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.

1. Introduction

The well-known variational inequality problem is to find x*∈C such that〈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: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 Question 1, very recently, Aoyama et al. [21] first considered the following generalized variational inequality problem in a Banach space.

Problem 1.

Let X be a smooth Banach space and C a nonempty closed convex subset of X. Let A be an accretive operator of C into X. Find a point x*∈C such that 〈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:
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.

Let X 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 fixed u∈C and arbitrarily given x0∈C, define a sequence {xn} iteratively byyn=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

Let X 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 〈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, 〈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, 〈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
〈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.

Let U={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, ‖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 limitlimt→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: 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 [<xref ref-type="bibr" rid="B11">25</xref>]).

Let q be a given real number with 1<q≤2, and let X be a q-uniformly smooth Banach space. Then,
‖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
Jq(x)={f∈X*:〈x,f〉=‖x‖q,‖f‖=‖x‖q-1},
for all x∈X.

Let D be a subset of C, and let Q be a mapping of C into D. Then, Q is said to be sunny if 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 [<xref ref-type="bibr" rid="B12">26</xref>]).

Let C 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
〈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 [<xref ref-type="bibr" rid="B8">21</xref>]).

Let C 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,
S(C,A)=F(QC(I-λA)),
where S(C,A)={x*∈C:〈Ax*,J(x-x*)〉≥0,forallx∈C}.

Lemma 2.7 (see [<xref ref-type="bibr" rid="B13">27</xref>]).

Let C 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 [<xref ref-type="bibr" rid="B14">28</xref>]).

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:
0<liminfn→∞αn≤limsupn→∞αn<1.
Suppose that
xn+1=αnxn+(1-αn)zn,n≥0,limsupn→∞(‖zn+1-zn‖-‖xn+1-xn‖)≤0.
Then, limn→∞∥zn-xn∥=0.

Lemma 2.9 (see [<xref ref-type="bibr" rid="B12">26</xref>]).

Assume that {an} is a sequence of nonnegative real numbers such that
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

∑n=0∞γn=∞,

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 an α-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.

Let X 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:

αn+βn+γn=1,foralln≥0;

limn→∞αn=0;

0<liminfn→∞βn≤limsupn→∞βn<1;

limn→∞(λn+1-λn)=0.

Then we have limn→∞∥xn+1-xn∥=0 and limn→∞∥Ayn-Axn∥=0.Proof.

First, we observe that I-λnA is nonexpansive. Indeed, for all x,y∈C, from Lemma 2.3, we have
‖(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.

Letting p∈S(C,A), it follows from Lemma 2.6 that p=QC(p-λnAp). Setting zn=QC(yn-λnAyn), from (3.1), we have
‖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
‖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
‖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
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
‖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
limsupn→∞(‖wn+1-wn‖-‖xn+1-xn‖)≤0.
Hence, by Lemma 2.8, we obtain ∥wn-xn∥→0 as n→∞. Consequently,
limn→∞‖xn+1-xn‖=limn→∞(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
limn→∞‖xn-zn‖=0.
For p∈S(C,A), from (3.1) and (3.2), we obtain
‖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
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
‖Axn-Ap‖⟶0asn⟶∞.
From the definition of zn and (3.1), we also have
‖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
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
‖Ayn-Ap‖⟶0asn⟶∞.
It follows from (3.12) and (3.15) that
‖Ayn-Axn‖⟶0asn⟶∞.
This completes the proof.

Now we state and study our main result.

Theorem 3.2.

Let X 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:

αn+βn+γn=1,foralln≥0;

limn→∞αn=0 and ∑n=0∞αn=∞;

0<liminfn→∞βn≤limsupn→∞βn<1;

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 Remark 2.1(2), we have that A is an (α/L2)-inverse-strongly accretive mapping. Then, from Lemma 3.1, we have
limn→∞‖Ayn-Axn‖=0.
On the other hand, we note that
‖Ayn-Axn‖≥α‖yn-xn‖,
which implies that
limn→∞‖yn-xn‖=0,
that is,
limn→∞‖QC(xn-λnAxn)-xn‖=0.
Next, we show that
limsupn→∞〈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
limsupn→∞〈u-Q′u,j(xn-Q′u)〉=limsupi→∞〈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
‖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))
limi→∞‖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
limsupn→∞〈u-Q′u,j(xn-Q′u)〉=limsupi→∞〈u-Q′u,j(xni-Q′u)〉=〈u-Q′u,j(z-Q′u)〉≤0.

Finally, from (1.5) and (3.2), we have
‖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
‖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 Theorem 3.2. We study the problem of finding a fixed point of a strictly pseudocontractive mapping.

A mapping T 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 〈Tx-Ty,j(x-y)〉≤σ‖x-y‖2.
This inequality can be written in the following form〈(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.

Let X 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:

αn+βn+γn=1,foralln≥0;

limn→∞αn=0 and ∑n=0∞αn=∞;

0<liminfn→∞βn≤limsupn→∞βn<1;

limn→∞(λn+1-λn)=0.

For fixed u∈C and arbitrarily given x0∈C, define a sequence {xn} iteratively by
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.

Putting A=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
‖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.

Acknowledgment

The paper is partially supported by the Program TH-1-3, Optimization Lean Cycle, of Sub-Projects TH-1 of Spindle Plan Four in Excellence Teaching and Learning Plan of Cheng Shiu University. The second author was partially supported by the Program TH-1-3, Optimization Lean Cycle, of Sub-Projects TH-1 of Spindle Plan Four inExcellence Teaching and Learning Plan of Cheng Shiu University and wassupported in part by NSC 100-2221-E-230-012.

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