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Integral-Type Operators from Spaces to Zygmund-Type Spaces on the Unit Ball

Abstract

Let denote the space of all holomorphic functions on the unit ball . This paper investigates the following integral-type operator with symbol , , , , where is the radial derivative of . We characterize the boundedness and compactness of the integral-type operators from general function spaces to Zygmund-type 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

(1.1)

stand for the radial derivative of . For , let , where is the Möbius transformation of satisfying , , and . For , , we say provided that and

(1.2)

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

(1.3)

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 Bloch-type spaces (or -Bloch space) , consists of all for which

(1.4)

The little Bloch-type space consists of all such that

(1.5)

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

(1.6)

Let denote the class of all such that

(1.7)

Write

(1.8)

With the norm , is a Banach space. is called the Zygmund space (see [8]). Let denote the class of all such that

(1.9)

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

(1.10)

It is easy to check that becomes a Banach space under the norm

(1.11)

and will be called the Zygmund-type space.

Let denote the class of holomorphic functions such that

(1.12)

and is called the little Zygmund-type space. When , from [8, page 261], we say that if and only if , and there exists a constant such that

(1.13)

for all and , where is the ball algebra on .

For , the following integral-type operator (so called extended Cesàro operator) is

(1.14)

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 Bloch-type spaces to Zygmund-type 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 Zygmund-type 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

(2.1)

Lemma 2.2 (see [9]).

For , if , then for any

(2.2)

Lemma 2.3 (see [15]).

For every , it holds .

Lemma 2.4 (see [10]).

Let . Suppose that for each , -variable functions satisfy , then

(2.3)

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

(2.4)

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 ,

(3.1)
(3.2)

(ii)for ,

(3.3)
(3.4)

Proof.

  1. (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,

(3.6)

For , set

(3.7)

Then, by [14] and .

Hence,

(3.8)

From (3.8), we have

(3.9)

On the other hand, we have

(3.10)

Combing (3.9) and (3.10), we get (3.1).

In order to prove (3.2), let and set

(3.11)

It is easy to see that , . We know that ; moreover, there is a positive constant such that . Hence,

(3.12)

From (3.1) and (3.12), we see that (3.2) holds.

  1. (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

(3.14)

then by [14].

By the boundedness of , it is easy to see that

(3.15)

By (3.14) and (3.15), in the same way as proving (3.1), we get that (3.3) holds.

Now, given any , set

(3.16)

then . Applying Lemma 2.4, we have that . Hence,

(3.17)

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)

  1. (i)

    for ,

(3.18)
(3.19)
  1. (ii)

    for ,

(3.20)
(3.21)

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

(3.22)

It is easy to see that for and uniformly on compact subsets of as . By Lemma 2.5, it follows that

(3.23)

By Lemma 2.3, we have

(3.24)

From (3.23) and (3.24), we obtain

(3.25)

which means that (3.18) holds.

Similarly, we take the test function

(3.26)

Then, for and uniformly on compact subsets of as . we obtain that (3.20) holds for the case .

For proving (3.19), we set

(3.27)

then , and converges to 0 uniformly on any compact subsets of as . By Lemma 2.5, it yields

(3.28)

Further, we have

(3.29)

From (3.25), (3.28), and (3.29), it follows that

(3.30)

which implies that (3.19) holds.

  1. (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

(3.32)

Hence, we have that

(3.33)

From (3.20), (3.32), and (3.33), it follows that

(3.34)

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

(3.35)

Note that (3.18) and (3.19) imply that

(3.36)

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

(3.37)

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

(3.38)

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

(3.39)

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

(3.40)

(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],

(3.41)

From (3.39), we have that for every , there is a , such that, for every ,

(3.42)

and from (3.39) that

(3.43)

Hence,

(3.44)

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 above-mentioned facts, , and since is an arbitrary positive number, we obtain

(3.45)

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

(3.46)

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

(3.47)

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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Correspondence to Congli Yang.

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Yang, C. Integral-Type Operators from Spaces to Zygmund-Type Spaces on the Unit Ball. J Inequal Appl 2010, 789285 (2011). https://doi.org/10.1155/2010/789285

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Keywords

  • Banach Space
  • Function Space
  • Compact Subset
  • Hardy Space
  • Besov Space