# On Certain Subclasses of Analytic Functions Defined by Differential Subordination

Authors: Hesam Mahzoon
Journal: International Journal of Mathematics and Mathematical Sciences
 (2011)

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## Abstract

We introduce and study certain subclasses of analytic functions which are defined by differential subordination. Coefficient inequalities, some properties of neighborhoods, distortion and covering theorems, radius of starlikeness, and convexity for these subclasses are given.


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## Body

## 1. Introduction

Let𝒯(j) be the class of analytic functions f of the form(1.1)f(z)=z-∑k=j+1∞akzk,(ak≥0,j∈N={1,2,…}),
defined in the open unit disc 𝒰={z∈ℂ:|z|<1}.LetΩ be the class of functions ω analytic in 𝒰 such that ω(0)=0, |ω(z)|<1.For any two functionsf and g in 𝒯(j), f is said to be subordinate to g that is denoted f≺g, if there exists an analytic function ω∈Ω such that f(z)=g(ω(z)) [1].Definition 1.1 (see [2]).
Forn∈ℕ and λ≥0, the Al-Oboudi operator Dλn:𝒯(j)→𝒯(j) is defined as Dλ0f(z)=f(z), Dλ1f(z)=(1-λ)f(z)+λzf′(z)=Dλf(z), and Dλnf(z)=Dλ(Dλn-1f(z)).Forλ=1, we get Sǎlǎgean differential operator [3].Further, iff(z)=z-∑k=j+1∞akzk, then(1.2)Dλnf(z)=z-∑k=j+1∞[1+(k-1)λ]nakzk(ak≥0).
For any function f∈𝒯(j) and δ≥0, the (j,δ)-neighborhood of f is defined as(1.3)Nj,δ(f)={g(z)=z-∑k=j+1∞bkzk∈T(j):∑k=j+1∞k|ak-bk|≤δ}.
In particular, for the identity function e(z)=z, we see that(1.4)Nj,δ(e)={g(z)=z-∑k=j+1∞bkzk∈T(j):∑k=j+1∞k|bk|≤δ}.
The concept of neighborhoods was first introduced by Goodman [4] and then generalized by Ruscheweyh [5].Definition 1.2.
A functionf∈𝒯(j) is said to be in the class 𝒯j(n,m,A,B,λ) if
(1.5)Dλn+mf(z)Dλnf(z)≺1+Az1+Bz,z∈U,
where n∈ℕ0, m∈ℕ, λ≥1, and -1≤B<A≤1.We observe that𝒯j(n,m,1-2α,-1,1)≡𝒯j(n,m,α)  [6],  𝒯j(0,1,1-2α,-1,1)≡𝒮j⋆(α)  [7], the class of starlike functions of order α and 𝒯j(1,1,1-2α,-1,1)≡𝒞j(α)  [7], the class of convex functions of order α.


## 2. Neighborhoods for the Class𝒯j(n,m,A,B,λ)

Theorem 2.1.
A functionf∈𝒯(j) belongs to the class 𝒯j(n,m,A,B,λ) if and only if
(2.1)∑k=j+1∞[1+(k-1)λ]n{(1-B)[1+(k-1)λ]m-(1-A)}ak≤A-B
for n∈ℕ0, m∈ℕ, λ≥1, and -1≤B<A≤1.Proof.
Letf∈𝒯j(n,m,A,B,λ). Then,
(2.2)Dλn+mf(z)Dλnf(z)≺1+Az1+Bz,z∈U.
Therefore,
(2.3)ω(z)=Dλnf(z)-Dλn+mf(z)BDλn+mf(z)-ADλnf(z).
Hence,
(2.4)|ω(z)|=|Dλnf(z)-Dλn+mf(z)BDλn+mf(z)-ADλnf(z)|=|∑k=j+1∞[1+(k-1)λ]n{[1+(k-1)λ]m-1}akzk(A-B)z+∑k=j+1∞[1+(k-1)λ]n{B[1+(k-1)λ]m-A}akzk|<1.
Thus,
(2.5)R{∑k=j+1∞[1+(k-1)λ]n{[1+(k-1)λ]m-1}akzk(A-B)z+∑k=j+1∞[1+(k-1)λ]n{B[1+(k-1)λ]m-A}akzk}<1.
Taking |z|=r, for sufficiently small r with 0<r<1, the denominator of (2.5) is positive and so it is positive for all r with 0<r<1, since ω(z) is analytic for |z|<1. Then, inequality (2.5) yields
(2.6)∑k=j+1∞[1+(k-1)λ]n{[1+(k-1)λ]m-1}akrk<(A-B)r+B∑k=j+1∞[1+(k-1)λ]n+makrk-A∑k=j+1∞[1+(k-1)λ]nakrk.
Equivalently,
(2.7)∑k=j+1∞[1+(k-1)λ]n{(1-B)[1+(k-1)λ]m-(1-A)}akrk≤(A-B)r,
and (2.1) follows upon letting r→1.
Conversely, for|z|=r, 0<r<1, we have rk<r. That is,
(2.8)∑k=j+1∞[1+(k-1)λ]n{(1-B)[1+(k-1)λ]m-(1-A)}akrk≤∑k=j+1∞[1+(k-1)λ]n{(1-B)[1+(k-1)λ]m-(1-A)}akr≤(A-B)r.
From (2.1), we have
(2.9)|∑k=j+1∞[1+(k-1)λ]n{[1+(k-1)λ]m-1}akzk|≤∑k=j+1∞[1+(k-1)λ]n{[1+(k-1)λ]m-1}akrk<(A-B)r+∑k=j+1∞{B[1+(k-1)λ]m-A}[1+(k-1)λ]nakrk<|(A-B)z+∑k=j+1∞{B[1+(k-1)λ]m-A}[1+(k-1)λ]nakzk|.
This proves that
(2.10)Dλn+mf(z)Dλnf(z)≺1+Az1+Bz,z∈U,
and hence f∈𝒯j(n,m,A,B,λ).Theorem 2.2.
If(2.11)δ=(A-B)(1+λj)n-1[(1-B)(1+λj)m-(1-A)],
then 𝒯j(n,m,A,B,λ)⊂Nj,δ(e).Proof.
It follows from (2.1) that if f∈𝒯j(n,m,A,B,λ), then
(2.12)(1+λj)n-1[(1-B)(1+λj)m-(1-A)]∑k=j+1∞kak≤(A-B),
which implies
(2.13)∑k=j+1∞kak≤(A-B)(1+λj)n-1[(1-B)(1+λj)m-(1-A)]=δ.
Using (1.4), we get the result.


## 3. Neighborhoods for the Classesℛj(n,A,B,λ) and 𝒫j(n,A,B,λ)

Definition 3.1.
A functionf∈𝒯(j) is said to be in the class ℛj(n,A,B,λ) if it satisfies
(3.1)(Dλnf(z))′≺1+Az1+Bz(z∈U),
where -1≤B<A≤1, λ≥1 and n∈𝒩0.Definition 3.2.
A functionf∈𝒯(j) is said to be in the class 𝒫j(n,A,B,λ) if it satisfies
(3.2)Dλnf(z)z≺1+Az1+Bz(z∈U),
where -1≤B<A≤1, λ≥1 and n∈𝒩0.Lemma 3.3.
A functionf∈𝒯(j) belongs to the class ℛj(n,A,B,λ) if and only if
(3.3)∑k=j+1∞(1-B)[1+(k-1)λ]n+1ak≤A-B.Lemma 3.4.
A functionf∈𝒯(j) belongs to the class 𝒫j(n,A,B,λ) if and only if
(3.4)∑k=j+1∞(1-B)[1+(k-1)λ]nak≤A-B.Theorem 3.5.
ℛj(n,A,B,λ)⊂𝒩j,δ(e), where
(3.5)δ=(A-B)[1+λj]n(1-B).Proof.
Iff∈ℛj(n,A,B,λ), we have
(3.6)[1+λj]n∑k=j+1∞(1-B)kak≤A-B,
which implies
(3.7)∑k=j+1∞kak≤(A-B)[1+λj]n(1-B)=δ.Theorem 3.6.
𝒫j(n,A,B,λ)⊂𝒩j,δ(e), where
(3.8)δ=(A-B)[1+λj]n-1(1-B).Proof.
Iff∈𝒫j(n,A,B,λ), we have
(3.9)[1+λj]n-1∑k=j+1∞(1-B)kak≤A-B,
which implies
(3.10)∑k=j+1∞kak≤(A-B)[1+λj]n-1(1-B)=δ.
Thus, in view of condition (1.4), we get the required result of Theorem 3.6.


## 4. Neighborhood of the Class𝒦jλ(n,m,A,B,C,D)

Definition 4.1.
A functionf∈𝒯(j) is said to be in the class 𝒦jλ(n,m,A,B,C,D) if it satisfies
(4.1)|f(z)g(z)-1|<A-B1-B,z∈U,
for -1≤B<A≤1, -1≤D<C≤1, λ≥1 and g∈𝒯j(n,m,C,D,λ).Theorem 4.2.
Forg∈𝒯j(n,m,C,D,λ), one has 𝒩j,δ(g)⊂𝒦jλ(n,m,A,B,C,D) and
(4.2)1-A1-B=1-[1+λj]n-1[(1-D)[1+λj]m-(1-C)]δ[1+λj]n[(1-D)[1+λj]m-(1-C)]-(C-D),
where
(4.3)δ≤(1-D)(1+λj)-(C-D)[1+λj]1-n{(1-D)[1+λj]m-(1-C)}-1.Proof.
Letf∈𝒩j,δ(g) for g∈𝒯j(n,m,C,D,λ). Then,
(4.4)∑k=j+1∞k|ak-bk|≤δ,∑k=j+1∞bk≤C-D[1+λj]n[(1-D)[1+λj]m-(1-C)].
Consider
(4.5)|f(z)g(z)-1|≤∑k=j+1∞|ak-bk|1-∑k=j+1∞bk≤δ(1+λj)[1+λj]n{(1-D)[1+λj]m-(1-C)}[1+λj]n{(1-D)[1+λj]m-(1-C)}-(C-D)=[1+λj]n-1{(1-D)[1+λj]m-(1-C)}δ[1+λj]n{(1-D)[1+λj]m-(1-C)}-(C-D)=A-B1-B.
This implies that f∈𝒦jλ(n,m,A,B,C,D).


## 5. Distortion and Covering Theorems

Theorem 5.1.
Iff∈𝒯j(n,m,A,B,λ), then
(5.1)r-A-B(1+jλ)n{(1-B)(1+jλ)m-(1-A)}rj+1≤|f(z)|≤r+A-B(1+jλ)n{(1-B)(1+jλ)m-(1-A)}rj+1(0<|z|=r<1),
with equality for
(5.2)f(z)=z-A-B(1+jλ)n{(1-B)(1+jλ)m-(1-A)}rj+1(z=±r).Proof.
In view of Theorem2.1, we have
(5.3)(1+jλ)n{(1-B)(1+jλ)m-(1-A)}∑k=j+1∞ak≤∑k=j+1∞[1+(k-1)λ]n{(1-B)[1+(k-1)λ]m-(1-A)}ak≤A-B.
Hence,
(5.4)|f(z)|≤r+∑k=j+1∞akrk≤r+rj+1∑k=j+1∞ak≤r+A-B(1+jλ)n{(1-B)(1+jλ)m-(1-A)}rj+1,|f(z)|≥r-∑k=j+1∞akrk≥r-rj+1∑k=j+1∞ak≥r-A-B(1+jλ)n{(1-B)(1+jλ)m-(1-A)}rj+1.
This completes the proof.Theorem 5.2.
Any functionf∈𝒯j(n,m,A,B,λ) maps the disk |z|<1 onto a domain that contains the disk
(5.5)|w|<1-A-B(1+jλ)n{(1-B)(1+jλ)m-(1-A)}.Proof.
The proof follows upon lettingr→1 in Theorem 5.1.Theorem 5.3.
Iff∈𝒯j(n,m,A,B,λ), then
(5.6)1-(A-B)(1+jλ)n-1{(1-B)(1+jλ)m-(1-A)}rj≤|f′(z)|≤1+A-B(1+jλ)n-1{(1-B)(1+jλ)m-(1-A)}rj(0<|z|=r<1),
with equality for
(5.7)f(z)=z-A-B(1+jλ)n-1{(1-B)(1+jλ)m-(1-A)}zj+1(z=±r).Proof.
We have(5.8)|f′(z)|≤1+∑k=j+1∞kak|z|k-1≤1+rj∑k=j+1∞kak.
In view of Theorem 2.1,
(5.9)∑k=j+1∞kak≤A-B(1+jλ)n-1{(1-B)(1+jλ)m-(1-A)}.
Thus,
(5.10)|f′(z)|≤1+A-B(1+jλ)n-1{(1-B)(1+jλ)m-(1-A)}rj.
On the other hand,
(5.11)|f′(z)|≥1-∑k=j+1∞kak|z|k-1≥1-rj∑k=j+1∞kak≥1-A-B(1+jλ)n-1{(1-B)(1+jλ)m-(1-A)}rj.
This completes the proof.


## 6. Radii of Starlikeness and Convexity

In this section, we find the radius of starlikeness of orderρ and the radius of convexity of order ρ for functions in the class 𝒯j(n,m,A,B,λ).Theorem 6.1.
Iff∈𝒯j(n,m,A,B,λ), then f is starlike of order ρ, (0≤ρ<1) in |z|<r1(n,m,A,B,λ,ρ), where
(6.1)r1(n,m,A,B,λ,ρ)=inf⁡k[[1+(k-1)λ]n{(1-B)[1+(k-1)λ]m-(1-A)}(1-ρ)(k-ρ)(A-B)]1/(k-1).Proof.
It is sufficient to show that|z(f′(z)/f(z))-1|≤1-ρ(0≤ρ<1) for |z|<r1(n,m,A,B,λ,ρ).
We have(6.2)|zf′(z)f(z)-1|≤∑k=j+1∞(k-1)ak|z|k-11-∑k=j+1∞ak|z|k-1.
Thus, |z(f′(z)/f(z))-1|≤1-ρ if
(6.3)∑k=j+1∞(k-ρ)ak|z|k-1(1-ρ)≤1.
Hence, by Theorem 2.1, (6.3) will be true if
(6.4)(k-ρ)|z|k-1(1-ρ)≤[1+(k-1)λ]n{(1-B)[1+(k-1)λ]m-(1-A)}(A-B)
or if
(6.5)|z|≤[[1+(k-1)λ]n{(1-B)[1+(k-1)λ]m-(1-A)}(1-ρ)(k-ρ)(A-B)]1/(k-1)(k≥j+1).
This completes the proof.Theorem 6.2.
Iff∈𝒯j(n,m,A,B,λ), then f is convex of order ρ, (0≤ρ<1) in |z|<r2(n,m,A,B,λ,ρ), where
(6.6)r2(n,m,A,B,λ,ρ)=inf⁡k[[1+(k-1)λ]n{(1-B)[1+(k-1)λ]m-(1-A)}(1-ρ)(k-ρ)(A-B)]1/(k-1).Proof.
It is sufficient to show that|z(f′′(z)/f′(z))|≤1-ρ(0≤ρ<1) for |z|<r1(n,m,A,B,λ,ρ).
We have(6.7)|zf′′(z)f′(z)|≤∑k=j+1∞k(k-1)ak|z|k-11-∑k=j+1∞kak|z|k-1.
Thus, |z(f′′(z)/f′(z))|≤1-ρ if
(6.8)∑k=j+1∞k(k-ρ)ak|z|k-1(1-ρ)≤1.
Hence, by Theorem 2.1, (6.8) will be true if
(6.9)k(k-ρ)|z|k-1(1-ρ)≤[1+(k-1)λ]n{(1-B)[1+(k-1)λ]m-(1-A)}(A-B)
or if
(6.10)|z|≤[[1+(k-1)λ]n{(1-B)[1+(k-1)λ]m-(1-A)}(1-ρ)k(k-ρ)(A-B)]1/(k-1)(k≥j+1).
This completes the proof.



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