Riemann-stieltjes Operators between Bergman-type Spaces and Α-bloch Spaces

We study the following integral operators: J g f (z)= z 0 f (ξ)g (ξ)dξ; I g f (z)= z 0 f (ξ)g(ξ)dξ, where g is an analytic function on the open unit disk in the complex plane. The bounded-ness and compactness of J g , I g between the Bergman-type spaces and the α-Bloch spaces are investigated.


Introduction
Let D be the open unit disk in the complex plane.Denote by H(D) the class of all analytic functions on D. An analytic function f in D is said to belong to the α-Bloch space Ꮾ α , or Bloch-type space, if (1.1) The expression f α defines a seminorm while the natural norm is given by f Ꮾ α = | f (0)| + f α .It makes Ꮾ α into a Banach space.
A positive continuous function φ on [0,1) is normal, if there exists 0 < s < t such that (see [7]) For 0 < p < ∞ and a normal function φ, let H(p, p,φ) denote the space of all analytic functions f on D such that Here dA denotes the normalized Lebesgue area measure on the unit disk D such that A(D) = 1.We call H(p, p,φ) the Bergman-type space.If 1 ≤ p < ∞, H(p, p,φ) is a Banach 2 Riemann-Stieltjes operators between Bergman and Bloch space equipped with the norm f H(p,p,φ) .When 0 < p < 1, H(p, p,φ) is a Fréchlet space.
In particular, if φ(r) = (1 − r) 1/p , then H(p, p,φ) is the Bergman space A p .For an analytic function f (z) on the unit disk D with the Taylor expansion f (z) = ∞ n=0 a n z n , the Cesàro operator acting on f is (1.4) The integral form of Ꮿ is taking simply as a path the segment joining 0 and z, we have that (1.6) The following operator: is closely related to the previous operator and on many spaces the boundedness of these two operators is equivalent.It is well known that Cesàro operator acts as a bounded linear operator on various analytic function spaces (see, e.g., [6,9,13,15,16,18,20], and the references therein).Suppose that g : D → C 1 is an analytic map, f ∈ H(D).A class of integral operator introduced by Pommerenke is defined by (see [11]) (1.8) The operator J g can be viewed as a generalization of the Cesàro operator which was called the Riemann-Stieltjes operator (see [21]).
In [11], Pommerenke showed that J g is a bounded operator on the Hardy space H 2 if and only if g ∈ BMOA.Aleman and Siskakis showed that J g is bounded (compact) on the Hardy space H p , 1 ≤ p < ∞, if and only if g ∈ BMOA (g ∈ VMOA), and that J g is bounded (compact) on the Bergman space A p if and only if g ∈ Ꮾ (g ∈ Ꮾ 0 ), see [2,3].Recently, J g acting on various function spaces, including the Bloch space, the weighted Bergman space, the BMOA, and VMOA spaces have been studied (see [1-3, 17, 22], and the related references therein).
Another integral operator has recently been defined as the following (see [22]): In this paper, we study the boundedness and compactness of the operators J g , I g between the Bergman-type space and the α-Bloch space.

Songxiao Li 3
Constants are denoted by C in this paper, they are positive and may differ from one occurrence to the other.a b means that there is a positive constant C such that a ≤ Cb.Moreover, if both a b and b a hold, then one says that a b.

J g , I
In this section, we consider the boundedness and compactness of J g , I g : H(p, p,φ) → Ꮾ α .First, let us state some useful lemmas.
Proof.Let β(z,w) denote the Bergman metric between two points z and w in D. It is given by For a ∈ D and r > 0, the set D(a,r) = {z ∈ D : β(a,z) < r} is a Bergman metric disk with center a and radius r.It is well known that (see [25]) when z ∈ D(a,r).For 0 < r < 1 and z ∈ D, by the subharmonicity of | f (z)| p and the normality of φ, we get from which we get the desired result.
The following lemma can be found in [ The following lemma can be found in [14].
Proof.By (1.8), it is easy to see that (J g f (2.10) Taking supremum over the unit disk in this inequality, we obtain (2.12) Songxiao Li 5 It is easy to see that By [12], we get Since φ is normal, by Lemma 2.4, Therefore, f w ∈ H(p, p,φ) (or see [23]).Moreover, there is a positive constant C such that for every z,w ∈ D.
Proof.Similar to the case of J g , we have ( (2.20) For w ∈ D, let f w (z) be defined by (2.12).From (2.3) and (2.13), that is, Taking supremum in the last inequality over the set 1/2 ≤ |w| < 1 and noticing that by the maximum modulus principle there is a positive constant Proof.First, we assume that (2.24) holds.In order to prove that J g is compact, by Lemma 2.5, it suffices to show that if { f n } is a bounded sequence in H(p, p,φ) that converges to 0 uniformly on compact subsects of D, then J g f n Ꮾ α → 0. Let { f n } be a sequence in H(p, p,φ) with f n H(p,p,φ) ≤ 1 and f n → 0 uniformly on compact subsets of D. By the assumption, for any > 0, there is a constant δ, Note that K is a compact subsect of D and φ is normal, we have where By the assumption and Theorem 2.6, we obtain J g f n Ꮾ α → 0 as n → ∞.Therefore, J g : Then f n ∈ H(p, p,φ) and f n converges to 0 uniformly on compact subsets of D (see [7]).Since J g is compact, by Lemma 2.5, J g f n Ꮾ α → 0 as n → ∞.In addition, from which the result follows.
8 Riemann-Stieltjes operators between Bergman and Bloch Theorem 2.9.Assume that 0 < p < ∞, α > 0, and φ is normal on [0,1).Then I g : H(p, p,φ) → Ꮾ α is compact if and only if Let f n (z) be defined by (2.28).Then from the proof of Theorem 2.8 and the compactness of I g , I g f n Ꮾ α → 0 as n → ∞.In addition, from which we get the desired result.
Assume (2.30) holds, in order to prove that I g is compact, it suffices to show that if { f n } is a bounded sequence in H(p, p,φ) that converges to 0 uniformly on compact subsects of D, then I g f n Ꮾ α → 0. Let { f n } be a sequence in H(p, p,φ) with f n H(p,p,φ) ≤ 1 and f n → 0 uniformly on compact subsets of D. By the assumption, for any > 0, there is a constant δ, 0 < δ < 1, such that δ < |z| < 1 implies (2.32) Similar to the proof of Theorem 2.8, we have (2.33) Therefore, I g : H(p, p,φ) → Ꮾ α is compact.
From the introduction, we can easily get the following corollary.
Corollary 2.10.In this section, we characterize the boundedness and compactness of J g ,I g : H(p, p,φ) → Ꮾ α 0 .For this purpose, we need Lemma 3.1.When α = 1, Lemma 3.1 was proved in [8].For the general case, the proof is similar to the proof of the case α = 1.We omit the details.
Proof.(i) It is clear to see that g ∈ Ꮾ α 0 and J g : H(p, p,φ) → Ꮾ α is bounded if J g : H(p, p,φ) → Ꮾ α 0 is bounded.Conversely, suppose that J g : H(p, p,φ) → Ꮾ α is bounded and g ∈ Ꮾ α 0 .For any polynomial p(z), since g ∈ Ꮾ α 0 and we know that J g p ∈ Ꮾ α 0 .For any f ∈ H(p, p,φ), there exists a sequence of polynomials {p n } such that f − p n H(p,p,φ) → 0 as n → ∞.Since Ꮾ α 0 is closed, we get In addition, J g : In fact, Then the following statements hold.
Then the following statements hold.

J g ,I
The following lemma is well known(e.g., see [19]).
The following lemma can be found in [10].
Songxiao Li 11 if and only if there is a positive constant C such that for all analytic functions f in D, in particular, for all f ∈ Ꮾ α .Let 0 < p < ∞, let μ be a positive Borel measure on D. Define Lemma 4.3.Let μ be a positive measure on D and 0 < p, α < ∞.Then the following statements are equivalent.
Theorem 4.5.Assume that 0 < α < ∞, 0 < p < ∞, and φ is normal on [0,1).Then the following statements are equivalent. (1) Proof.Since where 12 Riemann-Stieltjes operators between Bergman and Bloch By Lemma 4.3, we know that Theorem 4.6.Assume that α > 1, 0 < p < ∞, and φ is normal on [0,1).Then the following statements are equivalent. Proof.Since where Similar to the proof of Theorem 4.5, we get (i)⇔(iii) by Lemma 4.2.(ii)⇒(i) is clear.Next we prove that (iii)⇒(ii).Assume (iii) holds, we obtain that J g is bounded and so g ∈ H(p, p,φ).In addition to this, we also find that for any ε > 0, there is an r ∈ (0,1) such that |z|>r g (z) Let { f k } be any sequence in the unit ball of Ꮾ α and converges to 0 uniformly on compact subsets of D. For the above ε, there exists a Hence we have Songxiao Li 13 In other words, we obtain lim k→∞ J g f H(p,p,φ) = 0 and so J p : Ꮾ α → H(p, p,φ) is compact.
14 Riemann-Stieltjes operators between Bergman and Bloch we have f w Ꮾ ≤ ln2 + 2. By the subharmonicity, we have Therefore, we get the desired result.Remark 4.9.We use another method to prove the necessary condition of the boundedness of J g : Ꮾ → H(p, p,φ).
Since J g f ∈ H(p, p,φ), by Lemma 2.3, we have
16 Riemann-Stieltjes operators between Bergman and Bloch From (4.26), for any ε > 0, there exists an r, 0 < r < 1, such that as k > k 0 , from which we get the desired result.
Theorem 4.10.Assume that 0 < p < ∞ and φ is normal on [0,1).Then the following statements hold.(i)If the operator J g :Ꮾ → H(p, p,φ) is compact, then then J g : Ꮾ → H(p, p,φ) is compact.Proof.Suppose the operator J g : Ꮾ → H(p, p,φ) is compact.Let z n be a sequence in D such that |z n | → 1 as n → ∞.Take 24))Therefore, we get the desired result.