On Starlike Functions Using the Generalized Salagean Differential Operator

(1)

2016, Volume 3, e2895 ISSN Online: 2333-9721 ISSN Print: 2333-9705

On Starlike Functions Using the Generalized

Salagean Differential Operator

Saliu Afis

1

_{, Mashood Sidiq}

2

1_{Department of Mathematics, Gombe State University, Gombe,}_Nigeria 2_{Department of Mathematics, University of Ilorin, Ilorin, Nigeria}

Abstract

In this paper we investigate the new subclass of starlike functions in the unit disk

{

: 1

}

U= z∈ z < via the generalized salagean differential operator. Basic

proper-ties of this new subclass are also discussed.

Subject Areas

Mathematical Analysis

Keywords

Salagean Differential Operator, Starlike Functions, Unit Disk, Univalent Functions, Analytic Functions and Subordination

1. Introduction

Let  denote the class of functions:

( )

2

f z = +z a z +

(1)

which are analytic in the unit disk U=

{

z∈: z <1

}

. Denote by

( )

*

: zf z 0,

S f Re z U

f z

′

 

 

= ∈ > ∈ 

 

   the class of normalized univalent functions in U.

Let

( )

2 2

g z = +z b z + ∈ . We say that f z

( )

is subordinate to

g z

( )

(written

as f g) if there is a function w analytic in U, with w

( )

0 =0,w z

( )

<1, for all z∈U.

If g is univalent, then f g if and only if f

( )

0 =g

( )

0 and f U

( )

⊆g U

( )

[1].

Definition 1 ([2]). Let f∈,λ∈

(

0,1

]

and n∈. The operator D_λn is defined

by How to cite this paper: Afis, S. and Sidiq, M. (2016) On Starlike Functions Using the Generalized Salagean Differential Operator. Open Access Library Journal, 3: e2895.

http://dx.doi.org/10.4236/oalib.1102895

Received: August 19, 2016 Accepted: September 11, 2016 Published: September 14, 2016

This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).

http://creativecommons.org/licenses/by/4.0/

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( )

(

) ( )

( )

( ) (

)

( )

(

( )

)

(

( )

)

0

1

,

1 ,

1 , .

n n n n

D f z f z

D f z zf z D f z

D f z D f z z D f z D D f z z U

λ

λ λ

λ λ λ λ λ

λ λ

+

=

′ = − + =

′

= − + = ∈



(2)

Remark 1. If f∈ and f z

( )

z k ₂a zk k

∞ =

= +

∑

, then

( )

2 1

(

1

)

,

n

n k

k k

D f z_λ = +z

∑

∞₌ _ + k−

λ

_ a z z∈U.

Remark 2. For λ=1 in (2), we obtain the Salagean differential operator.

From (2), the following relations holds:

( )

(

n 1

)

(

n

( )

)

(

n

( )

)

D_λ+ f z ′= D f z_λ ′+z

λ

D f z_λ ′′

(3)

and from which, we get

( )

(

)

( )

(

)

( )

1

n n

D f z D f z

z

D f z D f z

λ λ

+ ′

= − + ₍₄₎

Definition 2 ([3]). Let f∈, and n∈0. Then

( )

(

( )

)

( )

(

)

1 1

0

2

1

d

1

,

1 1

z

n n n

k k n k

I f z I I f z z t I f t t

z a z

k

λ λ

λ λ λ _λ λ

λ

− −

∞

=

= =

= +

+ −

 

 

∫

∑

with 0

( )

I f zλ = f z .

This operator is a particular case of the operator defined in [3] and it is easy to see that for any f∈, I_λn

(

D f z_λn

( )

)

=D_λn

(

I f z_λn

( )

)

= f z

( )

.

Next, we define the new subclasses of *

S .

Definition 3. A function f ∈ belongs to the class

S

_λn if and only if

( )

(

)

(

]

1

1 , 0,1 .

n

n D f z Re

D f z

λ

λ λ

+

> − ∈

(5) Remark 3. 0 *

S

_λ

≡

S

.

Remark 4. n

f

∈

S

_λ if and only if n

( )

*

D f zλ ∈S .

Definition 4. Let u= +u1 u i2 , v= +v1 v i2 and Ψ, the set of functions

( )

u v

,

:

ψ

_{  }

× →

_satisfying_:

i)

ψ

( )

u v

,

is continuous in a domain Ω of  × , ii)

( )

1, 0

∈ Ω

and

Re

ψ

( )

1, 0

>

0

,

iii)

Re

ψ

(

u i v

2

,

1

)

≤

0

when

(

u i v

2

,

1

)

∈ Ω

and

(

)

2

1 2

1 1 2

v ≤ − +u for z∈U.

Several examples of members of the set Ψ_{have been mentioned in}_[4]_[5]_{and (}_[6]_,

p. 27).

2. Preliminary Lemmas

Let P denote the class of functions

( )

2 1 2

1

p z = +c z+c z + which are analytic in U

(3)

Lemma 1 ([5] [7]) Let ψ∈ Ψ_{with corresponding domain}Ω_{. If}

P

( )

Ψ

is de-fined as the set of functions

p z

( )

given as

( )

2

1 2

1

p z = +c z+c z + which are

regu-lar in U and satisfy: i)

(

p z

( )

,zp z′

( )

)

∈ Ω

ii) Re

ψ

(

p z

( )

,zp z′

( )

)

>0 when z∈U.Then

Re p z

( )

>

0

in U.

More general concepts were discussed in [4]-[6].

Lemma 2 ([8]). Let η and µ be complex constants and

h z

( )

a convex univalent function in U satisfying

h

( )

0 =

1

, and Re

(

η

h z

( )

+

µ

)

. Suppose p∈P satisfies the differential subordination:

( )

zp z

_{( )}

( )

, .

p z h z z U

p z

η µ ′

+ ∈

+  (6)

If the differential subordination:

( )

zq z

_{( )}

( )

,

( )

0 1.

q z h z q

q z

η µ ′

+ = =

+

(7)

has univalent solution

q z

( )

in U. Then

p z

( ) ( ) ( )



q z



h z

and

q z

( )

is the best

dominant in (6).

The formal solution of (6) is given as

( )

zF z

_{( )}

( )

q z

F z

′

=

₍₈₎

where

( )

1

( )

0 d

z

F z t H t t

z

η µ µ

µ

η µ+ −

=

_∫

and

( )

₀

( )

1

exp zh t d

H z z t

t

−

 

= _ _



∫



see [9][10].

Lemma 3 ([9]). Let η≠0 and µ be complex constants and

h z

( )

regular in U with

h′

( )

0 ≠

0

, then the solution

q z

( )

of (7) given by (8) is univalent in U if (i) Re

( )

{

G z =

η

h z +

µ

}

>0,(ii)

( )

* G z

Q z z S

G z

′

= ∈ (iii)

( )

Q z

( )

_{( )}

*

R z S

G z

= ∈ .

3 Main Results

Theorem 1. Let

λ ∈

(

0,1

]

and

h z

( )

0

1 h

=

, and Re 1

λ

h z

( )

λ

−

 ₊ 

 

 , z∈U . Let f ∈. If

( )

2

1 n

n

D f z

h z D f z

λ

+

+  , then

( )

1

n

n D f z

h z D f z

λ

+

 .

(4)

( )

( ) (

)

( )

(

)

( )

1 2

1 1 1 .

n n

D f z

z

D f z D f z

λ λ λ λ

λ

+ + + + ′ = − +

If we suppose 12

_{( )}

( )

n

D f z

h z D f z

λ

+

+  , we need to show that

( )

1

n

n D f z

h z D f z

λ

+

 . Using the above equation and (4) and Remark 4, it suffices to show that if

( )

(

)

( )

1 1 n n z D f z

h z D f z

λ

+

+ ′

 , then

(

_{( )}

( )

)

( )

n

n z D f z

h z D f z

λ λ ′  . Now, let

( )

(

)

( )

. n n z D f z p

D f z λ λ ′ = Then

( )

(

n

)

(

n

( )

)

=

( )

n

( )

(

n

( )

)

.

z D f z_λ ′′+ D f z_λ ′ p z D f z′ _λ + D f z_λ ′

By (2) and (3) we have

( )

(

)

( )

( ) (

(

)

) ( )

( )

1 2 1 1 1 . 1 n n

z D f z zp z p z p z

p z D f z

zp z p z

p z

λ

λ λ λ

λ λ λ λ + + ′ _′ _{+ −} ₊ = − + ′ = + − ₊ (9)

Applying Lemma 2 with η=1 and µ 1 λ

λ −

= _{, the proof is complete.}

Theorem 2. Let

λ ∈

(

0,1 2

]

and

h z

( )

0

1 h

=

, and Re 1

λ

h z

( )

0,z U

λ

−

 ₊ _> _∈

 

  . Let f ∈. If

n

f

∈

S

_λ, then

( )

1

n

n D f z

q z D f z

λ λ +  where

( )

(

)

(

)

(

)

(

)

2 0 0 1 1 1 1 1 1 k k k k k z k q z k z k λ λ ∞ = ∞ = + + + = + + +

∑

is the best dominant.

Proof. Let n1

f

∈

S

_λ+ , then by Remark 4,

( )

(

)

( )

1 1 1 . 1 n n

z D f z _z

z

D f z

λ λ + + ′ + − 

(5)

( )

1 ,

1 1

zp z z

p z

z p z

λ λ

′ +

+ ₋

−

+ 

where

( )

(

_{( )}

( )

)

.

n

n z D f z p z

D f z λ

λ

′ =

To show that

(

_{( )}

( )

)

( )

1 n

n z D f z

q z D f z

λ

+ ′

 , by Remark 4, it suffices to show that

( ) ( )

.

p z



q z

Now, considering the differential equation

( )

1

1 1

zq z z

q z

z q z

λ λ

′ +

+ ₋ =

− +

whose solution is obtained from (8). If we proof that

q z

( )

is univalent in U, our re-

sult follows trivially from Lemma 2. Setting µ 1 λ,η

λ

−

= _and

( )

1

1 z h z

z

+ =

− in Lemma

3, we have

i) ReG z

( )

=Re

(

µ

+h z

( )

)

>0,

ii)

( )

_{( )}

( )

2

₍

₎₍

₎

1 1

zG z z

Q z

G z λ βz z

′ = =

+ −

where β =2λ−1, so that by logarithmic differentiation, we have

( )

11 1 1 1.

zQ z

Q z z βz

′

= + − − +

Therefore,

_{( )}

( )

(

)(

2

)

1 2 1

0 2

zQ z Re

Q z

λ λ λ ′ − +

> > ,

iii)

( )

_{( )}

(

)

2

2 2

1

Q z z

R z

G z z

λ β

= =

+

so that

( )

11 0.

zR z Re

R z β µβ

′ _> − _{= >} +

Hence,

q z

( )

is univalent in U since it satisfies all the conditions of Lemma 3. This

completes the proof.

Theorem 3. n 1 n

S

_λ+

⊂

S

_λ. Proof. Let

_f

_S

n1

λ+

∈

_{. By Remark 4}

( )

(

)

( )

1

1 0.

n

z D f z

Re

D f z

λ

λ +

+

(6)

From (9), let

(

( )

)

( )

, :

1 zp z p z zp z p z

p z

ψ _λ

λ

′

′ = + ₋

+ with

( )

(

)

( )

n

n z D f z p z

D f z

λ

′

= _for

1

λ

  −  Ω =_ − −_ __×

 

  . Conditions (i) and (ii) of Lemma 1 are clearly satisfied by ψ .

Next,

(

)

1

2 1 2

2

, .

1 v u i v u i

u i

ψ _λ

λ

= + −

+ Then

(

)

1

2 1 2

2 2 1

, 0

1 v Re u i v

u

λ λ ψ

λ λ

−

= ≤

−  _{ +}

 

 

if

(

2

)

1 2

1 1 2

v ≤ − +u . Hence,

Re p z

( )

> 0.

Using Remark 4,

( )

1

1 n

n D f z Re

D f z

λ

+

> − which complete the proof.

Corollary 1. All functions in n

S

_λ are starlike univalent in U. Proof. The proof follows directly from Theorem 3 and Remark 4.

Corollary 2. The class 1

n

S

“clone” the analytic representation of convex functions. Proof. The proof is obvious from the above corollary and Definition 4.

The functions

( )

2 3

2! 3!

z z

f z = +z + + and

( )

2 3

2 2! 3 3!

z z

g z = −z + +

× ×  are

exam-ples of functions in 1

n

S

.

Theorem 4. The class n

S

_λ is preserve under the Bernardi integral transformation:

( )

1

( )

0 1

d , 1.

z c c c

F z t f t t c

z − +

=

_∫

> −

₍₁₀₎ Proof. let n

f

∈

S

_λ, then by Remark 4 n

( )

*

D f zλ ∈S . From (10) we get

(

c

+

1 ) ( )

f z

=

cF z

( )

+

zF z

′

( )

.

(11)

Applying

_D

n

λ on (10) and noting from Remark 1 that Dλn

(

zF z′

( )

)

=z D F z

(

λn

( )

)

′,

we have

( )

(

)

( )

(

)

(

( )

)

(

( )

)

( )

(

( )

)

2

1

= .

n n n

n

n n

z D f z c z D F z z D F z

D f z _{cD F z} _{z D F z}

λ λ λ

λ

λ λ

′ ₊ ′₊ ′′ ′ +

Let

( )

(

_{( )}

( )

)

n

n z D F z p z

D F z

λ

′

= and noting that

(

_{( )}

( )

)

( )

2

2 n

n z D F z

zp z p z p z

D f z

λ

′′ ′

= + − ,

we get

( )

(

)

( )

n

z D f z _{zp z}

p z

c p z

D f z λ

λ

′ _′

= + +

Let

(

p z

( )

,zp z

( )

)

:p z

( )

zp z

( )

_{( )}

c p z

ψ ′ = + ′

+ for Ω =− −

{ }

c ×. Then ψ satisfies

all the conditions of Lemma 1 and so

(

_{( )}

( )

)

0 n

n z D f z Re

D f z

λ

′

> ⇒

(

( )

)

( )

0. n

n z D F z Re

D F z

λ

′

(7)

Remark 4 n

F

∈

S

_λ.

Theorem 5. Let n

f

∈

S

_λ. Then f has integral representation:

( )

₀

( )

1

exp z d

n p t

f z I z t

t

λ

  − 

 

=   

  



∫



for some p∈P. Proof. Let n

f

∈

S

_λ. Then by Remark 4, n

( )

*

D F z_λ ∈S and so for some p∈P

( )

(

)

( )

.

n

n z D f z

p z D f z

λ

′ =

But d log

( )

1 d

n

D f z p z

z z z

λ

  −

=

 

  , so that

( )

₀

( )

1

exp z d .

n p t

D f z z t

t

λ

−

 

= _ _



∫



Applying the operator in Definition 2, we have the result.

With

( )

1

1 z p z

z

+ =

− , we have the extremal function for this new subclass of *

S

which is

( )

(

)

2

.

1 1

n k

k

f z z z

k λ

λ

∞

=

= +

+ −

 

 

∑

Theorem 6. Let n

f

∈

S

_λ. Then

(

)

(

1 1

)

, 2.

k n

k

a k

k λ

≤ ≥

+ −

The function n

( )

f_λ z given by (13) shows that the result is sharp.

Proof. Let

_f

_S

n λ

∈

_{, then by Remark 4,} n

( )

*

D f z_λ ∈S . Since it is well known that for

any *

f

∈

S

,

a

_k

≤

k k

,

≥

2

, then from Remark 1 we get the result.

Theorem 7. Let

_f

_S

n λ

∈

_{. Then}

(

1 n

)

( )

(

1 n

)

r −_λ < f z <r +_λ

and

( )

1−rRλn≤ f′ z ≤ +1 rRλn,

where

(

)

(

)

2

2 2

and .

1 1 1 1

n n

k k

R

k k

λ λ

∞ ∞

= =

+ − + −

   

   

∑



Proof. Let f ∈. Then by Theorem 6, we have

( )

(

)

2 2

1

1 1

k

k n

k k

k

f z z a z r

k

λ

∞ ∞

= =

 

 

≤ + = +

 __ ₊ ₋ __ 

 

(8)

and

( )

(

)

2 2

1

1 1

k

k n

k k

k

f z z a z r

k

λ

∞ ∞

= =

 

 

≥ − = −

 __ ₊ ₋ __ 

 

∑

for

z

= <

r

1

.

Also, upon differentiating f ∈, we get

( )

(

)

2 1

2 2

1 1

k

k n

k k

k

f z k a z r

k λ

∞ ∞

−

= =

′ ≤ + ≤ +

+ −

 

 

∑

and

( )

(

)

2 1

2 2

1 1

k

k n

k k

k

f z k a z r

k λ

∞ ∞

−

= =

′ ≥ − ≥ −

+ −

 

 

∑

for

z

= <

r

1

. This complete the proof.

Acknowledgements

The authors appreciates the immense role of Dr. K.O. Babalola (a senior lecturer at University of Ilorin, Ilorin, Nigeria) in their academic development.

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[9] Miller, S.S. and Mocanu, P.T. (1983) Univalent Solution of Briot-Bouquet Differential Equ-ations. Lecture Notes in Mathematics, 1013, Springer Berlin/Heidelberg, 292-310. [10] Srivastava, H.M. and Lashin A.Y. (1936) Some Applications of the Briot-Bouquet