1. Introduction

Let denote the class of functions of the form:

(1)

which are analytic and univalent in the open unit disk

Let denote the subclass of functions of the form:

(2)

The convolution of two power series given by(1) and

(3)

is defined as the following power series

Definition (1): A function is said to be in the class if it satisfies the condition:

(4)

where

2. Coefficient Estimates

In the following theorem, we obtain the sufficient and necessary condition to be the function in the class

Theorem(2.1): Let the function be defined by (2). Then if and only if

(5)

Proof: Suppose that . Then by the condition (4), we have

Since for all z, we have

Conversely, assume that the hypothesis(5) and then

by hypothesis. Then by Maximum modulus theorem, we have

Finally, the result is sharp for the function

Corollary (2.1): Let the function is in the class . Then

3. Subordination Theorems

Definition(2): Let and be analytic in the unit disk U. Then is said to be subordinate to f, written or if there exists a Schwarz function which is analytic in U with and such that Indeed it is Known that

In particular, if the function is univalent in U, we have the following equivalence([3], [4]):

Definition (3) [1]: The fractional derivative of order of a function is defined by

where is in Definition(1.1.14), and the multiplicity of is removed by requiring to be real, when

Definition (4): Under the hypothesis of Definition (3), the fractional derivative of order is defined, for a function by

It readily follows from Definition (3) that

(6)

We shall need the concept of Subordination between analytic functions and Subordination theorem of Littlewood [2]. (See also Duren [1])

Theorem (3.1): If the function and are analytic in U with Definition (2), then

(7)

Theorem (3.2): Let If and supposed that is defined by

If there exists an analytic function w defined by

Then, for and

(8)

Proof: Let

and

(9)

Then, we must show that

(10)

By Theorem (3.1), it suffices to show that

(11)

Set

(12)

From (12) and (5), we obtain

Next, the proof for the first derivative.

Theorem (3.3): Let If and

Then for and

(13)

Proof: It is suffices to show that

(14)

This follows because

Theorem (3.4): Let be of the form (3) and be of the form (2) and let for some where

Also, let for such the functions and be defined respectively by

(15)

Then, for and

Proof: Convolution of and is defined as:

Similarly, from (15), we obtain

To prove the theorem, we must show that for and

Thus, by applying Theorem (3.1), it would suffice to show that

(16)

If the subordination (16) holds true, then there exist an analytic function w with and such that

From the hypothesis of the theorem (3.2), there exists an analytic function given by

which readily yields Thus for such function using the hypothesis in the coefficient in equality for the class we get

Therefore, the subordination (16) holds true.

Now, we discuss the integral means inequalities for and h defined by

(17)

Theorem(3.5): Let and given by (17). If f satisfies

(18)

and there exists an analytic function such that

Then, for and

Proof: By putting and we see that

and

Applying Theorem (3.1), we have to show that

Let us define the function by

(19)

Since for

There exists an analytic function w in U such that

Next, we prove the analytic function w satisfies for the condition (18). By (19), we know that,

(20)

Letting in (20), we define the function by

If then we have for . Indeed we have

That is

Theorem (3.6): Let be given by

(21)

and suppose that

(22)

For or and where denotes the pochhammer symbol defined by

Then for

(23)

Proof: By means of the fractional derivative formula (6) and Definition (4), we find from (2) that

where

Since is adecreasing function of we have

Similarly, by using (21), (6) and Definition (4), we obtain

Thus, we have

For we must show that

By applying Theorem (2.1). It suffices to show that

By setting

We find that

Which readily yields . Therefore, we have

By means of the hypothesis (22) of theorem (3.6).

Theorem (3.7): If and

(24)

Then where

(25)

The result is sharp for the functions which is given by

(26)

Proof: Since Theorem (2.1) gives

For we have

Note that, we have to find the largest such that

The above inequality is true if

From the previous inequality, we obtain

That is,