International Journal of Mathematics and Mathematical Sciences
Volume 2009 (2009), Article ID 656534, 17 pages
doi:10.1155/2009/656534
Research Article

Fixed Points Approximation and Solutions of Some Equilibrium and Variational Inequalities Problems

Bashir Ali1,2

1Mathematics Stream, African University of Science and Technology, Abuja, Nigeria
2Department of Mathematical Sciences, Bayero University, Kano, P.M.B. 3011, Nigeria

Received 28 September 2009; Accepted 16 December 2009

Academic Editor: Asao Arai

Copyright © 2009 Bashir Ali. 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 prove a new strong convergence theorem for an element in the intersection of the set of common fixed points of a countable family of nonexpansive mappings, the set of solutions of some variational inequality problems, and the set of solutions of some equilibrium problems using a new iterative scheme. Our theorem generalizes and improves some recent results.

1. Introduction

Let 𝐻 be a real Hilbert space; a mapping 𝐵 ∶ 𝐷 ( 𝐵 ) → 𝐻 is said to be monotone if for all 𝑥 , 𝑦 ∈ 𝐷 ( 𝐵 )

⟨ 𝐵 𝑥 − 𝐵 𝑦 , 𝑥 − 𝑦 ⟩ ≥ 0 . ( 1 . 1 ) For some 𝜆 > 0 , the mapping 𝐵 is said to be 𝜆 -inverse strongly monotone if

⟨ 𝐵 𝑥 − 𝐵 𝑦 , 𝑥 − 𝑦 ⟩ ≥ 𝜆 ‖ 𝐵 𝑥 − 𝐵 𝑦 ‖ 2 . ( 1 . 2 ) A 𝜆 -inverse strongly monotone map is some time called 𝜆 -cocoercive. A map 𝐵 is said to be relaxed 𝜆 -cocoercive if there exists a constant 𝜆 > 0 such that

⟨ 𝐵 𝑥 − 𝐵 𝑦 , 𝑥 − 𝑦 ⟩ ≥ − 𝜆 ‖ 𝐵 𝑥 − 𝐵 𝑦 ‖ 2 ∀ 𝑥 , 𝑦 ∈ 𝐾 . ( 1 . 3 ) 𝐵 is said to be relaxed ( 𝜆 , 𝛾 ) -cocoercive, if there exist 𝜆 , 𝛾 > 0 such that

⟨ 𝐵 𝑥 − 𝐵 𝑦 , 𝑥 − 𝑦 ⟩ ≥ − 𝜆 ‖ 𝐵 𝑥 − 𝐵 𝑦 ‖ 2 + 𝛾 ‖ 𝑥 − 𝑦 ‖ 2 . ( 1 . 4 ) A map 𝐵 ∶ 𝐻 → 𝐻 is said to be 𝜆 -Lipschitzian if there exists a real number 𝜆 ≥ 0 such that

‖ 𝐵 𝑥 − 𝐵 𝑦 ‖ ≤ 𝜆 ‖ 𝑥 − 𝑦 ‖ ∀ 𝑥 , 𝑦 ∈ 𝐻 . ( 1 . 5 ) 𝐵 is a contraction map, if in the above inequality 𝜆 ∈ [ 0 , 1 ) and nonexpansive if 𝜆 = 1 .

Let 𝐾 be a nonempty, closed, and convex subset of a real Hilbert space 𝐻 . A variational inequality problem is searched for 𝑥 ∗ ∈ 𝐾 such that

⟨ 𝐵 𝑥 ∗ , 𝑦 − 𝑥 ∗ ⟩ ≥ 0 ∀ 𝑦 ∈ 𝐾 , ( 1 . 6 ) where 𝐵 is some nonlinear mapping of 𝐾 into 𝐻 . Inequality (1.6) is called the variational inequality.

Recall that for each 𝑥 ∈ 𝐻 there exists a unique nearest point in 𝐾 to 𝑥 denoted by 𝑃 𝐾 𝑥 . That is, ‖ 𝑥 − 𝑃 𝐾 𝑥 ‖ ≤ ‖ 𝑥 − 𝑦 ‖ for all 𝑦 ∈ 𝐾 . 𝑃 𝐾 is called a metric projection of 𝐻 onto 𝐾 . The mapping 𝑃 𝐾 is nonexpansive in this setting, that is, ‖ 𝑃 𝐾 𝑥 − 𝑃 𝐾 𝑦 ‖ ≤ ‖ 𝑥 − 𝑦 ‖ for all 𝑥 , 𝑦 ∈ 𝐻 . It is also known that 𝑃 𝐾 satisfies the following inequality ‖ 𝑃 𝐾 𝑥 − 𝑃 𝐾 𝑦 ‖ 2 ≤ ⟨ 𝑥 − 𝑦 , 𝑃 𝐾 𝑥 − 𝑃 𝐾 𝑦 ⟩ .

The solution set of the problem (1.6) is denoted by V I ( 𝐾 , 𝐵 ) . It is well known (see [1]) that 𝑥 ∗ ∈ V I ( 𝐾 , 𝐵 ) if and only if

𝑥 ∗ = 𝑃 𝐾 𝑥 ∗ − 𝜆 𝐵 𝑥 ∗ , ∀ 𝜆 > 0 . ( 1 . 7 ) A monotone map 𝐵 is said to be maximal if the graph Γ ( 𝐵 ) of 𝐵 is not properly contained in the graph of any other monotone map, where Γ ( 𝐵 ) = { ( 𝑥 , 𝑦 ) ∈ 𝐻 × 𝐻 ∶ 𝑦 ∈ 𝐵 𝑥 } for a multivalued map 𝐵 . It is also known that 𝐵 is maximal monotone if and only if for ( 𝑥 , 𝑓 ) ∈ 𝐻 × 𝐻 , ⟨ 𝑥 − 𝑦 , 𝑓 − 𝑔 ⟩ ≥ 0 for every ( 𝑦 , 𝑔 ) ∈ Γ ( 𝐵 ) implies 𝑓 ∈ 𝐵 𝑥 . Let 𝐵 be a monotone mapping defined from 𝐾 into 𝐻 and let 𝑁 𝐾 𝑞 be a normal cone to 𝐾 at 𝑞 ∈ 𝐾 , that is, 𝑁 𝐾 𝑞 = { 𝑝 ∈ 𝐻 ∶ ⟨ 𝑞 − 𝑢 , 𝑝 ⟩ ≥ , f o r a l l 𝑢 ∈ 𝐾 } . Define a map 𝑀 by

𝑀 𝑞 = 𝐵 𝑞 + 𝑁 𝐾 𝑞 , 𝑞 ∈ 𝐾 , ∅ , 𝑞 ∉ 𝐾 . ( 1 . 8 ) Then, 𝑀 is maximal monotone and 𝑥 ∗ ∈ 𝑀 − 1 ( 0 ) ⇔ 𝑥 ∗ ∈ V I ( 𝐾 , 𝐵 ) , see, for example, [2].

Let 𝐺 ∶ 𝐾 × 𝐾 → ℝ be a bifunction on a closed convex nonempty subset 𝐾 of a real Hilbert space 𝐻 . An equilibrium problem is searched for 𝑥 ∗ ∈ 𝐾 such that

𝐺 𝑥 ∗ , 𝑦 ≥ 0 ∀ 𝑦 ∈ 𝐾 . ( 1 . 9 ) The set of solutions of the equilibrium problem above is denoted by E P ( 𝐺 ) .

Several physical problems (such as the theories of lubrications, filterations and flows, moving boundary problems, see, e.g., [1, 3]) can be reduced to variational inequality or equilibrium problems. Consequently, these problems have solutions as the solutions of these resultant variational inequality or equilibrium problems.

Maingé [4] introduced a Halpern-type scheme and proved a strong convergence theorem for family of nonexpansive mappings in Hilbert space.

Recently, S. Takahashi and W. Takahashi [5] introduced an iterative scheme which they used to study the problem of approximating a common element of the set of solutions of an equilibrium problem and the set of fixed points of a nonexpansive mapping.

More recently, Kumam and Katchang [6], W. Kumam and P. Kumam [7], Li and Su [8] and many others (see, e.g. [9–12] and the references contained in them) studied the problem of fixed point approximations and solutions of some equilibrium and, or solutions of some variational inequalities problems.

In this paper we introduce a new iterative scheme for approximation of a common element in the intersection of the set of fixed points of some countable family of nonexpansive mappings, the set of solutions of some equilibrium problem, and the set of solutions of some variational inequality problem and prove a new theorem. Our theorem generalizes and improves some recent results.

2. Preliminaries

For a sequence { 𝑥 𝑛 } the notation 𝑥 𝑛 → 𝑥 ∗ and 𝑥 𝑛 ⇀ 𝑥 ∗ means that the sequence { 𝑥 𝑛 } converges strongly and weakly to 𝑥 ∗ , repectively. A Banach space 𝐸 is said to satisfy an Opial's condition (or in other words is an Opial's space) if for a sequence { 𝑥 𝑛 } in 𝐸 with 𝑥 𝑛 ⇀ 𝑥 ∗ , then

l i m i n f 𝑛 → ∞ ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ‖ < l i m i n f 𝑛 → ∞ ‖ ‖ 𝑥 𝑛 ‖ ‖ − 𝑦 f o r a n y 𝑦 ∈ 𝐸 𝑦 ≠ 𝑥 ∗ . ( 2 . 1 ) It is well known that Hilbert spaces are Opial’s spaces (see [13]).

In the sequel we shall make use of the following results.

Lemma 2.1 (see [14]). Let 𝐾 be a nonempty closed convex subset of 𝐻 and let 𝐺 be a bifunction of 𝐾 × 𝐾 into ℝ satisfying ( 𝐴 1 ) 𝐺 ( 𝑥 , 𝑥 ) = 0 f o r a l l 𝑥 ∈ 𝐾 ; ( 𝐴 2 ) 𝐺 is monotone, i.e. 𝐺 ( 𝑥 , 𝑦 ) + 𝐺 ( 𝑦 , 𝑥 ) ≤ 0 f o r a l l 𝑥 , 𝑦 ∈ 𝐾 ; ( 𝐴 3 ) f o r a l l 𝑥 , 𝑦 , 𝑧 ∈ 𝐾 , l i m s u p 𝑡 → 0 + 𝐺 ( 𝑡 𝑧 + ( 1 + 𝑡 ) 𝑥 , 𝑦 ) ≤ 𝐺 ( 𝑥 , 𝑦 ) ; ( 𝐴 4 ) f o r a l l 𝑥 ∈ 𝐾 , 𝐺 ( 𝑥 , . ) is convex and lower semicontinuous.Let 𝑟 > 0 and 𝑥 ∈ 𝐻 . Then there exists 𝑧 ∈ 𝐾 such that 1 𝐺 ( 𝑧 , 𝑦 ) + 𝑟 ⟨ 𝑦 − 𝑧 , 𝑧 − 𝑥 ⟩ ≥ 0 , ∀ 𝑦 ∈ 𝐾 . ( 2 . 2 )

Lemma 2.2 (see [15]). Let 𝐾 be a nonempty closed convex subset of 𝐻 and let 𝐺 be a bifunction of 𝐾 × 𝐾 into ℝ satisfying ( 𝐴 1 ) - ( 𝐴 4 ) . For 𝑟 > 0 and 𝑥 ∈ 𝐻 define a map 𝑇 𝑟 ∶ 𝐻 → 𝐾 by 𝑇 𝑟 1 𝑥 = 𝑧 ∈ 𝐾 ∶ 𝐺 ( 𝑧 , 𝑦 ) + 𝑟 . ⟨ 𝑦 − 𝑧 , 𝑧 − 𝑥 ⟩ ≥ 0 , ∀ 𝑦 ∈ 𝐾 ( 2 . 3 ) Then, the following holds: (1) 𝑇 𝑟 is single-valued;(2) 𝑇 𝑟 is firmly nonexpansive, that is, for any 𝑥 , 𝑦 ∈ 𝐻 ‖ 𝑇 𝑟 𝑥 − 𝑇 𝑟 𝑦 ‖ 2 ≤ ⟨ 𝑇 𝑟 𝑥 − 𝑇 𝑟 𝑦 , 𝑥 − 𝑦 ⟩ ; ( 2 . 4 ) (3) F i x ( 𝑇 𝑟 ) = E P ( 𝐺 ) ; (4) E P ( 𝐺 ) is closed and convex.

Lemma 2.3 (see [16]). Let 𝐻 be a real inner product space. Then, the following inequality holds: ‖ 𝑥 + 𝑦 ‖ 2 ≤ ‖ 𝑥 ‖ 2 + 2 ⟨ 𝑦 , 𝑥 + 𝑦 ⟩ ∀ 𝑥 , 𝑦 ∈ 𝐻 . ( 2 . 5 )

Lemma 2.4 (see [17]). Let { 𝑥 𝑛 } and { 𝑦 𝑛 } be bounded sequences in a Banach space 𝐸 and let { 𝛽 𝑛 } be a sequence in [ 0 , 1 ] with 0 < l i m i n f 𝛽 𝑛 ≤ l i m s u p 𝛽 𝑛 < 1 . Suppose 𝑥 𝑛 + 1 = 𝛽 𝑛 𝑦 𝑛 + ( 1 − 𝛽 𝑛 ) 𝑥 𝑛 for all integers 𝑛 ≥ 0 and l i m s u p ( ‖ 𝑦 𝑛 + 1 − 𝑦 𝑛 ‖ − ‖ 𝑥 𝑛 + 1 − 𝑥 𝑛 ‖ ) ≤ 0 . Then, l i m ‖ 𝑦 𝑛 − 𝑥 𝑛 ‖ = 0 .

Lemma 2.5 (see [18]). Let { 𝑎 𝑛 } be a sequence of nonnegative real numbers satisfying the following relation: 𝑎 𝑛 + 1 ≤ 1 − 𝛼 𝑛 𝑎 𝑛 + 𝛼 𝑛 𝜎 𝑛 + 𝛾 𝑛 , 𝑛 ≥ 0 , ( 2 . 6 ) Where (i) { 𝛼 𝑛 } ⊂ [ 0 , 1 ] , ∑ 𝛼 𝑛 = ∞ ; (ii) l i m s u p 𝜎 𝑛 ≤ 0 ; (iii) 𝛾 𝑛 ∑ 𝛾 ≥ 0 ; ( 𝑛 ≥ 0 ) , 𝑛 < ∞ . Then, 𝑎 𝑛 → 0 as 𝑛 → ∞ .

3. Main Results

In the sequel we assume that the sequences { 𝛼 𝑛 } , { 𝜎 𝑖 , 𝑛 } 𝑖 ⊂ ( 0 , 1 ) satisfy ∑ 𝑖 ≥ 1 𝜎 𝑖 , 𝑛 = 1 − 𝛼 𝑛 .

Theorem 3.1. Let 𝐾 be a nonempty, closed, and convex subset of a real Hilbert space 𝐻 . Let 𝐺 be a bifunction from 𝐾 × 𝐾 to ℝ which satisfies conditions ( 𝐴 1 ) - ( 𝐴 4 ) . Let { 𝑇 𝑖 , 𝑖 = 1 , 2 , 3 , … } be family of nonexpansive mappings of 𝐾 into 𝐻 and let 𝐵 be a 𝜇 -Lipschitzian, relaxed ( 𝜆 , 𝛾 ) -cocoercive map of 𝐾 into 𝐻 such that ⋂ 𝐹 ∶ = ∞ 𝑖 = 1 F i x ( 𝑇 𝑖 ) ∩ E P ( 𝐺 ) ∩ V I ( 𝐾 , 𝐵 ) ≠ ∅ . For an arbitrary but fixed 𝛿 ∈ ( 0 , 1 ) , let { 𝑥 𝑛 } and { 𝑦 𝑛 } be sequences generated by 𝑥 1 𝐺 𝑦 , 𝑢 ∈ 𝐻 , 𝑛 + 1 , 𝜂 𝑟 𝑛 ⟨ 𝜂 − 𝑦 𝑛 , 𝑦 𝑛 − 𝑥 𝑛 𝑥 ⟩ ≥ 0 , ∀ 𝜂 ∈ 𝐾 , 𝑛 + 1 = 𝛼 𝑛 𝑢 + ( 1 − 𝛿 ) 1 − 𝛼 𝑛 𝑥 𝑛 + 𝛿 𝑖 ≥ 1 𝜎 𝑖 𝑛 𝑇 𝑖 𝑃 𝐾 𝐼 − 𝑠 𝑛 𝐵 𝑦 𝑛 , 𝑛 ≥ 1 , ( 3 . 1 ) where { 𝛼 𝑛 } and { 𝜎 𝑖 𝑛 } are sequences in [ 0 , 1 ] and { 𝑟 𝑛 } and { 𝑠 𝑛 } are sequences in [ 0 , ∞ ) satisfying ( 𝐶 1 ) l i m 𝑛 → ∞ 𝛼 𝑛 = 0 , ( 𝐶 2 ) ∑ ∞ 𝑛 = 1 𝛼 𝑛 = ∞ , ( 𝐶 3 ) { 𝑠 𝑛 } ⊂ [ 𝑎 , 𝑏 ] for some 𝑎 , 𝑏 satisfying 0 ≤ 𝑎 ≤ 𝑏 ≤ 2 ( 𝛾 − 𝜆 𝜇 2 ) / 𝜇 2 , ( 𝐶 4 ) l i m 𝑛 → ∞ | 𝑠 𝑛 + 1 − 𝑠 𝑛 | = 0 , l i m 𝑛 → ∞ | 𝑟 𝑛 + 1 − 𝑟 𝑛 | = 0 , l i m 𝑛 → ∞ ∑ 𝑖 ≥ 1 | 𝜎 𝑖 , 𝑛 + 1 − 𝜎 𝑖 , 𝑛 | = 0 , ( 𝐶 5 ) l i m i n f 𝑛 → ∞ 𝑟 𝑛 > 0 . Then, both { 𝑥 𝑛 } and { 𝑦 𝑛 } converge strongly to 𝑃 𝐹 𝑢 .

Proof. First, we show that ( 𝐼 − 𝑠 𝑛 𝐵 ) is nonexpansive, actually, using the property of 𝐵 we have for 𝑥 , 𝑦 ∈ 𝐻 , ‖ 𝐼 − 𝑠 𝑛 𝐵 𝑥 − 𝐼 − 𝑠 𝑛 𝐵 𝑦 ‖ 2 = ‖ 𝑥 − 𝑦 − 𝑠 𝑛 ( 𝐵 𝑥 − 𝐵 𝑦 ) ‖ 2 = ‖ 𝑥 − 𝑦 ‖ 2 − 2 𝑠 𝑛 ⟨ 𝑥 − 𝑦 , 𝐵 𝑥 − 𝐵 𝑦 ⟩ + 𝑠 2 𝑛 ‖ 𝐵 𝑥 − 𝐵 𝑦 ‖ 2 ≤ ‖ 𝑥 − 𝑦 ‖ 2 − 2 𝑠 𝑛 − 𝜆 ‖ 𝐵 𝑥 − 𝐵 𝑦 ‖ 2 + 𝛾 ‖ 𝑥 − 𝑦 ‖ 2 + 𝑠 2 𝑛 ‖ 𝐵 𝑥 − 𝐵 𝑦 ‖ 2 ≤ ‖ 𝑥 − 𝑦 ‖ 2 + 2 𝑠 𝑛 𝜇 2 𝜆 ‖ 𝑥 − 𝑦 ‖ 2 − 2 𝑠 𝑛 𝛾 ‖ 𝑥 − 𝑦 ‖ 2 + 𝜇 2 𝑠 2 𝑛 ‖ 𝑥 − 𝑦 ‖ 2 = 1 + 2 𝑠 𝑛 𝜇 2 𝜆 − 2 𝑠 𝑛 𝛾 + 𝜇 2 𝑠 2 𝑛 ‖ 𝑥 − 𝑦 ‖ 2 ≤ ‖ 𝑥 − 𝑦 ‖ 2 , ( 3 . 2 ) and thus ( 𝐼 − 𝑠 𝑛 𝐵 ) is nonexpansive.
Let 𝑥 ∗ ∈ 𝐹 ; since 𝑦 𝑛 = 𝑇 𝑟 𝑛 𝑥 𝑛 , 𝑛 ∈ ℕ and the fact that 𝑇 𝑟 𝑛 is firmly nonexpansive (and hence nonexpansive) we have the following: ‖ 𝑦 𝑛 − 𝑥 ∗ ‖ = ‖ 𝑇 𝑟 𝑛 𝑥 𝑛 − 𝑇 𝑟 𝑛 𝑥 ∗ ‖ ≤ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ . ( 3 . 3 ) We claim that { 𝑥 𝑛 } satisfies ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ ≤ m a x ‖ 𝑢 − 𝑥 ∗ ‖ , ‖ 𝑥 1 − 𝑥 ∗ ‖ ∀ 𝑛 ≥ 1 . ( 3 . 4 ) We prove this by induction. Cearly the result is true for 𝑛 = 1 . Assume that the result holds for 𝑛 = 𝑘 for some 𝑘 ∈ ℕ . Then, for 𝑛 = 𝑘 + 1 we have ‖ 𝑥 𝑘 + 1 − 𝑥 ∗ ‖ = ‖ 𝛼 𝑘 𝑢 − 𝑥 ∗ + ( 1 − 𝛿 ) 1 − 𝛼 𝑘 𝑥 𝑘 − 𝑥 ∗ + 𝛿 𝑖 ≥ 1 𝜎 𝑖 𝑘 𝑇 𝑖 𝑃 𝐾 𝐼 − 𝑠 𝑘 𝐵 𝑦 𝑘 − 𝑥 ∗ ‖ ≤ 𝛼 𝑘 ‖ 𝑢 − 𝑥 ∗ ‖ + ( 1 − 𝛿 ) 1 − 𝛼 𝑘 ‖ 𝑥 𝑘 − 𝑥 ∗ ‖ + 𝛿 𝑖 ≥ 1 𝜎 𝑖 𝑘 ‖ 𝑇 𝑖 𝑃 𝐾 𝐼 − 𝑠 𝑘 𝐵 𝑦 𝑘 − 𝑇 𝑖 𝑃 𝐾 𝐼 − 𝑠 𝑘 𝐵 𝑥 ∗ ‖ ≤ 𝛼 𝑘 ‖ 𝑢 − 𝑥 ∗ ‖ + 1 − 𝛼 𝑘 ‖ 𝑥 𝑘 − 𝑥 ∗ ‖ ≤ m a x ‖ 𝑢 − 𝑥 ∗ ‖ , ‖ 𝑥 𝑘 − 𝑥 ∗ ‖ . ( 3 . 5 ) Hence the result, and so { 𝑥 𝑛 } is bounded. Furthermore, { 𝑦 𝑛 } , { 𝑇 𝑖 𝑃 𝐾 ( 𝑦 𝑛 − 𝑠 𝑛 𝐵 𝑦 𝑛 ) } and { 𝐵 𝑦 𝑛 } are each bounded.
We now show that l i m 𝑛 → ∞ ‖ 𝑥 𝑛 + 1 − 𝑥 𝑛 ‖ = 0 . Note that 𝑦 𝑛 = 𝑇 𝑟 𝑛 𝑥 𝑛 , 𝑦 𝑛 + 1 = 𝑇 𝑟 𝑛 + 1 𝑥 𝑛 + 1 , so that 𝐺 𝑦 𝑛 + 1 , 𝜂 𝑟 𝑛 ⟨ 𝜂 − 𝑦 𝑛 , 𝑦 𝑛 − 𝑥 𝑛 𝐺 𝑦 ⟩ ≥ 0 ∀ 𝜂 ∈ 𝐾 , 𝑛 + 1 + 1 , 𝜂 𝑟 𝑛 + 1 ⟨ 𝜂 − 𝑦 𝑛 + 1 , 𝑦 𝑛 + 1 − 𝑥 𝑛 + 1 ⟩ ≥ 0 ∀ 𝜂 ∈ 𝐾 . ( 3 . 6 ) Using (3.6) and ( 𝐴 2 ) , we have 𝑦 𝑛 + 1 − 𝑦 𝑛 , 𝑦 𝑛 − 𝑥 𝑛 𝑟 𝑛 − 𝑦 𝑛 + 1 − 𝑥 𝑛 + 1 𝑟 𝑛 + 1 ≥ 0 . ( 3 . 7 ) which implies that 𝑦 𝑛 + 1 − 𝑦 𝑛 , 𝑦 𝑛 − 𝑦 𝑛 + 1 + 𝑦 𝑛 + 1 − 𝑥 𝑛 − 𝑟 𝑛 𝑟 𝑛 + 1 𝑦 𝑛 + 1 − 𝑥 𝑛 + 1 ≥ 0 . ( 3 . 8 ) from which we get ‖ 𝑦 𝑛 + 1 − 𝑦 𝑛 ‖ 2 ≤ ‖ 𝑦 𝑛 + 1 − 𝑦 𝑛 ‖ ‖ 𝑥 𝑛 + 1 − 𝑥 𝑛 | | | | 𝑟 ‖ + 1 − 𝑛 𝑟 𝑛 + 1 | | | | ‖ 𝑦 𝑛 + 1 − 𝑥 𝑛 + 1 ‖ . ( 3 . 9 ) If, without loss of generality 𝑀 , 𝑚 are real numbers such that 𝑟 𝑛 > 𝑚 > 0 for all 𝑛 and 𝑀 ∶ = s u p 𝑛 , 𝑖 { ‖ 𝑦 𝑛 − 𝑥 𝑛 ‖ , ‖ 𝑇 𝑖 𝑃 𝐾 ( 𝐼 − 𝑠 𝑛 𝐵 ) 𝑦 𝑛 ‖ , ‖ 𝐵 𝑦 𝑛 ‖ } we then have ‖ 𝑦 𝑛 + 1 − 𝑦 𝑛 ‖ ≤ ‖ 𝑥 𝑛 + 1 − 𝑥 𝑛 | | | | 𝑟 ‖ + 1 − 𝑛 𝑟 𝑛 + 1 | | | | ‖ 𝑦 𝑛 + 1 − 𝑥 𝑛 + 1 ‖ ≤ ‖ 𝑥 𝑛 + 1 − 𝑥 𝑛 𝑀 ‖ + 𝑚 | | 𝑟 𝑛 + 1 − 𝑟 𝑛 | | . ( 3 . 1 0 ) Now, define two sequences { 𝛽 𝑛 } and { 𝑧 𝑛 } by 𝛽 𝑛 ∶ = ( 1 − 𝛿 ) 𝛼 𝑛 + 𝛿 and 𝑧 𝑛 ∶ = ( 𝑥 𝑛 + 1 − 𝑥 𝑛 + 𝛽 𝑛 𝑥 𝑛 ) / 𝛽 𝑛 . Then, 𝑧 𝑛 = 𝛼 𝑛 ∑ 𝑢 + 𝛿 𝑖 ≥ 1 𝜎 𝑖 , 𝑛 𝑇 𝑖 𝑃 𝐾 𝐼 − 𝑠 𝑛 𝐵 𝑦 𝑛 𝛽 𝑛 . ( 3 . 1 1 ) Observe that { 𝑧 𝑛 } is bounded and that ‖ 𝑧 𝑛 + 1 − 𝑧 𝑛 ‖ − ‖ 𝑥 𝑛 + 1 − 𝑥 𝑛 | | | | 𝛼 ‖ ≤ 𝑛 + 1 𝛽 𝑛 + 1 − 𝛼 𝑛 𝛽 𝑛 | | | | + 𝛿 ‖ 𝑢 ‖ 1 − 𝛼 𝑛 + 1 𝛽 𝑛 + 1 ‖ 𝑦 𝑛 + 1 − 𝑦 𝑛 ‖ − ‖ 𝑥 𝑛 + 1 − 𝑥 𝑛 ‖ + 𝛿 𝛽 𝑛 + 1 𝑖 ≥ 1 | | 𝜎 𝑖 , 𝑛 + 1 − 𝜎 𝑖 , 𝑛 | | ‖ 𝑇 𝑖 𝑃 𝐾 𝐼 − 𝑠 𝑛 + 1 𝐵 𝑦 𝑛 ‖ + 𝛿 𝛽 𝑛 + 1 𝛽 𝑛 | | 𝛽 𝑛 − 𝛽 𝑛 + 1 | | 𝑖 ≥ 1 𝜎 𝑖 , 𝑛 + 1 ‖ 𝑇 𝑖 𝑃 𝐾 𝐼 − 𝑠 𝑛 + 1 𝐵 𝑦 𝑛 ‖ + 𝛿 1 − 𝛼 𝑛 ‖ 𝐵 𝑦 𝑛 ‖ 𝛽 𝑛 | | 𝑠 𝑛 − 𝑠 𝑛 + 1 | | . ( 3 . 1 2 ) Using (3.10) we have that ‖ 𝑧 𝑛 + 1 − 𝑧 𝑛 ‖ − ‖ 𝑥 𝑛 + 1 − 𝑥 𝑛 | | | | 𝛼 ‖ ≤ 𝑛 + 1 𝛽 𝑛 + 1 − 𝛼 𝑛 𝛽 𝑛 | | | | + | | | | 𝛿 ‖ 𝑢 ‖ 1 − 𝛼 𝑛 + 1 𝛽 𝑛 + 1 | | | | − 1 ‖ 𝑥 𝑛 + 1 − 𝑥 𝑛 𝛿 ‖ + 1 − 𝛼 𝑛 + 1 𝑀 𝛽 𝑛 + 1 𝑚 | | 𝑟 𝑛 + 1 − 𝑟 𝑛 | | + 𝛿 𝑀 𝛽 𝑛 + 1 𝑖 ≥ 1 | | 𝜎 𝑖 , 𝑛 + 1 − 𝜎 𝑖 , 𝑛 | | 0 𝑥 0 2 0 0 𝑑 + 𝛿 𝑀 𝛽 𝑛 + 1 𝛽 𝑛 | | 𝛽 𝑛 − 𝛽 𝑛 + 1 | | + 𝛿 1 − 𝛼 𝑛 𝑀 𝛽 𝑛 | | 𝑠 𝑛 − 𝑠 𝑛 + 1 | | . ( 3 . 1 3 ) This implies l i m s u p 𝑛 → ∞ ‖ ‖ 𝑧 𝑛 + 1 − 𝑧 𝑛 ‖ ‖ − ‖ ‖ 𝑥 𝑛 + 1 − 𝑥 𝑛 ‖ ‖ ≤ 0 , ( 3 . 1 4 ) and by Lemma 2.4, l i m 𝑛 → ∞ ‖ 𝑧 𝑛 − 𝑥 𝑛 ‖ = 0 . Hence, ‖ ‖ 𝑥 𝑛 + 1 − 𝑥 𝑛 ‖ ‖ = 𝛽 𝑛 ‖ ‖ 𝑧 𝑛 − 𝑥 𝑛 ‖ ‖ ⟶ 0 a s 𝑛 ⟶ ∞ . ( 3 . 1 5 ) Using (3.10) we have ‖ ‖ 𝑦 𝑛 + 1 − 𝑦 𝑛 ‖ ‖ ⟶ 0 a s 𝑛 ⟶ ∞ . ( 3 . 1 6 ) We now have ‖ 𝑦 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 ∗ ⟩ = ⟨ 𝑦 𝑛 − 𝑥 ∗ , 𝑥 𝑛 − 𝑥 ∗ ⟩ = 1 2 ‖ 𝑦 𝑛 − 𝑥 ∗ ‖ 2 + ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ 2 − ‖ 𝑥 𝑛 − 𝑦 𝑛 ‖ 2 , ( 3 . 1 7 ) so that ‖ 𝑦 𝑛 − 𝑥 ∗ ‖ 2 ≤ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ 2 − ‖ 𝑥 𝑛 − 𝑦 𝑛 ‖ 2 . ( 3 . 1 8 ) But, ‖ 𝑥 𝑛 + 1 − 𝑥 ∗ ‖ 2 = ‖ ‖ ‖ ‖ 𝛼 𝑛 𝑢 − 𝑥 ∗ + ( 1 − 𝛿 ) 1 − 𝛼 𝑛 𝑥 𝑛 − 𝑥 ∗ + 𝛿 𝑖 ≥ 1 𝜎 𝑖 𝑛 𝑇 𝑖 𝑃 𝐾 𝐼 − 𝑠 𝑛 𝐵 𝑦 𝑛 − 𝑇 𝑖 𝑃 𝐾 𝐼 − 𝑠 𝑛 𝐵 𝑥 ∗ ‖ ‖ ‖ ‖ 2 ≤ 𝛼 𝑛 ‖ 𝑢 − 𝑥 ∗ ‖ 2 + ( 1 − 𝛿 ) 1 − 𝛼 𝑛 ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ 2 + 𝛿 𝑖 ≥ 1 𝜎 𝑖 𝑛 ‖ 𝑇 𝑖 𝑃 𝐾 𝐼 − 𝑠 𝑛 𝐵 𝑦 𝑛 − 𝑇 𝑖 𝑃 𝐾 𝐼 − 𝑠 𝑛 𝐵 𝑥 ∗ ‖ 2 ≤ 𝛼 𝑛 ‖ 𝑢 − 𝑥 ∗ ‖ 2 + ( 1 − 𝛿 ) 1 − 𝛼 𝑛 ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ 2 + 𝛿 1 − 𝛼 𝑛 ‖ 𝑦 𝑛 − 𝑥 ∗ ‖ 2 . ( 3 . 1 9 ) Putting (3.18) in (3.19), we have 𝛿 1 − 𝛼 𝑛 ‖ 𝑥 𝑛 − 𝑦 𝑛 ‖ 2 ≤ 𝛼 𝑛 ‖ 𝑢 − 𝑥 ∗ ‖ 2 + ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ 2 − ‖ 𝑥 𝑛 + 1 − 𝑥 ∗ ‖ 2 = 𝛼 𝑛 ‖ 𝑢 − 𝑥 ∗ ‖ 2 + ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ − ‖ 𝑥 𝑛 + 1 − 𝑥 ∗ ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ + ‖ 𝑥 𝑛 + 1 − 𝑥 ∗ ‖ ≤ 𝛼 𝑛 ‖ 𝑢 − 𝑥 ∗ ‖ 2 + ‖ 𝑥 𝑛 − 𝑥 𝑛 + 1 ‖ ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ + ‖ 𝑥 𝑛 + 1 − 𝑥 ∗ ‖ , ( 3 . 2 0 ) and hence, l i m 𝑛 → ∞ ‖ 𝑥 𝑛 − 𝑦 𝑛 ‖ = 0 . ( 3 . 2 1 ) Observe also that if 𝑤 𝑛 ∶ = 𝑃 𝐾 ( 𝐼 − 𝑠 𝑛 𝐵 ) 𝑦 𝑛 , then ‖ 𝑤 𝑛 − 𝑥 ∗ ‖ 2 = ‖ 𝑃 𝐾 𝐼 − 𝑠 𝑛 𝐵 𝑦 𝑛 − 𝑃 𝐾 𝐼 − 𝑠 𝑛 𝐵 𝑥 ∗ ‖ 2 ≤ ‖ 𝑦 𝑛 − 𝑥 ∗ − 𝑠 𝑛 𝐵 𝑦 𝑛 − 𝐵 𝑥 ∗ ‖ 2 = ‖ 𝑦 𝑛 − 𝑥 ∗ ‖ 2 − 2 𝑠 𝑛 ⟨ 𝑦 𝑛 − 𝑥 ∗ , 𝐵 𝑦 𝑛 − 𝐵 𝑥 ∗ ⟩ + 𝑠 2 𝑛 ‖ 𝐵 𝑦 𝑛 − 𝐵 𝑥 ∗ ‖ 2 ≤ ‖ 𝑦 𝑛 − 𝑥 ∗ ‖ 2 − 2 𝑠 𝑛 − 𝜆 ‖ 𝐵 𝑦 𝑛 − 𝐵 𝑥 ∗ ‖ 2 + 𝛾 ‖ 𝑦 𝑛 − 𝑥 ∗ ‖ 2 + 𝑠 2 𝑛 ‖ 𝐵 𝑦 𝑛 − 𝐵 𝑥 ∗ ‖ 2 = ‖ 𝑦 𝑛 − 𝑥 ∗ ‖ 2 + 2 𝑠 𝑛 𝜆 ‖ 𝐵 𝑦 𝑛 − 𝐵 𝑥 ∗ ‖ 2 − 2 𝑠 𝑛 𝛾 ‖ 𝑦 𝑛 − 𝑥 ∗ ‖ 2 + 𝑠 2 𝑛 ‖ 𝐵 𝑦 𝑛 − 𝐵 𝑥 ∗ ‖ 2 ≤ ‖ 𝑦 𝑛 − 𝑥 ∗ ‖ 2 + 2 𝑠 𝑛 𝜆 + 𝑠 2 𝑛 − 2 𝑠 𝑛 𝛾 𝜇 2 ‖ 𝐵 𝑦 𝑛 − 𝐵 𝑥 ∗ ‖ 2 . ( 3 . 2 2 ) Also, ‖ 𝑥 𝑛 + 1 − 𝑥 ∗ ‖ 2 = ‖ ‖ ‖ ‖ 𝛼 𝑛 𝑢 − 𝑥 ∗ + ( 1 − 𝛿 ) 1 − 𝛼 𝑛 𝑥 𝑛 − 𝑥 ∗ + 𝛿 𝑖 ≥ 1 𝜎 𝑖 𝑛 𝑇 𝑖 𝑃 𝐾 𝐼 − 𝑠 𝑛 𝐵 𝑦 𝑛 − 𝑇 𝑖 𝑃 𝐾 𝐼 − 𝑠 𝑛 𝐵 𝑥 ∗ ‖ ‖ ‖ ‖ 2 ≤ 𝛼 𝑛 ‖ 𝑢 − 𝑥 ∗ ‖ 2 + ( 1 − 𝛿 ) 1 − 𝛼 𝑛 ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ 2 + 𝛿 𝑖 ≥ 1 𝜎 𝑖 𝑛 ‖ 𝑇 𝑖 𝑃 𝐾 𝐼 − 𝑠 𝑛 𝐵 𝑦 𝑛 − 𝑇 𝑖 𝑃 𝐾 𝐼 − 𝑠 𝑛 𝐵 𝑥 ∗ ‖ 2 = 𝛼 𝑛 ‖ 𝑢 − 𝑥 ∗ ‖ 2 + ( 1 − 𝛿 ) 1 − 𝛼 𝑛 ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ 2 + 𝛿 𝑖 ≥ 1 𝜎 𝑖 𝑛 ‖ 𝑇 𝑖 𝑤 𝑛 − 𝑥 ∗ ‖ 2 ≤ 𝛼 𝑛 ‖ 𝑢 − 𝑥 ∗ ‖ 2 + ( 1 − 𝛿 ) 1 − 𝛼 𝑛 ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ 2 + 𝛿 1 − 𝛼 𝑛 ‖ 𝑤 𝑛 − 𝑥 ∗ ‖ 2 ≤ 𝛼 𝑛 ‖ 𝑢 − 𝑥 ∗ ‖ 2 + ( 1 − 𝛿 ) 1 − 𝛼 𝑛 ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ 2 + 𝛿 1 − 𝛼 𝑛 ‖ 𝑦 𝑛 − 𝑥 ∗ ‖ 2 + 2 𝑠 𝑛 𝜆 + 𝑠 2 𝑛 − 2 𝑠 𝑛 𝛾 𝜇 2 ‖ 𝐵 𝑦 𝑛 − 𝐵 𝑥 ∗ ‖ 2 ≤ 𝛼 𝑛 ‖ 𝑢 − 𝑥 ∗ ‖ 2 + 1 − 𝛼 𝑛 ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ 2 + 𝛿 1 − 𝛼 𝑛 2 𝑠 𝑛 𝜆 + 𝑠 2 𝑛 − 2 𝑠 𝑛 𝛾 𝜇 2 ‖ 𝐵 𝑦 𝑛 − 𝐵 𝑥 ∗ ‖ 2 , ( 3 . 2 3 ) which implies − 𝛿 1 − 𝛼 𝑛 2 𝑠 𝑛 𝜆 + 𝑠 2 𝑛 − 2 𝑠 𝑛 𝛾 𝜇 2 ‖ 𝐵 𝑦 𝑛 − 𝐵 𝑥 ∗ ‖ 2 ≤ 𝛼 𝑛 ‖ 𝑢 − 𝑥 ∗ ‖ 2 + ‖ 𝑥 𝑛 − 𝑥 ∗ ‖ 2 − ‖ 𝑥 𝑛 + 1 − 𝑥 ∗ ‖ 2 ( 3 . 2 4 ) so that l i m 𝑛 → ∞ ‖ 𝐵 𝑦 𝑛 − 𝐵 𝑥 ∗ ‖ = 0 . ( 3 . 2 5 )
We go further to prove that for each 𝑖 ∈ ℕ , l i m 𝑛 → ∞ ‖ 𝑇 𝑖 𝑃