R E S E A R C H
Open Access
The split equilibrium problem and its
convergence algorithms
Zhenhua He
**Correspondence:
[email protected] Department of Mathematics, Honghe University, Mengzi, Yunnan 661100, China
Abstract
The purpose of this paper is to introduce a split equilibrium problem (SEP) and find a solution of the equilibrium problem such that its image under a given bounded linear operator is a solution of another equilibrium problem. By using the iterative method, we construct some iterative algorithms to solve such problem in real Hilbert spaces and obtain some strong and weak convergence theorems. Finally, we point out that there exist many SEPs which need the use of new methods to solve them. Some examples are given to illustrate our results.
MSC: 47J25; 47H09; 65K10
Keywords: iterative algorithm; split equilibrium problem; strong and weak convergence
1 Introduction
Throughout this paper, the symbolsNandRare used to denote the sets of positive integers and real numbers, respectively.
In this paper, we propose a new equilibrium problem, which is called a split equilibrium problem (SEP). LetEandEbe two real Banach spaces. LetCbe a closed convex subset
ofE,K a closed convex subset ofE, andA:E→Ea bounded linear operator.f is a
bi-function fromC×CintoRandgis a bi-function fromK×KintoR. The SEP is
to find an elementp∈C such that f(p,y)≥, ∀y∈C, (.)
and
such thatu:=Ap∈K solves g(u,v)≥, ∀v∈K. (.)
If we consider only the problem (.), then (.) is a classical equilibrium problem. From (.) and (.), we can see that the SEP contains two equilibrium problems, and the im-age of a solution of one equilibrium problem under a given bounded linear operator is a solution of another equilibrium problem. Since many problems coming from physics, op-timization, and economics reduce to find a solution of the equilibrium problem (.) (see, for instance, [, ]), the equilibrium problem (.) is very important in the field of applied mathematics. Some authors have proposed some methods to find the solution of the equi-librium problem (.). As a generalization of the equiequi-librium problem (.), when finding a
common solution for some equilibrium problems, it has been considered in the same sub-set of the same space; see [–]. However, in general, some equilibrium problems always belong to different subsets of spaces, so the SEP is important and quite general. The SEP should enable us to split the solution between two different subsets of spaces so that the image of a solution point of one problem, under a given bounded linear operator, is a solu-tion point of another problem. A special case of the SEP is the split variasolu-tional inequality problem (SVIP); see [].
For convenience, in this paper letEP(f),EP(g) and={p∈EP(f) :Ap∈EP(g)}denote the solution set of (.), (.) and the SEP, respectively.
Example . LetE=E=R,C:= [, +∞) andK:= (–∞, –]. LetA(x) = –xfor allx∈R,
thenAis a bounded linear operator. Letf :C×C→R, andg:K×K→Rbe defined by f(x,y) =y–x,g(u,v) = (u–v), respectively. Clearly,EP(f) ={}andA() = –∈EP(g). So
={p∈EP(f) :Ap∈EP(g)} =∅.
Example . LetE=Rwith the standard norm| · |andE=Rwith the norm α =
(a +a)
for someα= (a,a)∈R.K:= [, +∞) andC:={α= (a,a)∈R|a–a≥}.
Define a bi-functionf(w,α) =w–w+a–a, wherew= (w,w),α= (a,a)∈C, then
f is a bi-function fromC×CintoRwithEP(f) ={p= (p,p)|p–p= }. For eachα=
(a,a)∈E, letAα=a–a, thenAis a bounded linear operator fromEintoE. In fact, it
is also easy to verify thatA(aα+bα) =aA(α) +bA(α) and A = √
for someα,α∈E
anda,b∈R. Now define another bi-functiong as follows:g(u,v) =v–ufor allu,v∈K. Thengis a bi-function fromK×KintoRwithEP(g) ={}.
Clearly, whenp∈EP(f), we haveAp= ∈EP(g). So={p∈EP(f) :Ap∈EP(g)} =∅. Remark . The SEP in Example . lies in two different subsets of the same space. While the SEP in Example . lies in two different subsets of the different space.
In this paper, we construct some iterative algorithms to solve the SEP. Some strong and weak convergence theorems are established. The results obtained in this paper can be reckoned as the new development of the equilibrium problem (.). Finally, we point out that there exist many SEPs which need the use of new methods to solve them. Some ex-amples are given to illustrate our results.
2 Preliminaries
We assume thatHis a real Hilbert space with zero vectorθwhose inner product and norm are denoted by·,·and · , respectively; and we use symbols→andto denote strong and weak convergence, respectively.
LetHandHbe two Hilbert spaces. The operatorAfromHintoHand the operator
BfromHintoHare two bounded linear operators.Bis called the adjoint operator ofA,
if for allz∈H,w∈H,BsatisfiesAz,w=z,Bw. Especially, ifH=H, thenBreduces
to the well-known adjoint operator ofA.
Remark . It is easy to verify that the operatorB, an adjoint operator ofA, has the fol-lowing characters:
Example . LetH=Rwith the standard norm| · |andH=Rwith the norm α =
(a +a)
for someα= (a,a)∈R.x,y=xydenotes the inner product ofHfor some
x,y∈Handα,β=ab+ab denotes the inner product ofH for someα= (a,a),
β= (b,b)∈H. LetAα=a–a, thenAis a bounded linear operator fromHintoH
with A =√. Forx∈H, letBx= (–x,x), thenBis a bounded linear operator fromH
intoHwith B = √
. Moreover, for anyα= (a,a)∈Handx∈H,Aα,x=α,Bx,
soBis an adjoint operator ofA.
Example . Let H =R with the norm α = (a +a)
for some α = (a,a)∈R
and H =R with the norm γ = (c +c+c)
for some γ = (c,c,c)∈ R. Let α,β=ab+abandγ,η=cd+cd+cddenote the inner product ofHandH,
respectively, whereα= (a,a),β = (b,b)∈H,γ = (c,c,c),η= (d,d,d)∈R. Let
Aα= (a,a,a–a) forα= (a,a)∈H, thenAis a bounded linear operator fromHinto
Hwith A = √
because (√ , –
√
, –
√
) ≤supα = Aα ≤√. Forγ = (c,c,c)∈
H, letBγ = (c+c,c–c), thenBis a bounded linear operator fromHintoH with
B =√. Moreover, for anyα= (a,a)∈H andγ = (c,c,c)∈H,Aα,γ=α,Bγ,
soBis an adjoint operator ofA.
LetK be a closed convex subset of a real Hilbert spaceH. For each pointx∈H, there exists a unique nearest point inK, denoted byPKx, such that
x–PKx ≤ x–y , ∀y∈K.
The mappingPKis called themetric projectionfromHontoK. It is well known thatPK
has the following characters:
(i) x–y,PKx–PKy ≥ PKx–PKy for everyx,y∈H.
(ii) Forx∈H, andz∈K,z=PK(x)⇔ x–z,z–y ≥,∀y∈K.
(iii) Forx∈Handy∈K,
y–PK(x)
+x–PK(x)
≤ x–y . (.)
A Banach space (X, · ) is said to satisfy Opial’s condition if, for each sequence{xn}in
Xwhich converges weakly to a pointx∈X, we have
lim inf
n→∞ xn–x <lim infn→∞ xn–y , ∀y∈X,y=x.
It is well known that each Hilbert space satisfies Opial’s condition. The following results are crucial to our main results.
Lemma .(see []) Let K be a nonempty closed convex subset of H and F be a bi-function of K×K intoRsatisfying the following conditions:
(A) F(x,x) = for allx∈K;
(A) Fis monotone, that is,F(x,y) +F(y,x)≤for allx,y∈K; (A) for eachx,y,z∈K,
lim sup
t↓
(A) for eachx∈K,y→F(x,y)is convex and lower semi-continuous.
Let r> and x∈H. Then, there exists z∈K such that
F(z,y) +
ry–z,z–x ≥, for all y∈K.
Lemma .(see []) Let K be a nonempty closed convex subset of H and let F be a bi-function of K×K intoRsatisfying (A)-(A). For r> and x∈H, define a mapping TrF : H→K as follows:
TrF(x) =
z∈K:F(z,y) +
ry–z,z–x ≥,∀y∈K
(.)
for all x∈H. Then the following hold:
(i) TF
r is single-valued; (ii) TF
r is firmly non-expansive, that is, for anyx,y∈H,
TF
rx–TrFy
≤TF
rx–TrFy,x–y
;
(iii) F(TrF) =EP(F)for∀r> ; (iv) EP(F)is closed and convex.
Lemma .(see []) Let H be a real Hilbert space. Then for any x,x, . . . ,xk∈H and
a,a, . . . ,ak∈[, ]with ki=ai= , k∈N, we have
k
i=
aixi
=
k
i=
ai xi – k–
i= k
j=i+
aiaj xi–xj .
In particular, we have
() αx+ ( –α)y =α x + ( –α) y –α( –α) x–y for allx,y∈Hand
α∈[, ];
() the mapf :H→Rdefined byf(x) = x is convex.
Lemma .(see,e.g., []) Let H be a real Hilbert space. Then the following hold:
(a) x+y ≤ y + x,x+yfor allx,y∈H; (b) x–y = x + y – x,yfor allx,y∈H.
Lemma . Let K be a nonempty closed convex subset of H. For x∈H, let the mapping TF
r(x)be the same as in Lemma .. Then for r,s> and x,y∈H,
TrF(x) –TsF(y)≤ y–x +|s–r| s T
F s(y) –y.
Proof Forr,s> andx,y∈H, by (i) of Lemma ., we can letz=TrF(x) andz=TsF(y).
By the definition ofTF
r, we have
F(z,u) +
and
F(z,u) +
su–z,z–y ≥, ∀u∈K. (.)
Takingu=zin (.) andu=zin (.), we have
F(z,z) +
rz–z,z–x ≥, (.)
and
F(z,z) +
sz–z,z–y ≥. (.)
Since the bi-functionFsatisfies the condition (A), from (.) and (.) we have
rz–z,z–x+
sz–z,z–y ≥,
which implies that
z–z,z–x–
z–z,r
z–y
s
≥.
Thus, we have
z–z,z–z+z–x–
r s(z–y)
≥,
this implies that
z–z ≤
z–z,z–x–
r s(z–y)
≤ z–z
z–x–
r s(z–y)
,
so
z–z ≤
z–x–
r s(z–y)
≤ y–x +
–r s
(z–y)
= y–x +|s–r| s T
F s(y) –y,
namely,
TrF(x) –TsF(y)≤ y–x +|s–r| s T
F s(y) –y.
The proof is completed.
3 Main results
In this section, we will solve the SEP which satisfies the conditions (A)-(A).
spaces.∧:={, , . . . ,k}denotes a finite index set. For any i∈ ∧, fi:C×C→Ris a
bi-function withki=EP(fi)=∅. Let A:H→Hbe a bounded linear operator with the
ad-joint B and g:K×K→Ra bi-function withEP(g)=∅. Let{xn}and{uin}(i∈ ∧) be
se-quences generated by
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
x∈C,
fi
uin,y+ rn
y–uin,uin–xn
≥, y∈C,i∈ ∧,
τn=
u
n+· · ·+ukn
k , g(wn,z) +
rn
z–wn,wn–Aτn ≥, z∈K,
xn+=PC
τn+μB(wn–Aτn)
, ∀n∈N,
(.)
where{rn} ⊂(, +∞)withlim infn→∞rn> , PCis a projection operator from Hinto C and
μ∈(,
B )is a constant. Suppose that={p∈
k
i=EP(fi) :Ap∈EP(g)} =∅, then the
sequences{xn}and{uin}(i∈ ∧) converge weakly to an element p∈, while{wn}converges
weakly to Ap∈EP(g).
Proof For eachi∈ ∧and eachr> , letTfi
r:H→Cbe defined by (.), thenuin=T fi rnxn
for alln∈Nby Lemma .. Again letTrg:H→Kbe defined by (.), thenwn=TrgnAτn
for alln∈N. So (.) can be rewritten as follows: ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
x∈C,
ui n=T
fi
rnxn, i∈ ∧,
τn=
u
n+· · ·+ukn
k , wn=TrgnAτn,
xn+=PC
τn+μB
Trgn–I
Aτn
, ∀n∈N.
(.)
Let x*∈C be a point such thatx*∈k
i=EP(fi) andAx*∈EP(g), namely,x*∈. By
Lemma . and Lemma ., it follows that
uin–x*=Tfi rnxn–T
fi rnx
*≤Tfi rnxn–T
fi rnx
*,x n–x*
= u
i n–x*
+xn–x*
–uin–xn
,
hence
uin–x*≤xn–x*
–uin–xn
. (.)
Applying Lemma ., we get
τn–x*≤
k
k
i=
(.) and (.) imply that
τn–x*
≤xn–x*
– k
k
i=
ui n–xn
. (.)
Again from Lemma ., we have
wn–Ax*=TrgnAτn–Ax
*≤Aτ
n–Ax* for eachn∈N. (.)
By (b) of Lemma . and (.), for eachn∈N, we have
μτn–x*,B
Trgn–IAτn
= μAτn–x*
+Trgn–IAτn–
Trgn–IAτn,
Trgn–IAτn
= μTrgnAτn–Ax*,
Trgn–IAτn
–Trgn–IAτn
= μ
T
g
rnAτn–Ax *+
T
g rn–I
Aτn
–
Aτn–Ax
*
–Trgn–IAτn
≤μ
T
g rn–I
Aτn–Trgn–I
Aτn
= –μTrgn–IAτn
. (.)
We also have
BTrgn–IAτn
≤ B Trgn–IAτn
. (.)
From (.), (.)-(.), we have
xn+–x* =PC
τn+μB
Trgn–IAτn
–PCx* ≤τn+μB
Trgn–IAτn) –x*
=τn–x*
+μBTg rn–I
Aτn
+ μτn–x*,B
Tg
rn–I
Aτn
≤τn–x*
+μ B Trgn–IAτn
–μTrgn–IAτn
=τn–x*–μ
–μ B Trgn–IAτn ≤xn–x*–μ
–μ B Trgn–IAτn. (.)
Noticeμ∈(, B),μ( –μ B ) > . It follows from (.) that
xn+–x*≤τn–x*≤xn–x* (.)
and
μ –μ B Trgn–IAτn
≤xn–x*
–xn+–x*
(.) implieslimn→∞ xn–x* exists. Further, from (.)-(.),
lim
n→∞xn–x
*= lim n→∞τn–x
*, lim
n→∞T
g rn–I
Aτn= . (.)
Again from (.), we have
lim
n→∞u i
n–xn= , i∈ ∧, (.)
which yields that
lim
n→∞T
fi rn–I
xn= lim n→∞T
fi
rnxn–xn=nlim→∞u i
n–xn= , i∈ ∧, (.)
and
τn–xn ≤un–xn+· · ·+ukn–xn→ asn→ ∞. (.)
Becauselimn→∞ xn–x* exists, which implies{xn}is bounded, hence{xn}has a weakly
convergence subsequence{xnj}. Assume thatxnjpfor somep∈C. Thenu i
njp,τnj
pandAτnjAp∈Kby (.) and (.).
Now we provep∈or, to be more precise, we provep∈ki=EP(fi) andAp∈EP(g).
By Lemma ., for anyr> ,EP(fi) =F(Trfi), i∈ ∧, andEP(g) =F(Trg). Fori∈ ∧, since
(I–Tfi
rn)xn→ by (.), we haveT fi
rp=pforr> . Otherwise, ifTrfip=pfor alli∈ ∧,
then by Opial’s condition, we have
lim inf
j→∞ xnj–p <lim inf j→∞ xnj–T
fi rp
=lim inf
j→∞ xnj–T fi rnjxnj+T
fi rnjxnj–T
fi rp ≤lim inf
j→∞
xnj–T fi
rnjxnj+T fi rnjxnj–T
fi
rp
=lim inf
j→∞
Tfi rnjxnj–T
fi rp
=lim inf
j→∞
Tfi
rp–Trfnji xnj
≤lim inf
j→∞
xnj–p + |rnj–r|
rnj
Tfi
rnjxnj–xnj
(by Lemma .)
=lim inf
j→∞ xnj–p ,
this is a contradiction. So this shows thatp∈ki=F(Tfi r) =
k
i=EP(fi). Similarly, we can
proveAp∈EP(g).
Finally, we prove{xn}and{uin}converge weakly top∈, while{wn}converges weakly
toAp∈EP(g). Firstly, if there exists other subsequence of{xn}which is denoted by{xnt}
such thatxntq∈withq=p, then, by Opial’s condition,
lim inf
t→∞ xnt–q <lim inft→∞ xnt–p <lim inft→∞ xnt–q .
On the other hand, by (.) we also have τnp. Notice that wn–Aτn = (Trgn–
I)Aτn → by (.), so we haveτnApandwnAp. We obtain the desired result.
Corollary . Let C be a nonempty closed convex subset of Hand K a nonempty closed
convex subset of H, where Hand Hare two real Hilbert spaces. f :C×C→Ris a
bi-function withEP(f)=∅. Let A:H→Hbe a bounded linear operator with the adjoint B
and g:K×K→Ra bi-function withEP(g)=∅. Let{xn}and{un}be sequences generated
by ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
x∈C,
f(un,y) +
rn
y–un,un–xn ≥, y∈C,
g(wn,z) +
rn
z–wn,wn–Aun ≥, z∈K,
xn+=PC
un+μB(wn–Aun)
, ∀n∈N,
(.)
where{rn} ⊂(, +∞)withlim infn→∞rn> , PC is a projection operator from H into C
andμ∈(,
B )is a constant. Suppose that={p∈EP(f) :Ap∈EP(g)} =∅, then the
sequences{xn}and{un}converge weakly to an element p∈, while{wn}converges weakly
to Ap∈EP(g).
Theorem .(Strong convergence theorem) Let C be a nonempty closed convex subset of H and K a nonempty closed convex subset of H, where Hand H are two real Hilbert
spaces.∧:={, , . . . ,k}denotes a finite index set. For any i∈ ∧, fi:C×C→Ris a
bi-function withki=EP(fi)=∅. Let A:H→Hbe a bounded linear operator with the
ad-joint B and g:K×K→Ra bi-function withEP(g)=∅. Let C=C and{xn}and{uin}
(i∈ ∧) be sequences generated by ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
x∈C,
fi
uin,y+ rn
y–uin,uin–xn
≥, y∈C,i∈ ∧,
τn=
u
n+· · ·+ukn
k , g(wn,z) +
rn
z–wn,wn–Aτn ≥, z∈K,
yn=PC
τn+μB(wn–Aτn)
,
Cn+=
v∈Cn: yn–v ≤ τn–v ≤ xn–v
,
xn+=PCn+(x), n∈N,
(.)
where{rn} ⊂(, +∞)withlim infn→∞rn> , PC is a projection operator from H into C
andμ∈(, B)is a constant. Suppose that={p∈
k
i=EP(fi) :Ap∈EP(g)} =∅, then the
sequences{xn}and{uin}(i∈ ∧) converge strongly to an element x*∈, while{wn}converges
strongly to Ax*∈EP(g).
Proof By Lemma .,ui n=T
fi
rnxnfor alli∈ ∧,n∈Nandwn=T g
rnAτnfor alln∈N. We
claimCn=∅forn∈N. In fact⊂Cnforn∈N. Indeed, letp∈, it follows from (.)
and (.) that
μτn–p,B
Trgn–IAτn
≤–μTrgn–IAτn
and
BTrgn–IAτn
≤ B Trgn–IAτn
. (.)
From (.)-(.) we have
yn–p ≤τn+μB
Trgn–IAτn–p
= τn–p +μB
Trgn–IAτn
+ μτn–p,B
Trgn–IAτn
≤ τn–p +μ B Trgn–I
Aτn
–μTrgn–IAτn
= τn–p –μ
–μ B Trgn–IAτn ≤ xn–p –μ
–μ B Trgn–IAτn
. (.)
Noticeμ∈(,
B ),μ( –μ B ) > . It follows from (.) that
yn–p ≤ τn–p ≤ xn–p for alln∈N, (.)
this showsp∈Cnfor alln∈N, so⊂CnandCn=∅forn∈N.
We want to proveCnis a closed convex set forn∈N. It is easy to verify thatCnis closed
forn∈N, so it suffices to verifyCnis convex forn∈N. In fact, letv,v∈Cn+, for each
λ∈(, ), we have yn–
λv+ ( –λ)v
=λ(yn–v) + ( –λ)(yn–v)
=λ yn–v + ( –λ) yn–v –λ( –λ) v–v ≤λ τn–v + ( –λ) τn–v –λ( –λ) v–v
=τn–
λv+ ( –λ)v
,
namely, yn– (λv+ ( –λ)v) ≤ τn– (λv+ ( –λ)v) . Similarly, we have τn– (λv+
( –λ)v) ≤ xn– (λv+ ( –λ)v) , this showsλv+ ( –λ)v∈Cn+andCn+is a convex
set forn∈N.
By (iv) of Lemma .,is a closed convex set, so there exists a unique element q= P(x)∈⊂Cn. Since xn=PCn(x), we have xn–x ≤ q–x , which shows that {xn} is bounded. So are{τn}and{yn}. Notice thatCn+⊂Cn andxn+=PCn+(x)⊂Cn,
then
xn+–x ≤ xn–x , n≥. (.)
It follows thatlimn→∞ xn–x exists.
For somem,n∈Nwithm>n, fromxm=PCm(x)⊂Cnand (.), we have
xn–xm + x–xm =xn–PCm(x)
+x–PCm(x)
≤ xn–x . (.)
Next we provex*∈. Firstly, byx
n+=PCn+(x)∈Cn+⊂Cn, from (.) we have
yn–xn ≤ yn–xn+ + xn+–xn ≤ xn+–xn →,
τn–xn ≤ τn–xn+ + xn+–xn ≤ xn+–xn →, (.)
yn–τn ≤ yn–xn + xn–τn →.
Again from (.), we have
Trgn–IAτn
≤
μ( –μ B )
xn–p – yn–p
≤
μ( –μ B ) xn–yn
xn–p + yn–p
→. (.)
So
lim
n→∞T
g rn–I
Aτn= . (.)
Noticeτn=u
n+···+ukn
k , hence from (.) and (.), we obtain
lim
n→∞T
fi
rnxn–xn=nlim→∞u i
n–xn= , i∈ ∧. (.)
Sincexn→x*, (.) and Lemma . imply that forr> ,
Tfi
rx*–x*≤Trfix*–Trfinxn+T fi
rnxn–xn+xn–x *
≤xn–x*+|
rn–r|
rn
Tfi
rnxn–xn+T fi
rnxn–xn+xn–x
*→,
which yields thatx*∈F(Tfi
r) for alli∈ ∧, furtherx*∈
k
i=EP(fi). SinceAis a bounded
linear operator, Axn–Ax* → byxn→x*. Then forr> , by (.) and Lemma . we
have
TrgAx*–Ax*≤TrgAx*–TrgnAxn+TrgnAxn–Axn+Axn–Ax *
≤Axn–Ax*+|
rn–r|
rn
TrgnAxn–Axn+TrgnAxn–Axn
+Axn–Ax*→,
hence,Ax*∈F(Tg
r) =EP(g) forr> . Thus we have provedx*∈, namely,{xn}converges
strongly tox*∈. Notice (.), we also have{ui
n}(i∈ ∧) converges strongly tox*∈.
Since τn–xn → by (.), we haveτn→x*byxn→x*. Again from (.) we have
lim
n→∞ wn–Aτn =nlim→∞T g rn–I
Aτn= ,
Corollary . Let C be a nonempty closed convex subset of Hand K a nonempty closed
convex subset of H, where Hand Hare two real Hilbert spaces. f :C×C→Ris a
bi-function withEP(f)=∅. Let A:H→Hbe a bounded linear operator with the adjoint B
and g:K×K→Ra bi-function withEP(g)=∅. Let C=C and{xn}and{un}be sequences
generated by ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
x∈C,
f(un,y) +
rn
y–un,un–xn ≥, y∈C,
g(wn,z) +
rn
z–wn,wn–Aun ≥, z∈K,
yn=PC
un+μB(wn–Aun)
,
Cn+=
v∈Cn: yn–v ≤ un–v ≤ xn–v
,
xn+=PCn+(x), n∈N,
(.)
where{rn} ⊂(, +∞)withlim infn→∞rn> ,μ∈(, B)is a constant. Suppose that= {p∈EP(f) :Ap∈EP(g)} =∅. Then the sequences {xn}and{un} converge strongly to an
element x*∈, while{w
n}converges strongly to Ax*∈EP(g).
IfC=HandK=Hin Theorem . and Theorem ., we have the following corollaries.
Corollary . Let Hand Hbe two real Hilbert spaces.∧:={, , . . . ,k}denotes a finite
index set. For any i∈ ∧, fi:H×H→Ris a bi-function with
k
i=EP(fi)=∅. Let A:H→
Hbe a bounded linear operator with the adjoint B and g:H×H→Ra bi-function with
EP(g)=∅. Let{xn}and{uin}be sequences generated by
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
x∈H,
fi
uin,y+ rn
y–uin,uin–xn
≥,y∈H, i∈ ∧,
τn=
u
n+· · ·+ukn
k , g(wn,z) +
rn
z–wn,wn–Aτn ≥, z∈H,
xn+=τn+μB(wn–Aτn), ∀n∈N,
(.)
where{rn} ⊂(, +∞)withlim infn→∞rn> ,μ∈(, B)is a constant. Suppose that= {p∈ki=EP(fi) :Ap∈EP(g)} =∅. Then the sequences{xn}and{uin}(i∈ ∧) converge weakly
to an element p∈, while{wn}converges weakly to Ap∈EP(g).
Corollary . Let Hand Hbe two real Hilbert spaces. f :H×H→Ris a bi-function
withEP(f)=∅. Let A:H→H be a bounded linear operator with the adjoint B and g:
H×H→Ra bi-function withEP(g)=∅. Let{xn}and{un}be sequences generated by
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
x∈H,
f(un,y) +
rn
y–un,un–xn ≥, y∈H,
g(wn,z) +
rn
z–wn,wn–Aun ≥, z∈H,
xn+=un+μB(wn–Aun), ∀n∈N,
where {rn} ⊂(, +∞)with lim infn→∞rn> , μ∈(, B) is a constant. Suppose that
={p∈EP(f) :Ap∈EP(g)} =∅. Then the sequences{xn} and{un} converge weakly to
an element p∈, while{wn}converges weakly to Ap∈EP(g).
Corollary . Let Hand Hbe two real Hilbert spaces.∧:={, , . . . ,k}denotes a finite
index set. For any i∈ ∧, fi:H×H→Ris a bi-function with
k
i=EP(fi)=∅. Let A:H→
Hbe a bounded linear operator with the adjoint B and g:H×H→Ra bi-function with
EP(g)=∅. Let C=Hand{xn}and{uin}(i∈ ∧) be sequences generated by
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
x∈H,
fi
uin,y+ rn
y–uin,uin–xn
≥, y∈H,i∈ ∧,
τn=
u
n+· · ·+ukn
k , g(wn,z) +
rn
z–wn,wn–Aτn ≥, z∈H,
yn=τn+μB(wn–Aτn),
Cn+=
v∈Cn: yn–v ≤ τn–v ≤ xn–v
,
xn+=PCn+(x), n∈N,
(.)
where {rn} ⊂(, +∞)with lim infn→∞rn> , μ∈(, B) is a constant. Suppose that
={p∈ki=EP(fi) :Ap∈EP(g)} =∅. Then the sequences{xn}and{uin}(i∈ ∧) converge
strongly to an element p∈, while{wn}converges strongly to Ax*∈EP(g).
Corollary . Let Hand Hbe two real Hilbert spaces. f :H×H→Ris a bi-function
withEP(f)=∅. Let A:H→H be a bounded linear operator with the adjoint B and g:
H×H→Ra bi-function withEP(g)=∅. Let C=H and{xn}and{un}be sequences
generated by ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
x∈H,
f(un,y) +
rn
y–un,un–xn ≥, y∈H,
g(wn,z) +
rn
z–wn,wn–Aun ≥, z∈H,
yn=un+μB(wn–Aun),
Cn+=
v∈Cn: yn–v ≤ un–v ≤ xn–v
,
xn+=PCn+(x), n∈N,
(.)
where{rn} ⊂(, +∞)withlim infn→∞rn> ,μ∈(, B)is a constant. Suppose that= {p∈EP(f) :Ap∈EP(g)} =∅. Then the sequences {xn}and{un} converge strongly to an
element x*∈, while{w
n}converges strongly to Ax*∈EP(g).
Remark . The results of this paper provide some solution algorithms for some SEPs; however, there are still some SEPs which cannot be solved by the results of this paper. The following examples belong to the case.
Example . LetH=RandH=Rwith the norm z = (x+y)
for somez= (x,y)∈
R.K:= [, +∞) andC:={z= (x,y)∈R|y–x≥}. Define a bi-functionf(w,z) =x +y+
y–x, wherew= (x,y),z= (x,y)∈C, thenf is a bi-function fromC×CintoRwith
EP(f) ={w= (x,y)|y–x≥,x+y≥–}. For eachz= (x,y)∈H, letAz=y–x, thenAis
a bounded linear operator fromHintoH. Now define another bi-functiongas follows:
g(u,v) =v–ufor allu,v∈K. Thengis a bi-function fromK×KintoRwithEP(g) ={}. Clearly, whenp= (x,y)∈EP(f) withy–x= andx+y≥–, we haveAp= ∈EP(g). So={p∈EP(f) :Ap∈EP(g)} =∅. However, because the bi-functionf does not satisfy the conditions (A)-(A), the SEP in this example cannot be solved by Corollary . or Corollary ..
Example . LetH, H,AandBbe the same as Example .. LetC:={α= (a,a)∈
H|a–a≥}andK:={γ = (c,c,c)∈H| γ ≤}. Define a bi-functionf(α,β) = (b–
b)– (a+a), whereα= (a,a),β= (b,b)∈C, thenf is a bi-function fromC×C
intoRwithEP(f) ={p= (p,p)|p–p≥≥p+p≥–}. Define another bi-function
g(γ,η) =c
+c– (c+d+d+d), whereγ= (c,c,c),η= (d,d,d)∈K, thenEP(g) = {u= (,u,u)∈K|u+u= }.
Clearly, whenp= (–, )∈EP(f), we haveAp= (, –, –)∈EP(g). So={p∈EP(f) : Ap∈EP(g)} =∅. However, since allf andg do not satisfy the conditions (A)-(A), we cannot use the results obtained in this paper to solve the SEP in this example.
4 Conclusion
There are still many SEPs which do not satisfy the conditions (A)-(A), so we need to develop some new methods to solve these problems in the future.
Competing interests
The author declares that he has no competing interests.
Acknowledgement
The author was supported by the Natural Science Foundation of Yunnan Province (2010ZC152).
Received: 1 April 2012 Accepted: 2 July 2012 Published: 20 July 2012
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