Open Access

Univalence of Certain Linear Operators Defined by Hypergeometric Function

Journal of Inequalities and Applications20092009:807943

Received: 11 January 2009

Accepted: 22 April 2009

Published: 2 June 2009


The main object of the present paper is to investigate univalence and starlikeness of certain integral operators, which are defined here by means of hypergeometric functions. Relevant connections of the results presented here with those obtained in earlier works are also pointed out.


Positive IntegerAnalytic FunctionIntegral OperatorUnit DiskFractional Derivative

1. Introduction and Preliminaries

Let denote the class of all analytic functions in the unit disk . For , a positive integer, let
with , where is referred to as the normalized analytic functions in the unit disc. A function is called starlike in if is starlike with respect to the origin. The class of all starlike functions is denoted by . For , we define
and it is called the class of all starlike functions of order . Clearly, for . For functions , given by
we define the Hadamard product (or convolution) of and by
An interesting subclass of (the class of all analytic univalent functions) is denoted by and is defined by

where and

The special case of this class has been studied by Ponnusamy and Vasundhra [1] and Obradović et al. [2].

For a,b,c C and c 0,-1,-2,, the Gussian hypergeometric series F(a,b;c;z) is defined as

where and . It is well-known that is analytic in . As a special case of the Euler integral representation for the hypergeometric function, we have
Now by letting
it is easily seen that
For , Owa and Srivastava [3] introduced the operator defined by
which is extensions involving fractional derivatives and fractional integrals. Using definition of we may write

This operator has been studied by Srivastava et al. [4] and Srivastava and Mishra [5].

Also for and , let us define the function by

This operator has been investigated by many authors such as Trimble [6], and Obradović et al. [7].

If we take
then we can rewrite operator defined by (1.11) as
From the definition of it is easy to check that
For with for all we define the transform by

where and

Also for with for all we define the transform by

where and

In this investigation we aim to find conditions on such that implies that the function to be starlike. Also we find conditions on for each the transforms and belong to and .

For proving our results we need the following lemmas.

Lemma 1.1 (cf. Hallenbeck and Ruscheweyh [8]).

Let be analytic and convex univalent in the unit disk with . Also let
be analytic in . If

and is the best dominant of (1.20).

Lemma 1.2 (cf. Ruscheweyh and Stankiewicz [8]).

If are analytic and are convex functions such that then .

Lemma 1.3 (cf. Ruscheweyh and Sheil-Small [9]).

Let and be univalent convex functions in . Then the Hadamard product is also univalent convex in .

2. Main Results

We follow the method of proof adopted in [1, 10].

Theorem 2.1.

Let n be positive integer with . Also let and . If belongs to , Then whenever , where


Let us define
Since , we have
where is an analytic function with and By Schwarz lemma, we have . By (2.3), it is easy to check that
We need to show that . To do this, according to a well-known result [9] and (2.5) it suffices to show that
which is equivalent to

Suppose that denote the class of all Schwarz functions such that and let

then, if . This observation shows that it suffices to find . First we notice that

Define by

Differentiating with respect to , weget

Case 1.

Let Then we see that has its only critical point in the positive real line at
Furthermore, we can see that for and for . Hence attains its maximum value at and

Case 2.

Let , then it is easy to see that and so attains its maximum value at and

Now the required conclusion follows from (2.13) and (2.14).

By putting in Theorem 2.1 we obtain the following result.

Corollary 2.2.

Let be the positive integer with . Also let and . If belongs to , then whenever .

Remark 2.3.

Taking in Theorem 2.1 and Corollary 2.2 we get results of [10].

We follow the method ofproofadopted in [11].

Theorem 2.4.

Let with and the function with be univalent convex in . If and defined by (1.8) satisfy the conditions

then the transform defined by (1.16) has the following:


(2) whenever


From the definition of we obtain
Differentiating shows that
It is easy to see that
From (1.9) and (2.19) we deduce that
Let us define
then is analytic in , with and Combining (2.18) with (2.21), one can obtain
Differentiating yields
In view of (2.21), (2.23), and (2.24), we obtain
Since and are convex and
by using Lemmas 1.2 and 1.3, from (2.26) we deduce that
It now follows from Lemma 1.1 that

and the result follows from the last subordination and Corollary 2.2.

It is well-known that (see, [12]) if and , then is univalent convex function in . So if we take in the Theorem 2.4, we obtain the following.

Corollary 2.5.

For and , let the function and defined by (1.8) satisfy the condition

Then the transform defined by (1.16) has the following:


(2) whenever

By putting on the (1.8), we get which is evidently convex. So by taking on Theorem 2.4 we have the following.

Corollary 2.6.

For with , let the function and defined by (1.8) satisfy the condition

Then the transform defined by (1.16) has the following:

(1) ;

(2) whenever

Remark 2.7.

Taking and on Corollary 2.6, we get a result of [11].

By putting and on Theorem 2.10 we obtain the following.

Corollary 2.8.

Let and with be univalent convex function in . Also let with and , satisfy
and let be the function which is defined by

then we have the following:


(2) whenever

Remark 2.9.

We note that if , then is convex function, and so we can replace with in Corollary 2.8 to get other new results.

In [13], Pannusamy and Sahoo have also considered the class for the case with

Theorem 2.10.

For let and defined by (1.13) satisfy the condition

Then the transform defined by (1.17) has the following:


(2) whenever


Let us define
then is analytic in , with and Using the same method as on Theorem 2.4 we get
Since and are convex,
Using Lemmas 1.2 and 1.3, from (2.42) it yields
It now follows from Lemma 1.1 that

and the result follows from (2.46) and Corollary 2.2.

Authors’ Affiliations

Department of Mathematics, Faculty of Science, Urmia University, Urmia, Iran


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© R. Aghalary and A. Ebadian. 2009

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