# General Higher-Order Lipschitz Mappings

Authors: Joseph Frank Gordon
Journal: Journal of Mathematics
 (2021)

---

## Abstract

In this paper, we introduce a new class of mappings and investigate their fixed point property. In the first direction, we prove a fixed point theorem for general higher-order contraction mappings in a given metric space and finally prove an approximate fixed point property for general higher-order nonexpansive mappings in a Banach space.


---

## Body

## 1. Introduction

Given a complete metric spaceX,d, the most well-studied types of self-maps are referred to as Lipschitz mappings (or Lipschitz maps, for short), which are given by the metric inequality:(1)dTx,Ty≤αdx,y,for all x,y∈X, where α≥0 is a real number, usually referred to as the Lipschitz constant of T. Now, for technical and historical reasons, we classify Lipschitz mappings into three categories, thus contraction mappings for the case where 0≤α<1, nonexpansive mappings for the case where α=1, and expansive mappings for the case where α>1.In 2007, Goebel and Japon Pineda [1] introduced the so-called mean nonexpansive mappings which are class of mappings more general than the class of nonexpansive mappings. Thus, given a Banach space X,⋅ and a nonempty subset C of X, a mapping T:C⟶C is called mean nonexpansive (or α-nonexpansive) if, for some α=α1,α2,…,αn with ∑k=1nαk=1,αk≥0 for all k, and α1,αn>0, we have(2)∑k=1nαkTkx−Tky≤x−y,for allx,y∈C.It is obvious that all nonexpansive mappings are mean nonexpansive mappings but the converse may not be true (see for instance [2], examples 2.3 and 2.4). Goebel and Japón Pineda further suggested the class of α,p-nonexpansive maps. A self-map T:C⟶C is called α,p-nonexpansive if, for some α=α1,α2,…,αn with ∑k=1nαk=1,αk≥0 for all k, and α1,αn>0 and for some p∈1,∞,(3)∑k=1nαkTkx−Tkyp≤x−yp,for all x,y∈C. It is easy to check that α,p-nonexpansive map for p>1 is also α-nonexpansive, but the converse does not hold (see [3] for details), whereas nonexpansive mappings are uniformly continuous and that continuous, mean nonexpansive mappings may not generally be continuous as seen in the following example taken from [2], example 2.2.Example 1.
Letf:0,1⟶0,1 be given by(4)fx≔1,x=0,0,x≠0.
Clearly,f is discontinuous but a mean nonexpansive mapping.In 2015, Ezearn [4] introduced a new class of mappings called higher-order Lipschitz mappings which are seen as a generalization of inequality (1). Thus, a mapping T:X⟶X on a metric space X,d is an rth-order Lipschitz mapping if(5)dTrx,Try≤∑k=0r−1ckdTkx,Tky∀x,y∈X,where r is a natural number and ck, for all 0≤k≤r−1, are nonnegative real numbers, whereas Lipschitz mappings are (uniformly) continuous mappings; this needs not be the case for higher-order Lipschitz mappings; they may not even be continuous as given by the following example taken from [4].Example 2.
LetT:ℝ⟶ℝ given by(6)Tx≔1−x,ifx<0,x,ifx≥0,with the metric induced by the usual absolute value on ℝ. Clearly, T is discontinuous at x=0 but a second-order Lipschitz (indeed, a second-order nonexpansive) mapping.In this paper, we introduce the following new class of mappings which generalizes both inequalities (3) and (5).Definition 1.
(general higher-order Lipschitz mappings). A mappingT:X⟶X on a metric space X,d is a r,p-general rth-order Lipschitz mapping if(7)∑k=l+1rαkdTkx,Tkyp≤∑k=0lαkdTkx,Tkyp∀x,y∈X,where p∈1,∞, r is a natural number, αk≥0 for all k, α0⋅αr≠0, and l∈0,…,r−1.It is clear that inequality (7) reduces to inequality (3) when l=0. Similarly, inequality (7) reduces to inequality (5) when p=1 and l=r−1. Generally, a (r,p)-general higher-order Lipschitz mapping may not be continuous as already seen in examples 1 and 2.Now, to everyr,p-general higher-order Lipschitz mapping, we associate a polynomial defined as(8)hz=∑k=l+1rαkzk−∑k=0lαkzk,just as Ezearn did in his paper [4]. Now, we classify r,p-general higher-order Lipschitz mappings into three main categories, thus(i)
T is r,p-general higher-order contraction mapping if h1>0.(ii)
T is r,p-general higher-order nonexpansive mapping if h1=0.(iii)
T is r,p-general higher-order expansive mapping if h1<0.Now, we provide some fixed point results as it is presently understood and known and their connections to the research focus of this paper.Theorem 1.
(Banach contraction mapping theorem). LetX,d be a complete metric space and let T:X⟶X be a contraction mapping. Then, T has a unique fixed point and limn⟶∞Tnx converges to this fixed point for any x∈X.Gordon [5] provided a version of Theorem 1 in a general Banach space using the concept of the normalized duality mappings and the generalized projection functional. He defined the following set of mappings.Definition 2.
(monotone contraction mapping). LetX be a smooth Banach space and let C be a closed subset of X. Then, the mapping T:C⟶C is said to be a monotone contraction mapping if there exists 0≤c<1 such that for all x,y∈C, the following two conditions are satisfied:(1)
ℜTx−Ty,JTx−JTy≤cℜx−y,Jx−Jy,(2)
ℜTm+1x−Tmy,JTn+1x−JTny≤0,where ℜ denotes the real part of a complex number and J is the normalized duality mapping for all m,n≥0 with m≠n.His result is summarized in the following theorem.Theorem 2.
(monotone contraction mapping theorem). LetC be a closed subset of a uniformly convex smooth Banach space X and let T:C⟶C be a monotone contraction mapping. Then, T has a unique fixed point, that is, FT=p and that the Picard iteration associated to T, that is, the sequence defined by xn=Txn−1=Tnx0 for all n≥1 converges to p for any initial guess x0∈X.Ezearn [4] extended the conclusion of the Banach contraction mapping theorem (Theorem 1) to higher-order contraction mappings as summarized in the following theorem.Theorem 3.
(higher-order contraction mapping theorem). LetX,d be a complete metric space and let T:X⟶X be an rth-order contraction mapping. Then, T has a unique fixed point and limn⟶∞Tnx converges to this fixed point for arbitrary x∈X.Ezearn later provided a remetrisation argument that relates his higher-order Lipschitz mappings to (first-order) Lipschitz mappings. His remetrisation of the original metric space does not necessarily result in a complete metric space, and as a result, he provided a completion of the remetrised space and an extension of the higher-order Lipschitz mapping into a complete remetrised space. Before stating this theorem, let us introduce the following: a new metric on the spaceX as already defined by Ezearn in his paper,(9)Dy,x=∑k=0r−1bkdTky,Tkx,wherebk=∑j=0kcjλj−k−1.His theorem (already stated as Theorem 3.5 in his paper) is stated as follows.Theorem 4.
Define the mapping,(10)T¯:X¯⟶X¯,xn⟶Txn.
Then, we have(11)D¯T¯yn,T¯xn≤λD¯yn,xn,where ynn≥1,xnn≥1 are Cauchy sequences in X,D and xn denotes the equivalence class of xnn≥1 in X,D. In particular, if X,d is complete, then T has a fixed point in X,d if and only if T¯ has a fixed point in X¯,D¯ where X¯,D¯ is the canonical completion of the metric space X,D; that is, D¯yn,xn:=limn⟶∞Dyn,xn.In light of Theorem4, this paper carefully uses ideas of the auxiliary results leading to the proof of Theorem 4 to establish entirely similar proof of Theorem 4 as far as r,p-general higher-order contraction mappings are concerned.These references [6–9] can be consulted for fixed point problems in metric spaces.An approximate fixed point sequence of a nonexpansive self-mapT on a closed convex subset C of Banach space X is any sequence xnn≥1 such that(12)limn⟶∞xn−Txn=0.WhenC is bounded, then such a sequence always exists. One of the results in this paper is to show that given any Banach space X and a closed bounded convex subset C containing the origin, an approximate fixed point sequence always exists for positively homogeneous r,p-general higher-order nonexpansive mappings.


## 2. Preliminaries

Theorem 5.
(Bolzano’s intermediate value [10]). If a continuous real function defined on an interval is sometimes positive and sometimes negative, then it must be 0 at some point in the interval.Theorem 6.
(Descartes’ rule of signs, see for instance, [11]). Let fz:=∑k=0rakzk be an rth degree polynomial over the real numbers ak. Then, the number of positive real roots of f is bounded above by the number of sign changes of the coefficients ak as one proceeds from k=0 to k=r (ignoring zero coefficients).Proposition 1.
Lethz=∑k=l+1rαkzk−∑k=0lαkzk, where αk≥0 is a polynomial.(i)
Ifh1>0, then there exists λ∈0,1, which is unique and positive if αk≠0 such that hλ=0.(ii)
Ifh1=0, then 1 is the only positive root of p.(iii)
Ifh1<0, then there exists a unique positive λ>1 such that hλ=0.Proof.
For (i): now, givenhz=∑k=l+1rαkzk−∑k=0lαkzk−α0, then h0=−α0<0 since α0≠0. So, with h1>0 and h0<0, by Theorem 5, we have that there exists λ∈0,1 such that hλ=0. The uniqueness of λ when α0≠0, in which case λ cannot be 0, follows from Theorem 6.For (ii): obviously, 1 is the only positive root which follows Theorem6.For (iii): here, we note that for real numberst, we have limt⟶∞ht=∞, and as a result, we can find t0≥1 such that ht0≥0. Since h1<0 and ht0≥0, then by Theorem 5, there exists λ∈1,t0 such that hλ=0. The uniqueness of λ follows from Theorem 6.


## 3. Main Result

We give the proof of the main result of this paper, which is accomplished in Theorems7 and 8. The following lemma, corollary, and proposition shall aid us in arriving at the conclusion of the main result.


### 3.1. Fixed Point Theorem for General Higher-Order Contraction Mappings

LetT be a r,p-general higher-order Lipschitz mapping on a complete metric space X,d as given in inequality (13) and let λ be the unique root of the polynomial hz=∑k=l+1rαkzk−∑k=0lαkzk as guaranteed by Proposition 1. Now, let us define the following on the space X:(13)Dpx,y=∑k=0r−1bkdTkx,Tkyp1/p,wherebk=∑j=0kαjλj−k−1sgnj,and p≥1 for all x,y∈X,0≤k≤r−1 and sgnj=1,ifj≤l,−1,ifj>l.Corollary 1.
Letbk be defined in equation (13). Then, bk is nonnegative.Proof.
To see this, ifk≤l, then λk+1bk=∑j=0kαjλj≥0. On the other hand, if k≥l+1, then(14)λk+1bk=∑j=0lαjλj−∑j=l+1kαjλj,=∑j=0lαjλj−∑j=l+1rαjλj+∑j=k+1rαjλj,=−hλ+∑j=k+1rαjλj,=0+∑j=k+1rαjλj≥0,and that completes the proof.Lemma 1.
LetDpx,y be defined in equation (13). Then, Dp is a metric on the space X.Proof.
By Corollary1, bk is nonnegative. Now, for p=1 is straightforward. For p>1, it is obvious that Dp is nonnegative since d and bk are nonnegative and it is subadditive since d is subadditive. We have also that Dpx,y=0 if and only if x=y since bk≠0 because of the fact that α0≠0. Also, Dpx,y=Dpy,x. Finally, without loss of generality, suppose x≠z. Then, we have the following evaluation:(15)Dpx,zp=∑k=0r−1bkdTkx,Tkzp,=∑k=0r−1bkdTkx,TkzdTkx,Tkzp−1,≤∑k=0r−1bk1/pdTkx,Tkybk1−1/pdTkx,Tkzp−1+∑k=0r−1bk1/pdTky,Tkzbk1−1/pdTkx,Tkzp−1.
Using the Hölder’s inequality on each of the terms on the right hand side, we have(16)Dpx,zp≤∑k=0r−1bkdTkx,Tkyp1/p∑k=0r−1bkdTkx,Tkzp1/q+∑k=0r−1bkdTky,Tkzp1/p∑k=0r−1bkdTkx,Tkzp1/q,=Dpx,y+Dpy,z∑k=0r−1bkdTkx,Tkzp1/p⋅p/q,=Dpx,y+Dpy,zDpx,zp/q.
From1/p+1/q=1,p−1/p=q⇒p−1q=p, we obtain(17)Dpx,z=Dpx,y+Dpy,z,and that completes the proof.Proposition 2.
Letbk be defined in equation (13). Then, the following recurrence relations hold:(18)α0=λb0,br−1−αr=0,λbk=bk−1+αksgnk,1≤k≤r−1.Proof.
b0=∑j=00αjλj−k−1sgnj=α0λ−1⇒α0=λb0.
Also,(19)bk−1+αksgnk=∑j=0k−1αjλj−ksgnj+αksgnk=∑j=0kαjλj−ksgnj,λ−1bk−1+αksgnk=λ−1∑j=0kαjλj−ksgnj=∑j=0kαjλj−k−1sgnj=bk,and multiplying through again by λ completes the proof.
Finally, fork=r−1, we have(20)λrbr−1=∑j=0lαjλj−∑j=l+1r−1αjλj,=∑j=0lαjλj−∑j=l+1r−1αjλj−αrλr+αrλr,=∑j=0lαjλj−∑j=l+1rαjλj+αrλr,=−hλ+αrλr.
Hence,λrbr−1=αrλr⇒br−1=αr.Lemma 2.
LetX,d be a (not necessarily complete) metric space and let T:X⟶X be a r,p-general higher-order Lipschitz mapping. Let Dp be the new metric defined in equation (13). Then,(21)DpTx,Ty≤λ1/pDpx,y.
Moreover, a sequencexnn≥1⊂X,Dp is Cauchy in X,Dp if and only if the sequence Tkxnn≥1⊂X,d is Cauchy in X,d for all 0≤k≤r−1.Proof.
(22)DpTx,Typ=∑k=0r−1bkdTk+1x,Tk+1yp,=∑k=1rbk−1dTkx,Tkyp,=∑k=l+1rαkdTkx,Tkyp+∑k=l+1rbk−1−αkdTkx,Tkyp+∑k=1lbk−1dTkx,Tkyp,≤∑k=0lαkdTkx,Tkyp+∑k=l+1rbk−1−αkdTkx,Tkyp+∑k=1lbk−1dTkx,Tkyp,=α0dx,yp+∑k=1lαkdTkx,Tkyp+∑k=l+1rbk−1−αkdTkx,Tkyp+∑k=1lbk−1dTkx,Tkyp,=α0dx,yp+∑k=1lαk+bk−1dTkx,Tkyp+∑k=l+1rbk−1−αkdTkx,Tkyp,=α0dx,yp+∑k=1rbk−1+αksgnkdTkx,Tkyp,=α0dx,yp+br−1−αrdTkx,Tkyp+∑k=1r−1bk−1+αksgnkdTkx,Tkyp,=λb0dx,yp+λ∑k=1r−1bkdTkx,Tkyp,=λ∑k=0r−1bkdTkx,Tkyp,=λDx,yp.
Now, given that the sequencexnn≥1⊂X,Dp is Cauchy, that is, limn,m⟶∞Dpxn,xm=0, then it is obvious that limn,m⟶∞dTkxn,Tkxm=0 for all 0≤k≤r−1. Similarly, if Tkxnn≥1⊂X,d is Cauchy in X,d, that is, limn,m⟶∞dTkxn,Tkxm=0 for all 0≤k≤r−1, then we have limn,m⟶∞Dpxn,xm=0 which completes the proof.As noted by Ezearn in his paper, Lemma2 does not imply that T is uniformly continuous or even continuous in X,d as seen in the introduction; rather, T is Lipschitz continuous and therefore uniformly continuous in X,Dp. We must also note that when λ<1, Theorem 1 cannot be used to conclude that T has a fixed point in X,Dp unless T is continuous in X,d. However, the following theorem remedies the case when T is discontinuous on X,d. Here, our proof follows exactly as in Ezearn’s Theorem 3.5, but we give the proof for the sake of completeness. To begin, let X¯,Dp¯ be the canonical completion of the metric space X,Dp; that is,(23)Dp¯yn,xn≔limn⟶∞Dpyn,xn,where ynn≥1,xnn≥1 are Cauchy sequences in X,Dp and xn,yn denote the equivalence class of xnn≥1 and ynn≥1 in X,Dp, where ynn≥1 is equivalent to xnn≥1 if limn⟶∞Dpyn,xn=0.Theorem 7.
Define the mapping,(24)T¯:X¯⟶X¯,xn⟶Txn.
Then, we have(25)Dp¯T¯yn,T¯xn≤λ1/pDp¯yn,xn.
In particular, ifX,d is complete, then T has a fixed point in X,d if and only if T¯ has a fixed point in X¯,Dp¯.Proof.
Sincexnn≥1 is Cauchy in X,Dp, then, by Lemma 2,(26)DpTxn,Txm≤λ1/pDpxn,xm,and so Txnn≥1 is Cauchy in X,Dp; thus, T¯ is well defined. Now, given Cauchy sequences xnn≥1,ynn≥1 in X,Dp, then we have(27)Dp¯T¯xn,T¯yn=Dp¯Tyn,Txn,=limn⟶∞DpTyn,Txn,≤λ1/plimn⟶∞Dpyn,xn,=λ1/pDp¯yn,xn.
Finally, ifx=Tx in X,d, then let x be the equivalence class of the constant sequence x,x,x,…∈X¯,Dp¯. Then, we have(28)T¯x=Tx=x.
On the other hand, ifxn=T¯xn=Txn in X¯,Dp¯, then, by Lemma 2, T is continuous at x:=limn⟶∞xn in X,d, hence(29)Tx=limn⟶∞Txn=limn⟶∞xn=x,which completes the proof.


### 3.2. Approximate Fixed Point Property for Positively Homogeneous General Higher-Order Nonexpansive Mappings

In this subsection, we prove the approximate fixed point property forr,p-general higher-order nonexpansive mapping in a Banach space, which is accomplished in the following theorem.Theorem 8.
LetC be a closed bounded convex subset containing the origin of a Banach space X and let T:C⟶C be a positively homogeneous general higher-order nonexpansive mapping. Then, T has an approximate fixed point sequence.Proof.
Now, for allγ∈0,1, define Tγ:C⟶C by(30)Tγx=γTx,∀x∈C.
Then,T2γx=γ2T2x,T3γx=γ3T3x,…,Tkγx=γkTkx since T is a homogeneous general higher-order nonexpansive map. By inequality (7), we have the following evaluation:(31)∑k=l+1rαkγ−kTkγx−Tkγyp=∑k=l+1rαkTkx−Tkyp,≤∑k=0lαkTkx−Tkyp,=∑k=0lαkγ−kTkγx−Tkγyp.
Hence, we have(32)∑k=l+1rαkγ−kTγkx−Tγkyp=∑k=0lαkγ−kTγkx−Tγkyp.
Now, we have that(33)∑k=l+1rαkγ−1k−∑k=0lαkγ−1k≥γ−l+1∑k=l+1rαk−γ−l∑k=0lαk.
SinceT is a general higher-order nonexpansive mapping, then, by Proposition 1, we have that ∑k=l+1rαk=∑k=0lαk. Now, let C=∑k=l+1rαk=∑k=0lαk. Then, inequality (33) becomes(34)∑k=l+1rαkγ−1k−∑k=0lαkγ−1k≥γ−l+1−γ−l⋅C>0,since αr⋅α0≠0. Hence, equation (32) is a general higher-order contraction mapping and thus, by Theorem 7, Tγ has a unique fixed point xγ in C. That is, Tγxγ=xγ. Let γ=1−1/n for all n≥1. Then,(35)xn−Txn=1nTxn≤1ndiamC.
In the limit asn⟶∞, it follows that xn−Txn⟶0 and that completes the proof.


## 3.1. Fixed Point Theorem for General Higher-Order Contraction Mappings

LetT be a r,p-general higher-order Lipschitz mapping on a complete metric space X,d as given in inequality (13) and let λ be the unique root of the polynomial hz=∑k=l+1rαkzk−∑k=0lαkzk as guaranteed by Proposition 1. Now, let us define the following on the space X:(13)Dpx,y=∑k=0r−1bkdTkx,Tkyp1/p,wherebk=∑j=0kαjλj−k−1sgnj,and p≥1 for all x,y∈X,0≤k≤r−1 and sgnj=1,ifj≤l,−1,ifj>l.Corollary 1.
Letbk be defined in equation (13). Then, bk is nonnegative.Proof.
To see this, ifk≤l, then λk+1bk=∑j=0kαjλj≥0. On the other hand, if k≥l+1, then(14)λk+1bk=∑j=0lαjλj−∑j=l+1kαjλj,=∑j=0lαjλj−∑j=l+1rαjλj+∑j=k+1rαjλj,=−hλ+∑j=k+1rαjλj,=0+∑j=k+1rαjλj≥0,and that completes the proof.Lemma 1.
LetDpx,y be defined in equation (13). Then, Dp is a metric on the space X.Proof.
By Corollary1, bk is nonnegative. Now, for p=1 is straightforward. For p>1, it is obvious that Dp is nonnegative since d and bk are nonnegative and it is subadditive since d is subadditive. We have also that Dpx,y=0 if and only if x=y since bk≠0 because of the fact that α0≠0. Also, Dpx,y=Dpy,x. Finally, without loss of generality, suppose x≠z. Then, we have the following evaluation:(15)Dpx,zp=∑k=0r−1bkdTkx,Tkzp,=∑k=0r−1bkdTkx,TkzdTkx,Tkzp−1,≤∑k=0r−1bk1/pdTkx,Tkybk1−1/pdTkx,Tkzp−1+∑k=0r−1bk1/pdTky,Tkzbk1−1/pdTkx,Tkzp−1.
Using the Hölder’s inequality on each of the terms on the right hand side, we have(16)Dpx,zp≤∑k=0r−1bkdTkx,Tkyp1/p∑k=0r−1bkdTkx,Tkzp1/q+∑k=0r−1bkdTky,Tkzp1/p∑k=0r−1bkdTkx,Tkzp1/q,=Dpx,y+Dpy,z∑k=0r−1bkdTkx,Tkzp1/p⋅p/q,=Dpx,y+Dpy,zDpx,zp/q.
From1/p+1/q=1,p−1/p=q⇒p−1q=p, we obtain(17)Dpx,z=Dpx,y+Dpy,z,and that completes the proof.Proposition 2.
Letbk be defined in equation (13). Then, the following recurrence relations hold:(18)α0=λb0,br−1−αr=0,λbk=bk−1+αksgnk,1≤k≤r−1.Proof.
b0=∑j=00αjλj−k−1sgnj=α0λ−1⇒α0=λb0.
Also,(19)bk−1+αksgnk=∑j=0k−1αjλj−ksgnj+αksgnk=∑j=0kαjλj−ksgnj,λ−1bk−1+αksgnk=λ−1∑j=0kαjλj−ksgnj=∑j=0kαjλj−k−1sgnj=bk,and multiplying through again by λ completes the proof.
Finally, fork=r−1, we have(20)λrbr−1=∑j=0lαjλj−∑j=l+1r−1αjλj,=∑j=0lαjλj−∑j=l+1r−1αjλj−αrλr+αrλr,=∑j=0lαjλj−∑j=l+1rαjλj+αrλr,=−hλ+αrλr.
Hence,λrbr−1=αrλr⇒br−1=αr.Lemma 2.
LetX,d be a (not necessarily complete) metric space and let T:X⟶X be a r,p-general higher-order Lipschitz mapping. Let Dp be the new metric defined in equation (13). Then,(21)DpTx,Ty≤λ1/pDpx,y.
Moreover, a sequencexnn≥1⊂X,Dp is Cauchy in X,Dp if and only if the sequence Tkxnn≥1⊂X,d is Cauchy in X,d for all 0≤k≤r−1.Proof.
(22)DpTx,Typ=∑k=0r−1bkdTk+1x,Tk+1yp,=∑k=1rbk−1dTkx,Tkyp,=∑k=l+1rαkdTkx,Tkyp+∑k=l+1rbk−1−αkdTkx,Tkyp+∑k=1lbk−1dTkx,Tkyp,≤∑k=0lαkdTkx,Tkyp+∑k=l+1rbk−1−αkdTkx,Tkyp+∑k=1lbk−1dTkx,Tkyp,=α0dx,yp+∑k=1lαkdTkx,Tkyp+∑k=l+1rbk−1−αkdTkx,Tkyp+∑k=1lbk−1dTkx,Tkyp,=α0dx,yp+∑k=1lαk+bk−1dTkx,Tkyp+∑k=l+1rbk−1−αkdTkx,Tkyp,=α0dx,yp+∑k=1rbk−1+αksgnkdTkx,Tkyp,=α0dx,yp+br−1−αrdTkx,Tkyp+∑k=1r−1bk−1+αksgnkdTkx,Tkyp,=λb0dx,yp+λ∑k=1r−1bkdTkx,Tkyp,=λ∑k=0r−1bkdTkx,Tkyp,=λDx,yp.
Now, given that the sequencexnn≥1⊂X,Dp is Cauchy, that is, limn,m⟶∞Dpxn,xm=0, then it is obvious that limn,m⟶∞dTkxn,Tkxm=0 for all 0≤k≤r−1. Similarly, if Tkxnn≥1⊂X,d is Cauchy in X,d, that is, limn,m⟶∞dTkxn,Tkxm=0 for all 0≤k≤r−1, then we have limn,m⟶∞Dpxn,xm=0 which completes the proof.As noted by Ezearn in his paper, Lemma2 does not imply that T is uniformly continuous or even continuous in X,d as seen in the introduction; rather, T is Lipschitz continuous and therefore uniformly continuous in X,Dp. We must also note that when λ<1, Theorem 1 cannot be used to conclude that T has a fixed point in X,Dp unless T is continuous in X,d. However, the following theorem remedies the case when T is discontinuous on X,d. Here, our proof follows exactly as in Ezearn’s Theorem 3.5, but we give the proof for the sake of completeness. To begin, let X¯,Dp¯ be the canonical completion of the metric space X,Dp; that is,(23)Dp¯yn,xn≔limn⟶∞Dpyn,xn,where ynn≥1,xnn≥1 are Cauchy sequences in X,Dp and xn,yn denote the equivalence class of xnn≥1 and ynn≥1 in X,Dp, where ynn≥1 is equivalent to xnn≥1 if limn⟶∞Dpyn,xn=0.Theorem 7.
Define the mapping,(24)T¯:X¯⟶X¯,xn⟶Txn.
Then, we have(25)Dp¯T¯yn,T¯xn≤λ1/pDp¯yn,xn.
In particular, ifX,d is complete, then T has a fixed point in X,d if and only if T¯ has a fixed point in X¯,Dp¯.Proof.
Sincexnn≥1 is Cauchy in X,Dp, then, by Lemma 2,(26)DpTxn,Txm≤λ1/pDpxn,xm,and so Txnn≥1 is Cauchy in X,Dp; thus, T¯ is well defined. Now, given Cauchy sequences xnn≥1,ynn≥1 in X,Dp, then we have(27)Dp¯T¯xn,T¯yn=Dp¯Tyn,Txn,=limn⟶∞DpTyn,Txn,≤λ1/plimn⟶∞Dpyn,xn,=λ1/pDp¯yn,xn.
Finally, ifx=Tx in X,d, then let x be the equivalence class of the constant sequence x,x,x,…∈X¯,Dp¯. Then, we have(28)T¯x=Tx=x.
On the other hand, ifxn=T¯xn=Txn in X¯,Dp¯, then, by Lemma 2, T is continuous at x:=limn⟶∞xn in X,d, hence(29)Tx=limn⟶∞Txn=limn⟶∞xn=x,which completes the proof.


## 3.2. Approximate Fixed Point Property for Positively Homogeneous General Higher-Order Nonexpansive Mappings

In this subsection, we prove the approximate fixed point property forr,p-general higher-order nonexpansive mapping in a Banach space, which is accomplished in the following theorem.Theorem 8.
LetC be a closed bounded convex subset containing the origin of a Banach space X and let T:C⟶C be a positively homogeneous general higher-order nonexpansive mapping. Then, T has an approximate fixed point sequence.Proof.
Now, for allγ∈0,1, define Tγ:C⟶C by(30)Tγx=γTx,∀x∈C.
Then,T2γx=γ2T2x,T3γx=γ3T3x,…,Tkγx=γkTkx since T is a homogeneous general higher-order nonexpansive map. By inequality (7), we have the following evaluation:(31)∑k=l+1rαkγ−kTkγx−Tkγyp=∑k=l+1rαkTkx−Tkyp,≤∑k=0lαkTkx−Tkyp,=∑k=0lαkγ−kTkγx−Tkγyp.
Hence, we have(32)∑k=l+1rαkγ−kTγkx−Tγkyp=∑k=0lαkγ−kTγkx−Tγkyp.
Now, we have that(33)∑k=l+1rαkγ−1k−∑k=0lαkγ−1k≥γ−l+1∑k=l+1rαk−γ−l∑k=0lαk.
SinceT is a general higher-order nonexpansive mapping, then, by Proposition 1, we have that ∑k=l+1rαk=∑k=0lαk. Now, let C=∑k=l+1rαk=∑k=0lαk. Then, inequality (33) becomes(34)∑k=l+1rαkγ−1k−∑k=0lαkγ−1k≥γ−l+1−γ−l⋅C>0,since αr⋅α0≠0. Hence, equation (32) is a general higher-order contraction mapping and thus, by Theorem 7, Tγ has a unique fixed point xγ in C. That is, Tγxγ=xγ. Let γ=1−1/n for all n≥1. Then,(35)xn−Txn=1nTxn≤1ndiamC.
In the limit asn⟶∞, it follows that xn−Txn⟶0 and that completes the proof.



---

*Source: 5570373-2021-04-01.xml*