Analytic and Harmonic Univalent FunctionsAbstractandAppliedAnalysis Analytic and Harmonic Univalent...
Transcript of Analytic and Harmonic Univalent FunctionsAbstractandAppliedAnalysis Analytic and Harmonic Univalent...
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Guest Editors: V. Ravichandran, Om P. Ahuja, and Rosihan M. Ali
Analytic and Harmonic Univalent Functions
Abstract and Applied Analysis
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Analytic and Harmonic Univalent Functions
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Abstract and Applied Analysis
Analytic and Harmonic Univalent Functions
Guest Editors: V. Ravichandran, Om P. Ahuja,and Rosihan M. Ali
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Copyright Β© 2014 Hindawi Publishing Corporation. All rights reserved.
This is a special issue published in βAbstract and Applied Analysis.β All articles are open access articles distributed under the CreativeCommons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the originalwork is properly cited.
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Editorial Board
Ravi P. Agarwal, USABashir Ahmad, Saudi ArabiaM. O. Ahmedou, GermanyNicholas D. Alikakos, GreeceDebora Amadori, ItalyDouglas R. Anderson, USAJan Andres, Czech RepublicGiovanni Anello, ItalyStanislav Antontsev, PortugalM. K. Aouf, EgyptNarcisa C. Apreutesei, RomaniaNatig M. Atakishiyev, MexicoFerhan M. Atici, USAIvan Avramidi, USASoohyun Bae, KoreaChuanzhi Bai, ChinaZhanbing Bai, ChinaDumitru Baleanu, TurkeyJoΜzef BanasΜ, PolandMartino Bardi, ItalyRoberto Barrio, SpainFeyzi BasΜ§ar, TurkeyAbdelghani Bellouquid, MoroccoDaniele Bertaccini, ItalyLucio Boccardo, ItalyIgor Boglaev, New ZealandMartin J. Bohner, USAGeraldo Botelho, BrazilElena Braverman, CanadaRomeo Brunetti, ItalyJanusz Brzdek, PolandDetlev Buchholz, GermanySun-Sig Byun, KoreaFabio M. Camilli, ItalyJinde Cao, ChinaAnna Capietto, ItalyJianqing Chen, ChinaWing-Sum Cheung, Hong KongMichel Chipot, SwitzerlandChangbum Chun, KoreaSoon-Yeong Chung, KoreaJaeyoung Chung, KoreaSilvia Cingolani, ItalyJean M. Combes, FranceMonica Conti, Italy
Juan Carlos CorteΜs LoΜpez, SpainGraziano Crasta, ItalyZhihua Cui, ChinaBernard Dacorogna, SwitzerlandVladimir Danilov, RussiaMohammad T. Darvishi, IranLuis F. Pinheiro de Castro, PortugalToka Diagana, USAJesuΜs I. DΔ±Μaz, SpainJosef DiblΜΔ±k, Czech RepublicFasma Diele, ItalyTomas Dominguez, SpainAlexander Domoshnitsky, IsraelMarco Donatelli, ItalyBoQing Dong, ChinaWei-Shih Du, TaiwanLuiz Duarte, BrazilRoman Dwilewicz, USAPaul W. Eloe, USAAhmed El-Sayed, EgyptLuca Esposito, ItalyKhalil Ezzinbi, MoroccoJulian F. Bonder, ArgentinaDashan Fan, USAAngelo Favini, ItalyMaΜrcia Federson, BrazilStathis Filippas, Equatorial GuineaAlberto Fiorenza, ItalyIlaria Fragala, ItalyXianlong Fu, ChinaMassimo Furi, ItalyJesuΜs G. Falset, SpainGiovanni P. Galdi, USAIsaac Garcia, SpainJ. A. G.a-R.guez, SpainLeszek Gasinski, PolandGyoΜrgy GaΜt, HungaryVladimir Georgiev, ItalyLorenzo Giacomelli, ItalyJaume GineΜ, SpainValery Y. Glizer, IsraelJean P. Gossez, BelgiumJose L. Gracia, SpainMaurizio Grasselli, ItalyLuca Guerrini, Italy
Yuxia Guo, ChinaQian Guo, ChinaC. P. Gupta, USAUno HaΜmarik, EstoniaMaoan Han, ChinaFerenc Hartung, HungaryJiaxin Hu, ChinaZhongyi Huang, ChinaChengming Huang, ChinaGennaro Infante, ItalyIvan Ivanov, BulgariaHossein Jafari, South AfricaJaan Janno, EstoniaAref Jeribi, TunisiaUncig Ji, KoreaZhongxiao Jia, ChinaLucas JoΜdar, SpainJong S. Jung, Republic of KoreaHenrik Kalisch, NorwayHamid R. Karimi, NorwayChaudry M. Khalique, South AfricaSatyanad Kichenassamy, FranceTero KilpelaΜinen, FinlandSung G. Kim, Republic of KoreaLjubisa Kocinac, SerbiaAndrei Korobeinikov, SpainPekka Koskela, FinlandVictor Kovtunenko, AustriaRen-Jieh Kuo, TaiwanPavel Kurasov, SwedenMilton C. L. Filho, BrazilMiroslaw Lachowicz, PolandKunquan Lan, CanadaRuediger Landes, USAIrena Lasiecka, USAMatti Lassas, FinlandChun-Kong Law, TaiwanMing-Yi Lee, TaiwanGongbao Li, ChinaElena Litsyn, IsraelShengqiang Liu, ChinaYansheng Liu, ChinaCarlos Lizama, ChileGuozhen Lu, USAJinhu LuΜ, China
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Grzegorz Lukaszewicz, PolandWanbiao Ma, ChinaNazim I. Mahmudov, TurkeyEberhard Malkowsky, TurkeySalvatore A. Marano, ItalyCristina Marcelli, ItalyPaolo Marcellini, ItalyJesuΜs MarΜΔ±n-Solano, SpainJose M. Martell, SpainM. MastyΕo, PolandMing Mei, CanadaTaras Melβnyk, UkraineAnna Mercaldo, ItalyStanislaw Migorski, PolandMihai MihaΜilescu, RomaniaFeliz MinhoΜs, PortugalDumitru Motreanu, FranceMaria Grazia Naso, ItalyGaston M. NβGuerekata, USAMicah Osilike, NigeriaMitsuharu OΜtani, JapanTurgut OΜzisΜ§, TurkeyNikolaos S. Papageorgiou, GreeceSehie Park, KoreaKailash C. Patidar, South AfricaKevin R. Payne, ItalyAdemir F. Pazoto, BrazilShuangjie Peng, ChinaAntonio M. Peralta, SpainSergei V. Pereverzyev, AustriaAllan Peterson, USAAndrew Pickering, SpainCristina Pignotti, ItalySomyot Plubtieng, ThailandMilan Pokorny, Czech RepublicSergio Polidoro, ItalyZiemowit Popowicz, PolandMaria M. Porzio, ItalyEnrico Priola, ItalyVladimir S. Rabinovich, Mexico
Irena RachuΜnkovaΜ, Czech RepublicMaria A. Ragusa, ItalySimeon Reich, IsraelAbdelaziz Rhandi, ItalyHassan Riahi, MalaysiaJuan P. RincoΜn-Zapatero, SpainLuigi Rodino, ItalyYuriy Rogovchenko, NorwayJulio D. Rossi, ArgentinaWolfgang Ruess, GermanyBernhard Ruf, ItalySatit Saejung, ThailandStefan G. Samko, PortugalMartin Schechter, USAJavier Segura, SpainValery Serov, FinlandNaseer Shahzad, Saudi ArabiaAndrey Shishkov, UkraineStefan Siegmund, GermanyAbdel-Maksoud A. Soliman, EgyptPierpaolo Soravia, ItalyMarco Squassina, ItalyHari M. Srivastava, CanadaSvatoslav StaneΜk, Czech RepublicStevo StevicΜ, SerbiaAntonio SuaΜrez, SpainWenchang Sun, ChinaWenyu Sun, ChinaRobert Szalai, UKSanyi Tang, ChinaChun-Lei Tang, ChinaGabriella Tarantello, ItalyNasser-Eddine Tatar, Saudi ArabiaGerd Teschke, GermanySergey Tikhonov, SpainClaudia Timofte, RomaniaThanh Tran, AustraliaJuan J. Trujillo, SpainGabriel Turinici, FranceMilan Tvrdy, Czech Republic
Mehmet nal, TurkeyCsaba Varga, RomaniaCarlos Vazquez, SpainJesus Vigo-Aguiar, SpainQing-WenWang, ChinaYushun Wang, ChinaShawn X. Wang, CanadaYouyu Wang, ChinaJing P. Wang, UKPeixuan Weng, ChinaNoemi Wolanski, ArgentinaNgai-Ching Wong, TaiwanPatricia J. Y. Wong, SingaporeYonghong Wu, AustraliaZili Wu, ChinaShi-Liang Wu, ChinaShanhe Wu, ChinaTiecheng Xia, ChinaXu Xian, ChinaYanni Xiao, ChinaGongnan Xie, ChinaFuding Xie, ChinaDaoyi Xu, ChinaZhenya Yan, ChinaXiaodong Yan, USANorio Yoshida, JapanBeong In Yun, KoreaAgacik Zafer, TurkeyJianming Zhan, ChinaWeinian Zhang, ChinaChengjian Zhang, ChinaZengqin Zhao, ChinaSining Zheng, ChinaYong Zhou, ChinaTianshou Zhou, ChinaChun-Gang Zhu, ChinaQiji J. Zhu, USAMalisa R. Zizovic, SerbiaWenming Zou, China
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Contents
Analytic and Harmonic Univalent Functions, V. Ravichandran, Om P. Ahuja, and Rosihan M. AliVolume 2014, Article ID 578214, 2 pages
ANote on Entire FunctionsThat Share Two Small Functions, Jun-Fan ChenVolume 2014, Article ID 601507, 9 pages
Radius Constants for Functions with the Prescribed Coefficient Bounds, Om P. Ahuja, Sumit Nagpal,and V. RavichandranVolume 2014, Article ID 454152, 12 pages
Third-Order Differential Subordination and Superordination Results for Meromorphically MultivalentFunctions Associated with the Liu-Srivastava Operator, Huo Tang, H. M. Srivastava, Shu-Hai Li,and Li-Na MaVolume 2014, Article ID 792175, 11 pages
Differential Subordinations for Nonanalytic Functions, Georgia Irina Oros and Gheorghe OrosVolume 2014, Article ID 251265, 9 pages
Upper Bound of Second Hankel Determinant for Certain Subclasses of Analytic Functions,Ming-Sheng Liu, Jun-Feng Xu, and Ming YangVolume 2014, Article ID 603180, 10 pages
On Certain Subclass of Harmonic Starlike Functions, A. Y. LashinVolume 2014, Article ID 467929, 7 pages
Initial Coefficients of Biunivalent Functions, See Keong Lee, V. Ravichandran, and Shamani SupramaniamVolume 2014, Article ID 640856, 6 pages
Starlikeness of Functions Defined byThird-Order Differential Inequalities and Integral Operators,R. Chandrashekar, Rosihan M. Ali, K. G. Subramanian, and A. SwaminathanVolume 2014, Article ID 723097, 6 pages
A Family of Minimal Surfaces and Univalent Planar Harmonic Mappings, Michael Dorff and Stacey MuirVolume 2014, Article ID 476061, 8 pages
Landau-TypeTheorems for Certain Biharmonic Mappings, Ming-Sheng Liu, Zhen-Xing Liu,and Jun-Feng XuVolume 2014, Article ID 925947, 7 pages
Some Connections between ClassU- and πΌ-Convex Functions, Edmond Aliaga and Nikola TuneskiVolume 2014, Article ID 692327, 4 pages
Meromorphic Solutions of Some Algebraic Differential Equations, Jianming Lin, Weiling Xiong,and Wenjun YuanVolume 2014, Article ID 796312, 5 pages
Differential Subordination Results for Analytic Functions in the Upper Half-Plane, Huo Tang,M. K. Aouf, Guan-Tie Deng, and Shu-Hai LiVolume 2014, Article ID 565727, 6 pages
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EditorialAnalytic and Harmonic Univalent Functions
V. Ravichandran,1 Om P. Ahuja,2 and Rosihan M. Ali3
1Department of Mathematics, University of Delhi, Delhi 110 007, India2Department of Mathematical Sciences, Kent State University, Burton, OH 44021, USA3School of Mathematical Sciences, Universiti Sains Malaysia, 11800 USM, Penang, Malaysia
Correspondence should be addressed to V. Ravichandran; [email protected]
Received 14 October 2014; Accepted 14 October 2014; Published 22 December 2014
Copyright Β© 2014 V. Ravichandran et al. This is an open access article distributed under the Creative Commons AttributionLicense, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properlycited.
Studies on analytic univalent functions became the focus ofintense researchwith the Bieberbach conjecture posed in 1916concerning the size of the moduli of the Taylor coefficientsof these functions. In efforts towards its resolution, theconjecture inspired the development of several ingeniouslydifferent mathematical techniques with powerful influence.These techniques include Lownerβs parametric representationmethod, the areamethod, Grunsky inequalities, andmethodsof variations. Despite the fact that the conjecture was affirma-tively settled by de Branges in 1985, complex function theorycontinued to remain a highly active relevant area of research.
Closely connected are harmonic univalent mappings,which arewidely known to have awealth of applications.Theyarise in the modelling of many physical problems, such as inthe study of fluid dynamics and elasticity problems, in theapproximation theory of plates subjected to normal loading,and in the investigations of Stokes flow in the engineering andbiological transport phenomena. Harmonic mappings arealso important to differential geometers because these mapsprovide isothermal (or conformal) parameters for minimalsurfaces. Indeed various properties of minimal surfaces suchas the Gauss curvature are studied more effectively throughplanar harmonic mappings.
Although a harmonic map provides a natural generaliza-tion to studies on analytic univalent functions, surprisinglyit fails to capture the interest of function theorists for quite aperiod of time. The defining moment came with the seminalpaper by Clunie and Sheil-Small in 1984. They introducedcomplex analytic approach in their studies and succeeded infinding viable analogues of the classical growth and distortion
theorems, covering theorems, and coefficient estimates inthe general setting of planar harmonic mappings. Althoughthere have been substantial steps forward in the studies ofharmonic mappings, yet many fundamental questions andconjectures remain unresolved. There is a great expectationthat the βharmonic Koebe functionβ will play the extremal rolein many of these problems, much akin to the role playedby the Koebe function in the classical theory of analyticunivalent functions.
This special issue aims to disseminate recent advances inthe studies of complex function theory, harmonic univalentfunctions, and their connections to produce deeper insightsand better understanding. These are crystallized in the formof original research articles or expository survey papers.
The response to this special issue was beyond ourexpectations. Forty papers were received in several areas ofresearch fields in analytic and harmonic univalent functions.All submitted papers went through a rigorous scrutiny oftwo or three peer-reviewed processes. Based on the reviewersβreports and editorsβ reviews, thirteen original research articleswere selected for inclusion in the special issue.
New concepts and techniques in the theory of first,second, and third order differential subordination and super-ordination for analytic functions (as well as nonanalyticfunctions) were introduced. These can be found in the threepapers entitled βThird-order differential subordination andsuperordination results for meromorphically multivalent func-tions associated with the Liu-Srivastava operatorβ (H. Tanget al.), βDifferential subordinations for nonanalytic functionsβ(G. I. Oros and G. Oros), and βDifferential subordination
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2 Abstract and Applied Analysis
results for analytic functions in the upper half-planeβ (H. Tanget al.). The paper βMeromorphic solutions of some algebraicdifferential equationsβ (J. Lin et al.) provides estimates on thegrowth order of meromorphic solutions of certain algebraicdifferential equations by means of the normal family theory.The paper βA note on entire functions that share two smallfunctionsβ by J.-F. Chen investigates a close linear relationshipbetween a non-constant entire function under certain condi-tions and its first derivative.
The paper βUpper bound of second Hankel determinantfor certain subclasses of analytic functionsβ (M.-S. Liu etal.) summarizes works done in the area of the Hankeldeterminant for univalent functions, while the paper βInitialcoefficients of biunivalent functionsβ (S. K. Lee et al.) givesestimates on the initial coefficients of the Taylor coefficientsof these functions.The paper βSome connections between classU- and πΌ-convex functionsβ (E. Aliaga and N. Tuneski) givessufficient conditions for an πΌ-convex function to be in aspecific class, while the paper βStarlikeness of functions definedby third-order differential inequalities and integral operatorsβ(R. Chandrashekar et al.) determines sufficient conditions foranalytic functions satisfying certain third-order differentialinequalities to be starlike.
Four papers in this issue deal with univalent harmonicmappings. These papers are βRadius constants for functionswith the prescribed coefficient boundsβ (O. P. Ahuja et al.),βOn certain subclass of harmonic starlike functionsβ (A. Y.Lashin), βA family of minimal surfaces and univalent planarharmonic mappingsβ (M. Dorff and S. Muir), and βLandau-type theorems for certain biharmonic mappingsβ (M.-S. Liuet al.). The first of these four papers, βRadius constants forfunctions with the prescribed coefficient bounds,β establishesa coefficient inequality for sense-preserving harmonic func-tions to compute the bounds for the radius of univalenceand radius of full starlikeness (and convexity) of positiveorder for functions with prescribed coefficient bound on theanalytic part; the second one, βOn certain subclass of harmonicstarlike functions,β discusses the geometric properties for anew class of harmonic univalent functions; the third one,βA family of minimal surfaces and univalent planar harmonicmappings,β presents a two-parameter family of minimalsurfaces constructed by lifting a family of planar harmonicmappings, while the fourth one, βLandau-type theoremsfor certain biharmonic mappings,β proves the Landau-typetheorems for biharmonic mappings connected with a linearcomplex operator.
We hope that the papers in this special issue will helpenrich our readers and stimulate further research.
Acknowledgment
Finally we take this opportunity to express our heartfeltgratitude to all contributing authors and to the reviewers forensuring the high standards of the issue.
V. RavichandranOm P. Ahuja
Rosihan M. Ali
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Research ArticleA Note on Entire Functions That Share Two Small Functions
Jun-Fan Chen
Department of Mathematics, Fujian Normal University, Fuzhou, Fujian 350007, China
Correspondence should be addressed to Jun-Fan Chen; [email protected]
Received 10 April 2014; Accepted 16 September 2014; Published 19 October 2014
Academic Editor: V. Ravichandran
Copyright Β© 2014 Jun-Fan Chen.This is an open access article distributed under the Creative CommonsAttribution License, whichpermits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
This note is to show that if π is a nonconstant entire function that shares two pairs of small functions ignoring multiplicities withits first derivative π, then there exists a close linear relationship between π and π. This result is a generalization of some resultsobtained by Rubel and Yang, Mues and Steinmetz, Zheng and Wang, and Qiu. Moreover, examples are provided to show that theconditions in the result are sharp.
1. Introduction and Main Result
Throughout this paper, we use standard notations in the Nev-anlinna theory (see, e.g., [1β4]). Let π(π§) be a meromorphicfunction. Here and in the following the word βmeromorphicβmeans meromorphic in the whole complex plane. We denoteby π(π, π) any real function of growth π(π(π, π)) as π β βoutside of a possible exceptional set of finite linear measure.The meromorphic function π is called a small function withrespect to π provided that π(π, π) = π(π, π).
Let π and π be two nonconstant meromorphic functions,and let π and π be two small functions with respect to π andπ. If the zeros of π β π and π β π coincide in locations andmultiplicities, then we say thatπ and π share the pair of smallfunctions (π, π) CM (counting multiplicities); if we do notconsider the multiplicities, then π and π are said to share thepair of small functions (π, π) IM (ignoring multiplicities). Wesee that π and π share the pair of small functions (π, π) CMif and only if π and π share the small function π CM, and πand π share the pair of small functions (π, π) IM if and onlyif π and π share the small function π IM.The same argumentapplies in the casewhen π and π are two values in the extendedplane.
Moreover, we introduce the following notations. Denotethe set of those points π§ β C by π
(π,π)(π1, π2) such that π§ is
a zero of π β π1of multiplicity π and a zero of π β π
2of
multiplicity π. The set π(π,π)
(π1, π2) can be similarly defined.
Now the notations π(π,π)
(π, 1/(π β π1)) and π
(π,π)(π, 1/(π β
π1)) denote the counting function and the reduced counting
function ofπwith respect to the set π(π,π)
(π1, π2), respectively.
The notationsπ(π,π)
(π, 1/(πβ π2)) andπ
(π,π)(π, 1/(π
β π2))
can be similarly defined.Many mathematicians have been interested in the value
distribution of different expressions of an entire or meromor-phic function and obtained a lot of fruitful and significantresults. When dealing with an entire function π and itsderivative π, Rubel and Yang [5] proved the following.
Theorem A. Let π be a nonconstant entire function, and let πand π be distinct finite complex numbers. If π and π share πand π CM, then π β‘ π.
Mues and Steinmetz [6] improved Theorem A andobtained the following.
Theorem B. Let π be a nonconstant entire function, and let πand π be distinct finite complex numbers. If π and π share πand π IM, then π β‘ π.
When the values π and πwere replaced by two small func-tions related to π, Zheng and Wang [7] proved the follow-ing.
Theorem C. Let π be a nonconstant entire function, and let πand π be distinct small functions with respect to π. If π and πshare π and π CM, then π β‘ π.
Recently, Qiu [8] proved the following result which wasan improvement of Theorem C.
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2 Abstract and Applied Analysis
TheoremD. Let π be a nonconstant entire function, and let πand π be distinct small functions with respect to π. If π and πshare π and π IM, then π β‘ π.
This paper is concerned with what can be said when theIM shared small function is replaced by the IM shared thepair of small functions in Theorem D. In fact, we prove thefollowing result by using themethod of [8], which generalizesthe above theorems from the point of view of shared pairs.
Theorem 1. Let π be a nonconstant entire function, and let π1,
π2, π1, and π
2be four small functions of π such that none of
them is identically equal to β and π1
ΜΈβ‘ π1, π2
ΜΈβ‘ π2. If π and
π share (π
1, π2) and (π
1, π2) IM, then (π
2βπ2)πβ (π
1βπ1)π+
π1π2β π2π1β‘ 0.
Remark 2. Let π1β‘ π2and π1β‘ π2.Then byTheorem 1 we can
get Theorem D.
Remark 3. Theorem 1 shows that a nonconstant entire func-tion sharing two pairs of small functions ignoring multiplic-ities with its first derivative implies that there exists a closelinear relationship between them.
Example 4 (see [9]). Let π = π½ + (π½ β πΌ)/(β β 1), where
πΌ = β
1
3
πβ2π§
β
1
2
πβπ§
, π½ = β
1
3
πβ2π§
+
1
2
πβπ§
, β = πβππ§
.
(1)
Set π = π½, π = πΌ. Then π(π, π) = π(π, π) and π(π, π) =π(π, π). It is easy to verify that
πβ π = π
2π§(π β π) (π β π½) ,
πβ π = π
2π§(π β π) (π β πΌ) .
(2)
Thus π and π share (π, π) and (π, π) IM, but π ΜΈβ‘ π. Thisshows that the conclusion inTheorem 1 is not valid generallyfor a meromorphic function π.
Example 5. Let π = π2π§ + π§, π1= 2π§, π
2= 2π§ + 1, π
1= π§, and
π2= π§ + 1. Then π and π share (π
1, π2) IM but do not share
(π1, π2) IM.Clearly, (π
2βπ2)πβ(π
1βπ1)π+π1π2βπ2π1
ΜΈβ‘ 0.Thisshows that the condition in Theorem 1 that π and π share(π1, π2) and (π
1, π2) IM cannot be weakened.
2. Some Lemmas
Lemma 1. Let π be a nonconstant entire function, and let π1,
π2, π1, and π
2be four small functions ofπ such that none of them
is identically equal to β and π1
ΜΈβ‘ π1, π2
ΜΈβ‘ π2. If π and π
share (π1, π2) and (π
1, π2) IM, then π(π, π) = π(π, π) := π(π).
Proof. Note thatπ andπ share (π1, π2) and (π
1, π2) IM. By the
second fundamental theorem, we get
π (π, π) β€ π(π,
1
π β π1
) + π(π,
1
π β π1
) + π (π, π)
= π(π,
1
πβ π2
) + π(π,
1
πβ π2
) + π (π, π)
β€ 2π (π, π) + π (π, π) ,
π (π, π) β€ π(π,
1
πβ π2
) + π(π,
1
πβ π2
) + π (π, π)
= π(π,
1
π β π1
) + π(π,
1
π β π1
) + π (π, π)
β€ 2π (π, π) + π (π, π) + π (π, π) ,
(3)
which implies from the definition of π(π, π) that π(π, π) =π(π, π
) and π(π, π) = π(π, π), respectively.
This completes the proof of Lemma 1.
Lemma 2. Let π be a nonconstant entire function, and let π1,
π2, π1, and π
2be four small functions of π such that none of
them is identically equal to β and π1
ΜΈβ‘ π1, π2
ΜΈβ‘ π2. If π and
π share (π
1, π2) and (π
1, π2) IM, then
π(π, π) = π(π,
1
π β π1
) + π(π,
1
π β π1
) + π (π) , (4)
provided that (π2β π2)π β (π
1β π1)π+ π1π2β π2π1
ΜΈβ‘ 0.
Proof. Note that
(π2β π2) π β (π
1β π1) π+ π1π2β π2π1
ΜΈβ‘ 0, (5)
(π2β π2) π β (π
1β π1) π+ π1π2β π2π1
β‘ (π2β π2) (π β π
1) β (π
1β π1) (πβ π2) ,
(6)
(π2β π2) π β (π
1β π1) π+ π1π2β π2π1
β‘ (π2β π2) (π β π
1) β (π
1β π1) (πβ π2) .
(7)
Since π and π share (π1, π2) and (π
1, π2) IM, from Lemma 1,
(5)β(7), and the condition that π is entire, we have
π(π,
1
π β π1
) + π(π,
1
π β π1
)
β€ π(π,
1
(π2β π2) π β (π
1β π1) π+ π1π2β π2π1
)
β€ π (π, (π2β π2) π β (π
1β π1) π) + π (π)
= π (π, (π2β π2) π β (π
1β π1) π)
+ π(π, (π2β π2) π β (π
1β π1) π) + π (π)
= π (π, (π2β π2) π β (π
1β π1) π) + π (π)
β€ π(π,
(π2β π2) π β (π
1β π1) π
π
) + π (π, π) + π (π)
β€ π (π, π) + π (π) .
(8)
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Abstract and Applied Analysis 3
On the other hand, by the second fundamental theorem,Lemma 1, and the condition that π is entire, we get
π(π, π) = π (π, π) β€ π(π,
1
π β π1
)
+ π(π,
1
π β π1
) + π (π) .
(9)
Now (8) and (9) imply
π(π, π) = π(π,
1
π β π1
) + π(π,
1
π β π1
) + π (π) . (10)
This completes the proof of Lemma 2.
Lemma 3. Let π be a nonconstant entire function, and let π1,
π2, π1, and π
2be four small functions of π such that none of
them is identically equal to β and π1
ΜΈβ‘ π1, π2
ΜΈβ‘ π2. Suppose
that π and π share (π1, π2) and (π
1, π2) IM. Set
πΌ = (π
1β π
1) (π β π
1) β (π
1β π1) (πβ π
1)
= (π
1β π
1) (π β π
1) β (π
1β π1) (πβ π
1) ,
(11)
π½ = (π
2β π
2) (πβ π2) β (π
2β π2) (πβ π
2)
= (π
2β π
2) (πβ π2) β (π
2β π2) (πβ π
2) ,
(12)
π =
πΌ [(π2β π2) π β (π
1β π1) π+ π1π2β π2π1]
(π β π1) (π β π
1)
, (13)
β =
π½ [(π2β π2) π β (π
1β π1) π+ π1π2β π2π1]
(πβ π2) (πβ π2)
, (14)
πΎπ= π2+ π (π2β π2) , (π = 1, 2) . (15)
If (π2β π2)π β (π
1β π1)π+ π1π2β π2π1
ΜΈβ‘ 0, then
(i) π(π, π) = π(π),(ii) π(π, β) β€ π(π, π) βπ(π, 1/(π β πΎ
π)) + π(π) for π = 1, 2.
Proof. Since π and π share (π1, π2) and (π
1, π2) IM, by
Lemma 1 we know π(π, π) = π(π, π) := π(π). Noting
πΌ
π β π1
=
(π
1β π
1) (π β π
1) β (π
1β π1) (πβ π
1)
π β π1
= π
1β π
1β (π1β π1)
πβ π
1
π β π1
,
πΌ
π β π1
=
(π
1β π
1) (π β π
1) β (π
1β π1) (πβ π
1)
π β π1
= π
1β π
1β (π1β π1)
πβ π
1
π β π1
,
1
(π β π1) (π β π
1)
=
1
π1β π1
(
1
π β π1
β
1
π β π1
) ,
π½
πβ π2
=
(π
2β π
2) (πβ π2) β (π
2β π2) (πβ π
2)
πβ π2
= π
2β π
2β (π2β π2)
πβ π
2
πβ π2
,
π½
πβ π2
=
(π
2β π
2) (πβ π2) β (π
2β π2) (πβ π
2)
πβ π2
= π
2β π
2β (π2β π2)
πβ π
2
πβ π2
,
1
(πβ π2) (πβ π2)
=
1
π2β π2
(
1
πβ π2
β
1
πβ π2
) ,
(16)
and the lemma of the logarithmic derivative, we obtain
π(π,
πΌ
π β π1
) = π (π) , π(π,
πΌ
π β π1
) = π (π) ,
π(π,
πΌ
(π β π1) (π β π
1)
) = π (π) ,
(17)
π(π,
π½
πβ π2
) = π (π) , π(π,
π½
πβ π2
) = π (π) ,
π(π,
π½
(πβ π2) (πβ π2)
) = π (π) .
(18)
Clearly, πΌ ΜΈβ‘ 0 and π½ ΜΈβ‘ 0. Otherwise from (11) and (12)we have π = π
1+ πΆ1(π1β π1) and π = π
2+ πΆ2(π2β π2) for
nonzero constants πΆ1, πΆ2, which implies that π(π, π) = π(π)
and π(π, π) = π(π), a contradiction. Then by using a similarmethod we can deduce that π ΜΈβ‘ 0 and β ΜΈβ‘ 0. It is easy to seeby (11) if any zero of π β π
1(π β π
1) of multiplicity π is not the
pole of π1β π1and is not the zero of π
1β π1, then it must be a
zero of πΌ of multiplicity π β 1 at least. Thus from (6), (7), (11),(13), the condition thatπ andπ share (π
1, π2) and (π
1, π2) IM,
and the condition that π is entire, we get
π(π, π) = π (π) . (19)
Likewise,
π(π, β) = π (π) . (20)
Now by (13) and (17) together with
πΌπ
(π β π1) (π β π
1)
=
πΌ
π β π1
+
π1πΌ
(π β π1) (π β π
1)
, (21)
it follows that
π(π, π) β€ π(π,
πΌ [(π2β π2) π β (π
1β π1) π]
(π β π1) (π β π
1)
)
+ π(π,
πΌ (π1π2β π2π1)
(π β π1) (π β π
1)
) + log 2
-
4 Abstract and Applied Analysis
β€ π(π,
πΌπ
(π β π1) (π β π
1)
)
+ π(π,
(π2β π2) π β (π
1β π1) π
π
) + π (π)
β€ π (π) .
(22)Thus from this and (19) we have
π (π, π) = π (π) , (23)implying (i). Next, it is easy to see that πΎ
πΜΈβ‘ π2and πΎ
πΜΈβ‘
π2(π = 1, 2). For π = 1, 2, by (14), (18), π
1ΜΈβ‘ π1, π2
ΜΈβ‘ π2, and
the condition that π is entire, we have
π(π, β) β€ π(π,
π½ [(π2β π2) π β (π
1β π1) π]
(πβ π2) (πβ π2)
)
+ π(π,
π½ (π1π2β π2π1)
(πβ π2) (πβ π2)
) + log 2
β€ π(π,
π½ (πβ πΎπ)
(πβ π2) (πβ π2)
Γ
(π2β π2) π β (π
1β π1) π
πβ πΎπ
) + π (π)
β€ π(π,
π½
πβ π2
) + π(π,
π½ (π2β πΎπ)
(πβ π2) (πβ π2)
)
+ π(π,
(π2β π2) π β (π
1β π1) π
πβ πΎπ
) + π (π)
β€ π(π,
((π2β π2) / (π1β π1)) π β π
πβ πΎπ
) + π (π)
β€ π(π,
((π2β π2) / (π1β π1)) π β πΎ
π
πβ πΎπ
β 1) + π (π)
β€ π(π,
πβ πΎπ
((π2β π2) / (π1β π1)) π β πΎ
π
)
+ π(π,
πβ πΎπ
((π2β π2) / (π1β π1)) π β πΎ
π
)
β π(π,
((π2β π2) / (π1β π1)) π β πΎ
π
πβ πΎπ
) + π (π)
β€ π(π,
1
(π2β π2) / (π1β π1)
Γ
πβ πΎπ
π β ((π1β π1) / (π2β π2)) πΎπ
)
+ π(π, πβ πΎπ)
+ π(π,
1
((π2β π2) / (π1β π1)) π β πΎ
π
)
β π(π,
π2β π2
π1β π1
π β πΎπ) β π(π,
1
πβ πΎπ
) + π (π)
β€ π(π,
πβ (((π
1β π1) / (π2β π2)) πΎπ)
π β ((π1β π1) / (π2β π2)) πΎπ
)
+ π(π,
(((π1β π1) / (π2β π2)) πΎπ)
β πΎπ
π β ((π1β π1) / (π2β π2)) πΎπ
)
+ π(π,
1
((π2β π2) / (π1β π1)) π β πΎ
π
)
β π(π,
1
πβ πΎπ
) + π (π)
β€ π(π,
1
π β ((π1β π1) / (π2β π2)) πΎπ
)
+ π(π,
1
((π2β π2) / (π1β π1)) π β πΎ
π
)
β π(π,
1
πβ πΎπ
) + π (π)
β€ π(π,
1
((π2β π2) / (π1β π1)) π β πΎ
π
)
+ π(π,
1
((π2β π2) / (π1β π1)) π β πΎ
π
)
β π(π,
1
πβ πΎπ
) + π (π)
β€ π (π, π) β π(π,
1
πβ πΎπ
) + π (π) .
(24)
Thus from (20) and (24) it follows that
π (π, β) β€ π (π, π) β π(π,
1
πβ πΎπ
) + π (π) , π = 1, 2.
(25)
This proves (ii) and completes the proof of Lemma 3.
Lemma 4 (see [10]; cf. [11, 12]). Let π be a nonconstantmeromorphic function, and let πππ(π) = π(π), where π(π)and π(π) are differential polynomials in π and the degree ofπ(π) is at most π. Then
π(π, π (π)) = π (π, π) . (26)
-
Abstract and Applied Analysis 5
Lemma 5. Let π be a nonconstant entire function, and let π1,
π2, π1, and π
2be four small functions of π such that none of
them is identically equal to β and π1
ΜΈβ‘ π1, π2
ΜΈβ‘ π2. Suppose
that π and π share (π1, π2) and (π
1, π2) IM. If
π(π, π) = π (π, π) + π (π, π) , (27)
then (π2β π2)π β (π
1β π1)π+ π1π2β π2π1β‘ 0.
Proof. Assume that (π2β π2)π β (π
1β π1)π+π1π2βπ2π1
ΜΈβ‘ 0.Let πΌ, π½, π, β, and πΎ
πbe defined by (11)β(15), respectively.Then
from the proof process of Lemma 3 we know πΌ ΜΈβ‘ 0, π½ ΜΈβ‘ 0,π ΜΈβ‘ 0, β ΜΈβ‘ 0, πΎ
πΜΈβ‘ π2, and πΎ
πΜΈβ‘ π2(π = 1, 2). Since π and π
share (π1, π2) and (π
1, π2) IM, by Lemmas 1, 2, and 3 it follows
that
π(π, π) = π(π,
1
π β π1
) + π(π,
1
π β π1
) + π (π) , (28)
π (π, π) = π (π) , (29)
π (π, β) β€ π (π, π) β π(π,
1
πβ πΎπ
) + π (π) , π = 1, 2.
(30)
Now from the second fundamental theorem, (27), (28), andthe assumption that π is entire, we deduce
2π (π, π) = 2π (π, π
)
β€ π(π,
1
πβ π2
)
+ π(π,
1
πβ π2
) + π(π,
1
πβ πΎπ
) + π (π)
β€ π(π,
1
π β π1
) + π(π,
1
π β π1
)
+ π(π,
1
πβ πΎπ
) + π (π)
= π (π, π) + π(π,
1
πβ πΎπ
) + π (π)
= π (π, π) + π(π,
1
πβ πΎπ
) + π (π) ,
(31)
which yields
π(π,
1
πβ πΎπ
) = π(π, π) + π (π) . (32)
Again by (27), (30), (32), and the assumption that π is entire,we obtain
π (π, β) = π (π) . (33)
For any π§0
β π(π,π)
(π1, π2) βͺ π(π,π)
(π1, π2), from (13) and
(14), we can get ππ(π§0) β πβ(π§
0) = 0.
If ππ β πβ β‘ 0, then by (13) and (14) we deduce
π(
πβ π
1
π β π1
β
πβ π
1
π β π1
) β‘ π(
πβ π
2
πβ π2
β
πβ π
2
πβ π2
) , (34)
which implies that
(
π β π1
π β π1
)
π
β‘ π1(
πβ π2
πβ π2
)
π
, (35)
where π1is a nonzero constant. If π ΜΈ= π, then from (35)
and the condition that π is entire, we obtain ππ(π, π) =ππ(π, π
) + π(π), which contradicts (27). If π = π, then we
get
π β π1
π β π1
β‘ π2(
πβ π2
πβ π2
) , (36)
where π2is a nonzero constant. We claim that π
2ΜΈ= 1. Indeed,
if π2= 1, then by (36) we deduce
π β π1
π β π1
β‘
πβ π2
πβ π2
, (37)
which leads to (π2β π2)π β (π
1β π1)π+ π1π2β π2π1β‘ 0. This
contradicts the assumption. Thus π2
ΜΈ= 1 and so from (36) wehave
π [(1 β π2) π+ π2π2β π2] = (π
1β π2π1) π+ π2π1π2β π2π1.
(38)
This and Lemma 4 yield
π(π, (1 β π2) π+ π2π2β π2) = π (π) , (39)
which givesπ(π, π) = π(π). From this and the condition thatπ is entire, it follows that π(π, π) = π(π), a contradiction.Hence ππ β πβ ΜΈβ‘ 0, for any positive integersπ and π.
Therefore by (29) and (33) we obtain
π(π,π)
(π,
1
π β π1
) + π(π,π)
(π,
1
π β π1
)
β€ π(π,
1
ππ β πβ
) + π (π)
β€ π (π, π) + π (π, β) + π (π)
= π (π) ,
(40)
-
6 Abstract and Applied Analysis
for any positive integers π and π. It follows from this,Lemma 1, the second fundamental theorem, and the condi-tion that π is entire that
π (π, π)
β€ π(π,
1
π β π1
) + π(π,
1
π β π1
) + π (π)
= β
π,π
(π(π,π)
(π,
1
π β π1
) + π(π,π)
(π,
1
π β π1
)) + π (π)
= β
π+πβ₯6
(π(π,π)
(π,
1
π β π1
) + π(π,π)
(π,
1
π β π1
)) + π (π)
β€ β
π+πβ₯6
1
6
(π(π,π)
(π,
1
π β π1
)
+π(π,π)
(π,
1
πβ π2
))
+ β
π+πβ₯6
1
6
(π(π,π)
(π,
1
π β π1
)
+π(π,π)
(π,
1
πβ π2
)) + π (π)
β€
1
6
(π(π,
1
π β π1
) + π(π,
1
πβ π2
))
+
1
6
(π(π,
1
π β π1
) + π(π,
1
πβ π2
)) + π (π)
β€
1
3
π (π, π) +
1
3
π (π, π) + π (π)
=
1
3
π (π, π) +
1
3
π (π, π) + π (π)
=
1
3
π (π, π) +
1
3
π(π,
π
π
π) + π (π)
β€
1
3
π (π, π) +
1
3
π (π, π) + π (π)
=
1
3
π (π, π) +
1
3
π (π, π) + π (π)
=
2
3
π (π, π) + π (π) ,
(41)
which implies that π(π, π) = π(π), a contradiction.Thus (π2β
π2)π β (π
1β π1)π+ π1π2β π2π1β‘ 0.
This completes the proof of Lemma 5.
3. Proof of Theorem 1
Suppose that (π2β π2)π β (π
1β π1)π+ π1π2β π2π1
ΜΈβ‘ 0.Since π and π share (π
1, π2) and (π
1, π2) IM, by Lemma 1
we have π(π, π) = π(π, π) := π(π). Let πΌ, π½, π, β, and πΎπbe
defined by (11)β(15), respectively.Then from the proof processof Lemma 3 we know πΌ ΜΈβ‘ 0, π½ ΜΈβ‘ 0, π ΜΈβ‘ 0, β ΜΈβ‘ 0, πΎ
πΜΈβ‘ π2,
and πΎπ
ΜΈβ‘ π2(π = 1, 2). Next we rewrite (13) as
[π β (π
1β π
1) (π2β π2)] π2
= π1ππ+ π2π + π3π+ π4π2
+ π5,
(42)
where π1= (π1β π1)(π
1+ π2β π2β π
1), π2= (π
1β π
1)(π1π2β
π2π1)+(π2β π2)(π1π
1β π
1π1) + (π1+ π1)π, π3= (π1β π1)(π2π1+
π
1π1β π1π2β π1π
1), π4= (π1βπ1)2, andπ
5= (π1π
1βπ
1π1)(π1π2β
π2π1) β π1π1π are all small functions with respect to π.
Now we divide into two cases.
Case 1. πβ (π1βπ
1)(π2βπ2) β‘ 0; that is, π β‘ (π
1βπ
1)(π2βπ2).
We again discuss the three subcases.
Subcase 1. π1
ΜΈβ‘ π2and π1
ΜΈβ‘ π2. Since π and π share (π
1, π2)
and (π1, π2) IM, the zeros of π β π
1and π β π
1of multiplicity
larger than one are the zeros π1β π2and π1β π2, respectively.
It then follows that
β
πβ₯2,πβ₯1
(π(π,π)
(π,
1
π β π1
) + π(π,π)
(π,
1
π β π1
))
β€ π(π,
1
π
1β π2
) + π(π,
1
π
1β π2
) + π (π)
β€ π (π, π
1) + π (π, π
2) + π (π, π
1)
+ π (π, π2) + π (π)
= π (π) ;
(43)
that is,
β
πβ₯2,πβ₯1
(π(π,π)
(π,
1
π β π1
) + π(π,π)
(π,
1
π β π1
)) = π (π) .
(44)
Let π§0β π(1,π)
(π1, π2). For π β₯ 2, from (13) we get
π (π§0) = (π
1(π§0) β π2(π§0)) (π2(π§0) β π2(π§0))
= (π
1(π§0) β π
1(π§0)) (π2(π§0) β π2(π§0)) ,
(45)
which implies that π2(π§0) β π
1(π§0) = 0 or π
2(π§0) β π2(π§0) = 0.
If π2β π
1β‘ 0, then by (13) we deduce
π=
π
1β π2
π1β π1
π +
π1π2β π
1π1
π1β π1
, (46)
which, in view of the condition that π is entire, implies thatπ(π, π) = π(π, π
) + π(π). From this and Lemma 5, it follows
that (π2βπ2)πβ(π
1βπ1)π+π1π2βπ2π1β‘ 0, contradicting the
assumption.Thus π2βπ
1ΜΈβ‘ 0. By the conditions inTheorem 1,
we know that π2β π2
ΜΈβ‘ 0.
-
Abstract and Applied Analysis 7
Hence
β
πβ₯2
π(1,π)
(π,
1
π β π1
)
β€ π(π,
1
π2β π
1
) + π(π,
1
π2β π2
) + π (π)
β€ 2π (π, π2) + π (π, π
2) + π (π, π
1) + π (π) = π (π) ;
(47)
that is,
β
πβ₯2
π(1,π)
(π,
1
π β π1
) = π (π) . (48)
Similarly,
β
πβ₯2
π(1,π)
(π,
1
π β π1
) = π (π) . (49)
It then follows from (44)β(49) and the second fundamentaltheorem that
π (π, π) β€ π(π,
1
π β π1
) + π(π,
1
π β π1
) + π (π)
= π(1,1)
(π,
1
π β π1
) + π(1,1)
(π,
1
π β π1
) + π (π) .
(50)
For any π§1
β π(1,1)
(π1, π2) βͺ π
(1,1)(π1, π2), from (13) and
(14), we can get π(π§1) β β(π§
1) = 0.
If π β β β‘ 0, then by (13) and (14) we have
πβ π
1
π β π1
β
πβ π
1
π β π1
β‘
πβ π
2
πβ π2
β
πβ π
2
πβ π2
, (51)
which implies that
π β π1
π β π1
β‘ π3
πβ π2
πβ π2
, (52)
where π3is a nonzero constant. We claim that π
3ΜΈ= 1. Indeed,
if π3= 1, then by (52) we have
π β π1
π β π1
β‘
πβ π2
πβ π2
, (53)
which leads to (π2β π2)π β (π
1β π1)π+ π1π2β π2π1β‘ 0. This
contradicts the assumption. Thus π3
ΜΈ= 1 and so from (52) weget
π [(1 β π3) π+ π3π2β π2] = (π
1β π3π1) π+ π3π1π2β π2π1.
(54)
This and Lemma 4 yield
π(π, (1 β π3) π+ π3π2β π2) = π (π) , (55)
which givesπ(π, π) = π(π). From this and the condition thatπ is entire, it follows that π(π, π) = π(π), a contradiction.Hence π β β ΜΈβ‘ 0.
Therefore by (50) and Lemma 3 we obtain
π (π, π) β€ π(1,1)
(π,
1
π β π1
) + π(1,1)
(π,
1
π β π1
) + π (π)
β€ π(π,
1
π β β
) + π (π)
β€ π (π, π) + π (π, β) + π (π)
β€ π (π, π) β π(π,
1
πβ πΎπ
) + π (π) ,
(56)
which implies that
π(π,
1
πβ πΎπ
) = π (π) , π = 1, 2. (57)
This is impossible by the second fundamental theorem.
Subcase 2. Either π1β‘ π2and π1
ΜΈβ‘ π2or π1
ΜΈβ‘ π2and π1β‘ π2.
Without loss of generality, we assume that π1β‘ π2and π1
ΜΈβ‘
π2. It is easy to see by (13) that the zeros of π β π
1and π β π
2
of multiplicity all larger than one are the zeros of π. Thus byLemma 3,
β
πβ₯2,πβ₯2
π(π,π)
(π,
1
π β π1
) β€ π(π,
1
π
) + π (π)
β€ π (π, π) + π (π) = π (π) ;
(58)
that is,
β
πβ₯2,πβ₯2
π(π,π)
(π,
1
π β π1
) = π (π) . (59)
By the discussion of Subcase 1, we see
π(π,
1
π β π1
) = π(1,1)
(π,
1
π β π1
) + π (π)
β€ π(π,
1
π β β
) + π (π)
β€ π (π, π) + π (π, β) + π (π)
β€ π (π, π) β π(π,
1
πβ πΎπ
) + π (π) ;
(60)
that is,
π(π,
1
π β π1
) β€ π (π, π) β π(π,
1
πβ πΎπ
) + π (π) ,
π = 1, 2.
(61)
Note that the zeros of π β π1of multiplicity larger than one
are all the zeros of π β π2
= πβ π
1. Since π and π share
(π1, π2) IM, it follows that
π(1,1)
(π,
1
π β π1
) = π (π) . (62)
-
8 Abstract and Applied Analysis
Now from (59)β(62) and the second fundamental theo-rem, we obtain
π (π, π) β€ π(π,
1
π β π1
) + π(π,
1
π β π1
) + π (π)
β€ π(2,1)
(π,
1
π β π1
) + π (π, π)
β π(π,
1
πβ πΎπ
) + π (π) ;
(63)
that is,
π(π,
1
πβ πΎπ
) β€ π(2,1)
(π,
1
π β π1
) + π (π) , π = 1, 2.
(64)
Let
π = 2
πβ π
2
πβ π2
β 2
πβ π
1
π β π1
+
π
1β π
1
π1β π1
β 2
π
2β π
2
π2β π2
. (65)
It is easily seen from (65) and the lemma of the logarithmicderivative that
π(π, π) = π (π) . (66)
Note that common simple zeros of π β π1and π β π
2are not
the poles of π. In terms of the discussion of Subcase 1, from(65) we know π(π, π) = π(π), which together with (66) givesthat
π (π, π) = π (π) . (67)
Let π§2
β π(2,1)
(π1, π2). Then by (65) and (13) we have
π(π§2) = 0.If π β‘ 0, then from (65) we derive
(πβ π2)
2
= π4
(π2β π2)2
π1β π1
(π β π1)2
, (68)
where π4is a nonzero constant. This, in view of the condition
that π is entire, implies that π(π, π) = π(π, π) + π(π). Fromthis and Lemma 5, it follows that (π
2β π2)π β (π
1β π1)π+
π1π2β π2π1β‘ 0, contradicting the assumption. Thus π ΜΈβ‘ 0.
Hence by (64) and (67) we obtain
π(π,
1
πβ πΎπ
) β€ π(2,1)
(π,
1
π β π1
) + π (π)
β€ π(π,
1
π
) + π (π)
β€ π (π, π) + π (π)
β€ π (π) ;
(69)
that is,
π(π,
1
πβ πΎπ
) = π (π) , π = 1, 2. (70)
This is also impossible by the second fundamental theorem.
Subcase 3. π1β‘ π2and π1β‘ π2. By the discussion of Subcase 2,
we see
π (π, π) β€ π(2,1)
(π,
1
π β π1
) + π(2,1)
(π,
1
π β π1
) + π (π) .
(71)
We claim that
π(2,1)
(π,
1
π β π1
) = π (π) , (72)
π(2,1)
(π,
1
π β π1
) = π (π) . (73)
Let
π = 2
πβ π
2
πβ π2
β
πβ π
1
π β π1
β 2
π
2β π
2
π2β π2
. (74)
It is easily known from (74) and the lemma of the logarithmicderivative that
π(π, π) = π (π) . (75)
Note that common zeros of π β π1of multiplicity two and
πβ π2of multiplicity one are not the poles of π. In terms of
the discussion of Subcase 2, we know π(π, π) = π(π), whichtogether with (75) gives that
π (π, π) = π (π) . (76)
Let π§3
β π(2,1)
(π1, π2). Then by (74) and (13) we have
π(π§3) = 0.If π ΜΈβ‘ 0, then from (76) we get
π(2,1)
(π,
1
π β π1
) β€ π(π,
1
π
) + π (π)
β€ π (π, π) + π (π) β€ π (π) ,
(77)
that is,
π(2,1)
(π,
1
π β π1
) = π (π) . (78)
If π β‘ 0, then by (74) we deduce
(πβ π2)
2
= π5(π2β π2)2
(π β π1) , (79)
where π5is a nonzero constant. This implies π
1(π§3) β π1(π§3) β
1/π5
= 0. Since π1
β‘ π2, π1
β‘ π2, and π
2ΜΈβ‘ π2, we obtain
π1β π1β 1/π5
ΜΈβ‘ 0. Thus
π(2,1)
(π,
1
π β π1
) β€ π(π,
1
π1β π1β 1/π5
) + π (π)
β€ π (π, π1) + π (π, π
1) + π (π) β€ π (π) ;
(80)
that is,
π(2,1)
(π,
1
π β π1
) = π (π) . (81)
Hence (72) follows. In the same manner as above, we canprove (73). The proof of the claim is complete. Now by (71)β(73) we get π(π, π) = π(π), a contradiction.
-
Abstract and Applied Analysis 9
Case 2. π β (π1β π
1)(π2β π2) ΜΈβ‘ 0. Then by (42) and (i) in
Lemma 3 we have
2π (π, π) β€ π(π,
1
π β (π
1β π
1) (π2β π2)
)
+ π(π, π1ππ+ π2π + π3π+ π4π2
+ π5)
β€ π(π, π(π1π+ π2+ π3
π
π
+ π4ππ
π
))
+ π (π, π5) + π (π)
β€ π (π, π) + π(π, π(π1+ π4
π
π
)) + π (π)
β€ π (π, π) + π (π, π) + π (π) ,
(82)
which implies that
π(π, π) β€ π (π, π) + π (π) . (83)
On the other hand,
π(π, π) β€ π(π, π
π
π
) β€ π (π, π) + π (π) . (84)
Combining (83) with (84) yields
π(π, π) = π (π, π) + π (π) . (85)
This andLemma 5 lead to (π2β π2)πβ(π
1β π1)π+ π1π2β π2π1β‘
0, contradicting the assumption.This completes the proof of Theorem 1.
Conflict of Interests
The author declares that there is no conflict of interestsregarding the publication of this paper.
Acknowledgments
Theauthor would like to thank the referees for their thoroughcomments and helpful suggestions. Project supported by theNational Natural Science Foundation of China (Grant no.11301076), theNatural Science Foundation of Fujian Province,China (Grant no. 2014J01004), the Education DepartmentFoundation of Fujian Province, China (Grant no. JB13018),and the Innovation Team of Nonlinear Analysis and itsApplications of Fujian Normal University, China (Grant no.IRTL1206).
References
[1] W. K. Hayman, Meromorphic Functions, Clarendon Press,Oxford, UK, 1964.
[2] L. Yang, Value Distribution Theory, Springer, Berlin, Germany,1993.
[3] C. C. Yang and H. X. Yi, Uniqueness Theory of MeromorphicFunctions, vol. 557, Kluwer Academic Publishers, Dordrecht,The Netherlands, 2003.
[4] I. Laine,NevanlinnaTheory andComplexDifferential Equations,Walter De Gruyter, Berlin, Germany, 1993.
[5] L. A. Rubel and C. C. Yang, βValues shared by an entire functionand its derivative,β inComplex Analysis, vol. 599 of Lecture Notesin Mathematics, pp. 101β103, Springer, Berlin, Germany, 1977.
[6] E. Mues and N. Steinmetz, βMeromorphe Funktionen, die mitihrer Ableitung Werte teilen,β Manuscripta Mathematica, vol.29, no. 2β4, pp. 195β206, 1979.
[7] J. H. Zheng and S. P. Wang, βOn the unicity of meromorphicfunctions and their derivatives,β Advances in Mathematics, vol.21, no. 3, pp. 334β341, 1992.
[8] G. Qiu, βUniqueness of entire functions that share some smallfunctions,β Kodai Mathematical Journal, vol. 23, no. 1, pp. 1β11,2000.
[9] P. Li, βUnicity of meromorphic functions and their derivatives,βJournal of Mathematical Analysis and Applications, vol. 285, no.2, pp. 651β665, 2003.
[10] J. Clunie, βOn integral and meromorphic functions,β Journal ofthe London Mathematical Society, vol. 37, pp. 17β27, 1962.
[11] E. Mues and N. Steinmetz, βThe theorem of Tumura-Clunie formeromorphic functions,β Journal of the London MathematicalSociety, vol. 23, no. 1, pp. 113β122, 1981.
[12] G. Weissenborn, βOn the theorem of Tumura and Clunie,βTheBulletin of the London Mathematical Society, vol. 18, no. 4, pp.371β373, 1986.
-
Research ArticleRadius Constants for Functions with the PrescribedCoefficient Bounds
Om P. Ahuja,1 Sumit Nagpal,2 and V. Ravichandran3
1 Department of Mathematics, Kent State University, Burton, OH 44021, USA2Department of Mathematics, Ramanujan College, University of Delhi, Delhi 110 019, India3 Department of Mathematics, University of Delhi, Delhi 110 007, India
Correspondence should be addressed to Om P. Ahuja; [email protected]
Received 19 June 2014; Accepted 16 August 2014; Published 9 September 2014
Academic Editor: Rosihan M. Ali
Copyright Β© 2014 Om P. Ahuja 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.
For an analytic univalent function π(π§) = π§ + ββπ=2
πππ§π in the unit disk, it is well-known that
ππ
β€ π for π β₯ 2. But the inequality
ππ
β€ π does not imply the univalence of π. This motivated several authors to determine various radii constants associated with
the analytic functions having prescribed coefficient bounds. In this paper, a survey of the related work is presented for analyticand harmonic mappings. In addition, we establish a coefficient inequality for sense-preserving harmonic functions to compute thebounds for the radius of univalence, radius of full starlikeness/convexity of order πΌ (0 β€ πΌ < 1) for functions with prescribedcoefficient bound on the analytic part.
1. Introduction
Let A denote the class of all analytic functions π defined inthe open unit disk D := {π§ β C : |π§| < 1} normalized byπ(0) = 0 = π
(0) β 1. For functions π of the form
π (π§) = π§ +
β
β
π=2
πππ§π (1)
belonging to the subclass S of A consisting of univalentfunctions, de Branges [1] proved the famous Bieberbachconjecture that |π
π| β€ π for π β₯ 2. However, the inequality
|ππ| β€ π (π β₯ 2) does not imply that π is univalent. A
function π given by (1) whose coefficients satisfy |ππ| β€ π for
π β₯ 2 is necessarily analytic inD by the usual comparison testand hence a member of A. But it need not be univalent. Forexample, the function
π (π§) = π§ β 2π§2β 3π§3β β β β = 2π§ β
π§
(1 β π§)2 (2)
satisfies the inequality |ππ| β€ π (π β₯ 2) but its derivative
vanishes inside D and so the function π is not univalent inD. It is therefore of interest to determine the largest subdisk
|π§| < π < 1 in which the functions π satisfying the inequality|ππ| β€ π are univalent. Motivated by this problem, various
radii problems associated with analytic as well as harmonicfunctions having prescribed coefficient bounds have beenstudied and we present a brief review of the research on thistopic. Recall that given two subsets F and G of A, the G-radius in F is the largest π such that, for every π β F,πβ1π(ππ§) β G for each π β€ π .
1.1. Analytic Case. Most of the classes in univalent functiontheory are characterized by the quantities π§π(π§)/π(π§) or 1 +π§π(π§)/π
(π§) lying in a given domain in the right half-plane.
For instance, the subclasses Sβ(πΌ) and K(πΌ) (0 β€ πΌ < 1)of S consisting of starlike functions of order πΌ and convexfunctions of order πΌ, respectively, are defined analytically bythe equivalences
π β Sβ
(πΌ) ββ Re(π§π(π§)
π (π§)
) > πΌ,
π β K (πΌ) ββ Re(π§π(π§)
π(π§)
+ 1) > πΌ.
(3)
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014, Article ID 454152, 12 pageshttp://dx.doi.org/10.1155/2014/454152
http://dx.doi.org/10.1155/2014/454152
-
2 Abstract and Applied Analysis
These classes were introduced by Robertson [2]. The classesSβ := Sβ(0) and K := K(0) are the familiar classes ofstarlike and convex functions, respectively. Goodman [3]introduced the class UCV of uniformly convex functionsπ β A, which map every circular arc πΎ contained in D withcenter π β D onto a convex arc. For π β A, RΓΈnning [4] andMa and Minda [5] independently proved that
π β UCV ββ Re(π§π(π§)
π(π§)
+ 1) >
π§π(π§)
π(π§)
(π§ β D) .
(4)
Closely related to the classUCV is the class Sπof parabolic
starlike functions, introduced by RΓΈnning [4] consisting offunctions π = π§π where π β UCV; that is, a function π βSπsatisfies
Re(π§π(π§)
π (π§)
) >
π§π(π§)
π (π§)
β 1
(π§ β D) . (5)
In 1970, Gavrilov [6] showed that the radius of univalencefor functions π β A satisfying |π
π| β€ π (π β₯ 2) is the real
root π0β 0.164 of the equation 2(1β π)3 β (1+ π) = 0. In 1982,
Yamashita [7] showed that the radius of univalence obtainedby Gavrilov [6] is also the radius of starlikeness for functionsπ β A satisfying |π
π| β€ π. Yamashita [7] also proved that the
radius of convexity for functions π β A satisfying |ππ| β€ π
(π β₯ 2) is the real root π0β 0.090 of the equation 2(1 β π)4 β
(1 + 4π + π2) = 0.
The inequality |ππ| β€ π holds for functions π β A
satisfying |π(π§)| β€ π. Gavrilov [6] proved that the radius ofunivalence for functions π β A satisfying |π
π| β€ π (π β₯ 2)
is 1 β βπ/(1 +π), which also turned out to be their radiusof starlikeness, a result proved by Yamashita [7]. The radiusof convexity for functions π β A satisfying |π
π| β€ π (π β₯ 2)
is the real root of the equation (π+1)(1β π)3 βπ(1+π) = 0.For 0 β€ π β€ 1, let A
πdenote the class of functions π
given by (1) with |π2| = 2π. Since the second coefficient of
normalized univalent functions determines their importantproperties such as Koebe-one-quarter theorem, growth anddistortion theorems, the last author [8] obtained the sharpSβ(πΌ),K(πΌ) (0 β€ πΌ < 1),UCV and S
πradii for functions
π β Aπsatisfying |π
π| β€ π, |π
π| β€ π, or |π
π| β€ π/π (π > 0)
for π β₯ 3. Observe that a function π β A with Reπ(π§) > 0satisfies |π
π| β€ 2/π for π β₯ 2. Indeed, Ravichandran [8]
proved the following theorem, which includes the results ofGavrilov [6] and Yamashita [7] as special cases.
Theorem 1 (see [8]). Let π β Aπbe given by (1) with |π
π| β€ π
for π β₯ 3. Then we have the following.
(i) π satisfies the inequality
π§π(π§)
π (π§)
β 1
< 1 β πΌ (6)
in |π§| < π0where π
0= π0(πΌ) is the real root in (0, 1)
of the equation 1 β πΌ + (1 + πΌ) π = 2(1 β πΌ + (2 βπΌ)(1 β π) π)(1 β π)
3. In particular, the number π0(πΌ)
is also the radius of starlikeness of order πΌ and thenumber π
0(1/2) is the radius of parabolic starlikeness
of the given functions.(ii) π satisfies the inequality
π§π(π§)
π(π§)
< 1 β πΌ (7)
in |π§| < π 0where π
0= π 0(πΌ) is the real root in (0, 1)
of the equation 2(1 β πΌ + 2(2 β πΌ)(1 β π) π)(1 β π)4 =1βπΌ+4π+(1+πΌ) π
2. In particular, the number π 0(πΌ) is
also the radius of convexity of order πΌ and the numberπ 0(1/2) is the radius of uniform convexity of the given
functions.
The results are sharp for the function
π0(π§) = 2π§ + 2 (1 β π) π§
2β
π§
(1 β π§)2
= π§ β 2ππ§2β 3π§3β 4π§4β β β β .
(8)
It is observed that [9] if a function π β A satisfiesRe(π(π§) + π§π(π§)) > 0 for π§ β D, then |π
π| β€ 2/π
2. Similarly,Reade [10] proved that a close-to-star functionπ β A satisfies|ππ| β€ π2 for π β₯ 2. However, the converse in both the cases is
not true, in general. Recently, Mendiratta et al. [11] obtainedsharp radii of starlikeness of order πΌ (0 β€ πΌ < 1), convexityof order πΌ (0 β€ πΌ < 1), parabolic starlikeness and uniformconvexity for the classA
πwhen |π
π| β€ π/π
2 or |ππ| β€ ππ
2
(π > 0) for π β₯ 3. Ali et al. [12] also worked in the similardirection and obtained similar radii constants.
1.2. Harmonic Case. In a simply connected domainΞ© β C, acomplex-valued harmonic function π has the representationπ = β + π, where β and π are analytic in Ξ©. We call thefunctions β and π the analytic and the coanalytic parts of π,respectively. LetH denote the class of all harmonic functionsπ = β + π in D normalized so that β and π take the form
β (π§) = π§ +
β
β
π=2
πππ§π,
π (π§) =
β
β
π=1
πππ§π.
(9)
Since the Jacobian of π is given by π½π
= |β|2β |π|2, by
a theorem of Lewy [13], π is sense-preserving if and onlyif |π| < |β|, or equivalently if β(π§) ΜΈ= 0 and the seconddilatation π€
π= π/β satisfies |π€
π(π§)| < 1 in D. Let Hsp be
the subclass of H consisting of sense-preserving functions.Then it is easy to see that |π
1| < 1 for functions in the class
Hsp. Set H0:= {π β H : π
1= 0} and H0sp := Hsp β© H
0.Finally, letS
π»andS0
π»be subclasses ofHsp andH
0
sp, respec-tively, consisting of univalent functions.
One of the important questions in the study of class S0π»
and its subclasses is related to coefficient bounds. In 1984,
-
Abstract and Applied Analysis 3
Clunie and Sheil-Small [14] conjectured that the Taylor coef-ficients of the series of β and π satisfy the inequality
ππ
β€
1
6
(2π + 1) (π + 1) ,
ππ
β€
1
6
(2π β 1) (π β 1) ,
βπ β₯ 2
(10)
and it is still open.These researchers proposed this coefficientconjecture because the harmonic Koebe functionπΎ = π»+πΊwhere
π»(π§) =
π§ β (1/2) π§2+ (1/6) π§
3
(1 β π§)3
= π§ +
1
6
β
β
π=2
(π + 1) (2π + 1) π§π,
πΊ (π§) =
(1/2) π§2+ (1/6) π§
3
(1 β π§)3
=
1
6
β
β
π=2
(π β 1) (2π β 1) π§π
(11)
is expected to play the extremal role in the classS0π». However,
this conjecture is proved for all functions π β S0π»with real
coefficients and all functionsπ β S0π»for which eitherπ(D) is
starlike with respect to the origin, close-to-convex, or convexin one direction (see [14β16]).
If π β S0π»for which π(D) is convex, Clunie and Sheil-
Small [14] proved that the Taylor coefficients of β andπ satisfythe inequalities
ππ
β€
π + 1
2
,ππ
β€
π β 1
2
, βπ β₯ 2, (12)
and equality occurs for the harmonic half-plane mapping
πΏ (π§) = π (π§) + π (π§),
π (π§) :=
π§ β (1/2) π§2
(1 β π§)2
,
π (π§) :=
β (1/2) π§2
(1 β π§)2.
(13)
LetK0π»and Sβ0
π»be subclasses of S0
π»consisting of func-
tions π for which π(D) is convex and π(D) is starlike withrespect to origin, respectively. Recall that convexity and star-likeness are not hereditary properties for univalent harmonicmappings (see [17β19]). Chuaqui et al. [19] introduced thenotion of fully starlike and fully convex harmonic functionsthat do inherit the properties of starlikeness and convex-ity, respectively. The last two authors [18] generalized thisconcept to fully starlike functions of order πΌ and fully con-vex harmonic functions of orderπΌ for 0 β€ πΌ < 1. LetFSβ
π»(πΌ)
and FKπ»(πΌ) (0 β€ πΌ < 1) be subclasses of S
π»consisting
of fully starlike functions of order πΌ and fully convex func-tions of order πΌ, with FSβ
π»:= FSβ
π»(0) and FK
π»:=
FKπ»(0).The functions in the classesFSβ
π»(πΌ) andFK
π»(πΌ)
are characterized by the conditions (π/ππ) argπ(ππππ) > πΌand (π/ππ)(arg{(π/ππ)π(ππππ)}) > πΌ for every circle |π§| = π,π§ = ππ
ππ, respectively, where 0 β€ π < 2π, 0 < π < 1.The radius of full convexity of the class K0
π»is β2 β 1
while the radius of full convexity of the class Sβ0π»
is 3 β β8(see [14, 16, 20]). The corresponding problems for the radiusof full starlikeness are still unsolved. However, Kalaj et al.[21] worked in this direction and determined the radius ofunivalence and full starlikeness of functions π = β+π whosecoefficients satisfy the conditions (10) and (12). This, in turn,provides a bound for the radius of full starlikeness for convexand starlike mappings inS0
π». These results are generalized in
context of fully starlike and fully convex functions of order πΌ(0 β€ πΌ < 1) in [18]. The authors [18] proved the followingresult.
Theorem 2 (see [18]). Let β and π have the form (9) with π1=
π(0) = 0 and 0 β€ πΌ < 1. Then we have the following.
(a) If the coefficients of the series satisfy the conditions (10),then π = β + π is univalent and fully starlike of orderπΌ in the disk |π§| < π
π, where π
π= ππ(πΌ) is the real root
in (0, 1) of the equation 2(1 β πΌ)(1 β π)4 + πΌ(1 β π)2 β(1 + π)
2= 0.
(b) If the coefficients of the series satisfy the conditions (12),then π = β+π is univalent and fully starlike of order πΌin the disk |π§| < π
π, where π
π= ππ(πΌ) is the real root in
(0, 1) of the equation (2βπΌ)(1βπ)3+πΌπ(1βπ)2β1βπ = 0.
Moreover, the results are sharp for each πΌ β [0, 1).
Theorem 2 gives the bounds for the radius of full starlike-ness of order πΌ (0 β€ πΌ < 1) for the classes Sβ0
π»and K0
π». In
addition, the authors in [18] also determined the bounds forthe radius of full convexity of order πΌ (0 < πΌ < 1) for theseclasses.
The analytic part of harmonic mappings plays a vital rolein shaping their geometric properties. For instance, if π =β+π β Hsp and β is convex univalent, thenπ β Sπ» andmapsD onto a close-to-convex domain (see [14, Theorem 5.17, p.20]). However, if π = β+π β Hsp where β and π are given by(9) and |π
π| β€ 1 for π β₯ 2, then π need not be even univalent;
for example, the function
π (π§) = π§ β
π§2
2
+
π§2
2
β
π§3
3
, π§ β D (14)
belongs to Hsp but is not univalent in D since π(π§0) =π(π§0) = 3/4 where π§
0= (3 + β3π)/4 β D. Note that a convex
univalent function π§ + π2π§2+ π3π§3+ β β β satisfies |π
π| β€ 1
for π = 2, 3, . . .. This paper aims to determine the coefficientinequalities and radius constants for certain subfamilies ofHsp with the prescribed coefficient bound on the analyticpart.
-
4 Abstract and Applied Analysis
A coefficient inequality for functions in the class Hsp isobtained in Section 2 which, in particular, improves the coef-ficient inequality proved by PolatogΜlu et al. [22] for perturbedharmonicmappings. Using this inequality, the bounds for theradius of univalence, full starlikeness, and full convexity oforder πΌ (0 β€ πΌ < 1) are obtained for functions π = β + π βH0sp where the coefficients of the analytic part β satisfy one ofthe conditions |π
π| β€ π, |π
π| β€ 1, or |π
π| β€ 1/π for π β₯ 2.
In addition, we will also discuss a case under which thesebounds can be improved.
In the third section, sharp bounds on π½ (depending uponπΌ and |π
1|) are determined for a function π = β + π β
H, where β and π are given by (9), satisfying either of thefollowing two conditions:β
β
π=2
πππ
+
β
β
π=1
πππ
β€ π½ or
β
β
π=2
π2 ππ
+
β
β
π=1
π2 ππ
β€ π½,
(15)
to be either fully starlike of order πΌ or fully convex of order πΌ.The obtained results are applied to hypergeometric functionsin Section 4.
2. A Coefficient Inequality andRadius Constants
Firstly, we will obtain a coefficient inequality for functions inthe classHsp.
Theorem 3. Let π = β + π β Hπ π, where β and π are given by
(9). Then
ππ
β€π1
ππ
+
(1 βπ1
2
)
π
πβ1
β
π=1
πππ
, (16)
for π β₯ 2, with π1= 1. In particular, one has
ππ
β€
1
π
π
β
π=1
πππ
, π = 2, 3, . . . . (17)
Proof. Since π β Hsp, the function π€(π§) = π(π§)/β(π§) =ββ
π=0π€ππ§π is analytic in D and |π€(π§)| < 1 in D. On equating
the coefficients of π§πβ1 in π(π§) = π€(π§) β(π§), we obtain
πππ= π1π€πβ1
+ 2π€πβ2
π2+ 3π€πβ3
π3
+ β β β + (π β 1)π€1ππβ1
+ ππ€0ππ,
(18)
where π1
= 1. Since |π€π| β€ 1 β |π€
0|2 (see [23, p. 172]), it
immediately follows that
πππ
β€ (1 β
π€0
2
)
πβ1
β
π=1
πππ
+ π
π€0
ππ
, (π
1= 1) . (19)
Since π€0= π(0)/β(0) = π
1, the desired result follows.
For specific choices of the analytic part β in a harmonicfunction π = β + π β Hsp, Theorem 3 yields the followingresult.
Corollary 4. Let π = β + π β Hπ π, where β and π are given
by (9). Then we have the following.
(i) If |ππ| β€ π or, in particular, β is univalent, then |π
π| β€
(2π + 1)(π + 1)/6, π = 2, 3, . . ..
(ii) If |ππ| β€ 1 or, in particular, β is convex univalent, then
|ππ| β€ (π + 1)/2, π = 2, 3, . . ..
(iii) If |ππ| β€ 1/π or, in particular, Re β(π§) > 0, then |π
π| β€
1, π = 2, 3, . . ..
Remark 5. PolatogΜlu et al. [22] determined the coefficientinequality for harmonic functions in a subclass of Hsp, forwhich the analytic part is a univalent function in D. Theyproved that if π = β + π β Hsp where β and π are givenby (9) and if β is univalent in D, then
ππ
β€
1
π
(2π6 β π2β 4π β 6) , π = 1, 2, . . . . (20)
It is evident that Corollary 4(i) improves this bound.
Now, we will determine the radius of univalence, radiusof full starlikeness/full convexity of order πΌ (0 β€ πΌ < 1) forthe classH0sp with specific choices of the coefficient bound onthe analytic part. We will make use of the following sufficientcoefficient conditions for a harmonic function to be in theclasses FSβ
π»(πΌ) and FK
π»(πΌ) (0 β€ πΌ < 1) that directly
follow from the corresponding results in [24, 25].
Lemma 6 (see [24, 25]). Let π = β + π, where β and π aregiven by (9) and let 0 β€ πΌ < 1. Then we have the following.
(i) If
β
β
π=2
π β πΌ
1 β πΌ
ππ
+
β
β
π=1
π + πΌ
1 β πΌ
ππ
β€ 1, (21)
then π β FSβπ»(πΌ).
(ii) If
β
β
π=2
π (π β πΌ)
1 β πΌ
ππ
+
β
β
π=1
π (π + πΌ)
1 β πΌ
ππ
β€ 1, (22)
then π β FKπ»(πΌ).
Theorem 7. Let π = β + π β H0π π, where β and π are given
by (9) with π1= π(0) = 0 and 0 β€ πΌ < 1. Then we have the
following.
(i) If |ππ| β€ π or, in particular, β is univalent, then π is
univalent and fully starlike of order πΌ in the disk |π§| <π1where π
1= π1(πΌ) is the real root of the equation
12 (1 β πΌ) π4+ (49πΌ β 48) π
3+ 8 (9 β 8πΌ) π
2
+ 3 (11πΌ β 18) π + 6 (1 β πΌ) = 0
(23)
in the interval (0, 1).
-
Abstract and Applied Analysis 5
(ii) If |ππ| β€ 1 or, in particular, β is convex univalent, then
π is univalent and fully starlike of order πΌ in the disk|π§| < π
2where π
2= π2(πΌ) is the real root of the equation
4 (1 β πΌ) π3+ (9πΌ β 12) π
2+ (12 β 7πΌ) π
β 2 (1 β πΌ) = 0
(24)
in the interval (0, 1).(iii) If |π
π| β€ 1/π or, in particular, Re β(π§) > 0, then π is
univalent and fully starlike of order πΌ in the disk |π§| <π3where π
3= π3(πΌ) is the real root of the equation
2 (1 β πΌ) π3+ (5πΌ β 4) π
2+ (1 β 3πΌ) π
β 2πΌ (1 β π)2 log (1 β π) = 0
(25)
in the interval (0, 1).
Proof. Since π = β + π β H0sp, we obtain
ππ
β€
1
π
πβ1
β
π=1
πππ
, (π β₯ 2; π
1= 1) , (26)
by applying Theorem 3. We will make use of (26) to obtainthe coefficient bounds for π
πin three different cases specified
in the theorem. For π β (0, 1), let ππ: D β C be defined
by
ππ(π§) :=
π (ππ§)
π
= π§ +
β
β
π=2
ππππβ1
π§π+
β
β
π=2
ππππβ1
π§π. (27)
We will show that ππβ FSβ
π»(πΌ). In view of Lemma 6(i), it
suffices to show that the sum
π =
β
β
π=2
π β πΌ
1 β πΌ
ππ
ππβ1
+
β
β
π=2
π + πΌ
1 β πΌ
ππ
ππβ1 (28)
is bounded above by 1 for 0 β€ π < ππfor π = 1, 2, 3.
(i) Since |ππ| β€ π, it is easy to deduce that |π
π| β€ (π β
1)(2π β 1)/6 by (26). Using these coefficient bounds in (28)and simplifying, we have
π β€
1
6 (1 β πΌ)
[2
β
β
π=2
π3ππβ1
+ (3 + 2πΌ)
β
β
π=2
π2ππβ1
+ (1 β 9πΌ)
β
β
π=2
πππβ1
+
πΌπ
1 β π
] .
(29)
Thus π β€ 1 if π satisfy the inequality
2
β
β
π=2
π3ππβ1
+ (2πΌ + 3)
β
β
π=2
π2ππβ1
+ (1 β 9πΌ)
β
β
π=2
πππβ1
+
πΌπ
1 β π
β€ 6 (1 β πΌ) .
(30)
By using the identities
π
(1 β π)2=
β
β
π=1
πππ,
π (1 + π)
(1 β π)3=
β
β
π=1
π2ππ,
π (π2+ 4π + 1)
(1 β π)4
=
β
β
π=1
π3ππ
(31)
the last inequality reduces to
2 (π2+ 4π + 1)
(1 β π)4
+ (2πΌ + 3)
1 + π
(1 β π)3
+
1 β 9πΌ
(1 β π)2+
πΌ
1 β π
β€ 12 (1 β πΌ)
(32)
or equivalently
2 (π2+ 4π + 1) + (2πΌ + 3) (1 β π
2)
+ (1 β 9πΌ) (1 β π)2+ πΌ(1 β π)
3
β€ 12 (1 β πΌ) (1 β π)4.
(33)
This gives
12 (1 β πΌ) π4+ (49πΌ β 48) π
3+ 8 (9 β 8πΌ) π
2
+ 3 (11πΌ β 18) π + 6 (1 β πΌ) β₯ 0.
(34)
Thus by Lemma 6(i), ππβ FSβ
π»(πΌ) for π β€ π
1where π
1is the
real root of (23) in (0, 1). In particular,π is univalent and fullystarlike of order πΌ in |π§| < π
1.
(ii) If |ππ| β€ 1 then (26) gives |π
π| β€ (π β 1)/2. These
coefficient bounds lead to the following inequality for the sum(28):
π β€
1
2 (1 β πΌ)
Γ [
β
β
π=2
π2ππβ1
+ (1 + πΌ)
β
β
π=2
πππβ1
β
3πΌπ
1 β π
] .
(35)
Therefore it follows that π β€ 1 if π satisfy the inequality
β
β
π=2
π2ππβ1
+ (1 + πΌ)
β
β
π=2
πππβ1
β
3πΌπ
1 β π
β€ 2 (1 β πΌ) . (36)
Making use of identities (31) in the last inequality, we obtain
1 + π
(1 β π)3+
1 + πΌ
(1 β π)2β
3πΌ
1 β π
β€ 4 (1 β πΌ) , (37)
which simplifies to
2 (1 β πΌ) + (7πΌ β 12) π + (12 β 9πΌ) π2β 4 (1 β πΌ) π
3β₯ 0.
(38)
-
6 Abstract and Applied Analysis
Lemma 6(i) shows that ππβ FSβ
π»(πΌ) for π β€ π
2where π
2is
the real root of (24) in (0, 1). In particular, π is univalent andfully starlike of order πΌ in |π§| < π
2.
(iii) Using (26), it is easily seen that |ππ| β€ (πβ1)/π. Using
the coefficient bounds for |ππ| and |π
π| in (28), it follows that
π β€
1
1 β πΌ
[
β
β
π=2
πππβ1
β 2πΌ
β
β
π=2
1
π
ππβ1
+
πΌπ
1 β π
] . (39)
The sum π β€ 1 if π satisfy the inequalityβ
β
π=2
πππβ1
β 2πΌ
β
β
π=2
1
π
ππβ1
+
πΌπ
1 β π
β€ 1 β πΌ. (40)
Using (31) and the identity β log(1β π) = π+ π2/2+ π3/3+ β β β ,the last inequality reduces to
1
(1 β π)2+
2πΌ
π
log (1 β π) + πΌ1 β π
β€ 2 (1 β πΌ) , (41)
which is equivalent to
2 (1 β πΌ) π3+ (5πΌ β 4) π
2+ (1 β 3πΌ) π
β 2πΌ(1 β π)2 log (1 β π) β₯ 0.
(42)
By applying Lemma 6(i), ππβ FSβ
π»(πΌ) for π β€ π
3where π
3is
the real root of (25) in (0, 1). In particular, π is univalent andfully starlike of order πΌ in |π§| < π
3. This completes the proof
of the theorem.
Remark 8. By (26), it follows that |π2| β€ 1/2 for all functions
π β H0sp. The bound 1/2 is sharp for the function π0(π§) =π§ + π§2/2 β H0sp. Since π0 is univalent in D, the coefficient
inequality |π2| β€ 1/2 remains sharp in the subclass S0
π».
Clunie and Sheil-Small [14] were the first to observe the sharpinequality |π