# Weighted Composition Operators fromF(p,q,s) Spaces to Hμ∞ Spaces

Authors: Xiangling Zhu
Journal: Abstract and Applied Analysis
 (2009)

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## Abstract

LetH(B) denote the space of all holomorphic functions on the
unit ball B. Let u∈H(B) and φ be a holomorphic
self-map of B. In this paper, we investigate the boundedness and
compactness of the weighted composition operator uCφ from
the general function space F(p,q,s) to the weighted-type space
Hμ∞ in the unit ball.


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## Body

## 1. Introduction

LetB be the unit
ball of ℂn. Let z=(z1,…,zn) and
let
w=(w1,…,wn) be points   in ℂn, we write(1.1)|z|=|z1|2+⋯+|zn|2,〈z,w〉=z1w¯1+⋯+znw¯n.Thus B={z∈ℂn:|z|<1}.  Let dv be the
normalized Lebesgue measure of B, that is, v(B)=1. Let H(B) be the space of
all holomorphic   functions  on B. For f∈H(B), let(1.2)ℜf(z)=∑j=1nzj∂f∂zj(z)represent the radial derivative
of f∈H(B). For a,z∈B, a≠0, let φa denote the
Möbius transformation of B taking 0 to a, which is defined by(1.3)φa(z)=a−Pa(z)−1−|z|2Qa(z)1−〈z,a〉,where Pa(z) is the
orthogonal projection of z onto the one
dimensional subspace of ℂn spanned by a, and Qa(z)=z−Pa(z).A positive continuous functionμ on [0,1) is called
normal, if there exist positive numbers s and t,0<s<t, and δ∈[0,1) such that(1.4)μ(r)(1−r)sisdecreasingon[δ,1),limr→1μ(r)(1−r)s=0;μ(r)(1−r)tisincreasingon[δ,1),limr→1μ(r)(1−r)t=∞ (see [1]).Letμ be a normal
function on [0,1). An f∈H(B) is said to
belong to the weighted-type space Hμ∞=Hμ∞(B), if(1.5)∥f∥Hμ∞=supz∈Bμ(|z|)|f(z)|<∞,where μ is normal on [0,1) (see, e.g.,
[2–4]). Hμ∞ is a Banach
space with the norm ∥⋅∥Hμ∞.  We denote by Hμ,0∞ the subspace of Hμ∞ consisting of
those f∈Hμ∞ such that(1.6)lim|z|→1μ(|z|)|f(z)|=0.When μ(r)=(1−r2)α, the induced spaces Hμ∞ and Hμ,0∞ become the
(classical) weighted space Hα∞ and Hα,0∞, respectively.Forα>0, recall that the α-Bloch space ℬα=ℬα(B) is the space of
all f∈H(B) for which (see
[5])(1.7)bα(f)=supz∈B(1−|z|2)α|ℜf(z)|<∞.Under the norm ∥f∥ℬα=|f(0)|+bα(f), ℬα is a Banach
space. When α=1, we get the classical Bloch space ℬ. For more information of the Bloch space and the α-Bloch space
(see, e.g., [5–8] and the references therein).Forp∈(0,∞), the weighted Bergman space Ap(B) is the space of
all holomorphic functions f on B for
which(1.8)∥f∥App=∫B|f(z)|pdv(z)<∞.The Hardy space Hp(B)(0<p<∞) on the unit
ball is defined by(1.9)Hp(B)={f∣f∈H(B),∥f∥Hp(B)=sup0≤r<1Mp(f,r)<∞},where(1.10)Mp(f,r)=(∫∂B|f(rζ)|pdσ(ζ))1/pand dσ is the
normalized surface measure on ∂B.For0<p<∞, the Qp space is
defined by (see [9])(1.11)Qp(B)={f∈H(B):∥f∥Qp2=supa∈B∫B|∇˜f(z)|2Gp(z,a)dv(z)(1−|z|2)n+1<∞}.Here ∇˜f(z)=∇(f∘φz)(0) denotes the
invariant gradient of f, and G(z,a) is the
invariant Green function defined by G(z,a)=g(φa(z)), where(1.12)g(z)=n+12n∫|z|1(1−t2)n−1t−2n+1dt.Let0<p,s<∞, −n−1<q<∞. A function f∈H(B) is said to
belong to F(p,q,s)=F(p,q,s)(B) (see [10–12]) if(1.13)∥f∥F(p,q,s)p=|f(0)|p+supa∈B∫B|∇f(z)|p(1−|z|2)qGs(z,a)dv(z)<∞.F(p,q,s) is called the
general function space since we can get many function spaces, such as Hardy
space, Bergman space, Bloch space, Qp space, if we
take special parameters of p,q,s. For example, F(2,1,0)=H2, F(p,p,0)=Ap, and F(2,0,s)=Qs. If q+s≤−1, then F(p,q,s) is the space of
constant functions. For the setting of the unit disk, see [13].Letu∈H(B) and φ be a
holomorphic self-map of B. For f∈H(B), the weighted composition operator uCφ is defined by(1.14)(uCφf)(z)=u(z)f(φ(z)),z∈B.The weighted composition
operator is the generalization of a multiplication operator and a composition
operator, which is defined by (Cφf)(z)=f(φ(z)). The main
subject in the study of composition operators is to describe operator theoretic
properties of Cφ in terms of
function theoretic properties of φ. The book [14] is a good reference for the theory of composition
operators. Recall that a linear operator is said to be bounded if the image of
a bounded set is a bounded set, while a linear operator is compact if it takes
bounded sets to sets with compact closure.In the setting of the unit ball, we studied the
boundedness and compactness of the weighted composition operator between
Bergman-type spaces andH∞ in [15]. More general results can
be found in [16]. Some
necessary and sufficient conditions for the weighted composition operator to be
bounded or compact between the Bloch space and H∞ are given in
[17]. In the setting
of the unit polydisk Dn, some necessary and sufficient conditions for a
weighted composition operator to be bounded or compact between the Bloch space
and H∞(Dn) are given in
[18, 19] (see, also [20] for the case of composition
operators). Some related results can be found, for example, in [2, 3, 6, 21–31].In the present paper, we are mainly concerned about
the boundedness and compactness of the weighted composition operator fromF(p,q,s) to the space Hμ∞. Some necessary and sufficient conditions for the
weighted composition operator uCφ to be bounded
and compact are given.Constants are denoted byC in this paper,
they are positive and may differ from one occurrence to the other. a⪯b means that
there is a positive constant C such that a≤Cb. Moreover, if both a⪯b and b⪯a hold, then one
says that a≍b.


## 2. Main Results and Proofs

In order to prove our results, we need some auxiliary
results which are incorporated in the following lemmas.Lemma 2.1 (see [12]).
For0<p,s<∞,   −n−1<q<∞,  q+s>−1, if  f∈F(p,q,s), then  f∈ℬ(n+1+q)/p
and (2.1)∥f∥ℬ(n+1+q)/p≤C∥f∥F(p,q,s).The following lemma can be found, for example, in
[32].Lemma 2.2.
Iff∈ℬα, then (2.2)|f(z)|≤C{∥f∥ℬα,0<α<1,∥f∥ℬαlne1−|z|2,α=1,∥f∥ℬα(1−|z|2)α−1,α>1,for some C independent of f.Lemma 2.3.
Assume thatμ  is normal. A
closed set  KinHμ,0∞is compact if
and only if it is bounded and satisfies(2.3)lim|z|→1supf∈Kμ(|z|)|f(z)|=0.The proof of Lemma2.3 is similar to the proof of Lemma
1 of   [33].  We omit the
details.Lemma 2.4.
Assume thatu∈H(B),  φis a
holomorphic self-map ofB,  μ
is a normal
function on  [0,1),  0<p,s<∞,  −n−1<q<∞,  q+s>−1. Then  uCφ:F(p,q,s)→Hμ∞
is compact if
and only if  uCφ:F(p,q,s)→Hμ∞is bounded and
for any bounded sequence(fk)k∈ℕinF(p,q,s)which converges
to zero uniformly on compact subsets ofBask→∞,  one has
∥uCφfk∥Hμ∞→0ask→∞.Lemma2.4  follows by standard arguments similar to those
outlined in [14, Proposition 3.11] (see, also, the corresponding lemmas in [20, 34, 35]). We omit the details.Note that whenp>n+1+q, the function in F(p,q,s) is indeed
Lipschitz continuous by Lemmas 2.1 and 2.2. By Arzela-Ascoli theorem, similarly to
the proof of Lemma 3.6 of  [26], we have the following result.Lemma 2.5.
Let0<p,s<∞,  −n−1<q<∞,  q+s>−1,  and  p>n+1+q. Let  (fk)be a bounded
sequence inF(p,q,s)
which converges
to 0 uniformly on compact subsets of  B, then (2.4)limk→∞supz∈B|fk(z)|=0.


## 2.1. Casep<n+1+q

In this subsection, we consider the casep<n+1+q. Our first result is the following theorem.Theorem 2.6.
Assume thatu∈H(B),  φis a
holomorphic self-map ofB,  μis a normal
function on[0,1),  0<p,s<∞,  −n−1<q<∞,  q+s>−1,  p<n+1+q. ThenuCφ:F(p,q,s)→Hμ∞is bounded if
and only if(2.5)supz∈Bμ(|z|)|u(z)|(1−|φ(z)|2)(n+1+q)/p−1<∞.Moreover, when uCφ:F(p,q,s)→Hμ∞ is bounded, the
following relationship(2.6)∥uCφ∥F(p,q,s)→Hμ∞≍supz∈Bμ(|z|)|u(z)|(1−|φ(z)|2)(n+1+q)/p−1holds.Proof.
Assume that (2.5) holds. For any f∈F(p,q,s), by Lemmas 2.1 and 2.2,(2.7)∥uCφf∥Hμ∞=supz∈Bμ(|z|)|(uCφf)(z)|=supz∈Bμ(|z|)|f(φ(z))||u(z)|≤C∥f∥ℬ(n+1+q)/psupz∈Bμ(|z|)|u(z)|(1−|φ(z)|2)(n+1+q)/p−1≤C∥f∥F(p,q,s)supz∈Bμ(|z|)|u(z)|(1−|φ(z)|2)(n+1+q)/p−1.Therefore, (2.5) implies that uCφ:F(p,q,s)→Hμ∞ is bounded.
Conversely, suppose thatuCφ:F(p,q,s)→Hμ∞ is bounded. For b∈B, let(2.8)fb(z)=1−|b|2(1−〈z,b〉)(n+1+q)/p.It is easy to see that(2.9)fφ(w)(φ(w))=1(1−|φ(w)|2)(n+1+q)/p−1,|ℜfφ(w)(φ(w))|=(n+1+q)|φ(w)|2p(1−|φ(w)|2)(n+1+q)/p.If φ(w)=0, then fφ(w)≡1 obviously
belongs to F(p,q,s). From [12], we know that fb∈F(p,q,s); moreover, there is a positive constant K such that supb∈B∥fb∥F(p,q,s)≤K. Therefore,(2.10)supz∈Bμ(|z|)|fφ(w)(φ(z))u(z)|=supz∈Bμ(|z|)|(uCφfφ(w))(z)|=∥uCφfφ(w)∥Hμ∞≤K∥uCφ∥F(p,q,s)→Hμ∞,for every w∈B, from which we get(2.11)supw∈Bμ(|w|)|u(w)|(1−|φ(w)|2)(n+1+q)/p−1≤K∥uCφ∥F(p,q,s)→Hμ∞<∞,that is, (2.5) follows. From (2.7)
and (2.11), we see that (2.6) holds. The proof of this theorem is finished.Theorem 2.7.
Assume thatu∈H(B),  φ  is a
holomorphic self-map of  B,  μ
is a normal
function on  [0,1),  0<p,s<∞,  −n−1<q<∞,  q+s>−1,  p<n+1+q.  Then  uCφ:F(p,q,s)→Hμ∞
is compact if
and only if  u∈Hμ∞
and (2.12)lim|φ(z)|→1μ(|z|)|u(z)|(1−|φ(z)|2)(n+1+q)/p−1=0.Proof.
First assume thatu∈Hμ∞ and the
condition in (2.12) hold. In order to
prove that uCφ is compact,
according to Lemma 2.4 it suffices to show that if (fk)k∈ℕ is bounded in F(p,q,s) and converges
to 0 uniformly on
compact subsets of B as k→∞, then ∥uCφfk∥Hμ∞→0 as k→∞.
Now assume that(fk)k∈ℕ is a sequence
in F(p,q,s) such that supk∈ℕ∥fk∥F(p,q,s)≤L and fk→0 uniformly on
compact subsets of B as k→∞. From (2.12), we
have that for every ε>0, there is a constant δ∈(0,1) such that(2.13)μ(|z|)|u(z)|(1−|φ(z)|2)(n+1+q)/p−1<ε,when δ<|φ(z)|<1. By Lemmas 2.1 and 2.2, we have(2.14)∥uCφfk∥Hμ∞=supz∈Bμ(|z|)|(uCφfk)(z)|=supz∈Bμ(|z|)|u(z)||fk(φ(z))|≤supφ(z)∈B(0,δ)¯μ(|z|)|u(z)||fk(φ(z))|+Csupφ(z)∈B∖B(0,δ)¯μ(|z|)|u(z)|(1−|φ(z)|2)(n+1+q)/p−1∥fk∥F(p,q,s)≤M1supφ(z)∈B(0,δ)¯|fk(φ(z))|+CLε,where M1:=supz∈Bμ(|z|)|u(z)|<∞. Using the fact that fk→0 uniformly on
compact subsets of B as k→∞, we obtain(2.15)M1lim supk→∞supφ(z)∈B(0,δ)¯|fk(φ(z))|=0.Therefore,(2.16)lim supk→∞∥uCφfk∥Hμ∞≤CLε.Since ε is an arbitrary
positive number, we have that limk→∞∥uCφfk∥Hμ∞=0, and, therefore, uCφ:F(p,q,s)→Hμ∞ is compact by
Lemma 2.4.
Conversely, supposeuCφ:F(p,q,s)→Hμ∞ is compact. Let (zk)k∈ℕ be a sequence
in B such that |φ(zk)|→1 as k→∞ (if such a
sequence does not exist that condition (2.12) is vacuously satisfied). Set(2.17)fk(z)=1−|φ(zk)|2(1−〈z,φ(zk)〉)(q+n+1)/p,k∈ℕ.From the proof of Theorem 2.6, we
see that fk∈F(p,q,s) for every k∈ℕ; moreover, supk∈ℕ∥fk∥F(p,q,s)≤C. Beside this, fk converges to 0
uniformly on compact subsets of B as k→∞. Since uCφ is compact, by
Lemma 2.4 we have that ∥uCφfk∥Hμ∞→0 as k→∞. Thus(2.18)(1−|φ(zk)|2)1−(q+n+1)/pμ(|zk|)|u(zk)|=μ(|zk|)|u(zk)||fk(φ(zk))|≤supz∈Bμ(|z|)|fk(φ(z))||u(z)|=supz∈Bμ(|z|)|(uCφfk)(z)|=∥uCφfk∥Hμ∞→0,as k→∞, from which we obtain (2.12), finishing the proof of the theorem.Theorem 2.8.
Assume thatu∈H(B),  φis a
holomorphic self-map ofB,  μis a normal
function on[0,1),  0<p,s<∞,  −n−1<q<∞,  q+s>−1,  p<n+1+q. ThenuCφ:F(p,q,s)→Hμ,0∞is compact if
and only if(2.19)lim|z|→1μ(|z|)|u(z)|(1−|φ(z)|2)(n+1+q)/p−1=0.Proof.
Suppose that (2.19) holds. From Lemma 2.3, we see that uCφ:F(p,q,s)→Hμ,0∞ is compact if
and only if(2.20)lim|z|→1sup∥f∥F(p,q,s)≤1μ(|z|)|(uCφf)(z)|=0.On the other hand, by Lemmas 2.1
and 2.2, we have that(2.21)μ(|z|)|(uCφf)(z)|≤C∥f∥F(p,q,s)μ(|z|)|u(z)|(1−|φ(z)|2)(n+1+q)/p−1.Taking the supremum in (2.21) over
the the unit ball in the space F(p,q,s), then letting |z|→1 and applying
(2.19) the result follows.
Conversely, suppose thatuCφ:F(p,q,s)→Hμ,0∞ is compact.
Then uCφ:F(p,q,s)→Hμ∞ is compact and
hence by Theorem 2.7,(2.22)lim|φ(z)|→1μ(|z|)|u(z)|(1−|φ(z)|2)(n+1+q)/p−1=0.In addition, uCφ:F(p,q,s)→Hμ,0∞ is bounded.
Taking f(z)=1, then employing the boundedness of uCφ:F(p,q,s)→Hμ,0∞, we get(2.23)lim|z|→1μ(|z|)|u(z)|=0.By (2.22), for every ε>0, there exists a δ∈(0,1),(2.24)μ(|z|)|u(z)|(1−|φ(z)|2)(n+1+q)/p−1<ε,when δ<|φ(z)|<1. By (2.23), for
above chosen ε, there exists r∈(0,1),(2.25)μ(|z|)|u(z)|≤ε(1−δ2)(n+1+q)/p−1,when r<|z|<1.
Therefore, whenr<|z|<1 and δ<|φ(z)|<1, we have that(2.26)μ(|z|)|u(z)|(1−|φ(z)|2)(n+1+q)/p−1<ε.If |φ(z)|≤δ and r<|z|<1, we obtain(2.27)μ(|z|)|u(z)|(1−|φ(z)|2)(n+1+q)/p−1≤1(1−δ2)(n+1+q)/p−1μ(|z|)|u(z)|<ε.Combing (2.26) with (2.27), we get
(2.19), as desired.


## 2.2. Casep=n+1+q

Theorem 2.9.
Assume thatu∈H(B),  φis a
holomorphic self-map ofB,  μis a normal
function on[0,1),  0<p,s<∞,  −n−1<q<∞,  q+s>−1,  p=n+1+q. ThenuCφ:F(p,q,s)→Hμ∞is bounded if
and only if(2.28)supz∈Bμ(|z|)|u(z)|lne1−|φ(z)|2<∞.Moreover, the following
relationship(2.29)∥uCφ∥F(p,q,s)→Hμ∞≍supz∈Bμ(|z|)|u(z)|lne1−|φ(z)|2holds.Proof.
Suppose that (2.28) holds. For any f∈F(p,q,s)⊆ℬ, by Lemmas 2.1 and 2.2, we have(2.30)∥uCφf∥Hμ∞=supz∈Bμ(|z|)|(uCφf)(z)|=supz∈Bμ(|z|)|f(φ(z))||u(z)|≤C∥f∥F(p,q,s)supz∈Bμ(|z|)|u(z)|lne1−|φ(z)|2.Therefore, (2.28) implies that uCφ is a bounded
operator from F(p,q,s) to Hμ∞.
Conversely, suppose thatuCφ:F(p,q,s)→Hμ∞ is bounded. For b∈B, let(2.31)fb(z)=lne1−〈z,b〉,z∈B.Then by [12] we see that fb∈F(p,q,s); moreover, there is a positive constant K such that supb∈B∥fb∥F(p,q,s)≤K. Hence(2.32)μ(|w|)|u(w)|lne1−|φ(w)|2=μ(|w|)|fφ(w)(φ(w))||u(w)|≤∥uCφfφ(w)∥Hμ∞,for every w∈B, that is, we get(2.33)supw∈Bμ(|w|)|u(w)|lne1−|φ(w)|2≤C∥uCφfφ(w)∥Hμ∞≤CK∥uCφ∥F(p,q,s)→Hμ∞<∞.From (2.30) and (2.33), (2.29)
follows. The proof is finished.Theorem 2.10.
Assume thatu∈H(B),  φis a
holomorphic self-map ofB,  μis a normal
function on[0,1),  0<p,s<∞,  −n−1<q<∞,  q+s>−1,  p=n+1+q. ThenuCφ:F(p,q,s)→Hμ∞is compact if
and only ifu∈Hμ∞and(2.34)lim|φ(z)|→1μ(|z|)|u(z)|lne1−|φ(z)|2=0.Proof.
Assume thatu∈Hμ∞ and
(2.34) hold, and that (fk)k∈ℕ is bounded in F(p,q,s) and converges
to 0 uniformly on compact subsets of B as k→∞. We have that, for every ε>0, there is a δ∈(0,1) such that(2.35)μ(|z|)|u(z)|lne1−|φ(z)|2<ε,when δ<|φ(z)|<1.
In addition(2.36)∥uCφfk∥Hμ∞=supz∈Bμ(|z|)|(uCφfk)(z)|=supz∈Bμ(|z|)|u(z)||fk(φ(z))|≤supφ(z)∈B(0,δ)¯μ(|z|)|u(z)fk(φ(z))|+C∥fk∥F(p,q,s)supφ(z)∈B∖B(0,δ)¯μ(|z|)|u(z)|lne1−|φ(z)|2≤Csupφ(z)∈B(0,δ)¯|fk(φ(z))|+Cε.Similar to the proof of Theorem
2.7, we obtain ∥uCφfk∥Hμ∞→0 as k→∞. Therefore, uCφ:F(p,q,s)→Hμ∞ is compact by
Lemma 2.4.
Conversely, suppose thatuCφ:F(p,q,s)→Hμ∞ is compact.
Assume that (zk)k∈ℕ is a sequence
in B such that |φ(zk)|→1 as k→∞ (if such a
sequence does not exist that condition (2.34) is vacuously satisfied). Set(2.37)fk(z)=(lne1−|φ(zk)|2)−1(lne1−〈z,φ(zk)〉)2,k∈ℕ.After some calculations or from
[12], we see that supk∈ℕ∥fk∥F(p,q,s)≤C for some
positive C independent of k, and fk converges to 0
uniformly on compact subsets of B as k→∞. Since uCφ is compact, by
Lemma 2.4, we have ∥uCφfk∥Hμ∞→0 as k→∞. Thus(2.38)μ(|zk|)|u(zk)|lne1−|φ(zk)|2≤supz∈Bμ(|z|)|fk(φ(zk))||u(zk)|=supz∈Bμ(|z|)|(uCφfk)(z)|=∥uCφfk∥Hμ∞→0,as k→∞, which is equivalent to (2.34). The proof of this
theorem is finished.Similarly to the proof of Theorem2.8, we can obtain the
following results. We omit the details of the proof.Theorem 2.11.
Assume thatu∈H(B),  φis a
holomorphic self-map ofB,  μis a normal
function on[0,1),  0<p,s<∞,  −n−1<q<∞,  q+s>−1,  p=n+1+q. ThenuCφ:F(p,q,s)→Hμ,0∞is compact if
and only if(2.39)lim|z|→1μ(|z|)|u(z)|lne1−|φ(z)|2<∞.


## 2.3. Casep>n+1+q

Theorem 2.12.
Assume thatu∈H(B),  φis a
holomorphic self-map ofB,  μis a normal
function on[0,1),  0<p,s<∞,  −n−1<q<∞,  q+s>−1,  p>n+1+q. Then the following statements are equivalent:(i)
uCφ:F(p,q,s)→Hμ∞ is bounded;(ii)
uCφ:F(p,q,s)→Hμ∞ is compact;(iii)
u∈Hμ∞.Proof.
(ii)⇒(i) This implication is obvious.
(i)⇒(iii) Taking f(z)=1, then using the boundedness of uCφ:F(p,q,s)→Hμ∞ the implication
follows.
(iii)⇒(ii) Suppose that u∈Hμ∞. For an f∈F(p,q,s), by Lemma 2.1, we see that f is continuous
on the closed unit ball and so is bounded in B. Therefore,(2.40)μ(|z|)|(uCφf)(z)|=μ(|z|)|f(φ(z))||u(z)|≤C∥f∥F(p,q,s)μ(|z|)|u(z)|.From the above inequality, we
see that uCφ:F(p,q,s)→Hμ∞ is bounded. Let (fk)k∈ℕ be any bounded
sequence in F(p,q,s) and fk→0 uniformly on
compact subsets of B as k→∞. Employing Lemma 2.5, we have(2.41)∥uCφfk∥Hμ∞=supz∈Bμ(|z|)|fk(φ(z))u(z)|≤∥u∥Hμ∞supz∈B|fk(φ(z))|→0,as k→∞. Then the result follows from Lemma 2.3.Theorem 2.13.
Assume thatu∈H(B),  φis a
holomorphic self-map ofB,  μis a normal
function on[0,1),  0<p,s<∞,  −n−1<q<∞,  q+s>−1,  p>n+1+q. Then the following statements are equivalent:(i)
uCφ:F(p,q,s)→Hμ,0∞  is bounded;(ii)
uCφ:F(p,q,s)→Hμ,0∞ is compact;(iii)
u∈Hμ,0∞.Proof.
(ii)⇒(i) It is obvious.
(i)⇒(iii) Taking f(z)=1, then using the boundedness of uCφ:F(p,q,s)→Hμ,0∞, we get the desired result.
(iii)⇒(ii) Suppose that u∈Hμ,0∞. For any f∈F(p,q,s) with ∥f∥F(p,q,s)≤1, we have(2.42)μ(|z|)|(uCφf)(z)|≤C∥f∥F(p,q,s)μ(|z|)|u(z)|≤Cμ(|z|)|u(z)|,from which we obtain(2.43)lim|z|→1sup∥f∥F(p,q,s)≤1μ(|z|)|(uCφf)(z)|≤Clim|z|→1μ(|z|)|u(z)|=0.Using Lemma 2.3, we see that uCφ:F(p,q,s)→Hμ,0∞ is compact, as
desired.



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*Source: 290978-2009-04-29.xml*