# Strong Convergence Theorems for Family of Nonexpansive Mappings and System of Generalized Mixed Equilibrium Problems and Variational Inequality Problems

Authors: Yekini Shehu
Journal: International Journal of Mathematics and Mathematical Sciences
 (2011)

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## Abstract

We introduce a new iterative scheme by hybrid method for finding a
common element of the set of common fixed points of infinite family of nonexpansive mappings,
the set of common solutions to a system of generalized mixed equilibrium problems, and the
set of solutions to a variational inequality problem in a real Hilbert space. We then prove strong
convergence of the scheme to a common element of the three sets. We give some applications of
our results. Our results extend important recent results.


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## Body

## 1. Introduction

LetK be a nonempty closed and convex subset of a real Hilbert space H. A mapping A:K→H is called monotone if(1.1)〈Ax-Ay,x-y〉≥0,∀x,y∈K.
The variational inequality problem is to find an x*∈K such that(1.2)〈y-x*,Ax*〉≥0,∀y∈K
(see, e.g., [1]). We will denote the set of solutions to the variational inequality problem (1.2) by VI(K,A).A mappingA:K→H is called inverse-strongly monotone (see, e.g., [2, 3]) if there exists a positive real number α such that 〈Ax-Ay,x-y〉≥α∥Ax-Ay∥2, for all x,y∈K. For such a case, A is called α-inverse-strongly monotone.A mappingT:K→K is said to be nonexpansive if(1.3)‖Tx-Ty‖≤‖x-y‖,
for all x,y∈K. A point x∈K is called a fixed point of T if Tx=x. The set of fixed points of T is the set F(T)∶={x∈K:Tx=x}.Letφ:K→ℝ be a real-valued function and A:K→H a nonlinear mapping. Suppose that F:K×K into ℝ is an equilibrium bifunction. That is, F(u,u)=0, for all u∈K. The generalized mixed equilibrium problem is to find x∈K (see, e.g., [4–6]) such that(1.4)F(x,y)+φ(y)-φ(x)+〈Ax,y-x〉≥0,
for all y∈K. We will denote the set of solutions of this generalized mixed equilibrium problem by GMEP(F,A,φ). Thus (1.5)GMEP(F,A,φ)∶={x*∈K:F(x*,y)+φ(y)-φ(x*)+〈Ax*,y-x*〉≥0,∀y∈K}.
If φ=0, A=0, then problem (1.4) reduces to equilibrium problem studied by many authors (see, e.g., [7–14]), which is to find x*∈K such that(1.6)F(x*,y)≥0,
for all y∈K. The set of solutions of (1.6) is denoted by EP(F).Ifφ=0, then problem (1.4) reduces to generalized equilibrium problem studied by many authors (see, e.g., [15–18]), which is to find x*∈K such that(1.7)F(x*,y)+〈Ax*,y-x*〉≥0,
for all y∈K. The set of solutions of (1.7) is denoted by GEP(F,A).IfA=0, then problem (1.4) reduces to mixed equilibrium problem considered by many authors (see, e.g., [19–21]), which is to find x*∈K such that(1.8)F(x*,y)+φ(y)-φ(x*)≥0,
for all y∈K. The set of solutions of (1.8) is denoted by MEP(F,φ).The generalized mixed equilibrium problems include fixed point problems, optimization problems, variational inequality problems, Nash equilibrium problems, and equilibrium problems as special cases (see, e.g., [22]). Numerous problems in physics, optimization, and economics reduce to find a solution of problem (1.4). Several methods have been proposed to solve the fixed point problems, variational inequality problems, and equilibrium problems in the literature. See, for example, [23–33].One of the iterative processes (see Halpern [34]) which is often used to approximate a fixed point of a nonexpansive mapping T is defined as follows. Take an initial guess x0∈K arbitrarily and define {xn} recursively by(1.9)xn+1=αnx0+(1-αn)Txn,n≥0,
where {αn} is a sequence in [0, 1]. The iteration process (1.9) has been proved to be strongly convergent both in Hilbert spaces [34–36] and uniformly smooth Banach spaces [37, 38] when the sequence {αn} satisfies the conditions(i)
limn→∞αn=0,(ii)
∑n=1∞αn=∞,(iii)
either∑n=1∞|αn+1-αn|<∞ or limn→∞αn/αn+1=1.Motivated by (1.9), Martinez-Yanes and Xu [39] introduced the following iterative scheme for a single nonexpansive mapping T in a Hilbert space:(1.10)x0∈K,yn=αnx0+(1-αn)Txn,Cn={z∈K:‖yn-z‖2≤‖xn-z‖2+αn(‖x0‖2+2〈xn-x0,z〉)},Qn={z∈K:〈xn-z,xn-x0〉≤0},xn+1=PCn∩Qnx0,
where PK denotes the metric projection of H onto a closed and convex subset K of H. They proved that if {αn}⊂(0,1) and limn→∞αn=0, then the sequence {xn} converges strongly to PF(T)x0.Furthermore, algorithm (1.10) has been modified by many authors for relatively nonexpansive mappings and quasi-ϕ-nonexpansive mappings in Banach spaces (see, e.g., [40–43]).Recently, Ceng and Yao [44] introduced a new iterative scheme of approximating a common element of the set of solutions to mixed equilibrium problem and set of common fixed points of finite family of nonexpansive mappings in a real Hilbert space H. In the proof process of their results, they imposed the following condition on a nonempty closed and convex subset K of H:(E)
A:K→ℝ is η-strongly convex and its derivative A′ is sequentially continuous from weak topology to the strong topology.We remark here that this condition (E) has been used by many authors for approximation of solution to mixed equilibrium problem in a real Hilbert space (see, e.g., [45, 46]). However, it is observed that condition (E) does not include the case A(x)=∥x∥2/2 and η(x,y)=x-y. Furthermore, Peng and Yao [19], Wangkeeree and Wangkeeree [47], and other authors replaced condition (E) with these conditions:(B1)
for eachx∈H and r>0, there exist a bounded subset Dx⊆K and yx∈K such that, for any z∈K∖Dx,
(1.11)F(z,yx)+φ(yx)-φ(z)+1r〈yx-z,z-x〉<0,
or(B2)
K is a bounded set.Consequently, conditions (B1) and (B2) have been used by many authors in approximating solution to generalized mixed equilibrium (mixed equilibrium) problems in a real Hilbert space (see, e.g., [19, 47]).In [48], Takahashi et al. proved the following convergence theorem using hybrid method.Theorem 1.1 (Takahashi et al. [48]).
LetK be a nonempty closed and convex subset of a real Hilbert space H. Let T be a nonexpansive mapping of K into itself such that F(T)≠∅. For C1=K, x1=PC1x0, define sequences {xn}n=0∞ and {yn}n=1∞ of K as follows:
(1.12)yn=αnxn+(1-αn)Txn,n≥1,Cn+1={z∈Cn:‖yn-z‖≤‖xn-z‖},n≥1,xn+1=PCn+1x0,n≥1.
Assume that {αn}n=1∞⊂[0,1) satisfies 0≤αn<α<1. Then, {xn}n=0∞ converges strongly to PF(T)x0.Motivated by the results of Takahashi et al. [48], Kumam [49] studied the problem of approximating a common element of set of solutions to an equilibrium problem, set of solutions to variational inequality problem, and set of fixed points of a nonexpansive mapping in a real Hilbert space. In particular, he proved the following theorem.Theorem 1.2 (Kumam [49]).
LetK be a nonempty closed convex subset of a real Hilbert space H. Let F be a bifunction from K×K satisfying (A1)–(A4), and let B be a β-inverse-strongly monotone mapping of K into H. Let T be a nonexpansive mapping of K into H such that F(T)∩EP(F)∩VI(K,B)≠∅. For C1=K, x1=PC1x0, define sequences {xn}n=0∞ and {zn}n=1∞ of K as follows:
(1.13)F(zn,y)+1rn〈y-zn,zn-xn〉≥0,∀y∈K,yn=αnxn+(1-αn)TPK(zn-λnBzn),n≥1,Cn+1={z∈Cn:‖yn-z‖≤‖xn-z‖},n≥1,xn+1=PCn+1x0,n≥1.
Assume that {αn}n=1∞⊂[0,1), {rn}n=1∞⊂(0,∞), and {λn}n=1∞⊂[0,2β] satisfy
(1.14)liminf⁡n→∞rn>0,0<c≤λn≤f<2β,lim⁡n→∞αn=0.
Then, {xn}n=0∞ converges strongly to PF(T)∩EP(F)∩VI(K,B)x0.Quite recently, Chantarangsi et al. [50] proved the following convergence theorem for approximation of fixed point of a nonexpansive mapping which is also a common solution to a system of generalized mixed equilibrium problems and variational inequality problem in a real Hilbert space.Theorem 1.3 (Chantarangsi et al. [50]).
LetK be a nonempty closed and convex subset of a real Hilbert space H. Let θ1, θ2 be bifunctions from K×K satisfying (A1)–(A4), Ψ1 an ξ-inverse-strongly monotone mapping of K into H, Ψ2 a β-inverse-strongly monotone mapping of K into H with assumption (B1) or (B2), and T : K→K a nonexpansive mapping. Let B be an ω-Lipschitz continuous and relaxed (υ,ν) co-coercive mapping of K into H, f:K→K a contraction mapping with coefficient η∈(0,1), and A a strongly positive linear bounded selfadjoint operator with coefficient γ¯>0 and 0<γ<γ¯/η. Suppose that F∶=F(T)∩GMEP(θ1,φ,Ψ1)∩GMEP(θ2,φ,Ψ2)∩VI(K,B)≠∅. Let {zn}n=1∞, {un}n=1∞, {yn}n=1∞, and {xn}n=1∞ be generated by
(1.15)un=Trn(θ1,φ)(xn-rnΨ1xn),νn=Tsn(θ2,φ)(un-snΨ1un),zn=PK(νn-αnBSνn),yn=εnγf(xn)+βnxn+((1-βn)I-εnA)zn,xn+1=γnxn+(1-γn)yn,∀n≥1,
where {rn}⊂[a,b]⊂[0,2ξ], {sn}⊂[c,d]⊂[0,2β], {γn}⊂[h,j]⊂(0,1), {γn}, {εn}, {βn} are three sequences in (0, 1) satisfying the following conditions:(C1)
limn→∞εn=0 and ∑n=1∞εn=∞,(C2)
0<liminfn→∞βn≤limsupn→∞βn<1,(C3)
0<liminfn→∞rn≤limsupn→∞rn<2ξ and limn→∞|rn+1-rn|=0,(C4)
0<liminfn→∞sn≤limsupn→∞sn<2β and limn→∞|sn+1-sn|=0,(C5)
{αn}⊂[e,g]⊂(0,(2(ν-υω2))/ω2),ν>υω2 and limn→∞|αn+1-αn|=0.Then, {xn} converges strongly to z=PF(γf+(I-A))(z).Motivated by the ongoing research and the above-mentioned results, we modify algorithm (1.10) and introduce a new iterative scheme for finding a common element of the set of fixed points of an infinite family of nonexpansive mappings, the set of common solutions to a system of generalized mixed equilibrium problems,and the set of solutions to a variational inequality problem in a real Hilbert space. Furthermore, we show that our new iterative scheme converges strongly to a common element of the three sets. In the proof process of our results, we use conditions (B1) and (B2) mentioned above. Our result extends many important recent results. Finally, we give some applications of our results.


## 2. Preliminaries

LetH be a real Hilbert space with inner product 〈·,·〉 and norm ∥·∥, and let K be a nonempty closed and convex subset of H. The strong convergence of {xn}n=1∞ to x is written xn→x as n→∞.For any pointu∈H, there exists a unique point PKu∈K such that(2.1)‖u-PKu‖≤‖u-y‖,∀y∈K.PK is called the metric projectionof H onto K. We know that PK is a nonexpansive mapping of H onto K. It is also known that PK satisfies(2.2)〈x-y,PKx-PKy〉≥‖PKx-PKy‖2,
for all x,y∈H. Furthermore, PKx is characterized by the properties PKx∈K and(2.3)〈x-PKx,PKx-y〉≥0,
for all y∈K and(2.4)‖x-PKx‖2≤‖x-y‖2-‖y-PKx‖2,∀x∈H,y∈K.
In the context of the variational inequality problem, (2.3) implies that (2.5)x*∈VI(K,A)⟺x*=PK(x*-λAx*),∀λ≥0.
If A is an α-inverse-strongly monotone mapping of K into H, then it is obvious that A is a (1/α)-Lipschitz continuous. We also have that, for all x,y∈K and r>0,(2.6)‖(I-rA)x-(I-rA)y‖2=‖x-y-r(Ax-Ay)‖2=‖x-y‖2-2r〈Ax-Ay,x-y〉+r2‖Ax-Ay‖2≤‖x-y‖2+r(r-2α)‖Ax-Ay‖2.
So, if r≤2α, then I-rA is a nonexpansive mapping of K into H.For solving the generalized mixed equilibrium problem for a bifunctionF : K×K→ℝ, let us assume that F, φ, and K satisfy the following conditions:(A1)
F(x,x)=0 for all x∈K,(A2)
F is monotone, that is, F(x,y)+F(y,x)≤0 for all x,y,∈K,(A3)
for eachx,y,z∈K, limt→0F(tz+(1-t)x,y)≤F(x,y),(A4)
for eachx∈K, y↦F(x,y) is convex and lower semicontinuous,(B1)
for eachx∈H and r>0 there exist a bounded subset Dx⊆K and yx∈K such that for any z∈K∖Dx,(2.7)F(z,yx)+φ(yx)-φ(z)+1r〈yx-z,z-x〉<0,(B2)
K is a bounded set.Then, we have the following lemma.Lemma 2.1 (Wangkeeree and Wangkeeree [47]).
Assume thatF:K×K→ℝ satisfies (A1)–(A4), and let φ:K→ℝ be a proper lower semicontinuous and convex function. Assume that either (B1) or (B2) holds. For r>0 and x∈H, define a mapping Tr(F,φ):H→K as follows:
(2.8)Tr(F,φ)(x)={z∈K:F(z,y)+φ(y)-φ(z)+1r〈y-z,z-x〉≥0,∀y∈K},
for all z∈H. Then, the following hold:(1)
for eachx∈H, Tr(F,φ)≠∅,(2)
Tr(F,φ) is single-valued,(3)
Tr(F,φ) is firmly nonexpansive, that is, for any x,y∈H,
(2.9)‖Tr(F,φ)x-Tr(F,φ)y‖2≤〈Tr(F,φ)x-Tr(F,φ)y,x-y〉,(4)
F(Tr(F,φ))=GMEP(F),(5)
GMEP(F) is closed and convex.We will also use the following lemma in our results.Lemma 2.2 (Baillon and Haddad [51]).
LetE be a Banach space, let f be a continuously Fréchet differentiable convex functional on E, and let ∇f be the gradient of f. If ∇f is (1/α)-Lipschitz continuous, then ∇f is α-inverse-strongly monotone.


## 3. Main Results

Theorem 3.1.
LetK be a nonempty closed and convex subset of a real Hilbert space H. For each m=1,2, let Fm be a bifunction from K×K satisfying (A1)–(A4), φm : K→ℝ∪{+∞} a proper lower semicontinuous and convex function with assumption (B1) or (B2), A be an α-inverse-strongly monotone mapping of K into H, and B a β-inverse-strongly monotone mapping of K into H, and, for each i=1,2,…, let Ti:K→K be a nonexpansive mapping such that ⋂i=1∞F(Ti)≠∅. Let D be a γ-inverse-strongly monotone mapping of K into H. Suppose that F∶=⋂i=1∞F(Ti)∩GMEP(F1,A,φ1)∩GMEP(F2,B,φ2)∩VI(K,D)≠∅. Let {zn}n=1∞, {un}n=1∞, {wn}n=1∞, {yn,i}n=1∞ (i=1,2,…), and {xn}n=0∞ be generated by x0∈K, C1,i=K, C1=⋂i=1∞C1,i, x1=PC1x0,
(3.1)zn=Trn(F1,φ1)(xn-rnAxn),un=Tλn(F2,φ2)(zn-λnBzn),wn=PK(un-snDun),yn,i=αn,ix0+(1-αn,i)Tiwn,Cn+1,i={z∈Cn,i:‖yn,i-z‖2≤‖xn-z‖2+αn,i(‖x0‖2+2〈xn-x0,z〉)},Cn+1=⋂i=1∞Cn+1,i,xn+1=PCn+1x0,n≥1.
Assume that {αn,i}n=1∞⊂(0,1) (i=1,2,…), {rn}n=1∞⊂[0,2α], and {λn}n=1∞⊂[0,2β] satisfy(i)
0<a≤rn≤b<2α,(ii)
0<c≤λn≤f<2β,(iii)
limn→∞αn,i=0,(iv)
0<h≤sn≤j<2γ. Then, {xn}n=0∞ converges strongly to PFx0.Proof.
Letx*∈F. Then,
(3.2)‖wn-x*‖2=‖PK(un-snDun)-PK(x*-snDx*)‖2≤‖(un-snDun)-(x*-snDx*)‖2≤‖un-x*‖2+sn(sn-2γ)‖Dun-Dx*‖2≤‖un-x*‖2.
Since both I-rnA and I-λnB are nonexpansive for each n≥1 and x*=Trn(F1,φ1)(x*-rnAx*), x*=Tλn(F2,φ2)(x*-λnBx*), from (2.6), we have that
(3.3)‖un-x*‖2=‖Tλn(F2,φ2)(zn-λnBzn)-x*‖2=‖Tλn(F2,φ2)(zn-λnBzn)-Tλn(F2,φ2)(x*-λnBx*)‖2≤‖(I-λnB)zn-(I-λnB)x*‖2≤‖zn-x*‖2+λn(λn-2β)‖Bzn-Bx*‖2≤‖zn-x*‖2(sinceλn<2β,∀n≥1),‖zn-x*‖2=‖Trn(F1,φ1)(xn-rnAxn)-x*‖2=‖Trn(F1,φ1)(xn-rnAxn)-Trn(F1,φ1)(x*-rnAx*)‖2≤‖(I-rnA)xn-(I-rnA)x*‖2≤‖xn-x*‖2+rn(rn-2α)‖Axn-Ax*‖2≤‖xn-x*‖2.
Therefore,
(3.4)‖un-x*‖≤‖xn-x*‖.
Let n=1, then C1,i=K is closed convex for each i=1,2,…. Now assume that Cn,i is closed convex for some n>1. Then, from definition of Cn+1,i, we know that Cn+1,i is closed convex for the same n>1. Hence, Cn,i is closed convex for n≥1 and for each i=1,2,…. This implies that Cn is closed convex for n≥1. Furthermore, we show that F⊂Cn. For n=1, F⊂K=C1,i. For n≥2, let x*∈F. Then,
(3.5)‖yn,i-x*‖2=‖αn,i(x0-x*)+(1-αn,i)(Tiwn-x*)‖2≤αn,i‖x0-x*‖2+(1-αn,i)‖wn-x*‖2≤‖wn-x*‖2+αn,i(‖x0-x*‖2-‖wn-x*‖2)≤‖xn-x*‖2+αn,i(‖x0‖2+2〈xn-x0,x*〉),
which shows that x*∈Cn,i, for all n≥2, for all i=1,2,…. Thus, F⊂Cn,i, for all n≥1, for all i=1,2,…. Hence, it follows that F⊂Cn, for all n≥1. Since xn=PCnx0, for all n≥1, and xn+1∈Cn+1⊂Cn, for all n≥1, we have that
(3.6)‖xn-x0‖≤‖xn+1-x0‖,∀n≥1.
Also, as F⊂Cn by (2.1), it follows that
(3.7)‖xn-x0‖≤‖z-x0‖,z∈F,∀n≥1.
From (3.6) and (3.7), we have that limn→∞∥xn-x0∥ exists. Hence, {xn}n=0∞ is bounded and so are {zn}n=1∞, {Axn}n=1∞, {un}n=1∞, {Dun}n=1∞, {Bzn}n=1∞, {wn}n=1∞, {Tiwn}n=1∞, and {yn,i}n=1∞, i=1,2,…. For m>n≥1, we have that xm=PCmx0∈Cm⊂Cn. By (2.4), we obtain
(3.8)‖xm-xn‖2≤‖xn-x0‖2-‖xm-x0‖2.
Letting m,n→∞ and taking the limit in (3.8), we have that xm-xn→0, m,n→∞, which shows that {xn}n=0∞ is Cauchy. In particular, limn→∞∥xn+1-xn∥=0. Since, {xn}n=0∞ is Cauchy, we assume that xn→z∈K.
Sincexn+1=PCn+1x0∈Cn+1, then
(3.9)‖yn,i-xn+1‖2≤‖xn-xn+1‖2+αn,i(‖x0‖2+2〈xn-x0,xn+1〉)⟶0,
and it follows that
(3.10)‖yn,i-xn‖≤‖yn,i-xn+1‖+‖xn-xn+1‖.
Thus,
(3.11)lim⁡n→∞‖yn,i-xn‖=0,i=1,2,….
Furthermore,
(3.12)‖yn,i-x*‖2≤αn,i‖x0-x*‖2+(1-αn,i)‖Tiwn-x*‖2≤αn,i‖x0-x*‖2+(1-αn,i)‖wn-x*‖2≤αn,i‖x0-x*‖2+(1-αn,i)‖un-x*‖2≤αn,i‖x0-x*‖2+(1-αn,i)‖Tλn(F2,φ2)(zn-λnBzn)-Tλn(F2,φ2)(x*-λnBx*)‖2≤αn,i‖x0-x*‖2+(1-αn,i)‖(zn-λnBzn)-(x*-λnBx*)‖2≤αn,i‖x0-x*‖2+(1-αn,i)[‖zn-x*‖2+λn(λn-2β)‖Bzn-Bx*‖2]≤αn,i‖x0-x*‖2+‖xn-x*‖2+λn(λn-2β)‖Bzn-Bx*‖2.
Since 0<c≤λn≤f<2β, we have that
(3.13)c(2β-f)‖Bzn-Bx*‖2≤‖xn-x*‖2-‖yn,i-x*‖2+αn,i‖x0-x*‖2≤‖yn,i-xn‖(‖xn-x*‖+‖yn,i-x*‖)+αn,i‖x0-x*‖2.
Hence, limn→∞∥Bzn-Bx*∥=0. From (3.1), we have that
(3.14)‖yn,i-x*‖2≤αn,i‖x0-x*‖2+(1-αn,i)‖Tiwn-x*‖2=αn,i‖x0-x*‖2+(1-αn,i)‖un-x*‖2≤αn,i‖x0-x*‖2+‖un-x*‖2.
On the other hand,
(3.15)‖un-x*‖2≤‖Tλn(F2,φ2)(zn-λnBzn)-Tλn(F2,φ2)(x*-λnBx*)‖2≤〈(zn-λnBzn)-(x*-λnBx*),un-x*〉=12[‖(zn-λnBzn)-(x*-λnBx*)‖2+‖un-x*‖2-‖(zn-λnBzn)-(x*-λnBx*)-(un-x*)‖2]≤12[‖zn-x*‖2+‖un-x*‖2-‖(zn-λnBzn)-(x*-λnBx*)-(un-x*)‖2]=12[‖zn-x*‖2+‖un-x*‖2-‖un-zn‖2+2λn〈zn-un,Bzn-Bx*〉-λn2‖Bzn-Bx*‖2],
and, hence,
(3.16)‖un-x*‖2≤‖zn-x*‖2-‖un-zn‖2+2λn〈zn-un,Bzn-Bx*〉-λn2‖Bzn-Bx*‖2≤‖zn-x*‖2-‖un-zn‖2+2λn‖zn-un‖‖Bzn-Bx*‖.
Putting (3.16) into (3.14), we have that
(3.17)‖yn,i-x*‖2≤‖zn-x*‖2-‖un-zn‖2+2λn‖zn-un‖‖Bzn-Bx*‖+αn,i‖x0-x*‖2.
It follows that
(3.18)‖zn-un‖2≤‖zn-x*‖2-‖yn,i-x*‖2+2λn‖zn-un‖‖Bzn-Bx*‖+αn,i‖x0-x*‖2≤‖xn-x*‖2-‖yn,i-x*‖2+2λn‖zn-un‖‖Bzn-Bx*‖+αn,i‖x0-x*‖2≤‖yn,i-xn‖(‖xn-x*‖+‖yn,i-x*‖)+2λn‖zn-un‖‖Bzn-Bx*‖+αn,i‖x0-x*‖2.
Therefore, limn→∞∥zn-un∥=0. Furthermore,
(3.19)‖yn,i-x*‖2≤αn,i‖x0-x*‖2+(1-αn,i)‖Tiwn-x*‖2≤αn,i‖x0-x*‖2+(1-αn,i)‖un-x*‖2≤αn,i‖x0-x*‖2+(1-αn,i)‖zn-x*‖2≤αn,i‖x0-x*‖2+(1-αn,i)‖Trn(F1,φ1)(xn-rnAxn)-Trn(F1,φ1)(x*-rnAx*)‖2≤αn,i‖x0-x*‖2+(1-αn,i)‖(xn-rnAxn)-(x*-rnAx*)‖2≤αn,i‖x0-x*‖2+(1-αn,i)[‖xn-x*‖2+rn(rn-2α)‖Axn-Ax*‖2]≤αn,i‖x0-x*‖2+‖xn-x*‖2+rn(rn-2α)‖Axn-Ax*‖2.
Since 0<a≤rn≤b<2α, we have that
(3.20)a(2α-b)‖Axn-Ax*‖2≤‖xn-x*‖2-‖yn,i-x*‖2+αn,i‖x0-x*‖2≤‖yn,i-xn‖(‖xn-x*‖+‖yn,i-x*‖)+αn,i‖x0-x*‖2.
Hence, limn→∞∥Axn-Ax*∥=0. From (3.1), we have that
(3.21)‖yn,i-x*‖2≤αn,i‖x0-x*‖2+(1-αn,i)‖Tiwn-x*‖2≤αn,i‖x0-x*‖2+(1-αn,i)‖un-x*‖2≤αn,i‖x0-x*‖2+‖zn-x*‖2.
On the other hand,
(3.22)‖zn-x*‖2≤‖Trn(F1,φ1)(xn-rnAxn)-Trn(F1,φ1)(x*-rnAx*)‖2≤〈(xn-rnAxn)-(x*-rnAx*),zn-x*〉=12[‖(xn-rnAxn)-(x*-rnAx*)‖2+‖zn-x*‖2-‖(xn-rnAxn)-(x*-rnAx*)-(zn-x*)‖2]≤12[‖xn-x*‖2+‖zn-x*‖2-‖(xn-rnAxn)-(x*-rnAx*)-(zn-x*)‖2]=12[‖xn-x*‖2+‖zn-x*‖2-‖zn-xn‖2+2rn〈xn-zn,Axn-Ax*〉-rn2‖Axn-Ax*‖2],
and, hence,
(3.23)‖zn-x*‖2≤‖xn-x*‖2-‖zn-xn‖2+2rn〈xn-zn,Axn-Ax*〉-rn2‖Axn-Ax*‖2≤‖xn-x*‖2-‖zn-xn‖2+2rn‖xn-zn‖‖Axn-Ax*‖.
Putting (3.23) into (3.21), we have that
(3.24)‖yn,i-x*‖2≤‖xn-x*‖2-‖zn-xn‖2+2rn‖xn-zn‖‖Axn-Ax*‖+αn,i‖x0-x*‖2.
It follows that
(3.25)‖xn-zn‖2≤‖xn-x*‖2-‖yn,i-x*‖2+2rn‖xn-zn‖‖Axn-Ax*‖+αn,i‖x0-x*‖2≤‖yn,i-xn‖(‖xn-x*‖+‖yn,i-x*‖)+2rn‖xn-zn‖‖Axn-Ax*‖+αn,i‖x0-x*‖2.
Therefore, limn→∞∥xn-zn∥=0. But yn,i=αn,ix0+(1-αn,i)Tiwn implies that
(3.26)‖yn,i-Tiwn‖=αn,i‖x0-Tiwn‖⟶0.
Furthermore, we have that
(3.27)‖xn-Tiwn‖≤‖yn,i-Tiwn‖+‖yn,i-xn‖⟶0.
Furthermore,
(3.28)‖yn,i-x*‖2≤αn,i‖x0-x*‖2+(1-αn,i)‖Tiwn-x*‖2≤αn,i‖x0-x*‖2+(1-αn,i)‖wn-x*‖2≤αn,i‖x0-x*‖2+(1-αn,i)‖PK(un-snDun)-PK(x*-snDx*)‖2≤αn,i‖x0-x*‖2+(1-αn,i)‖(un-snDun)-(x*-snDx*)‖2≤αn,i‖x0-x*‖2+(1-αn,i)[‖un-x*‖2+sn(sn-2γ)‖Dun-Dx*‖2]≤αn,i‖x0-x*‖2+‖xn-x*‖2+(1-αn,i)sn(sn-2γ)‖Dun-Dx*‖2.
Thus,
(3.29)(1-di)h(2γ-j)‖Dun-Dx*‖2≤(1-αn,i)sn(2γ-sn)‖Dun-Dx*‖2≤αn,i‖x0-x*‖2+‖xn-x*‖2-‖yn,i-x*‖2≤αn,i‖x0-x*‖2+‖yn,i-xn‖(‖xn-x*‖+‖yn,i-x*‖).
Since 0<h≤sn≤j<2γ, condition (iii) and ∥yn,i-xn∥→0 as n→∞, we have that limn→∞∥Dun-Dx*∥=0. Now, using (2.2), we obtain
(3.30)‖wn-x*‖2≤‖PK(un-snDun)-PK(x*-snDx*)‖2≤〈(un-snDun)-(x*-snDx*),wn-x*〉=12[‖(un-snDun)-(x*-snDx*)‖2+‖wn-x*‖2-‖(un-snDun)-(x*-snDx*)-(wn-x*)‖2]≤12[‖un-x*‖2+‖wn-x*‖2-‖(un-snDun)-(x*-snDx*)-(wn-x*)‖2]=12[‖xn-x*‖2+‖wn-x*‖2-‖wn-un‖2+2sn〈un-wn,Dun-Dx*〉-sn2‖Dun-Dx*‖2].
Thus,
(3.31)‖wn-x*‖2≤‖xn-x*‖2-‖wn-un‖2+2sn‖wn-un‖‖Dun-Dx*‖.
Using this last inequality, we obtain from (3.1)
(3.32)‖yn,i-x*‖2≤αn,i‖x0-x*‖2+(1-αn,i)‖Tiwn-x*‖2≤αn,i‖x0-x*‖2+(1-αn,i)‖wn-x*‖2≤αn,i‖x0-x*‖2+‖xn-x*‖2-(1-αn,i)‖wn-un‖2+2sn(1-αn,i)‖wn-un‖‖Dun-Dx*‖.
This implies that
(3.33)(1-αn,i)‖wn-un‖2≤αn,i‖x0-x*‖2+‖xn-x*‖2-‖yn,i-x*‖2+2sn(1-αn,i)‖wn-un‖‖Dun-Dx*‖≤αn,i‖x0-x*‖2+‖yn,i-xn‖(‖xn-x*‖+‖yn,i-x*‖)+2sn(1-αn,i)‖wn-un‖‖Dun-Dx*‖.
Since limn→∞αn,i=0, ∥yn,i-xn∥→0 as n→∞, and ∥Dun-Dx*∥→0 as n→∞, we have that limn→∞∥wn-un∥=0. Also since limn→∞∥wn-xn∥=0 and limn→∞∥xn-z∥=0, we have that limn→∞∥wn-z∥=0. Now,
(3.34)‖wn-Tiwn‖≤‖xn-Tiwn‖+‖wn-xn‖.
Hence, limn→∞∥wn-Tiwn∥=0, i=1,2,…. By limn→∞∥wn-z∥=0 and limn→∞∥wn-Tiwn∥=0, i=1,2,…, we have that z∈⋂i=1∞F(Ti).
Sincezn∶=Trn(F1,φ1)(xn-rnAxn), n≥1, we have, for any y∈K, that
(3.35)F1(zn,y)+φ1(y)-φ1(zn)+〈Axn,y-zn〉+1rn〈y-zn,zn-xn〉≥0.
Furthermore, replacing n by nj in the last inequality and using (A2), we obtain
(3.36)φ1(y)-φ1(znj)+〈Axnj,y-znj〉+1rnj〈y-znj,znj-xnj〉≥F1(y,znj).
Let zt∶=ty+(1-t)z for all t∈(0,1] and y∈K. This implies that zt∈K. Then, we have that
(3.37)〈zt-znj,Azt〉≥φ1(znj)-φ1(zt)+〈zt-znj,Azt〉-〈zt-znj,Axnj〉-〈zt-znj,znj-xnjrnj〉+F1(zt,znj)=φ1(znj)-φ1(zt)+〈zt-znj,Azt-Aznj〉+〈zt-znj,Aznj-Axnj〉-〈zt-znj,znj-xnjrnj〉+F1(zt,znj).
Since ∥xnj-znj∥→0, j→∞, we obtain ∥Axnj-Aznj∥→0, j→∞. Furthermore, by the monotonicity of A, we obtain 〈zt-znj,Azt-Aznj〉≥0. Then, by (A4), we obtain (noting that znj→z)
(3.38)〈zt-z,Azt〉≥φ1(z)-φ1(zt)+F1(zt,z),j⟶∞.
Using (A1), (A4), and (3.38), we also obtain
(3.39)0=F1(zt,zt)+φ1(zt)-φ1(zt)≤tF1(zt,y)+(1-t)F1(zt,z)+tφ1(y)+(1-t)φ1(z)-φ1(zt)≤t[F1(zt,y)+φ1(y)-φ1(zt)]+(1-t)〈zt-z,Azt〉=t[F1(zt,y)+φ1(y)-φ1(zt)]+(1-t)t〈y-z,Azt〉,
and, hence,
(3.40)0≤F1(zt,y)+φ1(y)-φ1(zt)+(1-t)〈y-z,Azt〉.
Letting t→0, we have, for each y∈K, that
(3.41)0≤F1(z,y)+φ1(y)-φ1(z)+〈y-z,Az〉.
This implies that z∈GMEP(F1,A,φ1). By following the same arguments, we can show that z∈GMEP(F2,B,φ2).
Following the arguments of [3, Theorem 3.1, pages 346-347], we can show that z∈VI(K,D). Therefore, z∈⋂i=1∞F(Ti)∩GMEP(F1,A,φ1)∩GMEP(F2,B,φ2)∩VI(K,D).
Noting thatxn=PCnx0, we have by (2.3),
(3.42)〈x0-xn,y-xn〉≤0,
for all y∈Cn. Since F⊂Cn and by the continuity of inner product, we obtain, from the above inequality,
(3.43)〈x0-z,y-z〉≤0,
for all y∈F. By (2.3) again, we conclude that z=PFx0. This completes the proof.Corollary 3.2.
LetK be a nonempty closed and convex subset of a real Hilbert space H. Let T:K→K be a nonexpansive mapping such that F(T)≠∅. Let {xn}n=0∞ be generated by
(3.44)x0∈K,yn=αnx0+(1-αn)Txn,Cn+1={z∈Cn:‖yn-z‖2≤‖xn-z‖2+αn(‖x0‖2+2〈xn-x0,z〉)},xn+1=PCn+1x0,n≥1.
Assume that {αn}n=1∞⊂(0,1) such that limn→∞αn=0. Then, {xn}n=0∞ converges strongly to PF(T)x0.Remark 3.3.
Corollary3.2 can be viewed as an improvement of Theorem 3.1 of Martinez-Yanes and Xu [39] because we relax the iterative step Qn in the algorithm of Theorem 3.1 of [39].


## 4. Applications

LetC be a nonempty closed and convex cone in H and D an operator of C into H. We define the polar of C in H to be the set (4.1)K*∶={y*∈H:〈x,y*〉≥0,∀x∈C}.
Then, the element u∈C is called a solution of the complementarity problem if (4.2)Du∈K*,〈u,Du〉=0.
The set of solutions of the complementarity problem is denoted by C(C,D). We will assume that D satisfies the following conditions:(E1)
D is γ-inverse strongly monotone,(E2)
C(C,D)≠∅.Also, we replace conditions (B1) and (B2) with(D1)
for eachx∈H and r>0, there exist a bounded subset Dx⊆C and yx∈C such that, for any z∈C∖Dx,
(4.3)F(z,yx)+φ(yx)-φ(z)+1r〈yx-z,z-x〉<0,(D2)
C is a bounded set.Theorem 4.1.
LetC be a nonempty closed and convex subset of a real Hilbert space H. For each m=1,2, let Fm be a bifunction from C×C satisfying (A1)–(A4), φm:C→ℝ∪{+∞} a proper lower semicontinuous and convex function with assumption (B1) or (B2), A an α-inverse-strongly monotone mapping of C into H, and B a β-inverse-strongly monotone mapping of C into H, and, for each i=1,2,…, let Ti:C→C be a nonexpansive mapping such that ⋂i=1∞F(Ti)≠∅. Let D be a γ-inverse-strongly monotone mapping of K into H. Suppose that F∶=⋂i=1∞F(Ti)∩GMEP(F1,A,φ1)∩GMEP(F2,B,φ2)∩C(C,D)≠∅. Let {zn}n=1∞, {un}n=1∞, {wn}n=1∞, {yn,i}n=1∞ (i=1,2,…), and {xn}n=0∞ be generated by x0∈C, C1,i=C, C1=⋂i=1∞C1,i, x1=PC1x0,
(4.4)zn=Trn(F1,φ1)(xn-rnAxn),un=Tλn(F2,φ2)(zn-λnBzn),wn=PC(un-snDun),yn,i=αn,ix0+(1-αn,i)Tiwn,Cn+1,i={z∈Cn,i:‖yn,i-z‖2≤‖xn-z‖2+αn,i(‖x0‖2+2〈xn-x0,z〉)},Cn+1=⋂i=1∞Cn+1,i,xn+1=PCn+1x0,n≥1.
Assume that {αn,i}n=1∞⊂(0,1) (i=1,2,…), {rn}n=1∞⊂[0,2α], and {λn}n=1∞⊂[0,2β] satisfy(i)
0<a≤rn≤b<2α,(ii)
0<c≤λn≤f<2β,(iii)
limn→∞αn,i=0,(iv)
0<h≤sn≤j<2γ.Then, {xn}n=0∞ converges strongly to PFx0.Proof.
Using Lemma 7.1.1 of [52], we have that VI(C,D)=C(C,D). Hence, by Theorem 3.1 we obtain the desired conclusion.Next we study the problem of finding a minimizer of a continuously Fréchet differentiable convex functional in a Hilbert space.Theorem 4.2.
For eachm=1,2, let Fm be a bifunction from H×H satisfying (A1)–(A4), φm:H→ℝ∪{+∞} a proper lower semicontinuous and convex function with assumption (B1) or (B2), A an α-inverse-strongly monotone mapping of H into itself, and B a β-inverse-strongly monotone mapping of H into itself, and, for each i=1,2,…, let Ti:H→H be a nonexpansive mapping such that ⋂i=1∞F(Ti)≠∅. Suppose that f is a functional on H which satisfies the following conditions:(1)
f is a continuously Fréchet differentiable convex functional on H and ∇f is (1/γ)-Lipschitz continuous,(2)
(∇f)-10={z∈H:f(z)=miny∈Hf(y)}≠∅.Suppose that F∶=⋂i=1∞F(Ti)∩GMEP(F1,A,φ1)∩GMEP(F2,B,φ2)∩(∇f)-10≠∅. Let {zn}n=1∞, {un}n=1∞, {wn}n=1∞, {yn,i}n=1∞ (i=1,2,…), and {xn}n=0∞ be generated by x0∈K, C1,i=K, C1=⋂i=1∞C1,i, x1=PC1x0,
(4.5)zn=Trn(F1,φ1)(xn-rnAxn),un=Tλn(F2,φ2)(zn-λnBzn),wn=un-sn∇f(un),yn,i=αn,iwn+(1-αn,i)Tiwn,Cn+1,i={z∈Cn,i:‖yn,i-z‖2≤‖xn-z‖2+αn,i(‖x0‖2+2〈xn-x0,z〉)},Cn+1=⋂i=1∞Cn+1,i,xn+1=PCn+1x0,n≥1.
Assume that {αn,i}n=1∞⊂(0,1) (i=1,2,…), {rn}n=1∞⊂[0,2α], and {λn}n=1∞⊂[0,2β] satisfy(i)
0<a≤rn≤b<2α,(ii)
0<c≤λn≤f<2β,(iii)
limn→∞αn,i=0,(iv)
0<h≤sn≤j<2γ.Then, {xn}n=0∞ converges strongly to PFx0.Proof.
We know from condition (i) and Lemma2.2 that ∇f is an γ-inverse-strongly monotone operator from H into H. Using Theorem 3.1, we have the desired conclusion.We now study a kind of multiobjective optimization problem with nonempty set of solutions:(4.6)min⁡h1(x),min⁡h2(x),x∈K,
where K is a nonempty closed convex subset of a real Hilbert space H, and hi:K→ℝ, i=1,2, is a convex and a lower semicontinuous functional. Let us denote the set of solutions to (4.6) by Ω and assume that Ω≠∅.We will denote the set of solutions of the following two optimization problems byΩ1 and Ω2, respectively: (4.7)min⁡x∈Kh1(x),min⁡x∈Kh2(x).
Clearly, if we find a solution x∈Ω1∩Ω2, then one must have x∈Ω.Now, for eachi=1,2, let Fi:K×K→ℝ be defined by Fi(x,y)∶=hi(y)-hi(x). Let us now find the following equilibrium problem: find x∈K such that(4.8)Fi(x,y)≥0,i=1,2,
for all y∈K. It is obvious that Fi satisfies conditions (A1)–(A4) and EP(Fi)=Ωi, i=1,2, where EP(Fi) is the set of solutions to (4.8). By Theorem 3.1, we have the following theorem.Theorem 4.3.
LetK be a nonempty closed and convex subset of a real Hilbert space H. For each i=1,2, let hi be a lower semicontinuous and convex function such that Ω1∩Ω2≠∅. Let {zn}n=1∞, {un}n=1∞, {yn}n=1∞, and {xn}n=0∞ be generated by x0∈K, C1=K, x1=PC1x0,
(4.9)h1(y)-h1(zn)+1rn〈y-zn,zn-xn〉≥0,∀y∈K,h(y)-h(un)+1λn〈y-un,un-zn〉≥0,∀y∈K,yn=αnx0+(1-αn)un,Cn+1={z∈Cn:‖yn-z‖2≤‖xn-z‖2+αn(‖x0‖2+2〈xn-x0,z〉)},xn+1=PCn+1x0,n≥1.
Assume that {αn}n=1∞⊂(0,1), {rn}n=1∞⊂(0,∞), and {λn}n=1∞⊂(0,∞) satisfy(i)
liminfn→∞rn>0,(ii)
liminfn→∞λn>0,(iii)
limn→∞αn=0.Then, {xn}n=0∞ converges strongly to PΩ1∩Ω2x0.Remark 4.4.
Our results in this paper also hold for infinite family of uniformly continuous quasi-nonexpansive mappings in a real Hilbert space.



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*Source: 734082-2011-04-17.xml*