# Sandwich-Type Theorems for a Class of Multiplier Transformations Associated with the Noor Integral Operators

Authors: Nak Eun Cho; Khalida Inayat Noor
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We obtain some subordination- and superordination-preserving properties for a class of multiplier transformations associated with Noor integral operators defined on the space of normalized analytic functions in the open unit disk. The sandwich-type theorems for these transformations are 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 and nonnegative integer n, let(1.1)H[a,n]={f∈H:f(z)=a+anzn+an+1zn+1+⋯}.
We also denote 𝒜 by the subclass of ℋ[a,1] with the usual normalization f(0)=f′(0)-1=0.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)(z∈U),
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 q̃ that satisfies q̃≺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))(z∈U),
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 q̃ that satisfies q≺q̃ for all subordinants q of (1.3) is said to be the best subordinant.Definition 1.3 (see [2]).
We denote by𝒬 the class of functions f that are analytic and injective on 𝕌¯∖E(f), where
(1.4)E(f)={ζ∈∂U:lim⁡z→ζ=∞},
and are such that f′(ζ)≠0 for ζ∈∂𝕌∖E(f).
Following Komatu [3], we introduce the integral operator ℒλ(c):𝒜→𝒜 defined by
(1.5)Lλ(c)f(z):=cλΓ(λ)∫01tc-2(log⁡1t)λ-1f(tz)dt(Re{c}>0;λ≥0;f∈A),
where the symbol Γ stands the Gamma function. We also note that the operator ℒλ(c)f(z) defined by (1.5) can be expressed by the series expansion as follows:
(1.6)Lλ(c)f(z)=z+∑k=2∞(cc+k-1)λakzk.
Obviously, we have, for λ,ν≥0,
(1.7)Lλ(c)(Lν(c)f(z))=Lλ+ν(c)f(z).
In particular, the operator ℒλ(2) is closely related to the multiplier transformation studied earlier by Flett [4]. Various interesting properties of the operator ℒλ(2) have been studied by Jung et al. [5] and Liu [6]. We also note from (1.6) that we can define the operator ℒλ(c) for any real number λ.
Let(1.8)fcλ(z)=z+∑k=2∞(cc+k-1)λzk(Re{c}>0;λ≥0)
and let fcλ,μ be defined such that
(1.9)fcλ(z)*fcλ,μ(z)=z(1-z)μ(μ>0;z∈U),
where the symbol * stands for the Hadamard product (or convolution). Then, motivated essentially by the Noor integral operator [7] (see also [8–11]), we now introduce the operator Iλ,μκ:𝒜→𝒜, which are defined here by
(1.10)Icλ,μf(z)=(fcλ,μ*f)(z)(f∈A;Re{c}>0;λ≥0;μ>0).
In view of (1.9) and (1.10), we obtain the following relations:
(1.11)z(Icλ,μf(z))′=cIcλ+1,μf(z)-(c-1)Icλ,μf(z),(1.12)z(Icλ,μf(z))′=μIcλ,μ+1f(z)-(μ-1)Icλ,μf(z).
Making use of the principle of subordination between analytic functions, Miller et al. [12] investigated some subordination theorems involving certain integral operators for analytic functions in 𝕌 (see, also [13]). Moreover, Miller and Mocanu [2] considered differential superordinations, as the dual concept of differential subordinations (see also [14]). In the present paper, we obtain the subordination- and superordination-preserving properties of the multiplier transformations Icλ,μ defined by (1.10) with the sandwich-type theorems.
The following lemmas will be required in our present investigation.Lemma 1.4 (see [15]).
Suppose that the functionH:ℂ2→ℂ satisfies the condition:
(1.13)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
(1.14)Re{H(p(z),zp′(z))}>0(z∈U),
then Re{p(z)}>0 in 𝕌.Lemma 1.5 (see [16]).
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:
(1.15)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 1.6 (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  q(𝕌r0)⊂p(𝕌),
(1.16)p(z0)=q(ζ0),z0p′(z0)=mζ0q′(ζ0)(m≥n).Lemma 1.7 (see [2]).
Letq∈ℋ[a,1], let  φ:ℂ2→ℂ, and set  φ(q(z),zq′(z))≡h(z). If  L(z,t)=φ(q(z),tzq′(z)) is a subordination chain and  p∈ℋ[a,1]∩𝒬, then
(1.17)h(z)≺φ(p(z),zp′(z))(z∈U)
implies that
(1.18)q(z)≺p(z)(z∈U).
Furthermore, if φ(q(z),zp′(z))=h(z) has a univalent solution q∈𝒬, then q is the best subordinant.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 z∈𝕌 and 0≤s<t.Lemma 1.8 (see [17]).
The functionL(z,t)=a1(t)z+⋯ with  a1(t)≠0 and  limt→∞|a1(t)|=∞. Suppose that  L(·;t) ia analytic in  𝕌 for all  t≥0,  L(z;·) is continuously differentiable on  [0,∞) for all  z∈𝕌. If  L(z;t) satisfies
(1.19)|L(z;t)|≤K0|a1(t)|(|z|<r0<1;0≤t<∞)
for some positive constants K0 and r0 and
(1.20)R{z∂L(z,t)/∂z∂L(z,t)/∂t}>0(z∈U;0≤t<∞),
then L(z;t) is a subordination chain.


## 2. Main Results

Firstly, we begin by proving the following subordination theorem involving the multiplier transformationIcλ,μ defined by (1.10).Theorem 2.1.
Letf,g∈𝒜. Suppose that
(2.1)Re{1+zϕ′′(z)ϕ′(z)}>-δ(ϕ(z):=(1-α)Icλ+1,μg(z)+αIcλ,μg(z);0≤α<1;z∈U),
where
(2.2)δ=(1-α)2+|c-1+α|2-|(1-α)2-(c-1+α)2|4(1-α)Re{c-1+α}(Re{c-1+α}>0).
Then the subordination:
(2.3)(1-α)Icλ+1,μf(z)+αIcλ,μf(z)≺ϕ(z)(z∈U)
implies that
(2.4)Icλ,μf(z)≺Icλ,μg(z)(z∈U).
Moreover, the function Icλ,μg(z) is the best dominant.Proof.
Let us define the functionsF and G, respectively, by
(2.5)F(z):=Icλ,μf(z),G(z):=Icλ,μg(z),
We first show that if the functionq is defined by
(2.6)q(z):=1+zG′′(z)G′(z)(z∈U),
then
(2.7)Re{q(z)}>0(z∈U).
Taking the logarithmic differentiation on both sides of the second equation in (2.5) and using (1.11) for g∈𝒜, we obtain
(2.8)cϕ(z)=(c-1+α)G(z)+(1-α)zG′(z),
which, in conjunction with (2.8), yields the relationship:
(2.9)1+zϕ′′(z)ϕ′(z)=1+zG′′(z)G′(z)+zq′(z)q(z)+(c-1+α)/(1-α)=q(z)+zq′(z)q(z)+(c-1+α)/(1-α)≡h(z).
From (2.1), we have
(2.10)Re{h(z)+c-1+α1-α}>0(z∈U),
and by using Lemma 1.5, we conclude that the differential equation (2.9) has a solution q∈ℋ(𝕌) with q(0)=h(0)=1.
Let us put(2.11)H(u,v)=u+vu+(c-1+α)/(1-α)+δ,
where δ is given by (2.2). From (2.1), (2.9), and (2.11), we obtain
(2.12)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 (2.11), we have
(2.13)Re{H(is,t)}=Re{is+tis+(c-1+α)/(1-α)+δ}=t(c-1+α)/(1-α)|(c-1+α)/(1-α)+is|2+δ≤-Eδ(s)2|(c-1+α)/(1-α)+is|2,
where
(2.14)Eδ(s):=(Re{c-1+α}(1-α)-2δ)s2-4δsIm⁡{c-1+α}(1-α)-2δ|(c-1+α)(1-α)|2+Re{c-1+α}(1-α).
For δ given by (2.2), we can prove easily that the expression Eδ(s) given by (2.14) is positive or equal to zero. Moreover, the quadratic expression by s in (2.14) is a perfect square for the assumed value of δ. Hence from (2.13), we see that Re{H(is,t)}≤0 for all real s and t≤-(1+s2)/2. Thus, by using Lemma 1.4, we conclude that Re{q(z)}>0 for all z∈𝕌. That is, q is convex in 𝕌.
Next, we prove that the subordination condition (2.3) implies that
(2.15)F(z)≺G(z)(z∈U)
for the functions F and G defined by (2.5). Without loss of generality, we can assume that G is analytic and univalent on 𝕌¯ and that G′(ζ)≠0(|ζ|=1). Now we consider the function L(z,t) given by
(2.16)L(z,t):=c-1+αcG(z)+(1-α)(1+t)czG′(z)(z∈U;0≤t<∞).
We note that
(2.17)∂L(z,t)∂z|z=0=G′(0)((c-1+α)+(1-α)(1+t)c)≠0(0≤t<∞;Re{c-1+α}>0).
This shows that the function
(2.18)L(z,t)=a1(t)z+⋯
satisfies the condition a1(t)≠0 for all t∈[0,∞). By using the well-known growth and distortion theorems for convex functions, it is easy to check that the first part of Lemma 1.8 is satisfied. Furthermore, we have
(2.19)Re{z∂L(z,t)/∂z∂L(z,t)/∂t}=Re{c-1+α1-α+(1+t)(1+zG′′(z)G′(z))}>0,
since G is convex and Re{c-1+α}>0. Therefore, by virtue of Lemma 1.8, L(z,t) is a subordination chain. We observe from the definition of a subordination chain that
(2.20)ϕ(z)=c-1+αcG(z)+1-αczG′(z)=L(z,0),L(z,0)≺L(z,t)(z∈U;0≤t<∞).
This implies that
(2.21)L(ζ,t)∉L(U,0)=ϕ(U)(ζ∈∂U;0≤t<∞).
Now suppose thatF is not subordinate to G, then by Lemma 1.6, there exists points z0∈𝕌 and ζ0∈∂𝕌 such that
(2.22)F(z0)=G(ζ0),z0F(z0)=(1+t)ζ0G′(ζ0)(0≤t<∞).
Hence we have
(2.23)L(ζ0,t)=c-1+αcG(ζ0)+(1-α)(1+t)cζ0G′(ζ0)=c-1+αcF(z0)+1-αcz0F′(z0)=(1-α)Icλ+1,μf(z0)+αIcλ,μf(z0)∈ϕ(U),
by virtue of the subordination condition (2.3). This contracts the above observation that L(ζ0,t)∉ϕ(𝕌). Therefore, the subordination condition (2.3) must imply the subordination given by (2.15). Considering F(z)=G(z), we see that the function G(z) is the best dominant. This evidently completes the proof of Theorem 2.1.Remark 2.2.
We note thatδ given by (2.2) in Theorem 2.1 satisfies the inequality 0<δ≤1/2.We next prove a dual problem of Theorem2.1, in the sense that the subordinations are replaced by superordinations.Theorem 2.3.
Letf,g∈𝒜. Suppose that
(2.24)Re{1+zϕ′′(z)ϕ′(z)}>-δ(ϕ(z):=(1-α)Icλ+1,μg(z)+αIcλ,μg(z);0≤α<1;z∈U),
where δ is given by (2.2), and the function Icλ+1,μf(z) is univalent in 𝕌 and Icλ,μf(z)∈𝒬. Then the superordination:
(2.25)ϕ(z)≺(1-α)Icλ+1,μf(z)+αIcλ,μf(z)(z∈U)
implies that
(2.26)Icλ,μg(z)z≺Icλ,μf(z)(z∈U).
Moreover, the function Icλ,μg(z) is the best subordinant.Proof.
Let us define the functionsF and G, respectively, by (2.5). We first note that, if the function q is defined by (2.6), by using (2.8), then we obtain
(2.27)ϕ(z)=c-1+αcG(z)+1-αczG′(z)=:φ(G(z),zG′(z)).
After a simple calculation, (2.27) yields the relationship:
(2.28)1+zϕ′′(z)ϕ′(z)=q(z)+zq′(z)q(z)+(c-1+α)/(1-α).
Then by using the same method as in the proof of Theorem 2.1, we can prove that Re{q(z)}>0 for all z∈𝕌. That is, G defined by (2.6) is convex (univalent) in 𝕌.
Next, we prove that the subordination condition (2.27) implies that
(2.29)F(z)≺G(z)(z∈U)
for the functions F and G defined by (2.5). Now consider the function L(z,t) defined by
(2.30)L(z,t):=c-1+αcG(z)+(1-α)tczG′(z)(z∈U;0≤t<∞).
Since G is convex and Re{c-1+α}>0, we can prove easily that L(z,t) is a subordination chain as in the proof of Theorem 2.1. Therefore according to Lemma 1.7, we conclude that the superordination condition (2.27) must imply the superordination given by (2.29). Furthermore, since the differential equation (2.27) has the univalent solution G, it is the best subordinant of the given differential superordination. Therefore we complete the proof of Theorem 2.3.If we combine this Theorems2.1 and 2.3, then we obtain the following sandwich-type theorem.Theorem 2.4.
Letf,gk∈𝒜(k=1,2). Suppose that
(2.31)Re{1+zϕk′′(z)ϕk′(z)}>-δ(ϕk(z):=(1-α)Icλ+1,μgk(z)+αIcλ,μgk(z);k=1,2;0≤α<1;z∈U),
where δ is given by (2.2), and the function Icλ+1,μf(z) is univalent in 𝕌 and Icλ,μf(z)∈𝒬. Then the subordination relation:
(2.32)ϕ1(z)≺Icλ+1,μf(z)≺ϕ2(z)(z∈U)
implies that
(2.33)Icλ,μgk(z)≺Icλ,μf(z)≺Icλ,μg2(z)(z∈U).
Moreover, the functions Icλ,μg1(z) and Icλ,μg2(z) are the best subordinant and the best dominant, respectively.The assumption of Theorem2.4, that the functions Icλ+1,μf(z)andIcλ,μf(z) need to be univalent in 𝕌, may be replaced by another conditions in the following result.Corollary 2.5.
Letf,gk∈𝒜(k=1,2). Suppose that the condition  (2.31) is satisfied and
(2.34)Re{1+zψ′′(z)ψ′(z)}>-δ(ψ(z):=(1-α)Icλ+1,μf(z)+αIcλ,μf(z);0≤α<1;z∈U;f∈Q),
where δ is given by (2.2). Then the subordination relation:
(2.35)ϕ1(z)≺ψ(z)≺ϕ2(z)(z∈U)
implies that
(2.36)Icλ,μgk(z)≺Icλ,μf(z)≺Icλ,μg2(z)(z∈U).
Moreover, the functions Icλ,μg1(z) and Icλ,μg2(z) are the best subordinant and the best dominant, respectively.Proof.
In order to prove Corollary2.5, we have to show that the condition (2.34) implies the univalence of ψ(z) and F(z):=Icλ,μf(z). Since 0<δ≤1/2 from Remark 2.2, the condition (2.34) means that ψ is a close-to-convex function in 𝕌 (see [18]) and hence ψ is univalent in 𝕌. Furthermore, by using the same techniques as in the proof of Theorem 2.4, we can prove the convexity (univalence) of F and so the details may be omitted. Therefore, by applying Theorem 2.4, we obtain Corollary 2.5.By settingc=2 and α=0 in Theorem 2.4, so that δ=1/2, we deduce the following consequence of Theorem 2.4.Corollary 2.6.
Letf,gk∈𝒜(k=1,2). Suppose that
(2.37)Re{1+zϕk′′(z)ϕk′(z)}>-12(ϕk(z):=I2λ+1,μgk(z);k=1,2;z∈U),
and the function I2λ+1,μf(z) is univalent functions in 𝕌 and I2λ,μf(z)∈𝒬. Then the subordination relation:
(2.38)I2λ+1,μg1(z)≺I2λ+1,μf(z)≺I2λ+1,μg2(z)(z∈U)
implies that
(2.39)I2λ,μg1(z)≺I2λ,μf(z)≺I2λ,μg2(z)(z∈U)
Moreover, the functions I2λ,μg1(z) and I2λ,μg2(z) are the best subordinant and the best dominant, respectively.If we takec=2+i and α=0 in Theorem 2.4, then we easily to lead to the following result.Corollary 2.7.
Letf,gk∈𝒜(k=1,2). Suppose that
(2.40)Re{1+zϕk′′(z)ϕk′(z)}>-3-54(ϕk(z):=I2+iλ,μgk(z);k=1,2;z∈U),
and the function I2+iλ+1,μf(z) is univalent functions in 𝕌 and I2+iλ,μf(z)∈𝒬. Then the subordination relation:
(2.41)I2+iλ+1,μg1(z)≺I2+iλ+1,μf(z)≺I2+iλ+1,μg2(z)(z∈U)
implies that
(2.42)I2+iλ,μg1(z)≺I2+iλ,μf(z)≺I2+iλ,μg2(z)(z∈U).
Moreover, the functions I2+iλ,μg1(z) and I2+iλ,μg2(z) are the best subordinant and the best dominant, respectively.The proof of Theorem2.8 below is similar to that of Theorem 2.4 by using (1.12) and so the details may be omitted.Theorem 2.8.
Letf,gk∈𝒜(k=1,2). Suppose that
(2.43)Re{1+zϕk′′(z)ϕk′(z)}>-δ(ϕk(z):=(1-α)Icλ,μ+1gk(z)+αIcλ,μgk(z);k=1,2;0≤α<1;z∈U),
where δ is given by (2.2) with c=μ, and the function Icλ,μ+1f(z) is univalent in 𝕌 and Icλ,μf(z)∈𝒬. Then the subordination relation:
(2.44)ϕ1(z)≺Icλ,μ+1f(z)≺ϕ2(z)(z∈U)
implies that
(2.45)Icλ,μgk(z)≺Icλ,μf(z)≺Icλ,μg2(z)(z∈U).
Moreover, the functions Icλ,μg1(z) and Icλ,μg2(z) are the best subordinant and the best dominant, respectively.By using a similar method given in the proof of Theorems2.4 and 2.8, we have the corresponding two theorems below.Theorem 2.9.
Letf,gk∈𝒜(k=1,2) with the additional condition  gk′′(z)≠0. Suppose that
(2.46)Re{1+zϕk′′(z)ϕk′(z)}>-δ(ϕk(z):=(1-α)Icλ+1,μgk(z)+αIcλ,μgk(z)z;k=1,2;0≤α<1;z∈U),
where δ is given by
(2.47)δ=(1-α)2+|c|2-|(1-α)2-c2|4(1-α)Re{c}(Re{c}>0),
and the function Icλ+1,μf(z) is univalent in 𝕌 and Icλ,μf(z)∈𝒬. Then the subordination relation:
(2.48)ϕ1(z)≺Icλ+1,μf(z)z≺ϕ2(z)(z∈U)
implies that
(2.49)Icλ,μg1(z)z≺Icλ,μf(z)z≺Icλ,μg2(z)z(z∈U).
Moreover, the functions Icλ,μg1(z)/z and Icλ,μg2(z)/z are the best subordinant and the best dominant, respectively.Theorem 2.10.
Letf,gk∈𝒜(k=1,2) with the additional condition  gk′′(z)≠0. Suppose that
(2.50)Re{1+zϕk′′(z)ϕk′(z)}>-δ(ϕk(z):=(1-α)Icλ,μ+1gk(z)+αIcλ,μgk(z)z;k=1,2;0≤α<1;z∈U),
where δ is given by (2.47) with c=μ, and the function Icλ,μ+1f(z)/z is univalent in 𝕌 and Icλ,μf(z)/z∈𝒬. Then the subordination relation:
(2.51)ϕ1(z)≺Icλ,μ+1f(z)z≺ϕ2(z)(z∈U)
implies that
(2.52)Icλ,μg1(z)z≺Icλ,μf(z)z≺Icλ,μg2(z)z(z∈U).
Moreover, the functions Icλ,μg1(z)/z and Icλ,μg2(z)/z are the best subordinant and the best dominant, respectively.



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