# Strong Convergence of a Hybrid Projection Algorithm for Equilibrium Problems, Variational Inequality Problems and Fixed Point Problems in a Banach Space

Authors: Wariam Chuayjan; Sornsak Thianwan
Journal: Abstract and Applied Analysis
 (2009)

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## Abstract

We introduce and study a new hybrid projection algorithm for finding a common element of the set of solutions of an equilibrium problem, the set of common fixed points of relatively quasi-nonexpansive mappings, and the
set of solutions of the variational inequality for an inverse-strongly-monotone operator in a Banach space. Under suitable assumptions, we show a strong convergence theorem. Using this result, we obtain some applications in a Banach space. The results obtained in this paper extend and improve the several recent results in this area.


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## Body

## 1. Introduction

LetE be a real Banach space with norm ∥·∥ and E* the dual space of E. Let C be a nonempty closed and convex subset of E, and A a monotone operator of C into E*. A mapping T:C→C is called nonexpansive if ∥Tx-Ty∥<∥x-y∥ for all x,y∈C. We denote by F(T)={x∈E:Tx=x} the set of fixed points of T. The classical variational inequality problem [1, 2], denoted by VI(C,A), is to find u∈C such that
(1.1)〈Au,v-u〉≥0
for all v∈C. One can see that the variational inequality problem (1.1) is connected with the convex minimization problem, the complementarity problem, the problem of finding a point u∈E satisfying 0=Au, and so on.Recall that an operatorA is called monotone if 〈Ax-Ay,x-y〉≥0 for all x,y∈C. An operator A of C into E* is said to be α-inverse-strongly-monotone [3–5] if each x,y∈C. We have 〈Ax-Ay,x-y〉≥α∥Ax-Ay∥2, for a constant α>0.Assume that the following hold:(C1)
A is α-inverse-strongly-monotone,(C2)
VI(C,A)≠∅,(C3)
∥Ay∥≤∥Ay-Au∥ for all y∈C and u∈VI(C,A).For finding a solution of the variational inequality problem for an operatorA that satisfies conditions  (C1)–(C3) in a 2-uniformly convex and uniformly smooth Banach space E, Iiduka and Takahashi [6] introduced and studied the following algorithm: x1=x∈C, define a sequence {xn} by
(1.2)xn+1=ΠCJ-1(Jxn-λnAxn)
for every n=1,2,…, where J is the duality mapping from E into E*, ΠC is the generalized projection from E onto C, and {λn} is a sequence of positive real numbers. They proved that under certain appropriate conditions imposed on {λn}, and J is weakly sequentially continuous, the sequence {xn} generated by (1.2) converges weakly to some element z in VI(C,A), where z=limn→∞ΠVI(C,A)(xn).In 2004, Matsushita and Takahashi [7] introduced the following iteration: x0∈C chosen arbitrarily,
(1.3)xn+1=ΠCJ-1(αnJxn+(1-αn)JTxn),
where {αn} is a real sequence in [0,1], T is a relatively nonexpansive mapping, and ΠC denotes the generalized projection from E onto a closed convex subset C of E. They prove that the sequence {xn} generated by (1.3) converges weakly to a fixed point of T.The problem of finding a common element of the set of the variational inequalities for monotone mappings in the framework of Hilbert spaces and Banach spaces has been intensively studied by many authors; see, for instance, [3–5] and the references cited therein.Letf be a bifunction of C×C into ℝ, where ℝ is the set of real numbers. The equilibrium problem for f:C×C→ℝ is to find x̂∈C such that
(1.4)f(x̂,y)≥0
for all y∈C. The set of solutions of (1.4) is denoted by EP(f), that is, EP(f)={x̂∈C:f(x̂,y)≥0forally∈C}. Given a mapping T:C→E*, let f(x,y)=〈Tx,y-x〉 for all x,y∈C. Then x̂∈EP(f) if and only if 〈Tx̂,y-x̂〉≥0 for all y∈C, that is, x̂ is a solution of the variational inequality. Many problems in physics, optimization, and economics reduce to finding a solution of (1.4). Equilibrium problems have been studied extensively; see, for instance, [8, 9]. Combettes and Hirstoaga [8] introduced an iterative scheme for finding the best approximation to the initial data when EP(f) is nonempty and proved a strong convergence theorem.For solving the equilibrium problem, let us assume that a bifunctionf 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 allx,y,z∈C,
(1.5)lim supt↓0f(tz+(1-t)x,y)≤f(x,y);(A4)
for allx∈C,f(x,·) is convex and lower semicontinuous.The problem of finding a common element of the set of fixed points and the set of solutions of an equilibrium problem in the framework of Hilbert spaces and Banach spaces has been studied by many authors; see [7, 8, 10–18].In 2008, Takahashi and Zembayashi [15] introduced the shrinking projection method which is the modification of (1.3) for a relatively nonexpansive mapping. It is given as follows:
(1.6)x0=x∈C,C0=C,yn=J-1(αnJxn+(1-αn)JTxn),un∈Csuchthatf(un,y)+1rn〈y-un,Jun-Jyn〉≥0,∀y∈C,Cn+1={z∈Cn:ϕ(z,un)≤ϕ(z,xn)},xn+1=ΠCn+1x0
for every n∈ℕ∪{0}, where J is the duality mapping on E,{αn}⊂[0,1] satisfying lim infn→∞αn(1-αn)>0 and {rn}⊂[s,∞) for some s>0. They proved that the sequence {xn} generated by (1.6) converges strongly to q=ΠF(T)∩EP(f)x0, where ΠF(T)∩EP(f) is the generalized projection of E onto F(T)∩EP(f).In the same year, Qin et al. [19] extend the iteration process (1.6) from a single relatively nonexpansive mapping to two relatively quasi-nonexpansive mappings: x0∈E,C1=C and x1=ΠC1x0,(1.7)yn=J-1(αnJxn+βnJTxn+γnJSxn),un∈Csuchthatf(un,y)+1rn〈y-un,Jun-Jyn〉≥0,∀y∈C,Cn+1={z∈Cn:ϕ(z,un)≤ϕ(z,xn)},xn+1=ΠCn+1x0
for every n∈ℕ∪{0}. Under appropriate conditions imposed on {αn},{βn},{γn}, and {rn}, they obtain that the sequence {xn} generated by (1.7) converges strongly to q=ΠF(T)∩F(S)∩EP(f)x0.In 2009, Wattanawitoon and Kumam [17] introduced the following iterative scheme which is the modification of (1.6) and (1.7) in a Banach space: x0∈E,C1=C, and x1=ΠC1x0,(1.8)yn=J-1(δnJxn+(1-δn)Jzn),zn=J-1(αnJxn+βnJTxn+γnJSxn),un∈Csuchthatf(un,y)+1rn〈y-un,Jun-Jzn〉≥0,∀y∈C,Cn+1={z∈Cn:ϕ(z,yn)≤ϕ(z,xn)},xn+1=ΠCn+1x0
for finding a common element of the set of solutions of an equilibrium problem and the set of common fixed points of two relatively quasi-nonexpansive mappings in a Banach space. They proved that under certain appropriate conditions imposed on {αn},{βn},{γn},{δn}, and {rn}, the sequences {xn} and {un} generated by (1.8) converge strongly to q∈F(T)∩F(S)∩EP(f), where q=ΠF(T)∩F(S)∩EP(f)x0.For finding common elements of the set of the equilibrium problem, the set of the variational inequality problem for an inverse-strongly-monotone operator and the set of common fixed points for relatively quasi-nonexpansive mappings. Cholamjiak [20] introduced an iterative scheme by using the new hybrid method in a Banach space. The scheme is defined as follows: x0∈E,C1=C and x1∈ΠC1x0,,
(1.9)zn=ΠCJ-1(Jxn-λnAxn),yn=J-1(αnJxn+βnJTxn+γnJSzn),un∈Csuchthatf(un,y)+1rn〈y-un,Jun-Jyn〉≥0,∀y∈C,Cn+1={z∈Cn:ϕ(z,un)≤ϕ(z,xn)},xn+1=ΠCn+1x0
for every n∈ℕ, where J is the duality mapping on E and {αn},{βn}, and {γn} are sequences in [0,1]. He proved that under certain appropriate conditions imposed on {αn},{βn},{γn},{rn}, and {λn}, the sequences {xn} and {un} generated by (1.9) converge strongly to q=ΠF(T)∩F(S)∩EP(f)∩VI(C,A)x0.Motivated by the recent works, we introduce an iterative scheme by a new hybrid method as follows:x0∈E,C1=C and x1=ΠC1x0,
(1.10)wn=ΠCJ-1(Jxn-λnAxn),yn=J-1(δnJxn+(1-δn)Jwn),zn=J-1(αnJxn+βnJTxn+γnJSyn),un∈Csuchthatf(un,y)+1rn〈y-un,Jun-Jzn〉≥0,∀y∈C,Cn+1={z∈Cn:ϕ(z,un)≤ϕ(z,xn)},xn+1=ΠCn+1x0
for every n∈ℕ, where J is the duality mapping on E, ΠC is the generalized projection from E onto a closed convex subset C of E, f is an equilibrium bifunction satisfying (A1)–(A4), A is an operator of C into E* satisfying (C1)–(C3), T,S:C→C are two closed relatively quasi-nonexpansive mappings, {αn},{βn,{γn}, and {δn} are sequences in [0,1] such that αn+βn+γn=1,lim supn→∞δn<1, lim infn→∞αnβn>0, and lim infn→∞αnγn>0,{rn}⊂[s,∞) for some s>0 and {λn}⊂[a,b] for some a,b with 0<a<b<c2α/2, where 1/c is the 2-uniformly convexity constant of E. We prove that the sequences {xn} and {un} generated by the above iterative scheme converge strongly to q=ΠF(T)∩F(S)∩EP(f)∩VI(C,A)x0.


## 2. Preliminaries

In this section, we recall some well know concepts and results.LetE be a real Banach space with dimension E≥2. The modulus of E is the function δE:(0,2]→[0,1] defined by
(2.1)δE(ε)=inf{1-∥x+y2∥:x,y∈E,∥x∥=∥y∥=1,∥x-y∥≥ε}.
Banach space E is uniformly convex if and only if δE(ε)>0 for all ε∈(0,2]. Let p be a fixed real number with p≥2. A Banach space E is said to be p-uniformly convex if there exists a constant c>0 such that δ(ε)≥cεp for all ε∈[0,2]; see [21–23] for more details. A Banach space E is said to be smooth if the limit
(2.2)limt→0∥x+ty∥-∥x∥t
exists for all x,y∈U, where U denotes the unit sphere of E (i.e., U={x∈E:∥x∥=1}). It is also said to be uniformly smooth if the limit (2.2) is attained uniformly for x,y∈U. One should note that no Banach space is p-uniformly convex for 1<p<2; see [23]. It is well known that a Hilbert space is 2-uniformly convex, uniformly smooth. For each p>1, the generalized duality mappingJp:E→2E* is defined by
(2.3)Jp(x)={x*∈E*:〈x,x*〉=∥x∥p,∥x*∥=∥x∥p-1}
for all x∈E. In particular, J=J2 is called the normalized duality mapping. If E is a Hilbert space, then J=I, where I is the identity mapping. It is also known that if E is uniformly smooth, then J is uniformly norm-to-norm continuous on each bounded subset of E. See [24, 25] for more details.LetE be a smooth, strictly convex, and reflexive Banach space and let C be a nonempty closed convex subset of E. We denote by ϕ the function defined by
(2.4)ϕ(x,y)=∥x∥2-2〈x,Jy〉+∥y∥2
for all x,y∈E. Following Alber [26], the generalized projection ΠC:E→C is a map that assigns to an arbitrary point x∈E the minimum point of the functional ϕ(x,y), that is, ΠCx=x¯, where x¯ is the solution to the minimization problem
(2.5)ϕ(x¯,x)=infy∈Cϕ(y,x).
The existence and uniqueness of the operator ΠC follows from the properties of the functional ϕ(x,y) and strict monotonicity of the mapping J (see, e.g., [24, 26–29]. In Hilbert space, ΠC=PC. It is obvious from the definition of function ϕ that
(2.6)(∥y∥-∥x∥)2≤ϕ(y,x)=(∥y∥+∥x∥)2
for all x,y∈E. If E is a Hilbert space, then ϕ(x,y)=∥x-y∥2.IfE is a reflexive, strictly convex, and smooth Banach space, then for x,y∈E,ϕ(x,y)=0 if and only if x=y. It is sufficient to show that if ϕ(x,y)=0, then x=y. From (2.6), we have ∥x∥=∥y∥. This implies that 〈x,Jy〉=∥x∥2=∥Jy∥2. From the definition of J, one has Jx=Jy. Therefore, we have x=y; see [24, 28] for more details.Recall that a pointp in a closed convex subset C of E is said to be an asymptotic fixed point of T [30] if C contains a sequence {xn} which converges weakly to p such that limn→∞∥xn-Txn∥=0. The set of asymptotic fixed point of T will be denoted by F̂(T). A mapping T:C→C is called relatively nonexpansive [11, 31–33] if T satisfies the following conditions:(1)
F(T)≠∅;(2)
ϕ(p,Tx)≤ϕ(p,x) for all p∈F(T) and x∈C;(3)
F̂(T)=F(T).The asymptotic behavior of a relatively nonexpansive mapping was studied in [31–33].A mappingT is said to be relatively quasi-nonexpansive if T satisfies conditions (1) and (2). It is easy to see that the class of relatively quasi-nonexpansive mappings is more general than the class of relatively nonexpansive mappings [11, 31–33] which requires the strong restriction: F(T)=F̂(T).We give some examples which are closed relatively quasi-nonexpansive; see [19].Example 2.1.
LetE be a uniformly smooth and strictly convex Banach space and A⊂E×E* be a maximal monotone mapping such that its zero set A-10≠∅. Then, Jr=(J+rA)-1J is a closed relatively quasi-nonexpansive mapping from E onto D(A) and F(Jr)=A-10.Example 2.2.
LetΠC be the generalized projection from a smooth, strictly convex, and reflexive Banach space E onto a nonempty closed convex subset C of E. Then, ΠC is a closed relatively quasi-nonexpansive mapping with F(ΠC)=C.An operatorA of C into E* is said to be hemicontinuous if for all x,y∈C, the mapping F of [0,1] into E* defined by F(t)=A(tx+(1-t)y) is continuous with respect to the weak* topology of E*. We define by NC(v) the normal cone for C at a point v∈C, that is,
(2.7)NC(v)={x*∈E*:〈v-y,x*〉≥0∀y∈C}.LetC be a nonempty, closed convex subset of a Banach space E and A a monotone, hemicontinuous operator of C into E*. Let Te⊂E×E* be an operator defined as follows:
(2.8)Tev={Av+NC(v),v∈C,∅,otherwise.
Then, Te is maximal monotone and Te-10=VI(C,A); see [34].In the sequel, the following lemmas are needed to prove our main results.Lemma 2.3 ([22, 35]).
Letp be a given real number with p≥2 and E a p-uniformly convex Banach space. Then, for all x,y∈E, jx∈Jp(x) and jy∈Jp(y),
(2.9)〈x-y,jx-jy〉≥cp2p-2p∥x-y∥p,
where Jp is the generalized duality mapping of E and 1/c is the p-uniformly convexity constant of E.Lemma 2.4 ([29]).
LetE be a uniformly convex and smooth Banach space and let {xn} and {yn} be two sequences of E. If ϕ(xn,yn)→0 and either {xn} or {yn} is bounded, then xn-yn→0.Lemma 2.5 ([26]).
LetC be a nonempty closed convex subset of a smooth Banach space E and x∈E. Then x0=ΠCx if and only if 〈x0-y,Jx-Jx0〉≥0 for all y∈C.Lemma 2.6 ([26]).
LetC be a nonempty closed convex subset of a reflexive, strictly convex, and smooth Banach space E and let x∈E. Then
(2.10)ϕ(y,ΠCx)+ϕ(ΠCx,x)≤ϕ(y,x),∀y∈C.Lemma 2.7 ([19]).
LetE be a uniformly convex, smooth Banach space, let C be a closed convex subset of E, let T be a closed and relatively quasi-nonexpansive mapping from C into itself. Then F(T) is a closed convex subset of C.Lemma 2.8 ([36]).
LetE be a uniformly convex Banach space and Br(0) be a closed ball of E. Then there exists a continuous strictly increasing convex function g:[0,∞)→[0,∞) with g(0)=0 such that
(2.11)∥αx+βy+γz∥2≤α∥x∥2+β∥y∥2+γ∥z∥2-αβg(∥x-y∥)
for all x,y,z∈Br(0) and α,β,γ∈[0,1] with α+β+γ=1.Lemma 2.9 ([10]).
LetC be a closed convex subset of a smooth, strictly convex, and reflexive Banach space E, let f be a bifunction from C×C to R satisfying (A1)–(A4), and let r>0 and x∈E. Then, there exists z∈C such that
(2.12)f(z,y)+1r〈y-z,Jz-Jx〉≥0,∀y∈C.Lemma 2.10 ([19]).
LetC be a closed convex subset of a uniformly smooth, strictly convex, and reflexive Banach space E, and let f be a bifunction from C×C to R satisfying (A1)–(A4). For all r>0 and x∈E, define a mapping Tr:E→C as follows:
(2.13)Trx={z∈C:f(z,y)+1r〈y-z,Jz-Jx〉≥0,∀y∈C}.Then, the following hold:(1)
Tr is single-valued;(2)
Tr is a firmly nonexpansive-type mapping [37], that is, for all x,y∈E,
(2.14)〈Trx-Try,JTrx-JTry〉≤〈Trx-Try,Jx-Jy〉;(3)
F(Tr)=EP(f);(4)
EP(f) is closed and convex.Lemma 2.11 ([16]).
LetC be a closed convex subset of a smooth, strictly, and reflexive Banach space E, let f be a bifucntion from C×C to R satisfying (A1)–(A4), let r>0. Then, for all x∈E and q∈F(Tr),
(2.15)ϕ(q,Trx)+ϕ(Trx,x)≤ϕ(q,x).We make use of the following mappingV studied in Alber [26]:
(2.16)V(x,x*)=∥x∥2-2〈x,x*〉+∥x*∥2
for all x∈E and x*∈E*, that is, V(x,x*)=ϕ(x,J-1(x*)).Lemma 2.12 ([26]).
LetE be a reflexive, strictly convex, smooth Banach space and let V be as in (2.16). Then
(2.17)V(x,x*)+2〈J-1(x*)-x,y*〉≤V(x,x*+y*)
for all x∈E and x*,y*∈E*.


## 3. Main Results

In this section, we prove strong convergence theorems by hybrid methods which solves the problem of finding a common element of the set of solutions of an equilibrium problem, the set of common fixed points of relatively quasi-nonexpansive mappings and the set of solutions of the variational inequality of anα-inverse-strongly-monotone mapping in a 2-uniformly convex, uniformly smooth Banach space.Theorem 3.1.
LetE be a 2-uniformly convex, uniformly smooth Banach space, C a nonempty closed convex subset of E, f a bifunction from C×C to ℝ which satisfies (A1)–(A4), A an operator of C into E* satisfying (C1)–(C3), and T,S two closed relatively quasi-nonexpansive mappings from C into itself such that the set F:=F(T)∩F(S)∩EP(f)∩VI(C,A)≠∅. For an initial point x0∈E with x1=ΠC1x0 and C1=C, define sequences {xn} and {un} of C as follows:
(3.1)wn=ΠCJ-1(Jxn-λnAxn),yn=J-1(δnJxn+(1-δn)Jwn),zn=J-1(αnJxn+βnJTxn+γnJSyn),un∈Csuchthatf(un,y)+1rn〈y-un,Jun-Jzn〉≥0,∀y∈C,Cn+1={z∈Cn:ϕ(z,un)≤ϕ(z,xn)},xn+1=ΠCn+1x0
for every n∈ℕ, where J is the duality mapping on E. Suppose that {αn},{βn},{γn}, and {δn} are sequences in [0,1] satisfying the restrictions:(B1)
αn+βn+γn=1;(B2)
lim supn→∞δn<1, lim infn→∞αnβn>0 and lim infn→∞αnγn>0;(B3)
{rn}⊂[s,∞) for some s>0;(B4)
{λn}⊂[a,b] for some a,b with 0<a<b<c2α/2, where 1/c is the 2-uniformly convexity constant of E.
Then,{xn} and {un} converge strongly to q=ΠFx0.Proof.
We divide the proof into eight steps.Step 1.
Show thatΠFx0 and ΠCn+1x0 are well defined.
It is obvious thatVI(C,A) is a closed convex subset of C. By Lemma 2.7, we know that F(T)∩F(S) is closed and convex. From Lemma 2.10 (4), we also have EP(f) is closed and convex. Hence, F:=F(T)∩F(S)∩EP(f)∩VI(C,A) is a nonempty, closed, and convex subset of C; consequently, ΠFx0 is well defined.
Next, we show thatCn is closed and convex for all n∈ℕ.
It is obvious that,C1=C is closed and convex. Suppose that Ck is closed and convex for some k∈ℕ. For all z∈Ck, we know that ϕ(z,uk)≤ϕ(z,xk) is equivalent to
(3.2)2〈z,Jxk-Juk〉≤∥xk∥2-∥uk∥2.
So, Ck+1 is closed and convex. Then, for any n∈ℕ, Cn is closed and convex. This implies that ΠCn+1x0 is well defined.Step 2.
We prove by induction thatF⊂Cn for all n∈ℕ.
Puttingvn=J-1(Jxn-λnAxn). First, we observe that un=Trnzn for all n∈ℕ and F⊂C1=C. Suppose that F⊂Ck for some k∈ℕ. Let u∈F⊂Ck. From Lemmas 2.6 and 2.12, we have
(3.3)ϕ(u,wk)=ϕ(u,ΠCvk)≤ϕ(u,vk)=ϕ(u,J-1(Jxk-λkAxk))=V(u,Jxk-λkAxk)≤V(u,(Jxk-λkAxk)+λkAxk)-2〈J-1(Jxk-λkAxk)-u,λkAxk〉=V(u,Jxk)-2λk〈vk-u,Axk〉=ϕ(u,xk)-2λk〈xk-u,Axk〉+2〈vk-xk,-λkAxk〉.
Using (C1) and u∈VI(C,A), we have
(3.4)-2λk〈xk-u,Axk〉=-2λk〈xk-u,Axk-Au+Au〉=-2λk〈xk-u,Axk-Au〉-2λk〈xk-u,Au〉≤-2αλk∥Axk-Au∥2.
By using Lemma 2.3 and (C3), we have
(3.5)2〈vk-xk,-λkAxk〉=2〈J-1(Jxk-λkAxk)-J-1(Jxk),-λkAxk〉≤2∥J-1(Jxk-λkAxk)-J-1(Jxk)∥∥λkAxk∥≤4c2∥JJ-1(Jxk-λkAxk)-JJ-1(Jxk)∥∥λkAxk∥=4c2∥Jxk-λkAxk-Jxk∥∥λkAxk∥=4c2λk2∥Axk∥2≤4c2λk2∥Axk-Au∥2.
Replacing (3.4) and (3.5) into (3.3) and using (B4), we get
(3.6)ϕ(u,wk)≤ϕ(u,xk)+2a(2c2b-α)∥Axk-Au∥2≤ϕ(u,xk).
By the convexity of∥·∥2 and (3.6), for each u∈F⊂Ck, we have
(3.7)ϕ(u,yk)=ϕ(u,J-1(δkJxk+(1-δk)Jwk))=V(u,δkJxk+(1-δk)Jwk)=∥u∥2-2〈u,δkJxk+(1-δk)Jwk〉+∥δkJxk+(1-δk)Jwk∥2≤∥u∥2-2δk〈u,Jxk〉-2(1-δk)〈u,Jwk〉+δk∥Jxk∥2+(1-δk)∥Jwk∥2=(δk+(1-δk))∥u∥2-2δk〈u,Jxk〉-2(1-δk)〈u,Jwk〉+δk∥Jxk∥2+(1-δk)∥Jwk∥2=δk(∥u∥2-2〈u,Jxk〉+∥Jxk∥2)+(1-δk)(∥u∥2-2〈u,Jwk〉+∥Jwk∥2)=δkϕ(u,xk)+(1-δk)ϕ(u,wk)≤δkϕ(u,xk)+(1-δk)ϕ(u,xk)=ϕ(u,xk),
and so
(3.8)ϕ(u,uk)=ϕ(u,Trkzk)≤ϕ(u,zk)=ϕ(u,J-1(αkJxk+βkJTxk+γkJSyk))=V(u,αkJxk+βkJTxk+γkJSyk)=∥u∥2-2αk〈u,Jxk〉-2βk〈u,JTxk〉-2γk〈u,JSyk〉+∥αkJxk+βkJTxk+γkJSyk∥2≤∥u∥2-2αk〈u,Jxk〉-2βk〈u,JTxk〉-2γk〈u,JSyk〉+αk∥Jxk∥2+βk∥JTxk∥2+γk∥JSyk∥2=(αk+βk+γk)∥u∥2-2αk〈u,Jxk〉-2βk〈u,JTxk〉-2γk〈u,JSyk〉+αk∥Jxk∥2+βk∥JTxk∥2+γk∥JSyk∥2=αk(∥u∥2-2〈u,Jxk〉+∥Jxk∥2)+βk(∥u∥2-2〈u,JTxk〉+∥JTxk∥2)+γk(∥u∥2-2〈u,JSyk〉+∥JSyk∥2)=αkϕ(u,xk)+βkϕ(u,Txk)+γkϕ(u,Syk)≤αkϕ(u,xk)+βkϕ(u,xk)+γkϕ(u,yk)≤αkϕ(u,xk)+βkϕ(u,xk)+γkϕ(u,xk)=(αk+βk+γk)ϕ(u,xk)=ϕ(u,xk).
This shows that u∈Ck+1 and hence F⊂Ck+1. This implies that F⊂Cn for all n∈ℕ.Step 3.
Show thatlimn→∞ϕ(xn,x0) exists.
Fromxn=ΠCnx0 and xn+1=ΠCn+1x0∈Cn+1⊂Cn, we have
(3.9)ϕ(xn,x0)≤ϕ(xn+1,x0),∀n∈ℕ.
Therefore, {ϕ(xn,x0)} is nondecreasing. From Lemma 2.6, we have
(3.10)ϕ(xn,x0)=ϕ(ΠCnx0,x0)≤ϕ(u,x0)-ϕ(u,xn)≤ϕ(u,x0).
Then the sequence {ϕ(xn,x0)} is bounded. It follows that limn→∞ϕ(xn,x0) exists.Step 4.
Show that{xn} is a Cauchy sequence in C.
Sincexm=ΠCmx0∈Cm⊂Cn for any positive integer m≥n, by Lemma 2.6, we also have
(3.11)ϕ(xm,xn)=ϕ(xm,ΠCnx0)≤ϕ(xm,x0)-ϕ(ΠCnx0,x0)=ϕ(xm,x0)-ϕ(xn,x0).
Letting m,n→∞ in (3.11), we have ϕ(xm,xn)→0. It follows from Lemma 2.4 that ∥xm-xn∥→0 as m,n→∞. Hence, {xn} is a Cauchy sequence. By the completeness of E and the closedness of C, one can assume that xn→q∈C as n→∞. Further, we obtain
(3.12)limn→∞ϕ(xn+1,xn)=0.
Since xn+1=ΠCn+1x0∈Cn+1, we have
(3.13)ϕ(xn+1,un)≤ϕ(xn+1,xn)→0(asn→∞).
Applying Lemma 2.4 to (3.12) and (3.13), we have
(3.14)∥un-xn∥=∥un-xn+1+xn+1-xn∥≤∥un-xn+1∥+∥xn+1-xn∥→0(asn→∞).
This implies that un→q as n→∞.  Since J is uniformly norm-to-norm continuous on bounded subsets of E, we also obtain
(3.15)limn→∞∥Jun-Jxn∥=0.Step 5.
Show thatxn→q∈F(T)∩F(S).
Letr=supn≥1{∥xn∥,∥Txn∥,∥Swn∥}. From (3.7) and Lemma 2.8, we obtain
(3.16)ϕ(u,un)=ϕ(u,Trnzn)≤ϕ(u,zn)=ϕ(u,J-1(αnJxn+βnJTxn+γnJSyn))=V(u,αnJxn+βnJTxn+γnJSyn)=∥u∥2-2αn〈u,Jxn〉-2βn〈u,JTxn〉-2γn〈u,JSyn〉+∥αnJxn+βnJTxn+γnJSyn∥2≤∥u∥2-2αn〈u,Jxn〉-2βn〈u,JTxn〉-2γn〈u,JSyn〉+αn∥Jxn∥2+βn∥JTxn∥2+γn∥JSyn∥2-αnβng(∥Jxn-JTxn∥)=(αn+βn+γn)∥u∥2-2αn〈u,Jxn〉-2βn〈u,JTxn〉-2γn〈u,JSyn〉+αn∥Jxn∥2+βn∥JTxn∥2+γn∥JSyn∥2-αnβng(∥Jxn-JTxn∥)=αnϕ(u,xn)+βnϕ(u,Txn)+γnϕ(u,Syn)-αnβng(∥Jxn-JTxn∥)≤αnϕ(u,xn)+βnϕ(u,xn)+γnϕ(u,yn)-αnβng(∥Jxn-JTxn∥)≤αnϕ(u,xn)+βnϕ(u,xn)+γnϕ(u,xn)-αnβng(∥Jxn-JTxn∥)=ϕ(u,xn)-αnβng(∥Jxn-JTxn∥).
This implies that
(3.17)αnβng(∥Jxn-JTxn∥)≤ϕ(u,xn)-ϕ(u,un)=∥xn∥2-∥un∥2-2〈u,Jxn-Jun〉≤∥xn-un∥(∥xn∥+∥un∥)+2∥u∥∥Jxn-Jun∥.
It follows from (3.14), (3.15), and (B2) that
(3.18)limn→∞g(∥Jxn-JTxn∥)=0.
Since g is strictly increasing and continuous at 0 with g(0)=0, it follows that
(3.19)limn→∞∥Jxn-JTxn∥=0.
Since J is uniformly norm-to-norm continuous on bounded sets, so is J-1. Then
(3.20)limn→∞∥xn-Txn∥=limn→∞∥J-1(Jxn)-J-1(JTxn)∥=0.
In the same manner, we can show that
(3.21)limn→∞∥xn-Syn∥=0.
In addition, ϕ(u,yn)≤δnϕ(u,xn)+(1-δn)ϕ(u,wn), using (3.6), we have
(3.22)ϕ(u,un)≤αnϕ(u,xn)+βnϕ(u,xn)+γnϕ(u,yn)≤αnϕ(u,xn)+βnϕ(u,xn)+γn(δnϕ(u,xn)+(1-δn)ϕ(u,wn))=αnϕ(u,xn)+βnϕ(u,xn)+γnδnϕ(u,xn)+γn(1-δn)ϕ(u,wn)≤αnϕ(u,xn)+βnϕ(u,xn)+γnδnϕ(u,xn)+γn(1-δn)(ϕ(u,xn)+2a(2c2b-α)∥Axn-Au∥2)=αnϕ(u,xn)+βnϕ(u,xn)+γnδnϕ(u,xn)+γn(1-δn)ϕ(u,xn)+γn(1-δn)2a(2c2b-α)∥Axn-Au∥2=(αn+βn+γn)ϕ(u,xn)-γn(1-δn)2a(α-2c2b)∥Axn-Au∥2=ϕ(u,xn)-γn(1-δn)2a(α-2c2b)∥Axn-Au∥2,
which leads to the following:
(3.23)γn(1-δn)2a(α-2c2b)∥Axn-Au∥2≤ϕ(u,xn)-ϕ(u,un).
Since lim supn→∞δn<1,0<lim infn→∞αnγn≤lim infn→∞γn and from (3.17), we observe
(3.24)limn→∞(ϕ(u,xn)-ϕ(u,un))=0,
which yields that
(3.25)limn→∞∥Axn-Au∥=0.
From Lemmas 2.6 and 2.12, and (3.5), we have
(3.26)ϕ(xn,wn)=ϕ(xn,ΠCvn)≤ϕ(xn,vn)=ϕ(xn,J-1(Jxn-λnAxn))=V(xn,Jxn-λnAxn)≤V(xn,(Jxn-λnAxn)+λnAxn)-2〈J-1(Jxn-λnAxn)-xn,λnAxn〉=ϕ(xn,xn)+2〈vn-xn,-λnAxn〉=2〈vn-xn,-λnAxn〉≤4c2b2∥Axn-Au∥2→0(asn→∞).
It follows from Lemma 2.4 and (3.25) that
(3.27)limn→∞∥xn-wn∥=0.
Hence wn→q as n→∞ and
(3.28)limn→∞∥Jxn-Jwn∥=0.
By using (3.26), we have
(3.29)ϕ(xn,yn)=ϕ(xn,J-1(δnJxn+(1-δn)Jwn))=V(u,δnJxn+(1-δn)Jwn)=∥xn∥2-2〈xn,δnJxn+(1-δn)Jwn〉+∥δnJxn+(1-δn)Jwn∥2≤∥xn∥2-2δn〈xn,Jxn〉-2(1-δn)〈xn,Jwn〉+δn∥Jxn∥2+(1-δn)∥Jwn∥2=(δn+(1-δn))∥xn∥2-2δn〈xn,Jxn〉-2(1-δn)〈xn,Jwn〉+δn∥Jxn∥2+(1-δn)∥Jwn∥2=δn∥xn∥2+(1-δn)∥xn∥2-2δn〈xn,Jxn〉-2(1-δn)〈xn,Jwn〉+δn∥Jxn∥2+(1-δn)∥Jwn∥2=δnϕ(xn,xn)+(1-δn)ϕ(xn,wn)≤ϕ(xn,wn)→0(asn→∞).
Applying Lemma 2.4, we get
(3.30)limn→∞∥xn-yn∥=0.
Hence, yn→q as n→∞.
In addition,(3.31)∥Syn-yn∥=∥Syn-xn+xn-yn∥≤∥Syn-xn∥+∥xn-yn∥.
It follows from (3.21), (3.30), and (3.31) that
(3.32)limn→∞∥Syn-yn∥=0.
From (3.20), (3.32) and by the closedness of T and S, we get q∈F(T)∩F(S).Step 6.
Show thatxn→q∈EP(f).
From (3.16), we have
(3.33)ϕ(u,zn)≤ϕ(u,xn).
Note that un=Trnzn. From (3.33) and Lemma 2.11, we have
(3.34)ϕ(un,zn)=ϕ(Trnzn,zn)≤ϕ(u,zn)-ϕ(u,Trnzn)≤ϕ(u,xn)-ϕ(u,Trnzn)=ϕ(u,xn)-ϕ(u,un).
Using (3.24), we have limn→∞ϕ(un,zn)=0. By Lemma 2.4, we obtain
(3.35)limn→∞∥un-zn∥=0.
Since rn≥s, we have
(3.36)∥Jun-Jzn∥rn→0asn→∞.
From un=Trnzn, we have
(3.37)f(un,y)+1rn〈y-un,Jun-Jzn〉≥0,∀y∈C.
By using (A2), we have
(3.38)∥y-un∥∥Jun-Jzn∥rn≥1rn〈y-un,Jun-Jzn〉≥-f(un,y)≥f(y,un),∀y∈C.
From (A4) and un→q, we get f(y,q)≤0 for all y∈C. For t with 0<t≤1 and y∈C, let yt=ty+(1-t)q. Since y∈C and q∈C, we have yt∈C and hence f(yt,q)≤0. So from (A1), we have 0=f(yt,yt)≤tf(yt,y)+(1-t)f(yt,q)≤tf(yt,y). That is, f(yt,y)≥0. It follows from (A3) that f(q,y)≥0 for all y∈C and hence q∈EP(f).Step 7.
Show thatxn→q∈VI(C,A).
DefineTe⊂E×E* be as in (2.8), which yields that Te is maximal monotone and Te-10=VI(C,A). Let (v,w)∈G(Te). Since w∈Tev=Av+NC(v), we get w-Av∈NC(v). From wn∈C, we have
(3.39)〈v-wn,w-Av〉≥0.
On the other hand, from wn=ΠCJ-1(Jxn-λnAxn) and Lemma 2.5, we have 〈v-wn,Jwn-(Jxn-λnAxn)〉≥0, and hence
(3.40)〈v-wn,Jxn-Jwnλn-Axn〉≤0.
It follows from (3.39) and (3.40) that
(3.41)〈v-wn,w〉≥〈v-wn,Av〉≥〈v-wn,Av〉+〈v-wn,Jxn-Jwnλn-Axn〉=〈v-wn,Av-Axn〉+〈v-wn,Jxn-Jwnλn〉=〈v-wn,Av-Awn〉+〈v-wn,Awn-Axn〉+〈v-wn,Jxn-Jwnλn〉≥-∥v-wn∥∥wn-xn∥α-∥v-wn∥∥Jxn-Jwn∥a≥-M(∥wn-xn∥α+∥Jxn-Jwn∥a),
where M=supn≥1{∥v-wn∥}. By taking the limit as n→∞ and from (3.27) and (3.28), we obtain 〈v-q,w〉≥0. By the maximality of Te, we have q∈Te-10 and hence q∈VI(C,A). That is, q∈F.Step 8.
Show thatq=ΠFx0.
Fromxn=ΠCnx0, we have
(3.42)〈xn-z,Jx0-Jxn〉≥0,∀z∈Cn.
Since F⊂Cn, we also have
(3.43)〈xn-u,Jx0-Jxn〉≥0,∀u∈F.
By taking the limit in (3.43), we obtain that
(3.44)〈q-u,Jx0-Jq〉≥0,∀u∈F.
By Lemma 2.5, we can conclude that q=ΠFx0. From (3.14), we have limn→∞∥un-xn∥=0, and it follows that limn→∞∥un-q∥=0. This completes the proof.Finally, we prove two strong convergence theorems in a 2-uniformly convex, uniformly smooth Banach space by using Theorem3.1.First, we consider the problem of finding a zero point of an inverse-strongly-monotone operator ofE into E*. Assume that A satisfies the conditions:(D1)
A is α-inverse-strongly monotone,(D2)
A-1(0)={u∈E:Au=0}≠∅.Theorem 3.2.
LetE be a 2-uniformly convex, uniformly smooth Banach space, f a bifunction from E×E to ℝ which satisfies (A1)–(A4), A an operator of E into E* satisfying (D1)–(D2), and T,S two closed relatively quasi-nonexpansive mappings from E into itself such that F:=F(T)∩F(S)∩EP(f)∩A-1(0)≠∅. For an initial point x0∈E with x1=ΠC1x0 and C1=E, define sequences {xn} and {un} as follows:
(3.45)wn=J-1(Jxn-λnAxn),yn=J-1(δnJxn+(1-δn)Jwn),zn=J-1(αnJxn+βnJTxn+γnJSyn),un∈Csuchthatf(un,y)+1rn〈y-un,Jun-Jzn〉≥0,∀y∈E,Cn+1={z∈Cn:ϕ(z,un)≤ϕ(z,xn)},xn+1=ΠCn+1x0
for every n∈ℕ, where J is the duality mapping on E. Suppose that {αn},{βn},{γn}, and {δn} are sequences in [0,1] satisfying conditions  (B1)–(B4) of Theorem 3.1.
Then,{xn} and {un} converge strongly to q=ΠFx0.Proof.
PuttingC=E in Theorem 3.1, we have ΠE=I. We also have VI(E,A)=A-1(0) and then condition (C3) of Theorem 3.1 holds for all y∈E and u∈A-1(0). So, we obtain the desired result.Next, letK be a nonempty closed convex cone in E and A an operator of K into E*. We define its polar in E* to be the set
(3.46)K*={y*∈E*:〈x,y*〉≥0∀x∈K}.
Then the element u∈K is called a solution of the complementarity problem if
(3.47)Au∈K*,〈u,Au〉=0.
The set of solutions of the complementarity problem is denoted by C(K,A).Assume thatA is an operator satisfying the conditions:(E1)
A is α-inverse-strongly-monotone,(E2)
C(K,A)≠∅,(E3)
∥Ay∥≤∥Ay-Au∥ for all y∈K and u∈C(K,A).Theorem 3.3.
LetE be a 2-uniformly convex, uniformly smooth Banach space, K a nonempty closed convex cone in E, f a bifunction from K×K to ℝ which satisfies (A1)–(A4), A an operator of K into E* satisfying (E1)–(E3), and T,S two closed relatively quasi-nonexpansive mappings from K into itself such that F:=F(T)∩F(S)∩EP(f)∩C(K,A)≠∅. For an initial point x0∈E with x1=ΠC1x0 and C1=K, define sequences {xn} and {un} as follows:
(3.48)wn=ΠKJ-1(Jxn-λnAxn),yn=J-1(δnJxn+(1-δn)Jwn),zn=J-1(αnJxn+βnJTxn+γnJSyn),un∈Csuchthatf(un,y)+1rn〈y-un,Jun-Jzn〉≥0,∀y∈K,Cn+1={z∈Cn:ϕ(z,un)≤ϕ(z,xn)},xn+1=ΠCn+1x0
for every n∈ℕ, where J is the duality mapping on E. Suppose that {αn},{βn},{γn}, and {δn} are sequences in [0,1] satisfying conditions  (B1)–(B4) of Theorem 3.1.
Then,{xn} and {un} converge strongly to q=ΠFx0.Proof.
From [24, Lemma 7.1.1], we have VI(K,A)=C(K,A). So by Theorem 3.1, we obtain the desired result.



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*Source: 613524-2009-10-26.xml*