# Toeplitz Operator and Carleson Measure on Weighted Bloch Spaces

Authors: Yanhua Zhang
Journal: Journal of Function Spaces
 (2019)

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## Abstract

In this paper, we consider Toeplitz operator acting on weighted Bloch spaces. Meanwhile, the inclusion map from weighted Bloch spaces into tent space is also investigated.


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## Body

## 1. Introduction

Denote the open unit disk of the complex planeC by D and the boundary of D by ∂D. Let H(D) denote the space of all functions analytic in D. For any a∈D, (1)φaz=a-z1-a¯z,z∈Dis the automorphism of D which exchanges 0 for a. Recall that (2)βz,a=12log1+φaz1-φazis the Bergman metric. For any 0<r<∞,a∈D, (3)Da,r=z∈D:βz,a<ris the Bergman disk. Let |D(a,r)| denote the normalized area of D(a,r). From [1], we see that |D(a,r)|≈(1-|a|2)2 when r is fixed.For0<p<∞ and α>-1, the weighted Bergman space Aαp is the space of all f∈H(D) such that (4)fAαpp≔∫Dfzp1-z2αdAz<∞.When α=0, Aαp is the classical Bergman space. We refer the readers to [1, 2] for more results on weighted Bergman spaces.Let0<α<∞. An f∈H(D) is said to belong to the weighted Bloch space, denoted by Bα, if (5)fBα=supz∈D1-z2αf′z<∞.The space Bα has been studied extensively in [3]. See [1, 4–8] for the study of some operators on weighted Bloch spaces.Letφ∈L∞(D). The Toeplitz operator Tφ with symbol φ is defined by (6)Tφfz=∫Dφwfw1-w¯z2+αdAαw,where dAα(w)=(1-|w|2)αdA(w). There are many results related to Tφ, see [1] and the references therein. Especially, some characterizations for the operator Tφ on Lα2 have been obtained by many authors. Since Bα⊆Aα1, it is nature to ask (7)Tφf∈Bα⇔?,f∈Bα.The following theorem is the first main result in this paper.Theorem 1.
Let0<α<∞ and φ∈L1(D) be harmonic. Then the following statements hold.
(1)Tφ:Bα+1→Bα+1 is bounded if and only if φ is bounded.
(2)Tφ:Bα+1→Bα+1 is compact if and only if φ=0.Given a positive Borel measureμ, the Toeplitz operator with the symbol μ is defined by (8)Tμfz=∫Dfw1-w¯z2+αdμw,f∈L1dAα.For the Toeplitz operatorTμ, we have the following result.Theorem 2.
Let0<α,r<∞ and μ be a positive Borel measure. Then the following statements hold.
(1)Tμ:Bα+1→Bα+1 is bounded if and only if (9)supa∈DμDa,r1-a22+α<∞.
(2)Tμ:Bα+1→Bα+1 is compact if and only if (10)lima→0μDa,r1-a22+α=0.ForI⊂∂D, |I|=1/2π∫I|dξ| is the normalized length of the subarc I and the corresponding Carleson square for I is defined as follows (see [9]). (11)SI=rξ∈D:r∈1-I,1,ξ∈I.For0<p<∞, a positive Borel measure μ on D is said to be a p-Carleson measure if(12)μp≕supI⊂∂DμSIIp1/2<∞.If p=1, p-Carleson measure is the classical Carleson measure. From Lemma 3.1.1 of [10], for p,q∈(0,∞) we know that μ is a p-Carleson measure if and only if (13)μp,q≔supz∈D∫D1-z2q1-z¯wp+qdμw1/2<∞.Moreover, μp,q≈μp.Let0<p,q<∞ and μ be a positive Borel measure on D. The tent space Tμp,q is the class of all f∈H(D) which satisfy (14)supI⊆∂D1Ip∫SIfzqdμz<∞.The tent space Tμp,2 was introduced by J. Xiao [11] to studied Carleson measure for Qs space. He proved that Qs space is continuously contained in Tμp,2 if and only if (15)supI⊆∂DμSIIslog⁡2I2<∞.J. Pau and R. Zhao [12] generalized the main results in [11]. In [13], J. Liu and Z.Lou studied Morrey spaces. They proved that an equivalent condition for Morrey spaces L2,s continuously contained in Tμp,2 is that μ is a Carleson measure. See [14, 15] for more information of the Morrey space.We state the last main result in this paper as follows.Theorem 3.
Let0<α<∞ and μ be a positive Borel measure. Then the following statements hold.
(1) The inclusion mapId:Bα+1→Tμ2+α,1 is bounded if and only if μ is a (2+2α)-Carleson measure.
(2) The inclusion mapId:Bα+1→Tμ2+α,1 is compact if and only if μ is a vanishing (2+2α)-Carleson measure.Throughout this paper, the letterC will denote constants and may differ from one occurrence to the other. The notation A≲B means that there is a positive constant C such that A≤CB. The notation A≈B means A≲B and B≲A.


## 2. Proofs of Main Results

To prove our main results in this paper, we need some auxiliary results. The following result can be found in [16, Theorem 3.8].Lemma 4.
Letp≥1, α>0, -1+pα<η<∞, and c>0. Then f∈Bα+1 if and only if (16)supz∈D∫Dfw-fzp1-z2c+pα1-z¯w2+c+ηdAηw<∞.From Lemma4, we can easily deduce the following result.Lemma 5.
Letp≥1, α>0, -1+pα<η<∞, and c>0. Then f∈Bα+1 if and only if (17)supa∈D∫Dfzp1-a2c+pα1-a¯z2+c+ηdAηz<∞.Proof.
First assume thatf∈Bα+1. It is clear that (18)fz≲fBα+11-z2α,z∈D.Thus,(19)∫Dfzp1-a2c+pα1-a¯z2+c+ηdAηz≲∫Dfz-fap1-a2c+pα1-a¯z2+c+ηdAηz+∫Dfap1-a2c+pα1-a¯z2+c+ηdAηz≲fBα+1p+fBα+1p∫D1-a2c1-a¯z2+c+ηdAηz≲fBα+1p.The proof of the inverse direction is similar to the above statements we omit the details. The proof is complete.Proof of Theorem1.
(1) First assume thatφ∈L∞(D). For f∈Bα+1, since (20)fBα+1≈supz∈D1-z2αfz,we obtain(21)TφfBα+1≈supz∈D1-z2αTφfz=supz∈D1-z2α∫Dφwfw1-w¯z2+αdAαw≲supz∈D1-z2α∫Dφwfw1-w¯z2+αdAαw≲φL∞DfBα+1supz∈D1-z2α∫D11-w¯z2+αdAw≲φL∞DfBα+1.Hence Tφ:Bα+1→Bα+1 is bounded.
Conversely, assume thatTφ:Bα+1→Bα+1 is bounded. For z∈D, set (22)fzw=1-z221-z¯w2+α∈Bα+1.It is easy to check that fzBα+1≈1. Using Lemma 5 with p=1, η=α and c=2+α, we get(23)∞>Tφ≳TφfzB1+α≳∫DTφfzw1-z22+2α1-z¯w22+αdAαw=1-z2α∫DTφfzφzwdAαw≳1-z2αTφfzφz0≳φz,which implies that φ∈L∞(D), as desired.
(2) Sufficiency. The result is obvious.
Necessity. For anyzn∈D, let (24)fznw=1-zn221-zn¯w2+α,w∈D.fzn∈B1+α. It is easy to check that fzn→0 uniformly on compact subsets on D as |zn|→1. From the fact that Tφ is compact on Bα+1→Bα+1 and the proof of (1), we have(25)0←TφfznBα+1≳∫DTφfznw1-zn22+2α1-zn¯w22+αdAαw≳1-z2αTφfznφzn0≳φzn.By the arbitrariness of zn and the Maximal Module Principle, we get φ=0. The proof is complete.Proof of Theorem2.
(1) First suppose thatTμ:Bα+1→Bα+1 is bounded. For any a∈D, from the proof of Theorem 1, we obtain that fa∈Bα+1 and faBα+1≲1. Thus,(26)∞>Tμ≳1-a2αTμfaa=1-a2α∫Dfaw1-w¯a2+αdμw=1-a2α∫D1-a221-w¯a4+2αdμw≥1-a2α∫Da,r1-a221-w¯a4+2αdμw≈μDa,r1-a22+α,as desired.
Conversely, suppose that(27)supa∈DμDa,r1-a22+α<∞.Then we can get that μ is a Carleson measure for L1(dAα). If g∈L1 and f∈Bα+1, we can easily obtain that fg∈L1(dAα). Using Fubini’s Theorem, we obtain(28)∫DgzTμfz¯1-z2αdAz=∫Dgz1-z2αdAz∫Dfw¯1-wz¯2+αdμw=∫Dfw¯dμw∫D1-z2αgz1-wz¯2+αdAz≤∫Dgwfw¯dμw≲∫DgwfwdAαw≲gL1fBα+1.Hence Tμ:Bα+1→Bα+1 is bounded.
(2) Suppose thatTμ:Bα+1→Bα+1 is compact. Let an∈D. Set (29)fanz=1-an221-an¯z2+α,z∈D.Then fan∈Bα+1 and fan→0 uniformly on compact subset on D as |an|→1. Thus,(30)TμfanBα+1≥1-an2αTμfanan=1-an2α∫Dfanw1-w¯an2+αdμw=1-an2α∫D1-an221-w¯an4+2αdμw≥1-an2α∫Dan,r1-an221-w¯an4+2αdμw≈μDan,r1-an22+α,which implies the desired result.
Conversely, assume that(31)lima→1μDa,r1-a22+α=0.We know that μ is a vanishing Carleson measure for L1(dAα). We want to show that Tμ is compact. Using Fubini’s Theorem we have (32)∫DTμfz¯gzdAαz=∫Dgzfz¯dμz.
Ifg∈L1 and f∈Bα+1, we can easily obtain fg∈L1(dAα). Therefore,(33)∫DTμfz¯gzdAαz≤∫Dgzfz¯dμz≲∫DgzfzdAαz≲gL1fBα+1.That is, (34)TμfBα+1≲fBα+1.If fn→0 weakly in Bα+1, it follows that TμfnBα+1→0. The proof of Theorem 2 is complete.Proof of Theorem3.
(1) Suppose thatId:Bα+1→Tμ2+α,1 is bounded. For a∈D, set (35)faz=1-a221-a¯z2+α,z∈D.Then fa∈Bα+1. For any I⊆∂D, we get(36)μSII2+2α≲1I2+α∫SIfazdμz<∞,as desired.
Conversely, assume thatμ is a (2+2α)-Carleson measure. Let f∈Bα+1. Using the well-known fact (37)fb≲11-b2α,b∈D,we have(38)1I2+α∫SIfwdμw≲1I2+α∫SIfw-fbdμw+1I2+α∫SIfbdμw≲1I2+α∫SIfw-fbdμw+μSII2+2α.Note that (39)μSII2+α≲μSII2+2α.Then μ is a Carleson measure for L1(dAα). Since Bα+1⊆L1(dAα), combined with Lemma 4, we obtain(40)1I2+α∫SIfw-fbdμw≈1-b22+α∫SIfw-fb1-b¯w4+2αdμw≲1-b22+α∫Dfw-fb1-b¯w4+2αdμw≲1-b22+α∫Dfw-fb1-b¯w4+2αdAαw≲∫Dfw-fb1-b22+α1-b¯w4+2αdAαw≲fBα+1.Hence Id:Bα+1→Tμ2+α,1 is bounded.
(2) Suppose thatId:Bα+1→Tμ2+α,1 is compact. Let an∈D such that |an|→1 as n→∞. We know that fan∈Bα+1 and fan→0 uniformly on compact subsets of D as n→∞. By Theorem 5.15 of [1] it follows that fan→0 weakly as n→∞. Hence for the compact operator Id:Bα+1→Tμ2+α,1, we have fanTμ2+α,1→0 as n→∞. Thus,(41)μSInIn2+2α≲11-an22+α∫SInfanzdμz≲fanTμ2+α,1→0as n→∞. Hence μ is a vanishing (2+2α)-Carleson measure.
Conversely, assume thatμ is a vanishing (2+2α)-Carleson measure. Let fn∈Bα+1, fnBα+1≲1, and fn→0  (n→∞) uniformly on compact subsets of D. Then it is easy to get that(42)1I2+α∫SIfnzdμz≲1I2+α∫SIfnzdμrz+1I2+α∫SIfnzdμ-μrz≲1I2+α∫SIfnzdμrz+μ-μr2fnBα+1.Let r→1- and n→∞; we get the desired result. The proof is complete.



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*Source: 4358959-2019-02-17.xml*