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Kyungpook Mathematical Journal 2023; 63(3): 451-461

Published online September 30, 2023 https://doi.org/10.5666/KMJ.2023.63.3.451

Copyright © Kyungpook Mathematical Journal.

Some Results on Generalized Asymptotically Nonexpansive Mappings in p-Hadamard Spaces

Kaewta Juanak, Aree Varatechakongka and Withun Phuengrattana∗

Department of Mathematics, Faculty of Science and Technology, Nakhon Pathom Rajabhat University, Nakhon Pathom 73000, Thailand
e-mail : ams_2413@hotmail.com, aree@npru.ac.th and withunph@webmail.npru.ac.th

Received: March 10, 2023; Revised: July 7, 2023; Accepted: July 20, 2023

In this paper, we study the fixed point property for generalized asymptotically nonexpansive mappings in the setting of p-Hadamard spaces, with p2. We prove the strong convergence of the sequence generated by the modified two-step iterative sequence for finding a fixed point of a generalized asymptotically nonexpansive mapping in p-Hadamard spaces.

Keywords: generalized asymptotically nonexpansive mappings, fixed point, existence theorems, convergence theorems, p-Hadamard spaces.

Let (X,d) be a metric space and x,yX. A geodesic joining x to y is a map γ from the closed interval [0,d(x,y)] to X such that γ(0)=x,γ(d(x,y))=y and d(γ(t1),γ(t2))=|t1t2| for all t1,t2[0,d(x,y)]. The image of γ is called a geodesic segment joining x and y. When it is unique, this geodesic segment is denoted by [x,y] and we write αx(1α)y for the unique point z in the geodesic segment joining from x to y such that d(x,z)=(1α)d(x,y) and d(y,z)=αd(x,y) for α[0,1]. The space (X,d) is said to be a geodesic metric space [2] if every two points of X are joined by a geodesic, and X is said to be uniquely geodesic if there is exactly one geodesic joining x and y for each x,yX.

A geodesic triangle (x1,x2,x3) in a geodesic metric space (X,d) consists of three points x1,x2,x3 in X and a geodesic segment between each pair of vertices. A comparison triangle for the geodesic triangle (x1,x2,x3) in X is a triangle ¯(x1,x2,x3):=( x ¯ 1, x ¯ 2, x ¯ 3) in 2 such that d2x¯i,x¯j=d(xi,xj) for i,j1,2,3. Let be a geodesic triangle in X and ¯ be a comparison triangle for in 2. Then is said to satisfy the CAT(0) inequality if for any x,y and their comparison points x¯,y¯¯, the following holds,

d(x,y)d2(x¯,y¯).

A geodesic metric space (X,d) is said to be a CAT(0) space if all geodesic triangles satisfy the CAT(0) inequality. It is well known that any complete, simply connected Riemannian manifold having nonpositive sectional curvature is a CAT(0) space; see more details in [2]. Other examples include Euclidean spaces, Hilbert spaces, the Hilbert ball [7], -trees [17], and many others.

In 2017, Khamsi and Shukri [9] introduced the concept of CATp(0) spaces based on the idea that comparison triangles belong to a general Banach space instead of the Euclidean plane as follows:

Definition 1.1. Let (E,) be a Banach space. A geodesic metric space (X,d) is said to be a CATE(0) space if for any geodesic triangle in X, there exists a comparison triangle ¯ in E such that the comparison axiom is satisfied, i.e., for all x,y and all comparison points x¯,y¯¯, we have

d(x,y)x¯y¯.

If E=lp, for p1, we say X is a CATp(0)

It is obvious that CAT2(0) space is exactly the classical CAT(0) space, which has been extensively studied.

Let x,y,z be in a CATp(0) space X, with p≥ 2, and xy2 is the midpoint of the geodesic [x,y]. Then the comparison axiom implies

dz,xy2p12d(z,x)p+12d(z,y)p12pd(x,y)p.

This inequality is the (CNp) inequality of Khamsi and Shukri [9]. Note that the (CNp) inequality coincides with the classical (CN) inequality [4] if p = 2. Below are some strong inequalities in a CATp(0) space.

Lemma 1.2.([1, 5]) Let (X,d) be a CATp(0) space, with p≥ 2. Then, for any x,y,z in X and α[0,1], we have

  • (i) d(z,αx(1α)y)αd(z,x)+(1α)d(z,y);

  • (ii) d(z,αx(1α)y)pαd(z,x)p+(1α)d(z,y)p12p1α(1α)d(x,y)p.

Let C be a nonempty subset of a CATp(0) space (X,d). A subset C of X is said to be convex if C includes every geodesic segment joining any two of its points, that is, for any x,yC, we have [x,y]C.

A complete CATp(0) space is called a p-Hadamard space. Throughout our work, we mainly focus on p-Hadamard spaces for p2.

Let T be a mapping of C into itself. An element x∈ C is called a fixed point of T if x = Tx. The set of all fixed points of T is denoted by F(T), that is, F(T)={xC:x=Tx}. A sequence {xn} in C is called approximate fixed point sequence for T (AFPS in short) if limnd(xn,Txn)=0.

The famous fixed point theorem for nonexpansive mappings in Banach spaces have first studied by Browder [3]and Göhde [8] in 1965 as follows:

Theorem 1.3. Let X be a uniformly convex Banach space and C be a nonempty bounded closed convex subset of X. Then, each nonexpansive mapping T:CC has a fixed point.

In 1972, Geobel and Kirk [6] extended their result to asymptotically nonexpansive mappings. Later in 2004, Kirk [10] obtained a similar result for complete CAT(0) spaces. In 2013, Phuengrattana and Suantai [12] extended those result to generalized asymptotically nonexpansive mappings and to complete uniformly convex metric spaces as follows:

Theorem 1.4. Let (X,d) be a complete uniformly convex metric space. Let C be a nonempty bounded closed convex subset of X and T:CC be a generalized asymptotically nonexpansive mapping whose graph G(T)={(x,y)C×C:y=Tx} is closed. Then, T has a fixed point.

In this work, we extend some known existence and convergence results for generalized asymptotically nonexpansive mappings in Hadamard spaces to the case of p-Hadamard spaces, for p2.

Let C be a nonempty subset of a CATp(0) space (X,d) and T:CC be a mapping. A mapping T is said to be generalized asymptotically nonexpansive [12, 15] if there exist sequences {kn}[1,) and {sn}[0,) with limnkn=1, limnsn=0 such that

dTnx,Tnyknd(x,y)+sn,

for all x,yC and n. In the case of sn=0 for all n, a mapping T is called an asymptotically nonexpansive mapping. In particular, if kn=1 and sn=0 for all n, a mapping T reduce to a nonexpansive mapping.

Remark 2.1. If T is a generalized asymptotically nonexpansive mapping, it is know that F(T) is not necessarily closed; see [13].

Recall that ϕ:X[0,) is called a type function if there exists a bounded sequence {xn} in X such that

ϕ(x)=limsupnd(xn,x),

for any x∈ X. A sequence {zn} in X is said to be a minimizing sequence of ϕ whenever

limnϕ(zn)=inf{ϕ(x):xX}.

Lemma 2.2.([9]) Let (X,d) be a p-Hadamard space, with p≥ 2. Let C be a nonempty bounded closed convex subset of X and ϕ be a type function defined on C. Then any minimizing sequence of ϕ is convergent and its limit z is the unique minimum point of ϕ, i.e., ϕ(z)=inf{ϕ(x):xC}.

The following results are needed for proving our convergence results.

Definition 2.3.([11]) A mapping T:CC is said to be semi-compact if C is closed and each bounded AFPS for T in C has a convergent subsequence.

Definition 2.4.([14]) A mapping T:CC is said to satisfy the condition (I) if there exists a nondecreasing function ρ:[0,)[0,) with ρ(0)=0 and ρ(r)>0 for all r(0,) such that

d(x,Tx)=ρ(dist(x,F(T)))

for all xC, where dist(x,F(T))=inf{d(x,z):zF(T)}.

Definition 2.5.([13]) Let {xn} be a sequence in a metric space (X,d) and FX. We say that {xn} is of monotone type (I) with respect to F if there exists sequences {δn} and {γn} of nonnegative real numbers such that n=1δn<, n=1γn<, and

d(xn+1,z)(1+δn)d(xn,z)+γn

for all n and zF.

Lemma 2.6.([13]) Let {xn} be a sequence in a complete metric space (X,d) and FX. If {xn} is of monotone type (I) with respect to F and liminfndist(xn,F)=0, then limnxn=z for some z∈ X satisfying dist(xn,F). In particular, if F is closed, then z∈ F.

Lemma 2.7.([18]) Let {an}, {bn} and {cn} be sequences of nonnegative real numbers satisfying

an+1(1+bn)an+cn,n,

where n=1bn< and n=1cn<. Then limnan exists.

In this section, we study the existence theorems for a generalized asymptotically nonexpansive mapping in p-Hadamard spaces.

We now state and prove our existence results.

Theorem 3.1. Let (X,d) be a p-Hadamard space, with p≥ 2. Let C be a nonempty bounded closed convex subset of X and T:CC be a generalized asymptotically nonexpansive mapping whose graph G(T)={(x,y)C×C:y=Tx} is closed. Then, T has a fixed point.

Proof. Fix xC. Consider the type function ϕ generated by the bounded sequence {Tnx}. Let z be the minimum point of ϕ which exists by using Lemma 2.2. Therefore,

dTn+mx,Tmzkmd(Tnx,z)+sm,

for any n,m. Taking n, we get

ϕ(Tmz)kmϕ(z)+sm=kminf{ϕ(x):xC}+sm,

for any m. Taking m, we get

limmϕ(Tmz)inf{ϕ(x):xC}.

Since z is the minimum point of ϕ, it implies that limmϕ(Tmz)=inf{ϕ(x):xC}. Thus, {Tmz} is a minimizing sequence of ϕ. By Lemma 2.2, we obtain that Tmzz as m, and so T(Tmz)=Tm+1zz as m. By the closedness of G(T), we have Tz=z. This completes the proof.

Theorem 3.2. Let (X,d) be a p-Hadamard space, with p2. Let C be a nonempty bounded closed convex subset of X and T:CC be a generalized asymptotically nonexpansive mapping whose graph G(T)={(x,y)C×C:y=Tx} is closed. Then, F(T) is nonempty closed and convex.

Proof. By Theorem 3.1, F(T) is nonempty. To show that F(T) is closed, we let {xn} be a sequence in F(T) such that limnxn=x. By the definition of T, we have

d(Tnx,x)d(Tnx,xn)+d(xn,x)(1+kn)d(xn,x)+sn.

Since limnkn=1 and limnsn=0, we get limnd(Tnx,x)=0. That is Tnxx as n, and so T(Tnx)=Tn+1x→ x as n. By the closedness of G(T), we have Tx=x. Hence xF(T) so that F(T) is closed.

In order to prove F(T) is convex, it is enough to prove that xy2F(T) whenever x,yF(T) with xy. Set z=xy2. By the (CNp) inequality and the definition of T, for any n, we have

d(Tnz,z)p=dTnz,xy2p    12d(Tnz,x)p+12d(Tnz,y)p12p d(x,y)p    =12d(Tnz,Tnx)p+12d(Tnz,Tny)p12p d(x,y)p    12knd(z,x)+snp+12knd(z,y)+snp12p d(x,y)p.

Since z=xy2, we get that d(z,x)=12d(x,y) and d(y,x)=12d(x,y). So, we have

d(Tnz,z)pkn2d(x,y)+snp12pd(x,y)p.

By limnkn=1 and limnsn=0, we get limnd(Tnz,z)p=0. This implies that Tnzz as n, and so T(Tnz)=Tn+1zz as n. From the closedness of G(T), we have Tz = z. Therefore, F(T) is convex. This completes the proof.

Next, we show that the existence of a fixed point of a generalized asymptotically nonexpansive mapping in a p-Hadamard space is equivalent to the existence of a bounded orbit at a point.

Theorem 3.3. Let (X,d) be a p-Hadamard space, with p2. Let C be a nonempty closed convex subset of X and T:CC be a generalized asymptotically nonexpansive mapping whose graph G(T)={(x,y)C×C:y=Tx} is closed. Then F(T) if and only if there exists an xC such that {Tnx} is bounded.

Proof. The necessity is obvious. Conversely, assume that x is an element in C such that {Tnx} is bounded. Consider the type function ϕ generated by {Tnx}. By the same step of the proof as in Theorem 3.1, we can conclude that F(T). This completes the proof.

Remark 3.4.

  • (i) If T is generalized asymptotically nonexpansive, F(T) is not necessarily closed. However, if G(T) is also closed, Theorem 3.2 guarantee that F(T) is always closed.

  • (ii) If T is continuous, then G(T) is always closed. Therefore, Theorems 3.1, 3.2 and 3.3 are obtained for a class of continuous generalized asymptotically nonexpansive mappings.

In this section, we study the strong convergence theorems for a generalized asymptotically nonexpansive mapping by the modified two-step iterative sequence for finding fixed points of such mapping in p-Hadamard spaces. We now introduce the modified two-step iterative sequence [16] as below:

Let C be a nonempty closed convex subset of a p-Hadamard space (X,d) and T:CC be a generalized asymptotically nonexpansive mapping. We generate the sequence {xn} in C by x1C and

yn=βnxn(1βn)Tnxn,xn+1=αnyn(1αn)Tnyn,n,

where {αn} and {βn} are sequences in [0,1].

Now we prove the strong convergence results.

Lemma 4.1. Let (X,d) be a p-Hadamard space, with p≥ 2. Let C be a nonempty bounded closed convex subset of X and T:CC be a uniformly continuous generalized asymptotically nonexpansive mapping with sequences {kn}[1,) and {sn}[0,) such that n=1(kn1)< and n=1sn<. Let x1C and {xn} be a sequence in C defined by \eqref{two-step} where {αn} and {βn} are sequences in (0,1) such that 0<aαn,βnb<1. Then, we have the following:

  • (i) there exit two sequences {δn} and {εn} of nonnegative real numbers such that n=1δn<, n=1εn<, and d(xn+1,z)(1+δn)d(xn,z)+εn for all n and zF(T);

  • (ii) limnd(xn,z) exists for all zF(T);

  • (iii) {xn} is an AFPS for T.

Proof. (i) : By the uniform continuity of T, we have G(T) is closed. It implies from Theorem 3.1 that F(T). Let zF(T). Since T is generalized asymptotically nonexpansive, by Lemma 1.2(i), we have

d(yn,z)βnd(xn,z)+(1βn)d(Tnxn,z)  =βnd(xn,z)+(1βn)d(Tnxn,Tnz)  βnd(xn,z)+(1βn)(knd(xn,z)+sn)  (βn+(1βn)kn)d(xn,z)+sn  =βnkn+(1βn)knd(xn,z)+sn.

Since 0βnkn+(1βn)1, we obtain

d(yn,z)knd(xn,z)+sn.

This implies that

d(xn+1,z)αnd(yn,z)+(1αn)d(Tnyn,z)    =αnd(yn,z)+(1αn)d(Tnyn,Tnz)    αnd(yn,z)+(1αn)(knd(yn,z)+sn)    (αn+(1αn)kn)d(yn,z)+sn    =αnkn+(1αn)knd(yn,z)+sn.

Since 0βnkn+(1βn)1, by (4.2), we have

d(xn+1,z)knd(yn,z)+sn    kn(knd(xn,z)+sn)+sn    kn2d(xn,z)+knsn+sn    =(1+( k n 1))2d(xn,z)+(1+(kn1))sn+sn    =(1+2(kn1)+( k n 1)2)d(xn,z)+(kn1)sn+2sn    =(1+δn)d(xn,z)+γn,

where δn=2(kn1)+(kn1)2 and γn=(kn1)sn+2sn. Since n=1(kn1)< and n=1sn<, it follows that n=1δn< and n=1γn<. Hence, we obtain the desired result.

(ii) : By (i) and Lemma 2.7, we obtain that limnd(xn,z) exists.

(iii) : By Lemma 1.2(ii) and (4.2), we have

d(xn+1,z)pαnd(y n,z)p+(1αn)d(T n y n,z)p αn (1αn )2 p1d(T n y n,y n)p    αn βn d ( xn ,z)p+(1βn )d ( Tn xn ,z)p βn (1βn ) 2 p1d ( Tn xn , xn )p      +(1αn)(k nd(y n,z)+s n)p αn (1αn )2 p1d(T n y n,y n)p    αnβnd(x n,z)p+αn(1βn)(k nd(x n,z)+s n)p      +(1αn)(k n2d(x n,z)+k ns n +s n)p αn βn (1βn )2 p1d(T n x n,x n)p       αn (1αn )2 p1d(T n y n,y n)p.

Since kn1 and sn0, we have

d(xn,z)knd(xn,z)+snkn2d(xn,z)+knsn+sn.

Then, by (4.3), we have

d(xn+1,z)p(kn2d(xn,z)+knsn+sn)p α n β n (1β n ) 2 p1d(Tn xn,xn)p       α n (1α n ) 2 p1d(Tn yn,yn)p.

This implies that

d(Tnxn,xn)p2p1a2(1b)(kn2d(xn,z)+knsn+sn)pd(xn+1,z)p,

and

d(Tnyn,yn)p2p1a(1b)(kn2d(xn,z)+knsn+sn)pd(xn+1,z)p.

By limnkn=1, limnsn=0 and limnd(xn,z) exists, we conclude that

limnd(Tnxn,xn)=0,

and

limnd(Tnyn,yn)=0.

By (4.4) and the uniform continuity of T, we have

limnd(Tn+1xn,Txn)=0.

By the definitions of xn+1 and yn, we obtain

d(xn,Txn)d(xn,xn+1)+d(xn+1,Tn+1xn+1)+d(Tn+1xn+1,Tn+1xn)  +d(Tn+1xn,Txn)(1+kn+1)d(xn,xn+1)+sn+1+d(xn+1,Tn+1xn+1)  +d(Tn+1xn,Txn)(1+kn+1)(αnd(xn,yn)+(1αn)d(xn,Tnyn))+sn+1  +d(xn+1,Tn+1xn+1)+d(Tn+1xn,Txn)(1+kn+1)(αnd(xn,yn)+(1αn)(d(xn,yn)+d(yn,Tnyn)))  +sn+1+d(xn+1,Tn+1xn+1)+d(Tn+1xn,Txn)(1+kn+1)(d(xn,yn)+(1αn)d(yn,Tnyn))+sn+1  +d(xn+1,Tn+1xn+1)+d(Tn+1xn,Txn)(1+kn+1)((1βn)d(xn,Tnxn)+(1αn)d(yn,Tnyn))  +sn+1+d(xn+1,Tn+1xn+1)+d(Tn+1xn,Txn).

Since limnkn=1, limnsn=0, by (4.4), (4.5), and (4.6), we conclude that

limnd(xn,Txn)=0.

Hence, we obtain the desired result.

Now, we prove a strong convergence theorem for a generalized asymptotically nonexpansive semi-compact mapping in p-Hadamard spaces.

Theorem 4.2. uppose that X, C, T, {xn}, {αn}, {βn} are as in Lemma 4.1. If Tm is semi-compact for some m, then {xn} converges strongly to a fixed point of T.

Proof. By Lemma 4.1(iii), limnd(xn,Txn)=0. Fix m, we have

d(xn,Tmxn)d(xn,Txn)+d(Txn,T2xn)++d(Tm1xn,Tmxn).

Since T is uniformly continuous, we have

limnd(xn,Tmxn)=0.

That is, {xn} is an AFPS for Tm. By the semi-compactness of Tm, there exist a subsequence {xnk} of {xn} and zC such that limkxnk=z. Again, by the uniform continuity of T, we have

d(Tz,z)d(Tz,Txnk)+d(Txnk,xnk)+d(xnk,z)0as k.

Then zF(T). By Lemma 4.1(ii), limnd(xn,z) exists, thus z is the strong limit of the sequence {xn} itself. This completes the proof.

Finally, we prove a strong convergence theorem for a generalized asymptotically nonexpansive mapping which satisfies condition (I) in p-Hadamard spaces.

Theorem 4.3. Suppose that X, C, T, {xn}, {αn}, {βn} are as in Lemma 4.1. If T satisfies condition (I), then {xn} converges strongly to a fixed point of T.

Proof. By condition (I), there exists a nondecreasing function ρ:[0,)[0,) with ρ(0)=0 and ρ(r)>0 for all r(0,) such that

d(xn,Txn)=ρ(dist(xn,F(T))).

It implies by Lemma 4.1(iii) that

limnρ(dist(xn,F(T)))=0.

Then we have

limndist(xn,F(T))=0.

By Lemma 4.1(i), we obtain that the sequence {xn} is of monotone type (I) with respect to F(T). This implies by Lemma 2.6 that the sequence {xn} converges strongly to a point zF(T). This completes the proof.

Remark 4.4. Any complete CAT(0) space is a 2-Hadamard space, therefore the results in this paper can be applied to any complete CAT(0) space.

The authors are thankful to the referees for careful reading and the useful comments and suggestions.

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