R E S E A R C H
Open Access
On variational inequality, fixed point and
generalized mixed equilibrium problems
Dong Feng Li
1*and Juan Zhao
2 *Correspondence: [email protected]1School of Information Engineering,
North China University of Water Resources and Electric Power, Zhengzhou, Henan 450011, China Full list of author information is available at the end of the article
Abstract
In this article, variational inequality, fixed point, and generalized mixed equilibrium problems are investigated based on an extragradient iterative algorithm. Weak convergence of the extragradient iterative algorithm is obtained in Hilbert spaces.
Keywords: fixed point; equilibrium problem; monotone mapping; nonexpansive
mapping; projection
1 Introduction
In this paper, we always assume thatHis a real Hilbert space with the inner product·,· and the norm · , andCis a nonempty, closed, and convex subset ofH.Ris denoted by the set of real numbers. LetFbe a bifunction ofC×CintoR. Consider the problem: find apsuch that
F(p,y)≥, ∀y∈C. (.)
In this paper, the solution set of the problem is denoted byEP(F),i.e.,
EP(F) =p∈C:F(p,y)≥,∀y∈C.
The above problem is first introduced by Ky Fan []. In the sense of Blum and Oettli [], the Ky Fan problem is also called an equilibrium problem.
Recently, the ‘so-called’ generalized mixed equilibrium problem has been investigated by many authors: The generalized mixed equilibrium problem is to find p∈C such that
F(p,y) +Ap,y–p+ϕ(y) –ϕ(p)≥, ∀y∈C, (.)
whereϕ:C→Ris a real valued function andA:C→His mapping. We useGMEP(F, A,ϕ) to denote the solution set of the equilibrium problem. That is,
GMEP(F,A,ϕ) :=p∈C:F(p,y) +Ap,y–p+ϕ(y) –ϕ(z)≥,∀y∈C.
Next, we give some special cases.
IfA= , then the problem (.) is equivalent to findp∈Csuch that
F(p,y) +ϕ(y) –ϕ(z)≥, ∀y∈C, (.)
which is called the mixed equilibrium problem.
IfF= , then the problem (.) is equivalent to findp∈Csuch that
Ap,y–p+ϕ(y) –ϕ(z)≥, ∀y∈C, (.)
which is called the mixed variational inequality of Browder type. Ifϕ= , then the problem (.) is equivalent to findp∈Csuch that
F(p,y) +Ap,y–p ≥, ∀y∈C, (.)
which is called the generalized equilibrium problem.
IfA= andϕ= , then the problem (.) is equivalent to (.).
For solving the above equilibrium problems, let us assume that the bifunctionF:C× C→Rsatisfies the following conditions:
(A) F(x,x) = ,∀x∈C;
(A) Fis monotone,i.e.,F(x,y) +F(y,x)≤,∀x,y∈C; (A)
lim sup t↓
Ftz+ ( –t)x,y≤F(x,y), ∀x,y,z∈C;
(A) for eachx∈C,y→F(x,y)is convex and weakly lower semicontinuous.
Equilibrium problems have intensively been studied. It has been shown that equilibrium problems cover fixed point problems, variational inequality problems, inclusion problems, saddle problems, complementarity problem, minimization problem, and Nash equilib-rium problem; see [–] and the references therein.
LetS:C→Cbe a mapping. In this paper, we useF(S) to stand for the set of fixed points. Recall that the mappingSis said to be nonexpansive if
Sx–Sy ≤ x–y, ∀x,y∈C.
Sis said to beκ-strictly pseudocontractive if there exists a constantκ∈[, ) such that
Sx–Sy≤ x–y+κx–y–Sx+Sy, ∀x,y∈C.
It is clear that the class ofκ-strictly pseudocontractive includes the class of nonexpansive mappings as a special case. The class ofκ-strictly pseudocontractive mappings was intro-duced by Browder and Petryshyn []; for existence and approximation of fixed points of the class of mappings, see [–] and the references therein.
LetA:C→Hbe a mapping. Recall thatAis said to be monotone if
Ais said to beκ-inverse strongly monotone if there exists a constantα> such that
Ax–Ay,x–y ≥κAx–Ay, ∀x,y∈C.
It is clear that theκ-inverse being strongly monotone is monotone and Lipschitz contin-uous.
A set-valued mappingT:H→His said to be monotone if, for allx,y∈H,f ∈Txand
g∈Tyimplyx–y,f –g> . A monotone mappingT:H→His maximal if the graph
G(T) ofT is not properly contained in the graph of any other monotone mapping. It is well known that a monotone mappingT is maximal if and only if, for any (x,f)∈H×H, x–y,f–g ≥ for all (y,g)∈G(T) impliesf ∈Tx. The class of monotone operators is one of the most important classes of operators. Within the past several decades, many authors have been devoting their efforts to the studies of the existence and convergence of zero points for maximal monotone operators.
LetF(x,y) =Ax,y–x,∀x,y∈C. We see that the problem (.) is reduced to the follow-ing classical variational inequality. Findx∈Csuch that
Ax,y–x ≥, ∀y∈C. (.)
It is well known thatx∈Cis a solution to (.) if and only ifxis a fixed point of the mapping PC(I–ρA), whereρ> is a constant, andIis the identity mapping. IfCis bounded, closed,
and convex, then the solution set of the variational inequality (.) is nonempty. In order to prove our main results, we need the following lemmas.
Lemma .[] Let S:C→C be aκ-strictly pseudocontractive mapping.Define St:C→
C by Stx=tx+ ( –t)Sx for each x∈C.Then,as t∈[κ, ),St is nonexpansive such that
F(St) =F(S).
Lemma .[] Let C be a nonempty,closed,and convex subset of H,and F:C×C→Ra bifunction satisfying(A)-(A).Then,for any r> and x∈H,there exists z∈C such that
F(z,y) +
ry–z,z–x ≥, ∀y∈C.
Further,define
Trx=
z∈C:F(z,y) +
ry–z,z–x ≥,∀y∈C
for all r> and x∈H.Then the following hold:
(a) Tris single-valued;
(b) Tris firmly nonexpansive,i.e.,for anyx,y∈H,
Trx–Try≤ Trx–Try,x–y;
(c) F(Tr) =EP(F);
Lemma .[] Let A be a monotone mapping of C into H and NCv the normal cone to C
at v∈C,i.e.,
NCv=
w∈H:v–u,w ≥,∀u∈C
and define a mapping T on C by
Tv=
⎧ ⎨ ⎩
Av+NCv, v∈C,
∅, v∈/C.
Then T is maximal monotone and∈Tv if and only ifAv,u–v ≥for all u∈C.
Lemma .[] Let{an}∞n=be real numbers in[, ]such that
∞
n=an= .Then we have
the following:
∞
i=
aixi
≤ ∞
i=
aixi
for any given bounded sequence{xn}∞n=in H.
Lemma .[] Let <p≤tn≤q< for all n≥.Suppose that{xn}and{yn}are
se-quences in H such that
lim sup
n→∞ xn ≤d, lim supn→∞ yn ≤d
and
lim
n→∞tnxn+ ( –tn)yn=d
hold for some r≥.Thenlimn→∞xn–yn= .
Lemma .[] Let C be a nonempty,closed,and convex subset of H,and S:C→C a strictly pseudocontractive mapping. If {xn} is a sequence in C such that xnx and limn→∞xn–Sxn= ,then x=Sx.
Lemma .[] Let{an},{bn},and{cn}be three nonnegative sequences satisfying the
fol-lowing condition:
an+≤( +bn)an+cn, ∀n≥n,
where n is some nonnegative integer,
∞
n=bn<∞ and
∞
n=cn<∞. Then the limit limn→∞anexists.
2 Main results
continuous and monotone mapping.Let Fmbe a bifunction from C×C toRwhich satisfies
(A)-(A),Bm:C→H a continuous and monotone mapping,ϕm:C→Ra lower
semicon-tinuous and convex function for each m≥.Assume thatF:=∞m=GMEP(Fm,Bm,ϕm)∩ VI(C,A)∩F(S)is not empty.Let{αn},{βn},and{δn,m}be real number sequences in(, ).
Let{λn},{rn,m}be positive real number sequences.Let{xn}be a sequence generated in the
following manner:
⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
x∈H,
xn+=αnxn+ ( –αn)(βnI+ ( –βn)S)ProjC(
∞
m=δn,mzn,m–λnAyn), n≥,
yn=ProjC(
∞
m=δn,mzn,m–λnA
∞
m=δn,mzn,m),
where zn,mis such that
Fm(zn,m,z)+Bmzn,m,z–zn,m+ϕm(z)–ϕm(zn,m)+
rn,m
z–zn,m,zn,m–xn ≥, ∀z∈C.
Assume that{αn},{βn},{δn,m},{λn},{rn,m}satisfy the following restrictions: (a) <a≤αn≤b< ;
(b) κ≤βn≤c< ;
(c) ∞m=δn,m= ,and <d≤δn,m≤;
(d) lim infn→∞rn,m> ande≤λn≤f,wheree,f ∈(, /L).
Then the sequence{xn}weakly converges to some pointx¯∈F.
Proof The proof is split into five steps.
Step . Show that the sequence{xn}is bounded.
DefineGm(p,y) =Fm(p,y) +Bmp,y–p+ϕm(y) –ϕm(p),∀p,y∈C. Next, we prove that
the bifunctionGmsatisfies the conditions (A)-(A). Therefore, generalized mixed
equi-librium problem is equivalent to the following equiequi-librium problem: findp∈Csuch that Gm(p,y)≥,∀y∈C. It is clear thatGm satisfies (A). Next, we proveGm is monotone.
SinceBmis a continuous and monotone operator, we find from the definition ofGthat
Gm(y,z) +Gm(z,y) =Fm(y,z) +Bmy,z–y+ϕm(z) –ϕm(y) +Fm(z,y)
+Bmz,y–z+ϕm(y) –ϕm(z)
=Fm(z,y) +Fm(y,z) +Bmz,y–z+Bmy,z–y
≤ Bmz–Bmy,y–z
≤.
Next, we showGmsatisfies (A), that is,
lim sup t↓
Gm
tz+ ( –t)x,y≤Gm(x,y), ∀x,y,z∈C.
SinceBmis continuous andϕmis lower semicontinuous, we have
lim sup t↓
Gm
tz+ ( –t)x,y=lim sup t↓
Fm
tz+ ( –t)x,y
+lim sup t↓
Bm
+lim sup t↓
ϕm(y) –ϕm
tz+ ( –t)x
≤Fm(x,y) +Bmx,y–x+ϕm(y) –ϕm(x)
=Gm(x,y).
Next, we show that, for eachx∈C,y→Gm(x,y) is a convex and lower semicontinuous.
For eachx∈C, for allt∈(, ) and for ally,z∈C, sinceFmsatisfies (A) andϕmis convex,
we have
Gm
x,ty+ ( –t)z
=Fm
x,ty+ ( –t)z+Bmx,ty+ ( –t)z–x
+ϕm
ty+ ( –t)z–ϕm(x)
≤tFm(x,y) +Bmx,y–x+ϕm(y) –ϕm(x)
+ ( –t)Fm(x,z) +Bmx,z–x+ϕm(z) –ϕm(x)
=tGm(x,y) + ( –t)Gm(x,z).
Thus,y→Gm(x,y) is convex. Similarly, we find thaty→Gm(x,y) is also lower
semicon-tinuous. Putun=ProjC( N
m=δn,mzn,m–λnAyn) andvn= N
m=δn,mzn,m. Lettingp∈F, we
see that
un–p≤ vn–λnAyn–p–vn–λnAyn–un
=vn–p–vn–un+ λn
Ayn–Ap,p–yn+Ap,p–yn
+Ayn,yn–un
≤ vn–p–vn–yn–yn–un+ vn–λnAyn–yn,un–yn.
Notice thatAisL-Lipschitz continuous andyn=ProjC(vn–λnAvn). It follows that
vn–λnAyn–yn,un–yn ≤λnLvn–ynun–yn.
It follows that
un–p≤ vn–p+
λnL– vn–yn. (.)
On the other hand, we have
vn–p≤
∞
m=
δn,mzn,m–p
≤ ∞
m=
δn,mTrn,mxn–p
≤ xn–p, (.)
whereTrn,m={z∈C:Gm(z,y) +
ry–z,z–x ≥,∀y∈C}. Substituting (.) into (.), we
obtain
un–p≤ xn–p+
PuttingSn=βnI+ ( –βn)S, we find from Lemma . thatSnis nonexpansive andF(Sn) =
F(S). It follows that
xn+–p≤αnxn–p+ ( –αn)Snun–p
≤αnxn–p+ ( –αn)un–p
≤αnxn–p+ ( –αn)
xn–p+
λnL– vn–yn
≤ xn–p+ ( –αn)
λnL– vn–yn
≤ xn–p. (.)
It follows from Lemma . that thelimn→∞xn–pexists. This shows that{xn}is bounded.
Since{xn}is bounded, we may assume that a subsequence{xni}of{xn}converges weakly toξ.
Step . Show thatξ∈VI(C,A)
From (.), we find thatβn( –λnL)vn–yn≤ xn–p–xn+–p. In view of the
restrictions (b) and (d), we see thatlimn→∞vn–yn= . Sinceyn–un ≤λLvn–yn,
we have thatlimn→∞yn–un= . It follows that
lim
n→∞vn–un= . (.)
Notice that
zn,m–p =Trn,mxn–Trn,mp
≤ Trn,mxn–Trn,mp,xn–p
=
zn,m–p+xn–p–zn,m–xn
.
This implies thatzn,m–p≤ xn–p–zn,m–xn. Sincevn=
∞
m=δn,mzn,m, where
∞
m=δn,m= , we find that
vn–p≤
∞
m=
δn,mzn,m–p
≤ xn–p–
∞
m=
δn,mzn,m–xn.
It follows that
xn+–p≤αnxn–p+ ( –αn)Snun–p
≤αnxn–p+ ( –αn)un–p
≤αnxn–p+ ( –αn)vn–p
≤ xn–p– ( –αn)
∞
m=
This implies that (–αn)δn,mzn,m–xn≤ xn–p–xn+–p. In view of the restrictions
(a) and (c), we find that
lim
n→∞zn,m–xn= . (.)
LetTbe the maximal monotone mapping defined by
Tx=
⎧ ⎨ ⎩
Ax+NCx, x∈C,
∅, x∈/C.
For any given (x,y)∈G(T), we havey–Ax∈NCx. So, we havex–m,y–Ax ≥, for all
m∈C. On the other hand, we haveun=ProjC(vn–λnAyn). We obtain
x–un,
un–vn λn
+Ayn
≥.
In view of the monotonicity ofA, we see that
x–uni,y ≥ x–uni,Ax
≥ x–uni,Ax–
x–uni,
uni–vni
λni
+Ayni
=x–uni,Ax–Auni+x–uni,Auni–Ayni
–
x–uni,
uni–vni
λni
≥ x–uni,Auni–Ayni–
x–uni,
uni–vni
λni
in view ofvn–xn ≤
∞
m=δn,mzn,m–xn. It follows from (.) thatlimn→∞vn–xn= .
Notice thatun–xn ≤ un–vn+vn–xn. It follows that
lim
n→∞un–xn= . (.)
This in turn implies thatuni ξ. It follows thatx–ξ,y ≥. Notice thatT is maximal monotone and hence ∈Tξ. This shows from Lemma . thatξ∈VI(C,A).
Step . Show thatξ∈GMEP(Fm,Bm,ϕm).
It follows from (.) that {zni,m} converges weakly toξ for eachm≥. Sincezn,m= Trn,mxn, we have
Gm(zn,m,z) +
rn,mz
–zn,m,zn,m–xn ≥, ∀z∈C.
From the assumption (A), we see that
z–zni,m,
zni,m–xni rni,m
≥Gm(z,zni,m), ∀z∈C.
In view of the assumption (A), we find from (.) thatGm(z,ξ)≤,∀z∈C. Fortm with
we haveztm∈C. It follows thatGm(ztm,ξ)≤. Notice that
=Gm(ztm,ztm)≤tmGm(ztm,z) + ( –tm)Gm(ztm,ξ)≤tmGm(ztm,z),
which yieldsGm(ztm,z)≥,∀z∈C. Lettingtm↓, one sees thatGm(ξ,z)≥,∀z∈C. This implies thatξ∈GMEP(Fm,Bm,ϕm) for eachm≥. This proves thatξ∈
∞
m=GMEP(Fm,
Bm,ϕm).
Step . Show thatξ∈F(S).
Sincelimn→∞xn–pexists, we putlimn→∞xn–p=d> . It follows that
lim
n→∞xn+–p=nlim→∞αn(xn–p) + ( –αn)(Snun–p)=d.
Notice thatlim supn→∞Snun–p ≤d. From Lemma ., we see that
lim
n→∞xn–Snun= . (.)
Since
Snxn–xn ≤ Snxn–Snun+Snun–xn
≤ xn–un+Snun–xn,
we find from (.) and (.) that
lim
n→∞xn–Snxn= . (.)
In view ofSxn–xn ≤ Sxn–Snxn+Snxn–xn, we find from (.) thatlimn→∞xn–
Sxn= . This implies from Lemma . thatξ∈F(S). This completes the proof thatξ∈F.
Step . Show that the whole sequence{xn}weakly converges toξ.
Let{xnj}be another subsequence of{xn}converging weakly toξ, whereξ=ξ. In the same way, we can show thatξ∈F. Since the spaceHenjoys Opial’s condition, we, there-fore, obtain
d=lim inf
i→∞ xni–ξ<lim infi→∞
xni–ξ
=lim inf j→∞
xj–ξ<lim inf
j→∞ xj–ξ=d.
This is a contradiction. Henceξ=ξ. This completes the proof.
3 Applications
In this section, we consider solutions of the mixed equilibrium problem (.), which in-cludes the Ky Fan inequality as a special case.
The so-called mixed equilibrium problem is to findp∈Csuch that
F(p,y) +ϕ(y) –ϕ(z)≥, ∀y∈C.
Theorem . Let C be a nonempty,closed,and convex subset of H,S:C→C aκ-strictly pseudocontractive mapping with a nonempty fixed point set,and A:C→H a L-Lipschitz continuous and monotone mapping.Let Fmbe a bifunction from C×C toRwhich satisfies
(A)-(A),andϕm:C→Ra lower semicontinuous and convex function for each m≥.
Assume thatF:=∞m=MEP(Fm,ϕm)∩VI(C,A)∩F(S)is not empty.Let{αn},{βn}and
{δn,m}be real number sequences in(, ).Let{λn},{rn,m}be positive real number sequences.
Let{xn}be a sequence generated in the following manner:
⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
x∈H,
xn+=αnxn+ ( –αn)(βnI+ ( –βn)S)ProjC(
∞
m=δn,mzn,m–λnAyn), n≥,
yn=ProjC(
∞
m=δn,mzn,m–λnA
∞
m=δn,mzn,m),
where zn,mis such that
Fm(zn,m,z) +ϕm(z) –ϕm(zn,m) +
rn,m
z–zn,m,zn,m–xn ≥, ∀z∈C.
Assume that{αn},{βn},{δn,m},{λn},{rn,m}satisfy the following restrictions: (a) <a≤αn≤b< ;
(b) κ≤βn≤c< ;
(c) ∞m=δn,m= ,and <d≤δn,m≤;
(d) lim infn→∞rn,m> ande≤λn≤f,wheree,f ∈(, /L).
Then the sequence{xn}weakly converges to some pointx¯∈F.
Proof IfBm= , we draw the desired conclusion immediately from Theorem ..
Further, ifSis nonexpansive, we find from Theorem . the following result.
Corollary . Let C be a nonempty,closed,and convex subset of H,S:C→C a nonexpan-sive mapping with a nonempty fixed point set,and A:C→H an L-Lipschitz continuous and monotone mapping.Let Fmbe a bifunction from C×C toRwhich satisfies(A)-(A),
andϕm:C→Ra lower semicontinuous and convex function for each m≥.Assume that F:=∞m=MEP(Fm,ϕm)∩VI(C,A)∩F(S)is not empty.Let{αn},{βn},and{δn,m}be real
number sequences in(, ).Let{λn},{rn,m}be positive real number sequences.Let{xn}be a
sequence generated in the following manner:
⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
x∈H,
xn+=αnxn+ ( –αn)SProjC(
∞
m=δn,mzn,m–λnAyn), n≥,
yn=ProjC(
∞
m=δn,mzn,m–λnA
∞
m=δn,mzn,m),
where zn,mis such that
Fm(zn,m,z) +ϕm(z) –ϕm(zn,m) +
rn,m
z–zn,m,zn,m–xn ≥, ∀z∈C.
(b) ∞m=δn,m= ,and <d≤δn,m≤;
(c) lim infn→∞rn,m> ande≤λn≤f,wheree,f∈(, /L).
Then the sequence{xn}weakly converges to some pointx¯∈F.
IfA= , we find from Theorem . the following result.
Theorem . Let C be a nonempty,closed,and convex subset of H,S:C→C aκ-strictly pseudocontractive mapping with a nonempty fixed point set.Let Fm be a bifunction from
C×C toRwhich satisfies(A)-(A),Bm:C→H a continuous and monotone mapping, ϕm:C→Ra lower semicontinuous and convex function for each m≥.Assume that F:=∞m=GMEP(Fm,Bm,ϕm)∩F(S)is not empty.Let{αn},{βn},and{δn,m}be real
num-ber sequences in(, ).Let{rn,m}be a positive real number sequence.Let{xn}be a sequence
generated in the following manner:
x∈H, xn+=αnxn+ ( –αn)
βnI+ ( –βn)S
∞
m=
δn,mzn,m, n≥,
where zn,mis such that
Fm(zn,m,z)+Bmzn,m,z–zn,m+ϕm(z)–ϕm(zn,m)+
rn,mz
–zn,m,zn,m–xn ≥, ∀z∈C.
Assume that{αn},{βn},{δn,m},and{rn,m}satisfy the following restrictions: (a) <a≤αn≤b< ;
(b) κ≤βn≤c< ;
(c) ∞m=δn,m= ,and <d≤δn,m≤; (d) lim infn→∞rn,m> .
Then the sequence{xn}weakly converges to some pointx¯∈F.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed equally to this manuscript. All authors read and approved the final manuscript.
Author details
1School of Information Engineering, North China University of Water Resources and Electric Power, Zhengzhou, Henan
450011, China.2School of Mathematics and Information Science, North China University of Water Resources and Electric
Power, Zhengzhou, Henan 450011, China.
Acknowledgements
The authors are grateful to the reviewers for useful suggestions which improved the contents of the article.
Received: 14 January 2014 Accepted: 8 May 2014 Published:22 May 2014
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Cite this article as:Li and Zhao:On variational inequality, fixed point and generalized mixed equilibrium problems.