# Fixed Points for Multivalued Mappings in Uniformly Convex Metric Spaces

Authors: A. Kaewcharoen; B. Panyanak
Journal: International Journal of Mathematics and Mathematical Sciences
 (2008)

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## Abstract

The purpose of this paper is to ensure the existence of fixed points for multivalued nonexpansive weakly inward nonself-mappings in uniformly convex metric spaces. This extends a result of Lim (1980) in Banach spaces. All results of Dhompongsa et al. (2005) and Chaoha and Phon-on (2006) are also extended.


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## Body

## 1. Introduction

In 1974,  Lim [1] developed a result concerning the existence of fixed
points for multivalued nonexpansive self-mappings in uniformly convex Banach
spaces. This result was extended to nonself-mappings satisfying the inwardness condition
independently by Downing and Kirk [2] and Reich [3]. This result was extended to weak inward mappings
independently by Lim [4] and Xu [5]. Recently, Dhompongsa et al. [6] presented an analog of
Lim-Xu's result in CAT(0) spaces. In this note, we extend the result to
uniformly convex metric spaces which improve results of both Lim-Xu and
Dhompongsa et al. In addition, we also give a new proof of a result of Lim
[7] by using Caristi's
theorem [8]. Finally,
we give some basic properties of fixed point sets for quasi-nonexpansive
mappings for these spaces.


## 2. Preliminaries

A concept of convexity in metric spaces was introduced
by Takahashi [9].
Definition 2.1.
Let(X,d) be a metric
space and I=[0,1]. A mapping W:X×X×I→X is said to be a  convex structure on X if for each (x,y,λ)∈X×X×I and z∈X,(2.1)d(z,W(x,y,λ))≤λd(z,x)+(1−λ)d(z,y).A metric space (X,d) together with a
convex structure W is called a                                    convex
metric space which will be denoted by (X,d,W).Definition 2.2.
A convex metric space(X,d,W) is said to be                                    uniformly
convex [10] if for
any ε>0, there exists δ=δ(ε)>0 such that for
all r>0 and x,y,z∈X with d(z,x)≤r,d(z,y)≤r and d(x,y)≥rε,(2.2)d(z,W(x,y,12))≤r(1−δ).Obviously, uniformly convex Banach spaces are uniformly convex metric spaces.
By using the (CN) inequality [11], it is easy to see that CAT(0) spaces are also
uniformly convex.Forx,y∈X,
C⊂X, and λ∈I, we denote W(x,y,λ):=λx⊕(1−λ)y, [x,y]:={λx⊕(1−λ)y:λ∈I}, (x,y]:=[x,y]∖{x}, and λx⊕(1−λ)C:={λx⊕(1−λ)z:z∈C}. So we can
define the                                    inward setIC(x) of x as follows:(2.3)IC(x):={x}∪{z:(x,z]∩C≠∅}.
Let C be a nonempty
subset of a metric space X. Then C is called
convex if for  x,y∈C,[x,y]⊂C. We will denote
by ℱ(C) the family of
nonempty closed subsets of C, by 𝒦(C) the family of
nonempty compact subsets of C, and by 𝒦𝒞(C) the family of
nonempty compact convex subsets of C. Let H(⋅,⋅) be the
Hausdorff distance on ℱ(X). That
is,(2.4)H(A,B)=max{supa∈Adist(a,B),supb∈Bdist(b,A)},A,B∈ℱ(X).Definition 2.3.
A multivalued mappingT:C→ℱ(X) is said to be                                    inward on C if for some p∈C,(2.5)λp⊕(1−λ)Tx⊂IC(x)∀x∈C,∀λ∈[0,1],and                                    weakly inward on C if for some p∈C,(2.6)λp⊕(1−λ)Tx⊂IC(x)¯∀x∈C,∀λ∈[0,1],where A¯ denotes the closure
of a subset A of X. In a Banach space setting, if C is convex, then
so is IC(x). Therefore, the conditions above can be replaced by Tx⊂IC(x) and Tx⊂IC(x)¯, respectively.Definition 2.4.
A multivalued mappingT:C→ℱ(X) satisfying(2.7)H(Tx,Ty)≤kd(x,y),x,y∈C,is called a                                    contraction if k∈[0,1) and                                    nonexpansive if k=1. A point x is a fixed
point of T if x∈Tx.Given a metric spaceX, one way to
describe a metric space ultrapower X˜ of X is to first
embed X as a closed
subset of a Banach space E (see, e.g.,
[12, page 129]). Let E˜ denote a Banach
space ultrapower of E relative to
some nontrivial ultrafilter 𝒰 (see, e.g.,
[13]). Then take(2.8)X˜:={x˜=[{xn}]∈E˜:xn∈X∀n}.One can then let d˜ denote the
metric on X˜ inherited from
the ultrapower norm ∥⋅∥𝒰 in E˜. If X is complete,
then so is X˜ since X˜ is a closed
subset of the Banach space E˜. In particular,
the metric d˜ on X˜ is given by(2.9)d˜(x˜,y˜)=lim𝒰∥xn−yn∥=lim𝒰d(xn,yn),with {un}∈[{xn}] if and only if lim𝒰∥xn−un∥=0.If(X,d,W) is a convex
metric space, we consider a metric space ultrapower (X˜,d˜) of (X,d) and define a
function W˜:X˜×X˜×I→X˜ by(2.10)W˜(x˜,y˜,λ)=[W(xn,yn,λ)].In order to show that the function W˜ is well
defined, we need the following condition. For each p,x,y∈X and λ∈[0,1],(2.11)d((1−λ)p⊕λx,(1−λ)p⊕λy)≤λd(x,y),which is equivalent to(2.12)d((1−λ)p⊕λx,(1−λ)q⊕λy)≤λd(x,y)+(1−λ)d(p,q),for all p,q,x,y∈X and λ∈[0,1].By using
condition (2.12), it is easy to see that W˜ is a convex
structure on X˜. This implies
that (X˜,d˜,W˜) is a convex
metric space.Example 2.5.
Every Banach space satisfies condition (2.11).Example 2.6.
Condition (2.11) is satisfied for spaces of hyperbolic type (for
more details of these spaces see [14]). This is also true for CAT(0) spaces and ℝ-trees.Example 2.7.
LetH be a
hyperconvex metric space. Then there exists a nonexpansive retract R:l∞(H)→H (see, e.g.,
[15] for more on
this). For any x,y∈H and t∈[0,1], we let(2.13)tx⊕(1−t)y=R(tx+(1−t)y).Since l∞(H) is a Banach
space, H also satisfies
condition (2.11).Let𝒰 be a nontrivial
ultrafilter on the natural number ℕ. If (X,d,W) is a uniformly
convex metric space satisfying condition (2.11), then the metric
space ultrapower   (X˜,d˜,W˜) relative to 𝒰 is also
uniformly convex. Indeed, let any ε>0 and let δ be a positive
number corresponding to the uniform convexity of X. Let r>0 and x˜,y˜,z˜∈X˜ be such that(2.14)d˜(z˜,x˜)≤r,d˜(z˜,y˜)≤r,d˜(x˜,y˜)≥rε.Then there are some representatives (xn) and (yn) of x˜ and y˜ and a set I∈𝒰 such
that(2.15)d(zn,xn)≤r,d(zn,yn)≤r,d(xn,yn)≥rε∀n∈I.For such n,d(zn,W(xn,yn,1/2))≤r(1−δ). This implies d˜(z˜,W˜(x˜,y˜,1/2))≤r(1−δ).Recall that a
subsetC of a metric
space X is said to be
(uniquely)                                    proximinal if each point x∈X has a (unique)
nearest point in C. A convex metric
space X is said to have                                    property (C) if every decreasing sequence of nonempty bounded closed
convex subsets of X has nonempty
intersection. In [10],
Shimizu and Takahashi proved that property (C) holds in complete uniformly
convex metric spaces. This implies that every nonempty closed convex subset of
a complete uniformly convex metric space is uniquely proximinal. Indeed, let C be a nonempty
closed convex subset of a complete uniformly convex metric space X, and x0∈X.LetN={x∈C:d(x0,x)=dist(x0,C)}. For each n, we define(2.16)Cn:={y∈C:d(x0,y)≤dist(x0,C)+1n}.Then (Cn) is a decreasing
sequence of nonempty bounded closed convex subsets of C. Moreover,N=⋂n=1∞Cn,which is nonempty by the above observation. The uniqueness follows from the
uniform convexity of X.


## 3. Main Results

We first establish the following lemma.Lemma 3.1.
LetX be a complete
uniformly convex metric space satisfying condition (2.11), C a nonempty
closed convex subset of X, x∈X, and p(x) the unique
nearest point of x in C. Then(3.1)d(x,p(x))<d(x,y)∀y∈IC(p(x))¯∖{p(x)}.Proof.
Lety∈IC(p(x))¯∖{p(x)}. Then there is a sequence (yn) in IC(p(x)) and yn→y. Choose n0∈ℕ such that (p(x),yn]∩C≠∅ for all n≥n0. For such n, let zn∈(p(x),yn]∩C and write zn=(1−αn)p(x)⊕αnyn,αn∈(0,1]. Then(3.2)d(x,p(x))≤d(x,zn)≤(1−αn)d(x,p(x))+αnd(x,yn).This implies(3.3)d(x,p(x))≤d(x,y).If d(x,p(x))=d(x,y), we let u=(1/2)p(x)⊕(1/2)y. By the uniform
convexity of X, we have d(x,u)<d(x,p(x)). On the other hand,
for each n≥n0, let un=(1/2)p(x)⊕(1/2)yn. We will show
that un∈IC(p(x)).
Case 1.
αn=1/2. We are done.
Case 2.
1/2<αn. Let vn=(1−1/2αn)p(x)⊕(1/2αn)zn. This implies(3.4)d(p(x),vn)=12αnd(p(x),zn)=12d(p(x),yn)=d(p(x),un),(3.5)d(vn,yn)≤d(vn,zn)+d(zn,yn)=(1−12αn)d(p(x),zn)+(1−αn)d(p(x),yn)=(αn−12)d(p(x),yn)+(1−αn)d(p(x),yn)=12d(p(x),yn)=d(un,yn).Therefore(3.6)d(vn,yn)≤d(un,yn).We claim that un=vn. If not, let wn=(1/2)un⊕(1/2)vn.
From (3.4), (3.6), and the uniform convexity of X, we have(3.7)d(p(x),wn)<d(p(x),un),d(wn,yn)<d(un,yn).This implies(3.8)d(p(x),yn)≤d(p(x),wn)+d(wn,yn)<d(p(x),un)+d(un,yn)=d(p(x),yn),which is a contradiction. Hence un=vn∈[p(x),zn] and so un∈IC(p(x)) by the
convexity of C.
Case 3.
αn<1/2. let vn=(1−2αn)p(x)⊕2αnun. By the same
arguments in the proof of Case 2, we can show that zn∈(p(x),un]. This means un∈IC(p(x)).
By condition (2.11), limnun=u, which implies u∈IE(p(x))¯∖{p(x)}. By the same arguments
in the first part of the proof, d(x,p(x))≤d(x,u) which is a
contradiction. Hence d(x,p(x))<d(x,y) as
desired.From  [6, Lemma 3.4], we observe
that the space is not necessarity assumed to be a CAT(0) space since the proof is
only involved with condition (2.11) which is weaker than the (CN) inequality (see
[11, Lemma 3]). Therefore, we
can obtain the following lemma.Lemma 3.2.
LetX be a complete
convex metric space satisfying condition (2.11), C a  nonempty
closed subset of X, and T:C→ℱ(X) a contraction
mapping satisfying, for all x∈C,(3.9)Tx⊂IC(x)¯.Then T has a fixed
point.This lemma was first proved in Banach spaces by Lim
[7], using transfinite
induction, while we apply directly Caristi's theorem. For completeness, we
include the details.Proof of Lemma3.2..
Let0≤k<1 be the
contraction constant of T and let ε>0 be such that ε+(k+2ε)(1+ε)<1. Let M={(x,z):z∈Tx,x∈C} and define a
metric ρ on M by ρ((x,z),(u,v))=max{d(x,u),d(z,v)}. It is easy to
see that (M,ρ) is a complete
metric space.
Now defineψ:M→[0,∞) by ψ(x,z)=d(x,z)/ε. Then ψ is continuous
on M. Suppose that T has no fixed
points, that is, dist(x,Tx)>0 for all x∈C. Let (x,z)∈M. By (3.9), we can find z′∈IC(x) satisfying d(z,z′)<εdist(x,Tx). Now choose u∈(x,z′]∩C and write u=(1−δ)x⊕δz′ for some 0<δ≤1. For such δ, we have(3.10)δε+(1−δ)+(k+2ε)δ(1+ε)<1.Since d(x,u)>0, we can find v∈Tu satisfying(3.11)d(z,v)≤H(Tx,Tu)+εd(x,u)≤(k+ε)d(x,u).
Now we define a mapping g:M→M by g(x,z)=(u,v) for all (x,z)∈M. We claim that g satisfies(3.12)ρ((x,z),g(x,z))<ψ(x,z)−ψ(g(x,z))∀(x,z)∈M.Caristi's
theorem [8] then
implies that g has a fixed
point, which contradicts to the strict inequality (3.12) and the proof is
complete. So it remains to prove (3.12). In fact, it is enough to show that(3.13)ρ((x,z),(u,v))<1ε(d(x,z)−d(u,v)).But d(z,v)≤d(x,u), it only needs
to prove that d(x,u)<(1/ε)(d(x,z)−d(u,v)).
Now,(3.14)d(x,u)=δd(x,z′)≤δ(d(x,z)+εdist(x,Tx))≤δ(d(x,z)+εd(x,z))≤δ(1+ε)d(x,z).Therefore(3.15)d(x,u)≤δ(1+ε)d(x,z).It follows that(3.16)d(z,v)≤(k+ε)d(x,u)≤(k+ε)δ(1+ε)d(x,z).We now let y=(1−δ)x⊕δz, then by the
condition (2.11),(3.17)d(u,v)≤d(u,y)+d(y,z)+d(z,v)≤δd(z,z′)+(1−δ)d(x,z)+(k+ε)δ(1+ε)d(x,z)≤δεd(x,z)+((1−δ)+(k+ε)δ(1+ε))d(x,z).Thus(3.18)d(u,v)≤(δε+(1−δ)+(k+ε)δ(1+ε))d(x,z).Inequalities (3.15), (3.18), and (3.10) imply that(3.19)εd(x,u)+d(u,v)≤εδ(1+ε)d(x,z)+(δε+(1−δ)+(k+ε)δ(1+ε))d(x,z)=(δε+(1−δ)+(k+2ε)δ(1+ε))d(x,z)<d(x,z).Therefore d(x,u)<(1/ε)(d(x,z)−d(u,v)) as
desired.By Lemmas3.1 and 3.2 with the same arguments in the proof of Theorem 3.3
of [6], we can obtain
the following theorem which  extends
[4, Theorem 8] by Lim and
[6, Theorem 3.3] by Dhompongsa et al.Theorem 3.3.
LetX be a complete
uniformly convex metric space satisfying condition  (2.11), C a nonempty
bounded closed convex subset of X, and T:C→𝒦(X) a nonexpansive
weakly inward mapping. Then T has a fixed
point.As an immediate
consequence of Theorem3.3, we obtain the following corollary.Corollary 3.4.
LetX be a complete
uniformly convex metric space satisfying condition (2.11), C a nonempty
bounded closed convex subset of X, and T:C→𝒦(C) a nonexpansive
mapping. Then T has a fixed
point.In fact, this corollary is a special case of [10, Theorem 2] in which
condition (2.11) was not assumed. An interesting question is whether
condition (2.11) in Theorem 3.3 can be dropped.LetC be a nonempty
subset of a metric space X. Recall that a
single-valued mapping t:C→C and
a
multivalued mapping T:C→2C∖∅ are said to be                                    commuting if ty∈Ttx for all y∈Tx and x∈C. If t:C→C is nonexpansive
with C being bounded closed convex and X complete
uniformly convex satisfying condition (2.11), then Fix(t) is nonempty by
the above corollary. Moreover, by a standard argument, we can show that it is
also closed and convex. So we can obtain a common fixed point theorem in
uniformly convex metric spaces as [6, Theorem 4.1] (see also [16, Theorem 4.2] for a related result in
Banach spaces).Theorem 3.5.
LetX be a complete
uniformly convex metric space satisfying condition  (2.11), let C be a nonempty
bounded closed convex subset of X, and let t:C→C and T:C→𝒦𝒞(C) be
nonexpansive. Assume that for some p∈Fix(t),(3.20)αp⊕(1−α)Txisconvex∀x∈C,∀α∈[0,1].If t and T are commuting,
then there exists a point z∈C such that tz=z∈Tz.


## 4. Fixed Point Sets of Quasi-Nonexpansive Mappings

LetX be a metric
space. Recall that a mapping f:X→X is said to be                                    quasi-nonexpansive if d(f(x),p)≤d(x,p) for all x∈X and p∈Fix(f). In this case,
we will assume that Fix(f)≠∅. In [17], Chaoha and Phon-on showed
that if X is a CAT(0)
space, then Fix(f) is closed
convex. Furthermore, they gave an explicit construction of a continuous
function defined on X whose fixed
point set is any prescribed closed subset of X. In this
section, we extend these results to uniformly convex metric spaces.We begin by proving the following lemma.Lemma 4.1.
LetX be a uniformly
convex metric space, and let x,y,z∈X for which(4.1)d(x,z)+d(z,y)=d(x,y).Then z∈[x,y].Proof.
Letu∈[x,y] be such that d(x,u)=d(x,z). Then d(x,y)=d(x,u)+d(u,y) and also d(z,y)=d(u,y) by (4.1). We will show
that z=u. Suppose not, we
let v=(1/2)z⊕(1/2)u and r=d(x,u)=d(x,z). Since d(z,u)>0, choose ε>0 so that d(z,u)>rε. By the uniform
convexity of X, there exists δ>0 such that(4.2)d(x,v)≤r(1−δ)<r=d(x,z).By using the same arguments, we can show that d(y,v)<d(y,z). Therefore(4.3)d(x,y)≤d(x,v)+d(y,v)<d(x,z)+d(y,z)=d(x,y),which is a contradiction.By using the
above lemma with the proof of Theorem 1.3 of [17], we obtain the following
result.Theorem 4.2.
LetX be a convex
subset of a uniformly convex metric space and f:X→X a
quasi-nonexpansive mapping whose fixed point set is nonempty. Then Fix(f) is closed
convex.In [17], the authors constructed a
continuous function defined on a CAT(0) space X whose fixed
point set is any prescribed closed subset of X by using the
following two implications of the (CN) inequality:(4.4)d((1−t)x⊕ty,(1−s)x⊕sy)=|t−s|d(x,y)∀x,y∈X,t,s∈[0,1],(4.5)d((1−t)x⊕ty,(1−t)x⊕tz)≤d(y,z)∀x,y,z∈X,t∈[0,1].
In fact,
condition (4.4) holds in uniformly convex metric spaces as the
following  lemma shows.Lemma 4.3.
Condition (4.4) holds in uniformly convex metric spaces.Proof.
We first note that the conclusion holds ifs=0 or t=0. We now let X be a uniformly
convex metric space, x,y∈X, and t,s∈(0,1]. Let u=(1−t)x⊕ty and z=(1−s)x⊕sy. Without loss of generality, we can assume that t<s. Let v=(1−t/s)x⊕(t/s)z, then(4.6)d(x,v)=tsd(x,z)=td(x,y),d(v,y)≤(1−ts)d(x,y)+tsd(z,y)=(1−t)d(x,y).If u≠v, we let w=(1/2)u⊕(1/2)v. Then by the
uniform convexity of X, we can show
that d(x,w)<d(x,u) and d(y,w)<d(y,u). This implies(4.7)d(x,y)<d(x,u)+d(u,y)=d(x,y),which is a contradiction, hence u=v. Therefore(4.8)d(z,u)=d(z,v)=(1−ts)d(x,z)=|s−t|d(x,y).It is unclear
that condition (4.5) holds for uniformly convex metric spaces. However,
the following theorem is a generalization of [17, Theorem 2.1], we omit the proof because
it is similar to the one given in [17].Theorem 4.4.
LetA be a nonempty
subset of a uniformly convex metric space X satisfying
condition (4.5). Then there exists a continuous function f:X→X such that Fix(f)=A¯.



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