# Maximal and Area Integral Characterizations of Bergman Spaces in the Unit Ball ofℂn

Authors: Zeqian Chen; Wei Ouyang
Journal: Journal of Function Spaces and Applications
 (2013)

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## Abstract

We present maximal and area integral characterizations of Bergman spaces in the unit ball ofℂn. The characterizations are in terms of maximal functions and area integral functions on Bergman balls involving the radial derivative, the complex gradient, and the invariant gradient. As an application, we obtain new maximal and area integral characterizations of Besov spaces. Moreover, we give an atomic decomposition of real-variable type with respect to Carleson tubes for Bergman spaces.


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## Body

## 1. Introduction and Main Results

Letℂ denote the set of complex numbers. Throughout the paper we fix a positive integer n, and let
(1)ℂn=ℂ×⋯×ℂ
denote the Euclidean space of complex dimension n. Addition, scalar multiplication, and conjugation are defined on ℂn component wise. For z=(z1,…,zn) and w=(w1,…,wn) in ℂn, we write
(2)〈z,w〉=z1w¯1+⋯+znw¯n,
where w¯k is the complex conjugate of wk. We also write
(3)|z|=|z1|2+⋯+|zn|2.
The open unit ball in ℂn is the set
(4)𝔹n={z∈ℂn:|z|<1}.
The boundary of 𝔹n will be denoted by 𝕊n and is called the unit sphere in ℂn, that is,
(5)𝕊n={z∈ℂn:|z|=1}.
Also, we denote by 𝔹¯n the closed unit ball, that is,
(6)𝔹¯n={z∈ℂn:|z|≤1}=𝔹n∪𝕊n.
The automorphism group of 𝔹n, denoted by Aut(𝔹n), consists of all biholomorphic mappings of 𝔹n. Traditionally, bi-holomorphic mappings are also called automorphisms.Forα∈ℝ, the weighted Lebesgue measure dvα on 𝔹n is defined by
(7)dvα(z)=cα(1-|z|2)αdv(z),
where cα=1 for α≤-1 and cα=Γ(n+α+1)/[n!Γ(α+1)] if α>-1, which is a normalizing constant so that dvα is a probability measure on 𝔹n. In the case of α=-(n+1), we denote the resulting measure by
(8)dτ(z)=dv(1-|z|2)n+1
and call it the invariant measure on 𝔹n, since dτ=dτ∘φ for any automorphism φ of 𝔹n.Forα>-1 and p>0, the (weighted) Bergman space 𝒜αp consists of holomorphic functions f in 𝔹n with
(9)∥f∥p,α=(∫𝔹n‍|f(z)|pdvα(z))1/p<∞,
where the weighted Lebesgue measure dvα on 𝔹n is defined by
(10)dvα(z)=cα(1-|z|2)αdv(z),
and cα=Γ(n+α+1)/[n!Γ(α+1)] is a normalizing constant so that dvα is a probability measure on 𝔹n. Thus,
(11)𝒜αp=ℋ(𝔹n)∩Lp(𝔹n,dvα),
where ℋ(𝔹n) is the space of all holomorphic functions in 𝔹n. When α=0, we simply write 𝒜p for 𝒜0p. These are the usual Bergman spaces. Note that for 1≤p<∞,𝒜αp is a Banach space under the norm ∥∥p,α. If 0<p<1, the space 𝒜αp is a quasi-Banach space with p-norm ∥f∥p,αp.Recall thatD(z,γ) denotes the Bergman metric ball at z(12)D(z,γ)={w∈𝔹n:β(z,w)<γ},
with γ>0, where β is the Bergman metric on 𝔹n. It is known that
(13)β(z,w)=12log1+|φz(w)|1-|φz(w)|,z,w∈𝔹n,
whereafter φz is the bijective holomorphic mapping in 𝔹n, which satisfies φz(0)=z, φz(z)=0, and φz∘φz=id.As is well known, maximal functions play a crucial role in the real-variable theory of Hardy spaces (cf. [1]). In this paper, we first establish a maximal-function characterization for the Bergman spaces. To this end, we define for each γ>0 and f∈ℋ(𝔹n):
(14)(Mγf)(z)=supw∈D(z,γ)|f(w)|,∀z∈𝔹n.
Then we have the following result.Theorem 1.
Supposeγ>0 and α>-1. Let 0<p<∞. Then for any f∈ℋ(𝔹n),f∈𝒜αp if and only if Mγf∈Lp(𝔹n,dvα). Moreover,
(15)∥f∥p,α≈∥Mγf∥p,α,
where “≈” depends only on γ,α,p, and n.The norm appearing on the right-hand side of (15) can be viewed as an analogue of the so-called nontangential maximal function in Hardy spaces. The proof of Theorem 1 is fairly elementary (see Section 2), using some basic facts and estimates on the Bergman balls.In order to state the area integral characterizations of the Bergman spaces, we require some more notation. For anyf∈ℋ(𝔹n) and z=(z1,…,zn)∈𝔹n, we define
(16)ℛf(z)=∑k=1n‍zk∂f(z)∂zk
and call it the radial derivative of f at z. The complex and invariant gradients of f at z are, respectively, defined as
(17)∇f(z)=(∂f(z)∂z1,…,∂f(z)∂zn),∇~f(z)=∇(f∘φz)(0).Now, for fixed1<q<∞ and γ>0, we define for each f∈ℋ(𝔹n) and z∈𝔹n:(1)
the radial area integral function(18)Aℛγ,q(f)(z)=(∫D(z,γ)‍|(1-|w|2)ℛf(w)|qdτ(w))1/q,(2)
the complex gradient area integral function(19)A∇γ,q(f)(z)=(∫D(z,γ)‍|(1-|w|2)∇f(w)|qdτ(w))1/q,(3)
the invariant gradient area integral function(20)A∇~γ,q(f)(z)=(∫D(z,γ)‍|∇~f(w)|qdτ(w))1q.We state the second main result of this paper as follows.Theorem 2.
Suppose1<q<∞,γ>0, and α>-1. Let 0<p<∞. Then, for any f∈ℋ(𝔹n), the following conditions are equivalent:(a)
f∈𝒜αp,(b)
Aℛγ,q(f) is in Lp(𝔹n,dvα),(c)
A∇γ,q(f) is in Lp(𝔹n,dvα),(d)
A∇~γ,q(f) is in Lp(𝔹n,dvα).
Moreover, the quantities(21)∥Aℛγ,q(f)∥p,α,∥A∇γ,q(f)∥p,α,∥A∇~γ,q(f)∥p,α
are all comparable to ∥f-f(0)∥p,α, where the comparable constants depend only on q,γ,α,p, and n.For0<p<∞ and -∞<α<∞, we fix a nonnegative integer k with pk+α>-1 and define the generalized Bergman space 𝒜αp as introduced in [2] to be the space of all f∈ℋ(𝔹n) such that (1-|z|2)kℛkf∈Lp(𝔹n,dvα). One then easily observes that 𝒜αp is independent of the choice of k and consistent with the traditional definition when α>-1. Let N be the smallest nonnegative integer such that pN+α>-1. Put
(22)∥f∥p,α=|f(0)|+(∫𝔹n‍(1-|z|2)pN|ℛNf(z)|pdvα(z))1/p,f∈𝒜αp.
Equipped with (22), 𝒜αp becomes a Banach space when p≥1 and a quasi-Banach space for 0<p<1.clized Bergman spaces𝒜αp covers most of the spaces of holomorphic functions in the unit ball of ℂn, such as the classical diagonal Besov space Bps and the Sobolev space Wk,βp, which has been extensively studied before in the literature under different names (see e.g., [2] for an overview).There are various characterizations forBps or Wk,βp involving complex-variable quantities in terms of radical derivatives, complex and invariant gradients, and fractional differential operators (for a review and details see [2] and references therein). However, as an application of Theorems 1 and 2, we obtain new maximal and area integral characterizations of the Besov spaces as follows, which can be considered as a unified characterization for such spaces involving real-variable quantities.Corollary 3.
Supposeγ>0 and α∈ℝ. Let 0<p<∞ and k a positive integer such that pk+α>-1. Then for any f∈ℋ(𝔹n),f∈𝒜αp if and only if Mγ(k)(f)∈Lp(𝔹n,dvα), where
(23)Mγ(k)(f)(z)=supw∈D(z,γ)|(1-|w|2)kℛkf(w)|,z∈𝔹n.
Moreover,
(24)∥f-f(0)∥p,α≈∥Mγ(ℛkf)∥p,α,
where “≈” depends only on γ,α,p,k, and n.Corollary 4.
Suppose1<q<∞,γ>0, and α∈ℝ. Let 0<p<∞ and k a nonnegative integer such that pk+α>-1. Then for any f∈ℋ(𝔹n),f∈𝒜αp if and only if Aℛk+1γ,q(f) is in Lp(𝔹n,dvα), where
(25)Aℛk+1γ,q(f)(z)=(∫D(z,γ)‍|(1-|w|2)k+1ℛk+1f(w)|qdτ(w))1/q.
Moreover,
(26)∥f-f(0)∥p,α≈∥Aℛk+1γ,q(f)∥p,α,
where “≈” depends only on q,γ,α,p,k, and n.To prove Corollaries3 and 4, one merely notices that f∈𝒜αp if and only if ℛkf∈Lp(𝔹n,dvα+pk) and applies Theorems 1 and 2, respectively, to ℛkf with the help of Lemma 5 below.The paper is organized as follows. In Section2 we will prove Theorems 1 and 2. An atomic decomposition of real-variable type with respect to Carleson tubes for Bergman spaces will be presented in Section 3 via duality method. Finally, in Section 4, we will prove Theorem 2 through using the real-variable atomic decomposition of Bergman spaces established in the preceding section.In what follows,C always denotes a constant depending (possibly) on n,q,p,γ, or α but not on f, which may be different in different places. For two nonnegative (possibly infinite) quantities X and Y, by X≲Y we mean that there exists a constant C>0 such that X≤CY and by X≈Y that X≲Y and Y≲X. Any notation and terminology not otherwise explained are as used in [3] for spaces of holomorphic functions in the unit ball of ℂn.


## 2. Proofs of Theorems1 and 2

For the sake of convenience, we collect some elementary facts on the Bergman metric and holomorphic functions in the unit ball ofℂn as follows.Lemma 5 (cf. [3, Lemma 2.20]).
For eachγ>0,
(27)1-|a|2≈1-|z|2≈|1-〈a,z〉|,
for all a and z in 𝔹n with β(a,z)<γ.Lemma 6 (cf. [3, Lemma 2.24]).
Supposeγ>0,p>0, and α>-1. Then there exists a constant C>0 such that for any f∈ℋ(𝔹n),
(28)|f(z)|p≤C(1-|z|2)n+1+α∫D(z,γ)‍|f(w)|pdvα(w),∀z∈𝔹n.Lemma 7 (cf. [3, Lemma 2.27]).
For eachγ>0,
(29)|1-〈z,u〉|≈|1-〈z,v〉|,
for all z in 𝔹-n and u,v in 𝔹n with β(u,v)<γ.


### 2.1. Proof of Theorem1

One needs the following result (cf. [4, Lemma 5]).Lemma 8.
For fixedγ>0, there exist a positive integer N and a sequence {ak} in 𝔹n such that(1)
𝔹n=⋃kD(ak,γ), and(2)
eachz∈𝔹n belongs to at most N of the sets D(ak,3γ).Proof of Theorem1.
Letp>0. By Lemmas 8, 6, and 5, one has
(30)∫𝔹n‍|Mγ(f)(z)|pdvα(z)≤∑k‍∫D(ak,γ)‍|Mγ(f)(z)|pdvα(z)=∑k‍∫D(ak,γ)‍supw∈D(z,γ)|f(w)|pdvα(z)≲∑k‍∫D(ak,γ)‍supw∈D(z,γ)1(1-|w|2)n+1+α×∫D(w,γ)‍|f(u)|pdvα(u)dvα(z)≲∑k‍∫D(ak,γ)‍(∫D(ak,3γ)‍1(1-|ak|2)n+1+α×∫D(ak,3γ)‍|f(u)|pdvα(u)1(1-|ak|2)n+1+α)dvα(z)≲∑k‍∫D(ak,3γ)‍|f(u)|pdvα(u)≲N∫𝔹n‍|f(u)|pdvα(u),
where N is the constant in Lemma 8 depending only on γ and n.


### 2.2. Proof of Theorem2

Recall thatℬ(𝔹n) is defined as the space of all f∈ℋ(𝔹n) so that
(31)∥f∥ℬ=supz∈𝔹n|∇~f(z)|<∞.ℬ(𝔹n) with the norm ∥f∥=|f(0)|+∥f∥ℬ is a Banach space and called the Bloch space. Then, the following interpolation result holds.Lemma 9 (cf. [3, Theorem 3.25]).
Let1<p<∞. Suppose α>-1 and
(32)1p=1-θp′,
for 0<θ<1 and 1≤p′<∞. Then
(33)𝒜αp(𝔹n)=[𝒜αp′(𝔹n),ℬ(𝔹n)]θ,
with equivalent norms.Moreover, to prove Theorem2 for the case 0<p≤1, one will use atom decomposition for Bergman spaces due to Coifman and Rochberg [5] (see also [3, Theorem 2.30]) as follows.Proposition 10.
Supposep>0,α>-1, and b>nmax{1,1/p}+(α+1)/p. Then there exists a sequence {ak} in 𝔹n such that 𝒜αp consists exactly of functions of the form
(34)f(z)=∑k=1∞‍ck(1-|ak|2)(pb-n-1-α)/p(1-〈z,ak〉)b,z∈𝔹n,
where {ck} belongs to the sequence space ℓp and the series converges in the norm topology of 𝒜αp. Moreover,
(35)∫𝔹n‍|f(z)|pdvα(z)≈inf{∑k‍|ck|p},
where the infimum runs over all the above decompositions.Also, we need a characterization of Carleson type measures for Bergman spaces as follows, which can be found in [2, Theorem 45].Proposition 11.
Supposen+1+α>0 and μ is a positive Borel measure on 𝔹n. Then, there exists a constant C>0 such that
(36)μ(Qr(ζ))≤Cr2(n+1+α),∀ζ∈𝕊n,r>0,
if and only if for each s>0, there exists a constant C>0 such that
(37)∫𝔹n‍(1-|z|2)s|1-〈z,w〉|n+1+α+sdμ(w)≤C,
for all z∈𝔹n.We are now ready to prove Theorem2. Note that for any f∈ℋ(𝔹n),
(38)(1-|z|2)|ℛf(z)|≤(1-|z|2)|∇f(z)|≤|∇~f(z)|,∀z∈𝔹n
(cf. [3, Lemma 2.14]). We have that (d) implies (c) and (c) implies (b) in Theorem 2. Then, it remains to prove that (b) implies (a) and (a) implies (d).Proof of(b)⇒(a).
Sinceℛf(z) is holomorphic, by Lemma 6 we have
(39)|ℛf(z)|q≤C(1-|z|2)n+1∫D(z,γ)‍|ℛf(w)|qdv(w)≤Cγ∫D(z,γ)‍|ℛf(w)|qdτ(w).
Then,
(40)(1-|z|2)|ℛf(z)|≤C(1-|z|2)(∫D(z,γ)‍|ℛf(w)|qdτ(w))1/q≤Cγ(∫D(z,γ)‍|(1-|w|2)ℛf(w)|qdτ(w))1/q=CγAℛγ,q(f)(z).
Hence, for any p>0, if Aℛγ,q(f)∈Lp(𝔹n,dvα) then (1-|z|2)|ℛf(z)| is in Lp(𝔹n,dvα), which implies that f∈𝒜αp (cf. [3, Theorem 2.16]).The proof of(a)⇒(d) is divided into two steps. We first prove the case 0<p≤1 using the atomic decomposition and then the remaining case via complex interpolation.Proof of(a)⇒(d)for0<p≤1.
To this end, we write(41)fk(z)=(1-|ak|2)(pb-n-1-α)/p(1-〈z,ak〉)b.
An immediate computation yields that
(42)∇fk(z)=ba¯k(1-|ak|2)(pb-n-1-α)/p(1-〈z,ak〉)b+1,ℛfk(z)=b〈z,ak〉(1-|ak|2)(pb-n-1-α)/p(1-〈z,ak〉)b+1.
Then we have
(43)|∇~fk(z)|2=(1-|z|2)(|∇fk(z)|2-|ℛfk(z)|2)=b2(1-|z|2)(1-|ak|2)2(pb-n-1-α)/p|ak|2-|〈z,ak〉|2|1-〈z,ak〉|2(b+1).
By Lemmas 5 and 7, one has
(44)A∇~γ,q(fk)(z)=(∫D(z,γ)‍|∇~fk(w)|qdτ(w))1/q≤b(1-|ak|2)(pb-n-1-α)/p×(∫D(z,γ)‍1|1-〈w,ak〉|qbdτ(w))1/q≤Cγb(1-|ak|2)(pb-n-1-α)/p|1-〈z,ak〉|b,
where we have used the fact v(D(z,γ))≈(1-|z|2)n+1. Note that vα(Qr)≈r2(n+1+α) (cf. [3, Corollary 5.24]); by Proposition 11 we have
(45)∫𝔹n|A∇~γ,q(fk)(z)|pdvα(z)≤Cbp∫𝔹n(1-|ak|2)(pb-n-1-α)|1-〈z,ak〉|pbdvα(z)≤Cp,α.
Hence, for 0<p≤1, we have for f=∑k=1∞‍ckfk with ∑k‍|ck|p<∞,
(46)∫𝔹n‍|A∇~γ,q(f)(z)|pdvα≤∑k=1∞‍|ck|p∫𝔹n‍|A∇~γ,q(fk)(z)|pdvα≤Cp,α∑k=1∞‍|ck|p.
This concludes that
(47)∫𝔹n‍|A∇~γ,q(f)(z)|pdvα≤Cp,αinf{∑k=1∞‍|ck|p}≤Cp,α∫𝔹n‍|f(z)|pdvα(z).
The proof is complete.Proof of(a)⇒(d) for p>1.
SetE=Lq(𝔹n,χD(0,γ)dτ;ℂn). Consider the operator
(48)T(f)(z,w)=(∇~f)(φz(w)),f∈ℋ(𝔹n).
Note that φz(D(0,γ))=D(z,γ) and the measure dτ is invariant under any automorphism of 𝔹n (cf. [3, Proposition 1.13]); we have
(49)∥T(f)(z)∥E=(∫𝔹n‍|(∇~f)(φz(w))|qχD(0,γ)(w)dτ(w))1/q=(∫𝔹n‍|∇~f(w)|qχD(z,γ)(w)dτ(w))1/q=A∇~γ,q(f)(z).
On the other hand,
(50)A∇~γ,q(f)(z)≤[Cγ(1-|z|2)-n-1v(D(z,γ))]1/2∥f∥ℬ≤C∥f∥ℬ.
This follows that T is bounded from ℬ into Lα∞(𝔹n,E). Notice that we have proved that T is bounded from 𝒜α1 to Lα1(𝔹n,E). Thus, by Lemma 9 and the well known fact that
(51)Lp(𝔹n,E)=(Lα1(𝔹n,E),Lα∞(𝔹n,E))θwithθ=1-1p,
we conclude that T is bounded from 𝒜αp into Lαp(𝔹n,E) for any 1<p<∞; that is,
(52)∥A∇~γ,q(f)∥p,α≤C∥f∥p,α,∀f∈𝒜αp,
where C depends only on q, γ, n, p, and α. The proof is complete.Remark 12.
From the proofs of that(b)⇒(a) and that (a)⇒(d) for p>1, we find that Theorem 2 still holds true for the Bloch space. That is, for any f∈ℋ(𝔹n),f∈ℬ if and only if one (or equivalently, all) of Aℛγ,q(f),A∇γ,q(f), and A∇~γ,q(f) is (or, are) in L∞(𝔹n). Moreover,
(53)∥f∥ℬ≈∥Aℛγ,q(f)∥L∞(𝔹n)≈∥A∇γ,q(f)∥L∞(𝔹n)≈∥A∇~γ,q(f)∥L∞(𝔹n),
where “≈” depends only on q,γ, and n.


## 2.1. Proof of Theorem1

One needs the following result (cf. [4, Lemma 5]).Lemma 8.
For fixedγ>0, there exist a positive integer N and a sequence {ak} in 𝔹n such that(1)
𝔹n=⋃kD(ak,γ), and(2)
eachz∈𝔹n belongs to at most N of the sets D(ak,3γ).Proof of Theorem1.
Letp>0. By Lemmas 8, 6, and 5, one has
(30)∫𝔹n‍|Mγ(f)(z)|pdvα(z)≤∑k‍∫D(ak,γ)‍|Mγ(f)(z)|pdvα(z)=∑k‍∫D(ak,γ)‍supw∈D(z,γ)|f(w)|pdvα(z)≲∑k‍∫D(ak,γ)‍supw∈D(z,γ)1(1-|w|2)n+1+α×∫D(w,γ)‍|f(u)|pdvα(u)dvα(z)≲∑k‍∫D(ak,γ)‍(∫D(ak,3γ)‍1(1-|ak|2)n+1+α×∫D(ak,3γ)‍|f(u)|pdvα(u)1(1-|ak|2)n+1+α)dvα(z)≲∑k‍∫D(ak,3γ)‍|f(u)|pdvα(u)≲N∫𝔹n‍|f(u)|pdvα(u),
where N is the constant in Lemma 8 depending only on γ and n.


## 2.2. Proof of Theorem2

Recall thatℬ(𝔹n) is defined as the space of all f∈ℋ(𝔹n) so that
(31)∥f∥ℬ=supz∈𝔹n|∇~f(z)|<∞.ℬ(𝔹n) with the norm ∥f∥=|f(0)|+∥f∥ℬ is a Banach space and called the Bloch space. Then, the following interpolation result holds.Lemma 9 (cf. [3, Theorem 3.25]).
Let1<p<∞. Suppose α>-1 and
(32)1p=1-θp′,
for 0<θ<1 and 1≤p′<∞. Then
(33)𝒜αp(𝔹n)=[𝒜αp′(𝔹n),ℬ(𝔹n)]θ,
with equivalent norms.Moreover, to prove Theorem2 for the case 0<p≤1, one will use atom decomposition for Bergman spaces due to Coifman and Rochberg [5] (see also [3, Theorem 2.30]) as follows.Proposition 10.
Supposep>0,α>-1, and b>nmax{1,1/p}+(α+1)/p. Then there exists a sequence {ak} in 𝔹n such that 𝒜αp consists exactly of functions of the form
(34)f(z)=∑k=1∞‍ck(1-|ak|2)(pb-n-1-α)/p(1-〈z,ak〉)b,z∈𝔹n,
where {ck} belongs to the sequence space ℓp and the series converges in the norm topology of 𝒜αp. Moreover,
(35)∫𝔹n‍|f(z)|pdvα(z)≈inf{∑k‍|ck|p},
where the infimum runs over all the above decompositions.Also, we need a characterization of Carleson type measures for Bergman spaces as follows, which can be found in [2, Theorem 45].Proposition 11.
Supposen+1+α>0 and μ is a positive Borel measure on 𝔹n. Then, there exists a constant C>0 such that
(36)μ(Qr(ζ))≤Cr2(n+1+α),∀ζ∈𝕊n,r>0,
if and only if for each s>0, there exists a constant C>0 such that
(37)∫𝔹n‍(1-|z|2)s|1-〈z,w〉|n+1+α+sdμ(w)≤C,
for all z∈𝔹n.We are now ready to prove Theorem2. Note that for any f∈ℋ(𝔹n),
(38)(1-|z|2)|ℛf(z)|≤(1-|z|2)|∇f(z)|≤|∇~f(z)|,∀z∈𝔹n
(cf. [3, Lemma 2.14]). We have that (d) implies (c) and (c) implies (b) in Theorem 2. Then, it remains to prove that (b) implies (a) and (a) implies (d).Proof of(b)⇒(a).
Sinceℛf(z) is holomorphic, by Lemma 6 we have
(39)|ℛf(z)|q≤C(1-|z|2)n+1∫D(z,γ)‍|ℛf(w)|qdv(w)≤Cγ∫D(z,γ)‍|ℛf(w)|qdτ(w).
Then,
(40)(1-|z|2)|ℛf(z)|≤C(1-|z|2)(∫D(z,γ)‍|ℛf(w)|qdτ(w))1/q≤Cγ(∫D(z,γ)‍|(1-|w|2)ℛf(w)|qdτ(w))1/q=CγAℛγ,q(f)(z).
Hence, for any p>0, if Aℛγ,q(f)∈Lp(𝔹n,dvα) then (1-|z|2)|ℛf(z)| is in Lp(𝔹n,dvα), which implies that f∈𝒜αp (cf. [3, Theorem 2.16]).The proof of(a)⇒(d) is divided into two steps. We first prove the case 0<p≤1 using the atomic decomposition and then the remaining case via complex interpolation.Proof of(a)⇒(d)for0<p≤1.
To this end, we write(41)fk(z)=(1-|ak|2)(pb-n-1-α)/p(1-〈z,ak〉)b.
An immediate computation yields that
(42)∇fk(z)=ba¯k(1-|ak|2)(pb-n-1-α)/p(1-〈z,ak〉)b+1,ℛfk(z)=b〈z,ak〉(1-|ak|2)(pb-n-1-α)/p(1-〈z,ak〉)b+1.
Then we have
(43)|∇~fk(z)|2=(1-|z|2)(|∇fk(z)|2-|ℛfk(z)|2)=b2(1-|z|2)(1-|ak|2)2(pb-n-1-α)/p|ak|2-|〈z,ak〉|2|1-〈z,ak〉|2(b+1).
By Lemmas 5 and 7, one has
(44)A∇~γ,q(fk)(z)=(∫D(z,γ)‍|∇~fk(w)|qdτ(w))1/q≤b(1-|ak|2)(pb-n-1-α)/p×(∫D(z,γ)‍1|1-〈w,ak〉|qbdτ(w))1/q≤Cγb(1-|ak|2)(pb-n-1-α)/p|1-〈z,ak〉|b,
where we have used the fact v(D(z,γ))≈(1-|z|2)n+1. Note that vα(Qr)≈r2(n+1+α) (cf. [3, Corollary 5.24]); by Proposition 11 we have
(45)∫𝔹n|A∇~γ,q(fk)(z)|pdvα(z)≤Cbp∫𝔹n(1-|ak|2)(pb-n-1-α)|1-〈z,ak〉|pbdvα(z)≤Cp,α.
Hence, for 0<p≤1, we have for f=∑k=1∞‍ckfk with ∑k‍|ck|p<∞,
(46)∫𝔹n‍|A∇~γ,q(f)(z)|pdvα≤∑k=1∞‍|ck|p∫𝔹n‍|A∇~γ,q(fk)(z)|pdvα≤Cp,α∑k=1∞‍|ck|p.
This concludes that
(47)∫𝔹n‍|A∇~γ,q(f)(z)|pdvα≤Cp,αinf{∑k=1∞‍|ck|p}≤Cp,α∫𝔹n‍|f(z)|pdvα(z).
The proof is complete.Proof of(a)⇒(d) for p>1.
SetE=Lq(𝔹n,χD(0,γ)dτ;ℂn). Consider the operator
(48)T(f)(z,w)=(∇~f)(φz(w)),f∈ℋ(𝔹n).
Note that φz(D(0,γ))=D(z,γ) and the measure dτ is invariant under any automorphism of 𝔹n (cf. [3, Proposition 1.13]); we have
(49)∥T(f)(z)∥E=(∫𝔹n‍|(∇~f)(φz(w))|qχD(0,γ)(w)dτ(w))1/q=(∫𝔹n‍|∇~f(w)|qχD(z,γ)(w)dτ(w))1/q=A∇~γ,q(f)(z).
On the other hand,
(50)A∇~γ,q(f)(z)≤[Cγ(1-|z|2)-n-1v(D(z,γ))]1/2∥f∥ℬ≤C∥f∥ℬ.
This follows that T is bounded from ℬ into Lα∞(𝔹n,E). Notice that we have proved that T is bounded from 𝒜α1 to Lα1(𝔹n,E). Thus, by Lemma 9 and the well known fact that
(51)Lp(𝔹n,E)=(Lα1(𝔹n,E),Lα∞(𝔹n,E))θwithθ=1-1p,
we conclude that T is bounded from 𝒜αp into Lαp(𝔹n,E) for any 1<p<∞; that is,
(52)∥A∇~γ,q(f)∥p,α≤C∥f∥p,α,∀f∈𝒜αp,
where C depends only on q, γ, n, p, and α. The proof is complete.Remark 12.
From the proofs of that(b)⇒(a) and that (a)⇒(d) for p>1, we find that Theorem 2 still holds true for the Bloch space. That is, for any f∈ℋ(𝔹n),f∈ℬ if and only if one (or equivalently, all) of Aℛγ,q(f),A∇γ,q(f), and A∇~γ,q(f) is (or, are) in L∞(𝔹n). Moreover,
(53)∥f∥ℬ≈∥Aℛγ,q(f)∥L∞(𝔹n)≈∥A∇γ,q(f)∥L∞(𝔹n)≈∥A∇~γ,q(f)∥L∞(𝔹n),
where “≈” depends only on q,γ, and n.


## 3. Atomic Decomposition for Bergman Spaces

We let(54)d(z,w)=|1-〈z,w〉|1/2,z,w∈𝔹¯n.
It is known that d satisfies the triangle inequality and the restriction of d to 𝕊n is a metric. As usual, d is called the nonisotropic metric.For anyζ∈𝕊n and r>0, the set
(55)Qr(ζ)={z∈𝔹n:d(z,ζ)<r}
is called a Carleson tube with respect to the nonisotropic metric d. We usually write Q=Qr(ζ) in short.As usual, we define the atoms with respect to the Carleson tube as follows: for1<q<∞,a∈Lq(𝔹n,dvα) is said to be a (1,q)α-atom if there is a Carleson tube Q such that(1)
a is supported in Q;(2)
∥a∥Lq(𝔹n,dvα)≤vα(Q)(1/q)-1;(3)
∫𝔹n‍a(z)dvα(z)=0.The constant function (1) is also considered to be a(1,q)α-atom. Note that for any (1,q)α-atom a,
(56)∥a∥1,α=∫Q‍|a|dvα≤vα(Q)1-1/q∥a∥q,α≤1.Recall thatPα is the orthogonal projection from L2(𝔹n,dvα) onto 𝒜α2, which can be expressed as
(57)Pαf(z)=∫𝔹n‍Kα(z,w)f(w)dvα(w),∀f∈L1(𝔹n,dvα),α>-1,
where
(58)Kα(z,w)=1(1-〈z,w〉)n+1+α,z,w∈𝔹n.Pα extends to a bounded projection from Lp(𝔹n,dvα) onto 𝒜αp (1<p<∞).We have the following useful estimates.Lemma 13.
Forα>-1 and 1<q<∞, there exists a constant Cq,α,n>0 such that
(59)∥Pα(a)∥1,α≤Cq,α,n,
for any (1,q)α-atom a.To prove Lemma13, we need first to show an inequality for reproducing kernel Kα associated with d, which is essentially borrowed from [6, Proposition 2.13].Lemma 14.
Forα>-1, there exists a constant δ>0 such that for all z,w∈𝔹n,ζ∈𝕊n satisfying d(z,ζ)>δd(w,ζ), we have
(60)|Kα(z,w)-Kα(z,ζ)|≤Cα,nd(w,ζ)d(z,ζ)2(n+1+α)+1.Proof.
Note that(61)Kα(z,w)-Kα(z,ζ)=∫01‍ddt(1(1-〈z,ζ〉-t〈z,w-ζ〉)n+1+α)dt.
We have
(62)|Kα(z,w)-Kα(z,ζ)|≤∫01‍(n+1+α)|〈z,w-ζ〉||1-〈z,ζ〉-t〈z,w-ζ〉|n+2+αdt.
Writez=z1+z2 and w=w1+w2, where z1 and w1 are parallel to ζ, while z2 and w2 are perpendicular to ζ. Then
(63)〈z,w〉-〈z,ζ〉=〈z2,w2〉-〈z1,w1-ζ〉
and so
(64)|〈z,w〉-〈z,ζ〉|≤|z2||w2|+|w1-ζ|.
Since |w1-ζ|=|1-〈w,ζ〉|,
(65)|z2|2=|z|2-|z1|2<1-|z1|2<(1+|z1|)(1+|z1|)≤|1-〈z1,ζ〉|=2|1-〈z,ζ〉|,
and similarly
(66)|w2|2≤2|1-〈w,ζ〉|,
we have
(67)|〈z,w〉-〈z,ζ〉|≤2|1-〈z,ζ〉|1/2|1-〈w,ζ〉|1/2+|1-〈w,ζ〉|=2d(w,ζ)[d(z,ζ)+d(w,ζ)]≤2(1+1δ)1δd2(z,ζ).
This concludes that there is δ>1 such that
(68)|〈z,w-ζ〉|<12|1-〈z,ζ〉|,∀z,w∈𝔹n,ζ∈𝕊n,
whenever d(z,ζ)>δd(w,ζ). Then, we have
(69)|1-〈z,ζ〉-t〈z,w-ζ〉|>|1-〈z,ζ〉|-t|〈z,ζ-w〉|>12|1-〈z,ζ〉|.
Therefore,
(70)|Kα(z,w)-Kα(z,ζ)|≤2n+3+α(n+1+α)(1+1/δ)d(w,ζ)d(z,ζ)|1-〈z,ζ〉|n+2+α≤Cα,nd(w,ζ)d(z,ζ)2(n+1+α)+1,
and the lemma is proved.Proof of Lemma13.
Whena is the constant function 1, the result is clear. Thus we may suppose that a is a (1,q)α-atom. Let a be supported in a Carleson tuber Qr(ζ) and δr≤2, where δ is the constant in Lemma 14. Since Pα is a bounded operator on Lq(𝔹n,dvα), we have
(71)∫Qδr‍|Pα(a)|dvα(z)≤vα(Qδr)1-(1/q)∥Pα(a)∥q,α≤∥Pα∥Lqvα(Qδr)1-(1/q)∥a∥q,α≤∥Pα∥Lq.
Next, if d(z,ζ)>δr then
(72)|∫𝔹na(w)(1-〈z,w〉)n+1+αdvα(w)|=|∫Qr(ζ)‍a(w)[1(1-〈z,w〉)n+1+α-1(1-〈z,ζ〉)n+1+α]dvα(w)∫Qr(ζ)‍a(w)[1(1-〈z,w〉)n+1+α|≤C∫Qr(ζ)‍|a(w)|d(w,ζ)d(z,ζ)2(n+1+α)+1dvα(w)≤Cr∫Qr(ζ)‍|a(w)|dvα(w)1d(z,ζ)2(n+1+α)+1≤Crd(z,ζ)2(n+1+α)+1.
Then
(73)∫d(z,ζ)>δr|Pα(a)|dvα(z)≤Cr∫d(z,ζ)>δr‍1d(z,ζ)2(n+1+α)+1dvα(z)=Cr∑k≥0‍∫2kδr<d(z,ζ)≤2k+1δr‍1d(z,ζ)2(n+1+α)+1dvα(z)≤Cr∑k≥0‍vα(Q2k+1δr)(2kδr)2(n+1+α)+1≤Cr∑k=0∞‍(2k+1δr)2(n+1+α)(2kδr)2(n+1+α)+1≤C,
where we have used the fact that vα(Qr)≈r2(n+1+α) in the third inequality (cf. [3, Corollary 5.24]). Thus, we get
(74)∫𝔹n‍|Pα(a)|dvα(z)=∫Qδr‍|Pα(a)|dvα(z)+∫d(z,ζ)>δr‍|Pα(h)|dvα(z)≤C,
where C depends only on q,n, and α.Now we turn to the atomic decomposition of𝒜α1(α>-1) with respect to the Carleson tubes. Recall that ∥a∥1,α≤1 for any (1,q)α-atom a. Then, we define 𝒜α1,q as the space of all f∈𝒜α1 which admits a decomposition
(75)f=∑i‍λiPαai,∑i‍|λi|≤Cq∥f∥1,α,
where for each i, ai is an (1,q)α-atom and λi∈ℂ so that ∑i‍|λi|<∞. We equip this space with the norm
(76)∥f∥𝒜α1,q=inf{∑i‍|λi|:f=∑i‍λiPαai},
where the infimum is taken over all decompositions of f described above.It is easy to see that𝒜α1,q is a Banach space. By Lemma 13, we have the contractive inclusion 𝒜α1,q⊂𝒜α1. We will prove in what follows that these two spaces coincide. That establishes the “real-variable” atomic decomposition of the Bergman space 𝒜α1. In fact, we will show the remaining inclusion 𝒜α1⊂𝒜α1,q by duality.Theorem 15.
Let1<q<∞ and α>-1. For every f∈𝒜α1 there exist a sequence {ai} of (1,q)α-atoms and a sequence {λi} of complex numbers such that
(77)f=∑i‍λiPαai,∑i‍|λi|≤Cq∥f∥1,α.
Moreover,
(78)∥f∥1,α≈inf∑i‍|λi|,
where the infimum is taken over all decompositions of f described above and “≈” depends only on α,n, and q.Recall that the dual space of𝒜α1 is the Bloch space ℬ (we refer to [3] for details). The Banach dual of 𝒜α1 can be identified with ℬ (with equivalent norms) under the integral pairing
(79)〈f,g〉α=limr→1-∫𝔹n‍f(rz)g(z)¯dvα(z),f∈𝒜α1,g∈ℬ.
(cf. [3, Theorem 3.17].)In order to prove Theorem15, we need the following result, which can be found in [4] (see also [3, Theorem 5.25]).Lemma 16.
Supposeα>-1 and 1≤p<∞. Then, for any f∈ℋ(𝔹n), f is in ℬ if and only if there exists a constant C>0 depending only on α and p such that
(80)1vα(Qr(ζ))∫Qr(ζ)‍|f-fα,Qr(ζ)|pdvα≤C,
for all r>0 and all ζ∈𝕊n, where
(81)fα,Qr(ζ)=1Qr(ζ)∫Qr(ζ)‍f(z)dvα(z).
Moreover,
(82)∥f∥ℬ≈supr>0,ζ∈𝕊(1vα(Qr(ζ))∫Qr(ζ)‍|f-fα,Qr(ζ)|pdvα)1/p,
where “≈” depends only on α,p, and n.As noted above, we will prove Theorem15 via duality. To this end, we first prove the following duality theorem.Proposition 17.
For any
1
<
q
<
∞
and
α
>
-
1,one has(𝒜α1,q)*=ℬisometrically. More precisely,(i)
everyg∈ℬdefines a continuous linear functionalφgon𝒜α1,qby(83)φg(f)=limr→1-∫𝔹n‍f(rz)g(z)¯dvα(z),∀f∈𝒜α1,q,(ii)
conversely, eachφ∈(𝒜α1,q)*is given as (83) by someg∈ℬ.
Moreover, we have(84)∥φg∥≈|g(0)|+∥g∥ℬ,∀g∈ℬ.Proof.
Letp be the conjugate index of q, that is, 1/p+1/q=1. We first show ℬ⊂(𝒜α1,q)*. Let g∈ℬ. For any (1,q)α-atom a, by Lemma 16 one has
(85)|∫𝔹n‍Pαa(z)g(z)¯dvα(z)|=|〈Pα(aj),g〉α|=|∫𝔹n‍ag¯dvα|=|∫𝔹n‍a(g-gQ)¯dvα|≤(∫Q‍|a|qdvα)1/q(∫Q‍|g-gQ|pdvα)1/p≤(1vα(Q)∫Q‍|g-gQ|pdvα)1/p≤C∥g∥ℬ.
On the other hand, for the constant function (1), we have Pα1=1 and so
(86)|∫𝔹n‍Pα1(z)g(z)¯dvα(z)|=|∫𝔹n‍g(z)dvα(z)|=|g(0)|.
Thus, we deduce that
(87)|∫𝔹n‍fg-dvα|≤C∥f∥𝒜α1,q(|g(0)|+∥g∥ℬ),
for any finite linear combination f of (1,q)α-atoms. Hence, g defines a continuous linear functional φg on a dense subspace of 𝒜α1,q, and φg extends to a continuous linear functional on 𝒜α1,q such that
(88)|φg(f)|≤C(|g(0)|+∥g∥ℬ)∥f∥𝒜α1,q,
for all f∈𝒜α1,q.
Next letφ be a bounded linear functional on 𝒜α1,q. Note that
(89)ℋq(𝔹n,dvα)=ℋ(𝔹n)∩Lq(𝔹n,dvα)⊂𝒜α1,q.
Then, φ is a bounded linear functional on ℋq(𝔹n,dvα). By duality there exists g∈ℋp(𝔹n,dvα) such that
(90)φ(f)=∫𝔹n‍fg-dvα,∀f∈ℋq(𝔹n,dvα).
Let Q=Qr(ζ) be a Carleson tube. For any f∈Lq(𝔹n,dvα) supported in Q, it is easy to check that
(91)af=(f-fQ)χQ[∥f∥Lqvα(Q)1/p]
is a (1,q)-atom. Then, |φ(Pαaf)|≤∥φ∥ and so
(92)|φ(Pα[(f-fQ)χQ])|≤∥φ∥∥f∥Lqvα(Q)1/p.
Hence, for any f∈Lq(𝔹n,dvα), we have
(93)|∫Q‍f(g-gQ)¯dvα|=|∫Q‍(f-fQ)g-dvα|=|∫𝔹n‍(f-fQ)χQg-dvα|=|∫𝔹n‍Pα[(f-fQ)χQ]g-dvα|=|φ(Pα[(f-fQ)χQ])|≤∥φ∥∥(f-fQ)χQ∥Lq(𝔹n,dvα)vα(Q)1/p≤2∥φ∥∥f∥Lq(Q,dvα)vα(Q)1/p.
This concludes that
(94)(1vα(Q)∫Q‍|g-gQ|pdvα)1/p≤2∥φ∥.
By Lemma 16, we have that g∈ℬ and ∥g∥ℬ≤C∥φ∥. Therefore, φ is given as (83) by g with |g(0)|+∥g∥ℬ≤C∥φ∥.Now we are ready to prove Theorem15.Proof of Theorem15.
By Lemma13 we know that 𝒜α1,q⊂𝒜α1. On the other hand, by Proposition 17 we have (𝒜α1)*=(𝒜α1,q)*. Hence, by duality we have ∥f∥1,q≈∥f∥𝒜α1,q.Remark 18.
(1) One would like to expect that when0<p<1,𝒜αp also admits an atomic decomposition in terms of atoms with respect to Carleson tubes. However, the proof of Theorem 15 via duality cannot be extended to the case 0<p<1. At the time of this writing, this problem is entirely open.
(2) The real-variable atomic decomposition of Bergman spaces should be known to specialists in the casep=1. Indeed, based on their theory of harmonic analysis on homogeneous spaces, Coifman and Weiss [7] claimed that the Bergman space 𝒜1 admits an atomic decomposition in terms of atoms with respect to (𝔹n,ϱ,dv), where
(95)ϱ(z,w)={||z|-|w||+|1-1|z||w|〈z,w〉|,ifz,w∈𝔹n∖{0},|z|+|w|,otherwise.
This also applies to 𝒜α1 because (𝔹n,ϱ,dvα) is a homogeneous space for α>-1 (see e.g. [6]). However, the approach of Coifman and Weiss is again based on duality and therefore not constructive and cannot be applied to the case 0<p<1. Recently, the present authors [8] extend this result to the case 0<p<1 through using a constructive method.


## 4. Area Integral Inequalities: Real-Variable Methods

In this section, we will prove the area integral inequality for the Bergman space𝒜α1 via atomic decomposition established in Section 3.Theorem 19.
Suppose1<q<∞,γ>0, and α>-1. Then,
(96)∥A∇~γ,q(f)∥1,α≲∥f∥1,α,∀f∈ℋ(𝔹n).This is the assertion(a)⇒(d) of Theorem 2 in the case p=1. The novelty of the proof here is to involve a real-variable method.The following lemma is elementary.Lemma 20.
Suppose1<q<∞,γ>0 and α>-1. If f∈𝒜αq, then
(97)∫𝔹n‍|A∇~γ,q(f)(z)|qdvα≈∫𝔹n‍|f(z)-f(0)|qdvα,
where “≈” depends only on q,γ,α, and n.Proof.
Note thatvα(D(z,γ))≈(1-|z|2)n+1+α. Then
(98)∫𝔹n‍|A∇~γ,q(f)(z)|qdvα=∫𝔹n‍∫D(z,γ)‍(1-|w|2)-1-n|∇~f(w)|qdv(w)dvα(z)=∫𝔹n‍vα(D(w,γ))(1-|w|2)-1-n|∇~f(w)|qdv(w)≈∫𝔹n‍|∇~f(w)|qdvα(w)≈∫𝔹n‍|f(w)-f(0)|qdvα(w).
In the last step, we have used [3, Theorem 2.16(b)].Proof of (96).
By Theorem15, it suffices to show that for 1<q<∞,γ>0, and α>-1, there exists C>0 such that
(99)∥A∇~γ,q(Pαa)∥1,α≤C,
for all (1,q)α-atoms a. Given an (1,q)α-atom a supported in Q=Qr(ζ), by Lemma 20, we have
(100)∫2Q‍A∇~γ,q(Pαa)dvα≤vα(2Q)1-(1/q)(∫2Q‍[A∇~γ,q(Pαa)]qdvα)1/q≤Cvα(Q)1-(1/q)(∫𝔹n‍|Pαa(z)-Pαa(0)|qdvα)1/q≤Cvα(Q)1-(1/q)∥a∥q,α≤C,
where 2Q=Q2r(ζ). On the other hand,(101)∫(2Q)cA∇~γ,q(Pαa)dvα=∫(2Q)c‍(∫D(z,γ)‍|∇~Pαa(w)|qdτ(w))1/qdvα(z)=∫(2Q)c‍(∫D(z,γ)‍|∫Q‍∇~w[Kα(w,u)-Kα(w,ζ)]a(u)dvα(u)|qdτ(w))1/qdvα(z)≤∥a∥q,α∫(2Q)c‍(∫D(z,γ)‍(∫Q‍|∇~w[Kα(w,u)-Kα(w,ζ)]|q/(q-1)dvα(u))q-1dτ(w))1/qdvα(z)≤∫(2Q)c‍(∫D(z,γ)‍supu∈Q|∇~w[Kα(w,u)-Kα(w,ζ)]|qdτ(w))1/qdvα(z),where (2Q)c=𝔹n∖2Q.
An immediate computation yields that(102)∇w[Kα(w,u)-Kα(w,ζ)]=(n+1+α)[u-(1-〈w,u〉)n+2+α-ζ-(1-〈w,ζ〉)n+2+α]=(n+1+α)u-(1-〈w,ζ〉)n+2+α-ζ-(1-〈w,u〉)n+2+α(1-〈w,u〉)n+2+α(1-〈w,ζ〉)n+2+α,ℛw[Kα(w,u)-Kα(w,ζ)]=(n+1+α)[〈w,u〉(1-〈w,u〉)n+2+α-〈w,ζ〉(1-〈w,ζ〉)n+2+α]=(n+1+α)×〈w,u〉(1-〈w,ζ〉)n+2+α-〈w,ζ〉(1-〈w,u〉)n+2+α(1-〈w,u〉)n+2+α(1-〈w,ζ〉)n+2+α.
Moreover,(103)|∇w[Kα(w,u)-Kα(w,ζ)]|2=(n+1+α)2×{|u|2|1-〈w,ζ〉|2(n+2+α)+|1-〈w,u〉|2(n+2+α)|1-〈w,u〉|2(n+2+α)|1-〈w,ζ〉|2(n+2+α)-(1-〈w,ζ〉)n+2+α(1-〈u,w〉)n+2+α〈ζ,u〉|1-〈w,u〉|2(n+2+α)|1-〈w,ζ〉|2(n+2+α)-(1-〈w,u〉)n+2+α(1-〈ζ,w〉)n+2+α〈u,ζ〉|1-〈w,u〉|2(n+2+α)|1-〈w,ζ〉|2(n+2+α)},|ℛw[Kα(w,u)-Kα(w,ζ)]|2=(n+1+α)2×{|〈w,u〉|2|1-〈w,ζ〉|2(n+2+α)+|〈w,ζ〉|2|1-〈w,u〉|2(n+2+α)|1-〈w,u〉|2(n+2+α)|1-〈w,ζ〉|2(n+2+α)-〈w,u〉〈ζ,w〉(1-〈w,ζ〉)n+2+α(1-〈u,w〉)n+2+α|1-〈w,u〉|2(n+2+α)|1-〈w,ζ〉|2(n+2+α)-〈w,ζ〉〈u,w〉(1-〈w,u〉)n+2+α(1-〈ζ,w〉)n+2+α|1-〈w,u〉|2(n+2+α)|1-〈w,ζ〉|2(n+2+α)}.Then, we have
(104)|∇w[Kα(w,u)-Kα(w,ζ)]|2-|ℛw[Kα(w,u)-Kα(w,ζ)]|2=(n+1+α)2|1-〈w,u〉|2(n+2+α)|1-〈w,ζ〉|2(n+2+α)×{(|u|2-|〈w,u〉|2)|1-〈w,ζ〉|2(n+2+α)+(1-|〈w,ζ〉|2)|1-〈w,u〉|2(n+2+α)+(〈w,u〉〈ζ,w〉-〈ζ,u〉)×(1-〈w,ζ〉)n+2+α(1-〈u,w〉)n+2+α+(〈w,ζ〉〈u,w〉-〈u,ζ〉)×(1-〈w,u〉)n+2+α(1-〈ζ,w〉)n+2+α(|u|2-|〈w,u〉|2)|1-〈w,ζ〉|2(n+2+α)}.
Note that for any f∈ℋ(𝔹n),
(105)|∇~f(z)|2=(1-|z|2)(|∇f(z)|2-|ℛf(z)|2),z∈𝔹n
(cf. [3, Lemma 2.13]). It is concluded that
(106)|∇~w[Kα(w,u)-Kα(w,ζ)]|≤(n+1+α)(1-|w|2)1/2|1-〈w,u〉|n+2+α|1-〈w,ζ〉|n+2+α×{(1-|〈w,u〉|2)|1-〈w,ζ〉|2(n+2+α)+(1-|〈w,ζ〉|2)|1-〈w,u〉|2(n+2+α)+[〈w,u-ζ〉〈ζ,w〉+(|〈w,ζ〉|2-1)+(1-〈ζ,u〉)]×(1-〈w,ζ〉)n+2+α(1-〈u,w〉)n+2+α+[〈w,ζ-u〉〈u,w〉+(|〈w,u〉|2-1)+(1-〈u,ζ〉)]×(1-〈w,u〉)n+2+α(1-〈ζ,w〉)n+2+α(1-|〈w,u〉|2)|1-〈w,ζ〉|2(n+2+α)}1/2≤(n+1+α)(1-|w|2)1/2(M1+M2+M3+M4)1/2|1-〈w,u〉|n+2+α|1-〈w,ζ〉|n+2+α,
where
(107)M1=|1-〈w,ζ〉|n+2+α|1-〈u,w〉|n+2+α×|〈w,u-ζ〉〈ζ,w〉+(1-〈ζ,u〉)|,M2=|1-〈w,u〉|n+2+α|1-〈ζ,w〉|n+2+α×|〈w,ζ-u〉〈u,w〉+(1-〈u,ζ〉)|,M3=(1-|〈w,u〉|2)|1-〈ζ,w〉|n+2+α×|(1-〈w,ζ〉)n+2+α-(1-〈w,u〉)n+2+α|,M4=(1-|〈w,ζ〉|2)|1-〈u,w〉|n+2+α×|(1-〈w,u〉)n+2+α-(1-〈w,ζ〉)n+2+α|,
for w∈D(z,γ),u∈Qr(ζ), and z∈𝔹n,ζ∈𝕊n.
Hence,(108)∫(2Q)c‍A∇~γ,q(Pαa)dvα≤∫(2Q)c‍(∫D(z,γ)‍supu∈Q|∇~w[Kα(w,u)-Kα(w,ζ)]|qdτ(w))1/qdvα(z)≤(n+1+α)∫(2Q)c‍(I1+I2+I3+I4)dvα(z),where(109)I1=(∫D(z,γ)‍supu∈Q(1-|w|2)q/2M1q/2|1-〈w,u〉|q(n+2+α)|1-〈w,ζ〉|q(n+2+α)dτ(w))1/q,I2=(∫D(z,γ)‍supu∈Q(1-|w|2)q/2M2q/2|1-〈w,u〉|q(n+2+α)|1-〈w,ζ〉|q(n+2+α)dτ(w))1/q,I3=(∫D(z,γ)‍supu∈Q(1-|w|2)q/2M3q/2|1-〈w,u〉|q(n+2+α)|1-〈w,ζ〉|q(n+2+α)dτ(w))1/q,I4=(∫D(z,γ)‍supu∈Q(1-|w|2)q/2M4q/2|1-〈w,u〉|q(n+2+α)|1-〈w,ζ〉|q(n+2+α)dτ(w))1/q.
We first estimateI1. Note that
(110)M1≤(|〈w,u-ζ〉|+|1-〈ζ,u〉|)×|1-〈w,ζ〉|n+2+α|1-〈w,u〉|n+2+α≤(2|1-〈u,ζ〉|1/2(|1-〈w,ζ〉|1/2+|1-〈u,ζ〉|1/2)+|1-〈ζ,u〉|(|1-〈w,ζ〉|1/2+|1-〈u,ζ〉|1/2))×|1-〈w,ζ〉|n+2+α|1-〈w,u〉|n+2+α≤(2|1-〈u,ζ〉|1/2(Cγ|1-〈z,ζ〉|1/2+12|1-〈z,ζ〉|1/2)+|1-〈ζ,u〉|12)×|1-〈w,ζ〉|n+2+α|1-〈w,u〉|n+2+α≤(Cγr|1-〈z,ζ〉|1/2+r2)×|1-〈w,ζ〉|n+2+α|1-〈w,u〉|n+2+α,
where the second inequality is the consequence of the following fact which has appeared in the proof of Lemma 14:
(111)|〈w,u-ζ〉|≤2|1-〈u,ζ〉|1/2(|1-〈w,ζ〉|1/2+|1-〈u,ζ〉|1/2);
the third inequality is obtained by Lemma 7, and the fact
(112)|1-〈u,ζ〉|1/2<r<12|1-〈z,ζ〉|1/2,
for u∈Q and z∈(2Q)c. Since
(113)|1-〈z,u〉|1/2≥|1-〈z,ζ〉|1/2-|1-〈u,ζ〉|1/2≥|1-〈z,ζ〉|1/2-12|1-〈z,ζ〉|1/2≥12|1-〈z,ζ〉|1/2,
by Lemmas 5 and 7, one has(114)I1≤(∫D(z,γ)‍supu∈Q(1-|w|2)q/2[Cr|1-〈z,ζ〉|1/2+r2]q/2|1-〈w,u〉|(q/2)(n+2+α)|1-〈w,ζ〉|(q/2)(n+2+α)dτ(w))1/q≤Cγ(∫D(z,γ)‍supu∈Q(1-|z|2)q/2(Cr|1-〈z,ζ〉|1/2+r2)q/2|1-〈z,u〉|(q/2)(n+2+α)|1-〈z,ζ〉|(q/2)(n+2+α)dτ(w))1/q≤Cγ((1-|z|2)q/2(r|1-〈z,ζ〉|1/2+r2)q/2|1-〈z,ζ〉|q(n+2+α))1/q≤Cγ(r|1-〈z,ζ〉|1/2+r2)1/2|1-〈z,ζ〉|n+(3/2)+α≤Cγ(r1/2d(z,ζ)2(n+1+α)+(1/2)+rd(z,ζ)2(n+1+α)+1).Hence,
(115)∫(2Q)c‍I1dvα(z)≲∫(2Q)c‍r1/2d(z,ζ)2(n+1+α)+(1/2)dvα(z)+∫(2Q)c‍rd(z,ζ)2(n+1+α)+1dvα(z).
The second term on the right hand side has been estimated in the proof of Lemma 13. The first term can be estimated as follows:
(116)∫d(z,ζ)>2rr1/2d(z,ζ)2(n+1+α)+(1/2)dvα(z)=r1/2∑k≥0‍∫2kr<d(z,ζ)≤2k+1r‍1d(z,ζ)2(n+1+α)+(1/2)dvα(z)≤r1/2∑k≥0‍vα(Q2k+1r)(2kr)2(n+1+α)+(1/2)≤Cr1/2∑k=0∞‍(2k+1r)2(n+1+α)(2kr)2(n+1+α)+(1/2)≤C,
where we have used the fact that vα(Qr)≈r2(n+1+α) in the third inequality (cf. [3, Corollary 5.24]).
By the same argument we can estimateI2 and omit the details.
Next, we estimateI3. Note that
(117)M3≤(1-|〈w,u〉|2)|1-〈w,ζ〉|n+2+α×|(1-〈w,ζ〉)n+2+α-(1-〈w,u〉)n+2+α|≤2|1-〈w,u〉||1-〈w,ζ〉|n+2+α×|∫01‍ddt(1-〈w,tζ+(1-t)u〉)n+2+αdt|=2(n+2+α)|1-〈w,u〉||1-〈w,ζ〉|n+2+α×|〈w,ζ-u〉∫01‍(1-〈w,tζ+(1-t)u〉)n+1+αdt|≤Cγ|1-〈w,u〉||1-〈w,ζ〉|n+2+αr|1-〈z,ζ〉|n+3/2+α,
where the last inequality is achieved by the following estimates:
(118)|1-〈w,tζ+(1-t)u〉|≤Cγ|1-〈z,tζ+(1-t)u〉|≤Cγ|1-〈z,u〉|+|〈z,ζ-u〉|≤Cγ|1-〈z,ζ〉|,|〈w,ζ-u〉|≤Cγr|1-〈z,ζ〉|1/2,
for any w∈D(z,γ) and u∈Qr(ζ). Thus, by Lemmas 5 and 7(119)I3≤Cγ(∫D(z,γ)‍supu∈Q(1-|w|2)q/2rq/2|1-〈z,ζ〉|(q/2)(n+(3/2)+α)|1-〈w,u〉|q(n+1+α)+(q/2)|1-〈w,ζ〉|(q/2)(n+2+α)dτ(w))1/q≤Cγ(∫D(z,γ)‍supu∈Q(1-|z|2)q/2rq/2|1-〈z,ζ〉|(q/2)(n+(3/2)+α)|1-〈z,u〉|q(n+1+α)+(q/2)|1-〈z,ζ〉|(q/2)(n+2+α)dτ(w))1/q≤Cγ((1-|z|2)q/2rq/2|1-〈z,ζ〉|q(n+1+α)+(3/4)q)1/q≤Cγr1/2d(z,ζ)2(n+1+α)+(1/2).Hence,
(120)∫(2Q)c‍I3dvα(z)≤Cγ∫(2Q)c‍r1/2d(z,ζ)2(n+1+α)+(1/2)dvα(z)≤Cγ,
as shown above.
Similarly, we can estimateI4 and omit the details. Therefore, combining the above estimates we conclude that
(121)∫(2Q)c‍Aγ(∇~Pαa)dvα≤C,
where C depends only on q,γ,n, and α.Remark 21.
We remark that whenever𝒜αp has an atomic decomposition in terms of atoms with respect to Carleson tubes for 0<p<1, the argument of Theorem 2 works as well in this case. However, as noted in Remark 18(1), the problem of the atomic decomposition of 𝒜αp with respect to Carleson tubes in 0<p<1 is entirely open.Remark 22.
The area integral inequality in case1<p<∞ can be also proved through using the method of vector-valued Calderón-Zygmund operators for Bergman spaces. This has been done in [9].



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*Source: 167514-2013-08-20.xml*