Abstract
This article aims to research iterative schemes for searching a solution of a quasimonotone variational inequality in a Hilbert space. For solving this quasimonotone variational inequality, we propose an iterative procedure which combines a self-adaptive rule and the extragradient algorithm. We demonstrate that the procedure weakly converges to the solution of the investigated quasimonotone variational inequality provided the considered operator satisfies several additional conditions.
1. Introduction
Variational inequality emerged in 1964 arising from the study of mechanics has many applications in engineering, economics, operations research, etc. ([1–4]). Variational inequality theory acts as a tool for solving many problems, such as equilibrium problems ([5, 6]), optimization problems ([7–9]), fixed point problems ([10–12]), and split problems ([13–17]). There are numerous iterative schemes for solving variational inequalities in the existing results; see [18–25]. Next, we briefly review several valuable iterative methods.
Throughout, suppose that is a Hilbert space. The symbols and are the inner and norm of , respectively. Let be a convex and closed set. For an operator , the variational inequality aims to seek a point satisfying
We use to indicate the set of solutions of (1).
A valuable algorithm for solving (1) is the projection algorithm ([26, 27]) which generates a procedure as follows: where stands for the orthogonal projection and means the step-size.
When is strongly monotone, strongly pseudomonotone, or inverse strongly monotone, iterative scheme (2) is convergent ([28, 29]).
Another powerful method is extragradient method studied by Korpelevich [30] which generates a procedure starting from an initial point :
Thereafter, (3) has been discussed extensively for solving (1); see, e.g., [30–34]. The main reason why the extragradient method attracts so much attention is that extragradient method can be used to find a solution of plain monotone operators. In fact, extragradient algorithm can be used to solve (1) if is pseudomonotone and sequentially weakly continuous ([35–37]).
Very recently, iterative methods for solving quasimonotone variational inequality have been investigated in the literature [24, 38, 39]. Especially, Salahuddin [40] utilized (3) for solving a Lipschitz quasimonotone variational inequality and achieve the following result.
Conclusion 1 ([40]). Assume that the operator satisfies (i) quasimonotone on ; (ii) sequentially weakly continuous on ; and (iii) Lipschitz continuous on . Suppose that and . Then, obtained from (3) weakly converges to .
In this article, we further utilize extragradient method (3) for solving quasimonotone variational inequality (1). For this task, we will make use of an auxiliary tool regarding the following dual variational inequality which is to find satisfying:
We use to indicate the set of solutions of (4).
Notice that is closed convex. At the same time, we have when is continuous and is convex. However, to acquire the convergence of the constructed sequence, one has to add the following extra condition which implies that
Note that the above condition (5) holds if is pseudomonotone. However, this condition (5) is not satisfied when is quasimonotone. Further, self-adaptive rule was applied for solving variational inequality problems, see [41–45]. In this paper, for solving quasimonotone variational inequality (1), we propose an iterative procedure which combines a self-adaptive method and extragradient method (3) without using condition (5). We show that the suggested iterative procedure is weakly convergent. Our result extends the above theorem (1) at two aspects: on the one hand “sequential weak continuity” imposed on can be replaced by a more general restriction and on the other hand a self-adaptive technique is used to relax Lipschitz condition of .
2. Notions and Lemmas
Throughout, suppose that is a Hilbert space and is convex and closed. A map is called (1)Monotone if(2)Pseudomonotone if(3)Quasimonotone if
By the above definition, we can deduce that if is pseudomonotone, then must be quasimonotone. However, the reverse conclusion may fail.
A map is called Lipschitz continuous if where is some positive constant. In this case, we call -Lipschitz. is called nonexpansive provided .
An orthogonal projection from onto , denoted by fulfills
possesses the following characteristic inequality:
3. Algorithms and Convergence Results
First, we declare several related conditions. Suppose that is a Hilbert space and is convex and closed. Suppose that the involved operator satisfies three restrictions:
(t1) is a quasimonotone operator
(t2) is -Lipschitz on
(t3) If with being a sequence in and , then
In the sequel, assume that and the set is finite.
Suppose that and are three constants in the open interval . Suppose that is a sequence in satisfying .
Next, we state our scheme for solving (1).
Algorithm 2. Select a fixed point in . Set .
Step 1. Assume is presented. Compute
where fulfills
Step 2. Compute . (2a) If , then stop. (2b) If , then calculate
where and compute
Let and return to Step 1.
Conclusion 3. Inequality (14) is well-defined. Moreover, or .
Proof. Since is -Lipschitz, we obtain
which equals to
This implies that (14) holds for all
It is obviously that
which implies that
Conclusion 4. If , then . (ii) If , then and (15) is well-defined.
Proof. (i)If , that is , by virtue of (12), we acquirewhich results in that .
(ii)Take . Notice thatAs a result of and , we have
As the same as (22), we obtain
due to .
In view of (21)-(23), we receive
Owing to , from (14), we get
Using (12) of and (14), we acquire
In the light of (24), (25) and (26), we achieve
If , then . Otherwise, by virtue of (12), which results in that and hence . This leads to a contradiction. So, . It follows from (27) that which yields that . Therefore, (15) is well-defined.
In this position, we prove a main theorem.
Theorem 5. defined by Algorithm 2 weakly converges to an element in .
Proof. Let . Since is nonexpansive and , from (16) and (21), we have Substituting (27) into (28) to derive Noting that , by (29), we acquire which leads to that exists. Then, is bounded and so is . From (13), we have Hence, and are bounded.
Taking into account (29), we gain
This leads to
Thanks to the boundedness of , by (33), we have
With the help of the Lipschitz continuity of , from (34), we deduce
Based on (15) and (16), we derive
In the light of (33) and (36), we achieve
So,
Observe that , , and are bounded. According to (34) and (39), we have
Owing to is bounded, there is fulfilling as . Taking into account (40), we attain
If , by and verifying (t1), we get that . Then, .
Now, we assume that . Then, there is an integer fulfilling for all . By virtue of (41), we attain
Let be a real number sequence fulfilling and as . Based on (42), there is of fulfilling and which results in that
Put for all . It is easily seen that for all . With the help of (44), we achieve
Owing to (45) and using the quasimonotonicity of , we acquire
As a result of Lipschitz continuity of and , we deduce as . In (46), letting , we receive
Thus, . Therefore, . Next, we prove has no more than one weak cluster point in . Suppose that and are two weak cluster points of . Then, there exist two subsequences and such that and .
It is obviously that
Letting on both sides of (48), we have that exists, denoted by . Therefore,
Note that and . By (45), we obtain which yields that . So, has no more than one weak cluster point in . Since the set is finite, we deduce that has only finite weak cluster points in . Let be unequal weak cluster points of in . Let and
For , there is of fulfilling when . Hence,
For , we have
Thanks to (52) and (53), there exists an integer satisfying for all ,
Let
Combining (54) with (55), we obtain
Next we demonstrate that if is large enough, . Suppose that there is of such that . According to the boundedness of , there exists a subsequence of , without loss of generality, still denoted by , which converges weakly to . Hence for any . Then, there is a subsequence of such that :
So, which results in that . It leads to a contradiction. Thus, there is a large enough integer such that for all .
Finally, we show that is singleton in . Suppose that . Taking into account (37), there is fulfilling when . Hence, there is fulfilling and , where and , that is
Thus, we have
Thanks to (59) and (60), we acquire
Note that
By virtue of (61) and (62), we receive which is impossible. Then, is singleton in . So, weakly converges to an element in .
Remark 6. A map is called weakly sequentially continuous, if , where is any sequence in .
To solve (1), many existing results have imposed the above “sequential weak-to-weak continuity” condition on ; see, [35, 36, 40]. We can check if satisfies sequential weak continuity and then satisfies condition (t3).
4. Conclusions
The main purpose of this paper is to investigate iterative algorithms for solving variational inequality (1). A powerful method to solve (1) is extragradient method (3) introduced by Korpelevich [30] where the involved operator is pseudomonotone monotone. Based on the corresponding result of Salahuddin [40], we further apply extragradient method (3) to solve quasimonotone variational inequality (1).
We propose an iterative algorithm (Algorithm 2) which combines a self-adaptive rule and the extragradient algorithm. In general, in order to show belongs to the solution set, should be sequentially weakly continuous. In this paper, we replace these conditions by a weaker condition (t3). We demonstrate that the procedure weakly converges to the solution of the investigated quasimonotone variational inequality under several additional conditions.
Data Availability
No data were used to support this 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 article. All authors read and approved the final manuscript.
Acknowledgments
Li-Jun Zhu was supported by the National Natural Science Foundation of China (grant number 61362033) and the Natural Science Foundation of Ningxia Province (grant number NZ17015).