Hindawi Publishing CorporationJournal of Inequalities and Applications Volume 2008, Article ID 143869, 4 pages doi:10.1155/2008/143869 Research Article A Convexity Property for an Integr
Trang 1Hindawi 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
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, Fnz, for analytic functions, fiz, in the open unit disk, U The object of this paper is to prove the convexity properties for the integral operator Fnz, on the class
S pβ.
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 a2z2 a3z3 · · · , 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
zfz
> 0, z ∈ U
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
zfz
fz 1 > α, z ∈ U
Consider the class S p β, which was introduced by Ronning 1 and which is defined by
zf fzz − 1
≤ Rezf fzz − β
where β is a real number with the property −1 ≤ β < 1.
Trang 22 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
f1t
t
α1
· · ·
f n t
t
α n
This integral operator was first defined by B Breaz and N Breaz2 It is easy to see that
F n z ∈ A.
2 Main results
Theorem 2.1 Let α i > 0, for i ∈ {1, , n}, let β 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
i1 α i β i − 1.
From1.4 we obtain
F nz
f1z
z
α1
· · ·
f n z
z
α n
,
F nz n
i1
α i
f i z
z
α i
zf iz − 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 nz
F nz α1
zf1z − f1z
zf1z
· · · α n
zf nz − f n z
zf n z
This relation is equivalent to
Fn z
Fn z α1
f
1z
f1z −
1
z
· · · α n
f nz
f n z−
1
z
If we multiply the relation2.4 with z, then we obtain
zF nz
F nz
n
i1
α i
zf iz
n
i1
α i zf
i z
f i z −
n
i1
The relation2.5 is equivalent to
zFn z
F nz 1
n
i1
α i
zf iz
n
i1
Trang 3Daniel Breaz 3 This relation is equivalent to
zF nz
F nz 1
n
i1
α i
zf iz
f i z − β i
n
i1
α i β i−n
i1
We calculate the real part from both terms of the above equality and obtain
Re
zF nz
F nz 1
n
i1
α iRe
zf iz
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 nz
F nz 1
>
n
i1
α i
zf iz
f i z − 1
n
i1
α i
Since α i |zf
i z/f i z − 1| > 0 for all i ∈ {1, , n}, we obtain that
Re
zF nz
F nz 1
>
n
i1
α i
So, F nis convex of order n
i1 α i β i− 1 1
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
i1 α i 1.
i1 α i 1, then
Re
zF nz
F nz 1
so F nis a convex function
Corollary 2.4 Let γ be a real number, γ > 0 Suppose that the functions f ∈ S p β and 0 < γ ≤ 1/1 − β In these conditions, the function F1z z
0ft/t γ dt is convex of order β − 1γ 1.
Corollary 2.5 Let f ∈ S p β and consider the integral operator of Alexander, Fz z
0ft/tdt.
In this condition, F is convex by the order β.
Proof We have
zFz
zfz
Trang 44 Journal of Inequalities and Applications From2.13, we have
Re
zFz
Re
zfz
β >
zf fzz − 1
β > β. 2.14
So, the relation2.14 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.
...i1
Trang 3Daniel Breaz This relation is equivalent to
zF nz...
Trang 44 Journal of Inequalities and Applications From2.13, we have
Re
zFz...
So, the relation2.14 implies that the Alexander operator is convex
References
1 F Ronning, “Uniformly convex functions and a corresponding class of starlike