# Integral-Type Operators from Bloch-Type Spaces toQK Spaces

Authors: Stevo Stević; Ajay K. Sharma
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

The boundedness and compactness of the integral-type operatorIφ,g(n)f(z)=∫0zf(n)(φ(ζ))g(ζ)dζ, where n∈ℕ0, φ is a holomorphic self-map of the unit disk 𝔻, and g is a holomorphic function on 𝔻, from α-Bloch spaces to QK spaces are characterized.


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## Body

## 1. Introduction

Let𝔻 be the open unit disk in the complex plane, ∂𝔻 be its boundary, D(w,r) be disk centered at w of radius r, and let H(𝔻) be the class of all holomorphic functions on 𝔻. Let
(1.1)ηa(z)=a-z1-a¯z,a∈D,
be the involutive Möbius transformation which interchanges points 0 and a. If X is a Banach space, then by BX we will denote the closed unit ball in X.Theα-Bloch space, ℬα(𝔻)=ℬα,α>0, consists of all f∈H(𝔻) such that
(1.2)sup⁡z∈D(1-|z|2)α|f′(z)|<∞.
The little α-Bloch space ℬ0α(𝔻)=ℬ0α consists of all functions f holomorphic on 𝔻 such that lim|z|→1(1-|z|2)α|f′(z)|=0. The norm on ℬα is defined by
(1.3)‖f‖Bα=|f(0)|+sup⁡z∈D(1-|z|2)α|f′(z)|.
With this norm, ℬα is a Banach space, and the little α-Bloch space ℬ0α is a closed subspace of the α-Bloch space. Note that ℬ1=ℬ is the usual Bloch space.Given a nonnegative Lebesgue measurable functionK on (0,1] the space QK consists of those f∈H(𝔻) for which
(1.4)bQK2(f)=sup⁡a∈D∫D|f′(z)|2K(1-|ηa(z)|2)dm(z)<∞,
where dm(z)=(1/π)dxdy=(1/π)rdrdθ is the normalized area measure on 𝔻 [1]. It is known that bQK is a seminorm on QK which is Möbius invariant, that is,
(1.5)bQK(f∘η)=bQK(f),η∈Aut(D),
where Aut(𝔻) is the group of all automorphisms of the unit disk 𝔻. It is a Banach space with the norm defined by
(1.6)‖f‖QK=|f(0)|+bQK(f).The spaceQK,0 consists of all f∈H(𝔻) such that
(1.7)lim⁡|a|→1∫D|f′(z)|2K(1-|ηa(z)|2)dm(z)=0.
It is known that QK,0 is a closed subspace of QK. For classical Q spaces, see [2].It is clear that eachQK contains all constant functions. If QK consists of just constant functions, we say that it is trivial. QK is nontrivial if and only if
(1.8)sup⁡t∈(0,1)∫01K(1-r)(1-t)2(1-tr2)3rdr<∞.
Throughout this paper, we assume that condition (1.8) is satisfied, so that the space QK is nontrivial. An important tool in the study of QK spaces is the auxiliary function λK defined by
(1.9)λK(s)=sup⁡0<t≤1K(st)K(t),0<s<∞,
where the domain of K is extended to (0,∞) by setting K(t)=K(1) when t>1. The next two conditions play important role in the study of QK spaces. (a)
There is a constantC>0 such that for allt>0(1.10)K(2t)≤CK(t).(b)
The auxiliary functionλK satisfies the following condition:
(1.11)∫01λK(s)sds<∞.LetΩ(0,∞) denote the class of all nondecreasing continuous functions on (0,∞) satisfying conditions (1.8), (1.10), and (1.11).A positive Borel measureμ on 𝔻 is called a K-Carleson measure [3] if
(1.12)sup⁡I∫S(I)K(1-|z||I|)dμ(z)<∞,
where the supermum is taken over all subarcs I⊂∂𝔻, |I| is the length of I, and S(I) is the Carleson box defined by
(1.13)S(I)={z:1-|I|<|z|<1,z|z|∈I}.A positive Borel measureμ is called a vanishing K-Carleson measure if
(1.14)lim⁡|I|→0∫S(I)K(1-|z||I|)dμ(z)=0.We also need the following results of Wulan and Zhu in [3], in which QK spaces are characterized in terms of K-Carleson measures.Theorem 1.1.
LetK∈Ω(0,∞). Then a positive Borel measure  μ on  𝔻 is a  K-Carleson measure if and only if
(1.15)sup⁡a∈D∫DK(1-|ηa(z)|2)dμ(z)<∞.
Also, μ is a vanishing K-Carleson measure if and only if
(1.16)lim⁡|a|→1∫DK(1-|ηa(z)|2)dμ(z)=0.From Theorem1.1 and the definition of the spaces QK and QK,0, we see that when K∈Ω(0,∞), then f∈QK if and only if the measure dμf=|f′(z)|2dm(z) is a K-Carleson measure, while f∈QK,0 if and only if this measure is a vanishing K-Carleson measure.Letφ∈S(𝔻) be the family of all holomorphic self-maps of 𝔻, g∈H(𝔻), and n∈ℕ0. We define an integral-type operator as follows:
(1.17)Iφ,g(n)f(z)=∫0zf(n)(φ(ζ))g(ζ)dζ,z∈D.
Operator (1.17) extends several operators which has been introduced and studied recently (see, e.g., [4–9]). For related operators in n-dimensional case, see, for example, [10–19]. For some classical operators see, for example, [20, 21] and the related references therein. For other product-type operators, see [22] and the references therein.Motivated by [23, 24] (see also [25–29]), we characterize when φ and g induce bounded and/or compact operators in (1.17) from α-Bloch to QK spaces.Throughout this paper, constants are denoted byC; they are positive and not necessarily the same at each occurrence. The notation A≍B means that there is a positive constant C such that B/C≤A≤CB.


## 2. Auxiliary Results

Here, we quote several lemmas which will be used in the proofs of the main results in this paper. The following lemma is folklore (see, e.g., [30]).Lemma 2.1.
For anyf∈H(𝔻) and  z∈𝔻, the following inequalities hold
(2.1)|f(z)|≤C{‖f‖Bα,if0<α<1,‖f‖Bαln⁡e1-|z|2,ifα=1,‖f‖Bα(1-|z|2)α-1,ifα>1,(2.2)|f(n)(z)|≤Csup⁡w∈D(z,(1-|z|)/2)(1-|w|2)α|f′(w)|(1-|z|2)α+n-1≤C‖f‖Bα(1-|z|2)α+n-1,ifn∈N.The next lemma is obtained in [31, 32].Lemma 2.2.
Letα>0. Then there are two functions  f1,f2∈ℬα such that
(2.3)|f1′(z)|+|f2′(z)|≥C(1-|z|2)α,z∈D.
Also, if α≠1, then there are two functions f3,f4∈ℬα and C>0, such that
(2.4)|f3(z)|+|f4(z)|≥C(1-|z|2)α-1,z∈D.The next Schwartz-type lemma [33] is proved in a standard way, so we omit the proof.Lemma 2.3.
Letα>0,K∈Ω(0,∞),φ∈S(𝔻),g∈H(𝔻), and  n∈ℕ0. Then  Iφ,g(n):ℬα(orℬ0α)→QK is compact if and only if for any bounded sequence  (fm)m∈ℕ in  ℬα converging to zero on compacts  of 𝔻, we have  limm→∞∥Iφ,g(n)fm∥QK=0.Lemma 2.4.
Letα>0,K∈Ω(0,∞),φ∈S(𝔻),g∈H(𝔻), and  n∈ℕ0. Then  Iφ,g(n):ℬ0α→QK(orQK,0) is weakly compact if and only if it is compact.Proof.
By a known theoremIφ,g(n):ℬ0α→QK(orQK,0) is weakly compact if and only if (Iφ,g(n))*:QK*(orQK,0*)→(ℬ0α)* is weakly compact. Since (ℬ0α)*≅A1 (the Bergman space) and A1 has the Schur property, it follows that it is equivalent to (Iφ,g(n))*:QK*(orQK,0*)→(ℬ0α)*, is compact, which is equivalent to Iφ,g(n):ℬ0α→QK(orQK,0), is compact, as claimed.Lemma 2.5.
Letα>0,K∈Ω(0,∞),φ∈S(𝔻),g∈H(𝔻), and  n∈ℕ0. Then  Iφ,g(n):ℬ0α→QK,0 is compact if and only if  Iφ,g(n):ℬα→QK,0 is bounded.Proof.
By Lemma2.4, Iφ,g(n):ℬ0α→QK,0 is compact if and only if it is weakly compact, which, by Gantmacher's theorem ([34]), is equivalent to (Iφ,g(n))**((ℬ0α)**)⊆QK,0. Since (ℬ0α)**=ℬα and by a standard duality argument (Iφ,g(n))**=Iφ,g(n) on ℬα, this can be written as Iφ,g(n)(ℬα)⊆QK,0, which by the closed graph theorem is equivalent to Iφ,g(n):ℬα→QK,0 is bounded.Fora∈𝔻, set
(2.5)Φφ,g,K(a)=∫DK(1-|ηa(z)|2)|g(z)|2(1-|φ(z)|2)2(1-α-n)dm(z).Lemma 2.6.
Letα>0,K∈Ω(0,∞),φ∈S(𝔻),g∈H(𝔻), and  n∈ℕ0. If  Φφ,g,K is finite at some point  a∈𝔻, then it is continuous on  𝔻.Proof.
We follow the lines of Lemma  2.3 in [24]. From the elementary inequality
(2.6)(1-|a|)(1-|a1|)4≤1-|ηa(z)|21-|ηa1(z)|2≤4(1-|a|)(1-|a1|),a,a1,z∈D,
and since K is nondecreasing and satisfies (1.10), we easily get
(2.7)K(1-|ηa1(z)|2)≤C[log⁡2(4/(1-|a|)(1-|a1|))]+1K(1-|ηa(z)|2).
From (2.7) and since Φφ,g,K(a) is finite, it follows that Φφ,g,K is finite at each point a1∈𝔻. Let a∈𝔻 be fixed, and let (al)l∈ℕ⊂𝔻 be a sequence converging to a.
We have(2.8)|Φφ,g,K(a)-Φφ,g,K(al)|≤∫D|g(z)|2|K(1-|ηa(z)|2)-K(1-|ηal(z)|2)|(1-|φ(z)|2)2(α+n-1)dm(z).
From (2.6), we have that for l such that 1-|al|≥(1-|a|)/2, say l≥l0, holds
(2.9)1-|ηal(z)|2≤8(1-|a|)2(1-|ηa(z)|2),
and consequently for l≥l0, it holds
(2.10)|K(1-|ηa(z)|2)-K(1-|ηal(z)|2)|≤(1+C[log⁡2(8/(1-|a|)2)]+1)K(1-|ηa(z)|2).
From this and sinceΦφ,g,K is finite at a, by the Lebesgue dominated convergence theorem, we get that the integral in (2.8) converges to zero as l→∞ which implies that Φφ,g,K(al)→Φφ,g,K(a) as l→∞, from which the lemma follows.


## 3. Boundedness and Compactness ofIφ,g(n):ℬα(orℬ0α)→QK(orQK,0)

In this section, we characterize the boundedness and compactness of the operatorsIφ,g(n):ℬα(orℬ0α)→QK(orQK,0). Let
(3.1)dμφ,g,n,α(z)=|g(z)|2(1-|φ(z)|2)2(1-α-n)dm(z).Theorem 3.1.
Letα>0,K∈Ω(0,∞),φ∈S(𝔻),g∈H(𝔻), and  n∈ℕ, or  n=0 and  α>1. Then the following statements are equivalent. (a)
Iφ,g(n):ℬα→QK is bounded.(b)
Iφ,g(n):ℬ0α→QK is bounded.(c)
M:=supa∈𝔻∫𝔻K(1-|ηa(z)|2)|g(z)|2(1-|φ(z)|2)2(1-α-n)dm(z)<∞.(d)
dμφ,g,n,α(z) is a K-Carleson measure.Moreover, if Iφ,g(n):ℬα→QK is bounded, then the next asymptotic relations hold
(3.2)‖Iφ,g(n)‖Bα→QK≍‖Iφ,g(n)‖B0α→QK≍M1/2.Proof.
By Theorem1.1, it is clear that (c) and (d) are equivalent.
(c)⇒ (a). Let f∈Bℬα. First note that Iφ,g(n)f(0)=0 for each f∈H(𝔹) and n∈ℕ0. From this and by Lemma 2.1, we have(3.3)‖Iφ,g(n)f‖QK2=sup⁡a∈D∫D|(Iφ,g(n)f)′(z)|2K(1-|ηa(z)|2)dm(z)=sup⁡a∈D∫D|f(n)(φ(z))|2|g(z)|2K(1-|ηa(z)|2)dm(z)≤C‖f‖Bα2sup⁡a∈D∫DK(1-|ηa(z)|2)|g(z)|2(1-|φ(z)|2)2(1-α-n)dm(z),
from which the boundedness of Iφ,g(n):ℬα→QK follows, and moreover
(3.4)‖Iφ,g(n)‖Bα→QK≤CM1/2.
(a)⇒(b). This implication is obvious.
(b)⇒(c). By Lemma 2.2, if n∈ℕ, there are two functions f1,f2∈ℬα such that (2.3) holds, and if n=0 and α>1 such that (2.4) holds. Let(3.5)h1(z)=f1(z)-∑k=1n-1f1(k)(0)k!zk,h2(z)=f2(z)-∑k=1n-1f2(k)(0)k!zk.
It is known (see [30]) that for each f∈H(𝔻) and n∈ℕ, we have
(3.6)(1-|z|2)α+n-1|f(n)(z)|+∑k=1n-1|f(k)(0)|≍(1-|z|2)α|f′(z)|.
From this, Lemma2.2, and since h1(k)(0)=h2(k)(0)=0, k=0,1,…,n-1, we have that there is a δ>0 such that(3.7)C(1-|z|2)-(α+n-1)≤|h1(n)(z)|+|h2(n)(z)|,for|z|>δ.
Now note that for any f∈ℬα, the functions fr(z)=f(rz), r∈(0,1) belong to ℬα, and moreover, sup0<r<1∥fr∥ℬα≤∥f∥ℬα.
Applying (3.7), using an elementary inequality, the boundedness of Iφ,g(n):ℬ0α→QK, and the last inequality, we obtain(3.8)∫|rφ(z)|>δr2nK(1-|ηa(z)|2)|g(z)|2(1-(r|φ(z)|)2)2(1-α-n)dm(z)≤C∫Dr2nK(1-|ηa(z)|2)|g(z)|2(|h1(n)(rφ(z))|2+|h2(n)(rφ(z))|2)dm(z)=C∫DK(1-|ηa(z)|2)|(Iφ,g(n)(h1)r)′(z)|2dm(z)+C∫DK(1-|ηa(z)|2)|(Iφ,g(n)(h2)r)′(z)|2dm(z)≤‖Iφ,g(n)‖B0α→QK2(‖h1‖Bα2+‖h2‖Bα2).
Letting r→1 in (3.8) and using the monotone convergence theorem, we get
(3.9)∫|φ(z)|>δK(1-|ηa(z)|2)|g(z)|2(1-|φ(z)|2)2(1-α-n)dm(z)≤C‖Iφ,g(n)‖B0α→QK2.
On the other hand, for f0(z)=zn/n!∈ℬ0α, we get Iφ,g(n)f0∈QK which implies
(3.10)sup⁡α∈D⁡∫|φ(z)|≤δK(1-|ηa(z)|2)|g(z)|2(1-|φ(z)|2)2(1-α-n)dm(z)≤‖Iφ,g(n)‖B0α→QK2‖f0‖Bα2(1-δ2)2(α+n-1).
From (3.9) and (3.10), (c) follows. Moreover we get M1/2≤C∥Iφ,g(n)∥ℬ0α→QK. From this, (3.4) and since ∥Iφ,g(n)∥ℬ0α→QK≤∥Iφ,g(n)∥ℬα→QK the asymptotic relations in (3.2) follow, finishing the proof of the theorem.Theorem 3.2.
Letα>0,K∈Ω(0,∞),φ∈S(𝔻),g∈H(𝔻), and  n∈ℕ, or  n=0 and  α>1. Let  Iφ,g(n):ℬα→QK be bounded. Then the following statements are equivalent.  (a)
Iφ,g(n):ℬα→QK is compact.(b)
Iφ,g(n):ℬ0α→QK is compact.(c)
Iφ,g(n):ℬ0α→QK is weakly compact.(d)
supa∈𝔻∫𝔻K(1-|ηa(z)|2)|g(z)|2dm(z)<∞, and(3.11)lim⁡r→1sup⁡a∈D∫|φ(z)|>rK(1-|ηa(z)|2)|g(z)|2(1-|φ(z)|2)2(1-α-n)dm(z)=0.Proof.
By Lemma2.4, we have that (b) is equivalent to (c).
(d)⇒ (a). Let (fl)l∈ℕ be a bounded sequence in ℬα, say by L, converging to zero uniformly on compacts of 𝔻. Then fl(n) also converges to zero uniformly on compacts of 𝔻. From (3.11) we have that for every ɛ>0 there is an r1∈(0,1) such that for r∈(r1,1)(3.12)sup⁡a∈D∫|φ(z)|>rK(1-|ηa(z)|2)|g(z)|2(1-|φ(z)|2)2(1-α-n)dm(z)<ɛ.
Therefore, by Lemma 2.1 and (3.12), we have that for r∈(r1,1)(3.13)‖Iφ,g(n)fl‖QK2=(∫|φ(z)|≤r+∫|φ(z)|>r)|fl(n)(φ(z))|2K(1-|ηa(z)|2)|g(z)|2dm(z)<sup⁡|φ(z)|≤r|fl(n)(φ(z))|2∫DK(1-|ηa(z)|2)|g(z)|2dm(z)+CL2ɛ.
Letting l→∞ in (3.13), using the first condition in (d) and sup|w|≤r|fl(n)(w)|→0 as l→∞, it follows that liml→∞∥Iφ,g(n)fl∥QK=0. Thus, by Lemma 2.3, Iφ,g(n):ℬα→QK is compact.
(a)⇒ (b). The implication is trivial since ℬ0α⊂ℬα.
(b)⇒ (d). By choosing f(z)=zn/n!∈ℬ0α, n∈ℕ0, we have that the first condition in (d) holds. Let fl(z)=zl/l, l∈ℕ. It is easy to see that (fl)l∈ℕ is a bounded sequence in ℬ0α converging to zero uniformly on compacts of 𝔻. Hence, by Lemma 2.3, it follows that ∥Iφ,g(n)(fl)∥QK→0 as l→∞. Thus, for every ɛ>0, there is an l0∈ℕ, l0>n such that forl≥l0(3.14)(∏j=1n-1(l-j))2sup⁡a∈D∫D|φ(z)|2(l-n)K(1-|ηa(z)|2)|g(z)|2dm(z)<ɛ.
From (3.14) we have that for each r∈(0,1) and l≥l0(3.15)r2(l-n)(∏j=1n-1(l-j))2sup⁡a∈D∫|φ(z)|>rK(1-|ηa(z)|2)|g(z)|2dm(z)<ɛ.
Hence, for r∈[(∏j=1n-1(l0-j))-1/(l0-n),1), we have that
(3.16)sup⁡a∈D∫|φ(z)|>rK(1-|ηa(z)|2)|g(z)|2dm(z)<ɛ.
Letf∈Bℬ0α, and let ft(z)=f(tz),0<t<1. Then sup0<t<1∥ft∥ℬα≤∥f∥ℬα, ft∈ℬ0α, t∈(0,1), and ft→f uniformly on compact subsets of 𝔻 as t→1. The compactness of Iφ,g(n):ℬ0α→QK implies(3.17)lim⁡t→1‖Iφ,g(n)ft-Iφ,g(n)f‖QK=0.
Hence, for every ɛ>0, there is a t∈(0,1) such that
(3.18)sup⁡a∈D∫D|ft(n)(φ(z))-f(n)(φ(z))|2K(1-|ηa(z)|2)|g(z)|2dm(z)<ɛ.
From this and (3.16), we have that for such t and each r∈[(∏j=1n-1(l0-j))-1/(l0-n),1)(3.19)sup⁡a∈D∫|φ(z)|>r|f(n)(φ(z))|2K(1-|ηa(z)|2)|g(z)|2dm(z)≤2sup⁡a∈D∫|φ(z)|>r|ft(n)(φ(z))-f(n)(φ(z))|2K(1-|ηa(z)|2)|g(z)|2dm(z)+2sup⁡a∈D∫|φ(z)|>r|ft(n)(φ(z))|2K(1-|ηa(z)|2)|g(z)|2dm(z)<2ɛ(1+‖ft(n)‖∞2).
From (3.19) we conclude that for every f∈Bℬ0α, there is a δ0∈(0,1) and δ0=δ0(f,ɛ) such that for r∈(δ0,1)(3.20)sup⁡a∈D∫|φ(z)|>r|f(n)(φ(z))|2K(1-|ηa(z)|2)|g(z)|2dm(z)<ɛ.
SinceIφ,g(n):ℬ0α→QK is compact, we have that for every ɛ>0 there is a finite collection of functions f1,f2,…,fk∈Bℬ0α such that, for each f∈Bℬ0α, there is a j∈{1,…,k}, such that(3.21)sup⁡a∈D∫D|f(n)(φ(z))-fj(n)(φ(z))|2K(1-|ηa(z)|2)|g(z)|2dm(z)<ɛ.
On the other hand, from (3.20), it follows that if δ̂:=max1≤j≤kδj(fj,ɛ), then for r∈(δ̂,1) and all j∈{1,…,k}, we have
(3.22)sup⁡a∈D∫|φ(z)|>r|fj(n)(φ(z))|2K(1-|ηa(z)|2)|g(z)|2dm(z)<ɛ.
From (3.21) and (3.22), we have that for r∈(δ̂,1) and every f∈Bℬ0α(3.23)sup⁡a∈D∫|φ(z)|>r|f(n)(φ(z))|2K(1-|ηa(z)|2)|g(z)|2dm(z)<4ɛ.
If we apply (3.23) to the delays of the functions in (3.5) which are normalized so that they belong to Bℬα and then use the monotone convergence theorem, we easily get that forr>max{δ,δ̂} where δ is chosen as in (3.7)
(3.24)sup⁡a∈D∫|φ(z)|>rK(1-|ηa(z)|2)|g(z)|2(1-|φ(z)|2)2(1-α-n)dm(z)<Cɛ,
from which (3.11) follows, as desired.Theorem 3.3.
Letα>0,K∈Ω(0,∞),φ∈S(𝔻), g∈H(𝔻) and  n∈ℕ, or  n=0 and  α>1. Then the next statements are equivalent. (a)
Iφ,g(n):ℬα→QK,0 is bounded.(b)
Iφ,g(n):ℬα→QK,0 is compact.(c)
Iφ,g(n):ℬ0α→QK,0 is compact.(d)
Iφ,g(n):ℬ0α→QK,0 is weakly compact.(e)
lim|a|→1∫𝔻K(1-|ηa(z)|2)|g(z)|2(1-|φ(z)|2)2(1-α-n)dm(z)=0.(f)
dμφ,g,n,α(z) is a vanishing K-Carleson measure.Proof.
By Theorem1.1, (e) and (f) are equivalent; by Lemma 2.4, (c) is equivalent to (d), while, by Lemma 2.5, (a) is equivalent to (c). Also (b) obviously implies (a).
(a)⇒ (e) Let h1 and h2 be as in (3.5). Then from (3.7) and an elementary inequality, we get(3.25)∫|φ(z)|>δK(1-|ηa(z)|2)(1-|φ(z)|2)2(1-α-n)|g(z)|2dm(z)≤C∫DK(1-|ηa(z)|2)|(Iφ,g(n)h1)′(z)|2dm(z)+C∫DK(1-|ηa(z)|2)|(Iφ,g(n)h2)′(z)|2dm(z).
For f0(z)=zn/n!∈ℬα, we get Iφ,g(n)f0∈QK,0. From this and since Iφ,g(n)(hj)∈QK,0,j=1,2, by letting |a|→1, we get that (e) holds.
(e)⇒ (b). We have that for every ɛ>0 there is a δ∈(0,1) so that for |a|>δ(3.26)Φφ,g,K(a)<ɛ.
On the other hand, by Lemma 2.6, Φφ,g,K is continuous on |a|≤δ, so uniformly bounded therein, which along with (3.26) gives the boundedness of Φφ,g,K on 𝔻. Hence, by Theorem 3.1, Iφ,g(n):ℬα→QK is bounded. By Lemma 2.1, we have
(3.27)lim⁡|a|→1sup⁡‖f‖Bα≤1⁡∫D|(Iφ,g(n)f)′(z)|2K(1-|ηa(z)|2)dm(z)≤Csup⁡‖f‖Bα≤1⁡‖f‖Bα2lim⁡|a|→1Φφ,g,K(a)=Clim⁡|a|→1Φφ,g,K(a)=0,
so Iφ,g(n):ℬα→QK,0 is bounded.
Now assume that(fl)l∈ℕ is a bounded sequence in ℬα, say by L, converging to zero uniformly on compacta of 𝔻 as l→∞. To show that the operator Iφ,g(n):ℬα→QK,0 is compact, it is enough to prove that there is a subsequence (flk)k∈ℕ of (fl)l∈ℕ such that Iφ,g(n)flk converges in QK,0 as k→∞. By Lemma 2.1 and Montel's theorem, it follows that there is a subsequence, which we may denote again by (fl)l∈ℕ converging uniformly on compacta of 𝔻 to an f∈ℬα, such that ∥f∥ℬα≤L. Since Iφ,g(n)(ℬα)⊆QK,0, then clearly Iφ,g(n)f∈QK,0. We show that(3.28)lim⁡l→∞‖Iφ,g(n)fl-Iφ,g(n)f‖QK=0.
From (3.26), Lemma 2.1, and some simple calculation, we obtain
(3.29)sup⁡δ<|a|<1∫D|(Iφ,g(n)fl(z)-Iφ,g(n)f(z))′|2K(1-|ηa(z)|2)dm(z)<4CL2ɛ.
Fora∈𝔻 and t∈(0,1), let(3.30)Ψt(a)=∫D∖tDK(1-|ηa(z)|2)|g(z)|2(1-|φ(z)|2)2(1-α-n)dm(z).
Lemma 2.6 essentially shows that Ψt is continuous on 𝔻. Hence, for each a∈𝔻, there is a t(a)∈(r,1) such that Ψt(a)(a)<ɛ/2. Moreover, there is a neighborhood 𝒪(a) of a such that, for every b∈𝒪(a), Ψt(a)(b)<ɛ. From this and since the set |a|≤δ is compact, it follows that there is a t0∈(0,1) such that Ψt0(a)<ɛ when |a|≤δ. This along with Lemma 2.1 implies that
(3.31)sup⁡|a|≤δ∫D∖t0D|(Iφ,g(n)fl(z)-Iφ,g(n)f(z))′|2K(1-|ηa(z)|2)dm(z)≤C‖fl-f‖Bα2sup⁡|a|≤δΨt0(a)<4CL2ɛ.
By the Weierstrass theoremfl(n)→f(n) uniformly on compacta as l→∞, from which along with (2.2) and since φ(t0𝔻) is compact, for r=supw∈φ(t0𝔻)|w|, it follows that(3.32)sup⁡|a|≤δ∫t0D|(Iφ,g(n)fl(z)-Iφ,g(n)f(z))′|2K(1-|ηa(z)|2)dm(z)≤Csup⁡|z|≤r|(fl-f)(n)(z)|2sup⁡|a|≤δΦφ,g,K(a)⟶0,asl⟶∞.
From (3.29)–(3.32) and since Iφ,g(n)f(0)=0 for each f∈H(𝔻), we easily get (3.28), from which (b) follows, finishing the proof of this theorem.Theorem 3.4.
Letα>0,K∈Ω(0,∞),φ∈S(𝔻),g∈H(𝔻), and  n∈ℕ, or  n=0 and  α>1.  Then the following statements are equivalent.(a)
Iφ,g(n):ℬ0α→QK,0 is bounded,(b)
supa∈𝔻∫𝔻|g(z)|2K(1-|ηa(z)|2)(1-|φ(z)|2)2(1-α-n)dm(z)<∞, and(3.33)lim⁡|a|→1∫D|g(z)|2K(1-|ηa(z)|2)dm(z)=0.Proof.
Suppose (b) holds andf∈ℬ0α. Then by Theorem 3.1, Iφ,g(n):ℬ0α→QK is bounded. We show Iφ,g(n)f∈QK,0, for every f∈ℬ0α. Since f∈ℬ0α, we have that, for every ɛ>0, there is an r∈(0,1) such that (see, e.g., the idea in [35, Lemma 2.4])
(3.34)|f(n)(φ(z))|2(1-|φ(z)|2)2(α+n-1)<ɛfor|φ(z)|>r.
Thus,
(3.35)sup⁡a∈D∫|φ(z)|>r|(Iφ,g(n)f(z))′|2K(1-|ηa(z)|2)dm(z)<ɛsup⁡a∈D∫DK(1-|ηa(z)|2)(1-|φ(z)|2)2(1-α-n)|g(z)|2dm(z).
We also have
(3.36)lim⁡|a|→1∫|φ(z)|≤r|(Iφ,g(n)f(z))′|2K(1-|ηa(z)|2)dm(z)≤C‖f‖Bα2(1-r2)2(α+n-1)lim⁡|a|→1∫|φ(z)|≤rK(1-|ηa(z)|2)|g(z)|2dm(z)≤C‖f‖Bα2(1-r2)2(α+n-1)lim⁡|a|→1∫DK(1-|ηa(z)|2)|g(z)|2dm(z)=0.
Combining (3.35) and (3.36), we get Iφ,g(n)f∈QK,0. Hence, Iφ,g(n):ℬ0α→QK,0 is bounded.
Conversely, ifIφ,g(n):ℬ0α→QK,0 is bounded, then Iφ,g(n):ℬ0α→QK is bounded too. Thus, by Theorem 3.1, we get the first condition in (b). For f0(z)=zn/n!∈ℬ0α, we get Iφ,g(n)f0∈QK,0, which is equivalent to (3.33), finishing the proof of the theorem.Ifn=0, we simply denote the operator Iφ,g(0) by Iφ,g.Theorem 3.5.
Letα∈(0,1),K∈Ω(0,∞),φ∈S(𝔻), and  g∈H(𝔻). Then the following statements are equivalent.  (a)
Iφ,g:ℬα→QK is bounded.(b)
Iφ,g:ℬ0α→QK is bounded.(c)
M1:=supa∈𝔻∫𝔻K(1-|ηa(z)|2)|g(z)|2dm(z)<∞.(d)
dμ1(z)=|g(z)|2dm(z) is a K-Carleson measure.(e)
Iφ,g:ℬα→QK is compact.(f)
Iφ,g:ℬ0α→QK is compact.(g)
Iφ,g:ℬ0α→QK is weakly compact.Moreover, if Iφ,g:ℬα→QK is bounded, then the next asymptotic relations hold
(3.37)‖Iφ,g‖Bα→QK≍‖Iφ,g‖B0α→QK≍M11/2.Proof.
The proof of the equivalence of statements (a)–(d) of this theorem is similar to the proof of Theorem3.1; moreover, the implication (b) ⇒ (c) is much simpler since it follows by using the test function f0(z)≡1. That (c) is equivalent to (e)–(g) is proved similarly as in Theorem 3.2, by using the well-known fact that if a bounded sequence (fl)l∈ℕ in ℬα, α∈(0,1) converges to zero uniformly on compacts of 𝔻, then it converges to zero uniformly on the whole 𝔻. The details are omitted.The proof of the next theorem is similar to the proofs of Theorems3.3 and 3.4 and will be omitted.Theorem 3.6.
Letα∈(0,1),K∈Ω(0,∞),φ∈S(𝔻), and g∈H(𝔻). Then the following statements are equivalent. (a)
Iφ,g:ℬ0α→QK,0 is bounded.(b)
Iφ,g:ℬα→QK,0 is bounded.(c)
Iφ,g:ℬα→QK,0 is compact.(d)
Iφ,g:ℬ0α→QK,0 is compact.(e)
Iφ,g:ℬ0α→QK,0 is weakly compact.(f)
lim|a|→1∫𝔻K(1-|ηa(z)|2)|g(z)|2dm(z)=0.(g)
dμ1(z)=|g(z)|2dm(z) is a vanishing K-Carleson measure.



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*Source: 698038-2011-10-09.xml*