# Extended Extragradient Methods for Generalized Variational Inequalities

Authors: Yonghong Yao; Yeong-Cheng Liou; Cun-Lin Li; Hui-To Lin
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

We suggest a modified extragradient method for solving the generalized variational inequalities in a Banach space. We prove some strong convergence results under some mild conditions on parameters. Some special cases are also discussed.


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## Body

## 1. Introduction

The well-known variational inequality problem is to findx*∈C such that(1.1)〈Ax*,x-x*〉≥0,∀x∈C,
where C is a nonempty closed convex subset of a real Hilbert space H and A:C→H is a nonlinear operator. This problem has been researched extensively due to its applications in industry, finance, economics, optimization, medical sciences, and pure and applied sciences; see, for instance, [1–19] and the reference contained therein. For solving the above variational inequality, Korpelevičh [20] introduced the following so-called extragradient method:(1.2)x0=x∈C,yn=PC(xn-λAxn),xn+1=PC(xn-λAyn)
for every n=0,1,2,…, where PC is the metric projection from Rn onto C and λ∈(0,1/k). He showed that the sequences {xn} and {yn} generated by (1.2) converge to the same point z∈VI(C,A). Since some methods related to extragradient methods have been considered in Hilbert spaces by many authors, please see, for example, [3, 5, 7, 14].This naturally brings us to the following questions.Question 1.
Could we extend variational inequality from Hilbert spaces to Banach spaces?Question 2.
Could we extend the extragradient methods from Hilbert spaces to Banach spaces?For solving Question1, very recently, Aoyama et al. [21] first considered the following generalized variational inequality problem in a Banach space.Problem 1.
LetXbe a smooth Banach space andCa nonempty closed convex subset ofX. LetAbe an accretive operator ofCintoX. Find a pointx*∈Csuch that(1.3)〈Ax*,J(x-x*)〉≥0,∀x∈C.
This problem is connected with the fixed point problem for nonlinear mapping, the problem of finding a zero point of an accretive operator, and so on. For the problem of finding a zero point of an accretive operator by the proximal point algorithm, please consult [22]. In order to find a solution of Problem 1, Aoyama et al. [21] introduced the following iterative scheme for an accretive operator A in a Banach space X:
(1.4)x1=x∈C,yn=QC(xn-λnAxn),xn+1=αnxn+(1-αn)yn,
for every n=1,2,…, where QC is a sunny nonexpansive retraction from X onto C. Then, they proved a weak convergence theorem in a Banach space which is generalized simultaneously by  theorems of [4, 23] as follows.Theorem 1.1.
LetX be a uniformly convex and 2-uniformly smooth Banach space, and let C be a nonempty closed convex subset of X. Let QC be a sunny nonexpansive retraction from X onto C, let α>0, and let A be an α-inverse-strongly accretive operator of C into X with S(C,A)≠∅. If {λn} and {αn} are chosen so that λn∈[a,α/K2] for some a>0 and αn∈[b,c] for some b,c with 0<b<c<1, then {xn} defined by (1.4) converges weakly to some element z of S(C,A):={x*∈C:〈Ax*,J(x-x*)〉≥0,forallx∈C}, where K is the 2-uniformly smoothness constant of X.In this paper, motivated by the ideas in the literature, we first introduce a new iterative method in a Banach space as follows.For fixedu∈C and arbitrarily given x0∈C, define a sequence {xn} iteratively by(1.5)yn=QC(xn-λnAxn),xn+1=αnu+βnxn+γnQC(yn-λnAyn),
for every n=1,2,…, where QC is a sunny nonexpansive retraction from X onto C, {αn}, {βn}, and {γn} are three sequences in (0,1), and {λn} is a sequence of real numbers. We prove some strong convergence results under some mild conditions on parameters.


## 2. Preliminaries

LetX be a real Banach space, and let X* denote the dual of X. Let C be a nonempty closed convex subset of X. A mapping A of C into X is said to be accretive if there exists j(x-y)∈J(x-y) such that (2.1)〈Ax-Ay,j(x-y)〉≥0,
for all x,y∈C, where J is called the duality mapping. A mapping A of C into X is said to be α-strongly accretive if, for α>0, (2.2)〈Ax-Ay,j(x-y)〉≥α‖x-y‖2,
for all x,y∈C. A mapping A of C into X is said to be α-inverse-strongly accretive if, for α>0, (2.3)〈Ax-Ay,j(x-y)〉≥α‖Ax-Ay‖2,
for all x,y∈C.Remark 2.1.
(1) Evidently, the definition of the inverse strongly accretive mapping is based on that of the inverse strongly monotone mapping.
(2) If A is an α-strongly accretive and L-Lipschitz continuous mapping of C into X, then
(2.4)〈Ax-Ay,j(x-y)〉≥α‖x-y‖2≥αL2‖Ax-Ay‖2,∀x,y∈C,
from which it follows that A must be (α/L2)-inverse-strongly accretive mapping.LetU={x∈X:∥x∥=1}. A Banach space X is said to be uniformly convex if, for each ϵ∈(0,2], there exists δ>0 such that, for any x,y∈U, (2.5)‖x-y‖≥ϵimplies‖x+y2‖≤1-δ.
It is known that a uniformly convex Banach space is reflexive and strictly convex. A Banach space X is said to be smooth if the limit(2.6)lim⁡t→0‖x+ty‖-‖x‖t
exists for all x,y∈U. It is also said to be uniformly smooth if the limit (2.6) is attained uniformly for x,y∈U. The norm of X is said to be Frechet differentiable if, for each x∈U, the limit (2.6) is attained uniformly for y∈U. And we define a function ρ:[0,∞)→[0,∞) called the modulus of smoothness of X as follows: (2.7)ρ(τ)=sup⁡⁡{12(‖x+y‖+‖x-y‖)-1:x,y∈X,‖x‖=1,‖y‖=τ}.
It is known that X is uniformly smooth if and only if limτ→0ρ(τ)/τ=0. Let q be a fixed real number with 1<q≤2. Then a Banach space X is said to be q-uniformly smooth if there exists a constant c>0 such that ρ(τ)≤cτq for all τ>0.Remark 2.2.
Takahashi et al. [24] remind us of the following fact: no Banach space is q-uniformly smooth for q>2. So, in this paper, we study a strong convergence theorem in a 2-uniformly smooth Banach space.We need the following lemmas for the proof of our main results.Lemma 2.3 (see [25]).
Letq be a given real number with 1<q≤2, and let X be a q-uniformly smooth Banach space. Then,
(2.8)‖x+y‖q≤‖x‖q+q〈y,Jq(x)〉+2‖Ky‖q,
for all x,y∈X, where K is the q-uniformly smoothness constant of X and Jq is the generalized duality mapping from X into 2X* defined by
(2.9)Jq(x)={f∈X*:〈x,f〉=‖x‖q,‖f‖=‖x‖q-1},
for all x∈X.LetD be a subset of C, and let Q be a mapping of C into D. Then, Q is said to be sunny if (2.10)Q(Qx+t(x-Qx))=Qx,
whenever Qx+t(x-Qx)∈C for x∈C and t≥0. A mapping Q of C into itself is called a retraction if Q2=Q. If a mapping Q of C into itself is a retraction, then Qz=z for every z∈R(Q), where R(Q) is the range of Q. A subset D of C is called a sunny nonexpansive retract of C if there exists a sunny nonexpansive retraction from C onto D. We know the following lemma concerning sunny nonexpansive retraction.Lemma 2.4 (see [26]).
LetC be a closed convex subset of a smooth Banach space X, D a nonempty subset of C, and Q a retraction from C onto D. Then, Q is sunny and nonexpansive if and only if
(2.11)〈u-Qu,j(y-Qu)〉≤0,
for all u∈C and y∈D.Remark 2.5.
(1) It is well known that, if X is a Hilbert space, then a sunny nonexpansive retraction QC is coincident with the metric projection from X onto C.
(2) Let C be a nonempty closed convex subset of a uniformly convex and uniformly smooth Banach space X, and let T be a nonexpansive mapping of C into itself with F(T)≠∅. Then, the set F(T) is a sunny nonexpansive retract of C.The following lemma is characterized by the set of solution Problem AIT by using sunny nonexpansive retractions.Lemma 2.6 (see [21]).
LetC be a nonempty closed convex subset of a smooth Banach space X. Let QC be a sunny nonexpansive retraction from X onto C, and let A be an accretive operator of C into X. Then, for all λ>0,
(2.12)S(C,A)=F(QC(I-λA)),
where S(C,A)={x*∈C:〈Ax*,J(x-x*)〉≥0,forallx∈C}.Lemma 2.7 (see [27]).
LetC be a nonempty bounded closed convex subset of a uniformly convex Banach space X, and let T be nonexpansive mapping of C into itself. If {xn} is a sequence of C such that xn→x weakly and xn-Txn→0 strongly, then x is a fixed point of T.Lemma 2.8 (see [28]).
Let{xn} and {zn} be bounded sequences in a Banach space X, and let {αn} be a sequence in [0,1] which satisfies the following condition:
(2.13)0<lim⁡inf⁡n→∞αn≤limsup⁡n→∞αn<1.
Suppose that
(2.14)xn+1=αnxn+(1-αn)zn,n≥0,limsup⁡n→∞(‖zn+1-zn‖-‖xn+1-xn‖)≤0.
Then, limn→∞∥zn-xn∥=0.Lemma 2.9 (see [26]).
Assume that{an} is a sequence of nonnegative real numbers such that
(2.15)an+1≤(1-γn)an+δn,n≥0,
where {γn} is a sequence in (0,1) and {δn} is a sequence in R such that  (i)
∑n=0∞γn=∞,(ii)
limsupn→∞δn/γn≤0 or ∑n=0∞|δn|<∞.Then, limn→∞an=0.


## 3. Main Results

In this section, we obtain a strong convergence theorem for finding a solution of Problem AIT for anα-strongly accretive and L-Lipschitz continuous mapping in a uniformly convex and 2-uniformly smooth Banach space. First, we assume that α>0 is a constant, L>0 a Lipschitz constant of A, and K>0 the 2-uniformly smoothness constant of X appearing in the following.In order to obtain our main result, we need the following lemma concerning(α/L2)-inverse-strongly accretive mapping.Lemma 3.1.
LetX be a uniformly convex and 2-uniformly smooth Banach space, and let C be a nonempty closed convex subset of X. Let QC be a sunny nonexpansive retraction from X onto C, and let A be an (α/L2)-inverse-strongly accretive mapping of C into X with S(C,A)≠∅. For given x0∈C, let the sequence {xn} be generated iteratively by (1.5), where {αn}, {βn}, and {γn} are three sequences in (0,1) and {λn} is a real number sequence in [a,b] for some a,b with 0<a<b<α/K2L2 satisfying the following conditions:  (i)
αn+βn+γn=1,foralln≥0;(ii)
limn→∞αn=0;(iii)
0<liminfn→∞βn≤limsupn→∞βn<1;(iv)
limn→∞(λn+1-λn)=0.Then we have limn→∞∥xn+1-xn∥=0 and limn→∞∥Ayn-Axn∥=0.Proof.
First, we observe thatI-λnA is nonexpansive. Indeed, for all x,y∈C, from Lemma 2.3, we have
(3.1)‖(I-λnA)x-(I-λnA)y‖2=‖(x-y)-λn(Ax-Ay)‖2≤‖x-y‖2-2λn〈Ax-Ay,j(x-y)〉+2K2λn2‖Ax-Ay‖2≤‖x-y‖2-2λnαL2‖Ax-Ay‖2+2K2λn2‖Ax-Ay‖2=‖x-y‖2+2λn(K2λn-αL2)‖Ax-Ay‖2.
If 0<a<λn<b<α/(K2L2), then I-λnA is a nonexpansive mapping.
Lettingp∈S(C,A), it follows from Lemma 2.6 that p=QC(p-λnAp). Setting zn=QC(yn-λnAyn), from (3.1), we have
(3.2)‖zn-p‖=‖QC(yn-λnAyn)-QC(p-λnAp)‖≤‖(yn-λnAyn)-(p-λnAp)‖≤‖yn-p‖=‖QC(xn-λnAxn)-QC(p-λnAp)‖≤‖(xn-λnAxn)-(p-λnAp)‖≤‖xn-p‖.
By (1.5) and (3.2), we have
(3.3)‖xn+1-p‖=‖αnu+βnxn+γnzn-p‖≤αn‖u-p‖+βn‖xn-p‖+γn‖zn-p‖=αn‖u-p‖+(1-αn)‖xn-p‖≤max⁡⁡{‖u-p‖,‖x0-p‖}.
Therefore, {xn} is bounded. Hence {zn}, {Axn}, and {Ayn} are also bounded. We observe that
(3.4)‖zn+1-zn‖=‖QC(yn+1-λn+1Ayn+1)-QC(yn-λnAyn)‖≤‖(yn+1-λn+1Ayn+1)-(yn-λnAyn)‖=‖(yn+1-λn+1Ayn+1)-(yn-λn+1Ayn)+(λn-λn+1)Ayn‖≤‖(yn+1-λn+1Ayn+1)-(yn-λn+1Ayn)‖+|λn-λn+1|‖Ayn‖≤‖yn+1-yn‖+|λn-λn+1|‖Ayn‖=‖QC(xn+1-λn+1Axn+1)-QC(xn-λnAxn)‖+|λn-λn+1|‖Ayn‖≤‖xn+1-xn‖+|λn-λn+1|(‖Axn‖+‖Ayn‖).
Setting xn+1=βnxn+(1-βn)wn for all n≥0 we obtain
(3.5)wn+1-wn=αn+1u+γn+1zn+11-βn+1-αnu+γnzn1-βn=(αn+11-βn+1-αn1-βn)u+γn+11-βn+1(zn+1-zn)+(γn+11-βn+1-γn1-βn)zn.
Combining (3.4) and (3.5), we have
(3.6)‖wn+1-wn‖-‖xn+1-xn‖≤|αn+11-βn+1-αn1-βn|‖u‖+γn+11-βn+1‖xn+1-xn‖+γn+11-βn+1|λn+1-λn|(‖Axn‖+‖Ayn‖)+|γn+11-βn+1-γn1-βn|‖zn‖-‖xn+1-xn‖≤|αn+11-βn+1-αn1-βn|(‖u‖+‖zn‖)+γn+11-βn+1|λn+1-λn|(‖Axn‖+‖Ayn‖);
this together with (ii) and (iv) implies that
(3.7)limsup⁡n→∞(‖wn+1-wn‖-‖xn+1-xn‖)≤0.
Hence, by Lemma 2.8, we obtain ∥wn-xn∥→0 as n→∞. Consequently,
(3.8)lim⁡n→∞‖xn+1-xn‖=lim⁡n→∞(1-βn)‖wn-xn‖=0.
From (1.5), we can write xn+1-xn=αn(u-xn)+γn(zn-xn) and note that 0<liminfn→∞γn≤limsupn→∞γn<1 and limn→∞αn=0. It follows from (3.8) that
(3.9)lim⁡n→∞‖xn-zn‖=0.
For p∈S(C,A), from (3.1) and (3.2), we obtain
(3.10)‖xn+1-p‖2=‖αnu+βnxn+γnzn-p‖2≤αn‖u-p‖2+βn‖xn-p‖2+γn‖zn-p‖2≤αn‖u-p‖2+βn‖xn-p‖2+γn‖yn-p‖2≤αn‖u-p‖2+βn‖xn-p‖2+γn{‖(xn-λnAxn)-(p-λnAp)‖2}≤αn‖u-p‖2+βn‖xn-p‖2+γn{‖xn-p‖2+2λn(K2λn-αL2)‖Axn-Ap‖2}≤αn‖u-p‖2+‖xn-p‖2+2γna(K2b-αL2)‖Axn-Ap‖2.
Therefore, we have
(3.11)0≤-2γna(K2b-αL2)‖Axn-Ap‖2≤αn‖u-p‖2+‖xn-p‖2-‖xn+1-p‖2=αn‖u-p‖2+(‖xn-p‖+‖xn+1-p‖)(‖xn-p‖-‖xn+1-p‖)≤αn‖u-p‖2+(‖xn-p‖+‖xn+1-p‖)‖xn-xn+1‖.
Since αn→0 and ∥xn-xn+1∥→0 as n→∞, from (3.11), we obtain
(3.12)‖Axn-Ap‖⟶0asn⟶∞.
From the definition of zn and (3.1), we also have
(3.13)‖zn-p‖2=‖QC(yn-λnAyn)-QC(p-λnAp)‖2≤‖(yn-λnAyn)-(p-λnAp)‖2≤‖yn-p‖2+2λn(K2λn-αL2)‖Ayn-Ap‖2.
From the above results and assumptions, we note that ∥yn-p∥≤∥xn-p∥, 0<a<b<α/(K2L2), ∥xn-p∥,∥zn-p∥ are bounded, and ∥xn-zn∥→0 as n→∞. Therefore, from (3.13), we have
(3.14)0≤-2a(K2b-αL2)‖Ayn-Ap‖2≤‖yn-p‖2-‖zn-p‖2≤‖xn-p‖2-‖zn-p‖2=(‖xn-p‖+‖zn-p‖)(‖xn-p‖-‖zn-p‖)≤(‖xn-p‖+‖zn-p‖)‖xn-zn‖⟶0,
which implies that
(3.15)‖Ayn-Ap‖⟶0asn⟶∞.
It follows from (3.12) and (3.15) that
(3.16)‖Ayn-Axn‖⟶0asn⟶∞.
This completes the proof.Now we state and study our main result.Theorem 3.2.
LetX be a uniformly convex and 2-uniformly smooth Banach space with weakly sequentially continuous duality mapping, and let C be a nonempty closed convex subset of X. Let QC be a sunny nonexpansive retraction from X onto C, and let A be an α-strongly accretive and L-Lipschitz continuous mapping of C into X with S(C,A)≠∅. Let {αn}, {βn}, and {γn} be three sequences in (0,1) and {λn} a real number sequence in [a,b] for some a,b with 0<a<b<α/(K2L2) satisfying the following conditions:  (i)
αn+βn+γn=1,foralln≥0;(ii)
limn→∞αn=0 and ∑n=0∞αn=∞;(iii)
0<liminfn→∞βn≤limsupn→∞βn<1;(iv)
limn→∞(λn+1-λn)=0.Then {xn} defined by (1.5) converges strongly to Q′u, where Q′ is a sunny nonexpansive retraction of C onto S(C,A).Proof.
From Remark2.1(2), we have that A is an (α/L2)-inverse-strongly accretive mapping. Then, from Lemma 3.1, we have
(3.17)lim⁡n→∞‖Ayn-Axn‖=0.
On the other hand, we note that
(3.18)‖Ayn-Axn‖≥α‖yn-xn‖,
which implies that
(3.19)lim⁡n→∞‖yn-xn‖=0,
that is,
(3.20)lim⁡n→∞‖QC(xn-λnAxn)-xn‖=0.
Next, we show that
(3.21)limsup⁡n→∞〈u-Q′u,j(xn-Q′u)〉≤0.
To show (3.21), since {xn} is bounded, we can choose a sequence {xni} of {xn} that converges weakly to z such that
(3.22)limsup⁡n→∞〈u-Q′u,j(xn-Q′u)〉=limsup⁡i→∞〈u-Q′u,j(xni-Q′u)〉.
We first prove z∈S(C,A). Since λn is in [a,b], it follows that {λni} is bounded, and so there exists a subsequence {λnij} of {λni} which converges to λ0∈[a,b]. We may assume, without loss of generality, that λni→λ0 as i→∞. Since QC is nonexpansive, it follows that
(3.23)‖QC(xni-λ0Axni)-xni‖≤‖QC(xni-λ0Axni)-QC(xni-λniAxni)‖+‖QC(xni-λniAxni)-xni‖≤‖(xni-λ0Axni)-(xni-λniAxni)‖+‖QC(xni-λniAxni)-xni‖≤|λni-λ0|‖Axni‖+‖QC(xni-λniAxni)-xni‖,
which implies that (noting that (3.20))
(3.24)lim⁡i→∞‖QC(I-λ0A)xni-xni‖=0.
By Lemma 2.7 and (3.24), we have z∈F(QC(I-λ0A)), and it follows from Lemma 2.6 that z∈S(C,A).
Now, from (3.22) and Lemma 2.4, we have
(3.25)limsup⁡n→∞〈u-Q′u,j(xn-Q′u)〉=limsup⁡i→∞〈u-Q′u,j(xni-Q′u)〉=〈u-Q′u,j(z-Q′u)〉≤0.
Finally, from (1.5) and (3.2), we have
(3.26)‖xn+1-z‖2=〈αnu+βnxn+γnzn-z,j(xn+1-z)〉=αn〈u-z,j(xn+1-z)〉+βn〈xn-z,j(xn+1-z)〉+γn〈zn-z,j(xn+1-z)〉≤12βn(‖xn-z‖2+‖xn+1-z‖2)+αn〈u-z,j(xn+1-z)〉+12γn(‖zn-z‖2+‖xn+1-z‖2)≤12(1-αn)(‖xn-z‖2+‖xn+1-z‖2)+αn〈u-z,j(xn+1-z)〉,
which implies that
(3.27)‖xn+1-z‖2≤(1-αn)‖xn-z‖2+2αn〈u-z,j(xn+1-z)〉.
Finally, by Lemma 2.9 and (3.27), we conclude that xn converges strongly to Q′u. This completes the proof.Remark 3.3.
From (3.1), we know that Q(I-λnA) is nonexpansive. If S(C,A)≠∅, it follows that there exists a sunny nonexpansive retraction Q′ of C onto F(Q(I-λnA))=S(C,A).


## 4. Application

In this section, we prove a strong convergence theorem in a uniformly convex and 2-uniformly smooth Banach space by using Theorem3.2. We study the problem of finding a fixed point of a strictly pseudocontractive mapping.A mappingT of C into itself is said to be strictly pseudocontractive if there exists 0≤σ<1 such that for all x,y∈C there exists j(x-y)∈J(x-y) such that (4.1)〈Tx-Ty,j(x-y)〉≤σ‖x-y‖2.
This inequality can be written in the following form(4.2)〈(I-T)x-(I-T)y,j(x-y)〉≥(1-σ)‖x-y‖2.
Now we give an application concerning a strictly pseudocontractive mapping.Theorem 4.1.
LetX be a uniformly convex and 2-uniformly smooth Banach space with weakly sequentially continuous duality mapping, and let C be a nonempty closed convex subset and a sunny nonexpansive retract of X. Let T be a strictly pseudocontractive and L-Lipschitz continuous mapping of C into itself with F(T)≠∅. Let {αn}, {βn}, and {γn} be three sequences in (0,1) and {λn} a real number sequence in [a,b] for some a,b with 0<a<b<1-σ/K2(L+1)2 satisfying the following conditions:  (i)
αn+βn+γn=1,foralln≥0;(ii)
limn→∞αn=0 and ∑n=0∞αn=∞;(iii)
0<liminfn→∞βn≤limsupn→∞βn<1;(iv)
limn→∞(λn+1-λn)=0.For fixed u∈C and arbitrarily given x0∈C, define a sequence {xn} iteratively by
(4.3)yn=(1-λn)xn+λnTxn,zn=(1-λn)yn+λnTyn,xn+1=αnu+βnxn+γnzn,
for every n=1,2,…. Then {xn} converges strongly to a fixed point of T.Proof.
PuttingA=I-T, we have from (4.2) that A is (1-σ)-strongly accretive. At the same time, since T is L-Lipschitz continuous, then we have
(4.4)‖Ax-Ay‖=‖(I-T)x-(I-T)y‖≤(L+1)‖x-y‖,
for all x,y∈C, that is, A is (L+1)-Lipschitz continuous mapping. It follows from Remark 2.1 (2) that A is (1-σ)/(L+1)2-inverse-strongly accretive mapping. It is easy to show that S(C,A)=S(C,I-T)=F(T)≠∅. Therefore, using Theorem 3.2, we can obtain the desired conclusion. This completes the proof.



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*Source: 237083-2011-11-19.xml*