# On a New Integral-Type Operator from the Weighted Bergman Space to

Authors: Stevo Stević
Journal: Discrete Dynamics in Nature and Society
 (2008)

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## Abstract

We introduce an integral-type operator, denoted byPφg,
on the space of holomorphic functions on the unit ball B⊂ℂn, which is an extension of the product of composition and
integral operators on the unit disk. The operator norm of
Pφg from the weighted Bergman space Aαp(B) to the
Bloch-type space ℬμ(B) or the little Bloch-type
space ℬμ,0(B) is calculated. The compactness
of the operator is characterized in terms of inducing functions
g and φ. Upper and lower bounds for the essential norm
of the operator Pφg:Aαp(B)→ℬμ(B),
when p>1, are also given.


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## Body

## 1. Introduction

Let𝔹 be the open unit ball in the complex vector
space ℂn,S=∂𝔹 its boundary, 𝔻 the open unit disk in the complex plane ℂ,dV(z) the Lebesgue measure on 𝔹,dVα(z)=cα(1−|z|2)αdV(z), where α>−1 and where the constant cα is chosen such that Vα(𝔹)=1,dσ the normalized rotation invariant measure on S (that is, σ(S)=1), H(𝔹) the class of all holomorphic functions on the
unit ball and H∞=H∞(𝔹) the space of all bounded holomorphic functions
on 𝔹 with the norm(1.1)∥f∥∞=sup⁡z∈B|f(z)|.Letz=(z1,…,zn) and w=(w1,…,wn) be points in ℂn,(1.2)〈z,w〉=∑k=1nzkw¯kand |z|=〈z,z〉.Forf∈H(𝔹) with the Taylor expansion f(z)=∑|β|≥0aβzβ,
let(1.3)ℜf(z)=∑|β|≥0|β|aβzβbe the radial derivative of f, where β=(β1,β2,…,βn) is a multi-index, |β|=β1+⋯+βn and zβ=z1β1⋯znβn.
It is well known (see, e.g., [1]) that(1.4)ℜf(z)=∑j=1nzj∂f∂zj(z).Forp>0 the Hardy space Hp=Hp(𝔹) consists of all f∈H(𝔹) such that(1.5)∥f∥pp=sup⁡0<r<1⁢∫S|f(rζ)|pdσ(ζ)<∞.It is well known that for every f∈Hp the radial limit(1.6)f*(ζ):=lim⁡r→1f(rζ)exists almost everywhere on ζ∈S.The weighted Bergman spaceAαp=Aαp(𝔹),p>0,α>−1,
consists of all f∈H(𝔹) such that(1.7)∥f∥Aαpp=∫B|f(z)|pdVα(z)<∞.When p≥1,
the weighted Bergman space with the norm ∥⋅∥Aαp becomes a Banach space. If p∈(0,1),
it is a Frechet space with the translation invariant metric(1.8)d(f,g)=∥f−g∥Aαpp.Since for everyf∈Hp(1.9)lim⁡α→−1+0∫B|f(z)|pdVα(z)=∫S|f*(ζ)|pdσ(ζ),we will also use the notation A−1p for the Hardy space Hp.A positive continuous functionϕ on [0,1) is called normal (see [2]) if there is δ∈[0,1) and a and b,0<a<b such that(1.10)ϕ(r)(1−r)ais⁢decreasing⁢on⁢[δ,1),lim⁡r→1ϕ(r)(1−r)a=0;ϕ(r)(1−r)bis⁢increasing⁢on⁢[δ,1),lim⁡r→1ϕ(r)(1−r)b=∞.From now on if we say that a
function μ:𝔹→[0,∞) is normal, we will also assume that it is
radial, that is, μ(z)=μ(|z|),z∈𝔹.The weighted spaceHμ∞=Hμ∞(𝔹) consists of all f∈H(𝔹) such that(1.11)∥f∥Hμ∞=sup⁡z∈B⁢μ(z)|f(z)|<∞,where μ is normal. For μ(z)=(1−|z|2)β, β>0,
we obtain the weighted space Hβ∞=Hβ∞(𝔹) (see, e.g., [3–5]).The little weighted spaceHμ,0∞=Hμ,0∞(𝔹) is a subspace of Hμ∞ consisting of all f∈H(𝔹) such that(1.12)lim⁡|z|→1μ(z)|f(z)|=0.The class of allf∈H(𝔹) such that(1.13)Bμ(f)=sup⁡z∈B⁢μ(z)|ℜf(z)|<∞,where μ is normal, is called the Bloch-type space, and
is denoted by ℬμ=ℬμ(𝔹).
With the norm(1.14)∥f∥ℬμ=|f(0)|+Bμ(f),the Bloch-type space becomes a
Banach space.The little Bloch-type spaceℬμ,0 is a subspace of ℬμ consisting of those f∈ℬμ such that(1.15)lim⁡|z|→1μ(z)|ℜf(z)|=0.The α-Bloch space ℬα is obtained for μ(z)=(1−|z|2)α,α∈(0,∞) (see, e.g., [6–11]). For α=1 the space ℬ1=ℬ becomes the classical Bloch space.Letφ be a holomorphic self-map of 𝔹. For any f∈H(𝔹), the composition operator is defined by(1.16)Cφf(z)=f(φ(z)),z∈B.It is of interest to provide
function theoretic characterizations when φ induces bounded or compact composition
operators on spaces of holomorphic functions. For some classical results in the
topic (see, e.g., [12]). For some recent results see, for example, [3–5, 7, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23] and the references therein.Letg∈H(𝔻) and φ be a holomorphic self-map of 𝔻. For f∈H(𝔻), products of integral-type and composition
operator are defined as follows:(1.17)CφJgf(z)=∫0φ(z)f(ζ)g′(ζ)dζ,JgCφf(z)=∫0zf(φ(ζ))g′(ζ)dζ.Whenφ(z)=z, operators in (1.17) are reduced to the integral operator introduced in [24].
For some other results on the operator; see, for example, [25, 26], and related references
therein. Some results on related integral-type operators on spaces of holomorphic
functions in ℂn can be found, for example, in [27–41]
(see also the references therein).In [42], among other results, we proved the following theorem
regarding the boundedness of the operator JgCφ:Aαp(𝔻)→ℬμ(𝔻).Theorem 1.1.
Assume thatp>0,α>−1,g∈H(𝔻),μ is normal, and φ is a holomorphic self-map of 𝔻.
Then JgCφ:Aαp(𝔻)→ℬμ(𝔻) is bounded if and only if(1.18)sup⁡z∈D⁢μ(z)|g′(z)|(1−|φ(z)|2)(α+2)/p<∞.One of the interesting questions is to extend
operators in (1.17) in the unit ball settings and to study their function
theoretic properties on spaces of holomorphic functions on the unit ball in
terms of inducing functions.Assume thatg∈H(𝔹),g(0)=0,
and φ is a holomorphic self-map of 𝔹.
We introduce the following important integral-type operator on the space of
holomorphic functions on 𝔹: (1.19)Pφg(f)(z)=∫01f(φ(tz))g(tz)dtt,f∈H(B),z∈B.First note that whenn=1,
the operator is reduced to an operator of the form as the second operator in
(1.17). Indeed, since g∈H(𝔻) and g(0)=0,
it follows that g(z)=zg0(z),z∈𝔻 for some g0∈H(𝔻). By using this fact and the change of variables ζ=tz,
we obtain(1.20)Pφgf(z)=∫01f(φ(tz))tzg0(tz)dtt=∫0zf(φ(ζ))g0(ζ)dζ.Hence operator (1.19) is a natural
extension of the second operator in (1.17).Now we formulate the following big research project
related to the operatorPφg.Research Project 1.
LetX and Y be two Banach spaces of holomorphic functions
on the unit ball in ℂn (e.g., the weighted Bergman space Aαp,
the Bloch-type space ℬμ,
the Hardy space Hp space, the weighted space Hμ∞,
the Besov space Bp,
BMOA etc.) Characterize the boundedness, compactness, essential norms, and
other operator theoretic properties of the operator Pφg:X→Y in terms of function theoretic properties of
inducing functions φ and g.Another interesting question is to find the exact
value of the norm of operators on spaces of holomorphic functions. Majority of
papers in the area only find asymptotics of the operator norm of certain linear
operators on some spaces of holomorphic functions. There are a few papers which
calculate the operator norm of these operators. Recently in [4] we calculated operator norm of the weighted
composition operator uCφ mapping the Bloch space ℬ to the weighted space Hμ∞, which motivates us to find the norms of
weighted composition and other closely related operators between various spaces
of holomorphic functions.Research Project 2.
LetX and Y be two Banach spaces of holomorphic functions
as in Research project 1. Calculate the operator norm of Pφg:X→Y in terms of inducing functions φ and g.In this paper, among other results, we will calculate
the operator norm ofPφg:Aαp(𝔹)→ℬμ(𝔹).
We will also characterize the boundedness, compactness, and the essential norm
of the operator. These results partially solve problems posed in the above
research projects.Throughout the paper,C denotes a positive constant not necessarily
the same at each occurrence. The notation A≍B means that there is a positive constant C such that A/C≤B≤CA.


## 2. Auxiliary Results

In this section, we give several auxiliary results,
which are used in the proofs of the main results.Lemma 2.1 (see [43, Corollary 3.5]).
Suppose thatp∈(0,∞) and α≥−1.
Then for all f∈Aαp(𝔹) and z∈𝔹,
the following inequality holds:(2.1)|f(z)|≤∥f∥Aαp(1−|z|2)(n+1+α)/p.The following criterion for the compactness follows by
standard arguments (see, e.g., [12, 20, 34–36]). Hence, we omit its proof.Lemma 2.2.
Suppose
that0<p<∞,α≥−1,g∈H(𝔹),μ is normal, and φ is a holomorphic self-map of 𝔹.
Then the operator Pφg:Aαp→ℬμ is compact if and only if Pφg:Aαp→ℬμ is bounded and for every bounded sequence (fk)k∈ℕ in Aαp converging to zero uniformly on compacts of 𝔹, one has  ∥Pφgfk∥ℬμ→0 as k→∞.The following result can be found in [44]. For closely related
results, see also [11, 45–52] and the references therein.Lemma 2.3.
Suppose
that0<p<∞,α>−1,
then(2.2)∥f∥Aαpp≍|f(0)|p+∫B|∇f(z)|p(1−|z|2)p+αdV(z),for every f∈Aαp (here ∇f=((∂f/∂z1),…,(∂f/∂zn))).The following lemma can be proved similar to [53, Lemma 1].Lemma 2.4.
Suppose
thatμ is normal. A closed set K in ℬμ,0 is compact if and only if it is bounded and(2.3)lim⁡|z|→1⁢sup⁡f∈K⁢μ(z)|ℜf(z)|=0.The following lemma is related to [32, Lemma 1] and [34, Lemma 2].Lemma 2.5.
Assume thatf,g∈H(𝔹) and g(0)=0.
Then(2.4)ℜPφg(f)(z)=f(φ(z))g(z).Proof.
Since the functionf(φ(z))g(z) is holomorphic and g(0)=0,
it has the Taylor expansion in the following form ∑α≠0aαzα. Then(2.5)ℜ[Pφg(f)](z)=ℜ∫01∑α≠0aα(tz)αdtt=ℜ(∑α≠0aα|α|zα)=∑α⁢≠0aαzα,as claimed.


## 3. The Norm of the OperatorPφg:Aαp→ℬμ

In this section, we calculate the norm∥Pφg∥Aαp→ℬμ.Theorem 3.1.
Assume thatp>0,α≥−1,g∈H(𝔹),μ is normal, φ is a holomorphic self-map of 𝔹,
and Pφg:Aαp(𝔹)→ℬμ(𝔹) is bounded. Then(3.1)∥Pφg∥Aαp→ℬμ=∥Pφg∥Aαp→ℬμ,0=sup⁡z∈B⁢μ(z)|g(z)|(1−|φ(z)|2)(n+1+α)/p=:M.Proof.
Iff∈Aαp, then by Lemmas 2.5 and 2.1 we obtain(3.2)∥Pφgf∥ℬμ=sup⁡z∈B⁢μ(z)|g(z)f(φ(z))|≤∥f∥Aαpsup⁡z∈B⁢μ(z)|g(z)|(1−|φ(z)|2)(n+1+α)/p,from which it follows that(3.3)∥Pφg∥Aαp→ℬμ≤M.
Now we prove the reverse inequality. Forw∈𝔹 fixed, set(3.4)fw(z)=(1−|w|2)(n+1+α)/p(1−〈z,w〉)2(n+1+α)/p,z∈B.We have that ∥fw∥Aαp=1, for each w∈𝔹. For α>−1 this fact is well known. The proof for the
case α=−1 could be less known, and we give a proof of it
for the lack of a specific reference and for the benefit of the reader. Let z=rζ, ζ∈S,
then we have (3.5)∥fw∥pp=sup⁡0<r<1⁢∫S(1−|w|2)n|1−〈z,w〉|2ndσ(ζ)=(1−|w|2)nsup⁡0<r<1⁢∫S|(1−〈z,w〉)−n|2dσ(ζ)=(1−|w|2)nsup⁡0<r<1⁢∫S|∑k=0∞Γ(n+k)Γ(k+1)Γ(n)rk〈ζ,w〉k|2dσ(ζ)=(1−|w|2)nsup⁡0<r<1⁢∫S∑k=0∞(Γ(n+k)Γ(k+1)Γ(n))2r2k|〈ζ,w〉|2kdσ(ζ)=(1−|w|2)nsup⁡0<r<1⁢∑k=0∞Γ(n+k)Γ(k+1)Γ(n)r2k|w|2k=(1−|w|2)nsup⁡0<r<1⁢1(1−r2|w|2)n=1,where we have used the following
formula (see, e.g., [1])(3.6)∫S|〈ζ,w〉|2kdσ(ζ)=Γ(k+1)Γ(n)Γ(n+k)|w|2k.
From this and the boundedness ofPφg:Aαp→ℬμ,
we have(3.7)∥Pφg∥Aαp→ℬμ=∥fφ(w)∥Aαp∥Pφg∥Aαp→ℬμ≥∥Pφg(fφ(w))∥ℬμ=sup⁡z∈B⁢μ(z)|g(z)||fφ(w)(φ(z))|≥μ(w)|g(w)||fφ(w)(φ(w))|=μ(w)|g(w)|(1−|φ(w)|2)(n+1+α)/p.Taking the supremum in (3.7) over w∈𝔹,
we obtain(3.8)∥Pφg∥Aαp→ℬμ≥M.From (3.3) and (3.8), it follows
that ∥Pφg∥Aαp→ℬμ=M.
Since(3.9)∥Pφg∥Aαp→ℬμ,0≤∥Pφg∥Aαp→ℬμand the proof of (3.8) does not
depend on the space ℬμ (we may replace it by ℬμ,0) the second equality in (3.1) also holds.Corollary 3.2.
Assume thatp>0,α≥−1,g∈H(𝔹),μ is normal, and φ is a holomorphic self-map of 𝔹.
Then Pφg:Aαp→ℬμ is bounded if and only if(3.10)sup⁡z∈B⁢μ(z)|g(z)|(1−|φ(z)|2)(n+1+α)/p<∞.Proof.
IfPφg:Aαp→ℬμ is bounded, then (3.10) follows from Theorem 3.1.
If (3.10) holds, then the boundedness of Pφg:Aαp→ℬμ follows from (3.3).


## 4. The Boundedness of the OperatorPφg:Aαp→ℬμ,0

Here we characterize the boundedness of the operatorPφg:Aαp→ℬμ,0.Theorem 4.1.
Assume thatp>0,α≥−1,g∈H(𝔹),μ is normal, and φ is a holomorphic self-map of 𝔹. Then Pφg:Aαp→ℬμ,0 is bounded if and only if Pφg:Aαp→ℬμ is bounded and g∈Hμ,0∞.Proof.
Assume thatPφg:Aαp→ℬμ is bounded and g∈Hμ,0∞. Then, for each polynomial p,
we have(4.1)μ(z)|ℜPφgp(z)|=μ(z)|g(z)p(φ(z))|≤μ(z)|g(z)|∥p∥∞⟶0,as⁢|z|⟶1from which it follows that Pφg(p)∈ℬμ,0.
Since the set of all polynomials is dense inAαp, we have that for every f∈Aαp there is a sequence of polynomials (pk)k∈ℕ such that(4.2)lim⁡k→∞∥f−pk∥Aαp=0.From this and since the operator Pφg:Aαp→ℬμ is bounded, it follows that(4.3)∥Pφgf−Pφgpk∥ℬμ≤∥Pφg∥Aαp→ℬμ∥f−pk∥Aαp⟶0,as k→∞. Hence Pφg(Aαp)⊂ℬμ,0. Since ℬμ,0 is a closed subset of ℬμ,
the boundedness of Pφg:Aαp→ℬμ,0 follows.
Now assume thatPφg:Aαp→ℬμ,0 is bounded. Then clearly Pφg:Aαp→ℬμ is bounded. Taking the test function f(z)=1∈Aαp,
we obtain g∈Hμ,0∞.


## 5. Compactness of the OperatorPφg:Aαp→ℬμ

This section is devoted to studying of the compactness
of the operatorPφg:Aαp→ℬμ.
We prove the following result.Theorem 5.1.
Assume thatp>0,α≥−1,g∈H(𝔹),μ is normal, φ is a holomorphic self-map of 𝔹,
and the operator Pφg:Aαp→ℬμ is bounded. Then the operator Pφg:Aαp→ℬμ is compact if and only if(5.1)lim⁡|φ(z)|→1μ(z)|g(z)|(1−|φ(z)|2)(n+1+α)/p=0.Proof.
First assume that the operatorPφg:Aαp→ℬμ is compact. If ∥φ∥∞<1, then condition (5.1) is vacuously satisfied.
Hence, assume that ∥φ∥∞=1 and assume to the contrary that (5.1) does not
hold. Then there is a sequence (zk)k∈ℕ satisfying the condition |φ(zk)|→1 as k→∞ and δ>0 such that(5.2)μ(zk)|g(zk)|(1−|φ(zk)|2)(n+1+α)/p≥δ,k∈ℕ.For w∈𝔹 fixed, set(5.3)Fk(z)=fφ(zk)(z),k∈ℕ,where fw is defined in (3.4). Recall that ∥fw∥Aαp=1, for each w∈𝔹. Then ∥Fk∥Aαp=1,k∈ℕ and it is easy to see that Fk→0 uniformly on compacts of 𝔹 as k→∞.
Hence, by Lemma 2.2, it follows that(5.4)lim⁡k→∞∥PφgFk∥ℬμ=0.
On the other hand, by Lemma2.5 and (5.2), we obtain(5.5)∥PφgFk∥ℬμ=sup⁡z∈B⁢μ(z)|g(z)||Fk(φ(z))|≥μ(zk)|g(zk)|(1−|φ(zk)|2)(n+1+α)/p≥δ>0,for every k∈ℕ, which contradicts with (5.4).
Now assume that (5.1) holds. Then for every ε>0 there is an r∈(0,1) such that when r<|φ(z)|<1,(5.6)μ(z)|g(z)|(1−|φ(z)|2)(n+1+α)/p<ε.
On the other hand, since the operatorPφg:Aαp→ℬμ is bounded, for f≡1∈Aαp,
we obtain ∥g∥Hμ∞<∞.
Assume that(hk)k∈ℕ is a bounded sequence in Aαp converging to zero uniformly on compacts of 𝔹 as k→∞. Let supk∈ℕ∥hk∥Aαp=M1. Then by Lemma 2.1 and (5.6), for r<|φ(z)|<1,
we obtain(5.7)μ(z)|g(z)||hk(φ(z))|≤sup⁡k∈ℕ⁢∥hk∥Aαpμ(z)|g(z)|(1−|φ(z)|2)(n+1+α)/p<M1ε.If |φ(z)|≤r, we have(5.8)μ(z)|g(z)||hk(φ(z))|≤∥g∥Hμ∞sup⁡|w|≤r⁢|hk(w)|⟶0,ask⟶∞.From (5.7) and (5.8), it follows
that ∥Pφghk∥ℬμ→0 as k→∞,
from which the compactness of the operator Pφg:Aαp→ℬμ follows.


## 6. Compactness of the OperatorPφg:Aαp→ℬμ,0

This section characterizes the compactness of the
operatorPφg:Aαp→ℬμ,0.Theorem 6.1.
Assumep>0,α≥−1,g∈H(𝔹),μ is normal, φ is a holomorphic self-map of 𝔹,
and the operator Pφg:Aαp→ℬμ,0 is bounded. Then the operator Pφg:Aαp→ℬμ,0 is compact if and only if(6.1)lim⁡|z|→1μ(z)|g(z)|(1−|φ(z)|2)(n+1+α)/p=0.Proof.
AssumePφg:Aαp→ℬμ,0 is compact. Then clearly Pφg:Aαp→ℬμ,0 is bounded and as in Theorem 4.1 we have that g∈Hμ,0∞.
Hence if∥φ∥∞<1,
then(6.2)lim⁡|z|→1μ(z)|g(z)|(1−|φ(z)|2)(n+1+α)/p≤lim⁡|z|→1μ(z)|g(z)|(1−∥φ∥∞2)(n+1+α)/p=0,from which the result follows in
this case.
Now assume∥φ∥∞=1.
By using the test functions Fk(z)=fφ(zk)(z), k∈ℕ,
defined in (5.3) we obtain that condition (5.1) holds, which implies that for
every ε>0,
there is an r∈(0,1) such that for r<|φ(z)|<1,
condition (5.6) holds.
Sinceg∈Hμ,0∞, there is σ∈(0,1) such that for σ<|z|<1(6.3)μ(z)|g(z)|<ε(1−r2)(n+1+α)/p.
Hence, if|φ(z)|≤r and σ<|z|<1,
we have(6.4)μ(z)|g(z)|(1−|φ(z)|2)(n+1+α)/p≤μ(z)|g(z)|(1−r2)(n+1+α)/p<ε.From (5.6) and (6.4), condition
(6.1) follows.
Now assume that condition (6.1) holds. Then the
quantity M in Theorem 3.1 is finite. Using this fact and
the following inequality(6.5)μ(z)|ℜPφgf(z)|≤μ(z)|g(z)f(φ(z))|≤∥f∥Aαpμ(z)|g(z)|(1−|φ(z)|2)(n+1+α)/p,it follows that the set Pφg({f:∥f∥Aαp≤1}) is bounded in ℬμ, moreover, in view of (6.1), it is bounded in ℬμ,0. Taking the supremum in the last inequality
over the unit ball in Aαp,
then letting |z|→1,
using condition (6.1) and employing Lemma 2.4, we obtain the compactness of the
operator Pφg:Aαp→ℬμ,0, as desired.


## 7. Essential Norm ofPφg:Aαp→ℬμ

LetX and Y be Banach spaces, and let L:X→Y be a bounded linear operator. The essential
norm of the operator L:X→Y,
denoted by ∥L∥e,X→Y,
is defined as follows:(7.1)∥L∥e,X→Y=inf⁡{∥L+K∥X→Y:K⁢is⁢compact⁢from⁢XtoY},where ∥⋅∥X→Y denote the operator norm.From this definition and since the set of all compact
operators is a closed subset of the set of bounded operators, it follows that
operatorL is compact if and only if ∥L∥e,X→Y=0.In this section, we study the essential norm of the
operatorPφg:Aαp→ℬμ for the case p>1.Theorem 7.1.
Assume thatp∈(1,∞),α≥−1,g∈H(𝔹),g(0)=0,φ is a holomorphic self-map of 𝔹,
and Pφg:Aαp→ℬμ is bounded. Then the following inequalities
hold:(7.2)limsup⁡|φ(z)|→1⁢μ(z)|g(z)|(1−|φ(z)|2)(n+1+α)/p≤∥Pφg∥e,Aαp→ℬμ≤2limsup⁡|φ(z)|→1⁢μ(z)|g(z)|(1−|φ(z)|2)(n+1+α)/p.Proof.
Assume that(φ(zk))k∈ℕ is a sequence in 𝔹 such that |φ(zk)|→1 as k→∞.
Note that the sequence (fφ(zk))k∈ℕ (where fw is defined in (3.4)) is such that ∥fφ(zk)∥Aαp=1 for each k∈ℕ and it converges to zero uniformly on compacts
of 𝔹. From this and by
[11, Theorems 2.12 and 4.50], it follows that fφ(zk)→0 weakly in Aαp,
as k→∞ (here we use the condition p>1). Hence, for every compact operator K:Aαp→ℬμ,
we have that ∥Kfφ(zk)∥ℬμ→0 as k→∞. Thus, for every such sequence and for every
compact operator K:Aαp→ℬμ,
we have that(7.3)∥Pφg+K∥Aαp→ℬμ≥limsup⁡k→∞⁢∥Pφgfφ(zk)∥ℬμ−∥Kfφ(zk)∥ℬμ∥fφ(zk)∥Aαp=limsup⁡k→∞⁢∥Pφgfφ(zk)∥ℬμ≥limsup⁡k→∞⁢μ(zk)|g(zk)fφ(zk)(φ(zk))|=limsup⁡n→∞⁢μ(zk)|g(zk)|(1−|φ(zk)|2)(n+1+α)/p.
Taking the infimum in (7.3) over the set of all compact
operators K:Aαp→ℬμ,
we obtain(7.4)∥Pφg∥e,Aαp→ℬμ≥limsup⁡n→∞⁢μ(zk)|g(zk)|(1−|φ(zk)|2)(n+1+α)/p,from which the first inequality
follows.
In the sequel we prove the second inequality. Assume
that(rl)l∈ℕ is a sequence which increasingly converges to
1. Consider the operators defined by(7.5)(Prlφgf)(z)=∫01g(tz)f(rlφ(tz))dtt,⁢l∈ℕ.We prove that these operators
are compact. Indeed, since |rlφ(z)|≤rl<1,
it follows that condition (5.1) in Theorem 5.1 is vacuously satisfied, from which
the claim follows.
Recall thatg∈Hμ∞. Let ρ∈(0,1) be fixed for a moment. Employing Lemma 2.1, and
using the fact(7.6)∥f−frl∥Aαp≤2∥f∥Aαp,l∈ℕ,which follows by using the
triangle inequality for the norm, the monotonicity of the integral means(7.7)Mpp(f,r)=∫S|f(rζ)|pdσ(ζ)and the polar coordinates, we
have (7.8)∥Pφg−Prlφg∥Aαp→ℬμ=sup⁡∥f∥Aαp≤1⁢sup⁡z∈B⁢μ(z)|g(z)||f(φ(z))−f(rlφ(z))|≤sup⁡∥f∥Aαp≤1⁢sup⁡|φ(z)|≤ρ⁢μ(z)|g(z)||f(φ(z))−f(rlφ(z))|+sup⁡∥f∥Aαp≤1⁢sup⁡|φ(z)|>ρ⁢μ(z)|g(z)||f(φ(z))−f(rlφ(z))|≤∥g∥Hμ∞sup⁡∥f∥Aαp≤1⁢sup⁡|φ(z)|≤ρ⁢|f(φ(z))−f(rlφ(z))|+2⁢sup⁡|φ(z)|>ρ⁢μ(z)|g(z)|(1−|φ(z)|2)(n+1+α)/p.
Let(7.9)Il:=sup⁡∥f∥Aαp≤1⁢sup⁡|φ(z)|≤ρ⁢|f(φ(z))−f(rlφ(z))|.If α>−1,
then by using the mean value theorem, the subharmonicity of the partial
derivatives of f and Lemma 2.3, we have (7.10)Il≤sup⁡∥f∥Aαp≤1⁢sup⁡|φ(z)|≤ρ⁢(1−rl)|φ(z)|sup⁡|w|≤ρ⁢|∇f(w)|≤Cρ(1−rl)sup⁡∥f∥Aαp≤1⁢(∫|w|≤(1+ρ)/2|∇f(w)|p(1−|w|2)p+αdV(w))1/p≤Cρ(1−rl)sup⁡∥f∥Aαp≤1⁢(∫B|f(w)|p(1−|w|2)αdV(w))1/p(7.11)≤Cρ(1−rl)⟶0⁢asl⟶∞.
Ifα=−1,
then applying in (7.10) the known fact that for each compact K⊂𝔹,(7.12)sup⁡w∈K⁢|∇f(w)|≤C∥f∥p, for some C independent of f (see [11]), we obtain that (7.11) also holds in this case.
Lettingl→∞ in (7.8), using (7.11), and then letting ρ→1,
the second inequality in (7.2) follows, finishing the proof of the theorem.Motivated by Theorem7.1, we leave the following open
problem.Open Problem 1.
Find the
exact value of the essential norm of the operatorPφg:Aαp→ℬμ.



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*Source: 154263-2008-09-14.xml*