Open Access

Strong convergence theorems for equilibrium problems and fixed point problems: A new iterative method, some comments and applications

Fixed Point Theory and Applications20112011:33

https://doi.org/10.1186/1687-1812-2011-33

Received: 2 April 2011

Accepted: 12 August 2011

Published: 12 August 2011

Abstract

In this paper, we introduce a new approach method to find a common element in the intersection of the set of the solutions of a finite family of equilibrium problems and the set of fixed points of a nonexpansive mapping in a real Hilbert space. Under appropriate conditions, some strong convergence theorems are established. The results obtained in this paper are new, and a few examples illustrating these results are given. Finally, we point out that some 'so-called' mixed equilibrium problems and generalized equilibrium problems in the literature are still usual equilibrium problems.

2010 Mathematics Subject Classification: 47H09; 47H10, 47J25.

Keywords

strong convergenceiterative methodequilibrium problemfixed point problem

1 Introduction and preliminaries

Throughout this paper, we assume that H is a real Hilbert space with zero vector θ, whose inner product and norm are denoted by 〈·, ·〉 and || · ||, respectively. The symbols and are used to denote the sets of positive integers and real numbers, respectively. Let K be a nonempty closed convex subset of H and T : KH be a mapping. In this paper, the set of fixed points of T is denoted by F(T). We use symbols → and to denote strong and weak convergence, respectively.

For each point x H, there exists a unique nearest point in K, denoted by P K x, such that
The mapping P K is called the metric projection from H onto K. It is well known that P K satisfies
for every x, y H. Moreover, P K x is characterized by the properties: for x H, and z K,
Let f be a bi-function from K × K into . The classical equilibrium problem is to find x K such that
(1.1)

Let EP(f) denote the set of all solutions of the problem (1.1). Since several problems in physics, optimization, and economics reduce to find a solution of (1.1) (see, e.g., [1, 2]), some authors had proposed some methods to find the solution of equilibrium problem (1.1); for instance, see [14]. We know that a mapping S is said to be nonexpansive mapping if for all x, y K, ||Sx - Sy|| ≤ ||x - y||. Recently, some authors used iterative method including composite iterative, CQ iterative, viscosity iterative etc. to find a common element in the intersection of EP(f) and F(S); see, e.g., [511].

Let I be an index set. For each i I, let f i be a bi-function from K × K into . The system of equilibrium problem is to find x K such that
(1.2)

We know that is the set of all solutions of the system of equilibrium problem (1.2).

For each i I, if f i (x, y) = 〈A i x, y - x〉, where A i : KK is a nonlinear operator, then the problem (1.2) becomes the following system of variational inequality problem:
(1.3)

It is obvious that the problem (1.3) is a special case of the problem (1.2).

The following Lemmas are crucial to our main results.

Lemma 1.1 (Demicloseness principle[12]) Let H be a real Hilbert space and K a closed convex subset of H. S : KH is a nonexpansive mapping. Then the mapping I - S is demiclosed on K, where I is the identity mapping, i.e., x n x in K and (I - S)x n y implies that × K and (I - S)x = y.

Lemma 1.2[13] 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 < lim infn→∞β n ≤ lim supn→∞β n < 1. Suppose xn+1= β n y n + (1 - β n )x n for all integers n ≥ 0 and lim supn→∞(||yn+1- y n || - ||xn+1- x n ||) ≤ 0, then limn→∞||y n - x n || = 0.

Lemma 1.3[5] Let H be a real Hilbert space. Then the following hold.
  1. (a)

    ||x + y||2 ≤ ||y||2 + 2〈x, x + yfor all x, y H;

     
  2. (b)

    ||αx + (1 - α)y||2 = α||x||2 + (1 - α) ||y||2 - α(1 - α) ||x - y||2 for all x, y H and α ;

     
  3. (c)

    ||x - y||2 = ||x||2 + ||y||2 - 2 〈x, yfor all x, y H.

     
Lemma 1.4. [14] Let {a n } be a sequence of nonnegative real numbers satisfying the following relation:
If
  1. (i)

    λ n [0,1], or, equivalently, ;

     
  2. (ii)

    or ,

     

then .

Lemma 1.5[1] Let K be a nonempty closed convex subset of H and F be a bi-function of K × K into satisfying the following conditions.

(A1) F(x, x) = 0 for all × K;

(A2) F is monotone, that is, F(x, y) + F(y, x) ≤ 0 for all x, y K;

(A3) for each x, y, z K,

(A4) for each × K, yF(x, y) is convex and lower semi-continuous.

Let r > 0 and × H. Then, there exists z K such that
Lemma 1.6[3] Let K be a nonempty closed convex subset of H and let F be a bi-function of K × K into R satisfying (A1) - (A4). For r > 0 and × H, define a mapping T r : HK as follows:
for all × H. Then the following hold:
  1. (i)

    T r is single-valued;

     
  2. (ii)
    T r is firmly nonexpansive, that is, for any x, y H,
     
  3. (iii)

    F(T r ) = EP (F);

     
  4. (iv)

    EP(F) is closed and convex.

     

2 Main results and their applications

Let I = {1, 2,..., k} be a finite index set, where k . For each i I, let f i be a bi-functions from K × K into satisfying the conditions (A1)-(A4). Denote by

For each (i, n) I × , applying Lemmas 1.5 and 1.6, is a firmly nonexpansive single-valued mapping such that is closed and convex. For each i I, let , n .

First, let us consider the following example.

Example A Let f i : [-1, 0]×[-1,0] be defined by f i (x, y) = (1+x2i)(x - y), i = 1, 2, 3. It is easy to see that for any i {1, 2, 3}, f i (x, y) satisfies the conditions (A1)-(A4) and . Let Sx = x3 and , x [-1, 0] Then g is a -contraction from K into itself and S : K → K is a nonexpansive mapping with . Let λ (0, 1), {r n } [1, + ∞) and {α n } (0,1) satisfy the conditions (i) limn→∞α n = 0, and (ii) , or equivalently, ; e.g., let , {α n } (0, 1) and {r n } [1, + ∞) be given by
Define a sequence {x n } by
(2.1)

Then the sequences {x n } and , i = 1, 2, 3, defined by (2.1) all strongly converge to 0.

Proof
  1. (a)

    By Lemmas 1.5 and 1.6, (2.1) is well defined.

     
  2. (b)
    Let K = [-1, 0]. For each i {1, 2, 3}, define
     
We claim that for each v K and any i {1, 2, 3}, there exists a unique z = 0 K such that
or, equivalently,
Obviously, z = 0 is a solution of the problem . On the other hand, there does not exist z [-1, 0) such that z - y ≤ 0 and . So z = 0 is the unique solution of the problem .
  1. (c)
    We notice that (2.1) is equivalent with (2.2), where
    (2.2)
     
It is easy to see that {x n } [-1, 0], so, by (b), for all n . We need to prove x n 0 as n → ∞. Since z n = 0 for all n , we have y n = (1 -λ)x n and
(2.3)
for all n . For any n , from (2.3), we have
(2.4)

Hence {|x n |} is a strictly deceasing sequence and |x n | ≥ 0 for all n . So exists.

On the other hand, for any n, m with n > m, using (2.4), we obtain

which implies . Therefore {x n } strongly converges to 0. □

In this paper, motivated by the preceding Example A, we introduce a new iterative algorithm for the problem of finding a common element in the set of solutions to the system of equilibrium problem and the set of fixed points of a nonexpansive mapping. The following new strong convergence theorem is established in the framework of a real Hilbert space H.

Theorem 2.1 Let K be a nonempty closed convex subset of a real Hilbert space H and I = {1, 2,..., k} be a finite index set. For each i I, let f i be a bi-function from K × K into satisfying (A1)-(A4). Let S : K → K be a nonexpansive mapping with . Let λ, ρ (0, 1) and g : K → K is a ρ-contraction. Let {x n } be a sequence generated in the following manner:

If the above control coefficient sequences {α n } (0, 1) and {r n } (0, +∞) satisfy the following restrictions:

(D1) , and ;

(D2) and .

then the sequences {x n } and , for all i I, converge strongly to an element c = PΩg(c) Ω. The following conclusion is immediately drawn from Theorem 2.1.

Corollary 2.1 Let K be a nonempty closed convex subset of a real Hilbert space H. Let f be a bi-function from K × K into satisfying (A1)-(A4) and S : KK be a nonexpansive mapping with Ω = EP(f) ∩F(S) ≠ . Let λ, ρ (0,1) and g : K → K is a ρ-contraction. Let {x n } be a sequence generated in the following manner:

If the above control coefficient sequences {α n } (0, 1) and {r n } (0, +∞) satisfy all the restrictions in Theorem 2.1, then the sequences {x n } and {u n } converge strongly to an element c = PΩg(c) Ω, respectively.

If f i (x, y) ≡ 0 for all (x, y) K × K in Theorem 2.1 and all i I, then, from the algorithm (D H ), we obtain , i I. So we have the following result.

Corollary 2.2 Let K be a nonempty closed convex subset of a real Hilbert space H. Let S : KK be a nonexpansive mapping with F(S) ≠ . Let λ, ρ (0, 1) and g : KK is a ρ-contraction. Let {x n } be a sequence generated in the following manner:

If the above control coefficient sequences {α n } (0, 1) satisfy , and , then the sequences {x n } converge strongly to an element c = PΩg(c) F (S).

As some interesting and important applications of Theorem 2.1 for optimization problems and fixed point problems, we have the following.

Application (I) of Theorem 2.1 We will give an iterative algorithm for the following optimization problem with a nonempty common solution set:

where h i (x), i {1, 2,..., k}, are convex and lower semi-continuous functions defined on a closed convex subset K of a Hilbert space H (for example, h i (x) = x i , x K := [0, 1], i {1, 2,..., k}).

If we put f i (x, y) = h i (y) - h i (x), i {1, 2,..., k}, then is the common solution set of the problem (OP), where denote the common solution set of the following equilibrium:
For i {1, 2,..., k}, it is obvious that the f i (x, y) satisfies the conditions (A1)-(A4). Let S = I (identity mapping), then from (D H ), we have the following algorithm
(2.5)

where x1 K, λ (0, 1), g : K → K is a ρ-contraction. From Theorem 2.1, we know that {x n } and , i {1,2,..., k}, generated by (2.5), strongly converge to an element of if the coefficients {α n } and {r n } satisfy the conditions of Theorem 2.1.

Application (II) of Theorem 2.1 Let H, K, I, λ, ρ, g be the same as Theorem 2.1. Let A1, A2,..., A k : KK be k nonlinear mappings with . For any i I, put f i (x, y) = 〈x - A i x, y - x〉, x, y K. Since , we have . Let S = I (identity mapping) in the algorithm (D H ). Then the sequences {x n } and , defined by the algorithm (D H ), converge strongly to a common fixed point of {A1, A2,..., A k }, respectively.

The following result is important in this paper.

Lemma 2.1 Let H be a real Hilbert space. Then for any x1, x2,... x k H and a1, a2,..., a k [0,1] with , k , we have
(2.6)
Proof It is obvious that (2.6) is true if a j = 1 for some j, so it suffices to show that (2.6) is true for a j ≠ 1 for all j. The proof is by mathematic induction on k. Clearly, (2.6) is true for k = 1. Let x1, x2 H and a1, a2 [0,1] with a1 + a2 = 1. By Lemma 1.3, we obtain
which means that (2.6) hold for k = 2. Suppose that (2.6) is true for k = l . Let x1, x2,..., x l , xl+1 H and a1, a2,..., a l , al+1 [0, 1) with . Let . Then applying the induction hypothesis we have

Hence, the equality (2.6) is also true for k = l + 1. This completes the induction. □

3 Proof of Theorem 2.1

We will proceed with the following steps.

Step 1: There exists a unique c Ω H such that PΩg(c) = c.

Since PΩg is a ρ-contraction on H, Banach contraction principle ensures that there exists a unique c H such that c = PΩg(c) Ω.

Step 2: We prove that the sequences {x n }, {y n }, {z n } and , i I, are all bounded.

First, we notice that (D H ) is equivalent with (Z H ), where
For each i I, we have
(3.1)
For any n , from (Z H ) we have
and
(3.2)
Since g is a ρ-contraction, it follows from (3.2) that
By induction, we obtain

which shows that {x n } is bounded. Also, we know that {y n }, {z n } and , i I, are all

bounded.

Step 3: We prove limn→∞||xn+1- x n || = 0.

For each i I, since , , from (Z H ), we have
(3.3)
and
(3.4)
By (3.3) and (3.4) and (A2),
which implies
(3.5)
It follows from (3.5) that
(3.6)
Let . For any n , since , by (3.6), we have
(3.7)
Set
(3.8)
where β n = 1 - (1 - λ)(1 - α n ), n . Then for each n ,
(3.9)
and
(3.10)
For any n , since
by (3.7), it follows that
From this and (D1), (D2), we get
(3.11)
By Lemma 1.2 and (3.11),
(3.12)
Owing to (3.9) and (3.12), we obtain
(3.13)

Step 4: We show .

By (3.6), (3.13) and (D2), we have
From (Z H ), we get
(3.14)
Since ||x n - y n || ≤ ||x n - xn+1|| + ||xn+1- y n ||, by (3.13) and (3.14),
which implies that
By Lemma 1.6,
which yields that
(3.15)
From (3.15) and Lemma 2.1,
Since
where
We have
(3.16)
Letting n → ∞ in the inequality (3.16), we obtain
(3.17)
Furthermore, it is easy to prove that
For any i I, since
it implies
(3.18)

Step 5: Prove lim supn→∞g(c) - q, x n - c〉 ≤ 0.

Take a subsequence of {x n } such that
(3.19)

Since is bounded, there exists a subsequence of which is still denoted by such that as → ∞. Notice that for each i I, by (3.17), so we also have as → ∞, i I.

We want to show z Ω. First, we show that z F(S). In fact, since and as → ∞, by Lemma 1.1, we have (I - S)z = θ or, equivalently, z F(S).

For each i I, since , y K, it follows from (A2) that
and hence
Applying (3.17) and (A4),
(3.20)
Let y K be given. Put y t = ty + (1 - t)z, t (0, 1). Then y t K and f i (y t , z) ≤ 0 for all i I. By (A1) and (A4), we get
For any i I, by (A3), we have
(3.21)
Hence, from (3.21), . Therefore, we proved . On the other hand, by (3.19), we obtain
(3.22)

Step 6: Finally, we prove {x n } and , for all i I, converge strongly to c = PΩg(c) Ω.

From (Z H ) and (a) of Lemma 1.3, we have
(3.23)
For any n , let
and
From (3.23), we have
It is easy to verify that all conditions of Lemma 1.4 are satisfied. Hence, applying Lemma 1.4, we obtain limn→∞a n = 0 which implies

or equivalence, {x n } strongly converges to c. By (3.17), we can prove that for any i I, strongly converges to c. The proof of Theorem 2.1 is completed. □

4 Further remarks

Let K be a nonempty closed convex subset of H and f be a bi-function of K × K into .

Remark 4.1 Recently, some authors introduced the following mixed equilibrium problem (MEP, for short) (see [1517] and references therein) and generalized equilibrium problem (GEP, for short) (see [1820] and references therein):
  1. (a)
    Mixed equilibrium problem [1517]:
     
where φ : C is a real-valued function.
  1. (b)
    Generalized equilibrium problem [1820]:
     

where A : CH is a nonlinear operator.

In [1517], the authors gave some iterative methods for finding the solution of MEP when the bi-function f(x, y) admits the conditions (A1)-(A4) and the real-valued function φ satisfies the following condition:

(A5) φ : C is a proper lower semi-continuous and convex function.

However, in this case, we argue that the problem MEP is still the equilibrium problem (1.1). In fact, if we put f1(x, y) = f(x, y), f2(x, y) = φ(y) - φ(x) and F(x, y) = f1(x, y) + f2(x, y) for each (x, y) C × C, then f1(x, y) satisfies the conditions (A1)-(A4), f2(x, y) satisfies the condition (A5) and the function φ must satisfy the conditions (A1)-(A4). This shows that for each (x, y) C × C, F(x, y) satisfies the conditions (A1)-(A4). So, when we study the solution of MEP, we only need to study the solution of the equilibrium (1.1). This also shows that some "so-called" mixed equilibrium problem studied in [1517] is still the equilibrium problem (1.1).

Remark 4. 2 Let us recall some well-known definitions. A mapping T : CC is said to be
  1. (1)

    v-expansive if ||Tx - Ty|| ≥ v||x - y|| for all x, y C. In particular, if v = 1, then T is called expansive.

     
  2. (2)
    v-strongly monotone if there exists a constant v > 0 such that
     
Clearly, any v-strongly monotone mapping is v-expansive.
  1. (3)
    u-inverse strongly monotone if there exists a constant u > 0 such that
     
  2. (4)

    L-Lipschitz continuous if ||Tx - Ty|| ≤ L||x - y|| for all x, y C. In particular, if L = 1, then T is called nonexpansive.

     

It is easy to see that a u-inverse strongly monotone operator is -Lipschitz continuous.

For the problem GEP, if the nonlinear operator A : CH is a u-inverse strongly monotone operator and the bi-function f(x, y) admits the conditions (A1)-(A4), we argue that the problem GEP is still the problem (1.1) and so it is indeed not a generalization. In fact, if A is a u-inverse strongly monotone operator from C into H, then A is a continuous operator. So, we obtain easily that the function (x, y) → <Ax, y - x〉, x, y C, satisfies the conditions (A1)-(A4). Hence, if we put F(x, y) = f(x, y) + 〈Ax, y - x〉 ≥ 0, then the problem GEP studied in [1820] is still the problem (1.1).

5 Conclusion

The problem MEP studied in [1517] and the problem GEP studied in [1820] are still the problem (1.1) studied in the literature [511, 2124] and others.

Declarations

Acknowledgements

Zhenhua He was supported by the Natural Science Foundation of Yunnan Province (2010ZC152) and the Scientific Research Foundation from Yunnan Province Education Committee (08Y0338); Wei-Shih Du was supported by the National Science Council of the Republic of China.

Authors’ Affiliations

(1)
Department of Mathematics, Honghe University
(2)
Department of Mathematics, National Kaohsiung Normal University

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Copyright

© He and Du; licensee Springer. 2011

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