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A new relaxed extragradient-like algorithm for approaching common solutions of generalized mixed equilibrium problems, a more general system of variational inequalities and a fixed point problem
Fixed Point Theory and Applications volume 2013, Article number: 126 (2013)
Abstract
In this paper, we introduce a new iterative algorithm by the relaxed extragradient-like method for finding a common element of the set of solutions of generalized mixed equilibrium problems, the set of solutions of a more general system of variational inequalities for finite inverse strongly monotone mappings and the set of solutions of a fixed point problem of a strictly pseudocontractive mapping in a Hilbert space. Then we prove strong convergence of the scheme to a common element of the three above described sets.
1 Introduction
In this paper, we assume that H is a real Hilbert space with the inner product and the induced norm and C is a nonempty closed convex subset of H. denotes the metric projection of H onto C and denotes the fixed points set of a mapping T. The sequence converges weakly to x which is denoted by .
Let be a real-valued function and let be a nonlinear mapping. Suppose that is a bifunction.
The generalized mixed equilibrium problem is to find (see, e.g., [1–6]) such that
The set of solutions of (1.1) is denoted by .
If , then problem (1.1) reduces to the generalized equilibrium problem, which is to find (see, e.g., [7–9]) such that
The set of solutions of (1.2) is denoted by .
If , then problem (1.1) reduces to the mixed equilibrium problem, which is to find (see, e.g., [10–13]) such that
The set of solutions of (1.3) is denoted by .
If , then problem (1.1) reduces to the mixed variational inequality of Browder type, which is to find (see, e.g., [3, 14]) such that
The set of solutions of (1.4) is denoted by .
If , , then problem (1.1) reduces to the equilibrium problem, which is to find (see, e.g., [15–17]) such that
The set of solutions of (1.5) is denoted by .
If , , then problem (1.1) reduces to the variational inequality, which is to find (see, e.g., [18–29]) such that
The set of solutions of (1.6) is denoted by .
If , , then problem (1.1) reduces to the minimized problem, which is to find (see, e.g., [18–28]) such that
The set of solutions of (1.7) is denoted by .
Let be two mappings. Ceng et al. [2] considered the following problem of finding such that
which is called a general system of variational inequalities where and are two constants. In particular, if , , then problem (1.8) reduces to the classical variational inequality problem (1.6).
In order to find the common element of the solutions of problem (1.8) and the set of fixed points of one nonexpansive mapping S, Ceng et al. [2] studied the following algorithm: fix , , and
Under appropriate conditions, they obtained one strong convergence theorem.
Let C be a nonempty closed convex subset of a real Hilbert space H. Let be a family of mappings. Cai and Bu [1] considered the following problem of finding such that
And (1.10) can be rewritten as
which is called a more general system of variational inequalities in Hilbert spaces, where for all . The set of solutions to (1.10) is denoted by Ω. In particular, if , , , , , , , then problem (1.10) reduces to problem (1.8).
In order to find a common element of the solutions of problem (1.10) and the common fixed points of a family of strictly pseudocontractive mappings, Cai and Bu [1] studied the following algorithm: pick any , set , , and
Under suitable conditions, they also obtained one strong convergence theorem.
In this paper, motivated and inspired by the above facts, we study a new iterative algorithm by the relaxed extragradient-like method for finding a common element of the set of solutions of generalized mixed equilibrium problems, the set of solutions of a more general system of variational inequalities for finite inverse strongly monotone mappings and the set of solutions of a fixed point problem of a strictly pseudocontractive mapping in a Hilbert space. Then we prove strong convergence of the scheme to a common element of the three above described sets.
2 Preliminaries
For solving the equilibrium problem, let us assume that the bifunction F satisfies the following conditions:
-
(A1)
for all ;
-
(A2)
F is monotone, i.e., for any ;
-
(A3)
F is weakly upper semicontinuous, i.e., for each ,
-
(A4)
is convex and lower semicontinuous for each ;
-
(B1)
For each and , there exists a bounded subset and such that for any ,
-
(B2)
C is a bounded set.
Let H be a real Hilbert space. It is well known that
and
for all .
Definition 2.1 Let C be a nonempty closed convex subset of a real Hilbert space H.
-
(1)
A mapping is said to be nonexpansive if
-
(2)
A mapping is said to be L-Lipschitzian if there exists such that
-
(3)
A mapping is said to be k-strictly pseudocontractive if there exists a constant such that
(2.3)It is obvious that , then the mapping T is nonexpansive;
-
(4)
A mapping is said to be monotone if
-
(5)
A mapping is said to be α-inverse-strongly monotone if there exists a positive real number α such that
It is obvious that any α-inverse-strongly monotone mapping T is monotone and -Lipschitz continuous.
Definition 2.2 is called a metric projection if for every point , there exists a unique nearest point in C, denoted by , such that
In order to prove our main results in the next section, we recall some lemmas.
Lemma 2.1 [30]
Let C be a nonempty closed convex subset of H and let be a k-strictly pseudocontractive mapping, then the following results hold:
-
(1)
equation (2.3) is equivalent to
(2.4) -
(2)
T is Lipschitz continuous with a constant , i.e.,
(2.5) -
(3)
(Demi-closed principle) is demi-closed on C, that is,
Lemma 2.2 [1]
Let C be a nonempty closed convex subset of H and let be an α-inverse-strongly monotone mapping, then for all and , we have
So, if , then is a nonexpansive mapping from C to H.
Lemma 2.3 Let C be a nonempty closed convex subset of H and let be a metric projection, then
-
(1)
, ;
-
(2)
moreover, is a nonexpansive mapping, i.e., , ;
-
(3)
, , ;
-
(4)
, , .
Lemma 2.4 [4]
Let C be a nonempty closed convex subset of H. Assume that satisfies (A1)-(A4) and let be a lower semicontinuous and convex function. Assume that either (B1) or (B2) holds. For and , define a mapping as follows:
for all . Then the following hold:
-
(1)
for each , and is single-valued;
-
(2)
is firmly nonexpansive, that is, for any ,
-
(3)
;
-
(4)
is closed and convex.
Lemma 2.5 [3]
Let C be a nonempty closed convex subset of H. Assume that satisfies (A1)-(A4), is a continuous monotone mapping and let be a lower semicontinuous and convex function. Assume that either (B1) or (B2) holds. For and , define a mapping as follows:
for all . Then the following hold:
-
(1)
for each , and is single-valued;
-
(2)
is firmly nonexpansive, that is, for any ,
-
(3)
;
-
(4)
is closed and convex.
Lemma 2.6 Let C be a nonempty closed convex subset of H. Let be a family of bifunctions from into ℝ satisfying (A1)-(A4), let be a family of lower semicontinuous functions from C into ℝ, and let be a family of -inverse-strongly monotone mappings from C into H. For and , , assume that either (B1) or (B2) holds. Let be a mapping defined by
Putting , where I is an identity mapping,
If and , , then
-
(1)
, ;
-
(2)
T is nonexpansive.
Proof (1) Since is a family of -inverse-strongly monotone mappings from C into H, so they are continuous monotone mappings. Observe that
By Lemma 2.5, we know that if then x is the fixed point of the mapping , , so we have
which implies that x is a fixed point of the mapping . Therefore we get
(2) Since is firmly nonexpansive, then it is obvious that is nonexpansive. And from Lemma 2.2, we have
which implies T is nonexpansive. □
Lemma 2.7 [1]
Let C be a nonempty closed convex subset of H. Let be -inverse-strongly monotone from C into H, respectively, where . Let be a mapping defined by
If , , then G is nonexpansive.
Proof Put , , and , where I is an identity mapping. Since is nonexpansive and from Lemma 2.2, we have
which implies G is nonexpansive. □
Lemma 2.8 Let C be a nonempty closed convex subset of H. Let be a nonlinear mapping, where . For given , , is a solution of problem (1.10) if and only if
that is,
Proof (⟸) From Lemma 2.3(3), it is obvious that (2.7) is the solution of problem (1.10).
(⟹) Since
Similarly, we get
Therefore we have
which completes the proof. □
From Lemma 2.8, we know that , that is, is a fixed point of the mapping G, where G is defined by Lemma 2.7. Moreover, if we find the fixed point , it is easy to solve the other points by (2.7).
Lemma 2.9 [31]
Let and be bounded sequences in a Banach space X and let be a sequence in with . Suppose that for all integers and . Then .
Lemma 2.10 [30]
Let be a nonexpansive mapping with . If and , then .
Lemma 2.11 [30]
Assume that is a sequence of nonnegative real numbers such that
where is a sequence in and is a sequence such that
-
(1)
;
-
(2)
or .
Then .
3 Main results
In this section, we state and verify our main results. We have the following theorem.
Theorem 3.1 Let C be a nonempty closed convex subset of a real Hilbert space H. Let be a family of bifunctions from into ℝ satisfying (A1)-(A4), let be a family of lower semicontinuous and convex functions and let be a family of -inverse-strongly monotone mappings from C into H. Let be -inverse-strongly monotone from C into H, respectively, where . Let S be a δ-strict pseudocontractive mapping from C into itself such that , where G is defined by Lemma 2.7. For and , , assume that either (B1) or (B2) holds. Pick any , let be a sequence generated by
where , , . satisfy the following conditions:
-
(i)
and for all ;
-
(ii)
and ;
-
(iii)
and ;
-
(iv)
;
-
(v)
, .
Then converges strongly to .
Proof Putting
and
, where I is the identity mapping on H. Then we have that and . From Lemma 2.6 and Lemma 2.7, it can be seen easily that and are nonexpansive, where , . We divide the proof into six steps.
Step 1. Firstly, we show that is bounded.
Indeed, take arbitrarily. Since , , . By Lemma 2.6, we have
It follows from Lemma 2.7 and (3.2) that
Furthermore, from (3.1), we have
Since , (2.3) and (2.4), we have
which implies that
From (3.2)-(3.5) it follows that
So, we have
By induction, we obtain that
Hence, is bounded. Consequently, we deduce immediately that , , are bounded.
Step 2. Next, we prove that .
Indeed, define for all . It follows that
Observe that
which implies that
Since
then we have
Hence it follows from (3.6), (3.7), (3.9) and that
Consequently, it follows from (3.10), conditions (ii), (iv) and , are bounded that
Hence, by Lemma 2.9, we get . Thus, from condition (iii), we have
Step 3. We show that , and , .
It follows from Lemma 2.6 that
By Lemma 2.3 and Lemma 2.2, we have
By induction, we get
From condition (i) and (3.5), we get
that is,
So, in terms of (3.12) and (3.13), we have
Therefore,
Since , , , , , , and is bounded, we have
and
Step 4. We prove that .
Indeed, utilizing firmly nonexpansive of and Lemma 2.2, we have
which implies
From (3.14), (3.3), Lemma 2.6 and (3.17), we have
It follows that
Since , , and , we conclude that
Therefore we get
From Lemma 2.3(1), we obtain
which implies that
By induction, we have
that is,
From (3.14) and (3.21),
It follows that
Since , , and , we conclude that
Therefore, we get
Thus from (3.19) and (3.23), we have
Observe that
Since , , and , we have
From (3.24) and (3.25), we conclude that
Step 5. In this step, we prove that , where .
Indeed, take a subsequence of such that
Since is bounded, there exists a subsequence of which converges weakly to . Without loss of generality, we may assume that . From (3.18) and (3.24), we have , , where . From (3.26) and Lemma 2.1, we have that is . Utilizing Lemma 2.7, we known that G is nonexpansive. And from (3.23), we obtain
According to Lemma 2.10, we obtain , that is, .
Next we prove that . Since
For all , we have
By (A2), we have