Analytic and Harmonic Univalent FunctionsAbstractandAppliedAnalysis Analytic and Harmonic Univalent Functions GuestEditors:V.Ravichandran,OmP.Ahuja, andRosihanM.Ali

Guest Editors: V. Ravichandran, Om P. Ahuja, and Rosihan M. Ali

Analytic and Harmonic Univalent Functions

Abstract and Applied Analysis

Abstract and Applied Analysis


Guest Editors: V. Ravichandran, Om P. Ahuja,and Rosihan M. Ali

Copyright © 2014 Hindawi Publishing Corporation. All rights reserved.

This is a special issue published in “Abstract and Applied Analysis.” 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.

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, TurkeyJózef Banaś, PolandMartino Bardi, ItalyRoberto Barrio, SpainFeyzi Başar, 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 Cortés López, SpainGraziano Crasta, ItalyZhihua Cui, ChinaBernard Dacorogna, SwitzerlandVladimir Danilov, RussiaMohammad T. Darvishi, IranLuis F. Pinheiro de Castro, PortugalToka Diagana, USAJesús I. Dı́az, SpainJosef Dibĺık, 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, ItalyMárcia Federson, BrazilStathis Filippas, Equatorial GuineaAlberto Fiorenza, ItalyIlaria Fragala, ItalyXianlong Fu, ChinaMassimo Furi, ItalyJesús G. Falset, SpainGiovanni P. Galdi, USAIsaac Garcia, SpainJ. A. G.a-R.guez, SpainLeszek Gasinski, PolandGyörgy Gát, HungaryVladimir Georgiev, ItalyLorenzo Giacomelli, ItalyJaume Giné, SpainValery Y. Glizer, IsraelJean P. Gossez, BelgiumJose L. Gracia, SpainMaurizio Grasselli, ItalyLuca Guerrini, Italy

Yuxia Guo, ChinaQian Guo, ChinaC. P. Gupta, USAUno Hämarik, 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 Jódar, SpainJong S. Jung, Republic of KoreaHenrik Kalisch, NorwayHamid R. Karimi, NorwayChaudry M. Khalique, South AfricaSatyanad Kichenassamy, FranceTero Kilpeläinen, 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 Lü, China

Grzegorz Lukaszewicz, PolandWanbiao Ma, ChinaNazim I. Mahmudov, TurkeyEberhard Malkowsky, TurkeySalvatore A. Marano, ItalyCristina Marcelli, ItalyPaolo Marcellini, ItalyJesús Maŕın-Solano, SpainJose M. Martell, SpainM. Mastyło, PolandMing Mei, CanadaTaras Mel’nyk, UkraineAnna Mercaldo, ItalyStanislaw Migorski, PolandMihai Mihǎilescu, RomaniaFeliz Minhós, PortugalDumitru Motreanu, FranceMaria Grazia Naso, ItalyGaston M. N’Guerekata, USAMicah Osilike, NigeriaMitsuharu Ôtani, JapanTurgut Ôziş, 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 Rachu̇nková, Czech RepublicMaria A. Ragusa, ItalySimeon Reich, IsraelAbdelaziz Rhandi, ItalyHassan Riahi, MalaysiaJuan P. Rincón-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 Staněk, Czech RepublicStevo Stević, SerbiaAntonio Suárez, 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

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 𝛼-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

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 © 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 Lowner’s 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 “harmonic Koebe function” 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 reviewers’reports and editors’ 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 “Third-order differential subordination andsuperordination results for meromorphically multivalent func-tions associated with the Liu-Srivastava operator” (H. Tanget al.), “Differential subordinations for nonanalytic functions”(G. I. Oros and G. Oros), and “Differential subordination

Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014, Article ID 578214, 2 pageshttp://dx.doi.org/10.1155/2014/578214

http://dx.doi.org/10.1155/2014/578214

2 Abstract and Applied Analysis

results for analytic functions in the upper half-plane” (H. Tanget al.). The paper “Meromorphic solutions of some algebraicdifferential equations” (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 “A note on entire functions that share two smallfunctions” by J.-F. Chen investigates a close linear relationshipbetween a non-constant entire function under certain condi-tions and its first derivative.

The paper “Upper bound of second Hankel determinantfor certain subclasses of analytic functions” (M.-S. Liu etal.) summarizes works done in the area of the Hankeldeterminant for univalent functions, while the paper “Initialcoefficients of biunivalent functions” (S. K. Lee et al.) givesestimates on the initial coefficients of the Taylor coefficientsof these functions.The paper “Some connections between classU- and 𝛼-convex functions” (E. Aliaga and N. Tuneski) givessufficient conditions for an 𝛼-convex function to be in aspecific class, while the paper “Starlikeness of functions definedby third-order differential inequalities and integral operators”(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 “Radius constants for functionswith the prescribed coefficient bounds” (O. P. Ahuja et al.),“On certain subclass of harmonic starlike functions” (A. Y.Lashin), “A family of minimal surfaces and univalent planarharmonic mappings” (M. Dorff and S. Muir), and “Landau-type theorems for certain biharmonic mappings” (M.-S. Liuet al.). The first of these four papers, “Radius constants forfunctions with the prescribed coefficient bounds,” 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, “On certain subclass of harmonicstarlike functions,” discusses the geometric properties for anew class of harmonic univalent functions; the third one,“A family of minimal surfaces and univalent planar harmonicmappings,” presents a two-parameter family of minimalsurfaces constructed by lifting a family of planar harmonicmappings, while the fourth one, “Landau-type theoremsfor certain biharmonic mappings,” 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

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 © 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 𝑓 is a nonconstant entire function that shares two pairs of small functions ignoring multiplicities withits first derivative 𝑓, then there exists a close linear relationship between 𝑓 and 𝑓. 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., [1–4]). Let 𝑓(𝑧) be a meromorphicfunction. Here and in the following the word “meromorphic”means meromorphic in the whole complex plane. We denoteby 𝑆(𝑟, 𝑓) any real function of growth 𝑜(𝑇(𝑟, 𝑓)) as 𝑟 → ∞outside of a possible exceptional set of finite linear measure.The meromorphic function 𝑎 is called a small function withrespect to 𝑓 provided that 𝑇(𝑟, 𝑎) = 𝑆(𝑟, 𝑓).

Let 𝑓 and 𝑔 be two nonconstant meromorphic functions,and let 𝑎 and 𝑏 be two small functions with respect to 𝑓 and𝑔. If the zeros of 𝑓 − 𝑎 and 𝑔 − 𝑏 coincide in locations andmultiplicities, then we say that𝑓 and 𝑔 share the pair of smallfunctions (𝑎, 𝑏) CM (counting multiplicities); if we do notconsider the multiplicities, then 𝑓 and 𝑔 are said to share thepair of small functions (𝑎, 𝑏) IM (ignoring multiplicities). Wesee that 𝑓 and 𝑔 share the pair of small functions (𝑎, 𝑎) CMif and only if 𝑓 and 𝑔 share the small function 𝑎 CM, and 𝑓and 𝑔 share the pair of small functions (𝑎, 𝑎) IM if and onlyif 𝑓 and 𝑔 share the small function 𝑎 IM.The same argumentapplies in the casewhen 𝑎 and 𝑏 are two values in the extendedplane.

Moreover, we introduce the following notations. Denotethe set of those points 𝑧 ∈ C by 𝑆

(𝑚,𝑛)(𝑎1, 𝑎2) such that 𝑧 is

a zero of 𝑓 − 𝑎1of multiplicity 𝑚 and a zero of 𝑓 − 𝑎

2of

multiplicity 𝑛. The set 𝑆(𝑚,𝑛)

(𝑏1, 𝑏2) can be similarly defined.

Now the notations 𝑁(𝑚,𝑛)

(𝑟, 1/(𝑓 − 𝑎1)) and 𝑁

(𝑚,𝑛)(𝑟, 1/(𝑓 −

𝑎1)) denote the counting function and the reduced counting

function of𝑓with respect to the set 𝑆(𝑚,𝑛)

(𝑎1, 𝑎2), respectively.

The notations𝑁(𝑚,𝑛)

(𝑟, 1/(𝑓− 𝑎2)) and𝑁

(𝑚,𝑛)(𝑟, 1/(𝑓

− 𝑎2))

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 𝑓 and itsderivative 𝑓, Rubel and Yang [5] proved the following.

Theorem A. Let 𝑓 be a nonconstant entire function, and let 𝑎and 𝑏 be distinct finite complex numbers. If 𝑓 and 𝑓 share 𝑎and 𝑏 CM, then 𝑓 ≡ 𝑓.

Mues and Steinmetz [6] improved Theorem A andobtained the following.

Theorem B. Let 𝑓 be a nonconstant entire function, and let 𝑎and 𝑏 be distinct finite complex numbers. If 𝑓 and 𝑓 share 𝑎and 𝑏 IM, then 𝑓 ≡ 𝑓.

When the values 𝑎 and 𝑏were replaced by two small func-tions related to 𝑓, Zheng and Wang [7] proved the follow-ing.

Theorem C. Let 𝑓 be a nonconstant entire function, and let 𝑎and 𝑏 be distinct small functions with respect to 𝑓. If 𝑓 and 𝑓share 𝑎 and 𝑏 CM, then 𝑓 ≡ 𝑓.

Recently, Qiu [8] proved the following result which wasan improvement of Theorem C.


http://dx.doi.org/10.1155/2014/601507


TheoremD. Let 𝑓 be a nonconstant entire function, and let 𝑎and 𝑏 be distinct small functions with respect to 𝑓. If 𝑓 and 𝑓share 𝑎 and 𝑏 IM, then 𝑓 ≡ 𝑓.

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 𝑓 be a nonconstant entire function, and let 𝑎1,

𝑎2, 𝑏1, and 𝑏

2be four small functions of 𝑓 such that none of

them is identically equal to ∞ and 𝑎1

̸≡ 𝑏1, 𝑎2

̸≡ 𝑏2. If 𝑓 and

𝑓 share (𝑎

1, 𝑎2) and (𝑏

1, 𝑏2) IM, then (𝑎

2−𝑏2)𝑓− (𝑎

1−𝑏1)𝑓+

𝑎1𝑏2− 𝑎2𝑏1≡ 0.

Remark 2. Let 𝑎1≡ 𝑎2and 𝑏1≡ 𝑏2.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 𝑓 = 𝛽 + (𝛽 − 𝛼)/(ℎ − 1), where

𝛼 = −

1

3

𝑒−2𝑧

−

1

2

𝑒−𝑧

, 𝛽 = −

1

3

𝑒−2𝑧

+

1

2

𝑒−𝑧

, ℎ = 𝑒−𝑒𝑧

.

(1)

Set 𝑎 = 𝛽, 𝑏 = 𝛼. Then 𝑇(𝑟, 𝑎) = 𝑆(𝑟, 𝑓) and 𝑇(𝑟, 𝑏) =𝑆(𝑟, 𝑓). It is easy to verify that

𝑓− 𝑎 = 𝑒

2𝑧(𝑓 − 𝑎) (𝑓 − 𝛽) ,

𝑓− 𝑏 = 𝑒

2𝑧(𝑓 − 𝑏) (𝑓 − 𝛼) .

(2)

Thus 𝑓 and 𝑓 share (𝑎, 𝑎) and (𝑏, 𝑏) IM, but 𝑓 ̸≡ 𝑓. Thisshows that the conclusion inTheorem 1 is not valid generallyfor a meromorphic function 𝑓.

Example 5. Let 𝑓 = 𝑒2𝑧 + 𝑧, 𝑎1= 2𝑧, 𝑎

2= 2𝑧 + 1, 𝑏

1= 𝑧, and

𝑏2= 𝑧 + 1. Then 𝑓 and 𝑓 share (𝑎

1, 𝑎2) IM but do not share

(𝑏1, 𝑏2) IM.Clearly, (𝑎

2−𝑏2)𝑓−(𝑎

1−𝑏1)𝑓+𝑎1𝑏2−𝑎2𝑏1

̸≡ 0.Thisshows that the condition in Theorem 1 that 𝑓 and 𝑓 share(𝑎1, 𝑎2) and (𝑏

1, 𝑏2) IM cannot be weakened.

2. Some Lemmas

Lemma 1. Let 𝑓 be a nonconstant entire function, and let 𝑎1,


2be four small functions of𝑓 such that none of them

is identically equal to ∞ and 𝑎1

̸≡ 𝑏1, 𝑎2

̸≡ 𝑏2. If 𝑓 and 𝑓

share (𝑎1, 𝑎2) and (𝑏

1, 𝑏2) IM, then 𝑆(𝑟, 𝑓) = 𝑆(𝑟, 𝑓) := 𝑆(𝑟).

Proof. Note that𝑓 and𝑓 share (𝑎1, 𝑎2) and (𝑏

1, 𝑏2) IM. By the

second fundamental theorem, we get

𝑇 (𝑟, 𝑓) ≤ 𝑁(𝑟,

1

𝑓 − 𝑎1

) + 𝑁(𝑟,

1

𝑓 − 𝑏1

) + 𝑆 (𝑟, 𝑓)

= 𝑁(𝑟,

1

𝑓− 𝑎2

) + 𝑁(𝑟,

1

𝑓− 𝑏2

) + 𝑆 (𝑟, 𝑓)

≤ 2𝑇 (𝑟, 𝑓) + 𝑆 (𝑟, 𝑓) ,

𝑇 (𝑟, 𝑓) ≤ 𝑁(𝑟,

1

𝑓− 𝑎2

) + 𝑁(𝑟,

1

𝑓− 𝑏2

) + 𝑆 (𝑟, 𝑓)

= 𝑁(𝑟,

1

𝑓 − 𝑎1

) + 𝑁(𝑟,

1

𝑓 − 𝑏1

) + 𝑆 (𝑟, 𝑓)

≤ 2𝑇 (𝑟, 𝑓) + 𝑆 (𝑟, 𝑓) + 𝑆 (𝑟, 𝑓) ,

(3)

which implies from the definition of 𝑆(𝑟, 𝑓) that 𝑆(𝑟, 𝑓) =𝑆(𝑟, 𝑓

) and 𝑆(𝑟, 𝑓) = 𝑆(𝑟, 𝑓), respectively.

This completes the proof of Lemma 1.





̸≡ 𝑏1, 𝑎2

̸≡ 𝑏2. If 𝑓 and

𝑓 share (𝑎

1, 𝑎2) and (𝑏

1, 𝑏2) IM, then

𝑚(𝑟, 𝑓) = 𝑁(𝑟,

1

𝑓 − 𝑎1

) + 𝑁(𝑟,

1

𝑓 − 𝑏1

) + 𝑆 (𝑟) , (4)

provided that (𝑎2− 𝑏2)𝑓 − (𝑎

1− 𝑏1)𝑓+ 𝑎1𝑏2− 𝑎2𝑏1

̸≡ 0.

Proof. Note that

(𝑎2− 𝑏2) 𝑓 − (𝑎

1− 𝑏1) 𝑓+ 𝑎1𝑏2− 𝑎2𝑏1

̸≡ 0, (5)

(𝑎2− 𝑏2) 𝑓 − (𝑎

1− 𝑏1) 𝑓+ 𝑎1𝑏2− 𝑎2𝑏1

≡ (𝑎2− 𝑏2) (𝑓 − 𝑎

1) − (𝑎

1− 𝑏1) (𝑓− 𝑎2) ,

(6)

(𝑎2− 𝑏2) 𝑓 − (𝑎

1− 𝑏1) 𝑓+ 𝑎1𝑏2− 𝑎2𝑏1

≡ (𝑎2− 𝑏2) (𝑓 − 𝑏

1) − (𝑎

1− 𝑏1) (𝑓− 𝑏2) .

(7)

Since 𝑓 and 𝑓 share (𝑎1, 𝑎2) and (𝑏

1, 𝑏2) IM, from Lemma 1,

(5)–(7), and the condition that 𝑓 is entire, we have

𝑁(𝑟,

1

𝑓 − 𝑎1

) + 𝑁(𝑟,

1

𝑓 − 𝑏1

)

≤ 𝑁(𝑟,

1

(𝑎2− 𝑏2) 𝑓 − (𝑎

1− 𝑏1) 𝑓+ 𝑎1𝑏2− 𝑎2𝑏1

)

≤ 𝑇 (𝑟, (𝑎2− 𝑏2) 𝑓 − (𝑎

1− 𝑏1) 𝑓) + 𝑆 (𝑟)

= 𝑚 (𝑟, (𝑎2− 𝑏2) 𝑓 − (𝑎

1− 𝑏1) 𝑓)

+ 𝑁(𝑟, (𝑎2− 𝑏2) 𝑓 − (𝑎

1− 𝑏1) 𝑓) + 𝑆 (𝑟)

= 𝑚 (𝑟, (𝑎2− 𝑏2) 𝑓 − (𝑎

1− 𝑏1) 𝑓) + 𝑆 (𝑟)

≤ 𝑚(𝑟,

(𝑎2− 𝑏2) 𝑓 − (𝑎

1− 𝑏1) 𝑓

𝑓

) + 𝑚 (𝑟, 𝑓) + 𝑆 (𝑟)

≤ 𝑚 (𝑟, 𝑓) + 𝑆 (𝑟) .

(8)

Abstract and Applied Analysis 3

On the other hand, by the second fundamental theorem,Lemma 1, and the condition that 𝑓 is entire, we get

𝑚(𝑟, 𝑓) = 𝑇 (𝑟, 𝑓) ≤ 𝑁(𝑟,

1

𝑓 − 𝑎1

)

+ 𝑁(𝑟,

1

𝑓 − 𝑏1

) + 𝑆 (𝑟) .

(9)

Now (8) and (9) imply

𝑚(𝑟, 𝑓) = 𝑁(𝑟,

1

𝑓 − 𝑎1

) + 𝑁(𝑟,

1

𝑓 − 𝑏1

) + 𝑆 (𝑟) . (10)






̸≡ 𝑏1, 𝑎2

̸≡ 𝑏2. Suppose

that 𝑓 and 𝑓 share (𝑎1, 𝑎2) and (𝑏

1, 𝑏2) IM. Set

𝛼 = (𝑎

1− 𝑏

1) (𝑓 − 𝑎

1) − (𝑎

1− 𝑏1) (𝑓− 𝑎

1)

= (𝑎

1− 𝑏

1) (𝑓 − 𝑏

1) − (𝑎

1− 𝑏1) (𝑓− 𝑏

1) ,

(11)

𝛽 = (𝑎

2− 𝑏

2) (𝑓− 𝑎2) − (𝑎

2− 𝑏2) (𝑓− 𝑎

2)

= (𝑎

2− 𝑏

2) (𝑓− 𝑏2) − (𝑎

2− 𝑏2) (𝑓− 𝑏

2) ,

(12)

𝑔 =

𝛼 [(𝑎2− 𝑏2) 𝑓 − (𝑎

1− 𝑏1) 𝑓+ 𝑎1𝑏2− 𝑎2𝑏1]

(𝑓 − 𝑎1) (𝑓 − 𝑏

1)

, (13)

ℎ =

𝛽 [(𝑎2− 𝑏2) 𝑓 − (𝑎

1− 𝑏1) 𝑓+ 𝑎1𝑏2− 𝑎2𝑏1]

(𝑓− 𝑎2) (𝑓− 𝑏2)

, (14)

𝛾𝑖= 𝑎2+ 𝑖 (𝑎2− 𝑏2) , (𝑖 = 1, 2) . (15)

If (𝑎2− 𝑏2)𝑓 − (𝑎

1− 𝑏1)𝑓+ 𝑎1𝑏2− 𝑎2𝑏1

̸≡ 0, then

(i) 𝑇(𝑟, 𝑔) = 𝑆(𝑟),(ii) 𝑇(𝑟, ℎ) ≤ 𝑇(𝑟, 𝑓) −𝑁(𝑟, 1/(𝑓 − 𝛾

𝑖)) + 𝑆(𝑟) for 𝑖 = 1, 2.

Proof. Since 𝑓 and 𝑓 share (𝑎1, 𝑎2) and (𝑏

1, 𝑏2) IM, by

Lemma 1 we know 𝑆(𝑟, 𝑓) = 𝑆(𝑟, 𝑓) := 𝑆(𝑟). Noting

𝛼

𝑓 − 𝑎1

=

(𝑎

1− 𝑏

1) (𝑓 − 𝑎

1) − (𝑎

1− 𝑏1) (𝑓− 𝑎

1)

𝑓 − 𝑎1

= 𝑎

1− 𝑏

1− (𝑎1− 𝑏1)

𝑓− 𝑎

1

𝑓 − 𝑎1

,

𝛼

𝑓 − 𝑏1

=

(𝑎

1− 𝑏

1) (𝑓 − 𝑏

1) − (𝑎

1− 𝑏1) (𝑓− 𝑏

1)

𝑓 − 𝑏1

= 𝑎

1− 𝑏

1− (𝑎1− 𝑏1)

𝑓− 𝑏

1

𝑓 − 𝑏1

,

1

(𝑓 − 𝑎1) (𝑓 − 𝑏

1)

=

1

𝑎1− 𝑏1

(

1

𝑓 − 𝑎1

−

1

𝑓 − 𝑏1

) ,

𝛽

𝑓− 𝑎2

=

(𝑎

2− 𝑏

2) (𝑓− 𝑎2) − (𝑎

2− 𝑏2) (𝑓− 𝑎

2)

𝑓− 𝑎2

= 𝑎

2− 𝑏

2− (𝑎2− 𝑏2)

𝑓− 𝑎

2

𝑓− 𝑎2

,

𝛽

𝑓− 𝑏2

=

(𝑎

2− 𝑏

2) (𝑓− 𝑏2) − (𝑎

2− 𝑏2) (𝑓− 𝑏

2)

𝑓− 𝑏2

= 𝑎

2− 𝑏

2− (𝑎2− 𝑏2)

𝑓− 𝑏

2

𝑓− 𝑏2

,

1

(𝑓− 𝑎2) (𝑓− 𝑏2)

=

1

𝑎2− 𝑏2

(

1

𝑓− 𝑎2

−

1

𝑓− 𝑏2

) ,

(16)

and the lemma of the logarithmic derivative, we obtain

𝑚(𝑟,

𝛼

𝑓 − 𝑎1

) = 𝑆 (𝑟) , 𝑚(𝑟,

𝛼

𝑓 − 𝑏1

) = 𝑆 (𝑟) ,

𝑚(𝑟,

𝛼

(𝑓 − 𝑎1) (𝑓 − 𝑏

1)

) = 𝑆 (𝑟) ,

(17)

𝑚(𝑟,

𝛽

𝑓− 𝑎2

) = 𝑆 (𝑟) , 𝑚(𝑟,

𝛽

𝑓− 𝑏2

) = 𝑆 (𝑟) ,

𝑚(𝑟,

𝛽

(𝑓− 𝑎2) (𝑓− 𝑏2)

) = 𝑆 (𝑟) .

(18)

Clearly, 𝛼 ̸≡ 0 and 𝛽 ̸≡ 0. Otherwise from (11) and (12)we have 𝑓 = 𝑎

1+ 𝐶1(𝑎1− 𝑏1) and 𝑓 = 𝑎

2+ 𝐶2(𝑎2− 𝑏2) for

nonzero constants 𝐶1, 𝐶2, which implies that 𝑇(𝑟, 𝑓) = 𝑆(𝑟)

and 𝑇(𝑟, 𝑓) = 𝑆(𝑟), a contradiction. Then by using a similarmethod we can deduce that 𝑔 ̸≡ 0 and ℎ ̸≡ 0. It is easy to seeby (11) if any zero of 𝑓 − 𝑎

1(𝑓 − 𝑏

1) of multiplicity 𝑙 is not the

pole of 𝑎1− 𝑏1and is not the zero of 𝑎

1− 𝑏1, then it must be a

zero of 𝛼 of multiplicity 𝑙 − 1 at least. Thus from (6), (7), (11),(13), the condition that𝑓 and𝑓 share (𝑎

1, 𝑎2) and (𝑏

1, 𝑏2) IM,

and the condition that 𝑓 is entire, we get

𝑁(𝑟, 𝑔) = 𝑆 (𝑟) . (19)

Likewise,

𝑁(𝑟, ℎ) = 𝑆 (𝑟) . (20)

Now by (13) and (17) together with

𝛼𝑓

(𝑓 − 𝑎1) (𝑓 − 𝑏

1)

=

𝛼

𝑓 − 𝑏1

+

𝑎1𝛼

(𝑓 − 𝑎1) (𝑓 − 𝑏

1)

, (21)

it follows that

𝑚(𝑟, 𝑔) ≤ 𝑚(𝑟,

𝛼 [(𝑎2− 𝑏2) 𝑓 − (𝑎

1− 𝑏1) 𝑓]

(𝑓 − 𝑎1) (𝑓 − 𝑏

1)

)

+ 𝑚(𝑟,

𝛼 (𝑎1𝑏2− 𝑎2𝑏1)

(𝑓 − 𝑎1) (𝑓 − 𝑏

1)

) + log 2


≤ 𝑚(𝑟,

𝛼𝑓

(𝑓 − 𝑎1) (𝑓 − 𝑏

1)

)

+ 𝑚(𝑟,

(𝑎2− 𝑏2) 𝑓 − (𝑎

1− 𝑏1) 𝑓

𝑓

) + 𝑆 (𝑟)

≤ 𝑆 (𝑟) .

(22)Thus from this and (19) we have

𝑇 (𝑟, 𝑔) = 𝑆 (𝑟) , (23)implying (i). Next, it is easy to see that 𝛾

𝑖̸≡ 𝑎2and 𝛾

𝑖̸≡

𝑏2(𝑖 = 1, 2). For 𝑖 = 1, 2, by (14), (18), 𝑎

1̸≡ 𝑏1, 𝑎2

̸≡ 𝑏2, and

the condition that 𝑓 is entire, we have

𝑚(𝑟, ℎ) ≤ 𝑚(𝑟,

𝛽 [(𝑎2− 𝑏2) 𝑓 − (𝑎

1− 𝑏1) 𝑓]

(𝑓− 𝑎2) (𝑓− 𝑏2)

)

+ 𝑚(𝑟,

𝛽 (𝑎1𝑏2− 𝑎2𝑏1)

(𝑓− 𝑎2) (𝑓− 𝑏2)

) + log 2

≤ 𝑚(𝑟,

𝛽 (𝑓− 𝛾𝑖)

(𝑓− 𝑎2) (𝑓− 𝑏2)

×

(𝑎2− 𝑏2) 𝑓 − (𝑎

1− 𝑏1) 𝑓

𝑓− 𝛾𝑖

) + 𝑆 (𝑟)

≤ 𝑚(𝑟,

𝛽

𝑓− 𝑏2

) + 𝑚(𝑟,

𝛽 (𝑎2− 𝛾𝑖)

(𝑓− 𝑎2) (𝑓− 𝑏2)

)

+ 𝑚(𝑟,

(𝑎2− 𝑏2) 𝑓 − (𝑎

1− 𝑏1) 𝑓

𝑓− 𝛾𝑖

) + 𝑆 (𝑟)

≤ 𝑚(𝑟,

((𝑎2− 𝑏2) / (𝑎1− 𝑏1)) 𝑓 − 𝑓

𝑓− 𝛾𝑖

) + 𝑆 (𝑟)

≤ 𝑚(𝑟,

((𝑎2− 𝑏2) / (𝑎1− 𝑏1)) 𝑓 − 𝛾

𝑖

𝑓− 𝛾𝑖

− 1) + 𝑆 (𝑟)

≤ 𝑚(𝑟,

𝑓− 𝛾𝑖

((𝑎2− 𝑏2) / (𝑎1− 𝑏1)) 𝑓 − 𝛾

𝑖

)

+ 𝑁(𝑟,

𝑓− 𝛾𝑖

((𝑎2− 𝑏2) / (𝑎1− 𝑏1)) 𝑓 − 𝛾

𝑖

)

− 𝑁(𝑟,

((𝑎2− 𝑏2) / (𝑎1− 𝑏1)) 𝑓 − 𝛾

𝑖

𝑓− 𝛾𝑖

) + 𝑆 (𝑟)

≤ 𝑚(𝑟,

1

(𝑎2− 𝑏2) / (𝑎1− 𝑏1)

×

𝑓− 𝛾𝑖

𝑓 − ((𝑎1− 𝑏1) / (𝑎2− 𝑏2)) 𝛾𝑖

)

+ 𝑁(𝑟, 𝑓− 𝛾𝑖)

+ 𝑁(𝑟,

1

((𝑎2− 𝑏2) / (𝑎1− 𝑏1)) 𝑓 − 𝛾

𝑖

)

− 𝑁(𝑟,

𝑎2− 𝑏2

𝑎1− 𝑏1

𝑓 − 𝛾𝑖) − 𝑁(𝑟,

1

𝑓− 𝛾𝑖

) + 𝑆 (𝑟)

≤ 𝑚(𝑟,

𝑓− (((𝑎

1− 𝑏1) / (𝑎2− 𝑏2)) 𝛾𝑖)

𝑓 − ((𝑎1− 𝑏1) / (𝑎2− 𝑏2)) 𝛾𝑖

)

+ 𝑚(𝑟,

(((𝑎1− 𝑏1) / (𝑎2− 𝑏2)) 𝛾𝑖)

− 𝛾𝑖

𝑓 − ((𝑎1− 𝑏1) / (𝑎2− 𝑏2)) 𝛾𝑖

)

+ 𝑁(𝑟,

1

((𝑎2− 𝑏2) / (𝑎1− 𝑏1)) 𝑓 − 𝛾

𝑖

)

− 𝑁(𝑟,

1

𝑓− 𝛾𝑖

) + 𝑆 (𝑟)

≤ 𝑚(𝑟,

1

𝑓 − ((𝑎1− 𝑏1) / (𝑎2− 𝑏2)) 𝛾𝑖

)

+ 𝑁(𝑟,

1

((𝑎2− 𝑏2) / (𝑎1− 𝑏1)) 𝑓 − 𝛾

𝑖

)

− 𝑁(𝑟,

1

𝑓− 𝛾𝑖

) + 𝑆 (𝑟)

≤ 𝑚(𝑟,

1

((𝑎2− 𝑏2) / (𝑎1− 𝑏1)) 𝑓 − 𝛾

𝑖

)

+ 𝑁(𝑟,

1

((𝑎2− 𝑏2) / (𝑎1− 𝑏1)) 𝑓 − 𝛾

𝑖

)

− 𝑁(𝑟,

1

𝑓− 𝛾𝑖

) + 𝑆 (𝑟)

≤ 𝑇 (𝑟, 𝑓) − 𝑁(𝑟,

1

𝑓− 𝛾𝑖

) + 𝑆 (𝑟) .

(24)

Thus from (20) and (24) it follows that

𝑇 (𝑟, ℎ) ≤ 𝑇 (𝑟, 𝑓) − 𝑁(𝑟,

1

𝑓− 𝛾𝑖

) + 𝑆 (𝑟) , 𝑖 = 1, 2.

(25)

This proves (ii) and completes the proof of Lemma 3.

Lemma 4 (see [10]; cf. [11, 12]). Let 𝑓 be a nonconstantmeromorphic function, and let 𝑓𝑛𝑃(𝑓) = 𝑄(𝑓), where 𝑃(𝑓)and 𝑄(𝑓) are differential polynomials in 𝑓 and the degree of𝑄(𝑓) is at most 𝑛. Then

𝑚(𝑟, 𝑃 (𝑓)) = 𝑆 (𝑟, 𝑓) . (26)






̸≡ 𝑏1, 𝑎2

̸≡ 𝑏2. Suppose

that 𝑓 and 𝑓 share (𝑎1, 𝑎2) and (𝑏

1, 𝑏2) IM. If

𝑚(𝑟, 𝑓) = 𝑚 (𝑟, 𝑓) + 𝑆 (𝑟, 𝑓) , (27)

then (𝑎2− 𝑏2)𝑓 − (𝑎

1− 𝑏1)𝑓+ 𝑎1𝑏2− 𝑎2𝑏1≡ 0.

Proof. Assume that (𝑎2− 𝑏2)𝑓 − (𝑎

1− 𝑏1)𝑓+𝑎1𝑏2−𝑎2𝑏1

̸≡ 0.Let 𝛼, 𝛽, 𝑔, ℎ, and 𝛾

𝑖be defined by (11)–(15), respectively.Then

from the proof process of Lemma 3 we know 𝛼 ̸≡ 0, 𝛽 ̸≡ 0,𝑔 ̸≡ 0, ℎ ̸≡ 0, 𝛾

𝑖̸≡ 𝑎2, and 𝛾

𝑖̸≡ 𝑏2(𝑖 = 1, 2). Since 𝑓 and 𝑓

share (𝑎1, 𝑎2) and (𝑏

1, 𝑏2) IM, by Lemmas 1, 2, and 3 it follows

that

𝑚(𝑟, 𝑓) = 𝑁(𝑟,

1

𝑓 − 𝑎1

) + 𝑁(𝑟,

1

𝑓 − 𝑏1

) + 𝑆 (𝑟) , (28)

𝑇 (𝑟, 𝑔) = 𝑆 (𝑟) , (29)

𝑇 (𝑟, ℎ) ≤ 𝑇 (𝑟, 𝑓) − 𝑁(𝑟,

1

𝑓− 𝛾𝑖

) + 𝑆 (𝑟) , 𝑖 = 1, 2.

(30)

Now from the second fundamental theorem, (27), (28), andthe assumption that 𝑓 is entire, we deduce

2𝑚 (𝑟, 𝑓) = 2𝑇 (𝑟, 𝑓

)

≤ 𝑁(𝑟,

1

𝑓− 𝑎2

)

+ 𝑁(𝑟,

1

𝑓− 𝑏2

) + 𝑁(𝑟,

1

𝑓− 𝛾𝑖

) + 𝑆 (𝑟)

≤ 𝑁(𝑟,

1

𝑓 − 𝑎1

) + 𝑁(𝑟,

1

𝑓 − 𝑏1

)

+ 𝑁(𝑟,

1

𝑓− 𝛾𝑖

) + 𝑆 (𝑟)

= 𝑚 (𝑟, 𝑓) + 𝑁(𝑟,

1

𝑓− 𝛾𝑖

) + 𝑆 (𝑟)

= 𝑚 (𝑟, 𝑓) + 𝑁(𝑟,

1

𝑓− 𝛾𝑖

) + 𝑆 (𝑟) ,

(31)

which yields

𝑁(𝑟,

1

𝑓− 𝛾𝑖

) = 𝑚(𝑟, 𝑓) + 𝑆 (𝑟) . (32)

Again by (27), (30), (32), and the assumption that 𝑓 is entire,we obtain

𝑇 (𝑟, ℎ) = 𝑆 (𝑟) . (33)

For any 𝑧0

∈ 𝑆(𝑚,𝑛)

(𝑎1, 𝑎2) ∪ 𝑆(𝑚,𝑛)

(𝑏1, 𝑏2), from (13) and

(14), we can get 𝑛𝑔(𝑧0) − 𝑚ℎ(𝑧

0) = 0.

If 𝑛𝑔 − 𝑚ℎ ≡ 0, then by (13) and (14) we deduce

𝑛(

𝑓− 𝑏

1

𝑓 − 𝑏1

−

𝑓− 𝑎

1

𝑓 − 𝑎1

) ≡ 𝑚(

𝑓− 𝑏

2

𝑓− 𝑏2

−

𝑓− 𝑎

2

𝑓− 𝑎2

) , (34)

which implies that

(

𝑓 − 𝑏1

𝑓 − 𝑎1

)

𝑛

≡ 𝑐1(

𝑓− 𝑏2

𝑓− 𝑎2

)

𝑚

, (35)

where 𝑐1is a nonzero constant. If 𝑛 ̸= 𝑚, then from (35)

and the condition that 𝑓 is entire, we obtain 𝑛𝑚(𝑟, 𝑓) =𝑚𝑚(𝑟, 𝑓

) + 𝑆(𝑟), which contradicts (27). If 𝑛 = 𝑚, then we

get

𝑓 − 𝑏1

𝑓 − 𝑎1

≡ 𝑐2(

𝑓− 𝑏2

𝑓− 𝑎2

) , (36)

where 𝑐2is a nonzero constant. We claim that 𝑐

2̸= 1. Indeed,

if 𝑐2= 1, then by (36) we deduce

𝑓 − 𝑏1

𝑓 − 𝑎1

≡

𝑓− 𝑏2

𝑓− 𝑎2

, (37)

which leads to (𝑎2− 𝑏2)𝑓 − (𝑎

1− 𝑏1)𝑓+ 𝑎1𝑏2− 𝑎2𝑏1≡ 0. This

contradicts the assumption. Thus 𝑐2

̸= 1 and so from (36) wehave

𝑓 [(1 − 𝑐2) 𝑓+ 𝑐2𝑏2− 𝑎2] = (𝑏

1− 𝑐2𝑎1) 𝑓+ 𝑐2𝑎1𝑏2− 𝑎2𝑏1.

(38)

This and Lemma 4 yield

𝑚(𝑟, (1 − 𝑐2) 𝑓+ 𝑐2𝑏2− 𝑎2) = 𝑆 (𝑟) , (39)

which gives𝑚(𝑟, 𝑓) = 𝑆(𝑟). From this and the condition that𝑓 is entire, it follows that 𝑇(𝑟, 𝑓) = 𝑆(𝑟), a contradiction.Hence 𝑛𝑔 − 𝑚ℎ ̸≡ 0, for any positive integers𝑚 and 𝑛.

Therefore by (29) and (33) we obtain

𝑁(𝑚,𝑛)

(𝑟,

1

𝑓 − 𝑎1

) + 𝑁(𝑚,𝑛)

(𝑟,

1

𝑓 − 𝑏1

)

≤ 𝑁(𝑟,

1

𝑛𝑔 − 𝑚ℎ

) + 𝑆 (𝑟)

≤ 𝑇 (𝑟, 𝑔) + 𝑇 (𝑟, ℎ) + 𝑆 (𝑟)

= 𝑆 (𝑟) ,

(40)


for any positive integers 𝑚 and 𝑛. It follows from this,Lemma 1, the second fundamental theorem, and the condi-tion that 𝑓 is entire that

𝑇 (𝑟, 𝑓)

≤ 𝑁(𝑟,

1

𝑓 − 𝑎1

) + 𝑁(𝑟,

1

𝑓 − 𝑏1

) + 𝑆 (𝑟)

= ∑

𝑚,𝑛

(𝑁(𝑚,𝑛)

(𝑟,

1

𝑓 − 𝑎1

) + 𝑁(𝑚,𝑛)

(𝑟,

1

𝑓 − 𝑏1

)) + 𝑆 (𝑟)

= ∑

𝑚+𝑛≥6

(𝑁(𝑚,𝑛)

(𝑟,

1

𝑓 − 𝑎1

) + 𝑁(𝑚,𝑛)

(𝑟,

1

𝑓 − 𝑏1

)) + 𝑆 (𝑟)

≤ ∑

𝑚+𝑛≥6

1

6

(𝑁(𝑚,𝑛)

(𝑟,

1

𝑓 − 𝑎1

)

+𝑁(𝑚,𝑛)

(𝑟,

1

𝑓− 𝑎2

))

+ ∑

𝑚+𝑛≥6

1

6

(𝑁(𝑚,𝑛)

(𝑟,

1

𝑓 − 𝑏1

)

+𝑁(𝑚,𝑛)

(𝑟,

1

𝑓− 𝑏2

)) + 𝑆 (𝑟)

≤

1

6

(𝑁(𝑟,

1

𝑓 − 𝑎1

) + 𝑁(𝑟,

1

𝑓− 𝑎2

))

+

1

6

(𝑁(𝑟,

1

𝑓 − 𝑏1

) + 𝑁(𝑟,

1

𝑓− 𝑏2

)) + 𝑆 (𝑟)

≤

1

3

𝑇 (𝑟, 𝑓) +

1

3

𝑇 (𝑟, 𝑓) + 𝑆 (𝑟)

=

1

3

𝑇 (𝑟, 𝑓) +

1

3

𝑚 (𝑟, 𝑓) + 𝑆 (𝑟)

=

1

3

𝑇 (𝑟, 𝑓) +

1

3

𝑚(𝑟,

𝑓

𝑓

𝑓) + 𝑆 (𝑟)

≤

1

3

𝑇 (𝑟, 𝑓) +

1

3

𝑚 (𝑟, 𝑓) + 𝑆 (𝑟)

=

1

3

𝑇 (𝑟, 𝑓) +

1

3

𝑇 (𝑟, 𝑓) + 𝑆 (𝑟)

=

2

3

𝑇 (𝑟, 𝑓) + 𝑆 (𝑟) ,

(41)

which implies that 𝑇(𝑟, 𝑓) = 𝑆(𝑟), a contradiction.Thus (𝑎2−

𝑏2)𝑓 − (𝑎

1− 𝑏1)𝑓+ 𝑎1𝑏2− 𝑎2𝑏1≡ 0.


3. Proof of Theorem 1

Suppose that (𝑎2− 𝑏2)𝑓 − (𝑎

1− 𝑏1)𝑓+ 𝑎1𝑏2− 𝑎2𝑏1

̸≡ 0.Since 𝑓 and 𝑓 share (𝑎

1, 𝑎2) and (𝑏

1, 𝑏2) IM, by Lemma 1

we have 𝑆(𝑟, 𝑓) = 𝑆(𝑟, 𝑓) := 𝑆(𝑟). Let 𝛼, 𝛽, 𝑔, ℎ, and 𝛾𝑖be

defined by (11)–(15), respectively.Then from the proof processof Lemma 3 we know 𝛼 ̸≡ 0, 𝛽 ̸≡ 0, 𝑔 ̸≡ 0, ℎ ̸≡ 0, 𝛾

𝑖̸≡ 𝑎2,

and 𝛾𝑖

̸≡ 𝑏2(𝑖 = 1, 2). Next we rewrite (13) as

[𝑔 − (𝑎

1− 𝑏

1) (𝑎2− 𝑏2)] 𝑓2

= 𝑑1𝑓𝑓+ 𝑑2𝑓 + 𝑑3𝑓+ 𝑑4𝑓2

+ 𝑑5,

(42)

where 𝑑1= (𝑎1− 𝑏1)(𝑏

1+ 𝑏2− 𝑎2− 𝑎

1), 𝑑2= (𝑎

1− 𝑏

1)(𝑎1𝑏2−

𝑎2𝑏1)+(𝑎2− 𝑏2)(𝑎1𝑏

1− 𝑎

1𝑏1) + (𝑎1+ 𝑏1)𝑔, 𝑑3= (𝑎1− 𝑏1)(𝑎2𝑏1+

𝑎

1𝑏1− 𝑎1𝑏2− 𝑎1𝑏

1), 𝑑4= (𝑎1−𝑏1)2, and𝑑

5= (𝑎1𝑏

1−𝑎

1𝑏1)(𝑎1𝑏2−

𝑎2𝑏1) − 𝑎1𝑏1𝑔 are all small functions with respect to 𝑓.

Now we divide into two cases.

Case 1. 𝑔− (𝑎1−𝑏

1)(𝑎2−𝑏2) ≡ 0; that is, 𝑔 ≡ (𝑎

1−𝑏

1)(𝑎2−𝑏2).

We again discuss the three subcases.

Subcase 1. 𝑎1

̸≡ 𝑎2and 𝑏1

̸≡ 𝑏2. Since 𝑓 and 𝑓 share (𝑎

1, 𝑎2)

and (𝑏1, 𝑏2) IM, the zeros of 𝑓 − 𝑎

1and 𝑓 − 𝑏

1of multiplicity

larger than one are the zeros 𝑎1− 𝑎2and 𝑏1− 𝑏2, respectively.

It then follows that

∑

𝑚≥2,𝑛≥1

(𝑁(𝑚,𝑛)

(𝑟,

1

𝑓 − 𝑎1

) + 𝑁(𝑚,𝑛)

(𝑟,

1

𝑓 − 𝑏1

))

≤ 𝑁(𝑟,

1

𝑎

1− 𝑎2

) + 𝑁(𝑟,

1

𝑏

1− 𝑏2

) + 𝑆 (𝑟)

≤ 𝑇 (𝑟, 𝑎

1) + 𝑇 (𝑟, 𝑎

2) + 𝑇 (𝑟, 𝑏

1)

+ 𝑇 (𝑟, 𝑏2) + 𝑆 (𝑟)

= 𝑆 (𝑟) ;

(43)

that is,

∑

𝑚≥2,𝑛≥1

(𝑁(𝑚,𝑛)

(𝑟,

1

𝑓 − 𝑎1

) + 𝑁(𝑚,𝑛)

(𝑟,

1

𝑓 − 𝑏1

)) = 𝑆 (𝑟) .

(44)

Let 𝑧0∈ 𝑆(1,𝑛)

(𝑎1, 𝑎2). For 𝑛 ≥ 2, from (13) we get

𝑔 (𝑧0) = (𝑎

1(𝑧0) − 𝑎2(𝑧0)) (𝑎2(𝑧0) − 𝑏2(𝑧0))

= (𝑎

1(𝑧0) − 𝑏

1(𝑧0)) (𝑎2(𝑧0) − 𝑏2(𝑧0)) ,

(45)

which implies that 𝑎2(𝑧0) − 𝑏

1(𝑧0) = 0 or 𝑎

2(𝑧0) − 𝑏2(𝑧0) = 0.

If 𝑎2− 𝑏

1≡ 0, then by (13) we deduce

𝑓=

𝑎

1− 𝑏2

𝑎1− 𝑏1

𝑓 +

𝑎1𝑏2− 𝑎

1𝑏1

𝑎1− 𝑏1

, (46)

which, in view of the condition that 𝑓 is entire, implies that𝑚(𝑟, 𝑓) = 𝑚(𝑟, 𝑓

) + 𝑆(𝑟). From this and Lemma 5, it follows

that (𝑎2−𝑏2)𝑓−(𝑎

1−𝑏1)𝑓+𝑎1𝑏2−𝑎2𝑏1≡ 0, contradicting the

assumption.Thus 𝑎2−𝑏

1̸≡ 0. By the conditions inTheorem 1,

we know that 𝑎2− 𝑏2

̸≡ 0.


Hence

∑

𝑛≥2

𝑁(1,𝑛)

(𝑟,

1

𝑓 − 𝑎1

)

≤ 𝑁(𝑟,

1

𝑎2− 𝑏

1

) + 𝑁(𝑟,

1

𝑎2− 𝑏2

) + 𝑆 (𝑟)

≤ 2𝑇 (𝑟, 𝑎2) + 𝑇 (𝑟, 𝑏

2) + 𝑇 (𝑟, 𝑏

1) + 𝑆 (𝑟) = 𝑆 (𝑟) ;

(47)

that is,

∑

𝑛≥2

𝑁(1,𝑛)

(𝑟,

1

𝑓 − 𝑎1

) = 𝑆 (𝑟) . (48)

Similarly,

∑

𝑛≥2

𝑁(1,𝑛)

(𝑟,

1

𝑓 − 𝑏1

) = 𝑆 (𝑟) . (49)

It then follows from (44)–(49) and the second fundamentaltheorem that

𝑇 (𝑟, 𝑓) ≤ 𝑁(𝑟,

1

𝑓 − 𝑎1

) + 𝑁(𝑟,

1

𝑓 − 𝑏1

) + 𝑆 (𝑟)

= 𝑁(1,1)

(𝑟,

1

𝑓 − 𝑎1

) + 𝑁(1,1)

(𝑟,

1

𝑓 − 𝑏1

) + 𝑆 (𝑟) .

(50)

For any 𝑧1

∈ 𝑆(1,1)

(𝑎1, 𝑎2) ∪ 𝑆

(1,1)(𝑏1, 𝑏2), from (13) and

(14), we can get 𝑔(𝑧1) − ℎ(𝑧

1) = 0.

If 𝑔 − ℎ ≡ 0, then by (13) and (14) we have

𝑓− 𝑏

1

𝑓 − 𝑏1

−

𝑓− 𝑎

1

𝑓 − 𝑎1

≡

𝑓− 𝑏

2

𝑓− 𝑏2

−

𝑓− 𝑎

2

𝑓− 𝑎2

, (51)

which implies that

𝑓 − 𝑏1

𝑓 − 𝑎1

≡ 𝑐3

𝑓− 𝑏2

𝑓− 𝑎2

, (52)

where 𝑐3is a nonzero constant. We claim that 𝑐

3̸= 1. Indeed,

if 𝑐3= 1, then by (52) we have

𝑓 − 𝑏1

𝑓 − 𝑎1

≡

𝑓− 𝑏2

𝑓− 𝑎2

, (53)

which leads to (𝑎2− 𝑏2)𝑓 − (𝑎

1− 𝑏1)𝑓+ 𝑎1𝑏2− 𝑎2𝑏1≡ 0. This

contradicts the assumption. Thus 𝑐3

̸= 1 and so from (52) weget

𝑓 [(1 − 𝑐3) 𝑓+ 𝑐3𝑏2− 𝑎2] = (𝑏

1− 𝑐3𝑎1) 𝑓+ 𝑐3𝑎1𝑏2− 𝑎2𝑏1.

(54)

This and Lemma 4 yield

𝑚(𝑟, (1 − 𝑐3) 𝑓+ 𝑐3𝑏2− 𝑎2) = 𝑆 (𝑟) , (55)

which gives𝑚(𝑟, 𝑓) = 𝑆(𝑟). From this and the condition that𝑓 is entire, it follows that 𝑇(𝑟, 𝑓) = 𝑆(𝑟), a contradiction.Hence 𝑔 − ℎ ̸≡ 0.

Therefore by (50) and Lemma 3 we obtain

𝑇 (𝑟, 𝑓) ≤ 𝑁(1,1)

(𝑟,

1

𝑓 − 𝑎1

) + 𝑁(1,1)

(𝑟,

1

𝑓 − 𝑏1

) + 𝑆 (𝑟)

≤ 𝑁(𝑟,

1

𝑔 − ℎ

) + 𝑆 (𝑟)

≤ 𝑇 (𝑟, 𝑔) + 𝑇 (𝑟, ℎ) + 𝑆 (𝑟)

≤ 𝑇 (𝑟, 𝑓) − 𝑁(𝑟,

1

𝑓− 𝛾𝑖

) + 𝑆 (𝑟) ,

(56)

which implies that

𝑁(𝑟,

1

𝑓− 𝛾𝑖

) = 𝑆 (𝑟) , 𝑖 = 1, 2. (57)

This is impossible by the second fundamental theorem.

Subcase 2. Either 𝑎1≡ 𝑎2and 𝑏1

̸≡ 𝑏2or 𝑎1

̸≡ 𝑎2and 𝑏1≡ 𝑏2.

Without loss of generality, we assume that 𝑎1≡ 𝑎2and 𝑏1

̸≡

𝑏2. It is easy to see by (13) that the zeros of 𝑓 − 𝑎

1and 𝑓 − 𝑎

2

of multiplicity all larger than one are the zeros of 𝑔. Thus byLemma 3,

∑

𝑚≥2,𝑛≥2

𝑁(𝑚,𝑛)

(𝑟,

1

𝑓 − 𝑎1

) ≤ 𝑁(𝑟,

1

𝑔

) + 𝑆 (𝑟)

≤ 𝑇 (𝑟, 𝑔) + 𝑆 (𝑟) = 𝑆 (𝑟) ;

(58)

that is,

∑

𝑚≥2,𝑛≥2

𝑁(𝑚,𝑛)

(𝑟,

1

𝑓 − 𝑎1

) = 𝑆 (𝑟) . (59)

By the discussion of Subcase 1, we see

𝑁(𝑟,

1

𝑓 − 𝑏1

) = 𝑁(1,1)

(𝑟,

1

𝑓 − 𝑏1

) + 𝑆 (𝑟)

≤ 𝑁(𝑟,

1

𝑔 − ℎ

) + 𝑆 (𝑟)

≤ 𝑇 (𝑟, 𝑔) + 𝑇 (𝑟, ℎ) + 𝑆 (𝑟)

≤ 𝑇 (𝑟, 𝑓) − 𝑁(𝑟,

1

𝑓− 𝛾𝑖

) + 𝑆 (𝑟) ;

(60)

that is,

𝑁(𝑟,

1

𝑓 − 𝑏1

) ≤ 𝑇 (𝑟, 𝑓) − 𝑁(𝑟,

1

𝑓− 𝛾𝑖

) + 𝑆 (𝑟) ,

𝑖 = 1, 2.

(61)

Note that the zeros of 𝑓 − 𝑎1of multiplicity larger than one

are all the zeros of 𝑓 − 𝑎2

= 𝑓− 𝑎

1. Since 𝑓 and 𝑓 share

(𝑎1, 𝑎2) IM, it follows that

𝑁(1,1)

(𝑟,

1

𝑓 − 𝑎1

) = 𝑆 (𝑟) . (62)


Now from (59)–(62) and the second fundamental theo-rem, we obtain

𝑇 (𝑟, 𝑓) ≤ 𝑁(𝑟,

1

𝑓 − 𝑎1

) + 𝑁(𝑟,

1

𝑓 − 𝑏1

) + 𝑆 (𝑟)

≤ 𝑁(2,1)

(𝑟,

1

𝑓 − 𝑎1

) + 𝑇 (𝑟, 𝑓)

− 𝑁(𝑟,

1

𝑓− 𝛾𝑖

) + 𝑆 (𝑟) ;

(63)

that is,

𝑁(𝑟,

1

𝑓− 𝛾𝑖

) ≤ 𝑁(2,1)

(𝑟,

1

𝑓 − 𝑎1

) + 𝑆 (𝑟) , 𝑖 = 1, 2.

(64)

Let

𝜙 = 2

𝑓− 𝑏

2

𝑓− 𝑏2

− 2

𝑓− 𝑏

1

𝑓 − 𝑏1

+

𝑎

1− 𝑏

1

𝑎1− 𝑏1

− 2

𝑎

2− 𝑏

2

𝑎2− 𝑏2

. (65)

It is easily seen from (65) and the lemma of the logarithmicderivative that

𝑚(𝑟, 𝜙) = 𝑆 (𝑟) . (66)

Note that common simple zeros of 𝑓 − 𝑏1and 𝑓 − 𝑏

2are not

the poles of 𝜙. In terms of the discussion of Subcase 1, from(65) we know 𝑁(𝑟, 𝜙) = 𝑆(𝑟), which together with (66) givesthat

𝑇 (𝑟, 𝜙) = 𝑆 (𝑟) . (67)

Let 𝑧2

∈ 𝑆(2,1)

(𝑎1, 𝑎2). Then by (65) and (13) we have

𝜙(𝑧2) = 0.If 𝜙 ≡ 0, then from (65) we derive

(𝑓− 𝑏2)

2

= 𝑐4

(𝑎2− 𝑏2)2

𝑎1− 𝑏1

(𝑓 − 𝑏1)2

, (68)

where 𝑐4is a nonzero constant. This, in view of the condition

that 𝑓 is entire, implies that 𝑚(𝑟, 𝑓) = 𝑚(𝑟, 𝑓) + 𝑆(𝑟). Fromthis and Lemma 5, it follows that (𝑎

2− 𝑏2)𝑓 − (𝑎

1− 𝑏1)𝑓+

𝑎1𝑏2− 𝑎2𝑏1≡ 0, contradicting the assumption. Thus 𝜙 ̸≡ 0.

Hence by (64) and (67) we obtain

𝑁(𝑟,

1

𝑓− 𝛾𝑖

) ≤ 𝑁(2,1)

(𝑟,

1

𝑓 − 𝑎1

) + 𝑆 (𝑟)

≤ 𝑁(𝑟,

1

𝜙

) + 𝑆 (𝑟)

≤ 𝑇 (𝑟, 𝜙) + 𝑆 (𝑟)

≤ 𝑆 (𝑟) ;

(69)

that is,

𝑁(𝑟,

1

𝑓− 𝛾𝑖

) = 𝑆 (𝑟) , 𝑖 = 1, 2. (70)

This is also impossible by the second fundamental theorem.

Subcase 3. 𝑎1≡ 𝑎2and 𝑏1≡ 𝑏2. By the discussion of Subcase 2,

we see

𝑇 (𝑟, 𝑓) ≤ 𝑁(2,1)

(𝑟,

1

𝑓 − 𝑎1

) + 𝑁(2,1)

(𝑟,

1

𝑓 − 𝑏1

) + 𝑆 (𝑟) .

(71)

We claim that

𝑁(2,1)

(𝑟,

1

𝑓 − 𝑎1

) = 𝑆 (𝑟) , (72)

𝑁(2,1)

(𝑟,

1

𝑓 − 𝑏1

) = 𝑆 (𝑟) . (73)

Let

𝜓 = 2

𝑓− 𝑏

2

𝑓− 𝑏2

−

𝑓− 𝑏

1

𝑓 − 𝑏1

− 2

𝑎

2− 𝑏

2

𝑎2− 𝑏2

. (74)

It is easily known from (74) and the lemma of the logarithmicderivative that

𝑚(𝑟, 𝜓) = 𝑆 (𝑟) . (75)

Note that common zeros of 𝑓 − 𝑏1of multiplicity two and

𝑓− 𝑏2of multiplicity one are not the poles of 𝜓. In terms of

the discussion of Subcase 2, we know 𝑁(𝑟, 𝜓) = 𝑆(𝑟), whichtogether with (75) gives that

𝑇 (𝑟, 𝜓) = 𝑆 (𝑟) . (76)

Let 𝑧3

∈ 𝑆(2,1)

(𝑎1, 𝑎2). Then by (74) and (13) we have

𝜓(𝑧3) = 0.If 𝜓 ̸≡ 0, then from (76) we get

𝑁(2,1)

(𝑟,

1

𝑓 − 𝑎1

) ≤ 𝑁(𝑟,

1

𝜓

) + 𝑆 (𝑟)

≤ 𝑇 (𝑟, 𝜓) + 𝑆 (𝑟) ≤ 𝑆 (𝑟) ,

(77)

that is,

𝑁(2,1)

(𝑟,

1

𝑓 − 𝑎1

) = 𝑆 (𝑟) . (78)

If 𝜓 ≡ 0, then by (74) we deduce

(𝑓− 𝑏2)

2

= 𝑐5(𝑎2− 𝑏2)2

(𝑓 − 𝑏1) , (79)

where 𝑐5is a nonzero constant. This implies 𝑎

1(𝑧3) − 𝑏1(𝑧3) −

1/𝑐5

= 0. Since 𝑎1

≡ 𝑎2, 𝑏1

≡ 𝑏2, and 𝑎

2̸≡ 𝑏2, we obtain

𝑎1− 𝑏1− 1/𝑐5

̸≡ 0. Thus

𝑁(2,1)

(𝑟,

1

𝑓 − 𝑎1

) ≤ 𝑁(𝑟,

1

𝑎1− 𝑏1− 1/𝑐5

) + 𝑆 (𝑟)

≤ 𝑇 (𝑟, 𝑎1) + 𝑇 (𝑟, 𝑏

1) + 𝑆 (𝑟) ≤ 𝑆 (𝑟) ;

(80)

that is,

𝑁(2,1)

(𝑟,

1

𝑓 − 𝑎1

) = 𝑆 (𝑟) . (81)

Hence (72) follows. In the same manner as above, we canprove (73). The proof of the claim is complete. Now by (71)–(73) we get 𝑇(𝑟, 𝑓) = 𝑆(𝑟), a contradiction.


Case 2. 𝑔 − (𝑎1− 𝑏

1)(𝑎2− 𝑏2) ̸≡ 0. Then by (42) and (i) in

Lemma 3 we have

2𝑚 (𝑟, 𝑓) ≤ 𝑚(𝑟,

1

𝑔 − (𝑎

1− 𝑏

1) (𝑎2− 𝑏2)

)

+ 𝑚(𝑟, 𝑑1𝑓𝑓+ 𝑑2𝑓 + 𝑑3𝑓+ 𝑑4𝑓2

+ 𝑑5)

≤ 𝑚(𝑟, 𝑓(𝑑1𝑓+ 𝑑2+ 𝑑3

𝑓

𝑓

+ 𝑑4𝑓𝑓

𝑓

))

+ 𝑚 (𝑟, 𝑑5) + 𝑆 (𝑟)

≤ 𝑚 (𝑟, 𝑓) + 𝑚(𝑟, 𝑓(𝑑1+ 𝑑4

𝑓

𝑓

)) + 𝑆 (𝑟)

≤ 𝑚 (𝑟, 𝑓) + 𝑚 (𝑟, 𝑓) + 𝑆 (𝑟) ,

(82)

which implies that

𝑚(𝑟, 𝑓) ≤ 𝑚 (𝑟, 𝑓) + 𝑆 (𝑟) . (83)

On the other hand,

𝑚(𝑟, 𝑓) ≤ 𝑚(𝑟, 𝑓

𝑓

𝑓

) ≤ 𝑚 (𝑟, 𝑓) + 𝑆 (𝑟) . (84)

Combining (83) with (84) yields

𝑚(𝑟, 𝑓) = 𝑚 (𝑟, 𝑓) + 𝑆 (𝑟) . (85)

This andLemma 5 lead to (𝑎2− 𝑏2)𝑓−(𝑎

1− 𝑏1)𝑓+ 𝑎1𝑏2− 𝑎2𝑏1≡

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, “Values shared by an entire functionand its derivative,” inComplex Analysis, vol. 599 of Lecture Notesin Mathematics, pp. 101–103, Springer, Berlin, Germany, 1977.

[6] E. Mues and N. Steinmetz, “Meromorphe Funktionen, die mitihrer Ableitung Werte teilen,” Manuscripta Mathematica, vol.29, no. 2–4, pp. 195–206, 1979.

[7] J. H. Zheng and S. P. Wang, “On the unicity of meromorphicfunctions and their derivatives,” Advances in Mathematics, vol.21, no. 3, pp. 334–341, 1992.

[8] G. Qiu, “Uniqueness of entire functions that share some smallfunctions,” Kodai Mathematical Journal, vol. 23, no. 1, pp. 1–11,2000.

[9] P. Li, “Unicity of meromorphic functions and their derivatives,”Journal of Mathematical Analysis and Applications, vol. 285, no.2, pp. 651–665, 2003.

[10] J. Clunie, “On integral and meromorphic functions,” Journal ofthe London Mathematical Society, vol. 37, pp. 17–27, 1962.

[11] E. Mues and N. Steinmetz, “The theorem of Tumura-Clunie formeromorphic functions,” Journal of the London MathematicalSociety, vol. 23, no. 1, pp. 113–122, 1981.

[12] G. Weissenborn, “On the theorem of Tumura and Clunie,”TheBulletin of the London Mathematical Society, vol. 18, no. 4, pp.371–373, 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 © 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 𝑓(𝑧) = 𝑧 + ∑∞𝑛=2

𝑎𝑛𝑧𝑛 in the unit disk, it is well-known that

𝑎𝑛

≤ 𝑛 for 𝑛 ≥ 2. But the inequality

𝑎𝑛

≤ 𝑛 does not imply the univalence of 𝑓. 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 𝛼 (0 ≤ 𝛼 < 1) for functions with prescribedcoefficient bound on the analytic part.

1. Introduction

Let A denote the class of all analytic functions 𝑓 defined inthe open unit disk D := {𝑧 ∈ C : |𝑧| < 1} normalized by𝑓(0) = 0 = 𝑓

(0) − 1. For functions 𝑓 of the form

𝑓 (𝑧) = 𝑧 +

∞

∑

𝑛=2

𝑎𝑛𝑧𝑛 (1)

belonging to the subclass S of A consisting of univalentfunctions, de Branges [1] proved the famous Bieberbachconjecture that |𝑎

𝑛| ≤ 𝑛 for 𝑛 ≥ 2. However, the inequality

|𝑎𝑛| ≤ 𝑛 (𝑛 ≥ 2) does not imply that 𝑓 is univalent. A

function 𝑓 given by (1) whose coefficients satisfy |𝑎𝑛| ≤ 𝑛 for

𝑛 ≥ 2 is necessarily analytic inD by the usual comparison testand hence a member of A. But it need not be univalent. Forexample, the function

𝑓 (𝑧) = 𝑧 − 2𝑧2− 3𝑧3− ⋅ ⋅ ⋅ = 2𝑧 −

𝑧

(1 − 𝑧)2 (2)

satisfies the inequality |𝑎𝑛| ≤ 𝑛 (𝑛 ≥ 2) but its derivative

vanishes inside D and so the function 𝑓 is not univalent inD. It is therefore of interest to determine the largest subdisk

|𝑧| < 𝜌 < 1 in which the functions 𝑓 satisfying the inequality|𝑎𝑛| ≤ 𝑛 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 𝑅 such that, for every 𝑓 ∈ F,𝑟−1𝑓(𝑟𝑧) ∈ G for each 𝑟 ≤ 𝑅.

1.1. Analytic Case. Most of the classes in univalent functiontheory are characterized by the quantities 𝑧𝑓(𝑧)/𝑓(𝑧) or 1 +𝑧𝑓(𝑧)/𝑓

(𝑧) lying in a given domain in the right half-plane.

For instance, the subclasses S∗(𝛼) and K(𝛼) (0 ≤ 𝛼 < 1)of S consisting of starlike functions of order 𝛼 and convexfunctions of order 𝛼, respectively, are defined analytically bythe equivalences

𝑓 ∈ S∗

(𝛼) ⇐⇒ Re(𝑧𝑓(𝑧)

𝑓 (𝑧)

) > 𝛼,

𝑓 ∈ K (𝛼) ⇐⇒ Re(𝑧𝑓(𝑧)

𝑓(𝑧)

+ 1) > 𝛼.

(3)


http://dx.doi.org/10.1155/2014/454152


These classes were introduced by Robertson [2]. The classesS∗ := S∗(0) and K := K(0) are the familiar classes ofstarlike and convex functions, respectively. Goodman [3]introduced the class UCV of uniformly convex functions𝑓 ∈ A, which map every circular arc 𝛾 contained in D withcenter 𝜁 ∈ D onto a convex arc. For 𝑓 ∈ A, Rønning [4] andMa and Minda [5] independently proved that

𝑓 ∈ UCV ⇐⇒ Re(𝑧𝑓(𝑧)

𝑓(𝑧)

+ 1) >

𝑧𝑓(𝑧)

𝑓(𝑧)

(𝑧 ∈ D) .

(4)

Closely related to the classUCV is the class S𝑃of parabolic

starlike functions, introduced by Rønning [4] consisting offunctions 𝑓 = 𝑧𝑔 where 𝑔 ∈ UCV; that is, a function 𝑓 ∈S𝑃satisfies

Re(𝑧𝑓(𝑧)

𝑓 (𝑧)

) >

𝑧𝑓(𝑧)

𝑓 (𝑧)

− 1

(𝑧 ∈ D) . (5)

In 1970, Gavrilov [6] showed that the radius of univalencefor functions 𝑓 ∈ A satisfying |𝑎

𝑛| ≤ 𝑛 (𝑛 ≥ 2) is the real

root 𝑟0≃ 0.164 of the equation 2(1− 𝑟)3 − (1+ 𝑟) = 0. In 1982,

Yamashita [7] showed that the radius of univalence obtainedby Gavrilov [6] is also the radius of starlikeness for functions𝑓 ∈ A satisfying |𝑎

𝑛| ≤ 𝑛. Yamashita [7] also proved that the

radius of convexity for functions 𝑓 ∈ A satisfying |𝑎𝑛| ≤ 𝑛

(𝑛 ≥ 2) is the real root 𝑟0≃ 0.090 of the equation 2(1 − 𝑟)4 −

(1 + 4𝑟 + 𝑟2) = 0.

The inequality |𝑎𝑛| ≤ 𝑀 holds for functions 𝑓 ∈ A

satisfying |𝑓(𝑧)| ≤ 𝑀. Gavrilov [6] proved that the radius ofunivalence for functions 𝑓 ∈ A satisfying |𝑎

𝑛| ≤ 𝑀 (𝑛 ≥ 2)

is 1 − √𝑀/(1 +𝑀), which also turned out to be their radiusof starlikeness, a result proved by Yamashita [7]. The radiusof convexity for functions 𝑓 ∈ A satisfying |𝑎

𝑛| ≤ 𝑀 (𝑛 ≥ 2)

is the real root of the equation (𝑀+1)(1− 𝑟)3 −𝑀(1+𝑟) = 0.For 0 ≤ 𝑏 ≤ 1, let A

𝑏denote the class of functions 𝑓

given by (1) with |𝑎2| = 2𝑏. 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 sharpS∗(𝛼),K(𝛼) (0 ≤ 𝛼 < 1),UCV and S

𝑃radii for functions

𝑓 ∈ A𝑏satisfying |𝑎

𝑛| ≤ 𝑛, |𝑎

𝑛| ≤ 𝑀, or |𝑎

𝑛| ≤ 𝑀/𝑛 (𝑀 > 0)

for 𝑛 ≥ 3. Observe that a function 𝑓 ∈ A with Re𝑓(𝑧) > 0satisfies |𝑎

𝑛| ≤ 2/𝑛 for 𝑛 ≥ 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 𝑓 ∈ A𝑏be given by (1) with |𝑎

𝑛| ≤ 𝑛

for 𝑛 ≥ 3. Then we have the following.

(i) 𝑓 satisfies the inequality

𝑧𝑓(𝑧)

𝑓 (𝑧)

− 1

< 1 − 𝛼 (6)

in |𝑧| < 𝑟0where 𝑟

0= 𝑟0(𝛼) is the real root in (0, 1)

of the equation 1 − 𝛼 + (1 + 𝛼) 𝑟 = 2(1 − 𝛼 + (2 −𝛼)(1 − 𝑏) 𝑟)(1 − 𝑟)

3. In particular, the number 𝑟0(𝛼)

is also the radius of starlikeness of order 𝛼 and thenumber 𝑟

0(1/2) is the radius of parabolic starlikeness

of the given functions.(ii) 𝑓 satisfies the inequality

𝑧𝑓(𝑧)

𝑓(𝑧)

< 1 − 𝛼 (7)

in |𝑧| < 𝑠0where 𝑠

0= 𝑠0(𝛼) is the real root in (0, 1)

of the equation 2(1 − 𝛼 + 2(2 − 𝛼)(1 − 𝑏) 𝑟)(1 − 𝑟)4 =1−𝛼+4𝑟+(1+𝛼) 𝑟

2. In particular, the number 𝑠0(𝛼) is

also the radius of convexity of order 𝛼 and the number𝑠0(1/2) is the radius of uniform convexity of the given

functions.

The results are sharp for the function

𝑓0(𝑧) = 2𝑧 + 2 (1 − 𝑏) 𝑧

2−

𝑧

(1 − 𝑧)2

= 𝑧 − 2𝑏𝑧2− 3𝑧3− 4𝑧4− ⋅ ⋅ ⋅ .

(8)

It is observed that [9] if a function 𝑓 ∈ A satisfiesRe(𝑓(𝑧) + 𝑧𝑓(𝑧)) > 0 for 𝑧 ∈ D, then |𝑎

𝑛| ≤ 2/𝑛

2. Similarly,Reade [10] proved that a close-to-star function𝑓 ∈ A satisfies|𝑎𝑛| ≤ 𝑛2 for 𝑛 ≥ 2. However, the converse in both the cases is

not true, in general. Recently, Mendiratta et al. [11] obtainedsharp radii of starlikeness of order 𝛼 (0 ≤ 𝛼 < 1), convexityof order 𝛼 (0 ≤ 𝛼 < 1), parabolic starlikeness and uniformconvexity for the classA

𝑏when |𝑎

𝑛| ≤ 𝑀/𝑛

2 or |𝑎𝑛| ≤ 𝑀𝑛

2

(𝑀 > 0) for 𝑛 ≥ 3. Ali et al. [12] also worked in the similardirection and obtained similar radii constants.

1.2. Harmonic Case. In a simply connected domainΩ ⊂ C, acomplex-valued harmonic function 𝑓 has the representation𝑓 = ℎ + 𝑔, where ℎ and 𝑔 are analytic in Ω. We call thefunctions ℎ and 𝑔 the analytic and the coanalytic parts of 𝑓,respectively. LetH denote the class of all harmonic functions𝑓 = ℎ + 𝑔 in D normalized so that ℎ and 𝑔 take the form

ℎ (𝑧) = 𝑧 +

∞

∑

𝑛=2

𝑎𝑛𝑧𝑛,

𝑔 (𝑧) =

∞

∑

𝑛=1

𝑏𝑛𝑧𝑛.

(9)

Since the Jacobian of 𝑓 is given by 𝐽𝑓

= |ℎ|2− |𝑔|2, by

a theorem of Lewy [13], 𝑓 is sense-preserving if and onlyif |𝑔| < |ℎ|, or equivalently if ℎ(𝑧) ̸= 0 and the seconddilatation 𝑤

𝑓= 𝑔/ℎ satisfies |𝑤

𝑓(𝑧)| < 1 in D. Let Hsp be

the subclass of H consisting of sense-preserving functions.Then it is easy to see that |𝑏

1| < 1 for functions in the class

Hsp. Set H0:= {𝑓 ∈ H : 𝑏

1= 0} and H0sp := Hsp ∩ H

0.Finally, letS

𝐻andS0

𝐻be subclasses ofHsp andH

0

sp, respec-tively, consisting of univalent functions.

One of the important questions in the study of class S0𝐻

and its subclasses is related to coefficient bounds. In 1984,


Clunie and Sheil-Small [14] conjectured that the Taylor coef-ficients of the series of ℎ and 𝑔 satisfy the inequality

𝑎𝑛

≤

1

6

(2𝑛 + 1) (𝑛 + 1) ,

𝑏𝑛

≤

1

6

(2𝑛 − 1) (𝑛 − 1) ,

∀𝑛 ≥ 2

(10)

and it is still open.These researchers proposed this coefficientconjecture because the harmonic Koebe function𝐾 = 𝐻+𝐺where

𝐻(𝑧) =

𝑧 − (1/2) 𝑧2+ (1/6) 𝑧

3

(1 − 𝑧)3

= 𝑧 +

1

6

∞

∑

𝑛=2

(𝑛 + 1) (2𝑛 + 1) 𝑧𝑛,

𝐺 (𝑧) =

(1/2) 𝑧2+ (1/6) 𝑧

3

(1 − 𝑧)3

=

1

6

∞

∑

𝑛=2

(𝑛 − 1) (2𝑛 − 1) 𝑧𝑛

(11)

is expected to play the extremal role in the classS0𝐻. However,

this conjecture is proved for all functions 𝑓 ∈ S0𝐻with real

coefficients and all functions𝑓 ∈ S0𝐻for which either𝑓(D) is

starlike with respect to the origin, close-to-convex, or convexin one direction (see [14–16]).

If 𝑓 ∈ S0𝐻for which 𝑓(D) is convex, Clunie and Sheil-

Small [14] proved that the Taylor coefficients of ℎ and𝑔 satisfythe inequalities

𝑎𝑛

≤

𝑛 + 1

2

,𝑏𝑛

≤

𝑛 − 1

2

, ∀𝑛 ≥ 2, (12)

and equality occurs for the harmonic half-plane mapping

𝐿 (𝑧) = 𝑀 (𝑧) + 𝑁 (𝑧),

𝑀 (𝑧) :=

𝑧 − (1/2) 𝑧2

(1 − 𝑧)2

,

𝑁 (𝑧) :=

− (1/2) 𝑧2

(1 − 𝑧)2.

(13)

LetK0𝐻and S∗0

𝐻be subclasses of S0

𝐻consisting of func-

tions 𝑓 for which 𝑓(D) is convex and 𝑓(D) is starlike withrespect to origin, respectively. Recall that convexity and star-likeness are not hereditary properties for univalent harmonicmappings (see [17–19]). 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 𝛼 and fully con-vex harmonic functions of order𝛼 for 0 ≤ 𝛼 < 1. LetFS∗

𝐻(𝛼)

and FK𝐻(𝛼) (0 ≤ 𝛼 < 1) be subclasses of S

𝐻consisting

of fully starlike functions of order 𝛼 and fully convex func-tions of order 𝛼, with FS∗

𝐻:= FS∗

𝐻(0) and FK

𝐻:=

FK𝐻(0).The functions in the classesFS∗

𝐻(𝛼) andFK

𝐻(𝛼)

are characterized by the conditions (𝜕/𝜕𝜃) arg𝑓(𝑟𝑒𝑖𝜃) > 𝛼and (𝜕/𝜕𝜃)(arg{(𝜕/𝜕𝜃)𝑓(𝑟𝑒𝑖𝜃)}) > 𝛼 for every circle |𝑧| = 𝑟,𝑧 = 𝑟𝑒

𝑖𝜃, respectively, where 0 ≤ 𝜃 < 2𝜋, 0 < 𝑟 < 1.The radius of full convexity of the class K0

𝐻is √2 − 1

while the radius of full convexity of the class S∗0𝐻

is 3 − √8(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 𝑓 = ℎ+𝑔 whosecoefficients satisfy the conditions (10) and (12). This, in turn,provides a bound for the radius of full starlikeness for convexand starlike mappings inS0

𝐻. These results are generalized in

context of fully starlike and fully convex functions of order 𝛼(0 ≤ 𝛼 < 1) in [18]. The authors [18] proved the followingresult.

Theorem 2 (see [18]). Let ℎ and 𝑔 have the form (9) with 𝑏1=

𝑔(0) = 0 and 0 ≤ 𝛼 < 1. Then we have the following.

(a) If the coefficients of the series satisfy the conditions (10),then 𝑓 = ℎ + 𝑔 is univalent and fully starlike of order𝛼 in the disk |𝑧| < 𝑟

𝑆, where 𝑟

𝑆= 𝑟𝑆(𝛼) is the real root

in (0, 1) of the equation 2(1 − 𝛼)(1 − 𝑟)4 + 𝛼(1 − 𝑟)2 −(1 + 𝑟)

2= 0.

(b) If the coefficients of the series satisfy the conditions (12),then 𝑓 = ℎ+𝑔 is univalent and fully starlike of order 𝛼in the disk |𝑧| < 𝑟

𝑆, where 𝑟

𝑆= 𝑟𝑆(𝛼) is the real root in

(0, 1) of the equation (2−𝛼)(1−𝑟)3+𝛼𝑟(1−𝑟)2−1−𝑟 = 0.

Moreover, the results are sharp for each 𝛼 ∈ [0, 1).

Theorem 2 gives the bounds for the radius of full starlike-ness of order 𝛼 (0 ≤ 𝛼 < 1) for the classes S∗0

𝐻and K0

𝐻. In

addition, the authors in [18] also determined the bounds forthe radius of full convexity of order 𝛼 (0 < 𝛼 < 1) for theseclasses.

The analytic part of harmonic mappings plays a vital rolein shaping their geometric properties. For instance, if 𝑓 =ℎ+𝑔 ∈ Hsp and ℎ is convex univalent, then𝑓 ∈ S𝐻 andmapsD onto a close-to-convex domain (see [14, Theorem 5.17, p.20]). However, if 𝑓 = ℎ+𝑔 ∈ Hsp where ℎ and 𝑔 are given by(9) and |𝑎

𝑛| ≤ 1 for 𝑛 ≥ 2, then 𝑓 need not be even univalent;

for example, the function

𝑓 (𝑧) = 𝑧 −

𝑧2

2

+

𝑧2

2

−

𝑧3

3

, 𝑧 ∈ D (14)

belongs to Hsp but is not univalent in D since 𝑓(𝑧0) =𝑓(𝑧0) = 3/4 where 𝑧

0= (3 + √3𝑖)/4 ∈ D. Note that a convex

univalent function 𝑧 + 𝑎2𝑧2+ 𝑎3𝑧3+ ⋅ ⋅ ⋅ satisfies |𝑎

𝑛| ≤ 1

for 𝑛 = 2, 3, . . .. This paper aims to determine the coefficientinequalities and radius constants for certain subfamilies ofHsp with the prescribed coefficient bound on the analyticpart.


A coefficient inequality for functions in the class Hsp isobtained in Section 2 which, in particular, improves the coef-ficient inequality proved by Polatoğlu et al. [22] for perturbedharmonicmappings. Using this inequality, the bounds for theradius of univalence, full starlikeness, and full convexity oforder 𝛼 (0 ≤ 𝛼 < 1) are obtained for functions 𝑓 = ℎ + 𝑔 ∈H0sp where the coefficients of the analytic part ℎ satisfy one ofthe conditions |𝑎

𝑛| ≤ 𝑛, |𝑎

𝑛| ≤ 1, or |𝑎

𝑛| ≤ 1/𝑛 for 𝑛 ≥ 2.

In addition, we will also discuss a case under which thesebounds can be improved.

In the third section, sharp bounds on 𝛽 (depending upon𝛼 and |𝑏

1|) are determined for a function 𝑓 = ℎ + 𝑔 ∈

H, where ℎ and 𝑔 are given by (9), satisfying either of thefollowing two conditions:∞

∑

𝑛=2

𝑛𝑎𝑛

+

∞

∑

𝑛=1

𝑛𝑏𝑛

≤ 𝛽 or

∞

∑

𝑛=2

𝑛2 𝑎𝑛

+

∞

∑

𝑛=1

𝑛2 𝑏𝑛

≤ 𝛽,

(15)

to be either fully starlike of order 𝛼 or fully convex of order 𝛼.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 𝑓 = ℎ + 𝑔 ∈ H𝑠𝑝, where ℎ and 𝑔 are given by

(9). Then

𝑏𝑛

≤𝑏1

𝑎𝑛

+

(1 −𝑏1

2

)

𝑛

𝑛−1

∑

𝑘=1

𝑘𝑎𝑘

, (16)

for 𝑛 ≥ 2, with 𝑎1= 1. In particular, one has

𝑏𝑛

≤

1

𝑛

𝑛

∑

𝑘=1

𝑘𝑎𝑘

, 𝑛 = 2, 3, . . . . (17)

Proof. Since 𝑓 ∈ Hsp, the function 𝑤(𝑧) = 𝑔(𝑧)/ℎ(𝑧) =∑∞

𝑛=0𝑤𝑛𝑧𝑛 is analytic in D and |𝑤(𝑧)| < 1 in D. On equating

the coefficients of 𝑧𝑛−1 in 𝑔(𝑧) = 𝑤(𝑧) ℎ(𝑧), we obtain

𝑛𝑏𝑛= 𝑎1𝑤𝑛−1

+ 2𝑤𝑛−2

𝑎2+ 3𝑤𝑛−3

𝑎3

+ ⋅ ⋅ ⋅ + (𝑛 − 1)𝑤1𝑎𝑛−1

+ 𝑛𝑤0𝑎𝑛,

(18)

where 𝑎1

= 1. Since |𝑤𝑛| ≤ 1 − |𝑤

0|2 (see [23, p. 172]), it

immediately follows that

𝑛𝑏𝑛

≤ (1 −

𝑤0

2

)

𝑛−1

∑

𝑘=1

𝑘𝑎𝑘

+ 𝑛

𝑤0

𝑎𝑛

, (𝑎

1= 1) . (19)

Since 𝑤0= 𝑔(0)/ℎ(0) = 𝑏

1, the desired result follows.

For specific choices of the analytic part ℎ in a harmonicfunction 𝑓 = ℎ + 𝑔 ∈ Hsp, Theorem 3 yields the followingresult.

Corollary 4. Let 𝑓 = ℎ + 𝑔 ∈ H𝑠𝑝, where ℎ and 𝑔 are given

by (9). Then we have the following.

(i) If |𝑎𝑛| ≤ 𝑛 or, in particular, ℎ is univalent, then |𝑏

𝑛| ≤

(2𝑛 + 1)(𝑛 + 1)/6, 𝑛 = 2, 3, . . ..

(ii) If |𝑎𝑛| ≤ 1 or, in particular, ℎ is convex univalent, then

|𝑏𝑛| ≤ (𝑛 + 1)/2, 𝑛 = 2, 3, . . ..

(iii) If |𝑎𝑛| ≤ 1/𝑛 or, in particular, Re ℎ(𝑧) > 0, then |𝑏

𝑛| ≤

1, 𝑛 = 2, 3, . . ..

Remark 5. Polatoğlu 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 𝑓 = ℎ + 𝑔 ∈ Hsp where ℎ and 𝑔 are givenby (9) and if ℎ is univalent in D, then

𝑏𝑛

≤

1

𝑛

(2𝑛6 − 𝑛2− 4𝑛 − 6) , 𝑛 = 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 𝛼 (0 ≤ 𝛼 < 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 FS∗

𝐻(𝛼) and FK

𝐻(𝛼) (0 ≤ 𝛼 < 1) that directly

follow from the corresponding results in [24, 25].

Lemma 6 (see [24, 25]). Let 𝑓 = ℎ + 𝑔, where ℎ and 𝑔 aregiven by (9) and let 0 ≤ 𝛼 < 1. Then we have the following.

(i) If

∞

∑

𝑛=2

𝑛 − 𝛼

1 − 𝛼

𝑎𝑛

+

∞

∑

𝑛=1

𝑛 + 𝛼

1 − 𝛼

𝑏𝑛

≤ 1, (21)

then 𝑓 ∈ FS∗𝐻(𝛼).

(ii) If

∞

∑

𝑛=2

𝑛 (𝑛 − 𝛼)

1 − 𝛼

𝑎𝑛

+

∞

∑

𝑛=1

𝑛 (𝑛 + 𝛼)

1 − 𝛼

𝑏𝑛

≤ 1, (22)

then 𝑓 ∈ FK𝐻(𝛼).

Theorem 7. Let 𝑓 = ℎ + 𝑔 ∈ H0𝑠𝑝, where ℎ and 𝑔 are given

by (9) with 𝑏1= 𝑔(0) = 0 and 0 ≤ 𝛼 < 1. Then we have the

following.

(i) If |𝑎𝑛| ≤ 𝑛 or, in particular, ℎ is univalent, then 𝑓 is

univalent and fully starlike of order 𝛼 in the disk |𝑧| <𝑟1where 𝑟

1= 𝑟1(𝛼) is the real root of the equation

12 (1 − 𝛼) 𝑟4+ (49𝛼 − 48) 𝑟

3+ 8 (9 − 8𝛼) 𝑟

2

+ 3 (11𝛼 − 18) 𝑟 + 6 (1 − 𝛼) = 0

(23)

in the interval (0, 1).


(ii) If |𝑎𝑛| ≤ 1 or, in particular, ℎ is convex univalent, then

𝑓 is univalent and fully starlike of order 𝛼 in the disk|𝑧| < 𝑟

2where 𝑟


4 (1 − 𝛼) 𝑟3+ (9𝛼 − 12) 𝑟

2+ (12 − 7𝛼) 𝑟

− 2 (1 − 𝛼) = 0

(24)

in the interval (0, 1).(iii) If |𝑎

𝑛| ≤ 1/𝑛 or, in particular, Re ℎ(𝑧) > 0, then 𝑓 is

univalent and fully starlike of order 𝛼 in the disk |𝑧| <𝑟3where 𝑟


2 (1 − 𝛼) 𝑟3+ (5𝛼 − 4) 𝑟

2+ (1 − 3𝛼) 𝑟

− 2𝛼 (1 − 𝑟)2 log (1 − 𝑟) = 0

(25)

in the interval (0, 1).

Proof. Since 𝑓 = ℎ + 𝑔 ∈ H0sp, we obtain

𝑏𝑛

≤

1

𝑛

𝑛−1

∑

𝑘=1

𝑘𝑎𝑘

, (𝑛 ≥ 2; 𝑎

1= 1) , (26)

by applying Theorem 3. We will make use of (26) to obtainthe coefficient bounds for 𝑏

𝑛in three different cases specified

in the theorem. For 𝑟 ∈ (0, 1), let 𝑓𝑟: D → C be defined

by

𝑓𝑟(𝑧) :=

𝑓 (𝑟𝑧)

𝑟

= 𝑧 +

∞

∑

𝑛=2

𝑎𝑛𝑟𝑛−1

𝑧𝑛+

∞

∑

𝑛=2

𝑏𝑛𝑟𝑛−1

𝑧𝑛. (27)

We will show that 𝑓𝑟∈ FS∗

𝐻(𝛼). In view of Lemma 6(i), it

suffices to show that the sum

𝑆 =

∞

∑

𝑛=2

𝑛 − 𝛼

1 − 𝛼

𝑎𝑛

𝑟𝑛−1

+

∞

∑

𝑛=2

𝑛 + 𝛼

1 − 𝛼

𝑏𝑛

𝑟𝑛−1 (28)

is bounded above by 1 for 0 ≤ 𝑟 < 𝑟𝑖for 𝑖 = 1, 2, 3.

(i) Since |𝑎𝑛| ≤ 𝑛, it is easy to deduce that |𝑏

𝑛| ≤ (𝑛 −

1)(2𝑛 − 1)/6 by (26). Using these coefficient bounds in (28)and simplifying, we have

𝑆 ≤

1

6 (1 − 𝛼)

[2

∞

∑

𝑛=2

𝑛3𝑟𝑛−1

+ (3 + 2𝛼)

∞

∑

𝑛=2

𝑛2𝑟𝑛−1

+ (1 − 9𝛼)

∞

∑

𝑛=2

𝑛𝑟𝑛−1

+

𝛼𝑟

1 − 𝑟

] .

(29)

Thus 𝑆 ≤ 1 if 𝑟 satisfy the inequality

2

∞

∑

𝑛=2

𝑛3𝑟𝑛−1

+ (2𝛼 + 3)

∞

∑

𝑛=2

𝑛2𝑟𝑛−1

+ (1 − 9𝛼)

∞

∑

𝑛=2

𝑛𝑟𝑛−1

+

𝛼𝑟

1 − 𝑟

≤ 6 (1 − 𝛼) .

(30)

By using the identities

𝑟

(1 − 𝑟)2=

∞

∑

𝑛=1

𝑛𝑟𝑛,

𝑟 (1 + 𝑟)

(1 − 𝑟)3=

∞

∑

𝑛=1

𝑛2𝑟𝑛,

𝑟 (𝑟2+ 4𝑟 + 1)

(1 − 𝑟)4

=

∞

∑

𝑛=1

𝑛3𝑟𝑛

(31)

the last inequality reduces to

2 (𝑟2+ 4𝑟 + 1)

(1 − 𝑟)4

+ (2𝛼 + 3)

1 + 𝑟

(1 − 𝑟)3

+

1 − 9𝛼

(1 − 𝑟)2+

𝛼

1 − 𝑟

≤ 12 (1 − 𝛼)

(32)

or equivalently

2 (𝑟2+ 4𝑟 + 1) + (2𝛼 + 3) (1 − 𝑟

2)

+ (1 − 9𝛼) (1 − 𝑟)2+ 𝛼(1 − 𝑟)

3

≤ 12 (1 − 𝛼) (1 − 𝑟)4.

(33)

This gives

12 (1 − 𝛼) 𝑟4+ (49𝛼 − 48) 𝑟

3+ 8 (9 − 8𝛼) 𝑟

2

+ 3 (11𝛼 − 18) 𝑟 + 6 (1 − 𝛼) ≥ 0.

(34)

Thus by Lemma 6(i), 𝑓𝑟∈ FS∗

𝐻(𝛼) for 𝑟 ≤ 𝑟

1where 𝑟

1is the

real root of (23) in (0, 1). In particular,𝑓 is univalent and fullystarlike of order 𝛼 in |𝑧| < 𝑟

1.

(ii) If |𝑎𝑛| ≤ 1 then (26) gives |𝑏

𝑛| ≤ (𝑛 − 1)/2. These

coefficient bounds lead to the following inequality for the sum(28):

𝑆 ≤

1

2 (1 − 𝛼)

× [

∞

∑

𝑛=2

𝑛2𝑟𝑛−1

+ (1 + 𝛼)

∞

∑

𝑛=2

𝑛𝑟𝑛−1

−

3𝛼𝑟

1 − 𝑟

] .

(35)

Therefore it follows that 𝑆 ≤ 1 if 𝑟 satisfy the inequality

∞

∑

𝑛=2

𝑛2𝑟𝑛−1

+ (1 + 𝛼)

∞

∑

𝑛=2

𝑛𝑟𝑛−1

−

3𝛼𝑟

1 − 𝑟

≤ 2 (1 − 𝛼) . (36)

Making use of identities (31) in the last inequality, we obtain

1 + 𝑟

(1 − 𝑟)3+

1 + 𝛼

(1 − 𝑟)2−

3𝛼

1 − 𝑟

≤ 4 (1 − 𝛼) , (37)

which simplifies to

2 (1 − 𝛼) + (7𝛼 − 12) 𝑟 + (12 − 9𝛼) 𝑟2− 4 (1 − 𝛼) 𝑟

3≥ 0.

(38)


Lemma 6(i) shows that 𝑓𝑟∈ FS∗


2where 𝑟

2is

the real root of (24) in (0, 1). In particular, 𝑓 is univalent andfully starlike of order 𝛼 in |𝑧| < 𝑟

2.

(iii) Using (26), it is easily seen that |𝑏𝑛| ≤ (𝑛−1)/𝑛. Using

the coefficient bounds for |𝑎𝑛| and |𝑏

𝑛| in (28), it follows that

𝑆 ≤

1

1 − 𝛼

[

∞

∑

𝑛=2

𝑛𝑟𝑛−1

− 2𝛼

∞

∑

𝑛=2

1

𝑛

𝑟𝑛−1

+

𝛼𝑟

1 − 𝑟

] . (39)

The sum 𝑆 ≤ 1 if 𝑟 satisfy the inequality∞

∑

𝑛=2

𝑛𝑟𝑛−1

− 2𝛼

∞

∑

𝑛=2

1

𝑛

𝑟𝑛−1

+

𝛼𝑟

1 − 𝑟

≤ 1 − 𝛼. (40)

Using (31) and the identity − log(1− 𝑟) = 𝑟+ 𝑟2/2+ 𝑟3/3+ ⋅ ⋅ ⋅ ,the last inequality reduces to

1

(1 − 𝑟)2+

2𝛼

𝑟

log (1 − 𝑟) + 𝛼1 − 𝑟

≤ 2 (1 − 𝛼) , (41)

which is equivalent to

2 (1 − 𝛼) 𝑟3+ (5𝛼 − 4) 𝑟

2+ (1 − 3𝛼) 𝑟

− 2𝛼(1 − 𝑟)2 log (1 − 𝑟) ≥ 0.

(42)

By applying Lemma 6(i), 𝑓𝑟∈ FS∗


3where 𝑟

3is

the real root of (25) in (0, 1). In particular, 𝑓 is univalent andfully starlike of order 𝛼 in |𝑧| < 𝑟

3. This completes the proof

of the theorem.

Remark 8. By (26), it follows that |𝑏2| ≤ 1/2 for all functions

𝑓 ∈ H0sp. The bound 1/2 is sharp for the function 𝑓0(𝑧) =𝑧 + 𝑧2/2 ∈ H0sp. Since 𝑓0 is univalent in D, the coefficient

inequality |𝑏2| ≤ 1/2 remains sharp in the subclass S0

𝐻.

Clunie and Sheil-Small [14] were the first to observe the sharpinequality |𝑏

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