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IntegralType Operators from Spaces to ZygmundType Spaces on the Unit Ball
Journal of Inequalities and Applications volume 2010, Article number: 789285 (2011)
Abstract
Let denote the space of all holomorphic functions on the unit ball . This paper investigates the following integraltype operator with symbol , , , , where is the radial derivative of . We characterize the boundedness and compactness of the integraltype operators from general function spaces to Zygmundtype spaces , where is normal function on .
1. Introduction
Let be the open unit ball of , let be its boundary, and let be the family of all holomorphic functions on . Let and be points in and .
Let
stand for the radial derivative of . For , let , where is the Möbius transformation of satisfying , , and . For , , we say provided that and
The space , introduced by Zhao in [1], is known as the general family of function spaces. For appropriate parameter values ,, and , coincides with several classical function spaces. For instance, let be the unit disk in , if (see [2]), where , consists of those analytic functions in for which
The space is the classical Bergman space (see [3]), is the classical Besov space , and, in particular, is just the Hardy space . The spaces are spaces, introduced by Aulaskari et al. [4, 5]. Further, , the analytic functions of bounded mean oscillation. Note that is the space of constant functions if . More information on the spaces can be found in [6, 7].
Recall that the Blochtype spaces (or Bloch space) , consists of all for which
The little Blochtype space consists of all such that
Under the norm introduced by is a Banach space and is a closed subspace of . If , we write and for and , respectively.
A positive continuous function on the interval [0,1) is called normal if there are three constants and such that
Let denote the class of all such that
Write
With the norm , is a Banach space. is called the Zygmund space (see [8]). Let denote the class of all such that
Let be a normal function on [0,1). It is natural to extend the Zygmund space to a more general form, for an , we say that belongs to the space if
It is easy to check that becomes a Banach space under the norm
and will be called the Zygmundtype space.
Let denote the class of holomorphic functions such that
and is called the little Zygmundtype space. When , from [8, page 261], we say that if and only if , and there exists a constant such that
for all and , where is the ball algebra on .
For , the following integraltype operator (so called extended Cesàro operator) is
where and . Stević [9] considered the boundedness of on Bloch spaces. Lv and Tang got the boundedness and compactness of from to Bloch spaces for all (see [10]). Recently, Li and Stević discussed the boundedness of from Blochtype spaces to Zygmundtype spaces in [11]. For more information about Zygmund spaces, see [12, 13].
In this paper, we characterize the boundedness and compactness of the operator from general analytic spaces to Zygmundtype spaces.
In what follows, we always suppose that , , , . Throughout this paper, constants are denoted by ; they are positive and may have different values at different places.
2. Some Auxiliary Results
In this section, we quote several auxiliary results which will be used in the proofs of our main results. The following lemma is according to Zhang [14].
Lemma 2.1.
If , then and
Lemma 2.2 (see [9]).
For , if , then for any
Lemma 2.3 (see [15]).
For every , it holds .
Lemma 2.4 (see [10]).
Let . Suppose that for each , variable functions satisfy , then
Lemma 2.5.
Assume that , , , , and is a normal function on , then is compact if and only if is bounded, and for any bounded sequence in which converges to zero uniformly on compact subsets of as , one has .
The proof of Lemma 2.5 follows by standard arguments (see, e.g., Lemma 3 in [16]). Hence, we omit the details.
The following lemma is similar to the proof of Lemma 1 in [17]. Hence, we omit it.
Lemma 2.6.
Let be a normal function. A closed set in is compact if and only if it is bounded and satisfies
3. Main Results and Proofs
Now, we are ready to state and prove the main results in this section.
Theorem 3.1.
Let , , , and let be normal, and , then is bounded if and only if
(i)for ,
(ii)for ,
Proof.

(i)
First, for , suppose that and . By Lemmas 2.1–2.3, we write . We have that
(3.5)
Hence, (3.1) and (3.2) imply that is bounded.
Conversely, assume that is bounded. Taking the test function , we see that , that is,
For , set
Then, by [14] and .
Hence,
From (3.8), we have
On the other hand, we have
Combing (3.9) and (3.10), we get (3.1).
In order to prove (3.2), let and set
It is easy to see that , . We know that ; moreover, there is a positive constant such that . Hence,
From (3.1) and (3.12), we see that (3.2) holds.

(ii)
If , then, by Lemmas 2.1 and 2.2, we have , for , we get
(3.13)
Applying (3.3) and (3.4) in (3.13), for the case , the boundedness of the operator follows.
Conversely, suppose that is bounded. Given any , set
then by [14].
By the boundedness of , it is easy to see that
By (3.14) and (3.15), in the same way as proving (3.1), we get that (3.3) holds.
Now, given any , set
then . Applying Lemma 2.4, we have that . Hence,
From (3.15) and (3.17), we see that (3.4) holds. The proof of this theorem is completed.
Theorem 3.2.
Let , , , and let be normal, and , then the following statements are equivalent:
(A) is compact;
(B) is compact;
(C)

(i)
for ,

(ii)
for ,
Proof.
(B) ⇒ (A). This implication is obvious.
(A) ⇒ (C). First, for the case .
Suppose that the operator is compact. Let be a sequence in such that . Denote , and set
It is easy to see that for and uniformly on compact subsets of as . By Lemma 2.5, it follows that
By Lemma 2.3, we have
From (3.23) and (3.24), we obtain
which means that (3.18) holds.
Similarly, we take the test function
Then, for and uniformly on compact subsets of as . we obtain that (3.20) holds for the case .
For proving (3.19), we set
then , and converges to 0 uniformly on any compact subsets of as . By Lemma 2.5, it yields
Further, we have
From (3.25), (3.28), and (3.29), it follows that
which implies that (3.19) holds.

(ii)
Second, for the case , take the test function
(3.31)
Then, by Lemma 2.4 and uniformly on any compact subset of . By Lemma 2.5 and condition (), we have
Hence, we have that
From (3.20), (3.32), and (3.33), it follows that
which implies that (3.21) holds.
(C) ⇒ (B). Suppose that (3.18) and (3.19) hold for . By Lemmas 2.1 and 2.2, we have that
Note that (3.18) and (3.19) imply that
Further, they also imply that (3.1) and (3.2) hold. From this and Theorem 3.1, it follows that set is bounded. Using these facts, (3.18), and (3.19), we have
Similarly, we obtain that (3.37) holds for the case by (3.20) and (3.21). Exploiting Lemma 2.6, the compactness of the operator follows. The proof of this theorem is completed.
Finally, we consider the case .
Theorem 3.3.
Let , , , and let be normal, , , then the following statements are equivalent:
(A) is bounded;
(B) and
The proof of Theorem 3.3 follows by the proof of Theorem 3.1. So, we omit the details here.
Theorem 3.4.
Let , , , and let be normal, and , then the following statements are equivalent:
(A) is compact;
(B) and
Proof.
(A) ⇒ (B). We assume that is compact. For , we obtain that . Exploiting the test function in (3.22), similarly to the proof of Theorem 3.2, we obtain that (3.39) holds. As a consequence, it follows that
(B) ⇒ (A). Assume that is a sequence in such that , and uniformly on compact of as . By Lemma 2.1 and [18, Lemma 4.2],
From (3.39), we have that for every , there is a , such that, for every ,
and from (3.39) that
Hence,
Since on compact subsets of by the Cauchy estimate, it follows that on compact subsets of , in particular on . Taking in (3.44), the supremum over , letting , using the abovementioned facts, , and since is an arbitrary positive number, we obtain
Hence, by Lemma 2.5, the compactness of the operator follows. The proof of this theorem is completed.
Theorem 3.5.
Let , , , and let be normal, and , then the following statements are equivalent:
(A) is compact;
(B) and
Proof.
(A) ⇒ (B). For , we obtain that . In the same way as in Theorem 3.4, we get that (3.46) holds.
(B) ⇒ (A). By Lemmas 2.1 and 2.2, we have that
This along with Theorem 3.2 implies that is bounded. Taking the supremum over the unit ball in , letting in (3.46), using the condition (), and finally by applying Lemma 2.6, we get the compactness of the operator . This completes the proof of the theorem.
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Acknowledgments
The author wishes to thank Professor Rauno Aulaskari for his helpful suggestions. This research was supported in part by the Academy of Finland 121281.
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Yang, C. IntegralType Operators from Spaces to ZygmundType Spaces on the Unit Ball. J Inequal Appl 2010, 789285 (2011). https://doi.org/10.1155/2010/789285
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DOI: https://doi.org/10.1155/2010/789285
Keywords
 Banach Space
 Function Space
 Compact Subset
 Hardy Space
 Besov Space