Journal of Inequalities and Applications
Volume 2010 (2010), Article ID 687374, 14 pages
doi:10.1155/2010/687374
Research Article

An Iterative Scheme with a Countable Family of Nonexpansive Mappings for Variational Inequality Problems in Hilbert Spaces

Yeol Je Cho1 and Shenghua Wang2,3

1Department of Mathematics Education and the RINS, Gyeongsang National University, Chinju 660-701, South Korea
2School of Applied Mathematics and Physics, North China Electric Power University, Baoding 071003, China
3Department of Mathematics, Gyeongsang National University, Chinju 660-701, South Korea

Received 13 December 2009; Accepted 8 February 2010

Academic Editor: Jong Kim

Copyright © 2010 Yeol Je Cho and Shenghua Wang. 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

We introduce a new iterative scheme with a countable family of nonexpansive mappings for the variational inequality problems in Hilbert spaces and prove some strong convergence theorems for the proposed schemes.

1. Introduction

Let 𝐻 be a Hilbert space and 𝐶 be a nonempty closed convex subset of 𝐻 . Let 𝐹 ∶ 𝐻 → 𝐻 be a nonlinear mapping. The classical variational inequality problem (for short, V I ( 𝐹 , 𝐶 ) ) is to find a point 𝑥 ∈ 𝐶 such that

𝐹 𝑥 ∗ , 𝑥 − 𝑥 ∗ ≥ 0 , ∀ 𝑥 ∗ ∈ 𝐶 . ( 1 . 1 )

This variational inequality was initially studied by Kinderlehrer and Stampacchia [1]. Since then, many authors have introduced and studied many kinds of the variational inequality problems (inclusions) and applied them to many fields.

It is well known that, if 𝐹 is a strongly monotone and Lipschitzian mapping on 𝐶 , then the V I ( 𝐹 , 𝐶 ) has a unique solution (see [2]).

Let 𝑇 ∶ 𝐻 → 𝐻 be a mapping. Recall that a mapping 𝑇 ∶ 𝐻 → 𝐻 is nonexpansive if

‖ 𝑇 𝑥 − 𝑇 𝑦 ‖ ≤ ‖ 𝑥 − 𝑦 ‖ , ∀ 𝑥 , 𝑦 ∈ 𝐻 . ( 1 . 2 ) The set of fixed points of 𝑇 is denoted by 𝐹 ( 𝑇 ) . Recently, the iterative methods for nonexpansive mappings and some kinds of nonlinear mappings have been applied to solve the convex minimization problems (see [3–7]).

A typical problem is to minimize a quadratic function over the set of the fixed points of a nonexpansive mapping on 𝐻 :

m i n 𝑥 ∈ 𝐶 1 2 ⟨ 𝐴 𝑥 , 𝑥 ⟩ − ⟨ 𝑥 , 𝑏 ⟩ , ( 1 . 3 ) where 𝐶 is the fixed point set of a nonexpansive mapping 𝑇 on H, 𝑏 is a given point in 𝐻 and 𝐴 is a strongly positive operator, that is, there is a constant 𝛾 > 0 such that

⟨ 𝐴 𝑥 , 𝑥 ⟩ ≥ 𝛾 ‖ 𝑥 ‖ 2 , ∀ 𝑥 ∈ 𝐻 . ( 1 . 4 )

Recently, for solving the variational inequality on 𝐴 , Marino and Xu [8] introduced the following general iterative scheme:

𝑥 𝑛 + 1 = 𝐼 − 𝛼 𝑛 𝐴 𝑇 𝑥 𝑛 + 𝛼 𝑛 𝑥 𝛾 𝑓 𝑛 , ∀ 𝑛 ≥ 0 , ( 1 . 5 ) where 𝐴 is a strongly positive linear bounded operator on 𝐻 , 𝑓 is a contraction on 𝐻 and { 𝛼 𝑛 } ⊂ ( 0 , 1 ) .

More precisely, they gave the following result.

Theorem 1 MX (see [8, Theorem 3.4]). Let { 𝑥 𝑛 } be generated by algorithm (1.5) with the sequence { 𝛼 𝑛 } satisfying the following conditions: (C1) l i m 𝑛 → ∞ 𝛼 𝑛 = 0 ,(C2) ∑ ∞ 𝑛 = 0 𝛼 𝑛 = ∞ ,(C3) either ∑ ∞ 𝑛 = 0 | 𝛼 𝑛 + 1 − 𝛼 𝑛 | < ∞ or l i m 𝑛 → ∞ 𝛼 𝑛 + 1 / 𝛼 𝑛 = 1 .Then the scheme { 𝑥 𝑛 } defined by (1.5) converges strongly to an element 𝑥 ∗ ∈ 𝐶 = 𝐹 ( 𝑇 ) which is the unique solution of the variational inequality ( for short, V I ( 𝐴 − 𝛾 𝑓 , 𝐶 ) ) : ⟨ ( 𝐴 − 𝛾 𝑓 ) 𝑥 ∗ , 𝑥 − 𝑥 ∗ ⟩ ≥ 0 , ∀ 𝑥 ∈ 𝐶 . ( 1 . 6 )

Let 𝑓 ∶ 𝐻 → 𝐻 be a contraction with coefficient 0 < 𝛼 < 1 and let 𝐴 , 𝐵 ∶ 𝐻 → 𝐻 be two strongly positive linear bounded operators with coefficients 𝛾 ∈ ( 0 , 1 ) and 𝛽 > 0 , respectively.

Motivated and inspired by the iterative sheme (1.5), Ceng et al. [9] introduced the following so-called hybrid viscosity-like approximation algorithms with variable parameters for nonexpansive mappings in Hilbert spaces.

Theorem 1 CGYone (see [9, Theorem 3.1]). Let 0 < 𝛾 𝛼 < 𝛽 and 𝛾 ∈ ( 0 , 1 ) . Let { 𝜆 𝑛 } be a sequence in ( 0 , 1 ) and { 𝜇 𝑛 } be a sequence in ( 0 , m i n { 1 , ‖ 𝐵 ‖ − 1 } ) . Starting with an arbitrary initial guess 𝑥 0 ∈ 𝐻 , generate a sequence { 𝑥 𝑛 } by the following iterative scheme: 𝑥 𝑛 + 1 = 𝐼 − 𝜆 𝑛 + 1 𝐴 𝑇 𝑥 𝑛 + 𝜆 𝑛 + 1 𝑇 𝑥 𝑛 − 𝜇 𝑛 + 1 𝐵 𝑇 𝑥 𝑛 𝑥 − 𝛾 𝑓 𝑛 , ∀ 𝑛 ≥ 0 . ( 1 . 7 ) Assume that (i) l i m 𝑛 → ∞ 𝜆 𝑛 = 0 ,(ii) ∑ ∞ 𝑛 = 1 𝜆 𝑛 = ∞ ,(iii)either ∑ ∞ 𝑛 = 1 | 𝜆 𝑛 + 1 − 𝜆 𝑛 | < ∞ or l i m 𝑛 → ∞ 𝜆 𝑛 / 𝜆 𝑛 + 1 = 1 ,(iv) ( 1 − 𝛾 ) / ( 𝛽 − 𝛾 𝛼 ) < l i m 𝑛 → ∞ 𝜇 𝑛 = 𝜇 < ( 2 − 𝛾 ) / ( 𝛽 − 𝛾 𝛼 ) .Then the scheme { 𝑥 𝑛 } defined by (1.7) converges strongly to an element 𝑥 ∗ ∈ 𝐶 = 𝐹 ( 𝑇 ) which is the unique solution of the variational inequality ( for short, V I ( 𝐴 − 𝐼 + 𝜇 ( 𝐵 − 𝛾 𝑓 ) , 𝐶 ) ) : ⟨ [ ] 𝑥 𝐴 − 𝐼 + 𝜇 ( 𝐵 − 𝛾 𝑓 ) ∗ , 𝑥 − 𝑥 ∗ ⟩ ≥ 0 , ∀ 𝑥 ∈ 𝐶 . ( 1 . 8 )

Theorem 1 CGYtwo (see [9, Theorem 3.2]). Let 0 < 𝛾 𝛼 < 𝛽 and 𝛾 ∈ ( 0 , 1 ) . Let { 𝜆 𝑛 } be a sequence in ( 0 , 1 ) and { 𝜇 𝑛 } be a sequence in ( 0 , m i n { 1 , ‖ 𝐵 ‖ − 1 } ) . Starting with an arbitrary initial guess 𝑥 0 ∈ 𝐻 , generate a sequence { 𝑥 𝑛 } by the following iterative scheme: 𝑥 𝑛 + 1 = 𝐼 − 𝜆 𝑛 + 1 𝐴 𝑇 [ 𝑛 + 1 ] 𝑥 𝑛 + 𝜆 𝑛 + 1 𝑇 [ 𝑛 + 1 ] 𝑥 𝑛 − 𝜇 𝑛 + 1 𝐵 𝑇 [ 𝑛 + 1 ] 𝑥 𝑛 𝑥 − 𝛾 𝑓 𝑛 , ∀ 𝑛 ≥ 0 . ( 1 . 9 ) Assume that (i) l i m 𝑛 → ∞ 𝜆 𝑛 = 0 ,(ii) ∑ ∞ 𝑛 = 1 𝜆 𝑛 = ∞ ,(iii)either ∑ ∞ 𝑛 = 1 | 𝜆 𝑛 + 𝑁 − 𝜆 𝑛 | < ∞ or l i m 𝑛 → ∞ 𝜆 𝑛 / 𝜆 𝑛 + 𝑁 = 1 ,(iv) ( 1 − 𝛾 ) / ( 𝛽 − 𝛾 𝛼 ) < l i m 𝑛 → ∞ 𝜇 𝑛 = 𝜇 < ( 2 − 𝛾 ) / ( 𝛽 − 𝛾 𝛼 ) .In addition, assume that 𝐶 = 𝑁 𝑖 = 1 𝐹 𝑇 𝑖 𝑇 = 𝐹 1 𝑇 2 ⋯ 𝑇 𝑁 𝑇 = 𝐹 𝑁 𝑇 1 ⋯ 𝑇 2 𝑇 = ⋯ = 𝐹 2 𝑇 3 ⋯ 𝑇 𝑁 𝑇 1 . ( 1 . 1 0 ) Then the scheme { 𝑥 𝑛 } defined by (1.9) converges strongly to an element 𝑥 ∗ ∈ 𝐶 = 𝐹 ( 𝑇 ) which is the unique solution of the variational inequality ( for short, V I ( 𝐴 − 𝐼 + 𝜇 ( 𝐵 − 𝛾 𝑓 ) , 𝐶 ) ) : ⟨ [ ] 𝑥 𝐴 − 𝐼 + 𝜇 ( 𝐵 − 𝛾 𝑓 ) ∗ , 𝑥 − 𝑥 ∗ ⟩ ≥ 0 , ∀ 𝑥 ∈ 𝐶 . ( 1 . 1 1 )

In this paper, motivated and inspired by the above research results, we introduce a new iterative process with a countable family of nonexpansive mappings for the variational inequality problem in Hilbert spaces.

More precisely, let 𝐻 be a Hilbert space and { 𝑇 𝑖 } ∞ 𝑖 = 1 be a countable family of nonexpansive mappings from 𝐻 to 𝐻 such that ⋂ 𝐶 = ∞ 𝑖 = 1 𝐹 ( 𝑇 𝑖 ) ≠ ∅ . Let 𝑓 ∶ 𝐻 → 𝐻 be a contraction with coefficient 0 < 𝛼 < 1 and 𝐴 , 𝐵 be strongly positive linear bounded operators with coefficients 𝜂 ∈ ( 0 , 1 ) and 𝛽 > 0 , respectively. Let { 𝜆 𝑛 } ∞ 𝑛 = 1 ⊂ ( 0 , 1 ) and { 𝛼 𝑛 } ∞ 𝑛 = 0 ⊂ ( 0 , 1 ] with 𝛼 0 = 1 . Take three fixed numbers 𝛾 , 𝜇 1 and 𝜇 2 such that 0 < 𝛾 𝛼 < 𝛽 , 𝜇 1 ∈ ( 0 , 1 ] and 𝜇 2 ∈ ( ( 1 − 𝜂 𝜇 1 ) / ( 𝛽 − 𝛾 𝛼 ) , m i n { 1 , ‖ 𝐵 ‖ − 1 , ( 2 − 𝜂 𝜇 1 ) / ( 𝛽 − 𝛾 𝛼 ) } ] . For any 𝑥 1 ∈ 𝐻 , generate the iterative scheme { 𝑥 𝑛 } by

𝑥 𝑛 + 1 = 𝛼 𝑛 𝐼 − 𝜆 𝑛 𝜇 1 𝐴 𝑥 𝑛 + 𝜆 𝑛 𝑥 𝑛 − 𝜇 2 𝐵 𝑥 𝑛 𝑥 − 𝛾 𝑓 𝑛 + 𝑛 𝑖 = 1 𝛼 𝑖 − 1 − 𝛼 𝑖 𝑇 𝑖 𝑥 𝑛 , ∀ 𝑛 ≥ 1 . ( 1 . 1 2 )

We prove that the iterative scheme { 𝑥 𝑛 } defined by (1.12) strongly converges to an element 𝑥 ∗ ∈ 𝐶 which is the unique solution of the variational inequality ( for short, V I ( 𝜇 1 𝐴 − 𝐼 + 𝜇 2 ( 𝐵 − 𝛾 𝑓 ) , 𝐶 ) ) :

𝜇 1 𝐴 − 𝐼 + 𝜇 2 𝑥 ( 𝐵 − 𝛾 𝑓 ) ∗ , 𝑥 − 𝑥 ∗ ≥ 0 , ∀ 𝑥 ∈ 𝐶 . ( 1 . 1 3 )

2. Preliminaries

Let 𝐻 be a Hilbert space and 𝑇 be a nonexpansive mapping of 𝐻 into itself such that 𝐹 ( 𝑇 ) ≠ ∅ . For all ̂ 𝑥 ∈ 𝐹 ( 𝑇 ) and 𝑥 ∈ 𝐻 , 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 ∈ 𝐻 . ( 2 . 2 )

Let { 𝑥 𝑛 } be a sequence in a Hilbert space 𝐻 and let 𝑥 ∈ 𝐻 . Throughout this paper, 𝑥 𝑛 → 𝑥 and 𝑥 𝑛 ⇀ 𝑥 denote that { 𝑥 𝑛 } strongly converges to 𝑥 ∈ 𝐻 and { 𝑥 𝑛 } converges weakly to a point 𝑥 ∈ 𝐻 , respectively.

Lemma 2.1 (see [10]). Let 𝐶 be a closed convex subset of a Hilbert space 𝐻 and 𝑇 be a nonexpansive mapping from 𝐶 into itself. Then 𝐼 − 𝑇 is demiclosed at zero, that is, 𝑥 𝑛 ⇀ 𝑥 , 𝑥 𝑛 − 𝑇 𝑥 𝑛 ⟶ 0 i m p l i e s 𝑥 = 𝑇 𝑥 . ( 2 . 3 )

The following lemma is an immediate consequence of the equality:

‖ 𝑥 + 𝑦 ‖ 2 = ‖ 𝑥 ‖ 2 + 2 ⟨ 𝑦 , 𝑥 + 𝑦 ⟩ − ‖ 𝑦 ‖ 2 , ∀ 𝑥 , 𝑦 ∈ 𝐻 . ( 2 . 4 )

Lemma 2.2. Let 𝐻 be a real Hilbert space. Then the following identity holds: ‖ 𝑥 + 𝑦 ‖ 2 ≤ ‖ 𝑥 ‖ 2 + 2 ⟨ 𝑦 , 𝑥 + 𝑦 ⟩ , ∀ 𝑥 , 𝑦 ∈ 𝐻 . ( 2 . 5 )

Lemma 2.3 (see [4, 11]). Let { 𝑠 𝑛 } , { 𝑐 𝑛 } be the sequences of nonnegative real numbers and let { 𝑎 𝑛 } ⊂ ( 0 , 1 ) . Suppose that { 𝑏 𝑛 } is a sequence of real numbers such that 𝑠 𝑛 + 1 ≤ 1 − 𝑎 𝑛 𝑠 𝑛 + 𝑏 𝑛 + 𝑐 𝑛 , ∀ 𝑛 ≥ 0 . ( 2 . 6 ) Assume that ∑ ∞ 𝑛 = 0 𝑐 𝑛 < ∞ . Then the following results hold. (1)If 𝑏 𝑛 ≤ 𝛽 𝑎 𝑛 , where ( 𝛽 ≥ 0 ) , then { 𝑠 𝑛 } is a bounded sequence.(2)If one has ∞ 𝑛 = 0 𝑎 𝑛 = ∞ , l i m s u p 𝑛 → ∞ 𝑏 𝑛 𝑎 𝑛 ≤ 0 , ( 2 . 7 ) then l i m 𝑛 → ∞ 𝑠 𝑛 = 0 .

Lemma 2.4 (see [8]). Let 𝐻 be a real Hilbert space, 𝑓 ∶ 𝐻 → 𝐻 be a contraction with coefficient 0 < 𝛼 < 1 and 𝐵 be a strongly positive linear bounded operator with coefficient 𝛽 > 0 . Then, for any 𝛾 with 0 < 𝛾 < 𝛽 / 𝛼 , ⟨ 𝑥 − 𝑦 , ( 𝐵 − 𝛾 𝑓 ) 𝑥 − ( 𝐵 − 𝛾 𝑓 ) 𝑦 ⟩ ≥ ( 𝛽 − 𝛾 𝛼 ) ‖ 𝑥 − 𝑦 ‖ 2 , ∀ 𝑥 , 𝑦 ∈ 𝐻 , ( 2 . 8 ) that is, 𝐵 − 𝛾 𝑓 is strongly monotone with coefficient 𝛽 − 𝛾 𝛼 .

Lemma 2.5. Assume 𝐴 is a strongly monotone linear bounded operator on a Hilbert space 𝐻 with coefficient 𝛼 > 0 . Take a fixed number 𝜌 such that 0 < 𝜌 ≤ ‖ 𝐴 ‖ − 1 . Then ‖ 𝐼 − 𝜌 𝐴 ‖ ≤ 1 − 𝜌 𝛼 .

Proof. The proof method is mainly from the idea of Marino and Xu [8, Lemma 2.5]. It is known that the norm of a linear bounded self-adjoint operator 𝑉 on 𝐻 is as follows: | | | | ‖ 𝑉 ‖ = s u p ⟨ 𝑉 𝑥 , 𝑥 ⟩ ∶ 𝑥 ∈ 𝐻 , ‖ 𝑥 ‖ = 1 . ( 2 . 9 ) Now, for all 𝑥 ∈ 𝐻 with ‖ 𝑥 ‖ = 1 , we see that (here 0 denotes zero point in 𝐻 ) } ‖ 𝐼 − 𝜌 𝐴 ‖ = s u p { ⟨ ( 𝐼 − 𝜌 𝐴 ) 𝑥 , 𝑥 ⟩ ∶ 𝑥 ∈ 𝐻 , ‖ 𝑥 ‖ = 1 = s u p { ⟨ ( 𝐼 − 𝜌 𝐴 ) 𝑥 − ( 𝐼 − 𝜌 𝐴 ) 0 , 𝑥 − 0 ⟩ ∶ 𝑥 ∈ 𝐻 , ‖ 𝑥 ‖ = 1 } = s u p { ⟨ ( 𝑥 − 0 ) − 𝜌 ( 𝐴 𝑥 − 𝐴 0 ) , 𝑥 − 0 ⟩ ∶ 𝑥 ∈ 𝐻 , ‖ 𝑥 ‖ = 1 } = s u p ‖ 𝑥 ‖ 2 − 𝜌 ⟨ ( 𝐴 𝑥 − 𝐴 0 ) , 𝑥 − 0 ⟩ ∶ 𝑥 ∈ 𝐻 , ‖ 𝑥 ‖ = 1 ≤ s u p 1 − 𝜌 𝛼 ‖ 𝑥 − 0 ‖ 2 ∶ 𝑥 ∈ 𝐻 , ‖ 𝑥 ‖ = 1 = 1 − 𝜌 𝛼 . ( 2 . 1 0 ) This completes the proof.

Remark 2.6. Lemma 2.5 still holds if 𝐴 is a strongly positive linear bounded operator (see [8, Lemma 2.5]). That is, Lemma 2.5 in this section and Lemma 2.5 in [8] both hold when 𝐴 is a strongly monotone linear bounded operator or a strongly positive linear bounded one because an operator on a Hilbert space is strongly monotone linear if and only if it is strongly positive linear.
In fact, if 𝐴 is a strongly monotone linear operator with coefficient 𝛼 > 0 on a Hilbert space 𝐻 , then, for all 𝑥 ∈ 𝐻 , ⟨ 𝐴 𝑥 , 𝑥 ⟩ = ⟨ 𝐴 𝑥 − 𝐴 0 , 𝑥 − 0 ⟩ ≥ 𝛼 ‖ 𝑥 − 0 ‖ 2 = 𝛼 ‖ 𝑥 ‖ 2 , ( 2 . 1 1 ) which shows that 𝐴 is strongly positive linear. Assume that 𝐴 is a strongly positive linear operator with coefficient 𝛼 > 0 on 𝐻 . Then, for all 𝑥 , 𝑦 ∈ 𝐻 , ⟨ 𝐴 𝑥 − 𝐴 𝑦 , 𝑥 − 𝑦 ⟩ = ⟨ 𝐴 ( 𝑥 − 𝑦 ) , 𝑥 − 𝑦 ⟩ ≥ 𝛼 ‖ 𝑥 − 𝑦 ‖ 2 , ( 2 . 1 2 ) which shows that 𝐴 is strongly monotone and linear.

3. Main Results

Let 𝐻 be a Hilbert space and 𝐶 be a nonempty closed and convex subset of 𝐻 . Let 𝑓 ∶ 𝐻 → 𝐻 be a contraction with coefficient 0 < 𝛼 < 1 . Let 𝐴 , 𝐵 ∶ 𝐻 → 𝐻 be strongly positive linear bounded operator with coefficient 𝜂 ∈ ( 0 , 1 ) and 𝛽 > 0 , respectively. Take a fixed number 𝛾 such that 0 ≤ 𝛾 𝛼 < 𝛽 . Then, from Lemma 2.4, it follows that 𝐵 − 𝛾 𝑓 is strongly monotone with coefficient 𝛽 − 𝛾 𝛼 > 0 . For any fixed numbers 𝜎 1 ∈ ( 0 , 1 ] and 𝜎 2 ∈ ( ( 1 − 𝜂 𝜎 1 ) / ( 𝛽 − 𝛾 𝛼 ) , ( 2 − 𝜂 𝜎 1 ) / ( 𝛽 − 𝛾 𝛼 ) ) , we have 𝜃 = 𝜂 𝜎 1 − 1 + 𝜎 2 ( 𝛽 − 𝛾 𝛼 ) ∈ ( 0 , 1 ) , which can be seen easily from the following:

𝜎 2 < 2 − 𝜂 𝜎 1 𝛽 − 𝛾 𝛼 ⟺ 𝜎 2 ( 𝛽 − 𝛾 𝛼 ) < 2 − 𝜂 𝜎 1 ⟺ 𝜃 = 𝜂 𝜎 1 − 1 + 𝜎 2 ( 𝛽 − 𝛾 𝛼 ) < 1 , ( 3 . 1 ) 1 − 𝜂 𝜎 1 𝛽 − 𝛾 𝛼 < 𝜎 2 ⟺ 𝜎 2 ( 𝛽 − 𝛾 𝛼 ) + 𝜂 𝜎 1 > 1 ⟺ 𝜃 = 𝜂 𝜎 1 − 1 + 𝜎 2 ( 𝛽 − 𝛾 𝛼 ) > 0 . ( 3 . 2 ) Moreover, observe that

‖ ‖ 𝜎 1 𝐴 − 𝐼 + 𝜎 2 𝜎 ( 𝐵 − 𝛾 𝑓 ) 𝑥 − 1 𝐴 − 𝐼 + 𝜎 2 𝑦 ‖ ‖ = ‖ ‖ 𝜎 ( 𝐵 − 𝛾 𝑓 ) 1 ( 𝐴 − 𝐼 𝑥 − 𝑦 ) + 𝜎 2 ( ‖ ‖ ≤ ‖ ‖ 𝜎 𝐵 − 𝛾 𝑓 ) ( 𝑥 − 𝑦 ) 1 ‖ ‖ 𝐴 − 𝐼 ‖ 𝑥 − 𝑦 ‖ + 𝜎 2 [ ] ≤ ‖ ‖ 𝜎 ‖ 𝐵 ( 𝑥 − 𝑦 ) ‖ + 𝛾 ‖ 𝑓 𝑥 − 𝑓 𝑦 ‖ 1 ‖ ‖ 𝐴 − 𝐼 + 𝜎 2 ( ‖ 𝐵 ‖ + 𝛾 𝛼 ) ‖ 𝑥 − 𝑦 ‖ , ( 3 . 3 ) which implies that 𝜎 1 𝐴 − 𝐼 + 𝜎 2 ( 𝐵 − 𝛾 𝑓 ) is Lipschitzian with coefficient ‖ 𝜎 1 𝐴 − 𝐼 ‖ + 𝜎 2 ( ‖ 𝐵 ‖ + 𝛾 𝛼 ) > 0 .

On the other hand, from Lemma 2.4, it follows that

𝜎 1 𝐴 − 𝐼 + 𝜎 2 𝜎 ( 𝐵 − 𝛾 𝑓 ) 𝑥 − 1 𝐴 − 𝐼 + 𝜎 2 ( 𝐵 − 𝛾 𝑓 ) 𝑦 , 𝑥 − 𝑦 = 𝜎 1 ⟨ 𝐴 𝑥 − 𝐴 𝑦 , 𝑥 − 𝑦 ⟩ + 𝜎 2 ⟨ ( 𝐵 − 𝛾 𝑓 ) 𝑥 − ( 𝐵 − 𝛾 𝑓 ) 𝑦 , 𝑥 − 𝑦 ⟩ − ‖ 𝑥 − 𝑦 ‖ 2 ≥ 𝜎 1 𝜂 ‖ 𝑥 − 𝑦 ‖ 2 + 𝜎 2 ( 𝛽 − 𝛾 𝛼 ) ‖ 𝑥 − 𝑦 ‖ 2 − ‖ 𝑥 − 𝑦 ‖ 2 = 𝜃 ‖ 𝑥 − 𝑦 ‖ 2 , ( 3 . 4 ) which implies that 𝜎 1 𝐴 − 𝐼 + 𝜎 2 ( 𝐵 − 𝛾 𝑓 ) is strongly monotone with coefficient 𝜃 > 0 . Hence the variational inequality (for short, V I ( 𝜎 1 𝐴 − 𝐼 + 𝜎 2 ( 𝐵 − 𝛾 𝑓 ) , 𝐶 ) )

⟨ 𝜎 1 𝐴 − 𝐼 + 𝜎 2 ( 𝐵 − 𝛾 𝑓 ) 𝑥 ∗ , 𝑥 − 𝑥 ∗ ⟩ ≥ 0 , ∀ 𝑥 ∈ 𝐶 ( 3 . 5 ) has the unique solution.

Let 𝑇 ∶ 𝐻 → 𝐻 be a nonexpansive mapping. Take two fixed numbers 𝜇 1 and 𝜇 2 such that 𝜇 1 ∈ ( 0 , 1 ] and 𝜇 2 ∈ ( 0 , m i n { 1 , ‖ 𝐵 ‖ − 1 } ] and, for all 𝜆 ∈ ( 0 , m i n { 1 , ‖ 𝐴 ‖ − 1 / 𝜇 1 } ) , define a mapping 𝑇 𝜆 ∶ 𝐻 → 𝐻 by

𝑇 𝜆 𝑥 = 𝐼 − 𝜆 𝜇 1 𝐴 𝑇 𝑥 + 𝜆 𝑇 𝑥 − 𝜇 2 ( 𝐵 𝑇 𝑥 − 𝛾 𝑓 ( 𝑥 ) ) , ∀ 𝑥 ∈ 𝐻 . ( 3 . 6 ) Then we have the following results.

Lemma 3.1. If 𝜇 2 ∈ ( ( 1 − 𝜂 𝜇 1 ) / ( 𝛽 − 𝛾 𝜆 ) , ( 2 − 𝜂 𝜇 1 ) / ( 𝛽 − 𝛾 𝜆 ) ) , then 𝑇 𝜆 is a contraction with coefficient 1 − 𝜆 𝜏 , where 𝜏 = 𝜂 𝜇 1 − 1 + 𝜇 2 ( 𝛽 − 𝛾 𝛼 ) ∈ ( 0 , 1 ) , that is, ‖ ‖ 𝑇 𝜆 𝑥 − 𝑇 𝜆 𝑦 ‖ ‖ ≤ ( 1 − 𝜆 𝜏 ) ‖ 𝑥 − 𝑦 ‖ , ∀ 𝑥 , 𝑦 ∈ 𝐻 . ( 3 . 7 )

Proof. From Lemma 2.5 and Remark 2.6, it follows that, for all 𝑥 , 𝑦 ∈ 𝐻 , ‖ ‖ 𝑇 𝜆 𝑥 − 𝑇 𝜆 𝑦 ‖ ‖ = ‖ ‖ 𝐼 − 𝜆 𝜇 1 𝐴 𝑇 𝑥 + 𝜆 𝑇 𝑥 − 𝜇 2 − ( 𝐵 𝑇 𝑥 − 𝛾 𝑓 ( 𝑥 ) ) 𝐼 − 𝜆 𝜇 1 𝐴 𝑇 𝑦 − 𝜆 𝑇 𝑦 − 𝜇 2 ‖ ‖ ≤ ‖ ‖ ( 𝐵 𝑇 𝑦 − 𝛾 𝑓 ( 𝑦 ) ) 𝐼 − 𝜆 𝜇 1 𝐴 𝑇 𝑥 − 𝐼 − 𝜆 𝜇 1 𝐴 ‖ ‖ ‖ ‖ 𝑇 𝑦 + 𝜆 𝑇 𝑥 − 𝜇 2 ( 𝐵 𝑇 𝑥 − 𝛾 𝑓 ( 𝑥 ) ) − 𝑇 𝑦 − 𝜇 2 ‖ ‖ ≤ ‖ ‖ ( 𝐵 𝑇 𝑦 − 𝛾 𝑓 ( 𝑦 ) ) 𝐼 − 𝜆 𝜇 1 𝐴 ‖ ‖ ‖ ‖ ‖ 𝑇 𝑥 − 𝑇 𝑦 ‖ + 𝜆 𝐼 − 𝜇 2 𝐵 𝑇 𝑥 − 𝐼 − 𝜇 2 𝐵 ‖ ‖ 𝑇 𝑦 + 𝜇 2 ≤ 𝛾 ‖ 𝑓 ( 𝑥 ) − 𝑓 ( 𝑦 ) ‖ 1 − 𝜆 𝜇 1 𝜂 ‖ ‖ ‖ 𝑥 − 𝑦 ‖ + 𝜆 𝐼 − 𝜇 2 𝐵 ‖ ‖ ‖ 𝑇 𝑥 − 𝑇 𝑦 ‖ + 𝜇 2 ≤ 𝛾 𝛼 ‖ 𝑥 − 𝑦 ‖ 1 − 𝜆 𝜇 1 𝜂 + 𝜆 1 − 𝜇 2 = 𝜇 ( 𝛽 − 𝛾 𝛼 ) ‖ 𝑥 − 𝑦 ‖ 1 − 𝜆 1 𝜂 − 1 + 𝜇 2 ‖ ( 𝛽 − 𝛾 𝛼 ) 𝑥 − 𝑦 ‖ = ( 1 − 𝜆 𝜏 ) ‖ 𝑥 − 𝑦 ‖ . ( 3 . 8 ) This completes the proof.

Let { 𝑇 𝑖 } ∞ 𝑛 = 1 be a countable family of nonexpansive mappings from 𝐻 into itself such that ⋂ 𝐶 = ∞ 𝑛 = 1 𝐹 ( 𝑇 𝑖 ) ≠ ∅ . Since each 𝐹 ( 𝑇 𝑖 ) is closed and convex, then 𝐶 is closed and convex.

Throughout this paper, let 𝑓 ∶ 𝐻 → 𝐻 be a contraction with coefficient 0 < 𝛼 < 1 . Let 𝐴 , 𝐵 ∶ 𝐻 → 𝐻 be strongly positive linear bounded mapping with coefficient 𝜂 ∈ ( 0 , 1 ) and 𝛽 > 0 , respectively. Take a fixed number 𝛾 such that 0 < 𝛾 𝛼 < 𝛽 . Suppose that 𝜇 1 ∈ ( 0 , 1 ] , 𝜇 2 ∈ ( ( 1 − 𝜂 𝜇 1 ) / ( 𝛽 − 𝛾 𝛼 ) , m i n { 1 , ‖ 𝐵 ‖ − 1 , ( 2 − 𝜂 𝜇 1 ) / ( 𝛽 − 𝛾 𝛼 ) } ) (assuming that ( 1 − 𝜂 𝜇 1 ) / ( 𝛽 − 𝛾 𝛼 ) < m i n { 1 , ‖ 𝐵 ‖ − 1 } such that ( ( 1 − 𝜂 𝜇 1 ) / ( 𝛽 − 𝛾 𝜆 ) , m i n { 1 , ‖ 𝐵 ‖ − 1 , ( 2 − 𝜂 𝜇 1 ) / ( 𝛽 − 𝛾 𝜆 ) } ] is nonempty), { 𝜆 𝑛 } ∞ 𝑛 = 1 ⊂ ( 0 , m i n { 1 , ‖ 𝐴 ‖ − 1 / 𝜇 1 } ) with l i m i n f 𝑛 → ∞ 𝜆 𝑛 > 0 and { 𝛼 𝑛 } ∞ 𝑛 = 0 ⊂ ( 0 , 1 ] with 𝛼 0 = 1 .

Now, we can rewrite the iterative scheme (1.12) as follows:

𝑥 𝑛 + 1 = 𝛼 𝑛 𝑇 𝜆 𝑛 𝑥 𝑛 + 𝑛 𝑖 = 1 𝛼 𝑖 − 1 − 𝛼 𝑖 𝑇 𝑖 𝑥 𝑛 , ∀ 𝑛 ≥ 1 , ( 3 . 9 ) where 𝑇 𝜆 𝑛 𝑥 𝑛 = ( 𝐼 − 𝜆 𝑛 𝜇 1 𝐴 ) 𝑥 𝑛 + 𝜆 𝑛 [ 𝑥 𝑛 − 𝜇 2 ( 𝐵 𝑥 𝑛 − 𝛾 𝑓 ( 𝑥 𝑛 ) ) ] . Then, by Lemma 3.1, for all 𝑥 , 𝑦 ∈ 𝐻 , we have

‖ ‖ 𝑇 𝜆 𝑛 𝑛 𝑥 − 𝑇 𝜆 𝑛 𝑛 𝑦 ‖ ‖ ≤ 1 − 𝜆 𝑛 𝜏 ‖ 𝑥 − 𝑦 ‖ , ∀ 𝑛 ≥ 1 , ( 3 . 1 0 ) where 𝜏 = 𝜂 𝜇 1 − 1 + 𝜇 2 ( 𝛽 − 𝛾 𝛼 ) ∈ ( 0 , 1 ) .

Lemma 3.2. If { 𝛼 𝑛 } is strictly decreasing, then the scheme { 𝑥 𝑛 } defined by (3.9) is bounded.

Proof. Since ‖ 𝑇 𝜆 𝑛 𝑝 − 𝑝 ‖ = 𝜆 𝑛 ‖ ( 𝜇 1 𝐴 − 𝐼 + 𝜇 2 ( 𝐵 − 𝛾 𝑓 ) ) 𝑝 ‖ , it follows from (3.10) that, for all 𝑝 ∈ 𝐶 , ‖ ‖ 𝑥 𝑛 + 1 ‖ ‖ = ‖ ‖ ‖ ‖ 𝛼 − 𝑝 𝑛 𝑇 𝜆 𝑛 𝑥 𝑛 + − 𝑝 𝑛 𝑖 = 1 𝛼 𝑖 − 1 − 𝛼 𝑖 𝑇 𝑖 𝑥 𝑛 ‖ ‖ ‖ ‖ − 𝑝 ≤ 𝛼 𝑛 ‖ ‖ 𝑇 𝜆 𝑛 𝑥 𝑛 ‖ ‖ + − 𝑝 𝑛 𝑖 = 1 𝛼 𝑖 − 1 − 𝛼 𝑖 ‖ ‖ 𝑇 𝑖 𝑥 𝑛 ‖ ‖ − 𝑝 ≤ 𝛼 𝑛 ‖ ‖ 𝑇 𝜆 𝑛 𝑥 𝑛 − 𝑇 𝜆 𝑛 𝑝 ‖ ‖ + 𝛼 𝑛 ‖ ‖ 𝑇 𝜆 𝑛 ‖ ‖ + 𝑝 − 𝑝 1 − 𝛼 𝑛 ‖ ‖ 𝑥 𝑛 ‖ ‖ − 𝑝 ≤ 𝛼 𝑛 1 − 𝜆 𝑛 𝜏 ‖ ‖ 𝑥 𝑛 ‖ ‖ − 𝑝 + 𝛼 𝑛 𝜆 𝑛 ‖ ‖ 𝜇 1 𝐴 − 𝐼 + 𝜇 2 𝑝 ‖ ‖ + ( 𝐵 − 𝛾 𝑓 ) 1 − 𝛼 𝑛 ‖ ‖ 𝑥 𝑛 ‖ ‖ = − 𝑝 1 − 𝛼 𝑛 𝜆 𝑛 𝜏 ‖ ‖ 𝑥 𝑛 ‖ ‖ − 𝑝 + 𝛼 𝑛 𝜆 𝑛 ‖ ‖ 𝜇 1 𝐴 − 𝐼 + 𝜇 2 𝑝 ‖ ‖ . ( 𝐵 − 𝛾 𝑓 ) ( 3 . 1 1 ) By induction, we obtain ‖ ‖ 𝑥 𝑛 + 1 ‖ ‖ ‖ ‖ 𝑥 − 𝑝 ≤ m a x 1 ‖ ‖ , 1 − 𝑝 𝜏 ‖ ‖ 𝜇 1 𝐴 − 𝐼 + 𝜇 2 𝑝 ‖ ‖ . ( 𝐵 − 𝛾 𝑓 ) ( 3 . 1 2 ) Hence { 𝑥 𝑛 } is bounded and so are { 𝑇 𝜆 𝑛 𝑥 𝑛 } and { 𝑇 𝑖 𝑥 𝑛 } for each 𝑖 ≥ 1 . This completes the proof.

Lemma 3.3. If { 𝛼 𝑛 } is strictly decreasing and the following conditions hold: ∞ 𝑛 = 1 𝛼 𝑛 = ∞ , ∞ 𝑛 = 1 | | 𝜆 𝑛 − 𝜆 𝑛 + 1 | | < ∞ , ( 3 . 1 3 ) then l i m 𝑛 → ∞ ‖ 𝑥 𝑛 + 1 − 𝑥 𝑛 ‖ = 0 .

Proof. By the iterative scheme (3.9), we have 𝑥 𝑛 + 1 − 𝑥 𝑛 = 𝛼 𝑛 𝑇 𝜆 𝑛 𝑥 𝑛 + 𝑛 𝑖 = 1 𝛼 𝑖 − 1 − 𝛼 𝑖 𝑇 𝑖 𝑥 𝑛 − 𝛼 𝑛 − 1 𝑇 𝜆 𝑛 − 1 𝑥 𝑛 − 1 + 𝑛 − 1 𝑖 = 1 𝛼 𝑖 − 1 − 𝛼 𝑖 𝑇 𝑖 𝑥 𝑛 − 1 = 𝛼 𝑛 𝑇 𝜆 𝑛 𝑥 𝑛 − 𝑇 𝜆 𝑛 𝑥 𝑛 − 1 + 𝑛 𝑖 = 1 𝛼 𝑖 − 1 − 𝛼 𝑖 𝑇 𝑖 𝑥 𝑛 − 𝑇 𝑖 𝑥 𝑛 − 1 + 𝑛 𝑖 = 1 𝛼 𝑖 − 1 − 𝛼 𝑖 𝑇 𝑖 𝑥 𝑛 − 1 − 𝑛 − 1 𝑖 = 1 𝛼 𝑖 − 1 − 𝛼 𝑖 𝑇 𝑖 𝑥 𝑛 − 1 + 𝛼 𝑛 𝑇 𝜆 𝑛 𝑥 𝑛 − 1 − 𝛼 𝑛 − 1 𝑇 𝜆 𝑛 − 1 𝑥 𝑛 − 1 = 𝛼 𝑛 𝑇 𝜆 𝑛 𝑥 𝑛 − 𝑇 𝜆 𝑛 𝑥 𝑛 − 1 + 𝑛 𝑖 = 1 𝛼 𝑖 − 1 − 𝛼 𝑖 𝑇 𝑖 𝑥 𝑛 − 𝑇 𝑖 𝑥 𝑛 − 1 + 𝛼 𝑛 − 1 − 𝛼 𝑛 𝑇 𝑛 𝑥 𝑛 − 1 + 𝛼 𝑛 − 1 𝜆 𝑛 − 1 − 𝛼 𝑛 𝜆 𝑛 𝜇 1 𝐴 − 𝐼 + 𝜇 2 𝑥 ( 𝐵 − 𝛾 𝑓 ) 𝑛 − 1 + 𝛼 𝑛 − 𝛼 𝑛 − 1 𝑥 𝑛 − 1 = 𝛼 𝑛 𝑇 𝜆 𝑛 𝑥 𝑛 − 𝑇 𝜆 𝑛 𝑥 𝑛 − 1 + 𝑛 𝑖 = 1 𝛼 𝑖 − 1 − 𝛼 𝑖 𝑇 𝑖 𝑥 𝑛 − 𝑇 𝑖 𝑥 𝑛 − 1 + 𝛼 𝑛 − 1 − 𝛼 𝑛 𝑇 𝑛 𝑥 𝑛 − 1 + 𝛼 𝑛 − 𝛼 𝑛 − 1 𝑥 𝑛 − 1 + 𝛼 𝑛 − 1 − 𝛼 𝑛 𝜆 𝑛 + 𝜆 𝑛 − 1 − 𝜆 𝑛 𝛼 𝑛 − 1 𝜇 1 𝐴 − 𝐼 + 𝜇 2 𝑥 ( 𝐵 − 𝛾 𝑓 ) 𝑛 − 1 ( 3 . 1 4 ) and hence ‖ ‖ 𝑥 𝑛 + 1 − 𝑥 𝑛 ‖ ‖ ≤ 𝛼 𝑛 ‖ ‖ 𝑇 𝜆 𝑛 𝑥 𝑛 − 𝑇 𝜆 𝑛 𝑥 𝑛 − 1 ‖ ‖ + 𝑛 𝑖 = 1 𝛼 𝑖 − 1 − 𝛼 𝑛 ‖ ‖ 𝑇 𝑖 𝑥 𝑛 − 𝑇 𝑖 𝑥 𝑛 − 1 ‖ ‖ + 𝛼 𝑛 − 1 − 𝛼 𝑛 ‖ ‖ 𝑇 𝑛 𝑥 𝑛 − 1 ‖ ‖ + 𝛼 𝑛 − 1 − 𝛼 𝑛 ‖ ‖ 𝑥 𝑛 − 1 ‖ ‖ + 𝛼 𝑛 − 1 − 𝛼 𝑛 𝜆 𝑛 + | | 𝜆 𝑛 − 1 − 𝜆