Fixed Point Theorems for an Elastic Nonlinear Mapping in Banach Spaces

Abstract

Let be a smooth Banach space with a norm . Let for any , where stands for the duality pair and is the normalized duality mapping. We define a -strongly nonexpansive mapping by . This nonlinear mapping is nonexpansive in a Hilbert space. However, we show that there exists a -strongly nonexpansive mapping with fixed points which is not nonexpansive in a Banach space. In this paper, we show a weak convergence theorem and strong convergence theorems for fixed points of this elastic nonlinear mapping and give the existence theorem.

1. Introduction

Let be a smooth Banach space with a norm and let be the dual space of . We denote by a duality pair on and let be the normalized duality mapping on . It is well known that is a continuous single-valued mapping in a smooth Banach space and a one-to-one mapping in a strictly convex Banach space (cf. [1]). We define a mapping by for all , where is a set of real numbers. It is obvious that . Let any be fixed, and then is a convex function because of convexity of . Many nonlinear mappings which are defined by using are studied (see [2–4]). We also defined a nonlinear mapping which is called a -strongly nonexpansive mapping in [5] as follows.

Definition 1. Let be a nonempty subset of a smooth Banach space . A mapping is called -strongly nonexpansive if there exists a constant such that for all where is the identity mapping on .

From this definition, it is obvious that the identity mapping is also a -strongly nonexpansive mapping. In a Hilbert space, it is trivial that this mapping is nonexpansive since and that any firmly nonexpansive mapping is a -strongly nonexpansive mapping with (see [5]). Moreover, we showed that if there exists a fixed point of a -strongly nonexpansive mapping , then is strongly nonexpansive with a Bregman distance in [5]. However, in Banach spaces, as we give an example in the later section, we find that there exists a -strongly nonexpansive mapping with fixed points which is not nonexpansive. We should point out that a guarantee of continuity of the -strongly nonexpansive mappings has not been given in a generalized Banach space yet.

In this paper, we prove a weak convergence theorem and strong convergence theorems for finding fixed points of a -strongly nonexpansive mapping in Banach spaces and show the existence theorem of fixed point with a dissipative property.

2. Preliminaries

In this section, at first we show the relationship between a -strongly nonexpansive mapping and other nonlinear mappings, in a Hilbert space. Secondly, we state some properties of -strongly nonexpansive mappings in a Banach space and give an example of a -strongly nonexpansive mapping which is not a quasinonexpansive mapping in a Banach space although has fixed points. We finally show some lemmas which are necessary in order to prove our theorems.

Let be a subset of a Banach space and let be a mapping. Then a point in the closure of is said to be an asymptotically fixed point of if contains a sequence which converges weakly to and the sequence converges strongly to 0. denotes the set of asymptotically fixed points of . In [6], Reich introduced a strongly nonexpansive mapping which is defined by using the Bregman distance .

Definition 2. Let be a Banach space. The Bregman distance corresponding to a function is defined by where is Gâteaux differentiable and stands for the derivative of at the point . Let be a nonempty subset of . We say that the mapping is strongly nonexpansive if and and if it holds that for a bounded sequence such that for any .

Taking the function as the convex, continuous, and Gâteaux differentiable function , we obtain the fact that the Bregman distance coincides with . In particular, in a Hilbert space, it is trivial that .

Proposition 3 (see [5]). In a Hilbert space, a -strongly nonexpansive mapping with is strongly nonexpansive.

Next we recall two mappings of other nonlinear mappings (cf. [6–9]). A firmly nonexpansive mapping and an -inverse strongly monotone mapping are defined as follows.

Definition 4. Let be a nonempty, closed, and convex subset of a Banach space . A mapping is said to be firmly nonexpansive if for all and some .

It is trivial that a firmly nonexpansive mapping is nonexpansive.

Definition 5. Let be a Hilbert space. A mapping is said to be -inverse strongly monotone if for all .

The relation among firmly nonexpansive mappings, -inverse strongly monotone mappings and -strongly nonexpansive mappings is shown in the following proposition.

Proposition 6 (see [5]). In a Hilbert space, the following hold.(a)A firmly nonexpansive mapping is -strongly nonexpansive with .(b)Let be an -inverse strongly monotone mapping for ; then is -strongly nonexpansive with .

The above (b) is obvious by showing that, for all , We will introduce some properties of -strongly nonexpansive mappings in [5].

Proposition 7 (see [5]). In a smooth Banach space , the following hold.(a)For , is -strongly nonexpansive. For , is -strongly nonexpansive for any . For , is -strongly nonexpansive for any .(b)If   is -strongly nonexpansive with , then, for any with , is also -strongly nonexpansive with .(c)If is -strongly nonexpansive with , then is -strongly nonexpansive with .(d)Suppose that is -strongly nonexpansive with and that satisfies . Then is -strongly nonexpansive with . Moreover, if  , then

Now we give an example of a -strongly nonexpansive mapping in a Banach space.

Example 8 (see [10]). Let such that . Let be a real Banach space with a norm defined by Then is smooth, and the normalized duality mapping is single-valued. is given by Hence, we have for that We define a mapping as follows: In a case of , we have shown that the mapping defined by (11) is a -strongly nonexpansive mapping (see [5]). We will show that is -strongly nonexpansive with any , for .

Proposition 9. Suppose that is defined by the formula (11) under the above situation. Then, is a -strongly nonexpansive mapping with any .

Proof. Case (a): suppose that with and .
Since , we have that Since we have that Hence, we obtain that Hölder’s inequality implies that Therefore, we obtain that
That is, the inequality (1) holds.
Case (b): suppose that with and .
Then we have that Hence, we have that As (a), we obtain from Hölder’s inequality that That is, the inequality (1) holds.
Case (c): suppose that with .
Then we have that Hence, we have that It is obvious that for any and . Thus, we have from Hölder’s inequality that That is, the inequality (1) holds.
It is clear that if then inequality (1) holds. Therefore, from Cases (a), (b), and (c), we obtain the conclusion that is -strongly nonexpansive for any .

Remark 10. When , we have given the result in [5]. When , we already know that is a Hilbert space and a -strongly nonexpansive mapping is nonexpansive.

Theorem 11. There exists a -strongly nonexpansive mapping with a nonempty subset of fixed points such that is not nonexpansive for some Banach space.

Proof. It is enough to show that the -strongly nonexpansive mapping which is given in the previous proposition is not nonexpansive.
Let . Suppose that satisfies that and . Then . Let and . We have that and . Then we obtain that . Then, we have that and since , we have that Therefore, we will show that since . Let be fixed. As , . Thus, we have for a sufficiently small that It is trivial that Let . For , we have that We obtain that and that Therefore, we obtain the conclusion.

We remark that the symbols and mean that converges strongly and weakly to , respectively. We will introduce the following important lemmas for proofs of our theorems.

Lemma 12. (a) For all , (b) Let be a sequence in such that there exists for some ; then is bounded.

Lemma 13 (see [3]). Let be a smooth and uniformly convex Banach space and a nonempty, convex, and closed subset of . Suppose that satisfies If a weakly convergent sequence satisfies that , it holds that .

Theorem 14 (see [1, 11]). Let be a compact subset of a topological vector space and let be a convex subset of   . Let be an operator such that, for each , is convex. Suppose that satisfies the following:(1) for each ,(2) for each ,(3) is open for each .
Then there exists a point such that .

Lemma 15 (see [12]). Let and let be a Banach space. Then is uniformly convex if and only if there exists a continuous, strictly increasing, and convex function , , such that for all and .

Lemma 16 (see [13]). Let be a smooth and uniformly convex Banach space. Then, there exists a continuous, strictly increasing, and convex function such that and, for each real number , for all .

Lemma 17 (see [13]). Let be a smooth and uniformly convex Banach space and and in . If   and either or is bounded, then .

3. Main Results

In this section, we prove a weak convergence theorem and strong convergence theorems for finding fixed points of a -strongly nonexpansive mapping in Banach spaces, and then we show the existence theorem for fixed points of with a dissipative property (cf. [10]).

Theorem 18. Let be a smooth and uniformly convex Banach space and a nonempty, closed, and convex subset of . Suppose that a mapping is -strongly nonexpansive with and that . One defines a Mann iterative sequence as follows: for any and , where and . Then for some .

Proof. Suppose that . Then we have from the convexity of that Since is -strongly nonexpansive with , we have that Hence, we have . From Lemma 12 (b), is bounded. Furthermore, we have that Since , we obtain that This means that converges strongly to . Hence, is also bounded, and there exists such that for all .
On the other hand, we have from Lemma 12 (a) that Hence, we obtain that . From Lemma 13, there exists a point such that and .

The duality mapping of a Banach space with Gâteaux differentiable norm is said to be weakly sequentially continuous if in implies that converges weak star to in (cf. [14]). This happens, for example, if is a Hilbert space, or finite-dimensional and smooth, or if (cf. [15]). Next we prove a strong convergence theorem.

Theorem 19. Let be a reflexive, smooth, and strictly convex Banach space. Suppose that the duality mapping of is weakly sequentially continuous. Suppose that is a nonempty, closed, and convex subset of , is -strongly nonexpansive with , and . One defines a Mann iterative sequence as follows: for any and , where and . If satisfies that then and for some .

Proof. As in the proof of Theorem 18, we obtain that and and for some . Furthermore, from Lemma 12 (a), we have that Hence, the assumptions imply that From Lemma 17, we have the conclusion that and .

Condition (44) is a definition of a linear dissipative mapping (cf. [16]). Moreover, we give a definition of a -dissipative mapping for nonlinear mappings in a Banach space.

Definition 20. Let be a single-valued duality mapping on and let be a nonempty subset of . Then a mapping is called -dissipative if it holds that for all .

In a Hilbert space, such a mapping is called dissipative. In Banach spaces, we remark that the -dissipative mapping is not equal to the dissipative mapping (cf. [17]). Next we give a characterization of -dissipative mappings by using .

Theorem 21. Let be a smooth Banach space, a nonempty subset of , and a mapping. Then, the following are equivalent.(a) is -dissipative.(b)For all ,

Proof. For any , is equal to From the definition of , this inequality is equivalent to

Furthermore, we have the following result by this theorem.

Lemma 22. Suppose that is a smooth and strictly convex Banach space and that is a nonempty convex subset. Assume that a mapping is J-dissipative. If there are fixed points of , then is singleton.

Proof. Assume that there exist and such that and . Since is -dissipative, we have by Theorem 21 that Thus, we have that . This implies that
Furthermore, we have and we have . Since is strictly convex and is one-to-one, we obtain that .

We give a result before proving an existence theorem for fixed points.

Theorem 23 (see [10]). Let be a smooth and uniformly convex Banach space, and let be a -strongly nonexpansive mapping with . Then, one has that for , where .

Proof. Since is a -strongly nonexpansive with , we have Thus, we obtain, for with , , From Lemma 16, we have that Therefore, we have from (57) that . From the definition of , we obtain that

Remark 24. If satisfies that for 0, the (57) implies that for in the neighborhood of .

We will prove the following existence theorem by using Theorem 14.

Theorem 25. Let be a reflexive, strictly convex, and smooth Banach space and a nonempty, bounded, closed, and convex subset of . Suppose is a -strongly nonexpansive and -dissipative mapping. Then, there exists a unique fixed point of .

Proof. At first, we will show that there exists such that Assume that, for all , Let and for all . Then, from the assumption, is nonempty for all . Since is -dissipative, Theorem 21 implies that for all . This means that for any . For any , let with , and suppose that and with . From the convexity of , we have Thus, we obtain that is convex for all . Since it is obvious that is open for each , Theorem 14 implies that there exists a point such that . This means that This is a contradiction. Thus, we have for some that This means that there exists such that for all .
Furthermore, we will show for all if satisfies (66). Let for any and . Since is convex, then . Thus, we obtain that From the convexity of for , and we have . From the definition of , we have that Therefore, we have, by Theorem 23 and the continuity of on a smooth Banach space, that and for all . Letting , we have that Hence, . This implies that and then we obtain that Thus, we have and we have by (72) that . Since is one-to-one on a strictly convex Banach space, implies that . Therefore, we have the conclusion.

Finally, we will prove a strong convergence theorem for finding fixed points of a -strongly nonexpansive mapping in a Banach space, without the assumption that .

Theorem 26. Let be a smooth and uniformly convex Banach space, and let be a nonempty, compact, and convex subset of . Suppose that is -dissipative and -strongly nonexpansive with . One defines a Mann iterative sequence as follows: for any and , where and . Then, there exists a unique fixed point such that and .

Proof. From Theorem 25, we have that . As in the proof of Theorem 18, we obtain that and that there exists a point such that and . Since is -dissipative, Theorem 21 implies that From , we have for that Since , we have that By Lemma 17, we obtain that and . We have the conclusion.

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The author would like to thank the reviewers of this paper for their enduring reading and Professor N. Hirano and Professor A. T. Lau for their encouragement. Finally, the author deeply appreciates her mother Ayako’s hearty support.