#### 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 [2â€“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 GÃ¢teaux 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 GÃ¢teaux 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. [6â€“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 â€‰ 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 â€‰, 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 HÃ¶lderâ€™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 HÃ¶lderâ€™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 HÃ¶lderâ€™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 ofâ€‰â€‰ . 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 . Ifâ€‰â€‰ 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 GÃ¢teaux 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 ,