# Weighted Differentiation Composition Operators to Bloch-Type Spaces

Authors: Junming Liu; Zengjian Lou; Ajay K. Sharma
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We characterized the boundedness and compactness of weighted differentiation composition operators from BMOA and the Bloch space to Bloch-type spaces. Moreover, we obtain new characterizations of boundedness and compactness of weighted differentiation composition operators.


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## Body

## 1. Introduction

Let𝔻 be the open unit disk in the complex plane ℂ, H(𝔻) the space of all functions holomorphic on 𝔻, dA(z)=(1/π)dxdy the normalized area measure on 𝔻, and H∞ the space of all bounded holomorphic functions with the norm ∥f∥∞=supz∈𝔻|f(z)|.Letα>0. The α-Bloch spaceℬα on 𝔻 is the space of all holomorphic functions f on 𝔻 such that
(1)supz∈𝔻(1-|z|2)α|f′(z)|<∞.
The littleα-Bloch spaceℬ0α consists of all f∈ℬα such that
(2)lim|z|→1(1-|z|2)α|f′(z)|=0.
Both spaces ℬα and ℬ0α are Banach spaces with the norm
(3)∥f∥ℬα=|f(0)|+supz∈𝔻(1-|z|2)α|f′(z)|,
and ℬ0α is a closed subspace of ℬα. If α=1, they become the classical Bloch space ℬ and little Bloch space ℬ0, respectively. For any α>0, the space 𝒜∞α consists of functions f∈H(𝔻) such that
(4)∥f∥𝒜∞α=supz∈𝔻(1-|z|2)α|f(z)|<∞.
For information of such spaces, see, for example, [1–4].Fora∈𝔻, let σa(z)=(a-z)/(1-a-z) be the automorphism of 𝔻 that interchanges 0 and a. Let the Green function in 𝔻 with logarithmic singularity at a be given by
(5)g(z,a)=log|1-a-za-z|=log1|σa(z)|.
The space BMOA consists of all f in the Hardy space H2 such that
(6)supa∈𝔻∥f∘σa-f(a)∥H2<∞.BMOA is a Banach space under following norm (see, e.g., [5]):
(7)∥f∥BMOA=|f(0)|+supa∈𝔻∥f∘σa-f(a)∥H2.Letφ and ψ be holomorphic maps on the open unit disk 𝔻 such that φ(𝔻)⊂𝔻. For a nonnegative integer n, we define a linear operator Dφ,ψn as follows:
(8)Dφ,ψnf=ψ·(f(n)∘φ),f∈H(𝔻).
We call it weighted differentiation composition operators, which was defined in [6, 7]. If n=0 and ψ≡1, Dφ,ψn becomes Cφ induced by φ, defined as Cφf=f∘φ, f∈H(𝔻). If ψ=1 and φ(z)=z, then Dφ,ψn is the differentiation operator defined as Dnf=f(n). If n=0, then we get the weighted composition operator ψCφ defined as ψCφf=ψ·(f∘φ). If n=1 and ψ(z)=φ′(z), then Dφ,ψn reduces to DCφ. When ψ≡1, then Dφ,ψn reduces to differentiation composition operator CφDn (also named as product of differentiation and composition operator). If we put φ(z)=z, then Dφ,ψn=MψDn, the product of multiplication and differentiation operator.The boundedness and compactness of differentiation composition operator between spaces of holomorphic functions have been studied extensively. For example, Hibschweiler; Portnoy and Ohno studied differentiation composition operatorCφD on Hardy and Bergman spaces in [8, 9]; Li; Stević and Ohno studied CφD on Bloch type spaces in [10–12]; Wu and Wulan gave a new compactness criterion of CφDm on the Bloch space in [13]. Recently, the weighted differentiation composition operator between different function spaces has also been investigated by several authors (see, for example, [14–21]).Boundedness, compactness, and essential norm of weighted composition operatorψCφ between Bloch-type spaces have been studied in [22–24]. Recently, Manhas and Zhao [25] and Hyvärinen and Lindström [26] gave a new characterization of boundedness and compactness of ψCφ in terms of the norm of φn (for the compactness of composition operator, see [27, 28]).Motivated by [13, 25, 26], we study the operator Dφ,ψn(n≥1) from BMOA and Bloch space to Bloch-type spaces.Throughout this paper, constants are denoted byC; they are positive and not necessarily the same at each occurrence. The notation A≲B means that there is a positive constant C such that A≤CB. When A≲B and B≲A, we write A≈B.


## 2. Some Lemmas

It is well known thatH∞⊂BMOA⊂ℬ. From the definition of the norm, we know
(9)∥f∥BMOA≲∥f∥∞,f∈H∞.
Indeed, Girela proved that
(10)∥f∥ℬ≤∥f∥BMOA1
in Corollary 5.2 of [5]. The following lemma is from Lemma 5 in [29] (see also Lemma 4.12 of [4]).Lemma 1.
Iff∈H(𝔻), then
(11)|f(0)|2≤2∫𝔻‍|f(z)|2log1|z|dA(z).The following lemma may be known, but we fail to find its reference; so we give a proof for the completeness of the paper.Lemma 2.
Letf∈H(𝔻). Then,
(12)∥f∥ℬ≤∥f∥BMOA.Proof.
Applying Littlewood-Paley identity(13)∥f∥H22=|f(0)|2+2∫𝔻‍|f′(z)|2log1|z|dA(z)
and Lemma 1, we have
(14)supa∈𝔻∥f∘σa-f(a)∥H2=supa∈𝔻(2∫𝔻‍|f′(σa(z))σa′(z)|2log1|z|dA(z))1/2≥supa∈𝔻(1-|a|2)|f′(a)|.
It follows from the definitions of Bloch space and BMOA space that
(15)∥f∥ℬ≤∥f∥BMOA.By Theorem 6.2 of [5] and the proof of Theorem 1 of [30], we have the following lemma.Lemma 3.
Letn be a fixed positive integer and f∈ℬ with f(0)=f′(0)=···=f(n-1)(0)=0. If
(16)supa∈𝔻∫𝔻‍|f(n)(z)|2(1-|z|2)2n-2(1-|σa(z)|2)dA(z)≲1,
then ∥f∥BMOA≲1.Lemma 4.
Suppose thatn is a fixed positive integer. Let k∈ℕ+, 0≤x≤1, and
(17)Hkn(x)={k(k-1)···(k-n+1)(1-x)nxk-nifk>nn!(1-x)nifk=n.
If k≥n, then there are two positive constants cn and Cn, depending only on n, such that
(18)cn≤Hkn(x)≤Cn,fork-nk≤x≤k-n+1k+1.Proof.
The proof is similar to that of Lemma 2.2 of [13] and is so omitted.


## 3. Boundedness ofDφ,ψn

In this section, we characterize the boundedness ofDφ,ψn from BMOA and the Bloch space to Bloch-type spaces.Theorem 5.
Letα>0, ψ∈H(𝔻), n∈ℕ+, and φ a holomorphic self-map of 𝔻. Then, the following statements are equivalent: (a)
Dφ,ψn:BMOA→ℬα is bounded.(b)
Dφ,ψ′n:BMOA→𝒜∞α and Dφ,ψφ′n+1:BMOA→𝒜∞α are bounded.(c)
Dφ,ψn:ℬ0→ℬα is bounded.(d)
Dφ,ψ′n:ℬ0→𝒜∞α and Dφ,ψφ′n+1:ℬ0→𝒜∞α are bounded.(e)
Dφ,ψn:ℬ→ℬα is bounded.(f)
Dφ,ψ′n:ℬ→𝒜∞α and Dφ,ψφ′n+1:ℬ→𝒜∞α are bounded.(g)
supz∈𝔻((1-|z|2)α/(1-|φ(z)|2)n)|ψ′(z)|<∞  andsupz∈𝔻((1-|z|2)α/(1-|φ(z)|2)n+1)|ψ(z)φ′(z)|<∞.(h)
supk∈ℕ∥Dφ,ψ′n(zk)∥𝒜∞α<∞ and supk∈ℕ∥Dφ,ψφ′n+1(zk)∥𝒜∞α<∞.Proof.
It is obvious that(f)⇒(b), (f)⇒(d), (e)⇒(c), and (e)⇒(a). Thus, we will prove the theorem according to the following steps. (I): (a)⇒(g), (c)⇒(g). (II): (b)⇒(g), (d)⇒(g). (III): (g)⇒(e), (g)⇒(f). (IV): (f)⇔(h).
(I):(a)⇒(g), (c)⇒(g). Suppose that (a) or (c) holds. We choose the test function g1(z)=zn. By Lemma 2, we get
(19)∥g1∥ℬ≤∥g1∥BMOA≲∥g1∥∞=1.
So
(20)supz∈𝔻(1-|z|2)α|ψ′(z)|≤∥Dφ,ψng1∥ℬα<∞.
Taking g2(z)=zn+1 and using the fact that |φ(z)|<1, we have
(21)supz∈𝔻(1-|z|2)α|ψ(z)φ′(z)|≤∥Dφ,ψng2∥ℬα+∥Dφ,ψng1∥ℬα<∞.
We now consider the function
(22)fλ(z)=(n+1)1-|φ(λ)|21-φ(λ)-z-(1-|φ(λ)|2)2(1-φ(λ)-z)2,λ∈𝔻.
It is easy to check that fλ∈ℬ0∩BMOA and ∥fλ∥BMOA≲∥fλ∥∞≲1. Moreover,
(23)fλ(n)(z)=(n+1)!(φ(λ)-)n×[1-|φ(λ)|2(1-φ(λ)-z)n+1-(1-|φ(λ)|2)2(1-φ(λ)-z)n+2].
Thus, fλ(n)(φ(λ))=0 and
(24)fλ(n+1)(φ(λ))=-(n+1)!(φ(λ)-)n+1(1-|φ(λ)|2)n+1.
We obtain
(25)∥Dφ,ψn∥≳∥Dφ,ψnfλ∥ℬα≳(1-|λ|2)α|ψ′(λ)fλ(n)(φ(λ))+ψ(λ)φ′(λ)fλ(n+1)(φ(λ))|≳(n+1)!(1-|λ|2)α(1-|φ(λ)|2)n+1|φ(λ)|n+1|ψ(λ)φ′(λ)|.
Thus, for any r0∈(0,1), we have
(26)supr0<|φ(λ)|<1(1-|λ|2)α(1-|φ(λ)|2)n+1|ψ(λ)φ′(λ)|<∞.
Using (21) yields
(27)sup|φ(λ)|≤r0(1-|λ|2)α(1-|φ(λ)|2)n+1|ψ(λ)φ′(λ)|≲1(1-r02)n+1supλ∈𝔻(1-|λ|2)α|ψ(λ)φ′(λ)|<∞.
Combining (26) with (27), we get
(28)supλ∈𝔻(1-|λ|2)α(1-|φ(λ)|2)n+1|ψ(λ)φ′(λ)|<∞.
We next consider the function
(29)gλ(z)=(n+2)1-|φ(λ)|21-φ(λ)-z-(1-|φ(λ)|2)2(1-φ(λ)-z)2,λ∈𝔻.
Similarly, we get gλ∈ℬ0∩BMOA and
(30)∥gλ∥BMOA≲∥gλ∥∞≲1.
Moreover,
(31)gλ(n)(z)=n!(φ(λ)-)n[(n+2)1-|φ(λ)|2(1-φ(λ)-z)n+1-(n+1)(1-|φ(λ)|2)2(1-φ(λ)-z)n+2].
So
(32)gλ(n)(φ(λ))=n!(φ(λ)-)n(1-|φ(λ)|2)n
and gλ(n+1)(φ(λ))=0. We have, as above,
(33)∥Dφ,ψn∥≳∥Dφ,ψngλ∥ℬα≳n!(1-|λ|2)α(1-|φ(λ)|2)n|φ(λ)|n|ψ′(λ)|.
Thus, for anys0∈(0,1),
(34)sups0<|φ(λ)|<1(1-|λ|2)α(1-|φ(λ)|2)n|ψ′(λ)|<∞.
Applying (20), we get
(35)sup|φ(λ)|≤s0(1-|λ|2)α(1-|φ(λ)|2)n|ψ′(λ)|<∞.
Combining (34) with (35) yields
(36)supλ∈𝔻(1-|λ|2)α(1-|φ(λ)|2)n|ψ′(λ)|<∞.
(II): (b)⇒(g) and (d)⇒(g). Suppose that Dφ,ψ′n:BMOA→𝒜∞α is bounded or Dφ,ψ′n:ℬ0→𝒜∞α is bounded. Set
(37)λ=supz∈𝔻(1-|z|2)α(1-|φ(z)|2)n|ψ′(z)|.
If λ=∞, then for any positive integer N, we can find b∈𝔻 such that
(38)(1-|b|2)α(1-|φ(b)|2)n|ψ′(b)|>N.
If φ(b)=0, then choose the test function g(z)=zn. It is clear that g∈ℬ0. From Lemma 2, we have
(39)∥g∥ℬ≤∥g∥BMOA≲∥g∥∞=1.
So(40)∥Dφ,ψ′n∥≳∥Dφ,ψ′ng∥𝒜∞α>(1-|b|2)α|ψ′(b)|>N.
If φ(b)≠0, consider the function
(41)g(z)=1a-n(1-|a|2)n(1-a-z)n≜∑j=0∞‍cjzj,
where a=φ(b). Let F(z)=∑j=n∞‍cjzj. Then, F(0)=F′(0)=···=F(n-1)(0)=0 and
(42)F(n)(z)=(1-|a|2(1-a-z)2)n.
It is easy to see that
(43)(1-|z|2)n|F(n)(z)|=(1-|σa(z)|2)n≤1.
So, by Theorems 5.4 and 5.13 of [4], we have F∈ℬ0 and ∥F∥ℬ≲1. By Lemma 1 of [31] and Lemma 3, we get ∥F∥BMOA≲1. We have
(44)∥Dφ,ψ′n∥≳∥Dφ,ψ′nF∥𝒜∞α>(1-|b|2)α(1-|φ(b)|2)n|ψ′(b)|>N.
Since N is arbitrary, we get ∥Dφ,ψ′n∥=∞. This contradicts the boundedness of Dφ,ψ′n:BMOA→𝒜∞α and that of Dφ,ψ′n:ℬ0→𝒜∞α.
Now, suppose thatDφ,ψφ′n+1:BMOA→𝒜∞α is bounded or Dφ,ψφ′n+1:ℬ0→𝒜∞α is bounded. Set
(45)η=supz∈𝔻(1-|z|2)α(1-|φ(z)|2)n+1|ψ(z)φ′(z)|.
If η=∞, then for any positive integer M, exists u∈𝔻 such that
(46)(1-|u|2)α(1-|φ(u)|2)n+1|ψ(u)φ′(u)|>M.
If φ(u)=0, then set g(z)=zn+1. The process as above gives
(47)∥Dφ,ψφ′n+1∥≳∥Dφ,ψφ′n+1g∥𝒜∞α>M.
If φ(u)≠0, consider the function
(48)g(z)=1a-n+1(1-|a|2)n+1(1-a-z)n+1≜∑j=0∞‍cjzj,
where a=φ(u). Let F(z)=∑j=n+1∞‍cjzj. Then, F(0)=F′(0)=···=F(n)(0)=0 and
(49)F(n+1)(z)=(1-|a|2(1-a-z)2)n+1,(1-|z|2)n+1|F(n+1)(z)|=(1-|σa(z)|2)n+1≤1.
Applying Theorems 5.4 and 5.13 of [4] again yields F∈ℬ0 and ∥F∥ℬ≲1. We get ∥F∥BMOA≲1 and
(50)∥Dφ,ψφ′n+1∥≳∥Dφ,ψφ′n+1F∥𝒜∞α∥Dφ,ψφ′n+1∥>(1-|u|2)α(1-|φ(u)|2)n+1|ψ(u)φ′(u)|>M.
Since M is arbitrary, we have ∥Dφ,ψφ′n+1∥=∞. This contradicts the boundedness of Dφ,ψφ′n+1.
(III):(g)⇒(e), (g)⇒(f). Note that
(51)∥Dφ,ψnf∥ℬα=supz∈𝔻(1-|z|2)α×|ψ(z)φ′(z)f(n+1)(φ(z))+ψ′(z)f(n)(φ(z))|≲∥f∥ℬ[supz∈𝔻(1-|z|2)α(1-|φ(z)|2)n+1|ψ(z)φ′(z)|+supz∈𝔻(1-|z|2)α(1-|φ(z)|2)n|ψ′(z)|],∥Dφ,ψφ′n+1f∥𝒜∞α=supz∈𝔻(1-|z|2)α×|ψ(z)φ′(z)f(n+1)(φ(z))|≲∥f∥ℬ[supz∈𝔻(1-|z|2)α(1-|φ(z)|2)n+1|ψ(z)φ′(z)|],∥Dφ,ψ′nf∥𝒜∞α=supz∈𝔻(1-|z|2)α|ψ′(z)f(n)(φ(z))|≲∥f∥ℬ[supz∈𝔻(1-|z|2)α(1-|φ(z)|2)n|ψ′(z)|].
The desired results follow.
(IV):(f)⇔(h). Suppose that (f) is true. It follows from Proposition 5.1 of [4] that ∥zk∥ℬ≤∥zk∥∞=1(k∈ℕ). So,
(52)supk∈ℕ∥Dφ,ψ′n(zk)∥𝒜∞α≤∥Dφ,ψ′n∥<∞,supk∈ℕ∥Dφ,ψφ′n+1(zk)∥𝒜∞α≤∥Dφ,ψφ′n+1∥<∞.
Conversely, assume that(h) is true. It is easy to see that
(53)supz∈𝔻(1-|z|2)α|ψ′(z)|≤∥Dφ,ψ′n(zn)∥𝒜∞α≤supk∈ℕ∥Dφ,ψ′n(zk)∥𝒜∞α<∞,supz∈𝔻(1-|z|2)α|ψ(z)φ′(z)|≤∥Dφ,ψφ′n+1(zn+1)∥𝒜∞α≤supk∈ℕ∥Dφ,ψφ′n+1(zk)∥𝒜∞α<∞.
If ∥φ∥∞<1, then
(54)supz∈𝔻(1-|z|2)α(1-|φ(z)|2)n|ψ′(z)|<1(1-∥φ∥∞)nsupz∈𝔻(1-|φ(z)|2)α|ψ′(z)|<∞,supz∈𝔻(1-|z|2)α|ψ(z)φ′(z)|(1-|φ(z)|2)n+1<1(1-∥φ∥∞)n+1supz∈𝔻(1-|φ(z)|2)α|ψ(z)φ′(z)|<∞.
Hence, (g) is true. From (g)⇒(f), we obtain that (f) is also true.
From now on, we assume that∥φ∥∞=1. For any integer k≥n, let
(55)Δkn={z∈𝔻:k-nk≤|φ(z)|≤k-n+1k+1}.
Let m with m≥n be the smallest positive integer such that Δmn≠⌀. Since Δkn is not empty for every integer k≥m and 𝔻=∪k=m∞Δkn. By Lemma 4, for f∈ℬ,
(56)∥Dφ,ψ′nf∥𝒜∞α=supz∈𝔻(1-|z|2)α|f(n)(φ(z))||ψ′(z)|=supk≥msupz∈Δkn(1-|z|2)α|f(n)(φ(z))ψ′(z)|=supk≥msupz∈Δkn(1-|φ(z)|)n|f(n)(φ(z))|×(1-|z|2)α|ψ′(z)|(Hkn(|φ(z)|)/(1-|φ(z)|)n)Hkn(|φ(z)|)≲1cn∥f∥ℬsupk∈ℕ∥Dφ,ψ′n(zk)∥𝒜∞α.
So, Dφ,ψ′n:ℬ→𝒜∞α is bounded. Similar argument implies
(57)∥Dφ,ψφ′n+1f∥𝒜∞α=supk≥m+1supz∈Δkn+1(1-|z|2)α|f(n+1)(φ(z))|×|ψ(z)φ′(z)|≲1cn+1∥f∥ℬsupk∈ℕ∥Dφ,ψφ′n+1(zk)∥𝒜∞α.
Thus, Dφ,ψφ′n+1:ℬ→𝒜∞α is bounded. Theorem 5 is proved.


## 4. Compactness ofDφ,ψn

The following criterion for the compactness is a useful tool and it follows from standard arguments, for example, Proposition 3.11 of [32] or Lemma 2.10 of [33].Lemma 6.
Letα>0, n∈ℕ+, and X=ℬ0,ℬ, or BMOA. Suppose that ψ and φ are in H(𝔻) such that φ(𝔻)⊂𝔻. Then, Dφ,ψn:X→ℬα is compact if and only if for any sequence {fm} in X with supm∥fm∥X<∞, which converges to zero locally uniformly on 𝔻; we have limm→∞∥Dφ,ψnfm∥ℬα=0.We now give the compactness ofDφ,ψn from BMOA and the Bloch space to Bloch-type spaces.Theorem 7.
Letα>0, ψ∈H(𝔻), n∈ℕ+, and φ a holomorphic self-map of 𝔻. Then, the following statements are equivalent: (a)
Dφ,ψn:BMOA→ℬα is compact.(b)
Dφ,ψ′n:BMOA→𝒜∞α is compact and Dφ,ψφ′n+1:BMOA→𝒜∞α is compact.(c)
Dφ,ψn:ℬ0→ℬα is compact.(d)
Dφ,ψ′n:ℬ0→𝒜∞α is compact and Dφ,ψφ′n+1:ℬ0→𝒜∞α is compact.(e)
Dφ,ψn:ℬ→ℬα is compact.(f)
Dφ,ψ′n:ℬ→𝒜∞α is compact and Dφ,ψφ′n+1:ℬ→𝒜∞α is compact.(g)
ψ∈ℬα, ψφ′∈𝒜∞α,
(58)lim|φ(z)|→1(1-|z|2)α(1-|φ(z)|2)n|ψ′(z)|=0,lim|φ(z)|→1(1-|z|2)α(1-|φ(z)|2)n+1|ψ(z)φ′(z)|=0.(h)
limsupk→∞∥Dφ,ψ′n(zk)∥𝒜∞α=0 and limsupk→∞∥Dφ,ψφ′n+1(zk)∥𝒜∞α=0.Proof.
The proof is a modification of that of Theorem5; so we give a sketch of the proof. We will prove the theorem according to the following steps. (I): (a)⇒(g), (c)⇒(g). (II): (b)⇒(g), (d)⇒(g). (III): (g)⇒(e), (g)⇒(f). (IV): (f)⇔(h).
(I):(a)⇒(g), (c)⇒(g). Suppose that (a) or (c) holds. Then by Theorem 5, we have
(59)supz∈𝔻(1-|z|2)α|ψ′(z)|≤supz∈𝔻(1-|z|2)α|ψ′(z)|(1-|φ(z)|2)n<∞,supz∈𝔻(1-|z|2)α|ψ(z)φ′(z)|≤supz∈𝔻(1-|z|2)α|ψ(z)φ′(z)|(1-|φ(z)|2)n+1<∞.
That is, ψ∈ℬα, ψφ′∈𝒜∞α.
Let{zj} be a sequence in 𝔻 such that |φ(zj)|→1 as j→∞. Now, we consider the function
(60)fj(z)=(n+1)1-|φ(zj)|21-φ(zj)-z-(1-|φ(zj)|2)2(1-φ(zj)-z)2.
Simple computation shows that fj∈ℬ0∩BMOA and
(61)∥fj∥BMOA≲∥fj∥∞≲1.
It is also easy to check that fj→0 uniformly on compact subsets of 𝔻 as j→∞. Moreover,
(62)fj(n)(z)=(n+1)!(φ(zj)-)n×[1-|φ(zj)|2(1-φ(zj)-z)n+1-(1-|φ(zj)|2)2(1-φ(zj)-z)n+2].
We have
(63)∥Dφ,ψnfj∥ℬα≳(n+1)!(1-|zj|2)α(1-|φ(zj)|2)n+1|φ(zj)|n+1|ψ(zj)φ′(zj)|.
By Lemma 6, we get
(64)lim|φ(zj)|→1(1-|zj|2)α(1-|φ(zj)|2)n+1|ψ(zj)φ′(zj)|=0.
We next consider the function
(65)gj(z)=(n+2)1-|φ(zj)|21-φ(zj)-z-(1-|φ(zj)|2)2(1-φ(zj)-z)2.
Similarly, we get gj∈ℬ0∩BMOA and
(66)∥gj∥BMOA≲∥gj∥∞≲1.
It is easy to see that gj converges to zero uniformly on compact subsets of 𝔻 as j→∞ and
(67)gj(n)(z)=n!(φ(zj)-)n×[(n+2)1-|φ(zj)|2(1-φ(zj)-z)n+1-(n+1)(1-|φ(zj)|2)2(1-φ(zj)-z)n+2].
Thus,
(68)∥Dφ,ψngj∥ℬα≳n!(1-|zj|2)α(1-|φ(zj)|2)n|φ(zj)|n|ψ′(zj)|.
Applying Lemma 6 again, we have
(69)lim|φ(zj)|→1(1-|zj|2)α(1-|φ(zj)|2)n|ψ′(zj)|=0.
Since zj∈𝔻 is arbitrary, we proved that (g) is true.
(II)(b)⇒(g), (d)⇒(g). Suppose that (b) or (d) holds. A similar argument to (I) shows that ψ∈ℬα, ψφ′∈𝒜∞α. Now, suppose that the equations in (g) are not true. Then, there exists a sequence {zj} in 𝔻 and δ>0 such that |φ(zj)|→1 as j→∞ and
(70)(1-|zj|2)α(1-|φ(zj)|2)n|ψ′(zj)|>δ,(1-|zj|2)α(1-|φ(zj)|2)n+1|ψ(zj)φ′(zj)|>δ.
Choose a subsequence of {zj} if necessary and suppose that infj|φ(zj)|>1/2. Let
(71)fj(z)=1-|φ(zj)|21-φ(zj)-z,z∈𝔻.
Then, it is easy to check that fj∈ℬ0∩BMOA, fj→0, uniformly on compact subsets of 𝔻 and
(72)fj(n)(z)=n!1-|φ(zj)|2(1-φ(zj)-z)n+1(φ(zj)-)n.
Thus,
(73)∥Dφ,ψ′nfj∥𝒜∞α≥n!(1-|zj|2)α(1-|φ(zj)|2)n|φ(zj)|n|ψ′(zj)|>n!δ2n,∥Dφ,ψφ′n+1fj∥𝒜∞α≥(n+1)!(1-|zj|2)α(1-|φ(zj)|2)n+1×|φ(zj)|n+1|ψ(zj)φ′(zj)|>(n+1)!δ2n+1.
Those contradict the compactness of Dφ,ψ′n and Dφ,ψφ′n+1.
(III)(g)⇒(e), (g)⇒(f). Let {fm} be a norm bounded sequence in ℬ that converges to zero uniformly on compact subsets of 𝔻. Let M=supm∥fm∥ℬ<∞. For ɛ>0, then there exists r0∈(0,1) such that for |φ(z)|>r0, we have
(74)(1-|z|2)α(1-|φ(z)|2)n|ψ′(z)|<ɛ,(1-|z|2)α(1-|φ(z)|2)n+1|ψ(z)φ′(z)|<ɛ.
Thus, for z∈𝔻, we have
(75)∥Dφ,ψnfm∥ℬα≲|ψ(0)fm(n)(φ(0))|+sup|φ(z)|≤r0(1-|z|2)α|fm(n)(φ(z))||ψ′(z)|+sup|φ(z)|>r0(1-|z|2)α|ψ′(z)|∥fm∥ℬ(1-|φ(z)|2)n+sup|φ(z)|≤r0(1-|z|2)α|fm(n+1)(φ(z))|×|ψ(z)φ′(z)|+sup|φ(z)|>r0(1-|z|2)α|ψ(z)φ′(z)|∥fm∥ℬ(1-|φ(z)|2)n+1≲|ψ(0)fm(n)(φ(0))|+K1sup|z|≤r0|fm(n)(z)|+K2sup|z|≤r0|fm(n+1)(z)|+2εM,
where K1=supz∈𝔻(1-|z|2)α|ψ′(z)| and K2=supz∈𝔻(1-|z|2)α|ψ(z)φ′(z)|. Since fm(n)→0 uniformly on compact subsets of 𝔻 as m→∞, we have ∥Dφ,ψnfm∥ℬα→0 as m→∞. It follows from Lemma 6 that Dφ,ψn:ℬ→ℬα is compact.
Similar as above, we know(76)∥Dφ,ψ′nfm∥𝒜∞α≲sup|φ(z)|≤r0(1-|z|2)α|fm(n)(φ(z))||ψ′(z)|+sup|φ(z)|>r0(1-|z|2)α|ψ′(z)|∥fm∥ℬ(1-|φ(z)|2)n≤K1sup|z|≤r0|fm(n)(z)|+ɛM,∥Dφ,ψφ′n+1fm∥𝒜∞α≲sup|φ(z)|≤r0(1-|z|2)α|fm(n+1)(φ(z))|×|ψ(z)φ′(z)|+sup|φ(z)|>r0(1-|z|2)α|ψ(z)φ′(z)|∥fm∥ℬ(1-|φ(z)|2)n+1≤K2sup|z|≤r0|fm(n+1)(z)|+ɛM.
From fm(n)→0 uniformly on compact subsets of 𝔻, we have ∥Dφ,ψ′nfm∥𝒜∞α→0 and ∥Dφ,ψφ′n+1fm∥𝒜∞α→0 as m→∞. So, Dφ,ψ′n, Dφ,ψφ′n+1:ℬ→𝒜∞α are compact.
(IV):(f)⇔(h). Suppose that (f) is true. Note that ∥zk∥ℬ≤∥zk∥∞=1 and zk→0 uniformly on compact subsets of 𝔻 as k→∞; by Lemma 6, we have
(77)limk→∞∥Dφ,ψ′n(zk)∥𝒜∞α=0,limk→∞∥Dφ,ψφ′n+1(zk)∥𝒜∞α=0.
Conversely, assume that (h)  is true. It is easy to see that
(78)supz∈𝔻(1-|z|2)α|ψ′(z)|≤∥Dφ,ψ′n(zn)∥𝒜∞α≤supk∈ℕ∥Dφ,ψ′n(zk)∥𝒜∞α<∞,supz∈𝔻(1-|z|2)α|ψ(z)φ′(z)|≤∥Dφ,ψφ′n+1(zn+1)∥𝒜∞α≤supk∈ℕ∥Dφ,ψφ′n+1(zk)∥𝒜∞α<∞.
If ∥φ∥∞<1, from (g)⇒(f), we get that (f) is true. If ∥φ∥∞=1, as in the proof of Theorem 5, let
(79)Δkn={z∈𝔻:k-nk≤|φ(z)|≤k-n+1k+1}.
And let m with m≥n be the smallest positive integer such that Δmn≠⌀. For given ɛ>0, there exists a large enough integer M1 with M1>m such that
(80)∥Dφ,ψ′n(zk)∥𝒜∞α<ɛ,∥Dφ,ψφ′n+1(zk)∥𝒜∞α<ɛ,
whenever k>M1. Let {fj} be a norm bounded sequence in ℬ that converges to zero uniformly on compact subsets of 𝔻 as j→∞. Denote M=supm∥fm∥ℬ<∞. We get
(81)∥Dφ,ψ′nfj∥𝒜∞α=supk≥msupz∈Δkn(1-|z|2)α|fj(n)(φ(z))||ψ′(z)|=(supm≤k≤M1+supk>M1)supz∈Δkn(1-|z|2)α×|fj(n)(φ(z))||ψ′(z)|=:I1+I2.
Then,
(82)I1=supm≤k≤M1supz∈Δkn(1-|z|2)α|fj(n)(φ(z))||ψ′(z)|≤supz∈𝔻(1-|z|2)α|ψ′(z)|sup|φ(z)|≤r|fj(n)(φ(z))|,
where
(83)r=M1-n+1M1+1,(84)I2=supk≥M1supz∈Δkn(1-|z|2)α|fj(n)(φ(z))||ψ′(z)|=supk≥M1supz∈Δkn(1-|φ(z)|)n|fj(n)(φ(z))|×(1-|z|2)α|ψ′(z)|(Hkn(|φ(z)|)/(1-|φ(z)|)n)Hkn(|φ(z)|)≲1cn∥fj∥ℬsupk>M1∥Dφ,ψ′n(zk)∥𝒜∞α≲1cnɛ.
Since fj(n)→0 uniformly on compact subsets of 𝔻, then ∥Dφ,ψ′nfj∥𝒜∞α→0 as j→∞. Thus, by Lemma 6, Dφ,ψ′n:ℬ→𝒜∞α is compact. Similar as above, we can prove that Dφ,ψφ′n+1:ℬ→𝒜∞α is compact. The proof is complete.



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*Source: 151929-2013-06-10.xml*