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`International Journal of AnalysisVolume 2014, Article ID 301709, 9 pageshttp://dx.doi.org/10.1155/2014/301709`
Research Article

## Weighted Fractional Differentiation Composition Operators from Mixed-Norm Spaces to Weighted-Type Spaces

Department of Mathematical Science, Gauhati University, Guwahati 781014, India

Received 3 December 2013; Accepted 10 January 2014; Published 27 February 2014

Academic Editor: Frédéric Robert

Copyright © 2014 D. Borgohain and S. Naik. 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

Let be an open unit disc in the complex plane and let as well as be analytic maps. For an analytic function on the weighted fractional differentiation composition operator is defined as , where , , and . In this paper, we obtain a characterization of boundedness and compactness of weighted fractional differentiation composition operator from mixed-norm space to weighted-type space .

#### 1. Introduction

The classical/Gaussian hypergeometric series is defined by the power series expansion Here , , are complex numbers such that , , and is Pochhammer’s symbol/shifted factorial defined by Appel’s symbol and for .

Obviously, . Many properties of the hypergeometric series including the Gauss and Euler transformations are found in standard textbooks such as [1, 2].

For any two analytic functions and represented by their power series expansion, in , the Hadamard product (or convolution) of and denoted by and is defined by in . Moreover, In particular, if , we have If and , then the fractional derivative (see ) of order is defined by In terms of convolution, we also have

For , we define

It is obvious to find that the fractional derivative and the ordinary derivative satisfy

Let be fixed and let   be nonconstant analytic self-map of . For and , we can define an operator on , called a weighted fractional differentiation composition operator, by We can regard this operator as a generalization of a multiplication operator and a weighted composition operator. In this paper we study the boundedness and the compactness of weighted fractional differentiation composition operator from mixed-norm spaces to weighted-type spaces. Recall that a positive continuous function on is called normal if there is and and with such that

Let be the normalized Lebesgue area measure on and let    be the space of all analytic functions on .

For , , and is normal we denote by the space of all functions such that where and where

When and , , then (classical weighted Bergman space), defined by Further, when then the natural limit to the weighted Bergman space is the Hardy space ; that is, . For more details on Hardy space, see .

Suppose is normal and radial; that is, . The weighted-type space consists of all such that A little version of is denoted by as the subset of consisting of all such that Note that is a closed subspace of . For the space , see .

For , we have the space of bounded analytic functions , where

For ,   equals the weighted composition operator defined by , , which reduces to the composition operator for . During the last century, composition operators were studied between different spaces of analytic functions. There are many papers about these operators. Among other things, they deal with boundedness and compactness of these operators. But we will discuss mostly the mixed-norm spaces. For some recent papers on weighted composition operators on some -type spaces, see .

If , we get the operator which for gives which for gives and gives . For particular choices of , , and , we obtain many operators which are product, addition, and composition of multiplication and differentiation operators. For more details of these types of operators see .

Recall that, in , Stević characterized the boundedness and compactness of weighted differentiation composition operators from mixed-norm spaces to the weighted-type spaces. Our results can be viewed as generalizations of their results.

Throughout this paper, denotes a positive constant which may vary for different lines. The notation means there is a positive constant such that .

#### 2. Some Lemmas

We collect some basic lemmas which are useful in the proof of the main results.

Lemma 1 (see [23, Lemma 6]). For and one has

Lemma 2 (see [24, Proposition ]). For one has

Lemma 3 (see [22, Lemma 1]). Assume , , is normal, and . Then, for every , there is a positive constant independent of such that

#### 3. Main Results

In the following results we show boundedness of the operator from mixed-norm space to weighted-type space . As a consequence we obtain many new results in the form of corollaries.

Lemma 3 is true for the th order derivative of . Here we will give a more general result which involves fractional derivative of .

Proposition 4. Assume , , is normal, and . Then, for every , there is a positive constant independent of such that

Theorem 5. For , , , , and are normal, and is an analytic self-map of . Then is bounded if and only if Moreover, if is bounded, then the following asymptotic relation holds:

Proof of Proposition 4. Let .
Using integral representation of convolution we have, for , Since , Minkowski’s inequality gives where .
For , we have Putting , we have, for any , Since therefore For Putting , we have By using the following asymptotic relations: we have A simple calculation using Cauchy’s integral formula gives The above inequality can be written as the following: When , from inequality (31) we have For we have and so . Therefore the proof is similar to the case, .

Proof of Theorem 5. Suppose (24) holds. We have Therefore,
Conversely, assume that is bounded. For a fixed , let where the constant is from the definition of the normality of the function : In view of the well-known Gauss identity , of hypergeometric function we can rewrite the previous equation to obtain the following:
We know the hypergeometric function is bounded if on (see ). The hypergeometric function is bounded on since ; therefore, using Lemma 2 gives Since is normal and using Lemma 1, we get . Now For every , we have
Therefore,
From (40) and (48) we get the asymptotic relation (25).

If we take in the proof of Theorem 5, we get the following result.

Theorem 6. For , , , , and are normal, and is an analytic self-map of . Then is bounded if and only if Moreover, if is bounded, then the following asymptotic relation holds:

Theorem 7. For , , , , and are normal, and is an analytic self-map of . Then is bounded if and only if is bounded and .

Proof. Suppose is bounded. Then it is clear that is also bounded, since Therefore, .
Conversely, assume that is bounded and . Let be any polynomial. Then, Let , where . Consider the following: Obviously, is bounded in . Also, ; therefore . Since the set of polynomials is dense in , therefore we have from the proof of Theorem 7 of  that from which the boundedness of follows.

As consequences of the above theorems we have the following important corollary.

Corollary 8. For , , , , and are normal, and is an analytic self-map of .(a)Then is bounded if and only if (b)When , one gets that is bounded if and only if (c)For , then is bounded if and only if (d)For , is bounded if and only if

A simple calculation shows that , , and . Thus the corollaries of Stević’s paper  on boundedness of these operators follow easily.

#### 4. Compactness of

In this section we will characterize when weighted composition operators acting between mixed-norm spaces and weighted-type space are compact. Before stating the results, we show the following lemma whose proof can be obtained by adapting the proof of Lemma 4 of .

Lemma 9. Suppose , , , and are normal, and is a holomorphic self-map of . Then the operator is compact if and only if is bounded and for any bounded sequence in which converges to zero uniformly on compact subsets of , we have as .

Theorem 10. Suppose , , , and are normal, and is a holomorphic self-map of . Then the operator is compact if and only if is bounded and

Proof. Suppose, is compact. Then it is bounded. Suppose (67) is not true. Then there is a sequence such that as and such that Let , .
It is obvious that As we have . From the above inequality we have as uniformly on compact subsets of . Therefore, But from our assumption we have when , which is a contradiction.
Suppose is bounded and (67) holds. Then for every there exist such that, when , Consider . Obviously, : Assume that is bounded sequence in , that is, , and converges to zero uniformly on compact subsets of as . For , Now, consider the case . We have Putting gives us where . Therefore, in compact subsets of . Therefore we have Since, on compact subsets of , uniformly as , therefore From (72) and (73) it follows that as . Therefore the operator is compact.

The following result is obvious as can be seen from the above theorem.

Theorem 11. Suppose , , , and are normal, and is a holomorphic self-map of . Then the operator is compact if and only if is bounded and

#### Conflict of Interests

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

#### Acknowledgment

The first author would like to acknowledge the financial support as Junior Research Fellow and the second author as Principal Investigator of the Project NBHM (National Board for Higher Mathematics, India) (2/48(9)/2012/NBHM (R.P.)/RD-II/2936).

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