# Carleson measures, Riemann-Stieltjes and multiplication operators on a general family of function spaces

aa r X i v : . [ m a t h . F A ] J u l CARLESON MEASURES, RIEMANN-STIELTJES AND MULTIPLICATIONOPERATORS ON A GENERAL FAMILY OF FUNCTION SPACES

JORDI PAU AND RUHAN ZHAOA

BSTRACT . Let µ be a nonnegative Borel measure on the unit disk of the complex plane.We characterize those measures µ such that the general family of spaces of analytic func-tions, F ( p, q, s ) , which contain many classical function spaces, including the Bloch space, BMOA and the Q s spaces, are embedded boundedly or compactly into the tent-typespaces T ∞ p,s ( µ ) . The results are applied to characterize boundedness and compactness ofRiemann-Stieltjes operators and multiplication operators on F ( p, q, s ) .

1. I

NTRODUCTION

The Carleson measure was introduced by Carleson [9] for studying the problem of inter-polation by bounded analytic functions and for solving the famous corona problem. Lateron variations of the Carleson measure have been also introduced and studied. It was foundthat Carleson type measures are usually closely related to certain function spaces. For ex-ample, the original Carleson measure is closely related to Hardy spaces H p . This featuremakes Carleson type measures important tools for the modern function theory and operatortheory.In this paper we study Carleson measures related to the general analytic function space F ( p, q, s ) on the unit disk of the complex plane, introduced by the second author in [27].Basically, we are going to characterize those nonnegative Borel measures µ such that theembedding from F ( p, q, s ) to a certain tent-type space is bounded or compact. The resultswill be applied to get characterizations for the bounded and compact Riemann-Stieltjesoperators and pointwise multiplication operators on F ( p, q, s ) .Our work involves several spaces of analytic functions. Let us ﬁrst review these spaces.Let D be the unit disk on the complex plane. Let H ( D ) be the space of all analyticfunctions on D . For α > − and p > , the weighted Bergman space A pα consists of allfunctions f ∈ H ( D ) such that k f k A pα = (cid:18)Z D | f ( z ) | p dA α ( z ) (cid:19) /p < ∞ . Here dA α ( z ) = ( α + 1)(1 − | z | ) α dA ( z ) , and dA is the normalized area measure on D .The weighted Dirichlet space D pα consists of all functions f ∈ H ( D ) such that k f k D pα = | f (0) | + (cid:18)Z D | f ′ ( z ) | p dA α ( z ) (cid:19) /p < ∞ . Key words and phrases.

Carleson measures, F ( p, q, s ) spaces, tent-type spaces, Riemann-Stieltjes opera-tors, multiplication operators.The ﬁrst author is supported by SGR grant SGR (Generalitat de Catalunya) and DGICYT grantMTM - - C - (MCyT/MEC). It is well-known that D = H , the Hardy space, and for α > p − , D pα = A pα − p , withequivalence of norms.Next, we recall the Bloch type spaces (also called α -Bloch spaces) B α . Let α > . The α -Bloch space B α consists of all analytic functions f on D such that k f k B α = sup z ∈ D | f ′ ( z ) | (1 − | z | ) α < ∞ . The little α -Bloch space B α consists of all analytic functions f on D for which lim | z |→ | f ′ ( z ) | (1 − | z | ) α = 0 . It is known that B α is a Banach space under the norm |k f k| B α = | f (0) | + k f k B α , and B α is a closed subspace of B α . As α = 1 , B = B , the Bloch space.Here are some related facts about B α . When < α < , B α = Lip − α , the Lipschitztype space, which contains analytic functions f on D for which there is a constant C > such that | f ( z ) − f ( w ) | ≤ C | z − w | − α for all z, w ∈ D . As a simple consequence, we know that, when < α < , B α ⊂ H ∞ ,the space of all bounded analytic functions f on D with k f k H ∞ = sup z ∈ D | f ( z ) | < ∞ . When α > , an analytic function f ∈ B α if and only if sup z ∈ D | f ( z ) | (1 − | z | ) α − < ∞ , and the above supremum is comparable to |k f k| B α . For all these results, see [31].Next, we introduce the spaces F ( p, q, s ) . For a point a ∈ D , let ϕ a ( z ) = ( a − z ) / (1 − ¯ az ) denote the M¨obius transformation of D that interchanges and a . An easy calculationshows ϕ ′ a ( z ) = − − | a | (1 − ¯ az ) . For < p < ∞ , − < q < ∞ , ≤ s < ∞ . The space F ( p, q, s ) is deﬁned as the spaceof all functions f ∈ H ( D ) such that k f k pF ( p,q,s ) = sup a ∈ D Z D | f ′ ( z ) | p (1 − | z | ) q (1 − | ϕ a ( z ) | ) s dA ( z ) < ∞ . It is known that, for p ≥ , F ( p, q, s ) is a Banach space under the norm |k f k| F ( p,q,s ) = | f (0) | + k f k F ( p,q,s ) . For < p < , the space F ( p, q, s ) is a complete metric space with the metric given by d ( f, g ) = |k f − g k| pF ( p,q,s ) . In other words, it is an F -space, in the terminology introduced by Banach [6]. The familyof spaces F ( p, q, s ) was introduced in [27]. It contains, as special cases, many classicalfunction spaces, such as the analytic Besov spaces, weighted Bergman spaces, weightedDirichlet spaces, the α -Bloch spaces, BMOA and the recently introduced Q s spaces.For convenience, we will write q = pα − , where α > . It is known that, for any α > , F ( p, pα − , s ) are subspaces of B α , the α -Bloch space. As s > , we actually have F ( p, pα − , s ) = B α . Also, it is known that F ( p, q, s ) contains only constant functions if ARLESON MEASURES ON FUNCTION SPACES 3 s + q ≤ − or s + pα ≤ when q = pα − (see Proposition 2.12 in [27]). Therefore, lateron we will assume that s + pα > , which guarantees that F ( p, pα − , s ) is nontrivial.Among these F ( p, pα − , s ) spaces, the case α = 1 is particularly interesting, sincethe spaces F ( p, p − , s ) are M¨obius invariant, in the sense that for any function f ∈ F ( p, p − , s ) and any a ∈ D , one has k f ◦ ϕ a k F ( p,p − ,s ) = k f k F ( p,p − ,s ) . As mentioned above, when s > we have that F ( p, p − , s ) = B , the Bloch space. As p = 2 , F ( p, p − , s ) = Q s , the Q s spaces, introduced in [4]. As p = 2 and s = 1 , F ( p, p − , s ) = BM OA , the space of analytic functions of bounded mean oscillation.See [27] for details of all of the above facts about F ( p, q, s ) spaces.The tent-type spaces used in this paper are deﬁned as follows. Let I be an arc on theunit circle ∂D . Denote by | I | the normalized arc length of I so that | ∂D | = 1 . Let S ( I ) be the Carleson box deﬁned by S ( I ) = { z : 1 − | I | < | z | < , z/ | z | ∈ I } . Let ≤ s < ∞ and < p < ∞ . For a nonnegative Borel measure µ on the unit disk D ,we deﬁne T ∞ p,s ( µ ) as the space of all µ -measurable functions f on D satisfying k f k pT ∞ p,s ( µ ) = sup I ⊂ ∂D | I | − s Z S ( I ) | f ( z ) | p dµ ( z ) < ∞ . By a standard argument we can show that for p ≥ , T ∞ p,s ( µ ) is a Banach space.Next, we introduce the Carleson measures needed in this paper. We say that a nonneg-ative Borel measure µ on D is a ( p, s ) -logarithmic Carleson measure if there is a constant C > such that sup I ⊂ ∂D | I | s (cid:18) log 2 | I | (cid:19) p µ ( S ( I )) ≤ C. We will denote the class of all ( p, s ) -logarithmic Carleson measures on D by LCM p,s , anddenote by k µ k LCM p,s = sup I ⊂ ∂D | I | s (cid:18) log 2 | I | (cid:19) p µ ( S ( I )) . When p = 0 , the (0 , s ) -logarithmic Carleson measures are called s -Carleson measures, andwe will denote by CM s = LCM ,s and k µ k CM s = k µ k LCM ,s . We say µ is a vanishing ( p, s ) -logarithmic Carleson measure if lim | I |→ | I | s (cid:18) log 2 | I | (cid:19) p µ ( S ( I )) = 0 . A vanishing (0 , s ) -logarithmic Carleson measure is also called a vanishing s -Carlesonmeasure.Logarithmic Carleson measures were ﬁrst introduced by the second author in [29]. Inthis paper, we are going to characterize the measures µ such that the identity operator I : F ( p, q, s ) → T ∞ p,s ( µ ) is bounded or compact. The results will be applied to givecharacterizations for the bounded and compact Riemann-Stieltjes operators and pointwisemultiplication operators on the F ( p, q, s ) spaces. Our results generalize some recent resultsby Xiao in [26] and by Pau and Pel´aez in [18].The paper is organized as follows. Section 2 is devoted to some preliminary results. Insection 3 we characterize boundedness and compactness of I : F ( p, p − , s ) → T ∞ p,s ( µ ) . Insection 4 we characterize boundedness and compactness of I : F ( p, pα − , s ) → T ∞ p,s ( µ ) J. PAU AND R. ZHAO for α = 1 . In section 5 we use these results to characterize bounded and compact Riemann-Stieltjes operators on F ( p, q, s ) , and in section 6 we characterize bounded and compactpointwise multiplication operators on F ( p, q, s ) .2. P RELIMINARY RESULTS

In this Section we state and prove some preliminary results needed for the rest of thepaper. Some of them may have independent interest. We begin with the following lemma.

Lemma 2.1.

Let < p < ∞ , − < q < ∞ , ≤ s < ∞ satisfy q + s > − .Let µ be a nonnegative Borel measure on D such that the point evaluation is a boundedfunctional on T ∞ p,s ( µ ) . Then I : F ( p, q, s ) → T ∞ p,s ( µ ) is a compact operator if and only if k f n k T ∞ p,s ( µ ) → whenever { f n } is a bounded sequence in F ( p, q, s ) that converges to uniformly on every compact subset of D . Using the fact that the point evaluations on F ( p, q, s ) and T ∞ p,s ( µ ) are bounded func-tionals (see Proposition 2.17 of [27] for the case of F ( p, q, s ) ) , the proof of Lemma 2.1 isstandard. See, for example, the proof of Proposition 3.11 in [10]. We omit the details here.We also need the following equivalent description of ( p, s ) -logarithmic Carleson mea-sures proved by O. Blasco (see Lemma . and Proposition . in [7]). Lemma 2.2.

Let s, t > and p ≥ . Let µ be a nonnegative Borel measure on D . Then µ is a ( p, s ) -logarithmic Carleson measure if and only if sup a ∈ D (cid:18) log 21 − | a | (cid:19) p Z D (1 − | a | ) t | − ¯ az | s + t dµ ( z ) < ∞ . Further, we have k µ k LCM p,s ≈ sup a ∈ D (cid:18) log 21 − | a | (cid:19) p Z D (1 − | a | ) t | − ¯ az | s + t dµ ( z ) . We also need the following equivalent deﬁnition for T ∞ p,s ( µ ) . Proposition 2.3.

Let µ be a nonnegative Borel measure on D . Let p > , s > , and t > . Then an analytic function f on D belongs to T ∞ p,s ( µ ) if and only if sup a ∈ D Z D (1 − | a | ) t | − ¯ az | s + t | f ( z ) | p dµ ( z ) < ∞ , and k f k pT ∞ p,s ( µ ) ≈ sup a ∈ D Z D (1 − | a | ) t | − ¯ az | s + t | f ( z ) | p dµ ( z ) . Proof.

Let dµ f ( z ) = | f ( z ) | p dµ ( z ) . Then k f k pT ∞ p,s ( µ ) = sup I ⊂ ∂D | I | s Z S ( I ) | f ( z ) | p dµ ( z ) = sup I ⊂ ∂D µ f ( S ( I )) | I | s . Applying Lemma 2.2 to dµ f ( z ) we immediately get the result. (cid:3) The following lemma is from [17].

ARLESON MEASURES ON FUNCTION SPACES 5

Lemma 2.4.

For s > − , r, t > with r + t − s − > , there is a constant C > suchthat Z D (1 − | z | ) s | − ¯ az | r | − ¯ bz | t dA ( z ) ≤ C | − ¯ ab | r + t − s − , if r, t < s,C (1 − | a | ) r − s − | − ¯ ab | t , if t < s < r,C (1 − | a | ) r − s − | − ¯ ab | t + C (1 − | b | ) t − s − | − ¯ ab | r , if r, t > s . Corollary 2.5.

For s > − , r, t > with < r + t − s − < r , we have I ( a, b ) = Z D (1 − | z | ) s | − ¯ az | r | − ¯ bz | t dA ( z ) ≤ C (1 − | a | ) r + t − s − . Proof.

From the condition < r + t − s − < r we know that t < s . If r < s or r > s then the result follows directly from Lemma 2.4.For the remaining case r = 2 + s , we cannot directly use Lemma 2.4. However, since s + 1 > and t > , we can choose a number δ such that < δ < min( t, s + 1) . Thus I ( a, b ) = Z D (1 − | z | ) s − δ | − ¯ az | r | − ¯ bz | t − δ · (1 − | z | ) δ | − ¯ bz | δ dA ( z ) ≤ C Z D (1 − | z | ) s − δ | − ¯ az | r | − ¯ bz | t − δ dA ( z ) . It is obvious that s − δ > − , r > , t − δ > , r + ( t − δ ) − ( s − δ ) − r + t − s − > .Since t < s and r = 2 + s we also have t − δ < s − δ ) = r − δ < r. Hence we can use the second inequality of Lemma 2.4 to get I ( a, b ) ≤ C (1 − | a | ) r − ( s − δ ) − | − ¯ ab | t − δ ≤ C (1 − | a | ) r + t − s − . The proof is complete. (cid:3)

Using this corollary, we are able to prove the following result.

Lemma 2.6.

Let < p < ∞ , < α < ∞ and < s < ∞ satisfy s + pα > . Let f b,α ( z ) = − α (1 − ¯ bz ) − α , if α = 1 , log 21 − ¯ bz , if α = 1 . Then sup b ∈ D k f b,α k F ( p,pα − ,s ) < ∞ . Proof.

A simple computation using the well-known identity − | ϕ a ( z ) | = (1 − | z | ) | ϕ ′ a ( z ) | J. PAU AND R. ZHAO shows k f b,α k pF ( p,pα − ,s ) = sup a ∈ D Z D (cid:12)(cid:12)(cid:12)(cid:12) ¯ b (1 − ¯ bz ) α (cid:12)(cid:12)(cid:12)(cid:12) p (1 − | z | ) pα − (1 − | ϕ a ( z ) | ) s dA ( z )= sup a ∈ D | b | p (1 − | a | ) s Z D (1 − | z | ) s + pα − | − ¯ az | s | − ¯ bz | pα dA ( z ) . Since s + pα − > − , pα > , s > , and < s + pα − ( s + pα − − s < s, we can use Corollary 2.5 to get Z D (1 − | z | ) s + pα − | − ¯ az | s | − ¯ bz | pα dA ( z ) ≤ C (1 − | a | ) s . Therefore, sup a ∈ D k f b,α k pF ( p,pα − ,s ) ≤ C < ∞ . The proof is complete. (cid:3)

Lemma 2.7.

Let < p < ∞ , and γ > − with γ > p − . For any f ∈ D pγ , and any a ∈ D , I ( a ) := Z D | f ( z ) − f (0) | p | − ¯ az | p (1 − | z | ) γ dA ( z ) ≤ C Z D | f ′ ( z ) | p (1 − | z | ) γ dA ( z ) . Proof. If p > , the result follows from Lemma . of [8]. If < p ≤ we use theatomic decomposition for D pγ (see [30, Theorem 32]): there is a sequence { a k } in D anda sequence of numbers { λ k } ∈ ℓ p such that g ( z ) = f ( z ) − f (0) = X k λ k (1 − | a k | ) b − (2+ γ − p ) /p (1 − ¯ a k z ) b , b > γ − pp and X k | λ k | p ≤ C k g k pD pγ = C Z D | f ′ ( z ) | p (1 − | z | ) γ dA ( z ) . Note that D pγ = A pγ − p in the notation of [30]. Now, since < p ≤ , we obtain I ( a ) ≤ X k | λ k | p (1 − | a k | ) pb − − γ + p Z D (1 − | z | ) γ | − ¯ a k z | pb | − ¯ az | p dA ( z ) ≤ C X k | λ k | p . The last inequality holds by Corollary 2.5. The proof is complete. (cid:3)

Proposition 2.8.

Let < p < ∞ , s > , and α ≥ with αp + s > . Let t = s +2 p ( α − .For any f ∈ F ( p, pα − , s ) and any a ∈ D , J ( a ) = (1 − | a | ) t Z D | f ( z ) − f ( a ) | p | − ¯ az | p + s + t (1 − | z | ) s + pα − dA ( z ) ≤ C k f k pF ( p,pα − ,s ) . ARLESON MEASURES ON FUNCTION SPACES 7

Proof.

Changing the variable z = ϕ a ( w ) , we get by Lemma 2.7, J ( a ) = (1 − | a | ) p ( α − Z D | f ◦ ϕ a ( w ) − f ◦ ϕ a (0) | p | − ¯ aw | p (1 − | w | ) s + pα − dA ( w ) ≤ C (1 − | a | ) p ( α − Z D (cid:12)(cid:12) ( f ◦ ϕ a ) ′ ( w ) (cid:12)(cid:12) p (1 − | w | ) s + pα − dA ( w )= C (1 − | a | ) p ( α − Z D | f ′ ( z ) | p (1 − | z | ) p − (1 − | ϕ a ( z ) | ) s + p ( α − dA ( z )= C (1 − | a | ) t Z D | f ′ ( z ) | p (1 − | z | ) s + pα − | − ¯ az | s + t dA ( z ) ≤ C k f k pF ( p,pα − ,s ) . The proof is complete. (cid:3)

Proposition 2.9.

Let ≤ p < ∞ and s ≥ . Let µ be a vanishing ( p, s ) -logarithmicCarleson measure on D . Then lim r → sup I ⊂ ∂D | I | s (cid:18) log 2 | I | (cid:19) p µ r ( S ( I )) = 0 . Proof.

Let ≤ p < ∞ and s ≥ . Since µ is a vanishing ( p, s ) -logarithmic Carlesonmeasure, for any ε > , there is an r , < r < , such that | I | s (cid:18) log 2 | I | (cid:19) p µ ( S ( I )) < ε for any arc I with | I | < − r . Now ﬁx the above r . Consider any arc I on the unit circle ∂D . Let α = 1 − r , and n = [ | I | /α ] . Then nα ≤ | I | < ( n + 1) α . Clearly, we cancover I by n + 1 arcs I , I , ..., I n +1 with | I k | = 1 − r = α for k = 1 , , ..., n + 1 . Let µ r = µ | D \ D r . Then | I | s (cid:18) log 2 | I | (cid:19) p µ r ( S ( I )) ≤ n +1 X k =1 | I | s (cid:18) log 2 | I | (cid:19) p µ ( S ( I k )) ≤ ε n +1 X k =1 | I | s (cid:18) log 2 | I | (cid:19) p | I k | s (cid:18) log 2 | I k | (cid:19) − p ≤ ε n +1 X k =1 nα ) s (cid:18) log 2 nα (cid:19) p α s (cid:18) log 2 α (cid:19) − p ≤ n − s ε (cid:20)(cid:18) log 2 nα (cid:19) (cid:14) (cid:18) log 2 α (cid:19)(cid:21) p ≤ Cε uniformly on I , when s ≥ . Hence, as s ≥ we have lim r → sup I ⊂ ∂D | I | s (cid:18) log 2 | I | (cid:19) p µ r ( S ( I )) = 0 . (cid:3)

3. E

MBEDDING FROM F ( p, p − , s ) TO T ∞ p,s ( µ ) In this section we describe the boundedness and compactness of the embedding fromthe M¨obius invariant space F ( p, p − , s ) to the tent-type space T ∞ p,s ( µ ) . The main resultin this section is the following. Theorem 3.1.

Let s > and < p < ∞ satisfy s + p > , and let µ be a nonnegativeBorel measure on D . Then J. PAU AND R. ZHAO (i) I : F ( p, p − , s ) → T ∞ p,s ( µ ) is bounded if and only if µ is a ( p, s ) -logarithmicCarleson measure. (ii) The following two conditions are equivalent: (a)

For any bounded sequence { f n } in F ( p, q, s ) satisfying f n ( z ) → uniformlyon every compact subset of D , lim n →∞ k f n k T ∞ p,s ( µ ) = 0 . (b) µ is a vanishing ( p, s ) -logarithmic Carleson measure.Remark. When p = 2 , the result is obtained by Xiao in [26] using techniques from [18].In order to prove Theorem 3.1, we need some results on Carleson measures for weightedDirichlet spaces D pα . A nonnegative Borel measure µ on D is said to be a Carleson measurefor D pα if there is a constant C > such that for every f ∈ D pα , Z D | f ( z ) | p dµ ( z ) ≤ C k f k pD pα , and is called a compact Carleson measure for D pα if lim n k f n k L p ( µ ) = 0 whenever { f n } is a bounded sequence in D pα converging to zero uniformly on compact subsets of D .Carleson measures for the weighted Bergman spaces A pα = D pα + p , with α > − and p > are described as those positive Borel measures on D such that sup I ⊂ ∂D µ ( S ( I )) | I | α < ∞ . This result was proved by several authors, including Oleinik and Pavlov [16] (for p > ),Stegenga [24] (for α = 0 ), and Hastings [12]. One can also ﬁnd a proof in [14]. The nextresult will be essential in order to prove Theorem 3.1. Lemma 3.2.

Let p > , s ≥ satisfy s + p > . Let µ be a nonnegative Borel measure on D . (a) If µ ∈ LCM p,s then µ is a Carleson measure for D ps + p − . Moreover (cid:13)(cid:13) I : D ps + p − → L p ( µ ) (cid:13)(cid:13) p ≤ C k µ k LCM p,s . (b) If µ is a vanishing ( p, s ) -logarithmic Carleson measure, then µ is a compact Car-leson measure for D ps + p − .Proof. The case p = 2 with < s < of part (a) was proved in [18] using the descriptionof Carleson measures for D ps + p − given in [5]. Here we will prove the result directly fromthe deﬁnition of Carleson measures for D ps + p − .We ﬁrst consider the case p > . Let p ′ denote the conjugate exponent of p . Since p ′ > is easy to see that Z D dA ( w ) | − ¯ wz | (cid:0) log −| w | (cid:1) p ′ ≤ C for some constant C independent of w ∈ D . Let α = s + p − , and f ∈ D pα with f (0) = 0 . Using the reproducing formula for the Bergman space A pα (see p.80 of [32]),and then integrating along the segment [0 , z ] it follows that f ( z ) = C Z D f ′ ( w ) K α ( w, z )(1 − | w | ) α dA ( w ) , ARLESON MEASURES ON FUNCTION SPACES 9 where K α ( w, z ) = 1 − (1 − ¯ wz ) α ¯ w (1 − ¯ wz ) α . It is easy to check that sup z,w ∈ D (cid:12)(cid:12)(cid:12)(cid:12) − (1 − ¯ wz ) α ¯ w (cid:12)(cid:12)(cid:12)(cid:12) ≤ C for some constant C > . Then | f ( z ) | p ≤ C (cid:18)Z D | f ′ ( w ) | (1 − | w | ) α | − ¯ wz | α dA ( w ) (cid:19) p ≤ C Z D | f ′ ( w ) | p (1 − | w | ) pα | − ¯ wz | pα +2 − p (cid:0) log 21 − | w | (cid:1) p dA ( w ) Z D (log −| w | ) − p ′ dA ( w ) | − ¯ wz | ! p − ≤ C Z D | f ′ ( w ) | p (1 − | w | ) pα | − ¯ wz | pα +2 − p (cid:0) log 21 − | w | (cid:1) p dA ( w ) . This inequality together with Fubini’s theorem and Lemma 2.2 yields Z D | f ( z ) | p dµ ( z ) ≤ C Z D | f ′ ( w ) | p (1 − | w | ) pα (cid:18) log 21 − | w | (cid:19) p Z D dµ ( z ) | − ¯ wz | pα +2 − p dA ( w ) ≤ C k µ k LCM p,s Z D | f ′ ( w ) | p (1 − | w | ) pα (1 − | w | ) − ( p − α dA ( w ) ≤ C k µ k LCM p,s k f k pD pα . Hence µ is a Carleson measure for D ps + p − .Now we consider the case < p ≤ . For r > , ﬁx an r -lattice { a n } in the Bergmanmetric. This means that the hyperbolic disks D ( a n , r ) = { z : β ( z, a n ) < r } cover the unitdisk D and β ( a i , a j ) ≥ r/ for all i and j with i = j . Here β ( z, w ) denotes the Bergmanor hyperbolic metric. If { a k } is an r -lattice in D , then it also has the following property:for any R > there exists a positive integer N (depending on r and R ) such that everypoint in D belongs to at most N sets in { D ( a k , R ) } . There are elementary constructionsof r -lattices in D . See [32, Chapter 4] for example. Note that by subharmonicity we have sup {| f ′ ( w ) | : w ∈ D ( a j , r ) } ≤ C − | a j | ) Z D ( a j , r ) | f ′ ( ζ ) | p dA ( ζ ) ! /p for all j = 1 , , . . . Let β be a sufﬁciently large number so that β ≥ s + p − and βp > s − p . It follows that | f ( z ) | p ≤ C (cid:18)Z D | f ′ ( w ) | (1 − | w | ) β | − ¯ wz | β dA ( w ) (cid:19) p ≤ C X j Z D ( a j ,r ) | f ′ ( w ) | (1 − | w | ) β | − ¯ wz | β dA ( w ) p ≤ C X j (1 − | a j | ) β | − ¯ a j z | β − | a j | ) Z D ( a j , r ) | f ′ ( ζ ) | p dA ( ζ ) ! /p p ≤ C X j (1 − | a j | ) p − βp | − ¯ a j z | p + βp Z D ( a j , r ) | f ′ ( ζ ) | p dA ( ζ ) Now, if µ ∈ LCM p,s , then clearly µ ∈ CM s with k µ k CM s ≤ C k µ k LCM p,s . Therefore,since − | a j | is comparable to − | ζ | if ζ ∈ D ( a j , r ) , an application of Lemma 2.2gives Z D | f ( z ) | p dµ ( z ) ≤ C X j (cid:18) (1 − | a j | ) p − βp Z D dµ ( z ) | − ¯ a j z | p + βp (cid:19) Z D ( a j , r ) | f ′ ( ζ ) | p dA ( ζ ) ≤ C k µ k CM s X j (1 − | a j | ) s + p − Z D ( a j , r ) | f ′ ( ζ ) | p dA ( ζ ) ≤ C k µ k CM s X j Z D ( a j , r ) | f ′ ( ζ ) | p (1 − | ζ | ) s + p − dA ( ζ ) ≤ C N k µ k CM s k f k pD ps + p − , where N is a positive integer such that each point of D belongs to at most N of the sets D ( a j , r ) . This ﬁnishes the proof of (a).To prove (b), we must show that if { f n } is a bounded sequence in D ps + p − convergingto zero uniformly on compact subsets of D , then k f n k L p ( µ ) → . Suppose ﬁrst that p > .Let α = s + p − . Since µ is a vanishing ( p, s ) -logarithmic Carleson measure on D , forany ε > , there is an r , < r < , such that sup | w | >r (cid:18) log 21 − | w | (cid:19) p (1 − | w | ) ( p − α Z D dµ ( z ) | − ¯ wz | pα +2 − p < ε p . Since { f n } converges to zero uniformly on compact sets, the same is true for the sequenceof its derivatives { f ′ n } , and hence there is a positive integer n such that for all n ≥ n sup | w |≤ r | f ′ n ( w ) | < ε. Now, we have k f n k pL p ( µ ) ≤ C Z D (cid:18)Z D | f ′ n ( w ) | (1 − | w | ) α | − ¯ wz | α dA ( w ) (cid:19) p dµ ( z )= C Z D Z | w |≤ r + Z | w | >r ! p dµ ( z ) ARLESON MEASURES ON FUNCTION SPACES 11

For the ﬁrst term, we have Z D Z | w |≤ r . . . ! p dµ ( z ) < ε p Z D (cid:18)Z D (1 − | w | ) α | − ¯ wz | α dA ( w ) (cid:19) p dµ ( z ) ≤ Cε p . For the second term, proceeding as in the proof of part (a), we have Z D Z | w | >r . . . ! p dµ ( z ) ≤ C Z | w | >r | f ′ n ( w ) | p (1 − | w | ) pα (cid:18) log 21 − | w | (cid:19) p Z D dµ ( z ) | − ¯ wz | pα +2 − p dA ( w ) < Cε p k f n k pD ps + p − ≤ Cε p . This ﬁnishes the proof of (b) for p > . For < p ≤ , we observe ﬁrst that the conditionon µ implies that µ is a vanishing s -Carleson measure. Therefore, given ε > , there is an r with < r < such that(1) sup | a | >r Z D (1 − | a | ) p + βp − s | − ¯ az | p + βp dµ ( z ) < ε, where β is a sufﬁciently large number so that β ≥ s + p − and βp > s − p .Let { f n } be a bounded sequence in D ps + p − converging to zero uniformly on compactsubsets of D , and for r > ﬁx an r -lattice { a k } in the Bergman metric. Since | a k | → ,there are only k points a k with | a k | ≤ r . Now, the same argument used in part (a) gives k f n k pL p ( µ ) ≤ C k µ k CM s k X k =1 (1 − | a k | ) s + p − Z D ( a k , r ) | f ′ n ( ζ ) | p dA ( ζ )+ C X | a k | >r (cid:18) (1 − | a k | ) p − βp Z D dµ ( z ) | − ¯ a k z | p + βp (cid:19) Z D ( a k , r ) | f ′ n ( ζ ) | p dA ( ζ )= ( I ) + ( II ) . Since { f n } converges to zero uniformly on compact sets, the same is true for the sequenceof its derivatives { f ′ n } , and hence there is a positive integer n such that for all n ≥ n sup (cid:8) | f ′ n ( ζ ) | p : ζ ∈ D ( a k , r ) (cid:9) < εk , ≤ k ≤ k . This gives ( I ) ≤ C k µ k CM s ε. Also, using (1), the same proof given in part (a) yields ( II ) ≤ C N ε k f n k pD ps + p − ≤ CN ε, where N is a positive integer such that each point of the unit disk belongs to at most N ofthe sets D ( a k , r ) . This ﬁnishes the proof of the lemma. (cid:3) Now we are ready to prove our main result in this section.

Proof of Theorem 3.1.

We ﬁrst prove (i). Suppose µ is a ( p, s ) -logarithmic Carlesonmeasure. Let s > and p > satisfy s + p > . By Lemma 3.2, µ is a Carleson measurefor D ps + p − .Given any subarc I of ∂D , let w = (1 − | I | ) ζ and ζ be the center on I . An easycomputation shows that, for any z ∈ S ( I ) , − | w | ≈ | − ¯ wz | ≈ | I | . Take any function f ∈ F ( p, p − , s ) . By Corollary 2.8 in [27], F ( p, p − , s ) ⊂ B . Thus,from a well-known estimate of Bloch functions (see, for example [32]), | f ( w ) | ≤ |k f k| B log 21 − | w | ≤ C |k f k| F ( p,p − ,s ) log 21 − | w | ≤ C |k f k| F ( p,p − ,s ) log 2 | I | . This, together with the fact that µ ∈ LCM p,s , gives | I | s Z S ( I ) | f ( z ) | p dµ ( z ) ≤ C | I | s Z S ( I ) | f ( z ) − f ( w ) | p dµ ( z ) + | f ( w ) | p µ ( S ( I )) ! ≤ C (1 − | w | ) s Z D (cid:12)(cid:12)(cid:12)(cid:12) f ( z ) − f ( w )(1 − ¯ wz ) s/p (cid:12)(cid:12)(cid:12)(cid:12) p dµ ( z ) + C k f k pF ( p,p − ,s ) k µ k LCM p,s . On the other hand, if we let f w ( z ) = (1 − | w | ) s/p f ( z ) − f ( w )(1 − ¯ wz ) s/p , then, applying Lemma 3.2, we get (1 − | w | ) s Z D (cid:12)(cid:12)(cid:12)(cid:12) f ( z ) − f ( w )(1 − ¯ wz ) s/p (cid:12)(cid:12)(cid:12)(cid:12) p dµ ( z ) = Z D | f w ( z ) | p dµ ( z ) ≤ C k µ k LCM p,s k f w k pD ps + p − ≤ C k µ k LCM p,s k f k pF ( p,p − ,s ) , since k f w k pD ps + p − ≤ C k f k pF ( p,p − ,s ) , with C being a positive constant independent of w .Indeed, notice ﬁrst that (1 −| w | ) s | f ( w ) − f (0) | p ≤ C k f k F ( p,p − ,s ) (1 −| w | ) s (cid:0) log 21 − | w | (cid:1) p ≤ C k f k F ( p,p − ,s ) , and this, together with Proposition 2.8, yields k f w k pD ps + p − = | f w (0) | p + Z D | f ′ w ( z ) | p (1 − | z | ) s + p − dA ( z ) ≤ (1 − | w | ) s | f ( w ) − f (0) | p + C (1 − | w | ) s Z D | f ′ ( z ) | p | − ¯ wz | s (1 − | z | ) s + p − dA ( z )+ C (1 − | w | ) s Z D | f ( z ) − f ( w ) | p | − ¯ wz | s + p (1 − | z | ) s + p − dA ( z ) ≤ C k f k pF ( p,p − ,s ) . ARLESON MEASURES ON FUNCTION SPACES 13

This shows that the identity operator I : F ( p, p − , s ) → T ∞ p,s ( µ ) is a bounded operator.Conversely, suppose that the identity operator I : F ( p, p − , s ) → T ∞ p,s ( µ ) is bounded.Given any arc I ⊂ ∂D , let a = (1 − | I | ) ζ , where ζ is the center of I , and consider thefunction f a ( z ) = log 21 − ¯ az . Then by Lemma 2.6, there is a constant

C > such that |k f a k| F ( p,p − ,s ) ≤ C. Since I : F ( p, p − , s ) → T ∞ p,s ( µ ) is bounded, we get | I | s Z S ( I ) | f a ( z ) | p dµ ( z ) ≤ C |k f a k| pF ( p,p − ,s ) ≤ C. It is easy to see that f a ( z ) ≈ log 2 | I | for any z ∈ S ( I ) . Combining the above two inequalities we get | I | s (cid:18) log 2 | I | (cid:19) p µ ( S ( I )) ≤ C. Thus µ is a ( p, s ) -logarithmic Carleson measure, completing the proof of (i).To prove (ii), ﬁrst let µ be a vanishing ( p, s ) -logarithmic Carleson measure and { f n } bea bounded sequence in F ( p, p − , s ) with f n ( z ) → uniformly on every compact subsetof D . We need to prove that k f n k T ∞ p,s ( µ ) → .The case s ≥ is easier. Let D r = { z ∈ D : | z | ≤ r } , and µ r = µ | D \ D r be therestriction of µ on D \ D r . By Proposition 2.9, if r → then sup I ⊂ ∂D | I | s (cid:18) log 2 | I | (cid:19) p µ r ( S ( I )) → . Now, by part (i), the fact that the limit k f n k T ∞ p,s ( µ ) → follows from Z S ( I ) | f n ( z ) | p dµ ( z ) ≤ C Z S ( I ) | f n ( z ) | p dµ D r ( z )+ C k f n k pF ( p,p − ,s ) (cid:18) log 2 | I | (cid:19) p µ r ( S ( I )) . For < s < we don’t have Proposition 2.9 at our disposal, so that the proof must followa different route that also works for s ≥ . For any arc I , let w = w I = (1 − | I | ) ζ , with ζ being the center of I . Then we have | I | s Z S ( I ) | f n ( z ) | p dµ ( z ) ≤ C | I | s Z S ( I ) | f n ( z ) − f n ( w ) | p dµ ( z ) + | f n ( w ) | p µ ( S ( I )) ! . Since µ is a vanishing ( p, s ) -logarithmic Carleson measure on D , for any ε > , there isan r , < r < , such that sup | I | < − r (cid:16) log 2 | I | (cid:17) p µ (cid:0) S ( I ) (cid:1) | I | s < ε. This, together with the facts that { f n } is a bounded sequence in F ( p, p − , s ) and theinequality | f n ( w ) | ≤ C |k f n |k F ( p,p − ,s ) log | I | gives sup | I | < − r | f n ( w ) | p µ (cid:0) S ( I ) (cid:1) | I | s < Cε. If | I | ≥ − r then | w | ≤ r , and since { f n } converges to zero uniformly on compactsubsets of D , there is a positive integer n such that sup | I |≥ − r | f n ( w ) | p µ (cid:0) S ( I ) (cid:1) | I | s < ε for n ≥ n . For the other term, we have J n := 1 | I | s Z S ( I ) | f n ( z ) − f n ( w ) | p dµ ( z ) ≤ C (1 − | w | ) s Z D (cid:12)(cid:12)(cid:12)(cid:12) f n ( z ) − f n ( w )(1 − ¯ wz ) s/p (cid:12)(cid:12)(cid:12)(cid:12) p dµ ( z )= C Z D | f n,w ( z ) | p dµ ( z ) , where f n,w ( z ) = (1 − | w | ) sp ( f n ( z ) − f n ( w ))(1 − ¯ wz ) sp . Fix a sufﬁciently large number β so that β ≥ s + p − and p (1 + β ) > s . The proof ofpart (b) in Lemma 3.2 gives k f n,w k pL p ( µ ) ≤ C | f n,w (0) | p µ ( D ) + C Z D Z | ζ |≤ r | f ′ n,w ( ζ ) | (1 − | ζ | ) β | − ¯ ζz | β dA ( ζ ) ! p dµ ( z )+ C ε k f n,w k pD ps + p − . Since { f n } is a bounded sequence in F ( p, p − , s ) , it follows from the proof of the bound-edness part that k f n,w k D ps + p − ≤ C k f n k F ( p,p − ,s ) ≤ C, with C independent of n and w .It is clear that f n,w (0) = (1 − | w | ) sp ( f n (0) − f n ( w )) . It is well-known F ( p, p − , s ) ⊂ B , the Bloch space. Hence { f n } is a bounded sequencein B and so there is a constant K > such that k f n k B ≤ K . Thus | f n,w (0) | p = (1 − | w | ) s | f n (0) − f n ( w ) | p ≤ K p (1 − | w | ) s (cid:18) log 21 − | w | (cid:19) p , and so for any ε > and any n ∈ N there is an r ∈ (0 , such that | f n,w (0) | p < ε whenever r < | w | < . On the other hand, since f n ( w ) → uniformly on compactsubsets of D , we know that there is an N > such that if n ≥ N then | f n,w (0) | p = (1 − | w | ) s | f n (0) − f n ( w ) | p < ε for all | w | ≤ r . Combining the above arguments we know that sup w ∈ D | f n,w (0) | p < ε if n is sufﬁciently large.It remains only to deal with the term A ( n, w ) := Z D Z | ζ |≤ r | f ′ n,w ( ζ ) | (1 − | ζ | ) β | − ¯ ζz | β dA ( ζ ) ! p dµ ( z ) . ARLESON MEASURES ON FUNCTION SPACES 15

We have that A ( n, w ) ≤ max(1 , p − ) (cid:16) A ( n, w ) + A ( n, w ) (cid:17) with A ( n, w ) := (1 − | w | ) s Z D Z | ζ |≤ r | f ′ n ( ζ ) | (1 − | ζ | ) β | − ¯ ζz | β | − ¯ wζ | sp dA ( ζ ) ! p dµ ( z ) , and A ( n, w ) := (1 − | w | ) s Z D Z | ζ |≤ r | f n ( ζ ) − f n ( w ) | (1 − | ζ | ) β | − ¯ ζz | β | − ¯ wζ | s + pp dA ( ζ ) ! p dµ ( z ) . Since { f ′ n } converges to zero uniformly on compact subsets of D , it is clear that sup w ∈ D A ( n, w ) < ε for n big enough. Similarly, if < r < is ﬁxed, it follows from the fact that { f n } converges uniformly to zero on compact subsets that lim n →∞ sup | w |≤ r A ( n, w ) = 0 . Since µ is a vanishing ( p, s ) -logarithmic Carleson measure, we can choose r so that sup | w | >r (cid:18) log 21 − | w | (cid:19) p Z D (1 − | w | ) s | − ¯ wz | s dµ ( z ) < ε. This together with Lemma 2.4 gives sup | w | >r (1 − | w | ) s | f n ( w ) | p Z D Z D (1 − | ζ | ) β | − ¯ ζz | β | − ¯ wζ | s + pp dA ( ζ ) ! p dµ ( z ) ≤ C sup | w | >r (1 − | w | ) s | f n ( w ) | p Z D dµ ( z ) | − ¯ wz | s ≤ C |k f n |k F ( p,p − ,s ) sup | w | >r (cid:18) log 21 − | w | (cid:19) p Z D (1 − | w | ) s | − ¯ wz | s dµ ( z ) < Cε. Now it is easy to deduce that lim n →∞ sup | w | >r A ( n, w ) = 0 completing this part of the proof.Conversely, suppose that for any bounded sequence { f n } in F ( p, p − , s ) with f n ( z ) → uniformly on every compact subset of D we have k f n k T ∞ p,s ( µ ) → . Let { I n } be asequence of subarcs of ∂D such that | I n | → . Let ζ n be the center of I n , w n = (1 −| I n | ) ζ n , and f n ( z ) = (cid:18) log 21 − | w n | (cid:19) − (cid:18) log 21 − ¯ w n z (cid:19) . Arguing as in Lemma 2.6, we easily see that { f n } is a bounded sequence on F ( p, p − , s ) ,and f n ( z ) → uniformly on every compact subset of D . Thus | I n | s (cid:18) log 2 | I n | (cid:19) p µ ( S ( I n )) ≤ C | I n | s Z S ( I n ) | f n ( z ) | p dµ ( z ) ≤ C k f n k pT ∞ p,s ( µ ) → , proving that µ is a vanishing ( p, s ) -logarithmic Carleson measure. The proof is complete. (cid:3) Using Lemma 2.1 we immediately get the following corollary of Theorem 3.1.

Corollary 3.3.

Let s > and p > satisfy s + p > , and let µ be a nonnegative Borelmeasure on D . Suppose that the point evaluation is a bounded functional on T ∞ p,s ( µ ) . Then I : F ( p, p − , s ) → T ∞ p,s ( µ ) is compact if and only if µ is a vanishing ( p, s ) -logarithmicCarleson measure.

4. E

MBEDDINGS FROM F ( p, pα − , s ) INTO T ∞ p,s ( µ ) WITH α = 1 In the previous section we studied the embedding I : F ( p, p − , s ) → T ∞ p,s ( µ ) . In thissection we consider the embedding I : F ( p, pα − , s ) → T ∞ p,s ( µ ) when α = 1 . First, welook at the case < α < . We need ﬁrst the following lemma from [15]. Lemma 4.1.

Let < α < and let T be a bounded linear operator from B α into anormed linear space Y . Then: T is compact if and only if k T f n k Y → , whenever { f n } is a bounded sequence in B α that converges to uniformly on D , the closure of D .Remark. It is clear from the proof given in [15] that the above lemma also holds for Y = T ∞ p,s ( µ ) in the case < p < . Theorem 4.2.

Let < α < , s > and p > satisfy s + pα > , and let µ be anonnegative Borel measure on D . Then the following conditions are equivalent: (i) I : F ( p, pα − , s ) → T ∞ p,s ( µ ) is bounded. (ii) I : F ( p, pα − , s ) → T ∞ p,s ( µ ) is compact. (iii) µ is an s -Carleson measure.Proof. (i) = ⇒ (iii). Suppose I : F ( p, pα − , s ) → T ∞ p,s ( µ ) is a bounded operator. Let f ( z ) = 1 . Then obviously f ∈ F ( p, pα − , s ) . Hence, | I | s µ ( S ( I )) = 1 | I | s Z S ( I ) | | p dµ ( z ) ≤ C |k k| pF ( p,pα − ,s ) ≤ C. Thus µ is an s -Carleson measure.(iii) = ⇒ (i). Suppose that µ is an s -Carleson measure. By Corollary 2.8 in [27], F ( p, pα − , s ) ⊂ B α . If we can prove that I : B α → T ∞ p,s ( µ ) is bounded then we are done. Since B α ⊂ H ∞ for < α < , a standard application of the Closed Graph Theorem showsthat k f k H ∞ ≤ C |k f k| B α . Hence, | I | s Z S ( I ) | f ( z ) | p dµ ( z ) ≤ k f k pH ∞ | I | s µ ( S ( I )) ≤ C |k f k| B α . Thus I : B α → T ∞ p,s ( µ ) is bounded. Consequently, I : F ( p, pα − , s ) → T ∞ p,s ( µ ) isbounded.(iii) = ⇒ (ii). We further prove that (iii) implies (ii). Since F ( p, pα − , s ) ⊂ B α (seeCorollary 2.8 in [27]), it is enough to prove that I : B α → T ∞ p,s ( µ ) is compact. Sincewe just proved that I : B α → T ∞ p,s ( µ ) is bounded, by Lemma 4.1, we need only provethat k f n k T ∞ p,s ( µ ) → , whenever { f n } is a bounded sequence in B α that converges to uniformly on D . Since { f n } converges to uniformly on D . for any given ε > , thereexists a positive integer N such that | f n ( z ) | < ε for any n > N . Hence, for any n > N we have k f n k pT ∞ p,s ( µ ) = sup I ⊂ ∂D | I | s Z S ( I ) | f n ( z ) | p dµ ( z ) ≤ ε sup I ⊂ ∂D | I | s Z S ( I ) dµ ( z ) ≤ Cε.

ARLESON MEASURES ON FUNCTION SPACES 17

Therefore lim n →∞ k f n k pT ∞ p,s ( µ ) = 0 , and so I : F ( p, pα − , s ) → T ∞ p,s ( µ ) is compact.(ii) = ⇒ (i). This is obvious. The proof is complete. (cid:3) Next, we consider the case α > . Theorem 4.3.

Let p > , s > and α > with s + pα > . Let µ be a ﬁnite positiveBorel measure on D . If µ is a Carleson measure for D ps + pα − , then the embedding I : F ( p, pα − , s ) → T ∞ p,s ( µ ) is bounded.Proof. Let β = s + p ( α − so that s + pα − β + p − . We prove ﬁrst that µ mustbe a β -Carleson measure. For each a ∈ D , consider the test functions f a ( z ) = (1 − | a | ) βp (1 − ¯ az ) βp , z ∈ D. Then sup a ∈ D k f a k D pβ + p − ≤ C , and since µ is a Carleson measure for D pβ + p − , one obtains Z D (1 − | a | ) β | − ¯ az | β dµ ( z ) = Z D | f a ( z ) | p dµ ( z ) ≤ C. Hence µ is a β -Carleson measure. Now, by Proposition 2.3, we have that k f k pT ∞ p,s ( µ ) iscomparable to the quantity sup a ∈ D Z D (1 − | a | ) t | − ¯ az | s + t | f ( z ) | p dµ ( z ) for all t > . Fix a ∈ D , and let t = s + 2 p ( α − , and f ∈ F ( p, pα − , s ) . Then Z D (1 − | a | ) t | − ¯ az | s + t | f ( z ) | p dµ ( z ) ≤ max(1 , p − ) (cid:0) I ( a ) + I ( a ) (cid:1) , where I ( a ) = | f ( a ) | p Z D (1 − | a | ) t | − ¯ az | s + t dµ ( z ) , and I ( a ) = Z D (1 − | a | ) t | − ¯ az | s + t | f ( z ) − f ( a ) | p dµ ( z ) . Since F ( p, pα − , s ) ⊂ B α , and α > , one has(2) | f ( a ) | p ≤ C |k f k| pF ( p,pα − ,s ) (1 − | a | ) − p ( α − . This, together with the fact that µ is an (cid:2) s + p ( α − (cid:3) -Carleson measure and Lemma 2.2,gives I ( a ) ≤ C |k f k| pF ( p,pα − ,s ) (1 − | a | ) t − p ( α − Z D dµ ( z ) | − ¯ az | s + t = C |k f k| pF ( p,pα − ,s ) (1 − | a | ) s + p ( α − Z D dµ ( z ) | − ¯ az | s +2 p ( α − ≤ C |k f k| pF ( p,pα − ,s ) . It remains to deal with the term I ( a ) . Set f a ( z ) = f ( z ) − f ( a )(1 − ¯ az ) s + tp . Since µ is a Carleson measure for D ps + pα − , we obtain I ( a ) =(1 − | a | ) t Z D | f a ( z ) | p dµ ( z ) ≤ C (1 − | a | ) t | f a (0) | p + C (1 − | a | ) t Z D (cid:12)(cid:12) ( f a ) ′ ( z ) (cid:12)(cid:12) p (1 − | z | ) s + pα − dA ( z ) . By (2), one has (1 − | a | ) t | f a (0) | p = (1 − | a | ) t | f ( a ) − f (0) | p ≤ C (1 − | a | ) s + p ( α − k f k pF ( p,pα − ,s ) ≤ C k f k pF ( p,pα − ,s ) . On the other hand, by Lemma 2.2 and Proposition 2.8, (1 − | a | ) t Z D (cid:12)(cid:12) ( f a ) ′ ( z ) (cid:12)(cid:12) p (1 − | z | ) s + pα − dA ( z ) ≤ C (1 − | a | ) t Z D | f ′ ( z ) | p (1 − | z | ) s + pα − | − ¯ az | s + t dA ( z )+ C (1 − | a | ) t Z D | f ( z ) − f ( a ) | p | − ¯ az | s + t + p (1 − | z | ) s + pα − dA ( z ) ≤ C k f k pF ( p,pα − ,s ) . All together gives I ( a ) ≤ C k f k pF ( p,pα − ,s ) , ﬁnishing the proof of the theorem. (cid:3) Theorem 4.4.

Let α > , s > , and p > with s + pα > , and let µ be a nonnegativeBorel measure on D . If < p ≤ ; or p > with s + p ( α − > ; or < p ≤ with s + p ( α −

1) = 1 , then (i)

The identity operator I : F ( p, pα − , s ) → T ∞ p,s ( µ ) is bounded if and only if µ isan [ s + p ( α − -Carleson measure. (ii) The following two conditions are equivalent: (a)

For any bounded sequence { f n } in F ( p, pα − , s ) satisfying f n ( z ) → uniformly on every compact subset of D , lim n →∞ k f n k T ∞ p,s ( µ ) = 0 . (b) µ is a vanishing [ s + p ( α − -Carleson measure.Proof. (i) Suppose ﬁrst that µ is an [ s + p ( α − -Carleson measure. In all the casesconsidered, this is equivalent to µ being a Carleson measure for D ps + αp − (see [25]. Notethat for < p ≤ , the implication needed here is proved in our proof of Lemma 3.2).Thus, by Theorem 4.3 it follows that I : F ( p, pα − , s ) → T ∞ p,s ( µ ) is bounded.Conversely, let I : F ( p, pα − , s ) → T ∞ p,s ( µ ) be bounded. Given any arc I ⊂ ∂D , let a = (1 − | I | ) ζ , where ζ is the center of I . Let f a ( z ) = (1 − ¯ az ) − α . By Lemma 2.6, there is a constant

C > such that |k f a k| F ( p,pα − ,s ) ≤ C. Since I : F ( p, pα − , s ) → T ∞ p,s ( µ ) is bounded, we get | I | s Z S ( I ) | f a ( z ) | p dµ ( z ) ≤ C |k f a k| pF ( p,pα − ,s ) ≤ C. ARLESON MEASURES ON FUNCTION SPACES 19

It is easy to see that f a ( z ) ≈ | I | − α for any z ∈ S ( I ) . Combining the above two inequalities we get | I | s + p ( α − µ ( S ( I )) ≤ C. Thus µ is an [ s + p ( α − -Carleson measure.Next, we prove (ii). Let µ be a vanishing [ s + p ( α − -Carleson measure and { f n } be a bounded sequence in F ( p, pα − , s ) with f n ( z ) → uniformly on every compactsubset of D . We ﬁrst deal with the cases considered with s + p ( α − ≥ . Let D r = { z ∈ D : | z | ≤ r } , and µ r = µ | D \ D r be the restriction of µ on D \ D r . By Proposition2.9, if r → then sup I ⊂ ∂D µ r (cid:0) S ( I ) (cid:1) | I | s + p ( α − → . Recall that, in the introduction we have seen that, when α > , if g ∈ B α then there is aconstant C > , independent of g , such that | g ( z ) | ≤ C k| g k| B α (1 − | z | ) − α . By Corollary 2.8 in [27], F ( p, pα − , s ) ⊂ B α . Thus, if f ∈ F ( p, pα − , s ) then | f ( z ) | ≤ C k| f k| B α (1 − | z | ) − α ≤ C k| f k| F ( p,pα − ,s ) (1 − | z | ) − α . Hence, given any arc I ⊂ ∂D with w = (1 −| I | ) ζ and ζ is the center of I , for our sequence { f n } we can ﬁnd a uniform constant C > such that | f n ( w ) | ≤ C k| f n k| F ( p,pα − ,s ) (1 − | w | ) α − = C k| f n k| F ( p,pα − ,s ) | I | α − . The fact that the limit k f n k T ∞ p,s ( µ ) → now follows from part (i) and Z S ( I ) | f n ( z ) | p dµ ( z ) ≤ C Z S ( I ) | f n ( z ) | p dµ D r ( z ) + C k| f n k| pF ( p,pα − ,s ) µ r ( S ( I )) | I | p ( α − . The proof for < p ≤ follows a similar argument as the corresponding one in Theorem3.1. We left the details to the interested reader.Conversely, suppose that for any bounded sequence { f n } in F ( p, pα − , s ) with f n ( z ) → uniformly on every compact subset of D we have k f n k T ∞ p,s ( µ ) → . Let { I n } be a se-quence of subarcs of ∂D such that | I n | → . Let ζ n be the center of I n , w n = (1 −| I n | ) ζ n ,and f n ( z ) = 1 − | w n | (1 − ¯ w n z ) α . By a proof similar to that of Lemma 2.6, we can easily see that { f n } is a bounded sequenceon F ( p, pα − , s ) , and f n ( z ) → uniformly on every compact subset of D . Clearly, forany z ∈ S ( I ) , | f n ( z ) | ≈ | I n | − α . Thus µ ( S ( I n )) | I n | s + p ( α − ≤ C | I n | s Z S ( I n ) | f n ( z ) | p dµ ( z ) ≤ C k f n k pT ∞ p,s ( µ ) → . Hence µ is a vanishing [ s + p ( α − -Carleson measure. The proof is complete. (cid:3) Using Lemma 2.1 we immediately get the following corollary of Theorem 4.4.

Corollary 4.5.

Let α > , s > and p > satisfy s + αp > , and let µ be a nonnegativeBorel measure on D . Suppose that the point evaluation is a bounded functional on T ∞ p,s ( µ ) .If < p ≤ ; or p > with s + p ( α − > ; or < p ≤ with s + p ( α −

1) = 1 , then I : F ( p, pα − , s ) → T ∞ p,s ( µ ) is compact if and only if µ is a vanishing [ s + p ( α − -Carleson measure.

5. R

IEMANN -S TIELTJES OPERATORS ON F ( p, q, s ) In this section we look at applications of our main theorems to Riemann-Stieltjes inte-gral operators. Recall that, for g ∈ H ( D ) , the Riemann-Stieltjes integral operator J g isdeﬁned by J g f ( z ) = Z z f ( ζ ) g ′ ( ζ ) dζ for f ∈ H ( D ) . The operators J g were ﬁrst used by Ch. Pommerenke in [20] to char-acterize BMOA functions. They were ﬁrst systematically studied by A. Aleman and A.G. Siskakis in [2]. They proved that J g is bounded on the Hardy space H p if and onlyif g ∈ BM OA . Thereafter there have been many works on these operators. See, [1],[3], [13], [19], and [22] for a few examples. Here we are considering boundedness andcompactness of these operators on F ( p, q, s ) .For < p < ∞ , − < q < ∞ , ≤ s < ∞ such that q + s > − , we deﬁne the space F L ( p, q, s ) , called logarithmic F ( p, q, s ) space , as the space of analytic functions f on D satisfying sup a ∈ D (cid:18) log 21 − | a | (cid:19) p Z D | f ′ ( z ) | p (1 − | z | ) q (1 − | ϕ a ( z ) | ) s dA ( z ) < ∞ . We also say f ∈ F L, ( p, q, s ) , if lim | a |→ (cid:18) log 21 − | a | (cid:19) p Z D | f ′ ( z ) | p (1 − | z | ) q (1 − | ϕ a ( z ) | ) s dA ( z ) = 0 . The following result establishes the boundedness of J g on F ( p, q, s ) . Theorem 5.1.

Let α > , s > and p > satisfy s + pα > . Then we have the followingresults. (i) As < α < , J g is bounded on F ( p, pα − , s ) if and only if g ∈ F ( p, pα − , s ) . (ii) J g is bounded on F ( p, p − , s ) if and only if g ∈ F L ( p, p − , s ) . (iii) Let α > and γ = s + p ( α − . For the cases < p ≤ ; or p > and γ > ;or < p ≤ and γ = 1 , we have that J g is bounded on F ( p, pα − , s ) if andonly if g ∈ F ( p, p − , γ ) .Proof. Since ( J g f ) ′ = f g ′ , the operator J g is bounded on F ( p, pα − , s ) if and only if Z D | f ( z ) | p | g ′ ( z ) | p (1 − | z | ) pα − (1 − | ϕ a ( z ) | ) s dA ( z ) ≤ C k| f k| F ( p,pα − ,s ) . Since − | ϕ a ( z ) | = (1 − | z | ) | ϕ ′ a ( z ) | , the above inequality is equivalent to(3) Z D | f ( z ) | p | ϕ ′ a ( z ) | s dµ g ( z ) ≤ C k| f k| F ( p,pα − ,s ) , where dµ g ( z ) = | g ′ ( z ) | p (1 − | z | ) s + pα − dA ( z ) . By Proposition 2.3, (3) means that I : F ( p, pα − , s ) → T ∞ p,s ( µ g ) is a bounded operator. ARLESON MEASURES ON FUNCTION SPACES 21 (i) As < α < , by Theorem 4.2, (3) is equivalent to that µ g is an s -Carleson measure.By Lemma 2.2, this means that sup a ∈ D Z D | ϕ ′ a ( z ) | s dµ g ( z ) < ∞ , which is the same as sup a ∈ D Z D | g ′ ( z ) | p (1 − | z | ) pα − (1 − | ϕ a ( z ) | ) s dA ( z ) < ∞ . Thus g ∈ F ( p, pα − , s ) .(ii) As α = 1 , by Theorem 3.1, (3) is equivalent to that µ g is a ( p, s ) -logarithmicCarleson measure. By Lemma 2.2 (or also by Theorem 2 in [29]), this is equivalent to sup a ∈ D (cid:18) log 21 − | a | (cid:19) p Z D | ϕ ′ a ( z ) | s dµ g ( z ) < ∞ , or sup a ∈ D (cid:18) log 21 − | a | (cid:19) p Z D | g ′ ( z ) | p (1 − | z | ) p − (1 − | ϕ a ( z ) | ) s dA ( z ) < ∞ , which means g ∈ F L ( p, p − , s ) .(iii). Let α > . By Theorem 4.4, in all the cases considered, (3) means that µ g is an [ s + p ( α − -Carleson measure. By Lemma 2.2, this is equivalent to sup a ∈ D Z D | ϕ ′ a ( z ) | s + p ( α − dµ g ( z ) < ∞ , or sup a ∈ D Z D | g ′ ( z ) | p (1 − | z | ) p − (1 − | ϕ a ( z ) | ) s + p ( α − dA ( z ) < ∞ . which means g ∈ F ( p, p − , s + p ( α − . The proof is complete. (cid:3) Remark.

Parts of the above results have been proved before. When p = 2 , α = 1 and s = 1 , it is known that F (2 , ,

1) =

BM OA , the space of analytic functions of boundedmean oscillation. In this case, the above result was proved by Siskakis and the secondauthor in [22]. When p = 2 , α = 1 , < s < , we know that F ( p, pα − , s ) = Q s .In this case, the result was proved by Xiao in [26]. Note also that in the case α > and γ = s + p ( α − > , by Theorem 1 in [28] (also see Theorem 1.3 in [27]), we know that F ( p, p − , γ ) = B , the Bloch space.We can also use our previous results to characterize compactness of J g on F ( p, q, s ) . Theorem 5.2.

Let α > , s > and p > satisfy s + pα > . Then we have the followingresults. (i) As < α < , J g is compact on F ( p, pα − , s ) if and only if g ∈ F ( p, pα − , s ) . (ii) J g is compact on F ( p, p − , s ) if and only if g ∈ F L, ( p, p − , s ) . (iii) Let α > and γ = s + p ( α − . For the cases < p ≤ ; or p > and γ > ;or < p ≤ and γ = 1 , we have that J g is compact on F ( p, pα − , s ) if andonly if g ∈ F ( p, p − , γ ) .Proof. As in the proof of Theorem 5.1, we know that the operator J g is compact on F ( p, pα − , s ) if and only if for any bounded sequence { f n } in F ( p, pα − , s ) with f n ( z ) → uniformly on compact subsets of D , we have lim n →∞ k J g ( f n ) k F ( p,pα − ,s ) = 0 . Thus we can apply Theorem 3.1, Theorem 4.2 and Theorem 4.4 to complete the proof. Theproof goes in the same way as the proof of Theorem 5.1, so we omit the details here. (cid:3)

Remark.

Recall that a function f analytic on D belongs to F ( p, q, s ) if f ∈ F ( p, q, s ) and lim | a |→ − Z D f ′ ( z ) | p (1 − | z | ) q (1 − | ϕ a ( z ) | ) s dA ( z ) = 0 . When α > and γ = s + p ( α − > , one has F ( p, p − , γ ) = B , the little Blochspace. 6. P OINTWISE MULTIPLICATION OPERATORS ON F ( p, q, s ) For an F -space X of analytic functions, we denote by M ( X ) the space of all pointwisemultipliers on X , that is, M ( X ) = { g ∈ H ( D ) : f g ∈ X for all f ∈ X } . Let M g denote the pointwise multiplication operator, given by M g f = f g . Since theClosed Graph Theorem is still available for F -spaces (see [11, Chap. II]), we know that g ∈ M ( X ) if and only if M g is bounded on X . We also denote M ( X ) = { g ∈ H ( D ) : M g is compact on X } . In this section, we give characterizations of M ( F ( p, q, s )) and M ( F ( p, q, s )) . To thisend, we ﬁrst study another integral operator. Let g be an analytic function on D . Deﬁne I g f ( z ) = Z z f ′ ( t ) g ( t ) dt, f ∈ H ( D ) . Proposition 6.1.

Let α > , s > and p > satisfy s + pα > . Then (i) I g is bounded on F ( p, pα − , s ) if and only if g ∈ H ∞ . (ii) I g is compact on F ( p, pα − , s ) if and only if g = 0 .Proof. (i) Let g ∈ H ∞ . Since ( I g f ) ′ = f ′ g , it is obvious that I g is bounded on F ( p, pα − , s ) .Conversely, suppose I g is bounded on F ( p, pα − , s ) . Let a ∈ D , and let h a ( z ) = 1 − | a | α (1 − ¯ az ) α . By a computation similar to the proof of Lemma 2.6, we can see that sup a ∈ D k h a k F ( p,pα − ,s ) < ∞ . Since F ( p, pα − , s ) ⊂ B α , we know that I g is bounded from F ( p, pα − , s ) to B α .Thus, k I g h a k B α ≤ C k h a k F ( p,pα − ,s ) ≤ C. Since ( I g h a ) ′ = h ′ a g , the above inequality is the same as(4) sup z ∈ D | h ′ a ( z ) || g ( z ) | (1 − | z | ) α ≤ C. Since h ′ a ( z ) = ¯ a (1 − | a | )(1 − ¯ az ) − α − , we get h ′ a ( a ) = ¯ a (1 − | a | ) − α . Letting z = a in(4) we get | a || g ( a ) | ≤ C. Since g is analytic on D , this implies that g ∈ H ∞ .(ii) Let g = 0 , then trivially, I g is compact on F ( p, pα − , s ) . ARLESON MEASURES ON FUNCTION SPACES 23

Conversely, suppose I g is compact on F ( p, pα − , s ) . Then I g is bounded on F ( p, pα − , s ) , and so by (i), g ∈ H ∞ . Since F ( p, pα − , s ) ⊂ B α , we know that I g is compactfrom F ( p, pα − , s ) to B α . Let { a n } be any sequence of points in D such that | a n | → ,and let h n ( z ) = 1 − | a n | α (1 − ¯ a n z ) α . As in the proof of (i), we know that sup n ≥ k h n k F ( p,pα − ,s ) < ∞ . Obviously, h n converges to uniformly on compact subsets of D . Hence lim n →∞ k I g h n k B α = 0 , which is equivalent to lim n →∞ sup z ∈ D | h ′ n ( z ) || g ( z ) | (1 − | z | ) α = 0 . Let z = a n we see lim n →∞ | h ′ n ( a n ) || g ( a n ) | (1 − | a n | ) α = 0 . Since h ′ n ( a n ) = ¯ a n (1 − | a n | ) − α , the above equation becomes lim n →∞ | a n || g ( a n ) | = 0 . Since g ∈ H ∞ , we must have g = 0 . The proof is complete. (cid:3) Theorem 6.2.

Let α > , s > and p > satisfy s + pα > . Then we have the followingresults. (i) If < α < , then M ( F ( p, pα − , s )) = F ( p, pα − , s ) . (ii) As α = 1 , we have that M ( F ( p, p − , s )) = H ∞ ∩ F L ( p, p − , s ) . (iii) Let α > and γ = s + p ( α − . Then (a) If < p ≤ , then M ( F ( p, pα − , s )) = F ( p, p − , γ ) . (b) If p > and γ > , then M ( F ( p, pα − , s )) = H ∞ ∩ B = H ∞ . (c) If < p ≤ and γ = 1 , then M ( F ( p, pα − , s )) = F ( p, p − , γ ) ∩ H ∞ .Proof. (i) If < α < , then F ( p, αp − , s ) ⊂ H ∞ , and it is easy to see that the space F ( p, αp − , s ) is an algebra.In order to prove (ii) and (iii), notice that(5) M g f ( z ) = f ( z ) g ( z ) = f (0) g (0) + I g f ( z ) + J g f ( z ) . Hence, if I g and J g are both bounded on F ( p, pα − , s ) , then M g is also bounded on F ( p, pα − , s ) . So the inclusions H ∞ ∩ F L ( p, p − , s ) ⊂ M ( F ( p, p − , s )) , for α = 1 , and, in all the cases considered, H ∞ ∩ F ( p, p − , γ ) ⊂ M ( F ( p, pα − , s )) , for α > , are direct consequences of Theorem 5.1 and (i) of Proposition 6.1.Now we are proving the inverse inclusions. Let α > and g ∈ M ( F ( p, pα − , s )) .For a ∈ D , let ψ a ( z ) = (1 − | a | ) (1 − ¯ az ) α +1 − − | a | (1 − ¯ az ) α . By a computation similar to the proof of Lemma 2.6, we get that sup a ∈ D k ψ a k F ( p,pα − ,s ) ≤ C < ∞ . It is also easy to check that ψ a ( a ) = 0 , ψ ′ a ( a ) = ¯ a (1 − | a | ) − α . Since M g is bounded on F ( p, pα − , s ) and F ( p, pα − , s ) ⊂ B α , we get that M g isbounded from F ( p, pα − s, s ) to B α . Hence, there is a constant C > such that C k ψ a k F ( p,pα − ,s ) > sup a ∈ D k M g ψ a k B α = sup a ∈ D sup z ∈ D | g ′ ( z ) ψ a ( z ) + g ( z ) ψ ′ a ( z ) | (1 − | z | ) α ≥ | g ′ ( a ) ψ a ( a ) + g ( a ) ψ ′ a ( a ) | (1 − | a | ) α = | g ( a ) || a | (1 − | a | ) − α (1 − | a | ) α = | g ( a ) || a | . Hence g ∈ H ∞ . By Proposition 6.1, I g is bounded on F ( p, pα − , s ) . Hence, from(5) we know that J g is also bounded on F ( p, pα − , s ) . The results now follow fromTheorem 2.3 and the fact that g ∈ H ∞ . When α > , in all the cases considered, we obtain M ( F ( p, αp − , s )) = H ∞ ∩ F ( p, p − , γ ) , and the statements of (iii) are consequencesof F ( p, p − , γ ) ⊂ H ∞ for < p ≤ , and the fact that H ∞ ⊂ B and F ( p, p − , γ ) = B for γ > . The proof is complete. (cid:3) Remark.

The above result for

BM OA = F (2 , , was obtained ﬁrst by Stegenga in [23](see also [17] and [29]). When p = 2 , α = 1 , < s < , we know F (2 , , s ) = Q s with < s < . In this case the result was proved by Pau and Pel´aez in [18]. Theorem 6.3.

Let α > , s > and p > satisfy s + pα > . Then M ( F ( p, pα − , s )) = { } .Proof. Let g ∈ M ( F ( p, pα − , s )) . Then obviously, g ∈ M ( F ( p, pα − , s )) . ByTheorem 6.2 we know that g ∈ H ∞ . Let { a n } be any sequence in D such that | a n | → as n → ∞ . Let ψ n ( z ) = (1 − | a n | ) (1 − ¯ a n z ) α +1 − − | a n | (1 − ¯ a n z ) α . Then, as in the proof of Theorem 6.2, we know sup n ≥ k ψ n k F ( p,pα − ,s ) ≤ C < ∞ , ψ n ( a n ) = 0 , ψ ′ n ( a n ) = ¯ a n (1 − | a n | ) − α . Since M g is compact on F ( p, pα − , s ) and F ( p, pα − , s ) ⊂ B α , we get that M g iscompact from F ( p, pα − s, s ) to B α . Thus ← lim n →∞ k M g ψ n k B α = lim n →∞ sup z ∈ D | g ′ ( z ) ψ n ( z ) + g ( z ) ψ ′ n ( z ) | (1 − | z | ) α ≥ lim n →∞ | g ′ ( a ) ψ n ( a ) + g ( a ) ψ ′ n ( a ) | (1 − | a | ) α = lim n →∞ | g ( a n ) || a n | (1 − | a n | ) − α (1 − | a n | ) α = lim n →∞ | g ( a n ) || a n | . Since g ∈ H ∞ , we must have g = 0 . The proof is complete. (cid:3) ARLESON MEASURES ON FUNCTION SPACES 25

7. A

N OPEN QUESTION

Finally, we want to mention a natural question that remains open. The question concernsthe embedding I : F ( p, pα − , s ) → T ∞ p,s ( µ ) in the case α > . From Theorem 4.3we know that µ being a Carleson measure for D ps + αp − is a sufﬁcient condition for theboundedness. On the other hand, it is easy to see that if the embedding is bounded, then µ must be an [ s + p ( α − -Carleson measure. In the cases < p ≤ ; or s + p ( α − > ;or < p ≤ and s + p ( α −

1) = 1 it is well known that the two conditions are equivalent,allowing to obtain a complete description in that case (see Theorem 4.4). However, it isknown that the two conditions are no longer equivalent in the remaining cases. So, what isthe criterion for the boundedness and compactness of the embedding I : F ( p, pα − , s ) → T ∞ p,s ( µ ) in these cases? Is the converse of Theorem 4.3 true? How about the boundednessof the Riemann-Stieltjes operator J g and the multiplication operator M g on F ( p, pα − , s ) for this case? R EFERENCES[1] A. Aleman and J. A. Cima, An integral operator on H p and Hardy’s inequality, J. Anal. Math. (2001),157–176.[2] A. Aleman and A. G. Siskakis, An integral operator on H p , Complex Variables (1995), 149–158.[3] A. Aleman and A. G. Siskakis, Integration operators on Bergman spaces, Indiana Univ. Math. J. (1997),337–356.[4] R. Aulaskari, J. Xiao and R. Zhao, On subspaces and subsets of BMOA and

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