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On some differential inequalities in the unit disk with applications

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.

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 (0α<1) if it satisfies

Re ( z f ( z ) f ( z ) ) >α(zU)
(1.2)

for some α (0α<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(zU).
(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 ζUE(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 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 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)=ap(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)=aC 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 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 ) ( 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+ρi (α,ρR), 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(ap(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)=aC 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 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 ) ) .
(2.7)

We have h( ζ 0 )=α+ρi (ρR); 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 +2βRe(a) [ ( Im ( a ) ) 2 + 2 α Re ( a ) ] + ( Re ( a ) α ) ( Im ( a ) ) 2 ,

with Re(a)>α (α0).

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 (α0).

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 (α0) 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].

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Acknowledgement

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

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Correspondence to Adel A Attiya.

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Attiya, A.A. On some differential inequalities in the unit disk with applications. J Inequal Appl 2014, 32 (2014). https://doi.org/10.1186/1029-242X-2014-32

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Keywords

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