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Abstract and Applied Analysis
Volume 2013 (2013), Article ID 836720, 6 pages
A Projection-Type Method for Multivalued Variational Inequality
Institute of Applied Mathematics, Chongqing University of Posts and Telecommunications, Chongqing 400065, China
Received 18 April 2013; Revised 5 August 2013; Accepted 19 August 2013
Academic Editor: Shawn X. Wang
Copyright © 2013 Changjie Fang et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
We propose a projection-type method for multivalued variational inequality. The iteration sequence generated by the algorithm is proven to be globally convergent to a solution, provided that the multivalued mapping is continuous with nonempty compact convex values. Moreover, we present a necessary and sufficient condition on the nonemptiness of the solution set. Preliminary computational experience is also reported.
We consider the following multivalued variational inequality, denoted by to find and such that where is a nonempty closed convex set in , is a multivalued mapping from into with nonempty values, and and denote the inner product and norm in , respectively.
Projection-type algorithms have been extensively studied in the literature; see [1–8] and the references therein. Reference  proposes a subgradient extragradient algorithm for solving single-valued variational inequality in which the next iterate is a projection onto a halfspace whose bounding hyperplane supports the feasible set at a certain point. Reference  proposes a projection method for variational inequality problems in which the hyperplane strictly separates the current iterate from the solution set of (1). Theory and algorithm of multivalued variational inequality have been much studied in the literature [1, 9–18]. Various algorithms for computing the solution of (1) are proposed. The well-known proximal point algorithm  requires the multivalued mapping to be monotone. Relaxing the monotonicity assumption,  proposed the double projection algorithm for solving (1); also see . Assume that is pseudomonotone;  described a combined relaxation method for solving (1); see also . Recently,  proposed an extragradient method for generalized variational inequality. In , the next iterate is the projection of the current iterate onto the feasible set ; also see .
In this paper, we introduce a projection-type method for multivalued variational inequality in which the next iterate is a projection of the initial iterate onto intersection of two halfspaces containing the solution set. We obtain a global convergence theorem, assuming that is pseudomonotone on with respect to the solution set; see (3) in the following. Moreover, we show that the iterative sequence diverges if and only if the solution set is empty. We also present numerical results of the proposed method. Now let us compare our algorithm with algorithms in [5, 12, 22, 23]. First, the next iterate in our method relates to the initial point. In [5, 12], the next iterate is the projection of the current iterate onto the intersection of the hyperplane and the feasible set . Secondly, the next iterate in our method is a projection of the initial point onto the intersection of the two hyperplanes and the feasible set . In addition, our Armijo-type linesearch procedure is also different from those in [12, 22, 23].
The organization of this paper is as follows. In the next section, we present the algorithm details and some lemmas. We prove several preliminary results for convergence analysis in Section 3. Numerical results are reported in the last section.
Let us recall the definition of continuous multivalued mapping. is said to be upper semicontinuous at if, for every open set containing , there is an open set containing such that for all . is said to be lower semicontinuous at if, given any sequence converging to and any , there exists a sequence that converges to . is said to be continuous at if, it is both upper semicontinuous and lower semicontinuous at . If is single valued, then both upper semicontinuity and lower semicontinuity reduce to the continuity of .
is called pseudomonotone on in the sense of Karamardian , if, for any ,
Let be the solution set of (1), that is, those points satisfying (1). Throughout this paper, we assume that the solution set of the problem (1) is nonempty and is continuous on with nonempty compact convex values satisfying the following property: The property (3) holds if is pseudomonotone on .
Let denote the projector onto , and let be a parameter.
Proposition 1. and solves the problem (1) if and only if
Algorithm 2. Choose and three parameters , and . Set .
Step 1. If for some , stop; else take arbitrarily .
Step 2. For every positive integer , let .
Step 3. Let be the smallest nonnegative integer satisfying Set and .
Step 4. Compute , where Let , and go to Step 1.
Remark 3. Let us compare the previous algorithm with Algorithm 3.1 in . First, the parameter is required to be strictly less than 1, and is assumed to be equal to 1 in their Algorithm 3.1. In our Algorithm 2, the parameter can take any positive scalar and . Secondly, since has compact convex values, has closed convex values. Therefore, in Step 2 of our algorithm is uniquely determined by . Hence, it is easy to compute the value of satisfying (5). In Step 1 of their Algorithm 3.1, since is a multivalued mapping, it is very difficult in practice to compute the value of satisfying and at the same time. In addition, we report numerical results concerning our algorithm, while  does not present numerical experiments for the proposed algorithm. Finally, we compare the performance of our algorithm and Algorithm 3.1 in  (see Table 4).
Lemma 4. The sequence generated in Step 2 has the following properties:
Proof. See Lemma 2.1 in .
Lemma 5. For every and ,
Proof. See Lemma 2.3 in .
We show that Algorithm 2 is well defined and implementable.
Lemma 6. If , there exists satisfying (5).
Lemma 7. Let be a closed convex subset of . For any and , the following statements hold:(i),(ii).
Proof. See .
Lemma 8. Let and . If , then the hyperplane strictly separates and the solution set .
Lemma 9. If , then .
Proof. It follows from Lemma 8 that . Next, it is sufficient to prove that for all . The proof will be given by induction.
Obviously, . Suppose that Then, .
Let . Since it follows from Lemma 7 (i) that Thus, . Therefore, we obtain that for all .
Lemma 10. Let be a nonempty bounded closed convex set, and let the mapping be lower semicontinuous with nonempty closed convex values; then, the solution set of is nonempty.
Proof. See Lemma 2.9 in .
The following lemma says that if the solution set is empty, then is a nonempty set, which implies the feasibility of Algorithm 2.
Lemma 11. Let be continuous with nonempty compact convex values on , and suppose that ; then, for all .
Proof. On the contrary, suppose that there exists such that . Then, there exists a positive number such that
Since is continuous with compact values, Proposition 3.11 in  implies that is a bounded set, and so is bounded. Without loss of generality, we assume that
Consider the variational inequality , where
It follows from Lemma 10 that the solution set of , denoted by , is nonempty. We denote the three sequences , and by , and , respectively, when Algorithm 2 is applied to with starting point . We claim that(i)the set has at least elements;(ii) for ;(iii) is not a solution of .
Items (i) and (iii) are obvious. Next we prove the item (ii). It is sufficient to prove that where .
Since where the second inequality follows from and Lemma 7 (ii), so , and hence, . Therefore,
Since , it follows from Lemma 9 that . Therefore, for , which contradicts the supposition that .
3. Main Results
Theorem 12. Let be continuous with nonempty compact convex values on satisfying condition (3). Suppose that Algorithm 2 generates an infinite sequence . If the solution set of is nonempty, then globally converges to a solution of satisfying .
Proof. Since , by Lemma 9 and the definition of projection, it follows that
Therefore, is a bounded sequence.
Since , . Since it follows that Thus, it follows from Lemma 7 (ii) that that is, Thus, the sequence is nondecreasing and bounded and hence convergent, which implies that On the other hand, since and since , we have where the second inequality follows from (5). Therefore, it follows from (27) that Since is continuous with compact values, Proposition 3.11 in  implies that is a bounded set, and so the sequences and are bounded. Thus, the continuity of implies that is a bounded set. Therefore, is bounded. It follows that By the boundedness of , there exists a convergent subsequence converging to .
If is a solution of the problem (1), we show next that the whole sequence converges to . Let . Since , by Lemma 9, Therefore, Thus, Letting in (34), we have where the last inequality follows from Lemma 7 (i) and the fact that and . Therefore, Thus, the sequence has a unique cluster point , which shows the global convergence of .
Suppose now that is not a solution of the problem (1). We show first that in Algorithm 2 cannot tend to . Since is continuous with compact values, Proposition in  implies that is a bounded set, and so the sequence is bounded. Therefore, there exists a subsequence converging to . Since is upper semicontinuous with compact values, Proposition in  implies that is closed, and so . By the definition of , we have If , then . The lower continuity of , in turn, implies the existence of such that converges to . Since , we have and . Therefore and Letting in (38), we have with being continuous. It follows from Lemma 5 that Thus, we obtain the contradiction because . Therefore, is bounded and so is .
By (31) and the boundedness of , we obtain that . Since is continuous and the sequences and are bounded, there exists an accumulation point of such that . This implies that solves the variational inequality (1). Similar to the preceding proof, we obtain that globally converges to .
Theorem 14. Let be continuous with nonempty compact convex values on satisfying condition (3). Suppose that Algorithm 2 generates an infinite sequence . Then the solution set of is empty if and only if the sequence generated by Algorithm 2 diverges.
Proof. In view of Theorem 12, it is sufficient to prove that if the solution set is empty, then the sequence generated by Algorithm 2 diverges. Since inequality (26) also holds in this case, the sequence is still nondecreasing. We claim that Otherwise, is bounded, and hence it follows from (26) that A similar discussion as in Theorem 12 would lead to the conclusion that every cluster point of is a solution of , which contradicts the emptiness of the solution set to .
4. Numerical Experiments
In this section, we present some numerical experiments for the proposed algorithm. The MATLAB codes are run on a PC (with CPU Intel P-T2390) under MATLAB Version 188.8.131.5204(R14) Service Pack 1. We compare the performance of our Algorithm 2 and the algorithms in [5, 11, 12, 22]. In Tables 1, 2, 3, and 4, “It.” denotes number of iteration, and “CPU" denotes the CPU time in seconds. The tolerance means that when , the procedure stops.
Example 15. Let , and be defined by Then, the set and the mapping satisfy the assumptions of Theorem 12, and is a solution of the multivalued variational inequality. Example 15 is tested in [5, 12, 22]. We choose , and for our algorithm and Algorithm 1 in ; , and for Algorithm 1 in ; , and for Algorithm 1 in ; for Algorithm 3.1 in . We use as the initial point (Tables 1– 4).
This work is partially supported by Natural Science Foundation Project of CQ CSTC (no. 2010BB9401), Science and Technology Project of Chongqing Municipal Education Committee of China (no. KJ110509), and Foundation of Chongqing University of Posts and Telecommunications for the Scholars with Doctorate (no. A2012-04).
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