1. Introduction

Let denote the class of functions of the following form:

(1.1)

which are analytic and multivalent in the open unit dics

Let denote the th-order ordinary differential operator for a function  , that is,

(1.2)

where ;     

Next, we define the differential operator as

(1.3)

In view of (1.3), it is clear that

(1.4)

If we take and for , then become the differential operator defined by Slgean [1].

Finally, in terms of a differential operator defined by (1.3) above, let denote the subclass ofconsisting of functionswhich satisfy the following inequality:

(1.5)

where , , , ; ;

For , , and , we define the next subclasses of

(1.6)

where ; ; ;

Remark 1.1.

() was studied by Nasr and Aouf [2] (also see Bulboac et al. [3]).

() and , were introduced by Srivastava et al. [4].

() and , were introduced by Silverman [5].

() and were introduced by Parvathan and Ponnusanny [6, pages 163-164].

() For and , the classes , , and are closely related with , , and which are defined by Owa and Sălăgean in [7].

In this paper we give relationships between the classes of , , and In the particular case when and , , and , we obtain the same results as in [8].

2. Main Results

Our main results are contained in

Theorem 2.1.

Let , and let ; then

(1)

(2)

(3)if , then

(2.1)

(4)if , then ;

(5)if , then

Proof.

() Let We prove that

(2.2)

If has the series expansion

(2.3)

then

(2.4)

We use the fact that for andthese imply

(2.5)

for

From (2.4) and (2.5), we deduce

(2.6)

By using the definition of from this last inequality we, obtain (2.2) which implies

(2.7)

hence

() Let be in Then (2.7) holds and, by using (2.3), this is equivalent to

(2.8)

For if , from (2.8) we obtain

(2.9)

which is equivalent to

(2.10)

Then multiplying the relation last inequality with we obtain

() if is a real positive number, then the definitions of and are equivalent, hence By using () and () from this theorem, we obtain ().

() We have the following two cases.

Case 1.

Let be defined by

(2.11)

and let We have

(2.12)

or

(2.13)

and then(see the definition of ).

Let now

(2.14)

Then, by a simple computation and by using the fact that

(2.15)

we obtain

(2.16)

where, , and

(2.17)

For we, have where is the disc with the center

(2.18)

and the radius

(2.19)

We have where and we deduce that for all does not hold.

We have obtained that for but and in this case

Case 2.

We consider the function defined by (2.11) for In this case, the inequality (2.13) holds too and this implies that

We also obtain that like in Case 1.

() Let be given by (2.11), whereand Then

(2.20)

which implies that

(2.21)

We have

(2.22)

where is given by (2.17).

From where and are given by (2.18) and (2.19), we obtain

(2.23)

If and, then

(2.24)

and if

(2.25)

then (2.24) also holds. By combining (2.24) with (2.23) and the definition of , we obtain that

(2.26)