# Kakeya Inequalities by Maximal Functions in Hardy Spaces

Authors: ZhuoRan Hu
Journal: Journal of Function Spaces
 (2022)

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## Abstract

In this paper, we will introduce and study several types of Kakeya inequalities by the maximal functions in Hardy spaces inℝn, n≥2, and we could obtain several inequalities associated with the Kakeya inequalities. We will show that MδSα,βtfp≲p,n,φ,ε1/δ5n/r+2εfp, when fx∈Lpℝn and suppf^ξ⊆B0,1.


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## Body

## 1. Introduction

In 1917, Kakeya [1] proposed a problem to determine the minimal area needed to continuously rotate a unit line segment in the plane by 180 degrees. In 1928, Besicovithch [2] proved the measure of such sets could be arbitrary small. Such sets are called Besicovitch sets or Kakeya sets. The Kakeya conjectures state that the Hausdorff dimension of any Besicovitch sets in ℝn is n. The case for n≥3 is still an open problem. The so-called maximal Kakeya conjecture (or maximal Nikodym conjecture) is actually a stronger one that involves the following Kakeya maximal function (or Nikodym maximal function):
(1)fδ∗ξ=supa∈ℝn1Tξδa∫Tξδafydy,where Tξδa is a 1×δ tube centered at a∈ℝn with the direction ξ∈Sn−1.
(2)fδ∗∗x=supx∈T1T∫Tfydy,where the supremum is taken over all 1×δ tubes T that contain x∈ℝn. Formula (1) is Kakeya maximal function, and Formula (2) is Nikodym maximal function. When n=2, in [3], Cordoba proved that for any ε>0,
(3)fδ∗L2S1≲εδ−εfL2ℝ2.The Kakeya maximal function conjecture is formulated by Bourgain [4] that
(4)fδ∗LpSn−1≲εδ−εfLpℝn holds for p≥n and n∈ℕ, and
(5)fδ∗LqSn−1≲εδ−n/p+1−εfLpℝn holds for 1<p≤n, q=n−1p′, and n∈ℕ. In 1983, Drury proved Formula (5) for p=d+1/2, q=n+1, in [5]. In 1991, Bourgain in [4] improved this result for each n≥3 to some pd∈d+1/2,d+2/2. By the interpolation theory, (see [6–8] and reference therein),
(6)fδ∗Lpℝn≲εδ−n−1/p−εfLpℝnholds for p≥n and n∈ℕ.


### 1.1. Main Result

Inspired by the Formulas (1), (2), and (4)–(6), we will consider maximal functions like MδSα,βfx and MδSα,βtfx in this paper. Notice that the classical case is δ=1, and then M1Sα,βfx and M1Sα,βtfx are classical maximal functions in Hardy spaces. And
(7)M1Sα,βfp≤Cf∗φ∇p,M1Sα,βfp≤Cfp,for some constant C>0.We will obtain several inequalities in Proposition6, Theorem 7, and Theorem 8. In Proposition 6, though the coefficient in Formula (71) is not better than the factor δ−n−1/p−ε in Formula (6), we use a way different to [3, 4] and [5]. And we could obtain Formula (70) which is different to the classical case δ=1. In Theorem 7, the coefficient is the same as the factor δ−ε in Formula (4) when suppf^ξ⊆B0,1. In Theorem 8, the coefficient is independent on δ when suppf^ξ⊆B0,t−1δ−1+4εc.


### 1.2. Notations

As usual, we usen to denote the dimension of ℝn. suppfx is the support set of fx. If x∈ℝn: x=x1,x2,⋯,xn, xe denotes xe=x12+x22+⋯+xn2. For α∈ℕn: α=α1,α2,⋯,αn, α denotes α=α1+α2+⋯+αn. We use .p to denote .Lpℝn and Oℝn to denote the n×n unit orthogonal matrix in ℝn: Oℝn=A:ATA=1.ATisthetransposedmatrixofA.Sℝn designates the space of C∞ functions on ℝn rapidly decreasing together with their derivatives. Sα,βℝn denotes Sα,βℝn=ϕ∈Sℝn:ϕα′,β′≤1,∀α′,β′∈ℕn,α′≤α,β′≤β and ε a positive fixed number (may be very small): ε>0.IfX and Y are two quantities, X≲Y or Y≳X denotes that X≤CY for some absolute constant C>0. More generally, given some parameters a1,⋯,ak, we use X≲a1,⋯,akY or Y≳a1,⋯,akXto denote the statement that X≤Ca1,⋯,akY for some constant Ca1,⋯,ak which can depend on the parameter a1,⋯,ak. We use X~Y to denote the statement X≲Y≲X, and similarly, X~a1,⋯,akY denotes X≲a1,⋯,akY≲a1,⋯,akX.


## 1.1. Main Result

Inspired by the Formulas (1), (2), and (4)–(6), we will consider maximal functions like MδSα,βfx and MδSα,βtfx in this paper. Notice that the classical case is δ=1, and then M1Sα,βfx and M1Sα,βtfx are classical maximal functions in Hardy spaces. And
(7)M1Sα,βfp≤Cf∗φ∇p,M1Sα,βfp≤Cfp,for some constant C>0.We will obtain several inequalities in Proposition6, Theorem 7, and Theorem 8. In Proposition 6, though the coefficient in Formula (71) is not better than the factor δ−n−1/p−ε in Formula (6), we use a way different to [3, 4] and [5]. And we could obtain Formula (70) which is different to the classical case δ=1. In Theorem 7, the coefficient is the same as the factor δ−ε in Formula (4) when suppf^ξ⊆B0,1. In Theorem 8, the coefficient is independent on δ when suppf^ξ⊆B0,t−1δ−1+4εc.


## 1.2. Notations

As usual, we usen to denote the dimension of ℝn. suppfx is the support set of fx. If x∈ℝn: x=x1,x2,⋯,xn, xe denotes xe=x12+x22+⋯+xn2. For α∈ℕn: α=α1,α2,⋯,αn, α denotes α=α1+α2+⋯+αn. We use .p to denote .Lpℝn and Oℝn to denote the n×n unit orthogonal matrix in ℝn: Oℝn=A:ATA=1.ATisthetransposedmatrixofA.Sℝn designates the space of C∞ functions on ℝn rapidly decreasing together with their derivatives. Sα,βℝn denotes Sα,βℝn=ϕ∈Sℝn:ϕα′,β′≤1,∀α′,β′∈ℕn,α′≤α,β′≤β and ε a positive fixed number (may be very small): ε>0.IfX and Y are two quantities, X≲Y or Y≳X denotes that X≤CY for some absolute constant C>0. More generally, given some parameters a1,⋯,ak, we use X≲a1,⋯,akY or Y≳a1,⋯,akXto denote the statement that X≤Ca1,⋯,akY for some constant Ca1,⋯,ak which can depend on the parameter a1,⋯,ak. We use X~Y to denote the statement X≲Y≲X, and similarly, X~a1,⋯,akY denotes X≲a1,⋯,akY≲a1,⋯,akX.


## 2. Preliminaries

Fort,ξ∈ℝn, f∈Sℝn, the Fourier transform of f is given by
(8)f^ξ=Ffξ=∫ℝnfte−2πi<ξ,t>dt,and thus fx=f∧∨x, where <ξ,t>=∑k=1nξktk and ∨ is the inversion of Fourier transform. For g∈Sℝn, gIx designates
(9)gIx=1δn−1gx1δ,x2δ,⋯,xn−1δ,xn.Letu=u1,u2,⋯,un=xA−1=x1,x2,⋯,xnA−1 where A is a variable (not fixed) matrix with A∈Oℝn, and then gAIx is given by
(10)gAIx=gIxA−1=gIu=1δn−1gu1δ,u2δ,⋯,un−1δ,un.IfA is a variable (not fixed) matrix with A∈Oℝn, let
(11)ζ=ζ1ζ2⋮ζn=Aξ=Aξ1ξ2⋮ξn,and thus
(12)FgIξ=Fgδξ1,δξ2,⋯,δξn−1,ξn,FgAIξ=FgIAξ=FgIζ=Fgδζ1,δζ2,⋯,δζn−1,ζn.In this paper, letφ∈Sℝn always to be a fixed radial function satisfying the following:
(13)φ^ξ=1,forξe≤1,φ^ξ=0,forξe≥2,φ^Aξ=φ^ξforA∈Oℝn.MYfx and MSα,βfx are given by MYfx=supt>0f∗Ytx,MSα,βfx=supY∈Sα,βℝnMYfx. And nontangential maximal functions f∗Y∇x are defined as usual: f∗Y∇x=supx−y≤tf∗Yty. The even larger tangential variant MYN∗∗ depending on a parameter N is given by
(14)MYN∗∗fx=supv∈ℝn,t>0∫ℝnfu1tnYx−u−vt1+vt−Ndu.Letf to be a distribution, and Hardy spaces Hpℝn are (c.f. [9]): fHpℝn=MSα,βfLpℝn, for 0<p<∞ with appropriate α and β depending on p. It is known that Hp=Lp for p>1: fHpℝn=fLpℝn.In this paper, the Kakeya type maximal functionMδSα,βfx is given by
(15)MδSα,βfx=supt>0,A∈Oℝn,Y∈Sα,βℝn∫fx−yYItyA−1dy,where YAIty=YItyA−1=1/tnYIyA−1/t. For some fixed t>0, MδSα,βtfx can be defined by
(16)MδSα,βtfx=supA∈Oℝn,Y∈Sα,βℝn∫fx−yYItyA−1dy.Lemma 1 (see [9]).
For anyψ∈Sℝn, 1<p<∞, f∈Lpℝn, and N>n/p, we could obtain
(17)MψN∗∗fp≲N,pf∗ψ∇p≲p,ψfp.Lemma 2 (see[10]).
Let0<C0<∞ and 0<r<∞. Then, there exist constants C1 and C2 (that depend only on n, C0, and r) such that for all t>0 and for all C1ℝn functions u on ℝn whose Fourier transform is supported in the ball ξe≤C0t and that satisfies uz≤B1+zen/r for some B>0, we have the estimation
(18)supz∈ℝn1t∇ux−z1+tzen/r≤C1supz∈ℝnux−z1+tzen/r≤C2Murx1/r,where M denotes the Hardy-Littlewood maximal operator. (The constants C1 and C2 are independent of B and u.)Lemma 3 (Phragmen-Lindelöf Lemma).
LetF be analytic in the open strip S=z∈ℂ:0<Rez<1, continuous, and bounded on its closure, such that Fz≤C0 when Rez=0 and Fz≤C1 when Rez=1. Then, Fz≤C01−θC1θ when Rez=θ for any 0<θ<1.


## 3. The Case when2≤δ−ε

When1≤δ−ε≤2, the case that δ~ε1 is trival for the Kakeya type inequalities; thus, we only want to discuss the case when 0<δ≪1. In the following of this paper, we will discuss under the assumption that 2≤δ−ε.


### 3.1. Decomposition of the Phase Space

In this section, we will decomposeℝn into a collection of regions:
(19)ξ∈ℝn:δ−k−1ε≤ξe≤δ−k+3εk≥1,k∈ℤandξ∈ℝn:ξe≤1.Then, we will give a decomposition of the regionξ∈ℝn:ξe≤1⋃ξ∈ℝn:1≤ξe≤δ−4ε into a collection of smaller ones:
(20)ξ∈ℝn:2k−1≤ξe≤2k+3s≥k≥1,k∈ℤandξ∈ℝn:ξe≤1.Let the functionsΦk^ξk for k∈ℤ,k≥0 to be defined as
(21)Φ0^ξ=φ^ξ,Φ0x=φx,Φk^ξ=φ^2−kξ−φ^21−kξ,Φkx=φ2−kx−φ2−k−1x,fork≥1.Then the functionsΦk^ξk for k∈ℤ,k≥0 are given by
(22)Φ0^ξ=Φ0^δξ1,δξ2,⋯,δξn−1,ξn,Φ0x=Φ0Ix,Φk^ξ=Φk^δξ1,δξ2,⋯,δξn−1,ξn,Φkx=ΦkIx,fork≥1.Thus, it is clear thatsuppΦk^ξ⊆ξ∈ℝn:2k−1≤ξe≤2k+1,fork≥1 and suppΦk^ξ⊆ξ∈ℝn:2k−1≤δξ12+⋯+δξn−12+ξn21/2≤2k+1fork≥1. Also, we could deduce that
(23)∑k=0∞Φk^ξ=1,withΦkx=ΦkIx.In the same way, we could define the functionsΨkxk and Ψkxk for k∈ℤ,k≥0 as
(24)Ψ0^ξ=φ^ξ,Ψ0x=φx,Ψk^ξ=φ^δkεξ−φ^δk−1εξ,Ψkx=φδkεx−φδk−1εx,fork≥1,Ψ0^ξ=Ψ0^δξ1,δξ2,⋯,δξn−1,ξn,Ψ0x=Ψ0IxΨk^ξ=Ψk^δξ1,δξ2,⋯,δξn−1,ξn,Ψkx=ΨkIx,fork≥1.Then, we could deduce thatsuppΨk^ξ⊆ξ∈ℝn:δ−k−1ε≤ξe≤δ−k+3εfork≥1 and suppΨk^ξ⊆ξ∈ℝn:δ−k−1ε≤δξ12+⋯+δξn−12+ξn21/2≤δ−k+3ε for k≥1 hold. Thus, we could have
(25)∑k=0∞Ψk^ξ=1withΨkx=ΨkIx.Notice thatδ1+k+3εξe≤1 for ξ∈suppΨk^ξ. Then, we could obtain
(26)φ^δ1+k+3εξ=1,forξ∈suppΨk^ξ,FYIξ=∑k=0∞Ψk^ξφ^δ1+k+3εξFYIξφ^δ1+k+3εξ.We setη1k^ξ, (for0≤k,k∈ℤ) as
(27)η1k^ξ=Ψk^ξφ^δ1+k+3εξFYIξ=Ψk^ξFYIξ.for0≤k,k∈ℤ.It is easy to see thatη1k^ξandη1kx∈Sℝn. ∃s∈ℕ, such that
(28)δ−4ε~2s,and∑k=0sΦk^ξ=1forξ∈suppΨ0^⋃suppΨ1^.We setη0k^ξ (k=0,1,2,⋯,s) as
(29)η0k^ξ=Ψ0^ξ+Ψ1^ξΦk^ξφ^2−k+1δξFYIξ.fork=0,1,2,⋯,s.Notice that2−k+1δξe≤1 holds, when ξ∈suppη0k^ξ. Thus, we could obtain
(30)φ^2−k+1δξ=1whenξ∈suppη0k^ξ.Thus,(31)η0k^ξ=Ψ0^ξ+Ψ1^ξΦk^ξFYIξfork=0,1,2,⋯,s.Thus, we could writeFYIξ and FYAIξφ^ as radial, and A is a variable (not fixed) matrix with A∈Oℝn) as
(32)FYIξ=∑k=0sη0k^ξφ^2−k+1δξ+∑k=2∞η1k^ξφ^δ1+k+3εξ,(33)FYAIξ=FYIAξ=∑k=0sη0k^Aξφ^2−k+1δξ+∑k=2∞η1k^Aξφ^δ1+k+3εξ,where2s~δ−4ε.


### 3.2. Two Lemmas

In this section, we will estimate the integrals (in Lemmas4 and 5) associated with η0kx and η1kx given in Formulas (32) and (33).Lemma 4.
ForN≥0,N∈ℝ, k∈ℕ, k≥2, and Y∈Sα,βℝn with appropriate α,β depending on ε,n,N, we have
(34)∫ℝn1+δ−k+3εδ−1xeNη1kxdx≲N,n,φ,εδkε.Proof.
First, we will prove that forl∈ℝ, l≥0, k∈ℕ, and k≥2, the following inequality holds:
(35)xel+2nη1kx≲l,n,φ,εδl+kε+k+3lε.
Notice that the following inequality holds for0<δ<1, for any m∈ℤ,m≥0:
(36)xe2m+2nη1kx≤x1δ2+x1δ2+⋯+xn−1δ2+xn2m+nη1kx.
Thus, by the formula of integration by parts, we could deduce the following for anym∈ℤ,m≥0:
(37)x1δ2+x1δ2+⋯+xn−1δ2+xn2m+nη1kx=∫ℝnC∂ξ1δ2+∂ξ2δ2+⋯+∂ξn−1δ2+∂ξn2m+nη1k^ξe2πi<x,ξ>dξ.
Make a variable substitution:(38)δξ1,δξ2⋯δξn−1,ξn⟶ξ1′,ξ2′⋯ξn−1′,ξn′.
We could write Formula (37) as
(39)x1δ2+x1δ2+⋯+xn−1δ2+xn2m+nη1kx=1δn−1∫ℝnCΔξ′n+mη1k^ξ′e2πi<x,ξ>dξ′,where Δξ′ is the Laplace operator: Δξ′=∂ξ′12+∂ξ′22+⋯+∂ξ′n2. We could also deduce that
(40)Δξ′n+mη1k^ξ′=Δξ′n+mΨk^ξ′FYξ′.
Thus,Δξ′n+mη1k^ξ′∈Sℝn, ξ′eα′Δξ′β′η1k^ξ′≲α′,β′1,forappropriateα′,β′, and
(41)suppΔξ′n+mη1k^ξ′⊆ξ′∈ℝn:δ−k−1ε≤ξ′e≤δ−k+3εfork≥2.
Whenδ−k−1ε≤ξ′e≤δ−k+3ε, k≥2, and δ−ε≥2, we could deduce that
(42)δ−k/2ε≤δ−k−1ε≤ξ′e≤δ−k+3ε≤δ−3kε.
That is,(43)ξ′e1/3≤δ−kε≤ξ′e2,δ−ε≤δ−kε≤ξ′e2.
Then, we could deduce that(44)xe2m+2nη1kx≲1δn−1∫ℝnΔξ′n+mη1k^ξ′dξ′≲δ2m+kε+2k+3mε∫ℝnξ′e2n+4m/ε+8mΔξ′n+mη1k^ξ′dξ′≲m,n,φ,εδ2m+kε+2k+3mε.
Thus, similar to Formula (44), we could obtain
(45)xe2mη1kx≲1δn−1∫ℝnΔξ′n+mη1k^ξ′dξ′≲m,n,φ,εδ2m+kε+2k+3mε,where k≥2, m∈ℤ,m≥0. By Lemma 3 and Formula (44), we could deduce Formula (35). Thus, we could obtain the following inequality for N≥0,N∈ℝ(46)δ−k+3εδxeNη1kx≲N,n,φ,εδkε1xe2n.
By Lemma3 and Formula (45), for l∈ℝ, l≥0, k∈ℕ, and k≥2, the following inequality holds:
(47)xelη1kx≲l,n,φ,εδl+kε+k+3lε.
Then, we could obtain the following inequality forN≥0,N∈ℝ, k∈ℕ, and k≥2,
(48)δ−k+3εδxeNη1kx≲N,n,φ,εδkε.
By Formulas (46) and (48), we could deduce that for N≥0,N∈ℝ, k∈ℕ, and k≥2, the following Formulas (49) and (50) hold:
(49)∫ℝnη1kxdx≲n,φ,εδkε,(50)∫ℝnδ−k+3εδ−1xeNη1kxdx≲N,n,φ,εδkε.
Then, we could obtain the Lemma4 directly from Formulas (49) and (50). This proves the Lemma.Lemma 5.
ForN≥0,N∈ℝ, k∈ℕ, and 0≤k≤s where 2s~δ−4ε, Y∈Sα,βℝn, with appropriate α,β depending on ε,n,N, the following two inequalities hold:
(51)∫ℝn1+2k+1δ−1xeNη0kxdx≲N,n,φ,εδ−4N+1εδ−N2−k,(52)∫ℝn1+2k+1xeNη0kxdx≲N,n,φ,εδ−4N+1ε2−k.Proof.
Notice that2s~δ−4ε, thus, for any k∈0,1,⋯,s and N≥0,N∈ℝ, we have
(53)1≲12k+1δ4εN.
Notice that2−k+1δξe≤1 holds, when ξ∈suppη0k^ξ. Thus, we could obtain
(54)φ^2−k+1δξ=1whenξ∈suppη0k^ξ.
Then, we could writeη0k^ξ as
(55)η0k^ξ=Ψ0^ξ+Ψ1^ξΦk^ξFYIξ.
It is clear that the following Formulas (56)–(60) hold:
(56)∂ξα1Ψ0∧ξ∨x=−2πixα1φIx,(57)∂ξα1Ψ1∧ξ∨x=−2πixα11δεnφIxδε−−2πixα1φIx,(58)∂ξβ1Φk∧ξ∨x=−2πixβ12knφI2kx−−2πixβ12k−1nφI2k−1xfork≥1,(59)∂ξβ1Φ0∧ξ∨x=−2πixβ1φIx,(60)∂ξγ1FYIξ∨x=−2πixγ1YIx.
By Young inequality, we could have(61)∫η0kxdx≤Ψ0+Ψ1∗Φk∗YI1≤Ψ0+Ψ11Φk1YI1≲φ1.
By the formula of integration by parts, we could deduce the following for anym∈ℕ:
(62)xe2n+2mη0kx=∫ℝnCΔξn+mη0k^ξe2πi<x,ξ>dξ=∑α1+β1+γ1=2m+2n∂ξα1Ψ1∧ξ+∂ξα1Ψ0∧ξ∨∗∂ξβ1Φk∧ξ∨∗∂ξγ1FYIξ∨x.
By Young Inequality, Formulas (62) and (56)–(60), we could obtain
(63)∫xe2n+2mη0kxdx≤∑α1+β1+γ1=2m+2n∂ξα1Ψ1∧ξ+∂ξα1Ψ0∧ξ∨1∂ξβ1Φk∧ξ∨1∂ξγ1FYIξ∨1≲φ,n,m1form∈ℕ.
By Lemma3 and Formulas (61) and (63), we could deduce the following Formula (64).
(64)∫xelη0kxdx≲φ,l1forl∈ℝ,l≥0.
By Formulas (61) and (64), the following two inequalities hold for N≥0, k∈0,1,⋯,s:
(65)∫ℝn1+2k+1δ−1xeNη0kxdx≲N,n,φ,εδ−N2k+1N,(66)∫ℝn1+2k+1xeNη0kxdx≲N,n,φ,ε2k+1N.
From Formulas (53), (65), and (66), we could obtain the Formula (51) and (52) together. This proves the Lemma.From Lemmas4 and 5, we could obtain the following inequalities (67)–(69). For N≥0,N∈ℝ, k∈ℕ, and k≥2, we have
(67)∫ℝn1+δ−k+3εδ−1xeNη1kxA−1dx≲N,n,φ,εδkε,where A is a variable (not fixed) matrix with A∈Oℝn. For N≥0,N∈ℝ, k∈ℕ, and 0≤k≤s where 2s~δ−4ε, we have Formulas (37) and (68)
(68)∫ℝn1+2k+1δ−1xeNη0kxA−1dx≲N,n,φ,εδ−4N+1εδ−N2−k,(69)∫ℝn1+2k+1xeNη0kxA−1dx≲N,n,φ,εδ−4N+1ε2−k,where A is a variable (not fixed) matrix with A∈Oℝn.


### 3.3. Main Results

From Formulas (67)–(69), we will obtain our main results in this section:Proposition 6.
Forp>1 with appropriate α, β depending on ε,n,p, we have
(70)MδSα,βfp≲p,n,φ,ε1δ4n/p+2εf∗φδ∇p,(71)MδSα,βfp≲p,n,φ,ε1δn/p+4n/p+3εfp,where f∗φδ∇x=supx−y≤tf∗φδty.Proof.
Notice thatf∈Lpℝn is a distribution. By Formula (33), we could obtain
(72)MδSα,βfx=supt>0,A∈Oℝn,Y∈Sα,βℝn∫fx−yYItyA−1dy≤∑k=2∞supt>0,A∈Oℝn∫ℝnfx−y∫ℝnt−nη1kuA−1tφδk+3εδty−ududy+∑k=0ssupt>0,A∈Oℝn∫ℝnfx−y∫ℝnt−nη0kuA−1tφ2−k+1δty−ududy≤∑k=2∞supt>0,A∈Oℝn∫ℝnt−nη1kuA−1tf∗φδk+3εδtx−udu+∑k=0ssupt>0,A∈Oℝn∫ℝnt−nη0kuA−1tf∗φ2−k+1δtx−udu≤∑k=2∞supt>0,A∈OℝnMφδN∗∗fx∫ℝnt−nη1kuA−1t1+ueδk+3εtNdu+∑k=0ssupt>0,A∈OℝnMφδN∗∗fx∫ℝnt−nη0kuA−1t1+ue2−k+1tNdu,where 2s~δ−4ε. Lemma 1 and Formulas (67), (69), and (72) yield to
(73)MδSα,βfp≲p,N,n,φ,ε1δ4N+2εf∗φδ∇pforp>1,N>np.
Similar to Formula (72), we could also obtain
(74)supt>0,A∈Oℝn∫fx−yYItyA−1dy≤∑k=2∞supt>0,A∈OℝnMφN∗∗fx∫ℝnt−nη1kuA−1t1+ueδk+3εδtNdu+∑k=0ssupt>0,A∈OℝnMφN∗∗fx∫ℝnt−nη0kuA−1t1+ue2−k+1δtNdu.
Notice that2s~δ−4ε; thus, Lemma 1 and Formulas (67), (68), and (74) yield to
(75)MδSα,βfp≲p,N,n,φ,ε1δN1δ4N+2εfpforp>1,N>np,N∈ℝ.
LetN be N=n/p+ε. From Formulas (73) and (75), we could prove the Proposition 6.Theorem 7.
For∞>p>r>1, 0<t≤δ−ε, fx∈Lpℝn, and suppf^ξ⊆B0,1. Then, with appropriate α, β depending on ε,n,r, we could obtain
(76)MδSα,βtfp≲p,n,φ,ε1δ5nr+2εfp.Proof.
By Formula (33), we could write MδSα,βtf as
(77)MδSα,βtfx≤∑k=2∞supu∈ℝnf∗φδk+3εδtx−u1+ue/δk+3εtn/rsupA∈Oℝn∫ℝnt−nη1kuA−1t1+ueδk+3εtn/rdu+∑k=0ssupu∈ℝnf∗φ2−k+1δtx−u1+ue/2−k+1tn/rsupA∈Oℝn∫ℝnt−nη0kuA−1t1+ue2−k+1tn/rdu,
where2s~δ−4ε.
By Holder inequality,f∗φδk+3εδtx and f∗φ2−k+1δtx are both bounded functions for any x∈ℝ, δ>0, k∈ℕ, and t>0. Thus, for some B>0, we could have
(78)f∗φδk+3εδtx≤B1+xen/rand,f∗φ2−k+1δtx≤B1+xen/r.
It is also clear thatsuppFf∗φδk+3εδtξ⊆B0,1, and suppFf∗φ2−k+1δtξ⊆B0,1. We could also deduce that f∗φδk+3εδtx∈C1ℝn,f∗φ2−k+1δtx∈C1ℝn. Thus, by Lemma 2, we could obtain
(79)supu∈ℝnf∗φδk+3εδtx−u1+ue/δk+3εtn/r≤supu∈ℝnf∗φδk+3εδtx−u1+uen/r≤C2Mf∗φδk+3εδtrx1/r,(80)supu∈ℝnf∗φ2−k+1δtx−u1+ue/2−k+1tn/r≤supu∈ℝnf∗φ2−k+1δtx−u1+uen/r≤C2Mf∗φ2−k+1δtrx1/r.
Notice that2s~δ−4ε, by Formulas (67), (69), (77), (79), and (80), we could deduce that for ∞>p>r>1, 0<t≤1(81)MδSα,βtfp≲p,n,φ,ε∑k=2∞δkε∫ℝnf∗φδk+3εδtpxdx1/p+∑k=0sδ−4n/r+1ε2−k∫ℝnf∗φ2−k+1δtpxdx1/p≲p,n,φ,ε1δ4n/r+2εfp.
When1<t≤δ−ε, notice that
(82)supu∈ℝnf∗φ2−k+1δtx−u1+ue/2−k+1tn/r≤tn/rsupu∈ℝnf∗φ2−k+1δtx−u1+ue/2−k+11/r≤tn/rsupu∈ℝnf∗φ2−k+1δtx−u1+uen/r≤C2δ−n/rεMf∗φ2−k+1δtrx1/r,(83)supu∈ℝnf∗φδk+3εδtx−u1+ue/δk+3εtn/r≤tn/rsupu∈ℝnf∗φδk+3εδtx−u1+ue/δk+3εn/r≤tn/rsupu∈ℝnf∗φδk+3εδtx−u1+uen/r≤C2δ−n/rεMf∗φδk+3εδtrx1/r
hold. From Formulas (67), (69), (77), (79), (80), (82), and (83), we could deduce that for ∞>p>r>1, 1<t≤δ−ε(84)MδSα,βtfp≲p,n,φ,ε∑k=2∞δ−n/rεδkε∫ℝnf∗φδk+3εδtpxdx1/p+∑k=0sδ−n/rεδ−4nr+1ε2−k∫ℝnf∗φ2−k+1δtpxdx1/p≲p,n,φ,ε1δ5nr+2εfp.
This proves the Theorem.Theorem 8.
For∞>p>1, 0<t≤δ−ε, fx∈Lpℝn, and suppf^ξ⊆B0,t−1δ−1+4εc, then with appropriate α, β depending on ε,n,p, we could obtain
(85)MδSα,βtfp≲p,n,φ,εfp.Proof.
By Formula (33), we could write MδSα,βtf as the following:
(86)MδSα,βtfx≤∑k=2∞supA∈Oℝn∫ℝnt−nη1kuA−1tf∗φδk+3εδtx−udu+∑k=0ssupA∈Oℝn∫ℝnt−nη0kuA−1tf∗φ2−k+1δtx−udu.
Notice thatsuppf^ξ⊆B0,t−1δ−1+4εc and suppFφ2−k+1δtξ⊆B0,t−1δ−1+4ε for 0≤k≤s,k∈ℤ; thus, we could deduce that f∗φ2−k+1δt=0. From Formula (86), we could obtain
(87)MδSα,βtfx≤∑k=2∞supA∈Oℝn∫ℝnt−nη1kuA−1tf∗φδk+3εδtx−udu.
By Formula (67) and Lemma 1, we could have
(88)∫ℝn∑k=2∞supA∈Oℝn∫ℝnt−nη1kuA−1tf∗φδk+3εδtx−udupdx1/p≤∫ℝn∑k=2∞supt>0,A∈Oℝn∫ℝnt−nη1kuA−1tf∗φδk+3εδtx−udupdx1/p≤∫ℝn∑k=2∞supt>0,A∈OℝnMφN∗∗fx∫ℝnt−nη1kuA−1t1+ueδk+3εδtNdupdx1/p≲p,N,n,φ,εfpforp>1,N>np,N∈ℝ.
LetN be N=n/p+ε, from Formulas (86)–(88), and we could prove the Theorem 8.


## 3.1. Decomposition of the Phase Space

In this section, we will decomposeℝn into a collection of regions:
(19)ξ∈ℝn:δ−k−1ε≤ξe≤δ−k+3εk≥1,k∈ℤandξ∈ℝn:ξe≤1.Then, we will give a decomposition of the regionξ∈ℝn:ξe≤1⋃ξ∈ℝn:1≤ξe≤δ−4ε into a collection of smaller ones:
(20)ξ∈ℝn:2k−1≤ξe≤2k+3s≥k≥1,k∈ℤandξ∈ℝn:ξe≤1.Let the functionsΦk^ξk for k∈ℤ,k≥0 to be defined as
(21)Φ0^ξ=φ^ξ,Φ0x=φx,Φk^ξ=φ^2−kξ−φ^21−kξ,Φkx=φ2−kx−φ2−k−1x,fork≥1.Then the functionsΦk^ξk for k∈ℤ,k≥0 are given by
(22)Φ0^ξ=Φ0^δξ1,δξ2,⋯,δξn−1,ξn,Φ0x=Φ0Ix,Φk^ξ=Φk^δξ1,δξ2,⋯,δξn−1,ξn,Φkx=ΦkIx,fork≥1.Thus, it is clear thatsuppΦk^ξ⊆ξ∈ℝn:2k−1≤ξe≤2k+1,fork≥1 and suppΦk^ξ⊆ξ∈ℝn:2k−1≤δξ12+⋯+δξn−12+ξn21/2≤2k+1fork≥1. Also, we could deduce that
(23)∑k=0∞Φk^ξ=1,withΦkx=ΦkIx.In the same way, we could define the functionsΨkxk and Ψkxk for k∈ℤ,k≥0 as
(24)Ψ0^ξ=φ^ξ,Ψ0x=φx,Ψk^ξ=φ^δkεξ−φ^δk−1εξ,Ψkx=φδkεx−φδk−1εx,fork≥1,Ψ0^ξ=Ψ0^δξ1,δξ2,⋯,δξn−1,ξn,Ψ0x=Ψ0IxΨk^ξ=Ψk^δξ1,δξ2,⋯,δξn−1,ξn,Ψkx=ΨkIx,fork≥1.Then, we could deduce thatsuppΨk^ξ⊆ξ∈ℝn:δ−k−1ε≤ξe≤δ−k+3εfork≥1 and suppΨk^ξ⊆ξ∈ℝn:δ−k−1ε≤δξ12+⋯+δξn−12+ξn21/2≤δ−k+3ε for k≥1 hold. Thus, we could have
(25)∑k=0∞Ψk^ξ=1withΨkx=ΨkIx.Notice thatδ1+k+3εξe≤1 for ξ∈suppΨk^ξ. Then, we could obtain
(26)φ^δ1+k+3εξ=1,forξ∈suppΨk^ξ,FYIξ=∑k=0∞Ψk^ξφ^δ1+k+3εξFYIξφ^δ1+k+3εξ.We setη1k^ξ, (for0≤k,k∈ℤ) as
(27)η1k^ξ=Ψk^ξφ^δ1+k+3εξFYIξ=Ψk^ξFYIξ.for0≤k,k∈ℤ.It is easy to see thatη1k^ξandη1kx∈Sℝn. ∃s∈ℕ, such that
(28)δ−4ε~2s,and∑k=0sΦk^ξ=1forξ∈suppΨ0^⋃suppΨ1^.We setη0k^ξ (k=0,1,2,⋯,s) as
(29)η0k^ξ=Ψ0^ξ+Ψ1^ξΦk^ξφ^2−k+1δξFYIξ.fork=0,1,2,⋯,s.Notice that2−k+1δξe≤1 holds, when ξ∈suppη0k^ξ. Thus, we could obtain
(30)φ^2−k+1δξ=1whenξ∈suppη0k^ξ.Thus,(31)η0k^ξ=Ψ0^ξ+Ψ1^ξΦk^ξFYIξfork=0,1,2,⋯,s.Thus, we could writeFYIξ and FYAIξφ^ as radial, and A is a variable (not fixed) matrix with A∈Oℝn) as
(32)FYIξ=∑k=0sη0k^ξφ^2−k+1δξ+∑k=2∞η1k^ξφ^δ1+k+3εξ,(33)FYAIξ=FYIAξ=∑k=0sη0k^Aξφ^2−k+1δξ+∑k=2∞η1k^Aξφ^δ1+k+3εξ,where2s~δ−4ε.


## 3.2. Two Lemmas

In this section, we will estimate the integrals (in Lemmas4 and 5) associated with η0kx and η1kx given in Formulas (32) and (33).Lemma 4.
ForN≥0,N∈ℝ, k∈ℕ, k≥2, and Y∈Sα,βℝn with appropriate α,β depending on ε,n,N, we have
(34)∫ℝn1+δ−k+3εδ−1xeNη1kxdx≲N,n,φ,εδkε.Proof.
First, we will prove that forl∈ℝ, l≥0, k∈ℕ, and k≥2, the following inequality holds:
(35)xel+2nη1kx≲l,n,φ,εδl+kε+k+3lε.
Notice that the following inequality holds for0<δ<1, for any m∈ℤ,m≥0:
(36)xe2m+2nη1kx≤x1δ2+x1δ2+⋯+xn−1δ2+xn2m+nη1kx.
Thus, by the formula of integration by parts, we could deduce the following for anym∈ℤ,m≥0:
(37)x1δ2+x1δ2+⋯+xn−1δ2+xn2m+nη1kx=∫ℝnC∂ξ1δ2+∂ξ2δ2+⋯+∂ξn−1δ2+∂ξn2m+nη1k^ξe2πi<x,ξ>dξ.
Make a variable substitution:(38)δξ1,δξ2⋯δξn−1,ξn⟶ξ1′,ξ2′⋯ξn−1′,ξn′.
We could write Formula (37) as
(39)x1δ2+x1δ2+⋯+xn−1δ2+xn2m+nη1kx=1δn−1∫ℝnCΔξ′n+mη1k^ξ′e2πi<x,ξ>dξ′,where Δξ′ is the Laplace operator: Δξ′=∂ξ′12+∂ξ′22+⋯+∂ξ′n2. We could also deduce that
(40)Δξ′n+mη1k^ξ′=Δξ′n+mΨk^ξ′FYξ′.
Thus,Δξ′n+mη1k^ξ′∈Sℝn, ξ′eα′Δξ′β′η1k^ξ′≲α′,β′1,forappropriateα′,β′, and
(41)suppΔξ′n+mη1k^ξ′⊆ξ′∈ℝn:δ−k−1ε≤ξ′e≤δ−k+3εfork≥2.
Whenδ−k−1ε≤ξ′e≤δ−k+3ε, k≥2, and δ−ε≥2, we could deduce that
(42)δ−k/2ε≤δ−k−1ε≤ξ′e≤δ−k+3ε≤δ−3kε.
That is,(43)ξ′e1/3≤δ−kε≤ξ′e2,δ−ε≤δ−kε≤ξ′e2.
Then, we could deduce that(44)xe2m+2nη1kx≲1δn−1∫ℝnΔξ′n+mη1k^ξ′dξ′≲δ2m+kε+2k+3mε∫ℝnξ′e2n+4m/ε+8mΔξ′n+mη1k^ξ′dξ′≲m,n,φ,εδ2m+kε+2k+3mε.
Thus, similar to Formula (44), we could obtain
(45)xe2mη1kx≲1δn−1∫ℝnΔξ′n+mη1k^ξ′dξ′≲m,n,φ,εδ2m+kε+2k+3mε,where k≥2, m∈ℤ,m≥0. By Lemma 3 and Formula (44), we could deduce Formula (35). Thus, we could obtain the following inequality for N≥0,N∈ℝ(46)δ−k+3εδxeNη1kx≲N,n,φ,εδkε1xe2n.
By Lemma3 and Formula (45), for l∈ℝ, l≥0, k∈ℕ, and k≥2, the following inequality holds:
(47)xelη1kx≲l,n,φ,εδl+kε+k+3lε.
Then, we could obtain the following inequality forN≥0,N∈ℝ, k∈ℕ, and k≥2,
(48)δ−k+3εδxeNη1kx≲N,n,φ,εδkε.
By Formulas (46) and (48), we could deduce that for N≥0,N∈ℝ, k∈ℕ, and k≥2, the following Formulas (49) and (50) hold:
(49)∫ℝnη1kxdx≲n,φ,εδkε,(50)∫ℝnδ−k+3εδ−1xeNη1kxdx≲N,n,φ,εδkε.
Then, we could obtain the Lemma4 directly from Formulas (49) and (50). This proves the Lemma.Lemma 5.
ForN≥0,N∈ℝ, k∈ℕ, and 0≤k≤s where 2s~δ−4ε, Y∈Sα,βℝn, with appropriate α,β depending on ε,n,N, the following two inequalities hold:
(51)∫ℝn1+2k+1δ−1xeNη0kxdx≲N,n,φ,εδ−4N+1εδ−N2−k,(52)∫ℝn1+2k+1xeNη0kxdx≲N,n,φ,εδ−4N+1ε2−k.Proof.
Notice that2s~δ−4ε, thus, for any k∈0,1,⋯,s and N≥0,N∈ℝ, we have
(53)1≲12k+1δ4εN.
Notice that2−k+1δξe≤1 holds, when ξ∈suppη0k^ξ. Thus, we could obtain
(54)φ^2−k+1δξ=1whenξ∈suppη0k^ξ.
Then, we could writeη0k^ξ as
(55)η0k^ξ=Ψ0^ξ+Ψ1^ξΦk^ξFYIξ.
It is clear that the following Formulas (56)–(60) hold:
(56)∂ξα1Ψ0∧ξ∨x=−2πixα1φIx,(57)∂ξα1Ψ1∧ξ∨x=−2πixα11δεnφIxδε−−2πixα1φIx,(58)∂ξβ1Φk∧ξ∨x=−2πixβ12knφI2kx−−2πixβ12k−1nφI2k−1xfork≥1,(59)∂ξβ1Φ0∧ξ∨x=−2πixβ1φIx,(60)∂ξγ1FYIξ∨x=−2πixγ1YIx.
By Young inequality, we could have(61)∫η0kxdx≤Ψ0+Ψ1∗Φk∗YI1≤Ψ0+Ψ11Φk1YI1≲φ1.
By the formula of integration by parts, we could deduce the following for anym∈ℕ:
(62)xe2n+2mη0kx=∫ℝnCΔξn+mη0k^ξe2πi<x,ξ>dξ=∑α1+β1+γ1=2m+2n∂ξα1Ψ1∧ξ+∂ξα1Ψ0∧ξ∨∗∂ξβ1Φk∧ξ∨∗∂ξγ1FYIξ∨x.
By Young Inequality, Formulas (62) and (56)–(60), we could obtain
(63)∫xe2n+2mη0kxdx≤∑α1+β1+γ1=2m+2n∂ξα1Ψ1∧ξ+∂ξα1Ψ0∧ξ∨1∂ξβ1Φk∧ξ∨1∂ξγ1FYIξ∨1≲φ,n,m1form∈ℕ.
By Lemma3 and Formulas (61) and (63), we could deduce the following Formula (64).
(64)∫xelη0kxdx≲φ,l1forl∈ℝ,l≥0.
By Formulas (61) and (64), the following two inequalities hold for N≥0, k∈0,1,⋯,s:
(65)∫ℝn1+2k+1δ−1xeNη0kxdx≲N,n,φ,εδ−N2k+1N,(66)∫ℝn1+2k+1xeNη0kxdx≲N,n,φ,ε2k+1N.
From Formulas (53), (65), and (66), we could obtain the Formula (51) and (52) together. This proves the Lemma.From Lemmas4 and 5, we could obtain the following inequalities (67)–(69). For N≥0,N∈ℝ, k∈ℕ, and k≥2, we have
(67)∫ℝn1+δ−k+3εδ−1xeNη1kxA−1dx≲N,n,φ,εδkε,where A is a variable (not fixed) matrix with A∈Oℝn. For N≥0,N∈ℝ, k∈ℕ, and 0≤k≤s where 2s~δ−4ε, we have Formulas (37) and (68)
(68)∫ℝn1+2k+1δ−1xeNη0kxA−1dx≲N,n,φ,εδ−4N+1εδ−N2−k,(69)∫ℝn1+2k+1xeNη0kxA−1dx≲N,n,φ,εδ−4N+1ε2−k,where A is a variable (not fixed) matrix with A∈Oℝn.


## 3.3. Main Results

From Formulas (67)–(69), we will obtain our main results in this section:Proposition 6.
Forp>1 with appropriate α, β depending on ε,n,p, we have
(70)MδSα,βfp≲p,n,φ,ε1δ4n/p+2εf∗φδ∇p,(71)MδSα,βfp≲p,n,φ,ε1δn/p+4n/p+3εfp,where f∗φδ∇x=supx−y≤tf∗φδty.Proof.
Notice thatf∈Lpℝn is a distribution. By Formula (33), we could obtain
(72)MδSα,βfx=supt>0,A∈Oℝn,Y∈Sα,βℝn∫fx−yYItyA−1dy≤∑k=2∞supt>0,A∈Oℝn∫ℝnfx−y∫ℝnt−nη1kuA−1tφδk+3εδty−ududy+∑k=0ssupt>0,A∈Oℝn∫ℝnfx−y∫ℝnt−nη0kuA−1tφ2−k+1δty−ududy≤∑k=2∞supt>0,A∈Oℝn∫ℝnt−nη1kuA−1tf∗φδk+3εδtx−udu+∑k=0ssupt>0,A∈Oℝn∫ℝnt−nη0kuA−1tf∗φ2−k+1δtx−udu≤∑k=2∞supt>0,A∈OℝnMφδN∗∗fx∫ℝnt−nη1kuA−1t1+ueδk+3εtNdu+∑k=0ssupt>0,A∈OℝnMφδN∗∗fx∫ℝnt−nη0kuA−1t1+ue2−k+1tNdu,where 2s~δ−4ε. Lemma 1 and Formulas (67), (69), and (72) yield to
(73)MδSα,βfp≲p,N,n,φ,ε1δ4N+2εf∗φδ∇pforp>1,N>np.
Similar to Formula (72), we could also obtain
(74)supt>0,A∈Oℝn∫fx−yYItyA−1dy≤∑k=2∞supt>0,A∈OℝnMφN∗∗fx∫ℝnt−nη1kuA−1t1+ueδk+3εδtNdu+∑k=0ssupt>0,A∈OℝnMφN∗∗fx∫ℝnt−nη0kuA−1t1+ue2−k+1δtNdu.
Notice that2s~δ−4ε; thus, Lemma 1 and Formulas (67), (68), and (74) yield to
(75)MδSα,βfp≲p,N,n,φ,ε1δN1δ4N+2εfpforp>1,N>np,N∈ℝ.
LetN be N=n/p+ε. From Formulas (73) and (75), we could prove the Proposition 6.Theorem 7.
For∞>p>r>1, 0<t≤δ−ε, fx∈Lpℝn, and suppf^ξ⊆B0,1. Then, with appropriate α, β depending on ε,n,r, we could obtain
(76)MδSα,βtfp≲p,n,φ,ε1δ5nr+2εfp.Proof.
By Formula (33), we could write MδSα,βtf as
(77)MδSα,βtfx≤∑k=2∞supu∈ℝnf∗φδk+3εδtx−u1+ue/δk+3εtn/rsupA∈Oℝn∫ℝnt−nη1kuA−1t1+ueδk+3εtn/rdu+∑k=0ssupu∈ℝnf∗φ2−k+1δtx−u1+ue/2−k+1tn/rsupA∈Oℝn∫ℝnt−nη0kuA−1t1+ue2−k+1tn/rdu,
where2s~δ−4ε.
By Holder inequality,f∗φδk+3εδtx and f∗φ2−k+1δtx are both bounded functions for any x∈ℝ, δ>0, k∈ℕ, and t>0. Thus, for some B>0, we could have
(78)f∗φδk+3εδtx≤B1+xen/rand,f∗φ2−k+1δtx≤B1+xen/r.
It is also clear thatsuppFf∗φδk+3εδtξ⊆B0,1, and suppFf∗φ2−k+1δtξ⊆B0,1. We could also deduce that f∗φδk+3εδtx∈C1ℝn,f∗φ2−k+1δtx∈C1ℝn. Thus, by Lemma 2, we could obtain
(79)supu∈ℝnf∗φδk+3εδtx−u1+ue/δk+3εtn/r≤supu∈ℝnf∗φδk+3εδtx−u1+uen/r≤C2Mf∗φδk+3εδtrx1/r,(80)supu∈ℝnf∗φ2−k+1δtx−u1+ue/2−k+1tn/r≤supu∈ℝnf∗φ2−k+1δtx−u1+uen/r≤C2Mf∗φ2−k+1δtrx1/r.
Notice that2s~δ−4ε, by Formulas (67), (69), (77), (79), and (80), we could deduce that for ∞>p>r>1, 0<t≤1(81)MδSα,βtfp≲p,n,φ,ε∑k=2∞δkε∫ℝnf∗φδk+3εδtpxdx1/p+∑k=0sδ−4n/r+1ε2−k∫ℝnf∗φ2−k+1δtpxdx1/p≲p,n,φ,ε1δ4n/r+2εfp.
When1<t≤δ−ε, notice that
(82)supu∈ℝnf∗φ2−k+1δtx−u1+ue/2−k+1tn/r≤tn/rsupu∈ℝnf∗φ2−k+1δtx−u1+ue/2−k+11/r≤tn/rsupu∈ℝnf∗φ2−k+1δtx−u1+uen/r≤C2δ−n/rεMf∗φ2−k+1δtrx1/r,(83)supu∈ℝnf∗φδk+3εδtx−u1+ue/δk+3εtn/r≤tn/rsupu∈ℝnf∗φδk+3εδtx−u1+ue/δk+3εn/r≤tn/rsupu∈ℝnf∗φδk+3εδtx−u1+uen/r≤C2δ−n/rεMf∗φδk+3εδtrx1/r
hold. From Formulas (67), (69), (77), (79), (80), (82), and (83), we could deduce that for ∞>p>r>1, 1<t≤δ−ε(84)MδSα,βtfp≲p,n,φ,ε∑k=2∞δ−n/rεδkε∫ℝnf∗φδk+3εδtpxdx1/p+∑k=0sδ−n/rεδ−4nr+1ε2−k∫ℝnf∗φ2−k+1δtpxdx1/p≲p,n,φ,ε1δ5nr+2εfp.
This proves the Theorem.Theorem 8.
For∞>p>1, 0<t≤δ−ε, fx∈Lpℝn, and suppf^ξ⊆B0,t−1δ−1+4εc, then with appropriate α, β depending on ε,n,p, we could obtain
(85)MδSα,βtfp≲p,n,φ,εfp.Proof.
By Formula (33), we could write MδSα,βtf as the following:
(86)MδSα,βtfx≤∑k=2∞supA∈Oℝn∫ℝnt−nη1kuA−1tf∗φδk+3εδtx−udu+∑k=0ssupA∈Oℝn∫ℝnt−nη0kuA−1tf∗φ2−k+1δtx−udu.
Notice thatsuppf^ξ⊆B0,t−1δ−1+4εc and suppFφ2−k+1δtξ⊆B0,t−1δ−1+4ε for 0≤k≤s,k∈ℤ; thus, we could deduce that f∗φ2−k+1δt=0. From Formula (86), we could obtain
(87)MδSα,βtfx≤∑k=2∞supA∈Oℝn∫ℝnt−nη1kuA−1tf∗φδk+3εδtx−udu.
By Formula (67) and Lemma 1, we could have
(88)∫ℝn∑k=2∞supA∈Oℝn∫ℝnt−nη1kuA−1tf∗φδk+3εδtx−udupdx1/p≤∫ℝn∑k=2∞supt>0,A∈Oℝn∫ℝnt−nη1kuA−1tf∗φδk+3εδtx−udupdx1/p≤∫ℝn∑k=2∞supt>0,A∈OℝnMφN∗∗fx∫ℝnt−nη1kuA−1t1+ueδk+3εδtNdupdx1/p≲p,N,n,φ,εfpforp>1,N>np,N∈ℝ.
LetN be N=n/p+ε, from Formulas (86)–(88), and we could prove the Theorem 8.



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*Source: 9821212-2022-06-15.xml*