# A Class of Nonlinear Integral Operators Preserving Double Subordinations

Authors: Oh Sang Kwon; Nak Eun Cho
Journal: Abstract and Applied Analysis
 (2008)

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## Abstract

The purpose of the present paper is to investigate some subordination- and superordination-preserving
properties of certain integral operators defined on the space of meromorphic functions in the punctured open unit disk. The sandwich-type theorem for these integral
operators is also considered.


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## Body

## 1. Introduction

Letℋ=ℋ(𝕌) denote the class of analytic functions in the
open unit disk 𝕌={z∈ℂ:|z|<1}.
For a∈ℂ,
let(1.1)ℋ[a,n]={f∈ℋ:f(z)=a+anzn+an+1zn+1+⋯}.Letf and F be members of ℋ.
The function f is said to be subordinate to F,
or F is said to be superordinate to f,
if there exists a function w analytic in 𝕌,
with w(0)=0 and |w(z)|<1,
and such that f(z)=F(w(z)). In such a case, we write f≺F or f(z)≺F(z). If the function F is univalent in 𝕌,
then we have f≺F if and only if f(0)=F(0) and f(𝕌)⊂F(𝕌) (cf. [1]).Definition 1.1 (see [1]).
Letϕ:ℂ2→ℂ and let h be univalent in 𝕌.
If p is analytic in 𝕌 and satisfies the differential subordination
(1.2)ϕ(p(z),zp′(z))≺h(z), then p is called a solution of the differential
subordination. The univalent function q is called a dominant of the solutions of the
differential subordination, or more simply a dominant if p≺q for all p satisfying (1.2). A dominant q˜ that satisfies q˜≺q for all dominants q of (1.2) is said to be the best dominant.Definition 1.2 (see [2]).
Letφ:ℂ2→ℂ and let h be analytic in 𝕌.
If p and φ(p(z),zp′(z)) are univalent in 𝕌 and satisfy the differential superordination
(1.3)h(z)≺φ(p(z),zp′(z)), then p is called a solution of the differential
superordination. An analytic function q is called a subordinant of the solutions of
the differential superordination, or more simply a subordinant if q≺p for all p satisfying (1.3). A univalent subordinant q˜ that satisfies q≺q˜ for all subordinants q of (1.3) is said to be the best
subordinant.Definition 1.3 (see [2]).
One
denotes by𝒬 the class of functions f that are analytic and injective on 𝕌¯∖E(f),
where
(1.4)E(f)={ζ∈∂𝕌:lim⁡z→ζf(z)=∞}, and are such
that f′(ζ)≠0 for ζ∈∂𝕌∖E(f).LetΣ denote the class of functions of the form(1.5)f(z)=1z+∑n=0∞anzn which are
analytic in the punctured open unit disk 𝔻=𝕌∖{0}.
Let Σ* and Σk be the subclasses of Σ consisting of all functions which are,
respectively, meromorphic starlike and meromorphic convex in 𝔻 (see, for details, [3–5]).For a functionf∈Σ,
we introduce the following integral operators Iβ,γ defined by(1.6)Iβ,γ(f)(z):=(γ−β+1zγ+1∫0ztγfβ(t)dt)1/β(f∈Σ;β,γ∈ℂ;β∈ℂ∖{0};Re{γ−β}>0). The integral
operator Iβ,γ(f) defined by (1.6) has been extensively studied
by many authors [6–10] with suitable restrictions on the parameters β and γ,
and for f belonging to some favored classes of
meromorphic functions. In particular, Bajpai [6] showed that the integral
operator I1,1(f) belongs to the classes Σ* and Σk,
whenever f belongs to the classes Σ* and Σk,
respectively. Moreover, the operator Iβ,γ for the case β=1 is related to the generalized Libera transform
introduced by Stević (see, e.g., [11–13]).Making use of the principle of subordination between
analytic functions, Miller et al. [14] obtained some subordination-preserving
properties for certain integral operators (see also [15]). Moreover, Miller and
Mocanu [2] considered differential superordinations as the dual concept of
differential subordinations (see also [16]). In the present paper, we obtain the
subordination- and superordination-preserving properties of the integral
operator Iβ,γ defined by (1.6) with the sandwich-type
theorem.Throughout this
paper, we denote the classΣβ,γ by(1.7)Σβ,γ:={f∈Σ:zf(z)≠0⁢andzIβ,γ(f)(z)≠0⁢(z∈𝕌;β⁢≠1)}, where Iβ,γ is the integral operator defined by (1.6). For
various interesting developments involving functions in the class Σβ,γ,
the reader may be referred, for example, to the work of Dwivedi et al. [9].


## 2. A Set of Lemmas

The following
lemmas will be required in our present investigation.Lemma 2.1 (see [17]).
Suppose that the functionH:ℂ2→ℂ satisfies the following condition:
(2.1)Re{H(is,t)}≤0 for all real s and t≥−n(1+s2)/2,
where n is a positive integer. If the function p(z)=1+pnzn+⋯ is analytic in 𝕌 and
(2.2)Re{H(p(z),zp′(z))}>0(z∈𝕌), then Re{p(z)}>0(z∈𝕌).Lemma 2.2 (see [5]).
Letβ,γ∈ℂ with β⁢≠0 and let h∈ℋ(𝕌) with h(0)=c.
If Re{βh(z)+γ}>0(z∈𝕌),
then the solution of the differential equation
(2.3)q(z)+zq′(z)βq(z)+γ=h(z)(z∈𝕌;q(0)=c) is analytic in 𝕌 and satisfies Re{βq(z)+γ}>0(z∈𝕌).Lemma 2.3 (see [1]).
Letp∈𝒬 with p(0)=a and let q(z)=a+anzn+⋯ be analytic in 𝕌 with q(z)≢a and n≥1.
If q is not subordinate to p,
then there exist points z0=r0eiθ∈𝕌 and ζ0∈∂𝕌∖E(f),
for which
(2.4)q(𝕌r0)⊂p(𝕌),q(z0)=p(ζ0),z0q′(z0)=mζ0p′(ζ0)(m≥n).A functionL(z,t) defined on 𝕌×[0,∞) is the subordination chain (or Löwner chain)
if L(⋅,t) is analytic and univalent in 𝕌 for all t∈[0,∞), L(z,⋅) is continuously differentiable on [0,∞) for all z∈𝕌, and L(z,s)≺L(z,t) for 0≤s<t.Lemma 2.4 (see [2]).
Letq∈ℋ[a,1],
set φ:ℂ2→ℂ, and let φ(q(z),zq′(z))≡h(z)(z∈𝕌).
If L(z,t)=φ(q(z),tzq′(z)) is a subordination chain and p∈ℋ[a,1]∩𝒬,
then
(2.5)h(z)≺φ(p(z),zp′(z)) implies that
(2.6)q(z)≺p(z). Furthermore, if φ(q(z),zp′(z))=h(z) has a univalent solution q∈𝒬,
then q is the best subordinant.Lemma 2.5 (see [18]).
The functionL(z,t)=a1(t)z+⋯ with a1(t)≠0 and lim⁡t→∞|a1(t)|=∞ is a subordination chain if and only if
(2.7)Re{z∂L(z,t)/∂z∂L(z,t)/∂t}>0(z∈𝕌;0≤t<∞).


## 3. Main Results

Subordination
theorem involving the integral operatorIβ,γ defined by (1.6) is contained in Theorem 3.1
below.Theorem 3.1.
Letf,g∈Σβ,γ.
Suppose that
(3.1)Re{1+zϕ′′(z)ϕ′(z)}>−δ(z∈𝕌;ϕ(z):=z(zg(z))β), where
(3.2)δ=1+|γ−β|2−|1−(γ−β)2|4Re{γ−β}(Re{γ−β}>0). Then, the
subordination
(3.3)z(zf(z))β≺z(zg(z))β implies that
(3.4)z(zIβ,γ(f)(z))β≺z(zIβ,γ(g)(z))β, where Iβ,γ is the integral operator defined by (1.6).
Moreover, the function z(zIβ,γ(g)(z))β is the best dominant.Proof.
Let us define the functionsF and G by
(3.5)F(z):=z(zIβ,γ(f)(z))β,G(z):=z(zIβ,γ(g)(z))β, respectively.
Without loss of generality, we can assume that G is analytic and univalent on 𝕌¯ and that
(3.6)G′(ζ)≠0(|ζ|=1).
We first show that if the functionq is defined by
(3.7)q(z):=1+zG′′(z)G′(z)(z∈𝕌), then
(3.8)Re{q(z)}>0(z∈𝕌). In terms of the
function ϕ involved in (3.1), the definition (1.6)
readily yields
(3.9)βz(Iβ,γ(g)(z))′Iβ,γ(g)(z)=−(γ+1)+(γ−β+1)ϕ(z)G(z). We also have
(3.10)βz(Iβ,γ(g)(z))′Iβ,γ(g)(z)=−(1+β)+zG′(z)G(z). By a simple
calculation with (3.9) and (3.10), we obtain the relationship
(3.11)1+zϕ′′(z)ϕ′(z)=q(z)+zq′(z)q(z)+γ−β≡h(z). From (3.1), we
note that
(3.12)Re{h(z)+γ−β}>0(z∈𝕌) and by using
Lemma 2.2, we conclude that the differential equation (3.11) has a solution q∈ℋ(𝕌) with
(3.13)q(0)=h(0)=1. Let us put
(3.14)H(u,v)=u+vu+γ−β+δ, where δ is given by (3.2).  From (3.1), (3.11), and (3.14), we obtain
(3.15)Re{H(q(z),zq′(z))}>0(z∈𝕌). Now, we proceed
to show that Re{H(is,t)}≤0 for all real s and t≤−(1+s2)/2.
From (3.14), we have(3.16)Re{H(is,t)}=Re{is+tis+γ−β+δ}=tRe{γ−β}|γ−β+is|2+δ≤−Eδ(s)2|γ−β+is|2, where(3.17)Eδ(s):=(Re{γ−β}−2δ)s2−4δ⁢Im{γ−β}s−2δ|γ−β|2+Re{γ−β}. For δ given by (3.2), we note that the coefficient
of s2 in the quadratic expression Eδ(s) given by (3.17) is positive or equal to zero.
Moreover, for the assumed value of δ given by (3.2), the quadratic expression Eδ(s) by s in (3.17) is a perfect square. Hence, from
(3.16), we see that Re{H(is,t)}≤0 for all real s and t≤−(1+s2)/2.
Thus, by using Lemma 2.1, we conclude that
(3.18)Re{q(z)}>0(z∈𝕌), that is, G is convex in 𝕌.
Next, we prove that the subordination condition (3.3)
implies that
(3.19)F(z)≺G(z) for the
functions F and G defined by (3.5). For this purpose, we
consider the function L(z,t) given by
(3.20)L(z,t):=γ−βγ−β+1G(z)+1+tγ−β+1zG′(z)(z∈𝕌;0≤t<∞). Since G is convex in 𝕌 and Re{γ−β}>0,
we obtain that
(3.21)∂L(z,t)∂z|z0=G′(0)(1+tγ−β+1)⁢≠0(z∈𝕌;0≤t<∞),Re{z∂L(z,t)/∂z∂L(z,t)/∂t}=Re{γ−β+(1+t)(1+zG′′(z)G′(z))}>0(z∈𝕌). Therefore, by
virtue of Lemma 2.5, L(z,t) is a subordination chain. We observe from the
definition of a subordination chain that
(3.22)ϕ(z)=γ−βγ−β+1G(z)+1γ−β+1zG′(z)=L(z,0),L(z,0)≺L(z,t)(0≤t<∞). This implies
that
(3.23)L(ζ,t)∉L(𝕌,0)=ϕ(𝕌)(ζ∈∂𝕌;0≤t<∞).
Now suppose thatF is not subordinate to G,
then by Lemma 2.3, there exist points z0∈𝕌 and ζ0∈∂𝕌 such that
(3.24)F(z0)=G(ζ0),z0F(z0)=(1+t)ζ0G′(ζ0)(0≤t<∞). Hence, we
have(3.25)L(ζ0,t)=γ−βγ−β+1G(ζ0)+1+tγ−β+1ζ0G′(ζ0)=γ−βγ−β+1F(ζ0)+1γ−β+1z0F′(z0)=z0(z0f(z0))β∈ϕ(𝕌), by virtue of
the subordination condition (3.3). This contradicts the above observation that L(ζ0,t)∉ϕ(𝕌).
Therefore, the subordination condition (3.3) must imply the subordination given
by (3.19). Considering F(z)=G(z),
we see that the function G(z) is the best dominant. This evidently completes
the proof of Theorem 3.1.Remark 3.2.
We note thatδ given by (3.2) in Theorem 3.1 satisfies the
inequality 0<δ≤1/2.We next prove a
dual problem of Theorem3.1 in the sense that the subordinations are replaced
by superordinations.Theorem 3.3.
Letf,g∈Σβ,γ.
Suppose that
(3.26)Re{1+zϕ′′(z)ϕ′(z)}>−δ(z∈𝕌;ϕ(z):=z(zg(z))β), where δ is given by (3.2), z(zf(z))β is univalent in 𝕌, and z(zIβ,γ(f)(z))β∈𝒬,
where Iβ,γ is the integral operator defined by (1.6).
Then, the superordination
(3.27)z(zg(z))β≺z(zf(z))β implies that
(3.28)z(zIβ,γ(g)(z))β≺z(zIβ,γ(f)(z))β. Moreover, the
function z(zIβ,γ(g)(z))β is the best subordinant.Proof.
The first part of the proof is
similar to that of Theorem3.1 and so we will use the same notation as in the
proof of Theorem 3.1. Now let us define the functions F and G,
respectively, by (3.5). We first note that by using (3.9) and
(3.10), we obtain(3.29)ϕ(z)=γ−βγ−β+1G(z)+1γ−β+1zG′(z):=φ(G′(z),zG′(z)). After a simple
calculation, (3.29) yields the relationship(3.30)1+zϕ′′(z)ϕ′(z)=q(z)+zq′(z)q(z)+γ−β, where the
function q is defined by (3.7). Then, by using the same
method as in the proof of Theorem 3.1, we can prove that
(3.31)Re{q(z)}>0(z∈𝕌), that is, G defined by (3.5) is convex(univalent) in 𝕌.
Next, we prove that the superordination condition
(3.27) implies that
(3.32)F(z)≺G(z)(z∈𝕌). Now, consider
the function L(z,t) defined by
(3.33)L(z,t):=γ−βγ−β+1G(z)+tγ−β+1zG′(z)(z∈𝕌;0≤t<∞). Since G is convex and Re{γ−β}>0,
we can prove easily that L(z,t) is a subordination chain as in the proof of
Theorem 3.1. Therefore, according to Lemma 2.4, we conclude that the
superordination condition (3.27) must imply the superordination given by
(3.32). Furthermore, since the differential equation (3.29) has the univalent
solution G,
it is the best subordinant of the given differential superordination.
Therefore, we complete the proof of Theorem 3.3.If we combine
Theorems3.1 and 3.3, then we obtain the following sandwich-type
theorem.Theorem 3.4.
Letf,gk∈Σβ,γ⁢(k=1,2).
Suppose that
(3.34)Re{1+zϕk′′(z)ϕk′(z)}>−δ(z∈𝕌;ϕk(z):=z(zgk(z))β;k=1,2), where δ is given by (3.2), z(zf(z))β is univalent in 𝕌,
and z(zIβ,γ(f)(z))β∈𝒬,
where Iβ,γ is the integral operator defined by (1.6).
Then, the subordination
(3.35)z(zg1(z))β≺z(zf(z))β≺z(zg2(z))β implies that
(3.36)z(zIβ,γ(g1)(z))β≺z(zIβ,γ(f)(z))β≺z(zIβ,γ(g2)(z))β. Moreover, the
functions z(zIβ,γ(g1)(z))β and z(zIβ,γ(g2)(z))β are the best subordinant and the best
dominant, respectively.The assumption
of Theorem3.4, that the functions z(zf(z))β and z(zIβ,γ(f)(z))β need to be univalent in 𝕌,
may be replaced by another condition in the
following result.Corollary 3.5.
Letf,gk∈Σβ,γ⁢(k=1,2).
Suppose that the condition (3.34) is satisfied and
(3.37)Re{1+zψ′′(z)ψ′(z)}>−δ(z∈𝕌;ψ(z):=z(zf(z))β;zf(z)∈𝒬), where δ is given by (3.2). Then, the subordination
(3.38)z(zg1(z))β≺z(zf(z))β≺z(zg2(z))β implies that
(3.39)z(zIβ,γ(g1)(z))β≺z(zIβ,γ(f)(z))β≺z(zIβ,γ(g2)(z))β, where Iβ,γ is the integral operator defined by (1.6).
Moreover, the functions z(zIβ,γ(g1)(z))β and z(zIβ,γ(g2)(z))β are the best subordinant and the best
dominant, respectively.Proof.
In order to prove Corollary3.5, we have to show that the condition
(3.37) implies the univalence of ψ(z) and
(3.40)F(z):=z(zIβ,γ(f)(z))β. Since 0<δ≤1/2 from Remark 3.2, the condition (3.37) means
that ψ is a close-to-convex function in 𝕌 (see [19]) and hence ψ is univalent in 𝕌.
Furthermore, by using the same techniques as in the proof of Theorem 3.4, we
can prove the convexity (univalence) of F and so the details may be omitted. Therefore,
by applying Theorem 3.4, we obtain Corollary 3.5.By settingγ−β=1 in Theorem 3.4, so that δ=1/2,
we deduce the following consequence of Theorem 3.4.Corollary 3.6.
Letf,gk∈Σβ,β+1⁢(k=1,2).
Suppose that
(3.41)Re{1+zϕk′′(z)ϕk′(z)}>−12(z∈𝕌;ϕk(z):=z(zgk(z))β;k=1,2),z(zf(z))β is univalent in 𝕌,
and z(zIβ,β+1(f)(z))β∈𝒬,
where Iβ,β+1 is the integral operator defined by (1.6) with γ=β+1.
Then, the subordination
(3.42)z(zg1(z))β≺z(zf(z))β≺z(zg2(z))β implies that
(3.43)z(zIβ,β+1(g1)(z))β≺z(zIβ,β+1(f)(z))β≺z(zIβ,β+1(g2)(z))β. Moreover, the
functions z(zIβ,β+1(g1)(z))β and z(zIβ,β+1(g2)(z))β are the best subordinant and the best
dominant, respectively.If we takeγ−β=1+i in Theorem 3.4, then we are easily led to the
following result.Corollary 3.7.
Letf,gk∈Σβ,β+1+i⁢(k=1,2).
Suppose that
(3.44)Re{1+zϕk′′(z)ϕk′(z)}>−3−54(z∈𝕌;ϕk(z):=z(zgk(z))β;k=1,2),(zf(z))β is univalent in 𝕌,
and (zIβ,β+1+i(f)(z))β∈𝒬,
where Iβ,β+1+i is the integral operator defined by (1.6) with γ=β+1+i.
Then, the subordination:
(3.45)z(zg1(z))β≺z(zf(z))β≺z(zg2(z))β implies that
(3.46)z(zIβ,β+1+i(g1)(z))β≺z(zIβ,β+1+i(f)(z))β≺z(zIβ,β+1+i(g2)(z))β. Moreover, the
functions z(zIβ,β+1+i(g1)(z))β and z(zIβ,β+1+i(g2)(z))β are the best subordinant and the best
dominant, respectively.



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