Volume 2010, Article ID 564361,15pages doi:10.1155/2010/564361
Research Article
An Iterative Algorithm for Mixed Equilibrium
Problems and Variational Inclusions Approach to
Variational Inequalities
Yeong-Cheng Liou
Department of Information Management, Cheng Shiu University, Kaohsiung 833, Taiwan
Correspondence should be addressed to Yeong-Cheng Liou,simplex [email protected]
Received 13 September 2009; Revised 10 November 2009; Accepted 10 January 2010
Academic Editor: Nanjing Huang
Copyrightq2010 Yeong-Cheng Liou. 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 present an iterative algorithm for finding a common elementx∗ of the set of solutions of a mixed equilibrium problem and the set of a variational inclusion in a real Hilbert space. Furthermore, we prove that the proposed iterative algorithms strongly converge tox∗which solves some variational inequality.
1. Introduction
LetCbe a nonempty closed convex subset of a real Hilbert spaceH. LetF :C → Hbe a nonlinear mapping, letϕ :C → Rbe a function, and letΘbe a bifunction ofC×CintoR. Now we consider the following mixed equilibrium problem:
Findu∈Csuch thatΘu, yϕy−ϕu Fu, y−u≥0, ∀y∈C. 1.1
The set of solution of problem1.1is denoted by EP.
If F 0, then the mixed equilibrium problem 1.1 becomes the following mixed equilibrium problem:
which was considered by Ceng and Yao1. Ifϕ 0, then the mixed equilibrium problem
1.1becomes the following equilibrium problem:
Find u∈Csuch thatΘu, yFu, y−u≥0, ∀y∈C, 1.3
which was studied by S. Takahashi and W. Takahashi2. Ifϕ0 andF 0, then the mixed equilibrium problem1.1becomes the following equilibrium problem:
Find u∈Csuch thatΘu, y≥0, ∀y∈C. 1.4
IfΘx, y 0 for allx, y ∈ C, the mixed equilibrium problem1.1becomes the following variational inequality problem:
Find u∈Csuch thatϕy−ϕu Fu, y−u≥0, ∀y∈C. 1.5
The mixed equilibrium problems include fixed point problems, optimization problems, variational inequality problems, Nash equilibrium problems, and the equilibrium problems as special cases; see, for example, 3–8. Some methods have been proposed to solve the mixed equilibrium problem and the equilibrium problem. In 1997, Flaim and Antipen4
introduced an iterative method of finding the best approximation to the initial data and proved a strong convergence theorem. Subsequently, S. Takahashi and W. Takahashi 9
introduced another iterative scheme for finding a common element of the set of solutions of the equilibrium problem1.2and the set of fixed point points of a nonexpansive mapping. Furthermore, Yao et al.10introduced some new iterative schemes for finding a common element of the set of solutions of the equilibrium problem 1.2 and the set of common fixed points of finitelyinfinitelynonexpansive mappings. Very recently, Ceng and Yao1
considered a new iterative scheme for finding a common element of the set of solutions of the mixed equilibrium problem and the set of common fixed points of finitely many nonexpansive mappings. Peng and Yao11developed a CQ method. They obtained some strong convergence results for finding a common element of the set of solutions of the mixed equilibrium problem1.1and the set of the variational inequality and the set of fixed points of a nonexpansive mapping. Their results extend and improve the corresponding results in
1,9,12.
Recall that a mappingB : C → Cis said to beβ-inverse strongly monotone if there exists a constantβ >0 such thatBx−By, x−y ≥β Bx−By 2,for allx, y∈C. A mappingAis strongly positive onHif there exists a constantμ >0 such thatAx, x ≥μ x 2 for allx∈H. LetB : H → H be a single-valued nonlinear mapping and letR : H → 2H be a set-valued mapping. Now we concern the following variational inclusion, which is to find a pointx∈Hsuch that
θ∈Bx Rx, 1.6
where θ is the zero vector in H. The set of solutions of problem 1.6 is denoted by
study of optimal solutions in many optimization related areas including mathematical programming, complementarity, variational inequalities, optimal control, mathematical economics, equilibria, and game theory. Also various types of variational inclusions problems have been extended and generalized. Recently, Zhang et al.15introduced a new iterative scheme for finding a common element of the set of solutions to the problem 1.6and the set of fixed points of nonexpansive mappings in Hilbert spaces. Peng et al.16introduced another iterative scheme by the viscosity approximate method for finding a common element of the set of solutions of a variational inclusion with set-valued maximal monotone mapping and inverse strongly monotone mappings, the set of solutions of an equilibrium problem, and the set of fixed points of a nonexpansive mapping. For some related works, please see
1,2,9–11,13–34and the references therein.
Inspired and motivated by the works in the literature, in this paper, we present an iterative algorithm for finding a common element x∗ of the set of solutions of a mixed equilibrium problem and the set of a variational inclusion in a real Hilbert space. Furthermore, we prove that the proposed iterative algorithms strongly converge tox∗which solves some variational inequality.
2. Preliminaries
LetHbe a real Hilbert space with inner product·,·and norm · . LetCbe a nonempty closed convex subset ofH. Then, for anyx ∈ H, there exists a unique nearest point inC, denoted byPCx, such that
x−PCx ≤x−y, ∀y∈C. 2.1
Such a PC is called the metric projection of H ontoC. We know thatPC is nonexpansive.
Further, forx∈Handx∗∈C,
x∗P
Cx⇐⇒x−x∗, x∗−y≥0, ∀y∈C. 2.2
A set-valued mappingT :H → 2His called monotone if, for allx, y∈H,f ∈Txand
g ∈Tyimplyx−y, f−g ≥ 0. A monotone mappingT :H → 2His maximal if its graph
GTis not properly contained in the graph of any other monotone mapping. It is known that a monotone mappingT is maximal if and only if, forx, f ∈H×H,x−y, f −g ≥0 for everyy, g∈GTimpliesf ∈Tx.
Let the set-valued mapping R : H → 2H be maximal monotone. We define the resolvent operatorJR,λassociated withRandλas follows:
JR,λ IλR−1x, x∈H, 2.3
whereλis a positive number. It is worth mentioning that the resolvent operatorJR,λis
single-valued, nonexpansive, and 1-inverse strongly monotone and that a solution of problem1.6
is a fixed point of the operatorJR,λI−λBfor allλ >0, see, for instance,25.
H1 Θx, x 0 for allx∈C;
H2 Θis monotone, that is,Θx, y Θy, x≤0 for allx, y∈C;
H3for eachy∈C,x→Θx, yis weakly upper semicontinuous;
H4for eachx∈C,y→Θx, yis convex and lower semicontinuous;
H5for eachx ∈Candr >0, there exists a bounded subsetDx ⊂ Candyx ∈ Csuch
that for anyz∈C\Dx,
Θz, yxϕyx−ϕz 1ryx−z, z−x<0. 2.4
Lemma 2.1see11. LetC be a nonempty closed convex subset of a real Hilbert spaceH. Let
Θ: C×C → Rbe a bifunction and letϕ :C → Rbe a proper lower semicontinuous and convex
function. Forr >0andx∈C, define a mappingSr :C → Cas follows:
Srx
z∈C:Θz, yϕy−ϕz 1
r
y−z, z−x≥0,∀y∈C
2.5
for allx∈C. Assume that the conditions (H1)–(H5) hold. Then one has the following results:
1for eachx∈C,Srx/∅andSr is single-valued;
2Sris firmly nonexpansive, that is, for anyx, y∈C,
Srx−Sry2≤Srx−Sry, x−y; 2.6
3FixSr EP;
4EPis closed and convex.
Lemma 2.2see24. LetR :H → 2Hbe a maximal monotone mapping and letB:H → Hbe
a Lipschitz-continuous mapping. Then the mappingRB:H → 2His maximal monotone.
Lemma 2.3see34. Assume taht{an}is a sequence of nonnegative real numbers such thatan1≤
1−γnanδnwhere{γn}is a sequence in0,1and{δn}is a sequence such that
1 ∞n1γn∞;
2lim supn→ ∞δn/γn≤0or ∞n1|δn|<∞.
Thenlimn→ ∞an0.
3. Main Results
LetCbe a nonempty closed convex subset of a real Hilbert spaceH. Letϕ :C → R be a lower semicontinuous and convex function and letΘ : H ×C → R be a bifunction satisfying conditionsH1–H5. LetAbe a strongly positive bounded linear operator with coefficient 0< μ <1 and letR:H → 2Hbe a maximal monotone mapping. Let the mappings
F, B:C → Cbeα-inverse strongly monotone andβ-inverse strongly monotone, respectively. Letr >0 andλ >0 be two constants such thatr <2αandλ <2β.
Now we introduce the following iteration algorithm.
Algorithm 3.1. For givenx0 ∈Carbitrarily, compute the sequences{xn}and{un}as follows:
Θun, yϕy−ϕun 1ry−un, un−xn−rFxn≥0, ∀y∈C,
xn1PCI−αnAJR,λI−λBun,
3.1
where{αn}is a real sequence in0,1.
Now we study the strong convergence of the algorithm3.1.
Theorem 3.2. Suppose thatΩ:EP∩IB, R/∅. Assume the following conditions are satisfied:
ilimn→ ∞αn0;
ii ∞n0αn∞;
iiilimn→ ∞αn1/αn 1.
Then the sequence{xn}generated by3.1converges strongly tox∗ ∈Ωwhich solves the following
variational inequality:
Ax, y−x≥0, ∀y∈Ω. 3.2
Proof. Takex∗∈Ω. It is clear that
Srx∗−rFx∗ JR,λx∗−λBx∗ x∗, n≥0. 3.3
We divide our proofs into the following five steps:
1the sequences{xn}and{un}are bounded;
2 xn1−xn → 0;
3 Fxn−Fx∗ → 0 and Bun−Bx∗ → 0;
4lim supn→ ∞Ax∗, xn−x∗ ≥0;
5xn → x∗.
Proof of (1).SinceFisα-inverse strongly monotone andBisβ-inverse strongly monotone, we
have
I−rFx−I−rFy2≤x−y2rr−2αFx−Fy2, 3.4
I−λBx−I−λBy2≤x−y2λλ−
It is clear that if 0≤r≤2αand 0≤λ≤2β, thenI−rFandI−λBare all nonexpansive. Set
ynJR,λun−λBun, n≥0. It follows that
yn−x∗ JR,λun−λBun−JR,λx∗−λBx∗ ≤ un−λBun−x∗−λBx∗ ≤ un−x∗ .
3.6
ByLemma 2.1, we haveunSrxn−rFxnfor alln≥0. Then, we have
un−x∗ 2 Srxn−rFxn−Srx∗−rFx∗ 2
≤ xn−rFxn−x∗−rFx∗ 2
≤ xn−x∗ 2rr−2α Fxn−Fx∗ 2
≤ xn−x∗ 2.
3.7
Hence, we have
yn−x∗≤ xn−x∗ . 3.8
SinceAis linear bounded self-adjoint operator onH, then
A sup{|Au, u|:u∈H, u 1}. 3.9
Observe that
I−αnAu, u1−αnAu, u ≥1−αn A ≥0, 3.10
that is to sayI−αnAis positive. It follows that
I−αnA sup{I−αnAu, u:u∈H, u 1}
sup{1−αnAu, u:u∈H, u 1} ≤1−αnμ.
3.11
From3.1, we deduce that
xn1−x∗ PCI−αnAyn−x∗
≤I−αnAyn−x∗−αnAx∗ ≤1−αnμyn−x∗αn Ax∗ ≤1−αnμ xn−x∗ αn Ax∗
≤max
x0−x∗ ,
Ax∗
μ
.
3.12
Proof of (2).From3.1, we have
xn1−xn PC
I−αnAyn−PCI−αn−1Ayn−1
≤I−αnAyn−I−αn−1Ayn−1
I−αnAyn−yn−1
αn−1−αnAyn−1
≤I−αnAyn−yn−1αn−1−αnAyn−1
≤1−αnμyn−yn−1|αn−αn−1|Ayn−1.
3.13
Note that
yn−yn−1 JR,λun−λBun−JR,λun−1−λBun−1 ≤ un−λBun−un−1−λBun−1
≤ un−un−1
Srxn−rFxn−Srxn−1−rFxn−1
≤ xn−rFxn−xn−1−rFxn−1
≤ xn−xn−1 .
3.14
Substituting3.14into3.13, we get
xn1−xn ≤
1−αnμ xn−xn−1 |αn−αn−1|Ayn−1
1−αnμ xn−xn−1 αnμ
1−αn−1
αn
μ1Ayn−1.
3.15
Notice that limn→ ∞|1−αn−1/αn| 0.This together with the last inequality andLemma 2.3
implies that
lim
n→ ∞ xn1−xn 0. 3.16
Proof of (3).From3.5and3.7, we get
yn−x∗2 JR,λun−λBun−JR,λx∗−λBx∗ 2 ≤ un−λBun−x∗−λBx∗ 2
≤ un−x∗ 2λλ−2β Bun−Bx∗ 2
≤ xn−x∗ 2rr−2α Fxn−Fx∗ 2λλ−2β Bun−Bx∗ 2.
By3.1, we obtain
xn1−x∗ 2PCI−αnAyn−x∗2
≤I−αnAyn−x∗2
yn−x∗−αnAyn2
yn−x∗2−2αnyn−x∗, Aynα2nAyn2
≤yn−x∗2αn
2yn−x∗AynAyn2
≤yn−x∗2αnM,
3.18
whereM >0 is some constant satisfying supn{2 yn−x∗ Ayn Ayn 2} ≤M. From3.17
and3.18, we have
xn1−x∗ 2≤ xn−x∗ 2rr−2α Fxn−Fx∗ 2
λλ−2β Bun−Bx∗ 2αnM.
3.19
Thus,
r2α−r Fxn−Fx∗ 2λ2β−λ Bun−Bx∗ 2
≤ xn−x∗ 2− xn1−x∗ 2αnM
≤ xn−x∗ xn1−x∗ xn1−xn αnM,
3.20
which imply that
lim
n→ ∞ Fxn−Fx
∗ 0, lim
n→ ∞ Bun−Bx
∗ 0.
Proof of (4).SinceSr is firmly nonexpansive, we have
un−x∗ 2 Srxn−rFxn−Srx∗−rFx∗ 2
≤ xn−rFxn−x∗−rFx∗, un−x∗
1
2
xn−rFxn−x∗−rFx∗ 2 un−x∗ 2
− xn−rFxn−x∗−rFx∗−un−x∗ 2
≤ 1
2
xn−x∗ 2 un−x∗ 2− xn−un−rFxn−Fx∗ 2
1
2
xn−x∗ 2 un−x∗ 2− xn−un 2
2rFxn−Fx∗, xn−un −r2 Fxn−Fx∗ 2
.
3.22
Hence, we have
un−x∗ 2≤ xn−x∗ 2− xn−un 22r Fxn−Fx∗ xn−un . 3.23
SinceJR,λis 1-inverse strongly monotone, we have
yn−x∗2 J
R,λun−λBun−JR,λx∗−λBx∗ 2 ≤un−λBun−x∗−λBx∗, yn−x∗
1
2
un−λBun−x∗−λBx∗ 2yn−x∗2
−un−λBun−x∗−λBx∗−yn−x∗2
≤ 1
2
un−x∗ 2yn−x∗2−un−yn−λBun−Bx∗2
1
2
un−x∗ 2yn−x∗2−un−yn2
2λBun−Bx∗, un−yn −λ2 Bun−Bx∗ 2
,
3.24
which implies that
yn−x∗2
Thus, by3.23and3.25, we obtain
yn−x∗2
≤ xn−x∗ 2− xn−un 22r Fxn−Fx∗ xn−un
−un−yn22λ Bun−Bx∗ un−yn.
3.26
Substitute3.26into3.18to get
xn1−x∗ 2≤ xn−x∗ 2− xn−un 22r Fxn−Fx∗ xn−un
−un−yn22λ Bun−Bx∗ un−ynαnM.
3.27
Then we derive
xn−un 2un−yn2
≤ xn−x∗ xn1−x∗ xn1−xn 2r Fxn−Fx∗ xn−un
2λ Bun−Bx∗ un−ynαnM.
3.28
So, we have
lim
n→ ∞ xn−un 0, nlim→ ∞un−yn0. 3.29
Proof of (5).We note thatPΩI−Ais a contraction. As a matter of fact,
PΩI−Ax−PΩI−Ay≤I−Ax−PΩI−Ay
≤ I−A x−y
≤1−μx−y
3.30
for allx, y ∈H. HencePΩI−Ahas a unique fixed point, sayx∗ ∈Ω. That is,x∗ PΩI−
Ax∗. This implies thatAx∗, y−x∗ ≥0 for ally∈Ω. Next, we prove that
lim sup
n→ ∞ Ax ∗, x
n−x∗ ≥0. 3.31
First, we note that there exists a subsequence{xni}of{xn}such that
lim sup
n→ ∞ Ax
∗, xn−x∗ lim
j→ ∞
Ax∗, xnj−x∗.
3.32
Since{xnj}is bounded, there exists a subsequence{xnji}of{xnj}which converges weakly to
We next show thatw∈EP. ByunSrxn−rFxn, we know that
Θun, yϕy−ϕun 1ry−un, un−xn−rFxn≥0, ∀y∈C. 3.33
It follows fromH2that
ϕy−ϕun 1ry−un, un−xn−rFxn≥Θy, un, ∀y∈C. 3.34
Hence,
ϕy−ϕuni
y−uni,uni−xnir−rFxni
≥Θy, uni, ∀y∈C. 3.35
Fort∈0,1andy∈H, letytty 1−tw. From3.35we have
yt−uni, Fyt ≥ yt−uni, Fyt −ϕytϕuni
−
yt−uni,uni−xnir−rFxni
Θyt, uni
yt−uni, Fyt−Funiyt−uni, Funi−Fxni−ϕyt
ϕuni−
yt−uni,uni−rxni
Θyt, uni.
3.36
Since uni −xni → 0, we have Funi −Fxni → 0. Further, from the inverse strongly
monotonicity ofF, we haveyt−uni, Fyt−Funi ≥ 0. So, fromH4and the weakly lower
semicontinuity ofϕ,uni−xni/r → 0 anduni → wweakly, we have
yt−w, Fyt≥ −ϕytϕw Θyt, w. 3.37
FromH1,H4, and3.37, we also have
0 Θyt, ytϕyt−ϕyt
≤tΘyt, y 1−tΘyt, wtϕy 1−tϕw−ϕyt
tΘyt, yϕy−ϕyt 1−tΘyt, wϕw−ϕyt
≤tΘyt, yϕy−ϕyt 1−tyt−w, Fyt
tΘyt, yϕy−ϕyt 1−tty−w, Fyt,
3.38
and hence
Lettingt → 0, we have, for eachy∈C,
Θw, yϕy−ϕw y−w, Fw≥0. 3.40
This implies thatw∈EP.
Next, we show thatw ∈IB, R. In fact, sinceBisβ-inverse strongly monotone,Bis Lipschitz continuous monotone mapping. It follows fromLemma 2.2thatRBis maximal monotone. Letv, g∈GRB, that is,g−Bv∈Rv. Again sinceyni JR,λuni −λBun−i,
we haveuni−λuni ∈IλRyni, that is,1/λuni−yni−λBuni∈Ryni. By virtue of the
maximal monotonicity ofRB, we have
v−yni, g−Bv− 1λuni−yni−λBuni≥0, 3.41
and so
v−yni, g ≥
v−yni, Bvλ1uni−yni−λBuni
v−yni, Bv−ByniByni−Buni1λuni−yni
≥v−yni, Byni−Buni
v−yni,λ1uni−yni.
3.42
It follows from un−yn → 0, Bun−Byn → 0 andyni wthat
lim
ni→ ∞
v−yni, gv−w, g≥0. 3.43
It follows from the maximal monotonicity ofBRthatθ ∈RBw, that is,w∈IB, R. Therefore,w∈Ω. It follows that
lim sup
n→ ∞ Ax
∗, xn−x∗ lim
j→ ∞
Ax∗, xnj−x∗
Ax∗, w−x∗
≥0.
3.44
Proof of (6).First, we note thatxn1PCI−αnAyn; then for allx∈C, we havexn1−I−
From3.1, we have
xn1−x∗ 2xn1−x∗, xn1−x∗
xn1−I−αnAyn I−αnAyn−x∗, xn1−x∗
xn1−I−αnAyn, xn1−x∗
I−αnAyn−x∗, xn1−x∗
≤I−αnAyn−x∗−αnAx∗, xn1−x∗
I−αnAyn−x∗, xn1−x∗
2αn−Ax∗, xn1−x∗
≤I−αnAyn−x∗ xn1−x∗
2αn−Ax∗, xn1−x∗
≤
1−αnμ
2
xn−x∗ 2 xn1−x∗ 2
2αn−Ax∗, xn1−x∗,
3.45
that is,
xn1−x∗ 2 ≤
1−αnμ xn−x∗ 2 2αn
1αnμ−Ax
∗, xn 1−x∗
1−δn xn−x∗ 2δnσn,
3.46
whereδn αnμandσn 2/1αnμμ−Ax∗, xn1−x∗. It is easy to see that ∞n1δn ∞ and lim supn→ ∞σn≤0. Hence, byLemma 2.3, we conclude that the sequence{xn}converges
strongly tox∗. This completes the proof.
Acknowledgment
The author was supported in part by NSC 98-2622-E-230-006-CC3 and NSC 98-2923-E-110-003-MY3.
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