Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 7, Issue 2, Article 49, 2006
ON CERTAIN CLASSES OF ANALYTIC FUNCTIONS
KHALIDA INAYAT NOOR
M ATHEMATICS D EPARTMENT
COMSATS I NSTITUTE OF I NFORMATION T ECHONOLGY
I SLAMABAD , PAKISTAN
khalidanoor@hotmail.com
Received 28 August, 2005; accepted 21 October, 2005
Communicated by Th.M. Rassias
P∞
A BSTRACT. Let A be the class of functions f : f (z) = z + n=2 an z n which are analytic
in the unit disk E. We introduce the class Bk (λ, α, ρ) ⊂ A and study some of their interesting
properties such as inclusion results and covering theorem. We also consider an integral operator
for these classes.
Key words and phrases: Analytic functions, Univalent, Functions with positive real part, Convex functions, Convolution, Integral operator.
2000 Mathematics Subject Classification. 30C45, 30C50.
1. I NTRODUCTION
Let A denote the class of functions
f : f (z) = z +
∞
X
an z n
n=2
which are analytic in the unit disk E = {z : |z| < 1} and let S ⊂ A be the class of functions
univalent in E.
Let Pk (ρ) be the class of functions p(z) analytic in E satisfying the properties p(0) = 1 and
Z 2π Re p(z) − ρ (1.1)
1 − ρ dθ ≤ kπ,
0
where z = rei θ, k ≥ 2 and 0 ≤ ρ < 1. This class has been introduced in [7]. We note that,
for ρ = 0, we obtain the class Pk defined and studied in [8], and for ρ = 0, k = 2, we have the
well known class P of functions with positive real part. The case k = 2 gives the class P (ρ) of
functions with positive real part greater than ρ.
ISSN (electronic): 1443-5756
c 2006 Victoria University. All rights reserved.
This research is supported by the Higher Education Commission, Pakistan, through grant No: 1-28/HEC/HRD/2005/90.
054-06
2
K HALIDA I NAYAT N OOR
From (1.1) we can easily deduce that p ∈ Pk (ρ) if, and only if, there exist p1 , p2 ∈ P (ρ) such
that, for E,
k 1
k 1
+
p1 (z) −
−
p2 (z).
(1.2)
p(z) =
4 2
4 2
P
P∞
m
m
Let f and g be analytic in E with f (z) = ∞
m=0 am z and g(z) =
m=0 bm z in E. Then the
convolution ? (or Hadamard Product) of f and g is defined by
(f ? g)(z) =
∞
X
am b m z m ,
m ∈ N0 = {0, 1, 2, . . .}.
m=0
Definition 1.1. Let f ∈ A. Then f ∈ Bk (λ, α, ρ) if and only if
α
α f (z)
zf 0 (z) f (z)
(1.3)
(1 − λ)
+λ
∈ Pk (ρ),
z
f (z)
z
z ∈ E,
where α > 0, λ > 0, k ≥ 2 and 0 ≤ ρ < 1. The powers are understood as principal values.
For k = 2 and with different choices of λ, α and ρ, these classes have been studied in [2, 3,
4, 10]. In particular B2 (1, α, ρ) is the class of Bazilevic functions studied in [1].
We shall need the following results.
Lemma 1.1 ([9]). If p(z) is analytic in E with p(0) = 1 and if λ is a complex number satisfying
Re λ ≥ 0, (λ 6= 0), then
Re[p(z) + λzp0 (z)] > β
(0 ≤ β < 1)
implies
Re p(z) > β + (1 − β)(2γ − 1),
where γ is given by
1
Z
(1 + tRe λ )−1 dt.
γ = γRe λ =
0
Lemma 1.2 ([5]). Let c > 0, λ > 0, ρ < 1 and p(z) = 1 + b1 z + b2 z 2 + · · · be analytic in E.
Let Re[p(z) + cλzp0 (z)] > ρ in E, then
1
Re[p(z) + czp0 (z)] ≥ 2ρ − 1 + 2(1 − ρ) 1 −
λ
1
cλ
1
Z
0
1 −1
u cλ
du.
1+u
This result is sharp.
2. M AIN R ESULTS
Theorem 2.1. Let λ, α > 0, 0 ≤ ρ < 1 and let f ∈ bk (λ, α, ρ). Then
ρ1 is given by
(2.1)
f (z)
z
α
∈ Pk (ρ1 ), where
ρ1 = ρ + (1 − ρ)(2γ − 1),
and
Z
γ=
1
λ
1 + tα
−1
dt.
0
J. Inequal. Pure and Appl. Math., 7(2) Art. 49, 2006
http://jipam.vu.edu.au/
O N C ERTAIN C LASSES OF A NALYTIC F UNCTIONS
3
Proof. Let
f (z)
z
α
= p(z) =
k 1
+
4 2
p1 (z) −
k 1
−
4 2
p2 (z).
Then p(z) = 1 + αa2 z + · · · is analytic in E, and
(f (z))α = z α p(z).
(2.2)
Differentiation of (2.2) and some computation give us
α
α
f (z)
zf 0 (z) f (z)
λ
+λ
= p(z) + zp0 (z).
(1 − λ)
z
f (z)
z
α
Since f ∈ Bk (λ, α, ρ), so {p(z) + αλ zp0 (z)} ∈ Pk (ρ) for z ∈ E. This implies that
λ 0
Re pi (z) + zpi (z) > ρ, i = 1, 2.
α
Using Lemma 1.1, we see that Re{pi (z)} > ρ1 , where ρ1 is given by (2.1). Consequently
p ∈ Pk (ρ1 ) for z ∈ E, and the proof is complete.
Corollary 2.2. Let f = zF10 and f ∈ B2 (λ, 1, ρ). Then F1 is univalent in E.
Proceeding as in Theorem 2.1 and using Lemma 1.2, we have the following.
Theorem 2.3. Let α > 0, λ > 0, 0 ≤ ρ < 1 and let f ∈ Bk (λ, α, ρ). Then
Pk (ρ2 ), where
α −1
Z
1−ρ
1 α 1 uλ
+ 2(1 − ρ) 1 −
du.
ρ2 = 2ρ − 1 +
λ
λ λ 0 1+u
zf 0 (z) f (z) α
( z )
f (z)
∈
This result is sharp.
For k = 2, we note that f is univalent, see [1].
Theorem 2.4. Let, for α > 0, λ > 0, 0 ≤ ρ < 1, f ∈ Bk (λ, α, ρ) and define I(f ) : A −→ A
as
α1
Z
1 α− λ1 z λ1 −1−α
α
(2.3)
I(f ) = F (z) =
z
t
(f (z)) dt , z ∈ E.
λ
0
Then F ∈ Bk (αλ, α, ρ1 ) for z ∈ E, where ρ1 is given by (2.1).
Proof. Differentiating (2.3), we have
α
α α
F (z)
zF 0 (z) F (z)
f (z)
(1 − αλ)
+ αλ
=
.
z
F (z)
z
z
Now, using Theorem 2.1, we obtain the required result.
Theorem 2.5. Let
f : f (z) = z +
∞
X
an z n ∈ Bk (λ, α, ρ).
n=2
Then
|an | ≤
J. Inequal. Pure and Appl. Math., 7(2) Art. 49, 2006
k(1 − ρ)
.
λ+α
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4
K HALIDA I NAYAT N OOR
The function fλ,α,ρ (z) defined as
α
Z α
fλ,α,ρ (z)
α 1
k 1
1 + (1 − 2ρ)uz
=
+
u λ −1
z
λ 0
4 2
1 − uz
α
k 1
1 − (1 − 2ρ)uz
−1
−
−
uλ
du
4 2
1 + uz
shows that this inequality is sharp.
Proof. Since f ∈ Bk (λ, α, ρ), so
!α
!
!α
∞
∞
∞
X
X
X
(1 − λ) 1 +
an z n−1
+λ 1+
nan z n−1
1+
an z n−1
n=2
n=2
n=2
= H(z) =
1+
∞
X
!
cn z n
∈ Pk (ρ).
n=1
It is known that |cn | ≤ k(1 − ρ) for all n and using this inequality, we prove the required
result.
Different choices of k, λ, α and ρ yield several known results.
Theorem 2.6 (Covering Theorem). Let λ > 0 and 0 < ρ < 1. Let f = zF10 ∈ B2 (λ, 1, ρ). If
D is the boundary of the image of E under F1 , then every point of D has a distance of at least
λ+1
from the origin.
(3+2λ−ρ)
Proof. Let F1 (z) 6= w0 , w0 6= 0. Then f1 (z) =
Let
∞
X
f (z) = z +
an z n ,
w0 F1 (z)
w0 +F1 (z)
is univalent in E since F1 is univalent.
F1 (z) = z +
n=2
∞
X
bn z n .
n=2
Then a2 = 2b2 . Also
1
f1 (z) = z + b2 +
w0
and so |b2 +
1
|
w0
≤ 2. Since, by Theorem 2.5, |b2 | ≤
z2 + · · · ,
1−ρ
,
1+λ
we obtain |w0 | ≥
λ+1
.
3+2λ−ρ
Theorem 2.7. For each α > 0, Bk (λ1 , α, ρ) ⊂ Bk (λ2 , α, ρ) for 0 ≤ λ2 < λ1 .
Proof. For λ2 = 0, the proof is immediate. Let λ2 > 0 and let f ∈ Bk (λ1 , α, ρ). Then there
exist two functions h1 , h2 ∈ Pk (ρ) such that, from Definition 1.1 and Theorem 2.1,
α
α
f (z)
zf 0 (z) f (z)
+ λ1
= h1 (z),
(1 − λ)
z
f (z)
z
and
f (z)
z
α
= h2 (z).
Hence
(2.4)
(1 − λ2 )
f (z)
z
α
zf 0 (z)
+ λ2
f (z)
f (z)
z
α
λ2
λ2
= h1 (z) + 1 −
h2 (z).
λ1
λ1
Since the class Pk (ρ) is a convex set, see [6], it follows that the right hand side of (2.4) belongs
to Pk (ρ) and this proves the result.
J. Inequal. Pure and Appl. Math., 7(2) Art. 49, 2006
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O N C ERTAIN C LASSES OF A NALYTIC F UNCTIONS
5
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J. Inequal. Pure and Appl. Math., 7(2) Art. 49, 2006
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