Volterra composition operators from generalized weighted weighted Bergman spaces to μ-Bloch spaces

Let φ be a holomorphic self-map and g be a fixed holomorphic function on the unit ball B . The boundedness and compactness of the operator

For any  > 0 and  ∈ ℝ, let  be the smallest nonnegative integer such that  +  > −1.We say that an  ∈ () belongs to the generalized weighted Bergman space    , if The generalized weighted Bergman space    is introduced by Zhao and Zhu (see, e.g., [15]).This space covers the traditional weighted Bergman space( > −1 ), the Besov space, the Hardy space  2 and the so-called Arveson space.For example, the space   0 is the classical Bergman space; the space  2  − is the so-called Arveson space; the space   −(+1) is the Besov space.See [15,16] for some basic facts on the weighted Bergman space.
Let  be a holomorphic self-map of  .The composition operator   is defined by The book [2] contains much information on this topic.
Suppose that  :  → ℂ 1 is a holomorphic map, the extended Cesàro operator, which was introduced in [4], is defined as following This operator is also called the Riemann-Stieltjes operator(see, e.g.[14]).
See [1,4,5,6,8,9,10,13,14] for more information of the operator   on various spaces in the unit ball.Motivated by the definition of operators   and   , we define a more general operator The operator  , will be called the Volterra composition operator.In the setting of the unit disk  , this operator has the following form which was first studied in [7].To the best of our knowledge, the operator  , in the unit ball is studied in the present paper for the first time.
In this paper we study the boundedness and compactness of Volterra composition operators  , from the generalized weighted Bergman space into ℬ  and ℬ ,0 .As some corollaries, we obtain characterizations of the extended Cesàro operator   from the generalized weighted Bergman space into ℬ  and ℬ ,0 .
Throughout the paper, constants are denoted by  , they are positive and may differ from one occurrence to the other.

Then every function in 𝐴 𝑝
is continuous on the closed unit ball and so is bounded.
(iii) Suppose  > 1 , 1/ + 1/ = 1 and  +  + 1 = 0. Then there exists a constant  > 0 such that Proof.The proof is similar to the proof of Lemma 1 in [11].We omit the details. □ The following criterion for compactness follows from standard arguments similar to those outlined in Proposition 3.11 of [2].We omit the details of the proof.
Especially, when  > 0 and  +  + 1 < 0 , we need the following criterion for compactness follows from arguments similar to those in Lemma 3.7 of [12].Lemma 4. Let  > 0 and  +  + 1 < 0 .Let  be a bounded linear operator from    into a normed linear space  .Then  is compact if and only if ∥   ∥  → 0 whenever (  ) is a norm-bounded sequence in    that converges to 0 uniformly on  .
Proof.The necessity is obvious.Now we prove the sufficiency part.Suppose that  is not compact.Then there is a bounded sequence (  ) in    such that (   ) has no convergent subsequence.Note that when  > 0 and  +  + 1 < 0,    are indeed Lipschitz continuous (see Theorem 66 of [15]).Similarly to the proof of Lemma 3.6 of [12], we see that every bounded sequence in    has a subsequence that converges uniformly on  by Lemma 1 and Arzela-Ascoli Theorem.Hence (  ) has a subsequence (  ) such that   →  uniformly on  .By Fatou's lemma we see that  ∈    .The sequence (  −  ) is bounded in    and converges to 0 uniformly on  .By assumption ∥   −  ∥  → 0 as  → ∞.This implies that the subsequence (   ) of (   ) converges in  (to   ), a contradiction.□

Case
Proof.Suppose that (2) holds.A calculation with (1) gives the following fundamental and useful formula(see, e.g.[4]) Then for arbitrary  ∈  and  ∈    , by Lemma 1 we have Using the condition (2), the boundedness of the operator  , :    → ℬ  follows by taking the supremum in (3) over  .
Conversely, suppose that  , : For  ∈  , set Then from Theorem 32 of [15] we see that from which we get (2).This completes the proof of Theorem 1.
(iii) ⇒ (ii).Suppose that  ∈ ℬ  .For an  ∈    , by Lemma 1 we see that  is continuous on the closed unit ball and so is bounded in  .Therefore From the above inequality we see that