R E S E A R C H
Open Access
Convergence theorems for finding zero
points of maximal monotone operators and
equilibrium problems in Banach spaces
Siwaporn Saewan
1*, Poom Kumam
2and Yeol Je Cho
3**Correspondence:
[email protected]; [email protected]
1Department of Mathematics and
Statistics, Faculty of Science, Thaksin University (TSU), Phatthalung 93110, Thailand
3Department of Mathematics
Education and RINS, Gyeongsang National University, Chinju, 660-701, Korea
Full list of author information is available at the end of the article
Abstract
In this paper, we construct a new hybrid projection method for approximating a common element of the set of zeroes of a finite family of maximal monotone operators and the set of common solutions to a system of generalized equilibrium problems in a uniformly smooth and strictly convex Banach space. We prove strong convergence theorems of the algorithm to a common element of these two sets. As application, we also apply our results to find common solutions of variational inequalities and zeroes of maximal monotone operators.
MSC: 47H05; 47H09; 47H10
Keywords: maximal monotone operators; hybrid projection method; system of generalized equilibrium problems
1 Introduction
LetE be a Banach space with the norm · and letE*denote the dual space ofE. Let
S={x∈E:x= }be the unit sphere ofE. A Banach spaceEis said to besmoothif the limit
lim
t→
x+ty–x
t (.)
exists for anyx,y∈S.Eis said to beuniformly smoothif the limit (.) is attained uniformly for (x,y) inS×S. A Banach spaceEis said to bestrictly convexifx+y< for allx,y∈E withx=y= andx=y(see [] for more details).
One of the major problems in the theory of monotone operators is as follows. Find a pointz∈Esuch that
∈Bz, (.)
whereBis an operator fromEintoE*. Suchz∈Eis called azero pointofB. We denote
the set of zeroes of the operatorBbyB–. An operatorB⊂E×E*is said to bemonotoneif
x–y,x*–y*≥, ∀x,x*,y,y*∈B.
A monotoneBis said to bemaximalif its graphG(B) ={(x,y*) :y*∈Bx}is not properly
contained in the graph of any other monotone operator. IfBis maximal monotone, then the solution setB– is closed and convex. Theresolvent of a monotone operatorBis
defined by
Jλ= (J+λB)–J, ∀λ> .
LetCbe a closed convex subset of a Banach spaceE, a mappingT:C→Cis said to be nonexpansiveif
Tx–Ty ≤ x–y, ∀x,y∈C.
Recall that a pointx∈Cis afixed pointofT providedTx=x. LetEbe a Banach space with dualE*and let·,·be the pairing betweenEandE*. Thenormalized duality mapping J:E→E*is defined by
J(x) =x*∈E*:x,x*=x,x*=x, ∀x∈E.
The Lyapunov functional is defined by
φ(x,y) =x– x,Jy+y forx,y∈E. (.)
It is obvious that
y–x≤φ(y,x)≤y+x, ∀x,y∈E. (.)
IfEis a Hilbert space, thenφ(x,y) =x–yfor allx,y∈E.
A pointpinCis said to be astrongly asymptotic fixed pointofT [] ifCcontains a sequence{xn}which converges strongly topsuch thatlimn→∞xn–Txn= . The set of
strong asymptotic fixed points ofTwill be denoted byF(T).
A mappingT fromCinto itself is said to be aweak relatively nonexpansive mappingif . F(T)is nonempty;
. φ(p,Tx)≤φ(p,x)for allx∈Candp∈F(T); . F(T) =F(T).
Kohasaka and Takahashi [] proved that ifEis a smooth strictly convex and reflexive Banach space andBis a continuous monotone operator withB–=∅, thenJλis a weak relatively nonexpansive mapping. By Takahashi [], we know thatF(Jλ) is closed and con-vex, whereF(Jλ) is the set of fixed points ofJλ.
LetBbe a maximal monotone operator in a Hilbert spaceH. The proximal point algo-rithm generates, for startingx=x∈H, a sequence{xn}inHby
xn+=Jλnxn, ∀n≥, (.)
where{λn} ⊂(,∞) andJλn= (I+λnB)–.
Also, Rockafellar [] proved that the sequence{xn}defined by (.) converges weakly to
LetCbe a nonempty closed convex subset ofEand letRbe the set of real numbers. Let fi:C×C→Rbe a bifunction andAi:C→E*be a nonlinear mapping fori= , , , . . . ,N.
Thesystem of generalized equilibrium problemsis as follows. Findu∈Csuch that for ally∈C,
⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩
f(u,y) +y–u,Au ≥,
f(u,y) +y–u,Au ≥,
· · ·
fN(u,y) +y–u,ANu ≥.
(.)
Iffi=fandAi=Ain (.), then from the problem (.) we have the followinggeneralized
equilibrium problemdenoted byGEP(f,A). Findu∈Csuch that
f(u,y) +y–u,Au ≥, ∀y∈C. (.)
The generalized equilibrium problems include fixed point problems, optimization lem, monotone inclusion problems, saddle point problems, variational inequality prob-lems, minimization probprob-lems, vector equilibrium probprob-lems, Nash equilibria in noncoop-erative games and equilibrium problems as special cases (see, for example, []). Also, some solution methods have been proposed to solve the equilibrium problem (see, for example, [–]) and numerous problems in physics, optimization and economics reduce to finding a solution of problem (.).
Recently, Li and Su [] introduced the hybrid iterative scheme for approximating a com-mon solution of the equilibrium problems and the variational inequality problems in a -uniformly convex real Banach space which is also uniformly smooth. In , Zegeye and Shahzad [] introduced the iterative process which converges strongly to a common solution of the variational inequality problems for two monotone mappings in Banach spaces.
Quite recently, Shehu [] introduced an iterative scheme by the hybrid method for ap-proximating a common element of the set of zeroes of a finite family ofα-inverse-strongly monotone operators and the set of common solutions of a system of generalized mixed equilibrium problems in a -uniformly convex real Banach space which is also uniformly smooth.
Motivated by the results of Shehu [], we prove some strong convergence theorems for finding a common zero of a finite family of continuous monotone mappings and a solution of the system of generalized equilibrium problems in a uniformly smooth and strictly convex real Banach space with the Kadec-Klee property.
2 Preliminaries
Throughout this paper, letEbe a Banach space with its dual spaceE*. For a sequence{xn}
ofEand a pointx∈E, the weak convergence of{xn}toxis denoted byxnxand the
strong convergence of{xn}toxis denoted byxn→x.
Thenormalized duality mapping J:E→E*is defined by
J(x) =x*∈E*:x,x*=x,x*=x, ∀x∈E,
Cioranescu [] proved the following properties:
() IfEis an arbitrary Banach space, thenJis monotone and bounded; () IfEis a strictly convex, thenJis strictly monotone;
() IfEis a smooth, thenJis single-valued and semi-continuous;
() IfEis uniformly smooth, thenJis uniformly norm-to-norm continuous on each bounded subset ofE;
() IfEis reflexive, smooth and strictly convex, then the normalized duality mappingJ is single-valued, one-to-one and onto;
() IfEis a reflexive, strictly convex and smooth Banach space andJis the duality mapping fromEintoE*, thenJ–is also single-valued, bijective and is also the duality mapping fromE*intoEand thusJJ–=IE*andJ–J=IE;
() IfEis uniformly smooth, thenEis smooth and reflexive; () Eis uniformly smooth if and only ifE*is uniformly convex.
A Banach spaceEhas the Kadec-Klee property [, ] if, for any sequence{xn} ⊂Eand
x∈Ewithxnxandxn → x, thenxn–x → asn→ ∞.
Consider the functionalφ:E×E→Rdefined by
φ(x,y) =x– x,Jy+y, ∀x,y∈E, (.)
whereJis the normalized duality mapping fromEto E*.
It is obvious from the definition of the functionφthat
y–x≤φ(y,x)≤y+x, ∀x,y∈E. (.)
IfEis a Hilbert space, thenφ(y,x) =y–x.
Remark . IfE is a reflexive, strictly convex and smooth Banach space, then, for any x,y∈E,φ(x,y) = if and only ifx=y. It is sufficient to show that ifφ(x,y) = , thenx=y. From (.) we havex=y. This implies thatx,Jy=x=Jy. From the definition
ofJ, one hasJx=Jy. Therefore, we havex=y(see [, ] for more details).
LetCbe a nonempty closed convex subset of a reflexive, strictly convex and smooth Banach spaceE. Thegeneralized projectionC:E→Cis a mapping that assigns to an
arbitrary pointx∈Ethe minimum point of the functionalφ(x,y), that is,Cx=x, where¯
¯
xis the solution to the minimization problem
φ(¯x,x) =inf
y∈Cφ(y,x). (.)
The existence and uniqueness of the operator C follows from the properties of the
functionalφ(y,x) and the strict monotonicity of the mappingJ(see, for example, [, – ]). IfEis a Hilbert space, thenCbecomes the metric projection ofEontoC.
Example .(Qinet al.[]) LetCbe the generalized projection from a smooth strictly
convex and reflexive Banach spaceEonto a nonempty closed convex subsetCofE. Then
Cis a closed relatively quasi-nonexpansive mapping fromEontoCwithF(C) =C.
Lemma .(Alber []) Let C be a nonempty closed convex subset of a smooth Banach space E and let x∈E.Then x=Cx if and only if
x–y,Jx–Jx ≥, ∀y∈C.
Lemma .(Alber []) Let E be a reflexive,strictly convex and smooth Banach space,let C be a nonempty closed convex subset of E and let x∈E.Then
φ(y,Cx) +φ(Cx,x)≤φ(y,x), ∀y∈C.
LetEbe a smooth strictly convex and reflexive Banach space,Cbe a nonempty closed convex subset ofEandB⊂E×E*be a monotone operator satisfying the following:
D(B)⊂C⊂J– λ>
R(J+λB)
.
Then theresolvent Jλ:C→D(B) ofBis defined by
Jλx=
z∈D(B) :Jx∈Jz+λBz,∀x∈C.
Jλis a single-valued mapping fromEtoD(B). For anyλ> , theYosida approximation Bλ:C→E*ofBis defined byBλx=Jx–λJJλx for allx∈C. We know thatBλx∈B(Jλx) for all
λ> andx∈E.
Lemma .(Kohsaka and Takahashi []) Let E be a smooth strictly convex and reflexive Banach space,let C be a nonempty closed convex subset of E and let B⊂E×E*be a
mono-tone operator satisfying D(B)⊂C⊂J–(
λ>R(J+λB)).For anyλ> ,let Jλand Bλbe the resolvent and the Yosida approximation of B,respectively.Then the following hold:
() φ(p,Jλx) +φ(Jλx,x)≤φ(p,x)for allx∈Candp∈B–; () (Jλx,Bλx)∈Bfor allx∈C;
() F(Jλ) =B–.
Lemma . (Rockafellar []) Let E be a reflexive strictly convex and smooth Banach space.Then an operator B⊂E×E* is maximal monotone if and only if R(J+λB) =E*
for allλ> .
For solving the equilibrium problem for a bifunctionf :C×C→R, assume thatf sat-isfies the following conditions:
(A) f(x,x) = for allx∈C;
(A) f is monotone,i.e.,f(x,y) +f(y,x)≤for allx,y∈C; (A) for eachx,y,z∈C,
lim
t↓f
tz+ ( –t)x,y≤f(x,y);
(A) for eachx∈C,y→f(x,y)is convex and lower semi-continuous. The following result is given in Blum and Oettli [].
(A)-(A).Then,for any r> and x∈E,there exists z∈C such that
f(z,y) +
ry–z,Jz–Jx ≥, ∀y∈C.
Lemma .(Takahashi and Zembayashi []) Let C be a closed convex subset of a uni-formly smooth strictly convex and reflexive Banach space E and let f be a bifunction from C×C toRsatisfying the conditions(A)-(A).For any r> and x∈E,define a mapping Tr:E→C as follows:
Trx=
z∈C:f(z,y) +
ry–z,Jz–Jx ≥,∀y∈C
, ∀x∈C.
Then the following hold: () Tris single-valued;
() Tris a firmly nonexpansive-type mapping for allx,y∈E,that is,
Trx–Try,JTrx–JTry ≤ Trx–Try,Jx–Jy;
() F(Tr) =EP(f);
() EP(f)is closed and convex.
Lemma .(Takahashi and Zembayashi []) Let C be a closed convex subset of a smooth, strictly convex and reflexive Banach space E,f be a bifunction from C×C toRsatisfying the conditions(A)-(A)and let r> .Then,for any x∈E and p∈F(Tr),
φ(p,Trx) +φ(Trx,x)≤φ(p,x).
Lemma . Let C be a nonempty closed convex subset of a smooth,strictly convex and reflexive Banach space E.Let A:C→E*be a continuous monotone mapping and f be a
bifunction from C×C toRsatisfying the conditions(A)-(A).Then,for any r> and x∈E,there exists z∈C such that
f(z,y) +y–z,Az+
ry–z,Jz–Jx ≥, ∀y∈C.
Proof Define a bifunction:C×C→Rby(x,y) =f(x,y) +y–x,Axfor allx,y∈C. We show thatsatisfies the conditions (A)-(A).
First, we show thatsatisfies the condition (A). Since
(x,x) =f(x,x) +x–x,Ax= , ∀x∈C,
the condition (A) is satisfied.
Next, we show thatsatisfies the condition (A). SinceAis a continuous monotone mapping andf satisfies the condition (A), for anyx,y∈C, we have
(x,y) +(y,x) =f(x,y) +f(y,x) +y–x,Ax+x–y,Ay
≤ +y–x,Ax–Ay ≤.
Thirdly, we show thatsatisfies the condition (A). Sincef satisfies the condition (A) andAis a continuous monotone mapping, for anyx,y,z∈C, we have
lim sup
t↓
x+t(z–x),y
=lim sup
t↓
fx+t(z–x),y+y–x+t(z–x),Ax+t(z–x)
≤lim sup
t↓
fx+t(z–x),y+lim sup
t↓
y–x+t(z–x),Ax+t(z–x)
≤f(x,y) +lim sup
t↓
y–x+t(z–x),Ax+t(z–x)
=f(x,y) +y–x,Ax
=(x,y).
The condition (A) is satisfied.
Finally, we show thatsatisfies the condition (A) sincey→ y–x,Axis convex and continuous; that is,y→ y–x,Axis convex and lower semi-continuous. Sincey→f(x,y) is convex and lower semi-continuous,y→(x,y) is convex and lower semi-continuous.
Therefore,(x,y) satisfies the conditions (A)-(A). From Lemma ., we can conclude that there existsz∈Csuch that
f(z,y) +y–z,Az+
ry–z,Jz–Jx ≥, ∀y∈C.
This completes the proof.
Lemma . Let C be a nonempty closed convex subset of a smooth,strictly convex and reflexive Banach space E.Let A:C→E*be a continuous and monotone mapping and f be
a bifunction from C×C toRsatisfying the conditions(A)-(A).Then,for any r> and x∈E,there exists z∈C such that
f(z,y) +y–z,Az+
ry–z,Jz–Jx ≥, ∀y∈C.
Define a mapping Kr:C→C as follows:
Kr(x) =
z∈C:f(z,y) +y–z,Az+
ry–z,Jz–Jx ≥,∀y∈C
, ∀x∈C. (.)
Then we have the following: () Kris single-valued;
() Kris firmly nonexpansive,i.e.,for allx,y∈E,
Krx–Kry,JKrx–JKry ≤ Krx–Kry,Jx–Jy;
() F(Kr) =GEP(f,A);
() GEP(f,A)is closed and convex;
Proof Define a bifunction:C×C→Rby(u,y) =f(u,y) +y–u,Aufor allu,y∈C. From Lemma ., it follows thatsatisfies the conditions (A)-(A). Now, we can rewrite the mappingKr:C→Cgiven in (.) as follows:
Kr(x) =
z∈C:(z,y) +
ry–z,Jz–Jx ≥,∀y∈C
, ∀x∈C. (.)
Thus, from Lemmas . and ., we obtain the conclusion. This completes the proof.
Throughout this paper, we define a mappingKi
ri (x) :C→Cby
Ki
ri (x) =
z∈C:i(z,y) +
riy
–z,Jz–Jx ≥,∀y∈C
, ∀x∈C, (.)
wherei(z,y) =fi(z,y) +y–z,Aizfor allz,y∈Candi= , , , . . . ,m.
3 Main results
Theorem . Let C be a nonempty closed and convex subset of a uniformly smooth and strictly convex Banach space E with the Kadec-Klee property.For any i= , , , . . . ,m,let fi
be a bifunction from C×C toRsatisfying the conditions(A)-(A)and let{Ai}be a finite
family of continuous and monotone mappings from C to E*.Let B
j⊂E×E*be maximal
monotone operators satisfying D(Bj)⊂C and J Bj
λj,n= (J+λj,nBj)
–J for allλ
j,n> and j=
, , . . . ,l.Assume that F:= (mi=GEP(fi,Ai))∩(
l
j=B–j )=∅.For arbitrary x∈C and
C=C,generate a sequence{xn}by
⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩
zn=JλBll,n◦J
Bl–
λl–,n◦ · · · ◦J
B λ,nxn,
un=Krmm,n◦K
m–
rm–,n◦ · · · ◦K
r,nzn,
Cn+={z∈Cn:φ(z,un)≤φ(z,xn)},
xn+=Cn+x, ∀n≥,
(.)
where{ri,n} ⊂[a,∞)for some a> for all i= , , . . . ,m andlim infn→∞λj,n> for all j=
, , . . . ,l.Then the sequence{xn}converges strongly to a point p∈F,where p=Fx.
Proof We split the proof into five steps as follows.
Step . We first show thatCn+is closed and convex for alln≥. Clearly,C=Cis closed
and convex. Suppose thatCnis closed and convex for alln≥. Since, for anyz∈Cn, we
know thatφ(z,un)≤φ(z,xn) is equivalent to the following:
z,Jxn–Jun ≤ xn–un.
ThusCn+is closed and convex for alln≥.
Step . We show thatF⊂Cnfor alln≥ and{xn}is well defined. SinceF⊂C=C,
sup-pose thatF⊂Cnfor somen≥. Letq∈F, from Lemma . and Lemma ., we have that
φ(q,un) =φ
q,Km
rm,n◦K
m–
rm–,n◦ · · · ◦K
r,nzn
≤φq,Km–
rm–,n◦ · · · ◦K
r,nzn
≤φq,K
r,nzn
≤φ(q,zn)
=φq,Jλll,n◦J
l–
λl–,n◦ · · · ◦J
λ,nxn
≤φq,Jλl–l–,n◦ · · · ◦J
λ,nxn
· · · ≤φq,Jλ,nxn
≤φ(q,xn). (.)
This shows thatq∈Cn+, which implies thatF⊂Cn+. HenceF⊂Cnfor alln≥. This
implies that the sequence{xn}is well defined.
Step . We show thatlimn→∞un–xn= andlimn→∞Jun–Jxn= . By the definition
ofCn+withxn=Cnxandxn+=Cn+x∈Cn+⊂Cn, it follows that
φ(xn,x)≤φ(xn+,x), ∀n≥, (.)
that is,{φ(xn,x)}is nondecreasing. By Lemma ., we get φ(xn,x) =φ(Cnx,x)
≤φ(q,x) –φ(q,xn)
≤φ(q,x), ∀q∈F. (.)
This implies that{φ(xn,x)}is bounded and solimn→∞φ(xn,x) exists. In particular, by
(.), the sequence{(xn–x)}is bounded. This implies{xn}is also bounded. So,{zn}
and{un}are also bounded. SinceEis reflexive andCnis closed and convex, without loss
of generality, we may assume that there existsp∈Cnsuch thatxnp.
Sincexn=Cnx, we have φ(xn,x)≤φ(p,x), ∀p∈Cn.
On the other hand, since
lim inf
n→∞ φ(xn,x) =lim infn→∞
xn– xn,Jx+x
≥ p– p,Jx
+x
=φ(p,x),
it follows that
φ(p,x)≤lim inf
n→∞ φ(xn,x)≤lim supn→∞ φ(xn,x)≤φ(p,x).
This implies thatlimn→∞φ(xn,x) =φ(p,x). Hence we getxn → pasn→ ∞. In view
of the Kadec-Klee property ofE, we obtain
lim
Now, we claim thatJun–Jxn → asn→ ∞. By the definition ofCnx, it follows that
φ(xn+,xn) =φ(xn+,Cnx)
≤φ(xn+,x) –φ(Cnx,x)
=φ(xn+,x) –φ(xn,x).
Sincelimn→∞φ(xn,x) exists, we obtain
lim
n→∞φ(xn+,xn) = . (.)
Since xn+ =Cn+x ∈ Cn+ ⊂ Cn and the definition of Cn+, we have φ(xn+,un) ≤
φ(xn+,xn). By (.) we also have
lim
n→∞φ(xn+,un) = . (.)
From (.) it follows that
un → p (n→ ∞). (.)
Since J is uniformly norm-to-norm continuous on bounded subsets of E, it follows that
Jun → Jp (n→ ∞). (.)
This implies that{Jun}is bounded inE*. Note thatEis reflexive andE*is also reflexive,
we can assume thatJunx*∈E*. SinceEis reflexive, we see thatJ(E) =E*. Hence there
existsx∈Esuch thatJx=x*, and we have
φ(xn+,un) =xn+– xn+,Jun+un
=xn+– xn+,Jun+Jun.
Takinglim infn→∞on both sides of the equation above, in view of the weak lower semi-continuity of the norm · , it follows that
≥ p– p,x*+x*
=p– p,Jx+Jx
=p– p,Jx+x
=φ(p,x).
From Remark . we havep=x, which implies thatx*=Jp, and soJu
nJp∈E*. From the
Kadec-Klee property, we have that
Note thatJ–:E*→Eis norm-weak*-continuous, that is,
unp asn→ ∞. (.)
From (.), (.) and the Kadec-Klee property ofE, it follows that
lim
n→∞un=p. (.)
Sincexn–un ≤ xn–p+p–un, it follows that
lim
n→∞xn–un= . (.)
SinceJis uniformly norm-to-norm continuous on bounded subsets ofE, we obtain
lim
n→∞Jun–Jxn= . (.)
Step . We show thatp∈F:= (mi=GEP(fi,Ai))∩(
l
j=B–j ). First, we show thatp∈
m
i=GEP(fi,Ai). From (.), (.) and (.), it follows that for anyq∈F,
lim
n→∞φ(q,zn) =φ(q,p). (.)
Denotejn:=J Bj
λj,n◦J
Bj–
λj–,n◦ · · · ◦J
B
λ,nxnfor eachj= , , . . . ,land
n=I. We have thatzn=
lnxnfor alln≥. From Lemma .(), it follows that
φ(zn,xn) =φ
lnxn,xn
≤φ(q,xn) –φ
q,lnxn
=φ(q,xn) –φ(q,zn).
Taking limit asn→ ∞on both sides of the inequality, we have
lim
n→∞φ(zn,xn) = .
From (.) it follows that (xn–zn)→. Sincexn → p, we have
zn → p (n→ ∞). (.)
SinceJis uniformly norm-to-norm continuous on bounded subsets ofE, it follows that
Jzn → Jp (n→ ∞). (.)
This implies that{Jzn}is bounded inE*andE*is reflexive, we can assume thatJzn
z*∈E*. In view ofJ(E) =E*, there existsz∈Esuch thatJz=z*, and so
φ(xn,zn) =xn– xn,Jzn+zn
Takinglim infn→∞on both sides of the equality above and in view of the weak lower
semi-continuity of the norm · , it follows that
≥ p– p,z*+z*
=p– p,Jz+Jz
=p– p,Jz+z
=φ(p,z).
From Remark ., we havep=z, which implies thatz*=Jpand soJz
nJp∈E*. From
(.) and the Kadec-Klee property ofE*, we haveJz
n→Jpasn→ ∞. Note thatJ–is
norm-weak*-continuous, that is,z
np. From (.) and the Kadec-Klee property ofE,
we have
lim
n→∞zn=p. (.)
For anyq∈F, we note that
φ(q,xn) –φ(q,un) =xn–un– q,Jxn–Jun
≤ xn–un
xn+un
+ qJxn–Jun.
Thus it follows fromxn–un → andJxn–Jun → that
φ(q,xn) –φ(q,un)→ (n→ ∞). (.)
Denote i n:=K
i
ri,n◦K
i–
ri–,n◦ · · · ◦K
r,nfor eachi= , , . . . ,mand n=I. We can rewriteun
asun= mnzn. It follows that for eachi= , , . . . ,m, we have
φ(q,un) =φ
q, mnzn
≤φq, mn–zn
≤φq, mn–zn
· · · ≤φq, inzn
. (.)
From Lemma .(), for eachi= , , . . . ,m, we have
φ inzn,zn
≤φ(q,zn) –φ
q, inzn
≤φ(q,xn) –φ
q, inzn
≤φ(q,xn) –φ(q,un). (.)
From (.) it follows thatφ( i
nzn,zn)→ asn→ ∞for eachi= , , . . . ,m, and so from
(.), that
i
nzn–zn
Sincezn → p, we also have, for eachi= , , . . . ,m,
i
nzn→ p (n→ ∞). (.)
Since{ inzn}is bounded for eachi= , , . . . ,mandEis reflexive, without loss of generality,
we may assume that inznhfor alli= , , . . . ,m. From the first step, sinceCnis closed
and convex for eachn≥, it is obvious thath∈Cn. Again, since
φ inzn,zn
= inzn–
i
nzn,Jzn
+zn,
takinglim infn→∞on both sides of the equality above, we have ≥ h– h,Jp+p=φ(h,p).
This implies thath=pfor eachi= , , . . . ,m, and so
i
nznp. (.)
From (.), (.) and the Kadec-Klee property, for eachi= , , . . . ,m, we have
lim
n→∞
i
nzn=p. (.)
By using the triangle inequality, for eachi= , , . . . ,m, we obtain
i
nzn– in–zn≤ inzn–p+p– in–zn.
Hence, for eachi= , , . . . ,m, we have
lim
n→∞ i
nzn– in–zn= . (.)
Since{ri,n} ⊂[a,∞) andJis uniformly norm-to-norm continuous on bounded subsets of
E, for eachi= , , , . . . ,m, we have
lim
n→∞
J i
nzn–J in–zn
ri,n
= . (.)
From Lemma . we have, for eachi= , , . . . ,m,
i
i
nzn,y
+ ri,n
y– inzn,J nizn–J in–zn
≥, ∀y∈C,
wherei(un,y) =fi(un,y) +y–un,Aiunfor allun,y∈C. From the condition (A), it
fol-lows that for eachi= , , . . . ,m,
ri,n
y– inzn,J nizn–J in–zn
≥i
y, inzn
, ∀y∈C.
From (.) and (.), for eachi= , , . . . ,m, we have
For anyt∈[, ] andy∈C, letyt=ty+ ( –t)p. Then we getyt∈C. From (.) it follows
that for eachi= , , . . . ,m,
i(yt,p)≤, ∀yt∈C. (.)
By the conditions (A) and (A), for eachi= , , . . . ,m, we have
= i(yt,yt)
≤ti(yt,y) + ( –t)i(yt,p)
≤ti(yt,y)
≤i(yt,y). (.)
From the condition (A), we get
≤i(yt,y) =i
ty+ ( –t)p,y.
Takingt→ in the equality above, for eachi= , , . . . ,m, we have
=lim
t→≤tlim→i
ty+ ( –t)p,y=i(p,y), ∀y∈C,
that is, fi(p,y) +y–p,Aip ≥ for all y∈C and i= , , . . . ,m. This implies that p∈
GEP(fi,Ai) for eachi= , , . . . ,m. Therefore,p∈
m
i=GEP(fi,Ai).
Next, we show thatp∈lj=B–
j . Letzn=lnxnfor eachn≥. For anyq∈F, it follows
that for eachj= , , . . . ,l,
φ(q,zn) =φ
q,lnxn
≤φq,ln–xn
≤φq,ln–xn
· · · ≤φq,jnxn
. (.)
By Lemma . we have, forj= , , , . . . ,m,
φjnxn,xn
≤φ(q,xn) –φ
q,jnxn
≤φ(q,xn) –φ(q,zn). (.)
Sincexn→pandzn→pasn→ ∞, we getφ(jnxn,xn)→ asn→ ∞forj= , , , . . . ,m.
From (.) it follows that
jnxn–xn
→.
Sincexn → p, we also have
This implies that for eachj= , , , . . . ,m,{jnxn}is bounded andEis reflexive, without
loss of generality, we assume thatjnxnk. We know thatCnis closed and convex for
eachn≥, it is obvious thatk∈Cn. Again, since
φjnxn,xn
=jnxn
– jnxn,Jxn
+xn,
takinglim infn→∞on both sides of equality above, we have
≥ k– k,Jp+p
=φ(k,p).
That is,k=p,∀j= , , , . . . ,l, it follows that
jnxnp. (.)
From (.), (.) and the Kadec-Klee property, it follows that
lim
n→∞ j
nxn=p, ∀j= , , , . . . ,m. (.)
We also have
lim
n→∞ j–
n xn=p, ∀j= , , , . . . ,m. (.)
It follows that
lim
n→∞
j
nxn–jn–xn= , ∀j= , , , . . . ,m. (.)
SinceJis uniformly norm-to-norm continuous on bounded subsets ofEand
lim inf
n→∞ λj,n>
for eachj= , , . . . ,l, we have
lim
n→∞
λj,n
Jjnxn–Jjn–xn= .
Letjnxn=Jλjj,n
j–
n xnfor eachj= , , . . . ,l. Then we have
lim
n→∞Bλj,n
j–
n xn= lim n→∞
λj,n
Jjnxn–Jjn–xn= .
For any (w,w*)∈G(Bj) and (njxn,Aλj,n
j–
n xn)∈G(Bj) for eachj= , , . . . ,l, it follows from
the monotonicity ofBjthat for alln≥,
w–jnxn,w*–Bλj,n
j–
n xn
Lettingn→ ∞in the inequality above, we getw–p,w* ≥. SinceB
jis maximal
mono-tone for eachj= , , . . . ,l, we obtainp∈lj=B–
j .
Step . We show thatp=Fx. Fromxn=Cnx, we haveJx–Jxn,xn–z ≥ for all
z∈Cn. SinceF⊂Cn, we also have
Jx–Jxn,xn–y ≥, ∀y∈F,
and so, taking limitn→ ∞, we get
Jx–Jp,p–y ≥, ∀y∈F.
Therefore, by Lemma ., we can conclude thatp=Fxandxn→pasn→ ∞. The proof
is completed.
Ifi= andj= , we have the following.
Corollary . Let C be a nonempty closed and convex subset of a uniformly smooth and strictly convex Banach space E with the Kadec-Klee property.Let f be a bifunction from C×C toRsatisfying the conditions(A)-(A)and let A:C→E*be a continuous and
monotone mapping.Let B⊂E×E*be a maximal monotone operator satisfying D(B)⊂C
and Jλn= (J+λnB)–J for allλn> .Assume that F:=GEP(f,A)∩B–=∅.For arbitrary
x∈C and C=C,generate a sequence{xn}by
⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩
zn=Jλnxn,
un=Krnzn,
Cn+={z∈Cn:φ(z,un)≤φ(z,xn)},
xn+=Cn+x, ∀n≥,
(.)
where{rn} ⊂[a,∞)for some a> andlim infn→∞λn> .Then the sequence{xn}converges
strongly to a point p∈F,where p=Fx.
4 Applications
In this section, we apply our result to find a common solution of the variational inequality problems and zeros of the maximal operators.
We need the following lemma for our result, which is a special case of Lemmas . and . of [].
Lemma . (Zegeye and Shahzad []) Let C be a closed convex subset of a uniformly smooth,strictly convex real Banach space E.Let A:C→E*be a continuous monotone
mapping.For any r> and x∈E,define a mapping Tr:E→C as follows:
Trx=
z∈C:y–z,Az+
ry–z,Jz–Jx ≥,∀y∈C
, ∀x∈E.
() Tris a firmly nonexpansive-type mapping,i.e.,for anyx,y∈E,
Trx–Try,JTrx–JTry ≤ Trx–Try,Jx–Jy;
() F(Tr) =VI(C,A);
() VI(C,A)is closed and convex;
() φ(p,Trx) +φ(Trx,x)≤φ(p,x)for anyp∈F(Tr).
Theorem . Let C be a nonempty closed and convex subset of a uniformly smooth and strictly convex Banach space E with the Kadec-Klee property.For each i= , , . . . ,m,let {Ai}be a finite family of continuous and monotone mappings C→E*.For rn> and x∈E,
define a mapping Tri,n:E→C by
Tri,nx:=
z∈C:y–z,Aiz+
rn
y–z,Jz–Jx ≥,∀y∈C
.
Let B⊂E×E* be a maximal monotone operator satisfying D(B)⊂C and JBj
λj,n = (J+ λj,nBj)–J for allλ> and j= , , . . . ,l.Assume that F:= (
m
i=VI(C,Ai))∩(
l
j=B–j )=∅.
For an initial point x∈E with C=C,define the sequence{xn}in C as follows:
⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩
zn=JλBll,n◦J
Bl–
λl–,n◦ · · · ◦J
B λ,nxn,
un=Trm,n◦Trm–,n◦ · · · ◦Tr,nzn,
Cn+={z∈Cn:φ(z,un)≤φ(z,xn)},
xn+=Cn+x, ∀n≥,
(.)
where{ri,n} ⊂[a,∞)for some a> for all i= , , . . . ,m andlim infn→∞λj,n> for all j=
, , . . . ,l.Then the sequence{xn}converges strongly to a point p∈F,where p=Fx.
Proof Takingfi(un,y) = for alli= , , . . . ,min Theorem ., we can get the desired
con-clusion.
By Theorem ., if we setBj≡ for eachj= , , . . . ,l, we obtain the following.
Corollary . Let C be a nonempty closed and convex subset of a uniformly smooth and strictly convex Banach space E with the Kadec-Klee property.For any i= , , . . . ,m, let {Ai}be a finite family of continuous and monotone mappings C→E*.Assume that F:=
m
i=VI(C,Ai)=∅.For an initial point x∈E with C=C,define the sequence{xn}in C as
follows:
⎧ ⎪ ⎨ ⎪ ⎩
un=Trm,n◦Trm–,n◦ · · · ◦Tr,nxn,
Cn+={z∈Cn:φ(z,un)≤φ(z,xn)},
xn+=Cn+x, ∀n≥,
(.)
where{ri,n} ⊂[a,∞)for some a> for each i= , , . . . ,m.Then the sequence{xn}converges
strongly to a point p∈F,where p=Fx.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed equally and significantly in writing this article. All authors read and approved the final manuscript.
Author details
1Department of Mathematics and Statistics, Faculty of Science, Thaksin University (TSU), Phatthalung 93110, Thailand. 2Department of Mathematics, Faculty of Science, King Mongkut’s University of Technology Thonburi (KMUTT), Bangkok,
Thailand.3Department of Mathematics Education and RINS, Gyeongsang National University, Chinju, 660-701, Korea.
Acknowledgements
This work was supported by Thaksin University Research Fund and was also supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (Grant Number: 2012-0008170).
Received: 23 October 2012 Accepted: 30 April 2013 Published: 16 May 2013
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doi:10.1186/1029-242X-2013-247
