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# Hybrid method for a class of accretive variational inequalities involving nonexpansive mappings

- Yaqin Wang
^{1}Email author and - Hong-kun Xu
^{2}

**2014**:217

https://doi.org/10.1186/1029-242X-2014-217

© Wang and Xu; licensee Springer. 2014

**Received:**21 March 2014**Accepted:**22 May 2014**Published:**29 May 2014

## Abstract

In this paper we use contractions to regularize a class of accretive variational inequalities and prove the strongly convergence in Banach spaces. We extend the result of Lu *et al.* (Nonlinear Anal. 71:1032-1041, 2009) to the framework of Banach spaces.

**MSC:**47H05, 47H09, 65J15.

## Keywords

- hybrid method
- nonexpansive mapping
- accretive mapping
- variational inequality

## 1 Introduction

*H*be a Hilbert space,

*C*be a nonempty closed convex subset of

*H*, and $F:C\to H$ a nonlinear mapping. The set of fixed points of

*F*is denoted by $Fix(F)$,

*i.e.*, $Fix(F)=\{x\in C:Fx=x\}$. A monotone variational inequality problem is to find a point ${x}^{\ast}$ with the property

where *F* is a monotone operator.

*et al.*[1] were concerned with a special class of variational inequalities in which the mapping

*F*is the complement of a nonexpansive mapping and the constraint set is the set of fixed points of another nonexpansive mapping. Namely, they considered the following type of monotone variational inequality (VI) problem:

where $T,V:C\to C$ are nonexpansive mappings and $Fix(T)\ne \mathrm{\varnothing}$.

*F*is Lipschitzian and strongly monotone. However, his methods do not apply to the variational inequality (2) since the mapping $I-V$ fails, in general, to be strongly monotone, though it is Lipschitzian. Therefore, other hybrid methods have to be sought. Recently, Moudafi and Mainge [3] studied VI (2) by regularizing the mapping $tS+(1-t)T$ and defined $\{{x}_{s,t}\}$ as the unique fixed point of the equation

*t*, the convergence of the scheme (3) is more complicated. Very recently, Lu

*et al.*[1] studied VI (2) by regularizing the mapping

*S*and defined $\{{x}_{s,t}\}$ as the unique fixed point of the equation

Note that Lu *et al.*’s regularization (4) no longer depends on *t*.

Motivated and inspired by the result of Lu *et al.* [1], we put forward a question: Can this implicit hybrid method [1] in Hilbert spaces be extended to the framework of Banach spaces? In this paper, we give a positive answer.

Throughout this paper, we always assume that *E* is a real Banach space. Let *C* be a nonempty closed convex subset of *E*. Let $F:C\to E$ be a nonlinear mapping.

where $S,T:C\to C$ are two nonexpansive mappings with the set of fixed point $Fix(T)\ne \mathrm{\varnothing}$. Let Ω denote the set of solutions of VI (5) and assume that Ω is nonempty.

## 2 Preliminaries

*E*be a real Banach space and

*J*be the normalized duality mapping from

*E*into ${2}^{{E}^{\ast}}$ given by

for all $x\in E$, where ${E}^{\ast}$ denotes the dual space of *E* and $\u3008\cdot ,\cdot \u3009$ the generalized duality pairing between *E* and ${E}^{\ast}$.

*C*be a nonempty closed convex subset of a real Banach space

*E*. Recall the following concepts of mappings.

- (i)A mapping $f:C\to C$ is a
*ρ*-contraction if $\rho \in [0,1)$ and the following property is satisfied:$\parallel f(x)-f(y)\parallel \le \rho \parallel x-y\parallel ,\phantom{\rule{1em}{0ex}}\mathrm{\forall}x,y\in C.$ - (ii)A mapping $T:C\to C$ is nonexpansive provided$\parallel Tx-Ty\parallel \le \parallel x-y\parallel ,\phantom{\rule{1em}{0ex}}\mathrm{\forall}x,y\in C.$
- (iii)A mapping $F:C\to E$ is
- (a)accretive if for any $x,y\in C$ there exists $j(x-y)\in J(x-y)$ such that$\u3008Fx-Fy,j(x-y)\u3009\ge 0;$
- (b)
strictly accretive if

*F*is accretive and the equality in (a) holds if and only if $x=y$; - (c)
*β*-strongly accretive if for any $x,y\in C$ there exists $j(x-y)\in J(x-y)$ such that$\u3008Fx-Fy,j(x-y)\u3009\ge \beta {\parallel x-y\parallel}^{2}$

- (a)

for some real constant $\beta >0$.

*φ*is called a gauge function. The duality mapping ${J}_{\phi}:E\to {E}^{\ast}$ associated with a gauge function

*φ*is defined by

In the case that $\phi (t)=t$, ${J}_{\phi}=J$, where *J* is the normalized duality mapping. Clearly, the relation ${J}_{\phi}(x)=\frac{\phi (\parallel x\parallel )}{\parallel x\parallel}J(x)$, $\mathrm{\forall}x\ne 0$ holds (see [4]).

*E*has a

*weakly continuous duality mapping*if there exists a gauge

*φ*for which the duality mapping ${J}_{\phi}(x)$ is single valued and weak-to-weak

^{∗}sequentially continuous (

*i.e.*, if $\{{x}_{n}\}$ is a sequence in

*E*weakly convergent to a point

*x*, then the sequence ${J}_{\phi}({x}_{n})$ converges weakly

^{∗}to ${J}_{\phi}(x)$). It is well known that ${l}^{p}$ has a weakly continuous duality mapping with a gauge function $\phi (t)={t}^{p-1}$ for all $1<p<\mathrm{\infty}$. Set

where *∂* denotes the sub-differential in the sense of convex analysis.

**Remark 2.1** If ${J}_{\phi}$ is weak-to-weak^{∗} sequentially continuous, then *J* is strong-to-weak^{∗} sequentially continuous.

^{∗}to ${J}_{\phi}(x)$ and $\phi (\parallel {x}_{n}\parallel )\to \phi (\parallel x\parallel )$ strongly. Since ${J}_{\phi}(x)\frac{\parallel x\parallel}{\phi (\parallel x\parallel )}=J(x)$, $\mathrm{\forall}x\ne 0$, for any $y\in E$, we have

*i.e.*, *J* is strong-to-weak^{∗} sequentially continuous.

**Lemma 2.1** ([[5], Lemma 2.1])

*Assume that a Banach space*

*E*

*has a weakly continuous duality mapping*${J}_{\phi}$

*with a gauge*

*φ*.

*For all*$x,y\in E$,

*the following inequality holds*:

*In particular*,

*for all*$x,y\in E$,

**Lemma 2.2** (see [6])

*Let*

*C*

*be a nonempty closed convex subset of a real Banach space*

*E*.

*Assume that*$F:C\to E$

*is accretive and weakly continuous along segments*;

*that is*$F(x+ty)\rightharpoonup F(x)$

*as*$t\to 0$.

*Then the variational inequality*

*is equivalent to the dual variational inequality*

## 3 Main results

*C*be a nonempty closed convex subset of a real Banach space

*E*. Let $f:C\to C$ be a contraction and $S,T:C\to C$ be two nonexpansive mappings. For $s,t\in (0,1)$, we define the following mapping:

**Theorem 3.1**

*Let*

*C*

*be a nonempty closed convex subset of a reflexive Banach space*

*E*

*which has a weakly continuous duality map*${J}_{\phi}(x)$

*with the gauge*

*φ*.

*Let*$f:C\to C$

*be a contraction with constant*$\rho >0$

*and*$S,T:C\to C$

*be two nonexpansive mappings with*$Fix(T)\ne \mathrm{\varnothing}$.

*Suppose that the solution set*Ω

*of VI*(5)

*is nonempty*.

*Let*,

*for each*$(s,t)\in {(0,1)}^{2}$, $\{{x}_{s,t}\}$

*be defined implicitly by*(6).

*Then*,

*for each fixed*$t\in (0,1)$,

*the net*$\{{x}_{s,t}\}$

*converges in norm*,

*as*$s\to 0$,

*to a point*${x}_{t}\in Fix(T)$.

*Moreover*,

*as*$t\to 0$,

*the net*$\{{x}_{t}\}$

*converges in norm to the unique solution*${x}^{\ast}$

*of the following variational inequality*:

*Hence*, *for each null sequence* $\{{t}_{n}\}$ *in* $(0,1)$, *there exists another null sequence* $\{{s}_{n}\}$ *in* $(0,1)$, *such that the sequence* ${x}_{{s}_{n},{t}_{n}}\to {x}^{\ast}$ *in norm as* $n\to \mathrm{\infty}$.

*Proof* Step 1. For each fixed $t\in (0,1)$, the net $\{{x}_{s,t}\}$ is bounded.

So for each fixed $t\in (0,1)$, $\{{x}_{s,t}\}$ is bounded, furthermore $\{f({x}_{s,t})\}$, $\{S({x}_{s,t})\}$ and $\{T({x}_{s,t})\}$ are all bounded.

Step 2. ${x}_{s,t}\to {x}_{t}\in Fix(T)$ as $s\to 0$.

Since ${J}_{\phi}$ is weakly continuous, it follows from (12) that $\mathrm{\Phi}(\parallel {x}_{{s}_{n},t}-{x}_{t}\parallel )\to 0$ as $n\to \mathrm{\infty}$, which implies that ${x}_{{s}_{n},t}\to {x}_{t}$ strongly. This has proved the relative norm compactness of the net $\{{x}_{s,t}\}$ as $s\to 0$.

*J*is strong-to-weak

^{∗}sequentially continuous. Let ${s}_{n}\to 0$ in the above inequality, we have

*i.e.*, ${x}_{t}={x}_{t}^{\prime}$. So the entire net $\{{x}_{s,t}\}$ converges in norm to ${x}_{t}\in Fix(T)$ as $s\to 0$.

Step 3. The net $\{{x}_{t}\}$ is bounded.

Step 4. The net ${x}_{t}\to {x}^{\ast}\in \mathrm{\Omega}$ which solves VI (7).

First, the uniqueness of the solution of VI (7) is obvious. We denote the unique solution by ${x}^{\ast}$.

*i.e.*, if $\{{t}_{n}\}$ is a null sequence in $(0,1)$ such that ${x}_{{t}_{n}}\to {x}^{\prime}$ weakly as $n\to \mathrm{\infty}$, then ${x}^{\prime}\in \mathrm{\Omega}$. Indeed, since $\{{x}_{t}\}\subset Fix(T)$, then ${x}^{\prime}\in Fix(T)$. Since $I-S$ is accretive, for any $z\in Fix(T)$ we have

*φ*is a continuous strictly increasing function, we have

*i.e.*, ${x}^{\prime}\in \mathrm{\Omega}$.

*φ*implies that ${x}_{{t}_{n}}\to {x}^{\prime}$ in norm. It follows from (15) and (17) that

*J*is strong-to-weak

^{∗}sequentially continuous and

*f*is a contraction, we have

So ${x}^{\prime}$ is the solution of VI (7). By uniqueness, we have ${x}^{\prime}={x}^{\ast}$. Therefore, ${x}_{t}\to {x}^{\ast}$ in norm as $t\to 0$. The proof is complete. □

## Declarations

### Acknowledgements

The first author was supported by Zhejiang Provincial Natural Science Foundation of China under Grant (no. LQ13A010007, no. LY14A010006) and the China Postdoctoral Science Foundation Funded Project (no. 2012M511928).

## Authors’ Affiliations

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## Copyright

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