Volume 2010, Article ID 895907,23pages doi:10.1155/2010/895907
Research Article
An Algorithm for Finding a Common Solution
for a System of Mixed Equilibrium Problem,
Quasivariational Inclusion Problem, and
Fixed Point Problem of Nonexpansive Semigroup
M. Liu, S. S. Chang, and P. Zuo
Department of Mathematics, Yibin University, Yibin, Sichuan 644007, China
Correspondence should be addressed to M. Liu,[email protected] S. S. Chang,[email protected]
Received 31 March 2010; Accepted 8 June 2010
Academic Editor: Vy Khoi Le
Copyrightq2010 M. Liu et al. 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 introduce a hybrid iterative scheme for finding a common element of the set of solutions for a system of mixed equilibrium problems, the set of common fixed point for nonexpansive semigroup, and the set of solutions of the quasi-variational inclusion problem with multivalued maximal monotone mappings and inverse-strongly monotone mappings in Hilbert space. Under suitable conditions, some strong convergence theorems are proved. Our results extend some recent results announced by some authors.
1. Introduction
Throughout this paper we assume thatHis a real Hilbert space, andCis a nonempty closed convex subset ofH.
In the sequel, we denote the set of fixed points ofSbyFS.
A bounded linear operatorA : H → His said to bestrongly positive, if there exists a constantγsuch that
Ax, x ≥γx2, ∀x∈H. 1.1
Let B : H → H be a single-valued nonlinear mapping and M : H → 2H
a multivalued mapping. The “so-called”quasi-variational inclusion problemsee, Chang1,2
is to find anu∈Hsuch that
A number of problems arising in structural analysis, mechanics, and economics can be studied in the framework of this kind of variational inclusionssee, e.g.,3.
The set of solutions of variational inclusion1.2is denoted by VIH, B, M.
Special Case
IfM∂δC, whereCis a nonempty closed convex subset ofH, andδC :H → 0,∞is the indicator function ofC, that is,
δC ⎧ ⎨ ⎩
0, x∈C,
∞, x /∈C, 1.3
then the variational inclusion problem1.2is equivalent to findu∈Csuch that
Bu, v−u ≥0, ∀v∈C. 1.4
This problem is calledHartman-Stampacchia variational inequality problemsee, e.g.,4. The set of solutions of1.4is denoted by VIC, B.
Recall that a mappingB : H → H is calledα-inverse strongly monotonesee5, if there exists anα >0 such that
Bx−By, x−y≥αBx−By2, ∀x, y∈H. 1.5
A multivalued mappingM:H → 2His calledmonotone, if for allx, y∈H,u∈Mx,
andv∈My, then it implies thatu−v, x−y ≥0. A multivalued mappingM:H → 2His
calledmaximal monotone, if it is monotone and if for anyx, u∈H×H
u−v, x−y≥0, ∀y, v ∈GraphM 1.6
the graph of mappingMimplies thatu∈Mx.
Proposition 1.1see5. LetB:H → Hbe anα-inverse strongly monotone mapping, then
aBis a1/α-Lipschitz continuous and monotone mapping;
bifλ is any constant in0,2α, then the mappingI −λBis nonexpansive, whereI is the identity mapping onH.
LetΘ :C×C → Rbe an equilibrium bifunction (i.e.,Θx, x 0, for allx ∈C), and let
ϕ:C → Rbe a real-valued function.
Recently, Ceng and Yao6introduced the followingmixed equilibrium problemMEP, that is, to findz∈Csuch that
The set of solutions of1.7is denoted by MEPΘ, ϕ, that is,
MEPΘ z∈C:Θz, y ϕy −ϕz≥0, ∀y∈C. 1.8
In particular, ifϕ 0, this problem reduces to theequilibrium problem, that is, to find
z∈Csuch that
EP :Θz, y ≥0, ∀y∈C. 1.9
Denote the set of solution of EP by EPΘ.
On the other hand, Li et al.7introduced two steps of iterative procedures for the approximation of common fixed point of a nonexpansive semigroup{Ts: 0≤s <∞}on a nonempty closed convex subsetCin a Hilbert space.
Very recently, Saeidi8 introduced a more general iterative algorithm for finding a common element of the set of solutions for a system of equilibrium problems and of the set of common fixed points for a finite family of nonexpansive mappings and a nonexpansive semigroup.
Recall that a family of mappings T {Ts : 0 ≤ s < ∞} : C → C is called a nonexpansive semigroup, if it satisfies the following conditions:
aTst TsTtfor alls, t≥0 andT0 I;
bTsx−Tsy ≤ x−y, for allx, y∈C.
cthe mappingT·xis continuous, for eachx∈C.
Motivated and inspired by Ceng and Yao6, Li et al.7, Saeidi8, and9–13, the purpose of this paper is to introduce a hybrid iterative scheme for finding a common element of the set of solutions for a system of mixed equilibrium problems, the set of common fixed point for a nonexpansive semigroup, and the set of solutions of the quasi-variational inclusion problem with multivalued maximal monotone mappings and inverse-strongly monotone mappings in Hilbert space. Under suitable conditions, some strong convergence theorems are proved. Our results extend the recent results in Zhang et al. 5, S. Takahashi and W. Takahashi14, Chang et al.15, Ceng and Yao6, Li et al.7and, Saeidi8.
2. Preliminaries
In the sequel, we use xn x and xn → x to denote the weak convergence and strong convergence of the sequence{xn}inH, respectively.
Definition 2.1. Let M : H → 2H be a multivalued maximal monotone mapping, then the
single-valued mappingJM,λ:H → Hdefined by
JM,λu IλM−1u, ∀u∈H 2.1
Proposition 2.2see5. aThe resolvent operatorJM,λassociated withMis single-valued and nonexpansive for allλ >0, that is,
JM,λx−JM,λ
y ≤x−y, ∀x, y∈H, ∀λ >0. 2.2
bThe resolvent operatorJM,λis 1-inverse-strongly monotone, that is,
JM,λx−JM,λy2≤
x−y, JM,λx−JM,λ
y , ∀x, y∈H. 2.3
Definition 2.3. A single-valued mappingP :H → His said to behemicontinuous, if for any
x, y∈H, the mappingt→Pxtyconverges weakly toP xast → 0. It is well known that every continuous mapping must be hemicontinuous.
Lemma 2.4see16. LetEbe a real Banach space,E∗the dual space ofE, T :E → 2E∗a maximal
monotone mapping, andP :E → E∗a hemicontinuous bounded monotone mapping withDP E, then the mappingSTP :E → 2E∗is a maximal monotone mapping.
For solving the equilibrium problem for bifunctionΘ:C×C → R,let us assume that
Θsatisfies the following conditions:
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 concave and upper semicontinuous.
H4for eachx∈C,y→Θx, yis convex.
A mapη:C×C → His called Lipschitz continuous, if there exists a constantL >0 such that
η
x, y ≤Lx−y, ∀x, y∈C. 2.4
A differentiable functionK:C → Ron a convex setCis called
iη-convex6if
Ky −Kx≥Kx, ηy, x , ∀x, y∈C, 2.5
whereKxis theFr´echetderivative ofKatx;
iiη-strongly convex6if there exists a constantμ >0 such that
Ky −Kx−Kx, ηy, x ≥μ
2
x−y2, ∀x, y∈C. 2.6
LetΘ:C×C → Rbe an equilibrium bifunction satisfying the conditionsH1–H4. Letrbe any given positive number. For a given pointx∈C, consider the followingauxiliary problem forMEPfor short, MEPx, rto findy∈Csuch that
Θy, z ϕz−ϕy 1 r
where η : C×C → H is a mapping, andKx is the Fr´echet derivative of a functional
K:C → Ratx. LetVrΘ:C → Cbe the mapping such that for eachx∈C,VrΘxis the set of solutions of MEPx, r, that is,
VrΘx
y∈C:Θy, z ϕz−ϕy
1
r
Ky −Kx, ηz, y ≥0, ∀z∈C
, ∀x∈C.
2.8
Then the following conclusion holds.
Proposition 2.5see6. Let Cbe a nonempty closed convex subset ofH, ϕ : C → R a lower semicontinuous and convex functional. LetΘ:C×C → Rbe an equilibrium bifunction satisfying conditions (H1)–(H4). Assume that
iη:C×C → His Lipschitz continuous with constantL >0such that
aηx, y ηy, x 0, for allx, y∈C,
bη·,·is affine in the first variable,
cfor each fixedy ∈C,x→ ηy, xis continuous from the weak topology to the weak topology;
iiK :C → Risη-strongly convex with constantμ >0, and its derivativeKis continuous from the weak topology to the strong topology;
iiifor each x ∈ C, there exists a bounded subsetDx ⊆ Cand zx ∈ Csuch that for any y∈C\Dx, one has
Θy, zx ϕzx−ϕ
y 1 r
Ky −Kx, ηzx, y <0. 2.9
Then the following hold:
iVrΘis single-valued;
iiVrΘis nonexpansive ifKis Lipschitz continuous with constantν >0such thatμ≥Lν;
iiiFVrΘ MEPΘ;
ivMEPΘis closed and convex.
Lemma 2.6see17. LetCbe a nonempty bounded closed convex subset ofH, and letI{Ts: 0≤s <∞}be a nonexpansive semigroup onC, then for anyh≥0
lim
t→ ∞supx∈C
1
t t
0
Tsxds−Th
1
t t
0
Lemma 2.7see7. LetCbe a nonempty bounded closed convex subset ofH, and letI{Ts: 0 ≤ s < ∞}be a nonexpansive semigroup onC. If{xn}is a sequence in Csuch thatxn zand
lim sups→ ∞lim supn→ ∞Tsxn−xn0, thenz∈FI.
3. The Main Results
In order to prove the main result, we first give the following lemma.
Lemma 3.1 see5. au ∈ H is a solution of variational inclusion 1.2 if and only ifu JM,λu−λBu, for allλ >0, that is,
VIH, B, M FJM,λI−λB, ∀λ >0. 3.1
bIfλ∈0,2α, then VIH, B, Mis a closed convex subset inH.
In the sequel, we assume thatH, C, M, A, B, f, T, F, ϕi, ηi, Ki i1,2, . . . , Nsatisfy the
following conditions:
1His a real Hilbert space,C⊂His a nonempty closed convex subset;
2A : H → His a strongly positive linear bounded operator with a coefficientγ >
0, f :H → His a contraction mapping with a contraction constanth0< h < 1, 0 < γ < γ/h,B : C → H is anα-inverse-strongly monotone mapping, andM :
H → 2His a multivalued maximal monotone mapping;
3T{Ts: 0≤s <∞}:C → Cis a nonexpansive semigroup;
4F {Θi : i 1,2, . . . , N} : C×C → Ris a finite family of bifunctions satisfying
conditionsH1–H4, andϕi : C → Ri 1,2, . . . , Nis a finite family of lower semicontinuous and convex functional;
5ηi :C×C → His a finite family of Lipschitz continuous mappings with constant Li>0i1,2, . . . , Nsuch that
aηix, y ηiy, x 0, for allx, y∈C,
bηi·,·is affine in the first variable,
cfor each fixedy ∈ C,x → ηiy, xis sequentially continuous from the weak topology to the weak topology;
6Ki : C → Ris a finite family ofηi-strongly convex with constantμi > 0, and its
In the sequel we always denote byFTthe set of fixed points of the nonexpansive semi-groupT, VIH, B, Mthe set of solutions to the variational inequality1.2, and MEPF the set of solutions to the followingauxiliary problem for a system of mixed equilibrium problems:
Θ1
yn1, x
φ1x−φ1
yn1
1
r1
Kyn1
−Kxn, η1
x, yn1
≥0, ∀x∈C,
Θ2
yn2, x
φ2x−φ2
yn2
1
r2
Kyn2
−Kyn1
, η2
x, yn2
≥0, ∀x∈C,
.. .
ΘN−1
ynN−1, x
φN−1x−φN−1
ynN−1
1
rN−1
KynN−1
−KynN−2
, ηN−1
x, ynN−1
≥0, ∀x∈C,
ΘN
yn, x φNx−φNyn 1 rN
Kyn −KynN−1
, ηNx, yn ≥0, ∀x∈C,
3.2
where
yn1 VrΘ11xn,
yni VrΘiiy
i−1
n VrΘiiV
Θi−1
ri−1y
i−2
n VrΘii· · ·V
Θ2
r2 y
1 n
VΘi ri · · ·V
Θ2
r2 V
Θ1
r1 xn, i2,3, . . . , N−1,
yn VrΘNN· · ·V
Θ2
r2 V
Θ1
r1 xn,
3.3
andVΘi
ri :C → C,i1,2, . . . , Nis the mapping defined by2.8.
In the sequel we denote byVlVΘl rl · · ·V
Θ2
r2 V
Θ1
r1 forl∈ {1,2, . . . , N}andV 0I.
Theorem 3.2. LetH, C, A, B, M, f, T, F, ϕi, ηi, Ki i1,2, . . . , Nbe the same as above. Letri i
1,2, . . . , N be a finite family of positive numbers, λ ∈ 0,2α,{αn},{βn} ⊂ 0,1, and {tn} ⊂
0,∞. IfG:FTMEPFVIH, B, M/∅and the following conditions are satisfied:
ifor each x ∈ C, there exists a bounded subsetDx ⊆ Cand zx ∈ Csuch that for any y∈C\Dx
Θi
y, zx ϕizx−ϕi
y 1 ri
Kiy −Kix, ηi
zx, y <0, 3.4
iilimn→ ∞αn 0,∞n1αn ∞, 0 < lim infn→ ∞βn ≤ lim supn→ ∞βn < 1, and
1for eachn≥1, there is a uniquexn∈Csuch that
xnαnγf
1
tn tn
0
Tsxnds
βnxn1−βn I−αnA 1 tn
tn
0
TsJM,λI−λB2VNxnds,
3.5
2the sequence{xn} converges strongly to some point x∗ ∈ G, provided thatVrΘii is
firmly nonexpansive;
3x∗is the unique solution of the following variational inequality
A−γf x∗, x∗−z≤0, ∀z∈ G. 3.6
Proof. We observe that from condition ii, we can assume, without loss of generality, that
αn≤1−βnA−1.
SinceAis a linear bounded self-adjoint operator onH, then
Asup{|Au, u|:u∈H,u1}. 3.7
Since
1−βn I−αnA u, u
1−βn−αnAu, u
≥1−βn−αnA ≥0, 3.8
this implies that1−βnI−αnAis positive. Hence we have
1−βn I−αnAsup1−βn I−αnA u, u:u∈H,u1
sup1−βn−αnAu, u:u∈H,u1
≤1−βn−αnγ <1.
3.9
For each givenn≥1, let us define the mapping
Wn:αnγf
1
tn tn
0
TsdsβnI
1−βn I−αnA
1
tn tn
0
TsJM,λI−λB2VNds.
Firstly we show that the mappingWn:C → Cis a contraction. Indeed, for anyx, y∈C, we have
Wnx−Wny
αnγf 1 tn tn 0
Tsxds
βnx1−βn I−αnA 1 tn
tn
0
TsJM,λI−λB2VNxds
−αnγf1
tn tn
0
Tsyds−βny−1−βn I−αnA 1 tn
tn
0
TsJM,λI−λB2VNyds ≤αnγ f 1 tn tn 0
Tsxds −f 1 tn tn 0
Tsyds
βnx−y
1−βn−αnγ
1
tn tn
0
TsJM,λI−λB2VNx−TsJM,λI−λB2VNyds
≤αnγhx−yβnx−y1−βn−αnγ x−y
1−αn
γ−γh x−y.
3.11
This implies thatWn:C → Cis a contraction mapping. Letxn∈Cbe the unique fixed point
ofWn. Thus,
xnαnγf
1
tn tn
0
Tsxnds
βnxn
1−βn I−αnA
1
tn tn
0
TsJM,λI−λB2VNxnds
3.12
is well defined.
LettingynVNxn,ξnJM,λI−λByn, andρnJM,λI−λBξn, then
xnαnγf
1
tn tn
0
Tsxnds
βnxn
1−βn I−αnA
1
tn tn
0
Tsρnds. 3.13
We divide the proof ofTheorem 3.2into 8 steps.
Step 1. First prove that the sequences{xn},{ρn},{ξn}, and{yn}are bounded.
aPickp∈G, sinceynVNxnandpVNp, we have
bSincep∈VIH, B, MandρnJM,λI−λBξn, we havepJM,λI−λBp, and so
ρn−pJM,λI−λBξn−JM,λI−λBp
≤I−λBξn−I−λBp≤ξn−p
JM,λI−λByn−JM,λI−λBp
≤yn−p≤xn−p.
3.15
Lettingun 1/tntn
0 Tsxnds, qn 1/tn
tn
0 Tsρnds, we have
un−p
1
tn tn
0
Tsxnds−p
≤ 1
tn tn
0
Tsxn−Tspds
≤xn−p.
3.16
Similarly, we have
qn−p≤ρn−p. 3.17
Form3.5,3.9,3.14,3.15,3.16, and3.17we have
xn−p
αnγfun βnxn1−βn I−αnA qn−p
αnγfun−fp βnxn−p 1−βn I−αnA qn−p αnγfp −Ap
≤αnγhun−pβnxn−p
1−βn −αnγ qn−pαnγf
p −Ap
≤αnγhxn−pβnxn−p1−βn −αnγ xn−pαnγfp −Ap.
3.18
So,xn−p ≤1/γ−γhγfp−Ap. This implies that{xn}is a bounded sequence inH. Therefore{yn},{ρn},{ξn},{γfun}, and{qn}are all bounded.
Step 2. Next we prove that
xn−Tsxn −→0, n−→ ∞. 3.19
Sincexnαnγfun βnxn 1−βnI−αnAqn, then
Hence
xn−qn≤ αn
1−βn
γfun−Aqn. 3.21
From conditionii, we have
xn−qn−→0. 3.22
LetK{w∈C:w−p ≤1/γ−γhγfp−Ap}, thenKis a nonempty bounded closed convex subset ofCandTs-invariant. Since{xn} ⊂KandKis bounded, there existsr >0 such thatK⊂Br; it follows fromLemma 2.6that
lim
n→ ∞qn−Tsqn−→0. 3.23
From3.22and3.23, we have
xn−Tsxnxn−qnqn−TsqnTsqn−Tsxn
≤xn−qnqn−TsqnTsqn−Tsxn
≤xn−qnqn−Tsqnqn−xn−→0.
3.24
Step 3. Next we prove that
i lim
n→ ∞
Vl1x
n− Vlxn0, ∀l∈ {0,1, . . . , N−1};
iiespecially, lim
n→ ∞
VNx
n−xnnlim→ ∞yn−xn0.
3.25
In fact, for any givenp ∈Gandl∈ {0,1, . . . , N−1}, sinceVΘl1
rl1 is firmly nonexpansive, we have
Vl1x
n−p
2
VΘl1
rl1 V
lx
n−VrΘll11p
2
≤VΘl1
rl1
Vlx n
−p,Vlxn−p
Vl1x
n−p,Vlxn−p
1
2
Vl1xn−p2Vlxn−p2−Vlxn− Vl1x
n
2
.
3.26
It follows that
Vl1x
n−p
2
≤xn−p2−Vlxn− Vl1xn
2
From3.5, we have
xn−p2αnγfun βnxn 1−βnI−αnAqn−p2
αnγfun−Ap βnxn−qn I−αnAqn−p2
≤I−αnAqn−p βnxn−qn22αn
γfun−Ap, xn−p
≤I−αnAqn−p βnxn−qn 22αnγfun−Ap, xn−p
≤1−αnγ ρn−pβnxn−qn22αn
γfun−Ap, xn−p
1−αnγ 2ρn−p2β2nxn−qn22
1−αnγ βnρn−p·xn−qn
2αnγfun−Ap·xn−p.
3.28
Since
ρn−p≤ξn−p≤VNxn−p≤Vl1xn−p, ∀l∈ {0,1, . . . , N−1}, 3.29
and this together with3.27and3.28, it yields
xn−p2
≤1−αnγ 2xn−p2−Vlxn− Vl1xn2
β2nxn−qn2
21−αnγ ·βnρn−p·xn−qn2αnγfun−Ap·xn−p
1−2αnγ
αnγ 2
xn−p2−
1−αnγ 2Vlxn− Vl1xn
2
βn2xn−qn2
21−αnγ βnρn−p·xn−qn2αnγfun−Ap·xn−p.
3.30
Simplifying it we have
1−αnγ 2Vlxn− Vl1xn
2
≤1αn
γ 2xn−p2−xn−p2
β2nxn−qn22
1−αnγ βnρn−p·xn−qn
2αnγfun−Ap·xn−p.
3.31
Sinceαn → 0 andxn−qn → 0, by conditionii, it yieldsVl1xn− Vlxn → 0.
Step 4. Now we prove that for any givenp∈G
lim
In fact, it follows from3.15that
ρn−p2≤
ξn−p2JM,λI−λByn−JM,λI−λBp2
≤I−λByn−I−λBp2
yn−p2−2λ
yn−p, Byn−Bp
λ2By
n−Bp2
≤yn−p2λλ−2αByn−Bp2
≤xn−p2λλ−2αByn−Bp2.
3.33
Substituting3.33into3.28, we obtain
xn−p2≤
1−αnγ 2xn−p2λλ−2αByn−Bp2βn2xn−qn2
21−αnγ βnρn−p·xn−qn2αnγfun−Ap·xn−p.
3.34
Simplifying it, we have
1−αnγ 2λ2α−λByn−Bp2
≤1αnγ 2xn−p2−xn−p2β2nxn−qn2
21−αnγ βnρn−p·xn−qn2αnγfun−Ap·xn−p
αnγ 2xn−p2β2nxn−qn2
21−αnγ βnρn−p·xn−qn2αnγfun−Ap·xn−p.
3.35
Sinceαn → 0,0<lim infn→ ∞βn≤lim supn→ ∞βn<1,xn−qn → 0, and{γfun−Ap},{xn}
are bounded, these imply thatByn−Bp → 0 n → ∞.
Step 5. Next we prove that
lim
n→ ∞yn−ρn0,
lim
n→ ∞xn−ρn0.
3.36
In fact, since
yn−ρn≤yn−ξnξn−ρn, 3.37
for the purpose, it is sufficient to prove
aFirst we prove thatyn−ξn → 0. In fact, since
ξn−p2
JM,λI−λByn−JM,λI−λBp2
≤yn−λByn−
p−λBp , ξn−p
1
2
yn−λByn−p−λBp 2ξn−p2−yn−λByn−p−λBp −ξn−p 2
≤ 1
2
yn−p2ξn−p2−yn−ξn−λByn−Bp2
≤ 1
2
yn−p2ξn−p2−yn−ξn22λyn−ξn, Byn−Bp−λ2Byn−Bp2,
3.39
we have
ξn−p2≤
yn−p2−yn−ξn22λyn−ξn, Byn−Bp−λ2Byn−Bp2. 3.40
Substituting3.40into3.28, it yields that
xn−p2≤
1−αnγ 2yn−p2−yn−ξn22λyn−ξn, Byn−Bp
−λ2Byn−Bp2β2nxn−qn2
21−αnγ βnρn−p·xn−qn2αnγfun−Ap·xn−p.
3.41
Simplifying it we have
1−αnγ 2yn−ξn2 ≤αnγ2xn−p221−αnγ2λyn−ξn, Byn−Bp
−1−αnγ 2λ2Byn−Bp2β2nxn−qn2
21−αnγ βnρn−p·xn−qn2αnγfun−Ap·xn−p.
3.42
Sinceαn → 0, 0<lim infn→ ∞βn≤lim supn→ ∞βn<1,xn−qn → 0,Byn−Bp → 0n →
∞, and{γfun−Ap},{xn},{ρn}are bounded, these imply thatyn−ξn → 0 n → ∞.
bNext we prove that
lim
n→ ∞ξn−ρn0. 3.43
In fact, sinceξn −ρn JM,λI −λByn −JM,λI −λBξn ≤ yn −ξn → 0, so
Step 6. Next we prove that there exists a subsequence{xnk}of{xn}such thatxnk x∗ ∈G,
andx∗is the unique solution of the variational inequality3.6.
a We first prove that x∗ ∈ FT. In fact, since {xn} is bounded, there exists a
subsequence {xnk} of {xn} such that {xnk} x∗. From Lemma 2.7and Step 2, we obtain x∗∈FT.
bNow we prove thatx∗∈ ∩N
l1MEPΘl, ϕl.
Sincexnk x∗ and noting Step 3, without loss of generality, we may assume that
Vlxn
k x∗, for alll∈ {0,1,2, . . . , N−1}. Hence for anyx∈Cand for anyl∈ {0,1,2, . . . , N−
1}, we have
Kl1Vl1xn
k −Kl1
Vlxn k rl1
, ηl1
x,Vl1xnk ≥ −Θl1
Vl1x
nk, x
−ϕl1xϕl1
Vl1x
nk
.
3.44
By the assumptions and by conditionH2we know that the functionϕi and the mapping x→ −Θl1x, yboth are convex and lower semicontinuous, hence they are weakly lower semicontinuous. These together withKl1Vl1xn
k−Kl1Vlxnk/rl1 → 0 andVl1xnk x∗, we have
0lim inf
k→ ∞
!
Kl1Vl1x
nk −Kl1
Vlx nk rl1
, ηl1
x,Vl1xnk "
≥lim inf
k→ ∞
−Θl1
Vl1xn k, x
−ϕl1x ϕl1
Vl1xn k
.
3.45
That is,
Θl1x∗, x ϕl1x−ϕl1x∗≥0 3.46
for allx∈Candl∈ {0,1, . . . , N−1}, hencex∗∈ ∩Nl1MEPΘl, ϕl.
cNow we prove thatx∗∈VIH, B, M.
In fact, sinceBisα-inverse-strongly monotone, it follows fromProposition 1.1thatBis a 1/α-Lipschitz continuous monotone mapping andDB HwhereDBis the domain of
B. It follows fromLemma 2.4thatMBis maximal monotone. Letν, g∈Graph MB, that is,g−Bν∈Mν. Sincexnk x∗and notingStep 3, without loss of generality, we may
assume thatVlxn
k x∗; in particular, we haveynk VNxnk x∗. Fromyn−ρn → 0, we
can prove thatρnk x∗. Again sinceρnk JM,λI−λBξnk, we have
ξnk−λBξnk ∈IλMρnk, that is,
1
λ
ξnk−ρnk−λBξnk ∈M
ρnk . 3.47
By virtue of the maximal monotonicity ofM, we have
#
ν−ρnk, g−Bν−
1
λ
ξnk −ρnk−λBξnk $
So,
ν−ρnk, g
≥
#
ν−ρnk, Bν
1
λ
ξnk −ρnk−λBξnk $
#
ν−ρnk, Bν−BρnkBρnk−Bξnk
1
λ
ξnk−ρnk $
≥0ν−ρnk, Bρnk−Bξnk
# ν−ρnk,
1
λ
ξnk−ρnk
$ .
3.49
Sinceξn−ρn → 0,Bξn−Bρn → 0, andρnk x∗, we have
lim
nk→ ∞
ν−ρnk, g
ν−x∗, g≥0. 3.50
SinceMBis maximal monotone, this implies thatθ∈MBx∗, that is,x∗∈VIH, B, M, and sox∗∈ G.
dNow we prove thatx∗is the unique solution of variational inequality3.6.
10We first prove that{x
nk} → x∗.
Since for allz∈G,
xn−z2 xn−z, xn−z
αnγfun βnxn1−βn I−αnA qn−z, xn−z
αnγfun−Az βnxn−z 1−βn I−αnA qn−z , xn−z
≤αn
γfun−Az, xn−z
βnxn−z2
1−βn−αnγ qn−z· xn−z
≤1−αnγ xn−z2αn
γfun−Az, xn−z
.
3.51
It follows that
xn−z2≤ 1
γ
γfun−Az, xn−z
1
γ
γfun−γfz γfz−Az, xn−z
≤ 1
γ
γhxn−z2γfz−Az, xn−z.
Therefore,
xn−z2≤ 1
γ−γh
γfz−Az, xn−z. 3.53
Now, replacingnin3.53withnkand lettingk → ∞andxnk x∗, we havexnk → x∗.
20Next we prove thatx∗is the unique solution of the variational inequality3.6. Since
xnαnγf
1
tn tn
0
Tsxnds
βnxn1−βn I−αnA 1 tn
tn
0
Tsρnds, 3.54
we have
αn
A−γf
1
tn tn
0
Tsxnds
−
!
1−βn
xn−
1
tn tn
0
Tsρnds "
αnA
1
tn tn
0
Tsxn−Tsρn ds
−1−βn
I− 1 tn
tn
0
TsJM,λI−λB2VNds
xnαnA
1
tn tn
0
Tsxn−Tsρn ds.
3.55
Hence for anyz∈Gwe have,
αn
A−γf
1
tn tn
0
Tsxnds
, xn−z
−1−βn
I− 1 tn
tn
0
TsJM,λI−λB2VNds
xn
−
I− 1
tn tn
0
TsJM,λI−λB2VNds
z, xn−z
αn A1 tn tn 0
Tsxn−Tsρn ds, xn−z ,
then
A−γf
1
tn tn
0
Tsxnds
, xn−z
−1−βn
αn
×
I− 1
tn tn
0
TsJM,λ2 I−λBVNds
xn
−
I− 1
tn tn
0
TsJM,λ2 I−λBVNds
z, xn−z
A1 tn tn 0
Tsxn−Tsρn ds, xn−z .
3.57
It is easily seen thatI−1/tntn
0 TsJM,λI−λB
2VNdsis monotone. Thus from3.57we
have that
A−γf
1
tn tn
0
Tsxnds
, xn−z ≤ A1 tn tn 0
Tsxn−Tsρn ds, xn−z . 3.58
Now, in3.58replacingnbynkand lettingk → ∞andxnk → x∗, from3.36, we have
1
tnk tnk
0
Tsxnk−Tsρnk ds−→0. 3.59
So, we have
A−γf x∗, x∗−z≤0 ∀z∈ G. 3.60
It follows from18, Theorem 3.2that the solution of the variational inequality3.6is unique, that is,x∗is a unique solution of3.6.
Step 7. Next we prove that
lim sup
n→ ∞
γfx∗−Ax∗, xn−x∗≤0. 3.61
aFirst, we prove that
lim sup
n→ ∞
1
tn tn
0
Indeed, there exists a subsequence{ρni}of{ρn}such that
lim sup
n→ ∞
1
tn tn
0
Tsρnds−x∗, γfx∗−Ax∗ lim
i→ ∞
1
tni tni
0
Tsρnids−x
∗, γfx∗−Ax∗ .
3.63
We may also assume thatρni w. This together with 3.22and 3.36 shows that qni
1/tni tni
0 Tsρnids w. Sincexn−qn → 0, we havexni w. Again by the same method
as given inStep 6we can prove thatw∈ G. So, we have
lim sup
n→ ∞
1
tn tn
0
Tsρnds−x∗, γfx∗−Ax∗
lim
i→ ∞
1
tni tni
0
Tsρnids−x∗, γfx∗−Ax∗
lim
i→ ∞
qni−x∗, γfx∗−Ax∗
w−x∗, γfx∗−Ax∗≤0.
3.64
bNow we prove that
lim sup
n→ ∞
γfx∗−Ax∗, xn−x∗
≤0. 3.65
Fromxn−qn → 0 anda, we have
lim sup
n→ ∞
γfx∗−Ax∗, xn−x∗
lim sup
n→ ∞
γfx∗−Ax∗, xn−qnqn−x∗
≤lim sup
n→ ∞
γfx∗−Ax∗, xn−qn
lim sup
n→ ∞
γfx∗−Ax∗, qn−x∗
≤0.
3.66
Step 8. Finally we prove that
Indeed, from3.5,3.15, and3.17, we have
xn−x∗2
αnγfun−Ax∗ βnxn−x∗ 1−βnI−αnAqn−x∗2
≤βnxn−x∗ 1−βnI−αnAqn−x∗22αn
γfun−Ax∗, xn−x∗
≤1−βn I−αnA qn−x∗ βnxn−x∗22αnγfun−fx∗, xn−x∗
2αn
γfx∗−Ax∗, xn−x∗
≤1−βn−αnγ ρn−x∗βnxn−x∗22αnγhxn−x∗2
2αnγfx∗−Ax∗, xn−x∗
1−αnγ 22αnγh
xn−x∗22αn
γfx∗−Ax∗, xn−x∗
.
3.68
This implies that
xn−x∗2 ≤ 2
2γ−γh −γ2
γfx∗−Ax∗, xn−x∗. 3.69
Combining3.61and3.69, we obtain thatxn → x∗.
This completes the proof ofTheorem 3.2.
Corollary 3.3. LetH, C, f, T, F, A, B, ϕi, ηi, Ki i1,2, . . . , Nbe the same as inTheorem 3.2. Let ri i 1,2, . . . , N be a finite family of positive parameter,λ ∈ 0,2α,{αn},{βn} ⊂ 0,1and
{tn} ⊂0,∞. IfG:FTMEPFVIH, B, M/∅and conditions (i) and (ii) inTheorem 3.2
are satisfied, then
1for each n≥1there is a uniquexn∈Csuch that
xnαnγf
1
tn tn
0
Tsxnds
βnxn
1−βn I−αnA 1 tn
tn
0
TsPCI−λB2VNxnd;
3.70
2the sequence{xn}converges strongly to some pointx∗ ∈ G, provided thatVΘi
ri is firmly
nonexpansive;
Proof. TakingM ∂δC :H → 2H inTheorem 3.2, whereδC :H → 0,∞is the indicator
function ofC, that is,
δC ⎧ ⎨ ⎩
0, x∈C,
∞, x /∈C, 3.71
then the variational inclusion problem1.2is equivalent to variational inequality1.4, that is, to findu∈Csuch that
Bu, v−u ≥0, ∀v∈C. 3.72
Again, sinceM∂δC, thenJM,λPC. Therefore we have
ρnPCI−λBξn, ξnPCI−λByn. 3.73
The conclusion ofCorollary 3.3can be obtained fromTheorem 3.2immediately.
4. Applications to Optimization Problem
Let H be a real Hilbert space, C a nonempty closed convex subset of H, A : H → H
a strongly positive linear bounded operator with a constant γ > 0, and T : C → C a nonexpansive mapping. In this section we will utilize the results presented inSection 3 to study the followingoptimization problem:
min
x∈FT
1
2Ax, x −hx, 4.1
whereFTis the set of fixed points ofT inCandh:C → Ris a potential function forγf
i.e.,hx γfx,x ∈ C, wheref : C → Cis a contractive mapping with a contractive constanth∈0,1. We have the following theorem.
Theorem 4.1. LetH, C, f, T, Abe the same as above. Let{αn},{βn}be sequences in0,1satisfying
condition (ii) inTheorem 3.2. IfFTis a nonempty compact subset ofC, then for eachn≥1there is a uniquexn∈Csuch that
xnαnγfTxn βnxn
1−βn I−αnA Txn, ∀n≥1, 4.2
and the sequence{xn}converges strongly to some pointx∗∈FTwhich is the unique minimal point of optimization problem4.1.
Proof. TakingΘi0,ϕi0,Ki0, ηi0, ri1i1,2, . . . , N,B0,TTinCorollary 3.3,
hence we haveF 0,VΘi
ri I,i 1,2, . . . , N, yn ξn ρn xn,1/tn tn
we know that the sequence{xn}defined by4.2converges strongly to some pointx∗∈FT
which is the unique solution of the following variational inequality:
A−γf x∗, x−x∗≥0, x∈FT. 4.3
SinceT is nonexpansive, thenFTis convex. Again by the assumption thatFTis compact, therefore it is a compact and convex subset ofC, and1/2Ax, x −hx : C → R is a continuous mapping. By virtue of the well-known Weierstrass theorem, there exists a point
y∗∈FTwhich is a minimal point of optimization problem4.1. As is known to all,4.3is the optimality necessary condition19for the optimization problem4.1. Therefore we also have
A−γf y∗, x−y∗≥0, ∀x∈FT. 4.4
Sincex∗is the unique solution of4.3, we havex∗y∗. This completes the proof ofTheorem 4.1.
Acknowledgment
The authors would like to express their thanks to the referees for their valuable suggestions and comments.
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