Volume 2011, Issue 1 459085
Research Article
Open Access

Common Fixed-Point Problem for a Family Multivalued Mapping in Banach Space

Zhanfei Zuo

Corresponding Author

Zhanfei Zuo

Department of Mathematics and Computer Science, Chongqing Three Gorges University, Wanzhou 404000, China sanxiau.edu.cn

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First published: 10 May 2011
Academic Editor: Vittorio Colao

Abstract

It is our purpose in this paper to prove two convergents of viscosity approximation scheme to a common fixed point of a family of multivalued nonexpansive mappings in Banach spaces. Moreover, it is the unique solution in F to a certain variational inequality, where stands for the common fixed-point set of the family of multivalued nonexpansive mapping {Tn}.

1. Introduction

Let X be a Banach space with dual X*, and let K be a nonempty subset of X. A gauge function is a continuous strictly increasing function φ : R+R+ such that φ(0) = 0 and lim tφ(t) = . The duality mapping Jφ : XX* associated with a gauge function φ is defined by Jφ(x): = {fX* : 〈x, f〉 = ∥x∥∥f∥, ∥f∥ = φ(∥x∥)}, xX, where 〈·, ·〉 denotes the generalized duality pairing. In the particular case φ(t) = t, the duality map J = Jφ is called the normalized duality map. We note that Jφ(x) = (φ(∥x∥)/∥x∥)J(x). It is known that if X is smooth, then Jφ is single valued and norm to weak* continuous (see [1]). When {xn} is a sequence in X, then xnx  (xnx, xnx) will denote strong (weak, weak*) convergence of the sequence {xn} to x. s

Following Browder [2], we say that a Banach space X has the weakly continuous duality mapping if there exists a gauge function φ for which the duality map Jφ is single valued and weak to weak* sequentially continuous, that is, if {xn} is a sequence in X weakly convergent to a point x, then the sequence {Jφ(xn)} converges weak* to Jφ(x). It is known that lp  (1 < p < 1) spaces have a weakly continuous duality mapping Jφ with a gauge φ(t) = tp−1. Setting
(1.1)
it is easy to see that Φ(t) is a convex function and Jφ(x) = Φ(∥x∥), for xX, where denotes the subdifferential in the sense of convex analysis. We will denote by 2X the family of all subsets of X, by CB(X) the family of all nonempty closed bounded subsets of X, and by C(X) the family of all nonempty compact subsets of X. A multivalued mapping T : K → 2X is said to be nonexpansive (resp., contractive) if
(1.2)
where H(·, ·) denotes the Hausdorff metric on CB(X) defined by
(1.3)
Since Banach′s contraction mapping principle was extended nicely to multivalued mappings by Nadler in 1969 (see [3]), many authors have studied the fixed-point theory for multivalued mappings.
In this paper, we construct two viscosity approximation sequences for a family of multivalued nonexpansive mappings in Banach spaces. Let K be a nonempty closed convex subset of Banach space X and let Tn : KC(K), n = 1,2… be a family of multivalued nonexpansive mapping, f : KK is a contraction mapping with constant α ∈ (0,1). Let αn ∈ (0,1), βn ∈ (0,1). For any given x0K, let y0T0x0 such that
(1.4)
From Nadler Theorem (see [3]), we can choose y1T1x1 such that
(1.5)
Inductively, we can get the sequence {xn} as follows:
(1.6)
where, for each nN, ynTnxn such that
(1.7)
Similarly, we also have the following multivalued version of the modified Mann iteration:
(1.8)
and ynTnxn such that ∥yn+1yn∥≤H(Tn+1xn+1, Tnxn). Then, {xn} is said to satisfy Condition (A) if for any subsequence and d(xn+1, Tn(xn)) → 0 implies that xF, where is the common fixed-point set of the family of multivalued mapping {Tn}. We give an example of a family of multivalued nonexpansive mappings with Condition (A) as follows.

Example 1.1. Take X = R and Tn = T (for all n ≥ 0), where T is defined by

(1.9)
Let f : R → {0} and αn = 1/n, n ≥ 2, then F = {0} and the iteration (1.6), reduced to
(1.10)
where ynTxn, and it satisfies Condition (A). In fact, if x0 ≤ 1, then (for all n, n > 0)xn = 0 and Condition (A) is automatically satisfied. If x0 > 1, then there exists an integer p ≥ 2, such that
(1.11)
Then, ypTxp−1 = {0}; hence, xn = 0  (for all np), from which we deduce that Condition (A) is satisfied.

2. Preliminaries

Let KX be a closed convex and Q a mapping of X onto K, then Q is said to be sunny if Q(Q(x) + t(xQ(x))) = Q(x) for all xX and t ≥ 0. A mapping Q of X into X is said to be a retraction if Q2 = Q. A subset K of X is said to be a sunny nonexpansive retract of X if there exists a sunny nonexpansive retraction of X onto K, and it is said to be a nonexpansive retract of X. If X = H, the metric projection P is a sunny nonexpansive retraction from H to any closed convex subset of H. The following Lemmas will be useful in this paper.

Lemma 2.1 (see [4].)Let K be a nonempty convex subset of a smooth Banach space X, let J : XX* be the (normalized) duality mapping of X, and let Q : XK be a retraction, then the following are equivalent:

  • (1)

    xPx, j(yPx)〉≤0 for all xX and yK,

  • (2)

    Q is both sunny and nonexpansive.

We note that Lemma 2.1 still holds if the normalized duality map J is replaced with the general duality map Jφ, where φ is a gauge function.

Lemma 2.2 (see [5].)Let {an} be a sequence of nonnegative real numbers satisfying the property

(2.1)
where {γn}⊂(0,1) and {βn} is a real number sequence such that
  • (i)

    ,

  • (ii)

    either limsup n(βn/γn) ≤ 0 or ,

then {an} converges to zero, as n.

Lemma 2.3 (see [1].)Let X be a real Banach space, then for all x, yX, one gets that

(2.2)

Lemma 2.4 (see [6].)Let {xn} and {yn} be bounded sequences in a Banach space X such that

(2.3)
where {γn} is a sequence in [0,1] such that
(2.4)
Assume that limsup n(∥yn+1yn∥−∥xn+1xn∥) ≤ 0, then lim nynxn∥ = 0.

3. Main Results

Theorem 3.1. Let X be a reflexive Banach space with weakly sequentially continuous duality mapping Jφ for some gauge φ, let K be a nonempty closed convex subset of X, and let Tn : KC(K), n = 0,1, 2…, be a family of multivalued nonexpansive mappings with F which is sunny nonexpansive retract of K with Q a nonexpansive retraction. Furthermore, Tn(p) = {p} for any fixed-point pF, {xn} is defined by (1.6), and αn ∈ (0,1) satisfies the following conditions:

  • (1)

    αn → 0 as n,

  • (2)

    ,

  • (3)

    {xn} satisfies Condition (A).

Then, {xn} converges strongly to a common fixed-point of a family Tn, n = 0,1, 2…, as n. Moreover, is the unique solution in F to the variational inequality
(3.1)

Proof. First, we show the uniqueness of the solution to the variational inequality (3.1) in X. In fact, let be another solution of (3.1) in F, then we have

(3.2)
From (3.2), we have that
(3.3)
We must have and the uniqueness is proved. Let pF, then, from iteration (1.6), we obtain that
(3.4)
Using an induction, we obtain ∥xnp∥≤max {∥x0p∥, (1/(1 − α))∥f(p) − p∥}, for all integers n, thus, {xn} is bounded and so are {Tnxn} and {f(xn)}. This implies that
(3.5)
Next, we will show that
(3.6)
Since X is reflexive and {xn} is bounded, we may assume that such that
(3.7)
From (3.5) and {xn} satisfying Condition (A), we obtain that qF. On the other hand, we notice that the assumption that the duality mapping Jφ is weakly continuous implies that X is smooth; from Lemma 2.1, we have
(3.8)
Finally, we will show that as n. From iteration (1.6) and Lemma 2.3, we get that
(3.9)
Lemma 2.2 gives that as n. Moreover, satisfying the variational inequality follows from the property of Q.

Let fuK in iteration (1.6) be a constant mapping, then . In fact, we have the following corollary.

Corollary 3.2. Let {xn} and Tn be as in Theorem 3.1, fuK, then {xn} converges strongly to a common fixed-point of a family Tn, n = 0,1, 2…, as n. Moreover, is the unique solution in F to the variational inequality

(3.10)
If X = H, then the condition that F is a sunny nonexpansive retract of K in Theorem 3.1 is not necessary, and one has the following Corollary.

Corollary 3.3. Let H be a Hilbert space with weakly sequentially continuous duality mapping Jφ for some gauge φ, and let {xn} and Tn be as in Theorem 3.1, then {xn} converges strongly to a common fixed-point of a family of Tn, n = 0,1, 2…, where PF is the metric projection from K onto F.

Proof. It is well known that H is reflexive; by Propositions 2.3 and 2.6(ii) of [7], we get that F is closed and convex, and hence the projection mapping PF is sunny nonexpansive retraction mapping, and the result follows from Theorem 3.1.

Corollary 3.4. Let X be a real smooth Banach space, let K be a nonempty compact subset of X, and let Tn and {xn} be as in Theorem 3.1, then {xn} converges strongly to a common fixed-point of a family of Tn, n = 0,1, 2…, as n. Moreover, is the unique solution in F to the variational inequality

(3.11)

Proof. Following the method of the proof of Theorem 3.1, we get that

(3.12)
Next, we will show that
(3.13)
Since K is compact and {xn} is bounded, we can assume that there exists a subsequence of {xn} such that ,
(3.14)
From (3.12) and {xn} satisfying Condition (A), we obtain that qF. On the other hand, from the fact that X is smooth, the duality being norm to weak* continuous, and the standard characterization of retraction on F, we obtain that
(3.15)
Now, following the method of the proof of Theorem 3.1, we get the required result.

Theorem 3.5. Let X be a reflexive Banach space with weakly sequentially continuous duality mapping Jφ for some gauge φ, let K be a nonempty closed convex subset of X, and let Tn : KC(K), n = 0,1, 2…, be a family of multivalued nonexpansive mappings with F which is sunny nonexpansive retract of K with Q a nonexpansive retraction. H(Tn+1x, Tny)≤∥xy∥ for arbitrary n. Furthermore, Tn(p) = {p} for any fixed-point pF. {xn} is defined by (1.8) and αn, βn satisfy the following conditions:

  • (i)

    βn → 0 as n,

  • (ii)

    ,

  • (iii)

    0 < liminf nαn ≤ lim sup nαn < 1.

If {xn} satisfies Condition (A), then {xn} converges strongly to a common fixed-point of a family of Tn, n = 0,1, 2…, as n. Moreover, is the unique solution in F to the variational inequality
(3.16)

Proof. We first show that the sequence {xn} defined by (1.8) is bounded. In fact, take pF, noting that Tn(p) = {p}, we have

(3.17)
It follows from induction that
(3.18)
so are {yn} and {f(xn)}. Thus, we have that
(3.19)
Next, we show that
(3.20)
Let λn = βn/(1 − αn) and zn = λnf(xn)+(1 − λn)yn, then
(3.21)
Therefore, we have for some appropriate constant M > 0 that the following inequality:
(3.22)
holds. Thus, limsup n(∥zn+1zn∥−∥xn+1xn∥) ≤ lim n(|λn+1λn | ∥f(xn+1) − f(xn)∥+(λn + λn+1)M) = 0. By Lemma 2.4, we obtain
(3.23)
Therefore, we have
(3.24)
Using (3.20) and {xn} satisfying Condition (A), we can use the same argumentation as Theorem 3.1 proves that and
(3.25)
Finally, we show that as n. In fact, from iteration (1.8) and Lemma 2.3, we have
(3.26)
From (ii) and (3.25), it then follows that
(3.27)
Apply Lemma 2.2 to conclude that .

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