Hindawi Publishing Corporation
International Journal of Mathematics and Mathematical Sciences
Volume 2007, Article ID 92439, 9 pages
doi:10.1155/2007/92439
Research Article
On Some Analytic Functions Defined by
a Multiplier Transformation
Khalida Inayat Noor
Received 26 July 2007; Accepted 19 November 2007
Recommended by Heinrich Begehr
We introduce and study a new class of analytic functions defined in the unit disc using a
certain multiplier transformation. Some inclusion results and other interesting properties
of this class are investigated.
Copyright © 2007 Khalida Inayat Noor. This is an open access article distributed under
the Creative Commons Attribution License, which permits unrestricted use, distribution,
and reproduction in any medium, provided the original work is properly cited.
1. Introduction
Let Pk (η) be the class of functions p(z) analytic in the unit disc E = {z : |z| < 1} satisfying
the properties p(0) = 1 and
2π Rep(z) − η 1 − η dθ ≤ kπ,
0
(1.1)
where z = reiθ , k ≥ 2, 0 ≤ η < 1. For η = 0, we obtain the class Pk defined by Pinchuk [1],
and for η = 0, k = 2, we have the class P of functions with positive real part, whereas
P2 (η) = P(η) is the class of functions with positive real part greater than η. We can write
(1.1) as
p(z) =
1
2
2π
0
1 + (1 − 2η)ze−it
dμ(t),
1 − ze−it
(1.2)
where μ(t) is a function with bounded variation on [0,2π] such that
2π
0
dμ(t) = 2,
2π
0
dμ(t) ≤ k.
(1.3)
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International Journal of Mathematics and Mathematical Sciences
We can also write (1.1), for p ∈ Pk (η) in E, if and only if
p(z) =
k 1
k 1
p1 (z) −
−
p2 (z),
+
4 2
4 2
p1 , p2 ∈ P(η).
(1.4)
It is known [2] that the class Pk (η) is a convex set. Let A be the class of functions f ,
defined by
f (z) = z +
∞
am z m ,
(1.5)
m=2
which are analytic in E. By S, K, S∗ , and C, we denote the subclasses of A which are
univalent, close-to-convex, starlike, and convex in E, respectively. The class A is closed
under the Hadamard product or convolution:
( f ∗g)(z) =
∞
am bm zm+1 ,
(1.6)
m=0
where
∞
f (z) =
am zm+1 ,
g(z) =
m=0
∞
bm zm+1 .
(1.7)
m=0
We define the following.
Definition 1.1. Let f ∈ A. Then, for α,β ≥ 0, 0 ≤ η < α + β ≤ 1, k ≥ 2, and z ∈ E, f ∈
Qk (α,β,η) if and only if
α f (z) + β z f (z)
∈ Pk (η).
(1.8)
We note that, for β = 0 and k = 2, f ∈ P(η) ⊂ P for z ∈ E and this implies that f
is univalent in E, see [3]. For any real number s, the multiplier transformations Iλs of
functions f ∈ A are defined by
fλs (z) = Iλs f (z) = z +
∞ m+λ s
m=2
1+λ
am z m
(λ > −1).
(1.9)
It is obvious that Iλs (Iλt f (z)) = Iλs+t f (z) for all real numbers s and t. The operator Iλs has
been studied by several authors for different choices of s and λ, see [4–7]. It is worth
noting that, for s any nonnegative integer and λ = 0, the operator Iλs is the differential
operator defined by Sălăgeam [8]. Also the operator Iλs is related rather closely to the
multiplier transformation discussed by Flett [9]. Using (1.9) and convolution, function
s
is defined as follows:
fλ,μ
fλs (z)∗ fλs (z) =
z
,
(1 − z)μ
z ∈ E, μ > 0.
(1.10)
Khalida Inayat Noor 3
Motivated essentially by Choi et al. operator [10] and Noor integral operator [11–14],
s
: A→A as
Cho and Kim [15] defined the operator Iλ,μ
s
s
f (z) = fλ,μ
(z)∗ f (z),
Iλ,μ
(1.11)
0
1
where s is real, λ > −1, μ > 0, and f ∈ A. In particular, I0,2
f (z) = z f (z), I0,2
f (z) = f (z).
From (1.10) and (1.11), we have
s+1
s
s+1
z Iλ,μ
f (z) = (λ + 1)Iλ,μ
f (z) − λIλ,μ
f (z),
s
s
s
f (z) = μIλ,μ+1
f (z) − (μ − 1)Iλ,μ
f (z).
z Iλ,μ
(1.12)
(1.13)
We now define the following.
Definition 1.2. Let f ∈ A. Then, for s real, λ > 1, μ > 0,
f ∈ Qks (λ,μ,α,β,η)
s
iff Iλ,μ
f (z) ∈ Qk (α,β,η) for z ∈ E.
(1.14)
2. Preliminary results
Lemma 2.1. If h(z) is analytic in E with h(0) = 1 and if λ1 is a complex number satisfying
Re λ1 ≥ 0 (λ1 = 0), then {h(z) + λ1 zh (z)} ∈ Pk (δ), 0 ≤ δ < 1, implies h(z) ∈ Pk (δ + (1 −
δ)(2γ − 1) and
γ=
1
0
1 + t Re λ1
−1
dt,
(2.1)
where γ is an increasing function of Re λ1 and 1/2 ≤ γ < 1. The estimate is sharp.
Proof. Let h(z) = (k/4 + 1/2)h1 (z) − (k/4 − 1/2)h2 (z), h(z) is analytic in E with h(0) =
1. Then, h(z) + λ1 zh (z) = (k/4 + 1/2)[h1 (z) + λ1 zh1 (z)] − (k/4 − 1/2)[h2 (z) + λ1 zh2 (z)].
Since [h(z) + λ1 zh (z)] ∈ Pk (δ), we use (1.4) to have [hi (z) + λ1 zhi (z)] ∈ P(δ), i = 1,2. We
now apply a lemma in [16] to conclude that hi ∈ P(δ 1 ), i = 1,2, and δ 1 = δ + (1 − δ)(2γ −
1), where γ is given by (2.1) and it is an increasing function of Re λ1 with 1/2 ≤ γ < 1.
Consequently h ∈ Pk (δ 1 ) in E.
Lemma 2.2 [17]. If p(z) is analytic in E with p(0) = 1, then, for any function F, analytic in
E, the function p∗F takes values in the convex hull of image of E under F.
Lemma 2.3. Let β1 < 1. If the function p is analytic in E, with p(0) = 1, then p ∈ Pk (β2 ),
β2 = (2β1 − 1) + 2(1 − β1 )ln 2, z ∈ E. This result is sharp.
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International Journal of Mathematics and Mathematical Sciences
Proof. The proof is immediate when we use (1.4) and a similar result for the class P(β2 )
in [18].
Lemma 2.4. For η1 ≤ 1 and η2 ≤ 1,Pk (η1 )∗Pk (η2 ) ⊂ Pk (1 − 2(1 − η1 )(1 − η2 )). This result
is sharp.
Proof. Let H ∈ Pk (η1 ), p ∈ Pk (η2 ). Then, using (1.4), we can write
(H ∗ p)(z) =
k 1 k 1 H1 ∗ p1 (z) −
−
H2 ∗ p2 (z) ,
+
4 2
4 2
Hi ∈ P η 1 ,
pi ∈ P η 2 ,
(2.2)
i = 1,2.
Now using a result from [19], we have, for i = 1,2,
η = 1 − 2 1 − η1 1 − η2 .
Hi ∗ pi ∈ P(η),
(2.3)
This result is shown to be sharp in [19] and consequently (H ∗ p) ∈ Pk (η).
3. Main results
Theorem 3.1. Qks (λ,μ,α,β,η) ⊂ Qks (λ,μ,1,0,σ) for
σ = σ 1 + 1 − σ 1 2σ 2 − 1 ,
σ2 =
Proof. Let f
1
0
∈ Qks (λ,μ,α,β,η).
s
α Iλ,μ
f
1 + t β/(α+β)
−1
dt,
σ1 =
with
η
,
α+β
(3.1)
1
≤ σ 2 ≤ 1.
2
Then, by definition it follows that
s
+ β z Iλ,μ
f
∈ Pk (η),
z ∈ E.
(3.2)
s
f (z)) = p(z). Then p is analytic in E with p(0) = 1 and for z ∈ E,
Set (Iλ,μ
s
s
α Iλ,μ
f (z) + β z Iλ,μ
f (z)
α+β−η
−η
α+β
η
β
p(z) +
zp (z) −
∈ Pk .
=
α+β−η
α+β−η
α+β−η
(3.3)
From (1.4) and (3.4), we have, for i = 1,2,
α+β
η
β
pi (z) +
zpi (z) −
= hi (z) ∈ P.
α+β−η
α+β−η
α+β−η
(3.4)
By putting σ 1 = η/(α + β), we see that
pi (z) +
β
zpi (z) = 1 − σ 1 hi (z) + σ 1 = Hi (z) ∈ P(σ 1 ).
α+β
(3.5)
s
f )
Now using Lemma 2.1, we obtain pi ∈ P(σ), where σ is given by (3.1). Therefore, (Iλ,μ
s
∈ Pk (σ) and consequently f ∈ Qk (λ,μ,1,0,σ) in E.
Khalida Inayat Noor 5
Remark 3.2. By writing σ 1 = η/(α + β), α1 = α/(α + β), we can deduce from Definition 1.2
that f ∈ Qks (λ,μ,α,β,η), if and only if, for 0 ≤ α1 ≤ 1,
s
α1 Iλ,μ
f
s
+ 1 − α1 z Iλ,μ
f
z ∈ E.
∈ Pk (σ 1 ),
(3.6)
In this case, we say that f ∈ Qks (λ,μ,α1 ,σ 1 ) in E.
Theorem 3.3. Let s be real, λ > −1, μ > 0. Then,
Qks λ,μ + 1,α1 ,σ 1 ⊂ Qks λ,μ,α1 ,δ 1 ⊂ Qks+1 λ,μ,α1 ,δ 2 ,
(3.7)
where α1 and σ 1 are as defined in Remark 3.2 and
δ 1 = σ 1 + 1 − σ 1 2η1 − 1 ,
η1 =
δ 2 = δ 1 + 1 − δ 1 (2η2 − 1 ,
η2 =
1
1 + t 1/μ
0
1
−1
0
1 + t 1/(λ+1)
dt,
−1
(3.8)
dt.
(3.9)
Proof. We first show that Qks (λ,μ + 1,α1 ,σ 1 ) ⊂ Qks (λ,μ,α1 ,δ 1 ).
Let f ∈ Qks (λ,μ + 1,α1 ,σ 1 ) and set
s
p(z) = α1 Iλ,μ
f (z)
+ 1 − α1
s
zIλ,μ
f (z)
.
(3.10)
From (1.13) and (3.10), we have, for z ∈ E,
s
α1 Iλ,μ+1
f (z) + 1 − α1 ) z Iλ+μ+1 f (z)
1
= p(z) + zp (z) ∈ Pk σ 1
μ
(3.11)
and, on using (1.4), it follows that Re { pi (z) + (1/μ)zpi (z)} > σ 1 , z ∈ E, i = 1,2.
Now, applying Lemma 2.1, we have Re pi (z) > δ 1 , i = 1,2, where δ 1 is given by (3.8).
This implies p ∈ Pk (δ 1 ) for z ∈ E and hence f ∈ Qks (λ,μ,α1 ,δ 1 ) in E. To prove Qks
(λ,μ,α1 ,δ 1 ) ⊂ Qks+1 (λ,μ,α1 ,δ 2 ), we proceed as follows. Set
s+1
s+1
α1 Iλ,μ
f (z) + 1 − α1 z Iλ,μ
f (z)
= h(z).
(3.12)
Then, using (1.12), we have
s
s
α1 Iλ,μ
f (z) + 1 − α1 z Iλ,μ
f (z)
= h(z) +
1
zh (z) ∈ Pk δ 1 .
λ+1
With similar argument as detailed above, we obtain the required result.
(3.13)
Theorem 3.4. The class Qks (λ,μ,α1 ,σ 1 ) is closed under the convolution with a convex function. That is, if f ∈ Qks (λ,μ,α1 ,σ 1 ) and φ ∈ C for z ∈ E, then (φ∗ f ) ∈ Qks (λ,μ,α1 ,σ 1 ).
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International Journal of Mathematics and Mathematical Sciences
Proof. Let f ∈ Qks (λ,μ,α1 ,σ 1 ). Consider
s
s
(φ∗ f )(z) + 1 − α1 ) z Iλ,μ
(φ∗ f )(z)
α1 Iλ,μ
s
s
= α1 fλ,μ
(z)∗(φ∗ f )(z) + 1 − α1 z fλ,μ
(z)∗(φ∗ f )(z)
s
s
= α1 φ(z)∗ fλ,μ
(z)∗ f (z) + 1 − α1 z φ(z)∗ fλ,μ
(z)∗ f (z)
=
s
φ(z) s
∗ α1 Iλ,μ f (z) + 1 − α1 z Iλ,μ
f (z)
z
=
k 1
+
4 2
φ(z)
k 1
∗h1 (z) −
−
z
4 2
(3.14)
φ(z)
∗h2 (z) ,
z
where φ(z)/z ∈ P(1/2) and hi ∈ P(σ 1 ). Using Lemma 2.2, we see that [(φ(z)/z)∗hi (z)] ∈
P(σ 1 ) and consequently h ∈ Pk (σ 1 ), which implies that φ∗ f ∈ Qks (λ,μ,α1 ,σ 1 ); the proof
is complete.
Corollary 3.5. The class Qks (λ,μ,α1 ,σ 1 ) is invariant under the following integral operators:
z
(i) f1 (z) = 0 ( f (t)/t)dt,
z
(ii) f2 (z) = (2/z) 0 f (t)dt (Libera’s operator [20]),
z
(iii) f3 (z) = 0 ( f (t) − f (xt)/(t − xt))dt, |x| ≤ 1, x = 1,
z
(iv) f4 (z) = ((1 + c)/zc ) 0 t c−1 f (t)dt,Re c > 0.
One may write (see [21, 22])
f1 (z) = f (z)∗φ1 (z),
f2 (z) = f (z)∗φ2 (z),
f3 (z) = f (z)∗φ3 (z),
f4 (z) = f (z)∗φ4 (z),
(3.15)
where φi , i = 1,2,3,4, are convex and
φ1 (z) = − log (1 − z) =
φ2 (z) =
∞
1
n =1
n
−2 z + log(1 − z)
z
zn ,
=
∞
2 n
z ,
n
+
1
n =1
(3.16)
∞
1 − xn
1 − xz
φ3 (z) =
zn ,
log
=
1−x
1−z
(1 − x)n
n =1
1
φ4 (z) =
∞
1+c
n+c
n =1
zn ,
|x| ≤ 1, x = 1,
Re c > 0.
Now, the result follows by applying Theorem 3.4. Let μ1 and μ2 be linear operators
defined on the class S as follows:
μ1 f (z) = z f (z),
f (z) + z f (z)
μ2 f (z) =
2
Livingston’s operator [23] .
(3.17)
Khalida Inayat Noor 7
Then, both of these operators can be written as a convolution operator [21], given by
μi ( f ) = hi ∗ f , i = 1,2, where
h1 (z) =
∞
nzn =
n =1
z
h2 (z) =
2,
(1 − z)
∞
n+1
n =1
2
zn =
z − z2 /2
.
(1 − z)2
(3.18)
√
It can easily be verified that the radius of convexity rc (h1 ) = 2 − 3 and rc (h2 ) = 1/2.
These facts together with Theorem 3.4 yield the following.
Theorem 3.6. Let f ∈ Qks (λ,μ,α1 ,σ 1 ). Then,
μ1 ( f ) = f ∗h1 ∈ Qks λ,μ,α1 ,σ 1 ,
μ2 ( f ) = f ∗h2 ) ∈ Qks (λ,μ,α1 ,σ 1 ,
√
for |z| < 2 − 3,
1
for |z| < .
2
(3.19)
Theorem 3.7. Let 0 ≤ α1 < α2 . Then, Qks (λ,μ,α1 ,σ 1 ) ⊂ Qks (λ,μ,α2 ,σ 1 ).
Proof. If α1 = 0, the result is obvious. Therefore, we assume that α1 > 0 and f ∈ Qks
s
(λ,μ,α2 ,σ 1 ). Let (Iλ,μ
f (z)) = H1 (z). Then, by Theorem 3.1, H1 ∈ Pk (σ 1 ). Also, let
s
s
α1 Iλ,
f (z) + 1 − α1 z Iλ,μ
f (z)
= H2 (z),
H2 ∈ Pk σ 1 in E.
(3.20)
Now,
s
s
α2 Iλ,μ
f (z) + 1 − α2 z Iλ,μ
f (z)
1 − α2
α2 − α1
H1 (z) + H2 (z)
=
1 − α1
1 − α1
α2 − α1
α − α1
H1 (z) + 1 − 2
H2 (z).
= 1 − α1
1 − α1
(3.21)
Since H1 ,H2 ∈ Pk (σ 1 ) and Pk (σ 1 ) is a convex set, see [2], we obtain the required result.
Theorem 3.8. Let fi ∈ Qks (λ,μ,α1 ,ζ i ), i = 1,2, and let Ψ = f1 ∗ f2 . Then, Ψ(z)/z ∈ Qks
(λ,μ,1,ζ) for z ∈ E, where ζ = 1 − δ(1 − δ 1 )(1 − δ 2 )(ln2 − 1)2 and
δ i = ζ i + 1 − ζ i (2m − 1).
(3.22)
Proof. Since fi ∈ Qks (λ,μ,α1 ,ζ i ), it follows from Theorem 3.1 that fi ∈ Qks (λ,μ,1,δ i ), δ i =
ζ i + (1 − ζ i )(2m − 1), and
m=
1
0
−1
(1 + t (1−α) ) dt.
(3.23)
Now,
s
z Iλ,μ
Ψ(z)
s
s
= Iλ,μ
(Ψ (z) + zψ (z) = z Iλ,μ
( f1 ∗ f2 )(z)
s
s
s
= Iλ,μ
f1 (z)∗ f2 (z) = Iλ,μ
f1 (z) ∗ Iλ,μ
f2 (z) .
(3.24)
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International Journal of Mathematics and Mathematical Sciences
Since fi ∈ Qks (λ,μ,1,δ i ), it follows, by Lemma 2.4, that {Ψ (z) + zΨ (z)} ∈ Qks (λ,μ,1,δ),
where
δ = 1 − 2 1 − δ1 1 − δ2 .
(3.25)
From (3.25) and Lemma 2.3, we have
Ψ (z) ∈ Qks λ,μ,1, 1 + 4(1 − δ 1 (1 − δ 2 (ln2 − 1
.
(3.26)
z ∈ E.
(3.27)
From (3.26) and Lemma 2.3, again, we have
2 Ψ(z)
∈ Qks λ,μ,1, 1 − δ 1 − δ 1 1 − δ 2 ln2 − 1
,
z
We now consider the converse case of Theorem 3.1 as follows.
Theorem 3.9. Let f ∈ Qks (λ,μ,1,σ). Then, f ∈ Qks (λ,μ,α1 ,σ), 0 < α1 ≤ 1, for |z| < rα1
(α1 = 1/2), where
1
rα1 = .
2(1 − α1 ) + 4α21 − 6α1 + 3
(3.28)
This result is sharp.
s
s
Proof. Let φα1 (z) = α1 (Iλ,μ
f (z)) + (1 − α1 )(z(Iλ,μ
f (z)) ) . Then,
φα1 (z) =
kα1 (z)
s
∗(Iλ,μ
f (z)) ,
z
where kα1 (z) = α1
z
z
.
+ 1 − α1
1−z
(1 − z)2
(3.29)
It is known [23] that the function kα1 is convex for |z| < rα1 , where rα1 is given by (3.28)
and this radius is sharp and consequently, for |z| < rα1 , by a well-known result, kα1 ∈
P(1/2). Thus, using Lemma 2.2, and the given fact that f ∈ Qks (λ,μ,1,σ), we obtain the
required result.
Acknowledgment
The author would like to thank Dr. S. M. Junaid Zaidi, (Rector CIIT), who generously
supports scientific research in all aspects.
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Khalida Inayat Noor 9
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Khalida Inayat Noor: Mathematics Department, COMSATS Institute of Information Technology,
Sector H-8/1, Islamabad 44000, Pakistan
Email addresses: khalidanoor@hotmail.com; khalidainayat@comsats.edu.pk