Nonlinear Analysis and Geometric Function Theory

Abstract and Applied Analysis Nonlinear Analysis and Geometric Function Theory

Guest Editors: Ljubomir B. C ´iric´ , Samuel Krushkal, Qamrul Hasan Ansari, David Kalaj, and Vesna Manojlovic´ Nonlinear Analysis and Geometric Function Theory Abstract and Applied Analysis

This is a special issue published in “Abstract and Applied Analysis.” All articles are open access articles distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Editorial Board

Ravi P. Agarwal, USA Jaeyoung Chung, Republic of Korea Lorenzo Giacomelli, Italy Bashir Ahmad, Saudi Arabia Silvia Cingolani, Italy Jaume Gine,´ Spain M. O. Ahmedou, Germany Jean M. Combes, France Valery Y. Glizer, Israel Nicholas D. Alikakos, Greece Monica Conti, Italy Laurent Gosse, Italy Debora Amadori, Italy Diego Cordoba,´ Spain Jean P. Gossez, Belgium Pablo Amster, Argentina Juan C. Cortes,´ Spain Jose L. Gracia, Spain Douglas R. Anderson, USA Graziano Crasta, Italy Maurizio Grasselli, Italy Jan Andres, Czech Republic Bernard Dacorogna, Switzerland Qian Guo, China Giovanni Anello, Italy Vladimir Danilov, Russia Yuxia Guo, China Stanislav Antontsev, Portugal Mohammad T. Darvishi, Iran Chaitan P. Gupta, USA Mohamed Kamal Aouf, Egypt Luis F. Pinheiro de Castro, Portugal Uno Hamarik,¨ Estonia Narcisa C. Apreutesei, Romania T. Diagana, USA Ferenc Hartung, Hungary Natig M. Atakishiyev, Mexico Jesus´ I. D´ıaz, Spain Behnam Hashemi, Iran Ferhan M. 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Samko, Portugal Tie-cheng Xia, China Changxing Miao, China Martin Schechter, USA Xu Xian, China Stanislaw Migorski, Poland Javier Segura, Spain Yanni Xiao, China Mihai Mihailescu,ˇ Romania Sigmund Selberg, Norway Fuding Xie, China Feliz Minhos,´ Portugal Valery Serov, Finland Gonang Xie, China Dumitru Motreanu, France Naseer Shahzad, Saudi Arabia Naihua Xiu, China Roberta Musina, Italy Andrey Shishkov, Ukraine Daoyi Xu, China G. M. N’Guer´ ekata,´ USA Stefan Siegmund, Germany Zhenya Yan, China Maria Grazia Naso, Italy Abdel-Maksoud A. Soliman, Egypt Xiaodong Yan, USA Sylvia Novo, Spain Pierpaolo Soravia, Italy Norio Yoshida, Japan Micah Osilike, Nigeria Marco Squassina, Italy Beong In Yun, Republic of Korea Mitsuharu Otani,ˆ Japan Svatoslav Stanek,ˇ Czech Republic Vjacheslav Yurko, Russia Turgut Ozis¨ ¸, Turkey Stevo Stevic, Serbia Agacik Zafer, Turkey Filomena Pacella, Italy Antonio Suarez,´ Spain Sergey V. Zelik, UK Nikolaos S. Papageorgiou, Greece Wenchang Sun, China Jianming Zhan, China Sehie Park, Republic of Korea Robert Szalai, UK Meirong Zhang, China Alberto Parmeggiani, Italy Sanyi Tang, China Weinian Zhang, China KailashC.Patidar,SouthAfrica Chun-Lei Tang, China Chengjian Zhang, China Kevin R. Payne, Italy Youshan Tao, China Zengqin Zhao, China Ademir Fernando Pazoto, Brazil Gabriella Tarantello, Italy Sining Zheng, China Josip E. Pecariˇ c,´ Croatia Nasser-eddine Tatar, Saudi Arabia Tianshou Zhou, China Shuangjie Peng, China Susanna Terracini, Italy Yong Zhou, China Sergei V.Pereverzyev, Austria Gerd Teschke, Germany Qiji J. Zhu, USA Maria Eugenia Perez, Spain Alberto Tesei, Italy Chun-Gang Zhu, China Josefina Perles, Spain Bevan Thompson, Australia Malisa R. Zizovic, Serbia Allan Peterson, USA Sergey Tikhonov, Spain Wenming Zou, China Andrew Pickering, Spain Claudia Timofte, Romania Cristina Pignotti, Italy Thanh Tran, Australia Contents

Nonlinear Analysis and Geometric Function Theory,LjubomirB.Ciri´ c,´ Samuel Krushkal, Qamrul Hasan Ansari, David Kalaj, and Vesna Manojlovic´ Volume 2014, Article ID 656957, 1 page

Distortion of Quasiregular Mappings and Equivalent Norms on Lipschitz-Type Spaces, Miodrag Mateljevic´ Volume 2014, Article ID 895074, 20 pages

Complete Self-Shrinking Solutions for Lagrangian Mean Curvature Flow in Pseudo-Euclidean Space, Ruiwei Xu and Linfen Cao Volume2014,ArticleID196751,9pages The Generalized Weaker (𝛼-𝜙-𝜑)-Contractive Mappings and Related Fixed Point Results in Complete Generalized Metric Spaces, Maryam A. Alghamdi, Chi-Ming Chen, and Erdal Karapınar Volume2014,ArticleID985080,10pages

Coupled and Tripled Coincidence Point Results with Application to Fredholm Integral Equations, Marwan Amin Kutbi, Nawab Hussain, Jamal Rezaei Roshan, and Vahid Parvaneh Volume 2014, Article ID 568718, 18 pages

Generalized Common Fixed Point Results with Applications, Marwan Amin Kutbi, Muhammad Arshad, Jamshaid Ahmad, and Akbar Azam Volume 2014, Article ID 363925, 7 pages

Convergence of Numerical Solution of Generalized Theodorsen’s Nonlinear Integral Equation, Mohamed M. S. Nasser Volume 2014, Article ID 213296, 11 pages

Implicit Approximation Scheme for the Solution of 𝐾-Positive Definite Operator Equation, Naseer Shahzad, Arif Rafiq, and Habtu Zegeye Volume2014,ArticleID683295,6pages

Caristi Fixed Point Theorem in Metric Spaces with a Graph, M. R. Alfuraidan and M. A. Khamsi Volume 2014, Article ID 303484, 5 pages

Applications of Differential Subordination for Argument Estimates of Multivalent Analytic Functions, Meng-Ting Lu, Ting Jia, Xing-Qian Ling, and Jin-Lin Liu Volume 2014, Article ID 634710, 4 pages Existence Theorems of 𝜖-Cone Saddle Points for Vector-Valued Mappings,TaoChen Volume2014,ArticleID478486,5pages

Fixed Point Theory in 𝛼-Complete Metric Spaces with Applications,N.Hussain,M.A.Kutbi,andP.Salimi Volume2014,ArticleID280817,11pages Algorithms of Common Solutions for Generalized Mixed Equilibria, Variational Inclusions, and Constrained Convex Minimization,Lu-ChuanCengandSulimanAl-Homidan Volume 2014, Article ID 132053, 25 pages

A New Iterative Method for Equilibrium Problems and Fixed Point Problems, Abdul Latif and Mohammad Eslamian Volume 2013, Article ID 178053, 9 pages Hindawi Publishing Corporation Abstract and Applied Analysis Volume 2014, Article ID 656957, 1 page http://dx.doi.org/10.1155/2014/656957

Editorial Nonlinear Analysis and Geometric Function Theory

Ljubomir B. SiriT,1 Samuel Krushkal,2 Qamrul Hasan Ansari,3 David Kalaj,4 and Vesna ManojloviT5

1 Faculty of Mechanical Engineering, University of Belgrade, Belgrade, Serbia 2 Department of Mathematics, Bar-Ilan University, 5290002 Ramat-Gan, Israel 3 Department of Mathematics, Aligarh Muslim University, Aligarh 202 002, India 4 Faculty of Natural Sciences and Mathematics, University of Montenegro, 81000 Podgorica, Montenegro 5 Mathematical Institute SANU and Faculty of Organizational Sciences, University of Belgrade, 11000 Belgrade, Serbia

Correspondence should be addressed to Ljubomir B. Ciri´ c;´ [email protected]

Received 30 September 2014; Accepted 30 September 2014; Published 10 November 2014

Copyright © 2014 Ljubomir B. Ciri´ c´ 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.

Sobolev mappings, quasiconformal mappings, or deforma- In connection with the general trend of the geometric tions, between subsets of Euclidean space or manifolds function theory in Euclidean space to generalize certain and harmonic maps between manifolds, may arise as the aspects of the analytic functions of one complex variable, solutions to certain optimization problems in the calculus there is another development related to maps of quasi- of variations, as stationary points which are the solutions to conformal type, which are solutions of nonlinear second- differential equations (particularly in conformal geometry). order elliptic equations or satisfy certain inequality related to The most recent developments in the theory of planar and Laplacian and gradient. space quasiconformal mappings are also related to the theory Variational analysis, fixed point theory, split feasibility to degenerate elliptic equations, which include p-Laplace problems, and so forth are also some important and vital equation whose solutions are p-harmonic maps. In particular, areas in nonlinear analysis. During the last two decades, Euclidean harmonic (2-harmonic) maps, which are critical several solution methods for these problems have been points of Dirichlet’s integral, are related to the theory of proposed and analyzed. These areas include application in minimal surfaces and are the primary object of the study optimization, medical sciences, image reconstruction, social in classical potential theory. The complex gradient of the p- sciences, engineering, management, and so forth. harmonic operator is quasiregular and it is general feature of In this special issue, thirteen articles were being pub- nonlinear PDEs. lished. All these articles have given important contributions Harmonic maps have also played a role in compact- to different parts of the above-discussed areas. Among them ifications and parametrizations of Teichmuller space, via stands out the very important work of Professor M. Matel- classical results of Mike Wolf, who gave a compactification in jevic,´ one of the most prominent scientists in the geometric terms of hyperbolic harmonic maps. Teichmuller spaces are function theory. In fact, this issue is devoted to his 65th central objects in geometry and complex analysis today, with birthday, as many authors emphasized in their articles. deep connections to quasiconformal mappings (in particular, to extremal problems of quasiconformal mappings), string Ljubomir B. Ciri´ ´c theory, and hyperbolic 3 manifolds. The study of connection Samuel Krushkal between extremal mappings of finite distortion and harmonic Qamrul Hasan Ansari mappings is also initiated. Since its introduction in the early David Kalaj 1980s, quasiconformal surgery has become a major tool in the Vesna Manojlovi´c development of the theory of holomorphic dynamics. Hindawi Publishing Corporation Abstract and Applied Analysis Volume 2014, Article ID 895074, 20 pages http://dx.doi.org/10.1155/2014/895074

Research Article Distortion of Quasiregular Mappings and Equivalent Norms on Lipschitz-Type Spaces

Miodrag MateljeviT

Faculty of Mathematics, University of Belgrade, Studentski Trg 16, 11000 Belgrade, Serbia

Correspondence should be addressed to Miodrag Mateljevic;´ [email protected]

Received 24 January 2014; Accepted 15 May 2014; Published 21 October 2014

Academic Editor: Ljubomir B. Ciri´ c´

Copyright © 2014 Miodrag Mateljevic.´ This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We prove a quasiconformal analogue of Koebe’s theorem related to the average Jacobian and use a normal family argument here to prove a quasiregular analogue of this result in certain domains in 𝑛-dimensional space. As an application, we establish that Lipschitz- type properties are inherited by a quasiregular function from its modulo. We also prove some results of Hardy-Littlewood type for Lipschitz-type spaces in several dimensions, give the characterization of Lipschitz-type spaces for quasiregular mappings by the average Jacobian, and give a short review of the subject.

1. Introduction Gehring [4,Theorems1.9and3.17](statedhereasLemma 5) to a quasiregular version (Lemma 33)involvingasomewhat 𝛼 The Koebe distortion theorem gives a series of bounds for a larger class of moduli of continuity than 𝑡 , 0<𝛼<1. univalent function and its derivative. A result of Hardy and In the paper [5] several properties of a domain which Littlewood relates Holder¨ continuity of analytic functions in satisfies the Hardy-Littlewood property with the inner length theunitdiskwithaboundonthederivative(werefertothis metricaregivenandalsosomeresultsontheHolder¨ result shortly as HL-result). continuity are obtained. Astala and Gehring [1]observedthatforcertaindis- The fact that Lipschitz-type properties are sometimes tortion property of quasiconformal mappings the function inherited by an analytic function from its modulus was 𝑎 |𝑓󸀠| 𝑓, defined in Section 2,playsanalogousroleas when firstdetectedin[6]. Later this property was considered for 𝑛=2 𝑓 and is conformal, and they establish quasiconformal different classes of functions and we will call shortly results version of the well-known result due to Koebe, cited here as of this type Dyk-type results. Theorem 22 yields a simple Lemma 4, and Hardy-Littlewood, cited here as Lemma 5. approach to Dyk-type result (the part (ii.1); see also [7]) In Section 2, we give a short proof of Lemma 4,using andestimateoftheaverageJacobianforquasiconformal 𝐽 𝑎 a version with the average Jacobian 𝑓 instead of 𝑓,and mappings in space. The characterization of Lipschitz-type we also characterize bi-Lipschitz mappings with respect to spaces for quasiconformal mappings by the average Jacobian quasihyperbolic metrics by Jacobian and the average Jaco- is established in Theorem 23 in space case and Theorem 24 bian; see Theorems 8, 9,and10 and Proposition 13.Gehring yields Dyk-type result for quasiregular mappings in planar and Martio [2] extended HL-result to the class of uniform case. domains and characterized the domains 𝐷 with the property In Section 4, we establish quasiregular versions of the that functions which satisfy a local Lipschitz condition in 𝐷 well-known result due to Koebe, Theorem 39 here, and use for some 𝛼 always satisfy the corresponding global condition this result to obtain an extension of Dyakonov’s theorem there. forquasiregularmappingsinspace(withoutDyakonov’s The main result of the Nolder paper [3] generalizes a hypothesis that it is a quasiregular local homeomorphism), quasiconformal version of a theorem, due to Astala and Theorem 40. The characterization of Lipschitz-type spaces 2 Abstract and Applied Analysis

𝑛 for quasiregular mappings by the average Jacobian is also R let 𝜌:𝐺 → [0,∞)be a continuous function. We say that established in Theorem 40. 𝜌 is a weight function or a metric density if, for every locally 𝑛 By R ={𝑥=(𝑥1,...,𝑥𝑛):𝑥1,...,𝑥𝑛 ∈ R} denote rectifiable curve 𝛾 in 𝐺,theintegral 𝑛 the real vector space of dimension 𝑛. For a domain 𝐷 in R with nonempty boundary, we define the distance function 𝑙𝜌 (𝛾) = ∫ 𝜌 (𝑥) 𝑑𝑠 (5) 𝑑=𝑑(𝐷)=dist(𝐷) by 𝑑(𝑥) = 𝑑(𝑥; 𝜕𝐷)= dist(𝐷)(𝑥) = 𝛾 󸀠 𝑛 inf{|𝑥 − 𝑦| : 𝑦 ∈,andif 𝜕𝐷} 𝑓 maps 𝐷 onto 𝐷 ⊂ R ,insome 𝑙 (𝛾) 𝜌 𝛾 settings, it is convenient to use short notation 𝑑∗ =𝑑𝑓(𝑥) for exists. In this case we call 𝜌 the -length of .Ametric 󸀠 𝑐 𝑐 𝑑 :𝐺×𝐺 → [0,∞) 𝑑(𝑓(𝑥); 𝜕𝐷 ).Itisclearthat𝑑(𝑥) = dist(𝑥, 𝐷 ),where𝐷 is density defines a metric 𝜌 as follows. For 𝑛 𝑎, 𝑏 ∈𝐺 the complement of 𝐷 in R . ,let 𝑛 𝑚 Let 𝐺 be an open set in R .Amapping𝑓:𝐺 → R 󸀠 𝑑𝜌 (𝑎,) 𝑏 = inf 𝑙𝜌 (𝛾) , is differentiable at 𝑥∈𝐺if there is a linear mapping 𝑓 (𝑥) : 𝛾 (6) 𝑛 𝑛 R → R , called the derivative of 𝑓 at 𝑥,suchthat wheretheinfimumistakenoveralllocallyrectifiablecurves 󸀠 𝑓 (𝑥+ℎ) −𝑓(𝑥) =𝑓 (𝑥) ℎ+|ℎ| 𝜀 (𝑥,) ℎ , (1) in 𝐺 joining 𝑎 and 𝑏. For the modern mapping theory, which also considers where 𝜀(𝑥,ℎ)→0as ℎ→0. For a vector-valued function 𝑛≥3 𝑛 𝑛 dimensions , we do not have a Riemann mapping 𝑓:𝐺 → R ,where𝐺⊂R is a domain, we define theorem and therefore it is natural to look for counterparts of 󵄨 󸀠 󵄨 󵄨 󸀠 󵄨 󸀠 󵄨 󸀠 󵄨 the hyperbolic metric. So-called hyperbolic type metrics have 󵄨𝑓 (𝑥)󵄨 = max 󵄨𝑓 (𝑥) ℎ󵄨 ,𝑙(𝑓(𝑥))=min 󵄨𝑓 (𝑥) ℎ󵄨 , 󵄨 󵄨 |ℎ|=1 󵄨 󵄨 |ℎ|=1 󵄨 󵄨 been the subject of many recent papers. Perhaps the most important metrics of these metrics are the quasihyperbolic (2) 𝑛 metric 𝜅𝐺 and the distance ratio metric 𝑗𝐺 of a domain 𝐺⊂R when 𝑓 is differentiable at 𝑥∈𝐺. (see [11, 12]). The quasihyperbolic metric 𝜅=𝜅𝐺 of 𝐺 is a In Section 3, we review some results from [7, 8]. For particular case of the metric 𝑑𝜌 when 𝜌(𝑥) = 1/𝑑(𝑥, 𝜕𝐺) (see [11, 12]). example, in [7] under some conditions concerning a majorant 𝑛 𝑛 𝜔,weshowedthefollowing. Given a subset 𝐸 of C or R ,afunction𝑓:𝐸 → C (or, 1 𝑛 𝑓 𝐸 C𝑚 R𝑚 Let 𝑓∈𝐶(𝐷, R ) and let 𝜔 be a continuous majorant more generally, a mapping from into or )issaid Λ (𝐸) such that 𝜔∗(𝑡) = 𝜔(𝑡)/𝑡 is nonincreasing for 𝑡>0. to belong to the Lipschitz space 𝜔 if there is a constant 𝐿 = 𝐿(𝑓) = 𝐿(𝑓; 𝐸) = 𝐿(𝑓,𝜔; 𝐸) Assume 𝑓 satisfies the following property (which we call , which we call Lipschitz Hardy-Littlewood (𝑐, 𝜔)-property): constant, such that 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󸀠 󵄨 󵄨𝑓 (𝑥) −𝑓(𝑦)󵄨 ≤𝐿𝜔(󵄨𝑥−𝑦󵄨) (HL (𝑐, 𝜔)) 𝑑 (𝑥) 󵄨𝑓 (𝑥)󵄨 ≤𝑐𝜔(𝑑 (𝑥)) , 󵄨 󵄨 󵄨 󵄨 (7) (3) 𝑥∈𝐷, 𝑑 (𝑥) = (𝑥, 𝐷𝑐). 𝑥, 𝑦 ∈𝐸 ‖𝑓‖ where dist for all .Thenorm Λ 𝜔(𝐸) is defined as the smallest 𝐿 in (7). Then There has been much work on Lipschitz-type properties ( Λ) 𝑓∈ Λ (𝐺) . of quasiconformal mappings. This topic was treated, among loc loc 𝜔 (4) other places, in [1–5, 7, 13–22]. As in most of those papers, we will currently restrict If, in addition, 𝑓 is harmonic in 𝐷 or, more generally, 𝛼 2 ourselves to the simplest majorants 𝜔𝛼(𝑡) := 𝑡 (0<𝛼≤1). 𝑓∈𝑂𝐶(𝐷),then(HL(𝑐, 𝜔)) is equivalent to (loc Λ).If𝐷 𝛼 The classes Λ 𝜔(𝐸) with 𝜔=𝜔𝛼 will be denoted by Λ (or by is a Λ 𝜔-extension domain, then (HL(𝑐, 𝜔)) is equivalent to Lip(𝛼; 𝐸) = Lip(𝛼, 𝐿;). 𝐸) 𝐿(𝑓) and 𝛼 are called, respectively, 𝑓∈Λ𝜔(𝐷). Lipschitz constant and exponent (of 𝑓 on 𝐸). We say that a In Section 3, we also consider Lipschitz-type spaces of 𝑛 domain 𝐺⊂𝑅 is uniform if there are constants 𝑎 and 𝑏 such pluriharmonic mappings and extend some results from [9]. that each pair of points 𝑥1,𝑥2 ∈𝐺canbejoinedbyrectifiable In Appendices A and B we discuss briefly distortion of arc 𝛾 in 𝐺 for which harmonic qc maps, background of the subject, and basic property of qr mappings, respectively. For more details on 󵄨 󵄨 𝑙(𝛾)≤𝑎󵄨𝑥1 −𝑥2󵄨 , related qr mappings we refer the interested reader to [10]. (8) min 𝑙(𝛾𝑗)≤𝑏𝑑(𝑥, )𝜕𝐺 2. Quasiconformal Analogue of Koebe’s 𝑥∈𝛾 𝑙(𝛾) 𝛾 𝛾 ,𝛾 Theorem and Applications for each ;here denotes the length of and 1 2 the components of 𝛾\{𝑥}. We define 𝐶(𝐺) = max{𝑎, .𝑏} Throughout the paper we denote by Ω, 𝐺,and𝐷 open subset 𝐶(𝐺) 𝑛 The smallest for which the previous inequalities hold of R , 𝑛≥1. 𝐺 𝑛 𝑛 𝑛−1 𝑛 is called the uniformity constant of and we denote it by Let 𝐵 (𝑥, 𝑟) = {𝑧∈ R :|𝑧−𝑥|<𝑟}, 𝑆 (𝑥, 𝑟) = 𝜕𝐵 (𝑥, 𝑟) 𝑐∗ =𝑐∗(𝐺) 𝑓 𝑛 𝑛−1 .Following[2, 17], we say that a function belongs (abbreviated 𝑆(𝑥,)andlet 𝑟) B , 𝑆 standfortheunitball Λ 𝜔(𝐺) 𝑛 to the local Lipschitz space loc if (7)holds,withafixed and the unit sphere in R , respectively. Sometimes we write 𝐶>0,whenever𝑥∈𝐺and |𝑦 − 𝑥| < (1/2)𝑑(𝑥,.Wesay 𝜕𝐺) 2 D and T instead of B and 𝜕D, respectively. For a domain 𝐺⊂ that 𝐺 is a Λ 𝜔-extension domain if Λ 𝜔(𝐺) = loc Λ 𝜔(𝐺).In Abstract and Applied Analysis 3

𝛼 particular if 𝜔=𝜔𝛼,wesaythat𝐺 is a Λ -extension domain; Gehring’s Theorem. For every 𝐾≥1and 𝑛≥2, there exists a 𝜃𝑛 : (0, 1) → R this class includes the uniform domains𝑛 mentioned above. function 𝐾 with the following properties: Suppose that Γ is a curve family in R . We denote by F(Γ) 𝜃𝑛 the family whose elements are nonnegative Borel-measurable (1) 𝐾 is increasing, 𝜌 ∫ 𝜌(𝑥)|𝑑𝑥| ≥1 𝜃𝑛 (𝑟) = 0 functions which satisfy the condition 𝛾 for (2) lim𝑟→0 𝐾 , 𝑛 every locally rectifiable curve 𝛾∈Γ,where𝑑𝑠 = |𝑑𝑥| denotes (3) lim𝑟→1𝜃𝐾(𝑟) = ∞, the arc length element. For 𝑝≥1,withthenotation 󸀠 𝑛 (4) Let Ω and Ω be proper subdomains of R and let 𝑓: Ω→Ω󸀠 𝐾 𝑥 𝑦 Ω 󵄨 󵄨𝑝 be a -qc. If and are points in such 𝐴𝑝 (𝜌) = ∫ 󵄨𝜌 (𝑥)󵄨 𝑑𝑉 (𝑥) , (9) 0 < |𝑦 − 𝑥| < 𝑑(𝑥, 𝜕Ω) R𝑛 that ,then 󵄨 󵄨 󵄨 󵄨 𝑑𝑉(𝑥) =𝑑𝑥 󵄨𝑓(𝑦)−𝑓(𝑥)󵄨 󵄨𝑦−𝑥󵄨 where denotes the Euclidean volume element 󵄨 󵄨 ≤𝜃𝑛 ( 󵄨 󵄨 ). 󸀠 𝐾 (14) 𝑑𝑥1𝑑𝑥2 ⋅⋅⋅𝑑𝑥𝑛, we define the 𝑝-modulus of Γ by 𝑑(𝑓(𝑥) ,𝜕Ω ) 𝑑 (𝑥, )𝜕Ω 𝑀 (Γ) = {𝐴 (𝜌) : 𝜌 ∈ F (Γ)}. 𝑝 inf 𝑝 (10) Introduce the quantity, mentioned in the introduction,

We will denote 𝑀𝑛(Γ) simply by 𝑀(Γ) and call it the mod- 1 ulus of Γ. 𝑎𝑓 (𝑥) =𝑎𝑓,𝐺 (𝑥) := exp ( 󵄨 󵄨 ∫ log 𝐽𝑓 (𝑧) 𝑑𝑧) , 𝑥 ∈𝐺, ∗ 𝑛 󵄨𝐵 󵄨 𝐵 Suppose that 𝑓:Ω → Ω is a homeomorphism. 󵄨 𝑥󵄨 𝑥 ∗ Consider a path family Γ in Ω and its image family Γ ={𝑓∘𝛾: (15) 𝛾∈Γ} . We introduce the quantities associated with a quasiconformal mapping 𝑓:𝐺 → 𝑓(𝐺)⊂ R𝑛 𝐽 𝑓 = 𝑀(Γ∗) 𝑀 (Γ) ;here 𝑓 is the Jacobian of , while B𝑥 B𝑥,𝐺 stands for 𝐾 (𝑓) = ,𝐾(𝑓) = , (11) the ball 𝐵(𝑥; 𝑑(𝑥, 𝜕𝐺)) and |B𝑥| for its volume. 𝐼 sup 𝑀 (Γ) 𝑂 sup 𝑀 (Γ∗) 󸀠 𝑛 Lemma 4 (see [4]). Suppose that 𝐺 and 𝐺 are domains in 𝑅 : where the suprema are taken over all path families Γ in Ω such 󸀠 ∗ If 𝑓:𝐺 →𝐺 is 𝐾-quasiconformal, then that 𝑀(Γ) and 𝑀(Γ ) are not simultaneously 0 or ∞. 󸀠 󸀠 ∗ 𝑑(𝑓(𝑥) ,𝜕𝐺 ) 𝑑(𝑓(𝑥) ,𝜕𝐺 ) Definition 1. Suppose that 𝑓:Ω → Ω is a homeomor- 1 ≤𝑎𝑓,𝐺 (𝑥) ≤𝑐 ,𝑥∈𝐺, phism; we call 𝐾𝐼(𝑓) the inner dilatation and 𝐾𝑂(𝑓) the 𝑐 𝑑 (𝑥, )𝜕𝐺 𝑑 (𝑥, )𝜕𝐺 outer dilatation of 𝑓.Themaximaldilatationof𝑓 is 𝐾(𝑓) = (16) max{𝐾𝑂(𝑓),𝐼 𝐾 (𝑓)}.If𝐾(𝑓)≤𝐾<∞,wesay𝑓 is 𝐾- quasiconformal (abbreviated qc). where 𝑐 is a constant which depends only on 𝐾 and 𝑛.

∗ 𝑛 1/(1−𝑛) Suppose that 𝑓:Ω → Ω is a homeomorphism and Set 𝛼0 =𝛼𝐾 =𝐾 . 𝑥∈Ω 𝑥 =∞̸ 𝑓(𝑥) =∞̸ , ,and . 𝑛 For each 𝑟>0such that 𝑆(𝑥; 𝑟) ⊂Ω we set 𝐿(𝑥, 𝑓,𝑟) = Lemma 5 (see [4]). Suppose that 𝐷 is a uniform domain in 𝑅 |𝑓(𝑦) − 𝑓(𝑥)| 𝑙(𝑥, 𝑓,𝑟) = |𝑓(𝑦) − 𝑓(𝑥)| and that 𝛼 and 𝑚 are constants with 0<𝛼≤1and 𝑚≥0.If max|𝑦−𝑥|=𝑟 , min|𝑦−𝑥|=𝑟 . 𝑛 𝑓 is 𝐾-quasiconformal in 𝐷 with 𝑓(𝐷) ⊂𝑅 and if 𝑓 𝑥 Definition 2. The linear dilatation of at is 𝛼−1 𝑎𝑓,𝐷 (𝑥) ≤𝑚𝑑(𝑥, 𝜕𝐷) for 𝑥∈𝐷, (17) 𝐿(𝑥,𝑓,𝑟) 𝐻(𝑥,𝑓)=lim sup . (12) 𝑟→0 𝑙(𝑥,𝑓,𝑟) then 𝑓 has a continuous extension to 𝐷 \ {∞} and 󵄨 󵄨 󵄨 󵄨 𝛼 Theorem 3 (the metric definition of quasiconformality). A 󵄨𝑓(𝑥1)−𝑓(𝑥2)󵄨 ≤𝐶(󵄨𝑥1 −𝑥2󵄨 +𝑑(𝑥1,𝜕𝐷)) (18) ∗ homeomorphism 𝑓:Ω → Ω is qc if and only if 𝐻(𝑥, 𝑓) is bounded on Ω. for 𝑥1,𝑥2 ∈ 𝐷 \ {∞},wheretheconstant𝐶=𝑐(𝐷)̂ depends ∗ ∗ only on 𝐾, 𝑛, 𝑎, 𝑚 and the uniformity constant 𝑐 =𝑐(𝐷) Ω 𝑅𝑛 𝑓:Ω →𝑛 𝑅 Let be a domain in and let be con- for 𝐷.Inthecase𝛼≤𝛼0, (18) can be replaced by the stronger tinuous. We say that 𝑓 is quasiregular (abbreviated qr) if conclusion that 𝑓 𝑊𝑛 (Ω) 󵄨 󵄨 󵄨 󵄨 𝛼 (1) belongs to Sobolev space 1,loc , 󵄨𝑓(𝑥1)−𝑓(𝑥2)󵄨 ≤𝐶(󵄨𝑥1 −𝑥2󵄨) (𝑥1,𝑥2 ∈ 𝐷\{∞}). (2) there exists 𝐾, 1≤𝐾<∞,suchthat (19) 󵄨 󵄨𝑛 󵄨𝑓󸀠 (𝑥)󵄨 ≤𝐾𝐽 (𝑥) . . 𝑎−1 1/(1−𝑛) 󵄨 󵄨 𝑓 a e (13) Example 6. The mapping 𝑓(𝑥) = |𝑥| 𝑥, 𝑎=𝐾 is 𝐾- qc with 𝑎𝑓 boundedintheunitball.Hence𝑓 satisfies the The smallest 𝐾 in (13) is called the outer dilatation 𝐾𝑂(𝑓). hypothesis of Lemma 5 with 𝛼=1.Since|𝑓(𝑥) − 𝑓(0)| ≤ 𝑎 𝑎 Aqrmappingisaqcifandonlyifitisahomeomorphism. |𝑥 − 0| =|𝑥|,weseethatwhen𝐾>1,thatis,𝑎<1=𝛼, First we need Gehring’s result on the distortion property of the conclusion (18)inLemma 5 cannot be replaced by the qc (see [23,page383],[24,page63]). stronger assertion |𝑓(𝑥1)−𝑓(𝑥2)| ≤ 𝑐𝑚|𝑥1 −𝑥2|. 4 Abstract and Applied Analysis

𝑛 + + Let Ω∈R and R =[0,∞)and 𝑓, 𝑔 : Ω→ R .Ifthere This pointwise result, combined with integration along is a positive constant 𝑐 such that 𝑓(𝑥) ≤ 𝑐𝑔(𝑥), 𝑥∈Ω,we curves, easily gives write 𝑓⪯𝑔on Ω.Ifthereisapositiveconstant𝑐 such that 𝑘𝐷󸀠 (ℎ (𝑧1) ,ℎ(𝑧2)) ≍𝑘𝐷 (𝑧1,𝑧2) ,𝑧1,𝑧2 ∈𝐷. (26) 1 𝑔 (𝑥) ≤𝑓(𝑥) ≤𝑐𝑔(𝑥) ,𝑥∈Ω, (20) 𝑐 Note that we do not use that log(1/𝐽ℎ) is a subharmonic function. we write 𝑓≈𝑔(or 𝑓≍𝑔)onΩ. 2 2 Let 𝐺⊂R be a domain and let 𝑓:𝐺 → R , 𝑓=(𝑓1,𝑓2) The following follows from the proof of Proposition 7: be a harmonic mapping. This means that 𝑓 is a map from 𝐺 2 Λ(ℎ, 𝑧) ≈ 𝑎 ≈ 𝜆(ℎ, 𝑧) √2 𝐽 ≈𝐽 into R and both 𝑓1 and 𝑓2 are harmonic functions, that is, (I) ℎ,𝐷 and 𝑓 𝑓;seebelowfor 𝐽 solutions of the two-dimensional Laplace equation the definition of average jacobian 𝑓. Δ𝑢 = 0. (21) When underlining a symbol (we also use other latex- symbols) we want to emphasize that there is a special meaning The above definition of a harmonic mapping extends ina of it; for example we denote by 𝑐 aconstantandby𝑐, 𝑐̂, 𝑐̃some natural way to the case of vector-valued mappings 𝑓:𝐺 → R𝑛 𝑓=(𝑓,...,𝑓) 𝐺⊂R𝑛 𝑛≥2 specific constants. , 1 𝑛 , defined on a domain , . Our next result concerns the quantity Let ℎ be a harmonic univalent orientation preserving 󸀠 mapping on a domain 𝐷, 𝐷 = ℎ(𝐷) and 𝑑ℎ(𝑧) = 𝑑(ℎ(𝑧), 1 󸀠 𝐸 (𝑥) := 󵄨 󵄨 ∫ 𝐽 (𝑧) 𝑑𝑧, 𝑥 ∈𝐺, 𝜕𝐷 ).Ifℎ=𝑓+𝑔 has the form, where 𝑓 and 𝑔 are analytic, 𝑓,𝐺 󵄨𝐵 󵄨 𝑓 (27) − 󸀠 󸀠 󵄨 𝑥󵄨 𝐵𝑥 we define 𝜆ℎ(𝑧) = 𝐷 (𝑧) = |𝑓 (𝑧)| − |𝑔 (𝑧)|,andΛ ℎ(𝑧) = + 󸀠 󸀠 𝐷 (𝑧) = |𝑓 (𝑧)| + |𝑔 (𝑧)|. associated with a quasiconformal mapping 𝑓:𝐺 → 𝑓(𝐺)⊂ 𝑛 R ;here𝐽𝑓 istheJacobianof𝑓,while𝐵𝑥 =𝐵𝑥,𝐺 stands for 2.1. Quasihyperbolic Metrics and the Average Jacobian. For the ball 𝐵(𝑥, 𝑑(𝑥, 𝜕𝐺)/2) and |𝐵𝑥| for its volume. harmonic qc mappings we refer the interested reader to [25– Define 28] and references cited therein. 𝐽 =𝐽 = √𝑛 𝐸 . 𝑓 𝑓,𝐺 𝑓,𝐺 (28) 󸀠 2 Proposition 7. Suppose 𝐷 and 𝐷 are proper domains in R . 󸀠 If ℎ:𝐷 → 𝐷 is 𝐾-qc and harmonic, then it is bi-Lipschitz Using the distortion property of qc (see [24,page63]) 󸀠 with respect to quasihyperbolic metrics on 𝐷 and 𝐷 . we give short proof of a quasiconformal analogue of Koebe’s theorem (related to Astala and Gehring’s results from [4], Results of this type have been known to the participants cited as Lemma 4 here). of Belgrade Complex Analysis seminar; see, for example, 󸀠 𝑛 [29, 30]andSection 2.2 (Proposition 13, Remark 14 and Theorem 8. Suppose that 𝐺 and 𝐺 are domains in R :If𝑓: 󸀠 Corollary 16). This version has been proved by Manojlovic´ 𝐺→𝐺is 𝐾-quasiconformal, then

[31] as an application of Lemma 4.In[8], we refine her 󸀠 󸀠 approach. 1 𝑑(𝑓(𝑥) ,𝜕𝐺 ) 𝑑(𝑓(𝑥) ,𝜕𝐺 ) ≤𝐽 (𝑥) ≤𝑐 ,𝑥∈𝐺, 𝑐 𝑑 (𝑥, )𝜕𝐺 𝑓,𝐺 𝑑 (𝑥, )𝜕𝐺 Proof. Let ℎ=𝑓+𝑔 be local representation on 𝐵𝑧. (29) Since ℎ is 𝐾-qc, then 󵄨 󵄨2 󵄨 󵄨2 𝑐 𝐾 𝑛 2 󵄨 󸀠󵄨 󵄨 󸀠󵄨 where is a constant which depends only on and . (1 − 𝑘 ) 󵄨𝑓 󵄨 ≤𝐽ℎ ≤𝐾󵄨𝑓 󵄨 (22) Proof. Bythedistortionpropertyofqc(see[23,page383], 𝐵 |𝑓󸀠(𝜁)| on 𝑧 and since log is harmonic, [24,page63]),therearetheconstants𝐶∗ and 𝑐∗ which depend 𝑛 𝐾 󵄨 󵄨 1 󵄨 󵄨2 on and only, such that 󵄨 󸀠 󵄨 󵄨 󸀠 󵄨 log 󵄨𝑓 (𝑧)󵄨 = 󵄨 󵄨 ∫ log 󵄨𝑓 (𝜁)󵄨 𝑑𝜉 𝑑𝜂. (23) 2 󵄨𝐵 󵄨 𝐵 󵄨 𝑧󵄨 𝑧 𝐵(𝑓(𝑥) ,𝑐∗𝑑∗)⊂𝑓(𝐵𝑥)⊂𝐵(𝑓(𝑥) ,𝐶∗𝑑∗), 𝑥∈𝐺, (30) Hence, 󸀠 1 1 󵄨 󵄨2 𝑑 (𝑥) := 𝑑(𝑓(𝑥)) = 𝑑(𝑓(𝑥), 𝜕𝐺 ) 𝑑(𝑥) := 𝑑(𝑥, 󵄨 󸀠 󵄨 where ∗ and log 𝑎ℎ,𝐷 (𝑧) ≤ log 𝐾+ 󵄨 󵄨 ∫ log 󵄨𝑓 (𝜁)󵄨 𝑑𝜉 𝑑𝜂 𝜕𝐺) 2 2 󵄨𝐵 󵄨 𝐵 .Hence 󵄨 𝑧󵄨 𝑧 (24) 󵄨 󵄨 = √𝐾 󵄨𝑓󸀠 (𝑧)󵄨 . 󵄨 󵄨 󵄨 󵄨 log 󵄨 󵄨 󵄨𝐵(𝑓(𝑥) ,𝑐∗𝑑∗)󵄨 ≤ ∫ 𝐽𝑓 (𝑧) 𝑑𝑧 ≤ 󵄨𝐵(𝑓(𝑥) ,𝐶∗𝑑∗)󵄨 (31) 𝐵𝑥 𝑎 (𝑧) ≤ √𝐾|𝑓󸀠(𝑧)| √1−𝑘|𝑓󸀠(𝑧)| ≤ 𝑎 (𝑧) Hence, ℎ,𝐷 and ℎ,𝐷 . and therefore we prove Theorem 8. Using Astala-Gehring result, we get We only outline proofs in the rest of this subsection. 󸀠 󸀠 𝑛 𝑑 (ℎ𝑧, 𝜕𝐷 ) Suppose that Ω and Ω are domains in R different from Λ (ℎ, 𝑧) ≍ ≍𝜆(ℎ, 𝑧) . (25) 𝑛 𝑑 (𝑧, 𝜕𝐷) R . Abstract and Applied Analysis 5

󸀠 1 󸀠 Theorem 9. Suppose 𝑓:Ω → Ω is a 𝐶 and the following Suppose 𝑓:Ω→Ωis onto qc mapping. We can hold. consider the following conditions:

(i1) 𝑓 is c-Lipschitz with respect to quasihyperbolic metrics (a.2) 𝑓 is bi-Lipschitz with respect to quasihyperbolic 󸀠 󸀠 on Ω and Ω ;then metrics on Ω and Ω ; √𝑛 𝐽 ≈𝑑 /𝑑 Ω 󸀠 (b.2) 𝑓 ∗ a.e. in ; (I1) 𝑑|𝑓 (𝑥0)| ≤ 𝑐𝑑∗(𝑥0) for every 𝑥0 ∈Ω. 𝑛 (c.2) √𝐽𝑓 ≈𝑎𝑓 a.e. in Ω; (i2) If 𝑓 is a qc 𝑐-quasihyperbolic-isometry, then √𝑛 𝐽 ≈𝐽 Ω (d.2) 𝑓 𝑓 a.e. in . 2 (I2) 𝑓 is a 𝑐 -qc mapping, |𝑓󸀠|≈𝑑 /𝑑 It seems that the above conditions are equivalent, but we (I3) ∗ . did not check details. 󸀠 𝑛 If Ω is a planar domain and 𝑓 harmonic qc, then we Proof. Since Ω and Ω are different from R ,therearequas- 󸀠 proved that (d.2) holds (see Proposition 18). ihyperbolic metrics on Ω and Ω .Thenforafixed𝑥0 ∈Ω,we have 2.2. Quasi-Isometry in Planar Case. For a function ℎ,weuse −1 󵄨 󵄨 𝜕ℎ = (1/2)(ℎ󸀠 −𝑖ℎ󸀠 ) 𝜕ℎ = (1/2)(ℎ󸀠 +𝑖ℎ󸀠 ) 𝑘(𝑥,𝑥0)=𝑑(𝑥0) 󵄨𝑥−𝑥0󵄨 (1+𝑜(1)) ,𝑥󳨀→𝑥0, notations 𝑥 𝑦 and 𝑥 𝑦 ; 󵄨 󵄨 we also use notations 𝐷ℎ and 𝐷ℎ instead of 𝜕ℎ and 𝜕ℎ, 󵄨𝑓 (𝑥) −𝑓(𝑥 )󵄨 󵄨 0 󵄨 respectively, when it seems convenient. Now we give another 𝑘(𝑓(𝑥) ,𝑓(𝑥0)) = (1+𝑜(1)) , (32) 𝑑∗ (𝑓 0(𝑥 )) proof of Proposition 7, using the following.

when 𝑥󳨀→𝑥0. Proposition 12 (see [29]). Let ℎ be an euclidean harmonic orientationpreservingunivalentmappingoftheunitdiscD into 𝑓 𝑐 Hence, (i1) implies (I1). If is a -quasihyperbolic-isom- C such that ℎ(D) contains a disc 𝐵𝑅 = 𝐵(𝑎; 𝑅) and ℎ(0) = 𝑎. 𝑑|𝑓󸀠(𝑥 )| ≤ 𝑐𝑑 𝑑𝑙(𝑓󸀠(𝑥 )) ≥𝑑 /𝑐 etry, then 0 ∗ and 0 ∗ .Hence, Then (I2) and (I3) follow. 𝑅 𝜕ℎ (0) ≥ . 󸀠 1 | | (34) Theorem 10. Suppose 𝑓:Ω→Ωis a 𝐶 qc homeomor- 4 phism. The following conditions are equivalent: If, in addition, ℎ is 𝐾-qc, then (a.1) 𝑓 is bi-Lipschitz with respect to quasihyperbolic metrics 1−𝑘 Ω Ω󸀠 𝜆 (0) ≥ 𝑅. (35) on and , ℎ 4 √𝑛 𝐽 ≈𝑑 /𝑑 (b.1) 𝑓 ∗ , For more details, in connection with material considered in this subsection, see also Appendix A,inparticular, √𝑛 𝐽 ≈𝑎 (c.1) 𝑓 𝑓, Proposition A.7. Let ℎ=𝑓+𝑔 be a harmonic univalent orientation √𝑛 𝐽 ≈𝐽 (d.1) 𝑓 𝑓, preserving mapping on the unit disk D, Ω=ℎ(D), 𝑑ℎ(𝑧) = − 󸀠 󸀠 𝑑(ℎ(𝑧), ,𝜕Ω) 𝜆ℎ(𝑧) = 𝐷 (𝑧) = |𝑓 (𝑧)| − |𝑔 (𝑧)|,andΛ ℎ(𝑧) = 󸀠 + 󸀠 󸀠 where 𝑑(𝑥) = 𝑑(𝑥, 𝜕Ω) and 𝑑∗(𝑥) = 𝑑(𝑓(𝑥), 𝜕Ω ). 𝐷 (𝑧) = |𝑓 (𝑧)| + |𝑔 (𝑧)|. By the harmonic analogue of the Koebe Theorem, then Proof. It follows from Theorem 9 that (a) is equivalent to (b) 1 − (see, e.g., [32, 33]). 𝐷 (0) ≤𝑑ℎ (0) (36) 𝐽 ≈𝑑 /𝑑 𝑎 ≈𝑑 /𝑑 16 Theorem 8 states that 𝑓 ∗ .ByLemma 4, 𝑓 ∗ and therefore (b) is equivalent to (c). The rest of the proof is and therefore straightforward. 2 1 − (1−|𝑧| ) 𝐷 (𝑧) ≤𝑑ℎ (𝑧) . (37) 1,1 16 Lemma 11. If 𝑓∈𝐶 is a 𝐾-quasiconformal mapping defined 𝑛 in a domain Ω⊂R (𝑛 ≥3) ,then If, in addition, ℎ is 𝐾-qc, then 𝐽 (𝑥) >0, 𝑥∈Ω 2 + 𝑓 (33) (1 − |𝑧| )𝐷 (𝑧) ≤ 16𝐾𝑑ℎ (𝑧) . (38)

𝑛−1 𝑛−1 provided that 𝐾<2 .Theconstant2 is sharp. Using Proposition 12,wealsoprove

2 1−𝑘 If 𝐺⊂Ω, then it is bi-Lipschitz with respect to Euclidean (1−|𝑧| ) 𝜆ℎ (𝑧) ≥ 𝑑ℎ (𝑧) . (39) 󸀠 4 and quasihyperbolic metrics on 𝐺 and 𝐺 = 𝑓(𝐺). It is a natural question whether is there an analogy of If we summarize the above considerations, we have 1 Theorem 10 if we drop the 𝐶 hypothesis. proved the part (a.3) of the following proposition. 6 Abstract and Applied Analysis

Proposition 13 (e-qch, hyperbolic distance version). (a.3). Let 𝐷 be a planar hyperbolic domain. Then if 𝐷 is simply Let ℎ=𝑓+𝑔 be a harmonic univalent orientation preserving connected, 𝐾 D Ω=ℎ(D) 𝑧∈D -qc mapping on the unit disk , .Then,for , 1 2 ≤𝜌≤ . 2 2𝑑 𝑑 (44) (1 − |𝑧| )Λℎ (𝑧) ≤ 16𝐾𝑑ℎ (𝑧) , (40) 𝑧→𝜕𝐷 2 1−𝑘 For general domain as , (1 − |𝑧| )𝜆ℎ (𝑧) ≥ 𝑑ℎ (𝑧) . 4 1+𝑜(1) 2 ≤𝜌≤ . (45) (b.3) If ℎ is a harmonic univalent orientation preserving 𝐾- 𝑑 ln (1/𝑑) 𝑑 󸀠 qc mapping of domain 𝐷 onto 𝐷 ,then 𝑑 (𝑧,󸀠 𝑧 )≤2𝜅 (𝑧,󸀠 𝑧 ) 𝑧,󸀠 𝑧 ∈𝐷 Hence hyp 𝐷 , . 𝑑 (𝑧) Λ ℎ (𝑧) ≤ 16𝐾𝑑ℎ (𝑧) , If Ω is a strongly hyperbolic domain, then there is a 𝜌 Ω 𝜌≈𝑑−1 1−𝑘 (41) hyperbolic density on the domain ,suchthat , 𝑑 (𝑧) 𝜆 (𝑧) ≥ 𝑑 (𝑧) , where 𝑑(𝑤) = 𝑑(𝑤, 𝜕Ω);see,forexample,[34]. Thus 𝑑 ;𝐷 ≈ ℎ 4 ℎ hyp 𝜅𝐷.Hence,wefindthefollowing. and Corollary 16. Every 𝑒-harmonic quasiconformal mapping of 1−𝑘 󸀠 󸀠 𝜅 (𝑧, 𝑧 )≤𝜅 󸀠 (ℎ𝑧, ℎ𝑧 ) the unit disc (more generally of a strongly hyperbolic domain) 4 𝐷 𝐷 (42) is a quasi-isometry with respect to hyperbolic distances. ≤ 16𝐾𝜅 (𝑧,󸀠 𝑧 ), 𝐷 Remark 17. Let 𝐷 be a hyperbolic domain, let ℎ be e-harmonic quasiconformal mapping of 𝐷 onto Ω, and let 𝜙:D → 󸀠 ∗ where 𝑧, 𝑧 ∈𝐷. 𝐷 be a covering and ℎ =ℎ∘𝜙. ∗ Remark 14. In particular, we have Proposition 7,butherethe (a.4) Suppose that 𝐷 is simply connected. Thus 𝜙 and ℎ proof of Proposition 13 isverysimpleandititisnotbasedon are one-to-one. Then Lemma 4. 2 1−𝑘 (1−|𝑧| ) 𝜆ℎ∗ (𝜁) ≥ 𝑑ℎ (𝑧) , (46) Proofof(b.3). Applying (a.3) to the disk 𝐵𝑧, 𝑧∈𝐷,weget(41). 4 It is clear that (41)implies(42). C 𝜌=𝜌 where 𝑧 = 𝜙(𝜁). For a planar hyperbolic domain in , we denote by 𝐷 󸀠 󸀠 𝑑 =𝑑 𝐷 𝜆 ∗ (𝜁) = 𝑙 (𝑧)|𝜙 (𝜁)| 𝜌 (𝑧)|𝜙 (𝜁)| = and hyp hyp;𝐷 the hyperbolic density and metric of , Hence, since ℎ ℎ and hyp;𝐷 respectively. 𝜌D(𝜁), We say that a domain 𝐷⊂C is strongly hyperbolic if 1−𝑘 it is hyperbolic and diameters of boundary components are 𝜆ℎ (𝑧) ≥ 𝑑ℎ (𝑧) 𝜌𝐷 (𝑧) . (47) uniformly bounded from below by a positive constant. 4

󸀠 Example 15. The Poincaremetricofthepunctureddisk´ D = Using Hall’s sharp result, one can also improve the {0<|𝑧|<1}is obtained by mapping its universal covering, constant in the second inequality in Propositions 13 and 12 − 1/4 𝜏 an infinitely-sheeted disk, on the half plane Π ={Re 𝑤<0} (i.e., the constant canbereplacedby 0;seebelowfor 𝑤 by means of 𝑤=ln 𝑧 (i.e., 𝑧=𝑒 ). The metric is more details): 2 |𝑑𝑤| |𝑑𝑧| (1 − |𝑧| )𝜆ℎ (𝑧) ≥ (1−𝑘) 𝜏0𝑑ℎ (𝑧) , (48) = . | 𝑤| |𝑧| (1/ |𝑧|) (43) Re ln √ where 𝜏0 =3 3/2𝜋. 0 Since a boundary component is the point ,thepunctured ∗ disk is not a strongly hyperbolic domain. Note also that 𝜌/𝑑 (b.4) Suppose that 𝐷 is not a simply connected. Then ℎ is and 1/ ln(1/𝑑) tend to 0 if 𝑧 tends to 0.Here𝑑=𝑑(𝑧)= not one-to-one and we cannot apply the procedure as 󸀠 󸀠 dist(𝑧, 𝜕D ) and 𝜌 is the hyperbolic density of D . Therefore in (a.4). one has the following: (c.4) It seems natural to consider whether there is an 󸀠 󸀠 󸀠 analogue in higher dimensions of Proposition 13. ( .1) 𝑧∈D 𝜅 󸀠 (𝑧, 𝑧 )= (1/𝑑(𝑧))𝑑 󸀠 (𝑧, 𝑧 ) A for , D ln hyp,D when 󸀠 𝑧 →𝑧. Proposition 18. For every 𝑒-harmonic quasiconformal map- 𝑓 There is no constant 𝑐 such that ping oftheunitdisc(moregenerallyofahyperbolicdomain 𝐷)thefollowingholds: 󸀠 󸀠 󸀠 − (A.2) 𝜅D󸀠 (𝑧, 𝑧 )≤𝑐𝜅Π− (𝑤, 𝑤 ) for every 𝑤, 𝑤 ∈Π ,where 𝑤 󸀠 𝑤󸀠 √𝐽 ≈𝑑 /𝑑 𝑧=𝑒 and 𝑧 =𝑒 . (e.2) 𝑓 ∗ .

𝑑 − ≈𝜅 − ( .2) 𝜅 󸀠 ≈ Since hyp,Π Π ,if A holds, we conclude that D In particular, it is a quasi-isometry with respect to quasihyper- 𝑑 󸀠 ( .1) hyp,D , which is a contradiction by A . bolic distances. Abstract and Applied Analysis 7

𝑎V 𝑧∈𝐷 𝑛 Proof. For , by the distortion property, and 𝐽𝑓 (𝐵; 𝑥) = √𝐸𝑓(𝐵; 𝑥);wealsousethenotation𝐽𝑓(𝐵; 𝑥) 𝑎V 󵄨 󵄨 and 𝐽𝑓(𝑥; 𝑟) instead of 𝐽𝑓 (𝐵; 𝑥). 󵄨𝑓(𝑧1)−𝑓(𝑧)󵄨 ≤𝑐𝑑(𝑓(𝑧)) for every 𝑧1 ∈𝐵𝑧. (49) 𝑛 For 𝑥, 𝑦 ∈ R ,wedefinetheEuclideaninnerproduct⟨⋅, ⋅⟩ 󸀠 Hence, by Schwarz lemma for harmonic maps, 𝑑|𝑓 (𝑧)| ≤ by 𝑐1𝑑(𝑓(𝑧)). Proposition 12 yields (e) and an application of ⟨𝑥, 𝑦⟩1 =𝑥 𝑦1 +⋅⋅⋅+𝑥𝑛𝑦𝑛. (55) Proposition 13 gives the proof. 𝑛 𝑛 By E we denote R with the Euclidean inner product and Recall 𝐵𝑥 stands for the ball 𝐵(𝑥; 𝑑(𝑥, 𝜕𝐺)/2) and |𝐵𝑥| for 𝑛 𝑚 call it Euclidean space 𝑛-space (space of dimension 𝑛). In this 𝑉 R 𝑢:𝑉 → R 𝑛 𝑛 its volume. If is a subset of and , we define paper, for simplicity, we will use also notation R for E .Then 𝑛 󵄨 󵄨 the Euclidean length of 𝑥∈R is defined by osc𝑉𝑢:=sup {󵄨𝑢 (𝑥) −𝑢(𝑦)󵄨 :𝑥,𝑦∈𝑉}, (50) 1/2 󵄨 󵄨 󵄨 󵄨 1/2 󵄨 󵄨2 󵄨 󵄨2 𝜔𝑢 (𝑟, 𝑥) := sup {󵄨𝑢 (𝑥) −𝑢(𝑦)󵄨 : 󵄨𝑥−𝑦󵄨 =𝑟}. |𝑥| = ⟨𝑥, 𝑥⟩ = (󵄨𝑥1󵄨 +⋅⋅⋅+󵄨𝑥𝑛󵄨 ) . (56)

𝑛 Suppose that 𝐺⊂R and 𝐵𝑥 = 𝐵(𝑥, 𝑑(𝑥)/2).Let The minimal analytic assumptions necessary for a viable 𝑂𝐶1(𝐺) = 𝑂𝐶1(𝐺; 𝑐 ) 𝑓∈𝐶1(𝐺) theory appear in the following definition. 1 denote the class of such 𝑛 𝑛 that Let Ω be a domain in R and let 𝑓:Ω → R be continuous. We say that 𝑓 has finite distortion if 󵄨 󸀠 󵄨 𝑑 (𝑥) 󵄨𝑓 (𝑥)󵄨 ≤𝑐 𝑓 󵄨 󵄨 1osc𝐵𝑥 (51) 1 (1) 𝑓 belongs to Sobolev space 𝑊1,loc(Ω); for every 𝑥∈𝐺. (2) the Jacobian determinant of 𝑓 is locally integrable and does not change sign in Ω; 󸀠 1 Proposition 19. 𝑓:Ω→Ω 𝐶 Suppose is a .Then (3) there is a measurable function 𝐾𝑂 =𝐾𝑂(𝑥) ≥1 ,finite 𝑓∈𝑂𝐶1(Ω; 𝑐) 𝑓 𝑐 ,ifandonlyif is -Lipschitz with respect to a.e., such that 𝐵 𝑓(𝐵) 𝐵⊂Ω quasihyperbolic metrics on and for every ball . 󵄨 󵄨𝑛 󵄨 󵄨 󵄨 󸀠 󵄨 󵄨 󵄨 󵄨𝑓 (𝑥)󵄨 ≤𝐾𝑂 (𝑥) 󵄨𝐽𝑓 (𝑥)󵄨 a.e. (57) 2.3. Dyk-Type Results. The characterization of Lipschitz-type spaces for quasiconformal mappings in space and planar The assumptions (1), (2),and(3) do not imply that 𝑓∈ 𝑛 quasiregular mappings by the average Jacobian are the main 𝑊1,loc(Ω),unlessofcourse𝐾𝑂 is a bounded function. results in this subsection. In particular, using the distortion If 𝐾𝑂 is a bounded function, then 𝑓 is qr. In this setting, property of qc mappings we give a short proof of a quasicon- the smallest 𝐾 in (57) is called the outer dilatation 𝐾𝑂(𝑓). formal version of a Dyakonov theorem which states: If 𝑓 is qr, also

𝛼 𝑛 󸀠 󸀠 𝑛 Suppose 𝐺 is a Λ -extension domain in R and 𝑓 is 𝐽𝑓 (𝑥) ≤𝐾𝑙(𝑓 (𝑥)) a.e. (58) 𝑛 a 𝐾-quasiconformal mapping of 𝐺 onto 𝑓(𝐺) ⊂ R . 𝛼 𝛼 󸀠 󸀠 󸀠 󸀠 Then 𝑓∈Λ (𝐺) if and only if |𝑓| ∈ Λ (𝐺). for some 𝐾 , 1≤𝐾 <∞,where𝑙(𝑓 (𝑥)) = inf{|𝑓 (𝑥)ℎ| : 󸀠 |ℎ| = 1}.Thesmallest𝐾 in (58) is called the inner dilatation This is Theorem B.3,inAppendix B below.It is convenient 𝐾𝐼(𝑓) and 𝐾(𝑓) = max(𝐾𝑂(𝑓),𝐼 𝐾 (𝑓)) is called the maximal to refer to this result as Theorem Dy; see also Theorems 23–24 dilatation of 𝑓.If𝐾(𝑓) ≤𝐾, 𝑓 is called 𝐾-quasiregular. and Proposition A, Appendix B. In a highly significant series of papers published in 1966– First we give some definitions and auxiliary results. 1969 Reshetnyak proved the fundamental properties of qr Recall, Dyakonov [15]usedthequantity mappings and in particular the main theorem concerning 1 topological properties of qr mappings: every nonconstant qr 𝐸 (𝑥) : 󵄨 󵄨 ∫ 𝐽 (𝑧) 𝑑𝑧, 𝑥 ∈𝐺, map is discrete and open; cf. [11, 35] and references cited there. 𝑓,𝐺 󵄨𝐵 󵄨 𝑓 (52) 󵄨 𝑥󵄨 𝐵𝑥 Lemma 20 (see Morrey’s Lemma, Lemma 6.7.1,[10,page 𝑓:𝐺 → 𝑓(𝐺)⊂ 1,𝑝 𝑛 associated with a quasiconformal mapping 170]). Let 𝑓 be a function of the Sobolev class 𝑊 (B, E ) in 𝑅𝑛 𝐽 𝑓 𝐵 =𝐵 ;here 𝑓 is the Jacobian of , while 𝑥 𝑥,𝐺 stands for the ball B =2𝐵=𝐵(𝑥0,2𝑅), 𝐵 =𝐵(𝑥0,𝑅), 1≤𝑝<∞,such 𝐵(𝑥; 𝑑(𝑥, 𝜕𝐺)/2) |𝐵 | the ball and 𝑥 for its volume. that 𝛼 =𝛼𝑛 =𝐾1/(1−𝑛) 𝐶 =𝜃𝑛 (1/2) 𝜃𝑛 (𝑐 )=1/2 Define 0 𝐾 , ∗ 𝐾 , 𝐾 ∗ , 1/𝑝 1 󵄨 󵄨𝑝 and 󵄨 󸀠󵄨 𝛾−1 𝐷𝑝 (𝑓,𝐵) =( ∫ 󵄨𝑓 󵄨 ) ≤𝑀𝑟 , (59) |𝐵| 𝐵 𝐽 = √𝑛 𝐸 . 𝑓,𝐺 𝑓,𝐺 (53) where 0<𝛾≤1,holdsforeveryball𝐵 = 𝐵(𝑎, 𝑟)⊂ B.Then𝑓 𝑛 𝑛 𝐵 𝛾 |𝑓(𝑥)− For a ball 𝐵=𝐵(𝑥,𝑟)⊂R and a mapping 𝑓:𝐵 → R , is Holder¨ continuous in with exponent ,andonehas 𝑓(𝑦)| ≤ 𝐶|𝑥𝛾 −𝑦| 𝑥, 𝑦 ∈𝐵 𝑐=4𝑀𝛾−12−𝛾 we define ,forall ,where .Here and in some places we omit to write the volume element 𝑑𝑥. 1 𝐸𝑓 (𝐵; 𝑥) := ∫ 𝐽𝑓 (𝑧) 𝑑𝑧, 𝑥 ∈𝐺, (54) |𝐵| 𝐵 We need a quasiregular version of this Lemma. 8 Abstract and Applied Analysis

𝑛 Lemma 21. Let 𝑓:B → R be a 𝐾-quasiregular mapping, Let 𝑥∈𝐺. An application of (vi.1) to 𝐵=𝐵𝑥 yields (i.1). 󸀠 󸀠 such that If |𝑥 − 𝑥 |≤𝑟,then𝑐∗𝑑 ≤|𝑦2|−|𝑦1| and therefore we have 𝛾−1 the following: 𝐽𝑓 (𝑎;) 𝑟 ≤𝑀𝑟 , (60) 󸀠 |𝑓(𝑥) − 𝑓(𝑥󸀠)| ≤ 𝐶 𝑑󸀠 ≤𝑐(|𝑦 |−|𝑦|) 𝑐 = 0<𝛾≤1 𝐵 = 𝐵(𝑎, 𝑟)⊂ 𝑓 (vi.1 ) ∗ 0 2 1 ,where 0 where ,holdsforeveryball B.Then 𝐶 /𝑐 is Holder¨ continuous in 𝐵 with exponent 𝛾,andonehas|𝑓(𝑥)− ∗ ∗. 𝛾 1/𝑛 −1 −𝛾 𝑓(𝑦)| ≤ 𝐶|𝑥 −𝑦| ,forall𝑥, 𝑦,where ∈𝐵 𝐶=4𝑀𝐾 𝛾 2 . If 𝑧1,𝑧2 ∈ 𝐵(𝑥,,then 𝑟) |𝑓(𝑧1)−𝑓(𝑧2)| ≤ |𝑓(𝑧1)−𝑓(𝑥)|+ 󸀠 |𝑓(𝑧2)−𝑓(𝑥)|.Hence,using(vi.1 ),wefind(ii.1). Proof of Proof. By hypothesis, 𝑓 satisfies (57) and therefore 𝐷𝑛(𝑓,𝐵) ≤ 1/𝑛 (iii.1). If the hypothesis (b) holds with multiplicative constant 𝐾 𝐽𝑓(𝑎; 𝑟) . An application of Lemma 20 to 𝑝=𝑛yields 󸀠 󸀠 𝛼 𝛼 𝐿,and|𝑥−𝑥|=𝑟,then𝑐∗𝑑 ≤|𝑦2|−|𝑦1|≤2𝐿𝑟 proof. 󸀠 󸀠 𝛼 and therefore |𝑓(𝑥) − 𝑓(𝑥 )| ≤ 𝐶∗𝑑 ≤𝑐1𝑟 ,where𝑐1 = 𝛼 𝛼 ̂ (𝐶∗/𝑐∗)2 𝐿.Hence𝑑1 ⪯𝑟and therefore, in particular, (a.0) By 𝐵(𝐺) denote a family of ball 𝐵 = 𝐵(𝑥) = 𝐵(𝑥, 𝑟) 𝛼 𝑅(𝑥) :=∗ 𝑐 𝑑∗(𝑥) ≤ 𝐿𝑑(𝑥) , 𝑥∈𝐺. such that B =𝐵(𝑥,2𝑟)⊂𝐺.For𝐵∈𝐵(𝐺)̂ , define 𝑑󸀠 =𝑑󸀠 = 𝑑(𝑓(𝑥), 𝜕 󸀠) 𝑅(𝑓(𝑥), 󸀠):=𝑐𝑑󸀠 It is clear, by the hypothesis (b), that there is a fixed con- 𝑓,B B and B ∗ , 𝐿 ( .1) |𝑓| (𝛼, 𝐿 ;𝐵) 󸀠 stant 1 such that vii belongs to Lip 1 for every B =𝑓(B) where . ball 𝐵=𝐵(𝑥,𝑟)such that 𝐵1 =𝐵(𝑥,2𝑟)⊂𝐺and, by (ii.1),we have the following: Define 𝜔loc(𝑓,𝑟)loc =𝜔 ,𝐺(𝑓,𝑟) = sup osc𝐵𝑓 and 𝑟0(𝐺) = sup 𝑟(𝐵), where supremum is taken over all balls 𝐵=𝐵(𝑎)= (viii.1) 𝐿(𝑓,𝐵) ≤𝐿2 for a fixed constant 𝐿2 and 𝑓∈loc 𝐵(𝑎, 𝑟) such that 𝐵1 =𝐵(𝑎,2𝑟)⊂𝐺and 𝑟(𝐵) denotes radius Lip(𝛼, 𝐿2;𝐺). of 𝐵. 𝛼 Hence, since, by the hypothesis (c.5), 𝐺 is a 𝐿 -extension Theorem 22. Let 0<𝛼≤1, 𝐾≥1.Supposethat(a.5) 𝐺 is 𝑛 𝑛 domain, we get (v.1). adomaininR and 𝑓:𝐺 → R is 𝐾-quasiconformal and 𝐺󸀠 = 𝑓(𝐺) Proof of (iv.1).Let𝐵 = 𝐵(𝑥) = 𝐵(𝑥, 𝑟) be a ball such that .Thenonehasthefollowing. 𝑛 𝑛 B =𝐵(𝑥,2𝑟)⊂𝐺.Then𝐸𝑔(𝐵; 𝑥)1 ≈𝑑 /𝑟 , 𝐽𝑓(𝐵; 𝑥)1 ≈𝑑 /𝑟, 𝑥∈𝐺 𝑥 ,𝑥 ∈𝐵 𝛼 (i.1) For every , there exists two points 1 2 𝑥 and, by (iii.1), 𝑑1 ⪯𝑟 on 𝐺.Hence such that |𝑦2|−|𝑦1|≥𝑅(𝑥),where𝑦𝑘 =𝑓(𝑥𝑘), 𝑘=1,2, and 𝑅(𝑥) :=∗ 𝑐 𝑑∗(𝑥). 𝛼−1 𝐽𝑓 (𝐵; 𝑥) ⪯𝑟 , (62) (ii.1) 𝜔loc,𝐺(𝑓,𝑟) 0≤ 2𝑐 𝜔loc,𝐺(|𝑓|, 𝑟), 0≤𝑟≤𝑟0(𝐺),where 𝑐 =2𝐶 /𝑐 0 ∗ ∗. on 𝐺. 𝛼 (iii.1) If, in addition, one supposes that (b.5) |𝑓| ∈ loc Λ (𝐺), Note that one can also combine (iii.1) and Lemma 33 (for then for all balls 𝐵 = 𝐵(𝑥) = 𝐵(𝑥, 𝑟) such that 𝐵1 = ( .1) 𝛼 details see proof of Theorem 40 below) to obtain v . 𝐵(𝑥, 2𝑟), ⊂𝐺 𝑑1 ⪯𝑟 and in particular 𝑅(𝑥) := 𝛼 𝑐∗𝑑∗(𝑥) ≤ 𝐿𝑑(𝑥) , 𝑥∈𝐺. Note that as an immediate corollary of (ii.1) we get a (iv.1) There is a constant 𝑐 such that simple proof of Dyakonov results for quasiconformal mappings (without appeal to Lemma 33 or Lemma 21,whichisa 𝛼−1 𝐽𝑓 (𝑥;) 𝑟 ≤𝑐𝑟 (61) version of Morrey’s Lemma). We enclose this section by proving Theorems 23 and 24 for every 𝑥0 ∈𝐺and 𝐵=𝐵(𝑥,𝑟)⊂𝐵𝑥 . 0 mentioned in the introduction; in particular, these results 𝛼 (v.1) If,inaddition,onesupposesthat(c.5) 𝐺 is a Λ - give further extensions of Theorem-Dy. 𝑛 extension domain in R ,then𝑓∈ Lip(𝛼, 𝐿2;𝐺). Theorem 23. Let 0<𝛼≤1, 𝐾≥1,and𝑛≥2.Suppose𝐺 𝛼 𝑛 Proof. Bythedistortionproperty(30), we will prove the fol- is a Λ -extension domain in R and 𝑓 is a 𝐾-quasiconformal 𝑛 lowing. mapping of 𝐺 onto 𝑓(𝐺) ⊂ R .Thefollowingareequivalent: 𝐵 = 𝐵(𝑥) = 𝐵(𝑥, 𝑟) 𝐵 = 𝐵(𝑥, 𝛼 (vi.1) For a ball such that 1 (i.2) 𝑓∈Λ (𝐺), 2𝑟) ⊂ 𝐺 there exist two points 𝑥1,𝑥2 ∈ 𝐵(𝑥) such that |𝑓| ∈𝛼 Λ (𝐺) |𝑦2|−|𝑦1| ≥𝑅(𝑥) ,where𝑦𝑘 =𝑓(𝑥𝑘), 𝑘=1,2. (ii.2) , 𝑐 Let 𝑙 be line throughout 0 and 𝑓(𝑥) which intersects the (iii.2) there is a constant 3 such that 𝑆(𝑓(𝑥), (𝑥)) 𝑦 𝑦 𝑥 =𝑓−1(𝑦 ) 𝑘=1,2 at points 1 and 2 and 𝑘 𝑘 , . 𝛼−1 By the left side of (30), 𝑥1,𝑥2 ∈ 𝐵(𝑥). We consider two cases: 𝐽𝑓 (𝑥;) 𝑟 ≤𝑐3𝑟 (63)

(a) if 0 ∉ 𝐵(𝑓(𝑥), 𝑅(𝑥)) and |𝑦2|≥|𝑦1|,then|𝑦2|−|𝑦1|= for every 𝑥0 ∈𝐺and 𝐵=𝐵(𝑥,𝑟)⊂𝐵𝑥 .If,inaddition, 2𝑅(𝑥) and |𝑦2|−|𝑦1| = 2𝑅(𝑥); 0 𝐺 𝛼≤𝛼 =𝐾1/(1−𝑛) 0 ∈ 𝐵(𝑓(𝑥), 𝑅(𝑥)) 0∈[𝑦, 𝑓(𝑥)] is a uniform domain and if 0 ,then (b) if , then, for example, 1 (i.2) and (ii.2) are equivalent to and if we choose 𝑥1 =𝑥,wefind|𝑦2| − |𝑓(𝑥)| = 𝑅(𝑥) 𝛼 and this yields (vi.1). (iv.2) |𝑓| ∈ Λ (𝐺, 𝜕𝐺). Abstract and Applied Analysis 9

Proof. Suppose (ii.2) holds, so that |𝑓| is 𝐿-Lipschitz in 𝐺. we find Then (iv.1) shows that (iii.2) holds. By Lemma 21, (iii.2) 𝛼 ( .2) 𝑟 implies i . 𝐽𝑓,𝐷 (𝑧0)≤𝑐2 𝐽𝑔,𝐺 (𝑧0) (68) 𝑑1 (𝜁0) We outline less direct proof that (ii.2) implies (ii.1). ( .2) 2 2 One can show first that ii implies (18)(ormoregen- and therefore, using 𝐸𝑔(𝐵; 𝑧)1 ≈𝑑 /𝑟 ,weget erally (iv.1);seeTheorem 22). Using 𝐴(𝑧) = 𝑎 +𝑅𝑧,we 𝑐∗(𝐵(𝑎, 𝑅))∗ =𝑐 (B) 𝑐(𝐵)̂ = 𝑐̂ 𝛼−1 conclude that and is a fixed 𝐽𝑓 (𝐵;) 𝑧 ≤𝑐3𝑟 . (69) constant (which depends only on 𝐾, 𝑛,1 𝑐 ), for all balls 𝐵⊂𝐺. Lemma 5 tells us that (18) holds, with a fixed constant, for all Thuswehave(iii.3)with𝑐=𝑐̌3. 𝐵⊂𝐺 𝑥 ,𝑥 ∈𝐵 balls and all pairs of points 1 2 .Further,we Proof of the implication (iii.4) ⇒ (i.4). 𝑥, 𝑦 ∈𝐺 |𝑥 − 𝑦| < 𝑑(𝑥, 𝜕𝐺) pick two points with and apply An application of Lemma 21 (see also a version of Astala- (18)with𝐷 = 𝐵(𝑥; |𝑥,letting −𝑦|) 𝑥1 =𝑥and 𝑥2 =𝑦.The 𝑓 𝛼 𝐵 Gehring lemma) shows that is -Holder¨ on 𝑧0 with a resulting inequality Lipschitz (multiplicative) constant which depends only on 𝛼 󵄨 󵄨 󵄨 󵄨𝛼 and it gives the result. 󵄨𝑓 (𝑥) −𝑓(𝑦)󵄨 ≤ const󵄨𝑥−𝑦󵄨 (64) 𝛼 𝛼 shows that 𝑓∈loc Λ (𝐺), and since 𝐺 is a Λ -extension 3. Lipschitz-Type Spaces of Harmonic 𝑓∈Λ𝛼(𝐺) domain, we conclude that . and Pluriharmonic Mappings The implication (ii.4) ⇒ (i.4)isthusestablished.Thecon- 𝛼≤𝛼 verse is clear. For the proof that (iii.4) implies (i.4) for 0, 3.1. Higher Dimensional Version of Schwarz Lemma. Before see [15]. 𝑛 𝑛 giving a proof of the higher dimensional version of the For a ball 𝐵=𝐵(𝑥,𝑟)⊂R and a mapping 𝑓:𝐵 → R , Schwarz lemma we first establish notation. 𝑛 𝑚 we define Suppose that ℎ:𝐵 (𝑎, 𝑟) → R is a continuous vector- 𝑛 1 valued function, harmonic on 𝐵 (𝑎, 𝑟),andlet 𝐸𝑓 (𝐵; 𝑥) := ∫ 𝐽𝑓 (𝑧) 𝑑𝑧, 𝑥 ∈𝐺, (65) |𝐵| 𝐵 ∗ 󵄨 󵄨 𝑛−1 𝑀𝑎 = sup {󵄨ℎ(𝑦)−ℎ(𝑎)󵄨 :𝑦∈𝑆 (𝑎,) 𝑟 }. (70)

𝑛 and 𝐽𝑓(𝐵; 𝑥) = √𝐸𝑓(𝐵; 𝑥). 1 2 𝑚 Let ℎ=(ℎ,ℎ ,...,ℎ ). A modification of the estimate in We use the factorization of planar quasiregular mappings [37,Equation(2.31)]gives to prove the following. 󵄨 𝑘 󵄨 ∗ 𝛼 𝑟 󵄨∇ℎ (𝑎)󵄨 ≤𝑛𝑀 , 𝑘=1,...,𝑚. (71) Theorem 24. Let 0<𝛼≤1, 𝐾≥1.Suppose𝐷 is a Λ - 󵄨 󵄨 𝑎 extension domain in C and 𝑓 is a 𝐾-quasiregular mapping of 𝐷 𝑓(𝐷) ⊂ C We next extend this result to the case of vector-valued onto . The following are equivalent: functions. See also [38]and[39, Theorem 6.16]. 𝑓∈Λ𝛼(𝐷) (i.3) , 𝑛 𝑚 𝛼 Lemma 25. Suppose that ℎ:𝐵 (𝑎, 𝑟) → R is a continuous |𝑓| ∈ Λ (𝐷) 𝑛 (ii.3) , mapping, harmonic in 𝐵 (𝑎, 𝑟).Then 𝑐̌ (iii.3) there is a constant such that 󵄨 󵄨 𝑟 󵄨ℎ󸀠 (𝑎)󵄨 ≤𝑛𝑀∗. 𝛼−1 󵄨 󵄨 𝑎 (72) 𝐽𝑓 (𝐵;) 𝑧 ≤ 𝑐𝑟̌ (𝑧) , (66) Proof. Without loss of generality, we may suppose that 𝑎=0 𝑧 ∈𝐷 𝐵 = 𝐵(𝑧, 𝑟)⊂𝐵 ℎ(0) = 0 for every 0 and 𝑧0 . and .Let 𝐷 𝑊 𝐶 𝑓:𝐷 →𝑊 𝑟2 − |𝑥|2 Proof. Let and be domains in and qr 𝐾(𝑥,𝑦)=𝐾 (𝑥) = , mapping. Then there is a domain 𝐺 and analytic function 𝜙 𝑦 󵄨 󵄨𝑛 (73) 𝑛𝜔𝑛𝑟󵄨𝑥−𝑦󵄨 on 𝐺 such that 𝑓=𝜙∘𝑔,where𝑔 is quasiconformal; see [36, 𝑛 page 247]. where 𝜔𝑛 is the volume of the unit ball in R .Hence,asin[8], Our proof will rely on distortion property of quasiconfor- for 1≤𝑗≤𝑛,wehave mal mappings. By the triangle inequality, (i.4) implies (ii.4). Now, we prove that (ii.4) implies (iii.4). 𝜕 𝜉 𝐾 (0, 𝜉) = 𝑗 . Let 𝑧0 ∈𝐷, 𝜁0 =𝑔(𝑧0), 𝐺0 =𝑔(𝐵𝑧 ), 𝑊0 =𝜙(𝐺0), 𝑑= 𝑛+1 (74) 0 𝜕𝑥𝑗 𝜔𝑛𝑟 𝑑(𝑧0)=𝑑(𝑧0,𝜕𝐷),and𝑑1 =𝑑1(𝜁0)=𝑑(𝜁0,𝜕𝐺).Let𝐵= 𝐵(𝑧, 𝑟)𝑧 ⊂𝐵 .Asinanalyticcasethereis𝜁1 ∈ 𝑔(𝐵) such 𝑛−1 0 Let 𝜂∈𝑆 be a unit vector and |𝜉| =.Forgiven 𝑟 𝜉, ||𝜙(𝜁 )| − |𝜙(𝜁 )|| ≥𝑐𝑑 (𝜁 )|𝜙󸀠(𝜁 )| 𝑧 =𝑔−1(𝜁 ) that 1 0 1 0 0 .If 1 1 ,then it is convenient to write 𝐾𝜉(𝑥) = 𝐾(𝑥, 𝜉) and consider 𝐾𝜉 as ||𝜙(𝜁1)|−|𝜙(𝜁0)|| = ||𝑓(𝑧1)|−|𝑓(𝑧0)||.Hence,if|𝑓| is 𝛼-Holder,¨ 𝑥 󸀠 𝛼 function of . then 𝑑1(𝜁)|𝜙 (𝜁)| ≤1 𝑐 𝑟 for 𝜁∈𝐺0. Hence, since Then

1 󵄨 󸀠 󵄨 󸀠 1 𝐸 (𝐵;) 𝑧 = 󵄨 󵄨 ∫ 󵄨𝜙 (𝜁)󵄨 𝐽 (𝑧) 𝑑𝑥 𝑑𝑦, 𝐾 (0) 𝜂= (𝜉, 𝜂) . 𝑓 󵄨𝐵 󵄨 󵄨 󵄨 𝑔 (67) 𝜉 𝑛+1 (75) 󵄨 𝑧󵄨 𝐵𝑧 𝜔𝑛𝑟 10 Abstract and Applied Analysis

Since |(𝜉, 𝜂)| ≤ |𝜉||𝜂|,weseethat =𝑟 functions 𝑓 for which there exists a constant 𝐶 such that, for all 𝑧 and 𝑤∈Ω(resp., 𝑧 and 𝑤∈𝜕Ω), 󵄨 󵄨 1 󵄨 󵄨 1 󵄨𝐾󸀠 (0) 𝜂󵄨 ≤ , 󵄨∇𝐾 (0)󵄨 ≤ . 󵄨 󵄨 󵄨 𝜉 󵄨 𝑛 and therefore 󵄨 𝜉 󵄨 𝑛 (76) 󵄨𝑓 (𝑧) −𝑓(𝑤)󵄨 ≤𝐶𝜔(|𝑧−𝑤|) . 𝜔𝑛𝑟 𝜔𝑛𝑟 󵄨 󵄨 (83)

This last inequality yields Using Lemma 25, one can prove the following. 󵄨 󵄨 𝜔 𝑓 󵄨 󸀠 󵄨 󵄨 𝑦 󵄨 󵄨 󵄨 Proposition 26. Let be a regular majorant and let be har- 󵄨ℎ (0) (𝜂)󵄨 ≤ ∫ 󵄨∇𝐾 (0)󵄨 󵄨ℎ(𝑦)󵄨 𝑑𝜎 (𝑦) Λ 𝑆𝑛−1(𝑎,𝑟) monic mapping in a simply connected 𝜔-extension domain 𝐷⊂𝑅𝑛 ℎ∈Λ (𝐷) |∇𝑓| ≤ 𝐶(𝜔(1−|𝑧|)/(1− (77) .Then 𝜔 if and only if 𝑀∗𝑛𝜔 𝑟𝑛−1 𝑀∗𝑛 |𝑧|)) ≤ 0 𝑛 = 0 , . 𝑛 𝜔𝑛𝑟 𝑟 It is easy to verify that the real part of any holomorphic where 𝑑𝜎(𝑦) is the surface element on the sphere, and the function is pluriharmonic. It is interesting that the converse proof is complete. is true in simply connected domains. 𝑛 Let C ={𝑧=(𝑧1,...,𝑧𝑛):𝑧1,...,𝑧𝑛 ∈ C} denote the Lemma 27. (i) Let 𝑢 be pluriharmonic in 𝐵0 = 𝐵(𝑎; 𝑟).Then complex vector space of dimension 𝑛.For𝑎=(𝑎1,...,𝑎𝑛)∈ there is an analytic function 𝑓 in 𝐵 such that 𝑓=𝑢+𝑖V. 𝑛 C , we define the Euclidean inner product ⟨⋅, ⋅⟩ by (ii) Let Ω be simply connected and 𝑢 be pluriharmonic in Ω. Then there is analytic function 𝑓 in Ω such that 𝑓=𝑢+𝑖V. ⟨𝑧,⟩ 𝑎 =𝑧1𝑎1 +⋅⋅⋅+𝑧𝑛𝑎𝑛, (78) Proof. (i) Let 𝐵0 = 𝐵(𝑎; 𝑟) ⊂Ω, 𝑃𝑘 =−𝑢𝑦 ,and𝑄=𝑢𝑥 ; 𝑎 (𝑘 ∈ {1,...,𝑛}) 𝑘 𝑘 where 𝑘 denotes the complex conjugate of define form 𝑎𝑘. Then the Euclidean length of 𝑧 is defined by 𝜔 =𝑃𝑑𝑥 +𝑄 𝑑𝑦 =−𝑢 𝑑𝑥 +𝑢 𝑑𝑦 , 𝑘 𝑘 𝑘 𝑘 𝑘 𝑦𝑘 𝑘 𝑥𝑘 𝑘 (84) 󵄨 󵄨2 󵄨 󵄨2 1/2 𝑧 = 𝑧, 𝑧 1/2 =(󵄨𝑧 󵄨 +⋅⋅⋅+󵄨𝑧 󵄨 ) . | | ⟨ ⟩ 󵄨 1󵄨 󵄨 𝑛󵄨 (79) (𝑥,𝑦) 𝜔=∑𝑛 𝜔 V(𝑥, 𝑦) =∫ 𝜔 𝑓 = 𝑘=1 𝑘 and (𝑥 ,𝑦 ) . Then (i) holds for 0 𝑛 󸀠 0 0 Denote a ball in C with center 𝑧 and radius 𝑟>0by 𝑢+𝑖V, which is analytic on 𝐵0. 󵄨 󵄨 (ii) If 𝑧∈Ω,thereisachain𝐶=(𝐵0,𝐵1,...,𝐵𝑛) in Ω B𝑛 (𝑧󸀠,𝑟)={𝑧∈C𝑛 : 󵄨𝑧−𝑧󸀠󵄨 <𝑟}. 󵄨 󵄨 (80) such that 𝑧 is center of 𝐵𝑛 and, by the lemma, there is analytic chain (𝐵𝑘,𝑓𝑘), 𝑘 = 0,...,𝑛. We define 𝑓(𝑧)𝑛 =𝑓 (𝑧).Asinin 𝑛 𝑛 𝑛−1 In particular, B denotes the unit ball B (0, 1) and S the the proof of monodromy theorem in one complex variable, 𝑛 1 sphere {𝑧 ∈ C :|𝑧|=1}.SetD = B , the open unit disk in one can show that this definition does not depend of chains 0 C,andlet𝑇=S be the unit circle in C. 𝐶 and that 𝑓=𝑢+𝑖V in Ω. A continuous complex-valued function 𝑓 definedina 𝑛 domain Ω⊂C is said to be pluriharmonic if, for fixed The following three theorems in9 [ ] are a generalization 𝑛−1 𝑧∈Ωand 𝜃∈S ,thefunction𝑓(𝑧 + 𝜃𝜁) is harmonic in of the corresponding one in [15]. {𝜁 ∈ C :|𝜃𝜁−𝑧|<𝑑Ω(𝑧)},where𝑑Ω(𝑧) denotes the distance Theorem 28. Let 𝜔 be a fast majorant, and let 𝑓=ℎ+𝑔 be from 𝑧 to the boundary 𝜕Ω of Ω.Itiseasytoverifythatthe a pluriharmonic mapping in a simply connected Λ 𝜔-extension real part of any holomorphic function is pluriharmonic; cf. domain Ω. Then the following are equivalent: [40]. 𝜔:[0,+∞)→ [0,+∞) 𝜔(0) = 0 Let with be a con- (1) 𝑓∈Λ𝜔(Ω); tinuous function. We say that 𝜔 is a majorant if (2) ℎ∈Λ𝜔(Ω) and 𝑔∈Λ𝜔(Ω); 𝜔(𝑡) (1) is increasing, (3) |ℎ| ∈ Λ 𝜔(Ω) and |𝑔| ∈ Λ 𝜔(Ω);

(2) 𝜔(𝑡)/𝑡 is nonincreasing for 𝑡>0. (4) |ℎ| ∈ Λ 𝜔(Ω, 𝜕Ω) and |𝑔| ∈ Λ 𝜔(Ω, 𝜕Ω),

If, in addition, there is a constant 𝐶>0depending only where Λ 𝜔(Ω, 𝜕Ω) denotes the class of all continuous functions on 𝜔 such that 𝑓 on Ω∪𝜕Ωwhich satisfy (83) with some positive constant 𝐶, whenever 𝑧∈Ωand 𝑤∈𝜕Ω. 𝛿 𝜔 (𝑡) ∫ 𝑑𝑡 ≤ 𝐶𝜔 (𝛿) , 0<𝛿<𝛿0, (81) 0 𝑡 Define 𝐷𝑓 = (𝐷 1𝑓,...,𝐷𝑛𝑓) and 𝐷𝑓𝐷 =( 1𝑓,...,𝐷𝑛𝑓), ∞ 𝑧 =𝑥 +𝑖𝑦 𝐷 𝑓 = (1/2)(𝜕 𝑓−𝑖𝜕 𝑓) 𝐷 𝑓= 𝜔 (𝑡) where 𝑘 𝑘 𝑘, 𝑘 𝑥𝑘 𝑦𝑘 and 𝑘 𝛿 ∫ 𝑑𝑡 ≤ 𝐶𝜔 (𝛿) , 0<𝛿<𝛿 (1/2)(𝜕 𝑓+𝑖𝜕 𝑓) 𝑘=1,2,...,𝑛 2 0 (82) 𝑥 𝑦 , . 𝛿 𝑡 𝑘 𝑘 Ω⊂C𝑛 𝑓 Ω for some 𝛿0,thenwesaythat𝜔 is a regular majorant.Amajor- Theorem 29. Let be a domain and analytic in . ant is called fast (resp., slow) if condition (81)(resp.,(82)) is Then fulfilled. 󵄨 󵄨 𝑑Ω (𝑧) 󵄨∇𝑓 (𝑧)󵄨 ≤4𝜔|𝑓| (𝑑Ω (𝑧)). (85) Given a majorant 𝜔, we define Λ 𝜔(Ω) (resp., Λ 𝜔(𝜕Ω))to be the Lipschitz-type space consisting of all complex-valued If |𝑓| ∈ Λ 𝜔(Ω),then𝑑Ω(𝑧)|∇𝑓(𝑧)| ≤ Ω4𝐶𝜔(𝑑 (𝑧)). Abstract and Applied Analysis 11

𝑛 If 𝑓=ℎ+𝑔 is a pluriharmonic mapping, where 𝑔 and ℎ Theorem 32. Suppose that 𝐷 is a domain in R , 0<𝛼≤1, 󸀠 󸀠 󸀠 󸀠 2 are analytic in Ω,then|𝑔 |≤|𝑓| and |ℎ |≤|𝑓|.Inparticular, and 𝑓 is harmonic (more generally 𝑂𝐶 (𝐷))in𝐷.Thenone 󸀠 |𝐷𝑖𝑔|, |𝐷𝑖ℎ| ≤ |𝑓 |≤|𝐷𝑔|+|𝐷ℎ|. has the following: 𝑓∈ (𝜔, 𝑐, 𝐷) Proof. Using a version of Koebe theorem for analytic func- (i.1) Lip implies tions (we can also use Bloch theorem), we outline a proof. |𝑓󸀠(𝑥)| ≤ 𝑐𝑑(𝑥)𝛼−1 𝑥∈𝐷 𝑛 (ii.1) , . Let 𝑧∈Ω, 𝑎=(𝑎1,...,𝑎𝑛)∈C , |𝑎| =, 1 𝐹(𝜆) = 𝐹(𝜆, 𝑎)= (ii.1) implies 𝑓(𝑧 +, 𝜆𝑎) 𝜆∈𝐵(0,𝑑Ω(𝑧)/2) ,and𝑢=|𝐹|. By the version of Koebe theorem for analytic functions, (iii.1) 𝑓∈loc Lip(𝛼, 𝐿1;𝐷). for every line 𝐿 which contains 𝑤=𝑓(𝑧),therearepoints 󸀠 𝑤1,𝑤2 ∈𝐿such that 𝑑Ω(𝑧)|𝐹 (0)| ≤ 4|𝑤𝑘 −𝑤|, 𝑘=1,2, 󸀠 4. Theorems of Koebe and Bloch Type for and 𝑤∈[𝑤1,𝑤2].Hence𝑑Ω(𝑧)|𝐹 (0)| ≤ 4𝜔𝑢(𝑑Ω(𝑧)) and 󸀠 Quasiregular Mappings 𝑑Ω(𝑧)|𝐹 (0)| ≤ 4𝜔|𝑓|(𝑑Ω(𝑧)). 𝑓󸀠(𝑧) = (𝑓󸀠 (𝑧),...𝑓󸀠 (𝑧)) 𝐹󸀠(0) = 𝐺⊂R𝑛 Define 𝑧1 𝑧𝑛 .Since We assume throughout that is an open connected set 𝑛 󸀠 󸀠 𝜕𝐺 𝐵(𝑥, 𝑅) ∑ 𝐷𝑘𝑓(𝑧)𝑘 𝑎 =(𝑓(𝑧), 𝑎),wefind|∇𝑓(𝑧)| = |𝑓 (𝑧)|,where whose boundary, ,isnonempty.Also is the open 𝑘=1 𝑥∈𝐺 𝑅 𝐵⊂R𝑛 𝑎=(𝑎1,...,𝑎𝑛).Hence ball centered at with radius .If is a ball, then 𝜎𝐵, 𝜎>0, denotes the ball with the same center as 𝐵 and with 󵄨 󵄨 𝜎 𝐵 𝑑Ω (𝑧) 󵄨∇𝑓 (𝑧)󵄨 ≤4𝜔𝑢 (𝑑Ω (𝑧)) ≤4𝜔|𝑓| (𝑑Ω (𝑧)) . (86) radius equal to times that of . The spherical (chordal) distance between two points 𝑎, 𝑛 Finally if |𝑓| ∈ Λ 𝜔(Ω),wehave𝑑Ω(𝑧)|∇𝑓(𝑧)| ≤ 𝑏∈R is the number 4𝐶𝜔(𝑑Ω(𝑧)). 󵄨 󵄨 𝑞 (𝑎,) 𝑏 = 󵄨𝑝 (𝑎) −𝑝(𝑏)󵄨 , (88) 𝐵𝑛 For the following result is proved in [9]. 𝑛 where 𝑝:R →𝑆(𝑒𝑛+1/2, 1/2) is stereographic projection, Theorem 30. Let 𝜔 be a regular majorant and let Ω be a simply defined by connected Λ 𝜔-extension domain. A function 𝑓 pluriharmonic 𝑥−𝑒 𝑝 (𝑥) =𝑒 + 𝑛+1 . in belongs to Λ 𝜔(Ω) if and only if, for each 𝑖 ∈ {1,2,...,𝑛}, 𝑛+1 󵄨 󵄨2 (89) 󵄨𝑥−𝑒𝑛+1󵄨 𝑑Ω(𝑧)|𝐷𝑖𝑓(𝑧)| ≤Ω 𝜔(𝑑 (𝑧)) and 𝑑Ω(𝑧) |𝐷𝑖𝑓(𝑧)| ≤Ω 𝜔(𝑑 (𝑧)) for some constant C depending only on 𝑓, 𝜔, Ω,and𝑛. Explicitly, if 𝑎 =∞̸ =𝑏̸ ,

𝑓=ℎ+𝑔 𝐷 𝑓=𝐷ℎ 2 −1/2 2 −1/2 Weonly outline a proof: let .Notethat 𝑖 𝑖 𝑞 (𝑎,) 𝑏 = |𝑎−𝑏| (1 + |𝑎| ) (1 + |𝑏| ) . (90) and 𝐷𝑖𝑓=𝐷𝑖𝑔. 𝑛 We can also use Proposition 26. When 𝑓:𝐺 → R is differentiable, we denote its 󸀠 Jacobi matrix by 𝐷𝑓 or 𝑓 and the norm of the Jacobi matrix 󸀠 2 󸀠 3.2. Lipschitz-Type Spaces. Let 𝑓:𝐺 →𝐺 be a 𝐶 function as a linear transformation by |𝑓 |.When𝐷𝑓 exists a.e. we 2 and 𝐵𝑥 = 𝐵(𝑥, 𝑑(𝑥)/2). We denote by 𝑂𝐶 (𝐺) the class of denote the local Dirichlet integral of 𝑓 at 𝑥∈𝐺by 𝐷𝑓(𝑥) = 1/𝑛 functions which satisfy the following condition: 𝐷 (𝑥) = (1/|𝐵| ∫ |𝑓󸀠|𝑛) 𝐵 =𝐵 =𝐵 𝑓,𝐺 𝐵 ,where 𝑥 𝑥,𝐺.Ifthereis 2 󵄨 󵄨 𝐺 = 𝑑 (𝑥) 󵄨Δ𝑓 (𝑥)󵄨 ≤𝑐osc𝐵 𝑓, no chance of confusion, we will omit the index .IfB B𝑥, sup 󵄨 󵄨 𝑥 (87) 𝐵𝑥 then 𝐷𝑓,𝐺 (𝑥) = 𝐷𝑓,B(𝑥) and if B is the unit ball, we write 𝐷(𝑓) instead of 𝐷 (0). 𝑥∈𝐺 𝑓,B for every . When the measure is omitted from an integral, as here, 𝑂𝐶2(𝐺) ⊂ 𝑂𝐶1(𝐺) It was observed in [8]that .In[7], we integration with respect to 𝑛-dimensional Lebesgue measure proved the following results. is assumed. A continuous increasing function 𝜔(𝑡) : [0, ∞)→ Theorem 31. Suppose that [0, ∞) is a majorant if 𝜔(0) = 0 and if 𝜔(𝑡1 +𝑡2)≤𝜔(𝑡1)+𝜔(𝑡2) 𝛼 𝑛 𝑡 ,𝑡 ≥0 (a1) 𝐷 is a Λ -extension domain in R , 0<𝛼≤1,and𝑓 is for all 1 2 . The main result of the paper [3] generalizes Lemma 5 to continuous on 𝐷 which is a 𝐾-quasiconformal mapping 𝑛 a quasiregular version involving a somewhat larger class of of 𝐷 onto 𝐺=𝑓(𝐷)⊂R ; 𝛼 moduli of continuity than 𝑡 , 0<𝛼<1. (a2) 𝜕𝐷 is connected; 𝐺 Λ 𝑓 ¨ 𝜕𝐷 𝛼 Lemma 33 (see [3]). Suppose that is 𝜔-extension domain (a3) is Holder on with exponent ; R𝑛 𝑓 𝐺 𝑓(𝐺) ⊂ R𝑛 2 in .If is K-quasiregular in with and if (a4) 𝑓∈𝑂𝐶(𝐷).Then𝑓 is Holder¨ on 𝐷 with exponent 𝛼 −1 𝐷𝑓 (𝑥) ≤𝐶1𝜔 (𝑑 (𝑥)) 𝑑(𝑥, 𝜕𝐷) 𝑓𝑜𝑟 𝑥 ∈𝐺, (91) The proof in [7] is based on Lemmas 3 and 8 in the paper of Martio and Nakki [18]. In the setting of Lemma 8, 𝑓 𝐷 \ {∞} 󸀠 𝛼 then has a continuous extension to and 𝑑|𝑓 (𝑦)| ≤ 𝑀𝑑̂ . In the setting of Lemma 3, using the fact that 󵄨 󵄨 󵄨 󵄨 𝜕𝐷 is connected, we get similar estimate for 𝑑 small enough. 󵄨𝑓(𝑥1)−𝑓(𝑥2)󵄨 ≤𝐶2𝜔(󵄨𝑥1 −𝑥2󵄨 +𝑑(𝑥1,𝜕𝐺)), (92) 12 Abstract and Applied Analysis

𝑥 ,𝑥 ∈ 𝐷 \ {∞} 𝐶 𝐾≥1 𝑚=𝑚(𝑛,𝐾) for 1 2 ,wheretheconstant 2 depends only on Lemma 38. For each there is 𝑛 with the 𝐾, 𝑛,1 𝜔,𝐶 ,and𝐺. following property. Let 𝜖>0,let𝐺⊂R be a domain, and ̂ If 𝐺 is uniform, the constant 𝐶2 = 𝑐(𝐺)̂ depends only on let 𝐹 denote the family of all qr mappings with max{𝐾𝑂(𝑥, 𝑓), ∗ ∗ 𝑛 𝐾, 𝑛,1 𝐶 and the uniformity constant 𝑐 =𝑐 (𝐺) for 𝐺. 𝐾𝐼(𝑥, 𝑓)} ≤𝐾 and 𝑓:𝐺 → R \{𝑎1,𝑓,𝑎2,𝑓,...,𝑎𝑚,𝑓} where 𝐶 𝑛 ̂ Conversely if there exists a constant 2 such that (92) holds 𝑎𝑖,𝑓 are points in R such that 𝑞(𝑎𝑖,𝑓,𝑎𝑗,𝑓)>𝜖, 𝑖 =𝑗̸ .Then𝐹 𝑥 ,𝑥 ∈𝐺 𝑥∈𝐺 𝐶 for all 1 2 ,then(91) holds for all with 1 forms a normal family in 𝐺. depending only on 𝐶2,𝐾,𝑛,𝜔,and𝐺. Now we prove Theorem 37. Now suppose that 𝜔=𝜔𝛼 and 𝛼≤𝛼0. Also in [3], Nolder, using suitable modification of a theo- Proof. If we suppose that this result is not true then there rem of Nakki¨ and Palka [41], shows that (92)canbereplaced is a sequence of positive numbers 𝑎𝑛,whichconvergesto by the stronger conclusion that zero, and a sequence of 𝐾-quasiregular functions 𝑓𝑛,such 𝑓 (B) [𝑎 ,+∞) 𝑛≥1 󵄨 󵄨 󵄨 󵄨 𝛼 that 𝑛 does not intersect 𝑛 , . Next, the 󵄨𝑓(𝑥 )−𝑓(𝑥 )󵄨 ≤𝐶(󵄨𝑥 −𝑥 󵄨) (𝑥 ,𝑥 ∈ 𝐷\{∞}). 𝑛 󵄨 1 2 󵄨 󵄨 1 2󵄨 1 2 functions 𝑔𝑛 =𝑓𝑛/𝑎𝑛 map B into 𝐺=R \[1,+∞)and (93) hence, by Lemma 38,thesequence𝑔𝑛 is equicontinuous and therefore forms normal family. Thus, there is a subsequence, Remark 34. Simple examples show that the term 𝑑(𝑥1,𝜕𝐺) which we denote again by 𝑔𝑛,whichconvergesuniformly 𝑎−1 cannot in general be omitted. For example, 𝑓(𝑥) = 𝑥|𝑥| on compact subsets of B to a quasiregular function 𝑔.Since 1/(1−𝑛) 𝐷(𝑔 ) 𝐷(𝑔) |𝐷(𝑔 )| = |𝐷(𝑓 )|/𝑎 with 𝑎=𝐾 is 𝐾-quasiconformal in 𝐵 = 𝐵(0; 1). 𝐷𝑓(𝑥) 𝑛 converges to and 𝑛 𝑛 𝑛 converges is bounded over 𝑥∈𝐵yet 𝑓∈Lip𝑎(𝐵);seeExample 6. to infinity, we have a contradiction by Lemma 35. 󸀠 𝑎−1 𝑖𝜃 If 𝑛=2,then|𝑓 (𝑧)| ≈ 𝜌 ,where𝑧=𝜌𝑒 ,andif𝐵𝑟 = 󸀠 𝑟 2𝜋 2(𝑎−1) 2𝑎 𝐵(0, 𝑟),then∫ |𝑓 (𝑧)|𝑑𝑥 𝑑𝑦 =∫ ∫ 𝜌 𝜌𝑑 𝜌𝑑𝜃 ≈𝑟 A path-connected topological space 𝑋 with a trivial fun- 𝐵𝑟 0 0 𝑎−1 damental group 𝜋1(𝑋) is said to be simply connected. We say and 𝐷𝑓,𝐵 (0) ≈ 𝑟 . 3 𝑟 that a domain 𝑉 in R is spatially simply connected if the fun- |𝑓| ∈𝛼 𝐿 (𝐺) Note that we will show below that if ,the damental group 𝜋2(𝑉) is trivial. conclusion (92)inLemma 33 canbereplacedbythestronger |𝑓(𝑥 )−𝑓(𝑥 )| ≤ 𝑐𝑚|𝑥 −𝑥 |𝛼 As an application of Theorem 37,weimmediatelyobtain assertion 1 2 1 2 . thefollowingresult,whichwecalltheKoebetheoremfor quasiregular mappings. Lemma 35 (see [3]). If 𝑓=(𝑓1,𝑓2,...,𝑓𝑛) is K-quasiregular 𝑛 in 𝐺 with 𝑓(𝐺) ⊂𝑅 and if 𝐵 is a ball with 𝜎𝐵 ⊂𝐺, 𝜎>1, 𝐾 𝐶 𝑛 Theorem 39 (second version of Koebe theorem for -qua- then there exists a constant , depending only on ,suchthat siregular functions). Let 𝐵 = 𝐵(𝑎;;let 𝑟) 𝑓 be 𝐾-quasiregular 𝑛 1/𝑛 1/𝑛 function on 𝐵⊂𝑅, 𝐷=𝑓(𝐵), 𝑓(𝑎) =𝑏,andletthe 1 󵄨 󸀠󵄨𝑛 𝜎 1 󵄨 󵄨𝑛 𝐷 𝐷𝑐 𝑑 = ( ∫ 󵄨𝑓 󵄨 ) ≤𝐶𝐾 ( ∫ 󵄨𝑓 −𝑎󵄨 ) (94) unbounded component ∞ of be not empty, and let ∞ |𝐵| 󵄨 󵄨 𝜎−1 |𝜎𝐵| 󵄨 𝑗 󵄨 𝐵 𝜎𝐵 𝑑∞(𝑏) = dist(𝑏,∞ 𝐷 ). Then there exists an absolute constant 𝑐: 𝑎∈R 𝑗 = 1,2,...,𝑛 for all and all .Hereandinsomeplaces (a.1) 𝑟|𝐷𝑓(𝑎)| ≤ 𝑐𝑑̃ ∞; we omit to write the volume and the surface element. 3 (a.2) if, in addition, 𝐷⊂R and 𝐷 is spatially simply con- Lemma 36 (see[42], second version of Koebe theorem for nected, then 𝐷 contains the disk 𝐵(𝑏, 𝜌) of radius 𝜌, 𝐵 = 𝐵(𝑎; 𝑟) 𝑓 analytic functions). Let ;let be holomorphic where 𝜌=𝜌𝑓(𝑎) = 𝑟|𝐷𝑓(𝑎)|/𝑐̃. function on 𝐵, 𝐷=𝑓(𝐵), 𝑓(𝑎) =𝑏, and let the unbounded 𝐷 𝐷 𝑑 =𝑑 (𝑏) = component ∞ of be not empty, and let ∞ ∞ (A.1) Now, using Theorem 39 and Lemma 20,wewill (𝑏, 𝐷 ) ( . ) 𝑟|𝑓󸀠(𝑎)| ≤ 4𝑑 ( . ) dist ∞ .Then a 1 ∞; b 1 if, in addition, establish the characterization of Lipschitz-type spaces for 𝐷 𝐷 𝐵(𝑏, 𝜌) is simply connected, then contains the disk of radius quasiregular mappings by the average Jacobian and in par- 𝜌 𝜌=𝜌 (𝑎) = 𝑟|𝑓󸀠(𝑎)|/4 ,where 𝑓 . ticular an extension of Dyakonov’s theorem for quasiregular mappings in space (without Dyakonov’s hypothesis that it The following result can be considered as a version of this is a quasiregular local homeomorphism). In particular, our lemma for quasiregular mappings in space. approachisbasedontheestimate(a.3) below. 𝑓 𝐾 Theorem 37. Suppose that is a -quasiregular mapping on Theorem 40 (Theorem-DyMa). Let 0<𝛼≤1, 𝐾≥1,and B 𝑓(0) = 0 |𝐷(𝑓)| ≥1 𝛼 𝑛 the unit ball , and . 𝑛≥2.Suppose𝐺 is a 𝐿 -extension domain in R and 𝑓 is a Then,thereexistsanabsoluteconstant𝛼 such that for every 𝐾 𝐺 𝑓(𝐺) ⊂ R𝑛 𝑥∈S 𝑦 Λ =Λ(0,𝑥)= -quasiregular mapping of onto .Thefollowing there exists a point on the half-line 𝑥 are equivalent: {𝜌𝑥 : 𝜌 ≥0} ,whichbelongsto𝑓(B),suchthat|𝑦| ≥2𝛼 . 𝑓 𝐾 𝛼 If is a -quasiconformal mapping, then there exists an (i.4) 𝑓∈Λ (𝐺), 𝜌 𝑓(B) B(0; 𝜌 ) absolute constant 1 such that contains 1 . 𝛼 (ii.4) |𝑓| ∈ Λ (𝐺), For the proof of the theorem, we need also the following 𝛼−1 result, Theorem 18.8.1 [10]. (iii.4) |𝐷𝑓(𝑟, 𝑥)| ≤ 𝑐𝑟 for every ball 𝐵=𝐵(𝑥,𝑟)⊂𝐺. Abstract and Applied Analysis 13

󸀠 󸀠 󸀠 Proof. Suppose that 𝑓 is a 𝐾-quasiregular mapping of 𝐺 onto afunctionℎ we denote by ℎ𝑟,ℎ𝑥 and ℎ𝑦 (or sometimes by 𝑓(𝐺) ⊂ R𝑛 𝐵=𝐵(𝑥,𝑟)⊂ .Wefirstestablishthatforeveryball 𝜕𝑟ℎ, 𝜕𝑥ℎ,and𝜕𝑥ℎ) partial derivatives with respect to 𝑟, 𝑥,and 𝐺,onehasthefollowing: 𝑦,respectively.Letℎ=𝑓+𝑔 be harmonic, where 𝑓 and 𝑔 (a.3) 𝑟|𝐷 (𝑟, 𝑥)| ≤ 𝑐𝜔̃ |𝑓|(𝑟, 𝑥),where𝐷 (𝑟, 𝑥) =𝐷 (𝑥). are analytic, and every complex valued harmonic function ℎ 𝑓 𝑓 𝑓,𝐵 󸀠 on simply connected set 𝐷 is of this form. Then 𝜕ℎ = 𝑓 , Let 𝑙 be line throughout 0 and 𝑓(𝑥) and denote by 𝐷∞,𝑟 󸀠 󸀠 󸀠 󸀠 𝑖𝜃 󸀠 2 󸀠 2 𝑐 ℎ =𝑓+ 𝑔󸀠 𝑓 =𝑓(𝑧)𝑒 𝐽 =|𝑓| −|𝑔| ℎ the unbounded component of 𝐷 ,where𝐷=𝑓(𝐵(𝑥,𝑟)). 𝑟 𝑟 𝑟, 𝑟 ,and ℎ .If is |𝑔󸀠|<|𝑓󸀠| |ℎ󸀠 |≤|𝑓󸀠|+|𝑔󸀠| Then, using a similar procedure as in the proof of Theorem 22, univalent, then and therefore 𝑟 𝑟 𝑟 and 󸀠 |ℎ󸀠 |<2|𝑓󸀠| the part (iii.1), one can show that there is 𝑧 ∈𝜕𝐷∞,𝑟 ∩𝑙such 𝑟 . 󸀠 󸀠 󸀠 󸀠 that 𝑑∞(𝑥 )≤|𝑧 − 𝑓(𝑥)| = ||𝑧 | − |𝑓(𝑥)||,where𝑥 = 𝑓(𝑥). After writing this paper and discussion with some col- −1 󸀠 󸀠 Take a point 𝑧∈𝑓 (𝑧 ).Then𝑧 = 𝑓(𝑧) and |𝑧 − 𝑥|. =𝑟 leagues (see Remark A.11 below), the author found out that it is useful to add this section. For origins of this section see Since ||𝑓(𝑧)| − |𝑓(𝑥)|||𝑓| ≤𝜔 (𝑥, 𝑟).ThenusingTheorem 39, we find (a.3). also [29]. Now we suppose (ii.4),thatis,|𝑓| ∈ Lip(𝛼, 𝐿;. 𝐺) 𝛼 Theorem A.1. Suppose that Thus we have 𝜔|𝑓|(𝑥, 𝑟) ≤ 𝐿𝑟 .Hence,weget 𝛼 (a) ℎ is an euclidean univalent harmonic mapping from an (a.4) 𝑟|𝐷𝑓(𝑟, 𝑥)| ≤ 𝑐𝑟 ,where𝑐=𝐿𝑐̃. open set 𝐷 which contains D into C; ( .4) It is clear that iii is denoted here as (a.4). The implica- ℎ(D) C tion (ii.4) ⇒ (iii.4) is thus established. (b) is a convex set in ; If we suppose (iii.4),thenanapplicationofLemma 20 (c) ℎ(D) contains a disc 𝐵(𝑎; 𝑅), ℎ(0) = 𝑎,andℎ(T) 𝛼 𝛼 shows that 𝑓∈loc Λ (𝐺), and since 𝐺 is a 𝐿 -extension belongs to the boundary of ℎ(D). 𝛼 domain, we conclude that 𝑓∈Λ(𝐺).Thus(iii.4) implies (i.4). Then ( .4)⇒( .4) 󸀠 𝑖𝜑 Finally, the implication i ii is a clear corollary (d) |ℎ (𝑒 )| ≥𝑅/2 , 0≤𝜑≤2𝜋. of the triangle inequality. 𝑟 (A.2) Now we give another outline that (i.4) is equivalent to A generalization of this result to several variables has been (ii.4). communicated at Analysis Belgrade Seminar, cf. [32, 33]. Here, we use approach as in [15]. In particular, (a.4) Proposition A.2. Suppose that implies that the condition (91)holds. 󸀠 ℎ We consider two cases: (a ) is an euclidean harmonic orientation preserving uni- 𝐷 D 𝑑(𝑥) ≤ 2|𝑥 −𝑦| valentmappingfromanopenset which contains (1) ; into C; (2) 𝑠=|𝑥−𝑦|≤𝑑(𝑥)/2. 󸀠 (b ) ℎ(D) is a convex set in C; 𝐺 𝐴=𝐵(𝑥,2𝑠) 󸀠 Then we apply Lemma 33 on and in Case (c ) ℎ(D) contains a disc 𝐵(𝑎; 𝑅) and ℎ(0) = 𝑎. (1) and Case (2),respectively. 𝛼 In more detail, if |𝑓| ∈ 𝐿 (𝐺),thenforeveryball𝐵(𝑥, 𝑟) ⊂ Then 𝛼−1 𝐺, by (a.3), 𝑟|𝐷 (𝑥)| ≤ 𝑐𝑟 and the condition (91)holds. 𝑓 󸀠 |𝜕ℎ(𝑧)| ≥𝑅/4 𝑧∈D Then Lemma 33 tellsusthat(92) holds, with a fixed constant, (d ) , . for all balls 𝐵⊂𝐺and all pairs of points 𝑥1,𝑥2 ∈𝐵. Next, we pick two points 𝑥, ∈ 𝐺 with |𝑥 − 𝑦| < 𝑑(𝑥, 𝜕𝐺) and apply By (d), we have (92)with𝐺=𝐴,where𝐴 = 𝐵(𝑥; |𝑥,letting −𝑦|) 𝑥1 =𝑥and 󸀠 (e) |𝑓 | ≥𝑅/4 on T. 𝑥2 =𝑦.Theresultinginequality 󵄨 󵄨 󵄨 󵄨𝛼 ℎ 𝑓󸀠 ≠ 󵄨𝑓 (𝑥) −𝑓(𝑦)󵄨 ≤ const󵄨𝑥−𝑦󵄨 (95) Since is an euclidean univalent harmonic mapping, 0 𝛼 𝛼 . Using (e) and applying Maximum Principle to the analytic shows that 𝑓∈loc Λ (𝐺), and since 𝐺 is a 𝐿 -extension 𝑓󸀠 =𝜕ℎ 𝛼 function ,weobtainProposition A.2. domain, we conclude that 𝑓∈Λ (𝐺). The implication (ii.4) ⇒ (i.4) is thus established. The Proof of Theorem A.1. Without loss of generality we can sup- converse being trivially true. pose that ℎ(0) = 0.Let0≤𝜑≤2𝜋be arbitrary. Since The consideration in (A.1) shows that (i.4) and (ii.4) are ℎ(D) is a bounded convex set in C there exists 𝜏∈[0,2𝜋] equivalent with (a.4). such that harmonic function 𝑢, defined by 𝑢=Re 𝐻,where 𝑖𝜏 𝑖𝜑 𝐻(𝑧) =𝑒 ℎ(𝑧),hasamaximumonD at 𝑒 . 𝑖𝜑 Appendices Define 𝑢0(𝑧) = 𝑢(𝑒 )−𝑢(𝑧),𝑧∈ D.Bythemeanvalue 2𝜋 𝑖𝜃 𝑖𝜑 theorem, (1/2𝜋) ∫ 𝑢0(𝑒 )𝑑𝜃= 𝑢(𝑒 )−𝑢(0)≥𝑅. A. Distortion of Harmonic Maps 0 Since Poisson kernel for D satisfies Recall by D and T =𝜕D we denote the unit disc and the unit 1−𝑟 𝑖𝜃 𝑃 (𝜃) ≥ , circle respectively, and we also use notation 𝑧=𝑟𝑒.For 𝑟 1+𝑟 (A.1) 14 Abstract and Applied Analysis

−1 using Poisson integral representation of the function 𝑢0(𝑧) = It is worthy to note that (A.4) holds (i.e., 𝑓 is Lipschitz) 𝑖𝜑 𝑢(𝑒 )−𝑢(𝑧), 𝑧∈D,weobtain under assumption that Ω is convex (without any smoothness 1−𝑟 𝑢(𝑒𝑖𝜑)−𝑢(𝑟𝑒𝑖𝜑)≥ (𝑢 𝑖𝜑(𝑒 )−𝑢(0)), hypothesis). 1+𝑟 (A.2) 2 Example A.5. 𝑓(𝑧) = (𝑧 −1) is univalent on D.Since ( ) 󸀠 󸀠 and hence d . 𝑓 (𝑧) = 2(𝑧 −1) it follows that 𝑓 (𝑧) tends 0 if 𝑧 tends to 1. Now we derive a slight generalization of Proposition A.2. This example shows that we cannot drop the hypothesis that 󸀠 𝑓(D) More precisely, we show that we can drop the hypothesis (a ) is a convex domain in Proposition A.4. 󸀠󸀠 and suppose weaker hypothesis (a ). 󸀠 Proof. Let 𝑐 = ((1 − 𝑘)/4)𝑅.Since 󸀠󸀠 󵄨 󵄨 Proposition A.3. Suppose that (𝑎 )ℎis an euclidean har- 󵄨 󵄨 󵄨𝐷ℎ (𝑧)󵄨 ≤𝑘 |𝐷ℎ (𝑧)| , (A.6) monic orientation preserving univalent mapping of the unit Ω Ω 𝐵(𝑎; 𝑅) disc onto convex domain .If contains a disc and it follows that 𝜆ℎ(𝑧) ≥𝑐 0 = ((1−𝑘)/4)𝑅 and therefore |ℎ(𝑧2)− ℎ(0) = 𝑎 󸀠 ,then ℎ(𝑧1)| ≥𝑐 |𝑧2 −𝑧1|. 𝑅 |𝜕ℎ (𝑧)| ≥ ,𝑧∈D. 4 (A.3) Hall, see [43,pages66–68],provedthefollowing.

A proof of the proposition can be based on Lemma A.6 (Hall Lemma). For all harmonic univalent map- Proposition A.2 and the hereditary property of convex pings 𝑓 of the unit disk onto itself with 𝑓(0) = 0, functions: (i) if an analytic function maps the unit disk 󵄨 󵄨2 󵄨 󵄨2 27 univalently onto a convex domain, then it also maps each 󵄨𝑎 󵄨 + 󵄨𝑏 󵄨 ≥𝑐 = , 󵄨 1󵄨 󵄨 1󵄨 0 4𝜋2 (A.7) concentric subdisk onto a convex domain. It seems that we can also use the approach as in the proof of Proposition A.7, 2 where 𝑎1 =𝐷𝑓(0), 𝑏1 = 𝐷𝑓(0),and𝑐0 = 27/4𝜋 = 0.6839 . . .. but an approximation argument for convex domain 𝐺,which we outline here, is interesting in itself: √ Set 𝜏0 = √𝑐0 =3 3/2𝜋. Now we derive a slight gener- (ii) approximation of convex domain 𝐺 with smooth alization of Proposition A.3.Moreprecisely,weshowthatwe convex domains. can drop the hypothesis that the image of the unit disc is 󸀠 convex. Let 𝜙 be conformal mapping of D onto 𝐺, 𝜙 (0) > 0, 𝐺𝑛 = 𝜙(𝑟𝑛D), 𝑟𝑛 =𝑛/(𝑛+1), ℎ is univalent mapping of the unit disc Proposition A.7. Let ℎ be an euclidean harmonic orientation −1 onto convex domain Ω and 𝐷𝑛 =ℎ (𝐺𝑛). preserving univalent mapping of the unit disc into C such that 𝑓(D) contains a disc 𝐵𝑅 = 𝐵(𝑎; 𝑅) and ℎ(0) = 𝑎. (iii) Let 𝜑𝑛 be conformal mapping of U onto 𝐷𝑛, 𝜑𝑛(0) = 0, 𝜑󸀠 (0) > 0 ℎ =ℎ∘𝜑 𝐷 ⊂𝐷 𝑛 ,and 𝑛 𝑛.Since 𝑛 𝑛+1 and (i.1) Then ∪𝐷 = D 𝜑 𝑛 , we can apply the Caratheodory´ theorem; 𝑛 𝑅 𝑧 𝜑󸀠 (𝑧) → |𝜕ℎ (0)| ≥ . tends to , uniformly on compacts, whence 𝑛 4 (A.8) 1 (𝑛→∞). By the hereditary property 𝐺𝑛 is convex. 𝐷 (iv) Since the boundary of 𝑛 is an analytic Jordan curve, (i.2) The constant 1/4 in inequality (A.8) can be replaced the mapping 𝜑𝑛 canbecontinuedanalyticallyacross with sharp constant 𝜏1 =3√3/2/2𝜋. T, which implies that ℎ𝑛 has a harmonic extension ℎ 𝐾 𝑘 = (𝐾 − 1)/(𝐾 + across T. (i.3) If in addition is -qc mapping and 1),then Thus we have the following. 𝜏0 (1−𝑘) (v) ℎ𝑛 are harmonic on D, 𝐺𝑛 =ℎ𝑛(D) are smooth convex 𝜆ℎ (0) ≥𝑅 . (A.9) √1+𝑘2 domains, and ℎ𝑛 tends to ℎ,uniformlyoncompacts subset of D. −1 Proof. Let 0<𝑟<𝑅and 𝑉=𝑉𝑟 =ℎ (𝐵𝑟) and let 𝜑 be a Using (v), an application of Proposition A.2 to ℎ𝑛,gives conformal mapping of the unit disc D onto 𝑉 such that 𝜑(0) = the proof. 0 and let ℎ𝑟 =ℎ∘𝜑. By Schwarz lemma As a corollary of Proposition A.3 we obtain (A.4). 󵄨 󵄨 󵄨 󸀠 󵄨 󵄨𝜑 (0)󵄨 ≤1. (A.10) Proposition A.4. Let ℎ be an euclidean harmonic orientation 𝐾 D preserving -qcmappingoftheunitdisc onto convex domain The function ℎ𝑟 has continuous partial derivatives on D.Since Ω Ω 𝐵(𝑎; 𝑅) ℎ(0) = 𝑎 󸀠 .If contains a disc and ,then 𝜕ℎ𝑟(0) = 𝜕ℎ(0)𝜑 (0),byProposition A.3,weget|𝜕ℎ𝑟(0)| = 󸀠 1−𝑘 |𝜕ℎ(0)||𝜑 (0)| ≥𝑟/4 .Hence,using(A.10)wefind|𝜕ℎ(0)| ≥ 𝜆 (𝑧) ≥ 𝑅, (A.4) ℎ 4 𝑟/4 and if 𝑟 tends to 𝑅,weget(A.8). 󵄨 󵄨 󸀠 󵄨 󵄨 𝑟 = {|𝑧| : 𝑧 ∈𝑉 } 𝑞 =1/𝑟 󵄨ℎ (𝑧2) −ℎ(𝑧1)󵄨 ≥𝑐 󵄨𝑧2 −𝑧1󵄨 ,𝑧1,𝑧2 ∈ D. (A.5) (i2) If 0 max 𝑅 and 0 0,then −1 (i.0) In particular, 𝑓 is Lipschitz on Ω. (vi) 𝑞0|𝜕ℎ𝑟(0)| ≤ |𝜕ℎ(0)|. Abstract and Applied Analysis 15

2 2 2 2 2 Hence, by the Hall lemma, 2|𝑎1| >|𝑎1| +|𝑏1| ≥𝑐0 and Thus Hall asserts the sharp form |𝑎1| +|𝑏1| ≥𝑐0 = 2 2 therefore 27/4𝜋 ,where𝑐0 = 27/4𝜋 = 0.6839 . . . and therefore if √ 𝑏1 =0,then|𝑎1|≥𝜏0,where𝜏0 = √𝑐0 =3 3/2𝜋. (vii) |𝑎1|≥𝜏1,where𝜏1 =3√3/2/2𝜋.Combining(vi) If we combine Hall’s sharp form with the Schwarz lemma and (vii), we prove (i2). for harmonic mappings, we conclude that if 𝑎1 is real, then √ 3 3/2𝜋 ≤1 𝑎 ≤1. (i3) If ℎ=𝑓+𝑔,where𝑓 and 𝑔 are analytic, then 𝜆ℎ(𝑧) = 󸀠 󸀠 󸀠 󸀠 Concerning general codomains, the author, using Hall’s |𝑓 (𝑧)| − |𝑔 (𝑧)| and 𝜆ℎ(𝑧) = |𝑓 (𝑧)| − |𝑔 (𝑧)| ≥(1 − 𝑘)|𝑓󸀠(𝑧)| 𝑎 =𝐷ℎ(0) 𝑏 = 𝐷ℎ(0) sharp result, communicated around 1990 at Seminar Uni- .Set 1 and 1 .BytheHall versity of Belgrade a proof of Corollary 16 and a version of |𝑎 |2 +|𝑏|2 ≥𝑐𝑅2 |𝑎 |2(1 + 𝑘2)≥ sharp form 1 1 0 we get 1 Proposition 13;cf.[32, 33]. 2 √ 2 𝑐0𝑅 ,then|𝑎1| ≥𝑅(𝜏 0/ 1+𝑘 ) and therefore (A.9). A.1. Characterization of Harmonic qc Mappings (See [25]). By 𝜒 we denote restriction of ℎ on R.Ifℎ∈𝐻𝑄𝐶0(H),itis 𝜒:R → R Also as a corollary of Proposition A.3 we obtain the fol- well-known that is a homeomorphism and ℎ=𝑃[𝜒] ℎ∈𝐻𝑄𝐶(H) lowing. Re . Now we give characterizations of 0 in terms of its boundary value 𝜒. Proposition A.8 (see [27, 44]). Let ℎ be an euclidean har- Suppose that ℎ is an orientation preserving diffeomor- monic diffeomorphism of the unit disc onto convex domain Ω. phism of H onto itself, continuous on H∪R such that ℎ(∞) = Ω 𝐵(𝑎; 𝑅) ℎ(0) = 𝑎 ∞,and𝜒 the is restriction of ℎ on R.Recallℎ∈𝐻𝑄𝐶0(H) if If contains a disc and ,then + and only if there is analytic function 𝜙:H →Π such that + 󸀠 ∗ 1 2 𝜙(H) Π 𝜒 (𝑥) = 𝜙 (𝑥) 𝑒 (ℎ)(𝑧) ≥ 𝑅 ,𝑧∈D, (A.11) is relatively compact subset of and Re 16 a.e.

2 2 We give similar characterizations in the case of the unit where 𝑒(ℎ)(𝑧) = |𝜕ℎ(𝑧)| +|𝜕ℎ(𝑧)| . disk and for smooth domains (see below).

The following example shows that Theorem A.1 and Prop- Theorem A.12. Let 𝜓 be a continuous increasing function on 𝑖𝜓(𝑡) ositions A.2, A.3, A.4, A.7,andA.8 are not true if we omit the R such that 𝜓(𝑡 + 2𝜋) − 𝜓(𝑡), =2𝜋 𝛾(𝑡) =𝑒 and ℎ=𝑃[𝛾]. condition ℎ(0) = 𝑎. Then ℎ isqcifandonlyifthefollowinghold:

󸀠 Example A.9. The mapping (1) ess inf 𝜓 >0; + 𝑧−𝑏 (2) there is analytic function 𝜙:D →Πsuch that 𝜑𝑏 (𝑧) = , |𝑏| <1, (A.12) + 󸀠 1−𝑏𝑧 𝜙(𝑈) isrelativelycompactsubsetof Π and 𝜓 (𝑥) = ∗ 𝑖𝑥 Re 𝜙 (𝑒 ) a.e. is a conformal automorphism of the unit disc onto itself and 𝜙 2 In the setting of this theorem we write ℎ=ℎ . The reader 󵄨 󸀠 󵄨 1−|𝑏| 󵄨𝜑 (𝑧)󵄨 = ,𝑧∈D. canusetheabovecharacterizationandfunctionsoftheform 󵄨 𝑏 󵄨 󵄨 󵄨2 (A.13) 󵄨1−𝑏𝑧󵄨 𝜙(𝑧) = 2 + 𝑀(𝑧),where𝑀 is an inner function, to produce 󵄨 󵄨 𝜙 examples of HQC mappings ℎ=ℎ of the unit disk onto itself 󸀠 2 ℎ In particular 𝜑𝑏(0) = 1 − |𝑏| . so the partial derivatives of have no continuous extension to certain points on the unit circle. In particular we can take Heinz proved (see [45]); that if ℎ is a harmonic diffeomor- 𝑀(𝑧) = exp((𝑧+1)/(𝑧−1)); for the subject of this subsections phism of the unit disc onto itself such that ℎ(0) = 0,then cf. [25, 28] and references cited therein. 1 𝑒 (ℎ)(𝑧) ≥ ,𝑧∈D. Remark A.13. Becauseoflackofspaceinthispaperwecould 2 (A.14) 𝜋 not consider some basic concepts related to the subject and Using Proposition A.3 we can prove another Heinz the- in particular further distortion properties of qc maps as orem. Gehring and Osgood inequality [12]. For an application of this inequality, see [25, 28]. Theorem A.10 (Heinz). There exists no euclidean harmonic diffeomorphism from the unit disc D onto C. B. Quasi-Regular Mappings Note that this result was a key step in his proof of the (A) The theory of holomorphic functions of one complex 3 Bernstein theorem for minimal surfaces in R . variable is the central object of study in complex analysis. It is one of the most beautiful and most useful parts of the whole Remark A.11. Professor Kalaj turned my attention to the fact mathematics. that in Proposition 12,theconstant1/4 canbereplacedwith Holomorphic functions are also sometimes referred to as sharp constant 𝜏1 =3√3/2/2𝜋 which is approximately analytic functions, regular functions, complex differentiable 0.584773. functions or conformal maps. 16 Abstract and Applied Analysis

This theory deals only with maps between two-dimen- Therefore the set 𝑆𝑓 of those points where it is not dif- sional spaces (Riemann surfaces). ferentiable has Lebesgue measure zero. Of course, 𝑆𝑓 may be For a function which has a domain and range in the com- nonempty in general and the behaviour of the mapping may plex plane and which preserves angles, we call a conformal be very interesting at the points of this set. Thus, there is a map. The theory of functions of several complex variables substantial difference between the two cases 𝐾>1and 𝐾= has a different character, mainly because analytic functions 1. This indicates that the higher dimensions theory of quasi- of several variables are not conformal. regular mappings is essentially different from the theory in Conformal maps can be defined between Euclidean the complex plane. There are several reasons for this: spaces of arbitrary dimension, but when the dimension is greater than 2, this class of maps is very small. A theorem of (a) there are neither general representation theorems of J.LiouvillestatesthatitconsistsofMobiustransformations Sto¨ılow’stype nor counterparts of power series expan- only; relaxing the smoothness assumptions does not help, as sions in higher dimensions; proved by Reshetnyak. This suggests the search of a gener- (b) the usual methods of function theory based on Cau- alization of the property of conformality which would give a chy’s Formula, Morera’s Theorem, Residue Theorem, rich and interesting class of maps in higher dimension. R𝑛 The Residue Calculus and Consequences, Laurent The general trend of the geometric function theory in Series, Schwarz’s Lemma, Automorphisms of the Unit is to generalize certain aspects of the analytic functions of one Disc, Riemann Mapping Theorem, and so forth, are complex variable. The category of mappings that one usu- notapplicableinthehigher-dimensionaltheory; ally considers in higher dimensions is the mappings with finitedistortion,thus,inparticular,quasiconformalandqua- (c) in the plane case the class of conformal mappings is siregular mappings. very rich, while in higher dimensions it is very small For the dimensions 𝑛=2and 𝐾=1,theclassof𝐾-quasi- (J. Liouville proved that for 𝑛≥3and 𝐾=1,suffi- regularmappingsagreeswiththatofthecomplex-analytic ciently smooth quasiconformal mappings are restric- functions. Injective quasi-regular mappings in dimensions tions of Mobius¨ transformations); 𝑛 𝑛≥2arecalledquasiconformal.If𝐺 is a domain in R , 𝑛 (d) for dimensions 𝑛≥3the branch set (i.e., the set 𝑛≥2, we say that a mapping 𝑓:𝐺 → R is discrete if of those points at which the mapping fails to be a thepreimageofapointisdiscreteinthedomain𝐺.Planar local homeomorphism) is more complicated than in quasiregular mappings are discrete and open (a fact usually the two-dimensional case; for instance, it does not proved via Sto¨ılow’s Theorem). contain isolated points. Theorem A (Sto¨ılow’s Theorem). For 𝑛=2and 𝐾≥1,a𝐾- 2 Injective quasiregular maps are called quasiconformal. quasi-regular mapping 𝑓:𝐺 → R can be represented in the 󸀠 Using the interaction between different coordinate systems, form 𝑓=𝜙∘𝑔,where𝑔:𝐺 →𝐺 is a 𝐾-quasi-conformal 󸀠 for example, spherical coordinates (𝜌, 𝜃, 𝜙)+ ∈𝑅 × [0, 2𝜋) × homeomorphism and 𝜙 is an analytic function on 𝐺 . [0, 𝜋] and cylindrical coordinates (𝑟, 𝜃, 𝑡),onecanconstruct There is no such representation in dimensions 𝑛≥3in certain qc maps. 𝑓 (𝜌,𝜃,𝜙)󳨃→(𝜙,𝜃, 𝜌) 𝑓 general, but there is representation of Sto¨ılow’s type for qua- Define by ln .Then maps the cone 𝑛 𝐶(𝜙 )={(𝜌,𝜃,𝜙):0≤𝜙<𝜙} 0<𝜙 ≤𝜋 𝑓 𝑛 S𝑛 = R 0 0 for 0 onto siregular mappings of the Riemann -sphere ,cf. {(𝑟, 𝜃, 𝑡) : 0 ≤𝑟<𝜙 } [46]. the infinite cylinder 0 .Weleaveitto 𝑛 𝑛 the reader as an exercise to check that the linear distortion 𝐻 Every quasiregular map 𝑓:S → S has a factorization 𝑛 𝑛 depends only on 𝜙 and that 𝐻(𝜌, 𝜃, 𝜙) =𝜙/ sin 𝜙≤𝜙0/ sin 𝜙0. 𝑓=𝜙∘𝑔,where𝑔:S → S is quasiconformal and 𝜙: 𝑛 𝑛 𝜙 =𝜋/2 S → S is uniformly quasiregular. For 0 we obtain a qc map of the half-space onto the cylinder with the linear distortion bounded by 𝜋/2. Gehring-Lehto Lemma: let 𝑓 be a complex, continuous, Ω Since the half space and ball are conformally equivalent, and open mapping of a plane domain which has finite par- wefindthatthereisaqcmapoftheunitballontotheinfinite tial derivatives a.e. in Ω.Then𝑓 is differentiable a.e. in Ω. 𝜋/2 𝑈 R𝑛 𝑓:𝑈 → R𝑚 cylinder with the linear distortion bounded by . Let be an open set in ,andlet be a 𝑓 mapping. Set A simple example of noninjective quasiregular map is 󵄨 󵄨 given in cylindrical coordinates in 3-space by the formula 󵄨𝑓 (𝑥+ℎ) −𝑓(𝑥)󵄨 𝐿(𝑥,𝑓)= . (B.1) (𝑟, 𝜃, 𝑧) 󳨃→(𝑟,2𝜃,𝑧). This map is two-to-one and it is quasi- lim sup 3 ℎ regularonanyboundeddomaininR whose closure does 󸀠 If 𝑓 is differentiable at 𝑥,then𝐿(𝑥, 𝑓) =|𝑓 (𝑥)|.Thetheorem not intersect the 𝑧-axis. The Jacobian 𝐽(𝑓) is different from 0 of Rademacher-Stepanov states that if 𝐿(𝑥, 𝑓) <∞ a.e., then except on the 𝑧-axis, and it is smooth everywhere except on 𝑓 is differentiable a.e. the 𝑧-axis. 2 2 2 3 Note that Gehring-Lehto Lemma is used in dimension Set 𝑆+ = {(𝑥, 𝑦, 𝑧) :𝑥 +𝑦 +𝑧 =1,𝑧≥0}and 𝐻 = 3 3 𝑛=2and Rademacher-Stepanov theorem to show the fol- {(𝑥, 𝑦, 𝑧) : 𝑧≥0} .TheZorichmap𝑍:R → R \{0}is a lowing. quasiregular analogue of the exponential function. It can be For all dimensions, 𝑛≥2, a quasiconformal mapping 𝑓: 𝑛 𝑛 defined as follows. 𝐺→R ,where𝐺 is a domain in R , is differentiable a.e. in 2 𝐺. (1) Choose a bi-Lipschitz map ℎ : [−𝜋/2, 𝜋/2] →𝑆+. Abstract and Applied Analysis 17

2 3 (2) Define 𝑍 : [−𝜋/2, 𝜋/2] ×𝑅 → 𝐻 by 𝑍(𝑥,𝑦,𝑧) = (f)Reshetnyakalsoprovedimportantconvergencethe- 𝑧 𝑒 ℎ(𝑥, .𝑦) orems for these mappings and several analytic prop- 3 (3) Extend 𝑍 to all of R by repeatedly reflecting in erties: they preserve sets of zero Lebesgue-measure, planes. are differentiable almost everywhere, and areolder H¨ continuous. The Reshetnyak theory (which uses the 𝑛 Quasiregular maps of R which generalize the sine and cosine phrase mapping with bounded distortion for “quasi- functions have been constructed by Drasin, by Mayer, and by regular mapping”) makes use of Sobolev spaces, BergweilerandEremenko,seeinFletcherandNickspaper potential theory, partial differential equations, cal- [47]. culus of variations, and differential geometry. Those (B) There are some new phenomena concerning quasireg- mappings solve important first-order systems of PDEs ular maps which are local homeomorphisms in dimensions analogous in many respects to the Cauchy-Riemann 𝑛≥3. A remarkable fact is that all smooth quasiregular equation. The solutions of these systems can be maps are local homeomorphisms. Even more remarkable is viewed as “absolute” minimizers of certain energy the following result of Zorich [48]. functionals. (g) We have mentioned that all pure topological results Theorem B.1. Every quasiregular local homeomorphism R𝑛 → R𝑛 𝑛≥3 about analytic functions (such as the Maximum Mod- ,where , is a homeomorphism. ulus Principle and Rouche’s´ theorem) extend to quasiregular maps. Perhaps the most famous result of this TheresultwasconjecturedbyM.A.Lavrentevin1938. sort is the extension of Picard’s theorem which is due The exponential function exp shows that there is no such 𝑛=2 to Rickman, cf. [35]: result for . Zorich’s theorem was generalized by Martio 𝑛 𝑛 et al., cf. [49](foraproofseealso,e.g.,[35, Chapter III, Section A 𝐾-quasiregular map R → R canomitatmostafinite 3]). set. When 𝑛=2, this omitted set can contain at most two Theorem B.2. There is a number 𝑟=𝑟(𝑛,𝐾)(which one points (this is a simple extension of Picard’s theorem). But calls the injectivity radius) such that every 𝐾-quasiregular local when 𝑛>2, the omitted set can contain more than two 𝑛 𝑛 homeomorphism 𝑔:B → R of the unit ball B ⊂𝑅 is points, and its cardinality can be estimated from above in actually homeomorphic on 𝐵(0; 𝑟). terms of 𝑛 and 𝐾. There is an integer 𝑞=𝑞(𝑛,𝐾)such that 𝑛 𝑛 every 𝐾-qr mapping 𝑓:R → R \{𝑎1,𝑎2,...,𝑎𝑞},where𝑎𝑗 An immediate corollary of this is Zorich’s result. are disjoint, is constant. It was conjectured for a while that This explains why in the definition of quasiregular maps 𝑞(𝑛, 𝐾). =2 Rickman gave a highly nontrivial example to it is not reasonable to restrict oneself to smooth maps: all show that it is not the case: for every positive integer 𝑝 there R𝑛 3 3 smooth quasiregular maps of to itself are quasiconformal. exists a nonconstant 𝐾-qr mapping 𝑓:R → R omitting 𝑝 𝑛≥3 𝜖 In each dimension there is a positive number 𝑛 points. such that every nonconstant quasiregular mapping 𝑓:𝐷 → 𝑛 (h) It turns out that these mappings have many properties R whose distortion function satisfies 𝐾𝛼,𝛽(𝑥, 𝑓) ≤ 1𝑛 +𝜖 for some 1≤𝛼, 𝛽≤𝑛−1is locally injective. similar to those of plane quasiconformal mappings. On the other hand, there are also striking differences. (C) Despite the differences between the two theories Probably the most important of these is that there described in parts (A) and (B), many theorems about geo- exists no analogue of the Riemann mapping theorem metric properties of holomorphic functions of one complex when 𝑛>2. This fact gives rise to the following two variable have been extended to quasiregular maps. These problems. Given a domain 𝐷 in Euclidean 𝑛-space, extensions are usually highly nontrivial. does there exist a quasiconformal homeomorphism 𝑛 (e) In a pioneering series of papers, Reshetnyak proved 𝑓 of 𝐷 onto the 𝑛-dimensional unit ball B ? Next, if in 1966–1969 that these mappings share the fun- such a homeomorphism 𝑓 exists, how small can the damental topological properties of complex-analytic dilatation of 𝑓 be? functions: nonconstant quasi-regular mappings are Complete answers to these questions are known when 𝑛= discrete, open, and sense-preserving, cf. [50]. Here we 2.Foraplanedomain𝐷 canbemappedquasiconformally state only the following. onto the unit disk 𝐵 if and only if 𝐷 is simply connected and has at least two boundary points. The Riemann mapping 𝑛 Open Mapping Theorem.If𝐷 is open in R and 𝑓 is a noncon- theorem then shows that if 𝐷 satisfies these conditions, 𝑛 stantqrfunctionfrom𝐷 to R ,wehavethat𝑓(𝐷) is open set. there exists a conformal homeomorphism 𝑓 of 𝐷 onto D. (Note that this does not hold for real analytic functions). The situation is very much more complicated in higher dimensions, and the Gehring-Vais¨ al¨ apaper[¨ 51]isdevotedto An immediate consequence of the open mapping theo- 𝑛=3 rem is the maximum modulus principle. It states that if 𝑓 is the study of these two questions in the case where . qr in a domain 𝐷 and |𝑓| achieves its maximum on 𝐷,then (D) We close this subsection with short review of Dya- 𝑓 is constant. This is clear. Namely, if 𝑎∈𝐷then the open konov’s approach [15]. mapping theorem says that 𝑓(𝑎) is an interior point of 𝑓(𝐷) The main result of Dyakonov’s papers [6](publishedin and hence there is a point in 𝐷 with larger modulus. Acta Math.) is as follows. 18 Abstract and Applied Analysis

Theorem B. Lipschitz-type properties are inherited from its We remark that a quasiregular mapping 𝑓 in dimension modulus by analytic functions. 𝑛≥3will be locally injective (or, equivalently, locally homeomorphic) if the dilatation 𝐾(𝑓) is sufficiently close A simple proof of this result was also published in Acta to 1. For more sophisticated local injectivity criteria, see the Math. by Pavlovic,´ cf. [52]. literature cited in [10, 15]. Here we only outline how to reduce the proof of (D1) In this item we shortly discuss the Lipschitz-type Theorem B.4 to qc case. properties for harmonic functions. The set of har- 𝑛≥3 monic functions on a given open set 𝑈 canbeseenas It is known that, by Theorem B.2,forgiven and 𝐾≥1 𝑟=𝑟(𝑛,𝐾) 𝐾 the kernel of the Laplace operator Δ and is therefore ,thereisanumber such that every - 𝑔:B →𝑅𝑛 avectorspaceoverR: sums, differences, and scalar quasiregular local homeomorphism of the unit B ⊂ R𝑛 𝐵(0; 𝑟) multiples of harmonic functions are again harmonic. ball is actually homeomorphic on . Applying 𝑔 (𝑤) = 𝑓(𝑥+𝑑(𝑥)𝑤) 𝑤∈B 𝑥∈ In several ways, the harmonic functions are real this to the mappings 𝑥 , ,where 𝐺 𝑑(𝑥) = 𝑑(𝑥, 𝜕𝐺) 𝑓 analogues to holomorphic functions. All harmonic and ,weseethat is homeomorphic (and 𝐵 (𝑥) := 𝐵(0; 𝑟𝑑(𝑥)) functions are real analytic; that is, they can be locally hence quasiconformal) on each ball 𝑟 .The 𝑟 = 𝑟(𝑛, 𝐾) ∈ (0,1) expressed as power series, they satisfy the mean value constant , coming from the preceding 𝑛 𝐾=𝐾(𝑓) 𝑥 theorem, there is Liouville’s type theorem for them, statement, depends on and ,butnoton . Note also the following. and so forth. 󸀠 󸀠 (i0) Define 𝑚𝑓(𝑥, 𝑟) = min{|𝑓(𝑥 )−𝑓(𝑥)|:|𝑥 −𝑥|=𝑟} Harmonic quasiregular (briefly, hqr) mappings in the 󸀠 󸀠 and 𝑀𝑓(𝑥, 𝑟) = max{|𝑓(𝑥 )−𝑓(𝑥)|: |𝑥 −𝑥|= 𝑟}.We planewerestudiedfirstbyMartioin[53]; for a review of this can express the distortion property of qc mappings in the subject and further results, see [25, 54] and the references following useful form. cited there. The subject has grown to include study of hqr 𝑛 maps in higher dimensions, which can be considered as a Proposition A. Suppose that 𝐺 is a domain in R and 𝑓: 𝑛 󸀠 natural generalization of analytic function in plane and good 𝐺→R is 𝐾-quasiconformal and 𝐺 = 𝑓(𝐺).Thereisa candidate for a generalization of Theorem B. constant 𝑐 such that 𝑀𝑓(𝑥, 𝑟) 𝑓≤ 𝑐𝑚 (𝑥, 𝑟) for 𝑥∈𝐺and 𝑟= For example, Chen et al. [19]andtheauthor[55]have 𝑑(𝑥)/2. This form shows in an explicit way that the maximal shown that Lipschitz-type properties are inherited from its dilatation of a qc mapping is controlled by minimal and it is modulo for 𝐾-quasiregular and harmonic mappings in planar convenient for some applications. For example, Proposition A case. These classes include analytic functions in planar case, also yields a simple proof of Theorem B.3. so this result is a generalization of Theorem A. Recall the following. (i1) Roughly speaking, quasiconformal 𝑛 and quasiregular mappings in R , 𝑛≥3, are natural gener- (D2) Quasiregular mappings. Dyakonov [15]madefurther alizations of conformal and analytic functions of one complex important step and roughly speaking showed that variable. Lipschitz-type properties are inherited from its modulus by qc mappings (see two next results). (i2) Dyakonov’s proof of Theorem 6.4 in15 [ ](aswehave indicated above) is reduced to quasiconformal case, but it seems Theorem B.3 (Theorem Dy, see15 [ ,Theorem4]).Let 0< likely that he wanted to consider whether the theorem holds 𝛼 𝛼≤1, 𝐾≥1,and𝑛≥2.Suppose𝐺 is a Λ -extension more generally for quasiregular mappings. 𝑛 domain in R and 𝑓 is a 𝐾-quasiconformal mapping of 𝐺 onto (i3) Quasiregular mappings are much more general than 𝑛 𝑓(𝐺) ⊂ R . Then the following conditions are equivalent: quasiconformal mappings and in particular analytic functions. 𝛼 (i.5) 𝑓∈Λ (𝐺); (E)Takingintoaccounttheabovediscussionitisnatural 𝛼 (ii.5) |𝑓| ∈ Λ (𝐺). to explore the following research problem. 𝑛 Question A. Is it possible to drop local homeomorphism If, in addition, 𝐺 is a uniform domain and if 𝛼≤𝛼0 =𝛼𝐾 = 1/(1−𝑛) hypothesis in Theorem B.4? 𝐾 , then (i.5) and (ii.5) are equivalent to 𝛼 It seems that using the approach from [15], we cannot (iii.5) |𝑓| ∈ Λ (𝐺, 𝜕𝐺). solve this problem and that we need new techniques. (i4) However, we establish the second version of Koebe An example in Section 2 [15] shows that the assumption 𝐾 𝛼≤𝛼 =𝛼𝑛 =𝐾1/(1−𝑛) 𝑛≥3 theorem for -quasiregular functions, Theorem 39,andthe 0 𝐾 cannot be dropped. For , characterization of Lipschitz-type spaces for quasiregular we have the following generalization—but also a consequence mappings by the average Jacobian, Theorem 40.Usingthese of Theorem B.3 dealing with quasiregular mappings that are theorems, we give a positive solution to Question A. local homeomorphisms (i.e., have no branching points).

Theorem B.4 (see [15,Proposition3]). Let 0<𝛼≤1, 𝐾≥1, 𝛼 𝑛 Conflict of Interests and 𝑛≥3.Suppose𝐺 is a Λ -extension domain in R and 𝑓 is a 𝐾-quasiregular locally injective mapping of 𝐺 onto 𝑓(𝐺) ⊂ The author declares that there is no conflict of interests 𝑛 𝛼 𝛼 R .Then𝑓∈Λ (𝐺) ifandonlyif|𝑓| ∈ Λ (𝐺). regarding the publication of this paper. Abstract and Applied Analysis 19

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Research Article Complete Self-Shrinking Solutions for Lagrangian Mean Curvature Flow in Pseudo-Euclidean Space

Ruiwei Xu and Linfen Cao

Department of Mathematics, Henan Normal University, Xinxiang, Henan 453007, China

Correspondence should be addressed to Linfen Cao; [email protected]

Received 14 January 2014; Accepted 15 June 2014; Published 6 July 2014

Academic Editor: Ljubomir B. Ciri´ c´

Copyright © 2014 R. Xu and L. Cao. 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.

2 𝑛 𝑛 Let 𝑓(𝑥) be a smooth strictly convex solution of det(𝜕 𝑓/𝜕𝑥𝑖𝜕𝑥𝑗)=exp {(1/2) ∑𝑖=1 𝑥𝑖(𝜕𝑓/𝜕𝑥𝑖)−𝑓}defined on a domain Ω⊂R ; 2𝑛 then the graph 𝑀∇𝑓 of ∇𝑓 is a space-like self-shrinker of mean curvature flow in Pseudo-Euclidean space R𝑛 with the indefinite metric ∑ 𝑑𝑥𝑖𝑑𝑦𝑖. In this paper, we prove a Bernstein theorem for complete self-shrinkers. As a corollary, we obtain if the Lagrangian 2𝑛 graph 𝑀∇𝑓 is complete in 𝑅𝑛 andpassesthroughtheoriginthenitisflat.

1. Introduction example, see [8–13] and so forth. But very little is known when self-shrinkers are complete not compact with respect 𝑀 𝑛 Let be an -dimensional submanifold immersed into to induced metric from pseudo-Euclidean space. In this R𝑛+𝑚 the Euclidean space .Meancurvatureflowisaone- paper, we will characterize self-shrinkers for Lagrangian 𝑋 = 𝑋(⋅, 𝑡) 𝑋 :𝑀 → parameter family 𝑡 of immersions 𝑡 mean curvature flow in the pseudo-Euclidean space from this R𝑛+𝑚 𝑀 =𝑋(𝑀) with corresponding images 𝑡 𝑡 such that aspect. 𝑑 Let (𝑥1,...,𝑥𝑛;𝑦1,...,𝑦𝑛) be null coordinates in 2𝑛- 𝑋 (𝑥,) 𝑡 =𝐻(𝑥,) 𝑡 ,𝑥∈𝑀, R2𝑛 𝑑𝑡 dimensional pseudo-Euclidean space 𝑛 . Then, the indef- (1) 2 𝑛 inite metric (cf. [14]) is defined by 𝑑𝑠 =∑𝑖=1 𝑑𝑥𝑖𝑑𝑦𝑖. 𝑋 (𝑥,) 0 =𝑋(𝑥) Suppose 𝑓(𝑥) is a smooth strictly convex function defined 𝑛 on domain Ω⊂R .Thegraph𝑀∇𝑓 of ∇𝑓 can be written as is satisfied, where 𝐻(𝑥, 𝑡) is the mean curvature vector of 𝑛+𝑚 (𝑥1,...,𝑥𝑛;𝜕𝑓/𝜕𝑥1,...,𝜕𝑓/𝜕𝑥𝑛).Then,theinducedRieman- 𝑀𝑡 at 𝑋(𝑥, 𝑡) in R . Self-similar solutions to the mean nian metric on 𝑀∇𝑓 is given by curvature flow play an important role in understanding the behavior of the flow and the types of singularities. They satisfy 𝑛 𝜕2𝑓 a system of quasilinear elliptic PDE of the second order as 𝐺=∑ 𝑑𝑥 𝑑𝑥 . 𝜕𝑥 𝜕𝑥 𝑖 𝑗 (3) follows: 𝑖=1 𝑖 𝑗 ⊥ 𝑋 𝑓 𝐻=− , (2) In particular, if function satisfies 2 2 𝑛 ⊥ 𝜕 𝑓 1 𝜕𝑓 where (⋅⋅⋅) stands for the orthogonal projection into the ( )= { ∑𝑥 −𝑓}, det 𝜕𝑥 𝜕𝑥 exp 2 𝑖 𝜕𝑥 (4) normal bundle 𝑁𝑀. 𝑖 𝑗 𝑖=1 𝑖 Self-shrinkers in the ambient Euclidean space have been 𝑀 ∇𝑓 studied by many authors; for example, see [1–6]andsoforth. then the graph ∇𝑓 of is a space-like self-shrinking 2𝑛 For recent progress and related results, see the introduction solution for mean curvature flow in R𝑛 . in [7]. When the ambient space is a pseudo-Euclidean space, Huang and Wang [12]andChauetal.[8]haveused there are many classification works about self-shrinkers; for different methods to investigate the entire solutions to the 2 Abstract and Applied Analysis above equation and showed that an entire smooth strictly For 𝑀, we choose the canonical relative normalization 𝑌= 𝑛 convex solution to (4)inR is the quadratic polynomial (0,0,...,1). Then, in terms of the language of the relative under the decay condition on Hessian of 𝑓.LaterDingand affine differential geometry, the Calabi metric Xinin[10] improve the previous ones in [8, 12]byremoving theadditionalassumptionandprovethefollowing. 𝐺=∑ 𝑓𝑖𝑗 𝑑𝑥𝑖𝑑𝑥𝑗 (7) 𝑌 Theorem 1. Anyspace-likeentiregraphicself-shrinkingsolu- istherelativemetricwithrespecttothenormalization .For 2𝑛 the position vector 𝑦=(𝑥1,...,𝑥𝑛,𝑓(𝑥1,...,𝑥𝑛)),wehave tion to Lagrangian mean curvature flow in 𝑅𝑛 with the indefinite metric ∑𝑖 𝑑𝑥𝑖𝑑𝑦𝑖 is flat. 𝑘 𝑦,𝑖𝑗 = ∑ 𝐴𝑖𝑗 𝑦𝑘 +𝑓𝑖𝑗 𝑌, (8) These rigidity results assume that the self-shrinker graphs where “,” denotes the covariant derivative with respect to the areentire.Namely,theyareEuclideancomplete.Here,wewill 𝐺 characterize the rigidity of self-shrinker graphs from another Calabi metric . We recall some fundamental formulas for the graph 𝑀; for details, see [19]. The Levi-Civita connection completenessandposethefollowingproblem. 𝐺 If a graphic self-shrinker is complete with respect to with respect to the metric has the Christoffel symbols 2𝑛 induced metric from ambient space 𝑅 ,thenisitflat? 1 𝑛 Γ𝑘 = ∑ 𝑓𝑘𝑙𝑓 . In this paper, we will use affine technique (see [15–18]) to 𝑖𝑗 2 𝑖𝑗𝑙 (9) prove the following Bernstein theorem. As a corollary, it gives a partial affirmative answer to the above problem. The Fubini-Pick tensor 𝐴𝑖𝑗𝑘 satisfies ∞ 1 Theorem 2. Let 𝑓(𝑥) be a 𝐶 strictly convex function defined 𝑛 𝐴𝑖𝑗𝑘 =− 𝑓𝑖𝑗𝑘 . (10) on a convex domain Ω⊆R satisfying the PDE (4).Ifthere 2 is a positive constant 𝛼 depending only on 𝑛 such that the 𝑛+1 Consequently, for the relative Pick invariant,wehave hypersurface 𝑀 = {(𝑥, 𝑓(𝑥))} in R is complete with respect 1 to the metric 𝐽= ∑ 𝑓𝑖𝑙𝑓𝑗𝑚𝑓𝑘𝑛𝑓 𝑓 . 4𝑛 (𝑛−1) 𝑖𝑗𝑘 𝑙𝑚𝑛 (11) 𝜕𝑓 𝜕2𝑓 𝐺=̃ {𝛼 ∑ 𝑥 } ∑ 𝑑𝑥 𝑑𝑥 , exp 𝑖 𝑖 𝑗 (5) The Gauss integrability conditions and the Codazzi equations 𝜕𝑥𝑖 𝜕𝑥𝑖𝜕𝑥𝑗 read 𝑓 𝑚ℎ then is the quadratic polynomial. 𝑅𝑖𝑗𝑘𝑙 = ∑ 𝑓 (𝐴𝑗𝑘𝑚𝐴ℎ𝑖𝑙 −𝐴𝑖𝑘𝑚𝐴ℎ𝑗𝑙 ), (12) 𝑓(𝑥) Remark 3. If is a strictly convex solution to (4), then the 𝐴𝑖𝑗𝑘,𝑙 =𝐴𝑖𝑗𝑙,𝑘. 2𝑛 (13) graph {(𝑥, ∇𝑓/2𝑛𝛼)} is a minimal manifold in 𝑅𝑛 endowed 2 with the conformal metric 𝑑𝑠 = exp{−𝛼𝑥 ⋅ 𝑦}𝑑𝑥. ⋅𝑑𝑦 From (12), we get the Ricci tensor 𝑅 = ∑ 𝑓𝑚ℎ𝑓𝑙𝑗 (𝐴 𝐴 −𝐴 𝐴 ) . As a direct application of Theorem 2,wehavethefollow- 𝑖𝑘 𝑖𝑚𝑙 ℎ𝑗𝑘 𝑖𝑚𝑘 ℎ𝑙𝑗 (14) ing. Introduce the Legendre transformation of 𝑓 𝑓 𝐶∞ Corollary 4. Let be a strictly convex -function defined 𝜕𝑓 Ω⊂R𝑛 𝑀 = {(𝑥, ∇𝑓(𝑥))} on a convex domain .Ifthegraph ∇𝑓 𝜉𝑖 = , 𝑖=1,2,...,𝑛, 2𝑛 𝜕𝑥𝑖 in R𝑛 is a complete space-like self-shrinker for mean curvature 𝑛 (15) flow and the sum ∑ 𝑥𝑖(𝜕𝑓/𝜕𝑥𝑖) has a lower bound, then 𝑀∇𝑓 𝜕𝑓 is flat. 𝑢(𝜉1,...,𝜉𝑛)=∑𝑥𝑖 −𝑓(𝑥) . 𝑖=1 𝜕𝑥𝑖 When the shrinker passes through the origin especially, Define the functions we have the following corollary. −1/(𝑛+2) 𝜕2𝑓 𝜕2𝑢 1/(𝑛+2) 𝑀 = {(𝑥, ∇𝑓(𝑥))} R2𝑛 𝜌:=[ ( )] =[ ( )] , Corollary 5. If the graph ∇𝑓 in 𝑛 is a det 𝜕𝑥 𝜕𝑥 det 𝜕𝜉 𝜕𝜉 complete space-like self-shrinker for mean curvature flow and 𝑖 𝑗 𝑖 𝑗 passes through the origin, then 𝑀∇𝑓 is flat. 󵄩 󵄩2 󵄩∇𝜌󵄩 Φ:=∑ 𝑓𝑖𝑗 ( 𝜌) ( 𝜌) = 󵄩 󵄩 , ln 𝑖 ln 𝑗 𝜌2 2. Preliminaries (16) ∞ Let 𝑓(𝑥1,...,𝑥𝑛) be a strictly convex 𝐶 -function defined on ‖⋅‖ Ω⊂R𝑛 here and later the norm is defined with respect to the adomain . Consider the graph hypersurface Calabi metric. From the PDE (4), we obtain 𝜕 𝜌 𝑀:={(𝑥, 𝑓 (𝑥)) |𝑥𝑛+1 =𝑓(𝑥1,...,𝑥𝑛) , (𝑥1,...,𝑥𝑛) ∈Ω} . ln 1 1 = (𝑓 − 𝑢)𝑖 = {𝑓𝑖 −𝑥𝑘𝑓𝑘𝑖}. (17) (6) 𝜕𝑥𝑖 2 (𝑛+2) 2 (𝑛+2) Abstract and Applied Analysis 3

That is, Because its calculation is standard as in [16], we will give its proof in the appendix. 𝑥 = ∑ 𝑓𝑖𝑘 (𝑓 −2(𝑛+2) ( 𝜌) ) . 𝑖 𝑘 ln 𝑘 (18) For affine hyperspheres, Calabi in [20]calculatedthe Laplacian of the Pick invariant 𝐽. Later, for a general convex Using (17)and(18), we can get function, Li and Xu proved the following lemma in [17]. 𝜌 𝜌 𝑓𝑘𝑙𝑓 𝑓𝜌 𝑖 𝑗 𝑘𝑙 𝑖𝑗𝑘 𝑙 Lemma 7. The Laplacian of the relative Pick invariant 𝐽 𝜌 = +𝑓 𝑓 𝜌 − . (19) 𝑖𝑗 𝜌 𝑖𝑗𝑘 𝑙 2 (𝑛+2) satisfies

𝑖𝑗 𝑛+2 𝑖𝑙 𝑗𝑚 𝑘𝑛 Put 𝜏 := (1/2) ∑ 𝑓 (𝜌𝑖/𝜌)𝑓𝑗.From(19), we have Δ𝐽 ≥ ∑ 𝑓 𝑓 𝑓 𝐴 ( 𝜌) 𝑛 (𝑛−1) 𝑖𝑗𝑘 ln ,𝑙𝑚𝑛 󵄩 󵄩2 (27) 𝑛 󵄩∇𝜌󵄩 2 (𝑛+2)4 Δ𝜌 = − +𝜏𝜌. (20) + ‖∇𝐴‖2 +2𝐽2 − Φ2, 2 𝜌 𝑛 (𝑛−1) 4

By (17), we get where “,” denotes the covariant derivative with respect to the Calabi metric. 2 2 󵄩 󵄩2 4(𝑛+2) Φ=‖∇𝑢‖ + 󵄩∇𝑓 󵄩 −2(𝑓+𝑢) , (21) Using Lemma 7,wegetthefollowingcorollary.Forthe and then proof, see the appendix. 󵄩 󵄩2 󵄩∇(𝑓 + 󵄩𝑢) =4(𝑛+2)2Φ+4(𝑓+𝑢). ∞ 󵄩 󵄩 (22) Corollary 8. Let 𝑓(𝑥1,...,𝑥𝑛) be a 𝐶 strictly convex function satisfying PDE (4);then Using (17)yields 2 8 2 1 2 Δ𝐽 ≥ 𝐽 −20(𝑛+2) Φ + ⟨∇𝐽, ∇ (𝑓 + 𝑢)⟩ Δ(𝑓+𝑢)=2𝑛+(𝑛+2) Φ. (23) 4 (28) √ 󵄩 󵄩 3/2 Define a conformal Riemannian metric 𝐺:=̃ exp{𝛼(𝑓+𝑢)}𝐺, +𝐽− 𝑛 (𝑛−1) 󵄩∇(𝑓+𝑢)󵄩 𝐽 . where 𝛼 is a constant. ̃ 3. Proof of Theorem 2 Conformal Ricci Curvature.Denoteby𝑅𝑖𝑗 the Ricci curvature with respect to the metric 𝐺̃;then It is our aim to prove Φ≡0; thus, from definition of 𝜌,

2 (𝑛−2) 𝛼 (𝑛−2) 𝛼 det (𝑓𝑖𝑗 )=const. (29) 𝑅̃ =𝑅 − (𝑓 + 𝑢) + (𝑓 + 𝑢) (𝑓 + 𝑢) 𝑖𝑗 𝑖𝑗 2 ,𝑖𝑗 4 ,𝑖 ,𝑗 everywhere on 𝑀.Asin[8], by Euler homogeneous theorem, 2 1 (𝑛−2) 𝛼 󵄩 󵄩2 we get Theorem 2. − (𝛼Δ (𝑓 + 𝑢)+ 󵄩∇(𝑓+𝑢)󵄩 )𝐺𝑖𝑗 , 2 2 Denote by 𝑠(𝑝0,𝑝) the geodesic distance function from ̃ (24) 𝑝0 ∈𝑀with respect to the metric 𝐺.Foranypositivenumber ̃ 𝑎,let𝐵𝑎(𝑝0, 𝐺) := {𝑝 ∈ 𝑀0 |𝑠(𝑝 ,𝑝)≤𝑎}.Denote where “,” again denotes the covariant derivation with respect 2 2 2 to the Calabi metric. A := max {(𝑎 −𝑠 ) exp {−𝛼(𝑓 + 𝑢)} Φ}, ̃ Usingtheaboveformulas,wecangetthefollowingcrucial 𝐵𝑎(𝑝0,𝐺) estimates. (30) 2 2 2 B := max {(𝑎 −𝑠 ) exp {−𝛼(𝑓 + 𝑢)} 𝐽}. ∞ 𝐵 (𝑝 ,𝐺)̃ Proposition 6. Let 𝑓(𝑥1,...,𝑥𝑛) be a 𝐶 strictly convex 𝑎 0 function satisfying PDE (4). Then, the following estimate holds: ∞ Lemma 9. Let 𝑓 be a strictly convex 𝐶 -function satisfying 1 ‖∇Φ‖2 the PDE (4). Then, there exist positive constants 𝛼 and 𝐶, ΔΦ ≥ 𝐴 ⟨∇Φ, ∇ 𝜌⟩ + ⟨∇ (𝑓 + 𝑢) , ∇Φ⟩ −𝐴 𝑛 1 ln 4 2 Φ depending only on ,suchthat 2 A ≤𝐶(𝑎2 +𝑎3). +𝐴3Φ +Φ, (31) (25) Proof. Step 1. We will prove that there exists a constant 𝐶 𝑛 where depending only on such that 6𝑛2 −𝑛+16 3𝑛2 + 32𝑛 A ≤𝐶(B1/2𝑎+𝑎2 +𝑎3). 𝐴 = ,𝐴= , (32) 1 (𝑛−1)(3𝑛 + 4) 2 8 (𝑛−1)(3𝑛 + 4) (26) To this end, consider the function 64𝑛3 − 72𝑛2 − 46𝑛 − 72 𝐴 = . 2 2 2 3 5𝑛 (𝑛−1)(3𝑛 + 4) 𝐹:=(𝑎 −𝑠 ) exp {−𝛼 (𝑓+𝑢)} Φ (33) 4 Abstract and Applied Analysis

̃ defined on 𝐵𝑎(𝑝0, 𝐺),where𝛼 is a positive constant to be here and later 𝐶 denotes positive constant depending only on determined later. Obviously, 𝐹 attains its supremum at some 𝑛. ∗ 2 2 ∗ ∗ ∗ interior point 𝑝 .Wemayassumethat𝑠 is a 𝐶 -function in a Denote 𝑎 =𝑠(𝑝0,𝑝 ).If𝑎 =0,from(39), it is easy to ∗ neighborhood of 𝑝 . Choose an orthonormal frame field on complete the proof of the lemma. In the following, we assume ∗ ∗ 2 2 𝑀 around 𝑝 with respect to the Calabi metric 𝐺.Then,at that 𝑎 >0.Now,wecalculatetheterm4𝑠Δ𝑠/(𝑎 −𝑠 ).Firstly, ∗ 𝑝 , we will give a lower bound of the Ricci curvature Ric(𝑀, 𝐺)̃ . Assume that Φ,𝑖 4𝑠𝑠,𝑖 −𝛼(𝑓+𝑢) − =0, (34) Φ ,𝑖 𝑎2 −𝑠2 max {exp {−𝛼(𝑓 + 𝑢)} Φ}= exp {−𝛼(𝑓 + 𝑢)} Φ( 𝑝)̃ , ̃ 𝐵𝑎∗ (𝑝0,𝐺) 2 ΔΦ ∑ (Φ ) 12𝑎2 {𝛼(𝑓 + 𝑢)} − ,𝑖 −𝛼Δ(𝑓+𝑢)− exp { {−𝛼(𝑓 + 𝑢)} 𝐽}= {−𝛼(𝑓 + 𝑢)} 𝐽( 𝑞)̃ . 2 2 2 2 max exp exp Φ Φ (𝑎 −𝑠 ) 𝐵 ∗ (𝑝 ,𝐺)̃ (35) 𝑎 0 4𝑠Δ𝑠 (40) − ≤0, 𝑎2 −𝑠2 ̃ For any 𝑝∈𝐵𝑎∗ (𝑝0, 𝐺), by a coordinate transformation, 𝑓 (𝑝) = 𝛿 𝑅 (𝑝) = 0 𝑖 =𝑗̸ 𝑝 where “,” denotes the covariant derivative with respect to the 𝑖𝑗 𝑖𝑗 and 𝑖𝑗 hold for .Then,at , 2 Calabi metric 𝐺 as before, and we used the fact ‖∇𝑠‖𝐺 = {𝛼(𝑓 + 𝑢)} 1 𝜕 (𝑛+2)2 exp . Inserting Proposition 6 into (35), we get 𝑅 ≥ (∑𝑓2 + (𝑛+2) ∑𝑓 𝜌) ≥ − Φ, 𝑖𝑖 4 𝑚𝑖𝑖 𝑚𝑖𝑖 𝜕𝑥 ln 16 2 𝑚 𝑚 𝑚 ∑(Φ ) 1 Φ −(1+𝐴 ) ,𝑖 +𝐴 Φ+ (𝑓 + 𝑢) ,𝑖 (𝑛−2) 𝛼 2 Φ2 3 4 ,𝑖 Φ (𝑓 + 𝑢) 2 ,𝑖𝑖 Φ 𝜌 +𝐴 ,𝑖 ,𝑖 +1−𝛼(2𝑛+(𝑛+2)2Φ) (𝑛−2) 𝛼 1 1 Φ 𝜌 (36) = (2 − 𝑓 (𝑓 + 𝑢) ) 2 2 𝑖𝑖𝑘 𝑘 2 12𝑎 {𝛼(𝑓 + 𝑢)} 4𝑠Δ𝑠 2 − exp − ≤0. (𝑛−2) 𝛼 󵄩 󵄩2 2 ≤ (𝑛−2) 𝛼+ 󵄩∇(𝑓 + 󵄩𝑢) +𝐶𝐽. (𝑎2 −𝑠2) 𝑎2 −𝑠2 4 󵄩 󵄩 (41) Combining (34)with(36) and using the Schwarz inequality, we have Then, using the Schwarz inequality and (22)–(24), we know 𝑝 1 Φ that at the point ∑ (𝑓 + 𝑢) ,𝑖 4 ,𝑖 Φ (𝑀, 𝐺)̃ 2 Ric 1 2 2 𝑎 {𝛼(𝑓 + 𝑢)} ≥ 𝛼 ∑ [(𝑓 + 𝑢) ] − exp , 8 ,𝑖 𝛼 (𝑎2 −𝑠2)2 ≥−exp {−𝛼(𝑓 + 𝑢)} ̃ Φ 𝜌 𝐴 𝐴2 Φ2 ×{𝐶Φ+𝐶𝐽+𝛼[3(𝑛−2) 𝛼(𝑓 + 𝑢) +2 (𝑛−1)]} 𝐺. 𝐴 ∑ ,𝑖 ,𝑖 ≥− 3 Φ− 1 ,𝑖 , 1 2 Φ 𝜌 4 𝐴3 Φ (42)

2 2 ∑(Φ ) 2 𝑎 {𝛼(𝑓 + 𝑢)} If 3(𝑛 − 2)𝛼(𝑓 + 𝑢) + 2(𝑛,then −1)≤0 ,𝑖 ≤2(𝛼2 ∑ [(𝑓 + 𝑢) ] +16 exp ). Φ2 ,𝑖 (𝑎2 −𝑠2)2 − exp {−𝛼 (𝑓+𝑢)} 𝛼 [3 (𝑛−2) 𝛼 (𝑓+𝑢) +2(𝑛−1)] ≥0. (37) (43) Choose 𝛼 small enough such that Otherwise, 𝐴2 1 𝐴 2(1+𝐴 + 1 )𝛼≤ ,𝛼(𝑛+2)2 ≤ 3 , exp {−𝛼 (𝑓+𝑢)} 𝛼 [3 (𝑛−2) 𝛼 (𝑓+𝑢) +2(𝑛−1)] ≤𝐶. 2 𝐴 16 4 3 (38) (44) 100𝑛𝛼≤ 1. ̃ ̃ Then,theRiccicurvatureRic(𝑀, 𝐺) on 𝐵𝑎∗ (𝑝0, 𝐺) is bounded Then, by substituting the three estimates above, we get from below by 2 ̃ 𝐴3 1 2 𝐶𝑎 Ric (𝑀, 𝐺) Φ+ 𝛼(𝑓 + 𝑢) − exp {𝛼(𝑓 + 𝑢)} 2 16 ,𝑖 (𝑎2 −𝑠2)2 Φ 𝐽 (39) ≥−𝐶( (𝑝)̃ + (𝑞)̃ + 1) 𝐺.̃ 4𝑠Δ𝑠 − ≤0, exp {𝛼(𝑓 + 𝑢)} exp {𝛼(𝑓 + 𝑢)} 𝑎2 −𝑠2 (45) Abstract and Applied Analysis 5

̃ By the Laplacian comparison theorem, we get defined on 𝐵𝑎(𝑝0, 𝐺),where𝛼 is the constant in (38). ∗ Obviously, 𝐻 attains its supremum at some interior point 𝑞 . ∗ 𝑠Δ𝑠 Choose an orthonormal frame field on 𝑀 around 𝑞 with ∗ 𝑎2 −𝑠2 respect to the Calabi metric 𝐺.Then,at𝑞 ,

𝑠Δ𝑠̃ (𝑛−2) 𝛼 𝑠(𝑓 + 𝑢) 𝑠 𝐽,𝑖 4𝑠𝑠,𝑖 = {𝛼(𝑓 + 𝑢)} − ,𝑖 ,𝑖 −𝛼(𝑓+𝑢) − =0, (52) exp 𝑎2 −𝑠2 2 𝑎2 −𝑠2 𝐽 ,𝑖 𝑎2 −𝑠2

∗ 2 2 exp {𝛼(𝑓 + 𝑢)} (𝑝 ) Δ𝐽 ∑(𝐽,𝑖) 12𝑎 exp {𝛼(𝑓 + 𝑢)} ≤𝐶3 − −𝛼Δ(𝑓+𝑢)− 𝑎2 −𝑠2 𝐽 𝐽2 (𝑎2 −𝑠2)2 (46) (53) ×(√ {−𝛼(𝑓 + 𝑢)} Φ( 𝑝)̃ 4𝑠Δ𝑠 exp − ≤0, 𝑎2 −𝑠2 +√exp {−𝛼(𝑓 + 𝑢)} 𝐽( 𝑞)̃ + 1) 𝑠 where “,” denotes the covariant derivative with respect to the Calabi metric 𝐺 as before. Inserting Corollary 8 into (53), we 2 𝛼 2 𝑎 exp {𝛼(𝑓 + 𝑢)} get + (𝑓 + 𝑢),𝑖 +𝐶 , 2 16 (𝑎2 −𝑠2)2 Φ2 1 𝐽 ∑ (𝐽 ) 𝐽−20(𝑛+2)8 + ∑ ,𝑖 (𝑓 + 𝑢) +1− ,𝑖 𝐽 4 𝐽 ,𝑖 𝐽2 where Δ̃ denotes the Laplacian with respect to the metric 𝐺̃. 2 󵄩 󵄩 1/2 Substituting (46)into(39)yields −𝛼(2𝑛+(𝑛+2) Φ) − √𝑛 (𝑛−1) 󵄩∇(𝑓+𝑢)󵄩 𝐽 (54) {−𝛼(𝑓 + 𝑢)} Φ 12𝑎2 {𝛼(𝑓 + 𝑢)} 4𝑠Δ𝑠 exp − exp − ≤0. (𝑎2 −𝑠2)2 𝑎2 −𝑠2 Φ ≤𝑎𝐶(√ (𝑝)̃ exp {𝛼(𝑓 + 𝑢)} Applying the Schwarz inequality, we have 2 1 𝐽,𝑖 𝛼 2 𝑎 exp {𝛼(𝑓 + 𝑢)} 𝐽 −1 (47) ∑ (𝑓 + 𝑢) ≥ ∑ [(𝑓 + 𝑢) ] −𝐶 , +√ (𝑞)̃ + 1) × 2(𝑎 −𝑠2) 4 𝐽 ,𝑖 8 ,𝑖 (𝑎2 −𝑠2)2 exp {𝛼(𝑓 + 𝑢)} 2 2 (𝐽 ) 2 32𝑎 {𝛼(𝑓 + 𝑢)} 𝐶𝑎2 ∑ ,𝑖 ≤2𝛼2 ∑ [(𝑓 + 𝑢) ] + exp , + . 𝐽2 ,𝑖 (𝑎2 −𝑠2)2 (𝑎2 −𝑠2)2 󵄩 󵄩 1/2 √𝑛 (𝑛−1) 󵄩∇(𝑓+𝑢)󵄩 𝐽 Note that 𝐽 2 2 2 2 ≤ +4𝑛(𝑛−1) ((𝑛+2) Φ + (𝑓 + 𝑢)) . A ≥[(𝑎−𝑠 ) exp {−𝛼(𝑓 + 𝑢)} Φ]( 𝑝)̃ 4 (55) 2 ≥(𝑎2 −𝑠2) (𝑝∗) {−𝛼(𝑓 + 𝑢)} ( 𝑝)̃ Φ (𝑝)̃ , exp Inserting these estimates into (54)yields (48) 2 2 2 3 Φ2 𝛼 B ≥[(𝑎−𝑠 ) exp {−𝛼(𝑓 + 𝑢)} 𝐽]( 𝑞)̃ 𝐽−20(𝑛+2)8 −𝐶Φ+ ∑ (𝑓 + 𝑢)2 4 𝐽 16 ,𝑖 2 2 2 ∗ (56) ≥(𝑎 −𝑠 ) (𝑝 ) exp {−𝛼(𝑓 + 𝑢)} ( 𝑞)̃ 𝐽 (𝑞)̃ . 𝑎2 {𝛼(𝑓 + 𝑢)} 4𝑠Δ𝑠 −𝐶 exp −𝐶(𝑓+𝑢)− ≤0, 2 2 2 𝑎2 −𝑠2 2 2 2 ∗ (𝑎 −𝑠 ) Multiplying by (𝑎 −𝑠 ) (𝑝 ), at both sides of (47), yields here and later 𝐶 denotes different positive constants depend- 1/2 1/2 2 3 A ≤𝐶𝑎(A + B )+𝐶(𝑎 +𝑎 ). (49) ing only on 𝑛. We discuss two subcases. Using the Schwarz inequality, we complete Step 1. Case 1.If Step 2. Wewill prove that there is a constant 𝐶 depending only 𝐽 ∗ Φ ∗ 𝑛 (𝑞 )≤ (𝑞 ), (57) on such that exp {𝛼(𝑓 + 𝑢)} exp {𝛼(𝑓 + 𝑢)} 2 4 B ≤𝐶(A +𝑎 +𝑎 ). (50) then B ≤ A. In this case, Step 2 is complete.

Consider Case 2. Now, assume that 𝐽 ∗ Φ ∗ 2 2 2 (𝑞 )> (𝑞 ). (58) 𝐻=(𝑎 −𝑠 ) exp {−𝛼 (𝑓+𝑢)} 𝐽 (51) exp {𝛼(𝑓 + 𝑢)} exp {𝛼(𝑓 + 𝑢)} 6 Abstract and Applied Analysis

∗ Then, 1 > (Φ/𝐽)(𝑞 ).Thus, Now, we assume that Φ(𝑝) =0̸ . Choose a local orthonormal 𝐺 𝑝 𝜌 (𝑝) = 2 frame field of the metric around such that ,1 3 𝛼 2 𝑎 {𝛼(𝑓 + 𝑢)} ‖∇𝜌‖(𝑝) >0 𝜌 (𝑝) = 0 𝑖>1 𝐽−𝐶Φ+ ∑ [(𝑓 + 𝑢) ] −𝐶 exp , ,𝑖 ,forall .Then, 4 16 ,𝑖 (𝑎2 −𝑠2)2 2 (59) (𝜌 ) 𝜌 𝜌 4𝑠Δ𝑠 ΔΦ = 2 (1−𝛿+𝛿) ∑ ,𝑖𝑗 +2∑ ,𝑗 ,𝑗𝑖𝑖 −𝐶(𝑓+𝑢)− ≤0. 2 2 𝑎2 −𝑠2 𝜌 𝜌 (67) The rest of the estimate is almost the same as in Step 1. The (𝜌 )2𝜌 ∗ −8 ,1 ,11 + (𝑛+6) Φ2 −2𝜏Φ, only difference is to deal with the term (𝑓+𝑢).If(𝑓+𝑢)(𝑞 )≤ 3 ∗ 𝜌 0,then−𝐶(𝑓 + 𝑢)(𝑞 )≥0.Wecandropthisterm. {−𝛼(𝑓 + 𝑢)}(𝑓 +𝑢) Otherwise, exp has a uniform upper where 1>𝛿>0is a constant to be determined later. Applying bound. (20), we obtain Using the same method as in Step 1, we can estimate the 4𝑠Δ𝑠/(𝑎2 −𝑠2) 2 term and finally get ∑(𝜌 ) 2 2 ,𝑖𝑗 2𝑛 (𝜌,11) ∑𝑖>1 (𝜌,1𝑖) 2 4 2 ≥ +4 B ≤𝐶(A +𝑎 +𝑎 ). (60) 𝜌2 𝑛−1 𝜌2 𝜌2

2 Then, combining the conclusion of Step 1, we get 2𝑛 (𝜌 ) 𝜌 𝑛2 + ,1 ,11 + Φ2 (68) 2 3 𝑛−1 𝜌3 2 (𝑛−1) A ≤𝐶(𝑎 +𝑎 ). (61) 4 𝜌 2 2𝑛 − ,11 𝜏+ 𝜏2 − Φ𝜏. This completes the proof of Lemma 9. 𝑛−1 𝜌 𝑛−1 𝑛−1 Proof of Theorem 2. For any point 𝑞∈𝑀,choosesufficient ̃ large constant 𝑅0 such that 𝑞∈𝐵𝑅 (𝑝0, 𝐺).Then,forall𝑎≥ An application of the Ricci identity shows that 𝑅 𝑞∈𝐵(𝑝 ) 0 0, 𝑎 0 .UsingLemma 9,weknow 2 4 2 (𝜌,1) 𝜌,11 (𝜌,1) 2 3 ∑𝜌 𝜌 =−2𝑛 +𝑛 𝐶 (𝑛) (𝑎 +𝑎 ) 𝜌2 ,𝑗 ,𝑗𝑖𝑖 𝜌3 𝜌4 exp {−𝛼(𝑓 + 𝑢)} Φ (𝑞)≤ . (62) (𝑎2 −𝑠2)2 (69) (𝜌 )2 𝜌 +2𝑅 ,1 +2 ,1 (𝜌𝜏) . Now, let 𝑎→+∞,andwehave 11 𝜌2 𝜌2 ,1 0≤exp {−𝛼(𝑓 + 𝑢)} Φ (𝑞) ≤0. (63) Substituting (68)and(69)into(67), we obtain

Consequently, 2 (𝜌 ) 2𝑛 (1−𝛿) 𝜕2𝑓 ΔΦ ≥ 2𝛿 ∑ ,𝑖𝑗 +(−2𝑛−8+ ) det ( )(𝑞)=const. (64) 𝜌2 𝑛−1 𝜕𝑥𝑖𝜕𝑥𝑗 (𝜌 )2𝜌 (𝜌 )2 This completes the proof of Theorem 2. × ,1 ,11 +2𝑅 ,1 𝜌3 11 𝜌2

4 4. Appendix 𝑛2 (1−𝛿) (𝜌 ) +( +2(𝑛+3)) ,1 4 Proof of Proposition 6. Let 𝑝∈𝑀, and we choose a local 2 (𝑛−1) 𝜌 orthonormal frame field of the metric 𝐺 around 𝑝.Then, (70) 𝜌,1 4𝑛 − 2 − 2𝑛𝛿 2 2 +2 (𝜌𝜏),1 − Φ𝜏 + (1−𝛿) ∑(𝜌 ) 𝜌 𝜌 ∑(𝜌 ) 𝜌2 𝑛−1 Φ= ,𝑗 ,Φ=2∑ ,𝑗 ,𝑗𝑖 −2𝜌 ,𝑗 , 2 ,𝑖 2 ,𝑖 3 2 2 𝜌 𝜌 𝜌 2𝑛 (𝜌 ) ∑ (𝜌 ) ×( ,11 +4 𝑖>1 ,1𝑖 2 𝑛−1 𝜌2 𝜌2 ∑(𝜌 ) 𝜌 𝜌 𝜌 𝜌 𝜌 (65) ΔΦ = 2 ,𝑗𝑖 +2∑ ,𝑗 ,𝑗𝑖𝑖 −8∑ ,𝑗 ,𝑖 ,𝑗𝑖 2 2 3 𝜌 𝜌 𝜌 4 𝜌 2 − ,11 𝜏+ 𝜏2). + (𝑛+6) Φ2 −2𝜏Φ, 𝑛−1 𝜌 𝑛−1 whereweused(20). In the case Φ(𝑝) = 0,itiseasytoget,at Note that 𝑝, 2 2 2 2 4 (𝜌 ) 1 (Φ ) ∑ (𝜌 ) (𝜌 ) 𝜌 (𝜌 ) 2 ,11 = ∑ ,𝑖 − 𝑖>1 ,1𝑖 +2 ,1 ,11 − ,1 . ∑ (𝜌,𝑖𝑗) 𝜌2 4 Φ 𝜌2 𝜌3 𝜌4 ΔΦ ≥ 2 . (66) 𝜌2 (71) Abstract and Applied Analysis 7

Then,70 ( )and(71)togethergiveus By the same method, as deriving (69), we have

2 2 (𝜌,𝑖𝑗) 𝑛 (1−𝛿) ∑(Φ ) 𝑛 2 ΔΦ ≥ 2𝛿 ∑ + ,𝑖 ∑ (𝐴 )2 ≥ ∑ (𝐴 )2 − 𝐴 ∑ 𝐴 𝜌2 2 (𝑛−1) Φ 𝑚𝑙1 𝑛−1 𝑖11 𝑛−1 111 𝑖𝑖1 (79) 2 2 1 2 6𝑛 (1−𝛿) (𝜌 ) 𝜌 (𝜌 ) + (∑ 𝐴 ) . +( −2(𝑛+4)) ,1 ,11 +2𝑅 ,1 𝑛−1 𝑖𝑖1 𝑛−1 𝜌3 11 𝜌2 ∑ 𝐴 = ((𝑛 + 2)/2)(𝜌 /𝜌) (𝑛2 −4𝑛)(1−𝛿) (𝜌 )4 Note that 𝑖𝑖1 1 . Therefore, by (14), (77), +[ +2(𝑛+3)] ,1 (78), and (79), we obtain 2 (𝑛−1) 𝜌4 2 2 3 (𝜌,1) 2 (𝜌,1) (𝜌,1) 1−𝛿 2 𝜌,11 4𝑛−2−2𝑛𝛿 2𝑅 =2∑ (𝐴 ) − (𝑛+2) 𝐴 + (2𝜏 −4 𝜏) − Φ𝜏 11 2 𝑘𝑗1 2 111 3 𝑛−1 𝜌 𝑛−1 𝜌 𝜌 𝜌 2 𝜌,1 𝑛 ∑(Φ −2(𝜌 𝑓 𝐴 /(𝑛 + 2)𝜌)) +2 (𝜌𝜏) . ≥ ,𝑖 ,1 𝑘 𝑘𝑖1 𝜌2 ,1 2 (𝑛−1) Φ (72) (𝑛+2)(𝑛+1) 𝜌 + ∑ Φ ,𝑖 Using the Schwarz inequality gives 2 (𝑛−1) ,𝑖 𝜌 2 (𝜌 ) 2 𝜌,11 7 ,𝑖𝑗 3 2 (𝜌 ) 2 2 𝜏≤ ∑ + 𝜏 . (73) 𝑛+1 ,1 (𝑛+2) 2 2 − 𝑓 𝐴 + Φ . 𝜌 3 𝜌 7 𝑛−1 𝑘 𝑘11 𝜌2 2 (𝑛−1) Using (80) 2 (𝜌 ) 𝜌 1 𝜌 ,1 ,11 = Φ ,𝑖 +Φ2, (74) On the other hand, we have 𝜌3 2 ,𝑖 𝜌 𝜌 1 𝜌 and choosing 𝛿 = 7/(3𝑛 +4),weget 2 ,1 (𝜌𝜏) =2Φ𝜏+ ∑ 𝐴 𝑓 𝑓 ,1 +Φ. 𝜌2 ,1 𝑛+2 1𝑖𝑘 𝑘 𝑖 𝜌 (81) 2 𝑛 (1−𝛿) ∑ (Φ ) 3𝑛 (1−𝛿) ΔΦ ≥ ,𝑖 +( − (𝑛+4)) 2 (𝑛−1) Φ 𝑛−1 Then, inserting80 ( )and(81)into(75), we get

2 𝜌,𝑖 (𝜌,1) 2 × ∑ Φ +2𝑅 2𝑛 − 𝑛𝛿 (Φ,𝑖) (𝑛+2)(𝑛−5) +6𝑛𝛿 ,𝑖 11 2 ΔΦ ≥ ∑ − 𝜌 𝜌 2 (𝑛−1) Φ 2 (𝑛−1) (75) 2 (𝑛 +8𝑛)(1−𝛿) 2 2 2 8 (1−𝛿) 2 𝜌 2(𝑛+2) −(𝑛 +8𝑛)𝛿 +[ −2]Φ + 𝜏 × ∑ Φ ,𝑖 +Φ+ Φ2 2 (𝑛−1) 7 (𝑛−1) ,𝑖 𝜌 2 (𝑛−1)

4𝑛−2−2𝑛𝛿 𝜌,1 8 (1−𝛿) 2𝑛 (1−𝛿) 1 − Φ𝜏 + 2 (𝜌𝜏) . + 𝜏2 − Φ𝜏 + 𝑛−1 𝜌2 ,1 7 (𝑛−1) 𝑛−1 𝑛+2

2 2 2 In the following, we will calculate the terms 𝑅11((𝜌,1) /𝜌 ) 𝜌,1 𝑛+1(𝜌,1) (𝜌 /𝜌2)(𝜌𝜏) × ∑ 𝐴 𝑓 𝑓 − 𝑓 𝐴 and ,1 ,1.Notethat(17) is invariant under an affine 1𝑖𝑘 𝑘 𝑖 𝜌 𝑛−1 𝜌2 𝑘 𝑘11 transformation of coordinates that preserved the origin. 𝑥 ,𝑥 ,...,𝑥 So, we can choose the coordinates 1 2 𝑛 such that 2𝑛 ∑Φ,𝑖𝑓𝑘𝐴𝑘𝑖1 2𝑛 𝑓 (𝑝) = 𝛿 𝜕𝜌/𝜕𝑥 =‖ 𝜌‖(𝑝) >0 (𝜕𝜌/𝜕𝑥 )(𝑝) = 0 − + 𝑖𝑗 𝑖𝑗 and 1 grad , 𝑖 , (𝑛−1)(𝑛+2) √Φ (𝑛−1)(𝑛+2)2 for all 𝑖>1.From(19), we easily obtain 2 𝜌 𝜌 𝐴 𝑓 𝜌 × ∑ (𝑓𝑘𝐴𝑘𝑖1) . 𝜌 =𝜌 +𝐴 𝜌 = ,𝑖 ,𝑗 −𝐴 𝜌 + 𝑖𝑗𝑘 𝑘 . ,𝑖𝑗 𝑖𝑗 𝑖𝑗1 ,1 𝜌 𝑖𝑗1 ,1 𝑛+2 (76) (82)

Thus, we get Using (77), we have 2 2 2𝜌,1𝜌,1𝑖 𝜌,𝑖(𝜌,1) (𝜌,1) 𝜌,1𝑓𝑘𝐴𝑘𝑖1 Φ = −2 =−2𝐴 +2 , 2 ,𝑖 𝜌2 𝜌3 𝑖11 𝜌2 (𝑛+2) 𝜌 1 𝜌 𝑛+1(𝜌 ) ∑ 𝐴 𝑓 𝑓 ,1 − ,1 𝑓 𝐴 (77) 𝑛+2 1𝑖𝑘 𝑘 𝑖 𝜌 𝑛−1 𝜌2 𝑘 𝑘11 (83) 3 2 2 𝜌 (𝜌 ) 𝑓 𝐴 (𝜌 ) 1 2 (𝜌 ) ∑ Φ ,𝑖 =−2𝐴 ,1 +2 𝑘 𝑘11 ,1 . (78) = 𝑓Φ − ,1 𝑓 𝐴 . ,𝑖 𝜌 111 𝜌3 𝑛+2 𝜌2 2 𝑖 ,𝑖 𝑛−1 𝜌2 𝑘 𝑘11 8 Abstract and Applied Analysis

𝑛+2 One observes that the Schwarz inequality gives ∑ 𝐴 𝐴 ( 𝜌) 𝑛 (𝑛−1) 𝑖𝑗𝑘 𝑖𝑗𝑙,𝑘 ln ,𝑙

2𝑛 ∑Φ,𝑖𝑓𝑘𝐴𝑘𝑖1 𝑛+2 = ∑ 𝐽 ( 𝜌) (𝑛−1)(𝑛+2) √Φ 2 ,𝑖 ln ,𝑖 2 2 1 2 𝑛+2 9𝑛 (Φ,𝑖) 8𝑛 ( ) ≤ ∑ + ≤ ∑ (𝐴𝑖𝑗𝑘,𝑙 ) + 𝐽Φ 8 (𝑛−1) Φ 2 𝑛 (𝑛−1) 4 9 (𝑛−1)(𝑛+2) (88) 4 2 1 2 1 2 (𝑛+2) 2 × ∑ (𝑓𝑘𝐴𝑘𝑖1) , ≤ ∑ (𝐴 ) + 𝐽 + Φ , 𝑛 (𝑛−1) 𝑖𝑗𝑘,𝑙 4 16 2 2 (𝜌,1) (84) 1 𝑓𝑘𝐴𝑘11 ∑ 𝐴 𝐴 𝐴 (𝑓 + 𝑢) 𝑛−1 𝜌2 𝑛 (𝑛−1) 𝑖𝑗𝑘 𝑗𝑖𝑙 𝑘𝑙𝑝 ,𝑝 9(𝑛+2)2 10𝑛 √ 󵄩 󵄩 3/2 ≤ Φ2 + ≤ 𝑛 (𝑛−1) 󵄩∇(𝑓+𝑢)󵄩 𝐽 . 10𝑛 (𝑛−1) 9 (𝑛−1)(𝑛+2)2

2 Thus, by inserting (88)intoLemma 7,weobtainCorollary 8. × ∑ (𝑓𝑘𝐴𝑘𝑖1) ,

2𝑛Φ𝜏2 ≤𝜏 +𝑛2Φ2. Conflict of Interests Note that by (17)wehave The authors declare that there is no conflict of interests 1 𝑛+2 1 regarding the publication of this paper. 𝑓𝑖𝑗 Φ 𝑓 = 𝑓𝑖𝑗 Φ ( 𝜌) + Φ 𝑥 4 𝑗 𝑖 2 𝑗 ln 𝑖 4 𝑗 𝑗 (85) Acknowledgments 𝑛+2 𝑖𝑗 1 𝑖𝑗 𝜕Φ 𝜕𝑢 = 𝑓 Φ𝑗(ln 𝜌)𝑖 + 𝑓 . 2 4 𝜕𝑥𝑖 𝜕𝑥𝑗 The first author is supported by Grants (nos. 11101129 and 11201318) of NSFC. Part of this work was done when the Then, inserting these estimates into82 ( )yieldsProposition 6. first author was visiting the University of Washington and he thanks Professor Yu Yuan and the support of China Scholarship Council. The second author is supported by Grants (nos. U1304101 and 11171091) of NSFC and NSF of Proof of Corollary 8. Now, we will calculate the term (ln 𝜌),𝑖𝑗𝑘. Henan Province (no. 132300410141). In particular, if 𝑓 satisfies PDE (4), choose the coordinate (𝑥1,𝑥2,...,𝑥𝑛) such that 𝑓𝑖𝑗 (𝑝) =𝑖𝑗 𝛿 ;thenwehave References 1 [1] T. H. Colding and I. Minicozzi, “Generic mean curvature flow (ln 𝜌),𝑖𝑗𝑘 = (𝐴𝑖𝑗𝑘 +𝐴𝑖𝑗𝑘,𝑝 𝑓,𝑝)−(ln 𝜌),𝑙𝐴𝑖𝑗𝑘,𝑙 𝑛+2 I: generic singularities,” Annals of Mathematics,vol.175,no.2, (86) 2 pp. 755–833, 2012. +𝐴 𝐴 (3( 𝜌) − 𝑓 ). 𝑖𝑗𝑙 𝑘𝑙𝑝 ln ,𝑝 𝑛+2 ,𝑝 [2] K. Ecker and G. Huisken, “Mean curvature evolution of entire graphs,” Annals of Mathematics,vol.130,no.3,pp.453–471,1989. Using (17), we have [3] G. Huisken, “Asymptotic behavior for singularities of the mean curvature flow,” Journal of Differential Geometry,vol.31,no.1, 2 1 pp. 285–299, 1990. 3( 𝜌) − 𝑓 =− (𝑓 + 𝑢) +( 𝜌) . ln ,𝑝 𝑛+2 ,𝑝 𝑛+2 ,𝑝 ln ,𝑝 (87) [4] G. Huisken, “Local and global behaviour of hypersurfaces moving by mean curvature,” in Differential Geometry: Partial BytheYounginequalityandtheSchwarzinequality,wehave Differential Equations on Manifolds,vol.54ofProceedings of Symposia in Pure Mathematics, American Mathematical 𝑛+2 Society, 1993. ∑ 𝐴 𝐴 𝐴 ( 𝜌) 𝑛 (𝑛−1) 𝑖𝑗𝑘 𝑖𝑗𝑙 𝑘𝑙ℎ ln ,ℎ [5] K. Smoczyk, “Self-shrinkers of the mean curvature flow in arbitrary codimension,” International Mathematics Research 1 ≤ 𝐽2 + 16𝑛2(𝑛−1)2(𝑛+2)4Φ2, Notices,no.48,pp.2983–3004,2005. 2 [6] L. Wang, “A Bernstein type theorem for self-similar shrinkers,” 1 Geometriae Dedicata,vol.151,pp.297–303,2011. ∑ 𝐴 𝐴 𝑓 𝑛 (𝑛−1) 𝑖𝑗𝑘 𝑖𝑗𝑙,𝑘 𝑙 [7] H. D. Cao and H. Li, “A gap theorem for self-shrinkers of the mean curvature flow in arbitrary codimension,” Calculus of 1 1 𝑛+2 = ∑ 𝐽 𝑓 = ⟨∇𝐽, ∇ (𝑓 + 𝑢)⟩ + ⟨∇𝐽, ∇ 𝜌⟩, Variations and Partial Differential Equations,vol.46,no.3-4, 2 ,𝑙 ,𝑙 4 2 ln pp. 879–889, 2013. Abstract and Applied Analysis 9

[8] A. Chau, J. Chen, and Y. Yuan, “Rigidity of entire self-shrinking solutions to curvature flows,” Journal fur¨ die Reine und Ange- wandte Mathematik,vol.664,pp.229–239,2012. [9] Q. Ding and Z. Z. Wang, On the self-shrinking systems in arbitrary codimension spaces, http://arxiv.org/abs/1012.0429. [10] Q. Ding and Y. L. Xin, “The Rigidity theorems for Lagrangian self-shrinkers,” Journal fur¨ die Reine und Angewandte Mathe- matik,2012. [11] K. Ecker, “Interior estimates and longtime solutions for mean curvature flow of noncompact spacelike hypersurfaces in Minkowski space,” JournalofDifferentialGeometry,vol.46,no. 3, pp. 481–498, 1997. [12] R. Huang and Z. Wang, “On the entire self-shrinking solutions to Lagrangian mean curvature flow,” Calculus of Variations and Partial Differential Equations, vol. 41, no. 3-4, pp. 321–339, 2011. [13] Y. L. Xin, “Mean curvature flow with bounded Gauss image,” Results in Mathematics,vol.59,no.3-4,pp.415–436,2011. [14] Y. L. Xin, Minimal Submanifolds and Related Topics,World Scientific Publishing, 2003. [15] A.-M. Li, R. Xu, U. Simon, and F. Jia, Affine Bernstein Problems and Monge-Ampere` Equations, World Scientific, Singapore, 2010. [16] A. M. Li and R. W. Xu, “A rigidity theorem for an affine Kahler-¨ Ricci flat graph,” Results in Mathematics,vol.56,no.1–4,pp.141– 164, 2009. [17] A.-M. Li and R. Xu, “A cubic form differential inequality with applications to affine Kahler-Ricci¨ flat manifolds,” Results in Mathematics,vol.54,no.3-4,pp.329–340,2009. [18] R. W. Xu and R. L. Huang, “On the rigidity theorems for Lagrangian translating solitons in pseudo-Euclidean space I,” ActaMathematicaSinica(EnglishSeries),vol.29,no.7,pp.1369– 1380, 2013. [19] A. V.Pogorelov, The Minkowski Multidimensional Problem,John Wiley&Sons,NewYork,NY,USA,1978. [20] E. Calabi, “Improper affine hyperspheres of convex type and a generalization of a theorem by K. Jorgens,”¨ The Michigan Mathematical Journal,vol.5,pp.105–126,1958. Hindawi Publishing Corporation Abstract and Applied Analysis Volume 2014, Article ID 985080, 10 pages http://dx.doi.org/10.1155/2014/985080

Research Article The Generalized Weaker (𝛼-𝜙-𝜑)-Contractive Mappings and Related Fixed Point Results in Complete Generalized Metric Spaces

Maryam A. Alghamdi,1 Chi-Ming Chen,2 and Erdal KarapJnar3,4

1 Department of Mathematics, Sciences Faculty for Girls, King Abdulaziz University, P.O. Box 4087, Jeddah 21491, Saudi Arabia 2 Department of Applied Mathematics, National Hsinchu University of Education, Taiwan 3 Department of Mathematics, Atilim University, Incek, 06836 Ankara, Turkey 4 Nonlinear Analysis and Applied Mathematics Research Group (NAAM), King Abdulaziz University, Jeddah, Saudi Arabia

Correspondence should be addressed to Maryam A. Alghamdi; [email protected]

Received 6 February 2014; Accepted 7 May 2014; Published 26 May 2014

Academic Editor: Qamrul Hasan Ansari

Copyright © 2014 Maryam A. Alghamdi et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We introduce the notion of generalized weaker (𝛼-𝜙-𝜑)-contractive mappings in the context of generalized metric space. We investigate the existence and uniqueness of fixed point of such mappings. Some consequences on existing fixed point theorems are also derived. The presented results generalize, unify, and improve several results in the literature.

1. Introduction and Preliminaries On the other hand, Samet et al. [5]introducedthenotion of 𝛼-𝜓 contraction mappings and proved the existence and In [1], Branciari introduced the notion of generalized metric uniqueness of such mappings in complete metric space. space by weakening the triangular inequality of metric The results of this paper are very impressive since several assumption with quadrilateral inequality. The author [1] existing results derived from the main theorem of Samet characterized and proved the analog of famous Banach fixed et al. [5] quiet easily. Later, a number of authors have point theorem in the setting of generalized metric space. appreciated these results and have used this technique to get Although the theorem of Branciari [1]iscorrect,theproofs further generalization via 𝛼-𝜓 contraction mappings; see, for had gaps [2] since the topology of generalized metric space example, [6–10]. is not strong enough as the topology of metric space. The In this paper, we introduce the generalized weaker 𝛼- disadvantages of generalized metric space can be listed as 𝜓 contraction mappings in the setting of generalized met- follows: ric spaces. Consequently, we investigate the existence and (𝑤1) generalized metric need not be continuous; uniqueness of fixed point by caring the problems (𝑤1)–(𝑤3) (𝑤2) a convergent sequence in generalized metric space mentioned above. need not be Cauchy; Let us recall basic definitions and notations and interest- (𝑤3) generalized metric space need not be Hausdorff, and ing results that will be in the sequel. Ψ 𝜓:[0,∞)→[0,∞) hence the uniqueness of limits cannot be guaranteed. Let be the family of functions satisfying the following conditions: Despite the weakness of the topology of generalized metric space, in [3, 4], the authors suggested some techniques to get a (unique) fixed point in such spaces. (i) 𝜓 is nondecreasing; 2 Abstract and Applied Analysis

(ii) there exist 𝑘0 ∈ N and 𝑎 ∈ (0, 1) and a convergent Proposition 5 (see [3]). Suppose that {𝑥𝑛} is a Cauchy ∞ series of nonnegative terms ∑𝑘=1 V𝑘 such that sequence in a GMS (𝑋, 𝑑) with lim𝑛→∞𝑑(𝑥𝑛,𝑢) =,where 0 𝑢∈𝑋.Thenlim𝑛→∞𝑑(𝑥𝑛,𝑧) = 𝑑(𝑢,𝑧) for all 𝑧∈𝑋.In 𝑘+1 𝑘 {𝑥 } 𝑧 𝑧 =𝑢̸ 𝜓 (𝑡) ≤𝑎𝜓 (𝑡) + V𝑘, (1) particular, the sequence 𝑛 does not converge to if .

+ Afunction𝜙:[0,∞)→[0,∞)is said to be a Meir- for 𝑘≥𝑘0 and any 𝑡∈R . Keeler function [16] if, for each 𝜂>0, there exists 𝛿>0 𝑡∈[0,∞) 𝜂≤𝑡<𝜂+𝛿 𝜙(𝑡) <𝜂 In the literature such functions are called either Bianchini- such that for with ,wehave . Grandolfi gauge functions (see, e.g.,11 [ –13]) or (𝑐)- Such mapping has been improved and used by several authors comparison functions (see, e.g., [14]). [17, 18]. In what follows we recall the notion of weaker Meir- Keeler function. Lemma 1 (see, e.g., [14]). If 𝜓∈Ψ, then the following hold: Definition 6 (see, e.g., [19]). We call 𝜙:[0,∞)→[0,∞) 𝑛 + 𝜂>0 (i) (𝜓 (𝑡))𝑛∈N converges to 0 as 𝑛→∞for all 𝑡∈R ; a weaker Meir-Keeler function if for each , there exists 𝛿>0 𝑡∈[0,∞) 𝜂≤𝑡<𝜂+𝛿 + such that for with , there exists 𝜓(𝑡) <𝑡 𝑡∈R 𝑛0 (ii) ,forany ; 𝑛0 ∈ N such that 𝜙 (𝑡) < 𝜂. (iii) 𝜓 is continuous at 0; Let Φ be the class of all nondecreasing function 𝜙: ∞ 𝑘 + (iv) the series ∑𝑘=1 𝜓 (𝑡) converges for any 𝑡∈R . [0, ∞) → [0, ∞) satisfying the following conditions: (𝜙 ) 𝜙 : [0, ∞) → [0, ∞) In the following, we recall the notion of generalized 1 is a weaker Meir-Keeler metric spaces. function; (𝜙2) 0 < 𝜙(𝑡) <𝑡 for all 𝑡>0, 𝜙(0) = 0; Definition 2 (see [1]). Let 𝑋 be a nonempty set and let 𝑑:𝑋× 𝑛 (𝜙3) for all 𝑡 ∈ (0, ∞), {𝜙 (𝑡)}𝑛∈N is decreasing; 𝑋→[0,∞]satisfy the following conditions for all 𝑥, 𝑦 ∈𝑋 (𝜙 ) 𝑡 =𝛾 𝜙(𝑡 )≤𝛾 and all distinct 𝑢, V ∈𝑋each of which is different from 𝑥 and 4 if lim𝑛→∞ 𝑛 ,thenlim𝑛→∞ 𝑛 . 𝑦 : Let Θ be the class of functions 𝜑 : [0, ∞) → [0, ∞) satisfying the following conditions: (GMS1) 𝑑(𝑥,𝑦)=0 iff 𝑥=𝑦, (𝜑1)𝜑is continuous; ( 2) 𝑑(𝑥,𝑦)=𝑑(𝑦,𝑥), GMS (2) (𝜑2) 𝜑(𝑡) >0 for 𝑡>0and 𝜑(0) = 0. (GMS3) 𝑑(𝑥,𝑦)≤𝑑(𝑥, 𝑢) +𝑑(𝑢, V) +𝑑(V,𝑦). By using the auxiliary functions, defined above, Chen and Sun [19] proved the following theorem. Then, the map 𝑑 is called generalized metric. Here, the pair (𝑋, 𝑑) is called a generalized metric space and abbreviated as Theorem 7. Let (𝑋, 𝑑) be a Hausdorff and complete general- GMS. ized metric space, and let 𝑓:𝑋 →𝑋be a function satisfying 𝑑 (𝑓𝑥, 𝑓𝑦) ≤𝜙(𝑑 (𝑥,)) 𝑦 −𝜑(𝑑 (𝑥,)) 𝑦 In the above definition, if 𝑑 satisfies only (GMS1) and (4) (GMS2), then it is called semimetric (see, e.g., [15]). for all 𝑥, 𝑦 ∈𝑋 and 𝜙∈Φ, 𝜑∈Θ.Then𝑓 has a periodic point 𝑝 The concepts of convergence, Cauchy sequence, and 𝜇 in 𝑋;thatis,thereexists𝜇∈𝑋such that 𝜇=𝑓 𝜇 for some completeness in a GMS are defined as follows. 𝑝∈N. {𝑥 } (𝑋, 𝑑) Definition 3. (1) A sequence 𝑛 in a GMS is GMS Another interesting auxiliary function, 𝛼-admissible, was 𝑥 𝑑(𝑥 ,𝑥) → 0 𝑛→∞ convergent to a limit if and only if 𝑛 as . defined by Samet et al. [5]. (2) A sequence {𝑥𝑛} in a GMS (𝑋, 𝑑) is GMS Cauchy if and only if for every 𝜀>0there exists positive integer 𝑁(𝜀) Definition 8 (see [5]). For a nonempty set 𝑋,let𝑇:𝑋 →𝑋 such that 𝑑(𝑥𝑛,𝑥𝑚)<𝜀for all 𝑛>𝑚>𝑁(𝜀). and 𝛼:𝑋×𝑋→ [0,∞)be mappings. We say that 𝑇 is (3) A GMS (𝑋, 𝑑) is called complete if every GMS Cauchy 𝛼-admissible if sequence in 𝑋 is GMS convergent. 𝛼(𝑥,𝑦)≥1󳨐⇒𝛼(𝑇𝑥,𝑇𝑦)≥1, (5) The following assumption was suggested by Wilson [15]to 𝑥, 𝑦 ∈𝑋 replace the triangle inequality with the weakened condition. for all . (𝑊) Foreachpairof(distinct)points𝑢, V there is a Example 9. Let 𝑋=[2,∞)and 𝑇:𝑋by →𝑋 𝑇𝑥 = (𝑥 + number 𝑟𝑢,V >0such that, for every 𝑧∈𝑋, 1)/(𝑥 − 1).Define𝛼(𝑥, 𝑦) : 𝑋 × 𝑋 →[0,∞) and 𝑥+1 𝑟𝑢,V <𝑑(𝑢,) 𝑧 +𝑑(𝑧, V) . (3) 𝑒 if 𝑥≥𝑦, 𝛼(𝑥,𝑦)={ (6) 0 if otherwise. Proposition 4 (see [3]). In a semimetric space, the assumption (𝑊) is equivalent to the assertion that limits are unique. Then 𝑇 is 𝛼-admissible. Abstract and Applied Analysis 3

Example 10. Let 𝑋=R and 𝑇:𝑋.Define →𝑋 𝛼(𝑥, 𝑦) : For the uniqueness, Karapınar [20](seealso[21]) added 𝑥+1 𝑋×𝑋 → [0,∞)by 𝑇𝑥=𝑒 and the following additional conditions.

2 𝑥 if 𝑥≥𝑦, (𝑈) 𝑥, 𝑦 ∈ (𝑇) 𝛼(𝑥, 𝑦) ≥1 𝛼(𝑥,𝑦)={ (7) For all Fix ,wehave ,where 0 if otherwise. Fix(𝑇) denotes the set of fixed points of 𝑇.

Then 𝑇 is 𝛼-admissible. (𝐻) For all 𝑥, 𝑦 ∈ Fix(𝑇), there exists 𝑧∈𝑋such that 𝛼(𝑥, 𝑧) ≥1 and 𝛼(𝑦, 𝑧). ≥1 Some interesting examples of such mappings were given in [5]. Theorem 15. Adding condition (𝑈) to the hypotheses of The notion of an 𝛼-𝜓 contractive mapping is defined in Theorem 13 (resp., Theorem 14), one obtains that 𝑢 is the unique the following way. fixed point of 𝑇.

Definition 11 (see [5]). Let (𝑋, 𝑑) be a metric space and let Theorem 16. Adding conditions (𝐻)and(𝑊)tothehypothe- 𝑇:𝑋be →𝑋 a given mapping. We say that 𝑇 is an 𝛼-𝜓 ses of Theorem 13 (resp., Theorem 14), one obtains that 𝑢 is the contractive mapping if there exist two functions 𝛼:𝑋×𝑋→ unique fixed point of 𝑇. [0, ∞) and 𝜓∈Ψsuch that Corollary 17. Adding condition (𝐻) to the hypotheses of 𝛼(𝑥,𝑦)𝑑(𝑇𝑥,𝑇𝑦)≤𝜓(𝑑(𝑥,𝑦)), ∀𝑥,𝑦∈𝑋. (8) Theorem 13 (resp., Theorem 14) and assuming that (𝑋, 𝑑) is Hausdorff, one obtains that 𝑢 istheuniquefixedpointof𝑇. Clearly, any contractive mapping, that is, a mapping 𝛼 𝜓 satisfying the Banach contraction, is an - contractive In this paper, we define the notion of weaker generalized mapping with 𝛼(𝑥, 𝑦) =1 for all 𝑥, 𝑦 ∈𝑋 and 𝜓(𝑡) =, 𝑘𝑡 𝛼 𝜓 𝑘 ∈ (0, 1) - contractive mappings and prove some fixed point results . in the setting of generalized metric spaces by using such Very recently, Karapınar [20]gavetheanalogofthe 𝛼 𝜓 mappings. We state some examples to illustrate the validity notion of an - contractive mapping, in the context of of the main results of this paper. generalized metric spaces as follows.

Definition 12. Let (𝑋, 𝑑) be a generalized metric space and let 2. Main Results 𝑇:𝑋be →𝑋 a given mapping. We say that 𝑇 is an 𝛼-𝜓 contractive mapping if there exist two functions 𝛼:𝑋×𝑋→ In this section, we will state and prove our main results. [0, ∞) and 𝜓∈Ψsuch that We give an extension of the notion of 𝛼-𝜓 contractive mappings, in the context of generalized metric space as 𝛼 (𝑥,) 𝑦 𝑑 (𝑇𝑥, 𝑇𝑦) ≤𝜓(𝑑 (𝑥,)) 𝑦 ,∀𝑥,𝑦∈𝑋. (9) follows.

Karapınar [20]alsostatedthefollowingfixedpoint Definition 18. Let (𝑋, 𝑑) be a generalized metric space and let theorems. 𝑇:𝑋be →𝑋 a given mapping. We say that 𝑇 is a (𝛼-𝜙- 𝜑)-contractive mapping of type I if there exist functions 𝛼: Theorem 13. Let (𝑋, 𝑑) be a complete generalized metric space 𝑋×𝑋 → [0,∞), 𝜑∈Θ,and𝜙∈Φsuch that and let 𝑇:𝑋be →𝑋 an 𝛼-𝜓 contractive mapping. Suppose that 𝛼 (𝑥, 𝑦) 𝑑 (𝑇𝑥, 𝑇𝑦) ≤ 𝜙 (𝑀 (𝑥, 𝑦))−𝜑(𝑀(𝑥, (10) (i) 𝑇 is 𝛼-admissible;

(ii) there exists 𝑥0 ∈𝑋such that 𝛼(𝑥0,𝑇𝑥0)≥1and 2 for all 𝑥, 𝑦 ∈𝑋,where 𝛼(𝑥0,𝑇 𝑥0)≥1; (iii) 𝑇 is continuous. 𝑀 (𝑥,) 𝑦 = max {𝑑 (𝑥,) 𝑦 ,𝑑(𝑥, 𝑇𝑥) ,𝑑(𝑦, 𝑇𝑦)} . (11) Then there exists a 𝑢∈𝑋such that 𝑇𝑢 =𝑢. Definition 19. Let (𝑋, 𝑑) be a generalized metric space and let Theorem 14. Let (𝑋, 𝑑) be a complete generalized metric space 𝑇:𝑋 →𝑋 𝑇 𝛼 𝜙 𝑇:𝑋 →𝑋 𝛼 𝜓 be a given mapping. We say that is a ( - - and let be an - contractive mapping. Suppose 𝜑)-contractive mapping of type II if there exist functions 𝛼: that 𝑋×𝑋 → [0,∞), 𝜑∈Θ,and𝜙∈Φsuch that (i) 𝑇 is 𝛼-admissible; 𝛼 (𝑥,) 𝑦 𝑑 (𝑇𝑥, 𝑇𝑦) ≤𝜙(𝑁 (𝑥,)) 𝑦 −𝜑(𝑁 (𝑥,)) 𝑦 (ii) there exists 𝑥0 ∈𝑋such that 𝛼(𝑥0,𝑇𝑥0)≥1and (12) 2 𝛼(𝑥0,𝑇 𝑥0)≥1;

(iii) if {𝑥𝑛} is a sequence in 𝑋 such that 𝛼(𝑥𝑛,𝑥𝑛+1)≥1for for all 𝑥, 𝑦 ∈𝑋,where all 𝑛 and 𝑥𝑛 →𝑥∈𝑋as 𝑛→∞,then𝛼(𝑥𝑛,𝑥)≥ 1 for all 𝑛. 𝑑 (𝑥, 𝑇𝑥) +𝑑(𝑦,𝑇𝑦) 𝑁 (𝑥,) 𝑦 = max {𝑑 (𝑥,) 𝑦 , } . (13) Then there exists a 𝑢∈𝑋such that 𝑇𝑢 =𝑢. 2 4 Abstract and Applied Analysis

Next, we introduce the notion of triangular 𝛼-admissible If 𝑀(𝑥𝑛,𝑥𝑛−1)=𝑑(𝑥𝑛,𝑥𝑛+1),thenby(15)andpropertyofthe as follows. function 𝜑,inequality(20)turnsinto

Definition 20. Let 𝑇:𝑋and →𝑋 𝛼:𝑋×𝑋→ [0,∞).The 𝑑(𝑥𝑛+1,𝑥𝑛)≤𝜙(𝑀(𝑥𝑛,𝑥𝑛−1)) − 𝜑 (𝑀𝑛 (𝑥 ,𝑥𝑛−1)) mapping 𝑇 is said to be weak triangular 𝛼-admissible if for all 𝑥∈𝑋,onehas =𝜙(𝑑(𝑥𝑛,𝑥𝑛+1)) − 𝜑 (𝑑𝑛 (𝑥 ,𝑥𝑛+1)) (22) 𝛼 (𝑥, 𝑇𝑥) ≥1, 𝛼(𝑇𝑥,𝑇2𝑥) ≥ 1 󳨐⇒ 𝛼 (𝑥,2 𝑇 𝑥) ≥ 1. <𝜙(𝑑(𝑥𝑛,𝑥𝑛+1)) . (14) 𝑛 Since {𝜙 (𝑡)} is decreasing, the inequality above yields a Now, we state the first fixed point theorem. contradiction. Hence, we conclude that 𝑀(𝑥𝑛,𝑥𝑛−1)= 𝑑(𝑥𝑛,𝑥𝑛−1) and (20)becomes Theorem 21. Let (𝑋, 𝑑) be a complete generalized metric space and let 𝑇:𝑋be →𝑋 a (𝛼-𝜙-𝜑)-contractive mapping of type 𝑑(𝑥𝑛+1,𝑥𝑛)≤𝜙(𝑑(𝑥𝑛,𝑥𝑛−1)) , (23) I. Suppose that (i) 𝑇 is weak triangular 𝛼-admissible; for all 𝑛≥1. Recursively, we derive that

(ii) there exists 𝑥0 ∈𝑋such that 𝛼(𝑥0,𝑇𝑥0)≥1; 𝑛 𝑑(𝑥𝑛+1,𝑥𝑛)≤𝜙 (𝑑 1(𝑥 ,𝑥0)) , ∀𝑛 ≥ 1. (24) (iii) 𝑇 is continuous. 𝑛 Then, 𝑇 has a fixed point 𝑢∈𝑋;thatis𝑇𝑢 =𝑢. Owing to the fact that the sequence {𝜙 (𝑑(𝑥0,𝑥1))}𝑛∈N is decreasing, it converges to some 𝜂≥0.Wewillshowthat𝜂= Proof. Due to statement (ii) of the theorem, there exists a 0.Suppose,onthecontrary,that𝜂>0. Taking the definition point 𝑥0 ∈𝑋such that 𝛼(𝑥0,𝑇𝑥0)≥1and 𝛼(𝑥0,𝑇𝑥0)≥1. of weaker Meir-Keeler function 𝜙 into account, there exists 𝑛+1 First, we define a sequence {𝑥𝑛} in 𝑋 by 𝑥𝑛+1 =𝑇𝑥𝑛 =𝑇 𝑥0 𝛿>0such that for 𝑥0,𝑥1 ∈𝑋with 𝜂≤𝑑(𝑥0,𝑥1)<𝛿+𝜂,and 𝑛0 for all 𝑛≥0.Noticethatif𝑥𝑛 =𝑥𝑛 +1 for some 𝑛0, then the there exists 𝑛0 ∈ N such that 𝜙 (𝑑(𝑥0,𝑥1)) < 𝜂. Regarding 0 0 𝑛 proof is completed. Indeed, we have 𝑢=𝑥𝑛 =𝑥𝑛 +1 =𝑇𝑥𝑛 = lim𝑛→∞𝜙 (𝑑(𝑥0,𝑥1)) = 𝜂, there exists 𝑝0 ∈ N such that 0 0 0 𝑝 𝑇𝑢.Thus,fortherestoftheproof,weassumethat 𝜂≤𝜙(𝑑(𝑥0,𝑥1)) < 𝛿 + 𝜂,forall𝑝≥𝑝0.Hence,wededuce 𝑝0+𝑛0 that 𝜙 (𝑑(𝑥0,𝑥1)) < 𝜂, which is a contradiction. Thus, 𝑥𝑛 =𝑥̸ 𝑛+1 ∀𝑛. (15) 𝑛 lim𝑛→∞𝜙 (𝑑(𝑥0,𝑥1)) = 0, and hence Owing to the fact that 𝑇 is 𝛼-admissible, we derive that 𝑑(𝑥 ,𝑥 )=0. 𝑛→∞lim 𝑛+1 𝑛 (25) 𝛼(𝑥0,𝑥1)=𝛼(𝑥0,𝑇𝑥0)≥1 (16) 󳨐⇒ 𝛼 ( 𝑇𝑥 0,𝑇𝑥1)=𝛼(𝑥1,𝑥2)≥1. Regarding (10)and(19), we deduce that

Utilizing the expression above, we find that 𝑑(𝑥𝑛+2,𝑥𝑛)=𝑑(𝑇𝑥𝑛+1,𝑇𝑥𝑛−1)

𝛼(𝑥𝑛,𝑥𝑛+1)≥1, ∀𝑛=0,1,.... (17) ≤𝛼(𝑥𝑛+1,𝑥𝑛−1)𝑑(𝑇𝑥𝑛+1,𝑇𝑥𝑛−1) Since 𝑇 is a weak triangular 𝛼-admissible mapping, we obtain ≤𝜙(𝑀(𝑥 ,𝑥 )) − 𝜑 (𝑀 (𝑥 ,𝑥 )) , that 𝑛+1 𝑛−1 𝑛+1 𝑛−1 (26) 𝛼 (𝑥0,𝑥2) ≥1. (18) 𝑥 2 for all 𝑛≥1,where Since 𝛼(𝑥0,𝑇0 )≥1and 𝛼(𝑇𝑥0,𝑇 𝑥0)=𝛼(𝑥0,𝑥2),iteratively, we conclude that 𝑀(𝑥𝑛+1,𝑥𝑛−1) 𝛼 (𝑥𝑛,𝑥𝑛+𝑘) ≥ 1, ∀𝑛, 𝑘 = 0, 1, . (19) = max {𝑑 𝑛+1(𝑥 ,𝑥𝑛−1),𝑑(𝑥𝑛+1,𝑇𝑥𝑛+1),𝑑(𝑥𝑛−1,𝑇𝑥𝑛−1)} Taking (10)and(17) into account, we observe that = {𝑑 (𝑥 ,𝑥 ),𝑑(𝑥 ,𝑥 ),𝑑(𝑥 ,𝑥 )} . 𝑑(𝑥 ,𝑥 )=𝑑(𝑇𝑥,𝑇𝑥 ) max 𝑛+1 𝑛−1 𝑛+1 𝑛+2 𝑛−1 𝑛 𝑛+1 𝑛 𝑛 𝑛−1 (27)

≤𝛼(𝑥𝑛,𝑥𝑛−1)𝑑(𝑇𝑥𝑛,𝑇𝑥𝑛−1) (20) If 𝑀(𝑥𝑛,𝑥𝑛−1)=𝑑(𝑥𝑛−1,𝑥𝑛+1) then inequality (26)turns ≤𝜙(𝑀(𝑥𝑛,𝑥𝑛−1)) − 𝜑 (𝑀𝑛 (𝑥 ,𝑥𝑛−1)) , into for all 𝑛≥1,where 𝑑(𝑥𝑛+2,𝑥𝑛)≤𝜙(𝑑(𝑥𝑛+1,𝑥𝑛−1)) − 𝜑 (𝑑𝑛+1 (𝑥 ,𝑥𝑛−1)) (28) 𝑀(𝑥𝑛,𝑥𝑛−1) ≤𝜙(𝑑(𝑥𝑛+1,𝑥𝑛−1)) = {𝑑 (𝑥 ,𝑥 ),𝑑(𝑥 ,𝑇𝑥 ),𝑑(𝑥 ,𝑇𝑥 )} max 𝑛 𝑛−1 𝑛 𝑛 𝑛−1 𝑛−1 𝑛≥1 (21) for all . By repeating the same argument, inequality (15) = max {𝑑 𝑛(𝑥 ,𝑥𝑛−1),𝑑(𝑥𝑛,𝑥𝑛+1),𝑑(𝑥𝑛−1,𝑥𝑛)} implies that 𝑛 = max {𝑑 𝑛(𝑥 ,𝑥𝑛−1),𝑑(𝑥𝑛,𝑥𝑛+1)} . 𝑑(𝑥𝑛+2,𝑥𝑛)≤𝜙 (𝑑 2(𝑥 ,𝑥0)) , ∀𝑛 ≥ 1. (29) Abstract and Applied Analysis 5

𝑛 Due to the fact that the sequence {𝜙 (𝑑(𝑥0,𝑥2))}𝑛∈N is If 𝑀(𝑥𝑚,𝑥𝑚−1)=𝑑(𝑥𝑚,𝑥𝑚+1),inequalities(34)and(23) decreasing, we conclude that become 𝑑(𝑥 ,𝑥 )=0, 𝑑(𝑥 ,𝑥 )=𝑑(𝑇𝑥,𝑥 )=𝑑(𝑇𝑥 ,𝑥 ) 𝑛→∞lim 𝑛+2 𝑛 (30) 𝑛+1 𝑛 𝑛 𝑛 𝑚 𝑚 =𝑑(𝑇𝑥 ,𝑇𝑥 ) by following the lines at the proof of (25). 𝑚 𝑚−1 If either 𝑀(𝑥𝑛,𝑥𝑛−1)=𝑑(𝑥𝑛−1,𝑥𝑛) or 𝑀(𝑥𝑛+1,𝑥𝑛+2)= ≤𝛼(𝑥𝑚,𝑥𝑚−1)𝑑(𝑇𝑥𝑚,𝑇𝑥𝑚−1) 𝑑(𝑥𝑛+1,𝑥𝑛+2),theninequality(26) becomes either (37) ≤ 𝜙 (𝑑𝑚 (𝑥 ,𝑥𝑚+1)) − 𝜑 (𝑑𝑚 (𝑥 ,𝑥𝑚+1)) 𝑑(𝑥𝑛+2,𝑥𝑛)≤𝜙(𝑑(𝑥𝑛−1,𝑥𝑛)) − 𝜑 (𝑑𝑛−1 (𝑥 ,𝑥𝑛)) (31) ≤ 𝜙 (𝑑𝑚 (𝑥 ,𝑥𝑚+1)) <𝜙(𝑑(𝑥𝑛−1,𝑥𝑛)) 𝑚−𝑛+1 or ≤𝜙 (𝑑 𝑛+1(𝑥 ,𝑥𝑛)) .

𝑑(𝑥𝑛+2,𝑥𝑛)≤𝜙(𝑑(𝑥𝑛+1,𝑥𝑛+2)) − 𝜑 (𝑑𝑛+1 (𝑥 ,𝑥𝑛+2)) Due to (𝜙2),inequalities(36)and(37)yieldthat (32) < 𝜙 (𝑑𝑛+1 (𝑥 ,𝑥𝑛+2)) , 𝑚−𝑛 𝑑(𝑥𝑛+1,𝑥𝑛)≤𝜙 (𝑑 𝑛+1(𝑥 ,𝑥𝑛)) < 𝑑 𝑛+1(𝑥 ,𝑥𝑛), for all 𝑛≥1.Letting𝑛→∞in any of the cases, (31)or(32), (38) 𝑑(𝑥 ,𝑥 )≤𝜙𝑚−𝑛+1 (𝑑 (𝑥 ,𝑥 )) < 𝑑 (𝑥 ,𝑥 ), together with (25), we have 𝑛+1 𝑛 𝑛+1 𝑛 𝑛+1 𝑛

𝑑(𝑥 ,𝑥 )≤ 𝜙(𝑑(𝑥 ,𝑥 )) ≤ 0 {𝑥𝑛} 𝑛→∞lim 𝑛+2 𝑛 𝑛→∞lim 𝑛−1 𝑛 which is a contradiction. Hence has no periodic point. In what follows we will prove that the sequence {𝑥𝑛} is or (33) Cauchy by standard technique. Suppose, on the contrary, that there exists 𝜀>0such that for any 𝑘∈N,thereare 𝑑 (𝑥 ,𝑥 ) ≤ 𝜙 (𝑑 (𝑥 ,𝑥 )) ≤0. 𝑚(𝑘), 𝑛(𝑘) ∈ N 𝑛(𝑘) > 𝑚(𝑘) >𝑘 𝑛→∞lim 𝑛+2 𝑛 𝑛→∞lim 𝑛+1 𝑛+2 with satisfying

Let 𝑥𝑛 =𝑥𝑚 for some 𝑚, 𝑛 ∈ N with 𝑚 =𝑛̸.Withoutloss 𝑑(𝑥𝑚(𝑘),𝑥𝑛(𝑘))≥𝜀. (39) 𝑚−𝑛 𝑛 of generality, assume that 𝑚>𝑛.Thus,𝑥𝑚 =𝑇 (𝑇 𝑥0)= 𝑛 𝑇 𝑥0 =𝑥𝑛. Regarding (15), we consider now Furthermore, corresponding to 𝑚(𝑘),onecanchoose𝑛(𝑘) in a way that it is the smallest integer 𝑛(𝑘) > 𝑚(𝑘) with 𝑑(𝑥 ,𝑥 )=𝑑(𝑇𝑥,𝑥 )=𝑑(𝑇𝑥 ,𝑥 ) 𝑛+1 𝑛 𝑛 𝑛 𝑚 𝑚 𝑑(𝑥𝑚(𝑘),𝑥𝑛(𝑘))≥𝜀.Consequently,wehave𝑑(𝑥𝑚(𝑘),𝑥𝑛(𝑘)−1)< 𝜀. Consider =𝑑(𝑇𝑥𝑚,𝑇𝑥𝑚−1) (34) 𝜀≤𝑑(𝑥𝑛(𝑘),𝑥𝑚(𝑘)) ≤𝛼(𝑥𝑚,𝑥𝑚−1)𝑑(𝑇𝑥𝑚,𝑇𝑥𝑚−1) ≤𝑑(𝑥𝑛(𝑘),𝑥𝑛(𝑘)−2)+𝑑(𝑥𝑛(𝑘)−2,𝑥𝑛(𝑘)−1)+𝑑(𝑥𝑛(𝑘)−1,𝑥𝑚(𝑘)) ≤𝜙(𝑀(𝑥𝑚,𝑥𝑚−1)) − 𝜑 (𝑀𝑚 (𝑥 ,𝑥𝑚−1)) , ≤𝑑(𝑥𝑛(𝑘),𝑥𝑛(𝑘)−2)+𝑑(𝑥𝑛(𝑘)−2,𝑥𝑛(𝑘)−1)+𝜀. where (40) 𝑀(𝑥𝑚,𝑥𝑚−1) Letting 𝑘→∞,wegetthat = max {𝑑 𝑚(𝑥 ,𝑥𝑚−1),𝑑(𝑥𝑚,𝑇𝑥𝑚),𝑑(𝑥𝑚−1,𝑇𝑥𝑚−1)} 𝑑 (𝑥𝑛(𝑘),𝑥𝑚(𝑘)) 󳨀→ 𝜀. (41) = max {𝑑 𝑚(𝑥 ,𝑥𝑚−1),𝑑(𝑥𝑚,𝑥𝑚+1),𝑑(𝑥𝑚−1,𝑥𝑚)} = {𝑑 (𝑥 ,𝑥 ),𝑑(𝑥 ,𝑥 )} . On the other hand, again by using the quadrilateral max 𝑚−1 𝑚 𝑚 𝑚+1 inequality, we find (35)

𝑑(𝑥𝑛(𝑘),𝑥𝑚(𝑘)) If 𝑀(𝑥𝑚,𝑥𝑚−1)=𝑑(𝑥𝑚−1,𝑥𝑚),thenfrom(34)and(23)we get that ≤𝑑(𝑥𝑛(𝑘),𝑥𝑛(𝑘)−1)+𝑑(𝑥𝑛(𝑘)−1,𝑥𝑚(𝑘)−1)

𝑑(𝑥𝑛+1,𝑥𝑛)=𝑑(𝑇𝑥𝑛,𝑥𝑛)=𝑑(𝑇𝑥𝑚,𝑥𝑚) +𝑑(𝑥𝑚(𝑘)−1,𝑥𝑚(𝑘)),

=𝑑(𝑇𝑥𝑚,𝑇𝑥𝑚−1) 𝑑(𝑥𝑛(𝑘)−1,𝑥𝑚(𝑘)−1) ≤𝛼(𝑥𝑚,𝑥𝑚−1)𝑑(𝑇𝑥𝑚,𝑇𝑥𝑚−1) ≤𝑑(𝑥 ,𝑥 )+𝑑(𝑥 ,𝑥 )+𝑑(𝑥 ,𝑥 ). (36) 𝑛(𝑘)−1 𝑛(𝑘) 𝑛(𝑘) 𝑚(𝑘) 𝑚(𝑘) 𝑚(𝑘)−1 ≤ 𝜙 (𝑑𝑚 (𝑥 ,𝑥𝑚−1)) − 𝜑 (𝑑𝑚 (𝑥 ,𝑥𝑚−1)) (42) 𝑛→∞ ≤ 𝜙 (𝑑𝑚 (𝑥 ,𝑥𝑚−1)) Letting , in the inequalities above, we get that 𝑚−𝑛 ≤𝜙 (𝑑 𝑛+1(𝑥 ,𝑥𝑛)) . 𝑑(𝑥𝑛(𝑘)−1,𝑥𝑚(𝑘)−1)󳨀→𝜀. (43) 6 Abstract and Applied Analysis

On account of (10), we have Proof. Following the proof of Theorem 21,weknowthatthe {𝑥 } 𝑥 =𝑇𝑥 𝑛≥0 𝑑(𝑥 ,𝑥 ) sequence 𝑛 defined by 𝑛+1 𝑛 for all converges 𝑛(𝑘) 𝑚(𝑘) for some 𝑢∈𝑋.Wewillshowthat𝑇𝑢 =𝑢.Suppose,on the contrary, that 𝑇𝑢 =𝑢̸ .From(17) and condition (iii), there =𝑑(𝑇𝑥𝑛(𝑘)−1,𝑇𝑥𝑚(𝑘)−1) exists a subsequence {𝑥𝑛(𝑘)} of {𝑥𝑛} such that 𝛼(𝑥𝑛(𝑘),𝑢)≥ 1 𝑘 ≤𝛼(𝑥𝑛(𝑘)−1,𝑥𝑚(𝑘)−1)𝑑(𝑇𝑥𝑛(𝑘)−1,𝑇𝑥𝑚(𝑘)−1) for all .Byapplyingthequadrilateralinequalitytogether with (10)and(15), for all 𝑘,wegetthat ≤𝜙(𝑀(𝑥𝑛(𝑘)−1,𝑥𝑚(𝑘)−1)) − 𝜑 (𝑀𝑛(𝑘)−1 (𝑥 ,𝑥𝑚(𝑘)−1)) , (44) 𝑑 (𝑢, 𝑇𝑢) where ≤𝑑(𝑢,𝑥𝑛(𝑘)+2)+𝑑(𝑥𝑛(𝑘)+2,𝑥𝑛(𝑘)+1)+𝑑(𝑥𝑛(𝑘)+1,𝑇𝑢)

𝑀(𝑥𝑛(𝑘)−1,𝑥𝑚(𝑘)−1) ≤𝑑(𝑢,𝑥𝑛(𝑘)+2)+𝑑(𝑥𝑛(𝑘)+2,𝑥𝑛(𝑘)+1)+𝑑(𝑇𝑥𝑛(𝑘),𝑇𝑢)

= max {𝑑 𝑛(𝑘)−1(𝑥 ,𝑥𝑚(𝑘)−1),𝑑(𝑥𝑛(𝑘)−1,𝑥𝑛(𝑘)), (45) ≤𝑑(𝑢,𝑥𝑛(𝑘)+2)+𝑑(𝑥𝑛(𝑘)+2,𝑥𝑛(𝑘)+1)

𝑑(𝑥𝑚(𝑘)−1,𝑥𝑚(𝑘))} . +𝛼(𝑥𝑛(𝑘),𝑢)𝑑(𝑇𝑥𝑛(𝑘),𝑇𝑢) Letting 𝑛→∞,in(44), and regarding definitions of ≤𝑑(𝑢,𝑥𝑛(𝑘)+2)+𝑑(𝑥𝑛(𝑘)+2,𝑥𝑛(𝑘)+1)+𝜙(𝑀(𝑥𝑛(𝑘),𝑢)), auxiliary functions 𝜙, 𝜑 and (45), we conclude that (49) 𝜀≤𝜀−𝜑(𝜀) , (46) where which yields that 𝜑(𝜀). =0 By definition of 𝜑, we derive that 𝜀=0 {𝑥𝑛} , which is a contradiction. Hence, we conclude that 𝑀(𝑥𝑛(𝑘),𝑢)=max {𝑑 𝑛(𝑘)(𝑥 ,𝑢),𝑑(𝑥𝑛(𝑘),𝑇𝑥𝑛(𝑘)),𝑑(𝑢, 𝑇𝑢)}. (𝑋, 𝑑) (𝑋, 𝑑) is a Cauchy sequence in .Since is complete, there (50) exists 𝑢∈𝑋such that 𝑑 (𝑥 ,𝑢) =0. Letting 𝑘→∞in the above equality and regarding that 𝑛→∞lim 𝑛 (47) the 𝜙 is an upper semicontinuous mapping, we find that Since 𝑇 is continuous, we obtain from (47)that 𝑑 (𝑢, 𝑇𝑢) ≤𝜙(𝑑 (𝑢, 𝑇𝑢)) . 𝑑(𝑥 ,𝑇𝑢)= 𝑑(𝑇𝑥 ,𝑇𝑢)=0. (51) 𝑛→∞lim 𝑛+1 𝑛→∞lim 𝑛 (48)

𝑛 It implies that from (𝜙2) From (47)and(48) we get immediately that lim𝑛→∞𝑇 𝑥0 = 𝑇𝑥 =𝑇𝑢 lim𝑛→∞ 𝑛 .TakingProposition 5 into account, we 𝑑 (𝑢, 𝑇𝑢) ≤𝜙(𝑑 (𝑢, 𝑇𝑢)) <𝑑(𝑢, 𝑇𝑢) , conclude that 𝑇𝑢 =𝑢. (52) The following result is deduced from the obvious inequal- which is a contradiction. Hence, we obtain that 𝑢 is a fixed ity 𝑁(𝑥, 𝑦) ≤ 𝑀(𝑥, 𝑦). point of 𝑇;thatis,𝑇𝑢 =𝑢.

Theorem 22. Let (𝑋, 𝑑) be a complete generalized metric space 𝑇:𝑋 →𝑋 𝛼 𝜙 𝜑 In the following theorem, we remove the semicontinuity and let be a ( - - )-contractive mapping of type 𝜙 II. Suppose that of by weakening the contractive mapping type. (i) 𝑇 is a weak triangular 𝛼-admissible; Theorem 24. Let (𝑋, 𝑑) be a complete generalized metric space and 𝑇:𝑋be →𝑋 a (𝛼-𝜙-𝜑)-contractive mapping of type II. (ii) there exists 𝑥0 ∈𝑋such that 𝛼(𝑥0,𝑇𝑥0)≥1; Suppose that (iii) 𝑇 is continuous. 𝑇 𝛼 Then there exists a 𝑢∈𝑋such that 𝑇𝑢 =𝑢. (i) is a weak triangular -admissible; (ii) there exists 𝑥0 ∈𝑋such that 𝛼(𝑥0,𝑇𝑥0)≥1; Theorem 23. Let (𝑋, 𝑑) be a complete generalized metric space and let 𝑇:𝑋be →𝑋 a (𝛼-𝜙-𝜑)-contractive mapping of type (iii) if {𝑥𝑛} is a sequence in 𝑋 such that 𝛼(𝑥𝑛,𝑥𝑛+1)≥1for I. Suppose that all 𝑛 and 𝑥𝑛 →𝑥∈𝑋as 𝑛→∞,then𝛼(𝑥𝑛,𝑥)≥ 1 for all 𝑛. (i) 𝑇 is a weak triangular 𝛼-admissible and 𝜙 is upper semicontinuous function; Then there exists a 𝑢∈𝑋such that 𝑇𝑢 =𝑢. (ii) there exists 𝑥0 ∈𝑋such that 𝛼(𝑥0,𝑇𝑥0)≥1; Proof. Following the proof of Theorem 21,weknowthatthe (iii) if {𝑥𝑛} is a sequence in 𝑋 such that 𝛼(𝑥𝑛,𝑥𝑛+1)≥1for sequence {𝑥𝑛} defined by 𝑥𝑛+1 =𝑇𝑥𝑛 for all 𝑛≥0converges all 𝑛 and 𝑥𝑛 →𝑥∈𝑋as 𝑛→∞,then𝛼(𝑥𝑛,𝑥)≥ 1 for some 𝑢∈𝑋.Wewillshowthat𝑇𝑢 =𝑢.Suppose,on for all 𝑛. the contrary, that 𝑇𝑢 =𝑢̸ .From(17) and condition (iii), there Then there exists a 𝑢∈𝑋such that 𝑇𝑢 =𝑢. exists a subsequence {𝑥𝑛(𝑘)} of {𝑥𝑛} such that 𝛼(𝑥𝑛(𝑘),𝑢)≥ 1 Abstract and Applied Analysis 7 for all 𝑘. By applying the quadrilateral inequality together Proof. The proof is analog of the proof of Theorem 25 with (10)and(15), for all 𝑘,wegetthat which will be concluded by using the reductio ad absurdum. Suppose, on the contrary, that V is another fixed point of 𝑇 𝑑 (𝑢, 𝑇𝑢) with V =𝑢̸ .Itisevidentthat𝛼(𝑢, V)=𝛼(𝑇𝑢,𝑇V). Now,dueto(12)and(𝜙2), we have ≤𝑑(𝑢,𝑥𝑛(𝑘)+2)+𝑑(𝑥𝑛(𝑘)+2,𝑥𝑛(𝑘)+1)+𝑑(𝑥𝑛(𝑘)+1,𝑇𝑢) 𝑑 (𝑢, V) ≤𝛼(𝑢, V) 𝑑 (𝑢, V) ≤𝑑(𝑢,𝑥𝑛(𝑘)+2)+𝑑(𝑥𝑛(𝑘)+2,𝑥𝑛(𝑘)+1)+𝑑(𝑇𝑥𝑛(𝑘),𝑇𝑢) =𝛼(𝑢, V) 𝑑 (𝑇𝑢,V 𝑇 ) ≤𝑑(𝑢,𝑥𝑛(𝑘)+2)+𝑑(𝑥𝑛(𝑘)+2,𝑥𝑛(𝑘)+1) ≤𝜙(𝑁 (𝑢, V)) −𝜑(𝑁 (𝑢, V)) +𝛼(𝑥 ,𝑢)𝑑(𝑇𝑥 ,𝑇𝑢) 𝑛(𝑘) 𝑛(𝑘) (58) =𝜙(𝑑 (𝑢, V)) −𝜑(𝑑 (𝑢, V)) ≤𝑑(𝑢,𝑥𝑛(𝑘)+2)+𝑑(𝑥𝑛(𝑘)+2,𝑥𝑛(𝑘)+1)+𝜙(𝑁(𝑥𝑛(𝑘),𝑢)), (53) =𝜙(𝑑 (𝑢, V)) where <𝑑(𝑢, V)

𝑁(𝑥𝑛(𝑘),𝑢) which is a contradiction, where

𝑑(𝑥𝑛(𝑘),𝑇𝑥𝑛(𝑘))+𝑑(𝑢, 𝑇𝑢) 𝑑 (𝑢, 𝑇𝑢) +𝑑(V,𝑇V) = {𝑑 (𝑥 ,𝑢), }. 𝑁 (𝑢, V) = max {𝑑 (𝑢, V) , }=𝑑(𝑢, V) . max 𝑛(𝑘) 2 2 (54) (59) 𝑢=V Letting 𝑘→∞intheaboveequalityandregarding(𝜙4), we Hence, . find that For the uniqueness, we can also consider the following 𝑑 (𝑢, 𝑇𝑢) 𝑑 (𝑢, 𝑇𝑢) condition. 𝑑 (𝑢, 𝑇𝑢) ≤𝜙( )≤ (55) 2 2 ∗ (𝐻 ) For all 𝑥, 𝑦 ∈ Fix(𝑇), there exists 𝑧∈𝑋such which is a contradiction. Hence, we obtain that 𝑢 is a fixed that 𝛼(𝑥, 𝑧) ≥1 and 𝛼(𝑦, 𝑧).Further, ≥1 point of 𝑇;thatis,𝑇𝑢 =𝑢. lim𝑛→∞𝑑(𝑧𝑛,𝑧𝑛+1)=0,where𝑧1 =𝑧and 𝑧𝑛+1 =𝑇𝑧𝑛 for 𝑛 = 1, 2, 3, .. Theorem 25. Adding condition (𝑈) to the hypotheses of ∗ Theorem 21 (resp., Theorem 23), one obtains that 𝑢 is the Theorem 27. Adding conditions (𝐻 )and(𝑊)tothehypothe- unique fixed point of 𝑇. ses of Theorem 21 (resp., Theorem 23), one obtains that 𝑢 is the unique fixed point of 𝑇. Proof. Inwhatfollowswewillshowthat𝑢 is a unique fixed ∗ point of 𝑇.Wewillusethereductio ad absurdum.LetV be Proof. Suppose that V is another fixed point of 𝑇.From(𝐻 ), another fixed point of 𝑇 with V =𝑢̸ . It is evident that 𝛼(𝑢, V)= there exists 𝑧∈𝑋such that 𝛼(𝑇𝑢,V 𝑇 ). 𝛼 (𝑢,) 𝑧 ≥1, 𝛼(V,𝑧) ≥1. Now,dueto(10)and(𝜙2), we have (60) 𝑑 (𝑢, V) ≤𝛼(𝑢, V) 𝑑 (𝑢, V) Since 𝑇 is 𝛼-admissible, from (60), we have 𝑛 𝑛 =𝛼(𝑢, V) 𝑑 (𝑇𝑢,V 𝑇 ) 𝛼(𝑢,𝑇 𝑧) ≥ 1, 𝛼V ( ,𝑇 𝑧) ≥ 1, ∀𝑛. (61) ≤𝜙(𝑀 (𝑢, V)) −𝜑(𝑀 (𝑢, V)) Define the sequence {𝑧𝑛} in 𝑋 by 𝑧𝑛+1 =𝑇𝑧𝑛 for all 𝑛≥0and (56) 𝑧 =𝑧 𝑛 =𝜙(𝑑 (𝑢, V)) −𝜑(𝑑 (𝑢, V)) 0 .From(61), for all ,wehave

≤𝜙(𝑑 (𝑢, V)) 𝑑(𝑢,𝑧𝑛+1)=𝑑(𝑇𝑢,𝑇𝑧𝑛)≤𝛼(𝑢,𝑧𝑛)𝑑(𝑇𝑢,𝑇𝑧𝑛) (62) <𝑑(𝑢, V) ≤𝜙(𝑀(𝑢,𝑧𝑛)) − 𝜑 (𝑀 (𝑢,𝑛 𝑧 )) , which is a contradiction, where where

𝑀 (𝑢, V) = max {𝑑 (𝑢, V) ,𝑑(𝑢, 𝑇𝑢) ,𝑑(V,𝑇V)} =𝑑(𝑢, V) . 𝑀(𝑢,𝑧𝑛)=max {𝑑 (𝑢,𝑛 𝑧 ),𝑑(𝑢, 𝑇𝑢) ,𝑑(𝑧𝑛,𝑇𝑧𝑛)} (57) (63) = max {𝑑 (𝑢,𝑛 𝑧 ),𝑑(𝑧𝑛,𝑇𝑧𝑛)} . Hence, 𝑢=V. If 𝑀(𝑢,𝑛 𝑧 )=𝑑(𝑧𝑛,𝑇𝑧𝑛) then by letting 𝑛→∞in (62)we Theorem 26. Adding condition (𝑈) to the hypotheses of get that Theorem 22 (resp., Theorem 24), one obtains that 𝑢 is the 𝑑(𝑧 ,𝑢)=0, unique fixed point of 𝑇. 𝑛→∞lim 𝑛 (64) 8 Abstract and Applied Analysis

due to the continuity of 𝜑, (𝜙4) and the fact that We omit the proof of Theorem 31, since it can be derived lim𝑛→∞𝑑(𝑧𝑛,𝑧𝑛+1)=0.If𝑀(𝑢,𝑛 𝑧 )=𝑑(𝑢,𝑧𝑛) then easily by following the lines in the proof of Theorem 21, (62)turnsinto analogously. 𝑑(𝑢,𝑧 )≤𝜙(𝑑(𝑢,𝑧)) − 𝜑 (𝑑 (𝑢, 𝑧 )) ≤ 𝜙 (𝑑 (𝑢, 𝑧 )) . 𝑛+1 𝑛 𝑛 𝑛 Theorem 32. Let (𝑋, 𝑑) be a complete generalized metric space (65) and let 𝑇:𝑋be →𝑋 a (𝛼-𝜙-𝜑)-contractive mapping of type Iteratively, by using inequality (62), we get that III. Suppose that 𝑛 𝑑 (𝑢,𝑛+1 𝑧 ) ≤𝜙 (𝑑 (𝑢,0 𝑧 )) , (66) (i) 𝑇 is 𝛼-admissible; for all 𝑛.Letting𝑛→∞in the above inequality, we obtain (ii) there exists 𝑥0 ∈𝑋such that 𝛼(𝑥0,𝑇x0)≥1and lim 𝑑(𝑧𝑛,𝑢)=0. 2 𝑛→∞ (67) 𝛼(𝑥0,𝑇 𝑥0)≥1; Similarly, one can show that (iii) if {𝑥𝑛} is a sequence in 𝑋 such that 𝛼(𝑥𝑛,𝑥𝑛+1)≥1for 𝑑 (𝑧 , V) =0. 𝑛 𝑥 →𝑥∈𝑋 𝑛→∞ 𝛼(𝑥 ,𝑥)≥ 1 𝑛→∞lim 𝑛 (68) all and 𝑛 as ,then 𝑛 for all 𝑛. Regarding (𝑊) together with (67)and(68), it follows that 𝑢= V.Thusweprovedthat𝑢 is the unique fixed point of 𝑇. Then there exists a 𝑢∈𝑋such that 𝑇𝑢 =𝑢. ∗ Theorem 28. Adding conditions (𝐻 )and(𝑊)tothehypothe- ses of Theorem 22 (resp., Theorem 24), one obtains that 𝑢 is the Proof. Following the proof of Theorem 21 (resp., {𝑥 } unique fixed point of 𝑇. Theorem 31), we know that the sequence 𝑛 defined by 𝑥𝑛+1 =𝑇𝑥𝑛 for all 𝑛≥0converges for some 𝑢∈𝑋.We The proof is the analog of the proof of Theorem 27;hence will show that 𝑇𝑢 =𝑢.Suppose,onthecontrary,that𝑇𝑢 =𝑢̸ . we omit it. From (17) and condition (iii), there exists a subsequence {𝑥𝑛(𝑘)} of {𝑥𝑛} such that 𝛼(𝑥𝑛(𝑘),𝑢)≥ 1for all 𝑘. By applying ∗ Corollary 29. Adding condition (𝐻 ) to the hypotheses of the quadrilateral inequality together with (10)and(15), for Theorem 21 (resp., Theorems 23, 22,and24) and assuming that all 𝑘,wegetthat (𝑋, 𝑑) is Hausdorff, one obtains that 𝑢 is the unique fixed point 𝑇 of . 𝑑 (𝑢, 𝑇𝑢) The proof is clear, and hence it is omitted. Indeed, ≤𝑑(𝑢,𝑥𝑛(𝑘)+2)+𝑑(𝑥𝑛(𝑘)+2,𝑥𝑛(𝑘)+1)+𝑑(𝑥𝑛(𝑘)+1,𝑇𝑢) Hausdorffness implies the uniqueness of the limit. Thus, the theorem above yields the conclusions. ≤𝑑(𝑢,𝑥𝑛(𝑘)+2)+𝑑(𝑥𝑛(𝑘)+2,𝑥𝑛(𝑘)+1)+𝑑(𝑇𝑥𝑛(𝑘),𝑇𝑢) ≤𝑑(𝑢,𝑥 )+𝑑(𝑥 ,𝑥 ) 3. Consequences 𝑛(𝑘)+2 𝑛(𝑘)+2 𝑛(𝑘)+1 +𝛼(𝑥 ,𝑢)𝑑(𝑇𝑥 ,𝑇𝑢) Now, we will show that many existing results in the literature 𝑛(𝑘) 𝑛(𝑘) canbededucedeasilyfromTheorems13 and 14. ≤𝑑(𝑢,𝑥𝑛(𝑘)+2)+𝑑(𝑥𝑛(𝑘)+2,𝑥𝑛(𝑘)+1)+𝜙(𝑑(𝑥𝑛(𝑘),𝑢)). Definition 30. Let (𝑋, 𝑑) be a generalized metric space and (70) let 𝑇:𝑋be →𝑋 a given mapping. We say that 𝑇 is a (𝛼- 𝜙-𝜑)-contractive mapping of type III if there exist functions 𝑘→∞ (𝜙 ) 𝛼:𝑋×𝑋 → [0,∞) 𝜑∈Θ 𝜙∈Φ Letting intheaboveequalityandregarding 4 , , ,and such that we find that 𝛼 (𝑥,) 𝑦 𝑑 (𝑇𝑥, 𝑇𝑦) ≤𝜙(𝑑 (𝑥,)) 𝑦 −𝜑(𝑑 (𝑥,)) 𝑦 (69) 𝑑 (𝑢, 𝑇𝑢) ≤ 𝜙 (𝑑 (𝑥 ,𝑢)) ≤0, for all 𝑥, 𝑦 ∈𝑋. 𝑛→∞lim 𝑛(𝑘) (71)

Now, we state the first fixed point theorem. which is a contradiction. Hence, we obtain that 𝑢 is a fixed 𝑇 𝑇𝑢 =𝑢 Theorem 31. Let (𝑋, 𝑑) be a complete generalized metric space point of ;thatis, . 𝑇:𝑋 →𝑋 𝛼 𝜙 𝜑 and let be a ( - - )-contractive mapping of type 𝑈 III. Suppose that Theorem 33. Adding condition ( ) to the hypotheses of Theorem 31 (resp., Theorem 32), one obtains that 𝑢 is the (i) 𝑇 is 𝛼-admissible; unique fixed point of 𝑇. (ii) there exists 𝑥0 ∈𝑋such that 𝛼(𝑥0,𝑇𝑥0)≥1and 2 𝑢 𝛼(𝑥0,𝑇 𝑥0)≥1; Proof. In what follows we will show that is a unique fixed point of 𝑇.Wewillusethereductio ad absurdum.LetV be (iii) 𝑇 is continuous. another fixed point of 𝑇 with V =𝑢̸ .Itisevidentthat𝛼(𝑢, V)= Then, 𝑇 has a fixed point 𝑢∈𝑋;thatis,𝑇𝑢 =𝑢. 𝛼(𝑇𝑢,V 𝑇 ). Abstract and Applied Analysis 9

Now,dueto(10)and(𝜙2), we have Authors’ Contribution 𝑑 (𝑢, V) ≤𝛼(𝑢, V) 𝑑 (𝑢, V) All authors contributed equally and significantly in writing this paper. All authors read and approved the final paper. =𝛼(𝑇𝑢,V 𝑇 ) 𝑑 (𝑇𝑢,V 𝑇 )

≤𝜙(𝑑 (𝑢, V)) −𝜑(𝑑 (𝑢, V)) (72) References

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[18] M. de la Sen, “Linking contractive self-mappings and cyclic Meir-Keeler contractions with Kannan self-mappings,” Fixed Point Theory and Applications, vol. 2010, Article ID 572057, 23 pages, 2010. [19] C. M. Chen and W.Y. Sun, “Periodic points and fixed points for the weaker (𝜙, 𝜑)-contractive mappings in complete generalized metric spaces,” Journal of Applied Mathematics,vol.2012,Article ID 856974, 7 pages, 2012. [20] E. Karapınar, “Discussion on 𝛼-𝜓 contractions on generalized metric spaces,” Abstract and Applied Analysis,vol.2014,Article ID 962784, 7 pages, 2014. [21] H. Aydi, E. Karapınar, and B. Samet, “Fixed points for generalized (alpha, psi)-contractions on generalized metric spaces,” Journal of Inequalities and Applications.Inpress. Hindawi Publishing Corporation Abstract and Applied Analysis Volume 2014, Article ID 568718, 18 pages http://dx.doi.org/10.1155/2014/568718

Research Article Coupled and Tripled Coincidence Point Results with Application to Fredholm Integral Equations

Marwan Amin Kutbi,1 Nawab Hussain,1 Jamal Rezaei Roshan,2 and Vahid Parvaneh3

1 DepartmentofMathematics,KingAbdulAzizUniversity,P.O.Box80203,Jeddah21589,SaudiArabia 2 Department of Mathematics, Qaemshahr Branch, Islamic Azad University, Qaemshahr, Iran 3 Department of Mathematics, Gilan-E-Gharb Branch, Islamic Azad University, Gilan-E-Gharb, Iran

Correspondence should be addressed to Vahid Parvaneh; [email protected]

Received 25 January 2014; Accepted 29 March 2014; Published 18 May 2014

Academic Editor: Ljubomir B. Ciri´ c´

Copyright © 2014 Marwan Amin Kutbi 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.

The aim of this paper is to define weak 𝛼-𝜓-𝜑-contractive mappings and to establish coupled and tripled coincidence point theorems forsuchmappingsdefinedon𝐺𝑏-metric spaces using the concept of rectangular 𝐺-𝛼-admissibility. As an application, we derive new coupled and tripled coincidence point results for weak 𝜓-𝜑-contractive mappings in partially ordered 𝐺𝑏-metric spaces. Our results are generalizations and extensions of some recent results in the literature. We also present an example as well as an application to nonlinear Fredholm integral equations in order to illustrate the effectiveness of our results.

1. Introduction and spaces (𝐺𝑏-metric spaces) and then they presented some basic 𝐺 Mathematical Preliminaries properties of 𝑏-metric spaces. The following is their definition of 𝐺𝑏-metric spaces. The concept of generalized metric space, ora 𝐺-metric space, was introduced by Mustafa and Sims. Definition 2 (see [2]). Let 𝑋 be a nonempty set and let 𝑠≥1be a given real number. Suppose that a mapping 𝐺 : 𝑋×𝑋×𝑋 → + Definition 1 (𝐺-metric space [1]). Let 𝑋 be a nonempty set R satisfies the following: + and let 𝐺 : 𝑋×𝑋×𝑋 →𝑅 be a function satisfying the (G 1) 𝐺(𝑥, 𝑦, 𝑧) =0 if 𝑥=𝑦=𝑧, following properties: b 0 < 𝐺(𝑥, 𝑥, 𝑦) 𝑥, 𝑦 ∈𝑋 𝑥 =𝑦̸ (Gb2) for all with , (G1) 𝐺(𝑥, 𝑦, 𝑧) =0 if and only if 𝑥=𝑦=𝑧; 𝐺(𝑥, 𝑥, 𝑦) ≤ 𝐺(𝑥, 𝑦,𝑧) 𝑥, 𝑦, 𝑧∈𝑋 𝑦 =𝑧̸ (Gb3) for all with , (G2) 0 < 𝐺(𝑥, 𝑥,,forall 𝑦) 𝑥, 𝑦 ∈𝑋 with 𝑥 =𝑦̸; 𝐺(𝑥, 𝑦, 𝑧) = 𝐺(𝑝{𝑥, 𝑦,𝑧}) 𝑝 (Gb4) ,where is a permutation (G3) 𝐺(𝑥, 𝑥, 𝑦) ≤ 𝐺(𝑥,,forall 𝑦,𝑧) 𝑥, 𝑦, 𝑧∈𝑋 with 𝑦 =𝑧̸; of 𝑥, 𝑦, 𝑧 (symmetry), 𝐺(𝑥, 𝑦, 𝑧) ≤ 𝑠[𝐺(𝑥, 𝑎, 𝑎)+𝐺(𝑎, 𝑦,𝑧)] 𝑥, 𝑦, 𝑧, 𝑎∈ (G4) 𝐺(𝑥, 𝑦, 𝑧) = 𝐺(𝑥, 𝑧, 𝑦) = 𝐺(𝑦, 𝑧,𝑥)=⋅ (symmetry (Gb5) for all 𝑋 in all three variables); (rectangle inequality). (G5) 𝐺(𝑥, 𝑦, 𝑧) ≤ 𝐺(𝑥, 𝑎, 𝑎)+𝐺(𝑎,,forall 𝑦,𝑧) 𝑥, 𝑦, 𝑧, 𝑎∈𝑋 Then 𝐺 is called a generalized 𝑏-metric and the pair (𝑋, 𝐺) 𝑏 𝐺 (rectangle inequality). is called a generalized -metric space or a 𝑏-metric space.

Then, the function 𝐺 is called a 𝐺-metric on 𝑋 and the Each 𝐺-metric space is a 𝐺𝑏-metric space with 𝑠=1. pair (𝑋, 𝐺) is called a 𝐺-metric space. Example 3 (see [2]). Let (𝑋, 𝐺) be a 𝐺-metric space and 𝑝 Recently,Aghajanietal.in[2] motivated by the concept of 𝐺∗(𝑥,𝑦,𝑧)=𝐺(𝑥,𝑦,𝑧) ,where𝑝>1is a real number. Then 𝑝−1 𝑏-metric [3] introduced the concept of generalized 𝑏-metric 𝐺∗ is a 𝐺𝑏-metric with 𝑠=2 . 2 Abstract and Applied Analysis

2 Example 4 (see [4]). Let 𝑋=R and 𝑑(𝑥, 𝑦) = |𝑥 −𝑦| . 𝐺𝑏-convergent sequences {𝑥𝑛} and {𝑦𝑛} converging to 𝑥 and 𝑦, We know that (𝑋, 𝑑) is a 𝑏-metric space with 𝑠=2.Let respectively, {𝐹(𝑥𝑛,𝑦𝑛)} is 𝐺𝑏-convergent to 𝐹(𝑥,. 𝑦) 𝐺(𝑥, 𝑦, 𝑧) = 𝑑(𝑥, 𝑦) + 𝑑(𝑦,,itiseasytosee 𝑧)+𝑑(𝑧,𝑥) 𝐺 𝐺(𝑥, 𝑦, 𝑧) 𝑠>1 that (𝑋, 𝐺) is not a 𝐺𝑏-metric space. Indeed, (𝐺𝑏3) is not In general, a 𝑏-metric function for is true for 𝑥=0, 𝑦=2,and𝑧=1.However,𝐺(𝑥, 𝑦, 𝑧)= not jointly continuous in all its variables. The following is an 𝐺 max{𝑑(𝑥, 𝑦), 𝑑(𝑦, 𝑧), 𝑑(𝑧,𝑥)} is a 𝐺𝑏-metric on R with 𝑠=2. example of a discontinuous 𝑏-metric. 𝑋=N∪{∞} 𝐷:𝑋×𝑋→ R Definition 5 (see [2]). A 𝐺𝑏-metric 𝐺 is said to be symmetric Example 14 (see [4]). Let and let if 𝐺(𝑥, 𝑦, 𝑦) = 𝐺(𝑦,,forall 𝑥,𝑥) 𝑥, 𝑦 ∈𝑋. be defined by 𝑋 𝐺 Proposition 6 (see [2]). Let be a b-metric space. Then for {0, if 𝑚=𝑛, 𝑥, 𝑦, 𝑧, 𝑎∈𝑋 { each it follows that {󵄨 1 1 󵄨 {󵄨 󵄨 {󵄨 − 󵄨 , if one of 𝑚, 𝑛 is even and the (1) if 𝐺(𝑥, 𝑦, 𝑧),then =0 𝑥=𝑦=𝑧, {󵄨𝑚 𝑛󵄨 𝐷 (𝑚, 𝑛) = { other is even or ∞, (2) 𝐺(𝑥, 𝑦, 𝑧) ≤ 𝑠(𝐺(𝑥, 𝑥, 𝑦), +𝐺(𝑥,𝑥,𝑧)) { {5, if one of 𝑚, 𝑛 is odd and the 𝐺(𝑥, 𝑦, 𝑦) ≤ 2𝑠𝐺(𝑦, 𝑥,𝑥) { (3) , { ( 𝑚 =𝑛̸) ∞, { other is odd and or (4) 𝐺(𝑥, 𝑦, 𝑧) ≤ 𝑠(𝐺(𝑥, 𝑎, 𝑧). +𝐺(𝑎,𝑦,𝑧)) {2, otherwise. (1) Definition 7 (see [2]). Let 𝑋 be a 𝐺𝑏-metric space. One defines 𝑑𝐺(𝑥,𝑦) = 𝐺(𝑥,𝑦,𝑦)+𝐺(𝑥,𝑥,𝑦),forall𝑥, 𝑦 ∈𝑋. Thenitiseasytoseethat,forall𝑚, 𝑛, 𝑝∈𝑋,wehave It is easy to see that 𝑑𝐺 defines a 𝑏-metric 𝑑 on 𝑋,whichone 𝑏 𝐺 calls the -metric associated with . 5 𝐷(𝑚,𝑝)≤ (𝐷 (𝑚, 𝑛) +𝐷(𝑛,𝑝)). 2 (2) Definition 8 (see [2]). Let 𝑋 be a 𝐺𝑏-metric space. A sequence {𝑥𝑛} in 𝑋 is said to be Thus, (𝑋, 𝐷) is a 𝑏-metric space with 𝑠=5/2(see [5]). 𝐺(𝑥, 𝑦, 𝑧)= {𝐷(𝑥, 𝑦), 𝐷(𝑦, 𝑧), 𝐷(𝑧,𝑥)} (1) 𝐺𝑏-Cauchy if, for each 𝜀>0, there exists a positive Let max .Itiseasy 𝐺 𝐺 𝑠=5/2 integer 𝑛0 such that, for all 𝑚,𝑛,𝑙≥𝑛0, 𝐺(𝑥𝑛,𝑥𝑚,𝑥𝑙)< to see that is a 𝑏-metric with which is not a 𝜀; continuous function. 𝐺 𝑥∈𝑋 𝜀>0 (2) 𝑏-convergent to a point if, for each ,there We will need the following simple lemma about the 𝐺𝑏- 𝑛 𝑚, 𝑛 ≥𝑛 exists a positive integer 0 such that, for all 0, convergent sequences in the proof of our main results. 𝐺(𝑥𝑛,𝑥𝑚,𝑥)<𝜀. Lemma 15 (see [4]). Let (𝑋, 𝐺) be a 𝐺𝑏-metric space with 𝑠> 𝑋 𝐺 Proposition 9 (see [2]). Let be a 𝑏-metric space. Then the 1 and suppose that {𝑥𝑛}, {𝑦𝑛},and{𝑧𝑛} are 𝐺𝑏-convergent to 𝑥, following are equivalent: 𝑦,and𝑧,respectively.Thenonehas {𝑥 } 𝐺 (1) the sequence 𝑛 is 𝑏-Cauchy, 1 𝐺(𝑥,𝑦,𝑧)≤lim inf 𝐺(𝑥𝑛,𝑦𝑛,𝑧𝑛) (2) for any 𝜀>0there exists 𝑛0 ∈ N such that 𝐺(𝑥𝑛,𝑥𝑚, 𝑠3 𝑛→∞ 𝑥𝑚)<𝜀for all 𝑚, 𝑛0 ≥𝑛 . ≤ lim sup𝐺(𝑥𝑛,𝑦𝑛,𝑧𝑛) (3) 𝑛→∞ Proposition 10 (see [2]). Let 𝑋 be a 𝐺𝑏-metric space. The following are equivalent. ≤𝑠3𝐺(𝑥,𝑦,𝑧).

(1) {𝑥𝑛} is 𝐺𝑏-convergent to 𝑥. In particular, if 𝑥=𝑦=𝑧,thenwehavelim𝑛→∞𝐺(𝑥𝑛,𝑦𝑛 𝐺(𝑥 ,𝑥 ,𝑥) → 0 𝑛→+∞ (2) 𝑛 𝑛 ,as . ,𝑧𝑛)=0. (3) 𝐺(𝑥𝑛,𝑥,𝑥),as →0 𝑛→+∞. The existence of fixed points, coupled fixed points, and Definition 11 (see [2]). A 𝐺𝑏-metric space 𝑋 is called 𝐺𝑏- tripled fixed points for contractive type mappings in partially complete if every 𝐺𝑏-Cauchy sequence is 𝐺𝑏-convergent in ordered metric spaces has been considered recently by several 𝑋. authors (see [6–28], etc.) Lakshmikantham and Ciri´ c[´ 17]introducedthenotionsof 󸀠 󸀠 Proposition 12. Let (𝑋, 𝐺) and (𝑋 ,𝐺 ) be two 𝐺𝑏-metric mixed 𝑔-monotone mapping and coupled coincidence point 󸀠 spaces. Then a function 𝑓:𝑋 →𝑋 is 𝐺𝑏-continuous at a and proved some coupled coincidence point and common point 𝑥∈𝑋if and only if it is 𝐺𝑏-sequentially continuous at coupled fixed point theorems in partially ordered complete 󸀠 𝑥; that is, whenever {𝑥𝑛} is 𝐺𝑏-convergent to 𝑥, {𝑓(𝑥𝑛)} is 𝐺𝑏- metric spaces. convergent to 𝑓(𝑥). Definition 16 (see [17]). Let (𝑋, ⪯) be a partially ordered set Proposition 13. Let (𝑋, 𝐺) be a 𝐺𝑏-metric space. A mapping and let 𝐹:𝑋×𝑋and →𝑋 𝑔:𝑋be →𝑋 two mappings. 𝐹:𝑋×𝑋→is 𝑋 said to be continuous if, for any two 𝐹 has the mixed 𝑔-monotone property, if 𝐹 is monotone Abstract and Applied Analysis 3

𝑔-nondecreasing in its first argument and is monotone 𝑔- Definition 22 (see [10]). Let (X,𝐺)be a generalized 𝑏-metric 2 nonincreasing in its second argument; that is, for all 𝑥1,𝑥2 ∈ space. Mappings 𝑓:X → X and 𝑔:X → X are called 𝑋, 𝑔𝑥1 ⪯𝑔𝑥2 implies 𝐹(𝑥1,𝑦)⪯ 𝐹(𝑥2,𝑦)for any 𝑦∈𝑋and compatible if for all 𝑦1,𝑦2 ∈𝑋, 𝑔𝑦1 ⪯𝑔𝑦2 implies 𝐹(𝑥,1 𝑦 )⪰𝐹(𝑥,𝑦2) for 𝑥∈𝑋 𝐺(𝑔𝑓(𝑥 ,𝑦 ),𝑓(𝑔𝑥 ,𝑔𝑦 ),𝑓(𝑔𝑥 ,𝑔𝑦 )) = 0, any . 𝑛→∞lim 𝑛 𝑛 𝑛 𝑛 𝑛 𝑛 (5) (𝑥, 𝑦) ∈ 𝑋×𝑋 𝐺(𝑔𝑓(𝑦 ,𝑥 ),𝑓(𝑔𝑦 ,𝑔𝑥 ),𝑓(𝑔𝑦 ,𝑔𝑥 )) = 0 Definition 17 (see [7, 17]). An element is called 𝑛→∞lim 𝑛 𝑛 𝑛 𝑛 𝑛 𝑛 (1) a coupled fixed point of mapping 𝐹:𝑋×𝑋if →𝑋 hold whenever {𝑥𝑛} and {𝑦𝑛} are sequences in X such that 𝑥=𝐹(𝑥,𝑦)and 𝑦 = 𝐹(𝑦,, 𝑥) 𝐹 : 𝑋×𝑋 → 𝑓(𝑥 ,𝑦 )= 𝑔𝑥 , (2) a coupled coincidence point of mappings 𝑛→∞lim 𝑛 𝑛 𝑛→∞lim 𝑛 𝑋 and 𝑔:𝑋if →𝑋 𝑔(𝑥) = 𝐹(𝑥, 𝑦) and 𝑔(𝑦) = (6) 𝐹(𝑦, 𝑥) 𝑓(𝑦 ,𝑥 )= 𝑔𝑦 . , 𝑛→∞lim 𝑛 𝑛 𝑛→∞lim 𝑛 (3) a common coupled fixed point of mappings 𝐹:𝑋× 𝑋→𝑋and 𝑔:𝑋if →𝑋 𝑥 = 𝑔(𝑥) = 𝐹(𝑥, 𝑦) and On the other hand, Berinde and Borcut [25]introduced 𝑦 = 𝑔(𝑦) = 𝐹(𝑦,. 𝑥) the concept of tripled fixed point and obtained some tripled fixed point theorems for contractive type mappings in par- Definition 18 (see [17]). Let 𝑋 be a nonempty set. We say that tially ordered metric spaces. For a survey of tripled fixed point the mappings 𝐹:𝑋×𝑋→and 𝑋 𝑔:𝑋are →𝑋 theorems and related topics we refer the reader to [25–28, 30]. commutative if 𝑔(𝐹(𝑥, 𝑦)) = 𝐹(𝑔𝑥, 𝑔𝑦) for all 𝑥, 𝑦 ∈𝑋. Definition 23 (see [25, 26]). Let (X,⪯)be a partially ordered 3 ChoudhuryandMaity[10] have established some coupled set, 𝑓:X → X,and𝑔:X → X. fixed point results for mappings with mixed monotone 3 (1) An element (𝑥,𝑦,𝑧) ∈ X is called a tripled fixed property in partially ordered 𝐺-metric spaces. They obtained point of 𝑓 if 𝑓(𝑥, 𝑦, 𝑧), =𝑥 𝑓(𝑦, 𝑥, 𝑦) =𝑦,and the following results. 𝑓(𝑧, 𝑦, 𝑥). =𝑧 3 Theorem 19 (see [10, Theorem 3.1]). Let (𝑋, ⪯) be a partially (2) An element (𝑥, 𝑦, 𝑧) ∈ X is called a tripled coinci- ordered set and let 𝐺 be a 𝐺-metric on 𝑋 such that (𝑋, 𝐺) is a dence point of the mappings 𝑓 and 𝑔 if 𝑓(𝑥, 𝑦, 𝑧)= complete 𝐺-metric space. Let 𝐹:𝑋×𝑋 →𝑋be a continuous 𝑔𝑥, 𝑓(𝑦, 𝑥, 𝑦) =𝑔𝑦,and𝑓(𝑧, 𝑦, 𝑥). =𝑔𝑧 mapping having the mixed monotone property on 𝑋. Assume 3 (3) An element (𝑥,𝑦,𝑧)∈X is called a tripled common that there exists 𝑘∈[0,1)such that fixed point of 𝑓 and 𝑔 if 𝑥=𝑔(𝑥)=𝑓(𝑥,𝑦,𝑧), 𝑦= 𝑘 𝑔(𝑦) = 𝑓(𝑦,,and 𝑥,𝑦) 𝑧=𝑔(𝑧)=𝑓(𝑧,𝑦,𝑥). 𝐺(𝐹(𝑥,𝑦),𝐹(𝑢, V) ,𝐹(𝑤,) 𝑧 )≤ [𝐺 (𝑥, 𝑢,) 𝑤 +𝐺(𝑦,V, 𝑧)] , 2 (4) One says that 𝑓 has the mixed 𝑔-monotone property (4) if 𝑓(𝑥, 𝑦, 𝑧) is 𝑔-nondecreasing in 𝑥, 𝑔-nonincreasing in 𝑦,and𝑔-nondecreasing in 𝑧;thatis,if,forany 𝑥⪯𝑢⪯𝑤 𝑦⪰V ⪰𝑧 𝑢 =𝑤̸ V =𝑧̸ for all and , where either or . 𝑥, 𝑦, 𝑧∈ X, If there exist 𝑥0,𝑦0 ∈𝑋such that 𝑥0 ⪯𝐹(𝑥0,𝑦0) and 𝑦0 ⪰ 𝐹(𝑦 ,𝑥 ) 𝐹 𝑋 0 0 ,then has a coupled fixed point in ; that is, there 𝑥1,𝑥2 ∈ X,𝑔𝑥1 ⪯𝑔𝑥2 󳨐⇒ 𝑓 (𝑥 1,𝑦,𝑧)⪯𝑓(𝑥2,𝑦,𝑧), exist 𝑥, 𝑦 ∈𝑋 such that 𝑥 = 𝐹(𝑥, 𝑦) and 𝑦 = 𝐹(𝑦,. 𝑥) 𝑦1,𝑦2 ∈ X,𝑔𝑦1 ⪯𝑔𝑦2 󳨐⇒ 𝑓 ( 𝑥 ,𝑦 1,𝑧)⪰𝑓(𝑥,𝑦2,𝑧), Theorem 20 (see [10, Theorem 3.2]). If, in the above theorem, in place of the continuity of 𝐹, one assumes the following 𝑧1,𝑧2 ∈ X,𝑔𝑧1 ⪯𝑔𝑧2 󳨐⇒ 𝑓 ( 𝑥 , 𝑦,𝑧 1)⪯𝑓(𝑥,𝑦,𝑧2). conditions, namely, (7) {𝑥 }→𝑥 𝑥 ⪯𝑥 (i) if a nondecreasing sequence 𝑛 ,then 𝑛 for Definition 24 (see [28]). Let X be a nonempty set. One says all 𝑛, 3 that the mappings 𝑓:X → X and 𝑔:X → X commute (ii) if a nonincreasing sequence {𝑦𝑛}→𝑦,then𝑦𝑛 ⪰𝑦for if 𝑔(𝑓(𝑥, 𝑦, 𝑧)) = 𝑓(𝑔𝑥,,forall 𝑔𝑦,𝑔𝑧) 𝑥, 𝑦, 𝑧∈ X. all 𝑛, In [26], Borcut obtained the following. then 𝐹 has a coupled fixed point. Theorem 25 (see [26,Corollary1]). Let (X,⪯)be a partially (𝑋, ⪯) Definition 21 (see [29]). Let be a partially ordered set orderedsetandsupposethereisametric𝑑 on X such that 𝐺 𝐺 𝑋 (𝑋, 𝐺, ⪯) 3 and let be a -metric on .Onesaysthat is regular (X,𝑑) is a complete metric space. Let 𝑓:X → X and if the following conditions hold. 𝑔:X → X be such that 𝑓 has the 𝑔-mixed monotone property. Assume that there exists 𝑘∈[0,1)such that (i) If {𝑥𝑛} is a nondecreasing sequence with 𝑥𝑛 →𝑥, 𝑥 ⪯𝑥 𝑛∈N then 𝑛 for all . 𝑑(𝑓(𝑥,𝑦,𝑧),𝑓(𝑢, V,𝑤)) (ii) If {𝑥𝑛} is a nonincreasing sequence with 𝑥𝑛 →𝑥,then (8) 𝑥𝑛 ⪰𝑥for all 𝑛∈N. ≤𝑘max {𝑑(𝑔𝑥,𝑔𝑢),𝑑(𝑔𝑦,𝑔V),𝑑(𝑔𝑧,𝑔𝑤)} 4 Abstract and Applied Analysis for all 𝑥, 𝑦, 𝑧, 𝑢, V,𝑤 ∈ X with 𝑔𝑥⪯𝑔𝑢, 𝑔𝑦 ⪰𝑔V,and Definition 27. Let 𝑇 be a self-mapping on 𝑋 and let 𝛼: 3 𝑔𝑧 ⪯.Suppose 𝑔𝑤 𝑓(X )⊆𝑔(X) and 𝑔 is continuous and 𝑋×𝑋 → [0,+∞)be a function. One says that 𝑇 is an 𝛼- commutes with 𝑓 and also suppose either admissible mapping if

(a) 𝑓 is continuous, or 𝑥, 𝑦 ∈ 𝑋, 𝛼 (𝑥, 𝑦) ≥ 1 󳨐⇒ 𝛼 (𝑇𝑥, 𝑇𝑦)≥1. (11) (b) X has the following properties: Definition 28 (see [32]). Let (𝑋, 𝐺) be a 𝐺-metric space, let 𝑇 𝑋 𝛼:𝑋3 →[0,+∞) 𝑥 →𝑥 𝑥 ⪯ be a self-mapping on ,andlet be a (i) if a nondecreasing sequence 𝑛 ,then 𝑛 function. One says that 𝑇 is an 𝐺-𝛼-admissible mapping if 𝑥 for all 𝑛, (ii) if a nonincreasing sequence 𝑦𝑛 →𝑦,then𝑦𝑛 ⪰𝑦 𝑥, 𝑦, 𝑧 ∈ 𝑋, 𝛼 (𝑥, 𝑦, 𝑧) ≥ 1 󳨐⇒ 𝛼 (𝑇𝑥,𝑇𝑦,𝑇𝑧)≥1. (12) for all 𝑛. Following the recent work in [33–35]wepresentthe 𝐺 If there exist 𝑥0,𝑦0,𝑧0 ∈ X such that 𝑔𝑥0 ⪯𝑓(𝑥0,𝑦0,𝑧0), following definition in the setting of -metric spaces. 𝑔𝑦0 ⪰𝑓(𝑦0,𝑥0,𝑧0),and𝑔𝑧0 ⪯𝑓(𝑧0,𝑦0,𝑥0),then𝑓 and 𝑔 Definition 29. Let (𝑋, 𝐺) be a 𝐺-metric space and let 𝑓, 𝑔 : have a tripled coincidence point. 3 𝑋→𝑋and 𝛼:𝑋 →[0,+∞).Onesaysthat𝑓 is a Definition 26. Let (X,𝐺) be a generalized 𝑏-metric space. rectangular 𝐺-𝛼-admissible mapping with respect to 𝑔 if 𝑓:X3 → X 𝑔:X → X Mappings and are called 𝛼(𝑔𝑥, 𝑔𝑦, 𝑔𝑧)≥1 𝛼(𝑓𝑥, 𝑓𝑦,𝑓𝑧) ≥1 𝑥, 𝑦, 𝑧∈ compatible if (R1) implies , 𝑋, 𝐺(𝑔𝑓(𝑥 ,𝑦 ,𝑧 ),𝑓(𝑔𝑥 ,𝑔𝑦 ,𝑔𝑧 ), {𝛼(𝑔𝑥, 𝑔𝑦, 𝑔𝑦) ≥ 1, 𝛼(𝑔𝑦, 𝑔𝑧,𝑔𝑧)≥1} 𝑛→∞lim 𝑛 𝑛 𝑛 𝑛 𝑛 𝑛 (R2) implies 𝛼(𝑔𝑥, 𝑔𝑧, 𝑔𝑧) ≥ 1, 𝑥,𝑦,𝑧∈𝑋. 𝑓(𝑔𝑥𝑛,𝑔𝑦𝑛,𝑔𝑧𝑛)) = 0, Lemma 30. Let 𝑓 be a rectangular 𝐺-𝛼-admissible mapping 𝐺(𝑔𝑓(𝑦 ,𝑥 ,𝑦 ),𝑓(𝑔𝑦 ,𝑔𝑥 ,𝑔𝑦 ), 𝑔 𝑓(𝑋) ⊆ 𝑔(𝑋) 𝑛→∞lim 𝑛 𝑛 𝑛 𝑛 𝑛 𝑛 with respect to such that . Assume that there (9) exists 𝑥0 ∈𝑋such that 𝛼(𝑔𝑥0,𝑓𝑥0,𝑓𝑥0)≥1. Define sequence 𝑓(𝑔𝑦𝑛,𝑔𝑥𝑛,𝑔𝑦𝑛)) = 0, {𝑦𝑛} by 𝑦𝑛 =𝑔𝑥𝑛 =𝑓𝑥𝑛−1.Then 𝐺(𝑔𝑓(𝑧 ,𝑦 ,𝑥 ),𝑓(𝑔𝑧 ,𝑔𝑦 ,𝑔𝑥 ), 𝛼(𝑦 ,𝑦 ,𝑦 )≥1 ∀𝑛,𝑚∈N 𝑤𝑖𝑡ℎ 𝑛 <𝑚. 𝑛→∞lim 𝑛 𝑛 𝑛 𝑛 𝑛 𝑛 𝑛 𝑚 𝑚 (13)

𝑓(𝑔𝑧𝑛,𝑔𝑦𝑛,𝑔𝑥𝑛)) = 0 Now, we prove the following coincidence point result. (𝑋, 𝐺) 𝑏 hold whenever {𝑥𝑛}, {𝑦𝑛},and{𝑧𝑛} are sequences in X such Theorem 31. Let be a generalized -metric space and that let 𝑓, 𝑔 : 𝑋 →𝑋 satisfy the following condition: 𝑓(𝑥 ,𝑦 ,𝑧 )= 𝑔𝑥 , 𝛼 (𝑔𝑥, 𝑔𝑦, 𝑔𝑧) 𝜓 (𝑠𝐺 (𝑓𝑥,𝑓𝑦,𝑓𝑧)) 𝑛→∞lim 𝑛 𝑛 𝑛 𝑛→∞lim 𝑛 (14) 𝑓(𝑦 ,𝑥 ,𝑦 )= 𝑔𝑦 , ≤ 𝜓 (𝐺 (𝑔𝑥, 𝑔𝑦, 𝑔𝑧)) − 𝜑 (𝐺(𝑔𝑥,𝑔𝑦,𝑔𝑧)) 𝑛→∞lim 𝑛 𝑛 𝑛 𝑛→∞lim 𝑛 (10) 𝑓(𝑧 ,𝑦 ,𝑥 )= 𝑔𝑧 . for all 𝑥, 𝑦, 𝑧∈ X,where𝜓, 𝜑 : [0, ∞) → [0, ∞) are two lim 𝑛 𝑛 𝑛 lim 𝑛 3 𝑛→∞ 𝑛→∞ altering distance mappings, 𝛼:𝑋 →[0,+∞),and𝑓 is a rectangular 𝐺-𝛼-admissible mapping with respect to 𝑔. Let 𝜓:[0,+∞)→ [0,+∞)satisfies the following: Then, maps 𝑓 and 𝑔 have a coincidence point if (i) 𝜓 is continuous and nondecreasing, (i) 𝑓(𝑋) ⊆ 𝑔(𝑋), (ii) 𝜓(𝑡) =0 if and only if 𝑡=0. (ii) there exists 𝑥0 ∈𝑋such that 𝛼(𝑔𝑥0,𝑓𝑥0,𝑓𝑥0)≥1, That is, 𝜓 is an altering distance function. (iii) 𝑓 and 𝑔 are continuous and compatible and (𝑋, 𝐺) is In this paper, we obtain some coupled and tripled coinci- complete, (𝜓, 𝜑) 󸀠 dence point theorems for nonlinear weakly contractive (iii ) one of 𝑓(𝑋) or 𝑔(𝑋) is complete and assume mappings which are 𝐺-𝛼-admissible with respect to another that whenever {𝑥𝑛} in 𝑋 is a sequence such that function in partially ordered 𝐺𝑏-metric spaces. These results 𝛼(𝑥𝑛,𝑥𝑛+1,𝑥𝑛+1)≥1for all 𝑛∈N∪{0}and 𝑥𝑛 →𝑥as generalize and modify several comparable results in the 𝑛→+∞,wehave𝛼(𝑥𝑛,𝑥,𝑥)≥1for all 𝑛∈N ∪{0}. literature. Proof. Let 𝑥0 ∈𝑋be such that 𝛼(𝑔𝑥0,𝑓𝑥0,𝑓𝑥0)≥1. {𝑦 } 𝑦 = 2. Main Results According to (i) one can define the sequence 𝑛 as 𝑛+1 𝑔𝑥𝑛+1 =𝑓𝑥𝑛 for all 𝑛 = 0, 1, 2, .. Samet et al. [31] defined the notion of 𝛼-admissible mapping As 𝛼(𝑔𝑥0,𝑔𝑥1,𝑔𝑥1)=𝛼(𝑔𝑥0,𝑓𝑥0,𝑓𝑥0)≥1and since as follows. 𝑓 is an 𝐺-𝛼-admissible mapping with respect to 𝑔,then Abstract and Applied Analysis 5

𝛼(𝑦1,𝑦2,𝑦2)=𝛼(𝑓𝑥0,𝑓𝑥1,𝑓𝑥1)≥1.Continuingthisprocess, Since 𝑓 is a rectangular 𝐺-𝛼-admissible mapping with respect we get 𝛼(𝑦𝑛,𝑦𝑛+1,𝑦𝑛+1)≥1for all 𝑛∈N ∪{0}. to 𝑔,thenfromLemma 30 𝛼(𝑦𝑛 ,𝑦𝑚 −1,𝑦𝑚 −1)≥1.Now, 𝑘 𝑘 k If 𝑦𝑛 =𝑦𝑛+1,then𝑥𝑛 is a coincidence point of 𝑓 and 𝑔. from (14)wehave Now, assume that 𝑦𝑛 =𝑦̸𝑛+1 for all 𝑛;thatis, 𝜓(𝑠𝐺(𝑦 ,𝑦 ,𝑦 )) 𝑛𝑘+1 𝑚𝑘 𝑚𝑘 𝐺 (𝑦𝑛,𝑦𝑛+1,𝑦𝑛+2) >0, (15) ≤𝛼(𝑦 ,𝑦 ,𝑦 )𝜓(𝑠𝐺(𝑦 ,𝑦 ,𝑦 )) 𝑛𝑘 𝑚𝑘−1 𝑚𝑘−1 𝑛𝑘+1 𝑚𝑘 𝑚𝑘 for all 𝑛.Let𝐺𝑛 =𝐺(𝑦𝑛,𝑦𝑛+1,𝑦𝑛+2).Then,from(14)weobtain =𝛼(𝑔𝑥𝑛 ,𝑔𝑥𝑚 −1,𝑔𝑥𝑚 −1)𝜓(𝑠𝐺(𝑓𝑥𝑛 ,𝑓𝑥𝑚 −1,𝑓𝑥𝑚 −1)) that 𝑘 𝑘 𝑘 𝑘 𝑘 𝑘

≤𝜓(𝐺(𝑦𝑛 ,𝑦𝑚 −1,𝑦𝑚 −1)) − 𝜑 (𝐺𝑛 (𝑦 ,𝑦𝑚 −1,𝑦𝑚 −1)) . 𝜓(𝑠𝐺(𝑦𝑛+1,𝑦𝑛+2,𝑦𝑛+3)) 𝑘 𝑘 𝑘 𝑘 𝑘 𝑘 (24) ≤𝛼(𝑦𝑛,𝑦𝑛+1,𝑦𝑛+2)𝜓(𝑠𝐺(𝑦𝑛+1,𝑦𝑛+2,𝑦𝑛+3)) (16) Using (Gb5) we obtain that =𝛼(𝑔𝑥𝑛,𝑔𝑥𝑛+1,𝑔𝑥𝑛+2)𝜓(𝑠𝐺(𝑓𝑥𝑛,𝑓𝑥𝑛+1,𝑓𝑥𝑛+2)) 𝐺(𝑦𝑛 ,𝑦𝑚 ,𝑦𝑚 ) ≤ 𝜓 (𝐺 (𝑦 ,𝑦 ,𝑦 )) − 𝜑 (𝐺 (𝑦 ,𝑦 ,𝑦 )) . 𝑘 𝑘 𝑘 𝑛 𝑛+1 𝑛+2 𝑛 𝑛+1 𝑛+2 (25) ≤𝑠𝐺(𝑦 ,𝑦 ,𝑦 )+𝑠𝐺(𝑦 ,𝑦 ,𝑦 ). 𝑛𝑘 𝑛𝑘+1 𝑛𝑘+1 𝑛𝑘+1 𝑚𝑘 𝑚𝑘 We prove that 𝐺𝑛+1 ≤𝐺𝑛 for each 𝑛∈N.If𝐺𝑛+1 >𝐺𝑛 for some 𝑛∈N,thenfrom(16)wehave𝜓(𝐺𝑛+1)≤𝜓(𝐺𝑛+1)− Taking the upper limit as 𝑘→∞and using (20)and(23)we 𝜑(𝐺𝑛) which implies that 𝐺𝑛 =0, a contradiction to (15). obtain that 0<𝐺 ≤𝐺 𝑛∈N Hence, we have 𝑛+1 𝑛 for each .Thus,the 𝜀 {𝐺 } 𝑟≥0 𝐺(𝑦 ,𝑦 ,𝑦 )≥ . sequence 𝑛 is nonincreasing and so there exists such lim sup 𝑛𝑘+1 𝑚𝑘 𝑚𝑘 (26) 𝑘→∞ 𝑠 that lim𝑛→∞𝐺𝑛 =𝑟≥0. Suppose that 𝑟>0.Thenfrom(16), taking the limit as Using (𝐺𝑏5)weobtainthat 𝑛→∞implies that

𝐺(𝑦𝑛 ,𝑦𝑚 ,𝑦𝑚 ) 𝜓 (𝑟) ≤𝜓(𝑟) −𝜑(𝑟) 𝑘 𝑘 𝑘 (17) (27) ≤𝑠𝐺(𝑦 ,𝑦 ,𝑦 )+𝑠𝐺(𝑦 ,𝑦 ,𝑦 ). 𝑛𝑘 𝑚𝑘−1 𝑚𝑘−1 𝑚𝑘−1 𝑚𝑘 𝑚𝑘 a contradiction. Hence, Taking the upper limit as 𝑘→∞and using (20)and(23)we 𝐺 = 𝐺 (𝑦 ,𝑦 ,𝑦 ) =0. 𝑛→∞lim 𝑛 𝑛→∞lim 𝑛 𝑛+1 𝑛+2 (18) obtain that 𝜀 𝐺 (𝑦 ,𝑦 ,𝑦 ) ≥ . 𝑦 =𝑦̸ 𝑛 lim inf 𝑛𝑘 𝑚𝑘−1 𝑚𝑘−1 (28) Since 𝑛+1 𝑛+2 for every ,sobyproperty(Gb3) we obtain 𝑘→∞ 𝑠 Taking the upper limit as 𝑘→∞in (24)andusing(23)and 𝐺(𝑦𝑛,𝑦𝑛+1,𝑦𝑛+1)≤𝐺(𝑦𝑛,𝑦𝑛+1,𝑦𝑛+2). (19) (26)weobtainthat Hence, 𝜓 (𝜖) ≤𝜓(𝑠 𝐺(𝑦 ,𝑦 ,𝑦 )) lim sup 𝑛𝑘+1 𝑚𝑘 𝑚𝑘 𝐺(𝑦 ,𝑦 ,𝑦 )=0. 𝑘→∞ 𝑛→∞lim 𝑛 𝑛+1 𝑛+1 (20) ≤𝜓( 𝐺(𝑦 ,𝑦 ,𝑦 )) lim sup 𝑛𝑘 𝑚𝑘−1 𝑚𝑘−1 Also, by part (3) of Proposition 6 we have 𝑘→∞ (29)

𝐺 (𝑦 ,𝑦 ,𝑦 ) =0. − lim inf𝜑(𝐺(𝑦𝑛 ,𝑦𝑚 −1,𝑦𝑚 −1)) 𝑛→∞lim 𝑛 𝑛 𝑛+1 (21) 𝑘→∞ 𝑘 𝑘 𝑘

Now, we prove that {𝑦𝑛} is a 𝐺𝑏-Cauchy sequence. Assume ≤𝜓(𝜖) −𝜑(lim inf𝐺(𝑦𝑛 ,𝑦𝑚 −1,𝑦𝑚 −1)) 𝑘→∞ 𝑘 𝑘 𝑘 on contrary that {𝑦𝑛} is not a 𝐺𝑏-Cauchy sequence. Then there 𝜀>0 {𝑦 } exists for which we can find subsequences 𝑚𝑘 and which implies that {𝑦 } {𝑦 } 𝑚 𝑚 > 𝑛𝑘 of 𝑛 such that 𝑘 is the smallest index for which 𝑘 𝑛𝑘 >𝑘and 𝜑( lim inf 𝐺(𝑦𝑛 ,𝑦𝑚 −1,𝑦𝑚 −1)) = 0, (30) 𝑘→∞ 𝑘 𝑘 𝑘 𝐺(𝑦 ,𝑦 ,𝑦 )≥𝜀. 𝑛𝑘 𝑚𝑘 𝑚𝑘 (22) 𝐺(𝑦 ,𝑦 ,𝑦 )=0 so lim𝑘→∞ inf 𝑛𝑘 𝑚𝑘−1 𝑚𝑘−1 , a contradiction to (28). It follows that {𝑦𝑛} is a 𝐺𝑏-Cauchy sequence in 𝑋. This means that Suppose first that (iii) holds. Then there exists

𝐺(𝑥 ,𝑥 ,𝑥 )<𝜀. lim 𝑓𝑥 𝑛 = lim 𝑔𝑥𝑛 =𝑧∈X. 𝑛𝑘 𝑚𝑘−1 𝑚𝑘−1 (23) 𝑛→∞ 𝑛→∞ (31) 6 Abstract and Applied Analysis

Further, since 𝑓 and 𝑔 are continuous and compatible, we get Proof. Define 𝛼:𝑋×𝑋×𝑋 → [0,+∞)by that 1, 𝑥⪯𝑦⪯𝑧, 𝛼(𝑥,𝑦,𝑧)={ if lim 𝑓𝑔 𝑥 𝑛 =𝑓𝑧, lim 𝑔𝑓𝑥𝑛 =𝑔𝑧, (37) 𝑛→∞ 𝑛→∞ 0, otherwise. (32) 𝐺(𝑓𝑔𝑥 ,𝑓𝑔𝑥 ,𝑔𝑓𝑥 )=0. 𝑛→∞lim 𝑛 𝑛 𝑛 First, we prove that 𝑓 is a rectangular 𝐺-𝛼-admissible mappingwithrespectto𝑔. Assume that 𝛼(𝑔𝑥, 𝑔𝑦, 𝑔𝑧)≥ We will show that 𝑓𝑧 = 𝑔𝑧. Indeed, we have 1. Therefore, we have 𝑔𝑥⪯𝑔𝑦⪯𝑔𝑧.Since𝑓 is 𝑔- nondecreasing with respect to ⪯,weget𝑓𝑥 ⪯ 𝑓𝑦 ⪯𝑓𝑧;thatis, 𝐺(𝑓𝑧,𝑔𝑧,𝑔𝑧)≤𝑠[𝐺(𝑓𝑧,𝑓𝑔𝑥𝑛,𝑓𝑔𝑥𝑛)+𝐺(𝑓𝑔𝑥𝑛,𝑔𝑧,𝑔𝑧)] 𝛼(𝑓𝑥, 𝑓𝑦,𝑓𝑧). ≥1 Also, let 𝛼(𝑥, 𝑧, 𝑧) ≥1 and 𝛼(𝑧, 𝑦, 𝑦), ≥1 and then 𝑥⪯𝑧and 𝑧⪯𝑦. Consequently, we deduce that ≤𝑠𝐺(𝑓𝑧,𝑓𝑔𝑥,𝑓𝑔𝑥 ) 𝑛 𝑛 𝑥⪯𝑦⪯𝑦;thatis,𝛼(𝑥, 𝑦, 𝑦).Thus, ≥1 𝑓 is a rectangular 2 𝐺-𝛼-admissible mapping with respect to 𝑔.Since +𝑠 [𝐺 (𝑓𝑔𝑥𝑛,𝑔𝑓𝑥𝑛,𝑔𝑓𝑥𝑛) 𝜓 (𝑠𝐺 (𝑓𝑥,𝑓𝑦, 𝑓𝑧)) +𝐺 (𝑔𝑓𝑥 , 𝑔𝑧, 𝑔𝑧)] 𝑛 (38) 2 2 ≤ 𝜓 (𝐺 (𝑔𝑥, 𝑔𝑦, 𝑔𝑧)) − 𝜑 (𝐺(𝑔𝑥,𝑔𝑦,𝑔𝑧)) 󳨀→𝑠⋅0+𝑠 ⋅0+𝑠 ⋅0=0 as (𝑛→∞) , (33) for all 𝑥, 𝑦,,with 𝑧∈𝑋 𝑔𝑥⪯𝑔𝑦⪯𝑔𝑧,then and it follows that 𝑓𝑧 = 𝑔𝑧.Itmeansthat𝑓 and 𝑔 have a 𝛼 (𝑔𝑥, 𝑔𝑦, 𝑔𝑧) 𝜓 (𝑠𝐺 (𝑓𝑥,𝑓𝑦,𝑓𝑧)) coincidence point. (39) 󸀠 In the case (iii ), if we assume that 𝑔(𝑋) is 𝐺𝑏-complete, ≤ 𝜓 (𝐺 (𝑔𝑥, 𝑔𝑦, 𝑔𝑧)) − 𝜑 (𝐺(𝑔𝑥,𝑔𝑦,𝑔𝑧)). then Moreover, from (ii) there exists 𝑥0 ∈𝑋such that 𝑔𝑥0 ⪯ 𝑓𝑥 = 𝑔𝑥 =𝑔𝑢=𝑧 𝑓𝑥 ⪯𝑓𝑥 𝛼(𝑔𝑥 ,𝑓𝑥 ,𝑓𝑥 )≥1 𝑛→∞lim 𝑛 𝑛→∞lim 𝑛 (34) 0 0;thatis, 0 0 0 .Hence,allthe conditions of Theorem 31 aresatisfiedandtherefore𝑓 and 𝑔 󸀠 have a coincidence point. for some 𝑢∈X. Also, from (iii )wehave𝛼(𝑔𝑥𝑛,𝑔𝑢,𝑔𝑢)≥1. 𝑥=𝑥 𝑦=𝑢 Applying (14)with 𝑛 and ,wehave If 𝛼(𝑥, 𝑦, 𝑧) =1 for all 𝑥, 𝑦, 𝑧∈𝑋 in Theorem 31,thenwe obtain the following coincidence point result. 𝜓(𝐺(𝑓𝑥𝑛,𝑓𝑢,𝑓𝑢)) (𝑋, 𝐺) 𝑏 ≤𝛼(𝑔𝑥,𝑔𝑢,𝑔𝑢)𝜓(𝑠𝐺(𝑓𝑥 ,𝑓𝑢,𝑓𝑢)) Theorem 33. Let be a generalized -metric space and 𝑛 𝑛 (35) let 𝑓, 𝑔 : 𝑋 →𝑋 satisfy the following condition: ≤𝜓(𝐺(𝑔𝑥,𝑔𝑢,𝑔𝑢))−𝜑(𝐺(𝑔𝑥 ,𝑔𝑢,𝑔𝑢)). 𝑛 𝑛 𝜓 (𝑠𝐺 (𝑓𝑥,𝑓𝑦, 𝑓𝑧)) (40) It follows that 𝐺(𝑓𝑥𝑛,𝑓𝑢,𝑓𝑢) →0when 𝑛→∞;thatis, ≤ 𝜓 (𝐺 (𝑔𝑥, 𝑔𝑦, 𝑔𝑧)) − 𝜑 (𝐺(𝑔𝑥,𝑔𝑦,𝑔𝑧)) 𝑓𝑥 𝑛 →𝑓𝑢. Uniqueness of the limit yields that 𝑓𝑢=𝑧=𝑔𝑢. 𝑓 𝑔 𝑢∈𝑋 Hence, and have a coincidence point . for all 𝑥, 𝑦, 𝑧∈ X,where𝜓, 𝜑 : [0, ∞) → [0, ∞) are two (𝑋, 𝐺, ⪯) 𝑏 altering distance functions. Theorem 32. Let be an ordered generalized -metric Then, maps 𝑓 and 𝑔 have a coincidence point if space and let 𝑓, 𝑔 : 𝑋 →𝑋 satisfy the following condition: (i) 𝑓(𝑋) ⊆ 𝑔(𝑋), 𝜓 (𝑠𝐺 (𝑓𝑥,𝑓𝑦, 𝑓𝑧)) (ii) 𝑓 and 𝑔 are continuous and compatible and (𝑋, 𝐺) is (36) ≤ 𝜓 (𝐺 (𝑔𝑥, 𝑔𝑦, 𝑔𝑧)) − 𝜑 (𝐺(𝑔𝑥,𝑔𝑦,𝑔𝑧)) complete, or 󸀠 (ii ) one of 𝑓(𝑋) or 𝑔(𝑋) is complete. for all 𝑥, 𝑦,,with 𝑧∈𝑋 𝑔𝑥⪯𝑔𝑦⪯𝑔𝑧,where𝜓, 𝜑 : [0, ∞)→ [0, ∞) are two altering distance functions. 3. Coupled Fixed Point Results Then, maps 𝑓 and 𝑔 have a coincidence point if We will use the following simple lemma in proving our next 𝑓 𝑔 ⪯ 𝑓(𝑋) ⊆ (i) is -nondecreasing with respect to and results. A similar case in the context of 𝑏-metric spaces can 𝑔(𝑋) ; be found in [24]. (ii) there exists 𝑥0 ∈𝑋such that 𝑔𝑥0 ⪯𝑓𝑥0; Lemma 34. Let (X,𝐺)be a generalized 𝑏-metric space (with 𝑓 𝑔 (𝑋, 𝐺) 2 (iii) and are continuous and compatible and is the parameter 𝑠)andlet𝑓:X → X and 𝑔:X → X. 2 2 complete, or Suppose that 𝐹:X → X is given by 󸀠 (𝑋, 𝐺, ⪯) 𝑓(𝑋) 𝑔(𝑋) (iii ) is regular and one of or is 2 complete. 𝐹𝑋=(𝑓(𝑥,𝑦),𝑓(𝑦,𝑥)), 𝑋=(𝑥,𝑦)∈X (41) Abstract and Applied Analysis 7

2 2 and 𝐻:X → X is defined by Theorem 35. Let (X,𝐺)be a generalized 𝑏-metric space with 𝑠 𝑓:X2 → X 𝑔:X → X 2 the parameter and let and . 𝐻𝑋=(𝑔𝑥,𝑔𝑦), 𝑋=(𝑥,𝑦)∈X . (42) Assume that 𝛼 ((𝑔𝑥, 𝑔𝑦) ,V (𝑔𝑢,𝑔 ) , (𝑔𝑧, 𝑔𝑤))

Ω𝑚 : X2 × X2 × X2 → R+ 𝑚 (a) If a mapping 2 is given by ×𝜓(𝑠𝑁𝑓 (𝑥,𝑦,𝑢,V,𝑧,𝑤)) Ω𝑚 (𝑋, 𝑈, 𝐴) = {𝐺 (𝑥, 𝑢,) 𝑎 ,𝐺(𝑦,V,𝑏)}, 𝑚 𝑚 2 max ≤𝜓(𝑁𝑔 (𝑥,𝑦,𝑢,V,𝑧,𝑤))−𝜑(𝑁𝑔 (𝑥,𝑦,𝑢,V,𝑧,𝑤)), (43) 𝑋=(𝑥,𝑦),( 𝑈= 𝑢, V) ,𝐴=(𝑎,) 𝑏 ∈ X2, (47) 𝑥, 𝑦, 𝑢, V,𝑧,𝑤∈ X 𝜓, 𝜑 : [0, ∞) → [0, ∞) 2 𝑚 for all ,where are (X ,Ω ) 𝑏 2 3 then 2 is a generalized -metric space (with the altering distance functions, 𝛼:(𝑋) →[0,∞),and𝑓 is a 𝑠 (X2,Ω𝑚) 𝐺 same parameter ). The space 2 is 𝑏-complete rectangular 𝐺-𝛼-admissible mapping with respect to 𝑔. (X,𝐺) 𝐺 if and only if is 𝑏-complete. Assume also that 2 𝑚 (b) If 𝑓 is continuous from (X ,Ω2 ) to (X,𝐺),then𝐹 is 2 2 𝑚 (1) 𝑓(X )⊆𝑔(X); continuous in (X ,Ω2 ). (2) there exist 𝑥0,𝑦0 ∈ X such that (c) If 𝑓 and 𝑔 are compatible, then 𝐹 and 𝐻 are compatible. 2 𝛼 ((𝑔𝑥 ,𝑔𝑦 ) , (𝑓 (𝑥 ,𝑦 ),𝑓(𝑦 ,𝑥 )) , (d) The mapping 𝑓:𝑋 →𝑋is 𝐺-𝛼-admissible with 0 0 0 0 0 0 respect to 𝑔;thatis, 𝑓(𝑥0,𝑦0),𝑓(𝑦0,𝑥0)) ≥ 1, 𝑥, 𝑦, 𝑢, V,𝑎,𝑏∈𝑋, (48) 𝛼 ((𝑔𝑦0,𝑔𝑥0) , (𝑓 0(𝑦 ,𝑥0),𝑓(𝑥0,𝑦0)) , 𝛼 ((𝑔𝑥, 𝑔𝑦)V ,(𝑔𝑢,𝑔 ) , (𝑔𝑎, 𝑔𝑏)) ≥1 𝑓(𝑦0,𝑥0),𝑓(𝑥0,𝑦0)) ≥ 1. 󳨐⇒𝛼(𝑓(𝑥,𝑦),𝑓(𝑦,𝑥)), (44) Also, suppose that either (𝑓 (𝑢, V) ,𝑓(V,𝑢)), (a) 𝑓 and 𝑔 are continuous, the pair (𝑓,𝑔) is compatible, ((𝑓 (𝑎,) 𝑏 ,𝑓(𝑏,) 𝑎 )) ≥ 1 and (X,𝐺)is 𝐺𝑏-complete, or

(b) (𝑔(X), 𝐺) is 𝐺𝑏-complete and assume that whenever 𝐹:X2 → X2 𝐺 𝛼 if and only if the mapping is - - {𝑥𝑛} and {𝑦𝑛} in 𝑋 are sequences such that admissible with respect to 𝐻;thatis, 𝛼((𝑥𝑛,𝑦𝑛),(𝑥𝑛+1,𝑦𝑛+1),(𝑥𝑛+1,𝑦𝑛+1)) ≥ 1, 𝑋, 𝑈, 𝐴 ∈𝑋2, (49) 𝛼 ((𝑦𝑛,𝑥𝑛),(𝑦𝑛+1,𝑥𝑛+1),(𝑦𝑛+1,𝑥𝑛+1)) ≥ 1 𝛼 (𝐻𝑋, 𝐻𝑈, 𝐻𝐴) ≥1 (45) 𝑛∈N ∪{0} 𝑥 →𝑥𝑦 →𝑦 𝑛→+∞ 󳨐⇒ 𝛼 (𝐹𝑋, 𝐹𝑈, 𝐹𝐴) ≥1, for all and 𝑛 , 𝑛 as ,we have 2 2 2 where 𝛼:𝑋 ×𝑋 ×𝑋 →[0,∞)is a function. 𝛼 ((𝑥𝑛,𝑦𝑛) , (𝑥, 𝑦) , (𝑥, 𝑦)) ≥1, 𝐺 𝛼 (50) (e) The statement (d) holds if we replace the - - 𝛼 ((𝑦 ,𝑥 ),(𝑦,𝑥),(𝑦,𝑥))≥1 admissibility by rectangular 𝐺-𝛼-admissibility. 𝑛 𝑛 𝑛∈N ∪{0} (X,𝐺) 𝑏 𝑓:X2 → X for all . Let be a generalized -metric space, , Then, 𝑓 and 𝑔 have a coupled coincidence point in X. and 𝑔:X → X. In the rest of this paper unless otherwise stated, for all 𝑥, 𝑦, 𝑢, V,𝑧,𝑤∈X,let 𝑚 2 Proof. Let Ω2 be the generalized 𝑏-metric on X defined in 𝐹, 𝐻 : X2 → X2 𝑁𝑚 (𝑥, 𝑦, 𝑢, V,𝑧,𝑤) Lemma 34. Also, define the mappings by 𝑓 𝐹𝑋= (𝑓(𝑥,𝑦),𝑓(𝑦,𝑥))and 𝐻𝑋 = (𝑔𝑥, 𝑔𝑦), 𝑋 = (𝑥, 𝑦) as 2 𝑚 = {𝐺 (𝑓 (𝑥, 𝑦),𝑓 (𝑢, V) ,𝑓(𝑧,) 𝑤 ), in Lemma 34.Then,(X ,Ω2 ) is a generalized 𝑏-metric space max 2 2 (with the same parameter 𝑠 as X), such that 𝐹(X )⊆𝐻(X ). 𝐺(𝑓(𝑦,𝑥),𝑓(V,𝑢) ,𝑓(𝑤,) 𝑧 )} , (46) Moreover, the contractive condition (47)impliesthat 𝑚 𝑚 𝑁𝑔 (𝑥, 𝑦, 𝑢, V,𝑧,𝑤) 𝛼 (𝐻𝑋, 𝐻𝑈, 𝐻𝐴) 𝜓(𝑠Ω2 (𝐹𝑋, 𝐹𝑈, 𝐹𝐴)) 𝑚 𝑚 (51) = max {𝐺 (𝑔𝑥, 𝑔𝑢, 𝑔𝑧) ,𝐺(𝑔𝑦,𝑔V,𝑔𝑤)}. ≤𝜓(Ω2 (𝐻𝑋, 𝐻𝑈, 𝐻𝐴))−𝜑(Ω2 (𝐻𝑋, 𝐻𝑈, 𝐻𝐴))

2 Now, we have the following coupled coincidence point holds for all 𝑋, 𝑈, 𝐴∈ X .Also,onecanshowthatall result. conditions of Theorem 31 aresatisfiedfor𝐹 and 𝐻 and we 8 Abstract and Applied Analysis

have proved in Theorem 31 that, under these conditions, it Similarly, we can show that {𝐻𝑈𝑛} is 𝐺𝑏-convergent to ∗ ∗ follows that 𝐹 and 𝐻 have a coincidence point 𝑋=(𝑥, 𝑦) ∈ 𝐻𝑋 . Since the limit is unique, it follows that 𝐻𝑋 = 𝐻𝑋 . 2 X which is obviously a coupled coincidence point of 𝑓 and The compatibility of 𝑓 and 𝑔 yields that 𝐹 and 𝐻 are 𝑔. compatible, and hence 𝐹 and 𝐻 are weak compatible. Since 𝐻𝑋 = 𝐹𝑋,wehave𝐻𝐻𝑋=𝐻𝐹𝑋=𝐹𝐻𝑋.Let𝐻𝑋=𝐴. In the following theorem, we give a sufficient condition Then, 𝐻𝐴 = 𝐹𝐴.Thus,𝐴 is another coincidence point of 𝐹 for the uniqueness of the common coupled fixed point (see and 𝐻.Then,𝐴=𝐻𝑋=𝐻𝐴. Therefore, 𝐴 = (𝑎, 𝑏) is a also [23]). coupled common fixed point of 𝑓 and 𝑔. To prove the uniqueness, assume that 𝑃 is another Theorem 36. In addition to the hypotheses of Theorem 35, common fixed point of 𝐹 and 𝐻.Then,𝑃=𝐻𝑃=𝐹𝑃and suppose that 𝑓 and 𝑔 are commutative and that, for all 𝐻𝑃 = 𝐻𝐴 𝑃=𝐻𝑃=𝐻𝐴=𝐴 ∗ ∗ 2 2 also .Thus, .Hence,thecoupled (𝑥, 𝑦) and (𝑥 ,𝑦 )∈X , there exists (𝑢, V)∈X , (𝑎, 𝑏) ∗ common fixed point is unique. Also, if is a common such that 𝛼((𝑔𝑥, 𝑔𝑦), (𝑔𝑢,𝑔V), (𝑔𝑢,V 𝑔 )) ≥ 1 and 𝛼((𝑔𝑥 , 𝑓 𝑔 (𝑏, 𝑎) ∗ coupled fixed point of and ,then is also a common 𝑔𝑦 ), (𝑔𝑢,V 𝑔 ), (𝑔𝑢,V 𝑔 )) ≥ 1.Then,𝑓 and 𝑔 have a unique coupledfixedpointof𝑓 and 𝑔. Uniqueness of the common common coupled fixed point of the form (𝑎, 𝑎). coupledfixedpointyieldsthat𝑎=𝑏.

Proof. Wewillusethenotationsasintheproofof 𝑎 2 2 2 + Let Ω : X × X × X → R be given by Theorem 35. It was proved in this theorem that the set of 2 coupled coincidence points of 𝑓 and 𝑔;thatis,thesetof 2 𝐺 (𝑥, 𝑢,) 𝑎 +𝐺(𝑦,V,𝑏) coincidence points of 𝐹 and 𝐻 in X is nonempty. We will Ω𝑎 (𝑋, 𝑈, 𝐴) = , ∗ 2 2 show that if 𝑋 and 𝑋 are coincidence points of 𝐹 and 𝐻, (56) that is, 𝑋=(𝑥,𝑦),( 𝑈= 𝑢, V) ,𝐴=(𝑎,) 𝑏 ∈ X2, 𝐻𝑋 = 𝐹𝑋, 𝐻𝑋∗ =𝐹𝑋∗, (52) 2 𝑎 and then (X ,Ω2) is a generalized 𝑏-metric space (with the ∗ then 𝐻𝑋=𝐻𝑋 . same parameter 𝑠). 2 2 Choose an element 𝑈 = (𝑢, V)∈X such Let (X,𝐺)be a generalized 𝑏-metric space, 𝑓:X → X, ∗ ∗ that 𝛼((𝑔𝑥, 𝑔𝑦), (𝑔𝑢,𝑔V), (𝑔𝑢,V 𝑔 )) ≥ 1 and 𝛼((𝑔𝑥 ,𝑔𝑦 ), and 𝑔:X → X.Forall𝑥, 𝑦, 𝑢, V,𝑧,𝑤∈X,let 2 (𝑔𝑢,V 𝑔 ), (𝑔𝑢,V 𝑔 )) ≥ 1.Let𝑈0 =𝑈and choose 𝑈1 ∈ X so 𝑎 that 𝐻𝑈1 =𝐹𝑈0. Then, we can inductively define a sequence 𝑁𝑓 (𝑥,𝑦,𝑢,V,𝑧,𝑤) {𝐻𝑈𝑛} such that 𝐻𝑈𝑛+1 =𝐹𝑈𝑛. 𝛼((𝑔𝑥, 𝑔𝑦) (𝑔𝑢,V 𝑔 ) (𝑔𝑢,V 𝑔 )) ≥ 1 𝑓 𝐺(𝑓(𝑥,𝑦),𝑓(𝑢, V) ,𝑓(𝑧,) 𝑤 ) As , , and is rect- = angular 𝐺-𝛼-admissible with respect to 𝑔,then𝛼((𝑓(𝑥, 𝑦), 2 𝑓(𝑦, 𝑥)), (𝑓(𝑢, V), 𝑓(V, 𝑢)), (𝑓(𝑢, V), 𝑓(V, 𝑢))) ≥ 1 ;thatis, 𝐺(𝑓(𝑦,𝑥),𝑓(V,𝑢) ,𝑓(𝑤,) 𝑧 ) 𝛼(𝐻𝑋,0 𝐻𝑈 ,𝐻𝑈0)≥1yields that + , 2 𝛼 (𝐹𝑋, 𝐹𝑈, 𝐹𝑈) =𝛼(𝐹𝑋,𝐹𝑈0,𝐹𝑈0) 𝑎 𝐺 (𝑔𝑥, 𝑔𝑢, 𝑔𝑧)V +𝐺(𝑔𝑦,𝑔 ,𝑔𝑤) 𝑁𝑔 (𝑥,𝑦,𝑢,V,𝑧,𝑤)= . =𝛼(𝐻𝑋,𝐻𝑈1,𝐻𝑈1) (53) 2 (57) ≥1. Remark 37. The result of Theorems 35 and 36 holds, if we Therefore, by the mathematical induction, we obtain that 𝑚 𝑚 𝑚 𝑎 𝑎 𝑎 replace Ω2 , 𝑁𝑓 ,and𝑁𝑔 by Ω2, 𝑁𝑓,and𝑁𝑔,respectively. 𝛼(𝐻𝑋,𝑛 𝐻𝑈 ,𝐻𝑈𝑛)≥1,forall𝑛≥0. Applying (47), one obtains that 4. Coupled Fixed Point Results in Partially 𝜓(𝑠Ω𝑚 (𝐹𝑋, 𝐻𝑈 ,𝐻𝑈 )) 2 𝑛+1 𝑛+1 Ordered Generalized 𝑏-Metric Spaces ≤𝛼(𝐻𝑋,𝐻𝑈,𝐻𝑈 )𝜓(𝑠Ω𝑚 (𝐹𝑋, 𝐹𝑈 ,𝐹𝑈 )) 𝑛 𝑛 2 𝑛 𝑛 We will use the following simple lemma in proving our 𝑚 𝑚 results. ≤𝜓(Ω2 (𝐻𝑋,𝑛 𝐻𝑈 ,𝐻𝑈𝑛)) − 𝜑 2(Ω (𝐻𝑋,𝑛 𝐻𝑈 ,𝐻𝑈𝑛))

𝑚 Lemma 38. Let (X,𝐺𝑏,⪯)be an ordered generalized 𝑏-metric ≤𝜓(Ω2 (𝐻𝑋,𝑛 𝐻𝑈 ,𝐻𝑈𝑛)) . 2 space (with the parameter 𝑠)andlet𝑓:X → X and 𝑔: (54) 2 2 X → X.Let𝐹:X → X be given by From the properties of 𝜓,wededucethatthesequence 𝑚 2 {Ω2 (𝐻𝑋,𝑛 𝐻𝑈 ,𝐻𝑈𝑛)} is nonincreasing. Hence, if we proceed 𝐹𝑋=(𝑓(𝑥,𝑦),𝑓(𝑦,𝑥)),X 𝑋=(𝑥,𝑦)∈ (58) as in Theorem 31,wecanshowthat 2 2 Ω𝑚 (𝐻𝑋, 𝐻𝑈 ,𝐻𝑈 )=0; and 𝐻:X → X is defined by 𝑛→∞lim 2 𝑛 𝑛 (55) 2 𝐻𝑋 = (𝑔𝑥, 𝑔𝑦) , 𝑋=(𝑥,𝑦)∈ X . (59) that is, {𝐻𝑈𝑛} is 𝐺𝑏-convergent to 𝐻𝑋. Abstract and Applied Analysis 9

2 (a) If a relation ⊑2 is defined on X by deduce that 𝑋⊑2 𝑈;thatis,𝛼(𝑋, 𝑈, 𝑈).Thus, ≥1 𝐹 is a 𝐺 𝛼 𝐻 2 rectangular - -admissible mapping with respect to . 𝑋⊑2 𝑈⇐⇒𝑥⪯𝑢∧𝑦⪰V, 𝑋=(𝑥,𝑦),( 𝑈= 𝑢, V) ∈ X , From (62) and the definition of 𝛼 and ⊑2, (60) 𝑚 𝜓(𝑠Ω2 (𝐹𝑋, 𝐹𝑈, 𝐹𝐴)) (X2,Ω𝑚,⊑ ) (X2,Ω𝑎,⊑ ) 𝑏 then 2 2 and 2 2 are ordered generalized - ≤𝜓(Ω𝑚 (𝐻𝑋, 𝐻𝑈, 𝐻𝐴))−𝜑(Ω𝑚 (𝐻𝑋, 𝐻𝑈, 𝐻𝐴)), metric spaces (with the same parameter 𝑠). 2 2 (65) (b) If the mapping 𝑓 has the 𝑔-mixed monotone property, 𝐹:X2 → X2 𝐺 2 then the mapping is -nondecreasing with for all 𝑋, 𝑈, 𝐴 ∈𝑋 with 𝐻𝑋 ⊑2 𝐻𝑈 ⊑2 𝐻𝐴. Moreover, from ⊑ 2 respect to 2;thatis, (2) there exists (𝑥0,𝑦0)∈𝑋 such that 𝐻𝑋 ⊑ 𝐻𝑈 󳨐⇒ 𝐹𝑋⊑ 𝐹𝑈. 2 2 (61) 𝐻(𝑥0,𝑦0)=(𝑔𝑥0,𝑔𝑦0)⊑2 ((𝑥0,𝑦0),𝑓(𝑦0,𝑥0)) (66) (X,𝐺 ,⪯) Theorem 39. Let 𝑏 be a partially ordered generalized =𝐹(𝑥0,𝑦0). 2 𝑏-metric space with the parameter 𝑠 and let 𝑓:X → X and 𝑔:X → X Hence, all the conditions of Theorem 32 are satisfied and so . Assume that 2 𝐹 and 𝐻 have a coincidence point 𝑋=(𝑥, 𝑦) ∈ X which is 𝑚 𝑓 𝑔 𝜓(𝑠𝑁𝑓 (𝑥,𝑦,𝑢,V,𝑧,𝑤)) a coupled coincidence point of and .

≤𝜓(𝑁𝑚 (𝑥,𝑦,𝑢,V,𝑧,𝑤))−𝜑(𝑁𝑚 (𝑥, 𝑦, 𝑢, V,𝑧,𝑤)), In the following theorem, we give a sufficient condition 𝑔 𝑔 for the uniqueness of the common coupled fixed point (see (62) also [25, 28, 30]). 𝑥, 𝑦, 𝑢, V,𝑧,𝑤 ∈ X 𝑔𝑥⪯𝑔𝑢⪯𝑔𝑧 𝑔𝑦 ⪰ for all with and Theorem 40. In addition to the hypotheses of Theorem 39, 𝑔V ⪰𝑔𝑤 𝜓, 𝜑 : [0, ∞) → [0, ∞) ∗ ∗ 2 ,where are altering distance suppose that, for all (𝑥, 𝑦) and (𝑥 ,𝑦 )∈X , there exists 2 functions. (𝑢, V)∈X ,suchthat(𝑔𝑢,V 𝑔 ) is comparable with (𝑔𝑥, 𝑔𝑦) ∗ ∗ Assume also that and (𝑔𝑥 ,𝑔𝑦 ).Then,𝑓 and 𝑔 have a unique common coupled 2 (𝑎, 𝑎) (1) 𝑓 has the mixed 𝑔-monotone property and 𝑓(X )⊆ fixed point of the form . 𝑔(X); Proof. It was proved in Theorem 39 that the set of coupled 𝑥 ,𝑦 ∈ X 𝑔𝑥 ⪯𝑓(𝑥,𝑦 ) (2) there exist 0 0 such that 0 0 0 and coincidence points of 𝑓 and 𝑔, that is, the set of coincidence 𝑔𝑦 ⪰𝑓(𝑦,𝑥 ) 2 0 0 0 . points of 𝐹 and 𝐻 in X ,isnonempty.Wewillshowthatif𝑋 𝑋∗ 𝐹 𝐻 Also, suppose that either and are coincidence points of and ,thatis, 𝐻𝑋 = 𝐹𝑋, 𝐻𝑋∗ =𝐹𝑋∗, (a) 𝑓 and 𝑔 are continuous, the pair (𝑓,𝑔) is compatible, (67) and (X,𝐺)is 𝐺𝑏-complete, or ∗ 2 then 𝐻𝑋 = 𝐻𝑋 . There exists (𝑢, V)∈X ,suchthat(𝑔𝑢,V 𝑔 ) (X,𝐺 ) (𝑔(X), 𝐺) 𝐺 ∗ ∗ (b) 𝑏 is regular and is 𝑏-complete. is comparable with (𝑔𝑥, 𝑔𝑦) and (𝑔𝑥 ,𝑔𝑦 ).Withoutanyloss of generality, we may assume that (𝑔𝑥, 𝑔𝑦)2 ⊑ (𝑔𝑢,V 𝑔 ) and Then, 𝑓 and 𝑔 have a coupled coincidence point in X. ∗ ∗ (𝑔𝑥 ,𝑔𝑦 )⊑2 (𝑔𝑢,V 𝑔 ). According to the definition of 𝛼 in (X2,Ω𝑚,⊑ ) the above theorem, 𝛼((𝑔𝑥, 𝑔𝑦), (𝑔𝑢,𝑔V), (𝑔𝑢,V 𝑔 )) ≥ 1 and Proof. By Lemma 38, 2 2 is an ordered generalized 𝛼((𝑔𝑥∗,𝑔𝑦∗), (𝑔𝑢,V 𝑔 ), (𝑔𝑢,V 𝑔 )) ≥ 1 𝑏-metric space (with the same parameter 𝑠). . 2 2 2 Now, following the proof of Theorem 36,onecanobtain Define 𝛼:𝑋 ×𝑋 ×𝑋 →[0,+∞)by ∗ that 𝐻𝑋=𝐻𝑋 . The remainder part of proof is analogous to 1, (𝑥, 𝑦) ⊑ (𝑢, V) ⊑ (𝑎,) 𝑏 , the proof of Theorem 36 and so we omit it. 𝛼 ((𝑥, 𝑦), (𝑢, V) , (𝑎,) 𝑏 )={ if 2 2 0, . otherwise Remark 41. In Theorem 39,wecanreplacethecontractive (63) condition (62)bythefollowing: First, we prove that 𝐹 is a rectangular 𝐺-𝛼-admissible 𝛼 ((𝑔𝑥, 𝑔𝑦)V ,(𝑔𝑢,𝑔 ) , (𝑔𝑧, 𝑔𝑤)) 𝐻 𝛼(𝐻𝑋, mapping with respect to . Hence, we assume that 𝑎 𝐻𝑈, 𝐻𝐴),where ≥1 𝑋 = (𝑥,, 𝑦) 𝑈=(𝑢,V),and𝐴 = (𝑎, 𝑏). ×𝜓(𝑠𝑁𝑓 (𝑥, 𝑦, 𝑢, V,𝑧,𝑤)) Therefore, we have 𝐻𝑋 ⊑2 𝐻𝑈2 ⊑ 𝐻𝐴.Since𝑓 has the mixed 𝑎 𝑎 𝑔-monotone property, then from Lemma 38,themapping ≤𝜓(𝑁𝑔 (𝑥,𝑦,𝑢,V,𝑧,𝑤))−𝜑(𝑁𝑔 (𝑥,𝑦,𝑢,V,𝑧,𝑤)). 2 2 𝐹:X → X is 𝐺-nondecreasing with respect to ⊑2;that (68) is, Remark 42. Theorem 39 provides conclusions of Theorems 𝐹𝑋 ⊑2 𝐹𝑈2 ⊑ 𝐹𝐴; (64) 3.1 and 3.2 of [4] for more general pair of compatible maps. that is, 𝛼(𝐹𝑋, 𝐹𝑈,.Also,let 𝐹𝐴)≥1 𝛼(𝑋, 𝐴, 𝐴) ≥1 and In Theorem 39,ifwetake𝜓(𝑡) =𝑡 for all 𝑡∈[0,∞),we 𝛼(𝐴, 𝑈, 𝑈);then ≥1 𝑋⊑2 𝐴 and 𝐴⊑2 𝑈.Consequently,we obtain the following result. 10 Abstract and Applied Analysis

Corollary 43. Let (X,𝐺,⪯) be a partially ordered 𝐺𝑏- (a) 𝑓 is continuous, or complete generalized 𝑏-metric space with the parameter 𝑠≥1 2 (X,𝐺) and let 𝑓:X → X. Assume that (b) is regular. 𝐺(𝑓(𝑥,𝑦),𝑓(𝑢, V) ,𝑓(𝑧,) 𝑤 ) Then, 𝑓 has a coupled fixed point in X. ∗ ∗ 2 2 In addition, suppose that, for all (𝑥, 𝑦) and (𝑥 ,𝑦 )∈X , 2 there exists (𝑢, V)∈X ,suchthat(𝑢, V) is comparable with 𝐺(𝑓(𝑦,𝑥),𝑓(V,𝑢) ,𝑓(𝑤,) 𝑧 ) ∗ ∗ + (𝑥, 𝑦) and (𝑥 ,𝑦 ).Then,𝑓 has a unique coupled fixed point 2 of the form (𝑎, 𝑎). (69) 1 𝐺 (𝑔𝑥, 𝑔𝑢, 𝑔𝑧)V +𝐺(𝑔𝑦,𝑔 ,𝑔𝑤) ≤ Now, we present an example to illustrate Theorem 39 and 𝑠 2 Remark 41. 1 𝐺 (𝑔𝑥, 𝑔𝑢, 𝑔𝑧)V +𝐺(𝑔𝑦,𝑔 ,𝑔𝑤) − 𝜑( ), Example 45. Let 𝑋=R be endowed with the usual 𝑠 2 ordering and let 𝐺𝑏-metric 𝐺 on 𝑋 be given by 𝐺(𝑥, 𝑦, 𝑧)= 2 2 2 𝑥, 𝑦, 𝑢, V,𝑧,𝑤 ∈ X 𝑥⪯𝑢⪯𝑧 𝑦⪰V ⪰𝑤 max{|𝑥 − 𝑦| ,|𝑦−𝑧| ,|𝑥−𝑧| },where𝑠=2.Define𝐹:𝑋× for all with and , 𝑋→𝑋 where 𝜑 : [0, ∞) → [0, ∞) is an altering distance function. as Assume also that 𝑥−𝑦 2 𝐹(𝑥,𝑦)= , (𝑥,𝑦)∈𝑋×𝑋. (71) (1) 𝑓 has the mixed 𝑔-monotone property and 𝑓(X )⊆ 9 𝑔(X); 𝜓, 𝜑 : [0, ∞) → [0, ∞) (2) there exist 𝑥0,𝑦0 ∈ X such that 𝑔𝑥0 ⪯𝑓(𝑥0,𝑦0) and We define by 𝑔𝑦0 ⪰𝑓(𝑦0,𝑥0). 𝑡+1 𝜓 (𝑡) = (𝑡+1) ,𝜑(𝑡) = ( ), Also, suppose that either ln ln 𝑐𝑡 +1 (72) (a) 𝑓 is continuous, or 𝑐 = 8/81 𝐹 (X,𝐺) where .Also, has mixed monotone property and (b) is regular. satisfies the condition (68). Indeed, for all 𝑥, 𝑦, 𝑢, V,𝑎,𝑏∈ 𝑋 Then, 𝑓 and 𝑔 have a coupled coincidence point in X. with 𝑥≤𝑢≤𝑎and 𝑦≥V ≥𝑏,wehave ∗ ∗ 2 In addition, suppose that, for all (𝑥, 𝑦) and (𝑥 ,𝑦 )∈X , 2 there exists (𝑢, V)∈X ,suchthat(𝑢, V) is comparable with 𝜓(2(𝐺(𝐹(𝑥,𝑦),𝐹(𝑢, V) ,𝐹(𝑎,) 𝑏 ))) ∗ ∗ (𝑥, 𝑦) and (𝑥 ,𝑦 ).Then,𝑓 and 𝑔 have a unique coupled 𝑥−𝑦 𝑢−V 2 𝑢−V 𝑎−𝑏 2 fixed point of the form (𝑎, 𝑎). = (2 {( − ) ,( − ) , ln max 9 9 9 9 In Corollary 43,ifwetake𝜑(𝑡) = (1−𝑘)𝑡 for all 𝑡∈[0,∞), 𝑘∈[0,1) 𝑎−𝑏 𝑥−𝑦 2 where , we obtain the following extension of the ( − ) }+1) results by Choudhury and Maity (Theorems 19 and 20). 9 9

2 2 Corollary 44. Let (X,𝐺,⪯) be a partially ordered 𝐺𝑏- 𝑥−𝑢+V −𝑦 𝑢−𝑎+𝑏−V = (2 {( ) ,( ) , complete generalized 𝑏-metric space with the parameter 𝑠≥1 ln max 9 9 2 and let 𝑓:X → X. Assume that 𝑎−𝑥+𝑦−𝑏 2 𝐺(𝑓(𝑥,𝑦),𝑓(𝑢, V) ,𝑓(𝑧,) 𝑤 ) ( ) }+1) 9 2 2 2 2 𝐺(𝑓(𝑦,𝑥),𝑓(V,𝑢) ,𝑓(𝑤,) 𝑧 ) = ln ( max {(𝑥−𝑢+V −𝑦) , (𝑢−𝑎+𝑏−V) , + (70) 81 2 2 𝑘 𝐺 (𝑥, 𝑢,) 𝑧 +𝐺(𝑦,V,𝑤) (𝑎−𝑥+𝑦−𝑏) }+1) ≤ , 𝑠 2 4 ≤ ( {(𝑥−𝑢)2 +(V −𝑦)2, (𝑢−𝑎)2 for all 𝑥, 𝑦, 𝑢, V,𝑧,𝑤∈ X with 𝑥⪯𝑢⪯𝑧and 𝑦⪰V ⪰𝑤,or ln 81 max 𝑧⪯𝑢⪯𝑥and 𝑤⪰V ⪰𝑦,where𝑘∈[0,1). Assume also that +(𝑏−V)2, (𝑎−𝑥)2 +(𝑦−𝑏)2}+1) 𝑓 (1) has the mixed monotone property; 4 ≤ ( [ {(𝑥−𝑢)2, (𝑢−𝑎)2, (𝑎−𝑥)2} (2) there exist 𝑥0,𝑦0 ∈ X such that 𝑥0 ⪯𝑓(𝑥0,𝑦0) and ln 81 max 𝑦0 ⪰𝑓(𝑦0,𝑥0). + {(V −𝑦)2, (𝑏−V)2,(𝑦−𝑏)2}] + 1) Also, suppose that either max Abstract and Applied Analysis 11

8 𝐺 (𝑥, 𝑢,) 𝑎 +𝐺(𝑦,V,𝑏) does not hold for all 𝑥, 𝑦, 𝑢, V,𝑎,𝑏∈ 𝑋such that 𝑥≥𝑢≥𝑎 = ln ( +1) 81 2 and 𝑦≤V ≤𝑏.Indeed,for𝑥=1, 𝑦=𝑢=V =𝑎=𝑏=0it reduces to 𝐺 (𝑥, 𝑢,) 𝑎 +𝐺(𝑦,V,𝑏) = (𝑐 +1) 𝜓 (𝐺 (𝐹 (1, 0) ,𝐹(0, 0) ,𝐹(0, 0))) ln 2 1 𝐺 (𝑥, 𝑢,) 𝑎 +𝐺(𝑦,V,𝑏) =𝜓( ) = 0.10536051565 = ( +1) 9 ln 2 ≰ 0.4054651081 − 0.35726300629 ((𝐺 (𝑥, 𝑢,) 𝑎 +𝐺(𝑦,V,𝑏))/2)+1 − ( ) 1 1 ln =𝜓( )−𝜑( ) (78) 𝑐 ((𝐺 (𝑥, 𝑢,) 𝑎 +𝐺(𝑦,V,𝑏))/2)+1 2 2 𝐺 (𝑥, 𝑢,) 𝑎 +𝐺(𝑦,V,𝑏) 𝐺 (1, 0, 0) +𝐺(0, 0, 0) =𝜓( ) =𝜓( ) 2 2

𝐺 (𝑥, 𝑢,) 𝑎 +𝐺(𝑦,V,𝑏) 𝐺 (1, 0, 0) +𝐺(0, 0, 0) −𝜑( ). −𝜑( ). 2 2 𝐺 (73) We conclude that, using a 𝑏-metric instead of the standard one, one has more possibilities for choosing a control func- Similarly, tion in order to get a coupled fixed point result. 𝜓(2(𝐺(𝐹(𝑦,𝑥),𝐹(V,𝑢)),𝐹(𝑏,) 𝑎 )) Remark 46. Theorems 3.5, 3.6, and 3.7 of[4]arespecialcases of Theorem 39. 𝐺 (𝑥, 𝑢,) 𝑎 +𝐺(𝑦,V,𝑏) ≤𝜓( ) 2 (74) 5. Tripled Coincidence Point Results 𝐺 (𝑥, 𝑢,) 𝑎 +𝐺(𝑦,V,𝑏) −𝜑( ). In this section we prove some tripled coincidence and tripled 2 common fixed point results.

So, Lemma 47. Let (X,𝐺,⪯)be an ordered generalized 𝑏-metric 3 space (with the parameter 𝑠)andlet𝑓:X → X and 𝑔: 𝐺(𝐹(𝑥,𝑦),𝐹(𝑢, V) ,𝐹(𝑎,) 𝑏 ) X → X. 𝜓(2( 3 2 (a) If a relation ⊑3 is defined on X by 𝐺(𝐹(𝑦,𝑥),𝐹(V,𝑢) ,𝐹(𝑏,) 𝑎 ) 𝑋⊑ 𝑈⇐⇒𝑥⪯𝑢∧𝑦⪰V ∧𝑧⪯𝑤, + )) 3 2 (79) 𝑋=(𝑥,𝑦,𝑧),( 𝑈= 𝑢, V,𝑤) ∈ X3, (75) 𝐺 (𝑥, 𝑢,) 𝑎 +𝐺(𝑦,V,𝑏) 𝑚 3 3 3 + ≤𝜓( ) and a mapping Ω : X × X × X → R is given by 2 3 𝑚 Ω3 (𝑋, 𝑈, 𝐴) = max {𝐺 (𝑥, 𝑢,) 𝑎 ,𝐺(𝑦,V,𝑏),𝐺(𝑧, 𝑤,) 𝑐 }, 𝐺 (𝑥, 𝑢,) 𝑎 +𝐺(𝑦,V,𝑏) −𝜑( ). (80) 2 3 for all 𝑋=(𝑥,𝑦,𝑧), 𝑈 = (𝑢, V,𝑤),and𝐴=(𝑎,𝑏,𝑐)∈X ,then 3 𝑚 Finally, there are obviously 𝑥0,𝑦0 ∈𝑋such that 𝑥0 ≤ (X ,Ω3 ,⊑3) is an ordered generalized 𝑏-metric space (with the 𝐹(𝑥 ,𝑦 ) 𝑦 ≥𝐹(𝑦,𝑥 ) 3 𝑚 0 0 and 0 0 0 .Thus,weconcludethatthe same parameter 𝑠). The space (X ,Ω3 ) is 𝐺𝑏-complete if and 𝐹 (0, 0) mapping has a coupled fixed point (which is ). only if (X,𝐺)is 𝐺𝑏-complete. Consider now the same example, but with the 𝐺-metric (b) If the mapping 𝑓 has the 𝑔-mixed monotone property, 3 3 󵄨 󵄨 󵄨 󵄨 then the mapping 𝐹:X → X given by 𝐺(𝑥,𝑦,𝑧)=max {󵄨𝑥−𝑦󵄨 , 󵄨𝑦−𝑧󵄨 , |𝑥−𝑧|} (76) 𝐹𝑋 = (𝑓 (𝑥, 𝑦, 𝑧) , 𝑓 (𝑦, 𝑥, 𝑦) ,𝑓(𝑧,𝑦,𝑥)) on 𝑋=R.Therespectivecontractivecondition (81) 𝑋=(𝑥,𝑦,𝑧)∈X3 𝜓(𝐺(𝐹(𝑥,𝑦),𝐹(𝑢, V) ,𝐹(𝑎,) 𝑏 )) is 𝐺-nondecreasing with respect to ⊑3;thatis, 𝐺 (𝑥, 𝑢,) 𝑎 +𝐺(𝑦,V,𝑏) ≤𝜓( ) 𝐻𝑋 ⊑3 𝐻𝑈 󳨐⇒3 𝐹𝑋⊑ 𝐹𝑈, (82) 2 (77) 𝐻:X3 → X3 𝐺 (𝑥, 𝑢,) 𝑎 +𝐺(𝑦,V,𝑏) where is defined by −𝜑( ) 3 2 𝐻𝑋 = (𝑔𝑥, 𝑔𝑦, 𝑔𝑧) ,𝑋=(𝑥,𝑦,𝑧)∈ X . (83) 12 Abstract and Applied Analysis

3 𝑚 (c) If 𝑓 is continuous from (X ,Ω3 ) to (X,𝐺),then𝐹 is (2) there exist 𝑥0,𝑦0,𝑧0 ∈ X such that 3 𝑚 continuous in (X ,Ω3 ). (d) If 𝑓 and 𝑔 are compatible, then 𝐹 and 𝐻 are compatible. 𝛼 ((𝑔𝑥 ,𝑔𝑦 ,𝑔𝑧 ), 3 0 0 0 (e) The mapping 𝑓:𝑋 →𝑋is 𝐺-𝛼-admissible with respect to 𝑔;thatis, (𝑓 0(𝑥 ,𝑦0,𝑧0),𝑓(𝑦0,𝑥0,𝑦0),𝑓(𝑧0,𝑦0,𝑥0)) ,

𝑥, 𝑦, 𝑧, 𝑢, V,𝑤,𝑎,𝑏,𝑐∈𝑋, (𝑓 0(𝑥 ,𝑦0,𝑧0),𝑓(𝑦0,𝑥0,𝑦0),𝑓(𝑧0,𝑦0,𝑥0))) ≥ 1,

𝛼 ((𝑔𝑥, 𝑔𝑦, 𝑔𝑧) , (𝑔𝑦, 𝑔𝑥, 𝑔𝑦) ,(𝑔𝑧,𝑔𝑦,𝑔𝑥))≥1 𝛼 ((𝑔𝑦0,𝑔𝑥0,𝑔𝑦0),

󳨐⇒ 𝛼 ((𝑓 (𝑥, 𝑦, 𝑧) , 𝑓 (𝑦, 𝑥, 𝑦),𝑓(𝑧,𝑦,𝑥)) (84) (𝑓 0(𝑦 ,𝑥0,𝑦0),𝑓(𝑥0,𝑦0,𝑧0),𝑓(𝑦0,𝑥0,𝑦0)) , (88)

(𝑓 (𝑢, V,𝑤) ,𝑓(V,𝑢,V) ,𝑓(𝑤, V,𝑢)), (𝑓 0(𝑦 ,𝑥0,𝑦0),𝑓(𝑥0,𝑦0,𝑧0),𝑓(𝑦0,𝑥0,𝑦0))) ≥ 1,

(𝑓 (𝑎, 𝑏,) 𝑐 ,𝑓(𝑏, 𝑎,) 𝑏 ,𝑓(𝑐, 𝑏, 𝑎))) ≥1 𝛼 ((𝑔𝑧0,𝑔𝑦0,𝑔𝑥0),

3 3 if and only if the mapping 𝐹:X → X is 𝐺-𝛼-admissible (𝑓 0(𝑧 ,𝑦0,𝑥0),𝑓(𝑦0,𝑥0,𝑦0),𝑓(𝑥0,𝑦0,𝑧0)) , with respect to 𝐻;thatis, (𝑓 0(𝑧 ,𝑦0,𝑥0),𝑓(𝑦0,𝑥0,𝑦0),𝑓(𝑥0,𝑦0,𝑧0))) ≥ 1. 𝑋, 𝑈, 𝐴 ∈𝑋3, 𝛼 (𝐻𝑋, 𝐻𝑈, 𝐻𝐴) ≥1 (85) Also, suppose that either 󳨐⇒ 𝛼 (𝐹𝑋, 𝐹𝑈, 𝐹𝐴) ≥1, (a) 𝑓 and 𝑔 are continuous, the pair (𝑓,𝑔) is compatible, 3 3 3 where 𝛼:𝑋 ×𝑋 ×𝑋 →[0,∞)is a function. and (X,𝐺)is 𝐺𝑏-complete, or

Let (X,𝐺,⪯) be an ordered generalized 𝑏-metric space, (b) (𝑔(X), 𝐺) is 𝐺𝑏-complete and assume that whenever 3 𝑓:X → X,and𝑔:X → X.Forall𝑥, 𝑦, 𝑧, 𝑢, V,𝑤,𝑎,𝑏,𝑐∈ {𝑥𝑛}, {𝑦𝑛}, {𝑧𝑛} in 𝑋 are sequences such that X,let 𝑚 𝛼((𝑥𝑛,𝑦𝑛,𝑧𝑛),(𝑥𝑛+1,𝑦𝑛+1,𝑧𝑛+1),(𝑥𝑛+1,𝑦𝑛+1,𝑧𝑛+1)) ≥ 1, 𝑀𝑓 (𝑥,𝑦,𝑧,𝑢,V,𝑤,𝑎,𝑏,𝑐) 𝛼 ((𝑦 ,𝑥 ,𝑦 ),(𝑦 ,𝑥 ,𝑦 ),(𝑦 ,𝑥 ,𝑦 )) ≥ 1, = max {𝐺 (𝑓 (𝑥, 𝑦, 𝑧),𝑓 (𝑢, V,𝑤) ,𝑓(𝑎, 𝑏,) 𝑐 ), 𝑛 𝑛 𝑛 𝑛+1 𝑛+1 𝑛+1 𝑛+1 𝑛+1 𝑛+1 𝛼((𝑧 ,𝑦 ,𝑥 ),(𝑧 ,𝑦 ,𝑥 ),(𝑧 ,𝑦 ,𝑥 )) ≥ 1 𝐺 (𝑓 (𝑦, 𝑥, 𝑦),𝑓 (V,𝑢,V) ,𝑓(𝑏, 𝑎,) 𝑏 ), 𝑛 𝑛 𝑛 𝑛+1 𝑛+1 𝑛+1 𝑛+1 𝑛+1 𝑛+1 (89) 𝐺 (𝑓 (𝑧, 𝑦, 𝑥),𝑓 (𝑤, V,𝑢) ,𝑓(𝑐, 𝑏, 𝑎))} ,

𝑀𝑚 (𝑥,𝑦,𝑧,𝑢,V,𝑤,𝑎,𝑏,𝑐) 𝑔 for all 𝑛∈N ∪{0}and 𝑥𝑛 →𝑥, 𝑦𝑛 →𝑦,and𝑧𝑛 →𝑧as 𝑛→+∞,wehave = max {𝐺(𝑔𝑥,𝑔𝑢,𝑔𝑎),𝐺(𝑔𝑦,𝑔V, 𝑔𝑏) , 𝐺 (𝑔𝑧, 𝑔𝑤, 𝑔𝑐)}. (86) 𝛼 ((𝑥𝑛,𝑦𝑛,𝑧𝑛) , (𝑥, 𝑦, 𝑧) , (𝑥, 𝑦, 𝑧))≥1, Theorem 48. Let (X,𝐺)be a generalized 𝑏-metric space with 3 𝛼((𝑦𝑛,𝑥𝑛,𝑦𝑛),(𝑦,𝑥,𝑦),(𝑦,𝑥,𝑦))≥1, (90) the parameter 𝑠 and let 𝑓:X → X and 𝑔:X → X. Assume that 𝛼 ((𝑧𝑛,𝑦𝑛,𝑥𝑛),(𝑧,𝑦,𝑥),(𝑧,𝑦,𝑥))≥1 𝛼 ((𝑔𝑥, 𝑔𝑦, 𝑔𝑧) ,(𝑔𝑢,𝑔V,𝑔𝑤),(𝑔𝑎,𝑔𝑏,𝑔𝑐)) 𝑛∈N ∪{0} 𝑚 for all . ×𝜓(𝑠𝑀𝑓 (𝑥,𝑦,𝑧,𝑢,V,𝑤,𝑎,𝑏,𝑐)) Then, 𝑓 and 𝑔 have a tripled coincidence point in X.

𝑚 (87) ≤𝜓(𝑀𝑔 (𝑥,𝑦,𝑧,𝑢,V,𝑤,𝑎,𝑏,𝑐)) Theorem 49. In addition to the hypotheses of Theorem 48, suppose that 𝑓 and 𝑔 are commutative and that, for all (𝑥,𝑦,𝑧) 𝑚 ∗ ∗ ∗ 3 3 −𝜑(𝑀𝑔 (𝑥,𝑦,𝑧,𝑢,V,𝑤,𝑎,𝑏,𝑐)), and (𝑥 ,𝑦 ,𝑧 )∈X , there exists (𝑢, V,𝑤)∈X ,suchthat 𝑥, 𝑦, 𝑧, 𝑢, V,𝑤,𝑎,𝑏,𝑐 ∈ X 𝜓, 𝜑 : [0, ∞)→ for all where 𝛼((𝑔𝑥,𝑔𝑦,𝑔𝑧),(𝑔𝑢,𝑔V,𝑔𝑤),(𝑔𝑢,𝑔V,𝑔𝑤))≥1, [0, ∞) 𝛼:(𝑋3)3 →[0,∞) are altering distance functions and (91) is a mapping such that 𝑓 is a rectangular 𝐺-𝛼-admissible 𝛼((𝑔𝑥∗,𝑔𝑦∗,𝑔𝑧∗),(𝑔𝑢,𝑔V,𝑔𝑤),(𝑔𝑢,𝑔V,𝑔𝑤))≥1. mapping with respect to 𝑔. Assume also that Then, 𝑓 and 𝑔 have a unique common tripled fixed point of the 3 (1) 𝑓(X )⊆𝑔(X); form (𝑎, 𝑎, 𝑎). Abstract and Applied Analysis 13

Let (b) (X,𝐺)is regular and (𝑔(X), 𝐺) is 𝐺𝑏-complete. 𝐺 (𝑥, 𝑢,) 𝑎 +𝐺(𝑦,V,𝑏)+𝐺(𝑧, 𝑤,) 𝑐 Ω𝑎 (𝑋, 𝑈, 𝐴) = , Then, 𝑓 and 𝑔 have a tripled coincidence point in X. 3 3 Theorem 52. In addition to the hypotheses of Theorem 48, 𝑋=(𝑥,𝑦,𝑧),( 𝑈= 𝑢, V,𝑤) ,𝐴=(𝑎, 𝑏,) 𝑐 ∈ X3, suppose that 𝑓 and 𝑔 are commutative and that, for all ∗ ∗ ∗ 3 3 𝑎 (𝑥, 𝑦, 𝑧) and (𝑥 ,𝑦 ,𝑧 )∈X , there exists (𝑢, V,𝑤) ∈ X , 𝑀𝑓 (𝑥,𝑦,𝑧,𝑢,V,𝑤,𝑎,𝑏,𝑐) such that (𝑔𝑢,V 𝑔 ,𝑔𝑤) is comparable with (𝑔𝑥, 𝑔𝑦, 𝑔𝑧) and ∗ ∗ ∗ 𝐺(𝑓(𝑥,𝑦,𝑧),𝑓(𝑢, V,𝑤) ,𝑓(𝑎, 𝑏,) 𝑐 ) (𝑔𝑥 ,𝑔𝑦 ,𝑔𝑧 ).Then,𝑓 and 𝑔 have a unique common tripled = (𝑎, 𝑎, 𝑎) 3 fixed point of the form . 𝐺 (𝑓 (𝑦, 𝑥, 𝑦),𝑓 (V,𝑢,V) ,𝑓(𝑏, 𝑎,) 𝑏 ) Remark 53. In Theorem 51,wecanreplacethecontractive + 3 condition (94)bythefollowing:

𝐺 (𝑓 (𝑧, 𝑦, 𝑥),𝑓 (𝑤, V,𝑢) ,𝑓(𝑐, 𝑏, 𝑎)) 𝜓(𝑠𝑀𝑎 (𝑥,𝑦,𝑧,𝑢,V,𝑤,𝑎,𝑏,𝑐)) + , 𝑓 3 𝑎 𝑎 ≤𝜓(𝑀𝑔 (𝑥,𝑦,𝑧,𝑢,V,𝑤,𝑎,𝑏,𝑐)) (95) 𝑀𝑔 (𝑥,𝑦,𝑧,𝑢,V,𝑤,𝑎,𝑏,𝑐) 𝑎 𝐺 (𝑔𝑥, 𝑔𝑢, 𝑔𝑎)V +𝐺(𝑔𝑦,𝑔 ,𝑔𝑏)+𝐺(𝑔𝑧,𝑔𝑤,𝑔𝑐) −𝜑(𝑀𝑔 (𝑥,𝑦,𝑧,𝑢,V,𝑤,𝑎,𝑏,𝑐)). = . 3 (92) Remark 54. Theorem 51 extends Theorem 2.1 of [36]toa compatible pair. Remark 50. In Theorem 48,wecanreplacethecontractive condition (87)bythefollowing: Remark 55. Corollaries 2.2, 2.3, 2.6, 2.7, and 2.8 of [36]are special cases of Theorem 51. 𝛼 ((𝑔𝑥, 𝑔𝑦, 𝑔𝑧) ,(𝑔𝑢,𝑔V,𝑔𝑤),(𝑔𝑎,𝑔𝑏,𝑔𝑐)) Remark 56. Theorem 25 is a special case of Theorems 51. 𝑎 ×𝜓(𝑠𝑀𝑓 (𝑥,𝑦,𝑧,𝑢,V,𝑤,𝑎,𝑏,𝑐)) (93) 𝑎 6. Application to Integral Equations ≤𝜓(𝑀𝑔 (𝑥,𝑦,𝑧,𝑢,V,𝑤,𝑎,𝑏,𝑐)) As an application of the (coupled) fixed point theorems estab- −𝜑(𝑀𝑎 (𝑥,𝑦,𝑧,𝑢,V,𝑤,𝑎,𝑏,𝑐)). 𝑔 lished in Section 4, we study the existence and uniqueness of a solution to a Fredholm nonlinear integral equation. The following tripled fixed point results in ordered metric In order to compare our results to the ones in [37, 38]we spacescanbeobtained. will consider the same integral equation; that is,

Theorem 51. Let (X,𝐺,⪯)be a partially ordered generalized 𝑏 3 𝑏-metric space with the parameter 𝑠 and let 𝑓:X → X and 𝑥 (𝑡) = ∫ (𝐾1 (𝑡, 𝑠) +𝐾2 (𝑡, 𝑠))(𝑓(𝑠, 𝑥 (𝑠)) +𝑔(𝑠, 𝑥 (𝑠)))𝑑𝑠 𝑔:X → X. Assume that 𝑎 +ℎ(𝑡) , 𝜓(𝑠𝑀𝑚 (𝑥,𝑦,𝑧,𝑢,V,𝑤,𝑎,𝑏,𝑐)) 𝑓 (96) 𝑚 ≤𝜓(𝑀 (𝑥,𝑦,𝑧,𝑢,V,𝑤,𝑎,𝑏,𝑐)) (94) 𝑔 where 𝑡 ∈ 𝐼 = [𝑎, 𝑏]. Θ 𝜃 : [0, ∞) → [0, ∞) −𝜑(𝑀𝑚 (𝑥,𝑦,𝑧,𝑢,V,𝑤,𝑎,𝑏,𝑐)), Let denote the set of all functions 𝑔 satisfying the following. for all 𝑥, 𝑦, 𝑧, 𝑢, V,𝑤,𝑎,𝑏,𝑐 ∈ X with 𝑔𝑥⪯𝑔𝑢⪯𝑔𝑎, 𝑔𝑦 ⪰ 𝑝 𝑝 (i𝜃)𝜃is nondecreasing and (𝜃(𝑟)) ≤𝜃(𝑟 ),forall𝑝≥1. 𝑔V ⪰𝑔𝑏,and𝑔𝑧 ⪯ 𝑔𝑤,where ⪯𝑔𝑐 𝜓, 𝜑 : [0, ∞) → [0, ∞) are altering distance functions. (ii𝜃) There exists 𝜑∈Φsuch that 2𝜃(𝑟) = (𝑟/2) −, 𝜑(𝑟/2) Assume also that for all 𝑟∈[0,∞). 3 (1) 𝑓 has the mixed 𝑔-monotone property and 𝑓(X )⊆ Θ is nonempty, as 𝜃1(𝑟) = 2𝑘𝑟 with 0≤4𝑘<1is an 𝑔(X); element of Θ.

(2) there exist 𝑥0,𝑦0,𝑧0 ∈ X such that 𝑔𝑥0 ⪯𝑓(𝑥0,𝑦0,𝑧0), Like in [38], we assume that the functions 𝐾1, 𝐾2, 𝑓,and 𝑔𝑦0 ⪰𝑓(𝑦0,𝑥0,𝑦0),and𝑔𝑧0 ⪯𝑓(𝑧0,𝑦0,𝑥0). 𝑔 fulfill the following conditions. Also, suppose that either Assumption 57. Consider the following: (a) 𝑓 and 𝑔 are continuous, the pair (𝑓,𝑔) is compatible, and (X,𝐺)is 𝐺𝑏-complete, or (i) 𝐾1(𝑡, 𝑠) ≥0 and 𝐾2(𝑡, 𝑠),forall ≤0 𝑡, 𝑠 ∈𝐼; 14 Abstract and Applied Analysis

(ii) there exist two positive numbers 𝜆 and 𝜇 and 𝜃∈Θ It is easy to see that (𝑋, 𝐺) is a complete 𝐺𝑏-metric space with 𝑝−1 such that, for all 𝑥, 𝑦 ∈ R with 𝑥≥𝑦,thefollowing 𝑠=2 (see Example 3). 2 Lipschitzian type conditions hold: Now define on 𝑋 the following partial order: (𝑥, 𝑦), (𝑢, V)∈𝑋2 0≤𝑓(𝑡, 𝑥) − 𝑓 (𝑡, 𝑦) ≤ 𝜆𝜃 (𝑥−𝑦), for all (97) (𝑥, 𝑦) ≤ (𝑢, V) ⇐⇒ 𝑥 (𝑡) ≤𝑢(𝑡) ,𝑦(𝑡) ≥ V (𝑡) ∀𝑡 ∈ 𝐼. −𝜇𝜃(𝑥−𝑦)≤𝑔(𝑡, 𝑥) −𝑔(𝑡,𝑦)≤0; (105)

(iii) 2 Obviously, for any (𝑥, 𝑦), (𝑧, 𝑡)∈𝑋 , the element (max 𝑝 𝑝 {𝐹(𝑥, 𝑦), 𝐹(𝑧, 𝑡)}, {𝐹(𝑦, 𝑥), 𝐹(𝑡, 𝑧)}) (max {𝜆 ,𝜇 }) min is comparable with (𝐹(𝑥, 𝑦), 𝐹(𝑦, 𝑥)) and (𝐹(𝑧, 𝑡), 𝐹(𝑡, 𝑧)). 𝑝 𝑝 𝑏 𝑏 1 Define now the mapping 𝐹:𝑋×𝑋by →𝑋 ⋅ [(∫ 𝐾 (𝑡, 𝑠) 𝑑𝑠) +(∫ −𝐾 (𝑡, 𝑠) 𝑑𝑠) ]≤ . sup 1 2 3𝑝−3 𝑡∈𝐼 𝑎 𝑎 2 𝑏 (98) 𝐹(𝑥,𝑦)(𝑡) = ∫ 𝐾1 (𝑡, 𝑠) [𝑓 (𝑠, 𝑥 (𝑠)) +𝑔(𝑠,𝑦(𝑠))] 𝑑𝑠 𝑎 2 Definition 58 (see [38]). A pair (𝛼, 𝛽) ∈𝑋 with 𝑋=𝐶(𝐼,R) 𝑏 𝑡∈𝐼 + ∫ 𝐾2 (𝑡, 𝑠) [𝑓 (𝑠, 𝑦 (𝑠))+𝑔(𝑠, 𝑥 (𝑠))]𝑑𝑠 is called a coupled lower-upper solution of (96)if,forall , 𝑎 𝑏 +ℎ(𝑡) ,∀𝑡∈𝐼. 𝛼 (𝑡) ≤ ∫ 𝐾1 (𝑡, 𝑠) [𝑓 (𝑠, 𝛼 (𝑠)) +𝑔(𝑠,𝛽(𝑠))] 𝑑𝑠 𝑎 (106) 𝑏 𝐹 + ∫ 𝐾2 (𝑡, 𝑠) [𝑓 (𝑠, 𝛽 (𝑠))+𝑔(𝑠, 𝛼 (𝑠))]𝑑𝑠+ℎ(𝑡) , It is not difficult to prove, like in[38], that has the mixed 𝑎 monotone property. Now for 𝑥, 𝑦, 𝑢, V,𝑎,𝑏∈𝑋with 𝑥≥𝑢≥ 𝑏 𝑎 and 𝑦≤V ≤𝑏,wehave 𝛽 (𝑡) ≥ ∫ 𝐾1 (𝑡, 𝑠) [𝑓 (𝑠, 𝛽 (𝑠))+𝑔(𝑠, 𝛼 (𝑠))]𝑑𝑠 𝑎 󵄨 󵄨𝑝 𝜌(𝐹(𝑥,𝑦),𝐹(𝑢, V))=sup󵄨𝐹(𝑥,𝑦)(𝑡) −𝐹(𝑢, V)(𝑡)󵄨 . 𝑏 𝑡∈𝐼 + ∫ 𝐾2 (𝑡, 𝑠) [𝑓 (𝑠, 𝛼 (𝑠)) +𝑔(𝑠,𝛽(𝑠))] 𝑑𝑠 +ℎ (𝑡) . (107) 𝑎 (99) Let us first evaluate the expression in the right hand side. According to the computations done by Berinde in [37], Theorem 59. Consider the integral equation (96) with 𝐹(𝑥,𝑦)(𝑡) −𝐹(𝑢, V)(𝑡) 𝐾1,𝐾2 ∈𝐶(𝐼×𝐼,R) ,ℎ∈𝐶(𝐼, R) . (100) 𝑏 (𝛼, 𝛽) Suppose that there exists a coupled lower-upper solution = ∫ 𝐾1 (𝑡, 𝑠) [𝑓 (𝑠, 𝑥 (𝑠)) +𝑔(𝑠,𝑦(𝑠))] 𝑑𝑠 of (96) with 𝛼≤𝛽and that Assumption 57 is satisfied. Then 𝑎 the integral equation (96) has a unique solution in 𝐶(𝐼, R). 𝑏 + ∫ 𝐾2 (𝑡, 𝑠) [𝑓 (𝑠, 𝑦 (𝑠))+𝑔(𝑠, 𝑥 (𝑠))]𝑑𝑠 Proof. Consider on 𝑋=𝐶(𝐼,R) the natural partial order 𝑎 relation; that is, for 𝑥, 𝑦 ∈ 𝐶(𝐼, R), 𝑏 − ∫ 𝐾1 (𝑡, 𝑠) [𝑓 (𝑠, 𝑢 (𝑠)) +𝑔(𝑠, V (𝑠))]𝑑𝑠 𝑥≤𝑦⇐⇒𝑥(𝑡) ≤𝑦(𝑡) , ∀𝑡∈𝐼. (101) 𝑎 𝑏 𝑋 It is well known that is a complete metric space with respect − ∫ 𝐾2 (𝑡, 𝑠) [𝑓 (𝑠, V (𝑠)) +𝑔(𝑠, 𝑢 (𝑠))]𝑑𝑠 to the sup metric: 𝑎 󵄨 󵄨 𝑏 𝑑(𝑥,𝑦)=sup 󵄨𝑥 (𝑡) −𝑦(𝑡)󵄨 ,𝑥,𝑦∈𝐶(𝐼, R) . = ∫ 𝐾 (𝑡, 𝑠) [𝑓 (𝑠, 𝑥 (𝑠)) −𝑓(𝑠, 𝑢 (𝑠)) +𝑔(𝑠,𝑦(𝑠)) 𝑡∈𝐼 (102) 1 𝑎 Now for 𝑝≥1we define −𝑔 (𝑠, V (𝑠))]𝑑𝑠 𝑝 𝑏 𝑝 󵄨 󵄨 𝜌(𝑥,𝑦)=𝑑(𝑥,𝑦) =( 󵄨𝑥 𝑡 −𝑦 𝑡 󵄨) sup 󵄨 ( ) ( )󵄨 + ∫ 𝐾2 (𝑡, 𝑠) [𝑓 (𝑠, 𝑦 (𝑠))−𝑓(𝑠, V (𝑠)) +𝑔(𝑠, 𝑥 (𝑠)) 𝑡∈𝐼 𝑎 (103) 󵄨 󵄨𝑝 −𝑔 (𝑠, 𝑢 (𝑠))]𝑑𝑠 = sup󵄨𝑥 (𝑡) −𝑦(𝑡)󵄨 ,𝑥,𝑦∈𝐶(𝐼, R) . 𝑡∈𝐼 𝑏 Define = ∫ 𝐾1 (𝑡, 𝑠) [(𝑓 (𝑠, 𝑥 (𝑠)) −𝑓(𝑠, 𝑢 (𝑠))) 𝑎 𝐺 (𝑥, 𝑦,) 𝑧 = max {𝜌 (𝑥,) 𝑦 ,𝜌(𝑦,) 𝑧 ,𝜌(𝑧, 𝑥)} . (104) −(𝑔(𝑠, V (𝑠)) −𝑔(𝑠,𝑦(𝑠)))] 𝑑𝑠 Abstract and Applied Analysis 15

𝑏 𝑏 𝑝 − ∫ 𝐾 𝑡, 𝑠 [(𝑓 (𝑠, 𝑦 𝑠 )−𝑓 𝑠, V 𝑠 ) 𝑝−1 2 ( ) ( ) ( ( )) ≤2 ((∫ 𝐾1 (𝑡, 𝑠) 𝑑𝑠) 𝑎 𝑎

−(𝑔(𝑠, 𝑢 (𝑠)) −𝑔(𝑠, 𝑥 (𝑠)))] 𝑑𝑠 ×2𝑝−1 (𝜆𝑝𝜃(𝑑 (𝑥, 𝑢))𝑝 +𝜇𝑝𝜃(𝑑 (V,𝑦))𝑝) 𝑏 𝑏 𝑝 ≤ ∫ 𝐾1 (𝑡, 𝑠) [𝜆𝜃 (𝑥 (𝑠) −𝑢(𝑠)) +𝜇𝜃(V (𝑠) −𝑦(𝑠))] 𝑑𝑠 𝑎 +(−∫ 𝐾2 (𝑡, 𝑠) 𝑑𝑠) 𝑎 𝑏 − ∫ 𝐾 (𝑡, 𝑠) [𝜆𝜃 (𝑦 (𝑠) − V (𝑠))+𝜇𝜃(𝑢 (𝑠) −𝑥(𝑠))]𝑑𝑠. 2 𝑝 𝑎 ×2𝑝−1 (𝜆𝑝𝜃(𝑑 (V,𝑦)) +𝜇𝑝𝜃(𝑑 (𝑥, 𝑢))𝑝)) (108) 𝑏 𝑝 𝑝−1 ≤2 ((∫ 𝐾1 (𝑡, 𝑠) 𝑑𝑠) Since the function 𝜃 is nondecreasing and 𝑥≥𝑢and 𝑦≤V, 𝑎 we have ×2𝑝−1 (𝜆𝑝𝜃(𝜌(𝑥, 𝑢))+𝜇𝑝𝜃(𝜌(V, 𝑦)))

𝑏 𝑝 +(−∫ 𝐾 (𝑡, 𝑠) 𝑑𝑠) 𝜃 (𝑥 (𝑠) −𝑢(𝑠)) ≤𝜃(sup |𝑥 (𝑡) −𝑢(𝑡)|)=𝜃(𝑑 (𝑥, 𝑢)) , 2 𝑡∈𝐼 𝑎

󵄨 󵄨 𝑝−1 𝑝 𝑝 𝜃(V (𝑠) −𝑦(𝑠))≤𝜃(sup 󵄨𝑦 (𝑡) − V (𝑡)󵄨)=𝜃(𝑑(𝑦,V)) . ×2 (𝜆 𝜃(𝜌(V,𝑦))+𝜇 𝜃(𝜌(𝑥, 𝑢))) ) 𝑡∈𝐼

(109) 𝑏 𝑝 2𝑝−2 𝑝 ≤2 [(𝜆 (∫ 𝐾1 (𝑡, 𝑠) 𝑑𝑠) 𝑎 𝐾2(𝑡, 𝑠) ≤0 Hence, by (108), in view of the fact that ,weobtain 𝑏 𝑝 that 𝑝 +𝜇 (− ∫ 𝐾2 (𝑡, 𝑠) 𝑑𝑠) )𝜃𝜌(𝑥, 𝑢) 𝑎 󵄨 󵄨 󵄨𝐹(𝑥,𝑦)(𝑡) −𝐹(𝑢, V)(𝑡)󵄨 𝑏 𝑝 󵄨 󵄨 𝑝 +(𝜇 (∫ 𝐾1 (𝑡, 𝑠) 𝑑𝑠) 𝑏 𝑎 ≤ ∫ 𝐾 (𝑡, 𝑠) [𝜆𝜃 (𝑑 (𝑥, 𝑢)) +𝜇𝜃(𝑑(𝑦,V))] 𝑑𝑠 1 𝑝 𝑎 (110) 𝑏 𝑝 +𝜆 (− ∫ 𝐾2 (𝑡, 𝑠) 𝑑𝑠) )𝜃𝜌(V,𝑦)]. 𝑏 𝑎 − ∫ 𝐾2 (𝑡, 𝑠) [𝜆𝜃 (𝑑 (𝑦, V)) + 𝜇𝜃 (𝑑 (𝑥, 𝑢))]𝑑𝑠, 𝑎 (111) So, we have as all quantities in the right hand side of (108) are nonnega- 󵄨 󵄨𝑝 󵄨𝐹(𝑥, 𝑦)(𝑡) −𝐹(𝑢, V)(𝑡)󵄨 tive. Now from (108)wehave 𝑏 𝑝 2𝑝−2 𝑝 ≤2 [(𝜆 (∫ 𝐾1 (𝑡, 𝑠) 𝑑𝑠) 𝑎 󵄨 󵄨𝑝 󵄨𝐹(𝑥,𝑦)(𝑡) −𝐹(𝑢, V)(𝑡)󵄨 𝑏 𝑝 𝑝 +𝜇 (− ∫ 𝐾2 (𝑡, 𝑠) 𝑑𝑠) )𝜃𝜌(𝑥, 𝑢) 𝑏 𝑎 ≤(∫ 𝐾1 (𝑡, 𝑠) [𝜆𝜃 (𝑑 (𝑥, 𝑢)) +𝜇𝜃(𝑑(𝑦,V))] 𝑑𝑠 𝑎 𝑏 𝑝 𝑝 𝑝 +(𝜇 (∫ 𝐾1 (𝑡, 𝑠) 𝑑𝑠) 𝑏 𝑎 − ∫ 𝐾2 (𝑡, 𝑠) [𝜆𝜃 (𝑑V ( , 𝑦)) + 𝜇𝜃(𝑑(𝑥, 𝑢))] 𝑑𝑠) 𝑎 𝑏 𝑝 +𝜆𝑝(− ∫ 𝐾 (𝑡, 𝑠) 𝑑𝑠) )𝜃𝜌(𝑦,V)]. 𝑏 𝑝 2 𝑝−1 𝑝 𝑎 ≤2 ((∫ 𝐾1 (𝑡, 𝑠) 𝑑𝑠) (𝜆𝜃 (𝑑 (𝑥, 𝑢)) +𝜇𝜃(𝑑(V, 𝑦))) 𝑎 (112)

𝑏 𝑝 Similarly, one can obtain that +(−∫ 𝐾2 (𝑡, 𝑠) 𝑑𝑠) 𝑝 𝑎 |𝐹 (𝑢, V)(𝑡) −𝐹(𝑧,)( 𝑡 𝑡)|

𝑏 𝑝 𝑝 2𝑝−2 𝑝 × (𝜆𝜃 (𝑑V ( ,𝑦))+𝜇𝜃(𝑑 (𝑥, 𝑢))) ) ≤2 [(𝜆 (∫ 𝐾1 (𝑡, 𝑠) 𝑑𝑠) 𝑎 16 Abstract and Applied Analysis

𝑏 𝑝 𝑝 Now, since 𝜃 is nondecreasing, we have +𝜇 (− ∫ 𝐾2 (𝑡, 𝑠) 𝑑𝑠) )𝜃𝜌(𝑢,) 𝑧 𝑎 𝜃 (𝐺 (𝑥, 𝑢,)) 𝑧 ≤𝜃(𝐺(𝑥, 𝑢,) 𝑧 +𝐺(𝑦,V,𝑡)), 𝑏 𝑝 𝑝 (115) +(𝜇 (∫ 𝐾1 (𝑡, 𝑠) 𝑑𝑠) 𝜃(𝐺(𝑦,V,𝑡))≤𝜃(𝐺(𝑥, 𝑢,) 𝑧 +𝐺(𝑦,V,𝑡)) 𝑎

𝑏 𝑝 and so 𝑝 +𝜆 (− ∫ 𝐾2 (𝑡, 𝑠) 𝑑𝑠) )𝜃𝜌(V,𝑡) ], 𝑎 𝜃 (𝐺 (𝑥, 𝑢,)) 𝑧 + 𝜃 (𝐺 (𝑦, V,𝑡)) 󵄨 󵄨𝑝 󵄨𝐹 (𝑧,)( 𝑡 𝑡) −𝐹(𝑥,𝑦)(𝑡)󵄨 ≤2𝜃(𝐺(𝑥, 𝑢,) 𝑧 +𝐺(𝑦,V,𝑡)) 𝑏 𝑝 𝐺 (𝑥, 𝑢,) 𝑧 +𝐺(𝑦,V,𝑡) 2𝑝−2 𝑝 (116) ≤2 [(𝜆 (∫ 𝐾1 (𝑡, 𝑠) 𝑑𝑠) = 𝑎 2

𝑏 𝑝 𝐺 (𝑥, 𝑢,) 𝑧 +𝐺(𝑦,V,𝑡) 𝑝 −𝜑( ), +𝜇 (− ∫ 𝐾2 (𝑡, 𝑠) 𝑑𝑠) )𝜃𝜌(𝑥,) 𝑧 2 𝑎

𝑏 𝑝 by the definition of 𝜃.Finally,from(114)wegetthat 𝑝 +(𝜇 (∫ 𝐾1 (𝑡, 𝑠) 𝑑𝑠) 𝑎 𝐺(𝐹(𝑥,𝑦),𝐹(𝑢, V) ,𝐹(𝑧,) 𝑡 ) 𝑏 𝑝 𝑝 1 𝐺 (𝑥, 𝑢,) 𝑧 +𝐺(𝑦,V,𝑡) +𝜆 (− ∫ 𝐾2 (𝑡, 𝑠) 𝑑𝑠) )𝜃𝜌(𝑦,𝑡)]. ≤ 𝑎 2𝑝−1 2 (117) (113) 1 𝐺 (𝑥, 𝑢,) 𝑧 +𝐺(𝑦,V,𝑡) − 𝜑( ). 2𝑝−1 2 Taking the supremum with respect to 𝑡 and using (98)weget

𝐺(𝐹(𝑥,𝑦),𝐹(𝑢, V) ,𝐹(𝑧,) 𝑡 ) Similarly, we can obtain that

𝐺(𝐹(𝑦,𝑥),𝐹(V,𝑢) ,𝐹(𝑡, 𝑧)) 󵄨 󵄨𝑝 = max {sup󵄨𝐹(𝑥,𝑦)(𝑡) −𝐹(𝑢, V)(𝑡)󵄨 , 𝑡∈𝐼 1 𝐺 (𝑥, 𝑢,) 𝑧 +𝐺(𝑦,V,𝑡) 𝑝 ≤ sup|𝐹 (𝑢, V)(𝑡) −𝐹(𝑧,)( 𝑡 𝑡)| , 2𝑝−1 2 (118) 𝑡∈𝐼 1 𝐺 (𝑥, 𝑢,) 𝑧 +𝐺(𝑦,V,𝑡) − 𝜑( ), 󵄨 󵄨𝑝 2𝑝−1 2 sup󵄨𝐹 (𝑧,)( 𝑡 𝑡) −𝐹(𝑥,𝑦)(𝑡)󵄨 } 𝑡∈𝐼 which is just the contractive condition (69)inCorollary 43. ≤22𝑝−2 ( {𝜆𝑝,𝜇𝑝}) 2 max Now, let (𝛼, 𝛽) ∈𝑋 be a coupled upper-lower solution of 𝑏 𝑝 𝑏 𝑝 (96). Then we have × sup [(∫ 𝐾1 (𝑡, 𝑠) 𝑑𝑠) +(∫ −𝐾2 (𝑡, 𝑠) 𝑑𝑠) ] 𝑡∈𝐼 𝑎 𝑎 𝛼 (𝑡) ≤𝐹(𝛼,𝛽)(𝑡) ,𝛽(𝑡) ≥𝐹(𝛽,𝛼)(𝑡) , (119) × {𝜃 (𝜌 (𝑥, 𝑢))+𝜃(𝜌(𝑦,V)) , 𝜃 (𝜌 (𝑢,) 𝑧 ) max for all 𝑡∈𝐼, which show that all hypotheses of Corollary 43 +𝜃 (𝜌 (V,𝑡)),𝜃(𝜌(𝑥,) 𝑧 )+𝜃(𝜌(𝑦,𝑡))} are satisfied. Thisprovesthat𝐹 has a coupled fixed point (𝑥∗,𝑦∗) in 2𝑝−2 𝑝 𝑝 𝑋2 ≤2 (max {𝜆 ,𝜇 }) . Since 𝛼≤𝛽,byCorollary 43 it follows that 𝑥∗ =𝑦∗;that 𝑏 𝑝 𝑏 𝑝 is, × sup [(∫ 𝐾1 (𝑡, 𝑠) 𝑑𝑠) +(∫ −𝐾2 (𝑡, 𝑠) 𝑑𝑠) ] 𝑡∈𝐼 𝑎 𝑎 𝑥∗ =𝐹(𝑥∗,𝑥∗), (120) ×𝜃(𝐺 (𝑥, 𝑢,)) 𝑧 +𝜃(𝐺(𝑦,V,𝑡)) and therefore 𝑥∗ ∈ 𝐶(𝐼, R) is the solution of the integral 22𝑝−2 equation (96). ≤ [𝜃 (𝐺 (𝑥, 𝑢,)) 𝑧 +𝜃(𝐺(𝑦,V,𝑡))] 23𝑝−3 1 Conflict of Interests = [𝜃 (𝐺 (𝑥, 𝑢,)) 𝑧 + 𝜃 (𝐺 (𝑦, V,𝑡))]. 2𝑝−1 The authors declare that there is no conflict of interests (114) regarding the publications of this paper. Abstract and Applied Analysis 17

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Research Article Generalized Common Fixed Point Results with Applications

Marwan Amin Kutbi,1 Muhammad Arshad,2 Jamshaid Ahmad,3 and Akbar Azam3

1 Department of Mathematics, King Abdulaziz University, Jeddah, Saudi Arabia 2 Department of Mathematics & Statistics, International Islamic University H-10, Islamabad, Pakistan 3 Department of Mathematics, COMSATS Institute of Information Technology, Chak Shahzad, Islamabad 44000, Pakistan

Correspondence should be addressed to Muhammad Arshad; marshad [email protected]

Received 10 February 2014; Accepted 23 March 2014; Published 27 April 2014

Academic Editor: Ljubomir B. Ciri´ c´

We obtained some generalized common fixed point results in the context of complex valued metric spaces. Moreover, we proved an existence theorem for the common solution for two Urysohn integral equations. Examples are presented to support our results.

1. Introduction and Preliminaries for mappings satisfying a rational inequality. The idea of complex valued metric spaces can be exploited to define Since the appearance of the Banach contraction mapping complex valued normed spaces and complex valued Hilbert principle, a number of papers were dedicated to the improve- spaces and then it will bring wonderful research activities in ment and generalization of that result. Most of these deal nonlinear analysis. with the generalizations of the contractive condition in metric In this paper we continue our investigations initiated by spaces. Azam et al. [15] and prove a common fixed point result for two Gahler¨ [1] generalized the idea of metric space and mappings and applied it to get the coincidence and common introduced a 2-metric space which was followed by a number fixed points of three and four mappings. of papers dealing with this generalized space. Plenty of We begin with listing some notations, definitions, and material is also available in other generalized metric spaces, basic facts on these topics that we will need to convey our such as, rectangular metric spaces, semimetric spaces, pseu- theorems. Let C be the set of complex numbers and 𝑧1,𝑧2 ∈ dometric spaces, probabilistic metric spaces, fuzzy metric C. Define a partial order ≾ on C as follows: spaces, quasimetric spaces, quasisemi metric spaces, 𝐷- metric spaces, 𝐺-metric space, partial metric space, and cone iff 𝑧1 ≾𝑧2, Re (𝑧1) ⩽ Re (𝑧2) , Im (𝑧1) ⩽ Im (𝑧2) . (1) metric spaces (see [2–14]). Azam et al. [15]improvedthe Banach contraction principle by generalizing it in complex It follows that valued metric space involving rational inequity which could 𝑧 ≾𝑧 not be handled in cone metric spaces [3, 5, 11, 15]due 1 2 (2) to limitations regarding product and quotient. Rouzkard if one of the following conditions is satisfied: and Imdad [16] extended the work of Azam et al. [15].

Sintunavarat and Kumam [17] obtained common fixed point (i) Re (𝑧1)=Re (𝑧2), Im (𝑧1)

𝑚 In particular, we will write 𝑧1⋨𝑧2 if 𝑧1 =𝑧̸ 2 and one of (i), (ii), Definition 4 (see [18]). Two families of self-mappings {𝑇𝑖}1 𝑛 and (iii) is satisfied and we will write 𝑧1 ≺𝑧2 if only (iii) is and {𝑆𝑖}1 are said to be pairwise commuting if: satisfied. Note that (1) 𝑇𝑖𝑇𝑗 =𝑇𝑗𝑇𝑖, 𝑖,𝑗∈{1,2,...,𝑚}; 󵄨 󵄨 󵄨 󵄨 0≾𝑧1⋨𝑧2 󳨐⇒ 󵄨𝑧1󵄨 < 󵄨𝑧2󵄨 , (2) 𝑆𝑘𝑆𝑙 =𝑆𝑙𝑆𝑘, 𝑘,𝑙∈{1,2,...,𝑛}; (4) (3) 𝑇𝑖𝑆𝑘 =𝑆𝑘𝑇𝑖, 𝑖∈{1,2,...,𝑚}, 𝑘∈{1,2,...,𝑛}. 𝑧1 ⪯𝑧2,𝑧2 ≺𝑧3 󳨐⇒ 𝑧 1 ≺𝑧3. Lemma 5 (see [19]). Let 𝑋 be a nonempty set and 𝑓:𝑋 →𝑋 Definition 1. Let 𝑋 be a nonempty set. Suppose that the self- 𝐸⊂𝑋 𝑓𝐸 = 𝑓𝑋 𝑑:𝑋×𝑋 → C a function. Then there exists a subset such that mapping satisfies: and 𝑓:𝐸 →𝑋is one to one. 0≾𝑑(𝑥,𝑦) 𝑥, 𝑦 ∈𝑋 𝑑(𝑥, 𝑦) =0 (1) ,forall and if and Lemma 6 (see [20]). Let 𝑋 beanonemptysetandthe 𝑥=𝑦 only if ; mappings 𝑆, 𝑇, 𝑓 : 𝑋→ have a unique point of coincidence V 𝑋 (𝑆, 𝑓) (𝑇, 𝑓) 𝑓V (2) 𝑑(𝑥, 𝑦) = 𝑑(𝑦, 𝑥) for all 𝑥, 𝑦 ∈𝑋; in .If and are weakly compatible, then is a unique common fixed point of 𝑆, 𝑇, 𝑓. (3) 𝑑(𝑥, 𝑦) ≾ 𝑑(𝑥, 𝑧),forall +𝑑(𝑧,𝑦) 𝑥, 𝑦,. 𝑧∈𝑋

Then 𝑑 is called a complex valued metric on 𝑋,and(𝑋, 𝑑) 2. Main Results is called a complex valued metric space. A point 𝑥∈𝑋is (𝑋, 𝑑) 𝐴⊆𝑋 Theorem 7. Let be a complete complex valued metric called interior point of a set whenever there exists space and 0≤ℎ<1.Iftheself-mappings𝑆, 𝑇 : 𝑋 →𝑋 satisfy 0≺𝑟∈C such that 𝑑 (𝑆𝑥, 𝑇𝑦) ≾ℎ𝐿(𝑥,) 𝑦 (8) 𝐵 (𝑥,) 𝑟 ={𝑦∈𝑋:𝑑(𝑥,𝑦)≺𝑟}⊆𝐴. (5) for all 𝑥, 𝑦 ∈𝑋,where Apoint𝑥∈𝑋is called a limit point of 𝐴 whenever for 𝐿(𝑥,𝑦) every 0≺𝑟∈C, 𝑑 (𝑥, 𝑆𝑥) 𝑑(𝑦,𝑇𝑦) ∈{𝑑(𝑥,𝑦),𝑑(𝑥, 𝑆𝑥) ,𝑑(𝑦,𝑇𝑦), } 𝐵 (𝑥,) 𝑟 ∩ (𝐴\{𝑥}) =𝜙.̸ (6) 1+𝑑(𝑥,𝑦) (9) 𝐴 is called open whenever each element of 𝐴 is an interior point of 𝐴.Moreover,asubset𝐵⊆𝑋is called closed then 𝑆 and 𝑇 have a unique common fixed point. whenever each limit point of 𝐵 belongs to 𝐵.Thefamily Proof. We will first show that fixed point of one map is afixed 𝑝=𝑇𝑝 𝐹={𝐵 (𝑥,) 𝑟 :𝑥∈𝑋,0≺𝑟} (7) pointoftheother.Supposethat .Thenfrom(8) 𝑑 (𝑆𝑝, 𝑝) = 𝑑 (𝑆𝑝, 𝑇𝑝) ≾ℎ𝐿(𝑝,𝑝). (10) isasubbasisforaHausdorfftopology𝜏 on 𝑋. Let 𝑥𝑛 be a sequence in 𝑋 and 𝑥∈𝑋.Ifforevery𝑐∈C Case 1 with 0≺𝑐there is 𝑛0 ∈ N such that for all 𝑛>𝑛0, 𝑑 (𝑆𝑝, 𝑝) ≾ ℎ𝑑 (𝑝, 𝑝) =0,𝑝𝑆𝑝. (11) 𝑑(𝑥𝑛,𝑥)≺𝑐,then{𝑥𝑛} is said to be convergent, {𝑥𝑛} converges 𝑥 𝑥 {𝑥 } to ,and is the limit point of 𝑛 . We denote this by Case 2 lim𝑛→∞𝑥𝑛 =𝑥,or𝑥𝑛 →𝑥,as𝑛→∞.Ifforevery 𝑐∈C with 0≺𝑐there is 𝑛0 ∈ N such that for all 𝑛>𝑛0, 𝑑 (𝑆𝑝, 𝑝) ≾ ℎ𝑑 (𝑝, 𝑆𝑝), (12) 𝑑(𝑥𝑛,𝑥𝑛+𝑚)≺𝑐,then{𝑥𝑛} is called a Cauchy sequence in 𝑝=𝑆𝑝 (𝑋, 𝑑). If every Cauchy sequence is convergent in (𝑋, 𝑑),then which yields that . (𝑋, 𝑑) is called a complete complex valued metric space. Let 𝑋 be a nonempty set and 𝑇, 𝑓 : 𝑋 →𝑋.Themappings𝑇, Case 3 𝑓 are said to be weakly compatible if they commute at their 𝑑(𝑆𝑝,𝑝)≾ℎ𝑑(𝑝,𝑇𝑝)=0, 𝑝=𝑆𝑝. (13) coincidence point (i.e., 𝑇𝑓𝑥 = 𝑓𝑇𝑥 whenever 𝑇𝑥 = 𝑓𝑥). A point 𝑦∈𝑋is called point of coincidence of 𝑇 and 𝑓 if there Case 4 𝑥∈𝑋 𝑦=𝑇𝑥=𝑓𝑥 exists a point such that .Werequirethe 𝑑 (𝑝, 𝑆𝑝) 𝑑 (𝑝, 𝑇𝑝) following lemmas. 𝑑(𝑆𝑝,𝑝)≾ℎ[ ], 1 + 𝑑 (𝑝, 𝑝) (14) Lemma 2 (see [15]). Let (𝑋, 𝑑) be a complex valued metric which implies that 𝑑(𝑆𝑝, 𝑝), ≾0 and hence 𝑝=𝑆𝑝.Ina space and let {𝑥𝑛} be a sequence in 𝑋.Then{𝑥𝑛} converges to 𝑥 similarmanneritcanbeshownthatanyfixedpointof𝑆 is if and only if |𝑑(𝑥𝑛,𝑥)| → 0as 𝑛→∞. also the fixed point of 𝑇.Let𝑥0 ∈𝑋and define Lemma 3 (see [15]). Let (𝑋, 𝑑) be a complex valued metric 𝑥 =𝑆𝑥 {𝑥 } 𝑋 {𝑥 } 2𝑛+1 2𝑛 space and let 𝑛 be a sequence in .Then 𝑛 is a Cauchy (15) sequence if and only if |𝑑(𝑥𝑛,𝑥𝑛+𝑚)| → 0 as 𝑛→∞. 𝑥2𝑛+2 =𝑇𝑥2𝑛+1,𝑛≥0. Abstract and Applied Analysis 3

We will assume that 𝑥𝑛 =𝑥̸ 𝑛+1 for each 𝑛.Otherwise,there Now for any 𝑚>𝑛, exists an 𝑛 such that 𝑥2𝑛 =𝑥2𝑛+1.Then𝑥2𝑛 =𝑆𝑥2𝑛 and 𝑥2𝑛 is a 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨𝑑(𝑥 ,𝑥 )󵄨 ≤ 󵄨𝑑(𝑥 ,𝑥 )󵄨 + 󵄨𝑑(𝑥 ,𝑥 )󵄨 fixed point of 𝑆, hence a fixed point of 𝑇.Similarly,if𝑥2𝑛+1 = 󵄨 𝑚 𝑛 󵄨 󵄨 𝑛 𝑛+1 󵄨 󵄨 𝑛+1 𝑛+2 󵄨 𝑥 𝑛 𝑥 𝑇 󵄨 󵄨 2𝑛+2 for some ,then 2𝑛+1 is common fixed point of and +⋅⋅⋅+󵄨𝑑(𝑥 ,𝑥 )󵄨 hence of 𝑆.From(8) 󵄨 𝑚−1 𝑚 󵄨 𝑛 𝑛+1 𝑚−1 󵄨 󵄨 (24) ≤[ℎ +ℎ +⋅⋅⋅+ℎ ] 󵄨𝑑(𝑥0,𝑥1)󵄨 𝑑(𝑥2𝑛+1,𝑥2𝑛+2)=𝑑(𝑆𝑥2𝑛,𝑇𝑥2𝑛+1)≤ℎ𝐿(𝑥2𝑛,𝑥2𝑛+1). (16) ℎ𝑛 󵄨 󵄨 ≤[ ] 󵄨𝑑(𝑥 ,𝑥 )󵄨 Case 1 1−ℎ 󵄨 0 1 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨𝑑(𝑥 ,𝑥 )󵄨 = 󵄨𝑑(𝑆𝑥 ,𝑇𝑥 )󵄨 ≤ℎ󵄨𝑑(𝑥 ,𝑥 )󵄨 . 󵄨 2𝑛+1 2𝑛+2 󵄨 󵄨 2𝑛 2𝑛+1 󵄨 󵄨 2𝑛 2𝑛+1 󵄨 and so (17) 𝑛 󵄨 󵄨 ℎ 󵄨 󵄨 󵄨𝑑(𝑥𝑚,𝑥𝑛)󵄨 ≤ 󵄨𝑑(𝑥0,𝑥1)󵄨 󳨀→ 0, as 𝑚, 𝑛 󳨀→ ∞. Case 2 1−ℎ (25) 󵄨 󵄨 󵄨 󵄨 󵄨𝑑(𝑥2𝑛+1,𝑥2𝑛+2)󵄨 = 󵄨𝑑(𝑆𝑥2𝑛,𝑇𝑥2𝑛+1)󵄨 This implies that {𝑥𝑛} is a Cauchy sequence. Since 𝑋 is 󵄨 󵄨 ≤ℎ󵄨𝑑(𝑥2𝑛,𝑆𝑥2𝑛)󵄨 (18) complete, there exists 𝑢∈𝑋such that 𝑥𝑛 →𝑢.Itfollows 󵄨 󵄨 that 𝑢=𝑆𝑢;otherwise𝑑(𝑢, 𝑆𝑢) =𝑧≻0 and we would then =ℎ󵄨𝑑(𝑥 ,𝑥 )󵄨 . 󵄨 2𝑛 2𝑛+1 󵄨 have

Case 3 𝑧≾𝑑(𝑢,𝑥2𝑛+2)+𝑑(𝑥2𝑛+2,𝑆𝑢) 󵄨 󵄨 󵄨 󵄨 ≾𝑑(𝑢,𝑥 )+𝑑(𝑇𝑥 ,𝑆𝑢) 󵄨𝑑(𝑥2𝑛+1,𝑥2𝑛+2)󵄨 = 󵄨𝑑(𝑆𝑥2𝑛,𝑇𝑥2𝑛+1)󵄨 2𝑛+2 2𝑛+1 (26) 󵄨 󵄨 ≾𝑑(𝑢,𝑥 ) + ℎ𝐿 (𝑢, 𝑥 ). ≤ℎ󵄨𝑑(𝑥2𝑛+1,𝑇𝑥2𝑛+1)󵄨 (19) 2𝑛+2 2𝑛+1 󵄨 󵄨 =ℎ󵄨𝑑(𝑥2𝑛+1,𝑥2𝑛+2)󵄨 , Case 1 󵄨 󵄨 󵄨 󵄨 |𝑧| ≤ 󵄨𝑑 (𝑢, 𝑥 )󵄨 +ℎ󵄨𝑑 (𝑢, 𝑥 )󵄨 . which implies that 󵄨 2𝑛+2 󵄨 󵄨 2𝑛+1 󵄨 (27) That is, |𝑧| ≤, 0 a contradiction and hence 𝑢=𝑆𝑢. 𝑥2𝑛+1 =𝑥2𝑛+2, (20) Case 2 a contradiction to our assumption. 󵄨 󵄨 |𝑧| ≤ 󵄨𝑑(𝑢,𝑥2𝑛+2)󵄨 +ℎ|𝑑 (𝑢, 𝑆𝑢)| . (28) Case 4 |𝑧| ≤ 0 𝑢=𝑆𝑢 󵄨 󵄨 󵄨 󵄨 That is, , a contradiction and hence . 󵄨𝑑(𝑥2𝑛+1,𝑥2𝑛+2)󵄨 = 󵄨𝑑(𝑆𝑥2𝑛,𝑇𝑥2𝑛+1)󵄨 󵄨 󵄨 Case 3 󵄨𝑑(𝑥 ,𝑆𝑥 )𝑑(𝑥 ,𝑇𝑥 )󵄨 ≤ℎ󵄨 2𝑛 2𝑛 2𝑛+1 2𝑛+1 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 |𝑧| ≤ 󵄨𝑑(𝑢,𝑥2𝑛+2)󵄨 +ℎ󵄨𝑑(𝑥2𝑛+1,𝑇𝑥2𝑛+1)󵄨 󵄨 1+𝑑(𝑥2𝑛,𝑥2𝑛+1) 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 ≤ 󵄨𝑑(𝑢,𝑥2𝑛+2)󵄨 +ℎ󵄨𝑑(𝑥2𝑛+1,𝑥2𝑛+2)󵄨 (29) 󵄨 𝑑(𝑥2𝑛,𝑥2𝑛+1) 󵄨 󵄨 󵄨 ≤ℎ󵄨 󵄨 󵄨𝑑(𝑥 ,𝑥 )󵄨 󵄨1+𝑑(𝑥 ,𝑥 )󵄨 󵄨 2𝑛+1 2𝑛+2 󵄨 󵄨 󵄨 2𝑛+2 󵄨 󵄨 󵄨 2𝑛 2𝑛+1 󵄨 ≤ 󵄨𝑑(𝑢,𝑥2𝑛+2)󵄨 +ℎ 󵄨𝑑(𝑥0,𝑥1)󵄨 . 󵄨 󵄨 ≤ℎ󵄨𝑑(𝑥 ,𝑥 )󵄨 . 󵄨 2𝑛+1 2𝑛+2 󵄨 This in turn gives us |𝑧| ≤, 0 a contradiction and hence 𝑢= (21) 𝑆𝑢.

∗ point. For this, assume that 𝑢 in 𝑋 is another common fixed As an application of Theorem 7, we prove the following ∗ ∗ ∗ point of 𝑆 and 𝑇.Then𝑑(𝑢, 𝑢 )=𝑑(𝑆𝑢,𝑇𝑢 )≾ℎ𝐿(𝑢,𝑢 ). theorem for two finite families of mappings.

𝑚 𝑛 Case 1 Theorem 11. If {𝑇𝑖}1 and {𝑆𝑖}1 are two finite pairwise com- mutingfinitefamiliesofself-mappingdefinedonacomplete 𝑑 (𝑢, 𝑢∗) ≾ℎ𝑑(𝑢,∗ 𝑢 ) . (31) complex valued metric space (𝑋, 𝑑) such that the mappings 𝑆 𝑇 𝑇=𝑇,𝑇 ,...,𝑇 𝑆=𝑆,𝑆 ,...,𝑆 Case 2 and (with 1 2 𝑚 and 1 2 𝑛)satisfy the contractive condition (8), then the component maps of the ∗ 𝑚 𝑛 𝑑(𝑢,𝑢 )≾ℎ𝑑(𝑢, 𝑆𝑢) ≾ℎ𝑑(𝑢, 𝑢) =0. (32) two families {𝑇𝑖}1 and {𝑆𝑖}1 have a unique common fixed point. ∗ This gives us 𝑢=𝑢 . Proof. From theorem we can say that the mappings 𝑇 and 𝑆 haveauniquecommonfixedpoint𝑧;thatis,𝑇𝑧 = 𝑆𝑧. =𝑧 Case 3 Now our requirement is to show that 𝑧 is a common fixed ∗ ∗ ∗ ∗ ∗ point of all the component mappings of both families. In view 𝑑(𝑢,𝑢 ) ≾ ℎ𝑑 (𝑢 ,𝑇𝑢 ) = ℎ𝑑 (𝑢 ,𝑢 )=0. (33) 𝑚 𝑛 of pairwise commutativity of the families {𝑇𝑖}1 and {𝑆𝑖}1,(for 1≤𝑘≤𝑚 𝑇 𝑧=𝑇𝑇𝑧 = 𝑇𝑇 𝑧 Case 4 every )wecanwrite 𝑘 𝑘 𝑘 and 𝑇𝑘𝑧=𝑇𝑘𝑆𝑧 =𝑘 𝑆𝑇 𝑧 which show that 𝑇𝑘𝑧 (for every 𝑘)isalso ℎ𝑑 (𝑢, 𝑆𝑢) 𝑑(𝑢∗,𝑇𝑢∗) 𝑑(𝑢,𝑢∗)≾ =0. a common fixed point of 𝑇 and 𝑆. By using the uniqueness of ∗ (34) 1+𝑑(𝑢, 𝑢 ) common fixed point, we can write 𝑇𝑘𝑧=𝑧(for every 𝑘)which 𝑧 {𝑇 }𝑚 𝑢∗ =𝑢 shows that is a common fixed point of the family 𝑖 1 . Hence, in all cases . This completes the proof of the Using the same argument one can also show that (for every theorem. 1≤𝑘≤𝑛) 𝑆𝑘𝑧=𝑧.Thuscomponentmapsofthetwofamilies 𝑚 𝑛 {𝑇𝑖} and {𝑆𝑖} haveauniquecommonfixedpoint. Corollary 8 (see [15]). Let (𝑋, 𝑑) be a complete complex 1 1 𝑆, 𝑇 : 𝑋 →𝑋 0<ℎ<1 valued metric space and let and . By setting 𝑇1 =𝑇2 = ⋅⋅⋅ = 𝑇𝑚 =𝐹and 𝑆1 =𝑆2 =⋅⋅⋅= 𝑆 𝑇 If the self-mappings , satisfy 𝑆𝑛 =𝐺,inTheorem 11,wegetthefollowingcorollary. 𝑑 (𝑆𝑥, 𝑇𝑦) ≾ ℎ𝐿 (𝑥,𝑦) (35) Corollary 12. If 𝐹 and 𝐺 are two commuting self-mappings (𝑋, 𝑑) for all 𝑥, 𝑦 ∈𝑋,where defined on a complete complex valued metric space satisfying the condition 𝑑 (𝑥, 𝑆𝑥) 𝑑(𝑦,𝑇𝑦) 𝑚 𝑛 𝐿 (𝑥, 𝑦) ∈ {𝑑 (𝑥, 𝑦), }, 𝑑 (𝐹 𝑥, 𝐺 𝑦) ≾ℎ𝐿(𝑥,) 𝑦 (41) 1+𝑑(𝑥,𝑦) (36) for all 𝑥, 𝑦 ∈𝑋 and 0≤ℎ<1,where then 𝑆 and 𝑇 have a unique common fixed point. 𝐿 (𝑥, 𝑦) ∈ {𝑑 (𝑥, 𝑦)𝑚 ,𝑑(𝑥,𝐹 𝑥),𝑑(𝑦,𝐺𝑛𝑦) , Corollary 9. Let (𝑋, 𝑑) be a complete complex valued metric space and 0≤ℎ<1.Iftheself-mapping𝑇:𝑋satisfies →𝑋 (42) 𝑑(𝑥,𝐹𝑚𝑥) 𝑑 (𝑦,𝑛 𝐺 𝑦) 𝑑(𝑇𝑥,𝑇𝑦)≾ℎ𝐿(𝑥,𝑦) }, (37) 1+𝑑(𝑥,𝑦) 𝑥, 𝑦 ∈𝑋 for all ,where then 𝐹 and 𝐺 have a unique common fixed point. 𝐿(𝑥,𝑦) Corollary 13. Let (𝑋, 𝑑) be a complete complex valued metric 𝑑 (𝑥, 𝑇𝑥) 𝑑(𝑦,𝑇𝑦) space and let 𝑇:𝑋be →𝑋 a self-mapping satisfying ∈{𝑑(𝑥,𝑦),𝑑 𝑥, 𝑇𝑥 ,𝑑(𝑦,𝑇𝑦), } ( ) 𝑛 𝑛 1+𝑑(𝑥,𝑦) 𝑑(𝑇 𝑥, 𝑇 𝑦) ≾ ℎ𝐿 (𝑥, 𝑦) (43) (38) for all 𝑥, 𝑦 ∈𝑋 and 0≤ℎ<1,where then 𝑇 has a unique fixed point. 𝐿(𝑥,𝑦)∈{𝑑(𝑥,𝑦),𝑑(𝑥,𝑇𝑚𝑥),𝑑(𝑦,𝑇𝑛𝑦) , Corollary 10 (see [15]). Let (𝑋, 𝑑) be a complete complex valued metric space and let 𝑇:𝑋and →𝑋 0≤ℎ<1.If (44) 𝑑(𝑥,𝑇𝑛𝑥) 𝑑 (𝑦,𝑛 𝑇 𝑦) the self-mapping 𝑇satisfies }. 1+𝑑(𝑥,𝑦) 𝑑 (𝑇𝑥, 𝑇𝑦) ≾ℎ𝐿(𝑥,) 𝑦 (39) Then 𝑇 has a unique fixed point. for all 𝑥, 𝑦 ∈𝑋,where Corollary 14 (see [15]). Let (𝑋, 𝑑) be a complete complex 𝑑 (𝑥, 𝑇𝑥) 𝑑(𝑦,𝑇𝑦) 𝑇:𝑋 →𝑋 0≤ℎ<1 𝐿 (𝑥, 𝑦) ∈ {𝑑 (𝑥, 𝑦), }, (40) valued metric space and and .The 1+𝑑(𝑥,𝑦) self-mapping 𝑇 satisfies 𝑛 𝑛 then 𝑇 has a unique fixed point. 𝑑(𝑇 𝑥, 𝑇 𝑦) ≾ ℎ𝐿 (𝑥, 𝑦) (45) Abstract and Applied Analysis 5 for all 𝑥, 𝑦 ∈𝑋,where Consider the Urysohn integral equations

𝑑(𝑥,𝑇𝑛𝑥) 𝑑 (𝑦,𝑛 𝑇 𝑦) 𝑏 𝐿 (𝑥, 𝑦) ∈ {𝑑 (𝑥, 𝑦), }. (46) 𝑥 (𝑡) = ∫ 𝐾1 (𝑡, 𝑠, 𝑥 (𝑠)) 𝑑𝑠 +𝑔 (𝑡) , (𝛼) 1+𝑑(𝑥,𝑦) 𝑎

Then 𝑇 has a unique fixed point. 𝑏 𝑥 (𝑡) = ∫ 𝐾2 (𝑡, 𝑠, 𝑥 (𝑠)) 𝑑𝑠 +ℎ (𝑡) , (𝛽) 𝑎 Our next example exhibits the superiority of Corollary 13 over Corollary 9. where 𝑡∈[𝑎,𝑏]⊂R, 𝑥, 𝑔,. ℎ∈𝑋

Example 15. Let 𝑋1 ={𝑧∈C :0≤Re 𝑧≤1,Im 𝑧=0}and 𝑛 𝑛 Suppose that 𝐾1,𝐾2 : [𝑎, 𝑏] × [𝑎, 𝑏]× R → R are such 𝑋2 ={𝑧∈C :0≤Im 𝑧≤1,Re 𝑧=0}and let 𝑋=𝑋1 ∪𝑋2. that 𝐹𝑥,𝐺𝑥 ∈𝑋for each 𝑥∈𝑋,where, Then with 𝑧=𝑥+𝑖𝑦,set𝑆=𝑇and define 𝑇:𝑋as →𝑋 follows: 𝑏 𝐹 (𝑡) = ∫ 𝐾 (𝑡, 𝑠, 𝑥 (𝑠)) 𝑑𝑠, (0, 0) 𝑥, 𝑦 ∈𝑄 𝑥 1 { if 𝑎 {(1, 0) 𝑥∈𝑄𝑐,𝑦∈𝑄 (51) 𝑇(𝑥,𝑦)= if 𝑏 {(0, 1) 𝑥∈𝑄,𝑦∈𝑄𝑐 (47) { if 𝐺𝑥 (𝑡) = ∫ 𝐾2 (𝑡, 𝑠, 𝑥 (𝑠)) 𝑑𝑠, ∀𝑡 ∈ [𝑎,] 𝑏 . 𝑐 {(1, 1) if 𝑥, 𝑦 ∈𝑄 . 𝑎 If there exists 0≤ℎ<1such that for every 𝑥, 𝑦 ∈𝑋 Consider a complex valued metric 𝑑:𝑋×𝑋→ C as follows:

󵄩 󵄩 𝑖 tan−1𝑎 2𝑖 󵄨 󵄨 󵄩𝐹 (𝑡) −𝐺 (𝑡) +𝑔(𝑡) −ℎ(𝑡)󵄩 √1+𝑎2𝑒 { 󵄨𝑥 −𝑥 󵄨 , 𝑧 ,𝑧 ∈𝑋 󵄩 𝑥 𝑦 󵄩∞ { 󵄨 1 2󵄨 if 1 2 1 { 3 (52) { { 𝑖 󵄨 󵄨 ≾ℎ𝐿(𝑥,𝑦)(𝑡) , { 󵄨𝑦 −𝑦󵄨 , 𝑧 ,𝑧 ∈𝑋 {3 󵄨 1 2󵄨 if 1 2 2 𝑑(𝑧1,𝑧2)={ (48) { 2 1 where {𝑖( 𝑥1 + 𝑦2), if 𝑧1 ∈𝑋1,𝑧2 ∈𝑋2 { 3 3 { 𝐿(𝑥,𝑦)(𝑡) { 1 2 {𝑖( 𝑦 + 𝑥 ) 𝑧 ∈𝑋,𝑧 ∈𝑋, 3 1 3 2 if 1 2 2 1 { ∈{𝐴(𝑥,𝑦)(𝑡) ,𝐵(𝑥,𝑦)(𝑡) ,𝐶(𝑥,𝑦)(𝑡) ,𝐷(𝑥,𝑦)(𝑡)}, 𝑧 =𝑥+𝑖𝑦 𝑧 =𝑥+𝑖𝑦 ∈𝑋 (𝑋, 𝑑) where 1 1 1, 2 2 2 .Then 󵄩 󵄩 −1 𝐴(𝑥,𝑦)(𝑡) = 󵄩𝑥(𝑡) − 𝑦(𝑡)󵄩 √1+𝑎2𝑒𝑖 tan 𝑎, is a complete complex valued metric space. By a routine 󵄩 󵄩∞ 2 calculation, one can verify that the map 𝑇 satisfies condition −1 𝜆 = (1/3)( ) 󵄩 󵄩 √ 2 𝑖 tan 𝑎 (43)with say . It is interesting to notice that this 𝐵(𝑥,𝑦)(𝑡) = 󵄩𝐹𝑥 (𝑡) +𝑔(𝑡)−𝑥(𝑡)󵄩∞ 1+𝑎 𝑒 , example cannot be covered by Corollary 9 as 𝑧1 = (1, 0), 𝑧2 = (1/2, 0) ∈ 𝑋 implies 󵄩 󵄩 −1 𝐶(𝑥,𝑦)(𝑡) = 󵄩𝐺 (𝑡) +ℎ(𝑡)−𝑦(𝑡)󵄩 √1+𝑎2𝑒𝑖 tan 𝑎, 󵄩 𝑦 󵄩∞ 2𝑖 𝑖 =𝑑(𝑇𝑧,𝑇𝑧 )≤𝑑(𝑧,𝑧 )= 3 1 2 1 2 3 (49) 𝐷(𝑥,𝑦)(𝑡)

󵄩 󵄩 󵄩 󵄩 a contradiction for every choice of 𝜆 which amounts to say 󵄩𝐹 (𝑡) +𝑔(𝑡) −𝑥(𝑡)󵄩 󵄩𝐺 (𝑡) +ℎ(𝑡) −𝑦(𝑡)󵄩 󵄩 𝑥 󵄩∞󵄩 𝑦 󵄩 that condition (37) is not satisfied. Notice that the point 0∈𝑋 = ∞ 2 1+ 𝐴(𝑥,𝑦) 𝑡 remains fixed under 𝑇 and 𝑇 and is indeed unique. max𝑡∈[𝑎,𝑏] ( )

−1 √ 2 𝑖 tan 𝑎 3. Application × 1+𝑎 𝑒 , (53) By providing the following result, we establish an existence theorem for the common solution for two Urysohn integral then the system of integral equations (𝛼) and (𝛽) has a unique equations. common solution. 𝑛 Theorem 16. Let 𝑋 = 𝐶([𝑎, 𝑏], R ), 𝑎>0,and𝑑:𝑋×𝑋 → C is defined as follows: Proof. Define 𝑆, 𝑇 : 𝑋 →𝑋 by

−1 󵄩 󵄩 √ 2 𝑖 tan 𝑎 𝑑(𝑥,𝑦)= max 󵄩𝑥 (𝑡) −𝑦(𝑡)󵄩∞ 1+𝑎 𝑒 . 𝑡∈[𝑎,𝑏] (50) 𝑆𝑥𝑥 =𝐹 +𝑔, 𝑇𝑥=𝐺𝑥 +ℎ. (54) 6 Abstract and Applied Analysis

Then where 󵄩 󵄩 𝑑(𝑆𝑥,𝑇𝑦)= max 󵄩𝐹𝑥 (𝑡) −𝐺𝑦 (𝑡) +𝑔(𝑡) −ℎ(𝑡)󵄩 𝑡∈[𝑎,𝑏]󵄩 󵄩∞ 𝐿 (𝑥, 𝑦) ∈ {𝑑 (𝑓𝑥, 𝑓𝑦), 𝑑 (𝑓𝑥,𝑔 (𝑓𝑥)) ,𝑑(𝑓𝑦,ℎ(𝑓𝑦)),

𝑖 tan−1𝑎 × √1+𝑎2𝑒 , 𝑑 (𝑓𝑥, 𝑔 (𝑓𝑥))𝑑 (𝑓𝑦,ℎ (𝑓𝑦)) }. 1+𝑑(𝑓𝑥,𝑓𝑦) 𝑑(𝑥,𝑦)= max 𝐴(𝑥,𝑦)(𝑡) , 𝑡∈[𝑎,𝑏] (60)

(55) By Theorem 7 as 𝑓𝐸 iscomplete,wededucethatthereexists 𝑑 (𝑥, 𝑆𝑥) = max 𝐵(𝑥,𝑦)(𝑡) , 𝑡∈[𝑎,𝑏] a unique common fixed point 𝑓𝑧 ∈ 𝑓𝐸 of 𝑔 and ℎ;thatis, 𝑓𝑧 = 𝑔(𝑓𝑧) =ℎ(𝑓𝑧).Thus,𝑧 is a coincidence point of 𝑆, 𝑇 𝑓 𝑆 𝑇 𝑓 𝑑(𝑦,𝑇𝑦)= max 𝐶(𝑥,𝑦)(𝑡) , ,and . Now we show that , ,and have unique point 𝑡∈[𝑎,𝑏] of coincidence. Now let 𝑥∈𝑋such that 𝑆𝑥=𝑇𝑥=𝑓𝑥, 𝑓𝑥 =𝑓𝑧̸ .Then𝑓𝑥 is another common fixed point of 𝑔 and 𝑑 (𝑥, 𝑆𝑥) 𝑑(𝑦,𝑇𝑦) ℎ, which is a contradiction, which implies that 𝑆, 𝑇,and𝑓 = max 𝐷(𝑥,𝑦)(𝑡) . 1+𝑑(𝑥,𝑦) 𝑡∈[𝑎,𝑏] have a unique point of coincidence. Since (𝑆, 𝑓) and (𝑇, 𝑓) are weakly compatible by Lemma 6,wededucethat𝑓𝑧 is a It is easily seen that 𝑑(𝑆𝑥, 𝑇𝑦) ≾ ,whereℎ𝐿(𝑥, 𝑦) unique common fixed point of 𝑆, 𝑇,and𝑓.

Theorem 19. Let (𝑋, 𝑑) be a complete complex valued metric space and 0≤ℎ<1.Let𝑆, 𝑇, 𝑓,𝑔 :𝑋→ by the self- 𝐿 (𝑥,) 𝑦 mappings such that 𝑆𝑋, 𝑇𝑋 ⊂ 𝑓𝑋 =𝑔𝑋. Assume that the following holds: 𝑑 (𝑥, 𝑆𝑥) 𝑑(𝑦,𝑇𝑦) ∈{𝑑(𝑥,𝑦),𝑑(𝑥, 𝑆𝑥) ,𝑑(𝑦,𝑇𝑦), } 𝑑(𝑆𝑥,𝑇𝑦)≾ℎ𝐿(𝑥,𝑦) 1+𝑑(𝑥,𝑦) (61) for all 𝑥, 𝑦 ∈𝑋,where (56) 𝐿(𝑥,𝑦)∈{𝑑(𝑓𝑥,𝑔𝑦),𝑑(𝑓𝑥,𝑆𝑥),𝑑(𝑔𝑦,𝑇𝑦), for every 𝑥, 𝑦.By ∈𝑋 Theorem 7,theUrysohnintegral equations (𝛼) and (𝛽) have a unique common solution. (62) 𝑑(𝑓𝑥,𝑆𝑥)𝑑(𝑔𝑦,𝑇𝑦) }. Remark 17. Now we will apply techniques of [6]toobtainthe 1+𝑑(𝑓𝑥,𝑔𝑦) common fixed points of three and four mappings by using a common fixed point result for two mappings. If (𝑆, 𝑓) and (𝑇, 𝑔) are weakly compatible and 𝑓𝑋 is closed in 𝑋,then𝑆, 𝑇, 𝑓,and𝑔 have a unique common fixed point in 𝑋. Theorem 18. Let (𝑋, 𝑑) be a complete complex valued metric 0≤ℎ<1 𝑆, 𝑇, 𝑓 : 𝑋→ space and .Let by the self-mappings Proof. By Lemma 5, there exists 𝐸1,𝐸2 ⊂𝑋such that 𝑓𝐸1 = 𝑆𝑋∪𝑇𝑋⊂𝑓𝑋 such that . Assume that the following holds: 𝑓𝑋 = 𝑔𝑋 =𝑔𝐸2, 𝑓:𝐸1 →𝑋,𝑔:𝐸2 →𝑋are one to one. 𝐴, 𝐵 : 𝑓𝐸 →𝑓𝐸 𝐴𝑓(𝑥) =𝑆𝑥 𝑑 (𝑆𝑥, 𝑇𝑦) ≾ℎ𝐿(𝑥,) 𝑦 (57) Now define the mappings 1 1 by and 𝐵𝑔(𝑥) =𝑇𝑥,respectively.Since𝑓, 𝑔 are one to one on 𝐸1 for all 𝑥, 𝑦 ∈𝑋,where and 𝐸2, respectively, then the mappings 𝐴, 𝐵 are well-defined. Now 𝐿(𝑥,𝑦)∈{𝑑(𝑓𝑥,𝑓𝑦),𝑑(𝑓𝑥,𝑆𝑥),𝑑(𝑓𝑦,𝑇𝑦), 𝑑 (𝑆𝑥, 𝑇𝑦) = 𝑑 (𝐴 (𝑓𝑥), 𝐵 (𝑔𝑦)) ≾ℎ𝐿(𝑥,𝑦), (63) (58) 𝑑(𝑓𝑥,𝑆𝑥)𝑑(𝑓𝑦,𝑇𝑦) where }. 1+𝑑(𝑓𝑥,𝑓𝑦) 𝐿(𝑥,𝑦) If (𝑆, 𝑓) and (𝑇, 𝑓) are weakly compatible and 𝑓𝑋 is closed, then 𝑆, 𝑇,and𝑓 have a unique common fixed point in 𝑋. ∈{𝑑(𝑓𝑥,𝑔𝑦),𝑑(𝑓𝑥,𝐴(𝑓𝑥)),𝑑(𝑔𝑦,𝐵(𝑔𝑦)), (64) 𝐸⊂𝑋 𝑓𝐸 = 𝑓𝑋 Proof. By Lemma 5, there exists such that 𝑑 (𝑓𝑥, 𝐴 (𝑓𝑥))𝑑 (𝑔𝑦, 𝐵(𝑔𝑦)) and 𝑓:𝐸 →𝑋is one to one. Now define the self-mappings } 1+𝑑(𝑓𝑥,𝑔𝑦) 𝑔, ℎ : 𝑓𝐸 →𝑓𝐸 by 𝑔(𝑓𝑥) =𝑆𝑥 and ℎ(𝑓𝑥) = 𝑇𝑥,respectively. 𝑓 𝐸 𝑔 ℎ Since is one to one on ,then , are well defined. Note 𝑓𝑥, 𝑔𝑦 ∈𝑓𝐸 𝑓𝐸 that for all 1.ByTheorem 7,as 1 is complete subspace of 𝑋, we deduce that there exists a unique common 𝑑 (𝑔 (𝑓𝑥) , ℎ (𝑓𝑥)) ≾ ℎ𝐿 (𝑥,𝑦), (59) fixed point 𝑓𝑧 ∈ 𝑓𝐸 1 of 𝐴 and 𝐵;thatis,𝐴(𝑓𝑧) = 𝐵𝑓(𝑧) =𝑓𝑧. Abstract and Applied Analysis 7

This implies that 𝑆𝑧 = 𝑓𝑧;letV ∈𝑋such that 𝑓𝑧 =𝑔V.We [9] Lj. Ciric,´ N. Hussain, and N. Cakic,´ “Common fixed points for have 𝐵(𝑔V)=𝑔V ⇒𝑇V =𝑔V.Weshowthat𝑆 and 𝑓 have a Cirictype´ 𝑓-weak contraction with applications,” Publicationes unique point of coincidence. If 𝑆𝑤 = 𝑓𝑤 then 𝑓𝑤 is a fixed Mathematicae Debrecen,vol.76,no.1-2,pp.31–49,2010. point of 𝐴.BytheproofofTheorem 7𝑓𝑤 is another common [10] Lj. Ciric,´ M. Abbas, R. Saadati, and N. Hussain, “Common fixed fixed point of 𝐴 and 𝐵 which is a contradiction. Hence, 𝑆 and points of almost generalized contractive mappings in ordered 𝑓 have a unique point of coincidence. By Lemma 6,itfollows metric spaces,” Applied Mathematics and Computation,vol.217, that 𝑓𝑧 is a unique common fixed point of 𝑆 and 𝑓.Similarly, no.12,pp.5784–5789,2011. 𝑔V istheuniquecommonfixedpointfor𝑇and 𝑔.Thisproves [11] L.-G. Huang and X. Zhang, “Cone metric spaces and fixed point that 𝑓𝑧 =𝑔V istheuniquecommonfixedpointfor𝑆, 𝑇, 𝑓, theorems of contractive mappings,” Journal of Mathematical and 𝑔. Analysis and Applications,vol.332,no.2,pp.1468–1476,2007. [12] C. Klin-eam and C. Suanoom, “Some common fixed-point theorems for generalized-contractive-type mappings on complex- Conflict of Interests valued metric spaces,” Abstract and Applied Analysis,vol.2013, ArticleID604215,6pages,2013. The authors declare that there is no conflict of interests [13] M. A. Kutbi, J. Ahmad, and A. Azam, “On fixed points of 𝛼- regarding the publication of this paper. 𝜓-contractive multivalued mappings in cone metric spaces,” Abstract and Applied Analysis,vol.2013,ArticleID313782,6 Authors’ Contribution pages, 2013. [14] M. A. Kutbi, A. Azam, J. Ahmad, and C. Di Bari, “Some com- All authors contributed equally and significantly in writing mon coupled fixed point results for generalized contraction in this paper. All authors read and approved the final paper. complex-valued metric spaces,” Journal of Applied Mathematics, vol. 2013, Article ID 352927, 10 pages, 2013. [15] A. Azam, B. Fisher, and M. Khan, “Common fixed point the- Acknowledgments orems in complex valued metric spaces,” Numerical Functional Analysis and Optimization, vol. 32, no. 3, pp. 243–253, 2011. This paper is dedicated to Professor Miodrag Mateljevicon´ [16] F. Rouzkard and M. Imdad, “Some common fixed point theoccasionofhis65thbirthday.Thefirstauthorgratefully theorems on complex valued metric spaces,” Computers & acknowledges the support from the Deanship of Scientific Mathematics with Applications,vol.64,no.6,pp.1866–1874, Research (DSR) at King Abdulaziz University (KAU) during 2012. this research. The third author gratefully acknowledges the [17] W. Sintunavarat and P. Kumam, “Generalized common fixed support from the Higher Education Commission of Pakistan. point theorems in complex valued metric spaces and applications,” Journal of Inequalities and Applications,vol.2012,article References 84, 2012. [18] M. Imdad, J. Ali, and M. Tanveer, “Coincidence and common [1] S. Gahler,¨ “2-metrische raume und ihre topologische strukture,” fixed point theorems for nonlinear contractions in Menger PM Mathematische Nachrichten,vol.26,pp.115–148,1963. spaces,” Chaos, Solitons & Fractals,vol.42,no.5,pp.3121–3129, 2009. [2] M. Abbas, B. Fisher, and T. Nazir, “Well-Posedness and periodic point property of mappings satisfying a rational inequality in [19] R. H. Haghi, Sh. Rezapour, and N. Shahzad, “Some fixed point an ordered complex valued metric space,” Numerical Functional generalizations are not real generalizations,” Nonlinear Analysis. Analysis and Optimization,vol.243,p.32,2011. Theory, Methods & Applications,vol.74,no.5,pp.1799–1803, A 2011. [3] M. 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Ume, “Common fixed point theorems for families of weakly compatible maps,” Computers & Mathematics with Applications,vol.55,no.11,pp. 2533–2543, 2008. [8] Lj. Ciric,´ “A generalization of Banach’s contraction principle,” Proceedings of the American Mathematical Society,vol.45,pp. 267–273, 1974. Hindawi Publishing Corporation Abstract and Applied Analysis Volume 2014, Article ID 213296, 11 pages http://dx.doi.org/10.1155/2014/213296

Research Article Convergence of Numerical Solution of Generalized Theodorsen’s Nonlinear Integral Equation

Mohamed M. S. Nasser1,2 1 Department of Mathematics, Faculty of Science, King Khalid University, P.O. Box 9004, Abha, Saudi Arabia 2 Department of Mathematics, Faculty of Science, Ibb University, P.O. Box 70270, Ibb, Yemen

Correspondence should be addressed to Mohamed M. S. Nasser; mms [email protected]

Received 7 February 2014; Accepted 24 February 2014; Published 2 April 2014

Academic Editor: Samuel Krushkal

Copyright © 2014 Mohamed M. S. Nasser. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We consider a nonlinear integral equation which can be interpreted as a generalization of Theodorsen’s nonlinear integral equation. This equation arises in computing the conformal mapping between simply connected regions. We present a numerical method for solving the integral equation and prove the uniform convergence of the numerical solution to the exact solution. Numerical results are given for illustration.

1. Introduction where 𝑆(𝑡) is the boundary correspondence function of the mapping function Ψ.Thefunction𝑆(𝑡) is a strictly increasing Numerical methods for conformal mapping from a simply function so that 𝑆(𝑡) −𝑡 is a 2𝜋-periodic function. connectedregionontoanothersimplyconnectedregionare The function 𝑆(𝑡) istheuniquesolutionofanonlinear availableonlywhenoneoftheregionisastandardregion, integral equation which can be interpreted as a generalization 𝐷 𝐺 Ω mostly the unit disk .Let and be bounded simply of Theodorsen’s nonlinear integral equation [1]. The proof 𝑧 𝑤 connected regions in the -plane and -plane, respectively, of the existence and the uniqueness of the solution of the Γ:=𝜕𝐺 𝐿:=𝜕Ω such that their boundaries and are smooth nonlinear integral equation was given in [1] for regions Ω of Ψ:𝐺 → Ω Jordan curves. Then the mapping is calculated which boundaries 𝐿=𝜕Ωsatisfy the so-called 𝜖-condition; 𝐺→𝐷→Ω as the composition of the maps . that is, Recently,anumericalmethodhasbeenproposedin[1]for direct approximation of the mapping Ψ:𝐺 →. Ω Assume 󵄨 󸀠 󵄨 󵄨𝑅 (𝑡)󵄨 that Γ and 𝐿 are star-like with respect to the origin and defined 𝜀:= max 󵄨 󵄨 <1. (3) 0≤𝑡≤2𝜋 󵄨 𝑅 𝑡 󵄨 by polar coordinates 󵄨 ( ) 󵄨

𝑡 𝑡 𝜂 (𝑡) =𝜌(𝑡) 𝑒i ,𝜁(𝑡) =𝑅(𝑡) 𝑒i , 0≤𝑡≤2𝜋, (1) In this paper, the nonlinear integral equation is solved by an iterative method. Each iteration of the iterative method 𝑛×𝑛 𝜌 𝑅 2𝜋 requires solving an linear system which is obtained respectively, such that both and are -periodic continu- by discretizing the integrals in the integral equation by the ously differentiable positive real functions with nonvanishing trapezoidal rule. The linear system is solved by a combination derivatives. By the Riemann-mapping theorem, there exists a of the generalized minimal residual (GMRES) method and unique conformal mapping function Ψ:𝐺 →normalized Ω 󸀠 the fast multipole method (FMM) in 𝑂(𝑛 ln 𝑛) operations. by Ψ(0) = 0, Ψ (0) > 0.Theboundaryvalueofthefunction Ψ 𝐿 Themainobjectiveofthispaperistoprovetheuniform is on the boundary and can be described as convergence of the numerical solution to the exact solution. We also study the properties of the generalized conjugation + Ψ (𝜂 (𝑡))=𝜁(𝑆 (𝑡)) , 0≤𝑡≤2𝜋, (2) operator. Numerical results are presented for illustration. 2 Abstract and Applied Analysis

2𝜋 ∫ 𝜃󸀠(𝑠)𝛾(𝑠)𝑑𝑠 =0 2. Auxiliary Materials Theorem 1. If 0 ,then 𝜃 𝜏 𝑤=𝑓(𝑧) 󵄩 󸀠 󵄩 2.1. The Functions and . Let be the mapping 󵄩 󵄩2 󵄩 󵄩 󵄩𝛾 󵄩 󵄩𝛾󵄩 ≤2𝜋󵄩𝛾󵄩 󵄩 󵄩 . function from the simply connected region 𝐺 onto the unit 󵄩 󵄩∞ 󵄩 󵄩𝜃󵄩 󸀠 󵄩 (13) 󸀠 󵄩𝜃 󵄩 disk 𝐷 with the normalization 𝑓(0) = 0 and 𝑓 (0) > 0.Then 𝜃 𝑓 the boundary value of the function is on the unit circle and Proof. Since 𝑠=𝜏(𝑡)and if 𝑡=𝜃(𝑠),wehave canbedescribedas 2𝜋 2𝜋 𝑓(𝜂(𝑡))=𝑒i𝜃(𝑡),0≤𝑡≤2𝜋. ∫ (𝛾 ∘ 𝜏) (𝑡) 𝑑𝑡 = ∫ 𝛾 (𝜏 (𝑡)) 𝑑𝑡 (4) 0 0 (14) 𝜃(𝑡) 2𝜋 The function is the boundary correspondence function 󸀠 of the mapping function 𝑓 where 𝜃(𝑡) −𝑡 is a 2𝜋-periodic = ∫ 𝜃 (𝑠) 𝛾 (𝑠) 𝑑𝑠 = 0. 󸀠 0 function and 𝜃 (𝑡) > 0 for all 𝑡 ∈ [0, 2𝜋].Let𝜏(𝑡) be the inverse of the function 𝜃(𝑡).Then𝜏(𝑡) is the boundary Thus, it follows from [2,page68]that correspondence function of the inverse mapping function 󵄩 󵄩2 󵄩 󵄩 󵄩 󸀠󵄩 −1 󵄩𝛾∘𝜏󵄩 ≤2𝜋󵄩𝛾∘𝜏󵄩 󵄩(𝛾 ∘ 𝜏) 󵄩 . 𝑧=𝑓 (𝑤) from 𝐷 onto 𝐺;thatis, 󵄩 󵄩∞ 󵄩 󵄩2󵄩 󵄩2 (15)

−1 𝑡 𝑓 (𝑒i )=𝜂(𝜏 (𝑡)) ,0≤𝑡≤2𝜋, (5) We have also 󵄩 󵄩2 1 2𝜋 󵄨 󵄨2 󸀠 󵄩 󸀠󵄩 󵄨 󸀠 󵄨 where 𝜏(𝑡) −𝑡 is a 2𝜋-periodic function and 𝜏 (𝑡) > 0 for all 󵄩(𝛾 ∘ 𝜏) 󵄩 = ∫ 󵄨(𝛾 ∘ 𝜏) (𝑡)󵄨 𝑑𝑡 2 2𝜋 0 𝑡∈[0,2𝜋]. 1 2𝜋 = ∫ 𝛾󸀠(𝜏 (𝑡))2𝜏󸀠(𝑡)2𝑑𝑡 2.2. The Norms. Let 𝐻 bethespaceofallrealHolder¨ 2𝜋 0 2𝜋 [0, 2𝜋] continuous -periodic functions on .Withtheinner 1 2𝜋 1 (16) product = ∫ 𝛾󸀠(𝑠)2 𝑑𝑠 󸀠 2𝜋 0 𝜃 (𝑠) 1 2𝜋 2 (𝛾, 𝜓) = ∫ 𝛾 (𝑠) 𝜓 (𝑠) 𝑑𝑠, (6) 1 2𝜋 𝛾󸀠 (𝑠) 2𝜋 0 = ∫ 𝜃󸀠 (𝑠) ( ) 𝑑𝑠. 󸀠 2𝜋 0 𝜃 (𝑠) the space 𝐻 is a pre-Hilbert space. We define the norm ‖⋅‖2 by Hence, 󵄩 󵄩 1/2 󵄩 󸀠 󵄩2 󵄩𝛾󵄩 := (𝛾,) 𝛾 . 󵄩 󸀠󵄩2 󵄩𝛾 󵄩 󵄩 󵄩2 (7) 󵄩(𝛾 ∘ 𝜏) 󵄩 = 󵄩 󵄩 . 󵄩 󵄩2 󵄩 󸀠 󵄩 (17) 󵄩𝜃 󵄩𝜃 Since 𝑠=𝜏(𝑡)and if 𝑡=𝜃(𝑠),wehave 𝜏(⋅) : [0, 2𝜋] → [0, 2𝜋] 󵄩 󵄩 2𝜋 Since is bijective, we have 󵄩 󸀠 1/2 󵄩 󸀠 2 󵄩(𝜃 ) 𝛾󵄩 = ∫ 𝜃 (𝑠) 𝛾(𝑠) 𝑑𝑠 󵄩 󵄩 󵄨 󵄨 󵄩 󵄩 󵄩𝛾∘𝜏󵄩 := max 󵄨𝛾 (𝜏 (𝑡))󵄨 󵄩 󵄩2 0 󵄩 󵄩∞ 0≤𝑡≤2𝜋 󵄨 󵄨 (8) 2𝜋 (18) 󵄩 󵄩 󵄨 󵄨 󵄩 󵄩 = ∫ 𝛾 𝜏 𝑡 2𝑑𝑡 = 󵄩𝛾∘𝜏󵄩 . = max 󵄨𝛾 (𝜏)󵄨 = 󵄩𝛾󵄩 . ( ( )) 󵄩 󵄩2 0≤𝜏≤2𝜋 󵄨 󵄨 󵄩 󵄩∞ 0 Hence, (15)and(18) imply that With the norm ‖⋅‖2, we define a norm ‖⋅‖𝜃 by 󵄩 󵄩 󵄩 󵄩 󵄩 󸀠󵄩 󵄩 󵄩 󵄩 1/2 󵄩 󵄩 󵄩 󵄩𝛾󵄩 ≤2𝜋󵄩𝛾∘𝜏󵄩 󵄩(𝛾∘𝜏) 󵄩 . 󵄩 󵄩 󵄩 󸀠 󵄩 󵄩 󵄩 󵄩 󵄩∞ 󵄩 󵄩2󵄩 󵄩2 (19) 󵄩𝛾󵄩𝜃 := 󵄩(𝜃 ) 𝛾󵄩 = 󵄩𝛾∘𝜏󵄩2. (9) 󵄩 󵄩2 Then (13)followsfrom(9), (17), and (19). We define also the maximum norm ‖⋅‖𝜃 by 󵄩 󵄩 󵄨 󵄨 2.3. The Operators K and J. The conjugation operator K is 󵄩𝛾󵄩 := max 󵄨𝛾 (𝑡)󵄨 . ∞ 0≤𝑡≤2𝜋 (10) defined by Since 𝜃(2𝜋) − 𝜃(0) =2𝜋,wehave 2𝜋 1 𝑠−𝑡 K𝜇=∫ cot 𝜇 (𝑡) 𝑑𝑡. (20) 2𝜋 0 2𝜋 2 󵄩 󵄩2 1 󸀠 2 󵄩𝛾󵄩𝜃 = ∫ 𝜃 (𝑡) 𝛾 (𝑡) 𝑑𝑡 2𝜋 0 Let J be the operator defined by (11) 2𝜋 2𝜋 󵄩 󵄩2 1 󸀠 󵄩 󵄩2 1 ≤ 󵄩𝛾󵄩∞ ∫ 𝜃 (𝑡) 𝑑𝑡 = 󵄩𝛾󵄩∞, J𝜇= ∫ 𝜇 (𝑡) 𝑑𝑡. (21) 2𝜋 0 2𝜋 0 which implies that Hence, the operators K and J satisfy [3] 󵄩 󵄩 󵄩 󵄩 2 󵄩𝛾󵄩𝜃 ≤ 󵄩𝛾󵄩∞. (12) JK =0, K =−I + J. (22) Abstract and Applied Analysis 3

3. The Generalized Conjugation Operator Lemma 4 (see [1]). If 𝛾∈𝐻and 𝜇=E𝛾,then𝛾=𝑐−E𝜇 with a real constant 𝑐=𝑓(0)where 𝑓 istheuniqueanalytic 𝐴 2𝜋 Let be the complex -periodic continuously differentiable function in 𝐺 with the boundary values 𝑓(𝜂(𝑡)) = 𝛾(𝑡) + i𝜇(𝑡) function: and Im 𝑓(0) = 0. 𝐴 (𝑠) := 𝜂 (𝑠) . (23) Lemma 5 (see [1]). The operator E has the following properties: 𝑀 𝑁 We define the real kernels and as real and imaginary 𝑁𝑢𝑙𝑙 (E) =𝑠𝑝𝑎𝑛{1} , parts: 3 E =−E, (32) 1 𝐴 (𝑠) 𝜂󸀠 (𝑡) 𝑀 (𝑠,) 𝑡 + 𝑁 (𝑠,) 𝑡 := . (24) i 𝜋 𝐴 (𝑡) 𝜂 (𝑡) −𝜂(𝑠) 𝜎 (E) = {0, ±i} .

The kernel 𝑁(𝑠, 𝑡) is called the generalized Neumann kernel Lemma 6. The operator E has the norm formed with 𝐴 and 𝜂.Thekernel𝑁(𝑠, 𝑡) is continuous and the ‖E‖𝜃 =1. (33) kernel 𝑀 has the representation 1 𝑠−𝑡 Proof. The operator E has the norm ‖E‖𝜃 ≤1[1]. Since i ∈ 𝑀 (𝑠,) 𝑡 =− +𝑀 (𝑠,) 𝑡 , 𝜎( ) 1=||≤‖ ‖ ≤1 2𝜋cot 2 1 (25) E ,hence i E 𝜃 .Hence,weobtain(33). with a continuous kernel 𝑀1.See[4] for more details. Lemma 7. The operators J𝜃 and E satisfy We define the Fredholm integral operators N and M1 and J𝜃E =0. (34) the singular integral operator M on 𝐻 by

2𝜋 Proof. For any 𝛾∈𝐻,let𝜇=E𝛾, 𝜇=𝜇∘𝜏̂ ,and𝛾=𝛾∘𝜏̂ . N𝜇=∫ 𝑁 (𝑠,) 𝑡 𝜇 (𝑡) 𝑑𝑡, Then, it follows from29 ( )that𝜇=̂ K𝛾̂.Thus, 0 2𝜋 J𝜃E𝛾 (𝑠) = J𝜃𝜇 (𝑠) 𝜇=∫ 𝑀 (𝑠,) 𝑡 𝜇 (𝑡) 𝑑𝑡, (26) M1 1 1 2𝜋 0 = ∫ 𝜃󸀠 (𝑡) 𝜇 (𝑡) 𝑑𝑡 2𝜋 2𝜋 0 𝜇=∫ 𝑀 (𝑠,) 𝑡 𝜇 (𝑡) 𝑑𝑡. M 1 2𝜋 0 = ∫ 𝜇 (𝜏 (𝑠)) 𝑑𝑠 2𝜋 We define an operator E on 𝐻 by 0 (35) 2𝜋 −1 1 E =−(I − N) M. (27) = ∫ 𝜇̂ (𝑠) 𝑑𝑠 2𝜋 0 𝐻 The operator E is singular but bounded on [1]. Finally, we = J𝜇̂ (𝑠) define an operator J𝜃 by = 𝛾̂(𝑠) , 2𝜋 JK 1 󸀠 J𝜃𝜇= ∫ 𝜃 (𝑡) 𝜇 (𝑡) 𝑑𝑡. (28) 2𝜋 0 which by (22)impliesthat 󸀠 𝛾 (𝑠) =0. Remark 2. When Γ reduces to the unit, then 𝜃 (𝑡) =, 1 J𝜃E (36) the operator J𝜃 reduces to the operator J,andtheoperator Since (36)holdsforallfunctions𝛾∈𝐻, the operator identity E reduces to the operator K;thatis,theoperatorE is a (34) follows. generalization of the well-known conjugation operator K (see [1] for more details). Lemma 8. The operator E satisfies The operator E is related to the operator K by [1] 2 E =−I + J𝜃. (37) 𝜇= 𝛾 𝜇∘𝜏= (𝛾 ∘ 𝜏) . E iff K (29) Proof. Let 𝛾∈𝐻and 𝜇=E𝛾.Then,byLemma 4, 𝛾=𝑐−E𝜇 with a real constant 𝑐. By the definition of the operator J𝜃,we Since 𝜇 = (𝜇 ∘ 𝜃) ∘𝜏 and 𝛾 = (𝛾 ∘ 𝜃) ∘𝜏,itfollowsfrom(29) have J𝜃𝑐=𝑐.SinceJ𝜃E =0,wehave that J𝜃𝛾=J𝜃𝑐−J𝜃E =𝑐. (38) 𝜇=K𝛾 iff 𝜇∘𝜃=E (𝛾 ∘ 𝜃) . (30) Hence, Lemma 3 (see [1]). Let 𝛾, 𝜇 ∈𝐻 be given functions. Then 2 𝑓(𝜂(𝑡)) = 𝛾(𝑡) + i𝜇(𝑡) is the boundary value of an analytic E 𝛾=E𝜇=−𝛾+𝑐=(−I + J𝜃)𝛾 (39) function in 𝐺 with Im 𝑓(0) = 0 if and only if holds for all 𝛾∈𝐻. Thus, the operator identities (37)follow. 𝜇=E𝛾. (31) 4 Abstract and Applied Analysis

Lemma 9. For all functions 𝛾∈𝐻,wehave That is, the approximate solutions 𝑆𝑘(𝑡) converge to 𝑆(𝑡) with ‖⋅‖ 𝜀<1 󵄩 󵄩 󵄩 󵄩 respect to the norm 𝜃 if . 󵄩 𝛾󵄩 ≤ 󵄩𝛾󵄩 , 󵄩E 󵄩𝜃 󵄩 󵄩𝜃 (40) In this section, we will prove the uniform convergence of the approximate solutions 𝑆𝑘(𝑡) to the exact solution 𝑆(𝑡).We 𝛾 𝛾=0 with equality for all with J𝜃 . will use the approach used in the proof of Proposition 1.5 in [2, page 69] related to Theodorsen’s integral equation. See also 𝛾∈𝐻 Proof. For all functions ,theinequality(40)follows [5, 6]. from (33). For all functions 𝛾 with J𝜃𝛾=0,wehavefrom(37)that 2 Lemma 11. Consider 𝛾=−E 𝛾.Hence E [ln 𝜌 (⋅)] (𝑡) =𝑡−𝜃(𝑡) . 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 (50) 󵄩𝛾󵄩 = 󵄩 2𝛾󵄩 ≤ 󵄩 𝛾󵄩 ≤ 󵄩𝛾󵄩 , 󵄩 󵄩𝜃 󵄩E 󵄩 󵄩E 󵄩𝜃 󵄩 󵄩𝜃 (41) 𝜃 Proof. The function 𝜃 is the boundary correspondence function of the conformal mapping 𝑓 from 𝐺 onto the unit disk. which means that ‖E𝛾‖𝜃 =‖𝛾‖𝜃. Hence, the function 𝜃(𝑡) −𝑡 satisfies [1] 𝛾∈𝐻 𝜇= 𝛾 Theorem 10. Let and E .Then 1 𝜃 (𝑡) −𝑡=E ln (𝑡) . (51) 𝜇󸀠 𝛾󸀠 𝜌 (⋅) = ( ). 󸀠 E 󸀠 (42) 𝜃 𝜃 Then (50) follows from (51). Proof. For 𝛾∈𝐻and 𝜇=E𝛾,wehavefrom(29)that𝜇∘𝜏= The previous lemma implies that (46)canberewrittenas K(𝛾 ∘ 𝜏).Then,itfollowsfrom[2,page64]that 𝑆 (𝑡) −𝜃(𝑡) = E [ln 𝑅 (𝑆 (⋅))] (𝑡) , (52) 󸀠 󸀠 (𝜇∘𝜏) = K(𝛾∘𝜏) . (43) and (47)canberewrittenas Hence, by (30), we have 𝑆𝑘 (𝑡) −𝜃(𝑡) = E [ln 𝑅(𝑆𝑘−1 (⋅))] (𝑡) . (53) (𝜇∘𝜏)󸀠 ∘𝜃= ((𝛾∘𝜏)󸀠 ∘𝜃) , E (44) Thus which implies that 𝑆𝑘 (𝑡) −𝑆(𝑡) = E [ln 𝑅(𝑆𝑘−1 (⋅))−ln 𝑅 (𝑆 (⋅))] (𝑡) . (54) 󸀠 󸀠 ((𝜇 ∘ 𝜏) ∘ 𝜃) ((𝛾 ∘ 𝜏) ∘𝜃) Lemma 12. Consider = ( ) . 󸀠 E 󸀠 (45) 𝜃 𝜃 󵄩 󵄩 󵄩 󵄩 𝑘 󵄩𝑆𝑘 −𝑆󵄩𝜃 ≤(𝜀+󵄩𝑆0 −𝜃󵄩𝜃)𝜀 . (55) Hence, we obtain (42). Proof. Let 𝑎 be such that 1 𝑅 (𝑡) 4. The Generalized Theodorsen Nonlinear ≤ ≤1+𝜀, ∀𝑡. 1+𝜀 𝑎 (56) Integral Equation Then The boundary correspondence function 𝑆(𝑡) is the unique 󵄨 󵄨 solution of the nonlinear integral equation 󵄨 𝑅 (𝑡)󵄨 󵄨ln 󵄨 ≤ ln (1+𝜀) <𝜀, ∀𝑡. (57) 󵄨 𝑎 󵄨 𝑅 (𝑆 (⋅)) 𝑆 (𝑡) −𝑡= (𝑡) E ln 𝜌 (⋅) (46) Hence, 󵄩 󵄩 󵄩 𝑅󵄩 which is a generalization of the well-known Theodorsen 󵄩 󵄩 <𝜀. 󵄩ln 𝑎 󵄩 (58) integral equation [1]. Nonlinear integral equation (46)canbe ∞ solved by the iterative method Thus 󵄩 󵄩 󵄩 󵄩 𝑅(𝑆 (⋅)) 󵄩 𝑅 (𝑆 (⋅)) 󵄩 󵄩 𝑅 (𝑆) 󵄩 𝑘−1 󵄩 [ ( )]󵄩 ≤ ‖ ‖ 󵄩 ( )]󵄩 𝑆𝑘 (𝑡) −𝑡=E ln (𝑡) , 𝑘=1,2,3,.... (47) 󵄩E ln 󵄩 E 𝜃󵄩ln 󵄩 𝜌 (⋅) 󵄩 𝑎 󵄩𝜃 󵄩 𝑎 󵄩𝜃 (59) 󵄩 󵄩 󵄩 𝑅 (𝑆) 󵄩 Thenwehave[1] ≤ 󵄩ln ( )]󵄩 <𝜀. 󵄩 𝑎 󵄩∞ 󵄩 󵄩 𝑘󵄩 󵄩 󵄩𝑆 −𝑆󵄩 ≤𝜀󵄩𝑆 −𝑆󵄩 . (48) 󵄩 𝑘 󵄩𝜃 󵄩 0 󵄩𝜃 Since 𝐿 𝜀 Thus, if the curve satisfies the -condition (3), then 𝑆 (𝑡) −𝑆0 (𝑡) =𝑆(𝑡) −𝜃(𝑡) +𝜃(𝑡) −𝑆0 (𝑡) 󵄩 󵄩 (60) 󵄩𝑆𝑘 −𝑆󵄩𝜃 󳨀→ 0. (49) = E [ln 𝑅 (𝑆 (⋅))] (𝑡) +𝜃(𝑡) −𝑆0 (𝑡) Abstract and Applied Analysis 5 and E(ln 𝑎) =,wehave 0 By (70)and(66), we have 𝑅 (𝑆 (⋅)) 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 𝑆 (𝑡) −𝑆 (𝑡) = [ ( )] (𝑡) +𝜃(𝑡) −𝑆 (𝑡) , 󵄩 𝑆󸀠 󵄩 󵄩 𝑆󸀠 󵄩 󵄩 𝑆󸀠 󵄩 0 E ln 0 (61) 󵄩 −1󵄩 ≤𝜀2󵄩 󵄩 =𝜀2 +𝜀2󵄩 −1󵄩 . 𝑎 󵄩 󸀠 󵄩 󵄩 󸀠 󵄩 󵄩 󸀠 󵄩 (71) 󵄩𝜃 󵄩𝜃 󵄩𝜃 󵄩𝜃 󵄩𝜃 󵄩𝜃 which implies that 󵄩 󵄩 Hence, 󵄩 󵄩 󵄩 𝑅 (𝑆 (⋅)) 󵄩 󵄩𝑆−𝑆 󵄩 ≤ 󵄩 [ ( )]󵄩 󵄩 󵄩2 󵄩 0󵄩𝜃 󵄩E ln 𝑎 󵄩 󵄩 𝑆󸀠 󵄩 𝜀2 󵄩 󵄩𝜃 (62) 󵄩 −1󵄩 ≤ . 󵄩 󸀠 󵄩 2 (72) 󵄩 󵄩 󵄩 󵄩 󵄩𝜃 󵄩𝜃 1−𝜀 + 󵄩𝜃−𝑆0󵄩𝜃 ≤𝜀+󵄩𝜃−𝑆0󵄩𝜃. Similarly, it follows from (69)that Hence (55)followsfrom(48). 󵄩 󸀠 󵄩 󵄩 󸀠 󵄩 󵄩𝑆𝑘 󵄩 󵄩𝑆𝑘−1 󵄩 Lemma 13. Consider 󵄩 −1󵄩 ≤𝜀󵄩 󵄩 , (73) 󵄩 𝜃󸀠 󵄩 󵄩 𝜃󸀠 󵄩 󵄩 󸀠 󸀠 󵄩 󵄩 󸀠 󵄩 󵄩 󵄩𝜃 󵄩 󵄩𝜃 󵄩𝑆𝑘 −𝑆 󵄩 𝜀 󵄩𝑆0 󵄩 󵄩 󵄩 ≤ (1+󵄩 󵄩 ) . 󵄩 󸀠 󵄩 2 󵄩 󸀠 󵄩 (63) which by (67)impliesthat 󵄩 𝜃 󵄩𝜃 √1−𝜀 󵄩 𝜃 󵄩𝜃 󵄩 󵄩2 󵄩 󵄩2 󵄩𝑆󸀠 󵄩 󵄩𝑆󸀠 󵄩 Proof. We have 󵄩 𝑘 −1󵄩 ≤𝜀2󵄩 𝑘−1 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󸀠 󵄩 󵄩 󸀠 󵄩 󵄩 󸀠 󸀠 󵄩 󵄩 󸀠 󸀠 󸀠 󸀠 󵄩 󵄩 𝜃 󵄩𝜃 󵄩 𝜃 󵄩𝜃 󵄩𝑆𝑘 −𝑆 󵄩 󵄩𝑆𝑘 −𝜃 +𝜃 −𝑆 󵄩 󵄩 󵄩 = 󵄩 󵄩 (74) 󵄩 󸀠 󵄩 󵄩 󸀠 󵄩 󵄩 󵄩2 󵄩 𝜃 󵄩𝜃 󵄩 𝜃 󵄩𝜃 󵄩 󸀠 󵄩 2 2󵄩𝑆𝑘−1 󵄩 (64) =𝜀 +𝜀 󵄩 −1󵄩 . 󵄩 󸀠 󵄩 󵄩 󸀠 󵄩 󵄩 𝜃󸀠 󵄩 󵄩𝑆𝑘 󵄩 󵄩 𝑆 󵄩 󵄩 󵄩𝜃 ≤ 󵄩 −1󵄩 + 󵄩 −1󵄩 . 󵄩 𝜃󸀠 󵄩 󵄩𝜃󸀠 󵄩 󵄩 󵄩𝜃 󵄩 󵄩𝜃 Hence, Since 󵄩 󸀠 󵄩2 󵄩 󸀠 󵄩2 󵄩𝑆𝑘 󵄩 2 2󵄩𝑆𝑘−1 󵄩 󵄩 󵄩2 2 󵄩 −1󵄩 ≤𝜀 +𝜀 󵄩 −1󵄩 󵄩 𝑆󸀠 󵄩 1 2𝜋 𝑆󸀠 󵄩 𝜃󸀠 󵄩 󵄩 𝜃󸀠 󵄩 󵄩 −1󵄩 = ∫ 𝜃󸀠 (𝑡) ( −1) 𝑑𝑡 󵄩 󵄩𝜃 󵄩 󵄩𝜃 󵄩 󸀠 󵄩 󸀠 󵄩𝜃 󵄩 2𝜋 0 𝜃 (75) 𝜃 󵄩 󸀠 󵄩2 2 4 2𝑘󵄩𝑆0 󵄩 2𝜋 󸀠 2 2𝜋 ≤⋅⋅⋅≤𝜀 +𝜀 +⋅⋅⋅+𝜀 󵄩 −1󵄩 , 1 󸀠 𝑆 1 󸀠 󵄩 𝜃󸀠 󵄩 = ∫ 𝜃 (𝑡) ( ) 𝑑𝑡 −2 ∫ 𝑆 (𝑡) 𝑑𝑡 󵄩 󵄩𝜃 󸀠 2𝜋 0 𝜃 2𝜋 0 which, in view of (67), implies that 2𝜋 1 󸀠 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 + ∫ 𝜃 (𝑡) 𝑑𝑡, 󵄩𝑆󸀠 󵄩 𝜀2 󵄩𝑆󸀠 󵄩 𝜀2 󵄩𝑆󸀠 󵄩 2𝜋 0 󵄩 𝑘 −1󵄩 ≤ (1 + 󵄩 0 −1󵄩 )= 󵄩 0 󵄩 . 󵄩 󸀠 󵄩 2 󵄩 󸀠 󵄩 2 󵄩 󸀠 󵄩 (65) 󵄩 𝜃 󵄩𝜃 1−𝜀 󵄩 𝜃 󵄩𝜃 1−𝜀 󵄩 𝜃 󵄩𝜃 2𝜋 2𝜋 (76) ∫ 𝑆󸀠(𝑡)𝑑𝑡 =2𝜋 ∫ 𝜃󸀠(𝑡)𝑑𝑡 =2𝜋 0 ,and 0 ,weobtain 󵄩 󵄩2 󵄩 󵄩2 󵄩 𝑆󸀠 󵄩 󵄩 𝑆󸀠 󵄩 󵄩 −1󵄩 = 󵄩 󵄩 −1. 󵄩 󸀠 󵄩 󵄩 󸀠 󵄩 (66) Then (63)followsfrom(64), (72), and (76). 󵄩𝜃 󵄩𝜃 󵄩𝜃 󵄩𝜃 𝜀<1 𝑆 Similarly, we have Theorem 14. If , then the approximate solution 𝑘 converges uniformly to the exact solution 𝑆 with 󵄩 󸀠 󵄩2 󵄩 󸀠 󵄩2 󵄩𝑆𝑘 󵄩 󵄩𝑆𝑘 󵄩 󵄩 −1󵄩 = 󵄩 󵄩 −1. (67) 2𝜋 󵄩 𝜃󸀠 󵄩 󵄩 𝜃󸀠 󵄩 󵄩 󵄩 󵄩 󵄩𝜃 󵄩 󵄩𝜃 󵄩𝑆𝑘 −𝑆󵄩 ≤ √ ∞ √1−𝜀2 In view of Theorem 10,itfollowsfrom(52)and(53)that (77) 󵄩 󸀠 󵄩 󸀠 󸀠 󸀠 󵄩𝑆 󵄩 𝑆 (𝑡) 𝑅 (𝑆 (⋅)) 𝑆 (⋅) √ 󵄩 󵄩 󵄩 0 󵄩 𝑘/2+1/2 −1= [ ] (𝑡) , × (𝜀 + 󵄩𝑆0 −𝜃󵄩 )(1+󵄩 󵄩 )𝜀 . 󸀠 E 󸀠 (68) 󵄩 󵄩𝜃 󵄩 󸀠 󵄩 𝜃 (𝑡) 𝑅 (𝑆 (⋅)) 𝜃 (⋅) 󵄩 𝜃 󵄩𝜃

𝑆󸀠 (𝑡) 𝑅󸀠 (𝑆 (⋅)) 𝑆󸀠 (⋅) Proof. In view of (54), Lemma 7 implies that 𝑘 −1= [ 𝑘−1 𝑘−1 ] (𝑡) . 󸀠 E 󸀠 (69) 2𝜋 𝜃 (𝑡) 𝑅(𝑆𝑘−1 (⋅)) 𝜃 (⋅) 󸀠 ∫ 𝜃 (𝑡) (𝑆𝑘 (𝑡) −𝑆(𝑡))𝑑𝑡=0. (78) Hence, it follows from (68)that 0 󵄩 󸀠 󵄩 󵄩 󸀠 󸀠 󵄩 󵄩 𝑆 󵄩 󵄩 𝑅 (𝑆 (⋅)) 𝑆 (⋅) 󵄩 Thus, we have from (13)that 󵄩 −1󵄩 = 󵄩 [ ]󵄩 󵄩𝜃󸀠 󵄩 󵄩E 𝑅 (𝑆 (⋅)) 𝜃󸀠 (⋅) 󵄩 󵄩 󸀠 󸀠 󵄩 󵄩 󵄩𝜃 󵄩 󵄩𝜃 󵄩 󵄩2 󵄩 󵄩 󵄩𝑆𝑘 −𝑆 󵄩 󵄩𝑆 −𝑆󵄩 ≤2𝜋󵄩𝑆 −𝑆󵄩 󵄩 󵄩 . (79) 󵄩 󵄩 󵄩 󵄩 (70) 󵄩 𝑘 󵄩∞ 󵄩 𝑘 󵄩𝜃󵄩 𝜃󸀠 󵄩 󵄩𝑅󸀠 (𝑆 (⋅)) 𝑆󸀠 (⋅) 󵄩 󵄩 𝑆󸀠 󵄩 󵄩 󵄩𝜃 ≤ ‖ ‖ 󵄩 󵄩 ≤𝜀󵄩 󵄩 . E 𝜃󵄩 󸀠 󵄩 󵄩 󸀠 󵄩 󵄩 𝑅 (𝑆 (⋅)) 𝜃 (⋅)󵄩𝜃 󵄩𝜃 󵄩𝜃 Hence (77) follows from (55)and(63). 6 Abstract and Applied Analysis

The following corollary follows from the previous theo- Since the kernel 𝑁(𝑠, 𝑡) is continuous on both variables and rem. since the function 𝛾(𝑡) is continuous, we have [9] 󵄩 󵄩 Corollary 15. If 󵄩(N − N𝑛)𝛾󵄩∞ 󳨀→ 0. (88) 𝑆 (𝑡) =𝜃(𝑡) =𝑡− [ 𝜌 (⋅)] (𝑡) , 0 E ln (80) The integral operator M1 will be discretized by the Nystrom¨ method as follows: then 2𝜋 𝑛 󵄩 󵄩 𝜋 𝑘/2+1 󵄩𝑆 −𝑆󵄩 ≤2√ 𝜀 . M1,𝑛𝛾 (𝑠) = ∑𝑀1 (𝑠,𝑗 𝑡 )[𝛾(𝑡𝑗)−𝛾(𝑠)]. 󵄩 𝑘 󵄩∞ (81) 𝑛 (89) √1−𝜀2 𝑗=1 Γ Remark 16. When reduces to the unit, then Since the kernel 𝑀1(𝑠, 𝑡) is continuous on both variables and since the function 𝛾(𝑡) is continuous, we have [9] 𝜌 (𝑡) =1, 𝜃(𝑡) =𝑡, E = K. (82) 󵄩 󵄩 󵄩(M1 − N1,𝑛)𝛾󵄩 Hence, the results presented in this section reduces to the 󵄩 󵄩∞ 󵄨 results presented in [2] for Theodorsen’s integral equation. 󵄨 2𝜋 󵄨 = max 󵄨 ∫ 𝑀1 (𝑠,) 𝑡 𝛾 (𝑡) 𝑑𝑡 𝑠∈[0,2𝜋] 󵄨 0 5. Discretizing (47) 󵄨 (90) 󵄨 𝑛 󵄨 In this paper, we will discretize (47) instead of (46). The 2𝜋 󵄨 − ∑𝑀 (𝑠, 𝑡 )𝛾(𝑡)󵄨 󳨀→ 0. numerical method used here is based on strict discretization 𝑛 1 𝑗 𝑗 󵄨 𝑗=1 󵄨 of the integrals in the operator E by the trapezoidal rule which gives accurate results since the integrals are over 2𝜋-periodic. 𝛾(𝑠) 𝑛 𝑛 To discretize the operator K , we first approximate the Let be a given even positive integer. We define equidistant function 𝛾(𝑠) by the interpolating trigonometric polynomial collocation points 𝑠𝑖 in the interval [0, 2𝜋] by of degree 𝑛/2 which interpolates 𝛾(𝑠) at the 𝑛 points 𝑡𝑗, 𝑗= 2𝜋 1,2,...,𝑛.Thatis, 𝑡𝑖 := (𝑖−1) , 𝑖=1,2,...,𝑛. (83) 𝑛 𝑛/2 𝑛/2−1 2𝜋 𝛾(𝑡) 𝛾 (𝑠) ≈ ∑𝑎𝑖 cos 𝑖𝑠 + ∑ 𝑏𝑖 sin 𝑖𝑠. (91) Then, for -periodic function ,thetrapezoidalrule 𝑖=0 𝑖=0 approximates the integral

2𝜋 Then K𝛾(𝑠) is approximated by 𝐼=∫ 𝛾 (𝑡) 𝑑𝑡 (84) 0 𝑛/2 𝑛/2−1 K𝑛𝛾 (𝑠) = ∑𝑎𝑖 sin 𝑖𝑠 − ∑ 𝑏𝑖 cos 𝑖𝑠, (92) by 𝑖=1 𝑖=0 2𝜋 𝑛 𝐼 = ∑𝛾(𝑡). where [6] 𝑛 𝑛 𝑖 (85) 𝑖=1 󵄩 󵄩 󵄩(K − K𝑛) 𝛾󵄩∞ 󳨀→ 0. (93) If the function 𝛾(𝑡) is continuous, then |𝐼−𝐼𝑛|→0.Ifthe integrand 𝛾(𝑡) is 𝑘 times continuously differentiable, then the The integral operator M is then discretized by 𝑘 rate of convergence of the trapezoidal rule is 𝑂(1/𝑛 ).For 𝑘 M𝑛 = M1,𝑛 − K𝑛. (94) analytic 𝛾(𝑡), the rate of convergence is better than 𝑂(1/𝑛 ) for any positive integer 𝑘 [7,page83].Seealso[8]. Then, it follows from90 ( )and(93)that For 𝛾∈𝐻,theintegraloperatorN will be discretized by the Nystrom¨ method as follows: 󵄩 󵄩 󵄩(M − M𝑛)𝛾󵄩∞ 󳨀→ 0. (95) 2𝜋 𝑛 𝛾 (𝑠) = ∑𝑁(𝑠,𝑡)𝛾(𝑡). The operator M𝑛 is bounded operator since the operator M1,𝑛 N𝑛 𝑛 𝑗 𝑗 (86) 𝑗=1 is bounded (𝑀1(𝑠, 𝑡) is continuous) and the operator K𝑛 is bounded operator (see [6]). Hence, we have Since the kernel 𝑁(𝑠, 𝑡) is continuous and 𝜆=1is not 𝑁(𝑠, 𝑡) − 󵄩 󵄩 an eigenvalue of the kernel [1], the operators I N𝑛 󵄩(N − N𝑛)𝛾󵄩∞ −1 are invertible and (I − N𝑛) areuniformlyboundedfor 󵄨 󵄨 𝑛 󵄨 2𝜋 𝑛 󵄨 sufficiently large [9]. Hence, we discretize the operator E 󵄨 2𝜋 󵄨 = max 󵄨∫ 𝑁 (𝑠,) 𝑡 𝛾 (𝑡) 𝑑𝑡 − ∑𝑁(𝑠,𝑡𝑗)𝛾(𝑡𝑗)󵄨 . by the bounded operator 𝑠∈[0,2𝜋] 󵄨 𝑛 󵄨 󵄨 0 𝑗=1 󵄨 −1 (87) E𝑛 := −(I − N𝑛) M𝑛. (96) Abstract and Applied Analysis 7

Lemma 17. If 𝛾∈𝐻,then Lemma 18. Consider 󵄩 󵄩 󵄩 󵄩 󵄩( − )𝛾󵄩 󳨀→ 0. 󵄩𝑆 −𝑆 󵄩 󳨀→ 0 𝑠𝑛󳨀→∞. 󵄩 E E𝑛 󵄩∞ (97) 󵄩 𝑘,𝑛 𝑘󵄩∞ a (106)

Proof. Let 𝜙:=E𝛾 and 𝜙𝑛 := E𝑛𝛾,then Proof. Let 𝛾(𝑠) := ln(𝑅(𝑆𝑘−1(𝑠)))/𝜌(𝑠).Then,wehave − 𝜙=− 𝛾, ( − ) 𝜙 =− 𝛾. (I N) M I N𝑛 𝑛 M𝑛 (98) 𝑆𝑘,𝑛 −𝑆𝑘 = E𝑛𝛾−E𝛾=(E𝑛 − E)𝛾. (107) ̂ Let also 𝜙𝑛 betheuniquesolutionofthediscretizedequation Hence, 󵄩 󵄩 󵄩 󵄩 ̂ 󵄩𝑆 −𝑆 󵄩 ≤ 󵄩( − ) 𝛾 󵄩 . (I − N𝑛) 𝜙𝑛 =−M𝛾. (99) 󵄩 𝑘,𝑛 𝑘󵄩∞ 󵄩 E E𝑛 𝑛󵄩∞ (108)

Thus, we have The lemma is then followed from (97). 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩( − )𝛾󵄩 = 󵄩𝜙−𝜙 󵄩 ≤ 󵄩𝜙−𝜙̂ 󵄩 The proof of the uniform convergence of the approximate 󵄩 E E𝑛 󵄩∞ 󵄩 𝑛󵄩∞ 󵄩 𝑛󵄩∞ solution 𝑆𝑘,𝑛 to the boundary correspondence function 𝑆 is 󵄩 󵄩 (100) + 󵄩𝜙̂ −𝜙 󵄩 . given in the following theorem. 󵄩 𝑛 𝑛󵄩∞ Theorem 19. If 𝜀<1,then Since the kernel 𝑁 is continuous and N𝑛 is the discretization 󵄩 󵄩 of N,thenitfollowsfrom[9,page108]that 󵄩𝑆 −𝑆󵄩 󳨀→ 0 𝑠𝑘,𝑛󳨀→∞. 󵄩 𝑘,𝑛 󵄩∞ a (109) 󵄩 󵄩 󵄩𝜙−𝜙̂ 󵄩 󳨀→ 0. 󵄩 𝑛󵄩∞ (101) Proof. We have 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ( − )−1 𝛾 󵄩𝑆 −𝑆󵄩 ≤ 󵄩𝑆 −𝑆 󵄩 + 󵄩𝑆 −𝑆󵄩 . Since I N𝑛 is bounded and is continuous, then (95) 󵄩 𝑘,𝑛 󵄩∞ 󵄩 𝑘,𝑛 𝑘󵄩∞ 󵄩 𝑘 󵄩∞ (110) implies that 𝜀<1 ‖𝑆 −𝑆‖ →0 𝑘→ 󵄩 󵄩 Since , it follows from (77)that 𝑘 ∞ as 󵄩𝜙̂ −𝜙 󵄩 ∞ 󵄩 𝑛 𝑛󵄩∞ . The theorem is then followed from (106). 󵄩 󵄩 = 󵄩( − )−1 ( − )𝛾󵄩 󵄩 I N𝑛 M M𝑛 󵄩∞ (102) 6. The Algebraic System 󵄩 −1󵄩 󵄩 󵄩 ≤ 󵄩( − ) 󵄩 󵄩( − )𝛾󵄩 󳨀→ 0, 󵄩 I N𝑛 󵄩∞󵄩 M M𝑛 󵄩∞ 𝑇 Let t be the 𝑛×1vector t := (𝑡1,𝑡2,...,𝑡𝑛) where 𝑇 denotes which with (100)and(101)implies(97). transposition. Then, for any function 𝛾(𝑡) defined on [0, 2𝜋], we define 𝛾(t) as the 𝑛×1vector obtained by componentwise 𝑆 𝑆 To calculate the function 𝑘 in (47)foragiven 𝑘−1,we evaluation of the function 𝛾(𝑡) at the points 𝑡𝑖, 𝑖=1,2,...,𝑛. replace the operator E in (47) by the approximate operator E𝑛 As in MATLAB, for any two vectors x and y, we define x⋅ ∗ y to obtain as the componentwise vector product of x and y.Ify𝑗 =0̸ ,for all 𝑗=1,2...,(𝑚+1)𝑛, we define x⋅/y as the componentwise 𝑅(𝑆𝑘−1 (⋅)) 𝑆 (𝑠) −𝑠= (𝑠) , vector division of x by y. For simplicity, we denote x⋅ ∗ y by 𝑘,𝑛 E𝑛 ln 𝜌 (⋅) (103) xy and x⋅/y by x/y. Let x𝑘−1 =𝑆𝑘−1(t)−t (given) and x𝑘 =𝑆𝑘,𝑛(t)−t where 𝑆𝑘,𝑛 is an approximation to 𝑆𝑘.Substituting𝑠=𝑡𝑖 and (unknown). Then system (105)canberewrittenas 𝑖=1,2,...,𝑛,in(103)weobtain 𝑅( + ) 𝑅(𝑆 (⋅)) x𝑘−1 t 𝑆 (𝑡 )−𝑡 = 𝑘−1 (𝑡 ), 𝑖=1,2...,𝑛. (𝐼−𝐵) x𝑘 =−𝐶ln , 𝑘=1,2,..., (111) 𝑘,𝑛 𝑖 𝑖 E𝑛 ln 𝜌 (⋅) 𝑖 (104) 𝜌 (t) 𝐼 𝑛×𝑛 𝐵 Equation (104)canberewrittenas where is the identity matrix, is the discretized matrix of the operator N, and 𝐶 is the discretized matrix of 𝑅(𝑆 (⋅)) the operator M [1]. Linear system (111) is uniquely solvable ( − )[𝑆 (𝑡 )−𝑡]=− 𝑘−1 (𝑡 ), I N𝑛 𝑘,𝑛 𝑖 𝑖 M𝑛 ln 𝜌 (⋅) 𝑖 [4, 10, 11]. (105) We start the iteration in (47)with𝑆𝑘(𝑡) = 𝑡 and 𝑖=1,2,...,𝑚, iterate until ‖𝑆𝑘 −𝑆𝑘−1‖∞ < tol where tol is a given tolerance; that is, we start the iteration in (111)with which represents an 𝑛×𝑛linear system for the unknown x0 = 0 and iterate until ‖x𝑘 − x𝑘−1‖∞ < tol. Each 𝑆𝑘,𝑛(𝑡1), 𝑆𝑘,𝑛(𝑡2),...,𝑆𝑘,𝑛(𝑡𝑛).Byobtaining𝑆𝑘,𝑛(𝑡𝑖) for 𝑖= iteration in (111)requiressolvingalinearsystemforx𝑘 1,2...,𝑛,thefunction𝑆𝑘,𝑛(𝑠) canbecalculatedfor𝑠∈ given x𝑘−1.Linearsystem(111)issolvedin𝑂(𝑛 ln 𝑛) oper- [0, 2𝜋] by the Nystrom¨ interpolating formula. In the fol- ationsbythefastmethodpresentedin[11, 12]whichis lowing lemma, we prove the uniform convergence of the based on a combination of the MATLAB function gmres approximate solution 𝑆𝑘,𝑛 ofdiscretizedequation(103)tothe and the MATLAB function zfmm2dpart in the MATLAB solution 𝑆𝑘 of (46). toolbox FMMLIB2D [13]. In the numerical results below, for 8 Abstract and Applied Analysis

Figure 1: The conformal mappings from 𝐺1 onto Ω1, Ω2,andΩ3.

Figure 2: The conformal mappings from 𝐺2 onto Ω1, Ω2,andΩ3. Abstract and Applied Analysis 9

Figure 3: The conformal mappings from 𝐺3 onto Ω1, Ω2,andΩ3.

function zfmm2dpart,weassumethatiprec =4which where the function 𝜌(𝑡) is given by −12 means that the tolerance of the FMM is 0.5 × 10 .Forthe function gmres, we choose the parameters restart =10, 1 =10−12 =10 Γ :𝜌(𝑡) =1+ 34𝑡, gmrestol ,andmaxit ,whichmeansthatthe 1 4cos GMRES method is restarted every 10 inner iterations, the −12 3 tolerance of the GMRES method is 10 ,andthemaximum (113) Γ2:𝜌(𝑡) =1+ cos 4𝑡, number of outer iterations of GMRES method is 10.See 4 [11, 12] for more details. cos 𝑡 2 sin 𝑡 2 Γ3:𝜌(𝑡) =𝑒 cos 2𝑡 + 𝑒 sin 2𝑡. By obtaining x𝑘,weobtainthevalues𝑆𝑘,𝑛(𝑡𝑖) for 𝑖= 1,2,...,𝑛.Then,thefunction𝑆𝑘,𝑛(𝑠) − 𝑠 can be calculated 𝑠∈[0,2𝜋] for by the Nystrom¨ interpolating formula. The boundaries 𝐿1, 𝐿2,and𝐿3 of the regions Ω1, Ω2,andΩ3 The convergence of 𝑆𝑘,𝑛(𝑠) to 𝑆(𝑠) follows from Theorem 19. are parameterized by Then the values of the mapping function Ψ can be computed from (2). The interior values of the mapping func- 𝜂 (𝑡) =𝑅(𝑡) 𝑒i𝑡,0≤𝑡≤2𝜋, tion can be computed by the Cauchy integral formula (114) which can be computed using the fast method presented in [12]. where the function 𝑅(𝑡) is given by

7. Numerical Examples 𝐿1:𝑅(𝑡) =1, 𝛼 In this section, we will compute the conformal mapping from 𝐿2:𝑅(𝑡) = , 𝐺 𝐺 𝐺 2 2 three simply connected regions 1, 2,and 3 onto three √1−(1−𝛼 ) cos 𝑡 simply connected regions Ω1, Ω2,andΩ3.TheboundariesΓ1, (115) Γ2,andΓ3 of the regions 𝐺1, 𝐺2,and𝐺3 are parameterized by 𝛼 = 0.6180339630899485, 1 𝑡 𝐿 :𝑅(𝑡) =1+ 8𝑡. 𝜂 (𝑡) =𝜌(𝑡) 𝑒i ,0≤𝑡≤2𝜋, (112) 3 10 cos 10 Abstract and Applied Analysis

100 100

10−3 10−3 ∞ ∞ ‖ ‖ k−1 10−6 k−1 10−6 −x −x k k ‖x ‖x

10−9 10−9

10−12 10−12 0 30 60 90 120 0 30 60 90 120 k k

Ω1 Ω3 Ω1 Ω3 Ω2 Ω2

(a) (b) 100

10−3 ∞ ‖

k−1 10−6 −x k ‖x

10−9

10−12 0 30 60 90 120 k

Ω1 Ω3 Ω2

(c)

Figure 4: The error norm ‖x𝑘 − x𝑘−1‖∞ versus the iteration number 𝑘 for (a) the conformal mapping from 𝐺1 onto Ω1, Ω2,andΩ3,(b)the conformal mapping from 𝐺2 onto Ω1, Ω2,andΩ3, and (c) the conformal mapping from 𝐺3 onto Ω1, Ω2,andΩ3.

The curves 𝐿1, 𝐿2,and𝐿3 satisfy the 𝜀-condition where versus the iteration number 𝑘 in (111)isshowninFigure 4. 󵄩 󵄩 It is clear from Figure 4 that the number of iterations in (111) 󵄩𝑅󸀠 (𝑡)󵄩 󵄩 󵄩 depends only on the boundary 𝐿 of the image region. More 𝐿1:𝜀=󵄩 󵄩 =0, 󵄩 𝑅 (𝑡) 󵄩∞ precisely, it depends on 𝜀.Theiterationsin(111)convergeonly 󵄩 󵄩 if 𝜀<1.Forsmall𝜀, a few number of iterations are required 󵄩𝑅󸀠 (𝑡)󵄩 󵄩 󵄩 for convergence. For values of 𝜀 close to 1,alargenumberof 𝐿2:𝜀=󵄩 󵄩 = 0.5 < 1, (116) 󵄩 𝑅 (𝑡) 󵄩∞ iterations are required for convergence. 󵄩 󵄩 󵄩𝑅󸀠 (𝑡)󵄩 󵄩 󵄩 𝐿3:𝜀=󵄩 󵄩 = 0.80403 < 1. 󵄩 𝑅 (𝑡) 󵄩∞ Conflict of Interests −12 The numerical results obtained with 𝑛 = 4096 and tol =10 The authors declare that there is no conflict of interests are shown in Figures 1, 2,and3. The error norm ‖x𝑘 − x𝑘−1‖∞ regarding the publication of this paper. Abstract and Applied Analysis 11

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Research Article Implicit Approximation Scheme for the Solution of 𝐾-Positive Definite Operator Equation

Naseer Shahzad,1 Arif Rafiq,2 and Habtu Zegeye3

1 Department of Mathematics, King Abdulaziz University, P.O. Box 80203, Jeddah 21859, Saudi Arabia 2 Department of Mathematics, Lahore Leads University, Lahore 54810, Pakistan 3 Departement of Mathematics, University of Botswana, Private Bag Box 00704, Gaborone, Botswana

Correspondence should be addressed to Naseer Shahzad; [email protected]

Received 5 December 2013; Accepted 9 February 2014; Published 23 March 2014

Academic Editor: Ljubomir B. Ciri´ c´

Copyright © 2014 Naseer Shahzad et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We construct an implicit sequence suitable for the approximation of solutions of K-positive definite operator equations in real Banach spaces. Furthermore, implicit error estimate is obtained and the convergence is shown to be faster in comparsion to the explicit error estimate obtained by Osilike and Udomene (2001).

1. Introduction Let 𝐸1 be a dense subspace of a Banach space 𝐸.An operator 𝑇 with domain 𝐷(𝑇)1 ⊇𝐸 is called continuously 𝐸1- Let 𝐸 be a real Banach space and let 𝐽 denote the normalized 𝑇 𝑅(𝑇) 𝑇 𝐸 𝐸∗ invertible if the range of , ,with in considered as an duality mapping from 𝐸 to 2 defined by operator restricted to 𝐸1, is dense in 𝐸 and 𝑇 has a bounded inverse on 𝑅(𝑇). ∗ ∗ ∗ 2 󵄩 ∗󵄩 𝐽 (𝑥) ={𝑓 ∈𝐸 :⟨𝑥,𝑓 ⟩=‖𝑥‖ , 󵄩𝑓 󵄩 = ‖𝑥‖}, (1) Let 𝐸 be a Banach space and let 𝐴 be a linear unbounded operator defined on a dense domain, 𝐷(𝐴),in𝐸.Anoperator ∗ where 𝐸 denotes the dual space of 𝐸 and ⟨⋅, ⋅⟩ denotes the 𝐴 will be called 𝐾 positive definite𝐾 ( pd) [1] if there exist a ∗ generalized duality pairing. It is well known that if 𝐸 is continuously 𝐷(𝐴)-invertible closed linear operator 𝐾 with strictly convex, then 𝐽 issinglevalued.Wewilldenotethe 𝐷(𝐴) ⊂ 𝐷(𝐾) and a constant 𝑐>0such that 𝑗(𝐾𝑥) ∈ ,𝐽(𝐾𝑥) 𝑗 single-valued duality mapping by . 2 Let 𝐸 be a Banach space. The modulus of smoothness of 𝐸 ⟨𝐴𝑥, 𝑗 (𝐾𝑥)⟩≥𝑐‖𝐾𝑥‖ ,∀𝑥∈𝐷(𝐴) . (4) is the function. 𝑐 ∈ (0, 1) 𝜌 : [0, ∞) → [0, ∞) Without loss of generality, we assume that . 𝐸 defined by In [1], Chidume and Aneke established the extension of 𝐾 1 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 pd operators of Martynjuk [2]andPetryshyn[3, 4]from 𝜌 (𝑡) = { (󵄩𝑥+𝑦󵄩 + 󵄩𝑥−𝑦󵄩)−1:‖𝑥‖ ≤1,󵄩𝑦󵄩 ≤𝑡}. 𝐸 sup 2 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 Hilbert spaces to arbitrary real Banach spaces. They proved (2) the following result. 𝐸 The Banach space 𝐸 is called uniformly smooth if Theorem 1. Let bearealseparableBanachspacewitha strictly convex dual 𝐸 and let 𝐴 be a 𝐾pd operator with 𝐷(𝐴) = 𝐷(𝐾) 𝜌𝐸 (𝑡) .Suppose lim =0. (3) 𝑡→0 𝑡 ⟨𝐴𝑥, 𝑗 (𝐾𝑦)⟩ = ⟨𝐾𝑥, 𝑗 (𝐴𝑦)⟩( ,∀𝑥,𝑦∈𝐷 𝐴) . (5) 𝐸 ABanachspace is said to be strictly convex if for two Then, there exists a constant 𝛼>0such that for all 𝑥∈𝐷(𝐴) elements 𝑥, 𝑦 ∈𝐸 which are linearly independent we have that ‖𝑥 + 𝑦‖ < ‖𝑥‖. +‖𝑦‖ ‖𝐴𝑥‖ ≤𝛼‖𝐾𝑥‖ . (6) 2 Abstract and Applied Analysis

Furthermore, the operator 𝐴 is closed, 𝑅(𝐴) =𝐸,andthe 2. Main Results equation 𝐴𝑥=𝑓has a unique solution for any given 𝑓∈𝐸. We need the following results. As the special case of Theorem 1 in which 𝐸=𝐿𝑝 (𝑙𝑝) 𝐸∗ spaces, 2≤𝑝<∞, Chidume and Aneke [1]introduced Lemma 3 (see [10]). If is uniformly convex then there exists 𝑏:[0,∞)→[0,∞) an iteration process which converges strongly to the unique a continuous nondecreasing function 𝑏(0) = 0 𝑏(𝛿𝑡) ≤ 𝛿𝑏(𝑡) 𝛿≥1 solution of the equation 𝐴𝑥=𝑓,where𝐴 and 𝐾 are such that , for all and commuting. Recently, Chidume and Osilike [5] extended the 󵄩 󵄩2 󵄩 󵄩 󵄩 󵄩 󵄩𝑥+𝑦󵄩 ≤ ‖𝑥‖2 +2⟨𝑦, 𝑗 (𝑥)⟩ + {‖𝑥‖ ,1} 󵄩𝑦󵄩 𝑏 (󵄩𝑦󵄩) , results of Chidume and Aneke [1] to the more general real 󵄩 󵄩 max 󵄩 󵄩 󵄩 󵄩 separable 𝑞-uniformly smooth Banach spaces, 1<𝑞<∞, (R) 𝐴 𝐾 by removing the commutativity assumption on and . 𝑥, 𝑦 ∈𝐸 Later on, Chuanzhi [6] proved convergence theorems for the for all . iterative approximation of the solution of the 𝐾pd operator Lemma 4 (see [22]). If there exists a positive integer 𝑁 such equation 𝐴𝑥 =𝑓 in more general separable uniformly that for all 𝑛≥𝑁, 𝑛∈N (the set of all positive integers), smooth Banach spaces. In [7], Osilike and Udomene proved the following result. 𝜌𝑛+1 ≤(1−𝜃𝑛)𝜌𝑛 +𝑏𝑛, (10)

Theorem 2. Let 𝐸 be a real separable Banach space with a then strictly convex dual and let 𝐴:𝐷(𝐴)⊆𝐸be →𝐸 a 𝜌 =0, 𝐾pd operator with 𝐷(𝐴) = 𝐷(𝐾).Suppose⟨𝐴𝑥, 𝑗(𝐾𝑦)⟩ = 𝑛→∞lim 𝑛 (11) 2 ⟨𝐾𝑥, 𝑗(𝐴𝑦)⟩ for all 𝑥, 𝑦 ∈ .Chooseany𝐷(𝐴) 𝜖1 ∈(0,𝑐/(1 + ∞ 2 where 𝜃𝑛 ∈ [0, 1), ∑𝑛=1 𝜃𝑛 =∞and 𝑏𝑛 = 𝑜(𝜃𝑛). 𝛼(1 − 𝑐) +𝛼 )) and define 𝑇𝜖 : 𝐷(𝐴) ⊆ 𝐸 →𝐸 by −1 −1 𝐾 𝐷(𝐴) 𝑇𝜖𝑥=𝑥+𝜖𝐾 𝑓−𝜖𝐾 𝐴𝑥. (7) Remark 5 (see [6]). Since is continuously invertible, there exists a constant 𝛽>0such that Then the Picard iteration scheme generated from an arbitrary ‖𝐾𝑥‖ ≥𝛽‖𝑥‖ ,∀𝑥∈𝐷(𝐾) =𝐷(𝐴) . 𝑥0 ∈ 𝐷(𝐴) by (12) 𝑛 𝑥𝑛+1 =𝑇𝜖𝑥𝑛 =𝑇𝜖 𝑥0 (8) In the continuation 𝑐 ∈ (0, 1), 𝛼 and 𝛽 are the constants appearing in (4), (6), and (12), respectively. Furthermore, 𝜖> converges strongly to the solution of the equation 𝐴𝑥=𝑓. 0 ∗ is defined by Moreover, if 𝑥 denotes the solution of the equation 𝐴𝑥 =𝑓, 𝑐−𝜂 then 𝜖= ,𝜂∈(0, 𝑐) . 𝛼(1−𝜂) (13) 󵄩 ∗󵄩 𝑛 −1 󵄩 ∗󵄩 󵄩𝑥𝑛+1 −𝑥 󵄩 ≤ (1−𝑐𝜖(1−𝑐)) 𝛽 󵄩𝐾𝑥0 −𝐾𝑥󵄩 . (9) With these notations, we now prove our main results. The most general iterative formula for approximating solutions of nonlinear equation and fixed point of nonlinear Theorem 6. Let 𝐸 be a real separable Banach space with a mapping is the Mann iterative method [8]whichproducesa 𝐴:𝐷(𝐴)⊆𝐸 →𝐸 {𝑥 } 𝑥 =𝛼𝑥 +(1− strictly convex dual and let be a sequence 𝑛 via the recursive approach 𝑛+1 𝑛 𝑛 𝐾pd operator with 𝐷(𝐴) = 𝐷(𝐾).Suppose⟨𝐴𝑥, 𝑗(𝐾𝑦)⟩ = 𝛼 )𝑇𝑥 𝑇:𝐶=𝐷(𝑇)→ 𝐶 ∗ 𝑛 𝑛, for nonlinear mapping ,where ⟨𝐾𝑥, 𝑗(𝐴𝑦)⟩ for all 𝑥, 𝑦 ∈ 𝐷(𝐴).Let𝑥 denote a solution of the initial guess 𝑥0 ∈𝐶is chosen arbitrarily. For convergence the equation 𝐴𝑥=𝑓. For arbitrary 𝑥0 ∈𝐸, define the sequence results of this scheme and related iterative schemes, see, for ∞ {𝑥𝑛}𝑛=0 in 𝐸 by example, [9–15]. −1 −1 In [16], Xu and Ori introduced the implicit iteration 𝑥𝑛 =𝑥𝑛−1 +𝜖𝐾 𝑓−𝜖𝐾 𝐴𝑥𝑛,𝑛≥0. (14) process {𝑥𝑛}, which is the modification of Mann, generated 𝑥 ∈𝐶 𝑥 =𝛼𝑥 +(1−𝛼 )𝑇 𝑥 𝑇 ,𝑖=1,2,...,𝑁 ∞ ∗ by 0 , 𝑛 𝑛 𝑛−1 𝑛 𝑛 𝑛,for 𝑖 , Then, {𝑥𝑛}𝑛=0 converges strongly to 𝑥 with nonexpansive mappings, and 𝑇𝑛 =𝑇𝑛 (mod𝑁) and {𝛼𝑛}⊂ 󵄩 ∗󵄩 𝑛 −1 󵄩 ∗󵄩 (0, 1). They proved the weak convergence of this process to 󵄩𝑥𝑛 −𝑥 󵄩 ≤𝜌𝛽 󵄩𝐾𝑥0 −𝐾𝑥󵄩 , (15) a common fixed point of the finite family of nonexpansive where 𝜌 = 1 − ((𝑐 − 𝜂)/(𝛼(1 − 𝜂) + 𝑐 −𝜂))𝜂∈(0,1).Thus,the mappings in Hilbert spaces. Since then fixed point problems 2 choice 𝜂=𝑐/2yields 𝜌=1−(𝑐/(4𝛼(1−𝑐/2)+2𝑐)).Moreover, and solving (or approximating) nonlinear equations based ∗ on implicit iterative processes have been considered by many 𝑥 is unique. authors (see, e.g., [17–21]). It is our purpose in this paper to introduce implicit Proof. The existence of the unique solution to the equation 𝐴𝑥=𝑓 scheme which converges strongly to the solution of the 𝐾pd comes from Theorem 1.From(4)wehave operator equation 𝐴𝑥=𝑓in a separable Banach space. Even ⟨𝐴𝑥−𝑐𝐾𝑥,𝑗(𝐾𝑥)⟩≥0, (16) though our scheme is implicit, the error estimate obtained indicates that the convergence of the implicit scheme is faster andfromLemma1.1ofKato[23], we obtain that in comparison to the explicit scheme obtained by Osilike and 󵄩 󵄩 Udomene [7]. ‖𝐾𝑥‖ ≤ 󵄩𝐾𝑥+𝛾(𝐴𝑥−𝑐𝐾𝑥)󵄩 , (17) Abstract and Applied Analysis 3

󵄩 ∗󵄩 2 󵄩 ∗󵄩 for all 𝑥∈𝐷(𝐴)and 𝛾>0.Now,from(14), linearity of 𝐾 and −𝜖(1−𝑐) 󵄩𝐾𝑥𝑛−1 −𝐾𝑥󵄩 −𝜖 (1−𝑐) 󵄩𝐴𝑥𝑛 −𝐴𝑥 󵄩 𝐴𝑥∗ =𝑓 the fact that we obtain that 󵄩 ∗󵄩 󵄩 ∗󵄩 ≥ (1+𝜖) 󵄩𝐾𝑥𝑛 −𝐾𝑥󵄩 −𝜖(1−𝑐) 󵄩𝐾𝑥𝑛−1 −𝐾𝑥󵄩 𝐾𝑥𝑛 =𝐾𝑥𝑛−1 +𝜖𝑓−𝜖𝐴𝑥𝑛 2 𝛼 󵄩 ∗󵄩 (18) −𝜖 (1−𝑐) 󵄩𝐾𝑥 −𝐾𝑥󵄩 , ∗ 1−𝛼𝜖󵄩 𝑛−1 󵄩 =𝐾𝑥𝑛−1 +𝜖𝐴𝑥 −𝜖𝐴𝑥𝑛, (24) which implies that which implies that 𝐾𝑥 =𝐾𝑥−𝜖𝐴𝑥∗ +𝜖𝐴𝑥 , 󵄩 ∗∗󵄩 𝑛−1 𝑛 𝑛 (19) 󵄩𝐾𝑥𝑛 −𝐾𝑥 󵄩 so that 1+𝜖(1−𝑐) +𝜖2 (1−𝑐)(𝛼/ (1−𝛼𝜖)) 󵄩 󵄩 ≤ 󵄩𝐾𝑥 −𝐾𝑥∗󵄩 ∗ ∗ ∗ 1+𝜖 󵄩 𝑛−1 󵄩 𝐾𝑥𝑛−1 −𝐾𝑥 =𝐾𝑥𝑛 −𝐾𝑥 −𝜖𝐴𝑥 +𝜖𝐴𝑥𝑛. (20) 󵄩 󵄩 =𝜌󵄩𝐾𝑥 −𝐾𝑥∗󵄩 , With the help of (14)andTheorem 1,wehavethefollowing 󵄩 𝑛−1 󵄩 estimate: (25) 󵄩 ∗󵄩 󵄩 ∗ 󵄩 󵄩 ∗ 󵄩 where 󵄩𝐴𝑥𝑛 −𝐴𝑥 󵄩 = 󵄩𝐴(𝑥𝑛 −𝑥 )󵄩 ≤𝛼󵄩𝐾(𝑥 𝑛 −𝑥 )󵄩 1+𝜖(1−𝑐) +𝜖2 (1−𝑐)(𝛼/ (1−𝛼𝜖)) 󵄩 ∗󵄩 𝜌= =𝛼󵄩𝐾𝑥𝑛 −𝐾𝑥󵄩 1+𝜖 (21) 󵄩 ∗ ∗ 󵄩 𝜖 𝛼 =𝛼󵄩𝐾𝑥𝑛−1 −𝐾𝑥 − 𝜖 (𝐴𝑥𝑛 −𝐴𝑥 )󵄩 =1− (𝑐 − 𝜖 (1−𝑐) ) 1+𝜖 1−𝛼𝜖 󵄩 ∗󵄩 󵄩 ∗󵄩 ≤𝛼󵄩𝐾𝑥𝑛−1 −𝐾𝑥󵄩 +𝛼𝜖󵄩𝐴𝑥𝑛 −𝐴𝑥 󵄩 , 𝜖 =1− 𝜂 (26) which gives 1+𝜖 𝑐−𝜂 󵄩 󵄩 𝛼 󵄩 󵄩 󵄩𝐴𝑥 −𝐴𝑥∗󵄩 ≤ 󵄩𝐾𝑥 −𝐾𝑥∗󵄩 . =1− 𝜂 󵄩 𝑛 󵄩 1−𝛼𝜖󵄩 𝑛−1 󵄩 (22) 𝛼(1−𝜂)+𝑐−𝜂 𝑐2 Furthermore, inequality (20)canberewrittenas =1− . ∗ 4𝛼 (1−𝑐/2) +2𝑐 𝐾𝑥𝑛−1 −𝐾𝑥 From (25)and(26), we have that ∗ = (1+𝜖) (𝐾𝑥𝑛 −𝐾𝑥) 󵄩 ∗󵄩 󵄩 ∗󵄩 𝑛 󵄩 ∗ 󵄩 󵄩𝐾𝑥𝑛 −𝐾𝑥󵄩 ≤𝜌󵄩𝐾𝑥𝑛−1 −𝐾𝑥󵄩 ≤⋅⋅⋅≤𝜌 󵄩𝐾(𝑥 0 −𝑥 )󵄩 . ∗ ∗ (27) + 𝜖 (𝐴𝑥𝑛 −𝐴𝑥 −𝑐(𝐾𝑥𝑛 −𝐾𝑥)) Hence by Remark 5,wegetthat −𝜖(1−𝑐) (𝐾𝑥 −𝐾𝑥∗) 𝑛 󵄩 ∗󵄩 󵄩𝑥𝑛 −𝑥 󵄩 ∗ = (1+𝜖) [𝐾𝑥𝑛 −𝐾𝑥 −1 󵄩 ∗󵄩 𝑛 −1 󵄩 ∗󵄩 ≤𝛽 󵄩𝐾𝑥𝑛 −𝐾𝑥󵄩 ≤⋅⋅⋅≤𝜌 𝛽 󵄩𝐾𝑥0 −𝐾𝑥󵄩 󳨀→ 0,

𝜖 ∗ ∗ (28) + (𝐴𝑥𝑛 −𝐴𝑥 −𝑐(𝐾𝑥𝑛 −𝐾𝑥))] ∗ 1+𝜖 as 𝑛→∞.Thus,𝑥𝑛 →𝑥 as 𝑛→∞. ∗ −𝜖(1−𝑐) (𝐾𝑥𝑛 −𝐾𝑥) In [6], Chuanzhi provided the following result.

∗ 𝐸 = (1+𝜖) [𝐾𝑥𝑛 −𝐾𝑥 Theorem 7. Let be a real uniformly smooth separable Banach space, and let 𝐴:𝐷(𝐴)⊆𝐸be →𝐸 a 𝐾pd operator 𝜖 with 𝐷(𝐴) = 𝐷(𝐾).Suppose⟨𝐴𝑥, 𝑗(𝐾𝑦)⟩ = ⟨𝐾𝑥, 𝑗(𝐴𝑦)⟩for + (𝐴𝑥 −𝐴𝑥∗ −𝑐(𝐾𝑥 −𝐾𝑥∗))] 𝑛 𝑛 all 𝑥, 𝑦 ∈ .𝐷(𝐴) For arbitrary 𝑓∈𝐸and 𝑥0 ∈ 𝐷(𝐴),define 1+𝜖 ∞ the sequence {𝑥𝑛}𝑛=0 by ∗ 2 ∗ −𝜖(1−𝑐) (𝐾𝑥𝑛−1 −𝐾𝑥)+𝜖 (1−𝑐) (𝐴𝑥𝑛 −𝐴𝑥 ). 𝑥𝑛+1 =𝑥𝑛 +𝑡𝑛𝛾𝑛, (23) 𝛾 =𝐾−1𝑓−𝐾−1𝐴𝑥 , In addition, from (17)and(22), we get that 𝑛 𝑛 󵄩 󵄩 1 󵄩𝐾𝑥 −𝐾𝑥∗󵄩 0≤𝑡 ≤ , 󵄩 𝑛−1 󵄩 𝑛 2𝑐 (29) 󵄩 󵄩 ∗ ≥ (1+𝜖) 󵄩𝐾𝑥 −𝐾𝑥 ∑ 𝑡 =0, 𝑡 =0, 󵄩 𝑛 𝑛 𝑛→∞lim 𝑛 󵄩 𝜖 ∗ ∗ 󵄩 2𝑐 + (𝐴𝑥 −𝐴𝑥 −𝑐(𝐾𝑥 −𝐾𝑥))󵄩 𝑏(𝛼𝑡 )≤ ,𝑛≥0, 1+𝜖 𝑛 𝑛 󵄩 𝑛 𝐵𝛼 4 Abstract and Applied Analysis

󵄩 󵄩2 󵄩 󵄩 󵄩 󵄩 where 𝑏(𝑡) is as in (R), 𝛼 is the constant appearing in inequality ≤(1−2𝑐𝑡𝑛) 󵄩𝐾𝛾𝑛−1󵄩 +2𝑡𝑛 󵄩𝐴𝛾𝑛−1 −𝐴𝛾𝑛󵄩 󵄩𝐾𝛾𝑛−1󵄩 𝑐 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 (6), is the constant appearing in inequality (4),and + {󵄩𝐾𝛾 󵄩 ,1}𝛼𝑡 󵄩𝐾𝛾 󵄩 𝑏(𝛼𝑡 󵄩𝐾𝛾 󵄩) 󵄩 󵄩 max 󵄩 𝑛−1󵄩 𝑛 󵄩 𝑛󵄩 𝑛 󵄩 𝑛󵄩 𝐵= {󵄩𝐾𝛾 󵄩 ,1}. max 󵄩 0󵄩 (30) 󵄩 󵄩2 ≤(1−2𝑐𝑡) 󵄩𝐾𝛾 󵄩 +2𝑀𝑡 𝜂 ∞ 𝑛 󵄩 𝑛−1󵄩 𝑛 𝑛 {𝑥𝑛} 𝐴𝑥 = Then, 𝑛=0 converges strongly to the unique solution of 2 2 𝑓. + max {𝑀,} 1 𝛼 𝑀 𝑡𝑛𝑏(𝑡𝑛), (39) However, its implicit version is as follows. where 𝐸 󵄩 󵄩 Theorem 8. Let be a real uniformly smooth separable 𝜂𝑛 = 󵄩𝐴𝛾𝑛−1 −𝐴𝛾𝑛󵄩 . (40) Banach space, and let 𝐴:𝐷(𝐴)⊆𝐸be →𝐸 a 𝐾pd operator By using (6)and(34)weobtainthat with 𝐷(𝐴) = 𝐷(𝐾).Suppose⟨𝐴𝑥, 𝑗(𝐾𝑦)⟩ = ⟨𝐾𝑥, 𝑗(𝐴𝑦)⟩for 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩𝐴𝛾 −𝐴𝛾󵄩 = 󵄩𝐴(𝛾 −𝛾)󵄩 ≤𝛼󵄩𝐾(𝛾 −𝛾)󵄩 all 𝑥, 𝑦 ∈ .𝐷(𝐴) For arbitrary 𝑓∈𝐸and 𝑥0 ∈ 𝐷(𝐴),define 󵄩 𝑛−1 𝑛󵄩 󵄩 𝑛−1 𝑛 󵄩 󵄩 𝑛−1 𝑛 󵄩 ∞ the sequence {𝑥𝑛}𝑛=0 by 󵄩 󵄩 =𝛼𝑡𝑛 󵄩𝐴𝛾𝑛󵄩 ≤𝑀𝛼𝑡𝑛 󳨀→ 0, as 𝑛󳨀→∞. 𝑥𝑛 =𝑥𝑛−1 +𝑡𝑛𝛾𝑛, (31) (41) −1 −1 Thus, 𝛾𝑛 =𝐾 𝑓−𝐾 𝐴𝑥𝑛, (32) 𝜂𝑛 󳨀→ 0 as 𝑛󳨀→∞. (42) ∑ 𝑡𝑛 =∞, lim 𝑡𝑛 =0, 𝑛≥0. 𝑛→∞ (33) Denote 󵄩 󵄩 𝜌 = 󵄩𝑥 −𝑝󵄩 , ∞ 𝑛 󵄩 𝑛 󵄩 Then, {𝑥𝑛}𝑛=0 converges strongly to the unique solution of 𝐴𝑥 = 𝑓. 𝜃𝑛 =2𝑐𝑡𝑛, (43) 𝜎 =2𝑀𝑡𝜂 + 𝑀, 1 𝛼2𝑀2𝑡 𝑏(𝑡 ). Proof. The existence of the unique solution to the equation 𝑛 𝑛 𝑛 max { } 𝑛 𝑛 𝐴𝑥=𝑓 comes from Theorem 1.Using(31)and(32)weobtain Condition (33) assures the existence of a rank 𝑛0 ∈ N such that 𝜃𝑛 =2𝑐𝑡𝑛 ≤1,forall𝑛≥𝑛0.Since𝑏(𝑡) is continuous, so 𝐾𝛾𝑛 =𝐾𝛾𝑛−1 −𝑡𝑛𝐴𝛾𝑛. (34) lim𝑛→∞𝑏(𝑡𝑛)=0(by condition (33)). Now with the help of Consider (33), (42), and Lemma 4,weobtainfrom(39)that 󵄩 󵄩 󵄩 󵄩2 󵄩𝐾𝛾 󵄩 =0. 󵄩𝐾𝛾𝑛󵄩 =⟨𝐾𝛾𝑛,𝑗(𝐾𝛾𝑛)⟩ = ⟨𝐾𝛾𝑛−1 −𝑡𝑛𝐴𝛾𝑛,𝑗(𝐾𝛾𝑛)⟩ 𝑛→∞lim 󵄩 𝑛󵄩 (44)

At last by Remark 5, 𝛾𝑛 →0as 𝑛→∞;thatis𝐴𝑥𝑛 →𝑓 =⟨𝐾𝛾𝑛−1,𝑗(𝐾𝛾𝑛)⟩ − 𝑡𝑛 ⟨𝐴𝛾𝑛,𝑗(𝐾𝛾𝑛)⟩ (35) as 𝑛→∞.Because𝐴 has bounded inverse, this implies that 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩2 𝑥 →𝐴−1𝑓 𝐴𝑥 =𝑓 ≤ 󵄩𝐾𝛾𝑛−1󵄩 󵄩𝐾𝛾𝑛󵄩 −𝑐𝑡𝑛󵄩𝐾𝛾𝑛󵄩 , 𝑛 , the unique solution of 𝑛 . This completes the proof. which implies that 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 (1) 󵄩𝐾𝛾 󵄩 ≤ 󵄩𝐾𝛾 󵄩 −𝑐𝑡 󵄩𝐾𝛾 󵄩 . Remark 9. According to the estimates (6–8) of Martynjuk 󵄩 𝑛󵄩 󵄩 𝑛−1󵄩 𝑛 󵄩 𝑛󵄩 (36) [2], we have {𝐾𝛾 }∞ 󵄩 ∗󵄩 Hence, 𝑛 𝑛=0 is bounded. Let 󵄩𝐾𝑥𝑛+1 −𝐾𝑥󵄩 󵄩 󵄩 𝑀 = 󵄩𝐾𝛾 󵄩 . 2 1 sup 󵄩 𝑛󵄩 (37) 1+𝜖1 (1−𝑐) +𝛼𝜖1 (1−𝑐+𝛼) 󵄩 ∗󵄩 𝑛≥0 ≤ 󵄩𝐾𝑥 −𝐾𝑥󵄩 (45) 1+𝜖 󵄩 𝑛 󵄩 ∞ 1 Also from (6)itcanbeeasilyseenthat{𝐴𝛾𝑛}𝑛=0 is also 󵄩 󵄩 =𝜃󵄩𝐾𝑥 −𝐾𝑥∗󵄩 , bounded. Let 󵄩 𝑛 󵄩 󵄩 󵄩 𝑀 = 󵄩𝐴𝛾 󵄩 . where 2 sup 󵄩 𝑛󵄩 (38) 𝑛≥0 1+𝜖 (1−𝑐) +𝛼𝜖2 (1−𝑐+𝛼) 𝜃= 1 1 𝑀=𝑀 +𝑀 𝑀<∞ Denote 1 2;then . 1+𝜖1 By using (34)andLemma 3,wehave 𝜖 =1− 1 (𝑐 − 𝛼 (1−𝑐+𝛼) 𝜖 ) (46) 󵄩 󵄩2 󵄩 󵄩2 1+𝜖 1 󵄩𝐾𝛾𝑛󵄩 = 󵄩𝐾𝛾𝑛−1 −𝑡𝑛𝐴𝛾𝑛󵄩 1 𝜖 󵄩 󵄩2 1 ≤ 󵄩𝐾𝛾 󵄩 −2𝑡 ⟨𝐴𝛾 ,𝑗(𝐾𝛾 )⟩ =1− 𝜂, 󵄩 𝑛−1󵄩 𝑛 𝑛 𝑛−1 1+𝜖1 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 𝜂 = 𝑐−𝛼(1−𝑐+𝛼)𝜖 𝜖 = (𝑐−𝜂)/𝛼(1−𝑐+𝛼) 𝜂 ∈ (0, 𝑐) + max {󵄩𝐾𝛾𝑛−1󵄩 ,1}󵄩𝑡𝑛𝐴𝛾𝑛󵄩 𝑏(󵄩𝑡𝑛𝐴𝛾𝑛󵄩) for 1 or 1 , . Thus, 󵄩 󵄩2 𝑐−𝜂 = 󵄩𝐾𝛾𝑛−1󵄩 −2𝑡𝑛 ⟨𝐴𝛾𝑛−1,𝑗(𝐾𝛾𝑛−1)⟩ 𝜃=1− 𝜂 𝛼 (1−𝑐+𝛼) +𝑐−𝜂 +2𝑡𝑛 ⟨𝐴𝛾𝑛−1 −𝐴𝛾𝑛,𝑗(𝐾𝛾𝑛−1)⟩ (47) 𝑐2 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 =1− . + max {󵄩𝐾𝛾𝑛−1󵄩 ,1}𝑡𝑛 󵄩𝐴𝛾𝑛󵄩 𝑏(𝑡𝑛 󵄩𝐴𝛾𝑛󵄩) 4𝛼 (1−𝑐+𝛼) +2𝑐 Abstract and Applied Analysis 5

Table 1 𝑛 12 3 4 5 6 7 8

𝑥𝑛 0.0922 0.0851 0.07859 0.07528 0.06947 0.06411 0.05916 0.05459

Table 2 𝑛 123 4 5 6 7 8

𝑥𝑛 0.0893 0.0798 0.07136 0.06376 0.05698 0.05092 0.04550 0.04066

Table 3 𝑛 12 3 4 5 6 7 8

𝑥𝑛 0.0098 0.0096 0.00949 0.00933 0.00917 0.00901 0.00885 0.00870

Table 4 𝑛 123 4 5 6 7 8

𝑥𝑛 0.0098 0.0096 0.00941 0.00923 0.00905 0.00887 0.00869 0.00852

(2) For 𝛼>𝑐/2,weobservethat It can be easily seen that for 𝑐≤1/2and 𝛼≥1/2,(4)and(6) are satisfied. Suppose 𝑐=1/4and 𝛼=3/5;then𝜂 = 0.125, 𝑐2 𝜌=1− 𝜖 = (𝑐 − 𝜂)/𝛼(1 − 𝜂), =0.23810 𝜖1 =(𝑐−𝜂)/𝛼(1−𝑐+𝛼)= 4𝛼 (1−𝑐/2) +2𝑐 0.15432, 𝜌 = 0.97596,and𝜃 = 0.983288 and so 𝜌<𝜃.Take 𝑥0 =0.1;thenfrom(52)wehaveTable 1 and for (54)weget 4𝛼𝑐2 𝑐 =𝜃− (𝛼 − ). Table 2. (4𝛼 (1−𝑐/2) +2𝑐)(4𝛼 (1−𝑐+𝛼) +2𝑐) 2 (48) Example 11. Let us take 𝐸=R, 𝐷(𝐴) = R+, 𝐴𝑥 = (1/4)𝑥, ∗ 𝐾𝑥=2𝑥(𝑥 =0is the solution of 𝐴𝑥=𝑓); then for the Thus, the relation between Martynjuk2 [ ] and our parameter explicit iterative scheme due to Osilike and Udomene [7]we of convergence, that is, between 𝜃 and 𝜌,respectively,isthe have following: 𝐾𝑥𝑛+1 =𝐾𝑥𝑛 −𝜖1𝐴𝑥𝑛, (55) 𝜌<𝜃. (49) which implies that 𝜖 Despite the fact that our scheme is implicit, inequality 2𝑥 =2𝑥 − 1 𝑥 , (49) shows that the results of Osilike and Udomene [7]are 𝑛+1 𝑛 4 𝑛 (56) improved in the sense that our scheme converges faster. and hence 𝜖 Example 10. Suppose 𝐸=R, 𝐷(𝐴) = R+, 𝐴𝑥 =𝑥, 𝐾𝑥= 1 ∗ 𝑥𝑛+1 =(1− )𝑥𝑛. (57) 2𝐼 (𝑥 =0is the solution of 𝐴𝑥 =𝑓); then for the explicit 8 iterative scheme due to Osilike and Udomene [7]wehave Alsofortheimplicititerativeschemewehavethat 𝐾𝑥 =𝐾𝑥−𝜖 𝐴𝑥 , 𝑛+1 𝑛 1 𝑛 (50) 𝐾𝑥𝑛 =𝐾𝑥𝑛−1 −𝜖𝐴𝑥𝑛, (58) which implies that which implies that 1 2𝑥𝑛+1 =2𝑥𝑛 −𝜖1𝑥𝑛, (51) 𝑥 = 𝑥 . 𝑛 1+𝜖/8 𝑛−1 (59) and hence Itcanbeeasilyseenthatfor𝑐≤1/8and 𝛼≥1/8,(4)and 𝜖1 (6) are satisfied. Suppose 𝑐 = 0.0625 and 𝛼 = 0.2;then𝜂= 𝑥𝑛+1 =(1− )𝑥𝑛. (52) 2 0.03125, 𝜖 = (𝑐 − 𝜂)/𝛼(1 − 𝜂), =0.16129 𝜖1 = (𝑐 − 𝜂)/𝛼(1 − 𝑐 + 𝛼) = 0.13736, 𝜌 = 0.99566,and𝜃 = 0.99623 and so 𝜌<𝜃. Also for the implicit iterative scheme we have that Take 𝑥0 =0.01;thenfrom(57)wehaveTable 3 and for (59) 𝐾𝑥 =𝐾𝑥 −𝜖𝐴𝑥 , we get Table 4. 𝑛 𝑛−1 𝑛 (53) Even though our scheme is implicit we observe that 𝐾 which implies that it converges strongly to the solution of the pd operator equation 𝐴𝑥 =𝑓 with the error estimate which is faster in 1 𝑥 = 𝑥 . comparison to the explicit error estimate obtained by Osilike 𝑛 1+𝜖/2 𝑛−1 (54) and Udomene [7]. 6 Abstract and Applied Analysis

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Research Article Caristi Fixed Point Theorem in Metric Spaces with a Graph

M. R. Alfuraidan1 and M. A. Khamsi2 1 Department of Mathematics and Statistics, King Fahd University of Petroleum and Minerals, Dhahran 31261, Saudi Arabia 2 Department of Mathematical Science, The University of Texas at El Paso, El Paso, TX 79968, USA

Correspondence should be addressed to M. R. Alfuraidan; [email protected]

Received 27 January 2014; Accepted 11 February 2014; Published 13 March 2014

Academic Editor: Qamrul Hasan Ansari

Copyright © 2014 M. R. Alfuraidan and M. A. Khamsi. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We discuss Caristi’s fixed point theorem for mappings defined on a metric space endowed with a graph. This work should be seenas a generalization of the classical Caristi’s fixed point theorem. It extends some recent works on the extension of Banach contraction principle to metric spaces with graph.

Dedicated to Rashed Saleh Alfuraidan and Prof. Miodrag Mateljevi’c for his 65th birthday

1. Introduction The proofs given to Caristi’s result vary and use different techniques (see [3, 6–8]). It is worth to mention that because This work was motivated by some recent works on the of Caristi’s result of close connection to the Ekeland’s [9] extension of Banach contraction principle to metric spaces variational principle, many authors refer to it as Caristi- with a partial order [1]oragraph[2]. Caristi’s fixed point Ekeland fixed point result. For more on Ekeland’s variational theorem is maybe one of the most beautiful extensions of principle and the equivalence between Caristi-Ekeland fixed Banach contraction principle [3, 4]. Recall that this theorem point result and the completeness of metric spaces, the reader 𝑇:𝑀 →𝑀 states the fact that any map has a fixed point isadvisedtoread[10]. provided that 𝑀 is a complete metric space and there exists a lower semicontinuous map 𝜙:𝑀 → [0,+∞)such that 2. Main Results 𝑑 (𝑥, 𝑇𝑥) ≤𝜙(𝑥) −𝜙(𝑇𝑥) , (1) Maybe one of the most interesting examples of the use of metric fixed point theorems is the proof of the existence of for every 𝑥∈𝑀.Recallthat𝑥∈𝑀is called a fixed point of 𝑇 solutions to differential equations. The general approach is to if 𝑇(𝑥) =𝑥.Thisgeneralfixedpointtheoremhasfoundmany convert such equations to integral equations which describes applications in nonlinear analysis. It is shown, for example, exactly a fixed point of a mapping. The metric spaces in which that this theorem yields essentially all the known inwardness such mapping acts are usually a function space. Putting a results [5] of geometric fixed point theory in Banach spaces. norm(inthecaseofavectorspace)oradistancegivesusa Recall that inwardness conditions are the ones which assert metric structure rich enough to use the Banach contraction that, in some sense, points from the domain are mapped principle or other known fixed point theorems. But one toward the domain. Possibly, the weakest of the inwardness structure naturally enjoyed by such function spaces is rarely conditions, the Leray-Schauder boundary condition, is the used. Indeed we have an order on the functions inherited assumption that a map points 𝑥 of 𝜕𝑀 anywhere except to from the order of R. In the classical use of Banach contraction the outward part of the ray originating at some interior point principle, the focus is on the metric behavior of the mapping. of 𝑀 and passing through 𝑥. The connection with the natural order is usually ignored. 2 Abstract and Applied Analysis

In [1, 11], the authors gave interesting examples where the Clearly, from this theorem, one may see that the contrac- order is used combined with the metric conditions. tive nature of the mapping 𝑇 is restricted to the comparable elements of (𝑋, ⪯) not to the entire set 𝑋. The detailed Example 1 (see [1]). Consider the periodic boundary value investigation of the example above shows that such mappings problem may exist which are not contractive on the entire set 𝑋. 𝑢󸀠 (𝑡) =𝑓(𝑡, 𝑢 (𝑡)) ,𝑡∈[0, 𝑇] , Therefore the classical Banach contraction principle will not (2) work in this situation. The analogue to Caristi’s fixed theorem 𝑢 (0) =𝑢(𝑇) , in this setting is the following result. where 𝑇>0and 𝑓:[0,𝑇]×R → R is a continuous function. Theorem 3. Let (𝑋, ⪯) be a partially ordered set and suppose Clearly any solution to this problem must be continuously that there exists a distance 𝑑 in 𝑋 such that (𝑋, 𝑑) is a differentiable on [0, 𝑇].Sothespacetobeconsideredforthis 𝑇:𝑋 →𝑋 1 complete metric space. Let be a continuous and problem is 𝐶 ([0, 𝑇], R).Theaboveproblemisequivalentto monotone increasing mapping. Assume that there exists a lower the integral problem semicontinuous function 𝜙:𝑋 → [0,+∞)such that 𝑇 𝑢 (𝑡) = ∫ 𝐺 (𝑡, 𝑠) [𝑓 (𝑠, 𝑢 (𝑠)) +𝜆𝑢(𝑠)]𝑑𝑠, (3) 𝑑 (𝑥, 𝑇 (𝑥)) ≤𝜙(𝑥) −𝜙(𝑇 (𝑥)) ,𝑤ℎ𝑒𝑛𝑒V𝑒𝑟 𝑇 (𝑥) ⪯𝑥. 0 (9) where 𝜆>0and the Green function is given by Then 𝑇 has a fixed point if and only if there exists 𝑥0 ∈𝑋,with 𝜆(𝑇+𝑠−𝑡) {𝑒 𝑇(𝑥0)⪯𝑥0. { 0≤𝑠<𝑡≤𝑇, { 𝑒𝜆𝑇 −1 𝐺 (𝑡, 𝑠) = 𝑥 𝑇 𝑇(𝑥 )=𝑥 { 𝜆(𝑠−𝑡) (4) Proof. Clearly, if 0 is a fixed point of ,thatis, 0 0, { 𝑒 0≤𝑡<𝑠≤𝑇. then we have 𝑇(𝑥0)⪯𝑥0. Assume that there exists 𝑥0 ∈𝑋 { 𝜆𝑇 𝑒 −1 such that 𝑇(𝑥0)⪯𝑥0.Since𝑇 is monotone increasing, we 𝑛+1 𝑛 Define the mapping A : 𝐶([0, 𝑇], R) → 𝐶([0, 𝑇], R) by have 𝑇 (𝑥0)⪯𝑇(𝑥0),forany𝑛≥1.Hence 𝑇 𝑑(𝑇𝑛 (𝑥 ),𝑇𝑛+1 (𝑥 )) A (𝑢)(𝑡) = ∫ 𝐺 (𝑡, 𝑠) [𝑓 (𝑠, 𝑢 (𝑠)) +𝜆𝑢(𝑠)]𝑑𝑠. (5) 0 0 0 (10) ≤𝜙(𝑇𝑛 (𝑥 )) − 𝜙 (𝑇𝑛+1 (𝑥 )), 𝑛=1,2,.... Note that if 𝑢(𝑡) ∈ 𝐶([0, 𝑇], R) is a fixed point of A,then 0 0 1 𝑢(𝑡) ∈𝐶 ([0, 𝑇], R) is a solution to the original boundary {𝜙(𝑇𝑛(𝑥 ))} value problem. Under suitable assumptions, the mapping A Hence 0 is a decreasing sequence of positive 𝜙 = 𝜙(𝑇𝑛(𝑥 )) 𝑛, ℎ ≥ 1 satisfies the following property: numbers. Let 0 lim𝑛→∞ 0 .Forany ,we have (i) if 𝑢(𝑡) ≤ V(𝑡),thenwehaveA(𝑢) ≤ A(V); ℎ−1 (ii) if 𝑢(𝑡) ≤ V(𝑡),then 𝑛 𝑛+ℎ 𝑛+𝑘 𝑛+𝑘+1 𝑑(𝑇 (𝑥0),𝑇 (𝑥0)) ≤ ∑𝑑(𝑇 (𝑥0),𝑇 (𝑥0)) ‖A (𝑢) − A (V)‖ ≤𝑘‖𝑢−V‖ , (6) 𝑘=0 𝑛 𝑛+ℎ for a constant 𝑘<1independent of 𝑢 and V. ≤𝜙(𝑇 (𝑥0)) − 𝜙 (𝑇 (𝑥0)) . The contractive condition is only valid for comparable func- (11) tions. It does not hold on the entire space 𝐶([0, 𝑇], R).This 𝑛 Therefore {𝑇 (𝑥0)} is a Cauchy sequence in 𝑋.Since𝑋 is condition led the authors in [1] to use a weaker version of 𝑛 complete, there exists 𝑥∈𝑋such that lim𝑛→∞𝑇 (𝑥0)=𝑥. the Banach contraction principle to prove the existence of Since 𝑇 is continuous, we conclude that 𝑇(𝑥) = 𝑥;thatis,𝑥 is the solution to the original boundary value problem, a result a fixed point of 𝑇. which was already known [12] using different techniques. 𝑇 (𝑋, ⪯) 𝑇:𝑋 →𝑋 The continuity assumption of may be relaxed if we Let be a partially ordered set and .We assume that 𝑋 satisfies the property (OSC). will say that 𝑇 is monotone increasing if (𝑋, ⪯) 𝑑 𝑥⪯𝑦󳨐⇒𝑇(𝑥) ⪯𝑇(𝑦), for any 𝑥, 𝑦 ∈ 𝑋. (7) Definition 4. Let be a partially ordered set. Let be a distance defined on 𝑋.Onesaysthat𝑋 satisfies the property The main result of [1] is the following theorem. (OSC) if and only if for any convergent decreasing sequence {𝑥𝑛} in 𝑋,thatis,𝑥𝑛+1 ⪯𝑥𝑛,forany𝑛≥1,onehas (𝑋, ⪯) Theorem 2 (see [1]). Let be a partially ordered set and lim𝑚→∞𝑥𝑚 = inf{𝑥𝑛,𝑛≥1}. suppose that there exists a distance 𝑑 in 𝑋 such that (𝑋, 𝑑) is a complete metric space. Let 𝑇:𝑋be →𝑋 a continuous and One has the following improvement to Theorem 3. monotone increasing mapping such that there exists 𝑘<1with Theorem 5. Let (𝑋, ⪯) be a partially ordered set and suppose 𝑑(𝑇(𝑥) , 𝑇 (𝑦)) ≤ 𝑘𝑑 (𝑥, 𝑦)V ,𝑤ℎ𝑒𝑛𝑒 𝑒𝑟 𝑦 ⪯𝑥. (8) that there exists a distance 𝑑 in 𝑋 such that (𝑋, 𝑑) is a complete If there exists 𝑥0 ∈𝑋,with𝑥0 ⪯𝑇(𝑥0),then𝑇 has a fixed metric space. Assume that 𝑋 satisfies the property (OSC). Let point. 𝑇:𝑋be →𝑋 a monotone increasing mapping. Assume that Abstract and Applied Analysis 3

there exists a lower semicontinuous function 𝜙:𝑋 → [0,+∞) then 𝑇(𝑥𝛼−1)⪯𝑇(𝑥𝛼−2));thatis,𝑥𝛼 ⪯𝑥𝛼−1.Otherwise,if such that 𝛼−2is an ordinal limit, then 𝑥𝛼−2 = inf{𝑥𝛾,𝛾<𝛼−2}.Since 𝑇 is monotone increasing, then we have 𝑑 (𝑥, 𝑇 (𝑥)) ≤𝜙(𝑥) −𝜙(𝑇 (𝑥)) , 𝑤ℎ𝑒𝑛𝑒V𝑒𝑟 𝑇 (𝑥) ⪯𝑥. (12) 𝑥𝛼−1 =𝑇(𝑥𝛼−2)⪯𝑥𝛾+1, for any 𝛾<𝛼−2, (15)

Then 𝑇 has a fixed point if and only if there exists 𝑥0 ∈𝑋,with which implies 𝑥𝛼 ⪯𝑥𝛾+2,forany𝛾 < 𝛼−2. Therefore we have 𝑇(𝑥0)⪯𝑥0. 𝑥𝛼 ⪯𝑥𝛽, which completes the proof of (3).Letusprove(4). Let 𝛽<𝛼. First assume that 𝛼−1exists. Then, in the proof of Proof. WeproceedaswedidintheproofofTheorem 3.Let (3) 𝑥 =𝑇(𝑥 )⪯𝑥 𝑇 𝑛 ,wesawthat 𝛼 𝛼−1 𝛼−1. Our assumption on 𝑥0 ∈𝑋such that 𝑇(𝑥0)⪯𝑥0.Write𝑥𝑛 =𝑇(𝑥0), 𝑛≥1.Then will then imply we have the fact that {𝑥𝑛} is decreasing and lim𝑛→∞𝑥𝑛 =𝑥𝜔 𝑑(𝑥 ,𝑥 )≤𝜙(𝑥 )−𝜙(𝑥 ). exists in 𝑋. Since we did not assume 𝑇 continuous, then 𝑥𝜔 𝛼−1 𝛼 𝛼−1 𝛼 (16) may not be a fixed point of 𝑇. The idea is to use transfinite 𝛽=𝛼−1 𝛽<𝛼−1 induction to build a transfinite orbit to help catch the fixed If , we are done. Otherwise, if ,thenweuse point. Note that since 𝑋 satisfies (OSC), then we have 𝑥𝜔 = the induction assumption to get inf{𝑥𝑛,𝑛 ≥ 1}.Since𝑇 is monotone increasing, then we will 𝑑(𝑥𝛽,𝑥𝛼−1)≤𝜙(𝑥𝛽)−𝜙(𝑥𝛼−1). (17) have 𝑇(𝑥𝜔)⪯𝑥𝜔. These basic facts so far will help us seek the transfinite orbit {𝑥𝛼}𝛼Γ,whereΓ is the set of all ordinals. This transfinite orbit must satisfy the following properties: The triangle inequality will then imply 𝑑(𝑥 ,𝑥 )≤𝜙(𝑥 )−𝜙(𝑥 ). (1) 𝑇(𝑥𝛼)=𝑥𝛼+1,forany𝛼∈Γ; 𝛽 𝛼 𝛽 𝛼 (18)

(2) 𝑥𝛼 = inf{𝑥𝛽,𝛽<𝛼},if𝛼 is a limit ordinal; Next we assume that 𝛼 is a limit ordinal. Then there exists 𝑥 ⪯𝑥 𝛽<𝛼 (3) 𝛼 𝛽,whenever ; an increasing sequence of ordinals {𝛽𝑛},with𝛽𝑛 <𝛼,such 𝑥 = 𝑥 𝛽<𝛼 (4) 𝑑(𝑥𝛼,𝑥𝛽)≤𝜙(𝑥𝛽)−𝜙(𝑥𝛼),whenever𝛽<𝛼. that 𝛼 lim𝑛→∞ 𝛽𝑛 .Given , assume that we have 𝛽𝑛 <𝛽,forall𝑛≥1.Inthiscase,wehaveseenthat𝑥𝛼 =𝑥𝛽. 𝛼∈ Clearly the above properties are satisfied for any Otherwise, let us assume that there exists 𝑛0 ≥1such that {0,1,...,𝜔}.Let𝛼 be an ordinal number. Assume that the 𝛽<𝛽 𝑛0 . In this case, from our induction assumption and the properties (1)–(4) aresatisfiedby{𝑥𝛽}𝛽<𝛼.Wehavetwocases triangle inequality, we get as follows. 𝑑(𝑥 ,𝑥 )≤𝜙(𝑥 )−𝜙(𝑥 ), 𝑛≥𝑛 . 𝛽 𝛽𝑛 𝛽 𝛽𝑛 0 (19) (i) If 𝛼=𝛽+1,thenset𝑥𝛼 =𝑇(𝑥𝛽).

(ii) Assume that 𝛼 is a limit ordinal. Set 𝜙0 = Using the lower semicontinuity of 𝜙,weconcludethat inf{𝜙(𝑥𝛽), 𝛽 < .𝛼} Then one can easily find an increasing sequence of ordinals {𝛽𝑛},with𝛽𝑛 <𝛼, 𝑑(𝑥𝛽,𝑥𝛼)≤𝜙(𝑥𝛽)−𝜙(𝑥𝛼), (20) such that lim𝑛→∞𝜙(𝑥𝛽 )=𝜙0.Property(4) will 𝑛 (4) force {𝑥𝛽 } to be Cauchy. Since 𝑋 is complete, then which completes the proof of . By the transfinite induction 𝑛 {𝑥 } lim𝑛→∞𝑥𝛽 = 𝑥 exists in 𝑋.Theproperty(OSC)will we conclude that the transfinite orbit 𝛼 exists which 𝑛 (1) (4) .6 then imply 𝑥=inf{𝑥𝛽 ,𝑛 ≥ 1}.Letusshowthat satisfies the properties – .UsingPropositionA ([13, 𝑛 𝛽 𝜙(𝑥 )=𝜙(𝑥 ) 𝑥=inf{𝑥𝛽,𝛽 < 𝛼}.Let𝛽<𝛼.If𝛽𝑛 <𝛽,forall page 284]), there exists an ordinal such that 𝛼 𝛽 , 𝑛≥1,thenwehave for any 𝛼≥𝛽.Inparticular,wehave𝜙(𝑥𝛼)=𝜙(𝑥𝛼+1),forany 𝛼≥𝛽.Property(4) will then force 𝑥𝛼+1 =𝑥𝛼;thatis,𝑇(𝑥𝛼)= 𝑑(𝑥 ,𝑥 )≤𝜙(𝑥 )−𝜙(𝑥 ), 𝑛=1,2,.... 𝑥 𝑇 𝛽 𝛽𝑛 𝛽𝑛 𝛽 (13) 𝛼. Therefore has a fixed point. One may wonder if Theorem 5 is truly an extension of the 𝜙(𝑥 )≥𝜙 = 𝜙(𝑥 )≥𝜙(𝑥 ) 𝜙(𝑥 )=𝜙 main results of [1, 2, 11]. The following example shows that it But 𝛽 0 lim𝑛→∞ 𝛽𝑛 𝛽 .Hence 𝛽 0 𝑥 =𝑥 𝑥 = 𝑥 is the case. which implies that lim𝑛→∞ 𝛽𝑛 𝛽.Hence 𝛽 . Assume otherwise that there exists 𝑛0 ≥1such that 𝛽<𝛽𝑛 .Hence 1 0 Example 6.Let𝑋=𝐿([0, 1], 𝑑𝑥) be the classical Banach 𝑥𝛽 ⪯𝑥𝛽 which implies 𝑥⪯𝑥𝛽.Inanycase,wehave𝑥⪯𝑥𝛽, 𝑛0 space with the natural pointwise order generated by R.Let 𝛽<𝛼 for any . Therefore we have 𝐶 = {𝑓 ∈ 𝑋, 𝑓(𝑡) ≥ 0𝑎.𝑒.} bethepositiveconeof𝑋.Define 𝑇:𝐶by →𝐶 𝑥=inf {𝑥𝛽 ,𝑛≥1}≤inf {𝑥𝛽,𝛽<𝛼}≤inf {𝑥𝛽 ,𝑛≥1}. 𝑛 𝑛 1 (14) {𝑓 (𝑡) 𝑓 (𝑡) > , { if 2 𝑥= {𝑥 ,𝛽 < 𝛼} 𝑥 = 𝑥 𝑇(𝑓)(𝑡) = { (21) Hence inf 𝛽 .Set 𝛼 .Letusprovethat { 1 {𝑥 ,𝛽≤𝛼} (1) (4) (1) (2) 0 𝑓 (𝑡) ≤ . 𝛽 satisfies all properties – .Clearly and { if 2 are satisfied. Let us focus on (3) and (4).Let𝛽<𝛼. We need to show that 𝑥𝛼 ⪯𝑥𝛽.If𝛼 is a limit ordinal, this is obvious. First note that 𝐶 is a closed subset of 𝑋.Hence𝐶 is Assume that 𝛼−1exists. We have two cases; if 𝛼−2exists, complete for the norm-1 distance. Also it is easy to check then we have 𝑥𝛼−1 ⪯𝑥𝛼−2.Since𝑇 is monotone increasing, that the property (OSC) holds in this case. Note that, for any 4 Abstract and Applied Analysis

2 𝑓∈𝐶,wehave0≤𝑇(𝑓)≤𝑓.Alsowehave𝑇 (𝑓) = Definition 7. One says that a mapping 𝑇:𝑋is →𝑋 𝐺-edge 𝑇(𝑇(𝑓)) = 𝑇(𝑓));thatis,𝑇(𝑓) is a fixed point of 𝑇 for any preserving if 𝑓∈𝐶.Notethatforany𝑓∈𝐶we have ∀𝑥,𝑦∈𝑋, (𝑥,𝑦)∈𝐸(𝐺) 󳨐⇒(𝑇𝑥,𝑇𝑦)∈𝐸(𝐺) . (25) 1 󵄨 󵄨 𝑑(𝑓,𝑇(𝑓))=∫ 󵄨𝑓 (𝑡) −𝑇(𝑓)(𝑡)󵄨 𝑑𝑡 𝑇 is said to be a Caristi 𝐺-mapping if there exists a lower 0 semicontinuous function 𝜙:𝑋 → [0,+∞)such that 1 1 󵄩 󵄩 󵄩 󵄩 𝑑 (𝑥, 𝑇𝑥) ≤𝜙(𝑥) −𝜙(𝑇𝑥) , (𝑇 (𝑥) ,𝑥) ∈𝐸(𝐺) . = ∫ 𝑓 (𝑡) 𝑑𝑡 − ∫ 𝑇(𝑓)(𝑡) 𝑑𝑡 = 󵄩𝑓󵄩 − 󵄩𝑇(𝑓)󵄩 . whenever 0 0 (26) (22) One can now give the graph theory versions of our two Therefore all the assumptions of Theorem 5 are satisfied. mean Theorems 3 and 5 as follows. But 𝑇 fails to satisfy the assumptions of [1, 2, 11]. Indeed if we take Theorem 8. Let 𝐺 be an oriented graph on the set 𝑋 with 𝐸(𝐺) containing all loops and suppose that there exists a distance 𝑑 1 1 1 in 𝑋 such that (𝑋, 𝑑) is a complete metric space. Let 𝑇:𝑋 → 𝑓 (𝑡) = ,𝑓𝑛 (𝑡) = + ,𝑛≥1, (23) 2 2 𝑛 𝑋 be continuous, 𝐺-edge preserving, and a 𝐺-Caristi mapping. Then 𝑇 has a fixed point if and only if there exists 𝑥0 ∈𝑋,with 𝑇(𝑓) =0 𝑇(𝑓 )=𝑓 𝑛≥1 then we have and 𝑛 𝑛,forany . (𝑇(𝑥0), 𝑥0)∈𝐸(𝐺). Therefore 𝑇 is not continuous since {𝑓𝑛} converges uniformly (and in norm-1 as well) to 𝑓. Note also that 𝑓≤𝑓𝑛,for𝑛≥1. Proof. 𝐺 hasalltheloops.Inparticular,if𝑥0 is a fixed point So any Lipschitz condition on the partial order of 𝐶 will not of 𝑇,thatis,𝑇(𝑥0)=𝑥0,thenwehave(𝑇(𝑥0), 𝑥0)∈𝐸(𝐺). be satisfied by 𝑇 in this case. Assume that there exists 𝑥0 ∈𝑋such that (𝑇(𝑥0), 𝑥0)∈𝐸(𝐺). 𝑛+1 𝑛 Since 𝑇 is 𝐺-edge preserving, we have (𝑇 (𝑥0), 𝑇 (𝑥0)) ∈ 𝐸(𝐺) 𝑛≥1 3. Caristi’s Theorem in Metric Spaces ,forany .Hence 𝑛 𝑛+1 with Graph 𝑑(𝑇 (𝑥0),𝑇 (𝑥0)) (27) It seems that the terminology of graph theory instead of ≤𝜙(𝑇𝑛 (𝑥 )) − 𝜙 (𝑇𝑛+1 (𝑥 )), 𝑛=1,2,.... partial ordering sets can give more clear pictures and yield 0 0 to generalize the theorems above. In this section, we give the 𝑛 Hence {𝜙(𝑇 (𝑥0))} is a decreasing sequence of positive graph versions of our two main results. 𝑛 numbers. Let 𝜙0 = lim𝑛→∞𝜙(𝑇 (𝑥0)).Forany𝑛, ℎ ≥,we 1 Throughout this section we assume that (𝑋, 𝑑) is a metric have space and 𝐺 is a directed graph (digraph) with set of vertices 𝑉(𝐺) =𝑋 𝐸(𝐺) ℎ−1 and set of edges containing all the loops; 𝑛 𝑛+ℎ 𝑛+𝑘 𝑛+𝑘+1 that is, (𝑥, 𝑥) ∈ 𝐸(𝐺) for any 𝑥∈𝑋. We also assume 𝑑(𝑇 (𝑥0),𝑇 (𝑥0)) ≤ ∑𝑑(𝑇 (𝑥0),𝑇 (𝑥0)) that 𝐺 has no parallel edges (arcs) and so we can identify 𝑘=0 𝐺 (𝑉(𝐺), 𝐸(𝐺)) with the pair . Our graph theory notations ≤𝜙(𝑇𝑛 (𝑥 )) − 𝜙 𝑛+ℎ(𝑇 (𝑥 )) . and terminology are standard and can be found in all graph 0 0 theory books, like [14, 15]. A digraph 𝐺 is called an oriented (28) graph; if whenever (𝑢, V)∈𝐸(𝐺),then(V,𝑢)∉𝐸(𝐺). 𝑛 Therefore {𝑇 (𝑥0)} is a Cauchy sequence in 𝑋.Since𝑋 is Let (𝑋, ⪯) be a partially ordered set. We define the 𝑛 complete, there exists 𝑥∈𝑋such that lim𝑛→∞𝑇 (𝑥0)=𝑥. oriented graph 𝐺⪯ on 𝑋 as follows. The vertices of 𝐺⪯ are the Since 𝑇 is continuous, we conclude that 𝑇(𝑥) = 𝑥;thatis,𝑥 is elements of 𝑋,andtwovertices𝑥, 𝑦 ∈𝑋 are connected by a a fixed point of 𝑇. directed edge (arc) if 𝑥⪯𝑦. Therefore, 𝐺⪯ has no parallel arcs as 𝑥⪯𝑦& 𝑦⪯𝑥⇒𝑥=𝑦. The following definition is needed to prove the analogue If 𝑥, 𝑦 are vertices of the digraph 𝐺,thenadirectedpath 𝑁 to Theorem 5. from 𝑥 to 𝑦 of length 𝑁 is a sequence {𝑥𝑖}𝑖=0 of 𝑁+1vertices such that Definition 9. Let 𝐺 be an acyclic oriented graph on the set 𝑋 with 𝐸(𝐺) containing all loops. One says that 𝐺 satisfies the 𝑥 =𝑥, 𝑥 =𝑦, 0 𝑁 property (OSC) if and only if (𝑋, ⪯𝐺) satisfies (OSC). (24) (𝑥 ,𝑥 )∈𝐸 𝐺 , 𝑖=0,1,...,𝑁. 𝑖 𝑖+1 ( ) The analogue to Theorem 5 may be stated as follows.

A closed directed path of length 𝑁>1from 𝑥 to 𝑦,that Theorem 10. Let 𝐺 be an oriented graph on the set 𝑋 with is, 𝑥=𝑦, is called a directed cycle. An acyclic digraph is a 𝐸(𝐺) containing all loops and suppose that there exists a metric digraph that has no directed cycle. 𝑑 in 𝑋 such that (𝑋, 𝑑) is a complete metric space. Assume that Given an acyclic digraph, 𝐺, we can always define a 𝐺 satisfies the property (OSC). Let 𝑇:𝑋be →𝑋 a 𝐺-edge partially order ⪯𝐺 on the set of vertices of 𝐺 by defining that preserving and a Caristi 𝐺-mapping. Then 𝑇 has a fixed point 𝑥⪯𝐺 𝑦 whenever there is a directed path from 𝑥 to 𝑦. ifandonlyifthereexists𝑥0 ∈𝑋,with(𝑇(𝑥0), 𝑥0)∈𝐸(𝐺). Abstract and Applied Analysis 5

The proof of Theorem 10 is similar to the proof of Theorem 5. In fact it is easy to check that these two theorems are equivalent.

Conflict of Interests The authors declare that there is no conflict of interests regarding the publication of this paper.

References

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Research Article Applications of Differential Subordination for Argument Estimates of Multivalent Analytic Functions

Meng-Ting Lu, Ting Jia, Xing-Qian Ling, and Jin-Lin Liu

Department of Mathematics, Yangzhou University, Yangzhou 225002, China

Correspondence should be addressed to Jin-Lin Liu; [email protected]

Received 21 November 2013; Accepted 20 January 2014; Published 26 February 2014

Academic Editor: David Kalaj

Copyright © 2014 Meng-Ting Lu 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.

By using the method of differential subordinations, we derive some properties of multivalent analytic functions. All results presented here are sharp.

ThispaperisdedicatedtoProfessorMiodragMateljevi´c on the occasion of his 65th birthday

1. Introduction Recently, a number of results for argument properties of analytic functions have been obtained by several authors 𝐴(𝑝) 𝑓(𝑧) Let denote the class of functions of the form (see, e.g., [1–5]).Theobjectiveofthepresentpaperisto ∞ derive some further interesting properties of multivalent 𝑝 𝑝+𝑛 analytic functions. The basic tool used here is the method of 𝑓 (𝑧) =𝑧 + ∑𝑎𝑝+𝑛𝑧 (𝑝 ∈ 𝑁 = {1, 2, 3, . . .}), (1) 𝑛=1 differential subordinations. To derive our results, we need the following lemmas. which are analytic in the open unit disk 𝐷={𝑧∈𝐶:|𝑧|<1}. ℎ(𝑧) Let 𝑓(𝑧) and 𝑔(𝑧) be analytic in 𝐷.Then,wesaythat𝑓(𝑧) Lemma 1 (see [6, Theorem 1, page 776]). Let be analytic 𝐷 ℎ(0) = 0 𝑔(𝑧) is subordinate to 𝑔(𝑧) in 𝐷,writtenas𝑓(𝑧) ≺ 𝑔(𝑧),ifthere and starlike univalent in with .If is analytic in 𝐷 𝑧𝑔󸀠(𝑧) ≺ ℎ(𝑧) exists an analytic function 𝑤(𝑧) in 𝐷,suchthat|𝑤(𝑧)| ≤ |𝑧| and ,then and 𝑓(𝑧) = 𝑔(𝑤(𝑧)).If (𝑧∈𝐷) 𝑔(𝑧) is univalent in 𝐷,then 𝑧 ℎ (𝑡) the subordination 𝑓(𝑧) ≺ 𝑔(𝑧) is equivalent to 𝑓(0) = 𝑔(0) 𝑔 (𝑧) ≺𝑔(0) + ∫ 𝑑𝑡. (5) 0 𝑡 and 𝑓(𝐷) ⊂ 𝑔(𝐷).Let𝑝(𝑧) = 11 +𝑝 𝑧+⋅⋅⋅ be analytic in 𝐷. Then, for −1≤𝐵<𝐴≤1,itisclearthat Lemma 2 (see [5,Theorem1,page1814]). Let 0<𝛼1 ≤1, 𝛽𝜋𝑖 0<𝛼2 ≤1, 𝛽=(𝛼1 −𝛼2)/(𝛼1 +𝛼2),and𝑐=𝑒 .Alsolet 1+𝐴𝑧 𝑝 (𝑧) ≺ (𝑧∈𝐷) (2) 1+𝐵𝑧 𝜆0𝑎≥0,( 𝜆 𝑏+2) ≥0, (𝑏+1) Re 𝜇≥0, 2 1 (6) if and only if |𝑏+1| ≤ , |𝑎−𝑏−1| ≤ . 𝛼1 +𝛼2 max {𝛼1,𝛼2} 󵄨 󵄨 󵄨 1−𝐴𝐵󵄨 𝐴−𝐵 󵄨𝑝 (𝑧) − 󵄨 < (−1<𝐵<𝐴≤1;𝑧∈𝐷) , 𝑞(𝑧) 𝐷 𝑞(0) = 1 󵄨 1−𝐵2 󵄨 1−𝐵2 If is analytic in with and (3) 𝑎 𝑏+2 𝑏+1 𝜆0(𝑞 (𝑧)) +𝜆(𝑞(𝑧)) +𝜇(𝑞(𝑧)) 1−𝐴 (7) 𝑝 (𝑧) > (𝐵=−1;𝑧∈𝐷) . 󸀠 𝑏 Re 2 (4) +𝑧𝑞 (𝑧) (𝑞 (𝑧)) ≺ℎ(𝑧)(𝑧∈𝐷) , 2 Abstract and Applied Analysis where Then, from the assumption of the theorem, we can see that 𝑎((𝛼 +𝛼 )/2) (1/2)(𝑏+1)(𝛼 +𝛼 ) 𝑝(𝑧) is analytic in 𝐷 with 𝑝(0) = 1 and 𝛼 + (1 − 𝛼)𝑝(𝑧) =0̸ 1+𝑐𝑧 1 2 1+𝑐𝑧 1 2 𝑧∈𝐷 ℎ (𝑧) =𝜆0( ) +( ) for all . Taking the logarithmic differentiations in both 1−𝑧 1−𝑧 sides of (17), we get (𝛼 +𝛼 )/2 1+𝑐𝑧 1 2 󸀠 󸀠 ×(𝜇+𝜆( ) 𝑧𝑓 (𝑧) (1−𝛼) 𝑧𝑝 (𝑧) (8) −𝑝= , (18) 1−𝑧 𝑓 (𝑧) 𝛼+(1−𝛼) 𝑝 (𝑧)

𝛼 +𝛼 𝑧 𝑐𝑧 𝑝 󸀠 󸀠 + 1 2 ( + )) 𝑧 𝑧𝑓 (𝑧) (1−𝛼) 𝑧𝑝 (𝑧) 2 1−𝑧 1+𝑐𝑧 ( −𝑝)= (19) 𝑓 (𝑧) 𝑓 (𝑧) (𝛼 + (1−𝛼) 𝑝 (𝑧))2 is (close-to-convex) univalent in 𝐷,then for all 𝑧∈𝐷.Thus,inequality(12)isequivalentto 𝜋 𝜋 − 𝛼 < (𝑞 (𝑧))< 𝛼 (𝑧∈𝐷) . 󸀠 2 2 arg 2 1 (9) (1−𝛼) 𝑧𝑝 (𝑧) ≺𝛿𝑧. (20) (𝛼 + (1−𝛼) 𝑝 (𝑧))2 The bounds 𝛼1 and 𝛼2 in (9) are sharp for the function 𝑞(𝑧) defined by By using Lemma 1,(20)leadsto (𝛼 +𝛼 )/2 1+𝑐𝑧 1 2 𝑧 (1−𝛼) 𝑝󸀠 (𝑡) 𝑞 (𝑧) =( ) . (10) ∫ 𝑑𝑡 ≺ 𝛿𝑧 1−𝑧 2 (21) 0 (𝛼 + (1−𝛼) 𝑝 (𝑡)) Remark 3 (see [5, Lemma 2, page 1813]). The function 𝑞(𝑧) or to defined by (10)isanalyticandunivalentconvexin𝐷 and 1 𝜋 𝜋 1− ≺𝛿𝑧. (22) 𝑞 (𝐷) ={𝑤:𝑤∈𝐶,− 𝛼 < 𝑤< 𝛼 }. 𝛼+(1−𝛼) 𝑝 (𝑧) 2 2 arg 2 1 (11) According to (16), (22)canbewrittenas 2. Main Results 1+(𝛼/ (1−𝛼)) 𝛿𝑧 𝑝 (𝑧) ≺ . (23) Our first result is contained in the following. 1−𝛿𝑧 𝐴 = (𝛼/(1 − 𝛼))𝛿 𝐵=−𝛿 𝛼 ∈ (0, 1/2] 𝛽 ∈ (0, 1) 𝑓(𝑧) ∈ 𝐴(𝑝) Now, by taking and in (2)and(3), Theorem 4. Let and .If we have satisfies 𝑓(𝑧) =0(0<|𝑧|<1)̸ and 󵄨 󵄨 󵄨 𝑓 (𝑧) 󵄨 󵄨 󵄨 󵄨 𝑝 󸀠 󵄨 󵄨arg ( −𝛼)󵄨 = 󵄨arg 𝑝 (𝑧)󵄨 󵄨 𝑧 𝑧𝑓 (𝑧) 󵄨 󵄨 𝑧𝑝 󵄨 󵄨 󵄨 󵄨 ( −𝑝)󵄨 <𝛿 (𝑧∈𝐷) , 󵄨 󵄨 󵄨𝑓 (𝑧) 𝑓 (𝑧) 󵄨 (12) (24) 󵄨 󵄨 𝛿 𝜋 < arcsin ( )= 𝛽 where 𝛿 isthesmallestpositiverootoftheequation 1−𝛼+𝛼𝛿2 2 𝜋𝛽 𝜋𝛽 for all 𝑧∈𝐷because of 𝑔(𝛿).Thisproves( =0 14). 𝛼 ( )𝑥2 −𝑥+(1−𝛼) ( )=0, sin 2 sin 2 (13) Next, we consider the function 𝑓(𝑧) defined by 𝑧𝑝 then 𝑓 (𝑧) = (25) 1−𝛿𝑧 󵄨 󵄨 󵄨 𝑓 (𝑧) 󵄨 𝜋 󵄨arg ( −𝛼)󵄨 < 𝛽 (𝑧∈𝐷) . (14) 𝑧∈𝐷 󵄨 𝑧𝑝 󵄨 2 for all .Itiseasytoseethat 󵄨 󵄨 󵄨 𝑧𝑝 𝑧𝑓󸀠 (𝑧) 󵄨 The bound 𝛽 is sharp for each 𝛼 ∈ (0, 1/2]. 󵄨 󵄨 󵄨 ( −𝑝)󵄨 = |𝛿𝑧| <𝛿 (26) 󵄨𝑓 (𝑧) 𝑓 (𝑧) 󵄨 Proof. Let for all 𝑧∈𝐷.Since 𝜋𝛽 2 𝜋𝛽 𝑔 (𝑥) =𝛼sin ( )𝑥 −𝑥+(1−𝛼) sin ( ). (15) 𝑓 (𝑧) 1+(𝛼/ (1−𝛼)) 𝛿𝑧 2 2 −𝛼=(1−𝛼) , (27) 𝑧𝑝 1−𝛿𝑧 We can see easily that (13)hastwopositiveroots.Since𝑔(0) > 0 𝑔(1) < 0 it follows from (3)that and ,wehave 󵄨 󵄨 󵄨 𝑓 (𝑧) 󵄨 𝛿 𝜋 𝛼 󵄨 ( −𝛼)󵄨 = ( ) = 𝛽. 0< 𝛿≤𝛿<1. sup 󵄨arg 𝑧𝑝 󵄨 arcsin 1−𝛼+𝛼𝛿2 2 (28) 1−𝛼 (16) 𝑧∈𝑈 󵄨 󵄨 𝛽 Put Hence, we conclude that the bound is the best possible for each 𝛼 ∈ (0, 1/2]. 𝑓 (𝑧) =𝛼+(1−𝛼) 𝑝 (𝑧) . (17) 𝑧𝑝 Next, we derive the following. Abstract and Applied Analysis 3

Theorem 5. If 𝑓(𝑧) ∈ 𝐴(𝑝) satisfies 𝑓(𝑧) =0(0<|𝑧|<1)̸ Theorem 6. Let 𝛼, 𝛾 ∈ (0,.If 1) 𝑓(𝑧) ∈ 𝐴(𝑝) satisfies and 𝑓(𝑧) =0(0<|𝑧|<1)̸ and 󸀠 󵄨 𝑧𝑝 𝑧𝑓 (𝑧) 󵄨 𝑓 (𝑧) 𝑧𝑓󸀠 (𝑧) 𝑓 (𝑧) { ( −𝑝)} <𝛾 (𝑧∈𝐷) , 󵄨 { (𝛾 ( −𝑝)+(1−𝛾) ) Re (29) 󵄨arg 𝑝 𝑝 𝑓 (𝑧) 𝑓 (𝑧) 󵄨 𝑧 𝑓 (𝑧) 𝑧 󵄨 (39) 󵄨 where 2 󵄨 −(1−𝛾)𝛼 }󵄨 <𝜋𝛿 1 󵄨 0<𝛾< , 2 2 (30) log for all 𝑧∈𝐷,where then √𝛾(2(1−𝛼) (1 − 𝛾) + 𝛾) 1 1 −1 𝑧𝑝 𝛿= + tan ( ), (40) >1−2𝛾 2 (𝑧∈𝐷) . 2 𝜋 2𝛼 (1 − 𝛾) Re 𝑓 (𝑧) log (31)

The bound in (31) is sharp. then 𝑓 (𝑧) >𝛼 (𝑧∈𝐷) . (41) Proof. Let Re 𝑧𝑝 𝑓 (𝑧) 𝛿 𝑝 (𝑧) = . (32) The bound in (39) is sharp. 𝑧𝑝 Proof. Define the function 𝑝(𝑧) by (17). For 𝛼, 𝛾 ∈ (0,,it 1) Then, from the assumption of the theorem we can see that follows from (17)and(18)that 𝑝(𝑧) is analytic in 𝐷 with 𝑝(0) = 1 and 𝑝(𝑧) =0̸ for all 𝑧∈𝐷. According to (32)and(29), we have immediately 1 𝑓 (𝑧) 𝑧𝑓󸀠 (𝑧) 𝑓 (𝑧) { (𝛾 ( −𝑝)+(1−𝛾) ) 𝛾 (1−𝛼) 𝑧𝑝 𝑓 (𝑧) 𝑧𝑝 𝑧𝑝󸀠 (𝑧) 1+𝑧 1− ≺ ; (33) 𝛾𝑝2 (𝑧) 1−𝑧 −(1−𝛾)𝛼2} that is, (1−𝛼) (1 − 𝛾) 2 2𝛼 (1 − 𝛾) 󸀠 1 󸀠 2𝛾𝑧 = 𝑝 (𝑧) + 𝑝 (𝑧) +𝑧𝑝 (𝑧) 𝑧( ) ≺ . (34) 𝛾 𝛾 𝑝 (𝑧) 1−𝑧 (42)

Now, by using Lemma 1,weobtain for all 𝑧∈𝐷.Putting 1 ≺1−2𝛾 (1−𝑧) . 𝑎=𝑏=𝜆0 =0, 𝛼1 =𝛼2 =1, 𝑝 (𝑧) log (35) (1−𝛼) (1 − 𝛾) 2𝛼 (1 − 𝛾) (43) 𝜆= ,𝜇= Since the function 1−2𝛾log(1 − 𝑧) is convex univalent in 𝐷 𝛾 𝛾 and in Lemma 2 and using (42), we see that if Re (1−2𝛾log (1−𝑧)) >1−2𝛾log 2 (𝑧∈𝐷) , (36) 1 𝑓 (𝑧) 𝑧𝑓󸀠 (𝑧) 𝑓 (𝑧) { (𝛾 ( −𝑝)+(1−𝛾) ) from (35), we get inequality (31). 𝛾 (1−𝛼) 𝑧𝑝 𝑓 (𝑧) 𝑧𝑝 To show that the bound in (31) cannot be increased, we (44) consider −(1−𝛾)𝛼2}≺ℎ(𝑧) , 𝑧𝑝 𝑓 (𝑧) = (𝑧∈𝐷) . (37) 1−2𝛾log (1−𝑧) where It is easy to verify that the function 𝑓(𝑧) satisfies inequality ℎ (𝑧) (29). On the other hand, we have 1+𝑧 (1−𝛼) (1 − 𝛾) 1+𝑧 2𝛼 (1 − 𝛾) =( )( ( )+ 𝑧𝑝 1−𝑧 𝛾 1−𝑧 𝛾 󳨀→ 1 − 2 𝛾 2 (45) Re 𝑓 (𝑧) log (38) 2𝑧 + ), as 𝑧→−1. Now, the proof of the theorem is complete. 1−𝑧2

Finally, we discuss the following theorem. then (41)holdstrue. 4 Abstract and Applied Analysis

Letting 0<𝜃<𝜋and 𝑥=cot(𝜃/2),wededucethat References

𝑖𝜃 arg ℎ(𝑒 ) [1] A. Gangadharan, V. Ravichandran, and T. N. Shanmugam, “Radii of convexity and strong starlikeness for some classes 𝜋 (1−𝛼) (1 − 𝛾) 2𝛼 (1 − 𝛾) of analytic functions,” Journal of Mathematical Analysis and = + { 𝑥𝑒𝜋𝑖/2 + Applications,vol.211,no.1,pp.301–313,1997. 2 arg 𝛾 𝛾 [2] J.-L. Liu, “The Noor integral and strongly starlike functions,” 𝑖 1 (46) Journal of Mathematical Analysis and Applications,vol.261,no. + (𝑥 + )} 2,pp.441–447,2001. 2 𝑥 [3] M. Nunokawa, S. Owa, E. Y. Duman, and M. Aydogan,ˇ “Some 2 properties of analytic functions relating to the Miller and 𝜋 −1 (2 (1−𝛼) (1 − 𝛾) + 𝛾) 𝑥 +𝛾 = + tan ( ). Mocanu result,” Computers & Mathematics with Applications, 2 4𝛼 (1 − 𝛾) 𝑥 vol.61,no.5,pp.1291–1295,2011. [4] N.-E. Xu, D.-G. Yang, and S. Owa, “On strongly starlike Making use of (46), we obtain that multivalent functions of order 𝛽 and type 𝛼,” Mathematische Nachrichten,vol.283,no.8,pp.1207–1218,2010. 󵄨 󵄨 inf 󵄨arg ℎ (𝑧)󵄨 [5] D.-G. Yang and J.-L. Liu, “Argument inequalities for certain |𝑧|=1(𝑧 =±1̸ ) 󵄨 󵄨 analytic functions,” Mathematical and Computer Modelling,vol. 𝑖𝜃 52, no. 9-10, pp. 1812–1821, 2010. = min arg ℎ(𝑒 ) 0<𝜃<𝜋 [6] T. J. Suffridge, “Some remarks on convex maps of the unit disk,” 2 Duke Mathematical Journal, vol. 37, no. 4, pp. 775–777, 1970. 𝜋 −1 (2 (1−𝛼) (1 − 𝛾) + 𝛾) 𝑥 +𝛾 = + min tan ( ) 2 𝑥>0 4𝛼 (1 − 𝛾) 𝑥

𝜋 √𝛾(2(1−𝛼) (1 − 𝛾) + 𝛾) = + −1 ( ) 2 tan 2𝛼 (1 − 𝛾)

=𝜋𝛿. (47)

Therefore, if 𝑓(𝑧) ∈ 𝐴(𝑝) satisfies (39), then the subordination (44)holds,and,thus,weobtain(41). For the function 𝑓 (𝑧) 1+(1−2𝛼) 𝑧 = , (48) 𝑧𝑝 1−𝑧 we find that 1 𝑓 (𝑧) 𝑧𝑓󸀠 (𝑧) 𝑓 (𝑧) { (𝛾 ( −𝑝)+(1−𝛾) ) 𝛾 (1−𝛼) 𝑧𝑝 𝑓 (𝑧) 𝑧𝑝 (49) −(1−𝛾)𝛼2}=ℎ(𝑧) , where ℎ(𝑧) is defined by (45). In view of (46)and(49), we conclude that the bound 𝛿 in (39) is the largest number such that (41)holdstrue.Thiscompletestheproof.

Conflict of Interests The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgment The authors would like to express their sincere thanks to the referees for careful reading and suggestions which helped them improve the paper. Hindawi Publishing Corporation Abstract and Applied Analysis Volume 2014, Article ID 478486, 5 pages http://dx.doi.org/10.1155/2014/478486

Research Article Existence Theorems of 𝜀-Cone Saddle Points for Vector-Valued Mappings

Tao Chen

College of Culture and Tourism, Yunnan Open University, Kunming 650223, China

Correspondence should be addressed to Tao Chen; [email protected]

Received 19 November 2013; Accepted 22 January 2014; Published 25 February 2014

Academic Editor: Qamrul Hasan Ansari

Copyright © 2014 Tao Chen. 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.

A new existence result of 𝜀-vector equilibrium problem is first obtained. Then, by using the existence theorem of 𝜀-vector equilibrium problem, a weakly 𝜀-cone saddle point theorem is also obtained for vector-valued mappings.

1. Introduction problem, or such an exact solution simply does not exist, for example, if the feasible set is not compact. Thus, it is Saddle point problems are important in the areas of opti- meaningful to look for an approximate solution instead. mization theory and game theory. As for optimization theory, Therearealsomanypaperstoinvestigatetheapproximate the main motivation of studying saddle point has been their solution problem, such as [19–21]. Kimura et al. [20]obtained connection with characterized solutions to minimax dual several existence results for 𝜀-vector equilibrium problem problems. Also, as for game theory, the main motivation has and the lower semicontinuity of the solution mapping of been the determination of two-person zero-sum games based 𝜀-vectorequilibriumproblem.AnhandKhanh[19]have on the minimax principle. considered two kinds of solution sets to parametric gener- In recent years, based on the development of vector alized 𝜀-vector quasiequilibrium problems and established optimization, a great deal of papers have been devoted the sufficient conditions for the Hausdorff semicontinuity (or to the study of cone saddle points problems for vector- Berge semicontinuity) of these solution mappings. X. B. Li valuedmappingsandset-valuedmappings,suchas[1–8]. andS.J.Li[21] established some semicontinuity results on Nieuwenhuis [5] introduced the notion of cone saddle points 𝜀-vector equilibrium problem. for vector-valued functions in finite-dimensional spaces and Theaimofthispaperistocharacterizethe𝜀-cone sad- obtained a cone saddle point theorem for general vector- dle point of vector-valued mappings. For this purpose, we valued mappings. Gong [2] established a strong cone sad- first establish an existence theorem for 𝜀-vector equilibrium dle point theorem of vector-valued functions. Li et al. [4] problem. Then, by this existence result, we obtain an existence obtained an existence theorem of lexicographic saddle point theorem for 𝜀-cone saddle point of vector-valued mappings. for vector-valued mappings. Bigi et al. [1] obtained a cone saddle point theorem by using an existence theorem of a vector equilibrium problem. Zhang et al. [9] established a 2. Preliminaries general cone loose saddle point for set-valued mappings. Let 𝑋 be a real Hausdorff topological vector space and let Zhang et al. [8] obtained a minimax theorem and an existence 𝑉 be a real local convex Hausdorff topological vector space. theorem of cone saddle points for set-valued mappings by Assume that 𝑆 is a pointed closed convex cone in 𝑉 with ∗ using Fan-Browder fixed point theorem. Some other types of nonempty interior int 𝑆 =0̸ .Let𝑉 be the topological dual ∗ existence results can be found in [3, 10–18]. space of 𝑉.Denotethedualconeof𝑆 by 𝑆 : On the other hand, in some situations, it may not ∗ ∗ ∗ ∗ be possible to find an exact solution for an optimization 𝑆 ={𝑠 ∈𝑉 :𝑠 (𝑠) ≥0,∀𝑠∈𝑆}. (1) 2 Abstract and Applied Analysis

Note that from Lemma 3.21 in [22]wehave where 𝑓:𝑋×𝑋is →𝑉 a vector-valued mapping, 𝐸 is a 𝑋, 𝜀∈ 𝑆 ∗ ∗ ∗ nonempty subset of and int . 𝑧∈𝑆⇐⇒{⟨𝑧 , 𝑧⟩ ≥ 0, ∀𝑧 ∈𝑆 }, If 𝑓(𝑥, 𝑦) = 𝑔(𝑦), −𝑔(𝑥) 𝑥, 𝑦,andif ∈𝐸 𝑥∈𝐸is (2) 𝑥∈𝐸 𝜀 𝑧∈ 𝑆⇐⇒{⟨𝑧∗, 𝑧⟩ > 0, ∀𝑧∗ ∈𝑆∗ \ {0}}. asolutionofVAEP,then is a solution of -vector int optimization of 𝑔,where𝑔 is a vector-valued mapping. Denote the 𝜀-solution set of (VAEP) by Definition 1 (see [7, 23]). Let 𝑓:𝑋 →𝑉be a vector-valued 𝑓 𝑆 𝑋 mapping. is said to be -upper semicontinuous on if and 𝑆 (𝜀) := {𝑥 ∈𝐸:𝑓(𝑥,) 𝑦 +𝜀∉−int 𝑆, ∀𝑦 ∈𝐸} . (8) only if, for each 𝑥∈𝑋and any 𝑠∈int 𝑆, there exists an open neighborhood 𝑈𝑥 of 𝑥 such that Lemma 6 (see [20]). Let 𝐸 be a nonempty subset of 𝑋.Suppose that 𝑓:𝑋×𝑋is →𝑉 a vector-valued mapping and the 𝑓 (𝑢) ∈𝑓(𝑥) +𝑠−int 𝑆, ∀𝑢𝑥 ∈𝑈 . (3) following conditions are satisfied:

𝑓 is said to be 𝑆-lower semicontinuous on 𝑋 if and only if −𝑓 (i) cl 𝐸 is a compact set; is 𝑆-upper semicontinuous on 𝑋. (ii) {𝑥 ∈ cl 𝐸:𝑓(𝑥,𝑦)∉−int 𝑆, ∀𝑦 ∈ cl 𝐸} =0̸ ; 𝑓 𝑆 𝐸× 𝐸 Lemma 2 (see [17]). Let 𝑓:𝑋×𝑋 →𝑉be a vector-valued (iii) is -lower semicontinuous on cl cl . ∗ ∗ mapping and 𝑠 ∈𝑆 \{0}.If𝑓 is 𝑆-lower semicontinuous, then 𝜀∈ 𝑆 𝑆(𝜀) =0̸ ∗ Then, for each int , . 𝑠 ∘𝑓is lower semicontinuous. Next, we give a sufficient condition for the condition (ii) Definition 3 (see [24]). Let 𝐴 and 𝐵 be nonempty subsets of in Lemma 6. 𝑋 and 𝑓:𝐴×𝐵 →𝑉be a vector-valued mapping. Lemma 7. Let 𝐸 be a nonempty subset of 𝑋.Supposethat𝑓: 𝑓 𝑆 𝐴 (i) is said to be -concavelike in its first variable on 𝑋×𝑋 →𝑉is a vector-valued mapping with 𝑓(𝑥, 𝑥) =0 for 𝑥 ,𝑥 ∈𝐴 𝑙∈[0,1] if and only if, for all 1 2 and ,there all 𝑥∈𝑋and the following conditions are satisfied: exists 𝑥∈𝐴such that (i) cl 𝐸 is a compact set; 𝑓(𝑥, 𝑦) ∈ 𝑙𝑓(𝑥 ,𝑦)+(1−𝑙) 𝑓(𝑥 ,𝑦)+𝑆, ∀𝑦∈𝐵. 1 2 (4) (ii) 𝑓 is 𝑆-concavelike-convexlike on cl 𝐸×cl 𝐸; 𝑥∈ 𝐸 𝑓(𝑥, ⋅) 𝑆 𝑓 𝑆 (iii) for each cl , is -lower semicontinuous on (ii) is said to be -convexlike in its second variable on cl 𝐸. 𝐵 if and only if, for all 𝑦1,𝑦2 ∈𝐵and 𝑙∈[0,1],there exists 𝑦∈𝐵such that Then, there exists 𝑥∈cl 𝐸 such that 𝑓 (𝑥,𝑦) ∉− 𝑆, ∀𝑦 ∈ 𝐸. 𝑓(𝑥,𝑦) ∈ 𝑙𝑓(𝑥,1 𝑦 )+(1−𝑙) 𝑓(𝑥,𝑦2)−𝑆, ∀𝑥∈𝐴. (5) int cl (9) ∗ ∗ Proof. For any 𝑡<0and 𝑠 ∈𝑆 \{0}, we define a multifunc- (iii) 𝑓 is said to be 𝑆-concavelike-convexlike on 𝐴×𝐵if cl 𝐸 tion 𝐺:cl 𝐸→2 by and only if 𝑓 is 𝑆-concavelike in its first variable and 𝑆 ∗ -convexlikeinitssecondvariable. 𝐺 (𝑥) ={𝑦∈cl 𝐸:𝑠 (𝑓 (𝑥, 𝑦)) ≤ 𝑡} ,∀𝑥∈ cl 𝐸. (10) Definition 4. Let 𝐴⊂𝑉be a nonempty subset and 𝜀∈int 𝑆. First, by assumptions, we must have (i) A point 𝑧∈𝐴is said to be a weak 𝜀-minimal point ⋂ 𝐺 (𝑥) =0. (11) of 𝐴 if and only if 𝐴 ∩(𝑧 −𝜀 int 𝑆) = 0 and Min𝜀𝐴 𝑥∈cl 𝐸 denotes the set of all weak 𝜀-minimal points of 𝐴. In fact, if there exists 𝑦∈cl 𝐸 such that 𝑦∈𝐺(𝑥),forall (ii) A point 𝑧∈𝐴is said to be a weak 𝜀-maximal point 𝑥∈cl 𝐸,then of 𝐴 if and only if 𝐴 ∩(𝑧 +𝜀 int 𝑆) = 0 and Max𝜀𝐴 𝜀 𝐴 ∗ denotes the set of all weak -maximal points of . 𝑠 (𝑓 (𝑥, 𝑦)) ≤𝑡, ∀𝑥∈cl 𝐸. (12) ∗ Definition 5. Let 𝑓:𝐴×𝐵 →𝑉be a vector-valued mapping Particularly, taking 𝑥=𝑦,wehave0=𝑠 (𝑓(𝑦, 𝑦)) ≤,which 𝑡 and 𝜀∈int 𝑆.Apoint(𝑎, 𝑏) ∈ 𝐴 ×𝐵 is said to be a weak 𝜀-𝑆- contradicts the assumption about 𝑡. saddle point of 𝑓 on 𝐴×𝐵if Then, by Lemma 2, 𝐺(𝑥) is a closed set, for each 𝑥∈cl 𝐸. By (11), for any 𝑦∈cl 𝐸,wehave 𝑓 (𝑎,) 𝑏 ∈ Max𝜀𝑓 (𝐴, 𝑏) ⋂ Min𝜀𝑓 (𝑎,) 𝐵 . (6) 𝑦∈𝑉\ ⋂ 𝐺 (𝑥) = ⋃ 𝑉\𝐺(𝑥) . (13) 𝑥∈cl 𝐸 𝑥∈cl 𝐸 3. Existence of 𝜀-Vector Equilibrium Problem Since cl 𝐸 is compact, there exists a finite point set {𝑥1,𝑥2, 𝜀 In this section, we deal with the following -vector equilib- ...,𝑥𝑛} in cl 𝐸 such that rium problem (for short VAEP). Find 𝑥∈𝐸such that 𝐸⊂ ⋃ 𝑉\𝐺(𝑥). cl 𝑖 (14) 𝑓(𝑥,𝑦)+𝜀∉−int 𝑆, ∀𝑦 ∈𝐸, (7) 1≤𝑖≤𝑛 Abstract and Applied Analysis 3

2 Namely, for each 𝑦∈cl 𝐸, there exists 𝑖 ∈ {1,2,...,𝑛}such Example 10. Let 𝑋=𝑅, 𝑉=𝑅,and𝐸 = [0, 1/3] ∪ [2/3, 1], that 𝑓 (𝑥, 𝑦) = {(𝑥𝑦, 𝑥𝑧)∈𝑅2 |𝑧=1−𝑦2}, 𝑥,𝑦∈𝑋, 𝑠∗ (𝑓 (𝑥 ,𝑦))>𝑡. 𝑖 (15) (23) 𝑆={(𝑥,𝑦)∈𝑅2 |𝑥≥0,𝑦≥0}. Now, we consider the set 𝑛+1 𝐸 𝐸 𝑀:={(𝑧1,𝑧2,...,𝑧𝑛,𝑟)∈𝑅 |∃𝑦∈cl 𝐸, Obviously, cl is a compact set. However, cl is not a convex (16) set. So, Theorem 3.2 in20 [ ] is not applicable. By the definition ∗ 𝑓 𝑓 𝑆 𝐸× 𝐸 𝑆 𝑠 (𝑓 𝑖(𝑥 ,𝑦))≤𝑟+𝑧𝑖,∀𝑖=1,2,...,𝑛}. of , is -concavelike-convexlike on cl cl and - lower semicontinuous on cl 𝐸×cl 𝐸.Thus,allconditionsof Obviously, by the condition (ii), 𝑀 is a convex set. By (15), we Theorem 8 hold. Indeed, for each 𝜀=(𝜀1,𝜀2)∈int 𝑆, have the fact that (0𝑅𝑛 ,𝑡)∉𝑀. 𝑓(0,𝑦)+𝜀=(𝜀,𝜀 )∉− 𝑆, ∀𝑦 ∈𝐸. By the separation theorem of convex sets, there exists 1 2 int (24) (𝜆1,𝜆2,...,𝜆𝑛, 𝑟) =0̸ 𝑅𝑛 such that Namely, 0∈𝑆(𝜀). 𝑛 ∑𝜆𝑖𝑧𝑖 + 𝑟𝑟 ≥ 𝑟𝑡, 1∀ (𝑧 ,𝑧2,...,𝑧𝑛,𝑟)∈𝑀. (17) 𝑖=1 4. Existence of 𝜀-Cone Saddle Points 𝑛+1 𝐸 𝑋 𝐸=𝐴×𝐵 Since 𝑀+𝑅 ⊂𝑀,wecanget𝜆𝑖 ≥0and 𝑟≥0,forall Lemma 11. Let be a nonempty subset of and .Let 𝑖=1,2,...,𝑛. By the definition of 𝑀,foreach𝑦∈cl 𝐸, 𝜀∈int 𝑆 and let 𝑓:𝑋×𝑋 →𝑉be a vector-valued mapping with 𝑓(𝑥, 𝑦) = 𝑔(𝑎, V)−𝑔(𝑢,𝑏),where𝑥 = (𝑎,, 𝑏) 𝑦 = (𝑢, V), ∗ 𝑎, 𝑢 ∈𝐴 V,𝑏∈𝐵 𝑥=(𝑎, 𝑏) ∈ 𝐸 (0𝑅𝑛 ,1+max 𝑠 (𝑓 𝑖(𝑥 , 𝑦))) ∈ int 𝑀, ,and . If there exists such that 1≤𝑖≤𝑛 (18) ∗ 𝑓 (𝑥,𝑦) +𝜀∉−int 𝑆, ∀𝑦 ∈𝐸, (25) (𝑠 (𝑓 1(𝑥 ,𝑦))−𝑟, ∗ ∗ (19) 𝑠 (𝑓 2(𝑥 ,𝑦))−𝑟,...,𝑠 (𝑓 𝑛(𝑥 ,𝑦))−𝑟,𝑟)∈𝑀. then (𝑎, 𝑏) ∈ 𝐴 ×𝐵 is a weak 𝜀-𝑆-saddle point of 𝑔 on 𝐴×𝐵.

By (18), 𝑟>0.Then,by(17)and(19), Proof. By assumptions, we have

𝑛 𝑛 𝜆𝑖 ∗ 𝜆𝑖 𝑓(𝑥, 𝑦) + 𝜀∉− int 𝑆, ∀𝑦 ∈𝐸. (26) ∑ 𝑠 (𝑓 𝑖(𝑥 ,𝑦))+𝑟(1−∑ )≥𝑡. (20) 𝑖=1 𝑟 𝑖=1 𝑟 Then, 𝑛 By (20), ∑ (𝜆𝑖/𝑟) =. 1 Thus, by the condition (ii), for each 𝑖=1 𝑔 (𝑎, V) −𝑔(𝑢, 𝑏) +𝜀∉− 𝑆, ∀ (𝑢, V) ∈𝐴×𝐵. 𝑦∈cl 𝐸, there exists 𝑥∈cl 𝐸 such that int (27) 𝑛 𝑢=𝑎 ∗ 𝜆𝑖 ∗ By (27), taking , 𝑠 (𝑓 (𝑥, 𝑦)) ≥ ∑ 𝑠 (𝑓 𝑖(𝑥 ,𝑦))≥𝑡. (21) 𝑖=1 𝑟 𝑔 (𝑎, V) −𝑔(𝑎, 𝑏) + 𝜀 ∉− int 𝑆, ∀V ∈𝐵, (28) ∗ By the assumption about 𝑡 and 𝑠 , there exists 𝑥∈cl 𝐸 such that which implies 𝑔(𝑎, 𝑏) ∈ Min𝜀𝑔(𝑎, 𝐵).Then,by(27), taking V = 𝑏, 𝑓 (𝑥,𝑦) ∉−int 𝑆, ∀𝑦 ∈ cl 𝐸. (22) 𝑔(𝑎, 𝑏) − 𝑔 (𝑢, 𝑏) + 𝜀 ∉− int 𝑆, ∀𝑢∈𝐴, (29) This completes the proof.

By Lemmas 6 and 7, we can get the following result. which implies 𝑔(𝑎, 𝑏) ∈ Max𝜀𝑔(𝐴, 𝑏).Thus,(𝑎, 𝑏) ∈ 𝐴 ×𝐵 is a weak 𝜀-𝑆-saddle point of 𝑔 on 𝐴×𝐵.Thiscompletesthe 𝐸 𝑋 Theorem 8. Let be a nonempty subset of .Supposethat proof. 𝑓:𝑋×𝑋 →𝑉is a vector-valued mapping with 𝑓(𝑥, 𝑥) =0 for all 𝑥∈𝑋and the following conditions are satisfied: Theorem 12. Let 𝐴 and 𝐵 be nonempty sets and 𝜀∈int 𝑆. Suppose that 𝑔 is a vector-valued mapping and the following (i) cl 𝐸 is a compact set; conditions are satisfied: (ii) 𝑓 is 𝑆-concavelike-convexlike on cl 𝐸×cl 𝐸; 𝐴 𝐵 (iii) 𝑓 is 𝑆-lower semicontinuous on cl 𝐸×cl 𝐸. (i) cl and cl are compact sets; (ii) 𝑔 is 𝑆-concavelike-convexlike on cl 𝐴×cl 𝐵; Then, for each 𝜀∈int 𝑆, 𝑆(𝜀) =0̸ . (iii) 𝑔 is 𝑆-upper semicontinuous on cl 𝐴×cl 𝐵; Remark 9. Note that the condition (i) does not require the (iv) 𝑔 is 𝑆-lower semicontinuous on cl 𝐴×cl 𝐵. fact that cl 𝐸 is a convex set. So Theorem 8 is different from Theorem 3.2 in [20]. The following example explains this case. Then, 𝑔 has a weak 𝜀-𝑆-saddle point on 𝐴×𝐵. 4 Abstract and Applied Analysis

Proof. Let 𝐴×𝐵=𝐸and 𝑓:cl 𝐸×cl 𝐸→𝑉be a vector- Conflict of Interests valued mappings by The author declares that there is no conflict of interests 𝑓(𝑥,𝑦)=𝑔(𝑎, V) −𝑔(𝑢,) 𝑏 ,∀𝑥=(𝑎,) 𝑏 ∈ cl 𝐸, regarding the publication of this paper. (30) 𝑦=(𝑢, V) ∈ cl 𝐸. Acknowledgments Next, we show that all assumptions of Theorem 8 are satisfied by 𝑔. The author would like to thank the anonymous referees Clearly, by the condition (i), cl 𝐸 is compact. Then, by the for their valuable comments and suggestions, which helped condition (ii), we have the fact that, for each 𝑎1,𝑎2 ∈ cl 𝐴 and to improve the paper. This paper is dedicated to Professor 𝑙∈[0,1], there exists 𝑎3 ∈ cl 𝐴 such that Miodrag Mateljevi’c on the occasion of his 65th birthday. 𝑔(𝑎 ,𝑏)∈𝑙𝑔(𝑎 ,𝑏)+(1−𝑙) 𝑔(𝑎 ,𝑏)+𝑆, ∀𝑏∈ 𝐵 3 1 2 cl (31) References 𝑏 ,𝑏 ∈ 𝐵 𝑙∈[0,1] 𝑏 ∈ 𝐵 and, for each 1 2 cl and , there exists 3 cl [1]G.Bigi,A.Capat˘ a,˘ and G. Kassay, “Existence results for 𝑔(𝑎,𝑏 )∈𝑙𝑔(𝑎,𝑏)+(1−𝑙) 𝑔(𝑎,𝑏 )−𝑆, ∀𝑎∈ 𝐴. strong vector equilibrium problems and their applications,” 3 1 2 cl Optimization,vol.61,no.5,pp.567–583,2012. (32) [2] X.-H. Gong, “The strong minimax theorem and strong saddle points of vector-valued functions,” Nonlinear Analysis. Theory, By (31)and(32), for each (𝑎1,𝑏1), (𝑎2,𝑏2)∈cl 𝐸 and 𝑙∈[0,1], Methods & Applications, vol. 68, no. 8, pp. 2228–2241, 2008. there exists (𝑎3,𝑏3)∈cl 𝐸 such that [3] X. Gong, “Strong vector equilibrium problems,” Journal of 𝑔(𝑎3,𝑏)−𝑔(𝑎,𝑏3)∈𝑙(𝑔(𝑎1,𝑏)−𝑔(𝑎,𝑏1)) Global Optimization,vol.36,no.3,pp.339–349,2006. [4] X. B. Li, S. J. Li, and Z. M. Fang, “Aminimax theorem for vector- + (1−𝑙) (𝑔 2(𝑎 ,𝑏)−𝑔(𝑎,𝑏2)) + 𝑆, valued functions in lexicographic order,” Nonlinear Analysis. Theory, Methods & Applications,vol.73,no.4,pp.1101–1108, ∀ (𝑎,) 𝑏 ∈ cl 𝐸, 2010. 𝑔(𝑎,𝑏 )−𝑔(𝑎,𝑏)∈𝑙(𝑔(𝑎,𝑏 )−𝑔(𝑎,𝑏)) [5] J. W. Nieuwenhuis, “Some minimax theorems in vector-valued 3 3 1 1 functions,” JournalofOptimizationTheoryandApplications,vol. 40,no.3,pp.463–475,1983. + (1−𝑙) (𝑔 (𝑎,2 𝑏 )−𝑔(𝑎2,𝑏))−𝑆, [6] T. Tanaka, “A characterization of generalized saddle points for ∀ (𝑎,) 𝑏 ∈ cl 𝐸. vector-valued functions via scalarization,” Nihonkai Mathemat- (33) ical Journal,vol.1,no.2,pp.209–227,1990. [7] T. Tanaka, “Generalized semicontinuity and existence theorems Namely, 𝑓 is 𝑆-concavelike-convexlike on cl 𝐸×cl 𝐸. forconesaddlepoints,”Applied Mathematics and Optimization, Now, we show that 𝑓 is 𝑆-lower semicontinuous on cl 𝐸× vol.36,no.3,pp.313–322,1997. cl 𝐸. By the condition (iii), for each (𝑎, V)∈cl 𝐴×cl 𝐵 and [8] Y. Zhang, S. J. Li, and S. K. Zhu, “Minimax problems for set- 𝑠∈int 𝑆, there exists an open neighborhood 𝑈𝑎 of 𝑎 and 𝑈V valued mappings,” Numerical Functional Analysis and Opti- of V such that mization,vol.33,no.2,pp.239–253,2012. 𝑠 [9] Q.-b. Zhang, M.-j. Liu, and C.-z. Cheng, “Generalized saddle 𝑔(𝑢 ,𝑢 )∈𝑔(𝑎, V) − + 𝑆, ∀𝑢 ∈𝑈,𝑢 ∈𝑈, 𝑎 V 2 int 𝑎 𝑎 V V (34) points theorems for set-valued mappings in locally generalized convex spaces,” Nonlinear Analysis. Theory, Methods & Applica- and, for each (𝑢, 𝑏) ∈ cl 𝐴×cl 𝐵 and 𝑠∈int 𝑆, there exists an tions,vol.71,no.1-2,pp.212–218,2009. open neighborhood 𝑈𝑢 of 𝑢 and 𝑈𝑏 of 𝑏 such that [10] Q.H.Ansari,I.V.Konnov,andJ.C.Yao,“Existenceofasolution 𝑠 and variational principles for vector equilibrium problems,” 𝑔(𝑢𝑢,𝑢𝑏)∈𝑔(𝑢,) 𝑏 + − int 𝑆, 𝑢∀𝑢 ∈𝑈𝑢,𝑢𝑏 ∈𝑈𝑏. JournalofOptimizationTheoryandApplications,vol.110,no.3, 2 pp.481–492,2001. (35) [11] Q. H. Ansari, I. V. Konnov, and J. C. Yao, “Characterizations of ((𝑎, 𝑏), (𝑢, V)) ∈ solutions for vector equilibrium problems,” Journal of Optimiza- By (34)and(35), we have the fact that, for any tion Theory and Applications,vol.113,no.3,pp.435–447,2002. cl 𝐸×cl 𝐸, [12] Q. H. Ansari, X. Q. Yang, and J.-C. Yao, “Existence and duality 𝑔(𝑢𝑎,𝑢V)−𝑔(𝑢𝑢,𝑢𝑏)∈𝑔(𝑎, V) −𝑔(𝑢,) 𝑏 −𝑠+int 𝑆, of implicit vector variational problems,” Numerical Functional (36) Analysis and Optimization,vol.22,no.7-8,pp.815–829,2001. ∀ ((𝑢𝑎,𝑢𝑏),(𝑢𝑢,𝑢V)) ∈ 𝑈𝑎 ×𝑈𝑏 ×𝑈𝑢 ×𝑈V. [13] J. W. Chen, Z. Wan, and Y. J. Cho, “The existence of solutions and well-posedness for bilevel mixed equilibrium problems in Namely, 𝑓 is 𝑆-lower semicontinuous on cl 𝐸×cl 𝐸. Therefore, Banach spaces,” Taiwanese Journal of Mathematics,vol.17,no.2, by Lemma 11, 𝑔 has a weak 𝜀-𝑆-saddle point on 𝐴×𝐵.This pp. 725–748, 2013. completes the proof. [14] Y. J. Cho, S. S. Chang, J. S. Jung, S. M. Kang, and X. Wu, “Minimax theorems in probabilistic metric spaces,” Bulletin of Remark 13. The conditions (iii) and (iv) of Theorem 12 do not the Australian Mathematical Society,vol.51,no.1,pp.103–119, imply that 𝑔 is continuous (see [23]). 1995. Abstract and Applied Analysis 5

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Research Article Fixed Point Theory in 𝛼-Complete Metric Spaces with Applications

N. Hussain,1 M. A. Kutbi,1 and P. Salimi2

1 Department of Mathematics, King Abdulaziz University, P.O. Box 80203, 21589 Jeddah, Saudi Arabia 2 Young Researchers and Elite Club, Rasht Branch, Islamic Azad University, Rasht, Iran

Correspondence should be addressed to P. Salimi; [email protected]

Received 28 November 2013; Accepted 14 December 2013; Published 6 February 2014

Academic Editor: Ljubomir B. Ciri´ c´

Copyright © 2014 N. Hussain 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.

The aim of this paper is to introduce new concepts of 𝛼-𝜂-complete metric space and 𝛼-𝜂-continuous function and establish fixed point results for modified 𝛼-𝜂-𝜓-rational contraction mappings in 𝛼-𝜂-complete metric spaces. As an application, we derive some Suzuki type fixed point theorems and new fixed point theorems for 𝜓-graphic-rational contractions. Moreover, some examples and an application to integral equations are given here to illustrate the usability of the obtained results.

ThispaperisdedicatedtoProfessorMiodragMateljevi´c on the occasion of his 65th birthday

1. Preliminaries Assume that there exists 𝑟∈[0,1)such that We know by the Banach contraction principle [1], which is 𝜃 (𝑟) 𝑑 (𝑥, 𝑇𝑥) ≤𝑑(𝑥,𝑦) 𝑖𝑚𝑝𝑙𝑖𝑒𝑠𝑑(𝑇𝑥,𝑇𝑦)≤𝑟𝑑(𝑥,𝑦) a classical and powerful tool in nonlinear analysis, that a self-mapping 𝑓 on a complete metric space (𝑋, 𝑑) such that (2) 𝑑(𝑓𝑥, 𝑓𝑦) ≤ 𝑐 𝑑(𝑥,𝑦) for all 𝑥, 𝑦,where ∈𝑋 𝑐∈[0,1),hasa unique fixed point. Since then, the Banach contraction prin- for all 𝑥, 𝑦. ∈𝑋 Then there exists a unique fixed point 𝑧 of 𝑇. 𝑛 ciple has been generalized in several directions (see [2–26] Moreover, lim𝑛→+∞𝑇 𝑥=𝑧for all 𝑥∈𝑋. and references cited therein). In 2008, Suzuki [21]provedthefollowingresultthatisan In 2012, Samet et al. [19] introduced the concepts of 𝛼- interesting generalization of the Banach contraction principle 𝜓-contractive and 𝛼-admissible mappings and established which also characterizes the metric completeness. variousfixedpointtheoremsforsuchmappingsdefinedon complete metric spaces. Afterwards Salimi et al. [16]and (𝑋, 𝑑) 𝑇 Theorem 1. Let be a complete metric space and let Hussain et al. [7] modified the notions of 𝛼-𝜓-contractive and be a self-mapping on 𝑋. Define a nonincreasing function 𝜃: 𝛼 [0,1) → (1/2,1] -admissiblemappingsandestablishedfixedpointtheorems by which are proper generalizations of the recent results in [12, 19]. (√5−1) { {1, 𝑖𝑓 0 ≤ 𝑟≤ , 𝑇 𝑋 { 2 Definition 2 (see [19]). Let be a self-mapping on and let { (√5−1) 𝛼:𝑋×𝑋 → [0,+∞)be a function. One says that 𝑇 is an 𝜃 (𝑟) = { −2 −1/2 (1) {(1−𝑟) 𝑟 ,𝑖𝑓 <𝑟<2 , 𝛼-admissible mapping if { 2 { (1+𝑟)−1,𝑖𝑓2−1/2 ≤𝑟<1. { 𝑥, 𝑦 ∈ 𝑋, 𝛼 (𝑥,) 𝑦 ≥1󳨐⇒𝛼(𝑇𝑥, 𝑇𝑦) ≥1. (3) 2 Abstract and Applied Analysis

Definition 3 (see [16]). Let 𝑇 be a self-mapping on 𝑋 and let where 𝛼(𝑥𝑛,𝑥𝑛+1)≥𝜂(𝑥𝑛,𝑥𝑛+1) for all 𝑛∈N,then{𝑥𝑛}⊆ 𝛼, 𝜂 : 𝑋 × 𝑋 → [0, +∞) be two functions. One says that 𝑇 is 𝑂(𝑤).Now,since𝑋 is an orbitally 𝑇-complete metric space, an 𝛼-admissible mapping with respect to 𝜂 if then {𝑥𝑛} converges in 𝑋.Thatis,(𝑋, 𝑑) is an 𝛼-𝜂-complete 𝛼(𝑥, 𝑦) ≥ 𝜂(𝑥, 𝑦) 𝑥, 𝑦 ∈ 𝑥, 𝑦 ∈ 𝑋, 𝛼 (𝑥, 𝑦) ≥ 𝜂 (𝑥, 𝑦) 󳨐⇒ 𝛼 (𝑇𝑥, 𝑇𝑦)≥𝜂(𝑇𝑥,. metric space. Also, suppose that ;then 𝑂(𝑤).Hence,𝑇𝑥, 𝑇𝑦 ∈ 𝑂(𝑤).Thatis,𝛼(𝑇𝑥, 𝑇𝑦) ≥ 𝜂(𝑇𝑥,. 𝑇𝑦) (4) Thus, 𝑇 is an 𝛼-admissible mapping with respect to 𝜂. Note that if we take 𝜂(𝑥, 𝑦), =1 then this definition reduces 𝛼(𝑥, 𝑦) =1 Definition 7. Let (𝑋, 𝑑) be a metric space. Let 𝛼, 𝜂 : 𝑋 ×𝑋→ to Definition 2.Also,ifwetake ,thenwesaythat [0, ∞) 𝑇:𝑋 →𝑋 𝑇 𝛼 𝜂 𝑇 is an 𝜂-subadmissible mapping. and .Onesays is an - -continuous mapping on (𝑋, 𝑑),ifforgiven𝑥∈𝑋and sequence {𝑥𝑛} with Here we introduce the notions of 𝛼-𝜂-complete metric 𝑥 󳨀→ 𝑥, 𝑛󳨀→∞, space and 𝛼-𝜂-continuous function and establish fixed point 𝑛 as 𝛼 𝜂 𝜓 𝛼 𝜂 results for modified - - -rational contractions in - - 𝛼(𝑥 ,𝑥 )≥𝜂(𝑥,𝑥 ), ∀𝑛∈N 󳨐⇒ 𝑇 𝑥 󳨀→ 𝑇 𝑥. complete metric spaces which are not necessarily complete. 𝑛 𝑛+1 𝑛 𝑛+1 𝑛 Asanapplication,wederivesomeSuzukitypefixedpointthe- (7) orems and new fixed point theorems for 𝜓-graphic-rational Example 8. Let 𝑋=[0,∞)and 𝑑(𝑥, 𝑦) = |𝑥 −𝑦| be a metric contractions. Moreover, some examples and an application to 𝑋 𝑇:𝑋 →𝑋 𝛼, 𝜂 : 𝑋 × 𝑋 → [0, +∞) integral equations are given here to illustrate the usability of on . Assume that and the obtained results. be defined by 𝑥5, 𝑥∈[0, 1] , 2. Main Results 𝑇𝑥 ={ if sin 𝜋𝑥 + 2, if (1, ∞) , First, we introduce the notions of 𝛼-𝜂-complete metric space 𝑥2 +𝑦2 +1, 𝑥, 𝑦 ∈ [0, 1] , and 𝛼-𝜂-continuous function. 𝛼(𝑥,𝑦)={ if (8) 0, otherwise, Definition 4. Let (𝑋, 𝑑) be a metric space and 𝛼,𝜂 : 𝑋×𝑋 → [0, +∞).Themetricspace𝑋 is said to be 𝛼-𝜂-complete if 𝜂(𝑥,𝑦)=𝑥2. and only if every Cauchy sequence {𝑥𝑛} with 𝛼(𝑥𝑛,𝑥𝑛+1)≥ 𝜂(𝑥𝑛,𝑥𝑛+1) for all 𝑛∈N converges in 𝑋.Onesays𝑋 is an 𝛼- Clearly, 𝑇 is not continuous, but 𝑇 is 𝛼-𝜂-continuous on 𝜂(𝑥, 𝑦) =1 𝑥, 𝑦 ∈𝑋 complete metric space when for all and (𝑋, 𝑑).Indeed,if𝑥𝑛 →𝑥as 𝑛→∞and 𝛼(𝑥𝑛,𝑥𝑛+1)≥𝜂(𝑥𝑛, (𝑋, 𝑑) 𝜂 𝛼(𝑥, 𝑦) = 5 one says is an -complete metric space when 𝑥𝑛+1),then𝑥𝑛 ∈ [0, 1] and so lim𝑛→∞𝑇𝑥𝑛 = lim𝑛→∞𝑥𝑛 = 1 𝑥, 𝑦 ∈𝑋 5 for all . 𝑥 =𝑇𝑥. 𝑋 = (0, ∞) 𝑑(𝑥, 𝑦) = |𝑥 −𝑦| Example 5. Let and be a metric Remark 9. Define (𝑋, 𝑑) and 𝛼, 𝜂 : 𝑋 × 𝑋 → [0, +∞) as in 𝑋 𝐴 𝑋 𝛼, 𝜂 : function on .Let be a closed subset of .Define Remark 6.Let𝑇:𝑋be →𝑋 a an orbitally continuous map 𝑋×𝑋 → [0,+∞) by on (𝑋, 𝑑).Then𝑇 is 𝛼-𝜂-continuous on (𝑋, 𝑑). Indeed if 𝑥𝑛 → 2 𝑥 as 𝑛→∞and 𝛼(𝑥𝑛,𝑥𝑛+1)≥𝜂(𝑥𝑛,𝑥𝑛+1) for all 𝑛∈N,so (𝑥 + 𝑦) , if 𝑥, 𝑦 ∈ 𝐴, 𝛼(𝑥,𝑦)={ 𝑥𝑛 ∈ 𝑂(𝑤) for all 𝑛∈N, then there exists sequence (𝑘𝑖)𝑖∈N 0, otherwise, 𝑘𝑖 (5) of positive integer such that 𝑥𝑛 =𝑇𝑤→𝑥as 𝑖→∞. 𝑇 (𝑋, 𝑑) 𝜂(𝑥,𝑦)=2𝑥𝑦. Now since is an orbitally continuous map on ,then 𝑘𝑖 𝑇𝑥𝑛 =𝑇(𝑇 𝑤) → 𝑇𝑥 as 𝑖→∞as required. Clearly, (𝑋, 𝑑) is not a complete metric space, but (𝑋, 𝑑) is 𝜓 : [0, ∞) → [0, ∞) an 𝛼-𝜂-complete metric space. Indeed, if {𝑥𝑛} is a Cauchy Afunction is called Bianchini- sequence in 𝑋 such that 𝛼(𝑥𝑛,𝑥𝑛+1)≥𝜂(𝑥𝑛,𝑥𝑛+1) for all Grandolfi gauge function13 [ , 14, 27]ifthefollowingcondi- 𝑛∈N,then𝑥𝑛 ∈𝐴for all 𝑛∈N.Now,since(𝐴, 𝑑) is a tions hold: ∗ complete metric space, then there exists 𝑥 ∈𝐴such that ∗ 𝜓 𝑥𝑛 →𝑥 as 𝑛→∞. (i) is nondecreasing; (ii) there exist 𝑘0 ∈ N and 𝑎 ∈ (0, 1) and a convergent Remark 6. Let 𝑇:𝑋be →𝑋 a self-mapping on metric space ∞ series of nonnegative terms ∑𝑘=1 V𝑘 such that 𝑋 and let 𝑋 be an orbitally 𝑇-complete. Define 𝛼,𝜂 : 𝑋×𝑋 → [0, +∞) by 𝑘+1 𝑘 𝜓 (𝑡) ≤𝑎𝜓 (𝑡) + V𝑘, (9) 3, if 𝑥, 𝑦 ∈𝑂 (𝑤) , 𝛼(𝑥,𝑦)={ + 0, , for 𝑘≥𝑘0 and any 𝑡∈R . otherwise (6) 𝜂(𝑥,𝑦)=1, In some sources, Bianchini-Grandolfi gauge function is known as (𝑐)—comparisonfunction(seee.g.,[2]).Wedenote where 𝑂(𝑤) isanorbitofapoint𝑤∈𝑋.Then(𝑋, 𝑑) is an 𝛼-𝜂- by Ψ the family of Bianchini-Grandolfi gauge functions. The complete metric space. Indeed, if {𝑥𝑛} be a Cauchy sequence, following lemma illustrates the properties of these functions. Abstract and Applied Analysis 3

2 2 Lemma 10 (see [2]). If 𝜓∈Ψ, then the following hold: that 𝛼(𝑥1,𝑥2)=𝛼(𝑇𝑥0,𝑇 𝑥0)≥𝜂(𝑇𝑥0,𝑇 𝑥0)=𝜂(𝑥1,𝑥2). 𝑛 + Continuing this process, we get (i) (𝜓 (𝑡))𝑛∈N converges to 0 as 𝑛→∞for all 𝑡∈R ; 𝛼(𝑥𝑛,𝑥𝑛+1)≥𝜂(𝑥𝑛,𝑥𝑛+1)=𝜂(𝑥𝑛,𝑇𝑥𝑛) (13) (ii) 𝜓(𝑡) <𝑡,forany𝑡 ∈ (0, ∞); for all 𝑛∈N ∪{0}. Now, by (a) we get (iii) 𝜓 is continuous at 0; ∞ 𝑘 + 𝑑 (𝑥𝑛,𝑥𝑛+1) =𝑑(𝑇𝑥𝑛−1,𝑇𝑥𝑛) ≤𝜓(𝑀 (𝑥𝑛−1,𝑥𝑛)) , (14) (iv) the series ∑𝑘=1 𝜓 (𝑡) converges for any 𝑡∈R . where Definition 11. Let (𝑋, 𝑑) be a metric space and let 𝑇 be a self- 𝑋 mapping on .Let 𝑀(𝑥𝑛−1,𝑥𝑛)=max {𝑑 𝑛−1(𝑥 ,𝑥𝑛), 𝑑 (𝑥, 𝑇𝑥) 𝑑(𝑦,𝑇𝑦) 𝑀(𝑥,𝑦)= {𝑑 (𝑥, 𝑦), , , 𝑑(𝑥 ,𝑇𝑥 ) 𝑑(𝑥 ,𝑇𝑥 ) max 1+𝑑(𝑥, 𝑇𝑥) 1+𝑑(𝑦,𝑇𝑦) 𝑛−1 𝑛−1 , 𝑛 𝑛 , 1+𝑑(𝑥𝑛−1,𝑇𝑥𝑛−1) 1+𝑑(𝑥𝑛,𝑇𝑥𝑛) 𝑑(𝑥,𝑇𝑦)+𝑑(𝑦,𝑇𝑥) }. 𝑑(𝑥 ,𝑇𝑥 )+𝑑(𝑥 ,𝑇𝑥 ) 2 𝑛−1 𝑛 𝑛 𝑛−1 } 2 (10)

𝑑(𝑥𝑛−1,𝑥𝑛) Then, = max {𝑑 𝑛−1(𝑥 ,𝑥𝑛), , 1+𝑑(𝑥𝑛−1,𝑥𝑛) 𝑇 𝛼 𝜂 𝜓 (a) we say is a modified - - -rational contraction 𝑑(𝑥 ,𝑥 ) 𝑑(𝑥 ,𝑥 ) mapping if 𝑛 𝑛+1 , 𝑛−1 𝑛+1 } 1+𝑑(𝑥𝑛,𝑥𝑛+1) 2 𝑥, 𝑦 ∈ 𝑋, (11) ≤ {𝑑 (𝑥 ,𝑥 ),𝑑(𝑥 ,𝑥 ), 𝜂 (𝑥, 𝑇𝑥) ≤ 𝛼 (𝑥, 𝑦) 󳨐⇒ 𝑑 (𝑇𝑥, 𝑇𝑦) ≤𝜓(𝑀(𝑥,𝑦)), max 𝑛−1 𝑛 𝑛 𝑛+1

𝜓∈Ψ 𝑑(𝑥 ,𝑥 )+𝑑(𝑥 ,𝑥 ) where ; 𝑛−1 𝑛 𝑛 𝑛+1 } (b) we say 𝑇 is a modified 𝛼-𝜓-rational contraction 2 mapping if = max {𝑑 𝑛−1(𝑥 ,𝑥𝑛),𝑑(𝑥𝑛,𝑥𝑛+1)} 𝑥,𝑦∈𝑋, 𝛼(𝑥,𝑦)≥1󳨐⇒𝑑(𝑇𝑥,𝑇𝑦)≤𝜓(𝑀(𝑥,𝑦)), (15)

(12) and so, 𝑀(𝑥𝑛−1,𝑥𝑛)≤max{𝑑(𝑥𝑛−1,𝑥𝑛), 𝑑(𝑥𝑛,𝑥𝑛+1)}.Now since 𝜓 is nondecreasing, so from (14), we have where 𝜓∈Ψ. 𝑑 (𝑥𝑛,𝑥𝑛+1) ≤𝜓(max {𝑑 (𝑥𝑛−1,𝑥𝑛) ,𝑑(𝑥𝑛,𝑥𝑛+1)}) . (16) The following is our first main result of this section. Now, if max{𝑑(𝑥𝑛−1,𝑥𝑛), 𝑑(𝑥𝑛,𝑥𝑛+1)} = 𝑑(𝑥𝑛,𝑥𝑛+1) for some 𝑛∈N Theorem 12. Let (𝑋, 𝑑) be a metric space and let 𝑇 be a self- ,then 𝑋 𝛼, 𝜂 : 𝑋 × 𝑋 → [0,∞) mapping on .Also,supposethat are 𝑑(𝑥𝑛,𝑥𝑛+1)≤𝜓(max {𝑑 𝑛−1(𝑥 ,𝑥𝑛),𝑑(𝑥𝑛,𝑥𝑛+1)}) two functions and 𝜓∈Ψ. Assume that the following assertions (17) hold true: = 𝜓 (𝑑𝑛 (𝑥 ,𝑥𝑛+1)) < 𝑑 𝑛(𝑥 ,𝑥𝑛+1) 𝑛∈N (i) (𝑋, 𝑑) is an 𝛼-𝜂-complete metric space; which is a contradiction. Hence, for all we have (ii) 𝑇 is an 𝛼-admissible mapping with respect to 𝜂; 𝑑 (𝑥𝑛,𝑥𝑛+1) ≤𝜓(𝑑 (𝑥𝑛−1,𝑥𝑛)) . (18) (iii) 𝑇 is modified 𝛼-𝜂-𝜓-rational contraction mapping on By induction, we have 𝑋; 𝑛 𝑑(𝑥𝑛,𝑥𝑛+1)≤𝜓 (𝑑 0(𝑥 ,𝑥1)) . (19) (iv) 𝑇 is an 𝛼-𝜂-continuous mapping on 𝑋; Fix 𝜖>0; there exists 𝑁∈N such that (v) there exists 𝑥0 ∈𝑋such that 𝛼(𝑥0,𝑇𝑥0)≥𝜂(𝑥0,𝑇𝑥0). ∑ 𝜓𝑛 (𝑑 (𝑥 ,𝑥 )) < 𝜖. 0 1 (20) Then 𝑇 has a fixed point. 𝑛≥𝑁 Let 𝑚, 𝑛 ∈ N with 𝑚>𝑛≥𝑁.Thenbytriangularinequality Proof. Let 𝑥0 ∈𝑋be such that 𝛼(𝑥0,𝑇𝑥0)≥𝜂(𝑥0,𝑇𝑥0). 𝑛 we get Define a sequence {𝑥𝑛} in 𝑋 by 𝑥𝑛 =𝑇𝑥0 =𝑇𝑥𝑛−1 for all 𝑛∈N 𝑥 =𝑥 𝑛∈N 𝑥=𝑥 𝑚−1 .If 𝑛+1 𝑛 for some ,then 𝑛 is a 𝑛 fixed point for 𝑇 and the result is proved. Hence, we suppose 𝑑(𝑥𝑛,𝑥𝑚)≤ ∑ 𝑑(𝑥𝑘,𝑥𝑘+1)≤ ∑ 𝜓 (𝑑 0(𝑥 ,𝑥1)) < 𝜖. 𝑘=𝑛 𝑛≥𝑁 that 𝑥𝑛+1 =𝑥̸ 𝑛 for all 𝑛∈N.Since𝑇 is 𝛼-admissible mapping with respect to 𝜂 and 𝛼(𝑥0,𝑇𝑥0)≥𝜂(𝑥0,𝑇𝑥0),wededuce (21) 4 Abstract and Applied Analysis

Consequently lim𝑚,𝑛, → +∞𝑑(𝑥𝑛,𝑥𝑚)=0.Hence{𝑥𝑛} is a (i) let 𝑥, 𝑦 ∈ [−1, 0) with 𝑥 =𝑦̸ ;then, Cauchy sequence. On the other hand from (13)weknowthat 1 𝛼(𝑥𝑛,𝑥𝑛+1)≥𝜂(𝑥𝑛,𝑥𝑛+1) for all 𝑛∈N.Nowsince𝑋 is an 𝛼- 𝑑(𝑇𝑥,𝑇𝑦)= {𝑥2,𝑦2} 4 max 𝜂-complete metric space, there is 𝑧∈𝑋such that 𝑥𝑛 →𝑧 as 𝑛→∞.Also,since𝑇 is an 𝛼-𝜂-continuous mapping, 1 󵄨 󵄨 ≤ max {|𝑥| , 󵄨𝑦󵄨}=𝜓(𝑑(𝑥,𝑦))≤𝜓(𝑀(𝑥,𝑦)); so 𝑥𝑛+1 =𝑇𝑥𝑛 →𝑇𝑧as 𝑛→∞.Thatis,𝑧=𝑇𝑧as 2 󵄨 󵄨 required. (24) 𝑋 = (−∞, −2)∪[−1, 1]∪(2, +∞) Example 13. Let .Weendow (ii) let 𝑥, 𝑦 ∈ (0, 1] with 𝑥 =𝑦̸ ;then 𝑋 with the metric 1 󵄨 󵄨 𝑑(𝑇𝑥,𝑇𝑦)= {|𝑥| , 󵄨𝑦󵄨} 4 max 󵄨 󵄨 󵄨 󵄨 max {|𝑥| , 󵄨𝑦󵄨}, if 𝑥 =𝑦,̸ 𝑑(𝑥,𝑦)={ 1 󵄨 󵄨 (22) ≤ {|𝑥| , 󵄨𝑦󵄨}=𝜓(𝑑(𝑥,𝑦))≤𝜓(𝑀(𝑥,𝑦)); 0, 𝑥=𝑦. 2 max 󵄨 󵄨 (25) 𝑇:𝑋 →𝑋𝛼, 𝜂 : 𝑋 × 𝑋 → [0,∞) 𝜓:[0, Define , ,and (iii) let 𝑥 ∈ (−1, 0) and 𝑦 ∈ (0, 1);then ∞) → [0, ∞) by 1 𝑑(𝑇𝑥,𝑇𝑦)= {𝑥2,𝑦} 4 max {√2𝑥2 −1, if 𝑥∈(−∞, −3] , { 1 󵄨 󵄨 { ≤ {|𝑥| , 󵄨𝑦󵄨}=𝜓(𝑑(𝑥,𝑦))≤𝜓(𝑀(𝑥,𝑦)) {𝑥3 −1, 𝑥∈(−3, −2) , 2 max 󵄨 󵄨 { if { (26) {1 2 { 𝑥 , if 𝑥∈[−1, 0] , 𝑇𝑥 = 4 { (iv) let 𝑥=𝑦∈[−1,0), 𝑥 = 𝑦 ∈ (0, 1] or let 𝑥=−1, 𝑦=1; {1 { 𝑥, 𝑥∈(0, 1] , then, 𝑇𝑥 = 𝑇𝑦.Thatis, {4 if { {5+ 𝜋𝑥, 𝑥∈(2, 4) , 𝑑(𝑇𝑥,𝑇𝑦)=0≤𝜓(𝑀(𝑥,𝑦)). { sin if (27) { 3 (23) {3𝑥 + ln 𝑥+1, if 𝑥∈[4, ∞) , Thus 𝑇 is a modified 𝛼-𝜂-𝜓-rational contraction mapping. 2 2 𝑇 𝑥 +𝑦 +1, if 𝑥, 𝑦 ∈ [−1, 1] , Hence all conditions of Theorem 12 aresatisfiedand has 𝛼(𝑥,𝑦)={ 2 𝑥=0 𝑇 𝑥 , otherwise, afixedpoint.Here, is fixed point of . 𝜂(𝑥, 𝑦) =1 𝑥, 𝑦 ∈𝑋 𝜂(𝑥,𝑦)=𝑥2 +𝑦2, By taking for all in Theorem 12,we obtain the following corollary. 1 𝜓 (𝑡) = 𝑡. (𝑋, 𝑑) 𝑇 2 Corollary 14. Let be a metric space and let be a self- mapping on 𝑋.Also,supposethat𝛼:𝑋×𝑋→ [0,∞)is a function and 𝜓∈Ψ. Assume that the following assertions hold Clearly, (𝑋, 𝑑) is not a complete metric space. However, it true: is an 𝛼-𝜂-complete metric space. In fact, if {𝑥𝑛} is a Cauchy (i) (𝑋, 𝑑) is an 𝛼-complete metric space; sequence such that 𝛼(𝑥𝑛,𝑥𝑛+1)≥𝜂(𝑥𝑛,𝑥𝑛+1) for all 𝑛∈N, 𝑇 𝛼 then {𝑥𝑛} ⊆ [−1, 1] for all 𝑛∈N.Now,since([−1, 1], 𝑑) is (ii) is an -admissible mapping; a complete metric space, then the sequence {𝑥𝑛} converges (iii) 𝑇 is a modified 𝛼-𝜓-rational contraction on 𝑋; [−1, 1] ⊆ 𝑋 𝛼(𝑥, 𝑦) ≥ 𝜂(𝑥, 𝑦) 𝑥, 𝑦 ∈ [−1, 1] in .Let ;then . (iv) 𝑇 is an 𝛼-continuous mapping on 𝑋; On the other hand, 𝑇𝑤 ∈ [−1, 1] for all 𝑤 ∈ [−1, 1].Then, 𝑥 ∈𝑋 𝛼(𝑥 ,𝑇𝑥 )≥1 𝛼(𝑇𝑥, 𝑇𝑦) ≥ 𝜂(𝑇𝑥,.Thatis, 𝑇𝑦) 𝑇 is an 𝛼-admissible mapping (v) there exists 0 such that 0 0 . 𝜂 {𝑥 } 𝑥 →𝑥 with respect to .Let 𝑛 be a sequence, such that 𝑛 Then 𝑇 has a fixed point. as 𝑛→∞and 𝛼(𝑥𝑛+1,𝑥𝑛)≥𝜂(𝑥𝑛,𝑥𝑛+1) for all 𝑛∈N. Then, {𝑥𝑛} ⊆ [−1, 1] for all 𝑛∈N.So,{𝑇𝑥𝑛} ⊆ [−1, 1] Theorem 15. Let (𝑋, 𝑑) be a metric space and let 𝑇 be a self- (since 𝑇𝑤 ∈ [−1, 1] for all 𝑤 ∈ [−1, 1]). Now, since 𝑇 is mapping on 𝑋.Also,supposethat𝛼, 𝜂 : 𝑋 × 𝑋 → [0,∞) are continuous on [−1, 1].Then,𝑇𝑥𝑛 →𝑇𝑥as 𝑛→∞.Thatis, two functions and 𝜓∈Ψ. Assume that the following assertions 𝑇 is an 𝛼-𝜂-continuous mapping. Clearly, 𝛼(0, 𝑇0) ≥ 𝜂(0, 𝑇0). hold true: Let 𝛼(𝑥, 𝑦) ≥ 𝜂(𝑥, 𝑇𝑥).Now,if𝑥 ∉ [−1, 1] or 𝑦 ∉ [−1, 1], 2 2 2 2 (i) (𝑋, 𝑑) is an 𝛼-𝜂-complete metric space; then 𝑥 ≥𝑥 +𝑦 +1which implies 𝑦 +1 ≤ 0which is 𝑇 𝛼 𝜂 a contradiction. Then, 𝑥, 𝑦 ∈ [−1, 1]. Now we consider the (ii) is an -admissible mapping with respect to ; following cases: (iii) 𝑇 is a modified 𝛼-𝜂-𝜓-rational contraction on 𝑋; Abstract and Applied Analysis 5

(iv) there exists 𝑥0 ∈𝑋such that 𝛼(𝑥0,𝑇𝑥0)≥𝜂(𝑥0,𝑇𝑥0); which is a contradiction. Hence, 𝑑(𝑧, 𝑇𝑧) =0 implies 𝑧= 𝑇𝑧 𝑧=𝑇𝑧 (v) if {𝑥𝑛} is a sequence in 𝑋 such that 𝛼(𝑥𝑛,𝑥𝑛+1)≥ . By the similar method we can show that if 𝜂(𝑥 ,𝑥 )≤𝛼(𝑥 ,𝑧) 𝑛∈N 𝜂(𝑥𝑛,𝑥𝑛+1) with 𝑥𝑛 →𝑥as 𝑛→∞, then either 𝑛+1 𝑛+2 𝑛+1 holds for all . 2 𝑋=(0,+∞) 𝑋 𝜂(𝑇𝑥𝑛,𝑇 𝑥𝑛)≤𝛼(𝑇𝑥𝑛,𝑥) Example 16. Let .Weendow with usual met- (28) ric. Define 𝑇:𝑋, →𝑋 𝛼, 𝜂 : 𝑋 × 𝑋 →,and [0,∞) 𝜓:[0, 2 3 2 ∞) → [0, ∞) 𝑜𝑟 𝜂 (𝑇 𝑥𝑛,𝑇 𝑥𝑛)≤𝛼(𝑇𝑥𝑛,𝑥) by √𝑥2 +1 holds for all 𝑛∈N. { , 𝑥∈(0, 1) , { 𝑥+ 𝑥+3 if 𝑇 {sin cos Then has a fixed point. { 1 𝑇𝑥 = 𝑥2 +1, 𝑥∈[1, 2] , {16 if Proof. Let 𝑥0 ∈𝑋be such that 𝛼(𝑥0,𝑇𝑥0)≥𝜂(𝑥0,𝑇𝑥0). { 𝑛 { 3 Define a sequence {𝑥𝑛} in 𝑋 by 𝑥𝑛 =𝑇 𝑥0 =𝑇𝑥𝑛−1 for all 𝑛∈ { 𝑥 +1 N 𝛼(𝑥 ,𝑥 )≥ , if 𝑥∈(2, ∞) , . Now as in the proof of Theorem 12 we have 𝑛+1 𝑛 {√𝑥2 +1 (33) 𝜂(𝑥𝑛+1,𝑥𝑛) for all 𝑛∈N and there exists 𝑧∈𝑋such that 𝑥 →𝑧 𝑛→∞ 𝑑(𝑧, 𝑇𝑧) =0̸ 1 𝑛 as .Let . From (v) either { , if 𝑥, 𝑦 ∈ [1, 2] , 𝛼(𝑥,𝑦)= 2 𝜂(𝑇𝑥 ,𝑇2𝑥 )≤𝛼(𝑇𝑥 ,𝑧) { 𝑛−1 𝑛−1 𝑛−1 0, , (29) { otherwise 2 3 2 or 𝜂(𝑇 𝑥𝑛−1,𝑇 𝑥𝑛−1)≤𝛼(𝑇𝑥𝑛−1,𝑧) 1 1 𝜂(𝑥,𝑦)= ,𝜓(𝑡) = 𝑡. holds for all 𝑛∈N.Then, 4 4 𝜂(𝑥 ,𝑥 )≤𝛼(𝑥,𝑧) Note that (𝑋, 𝑑) is not a complete metric space. But it is an 𝛼- 𝑛 𝑛+1 𝑛 𝜂 {𝑥 } (30) -complete metric space. Indeed, if 𝑛 is a Cauchy sequence 𝛼(𝑥 ,𝑥 )≥𝜂(𝑥,𝑥 ) 𝑛∈N {𝑥 }⊆ or 𝜂(𝑥𝑛+1,𝑥𝑛+2)≤𝛼(𝑥𝑛+1,𝑧) such that 𝑛 𝑛+1 𝑛 𝑛+1 for all ,then 𝑛 [1, 2] for all 𝑛∈N.Now,since([1, 2], 𝑑) is a complete metric 𝑛∈N 𝜂(𝑥 ,𝑥 )≤𝛼(𝑥,𝑧) holds for all .Let 𝑛 𝑛+1 𝑛 hold for all space, then the sequence {𝑥𝑛} converges in [1, 2] ⊆ 𝑋.Let 𝑛∈N. Now from (a) we get 𝛼(𝑥, 𝑦) ≥ 𝜂(𝑥,;then 𝑦) 𝑥, 𝑦 ∈ [1,. 2] On the other hand, 𝑇𝑤 ∈ [1, 2] for all 𝑤∈[1,2].Then,𝛼(𝑇𝑥, 𝑇𝑦) ≥ 𝜂(𝑇𝑥,.Thatis, 𝑇𝑦) 𝑑(𝑥𝑛 +1,𝑇𝑧) 𝑘 𝑇 is an 𝛼-admissible mapping with respect to 𝜂.If{𝑥𝑛} is a sequence in 𝑋 such that 𝛼(𝑥𝑛,𝑥𝑛+1)≥𝜂(𝑥𝑛,𝑥𝑛+1) with 𝑥𝑛 → =𝑑(𝑇𝑥𝑛 ,𝑇𝑧) 2 3 𝑘 𝑥 as 𝑛→∞.Then,𝑇𝑥𝑛,𝑇 𝑥𝑛,𝑇 𝑥𝑛 ∈ [1, 2] for all 𝑛∈N. That is, 𝑑(𝑥𝑛 ,𝑇𝑥𝑛 ) 𝑘 𝑘 2 ≤𝜓( {𝑑 𝑛(𝑥 ,𝑧), , max 𝑘 𝜂(𝑇𝑥𝑛,𝑇 𝑥𝑛)≤𝛼(𝑇𝑥𝑛,𝑥), 1+𝑑(𝑥𝑛 ,𝑇𝑥𝑛 ) 𝑘 𝑘 (34) 2 3 2 𝜂(𝑇 𝑥𝑛,𝑇 𝑥𝑛)≤𝛼(𝑇𝑥𝑛,𝑥), 𝑑 (𝑧,𝑇𝑧) 𝑑(𝑥𝑛 ,𝑇𝑧)+𝑑(𝑧,𝑇𝑥𝑛 ) , 𝑘 𝑘 }) 1+𝑑(𝑧,𝑇𝑧) 2 holds for all 𝑛∈N.Clearly,𝛼(0, 𝑇0) ≥ 𝜂(0, 𝑇0).Let,𝛼(𝑥, 𝑦) ≥ 𝜂(𝑥, 𝑇𝑥).Now,if𝑥∉[1,2]or 𝑦∉[1,2],then0≥1/4,which 𝑑(𝑥 ,𝑥 ) is a contradiction. So, 𝑥, 𝑦 ∈ [1,. 2] Therefore, 𝑛𝑘 𝑛𝑘+1 =𝜓(max {𝑑 𝑛(𝑥 ,𝑧), , 𝑘 1+𝑑(𝑥 ,𝑥 ) 1 󵄨 2 2󵄨 𝑛𝑘 𝑛𝑘+1 𝑑(𝑇𝑥,𝑇𝑦)= 󵄨𝑥 −𝑦 󵄨 16 󵄨 󵄨 𝑑(𝑥 ,𝑇𝑧)+𝑑(𝑧,𝑥 ) 𝑑 (𝑧,𝑇𝑧) 𝑛𝑘 𝑛𝑘+1 1 󵄨 󵄨 󵄨 󵄨 1 󵄨 󵄨 , }) = 󵄨𝑥−𝑦󵄨 󵄨𝑥+𝑦󵄨 ≤ 󵄨𝑥−𝑦󵄨 1+𝑑(𝑧,𝑇𝑧) 2 16 󵄨 󵄨 󵄨 󵄨 4 󵄨 󵄨 (35) 1 1 𝑑(𝑥𝑛 ,𝑥𝑛 +1) = 𝑑(𝑥,𝑦)≤ 𝑀 (𝑥, 𝑦) = 𝜓 (𝑀 (𝑥,𝑦)). < {𝑑 (𝑥 ,𝑧), 𝑘 𝑘 , 4 4 max 𝑛𝑘 1+𝑑(𝑥𝑛 ,𝑥𝑛 +1) 𝑘 𝑘 Therefore 𝑇 is a modified 𝛼-𝜂-𝜓-rational contraction mapping. Hence all conditions of Theorem 15 hold and 𝑇 has a 𝑑 (𝑧,𝑇𝑧) 𝑑(𝑥𝑛 ,𝑇𝑧)+𝑑(𝑧,𝑥𝑛 +1) , 𝑘 𝑘 }. fixed point. Here, 𝑥=8−2√14 is a fixed point of 𝑇. 1+𝑑(𝑧,𝑇𝑧) 2 𝜂(𝑥, 𝑦) =1 𝑥, 𝑦 ∈𝑋 (31) If in Theorem 15 we take for all ,then we obtain the following result. By taking limit as 𝑘→∞in the above inequality we get (𝑋, 𝑑) 𝑇 𝑑 (𝑧,𝑇𝑧) 𝑑 (𝑧,𝑇𝑧) Corollary 17. Let be a metric space and let be a self- 𝑑 (𝑧,𝑇𝑧) ≤ { , }<𝑑(𝑧,𝑇𝑧) mapping on 𝑋.Also,supposethat𝛼:𝑋×𝑋→ [0,∞)is a max 1+𝑑(𝑧,𝑇𝑧) 2 function and 𝜓∈Ψ. Assume that the following assertions hold (32) true: 6 Abstract and Applied Analysis

(i) (𝑋, 𝑑) is a 𝛼-complete metric space; Proof. Define 𝛼, 𝜂 : 𝑋 × 𝑋 → [0,∞) and 𝜓:[0,∞)→ [0, ∞) (ii) 𝑇 is an 𝛼-admissible mapping; by 𝑇 𝛼 𝜓 (iii) is a modified - -rational contraction mapping on 𝛼 (𝑥,) 𝑦 =𝑑(𝑥,) 𝑦 ,𝜂(𝑥,) 𝑦 =𝑑(𝑥,) 𝑦 , (42) 𝑋; 𝑥, 𝑦 ∈𝑋 𝜓(𝑡) = 𝑟𝑡 0≤𝑟<1 (iv) there exists 𝑥0 ∈𝑋such that 𝛼(𝑥0,𝑇𝑥0)≥1; for all and ,where .Clearly, 𝜂(𝑥, 𝑦) ≤ 𝛼(𝑥, 𝑦) for all 𝑥, 𝑦. ∈𝑋 That is, conditions (i)–(v) (v) if {𝑥𝑛} is a sequence in 𝑋 such that 𝛼(𝑥𝑛,𝑥𝑛+1)≥1with of Theorem 12 hold true. Let 𝜂(𝑥, 𝑇𝑥) ≤ 𝛼(𝑥, 𝑦).Then,𝑑(𝑥, 𝑥𝑛 →𝑥as 𝑛→∞, then either 𝑇𝑥) ≤ 𝑑(𝑥, 𝑦).Nowfrom(40)wehave𝑑(𝑇𝑥, 𝑇𝑦) ≤ 𝑟𝑀(𝑥, 2 𝑦) = 𝜓(𝑀(𝑥, 𝑦)) 𝑇 𝛼 𝜂 𝜓 𝛼(𝑇𝑥𝑛,𝑥)≥1 𝑜𝑟𝛼(𝑇 𝑥𝑛,𝑥)≥1 (36) .Thatis, is a modified - - -rational contraction mapping on 𝑋. Then all conditions of Theorem 12 holds for all 𝑛∈N. hold and 𝑇 has a fixed point. The uniqueness of the fixed point follows easily from (40). Then 𝑇 has a fixed point. Corollary 21. Let (𝑋, 𝑑) be a complete metric space and let 𝑇 Corollary 18. Let (𝑋, 𝑑) be a complete metric space and let 𝑇 𝑋 𝑋 𝑇 be a continuous self-mapping on . Assume that there exists be a continuous self-mapping on . Assume that is a modified 𝑟∈[0,1)such that rational contraction mapping, that is, 𝑑 (𝑥, 𝑇𝑥) ≤𝑑(𝑥,) 𝑦 𝑖𝑚𝑝𝑙𝑖𝑒𝑠 𝑑 (𝑇𝑥, 𝑇𝑦) ≤𝑟𝑑(𝑥,) 𝑦 ∀𝑥, 𝑦 ∈ 𝑋, 𝑑 (𝑇𝑥, 𝑇𝑦) ≤𝜓(𝑀 (𝑥,)) 𝑦 , (37) (43) where 𝜓∈Ψ.Then𝑇 has a fixed point. for all 𝑥, 𝑦 ∈𝑋.Then𝑇 has a unique fixed point. Corollary 19. Let (𝑋, 𝑑) be a complete metric space and let 𝑇 be a continuous self-mapping on 𝑋. Assume that 𝑇 satisfies the Now, we prove the following Suzuki type fixed point 𝑇 following rational inequality: theorem without continuity of . ∀𝑥, 𝑦 ∈ 𝑋, 𝑑 (𝑇𝑥, 𝑇𝑦) ≤𝑟𝑀(𝑥,) 𝑦 , (38) Theorem 22. Let (𝑋, 𝑑) be a complete metric space and let 𝑇 be a self-mapping on 𝑋. Define a nonincreasing function 𝜌: where 0≤𝑟<1and [0,1) → (1/2,1]by 𝑑 (𝑥, 𝑇𝑥) 1 𝑀(𝑥,𝑦)=max {𝑑 (𝑥, 𝑦), , 𝜌 (𝑟) = . 1+𝑑(𝑥, 𝑇𝑥) 1+𝑟 (44) 𝑑(𝑦,𝑇𝑦) 𝑑(𝑥,𝑇𝑦)+𝑑(𝑦,𝑇𝑥) , }. Assume that there exists 𝑟∈[0,1)such that 1+𝑑(𝑦,𝑇𝑦) 2 (39) 𝜌 (𝑟) 𝑑 (𝑥, 𝑇𝑥) ≤𝑑(𝑥,𝑦) 𝑖𝑚𝑝𝑙𝑖𝑒𝑠𝑑(𝑇𝑥,𝑇𝑦)≤𝑟𝑑(𝑥,𝑦) (45) Then 𝑇 has a fixed point. for all 𝑥, 𝑦 ∈𝑋.Then𝑇 has a unique fixed point. 3. Consequences Proof. Define 𝛼, 𝜂 : 𝑋 × 𝑋 → [0,∞) and 𝜓:[0,∞)→ [0, ∞) 3.1. Suzuki Type Fixed Point Results. From Theorem 12 we by deduce the following Suzuki type fixed point result. 𝛼 (𝑥,) 𝑦 =𝑑(𝑥,) 𝑦 ,𝜂(𝑥,) 𝑦 =𝜌(𝑟) 𝑑 (𝑥,) 𝑦 (46) Theorem 20. Let (𝑋, 𝑑) be a complete metric space and let 𝑇 𝑥, 𝑦 ∈𝑋 𝜓(𝑡) = 𝑟𝑡 0≤𝑟<1 be a continuous self-mapping on 𝑋. Assume that there exists for all and ,where .Now,since 𝜌(𝑟)𝑑(𝑥, 𝑦) ≤ 𝑑(𝑥, 𝑦) 𝑥, 𝑦 ∈𝑋 𝜂(𝑥, 𝑦) ≤ 𝛼(𝑥, 𝑦) 𝑟∈[0,1)such that for all , for all 𝑥, 𝑦. ∈𝑋 That is, conditions (i)–(iv) of Theorem 15 hold 𝑑 (𝑥, 𝑇𝑥) ≤𝑑(𝑥,𝑦) 𝑖𝑚𝑝𝑙𝑖𝑒𝑠𝑑(𝑇𝑥,𝑇𝑦)≤𝑟𝑀(𝑥,𝑦) true. Let {𝑥𝑛} be a sequence with 𝑥𝑛 →𝑥as 𝑛→∞.Since 2 2 (40) 𝜌(𝑟)𝑑(𝑇𝑥𝑛,𝑇 𝑥𝑛)≤𝑑(𝑇𝑥𝑛,𝑇 𝑥𝑛) for all 𝑛∈N,thenfrom (45)weget for all 𝑥, 𝑦 ∈𝑋,where 𝑑(𝑇2𝑥 ,𝑇3𝑥 )≤𝑟𝑑(𝑇𝑥,𝑇2𝑥 ) 𝑑 (𝑥, 𝑇𝑥) 𝑛 𝑛 𝑛 𝑛 (47) 𝑀(𝑥,𝑦)= {𝑑 (𝑥, 𝑦), , max 1+𝑑(𝑥, 𝑇𝑥) for all 𝑛∈N. 𝑛 ∈ N 𝑑(𝑦,𝑇𝑦) 𝑑(𝑥,𝑇𝑦)+𝑑(𝑦,𝑇𝑥) Assume there exists 0 such that , }. 1+𝑑(𝑦,𝑇𝑦) 2 𝜂(𝑇𝑥 ,𝑇2𝑥 )>𝛼(𝑇𝑥 ,𝑥), 𝑛0 𝑛0 𝑛0 (41) (48) 2 3 2 𝜂(𝑇 𝑥𝑛 ,𝑇 𝑥𝑛 )>𝛼(𝑇𝑥𝑛 ,𝑥); Then 𝑇 has a unique fixed point. 0 0 0 Abstract and Applied Analysis 7 then, That is, 𝑇 is a modified 𝛼-𝜓-rational contraction mapping. Let 2 {𝑥𝑛} be a sequence such that 𝛼(𝑥𝑛,𝑥𝑛+1)≥1with 𝑥𝑛 →𝑥as 𝜌 (𝑟) 𝑑(𝑇𝑥𝑛 ,𝑇 𝑥𝑛 )>𝑑(𝑇𝑥𝑛 ,𝑥), 0 0 0 𝑛→∞.So,{𝑥𝑛}⊆𝑂(𝑤). From (iii) we have 𝑥∈𝑂(𝑤).That (49) 2 3 2 is, 𝛼(𝑥𝑛,𝑥)≥1.Hence,allconditionsofCorollary17 hold and 𝜌 (𝑟) 𝑑(𝑇 𝑥𝑛 ,𝑇 𝑥𝑛 )>𝑑(𝑇𝑥𝑛 ,𝑥), 0 0 0 𝑇 has a fixed point. and so by (47)wehave Corollary 24. Let (𝑋, 𝑑) be a metric space and let 𝑇:𝑋 →𝑋 𝑑(𝑇𝑥 ,𝑇2𝑥 ) be a self-mapping on 𝑋. Suppose the following assertions hold: 𝑛0 𝑛0 (𝑋, 𝑑) 𝑇 2 (i) is an orbitally -complete metric space; ≤𝑑(𝑇𝑥𝑛 ,𝑥)+𝑑(𝑇 𝑥𝑛 ,𝑥) 0 0 (ii) there exists 𝑟∈[0,1)such that 2 2 3 <𝜌(𝑟) 𝑑(𝑇𝑥𝑛 ,𝑇 𝑥𝑛 )+𝜌(𝑟) 𝑑(𝑇 𝑥𝑛 ,𝑇 𝑥𝑛 ) 0 0 0 0 (50) 𝑑 (𝑇𝑥, 𝑇𝑦) ≤𝑟𝑀(𝑥,) 𝑦 (55) ≤𝜌(𝑟) 𝑑(𝑇𝑥 ,𝑇2𝑥 )+𝑟𝜌(𝑟) 𝑑(𝑇𝑥 ,𝑇2𝑥 ) 𝑥, 𝑦 ∈ 𝑂(𝑤) 𝑤∈𝑋 𝑛0 𝑛0 𝑛0 𝑛0 holds for all for some ,where =𝜌(𝑟)(1+𝑟) 𝑑(𝑇𝑥 ,𝑇2𝑥 )=𝑑(𝑇𝑥 ,𝑇2𝑥 ) 𝑑 (𝑥, 𝑇𝑥) 𝑛0 𝑛0 𝑛0 𝑛0 𝑀(𝑥,𝑦)= {𝑑 (𝑥, 𝑦), , max 1+𝑑(𝑥, 𝑇𝑥) which is a contradiction. Hence, either 𝑑(𝑦,𝑇𝑦) 𝑑(𝑥,𝑇𝑦)+𝑑(𝑦,𝑇𝑥) 𝜂(𝑇𝑥 ,𝑇2𝑥 )≤𝛼(𝑇𝑥,𝑥) , }; 𝑛 𝑛 𝑛 1+𝑑(𝑦,𝑇𝑦) 2 (51) 2 3 2 or 𝜂(𝑇 𝑥𝑛,𝑇 𝑥𝑛)≤𝛼(𝑇𝑥𝑛,𝑥) (56) holds for all 𝑛∈N. That is condition (v) of Theorem 15 holds. (iii) if {𝑥𝑛} is a sequence such that {𝑥𝑛}⊆𝑂(𝑤)with 𝑥𝑛 → Let, 𝜂(𝑥, 𝑇𝑥) ≤ 𝛼(𝑥, 𝑦).So,𝜌(𝑟)𝑑(𝑥, 𝑇𝑥) .≤ 𝑑(𝑥,𝑦) 𝑥 as 𝑛→∞,then𝑥∈𝑂(𝑤). Then from (45)weget𝑑(𝑇𝑥, 𝑇𝑦) ≤ 𝑟𝑑(𝑥, 𝑦) ≤𝑟𝑀(𝑥,𝑦)= 𝑇 𝜓(𝑀(𝑥,. 𝑦)) Hence, all conditions of Theorem 15 hold and 𝑇 Then has a fixed point. has a fixed point. The uniqueness of the fixed point follows easily from (45). 3.3. Fixed Point Results for Graphic Contractions. Consistent with Jachymski [11], let (𝑋, 𝑑) be a metric space and let Δ denote the diagonal of the Cartesian product 𝑋×𝑋. Consider 3.2. Fixed Point Results in Orbitally 𝑇-Complete Metric Spaces a directed graph 𝐺 such that the set 𝑉(𝐺) of its vertices 𝑋 𝐸(𝐺) Theorem 23. Let (𝑋, 𝑑) be a metric space and let 𝑇:𝑋 →𝑋 coincides with ,andtheset of its edges contains all 𝐸(𝐺) ⊇Δ 𝐺 be a self-mapping on 𝑋. Suppose the following assertions hold: loops; that is, . We assume that has no parallel edges, so we can identify 𝐺 with the pair (𝑉(𝐺), 𝐸(𝐺)). (𝑋, 𝑑) 𝑇 (i) is an orbitally -complete metric space; Moreover, we may treat 𝐺 as a weighted graph (see [11]) by (ii) there exists 𝜓∈Ψsuch that assigning to each edge the distance between its vertices. If 𝑥 and 𝑦 are vertices in a graph 𝐺,thenapathin𝐺 from 𝑥 to 𝑑(𝑇𝑥,𝑇𝑦)≤𝜓(𝑀(𝑥,𝑦)) (52) 𝑁 𝑦 of length 𝑁(𝑁 ∈ N) is a sequence {𝑥𝑖}𝑖=0 of 𝑁+1vertices 𝑥 =𝑥𝑥 =𝑦 (𝑥 ,𝑥 )∈𝐸(𝐺) 𝑖= holds for all 𝑥, 𝑦 ∈ 𝑂(𝑤) for some 𝑤∈𝑋,where such that 0 , 𝑁 and 𝑛−1 𝑛 for 1,...,𝑁.Agraph𝐺 is connected if there is a path between 𝑀(𝑥,𝑦) any two vertices. 𝐺 is weakly connected if 𝐺̃is connected (see for details [3, 6, 10, 11]). 𝑑 (𝑥, 𝑇𝑥) = max {𝑑 (𝑥, 𝑦), , Recently, some results have appeared providing sufficient 1+𝑑(𝑥, 𝑇𝑥) (53) conditions for a mapping to be a Picard operator if (𝑋, 𝑑) is endowed with a graph. The first result in this direction was 𝑑(𝑦,𝑇𝑦) 𝑑(𝑥,𝑇𝑦)+𝑑(𝑦,𝑇𝑥) , }; given by Jachymski [11]. 1+𝑑(𝑦,𝑇𝑦) 2 Definition 25 (see [11]). We say that a mapping 𝑇:𝑋is →𝑋 (iii) if {𝑥𝑛} is a sequence such that {𝑥𝑛}⊆𝑂(𝑤)with 𝑥𝑛 → a Banach 𝐺-contraction or simply 𝐺-contraction if 𝑇 pre- 𝑥 as 𝑛→∞,then𝑥∈𝑂(𝑤). serves edges of 𝐺;thatis, Then 𝑇 has a fixed point. ∀𝑥, 𝑦 ∈ 𝑋 ((𝑥, 𝑦)∈𝐸 (𝐺) 󳨐⇒ ( 𝑇 (𝑥) , 𝑇 (𝑦)) ∈𝐸 (𝐺)) (57) Proof. Define 𝛼:𝑋×𝑋→ [0,+∞)as in Remark 6.From Remark 6 we know that (𝑋, 𝑑) is an 𝛼-complete metric space and 𝑇 decreases weights of edges of 𝐺 in the following way: and 𝑇 is an 𝛼-admissiblemapping.Let𝛼(𝑥, 𝑦);then ≥1 𝑥, 𝑦 ∈ 𝑂(𝑤). Then from (ii) we have ∃𝛼∈(0, 1) ,∀𝑥,𝑦∈𝑋 (58) 𝑑 (𝑇𝑥, 𝑇𝑦) ≤𝜓(𝑀 (𝑥,)) 𝑦 . (54) ((𝑥, 𝑦) ∈𝐸 (𝐺) 󳨐⇒ 𝑑 (𝑇 (𝑥) , 𝑇 (𝑦)) ≤ 𝛼𝑑 (𝑥, 𝑦)). 8 Abstract and Applied Analysis

Definition 26 (see [11]). A mapping 𝑇:𝑋is →𝑋 called Theorem 28. Let (𝑋, 𝑑) be a complete metric space endowed 𝐺-continuous, if given 𝑥∈𝑋and sequence {𝑥𝑛} with a graph 𝐺 and let 𝑇 be a self-mapping on 𝑋.Supposethat the following assertions hold: 𝑥𝑛 󳨀→ 𝑥, as 𝑛󳨀→∞, (59) (i) for all 𝑥, 𝑦 ∈𝑋, (𝑥, 𝑦) ∈ 𝐸(𝐺) ⇒ (𝑇(𝑥), 𝑇(𝑦))∈ (𝑥 ,𝑥 )∈𝐸(𝐺) ,∀𝑛∈N 𝑇𝑥 󳨀→ 𝑇 𝑥. 𝑛 𝑛+1 implying 𝑛 𝐸(𝐺);

Theorem 27. Let (𝑋, 𝑑) be a metric space endowed with a (ii) there exists 𝑥0 ∈𝑋such that (𝑥0,𝑇𝑥0)∈𝐸(𝐺); 𝐺 𝑇 𝑋 graph and let be a self-mapping on . Suppose that the (iii) there exists 𝜓∈Ψsuch that following assertions hold: 𝑑 (𝑇𝑥, 𝑇𝑦) ≤𝜓(𝑀 (𝑥,)) 𝑦 (i) for all 𝑥, 𝑦 ∈𝑋, (𝑥, 𝑦) ∈ 𝐸(𝐺) ⇒ (𝑇(𝑥), 𝑇(𝑦))∈ (66) 𝐸(𝐺); for all (𝑥, 𝑦) ∈ 𝐸(𝐺),where (ii) there exists 𝑥0 ∈𝑋such that (𝑥0,𝑇𝑥0)∈𝐸(𝐺); (iii) there exists 𝜓∈Ψsuch that 𝑑 (𝑥, 𝑇𝑥) 𝑀(𝑥,𝑦)= {𝑑 (𝑥, 𝑦), , max 1+𝑑(𝑥, 𝑇𝑥) 𝑑(𝑇𝑥,𝑇𝑦)≤𝜓(𝑀(𝑥,𝑦)) (60) 𝑑(𝑦,𝑇𝑦) 𝑑(𝑥,𝑇𝑦)+𝑑(𝑦,𝑇𝑥) for all (𝑥, 𝑦) ∈ 𝐸(𝐺),where , }; 1+𝑑(𝑦,𝑇𝑦) 2 𝑑 (𝑥, 𝑇𝑥) 𝑀(𝑥,𝑦)= {𝑑 (𝑥, 𝑦), , (67) max 1+𝑑(𝑥, 𝑇𝑥)

𝑑(𝑦,𝑇𝑦) 𝑑(𝑥,𝑇𝑦)+𝑑(𝑦,𝑇𝑥) (iv) 𝑇 is 𝐺-continuous. , }; 1+𝑑(𝑦,𝑇𝑦) 2 Then 𝑇 has a fixed point. (61) As an application of Corollary 17,weobtain. (iv) 𝑇 is 𝐺-continuous; Theorem 29. Let (𝑋, 𝑑) be a metric space endowed with a (v) if {𝑥𝑛} is a Cauchy sequence in 𝑋 with (𝑥𝑛,𝑥𝑛+1)∈ graph 𝐺 and let 𝑇 be a self-mapping on 𝑋. Suppose that the 𝐸(𝐺) for all 𝑛∈N,then{𝑥𝑛} is convergent in 𝑋. following assertions hold: Then 𝑇 has a fixed point. (i) for all 𝑥, 𝑦 ∈𝑋, (𝑥, 𝑦) ∈ 𝐸(𝐺) ⇒ (𝑇(𝑥), 𝑇(𝑦))∈ 2 𝐸(𝐺) Proof. Define 𝛼:𝑋 →[0,+∞)by ; (ii) there exists 𝑥0 ∈𝑋such that (𝑥0,𝑇𝑥0)∈𝐸(𝐺); 1, if (𝑥, 𝑦) ∈𝐸 (𝐺) , 𝛼(𝑥,𝑦)={ (62) (iii) there exists 𝜓∈Ψsuch that 0, otherwise. 𝑑 (𝑇𝑥, 𝑇𝑦) ≤𝜓(𝑀 (𝑥,)) 𝑦 At first we prove that 𝑇 is an 𝛼-admissiblemapping.Let (68) 𝛼(𝑥, 𝑦);then ≥1 (𝑥, 𝑦) ∈ .From(i),wehave𝐸(𝐺) (𝑇𝑥, 𝑇𝑦) ∈ (𝑥, 𝑦) ∈ 𝐸(𝐺) 𝐸(𝐺).Thatis,𝛼(𝑇𝑥, 𝑇𝑦).Thus ≥1 𝑇 is an 𝛼-admissible for all ,where mapping. Let 𝑇 be 𝐺-continuous on (𝑋, 𝑑).Then, 𝑑 (𝑥, 𝑇𝑥) 𝑀(𝑥,𝑦)=max {𝑑 (𝑥, 𝑦), , 𝑥𝑛 󳨀→ 𝑥, as 𝑛󳨀→∞, 1+𝑑(𝑥, 𝑇𝑥) (63) (𝑥𝑛,𝑥𝑛+1)∈𝐸(𝐺) ,∀𝑛∈N implying 𝑇𝑥𝑛 󳨀→ 𝑇 𝑥. 𝑑(𝑦,𝑇𝑦) 𝑑(𝑥,𝑇𝑦)+𝑑(𝑦,𝑇𝑥) , }; 1+𝑑(𝑦,𝑇𝑦) 2 That is, (69) 𝑥𝑛 󳨀→ 𝑥, as 𝑛󳨀→∞, (64) {𝑥 } (𝑥 ,𝑥 )∈𝐸(𝐺) 𝛼(𝑥𝑛,𝑥𝑛+1) ≥ 1, ∀𝑛 ∈ N implying 𝑇𝑥𝑛 󳨀→ 𝑇 𝑥 (iv) if 𝑛 is a sequence such that 𝑛 𝑛+1 with 𝑥𝑛 →𝑥as 𝑛→∞, then either which implies that 𝑇 is 𝛼-continuous on (𝑋, 𝑑). From (ii) 𝑥 ∈𝑋 (𝑥 ,𝑇𝑥 )∈𝐸(𝐺) 2 there exists 0 such that 0 0 .Thatis, (𝑇𝑥𝑛,𝑥)∈𝐸(𝐺) 𝑜𝑟 (𝑇 𝑥𝑛,𝑥)∈𝐸(𝐺) (70) 𝛼(𝑥0,𝑇𝑥0)≥1.Let𝛼(𝑥, 𝑦);then ≥1 (𝑥, 𝑦) ∈ 𝐸(𝐺).Now, from (iii) we have 𝑑(𝑇𝑥, 𝑇𝑦) ≤ 𝜓(𝑀(𝑥,.Thatis, 𝑦)) holds for all 𝑛∈N;

𝛼(𝑥,𝑦)≥1󳨐⇒𝑑(𝑇𝑥,𝑇𝑦)≤𝜓(𝑀(𝑥,𝑦)). (65) (v) if {𝑥𝑛} is a Cauchy sequence in 𝑋 with (𝑥𝑛,𝑥𝑛+1)∈ 𝐸(𝐺) for all 𝑛∈N, then either {𝑥𝑛} is convergent in (𝑋, 𝑑) 𝛼 Condition (v) implies that is an -complete metric 𝑋 or (𝑋, 𝑑) is a complete metric space. space. Hence, all conditions of Corollary 14 are satisfied and 𝑇 has a fixed point. Then 𝑇 has a fixed point. Abstract and Applied Analysis 9

Let (𝑋,𝑑,⪯) be a partially ordered metric space. Define (iii) there exists 𝜓∈Ψsuch that the graph 𝐺 by 𝑑(𝑇𝑥,𝑇𝑦)≤𝜓(𝑀(𝑥,𝑦)) (76) 𝐸 (𝐺) := {(𝑥,) 𝑦 ∈𝑋×𝑋:𝑥⪯𝑦} . (71) for all 𝑥⪯𝑦,where For this graph, condition (i) in Theorem 27 means that 𝑇 is nondecreasing with respect to this order [5]. From 𝑑 (𝑥, 𝑇𝑥) 𝑀(𝑥,𝑦)= {𝑑 (𝑥, 𝑦), , Theorems 27–29 we derive the following important results in max 1+𝑑(𝑥, 𝑇𝑥) partially ordered metric spaces. 𝑑(𝑦,𝑇𝑦) 𝑑(𝑥,𝑇𝑦)+𝑑(𝑦,𝑇𝑥) Theorem 30. Let (𝑋, 𝑑, ⪯) be a partially ordered metric space , }; 1+𝑑(𝑦,𝑇𝑦) 2 and let 𝑇 be a self-mapping on 𝑋. Suppose that the following assertions hold: (77)

(i) 𝑇 is nondecreasing map; (iv) if {𝑥𝑛} is a sequence such that 𝑥𝑛 ⪯𝑥𝑛+1 with 𝑥𝑛 →𝑥 𝑥 ∈𝑋 𝑥 ⪯𝑇𝑥 (ii) there exists 0 such that 0 0; as 𝑛→∞, then either 𝜓∈Ψ (iii) there exists such that 2 𝑇𝑥𝑛 ⪯ 𝑥 𝑜𝑟 𝑇 𝑥𝑛 ⪯𝑥 (78) 𝑑 (𝑇𝑥, 𝑇𝑦) ≤𝜓(𝑀 (𝑥,)) 𝑦 (72) holds for all 𝑛∈N; for all 𝑥⪯𝑦,where (v) if {𝑥𝑛} is a Cauchy sequence in 𝑋 with 𝑥𝑛 ⪯𝑥𝑛+1 for all 𝑛∈N {𝑥 } 𝑋 (𝑋, 𝑑) 𝑑 (𝑥, 𝑇𝑥) , then either 𝑛 is convergent in or is a 𝑀(𝑥,𝑦)= {𝑑 (𝑥, 𝑦), , complete metric space. max 1+𝑑(𝑥, 𝑇𝑥) 𝑇 𝑑(𝑦,𝑇𝑦) 𝑑(𝑥,𝑇𝑦)+𝑑(𝑦,𝑇𝑥) Then has a fixed point. , }; 1+𝑑(𝑦,𝑇𝑦) 2 (73) 4. Application to Existence of Solutions of Integral Equations

(iv) either for a given 𝑥∈𝑋and sequence {𝑥𝑛} Fixed point theorems for monotone operators in ordered metric spaces are widely investigated and have found various 𝑥 󳨀→ 𝑥 , 𝑎 𝑠 𝑛󳨀→∞, 𝑛 applications in differential and integral equations (see28 [ – (74) 𝑥 ⪯𝑥 ,∀𝑛∈N,𝑜𝑛𝑒ℎ𝑎𝑠𝑇𝑥 󳨀→ 𝑇 𝑥 30] and references therein). In this section, we apply our 𝑛 𝑛+1 𝑛 result to the existence of a solution of an integral equation. Let 𝑋=𝐶([0,𝑇],R) be the set of real continuous functions or 𝑇 is continuous; defined on [0, 𝑇] and let 𝑑:𝑋×𝑋 → R+ be defined by (v) if {𝑥𝑛} is a Cauchy sequence in 𝑋 with 𝑥𝑛 ⪯𝑥𝑛+1 for all 𝑛∈N {𝑥 } 𝑋 (𝑋, 𝑑) 󵄩 󵄩 , then either 𝑛 is convergent in or is a 𝑑 (𝑥,) 𝑦 = 󵄩𝑥−𝑦󵄩∞ (79) complete metric space. for all 𝑥, 𝑦 ∈𝑋.Then(𝑋, 𝑑) is a complete metric space. Also, Then 𝑇 has a fixed point. assume this metric space endowed with a graph 𝐺. Corollary 31 (Ran and Reurings [15]). Let (𝑋,𝑑,⪯) be a Consider the integral equation as follows: 𝑇:𝑋 →𝑋 partially ordered complete metric space and let be 𝑇 𝑥 ⪯𝑇𝑥 a continuous nondecreasing self-mapping such that 0 0 𝑥 (𝑡) =𝑝(𝑡) + ∫ 𝑆 (𝑡, 𝑠) 𝑓 (𝑠, 𝑥 (𝑠)) 𝑑𝑠 (80) for some 𝑥0 ∈𝑋. Assume that 0

𝑑(𝑇𝑥,𝑇𝑦)≤𝑟𝑑(𝑥,𝑦) (75) and let 𝐹:𝑋be →𝑋 defined by holds for all 𝑥, 𝑦 ∈𝑋 with 𝑥⪯𝑦,where0≤𝑟<1.Then𝑇 has 𝑇 𝐹 (𝑥)(𝑡) =𝑝(𝑡) + ∫ 𝑆 (𝑡, 𝑠) 𝑓 (𝑠, 𝑥 (𝑠)) 𝑑𝑠. (81) afixedpoint. 0 (𝑋,𝑑,⪯) Theorem 32. Let be a partially ordered metric space We assume that and let 𝑇 be a self-mapping on 𝑋. Suppose that the following assertions hold: (A) 𝑓:[0,𝑇]×R → R is continuous; (i) 𝑇 is nondecreasing map; (B) 𝑝:[0,𝑇]→ R is continuous;

(ii) there exists 𝑥0 ∈𝑋such that 𝑥0 ⪯𝑇𝑥0; (C) 𝑆:[0,𝑇]×R →[0,+∞)is continuous; 10 Abstract and Applied Analysis

𝜓∈Ψ 𝑠∈[0,𝑇] 󵄩 󵄩 (D) there exists a such that for all ×(max { 󵄩𝑥 (𝑠) −𝑦(𝑠)󵄩 ,

󵄩 󵄩 ∀𝑥,𝑦∈𝑋 (𝑥,𝑦)∈𝐸(𝐺) 󳨐⇒ ( 𝐹 (𝑥) , 𝐹 (𝑦)) ∈𝐸 (𝐺) , ‖𝑥 (𝑠) −𝐹(𝑥 (𝑠))‖ 󵄩𝑦 (𝑠) −𝐹(𝑦(𝑠))󵄩 , 󵄩 󵄩, ∀𝑥,𝑦∈𝑋 (𝑥,𝑦)∈𝐸(𝐺) 󳨐⇒ 0 1+‖𝑥 (𝑠) −𝐹(𝑥 (𝑠))‖ 1+󵄩𝑦 (𝑠) −𝐹(𝑦(𝑠))󵄩 1 󵄩 󵄩 󵄩 󵄩 ≤𝑓(𝑠, 𝑥 (𝑠)) −𝑓(𝑠,𝑦(𝑠)) [󵄩𝑥 (𝑠) −𝐹(𝑦(𝑠))󵄩 + 󵄩𝑦 (𝑠) −𝐹(𝑥 (𝑠))󵄩]}). 2 󵄩 󵄩 󵄩 󵄩 󵄨 󵄨 󵄨𝑥 (𝑠) −𝑦(𝑠)󵄨 (84) ≤𝜓(max { 󵄨 󵄨, 1+󵄨𝑥 (𝑠) −𝑦(𝑠)󵄨 Then |𝑥 (𝑠) −𝐹(𝑥 (𝑠))| 󵄨 󵄨 󵄩 󵄩 , 󵄨𝑦 (𝑠) −𝐹(𝑦(𝑠))󵄨 , 󵄩𝐹𝑥−𝐹𝑦󵄩 1+|𝑥 (𝑠) −𝐹(𝑥 (𝑠))| 󵄨 󵄨 󵄩 󵄩∞ 󵄩 󵄩 1 󵄨 󵄨 󵄨 󵄨 ≤𝜓( { 󵄩𝑥 (𝑠) −𝑦(𝑠)󵄩 , [󵄨𝑥 (𝑠) −𝐹(𝑦(𝑠))󵄨 + 󵄨𝑦 (𝑠) −𝐹(𝑥 (𝑠))󵄨]}); max 󵄩 󵄩 2 󵄨 󵄨 󵄨 󵄨 󵄩 󵄩 (82) ‖𝑥 (𝑠) −𝐹(𝑥 (𝑠))‖ 󵄩𝑦 (𝑠) −𝐹(𝑦(𝑠))󵄩 , 󵄩 󵄩, 1+‖𝑥 (𝑠) −𝐹(𝑥 (𝑠))‖ 1+󵄩𝑦 (𝑠) −𝐹(𝑦(𝑠))󵄩 1 󵄩 󵄩 󵄩 󵄩 (E) there exists 𝑥0 ∈𝑋such that (𝑥0,𝐹(𝑥0)) ∈ 𝐸(𝐺); [󵄩𝑥 (𝑠) −𝐹(𝑦(𝑠))󵄩+󵄩𝑦 (𝑠) −𝐹(𝑥 (𝑠))󵄩]}). 2 󵄩 󵄩 󵄩 󵄩

(F) if {𝑥𝑛} is a sequence such that (𝑥𝑛,𝑥𝑛+1)∈𝐸(𝐺)with (85) 𝑥𝑛 →𝑥as 𝑛→∞, then either That is, (𝑥, 𝑦) ∈ 𝐸(𝐺) implies (𝐹𝑥 ,𝑥)∈𝐸(𝐺) 𝑜𝑟2 (𝐹 𝑥 ,𝑥)∈𝐸(𝐺) 󵄩 󵄩 𝑛 𝑛 (83) 󵄩𝐹𝑥−𝐹𝑦󵄩∞ 󵄩 󵄩 ≤𝜓( {󵄩𝑥−𝑦󵄩 , holds for all 𝑛∈N; max 󵄩 󵄩∞ 𝑇 󵄩 󵄩 ∫ 𝑆(𝑡, 𝑠)𝑑𝑠 ≤1 𝑡 󵄩𝑦 − 𝐹(𝑦)󵄩 (86) (G) 0 for all . ‖𝑥−𝐹(𝑥)‖∞ 󵄩 󵄩∞ , 󵄩 󵄩 , 1+‖𝑥−𝐹(𝑥)‖∞ 1+󵄩𝑦 − 𝐹(𝑦)󵄩 Theorem 33. Under assumptions (A)–(G), the integral equa- ∞ tion (80) has a solution in 𝑋=𝐶([0,𝑇],R). 1 󵄩 󵄩 󵄩 󵄩 [󵄩𝑥−𝐹(𝑦)󵄩 + 󵄩𝑦−𝐹(𝑥)󵄩 ]}). 2 󵄩 󵄩∞ 󵄩 󵄩∞ Proof. Consider the mapping 𝐹:𝑋defined →𝑋 by (81). Let (𝑥, 𝑦) ∈ 𝐸(𝐺). Then from (D) we deduce IteasilyshowsthatallthehypothesesofTheorem29 are satisfied and hence the mapping 𝐹 has a fixed point that is a 󵄨 󵄨 𝑋 = 𝐶([0, 𝑇], R) 󵄨𝐹 (𝑥)(𝑡) −𝐹(𝑦)(𝑡)󵄨 solution in of the integral equation (80). 󵄨 󵄨 󵄨 𝑇 󵄨 󵄨 󵄨 = 󵄨∫ 𝑆 (𝑡, 𝑠) [𝑓 (𝑠, 𝑥 (𝑠)) −𝑓(𝑠,𝑦(𝑠))] 𝑑𝑠󵄨 Conflict of Interests 󵄨 0 󵄨 The authors declare that there is no conflict of interests 𝑇 󵄨 󵄨 regarding the publication of this paper. ≤ ∫ 𝑆 (𝑡, 𝑠) 󵄨𝑓 (𝑠, 𝑥 (𝑠)) −𝑓(𝑠,𝑦(𝑠))󵄨 𝑑𝑠 0 𝑇 Acknowledgment ≤ ∫ 𝑆 (𝑡, 𝑠) 𝜓 0 ThispaperwasfundedbytheDeanshipofScientificResearch 󵄨 󵄨 (DSR), King Abdulaziz University, Jeddah. Therefore, the ×(max { 󵄨𝑥 (𝑠) −𝑦(𝑠)󵄨 , authors acknowledge with thanks DSR, KAU, for the financial support. 󵄨 󵄨 |𝑥 (𝑠) −𝐹(𝑥 (𝑠))| 󵄨𝑦 (𝑠) −𝐹(𝑦(𝑠))󵄨 , 󵄨 󵄨, 1+|𝑥 (𝑠) −𝐹(𝑥 (𝑠))| 1+󵄨𝑦 (𝑠) −𝐹(𝑦(𝑠))󵄨 References 1 󵄨 󵄨 󵄨 󵄨 [󵄨𝑥 (𝑠)−𝐹(𝑦(𝑠))󵄨+󵄨𝑦 (𝑠)−𝐹(𝑥 (𝑠))󵄨]})𝑑𝑠 [1] S. Banach, “Sur les operations´ dans les ensembles abstraits et 2 󵄨 󵄨 󵄨 󵄨 leur application aux equations´ integrales,”´ Fundamenta Mathe- maticae,vol.3,pp.133–181,1922. 𝑇 ≤(∫ 𝑆 (𝑡, 𝑠) 𝑑𝑠) 𝜓 [2] V. Berinde, Iterative Approximation of Fixed Points,Springer, 0 Berlin, Germany, 2007. Abstract and Applied Analysis 11

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Research Article Algorithms of Common Solutions for Generalized Mixed Equilibria, Variational Inclusions, and Constrained Convex Minimization

Lu-Chuan Ceng1 and Suliman Al-Homidan2

1 Department of Mathematics, Shanghai Normal University and Scientific Computing Key Laboratory of Shanghai Universities, Shanghai 200234, China 2 Department of Mathematics and Statistics, King Fahd University of Petroleum and Minerals, Dhahran, Saudi Arabia

Correspondence should be addressed to Lu-Chuan Ceng; [email protected]

Received 3 November 2013; Accepted 12 November 2013; Published 23 January 2014

Academic Editor: Qamrul Hasan Ansari

Copyright © 2014 L.-C. Ceng and S. Al-Homidan. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We introduce new implicit and explicit iterative algorithms for finding a common element of the set of solutions of the minimization problem for a convex and continuously Frechet´ differentiable functional, the set of solutions of a finite family of generalized mixed equilibrium problems, and the set of solutions of a finite family of variational inclusions in a real Hilbert space. Under suitable control conditions, we prove that the sequences generated by the proposed algorithms converge strongly to a common element of three sets, which is the unique solution of a variational inequality defined over the intersection of three sets.

1. Introduction The VIP (3) was first discussed by Lions2 [ ]. There are many applications of VIP (3) in various fields; see, for 𝐶 Let be a nonempty closed convex subset of a real Hilbert example, [3–6]. It is well known that if 𝐴 is a strongly 𝐻 𝑃 𝐻 𝐶 space and let 𝐶 be the metric projection of onto .Let monotone and Lipschitz continuous mapping on 𝐶,thenVIP 𝑆:𝐶 →𝐶 𝐶 (𝑆) be a self-mapping on . We denote by Fix the (3)hasauniquesolution.In1976,Korpelevic[̌7]proposedan 𝑆 set of fixed points of and by R thesetofallrealnumbers. iterative algorithm for solving the VIP (3) in Euclidean space 𝐴:𝐶 →𝐻 𝐿 𝑛 Amapping is called -Lipschitz continuous if R : there exists a constant 𝐿≥0such that 󵄩 󵄩 󵄩 󵄩 󵄩𝐴𝑥 − 𝐴𝑦󵄩 ≤𝐿󵄩𝑥−𝑦󵄩 ,∀𝑥,𝑦∈𝐶. (1) 𝑦𝑛 =𝑃𝐶 (𝑥𝑛 −𝜏𝐴𝑥𝑛), (4) In particular, if 𝐿=1,then𝐴 is called a nonexpansive 𝑥𝑛+1 =𝑃𝐶 (𝑥𝑛 −𝜏𝐴𝑦𝑛), ∀𝑛≥0, mapping [1]; if 𝐿∈[0,1),then𝐴 is called a contraction. Amapping𝑉 is called strongly positive on 𝐻 if there exists a constant 𝜇>0such that with 𝜏>0a given number, which is known as the extragradient method (see also [8]).TheliteratureontheVIP ⟨𝑉𝑥 ,𝑥 ⟩ ≥𝜇‖𝑥‖2,∀𝑥∈𝐻. (2) is vast and Korpelevich’s extragradient method has received Let 𝐴:𝐶be →𝐻 a nonlinear mapping on 𝐶.We great attention given by many authors, who improved it in consider the following variational inequality problem (VIP): various ways; see, for example, [9–24] and references therein, find a point 𝑥∈𝐶such that to name but a few. Let 𝜑:𝐶 → R be a real-valued function, 𝐴:𝐻 →𝐻 ⟨𝐴𝑥, 𝑦 − 𝑥⟩ ≥ 0, ∀𝑦 ∈𝐶. (3) a nonlinear mapping, and Θ:𝐶×𝐶→ R a bifunction. In The solution set of VIP3 ( ) is denoted by VI (𝐶, 𝐴). 2008, Peng and Yao [12] introduced the following generalized 2 Abstract and Applied Analysis mixed equilibrium problem (GMEP) of finding 𝑥∈𝐶such for all 𝑥∈𝐻. Then the following hold: that (Θ,𝜑) (i) for each 𝑥∈𝐻,𝑇 (𝑥) is nonempty and single- Θ(𝑥,𝑦)+𝜑(𝑦)−𝜑(𝑥) +⟨𝐴𝑥,𝑦−𝑥⟩≥0, ∀𝑦∈𝐶. 𝑟 (5) valued; (Θ,𝜑) We denote the set of solutions of GMEP (5)byGMEP (ii) 𝑇𝑟 is firmly nonexpansive; that is, for any 𝑥, 𝑦 ∈𝐻, (Θ, 𝜑, 𝐴).TheGMEP(5) is very general in the sense that 󵄩 (Θ,𝜑) (Θ,𝜑) 󵄩2 (Θ,𝜑) (Θ,𝜑) it includes, as special cases, optimization problems, varia- 󵄩𝑇 𝑥−𝑇 𝑦󵄩 ≤⟨𝑇 𝑥−𝑇 𝑦,𝑥−𝑦⟩; (10) tional inequalities, minimax problems, and Nash equilibrium 󵄩 𝑟 𝑟 󵄩 𝑟 𝑟 problems in noncooperative games. The GMEP is further considered and studied; see, for example, [11, 14, 23, 25–28]. If (Θ,𝜑) (iii) Fix (𝑇 ) = 𝑀𝐸𝑃(Θ, 𝜑); 𝜑=0and 𝐴=0,thenGMEP(5) reduces to the equilibrium 𝑟 problem (EP) which is to find 𝑥∈𝐶such that (iv) 𝑀𝐸𝑃(Θ, 𝜑) is closed and convex;

Θ (𝑥,) 𝑦 ≥0, ∀𝑦∈𝐶. (Θ,𝜑) (Θ,𝜑) 2 (Θ,𝜑) (Θ,𝜑) (6) (v) ‖𝑇𝑠 𝑥−𝑇𝑡 𝑥‖ ≤ ((𝑠 − 𝑡)/𝑠)⟨𝑇𝑠 𝑥−𝑇𝑡 𝑥, (Θ,𝜑) 𝑇 𝑥−𝑥⟩for all 𝑠, 𝑡 >0 and 𝑥∈𝐻. It is considered and studied in [29]. The set of solutions of 𝑠 EP is denoted by EP (Θ). It is worth mentioning that the EP Let 𝜆𝑛,1,𝜆𝑛,2,...,𝜆𝑛,𝑁 ∈ (0, 1], 𝑛≥1.Giventhe is a unified model of several problems, namely, variational nonexpansive mappings 𝑆1,𝑆2,...,𝑆𝑁 on 𝐻,foreach𝑛≥1, inequality problems, optimization problems, saddle point the mappings 𝑈𝑛,1,𝑈𝑛,2,...,𝑈𝑛,𝑁 are defined by problems,complementarityproblems,fixedpointproblems, Nashequilibriumproblems,andsoforth. 𝑈 =𝜆 𝑆 +(1−𝜆 )𝐼, Throughout this paper, it is assumed as in [12]thatΘ:𝐶× 𝑛,1 𝑛,1 1 𝑛,1 𝐶→R is a bifunction satisfying conditions (A1)–(A4) and 𝑈𝑛,2 =𝜆𝑛,2𝑆𝑛𝑈𝑛,1 +(1−𝜆𝑛,2)𝐼, 𝜑:𝐶 → R is a lower semicontinuous and convex function with restriction (B1) or (B2), where 𝑈𝑛,𝑛−1 =𝜆𝑛−1𝑇𝑛−1𝑈𝑛,𝑛 +(1−𝜆𝑛−1)𝐼, (11) (A1) Θ(𝑥, 𝑥) =0 for all 𝑥∈𝐶; . . (A2) Θ is monotone; that is, Θ(𝑥, 𝑦) + Θ(𝑦, 𝑥)≤0 for any 𝑥, 𝑦; ∈𝐶 𝑈𝑛,𝑁−1 =𝜆𝑛,𝑁−1𝑆𝑁−1𝑈𝑛,𝑁−2 +(1−𝜆𝑛,𝑁−1)𝐼, (A3) Θ is upper-hemicontinuous; that is, for each 𝑥, 𝑦, 𝑧∈ 𝑊 := 𝑈 =𝜆 𝑆 𝑈 +(1−𝜆 )𝐼. 𝐶, 𝑛 𝑛,𝑁 𝑛,𝑁 𝑁 𝑛,𝑁−1 𝑛,𝑁 Θ(𝑡𝑧+(1−𝑡) 𝑥, 𝑦) ≤ Θ (𝑥, 𝑦); lim sup The 𝑊𝑛 is called the 𝑊-mapping generated by 𝑆1,...,𝑆𝑁 𝑡→0+ (7) and 𝜆𝑛,1,𝜆𝑛,2,...,𝜆𝑛,𝑁. Note that the nonexpansivity of 𝑆𝑖 implies the nonexpansivity of 𝑊𝑛. Θ(𝑥, ⋅) (A4) is convex and lower semicontinuous for each In 2012, combining the hybrid steepest-descent method 𝑥∈𝐶 ; in [30] and hybrid viscosity approximation method in [31], (B1) for each 𝑥∈𝐻and 𝑟>0, there exists a bounded Ceng et al. [27] proposed and analyzed the following hybrid subset 𝐷𝑥 ⊂𝐶and 𝑦𝑥 ∈𝐶such that, for any 𝑧∈ iterative method for finding a common element of the set of 𝐶\𝐷𝑥, solutions of GMEP (5) and the set of fixed points of a finite 𝑁 family of nonexpansive mappings {𝑆𝑖} . 1 𝑖=1 Θ(𝑧,𝑦 )+𝜑(𝑦 )−𝜑(𝑧) + ⟨𝑦 −𝑧,𝑧−𝑥⟩<0; (8) 𝑥 𝑥 𝑟 𝑥 Theorem CGY (see [27, Theorem 3.1]). Let 𝐶 be a nonempty closed convex subset of a real Hilbert space 𝐻.LetΘ:𝐶× (B2) 𝐶 is a bounded set. 𝐶→R be a bifunction satisfying assumptions (A1)–(A4) and 𝜑:𝐶 → R alowersemicontinuousandconvex Next we list some elementary results for the MEP. function with restriction (B1) or (B2). Let the mapping 𝐴: 𝑁 𝐻→𝐻be 𝛿-inverse-strongly monotone and {𝑆𝑖}𝑖=1 afinite Proposition 1 (see [26]). Assume that Θ:𝐶×𝐶→ R 𝑁 family of nonexpansive mappings on 𝐻 such that ∩ Fix(𝑆𝑖)∩ satisfies (A1)–(A4) and let 𝜑:𝐶 → R be a proper lower 𝑖=1 𝐺𝑀𝐸𝑃(Θ, 𝜑,𝐴) =0̸ .Let𝐹:𝐻be →𝐻 a 𝜅-Lipschitzian semicontinuous and convex function. Assume that either (B1) and 𝜂-strongly monotone operator with constants 𝜅,𝜂>0 or (B2) holds. For 𝑟>0and 𝑥∈𝐻,defineamapping (Θ,𝜑) and 𝑉:𝐻a →𝐻 𝜌-Lipschitzian mapping with constant 𝑇 :𝐻 →as 𝐶 follows: 2 𝑟 𝜌≥0.Let0<𝜇<2𝜂/𝜅 and 0≤𝛾𝜌<𝜏,where𝜏= 2 (Θ,𝜑) 1−√1 − 𝜇(2𝜂 − 𝜇𝜅 ).Suppose{𝛼𝑛} and {𝛽𝑛} are two sequences 𝑇𝑟 (𝑥) = {𝑧 ∈ 𝐶 : Θ (𝑧, 𝑦) + 𝜑 (𝑦)−𝜑 (𝑧) (0, 1) {𝛾 } (0, 2𝛿] {𝜆 }𝑁 (9) in , 𝑛 is a sequence in ,and 𝑛,𝑖 𝑖=1 is a sequence 1 in [𝑎, 𝑏] with 0<𝑎≤𝑏<1.Forevery𝑛≥1,let𝑊𝑛 be the + ⟨𝑦 − 𝑧, 𝑧 − 𝑥⟩ ≥ 0, ∀𝑦 ∈𝐶}, 𝑟 𝑊-mapping generated by 𝑆1,...,𝑆𝑁 and 𝜆𝑛,1,𝜆𝑛,2,...,𝜆𝑛,𝑁. Abstract and Applied Analysis 3

∞ Given 𝑥1 ∈𝐻arbitrarily, suppose that the sequences {𝑥𝑛} and Let {𝑇𝑛}𝑛=1 be an infinite family of nonexpansive self- ∞ {𝑢𝑛} are generated iteratively by mappings on 𝐶 and {𝜆𝑛}𝑛=0 a sequence of nonnegative numbers in [0, 1].Forany𝑛≥1, define a self-mapping 𝑊𝑛 Θ(𝑢𝑛,𝑦)+𝜑(𝑦)−𝜑(𝑢𝑛)+⟨𝐴𝑥𝑛,𝑦−𝑢𝑛⟩ on 𝐶 as follows:

1 𝑈𝑛,𝑛+1 =𝐼, + ⟨𝑦 −𝑛 𝑢 ,𝑢𝑛 −𝑥𝑛⟩ ≥ 0, ∀𝑦 ∈ 𝐶, 𝑟𝑛 𝑈𝑛,𝑛 =𝜆𝑛𝑇𝑛𝑈𝑛,𝑛+1 +(1−𝜆𝑛)𝐼, 𝑥𝑛+1 =𝛼𝑛𝛾𝑉𝑥𝑛 +𝛽𝑛𝑥𝑛 +((1−𝛽𝑛)𝐼−𝛼𝑛𝜇𝐹)𝑛 𝑊 𝑢𝑛, 𝑈𝑛,𝑛−1 =𝜆𝑛−1𝑇𝑛−1𝑈𝑛,𝑛 +(1−𝜆𝑛−1)𝐼, ∀𝑛 ≥ 1, . (12) . where the sequences {𝛼𝑛}, {𝛽𝑛}, {𝑟𝑛} and the finite family of 𝑈𝑛,𝑘 =𝜆𝑘𝑇𝑘𝑈𝑛,𝑘+1 +(1−𝜆𝑘)𝐼, (15) 𝑁 sequences {𝜆𝑛,𝑖}𝑖=1 satisfy the conditions: 𝑈𝑛,𝑘−1 =𝜆𝑘−1𝑇𝑘−1𝑈𝑛,𝑘 +(1−𝜆𝑘−1)𝐼, ∞ (i) lim𝑛→∞𝛼𝑛 =0and ∑𝑛=1 𝛼𝑛 =∞; . (ii) 0

(iii) 0

∗ 𝑁 Such a mapping 𝑊𝑛 is called the 𝑊-mapping generated by Then both {𝑥𝑛} and {𝑢𝑛} converge strongly to 𝑥 ∈∩𝑖=1 Fix(𝑆𝑖)∩ ∗ 𝑇 ,𝑇 ,...,𝑇 𝜆 ,𝜆 ,...,𝜆 𝐺𝑀𝐸𝑃(Θ, 𝜑,𝐴) 𝑥 =𝑃𝑁 (𝐼−𝜇𝐹+ 𝑛 𝑛−1 1 and 𝑛 𝑛−1 1. ,where ∩ Fix (𝑆𝑖)∩𝐺𝑀𝐸𝑃(Θ,𝜑,𝐴) ∗ 𝑖=1 Whenever 𝐶=𝐻a real Hilbert space, Yao et al. [11]very 𝛾𝑓)𝑥 is a unique solution of the variational inequality problem recently introduced and analyzed an iterative algorithm for (VIP): finding a common element of the set of solutions of GMEP ⟨(𝜇𝐹 − 𝛾𝑉) 𝑥∗,𝑥∗ −𝑥⟩≤0, (5), the set of solutions of the variational inclusion (14), and the set of fixed points of an infinite family of nonexpansive 𝑁 (13) mappings. ∀𝑥 ∈ ⋂ Fix (𝑆𝑖)∩𝐺𝑀𝐸𝑃(Θ,𝜑,𝐴). 𝑖=1 Theorem YCL (see [11,Theorem3.2]). Let 𝜑:𝐻 → R be a lower semicontinuous and convex function and Θ:𝐻×𝐻 → Let 𝐵 be a single-valued mapping of 𝐶 into 𝐻 and 𝑅 a 𝑉 𝐷(𝑅) =𝐶 R a bifunction satisfying conditions (A1)–(A4) and (B1). Let multivalued mapping with . Consider the following be a strongly positive bounded linear operator with coefficient variational inclusion: find a point 𝑥∈𝐶such that 𝐻 𝜇>0and 𝑅:𝐻 →2 a maximal monotone mapping. Let 𝐴, 𝐵 : 𝐻 →𝐻 𝛼 0∈𝐵𝑥+𝑅𝑥. (14) the mappings be -inverse-strongly monotone and 𝛽-inverse-strongly monotone, respectively. Let 𝑓:𝐻 →𝐻 𝜌 𝑟>0𝛾>0 𝜆>0 We denote by 𝐼(𝐵, 𝑅) the solution set of the variational be a -contraction. Let , ,and be three 𝑟<2𝛼𝜆<2𝛽 0<𝛾<𝜇/𝜌 inclusion (14). In particular, if 𝐵=𝑅=0,then𝐼(𝐵, 𝑅). =𝐶 constants such that , ,and . {𝜆}∞ (0, 𝑏] If 𝐵=0,thenproblem(14) becomes the inclusion problem Let 𝑛=1 be a sequence of positive numbers in for some 𝑏 ∈ (0, 1) {𝑇 }∞ introduced by Rockafellar [32]. It is known that problem (14) and 𝑛 𝑛=1 an infinite family of nonexpansive self- 𝐻 Ω:=∩∞ (𝑇 )∩𝐺𝑀𝐸𝑃(Θ, 𝜑, 𝐴)∩ provides a convenient framework for the unified study of mappings on such that 𝑛=1 Fix 𝑛 𝐼(𝐵, 𝑅) =0̸ 𝑥 ∈𝐻 {𝑥 } optimal solutions in many optimization related areas includ- . For arbitrarily given 1 , let the sequence 𝑛 ing mathematical programming, complementarity problems, be generated by variational inequalities, optimal control, mathematical economics, and equilibria and game theory. Θ(𝑢 ,𝑦)+𝜑(𝑦)−𝜑(𝑢 )+⟨𝑦−𝑢 ,𝐴𝑥 ⟩ In 1998, Huang [33]studiedproblem(14)inthecase 𝑛 𝑛 𝑛 𝑛 where 𝑅 is maximal monotone and 𝐵 is strongly monotone 1 + ⟨𝑦 − 𝑢 ,𝑢 −𝑥 ⟩ ≥ 0, ∀𝑦 ∈ 𝐻, and Lipschitz continuous with 𝐷(𝑅)=𝐶=𝐻.Subsequently, 𝑟 𝑛 𝑛 𝑛 Zeng et al. [34] further studied problem (14)inthecasewhich is more general than Huang’s one [33]. Moreover, the authors 𝑥𝑛+1 =𝛼𝑛𝛾𝑓 (𝑥 𝑛)+𝛽𝑛𝑥𝑛 [34] obtained the same strong convergence conclusion as +[(1 − 𝛽 )𝐼 − 𝛼 𝑉]𝑊 𝐽 (𝑢 −𝜆𝐵𝑢 ), ∀𝑛 ≥ 1, in Huang’s result [33]. In addition, the authors also gave 𝑛 𝑛 𝑛 𝑅,𝜆 𝑛 𝑛 the geometric convergence rate estimate for approximate (16) solutions. Also, various types of iterative algorithms for solving variational inclusions have been further studied where {𝛼𝑛}, {𝛽𝑛} are two real sequences in [0, 1] and 𝑊𝑛 is the and developed; for more details, refer to [35–39]andthe 𝑊-mapping defined by (15) (with 𝑋=𝐻and 𝐶=𝐻). Assume references therein. that the following conditions are satisfied: 4 Abstract and Applied Analysis

∞ (C1) lim𝑛→∞𝛼𝑛 =0and ∑𝑛=1 𝛼𝑛 =∞; finite family of GMEPs, and the set of solutions of a finite family of variational inclusions for maximal monotone and (C2) 00 such that then the sequence {𝑥𝑛} defined by the GPA (20)convergesin norm to the unique minimizer of CMP (18). If the gradient 󵄩 󵄩2 ⟨𝐴𝑥−𝐴𝑦,𝑥−𝑦⟩ ≥𝜂󵄩𝑥−𝑦󵄩 ,∀𝑥,𝑦∈𝐶; (24) ∇𝑓 is only assumed to be Lipschitz continuous, then {𝑥𝑛} 󵄩 󵄩 canonlybeweaklyconvergentif𝐻 is infinite-dimensional (a counterexample is given in Section 5 of Xu [40]). 𝛼 ∇𝑓 (iii) -inverse-strongly monotone if there exists a constant Since the Lipschitz continuity of the gradient implies 𝛼>0such that that it is actually (1/𝐿)-inverse-strongly monotone (ism) [41], its complement can be an averaged mapping (i.e., it can be 󵄩 󵄩2 ⟨𝐴𝑥−𝐴𝑦,𝑥−𝑦⟩≥𝛼󵄩𝐴𝑥 − 𝐴𝑦󵄩 ,∀𝑥,𝑦∈𝐶. (25) expressed as a proper convex combination of the identity 󵄩 󵄩 mapping and a nonexpansive mapping). Consequently, the GPA can be rewritten as the composite of a projection and an It is obvious that if 𝐴 is 𝛼-inverse-strongly monotone, averaged mapping, which is again an averaged mapping. This then 𝐴 is monotone and (1/𝛼)-Lipschitz continuous. shows that averaged mappings play an important role in the The metric (or nearest point) projection from 𝐻 onto 𝐶 is GPA. Recently, Xu [40] used averaged mappings to study the the mapping 𝑃𝐶 :𝐻 →which 𝐶 assigns to each point 𝑥∈𝐻 convergence analysis of the GPA, which is hence an operator- the unique point 𝑃𝐶𝑥∈𝐶satisfying the property oriented approach. Motivated and inspired by the above facts, we in this 󵄩 󵄩 󵄩 󵄩 󵄩𝑥−𝑃𝐶𝑥󵄩 = inf 󵄩𝑥−𝑦󵄩 =: 𝑑 (𝑥,) 𝐶 . paper introduce new implicit and explicit iterative algorithms 𝑦∈𝐶 (26) forfindingacommonelementofthesetofsolutionsofthe CMP (18)foraconvexfunctional𝑓:𝐶 → R with 𝐿- Some important properties of projections are gathered in Lipschitz continuous gradient ∇𝑓 ,thesetofsolutionsofa the following proposition. Abstract and Applied Analysis 5

Proposition 2. For given 𝑥∈𝐻and 𝑧∈𝐶, Proposition 5 (see [42]). Let 𝑇:𝐻be →𝐻 a given mapping. (i) 𝑧=𝑃𝐶𝑥⇔⟨𝑥−𝑧,𝑦−𝑧⟩≤0,forall𝑦∈𝐶; 2 2 2 𝑇 𝐼−𝑇 (ii) 𝑧=𝑃𝐶𝑥 ⇔ ‖𝑥 −𝑧‖ ≤ ‖𝑥−𝑦‖ −‖𝑦−𝑧‖,forall (i) is nonexpansive if and only if the complement 𝑦∈𝐶; is (1/2)-ism. 2 𝑇 ] 𝛾>0,𝛾𝑇 (]/𝛾) (iii) ⟨𝑃𝐶𝑥−𝑃𝐶𝑦, 𝑥 − 𝑦⟩𝐶 ≥‖𝑃 𝑥−𝑃𝐶𝑦‖ ,forall𝑦∈𝐻. (ii) If is -ism, then, for is -ism. (iii) 𝑇 is averaged if and only if the complement 𝐼−𝑇 is ]-ism Consequently, 𝑃𝐶 is nonexpansive and monotone. for some ] >1/2. Indeed, for 𝛼 ∈ (0, 1), 𝑇 is 𝛼-averaged If 𝐴 is an 𝛼-inverse-strongly monotone mapping of 𝐶 into ifandonlyif𝐼−𝑇is (1/2𝛼)-ism. 𝐻,thenitisobviousthat𝐴 is (1/𝛼)-Lipschitz continuous. We 𝑆, 𝑇, 𝑉 : 𝐻→ also have that, for all 𝑢, V ∈𝐶and 𝜆>0, Proposition 6 (see [42, 43]). Let be given operators. ‖(𝐼−𝜆𝐴) 𝑢−(𝐼−𝜆𝐴) V‖2 = ‖(𝑢−V) −𝜆(𝐴𝑢 −𝐴V)‖2 (i) If 𝑇 = (1 − 𝛼)𝑆 +𝛼𝑉 for some 𝛼 ∈ (0, 1) and if 𝑆 is = ‖𝑢−V‖2 −2𝜆⟨𝐴𝑢 −𝐴V,𝑢−V⟩ averaged and 𝑉 is nonexpansive, then 𝑇 is averaged. 𝑇 +𝜆2‖𝐴𝑢 −𝐴V‖2 (ii) is firmly nonexpansive if and only if the complement 𝐼−𝑇is firmly nonexpansive. ≤ ‖𝑢−V‖2 +𝜆(𝜆−2𝛼) (iii) If 𝑇 = (1 − 𝛼)𝑆 +𝛼𝑉 for some 𝛼 ∈ (0, 1) and if 𝑆 is firmly nonexpansive and 𝑉 is nonexpansive, then 𝑇 is 2 × ‖𝐴𝑢 −𝐴V‖ . averaged. (27) (iv) The composite of finitely many averaged mappings {𝑇 }𝑁 So, if 𝜆≤2𝛼,then𝐼−𝜆𝐴is a nonexpansive mapping from 𝐶 is averaged. That is, if each of the mappings 𝑖 𝑖=1 𝑇 ⋅⋅⋅𝑇 to 𝐻. is averaged, then so is the composite 1 𝑁.In particular, if 𝑇1 is 𝛼1-averaged and 𝑇2 is 𝛼2-averaged, Definition 3. Amapping𝑇:𝐻is →𝐻 said to be where 𝛼1,𝛼2 ∈ (0, 1), then the composite 𝑇1𝑇2 is 𝛼- averaged, where 𝛼=𝛼1 +𝛼2 −𝛼1𝛼2. (a) nonexpansive [1]if 𝑁 󵄩 󵄩 󵄩 󵄩 (v) If the mappings {𝑇𝑖}𝑖=1 are averaged and have a 󵄩𝑇𝑥−𝑇𝑦󵄩 ≤ 󵄩𝑥−𝑦󵄩 , ∀𝑥, 𝑦 ∈ 𝐻; 󵄩 󵄩 󵄩 󵄩 (28) common fixed point, then

(b) firmly nonexpansive if 2𝑇 − 𝐼 is nonexpansive or, 𝑁 equivalently, if 𝑇 is 1-inverse-strongly monotone (1- ⋂ Fix (𝑇𝑖)=Fix (𝑇1 ⋅⋅⋅𝑇𝑁). (32) ism), 𝑖=1 󵄩 󵄩2 ⟨𝑥−𝑦,𝑇𝑥−𝑇𝑦⟩ ≥ 󵄩𝑇𝑥 − 𝑇𝑦󵄩 , ∀𝑥, 𝑦 ∈ 𝐻; 󵄩 󵄩 (29) The notation Fix(𝑇) denotes the set of all fixed points of the mapping 𝑇;thatis,Fix(𝑇) = {𝑥 ∈ 𝐻 : .𝑇𝑥 =𝑥} alternatively, 𝑇 is firmly nonexpansive if and only if 𝑇 can be expressed as We need some facts and tools in a real Hilbert space 𝐻 1 which are listed as lemmas below. 𝑇= (𝐼+𝑆) , 2 (30) Lemma 7. Let 𝑋 be a real inner product space. Then the 𝑆:𝐻 →𝐻 where is nonexpansive; projections are firmly following inequality holds: nonexpansive. 󵄩 󵄩2 2 󵄩𝑥+𝑦󵄩 ≤ ‖𝑥‖ +2⟨𝑦,𝑥+𝑦⟩, ∀𝑥,𝑦∈𝑋. (33) It can be easily seen that if 𝑇 is nonexpansive, then 𝐼−𝑇 is monotone. It is also easy to see that a projection Lemma 8. Let 𝐴:𝐶be →𝐻 a monotone mapping. In the 𝑃 1 𝐶 is -ism. Inverse-strongly monotone (also referred to as context of the variational inequality problem the characteriza- co-coercive) operators have been applied widely in solving tion of the projection (see Proposition 2(i)) implies practical problems in various fields. 𝑢∈VI (𝐶, 𝐴) ⇐⇒ 𝑢 = 𝑃𝐶 (𝑢−𝜆𝐴𝑢) , 𝑓𝑜𝑟 𝑠𝑜𝑚𝑒 𝜆>0. 𝑇:𝐻 →𝐻 Definition 4. Amapping is said to be an (34) averaged mapping if it can be written as the average of the 𝐼 identity and a nonexpansive mapping; that is, Lemma 9 (see [44, Demiclosedness principle]). Let 𝐶 be a 𝐻 𝑇≡(1−𝛼) 𝐼+𝛼𝑆, (31) nonempty closed convex subset of a real Hilbert space .Let 𝑇 be a nonexpansive self-mapping on 𝐶 with Fix(𝑇) =0̸ .Then where 𝛼 ∈ (0, 1) and 𝑆:𝐻is →𝐻 nonexpansive. More 𝐼−𝑇is demiclosed. That is, whenever {𝑥𝑛} is a sequence in 𝐶 precisely, when the last equality holds, we say that 𝑇 is 𝛼- weakly converging to some 𝑥∈𝐶and the sequence {(𝐼−𝑇)𝑥𝑛} averaged. Thus, firmly nonexpansive mappings (in particular, strongly converges to some 𝑦,itfollowsthat(𝐼 − 𝑇)𝑥 =𝑦.Here projections) are (1/2)-averaged mappings. 𝐼 is the identity operator of 𝐻. 6 Abstract and Applied Analysis

𝐻 Lemma 10 (see [45]). Let {𝑠𝑛} be a sequence of nonnegative Recall that a set-valued mapping 𝑅 : 𝐷(𝑅) ⊂𝐻→2 is numbers satisfying the conditions called monotone if for all 𝑥, 𝑦 ∈ ,𝐷(𝑅) 𝑓∈𝑅(𝑥)and 𝑔 ∈ 𝑅(𝑦) imply 𝑠𝑛+1 ≤ (1−𝛼𝑛) 𝑠𝑛 +𝛼𝑛𝛽𝑛,∀𝑛≥1, (35) where {𝛼𝑛} and {𝛽𝑛} are sequences of real numbers such that ⟨𝑓−𝑔,𝑥−𝑦⟩≥0. (43) ∞ (i) {𝛼𝑛}⊂[0,1]and ∑𝑛=1 𝛼𝑛 =∞or, equivalently, Aset-valuedmapping𝑅 is called maximal monotone if 𝑅 is ∞ 𝑛 monotone and (𝐼 + 𝜆𝑅)𝐷(𝑅) =𝐻 for each 𝜆>0,where𝐼 ∏ (1 − 𝛼 ):= ∏ (1 − 𝛼 )=0; 𝑛 𝑛→∞lim 𝑘 (36) is the identity mapping of 𝐻. We denote by 𝐺(𝑅) the graph 𝑛=1 𝑘=1 of 𝑅. It is known that a monotone mapping 𝑅 is maximal if ∞ and only if, for (𝑥, 𝑓) ∈ 𝐻 ×𝐻, ⟨𝑓−𝑔,𝑥−𝑦⟩≥0for every (ii) lim sup 𝛽𝑛 ≤0,or∑ |𝛼𝑛𝛽𝑛|<∞. 𝑛→∞ 𝑛=1 (𝑦, 𝑔) ∈ 𝐺(𝑅) implies 𝑓∈𝑅(𝑥). Then lim𝑛→∞𝑠𝑛 =0. Let 𝐴:𝐶be →𝐻 a monotone, 𝑘-Lipschitz continuous mapping and let 𝑁𝐶V be the normal cone to 𝐶 at V ∈𝐶;that {𝑥 } {𝑧 } Lemma 11 (see [46]). Let 𝑛 and 𝑛 be bounded sequences is, in a Banach space 𝑋 and {𝛽𝑛} a sequence in [0, 1] with 𝑁 V = {𝑤∈𝐻:⟨V −𝑢,𝑤⟩ ≥0,∀𝑢∈𝐶} . 0

Suppose that 𝑥𝑛+1 =(1−𝛽𝑛)𝑧𝑛 +𝛽𝑛𝑥𝑛 for each 𝑛≥1and Define 󵄩 󵄩 󵄩 󵄩 (󵄩𝑧 −𝑧 󵄩 − 󵄩𝑥 −𝑥 󵄩)≤0. lim sup 󵄩 𝑛+1 𝑛󵄩 󵄩 𝑛+1 𝑛󵄩 (38) 𝐴V +𝑁𝐶V, if V ∈𝐶, 𝑛→∞ 𝑇V ={ (45) 0, if V ∉𝐶. Then lim𝑛→∞‖𝑧𝑛 −𝑥𝑛‖=0. 𝑇 0∈𝑇V V ∈ The following lemma can be easily proven and, therefore, Then, is maximal monotone and if and only if 𝑉𝐼(𝐶, 𝐴);see[32]. we omit the proof. 𝐻 Assume that 𝑅 : 𝐷(𝑅) ⊂𝐻→2 is a maximal Lemma 12. Let 𝑉: 𝐻be →𝐻 an 𝑙-Lipschitzian mapping monotone mapping. Then, for 𝜆>0,associatedwith𝑅,the with constant 𝑙≥0,andlet𝐹:𝐻be →𝐻 a 𝜅- resolvent operator 𝐽𝑅,𝜆 can be defined as Lipschitzian and 𝜂-strongly monotone operator with positive 𝜅,𝜂>0 0≤𝛾𝑙<𝜇𝜂 −1 constants .Then,for , 𝐽𝑅,𝜆𝑥=(𝐼+𝜆𝑅) 𝑥, ∀𝑥 ∈𝐻. (46) 󵄩 󵄩2 ⟨(𝜇𝐹−𝛾𝑉)𝑥−(𝜇𝐹−𝛾𝑉)𝑦,𝑥−𝑦⟩≥(𝜇𝜂−𝛾𝑙)󵄩𝑥−𝑦󵄩 , In terms of Huang [33](seealso[34]), the following property ∀𝑥, 𝑦 ∈ 𝐻. holds for the resolvent operator 𝐽𝑅,𝜆 :𝐻 → 𝐷(𝑅). (39) Lemma 14. 𝐽𝑅,𝜆 is single-valued and firmly nonexpansive; that That is, 𝜇𝐹 − 𝛾𝑉 is strongly monotone with constant 𝜇𝜂 − 𝛾𝑙. is,

𝐶 󵄩 󵄩2 Let be a nonempty closed convex subset of a real Hilbert ⟨𝐽 𝑥−𝐽 𝑦,𝑥−𝑦⟩≥󵄩𝐽 𝑥−𝐽 𝑦󵄩 ,∀𝑥,𝑦∈𝐻. space 𝐻. We introduce some notations. Let 𝜆 be a number in 𝑅,𝜆 𝑅,𝜆 󵄩 𝑅,𝜆 𝑅,𝜆 󵄩 (0, 1] and let 𝜇>0. Associating with a nonexpansive mapping (47) 𝜆 𝑇:𝐶, →𝐻 we define the mapping 𝑇 :𝐶 →by 𝐻 Consequently, 𝐽𝑅,𝜆 is nonexpansive and monotone. 𝜆 𝑇 𝑥:=𝑇𝑥−𝜆𝜇𝐹(𝑇𝑥) ,∀𝑥∈𝐶, (40) Lemma 15 (see [39]). Let 𝑅 be a maximal monotone mapping where 𝐹:𝐻is →𝐻 an operator such that, for some with 𝐷(𝑅).Then,foranygiven =𝐶 𝜆>0, 𝑢∈𝐶is a solution positive constants 𝜅,𝜂>0, 𝐹 is 𝜅-Lipschitzian and 𝜂-strongly of problem (14) if and only if 𝑢∈𝐶satisfies monotone on 𝐻;thatis,𝐹 satisfies the conditions: 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩2 𝑢=𝐽 (𝑢 − 𝜆𝐵𝑢) . 󵄩𝐹𝑥−𝐹𝑦󵄩 ≤𝜅󵄩𝑥−𝑦󵄩 , ⟨𝐹𝑥−𝐹𝑦,𝑥−𝑦⟩ ≥𝜂󵄩𝑥−𝑦󵄩 , 𝑅,𝜆 (48) (41) Lemma 16 (see [34]). Let 𝑅 be a maximal monotone mapping for all 𝑥, 𝑦 ∈𝐻. with 𝐷(𝑅) =𝐶 and let 𝐵:𝐶be →𝐻 a strongly monotone, 𝑧∈𝐻 𝜆 continuous,andsingle-valuedmapping.Then,foreach , Lemma 13 (see [45, Lemma 3.1]). 𝑇 is a contraction provided 𝑧 ∈ (𝐵 + 𝜆𝑅)𝑥 𝑥 𝜆>0 2 the equation has a unique solution 𝜆 for . 0<𝜇<2𝜂/𝜅;thatis, 󵄩 󵄩 󵄩 󵄩 Lemma 17 (see [39]). Let 𝑅 be a maximal monotone mapping 󵄩𝑇𝜆𝑥−𝑇𝜆𝑦󵄩 ≤ (1−𝜆𝜏) 󵄩𝑥−𝑦󵄩 ,∀𝑥,𝑦∈𝐶, 󵄩 󵄩 󵄩 󵄩 (42) with 𝐷(𝑅) =𝐶 and let 𝐵:𝐶be →𝐻 a monotone, continuous, and single-valued mapping. Then (𝐼+𝜆(𝑅+𝐵))𝐶 =𝐻 for each 2 where 𝜏=1−√1 − 𝜇(2𝜂 − 𝜇𝜅 ) ∈ (0, 1]. 𝜆>0.Inthiscase,𝑅+𝐵is maximal monotone. Abstract and Applied Analysis 7

3. Implicit Iterative Algorithm and is ((2 + 𝜆𝑛𝐿)/4)-averaged for each 𝜆𝑛 ∈ (0, 2/𝐿). Therefore, Its Convergence Criteria we can write Wenowstateandprovethefirstmainresultofthispaper. 2−𝜆 𝐿 2+𝜆 𝐿 𝑃 (𝐼−𝜆 ∇𝑓 ) = 𝑛 𝐼+ 𝑛 𝑇 =𝑠𝐼+(1−𝑠 ) 𝑇 , 𝐶 𝑛 4 4 𝑛 𝑛 𝑛 𝑛 𝐶 Theorem 18. Let beanonemptyclosedconvexsubsetof (51) arealHilbertspace𝐻.Let𝑓:𝐶 → R be a convex functional with 𝐿-Lipschitz continuous gradient ∇𝑓 .Let𝑀, 𝑁 𝑇 𝑠 := 𝑠 (𝜆 )=(2−𝜆𝐿)/4 ∈ be two integers. Let Θ𝑘 be a bifunction from 𝐶×𝐶to R where 𝑛 is nonexpansive and 𝑛 𝑛 𝑛 𝑛 (0, 1/2) 𝜆 ∈ (0, 2/𝐿) satisfying (A1)–(A4) and let 𝜑𝑘 :𝐶 → R ∪{+∞}be a for each 𝑛 .Itisclearthat proper lower semicontinuous and convex function, where 𝑘∈ 𝐻 2 {1,2,...,𝑀}.Let𝑅𝑖 :𝐶 → 2 be a maximal monotone 𝜆𝑛 󳨀→ ⇐⇒ 𝑠 𝑛 󳨀→ 0. (52) mapping and let 𝐴𝑘 :𝐻 →and 𝐻 𝐵𝑖 :𝐶 →be 𝐻 𝜇𝑘- 𝐿 inverse-strongly monotone and 𝜂𝑖-inverse-strongly monotone, respectively, where 𝑘 ∈ {1,2,...,𝑀}, 𝑖 ∈ {1,2,...,𝑁}.Let Put 𝐹:𝐻be →𝐻 a 𝜅-Lipschitzian and 𝜂-strongly monotone operator with positive constants 𝜅,𝜂>0.Let𝑉: 𝐻 →𝐻be an 𝑘 (Θ𝑘,𝜑𝑘) (Θ𝑘−1,𝜑𝑘−1) Δ 𝑛 =𝑇𝑟 (𝐼 −𝑘,𝑛 𝑟 𝐴𝑘)𝑇𝑟 𝑙 𝑙≥0 0<𝜇<2𝜂/𝜅2 𝑘,𝑛 𝑘−1,𝑛 -Lipschitzian mapping with constant .Let (53) (Θ ,𝜑 ) and 0≤𝛾𝑙<𝜏,where𝜏=1−√1 − 𝜇(2𝜂 −2 𝜇𝜅 ). Assume that ×(𝐼−𝑟 𝐴 )⋅⋅⋅𝑇 1 1 (𝐼 − 𝑟 𝐴 )𝑥 , 𝑘−1,𝑛 𝑘−1 𝑟1,𝑛 1,𝑛 1 𝑛 𝑀 𝑁 Ω:=∩𝑘=1𝐺𝑀𝐸𝑃(Θ𝑘,𝜑𝑘,𝐴𝑘)∩∩𝑖=1𝐼(𝐵𝑖,𝑅𝑖)∩Γ=0̸ and that {𝑥 } either (B1) or (B2) holds. Let 𝑛 be a sequence generated by for all 𝑘∈{1,2,...,𝑀}and 𝑛≥1,

(Θ𝑀,𝜑𝑀) (Θ𝑀−1 ,𝜑𝑀−1 ) 𝑢𝑛 =𝑇 (𝐼−𝑟𝑀,𝑛𝐴𝑀)𝑇 𝑖 𝑟𝑀,𝑛 𝑟𝑀−1,𝑛 Λ =𝐽 (𝐼 − 𝜆 𝐵 )𝐽 𝑛 𝑅𝑖,𝜆𝑖,𝑛 𝑖,𝑛 𝑖 𝑅𝑖−1,𝜆𝑖−1,𝑛 (Θ ,𝜑 ) ×(𝐼−𝑟 𝐴 )⋅⋅⋅𝑇 1 1 (𝐼−𝑟 𝐴 )𝑥 , (54) 𝑀−1,𝑛 𝑀−1 𝑟1,𝑛 1,𝑛 1 𝑛 ×(𝐼−𝜆 𝐵 )⋅⋅⋅𝐽 (𝐼−𝜆 𝐵 ), 𝑖−1,𝑛 𝑖−1 𝑅1,𝜆1,𝑛 1,𝑛 1

V𝑛 =𝐽𝑅 ,𝜆 (𝐼−𝜆𝑁,𝑛𝐵𝑁)𝐽𝑅 ,𝜆 𝑁 𝑁,𝑛 𝑁−1 𝑁−1,𝑛 (49) 0 0 for all 𝑖 ∈ {1,2,...,𝑁}and 𝑛≥1,andΔ 𝑛 =Λ𝑛 =𝐼,where𝐼 ×(𝐼−𝜆𝑁−1,𝑛𝐵𝑁−1)⋅⋅⋅𝐽𝑅 ,𝜆 (𝐼 − 𝜆1,𝑛𝐵1)𝑢𝑛, 𝑀 1 1,𝑛 is the identity mapping on 𝐻.Thenwehavethat𝑢𝑛 =Δ𝑛 𝑥𝑛 𝑁 and V𝑛 =Λ𝑛 𝑢𝑛. 𝑥𝑛 =𝑠𝑛𝛾𝑉𝑥𝑛 +(𝐼−𝑠𝑛𝜇𝐹)𝑛 𝑇 V𝑛, Consider the following mapping 𝐺𝑛 on 𝐻 defined by ∀𝑛 ≥ 1, 𝐺 𝑥=𝑠𝛾𝑉𝑥+(𝐼−𝑠 𝜇𝐹) 𝑇 Λ𝑁Δ𝑀𝑥, ∀𝑥 ∈ 𝐻, 𝑛≥1, 𝑃 (𝐼−𝜆 ∇𝑓 ) = 𝑠 𝐼+(1−𝑠 )𝑇 𝑇 𝑛 𝑛 𝑛 𝑛 𝑛 𝑛 where 𝐶 𝑛 𝑛 𝑛 𝑛 (here 𝑛 is nonexpansive (55) and 𝑠𝑛 =(2−𝜆𝑛𝐿)/4 ∈ (0, 1/2) for each 𝜆𝑛 ∈ (0, 2/𝐿)). Assume that the following conditions hold: where 𝑠𝑛 =(2−𝜆𝑛𝐿)/4 ∈ (0, 1/2) for each 𝜆𝑛 ∈ (0, 2/𝐿).By (i) 𝑠𝑛 ∈ (0, 1/2) for each 𝜆𝑛 ∈ (0, 2/𝐿), lim𝑛→∞𝑠𝑛 =0(⇔ Proposition 1(ii) and Lemma 13 we obtain from (27)thatfor lim𝑛→∞𝜆𝑛 = 2/𝐿); all 𝑥, 𝑦 ∈𝐻

(ii) {𝜆𝑖,𝑛}⊂[𝑎𝑖,𝑏𝑖] ⊂ (0, 2𝜂𝑖),forall𝑖∈{1,2,...,𝑁}; 󵄩 󵄩 󵄩𝐺𝑛𝑥−𝐺𝑛𝑦󵄩 (iii) {𝑟𝑘,𝑛}⊂[𝑒𝑘,𝑓𝑘] ⊂ (0, 2𝜇𝑘),forall𝑘∈{1,2,...,𝑀}. 󵄩 󵄩 󵄩 ≤𝑠𝛾 󵄩𝑉𝑥−𝑉𝑦󵄩 + 󵄩(𝐼 − 𝑠 𝜇𝐹) 𝑇 Λ𝑁Δ𝑀𝑥 𝑛 󵄩 󵄩 󵄩 𝑛 𝑛 𝑛 𝑛 Then {𝑥𝑛} converges strongly as 𝜆𝑛 → 2/𝐿 (⇔𝑛 𝑠 →0)to a 𝑞∈Ω 󵄩 point ,whichisauniquesolutionoftheVIP: −(𝐼−𝑠 𝜇𝐹) 𝑇 Λ𝑁Δ𝑀𝑦󵄩 𝑛 𝑛 𝑛 𝑛 󵄩 ⟨(𝜇𝐹 − 𝛾𝑉) 𝑞, 𝑝 − 𝑞⟩ ≥0,∀𝑝∈Ω. 󵄩 󵄩 󵄩 󵄩 (50) ≤𝑠𝛾𝑙 󵄩𝑥−𝑦󵄩 +(1−𝑠 𝜏) 󵄩Λ𝑁Δ𝑀𝑥−Λ𝑁Δ𝑀𝑦󵄩 𝑛 󵄩 󵄩 𝑛 󵄩 𝑛 𝑛 𝑛 𝑛 󵄩 𝑞=𝑃 (𝐼 − 𝜇𝐹 + 𝛾𝑉)𝑞 󵄩 󵄩 󵄩 Equivalently, Ω . ≤𝑠𝛾𝑙 󵄩𝑥−𝑦󵄩 +(1−𝑠 𝜏) 󵄩(𝐼 − 𝜆 𝐵 )Λ𝑁−1Δ𝑀𝑥 𝑛 󵄩 󵄩 𝑛 󵄩 𝑁,𝑛 𝑁 𝑛 𝑛 Proof. Firstofall,letusshowthatthesequence{𝑥𝑛} is well 𝑁−1 𝑀 󵄩 −(𝐼−𝜆 𝐵 )Λ Δ 𝑦󵄩 defined. Indeed, since ∇𝑓 is 𝐿-Lipschitzian, it follows that ∇𝑓 𝑁,𝑛 𝑁 𝑛 𝑛 󵄩 is 1/𝐿-ism; see [41]. By Proposition 5(ii) we know that, for 𝜆> 󵄩 󵄩 󵄩 𝑁−1 𝑀 𝑁−1 𝑀 󵄩 ≤𝑠𝛾𝑙 󵄩𝑥−𝑦󵄩 +(1−𝑠 𝜏) 󵄩Λ Δ 𝑥−Λ Δ 𝑦󵄩 0, 𝜆∇𝑓 is (1/𝜆𝐿)-ism. So by Proposition 5(iii) we deduce that 𝑛 󵄩 󵄩 𝑛 󵄩 𝑛 𝑛 𝑛 𝑛 󵄩 𝐼 − 𝜆∇𝑓 (𝜆𝐿/2) 𝑃 is -averaged. Now since the projection 𝐶 is . (1/2)-averaged, it is easy to see from Proposition 6(iv) that . 𝑃 (𝐼 − 𝜆∇𝑓) (2 + 𝜆𝐿)/4 𝜆∈ the composite 𝐶 is -averaged for 󵄩 󵄩 󵄩 󵄩 ≤𝑠𝛾𝑙 󵄩𝑥−𝑦󵄩 +(1−𝑠 𝜏) 󵄩Λ0 Δ𝑀𝑥−Λ0 Δ𝑀𝑦󵄩 (0, 2/𝐿).Hence,weobtainthat,foreach𝑛≥1, 𝑃𝐶(𝐼 − 𝜆𝑛∇𝑓 ) 𝑛 󵄩 󵄩 𝑛 󵄩 𝑛 𝑛 𝑛 𝑛 󵄩 8 Abstract and Applied Analysis

󵄩 󵄩 󵄩 󵄩 =𝑠𝛾𝑙 󵄩𝑥−𝑦󵄩 +(1−𝑠 𝜏) 󵄩Δ𝑀𝑥−Δ𝑀𝑦󵄩 𝑛 󵄩 󵄩 𝑛 󵄩 𝑛 𝑛 󵄩 Similarly, we have 󵄩 󵄩 󵄩 𝑀−1 ≤𝑠𝛾𝑙 󵄩𝑥−𝑦󵄩 +(1−𝑠 𝜏) 󵄩(𝐼 − 𝑟 𝐴 )Δ 𝑥 𝑛 󵄩 󵄩 𝑛 󵄩 𝑀,𝑛 𝑀 𝑛 󵄩 󵄩 󵄩 𝑁−1 󵄩V𝑛 −𝑝󵄩 = 󵄩𝐽𝑅 ,𝜆 (𝐼−𝜆𝑁,𝑛𝐵𝑁)Λ 𝑢𝑛 󵄩 󵄩 󵄩 󵄩 𝑁 𝑁,𝑛 𝑛 −(𝐼−𝑟 𝐴 )Δ𝑀−1𝑦󵄩 𝑀,𝑛 𝑀 𝑛 󵄩 𝑁−1 󵄩 −𝐽𝑅 ,𝜆 (𝐼−𝜆𝑁,𝑛𝐵𝑁)Λ 𝑝󵄩 󵄩 󵄩 󵄩 󵄩 𝑁 𝑁,𝑛 𝑛 󵄩 ≤𝑠𝛾𝑙 󵄩𝑥−𝑦󵄩 +(1−𝑠 𝜏) 󵄩Δ𝑀−1𝑥−Δ𝑀−1𝑦󵄩 𝑛 󵄩 󵄩 𝑛 󵄩 𝑛 𝑛 󵄩 󵄩 󵄩 ≤ 󵄩(𝐼 − 𝜆 𝐵 )Λ𝑁−1𝑢 −(𝐼−𝜆 𝐵 )Λ𝑁−1𝑝󵄩 󵄩 𝑁,𝑛 𝑁 𝑛 𝑛 𝑁,𝑛 𝑁 𝑛 󵄩 . 󵄩 󵄩 . ≤ 󵄩Λ𝑁−1𝑢 −Λ𝑁−1𝑝󵄩 󵄩 𝑛 𝑛 𝑛 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 0 0 󵄩 ≤𝑠𝑛𝛾𝑙 󵄩𝑥−𝑦󵄩 +(1−𝑠𝑛𝜏) 󵄩Δ 𝑛𝑥−Δ𝑛𝑦󵄩 󵄩 󵄩 . 󵄩 󵄩 󵄩 󵄩 . =𝑠𝛾𝑙 󵄩𝑥−𝑦󵄩 +(1−𝑠 𝜏) 󵄩𝑥−𝑦󵄩 𝑛 󵄩 󵄩 𝑛 󵄩 󵄩 󵄩 󵄩 ≤ 󵄩Λ0 𝑢 −Λ0 𝑝󵄩 󵄩 󵄩 󵄩 𝑛 𝑛 𝑛 󵄩 =(1−𝑠 (𝜏 − 𝛾𝑙)) 󵄩𝑥−𝑦󵄩 . 𝑛 󵄩 󵄩 󵄩 󵄩 = 󵄩𝑢 −𝑝󵄩 . (56) 󵄩 𝑛 󵄩 (60)

Since 0<1−𝑠𝑛(𝜏 − 𝛾𝑙), <1 𝐺𝑛 :𝐻 →is 𝐻 a contraction. Therefore, by the Banach contraction principle, Combining (59)and(60), we have 𝐺𝑛 has a unique fixed point 𝑥𝑛 ∈𝐻, which uniquely solves the fixed point equation: 󵄩 󵄩 󵄩 󵄩 󵄩V𝑛 −𝑝󵄩 ≤ 󵄩𝑥𝑛 −𝑝󵄩 . (61) 𝑁 𝑀 𝑥𝑛 =𝑠𝑛𝛾𝑉𝑥𝑛 +(𝐼−𝑠𝑛𝜇𝐹)𝑛 𝑇 Λ 𝑛 Δ 𝑛 𝑥𝑛. (57) Since

This shows that the sequence {𝑥𝑛} is defined well. 0≤𝛾𝑙<𝜏 𝜇𝜂≥𝜏⇔𝜅≥𝜂 2 Note that and .Hence,by 𝑝=𝑃𝐶 (𝐼 − 𝜆𝑛∇𝑓 ) 𝑝 =𝑠 𝑛𝑝+(1−𝑠𝑛)𝑇𝑛𝑝, ∀𝜆𝑛 ∈(0, ), Lemma 12 we know that 𝐿 (62) 󵄩 󵄩2 ⟨(𝜇𝐹−𝛾𝑉)𝑥−(𝜇𝐹−𝛾𝑉)𝑦,𝑥−𝑦⟩≥(𝜇𝜂−𝛾𝑙)󵄩𝑥−𝑦󵄩 , where 𝑠𝑛 := 𝑠𝑛(𝜆𝑛)=(2−𝜆𝑛𝐿)/4 ∈ (0, 1/2),itisclearthat ∀𝑥, 𝑦 ∈ 𝐻. 𝑇𝑛𝑝=𝑝for each 𝜆𝑛 ∈ (0, 2/𝐿).Thus,utilizingLemma 13 and (58) the nonexpansivity of 𝑇𝑛,weobtainfrom(61)that

𝜇𝐹 − 𝛾𝑉 0≤𝛾𝑙<𝜏≤𝜇𝜂 󵄩 󵄩 󵄩 That is, is strongly monotone for . 󵄩𝑥𝑛 −𝑝󵄩 = 󵄩𝑠𝑛 (𝛾𝑉𝑥𝑛 −𝜇𝐹𝑝)+(𝐼−𝑠𝑛𝜇𝐹)𝑛 𝑇 V𝑛 Moreover, it is clear that 𝜇𝐹 − 𝛾𝑉 is Lipschitz continuous. So 󵄩 the VIP (50) has only one solution. Below we use 𝑞∈Ωto −(𝐼−𝑠𝑛𝜇𝐹)𝑛 𝑇 𝑝󵄩 denote the unique solution of the VIP (50). 󵄩 󵄩 ≤ 󵄩(𝐼 − 𝑠 𝜇𝐹) 𝑇 V −(𝐼−𝑠 𝜇𝐹) 𝑇 𝑝󵄩 Now,letusshowthat{𝑥𝑛} is bounded. In fact, take 𝑝∈Ω 󵄩 𝑛 𝑛 𝑛 𝑛 𝑛 󵄩 arbitrarily. Then from (27)andProposition 1(ii) we have 󵄩 󵄩 +𝑠𝑛 󵄩𝛾𝑉𝑥𝑛 −𝜇𝐹𝑝󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 (Θ𝑀,𝜑𝑀) 𝑀−1 ≤(1−𝑠𝑛𝜏) 󵄩V𝑛 −𝑝󵄩 󵄩𝑢𝑛 −𝑝󵄩 = 󵄩𝑇 (𝐼 −𝑀,𝑛 𝑟 𝐴𝑀)Δ 𝑥𝑛 󵄩 󵄩 󵄩 󵄩 󵄩 𝑟𝑀,𝑛 𝑛 󵄩 󵄩 󵄩 󵄩 󵄩 +𝑠𝑛 (𝛾 󵄩𝑉𝑥 𝑛 −𝑉𝑝󵄩 + 󵄩𝛾𝑉𝑝−󵄩 𝜇𝐹𝑝 ) (Θ𝑀,𝜑𝑀) 𝑀−1 󵄩 −𝑇 (𝐼−𝑟𝑀,𝑛𝐴𝑀)Δ 𝑝󵄩 𝑟𝑀,𝑛 𝑛 󵄩 󵄩 󵄩 ≤(1−𝑠𝜏) 󵄩𝑥 −𝑝󵄩 󵄩 𝑛 󵄩 𝑛 󵄩 ≤ 󵄩(𝐼 − 𝑟 𝐴 )Δ𝑀−1𝑥 󵄩 𝑀,𝑛 𝑀 𝑛 𝑛 󵄩 󵄩 󵄩 󵄩 +𝑠𝑛 (𝛾𝑙 󵄩𝑥𝑛 −𝑝󵄩 + 󵄩𝛾𝑉𝑝−󵄩 𝜇𝐹𝑝 ) 󵄩 −(𝐼−𝑟 𝐴 )Δ𝑀−1𝑝󵄩 𝑀,𝑛 𝑀 𝑛 󵄩 󵄩 󵄩 󵄩 󵄩 =(1−𝑠𝑛 (𝜏 − 𝛾𝑙)) 󵄩𝑥𝑛 −𝑝󵄩 +𝑠𝑛 󵄩𝛾𝑉𝑝−󵄩 𝜇𝐹𝑝 . 󵄩 󵄩 (59) ≤ 󵄩Δ𝑀−1𝑥 −Δ𝑀−1𝑝󵄩 (63) 󵄩 𝑛 𝑛 𝑛 󵄩

. ‖𝑥 −𝑝‖ ≤ ‖𝛾𝑉𝑝−𝜇𝐹𝑝‖/(𝜏−𝛾𝑙) {𝑥 } . This implies that 𝑛 .Hence, 𝑛 {𝑢 } 󵄩 󵄩 is bounded. So, according to (59)and(61)weknowthat 𝑛 , ≤ 󵄩Δ0 𝑥 −Δ0 𝑝󵄩 {V } {𝑇 V } {𝑉𝑥 } {𝐹𝑇 V } 󵄩 𝑛 𝑛 𝑛 󵄩 𝑛 , 𝑛 𝑛 , 𝑛 ,and 𝑛 𝑛 are bounded. Nextletusshowthat‖𝑢𝑛 −𝑥𝑛‖→0, ‖V𝑛 −𝑢𝑛‖→0,and 󵄩 󵄩 = 󵄩𝑥𝑛 −𝑝󵄩 . ‖𝑥𝑛 −𝑇𝑛𝑥𝑛‖→0as 𝑛→∞. Abstract and Applied Analysis 9

2󵄩 󵄩2 Indeed, from (27)itfollowsthatforall𝑖 ∈ {1,2,...,𝑁} = (1−𝑠𝑛𝜏) 󵄩V𝑛 −𝑝󵄩 +2𝑠𝑛𝛾 ⟨𝑉𝑥 𝑛 −𝑉𝑝,𝑥𝑛 −𝑝⟩ and 𝑘∈{1,2,...,𝑀} +2𝑠 ⟨𝛾𝑉𝑝− 𝜇𝐹𝑝, 𝑥 −𝑝⟩ 󵄩 󵄩2 𝑛 𝑛 󵄩 󵄩2 󵄩 𝑁 󵄩 󵄩V𝑛 −𝑝󵄩 = 󵄩Λ 𝑛 𝑢𝑛 −𝑝󵄩 󵄩 󵄩 2󵄩 󵄩2 󵄩 󵄩2 ≤(1−𝑠𝑛𝜏) 󵄩V𝑛 −𝑝󵄩 +2𝑠𝑛𝛾𝑙󵄩𝑥𝑛 −𝑝󵄩 󵄩 󵄩2 ≤ 󵄩Λ𝑖 𝑢 −𝑝󵄩 󵄩 𝑛 𝑛 󵄩 󵄩 󵄩 󵄩 󵄩 +2𝑠𝑛 󵄩𝛾𝑉𝑝−󵄩 𝜇𝐹𝑝 󵄩𝑥𝑛 −𝑝󵄩 󵄩 𝑖−1 = 󵄩𝐽 (𝐼 − 𝜆 𝐵 )Λ 𝑢 2 2 󵄩 𝑅𝑖,𝜆𝑖,𝑛 𝑖,𝑛 𝑖 𝑛 𝑛 󵄩 󵄩 ≤(1−𝑠𝑛𝜏) [󵄩𝑢𝑛 −𝑝󵄩 +𝜆𝑖,𝑛 (𝜆𝑖,𝑛 −2𝜂𝑖) 󵄩2 −𝐽𝑅 ,𝜆 (𝐼−𝜆𝑖,𝑛𝐵𝑖)𝑝󵄩 󵄩 󵄩2 𝑖 𝑖,𝑛 󵄩 ×󵄩𝐵 Λ𝑖−1𝑢 −𝐵𝑝󵄩 ] 󵄩 𝑖 𝑛 𝑛 𝑖 󵄩 󵄩 󵄩2 ≤ 󵄩(𝐼 − 𝜆 𝐵 )Λ𝑖−1𝑢 −(𝐼−𝜆 𝐵 )𝑝󵄩 󵄩 𝑖,𝑛 𝑖 𝑛 𝑛 𝑖,𝑛 𝑖 󵄩 󵄩 󵄩2 󵄩 󵄩 󵄩 󵄩 +2𝑠𝑛𝛾𝑙󵄩𝑥𝑛 −𝑝󵄩 +2𝑠𝑛 󵄩𝛾𝑉𝑝−󵄩 𝜇𝐹𝑝 󵄩𝑥𝑛 −𝑝󵄩 󵄩 𝑖−1 󵄩2 ≤ 󵄩Λ 𝑢 −𝑝󵄩 +𝜆 (𝜆 −2𝜂) 2 󵄩 𝑛 𝑛 󵄩 𝑖,𝑛 𝑖,𝑛 𝑖 2 󵄩 󵄩 ≤(1−𝑠𝑛𝜏) [ 󵄩𝑥𝑛 −𝑝󵄩 +𝑟𝑘,𝑛 (𝑟𝑘,𝑛 −2𝜇𝑘) 󵄩 󵄩2 × 󵄩𝐵 Λ𝑖−1𝑢 −𝐵𝑝󵄩 󵄩 𝑖 𝑛 𝑛 𝑖 󵄩 󵄩 𝑘−1 󵄩2 × 󵄩𝐴 Δ 𝑥 −𝐴 𝑝󵄩 󵄩 󵄩2 󵄩 𝑘 𝑛 𝑛 𝑘 󵄩 ≤ 󵄩𝑢𝑛 −𝑝󵄩 +𝜆𝑖,𝑛 (𝜆𝑖,𝑛 −2𝜂𝑖) 󵄩 𝑖−1 󵄩2 󵄩 󵄩2 +𝜆 (𝜆 −2𝜂) 󵄩𝐵 Λ 𝑢 −𝐵𝑝󵄩 ] × 󵄩𝐵 Λ𝑖−1𝑢 −𝐵𝑝󵄩 𝑖,𝑛 𝑖,𝑛 𝑖 󵄩 𝑖 𝑛 𝑛 𝑖 󵄩 󵄩 𝑖 𝑛 𝑛 𝑖 󵄩 󵄩 󵄩2 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩2 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 +2𝑠𝑛𝛾𝑙󵄩𝑥𝑛 −𝑝󵄩 +2𝑠𝑛 󵄩𝛾𝑉𝑝−󵄩 𝜇𝐹𝑝 󵄩𝑥𝑛 −𝑝󵄩 ≤ 󵄩𝑥𝑛 −𝑝󵄩 +𝜆𝑖,𝑛 (𝜆𝑖,𝑛 −2𝜂𝑖) 2 2 󵄩 󵄩2 󵄩 󵄩2 =[1−2𝑠 (𝜏−𝛾𝑙)+𝑠 𝜏 ] 󵄩𝑥 −𝑝󵄩 × 󵄩𝐵 Λ𝑖−1𝑢 −𝐵𝑝󵄩 , 𝑛 𝑛 󵄩 𝑛 󵄩 󵄩 𝑖 𝑛 𝑛 𝑖 󵄩 2 󵄩 𝑘−1 󵄩2 󵄩 󵄩2 󵄩 󵄩2 −(1−𝑠 𝜏) [𝑟 (2𝜇 −𝑟 ) 󵄩𝐴 Δ 𝑥 −𝐴 𝑝󵄩 󵄩𝑢 −𝑝󵄩 = 󵄩Δ𝑀𝑥 −𝑝󵄩 𝑛 𝑘,𝑛 𝑘 𝑘,𝑛 󵄩 𝑘 𝑛 𝑛 𝑘 󵄩 󵄩 𝑛 󵄩 󵄩 𝑛 𝑛 󵄩 󵄩 𝑖−1 󵄩2 󵄩 𝑘 󵄩2 +𝜆 (2𝜂 −𝜆 ) 󵄩𝐵 Λ 𝑢 −𝐵𝑝󵄩 ] ≤ 󵄩Δ 𝑥 −𝑝󵄩 𝑖,𝑛 𝑖 𝑖,𝑛 󵄩 𝑖 𝑛 𝑛 𝑖 󵄩 󵄩 𝑛 𝑛 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 +2𝑠 󵄩𝛾𝑉𝑝−󵄩 𝜇𝐹𝑝 󵄩𝑥 −𝑝󵄩 󵄩 (Θ𝑘,𝜑𝑘) 𝑘−1 𝑛 󵄩 󵄩 󵄩 𝑛 󵄩 = 󵄩𝑇 (𝐼−𝑟𝑘,𝑛𝐴𝑘)Δ 𝑥𝑛 󵄩 𝑟𝑘,𝑛 𝑛 󵄩 󵄩2 2 2󵄩 󵄩2 2 󵄩2 ≤ 󵄩𝑥𝑛 −𝑝󵄩 +𝑠𝑛𝜏 󵄩𝑥𝑛 −𝑝󵄩 −(1−𝑠𝑛𝜏) (Θ𝑘,𝜑𝑘) 󵄩 −𝑇𝑟 (𝐼 −𝑘,𝑛 𝑟 𝐴𝑘)𝑝󵄩 𝑘,𝑛 󵄩 󵄩 𝑘−1 󵄩2 ×[𝑟 (2𝜇 −𝑟 ) 󵄩𝐴 Δ 𝑥 −𝐴 𝑝󵄩 󵄩 󵄩2 𝑘,𝑛 𝑘 𝑘,𝑛 󵄩 𝑘 𝑛 𝑛 𝑘 󵄩 󵄩 𝑘−1 󵄩 ≤ 󵄩(𝐼 −𝑘,𝑛 𝑟 𝐴𝑘)Δ𝑛 𝑥𝑛 −(𝐼−𝑟𝑘,𝑛𝐴𝑘)𝑝󵄩 󵄩 󵄩 󵄩 󵄩2 󵄩 𝑖−1 󵄩 +𝜆𝑖,𝑛 (2𝜂𝑖 −𝜆𝑖,𝑛) 󵄩𝐵𝑖Λ 𝑛 𝑢𝑛 −𝐵𝑖𝑝󵄩 ] 󵄩 𝑘−1 󵄩2 󵄩 󵄩 ≤ 󵄩Δ 𝑥𝑛 −𝑝󵄩 +𝑟𝑘,𝑛 (𝑟𝑘,𝑛 −2𝜇𝑘) 󵄩 𝑛 󵄩 󵄩 󵄩 󵄩 󵄩 +2𝑠𝑛 󵄩𝛾𝑉𝑝−󵄩 𝜇𝐹𝑝 󵄩𝑥𝑛 −𝑝󵄩 , 󵄩 󵄩2 󵄩 󵄩 󵄩 󵄩 × 󵄩𝐴 Δ𝑘−1𝑥 −𝐴 𝑝󵄩 󵄩 𝑘 𝑛 𝑛 𝑘 󵄩 (65) 󵄩 󵄩2 ≤ 󵄩𝑥𝑛 −𝑝󵄩 +𝑟𝑘,𝑛 (𝑟𝑘,𝑛 −2𝜇𝑘) which implies that 󵄩 󵄩2 2 󵄩 𝑘−1 󵄩2 × 󵄩𝐴 Δ𝑘−1𝑥 −𝐴 𝑝󵄩 . (1 − 𝑠 𝜏) [𝑟 (2𝜇 −𝑟 ) 󵄩𝐴 Δ 𝑥 −𝐴 𝑝󵄩 󵄩 𝑘 𝑛 𝑛 𝑘 󵄩 𝑛 𝑘,𝑛 𝑘 𝑘,𝑛 󵄩 𝑘 𝑛 𝑛 𝑘 󵄩 (64) 󵄩 󵄩2 +𝜆 (2𝜂 −𝜆 ) 󵄩𝐵 Λ𝑖−1𝑢 −𝐵𝑝󵄩 ] (66) 𝑖,𝑛 𝑖 𝑖,𝑛 󵄩 𝑖 𝑛 𝑛 𝑖 󵄩 Thus, utilizing Lemma 7,from(49)and(64)wehave 2 2󵄩 󵄩2 󵄩 󵄩 󵄩 󵄩 2 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ≤𝑠𝑛𝜏 󵄩𝑥𝑛 −𝑝󵄩 +2𝑠𝑛 󵄩𝛾𝑉𝑝−󵄩 𝜇𝐹𝑝 󵄩𝑥𝑛 −𝑝󵄩 . 󵄩𝑥𝑛 −𝑝󵄩

󵄩 Since {𝜆𝑖,𝑛}⊂[𝑎𝑖,𝑏𝑖] ⊂ (0, 2𝜂𝑖) and {𝑟𝑘,𝑛}⊂[𝑒𝑘,𝑓𝑘] ⊂ (0, 2𝜇𝑘) = 󵄩𝑠𝑛 (𝛾𝑉𝑥𝑛 −𝜇𝐹𝑝)+(𝐼−𝑠𝑛𝜇𝐹)𝑛 𝑇 V𝑛 for all 𝑖 ∈ {1,2,...,𝑁}and 𝑘 ∈ {1,2,...,𝑀},from𝑠𝑛 →0 󵄩2 we conclude immediately that −(𝐼−𝑠𝑛𝜇𝐹)𝑛 𝑇 𝑝󵄩 󵄩 𝑘−1 󵄩 2 󵄩 󵄩 󵄩 󵄩 lim 󵄩𝐴𝑘Δ 𝑛 𝑥𝑛 −𝐴𝑘𝑝󵄩 =0, ≤ 󵄩(𝐼 −𝑛 𝑠 𝜇𝐹)𝑛 𝑇 V𝑛 −(𝐼−𝑠𝑛𝜇𝐹)𝑛 𝑇 𝑝󵄩 𝑛→∞󵄩 󵄩 󵄩 󵄩 (67) +2𝑠 ⟨𝛾𝑉𝑥 −𝜇𝐹𝑝,𝑥 −𝑝⟩ 󵄩𝐵 Λ𝑖−1𝑢 −𝐵𝑝󵄩 =0, 𝑛 𝑛 𝑛 𝑛→∞lim 󵄩 𝑖 𝑛 𝑛 𝑖 󵄩 2󵄩 󵄩2 ≤(1−𝑠𝑛𝜏) 󵄩V𝑛 −𝑝󵄩 +2𝑠𝑛 ⟨𝛾𝑉𝑥𝑛 −𝜇𝐹𝑝,𝑥𝑛 −𝑝⟩ for all 𝑘∈{1,2,...,𝑀}and 𝑖∈{1,2,...,𝑁}. 10 Abstract and Applied Analysis

1 󵄩 𝑖−1 󵄩2 Furthermore, by Proposition 1(ii) we obtain that for each = (󵄩(𝐼 − 𝜆 𝐵 )Λ 𝑢 −(𝐼−𝜆 𝐵 )𝑝󵄩 𝑘∈{1,2,...,𝑀} 2 󵄩 𝑖,𝑛 𝑖 𝑛 𝑛 𝑖,𝑛 𝑖 󵄩 󵄩 𝑖 󵄩2 󵄩 𝑖−1 󵄩 𝑘 󵄩2 + 󵄩Λ 𝑢 −𝑝󵄩 − 󵄩(𝐼 − 𝜆 𝐵 )Λ 𝑢 󵄩Δ 𝑥 −𝑝󵄩 󵄩 𝑛 𝑛 󵄩 󵄩 𝑖,𝑛 𝑖 𝑛 𝑛 󵄩 𝑛 𝑛 󵄩 󵄩 −(𝐼−𝜆 𝐵 )𝑝 󵄩 (Θ𝑘,𝜑𝑘) 𝑘−1 𝑖,𝑛 𝑖 = 󵄩𝑇 (𝐼 −𝑘,𝑛 𝑟 𝐴𝑘)Δ 𝑥𝑛 󵄩 𝑟𝑘,𝑛 𝑛 󵄩2 −(Λ𝑖 𝑢 −𝑝)󵄩 ) 󵄩2 𝑛 𝑛 󵄩 (Θ𝑘,𝜑𝑘) 󵄩 −𝑇 (𝐼 −𝑘,𝑛 𝑟 𝐴𝑘)𝑝󵄩 𝑟𝑘,𝑛 󵄩 1 󵄩 󵄩2 󵄩 󵄩2 ≤ (󵄩Λ𝑖−1𝑢 −𝑝󵄩 + 󵄩Λ𝑖 𝑢 −𝑝󵄩 𝑘−1 𝑘 󵄩 𝑛 𝑛 󵄩 󵄩 𝑛 𝑛 󵄩 ≤⟨(𝐼−𝑟𝑘,𝑛𝐴𝑘)Δ𝑛 𝑥𝑛 −(𝐼−𝑟𝑘,𝑛𝐴𝑘)𝑝,Δ𝑛𝑥𝑛 −𝑝⟩ 2 󵄩 𝑖−1 𝑖 𝑖−1 󵄩2 1 󵄩 𝑘−1 󵄩2 − 󵄩Λ 𝑢𝑛 −Λ 𝑢𝑛 −𝜆𝑖,𝑛 (𝐵𝑖Λ 𝑢𝑛 −𝐵𝑖𝑝)󵄩 ) = (󵄩(𝐼 − 𝑟 𝐴 )Δ 𝑥 −(𝐼−𝑟 𝐴 )𝑝󵄩 󵄩 𝑛 𝑛 𝑛 󵄩 2 󵄩 𝑘,𝑛 𝑘 𝑛 𝑛 𝑘,𝑛 𝑘 󵄩 1 󵄩 󵄩2 󵄩 𝑖 󵄩2 󵄩 󵄩2 󵄩 ≤ (󵄩𝑢 −𝑝󵄩 + 󵄩Λ 𝑢 −𝑝󵄩 + 󵄩Δ𝑘 𝑥 −𝑝󵄩 − 󵄩(𝐼 − 𝑟 𝐴 )Δ𝑘−1𝑥 󵄩 𝑛 󵄩 󵄩 𝑛 𝑛 󵄩 󵄩 𝑛 𝑛 󵄩 󵄩 𝑘,𝑛 𝑘 𝑛 𝑛 2 󵄩 󵄩2 −(𝐼−𝑟 𝐴 )𝑝 − 󵄩Λ𝑖−1𝑢 −Λ𝑖 𝑢 −𝜆 (𝐵 Λ𝑖−1𝑢 −𝐵𝑝)󵄩 ), 𝑘,𝑛 𝑘 󵄩 𝑛 𝑛 𝑛 𝑛 𝑖,𝑛 𝑖 𝑛 𝑛 𝑖 󵄩 󵄩2 −(Δ𝑘 𝑥 −𝑝)󵄩 ) (70) 𝑛 𝑛 󵄩

1 󵄩 𝑘−1 󵄩2 󵄩 𝑘 󵄩2 ≤ (󵄩Δ 𝑥 −𝑝󵄩 + 󵄩Δ 𝑥 −𝑝󵄩 which implies 2 󵄩 𝑛 𝑛 󵄩 󵄩 𝑛 𝑛 󵄩

󵄩 𝑘−1 𝑘 𝑘−1 󵄩2 −󵄩Δ 𝑥 −Δ 𝑥 −𝑟 (𝐴 Δ 𝑥 −𝐴 𝑝)󵄩 ), 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩 𝑛 𝑛 𝑛 𝑛 𝑘,𝑛 𝑘 𝑛 𝑛 𝑘 󵄩 󵄩Λ𝑖 𝑢 −𝑝󵄩 ≤ 󵄩𝑢 −𝑝󵄩 − 󵄩Λ𝑖−1𝑢 −Λ𝑖 𝑢 󵄩 𝑛 𝑛 󵄩 󵄩 𝑛 󵄩 󵄩 𝑛 𝑛 𝑛 𝑛 (68) 󵄩2 −𝜆 (𝐵 Λ𝑖−1𝑢 −𝐵𝑝)󵄩 which implies that 𝑖,𝑛 𝑖 𝑛 𝑛 𝑖 󵄩 󵄩 󵄩2 󵄩 󵄩2 󵄩 𝑖−1 𝑖 󵄩2 󵄩Δ𝑘 𝑥 −𝑝󵄩 = 󵄩𝑢 −𝑝󵄩 − 󵄩Λ 𝑢 −Λ 𝑢 󵄩 󵄩 𝑛 𝑛 󵄩 󵄩 𝑛 󵄩 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 󵄩 󵄩2 󵄩 𝑘−1 󵄩2 2 󵄩 𝑖−1 󵄩 ≤ 󵄩Δ 𝑥 −𝑝󵄩 −𝜆 󵄩𝐵𝑖Λ 𝑢𝑛 −𝐵𝑖𝑝󵄩 󵄩 𝑛 𝑛 󵄩 𝑖,𝑛󵄩 𝑛 󵄩 󵄩 󵄩2 +2𝜆 ⟨Λ𝑖−1𝑢 −Λ𝑖 𝑢 , (71) − 󵄩Δ𝑘−1𝑥 −Δ𝑘 𝑥 −𝑟 (𝐴 Δ𝑘−1𝑥 −𝐴 𝑝)󵄩 𝑖,𝑛 𝑛 𝑛 𝑛 𝑛 󵄩 𝑛 𝑛 𝑛 𝑛 𝑘,𝑛 𝑘 𝑛 𝑛 𝑘 󵄩 𝑖−1 󵄩 󵄩2 󵄩 󵄩2 𝐵 Λ 𝑢 −𝐵𝑝⟩ = 󵄩Δ𝑘−1𝑥 −𝑝󵄩 − 󵄩Δ𝑘−1𝑥 −Δ𝑘 𝑥 󵄩 𝑖 𝑛 𝑛 𝑖 󵄩 𝑛 𝑛 󵄩 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 󵄩 󵄩2 󵄩 𝑖−1 𝑖 󵄩2 󵄩 󵄩2 ≤ 󵄩𝑢 −𝑝󵄩 − 󵄩Λ 𝑢 −Λ 𝑢 󵄩 −𝑟2 󵄩𝐴 Δ𝑘−1𝑥 −𝐴 𝑝󵄩 󵄩 𝑛 󵄩 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 𝑘,𝑛󵄩 𝑘 𝑛 𝑛 𝑘 󵄩 󵄩 𝑖−1 𝑖 󵄩 +2𝜆 󵄩Λ 𝑢 −Λ 𝑢 󵄩 +2𝑟 ⟨Δ𝑘−1𝑥 −Δ𝑘 𝑥 ,𝐴 Δ𝑘−1𝑥 −𝐴 𝑝⟩ (69) 𝑖,𝑛 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 𝑘,𝑛 𝑛 𝑛 𝑛 𝑛 𝑘 𝑛 𝑛 𝑘 󵄩 󵄩 × 󵄩𝐵 Λ𝑖−1𝑢 −𝐵𝑝󵄩 . 󵄩 󵄩2 󵄩 󵄩2 󵄩 𝑖 𝑛 𝑛 𝑖 󵄩 ≤ 󵄩Δ𝑘−1𝑥 −𝑝󵄩 − 󵄩Δ𝑘−1𝑥 −Δ𝑘 𝑥 󵄩 󵄩 𝑛 𝑛 󵄩 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 󵄩 󵄩 󵄩 󵄩 +2𝑟 󵄩Δ𝑘−1𝑥 −Δ𝑘 𝑥 󵄩 󵄩𝐴 Δ𝑘−1𝑥 −𝐴 𝑝󵄩 𝑘,𝑛 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 󵄩 𝑘 𝑛 𝑛 𝑘 󵄩 Thus, utilizing Lemma 7,from(49), (69), and (71)wehave

󵄩 󵄩2 󵄩 𝑘−1 𝑘 󵄩2 ≤ 󵄩𝑥 −𝑝󵄩 − 󵄩Δ 𝑥 −Δ 𝑥 󵄩 󵄩 𝑛 󵄩 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 󵄩 󵄩2 󵄩𝑥 −𝑝󵄩 󵄩 󵄩 󵄩 𝑛 󵄩 +2𝑟 󵄩Δ𝑘−1𝑥 −Δ𝑘 𝑥 󵄩 𝑘,𝑛 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 󵄩 = 󵄩𝑠𝑛 (𝛾𝑉𝑥𝑛 −𝜇𝐹𝑝)+(𝐼−𝑠𝑛𝜇𝐹)𝑛 𝑇 V𝑛 󵄩 󵄩 × 󵄩𝐴 Δ𝑘−1𝑥 −𝐴 𝑝󵄩 . 󵄩 𝑘 𝑛 𝑛 𝑘 󵄩 󵄩2 −(𝐼−𝑠𝑛𝜇𝐹)𝑛 𝑇 𝑝󵄩 𝑖∈{1,2,...,𝑁} Also, by Lemma 14,weobtainthatforeach 󵄩 󵄩2 ≤ 󵄩(𝐼−𝑠𝑛𝜇𝐹)𝑛 𝑇 V𝑛 −(𝐼−𝑠𝑛𝜇𝐹)𝑛 𝑇 𝑝󵄩 󵄩 󵄩2 󵄩 𝑖 󵄩 󵄩Λ 𝑛𝑢𝑛 −𝑝󵄩 󵄩 󵄩 +2𝑠𝑛 ⟨𝛾𝑉𝑥𝑛 −𝜇𝐹𝑝,𝑥𝑛 −𝑝⟩ 󵄩 󵄩2 󵄩 𝑖−1 󵄩 2󵄩 󵄩2 󵄩 󵄩2 = 󵄩𝐽𝑅 ,𝜆 (𝐼 − 𝜆𝑖,𝑛𝐵𝑖)Λ 𝑢𝑛 −𝐽𝑅 ,𝜆 (𝐼 − 𝜆𝑖,𝑛𝐵𝑖)𝑝󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 𝑖 𝑖,𝑛 𝑛 𝑖 𝑖,𝑛 󵄩 ≤(1−𝑠𝑛𝜏) 󵄩V𝑛 −𝑝󵄩 +2𝑠𝑛𝛾𝑙󵄩𝑥𝑛 −𝑝󵄩 𝑖−1 𝑖 󵄩 󵄩 󵄩 󵄩 ≤⟨(𝐼−𝜆𝑖,𝑛𝐵𝑖)Λ𝑛 𝑢𝑛 −(𝐼−𝜆𝑖,𝑛𝐵𝑖)𝑝,Λ𝑛𝑢𝑛 −𝑝⟩ +2𝑠𝑛 󵄩𝛾𝑉𝑝−󵄩 𝜇𝐹𝑝 󵄩𝑥𝑛 −𝑝󵄩 Abstract and Applied Analysis 11

2󵄩 󵄩2 󵄩 󵄩2 󵄩 𝑖−1 𝑖 󵄩 󵄩 𝑖−1 󵄩 =(1−𝑠𝜏) 󵄩Λ𝑁𝑢 −𝑝󵄩 +2𝑠 𝛾𝑙󵄩𝑥 −𝑝󵄩 +2𝜆 󵄩Λ 𝑢 −Λ 𝑢 󵄩 󵄩𝐵 Λ 𝑢 −𝐵𝑝󵄩 𝑛 󵄩 𝑛 𝑛 󵄩 𝑛 󵄩 𝑛 󵄩 𝑖,𝑛 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 󵄩 𝑖 𝑛 𝑛 𝑖 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 +2𝑠 󵄩𝛾𝑉𝑝−󵄩 𝜇𝐹𝑝 󵄩𝑥 −𝑝󵄩 +2𝑠𝑛 󵄩𝛾𝑉𝑝−󵄩 𝜇𝐹𝑝 󵄩𝑥𝑛 −𝑝󵄩 𝑛 󵄩 󵄩 󵄩 𝑛 󵄩 󵄩 󵄩2 󵄩 󵄩2 2 2󵄩 𝑖 󵄩2 󵄩 󵄩2 󵄩 󵄩 2 2󵄩 󵄩 ≤(1−𝑠𝜏) 󵄩Λ 𝑢 −𝑝󵄩 +2𝑠 𝛾𝑙󵄩𝑥 −𝑝󵄩 ≤ 󵄩𝑥𝑛 −𝑝󵄩 +𝑠 𝜏 󵄩𝑥𝑛 −𝑝󵄩 −(1−𝑠𝑛𝜏) 𝑛 󵄩 𝑛 𝑛 󵄩 𝑛 󵄩 𝑛 󵄩 󵄩 󵄩 𝑛 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩2 󵄩 󵄩2 +2𝑠 󵄩𝛾𝑉𝑝−󵄩 𝜇𝐹𝑝 󵄩𝑥 −𝑝󵄩 ×(󵄩Δ𝑘−1𝑥 −Δ𝑘 𝑥 󵄩 + 󵄩Λ𝑖−1𝑢 −Λ𝑖 𝑢 󵄩 ) 𝑛 󵄩 󵄩 󵄩 𝑛 󵄩 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩 󵄩 󵄩 ≤(1−𝑠𝜏) [󵄩𝑢 −𝑝󵄩 − 󵄩Λ𝑖−1𝑢 −Λ𝑖 𝑢 󵄩 +2𝑟 󵄩Δ𝑘−1𝑥 −Δ𝑘 𝑥 󵄩 󵄩𝐴 Δ𝑘−1𝑥 −𝐴 𝑝󵄩 𝑛 󵄩 𝑛 󵄩 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 𝑘,𝑛 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 󵄩 𝑘 𝑛 𝑛 𝑘 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 +2𝜆 󵄩Λ𝑖−1𝑢 −Λ𝑖 𝑢 󵄩 +2𝜆 󵄩Λ𝑖−1𝑢 −Λ𝑖 𝑢 󵄩 󵄩𝐵 Λ𝑖−1𝑢 −𝐵𝑝󵄩 𝑖,𝑛 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 𝑖,𝑛 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 󵄩 𝑖 𝑛 𝑛 𝑖 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 × 󵄩𝐵 Λ𝑖−1𝑢 −𝐵𝑝󵄩 ] +2𝑠𝑛 󵄩𝛾𝑉𝑝−󵄩 𝜇𝐹𝑝 󵄩𝑥𝑛 −𝑝󵄩 . 󵄩 𝑖 𝑛 𝑛 𝑖 󵄩 (72) 󵄩 󵄩2 󵄩 󵄩 󵄩 󵄩 +2𝑠 𝛾𝑙󵄩𝑥 −𝑝󵄩 +2𝑠 󵄩𝛾𝑉𝑝−󵄩 𝜇𝐹𝑝 󵄩𝑥 −𝑝󵄩 𝑛 󵄩 𝑛 󵄩 𝑛 󵄩 󵄩 󵄩 𝑛 󵄩 It immediately follows that 󵄩 󵄩2 󵄩 󵄩2 2 󵄩 𝑀 󵄩 󵄩 𝑖−1 𝑖 󵄩 2 󵄩 𝑘−1 𝑘 󵄩2 󵄩 𝑖−1 𝑖 󵄩2 =(1−𝑠𝑛𝜏) [󵄩Δ 𝑥𝑛 −𝑝󵄩 − 󵄩Λ 𝑢𝑛 −Λ 𝑢𝑛󵄩 (1 − 𝑠 𝜏) (󵄩Δ 𝑥 −Δ 𝑥 󵄩 + 󵄩Λ 𝑢 −Λ 𝑢 󵄩 ) 󵄩 𝑛 󵄩 󵄩 𝑛 𝑛 󵄩 𝑛 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 󵄩 󵄩 󵄩 𝑖−1 𝑖 󵄩 2 2󵄩 󵄩2 󵄩 𝑘−1 𝑘 󵄩 +2𝜆𝑖,𝑛 󵄩Λ 𝑛 𝑢𝑛 −Λ𝑛𝑢𝑛󵄩 ≤𝑠𝜏 󵄩𝑥 −𝑝󵄩 +2𝑟 󵄩Δ 𝑥 −Δ 𝑥 󵄩 󵄩 󵄩 𝑛 󵄩 𝑛 󵄩 𝑘,𝑛 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 (73) 󵄩 𝑖−1 󵄩 󵄩 󵄩 󵄩 󵄩 × 󵄩𝐵 Λ 𝑢 −𝐵𝑝󵄩 ] × 󵄩𝐴 Δ𝑘−1𝑥 −𝐴 𝑝󵄩 +2𝜆 󵄩Λ𝑖−1𝑢 −Λ𝑖 𝑢 󵄩 󵄩 𝑖 𝑛 𝑛 𝑖 󵄩 󵄩 𝑘 𝑛 𝑛 𝑘 󵄩 𝑖,𝑛 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩2 󵄩 󵄩 󵄩 󵄩 × 󵄩𝐵 Λ𝑖−1𝑢 −𝐵𝑝󵄩 +2𝑠 󵄩𝛾𝑉𝑝−󵄩 𝜇𝐹𝑝 󵄩𝑥 −𝑝󵄩 . +2𝑠𝑛𝛾𝑙󵄩𝑥𝑛 −𝑝󵄩 +2𝑠𝑛 󵄩𝛾𝑉𝑝−󵄩 𝜇𝐹𝑝 󵄩𝑥𝑛 −𝑝󵄩 󵄩 𝑖 𝑛 𝑛 𝑖 󵄩 𝑛 󵄩 󵄩 󵄩 𝑛 󵄩

2 󵄩 𝑘 󵄩2 󵄩 𝑖−1 𝑖 󵄩2 {𝜆 }⊂[𝑎,𝑏] ⊂ (0, 2𝜂 ) {𝑟 }⊂[𝑒,𝑓] ⊂ (0, 2𝜇 ) ≤(1−𝑠𝜏) [󵄩Δ 𝑥 −𝑝󵄩 − 󵄩Λ 𝑢 −Λ 𝑢 󵄩 Since 𝑖,𝑛 𝑖 𝑖 𝑖 and 𝑘,𝑛 𝑘 𝑘 𝑘 𝑛 󵄩 𝑛 𝑛 󵄩 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 for all 𝑖 ∈ {1,2,...,𝑁}and 𝑘 ∈ {1,2,...,𝑀},from(67)and 󵄩 𝑖−1 𝑖 󵄩 𝑠𝑛 →0we deduce that +2𝜆 󵄩Λ 𝑢 −Λ 𝑢 󵄩 𝑖,𝑛 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 󵄩 󵄩 󵄩 󵄩 󵄩Δ𝑘−1𝑥 −Δ𝑘 𝑥 󵄩 =0, 󵄩Λ𝑖−1𝑢 −Λ𝑖 𝑢 󵄩 =0, 󵄩 󵄩 lim 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 lim 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 × 󵄩𝐵 Λ𝑖−1𝑢 −𝐵𝑝󵄩 ] 𝑛→∞ 𝑛→∞ 󵄩 𝑖 𝑛 𝑛 𝑖 󵄩 (74) 󵄩 󵄩2 󵄩 󵄩 󵄩 󵄩 𝑘∈{1,2,...,𝑀} 𝑖∈{1,2,...,𝑁} +2𝑠𝑛𝛾𝑙󵄩𝑥𝑛 −𝑝󵄩 +2𝑠𝑛 󵄩𝛾𝑉𝑝−󵄩 𝜇𝐹𝑝 󵄩𝑥𝑛 −𝑝󵄩 for all and .Hence,weget 󵄩 󵄩 󵄩 0 𝑀 󵄩 2 󵄩 󵄩2 󵄩 󵄩2 󵄩𝑥 −𝑢 󵄩 = 󵄩Δ 𝑥 −Δ 𝑥 󵄩 ≤(1−𝑠𝜏) [󵄩𝑥 −𝑝󵄩 − 󵄩Δ𝑘−1𝑥 −Δ𝑘 𝑥 󵄩 󵄩 𝑛 𝑛󵄩 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 𝑛 󵄩 𝑛 󵄩 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 󵄩 0 1 󵄩 󵄩 1 2 󵄩 󵄩 󵄩 ≤ 󵄩Δ 𝑥 −Δ 𝑥 󵄩 + 󵄩Δ 𝑥 −Δ 𝑥 󵄩 +2𝑟 󵄩Δ𝑘−1𝑥 −Δ𝑘 𝑥 󵄩 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 𝑘,𝑛 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 (75) 󵄩 𝑀−1 𝑀 󵄩 󵄩 󵄩 +⋅⋅⋅+󵄩Δ 𝑥 −Δ 𝑥 󵄩 󳨀→ 0 × 󵄩𝐴 Δ𝑘−1𝑥 −𝐴 𝑝󵄩 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 󵄩 𝑘 𝑛 𝑛 𝑘 󵄩 𝑛󳨀→∞, 󵄩 󵄩2 as − 󵄩Λ𝑖−1𝑢 −Λ𝑖 𝑢 󵄩 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 󵄩 󵄩 󵄩 󵄩 󵄩𝑢 − V 󵄩 = 󵄩Λ0 𝑢 −Λ𝑁𝑢 󵄩 󵄩 󵄩 󵄩 𝑛 𝑛󵄩 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 +2𝜆 󵄩Λ𝑖−1𝑢 −Λ𝑖 𝑢 󵄩 𝑖,𝑛 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 󵄩Λ0 𝑢 −Λ1 𝑢 󵄩 + 󵄩Λ1 𝑢 −Λ2 𝑢 󵄩 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 󵄩 𝑖−1 󵄩 × 󵄩𝐵 Λ 𝑢 −𝐵𝑝󵄩 ] 󵄩 󵄩 (76) 󵄩 𝑖 𝑛 𝑛 𝑖 󵄩 +⋅⋅⋅+󵄩Λ𝑁−1𝑢 −Λ𝑁𝑢 󵄩 󳨀→ 0 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 󵄩 󵄩2 +2𝑠 𝛾𝑙󵄩𝑥 −𝑝󵄩 𝑛 󵄩 𝑛 󵄩 as 𝑛󳨀→∞. 󵄩 󵄩 󵄩 󵄩 +2𝑠 󵄩𝛾𝑉𝑝−󵄩 𝜇𝐹𝑝 󵄩𝑥 −𝑝󵄩 𝑛 󵄩 󵄩 󵄩 𝑛 󵄩 So, taking into account that ‖𝑥𝑛 − V𝑛‖≤‖𝑥𝑛 −𝑢𝑛‖+‖𝑢𝑛 − V𝑛‖, 2 2 󵄩 󵄩2 we have ≤(1−2𝑠𝑛 (𝜏−𝛾𝑙)+𝑠𝑛𝜏 ) 󵄩𝑥𝑛 −𝑝󵄩 󵄩 󵄩 󵄩𝑥 − V 󵄩 =0. 𝑛→∞lim 󵄩 𝑛 𝑛󵄩 (77) 󵄩 󵄩2 −(1−𝑠 𝜏)2 ( 󵄩Δ𝑘−1𝑥 −Δ𝑘 𝑥 󵄩 𝑛 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 Thus, from (77)and𝑠𝑛 →0we have 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩2 󵄩V −𝑇V 󵄩 ≤ 󵄩V −𝑥 󵄩 + 󵄩𝑥 −𝑇V 󵄩 +󵄩Λ𝑖−1𝑢 −Λ𝑖 𝑢 󵄩 ) 󵄩 𝑛 𝑛 𝑛󵄩 󵄩 𝑛 𝑛󵄩 󵄩 𝑛 𝑛 𝑛󵄩 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 󵄩 󵄩 󵄩 󵄩 = 󵄩V𝑛 −𝑥𝑛󵄩 +𝑠𝑛 󵄩𝛾𝑉𝑥𝑛 −𝜇𝐹𝑇𝑛V𝑛󵄩 󳨀→ 0 (78) 󵄩 𝑘−1 𝑘 󵄩 󵄩 𝑘−1 󵄩 +2𝑟 󵄩Δ 𝑥 −Δ 𝑥 󵄩 󵄩𝐴 Δ 𝑥 −𝐴 𝑝󵄩 𝑘,𝑛 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 󵄩 𝑘 𝑛 𝑛 𝑘 󵄩 as 𝑛󳨀→∞. 12 Abstract and Applied Analysis

‖𝑥 −𝑇 𝑥 ‖→0 𝑛→∞ 𝑔−𝐵V ∈𝑅V Λ𝑚𝑢 =𝐽 (𝐼 − Now we show that 𝑛 𝑛 𝑛 as .Infact,from 𝑚 𝑚 . Again, since 𝑛 𝑛 𝑅𝑚,𝜆𝑚,𝑛 𝑚−1 𝑇𝑛 the nonexpansivity of ,wehave 𝜆𝑚,𝑛𝐵𝑚)Λ 𝑛 𝑢𝑛, 𝑛≥1, 𝑚∈{1,2,...,𝑁},wehave 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩𝑥𝑛 −𝑇𝑛𝑥𝑛󵄩 ≤ 󵄩𝑥𝑛 − V𝑛󵄩 + 󵄩V𝑛 −𝑇𝑛V𝑛󵄩 𝑚−1 𝑚−1 𝑚 Λ 𝑛 𝑢𝑛 −𝜆𝑚,𝑛𝐵𝑚Λ 𝑛 𝑢𝑛 ∈(𝐼+𝜆𝑚,𝑛𝑅𝑚)Λ𝑛 𝑢𝑛; (85) 󵄩 󵄩 + 󵄩𝑇𝑛V𝑛 −𝑇𝑛𝑥𝑛󵄩 (79) that is, 󵄩 󵄩 󵄩 󵄩 ≤2󵄩𝑥𝑛 − V𝑛󵄩 + 󵄩V𝑛 −𝑇𝑛V𝑛󵄩 . 1 𝑚−1 𝑚 𝑚−1 𝑚 (Λ 𝑛 𝑢𝑛 −Λ𝑛 𝑢𝑛 −𝜆𝑚,𝑛𝐵𝑚Λ 𝑛 𝑢𝑛) ∈𝑅𝑚Λ 𝑛 𝑢𝑛. (86) By (77)and(78), we get 𝜆𝑚,𝑛 󵄩 󵄩 󵄩𝑥 −𝑇𝑥 󵄩 =0. 𝑅 𝑛→∞lim 󵄩 𝑛 𝑛 𝑛󵄩 (80) In terms of the monotonicity of 𝑚,weget

From (78)itiseasytoseethat 𝑚 1 ⟨V −Λ𝑛 𝑢𝑛,𝑔−𝐵𝑚V − 󵄩 󵄩 𝜆𝑚,𝑛 󵄩𝑥 −𝑇V 󵄩 =0. 𝑛→∞lim 󵄩 𝑛 𝑛 𝑛󵄩 (81) (87) ×(Λ𝑚−1𝑢 −Λ𝑚𝑢 −𝜆 𝐵 Λ𝑚−1𝑢 )⟩≥0 Observe that 𝑛 𝑛 𝑛 𝑛 𝑚,𝑛 𝑚 𝑛 𝑛 󵄩 󵄩 󵄩𝑃 (𝐼 − 𝜆 ∇𝑓 ) V − V 󵄩 󵄩 𝐶 𝑛 𝑛 𝑛󵄩 and hence 󵄩 󵄩 = 󵄩𝑠𝑛V𝑛 +(1−𝑠𝑛)𝑇𝑛V𝑛 − V𝑛󵄩 𝑚 ⟨V −Λ𝑛 𝑢𝑛,𝑔⟩ 󵄩 󵄩 (82) =(1−𝑠) 󵄩𝑇 V − V 󵄩 𝑛 󵄩 𝑛 𝑛 𝑛󵄩 1 ≥⟨V −Λ𝑚𝑢 ,𝐵 V + 󵄩 󵄩 𝑛 𝑛 𝑚 𝜆 ≤ 󵄩𝑇𝑛V𝑛 − V𝑛󵄩 , 𝑚,𝑛 where 𝑠𝑛 =(2−𝜆𝑛𝐿)/4 ∈ (0, 1/2) for each 𝜆𝑛 ∈ (0, 2/𝐿). 𝑚−1 𝑚 𝑚−1 ×(Λ𝑛 𝑢𝑛 −Λ𝑛 𝑢𝑛 −𝜆𝑚,𝑛𝐵𝑚Λ 𝑛 𝑢𝑛)⟩ Hence, we have 󵄩 2 󵄩 󵄩 󵄩 𝑚 𝑚 𝑚 󵄩𝑃𝐶 (𝐼 − ∇𝑓 ) V𝑛 − V𝑛󵄩 =⟨V −Λ 𝑢 ,𝐵 V −𝐵 Λ 𝑢 +𝐵 Λ 𝑢 󵄩 𝐿 󵄩 𝑛 𝑛 𝑚 𝑚 𝑛 𝑛 𝑚 𝑛 𝑛 (88) 󵄩 󵄩 󵄩 2 󵄩 ≤ 󵄩𝑃 (𝐼 − ∇𝑓 ) V −𝑃 (𝐼−𝜆 ∇𝑓 ) V 󵄩 1 󵄩 𝐶 𝐿 𝑛 𝐶 𝑛 𝑛󵄩 −𝐵 Λ𝑚−1𝑢 + (Λ𝑚−1𝑢 −Λ𝑚𝑢 )⟩ 𝑚 𝑛 𝑛 𝜆 𝑛 𝑛 𝑛 𝑛 󵄩 󵄩 𝑚,𝑛 + 󵄩𝑃𝐶 (𝐼 − 𝜆𝑛∇𝑓 ) V𝑛 − V𝑛󵄩 𝑚 𝑚 𝑚−1 󵄩 󵄩 (83) ≥⟨V −Λ𝑛 𝑢𝑛,𝐵𝑚Λ 𝑛 𝑢𝑛 −𝐵𝑚Λ 𝑛 𝑢𝑛⟩ 󵄩 2 󵄩 ≤ 󵄩(𝐼 − ∇𝑓 ) V𝑛 −(𝐼−𝜆𝑛∇𝑓 ) V𝑛󵄩 󵄩 𝐿 󵄩 1 +⟨V −Λ𝑚𝑢 , (Λ𝑚−1𝑢 −Λ𝑚𝑢 )⟩ . 󵄩 󵄩 𝑛 𝑛 𝜆 𝑛 𝑛 𝑛 𝑛 + 󵄩𝑃𝐶 (𝐼 − 𝜆𝑛∇𝑓 ) V𝑛 − V𝑛󵄩 𝑚,𝑛 2 󵄩 󵄩 󵄩 󵄩 ≤( −𝜆 ) 󵄩∇𝑓 ( V )󵄩 + 󵄩𝑇 V − V 󵄩 . In particular, 𝐿 𝑛 󵄩 𝑛 󵄩 󵄩 𝑛 𝑛 𝑛󵄩 𝑚 ⟨V −Λ𝑛 𝑢𝑛 ,𝑔⟩ From the boundedness of {V𝑛}, 𝑠𝑛 → 0 (⇔ 𝜆𝑛 → 2/𝐿) and 𝑖 𝑖 ‖𝑇𝑛V𝑛 − V𝑛‖→0(due to (78)), it follows that 𝑚 𝑚 𝑚−1 ≥⟨V −Λ𝑛 𝑢𝑛 ,𝐵𝑚Λ 𝑛 𝑢𝑛 −𝐵𝑚Λ 𝑛 𝑢𝑛 ⟩ 󵄩 󵄩 𝑖 𝑖 𝑖 𝑖 𝑖 𝑖 (89) 󵄩 2 󵄩 lim 󵄩V𝑛 −𝑃𝐶 (𝐼 − ∇𝑓 ) V𝑛󵄩 =0. (84) 1 𝑛→∞󵄩 𝐿 󵄩 𝑚 𝑚−1 𝑚 +⟨V −Λ𝑛 𝑢𝑛 , (Λ 𝑛 𝑢𝑛 −Λ𝑛 𝑢𝑛 )⟩ . 𝑖 𝑖 𝜆 𝑖 𝑖 𝑖 𝑖 𝑚,𝑛𝑖 Further, we show that 𝜔𝑤(𝑥𝑛)⊂Ω.Indeed,since{𝑥𝑛} is bounded, there exists a subsequence {𝑥𝑛 } of {𝑥𝑛} which 𝑚 𝑚−1 𝑚 𝑖 Since ‖Λ 𝑛 𝑢𝑛 −Λ𝑛 𝑢𝑛‖→0(due to (74)) and ‖𝐵𝑚Λ 𝑛 𝑢𝑛 − converges weakly to some 𝑤.Notethatlim𝑛→∞‖𝑥𝑛 −𝑢𝑛‖=0 𝑚−1 𝐵𝑚Λ 𝑛 𝑢𝑛‖→0(due to the Lipschitz continuity of 𝐵𝑚), we (due to (75)). Hence, 𝑥𝑛 ⇀𝑤.Since𝐶 is closed and convex, 𝑚 𝑖 conclude from Λ 𝑛 𝑢𝑛 ⇀𝑤and condition (ii) that 𝐶 is weakly closed. So, we have 𝑤∈𝐶.From(74)and(75), 𝑖 𝑖 Δ𝑘 𝑥 ⇀𝑤Λ𝑚 𝑢 ⇀𝑤𝑢 ⇀𝑤V ⇀𝑤 we have that 𝑛𝑖 𝑛𝑖 , 𝑛𝑖 𝑛𝑖 , 𝑛𝑖 , 𝑛𝑖 , 𝑚 lim ⟨V −Λ𝑛 𝑢𝑛 ,𝑔⟩=⟨V −𝑤,𝑔⟩≥0. where 𝑘 ∈ {1,2,...,𝑀}, 𝑚 ∈ {1,2,...,𝑁}.First,weprove 𝑖→∞ 𝑖 𝑖 (90) 𝑁 that 𝑤∈∩𝑚=1𝐼(𝐵𝑚,𝑅𝑚). As a matter of fact, since 𝐵𝑚 is 𝜂𝑚- 𝐵 +𝑅 inverse-strongly monotone, 𝐵𝑚 is a monotone and Lipschitz It follows from the maximal monotonicity of 𝑚 𝑚 0∈(𝑅+𝐵)𝑤 𝑤∈𝐼(𝐵,𝑅 ) continuous mapping. It follows from Lemma 17 that 𝑅𝑚 +𝐵𝑚 that 𝑚 𝑚 ;thatis, 𝑚 𝑚 . 𝑁 is maximal monotone. Let (V,𝑔) ∈ 𝐺(𝑅𝑚 +𝐵𝑚);thatis, Therefore, 𝑤∈∩𝑚=1𝐼(𝐵𝑚,𝑅𝑚). Next we prove that Abstract and Applied Analysis 13

𝑀 𝑘 (Θ ,𝜑 ) 𝑤∈∩ (Θ ,𝜑 ,𝐴 ) Δ 𝑥 =𝑇𝑘 𝑘 (𝐼 − 𝑘=1 GMEP 𝑘 𝑘 𝑘 .Since 𝑛 𝑛 𝑟𝑘,𝑛 and hence 𝑘−1 𝑟𝑘,𝑛𝐴𝑘)Δ 𝑛 𝑥𝑛, 𝑛≥1, 𝑘∈{1,2,...,𝑀},wehave 0≤Θ𝑘 (𝑧𝑡,𝑦)+𝜑𝑘 (𝑦) −𝑘 𝜑 (𝑧𝑡)+(1−𝑡) ⟨𝑦 − 𝑤, 𝐴𝑘𝑧𝑡⟩. 𝑘 𝑘 Θ𝑘 (Δ 𝑛𝑥𝑛,𝑦)+𝜑𝑘 (𝑦) −𝜑𝑘 (Δ 𝑛𝑥𝑛) (96)

+⟨𝐴 Δ𝑘−1𝑥 ,𝑦−Δ𝑘 𝑥 ⟩ Letting 𝑡→0, we have, for each 𝑦∈𝐶, 𝑘 𝑛 𝑛 𝑛 𝑛 (91)

1 𝑘 𝑘 𝑘−1 0≤Θ𝑘 (𝑤, 𝑦) +𝜑𝑘 (𝑦) −𝜑𝑘 (𝑤) + ⟨𝑦−𝑤,𝐴𝑘𝑤⟩ . (97) + ⟨𝑦 − Δ 𝑛𝑥𝑛,Δ𝑛𝑥𝑛 −Δ𝑛 𝑥𝑛⟩≥0. 𝑟𝑘,𝑛 This implies that 𝑤∈GMEP(Θ𝑘,𝜑𝑘,𝐴𝑘) and hence 𝑤∈ 𝑀 By (A2), we have ∩𝑘=1 GMEP(Θ𝑘,𝜑𝑘,𝐴𝑘). Further, let us show that 𝑤∈Γ.Asa 𝑘 𝑘−1 𝑘 matter of fact, from (84), V𝑛 ⇀𝑤,andLemma 9,weconclude 𝜑 (𝑦) − 𝜑 (Δ 𝑥 )+⟨𝐴 Δ 𝑥 ,𝑦−Δ 𝑥 ⟩ 𝑖 𝑘 𝑘 𝑛 𝑛 𝑘 𝑛 𝑛 𝑛 𝑛 that 1 + ⟨𝑦 − Δ𝑘 𝑥 ,Δ𝑘 𝑥 −Δ𝑘−1𝑥 ⟩≥Θ (𝑦, Δ𝑘 𝑥 ). 2 𝑛 𝑛 𝑛 𝑛 𝑛 𝑛 𝑘 𝑛 𝑛 𝑤=𝑃𝐶 (𝐼 − ∇𝑓 ) 𝑤. (98) 𝑟𝑘,𝑛 𝐿 (92) 𝑀 So, 𝑤∈VI(𝐶, ∇𝑓). =Γ Therefore, 𝑤∈∩𝑘=1 GMEP(Θ𝑘, Let 𝑧𝑡 = 𝑡𝑦 + (1 − 𝑡)𝑤 for all 𝑡 ∈ (0, 1] and 𝑦∈𝐶. This implies 𝑁 𝜑𝑘,𝐴𝑘)∩∩𝑖=1𝐼(𝐵𝑖,𝑅𝑖)∩Γ=:Ω.Thisshowsthat𝜔𝑤(𝑥𝑛)⊂Ω. that 𝑧𝑡 ∈𝐶.Then,wehave Finally,letusshowthat𝑥𝑛 →𝑞as 𝑛→∞where 𝑞 is ⟨𝑧 −Δ𝑘 𝑥 ,𝐴 𝑧 ⟩ theuniquesolutionoftheVIP(50). Indeed, we note that, for 𝑡 𝑛 𝑛 𝑘 𝑡 𝑤∈Ω 𝑥 ⇀𝑤 with 𝑛𝑖 , 𝑘 𝑘 ≥𝜑𝑘 (Δ 𝑛𝑥𝑛)−𝜑𝑘 (𝑧𝑡)+⟨𝑧𝑡 −Δ𝑛𝑥𝑛,𝐴𝑘𝑧𝑡⟩ 𝑥𝑛 −𝑤=𝑠𝑛 (𝛾𝑉𝑥𝑛 −𝜇𝐹𝑤)+(𝐼−𝑠𝑛𝜇𝐹)𝑛 𝑇 V𝑛 𝑘 𝑘−1 (99) −⟨𝑧𝑡 −Δ𝑛𝑥𝑛,𝐴𝑘Δ 𝑛 𝑥𝑛⟩ −(𝐼−𝑠𝑛𝜇𝐹) 𝑤. 𝑘 𝑘−1 𝑘 Δ 𝑛𝑥𝑛 −Δ𝑛 𝑥𝑛 𝑘 −⟨𝑧𝑡 −Δ𝑛𝑥𝑛, ⟩+Θ𝑘 (𝑧𝑡,Δ𝑛𝑥𝑛) By (61)andLemma 13,weobtainthat 𝑟𝑘,𝑛 󵄩 󵄩2 𝑘 󵄩𝑥𝑛 −𝑤󵄩 =𝑠𝑛 ⟨𝛾𝑉𝑥𝑛 −𝜇𝐹𝑤,𝑥𝑛 −𝑤⟩ =𝜑𝑘 (Δ 𝑛𝑥𝑛)−𝜑𝑘 (𝑧𝑡) +⟨(𝐼−𝑠 𝜇𝐹) 𝑇 V −(𝐼−𝑠 𝜇𝐹) 𝑤, 𝑥 −𝑤⟩ 𝑘 𝑘 𝑛 𝑛 𝑛 𝑛 𝑛 +⟨𝑧𝑡 −Δ𝑛𝑥𝑛,𝐴𝑘𝑧𝑡 −𝐴𝑘Δ 𝑛𝑥𝑛⟩ =𝑠𝑛 ⟨𝛾𝑉𝑥𝑛 −𝜇𝐹𝑤,𝑥𝑛 −𝑤⟩ +⟨𝑧 −Δ𝑘 𝑥 ,𝐴 Δ𝑘 𝑥 −𝐴 Δ𝑘−1𝑥 ⟩ 𝑡 𝑛 𝑛 𝑘 𝑛 𝑛 𝑘 𝑛 𝑛 󵄩 󵄩 󵄩 󵄩 + 󵄩(𝐼 −𝑛 𝑠 𝜇𝐹)𝑛 𝑇 V𝑛 −(𝐼−𝑠𝑛𝜇𝐹)󵄩 𝑤 󵄩𝑥𝑛 −𝑤󵄩 𝑘 𝑘−1 𝑘 Δ 𝑛𝑥𝑛 −Δ𝑛 𝑥𝑛 𝑘 −⟨𝑧𝑡 −Δ𝑛𝑥𝑛, ⟩+Θ𝑘 (𝑧𝑡,Δ𝑛𝑥𝑛). ≤𝑠𝑛 ⟨𝛾𝑉𝑥𝑛 −𝜇𝐹𝑤,𝑥𝑛 −𝑤⟩ 𝑟𝑘,𝑛 󵄩 󵄩 󵄩 󵄩 (93) +(1−𝑠𝑛𝜏) 󵄩V𝑛 −𝑤󵄩 󵄩𝑥𝑛 −𝑤󵄩 ‖𝐴 Δ𝑘 𝑥 −𝐴Δ𝑘−1𝑥 ‖→0 󵄩 󵄩2 By (74), we have 𝑘 𝑛 𝑛 𝑘 𝑛 𝑛 (due ≤𝑠𝑛 ⟨𝛾𝑉𝑥𝑛 −𝜇𝐹𝑤,𝑥𝑛 −𝑤⟩+(1−𝑠𝑛𝜏) 󵄩𝑥𝑛 −𝑤󵄩 . to the Lipschitz continuity of 𝐴𝑘). Furthermore, by the 𝑘 𝑘 (100) monotonicity of 𝐴𝑘,weobtain⟨𝑧𝑡 −Δ𝑛𝑥𝑛,𝐴𝑘𝑧𝑡 −𝐴𝑘Δ 𝑛𝑥𝑛⟩≥ 0.Then,by(A4)weobtain Hence, it follows that ⟨𝑧 −𝑤,𝐴 𝑧 ⟩≥𝜑 (𝑤) −𝜑 (𝑧 )+Θ (𝑧 ,𝑤). 𝑡 𝑘 𝑡 𝑘 𝑘 𝑡 𝑘 𝑡 (94) 󵄩 󵄩2 1 󵄩𝑥𝑛 −𝑤󵄩 ≤ ⟨𝛾𝑉𝑥𝑛 −𝜇𝐹𝑤,𝑥𝑛 −𝑤⟩ Utilizing (A1), (A4), and (94), we obtain 𝜏 1 0=Θ (𝑧 ,𝑧)+𝜑 (𝑧 )−𝜑 (𝑧 ) = (𝛾 ⟨𝑉𝑥 −𝑉𝑤,𝑥 −𝑤⟩ 𝑘 𝑡 𝑡 𝑘 𝑡 𝑘 𝑡 𝜏 𝑛 𝑛

≤𝑡Θ𝑘 (𝑧𝑡,𝑦)+(1−𝑡) Θ𝑘 (𝑧𝑡,𝑤)+𝑡𝜑𝑘 (𝑦) +⟨𝛾𝑉𝑤−𝜇𝐹𝑤,𝑥𝑛 −𝑤⟩) + (1−𝑡) 𝜑 (𝑤) −𝜑 (𝑧 ) 𝑘 𝑘 𝑡 1 󵄩 󵄩2 ≤ (𝛾𝑙󵄩𝑥 −𝑤󵄩 +⟨𝛾𝑉𝑤−𝜇𝐹𝑤,𝑥 −𝑤⟩), 𝜏 󵄩 𝑛 󵄩 𝑛 ≤𝑡[Θ𝑘 (𝑧𝑡,𝑦)+𝜑𝑘 (𝑦) −𝑘 𝜑 (𝑧𝑡)] (95) (101)

+ (1−𝑡) ⟨𝑧𝑡 −𝑤,𝐴𝑘𝑧𝑡⟩ which hence leads to =𝑡[Θ (𝑧 ,𝑦)+𝜑 (𝑦) − 𝜑 (𝑧 )] 𝑘 𝑡 𝑘 𝑘 𝑡 ⟨𝛾𝑉𝑤 − 𝜇𝐹𝑤, 𝑥 −𝑤⟩ 󵄩 󵄩2 𝑛 󵄩𝑥𝑛 −𝑤󵄩 ≤ . (102) + (1−𝑡) 𝑡⟨𝑦−𝑤,𝐴𝑘𝑧𝑡⟩, 𝜏−𝛾𝑙 14 Abstract and Applied Analysis

In particular, we have Corollary 19. Let 𝐶 be a nonempty closed convex subset of a real Hilbert space 𝐻.Let𝑓:𝐶 → R be a convex functional 2 ⟨𝛾𝑉𝑤 − 𝜇𝐹𝑤, 𝑥 −𝑤⟩ 󵄩 󵄩 𝑛𝑖 with 𝐿-Lipschitz continuous gradient ∇𝑓 .LetΘ be a bifunction 󵄩𝑥𝑛 −𝑤󵄩 ≤ . (103) 󵄩 𝑖 󵄩 𝜏−𝛾𝑙 from 𝐶×𝐶to R satisfying (A1)–(A4) and let 𝜑:𝐶 → R ∪ {+∞} be a proper lower semicontinuous and convex function. 𝐻 𝑥𝑛 ⇀𝑤 𝑥𝑛 →𝑤 𝑖→∞ Since 𝑖 ,itfollowsfrom(103)that 𝑖 as . Let 𝑅𝑖 :𝐶 → 2 be a maximal monotone mapping and let 𝑤 𝑥 = Now we show that solves the VIP (50). Since 𝑛 𝐴:𝐻and →𝐻 𝐵𝑖 :𝐶 →be 𝐻 𝜁-inverse-strongly monotone 𝑠 𝛾𝑉𝑥 +(𝐼−𝑠 𝜇𝐹)𝑇 V 𝑛 𝑛 𝑛 𝑛 𝑛,wehave and 𝜂𝑖-inverse-strongly monotone, respectively, for 𝑖=1,2.Let 1 𝐹:𝐻be →𝐻 a 𝜅-Lipschitzian and 𝜂-strongly monotone (𝜇𝐹 − 𝛾𝑉) 𝑥 =− ((𝐼−𝑠 𝜇𝐹) 𝑥 − (𝐼−𝑠 𝜇𝐹) 𝑇 V ) . 𝜅,𝜂>0 𝑉:𝐻 →𝐻 𝑛 𝑠 𝑛 𝑛 𝑛 𝑛 𝑛 operator with positive constants .Let be an 𝑛 𝑙 𝑙≥0 0<𝜇<2𝜂/𝜅2 (104) -Lipschitzian mapping with constant .Let and 0≤𝛾𝑙<𝜏,where𝜏=1−√1 − 𝜇(2𝜂 − 𝜇𝜅2). Assume that It follows that, for each 𝑝∈Ω, Ω := 𝐺𝑀𝐸𝑃(Θ, 𝜑, 1𝐴) ∩𝐼(𝐵 ,𝑅1)∩𝐼(𝐵2,𝑅2)∩Γ=0̸ and that {𝑥 } ⟨(𝜇𝐹 − 𝛾𝑉)𝑛 𝑥 ,𝑥𝑛 −𝑝⟩ either (B1) or (B2) holds. Let 𝑛 be a sequence generated by

1 Θ(𝑢𝑛,𝑦)+𝜑(𝑦)−𝜑(𝑢𝑛)+⟨𝐴𝑥𝑛,𝑦−𝑢𝑛⟩ =− ⟨(𝐼 −𝑛 𝑠 𝜇𝐹)𝑛 𝑥 −(𝐼−𝑠𝑛𝜇𝐹)𝑛 𝑇 V𝑛,𝑥𝑛 −𝑝⟩ 𝑠𝑛 1 + ⟨𝑦 −𝑛 𝑢 ,𝑢𝑛 −𝑥𝑛⟩ ≥ 0, ∀𝑦 ∈ 𝐶, 1 𝑟 𝑛 (109) =− ⟨(𝐼 −𝑛 𝑠 𝜇𝐹)𝑛 𝑥 −(𝐼−𝑠𝑛𝜇𝐹)𝑛 𝑇 V𝑛,𝑥𝑛 −𝑝⟩ 𝑠𝑛 V =𝐽 (𝐼−𝜆 𝐵 )𝐽 (𝐼−𝜆 𝐵 )𝑢 , 𝑛 𝑅2,𝜆2,𝑛 2,𝑛 2 𝑅1,𝜆1,𝑛 1,𝑛 1 𝑛 1 =− ⟨(𝐼 − 𝑠 𝜇𝐹) 𝑥 −(𝐼−𝑠 𝜇𝐹) 𝑇 Λ𝑁Δ𝑀𝑥 ,𝑥 −𝑝⟩ 𝑛 𝑛 𝑛 𝑛 𝑛 𝑛 𝑛 𝑛 𝑥𝑛 =𝑠𝑛𝛾𝑉𝑥𝑛 +(𝐼−𝑠𝑛𝜇𝐹)𝑛 𝑇 V𝑛,∀𝑛≥1, 𝑠𝑛 where 𝑃𝐶(𝐼−𝜆𝑛∇𝑓 ) = 𝑠 𝑛𝐼+(1−𝑠𝑛)𝑇𝑛 (here 𝑇𝑛 is nonexpansive 1 𝑁 𝑀 𝑁 𝑀 =− ⟨(𝐼−𝑇Λ Δ )𝑥 −(𝐼−𝑇Λ Δ )𝑝,𝑥 −𝑝⟩ and 𝑠𝑛 =(2−𝜆𝑛𝐿)/4 ∈ (0, 1/2) for each 𝜆𝑛 ∈ (0, 2/𝐿)). Assume 𝑠 𝑛 𝑛 𝑛 𝑛 𝑛 𝑛 𝑛 𝑛 𝑛 that the following conditions hold: 𝑁 𝑀 𝑠 ∈ (0, 1/2) 𝜆 ∈ (0, 2/𝐿) 𝑠 =0(⇔ +⟨𝜇𝐹𝑥𝑛 −𝜇𝐹𝑇𝑛Λ 𝑛 Δ 𝑛 𝑥𝑛,𝑥𝑛 −𝑝⟩ (i) 𝑛 for each 𝑛 , lim𝑛→∞ 𝑛 lim𝑛→∞𝜆𝑛 = 2/𝐿); ≤⟨𝜇𝐹𝑥 −𝜇𝐹𝑇Λ𝑁Δ𝑀𝑥 ,𝑥 −𝑝⟩, 𝑛 𝑛 𝑛 𝑛 𝑛 𝑛 (ii) {𝜆𝑖,𝑛}⊂[𝑎𝑖,𝑏𝑖] ⊂ (0, 2𝜂𝑖) for 𝑖=1,2;

(105) (iii) {𝑟𝑛}⊂[𝑒,𝑓]⊂(0,2𝜁). 𝑁 𝑀 𝑁 𝑀 {𝑥 } 𝜆 → 2/𝐿 (⇔ 𝑠 →0) since 𝐼−𝑇𝑛Λ 𝑛 Δ 𝑛 is monotone (i.e., ⟨(𝐼 −𝑛 𝑇 Λ 𝑛 Δ 𝑛 )𝑥 − (𝐼 − Then 𝑛 converges strongly as 𝑛 𝑛 to a 𝑁 𝑀 𝑞∈Ω 𝑇𝑛Λ 𝑛 Δ 𝑛 )𝑦, 𝑥 − 𝑦⟩ ≥0 for all 𝑥, 𝑦 ∈𝐻. This is due to point ,whichisauniquesolutionoftheVIP: 𝑇 Λ𝑁Δ𝑀 ‖𝑥 −𝑇V ‖ = ‖(𝐼 − the nonexpansivity of 𝑛 𝑛 𝑛 ). Since 𝑛 𝑛 𝑛 ⟨(𝜇𝐹 − 𝛾𝑉) 𝑞, 𝑝 −𝑞⟩ ≥0, ∀𝑝∈Ω. (110) 𝑁 𝑀 𝑇𝑛Λ 𝑛 Δ 𝑛 )𝑥𝑛‖→0as 𝑛→∞,byreplacing𝑛 in (105)with Corollary 20. Let 𝐶 beanonemptyclosedconvexsubsetofa 𝑛𝑖 and letting 𝑖→∞,weget real Hilbert space 𝐻.Let𝑓:𝐶 → R be a convex functional ⟨(𝜇𝐹 − 𝛾𝑉) 𝑤, 𝑤−𝑝⟩ with 𝐿-Lipschitz continuous gradient ∇𝑓 .LetΘ be a bifunction from 𝐶×𝐶 to R satisfying (A1)–(A4) and let 𝜑:𝐶 → R∪{+∞} = lim ⟨(𝜇𝐹 − 𝛾𝑉)𝑛 𝑥 ,𝑥𝑛 −𝑝⟩ be a proper lower semicontinuous and convex function. Let 𝑖→∞ 𝑖 𝑖 (106) 𝐻 𝑅:𝐶 →2 be a maximal monotone mapping and let ≤ ⟨𝜇𝐹𝑥 −𝜇𝐹𝑇 V ,𝑥 −𝑝⟩=0. 𝐴:𝐻and →𝐻 𝐵:𝐶be →𝐻 𝜁-inverse-strongly lim 𝑛𝑖 𝑛𝑖 𝑛𝑖 𝑛𝑖 𝑖→∞ monotone and 𝜉-inverse-strongly monotone, respectively. Let 𝐹:𝐻 →𝐻 𝜅 𝜂 That is, 𝑤∈Ωis a solution of VIP (50). be a -Lipschitzian and -strongly monotone operator with positive constants 𝜅,𝜂>0.Let𝑉:𝐻 →𝐻be an Finally, in terms of the uniqueness of solutions of VIP 2 𝑤=𝑞 𝑥 →𝑞 𝑛→∞ 𝑙-Lipschitzian mapping with constant 𝑙≥0.Let0<𝜇<2𝜂/𝜅 (50), we deduce that and 𝑛𝑖 as .So,every {𝑥 } 2 weak convergence subsequence of 𝑛 converges strongly to and 0≤𝛾𝑙<𝜏,where𝜏=1−√1 − 𝜇(2𝜂 − 𝜇𝜅 ). Assume that 𝑞 {𝑥 } the unique solution of VIP (50). Therefore, 𝑛 converges Ω := 𝐺𝑀𝐸𝑃(Θ, 𝜑, 𝐴) ∩ 𝐼(𝐵, 𝑅)∩Γ =0̸ and that either (B1) or stronglytotheuniquesolution𝑞 of VIP (50). In addition, the (B2) holds. Let {𝑥𝑛} be a sequence generated by VIP (50)canberewrittenas Θ(𝑢𝑛,𝑦)+𝜑(𝑦)−𝜑(𝑢𝑛)+⟨𝐴𝑥𝑛,𝑦−𝑢𝑛⟩ ⟨(𝐼−𝜇𝐹+𝛾𝑉)𝑞−𝑞,𝑞−𝑝⟩≥0, ∀𝑝∈Ω. (107) 1 + ⟨𝑦 − 𝑢 ,𝑢 −𝑥 ⟩ ≥ 0, ∀𝑦 ∈ 𝐶, By Proposition 2(i), this is equivalent to the fixed point 𝑟 𝑛 𝑛 𝑛 equation: 𝑛 (111) 𝑥𝑛 =𝑠𝑛𝛾𝑉𝑥𝑛 +(𝐼−𝑠𝑛𝜇𝐹)𝑛 𝑇 𝐽𝑅,𝜌 (𝑢𝑛 −𝜌𝑛𝐵𝑢𝑛), 𝑃Ω (𝐼−𝜇𝐹+𝛾𝑉) 𝑞=𝑞. (108) 𝑛 ∀𝑛 ≥ 1, This completes the proof. Abstract and Applied Analysis 15

where 𝑃𝐶(𝐼−𝜆𝑛∇𝑓 ) = 𝑠 𝑛𝐼+(1−𝑠𝑛)𝑇𝑛 (here 𝑇𝑛 is nonexpansive Then {𝑥𝑛} converges strongly as 𝜆𝑛 → 2/𝐿 (⇔𝑛 𝑠 →0)to a and 𝑠𝑛 =(2−𝜆𝑛𝐿)/4 ∈ (0, 1/2) for each 𝜆𝑛 ∈ (0, 2/𝐿)). Assume point 𝑞∈Ω,whichisauniquesolutionofVIP(50). that the following conditions hold: Proof. Firstofall,repeatingthesameargumentsasin (i) 𝑠𝑛 ∈ (0, 1/2) for each 𝜆𝑛 ∈ (0, 2/𝐿), lim𝑛→∞𝑠𝑛 =0(⇔ Theorem 18,wecanwrite lim𝑛→∞𝜆𝑛 = 2/𝐿); {𝜌 } ⊂ [𝑎, 𝑏] ⊂ (0, 2𝜉) (ii) 𝑛 ; 2−𝜆 𝐿 2+𝜆 𝐿 {𝑟 }⊂[𝑒,𝑓]⊂(0,2𝜁) 𝑃 (𝐼−𝜆 ∇𝑓 ) = 𝑛 𝐼+ 𝑛 𝑇 =𝑠𝐼+(1−𝑠 ) 𝑇 , (iii) 𝑛 . 𝐶 𝑛 4 4 𝑛 𝑛 𝑛 𝑛 Then {𝑥𝑛} converges strongly as 𝜆𝑛 → 2/𝐿 (⇔𝑛 𝑠 →0)to a (114) point 𝑞∈Ω,whichisauniquesolutionoftheVIP:

⟨(𝜇𝐹 − 𝛾𝑉) 𝑞, 𝑝 − 𝑞⟩ ≥0,∀𝑝∈Ω. (112) where 𝑇𝑛 is nonexpansive and 𝑠𝑛 := 𝑠𝑛(𝜆𝑛)=(2−𝜆𝑛𝐿)/4 ∈ (0, 1/2) for each 𝜆𝑛 ∈ (0, 2/𝐿).Itisclearthat 4. Explicit Iterative Algorithm 2 and Its Convergence Criteria 𝜆 󳨀→ ⇐⇒ 𝑠 󳨀→ 0. 𝑛 𝐿 𝑛 (115) We next state and prove the second main result of this paper. Put Theorem 21. Let 𝐶 beanonemptyclosedconvexsubsetof 𝐻 𝑓:𝐶 → arealHilbertspace .Let R be a convex 𝑘 (Θ ,𝜑 ) (Θ ,𝜑 ) Δ =𝑇 𝑘 𝑘 (𝐼−𝑟 𝐴 )𝑇 𝑘−1 𝑘−1 functional with 𝐿-Lipschitz continuous gradient ∇𝑓 .Let𝑀, 𝑁 𝑛 𝑟𝑘,𝑛 𝑘,𝑛 𝑘 𝑟𝑘−1,𝑛 Θ 𝐶×𝐶 (116) be two integers. Let 𝑘 be a bifunction from to R (Θ ,𝜑 ) 𝜑 :𝐶 → ∪{+∞} ×(𝐼−𝑟 𝐴 )⋅⋅⋅𝑇 1 1 (𝐼 − 𝑟 𝐴 )𝑥 satisfying (A1)–(A4) and let 𝑘 R be a 𝑘−1,𝑛 𝑘−1 𝑟1,𝑛 1,𝑛 1 𝑛 proper lower semicontinuous and convex function, where 𝑘∈ {1,2,...,𝑀} 𝑅 :𝐶 →𝐻 2 .Let 𝑖 be a maximal monotone for all 𝑘∈{1,2,...,𝑀}and 𝑛≥1, mapping and let 𝐴𝑘 :𝐻 →and 𝐻 𝐵𝑖 :𝐶 →be 𝐻 𝜇𝑘- inverse-strongly monotone and 𝜂𝑖-inverse-strongly monotone, 𝑘 ∈ {1,2,...,𝑀} 𝑖 ∈ {1,2,...,𝑁} Λ𝑖 =𝐽 (𝐼 − 𝜆 𝐵 )𝐽 respectively, where , .Let 𝑛 𝑅𝑖,𝜆𝑖,𝑛 𝑖,𝑛 𝑖 𝑅𝑖−1,𝜆𝑖−1,𝑛 𝐹:𝐻be →𝐻 a 𝜅-Lipschitzian and 𝜂-strongly monotone (117) ×(𝐼−𝜆 𝐵 )⋅⋅⋅𝐽 (𝐼−𝜆 𝐵 ) operator with positive constants 𝜅,𝜂>0.Let𝑉: 𝐻 →𝐻be an 𝑖−1,𝑛 𝑖−1 𝑅1,𝜆1,𝑛 1,𝑛 1 2 𝑙-Lipschitzian mapping with constant 𝑙≥0.Let0<𝜇<2𝜂/𝜅 0≤𝛾𝑙<𝜏 𝜏=1−√1 − 𝜇(2𝜂 −2 𝜇𝜅 ) 0 0 and ,where . Assume that for all 𝑖∈{1,2,...,𝑁}and 𝑛≥1,andΔ 𝑛 =Λ𝑛 =𝐼,where𝐼 is 𝑀 𝑁 𝐻 𝑢 =Δ𝑀𝑥 Ω:=∩𝑘=1𝐺𝑀𝐸𝑃(Θ𝑘,𝜑𝑘,𝐴𝑘)∩∩𝑖=1𝐼(𝐵𝑖,𝑅𝑖)∩Γ=0̸ and that the identity mapping on .Thenwehavethat 𝑛 𝑛 𝑛 and 𝑁 either (B1) or (B2) holds. For arbitrarily given 𝑥1 ∈𝐻,let{𝑥𝑛} V𝑛 =Λ𝑛 𝑢𝑛. In addition, taking into consideration conditions be a sequence generated by (i) and (ii), we may assume, without loss of generality, that 𝑠 ≤1−𝛽 𝑛≥1 (Θ𝑀,𝜑𝑀) (Θ𝑀−1 ,𝜑𝑀−1 ) 𝑛 𝑛 for all . 𝑢𝑛 =𝑇𝑟 (𝐼 −𝑀,𝑛 𝑟 𝐴𝑀)𝑇𝑟 𝑀,𝑛 𝑀−1,𝑛 We divide the remainder of the proof into several steps.

(Θ1,𝜑1) ×(𝐼−𝑟𝑀−1,𝑛𝐴𝑀−1)⋅⋅⋅𝑇𝑟 (𝐼 −1,𝑛 𝑟 𝐴1)𝑥𝑛, 1,𝑛 Step 1. Let us show that ‖𝑥𝑛 −𝑝‖≤ max{‖𝑥1 − 𝑝‖, ‖𝛾𝑉𝑝 − 𝜇𝐹𝑝‖/(𝜏 − 𝛾𝑙)} for all 𝑛≥1and 𝑝∈Ω. Indeed, take 𝑝∈Ω V𝑛 =𝐽𝑅 ,𝜆 (𝐼 − 𝜆𝑁,𝑛𝐵𝑁)𝐽𝑅 ,𝜆 𝑁 𝑁,𝑛 𝑁−1 𝑁−1,𝑛 (113) arbitrarily. Repeating the same arguments as those of (59)– ×(𝐼−𝜆 𝐴 )⋅⋅⋅𝐽 (𝐼 − 𝜆 𝐵 )𝑢 , (61)intheproofofTheorem 18,weobtain 𝑁−1,𝑛 𝑁−1 𝑅1,𝜆1,𝑛 1,𝑛 1 𝑛

𝑥𝑛+1 =𝑠𝑛𝛾𝑉𝑥𝑛 +𝛽𝑛𝑥𝑛 + ((1 − 𝛽𝑛)𝐼−𝑠𝑛𝜇𝐹)𝑛 𝑇 V𝑛, 󵄩 󵄩 󵄩 󵄩 󵄩𝑢𝑛 −𝑝󵄩 ≤ 󵄩𝑥𝑛 −𝑝󵄩 , ∀𝑛 ≥ 1, 󵄩 󵄩 󵄩 󵄩 󵄩V𝑛 −𝑝󵄩 ≤ 󵄩𝑢𝑛 −𝑝󵄩 , (118) where 𝑃𝐶(𝐼−𝜆𝑛∇𝑓 ) = 𝑠 𝑛𝐼+(1−𝑠𝑛)𝑇𝑛 (here 𝑇𝑛 is nonexpansive 󵄩 󵄩 󵄩 󵄩 󵄩V −𝑝󵄩 ≤ 󵄩𝑥 −𝑝󵄩 . and 𝑠𝑛 =(2−𝜆𝑛𝐿)/4 ∈ (0, 1/2) for each 𝜆𝑛 ∈ (0, 2/𝐿)). Assume 󵄩 𝑛 󵄩 󵄩 𝑛 󵄩 that the following conditions hold:

(i) 𝑠𝑛 ∈ (0, 1/2) for each 𝜆𝑛 ∈ (0, 2/𝐿),andlim𝑛→∞𝑠𝑛 = Then from (118), 𝑇𝑛𝑝=𝑝,andLemma 13,wehave 0(⇔lim𝑛→∞𝜆𝑛 = 2/𝐿); {𝛽 } ⊂ (0, 1) 0< 𝛽 ≤ 󵄩 󵄩 (ii) 𝑛 and lim inf𝑛→∞ 𝑛 󵄩𝑥𝑛+1 −𝑝󵄩 lim sup 𝛽𝑛 <1; 𝑛→∞ 󵄩 = 󵄩𝑠𝑛 (𝛾𝑉𝑥𝑛 −𝜇𝐹𝑝)+𝛽𝑛 (𝑥𝑛 −𝑝) (iii) {𝜆𝑖,𝑛}⊂[𝑎𝑖,𝑏𝑖] ⊂ (0, 2𝜂𝑖) and lim𝑛→∞|𝜆𝑖,𝑛+1−𝜆𝑖,𝑛|=0 󵄩 for all 𝑖∈{1,2,...,𝑁}; +((1−𝛽𝑛)𝐼−𝑠𝑛𝜇𝐹)𝑛 𝑇 V𝑛 (iv) {𝑟𝑘,𝑛}⊂[𝑒𝑘,𝑓𝑘] ⊂ (0, 2𝜇𝑘) and lim𝑛→∞|𝑟𝑘,𝑛+1 −𝑟𝑘,𝑛|= 󵄩 0 for all 𝑘∈{1,2,...,𝑀}. − ((1 − 𝛽𝑛)𝐼−𝑠𝑛𝜇𝐹)𝑛 𝑇 𝑝󵄩 16 Abstract and Applied Analysis

󵄩 󵄩 󵄩 󵄩 𝑠𝑛+1 𝑠𝑛 ≤𝑠𝑛 󵄩𝛾𝑉𝑥𝑛 −𝜇𝐹𝑝󵄩 +𝛽𝑛 󵄩𝑥𝑛 −𝑝󵄩 = 𝛾𝑉𝑥 − 𝛾𝑉𝑥 +𝑇 V 1−𝛽 𝑛+1 1−𝛽 𝑛 𝑛+1 𝑛+1 󵄩 𝑛+1 𝑛 󵄩 𝑠 +(1−𝛽 ) 󵄩(𝐼 − 𝑛 𝜇𝐹) 𝑇 V 𝑠 𝑠 𝑛 󵄩 𝑛 𝑛 −𝑇V + 𝑛 𝜇𝐹𝑇 V − 𝑛+1 𝜇𝐹𝑇 V 󵄩 1−𝛽𝑛 𝑛 𝑛 𝑛 𝑛 𝑛+1 𝑛+1 1−𝛽𝑛 1−𝛽𝑛+1 󵄩 𝑠𝑛 󵄩 𝑠 −(𝐼− 𝜇𝐹) 𝑇 𝑝󵄩 𝑛+1 1−𝛽 𝑛 󵄩 = (𝛾𝑉𝑥𝑛+1 −𝜇𝐹𝑇𝑛+1V𝑛+1) 𝑛 󵄩 1−𝛽𝑛+1 󵄩 󵄩 󵄩 󵄩 𝑠 ≤𝑠𝑛 (󵄩𝛾𝑉𝑥𝑛 −𝛾𝑉𝑝󵄩 + 󵄩𝛾𝑉𝑝−󵄩 𝜇𝐹𝑝 ) 𝑛 + (𝜇𝐹𝑇𝑛V𝑛 −𝛾𝑉𝑥𝑛)+𝑇𝑛+1V𝑛+1 −𝑇𝑛V𝑛. 1−𝛽𝑛 󵄩 󵄩 𝑠𝑛𝜏 󵄩 󵄩 +𝛽𝑛 󵄩𝑥𝑛 −𝑝󵄩 +(1−𝛽𝑛)(1− ) 󵄩V𝑛 −𝑝󵄩 (122) 1−𝛽𝑛 󵄩 󵄩 󵄩 󵄩 Thus, it follows that ≤𝑠𝑛 (󵄩𝛾𝑉𝑥𝑛 −𝛾𝑉𝑝󵄩 + 󵄩𝛾𝑉𝑝−󵄩 𝜇𝐹𝑝 ) 󵄩 󵄩 𝑠𝑛+1 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩𝑧𝑛+1 −𝑧𝑛󵄩 ≤ (𝛾 󵄩𝑉𝑥 𝑛+1󵄩 +𝜇󵄩𝐹𝑇𝑛+1V𝑛+1󵄩) +𝛽𝑛 󵄩𝑥𝑛 −𝑝󵄩 +(1−𝛽𝑛 −𝑠𝑛𝜏) 󵄩𝑥𝑛 −𝑝󵄩 1−𝛽𝑛+1 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 𝑠𝑛 󵄩 󵄩 󵄩 󵄩 ≤𝑠𝑛𝛾𝑙 󵄩𝑥𝑛 −𝑝󵄩 +𝑠𝑛 󵄩𝛾𝑉𝑝−󵄩 𝜇𝐹𝑝 + (𝜇 󵄩𝐹𝑇 V 󵄩 +𝛾󵄩𝑉𝑥 󵄩) 1−𝛽 󵄩 𝑛 𝑛󵄩 󵄩 𝑛󵄩 󵄩 󵄩 𝑛 +(1−𝑠 𝜏) 󵄩𝑥 −𝑝󵄩 𝑛 󵄩 𝑛 󵄩 󵄩 󵄩 + 󵄩𝑇𝑛+1V𝑛+1 −𝑇𝑛V𝑛󵄩 . 󵄩 󵄩 =(1−𝑠𝑛 (𝜏 − 𝛾𝑙)) 󵄩𝑥𝑛 −𝑝󵄩 (123) 󵄩 󵄩 󵄩𝛾𝑉𝑝−󵄩 𝜇𝐹𝑝 On the other hand, since ∇𝑓 is (1/𝐿)-ism, 𝑃𝐶(𝐼 − 𝜆𝑛∇𝑓 ) is +𝑠 (𝜏 − 𝛾𝑙) 󵄩 󵄩 𝑛 𝜏−𝛾𝑙 nonexpansive for 𝜆𝑛 ∈ (0, 2/𝐿). So, it follows that, for any 𝑝∈Ω 󵄩 󵄩 given , 󵄩 󵄩 󵄩𝛾𝑉𝑝−󵄩 𝜇𝐹𝑝 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ≤ {󵄩𝑥 −𝑝󵄩 , 󵄩 󵄩}. 󵄩𝑃 (𝐼−𝜆 ∇𝑓 ) V 󵄩 ≤ 󵄩𝑃 (𝐼−𝜆 ∇𝑓 ) V −𝑝󵄩 + 󵄩𝑝󵄩 max 󵄩 𝑛 󵄩 𝜏−𝛾𝑙 󵄩 𝐶 𝑛+1 𝑛󵄩 󵄩 𝐶 𝑛+1 𝑛 󵄩 󵄩 󵄩 󵄩 (119) = 󵄩𝑃𝐶 (𝐼−𝜆𝑛+1∇𝑓 ) V𝑛 󵄩 󵄩 󵄩 −𝑃𝐶 (𝐼 − 𝜆𝑛+1∇𝑓 ) 𝑝 󵄩 + 󵄩𝑝󵄩 By induction, we have 󵄩 󵄩 󵄩 󵄩 ≤ 󵄩V𝑛 −𝑝󵄩 + 󵄩𝑝󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩𝛾𝑉𝑝−󵄩 𝜇𝐹𝑝 ≤ 󵄩V𝑛󵄩 +2󵄩𝑝󵄩 . 󵄩𝑥 −𝑝󵄩 ≤ {󵄩𝑥 −𝑝󵄩 , 󵄩 󵄩}, ∀𝑛≥1. 󵄩 𝑛 󵄩 max 󵄩 1 󵄩 𝜏−𝛾𝑙 (124) (120) This together with the boundedness of {V𝑛} implies that {𝑃𝐶(𝐼 − 𝜆𝑛+1∇𝑓 ) V𝑛} is bounded. Also, observe that 󵄩 󵄩 Hence, {𝑥𝑛} is bounded. According to (118), {𝑢𝑛}, {V𝑛}, {𝑇𝑛V𝑛}, 󵄩 󵄩 󵄩𝑇𝑛+1V𝑛 −𝑇𝑛V𝑛󵄩 {𝑉𝑥𝑛},and{𝐹𝑇𝑛V𝑛} are also bounded. 󵄩 󵄩4𝑃 (𝐼 − 𝜆 ∇𝑓 ) − ( 2 − 𝜆 𝐿) 𝐼 󵄩 𝐶 𝑛+1 𝑛+1 = 󵄩 V𝑛 Step 2. Let us show that ‖𝑥𝑛+1 −𝑥𝑛‖→0as 𝑛→∞.Tothis 󵄩 2+𝜆𝑛+1𝐿 end, define 󵄩 4𝑃 (𝐼 − 𝜆 ∇𝑓 ) − ( 2 − 𝜆 𝐿) 𝐼 󵄩 𝐶 𝑛 𝑛 󵄩 − V𝑛󵄩 2+𝜆𝑛𝐿 󵄩 𝑥𝑛+1 =𝛽𝑛𝑥𝑛 + (1−𝛽𝑛) 𝑧𝑛,∀𝑛≥1. (121) 󵄩 󵄩 󵄩4𝑃𝐶 (𝐼 − 𝜆𝑛+1∇𝑓 ) 4𝑃𝐶 (𝐼−𝜆𝑛∇𝑓 ) 󵄩 ≤ 󵄩 V − V 󵄩 󵄩 2+𝜆 𝐿 𝑛 2+𝜆 𝐿 𝑛󵄩 Observe that, from the definition of 𝑧𝑛, 󵄩 𝑛+1 𝑛 󵄩 󵄩 󵄩 󵄩2−𝜆 𝐿 2−𝜆 𝐿 󵄩 + 󵄩 𝑛 V − 𝑛+1 V 󵄩 󵄩 𝑛 𝑛󵄩 𝑧𝑛+1 −𝑧𝑛 󵄩2+𝜆𝑛𝐿 2+𝜆𝑛+1𝐿 󵄩 󵄩 𝑥 −𝛽 𝑥 𝑥 −𝛽 𝑥 = 󵄩 (4 (2 + 𝜆 𝐿) 𝑃 (𝐼 − 𝜆 ∇𝑓 ) V = 𝑛+2 𝑛+1 𝑛+1 − 𝑛+1 𝑛 𝑛 󵄩 𝑛 𝐶 𝑛+1 𝑛 1−𝛽𝑛+1 1−𝛽𝑛 −4 (2 + 𝜆𝑛+1𝐿)𝐶 𝑃 (𝐼−𝜆𝑛∇𝑓 ) V𝑛) 𝑠 𝛾𝑉𝑥 + ((1 − 𝛽 )𝐼−𝑠 𝜇𝐹) 𝑇 V 𝑛+1 𝑛+1 𝑛+1 𝑛+1 𝑛+1 𝑛+1 󵄩 = ×((2 + 𝜆 𝐿) (2 +𝜆 𝐿))−1󵄩 1−𝛽𝑛+1 𝑛+1 𝑛 󵄩 󵄨 󵄨 4𝐿 󵄨𝜆 −𝜆 󵄨 𝑠𝑛𝛾𝑉𝑥𝑛 + ((1 − 𝛽𝑛)𝐼−𝑠𝑛𝜇𝐹)𝑛 𝑇 V𝑛 󵄨 𝑛+1 𝑛󵄨 󵄩 󵄩 − + 󵄩V𝑛󵄩 1−𝛽𝑛 (2 + 𝜆𝑛+1𝐿) (2 +𝜆𝑛𝐿) Abstract and Applied Analysis 17

󵄩 󵄩 = 󵄩 (4𝐿 (𝜆 −𝜆 ) 𝑃 (𝐼−𝜆 ∇𝑓 ) V +4(2+𝜆 𝐿) ≤ 󵄩(𝐼−𝜆 𝐵 )Λ𝑁−1𝑢 󵄩 𝑛 𝑛+1 𝐶 𝑛+1 𝑛 𝑛+1 󵄩 𝑁,𝑛+1 𝑁 𝑛+1 𝑛+1 󵄩 ×(𝑃 (𝐼 − 𝜆 ∇𝑓 ) V −𝑃 (𝐼 − 𝜆 ∇𝑓 ) V )) −(𝐼−𝜆 𝐵 )Λ𝑁−1𝑢 󵄩 𝐶 𝑛+1 𝑛 𝐶 𝑛 𝑛 𝑁,𝑛 𝑁 𝑛+1 𝑛+1󵄩 −1󵄩 󵄩 ×((2 + 𝜆 𝐿) (2 +𝜆 𝐿)) 󵄩 + 󵄩(𝐼−𝜆 𝐵 )Λ𝑁−1𝑢 𝑛+1 𝑛 󵄩 󵄩 𝑁,𝑛 𝑁 𝑛+1 𝑛+1 󵄨 󵄨 𝑁−1 󵄩 4𝐿 󵄨𝜆𝑛+1 −𝜆𝑛󵄨 󵄩 󵄩 −(𝐼−𝜆 𝐵 )Λ 𝑢 󵄩 + 󵄩V 󵄩 𝑁,𝑛 𝑁 𝑛 𝑛󵄩 (2 + 𝜆 𝐿) (2 +𝜆 𝐿) 󵄩 𝑛󵄩 𝑛+1 𝑛 󵄩 󵄩 󵄨 󵄨 󵄩 𝑁−1 󵄩 󵄨 󵄨 󵄩 󵄩 ≤ 󵄨𝜆𝑁,𝑛+1 −𝜆𝑁,𝑛󵄨 󵄩𝐵𝑁Λ 𝑛+1 𝑢𝑛+1󵄩 4𝐿 󵄨𝜆 −𝜆 󵄨 󵄩𝑃 (𝐼 − 𝜆 ∇𝑓 ) V 󵄩 󵄩 󵄩 ≤ 󵄨 𝑛 𝑛+1󵄨 󵄩 𝐶 𝑛+1 𝑛󵄩 󵄩 󵄩 + 󵄩Λ𝑁−1𝑢 −Λ𝑁−1𝑢 󵄩 (2 + 𝜆𝑛+1𝐿) (2 +𝜆𝑛𝐿) 󵄩 𝑛+1 𝑛+1 𝑛 𝑛󵄩 󵄩 󵄨 󵄨 󵄩 󵄩 +(4(2+𝜆 𝐿) 󵄩𝑃 (𝐼−𝜆 ∇𝑓 ) V ≤ 󵄨𝜆 −𝜆 󵄨 󵄩𝐵 Λ𝑁−1𝑢 󵄩 𝑛+1 󵄩 𝐶 𝑛+1 𝑛 󵄨 𝑁,𝑛+1 𝑁,𝑛󵄨 󵄩 𝑁 𝑛+1 𝑛+1󵄩 󵄩 󵄨 󵄨 󵄩 󵄩 −𝑃𝐶 (𝐼−𝜆𝑛∇𝑓 ) V𝑛󵄩) + 󵄨𝜆 −𝜆 󵄨 󵄩𝐵 Λ𝑁−2𝑢 󵄩 󵄨 𝑁−1,𝑛+1 𝑁−1,𝑛󵄨 󵄩 𝑁−1 𝑛+1 𝑛+1󵄩 × ((2 + 𝜆 𝐿) (2 +𝜆 𝐿))−1 󵄩 󵄩 𝑛+1 𝑛 + 󵄩Λ𝑁−2𝑢 −Λ𝑁−2𝑢 󵄩 󵄩 𝑛+1 𝑛+1 𝑛 𝑛󵄩 󵄨 󵄨 4𝐿 󵄨𝜆 −𝜆 󵄨 󵄨 𝑛+1 𝑛󵄨 󵄩 󵄩 + 󵄩V𝑛󵄩 . (2 + 𝜆𝑛+1𝐿) (2 +𝜆𝑛𝐿) . 󵄨 󵄨 󵄩 󵄩 󵄨 󵄨 󵄩 󵄩 ≤ 󵄨𝜆 −𝜆 󵄨 [𝐿 󵄩𝑃 (𝐼−𝜆 ∇𝑓 ) V 󵄩 ≤ 󵄨𝜆 −𝜆 󵄨 󵄩𝐵 Λ𝑁−1𝑢 󵄩 󵄨 𝑛+1 𝑛󵄨 󵄩 𝐶 𝑛+1 𝑛󵄩 󵄨 𝑁,𝑛+1 𝑁,𝑛󵄨 󵄩 𝑁 𝑛+1 𝑛+1󵄩 󵄩 󵄩 󵄩 󵄩 󵄨 󵄨 󵄩 󵄩 +4 󵄩∇𝑓 ( V )󵄩 +𝐿󵄩V 󵄩] + 󵄨𝜆 −𝜆 󵄨 󵄩𝐵 Λ𝑁−2𝑢 󵄩 󵄩 𝑛 󵄩 󵄩 𝑛󵄩 󵄨 𝑁−1,𝑛+1 𝑁−1,𝑛󵄨 󵄩 𝑁−1 𝑛+1 𝑛+1󵄩 󵄨 󵄨 ̃ 󵄨 󵄨 󵄨 󵄨 󵄩 0 󵄩 ≤ 𝑀 󵄨𝜆𝑛+1 −𝜆𝑛󵄨 , +⋅⋅⋅+󵄨𝜆 −𝜆 󵄨 󵄩𝐵 Λ 𝑢 󵄩 󵄨 1,𝑛+1 1,𝑛󵄨 󵄩 1 𝑛+1 𝑛+1󵄩 (125) 󵄩 󵄩 + 󵄩Λ0 𝑢 −Λ0 𝑢 󵄩 󵄩 𝑛+1 𝑛+1 𝑛 𝑛󵄩

𝑁 {𝐿‖𝑃 (𝐼−𝜆 ∇𝑓 ) V ‖+4‖∇𝑓(V )‖ + 𝐿‖V ‖} ≤ 𝑀̃ 󵄨 󵄨 󵄩 󵄩 where sup𝑛≥1 𝐶 𝑛+1 𝑛 𝑛 𝑛 ≤ 𝑀̃ ∑ 󵄨𝜆 −𝜆 󵄨 + 󵄩𝑢 −𝑢 󵄩 , ̃ 0 󵄨 𝑖,𝑛+1 𝑖,𝑛󵄨 󵄩 𝑛+1 𝑛󵄩 for some 𝑀>0.So,by(125), we have that 𝑖=1 (127) 󵄩 󵄩 󵄩 󵄩 󵄩𝑇𝑛+1V𝑛+1 −𝑇𝑛V𝑛󵄩 ≤ 󵄩𝑇𝑛+1V𝑛+1 −𝑇𝑛+1V𝑛󵄩 𝑁 󵄩 󵄩 {∑ ‖𝐵 Λ𝑖−1 𝑢 ‖} ≤ 𝑀̃ 𝑀̃ >0 + 󵄩𝑇 V −𝑇V 󵄩 where sup𝑛≥1 𝑖=1 𝑖 𝑛+1 𝑛+1 0 for some 0 . 󵄩 𝑛+1 𝑛 𝑛 𝑛󵄩 Also, utilizing Proposition 1(v)wededucethat 󵄩 󵄩 ̃ 󵄨 󵄨 (126) ≤ 󵄩V𝑛+1 − V𝑛󵄩 + 𝑀 󵄨𝜆𝑛+1 −𝜆𝑛󵄨 󵄩 󵄩 ̃ 󵄩𝑢𝑛+1 −𝑢𝑛󵄩 󵄩 󵄩 4𝑀 ≤ 󵄩V𝑛+1 − V𝑛󵄩 + (𝑠𝑛+1 +𝑠𝑛). 󵄩 󵄩 𝐿 = 󵄩Δ𝑀 𝑥 −Δ𝑀𝑥 󵄩 󵄩 𝑛+1 𝑛+1 𝑛 𝑛󵄩 󵄩 󵄩 (Θ𝑀,𝜑𝑀) 𝑀−1 = 󵄩𝑇𝑟 (𝐼−𝑟𝑀,𝑛+1𝐴𝑀)Δ𝑛+1 𝑥𝑛+1 Note that 󵄩 𝑀,𝑛+1 󵄩 (Θ𝑀,𝜑𝑀) 𝑀−1 󵄩 −𝑇𝑟 (𝐼 −𝑀,𝑛 𝑟 𝐴𝑀)Δ𝑛 𝑥𝑛󵄩 󵄩 󵄩 𝑀,𝑛 󵄩 󵄩 󵄩 󵄩 𝑁 𝑁 󵄩 󵄩V𝑛+1 − V𝑛󵄩 = 󵄩Λ 𝑛+1𝑢𝑛+1 −Λ𝑛 𝑢𝑛󵄩 󵄩 󵄩 󵄩 󵄩 (Θ𝑀,𝜑𝑀) 𝑀−1 ≤ 󵄩𝑇𝑟 (𝐼−𝑟𝑀,𝑛+1𝐴𝑀)Δ𝑛+1 𝑥𝑛+1 󵄩 󵄩 𝑀,𝑛+1 = 󵄩𝐽 (𝐼 − 𝜆 𝐵 )Λ𝑁−1𝑢 󵄩 𝑅𝑁,𝜆𝑁,𝑛+1 𝑁,𝑛+1 𝑁 𝑛+1 𝑛+1 󵄩 (Θ𝑀,𝜑𝑀) 𝑀−1 󵄩 −𝑇𝑟 (𝐼 −𝑀,𝑛 𝑟 𝐴𝑀)Δ𝑛+1 𝑥𝑛+1󵄩 𝑁−1 󵄩 𝑀,𝑛 󵄩 −𝐽𝑅 ,𝜆 (𝐼 − 𝜆𝑁,𝑛𝐵𝑁)Λ 𝑢𝑛󵄩 𝑁 𝑁,𝑛 𝑛 󵄩 󵄩 󵄩 (Θ𝑀,𝜑𝑀) 𝑀−1 󵄩 + 󵄩𝑇 (𝐼 −𝑀,𝑛 𝑟 𝐴𝑀)Δ 𝑥𝑛+1 󵄩 𝑁−1 󵄩 𝑟𝑀,𝑛 𝑛+1 ≤ 󵄩𝐽𝑅 ,𝜆 (𝐼 − 𝜆𝑁,𝑛+1𝐵𝑁)Λ 𝑢𝑛+1 󵄩 𝑁 𝑁,𝑛+1 𝑛+1 󵄩 (Θ𝑀,𝜑𝑀) 𝑀−1 󵄩 𝑁−1 󵄩 −𝑇 (𝐼−𝑟𝑀,𝑛𝐴𝑀)Δ 𝑥𝑛󵄩 −𝐽 (𝐼−𝜆 𝐵 )Λ 𝑢 󵄩 𝑟𝑀,𝑛 𝑛 󵄩 𝑅𝑁,𝜆𝑁,𝑛 𝑁,𝑛 𝑁 𝑛+1 𝑛+1󵄩 󵄩 (Θ ,𝜑 ) 𝑀−1 󵄩 𝑁−1 ≤ 󵄩𝑇 𝑀 𝑀 (𝐼−𝑟 𝐴 )Δ 𝑥 + 󵄩𝐽 (𝐼−𝜆 𝐵 )Λ 𝑢 󵄩 𝑟𝑀,𝑛+1 𝑀,𝑛+1 𝑀 𝑛+1 𝑛+1 󵄩 𝑅𝑁,𝜆𝑁,𝑛 𝑁,𝑛 𝑁 𝑛+1 𝑛+1 󵄩 󵄩 󵄩 𝑁−1 󵄩 (Θ𝑀,𝜑𝑀) 𝑀−1 󵄩 −𝐽 (𝐼 − 𝜆 𝐵 )Λ 𝑢 󵄩 −𝑇𝑟 (𝐼 −𝑀,𝑛+1 𝑟 𝐴𝑀)Δ𝑛+1 𝑥𝑛+1󵄩 𝑅𝑁,𝜆𝑁,𝑛 𝑁,𝑛 𝑁 𝑛 𝑛󵄩 𝑀,𝑛 󵄩 18 Abstract and Applied Analysis 󵄩 󵄩 (Θ𝑀,𝜑𝑀) 𝑀−1 ̃ + 󵄩𝑇 (𝐼 −𝑀,𝑛+1 𝑟 𝐴𝑀)Δ 𝑥𝑛+1 where 𝑀1 >0isaconstantsuchthatforeach𝑛≥1 󵄩 𝑟𝑀,𝑛 𝑛+1 󵄩 (Θ𝑀,𝜑𝑀) 𝑀−1 󵄩 𝑀 −𝑇 (𝐼−𝑟𝑀,𝑛𝐴𝑀)Δ 𝑥𝑛+1󵄩 𝑟𝑀,𝑛 𝑛+1 󵄩 󵄩 𝑘−1 󵄩 1 ∑ [󵄩𝐴 Δ 𝑥 󵄩 + 󵄩 𝑘 𝑛+1 𝑛+1󵄩 𝑟 󵄩 󵄩 𝑘=1 𝑘,𝑛+1 + 󵄩(𝐼 − 𝑟 𝐴 )Δ𝑀−1𝑥 −(𝐼−𝑟 𝐴 )Δ𝑀−1𝑥 󵄩 󵄩 𝑀,𝑛 𝑀 𝑛+1 𝑛+1 𝑀,𝑛 𝑀 𝑛 𝑛󵄩 󵄩 󵄩 (Θ𝑘,𝜑𝑘) 𝑘−1 × 󵄩𝑇𝑟 (𝐼−𝑟𝑘,𝑛+1𝐴𝑘)Δ𝑛+1𝑥𝑛+1 (129) 󵄨 󵄨 󵄩 𝑘,𝑛+1 󵄨𝑟 −𝑟 󵄨 󵄩 󵄨 𝑀,𝑛+1 𝑀,𝑛󵄨 󵄩 (Θ𝑀,𝜑𝑀) 𝑀−1 ≤ 󵄩𝑇 (𝐼−𝑟𝑀,𝑛+1𝐴𝑀)Δ 𝑥𝑛+1 󵄩 󵄩 𝑟𝑀,𝑛+1 𝑛+1 𝑘−1 󵄩 𝑟𝑀,𝑛+1 −(𝐼−𝑟 𝐴 )Δ 𝑥 󵄩 ]≤𝑀̃ . 𝑘,𝑛+1 𝑘 𝑛+1 𝑛+1󵄩 1 󵄩 −(𝐼−𝑟 𝐴 )Δ𝑀−1𝑥 󵄩 𝑀,𝑛+1 𝑀 𝑛+1 𝑛+1󵄩 󵄨 󵄨 󵄩 󵄩 Combining (123)–(128), we get + 󵄨𝑟 −𝑟 󵄨 󵄩𝐴 Δ𝑀−1𝑥 󵄩 󵄨 𝑀,𝑛+1 𝑀,𝑛󵄨 󵄩 𝑀 𝑛+1 𝑛+1󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 + 󵄩Δ𝑀−1𝑥 −Δ𝑀−1𝑥 󵄩 󵄩𝑧 −𝑧 󵄩 − 󵄩𝑥 −𝑥 󵄩 󵄩 𝑛+1 𝑛+1 𝑛 𝑛󵄩 󵄩 𝑛+1 𝑛󵄩 󵄩 𝑛+1 𝑛󵄩 󵄨 󵄨 𝑠 = 󵄨𝑟 −𝑟 󵄨 𝑛+1 󵄩 󵄩 󵄩 󵄩 󵄨 𝑀,𝑛+1 𝑀,𝑛󵄨 ≤ (𝛾 󵄩𝑉𝑥 𝑛+1󵄩 +𝜇󵄩𝐹𝑇𝑛+1V𝑛+1󵄩) 1−𝛽𝑛+1 󵄩 󵄩 󵄩 𝑀−1 󵄩 𝑠 󵄩 󵄩 󵄩 󵄩 ×[󵄩𝐴𝑀Δ 𝑛+1 𝑥𝑛+1󵄩 𝑛 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 + (𝜇 󵄩𝐹𝑇𝑛V𝑛󵄩 +𝛾󵄩𝑉𝑥 𝑛󵄩) 1−𝛽𝑛 1 󵄩 󵄩 (Θ𝑀,𝜑𝑀) 𝑀−1 󵄩 󵄩 󵄩 󵄩 + 󵄩𝑇 (𝐼−𝑟𝑀,𝑛+1𝐴𝑀)Δ 𝑥𝑛+1 + 󵄩𝑇𝑛+1V𝑛+1 −𝑇𝑛V𝑛󵄩 − 󵄩𝑥𝑛+1 −𝑥𝑛󵄩 󵄩 𝑟𝑀,𝑛+1 𝑛+1 󵄩 󵄩 󵄩 󵄩 𝑟𝑀,𝑛+1 𝑠 󵄩 󵄩 󵄩 󵄩 ≤ 𝑛+1 (𝛾 󵄩𝑉𝑥 󵄩 +𝜇󵄩𝐹𝑇 V 󵄩) 𝑀−1 󵄩 󵄩 𝑛+1󵄩 󵄩 𝑛+1 𝑛+1󵄩 −(𝐼−𝑟 𝐴 )Δ 𝑥 󵄩 ] 1−𝛽𝑛+1 𝑀,𝑛+1 𝑀 𝑛+1 𝑛+1󵄩 𝑠 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 + 𝑛 (𝜇 󵄩𝐹𝑇 V 󵄩 +𝛾󵄩𝑉𝑥 󵄩) + 󵄩Δ𝑀−1𝑥 −Δ𝑀−1𝑥 󵄩 1−𝛽 󵄩 𝑛 𝑛󵄩 󵄩 𝑛󵄩 󵄩 𝑛+1 𝑛+1 𝑛 𝑛󵄩 𝑛 󵄩 󵄩 4𝑀̃ 󵄩 󵄩 . + 󵄩V − V 󵄩 + (𝑠 +𝑠 )−󵄩𝑥 −𝑥 󵄩 . 󵄩 𝑛+1 𝑛󵄩 𝐿 𝑛+1 𝑛 󵄩 𝑛+1 𝑛󵄩 󵄨 󵄨 ≤ 󵄨𝑟 −𝑟 󵄨 𝑠 󵄩 󵄩 󵄩 󵄩 󵄨 𝑀,𝑛+1 𝑀,𝑛󵄨 ≤ 𝑛+1 (𝛾 󵄩𝑉𝑥 󵄩 +𝜇󵄩𝐹𝑇 V 󵄩) 1−𝛽 󵄩 𝑛+1󵄩 󵄩 𝑛+1 𝑛+1󵄩 󵄩 󵄩 𝑛+1 ×[󵄩𝐴 Δ𝑀−1𝑥 󵄩 󵄩 𝑀 𝑛+1 𝑛+1󵄩 𝑠𝑛 󵄩 󵄩 󵄩 󵄩 + (𝜇 󵄩𝐹𝑇𝑛V𝑛󵄩 +𝛾󵄩𝑉𝑥 𝑛󵄩) 1−𝛽𝑛 1 󵄩 󵄩 (Θ𝑀,𝜑𝑀) 𝑀−1 + 󵄩𝑇 (𝐼−𝑟𝑀,𝑛+1𝐴𝑀)Δ 𝑥𝑛+1 𝑁 󵄩 𝑟𝑀,𝑛+1 𝑛+1 𝑟𝑀,𝑛+1 ̃ 󵄨 󵄨 󵄩 󵄩 + 𝑀0∑ 󵄨𝜆𝑖,𝑛+1 −𝜆𝑖,𝑛󵄨 + 󵄩𝑢𝑛+1 −𝑢𝑛󵄩 𝑖=1 󵄩 −(𝐼−𝑟 𝐴 )Δ𝑀−1𝑥 󵄩] 𝑀,𝑛+1 𝑀 𝑛+1 𝑛+1󵄩 ̃ 4𝑀 󵄩 󵄩 + (𝑠𝑛+1 +𝑠𝑛)−󵄩𝑥𝑛+1 −𝑥𝑛󵄩 󵄨 󵄨 𝐿 +⋅⋅⋅+󵄨𝑟1,𝑛+1 −𝑟1,𝑛󵄨 𝑠 󵄩 󵄩 󵄩 󵄩 ≤ 𝑛+1 (𝛾 󵄩𝑉𝑥 󵄩 +𝜇󵄩𝐹𝑇 V 󵄩) 󵄩 󵄩 1−𝛽 󵄩 𝑛+1󵄩 󵄩 𝑛+1 𝑛+1󵄩 ×[󵄩𝐴 Δ0 𝑥 󵄩 𝑛+1 󵄩 1 𝑛+1 𝑛+1󵄩 𝑠 󵄩 󵄩 󵄩 󵄩 + 𝑛 (𝜇 󵄩𝐹𝑇 V 󵄩 +𝛾󵄩𝑉𝑥 󵄩) 1 󵄩 1−𝛽 󵄩 𝑛 𝑛󵄩 󵄩 𝑛󵄩 󵄩 (Θ1,𝜑1) 0 𝑛 + 󵄩𝑇 (𝐼−𝑟1,𝑛+1𝐴1)Δ 𝑥𝑛+1 󵄩 𝑟1,𝑛+1 𝑛+1 𝑟1,𝑛+1 𝑁 𝑀 ̃ 󵄨 󵄨 ̃ 󵄨 󵄨 + 𝑀0∑ 󵄨𝜆𝑖,𝑛+1 −𝜆𝑖,𝑛󵄨 + 𝑀1 ∑ 󵄨𝑟𝑘,𝑛+1 −𝑟𝑘,𝑛󵄨 󵄩 −(𝐼−𝑟 𝐴 )Δ0 𝑥 󵄩] 𝑖=1 𝑘=1 1,𝑛+1 1 𝑛+1 𝑛+1󵄩 󵄩 󵄩 4𝑀̃ 󵄩 󵄩 󵄩 󵄩 + 󵄩𝑥 −𝑥 󵄩 + (𝑠 +𝑠 )−󵄩𝑥 −𝑥 󵄩 + 󵄩Δ0 𝑥 −Δ0 𝑥 󵄩 󵄩 𝑛+1 𝑛󵄩 𝑛+1 𝑛 󵄩 𝑛+1 𝑛󵄩 󵄩 𝑛+1 𝑛+1 𝑛 𝑛󵄩 𝐿 𝑠𝑛+1 󵄩 󵄩 󵄩 󵄩 𝑀 = (𝛾 󵄩𝑉𝑥 𝑛+1󵄩 +𝜇󵄩𝐹𝑇𝑛+1V𝑛+1󵄩) ̃ 󵄨 󵄨 󵄩 󵄩 1−𝛽𝑛+1 ≤ 𝑀1 ∑ 󵄨𝑟𝑘,𝑛+1 −𝑟𝑘,𝑛󵄨 + 󵄩𝑥𝑛+1 −𝑥𝑛󵄩 , 𝑘=1 𝑠𝑛 󵄩 󵄩 󵄩 󵄩 + (𝜇 󵄩𝐹𝑇𝑛V𝑛󵄩 +𝛾󵄩𝑉𝑥 𝑛󵄩) (128) 1−𝛽𝑛 Abstract and Applied Analysis 19

𝑁 𝑀 ̃ 󵄨 󵄨 ̃ 󵄨 󵄨 Also, from (27)itfollowsthatforall𝑖∈{1,2,...,𝑁}and 𝑘∈ + 𝑀0∑ 󵄨𝜆𝑖,𝑛+1 −𝜆𝑖,𝑛󵄨 + 𝑀1 ∑ 󵄨𝑟𝑘,𝑛+1 −𝑟𝑘,𝑛󵄨 {1,2,...,𝑀} 𝑖=1 𝑘=1 󵄩 󵄩2 󵄩 󵄩2 4𝑀̃ 󵄩V −𝑝󵄩 = 󵄩Λ𝑁𝑢 −𝑝󵄩 + (𝑠 +𝑠 ). 󵄩 𝑛 󵄩 󵄩 𝑛 𝑛 󵄩 𝐿 𝑛+1 𝑛 󵄩 󵄩2 ≤ 󵄩Λ𝑖 𝑢 −𝑝󵄩 (130) 󵄩 𝑛 𝑛 󵄩 󵄩 𝑖−1 = 󵄩𝐽 (𝐼 − 𝜆 𝐵 )Λ 𝑢 Thus, it follows from (130) and conditions (i)–(iv) that 󵄩 𝑅𝑖,𝜆𝑖,𝑛 𝑖,𝑛 𝑖 𝑛 𝑛 󵄩2 󵄩 󵄩 󵄩 󵄩 󵄩 (󵄩𝑧 −𝑧 󵄩 − 󵄩𝑥 −𝑥 󵄩)≤0. −𝐽𝑅 ,𝜆 (𝐼 − 𝜆𝑖,𝑛𝐵𝑖)𝑝󵄩 lim sup 󵄩 𝑛+1 𝑛󵄩 󵄩 𝑛+1 𝑛󵄩 (131) 𝑖 𝑖,𝑛 󵄩 𝑛→∞ 󵄩 ≤ 󵄩(𝐼 − 𝜆 𝐵 )Λ𝑖−1𝑢 󵄩 𝑖,𝑛 𝑖 𝑛 𝑛 Hence, by Lemma 11 we have 󵄩2 −(𝐼−𝜆 𝐵 )𝑝󵄩 󵄩 󵄩 𝑖,𝑛 𝑖 󵄩 󵄩𝑧 −𝑥 󵄩 =0. lim 󵄩 𝑛 𝑛󵄩 (132) 󵄩 󵄩2 𝑛→∞ ≤ 󵄩Λ𝑖−1𝑢 −𝑝󵄩 +𝜆 (𝜆 −2𝜂) 󵄩 𝑛 𝑛 󵄩 𝑖,𝑛 𝑖,𝑛 𝑖 󵄩 󵄩2 Consequently, × 󵄩𝐵 Λ𝑖−1𝑢 −𝐵𝑝󵄩 󵄩 𝑖 𝑛 𝑛 𝑖 󵄩 󵄩 󵄩 󵄩 󵄩 lim 󵄩𝑥𝑛+1 −𝑥𝑛󵄩 = lim (1 − 𝛽𝑛) 󵄩𝑧𝑛 −𝑥𝑛󵄩 =0, 󵄩 󵄩2 𝑛→∞ 𝑛→∞ (133) ≤ 󵄩𝑢𝑛 −𝑝󵄩 +𝜆𝑖,𝑛 (𝜆𝑖,𝑛 −2𝜂𝑖) 󵄩 󵄩2 × 󵄩𝐵 Λ𝑖−1𝑢 −𝐵𝑝󵄩 and, by (126)–(128), 󵄩 𝑖 𝑛 𝑛 𝑖 󵄩 󵄩 󵄩 󵄩 󵄩2 󵄩𝑢 −𝑢 󵄩 =0, ≤ 󵄩𝑥𝑛 −𝑝󵄩 +𝜆𝑖,𝑛 (𝜆𝑖,𝑛 −2𝜂𝑖) 𝑛→∞lim 󵄩 𝑛+1 𝑛󵄩 (139) 󵄩 󵄩2 󵄩 󵄩 󵄩 𝑖−1 󵄩 󵄩V − V 󵄩 =0, × 󵄩𝐵𝑖Λ 𝑛 𝑢𝑛 −𝐵𝑖𝑝󵄩 , 𝑛→∞lim 󵄩 𝑛+1 𝑛󵄩 (134) 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩2 󵄩 𝑀 󵄩2 󵄩𝑇 V −𝑇V 󵄩 =0. 󵄩𝑢 −𝑝󵄩 = 󵄩Δ 𝑥 −𝑝󵄩 𝑛→∞lim 󵄩 𝑛+1 𝑛+1 𝑛 𝑛󵄩 󵄩 𝑛 󵄩 󵄩 𝑛 𝑛 󵄩 󵄩 󵄩2 ≤ 󵄩Δ𝑘 𝑥 −𝑝󵄩 𝑘−1 󵄩 𝑛 𝑛 󵄩 Step 3. Let us show that ‖𝐴𝑘Δ 𝑛 𝑥𝑛 −𝐴𝑘𝑝‖→0and 𝑖−1 󵄩 ‖𝐵𝑖Λ 𝑛 𝑢𝑛 −𝐵𝑖𝑝‖ → 0 for all 𝑘 ∈ {1,2,...,𝑀} and 𝑖∈ 󵄩 (Θ𝑘,𝜑𝑘) 𝑘−1 = 󵄩𝑇𝑟 (𝐼 −𝑘,𝑛 𝑟 𝐴𝑘)Δ𝑛 𝑥𝑛 {1,2,...,𝑁}. 󵄩 𝑘,𝑛 Indeed, since 󵄩2 (Θ𝑘,𝜑𝑘) 󵄩 −𝑇 (𝐼−𝑟𝑘,𝑛𝐴𝑘)𝑝󵄩 𝑟𝑘,𝑛 󵄩 𝑥 =𝑠𝛾𝑉𝑥 +𝛽 𝑥 + ((1−𝛽 ) 𝐼−𝑠 𝜇𝐹) 𝑇 V , 𝑛+1 𝑛 𝑛 𝑛 𝑛 𝑛 𝑛 𝑛 𝑛 (135) 󵄩 󵄩2 ≤ 󵄩(𝐼 − 𝑟 𝐴 )Δ𝑘−1𝑥 −(𝐼−𝑟 𝐴 )𝑝󵄩 󵄩 𝑘,𝑛 𝑘 𝑛 𝑛 𝑘,𝑛 𝑘 󵄩 we have 󵄩 𝑘−1 󵄩2 ≤ 󵄩Δ 𝑥 −𝑝󵄩 +𝑟 (𝑟 −2𝜇 ) 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 𝑛 𝑛 󵄩 𝑘,𝑛 𝑘,𝑛 𝑘 󵄩𝑥 −𝑇V 󵄩 ≤ 󵄩𝑥 −𝑥 󵄩 + 󵄩𝑥 −𝑇V 󵄩 󵄩 𝑛 𝑛 𝑛󵄩 󵄩 𝑛 𝑛+1󵄩 󵄩 𝑛+1 𝑛 𝑛󵄩 󵄩 󵄩2 × 󵄩𝐴 Δ𝑘−1𝑥 −𝐴 𝑝󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 𝑘 𝑛 𝑛 𝑘 󵄩 ≤ 󵄩𝑥𝑛 −𝑥𝑛+1󵄩 +𝑠𝑛 󵄩𝛾𝑉𝑥𝑛 −𝜇𝐹𝑇𝑛V𝑛󵄩 (136) 󵄩 󵄩2 󵄩 󵄩 ≤ 󵄩𝑥𝑛 −𝑝󵄩 +𝑟𝑘,𝑛 (𝑟𝑘,𝑛 −2𝜇𝑘) +𝛽𝑛 󵄩𝑥𝑛 −𝑇𝑛V𝑛󵄩 ; 󵄩 󵄩2 × 󵄩𝐴 Δ𝑘−1𝑥 −𝐴 𝑝󵄩 . 󵄩 𝑘 𝑛 𝑛 𝑘 󵄩 that is, Furthermore, utilizing Lemma 7,wededucefrom(113)that 󵄩 󵄩 1 󵄩 󵄩 󵄩𝑥𝑛 −𝑇𝑛V𝑛󵄩 ≤ 󵄩𝑥𝑛 −𝑥𝑛+1󵄩 1−𝛽𝑛 󵄩 󵄩2 󵄩𝑥𝑛+1 −𝑝󵄩 𝑠 (137) 𝑛 󵄩 󵄩 󵄩 󵄩 󵄩 + (𝛾 󵄩𝑉𝑥 𝑛󵄩 +𝜇󵄩𝐹𝑇𝑛V𝑛󵄩). = 󵄩𝑠 (𝛾𝑉𝑥 −𝜇𝐹𝑝)+𝛽 (𝑥 −𝑇V ) 1−𝛽𝑛 󵄩 𝑛 𝑛 𝑛 𝑛 𝑛 𝑛 󵄩2 +(𝐼−𝑠𝑛𝜇𝐹)𝑛 𝑇 V𝑛 −(𝐼−𝑠𝑛𝜇𝐹)󵄩 𝑝 So, from 𝑠𝑛 →0, ‖𝑥𝑛+1 −𝑥𝑛‖→0, and condition (ii), it follows that 󵄩 󵄩2 ≤ 󵄩𝛽𝑛 (𝑥𝑛 −𝑇𝑛V𝑛)+(𝐼−𝑠𝑛𝜇𝐹)𝑛 𝑇 V𝑛 −(𝐼−𝑠𝑛𝜇𝐹)󵄩 𝑝 󵄩 󵄩 󵄩𝑥 −𝑇V 󵄩 =0. 𝑛→∞lim 󵄩 𝑛 𝑛 𝑛󵄩 (138) +2𝑠𝑛 ⟨𝛾𝑉𝑥𝑛 −𝜇𝐹𝑝,𝑥𝑛+1 −𝑝⟩ 20 Abstract and Applied Analysis

󵄩 󵄩 󵄩 󵄩 2 ≤ [𝛽𝑛 󵄩𝑥𝑛 −𝑇𝑛V𝑛󵄩 + 󵄩(𝐼−𝑠𝑛𝜇𝐹) 𝑇𝑛V𝑛 − (𝐼−𝑠𝑛𝜇𝐹) 𝑇𝑛𝑝󵄩] Since {𝜆𝑖,𝑛}⊂[𝑎𝑖,𝑏𝑖] ⊂ (0, 2𝜂𝑖) and {𝑟𝑘,𝑛}⊂[𝑒𝑘,𝑓𝑘] ⊂ (0, 2𝜇𝑘) 󵄩 󵄩 󵄩 󵄩 for all 𝑖 ∈ {1,2,...,𝑁}and 𝑘 ∈ {1,2,...,𝑀},by(133), (138), +2𝑠 󵄩𝛾𝑉𝑥 −𝜇𝐹𝑝󵄩 󵄩𝑥 −𝑝󵄩 𝑛 󵄩 𝑛 󵄩 󵄩 𝑛+1 󵄩 and (142)weconcludeimmediatelythat 󵄩 󵄩 󵄩 󵄩 2 󵄩 𝑘−1 󵄩 ≤[𝛽𝑛 󵄩𝑥𝑛 −𝑇𝑛V𝑛󵄩 +(1−𝑠𝑛𝜏) 󵄩V𝑛 −𝑝󵄩] 󵄩𝐴 Δ 𝑥 −𝐴 𝑝󵄩 =0, 󵄩 󵄩 󵄩 󵄩 𝑛→∞lim 󵄩 𝑘 𝑛 𝑛 𝑘 󵄩 󵄩 󵄩 󵄩 󵄩 +2𝑠 󵄩𝛾𝑉𝑥 −𝜇𝐹𝑝󵄩 󵄩𝑥 −𝑝󵄩 󵄩 󵄩 (143) 𝑛 󵄩 𝑛 󵄩 󵄩 𝑛+1 󵄩 󵄩𝐵 Λ𝑖−1𝑢 −𝐵𝑝󵄩 =0, 𝑛→∞lim 󵄩 𝑖 𝑛 𝑛 𝑖 󵄩 2󵄩 󵄩2 󵄩 󵄩2 =(1−𝑠𝜏) 󵄩V −𝑝󵄩 +𝛽2󵄩𝑥 −𝑇V 󵄩 𝑛 󵄩 𝑛 󵄩 𝑛󵄩 𝑛 𝑛 𝑛󵄩 for all 𝑘∈{1,2,...,𝑀}and 𝑖∈{1,2,...,𝑁}. 󵄩 󵄩 󵄩 󵄩 +2(1−𝑠𝑛𝜏)𝑛 𝛽 󵄩V𝑛 −𝑝󵄩 󵄩𝑥𝑛 −𝑇𝑛V𝑛󵄩 Step 4. Letusshowthat‖𝑥𝑛 −𝑢𝑛‖→0, ‖𝑢𝑛 − V𝑛‖→0,and 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ‖V𝑛 −𝑇𝑛V𝑛‖→0as 𝑛→∞. +2𝑠𝑛 󵄩𝛾𝑉𝑥𝑛 −𝜇𝐹𝑝󵄩 󵄩𝑥𝑛+1 −𝑝󵄩 . Indeed, by Proposition 1(iii) we obtain that for each 𝑘∈ (140) {1,2,...,𝑀}

󵄩 󵄩2 From (139)–(140), it follows that 󵄩Δ𝑘 𝑥 −𝑝󵄩 󵄩 𝑛 𝑛 󵄩 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩𝑥 −𝑝󵄩 ≤ 󵄩V −𝑝󵄩 +𝛽2󵄩𝑥 −𝑇V 󵄩 󵄩 𝑛+1 󵄩 󵄩 𝑛 󵄩 𝑛󵄩 𝑛 𝑛 𝑛󵄩 󵄩 󵄩2 󵄩 (Θ𝑘,𝜑𝑘) 𝑘−1 (Θ𝑘,𝜑𝑘) 󵄩 = 󵄩𝑇𝑟 (𝐼−𝑟𝑘,𝑛𝐴𝑘)Δ𝑛 𝑥𝑛 −𝑇𝑟 (𝐼−𝑟𝑘,𝑛𝐴𝑘)𝑝󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 𝑘,𝑛 𝑘,𝑛 󵄩 +2(1−𝑠𝑛𝜏)𝑛 𝛽 󵄩V𝑛 −𝑝󵄩 󵄩𝑥𝑛 −𝑇𝑛V𝑛󵄩 𝑘−1 𝑘 󵄩 󵄩 󵄩 󵄩 ≤⟨(𝐼−𝑟𝑘,𝑛𝐴𝑘)Δ𝑛 𝑥𝑛 −(𝐼−𝑟𝑘,𝑛𝐴𝑘)𝑝,Δ𝑛𝑥𝑛 −𝑝⟩ +2𝑠𝑛 󵄩𝛾𝑉𝑥𝑛 −𝜇𝐹𝑝󵄩 󵄩𝑥𝑛+1 −𝑝󵄩 󵄩 󵄩2 1 󵄩 𝑘−1 󵄩2 ≤ 󵄩𝑢 −𝑝󵄩 +𝜆 (𝜆 −2𝜂) = (󵄩(𝐼−𝑟 𝐴 )Δ 𝑥 −(𝐼−𝑟 𝐴 )𝑝󵄩 󵄩 𝑛 󵄩 𝑖,𝑛 𝑖,𝑛 𝑖 2 󵄩 𝑘,𝑛 𝑘 𝑛 𝑛 𝑘,𝑛 𝑘 󵄩 󵄩 󵄩2 󵄩 󵄩2 󵄩 𝑖−1 󵄩 2󵄩 󵄩 󵄩 𝑘 󵄩2 󵄩 𝑘−1 × 󵄩𝐵𝑖Λ 𝑛 𝑢𝑛 −𝐵𝑖𝑝󵄩 +𝛽𝑛󵄩𝑥𝑛 −𝑇𝑛V𝑛󵄩 + 󵄩Δ 𝑥 −𝑝󵄩 − 󵄩(𝐼 − 𝑟 𝐴 )Δ 𝑥 󵄩 󵄩 󵄩 𝑛 𝑛 󵄩 󵄩 𝑘,𝑛 𝑘 𝑛 𝑛 󵄩 󵄩 󵄩 󵄩 +2(1−𝑠 𝜏) 𝛽 󵄩V −𝑝󵄩 󵄩𝑥 −𝑇V 󵄩 𝑛 𝑛 󵄩 𝑛 󵄩 󵄩 𝑛 𝑛 𝑛󵄩 −(𝐼−𝑟𝑘,𝑛𝐴𝑘)𝑝 󵄩 󵄩 󵄩 󵄩 +2𝑠 󵄩𝛾𝑉𝑥 −𝜇𝐹𝑝󵄩 󵄩𝑥 −𝑝󵄩 󵄩2 𝑛 󵄩 𝑛 󵄩 󵄩 𝑛+1 󵄩 (141) −(Δ𝑘 𝑥 −𝑝)󵄩 ) 𝑛 𝑛 󵄩 󵄩 󵄩2 ≤ 󵄩𝑥𝑛 −𝑝󵄩 +𝑟𝑘,𝑛 (𝑟𝑘,𝑛 −2𝜇𝑘) 1 󵄩 𝑘−1 󵄩2 󵄩 𝑘 󵄩2 ≤ (󵄩Δ 𝑥 −𝑝󵄩 + 󵄩Δ 𝑥 −𝑝󵄩 󵄩 󵄩2 2 󵄩 𝑛 𝑛 󵄩 󵄩 𝑛 𝑛 󵄩 × 󵄩𝐴 Δ𝑘−1𝑥 −𝐴 𝑝󵄩 󵄩 𝑘 𝑛 𝑛 𝑘 󵄩 󵄩 󵄩2 −󵄩Δ𝑘−1𝑥 −Δ𝑘 𝑥 −𝑟 (𝐴 Δ𝑘−1𝑥 −𝐴 𝑝)󵄩 ), 󵄩 𝑖−1 󵄩2 󵄩 𝑛 𝑛 𝑛 𝑛 𝑘,𝑛 𝑘 𝑛 𝑛 𝑘 󵄩 +𝜆 (𝜆 −2𝜂) 󵄩𝐵 Λ 𝑢 −𝐵𝑝󵄩 𝑖,𝑛 𝑖,𝑛 𝑖 󵄩 𝑖 𝑛 𝑛 𝑖 󵄩 (144) 󵄩 󵄩2 +𝛽2󵄩𝑥 −𝑇V 󵄩 𝑛󵄩 𝑛 𝑛 𝑛󵄩 which implies that 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩2 󵄩 󵄩2 +2(1−𝑠𝑛𝜏)𝑛 𝛽 󵄩V𝑛 −𝑝󵄩 󵄩𝑥𝑛 −𝑇𝑛V𝑛󵄩 󵄩Δ𝑘 𝑥 −𝑝󵄩 ≤ 󵄩Δ𝑘−1𝑥 −𝑝󵄩 󵄩 𝑛 𝑛 󵄩 󵄩 𝑛 𝑛 󵄩 󵄩 󵄩 󵄩 󵄩 +2𝑠 󵄩𝛾𝑉𝑥 −𝜇𝐹𝑝󵄩 󵄩𝑥 −𝑝󵄩 , 󵄩 𝑛 󵄩 𝑛 󵄩 󵄩 𝑛+1 󵄩 − 󵄩Δ𝑘−1𝑥 −Δ𝑘 𝑥 󵄩 𝑛 𝑛 𝑛 𝑛 and so 󵄩2 −𝑟 (𝐴 Δ𝑘−1𝑥 −𝐴 𝑝)󵄩 󵄩 󵄩2 𝑘,𝑛 𝑘 𝑛 𝑛 𝑘 󵄩 𝑟 (2𝜇 −𝑟 ) 󵄩𝐴 Δ𝑘−1𝑥 −𝐴 𝑝󵄩 𝑘,𝑛 𝑘 𝑘,𝑛 󵄩 𝑘 𝑛 𝑛 𝑘 󵄩 󵄩 󵄩2 󵄩 󵄩2 = 󵄩Δ𝑘−1𝑥 −𝑝󵄩 − 󵄩Δ𝑘−1𝑥 −Δ𝑘 𝑥 󵄩 󵄩 󵄩2 󵄩 𝑛 𝑛 󵄩 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 +𝜆 (2𝜂 −𝜆 ) 󵄩𝐵 Λ𝑖−1𝑢 −𝐵𝑝󵄩 𝑖,𝑛 𝑖 𝑖,𝑛 󵄩 𝑖 𝑛 𝑛 𝑖 󵄩 󵄩 󵄩2 −𝑟2 󵄩𝐴 Δ𝑘−1𝑥 −𝐴 𝑝󵄩 󵄩 󵄩2 󵄩 󵄩2 2󵄩 󵄩2 𝑘,𝑛 󵄩 𝑘 𝑛 𝑛 𝑘 󵄩 ≤ 󵄩𝑥𝑛 −𝑝󵄩 − 󵄩𝑥𝑛+1 −𝑝󵄩 +𝛽𝑛󵄩𝑥𝑛 −𝑇𝑛V𝑛󵄩 𝑘−1 𝑘 𝑘−1 󵄩 󵄩 󵄩 󵄩 +2𝑟𝑘,𝑛 ⟨Δ 𝑛 𝑥𝑛 −Δ𝑛𝑥𝑛,𝐴𝑘Δ 𝑛 𝑥𝑛 −𝐴𝑘𝑝⟩ +2(1−𝑠𝑛𝜏)𝑛 𝛽 󵄩V𝑛 −𝑝󵄩 󵄩𝑥𝑛 −𝑇𝑛V𝑛󵄩 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩 󵄩 󵄩 (142) ≤ 󵄩Δ𝑘−1𝑥 −𝑝󵄩 − 󵄩Δ𝑘−1𝑥 −Δ𝑘 𝑥 󵄩 +2𝑠𝑛 󵄩𝛾𝑉𝑥𝑛 −𝜇𝐹𝑝󵄩 󵄩𝑥𝑛+1 −𝑝󵄩 󵄩 𝑛 𝑛 󵄩 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ≤(󵄩𝑥 −𝑝󵄩 + 󵄩𝑥 −𝑝󵄩) +2𝑟 󵄩Δ𝑘−1𝑥 −Δ𝑘 𝑥 󵄩 󵄩𝐴 Δ𝑘−1𝑥 −𝐴 𝑝󵄩 󵄩 𝑛 󵄩 󵄩 𝑛+1 󵄩 𝑘,𝑛 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 󵄩 𝑘 𝑛 𝑛 𝑘 󵄩 󵄩 󵄩 2󵄩 󵄩2 × 󵄩𝑥 −𝑥 󵄩 +𝛽 󵄩𝑥 −𝑇V 󵄩 󵄩 󵄩2 󵄩 𝑘−1 𝑘 󵄩2 󵄩 𝑛 𝑛+1󵄩 𝑛󵄩 𝑛 𝑛 𝑛󵄩 ≤ 󵄩𝑥 −𝑝󵄩 − 󵄩Δ 𝑥 −Δ 𝑥 󵄩 󵄩 𝑛 󵄩 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 󵄩 󵄩 󵄩 󵄩 +2(1−𝑠 𝜏) 𝛽 󵄩V −𝑝󵄩 󵄩𝑥 −𝑇V 󵄩 󵄩 󵄩 󵄩 󵄩 𝑛 𝑛 󵄩 𝑛 󵄩 󵄩 𝑛 𝑛 𝑛󵄩 +2𝑟 󵄩Δ𝑘−1𝑥 −Δ𝑘 𝑥 󵄩 󵄩𝐴 Δ𝑘−1𝑥 −𝐴 𝑝󵄩 . 𝑘,𝑛 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 󵄩 𝑘 𝑛 𝑛 𝑘 󵄩 󵄩 󵄩 󵄩 󵄩 +2𝑠𝑛 󵄩𝛾𝑉𝑥𝑛 −𝜇𝐹𝑝󵄩 󵄩𝑥𝑛+1 −𝑝󵄩 . (145) Abstract and Applied Analysis 21

2󵄩 󵄩2 󵄩 󵄩2 𝑖∈{1,2,...,𝑁} ≤(1−𝑠𝜏) 󵄩Λ𝑖 𝑢 −𝑝󵄩 +𝛽2󵄩𝑥 −𝑇V 󵄩 Also, by Lemma 14,weobtainthatforeach 𝑛 󵄩 𝑛 𝑛 󵄩 𝑛󵄩 𝑛 𝑛 𝑛󵄩 󵄩 󵄩2 󵄩Λ𝑖 𝑢 −𝑝󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 𝑛 𝑛 󵄩 +2(1−𝑠𝑛𝜏)𝑛 𝛽 󵄩V𝑛 −𝑝󵄩 󵄩𝑥𝑛 −𝑇𝑛V𝑛󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 𝑖−1 󵄩2 󵄩 󵄩 󵄩 󵄩 = 󵄩𝐽 (𝐼−𝜆 𝐵 )Λ 𝑢 −𝐽 (𝐼−𝜆 𝐵 )𝑝󵄩 +2𝑠𝑛 󵄩𝛾𝑉𝑥𝑛 −𝜇𝐹𝑝󵄩 󵄩𝑥𝑛+1 −𝑝󵄩 󵄩 𝑅𝑖,𝜆𝑖,𝑛 𝑖,𝑛 𝑖 𝑛 𝑛 𝑅𝑖,𝜆𝑖,𝑛 𝑖,𝑛 𝑖 󵄩 2 2 󵄩 󵄩2 𝑖−1 𝑖 󵄩 󵄩 󵄩 𝑖−1 𝑖 󵄩 ≤(1−𝑠𝑛𝜏) [󵄩𝑢𝑛 −𝑝󵄩 − 󵄩Λ 𝑢𝑛 −Λ 𝑢𝑛󵄩 ≤⟨(𝐼−𝜆𝑖,𝑛𝐵𝑖)Λ𝑛 𝑢𝑛 −(𝐼−𝜆𝑖,𝑛𝐵𝑖)𝑝,Λ𝑛𝑢𝑛 −𝑝⟩ 󵄩 󵄩 󵄩 𝑛 𝑛 󵄩 󵄩 𝑖−1 𝑖 󵄩 1 󵄩 𝑖−1 󵄩2 +2𝜆 󵄩Λ 𝑢 −Λ 𝑢 󵄩 = (󵄩(𝐼 − 𝜆 𝐵 )Λ 𝑢 −(𝐼−𝜆 𝐵 )𝑝󵄩 𝑖,𝑛 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 2 󵄩 𝑖,𝑛 𝑖 𝑛 𝑛 𝑖,𝑛 𝑖 󵄩 󵄩 𝑖−1 󵄩 󵄩 󵄩2 󵄩 × 󵄩𝐵 Λ 𝑢 −𝐵𝑝󵄩 ] + 󵄩Λ𝑖 𝑢 −𝑝󵄩 − 󵄩(𝐼 − 𝜆 𝐵 )Λ𝑖−1𝑢 󵄩 𝑖 𝑛 𝑛 𝑖 󵄩 󵄩 𝑛 𝑛 󵄩 󵄩 𝑖,𝑛 𝑖 𝑛 𝑛 2󵄩 󵄩2 󵄩 󵄩 +𝛽𝑛󵄩𝑥𝑛 −𝑇𝑛V𝑛󵄩 +2(1−𝑠𝑛𝜏)𝑛 𝛽 󵄩V𝑛 −𝑝󵄩 −(𝐼−𝜆𝑖,𝑛𝐵𝑖)𝑝 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩2 × 󵄩𝑥 −𝑇V 󵄩 +2𝑠 󵄩𝛾𝑉𝑥 −𝜇𝐹𝑝󵄩 󵄩𝑥 −𝑝󵄩 −(Λ𝑖 𝑢 −𝑝)󵄩 ) 󵄩 𝑛 𝑛 𝑛󵄩 𝑛 󵄩 𝑛 󵄩 󵄩 𝑛+1 󵄩 𝑛 𝑛 󵄩 󵄩 󵄩2 󵄩 󵄩2 2 󵄩 𝑀 󵄩 󵄩 𝑖−1 𝑖 󵄩 =(1−𝑠𝑛𝜏) [󵄩Δ 𝑛 𝑥𝑛 −𝑝󵄩 − 󵄩Λ 𝑛 𝑢𝑛 −Λ𝑛𝑢𝑛󵄩 1 󵄩 𝑖−1 󵄩2 󵄩 𝑖 󵄩2 󵄩 󵄩 󵄩 󵄩 ≤ (󵄩Λ 𝑢 −𝑝󵄩 + 󵄩Λ 𝑢 −𝑝󵄩 2 󵄩 𝑛 𝑛 󵄩 󵄩 𝑛 𝑛 󵄩 󵄩 󵄩 +2𝜆 󵄩Λ𝑖−1𝑢 −Λ𝑖 𝑢 󵄩 𝑖,𝑛 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 󵄩 󵄩2 −󵄩Λ𝑖−1𝑢 −Λ𝑖 𝑢 −𝜆 (𝐵 Λ𝑖−1𝑢 −𝐵𝑝)󵄩 ) 󵄩 𝑛 𝑛 𝑛 𝑛 𝑖,𝑛 𝑖 𝑛 𝑛 𝑖 󵄩 󵄩 󵄩 × 󵄩𝐵 Λ𝑖−1𝑢 −𝐵𝑝󵄩 ] 󵄩 𝑖 𝑛 𝑛 𝑖 󵄩 1 󵄩 󵄩2 󵄩 󵄩2 ≤ (󵄩𝑢 −𝑝󵄩 + 󵄩Λ𝑖 𝑢 −𝑝󵄩 󵄩 𝑛 󵄩 󵄩 𝑛 𝑛 󵄩 2󵄩 󵄩2 󵄩 󵄩 2 +𝛽𝑛󵄩𝑥𝑛 −𝑇𝑛V𝑛󵄩 +2(1−𝑠𝑛𝜏)𝑛 𝛽 󵄩V𝑛 −𝑝󵄩 󵄩 󵄩2 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 −󵄩Λ𝑖−1𝑢 −Λ𝑖 𝑢 −𝜆 (𝐵 Λ𝑖−1𝑢 −𝐵𝑝)󵄩 ), × 󵄩𝑥 −𝑇V 󵄩 +2𝑠 󵄩𝛾𝑉𝑥 −𝜇𝐹𝑝󵄩 󵄩𝑥 −𝑝󵄩 󵄩 𝑛 𝑛 𝑛 𝑛 𝑖,𝑛 𝑖 𝑛 𝑛 𝑖 󵄩 󵄩 𝑛 𝑛 𝑛󵄩 𝑛 󵄩 𝑛 󵄩 󵄩 𝑛+1 󵄩 (146) 󵄩 󵄩2 󵄩 󵄩2 ≤(1−𝑠𝜏)2 [󵄩Δ𝑘 𝑥 −𝑝󵄩 − 󵄩Λ𝑖−1𝑢 −Λ𝑖 𝑢 󵄩 𝑛 󵄩 𝑛 𝑛 󵄩 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 which implies 󵄩 𝑖−1 𝑖 󵄩 󵄩 󵄩2 󵄩 󵄩2 +2𝜆 󵄩Λ 𝑢 −Λ 𝑢 󵄩 󵄩Λ𝑖 𝑢 −𝑝󵄩 ≤ 󵄩𝑢 −𝑝󵄩 𝑖,𝑛 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 󵄩 𝑛 𝑛 󵄩 󵄩 𝑛 󵄩 󵄩 𝑖−1 󵄩 󵄩 󵄩2 × 󵄩𝐵 Λ 𝑢 −𝐵𝑝󵄩 ] − 󵄩Λ𝑖−1𝑢 −Λ𝑖 𝑢 −𝜆 (𝐵 Λ𝑖−1𝑢 −𝐵𝑝)󵄩 󵄩 𝑖 𝑛 𝑛 𝑖 󵄩 󵄩 𝑛 𝑛 𝑛 𝑛 𝑖,𝑛 𝑖 𝑛 𝑛 𝑖 󵄩 2󵄩 󵄩2 󵄩 󵄩 󵄩 󵄩2 +𝛽 󵄩𝑥 −𝑇V 󵄩 +2(1−𝑠 𝜏) 𝛽 󵄩V −𝑝󵄩 󵄩 󵄩2 󵄩 𝑖−1 𝑖 󵄩 𝑛󵄩 𝑛 𝑛 𝑛󵄩 𝑛 𝑛 󵄩 𝑛 󵄩 = 󵄩𝑢𝑛 −𝑝󵄩 − 󵄩Λ 𝑛 𝑢𝑛 −Λ𝑛𝑢𝑛󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 × 󵄩𝑥 −𝑇V 󵄩 +2𝑠 󵄩𝛾𝑉𝑥 −𝜇𝐹𝑝󵄩 󵄩𝑥 −𝑝󵄩 󵄩 󵄩2 󵄩 𝑛 𝑛 𝑛󵄩 𝑛 󵄩 𝑛 󵄩 󵄩 𝑛+1 󵄩 2 󵄩 𝑖−1 󵄩 −𝜆𝑖,𝑛󵄩𝐵𝑖Λ 𝑛 𝑢𝑛 −𝐵𝑖𝑝󵄩 󵄩 󵄩 2 󵄩 󵄩2 󵄩 󵄩2 ≤(1−𝑠𝜏) [󵄩𝑥 −𝑝󵄩 − 󵄩Δ𝑘−1𝑥 −Δ𝑘 𝑥 󵄩 𝑛 󵄩 𝑛 󵄩 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 +2𝜆 ⟨Λ𝑖−1𝑢 −Λ𝑖 𝑢 ,𝐵Λ𝑖−1𝑢 −𝐵𝑝⟩ 𝑖,𝑛 𝑛 𝑛 𝑛 𝑛 𝑖 𝑛 𝑛 𝑖 󵄩 󵄩 +2𝑟 󵄩Δ𝑘−1𝑥 −Δ𝑘 𝑥 󵄩 󵄩 󵄩2 󵄩 󵄩2 𝑘,𝑛 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 ≤ 󵄩𝑢 −𝑝󵄩 − 󵄩Λ𝑖−1𝑢 −Λ𝑖 𝑢 󵄩 󵄩 𝑛 󵄩 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 󵄩 󵄩 × 󵄩𝐴 Δ𝑘−1𝑥 −𝐴 𝑝󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 𝑘 𝑛 𝑛 𝑘 󵄩 +2𝜆 󵄩Λ𝑖−1𝑢 −Λ𝑖 𝑢 󵄩 󵄩𝐵 Λ𝑖−1𝑢 −𝐵𝑝󵄩 . 𝑖,𝑛 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 󵄩 𝑖 𝑛 𝑛 𝑖 󵄩 󵄩 󵄩2 − 󵄩Λ𝑖−1𝑢 −Λ𝑖 𝑢 󵄩 (147) 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 󵄩 󵄩 Thus, from (140), (145), and (147), we have +2𝜆 󵄩Λ𝑖−1𝑢 −Λ𝑖 𝑢 󵄩 𝑖,𝑛 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 󵄩 󵄩2 󵄩𝑥𝑛+1 −𝑝󵄩 󵄩 󵄩 × 󵄩𝐵 Λ𝑖−1𝑢 −𝐵𝑝󵄩 ] 󵄩 𝑖 𝑛 𝑛 𝑖 󵄩 2󵄩 󵄩2 2󵄩 󵄩2 ≤(1−𝑠𝑛𝜏) 󵄩V𝑛 −𝑝󵄩 +𝛽𝑛󵄩𝑥𝑛 −𝑇𝑛V𝑛󵄩 2󵄩 󵄩2 󵄩 󵄩 󵄩 󵄩 +𝛽 󵄩𝑥 −𝑇V 󵄩 +2(1−𝑠 𝜏) 𝛽 󵄩V −𝑝󵄩 󵄩𝑥 −𝑇V 󵄩 𝑛󵄩 𝑛 𝑛 𝑛󵄩 𝑛 𝑛 󵄩 𝑛 󵄩 󵄩 𝑛 𝑛 𝑛󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 +2(1−𝑠 𝜏) 𝛽 󵄩V −𝑝󵄩 󵄩𝑥 −𝑇V 󵄩 +2𝑠 󵄩𝛾𝑉𝑥 −𝜇𝐹𝑝󵄩 󵄩𝑥 −𝑝󵄩 𝑛 𝑛 󵄩 𝑛 󵄩 󵄩 𝑛 𝑛 𝑛󵄩 𝑛 󵄩 𝑛 󵄩 󵄩 𝑛+1 󵄩 󵄩 󵄩 󵄩 󵄩 +2𝑠 󵄩𝛾𝑉𝑥 −𝜇𝐹𝑝󵄩 󵄩𝑥 −𝑝󵄩 2󵄩 󵄩2 󵄩 󵄩2 𝑛 󵄩 𝑛 󵄩 󵄩 𝑛+1 󵄩 =(1−𝑠𝜏) 󵄩Λ𝑁𝑢 −𝑝󵄩 +𝛽2󵄩𝑥 −𝑇V 󵄩 𝑛 󵄩 𝑛 𝑛 󵄩 𝑛󵄩 𝑛 𝑛 𝑛󵄩 󵄩 󵄩2 󵄩 󵄩2 2󵄩 𝑘−1 𝑘 󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 󵄩𝑥𝑛 −𝑝󵄩 −(1−𝑠𝑛𝜏) 󵄩Δ 𝑛 𝑥𝑛 −Δ𝑛𝑥𝑛󵄩 +2(1−𝑠𝑛𝜏)𝑛 𝛽 󵄩V𝑛 −𝑝󵄩 󵄩𝑥𝑛 −𝑇𝑛V𝑛󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 +2𝑠 󵄩𝛾𝑉𝑥 −𝜇𝐹𝑝󵄩 󵄩𝑥 −𝑝󵄩 +2𝑟 󵄩Δ𝑘−1𝑥 −Δ𝑘 𝑥 󵄩 󵄩𝐴 Δ𝑘−1𝑥 −𝐴 𝑝󵄩 𝑛 󵄩 𝑛 󵄩 󵄩 𝑛+1 󵄩 𝑘,𝑛 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 󵄩 𝑘 𝑛 𝑛 𝑘 󵄩 22 Abstract and Applied Analysis

󵄩 󵄩2 −(1−𝑠 𝜏)2󵄩Λ𝑖−1𝑢 −Λ𝑖 𝑢 󵄩 𝑛 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 Also, observe that 󵄩 󵄩 󵄩 󵄩 +2𝜆 󵄩Λ𝑖−1𝑢 −Λ𝑖 𝑢 󵄩 󵄩𝐵 Λ𝑖−1𝑢 −𝐵𝑝󵄩 𝑖,𝑛 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 󵄩 𝑖 𝑛 𝑛 𝑖 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩𝑇𝑛V𝑛 − V𝑛󵄩 ≤ 󵄩𝑇𝑛V𝑛 −𝑥𝑛󵄩 + 󵄩𝑥𝑛 − V𝑛󵄩 . (154) 2󵄩 󵄩2 󵄩 󵄩 +𝛽𝑛󵄩𝑥𝑛 −𝑇𝑛V𝑛󵄩 +2(1−𝑠𝑛𝜏)𝑛 𝛽 󵄩V𝑛 −𝑝󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 Hence, we have from (138) × 󵄩𝑥𝑛 −𝑇𝑛V𝑛󵄩 +2𝑠𝑛 󵄩𝛾𝑉𝑥𝑛 −𝜇𝐹𝑝󵄩 󵄩𝑥𝑛+1 −𝑝󵄩 ; (148) 󵄩 󵄩 󵄩𝑇 V − V 󵄩 =0. 𝑛→∞lim 󵄩 𝑛 𝑛 𝑛󵄩 (155) that is, 󵄩 󵄩2 󵄩 󵄩2 (1 − 𝑠 𝜏)2 [󵄩Δ𝑘−1𝑥 −Δ𝑘 𝑥 󵄩 + 󵄩Λ𝑖−1𝑢 −Λ𝑖 𝑢 󵄩 ] ⟨(𝜇𝐹 − 𝛾𝑉)𝑞, 𝑞−𝑥 ⟩≤0 𝑛 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 Step 5. Letusshowthatlimsup𝑛→∞ 𝑛 , where 𝑞∈ΩisthesameasinTheorem 18;thatis,𝑞∈Ω 󵄩 󵄩2 󵄩 󵄩2 ≤ 󵄩𝑥𝑛 −𝑝󵄩 − 󵄩𝑥𝑛+1 −𝑝󵄩 is a unique solution of VIP (50). To show this inequality, we choose a subsequence {𝑥𝑛 } of {𝑥𝑛} such that 󵄩 󵄩 󵄩 󵄩 𝑖 +2𝑟 󵄩Δ𝑘−1𝑥 −Δ𝑘 𝑥 󵄩 󵄩𝐴 Δ𝑘−1𝑥 −𝐴 𝑝󵄩 𝑘,𝑛 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 󵄩 𝑘 𝑛 𝑛 𝑘 󵄩 󵄩 𝑖−1 𝑖 󵄩 󵄩 𝑖−1 󵄩 +2𝜆 󵄩Λ 𝑢 −Λ 𝑢 󵄩 󵄩𝐵 Λ 𝑢 −𝐵𝑝󵄩 lim sup ⟨(𝜇𝐹 − 𝛾𝑉)𝑛 𝑞,𝑞−𝑥 ⟩ 𝑖,𝑛 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 󵄩 𝑖 𝑛 𝑛 𝑖 󵄩 𝑛→∞ (156) 2󵄩 󵄩2 +𝛽𝑛󵄩𝑥𝑛 −𝑇𝑛V𝑛󵄩 = lim ⟨(𝜇𝐹 − 𝛾𝑉)𝑛 𝑞,𝑞−𝑥 ⟩. 󵄩 󵄩 𝑖→∞ 𝑖 󵄩 󵄩 󵄩 󵄩 +2(1−𝑠𝑛𝜏)𝑛 𝛽 󵄩V𝑛 −𝑝󵄩 󵄩𝑥𝑛 −𝑇𝑛V𝑛󵄩 (149) 󵄩 󵄩 󵄩 󵄩 Since {𝑥𝑛} is bounded, there exists a subsequence {𝑥𝑛 } of +2𝑠𝑛 󵄩𝛾𝑉𝑥𝑛 −𝜇𝐹𝑝󵄩 󵄩𝑥𝑛+1 −𝑝󵄩 𝑖𝑗 {𝑥𝑛 } which converges weakly to 𝑤. Without loss of generality, 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 𝑖 ≤(󵄩𝑥 −𝑝󵄩 + 󵄩𝑥 −𝑝󵄩) 󵄩𝑥 −𝑥 󵄩 𝑥 ⇀𝑤 󵄩 𝑛 󵄩 󵄩 𝑛+1 󵄩 󵄩 𝑛 𝑛+1󵄩 we may assume that 𝑛𝑖 .FromStep4,wehavethat 𝑘 𝑚 󵄩 󵄩 󵄩 󵄩 Δ 𝑛 𝑥𝑛 ⇀𝑤, Λ 𝑛 𝑢𝑛 ⇀𝑤, 𝑢𝑛 ⇀𝑤,andV𝑛 ⇀𝑤,where 󵄩 𝑘−1 𝑘 󵄩 󵄩 𝑘−1 󵄩 𝑖 𝑖 𝑖 𝑖 𝑖 𝑖 +2𝑟𝑘,𝑛 󵄩Δ 𝑛 𝑥𝑛 −Δ𝑛𝑥𝑛󵄩 󵄩𝐴𝑘Δ 𝑛 𝑥𝑛 −𝐴𝑘𝑝󵄩 󵄩 󵄩 󵄩 󵄩 𝑘 ∈ {1,2,...,𝑀}, 𝑚 ∈ {1,2,...,𝑁}.SinceV𝑛 −𝑇𝑛V𝑛 →0by 󵄩 󵄩 󵄩 󵄩 󵄩 𝑖−1 𝑖 󵄩 󵄩 𝑖−1 󵄩 Step 4, by the same arguments as in the proof of Theorem 18, +2𝜆𝑖,𝑛 󵄩Λ 𝑛 𝑢𝑛 −Λ𝑛𝑢𝑛󵄩 󵄩𝐵𝑖Λ 𝑛 𝑢𝑛 −𝐵𝑖𝑝󵄩 󵄩 󵄩 󵄩 󵄩 we get 𝑤∈Ω.Since𝑞=𝑃Ω(𝐼−𝜇𝐹+𝛾𝑉)𝑞,itfollowsthat 2󵄩 󵄩2 󵄩 󵄩 +𝛽𝑛󵄩𝑥𝑛 −𝑇𝑛V𝑛󵄩 +2(1−𝑠𝑛𝜏)𝑛 𝛽 󵄩V𝑛 −𝑝󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ⟨(𝜇𝐹 − 𝛾𝑉) 𝑞,𝑞−𝑥 ⟩= ⟨(𝜇𝐹 − 𝛾𝑉) 𝑞,𝑞−𝑥 ⟩ 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 lim sup 𝑛 lim 𝑛𝑖 × 󵄩𝑥𝑛 −𝑇𝑛V𝑛󵄩 +2𝑠𝑛 󵄩𝛾𝑉𝑥𝑛 −𝜇𝐹𝑝󵄩 󵄩𝑥𝑛+1 −𝑝󵄩 . 𝑛→∞ 𝑖→∞

So, from 𝑠𝑛 →0,(133), (138), and (143), we immediately get =⟨(𝜇𝐹−𝛾𝑉)𝑞,𝑞−𝑤⟩≤0. 󵄩 󵄩 󵄩 󵄩 (157) 󵄩Λ𝑖−1𝑢 −Λ𝑖 𝑢 󵄩 =0, 󵄩Δ𝑘−1𝑥 −Δ𝑘 𝑥 󵄩 =0, 𝑛→∞lim 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 𝑛→∞lim 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 (150) Step 6. Let us show that lim𝑛→∞‖𝑥𝑛 −𝑞‖=0,where𝑞∈Ωis 𝑞∈Ω for all 𝑖∈{1,2,...,𝑁}and 𝑘∈{1,2,...,𝑀}.Notethat thesameasinTheorem 18;thatis, is a unique solution 󵄩 󵄩 󵄩 󵄩 of VIP (50). From (113), we know that 󵄩𝑥 −𝑢 󵄩 = 󵄩Δ0 𝑥 −Δ𝑀𝑥 󵄩 󵄩 𝑛 𝑛󵄩 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 󵄩Δ0 𝑥 −Δ1 𝑥 󵄩 + 󵄩Δ1 𝑥 −Δ2 𝑥 󵄩 𝑥𝑛+1 −𝑞=𝑠𝑛 (𝛾𝑉𝑥𝑛 −𝜇𝐹𝑞)+𝛽𝑛 (𝑇𝑛V𝑛 −𝑞) 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 󵄩 󵄩 + ((1 − 𝛽 )𝐼−𝑠 𝜇𝐹) 𝑇 V +⋅⋅⋅+󵄩Δ𝑀−1𝑥 −Δ𝑀𝑥 󵄩 , 𝑛 𝑛 𝑛 𝑛 (158) 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 󵄩 󵄩 󵄩 󵄩 (151) − ((1 − 𝛽𝑛)𝐼−𝑠𝑛𝜇𝐹) 𝑞. 󵄩𝑢 − V 󵄩 = 󵄩Λ0 𝑢 −Λ𝑁𝑢 󵄩 󵄩 𝑛 𝑛󵄩 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 󵄩Λ0 𝑢 −Λ1 𝑢 󵄩 + 󵄩Λ1 𝑢 −Λ2 𝑢 󵄩 𝑇 𝑞=𝑞 ‖V − 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 Applying Lemmas 7 and 13 and noticing 𝑛 and 𝑛 𝑞‖ ≤ ‖𝑥 −𝑞‖ 𝑛≥1 󵄩 󵄩 𝑛 for all ,wehave +⋅⋅⋅+󵄩Λ𝑁−1𝑢 −Λ𝑁𝑢 󵄩 . 󵄩 𝑛 𝑛 𝑛 𝑛󵄩 󵄩 󵄩2 Thus, from (150)wehave 󵄩𝑥𝑛+1 −𝑞󵄩 󵄩 󵄩 󵄩 󵄩 󵄩𝑥 −𝑢 󵄩 =0, 󵄩𝑢 − V 󵄩 =0. 󵄩 𝑛→∞lim 󵄩 𝑛 𝑛󵄩 𝑛→∞lim 󵄩 𝑛 𝑛󵄩 (152) ≤ 󵄩𝛽𝑛 (𝑇𝑛V𝑛 −𝑞)+((1−𝛽𝑛)𝐼−𝑠𝑛𝜇𝐹)𝑛 𝑇 V𝑛 𝑛→∞ 󵄩2 It is easy to see that as −((1−𝛽𝑛)𝐼−𝑠𝑛𝜇𝐹) 𝑞󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩𝑥𝑛 − V𝑛󵄩 ≤ 󵄩𝑥𝑛 −𝑢𝑛󵄩 + 󵄩𝑢𝑛 − V𝑛󵄩 󳨀→ 0. 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 (153) +2𝑠𝑛 ⟨𝛾𝑉𝑥𝑛 −𝜇𝐹𝑞,𝑥𝑛+1 −𝑞⟩ Abstract and Applied Analysis 23

󵄩 󵄩 󵄩 ≤[𝛽 󵄩𝑇 V −𝑞󵄩 + 󵄩((1 − 𝛽 )𝐼−𝑠 𝜇𝐹) 𝑇 V 2(𝜏−𝛾𝑙) 󵄩 󵄩2 2(𝜏−𝛾𝑙)𝑠 𝑛 󵄩 𝑛 𝑛 󵄩 󵄩 𝑛 𝑛 𝑛 𝑛 ≤(1− 𝑠 ) 󵄩𝑥 −𝑞󵄩 + 𝑛 1−𝑠 𝛾𝑙 𝑛 󵄩 𝑛 󵄩 1−𝑠 𝛾𝑙 󵄩 2 𝑛 𝑛 − ((1 − 𝛽𝑛)𝐼−𝑠𝑛𝜇𝐹) 𝑞󵄩 ] 𝜏2𝑠 ×( 𝑛 𝑀̃ +2𝑠𝑛 ⟨𝛾𝑉𝑥𝑛 −𝛾𝑉𝑞,𝑥𝑛+1 −𝑞⟩ 2(𝜏−𝛾𝑙) 2 +2𝑠 ⟨𝛾𝑉𝑞 − 𝜇𝐹𝑞, 𝑥 −𝑞⟩ 𝑛 𝑛+1 1 + ⟨𝜇𝐹𝑞 − 𝛾𝑉𝑞, 𝑞−𝑥 ⟩) 𝜏−𝛾𝑙 𝑛+1 󵄩 󵄩 =[𝛽𝑛 󵄩𝑇𝑛V𝑛 −𝑞󵄩 +(1−𝛽𝑛) 󵄩 󵄩2 =(1−𝜎𝑛) 󵄩𝑥𝑛 −𝑞󵄩 +𝜎𝑛𝛿𝑛, 󵄩 󵄩 2 󵄩 𝑠 𝑠 󵄩 × 󵄩(𝐼 − 𝑛 𝜇𝐹) 𝑇 V −(𝐼− 𝑛 𝜇𝐹) 𝑇 𝑞󵄩] (160) 󵄩 𝑛 𝑛 𝑛 󵄩 󵄩 1−𝛽𝑛 1−𝛽𝑛 󵄩 ̃ 2 where 𝑀2 = sup𝑛≥1‖𝑥𝑛 −𝑞‖ , 𝜎𝑛 = (2(𝜏 − 𝛾𝑙)/(1𝑛 −𝑠 𝛾𝑙))𝑠𝑛, +2𝑠𝑛 ⟨𝛾𝑉𝑥𝑛 −𝛾𝑉𝑞,𝑥𝑛+1 −𝑞⟩ and 𝜏2𝑠 1 +2𝑠 ⟨𝛾𝑉𝑞 − 𝜇𝐹𝑞, 𝑥 −𝑞⟩ 𝛿 = 𝑛 𝑀̃ + ⟨𝜇𝐹𝑞−𝛾𝑉𝑞,𝑞−𝑥 ⟩. 𝑛 𝑛+1 𝑛 2(𝜏−𝛾𝑙) 2 𝜏−𝛾𝑙 𝑛+1 (161) 𝑠 𝜏 2 󵄩 󵄩 𝑛 󵄩 󵄩2 From condition (i) and Step 5, it is easy to see that 𝜎𝑛 →0, ≤[𝛽𝑛 󵄩V𝑛 −𝑞󵄩 +(1−𝛽𝑛)(1− ) 󵄩V𝑛 −𝑞󵄩 ] ∞ 1−𝛽𝑛 ∑𝑛=0 𝜎𝑛 =∞and lim sup𝑛→∞𝛿𝑛 ≤0.Hence,byLemma 10, we conclude that 𝑥𝑛 →𝑞as 𝑛→∞.Thiscompletesthe +2𝑠 ⟨𝛾𝑉𝑥 −𝛾𝑉𝑞,𝑥 −𝑞⟩ 𝑛 𝑛 𝑛+1 proof. +2𝑠 ⟨𝛾𝑉𝑞 − 𝜇𝐹𝑞, 𝑥 −𝑞⟩ 𝑛 𝑛+1 Corollary 22. Let 𝐶 beanonemptyclosedconvexsubsetofa 󵄩 󵄩 󵄩 󵄩 2 real Hilbert space 𝐻.Let𝑓:𝐶 → R be a convex functional =[𝛽 󵄩V −𝑞󵄩 +(1−𝛽 −𝑠 𝜏) 󵄩V −𝑞󵄩] 𝑛 󵄩 𝑛 󵄩 𝑛 𝑛 󵄩 𝑛 󵄩 with 𝐿-Lipschitz continuous gradient ∇𝑓 .LetΘ be a bifunction 𝐶×𝐶 𝜑:𝐶 → ∪ +2𝑠 ⟨𝛾𝑉𝑥 −𝛾𝑉𝑞,𝑥 −𝑞⟩ from to R satisfying (A1)–(A4) and let R 𝑛 𝑛 𝑛+1 {+∞} be a proper lower semicontinuous and convex function. 𝑅 :𝐶 →𝐻 2 +2𝑠𝑛 ⟨𝛾𝑉𝑞 − 𝜇𝐹𝑞,𝑛+1 𝑥 −𝑞⟩ Let 𝑖 be a maximal monotone mapping and let 𝐴:𝐻and →𝐻 𝐵𝑖 :𝐶 →be 𝐻 𝜁-inverse-strongly monotone 󵄩 󵄩 󵄩 󵄩 2 𝜂 𝑖=1,2 ≤[𝛽𝑛 󵄩𝑥𝑛 −𝑞󵄩 +(1−𝛽𝑛 −𝑠𝑛𝜏) 󵄩𝑥𝑛 −𝑞󵄩] and 𝑖-inverse-strongly monotone, respectively, for .Let 𝐹:𝐻be →𝐻 a 𝜅-Lipschitzian and 𝜂-strongly monotone +2𝑠𝑛 ⟨𝛾𝑉𝑥𝑛 −𝛾𝑉𝑞,𝑥𝑛+1 −𝑞⟩ operator with positive constants 𝜅,𝜂>0.Let𝑉:𝐻 →𝐻be an 2 𝑙-Lipschitzian mapping with constant 𝑙≥0.Let0<𝜇<2𝜂/𝜅 +2𝑠𝑛 ⟨𝛾𝑉𝑞 − 𝜇𝐹𝑞,𝑛+1 𝑥 −𝑞⟩ and 0≤𝛾𝑙<𝜏,where𝜏=1−√1 − 𝜇(2𝜂 − 𝜇𝜅2). Assume that 2󵄩 󵄩2 󵄩 󵄩 ≤(1−𝑠𝑛𝜏) 󵄩𝑥𝑛 −𝑞󵄩 +2𝑠𝑛𝛾𝑙 󵄩𝑥𝑛 −𝑞󵄩 Ω := 𝐺𝑀𝐸𝑃(Θ, 𝜑, 1𝐴) ∩𝐼(𝐵 ,𝑅1)∩𝐼(𝐵2,𝑅2)∩Γ=0̸ and that 𝑥 ∈𝐻 {𝑥 } 󵄩 󵄩 either (B1) or (B2) holds. For arbitrarily given 1 ,let 𝑛 × 󵄩𝑥𝑛+1 −𝑞󵄩 +2𝑠𝑛 ⟨𝛾𝑉𝑞 − 𝜇𝐹𝑞,𝑛+1 𝑥 −𝑞⟩ be a sequence generated by 2󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 Θ(𝑢𝑛,𝑦)+𝜑(𝑦)−𝜑(𝑢𝑛)+⟨𝐴𝑥𝑛,𝑦−𝑢𝑛⟩ ≤(1−𝑠𝑛𝜏) 󵄩𝑥𝑛 −𝑞󵄩 +𝑠𝑛𝛾𝑙󵄩 ( 𝑥𝑛 −𝑞󵄩 + 󵄩𝑥𝑛+1 −𝑞󵄩 ) 1 +2𝑠𝑛 ⟨𝛾𝑉𝑞 − 𝜇𝐹𝑞,𝑛+1 𝑥 −𝑞⟩. + ⟨𝑦 −𝑛 𝑢 ,𝑢𝑛 −𝑥𝑛⟩ ≥ 0, ∀𝑦 ∈ 𝐶, 𝑟𝑛 (159) V =𝐽 (𝐼−𝜆 𝐵 )𝐽 (𝐼 − 𝜆 𝐵 )𝑢 , (162) 𝑛 𝑅2,𝜆2,𝑛 2,𝑛 2 𝑅1,𝜆1,𝑛 1,𝑛 1 𝑛 This implies that 𝑥𝑛+1 =𝑠𝑛𝛾𝑉𝑥𝑛 +𝛽𝑛𝑥𝑛 + ((1 − 𝛽𝑛)𝐼−𝑠𝑛𝜇𝐹)𝑛 𝑇 V𝑛, 󵄩 󵄩2 󵄩𝑥𝑛+1 −𝑞󵄩 ∀𝑛 ≥ 1,

2 2 where 𝑃𝐶(𝐼−𝜆𝑛∇𝑓 ) = 𝑠 𝑛𝐼+(1−𝑠𝑛)𝑇𝑛 (here 𝑇𝑛 is nonexpansive 1−2𝜏𝑠 +𝜏 𝑠 +𝑠 𝛾𝑙󵄩 󵄩2 ≤ 𝑛 𝑛 𝑛 󵄩𝑥 −𝑞󵄩 𝑠 =(2−𝜆 𝐿)/4 ∈ (0, 1/2) 𝜆 ∈ (0, 2/𝐿) 1−𝑠 𝛾𝑙 󵄩 𝑛 󵄩 and 𝑛 𝑛 for each 𝑛 ). Assume 𝑛 that the following conditions hold: 2𝑠 𝑛 (i) 𝑠𝑛 ∈ (0, 1/2) for each 𝜆𝑛 ∈ (0, 2/𝐿),andlim𝑛→∞𝑠𝑛 = + ⟨𝛾𝑉𝑞 − 𝜇𝐹𝑞,𝑛+1 𝑥 −𝑞⟩ 1−𝑠𝑛𝛾𝑙 0(⇔lim𝑛→∞𝜆𝑛 = 2/𝐿); {𝛽 } ⊂ (0, 1) 0< 𝛽 ≤ 2(𝜏−𝛾𝑙)𝑠 𝜏2𝑠2 (ii) 𝑛 and lim inf𝑛→∞ 𝑛 𝑛 󵄩 󵄩2 𝑛 󵄩 󵄩2 𝛽 <1 =(1− ) 󵄩𝑥𝑛 −𝑞󵄩 + 󵄩𝑥𝑛 −𝑞󵄩 lim sup𝑛→∞ 𝑛 ; 1−𝑠𝑛𝛾𝑙 1−𝑠𝑛𝛾𝑙 (iii) {𝜆𝑖,𝑛}⊂[𝑎𝑖,𝑏𝑖] ⊂ (0, 2𝜂𝑖) and lim𝑛→∞|𝜆𝑖,𝑛+1−𝜆𝑖,𝑛|=0 2𝑠 for 𝑖=1,2; + 𝑛 ⟨𝛾𝑉𝑞 − 𝜇𝐹𝑞, 𝑥 −𝑞⟩ 𝑛+1 {𝑟 }⊂[𝑒,𝑓]⊂(0,2𝜁) |𝑟 −𝑟 |=0 1−𝑠𝑛𝛾𝑙 (iv) 𝑛 and lim𝑛→∞ 𝑛+1 𝑛 . 24 Abstract and Applied Analysis

Then {𝑥𝑛} converges strongly as 𝜆𝑛 → 2/𝐿 (⇔𝑛 𝑠 →0)to a References point 𝑞∈Ω,whichisauniquesolutionofVIP(110). [1] F. E. Browder and W. V. Petryshyn, “Construction of fixed Corollary 23. Let 𝐶 be a nonempty closed convex subset of a points of nonlinear mappings in Hilbert space,” Journal of real Hilbert space 𝐻.Let𝑓:𝐶 → R be a convex functional Mathematical Analysis and Applications,vol.20,pp.197–228, with 𝐿-Lipschitz continuous gradient ∇𝑓 .LetΘ be a bifunction 1967. from 𝐶×𝐶 to R satisfying (A1)–(A4) and let 𝜑:𝐶 → R∪{+∞} [2] J. L. Lions, Quelques Methodes´ de Resolution´ des Problemes` aux be a proper lower semicontinuous and convex function. Let Limites Non Lineaires, Dunod, Paris, France, 1969. 𝐻 𝑅:𝐶 →2 be a maximal monotone mapping and let [3] R. Glowinski, Numerical Methods for Nonlinear Variational Problems, Springer, New York, NY, USA, 1984. 𝐴:𝐻and →𝐻 𝐵:𝐶be →𝐻 𝜁-inverse-strongly monotone and 𝜉-inverse-strongly monotone, respectively. Let [4] W. Takahashi, Nonlinear Functional Analysis, Yokohama Pub- lishers, Yokohama, Japan, 2000. 𝐹:𝐻be →𝐻 a 𝜅-Lipschitzian and 𝜂-strongly monotone operator with positive constants 𝜅,𝜂>0.Let𝑉: 𝐻 →𝐻be an [5] J. T. Oden, Quantitative Methods on Nonlinear Mechanics, 2 Prentice-Hall, Englewood Cliffs, NJ, USA, 1986. 𝑙-Lipschitzian mapping with constant 𝑙≥0.Let0<𝜇<2𝜂/𝜅 [6] E. Zeidler, Nonlinear Functional Analysis and Its Applications, 2 and 0≤𝛾𝑙<𝜏,where𝜏=1−√1 − 𝜇(2𝜂 − 𝜇𝜅 ). Assume that Springer,NewYork,NY,USA,1985. Ω := 𝐺𝑀𝐸𝑃(Θ, 𝜑, 𝐴) ∩ 𝐼(𝐵, 𝑅)∩Γ =0̸ and that either (B1) or [7] G. M. Korpelevic,ˇ “An extragradient method for finding saddle (B2) holds. For arbitrarily given 𝑥1 ∈𝐻,let{𝑥𝑛} be a sequence points and for other problems,” Ekonomika` i Matematicheskie generated by Metody,vol.12,no.4,pp.747–756,1976. [8] F. Facchinei and J.-S. 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Assume andfixedpointproblems,”European Journal of Operational that the following conditions hold: Research,vol.212,no.2,pp.242–250,2011. (i) 𝑠𝑛 ∈ (0, 1/2) for each 𝜆𝑛 ∈ (0, 2/𝐿),andlim𝑛→∞𝑠𝑛 = [12] J.-W. Peng and J.-C. Yao, “A new hybrid-extragradient method 0(⇔lim𝑛→∞𝜆𝑛 = 2/𝐿); for generalized mixed equilibrium problems, fixed point prob- {𝛽 } ⊂ (0, 1) 0< 𝛽 ≤ lems and variational inequality problems,” Taiwanese Journal of (ii) 𝑛 and lim inf𝑛→∞ 𝑛 Mathematics,vol.12,no.6,pp.1401–1432,2008. lim sup 𝛽𝑛 <1; 𝑛→∞ [13] L.-C. Ceng, C.-Y. Wang, and J.-C. Yao, “Strong convergence (iii) {𝜌𝑛} ⊂ [𝑎, 𝑏] ⊂ (0, 2𝜉) and lim𝑛→∞|𝜌𝑛+1 −𝜌𝑛|=0; theorems by a relaxed extragradient method for a general system of variational inequalities,” Mathematical Methods of (iv) {𝑟𝑛}⊂[𝑒,𝑓]⊂(0,2𝜁)and lim𝑛→∞|𝑟𝑛+1 −𝑟𝑛|=0. Operations Research,vol.67,no.3,pp.375–390,2008. Then {𝑥𝑛} converges strongly as 𝜆𝑛 → 2/𝐿 (⇔𝑛 𝑠 →0)to a [14] L.-C. Ceng, Q. H. Ansari, and S. 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Research Article A New Iterative Method for Equilibrium Problems and Fixed Point Problems

Abdul Latif1 and Mohammad Eslamian2 1 Department of Mathematics, King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Saudi Arabia 2 Department of Mathematics, Mazandaran University of Science and Technology, Behshahr, Iran

Correspondence should be addressed to Abdul Latif; [email protected]

Received 31 October 2013; Accepted 15 December 2013

Academic Editor: Ljubomir B. Ciri´ c´

Copyright © 2013 A. Latif and M. Eslamian. 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.

Introducing a new iterative method, we study the existence of a common element of the set of solutions of equilibrium problems for a family of monotone, Lipschitz-type continuous mappings and the sets of fixed points of two nonexpansive semigroups in a real Hilbert space. We establish strong convergence theorems of the new iterative method for the solution of the variational inequality problem which is the optimality condition for the minimization problem. Our results improve and generalize the corresponding recent results of Anh (2012), Cianciaruso et al. (2010), and many others.

“Dedicated to Professor Miodrag Mateljevi’c on the occasion of his 65th birthday”

1. Introduction into 𝐻, then the Fan inequality problem (equilibrium problem) becomes the classical variational inequality problem 𝐻 ⟨⋅, ⋅⟩ ⋆ Let be a real Hilbert space with inner product and which is formulated as finding a point 𝑥 ∈𝐶such that norm ‖⋅‖.Let𝐶 be a nonempty closed convex subset of 𝐻, ⟨𝐹𝑥⋆,𝑦−𝑥⋆⟩ ≥∀𝑦∈𝐶. and let Proj𝐶 be a nearest point projection of 𝐻 into 𝐶;thatis, (2) for 𝑥∈𝐻,Proj𝐶𝑥 istheuniquepointin𝐶 with the property Variational inequalities were introduced and studied by ‖𝑥 − Proj𝐶𝑥‖ := inf{‖𝑥 − 𝑦‖ : 𝑦.Itiswellknownthat ∈𝐶} Stampacchia [5]. It is well known that this area covers 𝑦=Proj𝐶𝑥 iff ⟨𝑥 − 𝑦, 𝑦 − 𝑧⟩ ≥0 for all 𝑧∈𝐶. Let 𝑓 be a bifunction from 𝐶×𝐶into R,suchthat many branches of mathematics, such as partial differential 𝑓(𝑥, 𝑥) =0 for all 𝑥∈𝐶. Consider the Fan inequality [1]: equations, optimal control, optimization, mathematical pro- ⋆ find a point 𝑥 ∈𝐶such that gramming, mechanics, and finance; see [6–11]. ⋆ Here we recall some useful notions. 𝑓 (𝑥 ,𝑦) ≥0, ∀𝑦∈𝐶, (1) Amapping𝑇:𝐶is →𝐻 said to be 𝐿-Lipschitz on 𝐶 if 𝐿≥0 𝑥, 𝑦 ∈𝐶 where 𝑓(𝑥, ⋅) is convex and subdifferentiable on 𝐶 for every there exists a constant such that for each , 𝑥∈𝐶 󵄩 󵄩 󵄩 󵄩 . The set of solutions of this problem is denoted by 󵄩𝑇𝑥 − 𝑇𝑦󵄩 ≤𝐿󵄩𝑥−𝑦󵄩 . (3) Sol(𝑓,𝐶). In fact, the Fan inequality can be formulated as an equilibrium problem. Such problems arise frequently in In particular, if 𝐿∈[0,1[,then𝑇 is called a contraction on 𝐶; mathematics, physics, engineering, game theory, transporta- if 𝐿=1,then𝑇 is called a nonexpansive mapping on 𝐶.The tion, economics, and network. Due to importance of the set of fixed points of 𝑇 is denoted by 𝐹(𝑇). solutions of such problems, many researchers are working AfamilyT := {𝑇(𝑠) : 0 ≤ 𝑠 <∞} of mappings on a closed in this area and studying the existence of the solutions of convex subset 𝐶 of a Hilbert space 𝐻 is called a nonexpansive such problems; for example, see, [2–4]. Further, if 𝑓(𝑥, 𝑦) = semigroup if it satisfies the following: (i) 𝑇(0)𝑥=𝑥for all ⟨𝐹𝑥, 𝑦 −𝑥⟩ for every 𝑥, 𝑦,where ∈𝐶 𝐹 is a mapping from 𝐶 𝑥∈𝐶; (ii) 𝑇(𝑠+𝑡) = 𝑇(𝑠)𝑇(𝑡) for all 𝑠, 𝑡; ≥0 (iii) for all 𝑥∈𝐶, 2 Abstract and Applied Analysis

𝑠 → 𝑇(𝑠)𝑥 is continuous; (iv) ‖𝑇(𝑠)𝑥 − 𝑇(𝑠)𝑦‖ ≤‖𝑥−𝑦‖ for sequence {𝑥𝑛} generated by (8) converges strongly to the ⋆ all 𝑥, 𝑦 ∈𝐶 and 𝑠≥0. unique solution 𝑥 ∈ 𝐹(𝑇) of the variational inequality 𝐹(T) We use to denote the common fixed point set of the ⋆ ⋆ semigroup T;thatis,𝐹(T)={𝑥∈𝐶:𝑇(𝑠)𝑥=𝑥,∀𝑠≥ ⟨(𝐴 − 𝛾ℎ) 𝑥 ,𝑥−𝑥 ⟩ ≥ 0, ∀𝑥 ∈ 𝐹 (𝑇) , (9) 0}.Itiswellknownthat𝐹(T) isclosedandconvex[12]. A nonexpansive semigroup T on 𝐶 is said to be uniformly which is the optimality condition for the minimization asymptotically regular (in short, u.a.r.) on 𝐶 if for all ℎ≥0 problem 𝐵 𝐶 andanyboundedsubset of , 1 min ⟨𝐴𝑥, 𝑥⟩ −𝑔(𝑥) , (10) 𝑥∈𝐹(𝑇) 2 lim sup ‖𝑇 (ℎ)(𝑇 (𝑡) 𝑥) −𝑇(𝑡) 𝑥‖ =0. 𝑡→∞𝑥∈𝐵 (4) 󸀠 where 𝑔 is a potential function for 𝛾ℎ (i.e., 𝑔 (𝑥) = 𝛾ℎ(𝑥), 𝑡 ℎ≥0 𝜎 (𝑥) = (1/𝑡) ∫ 𝑇(𝑠)𝑥𝑑𝑠 ∀𝑥 ∈ 𝐻). For each , define 𝑡 0 .Then Iterative process for approximating common fixed points 󵄩 󵄩 󵄩𝑇 (ℎ) (𝜎 (𝑥))−𝜎 (𝑥)󵄩 =0, of a nonexpansive semigroup has been investigated by various 𝑡→∞lim sup 󵄩 𝑡 𝑡 󵄩 (5) 𝑥∈𝐵 authors (see [13, 14, 18–21]). Recently, Li et al. [19]introduced the following iterative procedure for the approximation of 𝐵 𝐶 Provided that isclosedboundedconvexsubsetof .Itis common fixed points of a nonexpansive semigroup T = {𝜎 (𝑥) : 𝑡 > 0} known that the set 𝑡 is a u.a.r. nonexpansive {𝑇(𝑠) : 0 ≤ 𝑠 <∞} on a closed convex subset 𝐶 of a Hilbert semigroup; see [13]. The other examples of u.a.r. operator space 𝐻: semigroup can be found in [14]. 𝑠 𝑓:𝐶×𝐶 → R 1 𝑛 A bifunction is said to be (i) strongly 𝑥 =𝛼𝛾ℎ (𝑥 )+(𝐼−𝛼 𝐴) ∫ 𝑇 (𝑠) 𝑥 𝑑𝑠, 𝑛 ≥1, 𝐶 𝛼>0 𝑓(𝑥, 𝑦) + 𝑓(𝑦, 𝑥)2 ≤−𝛼‖𝑥−𝑦‖ 𝑛 𝑛 𝑛 𝑛 𝑛 (11) monotone on with if , 𝑠𝑛 0 ∀𝑥, 𝑦; ∈𝐶 (ii) monotone on 𝐶 if 𝑓(𝑥, 𝑦)+𝑓(𝑦,, 𝑥)≤0 ∀𝑥, 𝑦 ∈ 𝐶; (iii) pseudomonotone on 𝐶 if 𝑓(𝑥, 𝑦) ≥ 0 ⇒ 𝑓(𝑦,, 𝑥)≤0 where 𝐴 is a strongly positive bounded linear operator on ∀𝑥, 𝑦;(iv) ∈𝐶 Lipschitz-type continuous on 𝐶 with constants 𝐻 and ℎ is a contraction on 𝐶.Imposingsomeappropriate 2 𝑐1 >0and 𝑐2 >0if 𝑓(𝑥, 𝑦) + 𝑓(𝑦, 𝑧) ≥1 𝑓(𝑥,𝑧)−𝑐 ‖𝑥−𝑦‖ − conditions on the parameters, they proved that the iterative 2 {𝑥 } 𝑐2‖𝑦 − 𝑧‖ , ∀𝑥, 𝑦, 𝑧. ∈𝐶 sequence 𝑛 generated by (11) converges strongly to the 𝑥⋆ ∈𝐹(T) Note that if 𝑇 is 𝐿-Lipschitz on 𝐶,thenforeach𝑥, 𝑦 ∈ unique solution of the variational inequality ⟨(𝛾ℎ − 𝐴)𝑥⋆,𝑧−𝑥⋆⟩≤0 ∀𝑧 ∈ 𝐹(T) 𝐶,thefunction𝑓(𝑥, 𝑦) = ⟨𝐹𝑥, 𝑦−𝑥⟩ is a Lipschitz-type , . (𝑓,𝐶) continuous with constants 𝑐1 =𝑐2 = 𝐿/2. For obtaining a common element of Sol and the set 𝑇 An operator 𝐴 on 𝐻 is called strongly positive if there is a of fixed points of a nonexpansive mapping , S. Takahashi constant 𝛾>0such that and W. Takahashi [9] first introduced an iterative scheme by the viscosity approximation method. They proved that under 2 ⟨𝐴𝑥, 𝑥⟩ ≥ 𝛾‖𝑥‖ ,∀𝑥∈𝐻. (6) certain conditions the iterative sequences converge strongly 𝑧= (ℎ(𝑧)) to Proj𝐹(𝑇)∩Sol(𝑓,𝐶) . Recently, iterative methods for nonexpansive mappings During last few years, iterative algorithms for finding a have been applied to solve convex minimization problems. common element of the set of solutions of Fan inequality and In [15], Xu defined an iterative sequence {𝑥𝑛} in 𝐶 which the set of fixed points of nonexpansive mappings in a real converges strongly to the unique solution of the minimization Hilbert space have been studied by many authors (see, e.g., problem under some suitable conditions. A well-known [2, 4, 22–28]). Recently, Anh [22] studied the existence of a typical problem is to minimize a quadratic function over the common element of the set of fixed points of a nonexpansive set of fixed points of a nonexpansive mapping 𝑇 on a real mapping and the set of solutions of Fan inequality for Hilbert space 𝐻: monotone and Lipschitz-type continuous bifunctions. He introduced the following new iterative process: 1 ⟨𝐴𝑥, 𝑥⟩ − ⟨𝑥,⟩ 𝑏 , min (7) 1󵄩 󵄩2 𝑥∈𝐹(𝑇) 2 𝑤 = {𝜆 𝑓(𝑥 ,𝑤)+ 󵄩𝑤−𝑥 󵄩 :𝑤∈𝐶}, 𝑛 argmin 𝑛 𝑛 2󵄩 𝑛󵄩 where 𝑏 is a given point in 𝐻 and 𝐴 is strongly positive 1󵄩 󵄩2 operator. 𝑧𝑛 = argmin {𝜆𝑛𝑓(𝑤𝑛,𝑧)+ 󵄩𝑧−𝑥𝑛󵄩 :𝑧∈𝐶}, 2󵄩 󵄩 (12) For solving the variational inequality problem, Marino and Xu [16] introduced the following general iterative process 𝑥𝑛+1 =𝛼𝑛ℎ(𝑥𝑛)+𝛽𝑛𝑥𝑛 +𝛾𝑛 (𝜇𝑆𝑛 (𝑥 )+(1−𝜇)𝑧𝑛), for nonexpansive mapping 𝑇 based on the viscosity approximation method (see [17]): ∀𝑛 ≥ 0, 𝜇∈]0,1[ 𝐶 𝑥𝑛+1 =𝑎𝑛𝛾ℎ 𝑛(𝑥 )+(𝐼−𝑎𝑛𝐴) 𝑇𝑥𝑛,∀𝑛≥0, (8) where , is nonempty, closed convex subset of a real Hilbert space 𝐻, 𝑓 is monotone, continuous, and Lipschitz- where 𝐴 is strongly positive bounded linear operator on 𝐻, type continuous bifunction, ℎ is self-contraction on 𝐶 with ℎ is contraction on 𝐻,and{𝑎𝑛}⊂]0,1[.Theyprovedthat, constant 𝑘∈]0,1[,and𝑆 is self nonexpansive mapping on under some appropriate conditions on the parameters, the 𝐶. He proved that, under some appropriate conditions over Abstract and Applied Analysis 3

positive sequences {𝛼𝑛}, {𝛽𝑛}, {𝛾𝑛},and{𝜆𝑛},thesequences In fact {𝑥𝑛}, {𝑤𝑛},and{𝑧𝑛} converge strongly to 𝑞∈𝐹(𝑆)∩Sol(𝑓,𝐶) which is a solution of the variational inequality ⟨(𝐼 − ℎ)𝑞, 𝑥− 𝜏 (𝑛) = max {𝑘≤𝑛:𝑡𝑘 <𝑡𝑘+1} . (17) 𝑞⟩ ≥, 0 ∀𝑥 ∈ 𝐹(𝑆) ∩ Sol(𝑓,𝐶). In this paper, we introduce a new iterative scheme based 𝐶 on the viscosity method and study the existence of a common Lemma 6 (see [2]). Let be a nonempty closed convex subset 𝐻 𝑓:𝐶×𝐶→ R element of the set of solutions of equilibrium problems for of a real Hilbert space and let be a family of monotone, Lipschitz-type continuous mappings a pseudomonotone and Lipschitz-type continuous bifunction 𝑐 ,𝑐 ≥0 𝑥∈𝐶 𝑓(𝑥, ⋅) and the sets of fixed points of two nonexpansive semigroups with constants 1 2 .Foreach ,let be convex 𝐶 𝑥 ∈𝐶 {𝑥 } {𝑧 } in a real Hilbert space. We establish strong convergence and subdifferentiable on .Let 0 and let 𝑛 , 𝑛 ,and {𝑤 } theorems of the new iterative scheme for the solution of 𝑛 be sequences generated by the variational inequality problem which is the optimality 1󵄩 󵄩2 condition for the minimization problem. Our results improve 𝑤 = {𝜆 𝑓(𝑥 ,𝑤)+ 󵄩𝑤−𝑥 󵄩 :𝑤∈𝐶}, and generalize the corresponding recent results of Anh [22], 𝑛 argmin 𝑛 𝑛 2󵄩 𝑛󵄩 Cianciaruso et al. [18], and many others. (18) 1󵄩 󵄩2 𝑧 = {𝜆 𝑓(𝑤 ,𝑧)+ 󵄩𝑧−𝑥 󵄩 :𝑧∈𝐶}. 𝑛 argmin 𝑛 𝑛 2󵄩 𝑛󵄩 2. Preliminaries 𝑥⋆ ∈ (𝑓,𝐶) In this section we collect some lemmas which are crucial for Then for each Sol , the proofs of our results. 󵄩 ⋆󵄩2 󵄩 ⋆󵄩2 󵄩 󵄩2 Let {𝑥𝑛} be a sequence in 𝐻 and 𝑥∈𝐻.Inthesequel, 󵄩𝑧 −𝑥 󵄩 ≤ 󵄩𝑥 −𝑥 󵄩 −(1−2𝜆 𝑐 ) 󵄩𝑥 −𝑤 󵄩 𝑥 ⇀𝑥 {𝑥 } 𝑥 𝑥 →𝑥 󵄩 𝑛 󵄩 󵄩 𝑛 󵄩 𝑛 1 󵄩 𝑛 𝑛󵄩 𝑛 denotes that 𝑛 weakly converges to and 𝑛 (19) {𝑥 } 𝑥 󵄩 󵄩2 denotes that 𝑛 weakly converges to . −(1−2𝜆𝑛 𝑐2) 󵄩𝑤𝑛 −𝑧𝑛󵄩 ,∀𝑛≥0. Lemma 1. Let 𝐻 be a real Hilbert space. Then the following inequality holds: 3. Main Results 󵄩 󵄩2 2 󵄩𝑥+𝑦󵄩 ≤ ‖𝑥‖ +2⟨𝑦, 𝑥⟩ +𝑦 ,∀𝑥,𝑦∈𝐻. (13) In this section, we prove the main strong convergence result which solves the problem of finding a common element Lemma 2 (see [15]). Assume that {𝑎𝑛} is a sequence of of three sets 𝐹(T), 𝐹(S),andSol(𝑓𝑖,𝐶) for finite family nonnegative real numbers such that of monotone, continuous, and Lipschitz-type continuous bifunctions 𝑓𝑖 in a real Hilbert space 𝐻. 𝑎𝑛+1 ≤ (1−𝜂𝑛) 𝑎𝑛 +𝜂𝑛𝛿𝑛,𝑛≥0, (14) Theorem 7. Let 𝐶 beanonemptyclosedconvexsubsetofareal {𝜂 } ]0, 1[ 𝛿 R where 𝑛 is a sequence in and 𝑛 is a sequence in such Hilbert space 𝐻 and let 𝑓𝑖 :𝐶×𝐶→ R (𝑖 = 1,2,...,𝑚) that be a finite family of monotone, continuous, and Lipschitz- ∞ type continuous bifunctions with constants 𝑐1,𝑖 and 𝑐2,𝑖.Let (i) ∑𝑛=1 𝜂𝑛 =∞, T = {𝑇(𝑢) : 𝑢 ≥0} and S ={𝑆(𝑢):𝑢≥0}be two 𝛿 ≤0 ∑∞ |𝜂 𝛿 |<∞ (ii) lim sup𝑛→∞ 𝑛 or 𝑛=1 𝑛 𝑛 . u.a.r. nonexpansive self-mapping semigroups on 𝐶 such that 𝑚 Ω=⋂𝑖=1 𝑆𝑜𝑙(𝑓𝑖,𝐶)⋂𝐹(T)⋂𝐹(S) =0̸. Assume that ℎ is a Then lim𝑛→∞𝑎𝑛 =0. 𝑘-contraction self-mapping of 𝐶 and 𝐴 is a strongly positive 𝐻 𝛾< Lemma 3 (see [16]). Let 𝐴 be a strongly positive linear bounded linear self-adjoint operator on with coefficient 1 0<𝛾<𝛾/𝑘 𝑥 ∈𝐶 {𝑥 } {𝑤 } {𝑧 } bounded self-adjoint operator on 𝐻 with coefficient 𝛾>0and and .Let 0 and let 𝑛 , 𝑛,𝑖 ,and 𝑛,𝑖 −1 0 < 𝜌 ≤ ‖𝐴‖ .Then‖𝐼−𝜌𝐴‖≤1−𝜌𝛾. be sequences generated by

Lemma 4 (see [29]). Let 𝐻 be a Hilbert space and 𝑥𝑖 ∈𝐻, 1󵄩 󵄩2 𝑚 𝑚 𝑤𝑛,𝑖 = argmin {𝜆𝑛,𝑖𝑓𝑖 (𝑥𝑛,𝑤)+ 󵄩𝑤−𝑥𝑛󵄩 :𝑤∈𝐶}, (1 ≤ 𝑖 ≤ 𝑚).Thenforanygiven{𝜆𝑖}𝑖=1 ⊂ ]0, 1[ with ∑𝑖=1 𝜆𝑖 = 2 1 and for any positive integer 𝑘, 𝑗 with 1≤𝑘<𝑗≤𝑚, 𝑖∈{1,2,...,𝑚} , 󵄩 𝑚 󵄩2 𝑚 󵄩 󵄩 󵄩 󵄩2 󵄩 󵄩2 1 󵄩∑𝜆 𝑥 󵄩 ≤ ∑𝜆 󵄩𝑥 󵄩 −𝜆 𝜆 󵄩𝑥 −𝑥󵄩 . (15) 󵄩 󵄩2 󵄩 𝑖 𝑖󵄩 𝑖󵄩 𝑖󵄩 𝑘 𝑗󵄩 𝑘 𝑗󵄩 𝑧𝑛,𝑖 = argmin {𝜆𝑛,𝑖𝑓𝑖 (𝑤𝑛,𝑖,𝑧)+ 󵄩𝑧−𝑥𝑛󵄩 :𝑧∈𝐶}, 󵄩𝑖=1 󵄩 𝑖=1 2 𝑖∈{1,2,...,𝑚} , Lemma 5 (see [30]). Let {𝑡𝑛} be a sequence of real numbers such that there exists a subsequence {𝑛𝑖} of {𝑛} such that 𝑡𝑛 < 𝑚 𝑡 𝑖∈N 𝑖 𝑛𝑖+1 for all . Then there exists a nondecreasing sequence 𝑦𝑛 =𝛼𝑛𝑥𝑛 + ∑𝛽𝑛,𝑖𝑧𝑛,𝑖 +𝛾𝑛𝑇(𝑡𝑛)𝑥𝑛 +𝜂𝑛𝑆(𝑡𝑛)𝑥𝑛, {𝜏(𝑛)} ⊂ N such that 𝜏(𝑛) →∞ and the following properties 𝑖=1 are satisfied by all (sufficiently large) numbers 𝑛∈N: 𝑥𝑛+1 =𝜃𝑛𝛾ℎ 𝑛(𝑥 )+(𝐼−𝜃𝑛𝐴)𝑛 𝑦 ,∀𝑛≥0, 𝑡𝜏(𝑛) ≤𝑡𝜏(𝑛)+1,𝑡𝑛 ≤𝑡𝜏(𝑛)+1. (16) (20) 4 Abstract and Applied Analysis

𝑚 where 𝛼𝑛 + ∑𝑖=1 𝛽𝑛,𝑖 +𝛾𝑛 +𝜂𝑛 =1and {𝑡𝑛}, {𝛼𝑛}, {𝛽𝑛,𝑖}, {𝛾𝑛}, It follows from Lemma 3 that {𝜂𝑛}, {𝜆𝑛,𝑖},and{𝜃𝑛} satisfy the following conditions: 𝑡 =∞ 󵄩 ⋆󵄩 (i) lim𝑛→∞ 𝑛 , 󵄩𝑥𝑛+1 −𝑥 󵄩 {𝜃 }⊂]0,1[ 𝜃 =0 ∑∞ 𝜃 =∞ (ii) 𝑛 , lim𝑛→∞ 𝑛 ,and 𝑛=1 𝑛 , 󵄩 ⋆ ⋆ 󵄩 = 󵄩𝜃𝑛 (𝛾ℎ (𝑥𝑛)−𝐴𝑥 )+(𝐼−𝜃𝑛𝐴) 𝑛(𝑦 −𝑥 )󵄩 (iii) {𝜆𝑛,𝑖} ⊂ [𝑎, 𝑏] ⊂]0, 1/𝐿[,where𝐿=max{2𝑐1,𝑖,2𝑐2,𝑖,1≤ 󵄩 ⋆󵄩 󵄩 󵄩 󵄩 ⋆󵄩 𝑖≤𝑚}, ≤𝜃𝑛 󵄩𝛾ℎ 𝑛(𝑥 )−𝐴𝑥 󵄩 + 󵄩𝐼−𝜃𝑛𝐴󵄩 󵄩𝑦𝑛 −𝑥 󵄩 󵄩 ⋆󵄩 󵄩 ⋆󵄩 (iv) {𝛼𝑛}, {𝛽𝑛,𝑖}, {𝛾𝑛}, {𝜂𝑛}⊂]0,1[, lim inf𝑛𝛼𝑛𝛽𝑛,𝑖 >0, 󵄩 󵄩 󵄩 󵄩 ≤𝜃𝑛 󵄩𝛾ℎ 𝑛(𝑥 )−𝐴𝑥 󵄩 +(1−𝜃𝑛𝛾) 󵄩𝑥𝑛 −𝑥 󵄩 lim inf𝑛𝛼𝑛𝛾𝑛 >0, lim inf𝑛𝛼𝑛𝜂𝑛 >0,and 𝛽 (1 − 2𝜆 𝑐 )>0 1≤𝑖≤𝑚 󵄩 ⋆ 󵄩 󵄩 ⋆ ⋆󵄩 lim inf𝑛 𝑛,𝑖 𝑛,𝑖 1,𝑖 for each . ≤𝜃𝑛𝛾 󵄩ℎ(𝑥𝑛)−ℎ(𝑥 )󵄩 +𝜃𝑛 󵄩𝛾ℎ (𝑥 )−𝐴𝑥 󵄩 ⋆ 󵄩 ⋆󵄩 Then, the sequence {𝑥𝑛} converges strongly to 𝑥 ∈ +(1−𝜃 𝛾) 󵄩𝑥 −𝑥 󵄩 𝑚 𝑛 󵄩 𝑛 󵄩 ⋂ 𝑆𝑜𝑙(𝑓𝑖,𝐶)⋂ 𝐹(T) ⋂ 𝐹(S) which solves the variational 𝑖=1 󵄩 ⋆󵄩 󵄩 ⋆ ⋆󵄩 inequality: ≤𝜃𝑛𝛾𝑘 󵄩𝑥𝑛 −𝑥 󵄩 +𝜃𝑛 󵄩𝛾ℎ (𝑥 )−𝐴𝑥 󵄩 󵄩 󵄩 ⋆ ⋆ +(1−𝜃 𝛾) 󵄩𝑥 −𝑥⋆󵄩 ⟨(𝐴 − 𝛾ℎ) 𝑥 ,𝑥−𝑥 ⟩ ≥ 0, ∀𝑥 ∈ Ω. (21) 𝑛 󵄩 𝑛 󵄩 󵄩 ⋆󵄩 󵄩 ⋆ ⋆󵄩 ≤(1−𝜃𝑛 (𝛾−𝛾𝑘))󵄩𝑥𝑛 −𝑥 󵄩 +𝜃𝑛 󵄩𝛾ℎ (𝑥 )−𝐴𝑥 󵄩 Proof. Since 𝐹(T), 𝐹(S),andSol(𝑓𝑖,𝐶) are closed and convex, Proj is well-defined. We claim that Proj (𝐼−𝐴+𝛾ℎ) 󵄩 ⋆󵄩 Ω Ω =(1−𝜃𝑛 (𝛾−𝛾𝑘))󵄩𝑥𝑛 −𝑥 󵄩 is a contraction from 𝐶 into itself. Indeed, for each 𝑥, 𝑦, ∈𝐶 󵄩 ⋆ ⋆󵄩 we have 󵄩𝛾ℎ (𝑥 )−𝐴𝑥 󵄩 +𝜃𝑛 (𝛾−𝛾𝑘) 󵄩 󵄩 𝛾−𝛾𝑘 󵄩ProjΩ (𝐼−𝐴+𝛾ℎ)(𝑥) − ProjΩ (𝐼 − 𝐴 + 𝛾ℎ)󵄩 (𝑦) 󵄩 ⋆ ⋆󵄩 󵄩 󵄩 󵄩 ⋆󵄩 󵄩𝛾ℎ𝑥 − A𝑥 󵄩 ≤ 󵄩(𝐼−𝐴+𝛾ℎ)(𝑥) −(𝐼−𝐴+𝛾ℎ)(𝑦)󵄩 ≤ {󵄩𝑥 −𝑥 󵄩 , }. max 󵄩 𝑛 󵄩 𝛾−𝛾𝑘 󵄩 󵄩 󵄩 󵄩 ≤ 󵄩(𝐼−𝐴) 𝑥−(𝐼−𝐴) 𝑦󵄩 +𝛾󵄩ℎ𝑥 − ℎ𝑦󵄩 (22) (25) 󵄩 󵄩 󵄩 󵄩 ≤(1−𝛾) 󵄩𝑥−𝑦󵄩 +𝛾𝑘󵄩𝑥−𝑦󵄩 󵄩 󵄩 {𝑥 } {𝑧 } {ℎ(𝑥 )} ≤(1−(𝛾−𝛾𝑘))󵄩𝑥−𝑦󵄩 . This implies that 𝑛 is bounded and so are 𝑛,𝑖 , 𝑛 , {𝑇(𝑡𝑛)𝑥𝑛},and{𝑆(𝑡𝑛)𝑥𝑛}. Next, we show that lim𝑛→∞‖𝑥𝑛 − 𝑇(𝑡 )𝑥 ‖= ‖𝑥 −𝑆(𝑡 )𝑥 ‖=0 Therefore, by the Banach contraction principle, there 𝑛 𝑛 lim𝑛→∞ 𝑛 𝑛 𝑛 .Indeed,byLemma 6, ⋆ ⋆ 1≤𝑖≤𝑚 exists a unique element 𝑥 ∈𝐶such that 𝑥 = ProjΩ(𝐼−𝐴+ for each ,wehave ⋆ 𝛾ℎ)𝑥 .Weshowthat{𝑥𝑛} is bounded. Since lim𝑛→∞𝜃𝑛 =0, −1 we can assume, with no loss of generality, that 𝜃𝑛 ∈ (0, ‖𝐴‖ ), 󵄩 ⋆󵄩2 󵄩 ⋆󵄩2 󵄩 󵄩2 for all 𝑛≥0.ByLemma 6,foreach1≤𝑖≤𝑚,wehave 󵄩𝑧𝑛,𝑖 −𝑥 󵄩 ≤ 󵄩𝑥𝑛 −𝑥 󵄩 −(1−2𝜆𝑛,𝑖𝑐1,𝑖) 󵄩𝑥𝑛 −𝑤𝑛,𝑖󵄩 (26) 󵄩 ⋆󵄩 󵄩 ⋆󵄩 󵄩 󵄩2 󵄩𝑧𝑛,𝑖 −𝑥 󵄩 ≤ 󵄩𝑥𝑛 −𝑥 󵄩 . (23) −(1−2𝜆𝑛,𝑖𝑐2,𝑖) 󵄩𝑤𝑛,𝑖 −𝑧𝑛,𝑖󵄩 .

This implies that 󵄩 󵄩 Applying Lemma 4 and inequality (26)for𝑘 ∈ {1,2,...,𝑚} 󵄩𝑦 −𝑥⋆󵄩 󵄩 𝑛 󵄩 we have that 󵄩 󵄩 󵄩 𝑚 󵄩 󵄩 ⋆󵄩 ≤ 󵄩𝛼𝑛𝑥𝑛 + ∑𝛽𝑛,𝑖𝑧𝑛,𝑖 +𝛾𝑛𝑇(𝑡𝑛)𝑥𝑛 +𝜂𝑛𝑆(𝑡𝑛)𝑥𝑛 −𝑥 󵄩 󵄩 󵄩 󵄩 ⋆󵄩2 󵄩 𝑖=1 󵄩 󵄩𝑦𝑛 −𝑥 󵄩 𝑚 󵄩 ⋆󵄩 󵄩 ⋆󵄩 󵄩 𝑚 󵄩2 ≤𝛼 󵄩𝑥 −𝑥 󵄩 + ∑𝛽 󵄩𝑧 −𝑥 󵄩 󵄩 󵄩 𝑛 󵄩 𝑛 󵄩 𝑛,𝑖 󵄩 𝑛,𝑖 󵄩 = 󵄩𝛼 𝑥 + ∑𝛽 𝑧 +𝛾𝑇(𝑡 )𝑥 +𝜂 𝑆(𝑡 )𝑥 −𝑥⋆󵄩 𝑖=1 󵄩 𝑛 𝑛 𝑛,𝑖 𝑛,𝑖 𝑛 𝑛 𝑛 𝑛 𝑛 𝑛 󵄩 󵄩 𝑖=1 󵄩 󵄩 ⋆󵄩 󵄩 ⋆󵄩 +𝛾𝑛 󵄩𝑇(𝑡𝑛)𝑥𝑛 −𝑥 󵄩 +𝜂𝑛 󵄩𝑆(𝑡𝑛)𝑥𝑛 −𝑥 󵄩 𝑚 󵄩 ⋆󵄩2 󵄩 ⋆󵄩2 𝑚 ≤𝛼𝑛󵄩𝑥𝑛 −𝑥 󵄩 + ∑𝛽𝑛,𝑖󵄩𝑧𝑛,𝑖 −𝑥 󵄩 󵄩 ⋆󵄩 󵄩 ⋆󵄩 𝑖=1 ≤𝛼𝑛 󵄩𝑥𝑛 −𝑥 󵄩 + ∑𝛽𝑛,𝑖 󵄩𝑥𝑛 −𝑥 󵄩 𝑖=1 󵄩 󵄩2 󵄩 󵄩2 +𝛾󵄩𝑇(𝑡 )𝑥 −𝑥⋆󵄩 +𝜂 󵄩𝑆(𝑡 )𝑥 −𝑥⋆󵄩 󵄩 󵄩 󵄩 󵄩 𝑛󵄩 𝑛 𝑛 󵄩 𝑛󵄩 𝑛 𝑛 󵄩 +𝛾 󵄩𝑥 −𝑥⋆󵄩 +𝜂 󵄩𝑥 −𝑥⋆󵄩 𝑛 󵄩 𝑛 󵄩 𝑛 󵄩 𝑛 󵄩 󵄩 󵄩2 󵄩 󵄩2 −𝛼𝑛𝛽𝑛,𝑘󵄩𝑥𝑛 −𝑧𝑛,𝑘󵄩 −𝛼𝑛𝛾𝑛󵄩𝑥𝑛 −𝑇(𝑡𝑛)𝑥𝑛󵄩 󵄩 ⋆󵄩 = 󵄩𝑥𝑛 −𝑥 󵄩 . 󵄩 󵄩2 (24) −𝛼𝑛𝜂𝑛󵄩𝑥𝑛 −𝑆(𝑡𝑛)𝑥𝑛󵄩 Abstract and Applied Analysis 5

𝑚 󵄩 ⋆󵄩2 󵄩 ⋆󵄩2 󵄩 ⋆󵄩2 By similar argument we can obtain that ≤𝛼𝑛󵄩𝑥𝑛 −𝑥 󵄩 + ∑𝛽𝑛,𝑖󵄩𝑥𝑛 −𝑥 󵄩 +𝛾𝑛󵄩𝑥𝑛 −𝑥 󵄩 󵄩 󵄩 󵄩 󵄩 𝑖=1 󵄩𝑥 −𝑤 󵄩 = 󵄩𝑥 −𝑇(𝑡 )𝑥 󵄩 𝑛→∞lim 󵄩 𝑛 𝑛,𝑘󵄩 𝑛→∞lim 󵄩 𝑛 𝑛 𝑛󵄩 󵄩 ⋆󵄩2 󵄩 󵄩2 +𝜂𝑛󵄩𝑥𝑛 −𝑥 󵄩 −𝛼𝑛𝛽𝑛,𝑘󵄩𝑥𝑛 −𝑧𝑛,𝑘󵄩 󵄩 󵄩 (32) = 󵄩𝑥 −𝑆(𝑡 )𝑥 󵄩 =0. 𝑛→∞lim 󵄩 𝑛 𝑛 𝑛󵄩 󵄩 󵄩2 −𝛽𝑛,𝑘 (1 − 2𝜆𝑛,𝑘𝑐1,𝑘) 󵄩𝑥𝑛 −𝑤𝑛,𝑘󵄩 Further, for all ℎ≥0and 𝑛≥0,weseethat 󵄩 󵄩2 󵄩 󵄩2 −𝛼𝑛𝛾𝑛󵄩𝑥𝑛 −𝑇(𝑡𝑛)𝑥𝑛󵄩 −𝛼𝑛𝜂𝑛󵄩𝑥𝑛 −𝑆(𝑡𝑛)𝑥𝑛󵄩 󵄩 󵄩 󵄩𝑥𝑛 −𝑇(ℎ) 𝑥𝑛󵄩 󵄩 ⋆󵄩2 󵄩 󵄩2 = 󵄩𝑥𝑛 −𝑥 󵄩 −𝛽𝑛,𝑘 (1 − 2𝜆𝑛,𝑘𝑐1,𝑘) 󵄩𝑥𝑛 −𝑤𝑛,𝑘󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 󵄩𝑥𝑛 −𝑇(𝑡𝑛)𝑥𝑛󵄩 + 󵄩𝑇(𝑡𝑛)𝑥𝑛 −𝑇(ℎ) 𝑇(𝑡𝑛)𝑥𝑛󵄩 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩 −𝛼𝑛𝛽𝑛,𝑘󵄩𝑥𝑛 −𝑧𝑛,𝑘󵄩 −𝛼𝑛𝛾𝑛󵄩𝑥𝑛 −𝑇(𝑡𝑛)𝑥𝑛󵄩 + 󵄩𝑇 (ℎ) 𝑇(𝑡 )𝑥 −𝑇(ℎ) 𝑥 󵄩 󵄩 𝑛 𝑛 𝑛󵄩 (33) 󵄩 󵄩2 󵄩 󵄩 −𝛼𝑛𝜂𝑛󵄩𝑥𝑛 −𝑆(𝑡𝑛)𝑥𝑛󵄩 . ≤2󵄩𝑥𝑛 −𝑇(𝑡𝑛)𝑥𝑛󵄩 (27) 󵄩 󵄩 + sup 󵄩𝑇(𝑡𝑛)𝑥𝑛 −𝑇(ℎ) 𝑇(𝑡𝑛)𝑥𝑛󵄩 . 𝑥∈{𝑥 } We now compute 𝑛 󵄩 ⋆󵄩2 Since {𝑇(𝑡)} is u.a.r. nonexpansive semigroup and 󵄩𝑥𝑛+1 −𝑥 󵄩 lim𝑛→∞𝑡𝑛 =∞,wehave 󵄩 󵄩2 = 󵄩𝜃 (𝛾ℎ (𝑥 )−𝐴𝑥⋆)+(𝐼−𝜃 𝐴) (𝑦 −𝑥⋆)󵄩 󵄩 󵄩 󵄩 𝑛 𝑛 𝑛 𝑛 󵄩 󵄩𝑥 −𝑇(ℎ) 𝑥 󵄩 =0. 𝑛→∞lim 󵄩 𝑛 𝑛󵄩 (34) 2󵄩 ⋆󵄩2 2󵄩 ⋆󵄩2 ≤𝜃𝑛󵄩𝛾ℎ 𝑛(𝑥 )−𝐴𝑥 󵄩 +(1−𝜃𝑛𝛾) 󵄩𝑦𝑛 −𝑥 󵄩 Similarly, for all ℎ≥0and 𝑛≥0,weobtainthat 󵄩 ⋆󵄩 󵄩 ⋆󵄩 +2𝜃 (1 − 𝜃 𝛾) 󵄩𝛾ℎ (𝑥 )−𝐴𝑥 󵄩 󵄩𝑦 −𝑥 󵄩 󵄩 󵄩 𝑛 𝑛 󵄩 𝑛 󵄩 󵄩 𝑛 󵄩 󵄩𝑥 −𝑆(ℎ) 𝑥 󵄩 =0. 𝑛→∞lim 󵄩 𝑛 𝑛󵄩 (35) 2󵄩 ⋆󵄩2 2󵄩 ⋆󵄩2 ≤𝜃𝑛󵄩𝛾ℎ 𝑛(𝑥 )−𝐴𝑥 󵄩 +(1−𝜃𝑛𝛾) 󵄩𝑥𝑛 −𝑥 󵄩 󵄩 󵄩 󵄩 󵄩 (28) Next, we show that +2𝜃 (1 − 𝜃 𝛾) 󵄩𝛾ℎ (𝑥 )−𝐴𝑥⋆󵄩 󵄩𝑥 −𝑥⋆󵄩 𝑛 𝑛 󵄩 𝑛 󵄩 󵄩 𝑛 󵄩 ⟨(𝐴 − 𝛾ℎ) 𝑥⋆,𝑥⋆ −𝑥 ⟩≤0. lim sup 𝑛 (36) 2 󵄩 󵄩2 𝑛→∞ −(1−𝜃𝑛𝛾) 𝛽𝑛,𝑘 (1 − 2𝜆𝑛,𝑘𝑐1,𝑘) 󵄩𝑥𝑛 −𝑤𝑛,𝑘󵄩 To show this inequality, we choose a subsequence {𝑥𝑛 } of {𝑥𝑛} 2 󵄩 󵄩2 𝑖 −(1−𝜃𝑛𝛾) 𝛼𝑛𝛽𝑛,𝑘󵄩𝑥𝑛 −𝑧𝑛,𝑘󵄩 such that 2 󵄩 󵄩2 ⋆ ⋆ −(1−𝜃 𝛾) 𝛼 𝛾 󵄩𝑥 −𝑇(𝑡 )𝑥 󵄩 lim ⟨(𝐴 − 𝛾ℎ) 𝑥 ,𝑥 −𝑥𝑛 ⟩ 𝑛 𝑛 𝑛󵄩 𝑛 𝑛 𝑛󵄩 𝑖→∞ 𝑖 (37) 2 󵄩 󵄩2 ⋆ ⋆ −(1−𝜃𝑛𝛾) 𝛼𝑛𝜂𝑛󵄩𝑥𝑛 −𝑆(𝑡𝑛)𝑥𝑛󵄩 . = lim sup ⟨(𝐴 − 𝛾ℎ) 𝑥 ,𝑥 −𝑥𝑛⟩. 𝑛→∞ Therefore, Since {𝑥𝑛 } is bounded, there exists a subsequence {𝑥𝑛 } of 2 󵄩 󵄩2 𝑖 𝑖𝑗 (1 − 𝜃𝑛𝛾) 𝛼𝑛𝛽𝑛,𝑘󵄩𝑥𝑛 −𝑧𝑛,𝑘󵄩 {𝑥 } 𝑥̃ 𝑛𝑖 which converges weakly to . Without loss of generality, we can assume that 𝑥𝑛 ⇀ 𝑥̃. Consider 󵄩 ⋆󵄩2 󵄩 ⋆󵄩2 𝑗 ≤ 󵄩𝑥𝑛 −𝑥 󵄩 − 󵄩𝑥𝑛+1 −𝑥 󵄩 +2𝜃𝑛 (1 − 𝜃𝑛𝛾) 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ⋆󵄩 󵄩 ⋆󵄩 2󵄩 ⋆󵄩2 󵄩𝑥𝑛 −𝑇(𝑡) 𝑥̃󵄩 ≤ 󵄩𝑥𝑛 −𝑇(𝑡) 𝑥𝑛 󵄩 + 󵄩𝑇 (𝑡) 𝑥𝑛 −𝑇(𝑡) 𝑥̃󵄩 × 󵄩𝛾ℎ 𝑛(𝑥 )−𝐴𝑥 󵄩 󵄩𝑥𝑛 −𝑥 󵄩 +𝜃𝑛󵄩𝛾ℎ 𝑛(𝑥 )−𝐴𝑥 󵄩 . 󵄩 𝑗 󵄩 󵄩 𝑗 𝑗 󵄩 󵄩 𝑗 󵄩 󵄩 󵄩 󵄩 󵄩 (29) 󵄩 󵄩 󵄩 󵄩 ≤ 󵄩𝑥𝑛 −𝑇(𝑡) 𝑥𝑛 󵄩 + 󵄩𝑥𝑛 − 𝑥̃󵄩 . 󵄩 𝑗 𝑗 󵄩 󵄩 𝑗 󵄩 𝑥 →𝑥⋆ 𝑛→∞ In order to prove that 𝑛 as ,weconsiderthe (38) following two cases.

⋆ Thus, we have Case 1. Assume that {‖𝑥𝑛 −𝑥‖} is a monotone sequence. ⋆ 󵄩 󵄩 󵄩 󵄩 In other words, for large enough 𝑛0, {‖𝑥𝑛 −𝑥‖}𝑛≥𝑛 is lim sup 󵄩𝑥𝑛 −𝑇(𝑡) 𝑥̃󵄩 ≤ lim sup 󵄩𝑥𝑛 − 𝑥̃󵄩 . ⋆ 0 𝑛→∞ 󵄩 𝑖 󵄩 𝑛→∞ 󵄩 𝑖 󵄩 (39) either nondecreasing or nonincreasing. Since {‖𝑥𝑛 −𝑥 ‖} is 𝜃 =0 {ℎ(𝑥 )} bounded, it is convergent. Since lim𝑛→∞ 𝑛 and 𝑛 By the Opial property of the Hilbert space 𝐻 we obtain {𝑥 } and 𝑛 are bounded, from (29), we have 𝑇(𝑡)𝑥=̃ 𝑥̃ for all 𝑡≥0.Similarlywehavethat𝑆(𝑡)𝑥=̃ 𝑥̃ 2 󵄩 󵄩2 for all 𝑡≥0. This implies that 𝑥∈𝐹(̃ T)⋂𝐹(S).Nowwe lim (1 − 𝜃𝑛𝛾) 𝛼𝑛𝛽𝑛,𝑘󵄩𝑥𝑛 −𝑧𝑛,𝑘󵄩 =0, (30) 𝑛→∞ show that 𝑥∈̃ Sol(𝑓𝑖,𝐶).Foreach1≤𝑖≤𝑚,since𝑓𝑖(𝑥, ⋅) is convex on 𝐶 for each 𝑥∈𝐶,weseethat and by assumption we get 1 󵄩 󵄩 󵄩 󵄩2 󵄩𝑥 −𝑧 󵄩 =0, 1≤𝑘≤𝑚. 𝑤𝑛,𝑖 = argmin {𝜆𝑛,𝑖 𝑓𝑖 (𝑥𝑛,𝑦)+ 󵄩𝑦−𝑥𝑛󵄩 :𝑦∈𝐶} (40) 𝑛→∞lim 󵄩 𝑛 𝑛,𝑘󵄩 (31) 2󵄩 󵄩 6 Abstract and Applied Analysis if and only if From Lemma 1,itfollowsthat

1󵄩 󵄩2 󵄩 󵄩 󵄩2 0∈𝜕2 (𝑓𝑖 (𝑥𝑛,𝑦)+ 󵄩𝑦−𝑥𝑛󵄩 )(𝑤𝑛,𝑖)+𝑁𝐶 (𝑤𝑛,𝑖), (41) 󵄩 ⋆󵄩 2 󵄩𝑥𝑛+1 −𝑥 󵄩 󵄩 ⋆ 󵄩2 ≤ 󵄩(𝐼 −𝑛 𝜃 𝐴) 𝑛(𝑦 −𝑥 )󵄩 where 𝑁𝐶(𝑥) is the (outward) normal cone of 𝐶 at 𝑥∈𝐶.This ⋆ ⋆ follows that +2𝜃𝑛 ⟨𝛾ℎ (𝑥𝑛)−𝐴𝑥 ,𝑥𝑛+1 −𝑥 ⟩ 2󵄩 ⋆󵄩2 ≤(1−𝜃𝑛𝛾) 󵄩𝑥𝑛 −𝑥 󵄩 0=𝜆𝑛,𝑖V +𝑤𝑛,𝑖 −𝑥𝑛 +𝑢𝑛, (42) ⋆ ⋆ +2𝜃𝑛𝛾⟨ℎ(𝑥𝑛)−ℎ(𝑥 ),𝑥𝑛+1 −𝑥 ⟩ where V ∈𝜕2𝑓𝑖(𝑥𝑛,𝑤𝑛,𝑖) and 𝑢𝑛 ∈𝑁𝐶(𝑤𝑛,𝑖). By the definition ⋆ ⋆ ⋆ +2𝜃𝑛 ⟨𝛾ℎ (𝑥 )−𝐴𝑥 ,𝑥𝑛+1 −𝑥 ⟩ of the normal cone 𝑁𝐶 we have 2󵄩 ⋆󵄩2 󵄩 ⋆󵄩 󵄩 ⋆󵄩 ≤(1−𝜃𝑛𝛾) 󵄩𝑥𝑛 −𝑥 󵄩 +2𝜃𝑛𝑘𝛾 󵄩𝑥𝑛 −𝑥 󵄩 󵄩𝑥𝑛+1 −𝑥 󵄩 ⟨𝑤 −𝑥 ,𝑦−𝑤 ⟩ ≥𝜆 ⟨V,𝑤 −𝑦⟩ ,∀𝑦∈𝐶. 𝑛,𝑖 𝑛 𝑛,𝑖 𝑛,𝑖 𝑛,𝑖 (43) ⋆ ⋆ ⋆ +2𝜃𝑛 ⟨𝛾ℎ (𝑥 )−𝐴𝑥 ,𝑥𝑛+1 −𝑥 ⟩ 2󵄩 ⋆󵄩2 Since 𝑓𝑖(𝑥𝑛,⋅) is subdifferentiable on 𝐶, by the well-known ≤(1−𝜃𝑛𝛾) 󵄩𝑥𝑛 −𝑥 󵄩 Moreau-Rockafellar theorem [31] (also see [6]), for V ∈ 𝜕 𝑓(𝑥 ,𝑤 ) 󵄩 ⋆󵄩2 󵄩 ⋆󵄩2 2 𝑖 𝑛 𝑛,𝑖 ,wehave +𝜃𝑛𝑘𝛾󵄩 ( 𝑥𝑛 −𝑥 󵄩 + 󵄩𝑥𝑛+1 −𝑥 󵄩 )

⋆ ⋆ ⋆ +2𝜃𝑛 ⟨𝛾ℎ (𝑥 )−𝐴𝑥 ,𝑥𝑛+1 −𝑥 ⟩ 𝑓𝑖 (𝑥𝑛,𝑦) −𝑓𝑖 (𝑥𝑛,𝑤𝑛,𝑖) ≥ ⟨V,𝑦−𝑤𝑛,𝑖⟩ ,∀𝑦∈𝐶.(44) 2 󵄩 ⋆󵄩2 ≤((1−𝜃𝑛𝛾) +𝜃𝑛𝑘𝛾) 󵄩𝑥𝑛 −𝑥 󵄩

Combining this with (43), we have 󵄩 ⋆󵄩2 ⋆ ⋆ ⋆ +𝜃𝑛𝛾𝑘󵄩𝑥𝑛+1 −𝑥 󵄩 +2𝜃𝑛 ⟨𝛾ℎ (𝑥 )−𝐴𝑥 ,𝑥𝑛+1 −𝑥 ⟩. (49) 𝜆𝑛,𝑖 (𝑓𝑖 (𝑥𝑛,𝑦)−𝑓𝑖 (𝑥𝑛,𝑤𝑛,𝑖)) ≥ ⟨𝑤𝑛,𝑖 −𝑥𝑛,𝑤𝑛,𝑖 −𝑦⟩, (45) ∀𝑦 ∈ 𝐶. This implies that

In particular, we have 󵄩 ⋆󵄩2 󵄩𝑥𝑛+1 −𝑥 󵄩

𝜆𝑛 ,𝑖 (𝑓𝑖 (𝑥𝑛 ,𝑦)−𝑓𝑖 (𝑥𝑛 ,𝑤𝑛,𝑖)) ≥ ⟨𝑤𝑛 ,𝑖 −𝑥𝑛 ,𝑤𝑛 ,𝑖 −𝑦⟩, 𝑗 𝑗 𝑗 𝑗 𝑗 𝑗 1−2𝜃 𝛾+(𝜃 𝛾)2 +𝜃 𝛾𝑘 𝑛 𝑛 𝑛 󵄩 ⋆󵄩2 ≤ 󵄩𝑥𝑛 −𝑥 󵄩 ∀𝑦 ∈ 𝐶. 1−𝜃𝑛𝛾𝑘 (46) 2𝜃𝑛 ⋆ ⋆ ⋆ + ⟨𝛾ℎ (𝑥 )−𝐴𝑥 ,𝑥𝑛+1 −𝑥 ⟩ 1−𝜃𝑛𝛾𝑘 𝑥 ⇀ 𝑥̃ 𝑤 ⇀ 𝑥̃ Since 𝑛𝑗 ,itfollowsfrom(32)that 𝑛𝑗,𝑖 .Andthus 2(𝛾−𝛾𝑘)𝜃 (𝜃 𝛾)2 we have 𝑛 󵄩 ⋆󵄩2 𝑛 󵄩 ⋆󵄩2 =(1− ) 󵄩𝑥𝑛 −𝑥 󵄩 + 󵄩𝑥𝑛 −𝑥 󵄩 1−𝜃𝑛𝛾𝑘 1−𝜃𝑛𝛾𝑘 𝑓 (𝑥,̃ 𝑦) ≥ 0, ∀𝑦 ∈𝐶. 𝑖 (47) 2𝜃𝑛 ⋆ ⋆ ⋆ + ⟨𝛾ℎ (𝑥 )−𝐴𝑥 ,𝑥𝑛+1 −𝑥 ⟩ 1−𝜃𝑛𝛾𝑘 𝑥∈̃ (𝑓,𝐶) 𝑥∈Ω̃ 𝑥⋆ = This implies that Sol 𝑖 and hence .Since 2(𝛾−𝛾𝑘)𝜃 2(𝛾−𝛾𝑘)𝜃 (𝐼−𝐴+𝛾ℎ)𝑥⋆ 𝑥∈Ω̃ 𝑛 󵄩 ⋆󵄩2 𝑛 ProjΩ and ,wehave ≤(1− ) 󵄩𝑥𝑛 −𝑥 󵄩 + 1−𝜃𝑛𝛾𝑘 1−𝜃𝑛𝛾𝑘

⋆ ⋆ (𝜃 𝛾2)𝑀 lim sup ⟨(𝐴 − 𝛾ℎ) 𝑥 ,𝑥 −𝑥𝑛⟩ 𝑛 1 ⋆ ⋆ ⋆ 𝑛→∞ ×( + ⟨𝛾ℎ (𝑥 )−𝐴𝑥 ,𝑥 −𝑥 ⟩) 2(𝛾−𝛾𝑘) 𝛾−𝛾𝑘 𝑛+1 ⋆ ⋆ = lim ⟨(𝐴 − 𝛾ℎ) 𝑥 ,𝑥 −𝑥𝑛 ⟩ (48) 𝑖→∞ 𝑖 󵄩 ⋆󵄩2 =(1−𝜂𝑛) 󵄩𝑥𝑛 −𝑥 󵄩 +𝜂𝑛𝛿𝑛, ⋆ ⋆ = ⟨(𝐴 − 𝛾ℎ) 𝑥 ,𝑥 − 𝑥⟩̃ ≤ 0. (50) Abstract and Applied Analysis 7 where Thesetofsolutionsofthisproblemisdenotedby𝑉𝐼(𝐹,. 𝐶) 𝑦 2(𝛾−𝛾𝑘)𝜃 In that particular case, the solution 𝑛 of the minimization 󵄩 ⋆󵄩2 𝑛 problem 𝑀=sup {󵄩𝑥𝑛 −𝑥 󵄩 :𝑛≥0},𝑛 𝜂 = , 1−𝜃𝑛𝛾𝑘 1󵄩 󵄩2 {𝜆 𝑓(𝑥 ,𝑦)+ 󵄩𝑦−𝑥 󵄩 :𝑦∈𝐶} 2 argmin 𝑛 𝑛 󵄩 𝑛󵄩 (59) (𝜃 𝛾 )𝑀 1 2 𝛿 = 𝑛 + ⟨𝛾ℎ𝑥⋆ −𝐴𝑥⋆,𝑥 −𝑥⋆⟩. 𝑛 2(𝛾−𝛾𝑘) 𝛾−𝛾𝑘 𝑛+1 can be expressed as (51) 𝑦𝑛 =𝑃𝐶 (𝑥𝑛 −𝜆𝑛𝐹 (𝑥𝑛)) . (60) ∞ It is easy to see that 𝜂𝑛 →0, ∑𝑛=1 𝜂𝑛 =∞,and Let 𝐹 be 𝐿-Lipschitz continuous on 𝐶.Then lim sup𝑛→∞𝛿𝑛 ≤0.Hence,byLemma 2 the sequence {𝑥𝑛} ⋆ 𝑓(𝑥,𝑦)+𝑓(𝑦,𝑧)−𝑓(𝑥,) 𝑧 =⟨𝐹(𝑥) − 𝐹 (𝑦) , 𝑦 −𝑧⟩, converges strongly to 𝑥 .From(32)wehavethat{𝑤𝑛,𝑖} and ⋆ (61) {𝑧𝑛,𝑖} converge strongly to 𝑥 . 𝑥, 𝑦, 𝑧 ∈𝐶. ⋆ Case 2. Assume that {‖𝑥𝑛 −𝑥 ‖} is not a monotone sequence. Therefore, Then, we can define an integer sequence {𝜏(𝑛)} for all 𝑛≥𝑛0 󵄨 󵄨 󵄩 󵄩 󵄩 󵄩 󵄨⟨𝐹 (𝑥) − 𝐹 (𝑦) , 󵄨𝑦 −𝑧⟩ ≤𝐿󵄩𝑥−𝑦󵄩 󵄩𝑦−𝑧󵄩 (for some large enough 𝑛0)by 󵄩 󵄩 󵄩 󵄩 𝐿 󵄩 󵄩2 󵄩 󵄩2 (62) 𝜏 (𝑛) := {𝑘 ∈ N,𝑘≤𝑛:󵄩𝑥 −𝑥⋆󵄩 < 󵄩𝑥 −𝑥⋆󵄩}. ≤ (󵄩𝑥−𝑦󵄩 + 󵄩𝑦−𝑧󵄩 ) max 󵄩 𝑘 󵄩 󵄩 𝑘+1 󵄩 2 󵄩 󵄩 󵄩 󵄩 (52) hence, 𝑓 satisfies Lipschitz-type continuous condition with Clearly, 𝜏 is a nondecreasing sequence such that 𝜏(𝑛) →∞ 𝑐1 =𝑐2 = 𝐿/2. as 𝑛→∞,andforall𝑛≥𝑛0, Using Theorem 7 we obtain the following convergence theorem. 󵄩 ⋆󵄩 󵄩 ⋆󵄩 󵄩𝑥𝜏(𝑛) −𝑥 󵄩 < 󵄩𝑥𝜏(𝑛)+1 −𝑥 󵄩 . (53) Theorem 8. Let 𝐶 beanonemptyclosedconvexsubsetof From (33)weobtainthat arealHilbertspace𝐻 and let 𝐹𝑖 : 𝐶 → 𝐻 (𝑖 = 1,2,...,𝑚) 1≤𝑖≤𝑚 󵄩 󵄩 󵄩 󵄩 be functions such that, for each , 󵄩𝑥𝜏(𝑛) −𝑧𝜏(𝑛),𝑖󵄩 = 󵄩𝑥𝜏(𝑛) −𝑤𝜏(𝑛),𝑖󵄩 𝑛→∞lim 󵄩 󵄩 𝑛→∞lim 󵄩 󵄩 𝐹𝑖 is monotone and 𝐿-Lipschitz continuous on 𝐶.LetT = {𝑇(𝑢) : 𝑢 ≥0} S ={𝑆(𝑢):𝑢≥0} 󵄩 󵄩 and be two = 󵄩𝑥 −𝑇(𝑡 )𝑥 󵄩 =0. 𝐶 𝑛→∞lim 󵄩 𝜏(𝑛) 𝜏𝑛 𝜏(𝑛)󵄩 u.a.r. nonexpansive self-mapping semigroups on such that Ω=⋂𝑚 𝑉𝐼(𝐹 ,𝐶)⋂ 𝐹(T) ⋂ 𝐹(S) =0̸ ℎ 𝑘 (54) 𝑖=1 𝑖 . Assume that is a - contraction of 𝐶 into itself and 𝐴 is a strongly positive bounded Following an argument similar to that in Case 1 we have linear self-adjoint operator on 𝐻 with coefficient 𝛾<1and 0<𝛾<𝛾/𝑘.Let𝑥0 ∈𝐶and let {𝑥𝑛}, {𝑤𝑛,𝑖},and{𝑧𝑛,𝑖} be 󵄩 ⋆󵄩2 󵄩 ⋆󵄩2 󵄩𝑥𝜏(𝑛)+1 −𝑥 󵄩 ≤(1−𝜂𝜏(𝑛)) 󵄩𝑥𝜏(𝑛) −𝑥 󵄩 +𝜂𝜏(𝑛)𝛿𝜏(𝑛), (55) sequences generated by

∞ 𝑤𝑛,𝑖 =𝑃𝐶 (𝑥𝑛 −𝜆𝑛,𝑖𝐹𝑖 (𝑥𝑛)) , 𝑖 ∈ {1,2,...,𝑚} , where 𝜂𝜏(𝑛) →0, ∑𝑛=1 𝜂𝜏(𝑛) =∞,andlimsup𝑛→∞𝛿𝜏(𝑛) ≤0. ⋆ Hence, by Lemma 2,weobtainlim𝑛→∞‖𝑥𝜏(𝑛) −𝑥 ‖=0and ⋆ 𝑧𝑛,𝑖 =𝑃𝐶 (𝑥𝑛 −𝜆𝑛,𝑖𝐹𝑖 (𝑤𝑛,𝑖)) , 𝑖 ∈ {1,2,...,𝑚} , lim𝑛→∞‖𝑥𝜏(𝑛)+1 −𝑥 ‖=0.NowLemma 5 implies that 𝑚 (63) 󵄩 ⋆󵄩 󵄩 ⋆󵄩 󵄩 ⋆󵄩 𝑦 =𝛼𝑥 + ∑𝛽 𝑧 +𝛾𝑇(𝑡 )𝑥 +𝜂 𝑆(𝑡 )𝑥 , 0≤󵄩𝑥𝑛 −𝑥 󵄩 ≤ max {󵄩𝑥𝜏(𝑛) −𝑥 󵄩 , 󵄩𝑥𝑛 −𝑥 󵄩} 𝑛 𝑛 𝑛 𝑛,𝑖 𝑛,𝑖 𝑛 𝑛 𝑛 𝑛 𝑛 𝑛 (56) 𝑖=1 󵄩 ⋆󵄩 ≤ 󵄩𝑥𝜏(𝑛)+1 −𝑥 󵄩 . 𝑥𝑛+1 =𝜃𝑛𝛾ℎ 𝑛(𝑥 )+(𝐼−𝜃𝑛𝐴)𝑛 𝑦 ,∀𝑛≥0, ⋆ 𝑚 Therefore, {𝑥𝑛} converges strongly to 𝑥 = ProjΩ(𝐼 − 𝐴 + ⋆ where 𝛼𝑛 +∑𝑖=1 𝛽𝑛,𝑖 +𝛾𝑛 +𝜂𝑛 =1and {𝛼𝑛}, {𝛽𝑛,𝑖}, {𝛾𝑛}, {𝜂𝑛}, 𝛾ℎ)𝑥 .Thiscompletestheproof. {𝜆𝑛,𝑖},and{𝜃𝑛} satisfy the following conditions:

(i) lim𝑛→∞𝑡𝑛 =∞, 4. Application ∞ (ii) {𝜃𝑛}⊂]0,1[, lim𝑛→∞𝜃𝑛 =0,and∑𝑛=1 𝜃𝑛 =∞,

In this section, we consider a particular Fan inequality (iii) {𝜆𝑛,𝑖} ⊂ [𝑎, 𝑏] ⊂ ]0, 1/𝐿[, corresponding to the function 𝑓 defined by the following: for (iv) {𝛼𝑛}, {𝛽𝑛,𝑖}, {𝛾𝑛}, {𝜂𝑛}⊂]0,1[, lim inf𝑛𝛼𝑛𝛽𝑛,𝑖 >0, every 𝑥, 𝑦 ∈𝐶 lim inf𝑛𝛼𝑛𝛾𝑛 >0, lim inf𝑛𝛼𝑛𝜂𝑛 >0,and 𝛽 (1 − 2𝜆 𝑐 )>0 1≤𝑖≤𝑚 𝑓 (𝑥,) 𝑦 = ⟨𝐹 (𝑥) ,𝑦−𝑥⟩ , (57) lim inf𝑛 𝑛,𝑖 𝑛,𝑖 1,𝑖 for each . ⋆ Then, the sequence {𝑥𝑛} converges strongly to 𝑥 ∈ 𝐹:𝐶 →𝐻 𝑚 where . Then, we obtain the classical variational ⋂𝑖=1 𝑉𝐼(𝐹𝑖,𝐶)⋂𝐹(T)⋂𝐹(S) which solves the variational inequality as follows. inequality ⋆ ⋆ Find 𝑧∈𝐶 such that ⟨𝐹 (𝑧) , 𝑦 − 𝑧⟩ ≥ 0, ∀𝑦 ∈𝐶. (58) ⟨(𝐴 − 𝛾ℎ) 𝑥 ,𝑥−𝑥 ⟩ ≥ 0, ∀𝑥 ∈ Ω. (64) 8 Abstract and Applied Analysis

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