# Properties Of Multivalent Holomorphic Functions And Some Results For Univalent Functions Defined By A Generalized Salagean Operator

DOI : 10.17577/IJERTV2IS4149

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#### Properties Of Multivalent Holomorphic Functions And Some Results For Univalent Functions Defined By A Generalized Salagean Operator

Dr. S. M. Khairnar1, R. A. Sukne2

1Professor and Dean (R & D)

MITS Maharashtra Academy of Engineering, Alandi, Pune-412105

2Assistant Professor in Mathematics

Dilkap Research Institute of Engineering and Management Studies, Neral, Tal. Karjat, Dist. Raigad.

Abstract

characterizations. For our convenience throughout this paper we are considering

In this paper we derived properties like differential

2 0, 0, 0, 0 1, (+)

= ,

Subordination, Hadamard product, Quasi-Hadamard product of Holomorphic Univalent and Multivalent functions with positive, negative Taylor series expansion. The results for Univalent functions are defined by a generalized Salagean operator.

Key Words Analytic function, Holomorphic function, Salagean operator, Subordination.

1. Introduction

(+)+

,

,

0, 0, 0, 0 < 1 , 0 < 1,

2

0, 0 < 1 , 1 < 1

2 2 2

+ + = , + = 2, + + =

1. Preliminary Lemma

1.1.1 Let () be of the form

=

=

= 2 2 2 , Then f(z) belongs to

2

2

, , , , , , , , if and only if

2 1 + 2 2 2 1

This paper contains the discussion of differential subordination and discussion of convolution and quasi-convolution. The differential subordination in

=2

Ã— 1+ 21 (+) 1+ 1+ 21 (+)

2 1

2

1.

the complex plane is the generalization of a

Proof: Since () , , , , , , , , And

differential inequality on real line. In convolution or Hadamard product of two power series the term

, , , , , , ,

2

, =

1 + +

convolution arises from the formula 2 =

: :

(, , )

1 + +

1 2 () ,

< 1. Convolution has

1 + +

2 0

Then

the algebraic properties of ordinary multiplication. The

1 + +

2 (1+ )2 2

1+ 2 1 +2

geometric series =

=

acts as the

= =2

=1

1

=2

(1+ 2 2

1+2

identity element under convolution = . A lot of literature on differential subordination is available in nuture for example Hardy, Littlewood and Polya [1], Protter and Weinberger [2], Walter [3],

Where = 1 + (2 1)( + )

Now, by definition of subordination, there exists ()

which is Holomorphic Function in U with 0 = 0,

= 1 in U such that

G. Goluzin [4],R. Robinson [5] S. S. Miller and P. T.

Mocanu [6] etc.The study of convolution has been

=2

2 (1+ )2 2

(1+ )2 2

1+2 1 + ()

1+2()

taken into consideration by number of authors,

=2

then by simple calculations, we obtain

Hayman [7], Epstein and Schoenberg [8], Loewner

=2

=2

=2

(21) (1+ 2 21

and Netanyahu [9], Suffridge [10] and Robertson [11]

(22 1)

2 2 1 (1+ )2 21

etc. We have also extended the concept of convolution to quasi-convolution and have obtained nice

Thus by noting |w (z)| < 1, we get

=

21 1+ 2 21

1 + (1 )1

=2

2 1

2 2 1 1+ 2 21

=2

= 1 1 , We obtain, 1 1

< 1

Letting 1, we get = 1 1 + (1 ) 1 1

=2

(21) (1+ )2

< 1 Differentiating both sides we get, 1 1

2 1 2 2 1 (1+ )2

=2

= 1 1 1 .

=2 2 1 1 +

2

=2

=2

< 2 1 2 2

Thus theorem holds true.

1 (1 + )2 Then

2 1 + 2 2 2(1)

Corollary 2.1.1 Let , 1 and

=2

1 (+)+1 is convex univalent function.

Ã— 1+(21) 1+ 1+(21)

2 (1)

2

1.

Then 1 1 is close-to-convex of order With respect to 1 1 .

1

2. Analysis and Main Results

Proof Since 1 1

1. Applications of Differential

Subordination Let , 1 denote the class of functions of the form

= 1 1 1 .

1 1 = 1

= +

+

0; (1)

=1

+

1 1 . We obtain

which are holomorphic in the open unit disc

= : < 1 . Let , , 1 , where

1

1

=

1 1 ()

+ 1 Since

= +

+

and

1 1

1 1 ()

=1

+

+

1

1

is a convex function,

=

+ =1 +

. Then the convolution

+

+

=

=1 + +

+

+

(2)

1

1 1

=

1 1

1 1 +

Let , , , , and , , be fixed real numbers.

A function , 1 belongs to the class , ,,, ,) ; , if it satisfies

> 1

Therefore, by definition of close-to-convex

, ,,

1+2

1+2

z U (3)

we get the required result.

Theorem 2.1.2 Let , , 1 ,

, ,,

and

1 2

,

1

(+)+ ()

, ,,, 1 1

, ,, 2 2

= 1 +

+ ( + )

where , are convex univalent in and if

1 2

where 1 = +

+)

+ .

(4)

0, > > 0 then

1(

)

=1 !

+

, ,, 1 1

Where is a gamma function. From above equation it

1

is follows that 1 =

( )1 . (5)

0

1 2 . 1 () 2()

This work is motivated by K. Piejko and J. Sokol [12]

Proof Since , ,, 1() 1() and

, ,, 2() 2() Then we have

and J. Patel and S. Rout [13], where we have used the

techniques of differential subordination to obtain

, ,, 1()

, ,, 2() 1 2 . And

several interesting properties. A holomorphic function f is said to be close-to-convex of order 0 < 1 . If there exists a convex function (1,1) and

by [13], the convolution of convex univalent functions is also the convex univalent function. Now, let

= , ,, 1(1 1)

()

1 1 (12)

1(12 )

such that Re

> .

= 1 +

()

Theorem 2.1.1 Let , 1 then

1 1 = 1

Then () is holomorphic function and 0 = 1 in .

since we have 1

1

1

= ( )1

[ + ] +

Proof We know

. (6)

+

+

= , ,, 1(1 1)

1( )

1 = ( )1 .

, ,, 1 1

= 1 1 1

1

+ 1 1 =

1

2

1

1

+ 1 1 1

( )

+ 1

.

1

2

= + (1 )1 . But owing to

+ 1 1 1

1 121 1,1; 2 + ; 12

. (7)

1

2

1 2

121

+ 1 1 1

1

2

Proof Since 1+1

and 1+2

are univalent convex

= 1 1 1 + 1

1+1

1+2

1

2

functions, 1+1 1+2 = 1 +

+ 1 +1 1 1

1+1

1+2

1 1 1+1

1

2

+ 1 1 + (1 )

1 + (2 2 ) 1+2

1

2

= 1 +

. Thus, by

Ã— 1 1

1 1 2

2 1+1 2

1

2

Theorem 2.1.2 we have

1 ()

+

1

1

1 1

2

, ,,

1 2

1

2

1 + (

)(

)

= 1 2

+

2

1 1 2

2

0 112

(+)+

1

(1 1 )(2 2 )

Ã— 1 +1

=

(+)

0

1 +

1

+ +

1

1

2

+

2

(1 )

1 2

1 1 2 2

1 1 2 2

= 1 + ( )( )

1 1

Ã— 1 1 1

Ã— 0

1 1 2 1 Hence we obtained the

1

2

required result. Putting

=

=

=

= 1 in

+ 2 2 1 1 .

1

Theorem 2.1.3, we have.

2 1 2

1

2

Now , ,, 1 , ,, 2 =

Corollary 2.1.2 Let , , 1 . Let

1 1 +

1 1

1 2

1+

1+

1

1

, ,, 1 1 and , ,, 2 1 then

1 1 +

1 1

1

2

2

1 + 4

.

2

, ,, 1 2

0 1+

= 1

1 1

Putting ( + ) = 1, ( + ) = 0 in above

1

2

 1 2 2 1 2 1 1 1+ and 0,0,0, 2 1+ then 1
 1 2 2 1 2 1 1 1+ and 0,0,0, 2 1+ then 1

+2 1

1 1 1

corollary 2.1.2 we have, next corollary.

+ 2 1 1 1 =

Corollary 2.1.3 Let 1 , 2 , 1 . Let

1

0,0,0, 1

1

1

2

1 + 4 1 .

+ 2 1 2 + 2 1

0, 1 2

0 1+

1

1

2

2

Consider the following integral transform

= + 1 = + ()

(8)

Ã—

1

2 +

0

=

+

Where () , 1 and + > 0. Now since

Ã— 1 1 . Then

1

2

1 =

= +

+ =

(1)

(+)

+ , Then we have 1

, ,, 1 , ,, 2 1() 2().

=1

!

+

(+)+

= + 1 1 . (9)

()

( +)

1 Theorem 2.1.4 Let , be real numbers ( )

0

1 2 1 () 2 () .

such that + > 0. If (), () (, 1) satisfy

1 2

1 1 2 () (1 1)(2 2)

then

Theorem 2.1.3 Let 1

, ,,, , (; 1, 1 )

11 2

and 2 , ,,, , (; 2, 2 ) that is

1 () () () (1 1 )(2 2 )

, ,, 1 1+1

1+

1+

1

and , ,, 2 1+2 .

1+

1+

2

Where () is defined as

112

where1 1 < 1 1; 1 2 < 2 1

= + 1 = + ()

( + ) + > ( + ) > 0 And

0 Then

0

=

+

1 =

=

(+)

+ .

1 () 1 + (

)(

)

1

=1

!

+

, ,, 1 2

1 1 2 2

() is defined as =

+ () and

1

= + 2

Ã—

0

112

= (). Where

= 1 + 1 1 +

1 2

1 2

++1

1 1

= 1 + (1 1)(2 2) Ã—

Ã—

21 1, 1; 2 + + ; 12 .

+

2 2 121

Proof Let =

1 ()

then () is

interesting generalized work of U. H. Naik and S. R. Kulkarni [15], also there are many researchers who have studied some proper -ties of univalent functions

holomorphic in with 0 = 1. Since we know

1 = + 1

with negative coefficients of type

= 2 like, Silverman and Berman

=2 2

1 . Hence

[16], Padmanabhan and Ganeshan [17] studied the

1 12

( )( )

convolution properties of univalent functions with

+

= 1 + 1 1 2 2

+

1

112

negative coefficients, Joshi and Kulkarni [18] also studied the properties of univalent functions with

missing coefficients. We shall make use of above

= + + +1 1+1 1+2

lemma,in our study.

0 1+1 1+2

=2

=2

1 + (1 1)(2 2)

11 2

Finally we obtain

Theorem 2.2.1 If =

2

2

= 1 + 1 1 +

and = 2 2 2 Where 2 0,

1 2 ++1 1 1

=

0 Such that

Ã—

21 1, 1; 2 + + ; 12

If we

2

2 2 121

, 2 , , , , , , , , , then

=

=

2

2

put = = 0, = 1 = 2 = 1, 1 = 2 = 1 = 2 2 2 2 , , , , , , 1, 1 ,

in above Theorem 2.1.4, then we have

With 1 2, +1 . Where

1 1

1 1

=

=

1

Corollary 2.1.4 Let + 1 > 0 where a real number. If (), () (, 1) and 1 2 () 1 + 4 then

1

3+2 4 2(1) 2 (1+3 )

Ã— 12 1 ()2

1 2

1

1+(13 ) 4()2(1)2

() () 1 + 4 . where

=

1

Proof Since and belong to,

=1 +

1 ,

, , , , , , , , therefore by lemma 1.1.1

= , 2

=1 + 2

=2 2 1 + 2 2 2 1

= 1 + 4 1 1 +1 21 1, 1; + 3 ;

Ã— 1+ 21 1+ 1+ 21

1 (20)

+2

1

2 1

2

=2

2 1 + 2 2 2 1

2. Convolution and Quasi-Convolution

Ã— 1 + 2 (1) 12 1 (21)

Properties Let T denote the subclass of A consisting

=2

2 1 + 2 2 2(1)

of functions of the form

=

=

= 2 ( 0) (10) Let (, , ) be the class of holomorphic functions

Ã— 1 + 2 (1) 12 1 (22) Where = 1 + 2 1 ( + ).

We contemplate to find 1, 1 , such that

in U that satisfies () 1+ 1 +

1+

Where

1 1 < 1 1 for

() , , , , , , 1 , 1 , , that is,

1 < 1 , 0 < 1.

2

2 1 + 2

(1)

2 2 =2 1 1

2

2

Consider as as subclass of T that consists of functions of the form = 2 2 2 (11)

Ã— 1 + 1 1 (1) 1 2 2 1 (23)

By using Cauchy-Schwarz inequality, we get

=

1 1 1

Define , , , , , , , , 2 2 2 2 2 2 2 2 2 2 1, (24)

=

=

=

+

+

= : : 1 + + (, , ) (12)

1 +

, , , , , , , ,

1 + +

= 2 1 + 2 2 2 1

Ã— 1 + 2 (1) 1 (25)

1

If 1 2 2 2 2 2 , then (23) is true, where

= : :

(13)

= 2 1 + 2 1

1 + +

1 1 1

(, , )

, , , , , , = , , 0, , , , , , (14)

Ã— 1 + Ã— 1 1 (1) 1 (26)

1

Or 1 2 2 2 ( = 2,3,4,)

, , , , , , , = , , 1, , , , , , (15)

, , , , , , = , , 0, , , , , , (16)

1

In view of (3.24), we have 2 2 2 1

(27)

2 2 2

, , , , , , , = , , 1, , , , , , (17)

Thus, to find 1 such that 1 =

(28)

2

() = +

2

1 + 1)( + )

2 1 + 2

1

=2 ( 1 1

() = + 1 + (2 1)( + )

2

(18)

Ã— 1 + 2 1 1 (1) (29)

=2

If =

2

and =

2

2 1 1 21 +(2)1 1+

2

=2

2

1 =

(1) 2 1+ (30)

=2 2

(2 , 2 0) then the convolution is

It is clear that 2 1 + for 1

=

=

defined by = 2 2 2 2 (19)

From (30) we can get

In the present work, we propose to give some

1 1 (21) 1+

2 (31)

21 (1+ )

(35)

11 (1) 2 1+

+1 (1) (1+ )

The right hand side of (31) decreases as k increases, then it has maximum for k = 2, then (31) is true if

1 1 1

The right hand side of (35) decreases as k increases and it has maximum for k = 2, then we obtain

6 1 ()

11 3+2 4 2 1 2 1+3

+1 3+2 4 2(1) 3+ 4 (1) (1+3 )

Ã—

Ã—

12 1 2

1+ 13 4 2 1 2

= (32)

1

Ã—

Ã—

1+(1+3 ) 2 ()(1)2

= . (36)

We can see that j < 1. Fixing A1 in (32), we have

1

1

B j+A1, (33)

1j

It is clear that < 1. Now fixing E

we get + , 1 1 2.

1

1 A1 < B1 1. Then the proof is complete.

Corollary 2.2.3: If , , , , , , And

2

2

=2

=2

2

2

Corollary 2.2.1 If f z = z

a2k

z2k

and

, , , , , , Then

+

= =2 2 2 where 2 0, 2 , , , , , , ,where 1 21 , 1

2

0 such that , , , , , , , , then

11

=2 2 2

=2 2 2

=

2

2

, , , , , ,

with 1

= 6 1 ()

3+2 4 2(1)

2 1 1

2 1 1

With, 1 < 1, 1 < 1 Ã— 1

1 1 2 2

3+ 4 (1) (1+3(+)) 2 ()(1)2

1 1 1

1 1 1

where 1 2 , 1+1 ,

=

=

11

This corollary is entirely new and not found in the literature. By putting = n = 0 in above Corollary, we

1

12 1 ()2

3+2 4 2(1) 2(1+3 ) 4()2(1)2

have the following result due to [15].

Corollary 2.2.4: If 0,0,0,0, , , 0 and

literature. By letting n == 0 in Corollary 2.2.1, we

2

0,0,0,0, , , 0 Th

have the following result due to [15].

2 en

0,0,0,0, , , 0 , where 1 22 , 2 +

Corollary 2.2.2 If =

2

2

and

12

2

=2

2

with

= 6 () .

= =2 2 where 2 0, 2 0

2 3+8 3+4 2 ()

such that f z , 0,0,0,0, , , 0 , then

=

2

2 0,0,0,0, , , 0 .

Corollary 2.2.5 If , , , , , , ,

=2 2

2 2 1 1

1 1

2

And , , , , , , , , then

Where 2 0, 2 0 , 2 < 2. With

2 +1

2

2

2

, , , , , , , .

1 1 22 , 1

1

, hence

Where 1 2 , 3 +

with

2

= .

= .

12()2

2 3+82 24()2

3

3 = 1

13

3+2 4 2(1) 3+ 4 (1)

Ã— 6 1 () . This corollary is

Theorem 2.2.2 If 2 , , , , , , , ,

(1+3 ) + 2 ()(1)2

and , , , , , , , , then

2

, , , , , , , , . Where 1

2 = = 0, =

, = 1, = 1, in above

1 2, +

1

with

2

Corollary, w have the following result due to [15].

=

=

6 1

3+2 4 2 1 3+ 4 1 1+3

Ã—

Ã—

1

1+(1+3 ) 2 ()(1)2

Corollary 2.2.6 If 0,1,1/2,1,1, , , 0

2

2

2

2

and 0,1,1/2,1,1, , , 0 . Then

0,1 /2,1,1, , , 0 . Where

2 ,1

Proof By making use of Theorem 2.2.1, and

1 2 , 4+

4

4

14

with

Lemma 1.1.1 we obtain,

2k F+1 1F+ 1 E Xn (1+Xm )

FE (1)

4

= 6 ()

4 3+82 3+4 2 ()

2 2+1 1+2+2 1 (1+ )

Theorem 2.2.3 Let =

2 Where,

2 (1)

0 , , , , , , ,

=2 2

2 +1 1++ 1 (1+ )

2 2

1 1 , and

Ã— (1) = , (34) Where = 1 + (2 1) , 0. Then by simple calculations we have

= 2 2 2 with 2 1, 1. Then

=

=

, , , , , , , , .

Proof By assumption, we have

=2

2 1 + 2 2 2(1)

Ã— 1+ 1+(21)

1 . And since

21 +2 2 2(1) ( 1+ ) 2

2 (1)

2

=2

2 (1) 2

2 1 1, then 1

1+(21) 1+ 1+(21)

.

21 +2 2 2(1) ( 1+ ) 2

=2

2 (1)

2

2

=2

2 (1) 2

2 1 + 2 2 2 1

=2

1+ 21 1+ 1+ 21

1.

1 (37)

Then we may write

2 1

That is

2

2

1 21 +2 2 2(1)

( 1+ ) 2

=

2

, , , , , , , ,

=2 2

2 (1)

=2 2 2

Ã— (2 2 + 2 2) 1 (38)

Corollary 2.2.7 Let =

2 .

Therefore, the inequality (38) holds if

Where

=2

0 , , . , , ,

2

21 + 2 1(1)1 ( 1+ )

2 2

and

11 (1)

=

=

= 2 2 2 With 2 1, 1.

1 21 +2 2 2(1) ( 1+ ) 2 2

Then , , , , , , .

2 2 (1) = 2 .

This corollary is entirely new and not found in the literature. By putting == 0 in above Corollary, we

And by simplification, the last inequality gives

1 1 2 21 (1+ )

(39)

have the following result due to [15].

Corollary 2.2.8 Let = 2 2 2 ,

1+1 1 22 (1+ )

The right hand side of (39) decreases as k increases

and if we put k = 2, we obtain

Where 0 0,0,0,0, , ,

=

and

1 1 24 1 2

2

2 0

2

1+1 3+2 4 2 1 2 1+3

= =2 2 With b2i 1, i 1. Then

Ã— 1 = (40)

0,0,0,0, , , 0

1+(1+3 ) 8()2(1)2

Now by fixing 1 in (40), we have 1 1 + and

Corollary 2.2.9 Let = =2 2 2 .

1

1 2. with j given in

1

2

2

Where 2 0 , , , , , , , . And

1 give 1

40 .

= 2 2 2

with 2 1, 1. then

Corollary 2.2.11 Let , , , , , , , ,

=

, , , , , , , .This corollary

2

then =

putting == 0 and = 1 , = = 1 in above

=2

(2

2 + 2

2)2 , , , , , ,

1 1 1

1 1 1

2 1 1

2 1 1

+

2

Corollary, we have the following result due to [15].

Where 1 2 and 1 1

=

=

.

.

11

With

Corollary 2.2.10

1

24 1 2

3+2 4 2(1) 2(1+3 ) 8()2 (1)2

Let 0,1, 1 2,1,1, , , 0 .

By putting == 0 in Corollary 2.2.11, we have the

2 2 following result due to [15].

= 2 2 2 Where 2 0 , and

2

2

=

=

2

2

=

2

2

Where 2

1, 1.

Corollary 2.2.12 Let , 0,0,0,0, , , 0 , then

Then 0,1,1/2,1,1, , , 0 .

=

2 2 2 + 2 2 2 0,0,0,0, 1, 1 , 0 .

Theorem 2.2.4 Let , , , , , , , , , ,

=

2

1 +2

2

then =

Where 1 1 22 and 1 12

24 2

with

2 2 2 + 2 2 , , , , , , , , .

2 =

2 2 . Let the class T (n, p) of

= 2

=+

=+

3+82 8 ()

1 1

1 1

where 1 2 1 +

1

with

functions of the form =

=

=

1

3+2 4 2(1) 2 (1+3 )

Ã—

Ã—

.

.

24 1 2

1+(1+3 ) 8()2(1)2

where , , 0 (41)

will be Holomorphic and multivalent in the unit

disk = : < 1 . Consider the generalized Ruscheweyh derivative , ,,,, , () defined as

Proof By assumption, we have

21 +2 2 2(1) ( 1+ )

, ,,,, ,

, ,,,, ,

2

2

=

2 (1)

21 +2 2 2(1) ( 1+ )

2 1

= +

=

=

, ,, , , =

()

(42)

=2

2 (1) 2 1

Where = 1 + (2 1), thus

+1+2 (+2+ )(++2)

+1 (+2+2+ )(+2)(1+)

(43)

21 +2 2 2(1) ( 1+ ) 2

, , , , = , = , = and = = 1

=2

1.

2 (1)

2

We have Ruscheweyh derivative to univalent function. Now we define a class

(, , , , , , , , , ) consisting of functions

f (z) o f the form (41) satisfying the condition

Corollary 2.2.13 Let

, , , , , , , , , , then

1 1 1 + +1 ,

, , ,, ,)()

1 1 (+) +1 +(+)+1 , ,,, , (+)

1 2 , , ,, ,) + , , ,, ,)() +, , ,, ,)()

k n + p. (47)

>

1

1

(44)

Also, consider the class , , , , , , , , ,

consisting of all functions (, ) such that

z u, 0 < , 0 < and

2 1 p p1 +p+1

p

p

,,,,,f(z) as defined in (42). Also Let

, , , , , , , , , .

=

=

= + ,

( = 1,2)

Theorem 2.2.6 The function

, , , , , , , , , if and only if

be in the class (, ) consisting of the family of functions that are holomorphic in u. Then the quasi-

=+

1 + + 1 1

convolution 1 2 of the functions

, , is defined by

Ã— 2 + 2 1

Ã— , ,, , , ()

1 2, 1 2

= = + 1 , ,

(45)

1 1 1 + + 1 .

Where

=

,

.

1 1

=1

, = ,1

,2

,

Where 0 < , 0 < ,

1 1 + +1 2

and , ,,, , as defined in (43)

Theorem 2.2.5 Let , . Then

() (, , , , , , , , , ) if and only if

=+

1 1 k k 1 + k + 1 ,,,,, k ak

Corollary 2.2.14 Let

, , , , , , , , , , then

p

p

< 1 1 1 + + 1 (46)

1

0 < 1

1 1 + +1

and , ,,, , ()

(+) 1 1 (+) +1 + ()+1

as defined in (43). The result holds true.

Proof Let , . and suppose that

, , , , , , , , , , then

Ã— 1 1 1 + +1

, ,,, , (+)

( 2+2) .

(+) 2 (+)2+2 , ,,, , (+)

, , ,, , ()

Theorem 2.2.7: Let , , , , , , , , ,

1 2 , , ,, , + , , ,, , () +, , ,, , ()

then 1 1 1 + +1

+

> ( ).

1 1 (+) +1 +(+)+1

, ,,, , () +

1 ,,,, , 1 2

Ã— , ,,, , , ,, , , ()

> 0.

1 1 1 + +1

1 1 (+) +1 +(+)+1

+ (48)

, ,, ,

1 (+) 1 1 1 + +1

+1

By using the definition of

,

(), we obtain

1 1 (+) +1 +(+)+1

1 1 1 + + 1 1 1 1 + + 1

, ,,, , 1 +

=+

Ã— , ,,, , ()

> 0

(+) 1 1 1 + +1

1 1 (+) +1 +(+)+1

+1 (49)

Where 0 < 1 , 0 < 1 ,

Letting 1

on real values yields

2 1 1 + +1

=+

1 1 1 + + 1

Ã— , ,, , , ()

+ > , , , , , .

, ,, , , + =

< 1 1 1 + + 1 Where

, ,, , , =

2+1+ (+2+ )(++2)

+1 (+2+2+ )(+2)(1+ )

+1+ (+2+ )(++2) .

+1 (++2+ )(+2)(1+)

Proof For , , , , , , , , , , we have

, ,, , , ()

Conversely, suppose (46) holds true, then

=+

, ,,, , ()

=

1 1 1 + +1

. Therefore

1 2 , , , , , + , , , ,, () +, ,,, , ()

1 1 (+) +1 +(+)+1

, ,,, , () =

, ,,, , ()

Thus by simple calculations we get the required result.

=+

For sharpness the function () is given by following

=

+ + = + , ,,, , ()

1 1 1 + +1

+

1 1 1 + +1

1 1 (+) +1 +(+)+1 , ,,, , (+)

+ .

1 1 (+) +1 +(+)+1

And , ,, , , ()

1+1 1 + +1 + + +1 , ,,, , ()

Ã—

1,

1 1 1 + +1

1 1 (+) +1 +(+)+1

Similarly, we can prove the relation (49).

Theorem 2.2.8 Let

+ .

1+1 1 1 + +1

then by mathematical induction, we obtain the result

which is true for any positive integer. We want to prove that 1 2 = + + . Where An+p =

() , , , , , , , , , then 1 1 1 + +1 1

1 2 () , , , , , , , , , =1 1 1 + +1 + + +1 , , ,, , (+)

and 0 < 1

. Where

0 < 1

. This is the required

1 1 + +1 1(+,)

1 1 + +1

=

=

= + , = 1,2,, .

condition and this completes the proof of theorem. For

+ , =

1 1 + +1 + + +1

sharpness take the function =

1 =1

1 1 1 + +1

1 1 1 + +1

+

, ,, , , + 1 For 0 < 1 .

1 1 + +1 + + +1 , ,,, , (+)

Thus result holds true.

1 1 + +1

Similarly we can prove the result for

, , , , , , , , , in next Theorem.

Proof By induction on For = 1, the result is true. For = 2, we have

1

1

1 1 1 + +1 , ,,, , ()

=+ 11 1 1 + +1 ,1 1.

Theorem 2.2.9 If

, , , , , , , , , for each

( = 1,2,, ) Then

, , , , , , , , ,

12 1 1 + +1 ( +),(+)()

1 2

=+ ,2

=+ ,2

& 1

12 1 1 + +1

0 < 1

1 1 + +1

2(+, )

By Cauchy-Schwarz inequality we have

2

2 + , =

2

1 1 1 + +1

(+), ,,, , (+)

=+

=1 1 1 1 + +1

,

Ã— , ,, , , () 1. We have only to find the largest

such that

=1 1 1 1 + +1

Ã— 1 1 + + 1 + + + 1

1. The result is sharp for the functions: for

1 1 1 + +1 , ,,, , ()

=+ 1 1 1 + +1 ,1,2

1 Such that

1 1 1 + +1

( = 1,2,, ) given by

=

1 1 1 + +1 + (+) 1 1 + +1 + + +1 , , ,, , (+)

1 1 1 + +1

,1

,2

1

Put = ( = 1,2,, ) in Them. 2.2.8, we get

2

1 1 1 + +1 2 . Consequently, we

=1 1 1 1 + +1

have to find such that 1 1 1 + +1

1 1 1 + +1

Corollary 2.2.15 If

, , , , , , , , , = 1,2,, .

, , , , , , , , , .

, ,,, , 2 1 1 1 + +1 .

1 2

1

=1 1 1 1 + +1 Where =

Thus , , , , , , , , , ,

1 1 + +1

1 2

1

for 0 <

Where

, ,,, ,

1 =

1 1 + +1

1()

1 1 ( +) +1 +( + )+1

1 1 1 + +1

( +)

1

, ,, , , () 2 1 1 1 + +1

1.

0 < 1

Then result is sharp for the

=1 1 1 1 + +1

1 1 + +1

So for + we get

0 < 1 . Where

function () for all = 1,2,, given by

=

1 1 + +1 1(+)

1 1 1 + +1

+ .

1 + = , ,,, , ( + )

1 1 + +)1 + ++1 , ,,, , (+)

Ã—

Ã—

2

=1

1 1 + +1 + + +1 1.

1 1 1 + +1

Put = for ( = 1,2,, ) in Theorem 2.2.9, then next corollary.

Now suppose the result is true for any N . Then we

must show that

1 2 , , , , , , , , , ,

where 0 < 1 n and

Corollary 2.2.16: If ( , , , , , , ,

, , for i = 1,2,, N . Then

() , , , , , , , , ,

1 p p1 +p+1 M1 (n+p,+1)

1 2

+ , + 1 = 1 1 + +1 + + +1

1 1 1 1 + +1

where = 1

1 1 1 + +1 , ,,, ,

1 1 + +1

=+ 1 1 1 + 1

. ++ (+)

, ,,, ,

Ã—

+ (++)

( +) 1 1 ( +) +1 +( +)+1

( +)

1 1 1 + +1

, ,,, ,

1 1 1 + +1

+

1.

=

1 1 1 + +1

Since ++ (+) 1 for + , then

Theorem 2.2.10: Let 1 , 2 , 3 ,, ()

+ (++)

=

=

defined by = + ,

, , , , , , , , , . Where = 1,2,,

. Then arithmetic mean of fi = 1,2,,

() (, , , , , , , , , )

Theorem 2.2.13: Let () be defined, having Taylor

defined by = 1 (), is also in series expansion of the form, =

=1

, , , , , , , , , = 1,2,, .

++ (+) Then () is starlike of

=+ + (++)

Proof: By definition of () we have

order 0 < . in

1

1 , , , , , , , , , , =

= =1

=+ ,

=

() ++ (+) 1 1 1 + +1 1

1

.

() + (++) 1 1 1 + +1

=+

=1

,

Using Theorem 2.2.5,

+

1

=+

1 1 1 + + 1

Ã— , ,, , ,

.

Ã— , ,, , , 1

, =

=1

1

1 1 1 + + 1

Proof: We must show that

=1

=+

()

Ã— , ,, , ,

< . Or

1

,

1 1 1 + + 1 ,

()

++ (+) ( )

=1

=+ + (++)

Then we obtain (, , , , , , , , , ).

< ( ) 1

++ (+)

Theorem 2.2.11: Let () and () be in the class

(, , , , , , , , p, n), then ()) = () + (1 t) (), 0 t 1, also belongs to

, , , , , , , , , .

Proof: By definition of h(z) we have

=+ + (++)

then ++ +

+ ++

Ã— 1 1 1 + +1 , ,,, , . Hence proof.

1 1 1 + +1

3. References

=

=

= + + (1 ) ,

where

1. G. H. Hardy, J. E. Littlewood and G. Polya,

=

=

= +

And =

Inequalities, Sec. Edit., Cambridge University Press,

= + ,

( , 0).Using theorem 2.2.5

(1952).

=+

1 1 1 + +1 , ,,, , 1 1 1 + +1

2. M. H. Protter and H. E. Weinberger, Maximum

Principles in Differential Equations, Prentice Hall, Englewood Cliffs, New Jersey, (1967).

Ã— + (1 )

1 1 1 + +1 , , ,, ,

= + +

3. W. Walter, Differential and Integral Inequalities, Springer, New York (1970).

4. G. M. Goluzin, On the majorization principle in

=

1 1 1 + +1

1 1 1 + +1 , , ,, ,

function theory (Russian), Dockl.Akad.Nauk SSSR, 42

=+

=+

(1 ) 1

1 1 1 + +1

then (, , , , , , , , , ). consider the generalized Jung-Kim-Srivastava integral operator

(1935), 647-650.

5. R. M. Robinson, Univalent majorants, Trans. Amer. Math. Soc., 61 (1947), 1-35.

6. S. S. Miller and P. T. Mocanu, Differential

=

++ + . Where

subordinations and univalent functions, Michigan Math. J.

=+ + ++

0, > 1. [19], then we have the next theorem.

Theorem 2.2.12: Let (, , , , , , , , , )

be defined by (41) and 0, > 1 then ()

defined above also belongs to

(, , , , , , , , , ).

.

Proof: By Theorem 3.3.5, we have

28 (1981), 157-171.

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9. C. Loewner and E. Netanyahu, On some compositions of Hadamard type in classes at analytic functions, Bull. Amer. Math. Soc., 65 (1959), 284-286.

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17. K. S. Padmanabhan and M. S. Ganeshan, Convolution of certain classes of univalent functions with negative coefficients, Indian J. Pure Appl. Math., 19 (9) (1988), 880-889.

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Dr. S. M. Khairnar1

1Professor and Dean (R & D)

MITS Maharashtra Academy of Engineering, Alandi, Pune-412105 Smkhairnar2007@gmail.com

R. A. Sukne2

2Assistant Professor in Mathematics Dilkap Research Institute of Engineering and

Management Studies, Neral, Tal. Karjat, Dist. Raigad.

rasukne@gmail.com