on sandwich theorem for certain subclasses of symmetric ... · symmetric analytic functions in the...

International Journal of Science and Research (IJSR) ISSN (Online): 2319-7064

Index Copernicus Value (2013): 6.14 | Impact Factor (2015): 6.391

Volume 5 Issue 5, May 2016

www.ijsr.net Licensed Under Creative Commons Attribution CC BY

On Sandwich Theorem for Certain Subclasses of

Symmetric Analytic Functions Associated with

Noor Integral Operator

C. Selvaraj1, T. R. K. Kumar

2

1Department of Mathematics, Presidency College, Chennai, Tamilnadu, India

2Department of Mathematics, R. M. K. Engineering College, Tamilnadu, India

Abstract: In this paper, we obtain some interesting properties of differential subordination and superordination for the classes of

symmetric analytic functions in the unit disk, by applying Noor integral operator. We investigate several sandwich theorems on basis of

this theory.

Keywords: Convex functions, Differential subordination and superordination, Noor integral operator, Best dominant

1. Introduction

Let H(U) denote the class of analytic functions in the open

unit disk 1}|<:|{= zzU and let ,1][aH denote the

subclass of the functions H(U)f of the form:

)(=)( 2

21 C azazaazf

Also, let A be the class of functions H(U)f of the

form

,=)(2=

n

n

n

zazzf

(1)

For two functions )(zf given by (1) and

n

n

n

zbzzg

2=

=

The Hadamard product (or convolution) of f and g is

defined by

.==2=

zfgzbazzgf n

nn

n

Let H(U)gf , , we say that the function f is

subordinate to g , if there exist a Schwarz function w ,

analytic in U , with 0=(0)w and )(1|<)(| Uzzw ,

such that ))((=)( zwgzf for all Uz .

This subordination is denoted by gf or )()( zgzf .

It is well known that, if the function g is univalent in U ,

then )()( zgzf if and only if (0)=(0) gf and

)()( UU gf .

Let H(U))(),( zhzp , and let

CUC 3:);,,( ztsr . If )(zp and

));(),(),(( 2 zzpzzpzzp are univalent functions, and

if )(zp satisfies the second-order superordination

));(),(),(()( 2 zzpzzpzzpzh (2)

then )(zp is called to be a solution of the differential

superordination (2). (If )(zf is subordinatnate to )(zg ,

then )(zg is called to be superordinate to )(zf ). An

analytic function )(zq is called a subordinant if

)()( zpzq for all )(zp satisfies (2). An univalent

subordinant )(~ zq that satisfies )(~)( zqzq for all

subordinants )(zq of (2) is said to be the best subordinant.

Recently, Miller amd Mocanu [13] obtained conditions on

)(),( zqzh and for which the following implication

holds true:

)()());(),(),(()( 2 zpzqzzpzzpzzpzh

Using these results, the authors in [3] considered certain

classes of first- order differential superordinations, see also

[7], as well as superordination-preserving integral operators

[6]. Aouf et al. [3, 4], obtained sufficient conditions for

certain normalized analytic functions )(zf to satisfy

)()(

)()( 21 zq

zf

zfzzq

where )(1 zq and )(2 zq are given univalent functions in

U with 1=(0)1q and 1=(0)2q .

In [18], Sakaguchi defined the class of starlike functions with

respect to symmetrical points as follows:

Let Af . Then f is said to be starlike with respect to

symmetrical points in U if, and only if,

).(0,> UR

zzfzf

zzf '

Obviously, it forms a subclass of close-to-convex functions

and hence univalent. Moreover, this class includes the class

Paper ID: NOV163819 1657





of convex functions and odd starlike functions with respect

to the origin, see [18].

Let A denote by AA :D the operator defined by

1>1

=1

zfz

zzfD

or equiavalently,

0,1,2,=!

= 0

1

N

kk

zfzzzfD

kkk

where the symbol )( stands for the Hadamard product (or

Convolution). We note that )(=)(0 zfzfD and

)(=)(1 zzfzfD '. The operator fDk

is called the

Ruscheweyh derivative of kth order of f , see [17].

Analogous to fDk, Noor [14] and Noor et al. [15] defined

an integral operator AA :kI as follows.

Let 01

,1

=)( N

k

z

zzf

kk , and let

kf be defined

such that

21

=)(z

zzfzf kk

Then

.1

=)(=1

zfz

zzfzfzfI

kkkk

(3)

From (3) it is easy to verify that

.1= 11 zfkIzfIkzfIz kk

'

k

(4)

We note that zzfzfI '=0 and zfzfI =1 . The

operator zfIk defined by (3) is called the Noor Integral

operator of kth order of f , see [8]. Moreover, Liu [8]

introduced some new subclasses of strongly starlike

functions defined by using the Noor integral operator and

studied their properties. Liu and Noor [9] investigated some

interesting properties of the Noor integral operator.

Definition 1.1 A function Af is said to be in the class

BA,,B , if it satisfies the following subordination

condition:

z

zfIzfI

zfIzfI

zfIzfIz

z

zfIzfI kk

kk

kkkk

221 11

11

11

Bz

Az

1

1

(5)

where and throughout this paper unless otherwise mention

the parameters , , A and B are constrained as follows:

,,1,1:0>)(: RABAB RC

and all powers are understood as principal values.

In this paper, we prove such results as subordination and

superordination properties, convolution properties, distortion

theorems, and inequality properties of the class BA,,B .

For interested readers see the work done by the authors [1,

5].

2. Preliminary Results

Definition 2.1 Let Q be the set of all functions f that are

analytic and injective on )(\ fEU , where

,=lim:=)(

zffEz

U

and are such that 0)( 'f for )(\ fEU .

To establish our main results we need the following

Lemmas.

Lemma 2.1 (Miller and Mocanu [12, 13]). Let the function

)(zh be analytic and convex (univalent) in U with

1=(0)h . Suppose also that the function )(z given b

2

211=)( zczcz (6)

is analytic in U ,

0),0;,()()(

)(

RUzzhzz

z'

(7)

then 1

0( ) ( ) = ( ) ( ),

z

z z t h t dt h zz

( ),z U and )(z is the best dominant of (1).

Lemma 2.2 (Shanmugam et al. [19]). Let

0\=, CCC and let q be a convex univalent

function in U with

),(,0;>)(

)(1 URR

zmax

zq

zzq'

''

If p is analytic in U and

)()()()( zzqzqzzpzp '' (8)

then )()( zqzp , and q is the best dominant of (3).

Lemma 2.3 ([13]). let )(zq be a convex univalent function

in U and let 0>, mm C . Further assume that

0>mR . If ,(0),1)( Qqzg H and

),()()()( zmzgzgzmzqzg ''

implies )()( zgzq , and )(zq is the best subordinant.

Lemma 2.4 ([10]). let F be a analytic and convex in U . If

Agf , and Fgf , Then

1 , (0 1).f g F

Lemma 2.5 ([16]). let n

nnzazf

1=1=)( be analytic






and convex in U and n

nnzbzg

1=1=)( be analytic

and convex in U . If )()( zgzf , then

)(,< 1 Nnban .

3. Main Results

Theorem 3.1 Let BAzf ,)( ,B with 0>R . Then

11 1

1 1

0

1 1( ) =

2 1

k

k kI f z I f z k Azuz u du

z Bzu

1

1

Az

Bz

(9)

and )(z is the best dominant.

Proof. Set

).(),(=

2

11 U

zzhz

zfIzfI kk

(10)

Then )(zh is analytic in U with 1=(0)h . Logarithmic

differentiation of (5) and simple computations yield

z

zfIzfI

zfIzfI

zfIzfIz

z

zfIzfI kk

kk

kkkk

221 11

11

11

.1

1

1=

Bz

Azzzh

kzh '

(11)

Applying Lemma 2.2 to (11) with

1=

n, we have

1 1( )

2

k kI f z I f zz

z

1 1

1

0

1 1=

1

k kzk At

z t dtBt

,1

1

1

11=

11

1

0 Bz

Azduu

Bzu

Azukn

(12)

and )(z is the best dominant. This completes the proof.

Theorem 3.2 Let )(zq be univalent in CU , . Suppose

also that )(zq satisfies the following inequality:

.10;>)(

)(1

RR kmax

zq

zzq'

''

(13)

If Af satisfies the following subordination:

z

zfIzfI

zfIzfI

zfIzfIz

z

zfIzfI kk

kk

kkkk

221 11

11

11

,1

zzqk

zq '

(14)

then

1 1( ),

2

k kI f z I f zq z

z

( ).z U

and )(zq is the best dominant.

Proof. Let the function )(zh be defined by (10). We know

that the first part of (11) holds true. Combining (11) and

(14), we have

.11

zzqk

zqzzhk

zh ''

(15)

By using Lemma 2.3 and (7), we easily get the assertion of

Theorem 3.2.

Corollary 3.3 Let C and 1<1 AB . Suppose

also that

.10;>1

1

RR kmax

Bz

Bz

If Af satisfies the following subordination:

z

zfIzfI

zfIzfI

zfIzfIz

z

zfIzfI kk

kk

kkkk

221 11

11

11

,11

12

Bz

zBA

Bz

Az

then

1 1 1, ( ).

2 1

k kI f z I f z Azz U

z Bz

and Bz

Az

1

1 is the best dominant.

If f is subordinate to F , then F is superordinate to f .

We now derive the following superordination result for the

class BA,,B .

Theorem 3.4 let )(zq be convex univalent function in U

and let C with 0>R . Also let

,(0),1

2

11 QH

qz

zfIzfI kk

and

z

zfIzfI

zfIzfI

zfIzfIz

z

zfIzfI kk

kk

kkkk

221 11

11

11

be univalent in U . If

zzqk

zq '

1

z

zfIzfI

zfIzfI

zfIzfIz

z

zfIzfI kk

kk

kkkk

221 11

11

11

then






z

zfIzfIzq kk

2

11

and q is the best subordinant.

Proof. Let the function )(zh be defined by (10). Then

zzqk

zq '

1

z

zfIzfI

zfIzfI

zfIzfIz

z

zfIzfI kk

kk

kkkk

221 11

11

11

.1

= zzhk

zh '

An application of Lemma 2.4 yields the assertion of

Theorem 3.4.

Taking Bz

Azzq

1

1= in Theorem 3.4, we obtain the

following corollary.

Corollary 3.5 let )(zq be convex univalent function in U

and let C 1,<1 AB with 0>R . Also let

,(0),1

20 11 Qq

z

zfIzfI kk

H

and

z

zfIzfI

zfIzfI

zfIzfIz

z

zfIzfI kk

kk

kkkk

221 11

11

11

be univalent in U . If

211

1

Bz

zBA

Bz

Az

,22

1 11

11

11

z

zfIzfI

zfIzfI

zfIzfIz

z

zfIzfI kk

kk

kkkk

then

1 11, ( ).

1 2

k kI f z I f zAzz U

Bz z

and Bz

Az

1

1 is the best subordinant.

Combining the above results of subordination and

superordination, we easily get the following sandwich- type

result.

Corollary 3.6 let 1q be convex univalent and let 2q be

univalent in U , C with 0>R . Let 2q satisfy(12 ).

If

,(0),1

20 11 Qq

z

zfIzfI kk

H

and

z

zfIzfI

zfIzfI

zfIzfIz

z

zfIzfI kk

kk

kkkk

221 11

11

11

be univalent function in U , also

zzqn

zq '

111

z

zfIzfI

zfIzfI

zfIzfIz

z

zfIzfI kk

kk

kkkk

221 11

11

11

zzqk

zq '

221

then

,2

211

1 zqz

zfIzfIzq kk

and 1q and 2q are, respectively, the best subordinant and

dominant.

Theorem 3.7 If 0>. C and 0,( ) 1 2 , 1f z B ,

(0 < 1) , then 1,21)( , Bzf for Rz < ,

where

11

1=

2

kkR

(16)

The bound R is the best possible.

Proof. Set

1).<(0,,)()(1=

2

11

Uzzhz

zfIzfI kk

(17)

Then, clearly the function )(zh is analytic in U with

1=(0)h . Proceeding as an Theorem 3.1, we have

z

zfIzfI kk

21

1

1 11

z

zfIzfI

zfIzfI

zfIzfIz kk

kk

kk

2

11

11

.1

= zzhk

zh '

(18)

Using the following well-known estimate, see [11]

1<=

1

22

rzr

r

zh

zh'

R

in (18), we obtain that

z

zfIzfI kk

21

1

1 11R

z

zfIzfI

zfIzfI

zfIzfIz kk

kk

kk

2

11

11

.11

21

2

rk

rzh

R

(19)






Right hand side of (19) is positive, provided that Rr < ,

where R is given by (16).

In order to show that the bound R is best possible, we

consider the function A)(zf defined by

.1<0,

1

11=

2

11

Uzz

z

z

zfIzfI kk

We note that

z

zfIzfI kk

21

1

1 11

z

zfIzfI

zfIzfI

zfIzfIz kk

kk

kk

2

11

11

0,=

11

2

1

1=

2zk

z

z

z

for Rz = , we conclude that the bound is the best possible

and this proves the theorem.

Theorem 3.8 If 210 and

1<1 1221 AABB . Then

.,, 11

,1

22

,2 BABA

BB

(20)

Proof. Suppose that 22

,2 , BAf

B . We know that

z

zfIzfI

zfIzfI

zfIzfIz

z

zfIzfI kk

kk

kkkk

221 11

11

211

2

.1

1

2

2

zB

zA

Since 1<1 1221 AABB , we easily find that

z

zfIzfI

zfIzfI

zfIzfIz

z

zfIzfI kk

kk

kkkk

221 11

11

211

2

,1

1

1

1

1

1

2

2

zB

zA

zB

zA

(21)

this is 11

,1 , BAf

B . Thus the assertion (20) holds

true for 21 =0 . If 12 > , by Theorem 3.1 and (21),

we know that , 22

0, , BAf B that is,

.

1

1

2 1

111

zB

zA

z

zfIzfI kk

(22)

At the same time, we have

z

zfIzfI

zfIzfI

zfIzfIz

z

zfIzfI kk

kk

kkkk

221 11

11

111

1

zfIzfI

zfIzfIz

z

zfIzfI

kk

kkkk

11

211

2

1

21=

1 1 1 11

2

1 .2 2

k k k kI f z I f z I f z I f z

z z

(23)

Moreover,

1,<02

1

and the function zB

zA

1

1

1

1

, U zAB 1,<1 11 is

analytic and convex in U . Combining (21 - 23) and Lemma

2.4, we find that

z

zfIzfI

zfIzfI

zfIzfIz

z

zfIzfI kk

kk

kkkk

221 11

11

111

1

.1

1

1

1

zB

zA

this is 11

,1 , BAf

B , which implies that the assertion

(12) of Theorem 3.8 holds and this completes the proof.

Theorem 3.9 Let BAzf ,)( ,B with 0> and

1<1 11 AB . Then

duuBu

Aukk

11

1

0 1

11

duuBu

Auk

z

zfIzfIk

kk1

11

0

11

1

11<

2<

R

(24)

The extremal function of (24) is defined by

.1

112=

1

11

1

0,,,

duuBzu

AzukzzF

k

BA

(25)

Proof. Let BAzf ,)( ,B with 0> . From

Theorem 1, we know that (1) holds, which implies that

duuBzu

Azuk

z

zfIzfIk

z

kk1

11

0

11

1

11sup<

2

RR

U

duu

Bzu

Azukk

z

11

1

0 1

1sup

1

R

U

,1

11<

11

1

0duu

Bu

Aukk

(26)

and

duuBzu

Azuk

z

zfIzfIn

z

kk1

11

0

11

1

11inf>

2

RR

U

duu

Bzu

Azukk

z

11

1

0 1

1inf

1

R

U

.1

11>

11

1

0duu

Bu

Aukk

(27)

Combining (26) and (27), we obtain (20). Noting that the

function zF BA,,, defined by (25) belongs to the class






BA,,B , we get that inequality (24) is sharp. This

completes the proof.

In view of Theorem 9, we have the following distortion

theorems for the class BA,,B .

Corollary 3.10 Let BAzf ,)( ,B with 0> and

1<1 11 AB . Then for 1<= rz , we have

1

11

1

0 1

112

duuBur

Aurkr

k

.1

112<<

1

11

1

011

duuBur

AurkrzfIzfI

k

kk

(28)

The extremal function of (28) is defined by (25).

By noting that

0.;,2

12

1

2

1

vvvvv RCRR

From Theorem 9, we can easily derive the following result.

Corollary 3.11 Let BAzf ,)( ,B with 0> and

1<1 11 AB . Then

2

1

11

1

0 1

11

duuBu

Aukk

.1

11<

2<

2

1

11

1

0

211

duuBu

Auk

z

zfIzfIk

kk

R

Theorem 3.12 Let BAzf ,)( ,B with 0> and

1<1 11 AB . Then

.12

21

k

BAa

(29)

The inequality (29) is sharp, with the extremal function

defined by (25).

Proof. Combining (1) and (5), we have

z

zfIzfI

zfIzfI

zfIzfIz

z

zfIzfI kk

kk

kkkk

221 11

11

11

Bz

Azzak

k

1

11

1211= 1

= zBA1

(30)

An application of Lemma 2.4 to (30) yields

.1

121 1 BAak

k

(31)

Thus, from (31), we easily arrive at (29) asserted by

Theorem 3.12.

Theorem 3.13 Let ,0)( , Azf B with 0>R and

0>A and

1>1

1

kA

k

R . Then

1

11

11

1

<111

kAk

kk

A

zfIzfI

zfIzfIz

kk

kk

R

R

Proof. let zh be defined by (9). It follows from (10) that

,1=1

zAwzzhk

zh '

(32)

where

,=1=

nn

n

zwzw

is analytic in U with Uzzw 1,<)( . From (32), we can

get

dttzwtk

Azh

k1

11

0

11=

.1

111=

1=

n

n

n

zwk

n

kA

(33)

It follows from (33) that

.

1

111=

1=

n

n

n

'zw

kn

nkAzzh

n

n

n

zwk

n

kA

1

111=

1=

1

1 1

0

1 1.

kk k

A w z t w tz dt

(34)

We now find from (33) and (34) that

.11

=1

11

0

dttzwtk

zwk

Azzh

k'

(35)

Combining (33) and (35), we can get

11

1

11

1

<

kAk

kk

A

zh

zzh'

R

R

(36)

Thus, from (10) and (36), we easily arrive at the assertion of

Theorem 3.13.

Remark 3.1 If 1= , we obtain the results of [2], Theorems

3.1, 4.1 and 4.4.






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