# Strong Convergence to Solutions of Generalized Mixed Equilibrium Problems with Applications

Authors: Prasit Cholamjiak; Suthep Suantai; Yeol Je Cho
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

We introduce a Halpern-type iteration for a generalized mixed equilibrium
problem in uniformly smooth and uniformly convex Banach spaces. Strong convergence
theorems are also established in this paper. As applications, we apply
our main result to mixed equilibrium, generalized equilibrium, and mixed variational
inequality problems in Banach spaces. Finally, examples and numerical
results are also given.


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## Body

## 1. Introduction

LetE be a real Banach space, C a nonempty, closed, and convex subset of E, and E* the dual space of E. Let T:C→C be a nonlinear mapping. The fixed points set of T is denoted by F(T), that is, F(T)={x∈C:x=Tx}.One classical way often used to approximate a fixed point of a nonlinear self-mappingT on C was firstly introduced by Halpern [1] which is defined by x1=x∈C and
(1.1)xn+1=αnx+(1-αn)Txn,∀n≥1,
where {αn} is a real sequence in [0,1]. He proved, in a real Hilbert space, a strong convergence theorem for a nonexpansive mapping T when αn=n-a for any a∈(0,1).Subsequently, motivated by Halpern [1], many mathematicians devoted time to study algorithm (1.1) in different styles. Several strong convergence results for nonlinear mappings were also continuously established in some certain Banach spaces (see also [2–9]).Letf:C×C→ℝ be a bifunction, A:C→E* a mapping, and φ:C→ℝ a real-valued function. The generalized mixed equilibrium problem is to find x̂∈C such that
(1.2)f(x̂,y)+〈Ax̂,y-x̂〉+φ(y)≥φ(x̂),∀y∈C.
The solutions set of (1.2) is denoted by GMEP(f,A,φ) (see Peng and Yao [10]).IfA≡0, then the generalized mixed equilibrium problem (1.2) reduces to the following mixed equilibrium problem: finding x̂∈C such that
(1.3)f(x̂,y)+φ(y)≥φ(x̂),∀y∈C.
The solutions set of (1.3) is denoted by MEP(f,φ) (see Ceng and Yao [11]).Iff≡0, then the generalized mixed equilibrium problem (1.2) reduces to the following mixed variational inequality problem: finding x̂∈C such that
(1.4)〈Ax̂,y-x̂〉+φ(y)≥φ(x̂),∀y∈C.
The solutions set of (1.4) is denoted by VI(C,A,φ) (see Noor [12]).Ifφ≡0, then the generalized mixed equilibrium problem (1.2) reduces to the following generalized equilibrium problem: finding x̂∈C such that
(1.5)f(x̂,y)+〈Ax̂,y-x̂〉≥0,∀y∈C.
The solutions set of (1.5) is denoted by GEP(f,A) (see Moudafi [13]).Ifφ≡0, then the mixed equilibrium problem (1.3) reduces to the following equilibrium problem: finding x̂∈C such that
(1.6)f(x̂,y)≥0,∀y∈C.
The solutions set of (1.6) is denoted by EP(f) (see Combettes and Hirstoaga [14]).Iff≡0, then the mixed equilibrium problem (1.3) reduces to the following convex minimization problem: finding x̂∈C such that
(1.7)φ(y)≥φ(x̂),∀y∈C.
The solutions set of (1.7) is denoted by CMP(φ).Ifφ≡0, then the mixed variational inequality problem (1.4) reduces to the following variational inequality problem: finding x̂∈C such that
(1.8)〈Ax̂,y-x̂〉≥0,∀y∈C.
The solutions set of (1.8) is denoted by VI(C,A) (see Stampacchia [7]).The problem (1.2) is very general in the sense that it includes, as special cases, optimization problems, variational inequalities, minimax problems, the Nash equilibrium problem in noncooperative games, and others. For more details on these topics, see, for instance, [14–34].For solving the generalized mixed equilibrium problem, let us assume the following [25]:(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, limsupt↓0f(tz+(1-t)x,y)≤f(x,y);(A4)
for allx∈C,f(x,·) is convex and lower semicontinuous.The purpose of this paper is to investigate strong convergence of Halpern-type iteration for a generalized mixed equilibrium problem in uniformly smooth and uniformly convex Banach spaces. As applications, our main result can be deduced to mixed equilibrium, generalized equilibrium, mixed variational inequality problems, and so on. Examples and numerical results are also given in the last section.


## 2. Preliminaries and Lemmas

In this section, we need the following preliminaries and lemmas which will be used in our main theorem.LetE be a real Banach space and let U={x∈E:∥x∥=1} be the unit sphere of E. A Banach space E is said to be strictly convex if, for any x,y∈U,
(2.1)x≠yimplies‖x+y2‖<1.
It is also said to be uniformly convex if, for any ε∈(0,2], there exists δ>0 such that, for any x,y∈U,
(2.2)‖x-y‖≥εimplies‖x+y2‖<1-δ.
It is known that a uniformly convex Banach space is reflexive and strictly convex. Define a function δ:[0,2]→[0,1] called the modulus of convexity of E as follows:
(2.3)δ(ε)=inf⁡{1-‖x+y2‖:x,y∈E,‖x‖=‖y‖=1,‖x-y‖≥ε}.
Then E is uniformly convex if and only if δ(ε)>0 for all ε∈(0,2]. A Banach space E is said to be smooth if the limit
(2.4)lim⁡t→0‖x+ty‖-‖x‖t
exists for all x,y∈U. It is also said to be uniformly smooth if the limit (2.4) is attained uniformly for x,y∈U. The normalized duality mappingJ:E→2E* is defined by
(2.5)J(x)={x*∈E*:〈x,x*〉=‖x‖2=‖x*‖2}
for all x∈E. 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 [35]).LetE be a smooth Banach space. The function ϕ:E×E→ℝ is defined by
(2.6)ϕ(x,y)=‖x‖2-2〈x,Jy〉+‖y‖2,∀x,y∈E.Remark 2.1.
We know the following: for anyx,y,z∈E,(1)
(∥x∥-∥y∥)2≤ϕ(x,y)≤(∥x∥+∥y∥)2;(2)
ϕ(x,y)=ϕ(x,z)+ϕ(z,y)+2〈x-z,Jz-Jy〉;(3)
ϕ(x,y)=∥x-y∥2 in a real Hilbert space.Lemma 2.2 (see [36]).
LetE be a uniformly convex and smooth Banach space and let {xn} and {yn} be sequences of E such that {xn} or {yn} is bounded and limn→∞ϕ(xn,yn)=0. Then limn→∞∥xn-yn∥=0.LetE be a reflexive, strictly convex, and smooth Banach space and let C be a nonempty closed and convex subset of E. The generalized projection mapping, introduced by Alber [37], is a mapping ΠC:E→C, that assigns to an arbitrary point x∈E the minimum point of the functional ϕ(y,x), that is, ΠCx=x¯, where x¯ is the solution to the minimization problem:
(2.7)ϕ(x¯,x)=min⁡{ϕ(y,x):y∈C}.In fact, we have the following result.Lemma 2.3 (see [37]).
LetC be a nonempty, closed, and convex subset of a reflexive, strictly convex, and smooth Banach space E and let x∈E. Then there exists a unique element x0∈C such that ϕ(x0,x)=min{ϕ(z,x):z∈C}.Lemma 2.4 (see [36, 37]).
LetC be a nonempty closed and convex subset of a reflexive, strictly convex, and smooth Banach space E, x∈E, and z∈C. Then z=ΠCx if and only if
(2.8)〈Jx-Jz,y-z〉≤0,∀y∈C.Lemma 2.5 (see [36, 37]).
LetC be a nonempty closed and convex subset of a reflexive, strictly convex, and smooth Banach space E and let x∈E. Then
(2.9)ϕ(y,ΠCx)+ϕ(ΠCx,x)≤ϕ(y,x),∀y∈C.Lemma 2.6 (see [38]).
LetE be a uniformly convex and uniformly smooth Banach space and C a nonempty, closed, and convex subset of E. Then ΠC is uniformly norm-to-norm continuous on every bounded set.We make use of the following mappingV studied in Alber [37]:
(2.10)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 2.7 (see [39]).
LetE be a reflexive, strictly convex, smooth Banach space. Then
(2.11)V(x,x*)+2〈J-1x*-x,y*〉≤V(x,x*+y*)
for all x∈E and x*,y*∈E*.Lemma 2.8 (see [25]).
LetC be a closed and convex subset of a smooth, strictly convex, and reflexive Banach space E, let f be a bifunction from C×C to ℝ which satisfies conditions (A1)–(A4), and let r>0 and x∈E. Then there exists z∈C such that
(2.12)f(z,y)+1r〈Jz-Jx,y-z〉≥0,∀y∈C.Following [25, 40], we know the following lemma.Lemma 2.9 (see [41]).
LetC be a nonempty closed and convex subset of a smooth, strictly convex, and reflexive Banach space E. Let A:C→E* be a continuous and monotone mapping, let f be a bifunction from C×C to ℝ satisfying (A1)–(A4), and let φ be a lower semicontinuous and convex function from C to ℝ. For all r>0 and x∈E, there exists z∈C such that
(2.13)f(z,y)+〈Az,y-z〉+φ(y)+1r〈Jz-Jx,y-z〉≥φ(z),∀y∈C.
Define the mapping Tr:E→2C as follows:
(2.14)Tr(x)={z∈C:f(z,y)+〈Az,y-z〉+φ(y)+1r〈Jz-Jx,y-z〉≥φ(z),∀y∈C}.
Then, the followings hold:(1)
Tr is single-valued;(2)
Tr is firmly nonexpansive-type mapping [42], that is, for all x,y∈E,
(2.15)〈Trx-Try,JTrx-JTry〉≤〈Trx-Try,Jx-Jy〉;(3)
F(Tr)=GMEP(f,A,φ);(4)
GMEP(f,A,φ) is closed and convex.Remark 2.10.
It is known thatT is of firmly nonexpansive type if and only if
(2.16)ϕ(Tx,Ty)+ϕ(Ty,Tx)+ϕ(Tx,x)+ϕ(Ty,y)≤ϕ(Tx,y)+ϕ(Ty,x)
for all x,y∈ dom T (see [42]).The following lemmas give us some nice properties of real sequences.Lemma 2.11 (see [43]).
Assume that{an} is a sequence of nonnegative real numbers such that
(2.17)an+1≤(1-αn)an+bn,∀n≥1,
where {αn} is a sequence in (0,1) and {bn} is a sequence such that(a)
∑n=1∞αn=+∞;(b)
limsupn→∞bn/αn≤0 or ∑n=1∞|bn|<+∞.Then limn→∞an=0.Lemma 2.12 (see [44]).
Let{γn} be a sequence of real numbers such that there exists a subsequence {γnj} of {γn} such that γnj<γnj+1 for all j≥1. Then there exists a nondecreasing sequence {mk} of ℕ such that limk→∞mk=∞ and the following properties are satisfied by all (sufficiently large) numbers k≥1:
(2.18)γmk≤γmk+1,γk≤γmk+1.
In fact, mk is the largest number n in the set {1,2,…,k} such that the condition γn<γn+1 holds.


## 3. Main Results

In this section, we prove our main theorem in this paper. To this end, we need the following proposition.Proposition 3.1.
LetC be a nonempty closed and convex subset of a reflexive, strictly convex, and uniformly smooth Banach space E. Let f be a bifunction from C×C to ℝ satisfying (A1)–(A4), A:C→E* a continuous and monotone mapping, and φ a lower semicontinuous and convex function from C to ℝ such that GMEP(f,A,φ)≠∅. Let {rn}⊂(0,∞) be such that liminfn→∞rn>0. For each n≥1, let Trn be defined as in Lemma 2.9. Suppose that x∈C and {xn} is a bounded sequence in C such that limn→∞∥xn-Trnxn∥=0. Then
(3.1)limsup⁡n→∞〈Jx-Jp,xn-p〉≤0,
where p=ΠGMEP(f,A,φ)x and ΠGMEP(f,A,φ) is the generalized projection of C onto GMEP(f,A,φ).Proof.
Letx∈C and put p=ΠGMEP(f,A,φ)x. Since E is reflexive and {xn} is bounded, there exists a subsequence {xnk} of {xn} such that xnk⇀v∈C and
(3.2)limsup⁡n→∞〈Jx-Jp,xn-p〉=〈Jx-Jp,v-p〉.
Put yn=Trnxn. Since limk→∞∥xnk-ynk∥=0, we have ynk⇀v. On the other hand, since E is uniformly smooth, J is uniformly norm-to-norm continuous on bounded subsets of E. So we have
(3.3)lim⁡k→∞‖Jxnk-Jynk‖=0.
Since liminfk→∞rnk>0,
(3.4)lim⁡k→∞‖Jxnk-Jynk‖rnk=0.
By the definition of Trnk, for any y∈C, we see that
(3.5)f(ynk,y)+〈Aynk,y-ynk〉+φ(y)+1rnk〈Jynk-Jxnk,y-ynk〉≥φ(ynk).
By (A2), for each y∈C, we obtain
(3.6)f(y,ynk)+φ(ynk)≤-f(ynk,y)+φ(ynk)≤〈Aynk,y-ynk〉+φ(y)+1rnk〈Jynk-Jxnk,y-ynk〉.
For any t∈(0,1) and y∈C, we define yt=ty+(1-t)v. Then yt∈C. It follows by the monotonicity of A that
(3.7)f(yt,ynk)+φ(ynk)≤〈Aynk-Ayt,yt-ynk〉+〈Ayt,yt-ynk〉+φ(yt)+1rnk〈Jynk-Jxnk,yt-ynk〉≤〈Ayt,yt-ynk〉+φ(yt)+1rnk〈Jynk-Jxnk,yt-ynk〉.
By (A4), (3.4), and the weakly lower semicontinuity of φ, letting k→∞, we obtain
(3.8)f(yt,v)+φ(v)≤〈Ayt,yt-v〉+φ(yt).
By (A1), (A4), and the convexity of φ, we have
(3.9)0=f(yt,yt)+φ(yt)-φ(yt)≤tf(yt,y)+(1-t)f(yt,v)+tφ(y)+(1-t)φ(v)-φ(yt)=t(f(yt,y)+φ(y)-φ(yt))+(1-t)(f(yt,v)+φ(v)-φ(yt))≤t(f(yt,y)+φ(y)-φ(yt))+(1-t)〈Ayt,yt-v〉=t(f(yt,y)+φ(y)-φ(yt))+(1-t)t〈Ayt,y-v〉.
It follows that
(3.10)f(yt,y)+φ(y)-φ(yt)+(1-t)〈Ayt,y-v〉≥0.
By (A3), the weakly lower semicontinuity of φ, and the continuity of A, letting t→0, we obtain
(3.11)f(v,y)+φ(y)-φ(v)+〈Av,y-v〉≥0,∀y∈C.
This shows that v∈GMEP(f,A,φ). By Lemma 2.4, we have
(3.12)limsup⁡n→∞〈Jx-Jp,xn-p〉=〈Jx-Jp,v-p〉≤0.
This completes the proof.Theorem 3.2.
LetC be nonempty, closed, and convex subset of a uniformly smooth and uniformly convex Banach space E. Let f be a bifunction from C×C to ℝ satisfying (A1)–(A4), A:C→E* a continuous and monotone mapping, and φ a lower semicontinuous and convex function from C to ℝ such that GMEP(f,A,φ)≠∅. Define the sequence {xn} as follows: x1=x∈C and
(3.13)f(yn,y)+〈Ayn,y-yn〉+φ(y)+1rn〈Jyn-Jxn,y-yn〉≥φ(yn),∀y∈C,xn+1=ΠCJ-1(αnJx+(1-αn)Jyn),∀n≥1,
where {αn}⊂(0,1) and {rn}⊂(0,∞) satisfy the following conditions:(a)
limn→∞αn=0;(b)
∑n=1∞αn=∞;(c)
liminfn→∞rn>0.Then {xn} converges strongly to ΠGMEP(f,A,φ)x, where ΠGMEP(f,A,φ) is the generalized projection of C onto GMEP(f,A,φ).Proof.
From Lemma2.9(4), we know that GMEP(f,A,φ) is closed and convex. Let p=ΠGMEP(f,A,φ)x. Put yn=Trnxn and zn=J-1(αnJx+(1-αn)Jyn) for all n∈ℕ. So, by Lemma 2.5, we have
(3.14)ϕ(p,xn+1)≤ϕ(p,zn)≤αnϕ(p,x)+(1-αn)ϕ(p,yn)≤αnϕ(p,x)+(1-αn)ϕ(p,xn).
By induction, we can show that ϕ(p,xn)≤ϕ(p,x) for each n∈ℕ. Hence {ϕ(p,xn)} is bounded and thus {xn} is also bounded.
We next show that if there exists a subsequence{xnk} of {xn} such that
(3.15)lim⁡k→∞(ϕ(p,xnk+1)-ϕ(p,xnk))=0,
then
(3.16)lim⁡k→∞(ϕ(p,ynk)-ϕ(p,xnk))=0.
Since αnk→0,
(3.17)lim⁡k→∞‖Jznk-Jynk‖=lim⁡k→∞αnk‖Jx-Jynk‖=0.
Since J is uniformly norm-to-norm continuous on bounded subsets of E, so is J-1. It follows that
(3.18)lim⁡k→∞‖znk-ynk‖=0.
Since E is uniformly smooth and uniformly convex, by Lemma 2.6, ΠC is uniformly norm-to-norm continuous on bounded sets. So we obtain
(3.19)lim⁡k→∞‖xnk+1-ynk‖=lim⁡k→∞‖ΠCznk-ΠCynk‖=0,
and hence
(3.20)lim⁡k→∞‖Jxnk+1-Jynk‖=0.
Furthermore, limk→∞ϕ(xnk+1,ynk)=0. Indeed, by the definition of ϕ, we observe that
(3.21)ϕ(xnk+1,ynk)=‖xnk+1‖2-2〈xnk+1,Jynk〉+‖ynk‖2=〈xnk+1,Jxnk+1-Jynk〉+〈ynk-xnk+1,Jynk〉.
It follows from (3.19) and (3.20) that limk→∞ϕ(xnk+1,ynk)=0. On the other hand, from Remark 2.1(2), we have
(3.22)ϕ(p,ynk)-ϕ(p,xnk)=(ϕ(p,xnk+1)-ϕ(p,xnk))+(ϕ(p,ynk)-ϕ(p,xnk+1))=(ϕ(p,xnk+1)-ϕ(p,xnk))+ϕ(xnk+1,ynk)+2〈p-xnk+1,Jxnk+1-Jynk〉.
It follows from (3.20) and (3.21) thatlimk→∞(ϕ(p,ynk)-ϕ(p,xnk))=0.
We next consider the following two cases.
Case 1.
ϕ(p,xn+1)≤ϕ(p,xn) for all sufficiently large n. Hence the sequence {ϕ(p,xn)} is bounded and nonincreasing. So limn→∞ϕ(p,xn) exists. This shows that limn→∞(ϕ(p,xn+1)-ϕ(p,xn))=0 and hence
(3.23)lim⁡n→∞(ϕ(p,yn)-ϕ(p,xn))=0.
Since Trn is of firmly nonexpansive type, by Remark 2.10, we have
(3.24)ϕ(yn,p)+ϕ(p,yn)+ϕ(yn,xn)+ϕ(Trnp,p)≤ϕ(yn,p)+ϕ(p,xn),
which implies
(3.25)ϕ(p,yn)+ϕ(yn,xn)≤ϕ(p,xn).
Hence
(3.26)ϕ(yn,xn)≤ϕ(p,xn)-ϕ(p,yn)⟶0
as n→∞. By Lemma 2.2, we obtain
(3.27)lim⁡n→∞‖xn-yn‖=0.
Proposition 3.1 yields that
(3.28)limsup⁡n→∞〈Jx-Jp,xn-p〉≤0.
It also follows that
(3.29)limsup⁡n→∞〈Jx-Jp,yn-p〉≤0.
Finally, we show thatxn→p. Using Lemma 2.7, we see that
(3.30)ϕ(p,xn+1)≤ϕ(p,zn)=V(p,αnJx+(1-αn)Jyn)≤V(p,αnJx+(1-αn)Jyn-αn(Jx-Jp))+〈αn(Jx-Jp),zn-p〉=V(p,αnJp+(1-αn)Jyn)+αn〈Jx-Jp,zn-p〉≤αnV(p,Jp)+(1-αn)V(p,Jyn)+αn〈Jx-Jp,zn-p〉=(1-αn)ϕ(p,yn)+αn〈Jx-Jp,zn-p〉≤(1-αn)ϕ(p,xn)+αn〈Jx-Jp,zn-p〉=(1-αn)ϕ(p,xn)+αn(〈Jx-Jp,zn-yn〉+〈Jx-Jp,yn-p〉).
Set an=ϕ(p,xn) and bn=αn(〈Jx-Jp,zn-yn〉+〈Jx-Jp,yn-p〉). We see that limsupn→∞bn/αn≤0. By Lemma 2.11, since ∑n=1∞αn=+∞, we conclude that limn→∞ϕ(p,xn)=0. Hence xn→p as n→∞.Case 2.
There exists a subsequence{ϕ(p,xnj)} of {ϕ(p,xn)} such that ϕ(p,xnj)<ϕ(p,xnj+1) for all j∈ℕ. By Lemma 2.12, there exists a strictly increasing sequence {mk} of positive integers such that the following properties are satisfied by all numbers k∈ℕ:
(3.31)ϕ(p,xmk)≤ϕ(p,xmk+1),ϕ(p,xk)≤ϕ(p,xmk+1).
So we have
(3.32)0≤lim⁡k→∞(ϕ(p,xmk+1)-ϕ(p,xmk))≤limsup⁡n→∞(ϕ(p,xn+1)-ϕ(p,xn))≤limsup⁡n→∞(ϕ(p,zn)-ϕ(p,xn))≤limsup⁡n→∞(αnϕ(p,x)+(1-αn)ϕ(p,yn)-ϕ(p,xn))=limsup⁡n→∞(αn(ϕ(p,x)-ϕ(p,yn))+(ϕ(p,yn)-ϕ(p,xn)))≤limsup⁡n→∞αn(ϕ(p,x)-ϕ(p,yn))=0.
This shows that
(3.33)lim⁡k→∞(ϕ(p,xmk+1)-ϕ(p,xmk))=0.
Following the proof line in Case 1, we can show that
(3.34)limsup⁡k→∞〈Jx-Jp,ymk-p〉≤0,ϕ(p,xmk+1)≤(1-αmk)ϕ(p,xmk)+αmk(〈Jx-Jp,zmk-ymk〉+〈Jx-Jp,ymk-p〉).
This implies
(3.35)αmkϕ(p,xmk)≤ϕ(p,xmk)-ϕ(p,xmk+1)+αmk(〈Jx-Jp,zmk-ymk〉+〈Jx-Jp,ymk-p〉)≤αmk(〈Jx-Jp,zmk-ymk〉+〈Jx-Jp,ymk-p〉).
Hence limk→∞ϕ(p,xmk)=0. Using this and (3.33) together, we conclude that
(3.36)limsup⁡k→∞ϕ(p,xk)≤lim⁡k→∞ϕ(p,xmk+1)=0.
This completes the proof.As a direct consequence of Theorem3.2, we obtain the following results.Corollary 3.3.
LetC be nonempty closed and convex subset of a uniformly smooth and uniformly convex Banach space E. Let f be a bifunction from C×C to ℝ satisfying (A1)–(A4) and φ a lower semicontinuous and convex function from C to ℝ such that MEP(f,φ)≠∅. Define the sequence {xn} as follows: x1=x∈C and
(3.37)f(yn,y)+φ(y)+1rn〈Jyn-Jxn,y-yn〉≥φ(yn),∀y∈C,xn+1=ΠCJ-1(αnJx+(1-αn)Jyn),∀n≥1,
where {αn}⊂(0,1) and {rn}⊂(0,∞) satisfy the following conditions:(a)
limn→∞αn=0;(b)
∑n=1∞αn=∞;(c)
liminfn→∞rn>0.Then {xn} converges strongly to ΠMEP(f,φ)x, where ΠMEP(f,φ) is the generalized projection of C onto MEP(f,φ).Corollary 3.4.
LetC be nonempty, closed, and convex subset of a uniformly smooth and uniformly convex Banach space E. Let f be a bifunction from C×C to ℝ satisfying (A1)–(A4), and A:C→E* a continuous and monotone mapping such that GEP(f,A)≠∅. Define the sequence {xn} as follows: x1=x∈C and
(3.38)f(yn,y)+〈Ayn,y-yn〉+1rn〈Jyn-Jxn,y-yn〉≥0,∀y∈C,xn+1=ΠCJ-1(αnJx+(1-αn)Jyn),∀n≥1,
where {αn}⊂(0,1) and {rn}⊂(0,∞) satisfy the following conditions:(a)
limn→∞αn=0;(b)
∑n=1∞αn=∞;(c)
liminfn→∞rn>0.Then {xn} converges strongly to ΠGEP(f,A)x, where ΠGEP(f,A) is the generalized projection of C onto GEP(f,A).Corollary 3.5.
LetC be nonempty, closed, and convex subset of a uniformly smooth and uniformly convex Banach space E. Let A:C→E* be a continuous and monotone mapping, and φ a lower semicontinuous and convex function from C to ℝ such that VI(C,A,φ)≠∅. Define the sequence {xn} as follows: x1=x∈C and
(3.39)〈Ayn,y-yn〉+φ(y)+1rn〈Jyn-Jxn,y-yn〉≥φ(yn),∀y∈C,xn+1=ΠCJ-1(αnJx+(1-αn)Jyn),∀n≥1,
where {αn}⊂(0,1) and {rn}⊂(0,∞) satisfy the following conditions:(a)
limn→∞αn=0;(b)
∑n=1∞αn=∞;(c)
liminfn→∞rn>0.Then {xn} converges strongly to ΠVI(C,A,φ)x, where ΠVI(C,A,φ) is the generalized projection of C onto VI(C,A,φ).


## 4. Examples and Numerical Results

In this section, we give examples and numerical results for our main theorem.Example 4.1.
LetE=ℝandC=[-1,1]. Let f(x,y)=-9x2+xy+8y2,φ(x)=3x2, and Ax=2x. Find x̂∈[-1,1] such that
(4.1)f(x̂,y)+〈Ax̂,y-x̂〉+φ(y)≥φ(x̂),∀y∈[-1,1].Solution 4.
It is easy to check thatf, φ, and A satisfy all conditions in Theorem 3.2. For each r>0 and x∈[-1,1], Lemma 2.9 ensures that there exists z∈[-1,1] such that, for any y∈[-1,1],
(4.2)f(z,y)+〈Az,y-z〉+φ(y)+1r〈z-x,y-z〉≥φ(z)⟺-9z2+yz+8y2+2z(y-z)+3y2+1r(z-x)(y-z)≥3z2⟺11ry2+(3rz+z-x)y-(14rz2+z2-xz)≥0.
Put G(y)=11ry2+(3rz+z-x)y-(14rz2+z2-xz). Then G is a quadratic function of y with coefficient a=11r, b=(3rz+z-x), and c=-(14rz2+z2-xz). We next compute the discriminant Δ of G as follows:
(4.3)Δ=b2-4ac=[(3r+1)z-x]2+44r(14rz2+z2-xz)=x2-2(3r+1)xz+(3r+1)2z2+616r2z2+44rz2-44rxz=x2-50rxz-2xz+625r2z2+50rz2+z2=x2-2(25rz+z)x+(625r2z2+50rz2+z2)=[x-(25rz+z)]2.
We know that G(y)≥0 for all y∈[-1,1] if it has at most one solution in [-1,1]. So Δ≤0 and hence x=25rz+z. Now we have z=Trx=x/(25r+1).
Let{xn}n=1∞ be the sequence generated by x1=x∈[-1,1] and
(4.4)f(yn,y)+〈Ayn,y-yn〉+φ(y)+1rn〈yn-xn,y-yn〉≥φ(yn),∀y∈[-1,1],xn+1=αnx+(1-αn)yn,∀n≥1,
and, equivalently,
(4.5)xn+1=αnx+(1-αn)Trnxn,∀n≥1.We next give two numerical results for algorithm (4.5).Algorithm 4.2.
Letαn=1/80n and rn=n/(n+1). Choose x1=x=1. Then algorithm (4.5) becomes
(4.6)xn+1=180n+(1-180n)(n+126n+1)xn,∀n≥1.Numerical Result I
See Table1.Table 1nxn11.000020.085630.011140.004750.0033⋮⋮2610.00012620.0000Algorithm 4.3.
Letαn=1/100n and rn=(n+1)/2n. Choose x1=x=-1. Then algorithm (4.5) becomes
(4.7)xn+1=-1100n+(1-1100n)(2n27n+25)xn,∀n≥1.Numerical Result II
See Table2.Table 2nxn1−1.00002−0.04813−0.00744−0.00385−0.0027⋮⋮217−0.00012180.0000


## 5. Conclusion

Tables1 and 2 show that the sequence {xn} converges to 0 which solves the generalized mixed equilibrium problem. On the other hand, using Lemma 2.9(3), we can check that GMEP(f,A,φ)=F(Tr)={0}.Remark 5.1.
In the view of computation, our algorithm is simple in order to get strong convergence for generalized mixed equilibrium problems.



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*Source: 308791-2012-01-16.xml*