Abstract and Applied Analysis

Volume 2014, Article ID 692327, 4 pages

http://dx.doi.org/10.1155/2014/692327

## Some Connections between Class *𝒰*- and *α*-Convex Functions

*𝒰*

*α*

^{1}Department of Mathematics and Computer Sciences, Faculty of Mathematical and Natural Sciences, University of Prishtina, M. Tereza n.n., 10000 Prishtina, Kosovo^{2}Faculty of Mechanical Engineering, Ss. Cyril and Methodius University in Skopje, Karpoš II b.b., 1000 Skopje, Macedonia

Received 19 November 2013; Accepted 12 February 2014; Published 13 March 2014

Academic Editor: V. Ravichandran

Copyright © 2014 Edmond Aliaga and Nikola Tuneski. 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

The class *𝒰* of normalized analytic functions that satisfy for all *z* in the open unit disk is studied and sufficient conditions for an *α*-convex function to be in *𝒰* are given.

#### 1. Introduction

Let denote the class of functions which are analytic in the unit disk with the normalization and .

For a function , we say that is starlike of order , , if and only if We denote by the class of all such functions. Also, we denote by the class of convex functions of order , , that is, the class of functions for which For , we have the classes of and of starlike and convex functions, respectively. All of the above classes are subclasses of the class of univalent functions in and even more, . For details see [1].

Further, for and , let us define the operator and the class This class and its special cases and are widely studied in the past decades ([2–12]). It is known [2, 12] that functions in are univalent if , but not necessarily univalent if . Further, Fournier and Ponnusamy [3] proved that assuming the following equivalency holds: that is, in general case, is not a subset of . In particular, that is, , which can be also verified by the function Finally, let us consider the classes where , , and . These classes make a “bridge” between the classes of starlike and convex functions (of some order). The class is in fact the class of -convex functions of order and , in the case when is a subclass of . Further, -convex functions of some order are also starlike ([13], page 10). Therefore, it gives rise to the question (which is studied in this paper) of finding the sufficient conditions for

Let and be analytic in the unit disk. We say that is subordinate to , and we write ; if is univalent in , then and . Further, we use the method of differential subordination introduced by Miller and Mocanu [14]. In fact, if ( is the complex plane) is analytic in domain , if is univalent in , and if is analytic in with , when , then we say that satisfies a first-order differential subordination if The univalent function is called a dominant of the differential subordination (11) if for all satisfying (11). If is a dominant of (11) and for all dominants of (11), then we say that is the best dominant of the differential subordination (11).

We will make use of the following lemma.

Lemma 1 (see [15]). *Let be univalent in the unit disk , and let and be analytic in a domain containing , with when . Set , and suppose that *(i)*is starlike in the unit disk ;*(ii)*, .** If is analytic in , with , , and
**
then , and is the best dominant of (12).*

#### 2. Main Results and Consequences

Now we will prove the following theorem that will further lead to connections between class and classes and .

Theorem 2. *Let , and . If for all , and if
**
then
**
that is, , and is the best dominant of (13).*

*Proof. *Let , , , and , where . Then is univalent in and are analytic in domain which contains ( and is the unit disk), and when . On the other hand, let
Then
The last inequality holds since for all and does not contain the origin. So, conditions (i) and (ii) from Lemma 1 are satisfied.

Further, is analytic in and . Also, for all ; that is, , since for all (condition of the theorem); for (because ) and has no poles on . Hence from Lemma 1 and the fact that
we receive that , that is, relation (14), and we also receive that is the best dominant of (13).

Applying the definition of subordination to Theorem 2, we receive the following.

Corollary 3. *Let , , and . Also, let and for all . If
**
then ; that is,
**
This result is sharp; that is, the constant in inequality (19) cannot be replaced by a smaller one such that the implication still holds.*

*Proof. *First, let us note that the function defined by expression (13) is univalent in the unit disk such that
So, the disk is contained in which, having in mind the definition of subordination, means that inequality (18) implies subordination (13). Further, from Theorem 2 follows subordination (14), which is equivalent to the inequality (19). Even more, Theorem 2 says that is the best dominant of (13).

In order to prove the sharpness of the result let us assume the opposite; that is, there exists , , such that inequality (18) implies
that is,
On the other hand, inequality (18) implies subordination (13) with best dominant , meaning that . This is a contradiction to the assumption which proves the sharpness of the result.

Previous corollary can be written in the following, equivalent, form that gives conditions for inclusion of the class into the class .

Corollary 4. *Let , , , and . Also, let and for all . Then
**
The constant , for the class , cannot be replaced by a smaller one such that the inclusion still holds.*

Next result gives the connection between classes and .

Corollary 5. *Let , , , and
**
Also, let and for all . If subordination (13) holds, then
*

*Proof. *It is easy to check that all conditions of Theorem 2 are fulfilled; hence, .

It remains to verify that ; that is, subordination (13) implies
Having in mind the definition of subordination and the fact that is univalent, it is enough to show that for all . The last is true because

*Remark 6. *The case and is not covered by the previous corollary since then .

*3. Examples*

*Now we will apply the results from the previous section on specific functions and receive interesting conclusions.*

*Example 1. *Let , , and . Consider the following.(i)Let
If , , , , and , then
(ii)If and one of the following two sets of conditions holds:
or
then
In both cases, power is taken by its principal value.

*Proof. *(i) For the function , we have and

Condition guarantees that for all ; hence, is an analytic function and . For the function , it is easy to verify that

Further, from the definition of subordination, we have that when and ; that is, both are positive or both negative. Therefore,

So, all conditions of Corollary 5 bring us to the conclusion that .

(ii) It is easy to verify that and that

Therefore,

Further, if one of the conditions (30) or (31) holds, then ; that is, we receive inequality (18). Finally, we have shown that all conditions of Corollary 3 are fulfilled, which leads to .

*The following example exhibits some concrete conclusions that can be obtained from the results of the previous section by specifying the values of and/or .*

*Example 2. *Let , and let and for all . Consider the following: (i) ( in Corollary 4);(ii) ( and in Corollary 5).

*Conflict of Interests*

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

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