# On Picard–Krasnoselskii Hybrid Iteration Process in Banach Spaces

Authors: Thabet Abdeljawad; Kifayat Ullah; Junaid Ahmad
Journal: Journal of Mathematics
 (2020)

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## 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.


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## Body

## 1. Introduction

A mappingF on a subset G of a Banach space X is called nonexpansive if(1)Fa−Fb≤a−b,for alla,b∈G.A pointq∈G is called a fixed point of F provided that q=Fq. We denote the fixed point set of F by fF. The set fF is nonempty if G is nonempty closed convex bounded and X is uniformly convex (cf. [1–3]).In 2008, Suzuki [4] generalized this concept as follows. The mapping F on G is said to satisfy condition (C) (or Suzuki mapping) if for all a,b∈G,(2)1/2a−Fa≤a−b⇒Fa−Fb≤a−b.Recently, Patir et al. [5] extended the (C) condition as follows. A mapping F on G is said to satisfy condition Bγ,μ if there exists γ∈0,1 and μ∈0,1/2 satisfying 2μ≤γ such that for each a,b∈G,(3)γa−Fa≤a−b+μb−Fb,implyingFa−Fb≤1−γa−b+μa−Fb+b−Fa.They also showed that if a mapping satisfies the (C) condition, then it satisfies theBγ,μ condition but the converse does not hold in general.LetG be a nonempty closed convex subset of a Banach space X, F:G⟶G be a mapping, λ∈0,1, and n≥1.The well-known Picard [6] and Krasnoselskii [7] iteration processes are, respectively, defined as(4)a1=a∈G,an+1=Fan,(5)a1=a∈G,an+1=1−λan+λFan.In 2017, Okeke and Abbas [8] considered the Picard–Krasnoselskii hybrid iteration process as follows:(6)a1=a∈G,bn=1−λan+λFan,an+1=Fbn.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 Bγ,μ. In the last section, we give an example of mapping F which satisfies condition Bγ,μ but not (C) and compare the rate of convergence of Picard–Krasnoselskii iteration, Picard iteration, and Krasnoselskii iteration.


## 2. Preliminaries

LetX be a Banach space. We say that X is uniformly convex [9] provided that for any r∈0,2, there is a d>0 such that for any a,b∈X with a≤1, b≤1, and a+b>r; it follows that(7)12a+b≤1−d.We say thatX satisfies Opial’s property [10] provided that for any an in X which weakly converges to a∈X and for all b∈X−a, someone has(8)limsupn⟶∞an−a<limsupn⟶∞an−b.Definition 1.
LetG be a nonempty subset of a Banach space X, an in X be bounded, and ra,an=limsupn⟶∞an−a. The asymptotic radius of an relative to G is the set rG,an=infra,an:a∈G. Moreover, the asymptotic center of an relative to G is the set AG,an=a∈G:ra,an=rG,an.
We know that the setAG,an is singleton whenever the underlying space is uniformly convex Banach. Also, AG,an is nonempty as well as convex whenever G is weakly compact and convex (see, e.g., [11, 12]).Lemma 1.
(see [5]). Let G be a nonempty subset of a Banach space X and F:G⟶G satisfies Bγ,μ condition. If q is a fixed point of F:G⟶G, then for each a∈G,(9)q−Fa≤q−a.
From Lemma1, we obtain the following facts.Lemma 2.
LetG be a nonempty subset of a Banach space X. Let F:G⟶G satisfy condition Bγ,μ. Then, the set fF is closed. Moreover, if X is strictly convex and G is convex, then fF is also convex.Theorem 1.
(see [5]). Let G be a nonempty subset of a Banach space X having Opial property. Let F:G⟶G satisfy condition Bγ,μ. If an is sequence in X such that(i)
an converges weakly to a,(ii)
limn⟶∞an−Fan=0,then Fa=a.Proposition 1.
(see [5]). Let G be a nonempty subset of a Banach space X. If F:G⟶G satisfies condition Bγ,μ on G, then for all a,b∈G and c∈0,1,(i)
Fa−F2a≤a−Fa(ii)
At least (g) and (h) hold:(g)
c/2a−Fa≤a−b(h)
c/2Fa−F2a≤Fa−bThe Condition (g) implies Fa−Fb≤1−c/2a−b+μa−Fb+b−Fa and condition (h) implies F2a−Fb≤1−c/2Fa−b+μFa−Fb+b−F2a(iii)
a−Fb≤3−ca−Fa+1−c/2a−b+μ2a−Fa+a−Fb+b−Fa+2Fa−F2a
We need the following useful lemma from [13].Lemma 3.
LetX be a uniformly convex Banach space and 0<g≤θn≤h<1 for every n≥1. If sn and tn are two sequences in X such that limsupn⟶∞sn≤r, limsupn⟶∞tn≤r, and limn⟶∞θnsn+1−θntn=r for some r≥0, then limn⟶∞sn−tn=0.


## 3. Main Results

The following lemma will be used in the upcoming results.Lemma 4.
LetG be a nonempty closed convex subset of a Banach space X. Suppose that F:G⟶G satisfies condition Bγ,μ and fF≠∅. If an is a sequence defined by (6), then limn⟶∞an−q exists for each q∈fF.Proof.
Suppose thatq∈fF. By Lemma 1, we have(10)bn−q=1−λan+λFan−q≤1−λan−q+λFan−q≤1−λan−q+λan−q≤an−q,which implies that(11)an+1−q=Fbn−q≤bn−q≤an−q.
Thus,an−q is bounded as well as nonincreasing. Hence, limn⟶∞an−q exists for each q∈fF.Theorem 2.
LetG be a nonempty closed convex subset of a uniformly convex Banach space X. Suppose that F:G⟶G satisfies condition Bγ,μ and let an be a sequence defined by (6). Then, fF≠∅ if and only if an is bounded and limn⟶∞an−Fan=0.Proof.
We assumeq∈AG,an. By Proposition 1 (iii), for γ=c/2, c∈0,1,(12)an−Fq≤3−can−Fan+1−c2an−q+μ2an−Fan+an−Fq+q−Fan+2Fan−F2an≤3−can−Fan+1−c2an−q+μ2an−Fan+an−Fq+an−q+an−Fan+2an−Fan,(by Proposition 1 (i))(13)⇒1−μlimsupn⟶∞an−Fq≤1−c2+μlimsupn⟶∞an−q⇒limsupn⟶∞an−Fq≤1−c/2+μ1−μlimsupn⟶∞an−q≤limsupn⟶∞an−qas1−c/2+μ1−μ≤1,for2μ≤γ=c2,⇒rFq,an≤rq,an.
So,Fq∈AG,an. But AG,an is singleton, and we have Fq=q. Hence, fF≠∅.
Conversely, letq∈fF. By Lemma 4, limn⟶∞an−q exists. Assume that limn⟶∞an−q=r. We first prove that limn⟶∞bn−q=r. By the proof of Lemma 4, an+1−q≤bn−q; therefore,(14)liminfn⟶∞an+1−q≤liminfn⟶∞bn−q,and so r≤liminfn⟶∞bn−q.
Again by the proof of Lemma4, bn−q≤an−q. Hence, limsupn⟶∞bn−q≤r. Therefore, we obtain limn⟶∞bn−q=r. Also, by Lemma 1, Fan−q≤an−q. It follows that limsupn⟶∞Fan−q≤r. By Lemma 3, we obtain(15)limn⟶∞Fan−an=0.
Now, we are in the position to establish a weak convergence ofan defined by (6) for the class of mappings with condition Bγ,μ.Theorem 3.
LetG a nonempty closed convex subset of a uniformly convex Banach space X with the Opial property. Suppose that F:G⟶G satisfies condition Bγ,μ and fF≠∅. Then, an defined by (6) converges weakly to an element of fF.Proof.
By Theorem2, an is bounded and limn⟶∞an−Fan=0. Since X is uniformly convex, X is reflexive. Thus, we can find a subsequence ant of an such that ant converges weakly to some u1∈G. By Theorem 1, we obtain u1∈fF. It is sufficient to prove that an converges weakly to u1. Indeed, if an does not converge weakly to u1, then we can find a subsequence ans of an and u2∈G such that ans converges weakly to u2 and u2≠u1. Hence, u2∈fF by Theorem 1. Opial condition and Lemma 4 give us(16)limn⟶∞an−u1=limt⟶∞ant−u1<limt⟶∞ant−u2=limn⟶∞an−u2=lims⟶∞ans−u2<lims⟶∞ans−u1=limn⟶∞an−u1.
This is a contradiction. So,u1=u2.Theorem 4.
LetG be a nonempty closed convex subset of a uniformly convex Banach space X. Suppose that F:G⟶G satisfies condition Bγ,μ. If fF≠∅ and liminfn⟶∞distan,fF=0, where an be the sequence defined by (6), then an converges strongly to a fixed point of F.Proof.
By Lemma4, limn⟶∞an−q exists, for each q∈fF. Thus, limn⟶∞distan,fF exists. Hence,(17)limn⟶∞distan,fF=0.
We can find a subsequenceant of an and pt in fF with ant−pt≤1/2t, n≥1. Moreover, an is nonincreasing by the proof of Lemma 4. Hence,(18)ant+1−pt≤ant−pt≤12t.
We shall prove thatpt is a Cauchy in fF.(19)pt+1−pt≤pt+1−ant+1+ant+1−pt≤12t+1+12t≤12t−1⟶0,ast⟶∞.
This shows that the sequencept is Cauchy in fF. By Lemma 2, fF is closed. Hence, pt⟶ω for some ω∈fF. By Lemma 4, limn⟶∞an−ω exists. So, the proof is finished.
Finally, we prove the following strong convergence theorem for the sequencean defined by (6) with the help of condition (I).Definition 2.
Recall that a self-mappingF on G subset of a Banach space is said to satisfy condition I [14] if and only if there exists ρ:0,∞⟶0,∞ satisfying ρ0=0 and ρu>0 for every u>0 such that(20)a−Fa≥ρdista,fF,for alla∈G.Theorem 5.
LetG be a nonempty closed convex subset of a uniformly convex Banach space X. Suppose that F:G⟶G satisfies condition Bγ,μ and fF≠∅. If F satisfies condition I, then an defined by (6) converges strongly to a fixed point of F.Proof.
By Theorem2, we have(21)liminfn⟶∞an−Fan=0.
By condition (I), we obtain(22)liminfn⟶∞distan,fF=0.
The conclusion follows from Theorem4.


## 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.
LetG=0,1 be endowed with the usual norm. Set Fa=0 if a∈A=0,1/100 and Fa=1/2a if a∈B=1/100,1. If a=1/160 and b=1/100, then 1/2a−Fa<a−b but Fa−Fb>a−b. Hence, F does not satisfy condition (C). However, F satisfies condition B1,1/2. The case when a,b∈A is trivial and hence omitted. We consider the following two nontrivial cases.
Whena,b∈B, we have(23)1−γa−b+μa−Fb+b−Fa=12a−b2+b−a2≥123a2−3b2=34a−b≥12a−b=Fa−Fb.
Whena∈B and b∈A, we have(24)1−γa−b+μa−Fb+b−Fa=12a+b−a2=12a+12b−a2≥12a=Fa−Fb.
Chooseλ=1/2∈0,1; the strong convergence of the sequence an defined by Picard–Krasnoselskii hybrid process (6) to q=0 can be seen in Table 1.Table 1
Convergence of Picard–Krasnoselskii hybrid, Picard, and Krasnoselskii iterations to the fixed pointq=0.
nPicard–KrasnoselskiiPicardKrasnoselskii10.80.80.820.30000000000000.40000000000000.600000000000030.11250000000000.20000000000000.450000000000040.04218750000000.10000000000000.337500000000050.01582031250000.05000000000000.253125000000060.00593261718750.02500000000000.1898437500000700.01250000000000.1423828125000800.00625000000000.10678710937509000.080090332031210000.0059326171875Remark 1.
From Table1 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 Bγ,μ.Figure 1
Convergence behavior of Picard–Krasnoselskii, Picard, and Krasnoselskii iterates for mappingF defined in Example 1 where a1=0.8.



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