Journal of Complex Analysis

Volume 2014 (2014), Article ID 473069, 10 pages

http://dx.doi.org/10.1155/2014/473069

## Integral Transforms of Functions to Be in a Class of Analytic Functions Using Duality Techniques

Department of Mathematics, Indian Institute of Technology, Roorkee, Uttarakhand 247 667, India

Received 26 February 2014; Accepted 13 May 2014; Published 1 July 2014

Academic Editor: Hari M. Srivastava

Copyright © 2014 Satwanti Devi and A. Swaminathan. 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 , denote the class of all normalized analytic functions in the unit disc such that for some with , and . Let , , denote the Pascu class of -convex functions given by the analytic condition which unifies the classes of starlike and convex functions. The aim of this paper is to find conditions on so that the integral transform of the form carry functions from into . As for the applications, for specific values of , it is found that several known integral operators carry functions from into . The results for a more generalized operator related to are also given.

#### 1. Introduction

Let denote the class of all functions analytic in the open unit disc with the normalization , and let be the class of functions that are univalent in . A function is said to be starlike or convex , if maps conformally onto the domains, respectively, starlike with respect to origin and convex. Note that in , if follows from the well-known Alexander theorem (see [1] for details). A useful generalization of the class is the class that has the analytic characterization and . Various generalizations of classes and are abundant in the literature. One such generalization is the following.

A function is said to be in the Pascu class of -convex functions of order if [2] or in other words This class is denoted by . Even though this class is known as Pascu class of -convex functions of order , since we use the parameter for another important class, we denote this class by , , and we remark that, in the sequel, we only consider the class . Clearly, and , which implies that this class is a smooth passage between the classes of starlike and convex functions.

The main objective of this work is to find conditions on the nonnegative real valued integrable function satisfying , such that the operator is in the class . Note that this operator was introduced in [3]. To investigate this admissibility property, the class to which the function belongs is important. Let , where , , and , denote the class of all normalized analytic functions in the open unit disc such that for some . This class and its particular cases were considered by many authors so that the corresponding operator given by (3) is univalent and in for some particular values of , , , and . This work was motivated in [3] by studying the conditions under which and was generalized in [4] by studying the case . Similar situation for the convex case, namely, , was initiated in [5]. After several generalizations by many authors, recently, the conditions under which were obtained in [6] and the corresponding results for the convex case so that were obtained in [7]. Applications involving several well-known integral transforms were studied in [6, 7] (see also [8]). For all of the literatures involving the complete study in this direction so far, we refer to [6–9] and the references therein.

In this work, we find conditions on so that using duality techniques which are presented in Section 2. As applications, in Section 3, we consider particular values for in (3) so that results for some of the well-known integral operators can be deduced. A more generalized operator introduced in [5] is considered in Section 4 for similar type of results.

First, we underline some preliminaries that are useful for our discussion. We introduce two constants and satisfying [6, 10]

When , then is chosen to be , in which case . When , (5) yields , or , and leads to two cases as follows.(i)For , then, choosing gives .(ii)For , then, and .

*Note.* Since the case is considered in [9], we only consider results for the case , except for Theorem 12 (see Remark 11).

Next, we introduce two known auxiliary functions [6]. Let

Here, denotes the convolution inverse of such that . By , we mean the following: if and are in with the power series expansions and , respectively, then the convolution or Hadamard product of and is given by .

Since , , when , making the change of variables , in (7) results in writing as

Now, let be the solution of the initial value-problem satisfying . The series solution is given by Let be the solution of the differential equation satisfying . The series solution of is given by Note that also satisfies .

Our main result is the generalization of the following results given in [6, 7]. The necessary and sufficient conditions under which the operator carries the function from to the classes and , respectively, are given in the next two results.

Theorem 1 (see [6]). *Consider , given by (5) and let satisfy
**
where is the solution of the initial value-problem (9), and let . Assume that , and as . Then, is in if and only if
**
where
*

*Theorem 2 (see [7]).
Let , let , satisfy (5), and let be given by
where is given by (11). Further, and are given in (15) and (16), and assume that and as . Then,
if and only if is in .*

*It is difficult to verify conditions (14) and (19). Hence, the following results involving sufficient conditions are useful for finding applications.*

*Theorem 3 (see [6]). Let and be defined in (15) and (16). Assume that both and are integrable on and positive on . Assume further that and
is decreasing on . If satisfy (13) and , then .*

*Theorem 4 (see [7]). Let and be defined in (15) and (16). Assume that both are integrable on and positive on . Assume further that and
is decreasing on . If satisfy (18) and , then .*

*2. Main Results*

*We start with a result that gives both necessary and sufficient conditions for an integral transform that satisfies the admissibility property of the class , which contains nonunivalent functions also, to the Pascu class .*

*Theorem 5.
Let , satisfy (5), and let satisfy
where and are defined by the differential equations given in (9) and (11), respectively. Assume that and as . Then,
or
where
The value of is sharp.*

*Proof. *Let
Since , [11] we may assume that , . Further, assuming that , from (5) we get
using (6) and (7). This gives
Since implies that , by the well-known result [11, page 94] of convolution theory we get
Thus,
This condition holds if and only if [11, page 23]
Using the results (14) and (19) from Theorems 1 and 2, respectively, we obtain
which is the required result.

To verify sharpness, let be the solution of the differential equation
where satisfies (22). Further simplification using (10) and (12) gives
where .

Clearly, . Using and the series expansion of , we get

This means that
Using (37), a simple computation gives
This means that
Hence, for some , so is not even locally univalent in . This shows that the result is sharp for .

*Remark 6. *This result generalizes various results known in this direction. For example, gives Theorem 1 [6, Theorem 3.1], and gives Theorem 2 [7, Theorem ]. For other particular cases with or , we refer to [6, 7] and the references therein.

*Theorem 7. Let and be defined as in (15) and (16), respectively, with both of them being integrable on and positive on . Further, assume that , , and
is increasing on . Then, for satisfying (22), .*

*Proof. *Consider the following:
Integration by parts gives
by a simple computation. It is easy to see that, from Theorems 3 and 4, only when is decreasing on which is nothing but (40), and the proof is complete by applying Theorem 5.

*Remark 8. *Even though we did not consider the case , even at , Theorem 7 does not reduce to a similar result given in [9]. This is due to the fact that our condition (40) has the term in the denominator, whereas the corresponding result in [9] has the term and hence has different condition.

*3. Applications*

*It is difficult to check the condition given in Section 2, for . In order to find applications, simplified conditions are required. For this purpose, from (40), it is enough to show that
is increasing on which is equivalent to having the fact that
is decreasing on , where and are defined in (15) and (16). It is enough to have .*

*Let , where . So to satisfy the above condition, we need to have
Since and , we get . This implies that it suffices to have that is increasing on , which means that
The above equation also holds if which is equivalent to the condition
This inequality can further be reduced to
if the inequality
is true. So, in order to obtain further results, we check conditions (48) and (49). Note that, whenever , from (47) we see that it is sufficient to check (48) as there is no necessity for the condition given by (49).*

*As the first application, we consider , so that the Bernardi operator of the function in is in .*

*Theorem 9. Let , , and let satisfy (22). If , then the function, given by the Bernardi operator,
belongs to the class if
*

*Proof. *With , using (51), we get
So, inequality (49) is satisfied. If inequality (48) holds, then , which means that
On further simplification, this inequality reduces to
Clearly, . Hence, using and substituting in (54), we get
which is true by the hypothesis.

*Remark 10. *(1) When , then , whereas, in [6], the range for is given as .

(2) For , we have . The result obtained in [7], for , is .

*So in both cases the result (Theorem 9) obtained for admissibility property of Bernardi operator in class differs from [6, Theorem 5.1] and [7, Theorem 5.1]. But our result is true for also.*

*Remark 11. *We recall that, throughout this paper, we use , as case is considered in [9]. But we note that, for the Bernardi operator, the result given in [9] uses the fact , which is not true. Hence, in order to make completion of the work for in this direction, we give the result related to Bernardi operator for . Since the condition for given in [9] is different from the one given by (49) and (48), we explicitly prove this result.

*Theorem 12. Let , , and let satisfy (22). If , then the function
belongs to the class if and for .*

*Proof. *Note that, for the Bernardi operator, the case is considered in [6, Theorem 5.1] and is considered in [7, Theorem 5.1]. We consider only the case .

For the case , the integral operator is in if and only if
which can be easily obtained as in Theorem 7.

Let , where
for the integral operator to be in , .

Substituting the value of and in the above equation, we have
Consider the case when , where . So inequality (59) on simplification reduces to
If and , then, for , (60) is satisfied.

For , , which clearly implies that if is decreasing on , that is, , then also inequality (60) holds. Further simplifying (59), for Bernardi operator, we need to show that
On substituting the values of , , and in (61), we get
To satisfy (62), should hold:
Since , we get
satisfying the hypothesis of the theorem which gives that given by (56) is in .

*Theorem 13. Let and . If satisfies
in and let satisfy
where and are given by (10) and (12), respectively, then .*

*Proof. *Let ; then, . Therefore,
When , the hypothesis satisfies (51), and, hence, from Theorem 9, the required result follows.

*Example 14. *If , , then . In this case, (10) and (12) yield
Thus, .

*Remark 15. *(1) For [6, Remark 5.2], we get , such that .

(2) For [7, Example 5.2], we get , such that .

*Theorem 16. Let , , , , and , and let satisfy (18). If , then the function
belongs to the class if
*

*Proof. *For to be in , should satisfy (47), where

Since , and hold. Hence it is enough to check (48). If we substitute the values of , , and in (48), a simple computation shows that (48) is equivalent to
where
with
with

Since , , and (see also [6, 7, 9]), proving for gives and which will imply the required result for .

Now, , which clearly holds since and for all .

Similarly, to prove , it is enough to prove
as the other term in is positive on . Since and from the hypothesis, the term involving is nonnegative. Given that , since , we have and , which gives for .

Now, proving that is equivalent to proving that .

Since , by hypothesis, and for and , .

*Remark 17. *(1) For the particular value of , Theorem 16 yields a result with a smaller range for the parameters than the result given in [6, Theorem 5.5].

(2) For the case , the result of Theorem 16 coincides with the result given in [7, Theorem 5.8]

*Theorem 18. Let , , and , and let satisfy (18). If , then the function
where is given by
belongs to the class if , , and satisfy one of the following conditions: (i), , and ,(ii), , and ,(iii) and , which for (as in Theorem 7) gives .*

*Proof. *To prove the required result, we need to show that (47) holds.

For the case , , , and . So, it is enough to show inequality (48). On substituting the values for , , and , we need to show that , where
So, it is enough to prove and for .*Case i *. Since and , so . We need only to show that , which clearly holds since . Hence, for all . Now, for to be positive, it is enough to show that , which is satisfied by the given condition on .*Case ii *. Since and , so . We need only to show that , which is true since . Now, for to be positive, it is enough to show that
which is satisfied by the given condition on .*Case iii *. We change inequality (48), which is true for . Hence, we only consider the situation . Substituting the values of , , and in (48), an easy computation shows that, for , it suffices to show that the expressions
are nonnegative. Since is positive, the nonnegativity of the first expression follows from hypothesis (iii) of the theorem. Similar observation shows that the second expression reduces to and using is positive. These two inequalities, for , give , which is hypothesis (iii). The proof of Theorem 18 is completed.

*Theorem 19. Let and with . Further, suppose that and let satisfy
where satisfies (11). Then, for the function ,
belongs to the class .*

*Proof. *Choosing , we take and so that takes the form
We complete the proof by applying Theorem 16 and using a simple computation to obtain .

*4. A Generalized Integral Operator*

*4. A Generalized Integral Operator*

*In this section, for the functions , we consider another integral operator introduced in [5] and find the admissibility conditions to be in the class .*

*Theorem 20. Let , satisfy (5); let , and satisfies
where and are given in (10) and (12), respectively. Assume that, for ,
Then, .*

*Proof. *The proof follows similar lines of the proof of Theorem 5. Hence, we omit the details.

*Note. * and, hence, this operator generalizes the operator given in (3).

*For finding applications of the operator , by virtue of (25), Theorem 12 is sufficient. This means that we use the conditions given in Section 3. Hence, we state the following results without giving their proofs as they can be obtained in a similar fashion as in the results of Section 3.*

*Corollary 21. Let , , and satisfy (22). Let , and satisfies
where and are given by (10) and (12), respectively. Assume that, for , the function belongs to the class provided
*

*Corollary 22. Let , , and satisfy (22). Let , and satisfies
where and are given by (10) and (12), respectively. Assume that, for , the function belongs to the class provided
*

*Corollary 23. Let , , and . Let , and satisfies
where and are given by (10) and (12), respectively, and is given by
Assume that, for *