# Subordination Properties for Certain Analytic Functions

Authors: A. A. Attiya; Nak Eun Cho; M. A. Kutbi
Journal: International Journal of Mathematics and Mathematical Sciences
 (2008)

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## Abstract

The purpose of the present paper is to derive a subordination result for functions in the classHn*(α,λ,b) of normalized analytic functions in the
open unit disk U. A number of interesting applications of the subordination result are also considered.


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## Body

## 1. Introduction

LetA denote the class of functions of the
form (1.1)f(z)=z+∑n=2∞anzn  which are analytic in the unit disc 𝕌={z:|z|<1}. We also denote by K the class of
functions f∈A that are
convex in 𝕌.Given two functionsf,g∈A, where f is given by
(1.1) and g is defined by (1.2)g(z)=z+∑n=2∞bnzn,  the Hadamard
product (or convolution) f∗g is defined by (1.3)(f∗g)(z)=z+∑n=2∞anbnzn(z∈𝕌).By using the Hadamard product, Ruscheweyh [1] defined (1.4)Dαf(z)=z(1−z)α+1∗f(z)(α≥−1).  From the definition of (1.4), we observe that (1.5)Dnf(z)=z(zn−1f(z))(n)n!,  when n=α∈ℕ0=ℕ∪{0}={0,1,2,…}. The symbol Dnf(z)  (n∈ℕ0) was called the n-th order
Ruscheweyh derivative of f by Al-Amiri [2]. We also note that D0f(z)=f(z) and D1f(z)=zf′(z).Definition 1.1.
Suppose
thatf∈A. Then the function f is said to be a
member of the class Hn(α,λ,b) if it satisfies (1.6)|λ(n+1)(Dn+1f(z)/z)+[1−λ(n+1)](Dnf(z)/z)−1λ(n+1)(Dn+1f(z)/z)+[1−λ(n+1)](Dnf(z)/z)+2b(1−α)−1|<1(z∈𝕌;0≤α<1;λ≥0;b∈ℂ∖{0};n∈ℕ0).    By
specializing α, λ, b, and n, one can obtain various subclasses studied by many
authors (see, e.g., [3–11]).Definition 1.2.
Letg be analytic and
univalent in 𝕌. If f is analytic in 𝕌, f(0)=g(0), and f(𝕌)⊂g(𝕌), then one says that f is subordinate
to g in 𝕌, and one writes f≺g or f(z)≺g(z). One also says that g is superordinate
to f in 𝕌.Definition 1.3.
An infinite sequence{bk}k=1∞ of complex
numbers will be called a subordinating factor sequence if for every univalent
function f in K, one has (1.7)∑k=1∞bkakzk≺f(z)(z∈𝕌;a1=1).Lemma 1.4. (see [12]).
The sequence{bk}k=1∞ is a
subordinating factor sequence if and only if (1.8)Re{1+2∑k=1∞bkzk}>0(z∈𝕌).Now, we prove
the following lemma which gives a sufficient condition for functions belonging
to the classHn(α,λ,b).Lemma 1.5.
If the
functionf which is
defined by (1.1) satisfies the following condition: (1.9)∑k=2∞[1+λ(k−1)]C(n,k)|ak|≤(1−α)|b|(0≤α<1;λ≥0;b∈ℂ∖{0};n∈ℕ0),    where (1.10)C(n,k)=∏j=2k(j+n−1)(k−1)!(k=2,3,…),  then f∈Hn(α,λ,b).Proof.
Suppose that the inequality (1.9) holds.
Using the identity (1.11)z(Dnf(z))′=(n+1)Dn+1f(z)−nDnf(z),  we have for z∈𝕌, (1.12)|(1−λ)Dnf(z)z+λ(Dnf(z))′−1|−|2b(1−α)+(1−λ)Dnf(z)z+λ(Dnf(z))′−1|=|∑k=2∞[1+λ(k−1)]C(n,k)akzk−1|−|2b(1−α)+∑k=2∞[1+λ(k−1)]C(n,k)akzk−1|≤∑k=2∞[1+λ(k−1)]C(n,k)|ak||z|k−1−{2|b|(1−α)−∑k=2∞[1+λ(k−1)]C(n,k)|ak||z|k−1}≤2{∑k=2∞[1+λ(k−1)]C(n,k)|ak|−|b|(1−α)}≤0, which shows
that f belongs to Hn(α,λ,b).LetHn∗(α,λ,b) denote the
class of functions f in A whose
Taylor-Maclaurin coefficients ak satisfy the
condition (1.9).We note that(1.13)Hn∗(α,λ,b)⊆Hn(α,λ,b).Example 1.6.
(i) For0≤α<1, λ>0, b∈ℂ∖{0}, and n∈ℕ0, the following function defined by:
(1.14)f0(z)=z+2b(1−α)(n+1)(λ+1)z23F2(1,2,1+1λ;2+1λ,n+2;z)(z∈𝕌),
is in the class Hn(α,λ,b).
(ii) For0≤α<1, λ>0, b∈ℂ∖{0}, and n∈ℕ0, the following functions defined by:
(1.15)f1(z)=z±(1−α)|b|(λ+1)(n+1)z2(z∈𝕌),f2(z)=z±(1−α)|b|(2λ+1)(n+1)(n+2)z3(z∈𝕌),f3(z)=z±1(λ+1)(n+1)z2±2[(1−α)|b|−1](2λ+1)(n+1)(n+2)z3(z∈𝕌)     are in the class Hn∗(α,λ,b).In this paper, we obtain a sharp subordination result
associated with the classHn∗(α,λ,b) by using the
same techniques as in [13] (see also [14–16]). Some applications of the
main result which give important results of analytic functions are also
investigated.


## 2. Main Theorem

Theorem 2.1.
Letf∈Hn∗(α,λ,b). Then (2.1)(λ+1)(n+1)2[(λ+1)(n+1)+|b|(1−α)](f∗g)(z)≺g(z)(z∈𝕌)  for every
function g in K, and (2.2)Ref(z)>−(λ+1)(n+1)+|b|(1−α)(λ+1)(n+1).  The constant (λ+1)(n+1)/2[(λ+1)(n+1)+|b|(1−α)] cannot be
replaced by a larger one.Proof.
Letf∈Hn∗(α,λ,b) and let (2.3)g(z)=z+∑k=2∞ckzk be any function
in the class K. Then we readily have (2.4)(λ+1)(n+1)2[(λ+1)(n+1)+|b|(1−α)](f∗g)(z)=(λ+1)(n+1)2[(λ+1)(n+1)+|b|(1−α)](z+∑k=2∞akckzk).  Thus, by
Definition 1.2, the subordination result (2.1) will hold true if the sequence (2.5){(λ+1)(n+1)ak2[(λ+1)(n+1)+|b|(1−α)]}k=1∞  is a
subordinating factor sequence, with a1=1. In view of
Lemma 1.4, this is equivalent to the following inequality: (2.6)Re{1+∑k=1∞(λ+1)(n+1)[(λ+1)(n+1)+|b|(1−α)]akzk}>0(z∈𝕌).  Now, since (2.7)[1+λ(k−1)]C(n,k)(λ≥0,n∈ℕ0) is an
increasing function of k, we have (2.8)Re{1+∑k=1∞(λ+1)(n+1)[(λ+1)(n+1)+|b|(1−α)]akzk}=Re{1+(λ+1)(n+1)[(λ+1)(n+1)+|b|(1−α)]z+1[(λ+1)(n+1)+|b|(1−α)]∑k=2∞(λ+1)(n+1)akzk}>1−(λ+1)(n+1)[(λ+1)(n+1)+|b|(1−α)]r−1[(λ+1)(n+1)+|b|(1−α)]∑k=2∞[1+λ(k−1)]C(n,k)|ak|rk>1−(λ+1)(n+1)[(λ+1)(n+1)+|b|(1−α)]r−|b|(1−α)[(λ+1)(n+1)+|b|(1−α)]r>0(|z|=r). This proves the inequality (2.6), and hence also
the subordination result (2.1) asserted by Theorem 2.1. The inequality (2.2)
follows from (2.1) by taking (2.9)g(z)=z1−z∈K.
Next, we consider the function(2.10)f1(z)=z−|b|(1−α)(λ+1)(n+1)z2(0≤α<1;λ≥0;b∈ℂ∖{0};n∈ℕ0)  which is a
member of the class Hn∗(α,λ,b). Then by using
(2.1), we have (2.11)(λ+1)(n+1)2[(λ+1)(n+1)+|b|(1−α)]f1(z)≺z1−z(z∈𝕌).  It can be
easily verified for the function f1(z) defined by
(2.10) that (2.12)infz∈𝕌{Re((λ+1)(n+1)2[(λ+1)(n+1)+|b|(1−α)]f1(z))}=−12(z∈𝕌)  which completes the proof of Theorem 2.1.


## 3. Some Applications

Takingn=0 in Theorem
2.1, we obtain the following.Corollary 3.1.
If the functionf defined by
(1.1) satisfies (3.1)∑k=2∞[1+λ(k−1)]|ak|≤m(λ≥0,m>0)  then  for every function g in K, one has (3.2)(λ+1)2(λ+m+1)(f∗g)(z)≺g(z)(z∈𝕌),Ref(z)>−(1+mλ+1).     The constant (λ+1)/2(λ+m+1) cannot be
replaced by larger one.Puttingλ=0 in  Theorem
2.1, we have the following corollary.Corollary 3.2.
If the functionf defined by
(1.1) satisfies (3.3)∑k=2∞C(n,k)|ak|≤m,m>0,n∈ℕ0,  where C(n,k) is defined by
(1.10), then for every function g in K, one has (3.4)(n+1)2(n+m+1)(f∗g)(z)≺g(z)(z∈𝕌),Ref(z)>−(1+mn+1).     The constant (n+1)/2(n+m+1) cannot be
replaced by larger one.Next lettingλ=1 and n=0, in Theorem
2.1, we obtain the following corollary.Corollary 3.3.
If the functionf satisfies (3.5)∑k=2∞k|ak|≤m(m>0),  then for every
function g in K, one has (3.6)1(m+2)(f∗g)(z)≺g(z)(z∈𝕌),Ref(z)>−(1+m2).     The constant 1/(m+2) cannot be
replaced by a larger one.Remark 3.4.
Puttingλ=1, m=1, and n=0, in Theorem 2.1, we obtain the result due to Singh
[17].Also, by takingλ=0 and n=0, in Theorem
2.1, we have the following.Corollary 3.5.
If the functionf satisfies (3.7)∑k=2∞|ak|≤m(m>0),  then for every
function g in K, one has (3.8)12(m+1)(f∗g)(z)≺g(z)(z∈𝕌),Ref(z)>−(1+m).     The constant 1/2(m+1) cannot be
replaced by a larger one.It is clearly from the proof of Theorem2.1 that the
function f(z)=z−mz2 (m>0,z∈𝕌) is the extremal
function of  Corollary 3.5. Also, the following example gives a
nonpolynomial extremal function for the same corollary.Example 3.6.
Let the functionh be defined by (3.9)h(z)=(m+1)z(m+1)+mz(m>0,z∈𝕌),  the above
function is analytic in 𝕌 and it is
equivalent to (3.10)h(z)=z+∑k=2∞(−mm+1)k−1zk. Then we have (3.11)∑k=2∞|(−mm+1)k−1|=m. Therefore, the Taylor-Maclaurin coefficients of the
function h satisfy the
condition in Corollary 3.5. Moreover, it can be easily verified that (3.12)infz∈𝕌Reh(z)=h(−1)=−(m+1). Then, the
constant −(m+1) cannot be
replaced by  a larger one. Therefore, the function h is the extremal
function of  Corollary 3.5.



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