# Anisotropic Weak Hardy Spaces and Wavelets

Authors: B. Barrios; J. J. Betancor
Journal: Journal of Function Spaces and Applications
 (2012)

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## Abstract

We characterize the anisotropic weak Hardy spacesHAp,∞(ℝn) associated with an expansive matrix A by using square functions involving wavelets coefficients.


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## Body

## 1. Introduction

Bownik, in a series of papers [1–5], studied anisotropic function spaces associated with dilations. In the monography [1] he investigated anisotropic Hardy spaces. Suppose that A is an expansive matrix (also called dilation) in ℝn, that is, A is an n×n-matrix all of whose eigenvalues λ satisfy |λ|>1. If ϕ is a function in the Schwartz class S(ℝn) such that ∫ℝnϕ(x)dx≠0 and f is a distribution in S(ℝn)′, the dual space of S(ℝn), the radial maximal function Mϕ(f) is defined by(1.1)Mϕ(f)=sup⁡k∈Z|f*ϕk|,
where ϕk(x)=|detA|-kϕ(A-kx),x∈ℝn, and k∈ℤ. For every 0<p<∞, the Hardy space HAp(ℝn) associated with A consists of all those f∈S(ℝn)′ such that Mϕ(f)∈Lp(ℝn). The space HAp(ℝn) does not depend on the function ϕ and the ∥·∥HAp(ℝn)-norm is defined by(1.2)‖f‖HAp(R)n=∥Mϕ(f)∥p,f∈HAp(Rn).Like in the classical case [6], the anisotropic Hardy space HAp(ℝn) can be characterized by nontangential or grand maximal functions [1, Theorem 7.1, page 42]. Also, atomic representations of the distributions in HAp(ℝn) are obtained in [1, Theorem 6.5, page 39]. Wavelets for a dilation A are studied in [1, Chapter 2]. The author proved that r-wavelets associated with the expansive matrix A form an unconditional basis for the anisotropic Hardy spaces defined by A. Recently, Bownik et al. [7, 8] have investigated weighted anisotropic Hardy spaces.The weak anisotropic Hardy spaceHAp,∞(ℝn),0<p≤1, was introduced by Ding and Lan [9]. A distribution f∈S(ℝn)′ is in HAp,∞(ℝn) if and only if Mϕ(f)∈Lp,∞(ℝn), where Lp,∞(ℝn) denotes the weak Lp-space. We define(1.3)‖f‖HAp,∞(Rn)=‖Mϕf‖Lp,∞(Rn),f∈HAp,∞(Rn).
As the case HAp(ℝn), the space HAp,∞(ℝn) does not depend on the election of ϕ and it can be described by nontangential and grand maximal functions. Atomic representations of distributions in HAp,∞(ℝn) were established in [9, Theorem 1.1]. In this paper we study wavelets for HAp,∞(ℝn). We characterize in Theorem 2.2 bellow the distributions in HAp,∞(ℝn) by square functions involving wavelets coefficients. Our result can be seen as an anisotropic version of the one showed in [10] (see also [11]).This paper is organized as follows. In Section2 we recall the main definitions and properties about the anisotropic setting that we need throughout the paper. We also state our result (Theorem 2.2). The proof of Theorem 2.2 is presented in Section 3.


## 2. Preliminaries and Results

We now recall the main definitions and properties concerning the analysis in the anisotropic setting. We refer the reader to [1] where the anisotropic theory associated with expansive matrixes was developed. Suppose that A is an expansive matrix in ℝn. We denote by Δ an ellipsoid for A such that the Lebesgue measure |Δ| of Δ is equal to 1. For every k∈ℤ, we define Bk=AkΔ. We consider the mapping ρ:ℝn→[0,∞) given by(2.1)ρ(x)={|det⁡⁡A|j,x∈Bj+1∖Bj,j∈Z,0,x=0.
Thus, ρ is a homogeneous quasinorm associated with A in the sense of [1, Definition 2.3, page 6]. This means that ρ satisfies the following properties:(a)
ρ(x)>0, x∈ℝn∖{0},(b)
ρ(Ax)=|detA|ρ(x), x∈ℝn,(c)
ρ(x+y)≤Hρ(ρ(x)+ρ(y)), x,y∈ℝn,whereHρ≥1. In [1, Lemma 2.4, page 6] it was proved that if ρ1 and ρ2 are two homogeneous quasinorms associated with A then,(2.2)1Cρ1(x)≤ρ2(x)≤Cρ1(x),x∈Rn,
for a certain C>0. As it was mentioned above a tempered distribution f∈HAp,∞(ℝn),  0<p≤1, when Mϕ(f)∈Lp,∞(ℝn), where ϕ∈S(ℝn) such that ∫ℝnϕ(x)dx≠0. In [9, Theorem 1.1] Ding and Lan proved the following atomic representation for the distributions in HAp,∞(ℝn) that will be useful in the sequel.Theorem 2.1 (see [9, Theorem 1.1]).
Suppose thatf∈HAp,∞(ℝn), 0<p≤1, and s≥[(1/p-1)log|detA|/logm1], where m1=min{|λ|:λisaneigenvalueofA}. Then, there exists a sequence of bounded functions {fk}k∈ℤ satisfying, for a certain C>0,(a)
f=∑k=-∞∞fk in S(ℝn)′, and ∥fk∥∞≤C2k, for every k∈ℤ,(b)
for everyk∈ℤ, fk=∑i=0∞βik, in S(ℝn)′, where, for each i∈ℕ,
(i)
there existxi,k∈ℝn and ji,k∈ℤ such that suppβik⊂B̃ik=xi,k+Aji,kΔ, and
(2.3)∑i=0∞|B̃ik|≤C2-kp‖f‖HAp,∞(Rn)p,(ii)
∥βik∥∞≤C2k,(iii)
∫ℝnβik(x)P(x)dx=0, for every polynomial P whose degree is less or equal to s.Conversely, if f∈S(ℝn)′ satisfying (a) and (b) where in (i)C∥f∥HAp,∞(ℝn)p is replacing by M>0, then f∈HAp,∞(ℝn) and M:∥f∥HAp,∞(ℝn)p.Theorem2.1 is an anisotropic version of the isotropic result established by Fefferman and Soria [12, Proposition, page 8].Letψ∈L2(ℝn). For every j∈ℤ and k∈ℤn, we define(2.4)ψj,k(x)=|det⁡⁡A|j/2ψ(Ajx-k),x∈Rn.We say thatψ is a Bessel A-wavelet if there exists C>0 such that, for every f∈L2(ℝn),(2.5)∑j∈Z,k∈Zn|〈f,ψj,k〉|2≤C‖f‖22.
ψ is a frame A-wavelet when, for a certain C>0,(2.6)1C‖f‖22≤∑j∈Z,k∈Zn|〈f,ψj,k〉|2≤C‖f‖22,f∈L2(Rn).
We say that ψ is a tight frame A-wavelet when (2.6) holds with C=1.As usual ifg∈L2(ℝn) we denote by ĝ the Fourier transform of g.We now establish our result where the distributions inHAp(ℝn), 0<p≤1, are characterized by using square functions involving A-wavelets.Theorem 2.2.
Letf∈S(ℝn)′ and 0<p≤1. Assume that ψ∈S(ℝn) is a tight frame A-wavelet where A is an expansive matrix in ℝn, such that ψ(0)≠0, suppψ̂ is compact, and 0∉suppψ̂. Then, the following properties are equivalent:(a)
W(f)=(∑j∈ℤ,k∈ℤn|〈f,ψj,k〉|2|ψj,k|2)1/2∈Lp,∞(ℝn),(b)
G(f)=(∑j∈ℤ,k∈ℤn|〈f,ψj,k〉|2χQj,k|Qj,k|-1)1/2∈Lp,∞(ℝn),whereQj,k=A-j([0,1]n+k), j∈ℤ, and k∈ℤn,(c)
S(f)=(∑j∈ℤ,k∈ℤn|〈f,ψj,k〉|2χRj,k|Qj,k|-1)1/2∈Lp,∞(ℝn),where for everyj∈ℤ and k∈ℤn, Rj,k is a measurable set such that Rj,k⊂Qj,k and |Rj,k|≥γ|Qj,k|, for a certain γ∈(0,1),(d)
the distributionf=∑j∈ℤ,k∈ℤn〈f,ψj,k〉ψj,k is in HAp,∞(ℝn).Moreover, if one of (and then all) the above conditions is satisfied, then
(2.7)‖f‖HAp,∞(Rn)~‖W(f)‖Lp,∞(Rn)~‖G(f)‖Lp,∞(Rn)~‖S(f)‖Lp,∞(Rn).Bownik proved in [1, Theorem 4.2, page 94] that there exists a tight frame A-wavelet ψ∈S(ℝn) satisfying the conditions in Theorem 2.2 such that ∫ℝnf(x)P(x)dx=0, for every polynomial P in ℝn.


## 3. Proof of Theorem2.2

In this section we present a proof of Theorem2.2. Throughout this section with C we denote a positive constant that can change in each occurrence.(a)⇒(c) According to [1, equation (2.2), page 5], Ajx→0, as j→-∞, uniformly in x∈[0,1]n. Moreover, there exist 0<η<1 and δ>0 for which(3.1)|ψ(x)|≥δ,x∈[0,η]n.
Then, for every j∈ℤ and k∈ℤn, we have(3.2)|ψj,k(x)|≥δ|Qj,k|,x∈Rj,k,
where Rj,k=A-j([0,η]n+k)⊆Qj,k. Hence, we can write
(3.3)(∑j∈Z,k∈Zn|〈f,ψj,k〉|2|ψj,k(x)|2)1/2≥δ(∑j∈Z,k∈Zn|〈f,ψj,k〉|21|Qj,k|χRj,k(x))1/2.
Note that(3.4)|Rj,k|=|A-j[0,η]n|=|det⁡⁡A|-jηn=|Qj,k|ηn.
Thus we show that (a) implies (b).(b)⇒(c) It is clear.(c)⇒(b) Assume that (c) holds. For every k∈ℤ, we define(3.5)Ek={x∈Rn:Sf(x)>2k}.
According to our assumption we have that supk∈ℤ2kp|Ek|<∞. Fix β∈(0,γ) and k∈ℕ. We denote by Dk the set defined by(3.6)Dk={(j,l)∈Z×Zn:|Q-j,l∩Ek|≥β|Q-j,l|}.
By Dk,max we represent the set that consists of all (j,l)∈Dk such that Q-j,l is maximal with respect to the order in ≤Dk introduced in [1, Definition 6.4, page 105].It is clear that(3.7)|Q-j,l|≤1β|Q-j,l∩Ek|≤C2-kp,(j,l)∈Dk.Moreover,|Q-j,l|=|detA|j,j∈ℤ, and l∈ℤn. Hence, there exists j0∈ℤ such that j≤j0 provided that (j,l)∈Dk. Then, for every (j,l)∈Dk, there exists (j1,l1)∈Dk,max for which Q-j,l≤DkQ-j1,l1 (see [1, page 105]).For every(j,l)∈Dk,max we define(3.8)Dk(j,l)={(m,s)∈Dk:Q-m,s≤DkQ-j,l}.
By [1, Lemma 6.5, page 105], we can find η∈ℕ such that for every (j,l)∈Dk,max we have(3.9)⋃(m,s)∈Dk(j,l)Q-m,s⊆⋃|l-l1|<ηQ-j,l1.
If Ek*=⋃(j,l)∈DkQ-j,l, it follows that(3.10)|Ek*|=|⋃(j,l)∈Dk,max⁡⋃(m,s)∈Dk(j,l)Q-m,s|≤∑(j,l)∈Dk,max⁡|⋃(m,s)∈Dk(j,l)Q-m,s|≤∑(j,l)∈Dk,max⁡|⋃|l-l1|<ηQ-j,l1|≤(2η+1)n∑(j,l)∈Dk,max⁡|det⁡⁡A|j=(2η+1)n∑(j,l)∈Dk,max⁡|Q-j,l|≤(2η+1)nβ∑(j,l)∈Dk,max⁡|Q-j,l∩Ek|≤(2η+1)nβ|Ek|.
In the last inequality we have used |Q-j,l∩Q-j1,l1|=0 provided that (j,l),  (j1,l1)∈Dk,max, (j,l)≠(j1,l1).By proceeding as in [1, page 107] by induction on (j,l)∈Dk,max we find, for every (j,l)∈Dk,max, a set(3.11)I(j,l)⊆Dk(j,l),
satisfying the following properties:(3.12)Dk=⋃(j,l)∈Dk,max⁡I(j,l),I(j,l)∩I(j1,l1)=∅,(j,l),(j1,l1)∈Dk,max⁡,(j,l)≠(j1,l1).
Note that Dk+1⊆Dk. We can consider the sets Zk=Dk∖Dk+1,Zk(j,l)=Zk∩I(j,l) and Mk(j,l)=⋃(m,s)∈Zk(j,l)Q-m,s, for every (j,l)∈Dk,max.Let(j,l)∈Dk,max. We can write(3.13)∫Mk(j,l)∖Ek+1|S(f)(x)|2dx=∫Mk(j,l)∖Ek+1∑m∈Z,s∈Zn|〈f,ψ-m,s〉|21|Q-m,s|χR-m,s(x)dx≥∫Mk(j,l)∖Ek+1∑(m,s)∈Zk(j,l)|〈f,ψ-m,s〉|21|Q-m,s|χR-m,s(x)dx≥∑(m,s)∈Zk(j,l)|〈f,ψ-m,s〉|21|Q-m,s||R-m,s∩Ek+1c|.
Also, for every (m,s)∈Zk(j,l), we get(3.14)|R-m,s∩Ek+1c|=|R-m,s|-|R-m,s∩Ek+1|≥(γ-β)|Q-m,s|,
because |R-m,s|≥γ|Q-m,s|, and, since (m,s)∉Dk+1,(3.15)|R-m,s∩Ek+1|≤|Q-m,s∩Ek+1|<β|Q-m,s|.
Hence(3.16)∫Mk(j,l)∖Ek+1|S(f)(x)|2dx≥∑(m,s)∈Zk(j,l)|〈f,ψ-m,s〉|2(γ-β).
On the other hand, we have that(3.17)∫Mk(j,l)∖Ek+1|S(f)(x)|2dx≤4k+1|Mk(j,l)∖Ek+1|.
We conclude that(3.18)∑(m,s)∈Zk(j,l)|〈f,ψ-m,s〉|2≤4k+1γ-β|Mk(j,l)|.
Again according to [1, Lemma 6.5] (see (3.9)) it follows that(3.19)Mk(j,l)⊆⋃|s-l|<ηQ-j,s.
Then(3.20)|Mk(j,l)|≤∑|s-l|<η|Q-j,s|≤(2η+1)n|det⁡⁡A|j.
Therefore(3.21)∑(m,s)∈Zk(j,l)|〈f,ψ-m,s〉|2≤(2η+1)nγ-β4k+1|Q-j,l|.
Let ν>0. We choose k0∈ℕ such that 2k0≤ν<2k0+1 and we define the functions G1 and G2 by(3.22)G1(f)=(∑k=-∞k0∑(j,l)∈Dk,max⁡∑(m,s)∈Zk(j,l)|〈f,ψ-m,s〉|21|Q-m,s|χQ-m,s)1/2,G2(f)=(∑k=k0+1∞∑(j,l)∈Dk,max⁡∑(m,s)∈Zk(j,l)|〈f,ψ-m,s〉|21|Q-m,s|χQ-m,s)1/2.
Note that if (m,s)∈ℤ×ℤn and 〈f,ψ-m,s〉≠0 then there exists k1∈ℤ such that R-m,s⊆Ek1. Hence (m,s)∈Dk, for every k∈ℤ,k≤k1. Moreover, there exists k2∈ℤ such that (m,s)∉Dk, when k≥k2. We deduce that G2=G12+G22. By (3.10) and (3.21), since ∑(j,l)∈Dk,max|Q-j,l|≤|Ek*|, we infer(3.23)‖G1f‖22≤(2η+1)nγ-β∑k=-∞k0∑(j,l)∈Dk,max⁡4k+1|Q-j,l|≤(2η+1)nγ-β∑k=-∞k04k+1|Ek*|≤(2η+1)2nβ(γ-β)∑k=-∞k04k+1|Ek|≤4(2η+1)2nβ(γ-β)∑k=-∞k02(2-p)k≤4(2η+1)2nβ(γ-β)2(2-p)k01-2-(2-p)≤Cν2-p.
Moreover, since G2(f)(x)=0 when x∉⋃k≥k0+1Ek*, by (3.10), it follows that(3.24)|{x∈Rn:G2(f)(x)≠0}|≤∑k=k0+1∞|Ek*|≤(2η+1)nβ∑k=k0+1∞|Ek|≤C(2η+1)nβ∑k=k0+1∞2-kp≤Cν-p.
Then, we conclude that(3.25)|{x∈Rn:G(f)(x)>ν}|≤|{x∈Rn:G1(f)(x)>ν2}|+|{x∈Rn:G2(f)(x)≠0}|≤∫{x∈Rn:G1(f)(x)>ν/2}(2G1(f)(x)ν)2dx+Cν-p≤Cν-p.
Thus (b) is proved.(c)⇒(d) Suppose that (c) holds. We keep the notation that it was used in the proof of (c)⇒(b).Letν>0. We choose k0∈ℤ such that 2k0≤ν<2k0+1 and we define(3.26)f1=∑k=-∞k0∑(j,l)∈Dk,max⁡∑(-m,s)∈Zk(j,l)〈f,ψ-m,s〉ψ-m,s.
By proceeding as in the proof of (c)⇒(b) we get(3.27)∑k=-∞k0∑(j,l)∈Dk,max⁡∑(m,s)∈Zk(j,l)|〈f,ψ-m,s〉|2≤Cν2-p.
Then, since ψ is a tight frame A-wavelet, by using duality, we obtain that
(3.28)‖f1‖22≤C∑k=-∞k0∑(j,l)∈Dk,max⁡∑(m,s)∈Zk(j,l)|〈f,ψ-m,s〉|2≤C2-p.
We now consider(3.29)f2=∑k=k0+1∞∑(j,l)∈Dk,max⁡∑(m,s)∈Zk(j,l)〈f,ψ-m,s〉ψ-m,s.
We choose φ∈S(ℝn) such that suppφ̂ is compact and bounded away from the origin and that ∑l∈ℤφ̂((A*)ly)=1,y∈ℝn∖{0}. We are going to prove that
(3.30)|{x∉R2:sup⁡α∈Z⁡|φα*f2(x)|>ν}|≤Cνp.
For every (j,l)∈Dk,max, we define(3.31)C(j,l)=⋃|l-l1|≤bηQ-j,l1,
where η>0 is the one appeared in (3.9) and b>2 will be chosen later. We also consider(3.32)Ω=⋃k=k0+1∞⋃(j,l)∈Dk,max⁡C(j,l).
If (j,l)∈Dk,max, then(3.33)|C(j,l)|≤∑|l-l1|≤bη|Q-j,l1|≤(2bη+1)n|Q-j,l|.
Hence, since ∑(j,l)∈Dk,max|Q-j,l|≤|Ek*|, by (3.10) it follows that(3.34)|Ω|=|⋃k=k0+1∞⋃(j,l)∈Dk,max⁡C(j,l)|≤(2bη+1)n∑k=k0+1∞|Ek*|≤C∑k=k0+1∞|Ek|≤C∑k=k0+1∞2-kp≤C2-k0p≤Cνp.
By (3.34), (3.30) will be proved when we obtain that(3.35)|{x∉Ω:sup⁡α∈Z|φα*f2(x)|>ν}|≤Cνp.
Since suppψ̂ and suppφ̂ are compact and 0∉suppψ̂∪suppφ̂, there exists M>0 such that suppψ̂-m,s∩suppφ̂α=∅, provided that |m-α|>M,m,α∈ℤ, and s∈ℤn.Letα∈ℤ. We can write(3.36)φα*f2=∑k=k0+1∞∑(j,l)∈Dk,max⁡∑m=α-Mα+M∑s:(m,s)∈Zk(j,l)〈f,ψ-m,s〉φα*ψ-m,s.
Using Hölder’s inequality we get(3.37)|∑m=α-Mα+M∑s:(m,s)∈Zk(j,l)〈f,ψ-m,s〉φα*ψ-m,s(x)|≤(∑m=α-Mα+M∑s:(m,s)∈Zk(j,l)|〈f,ψ-m,s〉|2)1/2(∑m=α-Mα+M∑s:(m,s)∈Zk(j,l)|(φα*ψ-m,s)(x)|2)1/2.
From (3.21) we deduce that(3.38)(∑m=α-Mα+m∑s:(m,s)∈Zk(j,l)|〈f,ψ-m,s〉|2)1/2≤(∑(m,s)∈Zk(j,l)|〈f,ψ-m,s〉|2)1/2≤C2k|Q-j,l|1/2.
According to [13, page 1482] for any λ>1 there exists C=C(λ)>0 for which(3.39)|(φβ*ψ-m,s)(x)|≤C|det⁡⁡A|-m/2(1+ρA(A-mx-s))-λ,x∈Rn,
when |β-m|≤M.Letλ>0 that will be fixed later. We can write(3.40)(∑m=α-Mα+M∑s:(m,s)∈Zk(j,l)|(φα*ψ-m,s)(x)|2)1/2≤C(∑m=α-Mα+M∑s:(m,s)∈Zk(j,l)|det⁡⁡A|-m(1+ρA(A-mx-s))2λ)1/2≤C∑m=α-Mα+M∑s:(m,s)∈Zk(j,l)|det⁡⁡A|-m/2(1+ρA(A-mx-s))λ,x∈Rn.
Let k∈ℤ,k>k0, and (j,l)∈Dk,max. Assume that x∉Ω. Then x∉C(j,l). By (3.9) we know that(3.41)⋃(m,s)∈Zk(j,l)Q-m,s⊆⋃|l-l1|<ηQ-j,l1.
Let (m,s)∈Zk(j,l). We define(3.42)x′=Ajx,xm,s=Ams,xm,s′=Ajxm,s,xj,l=A-jl,xj,l′=l.
Note that(3.43)x′∉⋃|l-l1|<bη([0,1]n+l1),xm,s′∈⋃|l-l1|<η([0,1]n+l1),xj,l′∈⋃|l-l1|<η([0,1]n+l1).
We have that(3.44)ρA(x′-xm,s′)≥1HρA(x′-xj,l′)-ρA(xm,s′-xj,l′).
By [1, Lemma 3.2] it follows that, for certain γ>0,(3.45)ρA(xm,s′-xj,l′)≤Cηγ,
and, for some γ′>0,(3.46)ρA(x′-xj,l′)≥C(bη)γ′.
Choosing b large enough we get that(3.47)ρA(x′-xm,s′)≥12HρA(x′-xj,l′).
Then,(3.48)ρA(x-xm,s)=ρA(A-j(x′-xm,s′))=|det⁡⁡A|-jρA(x′-xm,s′)≥|det⁡⁡A|-j2HρA(x′-xj,l′)=12HρA(x′-xj,l′).
By (3.40) and (3.48) with λ>3/2 we obtain that(3.49)(∑m=α-Mα+M∑s:(m,s)∈Zk(j,l)|(φα*ψ-m,s)(x)|2)1/2≤CρA(x-xj,l)λ∑m=α-Mα+M∑s:(m,s)∈Zk(j,l)|det⁡⁡A|-m(1/2-λ)≤CρA(x-xj,l)λ∑m=α-Mα+M(∑s:(m,s)∈Zk(j,l)|det⁡⁡A|m)-1/2+λ≤CρA(x-xj,l)λ∑m=α-Mα+M|⋃s:(m,s)∈Zk(j,l)Q-m,s|-1/2+λ≤C|Q-j,l|-1/2+λρA(x-xj,l)λ.
In the last inequality we have used (3.20). From (3.37), (3.38), and (3.49) it follows that(3.50)|∑m=α-Mα+M∑s:(m,s)∈Zk(j,l)〈f,ψ-m,s〉φα*ψ-m,s(x)|≤C2k|Q-j,l|λρA(x-xj,l)λ,λ≥32.
Whence(3.51)|{x∉Ω:sup⁡α∈Z|φα*f2(x)|>ν}|≤|{x∉Ω:∑k=k0+1∞∑(j,l)∈Dk,max⁡C2k|Q-j,l|λρA(x-xj,l)λ>ν}|≤|{x∉Ω:∑k=k0+1∞∑(j,l)∈Dk,max⁡cj,lkgj,lk>ν}|,
where λ>3/2,cj,lk=C2k|Q-j,l|λ and(3.52)gj,lk(x)=χΩc(x)ρA(x-xj,l)λ,x∈Rn.
By [2, equation (2.4)] we know that |BρA(a,r)|≈r,a∈ℝn, and r>0. Therefore,(3.53)|{x∈Rn:|gj,lk(x)|>ω}|=|{x∈Ωc:ρA(x-xj,l)<ω-1/λ}|≤Cω-1/λ,ω>0.
Hence, according to [9, Lemma 2.1], by taking λ>1/p, we obtain(3.54)|{x∉Ω:∑k=k0+1∞∑(j,l)∈Dk,max⁡C2k|Q-j,l|λρA(x-xj,l)λ>ν}|≤C1ν1/λ∑k=k0+1∞∑(j,l)∈Dk,max⁡2k/λ|Q-j,l|≤C1ν1/λ∑k=k0+1∞2k/λ|Ek*|≤C1ν1/λ∑k=k0+1∞2k/λ|Ek|≤C1ν1/λ∑k=k0+1∞2k(1/λ-p)≤C1ν1/λ2k0(1/λ-p)1-21/λ-p≤Cνp.
Hence,(3.55)|{x∉Ω:sup⁡α∈Z⁡|φα*f2(x)|>ν}|≤Cνp.
Combining (3.28), (3.34), and (3.55) we get(3.56)|{x∈Rn:sup⁡α∈Z|φα*f(x)|>ν}|≤|{x∈Rn:sup⁡α∈Z|φα*f1(x)|>ν2}|+|Ω|+|{x∉Ω:sup⁡α∈Z|φα*f2(x)|>ν2}|≤4ν2‖sup⁡α∈Z|φα*f1(x)|‖22+Cνp≤C(1ν2‖f1‖22+1νp)≤Cνp.
Thus we conclude that f∈HAp,∞(ℝn) and, therefore, (d) is established.(d)⇒(a) We define the sublinear operator W as follows:(3.57)Wf=(∑j∈Z,k∈Zn|〈f,ψj,k〉|2|ψj,k(x)|2)1/2,f∈L2(Rn).
Since ψ is a tight frame A-wavelet, we have that(3.58)‖Wf‖22=∫Rn∑j∈Z,k∈Zn|〈f,ψj,k〉|2|ψj,k(x)|2dx=∑j∈Z,k∈Zn|〈f,ψj,k〉|2|det⁡⁡A|j∫Rn|ψ(Ajx-k)|2dx=∫Rn|ψ(x)|2dx∑j∈Z,k∈Zn|〈f,ψj,k〉|2=‖f‖22∫Rn|ψ(x)|2dx,f∈L2(Rn).
Hence, W is a bounded operator from L2(ℝn) into itself.According to [1, Theorem 6.7, page 109], for every 0<q≤1 and f∈HAq(ℝn),(3.59)‖f‖HAq(Rn)~‖(∑j∈Z,k∈Zn|〈f,ψj,k〉|2|ψj,k(x)|2)1/2‖q.
Then, the operator W defined as above is also bounded from the anisotropic Hardy space HAq(ℝn) into Lq(ℝn), for every 0<q≤1. Our proof will be finished when we see that the operator T is bounded from HAp,∞(ℝn) into Lp,∞(ℝn). In order to do this we proceed interpolating by using the ideas developed in the proof of [9, Theorem 3.4].Letf∈HAp,∞(ℝn) and ν>0. We choose k0∈ℤ such that 2k0≤ν<2k0+1. We take the atomic decomposition of f given by f=∑k=-∞∞fk. Here fk=∑i=0∞βik, k∈ℤ, as in Theorem 2.1 with s=[(2/p-1)((log|detA|)/logm)], where m=min{|λ|:λisaneigenvalueofA}. We decompose f as follows:(3.60)f=∑k=-∞k0fk+∑k=k0+1∞fk=F1+F2.
We have that(3.61)‖F1‖2≤∑k=-∞k0‖fk‖2≤C∑k=-∞k02k(∑i=0∞|B̃ik|)1/2≤C∑k=-∞k02k(1-p/2)‖f‖HAp,∞(Rn)p/2≤C2k0(1-p/2)‖f‖HAp,∞(Rn)p/2.
Then,(3.62)|{x∈Rn:|WF1(x)|>ν}|≤Cν2‖F1‖22≤C2k0(2-p)ν2‖f‖HAp,∞(Rn)p≤Cνp‖f‖HAp,∞(Rn)p.According to Theorem2.1, there exists C>0 such that, for every k∈ℤ and i∈ℕ, C2-k|B̃ik|-2/pβik is a (p/2,∞,s)HAp/2(ℝn)-atom. Hence, for every k∈ℤ, fk∈HAp/2(ℝn) and(3.63)‖fk‖HAp/2(Rn)p/2≤C∑i=0∞2kp/2|B̃ik|≤C2-kp/2‖f‖HAp,∞(Rn)p.
Since W is bounded from HAp/2(ℝn) into Lp/2(ℝn), it follows that(3.64)|{x∈Rn:|W(fk)(x)|>ν}|≤ν-p/2‖W(fk)‖p/2p/2≤Cν-p/2‖fk‖HAp/2(Rn)p/2,k∈Z.Therefore, using [9, Lemma 2.1] we get(3.65)|{x∈Rn:|W(F2)(x)|>ν}|≤|{x∈Rn:∑k=k0+1∞|W(fk)(x)|>ν}|=|{x∈Rn:∑k=k0+1∞‖fk‖HAp/2(Rn)|W(fk‖fk‖HAp/2(Rn))(x)|>ν}|≤Cν-p/2∑k=k0+1∞‖fk‖HAp/2(Rn)p/2≤Cν-p/2∑k=k0+1∞2-kp/2‖f‖HAp,∞(Rn)p≤Cν-p/22-k0p/2‖f‖HAp,∞(Rn)p≤Cνp‖f‖HAp,∞(Rn)p.
Putting together the above estimates we obtain(3.66)|{x∈Rn:|W(f)(x)|>ν}|≤|{x∈Rn:|W(F1)(x)|>ν2}|+|{x∈Rn:|W(F2)(x)|>ν2}|≤Cνp‖f‖HAp,∞(Rn)p.



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*Source: 809121-2012-05-03.xml*