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Volume 2008, Article ID 619525,14pages doi:10.1155/2008/619525

Research Article

Weighted Composition Operators from

Generalized Weighted Bergman Spaces to

Weighted-Type Spaces

Dinggui Gu

Department of Mathematics, JiaYing University, Meizhou, GuangDong 514015, China

Correspondence should be addressed to Dinggui Gu,gudinggui@163.com

Received 3 November 2008; Revised 22 November 2008; Accepted 24 November 2008

Recommended by Kunquan Lan

Letϕbe a holomorphic self-map and let ψ be a holomorphic function on the unit ballB. The boundedness and compactness of the weighted composition operatorψCϕfrom the generalized weighted Bergman space into a class of weighted-type spaces are studied in this paper.

Copyrightq2008 Dinggui Gu. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

1. Introduction

LetBbe the unit ball ofCnand letHBbe the space of all holomorphic functions onB. For

fHB, let

Rfz

n

j1 zj

∂f

∂zjz

1.1

represent the radial derivative offHB. We writeRmfRRm−1f.

For anyp >0 andα∈R, letNbe the smallest nonnegative integer such thatpNα >

−1. The generalized weighted Bergman spaceApαis defined as follows:

Apα

fHB| fApα f0

B

RNfzp

1− |z|2pNαdvz 1/p

<

. 1.2

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classicalweighted Bergman spaceα > −1, the Besov spaceApn1, and the Hardy space

H2. See1,2for some basic facts on the weighted Bergman space.

Letμbe a positive continuous function on0,1. We say thatμis normal if there exist positive numbersαandβ, 0< α < β,andδ∈0,1such thatsee3

μr

1− is decreasing on δ,1, limr→1 μr

1− 0; μr

1− is increasing onδ,1, limr→1 μr

1−.

1.3

AnfHBis said to belong to the weighted-type space, denoted byB,

if

fHμ∞ sup zB

μ|z|fz<, 1.4

whereμis normal on0,1. The little weighted-type space, denoted byHμ,0, is the subspace of∞consisting of thosef∞such that

lim |z| →1μ

|z|fz0. 1.5

See4,5for more information on∞.

Let ϕ be a holomorphic self-map of B. The composition operator is defined as

follows:

Cϕf

z fϕz, fHB. 1.6

LetψHB. ForfHB, the weighted composition operatorψCϕis defined by

ψCϕf

z ψzfϕz, zB. 1.7

The book6contains a plenty of information on the composition operator and the weighted composition operator.

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In this paper, we study the weighted composition operatorψCϕ from the generalized

weighted Bergman space to the spaces ∞ and Hμ,∞0. Some necessary and sufficient

conditions for the weighted composition operator ψCϕ to be bounded and compact are

given.

Throughout the paper, constants are denoted byC, they are positive and may differ from one occurrence to the other.

2. Main results and proofs

Before we formulate our main results, we state several auxiliary results which will be used in the proofs. They are incorporated in the lemmas which follow.

Lemma 2.1see1. iSuppose thatp >0andαn1>0. Then there exists a constantC >0

such that

fz CfApα

1− |z|21/p 2.1

for allfApαandzB.

iiSuppose thatp >0andαn1<0or0< p≤1andαn10. Then every function inApαis continuous on the closed unit ball. Moreover, there is a positive constantCsuch that

f∞≤CfApn1, 2.2

for everyfApn1.

iiiSuppose thatp >1,1/p1/q1,andαn10. Then there exists a constantC >0

such that

fzCln e 1− |z|2

1/q

2.3

for allfApαandzB.

The following criterion for compactness of weighted composition operators follows from standard arguments similar to those outlined in6, Proposition 3.11 see also12, proof of Lemma 2. We omit the details of the proof.

Lemma 2.2. Assume thatψHB,ϕis a holomorphic self-map ofB,andμis a normal function on0,1. ThenψCϕ :Apαis compact if and only ifψCϕ :Apαis bounded and for

any bounded sequencefkk∈NinApα which converges to zero uniformly on compact subsets ofBas

k → ∞, one hasψCϕfkHμ∞ → 0ask → ∞.

Note that whenp > 0 andαn1<0, the functions inApαare Lipschitz continuous

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Lemma 2.3. Letp >0andαn1<0. Letfkbe a bounded sequence inApαwhich converges to0

uniformly on compact subsets ofB, then

lim

k→ ∞supzBfkz0. 2.4

The following lemma is from21 one-dimensional case is20, Lemma 2.1.

Lemma 2.4. Assume thatμis a normal function on0,1. A closed setKinHμ,0is compact if and only if it is bounded and satisfies

lim |z| →1supfKμ

|z|fz0. 2.5

We will consider three cases:n1α >0,n1α0,andn1α <0.

2.1. Casen1α >0

Theorem 2.5. Assume thatp > 0,αis a real number such thatnα1 > 0HB,ϕis a holomorphic self-map ofB,andμis a normal function on0,1. ThenψCϕ :Apαis bounded

if and only if

M:sup

zB

μ|z|ψz

1−ϕz2n1α/p

<. 2.6

Proof. Assume thatψCϕ:Apα∞is bounded. Let

t > nmax

1,1 p

α1

p . 2.7

ForaB, set

faz

1− |a|2tn1α/p

1− z, a t . 2.8

It follows from1, Theorem 32thatfaApαand supaBfaApα <.Hence

CψCϕAp

αHμψCϕfϕbHμ

sup

zB

μ|z|ψCϕfϕbz

μ

|b|ψb

1−ϕb2n1α/p ,

2.9

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Conversely, suppose that 2.6 holds. Then for arbitrary zB and fApα, by

Lemma 2.1we have

μ|z|ψCϕf

|z|fϕzψzCfApα

μ|z|ψz

1−ϕz2n1α/p

. 2.10

In light of condition2.6, the boundedness of the operatorψCϕ : Apα∞follows from

2.10by taking the supremum overB. This proof is completed.

Theorem 2.6. Assume thatp > 0,αis a real number such thatnα1 > 0HB,ϕis a holomorphic self-map ofB,andμis a normal function on0,1. ThenψCϕ:Apαis compact if

and only ifψand

lim |ϕz| →1

μ|z|ψz

1−ϕz2n1α/p

0. 2.11

Proof. Assume thatψCϕ :Apα∞is compact, thenψCϕ :Apα∞is bounded. Taking

fz≡1, we get thatψ∞. Letzkk∈Nbe a sequence inBsuch that|ϕzk| → 1 ask → ∞

if such a sequence does not exist that condition2.11is vacuously satisfied. Set

fkz

1−ϕzk2

t1/p

1−z, ϕzk

t , k∈N, 2.12

wheretsatisfies2.7. From1, Theorem 32, we see thatfkk∈Nis a bounded sequence in

Apα. Moreover, it is easy to see thatfkconverges to zero uniformly on compact subsects ofB.

ByLemma 2.2, lim supk→ ∞ψCϕfkHμ∞ 0.On the other hand, we have

ψCϕfk

∞ sup zB

μ|z|ψCϕfk

z| ≥ μzkψ

zk

1−ϕzk2

n1α/p. 2.13

Hence

lim sup

k→ ∞

μzkψ

zk

1−ϕzk2

n1α/p 0, 2.14

from which2.11follows.

Conversely, assume thatψ∞and2.11holds. Then, it is easy to check that2.6

holds. HenceψCϕ : Apα∞is bounded. According to2.11, for givenε > 0, there is a

constantδ∈0,1such that

sup {zB:δ<|ϕz|<1}

μ|z|ψz

1−ϕz2n1α/p

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LetfkkNbe a bounded sequence inApαsuch thatfk → 0 uniformly on compact subsets of

Bask → ∞. LetδD{wB:|w| ≤δ}. From2.15andψ∞, we have

ψCϕfk

∞ sup zB

μ|z|fk

ϕzψz

sup

{zB:|ϕz|≤δ}{zB:δ<sup|ϕz|<1}

μ|z|ψzfk

ϕz

ψHμ sup wδD

fkwCfk

Apα sup

{zB:δ<|ϕz|<1}

μ|z|ψz

1−ϕz2n1α/p

ψHμ sup wδD

fkwCε.

2.16

SinceδDis a compact subset ofB, we have limk→ ∞supwδD|fkw| 0. Using this fact and

lettingk → ∞in2.16, we obtain

lim sup

k→ ∞

ψCϕfk

∞≤Cε. 2.17

Sinceεis an arbitrary positive number, we obtain lim supk→ ∞ψCϕfkHμ∞ 0.ByLemma 2.2, the implication follows.

Theorem 2.7. Assume thatp > 0,αis a real number such thatnα1 > 0HB,ϕis a holomorphic self-map ofB,andμis a normal function on0,1. ThenψCϕ:ApαHμ,∞0is bounded

if and only ifψCϕ :Apαis bounded andψHμ,∞0.

Proof. Assume thatψCϕ : ApαHμ,∞0 is bounded. Then it is clear thatψCϕ :Apα∞is

bounded. Takingfz 1 and employing the boundedness ofψCϕ :ApαHμ,∞0, we see that ψHμ,0.

Conversely, assume thatψCϕ : Apα∞ is bounded andψHμ,∞0. Suppose that fApαwithfApαL, using polynomial approximations we obtainsee, e.g.,1

lim |z| →1

1− |z|2n1α/pfz0. 2.18

From the above equality andψHμ,0, we have that for everyε > 0, there exists aδ ∈0,1 such that whenδ <|z|<1,

1− |z|2n1α/pfz< ε

M, 2.19

μ|z|ψz< ε

1−δ2n1α/p

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whereMis defined in2.6. Therefore, ifδ <|z|<1 andδ <|ϕz|<1, from2.6and2.19 we have

μ|z|ψCϕf

z μ

|z|ψz

1−ϕz2n1α/p

1−ϕz2n1α/pfϕz

M1−ϕz2n1α/pfϕz< ε.

2.21

Ifδ <|z|<1 and|ϕz| ≤δ, usingLemma 2.1and2.20we have

μ|z|ψCϕf

z μ

|z|ψz

1−ϕz2n1α/p

1−ϕz2n1α/pfϕz

CfApα

μ|z|ψz

1−ϕz2n1α/p

CfApα

1−δ2n1α/pμ

|z|ψz< ε.

2.22

Combining2.21and2.22, we obtain thatψCϕfHμ,∞0. Sincef is an arbitrary element of Apαwe see that

ψCϕ

Apα

Hμ,∞0, 2.23

which, along with the boundedness ofψCϕ :Apα∞, implies the result.

Theorem 2.8. Assume thatp > 0,αis a real number such thatnα1 > 0HB,ϕis a holomorphic self-map ofB,andμis a normal function on0,1. ThenψCϕ:ApαHμ,∞0is compact

if and only if

lim |z| →1

μ|z|ψz

1−ϕz2n1α/p

0. 2.24

Proof. Assume that2.24holds. For anyfApαwithfApα ≤1, by2.10we have

μ|z|ψCϕf

zCfApα

μ|z|ψz

1−ϕz2n1α/p

(8)

Using2.24, we get

lim |z| →1fsup

Apα≤1

μ|z|ψCϕf

zC lim |z| →1

μ|z|ψz

1−ϕz2n1α/p

0. 2.26

From this andLemma 2.4, we see thatψCϕ :ApαHμ,∞0is compact.

Conversely, assume that ψCϕ : ApαHμ,∞0 is compact. ThenψCϕ : A

p

αHμ,∞0 is

bounded andψCϕ :Apα∞is compact. By Theorems2.6and2.7, we obtain

lim |ϕz| →1

μ|z|ψz

1−ϕz2n1α/p

0, 2.27

lim |z| →1μ

|z|ψz0. 2.28

Ifϕ<1, it holds that

lim |z| →1

μ|z|ψz

1−ϕz2n1α/p

1

1− ϕ2

n1α/p lim |z| →1μ

|z|ψz0, 2.29

from which the result follows in this case.

Hence, assume thatϕ∞1. In terms of2.27, for everyε >0, there exists aδ∈0,1, such that whenδ <|ϕz|<1,

μ|z|ψz

1−ϕz2n1α/p

< ε. 2.30

According to2.28, for the aboveε, there exists anr∈0,1, such that whenr <|z|<1,

μ|z|ψz< ε1−δ2n1α/p. 2.31

Therefore, whenr <|z|<1 andδ <|ϕz|<1, we have that

μ|z|ψz

1−ϕz2n1α/p

< ε. 2.32

Ifr <|z|<1 and|ϕz| ≤δ, we obtain

μ|z|ψz

1−ϕz2n1α/p

1

1−δ2n1α/pμ

|z|ψz< ε. 2.33

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2.2. Casen1α0

Theorem 2.9. Assume thatp > 1,αis a real number such thatnα1 0HB,ϕis a holomorphic self-map ofB,andμis a normal function on0,1. ThenψCϕ :Apαis bounded

if and only if

M1:sup

zB

μ|z|ψz

ln e

1−ϕz2

1−1/p

<. 2.34

Proof. Assume that2.34holds. Then for arbitraryzBandfApα, byLemma 2.1we have

μ|z|ψCϕf

|z|fϕzψz

CfApαμ

|z|ψz

ln e

1−ϕz2

1−1/p

. 2.35

From2.34and2.35, the boundedness ofψCϕ :Apα∞follows.

Now assume thatψCϕ :Apα∞is bounded. ForaB, set

faz

ln e

1− |a|2

−1/p

ln e

1− z, a

. 2.36

By using2, Theorem 1.12, we easily check thatfaApn1. Therefore,

CψCϕAp

α∞ ≥ψCϕfϕbHμ

sup

zB

μ|z|ψCϕfϕbz

μ|b|ψb

ln e

1−ϕb2

1−1/p

.

2.37

From the last inequality, we get the desired result.

Theorem 2.10. Assume thatp >1,αis a real number such thatnα1 0HB,ϕis a holomorphic self-map ofB,andμis a normal function on0,1. ThenψCϕ:Apαis compact if

and only ifψand

lim |ϕz| →1μ

|z|ψz

ln e

1−ϕz2

1−1/p

0. 2.38

Proof. First assume that2.38holds andψ∞. In this case, the proof ofTheorem 2.6still

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Now we assume thatψCϕ :Apα∞is compact, then it is clear thatψCϕ :Apα

∞is bounded. Similarly to the proof ofTheorem 2.6, we see thatψ∞. Letzkk∈Nbe

a sequence inBsuch that |ϕzk| → 1 ask → ∞if such a sequence does not exist that

condition2.38is vacuously satisfied. Set

fkz

ln e

1−ϕzk2

−1/p

ln e

1−z, ϕzk

, k∈N. 2.39

From2, Theorem 1.12, we see thatfkk∈Nis a bounded sequence inApα. Moreover,fk → 0

uniformly on compact subsets ofBask → ∞. It follows fromLemma 2.2thatψCϕfkHμ∞ → 0 ask → ∞.Because

ψCϕfk

∞ sup zB

μ|z|ψCϕfk

z

μzkψ

zk

ln e

1−ϕzk2

1−1/p

,

2.40

we obtain

lim

k→ ∞μzkψ

zk

ln e

1−ϕzk2

1−1/p

0, 2.41

from which we get the desired result. The proof is completed.

Theorem 2.11. Assume thatp >1,αis a real number such thatnα1 0HB,ϕis a holomorphic self-map ofB,andμis a normal function on0,1. ThenψCϕ:ApαHμ,0is bounded

if and only ifψCϕ :Apαis bounded andψHμ,∞0.

Proof. First assume thatψCϕ : ApαHμ,∞0 is bounded. Then clearly ψCϕ : Apα∞ is

bounded. Takingfz 1, then employing the boundedness ofψCϕ :ApαHμ,∞0, we have

thatψHμ,0, as desired.

Conversely, assume that ψCϕ : Apα∞ is bounded and ψHμ,∞0. For each

polynomialp, we have

μ|z|ψCϕp

|z|pϕzψzpμ|z|ψz, 2.42

from which we have thatψCϕpHμ,∞0.

Since the set of all polynomials is dense inApαsee2, for everyfApα there is a

sequence of polynomialspkkNsuch that

pkf

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From the boundedness ofψCϕ:Apα∞, we have that

ψCϕpkψCϕf

Hμ

ψCϕpkf

Apα −→0 ask−→ ∞. 2.44

SinceHμ,0is a closed subset of∞, we obtain

ψCϕf lim

k→ ∞ψCϕpkH

μ,0. 2.45

Therefore,ψCϕ:ApαHμ,0is bounded.

Using Theorems 2.10 and 2.11, similarly to the proof of Theorem 2.8we obtain the following result. We omit the proof.

Theorem 2.12. Assume thatp >1,αis a real number such thatnα1 0HB,ϕis a holomorphic self-map ofB,andμis a normal function on0,1. ThenψCϕ:ApαHμ,∞0is compact

if and only if

lim |z| →1μ

|z|ψz

ln e

1−ϕz2

1−1/p

0. 2.46

Theorem 2.13. Assume that0< p≤1,αis a real number such thatnα10HB,ϕis a holomorphic self-map ofB,andμis a normal function on0,1. ThenψCϕ :Apαis bounded

if and only ifψ.

Proof. Assume thatψ∞. For anyfA p

α, byLemma 2.1we have

sup

zB

μ|z|ψCϕf

zCfApαsup zB

μ|z|ψz. 2.47

From the above inequality, we obtain thatψCϕ:Apα∞is bounded.

Conversely, assume thatψCϕ :Apα∞is bounded. Takingfz 1 and using the

boundedness ofψCϕ:Apα∞, we getψ∞, as desired.

Theorem 2.14. Assume that0< p≤1,αis a real number such thatnα10HB,ϕis a holomorphic self-map ofB,andμis a normal function on0,1. ThenψCϕ :Apαis compact

if and only ifψand

lim |ϕz| →1μ

|z|ψz0. 2.48

Proof. First assume that ψ∞ and 2.48 holds. In this case, the proof is similar to the

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Now we suppose thatψCϕ :Apα∞is compact. It follows fromTheorem 2.13and

the boundedness ofψCϕ :Apα∞thatψ∞. Letzkk∈Nbe a sequence inBsuch that

|ϕzk| → 1 ask → ∞. Set

fkz

1−ϕzk2

1−z, ϕzk

, k∈N. 2.49

From 2, Theorem 6.6, we see that fkkN is a bounded sequence in Apα. Moreover, fk

converges to zero uniformly on compact subsets ofB. Hence byLemma 2.2it follows that

lim sup

k→ ∞

ψCϕfk

∞ 0. 2.50

On the other hand, we obtain ψCϕfk

∞ sup zB

μ|z|ψCϕfk

zμzkψ

zk. 2.51

Combining2.50with2.51we obtain that2.48holds.

Theorem 2.15. Assume that0< p≤1,αis a real number such thatnα10HB,ϕis a holomorphic self-map ofB,andμis a normal function on0,1. Then the following statements are

equivalent:

iψCϕ :ApαHμ,∞0is bounded;

iiψCϕ :ApαHμ,∞0is compact;

iiiψHμ,0.

Proof. ii⇒i. This implication is clear.

i⇒iii. Takingfz 1 and employing the boundedness ofψCϕ :ApαHμ,∞0,we

obtain thatψHμ,0.

iii⇒ii. For anyfAαpwithfApα ≤1, we have

μ|z|ψCϕf

zCfAp αμ

|z|ψz|z|ψz, 2.52

from which we obtain

lim |z| →1fsup

Apα≤1

μ|z|ψCϕf

zC lim |z| →1μ

|z|ψz0. 2.53

UsingLemma 2.4, we obtain thatψCϕ :ApαHμ,0is compact.

2.3. Casen1α <0

Theorem 2.16. Assume thatp > 0,αis a real number such thatnα1 < 0HB,ϕis a holomorphic self-map ofB,andμis a normal function on0,1. Then the following statements are

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iψCϕ :Apαis bounded;

iiψCϕ :Apαis compact;

iiiψ.

Proof. ii⇒i. This implication is obvious.

i⇒iii. Takingfz 1, then using the boundedness ofψCϕ :Apα,we obtain

thatψ∞.

iii⇒ii. IffApα, byLemma 2.1we obtain

μ|z|ψCϕf

zCfApαμ

|z|ψz, 2.54

from which it follows thatψCϕ:Apα∞is bounded. LetfkkNbe any bounded sequence

inApαandfk → 0 uniformly onBask → ∞. ByLemma 2.3, we have

ψCϕfk

Hμ sup zB

μ|z|fk

ϕzψzψHμ sup zB

fk

ϕz−→0, 2.55

ask → ∞. The result follows fromLemma 2.2.

Similarly to the proof ofTheorem 2.15, we have the following result. We omit the proof here.

Theorem 2.17. Assume thatp > 0,αis a real number such thatnα1 < 0HB,ϕis a holomorphic self-map ofB,andμis a normal function on0,1. Then the following statements are

equivalent:

iψCϕ :ApαHμ,0is bounded;

iiψCϕ :ApαHμ,0is compact;

iiiψHμ,0.

Acknowledgment

The author is supported partly by the NSF of Guangdong Province of Chinano. 07006700.

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Their potential role in controlled release is being globally exploited by the scientists with high rate of attainment, thus concept of administration of drug

Dekan: Prof.. Meiner Familie gewidmet.. Bedeutung für die Medizin ... Material und Methoden ... Einteilung des Gesamtkollektivs ... Begutachtungswesen im Allgemeinen

A recent study by Abdur Rehman and others evaluated allergic fungal sinusitis occurrence in patients with nasal polyps and recorded that allergic fungal sinusitis (AFS)

Brake specific fuel consumption for all four tests with and without fuel modification at 1500 rpm For better identification of the test results, charts showing relation of