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On some differential inequalities in the unit disk with applications

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R E S E A R C H

Open Access

On some differential inequalities in the unit

disk with applications

Adel A Attiya

*

*Correspondence:

[email protected] Department of Mathematics, Faculty of Science, University of Mansoura, Mansoura, 35516, Egypt

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: Primary 30C45; 30C80

Keywords: analytic functions; starlike functions; convex functions; spiral-like functions; Carathéodory functions

1 Introduction

LetAdenote the class of functions of the form

f(z) =z+

n=

anzn (.)

which are analytic in the unit discU={z:|z|< }. Also, we denote byKthe class of func-tionsf(z)∈Athat are convex inU.

A functionf(z) in the classAis said to be in the classS∗(α) of starlike functions of order

α(≤α< ) if it satisfies

Re

zf(z)

f(z)

>α (z∈U) (.)

for someα(≤α< ). Also, we writeS() =S∗, the class of starlike functions inU. A functionf(z)∈Ais in(|λ|<π

), the class ofλ-spiral-like functions, if it satisfies

Re

eiλzf

(z)

f(z)

>  (z∈U). (.)

Definition . Letf(z) andF(z) be analytic functions. The functionf(z) is said to be sub-ordinate to F(z), writtenf(z)≺F(z), if there exists a functionw(z) analytic inU, with

w() =  and|w(z)| ≤, and such thatf(z) =F(w(z)). IfF(z) is univalent, thenf(z)≺F(z)

if and only if f() =F() andf(U)⊂F(U).

LetDbe the set of analytic functionsq(z) injective onU\E(q), where

E(q) =ζU:lim

zζq(z) =∞

andq(ζ)=  forζU\E(q). Further, letDa={q(z)∈D:q() =a}.

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In this paper we obtain some interesting relations associated with some differential in-equalities inU. These relations extend and generalize the Carathéodory functions inU which have been studied by many authorse.g., see [–].

2 Main results

To prove our results, we need the following lemma due to Miller and Mocanu [, p.].

Lemma . Let q(z)∈Daand let p(z) =b+bnzn+· · ·

be analytic inUwith p(z)=b.If p(z)⊀q(z),then there exist points z∈Uandζ∈U\E(q)

and on mn≥for which

(i) p(z) =q(ζ), (ii) zp(z) =q(ζ).

Theorem . Let

P:U→C

with

ReaP(z)>  (a∈C).

If p is a function analytic inUwith p() = and

Rep(z) +P(z)zp(z)> E

|a|Re(aP(z)), (.)

then

Reap(z)>α,

where

E= –Re(a) –αReaP(z)

+ ReaP(z) Im(a)+ αRe(a)

+Re(a) –αIm(a), (.)

withRe(a) >α.

Proof Let us defineq(z) andh(z) as follows:

q(z) =ap(z) and

h(z) = a– (αa)z  –z

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The functionsqandhare analytic inUwithq() =h() =a∈Cwith

h(U) =w:Re(w) >α.

Now, we suppose thatq(z)⊀h(z). Therefore, by using Lemma ., there exist points

z∈U and ζ∈U\{}

such thatq(z) =h(ζ) andzq(z) =h(ζ),mn≥.

We note that

ζ=h–

q(z)

= q(z) –a

q(z) – (αa)

(.)

and

ζh(ζ) =

–|q(z) –a|

Re(aq(z))

. (.)

We haveh(ζ) =α+ρi(α,ρ∈R), therefore

Rep(z) +P(z)zp(z)

=Re

ah(ζ) +

aP(z)h

(ζ )

=Re

α+ρi

a

m|α+ρia|

Re(aα) Re

P(z)

a

≤Re

α+ρi

a

–|α+ρia|

Re(aα) Re

P(z)

a

=++C

=g(ρ), (.)

where

A= – Re(aP(z)) |a|Re(aα),

B=Im(a)

|a|

 +Re(aP(z))

Re(a) –α

and

C= 

|a|

αRe(a) –α

+|a|– αRe(a)Re(aP(z ))

(Re(a) –α)

.

We can see that the functiong(ρ) in (.) takes the maximum value atρgiven by

ρ=Im(a)

 + Re(a) –α

Re(aP(z))

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Hence, we have

Rep(z) +P(z)zp(z)

g(ρ)

= E |a|Re(aP(z)),

whereEis defined by (.). This is in contradiction to (.). Then we obtainRe(ap(z)) >α.

Theorem . Let p(z)a nonzero analytic function inUwith p() = .If

p(z) +zp

(z)

p(z) – 

<Re(aα)

|a| p(z), (.) then

Re

a p(z)

>α,

whereRe(a) >α.

Proof Let us define bothq(z) andh(z) as follows:

q(z) =a/p(z) and

h(z) =a– (αa)z  –z

Re(a) >α.

The functionsqandhare analytic inUwithq() =h() =a∈Cwith

h(U) =w:Re(w) >α.

Now, we suppose thatq(z)⊀h(z). Therefore, by using Lemma ., there exist points

z∈U and ζ∈U\{}

such thatq(z) =h(ζ) andzq(z) =h(ζ),mn≥.

We note that

ζh(ζ) =

–|q(z) –a|

Re(aq(z))

. (.)

We haveh(ζ) =α+ρi(ρ∈R); therefore,

|p(z) +zp

(z) p(z) – | |p(z)|

=α+ρi

am

a

|aα|

Re(aα) – 

|

a|

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≥  |a|

|aα|

Re(aα) +Re(aα)

≥ 

|a|Re(aα)

Re(aα)+Im(a) –ρ ≥Re(aα)

|a| .

This is in contradiction to (.). Then we obtainRe(pa(z)) >α.

3 Applications and examples

PuttingP(z) =β(β> ; real) in Theorem . we have the following corollary.

Corollary . If p is a function analytic inUwith p() = and

Rep(z) +βzp(z)> Eβ|a|Re(a),

then

Reap(z)>α,

where

E= –Re(a) –αβRe(a)+ βRe(a) Im(a)+ αRe(a)+Re(a) –αIm(a),

withRe(a) >α(α≥).

Puttingβ=  in Corollary ., we obtain the following corollary.

Corollary . If p is a function analytic inUwith p() = and

Rep(z) +zp(z)>  Re(a)

Re(a) –α–Re(a)

|a|

Re(a) – α,

then

Reap(z)>α,

withRe(a) >α(α≥).

Corollary . Let f(z)∈A, (g(z))aSand

Re

f(z)

g(z)

> E |a|Re(ag(z)

zg(z))

,

then

Re

af(z) g(z)

>α,

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Proof Puttingp(z) = fg((zz)) andP(z) =zgg(z(z))in Theorem ., we have

Rep(z) +P(z)zp(z)=Re

f(z)

g(z)

.

Since (g(z))aS, which givesRe(azg(z)

g(z) ) > , therefore,Re(aP(z)) > . This completes the

proof of the corollary.

Example . Letf(z)∈Aand

Ref(z)>  Re(a)

Re(a) –α–Re(a)

|a|

Re(a) – α,

then

Re

af(z) z

>α,

whereRe(a) >α.

Example . Letf(z)∈Aand

Re

 +zf

(z)

f(z) –

zf(z)

f(z)

zf(z)

f(z)

>  Re(a)

Re(a) –α–Re(a)

|a|

Re(a) – α,

then

Re

azf

(z)

f(z)

>α,

whereRe(a) >α.

() Puttinga=eiλ(|λ|<π

) andα= in Theorem ., we have Theorem  due to Kim and Cho [].

() Puttinga=eiλ(|λ|<π

),P(z) =β(β> ; real) andα= in Theorem ., we have Corollary  due to Kim and Cho [].

() Puttinga=α= andP(z) = in Theorem ., we have the result due to Nunokawa

et al.[].

() Puttinga=eiλ(|λ|<π

),P(z) = andα= in Theorem ., we have Corollary  due to Kim and Cho [].

Puttingp(z) =zff((zz)) in Theorem ., we have the following corollary.

Corollary . Let p(z)a nonzero analytic function in U with p() = .If

zff((zz))

<Re(aα) |a|

zff((zz)),

then

Re

a zf(z)

f(z)

>α,

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Remark

() Puttinga= andα= in Corollary ., we have the result due to Attiya and Nasr [].

() Puttinga= andα= in Corollary ., we have the result due to Kim and Cho [].

Competing interests

The author declares that he has no competing interests.

Acknowledgement

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

Received: 13 December 2013 Accepted: 6 January 2014 Published:24 Jan 2014

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