A Viscosity Hybrid Steepest Descent Method for Generalized Mixed Equilibrium Problems and Variational Inequalities for Relaxed Cocoercive Mapping in Hilbert Spaces
We present an iterative method for fixed point problems, generalized mixed equilibrium problems,
and variational inequality problems. Our method is based on the so-called viscosity hybrid steepest descent
method. Using this method, we can find the common element of the set of fixed points of a nonexpansive mapping, the
set of solutions of generalized mixed equilibrium problems, and the set of solutions of variational inequality problems for
a relaxed cocoercive mapping in a real Hilbert space. Then, we prove the strong convergence of the proposed iterative
scheme to the unique solution of variational inequality. The results presented in this paper generalize and extend some
well-known strong convergence theorems in the literature.
1. Introduction
Throughout this paper, unless otherwise specified, we consider to be a real Hilbert space with inner product and its induced norm . Let be a nonempty closed convex subset of and let be the metric projection of onto the closed convex subset . Let be a nonexpansive mapping, that is, for all . The fixed point set of is defined by
If is nonempty, bounded, closed, and convex and is a nonexpansive mapping of into itself, then is nonempty; see, for example, [1, 2]. A mapping is a contraction on if there exists a constant such that for all In addition, let be a nonlinear mapping. Let be a real-valued function and let be a bifunction such that , where is the set of real numbers and .
The generalized mixed equilibrium problem for finding
The set of solutions of (1.2) is denoted by , that is,
We see that if is a solution of a problem (1.2), then .
Special Examples (1)If , then the problem (1.2) is reduced into the mixed equilibrium problem for finding such that
The set of solutions of (1.4) is denoted by .(2)If , then the problem (1.2) is reduced into the generalized equilibrium problem for finding such that
The set of solutions of (1.5) is denoted by .(3)If and , then the problem (1.2) is reduced into the equilibrium problem for finding such that
The set of solutions of (1.6) is denoted by .(4)If and , then the problem (1.2) is reduced into the variational inequality problem for finding such that
The set of solutions of (1.7) is denoted by .
The generalized mixed equilibrium problem is very general in the sense that it includes, as special cases, fixed point problems, variational inequality problems, optimization problems, Nash equilibrium problems in noncooperative games, the equilibrium problem, and Numerous problems in physics, economics, and others. Some methods have been proposed to solve problem (1.2); see, for instance, [3, 4] and the references therein.
Let be a nonlinear mapping. Now, we recall the following definitions.(d1) is said to be monotone if for each (d2) is said to be -strongly monotone if there exists a positive real number such that
(d3) is said to be -Lipschitz continuous if there exists a positive real number such that
(d4) is said to be -inverse-strongly monotone if there exists a constant such that
(d5) is said to be relaxed -cocoercive if there exist positive real numbers such that
(d6) A set-valued mapping is called monotone if for all , and imply .(d7) A monotone mapping is called maximal if the graph of is not properly contained in the graph of any other monotone mapping. It is well known that a monotone mapping is maximal if and only if for , for every implies .
For finding a common element of the set of fixed points of a nonexpansive mapping and the set of solutions of variational inequalities for a -inverse-strongly monotone mapping, Takahashi and Toyoda [5] introduced the following iterative scheme:
where is a -inverse-strongly monotone mapping, is a sequence in , and is a sequence in . They showed that if is nonempty, then the sequence generated by (1.13) converges weakly to some .
For finding an element of , Iiduka et al. [6] introduced the following iterative scheme:
where is a -inverse-strongly monotone mapping, is a sequence in , and is a sequence in . They showed that if is nonempty, then the sequence generated by (1.14) converges weakly to some .
For finding a common element of , let be a nonexpansive mapping. Yamada [7] introduced the following iterative scheme called the hybrid steepest descent method:
where is a strongly monotone and Lipschitz continuous mapping, and is a positive real number. He proved that the sequence generated by (1.15) converges strongly to the unique solution of the .
The hybrid steepest descent method is constructed by blending important ideas in the steepest descent method and in the fixed point theory. The remarkable applicability of this method to the convexly constrained generalized pseudoinverse problem as well as to the convex feasibility problem is demonstrated by constructing nonexpansive mappings whose fixed point sets are the feasible sets of the problems.
On the other hand, Shang et al. [8] introduced a new iterative process for finding a common element of the set of fixed points of a nonexpansive mapping and the set of solutions of the variational inequalities for relaxed -cocoercive mappings in a real Hilbert space by using viscosity approximation method. Let be a nonexpansive mapping and let be a contraction mapping. Starting with arbitrary initial and define sequences recursively by
They proved that under certain appropriate conditions imposed on , and , the sequence converges strongly to , where
For finding a common element of , let be a nonempty closed convex subset of a real Hilbert space . Let be a -inverse-strongly monotone mapping of into and let be a nonexpansive mapping of into itself. S. Takahashi and W. Takahashi [9] introduced the following iterative scheme:
where , and satisfy some parameters controlling conditions. They proved that the sequence defined by (1.17) converges strongly to a common element of .
Iterative methods for nonexpansive mappings have recently been applied to solve convex minimization problems; see, for example, [7, 10–12] and the references therein. Convex minimization problems have a great impact and influence in the development of almost all branches of pure and applied sciences.
A typical problem is to minimize a quadratic function over the set of fixed points of a nonexpansive mapping defined on a real Hilbert space :
where is the fixed point set of a nonexpansive mapping defined on and is a given point in .
A linear bounded operator is strongly positive if there exists a constant with the property
Recently, Marino and Xu [13] introduced a new iterative scheme by the viscosity approximation method:
They proved that the sequence generated by (1.20) converges strongly to the unique solution of the variational inequality:
which is the optimality condition for the minimization problem:
where is a potential function for .
In 2008, Qin et al. [14] proposed the following iterative algorithm:
where is a strongly positive linear bounded operator and is a relaxed cocoercive mapping of into . They proved that if the sequences , and of parameters satisfy appropriate condition, then the sequence defined by (1.23) converges strongly to the unique solution of the variational inequality:
which is the optimality condition for the minimization problem:
where is a potential function for .
In this paper, we introduce an iterative scheme by using a viscosity hybrid steepest descent method for finding a common element of the set of solutions of a generalized mixed equilibrium problem, the set of fixed points of a nonexpansive mapping, and the set of solutions of variational inequality problem for a relaxed cocoercive mapping in a real Hilbert space. The results shown in this paper improve and extend the recent ones announced by many others.
2. Preliminaries
Throughout this paper, we always assume that is a real Hilbert space and is a nonempty closed convex subset of . For a sequence , the notation of and means that the sequence converges weakly and strongly to , respectively.
The following lemmata give some characterizations and useful properties of the metric projection in a real Hilbert space. The metric (or nearest point) projection from onto is the mapping which assigns to each point the unique point satisfying the following property:
Lemma 2.1. It is well known that the metric projection has the following properties: (m1) for each and ,
(m2) is nonexpansive, that is,
(m3) is firmly nonexpansive, that is,
In order to prove our main results, we also need the following lemmata.
Lemma 2.2 (see [2]). Let be a Hilbert space, let be a nonempty closed convex subset of , and let be a mapping of into . Let . Then, for ,
that is,
where is the metric projection of onto .
Lemma 2.3 (see [15]). Let be a monotone mapping of into and let be the normal cone to at , that is,
and define a mapping on by
Then is maximal monotone and if and only if for all .
Lemma 2.4 (see [16]). Each Hilbert space satisfies Opials condition; that is, for any sequence with , the inequality
holds for each with .
Lemma 2.5 (see [13]). Let be a nonempty closed convex subset of , let be a contraction of into itself with coefficient , and let be a strongly positive linear bounded operator on with coefficient . Then, for ,
That is, is strongly monotone with coefficient .
Lemma 2.6 (see [13]). Assume that is a strongly positive linear bounded operator on with coefficient and . Then .
Lemma 2.7 (see [17]). Let and be bounded sequences in a Banach space and let be a sequence in with
Suppose
Then,
Lemma 2.8 (see [18]). Assume that is a sequence of nonnegative real numbers such that
where is a sequence in and is a sequence in such that (1), (2) or Then
For solving the generalized mixed equilibrium problem and the mixed equilibrium problem, let us give the following assumptions for the bifunction , the function , and the set : (H1)(H2) is monotone, that is, (H3) for each is weakly upper semicontinuous; (H4) for each is convex; (H5) for each is lower semicontinuous; (B1) for each and , there exist abounded subsets and such that for any ,
(B2) is a bounded set.
Lemma 2.9 (see [19]). Let be a nonempty closed convex subset of . Let be a bifunction that satisfies (H1)–(H5) and let be a proper lower semicontinuous and convex function. Assume that either (B1) or (B2) holds. For and , define a mapping as follows:
for all . Then, the following properties hold: (i)for each ; (ii) is single-valued; (iii) is firmly nonexpansive; that is, for any (iv)(v) is closed and convex.
Remark 2.10. If , then is rewritten as .
Lemma 2.11 (see [9]). Let , and be as in Remark 2.10. Then the following holds:
for all and .
The following lemma is an immediate consequence of an inner product.
Lemma 2.12. Let be a real Hilbert space, let and be elements in , and let . Then (1), (2).
3. Main Results
In this section, we will introduce an iterative scheme by using a viscosity hybrid steepest descent method for finding a common element of the set of fixed points for nonexpansive mappings, the set of solutions of a generalized mixed equilibrium problem, and the set of solutions of variational inequality problem for a relaxed cocoercive mapping in a real Hilbert space. We show that the iterative sequence converges strongly to a common element of the three sets.
In order to prove our main results, we first prove the following lemmata.
Lemma 3.1. Let and be as in Lemma 2.9. Then the following holds:
for all and .
Proof. By similar argument as in the proof of Lemma 2.11 in [9], for and . Observing that and , we have
Putting in (3.2) and in (3.3), we obtain
So, summing up these two equalities and using the monotonicity of (H2), we get
and hence
We derive from (3.6) that
and so
This indicates that
In other words,
and thus the claim holds.
Lemma 3.2. Let be a real Hilbert space, let be a nonempty closed convex subset of , let be a nonexpansive mapping, and let be an -Lipschitz continuous and relaxed -cocoercive mappings with . If , then is a nonexpansive mapping in .
Proof. Let . Then, for every , we have
Now, since , thus . Thus, is a nonexpansive mapping of into .
Now we can prove that a strong convergence theorem is a real Hilbert space.
Theorem 3.3. Let be a nonempty closed convex subset of a real Hilbert space . Let and be two bifunctions from to satisfying (H1)–(H5) and let be a proper lower semicontinuous and convex function with assumption (B1) or (B2). Let (i) be a -inverse-strongly monotone mapping, (ii) be a -inverse-strongly monotone mapping, (iii) be an -Lipschitz continuous and relaxed -cocoercive mappings, (iv) be a contraction mapping with coefficient and let be a strongly positive linear bounded self-adjoint operator with the coefficient and . Let be a nonexpansive mapping with . Assume that
Let , , , , and be the sequences generated by
where , and , and are three sequences in satisfying the following conditions: (C1) and , (C2), and , (C3), and , (C4), and , (C5), and . Then, converges strongly to , which is the unique solution of the variational inequality:
Proof. From the restrictions on control sequences, we may assume, without loss of generality, that for all . Since is a strongly positive linear bounded self-adjoint operator on , we have
Observe that
That is, is positive. It follows that
We will split the proof of Theorem 3.3 into six steps.Step 1. We claim that the sequence is bounded. Indeed, let , by Lemmas 2.2 and 2.9, we obtain
Since is -inverse-strongly monotone, and , we know that, for any , we have
Similarly, from (3.19), , and , we can prove that
and hence
Let , and form Lemma 3.2 is a nonexpansive mapping and from Lemma 2.2 , we have
which yields that
By mathematical induction, putting , we have that for all . Indeed, we can easily see that . Suppose that for some positive integral . Then we have that
This shows that is bounded in . From (3.21), we know that and are bounded in and so , , , , , and are bounded sequence in .Step 2. We claim that Since is nonexpansive, we have
Next, we estimate . Observing that and , we have
Similarly, we can prove that
Substitution (3.26) into (3.27), we derive
Since , and are bounded, is an appropriate constant such that
Substitution (3.28) into (3.25), we obtain
From (3.13), we have
Simple calculations show that
which yields that
Substitution (3.30) into (3.33) yields that
where is an appropriate constant such that
Since and
next, we estimate
Note that ; there exists a constant such that for all . From Lemma 3.1, we get
It follows that
Since , we obtain
Similarly, we can prove that
Consequently, from (3.40), (3.41), and conditions in Theorem 3.3, we obtain
It follows from Lemma 2.7 that
In view of (3.13), we see that
which, combining with (3.43) and , yields that
Step 3. We claim that Observing that , we have
which, combining with the conditions (C1) and (C2), gives
From (3.43) and (3.47), we have
For any , we see that
Furthermore, from (3.13) and Lemma 2.12(1), we have
It follows that
From the condition (C1) and (3.45), we arrive at
On the other hand, we have
which yields that
Substituting (3.55) into (3.50), we have
Using (3.51) and (3.56), we have
It follows that
From condition (C1), (3.45), and (3.53), we obtain
Consequently, from (3.50) we derive that
From (3.51) and (3.60), we obtain
So, we obtain
Since and , we obtain
Similarly, we can prove that
In addition, from the firmly nonexpansivity of , we have
So, we obtain
Similarly, we can prove that
Substituting (3.66) into (3.50), we have
Using (3.51) and (3.68), we have
It follows that
Since , and , we obtain
Similarly, we can prove that
Furthermore, by the triangular inequality, we also have
Applying (3.71) and (3.72), we have
Since
so we get
Also, observe that
Consequently, we obtain
Step 4. We prove that the mapping has a unique fixed point. Since is a contraction of into itself with coefficient , then, we have
Since , it follows that is a contraction of into itself. Therefore by the Banach Contraction Mapping Principle, it has a unique fixed point, say , that is,
Step 5. We claim that , where is the unique solution of the variational inequality , for all Since is a unique solution of the variational inequality (3.14), to show this inequality, we choose a subsequence of such that
Correspondingly, there exists a subsequence of such that
Since is bounded, there exists a subsequence of which converges weakly to . Without loss of generality, we can assume that From we obtain . Next, we show that . First, we show that . Assume Since and it follows by Opial's condition (Lemma 2.4) that
This is a contradiction. Thus, we have . Next, we prove that For any , we have
which yield that
It follows from Lemma 2.9 that for all Thus, we conclude that is equivalent to
From (H2), we also have
Replacing by , we obtain
Let for all and . Since and we obtain So, from (3.88) we have
Since , we have . Further, from the inverse strong monotonicity of , we have
So, from (H4), (H5), and the weak lower semicontinuity of , and , we have
From (H1), (H4), and (3.91), we also get
Dividing by , we get
Letting in the above inequality, we arrive that
Thus, Similarly, we can prove that Finally, now we prove that We define the maximal monotone operator:
Since is relaxed -cocoercive and condition (C5), we have
which yields that is monotone. Thus, is maximal monotone. Let Since and we have
On the other hand, from we have
and hence
It follows that
which implies that
Since is maximal monotone, we obtain that . From Lemma 2.3, we get that That is, . Since , we have
On the other hand, we have
From (3.43) and (3.102), we obtain that
Step 6. Finally, we show that converges strongly to . Indeed, by (3.13) and using Lemmas 2.6 and 2.12(2), we observe that
which implies that
On the other hand, we have
Substituting (3.106) into (3.107) yields that
Taking
then, we can rewrite (3.108) as
It follows from condition (C1) and (3.104) that
Since
and is bounded, we have
Applying Lemma 2.8 to (3.110), we conclude that converges strongly to in norm. This completes the proof.
Corollary 3.4. Let be a nonempty closed convex subset of a real Hilbert space . Let and be two bifunctions from to satisfying (H1)–(H5) and let be a proper lower semicontinuous and convex function with assumption (B1) or (B2). Let (i) be a -inverse-strongly monotone mapping, (ii) be a -inverse-strongly monotone mapping, (iii) be a contraction mapping with coefficient . Let be a nonexpansive mapping with . Assume that
Let , and be the sequences generated by
where , and , and are three sequences in satisfying the following conditions: (C1), (C2) and , (C3) and , (C4) and , (C5) and . Then, converges strongly to , which is the unique solution of the variational inequality:
Proof. In Theorem 3.3, put , and . Let in Theorem 3.3; then we have and
and we can obtain the desired conclusion from Theorem 3.3 immediately.
Corollary 3.5. Let be a nonempty closed convex subset of a real Hilbert space . Let and be two bifunctions from to satisfying (H1)–(H5) and let be a proper lower semicontinuous and convex function with assumption (B1) or (B2). Let be a contraction mapping with coefficient and let be a nonexpansive mapping with . Assume that
Let , and be the sequences generated by
where , and , and are three sequences in satisfying the following conditions: (C1), (C2) and , (C3) and , (C4) and , (C5) and . Then, converges strongly to , which is the unique solution of the variational inequality:
Proof. In Theorem 3.3, put to be equivalent to
and put to be equivalent to
Now, put . Then, it follows that
Observe that for all , we see that
Thus, let be a sequence satisfying the restriction: , where with Similarly, we obtain , where with and we obtain the desired result by Corollary 3.4.
Acknowledgments
The first author was supported by Kasetsart University Research and Development Institute (KURDI). The second author was supported by Rajamangala University of Technology Rattanakosin Research and Development Institute. The third author was supported by the Thailand Research Fund and the Commission on Higher Education under Grant No. MRG5380044.
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