About this Journal Submit a Manuscript Table of Contents 
Journal of Applied Mathematics
Volume 2012 (2012), Article ID 572326, 18 pages
doi:10.1155/2012/572326
Research Article

Hybrid Iterative Scheme by a Relaxed Extragradient Method for Equilibrium Problems, a General System of Variational Inequalities and Fixed-Point Problems of a Countable Family of Nonexpansive Mappings

Qiao-Li Dong,1 Yan-Ni Guo,1 and Fang Su2

1College of Science, Civil Aviation University of China, Tianjin 300300, China
2College of Science, National University of Defense Technology, Changsha 410073, China

Received 29 November 2011; Accepted 5 December 2011

Academic Editor: Yonghong Yao

Copyright © 2012 Qiao-Li Dong 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.

Abstract

Based on the relaxed extragradient method and viscosity method, we introduce a new iterative method for finding a common element of solution of equilibrium problems, the solution set of a general system of variational inequalities, and the set of fixed points of a countable family of nonexpansive mappings in a real Hilbert space. Furthermore, we prove the strong convergence theorem of the studied iterative method. The results of this paper extend and improve the results of Ceng et al., (2008), W. Kumam and P. Kumam, (2009), Yao et al., (2010) and many others.

1. Introduction

Let 𝐻 be a real Hilbert space with the inner product ⟨ ⋅ , ⋅ ⟩ and the norm ‖ ⋅ ‖ . Let 𝐶 be a closed convex subset of 𝐻 . Let 𝐹 be a bifunction of 𝐶 × 𝐶 into ℝ , where ℝ is the set of real numbers. The equilibrium problem for 𝐹 ∶ 𝐶 × 𝐶 → ℝ is to find 𝑥 ∈ 𝐶 such that 𝐹 ( 𝑥 , 𝑦 ) ≥ 0 , ∀ 𝑦 ∈ 𝐶 . ( 1 . 1 ) The set of solutions of (1.1) is denoted by E P ( 𝐹 ) . The equilibrium problems covers, as special cases, monotone inclusion problems, saddle point problems, minimization problems, optimization problems, variational inequality problems, Nash equilibria in noncooperative games, and various forms of feasibility problems (see [1–4] and the references therein).

A mapping 𝐴 ∶ 𝐶 → 𝐻 is called 𝛼 -inverse-strongly monotone if there exists a positive real number 𝛼 such that ⟨ 𝐴 𝑢 − 𝐴 𝑣 , 𝑢 − 𝑣 ⟩ ≥ 𝛼 ‖ 𝐴 𝑢 − 𝐴 𝑣 ‖ 2 , ∀ 𝑢 , 𝑣 ∈ 𝐶 . ( 1 . 2 ) It is obvious that any 𝛼 -inverse-strongly monotone mapping 𝐴 is monotone and Lipschitz continuous. A mapping 𝑇 ∶ 𝐶 → 𝐶 is said to be nonexpansive if ‖ 𝑇 𝑥 − 𝑇 𝑦 ‖ ≤ ‖ 𝑥 − 𝑦 ‖ , ∀ 𝑥 , 𝑦 ∈ 𝐶 . ( 1 . 3 ) We denote by 𝐹 ( 𝑇 ) the set of fixed points of 𝑇 . Recently, Wang and Guo [5] introduced an iterative scheme for a countable family of nonexpansive mappings.

Let 𝐶 be a nonempty closed convex subset of a real Hilbert space 𝐻 . For a given nonlinear operator 𝐴 ∶ 𝐶 → 𝐻 , consider the following variational inequality problem of finding 𝑥 ∗ ∈ 𝐶 such that ⟨ 𝐴 𝑥 ∗ , 𝑥 − 𝑥 ∗ ⟩ ≥ 0 , 𝑥 ∈ 𝐶 . ( 1 . 4 ) The set of solutions of the variational inequality (1.4) is denoted by V I ( 𝐶 , 𝐴 ) (see [6–9] and the references therein).

Let 𝐴 , 𝐵 ∶ 𝐶 → 𝐻 be two mappings. Consider the following problem of finding ( 𝑥 ∗ , 𝑦 ∗ ) ∈ 𝐶 × 𝐶 such that ⟨ 𝜆 𝐴 𝑦 ∗ + 𝑥 ∗ − 𝑦 ∗ , 𝑥 − 𝑥 ∗ ⟩ ≥ 0 , ∀ 𝑥 ∈ 𝐶 , ⟨ 𝜇 𝐵 𝑥 ∗ + 𝑦 ∗ − 𝑥 ∗ , 𝑥 − 𝑦 ∗ ⟩ ≥ 0 , ∀ 𝑥 ∈ 𝐶 , ( 1 . 5 ) which is called a general system of variational inequalities, where 𝜆 > 0 and 𝜇 > 0 are two constants. The set of solutions of (1.5) is denoted by GSVI( 𝐴 , 𝐵 , a n d 𝐶 ). In particular, if 𝐴 = 𝐵 , then problem (1.5) reduces to finding ( 𝑥 ∗ , 𝑦 ∗ ) ∈ 𝐶 × 𝐶 such that ⟨ 𝜆 𝐴 𝑦 ∗ + 𝑥 ∗ − 𝑦 ∗ , 𝑥 − 𝑥 ∗ ⟩ ≥ 0 , ∀ 𝑥 ∈ 𝐶 , ⟨ 𝜇 𝐴 𝑥 ∗ + 𝑦 ∗ − 𝑥 ∗ , 𝑥 − 𝑦 ∗ ⟩ ≥ 0 , ∀ 𝑥 ∈ 𝐶 , ( 1 . 6 ) which is defined by Verma [7] (see also [10]) and is called the new system of variational inequalities. Further, if we add up the requirement that 𝑥 ∗ = 𝑦 ∗ , then problem (1.6) reduces to the classical variational inequality problem (1.4). Recently, Yao et al. [11] presented system of variational inequalities in Banach space. For solving problem (1.5), recently, Ceng et al. [12] introduced and studied a relaxed extragradient method. Based on the relaxed extragradient method and the viscosity approximation method, W. Kumam and P. Kumam [13] constructed a new viscosity-relaxed extragradient approximation method. Very recently, based on the extragradient method, Yao et al. [14] proposed an iterative method for finding a common element of the set of a general system of variational inequalities and the set of fixed points of a strictly pseudocontractive mapping in a real Hilbert space.

Motivated and inspired by the above works, in this paper, we introduce an iterative method based on the extragradient method and viscosity method for finding a common element of solution of equilibrium problems, the solution set of a general system of variational inequalities, and the set of fixed points of a countable family of nonexpansive mappings in a real Hilbert space. Furthermore, we prove the strong convergence theorem of the proposed iterative method.

2. Preliminaries

Let 𝐶 be a closed convex subset of 𝐻 , and let 𝑇 ∶ 𝐶 → 𝐶 be nonexpansive such that 𝐹 ( 𝑇 ) ≠ ∅ . For all ̂ 𝑥 ∈ 𝐹 ( 𝑇 ) and all 𝑥 ∈ 𝐶 , we have ‖ 𝑥 − ̂ 𝑥 ‖ 2 ≥ ‖ 𝑇 𝑥 − 𝑇 ̂ 𝑥 ‖ 2 = ‖ 𝑇 𝑥 − ̂ 𝑥 ‖ 2 ‖ = ‖ ( 𝑇 𝑥 − 𝑥 ) + ( 𝑥 − ̂ 𝑥 ) 2 = ‖ 𝑇 𝑥 − 𝑥 ‖ 2 + ‖ 𝑥 − ̂ 𝑥 ‖ 2 + 2 ⟨ 𝑇 𝑥 − 𝑥 , 𝑥 − ̂ 𝑥 ⟩ , ( 2 . 1 ) and hence ‖ 𝑇 𝑥 − 𝑥 ‖ 2 ≤ 2 ⟨ 𝑥 − 𝑇 𝑥 , 𝑥 − ̂ 𝑥 ⟩ , ∀ ̂ 𝑥 ∈ 𝐹 ( 𝑇 ) , ∀ 𝑥 ∈ 𝐶 . ( 2 . 2 )

Remark 2.1. Let A be 𝛼 -inverse-strongly monotone. For all 𝑥 , 𝑦 ∈ 𝐶 , 𝜆 > 0 , we have ‖ ‖ ( 𝐼 − 𝜆 𝐴 ) 𝑥 − ( 𝐼 − 𝜆 𝐴 ) 𝑦 2 = ‖ 𝑥 − 𝑦 ‖ 2 − 2 𝜆 ⟨ 𝐴 𝑥 − 𝐴 𝑦 , 𝑥 − 𝑦 ⟩ + 𝜆 2 ‖ 𝐴 𝑥 − 𝐴 𝑦 ‖ 2 ≤ ‖ 𝑥 − 𝑦 ‖ 2 + 𝜆 ( 𝜆 − 2 𝛼 ) ‖ 𝐴 𝑥 − 𝐴 𝑦 ‖ 2 . ( 2 . 3 ) So, if 𝜆 ≤ 2 𝛼 , then 𝐼 − 𝜆 𝐴 is a nonexpansive mapping from 𝐶 to 𝐻 .

Recall that the (nearest point) projection 𝑃 𝐶 from 𝐻 onto 𝐶 assigns to each 𝑥 ∈ 𝐻 the unique point 𝑃 𝐶 𝑥 ∈ 𝐶 satisfying the property ‖ ‖ 𝑥 − 𝑃 𝐶 𝑥 ‖ ‖ = m i n 𝑦 ∈ 𝐶 ‖ 𝑥 − 𝑦 ‖ . ( 2 . 4 ) The following characterizes the projection 𝑃 𝐶 .

Lemma 2.2. Given that 𝑥 ∈ 𝐻 and 𝑦 ∈ 𝐶 , then 𝑃 𝐶 𝑥 = 𝑦 if and only if there holds the inequality ⟨ 𝑥 − 𝑦 , 𝑦 − 𝑧 ⟩ ≥ 0 ∀ 𝑧 ∈ 𝐶 . ( 2 . 5 )

Lemma 2.3 (see [15]). Let 𝐻 be a Hilbert space, 𝐶 a closed convex subset of 𝐻 , and 𝑇 ∶ 𝐶 → 𝐶 a nonexpansive mapping with 𝐹 ( 𝑇 ) ≠ ∅ . If { 𝑥 𝑛 } is a sequence in 𝐶 weakly converging to 𝑥 ∈ 𝐶 and if { ( 1 − 𝑇 ) 𝑥 𝑛 } converges strongly to 𝑦 , then ( 1 − 𝑇 ) 𝑥 = 𝑦 .

Lemma 2.4 (see [16]). Assume that { 𝑎 𝑛 } is a sequence of nonnegative real numbers such that 𝛼 𝑛 + 1 ≤ 1 − 𝛾 𝑛 𝛼 𝑛 + 𝛿 𝑛 , ∀ 𝑛 ≥ 0 , ( 2 . 6 ) where two sequences { 𝛾 𝑛 } ⊂ ( 0 , 1 ) and { 𝛿 𝑛 } satisfy(1) ∑ ∞ 𝑛 = 1 𝛾 𝑛 = ∞ ; (2) l i m s u p 𝑛 → ∞ 𝛿 𝑛 / 𝛾 𝑛 ∑ ≤ 0 𝑜 𝑟 ∞ 𝑛 = 1 | 𝛿 𝑛 | < ∞ . Then l i m 𝑛 → ∞ 𝛼 𝑛 = 0 .

Lemma 2.5 (see [12]). For given 𝑥 ∗ , 𝑦 ∗ ∈ 𝐶 , ( 𝑥 ∗ , 𝑦 ∗ ) is a solution of problem (1.5) if and only if 𝑥 ∗ is a fixed point of the mapping 𝐺 ∶ 𝐶 → 𝐶 defined by 𝐺 ( 𝑥 ) = 𝑃 𝐶 𝑃 𝐶 ( 𝑥 − 𝜇 𝐵 𝑥 ) − 𝜆 𝐴 𝑃 𝐶 ( 𝑥 − 𝜇 𝐵 𝑥 ) , ∀ 𝑥 ∈ 𝐶 , ( 2 . 7 ) where 𝑦 ∗ = 𝑃 𝐶 ( 𝑥 ∗ − 𝜇 𝐵 𝑥 ∗ ) .

Note that the mapping 𝐺 is nonexpansive provided that 𝜆 ∈ ( 0 , 2 𝛼 ) and 𝜇 ∈ ( 0 , 2 𝛽 ) .

Throughout this paper, the set of fixed points of the mapping 𝐺 is denoted by Γ .

Lemma 2.6 (see [1]). Let 𝐶 be a nonempty closed convex subset of 𝐻 and 𝐹 ∶ 𝐶 × 𝐶 → ℝ satisfy following conditions: (A1) 𝐹 ( 𝑥 , 𝑥 ) = 0 , ∀ 𝑥 ∈ 𝐶 ; (A2) 𝐹 is monotone, that is, 𝐹 ( 𝑥 , 𝑦 ) + 𝐹 ( 𝑦 , 𝑥 ) ≤ 0 , ∀ 𝑥 , 𝑦 ∈ 𝐶 ;(A3) l i m s u p 𝑡 → 0 + 𝐹 ( 𝑡 𝑧 + ( 1 − 𝑡 ) 𝑥 , 𝑦 ) ≤ 𝐹 ( 𝑥 , 𝑦 ) , ∀ 𝑥 , 𝑦 , 𝑧 ∈ 𝐶 ;(A4)for each 𝑥 ∈ 𝐶 , 𝐹 ( 𝑥 , ⋅ ) is convex and lower semicontinuous.For 𝑥 ∈ 𝐶 and 𝑟 > 0 , set 𝑇 𝐹 𝑟 ∶ 𝐻 → 𝐶 to be 𝑇 𝐹 𝑟 1 ( 𝑥 ) = 𝑧 ∈ 𝐶 ∶ 𝐹 ( 𝑧 , 𝑦 ) + 𝑟 ⟨ 𝑦 − 𝑧 , 𝑧 − 𝑥 ⟩ ≥ 0 , ∀ 𝑦 ∈ 𝐶 . ( 2 . 8 ) Then 𝑇 𝐹 𝑟 is well defined and the following holds: (1) 𝑇 𝐹 𝑟 is single valued;(2) 𝑇 𝐹 𝑟 is firmly nonexpansive [17], that is, for any 𝑥 , 𝑦 ∈ 𝐸 , ‖ ‖ 𝑇 𝐹 𝑟 𝑥 − 𝑇 𝐹 𝑟 𝑦 ‖ ‖ 2 ≤ 𝑇 𝐹 𝑟 𝑥 − 𝑇 𝐹 𝑟 𝑦 , 𝑥 − 𝑦 ; ( 2 . 9 ) (3) 𝐹 ( 𝑇 𝐹 𝑟 ) = 𝐸 𝑃 ( 𝐹 ) ; (4) 𝐸 𝑃 ( 𝐹 ) is closed and convex.

By the proof of Lemma  5 in [2] (also see [3]), we have following lemma.

Lemma 2.7. Let 𝐶 be a nonempty closed convex subset of a Hilbert space 𝐻 and 𝐹 ∶ 𝐶 × 𝐶 → ℝ be a bifunction. Let 𝑥 ∈ 𝐶 and 𝑟 1 , 𝑟 2 ∈ ( 0 , ∞ ) . Then ‖ ‖ 𝑇 𝑟 1 𝑥 − 𝑇 𝑟 2 𝑥 ‖ ‖ ≤ | | | | 𝑟 1 − 2 𝑟 1 | | | | ‖ ‖ 𝑇 𝑟 1 𝑥 ‖ ‖ + ‖ 𝑥 ‖ . ( 2 . 1 0 )

3. Main Results

Theorem 3.1. Let 𝐶 be a nonempty closed convex subset of a real Hilbert space 𝐻 . Let the mappings 𝐴 , 𝐵 ∶ 𝐶 → 𝐻 be 𝛼 -inverse strongly monotone and 𝛽 -inverse strongly monotone, respectively. Let 𝐹 be a bifunction from 𝐶 × 𝐶 to ℝ satisfying (A1)–(A4) and { 𝑇 𝑛 } ∞ 𝑛 = 1 ∶ 𝐶 → 𝐶 be a countable family of nonexpansive mappings such that ⋂ Ω ∶ = ∞ 𝑛 = 1 𝐹 ( 𝑇 𝑛 ) ∩ 𝐸 𝑃 ( 𝐹 ) ∩ Γ ≠ ∅ . Let 𝑓 ∶ 𝐶 → 𝐶 be a contraction with coefficient 𝜌 ∈ ( 0 , 1 / 2 ) . Set 𝛽 0 = 1 . For given 𝑥 1 ∈ 𝐶 arbitrarily, let the sequences { 𝑥 𝑛 } , { 𝑦 𝑛 } , { 𝑧 𝑛 } , and { 𝑢 𝑛 } be generated by 𝐹 𝑢 𝑛 + 1 , 𝑦 𝑟 𝑛 ⟨ 𝑦 − 𝑢 𝑛 , 𝑢 𝑛 − 𝑥 𝑛 𝑧 ⟩ ≥ 0 , ∀ 𝑦 ∈ 𝐶 , 𝑛 = 𝑃 𝐶 𝑢 𝑛 − 𝜇 𝐵 𝑢 𝑛 , 𝑦 𝑛 = 𝛼 𝑛 𝑓 𝑥 𝑛 + 1 − 𝛼 𝑛 𝑃 𝐶 𝑧 𝑛 − 𝜆 𝐴 𝑧 𝑛 , 𝑥 𝑛 + 1 = 𝛽 𝑛 𝑥 𝑛 + 𝜎 𝑛 𝑛 𝑖 = 1 𝛽 𝑖 − 1 − 𝛽 𝑖 𝑇 𝑖 𝑦 𝑛 + 1 − 𝛽 𝑛 1 − 𝜎 𝑛 𝑃 𝐶 𝑧 𝑛 − 𝜆 𝐴 𝑧 𝑛 , ( 3 . 1 ) where 𝜆 ∈ ( 0 , 2 𝛼 ) , 𝜇 ∈ ( 0 , 2 𝛽 ) , and sequences { 𝛼 𝑛 } ⊂ [ 0 , 1 ] , { 𝛽 𝑛 } ⊂ [ 0 , 1 ] , { 𝜎 𝑛 } ⊂ [ 0 , 1 ] , and { 𝑟 𝑛 } ⊂ ( 𝑟 , ∞ ) , 𝑟 > 0 , are such that
(i) { 𝛽 𝑛 } is strictly decreasing, (ii) 0 < l i m i n f 𝑛 → ∞ 𝛽 n ≤ l i m s u p 𝑛 → ∞ 𝛽 𝑛 < 1 , (iii) l i m 𝑛 → ∞ 𝛼 𝑛 = 0 and ∑ ∞ 𝑛 = 1 𝛼 𝑛 = ∞ ,(iv) 𝜎 𝑛 ∑ > 1 / 2 ( 1 − 𝜌 ) , ∞ 𝑛 = 1 | 𝜎 𝑛 − 𝜎 𝑛 − 1 | < ∞ , (v) ∑ ∞ 𝑛 = 1 | 𝑟 𝑛 − 𝑟 𝑛 − 1 | < ∞ . Then the sequence { 𝑥 𝑛 } generated by (3.1) converges strongly to 𝑥 ∗ = 𝑃 Ω ⋅ 𝑓 ( 𝑥 ∗ ) , and ( 𝑥 ∗ , 𝑦 ∗ ) is a solution of the general system of variational inequalities (1.5), where 𝑦 ∗ = 𝑃 𝐶 ( 𝑥 ∗ − 𝜇 𝐵 𝑥 ∗ ) .

Proof. The proof is divided into several steps.
Step 1. The sequence { 𝑥 𝑛 } defined by (3.1) is bounded.
For each 𝑥 ∗ ∈ Ω , from Lemma 2.6, we have 𝑢 𝑛 = 𝑇 𝑟 𝑛 𝑥 𝑛 and hence ‖ ‖ 𝑢 𝑛 − 𝑥 ∗ ‖ ‖ = ‖ ‖ 𝑇 𝑟 𝑛 𝑥 𝑛 − 𝑇 𝑟 𝑛 𝑥 ∗ ‖ ‖ ≤ ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ . ( 3 . 2 ) Since 𝑓 is a 𝜌 -contraction mapping, using (2.3) and (3.2), we have ‖ ‖ 𝑧 𝑛 − 𝑦 ∗ ‖ ‖ 2 = ‖ ‖ 𝑃 𝐶 𝑢 𝑛 − 𝜇 𝐵 𝑢 𝑛 − 𝑃 𝐶 𝑥 ∗ − 𝜇 𝐵 𝑥 ∗ ‖ ‖ 2 ≤ ‖ ‖ 𝑢 𝑛 − 𝜇 𝐵 𝑢 𝑛 − 𝑥 ∗ − 𝜇 𝐵 𝑥 ∗ ‖ ‖ 2 ≤ ‖ ‖ 𝑢 𝑛 − 𝑥 ∗ ‖ ‖ 2 ‖ ‖ + 𝜇 ( 𝜇 − 2 𝛽 ) 𝐵 𝑢 𝑛 − 𝐵 𝑥 ∗ ‖ ‖ 2 ≤ ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 ‖ ‖ + 𝜇 ( 𝜇 − 2 𝛽 ) 𝐵 𝑢 𝑛 − 𝐵 𝑥 ∗ ‖ ‖ 2 ≤ ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 . ( 3 . 3 ) Set 𝑣 𝑛 = 𝑃 𝐶 ( 𝑧 𝑛 − 𝜆 𝐴 𝑧 𝑛 ) . Since 𝑃 𝐶 is nonexpansive, from (2.3), we have ‖ ‖ 𝑣 𝑛 − 𝑥 ∗ ‖ ‖ 2 = ‖ ‖ 𝑃 𝐶 𝑧 𝑛 − 𝜆 𝐴 𝑧 𝑛 − 𝑃 𝐶 𝑦 ∗ − 𝜆 𝐴 𝑦 ∗ ‖ ‖ 2 ≤ ‖ ‖ 𝑧 𝑛 − 𝑦 ∗ ‖ ‖ 2 ‖ ‖ + 𝜆 ( 𝜆 − 2 𝛼 ) 𝐴 𝑧 𝑛 − 𝐴 𝑦 ∗ ‖ ‖ 2 ≤ ‖ ‖ 𝑧 𝑛 − 𝑦 ∗ ‖ ‖ 2 . ( 3 . 4 ) Hence we get ‖ ‖ 𝑦 𝑛 − 𝑥 ∗ ‖ ‖ = ‖ ‖ 𝛼 𝑛 𝑓 𝑥 𝑛 + 1 − 𝛼 𝑛 𝑣 𝑛 − 𝑥 ∗ ‖ ‖ ≤ 𝛼 𝑛 ‖ ‖ 𝑓 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ + 1 − 𝛼 𝑛 ‖ ‖ 𝑧 𝑛 − 𝑦 ∗ ‖ ‖ ≤ 𝛼 𝑛 ‖ ‖ 𝑓 𝑥 𝑛 𝑥 − 𝑓 ∗ ‖ ‖ + ‖ ‖ 𝑓 𝑥 ∗ − 𝑥 ∗ ‖ ‖ + 1 − 𝛼 𝑛 ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ ≤ 1 − 𝛼 𝑛 ( ‖ ‖ 𝑥 1 − 𝜌 ) 𝑛 − 𝑥 ∗ ‖ ‖ + 𝛼 𝑛 ‖ ‖ 𝑓 𝑥 ∗ − 𝑥 ∗ ‖ ‖ . ( 3 . 5 ) From (3.1) and (3.3)–(3.5), we get ‖ ‖ 𝑥 𝑛 + 1 − 𝑥 ∗ ‖ ‖ = ‖ ‖ ‖ ‖ 𝛽 𝑛 𝑥 𝑛 + 𝜎 𝑛 𝑛 𝑖 = 1 𝛽 𝑖 − 1 − 𝛽 𝑖 𝑇 𝑖 𝑦 𝑛 + 1 − 𝛽 𝑛 1 − 𝜎 𝑛 𝑣 𝑛 − 𝑥 ∗ ‖ ‖ ‖ ‖ ≤ 𝛽 𝑛 ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ + 𝜎 𝑛 𝑛 𝑖 = 1 𝛽 𝑖 − 1 − 𝛽 𝑖 ‖ ‖ 𝑇 𝑖 𝑦 𝑛 − 𝑥 ∗ ‖ ‖ + 1 − 𝛽 𝑛 1 − 𝜎 𝑛 ‖ ‖ 𝑣 𝑛 − 𝑥 ∗ ‖ ‖ ≤ 𝛽 𝑛 ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ + 𝜎 𝑛 𝑛 𝑖 = 1 𝛽 𝑖 − 1 − 𝛽 𝑖 ‖ ‖ 𝑦 𝑛 − 𝑥 ∗ ‖ ‖ + 1 − 𝛽 𝑛 1 − 𝜎 𝑛 ‖ ‖ 𝑧 𝑛 − 𝑦 ∗ ‖ ‖ ≤ 1 − 𝜎 𝑛 1 − 𝛽 𝑛 ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ + 𝜎 𝑛 1 − 𝛽 𝑛 ‖ ‖ 𝑦 𝑛 − 𝑦 ∗ ‖ ‖ ≤ 1 − 𝜎 𝑛 1 − 𝛽 𝑛 ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ + 𝜎 𝑛 1 − 𝛽 𝑛 1 − 𝛼 𝑛 ‖ ‖ 𝑥 ( 1 − 𝜌 ) 𝑛 − 𝑥 ∗ ‖ ‖ + 𝛼 𝑛 ‖ ‖ 𝑓 𝑥 ∗ − 𝑥 ∗ ‖ ‖ ≤ 1 − 𝜎 𝑛 1 − 𝛽 𝑛 𝛼 𝑛 ‖ ‖ 𝑥 ( 1 − 𝜌 ) 𝑛 − 𝑥 ∗ ‖ ‖ + 𝜎 𝑛 1 − 𝛽 𝑛 𝛼 𝑛 1 ( 1 − 𝜌 ) ‖ ‖ 𝑓 𝑥 1 − 𝜌 ∗ − 𝑥 ∗ ‖ ‖ ≤ 𝑀 , ( 3 . 6 ) where 𝑀 = m a x { ‖ 𝑥 1 − 𝑥 ∗ ‖ , 1 / ( 1 − 𝜌 ) ‖ 𝑓 ( 𝑥 ∗ ) − 𝑥 ∗ ‖ } . Hence, { 𝑥 𝑛 } is bounded and therefore { 𝑢 𝑛 } , { 𝑧 𝑛 } , { 𝑣 𝑛 } , and { 𝑦 𝑛 } are also bounded.Step 2. l i m 𝑛 → ∞ ‖ 𝑥 𝑛 + 1 − 𝑥 𝑛 ‖ = 0 . Since 𝑢 𝑛 = 𝑇 𝑟 𝑛 𝑥 𝑛 and 𝑢 𝑛 − 1 = 𝑇 𝑟 𝑛 − 1 𝑥 𝑛 − 1 , using Lemma 2.7, we have ‖ ‖ 𝑢 𝑛 − 𝑢 𝑛 − 1 ‖ ‖ = ‖ ‖ 𝑇 𝑟 𝑛 𝑥 𝑛 − 𝑇 𝑟 𝑛 − 1 𝑥 𝑛 − 1 ‖ ‖ ≤ ‖ ‖ 𝑇 𝑟 𝑛 𝑥 𝑛 − 𝑇 𝑟 𝑛 𝑥 𝑛 − 1 ‖ ‖ + ‖ ‖ 𝑇 𝑟 𝑛 𝑥 𝑛 − 1 − 𝑇 𝑟 𝑛 − 1 𝑥 𝑛 − 1 ‖ ‖ ≤ ‖ ‖ 𝑥 𝑛 − 𝑥 𝑛 − 1 ‖ ‖ + | | | | 𝑟 1 − 𝑛 𝑟 𝑛 − 1 | | | | ‖ ‖ 𝑢 𝑛 − 1 ‖ ‖ + ‖ ‖ 𝑥 𝑛 − 1 ‖ ‖ ≤ ‖ ‖ 𝑥 𝑛 − 𝑥 𝑛 − 1 ‖ ‖ + 1 𝑟 | | 𝑟 𝑛 − 1 − 𝑟 𝑛 | | 𝐿 , ( 3 . 7 ) where 𝐿 = s u p { ‖ 𝑢 𝑛 ‖ + ‖ 𝑥 𝑛 ‖ ∶ 𝑛 = 1 , 2 , … } . From (3.1) and (3.7), it follows that ‖ ‖ 𝑧 𝑛 − 𝑧 𝑛 − 1 ‖ ‖ = ‖ ‖ 𝑃 𝐶 ( 𝐼 − 𝜇 𝐵 ) 𝑢 𝑛 − 𝑃 𝐶 ( 𝐼 − 𝜇 𝐵 ) 𝑢 𝑛 − 1 ‖ ‖ ≤ ‖ ‖ 𝑢 𝑛 − 𝑢 𝑛 − 1 ‖ ‖ ≤ ‖ ‖ 𝑥 𝑛 − 𝑥 𝑛 − 1 ‖ ‖ + 1 𝑟 | | 𝑟 𝑛 − 1 − 𝑟 𝑛 | | 𝐿 . ( 3 . 8 ) From (3.1) and (3.8), we have ‖ ‖ 𝑦 𝑛 − 𝑦 𝑛 − 1 ‖ ‖ ≤ 𝛼 𝑛 ‖ ‖ 𝑓 𝑥 𝑛 𝑥 − 𝑓 𝑛 − 1 ‖ ‖ + 1 − 𝛼 𝑛 ‖ ‖ 𝑃 𝐶 ( 𝐼 − 𝜆 𝐴 ) 𝑧 𝑛 − 𝑃 𝐶 ( 𝐼 − 𝜆 𝐴 ) 𝑧 𝑛 − 1 ‖ ‖ ≤ 𝛼 𝑛 𝜌 ‖ ‖ 𝑥 𝑛 − 𝑥 𝑛 − 1 ‖ ‖ + 1 − 𝛼 𝑛 ‖ ‖ 𝑧 𝑛 − 𝑧 𝑛 − 1 ‖ ‖ ≤ 𝛼 𝑛 𝜌 ‖ ‖ 𝑥 𝑛 − 𝑥 𝑛 − 1 ‖ ‖ + 1 − 𝛼 𝑛 ‖ ‖ 𝑥 𝑛 − 𝑥 𝑛 − 1 ‖ ‖ + 1 𝑟 | | 𝑟 𝑛 − 1 − 𝑟 𝑛 | | 𝐿 ≤ 1 − 𝛼 𝑛 ‖ ‖ 𝑥 ( 1 − 𝜌 ) 𝑛 − 𝑥 𝑛 − 1 ‖ ‖ + 1 − 𝛼 𝑛 1 𝑟 | | 𝑟 𝑛 − 1 − 𝑟 𝑛 | | 𝐿 . ( 3 . 9 ) By definition of scheme (3.1), we have 𝑥 𝑛 + 1 − 𝑥 𝑛 = 𝛽 𝑛 𝑥 𝑛 − 𝑥 𝑛 − 1 + 𝛽 𝑛 − 𝛽 𝑛 − 1 𝑥 𝑛 − 1 + 𝜎 𝑛 𝑛 𝑖 = 1 𝛽 𝑖 − 1 − 𝛽 𝑖 𝑇 𝑖 𝑦 𝑛 − 𝑇 𝑖 𝑦 𝑛 − 1 + 𝜎 𝑛 − 𝜎 𝑛 − 1 𝑛 − 1 𝑖 = 1 𝛽 𝑖 − 1 − 𝛽 𝑖 𝑇 𝑖 𝑦 𝑛 − 1 + 𝜎 𝑛 𝛽 𝑛 − 1 − 𝛽 𝑛 𝑇 𝑛 𝑦 𝑛 − 1 + 1 − 𝛽 𝑛 1 − 𝜎 𝑛 𝑃 𝐶 𝑧 𝑛 − 𝜆 𝐴 𝑧 𝑛 − 𝑃 𝐶 𝑧 𝑛 − 1 − 𝜆 𝐴 𝑧 𝑛 − 1 − 1 − 𝛽 𝑛 − 1 1 − 𝜎 𝑛 − 1 𝑃 𝐶 𝑧 𝑛 − 1 − 𝜆 𝐴 𝑧 𝑛 − 1 + 1 − 𝛽 𝑛 1 − 𝜎 𝑛 𝑃 𝐶 𝑧 𝑛 − 1 − 𝜆 𝐴 𝑧 𝑛 − 1 . ( 3 . 1 0 ) Thus, from (3.8)–(3.10), ‖ ‖ 𝑥 𝑛 + 1 − 𝑥 𝑛 ‖ ‖ ≤ 𝛽 𝑛 ‖ ‖ 𝑥 𝑛 − 𝑥 𝑛 − 1 ‖ ‖ + 𝛽 𝑛 − 1 − 𝛽 𝑛 ‖ ‖ 𝑥 𝑛 − 1 ‖ ‖ + 𝜎 𝑛 𝑛 𝑖 = 1 𝛽 𝑖 − 1 − 𝛽 𝑖 ‖ ‖ 𝑇 𝑖 𝑦 𝑛 − 𝑇 𝑖 𝑦 𝑛 − 1 ‖ ‖ + | | 𝜎 𝑛 − 𝜎 𝑛 − 1 | | 𝑛 − 1 𝑖 = 1 𝛽 𝑖 − 1 − 𝛽 𝑖 ‖ ‖ 𝑇 𝑖 𝑦 𝑛 − 1 ‖ ‖ + 𝜎 𝑛 𝛽 𝑛 − 1 − 𝛽 𝑛 ‖ ‖ 𝑇 𝑛 𝑦 𝑛 − 1 ‖ ‖ + 1 − 𝛽 𝑛 1 − 𝜎 𝑛 ‖ ‖ 𝑃 𝐶 𝑧 𝑛 − 𝜆 𝐴 𝑧 𝑛 − 𝑃 𝐶 𝑧 𝑛 − 1 − 𝜆 𝐴 𝑧 𝑛 − 1 ‖ ‖ + | | 1 − 𝛽 𝑛 1 − 𝜎 𝑛 − 1 − 𝛽 𝑛 − 1 1 − 𝜎 𝑛 − 1 | | ‖ ‖ 𝑃 𝐶 𝑧 𝑛 − 1 − 𝜆 𝐴 𝑧 𝑛 − 1 ‖ ‖ ≤ 𝛽 𝑛 ‖ ‖ 𝑥 𝑛 − 𝑥 𝑛 − 1 ‖ ‖ + 𝛽 𝑛 − 1 − 𝛽 𝑛 ‖ ‖ 𝑥 𝑛 − 1 ‖ ‖ + 𝜎 𝑛 𝑛 𝑖 = 1 𝛽 𝑖 − 1 − 𝛽 𝑖 ‖ ‖ 𝑦 𝑛 − 𝑦 𝑛 − 1 ‖ ‖ + | | 𝜎 𝑛 − 𝜎 𝑛 − 1 | | 𝑛 − 1 𝑖 = 1 𝛽 𝑖 − 1 − 𝛽 𝑖 ‖ ‖ 𝑇 𝑖 𝑦 𝑛 − 1 ‖ ‖ + 𝛽 𝑛 − 1 − 𝛽 𝑛 ‖ ‖ 𝑇 𝑛 𝑦 𝑛 − 1 ‖ ‖ + 1 − 𝛽 𝑛 1 − 𝜎 𝑛 ‖ ‖ 𝑧 𝑛 − 𝑧 𝑛 − 1 ‖ ‖ + 𝛽 𝑛 − 1 − 𝛽 𝑛 1 − 𝜎 𝑛 ‖ ‖ 𝑃 𝐶 ( 𝐼 − 𝜆 𝐴 ) 𝑧 𝑛 − 1 ‖ ‖ + 1 − 𝛽 𝑛 − 1 | | 𝜎 𝑛 − 𝜎 𝑛 − 1 | | ‖ ‖ 𝑃 𝐶 ( 𝐼 − 𝜆 𝐴 ) 𝑧 𝑛 − 1 ‖ ‖ ≤ 𝛽 𝑛 ‖ ‖ 𝑥 𝑛 − 𝑥 𝑛 − 1 ‖ ‖ + 𝛽 𝑛 − 1 − 𝛽 𝑛 ‖ ‖ 𝑥 𝑛 − 1 ‖ ‖ + 𝜎 𝑛 1 − 𝛽 𝑛 × 1 − 𝛼 𝑛 ‖ ‖ 𝑥 ( 1 − 𝜌 ) 𝑛 − 𝑥 𝑛 − 1 ‖ ‖ + 1 − 𝛼 𝑛 1 𝑟 | | 𝑟 𝑛 − 1 − 𝑟 𝑛 | | 𝐿 + | | 𝜎 𝑛 − 𝜎 𝑛 − 1 | | 𝑛 − 1 𝑖 = 1 𝛽 𝑖 − 1 − 𝛽 𝑖 ‖ ‖ 𝑇 𝑖 𝑦 𝑛 − 1 ‖ ‖ + 𝛽 𝑛 − 1 − 𝛽 𝑛 ‖ ‖ 𝑇 𝑛 𝑦 𝑛 − 1 ‖ ‖ + 1 − 𝛽 𝑛 1 − 𝜎 𝑛 ‖ ‖ 𝑥 𝑛 − 𝑥 𝑛 − 1 ‖ ‖ + 1 𝑟 | | 𝑟 𝑛 − 1 − 𝑟 𝑛 | | 𝐿 + 𝛽 𝑛 − 1 − 𝛽 𝑛 ‖ ‖ 𝑃 𝐶 ( 𝐼 − 𝜆 𝐴 ) 𝑧 𝑛 − 1 ‖ ‖ + | | 𝜎 𝑛 − 𝜎 𝑛 − 1 | | ‖ ‖ 𝑃 𝐶 ( 𝐼 − 𝜆 𝐴 ) 𝑧 𝑛 − 1 ‖ ‖ ≤ 1 − 𝜎 𝑛 1 − 𝛽 𝑛 𝛼 𝑛 ( ‖ ‖ 𝑥 1 − 𝜌 ) 𝑛 − 𝑥 𝑛 − 1 ‖ ‖ + 𝛽 𝑛 − 1 − 𝛽 𝑛 𝑀 + 1 𝑟 | | 𝑟 𝑛 − 1 − 𝑟 𝑛 | | | | 𝜎 𝐿 + 𝑛 − 𝜎 𝑛 − 1 | | 𝑛 − 1 𝑖 = 1 𝛽 𝑖 − 1 − 𝛽 𝑖 𝛽 𝑀 + 𝑛 − 1 − 𝛽 𝑛 𝑀 + 𝛽 𝑛 − 1 − 𝛽 𝑛 + | | 𝜎 𝑛 − 𝜎 𝑛 − 1 | | 𝑀 ≤ 1 − 𝜎 𝑛 1 − 𝛽 𝑛 𝛼 𝑛 ‖ ‖ 𝑥 ( 1 − 𝜌 ) 𝑛 − 𝑥 𝑛 − 1 ‖ ‖ + 1 𝑟 | | 𝑟 𝑛 − 1 − 𝑟 𝑛 | | 𝐿 𝛽 + 3 𝑛 − 1 − 𝛽 𝑛 | | 𝜎 𝑀 + 2 𝑛 − 𝜎 𝑛 − 1 | | 𝑀 , ( 3 . 1 1 ) where 𝑀 = m a x { s u p 𝑛 ≥ 1 ‖ 𝑥 𝑛 ‖ , s u p 𝑖 ≥ 1 , 𝑛 ≥ 1 ‖ 𝑇 𝑖 𝑦 𝑛 ‖ , s u p 𝑛 ≥ 1 ‖ 𝑃 𝐶 ( 𝐼 − 𝜆 𝐴 ) 𝑧 𝑛 ‖ } . Since { 𝛽 𝑛 } is strictly decreasing, we have ∑ ∞ 𝑛 = 2 ( 𝛽 𝑛 − 1 − 𝛽 𝑛 ) = 𝛽 1 < ∞ . Further, by assumption conditions (iv)-(v), we have ∞ 𝑛 = 2 1 𝑟 | | 𝑟 𝑛 + 1 − 𝑟 𝑛 | | 𝛽 𝐿 + 3 𝑛 − 1 − 𝛽 𝑛 | | 𝜎 𝑀 + 2 𝑛 − 𝜎 𝑛 − 1 | | 𝑀 < ∞ . ( 3 . 1 2 ) Thus, using Lemma 2.4, we have l i m 𝑛 → ∞ ‖ 𝑥 𝑛 + 1 − 𝑥 𝑛 ‖ = 0 .Step 3. l i m 𝑛 → ∞ ‖ 𝑥 𝑛 − 𝑢 𝑛 ‖ = 0 .
For any 𝑥 ∗ ∈ Ω , it follows from Lemma 2.6 that ‖ ‖ 𝑢 𝑛 − 𝑥 ∗ ‖ ‖ 2 = ‖ ‖ 𝑇 𝑟 𝑛 𝑥 𝑛 − 𝑇 𝑟 𝑛 𝑥 ∗ ‖ ‖ 2 ≤ 𝑇 𝑟 𝑛 𝑥 𝑛 − 𝑇 𝑟 𝑛 𝑥 ∗ , 𝑥 𝑛 − 𝑥 ∗ = ⟨ 𝑢 𝑛 − 𝑥 ∗ , 𝑥 𝑛 − 𝑥 ∗ ⟩ = 1 2 ‖ ‖ 𝑢 𝑛 − 𝑥 ∗ ‖ ‖ 2 + ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 − ‖ ‖ 𝑢 𝑛 − 𝑥 𝑛 ‖ ‖ 2 . ( 3 . 1 3 ) From (2.3), (3.1)–(3.4), we can get ‖ ‖ 𝑦 𝑛 − 𝑥 ∗ ‖ ‖ 2 ≤ 𝛼 𝑛 ‖ ‖ 𝑓 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 + 1 − 𝛼 𝑛 ‖ ‖ 𝑣 𝑛 − 𝑥 ∗ ‖ ‖ 2 ≤ 𝛼 𝑛 ‖ ‖ 𝑓 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 + 1 − 𝛼 𝑛 ‖ ‖ 𝑧 𝑛 − 𝑦 ∗ ‖ ‖ 2 ‖ ‖ + 𝜆 ( 𝜆 − 2 𝛼 ) 𝐴 𝑧 𝑛 − 𝐴 𝑦 ∗ ‖ ‖ 2 ≤ 𝛼 𝑛 ‖ ‖ 𝑓 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 + 1 − 𝛼 𝑛 ‖ ‖ 𝑧 𝑛 − 𝑦 ∗ ‖ ‖ 2 ≤ 𝛼 𝑛 ‖ ‖ 𝑓 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 + 1 − 𝛼 𝑛 ‖ ‖ 𝑢 𝑛 − 𝑥 ∗ ‖ ‖ 2 . ( 3 . 1 4 ) From (3.1), (3.3)-(3.4), and (3.13)-(3.14), it follows that ‖ ‖ 𝑥 𝑛 + 1 − 𝑥 ∗ ‖ ‖ 2 = ‖ ‖ ‖ ‖ 𝛽 𝑛 𝑥 𝑛 + 𝜎 𝑛 𝑛 𝑖 = 1 𝛽 𝑖 − 1 − 𝛽 𝑖 𝑇 𝑖 𝑦 𝑛 + 1 − 𝛽 𝑛 1 − 𝜎 𝑛 𝑣 𝑛 − 𝑥 ∗ ‖ ‖ ‖ ‖ 2 ≤ 𝛽 𝑛 ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 + 1 − 𝛽 𝑛 ‖ ‖ ‖ ‖ 𝜎 𝑛 1 1 − 𝛽 𝑛 𝑛 𝑖 = 1 𝛽 𝑖 − 1 − 𝛽 𝑖 𝑇 𝑖 𝑦 𝑛 + 1 − 𝜎 𝑛 𝑣 𝑛 − 𝑥 ∗ ‖ ‖ ‖ ‖ 2 ≤ 𝛽 𝑛 ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 + 1 − 𝛽 𝑛 𝜎 𝑛 ‖ ‖ ‖ ‖ 1 1 − 𝛽 𝑛 𝑛 𝑖 = 1 𝛽 𝑖 − 1 − 𝛽 𝑖 𝑇 𝑖 𝑦 𝑛 − 𝑥 ∗ ‖ ‖ ‖ ‖ 2 + 1 − 𝛽 𝑛 1 − 𝜎 𝑛 ‖ ‖ 𝑣 𝑛 − 𝑥 ∗ ‖ ‖ 2 ≤ 𝛽 𝑛 ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 + 1 − 𝛽 𝑛 𝜎 𝑛 1 1 − 𝛽 𝑛 𝑛 𝑖 = 1 𝛽 𝑖 − 1 − 𝛽 𝑖 ‖ ‖ 𝑇 𝑖 𝑦 𝑛 − 𝑥 ∗ ‖ ‖ 2 + 1 − 𝛽 𝑛 1 − 𝜎 𝑛 ‖ ‖ 𝑣 𝑛 − 𝑥 ∗ ‖ ‖ 2 ≤ 𝛽 𝑛 ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 + 1 − 𝛽 𝑛 𝜎 𝑛 ‖ ‖ 𝑦 𝑛 − 𝑥 ∗ ‖ ‖ 2 + 1 − 𝛽 𝑛 1 − 𝜎 𝑛 ‖ ‖ 𝑧 𝑛 − 𝑦 ∗ ‖ ‖ 2 ≤ 𝛽 𝑛 ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 + 1 − 𝛽 𝑛 𝜎 𝑛 1 − 𝛼 𝑛 ‖ ‖ 𝑢 𝑛 − 𝑥 ∗ ‖ ‖ 2 + 𝛼 𝑛 ‖ ‖ 𝑓 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 + 1 − 𝛽 𝑛 1 − 𝜎 𝑛 ‖ ‖ 𝑢 𝑛 − 𝑥 ∗ ‖ ‖ 2 ≤ ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 + 1 − 𝛽 𝑛 𝜎 𝑛 𝛼 𝑛 ‖ ‖ 𝑓 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 − ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 − 1 − 𝛽 𝑛 1 − 𝜎 𝑛 𝛼 𝑛 ‖ ‖ 𝑢 𝑛 − 𝑥 𝑛 ‖ ‖ 2 ( 3 . 1 5 ) which implies that 1 − 𝛽 𝑛 1 − 𝜎 𝑛 𝛼 𝑛 ‖ ‖ 𝑢 𝑛 − 𝑥 𝑛 ‖ ‖ 2 ≤ ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ + ‖ ‖ 𝑥 𝑛 + 1 − 𝑥 ∗ ‖ ‖ ‖ ‖ 𝑥 𝑛 − 𝑥 𝑛 + 1 ‖ ‖ + 𝛼 𝑛 ‖ ‖ 𝑓 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 . ( 3 . 1 6 ) Since { 𝑥 𝑛 } is both bounded, using Step 2 and conditions (ii)–(iv), we conclude the result.Step 4. l i m 𝑛 → ∞ ‖ 𝐴 𝑧 𝑛 − 𝐴 𝑦 ∗ ‖ = 0 and l i m 𝑛 → ∞ ‖ 𝐵 𝑥 𝑛 − 𝐵 𝑥 ∗ ‖ = 0 .Using (3.3), (3.14), and (3.15), we have ‖ ‖ 𝑥 𝑛 + 1 − 𝑥 ∗ ‖ ‖ 2 ≤ 𝛽 𝑛 ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 + 1 − 𝛽 𝑛 𝜎 𝑛 ‖ ‖ 𝑦 𝑛 − 𝑥 ∗ ‖ ‖ 2 + 1 − 𝛽 𝑛 1 − 𝜎 𝑛 ‖ ‖ 𝑧 𝑛 − 𝑦 ∗ ‖ ‖ 2 ≤ 𝛽 𝑛 ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 + 1 − 𝛽 𝑛 𝜎 𝑛 × 𝛼 𝑛 ‖ ‖ 𝑓 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 + 1 − 𝛼 𝑛 ‖ ‖ 𝑧 𝑛 − 𝑦 ∗ ‖ ‖ 2 ‖ ‖ + 𝜆 ( 𝜆 − 2 𝛼 ) 𝐴 𝑧 𝑛 − 𝐴 𝑦 ∗ ‖ ‖ 2 + 1 − 𝛽 𝑛 1 − 𝜎 𝑛 ‖ ‖ 𝑧 𝑛 − 𝑦 ∗ ‖ ‖ 2 ≤ 𝛽 𝑛 ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 + 1 − 𝛽 𝑛 𝜎 𝑛 1 − 𝛼 𝑛 𝜆 ‖ ‖ ( 𝜆 − 2 𝛼 ) 𝐴 𝑧 𝑛 − 𝐴 𝑦 ∗ ‖ ‖ 2 + 𝛼 𝑛 ‖ ‖ 𝑓 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 + 1 − 𝛽 𝑛 ‖ ‖ 𝑧 𝑛 − 𝑦 ∗ ‖ ‖ 2 ≤ 𝛽 𝑛 ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 ‖ ‖ + 𝜆 ( 𝜆 − 2 𝛼 ) 𝐴 𝑧 𝑛 − 𝐴 𝑦 ∗ ‖ ‖ 2 + 𝛼 𝑛 ‖ ‖ 𝑓 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 + 1 − 𝛽 𝑛 ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 ‖ ‖ + 𝜇 ( 𝜇 − 2 𝛽 ) 𝐵 𝑢 𝑛 − 𝐵 𝑥 ∗ ‖ ‖ 2 ≤ ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 ‖ ‖ + 𝜆 ( 𝜆 − 2 𝛼 ) 𝐴 𝑧 𝑛 − 𝐴 𝑦 ∗ ‖ ‖ 2 + 𝛼 𝑛 ‖ ‖ 𝑓 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 ‖ ‖ + 𝜇 ( 𝜇 − 2 𝛽 ) 𝐵 𝑢 𝑛 − 𝐵 𝑥 ∗ ‖ ‖ 2 . ( 3 . 1 7 ) Therefore, ‖ ‖ 𝜆 ( 2 𝛼 − 𝜆 ) 𝐴 𝑧 𝑛 − 𝐴 𝑦 ∗ ‖ ‖ 2 ‖ ‖ + 𝜇 ( 2 𝛽 − 𝜇 ) 𝐵 𝑢 𝑛 − 𝐵 𝑥 ∗ ‖ ‖ 2 ≤ ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ + ‖ ‖ 𝑥 𝑛 + 1 − 𝑥 ∗ ‖ ‖ ‖ ‖ 𝑥 𝑛 − 𝑥 𝑛 + 1 ‖ ‖ + 𝛼 𝑛 ‖ ‖ 𝑓 𝑥 𝑛 − 𝑦 ∗ ‖ ‖ 2 . ( 3 . 1 8 ) From Step 2, using condition (iii), we get l i m 𝑛 → ∞ ‖ 𝐴 𝑧 𝑛 − 𝐴 𝑦 ∗ ‖ = 0 and l i m 𝑛 → ∞ ‖ 𝐵 𝑢 𝑛 − 𝐵 𝑥 ∗ ‖ = 0 . From the fact that the 𝐵 is 𝛽 -inverse strongly monotone operator, it follows that ‖ ‖ 𝐵 𝑥 𝑛 − 𝐵 𝑥 ∗ ‖ ‖ ≤ ‖ ‖ 𝐵 𝑥 𝑛 − 𝐵 𝑢 𝑛 ‖ ‖ + ‖ ‖ 𝐵 𝑢 𝑛 − 𝐵 𝑥 ∗ ‖ ‖ ≤ 1 𝛽 ‖ ‖ 𝑥 𝑛 − 𝑢 𝑛 ‖ ‖ + ‖ ‖ 𝐵 𝑢 𝑛 − 𝐵 𝑥 ∗ ‖ ‖ . ( 3 . 1 9 ) Applying Step 3, we have l i m 𝑛 → ∞ ‖ 𝐵 𝑥 𝑛 − 𝐵 𝑥 ∗ ‖ = 0 .Step 5. l i m 𝑛 → ∞ ‖ 𝑥 𝑛 − 𝑇 𝑖 𝑥 𝑛 ‖ = 0 , 𝑓 𝑜 𝑟 𝑎 𝑙 𝑙 𝑖 = 1 , 2 , … .
Noting that 𝑃 𝐶 is firmly nonexpansive, we have ‖ ‖ 𝑧 𝑛 − 𝑦 ∗ ‖ ‖ 2 = ‖ ‖ 𝑃 𝐶 𝑢 𝑛 − 𝜇 𝐵 𝑢 𝑛 − 𝑃 𝐶 𝑥 ∗ − 𝜇 𝐵 𝑥 ∗ ‖ ‖ 2 ≤ 𝑢 𝑛 − 𝜇 𝐵 𝑢 𝑛 − 𝑥 ∗ − 𝜇 𝐵 𝑥 ∗ , 𝑧 𝑛 − 𝑦 ∗ = 1 2 ‖ ‖ 𝑢 𝑛 − 𝑥 ∗ − 𝜇 𝐵 𝑢 𝑛 − 𝐵 𝑥 ∗ ‖ ‖ 2 + ‖ ‖ 𝑧 𝑛 − 𝑦 ∗ ‖ ‖ 2 − ‖ ‖ 𝑢 𝑛 − 𝑧 𝑛 − 𝜇 𝐵 𝑢 𝑛 − 𝐵 𝑥 ∗ − 𝑥 ∗ − 𝑦 ∗ ‖ ‖ 2 ≤ 1 2 ‖ ‖ 𝑢 𝑛 − 𝑥 ∗ ‖ ‖ 2 + ‖ ‖ 𝑧 𝑛 − 𝑦 ∗ ‖ ‖ 2 − ‖ ‖ 𝑢 𝑛 − 𝑧 𝑛 − 𝑥 ∗ − 𝑦 ∗ ‖ ‖ 2 𝑢 + 2 𝜇 𝑛 − 𝑧 𝑛 − 𝑥 ∗ − 𝑦 ∗ , 𝐵 𝑢 𝑛 − 𝐵 𝑥 ∗ − 𝜇 2 ‖ ‖ 𝐵 𝑢 𝑛 − 𝐵 𝑥 ∗ ‖ ‖ 2 ≤ 1 2 ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 + ‖ ‖ 𝑧 𝑛 − 𝑦 ∗ ‖ ‖ 2 − ‖ ‖ 𝑢 𝑛 − 𝑧 𝑛 − 𝑥 ∗ − 𝑦 ∗ ‖ ‖ 2 𝑢 + 2 𝜇 𝑛 − 𝑧 𝑛 − 𝑥 ∗ − 𝑦 ∗ , 𝐵 𝑢 𝑛 − 𝐵 𝑥 ∗ , ‖ ‖ 𝑣 𝑛 − 𝑥 ∗ ‖ ‖ 2 = ‖ ‖ 𝑃 𝐶 𝑧 𝑛 − 𝜆 𝐴 𝑧 𝑛 − 𝑃 𝐶 𝑦 ∗ − 𝜆 𝐴 𝑦 ∗ ‖ ‖ 2 ≤ 𝑧 𝑛 − 𝜆 𝐴 𝑧 𝑛 − 𝑦 ∗ − 𝜆 𝐴 𝑦 ∗ , 𝑣 𝑛 − 𝑥 ∗ = 1 2 ‖ ‖ 𝑧 𝑛 − 𝑦 ∗ − 𝜆 𝐴 𝑧 𝑛 − 𝐴 𝑦 ∗ ‖ ‖ 2 + ‖ ‖ 𝑣 𝑛 − 𝑥 ∗ ‖ ‖ 2 − ‖ ‖ 𝑧 𝑛 − 𝑣 𝑛 − 𝜆 𝐴 𝑧 𝑛 − 𝐴 𝑦 ∗ − 𝑦 ∗ − 𝑥 ∗ ‖ ‖ 2 ≤ 1 2 ‖ ‖ 𝑧 𝑛 − 𝑦 ∗ ‖ ‖ 2 + ‖ ‖ 𝑣 𝑛 − 𝑥 ∗ ‖ ‖ 2 − ‖ ‖ 𝑧 𝑛 − 𝑣 𝑛 − 𝑦 ∗ − 𝑥 ∗ ‖ ‖ 2 𝑧 + 2 𝜆 𝑛 − 𝑣 𝑛 − 𝑦 ∗ − 𝑥 ∗ , 𝐴 𝑧 𝑛 − 𝐴 𝑦 ∗ − 𝜆 2 ‖ ‖ 𝐴 𝑧 𝑛 − 𝐴 𝑦 ∗ ‖ ‖ 2 ≤ 1 2 ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 + ‖ ‖ 𝑣 𝑛 − 𝑥 ∗ ‖ ‖ 2 − ‖ ‖ 𝑧 𝑛 − 𝑣 𝑛 − 𝑦 ∗ − 𝑥 ∗ ‖ ‖ 2 𝑧 + 2 𝜆 𝑛 − 𝑣 𝑛 − 𝑦 ∗ − 𝑥 ∗ 𝐴 𝑧 𝑛 − 𝐴 𝑦 ∗ . ( 3 . 2 0 ) It follows that ‖ ‖ 𝑧 𝑛 − 𝑦 ∗ ‖ ‖ 2 ≤ ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 − ‖ ‖ 𝑢 𝑛 − 𝑧 𝑛 − 𝑥 ∗ − 𝑦 ∗ ‖ ‖ 2 𝑢 + 2 𝜇 𝑛 − 𝑧 𝑛 − 𝑥 ∗ − 𝑦 ∗ , 𝐵 𝑢 𝑛 − 𝐵 𝑥 ∗ , ‖ ‖ 𝑣 ( 3 . 2 1 ) 𝑛 − 𝑥 ∗ ‖ ‖ 2 ≤ ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 − ‖ ‖ 𝑧 𝑛 − 𝑣 𝑛 − 𝑦 ∗ − 𝑥 ∗ ‖ ‖ 2 𝑧 + 2 𝜆 𝑛 − 𝑣 𝑛 − 𝑦 ∗ − 𝑥 ∗ , 𝐴 𝑧 𝑛 − 𝐴 𝑦 ∗ . ( 3 . 2 2 ) By (3.3), (3.14)-(3.15), and (3.22), we have ‖ ‖ 𝑥 𝑛 + 1 − 𝑥 ∗ ‖ ‖ 2 ≤ 1 − 1 − 𝛽 𝑛 𝜎 𝑛 ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 + 1 − 𝛽 𝑛 𝜎 𝑛 ‖ ‖ 𝑦 𝑛 − 𝑥 ∗ ‖ ‖ 2 ≤ 1 − 1 − 𝛽 𝑛 𝜎 𝑛 ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 + 1 − 𝛽 𝑛 𝜎 𝑛 𝛼 𝑛 ‖ ‖ 𝑓 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 + ‖ ‖ 𝑣 𝑛 − 𝑥 ∗ ‖ ‖ 2 ≤ 1 − 1 − 𝛽 𝑛 𝜎 𝑛 ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 + 1 − 𝛽 𝑛 𝜎 𝑛 𝛼 𝑛 ‖ ‖ 𝑓 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 + ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 − ‖ ‖ 𝑧 𝑛 − 𝑣 𝑛 − 𝑦 ∗ − 𝑥 ∗ ‖ ‖ 2 𝑧 + 2 𝜆 𝑛 − 𝑣 𝑛 − 𝑦 ∗ − 𝑥 ∗ 𝐴 𝑧 𝑛 − 𝐴 𝑦 ∗ = ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 + 𝛼 𝑛 ‖ ‖ 𝑓 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 + 1 − 𝛽 𝑛 1 − 𝜎 𝑛 × − ‖ ‖ 𝑧 𝑛 − 𝑣 𝑛 − 𝑦 ∗ − 𝑥 ∗ ‖ ‖ 2 𝑧 + 2 𝜆 𝑛 − 𝑣 𝑛 − 𝑦 ∗ − 𝑥 ∗ , 𝐴 𝑧 𝑛 − 𝐴 𝑦 ∗ . ( 3 . 2 3 ) It follows that 1 − 𝛽 𝑛 𝜎 𝑛 ‖ ‖ 𝑣 𝑛 𝑥 − 𝑧 − ∗ − 𝑦 ∗ ‖ ‖ 2 ≤ ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ + ‖ ‖ 𝑥 𝑛 + 1 − 𝑥 ∗ ‖ ‖ ‖ ‖ 𝑥 𝑛 − 𝑥 𝑛 + 1 ‖ ‖ + 𝛼 𝑛 ‖ ‖ 𝑓 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 + 2 𝜆 1 − 𝛽 𝑛 𝜎 𝑛 𝑧 𝑛 − 𝑣 𝑛 − 𝑦 ∗ − 𝑥 ∗ , 𝐴 𝑧 𝑛 − 𝐴 𝑦 ∗ . ( 3 . 2 4 ) From conditions (ii)–(iv), Steps 2 and 4, we get the following: l i m 𝑛 ⟶ ∞ ‖ ‖ 𝑣 𝑛 − 𝑧 𝑛 − 𝑥 ∗ − 𝑦 ∗ ‖ ‖ = 0 . ( 3 . 2 5 ) On the other hand, from (3.14)–(3.21), we have ‖ ‖ 𝑥 𝑛 + 1 − 𝑥 ∗ ‖ ‖ 2 ≤ 𝛽 𝑛 ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 + 1 − 𝛽 𝑛 𝜎 𝑛 𝛼 𝑛 ‖ ‖ 𝑓 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 + ‖ ‖ 𝑧 𝑛 − 𝑦 ∗ ‖ ‖ 2 + 1 − 𝛽 𝑛 1 − 𝜎 𝑛 ‖ ‖ 𝑧 𝑛 − 𝑦 ∗ ‖ ‖ 2 ≤ 𝛽 𝑛 ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 + 𝛼 𝑛 ‖ ‖ 𝑓 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 + 1 − 𝛽 𝑛 ‖ ‖ 𝑧 𝑛 − 𝑦 ∗ ‖ ‖ 2 ≤ 𝛽 𝑛 ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 + 𝛼 𝑛 ‖ ‖ 𝑓 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 + 1 − 𝛽 𝑛 × ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 − ‖ ‖ 𝑢 𝑛 − 𝑧 𝑛 − 𝑥 ∗ − 𝑦 ∗ ‖ ‖ 2 𝑢 + 2 𝜇 𝑛 − 𝑧 𝑛 − 𝑥 ∗ − 𝑦 ∗ , 𝐵 𝑢 𝑛 − 𝐵 𝑥 ∗ = ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 + 𝛼 𝑛 ‖ ‖ 𝑓 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 + 1 − 𝛽 𝑛 − ‖ ‖ 𝑢 𝑛 − 𝑧 𝑛 − 𝑥 ∗ − 𝑦 ∗ ‖ ‖ 2 𝑢 + 2 𝜇 𝑛 − 𝑧 𝑛 − 𝑥 ∗ − 𝑦 ∗ , 𝐵 𝑢 𝑛 − 𝐵 𝑥 ∗ , ( 3 . 2 6 ) which implies that ‖ ‖ 1 − 𝛽 𝑛 𝑢 𝑛 − 𝑧 𝑛 − 𝑥 ∗ − 𝑦 ∗ ‖ ‖ 2 ≤ ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ + ‖ ‖ 𝑥 𝑛 + 1 − 𝑥 ∗ ‖ ‖ ‖ ‖ 𝑥 𝑛 − 𝑥 𝑛 + 1 ‖ ‖ + 𝛼 𝑛 ‖ ‖ 𝑓 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 + 2 𝜇 1 − 𝛽 𝑛 × 𝑢 𝑛 − 𝑧 𝑛 − 𝑥 ∗ − 𝑦 ∗ , 𝐵 𝑢 𝑛 − 𝐵 𝑥 ∗ . ( 3 . 2 7 ) From (ii)-(iii), Steps 2 and 4, we get the following: l i m 𝑛 ⟶ ∞ ‖ ‖ 𝑢 𝑛 − 𝑧 𝑛 − 𝑥 ∗ − 𝑦 ∗ ‖ ‖ = 0 . ( 3 . 2 8 ) Combining (3.25) and (3.28), we get the following: l i m 𝑛 ⟶ ∞ ‖ ‖ 𝑢 𝑛 − 𝑣 𝑛 ‖ ‖ = 0 . ( 3 . 2 9 ) Using Step 3, we obtain the following: l i m 𝑛 ⟶ ∞ ‖ ‖ 𝑥 𝑛 − 𝑣 𝑛 ‖ ‖ = 0 . ( 3 . 3 0 ) This together with ‖ 𝑦 𝑛 − 𝑣 𝑛 ‖ = 𝛼 𝑛 ‖ 𝑓 ( 𝑥 𝑛 ) − 𝑣 𝑛 ‖ → 0 implies that l i m 𝑛 ⟶ ∞ ‖ ‖ 𝑥 𝑛 − 𝑦 𝑛 ‖ ‖ = 0 . ( 3 . 3 1 ) For any 𝑥 ∗ ∈ Ω , we have from (3.1) that 𝜎 𝑛 𝑛 𝑖 = 1 𝛽 𝑖 − 1 − 𝛽 𝑖 ⟨ 𝑥 𝑛 − 𝑇 𝑖 𝑥 𝑛 , 𝑥 𝑛 − 𝑥 ∗ ⟩ + ⟨ 𝑇 𝑖 𝑥 𝑛 − 𝑇 𝑖 𝑦 𝑛 , 𝑥 𝑛 − 𝑥 ∗ ⟩ = 𝜎 𝑛 𝑛 𝑖 = 1 𝛽 𝑖 − 1 − 𝛽 𝑖 ⟨ 𝑥 𝑛 − 𝑇 𝑖 𝑦 𝑛 , 𝑥 𝑛 − 𝑥 ∗ ⟩ = 𝑥 𝑛 − 𝑥 𝑛 + 1 , 𝑥 𝑛 − 𝑥 ∗ + 1 − 𝛽 𝑛 1 − 𝜎 𝑛 ⟨ 𝑣 𝑛 − 𝑥 𝑛 , 𝑥 𝑛 − 𝑥 ∗ ⟩ . ( 3 . 3 2 ) Since each 𝑇 𝑖 is nonexpansive, from (2.2), we have ‖ ‖ 𝑇 𝑖 𝑥 𝑛 − 𝑥 𝑛 ‖ ‖ 2 ≤ 2 ⟨ 𝑥 𝑛 − 𝑇 𝑖 𝑥 𝑛 , 𝑥 𝑛 − 𝑥 ∗ ⟩ . ( 3 . 3 3 ) Hence, combining this inequality with (3.32), we get the following: 1 2 𝜎 𝑛 𝑛 𝑖 = 1 𝛽 𝑖 − 1 − 𝛽 𝑖 ‖ ‖ 𝑇 𝑖 𝑥 𝑛 − 𝑥 𝑛 ‖ ‖ 2 ≤ − 𝜎 𝑛 𝑛 𝑖 = 1 𝛽 𝑖 − 1 − 𝛽 𝑖 ⟨ 𝑇 𝑖 𝑥 𝑛 − 𝑇 𝑖 𝑦 𝑛 , 𝑥 𝑛 − 𝑥 ∗ ⟩ + 𝑥 𝑛 − 𝑥 𝑛 + 1 , 𝑥 𝑛 − 𝑥 ∗ + 1 − 𝛽 𝑛 1 − 𝜎 𝑛 × ⟨ 𝑣 𝑛 − 𝑥 𝑛 , 𝑥 𝑛 − 𝑥 ∗ ⟩ , ( 3 . 3 4 ) that is (noting that 𝛽 𝑛 is strictly decreasing), ‖ ‖ 𝑇 𝑖 𝑥 𝑛 − 𝑥 𝑛 ‖ ‖ 2 ≤ 2 1 − 𝛽 𝑛 𝛽 𝑖 − 1 − 𝛽 𝑖 ‖ ‖ 𝑥 𝑛 − 𝑦 𝑛 ‖ ‖ ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ + 2 𝜎 𝑛 𝛽 𝑖 − 1 − 𝛽 𝑖 ‖ ‖ 𝑥 𝑛 − 𝑥 𝑛 + 1 ‖ ‖ ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ + 2 1 − 𝛽 𝑛 1 − 𝜎 𝑛 𝜎 𝑛 𝛽 𝑖 − 1 − 𝛽 𝑖 ‖ ‖ 𝑣 𝑛 − 𝑥 𝑛 ‖ ‖ ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ . ( 3 . 3 5 ) Now from (3.30)-(3.31) and Step 2, we conclude that l i m 𝑛 ⟶ ∞ ‖ ‖ 𝑇 𝑖 𝑥 𝑛 − 𝑥 𝑛 ‖ ‖ = 0 , ∀ 𝑖 = 1 , 2 , … , ( 3 . 3 6 ) which completes the proof.Step 6. l i m s u p 𝑛 → ∞ ⟨ 𝑓 ( 𝑥 ∗ ) − 𝑥 ∗ , 𝑥 𝑛 − 𝑥 ∗ ⟩ ≤ 0 , where 𝑥 ∗ = 𝑃 Ω ⋅ 𝑓 ( 𝑥 ∗ ) .
As { 𝑥 𝑛 } is bounded, there exists a subsequence { 𝑥 𝑛 𝑖 } of { 𝑥 𝑛 } such that { 𝑥 𝑛 𝑖 } → ̃ 𝑥 weakly. First, it is clear from Step 5 and Lemma 2.3 that ⋂ ̃ 𝑥 ∈ ∞ 𝑛 = 1 𝐹 ( 𝑇 𝑛 ) . Next, we prove that ̃ 𝑥 ∈ Γ . From (iii), Step 3 and (3.31), we note that ‖ ‖ 𝑦 𝑛 𝑦 − 𝐺 𝑛 ‖ ‖ ≤ 𝛼 𝑛 ‖ ‖ 𝑓 𝑥 𝑛 𝑦 − 𝐺 𝑛 ‖ ‖ + 1 − 𝛼 𝑛 ‖ ‖ 𝑃 𝐶 𝑃 C 𝑢 𝑛 − 𝜇 𝐵 𝑢 𝑛 − 𝜆 𝐴 𝑃 𝐶 𝑢 𝑛 − 𝜇 𝐵 𝑢 𝑛 𝑦 − 𝐺 𝑛 ‖ ‖ = 𝛼 𝑛 ‖ ‖ 𝑓 𝑥 𝑛 𝑦 − 𝐺 𝑛 ‖ ‖ + 1 − 𝛼 𝑛 ‖ ‖ 𝐺 𝑢 𝑛 𝑦 − 𝐺 𝑛 ‖ ‖ ≤ 𝛼 𝑛 ‖ ‖ 𝑓 𝑥 𝑛 𝑦 − 𝐺 𝑛 ‖ ‖ + 1 − 𝛼 𝑛 ‖ ‖ 𝑢 𝑛 − 𝑦 𝑛 ‖ ‖ ⟶ 0 . ( 3 . 3 7 ) According to (3.31) and Lemma 2.3, we obtain that ̃ 𝑥 ∈ Γ . Next, we show that ̃ 𝑥 ∈ E P ( 𝐹 ) . Indeed, by 𝑢 𝑛 = 𝑇 𝑟 𝑛 𝑥 𝑛 , we have 𝐹 𝑢 𝑛 + 1 , 𝑦 𝑟 𝑛 ⟨ 𝑦 − 𝑢 𝑛 , 𝑢 𝑛 − 𝑥 𝑛 ⟩ ≥ 0 , ∀ 𝑦 ∈ 𝐶 . ( 3 . 3 8 ) From (A2), we have also 1 𝑟 𝑛 ⟨ 𝑦 − 𝑢 𝑛 , 𝑢 𝑛 − 𝑥 𝑛 𝑢 ⟩ ≥ − 𝐹 𝑛 , 𝑦 ≥ 𝐹 𝑦 , 𝑢 𝑛 , ( 3 . 3 9 ) and hence, 𝑦 − 𝑢 𝑛 𝑘 , 𝑢 𝑛 𝑘 − 𝑥 𝑛 𝑘 𝑟 𝑛 𝑘 ≥ 𝐹 𝑦 , 𝑢 𝑛 𝑘 . ( 3 . 4 0 ) According to 𝑟 𝑛 > 𝑟 > 0 and 𝑢 𝑛 − 𝑥 𝑛 → 0 , we conclude that ( 𝑢 𝑛 𝑖 − 𝑥 𝑛 𝑖 ) / 𝑟 𝑛 𝑖 → 0 and 𝑢 𝑛 𝑘 ⇀ ̃ 𝑥 . From (A4), we obtain the following: 𝐹 ( 𝑦 , ̃ 𝑥 ) ≤ 0 , ∀ 𝑦 ∈ 𝐶 . ( 3 . 4 1 ) For 𝑡 with 0 < 𝑡 ≤ 1 and 𝑦 ∈ 𝐶 , let 𝑦 𝑡 = 𝑡 𝑦 + ( 1 − 𝑡 ) ̃ 𝑥 . Since 𝑦 ∈ 𝐶 and ̃ 𝑥 ∈ 𝐶 (due to 𝑢 𝑛 𝑖 → ̃ 𝑥 as 𝑖 → ∞ ), we have 𝑦 𝑡 ∈ 𝐶 and hence 𝐹 ( 𝑦 𝑡 , ̃ 𝑥 ) ≤ 0 . So, from (A1) and (A4), we have 𝑦 𝑡 𝐹 𝑡 𝑦 , 𝑦 ≥ 𝑡 𝐹 𝑡 + 𝑦 , 𝑦 ( 1 − 𝑡 ) 𝐹 𝑡 𝑦 , ̃ 𝑥 ≥ 𝑡 𝐹 𝑡 , 𝑦 𝑡 = 0 ( 3 . 4 2 ) and 𝐹 ( 𝑦 𝑡 , 𝑦 ) ≥ 0 , for all 𝑡 ∈ ( 0 , 1 ) and 𝑦 ∈ 𝐶 . From (A3), we obtain the following: 𝐹 ( ̃ 𝑥 , 𝑦 ) ≥ 0 ∀ 𝑦 ∈ 𝐶 , ( 3 . 4 3 ) and hence ̃ 𝑥 ∈ E P ( 𝐹 ) . Therefore, we obtain that ̃ 𝑥 ∈ Ω . Hence, it follows from Lemma 2.2 that l i m s u p 𝑛 → ∞ 𝑓 𝑥 ∗ − 𝑥 ∗ , 𝑥 𝑛 − 𝑥 ∗ = l i m 𝑖 → ∞ 𝑓 𝑥 ∗ − 𝑥 ∗ , 𝑥 𝑛 𝑖 − 𝑥 ∗ = 𝑓 𝑥 ∗ − 𝑥 ∗ , ̃ 𝑥 − 𝑥 ∗ ≤ 0 . ( 3 . 4 4 ) Step 7. l i m 𝑛 → ∞ 𝑥 𝑛 = 𝑥 ∗ .
From (3.1), (3.3)-(3.4), and the convexity of ‖ ⋅ ‖ , we have ‖ ‖ 𝑦 𝑛 − 𝑥 ∗ ‖ ‖ 2 = ‖ ‖ 𝛼 𝑛 𝑓 𝑥 𝑛 + 1 − 𝛼 𝑛 𝑣 𝑛 − 𝑥 ∗ ‖ ‖ 2 ≤ 1 − 𝛼 𝑛 ‖ ‖ 𝑣 𝑛 − 𝑥 ∗ ‖ ‖ 2 + 2 𝛼 𝑛 𝑓 𝑥 𝑛 − 𝑥 ∗ , 𝑦 𝑛 − 𝑥 ∗ ≤ 1 − 𝛼 𝑛 ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 + 2 𝛼 𝑛 𝑓 𝑥 𝑛 − 𝑥 ∗ , 𝑦 𝑛 − 𝑥 ∗ . ( 3 . 4 5 ) We can also get ‖ ‖ 𝑥 𝑛 + 1 − 𝑥 ∗ ‖ ‖ 2 = ‖ ‖ ‖ ‖ 𝛽 𝑛 𝑥 𝑛 − 𝑥 ∗ + 𝜎 𝑛 𝑛 𝑖 = 1 𝛽 𝑖 − 1 − 𝛽 𝑖 𝑇 𝑖 𝑦 𝑛 − 𝑥 ∗ + 1 − 𝛽 𝑛 1 − 𝜎 𝑛 𝑦 𝑛 − 𝑥 ∗ + 1 − 𝛽 𝑛 1 − 𝜎 𝑛 𝛼 𝑛 𝑣 𝑛 𝑥 − 𝑓 𝑛 ‖ ‖ ‖ 2 ≤ ‖ ‖ ‖ ‖ 𝛽 𝑛 𝑥 𝑛 − 𝑥 ∗ + 𝜎 𝑛 𝑛 𝑖 = 1 𝛽 𝑖 − 1 − 𝛽 𝑖 𝑇 𝑖 𝑦 𝑛 − 𝑥 ∗ + 1 − 𝛽 𝑛 1 − 𝜎 𝑛 𝑦 𝑛 − 𝑥 ∗ ‖ ‖ ‖ ‖ 2 + 2 1 − 𝛽 𝑛 1 − 𝜎 𝑛 𝛼 𝑛 𝑣 𝑛 𝑥 − 𝑓 𝑛 , 𝑥 𝑛 + 1 − 𝑥 ∗ ≤ 𝛽 𝑛 ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 + 1 − 𝛽 𝑛 ‖ ‖ 𝑦 𝑛 − 𝑥 ∗ ‖ ‖ 2 + 2 1 − 𝛽 𝑛 1 − 𝜎 𝑛 𝛼 𝑛 𝑣 𝑛 − 𝑥 ∗ , 𝑥 𝑛 + 1 − 𝑥 ∗ + 2 1 − 𝛽 𝑛 1 − 𝜎 𝑛 𝛼 𝑛 𝑥 ∗ 𝑥 − 𝑓 𝑛 , 𝑥 𝑛 + 1 − 𝑥 ∗ . ( 3 . 4 6 ) By (3.45), we have ‖ ‖ 𝑥 𝑛 + 1 − 𝑥 ∗ ‖ ‖ 2 ≤ 𝛽 𝑛 ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 + 1 − 𝛽 𝑛 1 − 𝛼 𝑛 ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 + 2 𝛼 𝑛 𝑓 𝑥 𝑛 − 𝑥 ∗ , 𝑦 𝑛 − 𝑥 ∗ + 2 1 − 𝛽 𝑛 1 − 𝜎 𝑛 𝛼 𝑛 ‖ ‖ 𝑣 𝑛 − 𝑥 ∗ ‖ ‖ ‖ ‖ 𝑥 𝑛 + 1 − 𝑥 ∗ ‖ ‖ + 2 1 − 𝛽 𝑛 𝛼 𝑛 1 − 𝜎 𝑛 𝑥 ∗ 𝑥 − 𝑓 𝑛 , 𝑥 𝑛 + 1 − 𝑥 ∗ ≤ 1 − 1 − 𝛽 𝑛 𝛼 𝑛 ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 + 2 1 − 𝛽 𝑛 1 − 𝜎 𝑛 𝛼 𝑛 𝑓 𝑥 𝑛 − 𝑥 ∗ , 𝑦 𝑛 − 𝑥 𝑛 + 1 + 2 1 − 𝛽 𝑛 1 − 𝜎 𝑛 𝛼 𝑛 ‖ ‖ 𝑧 𝑛 − 𝑦 ∗ ‖ ‖ ‖ ‖ 𝑥 𝑛 + 1 − 𝑥 ∗ ‖ ‖ + 2 1 − 𝛽 𝑛 𝛼 𝑛 𝜎 𝑛 𝑓 𝑥 𝑛 − 𝑥 ∗ , 𝑦 𝑛 − 𝑥 ∗ ≤ 1 − 1 − 𝛽 𝑛 𝛼 𝑛 ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ 2 + 2 1 − 𝛽 𝑛 1 − 𝜎 𝑛 𝛼 𝑛 ‖ ‖ 𝑓 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ ‖ ‖ 𝑥 𝑛 + 1 − 𝑦 𝑛 ‖ ‖ + 2 1 − 𝛽 𝑛 1 − 𝜎 𝑛 𝛼 𝑛 ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ ‖ ‖ 𝑥 𝑛 + 1 − 𝑥 ∗ ‖ ‖ + 2 1 − 𝛽 𝑛 𝛼 𝑛 𝜎 𝑛 𝑓 𝑥 𝑛 − 𝑥 ∗ , 𝑦 𝑛 − 𝑥 𝑛 + 2 1 − 𝛽