1, 410087 Oradea, Romania 2 Department of Mathematics and Computer Science, 1 Decembrie 1918 University of Alba Iulia, str.. 11–13, 510009 Alba Iulia, Romania Correspondence should be ad
Trang 1Volume 2008, Article ID 127645, 7 pages
doi:10.1155/2008/127645
Research Article
Sufficient Conditions for Univalence of
an Integral Operator
Georgia Irina Oros, 1 Gheorghe Oros, 1 and Daniel Breaz 2
1 Department of Mathematics, University of Oradea, str Universitatii nr 1, 410087 Oradea, Romania
2 Department of Mathematics and Computer Science, 1 Decembrie 1918 University of Alba Iulia,
str N Iorga, no 11–13, 510009 Alba Iulia, Romania
Correspondence should be addressed to Daniel Breaz, dbreaz@uab.ro
Received 27 February 2008; Accepted 24 May 2008
Recommended by Ulrich Abel
In this paper we have introduced an integral general operator For this general operator which
is a generalization of more known integral operators we have demonstrated some univalence properties.
Copyright q 2008 Georgia Irina Oros et al 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 and preliminaries
Let U be the unit disk of the complex plane:
U
z ∈ C : |z| < 1
LetHU be the space of holomorphic functions in U,
A nf ∈ HU, fz z a n1 z n1 · · · , z ∈ U 1.2
with A1 A, and
S
f ∈ A : f is univalent in U
Lemma 1.1 see 1 If the function f is regular in the unit disc U,
fz z a2z2 · · · ,
1− |z|2zfz
fz
≤ 1 ∀z ∈ U, 1.4
then the function f is univalent in U.
Trang 2Definition 1.2St Ruscheweyh 2 For f ∈ A, n ∈ N ∪ {0}, let R nbe the operator defined by
R n : A → A,
R0fz fz,
R1fz zfz
n 1R n1 fz z
R n fz
nR n fz, z ∈ U.
1.5
Remark 1.3 If f ∈ A
fz z
∞
j2
then
R n fz z
∞
j1
with
R n f0 0,
R n f0
Lemma 1.4 3, Schwarz’s lemma, 4, Lemma 4.26, page 103 If the analytic function fz is
regular in U with f0 0 and |fz| < 1 for all z ∈ U, then
fz ≤ |z|, ∀z ∈ U, 1.9
and |f0| ≤ 1.
The equality holds if and only if fz cz, z ∈ U, |c| 1.
2 Main results
By using the Ruscheweyh differential operator given by Definition 1.2, we introduce the following integral operator
Definition 2.1 Let n, m ∈ N ∪ {0}, i ∈ {1, 2, 3, , m}, α i ∈ C Define the integral operator
If1, f2, , f m : A m → A,
I
f1, f2, , f m
z z
0
R n f1t
t
α1
· · · R n f m t
t
α m
where f i z ∈ A and R nis the Ruscheweyh differential operator
Remark 2.2 i For n 0, m 1, α1 1, α2 α3 · · · α m 0,
Trang 3we obtain Alexander integral operator introduced in 1915 in5:
Iz
z
0
ft
ii For n 0, m 1, α1 α ∈ 0, 1, α2 α3 · · · α m 0, R0fz fz ∈ S, and we
obtain the integral operator
I α z z
0
ft
t
α
studied in6
iii For n 1, m 1, α1 γ ∈ C, |γ| ≤ 1/4, α2 · · · α m 0, R1fz zfz ∈ S, we
obtain the integral operator
F γ z z
0
ftγ
studied in7,8
iv For n 0, m ∈ N ∪ {0}, α i ∈ C, i ∈ {1, 2, , m}, R0fz fz ∈ S, and we obtain
the integral operator
Fz
z
0
f1t
t
α1
· · ·
f m t
t
α m
studied in9
v For n, m ∈ N ∪ {0}, i ∈ {1, 2, , m}, α i > 0, we obtain the integral operator F m :
A m → A,
F m
f1, f2, , f m
z z
0
R n f1t
t
α1
· · ·
R n f m t
t
α m
studied in10
vi For n 0, m 1, α1 γ, α2 · · · α m 0, R0fz fz, and we obtain the
integral operator
F γ z z
0
ft
t
γ
studied in11,12
Theorem 2.3 Let n, m ∈ N ∪ {0}, i ∈ {1, 2, , m}, α i ∈ C, f i ∈ A If
zR n f i z
R n f i z − 1
≤ 1, α1 α2 ··· α m ≤ 1, z ∈ U, 2.9
then If1, f2, , f m z given by 2.1 is univalent.
Trang 4Proof Since f i ∈ A, i ∈ {1, 2, , m}, fromRemark 1.3we have
R n f i z
z z
∞
j2 C n nj−1 a j,i z j
z 1 ∞
j2
C nj−1 n a j,i z j−1 ,
R n f i z
z / 0, z ∈ U.
2.10
For z 0, we have
R n f1z
z
α1
· · ·
R n f m z
z
α m
By differentiating 2.1, we obtain
I
f1, f2, , f m
z
R n f1z
z
α1
· · ·
R n f m z
z
α m
, z ∈ U,
I
f1, f2, , f m
0 1.
2.12 Using2.12, we obtain
log I
f1, f2, , f m
z α1
log R n f1z − log z · · · α mlog R n f m z − log z, z ∈ U.
2.13
By differentiating 2.13, we have
I
f1, f2, , f m
z
I
f1, f2, , f m
z α1
R n f
1z
R n f1z −
1
z
· · · α m
R n f
m z
R n f m z −
1
z
, z ∈ U 2.14 and after a short calculus we obtain
zI
f1, f2, , f m
z
I
f1, f2, , f m
z α1 z
R n f1z
R n f1z − 1
· · · α m z
R n f m z
R n f m z − 1
, z ∈ U.
2.15
We multiply the modulus of2.15 by 1 − |z|2 and we obtain
1− |z|2zI
f1, f2, , f m
z
I
f1, f2, , f m
z
1− |z|2α1 z
R n f1z
R n f1z − 1
· · · α m
zR n f
m z
R n f m z − 1
≤1− |z|2 α1 z
R n f1z
R n f1z − 1
··· α m z
R n f m z
R n f m z − 1
≤α1 ··· α m1−z2 ≤ α1 ··· α m ≤ 1.
2.16
From Lemma A, we have If1, f2, , f m z ∈ S.
Trang 5Remark 2.4 i For n 0, R n f i z f i z ∈ S, we obtainTheorem 2.3from9.
ii For α i ∈ R, α i > 0,Theorem 2.3can be rewritten as follows
Corollary 2.5 Let n, m ∈ N ∪ {0}, i ∈ {1, 2, , m}, α i > 0 with α1 α2 · · · α m ≤ 1 If f i ∈ A
satisfy
zR n f i z
R n f i z − 1
≤ 1, z ∈ U, 2.17
then the integral operator given by2.1 is univalent.
Theorem 2.6 Let n, m ∈ N ∪ {0}, i ∈ {1, 2, , m}, α i ∈ C If f i ∈ A satisfy
i |α1| · · · |α m | ≤ 1/3,
ii |R n f i z| ≤ 1,
iii |z2R n f i z/R n i f i z2− 1| < 1
for all z ∈ U, then the integral operator given by 2.1 is univalent.
Proof Using2.14, we obtain
z
I
f1, , f m
z
If1, , f m
z
α1
z
R n f1z
R n f1z − 1
α m
z
R n f m z
R n f m z − 1
. 2.18
We multiply2.18 by 1 − |z|2, use Schwarz’s lemma, and obtain
1− |z|2zTz
Tz
1− |z|2α1 zR n f1z
R n f1z − 1
··· 1− |z|2α m z
R n f m z
R n f1z − 1
1− |z|2α1 z
R n f1z
R n f1z
1− |z|2α1 ··· 1 − |z|2α m z
R n f m z
R n f1z
1− |z|2α m
1− |z|2α1 z
R n f1z
R n f1z
··· zR n f m z
R n f1z
1− |z|2α1 ··· α m
1−z2 α1z2
R n f1z
R n f1z2
R n f1
|z| · · · α mz2
R n f m z
R n f m z2
R n f m
|z|
1− |z|2α1 ··· α m
Trang 6≤1− |z|2 α1z2
R n f1z
R n f1z2
··· α mz2
R n f m z
R n f m z2
1− |z|2α1 ··· α m
1− |z|2 α1z2
R n f1z
R n f1z2
−α1 α1
· · · 1− |z|2 α m z2
R n f m z
R n f m z2
−α m α m
1− |z|2α1 ··· α m
1− |z|2 α1z2
R n f1z
R n f1z2 − 1
··· α mz2
R n f m z
R n f m z2 − 1
1− |z|2α1 ··· α m 1 − |z|2α1 ··· α m
≤1− |z|2α1 α1 ··· α m 21 − |z|2α1 ··· α m
31− |z|2α1 ··· α m
≤ 3α1 ··· α m.
2.19 From2.19 and condition i, we have
1− |z|2zFz
Fz
for all z ∈ U.
By Lemma A, it follows that the integral operator If1, f2, , f m z is univalent.
Remark 2.7 For n 0, m 1, α1 α ∈ C, |α| ≤ 1/3, α2 · · · α m 0, the result was obtained
in11, Theorem 1
For α i ∈ R, α i > 0,Theorem 2.6can be rewritten as follows
Corollary 2.8 Let n, m ∈ N ∪ {0}, i ∈ {1, 2, , m}, α i > 0 If f i ∈ A satisfy
i α1 α2 · · · α n ≤ 1/3,
ii |R n f i z| ≤ 1,
iii |z2R n f i z/R n f i z2− 1| < 1
for all z ∈ U, then the integral operator given by 2.1 is univalent.
References
1 J Becker, “L¨ownersche Differentialgleichung und quasikonform fortsetzbare schlichte Funktionen,”
Journal f ¨ur die reine und angewandte Mathematik, vol 255, pp 23–43, 1972.
2 St Ruscheweyh, “New criteria for univalent functions,” Proceedings of the American Mathematical
Society, vol 49, no 1, pp 109–115, 1975.
Trang 73 Z Nehari, Conformal Mapping, Dover, New York, NY, USA, 1975.
4 P Hamburg, P Mocanu, and N Negoescu, Analiz˘a matematic˘a (Funct¸ii complexe), Editura Didactic˘a s¸i
Pedagogic˘a, Bucures¸ti, Romania, 1982.
5 J W Alexander, “Functions which map the interior of the unit circle upon simple regions,” Annals of
Mathematics, vol 17, no 1, pp 12–22, 1915.
6 S S Miller, P T Mocanu, and M O Reade, “Starlike integral operators,” Pacific Journal of Mathematics,
vol 79, no 1, pp 157–168, 1978.
7 Y J Kim and E P Merkes, “On an integral of powers of a spirallike function,” Kyungpook Mathematical
Journal, vol 12, pp 249–252, 1972.
8 N N Pascu and V Pescar, “On the integral operators of Kim-Merkes and Pfaltzgraff,” Mathematica,
vol 3255, no 2, pp 185–192, 1990.
9 D Breaz and N Breaz, “Two integral operators,” Studia Universitatis Babes¸-Bolyai Mathematica, vol 47,
no 3, pp 13–19, 2002.
10 G I Oros and G Oros, “A convexity property for an integral operator F m,” in preparation.
11 V Pescar and S Owa, “Sufficient conditions for univalence of certain integral operators,” Indian Journal
of Mathematics, vol 42, no 3, pp 347–351, 2000.
12 V Pescar, “On some integral operations which preserve the univalence,” The Punjab University Journal
of Mathematics, vol 30, pp 1–10, 1997.
... · · α m 0, Trang 3we obtain Alexander integral operator introduced in 1915 in5:
Iz... given by 2.1 is univalent.
Trang 4Proof Since f i ∈ A, i ∈ {1, 2, ,... m z ∈ S.
Trang 5Remark 2.4 i For n 0, R n f i