Abstract
In this research, we prove strong and weak convergence results for a class of mappings which is much more general than that of Suzuki nonexpansive mappings on Banach space through the Picard–Krasnoselskii hybrid iteration process. Using a numerical example, we prove that the Picard–Krasnoselskii hybrid iteration process converges faster than both of the Picard and Krasnoselskii iteration processes. Our results are the extension and improvement of many well-known results of the literature.
1. Introduction
A mapping on a subset of a Banach space is called nonexpansive if
A point is called a fixed point of provided that . We denote the fixed point set of by . The set is nonempty if is nonempty closed convex bounded and is uniformly convex (cf. [1–3]).
In 2008, Suzuki [4] generalized this concept as follows. The mapping on is said to satisfy condition (C) (or Suzuki mapping) if for all ,
Recently, Patir et al. [5] extended the (C) condition as follows. A mapping on is said to satisfy condition if there exists and satisfying such that for each ,
They also showed that if a mapping satisfies the (C) condition, then it satisfies the condition but the converse does not hold in general.
Let be a nonempty closed convex subset of a Banach space , be a mapping, , and .
The well-known Picard [6] and Krasnoselskii [7] iteration processes are, respectively, defined as
In 2017, Okeke and Abbas [8] considered the Picard–Krasnoselskii hybrid iteration process as follows:
They proved that the Picard–Krasnoselskii hybrid iteration (6) converges faster than all of Picard (4) and Krasnoselskii (5) processes for contraction mappings. We study this process in the general setting of mappings. We establish some weak and strong convergence results for mappings with condition . In the last section, we give an example of mapping which satisfies condition but not (C) and compare the rate of convergence of Picard–Krasnoselskii iteration, Picard iteration, and Krasnoselskii iteration.
2. Preliminaries
Let be a Banach space. We say that is uniformly convex [9] provided that for any , there is a such that for any with , , and ; it follows that
We say that satisfies Opial’s property [10] provided that for any in which weakly converges to and for all , someone has
Definition 1. Let be a nonempty subset of a Banach space , in be bounded, and . The asymptotic radius of relative to is the set . Moreover, the asymptotic center of relative to is the set .
We know that the set is singleton whenever the underlying space is uniformly convex Banach. Also, is nonempty as well as convex whenever is weakly compact and convex (see, e.g., [11, 12]).
Lemma 1. (see [5]). Let be a nonempty subset of a Banach space and satisfies condition. If is a fixed point of , then for each ,From Lemma 1, we obtain the following facts.
Lemma 2. Let be a nonempty subset of a Banach space . Let satisfy condition . Then, the set is closed. Moreover, if is strictly convex and is convex, then is also convex.
Theorem 1. (see [5]). Let be a nonempty subset of a Banach space having Opial property. Let satisfy condition . If is sequence in such that(i) converges weakly to ,(ii),then .
Proposition 1. (see [5]). Let be a nonempty subset of a Banach space . If satisfies condition on , then for all and ,(i)(ii)At least () and (h) hold:(g)(h) The Condition () implies and condition (h) implies (iii)We need the following useful lemma from [13].
Lemma 3. Let be a uniformly convex Banach space and for every . If and are two sequences in such that , , and for some , then .
3. Main Results
The following lemma will be used in the upcoming results.
Lemma 4. Let be a nonempty closed convex subset of a Banach space . Suppose that satisfies condition and . If is a sequence defined by (6), then exists for each .
Proof. Suppose that . By Lemma 1, we havewhich implies thatThus, is bounded as well as nonincreasing. Hence, exists for each .
Theorem 2. Let be a nonempty closed convex subset of a uniformly convex Banach space . Suppose that satisfies condition and let be a sequence defined by (6). Then, if and only if is bounded and .
Proof. We assume . By Proposition 1 (iii), for , ,(by Proposition 1 (i))So, . But is singleton, and we have . Hence, .
Conversely, let . By Lemma 4, exists. Assume that . We first prove that . By the proof of Lemma 4, ; therefore,and so .
Again by the proof of Lemma 4, . Hence, . Therefore, we obtain . Also, by Lemma 1, . It follows that . By Lemma 3, we obtainNow, we are in the position to establish a weak convergence of defined by (6) for the class of mappings with condition .
Theorem 3. Let a nonempty closed convex subset of a uniformly convex Banach space with the Opial property. Suppose that satisfies condition and . Then, defined by (6) converges weakly to an element of .
Proof. By Theorem 2, is bounded and . Since is uniformly convex, is reflexive. Thus, we can find a subsequence of such that converges weakly to some . By Theorem 1, we obtain . It is sufficient to prove that converges weakly to . Indeed, if does not converge weakly to , then we can find a subsequence of and such that converges weakly to and . Hence, by Theorem 1. Opial condition and Lemma 4 give usThis is a contradiction. So, .
Theorem 4. Let be a nonempty closed convex subset of a uniformly convex Banach space . Suppose that satisfies condition . If and , where be the sequence defined by (6), then converges strongly to a fixed point of .
Proof. By Lemma 4, exists, for each . Thus, exists. Hence,We can find a subsequence of and in with , . Moreover, is nonincreasing by the proof of Lemma 4. Hence,We shall prove that is a Cauchy in .This shows that the sequence is Cauchy in . By Lemma 2, is closed. Hence, for some . By Lemma 4, exists. So, the proof is finished.
Finally, we prove the following strong convergence theorem for the sequence defined by (6) with the help of condition (I).
Definition 2. Recall that a self-mapping on subset of a Banach space is said to satisfy condition [14] if and only if there exists satisfying and for every such that
Theorem 5. Let be a nonempty closed convex subset of a uniformly convex Banach space . Suppose that satisfies condition and . If satisfies condition , then defined by (6) converges strongly to a fixed point of .
Proof. By Theorem 2, we haveBy condition (I), we obtainThe conclusion follows from Theorem 4.
4. Numerical Example
In this section, we compare the rate of convergence of Picard–Krasnoselskii hybrid iteration process with Picard and Krasnoselskii iterations in general setting of mappings.
Example 1. Let be endowed with the usual norm. Set if and if . If and , then but . Hence, does not satisfy condition (C). However, satisfies condition . The case when is trivial and hence omitted. We consider the following two nontrivial cases.
When , we haveWhen and , we haveChoose ; the strong convergence of the sequence defined by Picard–Krasnoselskii hybrid process (6) to can be seen in Table 1.
Remark 1. From Table 1 and Figure 1, we observe that the Picard–Krasnoselskii hybrid iteration process converges faster than Picard and Krasnoselskii iterations in the class of mappings with condition .
Data Availability
No data were used to support this study.
Conflicts of Interest
The authors declare that they have no conflicts of interest.
Authors’ Contributions
All authors contributed equally and significantly in writing this article. All authors read and approved the final manuscript.
Acknowledgments
Thabet Abdeljawad would like to thank Prince Sultan University for funding this work through research group Nonlinear Analysis Methods in Applied Mathematics (NAMAM) (group no. RG-DES-2017-01-17).