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Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2012, Article ID 383592, 10 pagesdoi:10.1155/2012/383592
Research Articlen-Bazilevic Functions
F. M. Al-Oboudi
Department of Mathematical Sciences, College of Science, Princess Nora bint Abdul Rahman University,P.O. Box 4384, Riyadh 11491, Saudi Arabia
Correspondence should be addressed to F. M. Al-Oboudi, [email protected]
Received 31 December 2011; Revised 2 March 2012; Accepted 3 March 2012
Academic Editor: Khalida Inayat Noor
Copyright q 2012 F. M. Al-Oboudi. This is an open access article distributed under the CreativeCommons Attribution License, which permits unrestricted use, distribution, and reproduction inany medium, provided the original work is properly cited.
The aim of this paper is to define and study a class of Bazilevic functions using the generalizedSalagean operator. Some properties of this class are investigated: inclusion relation, someconvolution properties, coefficient bounds, and other interesting results.
1. Introduction
Let H be the set of analytic functions in the open unit disc E = {z : |z| < 1}. Let A be theset of functions f ∈ H, with f(z) = z +
∑∞k=2 akz
k, and let A0 be the set of functions f ∈ H,with f(0) = 1. Let S be the class of functions f ∈ A, which are univalent in E. Denote byST(γ)(CV (γ))(K(γ)), γ < 1, the class of starlike (convex)(close-to-convex) functions of orderγ . Note that when 0 ≤ γ < 1, then ST(γ)(CV (γ))(K(γ)) ⊂ S, let ST(0)(CV (0))(K(0)) ≡ST(CV )(K). A function F ∈ A0, where F /= 0 belongs to the Kaplan class K(α, β), α ≥ 0, β ≥ 0,[1] if
−απ +12(α − β
)(θ2 − θ1) ≤ argF
(reiθ2
)− argF
(reiθ1
)≤ 1
2(α − β
)(θ2 − θ1) + βπ, (1.1)
for θ1 < θ2 < θ1 + 2π and 0 < r < 1.The Dual of ν ⊂ A0 is defined as
ν� ={g ∈ A0 : f ∗ g /= 0 in Δ, f ∈ ν
}, (1.2)
where ∗ denotes Hadamard product (convolution).
2 Abstract and Applied Analysis
A set ν ⊂ A0 is called a test set for u ⊂ A (denoted by ν � u) if ν ⊂ u ⊂ ν��. Note thatif ν � u, then ν� ⊂ u�.
Denote by P the class of functions p ∈ A0, such that Re p > 0, in E, and let Pα = {fα ∈A0, f ∈ P}. Note that for 0 ≤ α ≤ β,
K(α, β)= Pα ·K(0, β − α
), (1.3)
and that f ∈ ST(α), α < 1, if, and only if, f/z ∈ K(0, 2 − 2α).For α ≥ 0 and β ≥ 0, define the class T(α, β) as
T(α, β)=
{(1 + xz)[α]
(1 + yz
)α−[α]
(1 + uz)β: |x| = ∣∣y∣∣ = |u| = 1
}
. (1.4)
Note that T(α, β) � K(α, β), α ≥ 1, β ≥ 1.A function f ∈ A0, is called prestarlike of order α, α ≤ 1, (denoted by R(α)) if and only
if f/z ∈ T(0, 3 − 2α)�, or f ∈ R(α) if and only if
f ∗ z
(1 − z)2−2α∈ ST(α) α < 1,
Ref
z>
12, z ∈ E α = 1.
(1.5)
Let B(α, β), α > 0, β ∈ R, denote the class of Bazilevic functions in E, introduced byBazilevic [2], f ∈ B(α, β), α > 0, β ∈ R, if and only if there exists a g ∈ ST(1 − α), such that forz ∈ E
zf ′
g
(f
z
)α+iβ−1∈ P, (1.6)
where (f/z)α+iβ = 1 at z = 0. We denote B(α, 0) by B(α). Bazilevic shows that B(α, β) ⊂ S, forα > 0, β ∈ R. Note that
CV ⊂ ST ⊂ K ⊂ B(α, β). (1.7)
For further information, see [3–7].The generalized Salagean operator Dn
λf : A −→ A, n ∈ N0 = N ∪ {0}, λ ≥ 0, is defined
[8] as
Dnλf(z) = hn,λ(z) ∗ f(z), (1.8)
Abstract and Applied Analysis 3
where
hn,λ(z) = hλ(z) ∗ · · · ∗ hλ(z)︸ ︷︷ ︸
(n-times)
,
hλ(z) =z − (1 − λ)z2
(1 − z)2= z +
∞∑
k=2
[1 + λ(k − 1)]zk.
(1.9)
The Operator Dnλf satisfies the following identity:
Dn+1λ f(z) = (1 − λ)Dn
λf(z) + λz(Dn
λf(z))′. (1.10)
Not that for λ = 1, Dn1f(z) ≡ Dnf(z), Salagean differential operator [9].
Let
kn,λ(z) = h(−1)n,λ (z) = z +
∞∑
k=2
zk
[1 + λ(k − 1)]n, λ > 0, (1.11)
we mean by f (−1), the solution of f ∗ f (−1) = z/(1 − z). Hence
kn,λ(z) = kλ(z) ∗ · · · ∗ kλ(z)︸ ︷︷ ︸
(n-times)
, (1.12)
where kλ(z) = h(−1)λ
(z). It is known [10] that kλ/z ∈ T(1, 1+ 1/λ)∗, hence kλ ∈ R(1− 1/λ), andthat
kλ(z) ∗ z
(1 − z)(1/λ)+1=
z
(1 − z)1/λ. (1.13)
The class STn(γ), γ ≤ 1 is defined as f ∈ STn(γ) if and only if Dnλf ∈ ST(γ). For λ = 1,
we get Salagean-type n-starlike functions [9].The operator Dn
λf is now called “Al-Oboudi Operator” and has been extensivelystudied latly, [5, 11, 12].
In this paper we define and study a class of Bazilevic functions using the operatorDnλf
and study some of its basic properties, inclusion relation, convolution properties coefficientbounds, and other interesting results.
2. Definition and Preliminaries
In this section, the class of n-Bazilevic functions Bnλ , λ > 0, where B0
λ ≡ B(1/λ, 0), is definedand some preliminary lemmas are given.
4 Abstract and Applied Analysis
2.1. Definition
Let f ∈ A. Then f ∈ Bnλ, n ∈ N0, λ > 0, if and only if there exists a g ∈ STn(1 − 1/λ), such that
Dn+1λ z(f/z)1/λ
Dnλg(z)
∈ P, z ∈ E, (2.1)
where the power (f/z)1/λ is chosen as a principal one.Denote by Bn
1λ the class of functions f ∈ Bnλ, where g ≡ z.
Using (1.10), we see that Dn+1λ z(f/z)1/λ = f ′(z)Dn
λz(f/z)1/λ−1, from which the
following special cases are clear.
2.1.1. Special Cases
(1) For n = 0, B01/α ≡ B(α), α > 0, Bazilevic [2].
(2) For λ = 1, Bn1 ≡ Kn(0), Salagean type close to convex functions, Blezu [13].
(3) For n = 0, λ = 1, B01 ≡ K, Kaplan [14].
(4) For λ = 1, Bn11 ≡ Bn+1(1), Abdul Halim [15] and Bn
11 ≡ T1n+1(0), Opoola [16].
2.2. Lemmas
The following lemmas are needed to prove our results.
Lemma 2.1 (see [10]). Let α, β ≥ 1 and f ∈ T(α, β)∗, g ∈ K(α − 1, β − 1). Then for F ∈ A
(ϕ ∗ g F
)
(ϕ ∗ g) (E) ⊂ CO (F(E)). (2.2)
Lemma 2.2. If Dn+1λ f ∈ ST(1 − 1/λ), λ > 0, then Dn
λf ∈ ST(1 − 1/λ).
Proof. Since Dnλf = kλ ∗Dn+1
λf , we will show that (kλ ∗Dn+1
λf) ∈ ST(1 − 1/λ). Now Dn+1
λf ∈
ST(1 − 1/λ) ⊂ ST(1/2 − 1/2λ), implies
(z
(1 − z)1/λ+1
)(−1)∗Dn+1
λ f(z) ∈ R
(12− 12λ
)
⊂ R
(
1 − 12λ
)
. (2.3)
Hence
(z
(1 − z)1/λ
)(z
(1 − z)1/λ+1
)(−1)∗Dn+1
λ f(z) ∈ ST
(
1 − 12λ
)
⊂ ST
(
1 − 1λ
)
. (2.4)
From (1.13), we get the required result.
Abstract and Applied Analysis 5
Lemma 2.3 (see [1]). Let α, β ≥ 1 and f ∈ T(α, β)∗, g ∈ K(α, β). Then f ∗ g ∈ K(α, β).
For X ⊂ A, let rc(X) denote the largest positive number so that every f ∈ X is convexin |z| < rc(X). The following result is due to Al-Amiri [17].
Lemma 2.4. One has
rc(hλ) = rc =1
1 + |c| +√
1 + |c| + |c|2, c = 2λ − 1, 0 < λ ≤ 1. (2.5)
Lemma 2.5 (see [18]). Let f, g ∈ H, with f(0) = g(0) = 0 and f ′(0)g ′(0)/= 0. Let ϕ ∈ ν� in|z| < r < 1, where
ν ={1 + xz
1 + yzg(z) : |x| = ∣∣y∣∣ = 1
}
. (2.6)
Then for each F ∈ H,
(ϕ ∗ Fg)(ϕ ∗ g) (|z| < r) ⊂ CO (F(E)), (2.7)
where CO stands for closed convex hull.
Remark 2.6. In [1], it was shown that condition (2.6) is satisfied for all z in E whenever ϕ is inCV and g is in ST .
Lemma 2.7 (see [10]). Let α, β, γ, δ, μ, ν ∈ R be such that
0 ≤ γ ≤ α − 1, 0 ≤ δ ≤ β − 1, 0 ≤ μ ≤ α − γ, 0 ≤ γ ≤ β − δ, (2.8)
and let f ∈ T(α, β)∗, g ∈ K(γ, δ), F ∈ K(μ, ν). Then
(f ∗ gF)(f ∗ g) ∈ Pmax{μ,ν}. (2.9)
From (1.12) and (1.13), we immediately have;
Lemma 2.8. One has
kn+1,λ =z
(1 − z)(1/λ)−n∗(
z
(1 − z)(1/λ)+1
)(−1), (2.10)
6 Abstract and Applied Analysis
Lemma 2.9 (see [10]). Let f ∈ K(α, β), α, β ≥ 1. Then
f � (1 + z)α
(1 − z)β, (2.11)
where� stands for coefficient majorization.
3. Main Results
Theorem 3.1. One has
Bn+1λ ⊂ Bn
λ, λ > 0. (3.1)
Proof. Let f ∈ Bn+1λ . Then there exists g ∈ STn+1(1 − 1/λ), such that
Dn+2λ z
(f
z
)1/λ
= Dn+1λ g(z) · p(z), p ∈ K(1, 1). (3.2)
Hence
(f
z
)1/λ
=kn+1,λz
∗ kλz
∗ Dn+1λ g
z· p, p ∈ K(1, 1). (3.3)
Since Dn+1λ
g/z ∈ K(0, 2/λ), and kλ/z ∈ T(1, 1 + 1/λ)∗, application of Lemma 2.3 gives
kλz
∗ Dn+1λ
g
z· p =
(kλz
∗ Dn+1λ
g
z
)
p0, p0 ∈ K(1, 1) (3.4)
hence
(f
z
)1/λ
=kn+1,λz
∗ Dnλg
zp0, p0 ∈ K(1, 1) (3.5)
Using Lemma 2.2 we deduce that f ∈ Bnλ .
As a consequence of (3.1) we immediately have the following.
Abstract and Applied Analysis 7
Corollary 3.2. One has
Bnλ ⊂ S. (3.6)
Corollary 3.3. If f ∈ Bnλ, n ∈ N0, λ > 0, then, for z ∈ E
(f
z
)1/λ
∈ K
(
1, 1 +2λ
)
. (3.7)
Proof. Since f ∈ Bnλ , there exists a g ∈ STn(1 − 1/λ) or Dn
λg/z ∈ K(0, 2/λ) such that
(f
z
)1/λ
=kn+1,λz
∗ Dnλg
z· p, p ∈ K(1, 1), (3.8)
=kn+1,λz
∗ F, F ∈ K
(
1, 1 +2λ
)
, (3.9)
using (1.3). From (3.6), we conclude that
(f
z
)1/λ
=kn+1,λz
∗ F /= 0, 0 < |z| < 1, (3.10)
which implies that
kn+1,λz
∈ K
(
1, 1 +2λ
)∗≡ T
(
1, 1 +2λ
)∗, (3.11)
Applying Lemma 2.3 to (3.9), we get the result.
Theorem 3.4. Let f ∈ Bnλ . Then
⎛
⎝Dn+1
λz(f/z)1/λ
z
⎞
⎠
λ
∈ K(λ,λ + 2). (3.12)
Proof. From (2.1), we see that
Dn+1λ z
(f
z
)1/λ
= Dnλg(z) · p(z), p ∈ K(1, 1). (3.13)
8 Abstract and Applied Analysis
Since (Dnλg/z)
λ ∈ K(0, 2), and p(z)λ ∈ K(λ, λ), then
⎛
⎝Dn+1
λz(f/z)1/λ
z
⎞
⎠
λ
=
(Dn
λg
z
)λ
p(z)λ ∈ K(λ,λ + 2), (3.14)
which is the required result.
In the following we prove the converse of Theorem 3.1, for 0 < λ ≤ 1.
Theorem 3.5. Let f ∈ Bnλ , 0 < λ ≤ 1. Then f ∈ Bn+1
λ in |z| < rc, where rc is given by (2.5)
Proof. f ∈ Bnλimplies (2.1), where Dn
λg ∈ ST(1 − 1/λ) ⊂ ST .
Now
Dn+2λ
z(f/z)1/λ
Dn+1λ
g=
hλ ∗Dn+1λ
z(f/z)1/λ
hλ ∗Dnλg
. (3.15)
Using Lemma 2.4, we see that hλ ∈ CV in |z| < rc, for 0 < λ ≤ 1, where rc is given by (2.5).From Remark 2.6, we conclude
hλ ∗{1 + xz
1 + yzDn
λg : |x| = ∣∣y∣∣ = 1}
/= 0. (3.16)
Applying Lemma 2.5, we deduce
⎛
⎜⎝
hλ ∗(Dn+1
λ z(f/z)1/λ
/Dnλg)Dn
λg
hλ ∗Dnλg
⎞
⎟⎠(|z| < rc) ⊂ CO
⎛
⎝Dn+1
λz(f/z)1/λ
Dnλg
⎞
⎠(E), (3.17)
hence Dn+2λ
z(f/z)1/λ/Dn+1λ
g ∈ P in |z| < rc, as required.
Corollary 3.3 can be improved for 0 < λ ≤ 1, as follows.
Theorem 3.6. Let f ∈ Bnλ , 0 < λ ≤ 1. Then
f
z∈ K(1, 2). (3.18)
Proof. Wewill use Ruscheweyh’s.method of proof [10]. f ∈ Bnλimplies (3.8), whereDn+1
λg/z ∈
K(0, 2/λ), kn+1,λ/z ∈ T(1, 1 + 2/λ)∗.Let Dn
λg/z = l ·m, where l = (Dn
λg/z)(1+λ)/2 and m = (Dn
λg/z)(1−λ)/2.
Then l ∈ K(0, 1/λ + 1) ·m ∈ K(0, 1/λ − 1) andm · p = F ∈ K(1, 1/λ). This implies
(f
z
)1/λ
=kn+1,λz
∗ lF. (3.19)
Abstract and Applied Analysis 9
Using Lemma 2.7, we get
(kn+1,λ/z) ∗ lF(kn+1,λ/z) ∗ l = F0 ∈ K
(1λ,1λ
)
, 0 < λ ≤ 1. (3.20)
Hence
(f
z
)1/λ
=(kn+1,λz
∗ l)
F0, F0 ∈ K
(1λ,1λ
). (3.21)
To prove that (f/z)1/λ ∈ K(1/λ, 2/λ), we have to show that ((kn+1,λ/z) ∗ l) ∈ K(0, 1/λ), orequivalently kn+1,λ ∗ z l ∈ ST(1 − 1/2λ).
Since z l ∈ ST((λ − 1)/2λ), then from (1.5)
(z
(1 − z)1/λ+1
)(−1)∗ z l ∈ R
(λ − 12λ
)
⊂ R
((n + 2)λ − 1
2λ
)
,
z
(1 − z)1/λ−n∗(
z
(1 − z)1/λ+1
)(−1)∗ z l ∈ ST
((n + 2)λ − 1
2λ
)
⊂ ST
(
1 − 12λ
)
.
(3.22)
From Lemma 2.8, (1.13), and (3.22), we see that ((kn+1,λ/z) ∗ l) ∈ K(0, 1/λ). Using (3.21), weobtain (f/z)1/λ ∈ K(1/λ, 2/λ). From (1.1) we get the required result.
Remark 3.7. For n = 0, λ = 1/α Theorem 3.6 and other stronger results depending on α, areproved by Sheil-Small [7].
For the coefficient bounds of f ∈ Bnλ, Theorem 3.6 is not strong enough to settle this
problem for 0 < λ < 1, In 1962, Zamorski [19] proved the Bieberbach conjecture for f ∈ B(α),when α = 1, 1/2, 1/3, . . ., in the following we prove this result for f ∈ Bn
λ , using the extremepoints of Kaplan class K(α, β).
Theorem 3.8. For f ∈ Bnλ, λ = m ∈ N,
f � z
(1 − z)2−mn. (3.23)
Proof. From (3.9) and Lemma 2.9, we get
(f
z
)1/λ
� kn+1,λz
∗ 1 + z
(1 − z)1+2/λ,
=1
(1 − z)2/λ−n,
(3.24)
using (2.10). Raising both sides of (3.24) to the mth power, where λ = m ∈ N, we get therequired result.
10 Abstract and Applied Analysis
Remark 3.9. For n = 0, we get the result of Zamorski [19], and the result of Sheil-Small [7],from which we get the idea of proof.
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