# Weighted Composition Operators from Hardy to Zygmund Type Spaces

Authors: Shanli Ye; Zhengyuan Zhuo
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

This paper aims at studying  the boundedness and compactness of weighted composition operator between  spaces of analytic functions. We characterize boundedness and compactness of  the weighted composition operatoruCϕ from the Hardy spaces Hp to the Zygmund type spaces 𝒵α={f∈H(D):supz∈D(1−|z|2)α|f″(z)|<∞} and the little Zygmund type spaces 𝒵α,0 in terms of function theoretic properties of the symbols u and ϕ.


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## Body

## 1.  Introduction

LetD={z:|z|<1} be the open unit disk in the complex plane C and T={z:|z|=1} its boundary, and H(D) denotes the set of all analytic functions on D. An analytic self-map φ:D→D induces the composition operator Cφ on H(D), defined by Cφ(f)=f(φ(z)) for f analytic on D. It is a well-known consequence of Littlewood’s subordination principle that the composition operator Cφ is bounded on the classical Hardy Hp  (0<p≤∞) spaces, Bergman Ap  (0<p≤∞) spaces, and Bloch spaces (see, e.g., [1–4]).Letu be a fixed analytic function on the open unit disk. Define a linear operator uCφ on the space of analytic functions on D, called a weighted composition operator, by uCφf=u·(f∘φ), where f is an analytic function on D. We can regard this operator as a generalization of a multiplication operator and a composition operator. In recent years the weighted composition operator has received much attention and appears in various settings in the literature. For example, it is known that isometries of many analytic function spaces are weighted composition operators (see [5], for instance). Their boundedness and compactness have been studied on various Banach spaces of analytic functions, such as Hardy, Bergman, BMOA, Bloch-type, and Zygmund spaces; see, for example, [6–11]. Also, it has been studied from one Banach space of analytic functions to another; one may see [12–23].The purpose of this paper is to consider the weighted composition operators from the Hardy spaceHp  (0<p<∞) to the Zygmund type spaces 𝒵α. Our main goal is to characterize boundedness and compactness of the operators uCφ from Hp to 𝒵α in terms of function theoretic properties of the symbols u and φ.Now we give a detailed definition of these spaces. For0≤r<1, f(z)∈H(D), we set
(20)Mp(r,f)=(12π∫02π‍|f(reiθ)|pdθ)1/p,0<p<∞,M∞(r,f)=max0≤θ≤2π|f(reiθ)|.
For 0<p≤∞, the Hardy space Hp consists of those functions f∈H(D), for which
(2)∥f∥p=sup0≤r<1Mp(r,f)<∞.
It is well known that with norm (2) the Hp space is a Banach space if 1≤p≤∞, for 0<p<1, Hp space is a nonlocally convex topological vector space, and d(f,g)=∥f-g∥pp is a complete metric for it. For more information about the Hp space, one may see these books, for example, [24, 25].Forα>0 the α-Bloch space βα consists of all analytic functions f defined on D such that
(3)∥f∥βα=sup{(1-|z|2)α|f′(z)|:z∈D},0<α<+∞.
The space 𝒵α consists of all analytic functions f defined on D such that
(4)∥f∥𝒵α=sup{(1-|z|2)α|f′′(z)|:z∈D},0<α<+∞.
When α=1, it is called the Zygmund space. From a theorem by Zygmund (see [26, vol. I, p. 263] or [24, Theorem 5.3]), we see that f∈𝒵1 if and only if f is continuous in the close unit disk D¯={z:|z|≤1} and the boundary function f(eiθ) such that
(5)|f(ei(θ+h))+f(ei(θ-h))-2f(eiθ)|h<∞.
When α>1, from Proposition 8 of [27], we know that 𝒵α=βα-1. Then the space 𝒵α is called a Zygmund type space if 0<α≤1. However, all results in this paper are valid for all 𝒵α spaces (α>0). An analytic function f∈H(D) is said to belong to the little Zymund type space 𝒵α,0 which consists of all f∈𝒵α satisfying lim|z|→1-(1-|z|2)α|f′′(z)|=0. It can be easily proved that 𝒵α is a Banach space under the norm
(6)∥f∥*=|f(0)|+|f′(0)|+∥f∥𝒵α.
And the polynomials are norm-dense in closed subspace 𝒵α,0. For some other information on this space and some operators on it, see, for example, [28–31].Throughout this paper, constants are denoted byC, C(p), they are positive, and C(p) are only depending on p and may differ from one occurrence to the another.


## 2. Auxiliary Results

In order to prove the main results of this paper. We need some auxiliary results. The first lemma is well known.Lemma 1 (see [24, p. 65]).
Forp>1, there exists a constant C(p) such that
(7)∫02π‍dθ|1-z|p≤C(p)(1-|z|2)p-1foreveryz∈D.Lemma 2.
Suppose that0<p<∞, f∈Hp; then
(8)|f(n)(z)|≤∥f∥p(1-|z|2)1/p+n
for every z∈D and all nonnegative integer n=0,1,2,….Proof.
We use induction onn. The case n=0 holds because it is Exercise  5 in [25, p. 85]. Assume the case n=k holds. Fix 0<r<1 and let g(z)=f(k)(rz). Then g(z) is in H∞⊂β1, and ∥g∥β1≤∥g∥∞. It follows that
(9)(1-|z|2)|g′(z)|=(1-|z|2)|rf(k+1)(rz)|≤∥f∥p(1-|rz|2)1/p+k≤∥f∥p(1-|z|2)1/p+k.
Let r→1-; we have
(10)|f(k+1)(z)|≤∥f∥p(1-|z|2)1/p+k+1.
Then the case n=k+1 holds. Hence (8) holds.Lemma 3.
For0<p<∞, suppose uCφ:Hp→𝒵α,0 is a bounded operator. Then uCφ:Hp→𝒵α is a bounded operator.This is obvious.


## 3. Boundedness ofuCφ from Hp(0<p<∞) to 𝒵α and 𝒵α,0

In this section we characterize bounded weighted composition operators from the Hardy spaceHp  (0<p<∞) to the Zygmund spaces 𝒵α.Theorem 4.
Letα>0,  0<p<∞,  and u be an analytic function on the unit disc D and φ an analytic self-map of D. Then uCφ is a bounded operator from Hp to the Zygmund spaces 𝒵α if and only if the following are satisfied:
(11)supz∈D(1-|z|2)α|u′′(z)|(1-|φ(z)|2)1/p<∞,(12)supz∈D(1-|z|2)α|u(z)(φ′(z))2|(1-|φ(z)|2)1/p+2<∞,(13)supz∈D(1-|z|2)α|2φ′(z)u′(z)+φ′′(z)u(z)|(1-|φ(z)|2)1/p+1<∞.Proof .
SupposeuCφ is bounded from Hp to the Zygmund spaces 𝒵α. Then we can easily obtain the following results by taking f(z)=1 and f(z)=z in Hp, respectively:
(14)u∈𝒵α,supz∈D(1-|z|2)α|2φ′(z)u′(z)+φ′′(z)u(z)+φ(z)u′′(z)|<+∞.
By (14) and the boundedness of the function φ(z), we get
(15)K1=supz∈D(1-|z|2)α|2φ′(z)u′(z)+φ′′(z)u(z)|<+∞.
Let f(z)=z2 in Hp again; in the same way we have
(16)supz∈D(1-|z|2)α|∫4φ(z)φ′(z)u′(z)+φ2(z)u′′(z)supz∈D(1-|z|2)α+2u(z)(φ(z)φ′′(z)+(φ′(z))2)∫|<∞.
Using these facts and the boundedness of the function φ(z) again, we get
(17)K2=supz∈D(1-|z|2)α|(φ′(z))2u(z)|<+∞.
Fix a∈D; we take the test functions
(18)fa(z)=-1-|a|2(1-a¯z)1+1/p+2(1-|a|2)2(1-a¯z)2+1/p-(1-|a|2)3(1-a¯z)3+1/p
for z∈D. From Lemma 1 we obtain that fa∈Hp and supa∥fa∥p≤C(p)<∞ with a direct calculation. Since fa(a)=0,  fa′(a)=0,  and fa′′(a)=-2a¯2/(1-|a|2)2+1/p, it follows that, for all λ∈D with |φ(λ)|>1/2, we have
(19)C∥fa∥*≥∥uCφfa∥*≥supz∈D(1-|z|2)α|(uCφfa)′′(z)|=supz∈D(1-|z|2)α×|∫(2φ′(z)u′(z)+φ′′(z)u(z))fa′(φ(z))+fa′′(φ(z))(φ′(z))2u(z)+u′′(z)fa(φ(z))∫|.
Let a=φ(λ); it follows that
(1)C∥fa∥*≥(1-|λ|2)α×|∫(2φ′(λ)u′(λ)+φ′′(λ)u(λ))fφ(λ)′(φ(λ))+fφ(λ)′′(φ(λ))(φ′(λ))2u(λ)+u′′(λ)fφ(λ)(φ(λ))∫|=(1-|λ|2)α|(φ′(λ))2u(λ)2(φ(λ)¯)2(1-|φ(λ)|2)2+1/p|≥12(1-|λ|2)α|u(λ)(φ′(λ))2|(1-|φ(λ)|2)2+1/p.
For all λ∈D with |φ(λ)|≤1/2, by (17), we have
(21)supλ∈D(1-|λ|2)α|u(λ)(φ′(λ))2|1-|φ(λ)|1/p+2≤(43)1/p+2supλ∈D(1-|λ|2)α|u(λ)(φ′(λ))2|<+∞.
Hence (12) holds.
Next, fixa∈D; we take another test functions
(22)ga(z)=-31-|a|2(1-a¯z)1/p+1+5(1-|a|2)2(1-a¯z)1/p+2-2(1-|a|2)3(1-a¯z)1/p+3
for z∈D. From Lemma 1 we obtain that ga∈Hp and supa∥ga∥p≤C(p)<∞ with a direct calculation. Since ga(a)=0,  ga′′(a)=0,  and ga′(a)=a¯/(1-|a|2)1/p+1, it follows that, for all λ∈D with |φ(λ)|>1/2, we obtain that
(23)C∥ga∥*≥∥uCφga(z)∥*≥supz∈D(1-|z|2)α|(uCφga)′′(z)|=supz∈D(1-|z|2)α×|∫(2φ′(z)u′(z)+φ′′(z)u(z))ga′(φ(z))+ga′′(φ(z))(φ′(z))2u(z)+u′′(z)ga(φ(z))∫|≥(1-|λ|2)α×|∫(2φ′(λ)u′(λ)+φ′′(λ)u(λ))gφ(λ)′(φ(λ))+gφ(λ)′′(φ(λ))(φ′(λ))2u(λ)+u′′(λ)gφ(λ)(φ(λ))∫|=(1-|λ|2)α×|φ(λ)¯(1-|φ(λ)|2)1/p+1(2φ′(λ)u′(λ)+φ′′(λ)u(λ))×φ(λ)¯(1-|φ(λ)|2)1/p+1|≥12(1-|λ|2)α|2φ′(λ)u′(λ)+φ′′(λ)u(λ)|(1-|φ(λ)|2)1/p+1.
For all λ∈D with |φ(λ)|≤1/2, by (15), we have
(24)sup|φ(λ)|≤1/2(1-|λ|2)α|2φ′(λ)u′(λ)+φ′′(λ)u(λ)|(1-|φ(λ)|2)1/p+1≤(43)1/p+1sup|φ(λ)|≤1/2(1-|λ|2)α×|2φ′(λ)u′(λ)+φ′′(λ)u(λ)|<∞.
Hence (13) holds.
Finally, fixa∈D, and, for all z∈D, let
(25)ha(z)=-p+3p+11-|a|2(1-a¯z)1/p+1+2(p+3)p+2(1-|a|2)2(1-a¯z)1/p+2-(1-|a|2)3(1-a¯z)1/p+3.
From Lemma 1 we obtain that ha∈Hp and supa∥ha∥p≤C(p)<∞ with a direct calculation. Since ha′(a)=0,  ha′′(a)=0,  and ha(a)=-2/(p+1)(p+2)(1-|a|2)1/p, it follows that, for all λ∈D, we obtain that
(26)C∥ha∥*≥∥uCφha(z)∥*≥supz∈D(1-|z|2)α|(uCφha)′′(z)|=supz∈D(1-|z|2)α×|∫(2φ′(z)u′(z)+φ′′(z)u(z))ha′(φ(z))+ha′′(φ(z))(φ′(z))2u(z)+u′′(z)ha(φ(z))∫|≥(1-|λ|2)α×|∫(2φ′(λ)u′(λ)+φ′′(λ)u(λ))hφ(λ)′(φ(λ))+hφ(λ)′′(φ(λ))(φ′(λ))2u(λ)+u′′(λ)hφ(λ)(φ(λ))∫|=(1-|λ|2)α|u′′(λ)hφ(λ)(φ(λ))|=2(p+1)(p+2)(1-|λ|2)α|u′′(λ)|(1-|φ(λ)|2)1/p.
Then (11) holds.
Conversely, suppose that (11), (12), and (13) hold. For f∈Hp, by Lemma 2, we have the following inequality:
(27)(1-|z|2)α|(uCφf)′′(z)|=(1-|z|2)α×|∫(2φ′(z)u′(z)+φ′′(z)u(z))f′(φ(z))+f′′(φ(z))(φ′(z))2u(z)+u′′(z)f(φ(z))∫|≤(1-|z|2)α×|(2φ′(z)u′(z)+φ′′(z)u(z))f′(φ(z))|+(1-|z|2)α|f′′(φ(z))(φ′(z))2u(z)|+(1-|z|2)|u′′(z)f(φ(z))|≤(1-|z|2)α|2φ′(z)u′(z)+φ′′(z)u(z)|(1-|φ(z)|2)1/p+1∥f∥p+(1-|z|2)α|(φ′(z))2u(z)|(1-|φ(z)|2)1/p+2∥f∥p+(1-|z|2)α|u′′(z)|(1-|φ(z)|2)1/p∥f∥p≤C∥f∥p,|u(0)f(φ(0))|+|u′(0)f(φ(0))|+|u(0)f′(φ(0))|≤(|u(0)|(1-|φ(0)|2)1/p+|u′(0)|(1-|φ(0)|2)1/p+|u(0)|(1-|φ(0)|2)1/p+1)∥f∥p.
This shows that uCφ is bounded. This completes the proof of Theorem 4.Theorem 5.
Letα>0,  0<p<∞, and u be an analytic function on the unit disc D and φ an analytic self-map of D. Then uCφ:Hp→𝒵α,0 is a bounded operator provided that the following are satisfied:
(28)lim|z|→1-(1-|z|2)α|u′′(z)|(1-|φ(z)|2)1/p=0,(29)lim|z|→1-(1-|z|2)α|u(z)(φ′(z))2|(1-|φ(z)|2)1/p+2=0,(30)lim|z|→1-(1-|z|2)α|2φ′(z)u′(z)+φ′′(z)u(z)|(1-|φ(z)|2)1/p+1=0.
Conversely, if uCφ:Hp→𝒵α,0 is a bounded operator, then u∈𝒵α,0, (11), (12), and (13) hold, and the following are satisfied:
(31)lim|z|→1-(1-|z|2)α|u(z)(φ′(z))2|=0,(32)lim|z|→1-(1-|z|2)α|2φ′(z)u′(z)+φ′′(z)u(z)|=0.Proof.
Assume that (28), (29), and (30) hold. Then for any ϵ>0, there is a constant δ,  0<δ<1, such that δ<|z|<1 implies
(33)(1-|z|2)α|u(z)(φ′(z))2|(1-|φ(z)|2)1/p+2<ϵ,(1-|z|2)α|u′′(z)|(1-|φ(z)|2)1/p<ϵ,(1-|z|2)α|2φ′(z)u′(z)+φ′′(z)u(z)|(1-|φ(z)|2)1/p+1<ϵ.
Then, for any f∈Hp, from Lemma 2 we obtain that
(34)(1-|z|2)α|(uCφf)′′(z)|=(1-|z|2)α×|∫(2φ′(z)u′(z)+φ′′(z)u(z))f′(φ(z))+f′′(φ(z))(φ′(z))2u(z)+u′′(z)f(φ(z))∫|≤(1-|z|2)α|2φ′(z)u′(z)+φ′′(z)u(z)|×1(1-|φ(z)|2)1/p+1∥f∥p+(1-|z|2)α|(φ′(z))2u(z)|1(1-|φ(z)|2)1/p+2∥f∥p+(1-|z|2)|u′′(z)|(1-|φ(z)|2)1/p∥f∥p≤3∥f∥pϵ.
Hence uCφf∈𝒵α,0 for all f∈𝒵α,0. On the other hand, (25), (28), and (29) imply that (11), (12), and (13) hold; then uCφ:Hp→𝒵α is bounded by Theorem 4. So uCφ:Hp→𝒵α,0 is bounded.
Conversely, assume thatuCφ is bounded from Hp to the little Zygmund type space 𝒵α,0. Then u=uCφ1∈𝒵α,0. Also uφ=uCφz∈𝒵α,0; thus
(35)(1-|z|2)α|2φ′(z)u′(z)+φ′′(z)u(z)+φ(z)u′′(z)|⟶0(|z|⟶1-).
Since |φ|≤1 and u∈𝒵α,0, we have lim|z|→1-(1-|z|2)α|2φ′(z)u′(z)+φ′′(z)u(z)|=0. Hence (32) holds.
Similarly,uCφz2∈𝒵α,0; then
(36)(1-|z|2)α×|(φ(z)φ′′(z)+(φ′(z))2)4φ(z)φ′(z)u′(z)+φ2(z)u′′(z)+2u(z)(φ(z)φ′′(z)+(φ′(z))2)∫|⟶0(|z|⟶1-).
By (32), |φ|≤1, and u∈𝒵α,0, we get that lim|z|→1-(1-|z|2)α|u(z)(φ′(z))2|=0; that is, (31) holds.
On the other hand, from Lemma3 and Theorem 4, we obtain that (11), (12), and (13) hold.


## 4.  Compactness ofuCφ

In order to prove the compactness ofuCφ from Hp to the Zygmund spaces 𝒵α, we require the following lemmas.Lemma 6.
Letα>0,  0<p<∞, and u be an analytic function on the unit disc D and φ an analytic self-map of D. Suppose that uCφ is a bounded operator from Hp to 𝒵α. Then uCφ is compact if and only if, for any bounded sequence {fn} in Hp which converges to 0 uniformly on compact subsets of D, one has ∥uCφ(fn)∥*→0 as n→∞.The proof is similar to that of Proposition  3.11 in [32]. The details are omitted.Theorem 7.
Letα>0,0<p<∞, u be an analytic function on the unit disc D and φ an analytic self-map of D. Then uCφ is a compact operator from Hp to 𝒵α if and only if uCφ is a bounded operator and the following are satisfied:
(37)(i)lim|φ(z)|→1-(1-|z|2)α|u′′(z)|(1-|φ(z)|2)1/p=0,(ii)lim|φ(z)|→1-(1-|z|2)α|u(z)(φ′(z))2|(1-|φ(z)|2)1/p+2=0,(iii)lim|φ(z)|→1-(1-|z|2)α|2φ′(z)u′(z)+φ′′(z)u(z)|(1-|φ(z)|2)1/p+1=0.Proof.
Suppose thatuCφ is compact from Hp to the Zygmund type space 𝒵α. Let {zn} be a sequence in D such that |φ(zn)|→1 as n→∞. If such a sequence does not exist, then (37) are automatically satisfied. Without loss of generality we may suppose that |φ(zn)|>1/2 for all n. We take the test functions
(38)fn(z)=-1-|φ(zn)|2(1-φ(zn)¯z)1/p+1+2(1-|φ(zn)|2)2(1-φ(zn)¯z)1/p+2-(1-|φ(zn)|2)3(1-φ(zn)¯z)1/p+3.
By a direct calculation, we may easily prove that {fn} converges to 0 uniformly on compact subsets of D and supn∥fn∥p≤C(p)<∞. Then {fn} is a bounded sequence in Hp which converges to 0 uniformly on compact subsets of D. Then limn→∞∥uCφ(fn)∥*=0 by Lemma 6. Note that
(39)fn(φ(zn))=0,fn′(φ(zn))=0,fn′′(φ(zn))=-2φ(zn)¯2(1-|φ(zn)|2)1/p+2.
It follows that
(40)∥uCφfn∥*≥∥uCφfn∥𝒵α≥(1-|zn|2)α×|∫(2u′(zn)φ′(zn)+φ′′(zn)u(zn))×fn′(φ(zn))+u(zn)fn′′(φ(zn))(φ′(zn))2+u′′(zn)fn(φ(zn))∫|=(1-|zn|2)α×|(φ′(zn))2u(zn)2φ(zn)¯2(1-|φ(zn)|2)1/p+2|.
Then
(41)limn→∞(1-|zn|2)α|u(zn)(φ′(zn))2|(1-|φ(zn)|2)1/p+2=0.
Next, let(42)gn(z)=-31-|φ(zn)|2(1-φ(zn)¯z)1/p+1+5(1-|φ(zn)|2)2(1-φ(zn)¯z)1/p+2-2(1-|φ(zn)|2)3(1-φ(zn)¯z)1/p+3.
By a direct calculation we obtain that gn⇉0  (n→∞) on compact subsets of D and supn∥gn∥p≤C(p)<+∞. Consequently, {gn} is a bounded sequence in Hp which converges to 0 uniformly on compact subsets of D. Then limn→∞∥uCφ(gn)∥*=0 by Lemma 6. Note that gn(φ(zn))≡0, gn′′(φ(zn))≡0 and gn′(φ(zn))=φ(zn)¯/(1-|φ(zn)|2)1/p+1; it follows that
(43)∥uCφgn∥*≥∥uCφgn∥𝒵α≥(1-|zn|2)α×|∫(2u′(zn)φ′(zn)+φ′′(zn)u(zn))×gn′(φ(zn))+u(zn)gn′′(φ(zn))(φ′(zn))2+u′′(zn)gn(φ(zn))∫|=(1-|zn|2)α×|φ(zn)¯(1-|φ(zn)¯|2)1/p+1(2u′(zn)φ′(zn)+φ′′(zn)u(zn))×φ(zn)¯(1-|φ(zn)¯|2)1/p+1|.
Then limn→∞(1-|zn|2)α(2u′(zn)φ′(zn)+φ′′(zn)u(zn))/(1-|φ(zn)|2)1/p+1=0.
Finally, let(44)hn(z)=-p+3p+11-|φ(zn)|2(1-φ(zn)¯z)1/p+1+2(p+3)p+2(1-|φ(zn)|2)2(1-φ(zn)¯z)1/p+2-(1-|φ(zn)|2)3(1-φ(zn)¯z)1/p+3.
By a direct calculation we obtain that hn⇉0  (n→∞) on compact subsets of D and supn∥hn∥p<∞. Consequently, {hn} is a bounded sequence in Hp which converges to 0 uniformly on compact subsets of D. Then limn→∞∥uCφ(hn)∥*=0 by Lemma 6. Note that hn(φ(zn))=2/(p+1)(p+2)(1-|φ(zn)|2)1/p,  hn′(φ(zn))≡0, and hn′′(φ(zn))≡0; it follows that
(45)∥uCφhn∥*≥∥uCφhn∥𝒵α≥(1-|zn|2)α×|∫(2u′(zn)φ′(zn)+φ′′(zn)u(zn))×hn′(φ(zn))+u(zn)hn′′(φ(zn))(φ′(zn))2+u′′(zn)hn(φ(zn))∫|=2(1-|zn|2)α|u′′(zn)|(p+1)(p+2)(1-|φ(zn)|2)1/p.
Then limn→∞((1-|zn|2)α|u′′(zn)|/(1-|φ(zn)|2)1/p)=0. The proof of the necessary is completed.
Conversely, Suppose that (37) hold. Since uCφ is a bounded operator, from Theorem 4, we have
(46)M1≜supz∈D(1-|z|2)α|u(z)(φ′(z))2|<∞,M2≜supz∈D(1-|z|2)α|u′′(z)|<∞,M3≜supz∈D(1-|z|2)α|2φ′(z)u′(z)+φ′′(z)u(z)|<∞.
Let {fn} be a bounded sequence in Hp with ∥fn∥p≤1 and fn→0 uniformly on compact subsets of D. We only prove limn→∞∥uCφ(fn)∥*=0 by Lemma 6. By the assumption, for any ϵ>0, there is a constant δ,  0<δ<1, such that δ<|φ(z)|<1 implies
(47)(1-|z|2)α|u(z)(φ′(z))2|(1-|φ(z)|2)1/p+2<ϵ,(1-|z|2)α|u′′(z)|(1-|φ(z)|2)1/p<ϵ,(1-|z|2)α|2φ′(z)u′(z)+φ′′(z)u(z)|(1-|φ(z)|2)1/p+1<ϵ.
Let K={w∈D:|w|≤δ}. Note that K is a compact subset of D. Then from Lemma 2 it follows that
(48)∥uCφfn∥𝒵α=supz∈D(1-|z|2)α|(uCφfn)′′(z)|=(1-|z|2)α×|∫(2φ′(z)u′(z)+φ′′(z)u(z))fn′(φ(z))+fn′′(φ(z))(φ′(z))2u(z)+u′′(z)fn(φ(z))∫|≤supz∈D(1-|z|2)α×|(2φ′(z)u′(z)+φ′′(z)u(z))fn′(φ(z))|+supz∈D(1-|z|2)α|fn′′(φ(z))(φ′(z))2u(z)|+supz∈D(1-|z|2)|u′′(z)fn(φ(z))|≤3ϵ+sup|φ(z)|≤δ(1-|z|2)α×|(2φ′(z)u′(z)+φ′′(z)u(z))fn′(φ(z))|+sup|φ(z)|≤δ(1-|z|2)α|fn′′(φ(z))(φ′(z))2u(z)|+sup|φ(z)|≤δ(1-|z|2)α|u′′(z)fn(φ(z))|≤3ϵ+M3supw∈K|fn′(w)|+M1supw∈K|fn′′(w)|+M2supw∈K|fn(w)|.
As n→∞,
(49)∥uCφfn∥𝒵α⟶0.
Hence uCφ is compact. This completes the proof of Theorem 7.In order to prove the compactness ofuCφ on the little Zygmund spaces 𝒵α,0, we require the following lemma.Lemma 8.
LetU⊂𝒵α,0. Then U is compact if and only if it is closed, bounded and satisfies
(50)lim|z|→1-supf∈U(1-|z|2)|f′′(z)|=0.The proof is similar to that of Lemma  1 in [1], but we omit it.Theorem 9.
Letα>0, 0<p<∞, u be an analytic function on the unit disc D and φ an analytic self-map of D. Then uCφ is compact from Hp to the little Zygmund type spaces 𝒵α,0 if and only if (28), (29), and (30) hold.Proof.
Assume that (28), (29), and (30) hold. By Theorem 5, we know that uCφ is bounded from Hp to the little Zygmund type spaces 𝒵α,0. Suppose that f∈Hp with ∥f∥p≤1. From Lemmas 1 and 2 we obtain that
(51)(1-|z|2)α|(uCφf)′′(z)|≤(1-|z|2)α|(2φ′(z)u′(z)+φ′′(z)u(z))f′(φ(z))|+(1-|z|2)α|f′′(φ(z))(φ′(z))2u(z)|+(1-|z|2)|u′′(z)f(φ(z))|≤(1-|z|2)α|2φ′(z)u′(z)+φ′′(z)u(z)|×1(1-|φ(z)|2)1/p+1∥f∥p+(1-|z|2)α|(φ′(z))2u(z)|(1-|φ(z)|2)1/p+2∥f∥p+(1-|z|2)α|u′′(z)|(1-|φ(z)|2)1/p∥f∥p,
thus
(52)sup{|(1-|z|2)α(uCφf)′′(z)|:f∈Hp,∥f∥p≤1}≤(1-|z|2)α|2φ′(z)u′(z)+φ′′(z)u(z)|11-|φ(z)|2+(1-|z|2)α|(φ′(z))2u(z)|(1-|φ(z)|2)1/p+2+(1-|z|2)α|u′′(z)|(1-|φ(z)|2)1/p,
and it follows that
(53)lim|z|→1-sup{|(1-|z|2)α(uCφf)′′(z)|:f∈Hp,∥f∥p≤1}=0.
Hence uCφ:Hp→𝒵α,0 is compact by Lemma 8.
Conversely, suppose thatuCφ:Hp→𝒵α,0 is compact.
Firstly, it is obviousuCφ is bounded, from Theorem 5 we have u∈𝒵α,0, and (31), (32) hold. On the other hand, we have
(54)lim|z|→1-sup{|(1-|z|2)α(uCφf)′′(z)|:f∈Hp,∥f∥p≤M}=0,
for some M>0 by Lemma 6.
Next, note that the proof of Theorem4 and the fact that the functions given in (18) are in Hp and have norms bounded independently of a; we obtain that
(55)lim|z|→1-(1-|z|2)α|u(z)(φ′(z))2|(1-|φ(z)|2)1/p+2=0
for |φ(z)|>1/2. However, if |φ(z)|≤1/2, by (31), we easily have
(56)lim|z|→1-(1-|z|2)α|u(z)(φ′(z))2|(1-|φ(z)|2)1/p+2≤lim|z|→1-(43)1/p+2(1-|z|2)α|u(z)(φ′(z))2|=0.
Similarly, note that the functions given in (22) and (25) are in Hp and have norms bounded independently of a, we obtain that
(57)lim|z|→1-(1-|z|2)α|2φ′(z)u′(z)+φ′′(z)u(z)|(1-|φ(z)|2)1/p+1=0,lim|z|→1-(1-|z|2)α|u′′(z)|(1-|φ(z)|2)1/p=0
for |φ(z)|>1/2. However, if |φ(z)|≤1/2, from u∈𝒵α,0 and (32), we easily have
(58)lim|z|→1-(1-|z|2)α|2φ′(z)u′(z)+φ′′(z)u(z)|(1-|φ(z)|2)1/p+1≤(43)1/p+1lim|z|→1-(1-|z|2)α×|2φ′(z)u′(z)+φ′′(z)u(z)|=0,lim|z|→1-(1-|z|2)α|u′′(z)|(1-|φ(z)|2)1/p≤lim|z|→1-(43)1/p(1-|z|2)α|u′′(z)|=0.
This completes the proof of Theorem 9.Remark 10.
From Theorems5 and 9, we conjecture that uCφ:Hp→𝒵α,0 is compact if and only if uCφ:Hp→𝒵α,0 is bounded.Takingu(z)=1 from Theorems 4, 7, and 9, we obtain the following results about the characterization of the boundedness and compactness of the composition operator Cφ:Hp→𝒵α(or𝒵α,0).Corollary 11.
Letα>0,  0<p<∞,  and  φ be an analytic self-map of D. Then Cφ:Hp→𝒵α is a bounded operator if and only if the following are satisfied:
(59)supz∈D(1-|z|2)α|(φ′(z))2|(1-|φ(z)|2)1/p+2<∞,supz∈D(1-|z|2)α|φ′′(z)|(1-|φ(z)|2)1/p+1<∞.Corollary 12.
Letα>0,  0<p<∞,  and  φ be an analytic self-map of D. Then Cφ:Hp→𝒵α is a compact operator if and only if Cφ is bounded and the following are satisfied:
(60)lim|φ(z)|→1-(1-|z|2)α|(φ′(z))2|(1-|φ(z)|2)1/p+2=0,lim|φ(z)|→1-(1-|z|2)α|φ′′(z)|(1-|φ(z)|2)1/p+1=0.Corollary 13.
Letα>0,  0<p<∞,  and  φ be an analytic self-map of D. Then Cφ:Hp→𝒵α,0 is a compact operator if and only if
(61)lim|z|→1-(1-|z|2)α|(φ′(z))2|(1-|φ(z)|2)1/p+2=0,lim|z|→1-(1-|z|2)α|φ′′(z)|(1-|φ(z)|2)1/p+1=0.In the formulation of corollary, we use the notationMu on H(D) defined by Muf=uf for f∈H(D). Taking φ(z)=z from Theorems 4, 5, 7, and 9, we obtain the following results about the characterization of the boundedness and compactness of pointwise multiplier Mu:Hp→𝒵α(or𝒵α,0).Corollary 14.
Letα>0,  0<p<∞. Then the pointwise multiplier Mu:Hp→𝒵α is a bounded operator if and only if(i)
u=0 if α<2+1/p;(ii)
u∈H∞ if α=2+1/p;(iii)
supz∈D(1-|z|2)α-2-1/p|u(z)|<∞ if α>2+1/p.Corollary 15.
Letα>0,  0<p<∞. Then the pointwise multiplier Mu:Hp→𝒵α,0 is a bounded operator if and only if Mu:Hp→𝒵α,0 is a compact operator if and only if Mu:Hp→𝒵α is a compact operator if and only if(i)
u=0 if α≤2+1/p,(ii)
lim|z|→1-(1-|z|2)α-2-1/p|u(z)|=0 if α>2+1/p.



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*Source: 365286-2013-04-18.xml*