# Weighted Composition Operators on Some Weighted Spaces in the Unit Ball

Authors: Xiaohong Fu; Xiangling Zhu
Journal: Abstract and Applied Analysis
 (2008)

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## Abstract

Let𝔹n be the unit ball of ℂn, H(𝔹n) the space of all holomorphic functions  in 𝔹n. Let u𝞊H(𝔹n) and α be a holomorphic self-map of 𝔹n. For  f𝞊H(𝔹n), the weigthed composition operator uCα is defined by (uCαf)(z)=u(z)f(α(z)),z𝞊𝔹n. The boundedness and compactness of the weighted composition operator
on some weighted spaces on the unit ball are studied in this paper.


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## Body

## 1. Introduction

Let𝔹n be the unit ball of ℂn, H(𝔹n) the space of all holomorphic functions in 𝔹n,
and H∞=H∞(𝔹n) the space of all bounded holomorphic functions
in the unit ball. For f∈H(𝔹n),
let(1.1)ℜf(z)=∑j=1nzj∂f∂zj(z)be the radial derivative of f.A positive continuous functionμ on [0,1) is called normal if there exist positive
numbers α and β,0<α<β, and δ∈[0,1) such that (see, e.g., [1, 2])(1.2)μ(r)(1−r)αisdecreasingon[δ,1),lim⁡r→1μ(r)(1−r)α=0,μ(r)(1−r)βisincreasingon[δ,1),lim⁡r→1μ(r)(1−r)β=∞.Anf∈H(𝔹n) is said to belong to the weighted-type space,
denoted by Hμ∞=Hμ∞(𝔹n),
if(1.3)∥f∥Hμ∞=sup⁡z∈Bnμ(|z|)|f(z)|<∞,where μ is normal on [0,1) (see [3]). Hμ∞ is a Banach space with the norm ∥⋅∥Hμ∞.The little weighted-type space, denoted byHμ,0∞,
is the subspace of Hμ∞ consisting of those f∈Hμ∞ such that(1.4)lim⁡|z|→1μ(|z|)|f(z)|=0.
When μ(r)=(1−r2)α,
the induced spaces Hμ∞ and Hμ,0∞ become the (classical) weighted
spaces
Hα∞ and Hα,0∞ respectively.Anf∈H(𝔹n) is said to belong to the logarithmic-type
space Hlog∞ if(1.5)∥f∥Hlog⁡∞=sup⁡z∈Bn⁢|f(z)|log⁡(e/(1−|z|2))<∞.It is easy to see that Hlog∞ becomes a Banach space under the norm ∥⋅∥Hlog∞,
and that the following inclusions hold:(1.6)H∞⊂ℬ⊂Hlog⁡∞⊂Hα∞,⁢α>0,where ℬ is the Bloch space defined by(1.7)ℬ={f∈H(Bn):sup⁡z∈Bn(1−|z|2)|ℜf(z)|<∞}.For some information on the
Bloch and related spaces see, for example, [4–13] and the references therein.
For some information on the space Hlog∞ in the unit disk see [14].Letu∈H(𝔹n),
and let φ be a holomorphic self-map of 𝔹n.
For f∈H(𝔹n),
the weighted composition operator uCφ is defined by(1.8)(uCφf)(z)=u(z)f(φ(z)),z∈Bn.The weighted composition
operator can be regarded as a generalization of a multiplication operator and a
composition operator, which is defined by (Cφf)(z)=f(φ(z)). The work in [15] contains much information on
this topic.In the
setting of the unit ball, Zhu studied the
boundedness and compactness of the weighted composition operator between
Bergman-type spaces andH∞ in [16]. More general results can be found in [17, 18]. Some necessary and
sufficient conditions for the weighted composition operator to be bounded and
compact between the Bloch space and H∞ are given in [19]. In the setting of the unit
polydisk, some necessary and sufficient conditions for a weighted composition
operator to be bounded and compact between the Bloch space and H∞(𝔹1n) are given in [20, 21] (see also [22] for the case of composition
operators). Other related results can be found, for example, in [3, 23–32].In this paper, we study the weighted composition
operator fromHlog∞ to the spaces Hμ∞ and Hlog∞.
Some necessary and sufficient conditions for the weighted composition operator uCφ to be bounded and compact are given.Throughout the paper, constants are denoted byC;
they are positive and may not be the same in every occurrence.


## 2. Main Results and Proofs

In this section, we give our main results and their
proofs. Before stating these results, we need some auxiliary results, which are
incorporated in the lemmas which follow.Lemma 2.1.
Assume thatu∈H(𝔹n), φ is a holomorphic self-map of 𝔹n,
and μ is a normal function on [0,1).
Then, uCφ:Hlog∞→Hμ∞ is compact if and only if uCφ:Hlog∞→Hμ∞ is bounded, and for any bounded sequence (fk)k∈ℕ in Hlog∞ which converges to zero uniformly on compact
subsets of 𝔹n as k→∞,
one has
∥uCφfk∥Hμ∞→0 as k→∞.The proof of Lemma2.1
follows by standard arguments (see, e.g., [15, Proposition 3.11] as
well as the proofs of the corresponding results in [7, 22, 33, 34]). Hence, we omit the
details.Lemma 2.2.
Assume thatμ is normal. A closed set K in Hμ,0∞ is compact if and only if it is bounded and
satisfies (2.1)lim⁡|z|→1⁢sup⁡f∈K⁢μ(|z|)|f(z)|=0.The proof of Lemma2.2 is similar to the proof of
Lemma
1 in [35]. We omit the
details.Now, we are in a position to state and prove our main
results.Theorem 2.3.
Assume thatu∈H(𝔹n), φ is a holomorphic self-map of 𝔹n,
and μ is normal on [0,1).
Then, uCφ:Hlog∞→Hμ∞ is bounded if and only if (2.2)M=sup⁡z∈Bnμ(|z|)|u(z)|log⁡e1−|φ(z)|2<∞.Proof.
Assume thatuCφ:Hlog∞→Hμ∞ is bounded. For a∈𝔹n,
set(2.3)fa(z)=log⁡e1−〈z,a〉.It is easy to see that fa∈Hlog∞ and supa∈𝔹n∥fa∥Hlog∞<∞.
For anyb∈𝔹n,
we have(2.4)∞>∥uCφfφ(b)∥Hμ∞=sup⁡z∈Bnμ(|z|)|(uCφfφ(b))(z)|=sup⁡z∈Bnμ(|z|)|u(z)||fφ(b)(φ(z))|≥μ(|b|)|u(b)|log⁡e1−|φ(b)|2,which implies (2.2).
Conversely, assume that (2.2) holds. Then, for any f∈Hlog∞,
we have(2.5)μ(|z|)|(uCφf)(z)|=μ(|z|)|u(z)||f(φ(z))|≤μ(|z|)|u(z)|log⁡e1−|φ(z)|2∥f∥Hlog⁡∞.Taking the supremum in (2.5) over 𝔹n and using condition (2.2), the boundedness of
the operator uCφ:Hlog∞→Hμ∞ follows, as desired.Theorem 2.4.
Assume thatu∈H(𝔹n), φ is a holomorphic self-map of 𝔹n,
and μ is a normal function on [0,1).
Then, uCφ:Hlog∞→Hμ∞ is compact if and only if u∈Hμ∞ and (2.6)lim⁡|φ(z)|→1μ(|z|)|u(z)|log⁡e1−|φ(z)|2=0.Proof.
Assume thatuCφ:Hlog∞→Hμ∞ is compact. Then, it is obvious that uCφ:Hlog∞→Hμ∞ is bounded. Taking the function f(z)=1∈Hlog∞,
we see that u∈Hμ∞.
Let (φ(zk))k∈ℕ be a sequence in 𝔹n such that limk⇀∞|φ(zk)|=1.
Set(2.7)fk(z)=(log⁡e1−〈z,φ(zk)〉)2(log⁡e1−|φ(zk)|2)−1,k∈ℕ.It is easy to see that supk∈ℕ∥fk∥Hlog∞<∞.
Moreover, fk→0 uniformly on compact subsets of 𝔹n as k→∞.
By Lemma 2.1,(2.8)lim⁡k→∞∥uCφfk∥Hμ∞=0.We have(2.9)∥uCφfk∥Hμ∞=sup⁡z∈Bn⁢μ(|z|)|u(z)fk(φ(z))|≥μ(|zk|)|u(zk)|log⁡e1−|φ(zk)|2,which together with (2.8) implies
that(2.10)lim⁡k→∞μ(|zk|)|u(zk)|log⁡e1−|φ(zk)|2=0.This proves that (2.6) holds.
Conversely, assume thatu∈Hμ∞ and (2.6) holds. From this, it follows that (2.2)
holds; hence uCφ:Hlog∞→Hμ∞ is bounded. In order to prove that uCφ:Hlog∞→Hμ∞ is compact, according to Lemma 2.1, it suffices
to show that if (fk)k∈ℕ is a bounded sequence in Hlog∞ converging to 0 uniformly on compact subsets
of 𝔹n,
then(2.11)lim⁡k→∞∥uCφfk∥Hμ∞=0.Let (fk)k∈ℕ be a bounded sequence in Hlog∞ such that fk→0 uniformly on compact subsets of 𝔹n as k→∞. By (2.6), we have that for any ε>0,
there is a constant δ∈(0,1) such that(2.12)μ(|z|)|u(z)|log⁡e1−|φ(z)|2<εwhenever δ<|φ(z)|<1.
Let K={w∈𝔹n:|w|≤δ}.
Equation (2.12) along with the fact that u∈Hμ∞ implies(2.13)∥uCφfk∥Hμ∞=sup⁡z∈Bn⁢μ(|z|)|(uCφfk)(z)|=sup⁡z∈Bn⁢μ(|z|)|u(z)fk(φ(z))|≤(sup⁡{z∈Bn:|φ(z)|≤δ}+sup⁡{z∈Bn:δ≤|φ(z)|<1})μ(|z|)|u(z)||fk(φ(z))|≤∥u∥Hμ∞sup⁡w∈K|fk(w)|+sup⁡{z∈Bn:δ≤|φ(z)|<1}μ(|z|)|u(z)|log⁡e1−|φ(z)|2∥fk∥Hlog⁡∞≤∥u∥Hμ∞sup⁡w∈K|fk(w)|+Cε.Observe that K is a compact subset of 𝔹n so that(2.14)lim⁡k→∞⁢sup⁡w∈K|fk(w)|=0.With the aid of the above
inequality, we can deduce that(2.15)limsup⁡k→∞∥uCφfk∥Hμ∞≤Cεby letting k→∞.
Since ε is an arbitrary positive number, it follows
that the last limit is equal to zero. Therefore, uCφ:Hlog∞→Hμ∞ is compact. The proof is complete.Theorem 2.5.
Assume thatu∈H(𝔹n), φ is a holomorphic self-map of 𝔹n,
and μ is a normal function on [0,1).
Then, uCφ:Hlog∞→Hμ,0∞ is compact if and only if u∈Hμ,0∞ and (2.16)lim⁡|φ(z)|→1μ(|z|)|u(z)|log⁡e1−|φ(z)|2=0.Proof.
Assume thatuCφ:Hlog∞→Hμ,0∞ is compact. Then, it is clear that uCφ:Hlog∞→Hμ∞ is compact, and hence (2.16) holds. In addition,
taking the function given by f(z)=1,
we get u∈Hμ,0∞.
Conversely, suppose thatu∈Hμ,0∞ and (2.16) holds. In the proof of the
implication we follow the lines, for example, of the proof of Lemma 4.2 in
[24]. From (2.16), it
follows that for every ε>0,
there exists a δ∈(0,1) such that(2.17)μ(|z|)|u(z)|log⁡e1−|φ(z)|2<εwhen δ<|φ(z)|<1. From the assumption u∈Hμ,0∞,
we have that for the above ε,
there exists an r∈(0,1) such that when r<|z|<1,(2.18)μ(|z|)|u(z)|<εlog⁡(e/(1−δ2)).Therefore, if r<|z|<1 and δ<|φ(z)|<1,
we obtain(2.19)μ(|z|)|u(z)|log⁡e1−|φ(z)|2<ε.If |φ(z)|≤δ and r<|z|<1, we have that(2.20)μ(|z|)|u(z)|log⁡e1−|φ(z)|2≤μ(|z|)|u(z)|log⁡e1−δ2<ε.Combining (2.19) with
(2.20), we
get(2.21)lim⁡|z|→1μ(|z|)|u(z)|log⁡e1−|φ(z)|2=0. On the other hand, from (1.5) we
have thatμ(|z|)|(uCφf)(z)|≤μ(|z|)|u(z)|log⁡e1−|φ(z)|2∥f∥Hlog⁡∞.Taking the supremum in the above
inequality over all f∈Hlog∞ such that ∥f∥Hlog∞≤1, then letting |z|→1,
by (2.21) it follows that(2.22)lim⁡|z|→1⁢sup⁡∥f∥Hlog⁡∞≤1μ(|z|)|(uCφ(f))(z)|=0.From this and by employing Lemma
2.2, we see that uCφ:Hlog∞→Hμ,0∞ is compact. The proof is complete.Similar to the proofs of Theorems2.3 and
2.4, we easily
get the following two results. We also omit their proofs.Theorem 2.6.
Assume
thatu∈H(𝔹n) and φ is a holomorphic self-map of 𝔹n.
Then, the following statements hold.
(a)uCφ:Hlog∞→Hlog∞ is bounded if and only if(2.23)sup⁡z∈Bn|u(z)|log(e/(1−|φ(z)|2))⁡log(e/(1−|z|2))⁡<∞.
(b)uCφ:Hlog∞→Hlog∞ is compact if and only if u∈Hlog∞ and(2.24)lim⁡|φ(z)|→1|u(z)|log(e/(1−|φ(z)|2))⁡log(e/(1−|z|2))⁡=0.Theorem 2.7.
Assume thatu∈H(𝔹n) and φ is a holomorphic self-map of 𝔹n.
Then, the following statements hold.
(a)uCφ:Hlog∞→H∞ is bounded if and only if(2.25)sup⁡z∈Bn|u(z)|log⁡e1−|φ(z)|2<∞.
(b)uCφ:Hlog∞→H∞ is compact if and only if u∈H∞ and(2.26)lim⁡|φ(z)|→1|u(z)|log⁡e1−|φ(z)|2=0.



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