Table of Contents
ISRN Geometry
VolumeΒ 2011, Article IDΒ 151474, 8 pages
http://dx.doi.org/10.5402/2011/151474
Research Article

Criteria for Strongly Starlike Functions

Department of Mathematics, Changshu Institute of Technology, Jiangsu, Changshu 215500, China

Received 5 April 2011; Accepted 11 May 2011

Academic Editors: A.Β Belhaj and A. A.Β Ungar

Copyright Β© 2011 Neng Xu. 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.

Abstract

Let 𝑓(𝑧) be analytic in the unit disk π‘ˆ={π‘§βˆΆ|𝑧|<1} with 𝑓(0)=π‘“ξ…ž(0)βˆ’1=0 and (𝑓(𝑧)/𝑧)π‘“ξ…ž(𝑧)β‰ 0. By using the method of differential subordinations, we determine the largest number 𝛼(𝛽,πœ†,πœ‡,π‘š) such that, for some 𝛽,πœ†,πœ‡, and π‘š, the differential subordination πœ†(π‘§π‘“ξ…ž(𝑧)/𝑓(𝑧))1βˆ’πœ‡(1+(π‘§π‘“ξ…žξ…ž(𝑧)/π‘“ξ…ž(𝑧))βˆ’π‘§π‘“ξ…ž(𝑧)/𝑓(𝑧))+(π‘§π‘“ξ…ž(𝑧)/𝑓(𝑧))π‘šβ‰Ί(1+𝑧/1βˆ’π‘§)𝛼(𝛽,πœ†,πœ‡,π‘š)(π‘§βˆˆπ‘ˆ) implies π‘§π‘“ξ…ž(𝑧)/𝑓(𝑧)β‰Ί(1+𝑧/1βˆ’π‘§)𝛽. Some useful consequences of this result are also given.

1. Introduction

Let 𝐴 be the class of functions of the form 𝑓(𝑧)=𝑧+βˆžξ“π‘›=2π‘Žπ‘›π‘§π‘›,(1.1) which are analytic in π‘ˆ={π‘§βˆΆ|𝑧|<1}. A function 𝑓(𝑧)∈𝐴 is said to be starlike of order 𝛼 if Reπ‘§π‘“ξ…ž(𝑧)𝑓(𝑧)>𝛼(π‘§βˆˆπ‘ˆ)(1.2) for some 𝛼(0≀𝛼<1). We denote this class by π‘†βˆ—(𝛼). A function 𝑓(𝑧)∈𝐴 is said to be convex of order 𝛼 if ξ‚»Re1+π‘§π‘“ξ…žξ…ž(𝑧)π‘“ξ…žξ‚Ό(𝑧)>𝛼(π‘§βˆˆπ‘ˆ)(1.3) for some 𝛼(0≀𝛼<1). We denote this class by 𝐢(𝛼). Further, a function 𝑓(𝑧)∈𝐴 is said to be strongly starlike of order 𝛽 if ||||ξ‚»argπ‘§π‘“ξ…ž(𝑧)ξ‚Ό||||<πœ‹π‘“(𝑧)2𝛽(π‘§βˆˆπ‘ˆ)(1.4) for some 𝛽(0<𝛽≀1). Also we denote this class by ξ‚π‘†βˆ—(𝛽)(0<𝛽≀1). Clearly π‘†βˆ—(𝛼)βŠ‚π‘†βˆ—(0)β‰‘π‘†βˆ— for 𝑆0≀𝛼<1,βˆ—(1)=π‘†βˆ—.

Let 𝑓(𝑧) and 𝑔(𝑧) be analytic in π‘ˆ. Then the function 𝑓(𝑧) is said to be subordinate to 𝑔(𝑧), written as 𝑓(𝑧)≺𝑔(𝑧), if there exists an analytic function 𝑀(𝑧) with 𝑀(0)=0 and |𝑀(𝑧)|<1(π‘§βˆˆπ‘ˆ) such that 𝑓(𝑧)=𝑔(𝑀(𝑧)) for π‘§βˆˆπ‘ˆ. If 𝑔(𝑧) is univalent in π‘ˆ, then 𝑓(𝑧)≺𝑔(𝑧) is equivalent to 𝑓(0)=𝑔(0) and 𝑓(π‘ˆ)βŠ‚π‘”(π‘ˆ). It is easy to see that a function 𝑆𝑓(𝑧)βˆˆβˆ—(𝛽)(π‘œ<𝛽≀1) if and only if π‘§π‘“ξ…ž(𝑧)≺𝑓(𝑧)1+𝑧1βˆ’π‘§π›½(π‘§βˆˆπ‘ˆ).(1.5)

A number of results for strongly starlike functions in π‘ˆ have been obtained by several authors (see, e.g., [1–7]). In this paper, by using the method of differential subordinations, we determine the largest number 𝛼(𝛽,πœ†,πœ‡,π‘š) such that, for some 𝛽,πœ†,πœ‡, and π‘š, the differential subordination πœ†ξ‚΅π‘§π‘“ξ…ž(𝑧)𝑓(𝑧)1βˆ’πœ‡ξ‚΅1+π‘§π‘“ξ…žξ…ž(𝑧)π‘“ξ…žβˆ’(𝑧)π‘§π‘“ξ…ž(𝑧)ξ‚Ά+𝑓(𝑧)π‘§π‘“ξ…ž(𝑧)𝑓(𝑧)π‘šβ‰Ίξ‚€1+𝑧1βˆ’π‘§π›Ό(𝛽,πœ†,πœ‡,π‘š)(π‘§βˆˆπ‘ˆ)(1.6) implies π‘§π‘“ξ…ž(𝑧)≺𝑓(𝑧)1+𝑧1βˆ’π‘§π›½.(1.7) Our results improved or extended the above results.

To prove our results, we need the following lemma due to Miller and Mocanu [3].

Lemma 1.1. Let 𝑔(𝑧) be analytic and univalent in U, and let πœƒ(𝑀) and πœ‘(𝑀) be analytic in a domain D containing 𝑔(π‘ˆ), with πœ‘(𝑀)β‰ 0 when π‘€βˆˆπ‘”(π‘ˆ). Set 𝑄(𝑧)=π‘§π‘”ξ…ž(𝑧)πœ‘(𝑔(𝑧)),β„Ž(𝑧)=πœƒ(𝑔(𝑧))+𝑄(𝑧)(1.8) and suppose that (i)𝑄(𝑧) is starlike univalent in U, (ii)Reπ‘§β„Žξ…ž(𝑧)/𝑄(𝑧)=Re{πœƒβ€²(𝑔(𝑧))/πœ‘(𝑔(𝑧))+π‘§π‘„ξ…ž(𝑧)/𝑄(𝑧)}>0(π‘§βˆˆπ‘ˆ).
If 𝑝(𝑧) is analytic in U, with 𝑝(0)=𝑔(0),𝑝(π‘ˆ)βŠ‚π·, πœƒ(𝑝(𝑧))+π‘§π‘ξ…ž(𝑧)πœ‘(𝑝(𝑧))β‰Ίπœƒ(𝑔(𝑧))+π‘§π‘”ξ…ž(𝑧)πœ‘(𝑔(𝑧))=β„Ž(𝑧),(1.9) then 𝑝(𝑧)≺𝑔(𝑧) and 𝑔(𝑧) is the best dominant of (1.9).

2. Main Results

Theorem 2.1. Let 0<𝛽<1,πœ‡βˆˆ{0,1,2},π‘š an integer, |π‘š+πœ‡βˆ’1|𝛽<1,π‘šπœ†β‰₯0, and πœ†β‰ 0. If 𝑓(𝑧)∈𝐴 satisfies (𝑓(𝑧)/𝑧)π‘“ξ…ž(𝑧)β‰ 0(π‘§βˆˆπ‘ˆ) and πœ†ξ‚΅π‘§π‘“ξ…ž(𝑧)𝑓(𝑧)1βˆ’πœ‡ξ‚΅1+π‘§π‘“ξ…žξ…ž(𝑧)π‘“ξ…žβˆ’(𝑧)π‘§π‘“ξ…ž(𝑧)ξ‚Ά+𝑓(𝑧)π‘§π‘“ξ…ž(𝑧)𝑓(𝑧)π‘šβ‰Ίξ‚€1+𝑧1βˆ’π‘§π›Ό(𝛽,πœ†,πœ‡,π‘š)(π‘§βˆˆπ‘ˆ),(2.1) where =2𝛼(𝛽,πœ†,πœ‡,π‘š)πœ‹Γ—ξ‚΅arctantan(π‘š+πœ‡βˆ’1)π›½πœ‹2+πœ†π›½(1βˆ’(π‘š+πœ‡βˆ’1)𝛽)1βˆ’(π‘š+πœ‡βˆ’1)𝛽/2(1+(π‘š+πœ‡βˆ’1)𝛽)1+(π‘š+πœ‡βˆ’1)𝛽/2ξ‚Άcos(π‘š+πœ‡βˆ’1)π›½πœ‹/2+(1βˆ’πœ‡)𝛽,(2.2) then π‘§π‘“ξ…ž(𝑧)≺𝑓(𝑧)1+𝑧1βˆ’π‘§π›½(2.3) and 𝛼(𝛽,πœ†,πœ‡,π‘š) given by (2.2) is the largest number such that (2.3) holds.

Proof. Let 𝑓(𝑧)∈𝐴 with (𝑓(𝑧)/𝑧)π‘“ξ…ž(𝑧)β‰ 0(π‘§βˆˆπ‘ˆ), and define the function 𝑝(𝑧) in π‘ˆ by 𝑝(𝑧)=π‘§π‘“ξ…ž(𝑧)𝑓(𝑧).(2.4) Then 𝑝(𝑧) is analytic in π‘ˆ and πœ†ξ‚΅π‘§π‘“ξ…ž(𝑧)𝑓(𝑧)1βˆ’πœ‡ξ‚΅1+π‘§π‘“ξ…žξ…ž(𝑧)π‘“ξ…žβˆ’(𝑧)π‘§π‘“ξ…ž(𝑧)ξ‚Ά+𝑓(𝑧)π‘§π‘“ξ…ž(𝑧)𝑓(𝑧)π‘š=πœ†π‘§π‘ξ…ž(𝑧)π‘πœ‡(𝑧)+π‘π‘š(𝑧).(2.5) Let 0<𝛽<1,πœ‡βˆˆ{0,1,2},π‘š an integer, |π‘š+πœ‡βˆ’1|𝛽<1,π‘šπœ†β‰₯0,πœ†β‰ 0, and 𝐷=𝐢,πœ‡=0,π‘šβ‰₯0,𝐢⧡{0},πœ‡=1,2orπ‘š<0,(2.6) and choose 𝑔(𝑧)=1+𝑧1βˆ’π‘§π›½,πœƒ(𝑀)=π‘€π‘šπœ†,πœ‘(𝑀)=π‘€πœ‡.(2.7) Then 𝑔(𝑧) is analytic and univalent in π‘ˆ,𝑝(0)=𝑔(0)=1,𝑝(π‘ˆ)βŠ‚π·,andπœƒ(𝑀) and πœ‘(𝑀) satisfy the conditions of the lemma. The function 𝑄(𝑧)=π‘§π‘”ξ…ž(𝑧)πœ‘(𝑔(𝑧))=2πœ†π›½π‘§(1+𝑧)1+(πœ‡βˆ’1)𝛽(1βˆ’π‘§)1βˆ’(πœ‡βˆ’1)𝛽(2.8) is univalent and starlike in π‘ˆ because Reπ‘§π‘„ξ…ž(𝑧)π‘„ξ‚€βˆ’π‘§(𝑧)=1+(1+(πœ‡βˆ’1)𝛽)Re𝑧1+𝑧+(1βˆ’(πœ‡βˆ’1)𝛽)Re1βˆ’π‘§>0(π‘§βˆˆπ‘ˆ)(2.9) for |(πœ‡βˆ’1)|𝛽<1. Further, we have that ξ‚€πœƒ(𝑔(𝑧))+𝑄(𝑧)=1+𝑧1βˆ’π‘§π‘šπ›½+2πœ†π›½π‘§(1+𝑧)1+(πœ‡βˆ’1)𝛽(1βˆ’π‘§)1βˆ’(πœ‡βˆ’1)𝛽=β„Ž(𝑧)(say),(2.10)π‘§β„Žξ…ž(𝑧)𝑄=π‘š(𝑧)πœ†ξ‚€1+𝑧1βˆ’π‘§(π‘š+πœ‡βˆ’1)𝛽+π‘§π‘„ξ…ž(𝑧)𝑄(𝑧)(π‘§βˆˆπ‘ˆ).(2.11) Since 0≀|π‘š+πœ‡βˆ’1|𝛽<1, we have that ||||ξ‚€arg1+𝑧1βˆ’π‘§(π‘š+πœ‡βˆ’1)𝛽||||<πœ‹2(π‘§βˆˆπ‘ˆ)(2.12) and so Reπ‘§β„Žξ…ž(𝑧)=π‘šπ‘„(𝑧)πœ†ξ‚»ξ‚€Re1+𝑧1βˆ’π‘§(π‘š+πœ‡βˆ’1)𝛽+Reπ‘§π‘„ξ…ž(𝑧)𝑄(𝑧)>0(π‘§βˆˆπ‘ˆ).(2.13) Inequality (2.13) shows that the function β„Ž(𝑧) is close-to-convex and univalent in π‘ˆ. Letting 0<πœƒ<πœ‹ and π‘₯=cot(πœƒ/2)>0, then β„Žξ€·π‘’π‘–πœƒξ€Έ=ξ‚΅π‘₯π‘šπ›½π‘’(π‘š+πœ‡βˆ’1)π›½πœ‹π‘–/2+πœ†π›½π‘–2ξ‚€π‘₯1βˆ’(πœ‡βˆ’1)𝛽+1π‘₯1+(πœ‡βˆ’1)𝛽𝑒(1βˆ’πœ‡)π›½πœ‹π‘–/2(2.14) and so 𝑒argβ„Žπ‘–πœƒξ€Έ=arctan𝑀(π‘₯)+(1βˆ’πœ‡)π›½πœ‹2,(2.15) where 𝑀(π‘₯)=tan(π‘š+πœ‡βˆ’1)π›½πœ‹2+πœ†π›½ξ‚€π‘₯2cos(π‘š+πœ‡βˆ’1)π›½πœ‹/21βˆ’(π‘š+πœ‡βˆ’1)𝛽+1π‘₯1+(π‘š+πœ‡βˆ’1)𝛽.(2.16) It is easy to know that 𝑀(π‘₯) takes its minimum value at √π‘₯=1+(π‘š+πœ‡βˆ’1)𝛽/1βˆ’(π‘š+πœ‡βˆ’1)𝛽. Hence, in view of β„Ž(π‘’βˆ’π‘–πœƒ)=β„Ž(π‘’π‘–πœƒ), we deduce that inf|𝑧|(𝑧≠1)||||βŽ›βŽœβŽœβŽπ‘€βŽ›βŽœβŽœβŽξƒŽargβ„Ž(𝑧)=arctan1+(π‘š+πœ‡βˆ’1)π›½βŽžβŽŸβŽŸβŽ βŽžβŽŸβŽŸβŽ +1βˆ’(π‘š+πœ‡βˆ’1)𝛽(1βˆ’πœ‡)π›½πœ‹2πœ‹=𝛼(𝛽,πœ†,πœ‡,π‘š)2,(2.17) where 𝛼(𝛽,πœ†,πœ‡,π‘š) is given by (2.2).
Now, if 𝑓(𝑧) satisfies (2.1), it follows from (2.17) that the subordinationπœ†ξ‚΅π‘§π‘“β€²(𝑧)𝑓(𝑧)1βˆ’πœ‡ξ‚΅1+π‘§π‘“ξ…žξ…ž(𝑧)π‘“ξ…žβˆ’(𝑧)π‘§π‘“ξ…ž(𝑧)ξ‚Ά+𝑓(𝑧)π‘§π‘“ξ…ž(𝑧)𝑓(𝑧)π‘šβ‰Ίβ„Ž(𝑧)(π‘§βˆˆπ‘ˆ)(2.18) holds. Hence it follows from (2.5), (2.7), (2.10), and (2.18) that πœƒ(𝑝(𝑧))+π‘§π‘ξ…ž(𝑧)πœ‘(𝑝(𝑧))β‰Ίπœƒ(𝑔(𝑧))+π‘§π‘”ξ…ž(𝑧)πœ‘(𝑔(𝑧))=β„Ž(𝑧)(2.19) holds. Therefore, by virtue of the lemma, we conclude that 𝑝(𝑧)≺𝑔(𝑧), that is, (2.3) holds.
Next we consider the extremal functionξ‚»ξ€œπ‘“(𝑧)=𝑧exp𝑧01𝑑1+𝑑1βˆ’π‘‘π›½ξ‚Άξ‚Όβˆ’1𝑑𝑑.(2.20) Then 𝑓(𝑧) satisfies πœ†ξ‚΅π‘§π‘“ξ…ž(𝑧)𝑓(𝑧)1βˆ’πœ‡ξ‚΅1+π‘§π‘“ξ…žξ…ž(𝑧)π‘“ξ…žβˆ’(𝑧)π‘§π‘“ξ…ž(𝑧)ξ‚Ά+𝑓(𝑧)π‘§π‘“ξ…ž(𝑧)𝑓(𝑧)π‘š=ξ‚€1+𝑧1βˆ’π‘§π‘šπ›½+2πœ†π›½π‘§(1+𝑧)1+(πœ‡βˆ’1)𝛽(1βˆ’π‘§)1βˆ’(πœ‡βˆ’1)𝛽=β„Ž(𝑧),(2.21) and it follows from (2.17) that the bound 𝛼(𝛽,πœ†,πœ‡,π‘š) in (2.1) is sharp. The proof of the theorem is complete.

Making use of the theorem, we can obtain a number of interesting results.

Letting πœ‡=0 and π‘š=2 in the theorem, we have the following corollary.

Corollary 2.2. Let 0<𝛽<1 and πœ†>0. If 𝑓(𝑧)∈𝐴 satisfies 𝑓(𝑧)β‰ 0(0<|𝑧|<1) and π‘§π‘“ξ…ž(𝑧)ξ‚΅πœ†ξ‚΅π‘“(𝑧)1+π‘§π‘“ξ…žξ…ž(𝑧)π‘“ξ…žξ‚Ά(𝑧)+(1βˆ’πœ†)π‘§π‘“ξ…ž(𝑧)≺𝑓(𝑧)1+𝑧1βˆ’π‘§π›Ό3(𝛽,πœ†)(π‘§βˆˆπ‘ˆ),(2.22) where 𝛼32(𝛽,πœ†)=πœ‹ξ‚΅arctantanπ›½πœ‹2+πœ†π›½(1βˆ’π›½)(1βˆ’π›½)/2(1+𝛽)(1+𝛽)/2ξ‚Άcosπ›½πœ‹/2+𝛽,(2.23) then π‘§π‘“ξ…ž(𝑧)≺𝑓(𝑧)1+𝑧1βˆ’π‘§π›½(2.24) and 𝛼3(𝛽,πœ†) given by (2.23) is the largest number such that (2.24) holds.

Remark 2.3. Ramesha et al. [6] have proved that, if 𝑓(𝑧)∈𝐴 satisfies 𝑓(𝑧)β‰ 0(0<|𝑧|<1) and ξ‚»Reπ‘§π‘“ξ…ž(𝑧)𝑓(𝑧)1+π‘§π‘“ξ…žξ…ž(𝑧)π‘“ξ…ž(𝑧)ξ‚Άξ‚Ό>0(π‘§βˆˆπ‘ˆ),(2.25) then 𝑆𝑓(𝑧)βˆˆβˆ—(1/2).
For 𝛽=1/2 and πœ†>0, it follows from (2.23) that 𝛼3(1/2,πœ†)>1. Hence, the image of 𝑀=((1+𝑧)/(1βˆ’π‘§))𝛼3(1/2,πœ†)(π‘§βˆˆπ‘ˆ) is a region which properly contains the right half plane. Thus we conclude that Corollary 2.2 with 𝛽=1/2 and πœ†=1 is better than the result of Ramesha et al. [6].
Letting 𝛼3(𝛽,πœ†)=1 in Corollary 2.2, we have the following corollary.

Corollary 2.4. Let πœ†>0. If 𝑓(𝑧)∈𝐴 satisfies 𝑓(𝑧)β‰ 0(0<|𝑧|<1) and ξ‚»Reπ‘§π‘“ξ…ž(𝑧)ξ‚΅πœ†ξ‚΅π‘“(𝑧)1+π‘§π‘“ξ…žξ…ž(𝑧)π‘“ξ…žξ‚Ά(𝑧)+(1βˆ’πœ†)π‘§π‘“ξ…ž(𝑧)𝑓(𝑧)ξ‚Άξ‚Ό>0(π‘§βˆˆπ‘ˆ),(2.26) then 𝑆𝑓(𝑧)βˆˆβˆ—(𝛽), where π›½βˆˆ(0,1) is the root of the equation 2πœ‹ξ‚΅arctantanπ›½πœ‹2+πœ†(1βˆ’π›½)(1βˆ’π›½)/2(1+𝛽)(1+𝛽)/2ξ‚Άcosπ›½πœ‹/2+𝛽=1.(2.27)

Remark 2.5. For πœ†=1, Corollary 2.4 reduces to a main result of Nunokawa et al. [5, Theorem 1] by a different method.

Remark 2.6. Note that lim𝛽→1𝛼3(𝛽,1)=2. Hence it follows from Corollary 2.2 that ||||ξ‚»argπ‘§π‘“ξ…ž(𝑧)𝑓(𝑧)1+π‘§π‘“ξ…žξ…ž(𝑧)π‘“ξ…ž||||(𝑧)ξ‚Άξ‚Ό<πœ‹(π‘§βˆˆπ‘ˆ)(2.28) implies 𝑓(𝑧)βˆˆπ‘†βˆ—.(2.29)
Letting πœ‡=π‘š=1 in the theorem, we have the following corollary.

Corollary 2.7. Let 0<𝛽<1,πœ†>0. If 𝑓(𝑧)∈𝐴 satisfies (𝑓(𝑧)/𝑧)π‘“ξ…ž(𝑧)β‰ 0(π‘§βˆˆπ‘ˆ) and πœ†ξ‚΅1+π‘§π‘“ξ…žξ…ž(𝑧)π‘“ξ…žξ‚Ά(𝑧)+(1βˆ’πœ†)π‘§π‘“ξ…ž(𝑧)≺𝑓(𝑧)1+𝑧1βˆ’π‘§π›Ό1(𝛽,πœ†)(π‘§βˆˆπ‘ˆ),(2.30) where 𝛼12(𝛽,πœ†)=πœ‹ξ‚΅arctantanπ›½πœ‹2+πœ†π›½(1βˆ’π›½)(1βˆ’π›½)/2(1+𝛽)(1+𝛽)/2ξ‚Άcosπ›½πœ‹/2,(2.31) then π‘§π‘“ξ…ž(𝑧)≺𝑓(𝑧)1+𝑧1βˆ’π‘§π›½(2.32) and 𝛼1(𝛽,πœ†) given by (2.31) is the largest number such that (2.32) holds.

Remark 2.8. Marjono and Thomas [2] proved the above result by a different method. For πœ†=𝛿/𝛾(𝛿>0,𝛾>0) Corollary 2.2 reduces to a result of Darus [1]. For πœ†=1, Corollary 2.2 reduces to a result of Nunokawa and Thomas [4].
Letting πœ‡=2 and π‘š=0 in the theorem, we have the following corollary.

Corollary 2.9. Let 0<𝛽<1 and πœ†β‰ 0. If 𝑓(𝑧)∈𝐴 satisfies π‘“ξ…ž(𝑧)β‰ 0(0<|𝑧|<1) and πœ†ξ€·1+π‘§π‘“ξ…žξ…žξ€Έ(𝑧)/π‘“ξ…ž(𝑧)+(1βˆ’πœ†)π‘§π‘“ξ…ž(𝑧)/𝑓(𝑧)π‘§π‘“ξ…žβ‰Ίξ‚€(𝑧)/𝑓(𝑧)1+𝑧1βˆ’π‘§π›Ό2(𝛽,πœ†)(π‘§βˆˆπ‘ˆ),(2.33) where 𝛼22(𝛽,πœ†)=πœ‹ξ‚΅arctantanπ›½πœ‹2+πœ†π›½(1βˆ’π›½)(1βˆ’π›½)/2(1+𝛽)(1+𝛽)/2ξ‚Άcosπ›½πœ‹/2βˆ’π›½,(2.34) then π‘§π‘“ξ…ž(𝑧)≺𝑓(𝑧)1+𝑧1βˆ’π‘§π›½(2.35) and 𝛼2(𝛽,πœ†) given by (2.34) is the largest number such that (2.35) holds.

Remark 2.10. For πœ†=1, Corollary 2.9 reduces to a result of Ravichandran and Darus [7].

Acknowledgments

This work was partially supported by the National Natural Science Foundation of China (Grant no. 10871094) and Natural Science Foundation of Universities of Jiangsu Province (Grant no. 08KJB110001).

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