A Cyclic Method for Solutions of a Class of Split Variational Inequality Problem in Banach Space

Ali, Bashir; Adam, Aisha A.; Ibrahim, Yusuf

doi:https://doi.org/10.1155/2020/7151953

Abstract and Applied Analysis

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Abstract Introduction Preliminaries Numerical Example Data Availability Conflicts of Interest Authors’ Contributions Acknowledgments References Copyright Related Articles

Research Article | Open Access

Volume 2020 | Article ID 7151953 | https://doi.org/10.1155/2020/7151953

A Cyclic Method for Solutions of a Class of Split Variational Inequality Problem in Banach Space

Bashir Ali,¹Aisha A. Adam,²and Yusuf Ibrahim^1,3

Academic Editor: Micah Osilike

Received15 Jan 2020

Revised26 Feb 2020

Accepted10 Mar 2020

Published10 Apr 2020

Abstract

In this paper, a cyclic algorithm for approximating a class of split variational inequality problem is introduced and studied in some Banach spaces. A strong convergence theorem is proved. Some applications of the theorem are presented. The results presented here improve, unify, and generalize certain recent results in the literature.

1. Introduction

Let be a nonempty closed and convex subset of a real Banach space , with dual . Then, a mapping is said to be(1)Nonexpansive if .(2)Demiclosed at zero if whenever a sequence in converges weakly to and converges strongly to 0, then .(3)L-Lipschitz continuous on if there exists such that A mapping is said to be(1)Monotone if(2)-inverse strongly monotone if(3)Strongly monotone if

Problem of the type finding such thatis called a variational inequality problem, and the set of solution of such problem is denoted by .

Variational inequality problems have played a crucial role in the study of several problems arising in physics, finance, economics, network analysis, optimization, medical image and structural analysis, and so on (see, for example, [1–5]). Variational inequality problems were formulated in the late 1960’s by Lions and Stampacchia [6]. Since then, various iterative algorithms for approximating solutions of such problems have been proposed by numerous researchers (see, for example, [7–12, 27]) and the references therein.

In 1976, Korpelevch [14] introduced the following extragradient method for solving the variational inequality problem when the operator is monotone and -Lipschitz continuous in a finite dimensional Euclidean space ,for each , .

The split feasibility problem in the finite dimensional Hilbert space was first introduced by Censor and Elfving [15] for modeling inverse problems which arise from phase retrievals and in medical image reconstruction.

Let and be two real Hilbert spaces. Let and be two nonempty closed convex subset of and , respectively. The split feasibility problem is to find

Assuming that the split feasibility problem is consistent (i.e., (7) has a solution), it is easy to see that solves (7) if and only if it solves the fixed point equation:where and are the orthogonal projections onto and , respectively, , and is the adjoint of . To solve (8), Byrne [16] proposed the algorithm which generates a sequence byfor each , where , being the spectral radius of the operator .

In 2010, Censor et al. [17] considered a new variational problem called split variational inequality problem (). It entails finding a solution of one variational inequality problem whose image under a bounded linear transformation is a solution of another variational inequality problem. The is formulated aswhere and are the nonempty closed convex subsets of real Hilbert spaces and , respectively, and is a bounded linear operator. They constructed the following iterative algorithm to solve such problem and proved a strong convergence theorem in a Hilbert space:where , is the spectral radius of the operator , and is the adjoint of .

One can easily observe that split variational inequality has the split feasibility problem as a special case.

Recently, Tian and Jiang [12], based on the work of Censor et al. [17], considered a class of which is to findwhere is a nonempty closed convex subset of a real Hilbert space , is a monotone and -Lipschitz continuous map, is a bounded linear map, and is a nonexpansive map. They proposed the following algorithm by combining the Korpelevich extragradient method and Byrne algorithm:

They obtained the following result.

Theorem 1. (see [12]). Let and be real Hilbert spaces. Let be a nonempty, closed and convex subset of , be a bounded linear operator such that , be a monotone and -Lipschitz continuous map, and be a nonexpansive map. Setting , assume that . Let the sequence be generated by (13), where , , and . Then, the sequence converges weakly to a point .

Remark 1. Inspired by the results of Tian and Jiang [12], the authors raised the following motivational questions: Q1. Can the result of Tian and Jiang hold in a more general setting of Banach space than Hilbert? Q2. Can the result also be proved for a common fixed point of finite family of nonexpansive mapping? Q3. Can strong convergence theorem be proved?In this paper, the above questions are answered in affirmative. We study a cyclic algorithm in the setting of uniformly smooth which is also 2-uniformly convex real Banach space and 2-uniformly smooth real Banach space and prove its strong convergence to a solution of a variational inequality problem for a monotone -Lipschitz continuous map whose image under a bounded linear operator is a common fixed point of a finite family of nonexpansive maps. Our theorems improve and extend the results of Tian and Jiang [12].

2. Preliminaries

The duality map of a Banach space has the following properties:(1)If is a reflexive, strictly convex, and smooth real Banach space, then is single-valued and bijective. In this case, the inverse is given by with being the duality mapping of .(2)In a Hilbert space , the duality map and its inverse are the identity maps on .(3)If is uniformly smooth and uniformly convex, then the dual space is also uniformly smooth and uniformly convex and the normalized duality map and its inverse, , are both uniformly continuous on bounded sets.

Let be a smooth real Banach space and be defined by

It is easy to see from the definition of that, in a real Hilbert space , equation (14) reduces to .

Furthermore, given and , we have the following properties (see, for example, [18]): P1: , P2: , P3:

Definition 1. Let be a smooth, strictly convex, and reflexive real Banach space and let be a nonempty, closed, and convex subset of . The map defined by such that is called the generalized projection of onto . Clearly, in a real Hilbert space , the generalized projection coincides with the metric projection from onto .

Definition 2. Let and be two reflexive, strictly convex, and smooth Banach spaces. The collection of mappings is linear, and continuous is a normed linear space with norm defined by . The dual operator defined by , is called the adjoint operator of A. The adjoint operator has the property .

Lemma 1 (see [19]). Let be a nonempty closed and convex subset of a smooth, strictly convex, and reflexive real Banach space . Then,(1)If , then if and only if , for all ,(2).

Lemma 2. (see [20]). Let be -uniformly smooth Banach space. Then, there exists a constant such that

Lemma 3. (see [21]). Let be a 2-uniformly convex and smooth real Banach space. Then, there exists a positive constant such that

Lemma 4. (see [22]). Let be a nonempty closed and convex subset of a reflexive space and , a monotone, and hemicontinuous map of into . Let be an operator defined bywhere is defined asThen, is maximal monotone and .

Lemma 5. (see [23]). Let be a uniformly convex and smooth real Banach space, and let and be two sequences of . If either or is bounded and , then .

3. Main Results

Theorem 2. Let be a uniformly smooth and 2-uniformly convex real Banach space and a 2-uniformly smooth real Banach space with smoothness constant . Let be a nonempty, closed, and convex subset of . Let be a monotone and -Lipschitz continuous map, and be a bounded linear operator with its adjoint such that . Let , be nonexpansive mappings. Setting and assuming . Let a sequence be generated bywhere , , being the smoothness constant of as in Lemma 2, and , being a positive constant as in Lemma 3. Then, the sequence converges to a point .

Proof. We divide the proof into five steps.

Step 1. We show that is closed and convex for any .
Since , is closed and convex.
Assume is closed and convex for some . Since for any ,we have that is closed and convex. Therefore, is closed and convex for any .

Step 2. We prove for any .
For , .
Assume for some . Let , thenFrom the fact that is 2-uniformly smooth, we haveUsing the fact that is 2-uniformly smooth and being nonexpansive, we haveFrom (22) and (23), we getAlso by Lemma 1, we haveBy the fact that and using property P2, we haveAlso from the fact that , , the Lipschitz continuity of , Lemma 1, and Lemma 3, we obtain thatThus,Hence, for any .

Step 3. We shall show that is a Cauchy sequence.
Since and , then by Lemma 1, we have that and also . Hence, is nondecreasing. So, exists. By property , is bounded. Also, it follows from (24), (28), and the fact that is a bounded linear operator that , , and are bounded.
From Lemma 1, we have that
as . Hence, is a Cauchy sequence.

Step 4. We show thatSince ,Taking limit as , we have .
Similarly,Taking limit as , we have .
Since is bounded, it follows from Lemma 5 that and .
Now,Taking limit as , we have .
In a similar way, we also have .
From (24), we obtainThus,From (24) and (28),(35) and (36), respectively, implies thatSince and are bounded, by Lemma 5, we getThus, (39) and (40) imply

Step 5. We show that converges to an element of
Since is a Cauchy sequence, we may assume that .
From the fact that we obtain that , , and . Since is a bounded linear operator, we have that .
From (35), we haveThus, for , .
Since is nonexpansive for each , we have that is demiclosed at 0 for . And therefore, .
Thus, . Next, we show that
DefineBy Lemma 4, is a maximal monotone and if and only if .
Let . .
Thus, . Since and , we have by Lemma 4 that
Thus,Using the fact that and , we haveUsing the fact that is uniformly continuous on bounded sets and is Lipschitz continuous, as , we haveSince is a maximal monotone, , and hence , therefore .

Corollary 1. Let be a uniformly smooth and 2-uniformly convex real Banach space and a 2-uniformly smooth real Banach space with smoothness constant . Let be a nonempty, closed, and convex subset of . Let be a monotone and -Lipschitz continuous map and be a bounded linear operator with its adjoint such that . Let be a nonexpansive map. Let . Let a sequence be generated bywhere , , being the smoothness constant of as in Lemma 2, and , being a positive constant as in Lemma 3. Then, the sequence converges to a point .

Proof. The result followed from Theorem 2 by setting .

Corollary 2. Let and be real Hilbert spaces. Let be a nonempty, closed, and convex subset of . be a bounded linear operator such that , be a monotone and -Lipschitz continuous map, and , , be a nonexpansive map. Let . Let a sequence be generated bywhere , , and . Then, the sequence converges strongly to a point .

Proof. Setting and in Theorem 2, the result is as follows.

Remark 2. Corollary 2 complements Theorem 1 (the result of Tian and Jiang [12]) in the sense that strong convergence of the sequence generated is obtained to a solution of the split variational inequality problem involving common fixed points of finite family of nonexpansive mappings, while weak convergence of the scheme is obtained in Theorem 1 involving only a single nonexpansive map. Though, additional condition of projecting the iterates on the half spaces at each step is imposed.

Lemma 6. Let be a real Hilbert space and be a nonempty closed convex subset of . Let be an - inverse strongly monotone mapping, that is, . Let . Then, the mapping is nonexpansive.

Proof. Let:Hence, is nonexpansive.

Corollary 3. Let and be real Hilbert spaces. Let and be two nonempty, closed, and convex subset of and , respectively. Let be a bounded linear operator such that , be a monotone and -Lipschitz continuous map, and be an inverse strongly monotone mapping. Setting . Let a sequence be defined bywhere , , , and . Then, the sequence converges to a point .

Proof. Since is nonexpansive for and if and only if for , putting in Corollary 1, we get the desired result.

4. Application to Equilibrium Problem

Let be a nonempty closed convex subset of a real Banach space and let be a bifunction.

The equilibrium problem with respect to and is to find such that

The set of solutions of the equilibrium problem mentioned above is denoted by . For solving the equilibrium problem, we assume that satisfies the following conditions: (A1) for all ; (A2) is monotone, i.e., ; (A3) for each , ; (A4) for each , is convex and lower semicontinous.

Lemma 7. (see [24]). Let be a reflexive, strictly convex, and uniformly smooth Banach space and be a nonempty closed convex subset of . Let be a bifunction satisfying conditions (A1)–(A4), then for any and , there exists a unique point such that

Lemma 8. (see [24]). Let be a reflexive, strictly convex, and smooth Banach space and be a nonempty closed convex subset of . Let be a bifunction satisfying conditions (A1)–(A4), then for any and , define a mapping byThen, the following holds:(1) is single valued;(2) is a firmly nonexpansive type, i.e.,(3);(4) is closed and convex.

Theorem 3. Let be a uniformly smooth and 2-uniformly convex real Banach space and a 2-uniformly smooth real Banach space with smoothness constant . Let and be two nonempty, closed, and convex subsets of and , respectively. Let be a bounded linear operator with its adjoint such that , be a monotone and -Lipschitz continuous map, and be a bifunction satisfying conditions (A1)–(A4). Let . Let a sequence be generated bywhere , , is the smoothness constant of as in Lemma 2, , being a positive constant as in Lemma 3, and is the resolvent of for . Then, the sequence converges to a point .

Proof. Putting in Corollary 1, we get the desired result.

5. Application to Maximal Monotone Operator

A set valued mapping with domain and range is said to be monotone if whenever . A monotone operator is said to be maximal monotone if its graph is not properly contained in the graph of any other monotone mapping. We know that if is a maximal monotone, then the zero of , is closed and convex. If is a smooth, strictly convex, and reflexive Banach space, then a monotone operator is maximal if and only if for each . Let be a smooth, strictly convex, and reflexive Banach space, be a nonempty closed convex subset of , and be a monotone operator satisfyingfor all . If is a maximal monotone, then (56) holds for and we can define the resolvent byfor all , i.e., . We know the following (see [23, 25–27]):(1) is single valued,(2), where is the set of fixed points of ,(3) is a firmly nonexpansive type, i.e.,

Theorem 4. Let be a uniformly smooth and 2-uniformly convex real Banach space and a 2-uniformly smooth real Banach space with smoothness constant . Let and be two nonempty, closed, and convex subsets of and , respectively; be a bounded linear operator with its adjoint such that ; be a monotone and -Lipschitz continuous map; and be a maximal monotone operator. Let . Let a sequence be generated bywhere , , being the smoothness constant of as in Lemma 2, , being a positive constant as in Lemma 3, and is the resolvent of for . Then, the sequence converges to a point .

Proof. Putting in Corollary 1, we get the desired result.

6. Application to Constrained Convex Minimization Problem

The problem of finding such thatwhere is a nonempty closed convex subset of and is a real-valued convex function is called constraint convex minimization problem. We denote the set of solution of the constraint convex minimization problem by .

Lemma 9. (see [12]). Let be a real Hilbert space and be a nonempty closed convex subset of . Let be a convex function of into . If is differentiable, then is a solution of the constraint convex minimization problem if and only if .

Theorem 5. Let and be real Hilbert spaces. Let and be two nonempty, closed, and convex subset of and , respectively. Let be a bounded linear operator such that , and be a monotone and -Lipschitz continuous. Let be a differentiable convex function and suppose is -inverse strongly monotone mapping. Setting , assume that . Let a sequencewhere , , , and . Then, the sequence converges to a point .

Proof. Putting in Corollary 3, by Lemma 9 we get the desired result.

7. Application to Split Minimization Problem

Let and be real Hilbert spaces. Let and be two nonempty, closed, and convex subset of and , respectively, and , be two convex functions. Let be a bounded linear operator.

The problem of finding satisfying the conditionsis called the split minimization problem.

Theorem 6. Let and be real Hilbert spaces. Let and be two nonempty, closed, and convex subset of and , respectively. Let be a bounded linear operator such that , and and be two differentiable convex functions. Suppose that is -Lipschitz continuous and is -inverse strongly monotone mapping, let . Let a sequencewhere , , , and . Then, the sequence converges to a point .

Proof. Since is convex, we have is monotone. Putting and in Corollary 3 and by Lemma 9, we get the desired result.

8. Numerical Example

In this section, we present a numerical example to show the convergence of a sequence generated by our algorithm. Let , .

Set and Let be defined by

Then, and are nonexpansive.

Let be defined by

Then, is monotone and .

Let be defined by

Then, is a bounded linear operator and , .

When , .

So, .

Clearly, . Taking and . It follows from Theorem 2 that a sequence is generated by the following algorithm:converges strongly to (see Figure 1).

Data Availability

Data sharing not applicable to this article as no datasets were generated or analysed during the current study.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Authors’ Contributions

All authors contributed equally and significantly in writing this paper. All authors read and approved the final manuscript.

Acknowledgments

This work was supported by ACBF and AfDB research grant funds to AUST. The authors appreciate the support of their institution. They thank ACBF and AfDB for their financial support.

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