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Viscosity approximation methods for hierarchical optimization problems in spaces
Journal of Inequalities and Applicationsvolume 2013, Article number: 471 (2013)
This paper aims at investigating viscosity approximation methods for solving a system of variational inequalities in a space. Two algorithms are given. Under certain appropriate conditions, we prove that the iterative schemes converge strongly to the unique solution of the hierarchical optimization problem. The result presented in this paper mainly improves and extends the corresponding results of Shi and Chen (J. Appl. Math. 2012:421050, 2012, doi:10.1155/2012/421050), Wangkeeree and Preechasilp (J. Inequal. Appl. 2013:93, 2013, doi:10.1186/1029-242X-2013-93) and others.
The concept of variational inequalities plays an important role in various kinds of problems in pure and applied sciences (see, for example, [1–11]). Moreover, the rapid development and the prolific growth of the theory of variational inequalities have been made by many researchers.
In a space, Saejung  studied the convergence theorems of the following Halpern iterations for a nonexpansive mapping T. Let u be fixed and be the unique fixed point of the contraction ; i.e.,
where and are arbitrarily chosen and
where . It is proved that converges strongly as to such that , and converges strongly as to under certain appropriate conditions on , where is a metric projection from X onto C.
In 2012, Shi and Chen  studied the convergence theorems of the following Moudafi viscosity iterations for a nonexpansive mapping T: For a contraction f on C and , let be the unique fixed point of the contraction ; i.e.,
and is arbitrarily chosen and
where . They proved that defined by (1.3) converges strongly as to such that in the framework of a space satisfying the property , i.e., if for ,
Furthermore, they also obtained that defined by (1.4) converges strongly as to under certain appropriate conditions imposed on .
By using the concept of quasilinearization, which was introduced by Berg and Nikolaev , Wangkeeree and Preechasilp  studied the strong convergence theorems of iterative schemes (1.3) and (1.4) in spaces without the property . They proved that iterative schemes (1.3) and (1.4) converge strongly to such that , which is the unique solution of the variational inequality (VIP)
In this paper, we are interested in the following so-called hierarchical optimization problems (HOP). More precisely, let be two contractions with coefficient , and let be two nonexpansive mappings such that and are nonempty. The class of hierarchical optimization problems (HOP) consists in finding such that the following inequalities hold:
For this purpose, we introduce the following iterative schemes:
where , and
(H3) either or .
We prove that iterative schemes (1.7) and (1.8) converge strongly to such that and , which is the unique solution of (1.6).
Let be a metric space. A geodesic path joining to (or, more briefly, a geodesic from x to y) is a map such that , , and for all . In particular, c is an isometry and . The image of c is called a geodesic segment joining x and y. When it is unique, this geodesic segment is denoted by . The space is said to be a geodesic space if every two points of X are joined by a geodesic, and X is said to be uniquely geodesic if there is exactly one geodesic joining x and y for each . A subset is said to be convex if Y includes every geodesic segment joining any two of its points.
A geodesic triangle in a geodesic metric space consists of three points , , and in X (the vertices of △) and a geodesic segment between each pair of vertices (the edges of △). A comparison triangle for the geodesic triangle in is a triangle in the Euclidean plane such that for .
A geodesic space is said to be a space if all geodesic triangles satisfy the following comparison axiom.
: Let △ be a geodesic triangle in X, and let be a comparison triangle for △. Then △ is said to satisfy the inequality if for all and all comparison points ,
Let by [, Lemma 2.1(iv)] for each , then there exists a unique point such that
From now on, we will use the notation for the unique point z satisfying (2.1).
We now collect some elementary facts about spaces which will be used in the proofs of our main results.
Lemma 2.1 Let X be a space. Then
(see [, Lemma 2.4]) for each and , one has(2.2)
(see ) for each and , one has(2.3)
(see ) for each and , one has(2.4)
(see ) for each and , one has(2.5)
(see ) for each and , one has(2.6)
Let C be a nonempty subset of a complete space X. Recall that a self-mapping is a nonexpansion on C iff for all . A point is called a fixed point of T if . We denote by the set of all fixed points of T. A self-mapping is a contraction on C if there exists a constant such that . Banach’s contraction principle  guarantees that f has a unique fixed point when C is a nonempty closed convex subset of a complete metric space.
Fixed-point theory in spaces was first studied by Kirk (see [19, 21]). He showed that every nonexpansive (single-valued) mapping defined on a bounded closed convex subset of a complete space always has a fixed point. Since then, the fixed-point theory for single-valued and multivalued mappings in spaces has been rapidly developed.
Berg and Nikolaev  introduced the concept of quasilinearization as follows.
Let be a metric space. Let us formally denote a pair by and call it a vector. Then quasilinearization is defined as a mapping defined by
It is easily seen that , and for all .
We say that X satisfies the Cauchy-Schwarz inequality if
for all .
It is known [, Corollary 3] that a geodesically connected metric space is a space if and only if it satisfies the Cauchy-Schwarz inequality.
Recently, Dehghan and Rooin  presented a characterization of a metric projection in spaces as follows.
Lemma 2.2 Let C be a nonempty closed and convex subset of a complete space X, and . Then if and only if for all .
Let be a bounded sequence in a space X. For , we set
The asymptotic radius of is given by
and the asymptotic center of is the set
It is known from Proposition 7 of  that for each bounded sequence in a complete space, consists of exactly one point. A sequence is said to △-converge to if for every subsequence of . We use to denote that △-converges to x. The uniqueness of an asymptotic center implies that the space X satisfies Opial’s property, i.e., for given such that , then for any given with , the following holds:
Lemma 2.3 
Assume that X is a complete space. Then:
Every bounded sequence in X always has a △-convergent subsequence.
If C is a closed convex subset of X and is a nonexpansive mapping, then the conditions and imply and .
The following lemma shows a characterization of △-convergence.
Lemma 2.4 
Let X be a complete space, be a sequence in X, and . Then if and only if for all .
Lemma 2.5 
Let be a sequence of nonnegative real numbers satisfying the property
where and such that
Then converges to zero as .
Lemma 2.6 
Let X be a complete space. Then,
for each , one has(2.9)
for any and , letting for all , we have:
3 Main results
Now we are ready to give our main results in this paper.
Let be a metric space. Define a mapping by
for all . Then it is easy to verify that is a metric space, and is complete if and only if is complete.
Lemma 3.1 Let C be a closed convex subset of a complete space. Let be two contractions with coefficient , and let be two nonexpansive mappings. For any , define another mapping by
Then is a contraction on .
Proof For any and , we have
This implies that is a contraction mapping. Therefore there exists a unique fixed point of such that
Theorem 3.2 Let C be a closed convex subset of a complete space X, and let be two nonexpansive mappings such that and are nonempty. Let f, g be two contractions on C with coefficient . For each , let and be given by (1.7). Then and as such that , which is the unique solution of HOP (1.6).
Proof We first show that and are bounded. Indeed, take to derive that
After simplifying, we have
Hence and are bounded, so are , , and . Consequently,
In particular, we have
Next we prove that is relatively compact as .
In fact, let be any subsequence such that as . Put and . Now we prove that contains a subsequence converging strongly to where , and it is a solution of HOP (1.6).
In fact, since and are both bounded, by Lemma 2.3(i), (ii) and (3.1), we may assume that and , and , . Hence it follows from Lemma 2.6 that
After simplifying, we have
Since and , by Lemma 2.4,
Hence we have
It follows from (3.3) that . Hence and .
Next we show that , which solves HOP (1.6).
Indeed, for each , , we have
This implies that
Letting and taking the limit and noting that and , we have
It is similar to proving that
That is, solves inequality (1.6).
Finally, we show that the entire net converges to , and converges to . In fact, for any subsequence such that (as ), assume that and . By the same argument as above, we get that and solves inequality (1.6). Hence we have
Adding up (3.5) and (3.6), we get that
Since , we have that , and so , . Hence the entire net converges to and converges to , which solves HOP (1.6). This completes the proof of Theorem 3.2. □
Theorem 3.3 Let C be a closed convex subset of a complete space X, and let be two nonexpansive mappings such that and are nonempty. Let f, g be two contractions on C with coefficient . Let and be the sequences defined by (1.8). If conditions (H1)-(H3) are satisfied, then and as , where , , which solves HOP (1.6).
Proof First we show that and are bounded. Indeed, taking , it follows that
By induction, we can prove that
for all . This implies that and are bounded, so are , , and .
We claim that and . Indeed, we have (for some appropriate constant )
By conditions (H2) and (H3) and Lemma 2.5, we have
and thus , .
Consequently, by condition (H1), we have
This implies that
Let and be two nets in C such that
By Theorem 3.2, we have that and as such that , , which solves the variational inequality (1.6). Now, we claim that
From Lemma 2.5, we have
where . Simplifying this, we have
Taking the limit as first and then letting on both sides of (3.11), we have
Since and as , by the continuity of a metric d, it follows that
This implies that, for any , there exists such that for any , we have
First letting and taking limit, and then letting and taking the upper limit on (3.12), we obtain
Since is arbitrary, we have
Finally, we prove that and as . Indeed, taking , for any , it follows from Lemma 2.6(i) that
which implies that
where . Thus,
Applying Lemma 2.6, we have . Hence and as . This completes the proof of Theorem 3.3. □
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The authors would like to express their thanks to the referees for their helpful suggestions and comments. This work was supported by the Scientific Research Fund of Sichuan Provincial Education Department (11ZA221) and the Scientific Research Fund of Science Technology Department of Sichuan Province 2011JYZ010.
The authors declare that they have no competing interests.
All the authors participated in this article’s design and coordination. And they read and approved the final manuscript.