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Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2008, Article ID 143869, 4 pages
doi:10.1155/2008/143869
Research Article
A Convexity Property for an Integral Operator
on the Class S
P
β
Daniel Breaz
Department of Mathematics, “1 Decembrie 1918” University, Alba Iulia 510009, Romania
Correspondence should be addressed to Daniel Breaz,
Received 30 October 2007; Accepted 30 December 2007
Recommended by Narendra Kumar K. Govil
We consider an integral operator, F
n
z, for analytic functions, f
i
z, in the open unit disk, U.The
object of this paper is to prove the convexity properties for the integral operator F
n
z,ontheclass
S
p
β.
Copyright q 2008 Daniel Breaz. 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 U {z ∈ C, |z| < 1} be the unit disc of the complex plane and denote by HU the class of
the holomorphic functions in U.LetA {f ∈ HU,fzz a
2
z
2
a
3
z
3
···,z∈ U} be the
class of analytic functions in U and S {f ∈ A : f is univalent in U}.
Denote with K the class of convex functions in U, defined by
K
f ∈ A : Re
zf
z
f
z
1
> 0,z∈ U
. 1.1
A function f ∈ S is the convex function of order α, 0 ≤ α<1, and denote this class by
Kα if f verifies the inequality
Re
zf
z
f
z
1 >α, z∈ U
. 1.2
Consider the class S
p
β, which was introduced by Ronning 1 and which is defined by
f ∈ S
p
β ⇐⇒
zf
z
fz
− 1
≤ Re
zf
z
fz
− β
, 1.3
where β is a real number with the property −1 ≤ β<1.
2 Journal of Inequalities and Applications
For f
i
z ∈ A and α
i
> 0,i∈{1, ,n}, we define the integral operator F
n
z given by
F
n
z
z
0
f
1
t
t
α
1
·····
f
n
t
t
α
n
dt. 1.4
This integral operator was first defined by B. Breaz and N. Breaz 2. It is easy to see that
F
n
z ∈ A.
2. Main results
Theorem 2.1. Let α
i
> 0,fori ∈{1, ,n},letβ
i
be real numbers with the property −1 ≤ β
i
< 1,and
let f
i
∈ S
p
β
i
for i ∈{1, ,n}.
If
0 <
n
i1
α
i
1 − β
i
≤ 1, 2.1
then the function F
n
given by 1.4 is convex of order 1
n
i1
α
i
β
i
− 1.
Proof. We calculate for F
n
the derivatives of first and second orders.
From 1.4 we obtain
F
n
z
f
1
z
z
α
1
·····
f
n
z
z
α
n
,
F
n
z
n
i1
α
i
f
i
z
z
α
i
zf
i
z − f
i
z
zf
i
z
n
j1
j
/
i
f
j
z
z
α
j
.
2.2
After some calculus, we obtain that
F
n
z
F
n
z
α
1
zf
1
z − f
1
z
zf
1
z
··· α
n
zf
n
z − f
n
z
zf
n
z
. 2.3
This relation is equivalent to
F
n
z
F
n
z
α
1
f
1
z
f
1
z
−
1
z
··· α
n
f
n
z
f
n
z
−
1
z
. 2.4
If we multiply the relation 2.4 with z,thenweobtain
zF
n
z
F
n
z
n
i1
α
i
zf
i
z
f
i
z
− 1
n
i1
α
i
zf
i
z
f
i
z
−
n
i1
α
i
. 2.5
The relation 2.5 is equivalent to
zF
n
z
F
n
z
1
n
i1
α
i
zf
i
z
f
i
z
−
n
i1
α
i
1. 2.6
Daniel Breaz 3
This relation is equivalent to
zF
n
z
F
n
z
1
n
i1
α
i
zf
i
z
f
i
z
− β
i
n
i1
α
i
β
i
−
n
i1
α
i
1. 2.7
We calculate the real part from both terms of the above equality and obtain
Re
zF
n
z
F
n
z
1
n
i1
α
i
Re
zf
i
z
f
i
z
− β
i
n
i1
α
i
β
i
−
n
i1
α
i
1. 2.8
Because f
i
∈ S
p
β
i
for i {1, ,n}, we apply in the above relation inequality 1.3 and
obtain
Re
zF
n
z
F
n
z
1
>
n
i1
α
i
zf
i
z
f
i
z
− 1
n
i1
α
i
β
i
− 1
1. 2.9
Since α
i
|zf
i
z/f
i
z − 1| > 0 for all i ∈{1, ,n},weobtainthat
Re
zF
n
z
F
n
z
1
>
n
i1
α
i
β
i
− 1
1. 2.10
So, F
n
is convex of order
n
i1
α
i
β
i
− 11.
Corollary 2.2. Let α
i
,i∈{1, ,n} be real positive numbers and f
i
∈ S
p
β for i ∈{1, ,n}.
If
0 <
n
i1
α
i
≤
1
1 − β
, 2.11
then the function F
n
is convex of order β − 1
n
i1
α
i
1.
Proof. In Theorem 2.1, we consider β
1
β
2
··· β
n
β.
Remark 2.3. If β 0and
n
i1
α
i
1, then
Re
zF
n
z
F
n
z
1
> 0, 2.12
so F
n
is a convex function.
Corollary 2.4. Let γ bearealnumber,γ>0. Suppose that the functions f ∈ S
p
β and 0 <γ≤
1/1 − β. In these conditions, the function F
1
z
z
0
ft/t
γ
dt is convex of order β − 1γ 1.
Proof. In Corollary 2.2, w e consider n 1.
Corollary 2.5. Let f ∈ S
p
β and consider the integral operator of Alexander, Fz
z
0
ft/tdt.
In this condition, F is convex by the order β.
Proof. We have
zF
z
F
z
zf
z
fz
− 1. 2.13
4 Journal of Inequalities and Applications
From 2.13,wehave
Re
zF
z
F
z
1
Re
zf
z
fz
− β
β>
zf
z
fz
− 1
β>β. 2.14
So, the relation 2.14 implies that the Alexander operator is convex.
References
1 F. Ronning, “Uniformly convex functions and a corresponding class of starlike functions,” Proceedings
of the American Mathematical Society, vol. 118, no. 1, pp. 189–196, 1993.
2 D. Breaz and N. Breaz, “Two integral operators,” Studia Universitatis Babes¸-Bolyai, Mathematica, vol. 47,
no. 3, pp. 13–19, 2002.
