An Iterative Method for Solving Split Monotone Variational Inclusion Problems and Finite Family of Variational Inequality Problems in Hilbert Spaces
The purpose of this paper is to study the convergence analysis of an intermixed algorithm for finding the common element of the set of solutions of split monotone variational inclusion problem (SMIV) and the set of a finite family of variational inequality problems. Under the suitable assumption, a strong convergence theorem has been proved in the framework of a real Hilbert space. In addition, by using our result, we obtain some additional results involving split convex minimization problems (SCMPs) and split feasibility problems (SFPs). Also, we give some numerical examples for supporting our main theorem.
Let and be real Hilbert spaces whose inner product and norm are denoted by and , respectively, and let C, Q be nonempty closed convex subsets of and , respectively. For a mapping , we denoted by the set of fixed points of (i.e., ). Let be a nonlinear mapping. The variational inequality problem (VIP) is to find such thatand the solution set of problem (1) is denoted by . It is known that the variational inequality, as a strong and great tool, has already been investigated for an extensive class of optimization problems in economics and equilibrium problems arising in physics and many other branches of pure and applied sciences. Recall that a mapping is said to be -inverse strongly monotone if there exists such that
A multivalued mapping is called monotone if for all , , for any and . A monotone mapping is maximal if the graph for is not properly contained in the graph of any other monotone mapping. It is generally known that is maximal if and only if for for all implies . Let be a multivalued maximal monotone mapping. The resolvent mapping associated with is defined bywhere stands for the identity operator on . We note that for all , the resolvent is single-valued, nonexpansive, and firmly nonexpansive.
In 2011, Moudafi  introduced the following split monotone variational inclusion problem (SMVI):where is the zero vector in and , and are multivalued mappings on and , and are two given single-valued mappings, and is a bounded linear operator with adjoint of . We note that if (4) and (5) are considered separately, we have that (4) is a variational inclusion problem with its solution set and (5) is a variational inclusion problem with its solution set . We denoted the set of all solutions of (SMVI) by .
It is worth noticing that by taking normal cones to closed convex sets and , then (SMVI) (4) and (5) reduce to the split variational inequality problem (SVIP) that was introduced by Censor et at. . In , they mentioned that (SMVI) (4) and (5) contain many special cases, such as split minimization problem (SMP), split minimax problem (SMMP), and split equilibrium problem (SEP). Some related works can be found in [1, 3–10].
For solving (SMVI) (4) and (5), Modafi  proposed the following algorithm.
Algorithm 1. Let , , and the sequence be generated bywhere with being the spectral radius of the operator .
He obtained the following weak convergence theorem for algorithm (6).
Theorem 1. (see ). Let , be real Hilbert spaces. Let be a bounded linear operator with adjoint . For , let be -inverse strongly monotone with and let be two maximal monotone operators. Then, the sequence generated by (6) converges weakly to an element provided that , , and with being the spectral radius of the operator .
Since then, because of a lot of applications of (SMVI), it receives much attention from many authors. They presented many approximation methods for solving (SMVI) (4) and (5). Also the iterative methods for solving (SMVIP) (4) and (5) and fixed-point problems of some nonlinear mappings have been investigated (see [11–19]).
On the other hand, Yao et al.  presented an intermixed Algorithm 1.3 for two strict pseudo-contractions in real Hilbert spaces. They also showed that the suggested algorithms converge strongly to the fixed points of two strict pseudo-contractions, independently. As a special case, they can find the common fixed points of two strict pseudo-contractions in Hilbert spaces (i.e., a mapping is said to be -strictly pseudo-contractive if there exists a constant such that ).
Algorithm 2. For arbitrarily given , let the sequences and be generated iteratively bywhere and are two sequences of the real number in , are -strictly pseudo-contractions, is a -contraction, is a -contraction, and is a constant.
Under some control conditions, they proved that the sequence converges strongly to and converges strongly to , respectively, where , , and are the metric projection of onto and , respectively. After that, many authors have developed and used this algorithm to solve the fixed-point problems of many nonlinear operators in real Hilbert spaces (see for example [21–27]). Question: can we prove the strong convergence theorem of two sequences of split monotone variational inclusion problems and fixed-point problems of nonlinear mappings in real Hilbert spaces?
The purpose of this paper is to modify an intermixed algorithm to answer the question above and prove a strong convergence theorem of two sequences for finding a common element of the set of solutions of (SMVI) (4) and (5) and the set of solutions of a finite family of variational inequality problems in real Hilbert spaces. Furthermore, by applying our main result, we obtain some additional results involving split convex minimization problems (SCMPs) and split feasibility problems (SFPs). Finally, we give some numerical examples for supporting our main theorem.
Let be a real Hilbert space and be a nonempty closed convex subset of . We denote the strong convergence of to and the weak convergence of to by notations “ as ” and “ as ,” respectively. For each and with , we have
Definition 1. Let be a real Hilbert space and be a closed convex subset of . Let be a mapping. Then, is said to be(1)Monotone, if (2)Firmly nonexpansive, if (3)Lipschitz continuous, if there exists a constant such that (4)Nonexpansive, if =
It is well known that if is -inverse strongly monotone, then it is -Lipschitz continuous and every nonexpansive mapping is 1-Lipschitz continuous. We note that if is a nonexpansive mapping, then it satisfies the following inequality (see Theorem 3 in  and Theorem 1 in ):Particularly, for every and , we have
For every , there is a unique nearest point in such that
Such an operator is called the metric projection of onto .
Lemma 1. (see ). For a given and ,Furthermore, is a firmly nonexpansive mapping of onto and satisfies
Moreover, we also have the following lemma.
Lemma 2. (see ). Let be a real Hilbert space, let be a nonempty closed convex subset of , and let be a mapping of into . Let . Then, for ,where is the metric projection of onto .
Lemma 3. Let be a nonempty closed and convex subset of a real Hilbert space . For every , let be the -inverse strongly monotone with . If , thenwhere for all and . Moreover, is a nonexpansive mapping for all .
Proof. By Lemma 4.3 of , we have that . Let and let . As the same argument as in the proof of Lemma 8 in , we have as nonexpansive.
Lemma 4. (see ). Let be a real Hilbert space, be a single-valued nonlinear mapping, and be a set-valued mapping. Then, a point is a solution of variational inclusion problem if and only if , i.e.,Furthermore, if is -inverse strongly monotone and , then is a closed convex subset of .
Lemma 5. (see ). The resolvent operator associated with is single-valued, nonexpansive, and 1-inverse strongly monotone for all .
The following two lemmas are the particular case of Lemmas 7 and 8 in .
Lemma 6. (see ). For every , let be real Hilbert spaces, let be a multivalued maximal monotone mapping, and let be an -inverse strongly monotone mapping. Let be a bounded linear operator with adjoint of , and let be a mapping defined by , for all . Then, , for all , where is the spectral radius of the operator , , and . Furthermore, if , then is a nonexpansive mapping.
Lemma 7. (see ). Let and be Hilbert spaces. For , let be a multivalued maximal monotone mapping and let be an -inverse strongly monotone mapping. Let be a bounded linear operator with adjoint . Assume that . Then, if and only if , where is a mapping defined byfor all , , and is the spectral radius of the operator .
Next, we give an example to support Lemma 7.
Example 1. Let be a set of real number and , and let be inner product defined by , for all and and the usual norm given by , for all . Let be defined by for all and be defined by for all . Let be defined by , respectively, for all . Let the mapping be defined by , respectively, for all . Then, is a fixed point of . That is, .
Proof. It is obvious to see that , is 2-inverse strongly monotone, and is 3-inverse strongly monotone. Choose . Since and the resolvent of for all , we obtain thatChoose . Since and the resolvent of for all , we obtain thatSince the spectral radius of the operator is 4, we choose . Then, from (18) and (19), we get thatfor all . Then, by Lemma 7, we have that .
Lemma 8. (see ). Let be a sequence of nonnegative real numbers satisfying where is a sequence in and is a sequence in such that(1).(2) or . Then .
3. Main Results
In this section, we introduce an iterative algorithm of two sequences which depend on each other by using the intermixed method. Then, we prove a strong convergence theorem for solving two split monotone variational inclusion problems and a finite family of variational inequality problems.
Theorem 2. Let and be Hilbert spaces, and let be a nonempty closed convex subset of . Let be a bounded linear operator, and let be -contraction mappings with . For , let be multivalued maximal monotone mappings and let be inverse strongly monotone mappings, respectively. For let be inverse strongly monotone mappings, respectively, , and . Let be defined by , and , respectively, where , and with being a spectral radius of . Assume that and . Let and be sequences generated by andfor all where with , , and . Assume the following condition holds:(1), and for some .(2).(3).(4), for all , for some .(5). Then, converges strongly to and converges strongly to .
Proof. We divided the proof into five steps.
Step 1. We will show that and are bounded. Let and . Then, from Lemma 7 and Lemma 6, we getFrom (21), Lemma 3, and (22), we haveSimilarly, from definition of , we haveHence, from (23) and (24), we obtainBy induction, we havefor every . Thus, and are bounded.
Step 2. We will show that . Put , , , and , for all . From Lemma 6, we haveBy applying Lemma 3, we get thatFrom the definition of , (27), and (28), we haveBy the same argument as in (27) and (29), we also haveFrom (29) and (31), we obtain thatFrom (32), conditions , , and , and Lemma 8, we obtain that
Step 3. We show that