Abstract
Multistep composite implicit and explicit extragradient-like schemes are presented for solving the minimization problem with the constraints of variational inclusions and generalized mixed equilibrium problems. Strong convergence results of introduced schemes are given under suitable control conditions.
Keywords:
extragradient-like method; minimization problem; variational inclusion; generalized mixed equilibrium problem MSC:
49J30; 47H09; 47J20; 49M05
1. Introduction
Let H be a real Hilbert space and be a closed convex set. Let be a nonlinear operator. Let us consider the variational inequality problem (VIP) which aims to find verifying
We denote the solution set of VIP (1) by .
In [1], Korpelevich suggested an extragradient method for solving VIP (1). Korpelevich’s method has been studied by many authors, see e.g., references [2,3,4,5,6,7,8,9,10,11,12,13] and references therein. In 2011, Ceng, Ansari and Yao [14] considered the following algorithm
They showed that strongly converges to which solves the following VIP
Ceng, Guu and Yao [15] presented a composite implicit scheme
and another composite explicit scheme
Ceng, Guu and Yao proved that and converge strongly to the same point , which solves the following VIP
In [2], Peng and Yao considered the generalized mixed equilibrium problem (GMEP) which aims to find verifying
where is a real-valued function and is a bifunction. The set of solutions of GMEP (7) is denoted by .
It is clear that GMEP (7) includes optimization problems, VIP and Nash equilibrium problems as special cases.
Special cases:
GMEP (7) has been discussed extensively in the literature; see e.g., [6,7,20,21,22,23,24,25,26,27,28,29,30,31,32].
Now, we consider the minimization problem (CMP):
where is a convex and continuously Fréchet differentiable functional. Use to denote the set of minimizers of CMP (8).
For solving (8), an efficient algorithm is the gradient-projection algorithm (GPA) which is defined by: for given the initial guess ,
or a general form,
It is known that is a contractive operator if is -strongly monotone and L-Lipschitz and . In this case, GPA (9) has strong convergence.
Recall that the variational inclusion problem is to find a point verifying
where is a single-valued mapping and R is a multivalued mapping with . Use to denote the solution set of (11).
Taking in (11), it reduces to the problem introduced by Rockafellar [33]. Let be a maximal monotone operator. The resolvent operator is defined by In [34], Huang considered problem (11) under the assumptions that B is strongly monotone Lipschitz operator and R is maximal monotone. Zeng, Guu and Yao [35] discussed problem (11) under more general cases. Related work, please refer to [36,37] and the references therein.
In the present paper, we introduce two composite schemes for finding a solution of the CMP (8) with the constraints of finitely many GMEPs and finitely many variational inclusions for maximal monotone and inverse strongly monotone mappings in a real Hilbert space H. Strong convergence of the suggested algorithms are given. Our theorems complement, develop and extend the results obtained in [6,15,38], having as background [39,40,41,42,43,44,45,46].
2. Preliminaries
Let C be a nonempty closed convex subset of a real Hilbert space H. Recall that the metric projection is defined by . A mapping is called strongly positive if for . A mapping is called Lipschitz if for some . F is called nonexpansive if and it is called contractive provided .
Proposition 1.
Let and . Then, the following hold
- ;
- ;
- .
Definition 1.
A mapping is called
- monotone if ;
- η-strongly monotone if for some ;
- α-inverse-strongly monotone (ism) if for some . In this case, we have for all ,where is a constant.
Definition 2.
A mapping is called firmly nonexpansive if is nonexpansive. Thus, T is firmly nonexpansive iff , where is nonexpansive.
Let be a bifunction satisfying the following assumptions:
- (A1)
- for all ;
- (A2)
- for all ;
- (A3)
- for , ;
- (A4)
- is convex and lower semicontinuous for each .
Let be a lower semicontinuous and convex function satisfying the following conditions:
- (B1)
- for each and , there exists a bounded subset and such that for any ,
- (B2)
- C is a bounded set.
Let . Define a mapping by the following form: for any ,
If , then .
Proposition 2
([17]). Let a bifunction satisfying assumptions (A1)–(A4). Let a convex function be a proper lower semi-continuous. Assume conditions (B1) or (B2) holds. Then, we have
- , is single-valued;
- is firmly nonexpansive;
- ;
- is convex and closed;
- , for .
Proposition 3
([41]). We have the following conclusions:
- A mapping T is nonexpansive iff the complement is -ism.
- If a mapping T is α-ism, then is -ism where .
- A mapping T is averaged iff the complement is α-ism for some .
Lemma 1.
Let E be a real inner product space. Then,
Lemma 2.
In a real Hilbert space H, we have the following results
- , ;
- , and with ;
- is a sequence satisfying . Then,
Lemma 3
([45]). Let H be a real Hilbert space and be a closed convex set. Let be a k-strictly pseudocontractive mapping. Then,
- ,.
- is demiclosed at 0.
- of T is closed and convex.
Lemma 4
([39]). Let H be a real Hilbert space and be a closed convex set. Let be a k-strictly pseudocontractive mapping. Then,
where and with .
Let be a nonexpansive operator and the operator be -Lipschitzian and -strongly monotone. Define an operator by , where and are two constants.
Lemma 5
([42]). Let . Then, for , we have , where .
Lemma 6
([46]). Let , , and be four sequences. If
- (i)
- and ;
- (ii)
- either or ;
- (iii)
- , .
Then, .
Lemma 7
([44]). Let bounded linear operator be -strongly positive. Then, provided .
Let be a Banach limit. Then, we have following properties:
- ⇒;
- , where N is a fixed positive integer;
- , .
Lemma 8
([40]). Assume that the sequence satisfy , where M is a constant. If , then .
Let the operator be maximal monotone. Let be two constants.
Lemma 9
([43]). There holds the resolvent identity
Remark 1.
The resolvent has the following property
Lemma 10
([34,35]). satisfies
3. Main Results
Let H be a real Hilbert space and be a closed convex set. In what follows, we assume:
is a convex functional with gradient being L-Lipschitz, the operator is -strongly monotone and -Lipschitzian, is a maximal monotone mapping and is -ism, satisfies (A1)–(A4), is a proper convex and lower semicontinuous satisfying (B1) or (B2), and is -ism for each ;
is a nonexpansive mapping and is a -strongly positive bounded linear operator such , and with ;
where is nonexpansive, and with ;
The operator is defined by , for ;
The operator is defined by with and , for each ;
The operator is defined by , for ;
The operator is defined by with and , for each ;
;
and .
Next, set
and
for and , and .
By Proposition 3, is -ism for . In addition, hence, is -averaged. It is clear that is -averaged for each . Thus, , where is nonexpansive and for each . Similarly, for each , is -averaged for each and . Please note that . Since , by (12) and Lemma 10, we deduce
On the other hand, since , according to (12) and Proposition 2, we get
Next, we present the following net :
We prove the strong convergence of as to a point which solves
Let , define a sequence as follows
We will show the convergence of as to , which solves (15).
Now, for , and , let be defined by . By Lemmas 5 and 7, we have
It is easy to check that . Therefore, is contraction and has a unique fixed point, denoted by .
Proposition 4.
Let be defined via (14). Then
- (i)
- is bounded for ;
- (ii)
- and provided ;
- (iii)
- are locally Lipschitzian.
Proof.
(i) Pick up . Noting that and for all and , by the nonexpansivity of and and Lemmas 5 and 7 we get
Thus,
Hence is bounded and so are and .
(ii) From (14), we obtain
Simple calculations show that
where . Then, by the nonexpansivity of , Proposition 1, and Lemmas 5 and 7, from (18)–(20) we obtain that
which together with and , implies that
Since and is bounded, we have
According to Proposition 2, we have
which implies that
Also, by Lemma 10, we obtain that for each
which implies
Since and (due to (22)), we deduce from the boundedness of and that
Hence we get
and
So, taking into account that , we have
In the meantime, from the nonexpansivity of and , it is easy to see that
and
(iii) Since is -ism, is nonexpansive for . Then,
Utilizing (12) and (13), we obtain that
where
for some and for some . Also, utilizing Proposition 2, we deduce that
where is a constant and
This immediately implies that
Since are locally Lipschitzian, we conclude that is locally Lipschitzian. □
Proof.
Let be the unique solution of (15). Let . Then, we have
where . Then, by Proposition 1 and the nonexpansivity of , we obtain from (18) that
Hence,
It follows that , where such that . By Proposition 4, we get . Observe that
where for . Hence we have
By the boundedness of () and , we deduce
Therefore, .
Furthermore, from (25), (26) and (28), we have that and . First, we prove that . Please note that is maximal monotone. Let , i.e., . Noting that , we have
that is,
According to the monotonicity of , we have
and hence
Since (due to (25)) and , we deduce from and that
By the maximal monotonicity of , we derive , i.e., . Thus, . Next we prove that . Since , we have
By (A2), we have
Let and . Set . Thus, . Hence,
By (25), we have as . Please note that . Then,
Thus,
So, and . Therefore, .
Next, we prove that as . First, let us assert that is a solution of the VIP (15). As a matter of fact, since , we have
Since and are nonexpansive mappings, is monotone. By the monotonicity of , we have
Hence,
Thus, is a solution of (15); hence by uniqueness. Consequently, as . □
Theorem 2.
Assume that the sequences and satisfy: and . Let the sequence be defined by (16). Then , where and is defined by
where are defined by and for and .
Proof.
We assume, without loss of generality, that for all . Let and is the unique solution of (15). By Proposition 4, we deduce that the nets , and are bounded.
Let . Then and . Applying Lemmas 5 and 7, we obtain that
By induction
Thus, , and are all bounded. By (C1), we obtain
By (31), we also have
where for some . According to (32), we have
where
for some and for some . In terms of (33), we have
where for some . In terms of (39)–(41) we calculate
It is clear that
where as . Please note that
Furthermore, for simplicity, we write . By (38), we get and
Applying Lemma 1 and Proposition 1, we have
Applying the Banach limit to (45), we have
Using the property , we have
This completes the proof. □
Theorem 3.
Proof.
We assume, without loss of generality, that and for all . Let be defined by (38). Then, (due to Theorem 1). We divide the rest of the proof into several steps.
Step 1. It is easy to show that
Hence , , and are all bounded.
Step 2. We show that . To this end, let
According to Theorem 2, we deduce . Choose a subsequence of such that
and . This indicates that . Hence,
and therefore
Then, by Lemma 8, we derive
Step 3. We show that . Set for all . Then . Utilizing (16) and , we have
Thus, utilizing Proposition 1 and Lemma 1, we get
where and
It is easy to check that and . By Lemma 6 with , we deduce that . This completes the proof. □
Author Contributions
All the authors have contributed equally to this paper. All the authors have read and approved the final manuscript.
Funding
This research was partially supported by the Innovation Program of Shanghai Municipal Education Commission (15ZZ068), Ph.D. Program Foundation of Ministry of Education of China (20123127110002) and Program for Outstanding Academic Leaders in Shanghai City (15XD1503100).
Conflicts of Interest
The authors declare no conflict of interest.
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