# Sufficient Conditions for Non-Bazilevič Functions

Authors: Qun-Fa Cui; Zhi-Gang Wang; Xiao-Hong Chen; Feng-Hua Wen
Journal: Abstract and Applied Analysis
 (2013)

---

## Abstract

The main purpose of this paper is to derive some sufficient conditions for analytic functions to be of non-Bazilevič type.


---

## Body

## 1. Introduction

Let𝒜 denote the class of functions of the form
(1)f(z)=z+∑j=2∞‍ajzj,
which are analytic in the open unit disk:
(2)𝕌:={z:z∈ℂ,|z|<1}.For0≦α<1 and 0<μ<1, a function f∈𝒜 is said to be in the class 𝒩(μ,α) if it satisfies the condition
(3)ℜ(f′(z)(zf(z))1+μ)>α,(z∈𝕌).
As usual, the class 𝒩(μ,α) is said to be non-Bazilevič functions of order α (see [1]).For some recent investigations of non-Bazilevič functions, see, for example  the works of [2–6] and the references cited therein.For two functionsf and g, analytic in 𝕌, we say that the function f is subordinate to g in 𝕌  and write
(4)f(z)≺g(z),(z∈𝕌),
if there exists a Schwarz function ω, which is analytic in 𝕌 with
(5)ω(0)=0,|ω(z)|<1,(z∈𝕌),
such that
(6)f(z)=g(ω(z)),(z∈𝕌).
Indeed, it is known that
(7)f(z)≺g(z),(z∈𝕌)⟹f(0)=g(0),f(𝕌)⊂g(𝕌).
Furthermore, if the function g is univalent in 𝕌, then we have the following equivalence:
(8)f(z)≺g(z),(z∈𝕌)⟹f(0)=g(0),f(𝕌)⊂g(𝕌).To derive our main results, we need the following lemmas.Lemma 1 (see [7]).
Let𝔭(z)=1+b1z+b2z2+⋯ be analytic in 𝕌 and let  𝔥 be analytic and starlike (with respect to the origin) univalent in 𝕌 with 𝔥(0)=0. If
(9)z𝔭′(z)≺𝔥(z),
then
(10)𝔭(z)≺1+∫0z‍𝔥(t)tdt.Lemma 2 (see [8]).
Letq be univalent in 𝕌. Also let ϕ be analytic in the domain 𝔻 containing q(𝕌) with ϕ(ω)≠0 when ω∈q(𝕌). Set
(11)Q(z)=zq′(z)ϕ(q(z)),h(z)=θ(q(z))+Q(z).
Suppose that  (1)
Q(z) is starlike univalent in 𝕌;(2)
ℜ(zh′(z)/Q(z))=ℜ((θ′(q(z))/ϕ(q(z)))+(zQ′(z)/Q(z)))>0 for z∈𝕌.
Ifp is analytic in 𝕌 with p(0)=q(0), p(𝕌)⊂𝔻 and
(12)θ(p(z))+zp′(z)ϕ(p(z))≺θ(q(z))+zq′(z)ϕ(q(z)),
then p≺q, and q is the best dominant.Lemma 3 (see [9]).
LetΩ be a set in the complex plane ℂ and suppose that Φ is a mapping from ℂ2×𝕌 to ℂ which satisfies Φ(ix,y;z)∉Ω for z∈𝕌  and for all real x,y such that y≦-(1+x2)/2.
If the functionp(z)=1+c1z+c2z2+⋯ is analytic in 𝕌 and Φ(p(z),zp′(z);z)∈Ω for all z∈𝕌, then  ℜ(p(z))>0.In this paper, we aim at proving some sufficient conditions for analytic functions to be of non-Bazilevič type.


## 2. Main Results

Our first main result is given by Theorem4.Theorem 4.
Suppose thath(z) is starlike in 𝕌 with h(0)=0. If
(13)zf′′(z)f′(z)+(1+μ)(1-zf′(z)f(z))≺h(z),(0<μ<1),
then
(14)f′(z)(zf(z))1+μ≺exp(1+∫0z‍h(t)tdt).Proof.
We define the functionp by
(15)p(z):=f′(z)(zf(z))1+μ,(z∈𝕌;0<μ<1).
Then p is analytic in 𝕌 with p(0)=1. It follows from (15) that
(16)z(log(p(z)))′=zf′′(z)f′(z)+(1+μ)(1-zf′(z)f(z)),(0<μ<1).
Combining (13) and (16), we find that
(17)z(log(p(z)))′≺h(z).
By Lemma 1, we deduce that
(18)log(p(z))≺1+∫0z‍h(t)tdt.
From (15) and (18), we readily get the assertion (14) of Theorem 4.Theorem 5.
Iff∈𝒜 satisfies the inequality(19)|×(f′(z)(zf(z))1+μ)-1[zf′′(z)f′(z)+(1+μ)(1-zf′(z)f(z))]×(f′(z)(zf(z))1+μ)-1|<ν,(0<μ,ν<1),
then f∈𝒩(μ,1/(1+ν)).Proof.
Suppose that the functionp is defined by (15). It follows that(20)z(1p(z))′=-[zf′′(z)f′(z)+(1+μ)(1-zf′(z)f(z))]×(f′(z)(zf(z))1+μ)-1.
Combining (19) and (20), we know that
(21)z(1p(z))′≺νz.
An application of Lemma 1 to (21) yields
(22)p(z)≺11+νz=:q(z).
By noting that
(23)ℜ(1+zq′′(z)q′(z))=ℜ(1-νz1+νz)≧1-ν1+ν>0,(0<ν<1;z∈𝕌),
which implies that the region q(𝕌) is symmetric with respect to the real axis and q is convex univalent in 𝕌  therefore, we have
(24)ℜ(q(z))≧q(1)≧0,(z∈𝕌).
Combining (15), (22), and (24), we conclude that
(25)ℜ(f′(z)(zf(z))1+μ)>11+ν,(0<ν<1;z∈𝕌).
This completes the proof of Theorem 5.Theorem 6.
Suppose thatq is convex in 𝕌 with q(0)=1. If
(26)ℜ(λq(z))>0,(z∈𝕌;λ∈ℂ),(27)[zf′′(z)f′(z)+(1+μ)(1-zf′(z)f(z))]f′(z)(zf(z))1+μ+λ(f′(z)(zf(z))1+μ)2≺zq′(z)+λq2(z),
then
(28)f′(z)(zf(z))1+μ≺q(z),
and q is the best dominant.Proof.
Suppose that the functionp is defined by (15). It follows that(29)zp′(z)+λp2(z)=[zf′′(z)f′(z)+(1+μ)(1-zf′(z)f(z))]×f′(z)(zf(z))1+μ+λ(f′(z)(zf(z))1+μ)2.
We now assume that(30)θ(ω)=λω2,ϕ(ω)=1.
Obviously, θ(ω) and ϕ(ω) are analytic in the ω plane. By noting that the function
(31)Q(z)=zp′(z)ϕ(p(z))=zp′(z)
is starlike in 𝕌 and
(32)χ(z)=θ(p(z))+Q(z)=λp2(z)+zp′(z),
it follows from (26) that (33)ℜ(zχ′(z)Q(z))=ℜ(2λp(z)+zQ′(z)Q(z))>0.
Combining (27), (29), and Lemma 2, we get the assertion of Theorem 6.Remark 7.
By taking suitableh(z) and q(z) in Theorems 4 and 6, respectively, we can get some useful consequences. Here we choose to omit the details.Theorem 8.
Iff∈𝒜 satisfies the condition(34)f1+μ(z)zμf′(z)(f′(z)(zf(z))1+μ)′>{γ2(γ-1),(0≦γ≦12),γ-12γ,(12≦γ<1),
then f∈𝒩(μ,γ).Proof.
Suppose that(35)ψ(z):=f′(z)(z/f(z))1+μ-γ1-γ,(0≦γ<1;z∈𝕌).
Then ψ is analytic in 𝕌. It follows from (35) that
(36)f1+μ(z)zμf′(z)(f′(z)(zf(z))1+μ)′=(1-γ)zψ′(z)γ+(1-γ)ψ(z)=Φ(ψ(z),zψ′(z);z),
where
(37)Φ(r,s;t)=(1-γ)sγ+(1-γ)r.
For all real x and y satisfying y≦-(1+x2)/2, we have
(38)ℜ(Φ(ix,y;z))=(1-γ)γyγ2+(1-γ)2x2≦-(1-γ)γ2·1+x2γ2+(1-γ)2x2≦{-(1-γ)γ2·1(1-γ)2,(0≦γ≦12),-(1-γ)γ2·1γ2,(12≦γ<1).
We now put(39)Ω={ξ:ℜ(ξ)>{γ2(γ-1)(0≦γ≦12)γ-12γ(12≦γ<1)}.
Then Φ(ix,y;z)∉Ω for all real x,y such that y≦-(1+x2)/2. Moreover, in view of (34), we know that Φ(ψ(z),zψ′(z);z)∈Ω. Thus, by Lemma 3, we deduce that
(40)ℜ(ψ(z))>0,(z∈𝕌),
which shows that the desired assertion of Theorem 8 holds.



---

*Source: 154912-2013-10-31.xml*