# Subordination and Superordination Results for a Class of Analytic Multivalent Functions

Authors: S. P. Goyal; Pranay Goswami; H. Silverman
Journal: International Journal of Mathematics and Mathematical Sciences
 (2008)

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## Abstract

We derive subordination and superordination results for a family of
normalized analytic functions in the open unit disk defined by integral operators. We
apply this to obtain sandwich results and generalizations of some known results.


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## Body

## 1. Introduction

Letℋ denote the class of  analytic functions in the
unit disk Δ:={z:|z|<1},
and let ℋ[a,p] be the subclass of ℋ of the form(1.1)f(z)=a+apzp+ap+1zp+1+⋯,p∈N={1,2,3,…}.Let𝒜(p) be the subclass of ℋ of the form(1.2)f(z)=zp+∑k=p+1∞akzk,p∈N.Iff and g are analytic and there exists a Schwarz
function w(z),
analytic in Δ with(1.3)w(0)=0,|w(z)|<1,z∈Δ,such that f(z)=g(w(z)),
then the function f is called subordinate to g and is denoted by(1.4)f≺gorf(z)≺g(z),z∈Δ.In particular, if the function g is univalent in Δ,
the above subordination is equivalent to(1.5)f(0)=g(0),f(Δ)⊂g(Δ).Supposeh and k are analytic functions in Δ and ϕ(r,s,t;z):C3×Δ→C.
If h and ϕ(h(z),zh′(z),z2h′′(z);z) are univalent and if h satisfies the second-order superordination(1.6)k(z)≺ϕ(h(z),zh′(z),z2h′′(z);z), then h is a solution of the differential
superordination (1.6). Note that if f is subordinate to g,
then g is superordinate to f.
An analytic function q is called subordinant if q≺h for all h satisfying (1.6). A univalent
subordinant
q˜ that satisfies q≺q˜ for all subordinants q of (1.6) is said to be the bestsubordinant.
Miller and Mocanu [1] have obtained conditions on k, q,
and ϕ for which the following implication holds:(1.7)k(z)≺ϕ(h(z),zh′(z),z2h′′(z);z)⇒q(z)≺h(z).Ali et al. [2] have obtained sufficient conditions for certain
normalized analytic functions f(z) to satisfy(1.8)q1(z)≺zf′(z)f(z)≺q2(z),where q1 and q2 are given univalent functions in Δ with q1(0)=1 and q2(0)=1.Recently,
Shanmugam et al. [3, 4] have also obtained sandwich results for certain
classes of analytic functions. Further subordination results can be found in
[5–8].


## 2. Definitions and Preliminaries

Definition 2.1.
Forf(z)∈𝒜(p),
Shams et al. [9] defined the following integral operator:(2.1)ℐσf(z)=(p+1)σzΓ(σ)∫0z(logzt)σ−1f(t)dt(2.2)=zp+∑n=p+1∞(p+1n+1)σanzn,σ>0.For the operator, one easily
gets(2.3)z[ℐσf(z)]'=(p+1)ℐσ−1f(z)−ℐσf(z).Also for −1≤B<A≤1 and λ≥0,
Shams et al. [9] defined a class Ωpσ(A,B;λ) of functions f(z)∈𝒜(p),
so that(2.4)λp(ℐσ−1f(z)zp)+p−λp(ℐσf(z)zp)≺1+Az1+Bz.The family Ωpσ(A;B;λ) is a general family containing various new and
known classes of analytic functions (see, e.g., [10, 11]).Definition 2.2 (see [1]).
Denote byQ the set of all functions f(z) that are analytic and injective on Δ¯−E(f),
where(2.5)E(f)={ζ∈∂Δ:limz→ζf(z)=∞},and are such that f′(ζ)≠0 for ζ∈∂Δ−E(f).We will
require certain results due to Miller and Mocanu [1, 12], Bulboacă [13], and Shanmugam et al. [4] contained in
the following lemmas.Lemma 2.3 (see [12]).
Letq(z) be univalent in the unit disk Δ,
and let θ and ϕ be analytic in the domain D containing q(Δ) with ϕ(w)≠0 when w∈q(Δ).
Set Q(z)=zq′(z)ϕ(q(z)),h(z)=θ(q(z))+Q(z). Suppose that
(i)
Q(z) is starlike univalent in Δ;(ii)
Re(zh′(z)/Q(z))>0 for z∈Δ.
If p(z) is analytic in Δ,
with p(0)=q(0), p(Δ)∈D,
and(2.6)θ(p(z))+zp′(z)ϕ(p(z))≺θ(q(z))+zq′(z)ϕ(q(z)),then p(z)≺q(z) and q(z) is the best dominant.Lemma 2.4 (see [4]).
Letq(z) be a convex univalent function in Δ and ψ,γ∈𝒞 with Re(1+(zq′′(z)/q′(z)))>max{0,−Re(ψ/γ)}. If p(z) is analytic in Δ and(2.7)ψp(z)+γzp′(z)≺ψq(z)+γzq′(z),then p(z)≺q(z) and q(z) is the best dominant.Lemma 2.5 (see [12]).
Letq(z) be univalent in Δ,
and let ϕ(z) be analytic in a domain containing q(Δ).
If zq′(z)/φ(q(z)) is starlike and(2.8)zp′(z)φ(p(z))≺zq′(z)φ(q(z)),then p(z)≺q(z) and q(z) is the best dominant.Lemma 2.6 (see [13]).
Letq(z) be convex univalent in the unit disk Δ,
and let ϑ and φ be analytic in a domain D containing q(Δ).
Suppose that
(i)
Re[ϑ′(q(z))/φ(q(z))]>0 for z∈Δ;(ii)
zq′(z)φ(q(z)) is starlike univalent in z∈Δ.
If p(z)∈ℋ[q(0),1]∩Q,
with p(Δ)⊆D,
and if ϑ(p(z))+zp′(z)ϕ(p(z)) is univalent in Δ and(2.9)ϑ(q(z))+zq′(z)φ(q(z))≺ϑ(p(z))+zp′(z)φ(p(z)),then q(z)≺p(z) and q(z) is the best subordinant.Lemma 2.7 (see [1]).
Letq(z) be convex univalent in Δ and γ∈𝒞.
Further assume that Re(γ)>0.
If p(z)∈ℋ[q(0),1]∩Q and p(z)+γzp′(z) is univalent in Δ,
then(2.10)q(z)+γzq′(z)≺p(z)+γzp′(z),which implies that q(z)≺p(z) and q(z) is the best subordinant.The main
object of this paper is to apply a method based on the differential
subordination in order to derive several subordination results.


## 3. Subordination for Analytic Functions

Theorem 3.1.
Letq(z) be univalent in the unit disk Δ, λ∈𝒞,
and(3.1)Re(1+zq′′(z)q′(z))>max{0,−Re(p(p+1)λ)},λ≠0(p∈N).If f(z)∈𝒜(p) satisfies the subordination
(3.2)λp(ℐσ−1f(z)zp)+p−λp(ℐσf(z)zp)≺q(z)+λzq′(z)p(p+1),where ℐσf(z) is defined by (2.1), then(3.3)(ℐσf(z)zp)≺q(z)and q(z) is the best dominant.Proof.
Consider(3.4)h(z):=(ℐσf(z)zp).Differentiating (3.4) with
respect to z logarithmically, we get(3.5)zh′(z)h(z)=z[ℐσf(z)]′ℐσf(z))−p.Now, in view of (2.3),  we obtain
from (3.5) the following subordination:(3.6)h(z)+λzh′(z)p(p+1)≺q(z)+λzq′(z)p(p+1).An application of Lemma 2.4,
with γ=λ/p(p+1) and ψ=1,
leads to (3.3).Takingq(z)=(1+Az)/(1+Bz) in Theorem 3.1, we arrive at the following.Corollary 3.2.
Let−1≤B<A≤1 and Re((1−Bz)/(1+Bz))>max{0,−Re(p(p+1)/λ)}
(λ≠0),p∈N.
If f∈𝒜(p) and(3.7)λp(ℐσ−1f(z)zp)+p−λp(ℐσf(z)zp)≺1+Az1+Bz+λp(p+1)(A−B)z(1+Bz)2,then(3.8)ℐσf(z)zp≺1+Az1+Bzand (1+Az)/(1+Bz) is the best dominant.Puttingp=1 and q(z)=(1+z)/(1−z) in Theorem 3.1,  we get the
following corollary.Corollary 3.3.
LetRe((1+z)/(1−z))>max{0,−Re(2/λ)} and λ≠0.
If f∈𝒜(1) and(3.9)λℐσ−1f(z)z+(1−λ)ℐσf(z)z≺1+z1−z+λz(1−z)2,then(3.10)ℐσf(z)z≺1+z1−zand (1+z)/(1−z) is the best dominant.Theorem 3.4.
Letq(z) be univalent in Δ and 0≠γ,μ∈𝒞,
and α,β∈𝒞 such that α+β≠0.
Let f∈𝒜(p) and suppose that q satisfies(3.11)Re{1+zq′′(z)q′(z)−zq′(z)q(z)}>0.If(3.12)1+γμ[αz[ℐσ−1f(z)]'+βz[ℐσf(z)]'αℐσ−1f(z)+βℐσf(z)−p]≺1+γzq′(z)q(z),then(3.13)[αℐσ−1f(z)+βℐσf(z)(α+β)zp]μ≺q(z),and q(z) is the best dominant.Proof.
Let us consider a functionh(z) defined by(3.14)h(z):=[αℐσ−1f(z)+βℐσf(z)(α+β)zp]μ,μ≠0,α+β≠0.Now, differentiating (3.14)
logarithmically, we get(3.15)zh′(z)h(z)=μ[αz[ℐσ−1f(z)]'+βz[ℐσf(z)]'αℐσ−1f(z)+βℐσf(z)−p].By setting(3.16)θ(w)=1,ϕ(w)=γw,it can be easily observed that θ(w) is analytic in 𝒞 and that ϕ(w)≠0 is analytic in 𝒞/{0}.
Also, we let(3.17)Q(z)=zq′(z)ϕ(q(z))=γzq′(z)q(z),(3.18)p(z)=θ(q(z))+Q(z)=1+γzq′(z)q(z).From (3.11) we see that Q(z) is starlike univalent in the unit disk Δ,
and from (3.18) we get(3.19)Re(zp′(z)Q(z))=Re{1+zq′′(z)q′(z)−zq′(z)q(z)}>0.An application of Lemma 2.3 to
(3.12) yields the result.Puttingα=0,β=1,γ=1,
and q(z)=(1+Az)/(1+Bz) in Theorem 3.4, we obtain the
following corollary.Corollary 3.5.
Iff(z)∈𝒜(p) and for −1≤A<B≤1,μ≠0,(3.20)1+μ[z[ℐσf(z)]′ℐσf(z)−p]≺1+(A−B)z(1+Az)(1+Bz),then(3.21)[ℐσf(z)zp]μ≺1+Az1+Bzand (1+Az)/(1+Bz) is the best dominant.By settingα=0,β=1,γ=1,σ=0,p=1,
and q(z)=(1+Bz)μ(A−B)/B in Theorem
3.4, we get the following corollary.Corollary 3.6.
Supposef(z)∈𝒜(1)
and let −1≤B<A≤1 and B≠0. If(3.22)1+μ[zf′(z)f(z)−1]≺1+Az1+Bz,then(3.23)[f(z)z]μ≺(1+Bz)μ(A−B)/Band (1+Bz)μ(A−B)/B is the best dominant.Remark 3.7.
q(z)=(1+Bz)μ(A−B)/B is univalent if and only if |(μ(A−B)/B)−1|≤1 or |(μ(A−B)/B)+1|≤1 (see [5]).Again by
settingβ=1,μ=1,α=0,γ=1/b,p=1,
and σ=0,
and by q(z)=1/(1−z)2b(b∈𝒞∖{0}) in Theorm 3.4, we get the
following corollary.Corollary 3.8.
Supposef(z)∈𝒜(1) and b is a nonzero complex number for which(3.24)1+1b[zf′(z)f(z)−1]≺1+z1−z.Then,(3.25)f(z)z≺1(1−z)2b
and 1/(1−z)2b is the best dominant.The result contained in Corollary3.8
was earlier given by Srivastava and Lashin [7].Theorem 3.9.
Letq be univalent in the unit disk Δ,
and let μ,γ≠0,η,δ,α,β∈𝒞,
and f(z)∈𝒜(p).
Suppose that q satisfies(3.26)Re{1+zq′′(z)q′(z)}>max{0,−Re(ηγ)}.Let(3.27)ψ(z)=[αℐσ−1f(z)+βℐσf(z)(α+β)zp]μ{η+γμ(αz[ℐσ−1f(z)]′+βz[ℐσf(z)]′αℐσ−1f(z)+βℐσf(z)−p)}+δ.If(3.28)ψ(z)≺ηq(z)+δ+γzq′(z),then(3.29)[αℐσ−1f(z)+βℐσf(z)(α+β)zp]μ≺q(z),α+β≠0,and q(z) is the best dominant.Proof.
Define a functionh(z) by(3.30)h(z):=[αℐσ−1f(z)+βℐσf(z)(α+β)zp]μ.Then, a computation shows that(3.31)zh′(z)h(z)=μ{αz[ℐσ−1f(z)]′+βz[ℐσf(z)]′αℐσ−1f(z)+βℐσf(z)−p}and hence(3.32)zh′(z)=μh(z)(z[αℐσ−1f(z)]′+zβ[ℐσf(z)]′αℐσ−1f(z)+βℐσf(z)−p).Set(3.33)θ(w)=ηw+δ,ϕ(w)=γ,and let(3.34)Q(z)=zq′(z)ϕ(q(z))=γzq′(z),p(z)=θ(q(z))+Q(z)=ηq(z)+δ+γzq′(z).
From (3.26), we see that Q(z) is starlike in Δ and that(3.35)Re{zp′(z)Q(z)}=Re{ηγ+1+zq′′(z)q′(z)}>0,by the hypothesis (3.26) of
Theorem 3.9. Thus, applying Lemma 2.3, the proof of Theorem 3.9 is completed.By settingβ=1,γ=1,α=0,andq(z)=(1+Az)/(1+Bz),
we obtain the following corollary.Corollary 3.10.
Letf(z)∈𝒜(p) and Re(η)>0.
Suppose that(3.36)Re{1−Bz1+Bz}>max{0,−Re(η)}.If(3.37)[ℐσf(z)zp]μ{η+μ(z[ℐσf(z)]′ℐσf(z)−p)}+δ≺η1+Az1+Bz+δ+z(A−B)(1+Bz)2,then(3.38)[ℐσf(z)zp]μ≺1+Az1+Bzand (1+Az)/(1+Bz) is the best dominant.Again by settingβ=1,γ=1,α=0,p=1,
and σ=0,
and by q(z)=(1+z)/(1−z),
we get the
following
corollary.Corollary 3.11.
Letf(z)∈𝒜(1) and(3.39)[f(z)z]μ{η+μ(zf′(z)f(z)−1)}+δ≺η1+z1−z+δ+2z(1−z)2,then(3.40)[f(z)z]μ≺1+z1−zand (1+z)/(1−z) is the best dominant.


## 4. Superordination for Analytic Functions

Theorem 4.1.
Letq be convex univalent in the unit disk Δ,
and λ∈𝒞.
Suppose λ satisfies Re{λ}>0 and ℐσf(z)/zp∈ℋ(q(0),1)∩Q.
Suppose that(4.1)λp(ℐσ−1f(z)zp)+p−λp(ℐσf(z)zp)is univalent in the unit disk Δ.
If(4.2)q(z)+λzq′(z)p(p+1)≺λp(ℐσ−1f(z)zp)+p−λp(ℐσf(z)zp),then(4.3)q(z)≺ℐσf(z)zpand q(z) is the best subordinant.Proof.
Let(4.4)p(z)=ℐσf(z)zp,z≠0.Differentiating logarithmically,
we get(4.5)zp′(z)p(z)=z[ℐσf(z)]′ℐσf(z)−p.After some computation, we get(4.6)p(z)+λzp′(z)p(p+1)=λp(ℐσ−1f(z)zp)+p−λp(ℐσf(z)zp).Now, using Lemma 2.7, we get the
desired result (4.3).Corollary 4.2.
Letq be convex univalent in Δ,
and λ∈𝒞.
Suppose λ satisfies R{λ}>0 and ℐσf(z)/zp∈ℋ(q(0),1)∩Q. Let(4.7)λp(ℐσ−1f(z)zp)+p−λp(ℐσf(z)zp)be univalent in the unit disk Δ.
If(4.8)λ(A−B)zp(p+1)(1+Bz)2+1+Az1+Bz≺λp(ℐσ−1f(z)zp)+p−λp(ℐσf(z)zp),then(4.9)1+Az1+Bz≺ℐσf(z)zpand (1+Az)/(1+Bz) is the best subordinant.Since the
proofs of Theorems4.3 and 4.4 are similar to the proofs of the previous
theorems, we only give statements of these theorems without proofs.Theorem 4.3.
Letq(z) be convex univalent in Δ,
and 0≠γ,μ∈𝒞,
and α,β∈𝒞 such that α+β≠0.
Let f(z)∈𝒜(p).
Suppose that [(αℐσ−1f(z)+βℐσf(z))/(α+β)zp]μ∈ℋ(q(0),1)∩Q, and(4.10)1+γμ[αz[ℐσ−1f(z)]'+βz[ℐσf(z)]'αℐσ−1f(z)+βℐσf(z)−p]is univalent in Δ.
If(4.11)1+γzq′(z)q(z)≺1+γμ[αz[ℐσ−1f(z)]'+β[ℐσf(z)]'αℐσ−1f(z)+βℐσf(z)−p],then(4.12)q(z)≺[αℐσ−1f(z)+βℐσf(z)(α+β)zp]μand q(z) is the best subordinant.Theorem 4.4.
Letq be convex univalent in the unit disk Δ,
and let γ≠0∈𝒞,η,δ,α,β∈𝒞 with α+β≠0,
and f(z)∈𝒜(p).
Suppose that ℐσf(z)/zp∈ℋ(q(0),1)∩Q,  and(4.13)Re{ηq′(z)γ}>0.If(4.14)ηq(z)+δ+γzq′(z)≺[αℐσ−1f(z)+βℐσf(z)(α+β)zp]μ{η+γμ(zα[ℐσ−1f(z)]′+zβ[ℐσf(z)]′αℐσ−1f(z)+βℐσf(z)−p)}+δ,then(4.15)q(z)≺[αℐσ−1f(z)+βℐσf(z)(α+β)zp]μand q(z) is the best subordinant.


## 5. Sandwich Results

Combining
results of differential subordinations and superordinations, we arrive at the
following “sandwich results.”Theorem 5.1.
Letq1(z) be convex univalent, and let q2(z) be univalent in Δ,
and λ∈𝒞.
Suppose q1 satisfies Re{λ}>0 and q2 satisfies (3.1). If ℐσf(z)/zp∈ℋ(q(0),1)∩Q and(5.1)[αℐσ−1f(z)+βℐσf(z)(α+β)zp]μ,α+β≠0,is univalent in Δ,
and if(5.2)q1(z)+λzq1′(z)p(p+1)≺λℐσ−1f(z)zp+(p−λ)ℐσf(z)zp≺q2(z)+λzq2′(z)p(p+1),then(5.3)q1(z)≺(ℐσf(z)zp)≺q2(z)and q1(z) and q2(z) are, respectively, the best subordinant and
the best dominant.Theorem 5.2.
Letq1(z) be convex univalent, and let q2(z) be univalent in Δ,
and λ∈𝒞.
Suppose that q2 satisfies (3.11). Further suppose that [(αℐσ−1f(z)+βℐσf(z))/(α+β)zp]μ∈ℋ(q(0),1)∩Q and 1+γμ[(αz[ℐσ−1f(z)]'+βz[ℐσf(z)]')/(αℐσ−1f(z)+βℐσf(z))−p] is univalent in Δ.
If(5.4)1+γzq1'(z)q1(z)≺1+γμ[αz[ℐσ−1f(z)]'+βz[ℐσf(z)]'αℐσ−1f(z)+βℐσf(z)−p]≺1+γzq2'(z)q2(z),then(5.5)q1(z)≺[αℐσ−1f(z)+βℐσf(z)(α+β)zp]μ≺q2(z),α+β≠0,and q1(z) and q2(z) are, respectively, the best subordinant and
the best dominant.Theorem 5.3.
Letq1(z) be convex univalent, and let q2(z) be univalent in Δ,
and λ∈𝒞.
Suppose that q1(z) satisfies (4.13) and q2(z) satisfies (3.28). Further suppose that [(αℐσ−1f(z)+βℐσf(z))/(α+β)zp]μ∈ℋ(q(0),1)∩Q with α+β≠0,
and that(5.6)[αℐσ−1f(z)+βℐσf(z)(α+β)zp]μ{η+γμ(zα[ℐσ−1f(z)]′+zβ[ℐσf(z)]′αℐσ−1f(z)+βℐσf(z)−p)}+δis univalent in Δ.
If(5.7)ηq1(z)+δ+γzq1'(z)≺[αℐσ−1f(z)+βℐσf(z)(α+β)zp]μ×{η+γμ(zα[ℐσ−1f(z)]′+zβ[ℐσf(z)]′αℐσ−1f(z)+βℐσf(z)−p)}+δ≺ηq2(z)+δ+γzq2′(z), then(5.8)q1(z)≺[αℐσ−1f(z)+βℐσf(z)(α+β)zp]μ≺q2(z)and q1(z) and q2(z) are, respectively, the best subordinant and
the best dominant.



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