Abstract and Applied Analysis
Volume 2013, Issue 1 619762
Research Article
Open Access

Strong Convergence Iterative Algorithms for Equilibrium Problems and Fixed Point Problems in Banach Spaces

Shenghua Wang, 

Shenghua Wang

Department of Applied Mathematics and Physics, North China Electric Power University, Baoding 071003, China ncepu.edu.cn

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Shin Min Kang, 

Corresponding Author

Shin Min Kang

Department of Mathematics and RINS, Gyeongsang National University, Jinju 660-701, Republic of Korea gnu.ac.kr

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First published: 14 April 2013
https://doi.org/10.1155/2013/619762
Academic Editor: Yisheng Song

Abstract

We first introduce the concept of Bregman asymptotically quasinonexpansive mappings and prove that the fixed point set of this kind of mappings is closed and convex. Then we construct an iterative scheme to find a common element of the set of solutions of an equilibrium problem and the set of common fixed points of a countable family of Bregman asymptotically quasinonexpansive mappings in reflexive Banach spaces and prove strong convergence theorems. Our results extend the recent ones of some others.

1. Introduction

Let E be a real reflexive Banach space with norm ∥·∥ and E* the dual space of E equipped with the inducted norm ∥·∥*. Throughout this paper, f : E → (−∞, +∞] is a proper, lower semicontinuous, and convex function and the Fenchel conjugate of f is the function f* : E* → (−∞, +∞] defined by
()

We denote by dom f the domain of f, that is, the set {x ∈ E : f(x)<+∞}.

Let C be a nonempty, closed, and convex subset of E and T : C → C a nonlinear mapping. The fixed points set of T is denoted by
()
Recall that a mapping T : C → C is said to be nonexpansive if, for each x, y ∈ C,
()
Nakajo-Takahashi [1] introduced the following hybrid method which is the so-called CQ-method for a nonexpansive mapping T in a Hilbert space H:
()
where {αn}⊂[0,1] and PK is the metric projection from H onto a closed and convex subset K of H. They proved that the sequence {xn} generated by (4) converges strongly to a fixed point of T under suitable conditions.
Takahashi et al. [2] introduced a new hybrid iterative scheme called the shrinking projection method for a nonexpansive mapping T in a Hilbert space H as follows:
()
where {αn}⊂[0,1], and they proved that the sequence {xn} generated by (5) converges strongly to a fixed point of T under suitable conditions.
In 2010, Reich and Sabach [3] introduced the following two hybrid iterative schemes for Bregman strongly nonexpansive mappings Ti : E → E  (i = 1,2, …, N) in a reflexive Banach space E:
()
()
where is the Bregman projection with respect to f from E onto a closed and convex subset K of E. They proved that the sequence {xn} generated by both (6) and (7) converges strongly to a common fixed point of .

The construction of fixed points for Bregman-type mappings via iterative processes has been investigated in, for example, [4–8].

In this paper, we design a new hybrid iterative scheme for finding a common element of the set of solutions of an equilibrium problem and the set of common fixed points of a countable family of Bregman asymptotically quasinonexpansive mappings in reflexive Banach spaces and prove some strong convergence theorems. Our results extend the recent one of Reich and Sabach [3].

2. Preliminaries

Let E be a real Banach space. For any x ∈ int  dom  f and y ∈ E, we define the right-hand derivative of f at x in the direction y by
()

The function f is said to be Gâteaux differentiable at x if exists for any y. In this case, fo(x, y) coincides with ∇f(x), the value of the gradient ∇f of f at x. The function f is said to be Gâteaux differentiable if it is Gâteaux differentiable for any x∈ int dom f. The function f is said to be Fréchet differentiable at x if this limit is attained uniformly in ∥y∥ = 1. Finally, f is said to be uniformly Fréchet differentiable on a subset C of E if the limit is attained uniformly for x ∈ C and ∥y∥ = 1.

Let E be a reflexive Banach space. The Legendre function is defined from a general Banach space E into (−∞, +∞] (see [9]). According to [9], the function f is Legendre if and only if it satisfies the following conditions
  • (L1)

    The interior of the domain of f (denoted by int  dom f) is nonempty; f* is Gâteaux differentiable on int  dom f and dom ∇f = int  dom f.

  • (L2)

    The interior of the domain f* (denoted by int domf*) is nonempty; f* is Gâteaux differentiable on int domf* and dom ∇f* = int  dom f*.

Since E is reflexive, we always have (∂f) −1 = ∂f* (see [10]). This fact, when combined with conditions (L1) and (L2), implies the following equalities:
()

Also, conditions (L1) and (L2), in conjunction with [9], imply that the functions f and f* are strictly convex on the interior of their respective domains. Several interesting examples of the Legendre functions are presented in [9, 11]. Especially, the functions (1/s)∥·∥s with s ∈ (1, ∞) are Legendre, where the Banach space E is smooth and strictly convex and, in particular, a Hilbert space.

Let f : E → (−∞, +∞] be a convex and Gâteaux differentiable function. The function Df : dom f × int  dom f → [0, +∞) defined as
()
is called the Bregman distance with respect to f [12].
By the definition, we know the following property (the three point identity): for any x ∈ dom f and y, z ∈ int  dom f,
()
Recall that the Bregman projection [13] of x ∈ int  dom f onto the nonempty, closed, and convex subset C of dom f is the necessarily unique vector satisfying
()
Let f : E → (−∞, +∞] be a convex and Gâteaux differentiable function. The function f is said to be totally convex at x ∈ int  dom f if its modulus of total convexity at x, that is, the function νf : int  dom f × [0, +∞)→[0, +∞] defined by
()
is positive whenever t > 0. The function f is said to be totally convex when it is totally convex at every point x ∈ int  dom f. In addition, the function f is said to be totally convex on bounded sets if νf(B, t) is positive for any nonempty bounded subset B of E and t > 0, where the modulus of total convexity of the function f on the set B is the function νf : int  dom f × [0, +∞)→[0, +∞] defined by
()

Some examples of the totally convex functions can be found in [14, 15].

Recall that the function f is said to be sequentially consistent [15] if, for any two sequences {xn} and {yn} in E such that the first is bounded,
()
Let C be a nonempty, closed, and convex subset of E and g : C × C → ℝ a bifunction that satisfies the following conditions:
  • (C1)

    g(x, x) = 0 for all x ∈ C;

  • (C2)

    g is monotone, that is, g(x, y) + g(y, x) ≤ 0 for all x, y ∈ C;

  • (C3)

    limsup t↓0g(tz + (1 − t)x, y) ≤ g(x, y) for all x, y, z ∈ C;

  • (C4)

    for all x ∈ C,   g(x, ·) is convex and lower semicontinuous.

The equilibrium problem with respect to g is as follows: find such that
()
The set of all solutions of (16) is denoted by EP (g). The resolvent of a bifunction g : C × C → ℝ [16] is the operator denoted by
()
For any x ∈ E, there exists z ∈ C such that ; see [3].
Let K be a convex subset of int  dom f and T : K → K a mapping. A point p in the closure of K is said to be an asymptotic fixed point of T [17, 18] if K contains a sequence {xn} which converges weakly to p such that the strong lim n→∞(xn − Txn) = 0. The set of asymptotic fixed points of T will be denoted by . The mapping T is called Bregman quasi-nonexpansive if F(T) ≠ ∅ and
()
T is said to be Bregman (quasi)-strongly nonexpansive [6] with respcet to a nonempty if
()
for all and x ∈ K, and if whenever {xn} ⊂ K is bounded, , and
()
it follows that
()
The mapping T is called Bregman firmly nonexpansive if
()
for all x, y ∈ K.
Next, we introduce a new mapping that is called Bregman asymptotically quasinonexpansive mapping which is a natural extension of Bregman quasinonexpansive mapping introduced by Reich and Sabach [3]. The mapping T : K → K is said to be Bregman asymptotically quasi-nonexpansive if there exists a sequence {kn}⊂[1, ∞) satisfying lim n→∞kn = 1 such that, for every n ≥ 1,
()
Obviously, every Bregman quasinonexpansive mapping is a Bregman asymptotically quasi-nonexpansive one with kn = 1.
Let E be a Banach space and C a nonempty subset of E. The mapping T : C → C is said to be uniformly asymptotically regular on C if
()
The mapping T is said to be closed if, for any sequence {xn} in C such that lim n→∞xn = x0 and lim n→∞Txn = y0, Tx0 = y0.

The following is an important result which will be used in the next section.

Lemma 1. Let E be a reflexive Banach space and f : E → (−∞, +∞) a Gâteaux differentiable and Legendre function which is totally convex on bounded sets. Let K be a nonempty, closed and convex subset of int  dom f and T : K → K a closed Bergman asymptotically quasi-nonexpansive mapping with the sequence {kn}⊂[1, +∞) such that kn → 1 as n → ∞. Then F(T) is closed and convex.

Proof. The closedness of F(T) comes directly from the closedness of T. Now, for arbitrary p1, p2 ∈ F(T), t ∈ (0,1), put p3 = tp1 + (1 − t)p2. We prove that Tp3 = p3. Indeed, from the definition of Df, we see that

()
This implies that lim n→∞Df(p3, Tnp3) = 0. It follows from Lemma 3 below that
()
that is, TTnp3 − p3 → 0 as n → ∞. In view the closedness of T, we can obtain the desired conclusion. This completes the proof.

Finally, we state some lemmas that will used in the proof of main results in next section.

Lemma 2 (see [7].)If f : E → ℝ is uniformly Fréchet differentiable and bounded on bounded subsets of E, then ∇f is uniformly continuous on bounded subsets of E from the strong topology of E to the strong topology of E*.

Lemma 3 (see [14].)The function f is totally convex on bounded sets if and only if it is sequentially consistent.

Lemma 4 (see [15].)Suppose that f is Gâteaux differentiable and totally convex on int dom f. Let x ∈ int  dom f and C a nonempty, closed, and convex subset of int dom f. If , then the following conditions are equivalent.

  • (i)

    The vector is the Bregman projection of x onto C with respect to f.

  • (ii)

    The vector is the unique solution of the variational inequality.

    ()

  • (iii)

    The vector is the unique solution of the inequality

    ()

Lemma 5 (see [6].)Let f : E → ℝ be a Gâteaux differentiable and totally convex function. If x0 ∈ E and the sequence is bounded, then the sequence is also bounded.

Lemma 6 (see [3].)Let f : E → (−∞, +∞) be a coercive (i.e., lim ∥x ∥→∞(f(x)/∥x∥) = +∞) and Gâteaux differentiable function. Let C be a closed and convex subset of E. If the bifunction g : C × C → ℝ satisfies conditions (C1)–(C4), then

  • (1)

    is single-valued;

  • (2)

    is a Bregman firmly nonexpansive mapping;

  • (3)

    the set of fixed points of is the solution set of the equilibrium problem, that is, ;

  • (4)

    EP (g) is a closed and convex subset of C;

  • (5)

    for all x ∈ E and , one has

    ()

3. Main Results

Now, we give our main theorems.

Theorem 7. Let E be a reflexive Banach space and f : E → ℝ a coercive Legendre function which is bounded, uniformly Fréchet differentiable, and totally convex on bounded subsets of E. Let K be a nonempty, closed, and convex subset of int  dom f and a countable family of closed Bregman asymptotically quasi-nonexpansive mappings with the sequences {ki,n}⊂[1, ∞) such that lim n→∞ki,n = 1 for every i ≥ 1. Let kn = sup {ki,n : i ≥ 1} and suppose that lim n→∞kn = 1. Let g : K × K → ℝ be a bifunction satisfying conditions (C1)–(C4). Assume that each Ti(i ≥ 1) is uniformly asymptotically regular and is nonempty and bounded. Let {αi,n} be a real sequence in (0,1) with for every n ≥ 1 and liminf n→∞αi,n > 0 for every i ≥ 1. Let {xn} be a sequence generated by the following manner:

()
where Mn = sup {Df(v, xn) : v ∈ Ω} for each n ≥ 1. Then, {xn} defined by (30) converges strongly to as n → ∞.

Proof. First, we prove that the sequence {xn} is well defined. Note that

()
is
()
that is,
()
This shows that Cn is closed and convex for every n ≥ 1. From the definition of Dn, it is easy to see that Dn is closed and convex for every n ≥ 1. For every i ≥ 1 and n ≥ 1, Lemma 6 shows that and for any v ∈ Ω and y ∈ E. Hence,
()

Since for every n ≥ 1, we have

()
This shows that v ∈ Cn for every n ≥ 1. Thus Ω ⊂ Cn for every n ≥ 1. Further, we have Ω ⊂ Dn for every n ≥ 1. Thus the sequence {xn} is well defined.

From , by Lemma 4(iii) we have

()
for any v ∈ Ω. Hence the sequence Df(xn, x) is bounded. Therefore by Lemma 5 the sequence {xn} is bounded.

On the other hand, in view of and , from Lemma 4(iii) we have

()
that is,
()
Therefore the sequence {Df(xn, x)} is increasing, and since it is also bounded, lim n→∞Df(xn, x) exists. By the construction of Dn, we have that Dm ⊂ Dn and for any positive integer m ≥ n. It follows that
()
Letting m, n → ∞ in (39), we see that Df(xm, xn) → 0. It follows from Lemma 3 that xm − xn → 0 as m, n → ∞. Hence, {xn} is a Cauchy sequence. Since E is a Banach space and K is closed and convex, we can assume that
()
By taking m = n + 1 in (39), we see that
()
Lemma 3 implies that
()
Since , we have
()
Then (41) implies that
()
Note that and liminf n→∞αi,n > 0, we have
()
for every i ≥ 1. It follows from Lemma 3 that
()
for every i ≥ 1. Note that
()
Combining (42) with (46), we see that
()
for every i ≥ 1. This means that the sequence {ui,n} is bounded. Since f is uniformly Fréchet differentiable, it follows from Lemma 2 that
()
Since f is uniformly Fréchet differentiable, it is also uniformly continuous (see [19, Theorem 1.8, p.13]) and therefore
()
From the definition of the Bregman distance, we obtain that for every
()
for any v ∈ Ω. Since every sequence {ui,n} is bounded, {∇f(ui,n)} is also bounded for every i ≥ 1. Now from (48)–(50), we have
()
for any v ∈ Ω and for every i ≥ 1.

In view of , by Lemma 6 (5) we have

()
Note that Mn is bounded and kn → 1 as n → ∞. It follows from (52) that
()
for every i ≥ 1. Lemma 3 shows that
()
Note that . From (48) and (55) we get
()
for every i ≥ 1. Note that
()

It follows from (40) and (56) that

()
for every i ≥ 1. On the other hand, we have
()
Since every Ti is uniformly asymptotically regular and (58), we obtain that, for every i ≥ 1,
()
that is, as n → ∞. From the closedness of Ti, we see that x* ∈ F(Ti) for every i ≥ 1. Thus .

Next we prove that x* ∈ EP (g) for every i ≥ 1. Since f is uniformly Fréchet differentiable, ∇f is uniformly continuous. Thus, by (55) we have

()
Since , we have
()
We have from (C2) that
()
Letting n → ∞, we have from (61) and (C4) that
()
For t with 0 < t ≤ 1 and y ∈ K, let yt = ty + (1 − t)x*. Since y ∈ K and x* ∈ K, we have yt ∈ K and hence g(yt, x*) ≤ 0. So, from (C1) we have
()
Dividing by t, we have
()
Letting t ↓ 0, from (C3) we have
()
Therefore, x* ∈ EP (g). Thus .

Finally, we show that x* = projΩx. Since Ω ⊂ Dn for every n ≥ 1, by Lemma 4(ii) we arrive at

()
Taking the limit as n → ∞ in (68), we obtain that
()
and hence x* = projΩx by Lemma 4(ii). This completes the proof.

Corollary 8. Let E be a reflexive Banach space and f : E → ℝ a coercive Legendre function which is bounded, uniformly Fréchet differentiable and totally convex on bounded subsets of E. Let K be a nonempty, closed, and convex subset of int  dom f and T : K → K a closed Bregman asymptotically quasi-nonexpansive mapping with the sequence {kn}⊂[1, ∞) such that lim n→∞kn = 1. Let g : K × K → ℝ be a bifunction satisfying conditions (C1)–(C4). Assume that T is uniformly asymptotically regular and Ω = F(T)∩EP (g) is nonempty and bounded. Let {xn} be a sequence generated by the following manner:

()
where Mn = sup {Df(v, xn) : v ∈ Ω} for each n ≥ 1. Then, {xn} defined by (70) converges strongly to as n → ∞.

Since every Bregman quasi-nonexpansive mapping is Bregman quasi-asymptotically nonexpansive, we have the following results.

Corollary 9. Let E be a reflexive Banach space and let f : E → ℝ be a coercive Legendre function which is bounded, uniformly Fréchet differentiable, and totally convex on bounded subsets of E. Let K be a nonempty, closed, and convex subset of int  dom f. Let be a countable family of closed Bregman quasi-nonexpansive mappings and g : K × K → ℝ a bifunction satisfying conditions (C1)–(C4). Assume that . Let {αi,n} be a real sequence in (0,1) with and liminf n→∞αi,n > 0 for every i ≥ 1. Let {xn} be a sequence generated by the following manner:

()
Then, {xn} defined by (71) converges strongly to as n → ∞.

Corollary 10. Let E be a reflexive Banach space and let f : E → ℝ be a coercive Legendre function which is bounded, uniformly Fréchet differentiable, and totally convex on bounded subsets of E. Let K be a nonempty, closed, and convex subset of int  dom f. Let T : K → K be a closed Bregman quasi-nonexpansive mapping and g : K × K → ℝ a bifunction satisfying conditions (C1)–(C4). Assume that Ω = F(T)∩EP (g) ≠ ∅. Let {xn} be a sequence generated by the following manner:

()
Then, {xn} defined by (72) converges strongly to as n → ∞

Remark 11. Set αn,i = 1/i(i + 1) + 1/n(n + 1) for each n ≥ 1 and i = 1,2, …, n and ki,n = 1 + 1/in for each n ≥ 1 and i ≥ 1. Then and liminf n→∞αi,n = 1/i(i + 1) > 0. Also, kn = sup {ki,n : i ≥ 1} = 1 for every n ≥ 1. Hence, {αi,n} and {ki,n} satisfy the conditions of Theorem 7.

Remark 12. It needs to notice that Corollaries 9 and 10 still hold if we replace the closedness of the mappings with .

In the equilibrium problem, the bifunction g is usually required to satisfy conditions (C1)–(C4). But, if the condition (C3) is replaced with the following condition:
  •  

    (C3′) for every fixed y ∈ C,  g(·, y) is continuous, then we have the following result:

Lemma 13. Let f : E → (−∞, +∞) be a coercive (i.e., lim  ∥x∥ →∞(f(x)/ ∥x∥ ) = +∞) and Gâteaux differentiable function. Let C be a closed and convex subset of E. If the bifunction g : C × C → ℝ satisfies conditions (C1), (C2), (C3′), and (C4), then the mapping defined by (2.2) is closed.

Proof. Let {xn} ⊂ E converge to x′ and to . To end the conclusion, we need to prove that . Indeed, for each xn, Lemma 6 shows that there exists a unique zn ∈ C such that , that is,

()

Since f is uniformly Fréchet differentiable, ∇f is uniformly continuous. So, taking the limit as n → ∞ in (73), by using (C3′) we get

()
which implies that . This completes the proof.

If the bifunction g satisfies conditions (C1), (C2), (C3′), and (C4) instead of (C1)–(C4), then we have a simple method to prove that x* ∈ EP (g) in the proof of Theorem 7. Indeed, from the proof of Theorem 7, we see that
()

Note that xn → x* as n → ∞. This shows that as n → ∞ for every i ≥ 1. It follows from the closedness of that . Lemma 6 shows that x* ∈ EP (g).

Remark 14. Obviously, the proof process of x* ∈ EP (g) is simple if we replace condition (C3) with (C3′) which is such that is closed. In fact, although condition (C3′) is stronger than (C3), it is not easier to verify condition (C3) than to verify the condition (C3′). Hence, from this viewpoint, the condition (C3′) is acceptable.

Acknowledgments

This work is supported by the Fundamental Research Fundsfor the Central Universities (Grant Number: 13MS109) and the HeBei Education 4 Department (Grant Number: 936101101).

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