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On sufficient conditions for Carathéodory functions with applications

Abstract

In the present paper, we derive some interesting relations associated with the Carathéodory functions which yield sufficient conditions for the Carathéodory functions in the open unit disk U={z:|z|<1}. Some interesting applications of the main results are also obtained.

MSC:30C45, 30C80.

1 Introduction

Let P denote the class of functions of the form

p(z)= n = 0 p n z n ,

which are analytic in the unit disc U={z:|z|<1}. The function p(z) is called a Carathéodory function if it satisfies the condition

Re ( p ( z ) ) >0.

Moreover, let A denote the class of functions of the form

f(z)=z+ n = 2 a n z n ,
(1.1)

which are analytic in the unit disc U.

A function f(z)A is in K, the class of convex functions, if it satisfies

Re ( 1 + z f ( z ) f ( z ) ) >0(zU).
(1.2)

Also, a function f(z)A is in S λ (|λ|< π 2 ), the class of λ-spirallike functions, if it satisfies

Re ( e i λ z f ( z ) f ( z ) ) >0(zU).
(1.3)

Moreover, we denote by S = S 0 the class of starlike functions in U.

Definition 1.1 Let f(z) and F(z) be analytic functions. The function f(z) is said to be subordinate to F(z), written f(z)F(z), if there exists a function w(z) analytic in U, with w(0)=0 and |w(z)|1, and such that f(z)=F(w(z)). If F(z) is univalent, then f(z)F(z) if and only if f(0)=F(0) and f( U ) F(U).

Definition 1.2 Let D be the set of analytic functions q(z) and injective on U ¯ E(q), where

E(q)= { ζ U : lim z ζ q ( z ) = }

and q (ζ)0 for ζUE(q). Further, let D a ={q(z)D:q(0)=a}.

Many authors have obtained several relations of Carathéodory functions, e.g., see ([113]).

In the present paper, we derive some relations associated with the Carathéodory functions which yield the sufficient conditions for Carathéodory functions in U. Some applications of the main results are also obtained.

2 Main results

To prove our results, we need the following lemma due to Miller and Mocanu [[14], p.24]

Lemma 2.1 Let q(z) D a and let

p(z)=b+ b n z n +
(2.1)

be analytic in U with p(z)b. If p(z)q(z), then there exist points z 0 U and ζ 0 UE(q) and on mn1 for which

  1. (i)

    p( z 0 )=q( ζ 0 ),

  2. (ii)

    z 0 p ( z 0 )=m ζ 0 q ( ζ 0 ).

Theorem 2.1 Let

P:UC

with

Re ( a ¯ P ( z ) ) >0(aC).

If p(z) is an analytic function in U with p(0)=1 and

Re ( p ( z ) + P ( z ) z p ( z ) ) > E 2 | a | 2 Re ( a ¯ P ( z ) ) ,
(2.2)

then

Re ( a p ( z ) ) >0,

where

E= ( Re ( a ) ) ( Re ( a ¯ P ( z ) ) ) 2 +2Re ( a ¯ P ( z ) ) ( Im ( a ) ) 2 + ( Re ( a ) ) ( Im ( a ) ) 2
(2.3)

with Re(a)>0.

Proof Let us define both q(z) and h(z) as follows:

q(z)=ap(z)

and

h(z)= a + a ¯ z 1 z ( Re ( a ) > 0 ) ,

where p(z) is defined by (2.1) since q(z) and h(z) are analytic functions in U with q(0)=h(0)=aC with

h(U)= { w : Re ( w ) > 0 } .

Now, we suppose that q(z)h(z). Therefore, by using Lemma 2.1, there exist points

z 0 Uand ζ 0 U{1}

such that q( z 0 )=h( ζ 0 ) and z 0 q ( z 0 )=m ζ 0 h ( ζ 0 ), mn1.

We note that

ζ 0 = h 1 ( q ( z 0 ) ) = q ( z 0 ) a q ( z 0 ) + a ¯
(2.4)

and

ς 0 h ( ς 0 )= | q ( z 0 ) a | 2 2 Re ( a q ( z 0 ) ) .
(2.5)

We have h( ζ 0 )=ρi (ρR); therefore,

(2.6)

where

and

C= Re ( a ¯ p ( z 0 ) ) 2 Re ( a ) .

We can see that the function g(ρ) in (2.6) takes the maximum value at ρ 1 given by

ρ 1 =Im(a) ( 1 + Re ( a ) Re ( a ¯ p ( z 0 ) ) ) .

Hence, we have

Re ( p ( z 0 ) + P ( z 0 ) z p ( z 0 ) ) g ( ρ 1 ) = E 2 | a | 2 Re ( a ¯ P ( z ) ) ,

where E is defined by (2.3). This is a contradiction to (2.2). Then we obtain Re(ap(z))>0. □

Theorem 2.2 Let p(z) be a nonzero analytic function in U and p(0)=1. If

γ 1 <Im ( p ( z ) + z p ( z ) p ( z ) ) < γ 2 ,
(2.7)

where

γ 1 = | a | 2 + 2 ( Re ( a ) ) 2 Im ( a ) Re a

and

γ 2 = | a | 2 + 2 ( Re ( a ) ) 2 + Im ( a ) Re ( a ) ,

then

Re ( a p ( z ) ) >0,

where Re(a)>0.

Proof Let us define both q(z) and h(z) as follows:

q(z)=ap(z)

and

h(z)= a + a ¯ z 1 z ( Re ( a ) > 0 ) ,

where p(z) is defined by (2.1) since q(z) and h(z) are analytic functions in U with q(0)=h(0)=aC with

h(U)= { w : Re ( w ) > 0 } .

Now, we suppose that q(z)h(z). Therefore, by using Lemma 2.1, there exist points

z 0 Uand ζ 0 U{1}

such that q( z 0 )=h( ζ 0 ) and z 0 q ( z 0 )=m ζ 0 h ( ζ 0 ), mn1.

We note that

ζ 0 h ( ζ 0 )= | q ( z 0 ) a | 2 2 Re ( a q ( z 0 ) ) .
(2.8)

We have h( ζ 0 )=ρi (ρR); therefore,

Im ( p ( z 0 ) + z 0 p ( z 0 ) p ( z 0 ) ) = Im ( q ( z 0 ) + z 0 q ( z 0 ) q ( z 0 ) ) = Im ( h ( ζ 0 ) a + m ζ 0 h ( ζ 0 ) h ( ζ 0 ) ) = Im ( ρ i a m | ρ i a | 2 2 Re ( a ) ρ i ) = ρ | a | 2 Re ( a ) + m | ρ i a | 2 2 ρ Re ( a ) .

For the case ρ>0, we obtain

Im ( p ( z 0 ) + z 0 p ( z 0 ) p ( z 0 ) ) ρ | a | 2 Re ( a ) + | ρ i a | 2 2 ρ Re ( a ) = 1 2 ρ Re ( a ) [ ( 1 + 2 ( Re ( a ) | a | ) 2 ) ρ 2 + 2 Im ( a ) ρ + | a | ] = g ( ρ ) .
(2.9)

We can see that the function g(ρ) in (2.9) takes the minimum value at ρ 1 given by

ρ 1 = | a | 2 | a | 2 + 2 ( Re ( a ) ) 2 .

Hence, we have

Im ( p ( z 0 ) + z 0 p ( z 0 ) p ( z 0 ) ) g ( ρ 1 ) = γ 2 .

This is a contradiction to (2.7). Then we obtain Re(ap(z))>0.

For the case ρ<0, we obtain

Im ( p ( z 0 ) + z 0 p ( z 0 ) p ( z 0 ) ) ρ | a | 2 Re ( a ) + | ρ i a | 2 2 ρ Re ( a ) = 1 2 ρ Re ( a ) [ ( 1 + 2 ( Re ( a ) | a | ) 2 ) ρ 2 + 2 Im ( a ) ρ + | a | 2 ] = g ( ρ ) .
(2.10)

We can see that the function g(ρ) in (2.10) takes the maximum value at ρ 2 given by

ρ 2 = | a | 2 | a | 2 + 2 ( Re ( a ) ) 2 .

Hence, we have

Im ( p ( z 0 ) + z 0 p ( z 0 ) p ( z 0 ) ) g ( ρ 2 ) = γ 1 .

This is a contradiction to (2.7). Then we obtain Re(ap(z))>0. □

Theorem 2.3 Let p(z) be a nonzero analytic function in U with p(0)=1. If

|p(z)+ z p ( z ) p ( z ) 1|< 3 Re ( a ) 2 | a | ,

then

Re ( a p ( z ) ) >0,

where Re(a)>0.

Proof Let us define both q(z) and h(z) as follows:

q(z)=ap(z)

and

h(z)= a + a ¯ z 1 z ( Re ( a ) > 0 ) ,

where p(z) is defined by (2.1) since q(z) and h(z) are analytic functions in U with q(0)=h(0)=aC with

h(U)={w:Rew>0}.

Now, we suppose that q(z)h(z). Therefore, by using Lemma 2.1, there exist points

z 0 Uand ζ 0 U{1}

such that q( z 0 )=h( ζ 0 ) and z 0 q ( z 0 )=m ζ 0 h ( ζ 0 ), mn1.

We note that

ζ 0 h ( ζ 0 )= | q ( z 0 ) a | 2 2 Re ( a q ( z 0 ) ) .
(2.11)

We have h( ζ 0 )=ρi (ρR).

Therefore,

| p ( z 0 ) + z p ( z 0 ) p ( z 0 ) 1 | | p ( z 0 ) | = | ρ i a m a | a i ρ | 2 2 Re ( a ) 1 | 1 | a | | m | a i ρ | 2 2 Re ( a ) + Re ( a ) | 1 | a | ( | a i ρ | 2 2 Re ( a ) + Re ( a ) ) 1 2 | a | Re ( a ) ( 3 ( Re ( a ) ) 2 + ( Im ( a ) ρ ) 2 ) 3 Re ( a ) 2 | a | .

This is a contradiction to (2.7). Then we obtain Re( a p ( z ) )>0. □

3 Applications and examples

Putting P(z)=β (β>0; real) in Theorem 2.1, we have the following corollary.

Corollary 3.1 If p(z) is an analytic function in U with p(0)=1 and

Re ( p ( z ) + β z p ( z ) ) > E 2 β | a | 2 Re ( a ) ,

then

Re ( a p ( z ) ) >0,

where

E= ( Re ( a ) ) [ β 2 ( Re ( a ) ) 2 + ( 1 + 2 β ) ( Im ( a ) ) 2 ]

with Re(a)>0.

Putting β=1 in Corollary 3.1, we obtain the following corollary.

Corollary 3.2 If p(z) is an analytic function in U with p(0)=1 and

Re ( p ( z ) + z p ( z ) ) > 3 2 2 ( Re ( a ) | a | ) 2 ,

then

Re ( a p ( z ) ) >0,

where Re(a)>0.

Putting p(z)= f ( z ) g ( z ) and P(z)= g ( z ) z g ( z ) in Theorem 2.1, we have the following corollary.

Corollary 3.3 Let f(z)A, g(z) S and

Re ( f ( z ) g ( z ) ) > 3 2 2 ( Re ( a ) | a | ) 2 .

Then

Re ( a f ( z ) g ( z ) ) >0,

where Re(a)>0.

Example 3.1 Let f(z)A satisfy the following relation:

Re ( f ( z ) ) > 3 2 2 ( Re ( a ) | a | ) 2 .

Then

Re ( a f ( z ) z ) >0,

where Re(a)>0.

Example 3.2 Let f(z)A satisfy the following relation:

Re ( ( 2 + z f ( z ) f ( z ) z f ( z ) f ( z ) ) z f ( z ) f ( z ) ) > 3 2 2 ( Re ( a ) | a | ) 2 .

Then

Re ( a z f ( z ) f ( z ) ) >0,

where Re(a)>0.

Remark 3.1

  1. (i)

    Putting a= e i λ (|λ|< π 2 ) in Theorem 2.1, we have Theorem 1 due to Kim and Cho [3].

  2. (ii)

    Putting a= e i λ (|λ|< π 2 ), P(z)=β (β>0; real) in Theorem 2.1, we have Corollary 1 due to Kim and Cho [3].

  3. (iii)

    Putting a=0 and P(z)=1 in Theorem 2.1, we have the result due to Nunokawa et al. [15].

  4. (iv)

    Putting a= e i λ (|λ|< π 2 ), P(z)=1 in Theorem 2.1, we have Corollary 2 due to Kim and Cho [3].

Putting p(z)= z f ( z ) f ( z ) in Theorem 2.2, we have the following corollary.

Corollary 3.4 Let f(z)A. If

γ 1 <Im ( 1 + z f ( z ) f ( z ) ) < γ 2 ,

where

γ 1 = | a | 2 + 2 ( Re ( a ) ) 2 Im ( a ) Re ( a )

and

γ 2 = | a | 2 + 2 ( Re ( a ) ) 2 + Im ( a ) Re ( a ) ,

then

Re ( a z f ( z ) f ( z ) ) >0,

where Re(a)>0.

Putting p(z)= z f ( z ) f ( z ) in Theorem 2.3, we have the following corollary.

Corollary 3.5 Let p(z) be a nonzero analytic function in U with p(0)=1. If

| z f ( z ) f ( z ) |< 3 Re ( a ) 2 | a | ,

then

Re ( 1 a z f ( z ) f ( z ) ) >0,

where Re(a)>0.

Remark 3.2 Putting a= e i λ (|λ|< π 2 ) in Corollary 3.5, we have the result due to Kim and Cho [3].

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Acknowledgements

Dedicated to Professor Hari M Srivastava.

The authors would like to express their gratitude to the referees for the valuable advices to improve this paper.

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Attiya, A.A., Nasr, M.A. On sufficient conditions for Carathéodory functions with applications. J Inequal Appl 2013, 191 (2013). https://doi.org/10.1186/1029-242X-2013-191

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Keywords

  • analytic functions
  • starlike functions
  • convex functions
  • spirallike functions
  • Carathéodory functions