1. Trang chủ
  2. » Khoa Học Tự Nhiên

Hindawi Publishing Corporation Journal of Inequalities and Applications Volume 2010, Article ID docx

11 258 0
Tài liệu đã được kiểm tra trùng lặp

Đang tải... (xem toàn văn)

Tài liệu hạn chế xem trước, để xem đầy đủ mời bạn chọn Tải xuống

THÔNG TIN TÀI LIỆU

Thông tin cơ bản

Định dạng
Số trang 11
Dung lượng 484,26 KB

Các công cụ chuyển đổi và chỉnh sửa cho tài liệu này

Nội dung

Volume 2010, Article ID 175369, 11 pagesdoi:10.1155/2010/175369 Research Article Some Starlikeness Criterions for Analytic Functions Gejun Bao,1 Lifeng Guo,1 and Yi Ling2 1 Department of

Trang 1

Volume 2010, Article ID 175369, 11 pages

doi:10.1155/2010/175369

Research Article

Some Starlikeness Criterions for

Analytic Functions

Gejun Bao,1 Lifeng Guo,1 and Yi Ling2

1 Department of Mathematics, Harbin Institute of Technology, Harbin 150001, China

2 Department of Mathematical Science, Delaware State University, Dover, DE 19901, USA

Correspondence should be addressed to Gejun Bao,baogj@hit.edu.cnand

Yi Ling,yiling@desu.edu

Received 26 October 2010; Accepted 16 December 2010

Academic Editor: Vijay Gupta

Copyrightq 2010 Gejun Bao 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

We determine the condition onα, μ, β, and λ for which |1−αz/fz μ αzfz/fzz/fz μ− 1| < λ implies fz ∈ S∗β, where Sβ is the class of starlike functions of order β Some results of

Obradovi´c and Owa are extended We also obtain some new results on starlikeness criterions

1 Introduction

Letn be a positive integer, and let H ndenote the class of function

fz  z ∞

kn

that are analytic in the unit diskU  {z : |z| < 1} For 0 ≤ β < 1, let

S∗

β



f ∈ H1: Rezfz

fz > β, z ∈ U



1.2

denote the class of starlike function of orderβ and S0  S

Letfz and Fz be analytic in U; then we say that the function fz is subordinate

toFz in U, if there exists an analytic function wz in U such that |wz| ≤ |z|, and fz ≡ Fwz, denoted that f ≺ F or fz ≺ Fz If Fz is univalent in U, then the subordination

is equivalent tof0  F0 and fU ⊂ FU 1

Trang 2

S λf ∈ H1:fz − 1 < λ,z ∈ U 1.3

Singh2 proved that S λ ⊂ S∗if 0< λ ≤ 2/√5 More recently, Fournier3,4 proved that

S λ ⊂ S⇐⇒ 0 ≤ λ ≤ √2

5,

ρ λ

1 − λ1 − λ/2

1− λ2/4 , if 0≤ λ ≤

2

3,

1/21− 5/4λ2

1− λ2/4 , if

2

3 ≤ λ ≤ 1,

1.4

is the order of starlikeness ofS λ Now, we define

Uλ, μ, n



f ∈ H n:

zf fzzfz z μ− 1

 < λ, z ∈ U. 1.5 Clearly,Uλ, −1, 1  S λ In 1998, Obradovi´c5 proved that

Uλ, μ, 1⊂ S∗ 1.6

if 0< μ < 1 and 0 < λ ≤ 1 − μ/1 − μ2 μ2 Recently, Obradovi´c and Owa6 proved that

Uλ, μ, n⊂ S∗ 1.7

if 0< μ < 1 and 0 < λ ≤ n − μ/n − μ2 μ2

In this paper we find a condition onα, μ, β, and λ for which



1 − αfz z μ  α zfz

fz

 z

fz

μ

− 1

impliesfz ∈ Sβ and extend some results of Obradovi´c and Owa 5,6 Also, we obtain some new results on starlikeness criterions

Trang 3

2 Main Results

For our results we need the following lemma

Lemma 2.1 see 6 Let pz  1  p n z n  p n1 z n1  · · · be analytic in U, n ≥ 1, and satisfy the

condition

pz −1μ zpz ≺ 1  λz, 0 < μ < 1, 0 < λ ≤ 1. 2.1

Then

pz ≺ 1  λ1z, 2.2

where

Theorem 2.2 Let 0 ≤ μ < 1, nα > μ, 0 ≤ β < 1, and

M n

α, β, μ

αnα − μ1− β

αn  μ − μβ− 2μ , if α ≥ α2,



nα − μ2α1− β− 1



n2α2 2μ2

1− β− nμα , if α1< α < α2,



nα − μ1− β

n − μ1− β  , if 0 < α ≤ α1,

2.4

where

α1  n − μ



1− β

n1− β  ,

α2 n  3μ



1− β

n  3μ1− β2− 8nμ1− β

2.5

If pz  1  p n z n  p n1 z n1  · · · and qz  1  q n z n  q n1 z n1  · · · are analytic in U, satisfy

qz ≺ 1  nα − μ μλ z, 2.6

Trang 4

where 0 < λ ≤ M n α, β, μ, then

Repz > β, for z ∈ U. 2.8

Proof If μ  0, it is easy to see the result is true Now, assume μ > 0 Let

If there existsz0∈ U, such that Re pz0  β, then we will show that

qz01− α  αpz0− 1 ≥ λ 2.10 for 0< λ ≤ M n α, β, μ Note that |qz0 − 1| ≤ N for z ∈ U; it is sufficient to show that

α pz0 − 1 − N1 − α  αpz0 ≥ λ 2.11 for 0< λ ≤ M n α, β, μ Let pz0  β  iy, y ∈ R; then, the left-hand side of 2.11 is

α

β − 12 y2− N

αβ  1 − α2 α2y2

 αβ2 y2 1 − 2β − Nα2β2 α2y2 2α1 − αβ  1 − α2.

2.12

Suppose thatx  β2 y2and note thatnα − μN  μλ; then inequality 2.11 is equivalent to



x  1 − 2β

nα − μ  μα2x  2α1 − αβ  1 − α2 2.13 for allx ≥ β2and 0< λ ≤ M n α, β, μ Now, if we define

ϕx 



x  1 − 2β

nα − μ  μα2x  2α1 − αβ  1 − α2, x ≥ β2, 2.14

then we have

ϕx 



nα − μψx  μ1− 2α1− β

2ψxx  1 − 2βnα − μ μψx2, x > β2, 2.15

where

ψx α2x  2α1 − αβ  1 − α2. 2.16

Trang 5

ψx α2x2 2α1 − αβ  1 − α2>1− α

1− β, for x > β2, 2.17

the denominator ofϕx is positive Further, let

Tx nα − μψx  μ1− 2α1− β, x ≥ β2. 2.18

We have

Tx ≥nα − μ1− α

1− β  μ1 − 2α1 − β. 2.19 If

1

we get

Tx ≥ nα2

1− β−n  3μ1− βα  2μ

 n1− βα − r1α − r2, 2.21

where

r1  n  3μ



1− β−

n  3μ1− β2− 8nμ1− β

r2  n  3μ



1− β

n  3μ1− β2− 8nμ1− β

2.22

Note that

r1< 1

1− β < r2. 2.23

We obtain

Tx ≥ 0 for α ≥ α2 r2. 2.24 If

1

2

1− β  ≤ α <

1

Trang 6

we have

Tx ≥ αn − μ1− β− n1− βα. 2.26 Hence we obtain

Tx ≥ 0 for 1

2

where

α1 n − μ



1− β

n1− β  <

1

If

0< α < 1

2

we have 1− 2α1 − β > 0 It follows that Tx > 0.

Therefore we obtainϕx ≥ 0 for x > β2if 0< α ≤ α1orα ≥ α2 It follows that

min

x≥β2ϕx  ϕβ2





1− β

αn  μ − μβ− 2μ , if α ≥ α2,



1− β

αn − μ1− β , if 0< α ≤ α1.

2.30

Ifα1< α < α2, we have

lim

x →β2Tx  Tβ2

nα − μ1− α

1− β  μ1 − 2α1 − β < 0 2.31

by2.13 and 2.21 for 1/1 − β ≤ α < α2and by2.23 for α1< α < 1/1 − β Note that Tx

is an continuous increasing function forx ≥ β2, and

lim

Then there exists a uniquex0 ∈ β2, ∞, such that

Thus,x0is the global minimum point ofϕx on β2, ∞ It follows from 2.33 that



nα − μα2x0 2α1 − αβ  1 − α2 μ2α1− β− 1, 2.34

Trang 7

x0 α12



μ2

2α1− β− 12



nα − μ2 − 2α1 − αβ − 1 − α2



By a simple calculation, we may obtain

min

x≥β2ϕx  ϕx0 



2α1− β− 1

αn2α2 2μ2

forα1< α < α2 It follows from2.30 and 2.36 that that inequality 2.13 holds This shows that inequality2.10 holds, which contradicts with 2.7 Hence we must have

Repz > β, z ∈ U. 2.37

Theorem 2.3 Let α, μ, β, λ and M n α, β, μ be defined as in Theorem 2.2 If fz ∈ H n satisfies



1 − αfz z μ  α zfz

fz

 z

fz

μ

− 1

where 0 < λ ≤ M n α, β, μ, then fz ∈ Sβ.

Proof If μ  0, M n α, β, 0  α1 − β and the result is trivial Now, assume μ > 0 If we put

qz 

fz

μ

then by some transformations and2.38 we get

qz − α μ zqz ≺ 1  λz. 2.40

qz ≺ 1  nα − μ μλ z. 2.41 Let

pz  zf fzz 2.42

Trang 8

Then we have

Repz > β, z ∈ U. 2.44

It follows thatfz ∈ Sβ.

Forβ  0, we get the following corollary.

Corollary 2.4 Let 0 ≤ μ < 1, nα > μ, and let

M nα, μ

αnα − μ

αn  μ− 2μ , if α ≥ α2,



nα − μ√2α − 1



n2α2 2μ2− nμα , if α1 ≤ α < α2,



nα − μ

n − μ , if 0 < α < α1,

2.45

where

α1  n − μ n ,

α2 n  3μ 



n  3μ2− 8nμ

2.46

If fz ∈ H n satisfies



1 − αfz z μ  α zfz

fz

 z

fz

μ

− 1

where 0 < λ ≤ M n α, μ, then fz ∈ S.

Corollary 2.5 Let 0 ≤ μ < 1, 0 ≤ β < 1, and let

M n

β, μ



n − μ1− β

n − μ1 β  , if 1 > β ≥

μ

n  μ ,



n − μ1− 2β



n2 2μ2

1− β− nμ , if 0 ≤ β <

μ

n  μ .

2.48

Trang 9

If fz ∈ H n satisfies



zf fzzfz z μ− 1

where 0 < λ ≤ M n β, μ, then fz ∈ Sβ.

Proof Note that

α1 n − μ



1− β

n1− β  ≥ 1, for β ≥

μ

n  μ ,

α1 n − μ



1− β

n1− β  < 1, for β <

μ

n  μ ,

α2 n  3μ



1− β

n  3μ1− β2− 8nμ1− β

2.50

Puttingα  1 inTheorem 2.3, we obtain the above corollary

Remark 2.6 Our results extend the results given by Obradovi´c5, and Obradovi´c and Owa

6

Theorem 2.7 Let 0 < μ < 1, 0 ≤ β < 1, Re{c} > −μ, and let

β nβ, μ



n − μ1− βn  c − μ



n − μ1− βc − μ , if β ≥ n  μ μ ,



n − μ1− 2βn  c − μ



n2 2μ2

1− β− nμc − μ , if β < n  μ μ .

2.51

If fz ∈ H n satisfies



zf fzzfz z μ− 1

where 0 < λ ≤ β n β, μ, and

Fz  z c − μ

z c−μ

z

0

 t

ft

μ

t c−μ−1 dt

−1/μ

then Fz ∈ Sβ.

Trang 10

Proof Let

Qz  Fz

Fz

1μ

Then from2.52 and 2.53 we obtain

Qz  c − μ1 Qz  fz

fz

1μ

Hence, by using Theorem 1 given by Hallenbeck and Ruscheweyh7, we have that

Qz ≺ 1  λ1z, λ1 c − μλ

and the desired result easily follows fromCorollary 2.5

Forc  μ  1, we have the following corollary.

Corollary 2.8 Let 0 < μ < 1, 0 ≤ β < 1, and let

β nβ, μ



n − μ1− βn  1



n − μ1− β , if 1 > β ≥

μ

n  μ ,



n − μ1− 2βn  1



n2 2μ2

1− β− nμ , if 0 ≤ β <

μ

n  μ .

2.57

If fz ∈ H n satisfies



zf fzzfz z μ− 1

where 0 < λ ≤ β n β, μ, and

Fz  z

 1

z

z

0

 t

ft

μ

dt

−1/μ

then Fz ∈ Sβ.

Acknowledgment

The authors would like to thank the referee for giving them thoughtful suggestions which greatly improved the presentation of the paper Bao Gejun was supported by NSF of P.R.Chinano 11071048

Trang 11

1 C Pommerenke, Univalent Functions with a Chapter on Quadratic Differentials by Gerd Jensen, vol 20 of

Studia Mathematica/Mathematische Lehrb ¨ucher, Vandenhoeck & Ruprecht, G ¨ottingen, Germany, 1975.

2 V Singh, “Univalent functions with bounded derivative in the unit disc,” Indian Journal of Pure and

Applied Mathematics, vol 8, no 11, pp 1370–1377, 1977.

3 R Fournier, “On integrals of bounded analytic functions in the closed unit disc,” Complex Variables.

Theory and Application, vol 11, no 2, pp 125–133, 1989.

4 R Fournier, “The range of a continuous linear functional over a class of functions defined by

subordination,” Glasgow Mathematical Journal, vol 32, no 3, pp 381–387, 1990.

5 M Obradovi´c, “A class of univalent functions,” Hokkaido Mathematical Journal, vol 27, no 2, pp 329–

335, 1998

6 M Obradovi´c and S Owa, “Some sufficient conditions for strongly starlikeness,” International Journal

of Mathematics and Mathematical Sciences, vol 24, no 9, pp 643–647, 2000.

7 D J Hallenbeck and S Ruscheweyh, “Subordination by convex functions,” Proceedings of the American

Mathematical Society, vol 52, pp 191–195, 1975.

Ngày đăng: 21/06/2014, 06:20

TỪ KHÓA LIÊN QUAN

TÀI LIỆU CÙNG NGƯỜI DÙNG

TÀI LIỆU LIÊN QUAN

🧩 Sản phẩm bạn có thể quan tâm