# A Viscosity Approximation Scheme for Finding Common Solutions of Mixed Equilibrium Problems, a Finite Family of Variational Inclusions, and Fixed Point Problems in Hilbert Spaces

Authors: Bin-Chao Deng; Tong Chen; Baogui Xin
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

We introduce an iterative method for finding a common
element of set of fixed points of nonexpansive mappings, the set of
solutions of a finite family of variational inclusion with
set-valued maximal monotone mappings and inverse strongly monotone
mappings, and the set of solutions of a mixed equilibrium problem in
Hilbert spaces. Under suitable conditions, some strong convergence
theorems for approximating this common elements are proved. The
results presented in the paper improve and extend the main results
of Plubtemg and Sripard and many others.


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## Body

## 1. Introduction

LetH be a real Hilbert space whose inner product and norm are denoted by 〈·,·〉 and ∥·∥, respectively. Let C be a nonempty closed convex subset of H, and let F be a bifunction of C×C into ℝ which is the set of real numbers. The equilibrium problem for F:C×C→ℝ is to find x∈C such that
(1.1)F(x,y)≥0,∀y∈C.
The set of solutions of (1.1) is denoted by EP(F). The mixed equilibrium problem for two bifunction of F1,F2:C×C→ℝ is to find x∈C such that
(1.2)F1(x,y)+F2(x,y)+〈Ax,x-y〉≥0,∀y∈C.
In the sequel we will indicate by MEP(F1,F2,A) the set of solution of our mixed equilibrium problem. If A=0 we denote MEP(F1,F2,0) with MEP(F1,F2).In 2005, Combettes and Hirstoaga [1] introduced an iterative scheme of finding the best approximation to the initial data when EP(F) is nonempty and proved a strong convergence theorem. Let A:C→H be a nonlinear mapping. The classical variational inequality which is denoted by VI(A,C) is to find x∈C such that
(1.3)〈Ax,y-x〉≥0,∀y∈C.
The variational inequality has been extensively studied in the literature; see, for example, [2, 3] and the reference therein. Recall that mapping T of C into itself is called nonexpansive if
(1.4)‖Tx-Ty‖≤‖x-y‖,∀x,y∈C.
A mapping f:C→C is called contractive if there exists a constant β∈(0,1) such that(1.5)‖fx-fy‖≤β‖x-y‖,∀x,y∈C.
We denote by Fix(T) the set of fixed points of T.Some methods have been proposed to solve the equilibrium problem and fixed point problem of nonexpansive mapping; see, for instance, [2, 4–6] and the references therein. In 2007, Plubtieng and Punpaeng [6] introduced the following iterative scheme. Let x1∈H and let {xn} and {un} be sequences generated by
(1.6)F(un,y)+1rn〈y-un,un-xn〉≥0,∀y∈H,xn+1=αnγf(xn)+(I-αnA)Tun,∀n∈N.
They proved that if the sequences {αn} and {rn} of parameters satisfy appropriate conditions, then the sequences {xn} and {un} both converge strongly to the unique solution of the variational inequality
(1.7)〈(A-γf)z,z-x〉≥0,∀x∈Fix(T)∩EP⁡(F),
which is the optimality condition for the minimization problem
(1.8)min⁡x∈Fix(T)∩EP⁡(F)12〈Ax,x〉-h(x),
where h is a potential function for γf.LetA:H→H be a single-valued nonlinear mapping, and let M:H→2H be a set-valued mapping. We consider the following variational inclusion, which is to find a point u∈H such that
(1.9)θ∈A(u)+M(u),
where θ is the zero vector in H. The set of solutions of problem (1.9) is denoted by I(A,M). Let Ai:H→H,i=1,2,…N be single-valued nonlinear mappings, and let Mi:H→2H,i=1,2,…N, be set-valued mappings. If A≡0, then problem (1.9) becomes the variational inclusion problem introduced by Rockafellar [7]. If M=∂δC, where C is a nonempty closed convex subset of H and δC:H→[0,∞] is the indicator function of C, that is,
(1.10)δC(x)={0,x∈C,+∞,x∉C,
then the variational inclusion problem (1.9) is equivalent to variational inequality problem (1.3). It is known that (1.9) provides a convenient framework for the unified study of optimal solutions in many optimization-related areas including mathematical programming, complementarity, variational inequalities, optimal control, mathematical economics, equilibria, and game theory. Also various types of variational inclusions problems have been extended and generalized (see [8] and the references therein). We introduce following a finite family of variational inclusions, which are to find a point u∈H such that
(1.11)θ∈Ai(u)+Mi(u),i=1,2,…N,
where θ is the zero vector in H. The set of solutions of problem (1.11) is denoted by ⋂i=1NI(Ai,Mi). The formulation (1.11) extends this formalism to a finite family of variational inclusions covering, in particular, various forms of feasibility problems (see, e.g., [9]).In 2009, Plubtemg and Sripard [10] introduced the following iterative scheme for finding a common element of set of solutions to the problem (1.9) with multivalued maximal monotone mapping and inverse-strongly monotone mapping, the set solutions of an equilibrium problem and the set of fixed points of a nonexpansive mapping in Hilbert spaces. Starting with an arbitrary x1∈H, define sequence {xn}, {yn}, and {un} by
(1.12)F(un,y)+1rn〈y-un,un-xn〉≥0,∀y∈H,yn=JM,λ(un-λAun),∀n>0,xn+1=αnγf(xn)+(I-αnB)Snyn,
for all n∈N, where λ∈(0,2α], {αn}⊂[0,1], and {rn}⊂(0,∞); B is a strongly positive bounded linear operator on H, and {Sn} is a sequence of nonexpansive mappings on H. They proved that under certain appropriate conditions imposed on {αn} and {rn}, the sequence {xn}, {yn}, and {un} generated by (1.12) converge strongly to z∈⋂i=1∞Fix(Si)∩I(A,M)∩EP(F), where z=P⋂i=1∞Fix(Si)∩I(A,M)∩EP(F)f(z).In 2011, Yao et al. [11] considered the following iterative method for finding a common element of set of solutions to the problem (1.9) with multi-valued maximal monotone mapping and inverse-strongly monotone mapping, the set solutions of a mixed equilibrium problem, and the set of fixed points of an infinite family of nonexpansive mappings in Hilbert spaces. Let F:H×H→ℝ be a bifunction, A be a strongly positive bounded linear operator, and B1,B2:H→H be inverse strongly monotone and let inverse strongly monotone, and φ:H→ℝ be a lower semicontinuous and convex function. For any initial x0 is selected in H arbitrarily
(1.13)F(un,y)+φ(y)-φ(un)+1r〈y-un,un-(xn-rB1xn)〉≥0,∀y∈H,xn+1=αnγf(xn)+βnxn+((1-βn)I-αnB)WnJR,λ(un-λB2un),n≥0,
where {αn} and {βn} are two sequences in [0,1] and Wn is an infinite family of nonexpansive mappings. They study the strong convergence of the iterative algorithm (1.13).Motivated and inspired by Saeidi [12], Aoyama et al. [13], Plubieng and Punpaeng [6], Plubtemg and Sripard [10], Peng et al. [14], and Yao et al. [11], we introduce an iterative scheme for finding a common element of the set of solutions of a finite family of variational inclusion problems (1.11) with multi-valued maximal monotone mappings and inverse-strongly monotone mappings, the set of solutions of a mixed equilibrium problem, and the set of fixed points of nonexpansive mappings in Hilbert space. Starting with an arbitrary x1∈H, define sequence {xn}, {yn} and {un} by
(1.14)F1(un,y)+F2(un,y)+1rn〈y-un,un-xn〉≥0,∀y∈H,yn=JMN,λN,n(I-λN,nAN)⋯JM1,λ1,n(I-λ1,nA1)un,xn+1=ϵnγf(xn)+βnxn+((1-βn)I-ϵnB)Snyn,
for all n∈N, where λi,n∈(0,2αi],i∈{1,2,…,N},{ϵn}⊂[0,1], and {rn}⊂(0,∞), B is a strongly positive bounded linear operator on H, and {Sn} is a sequence of nonexpansive mappings on H. Under suitable conditions, some strong convergence theorems for approximating to this common elements are proved. Our results extend and improve some corresponding results in [10, 11, 14] and the references therein.


## 2. Preliminaries

This section collects some lemmas which will be used in the proofs for the main results in next section.LetH be a real Hilbert space with inner product 〈·,·〉 and norm ∥·∥, respectively. It is well known that for all x,y∈H and λ∈[0,1], there holds
(2.1)‖λx+(1-λ)y‖2=λ‖x‖2+(1-λ)‖y‖2-λ(1-λ)‖x-y‖2.LetC be a nonempty closed convex subset of H. Then, for any x∈H, there exists a unique nearest point of C, denoted by PCx. such that ∥x-PCx∥≤∥x-y∥ for all y∈C. Such a PC is called the metric projection from H into C. We know that PC is nonexpansive. It is also known that PCx∈C and
(2.2)〈x-PCx,PCx-z〉≥0,∀x∈H,z∈C.
It is easy to see that (2.2) is equivalent to
(2.3)‖x-z‖2≥‖x-PCx‖2+‖PCx-z‖2,∀x∈H,z∈C.For solving the mixed equilibrium problem for a bifunctionF:C×C→ℝ, let us assume that F satisfies the following conditions:(A1)
F(x,x)=0 for all x∈C;(A2)
F is monotone, that is, F(x,y)+F(y,x)≤0 for all x,y∈C;(A3)
for eachx,y,z∈C,
(2.4)lim⁡t→0F(tz+(1-t)x,y)≤F(x,y);(A4)
for eachx∈C,y↦F(x,y) is convex and lower semicontinuous.Lemma 2.1 (see [15]).
LetC be a convex closed subset of a Hilbert spaces  H.
LetF1:C×C→R be a bifunction such that
(f1)
F1(x,x)=0 for all x∈C;(f2)
F1 is monotone and upper hemicontinuous in the first variable;(f3)
F1 is lower semicontinuous and convex in the second variable.Let F2:C×C→R be a bifunction such that(h1)
F2(x,x)=0 for all x∈C;(h2)
F2 is monotone and weakly upper semicontinuous in the first variable;(h3)
F2 is convex in the second variable. Moreover let us suppose that(H)
for fixedr>0 and x∈C there exists a bounded set K⊂C and a∈K such that for all z∈C∖K,-F1(a,z)+F2(z,a)+(1/r)〈a-z,z-x〉<0. For r>0 and x∈H, let Tr:H→C be a mapping defined by
(2.5)Tr(x)={y∈C:F1(z,y)+F1(z,y)+1r〈y-z,z-x〉≥0,∀z∈C}.
Called resolvent of F1 and F2.
Then,(1)
Trx≠∅;(2)
Tr is a single value;(3)
Tr is firmly nonexpansive;(4)
MEP(F1,F2)=Fix(Trx) and it is closed and convex.Recall that a mappingA:H→H is called α-inverse-strongly monotone, if there exists a positive number α such that
(2.6)〈Au-Av,u-v〉≥α‖Au-Av‖2,∀u,v∈H.LetI be the identity mapping on H. It is well known that if A:H→H is α-inverse-strongly monotone, then A is 1/α-Lipschitz continuous and monotone mapping. In addition, if 0<λ≤2α, then I-λA is a nonexpansive mapping.A set-valuedM:H→2H is called monotone, if for all x,y∈H,f∈Mx, and g∈My imply 〈x-y,f-g〉≥0. A monotone mapping M:H→2H is maximal if its graph G(M):{(x,f)∈H×H∣f∈M(x)} of M is not properly contained in the graph of any other monotone mapping. It is known that a monotone mapping M is maximal if and only if for (x,f)∈H×H,〈x-y,f-g〉≥0 for every (y,g)∈G(H) implies f∈Mx.Let the set-valuedM:H→2H be maximal monotone. we define the resolvent operator JM,λ associated with M and λ as follows:
(2.7)JM,λ(u)=(I+λM)-1(u),∀u∈H,
where λ is a positive number. It is worth mentioning that the resolvent operator JM,λ is single-valued, nonexpansive, and 1-inverse-strongly monotone, see for example [16] and that a solution of problem (1.9) is a fixed point of the operator JM,λ(I-λA) for all λ>0, see for instance, [17]. Furthermore, a solution of a finite family of variational inclusion problems (1.11) is a common fixed point of JMk,λ(I-λAk), k∈{1,…,N}, λ>0.Lemma 2.2 (see [16]).
LetM:H→2H be a maximal monotone mapping and  A:H→H a Lipschitz-continuous mapping. Then the mapping  S=M+A:H→2H is a maximal monotone mapping.Lemma 2.3 (see [18]).
LetH be a Hilbert space,  C a nonempty closed subset of  H,  f:H→H a contraction with coefficient  0<α<1, and  B a strongly positive linear bounded operator with coefficient  γ̅>0. Then, (1)
if0<γ<γ̅/α, then 〈x-y,(B-γf)x-(B-γf)y〉≥(γ̅-γα)∥x-y∥2,x,y∈H;(2)
if0<ρ<∥B∥-1, then ∥I-ρB∥≤1-ργ̅.Lemma 2.4.
For allx,y∈H, there holds the inequality(2.8)‖x+y‖2≤‖x‖2+2〈y,x+y〉.Lemma 2.5 (the resolvent identity).
LetE be a Banach space, for  λ>0,μ>0 and  x∈E,
(2.9)Jλx=Jμ(μλx+(1-μλ)Jλx).Lemma 2.6.
LetH be a Hilbert space. Let  Ai:H→H,i=1,2,…,N be  αi-inverse-strongly monotone mappings,  Mi:H→2H,i=1,2,…,N maximal monotone mappings,  and {ωn} be a bounded sequence in  H. Assume  λj,n>0,j=1,2,…,N, satisfy (H1)
limn→∞∑n=1∞|λj,n-λj,n+1|<∞,(H2)
liminfn→∞λj,n>0.Set Θnk=JMk,λk,n(I-λk,nAk)⋯JM1,λ1,n(I-λ1,nA1) for k∈{1,2,…,N} and Θn0=I for all n. Then, for k∈{1,2,…,N},
(2.10)∑i=1∞‖Θi+1kωi-Θikωi‖<∞.Proof.
From Lemma2.5, we have, for all k∈{1,2,…,N},
(2.11)‖JMk,λk,n+1(I-λk,n+1Ak)ωn-JMk,λk,n(I-λk,nAk)ωn‖≤|1-λk,nλk,n+1|(‖JMk,λk,n+1(I-λk,n+1Ak)ωn‖+‖ωn‖).
Furthermore, from the definition of Θnk, it follows
(2.12)Θnk=JMk,λk,n(I-λk,nAk)Θnk-1.
Combining (2.11) and (2.12), we obtain
(2.13)‖Θn+1kωn-Θnkωn‖≤‖JMk,λk,n+1(I-λk,n+1Ak)Θn+1k-1ωn-JMk,λk,n(I-λk,nAk)Θnk-1ωn‖≤‖JMk,λk,n+1(I-λk,n+1Ak)Θn+1k-1ωn-JMk,λk,n+1(I-λk,n+1Ak)Θnk-1ωn‖+‖JMk,λk,n+1(I-λk,n+1Ak)Θnk-1ωn-JMk,λk,n(I-λk,nAk)Θnk-1ωn‖≤‖Θn+1k-1ωn-Θnk-1ωn‖+|1-λk,nλk,n+1|×(‖JMk,λk,n+1(I-λk,n+1Ak)Θnk-1ωn‖+‖ωn‖)≤‖Θn+1k-1ωn-Θnk-1ωn‖+|1-λk,nλk,n+1|M1≤⋯≤‖Θn+10ωn-Θn0ωn‖+∑l=1k|1-λl,nλl,n+1|M1=∑l=1k|1-λl,nλl,n+1|M1,
where M1=sup{∥ωn∥+∑k=1N∥JMk,λk,n+1(I-λk,n+1Ak)Θnk-1ωn∥}. According to (H1) and (H2), then (2.11) holds.Lemma 2.7 (see [19]).
Let{xn} and  {zn} be bounded sequences in a Banach space  X, and let  {βn} be a sequence in  [0,1] with  0<liminfn→∞βn and limsupn→∞βn<1. Suppose
(2.14)xn+1=βnxn+(1-βn)zn
for all n≥0 and
(2.15)limsup⁡n→∞(‖zn+1-zn‖-‖xn+1-xn‖)≤0.
Then limn→∞∥xn-zn∥=0.Lemma 2.8 (see [20]).
Assume{αn} is a sequence of nonnegative real numbers such that
(2.16)αn+1≤(1-γn)αn+δn,n≥0,
where {αn} is a sequence in (0,1) and {δn} is a sequence in ℝ such that(i)
∑n=1∞γn=∞,(ii)
limsupn→∞(δn/γn)≤0 or ∑n=1∞|δn|<∞.Then limn→∞αn=0.


## 3. Main Results

Theorem 3.1.
LetH be a real Hilbert space, let  F1 and  F2 be bifunction  H×H→ℝ satisfying (A1)–(A4), and let  {Sn} be a sequence of nonexpansive mappings on  H. Let  Ai:H→H,i=1,2,…,N, be  αi-inverse-strongly monotone mappings, and let Mi:H→2H,i=1,2,…,N be maximal monotone mappings such that  Ω:=(⋂n=1∞Fix(Sn))∩MEP(F1,F2)∩(⋂i=1NI(Ai,Mi))≠∅. Let  f be a contraction of  H into itself with a constant  α∈(0,1), and let  B be a strongly positive bounded linear operator on  H with coefficient  γ̅>0 and  0<γ<γ̅/α. Let  {xn},  {yn}, and  {un} be sequences generated by  x1∈H and
(3.1)F1(un,y)+F2(un,y)+1rn〈y-un,un-xn〉≥0,∀y∈H,yn=JMN,λN,n(I-λN,nAN)⋯JM1,λ1,n(I-λ1,nA1)un,xn+1=ϵnγf(xn)+βnxn+((1-βn)I-ϵnB)Snyn,
for all n∈ℕ, where βn∈(0,1), λi,n∈(0,2αi], i∈{1,2,…,N}, satisfy (H1)-(H2) and {ϵn}⊂[0,1] and {rn}⊂(0,∞) satisfy(C1)
limn→∞ϵn=0;(C2)
∑n=1∞ϵn=∞;(C3)
∑n=1∞|ϵn+1-ϵn|<∞;(C4)
liminfn→∞rn>0;(C5)
∑n=1∞|rn+1-rn|<∞;(C6)
limn→∞βn=0. Suppose that ∑n=1∞sup{∥Sn+1z-Snz∥:z∈K}<∞ for any bounded subset K of H. Let S be a mapping of H into itself defined by Sx=limn→∞Snx, for all x∈H and suppose that Fix(S)=⋂n=1∞Fix(Sn). Then, {xn}, {yn}, and {un} converge strongly to z, where z=PΩ(I-B+γf)(z) is a unique solution of the variational inequality
(3.2)〈(B-γf)z,z-x〉≤0x∈Ω.Proof.
Sinceϵn→0, we will assume that ϵn≤(1-βn)∥B∥-1 and 1-ϵn(γ̅-αγ)>0. Observe that, if ∥u∥=1, then
(3.3)〈((1-βn)I-ϵnB)u,u〉=(1-βn)-ϵn〈Bu,u〉≥(1-βn-ϵn‖B‖)≥0.
By Lemma 2.3 we have
(3.4)‖(1-βn)I-ϵnB‖≤(1-βn)-ϵnγ̅.
Moreover, using the definition of Θnk in Lemma 2.6, we have yn=ΘnNun. We divide the proof into several steps.Step 1.
The sequence{xn} is bounded.
Sinceϵn→0, we may assume that ϵn≤∥B∥-1 for all n. Let p∈Ω. Using the fact that JMk,λk,n(I-λk,nAk), k∈{1,2,…,N}, is nonexpansive and p=JMk,λk,n(I-λk,nAk)p, we have
(3.5)‖yn-p‖=‖ΘnNun-ΘnNp‖≤‖un-p‖≤‖Trxn-Trp‖≤‖xn-p‖,
for all n≥1. Then, we have
(3.6)‖xn+1-p‖=‖((1-βn)I-ϵnB)(SnΘnNun-SnΘnNp)+ϵn(f(xn)-f(p))+ϵn(γf(p)-Bp)+βn(xn-p)SnΘnNun‖≤(1-ϵn(γ̅-αγ))‖xn-p‖+ϵn(γ̅-αγ)‖γf(p)-Bp‖γ̅-αγ.
It follow from (3.6) and induction that
(3.7)‖xn-p‖≤max⁡{‖x1-p‖,‖γf(p)-Ap‖γ̅-αγ},∀n≥1.
Hence {xn} is bounded and therefore {un}, {yn}, {f(xn)}, and {Snyn} are also bounded.Step 2.
We show that∥xn+1-xn∥→0.
Definexn+1=βnxn+(1-βn)vn for each n≥0. From the definition of vn, we obtain
(3.8)vn+1-vn=11-βn+1(xn+2-βn+1xn+1)-11-βn(xn+1-βnxn)=ϵn+1γf(xn+1)+((1-βn+1)I-ϵn+1B)Sn+1yn+11-βn+1-ϵnγf(xn)+((1-βn)I-ϵnB)Snyn1-βn=ϵn+1γf(xn+1)1-βn+1-ϵnγf(xn)1-βn+Sn+1yn+1-Snyn+ϵn1-βnBSnyn-ϵn+11-βn+1BSn+1yn+1=ϵn+11-βn+1(γf(xn+1)-BSn+1yn+1)+ϵn1-βn(γf(xn)-BSnyn)+Sn+1yn+1-Sn+1yn+Sn+1yn-Snyn.
It follows that
(3.9)‖vn+1-vn‖-‖xn+1-xn‖≤ϵn+11-βn+1(‖γf(xn+1)‖+‖BSn+1yn+1‖)+ϵn1-βn(‖γf(xn)‖+‖BSnyn‖)+‖Sn+1yn+1-Sn+1yn‖+‖Sn+1yn-Snyn‖-‖xn+1-xn‖≤ϵn+11-βn+1(‖γf(xn+1)‖+‖BSn+1yn+1‖)+ϵn1-βn(‖γf(xn)‖+‖BSnyn‖)+‖yn+1-yn‖+‖Sn+1yn-Snyn‖-‖xn+1-xn‖.
By the suppose of ∑n=1∞sup{∥Sn+1z-Snz∥:z∈K}<∞, we obtain
(3.10)lim⁡n→∞‖Sn+1yn-Snyn‖=0.
From Lemma 2.6, we obtain
(3.11)lim⁡n→∞∥Θn+1Nun+1-ΘnNun+1∥=0.
By ΘnN and Tr being nonexpansive, we have
(3.12)‖yn+1-yn‖=‖Θn+1Nun+1-ΘnNun‖≤‖Θn+1Nun+1-ΘnNun+1‖+‖ΘnNun+1-ΘnNun‖≤‖Θn+1Nun+1-ΘnNun+1‖+‖un+1-un‖≤‖Θn+1Nun+1-ΘnNun+1‖+‖Trxn+1-Trxn‖≤‖Θn+1Nun+1-ΘnNun+1‖+‖xn+1-xn‖.
Substituting (3.12) into (3.9), we get
(3.13)‖vn+1-vn‖-‖xn+1-xn‖≤ϵn+11-βn+1(‖γf(xn+1)‖+‖BSn+1yn+1‖)+ϵn1-βn(‖γf(xn)‖+‖BSnyn‖)+‖Θn+1Nun+1-ΘnNun+1‖+‖Sn+1yn-Snyn‖.
By (3.10), (3.11), and the conditions (C1) and (C6), we imply that
(3.14)lim⁡n→∞(‖vn+1-vn‖-‖xn+1-xn‖)=0.
Hence, by Lemma 2.7, we have limn→∞∥vn-xn∥=0. Consequently, it follows that
(3.15)lim⁡n→∞‖xn+1-xn‖=lim⁡n→∞(1-βn)‖vn-xn‖=0.
From (3.11), (3.12), and (3.15), we also imply that
(3.16)lim⁡n→∞‖yn+1-yn‖=0.Step 3.
We now show that(3.17)lim⁡n→∞‖Θnkun-Θnk-1un‖=0,k=1,2,…,N.
Indeed, letp∈Ω. It follows from the firmly nonexpansiveness of JMk,λk,n(I-λk,nAk) that
(3.18)‖Θnkun-p‖2=‖JMk,λk,n(I-λk,nAk)Θnk-1un-JMk,λk,n(I-λk,nAk)p‖2≤〈Θnkun-p,Θnk-1un-p〉=12(‖Θnkun-p‖2+‖Θnk-1un-p‖2‖Θnkun-Θnk-1un‖2),
for each k∈{1,2,…,N}. Thus we get
(3.19)‖Θnkun-p‖2≤‖Θnk-1un-p‖2-‖Θnkun-Θnk-1un‖2,
which implies that, for each k∈{1,2,…,N},
(3.20)‖yn-p‖2=‖ΘnNun-p‖2≤‖Θn0un-p‖2-∑k=1N‖Θnkun-Θnk-1un‖2≤‖un-p‖2-‖Θnkun-Θnk-1un‖2≤‖xn-p‖2-‖Θnkun-Θnk-1un‖2.
Set θn=γf(xn)-BSnyn, and let λ>0 be a constant such that
(3.21)λ>sup⁡n,k{‖θn‖,‖xk-p‖}.
Using Lemma 2.2 and noting that ∥·∥2 is convex, we derive, from (3.20),
(3.22)‖xn+1-p‖2=‖(1-βn)(Snyn-p)+βn(xn-p)+ϵnθn‖2≤‖(1-βn)(Snyn-p)+βn(xn-p)‖2+2ϵn〈θn,xn+1-p〉≤(1-βn)‖Snyn-p‖2+βn‖xn-p‖2+2λ2ϵn≤(1-βn)‖yn-p‖2+βn‖xn-p‖2+2λ2ϵn≤(1-βn)(‖xn-p‖2-‖Θnkun-Θnk-1un‖2)+βn‖xn-p‖2+2λ2ϵn≤‖xn-p‖2-(1-βn)‖Θnkun-Θnk-1un‖2+2λ2ϵn.
It follows, by Step 2 and condition (C1), that
(3.23)‖Θnkun-Θnk-1un‖2≤11-βn(‖xn-p‖2-‖xn+1-p‖2+2λ2ϵn)≤11-βn(2λ‖xn-xn+1‖+2λ2ϵn)⟶0,asn⟶∞.Step 4.
We will provelimn→∞∥un-xn∥=0.
We note from (3.1)
(3.24)‖xn-Snyn‖≤‖xn-Sn-1yn-1‖+‖Sn-1yn-1-Sn-1yn‖+‖Sn-1yn-Snyn‖≤ϵn-1‖γf(xn-1)-BSn-1yn-1‖+‖yn-1-yn‖+βn‖xn-1-Sn-1yn-1‖+sup⁡{‖Sn+1z-Snz‖:z∈‖yn‖}.
Since ϵn→0, βn→0 as n→∞, limn→∞∥yn+1-yn∥=0 and sup{∥Sn+1z-Snz∥:z∈{yn}}→0, we get
(3.25)‖xn-Snyn‖⟶0asn⟶∞.
Let v∈Ω. Since un=Trnxn, it follows from Lemma 2.1 that
(3.26)‖un-v‖2=‖Trnxn-Trnv‖2≤〈Trnxn-Trnv,xn-v〉=〈un-v,xn-v〉≤12(‖un-v‖2+‖xn-v‖2-‖un-xn‖2),
and hence ∥un-v∥2≤∥xn-v∥2-∥un-xn∥2. Therefore, using Lemma 2.6 and (3.22), we have
(3.27)‖xn+1-v‖2≤(1-βn)‖yn-p‖2+βn‖xn-p‖2+2λ2ϵn≤(1-βn)‖un-p‖2+βn‖xn-p‖2+2λ2ϵn≤(1-βn)(‖xn-v‖2-‖un-xn‖2)+βn‖xn-p‖2+2λ2ϵn≤‖xn-v‖2-(1-βn)‖un-xn‖2+2λ2ϵn,
and hence
(3.28)‖un-xn‖2≤11-βn(‖xn-v‖2-‖xn+1-v‖2+2λ2ϵn)≤11-βn{‖xn-xn+1‖(‖xn-v‖-‖xn+1-v‖)+2λ2ϵn}.
Since {xn} is bounded, limn→∞ϵn=0, limn→∞∥xn-xn+1∥=0, and λ>supn,k{∥θn∥,∥xk-p∥}, it follows that
(3.29)lim⁡n→∞∥xn-un∥=0.
Next we will prove limn→∞∥un-yn∥=0. 									(3.30)‖un-yn‖=‖ΘnNun-un‖≤‖ΘnNun-ΘnN-1un‖+‖ΘnN-1un-ΘnN-2un‖+⋯+‖Θn2un-Θn1un‖+‖Θn1un-Θn0un‖+‖un-un‖.
From (2.10), we obtain
(3.31)lim⁡n→∞‖un-yn‖=0.
In addition, according to ∥xn-yn∥≤∥xn-un∥+∥un-yn∥, we have
(3.32)lim⁡n→∞‖xn-yn‖=0.
It follows from (3.25), (3.32), and the inequality ∥yn-Snyn∥≤∥yn-xn∥+∥xn-Snyn∥ that limn→∞∥yn-Snyn∥=0. Since
(3.33)‖Syn-yn‖≤‖Syn-Snyn‖+‖Snyn-yn‖≤sup⁡{‖Sz-Snz‖:z∈{yn}}+‖Snyn-yn‖,
for all n∈ℕ, it follows that
(3.34)lim⁡n→∞∥Syn-yn∥=0.Step 5.
We showω∈(⋂n=1∞Fix(Sn))∩MEP(F1,F2)∩(⋂i=1NI(Ai,Mi)).
Since{xn} is bounded, there exists a subsequence {xni} of {xn} which converges weakly to ω. From (3.29), we obtain {uni} which converges weakly to ω. From (3.32), it follows that yni⇀ω. We show ω∈MEP(F1,F2). According to (3.1) and (A2), we obtain
(3.35)1rn〈y-un,un-xn〉≥F1(y,un)+F2(y,un),
and hence
(3.36)〈y-uni,uni-xnirni〉≥F1(y,un)+F2(y,un).
Since (uni-xni)/rni→0 and uni⇀ω, from (A4), it follows that 0≥F1(y,ω)+F2(y,ω) for all y∈H. For t with 0<t≤1 and y∈H, let yt=ty+(1-t)ω, then we obtain 0≥F1(yt,ω)+F2(yt,ω). So, from (A1) and (A4) we have
(3.37)0=F1(yt,yt)+F2(yt,yt)≤tF1(yt,y)+(1-t)F1(yt,ω)+tF2(yt,y)+(1-t)F2(yt,ω)≤F1(yt,y)+F2(yt,y)
and hence 0≤F1(yt,y)+F2(yt,y). From (A3), we have 0≤F1(ω,y)+F2(ω,y) for all y∈H. Therefore, ω∈MEP(F1,F2).
Next, we showω∈⋂n=1∞Fix(Sn). Assume ω∉⋂n=1∞Fix(Sn); then we have ω≠Sω. It follows by the Opial’s condition and (3.34) that
(3.38)lim⁡inf⁡n→∞∥yn-ω∥<lim⁡inf⁡⁡n→∞‖yn-Sω‖≤lim⁡inf⁡n→∞{‖yn-Syn‖+‖Syn-Sω‖}≤lim⁡inf⁡n→∞‖yn-ω‖.
This is a contradiction. Hence ω∈⋂n=1∞Fix(Sn).
We now show thatω∈⋂i=1NI(Ai,Mi). In fact, since Ai is αi-inverse-strongly monotone, Ai, i=1,2,…,N, is a 1/αi-Lipschitz continuous monotone mapping and D(Ai)=H, i=1,2,…,N. It follows from Lemma 2.2 that Mi+Ai, i=1,2,…,N, is maximal monotone. Let (p,g)∈G(Mi+Ai), i=1,2,…,N, that is, g-Aip∈(Mip), i=1,2,…,N. Since Θnkun=JMk,λk,n(I-λk,nAk)Θnk-1un, we have Θnk-1un-λk,nAkΘnk-1un∈(I+λk,nMk)(Θnkun), that is,
(3.39)1λn,k(Θnk-1un-Θnkun-λN,nAkΘnk-1un)∈Mk(Θnkun).
By the maximal monotonicity of Mi+Aii=1,2,…,N, we have
(3.40)〈p-Θnkun,g-Akp-1λk,n(Θnk-1un-Θnkun-λk,nAkΘnk-1un)〉≥0,
which implies
(3.41)〈p-Θnkun,g〉≥〈p-Θnkun,Akp+1λn,k(Θnk-1un-Θnkun-λk,nAkΘnk-1un)〉=〈p-Θnkun,Akp-AkΘnkun+AkΘnkun-AkΘnk-1un+1λk,n(Θnk-1un-Θnkun)〉≥0+〈p-Θnkun,AkΘnkun-AkΘnk-1un〉+〈p-Θnkun,1λk,n(Θnk-1un-Θnkun)〉,
for k∈{1,2,…,N}. From (3.17), it follows limn→∞∥Θnkun-Θnk-1un∥=0, especially, Θnikuni⇀ω. Since Ak, k=1,…,N, are Lipschitz continuous operators, we have ∥AkΘnk-1un-AkΘnkun∥→0. So, from (3.41), we have
(3.42)lim⁡i→∞〈p-Θnikuni,g〉=〈p-ω,g〉≥0.
Since Ak+Mk, k∈{1,2,…,N} is maximal monotone, this implies that 0∈(Mk+Ak)(ω), k∈{1,2,…,N}, that is, ω∈⋂i=1NI(Ai,Mi). So, we obtain result.Step 6.
We show that(3.43)limsup⁡n→∞〈(B-γf)z,z-xn〉≤0,
where z=PΩ(I-B+γf)(z) is unique solution of the variational inequality (3.2).
To show this, we choose a subsequence{xni} of {xn} such that
(3.44)lim⁡i→∞〈(B-γf)z,z-xni〉=limsup⁡n→∞〈(B-γf)z,z-xn〉.
By the proof of Step 5, we obtain that
(3.45)limsup⁡n→∞〈(B-γf)z,z-xn〉=lim⁡i→∞〈(B-γf)z,z-xni〉=〈(B-γf)z,z-ω〉≤0.Step 7.
We prove thatxn→ω.
By using Lemmas2.3 and 2.4, we have
(3.46)‖xn+1-ω‖2=‖((1-βn)I-ϵnB)(Snyn-ω)+βn(xn-ω)+ϵn(γf(xn)-Bω)‖2≤‖((1-βn)I-ϵnB)(Snyn-ω)+βn(xn-ω)‖2+2ϵn〈γf(xn)-Bω,xn+1-ω〉≤‖(1-βn)I-ϵnB)(Snyn-ω)‖2+βn‖xn-ω‖2+2ϵn〈γf(xn)-f(ω),xn+1-ω〉+2ϵn〈γf(ω)-Bω,xn+1-ω〉≤((1-βn)-ϵnγ̅)‖Snyn-ω‖2+βn‖xn-ω‖2+2ϵnγα‖xn-ω‖‖xn+1-ω‖+2ϵn〈γf(ω)-Bω,xn+1-ω〉≤((1-βn)-ϵnγ̅)‖xn-ω‖2+βn‖xn-ω‖2+ϵnγα(‖xn-ω‖2+‖xn+1-ω‖2)+2ϵn〈γf(ω)-Bω,xn+1-ω〉≤(1-ϵn(γ̅-γα))‖xn-ω‖2+ϵnγα‖xn+1-ω‖2+2ϵn〈γf(ω)-Bω,xn+1-ω〉.
It follows that
(3.47)‖xn+1-ω‖2≤(1-(γ̅-αγ)ϵn1-αγϵn)‖xn-ω‖2+2ϵn1-αγϵn〈γf(ω)-Bω,xn+1-ω〉.
Now, from conditions (C1), (C2), and (C6), Step 6 and Lemma 2.8, we obtain limn→∞∥xn-ω∥=0. Namely, xn→ω in norm.Corollary 3.2.
LetH be a real Hilbert space, let F be a bifunction H×H→ℝsatisfying (A1)–(A4), and let {Sn} be a sequence of nonexpansive mappings onH. Let  Ai:H→H,i=1,2,…,N be  αi- inverse-strongly monotone mappings and Mi:H→2H,i=1,2,…,N, maximal monotone mappings such thatΩ:=(⋂n=1∞Fix(Sn))∩EP(F)∩(⋂i=1NI(Ai,Mi))≠∅.  Letf be a contraction of H into itself with a constantα∈(0,1), and let B be a strongly positive bounded linear operator on H with coefficient γ̅>0 and 0<γ<γ̅/α.  Let {xn}, {yn}, and {un}be sequences generated by x1∈H and
(3.48)F(un,y)+1rn〈y-un,un-xn〉≥0,∀y∈H,yn=JMN,λN,n(I-λN,nAN)⋯JM1,λ1,n(I-λ1,nA1)un,xn+1=ϵnγf(xn)+βnxn+((1-βn)I-ϵnB)Snyn,
for all n∈ℕ, where βn∈(0,1)λi,n∈(0,2αi], i∈{1,2,…,N}, satisfy (H1)-(H2) and {ϵn}⊂[0,1], and {rn}⊂(0,∞) satisfy(C1)
limn→∞ϵn=0;(C2)
∑n=1∞ϵn=∞;(C3)
∑n=1∞|ϵn+1-ϵn|<∞;(C4)
liminfn→∞rn>0;(C5)
∑n=1∞|rn+1-rn|<∞;(C6)
limn→∞βn=0. Suppose that ∑n=1∞sup{∥Sn+1z-Snz∥:z∈K}<∞ for any bounded subset K of H. Let S be a mapping of H into itself defined by Sx=limn→∞Snx, for all x∈H, and suppose that Fix(S)=⋂n=1∞Fix(Sn). Then, {xn}, {yn}, and {un} converge strongly to z, where z=PΩ(I-B+γf)(z) is a unique solution of the variational inequality
(3.49)〈(B-γf)z,z-x〉≤0x∈Ω.



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*Source: 152023-2012-06-10.xml*