# 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 ($0≤α<1$) 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 .

## 2 Main results

To prove our results, we need the following lemma due to Miller and Mocanu [, 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:

$q(z)=ap(z)$

and

$h(z)= a − ( 2 α − a ¯ ) z 1 − z ( Re ( a ) > α ) .$

The functions q and h are analytic in with $q(0)=h(0)=a∈C$ with

$h(U)= { w : Re ( w ) > α } .$

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

$z 0 ∈Uand ζ 0 ∈∂U∖{1}$

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

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 $h( ζ 0 )=α+ρi$ ($α,ρ∈R$), 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 $Re(ap(z))>α$. □

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 $q(0)=h(0)=a∈C$ with

$h(U)= { w : Re ( w ) > α } .$

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

$z 0 ∈Uand ζ 0 ∈∂U∖{1}$

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

We note that

$ζ 0 h ′ ( ζ 0 )= − | q ( z 0 ) − a | 2 2 Re ( a − q ( z 0 ) ) .$
(2.7)

We have $h( ζ 0 )=α+ρi$ ($ρ∈R$); 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

$E=− ( Re ( a ) − α ) β 2 ( Re ( a ) ) 2 +2βRe(a) [ ( Im ( a ) ) 2 + 2 α Re ( a ) ] + ( Re ( a ) − α ) ( Im ( a ) ) 2 ,$

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 .

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 .

3. (3)

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

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 .

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 .

2. (2)

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

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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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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 