# Double Integral Operators Concerning Starlike of Orderβ

Authors: Kazuo Kuroki; Shigeyoshi Owa
Journal: International Journal of Differential Equations
 (2009)

---

## Abstract

Double integral operators which were considered by S. S. Miller and P. T. Mocanu
(Integral Transform. Spec. Funct.19(2008), 591–597) are discussed. In order to show the analytic function f(z) is starlike of order β in the open unit disk 𝕌, the theory of differential subordinations for analytic functions is applied. The object of
the present paper is to discuss some interesting conditions for f(z) to be starlike
of order β in 𝕌 concerned with second-order differential inequalities and double
integral operators.


---

## Body

## 1. Introduction, Definition and Preliminaries

Letℋ=ℋ(𝕌) denote the class of functions f(z) which are analytic in the open unit disk
(1.1)𝕌={z:z∈ℂand|z|<1}.
For a positive integer n and a∈ℂ, we define the following classes of analytic functions:
(1.2)ℋ[a,n]={f∈ℋ:f(z)=a+anzn+an+1zn+1+⋯,z∈𝕌}and(1.3)𝒜n={f∈ℋ:f(z)=z+an+1zn+1+an+2zn+2+⋯,z∈𝕌}
with 𝒜1=𝒜.A functionf(z)∈𝒜 is said to be starlike of order β in 𝕌 if it satisfies
(1.4)Re(zf′(z)f(z))>β(z∈𝕌),
for some β(0≦β<1). We denote by 𝒮*(β) the  subclass of 𝒜 consisting of all functions f(z) which are starlike of order β in 𝕌.By the familiar principle of differential subordinations between analytic functionsf(z) and g(z) in 𝕌, we say that f(z) is subordinate to g(z) in 𝕌 if there exists an analytic function w(z) with w(0)=0 and |w(z)|<1 such that f(z)=g(w(z))(z∈𝕌).We denote this subordination by(1.5)f(z)≺g(z)(z∈𝕌).
In particular, if g(z) is univalent in 𝕌, then it is known that
(1.6)f(z)≺g(z),(z∈𝕌)⟺f(0)=g(0)andf(𝕌)⊂g(𝕌).To obtain some results of this paper, we need the following two lemmas concerning the differential subordinations.Lemma 1.1 (see [1, Hallenbeck and Ruscheweyh]).
Leth(z)be a convex function with h(0)=aand letReγ>0. If  p(z)∈ℋ[a,n] and
(1.7)p(z)+zp′(z)γ≺h(z),
then
(1.8)p(z)≺q(z)≺h(z),
where
(1.9)q(z)=γnzγ/n∫0zh(t)tγ/n-1dt.
This result is sharp.Lemma 1.2 (see [2, Al-Amiri and Mocanu]).
Letn be a positive integer, and let α be real with0≦α<n. Let  q(z)∈ℋ with  q(0)=0,q′(0)≠0 and
(1.10)Rezq′′(z)q′(z)+1>αn.
If p∈ℋ[0,n] satisfies
(1.11)zp′(z)-αp(z)≺nzq′(z)-αq(z),
then p(z)≺q(z) and this result is sharp.By making use of these lemmas, Miller and Mocanu [3] have investigated some second-order differential inequality that implies starlikeness and deduced the following lemma.Lemma 1.3.
Letf(z)∈𝒜n and let 0≦α<n. If f(z) satisfies
(1.12)|zf′′(z)-α(f′(z)-1)|<n-α,
then f(z) is starlike.Furthermore, by using Lemma1.3, Miller and Mocanu [3] obtained some result concerning the double integral starlike operator as follows.Lemma 1.4.
Let0≦α<n and let  g(z)∈ℋ. If g(z) satisfies
(1.13)|g(z)|≦n-α,
then the function f(z) given by
(1.14)f(z)=z+zn+1∬01g(rsz)rn-α-1sndrds
is starlike.


## 2. Some Second-Order Differential Inequalities for Starlike of Orderβ

In this section, we deduced some conditions concerning the second-order differential inequality to show thatf(z) is starlike of order β in 𝕌.Theorem 2.1.
Letf(z)∈𝒜n and let 0≦α<n and 0≦β<1. If f(z) satisfies
(2.1)|zf′′(z)-α(f′(z)-1)|<(n+1)(1-β)(n-α)n+1-β,
then f(z) is starlike of order β in 𝕌.Proof.
We can rewrite the inequality (2.1) in terms of subordination as
(2.2)zf′′(z)-α(f′(z)-1)≺(n+1)(1-β)(n-α)n+1-βz.
If we set
(2.3)P(z)=f′(z)-(1+α)f(z)z=-α+(n-α)an+1zn+(n+1-α)an+2zn+1+⋯∈ℋ[-α,n],
then the subordination (2.2) becomes
(2.4)P(z)+zP′(z)≺-α+(n+1)(1-β)(n-α)n+1-βz.
Applying Lemma 1.1 to this first-order differential subordination, we obtain
(2.5)P(z)≺1nz1/n∫0z{-α+(n+1)(1-β)(n-α)n+1-βt}t1/n-1dt=-α+(1-β)(n-α)n+1-βz,
or equivalently
(2.6)f′(z)-(1+α)f(z)z≺-α+(1-β)(n-α)n+1-βz.
If we consider
(2.7)q(z)=1-βn+1-βz(q(0)=0,q′(0)=1-βn+1-β≠0)
and
(2.8)p(z)=f(z)z-1=an+1zn+an+2zn+1+⋯∈ℋ[0,n],
then the subordination (2.6) can be written as
(2.9)zp′(z)-αp(z)≺(1-β)(n-α)n+1-βz=nzq′(z)-αq(z).
Since 0≦α<n and the function q(z) satisfies
(2.10)Rezq′′(z)q′(z)+1=1>αn,
using Lemma 1.2, we obtain the subordination p(z)≺q(z), or
(2.11)f(z)z-1≺1-βn+1-βz.
It follows from the subordination (2.6) that
(2.12)|f′(z)-(1+α)f(z)z|<α+(1-β)(n-α)n+1-β=n(1+α-β)n+1-β,
while from the subordination (2.11) that
(2.13)|f(z)z|>1-1-βn+1-β=nn+1-β.
Combining these last two inequalities, we see that
(2.14)nn+1-β|zf′(z)f(z)-(1+α)|<|f(z)z||zf′(z)f(z)-(1+α)|=|f′(z)-(1+α)f(z)z|<n(1+α-β)n+1-β,
which simplifies to
(2.15)|zf′(z)f(z)-(1+α)|<n(1+α-β)n+1-β·n+1-βn=1+α-β.
This gives us that
(2.16)Rezf′(z)f(z)>(1+α)-(1+α-β)=β,
which proves that f(z) is starlike of order β in 𝕌.We introduce the following example for Theorem2.1.Example 2.2.
For the functionf(z)=z+((1-β)/(n+1-β))zn+1(0≦β<1), we have
(2.17)|zf′′(z)-α(f′(z)-1)|=|n(n+1)(1-β)n+1-βzn-α(n+1)(1-β)n+1-βzn|=|(n+1)(1-β)(n-α)n+1-β||z|n<(n+1)(1-β)(n-α)n+1-β.
Furthermore, we see that
(2.18)Rezf′(z)f(z)=Re(1+((n+1)(1-β)/(n+1-β))zn1+((1-β)/(n+1-β))zn)>1-(n+1)(1-β)/(n+1-β)1-((1-β)/(n+1-β))=β(z∈𝕌).Remark 2.3.
Lettingβ=0 in Theorem 2.1, we obtain Lemma 1.3 given by Miller and Mocanu [3].Also, settingα=0 in Theorem 2.1, we have the following.Corollary 2.4.
Letf(z)∈𝒜n and let0≦β<1. If  f(z) satisfies
(2.19)|zf′′(z)|<n(n+1)(1-β)n+1-β,
then f(z) is starlike of order β in 𝕌.For the casen=1 in the above corollary, we find the following.Remark 2.5.
Forf(z)∈𝒜, we have that
(2.20)|zf′′(z)|<2(1-β)2-β⇒f(z)∈𝒮*(β).
The case β=0 was first discussed by Obradović [4].Next, by making use of Theorem2.1, we obtain the following result concerning the double, integral operator for starlike of order β.Theorem 2.6.
Let a functiong(z)∈ℋ satisfy
(2.21)|g(z)|≦(n+1)(1-β)(n-α)n+1-β,
for some 0≦α<n and 0≦β<1. Then the function f(z) given by
(2.22)f(z)=z+zn+1∬01g(rsz)rn-α-1sndrds
is starlike of order β in 𝕌.Proof.
We first consider the functionf(z)∈𝒜n satisfying the differential equation
(2.23)zf′′(z)-α(f′(z)-1)=zng(z).
Then, it is clear that
(2.24)|zf′′(z)-α(f′(z)-1)|=|z|n|g(z)|<(n+1)(1-β)(n-α)n+1-β(z∈𝕌).
Thus, from Theorem 2.1, we see that the solution of the differential equation (2.23) must be starlike of order β. The solution of (2.23) can be obtained in two integrations. If we set φ(z)=f′(z)-1, then the equation (2.23) can be simplified to
(2.25)zφ′(z)-αφ(z)=zng(z),
which has the solution φ(z) given by
(2.26)φ(z)=zα∫0zg(ζ)ζn-α-1dζ=zn∫01g(rz)rn-α-1dr.
Since φ(z)=f′(z)-1, we have
(2.27)f′(z)-1=zn∫01g(rz)rn-α-1dr,
that is,
(2.28)f(z)=z+∫0zζn(∫01g(rζ)rn-α-1dr)dζ=z+zn+1∬01g(rsz)rn-α-1sndrds.Remark 2.7.
Takingβ=0 in Theorem 2.6, we find Lemma 1.4 given by Miller and Mocanu [3]. However, (1.14) and (2.22) are double integral operators of the same form.Moreover, makingn=1 and α=0 in Theorem 2.6, we have the following.Corollary 2.8.
Ifg(z)∈ℋ and
(2.29)|g(z)|≦2(1-β)2-β(z∈𝕌)
for some 0≦β<1, then
(2.30)f(z)=z+z2∬01g(rsz)sdrds∈𝒮*(β).As examples of Corollary2.8, we get the following.Example 2.9.
For the functiong(z)=2(1-β)/(2-β)(0≦β<1), we find
(2.31)f(z)=z+1-β2-βz2∈𝒮*(β),
because
(2.32)Rezf′(z)f(z)=Re(1+(2(1-β)/(2-β))z1+((1-β)/(2-β))z)>1-(2(1-β)/(2-β))1-((1-β)/(2-β))=β(z∈𝕌).Example 2.10.
For the functiong(z)=(2(1-β)/(2-β))z(0≦β<1), we have
(2.33)f(z)=z+1-β3(2-β)z3.
Then, we see that
(2.34)Rezf′(z)f(z)=Re(1+((1-β)/(2-β))z21+((1-β)/3(2-β))z2)>35-2β>β,(z∈𝕌),
that is,
(2.35)f(z)=z+1-β3(2-β)z3∈𝒮*(35-2β)⊂𝒮*(β).


## 3. Other Result for Starlikeness of Orderβ

To obtain thatf(z) is starlike of order β in Theorem 2.1, we showed that
(3.1)|zf′(z)f(z)-(1+α)|<1+α-β(z∈𝕌).
In this section, to obtain that f(z) is starlike of order β, we consider some second-order differential inequality concerning the order β and show the following inequality:
(3.2)|f(z)zf′(z)-12β|<12β(z∈𝕌)
for some 0<β<1.Remark 3.1.
Foe someβ(0<β<1), we see that
(3.3)|f(z)zf′(z)-12β|<12β⟺Rezf′(z)f(z)>β.Now, we consider the following theorem.Theorem 3.2.
Letf(z)∈𝒜n and let0<β<1. If f(z) satisfies
(3.4)|zf′′(z)+(1-2β)(f′(z)-1)|<{β(n+1)(n+1-2β)n+1-β(0<β<12)(1-β)(n+1)(n+1-2β)n+1-β,(12≦β<1),
then
(3.5)|f(z)zf′(z)-12β|<12β(z∈𝕌),
or f(z) is starlike of order β in 𝕌.Proof.
(i) For the case0<β<1/2, the inequality (3.4) can be written as follows:
(3.6)zf″(z)+(1-2β)(f′(z)-1)≺β(n+1)(n+1-2β)n+1-βz.
If we set
(3.7)P(z)=f′(z)-2βf(z)z=(1-2β)+(n+1-2β)an+1zn+(n+2-2β)an+2zn+1+⋯∈ℋ[1-2β,n],
then the subordination (3.6) becomes
(3.8)P(z)+zP′(z)≺(1-2β)+β(n+1)(n+1-2β)n+1-βz.
Applying Lemma 1.1 as well as the proof of Theorem 2.1, we obtain that
(3.9)P(z)≺1nz1/n∫0z{(1-2β)+β(n+1)(n+1-2β)n+1-βt}t1/n-1dt=(1-2β)+β(n+1-2β)n+1-βz,
or equivalently, that
(3.10)f′(z)-2βf(z)z≺(1-2β)+β(n+1-2β)n+1-βz.
Also, if the function p(z) is defined by
(3.11)p(z)=f′(z)-1=(n+1)an+1zn+(n+2)an+2zn+1+⋯∈ℋ[0,n],
then the subordination (3.6) becomes
(3.12)zp′(z)+(1-2β)p(z)≺β(n+1)(n+1-2β)n+1-βz,
namely,
(3.13)p(z)+zp′(z)1-2β≺β(n+1)(n+1-2β)(1-2β)(n+1-β)z.
Since 0<1-2β<1, we can use Lemma 1.1 as γ=1-2β and obtain
(3.14)p(z)≺1-2βnz(1-2β)/n∫0zβ(n+1)(n+1-2β)(1-2β)(n+1-β)t(1-2β)/ndt=β(n+1)n+1-βz,
that is,
(3.15)f′(z)-1≺β(n+1)n+1-βz.
From the subordination (3.10), we find
(3.16)|f′(z)-2βf(z)z|<(1-2β)+β(n+1-2β)n+1-β=n(1-β)+(1-2β)n+1-β,
while, from the subordination (3.15), we get
(3.17)|f′(z)|>1-β(n+1)n+1-β=n(1-β)+(1-2β)n+1-β.
By combining these last two inequalities, we obtain that
(3.18)n(1-β)+(1-2β)n+1-β|f(z)zf′(z)-12β|<|f′(z)||f(z)zf′(z)-12β|=12β|f′(z)-2βf(z)z|<n(1-β)+(1-2β)2β(n+1-β)(z∈𝕌),
which simplifies to
(3.19)|f(z)zf′(z)-12β|<12β(z∈𝕌).
(ii) For the case 1/2≦β<1, we can rewrite (3.4) in terms of the subordination as
(3.20)zf′′(z)+(1-2β)(f′(z)-1)≺(1-β)(n+1)(n+1-2β)n+1-βz.
Applying Lemma 1.1 in similar to case (i), we can obtain that
(3.21)f(z)-2βf(z)z≺(1-2β)+(1-β)(n+1-2β)n+1-βz.
Furthermore, if we set
(3.22)q(z)=(1-β)(n+1)n+1-βzandp(z)=f′(z)-1∈ℋ[0,n],
then the subordination (3.20) can be written as
(3.23)zp′(z)-(2β-1)p(z)≺nzq′(z)-(2β-1)q(z).
Since 0≦2β-1<1≦n and the function q(z) satisfies
(3.24)q(0)=0,q′(0)=(1-β)(n+1)n+1-β≠andRezf′′(z)f′(z)+1=1>2β-1n,
we can use Lemma 1.2 and obtain the following fact:
(3.25)zp′(z)-(2β-1)p(z)≺nzq′(z)-(2β-1)q(z)⇒p(z)≺q(z).
This implies that
(3.26)f′(z)-1≺(1-β)(n+1)n+1-βz.
From subordinations (3.21) and (3.26), we find
(3.27)|f′(z)-2βf(z)z|<|(1-2β)-(1-β)(n+1-2β)n+1-β|=βnn+1-β(z∈𝕌)
and
(3.28)|f′(z)|>1-(1-β)(n+1)n+1-β=βnn+1-β(z∈𝕌).
Therefore, we obtain
(3.29)βnn+1-β|f(z)zf′(z)-12β|<|f′(z)||f(z)zf′(z)-12β|=12β|f′(z)-2βf(z)z|<βn2β(n+1-β)(z∈𝕌),
which simplifies to
(3.30)|f(z)zf′(z)-12β|<12β(z∈𝕌).
This completes the proof of this theorem.Making use of Theorem3.2, we obtain the following result concerning the double integral operator for starlike of order β.Theorem 3.3.
Ifg(z)∈ℋ satisfies
(3.31)|g(z)|≦{β(n+1)(n+1-2β)n+1-β(0<β<12)(1-β)(n+1)(n+1-2β)n+1-β(12≦β<1),
for some 0<β<1, then
(3.32)f(z)=z+zn+1∬01g(rsz)rn-2βsndrds,
is a starlike function of order β.This theorem can be shown as well as the proof of Theorem2.6.Makingn=1 in Theorem 3.3, we haveCorollary 3.4.
Ifg(z)∈ℋ satisfies
(3.33)|g(z)|≦{4β(1-β)2-β(0<β<12)4(1-β)22-β(12≦β<1),
for some 0<β<1, then
(3.34)f(z)=z+z2∬01g(rsz)r1-2βsdrds∈𝒮*(β).Let us consider an example for Corollary3.4.Example 3.5.
If we considerg(z)=4β(1-β)/(2-β)(0<β<1/2), then we find
(3.35)f(z)=z+β2-βz2∈𝒮*(2-3β2(1-β))⊂𝒮*(β),
because
(3.36)Rezf′(z)f(z)=Re(1+(2β/(2-β))z1+(β/(2-β))z)>(2-3β)/2(1-β)>β(z∈𝕌).
Further, if we take g(z)=4(1-β)2/(2-β)(1/2≦β<1), then we see that
(3.37)f(z)=z+1-β2-βz2∈𝒮*(β).



---

*Source: 737129-2009-07-09.xml*