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Hybrid method for a class of accretive variational inequalities involving nonexpansive mappings
Journal of Inequalities and Applications volume 2014, Article number: 217 (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. 71:10321041, 2009) to the framework of Banach spaces.
MSC:47H05, 47H09, 65J15.
1 Introduction
Let 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.
Recently, Lu 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}.
Hybrid methods for solving VI (1) were studied by Yamada [2], where F is Lipschitzian and strongly monotone. However, his methods do not apply to the variational inequality (2) since the mapping IV 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+(1t)T and defined \{{x}_{s,t}\} as the unique fixed point of the equation
Since Moudafi and Mainge’s regularization depends on 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.
In this paper, we consider the following type of accretive variational inequality problem:
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
Let 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}.
Let 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 ρcontraction if \rho \in [0,1) and the following property is satisfied:
\parallel f(x)f(y)\parallel \le \rho \parallel xy\parallel ,\phantom{\rule{1em}{0ex}}\mathrm{\forall}x,y\in C. 
(ii)
A mapping T:C\to C is nonexpansive provided
\parallel TxTy\parallel \le \parallel xy\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(xy)\in J(xy) such that
\u3008FxFy,j(xy)\u3009\ge 0; 
(b)
strictly accretive if F is accretive and the equality in (a) holds if and only if x=y;

(c)
βstrongly accretive if for any x,y\in C there exists j(xy)\in J(xy) such that
\u3008FxFy,j(xy)\u3009\ge \beta {\parallel xy\parallel}^{2}

(a)
for some real constant \beta >0.
Let \phi :[0,\mathrm{\infty}):={R}^{+}\to {R}^{+} be a continuous strictly increasing function such that \phi (0)=0 and \phi (t)\to \mathrm{\infty} as t\to \mathrm{\infty}. This function 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]).
Following Browder [4], we say that a Banach space 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 weaktoweak^{∗} 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}^{p1} for all 1<p<\mathrm{\infty}. Set
then
where ∂ denotes the subdifferential in the sense of convex analysis.
Remark 2.1 If {J}_{\phi} is weaktoweak^{∗} sequentially continuous, then J is strongtoweak^{∗} sequentially continuous.
Indeed, if {x}_{n}\to x strongly, then {x}_{n}\to x weakly, {J}_{\phi}({x}_{n}) converges weakly^{∗} 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
Letting n\to \mathrm{\infty}, we have
i.e., J is strongtoweak^{∗} 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
In this section, we introduce an implicit algorithm and prove this algorithm converges strongly to {x}^{\ast} which solves VI (5). Let 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:
It is obvious that {W}_{s,t}:C\to C is a contraction. So the contraction {W}_{s,t} has a unique fixed point which is denoted {x}_{s,t}. Namely,
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.
For any z\in Fix(T), we have
Combining the above inequality and Lemma 2.1, we obtain
which implies that
Taking \phi (t)=t, then {J}_{\phi}=J and \mathrm{\Phi}(t)=\frac{{t}^{2}}{2}, from (8) we have
which implies that
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.
From (6) and the boundedness of the sequences \{f({x}_{s,t})\}, \{S({x}_{s,t})\} and \{T({x}_{s,t})\}, for each fixed t\in (0,1) we have
Assume that \{{s}_{n}\}\subset (0,1) is such that {s}_{n}\to 0 (n\to \mathrm{\infty}). From (8), for any z\in Fix(T), we have
Since \{{x}_{{s}_{n},t}\} is bounded, without loss of generality, we may assume that \{{x}_{{s}_{n},t}\} converges weakly to a point {x}_{t} as n\to \mathrm{\infty}. This together with (10) implies that {x}_{t}\in Fix(T). Taking z={x}_{t} in (11), we have
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.
Taking s={s}_{n} in (9), we have
Since {J}_{\phi} is weakly continuous, then by Remark 2.1, J is strongtoweak^{∗} sequentially continuous. Let {s}_{n}\to 0 in the above inequality, we have
Hence we obtain
This together with Lemma 2.2, we have
Next, we prove that the entire net \{{x}_{s,t}\} converges strongly to {x}_{t} as s\to 0. We assume that {x}_{{s}_{n}^{\prime},t}\to {x}_{t}^{\prime} where {s}_{n}^{\prime}\to 0. Similar to the above proof, we have {x}_{t}^{\prime}\in Fix(T) and
Taking z={x}^{\prime}(t) and z={x}_{t} in (13) and (14), respectively, we have
Adding up the above two inequalities yields
Since
we obtain
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.
For any y\in \mathrm{\Omega}, taking z=y in (13), we have
which together with the fact of y\in \mathrm{\Omega} implies that
Since If is strongly accretive and IS is accretive, we obtain
It follows from (15)(17) that
Hence we have
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}.
Next we prove that {\omega}_{w}({x}_{t})\subset \mathrm{\Omega}, 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 IS is accretive, for any z\in Fix(T) we have
It follows from (13) that
By virtue of (19) and (20), we have
furthermore, we get
Letting t={t}_{n}\to 0 (n\to \mathrm{\infty}) in the above inequality, since \{{x}_{t}\} is bounded and φ is a continuous strictly increasing function, we have
This implies that
hence from the above inequality and Lemma 2.2, we have
i.e., {x}^{\prime}\in \mathrm{\Omega}.
Next we show that {x}^{\prime} is the solution of VI (7). Taking y={x}^{\prime} and t={t}_{n} in (18), we obtain
which implies that
Since {x}_{{t}_{n}}\to {x}^{\prime} weakly and {J}_{\phi} is weakly continuous, let {t}_{n}\to 0 in (21), we get
which together with the property of φ implies that {x}_{{t}_{n}}\to {x}^{\prime} in norm. It follows from (15) and (17) that
Since J is strongtoweak^{∗} sequentially continuous and f is a contraction, we have
Letting t={t}_{n}\to 0 (n\to \mathrm{\infty}) in (22) and combining (23) 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. □
References
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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).
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Wang, Y., Xu, Hk. Hybrid method for a class of accretive variational inequalities involving nonexpansive mappings. J Inequal Appl 2014, 217 (2014). https://doi.org/10.1186/1029242X2014217
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DOI: https://doi.org/10.1186/1029242X2014217