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 SP β
Daniel Breaz
Department of Mathematics, “1 Decembrie 1918” University, Alba Iulia 510009, Romania
Correspondence should be addressed to Daniel Breaz, dbreaz@uab.ro
Received 30 October 2007; Accepted 30 December 2007
Recommended by Narendra Kumar K. Govil
We consider an integral operator, Fn z, for analytic functions, fi z, in the open unit disk, U. The
object of this paper is to prove the convexity properties for the integral operator Fn z, on the class
Sp 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. Let A {f ∈ HU, fz z a2 z2 a3 z3 · · · , 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
zf z
K f ∈ A : Re
1
>
0,
z
∈
U
.
1.1
f z
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
zf z
Re
1 > α, z ∈ U .
1.2
f z
Consider the class Sp β, which was introduced by Ronning 1 and which is defined by
zf z
zf z
− 1 ≤ Re
−β ,
1.3
f ∈ Sp β ⇐⇒ fz
fz
where β is a real number with the property −1 ≤ β < 1.
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Journal of Inequalities and Applications
For fi z ∈ A and αi > 0, i ∈ {1, . . . , n}, we define the integral operator Fn z given by
Fn z z 0
f1 t
t
α1
·····
fn t
t
αn
dt.
1.4
This integral operator was first defined by B. Breaz and N. Breaz 2. It is easy to see that
Fn z ∈ A.
2. Main results
Theorem 2.1. Let αi > 0, for i ∈ {1, . . . , n}, let βi be real numbers with the property −1 ≤ βi < 1, and
let fi ∈ Sp βi for i ∈ {1, . . . , n}.
If
0<
n
i1
αi 1 − βi ≤ 1,
then the function Fn given by 1.4 is convex of order 1 2.1
n
i1 αi βi
− 1.
Proof. We calculate for Fn the derivatives of first and second orders.
From 1.4 we obtain
Fn z
Fn z f1 z
z
n
i1
αi
α1
fn z αn
·····
,
z
fi z
z
αi n fj z αj
zfi z − fi z .
zfi z
z
j1
2.2
j/
i
After some calculus, we obtain that
zf1 z − f1 z
zfn z − fn z
Fn z
α
·
·
·
α
.
1
n
zf1 z
zfn z
Fn z
2.3
This relation is equivalent to
f1 z 1
fn z 1
Fn z
−
·
·
·
α
−
.
α
1
n
f1 z z
fn z z
Fn z
2.4
If we multiply the relation 2.4 with z, then we obtain
n
n
n
zfi z zfi z
zFn z α
α
αi .
−
1
−
i
i
fi z
fi z
Fn z
i1
i1
i1
2.5
The relation 2.5 is equivalent to
n
n
zf z zFn z
−
1
αi i
αi 1.
fi z
Fn z
i1
i1
2.6
Daniel Breaz
3
This relation is equivalent to
n
n
n
zfi z
zFn z
1
α
α
β
−
αi 1.
−
β
i
i
i
i
fi z
Fn z
i1
i1
i1
2.7
We calculate the real part from both terms of the above equality and obtain
n
n
n
zfi z
zFn z
1
α
Re
α
β
−
αi 1.
Re
−
β
i
i
i
i
fi z
Fn z
i1
i1
i1
2.8
Because fi ∈ Sp βi for i {1, . . . , n}, we apply in the above relation inequality 1.3 and
obtain
n
n
zfi z
zFn z
Re
−
1
1
>
α
αi βi − 1 1.
2.9
i
f
z
Fn z
i
i1
i1
Since αi |zfi z/fi z − 1| > 0 for all i ∈ {1, . . . , n}, we obtain that
zFn z
1
Re
Fn z
So, Fn is convex of order
n
i1 αi βi
>
n
i1
αi βi − 1 1.
2.10
− 1 1.
Corollary 2.2. Let αi , i ∈ {1, . . . , n} be real positive numbers and fi ∈ Sp β for i ∈ {1, . . . , n}.
If
0<
n
αi ≤
i1
n
then the function Fn is convex of order β − 1
i1 αi
1
,
1−β
2.11
1.
Proof. In Theorem 2.1, we consider β1 β2 · · · βn β.
Remark 2.3. If β 0 and ni1 αi 1, then
zFn z
1
> 0,
Re
Fn z
2.12
so Fn is a convex function.
Corollary 2.4. Let γ be a real number, γ > 0. Suppose
z that the functions f ∈ Sp β and 0 < γ ≤
1/1 − β. In these conditions, the function F1 z 0 ft/tγ dt is convex of order β − 1γ 1.
Proof. In Corollary 2.2, we consider n 1.
Corollary 2.5. Let f ∈ Sp β and consider the integral operator of Alexander, Fz In this condition, F is convex by the order β.
z
0
ft/tdt.
Proof. We have
zF z zf z
− 1.
F z
fz
2.13
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Journal of Inequalities and Applications
From 2.13, we have
Re
zF z
1
F z
Re
zf z
zf z
− β β > − 1 β > β.
fz
fz
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 Babeş-Bolyai, Mathematica, vol. 47,
no. 3, pp. 13–19, 2002.