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Journal of Applied Mathematics
Volume 2012 (2012), Article ID 812783, 10 pages
doi:10.1155/2012/812783
Research Article

Approximation of Solutions of an Equilibrium Problem in a Banach Space

Hecai Yuan1 and Guohong Shi2

1School of Mathematics and Information Science, North China University of Water Resources and Electric Power, Zhengzhou 450011, China
2College of Science, Hebei University of Engineering, Handan 056038, China

Received 15 December 2011; Accepted 4 February 2012

Academic Editor: Rudong Chen

Copyright © 2012 Hecai Yuan and Guohong Shi. 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

An equilibrium problem is investigated based on a hybrid projection iterative algorithm. Strong convergence theorems for solutions of the equilibrium problem are established in a strictly convex and uniformly smooth Banach space which also enjoys the Kadec-Klee property.

1. Introduction

Equilibrium problems which were introduced by Fan [1] and Blum and Oettli [2] have had a great impact and influence on the development of several branches of pure and applied sciences. It has been shown that the equilibrium problem theory provides a novel and unified treatment of a wide class of problems which arise in economics, finance, image reconstruction, ecology, transportation, network, elasticity, and optimization. It has been shown [3–8] that equilibrium, problems include variational inequalities, fixed point, the Nash equilibrium, and game theory as special cases. A number of iterative algorithms have recently been studying for fixed point and equilibrium problems, see [9–26] and the references therein. However, there were few results established in the framework of the Banach spaces. In this paper, we suggest and analyze a projection iterative algorithm for finding solutions of equilibrium in a Banach space.

2. Preliminaries

In what follows, we always assume that 𝐸 is a Banach space with the dual space 𝐸 ∗ . Let 𝐶 be a nonempty, closed, and convex subset of 𝐸 . We use the symbol 𝐽 to stand for the normalized duality mapping from 𝐸 to 2 𝐸 ∗ defined by 𝑓 𝐽 𝑥 = ∗ ∈ 𝐸 ∗ ∶ ⟨ 𝑥 , 𝑓 ∗ ⟩ = ‖ 𝑥 ‖ 2 = ‖ 𝑓 ∗ ‖ 2 , ∀ 𝑥 ∈ 𝐸 , ( 2 . 1 )

where ⟨ ⋅ , ⋅ ⟩ denotes the generalized duality pairing of elements between 𝐸 and 𝐸 ∗ .

Let 𝑈 𝐸 = { 𝑥 ∈ 𝐸 ∶ ‖ 𝑥 ‖ = 1 } be the unit sphere of 𝐸 . 𝐸 is said to be strictly convex if ‖ ( 𝑥 + 𝑦 ) / 2 ‖ < 1 for all 𝑥 , 𝑦 ∈ 𝑈 𝐸 with 𝑥 ≠ 𝑦 . It is said to be uniformly convex if for any 𝜖 ∈ ( 0 , 2 ] there exists 𝛿 > 0 such that for any 𝑥 , 𝑦 ∈ 𝑈 𝐸 , ‖ ‖ ‖ ‖ 𝑥 − 𝑦 ‖ ≥ 𝜖 i m p l i e s 𝑥 + 𝑦 2 ‖ ‖ ‖ ≤ 1 − 𝛿 . ( 2 . 2 )

It is known that a uniformly convex Banach space is reflexive and strictly convex; for details see [27] and the references therein.

Recall that a Banach space 𝐸 is said to have the Kadec-Klee property if a sequence { 𝑥 𝑛 } of 𝐸 satisfies that 𝑥 𝑛 ⇀ 𝑥 ∈ 𝐶 , where ⇀ denotes the weak convergence, and ‖ 𝑥 𝑛 ‖ → ‖ 𝑥 ‖ , where → denotes the strong convergence, and then 𝑥 𝑛 → 𝑥 . It is known that if 𝐸 is uniformly convex, then 𝐸 enjoys the Kadec-Klee property; for details see [26] and the references therein.

𝐸 is said to be smooth provided l i m 𝑡 → 0 ( ‖ 𝑥 + 𝑡 𝑦 ‖ − ‖ 𝑥 ‖ ) / 𝑡 exists for all 𝑥 , 𝑦 ∈ 𝑈 𝐸 . It is also said to be uniformly smooth if the limit is attained uniformly for all 𝑥 , 𝑦 ∈ 𝑈 𝐸 .

It is well known that if 𝐸 ∗ is strictly convex, then 𝐽 is single valued; if 𝐸 ∗ is reflexive, and smooth, then 𝐽 is single valued and demicontinuous; for more details see [27, 28] and the references therein.

It is also well known that if 𝐷 is a nonempty, closed, and convex subset of a Hilbert space 𝐻 , and 𝑃 𝐷 ∶ 𝐻 → 𝐷 is the metric projection from 𝐻 onto 𝐷 , then 𝑃 𝐷 is nonexpansive. This fact actually characterizes the Hilbert spaces, and consequently, it is not available in more general Banach spaces. In this connection, Alber [29] introduced a generalized projection operator Π 𝐷 in the Banach spaces which is an analogue of the metric projection in the Hilbert spaces.

Let 𝐸 be a smooth Banach space. Consider the functional defined by 𝜙 ( 𝑥 , 𝑦 ) = ‖ 𝑥 ‖ 2 − 2 ⟨ 𝑥 , 𝐽 𝑦 ⟩ + ‖ 𝑦 ‖ 2 , ∀ 𝑥 , 𝑦 ∈ 𝐸 . ( 2 . 3 ) Notice that, in a Hilbert space 𝐻 , (2.3) is reduced to 𝜙 ( 𝑥 , 𝑦 ) = ‖ 𝑥 − 𝑦 ‖ 2 for all 𝑥 , 𝑦 ∈ 𝐻 . The generalized projection Π 𝐶 ∶ 𝐸 → 𝐶 is a mapping that is assigned to an arbitrary point 𝑥 ∈ 𝐸 , the minimum point of the functional 𝜙 ( 𝑥 , 𝑦 ) ; that is, Π 𝐶 𝑥 = 𝑥 , where 𝑥 is the solution to the following minimization problem: 𝜙 𝑥 , 𝑥 = m i n 𝑦 ∈ 𝐶 𝜙 ( 𝑦 , 𝑥 ) . ( 2 . 4 )

The existence and uniqueness of the operator Π 𝐶 follow from the properties of the functional 𝜙 ( 𝑥 , 𝑦 ) and the strict monotonicity of the mapping 𝐽 ; see, for example, [27, 28]. In the Hilbert spaces, Π 𝐶 = 𝑃 𝐶 . It is obvious from the definition of the function 𝜙 that ( ) ‖ 𝑦 ‖ − ‖ 𝑥 ‖ 2 ) ≤ 𝜙 ( 𝑦 , 𝑥 ) ≤ ( ‖ 𝑦 ‖ + ‖ 𝑥 ‖ 2 , ∀ 𝑥 , 𝑦 ∈ 𝐸 , ( 2 . 5 ) 𝜙 ( 𝑥 , 𝑦 ) = 𝜙 ( 𝑥 , 𝑧 ) + 𝜙 ( 𝑧 , 𝑦 ) + 2 ⟨ 𝑥 − 𝑧 , 𝐽 𝑧 − 𝐽 𝑦 ⟩ , ∀ 𝑥 , 𝑦 , 𝑧 ∈ 𝐸 . ( 2 . 6 )

Let 𝑇 ∶ 𝐶 → 𝐶 be a mapping. Recall that a point 𝑝 in 𝐶 is said to be an asymptotic fixed point of 𝑇 if 𝐶 contains a sequence { 𝑥 𝑛 } which converges weakly to 𝑝 such that l i m 𝑛 → ∞ ‖ 𝑥 𝑛 − 𝑇 𝑥 𝑛 ‖ = 0 . The set of asymptotic fixed points of 𝑇 will be denoted by 𝐹 ( 𝑇 ) . 𝑇 is said to be relatively nonexpansive if 𝐹 ( 𝑇 ) = 𝐹 ( 𝑇 ) ≠ ∅ , 𝜙 ( 𝑝 , 𝑇 𝑥 ) ≤ 𝜙 ( 𝑝 , 𝑥 ) , ∀ 𝑥 ∈ 𝐶 , ∀ 𝑝 ∈ 𝐹 ( 𝑇 ) . ( 2 . 7 )

The asymptotic behavior of a relatively nonexpansive mapping was studied in [27, 29, 30].

Let 𝑓 be a bifunction from 𝐶 × 𝐶 to ℝ , where ℝ denotes the set of real numbers. In this paper, we consider the following equilibrium problem. Find 𝑝 ∈ 𝐶 such that 𝑓 ( 𝑝 , 𝑦 ) ≥ 0 , ∀ 𝑦 ∈ 𝐶 . ( 2 . 8 ) We use E P ( 𝑓 ) to denote the solution set of the equilibrium problem (2.3). That is, E P ( 𝑓 ) = { 𝑝 ∈ 𝐶 ∶ 𝑓 ( 𝑝 , 𝑦 ) ≥ 0 , ∀ 𝑦 ∈ 𝐶 } . ( 2 . 9 )

Given a mapping 𝑄 ∶ 𝐶 → 𝐸 ∗ , let 𝑓 ( 𝑥 , 𝑦 ) = ⟨ 𝑄 𝑥 , 𝑦 − 𝑥 ⟩ , ∀ 𝑥 , 𝑦 ∈ 𝐶 . ( 2 . 1 0 )

Then 𝑝 ∈ E P ( 𝑓 ) if and only if 𝑝 is a solution of the following variational inequality. Find 𝑝 such that ⟨ 𝑄 𝑝 , 𝑦 − 𝑝 ⟩ ≥ 0 , ∀ 𝑦 ∈ 𝐶 . ( 2 . 1 1 )

To study the equilibrium problem (2.8), we may assume that 𝑓 satisfies the following conditions:

(A1) 𝑓 ( 𝑥 , 𝑥 ) = 0 , f o r a l l 𝑥 ∈ 𝐶 ;

(A2) 𝑓 is monotone, that is, 𝑓 ( 𝑥 , 𝑦 ) + 𝑓 ( 𝑦 , 𝑥 ) ≤ 0 , f o r a l l 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 ; ( 2 . 1 2 )

(A4) for each 𝑥 ∈ 𝐶 , 𝑦 ↦ 𝑓 ( 𝑥 , 𝑦 ) is convex and weakly lower semicontinuous.

In this paper, we study the problem of approximating solutions of equilibrium problem (2.8) based on a hybrid projection iterative algorithm in a strictly convex and uniformly smooth Banach space which also enjoys the Kadec-Klee property. To prove our main results, we need the following lemmas.

Lemma 2.1. Let 𝐸 be a strictly convex and uniformly smooth Banach space and 𝐶 a nonempty, closed, and convex subset of 𝐸 . Let 𝑓 be a bifunction from 𝐶 × 𝐶 to ℝ satisfying (A1)–(A4). Let 𝑟 > 0 and 𝑥 ∈ 𝐸 . Then (a)(see [2]). There exists 𝑧 ∈ 𝐶 such that 1 𝑓 ( 𝑧 , 𝑦 ) + 𝑟 ⟨ 𝑦 − 𝑧 , 𝐽 𝑧 − 𝐽 𝑥 ⟩ ≥ 0 , ∀ 𝑦 ∈ 𝐶 . ( 2 . 1 3 ) (b)(see [31]). Define a mapping 𝑇 𝑓 𝑟 ∶ 𝐸 → 𝐶 by 𝑇 𝑓 𝑟 1 𝑥 = 𝑧 ∈ 𝐶 ∶ 𝑓 ( 𝑧 , 𝑦 ) + 𝑟 . ⟨ 𝑦 − 𝑧 , 𝐽 𝑧 − 𝐽 𝑥 ⟩ , ∀ 𝑦 ∈ 𝐶 ( 2 . 1 4 ) Then the following conclusions hold:(1) 𝑇 𝑓 𝑟 is single valued;(2) 𝑇 𝑓 𝑟 is a firmly nonexpansive-type mapping; that is, for all 𝑥 , 𝑦 ∈ 𝐸 , 𝑇 𝑓 𝑟 𝑥 − 𝑇 𝑓 𝑟 𝑦 , 𝐽 𝑇 𝑓 𝑟 𝑥 − 𝐽 𝑇 𝑓 𝑟 𝑦 ≤ 𝑇 𝑓 𝑟 𝑥 − 𝑇 𝑓 𝑟 𝑦 , 𝐽 𝑥 − 𝐽 𝑦 ; ( 2 . 1 5 ) (3) 𝐹 ( 𝑇 𝑓 𝑟 ) = E P ( 𝑓 ) ; (4) E P ( 𝑓 ) is closed and convex;(5) 𝑇 𝑓 𝑟 is relatively nonexpansive.

Lemma 2.2 (see [29]). Let 𝐸 be a reflexive, strictly convex, and smooth Banach space and 𝐶 a nonempty, closed, and convex subset of 𝐸 . Let 𝑥 ∈ 𝐸 , and 𝑥 0 ∈ 𝐶 . Then 𝑥 0 = Π 𝐶 𝑥 if and only if

⟨ 𝑥 0 − 𝑦 , 𝐽 𝑥 − 𝐽 𝑥 0 ⟩ ≥ 0 , ∀ 𝑦 ∈ 𝐶 . ( 2 . 1 6 )

Lemma 2.3 (see [29]). Let 𝐸 be a reflexive, strictly convex, and smooth Banach space and 𝐶 a nonempty, closed, and convex subset of 𝐸 , and 𝑥 ∈ 𝐸 . Then

𝜙 𝑦 , Π 𝐶 𝑥 Π + 𝜙 𝐶 𝑥 , 𝑥 ≤ 𝜙 ( 𝑦 , 𝑥 ) , ∀ 𝑦 ∈ 𝐶 . ( 2 . 1 7 )

Lemma 2.4 (see [27]). Let 𝐸 be a reflexive, strictly convex, and smooth Banach space. Then one has the following

𝜙 ( 𝑥 , 𝑦 ) = 0 ⟺ 𝑥 = 𝑦 , ∀ 𝑥 , 𝑦 ∈ 𝐸 . ( 2 . 1 8 )

3. Main Results

Theorem 3.1. Let 𝐸 be a strictly convex and uniformly smooth Banach space which also enjoys the Kadec-Klee property and 𝐶 a nonempty, closed, and convex subset of 𝐸 . Let 𝑓 be a bifunction from 𝐶 × 𝐶 to ℝ satisfying (A1)–(A4) such that E P ( 𝑓 ) ≠ ∅ . Let { 𝑥 𝑛 } be a sequence generated by the following manner: 𝑥 0 C ∈ 𝐸 c h o s e n a r b i t r a r i l y , 1 𝑥 = 𝐶 , 1 = Π 𝐶 1 𝑥 0 , 𝑦 𝑛 𝑦 ∈ 𝐶 , s u c h t h a t 𝑓 𝑛 + 1 , 𝑢 𝑟 𝑛 ⟨ 𝑢 − 𝑦 𝑛 , 𝐽 𝑦 𝑛 − 𝐽 𝑥 𝑛 𝐶 ⟩ ≥ 0 , ∀ 𝑢 ∈ 𝐶 , 𝑛 + 1 = 𝑢 ∈ 𝐶 𝑛 ∶ 2 ⟨ 𝑥 𝑛 − 𝑢 , 𝐽 𝑥 𝑛 − 𝐽 𝑦 𝑛 𝑥 ⟩ ≥ 𝜙 𝑛 , 𝑦 𝑛 , 𝑥 𝑛 + 1 = Π 𝐶 𝑛 + 1 𝑥 0 , ∀ 𝑛 ≥ 1 , ( 3 . 1 ) where { 𝑟 𝑛 } is a real number sequence in [ 𝑟 , ∞ ) , where 𝑟 is some positive real number. Then the sequence { 𝑥 𝑛 } converges strongly to 𝑥 = Π E P ( 𝑓 ) 𝑥 0 .

Proof. In view of Lemma 2.1, we see that E P ( 𝑓 ) is closed and convex. Next, we show that 𝐶 𝑛 is closed and convex. It is not hard to see that 𝐶 𝑛 is closed. Therefore, we only show that 𝐶 𝑛 is convex. It is obvious that 𝐶 1 = 𝐶 is convex. Suppose that 𝐶 ℎ is convex for some ℎ ∈ ℕ . Next, we show that 𝐶 ℎ + 1 is also convex for the same ℎ . Let 𝑎 , 𝑏 ∈ 𝐶 ℎ + 1 and 𝑐 = 𝑡 𝑎 + ( 1 − 𝑡 ) 𝑏 , where 𝑡 ∈ ( 0 , 1 ) . It follows that 𝜙 𝑥 ℎ , 𝑦 ℎ ≤ 2 ⟨ 𝑥 ℎ − 𝑎 , 𝐽 𝑥 ℎ − 𝐽 𝑦 ℎ 𝑥 ⟩ , 𝜙 ℎ , 𝑦 ℎ ≤ 2 ⟨ 𝑥 ℎ − 𝑏 , 𝐽 𝑥 ℎ − 𝐽 𝑦 ℎ ⟩ , ( 3 . 2 ) where 𝑎 , 𝑏 ∈ 𝐶 ℎ . From the above two inequalities, we can get that 𝜙 𝑥 ℎ , 𝑦 ℎ ≤ 2 ⟨ 𝑥 ℎ − 𝑐 , 𝐽 𝑥 ℎ − 𝐽 𝑦 ℎ ⟩ , ( 3 . 3 ) where 𝑐 ∈ 𝐶 ℎ . It follows that 𝐶 ℎ + 1 is closed and convex. This completes the proof that 𝐶 𝑛 is closed, and convex.
Next, we show that E P ( 𝑓 ) ⊂ 𝐶 𝑛 . It is obvious that E P ( 𝑓 ) ⊂ 𝐶 = 𝐶 1 . Suppose that E P ( 𝑓 ) ⊂ 𝐶 ℎ for some ℎ ∈ ℕ . For any 𝑧 ∈ E P ( 𝑓 ) ⊂ 𝐶 ℎ , we see from Lemma 2.1 that 𝜙 𝑧 , 𝑦 ℎ ≤ 𝜙 𝑧 , 𝑥 ℎ . ( 3 . 4 ) On the other hand, we obtain from (2.6) that 𝜙 𝑧 , 𝑦 ℎ = 𝜙 𝑧 , 𝑥 ℎ 𝑥 + 𝜙 ℎ , 𝑦 ℎ + 2 ⟨ 𝑧 − 𝑥 ℎ , 𝐽 𝑥 ℎ − 𝐽 𝑦 ℎ ⟩ . ( 3 . 5 ) Combining (3.4) with (3.5), we arrive at 2 ⟨ 𝑥 ℎ − 𝑧 , 𝐽 𝑥 ℎ − 𝐽 𝑦 ℎ 𝑥 ⟩ ≥ 𝜙 ℎ , 𝑦 ℎ ( 3 . 6 ) which implies that 𝑧 ∈ 𝐶 ℎ + 1 . This shows that E P ( 𝑓 ) ⊂ 𝐶 ℎ + 1 . This completes the proof that E P ( 𝑓 ) ⊂ 𝐶 𝑛 .
Next, we show that { 𝑥 𝑛 } is a convergent sequence and strongly converges to 𝑥 , where 𝑥 ∈ E P ( 𝑓 ) . Since 𝑥 𝑛 = Π 𝐶 𝑛 𝑥 0 , we see from Lemma 2.2 that ⟨ 𝑥 𝑛 − 𝑧 , 𝐽 𝑥 0 − 𝐽 𝑥 𝑛 ⟩ ≥ 0 , ∀ 𝑧 ∈ 𝐶 𝑛 . ( 3 . 7 ) It follows from E P ( 𝑓 ) ⊂ 𝐶 𝑛 that ⟨ 𝑥 𝑛 − 𝑤 , 𝐽 𝑥 0 − 𝐽 𝑥 𝑛 ⟩ ≥ 0 , ∀ 𝑤 ∈ E P ( 𝑓 ) . ( 3 . 8 ) By virtue of Lemma 2.3, we obtain that 𝜙 𝑥 𝑛 , 𝑥 0 Π = 𝜙 𝐶 𝑛 𝑥 0 , 𝑥 0 Π ≤ 𝜙 E P ( 𝑓 ) 𝑥 0 , 𝑥 0 Π − 𝜙 E P ( 𝑓 ) 𝑥 0 , 𝑥 𝑛 Π ≤ 𝜙 E P ( 𝑓 ) 𝑥 0 , 𝑥 0 . ( 3 . 9 ) This implies that the sequence { 𝜙 ( 𝑥 𝑛 , 𝑥 0 ) } is bounded. It follows from (2.5) that the sequence { 𝑥 𝑛 } is also bounded. Since the space is reflexive, we may assume that 𝑥 𝑛 ⇀ 𝑥 . Since 𝐶 𝑛 is closed and convex, we see that 𝑥 ∈ 𝐶 𝑛 . On the other hand, we see from the weakly lower semicontinuity of the norm that 𝜙 𝑥 , 𝑥 0 = ‖ ‖ 𝑥 ‖ ‖ 2 − 2 𝑥 , 𝐽 𝑥 0 + ‖ ‖ 𝑥 0 ‖ ‖ 2 ≤ l i m i n f 𝑛 → ∞ ‖ ‖ 𝑥 𝑛 ‖ ‖ 2 − 2 ⟨ 𝑥 𝑛 , 𝐽 𝑥 0 ‖ ‖ 𝑥 ⟩ + 0 ‖ ‖ 2 = l i m i n f 𝑛 → ∞ 𝜙 𝑥 𝑛 , 𝑥 0 ≤ l i m s u p 𝑛 → ∞ 𝜙 𝑥 𝑛 , 𝑥 0 ≤ 𝜙 𝑥 , 𝑥 0 , ( 3 . 1 0 ) which implies that 𝜙 ( 𝑥 𝑛 , 𝑥 0 ) → 𝜙 ( 𝑥 , 𝑥 0 ) as 𝑛 → ∞ . Hence, ‖ 𝑥 𝑛 ‖ → ‖ 𝑥 ‖ as 𝑛 → ∞ . In view of the Kadec-Klee property of 𝐸 , we see that 𝑥 𝑛 → 𝑥 as 𝑛 → ∞ . Notice that 𝑥 𝑛 + 1 = Π E P ( 𝑓 ) 𝑥 0 ∈ 𝐶 𝑛 + 1 ⊂ 𝐶 𝑛 . It follows that 𝜙 𝑥 𝑛 + 1 , 𝑥 𝑛 𝑥 = 𝜙 𝑛 + 1 , Π 𝐶 𝑛 𝑥 0 𝑥 ≤ 𝜙 𝑛 + 1 , 𝑥 0 Π − 𝜙 𝐶 𝑛 𝑥 0 , 𝑥 0 𝑥 = 𝜙 𝑛 + 1 , 𝑥 0 𝑥 − 𝜙 𝑛 , 𝑥 0 . ( 3 . 1 1 ) Since 𝑥 𝑛 = Π 𝐶 𝑛 𝑥 0 and 𝑥 𝑛 + 1 = Π 𝐶 𝑛 + 1 𝑥 0 ∈ 𝐶 𝑛 + 1 ⊂ 𝐶 𝑛 , we arrive at 𝜙 ( 𝑥 𝑛 , 𝑥 0 ) ≤ 𝜙 ( 𝑥 𝑛 + 1 , 𝑥 0 ) . This shows that { 𝜙 ( 𝑥 𝑛 , 𝑥 0 ) } is nondecreasing. It follows from the boundedness that l i m 𝑛 → ∞ 𝜙 ( 𝑥 , 𝑥 0 ) exists. It follows that l i m 𝑛 → ∞ 𝜙 𝑥 𝑛 + 1 , 𝑥 𝑛 = 0 . ( 3 . 1 2 ) By virtue of 𝑥 𝑛 + 1 = Π 𝐶 𝑛 + 1 𝑥 0 ∈ 𝐶 𝑛 + 1 , we find that 𝜙 𝑥 𝑛 , 𝑦 𝑛 𝑥 ≤ 2 𝑛 − 𝑥 𝑛 + 1 , 𝐽 𝑥 𝑛 − 𝐽 𝑦 𝑛 . ( 3 . 1 3 ) It follows that l i m 𝑛 → ∞ 𝜙 𝑥 𝑛 , 𝑦 𝑛 = 0 . ( 3 . 1 4 ) In view of (2.5), we see that l i m 𝑛 → ∞ ‖ ‖ 𝑥 𝑛 ‖ ‖ − ‖ ‖ 𝑦 𝑛 ‖ ‖ = 0 . ( 3 . 1 5 ) Since 𝑥 𝑛 → 𝑥 , we find that l i m 𝑛 → ∞ ‖ ‖ 𝑦 𝑛 ‖ ‖ = ‖ ‖ 𝑥 ‖ ‖ . ( 3 . 1 6 ) It follows that l i m 𝑛 → ∞ ‖ ‖ 𝐽 𝑦 𝑛 ‖ ‖ = ‖ ‖ 𝐽 𝑥 ‖ ‖ . ( 3 . 1 7 ) This implies that { 𝐽 𝑦 𝑛 } is bounded. Note that both 𝐸 and 𝐸 ∗ are reflexive. We may assume that 𝐽 𝑦 𝑛 ⇀ 𝑦 ∗ ∈ 𝐸 ∗ . In view of the reflexivity of 𝐸 , we see that there exists an element 𝑦 ∈ 𝐸 such that 𝐽 𝑦 = 𝑦 ∗ . It follows that 𝜙 𝑥 𝑛 , 𝑦 𝑛 = ‖ ‖ 𝑥 𝑛 ‖ ‖ 2 − 2 ⟨ 𝑥 𝑛 , 𝐽 𝑦 𝑛 ‖ ‖ 𝑦 ⟩ + 𝑛 ‖ ‖ 2 = ‖ ‖ 𝑥 𝑛 ‖ ‖ 2 − 2 ⟨ 𝑥 𝑛 , 𝐽 𝑦 𝑛 ‖ ‖ ⟩ + 𝐽 𝑦 𝑛 ‖ ‖ 2 . ( 3 . 1 8 ) Taking l i m i n f 𝑛 → ∞ on the both sides of the equality above yields that ‖ ‖ 0 ≥ 𝑥 ‖ ‖ 2 − 2 ⟨ 𝑥 , 𝑦 ∗ ⟩ + ‖ 𝑦 ∗ ‖ 2 = ‖ ‖ 𝑥 ‖ ‖ 2 − 2 ⟨ 𝑥 , 𝐽 𝑦 ⟩ + ‖ 𝐽 𝑦 ‖ 2 = ‖ ‖ 𝑥 ‖ ‖ 2 − 2 ⟨ 𝑥 , 𝐽 𝑦 ⟩ + ‖ 𝑦 ‖ 2 = 𝜙 . 𝑥 , 𝑦 ( 3 . 1 9 ) That is, 𝑥 = 𝑦 , which in turn implies that 𝑦 ∗ = 𝐽 𝑥 . It follows that 𝐽 𝑦 𝑛 ⇀ 𝐽 𝑥 ∈ 𝐸 ∗ . Since 𝐸 ∗ enjoys the Kadec-Klee property, we obtain from (3.17) that l i m 𝑛 → ∞ 𝐽 𝑦 𝑛 = 𝐽 𝑥 . Since 𝐽 − 1 ∶ 𝐸 ∗ → 𝐸 is demicontinuous, we find that 𝑦 𝑛 ⇀ 𝑥 . This implies from (3.16) and the Kadec-Klee property of 𝐸 that l i m 𝑛 → ∞ 𝑦 𝑛 = 𝑥 . This in turn implies that l i m 𝑛 → ∞ ‖ 𝑦 𝑛 − 𝑥 𝑛 ‖ = 0 . Since 𝐽 is uniformly norm-to-norm continuous on any bounded sets, we find that l i m 𝑛 → ∞ ‖ ‖ 𝐽 𝑦 𝑛 − 𝐽 𝑥 𝑛 ‖ ‖ = 0 . ( 3 . 2 0 ) Next, we show that 𝑥 ∈ 𝐸 𝐹 ( 𝑓 ) . In view of Lemma 2.1, we find from 𝑦 𝑛 = 𝑇 𝑓 𝑟 𝑛 𝑥 𝑛 that 𝑓 𝑦 𝑛 + 1 , 𝑢 𝑟 𝑛 ⟨ 𝑢 − 𝑦 𝑛 , 𝐽 𝑦 𝑛 − 𝐽 𝑥 𝑛 ⟩ ≥ 0 , ∀ 𝑢 ∈ 𝐶 . ( 3 . 2 1 ) It follows from condition (A2) and (3.20) that 1 𝑟 𝑛 ‖ ‖ 𝑢 − 𝑦 𝑛 ‖ ‖ ‖ ‖ 𝐽 𝑦 𝑛 − 𝐽 𝑥 𝑛 ‖ ‖ ≥ 𝑓 𝑢 , 𝑦 𝑛 , ∀ 𝑢 ∈ 𝐶 . ( 3 . 2 2 ) In view of condition (A4), we obtain from (3.17) that 𝑓 𝑢 , 𝑥 ≤ 0 , ∀ 𝑢 ∈ 𝐶 . ( 3 . 2 3 ) For 0 < 𝑡 < 1 and 𝑢 ∈ 𝐶 , define 𝑢 𝑡 = 𝑡 𝑢 + ( 1 − 𝑡 ) 𝑥 . It follows that 𝑢 𝑡 ∈ 𝐶 , which yields that 𝑓 ( 𝑢 𝑡 , 𝑥 ) ≤ 0 . It follows from conditions (A1) and (A4) that 𝑢 0 = 𝑓 𝑡 , 𝑢 𝑡 𝑢 ≤ 𝑡 𝑓 𝑡 + 𝑢 , 𝑢 ( 1 − 𝑡 ) 𝑓 𝑡 , 𝑥 𝑢 ≤ 𝑡 𝑓 𝑡 , 𝑢 . ( 3 . 2 4 ) That is, 𝑓 𝑢 𝑡 , 𝑢 ≥ 0 . ( 3 . 2 5 ) Letting 𝑡 ↓ 0 , we find from condition (A3) that 𝑓 ( 𝑥 , 𝑢 ) ≥ 0 , f o r a l l 𝑢 ∈ 𝐶 . This implies that 𝑥 ∈ E P ( 𝑓 ) . This shows that 𝑥 ∈ E P ( 𝑓 ) .
Finally, we prove that 𝑥 = Π E P ( 𝑓 ) 𝑥 0 . Letting 𝑛 → ∞ in (3.8), we see that 𝑥 − 𝑤 , 𝐽 𝑥 0 − 𝐽 𝑥 ≥ 0 , ∀ 𝑤 ∈ E P ( 𝑓 ) . ( 3 . 2 6 ) In view of Lemma 2.2, we can obtain that 𝑥 = Π E P ( 𝑓 ) 𝑥 0 . This completes the proof.

In the framework of the Hilbert spaces, we have the following.

Corollary 3.2. Let 𝐸 be a Hilbert space and 𝐶 a nonempty, closed, and convex subset of 𝐸 . Let 𝑓 be a bifunction from 𝐶 × 𝐶 to ℝ satisfying (A1)–(A4) such that E P ( 𝑓 ) ≠ ∅ . Let { 𝑥 𝑛 } be a sequence generated by the following manner: 𝑥 0 𝐶 ∈ 𝐸 c h o s e n a r b i t r a r i l y , 1 𝑥 = 𝐶 , 1 = 𝑃 𝐶 1 𝑥 0 , 𝑦 𝑛 𝑦 ∈ 𝐶 , s u c h t h a t 𝑓 𝑛 + 1 , 𝑢 𝑟 𝑛 ⟨ 𝑢 − 𝑦 𝑛 , 𝑦 𝑛 − 𝑥 𝑛 𝐶 ⟩ ≥ 0 , ∀ 𝑢 ∈ 𝐶 , 𝑛 + 1 = 𝑢 ∈ 𝐶 𝑛 ∶ 2 ⟨ 𝑥 𝑛 − 𝑢 , 𝑥 𝑛 − 𝑦 𝑛 ‖ ‖ 𝑥 ⟩ ≥ 𝑛 − 𝑦 𝑛 ‖ ‖ 2 , 𝑥 𝑛 + 1 = 𝑃 𝐶 𝑛 + 1 𝑥 0 , ∀ 𝑛 ≥ 1 , ( 3 . 2 7 ) where { 𝑟 𝑛 } is a real number sequence in [ 𝑟 , ∞ ) , where 𝑟 is some positive real number. Then the sequence { 𝑥 𝑛 } converges strongly to 𝑥 = 𝑃 E P ( 𝑓 ) 𝑥 0 .

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