R E S E A R C H
Open Access
Strong convergence of a splitting algorithm
for treating monotone operators
Sun Young Cho
1, Xiaolong Qin
2,3and Lin Wang
4* *Correspondence:[email protected] 4College of Statistics and
Mathematics, Yunnan University of Finance and Economics, Kunming, 650221, China
Full list of author information is available at the end of the article
Abstract
In this paper, we investigate a splitting algorithm for treating monotone operators. Strong convergence theorems are established in the framework of Hilbert spaces.
Keywords: maximal monotone operator; fixed point; nonexpansive mapping;
proximal point algorithm; zero point
1 Introduction and preliminaries
In this article, we always assume thatHis a real Hilbert space with inner product·,·and norm · andCis a nonempty, closed, and convex subset ofH.
Let S:C→Cbe a mapping.F(S) stands for the fixed point set ofS. Sis said to be contractiveiff there exists a constantα∈(, ) such that
Sx–Sy ≤αx–y, ∀x,y∈C.
It is well known that every contractive mapping has a unique fixed point in metric spaces. Sis said to benonexpansiveiff
Sx–Sy ≤ x–y, ∀x,y∈C.
IfCis a bounded, closed, and convex subset ofH, thenF(S) is not empty, closed, and convex; see [] and the references therein.Sis said to bestrictly pseudocontractiveiff there exists a constantκ∈[, ) such that
Sx–Sy≤ x–y+κx–y–Sx+Sy, ∀x,y∈C.
The class of strictly pseudocontractive mapping was introduced by Browder and Petryshyn []. It is clear that the class of strictly pseudocontractive mappings include the class of nonexpansive mappings as a special case. It is also not hard to see that strictly pseudocon-tractive mapping is continuous.
LetA:C→Hbe a mapping. Recall thatAis said to bemonotoneiff
Ax–Ay,x–y ≥, ∀x,y∈C.
Ais said to bestrongly monotoneiff there exists a constantκ> such that
Ax–Ay,x–y ≥κx–y, ∀x,y∈C.
Ais said to beinverse-strongly monotoneiff there exists a constantκ> such that
Ax–Ay,x–y ≥κAx–Ay, ∀x,y∈C.
Ais inverse-strongly monotone iff the inverse ofAis strongly monotone. It is not hard to see that every inverse-strongly monotone mapping is monotone and continuous. LetIbe the identity mapping onH. From [], we know thatI–Sis inverse-strongly monotone iff Sis strictly pseudocontractive; for more details, see [] and the references therein.
The classical variational inequality problem is formulated as finding a pointx∈Csuch that
y–x,Ax ≥, ∀y∈C.
Such a pointx∈Cis called a solution of the variational inequality. In this paper, we use VI(C,A) to denote the solution set of the variational inequality. It is known thatx is a solution of the variational inequality iffxis a fixed point of the mappingProjC(I–rA), whereProjCis the metric projection fromHontoC,Iis the identity andris some posi-tive real number. Recently, many authors studied solutions of inverse-strongly monotone variational inequalities based on the equivalence; see [–].
Recall that a set-valued mappingB:H⇒His said to bemonotoneiff, for allx,y∈H, f ∈Bxandg∈Byimplyx–y,f –g> . In this paper, we useB–() to stand for the zero
point ofB. A monotone mappingB:H⇒H ismaximaliff the graphGraph(B) ofBis not properly contained in the graph of any other monotone mapping. It is known that a monotone mappingBis maximal if and only if, for any (x,f)∈H×H,x–y,f –g ≥, for all (y,g)∈Graph(B) impliesf ∈Bx. For a maximal monotone operatorBonH, and r> , we may define the single-valued resolventJr:H→Dom(B), whereDom(B) denote
the domain ofB. It is known thatJris firmly nonexpansive, andB–() =F(Jr).
One of the most important techniques for solving zero point problem of monotone op-erators goes back to the work of Browder []. Many important problems have reformula-tions which require finding zero points, for instance, evolution equareformula-tions, complementar-ity problems, mini-max problems, variational inequalities and fixed point problems. It is well known that minimizing a convex functionf can be reduced to finding zero points of the subdifferential mappingA=∂f. One of the basic ideas in the case of a Hilbert spaceHis reducing the above inclusion problem to a fixed point problem of the operatorRAdefined
byRA= (I+A)–, which is called the classical resolvent ofA. IfAhas some monotonicity
conditions, the classical resolvent ofAis with full domain and firmly nonexpansive. The property of the resolvent ensures that the Picard iterative algorithmxn+=RAxnconverge
weakly to a fixed point ofRA, which is necessarily a zero point ofA. Rockafellar introduced
this iteration method and call it the proximal point algorithm (PPA); for more details, see [] and [] and the references therein.
or of the same difficult as the original zero point problem. Therefore, one of the most inter-esting and important problems in the theory of monotone operators is to find an efficient iterative algorithm to compute their zero points. In many disciplines, including economics [], image recovery [], quantum physics [], and control theory [], problems arises in infinite dimension spaces. In such problems, strong convergence (norm convergence) is often much more desirable than weak convergence, for it translates the physically tangible property that the energyxn–xof the error between the iteratexnand the solutionx
eventually becomes arbitrarily small. The important of strong convergence is also under-lined in [], where a convex functionf is minimized via the proximal point algorithm: it is shown that the rate of convergence of the value sequence{f(xn)}is better when{xn}
converges strongly that it converges weakly. Such properties have a direct impact when the process is executed directly in the underlying infinite dimensional space.
To improve the weak convergence of PPA, many authors considered lots of different modifications; see [–] the references therein. One of the classic results was estab-lished by Solodov and Svaiter []. They obtained strong convergence theorems in Hilbert space without any compact assumption but with the aid of the metric projection.
In this paper, we are concerned with the problem of finding an element in the zero point set of the sum of two operators which are inverse-strongly monotone and a maxi-mal monotone and in the fixed point set of a mapping which is strictly pseudocontractive. Strong convergence theorems are established without the aid of the metric projections. The organization of this paper is as follows. In Section , we provide an introduction and some necessary preliminaries. In Section , a regularization iterative algorithm is investi-gated. A strong convergence theorem is established without the aid of metric projections. In Section , applications of the main results are discussed.
In order to prove our main results, we also need the following lemmas.
Lemma .[] Let A:C→H be a mapping,and B:H⇒H a maximal monotone oper-ator.Then F(Jr(I–rB)) = (A+B)–().
Lemma .[] Let E be a Banach space and let A be an m-accretive operator.Forλ> , μ> ,and x∈E,we have Jλx=Jμ(μλx+ ( –μλ)Jλx),where Jλ= (I+λA)–and Jμ= (I+μA)–.
Lemma .[] Let{xn}and{yn}be bounded sequences in a Banach space E,and{βn}
be a sequence in (, )with <lim infn→∞βn≤lim supn→∞βn< .Suppose that xn+=
( –βn)yn+βnxn,∀n≥and
lim sup
n→∞
yn+–yn–xn+–xn
≤.
Thenlimn→∞yn–xn= .
Lemma .[] Let{an}be a sequence of nonnegative numbers satisfying the condition
an+≤( –tn)an+tnbn+cn,∀n≥,where{tn}is a number sequence in(, )such that
limn→∞tn= and
∞
n=tn=∞,{bn}is a number sequence such thatlim supn→∞bn≤,
and{cn}is a positive number sequence such that∞n=cn<∞.Thenlimn→∞an= .
T:C→C by Tx:=ax+ ( –a)Sx for each x∈C.Then,as a∈[κ, ),T is nonexpansive such that F(S) =F(T).
2 Convergence analysis
Theorem . Let A:C→H be anα-inverse-strongly monotone mapping and let B be a maximal monotone operator on H.Let S:C→C be a strictly pseudocontractive mapping with the constantκ∈[, )and let f :C→C be a contractive mapping with the constant β∈[, ).Assume thatDom(B)⊂C and F(S)∩(A+B)–()is not empty.Let J
rn= (I+rnB)–
and let{xn}be a sequence generated in the following process:x∈C and ⎧
⎪ ⎪ ⎨ ⎪ ⎪ ⎩
zn=αnf(xn) + ( –αn)xn,
yn=Jrn(zn–rnAzn+en),
xn+=βnxn+ ( –βn)(γnyn+ ( –γn)Syn), ∀n≥,
where{αn}, {βn}and{γn}are real number sequences in(, )and{rn}is a positive real
number sequence in(, α).Assume that the control sequences satisfy the following restric-tions:
(a) limn→∞αn= ,
∞
n=αn=∞;
(b) <lim infn→∞βn≤lim supn→∞βn< ;
(c) κ≤γn≤a< ,limn→∞|γn+–γn|= ;
(d) <b≤rn≤c< αand
∞
n=|rn–rn–|<∞; (e) ∞n=en<∞,
where a,b and c are three real numbers.Then{xn}converges strongly to a pointx¯∈F(S)∩
(A+B)–(),wherex¯=Proj
F(S)∩(A+B)–()f(¯x).
Proof First, we show that{xn}is bounded. Notice thatI–rnAis nonexpansive. Indeed, we
have
(I–rnA)x– (I–rnA)y
=x–y– rnx–y,Ax–Ay+rnAx–Ay
≤ x–y–rn(α–rn)Ax–Ay.
In view of the restriction (d), we find thatI–rnAis nonexpansive. Fixingp∈F(S)∩(A+
B)–(), we find that
zn–p ≤αn f(xn) –p + ( –αn)xn–p
≤ –αn( –β)
xn–p+αn f(p) –p .
PuttingTnx:=γnx+ ( –γn)Sxfor eachx∈C, we see from Lemma . thatTnis
nonex-pansive withF(Tn) =F(S) for eachn≥. It follows that
xn+–p ≤βnxn–p+ ( –βn) TnJrn(zn–rnAzn+en) –p
≤βnxn–p+ ( –βn)zn–p+ ( –βn)en
≤βnxn–p+
–αn( –β)
+αn( –βn) f(p) –p +en
≤ –αn( –β)( –βn)
xn–p+αn( –βn) f(p) –p +en
≤max
xn–p,
f(p) –p –β
+en
≤max
xn––p,
f(p) –p –β
+en–+en
.. .
≤max
x–p,f
(p) –p –β
+ n i= ei ≤max
x–p,f
(p) –p –β
+ ∞
i=
ei<∞.
This proves that the sequence{xn}is bounded, so are{yn}and{zn}. Notice that
zn–zn– ≤
–αn( –β)
xn–xn–+|αn–αn–| f(xn–) –xn– .
Puttingρn=zn–rnAzn+en, we find that
ρn–ρn– ≤ zn–zn–+rn–rn–Azn–+en+en–
≤ –αn( –β)
xn–xn–+|αn–αn–| f(xn–) –xn–
+|rn–rn–|Azn–+en+en–.
It follows from Lemma . that
yn–yn–= Jrn–
rn–
rn
ρn+
–rn– rn
Jrnρn
–Jrn–ρn–
≤ rn–
rn
(ρn–ρn–) +
–rn– rn
(Jrnρn–ρn–)
≤ (ρn–ρn–) +
–rn– rn
(Jrnρn–ρn)
≤ ρn–ρn–+|rn
–rn–|
b Jrnρn–ρn
≤ –αn( –β)
xn–xn–+fn
≤ xn–xn–+fn,
where
fn=|αn–αn–| f(xn–) –xn– +|rn–rn–|
Azn–+
Jrnρn–ρn
b
+en+en–.
This implies that
Tnyn–Tn–yn– ≤ Tnyn–Tnyn–+Tnyn––Tn–yn–
In view of the restrictions (a), (c), (d), and (e), we find that
lim sup
n→∞
Tnyn–Tn–yn––xn–xn–
≤.
It follows from Lemma . that
lim
n→∞Tnyn–xn= . (.)
This in turn implies that
lim
n→∞xn+–xn= . (.)
Notice that
xn+–p≤βnxn–p+ ( –βn)TnJrnρn–p
≤βnxn–p+ ( –βn) Jrn(zn–rnAzn+en) –p
≤βnxn–p+ ( –βn) (zn–rnAzn) – (I–rnA)p
+en
en+ (zn–rnAzn) – (I–rnA)p
≤βnxn–p+ ( –βn)zn–p–rn( –βn)(α–rn)Azn–Ap
+en
en+ (zn–rnAzn) – (I–rnA)p
≤βnxn–p+αn( –βn) f(xn) –p
+ ( –βn)( –αn)xn–p
–rn( –βn)(α–rn)Azn–Ap+gn,
wheregn=en(en+ (zn–rnAzn) – (I–rnA)p). It follows that
rn( –βn)(α–rn)Azn–Ap
≤ xn–p–xn+–p+αn( –βn) f(xn) –p
+gn
≤xn–p+xn+–p
xn+–xn+αn f(xn) –p
+gn.
In view of the restrictions (a), (b), (c), (d), and (e), we find from (.) that
lim
n→∞Azn–Ap= . (.)
SinceJrnis firmly nonexpansive, we find that
Jrnρn–p
≤J
rnρn–p, (zn–rnAzn+en) – (p–rnAp)
=
Jrnρn–p+ (zn–rnAzn+en) – (p–rnAp)
– (Jrnρn–p) –
(zn–rnAzn+en) – (p–rnAp)
≤
Jrnρn–p
+ (I–r
nA)zn– (I–rnA)p
–Jrnρn–zn–en+rnAzn–rnAp
≤
Jrnρn–p+zn–p+gn–Jrnρn–zn–en
–rnAzn–rnAp+ rnAzn–ApJrnρn–zn–en
.
It follows that
Jrnρn–p≤ zn–p+gn–Jrnρn–zn–en
–rnAzn–rnAp+ enJrnρn–zn+rnAzn–rnAp
≤αn f(xn) –p + ( –αn)xn–p+gn–Jrnρn–zn–en
+ rnAzn–ApJrnρn–zn–en.
This implies that
xn+–p≤βnxn–p+ ( –βn)TnJrnρn–p
≤βnxn–p+ ( –βn)Jrnρn–p
≤ xn–p+αn f(xn) –p
+gn– ( –βn)Jrnρn–zn–en
+ rnAzn–ApJrnρn–zn–en.
It follows that
( –βn)Jrnρn–zn–en
≤x
n–p+xn+–p
xn–xn++αn f(xn) –p
+gn+ rnAzn–ApJrnρn–zn–en.
In view of the restrictions (a), (b), and (e), we find from (.) and (.) thatlimn→∞Jrnρn–
zn–en= . This in turn implies that
lim
n→∞Jrnρn–zn= . (.)
Notice that
yn–xn ≤ yn–zn+zn–xn
≤ yn–zn+αn f(xn) –xn .
It follows from (.) that
lim
n→∞yn–xn= . (.)
On the other hand, we have
Tnxn–xn ≤ γnxn+ ( –γn)Sxn
–γnyn+ ( –γn)Syn
+ γnyn+ ( –γn)Syn
–xn
SinceSis Lipschitz continuous, we find from (.) and (.) that
lim
n→∞Tnxn–xn= . (.)
Notice that
Syn–yn ≤ Syn–Tnyn+Tnyn–Tnxn+Tnxn–xn+xn–yn
≤γnyn–Syn+ yn–xn+Tnxn–xn.
That is,
( –γn)Syn–yn ≤yn–xn+Tnxn–xn.
In view of (.) and (.), we find from the restriction (c) that
lim
n→∞Syn–yn= . (.)
SinceProjF(S)∩(A+B)–()f is contractive, we see that there exists a unique fixed point, sayx.¯
Next, we show thatlim supn→∞f(x) –¯ x,¯ zn–x¯ ≤. To show it, we can choose a
subse-quence{zni}of{zn}such that
lim sup
n→∞
f(x) –¯ x,¯ zn–x¯
= lim
i→∞
f(x) –¯ x,¯ zni–x¯
.
Since{zni}is bounded, we can choose a subsequence{znij}of{zni}which converges weakly
to some pointx. We may assume, without loss of generality, thatzniconverges weakly tox.
In view of (.), we find thatynialso converges weakly tox. It follows from Lemma . that
x∈F(S).
Now, we are in a position to show thatx∈(A+B)–(). Notice thaty
n=Jrn(zn–rnAzn+
en). It follows that
zn–rnAzn+en∈(I+rnB)yn.
That is,
zn–yn
rn
–Azn+en∈Byn.
SinceBis monotone, we get, for any (μ,ν)∈B,
yn–μ,
zn–yn
rn
–Azn+en–ν
≥.
Replacingnbyniand lettingi→ ∞, we obtain from (.) that
This gives –Ax∈Bx, that is, ∈(A+B)(x). This proves thatx∈(A+B)–(). This complete
the proof thatx∈F(S)∩(A+B)–(). It follows that
lim sup
n→∞
f(¯x) –x,¯ zn–x¯
≤.
Finally, we show thatxn→ ¯x. Notice that
zn–x¯ ≤αn
f(xn) –x,¯ zn–x¯
+ ( –αn)xn–xz¯ n–x¯
≤αn f(xn) –f(¯x) zn–x¯ +αn
f(¯x) –x,¯ zn–x¯
+ ( –αn)xn–xz¯ n–x¯
≤ –αn( –β)
xn–x¯ +zn–x¯
+αn
f(¯x) –x,¯ zn–x¯
.
This implies that
zn–x¯≤
–αn( –β)
xn–x¯+ αn
f(x) –¯ x,¯ zn–x¯
.
It follows that
yn–x¯ ≤ (zn–rnAzn) – (¯x–rnA¯x) +en
≤ (zn–rnAzn) – (¯x–rnA¯x)
+en
+ en (zn–rnAzn) – (¯x–rnA¯x)
≤ zn–x¯ +hn
≤ –αn( –β)
xn–x¯+ αn
f(¯x) –x,¯ zn–x¯
+hn, (.)
wherehn=en(en+ (zn–rnAzn) – (I–rnA)¯x). It follows from (.) that
xn+–x¯ ≤βnxn–x¯ + ( –βn)Tnyn–x¯
≤βnxn–x¯ + ( –βn)yn–x¯
≤ –αn( –βn)( –β)
xn–x¯
+ αn( –βn)
f(¯x) –x,¯ zn–x¯
+hn.
In view of the restriction (a), (b), and (e), we find from Lemma . thatxn→ ¯x. This
com-pletes the proof.
IfSis nonexpansive andγn≡, then we have the following result immediately.
Corollary . Let A:C→H be anα-inverse-strongly monotone mapping and let B be a maximal monotone operator on H.Let S:C→C be a nonexpansive mapping and let f :C→C be a contractive mapping with the constantβ∈[, ).Assume thatDom(B)⊂C and F(S)∩(A+B)–()is not empty.Let J
in the following process:x∈C and ⎧
⎨ ⎩
yn=αnf(xn) + ( –αn)xn,
xn+=βnxn+ ( –βn)SJrn(yn–rnAyn+en), ∀n≥,
where{αn}and{βn}are real number sequences in(, )and{rn}is a positive real number
sequence in(, α).Assume that the control sequences satisfy the following restrictions:
(a) limn→∞αn= ,
∞
n=αn=∞;
(b) <lim infn→∞βn≤lim supn→∞βn< ;
(c) <b≤rn≤c< αand
∞
n=|rn–rn–|<∞; (d) ∞n=en<∞,
where b and c are two real numbers.Then{xn}converges strongly to a pointx¯∈F(S)∩(A+
B)–(),wherex¯=Proj
F(S)∩(A+B)–()f(x).¯
3 Applications
Many nonlinear problems arising in applied areas such as image recovery, signal process-ing, and machine learning are mathematically modeled as a nonlinear operator equation and this operator is decomposed as the sum of two nonlinear operators. The central prob-lem is to iteratively find a zero point of the sum of two monotone operators, that is,
∈(A+B)(x).
Many real word problems can be formulated as a problem of the above form. For instance, a stationary solution to the initial value problem of the evolution equation
⎧ ⎨ ⎩
∈Fu+∂∂ut, u=u()
can be recast as the inclusion problem when the governing maximal monotoneFis of the formF=A+B; for more details, see [] and the references therein.
First, we give the following result.
Theorem . Let A:C→H be anα-inverse-strongly monotone mapping and let B be a maximal monotone operator on H.Let f :C→C be a contractive mapping with the constantβ∈[, ).Assume thatDom(B)⊂C and(A+B)–()is not empty.Let J
rn= (I+
rnB)–and let{xn}be a sequence generated in the following process:x∈C and ⎧
⎨ ⎩
yn=αnf(xn) + ( –αn)xn,
xn+=βnxn+ ( –βn)Jrn(yn–rnAyn+en), ∀n≥,
where{αn}and{βn}are real number sequences in(, )and{rn}is a positive real number
sequence in(, α).Assume that the control sequences satisfy the following restrictions:
(a) limn→∞αn= ,
∞
n=αn=∞;
(b) <lim infn→∞βn≤lim supn→∞βn< ;
(c) <b≤rn≤c< αand
∞
where b and c are two real numbers.Then{xn}converges strongly to a pointx¯∈(A+B)–(),
wherex¯=Proj(A+B)–()f(¯x).
Proof PutS=I, the identity onH. The desired conclusion can be obtained immediately. LetHbe a Hilbert space andf :H→(–∞, +∞] a proper convex lower semicontinuous function. Then the subdifferential∂f off is defined as follows:
∂f(x) =y∈H:f(z)≥f(x) +z–x,y,z∈H, ∀x∈H.
From Rockafellar [], we know that ∂f is maximal monotone. It is easy to verify that ∈∂f(x) if and only iff(x) =miny∈Hf(y). LetICbe the indicator function ofC,i.e.,
IC(x) =
⎧ ⎨ ⎩
, x∈C,
+∞, x∈/C. (.)
SinceICis a proper lower semicontinuous convex function onH, we see that the
subdif-ferential∂ICofICis a maximal monotone operator. Theny= (I+λ∂IC)–x⇐⇒y=ProjCx,
∀x∈H,y∈C.
Theorem . Let A:C→H be anα-inverse-strongly monotone mapping.Let S:C→C be a strictly pseudocontractive mapping with the constantκ ∈[, )and let f:C→C be a contractive mapping with the constantβ∈[, ).Assume that F(S)∩VI(C,A)is not empty. Let{xn}be a sequence generated in the following process:x∈C and
⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
zn=αnf(xn) + ( –αn)xn,
yn=ProjC(zn–rnAzn+en),
xn+=βnxn+ ( –βn)(γnyn+ ( –γn)Syn), ∀n≥,
where{αn}, {βn}and{γn}are real number sequences in(, )and{rn}is a positive real
number sequence in(, α).Assume that the control sequences satisfy the following restric-tions:
(a) limn→∞αn= ,
∞
n=αn=∞;
(b) <lim infn→∞βn≤lim supn→∞βn< ;
(c) κ≤γn≤a< ,limn→∞|γn+–γn|= ;
(d) <b≤rn≤c< αand
∞
n=|rn–rn–|<∞; (e) ∞n=en<∞,
where a,b,and c are three real numbers.Then{xn}converges strongly to a pointx¯∈F(S)∩
(A+B)–(),wherex¯=Proj
F(S)∩(A+B)–()f(¯x).
Proof PutBx=∂IC. Next, we show thatVI(C,A) = (A+∂IC)–(). Notice that
x∈(A+∂IC)–() ⇐⇒ ∈Ax+∂ICx
⇐⇒ –Ax∈∂ICx
⇐⇒ Ax,y–x ≥ ⇐⇒ x∈VI(C,A).
IfS=I, the identity onH, then we find from Theorem . the following result immedi-ately.
Corollary . Let A:C→H be anα-inverse-strongly monotone mapping.Let S:C→C be a strictly pseudocontractive mapping with the constantκ∈[, )and let f :C→C be a contractive mapping with the constantβ∈[, ).Assume that F(S)∩VI(C,A)is not empty. Let{xn}be a sequence generated in the following process:x∈C and
⎧ ⎨ ⎩
yn=αnf(xn) + ( –αn)xn,
xn+=βnxn+ ( –βn)ProjC(yn–rnAyn+en), ∀n≥,
where{αn}and{βn}are real number sequences in(, )and{rn}is a positive real number
sequence in(, α).Assume that the control sequences satisfy the following restrictions:
(a) limn→∞αn= ,
∞
n=αn=∞;
(b) <lim infn→∞βn≤lim supn→∞βn< ;
(c) <b≤rn≤c< αand
∞
n=|rn–rn–|<∞; (d) ∞n=en<∞,
where b and c are three real numbers.Then{xn}converges strongly to a pointx¯∈VI(C,A),
wherex¯=ProjVI(C,A)f(¯x).
LetFbe a bifunction ofC×CintoR, whereRdenotes the set of real numbers. Recall the following equilibrium problem:
Findx∈Csuch thatF(x,y)≥, ∀y∈C. (.)
In this paper, we useEP(F) to denote the solution set of the equilibrium problem (.). To study the equilibrium problems (.), we may assume thatFsatisfies the following conditions:
(A) F(x,x) = for allx∈C;
(A) Fis monotone,i.e.,F(x,y) +F(y,x)≤for allx,y∈C; (A) for eachx,y,z∈C,
lim sup
t↓
Ftz+ ( –t)x,y≤F(x,y);
(A) for eachx∈C,y→F(x,y)is convex and weakly lower semicontinuous.
PuttingF(x,y) =Ax,y–xfor everyx,y∈C, we see that the equilibrium problem (.) is reduced to a variational inequality.
Lemma .[, ] Let C be a nonempty closed convex subset of H and let F:C×C→R be a 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);
(d) EP(F)is closed and convex.
Lemma .[] Let C be a nonempty closed convex subset of a real Hilbert space H,F a bifunction from C×C toRwhich satisfies(A)-(A)and AFa multivalued mapping of H
into itself defined by
AFx=
⎧ ⎨ ⎩
{z∈H:F(x,y)≥ y–x,z,∀y∈C}, x∈C,
∅, x∈/C.
(.)
Then AFis a maximal monotone operator with the domain D(AF)⊂C,EP(F) =A–F ()and
Trx= (I+rAF)–x, ∀x∈H,r> ,
where Tris defined as in(.).
Theorem . Let A:C→H be anα-inverse-strongly monotone mapping and let FBbe a
bifunction from C×C toRwhich satisfies(A)-(A).Let S:C→C be a strictly pseudocon-tractive mapping with the constantκ∈[, )and let f :C→C be a contractive mapping with the constantβ∈[, ).Assume that F(S)∩EP(F)is not empty.Let{xn}be a sequence
generated in the following process:x∈C and ⎧
⎪ ⎪ ⎨ ⎪ ⎪ ⎩
zn=αnf(xn) + ( –αn)xn,
yn=Trn(zn–rnAzn+en),
xn+=βnxn+ ( –βn)(γnyn+ ( –γn)Syn), ∀n≥,
where{αn},{βn},and{γn}are real number sequences in(, )and{rn}is a positive real
number sequence in(, α).Assume that the control sequences satisfy the following restric-tions:
(a) limn→∞αn= ,
∞
n=αn=∞;
(b) <lim infn→∞βn≤lim supn→∞βn< ;
(c) κ≤γn≤a< ,limn→∞|γn+–γn|= ;
(d) <b≤rn≤c< αand
∞
n=|rn–rn–|<∞; (e) ∞n=en<∞,
where a,b,and c are three real numbers.Then{xn}converges strongly to a pointx¯∈F(S)∩
EP(F),wherex¯=ProjF(S)∩EP(F)f(¯x).
Corollary . Let A:C→H be anα-inverse-strongly monotone mapping and let FBbe
a bifunction from C×C toRwhich satisfies(A)-(A).Let f :C→C be a contractive mapping with the constantβ∈[, ).Assume that EP(F)is not empty.Let{xn}be a sequence
generated in the following process:x∈C and ⎧
⎨ ⎩
yn=αnf(xn) + ( –αn)xn,
xn+=βnxn+ ( –βn)Trn(yn–rnAyn+en), ∀n≥,
where{αn}and{βn}are real number sequences in(, )and{rn}is a positive real number
sequence in(, α).Assume that the control sequences satisfy the following restrictions:
(a) limn→∞αn= ,
∞
n=αn=∞;
(b) <lim infn→∞βn≤lim supn→∞βn< ;
(c) <b≤rn≤c< αandn∞=|rn–rn–|<∞; (d) ∞n=en<∞,
where b and c are two real numbers.Then{xn} converges strongly to a pointx¯∈EP(F),
wherex¯=ProjEP(F)f(x).¯
Competing interests
The authors declare that there is no competing interests.
Authors’ contributions
The main idea of this paper was proposed by SYC. XQ and LW participate the research and performed some steps of the proof in this research. All authors read and approved the final manuscript.
Author details
1Department of Mathematics, Gyengsang National University, Jinju, 660-701, Korea.2Department of Mathematics,
Hangzhou Normal University, Hangzhou, 310036, China.3Department of Mathematics, Faculty of Science, King Abdulaziz University, Jeddah, Saudi Arabia.4College of Statistics and Mathematics, Yunnan University of Finance and Economics, Kunming, 650221, China.
Acknowledgements
The authors are grateful to the referees for useful suggestions which improved the contents of the paper.
Received: 5 January 2014 Accepted: 24 March 2014 Published:09 Apr 2014
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