{2,γ′}+1/β,1/2)2 and (max⁡2,γ′/2β+(1/2γ)(1-max⁡{2,γ′}/β), max⁡{2,γ′}/2β+(1-1/2γ)(1-max⁡{2,γ′}/β))2.
( i i ) Then, for α∈R, q∈(1,∞), and 1/p-1/2<1/max⁡{2,γ′}-1/β, there exists a constant C>0 such that (18)Th,Ω,ΓfB˙αp,qRn≤CfB˙αp,qRnfor all f∈B˙αp,q(Rn), where C=Cn,α,p,q,γ,β is independent of the coefficients of {Pj}j=1n.Applying a switched method following from [12], Theorem 4 yields the following more general result.Theorem 5.
LetΓ(y)=(P1(φ(y))y1′,…,Pn(φ(y))yn′) with each Pj being a real-valued polynomial on R with Pj(0)=0 and φ∈G. Here G is the set of all nonnegative (or nonpositive) and monotonic C1(R+) functions φ satisfying Υ(t)≔φ(t)/tφ′(t) with Υt≤C, where C>0 depends only on φ. Suppose that h∈Δγ(R+) for some γ>1 and Ω∈WFβ(Sn-1) for some β>max⁡2,γ′ satisfying (1). Let Rγ,β be given as in Theorem 4.
( i ) Then, for α∈R and (1/p,1/q)∈Rγ,β, there exists a constant C>0 such that (19)Th,Ω,ΓfF˙αp,qRn≤CfF˙αp,qRnfor all f∈F˙αp,q(Rn), where C=Cn,α,p,q,γ,β,φ is independent of the coefficients of {Pj}j=1n.
( i i ) Then, for α∈R, q∈(1,∞), and 1/p-1/2<1/max⁡{2,γ′}-1/β, there exists a constant C>0 such that (20)Th,Ω,ΓfB˙αp,qRn≤CfB˙αp,qRnfor all f∈B˙αp,q(Rn), where C=Cn,α,p,q,γ,β,φ is independent of the coefficients of {Pj}j=1n.Remark 6.
( i ) It is not difficult to see that Rγ1,β⊊Rγ2,β when 1<γ1<γ2≤∞ and (21)Rγ,β⊊p,q;max⁡1p-12,1q-12<1max⁡2,γ′-1β.
( i i ) If φ∈G, then limt→0φ(t)=0 and limt→∞φt=∞ if φ is nonnegative and increasing, or nonpositive and decreasing; limt→0φt=∞ and limt→∞φ(t)=0 if φ is nonnegative and decreasing or nonpositive and increasing (see [12]).
(iii) It follows from Theorem5 that Th,Ω,Γ is bounded on Lp(Rn) for β>max⁡{2,γ′} and 1/p-1/2<1/max⁡{2,γ′}-1/β under the same conditions of Theorem 5.It is well known that the operators defined in (3) have their roots in the classical Marcinkiewicz integral operators MΩ, corresponding to ρ=1, h(t)≡1, and Γ(y)=y. The Lp bounds for parametric Marcinkiewicz integrals have been extensively studied by many authors (see [13–15], etc.). In recent years, the investigation of boundedness for parametric Marcinkiewicz integral operators on the Triebel-Lizorkin spaces has also attracted the attention of many authors (see [16–19] for examples). Particularly, Yabuta [18] proved the following result.Theorem B (see [18]).
Letρ>0 and Γ(y)=φyy′ with φ∈F, where F is the set of all functions ϕ which satisfy the following conditions:
( a )   ϕ is a positive increasing function on R+ such that tδϕ′(t) is monotonic on R+ for some δ∈R;
( b ) there exist positive constants Cϕ and cϕ such that tϕ′(t)≥Cϕϕ(t) and ϕ(2t)≤cϕϕ(t) for all t>0.
Suppose thath∈Δγ(R+) for some γ>1 and Ω∈WFβ(Sn-1) for some β>max⁡{2,γ′} satisfying (1). Let Rγ,β be given as in Theorem 4. Then
( i )   M h , Ω , Γ , ρ is bounded on F˙αp,q(Rn) for α∈(0,1) and (1/p,1/q)∈Rγ,β;
( i i )   M h , Ω , Γ , ρ is bounded on B˙αp,q(Rn) for α∈(0,1), q∈(1,∞), and 1/p-1/2<1/max⁡{2,γ′}-1/β.Remark 7.
We notice thatF⊊G. There are some examples for the class F, such as tα(α>0), tβln⁡(1+t)(β≥1), tln⁡ln⁡(e+t), and real-valued polynomials P on R with positive coefficients and P(0)=0. It should be pointed out that there exists Bφ>1 such that φ(2t)≥Bφφ(t) for any φ∈F (see [16]).The second one of main results is listed as follows.Theorem 8.
LetΓ(y)=P1φyy1′,…,Pnφyyn′ with each Pj being a real-valued polynomial on R satisfying Pj(0)=0 and φ∈F. Suppose that h∈Δγ(R+) for some γ>1 and Ω∈WFβ(Sn-1) for some β>max⁡{2,γ′} satisfying (1). Let Rγ,β be given as in Theorem 4.
( i ) Then for α∈(0,1) and (1/p,1/q)∈Rγ,β, there exists a constant C>0 such that (22)Mh,Ω,Γ,ρfF˙αp,qRn≤CfF˙αp,qRnfor all f∈F˙αp,q(Rn), where C=Cn,ς,α,p,q,γ,β,φ is independent of the coefficients of {Pj}j=1n.
( i i )   M h , Ω , Γ , ρ is continuous from Fαp,q(Rn) to F˙αp,q(Rn) for α∈(0,1) and (1/p,1/q)∈Rγ,β.
( i i i ) Then, for α∈(0,1), q∈(1,∞), and 1/p-1/2<1/max⁡{2,γ′}-1/β, there exists a constant C>0 such that (23)Mh,Ω,Γ,ρfB˙αp,qRn≤CfB˙αp,qRnfor all f∈B˙αp,q(Rn), where C=Cn,ς,α,p,q,γ,β,φ is independent of the coefficients of {Pj}j=1n.
( i v )   M h , Ω , Γ , ρ is continuous from Bαp,q(Rn) to B˙αp,q(Rn) for α∈(0,1), q∈(1,∞), and 1/p-1/2<1/max⁡{2,γ′}-1/β.Remark 9.
Parts (i) and (iii) in Theorem8 extend Theorem B, which corresponds to the case P1(t)=P2(t)=⋯=Pn(t)=t. Comparing with the singular integral operators, the continuity of the singular integral operators on the Triebel-Lizorkin spaces and Besov spaces can be obtained automatically by the corresponding boundedness since the singular integral operators are linear. However, the continuity of the Marcinkiewicz integral operators on the above function spaces is nontrivial. The reason for this is twofold. First, the Marcinkiewicz integral operators are not linear. Second, f≤g can not imply fF˙αp,q(Rn)≤gF˙αp,q(Rn) and fB˙αp,q(Rn)≤gB˙αp,q(Rn).Remark 10.
By employing the method in the proof of [20, Theorem 1.4] and applying some estimates about Fourier transforms of measures appearing in the proof of Theorem 8, one can obtain that Mh,Ω,Γ,ρ is bounded on Lp(Rn) for 1/p-1/2<1/max⁡{2,γ′}-1/β if h,Ω,Γ,β are given as in Theorem 8.The study of integral operators in form (4) is motivated by the early work of Fefferman on singular integral operators with rough kernels multiplied by bounded radials functions [6] and was introduced by Chen and Lin [21]. Recently, the Triebel-Lizorkin space and Besov space bounds for maximal operators have also been investigated by many authors. For example, see [22, 23] for the Hardy-Littlewood maximal operator and [24, 25] for the maximal functions related to rough singular integrals and Marcinkiewicz integrals. Motivated by the above works, we shall establish the following theorem.Theorem 11.
LetΓ(y)=(P1(φ(y))y1′,…,Pn(φ(y))yn′) with each Pj being a real-valued polynomial on R satisfying Pj(0)=0 and φ∈F. Suppose that Ω∈WFβ(Sn-1) for some β>2 satisfying (1).
( i ) Then, for α∈(0,1) and 1/p,1/q∈Gβ, there exists a constant C>0 such that (24)SΩ,ΓfF˙αp,qRn+MΩ,Γ,ρfF˙αp,qRn≤CfF˙αp,qRnfor all f∈F˙αp,q(Rn), where C=Cn,α,p,q,β,φ is independent of the coefficients of {Pj}j=1n. Here Gβ is the set of all interiors of the convex hull of two squares (1/β,1/2)2 and (1/2,1-1/β)2.
( i i )   S Ω , Γ and MΩ,Γ,ρ are continuous from Fαp,q(Rn) to F˙αp,q(Rn) for α∈(0,1), p∈[2,β), and 1/q∈(1/β,1/p+1/2-1/β).
( i i i ) Then, for α∈(0,1), p∈[2,β), and q∈(1,∞), there exists a constant C>0 such that (25)SΩ,ΓfB˙αp,qRn+MΩ,Γ,ρfB˙αp,qRn≤CfB˙αp,qRnfor all f∈B˙αp,q(Rn), where C=Cn,α,p,q,β,φ is independent of the coefficients of {Pj}j=1n.
( i v )   S Ω , Γ and MΩ,Γ,ρ are continuous from Bαp,q(Rn) to B˙αp,q(Rn) for α∈(0,1), p∈[2,β), and q∈(1,∞).Remark 12.
Note thatGβ=R2,β. By using the estimates of measures appearing in the proof of Theorem 11 and the arguments similar to those used in deriving [7, Theorem 1.9], we can obtain that SΩ,Γ is bounded on Lp for p∈[2,β) under the conditions of Theorem 11.Applying (15)-(16), Theorems 4, 5, 8, and 11, and Remarks 6, 9, 10, and 12, we can obtain the following result immediately.Theorem 13.
Under the same conditions of Theorems4, 5, 8, and 11 and Remarks 6, 9, 10, and 12 with α>0, these operators are bounded and continuous on Fαp,q(Rn) and Bαp,q(Rn), respectively.Due to the fact thatFβ(S1)⊂WFβ(S1), Theorem 13 may yield directly the following conclusion.Theorem 14.
Letn=2 and Γ(y)=(P1(φ(y))y1′,…,P2(φ(y))yn′) with each Pj being a real-valued polynomial on R satisfying Pj(0)=0 and φ∈F.
( i ) If h∈Δγ(R+) for some γ>1 and Ω∈Fβ(S1) for some β>max⁡{2,γ′} satisfying (1), let Rγ,β be given as in Theorem 4. Then
( a )   T h , Ω , Γ is bounded and continuous on Fαp,q(R2) for α∈(0,∞) and (1/p,1/q)∈Rγ,β. Th,Ω,Γ is also bounded and continuous on Bαp,q(R2) for α∈R, q∈(1,∞), and 1/p-1/2<1/max⁡{2,γ′}-1/β;
( b )   M h , Ω , Γ , ρ is bounded and continuous on Fαp,q(R2) for α∈(0,1) and (1/p,1/q)∈Rγ,β. μh,Ω,Γ,ρ is also bounded and continuous on Bαp,q(R2) for α∈(0,1), q∈(1,∞), and 1/p-1/2<1/max⁡{2,γ′}-1/β.
( i i ) If Ω∈Fβ(S1) for some β>2 satisfying (1), then SΩ,Γ and MΩ,Γ,ρ are bounded and continuous on Fαp,q(R2) for α∈(0,1), p∈[2,β), and 1/q∈(1/β,1/p+1/2-1/β). SΩ,Γ and MΩ,Γ,ρ are also bounded and continuous on Bαp,q(R2) for α∈(0,1), p∈[2,β), and q∈(1,∞).The paper is organized as follows. Section2 contains some known results, which play key roles in the proofs of main results. In Section 3, we will present some criterions on the boundedness and continuity of several operators on Triebel-Lizorkin spaces and Besov spaces, which are the main ingredients of our proofs. The proofs of main results will be given in Section 4. We remark that the methods employed in this paper follow from a combination of ideas and arguments in [10, 12, 17, 18, 23, 24, 26, 27], among others. It should be also point out that our methods can be used to deal with other integral operators, such as singular integrals, Marcinkiewicz integrals, and related maximal functions associated with other surfaces with other rough kernels.Throughout the paper, we denotep′ by the conjugate index of p, which satisfies 1/p+1/p′=1. The letter C or c, sometimes with certain parameters, will stand for positive constants not necessarily the same one at each occurrence but are independent of the essential variables. In what follows, we set Rn={ξ∈Rn:1/2<ξ≤1}. We denote by △ζ the difference of f for an arbitrary function f defined on Rn and ζ∈Rn; that is, △ζf(x)=f(x+ζ)-f(x). We also set ∑j∈∅aj=0 and ∏j∈∅aj=1.


## 2. Preliminary Lemmas

This section is devoted to recalling some known lemmas, which plays key roles in the proofs of main theorems. Let us begin with the following lemma of van der Corput type.Lemma 15 (see [28]).
LetΦ(t)=tα1+μ2tα2+⋯+μntαn and Ψ∈C1([0,1]), where μ2,…,μn are real parameters, and α1,…,αn are distinct positive (not necessarily integer) exponents. Then, for λ≠0, the following holds: (26)∫abexp⁡iλΦtΨtdt≤Cλ-ϵsupa≤t≤b⁡Ψt+∫abΨ′tdt,where ϵ=min⁡{1/α1,1/n} and C does not depend on μ2,…,μn as long as 0≤a<b≤1.Applying Lemma15 and the arguments similar to those used in the proof of [16, Lemma 2.2], we can obtain the following result.Lemma 16.
LetΦ(t)=tα1+μ2tα2+⋯+μntαn, where μ2,…,μn are real parameters, and α1,…,αn are distinct positive (not necessarily integer) exponents. Suppose that φ∈F satisfying tδφ′(t) is monotonic on R+ for some δ∈R. Then, for any r>0 and λ≠0, the following holds: (27)∫r/2rexp⁡iλΦφtdtt≤Cλφrα1-ϵ,with ϵ=min⁡{1/α1,1/n}, where C is independent of μ2,…,μn, but may depend on ϱ, φ, and δ.Proof.
By the change of variables, we have(28)∫r/2rexp⁡iλΦφtdtt=∫φr/2φrexp⁡iλΦtdtφ′φ-1tφ-1t=∫φr/2φrexp⁡iλΦtφ-1tδ-1dtφ-1tδφ′φ-1t=φr∫φr/2/φr1exp⁡iλΦφrtφ-1φrtδ-1dtφ-1φrtδφ′φ-1φrt=φr∫ς1exp⁡iλΦφrtgr,φtψtdt,where ς=φ(r/2)/φ(r), gr,φ(t)=1/(φ-1φrt)δφ′(φ-1(φ(r)t)), and ψ(t)=(φ-1(φ(r)t))δ-1. We can also write (29)∫r/2rexp⁡iλΦφtdtt=φr∫ς1ψtdJt,where (30)Jt=∫ςtexp⁡iλΦφrsgr,φsds,ς≤t≤1.Since gr,φ∈C1([0,1]) is monotonic, applying Lemma 15, we obtain (31)Jt≤Cλφrα1-ϵrδφ′r-1+r2δφ′r2-1for ς≤t≤1, where ϵ=min⁡{1/α1,1/n}. By (29) and the integration by parts, we have (32)∫r/2rexp⁡iλΦφtdtt≤φrJ1ψ1+∫ς1Jtψ′tdt≤Cφrλφrα1-ϵrδφ′r-1+r2δφ′r2-1rδ-1+r2δ-1≤Cφ,δλφrα1-ϵ.This proves Lemma 16.Lemmas17–19 are known and will play key roles in the proofs of Theorems 4, 8, and 11, respectively.Lemma 17 (see [26]).
LetΓ(y)=(P1(y)a1(y/y),…,Pn(y)an(y/y)), where P1,…,Pn are real-valued polynomials defined on R and a1,…,an are arbitrary functions defined on Sn-1. Suppose that Ω∈L1(Sn-1) is a homogeneous function of degree zero and h∈Δγ(R+) for some γ>1. Define the measures {σk,Γ,Ω}k∈Z by (33)∫Rnfxdσk,Γ,Ωx=∫2k≤x<2k+1fΓxhxΩxxndx.If (1/p,1/q) belongs to the interior of the convex hull of three squares (1/2,1/2+1/max⁡{2,γ′})2, (1/2-1/max⁡{2,γ′},1/2)2, and (1/2γ,1-1/2γ)2, then, for arbitrary functions {gk,j}k,j∈Z∈Lp(lq(l2),Rn), there exists C>0 independent of γ such that (34)∑j∈Z∑k∈Zσk,Γ,Ω∗gk,j2q/21/qLpRn≤CΩL1Sn-1hΔγR+∑j∈Z∑k∈Zgk,j2q/21/qLpRn.The constant C is independent of Ω and the coefficients of {Pj}j=1n.Lemma 18 (see [17]).
LetΓ(y)=(P1(φ(y))a1(y),…,Pn(φ(y))an(y)), where φ∈F, P1,…,Pn are real-valued polynomials on R and a1(y),…,an(y) are arbitrary functions independent of y. Define the family of measures {σt,Γ}t∈R+ on Rn by (35)∫Rnfxdσt,Γx=1tρ∫t/2<x≤tfΓxhxΩxxn-ρdx.
Suppose thath∈Δγ(R+) for some γ>1 and Ω∈L1(Sn-1). If (1/p,1/q,1/r) belongs to the interior of the convex hull of three cubes (1/2,1/2+1/max⁡{2,γ′})3, (1/2-1/max⁡{2,γ′},1/2)3, and (1/2γ,1-1/2γ)3, then, for arbitrary functions {gj,ζ,k}j,ζ,k∈Lp(lq(Lr(l2)),Rn), there exists C>0 such that (36)
∑
j
∈
Z
∫
R
n
∑
k
∈
Z
∫
2
k
2
k
+
1
σ
t
,
Γ
∗
g
j
,
ζ
,
k
2
d
t
t
1
/
2
d
ζ
q
1
/
q
L
p
R
n
≤
C
Ω
L
1
S
n
-
1
∑
j
∈
Z
∑
k
∈
Z
g
j
,
ζ
,
k
2
1
/
2
L
r
R
n
q
1
/
q
L
p
R
n
.The constant C>0 is independent of Ω and the coefficients of {Pj}j=1n.Lemma 19 (see [24]).
LetΓ(y)=(P1(φ(y))a1(y/y),…,Pn(φ(y))an(y/y)), where φ∈F and P1,P2,…,Pn are real-valued polynomials on R+, and a1,a2,…,an are arbitrary functions defined on Sn-1. Suppose that Ω∈L1(Sn-1). Define the measures {σt,Γ}t∈R+ by (37)σt,Γ^x=∫Sn-1e-2πiΓty′·xΩy′dσy′.If (1/p,1/q,1/r) belongs to the interior of the convex hull of two cubes (0,1/2)3 and (1/2,1)3, then, for arbitrary functions {gj,ζ,k}j,ζ,k∈Lp(lq(Lr(l2)),Rn), there exists C>0 independent of Ω and the coefficients of {Pj}j=1n such that (38)
∑
j
∈
Z
∫
R
n
∑
k
∈
Z
∫
2
k
2
k
+
1
σ
t
,
Γ
∗
g
j
,
ζ
,
k
2
d
t
t
1
/
2
d
ζ
q
1
/
q
L
p
R
n
≤
C
Ω
L
1
S
n
-
1
∑
j
∈
Z
∑
k
∈
Z
g
j
,
ζ
,
k
2
1
/
2
L
r
R
n
q
1
/
q
L
p
R
n
.Below is the vector-valued inequality of the Hardy-Littlewood maximal functions, which is one of the main ingredients of our proofs.Lemma 20 (see [16]).
LetM(n) be the Hardy-Littlewood maximal operator on Rn. Then (39)∑j∈Z∑k∈ZMngj,ζ,k21/2LrRnq1/qLpRn≤C∑j∈Z∑k∈Zgj,ζ,k21/2LrRnq1/qLpRnfor any 1<p, q,r<∞.In order to deal with Marcinkiewicz integrals and maximal functions, we need a useful characterization of Triebel-Lizorkin spaces and Besov spaces.Lemma 21 (see [18]).
Let0<α<1.
( i ) If 1<p<∞, 1<q≤∞, and 1≤r<min⁡{p,q}, then (40)fF˙αp,qRn~∑k∈Z2kqα∫Rn△2-kζfrdζq/r1/qLpRn.
( i i ) If 1≤p<∞, 1≤q≤∞, and 1≤r≤p, then (41)fB˙αp,qRn~∑k∈Z2kqα∫Rn△2-kζfrdζ1/rLpRdq1/q.To prove Theorem5, we need the following results.Lemma 22 (see [12]).
LetΥ,φ be given as in Theorem 5. Suppose that h∈Δγ(R+) for some γ>1, then h(φ-1)Υ(φ-1)∈Δγ(R+).The following lemma is a key switched result about singular integrals associated with compound surfaces.Lemma 23 (see [29]).
Letφ∈G and Υ be given as in Theorem 5. Let Th,Ω,Γ be defined as in (2) and Ω~(y)=Ω(-y). Define the operator Th,Ω,Γ,φ by (42)Th,Ω,Γ,φfx=p.v.∫Rnfx-Γφyy′hyΩyyndy.
( i ) If φ is nonnegative and increasing, then Th,Ω,Γ,φf=Th(φ-1)Υ(φ-1),Ω,Γf.
( i i ) If φ is nonnegative and decreasing, then Th,Ω,φf=-Th(φ-1)Υ(φ-1),Ω,Γf.
( i i i ) If φ is nonpositive and decreasing, then Th,Ω,Γ,φf=Th(φ-1)Υ(φ-1),Ω~,Γf.
( i v ) If φ is nonpositive and increasing, then Th,Ω,Γ,φf=-Th(φ-1)Υ(φ-1),Ω~,Γf.


## 3. Some Criterions

To prove Theorem4, we need the following criterion on the boundedness of the convolution operators on Triebel-Lizorkin spaces, which is a variant of [15, Theorem 1.10]. This can be proved by making some minor modifications in the proof of [15, Theorem 1.10]. We omit the details.Proposition 24.
Letl∈N∖{0} and {σs,k:0≤s≤landk∈Z} be a family of measures on Rn. For 1≤s≤l, let {ak,s}k∈Z be some sequences of positive real numbers with satisfying (43)infk∈Z⁡ak+1,sak,s≥ηsfor some η>1. For 1≤s≤l, let ls∈N∖{0} and Ls:Rn→Rls be linear transformations. Suppose that there exists constants C>0 and β>1 such that
i   σ 0 , k = 0 for every k∈Z;
( i i )   σ s , k ≤ C for every k∈Z and 1≤s≤l;
( i i i )   σ s , k ^ ξ ≤ c log ⁡ ( a k , s L s ξ ) - β if ak,sLsξ>1 for ξ∈Rn, k∈Z, and 1≤s≤l;
( i v )   σ s , k ^ ξ - σ s - 1 , k ^ ξ ≤ c a k , s L s ξ for ξ∈Rn, k∈Z, and 1≤s≤l;
( v ) for any 1≤s≤l and arbitrary functions {gk,j}k,j∈Lp(lq(l2),Rn), there exists a positive constant C which is independent of {Ls}s=1l such that (44)∑j∈Z∑k∈Zσs,k∗gk,j2q/21/qLpRn≤C∑j∈Z∑k∈Zgk,j2q/21/qLpRnfor some p0,q0∈(1,∞) with p0≠2 and q0≠2.
LetP1P2 be the line segment from P1 to P2 with P1=(1/2,1/2) and P2=(1/2β+(1/p0)(1-1/β),1/2β+(1/q0)(1-1/β). Then there exists a positive constant C such that (45)∑k∈Zσl,k∗fF˙αp,qRn≤CfF˙αp,qRn holds for any α∈R and (1/p,1/q)∈P1P2∖{P2}.To establish the Triebel-Lizorkin space boundedness parts in Theorems8 and 11, we will give the following lemma, which is the heart of the proofs of Theorems 8 and 11.Proposition 25.
LetΛ∈N∖{0} and {σs,t:t>0,1≤s≤Λ} be a family of Borel measures on Rn. Let σs,t be the total variation of σs,t. For 1≤s≤Λ, let {ak,s}k∈Z be some sequences of positive real numbers with satisfying (46)δs≥infk∈Z⁡ak+1,sak,s≥ηs>1for some ηs,δs>1. For 1≤s≤Λ, let ls∈N∖{0} and Ls:Rn→RMs be linear transformations. Suppose that there exist p0,q0>1,1<r0<min⁡{p0,q0}, β>1, and C>0 such that the following conditions hold for 1≤s≤Λ, t>0, ξ∈Rn, and {gl,ζ,k}∈Lp0(lq0(Lr0(l2,Rn)),Rn):
( i )   σ 0 , t = 0;
( i i )   ( ∫ 2 k 2 k + 1 σ s , t ^ ξ - σ s - 1 , t ^ ξ 2 ( d t / t ) ) 1 / 2 ≤ C min ⁡ 1 , a k + 1 , s L s ξ;
( i i i )   ( ∫ 2 k 2 k + 1 σ s , t ^ ξ ( d t / t ) ) 1 / 2 ≤ C ( log ⁡ a k , s L s ξ ) - β if ak,sLsξ>1;
( i v )  (47) ∑ l ∈ Z ∫ R n ∑ k ∈ Z ∫ 2 k 2 k + 1 σ s , t ∗ g l , ζ , k 2 d t t 1 / 2 d ζ q 0 1 / q 0 L p 0 R n ≤ C ∑ l ∈ Z ∑ k ∈ Z g l , ζ , k 2 1 / 2 L r 0 R n q 0 1 / q 0 L p 0 R n . Let P1P2 be the line segment from P1 to P2 with P1=(1/2,1/2) and P2=(1/2β+(1/p0)(1-1/β),1/2β+(1/q0)(1-1/β)). Then there exists a positive constant C such that (48)∑l∈Z2lqα∫Rn∫0∞σΛ,t∗△2-lζf2dtt1/2dζq1/qLpRn≤CfF˙αp,qRnholds for any α∈(0,1) and (1/p,1/q)∈P1P2∖{P2}.Proof.
For any1≤s≤Λ, let ls=rank(Ls)≤min⁡{n,ls}. By [5, Lemma 6.1], there are two nonsingular linear transformations Hs:Rls→Rls and Gs:Rn→Rn such that (49)HsπlsnGsξ≤Lsξ≤MsHsπlsnGsξ,ξ∈Rn.For t>0 and 1≤s≤Λ, we define the family of measures {τs,t}t>0 by (50)τs,t^ξ=σs,t^ξ∏j=s+1Λψak,jHjπljnGjξ-σs-1,t^ξ∏j=sΛψak,jHjπljnGjξ,where ψ∈C0∞(R) such that ψ(t)≡1 for t≤1/2 and ψ(t)≡0 for t>1. Then (50) together with assumption (i) implies that (51)σΛ,t=∑s=1Λτs,t.It follows that (52)∑l∈Z2lqα∫Rn∫0∞σΛ,t∗△2-lζf2dtt1/2dζq1/qLpRn≤∑s=1Λ∑l∈Z2lqα∫Rn∫0∞τs,t∗△2-lζf2dtt1/2dζq1/qLpRn.Therefore, to prove (48), it suffices to show that there exists C>0 such that (53)∑l∈Z2lqα∫Rn∫0∞τs,t∗△2-lζf2dtt1/2dζq1/qLpRn≤CfF˙αp,qRnfor any 1≤s≤Λ, α∈(0,1), and (1/p,1/q)∈P1P2∖{P2}.
We now prove (53). By our assumptions (ii)-(iii), we have (54)∫2k2k+1τs,t^ξdtt1/2≤Cmin⁡1,ak+1,sLsξ;∫2k2k+1τs,t^ξdtt1/2≤Clog⁡ak,sLsξ-βifak,sLsξ>Ms.Let {υk,s}k∈Z be a collection of C∞ functions on (0,∞) with the following properties: (55)suppυk,s⊂ak+1,s-1,ak-1,s-1;0≤υk,st≤1;∑k∈Zυk,st=1.Define the multiplier operator Sk,s on Rn by (56)Sk,sf^ξ=υk,sHsπlsnQsξf^ξ.Note that δs≥ak+1,s/ak,s≥ηs>1. By [16, Lemma 2.5] we obtain (57)∑l∈Z∑k∈ZSk,sfζ,l21/2LrRnq1/qLpRn≤C∑l∈Zfζ,lLrRnq1/qLpRn.By Minkowski’s inequality we have (58)∑l∈Z2lqα∫Rn∫0∞τs,t∗△2-lζf2dtt1/2dζq1/qLpRn=∑l∈Z2lqα∫Rn∑k∈Z∫2k2k+1τs,t∗∑j∈ZSj-k,s△2-lζf2dtt1/2dζq1/qLpRn≤∑j∈Z∑l∈Z2lqα∫Rn∑k∈Z∫2k2k+1τs,t∗Sj-k,s△2-lζf2dtt1/2dζq1/qLpRn.Define the mixed norm ·Eαp,q for measurable functions on Rn×Rn×Z×Z×R+ by (59)gEαp,q≔∑l∈Z2lqα∫Rn∑k∈Z∫0∞gx,ζ,l,k,t2dtt1/2dζq1/qLpRn.For any j∈Z, let (60)Vj,sfx,ζ,l,k,t≔τs,t∗Sj-k,s△2-lζfxχ2k,2k+1t.Then we have (61)∑l∈Z2lqα∫Rn∑k∈Z∫2k2k+1τs,t∗△2-lζf2dtt1/2dζq1/qLpRn≤∑j∈ZVj,sfEαp,q.By (54), Hölder’s inequality, Minkowski’s inequality, Fubini’s theorem, Plancherel’s theorem, and Lemma 21(ii), we have (62)Vj,sfEα2,22=∑l∈Z22lα∫Rn∑k∈Z∫2k2k+1τs,t∗Sj-k,s△2-lζf2dtt1/2dζ21/2L2Rn2=∫Rn∑l∈Z22lα∫Rn∑k∈Z∫2k2k+1τs,t∗Sj-k,s△2-lζfx2dtt1/2dζ2dx≤C∑l∈Z22lα∫Rn∑k∈Z∫2k2k+1∫Rnτs,t∗Sj-k,s△2-lζfx2dxdttdζ≤C∑l∈Z22lα∫Rn∑k∈Z∫Ej-k,s∫2k2k+1τs,t^x2dtt△2-lζf^x2dxdζ≤CBj,s2fB˙α2,2Rn2,where Bj,s=Msηs-j+1χ{j≥ks}(j)+j-βχ{j<ks}(j) with ks=max⁡{k∈Z:k<-1-logηsMs} and (63)Ej+k,s=x∈Rn:aj+k+1,s-1≤Lsξ≤Msaj+k-1,s-1.It follows from (62) and (14) that (64)Vj,sfEα2,2≤CBj,sfF˙α2,2Rn.
We now prove(65)Vj,sfEαp0,q0≤CfF˙αp0,q0Rn.For 1≤s≤Λ, let Φs be a radial function in S(Rls) defined by Φs^(x)=ψ(x), where x∈Rls and ψ is given as in (50). Define Js and Xs by (66)Jsfx≔fGstHst⊗idRn-lsx,Xsfx=supk∈Z⁡supt∈2k,2k+1⁡Xk,t;sfx,where Xk,t;sf(x)=Js-1((Φk,t;s⊗δRn-ls)∗Jsf)(x) and Φk,t;s(x0)=(φ(t)γs)-lsΦs(φ(t)-γsx0) with x0∈Rls. It is easy to check that (67)Xsfx≤CsJs-1∘Mls⊗idRn-ls∘Jsfx,where x=(x0,x1)∈Rls×Rn-ls. (67) together with Lemma 20 yields that (68)∑l∈Z∑k∈ZXsgl,ζ,k21/2LrRnq1/qLpRnp≤C∑l∈Z∑k∈ZJs-1∘Mls⊗idRn-ls∘Jsgl,ζ,k21/2LrRnq1/qLpRnp≤CJs∫Rn-ls∫Rls∑l∈Z∑k∈ZMlsJsgl,ζ,k·,x1x021/2LrRnqp/qdx0dx1≤C∑l∈Z∑k∈Zgl,ζ,k21/2LrRnq1/qLpRnpfor any 1≤s≤Λ and 1<p, q,r<∞. Define Xsf=Xs∘Xs+1∘⋯∘XΛf for 1≤s≤Λ. We get from (68) that, for any 1≤s≤Λ and 1<p, q,r<∞,(69)∑l∈Z∑k∈ZXsgl,ζ,k21/2LrRnq1/qLpRnp≤C∑l∈Z∑k∈Zgl,ζ,k21/2LrRnq1/qLpRnp.On the other hand, by the definition of Xk,t;s we have (70)τs,t∗f=σs,t∗Xk,t;s+1∘Xk,t;s+2∘⋯∘Xk,t;Λf-σs-1,t∗Xk,t;s∘Xk,t;s+1∘⋯∘Xk,t;Λf.It follows that (71)∫2k2k+1τs,t∗f2dtt≤2∫2k2k+1σs,t∗Xs+1f2dtt+∫2k2k+1σs-1,t∗Xsf2dtt.By (69), (71), and assumption (iv), we have (72)∑l∈Z∫Rn∑k∈Z∫2k2k+1τs,t∗gl,ζ,k2dtt1/2dζq01/q0Lp0Rn≤C∑l∈Z∑k∈Zgl,ζ,k21/2Lr0Rnq01/q0Lp0Rnfor arbitrary functions {gl,ζ,k}∈Lp0(lq0(Lr0(l2,Rd)),Rn) and 1≤s≤Λ. Then (72) together with Lemma 21(i) and (57) leads to (73)Vj,sfEαp0,q0=∑l∈Z2lq0α∫Rn∑k∈Z∫2k2k+1τs,t∗Sj-k,s△2-lζf2dtt1/2dζq01/q0Lp0Rn≤C∑l∈Z2lq0α∑k∈ZSj-k,s△2-lζf21/2Lr0Rnq01/q0Lp0Rn≤C∑l∈Z2lq0α△2-lζfLr0Rnq01/q0Lp0Rn≤CBφδsBφδs-1n+2fF˙αp0,q0Rn.This proves (65).
By the interpolation between (64) and (65), we obtain that, for α∈(0,1) and (1/p,1/q)∈P1P2∖{P2}, there exists θ∈(1/β,1] such that (74)Vj,sfEαp,q≤CBj,sθfF˙αp,qRn.Then, (74) together with (61) yields (48) and completes the proof of Proposition 25.The following result is a criterion on the boundedness and continuity of several operators on Besov spaces, which can be used to prove the boundedness and continuity result on Besov spaces in Theorems8 and 11.Proposition 26 (see [23]).
Assume thatT:Lp(Rn)→Lp(Rn) for some p∈(1,∞). If (75)△ζTfx≤T△ζfxfor any x,ζ∈Rn. Then T is bounded on B˙sp,q(Rn) for any s∈(0,1) and q∈(1,∞). Particularly, if T satisfies (76)Tf-Tg≤Tf-gfor arbitrary functions f,g defined on Rn, then T is continuous from Bsp,q(Rn) to B˙sp,q(Rn) for any s∈(0,1) and q∈(1,∞).To establish the Triebel-Lizorkin space continuity parts in Theorems8 and 11, we will give the following criterion of continuity for several sublinear operators on Triebel-Lizorkin spaces.Proposition 27.
Assume thatT is a sublinear operator and the following conditions hold.
( i )   T : L p ( R d ) → L p ( R d ) for some p∈(1,∞).
( i i ) For all x,ζ∈Rn, the following holds: (77)△ζTfx≤T△ζfx.
( i i i ) For arbitrary functions f,g defined on Rn, the following holds:(78)Tf-Tg≤Tf-g.
( i v ) There exist α∈(0,1) and q∈(1,∞) such that (79)∑l∈Z2lqα∫RnT△2-lζfdζq1/qLpRn≤CfF˙αp,qRn.
ThenT is continuous from Fsp,q(Rn) to F˙sp,q(Rn).Proof.
Letfj→f in Fαp,q(Rn) as j→∞. By (15) we see that fj→f in F˙αp,q(Rn) and in Lp(Rn) as j→∞. Since fj-fLp(Rn)→0 as j→∞, by assumptions (i) and (iii) we obtain that Tfj→Tf in Lp(Rn) as j→∞. Therefore, it suffices to show that Tfj→Tf in F˙αp,q(Rn) as j→∞.
We shall prove this claim by contradiction. Without loss of generality we may assume that there existsc>0 such that (80)Tfj-TfF˙αp,qRn>cfor every j. Since Tfj→Tf in Lp(Rn) as j→∞, by extracting a subsequence we may assume that Tfjx-Tfx→0 as j→∞ for almost every x∈Rn. It follows that △2-lζ(Tfj-Tf)(x)→0 as j→∞ for every (l,ζ)∈Z×Rn and almost every x∈Rn. We get from assumption (ii) and the sublinearity of T that (81)△2-lζTfj-Tfx≤2T△2-lζfx+T△2-lζfj-fxfor (x,l,ζ)∈Rn×Z×Rn. For convenience, we set (82)gp,q,α≔∑l∈Z2lqα∫Rngx,l,ζdζq1/qLpRnfor α∈R and (p,q)∈(1,∞)2. It follows from Lemma 21(i) that fF˙αp,q(Rn)~△2-lζfp,q,α for α∈(0,1) and (p,q)∈(1,∞)2. By assumption (iv) we obtain (83)T△2-lζfp,q,α≤C∑l∈Z2lqα∫RnT△2-lζfdζq1/qLpRn≤CfF˙αp,qRn.It follows that T(△2-lζ(fj-f))p,q,α≲fj-fF˙αp,q(Rn)→0 as j→∞. One can extract a subsequence such that ∑j=1∞T(△2-lζ(fj-f))p,q,α<∞. Define a function G:Rn×Z×Rn→R by (84)Gx,l,ζ=∑j=1∞T△2-lζfj-fx+2T△2-lζfx.One can easily check that Gp,q,α<∞ and (85)△2-lζTfj-Tfx≤Gx,l,ζfor  almost  everyx,l,ζ∈Rn×Z×Rn.Since Gp,q,α<∞, we have that ∫RdG(x,k,ζ)dζ<∞ for every k∈Z and almost every x∈Rn. (85) together with the dominated convergence theorem leads to (86)∫Rn△2-lζTfj-Tfxdζ⟶0asj⟶∞for every l∈Z and almost every x∈Rn. By the fact Gp,q,α<∞ again, we have (87)∑l∈Z2lqα∫RnGx,l,ζdζq1/q<∞for almost every x∈Rn. Using (85) we obtain (88)∫Rn△2-lζTfj-Tfxdζ≤∫RnGx,l,ζdζfor almost every x∈Rd and l∈Z. It follows from (86)–(88) and the dominated convergence theorem that (89)∑l∈Z2lqα∫Rn△2-lζTfj-Tfxdζq1/q⟶0asj⟶∞for almost every x∈Rn. By (85) again, it holds that (90)∑l∈Z2lqα∫Rn△2-lζTfj-Tfxdζq1/q≤∑l∈Z2lqα∫RnGx,l,ζdζq1/qfor almost every x∈Rn. By (89)-(90), the fact Gp,q,α<∞, and the dominated convergence theorem, we obtain (91)△2-lζTfj-Tfp,q,α⟶0asj⟶∞.This yields that Tfj-TfF˙αp,q(Rn)→0 as j→∞ and gives a contradiction.


## 4. Proofs of Theorems4, 5, and 11

In this section we shall prove Theorems4–11. In what follows, let deg(P)=max1≤j≤n⁡deg(Pj). For 1≤j≤n, we set Pl(t)=∑i=1deg(P)ai,jti. Then there are integers 0<l1<l2<⋯<lΛ≤deg(P) such that Pj(t)=∑i=1Λali,jtli for any 1≤j≤n and (ali,1,ali,2,…,ali,n)≠(0,0,…,0)∈Rn for all 1≤i≤Λ. For 1≤j≤n and 0≤s≤Λ, we set Pj(s)(t)=∑i=1sali,jtli for 1≤s≤Λ and Pj(0)(t)=(0,…,0). For 1≤s≤Λ, we define the linear transformation Li:Rn→Rn by (92)Liξ=ali,1ξ1,ali,2ξ2,…,ali,nξn.We now turn to prove Theorems4, 5, and 11Proof of Theorem4.
DefineΦss=0Λ by (93)Φsy=P1syy1′,…,Pnsyyn′,0≤s≤Λ.It is clear to see that (94)Φsx·ξ=∑j=1nPjsxx′·ξl=∑i=1sxliLiξ·x′for any x,ξ∈Rn and 1≤s≤Λ. For 0≤s≤Λ and ξ∈Rn, we define the measures {σk,s}k∈Z by (95)σk,s^ξ=∫2k<y≤2k+1e-2πiΦsyy′·ξΩyhyyndy.It is clear that (96)Th,Ω,Γf=∑k∈Zσk,Λ∗f.By the change of the variables, we have (97)σk,s^ξ-σk,s-1^ξ=∫2k2k+1∫Sn-1Ωy′e-2πiΦsty′·ξ-e-2πiΦs-1ty′·ξdσy′htdtt≤CΩL1Sn-1hΔγR+2k+1sLsξ.On the other hand, it is easy to check that (98)σk,s≤CΩL1Sn-1hΔγR+;σk,0=0.By the change of the variables and Hölder’s inequality, we have (99)σk,s^ξ=∫2k2k+1∫Sn-1Ωy′e-2πiΦsty′·ξdσy′htdtt≤ChΔγR+∫2k2k+1∫Sn-1Ωy′e-2πiΦsty′·ξdσy′γ′dtt1/γ′≤C∫2k2k+1∫Sn-1Ωy′e-2πiΦsty′·ξdσy′2dtt1/max2,γ′≤C∫2k2k+1∬Sn-1×Sn-1Ωy′Ωθ¯e-2πiΦsty′-θ·ξdσy′dσθdtt1/max2,γ′≤C∬Sn-1×Sn-1Hkξ,y′,θΩy′Ωθ¯dσy′dσθ1/max2,γ′,where (100)Hkξ,y′,θ=∫2k2k+1e-2πiΦsty′-θ·ξdtt.By the Van der Corput lemma, there exists a constant C>0, such that (101)Hkξ,y′,θ≤Cmin⁡1,2k+1sLsξ·y′-θ-1/s.When 2k+1sLsξ>1, since t/(log⁡t)β is increasing in (eβ,∞), we have (102)Hkξ,y′,θ≤Clog⁡2eβλη·y′-θ-1βlog⁡2k+1sLsξβ,where η=Ls(ξ)/Lsξ. Combining (99), (102) with the fact that Ω∈WFβ(Sn-1) yields that (103)σk,s^ξ≤Clog⁡2k+1sLsξ-β/max⁡2,γ′,if2k+1sLsξ>1.On the other hand, Lemma 17 yields that (104)∑j∈Z∑k∈Zσk,s∗gk,j2q/21/qLpRn≤C∑j∈Z∑k∈Zgk,j2q/21/qLpRnfor (1/p,1/q) belonging to the interior of the convex hull of three squares (1/2,1/2+1/max⁡{2,γ′})2, (1/2-1/max⁡{2,γ′},1/2)2, and (1/2γ,1-1/2γ)2. Here C>0 is independent of the coefficients of {Pj}j=1n.
Takeak,s=2ks. By (96)–(98), (103)-(104) and Proposition 24, we obtain (105)Th,Ω,ΓfF˙αp,qRn≤CfF˙αp,qRnfor β>max⁡{2,γ′}, α∈R, and all (1/p,1/q)∈Rγ,β, where Rγ,β is given as in Theorem 4. This proves Theorem 4(i).
On the other hand, it follows from Theorem4(i) and (13) that Th,Ω,Γ is bounded on Lp(Rn) for β>max⁡{2,γ′} and 1/p-1/2<1/max⁡{2,γ′}-1/β. This together with the arguments similar to those used in deriving [30, Theorem 1.2] yields Theorem 4(ii).Proof of Theorem5.
Theorem5 follows from Theorem 4 and Lemmas 22 and 23.Proof of Theorem8.
DefineΦ0,Φ1,…,ΦΛ by (106)Φsy=P1sφyy1′,…,Pnsφyyn′,0≤s≤Λ.Clearly, (107)Φsx·ξ=∑j=1nPjsφxx′·ξl=∑i=1sφxliLiξ·x′,for any x,ξ∈Rn and 1≤s≤Λ. For 0≤s≤Λ, define the family of measures {σt,s}t∈R+ by (108)∫Rnfxdσt,sx=1tρ∫t/2<x≤tfΦsxhxΩxxn-ρdx,where σt,s is defined in the same way as σt,s, but with h and Ω replaced by h and Ω, respectively. By the change of variables and Minkowski’s inequality, we have (109)Mh,Ω,Γ,ρfx=∫0∞∑k=-∞02kρσ2kt,Λ∗fx2dtt1/2≤∑k=-∞02kς∫0∞σ2kt,Λ∗fx2dtt1/2≤11-2-ς∫0∞σt,Λ∗fx2dtt1/2.By Lemma 18, we obtain (110)∑j∈Z∫Rn∑k∈Z∫2k2k+1σt,s∗gj,ζ,k2dtt1/2dζq1/qLpRn≤CΩL1Sn-1∑j∈Z∑k∈Zgj,ζ,k21/2LrRnq1/qLpRnfor (1/p,1/q,1/r) belonging to the interior of the convex hull of three cubes (1/2,1/2+1/max⁡{2,γ′})3, (1/2-1/max⁡{2,γ′},1/2)3, and (1/2γ,1-1/2γ)3. Here C>0 is independent of the coefficients of {Pj}j=1n.
One can easily check that(111)σt,0=0;(112)σt,s^ξ-σt,s-1^ξ≤Cmin⁡1,φtlsLsξ.It follows from (112) that (113)∫2k2k+1σt,s^ξ-σt,s-1^ξ2dtt1/2≤Cmin⁡1,φ2k+1lsLsξ.By a change of variable, we have (114)σt,s^ξ=1tρ∫t/2t∫Sn-1exp⁡-2πi∑j=1sLjξ·y′φrljΩy′dσy′hrdrr1-ρ≤∫t/2t∫Sn-1exp⁡-2πi∑j=1sLjξ·y′φrljΩy′dσy′hrdrr≤ChΔγR+ΩL1Sn-1max⁡0,1-2/γ′×∫t/2t∫Sn-1exp⁡-2πi∑j=1sLjξ·y′φrljΩy′dσy′2drr1/max2,γ′.Since (115)∫t/2t∫Sn-1exp⁡-2πi∑j=1sLjξ·y′φrljΩy′dσy′2drr≤∫t/2t∬Sn-1×Sn-1exp⁡-2πi∑j=1sLjξ·y′-u′φrljΩy′Ωu′¯dσy′dσu′drr≤∬Sn-1×Sn-1Ωy′Ωu′¯∫t/2texp⁡-2πi∑j=1sLjξ·y′-u′φrljdrrdσy′dσu′,by Lemma 16, we have (116)∫t/2texp⁡-2πi∑j=1sLjξ·y′-u′φrljdrr≤Cmin⁡1,φtlsLsξ·y′-u′-1/s.For φ(t)lsLs(ξ)>1, since r/(log⁡r)β is increasing in (eβ,∞), we have (117)∫t/2texp⁡-2πi∑j=1sLjξ·y′-u′φrljdrr≤Clog⁡2eβsη·y′-θ-1βlog⁡φtlsLsξβ,where η=Ls(ξ)/Lsξ. Combining (114), (115), and (117) with the fact that Ω∈WFβ(Sn-1) yields that (118)σt,s^ξ≤Clog⁡φtlsLsξ-β/max2,γ′when φ(t)lsLs(ξ)>1. It follows from (118) that (119)∫2k2k+1σt,s^ξ2dtt1/2≤Clog⁡φ2klsLsξ-β/max⁡2,γ′when φ(2k)lsLs(ξ)>1.
Takeak,s=φ(2k)ls. By Remark 7 we have that cφls≥ak+1,s/ak,s≥Bφls>1. It follows from (110)-(111), (113), (119), and Proposition 25 that (120)∑l∈Z2lqα∫Rn∫0∞σt,Λ∗△2-lζf2dtt1/2dζq1/qLpRn≤CfF˙αp,qRnholds for β>max⁡{2,γ′}, any α∈(0,1), and all (1/p,1/q)∈Rγ,β. Thus (120) together with (109) yields that (121)∑l∈Z2lqα∫RnMh,Ω,Γ,ρ△2-lζfdζq1/qLpRn≤CfF˙αp,qRnholds for β>max⁡{2,γ′}, any α∈(0,1), and all (1/p,1/q)∈Rγ,β. Here C=Cn,ς,α,p,q,γ,β,φ is independent of the coefficients of {Pj}j=1n. On the other hand, one can easily check that (122)△ζMh,Ω,Γ,ρfx≤Mh,Ω,Γ,ρ△ζfxfor any x,ζ∈Rn and (123)Mh,Ω,Γ,ρf-Mh,Ω,Γ,ρg≤Mh,Ω,Γ,ρf-gfor arbitrary functions f and g defined on Rn. By Lemma 21(i), we have (124)Mh,Ω,Γ,ρfF˙αp,qRn≤C∑l∈Z2lqα∫Rn△2-lζMh,Ω,Γ,ρfdζq1/qLpRnfor all α∈(0,1) and (p,q)∈(1,∞)2. Here C=Cn,ς,α,p,q,γ,β,φ is independent of the coefficients of {Pj}j=1n. And (122) and (124) yield Theorem 8(i). We get from Remark 10 that Mh,Ω,Γ,ρ is bounded on Lp(Rn) for β>max⁡{2,γ′} and 1/p-1/2<1/max⁡{2,γ′}-1/β. Note that Mh,Ω,Γ,ρ is a sublinear operator. These facts together with (121)–(123) yield Theorem 8(ii). Theorem 8(iii) and (iv) follow from the Lp bounds for μh,Ω,Γ,ρ and (122)-(123).Proof of Theorem11.
We first consider the operatorSΩ,Γ. One can easily check that (125)△ζSΩ,Γfx≤SΩ,Γ△ζfx∀x,ζ∈Rd.By Lemma 21(i) and (125) we have (126)SΩ,ΓfF˙αp,qRn≤C∑l∈Z2lqα∫Rn△2-lζSΩ,Γfdζq1/qLpRn≤C∑l∈Z2lqα∫RnSΩ,Γ△2-lζfdζq1/qLpRnfor α∈(0,1) and (p,q)∈(1,∞)2. Therefore, to prove Theorem 11(i) for SΩ,Γ, it suffices to show that (127)∑l∈Z2lqα∫RnSΩ,Γ△2-lζfdζq1/qLpRn≤CfF˙αp,qRnfor α∈(0,1) and (1/p,1/q) belonging to the set of all interiors of the convex hull of two squares (1/β,1/2)2 and (1/2,1-1/β)2. Here C=Cn,α,p,q,β,φ is independent of the coefficients of {Pj}j=1n.
LetΦs,Ls, and Λ be given as in the proof of Theorem 8. Define the family of measures {σt,s}t∈R+ and {σt,s}t∈R+ on Rn by (128)σt,s^x=∫Sn-1e-2πiΦsty′·xΩy′dσy′;σt,s^x=∫Sn-1e-2πiΦsty′·xΩy′dσy′.By duality we have (129)SΩ,Γfx=∫0∞σt,Λ∗fx2dtt1/2.One can easily check that (130)σt,0=0,∫2k2k+1σt,s^ξ-σt,s-1^ξ2dtt1/2≤∫2k2k+1max⁡1,φtlsLsξ2dtt1/2≤Cmin⁡1,φ2k+1lsLsξ.On the other hand, (131)σt,s^ξ2=∫Sn-1e-2πiΦsty′·ξΩy′dσy′2=∫Sn-1exp⁡-2πi∑i=1sφtliLiξ·y′Ωy′dσy′2=∬Sn-1×Sn-1exp⁡-2πi∑i=1sφtliLiξ·y′-θΩy′Ωθ¯dσy′dσθ. It follows that (132)∫2k2k+1σt,s^ξ2dtt≤∬Sn-1×Sn-1∫2k2k+1exp⁡-2πi∑i=1sφtliLiξ·y′-θdttΩy′Ωθ¯dσy′dσθ.Invoking Lemma 16 we obtain (133)∫2k2k+1exp⁡-2πi∑i=1sφtliLiξ·y′-θdtt≤min⁡1,φ2k+1lsLsξ·y′-θ-1/s.When φ(2k+1)lsLs(ξ)>1, since t/(log⁡t)β is increasing in (eβ,∞), we have (134)∫2k2k+1exp⁡-2πi∑i=1sφtliLiξ·y′-θdtt≤Clog⁡2eβλη·y′-θ-1βlog⁡φ2k+1lsLsξβ,where η=Ls(ξ)/Lsξ. Combining (132), (134) with the fact that Ω∈WFβ(Sn-1) yields that (135)∫2k2k+1σt,s^ξ2dtt1/2≤Clog⁡φ2k+1lsLsξ-β/2if φ(2k+1)lsLs(ξ)>1. By Lemma 19 we have (136)∑j∈Z∫Rn∑k∈Z∫2k2k+1σt,s∗gj,ζ,k2dtt1/2dζq1/qLpRn≤CΩL1Sn-1∑j∈Z∑k∈Zgj,ζ,k21/2LrRnq1/qLpRnfor (1/p,1/q,1/r) belonging to the interior of the convex hull of two cubes (0,1/2)3 and (1/2,1)3. Here C>0 is independent of the coefficients of {Pj}j=1n.
Takeak,s=φ(2k)ls. By Remark 7 we have that cφls≥ak+1,s/ak,s≥Bφls>1. By (130), (135)-(136), and Proposition 25 we obtain that (137)∑l∈Z2lqα∫Rn∫0∞σt,Λ∗△2-lζf2dtt1/2dζq1/qLpRn≤CfF˙αp,qRnfor β>2, α∈(0,1) and (1/p,1/q) belonging to the interior of the convex hull of two squares (1/β,1/2)2 and (1/2,1-1/β)2. Here C=Cn,α,p,q,β,φ is independent of the coefficients of {Pj}j=1n. Equation (137) together with (129) yields (127). By arguments similar to those used in deriving (3.14) and (3.16) in [31], one can obtain (138)MΩ,Γ,ρfx≤CςSΩ,Γfx∀x∈Rn.Thus (138) together with (127) yields that (139)∑l∈Z2lqα∫RnMΩ,Γ,ρ△2-lζfdζq1/qLpRn≤CfF˙αp,qRnfor α∈(0,1) and (1/p,1/q) belonging to the set of all interiors of the convex hull of two squares (1/β,1/2)2 and (1/2,1-1/β)2. Here C=Cn,ς,α,p,q,β,φ is independent of the coefficients of {Pj}j=1n. One can easily check that (140)△ζMΩ,Γ,ρfx≤MΩ,Γ,ρ△ζfx∀x,ζ∈Rn.By Lemma 21(i) and (140) we have (141)MΩ,Γ,ρfF˙αp,qRn≤C∑l∈Z2lqα∫Rn△2-lζMΩ,Γ,ρfdζq1/qLpRn≤C∑l∈Z2lqα∫RnMΩ,Γ,ρ△2-lζfdζq1/qLpRnfor α∈(0,1) and (p,q)∈(1,∞)2. Then Theorem 11(i) follows from (126)-(127), (139), and (141).
It is known that bothSΩ,Γ and MΩ,Γ,ρ are sublinear operators. Moreover, one can easily check that (142)SΩ,Γf-SΩ,Γg≤SΩ,Γf-g,MΩ,Γ,ρf-MΩ,Γ,ρg≤MΩ,Γ,ρf-gfor arbitrary functions f,g defined on Rn. It follows from Remark 12 that (143)max⁡SΩ,ΓfLpRn,MΩ,Γ,ρfLpRn≤CfLpRnfor p∈[2,β). It follows from (142)-(143), (127), (139), and Proposition 27 that Theorem 11(ii) holds. Theorem 11(iii)-(iv) follows from (125), (140), (142)-(143), and Proposition 26.



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