New Iterative Approximation Methods for a Countable Family of Nonexpansive Mappings in Banach Spaces

Abstract

We introduce new general iterative approximation methods for finding a common fixed point of a countable family of nonexpansive mappings. Strong convergence theorems are established in the framework of reflexive Banach spaces which admit a weakly continuous duality mapping. Finally, we apply our results to solve the the equilibrium problems and the problem of finding a zero of an accretive operator. The results presented in this paper mainly improve on the corresponding results reported by many others.

1. Introduction

In recent years, the existence of common fixed points for a finite family of nonexpansive mappings has been considered by many authors (see [1–4] and the references therein). The well-known convex feasibility problem reduces to finding a point in the intersection of the fixed point sets of a family of nonexpansive mappings (see [5, 6]). The problem of finding an optimal point that minimizes a given cost function over the common set of fixed points of a family of nonexpansive mappings is of wide interdisciplinary interest and practical importance (see [2, 7]). A simple algorithmic solution to the problem of minimizing a quadratic function over the common set of fixed points of a family of nonexpansive mappings is of extreme value in many applications including set theoretic signal estimation (see [7, 8]).

Let be a normed linear space. Recall that a mapping is called nonexpansive if

(11)

We use to denote the set of fixed points of , that is, . A self mapping is a contraction on if there exists a constant such that

(12)

One classical way to study nonexpansive mappings is to use contractions to approximate a nonexpansive mapping [9–11]. More precisely, take and define a contraction by

(13)

where is a fixed point. Banach's contraction mapping principle guarantees that has a unique fixed point in . It is unclear, in general, what is the behavior of as , even if has a fixed point. However, in the case of having a fixed point, Browder [9] proved that if is a Hilbert space, then converges strongly to a fixed point of . Reich [10] extended Browder's result to the setting of Banach spaces and proved that if is a uniformly smooth Banach space, then converges strongly to a fixed point of and the limit defines the (unique) sunny nonexpansive retraction from onto . Xu [11] proved Reich's results hold in reflexive Banach spaces which have a weakly continuous duality mapping.

The iterative methods for nonexpansive mappings have recently been applied to solve convex minimization problems; see, for example, [12–14] and the references therein. Let be a real Hilbert space, whose inner product and norm are denoted by and , respectively. Let be a strongly positive bounded linear operator on ; that is, there is a constant with property

(14)

A typical problem is to minimize a quadratic function over the set of the fixed points of a nonexpansive mapping on a real Hilbert space

(15)

where is a given point in . In 2003, Xu [13] proved that the sequence defined by the iterative method below, with the initial guess chosen arbitrarily

(16)

converges strongly to the unique solution of the minimization problem (1.5) provided the sequence satisfies certain conditions. Using the viscosity approximation method, Moudafi [15] introduced the following iterative process for nonexpansive mappings (see [16] for further developments in both Hilbert and Banach spaces). Let be a contraction on . Starting with an arbitrary initial , define a sequence recursively by

(17)

where is a sequence in . It is proved [15, 16] that under certain appropriate conditions imposed on , the sequence generated by (1.7) strongly converges to the unique solution in of the variational inequality

(18)

Recently, Marino and Xu [17] mixed the iterative method (1.6) and the viscosity approximation method (1.7) and considered the following general iterative method:

(19)

where is a strongly positive bounded linear operator on . They proved that if the sequence of parameters satisfies the following conditions:

(C1),

(C2),

(C3),

then the sequence generated by (1.9) converges strongly to the unique solution in of the variational inequality

(110)

which is the optimality condition for the minimization problem: , where is a potential function for for ).

On the other hand, in order to find a fixed point of nonexpansive mapping , Halpern [18] was the first who introduced the following iteration scheme which was referred to as Halpern iteration in a Hilbert space: ,

(111)

He pointed out that the control conditions (C1) and (C2) are necessary for the convergence of the iteration scheme (1.11) to a fixed point of . Furthermore, the modified version of Halpern iteration was investigated widely by many mathematicians. Recently, for the sequence of nonexpansive mappings with some special conditions, Aoyama et al. [1] studied the strong convergence of the following modified version of Halpern iteration for :

(112)

where is a nonempty closed convex subset of a uniformly convex Banach space whose norm is uniformly Gáteaux differentiable, is a sequence in satisfying (C1) , (C2) , and either (C3) or (C3^') for all and . Very recently, Song and Zheng [19] also introduced the conception of the condition on a countable family of nonexpansive mappings and proved strong convergence theorems of the modified Halpern iteration (1.12) and the sequence defined by

(113)

in a reflexive Banach space with a weakly continuous duality mapping and in a reflexive strictly convex Banach space with a uniformly Gáteaux differentiable norm.

Other investigations of approximating common fixed points for a countable family of nonexpansive mappings can be found in [1, 20–24] and many results not cited here.

In a Banach space having a weakly continuous duality mapping with a gauge function , an operator is said to be strongly positive [25] if there exists a constant with the property

(114)

(115)

where is the identity mapping. If is a real Hilbert space, then the inequality (1.14) reduces to (1.4).

In this paper, motivated by Aoyama et al. [1], Song and Zheng [19], and Marino and Xu [17], we will combine the iterative method (1.12) with the viscosity approximation method (1.9) and consider the following three new general iterative methods in a reflexive Banach space which admits a weakly continuous duality mapping with gauge :

(116)

(117)

where is strongly positive defined by (1.15), is a countable family of nonexpansive mappings, and is an -contraction. We will prove in Section 3 that if the sequence of parameters satisfies the appropriate conditions, then the sequences , and converge strongly to the unique solution of the variational inequality

(118)

Finally, we apply our results to solve the the equilibrium problems and the problem of finding a zero of an accretive operator.

2. Preliminaries

Throughout this paper, let be a real Banach space, and be its dual space. We write (resp., ) to indicate that the sequence weakly (resp., weak*) converges to ; as usual will symbolize strong convergence. Let . A Banach space is said to uniformly convex if, for any , there exists such that, for any , implies . It is known that a uniformly convex Banach space is reflexive and strictly convex (see also [26]). A Banach space is said to be smooth if the limit exists for all . It is also said to be uniformly smooth if the limit is attained uniformly for .

By a gauge function , we mean a continuous strictly increasing function such that and as . Let be the dual space of . The duality mapping associated to a gauge function is defined by

(21)

In particular, the duality mapping with the gauge function , denoted by , is referred to as the normalized duality mapping. Clearly, there holds the relation for all (see [27]). Browder [27] initiated the study of certain classes of nonlinear operators by means of the duality mapping . Following Browder [27], we say that a Banach space has a  weakly  continuous  duality  mapping  if there exists a gauge for which the duality mapping is single valued and continuous from the weak topology to the weak* topology, that is, for any with , the sequence converges weakly* to . It is known that has a weakly continuous duality mapping with a gauge function for all . Set

(22)

then

(23)

where denotes the subdifferential in the sense of convex analysis.

Now, we collect some useful lemmas for proving the convergence result of this paper.

The first part of the next lemma is an immediate consequence of the subdifferential inequality and the proof of the second part can be found in [28].

Lemma 2.1 (see [28]).

Assume that a Banach space has a weakly continuous duality mapping with gauge .

(i)For all , the following inequality holds:

(24)

In particular, for all ,

(25)

(ii)Assume that a sequence in converges weakly to a point ,

then the following identity holds:

(26)

Lemma 2.2 (see [1, Lemma  2.3]).

Let be a sequence of nonnegative real numbers such that satisfying the property

(27)

where satisfying the restrictions

(i);    (ii);    (iii).

Then, .

Definition 2.3 (see [1]).

Let be a family of mappings from a subset of a Banach space into with . We say that satisfies the AKTT-condition if for each bounded subset of ,

(28)

Remark 2.4.

The example of the sequence of mappings satisfying AKTT-condition is supported by Lemma 4.6.

Lemma 2.5 (see [1, Lemma  3.2]).

Suppose that satisfies AKTT-condition, then, for each , converses strongly to a point in . Moreover, let the mapping be defined by

(29)

Then, for each bounded subset of , .

The next valuable lemma was proved by Wangkeeree et al. [25]. Here, we present the proof for the sake of completeness.

Lemma 2.6.

Assume that a Banach space has a weakly continuous duality mapping with gauge . Let be a strongly positive bounded linear operator on with coefficient and , then .

Proof.

From (1.15), we obtain that . Now, for any with , we see that

(210)

That is, is positive. It follows that

(211)

Let be a Banach space which admits a weakly continuous duality with gauge such that is invariant on that is, . Let be a nonexpansive mapping, , an -contraction, and a strongly positive bounded linear operator with coefficient and . Define the mapping by

(212)

Then, is a contraction mapping. Indeed, for any ,

(213)

Thus, by Banach contraction mapping principle, there exists a unique fixed point in , that is

(214)

Remark 2.7.

We note that space has a weakly continuous duality mapping with a gauge function for all . This shows that is invariant on .

Lemma 2.8 (see [25, Lemma  3.3]).

Let be a reflexive Banach space which admits a weakly continuous duality mapping with gauge such that is invariant on . Let be a nonexpansive mapping with , an -contraction, and a strongly positive bounded linear operator with coefficient and . Then, the net defined by (2.14) converges strongly as to a fixed point of which solves the variational inequality

(215)

3. Main Results

We now state and prove the main theorems of this section.

Theorem 3.1.

Let be a reflexive Banach space which admits a weakly continuous duality mapping with gauge such that is invariant on . Let be a countable family of nonexpansive mappings satisfying . Let be an -contraction and a strongly positive bounded linear operator with coefficient and . Let the sequence be generated by (1.16), where is a sequence in satisfying the following conditions:

(C1),

(C2),

(C3).

Suppose that satisfies the AKTT-condition. Let be a mapping of into itself defined by for all , and suppose that . Then, converges strongly to which solves the variational inequality

(31)

Proof.

Applying Lemma 2.8, there exists a point which solves the variational inequality (3.1). Next, we observe that is bounded. Indeed, pick any to obtain

(32)

It follows from induction that

(33)

Thus, is bounded, and hence so are and . Now, we show that

(34)

We observe that

(35)

for all , where is a constant satisfying . Putting . From AKTT-condition and (C3), we have

(36)

Therefore, it follows from Lemma 2.2 that . Since , we obtain

(37)

Using Lemma 2.5, we obtain

(38)

Next, we prove that

(39)

Let be a subsequence of such that

(310)

If follows from reflexivity of and the boundedness of a sequence that there exists which is a subsequence of converging weakly to as . Since is weakly continuous, we have by Lemma 2.1 that

(311)

Let

(312)

It follows that

(313)

Then, from , we have

(314)

On the other hand, however,

(315)

It follows from (3.14) and (3.15) that

(316)

Therefore, , and hence . Since the duality map is single valued and weakly continuous, we obtain, by (3.1), that

(317)

Next, we show that as . In fact, since , and is a gauge function, then for , and

(318)

Finally, we show that as . Following Lemma 2.1, we have

(319)

It then follows that

(320)

where . Put

(321)

It follows that from condition (C1), and (3.9) that

(322)

Applying Lemma 2.2 to (3.20), we conclude that as ; that is, as . This completes the proof.

Setting , where is the identity mapping and for all in Theorem 3.1, we have the following result.

Corollary 3.2.

Let be a reflexive Banach space which admits a weakly continuous duality mapping with gauge . Suppose that is a countable family of nonexpansive mappings satisfying . Assume that is defined by, for ,

(323)

where is a sequence in satisfying the following conditions:

(C1),

(C2),

(C3).

Suppose that satisfies the AKTT-condition. Let be a mapping of into itself defined by for all , and suppose that , then converges strongly to of which solves the variational inequality

(324)

Applying Theorem 3.1, we can obtain the following two strong convergence theorems for the iterative sequences and defined by (1.17).

Theorem 3.3.

Let be a reflexive Banach space which admits a weakly continuous duality mapping with gauge such that is invariant on . Let be a countable family of nonexpansive mappings satisfying . Let be an -contraction and a strongly positive bounded linear operator with coefficient and . Let the sequence be generated by (1.17), where is a sequence in satisfying the following conditions:

(C1),

(C2),

(C3).

Suppose that satisfies the AKTT-condition. Let be a mapping of into itself defined by for all , and suppose that , then converges strongly to which solves the variational inequality (3.1).

Proof.

Let be the sequence given by and

(325)

Form Theorem 3.1, . We claim that . Applying Lemma 2.6, we estimate

(326)

It follows from , , and Lemma 2.2 that . Consequently, as required.

Theorem 3.4.

Let be a reflexive Banach space which admits a weakly continuous duality mapping with gauge such that is invariant on . Let be a countable family of nonexpansive mappings satisfying . Let be an -contraction and a strongly positive bounded linear operator with coefficient and . Let the sequence be generated by (1.17), where is sequence in satisfying the following conditions:

(C1),

(C2),

(C3).

Suppose that satisfies the AKTT-condition. Let be a mapping of into itself defined by for all , and suppose that , then converges strongly to which solves the variational inequality (3.1).

Proof.

Let the sequences and be given by

(327)

Taking , we have

(328)

It follows from induction that

(329)

Thus, both and are bounded. We observe that

(330)

Thus, Theorem 3.1 implies that converges strongly to some point . In this case, we also have

(331)

Hence, the sequence converges strongly to . This competes the proof.

Setting , where is the identity mapping and for all in Theorem 3.4, we have the following result.

Corollary 3.5.

Let be a reflexive Banach space which admits a weakly continuous duality mapping with gauge . Suppose that is a countable family of nonexpansive mappings satisfying . Assume that is defined by for ,

(332)

where is a sequence in satisfying the following conditions:

(C1),

(C2),

(C3).

Suppose that satisfies the AKTT-condition. Let be a mapping of into itself defined by for all , and suppose that , then converges strongly to of which solves the variational inequality

(333)

4. Applications

4.1. -Mappings

Let be infinite mappings of into itself, and let be a nonnegative real sequence with , . For any , define a mapping of into itself as follows:

(41)

Nonexpansivity of each ensures the nonexpansivity of . The mapping is called a -mapping generated by and .

Throughout this section, we will assume that . Concerning defined by (4.1), we have the following useful lemmas.

Lemma 4.1 (see [4]).

Let be a nonempty closed convex subset of a a strictly convex, reflexive Banach space , a family of infinitely nonexpansive mapping with , and a real sequence such that , , then:

(1) is nonexpansive and for each ;

(2)for each and for each positive integer , the limit exists;

(3)the mapping define by

(42)

is a nonexpansive mapping satisfying , and it is called the -mapping generated by and .

From Remark 3.1 of Peng and Yao [29], we obtain the following lemma.

Lemma 4.2.

Let be a strictly convex, reflexive Banach space, a family of infinitely nonexpansive mappings with , and a real sequence such that , . Then sequence satisfies the -condition.

Applying Lemma 4.2 and Theorem 3.1, we obtain the following result.

Theorem 4.3.

Let be a reflexive Banach space which admits a weakly continuous duality mapping with gauge such that is invariant on . Let be a countable family of nonexpansive mappings with and an -contraction and a strongly positive bounded linear operator with coefficient and . Let the sequence be generated by the following:

(43)

where is defined by (4.1) and is a sequence in satisfying the conditions (C1), (C2), and (C3). Then converges strongly to in .

Applying Lemma 4.2 and Theorem 3.3, we obtain the following result.

Theorem 4.4.

Let , , , , , and be as in Theorem 4.3. Let the sequence be generated by the following:

(44)

then converges strongly to in .

Applying Lemma 4.2 and Theorem 3.4, we obtain the following result.

Theorem 4.5.

Let , , , , , and be as in Theorem 4.3. Let the sequence be generated by the following:

(45)

then converges strongly to in .

4.2. Accretive Operators

We consider the problem of finding a zero of an accretive operator. An operator is said to be accretive if for each and , there exists such that . An accretive operator is said to satisfy the range condition if for all , where is the domain of ,   is the identity mapping on , is the range of , and is the closure of . If is an accretive operator which satisfies the range condition, then we can define, for each , a mapping by , which is called the resolvent of . We know that is nonexpansive and for all . We also know the following [30]: for each and , it holds that

(46)

From the Resolvent identity, we have the following lemma.

Lemma 4.6.

Let be a Banach space and a nonempty closed convex subset of . Let be an accretive operator such that and . Suppose that is a sequence of such that and , then

(i)the sequence satisfies AKTT-condition,

(ii) for all and , where as .

Proof.

By the proof of Theorem 4.3 in [1] and applying Lemma 4.6 and Theorem 3.1, we obtain the following result.

Theorem 4.7.

Let be a reflexive Banach space which admits a weakly continuous duality mapping with gauge such that is invariant on . Let be an accretive operator such that . Assume that is a nonempty closed convex subset of such that and is an -contraction. Let be a strongly positive bounded linear operator with coefficient and . Suppose that is a sequence of such that . Let the sequence be generated by the following:

(47)

where is a sequence in satisfying the following conditions (C1), (C2), and (C3), then converges strongly to in .

Applying Lemma 4.6 and Theorem 3.3, we obtain the following result.

Theorem 4.8.

Let , , , , , , and be as in Theorem 4.7. Let be generated by the following:

(48)

then converges strongly to in .

Applying Lemma 4.6 and Theorem 3.4, we obtain the following result.

Theorem 4.9.

Let , , , , , , and be as in Theorem 4.7. Let be generated by the following:

(49)

Then converges strongly to in .

4.3. The Equilibrium Problems

Let be a real Hilbert space, and let   be a bifunction of , where is the set of real numbers. The equilibrium problem for is to find such that

(410)

The set of solutions of (4.10) is denoted by . Given a mapping , let for all . Then, if and only if for all , that is, is a solution of the variational inequality. Numerous problems in physics, optimization, and economics reduce to find a solution of (4.10). Some methods have been proposed to solve the equilibrium problem; see, for instance, Blum and Oettli [31] and Combettes and Hirstoaga [32]. For the purpose of solving the equilibrium problem for a bifunction , let us assume that satisfies the following conditions:

(A1) for all ,

(A2) is monotone, that is, for all ,

(A3) for each ,

(A4) for each is convex and lower semicontinuous.

The following lemmas were also given in [31, 32], respectively.

Lemma 4.10 (see [31, Corollary  1]).

Let be a nonempty closed convex subset of , and let be a bifunction of satisfying . Let and , then there exists such that .

Lemma 4.11 (see [32, Lemma  2.12]).

Assume that satisfies . For and , define a mapping as follows:

(411)

then, the following hold:

(1) is single valued,

(2) is firmly nonexpansive, that is, for any ,

(3),

(4) is closed and convex.

Theorem 4.12.

Let be a real Hilbert space. Let be a bifunction from satisfying (A1)–(A4) and . Let be an -contraction, a strongly positive bounded linear operator with coefficient and . Let the sequences be generated by and

(412)

for all , where is a sequence in and satisfying the following conditions:

(C1),

(C2),

(C3),

(C4) and .

then and converge strongly to .

Proof.

Following the proof technique of Theorem 3.1, we only need, show that , for all . From (4.12), it follows that

(413)

On the other hand, from the definition of we have

(414)

Putting and in (4.14), we have

(415)

So, from (A2), we have

(416)

and hence,

(417)

then we have

(418)

and hence,

(419)

where is a constant satisfying . Substituting (4.19) in (4.13) yields

(420)

for some with (the definition ). By the assumptions on and and using Lemma 2.2, we conclude that

(421)

From the definition of and , it follows that

(422)

Combining (4.21) and (4.22), we have

(423)

From the definition of , it follows that

(424)

Putting in (4.14) and in (4.24), we have

(425)

So, from (A2), we have

(426)

and hence, for each ,

(427)

then

(428)

Since

(429)

then for each , we have from (4.23)

(430)

This completes the proof.

Applying Theorem 4.12, we can obtain the following result.

Corollary 4.13.

Let be a real Hilbert space. Let be a bifunction from satisfying (A1)–(A4) and . Let be an -contraction, a strongly positive bounded linear operator with coefficient and . Let the sequences be generated by and

(431)

for all , where is a sequence in and satisfying the following conditions:

(C1),

(C2),

(C3),

(C4) and ,

then and converge strongly to .

Proof.

We observe that for all . Then we rewrite the iterative sequence (4.31) by the following:

(432)

Let be the sequence given by and

(433)

Form Theorem 4.12, in . We claim that . Applying Lemma 2.6, we estimate

(434)

It follows from , , and Lemma 2.2 that as . Consequently, as required.