 Open Access
 Total Downloads : 327
 Authors : Dr. S. M. Khairnar, R. A. Sukne
 Paper ID : IJERTV2IS4149
 Volume & Issue : Volume 02, Issue 04 (April 2013)
 Published (First Online): 04042013
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
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, Pune412105
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, QuasiHadamard 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.

Introduction
(+)+
,
,
0, 0, 0, 0 < 1 , 0 < 1,
2
0, 0 < 1 , 1 < 1
2 2 2
+ + = , + = 2, + + =

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 quasiconvolution. 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 quasiconvolution 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 closetoconvex of order With respect to 1 1 .
1


Analysis and Main Results
Proof Since 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 closetoconvex
, ,,
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 closetoconvex 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 the1 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

Convolution and QuasiConvolution
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 CauchySchwarz 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].
This corollary is entirely new and not found in the
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
entirely new and not found in the literature. Letting
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 =
is entirely new and not found in the literature. By
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 CauchySchwarz 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


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Dr. S. M. Khairnar1
1Professor and Dean (R & D)
MITS Maharashtra Academy of Engineering, Alandi, Pune412105 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