1. Introduction and Preliminaries

Let denote the class of all analytic functions in the unit disk . For , a positive integer, let

(1.1)

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

(1.2)

and it is called the class of all starlike functions of order . Clearly, for . For functions , given by

(1.3)

we define the Hadamard product (or convolution) of and by

(1.4)

An interesting subclass of (the class of all analytic univalent functions) is denoted by and is defined by

(1.5)

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

(1.6)

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

(1.7)

Now by letting

(1.8)

it is easily seen that

(1.9)

For , Owa and Srivastava [3] introduced the operator defined by

(1.10)

which is extensions involving fractional derivatives and fractional integrals. Using definition of we may write

(1.11)

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

Also for and , let us define the function by

(1.12)

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

If we take

(1.13)

then we can rewrite operator defined by (1.11) as

(1.14)

From the definition of it is easy to check that

(1.15)

For with for all we define the transform by

(1.16)

where and

Also for with for all we define the transform by

(1.17)

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

(1.18)

be analytic in . If

(1.19)

then

(1.20)

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

(2.1)

Proof.

Let us define

(2.2)

Since , we have

(2.3)

where is an analytic function with and By Schwarz lemma, we have . By (2.3), it is easy to check that

(2.4)

Therefore

(2.5)

We need to show that . To do this, according to a well-known result [9] and (2.5) it suffices to show that

(2.6)

which is equivalent to

(2.7)

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

(2.8)

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

(2.9)

Define by

(2.10)

Differentiating with respect to , weget

(2.11)

Case 1.

Let Then we see that has its only critical point in the positive real line at

(2.12)

Furthermore, we can see that for and for . Hence attains its maximum value at and

(2.13)

Case 2.

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

(2.14)

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

(2.15)

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

(1)

(2)whenever

(2.16)

Proof.

From the definition of we obtain

(2.17)

Differentiating shows that

(2.18)

It is easy to see that

(2.19)

From (1.9) and (2.19) we deduce that

(2.20)

or

(2.21)

Let us define

(2.22)

then is analytic in , with and Combining (2.18) with (2.21), one can obtain

(2.23)

Differentiating yields

(2.24)

In view of (2.21), (2.23), and (2.24), we obtain

(2.25)

Hence

(2.26)

Since and are convex and

(2.27)

by using Lemmas 1.2 and 1.3, from (2.26) we deduce that

(2.28)

It now follows from Lemma 1.1 that

(2.29)

Therefore

(2.30)

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

(2.31)

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

(1)

(2) whenever

(2.32)

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

(2.33)

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

(1);

(2)whenever

(2.34)

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

(2.35)

and let be the function which is defined by

(2.36)

If

(2.37)

then we have the following:

(1)

(2) whenever

(2.38)

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

(2.39)

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

(1)

(2) whenever

(2.40)

Proof.

Let us define

(2.41)

then is analytic in , with and Using the same method as on Theorem 2.4 we get

(2.42)

Since and are convex,

(2.43)

Using Lemmas 1.2 and 1.3, from (2.42) it yields

(2.44)

It now follows from Lemma 1.1 that

(2.45)

Therefore

(2.46)

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