Journal of Function Spaces
Volume 2015 (2015), Article ID 362681, 6 pages
http://dx.doi.org/10.1155/2015/362681

Research Article

The Improvement on the Boundedness and Norm of a Class of Integral Operators on Space

Lifang Zhou and Jin Lu

Department of Mathematics, Huzhou University, Huzhou, Zhejiang 313000, China

Received 14 July 2014; Accepted 27 August 2014

Academic Editor: Janusz Matkowski

Copyright © 2015 Lifang Zhou and Jin Lu. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract.

We prove the condition “ is neither nor a negative integer” can be dropped on the boundedness of a class of integral operators on space, which improves the result by Krues and Zhu. Besides, the exact norm of on space is also obtained under the assumption .

1. Introduction

Let be the open unit ball in the complex space . The measure,

denotes the weighted Lebesgue measure on , where is real parameter and is the normalized Lebesgue measure on such that . It is easy to know is finite if and only if . Suppose ; to simplify the notation, we write for the weighted -space under the measure on and for the usual -space under the measure .

Suppose are real numbers, and a class of integral operators is defined by

The class of integral operators is introduced by Kures and Zhu [1]. And it is closely related to “maximal Bergman projection” and Berezin transform. In fact, the boundedness of Bergman projection on comes from the boundedness of the operator

on ; see [2]. Therefore, we can call by “maximal Bergman projection,” which is the particular case of . Berezin transforms, whatever the case of the unit disk [3, page 141] or the case of unit ball ([4, page 76], [5, page 383]), are all concluded in the form of with special .

In [1], Krues and Zhu gave the sufficient and necessary conditions of the boundedness of operator .

Theorem A (see [1]). Suppose is neither 0 nor a negative integer.

(1)The operator is bounded on if and only if .(2)The operator is bounded on if and only if or .

The main purposes of this note contain two parts. One part is to prove the condition “ is neither nor a negative integer” in Theorem can be removed; see Section 3. The other part is to give the accurate norm of the operator on under the assumption , which can be seen from the following two theorems.

Theorem 1. Suppose . If and , then

Else, we also give the sufficient and necessary conditions of the operator on and the accurate norm under of this case, where denotes the set of all essentially bounded and measurable functions under the measure on .

Theorem 2. The operator is bounded on if and only if , , and or , , and . Moreover, when , we have

Notice is the generalization of “maximal Bergman projection” and Berezin transform which was first introduced by Berezin [6]. The boundedness of Berezin transform of is a well-known fact; see [7, Proposition 2.2]. But the norm of it was not calculated out until 2008 by Dostanić; see [8, Corollary 2]. Recently, the result by Dostanić has been extended to several complex variables in [9, Theorem 1.1]. Thus, Theorems 1 and 2 promote the main results in [8, 9]. And they also imply the following corollary.

Corollary 3. Suppose , , and the norm of on can be

which implies grows at most like as .

Next, we will see that the boundedness of an operator called Berezin-type transform on can also be obtained from our main results. The Berezin-type transform is defined by

where

and , , , and . The transform was introduced by Li and Liu [10] when they discuss whether the mean-value property implies -harmonicity for integrable functions on the unit ball in . Notice that

with , , and . And as . Therefore, the boundedness of Berezin-type transform on comes from the boundedness of the operator on . Thus, we have the following result, which extends Propositions 3.3 and 3.4 in [10] combining the fact of Lemma 2.4 in [10] therein.

Corollary 4. If such that , then the Berezin-type is bounded on and

where

Moreover, the Berezin-type transform is bounded on , and

2. Preliminaries

A number of hypergeometric functions will appear throughout. We use the classical notation to denote

with , where

And the hypergeometric series in (13) converges absolutely for all the value of . Moreover, as , it is easy to know that

where represents the ratio and has a positive finite limit as . Now we list a few formulas for easy reference (see [11, Chapter II]):

Lemma 5. Suppose and . Then

Proof. Note that, under the assumption of the lemma, both sides of (18) are continuous at . The lemma then follows by letting in (18) and applying (16).

The following integral formulae concerning the hypergeometric function are significant for our main results. And all these formulae are contained in [12]. Now we list them.

Lemma 6 (see [12, Corollary 2.4]). For and , we have

Lemma 6 is also contained implicitly in the proof of Theorem 1.4.10 in [13] (see the formula in page 19, line 5 of [13]).

Lemma 7 (see [12, Corollary 2.5]). Suppose that , , and . Then

Proof. Using Lemma 6 in the inner integral, we have

Then (19) gives the result.

The following result, usually called Schur’s test, is a very effective tool in proving the -boundedness of integral operators. See, for example, [3].

Lemma 8. Suppose that is a -finite measure space, is a nonnegative measurable function on , and is the associated integral operator:

Let and . If there exist a positive constant and a positive measurable function on such that

for almost every in , and

for almost every in , then is bounded on with .

3. The Improvement

The section mainly proposes the condition “ is neither nor a negative integer” can be omitted in Theorem . Notice the condition is only used to give while proving the necessity for the boundedness of the operator on ; see [1, lemma 12]. Now we will give a new proof of the necessity for the boundedness of on in Propositions 9 and 11 to introduce the condition can be put off.

Proposition 9. Suppose the operator is bounded on , and then .

Proof. Let be the number such that . For any fixed , define

where

Easy calculation shows . Notice the fact

Then the boundedness of the operator on leads to the integral

Hence, using Lemma 7 with , , and , we can conclude that

Then the arbitrariness of gives

Now, we will give the proof by dividing into the following two cases.
When , by Lemma 7, the integral in (30) equals

Then letting , by (30), we can know the limits

Then the boundedness of the operator gives .
When , take the function

with . The condition (32) implies the function . And using Lemma 6, we have

According to (15), we can obtain that . Thus the boundedness of the operator on gives that ; that is, . Now we consider the adjoint operator of the operator ; that is,

The boundedness of on implies the boundedness of on . With the similar discussion above, we can obtain that ; that is, .

When , (34) implies the following result.

Corollary 10. Suppose and , , and then

Proposition 11. The operator is bounded on if and only if or . And when , we have

When , we have

where .

Proof. By Lemma 6, we have

where denotes the adjoint operator of . Then, using (15), we can obtain that the operator is bounded on if and only if

or

which gives the first part of the proposition.
Now we will give the second part. When and , the hypergeometric function in (41) is increasing since its Taylor coefficients are all positive. Applying (16), we have (40). When , (17) gives

Thus (41), the increase of the last hypergeometric function, and (16) lead to

4. The Proof of Theorems 1 and 2

Proof of Theorems 1 and 2. Since

therefore Theorem 2 comes out as the same discussion as Proposition 11.
Next, we will concentrate on the proof of Theorem 1. Remember the hypothesis throughout the following proof. Since (39) gives the case of , for the case , Corollary 10 gives the lower bound of . Thus we only show the fact

To this end, we will use Schur’s test (Lemma 8) with

Set

where is the conjugate exponent of such that . It then suffices to show

for all , and

for all . We only prove (50), since (51) comes from the same way as (50). Applying Lemma 6 and (17), we have

By (16), the last hypergeometric function is bounded from the above by

since it is increasing on the interval . This proves (50), which in turn implies (47). The proof is completed.

5. Remark

The topic on the exact norm of an operator is an interesting but difficult problem. In this note, we only give the accurate norm of the generalized operator on under . But for other cases, except the particular case (40), we can give an upper bound of by Theorem 1 according to the fact

and a lower bound for one fixed by (30) and Lemma 7; thus the problem of the norm of other cases may be left as an open problem to consider.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The authors thank the referee for his/her careful reading and corrections. This work is supported by the National Natural Science Foundation of China (nos. 11271124, 11201141, 11301136, 11261022, and 61473332), the Natural Science Foundation of Zhejiang Province (nos. LQ13A010005 and LY14A010017), and the Scientific Research and Teachers Project of Huzhou Teachers College (nos. KX21058 and RP21028).

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