Open Access

On some differential inequalities in the unit disk with applications

Journal of Inequalities and Applications20142014:32

https://doi.org/10.1186/1029-242X-2014-32

Received: 13 December 2013

Accepted: 6 January 2014

Published: 24 January 2014

Abstract

In this paper we obtain a number of interesting relations associated with some differential inequalities in the open unit disk, U = { z : | z | < 1 } . Some applications of the main results are also obtained.

MSC:30C45, 30C80.

Keywords

analytic functions starlike functions convex functions spiral-like functions Carathéodory functions

1 Introduction

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 = { z : | z | < 1 } . Also, we denote by K the class of functions f ( z ) A that are convex in .

A function f ( z ) in the class A is said to be in the class S ( α ) of starlike functions of order α ( 0 α < 1 ) if it satisfies
Re ( z f ( z ) f ( z ) ) > α ( z U )
(1.2)

for some α ( 0 α < 1 ). Also, we write S ( 0 ) = S , the class of starlike functions in .

A function f ( z ) A is in S λ ( | λ | < π 2 ), the class of λ-spiral-like functions, if it satisfies
Re ( e i λ z f ( z ) f ( z ) ) > 0 ( z U ) .
(1.3)

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 , 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 ) .

Let be the set of analytic functions q ( z ) injective on U ¯ E ( q ) , where
E ( q ) = { ζ U : lim z ζ q ( z ) = }

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

In this paper we obtain some interesting relations associated with some differential inequalities in . These relations extend and generalize the Carathéodory functions in which have been studied by many authors e.g., see [114].

2 Main results

To prove our results, we need the following lemma due to Miller and Mocanu [[15], p.24].

Lemma 2.1 Let q ( z ) D a and let
p ( z ) = b + b n z n +
be analytic in with p ( z ) b . If p ( z ) q ( z ) , then there exist points z 0 U and ζ 0 U E ( q ) and on m n 1 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 : U C
with
Re ( a ¯ P ( z ) ) > 0 ( a C ) .
If p is a function analytic in with p ( 0 ) = 1 and
Re ( p ( z ) + P ( z ) z p ( z ) ) > E 2 | a | 2 Re ( a ¯ P ( z ) ) ,
(2.1)
then
Re ( a p ( z ) ) > α ,
where
E = ( Re ( a ) α ) ( Re ( a ¯ P ( z ) ) ) 2 + 2 Re ( a ¯ P ( z ) ) [ ( Im ( a ) ) 2 + 2 α Re ( a ) ] + ( Re ( a ) α ) ( Im ( a ) ) 2 ,
(2.2)

with Re ( a ) > α .

Proof Let us define q ( z ) and h ( z ) as follows:
q ( z ) = a p ( z )
and
h ( z ) = a ( 2 α a ¯ ) z 1 z ( Re ( a ) > α ) .
The functions q and h are analytic in with q ( 0 ) = h ( 0 ) = a C with
h ( U ) = { w : Re ( w ) > α } .
Now, we suppose that q ( z ) h ( z ) . Therefore, by using Lemma 2.1, there exist points
z 0 U and ζ 0 U { 1 }

such that q ( z 0 ) = h ( ζ 0 ) and z 0 q ( z 0 ) = m ζ 0 h ( ζ 0 ) , m n 1 .

We note that
ζ 0 = h 1 ( q ( z 0 ) ) = q ( z 0 ) a q ( z 0 ) ( 2 α a ¯ )
(2.3)
and
ζ 0 h ( ζ 0 ) = | q ( z 0 ) a | 2 2 Re ( a q ( z 0 ) ) .
(2.4)
We have h ( ζ 0 ) = α + ρ i ( α , ρ R ), therefore
Re ( p ( z 0 ) + P ( z 0 ) z 0 p ( z 0 ) ) = Re ( 1 a h ( ζ 0 ) + 1 a P ( z 0 ) m ζ 0 h ( ζ 0 ) ) = Re ( α + ρ i a ) m | α + ρ i a | 2 2 Re ( a α ) Re ( P ( z 0 ) a ) Re ( α + ρ i a ) | α + ρ i a | 2 2 Re ( a α ) Re ( P ( z 0 ) a ) = A ρ 2 + B ρ + C = g ( ρ ) ,
(2.5)
where
A = Re ( a ¯ P ( z 0 ) ) 2 | a | 2 Re ( a α ) , B = Im ( a ) | a | 2 ( 1 + Re ( a ¯ P ( z 0 ) ) Re ( a ) α )
and
C = 1 | a | 2 ( α Re ( a ) α 2 + | a | 2 2 α Re ( a ) Re ( a ¯ P ( z 0 ) ) 2 ( Re ( a ) α ) ) .
We can see that the function g ( ρ ) in (2.5) takes the maximum value at ρ 1 given by
ρ 1 = Im ( a ) ( 1 + Re ( a ) α 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 (2.2). This is in contradiction to (2.1). Then we obtain Re ( a p ( z ) ) > α . □

Theorem 2.2 Let p ( z ) a nonzero analytic function in with p ( 0 ) = 1 . If
| p ( z ) + z p ( z ) p ( z ) 1 | < 3 Re ( a α ) 2 | a | | p ( z ) | ,
(2.6)
then
Re ( a p ( z ) ) > α ,

where Re ( a ) > α .

Proof Let us define both q ( z ) and h ( z ) as follows:
q ( z ) = a / p ( z )
and
h ( z ) = a ( 2 α a ¯ ) z 1 z ( Re ( a ) > α ) .
The functions q and h are analytic in with q ( 0 ) = h ( 0 ) = a C with
h ( U ) = { w : Re ( w ) > α } .
Now, we suppose that q ( z ) h ( z ) . Therefore, by using Lemma 2.1, there exist points
z 0 U and ζ 0 U { 1 }

such that q ( z 0 ) = h ( ζ 0 ) and z 0 q ( z 0 ) = m ζ 0 h ( ζ 0 ) , m n 1 .

We note that
ζ 0 h ( ζ 0 ) = | q ( z 0 ) a | 2 2 Re ( a q ( z 0 ) ) .
(2.7)
We have h ( ζ 0 ) = α + ρ i ( ρ R ); therefore,
| p ( z 0 ) + z p ( z 0 ) p ( z 0 ) 1 | | p ( z 0 ) | = | α + ρ i a m a | a α i ρ | 2 2 Re ( a α ) 1 | 1 | a | | m | a α i ρ | 2 2 Re ( a α ) + Re ( a α ) | 1 | a | ( | a α i ρ | 2 2 Re ( a α ) + Re ( a α ) ) 1 2 | a | Re ( a α ) ( 3 ( Re ( a α ) ) 2 + ( Im ( a ) ρ ) 2 ) 3 Re ( a α ) 2 | a | .

This is in contradiction to (2.6). Then we obtain Re ( a p ( z ) ) > α . □

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 is a function analytic in with p ( 0 ) = 1 and
Re ( p ( z ) + β z p ( z ) ) > E 2 β | a | 2 Re ( a ) ,
then
Re ( a p ( z ) ) > α ,
where
E = ( Re ( a ) α ) β 2 ( Re ( a ) ) 2 + 2 β Re ( a ) [ ( Im ( a ) ) 2 + 2 α Re ( a ) ] + ( Re ( a ) α ) ( Im ( a ) ) 2 ,

with Re ( a ) > α ( α 0 ).

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

Corollary 3.2 If p is a function analytic in with p ( 0 ) = 1 and
Re ( p ( z ) + z p ( z ) ) > 1 2 Re ( a ) ( 3 Re ( a ) α ) Re ( a ) | a | 2 ( 2 Re ( a ) 3 α ) ,
then
Re ( a p ( z ) ) > α ,

with Re ( a ) > α ( α 0 ).

Corollary 3.3 Let f ( z ) A , ( g ( z ) ) a S and
Re ( f ( z ) g ( z ) ) > E 2 | a | 2 Re ( a ¯ g ( z ) z g ( z ) ) ,
then
Re ( a f ( z ) g ( z ) ) > α ,

where Re ( a ) > α ( α 0 ) and E is defined by (2.2) with P ( z ) = g ( z ) z g ( z ) .

Proof Putting p ( z ) = f ( z ) g ( z ) and P ( z ) = g ( z ) z g ( z ) in Theorem 2.1, we have
Re ( p ( z ) + P ( z ) z p ( z ) ) = Re ( f ( z ) g ( z ) ) .

Since ( g ( z ) ) a S , which gives Re ( a z g ( z ) g ( z ) ) > 0 , therefore, Re ( a ¯ P ( z ) ) > 0 . This completes the proof of the corollary. □

Example 3.1 Let f ( z ) A and
Re ( f ( z ) ) > 1 2 Re ( a ) ( 3 Re ( a ) α ) Re ( a ) | a | 2 ( 2 Re ( a ) 3 α ) ,
then
Re ( a f ( z ) z ) > α ,

where Re ( a ) > α .

Example 3.2 Let f ( z ) A and
Re ( ( 2 + z f ( z ) f ( z ) z f ( z ) f ( z ) ) z f ( z ) f ( z ) ) > 1 2 Re ( a ) ( 3 Re ( a ) α ) Re ( a ) | a | 2 ( 2 Re ( a ) 3 α ) ,
then
Re ( a z f ( z ) f ( z ) ) > α ,

where Re ( a ) > α .

  1. (1)

    Putting a = e i λ ( | λ | < π 2 ) and α = 0 in Theorem 2.1, we have Theorem 1 due to Kim and Cho [4].

     
  2. (2)

    Putting a = e i λ ( | λ | < π 2 ), P ( z ) = β ( β > 0 ; real) and α = 0 in Theorem 2.1, we have Corollary 1 due to Kim and Cho [4].

     
  3. (3)

    Putting a = α = 0 and P ( z ) = 1 in Theorem 2.1, we have the result due to Nunokawa et al. [16].

     
  4. (4)

    Putting a = e i λ ( | λ | < π 2 ), P ( z ) = 1 and α = 0 in Theorem 2.1, we have Corollary 2 due to Kim and Cho [4].

     

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

Corollary 3.4 Let p ( z ) a nonzero analytic function in U with p ( 0 ) = 1 . If
| z f ( z ) f ( z ) | < 3 Re ( a α ) 2 | a | | z f ( z ) f ( z ) | ,
then
Re ( 1 a z f ( z ) f ( z ) ) > α ,

where Re ( a ) > α .

Remark
  1. (1)

    Putting a = 1 and α = 0 in Corollary 3.4, we have the result due to Attiya and Nasr [1].

     
  2. (2)

    Putting a = 1 and α = 0 in Corollary 3.4, we have the result due to Kim and Cho [4].

     

Declarations

Acknowledgement

The author would like to express his gratitude to the referee(s) for the valuable advices to improve this paper.

Authors’ Affiliations

(1)
Department of Mathematics, Faculty of Science, University of Mansoura

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© Attiya; licensee Springer. 2014

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