R E S E A R C H
Open Access
Remarks on endpoints of multivalued
mappings in geodesic spaces
Satit Saejung
**Correspondence:
[email protected] Department of Mathematics, Faculty of Science, Khon Kaen University, Khon Kaen, 40002, Thailand
Research Center for Environmental and Hazardous Substance Management, Khon Kaen University, Khon Kaen, 40002, Thailand
Abstract
We discuss Panayanak’s results on the existence of an endpoint of a multivalued nonexpansive mapping. We show that all of his results can be extended and some can be established in a wider class of mappings. Out of his three open questions, two of them are solved in affirmative.
1 Introduction
Let (X,d) be a metric space. Thedistancefrom an elementx∈Xto a nonempty subset E⊂Xis defined by
d(x,E) :=infd(x,y) :y∈E.
For a nonempty subsetE⊂X, we denote by BC(E) (K(E), respectively) the family of nonempty bounded and closed subsets (nonempty compact subsets, respectively) ofE. Note thatK(E)⊂BC(E) and the inclusion may be proper. ThePompeiu-Hausdorff dis-tanceonBC(X) is defined by
H(A,B) :=max
sup a∈A
d(a,B),sup b∈B
d(b,A)
for allA,B∈BC(X).
An elementx∈Eis afixed pointof a multivalued mappingT:E→BC(X) ifxbelongs to the setTx. Moreover, if{x}=Tx, thenxis called anendpointofT(or astationary point ofT). The set of all endpoints (all fixed points, respectively) ofT is denoted byEnd(T) (Fix(T), respectively). It is clear thatEnd(T)⊂Fix(T) and the inclusion may be proper (see Example ). In the case of single-valued mappings, both notions coincide.
The existence of an endpoint and of a fixed point of a multivalued mapping has been widely investigated by many researchers (see,e.g., [–]). Corley [] proved that a max-imization with respect to a cone, which subsumes ordinary and Pareto optmax-imization, is equivalent to the problem of finding an endpoint of certain multivalued mapping. Note that the results in multivalued case are suggested but do not follow directly from the one in the single-valued case. In spite of the Michael selection theorem, which gives a contin-uous selection for multivalued upper semicontinuos mappings, almost nothing is known
about obtaining a nonexpansive selection. In our problem studied below, we do not know how they can be proved via the classical results for single-valued mappings.
First, let us recall the following simple example of Ko []. It suggests that fixed point re-sults in the single-valued case should be extended to endpoint rere-sults in the corresponding multivalued one.
Example LetXbe the two-dimensional Euclidean space (R,·), andE= [, ]×[, ]. LetT:E→K(E) be defined by
T(a,b) := the closed convex hull of(, ), (a, ), (,b).
Note thatH(T(a,b),T(c,d))≤ (a,b) – (c,d)for all (a,b), (c,d)∈E. Moreover, we remark the following facts.
• It is clear thatFix(T) ={(a,b)∈E:ab= }andEnd(T) ={(, )}. Hence,End(T)is convex, butFix(T)is not.
• This example shows that even in a Hilbert space the class of-type mappings, which is given analogously to Bruck’s theorem does not include all nonexpansive mappings (see [], Example ., for more details).
• For a fixed element(a,b)∈Eandt∈(, ), defineSt:E→K(E)by
St(a,b) := ( –t)(a,b) +tT(a,b).
It follows that, ast↑, the net{Fix(St)}t∈(,)does not converge to the fixed point ofT
nearest to(a,b)even in the weaker convergence of sets (see []).
Recall that a multivalued mappingT:E→BC(X) has theapproximate endpoint prop-ertyif
inf
x∈Eysup∈Txd(x,y) = .
Obviously, ifEnd(T)=∅, thenThas the approximate endpoint property. In this paper, we investigate the following statement:
Thas the approximate endpoint property ⇒ End(T)=∅. (A)
This research problem has been investigated by many mathematicians (see [–] for the single-valued case and [, , ] for the multivalued one).
Recently, Panyanak [] showed that (A) holds in some situations. We first quote all main results recently proved in his paper. Some relevant definitions and concept will be given in the next section.
Theorem P Suppose that E is a nonempty bounded closed convex subset of a uniformly convex Banach space X.If T:E→K(E)is a nonexpansive mapping,then(A)holds.
Theorem P Suppose that E is a nonempty bounded closed convex subset of a complete
CAT()space X.If T:E→K(X)is a nonexpansive mapping,then(A)holds.
Note that the proofs of Theorems P, P, and P are based on the technique of asymp-totic center introduced by Lim [].
Theorem P Suppose that E is a nonempty convex subset of a completeCAT()space X. If T:E→K(X)is a nonexpansive mapping,thenEnd(T)is convex.
We also quote some of his questions.
Question P LetEbe a nonempty bounded closed convex subset of a uniformly convex Banach spaceX, andT:E→K(X) be a nonexpansive mapping. Does (A) hold?
Question P LetE be a nonempty closed convex subset of a uniformly convex Banach spaceX, andT:E→BC(X) be a nonexpansive mapping. IsEnd(T) convex?
The purpose of this paper is to give significant extensions of all theorems above. We also give affirmative answers to the preceding two questions. It should be noted that our proofs are different from those in Panyanak’s paper. Moreover, many results are established under weaker assumptions.
2 Main results
For an elementxin a metric spaceX:= (X,d) and for a nonempty bounded subsetEofX, we write
D(x,E) :=supd(x,y) :y∈E.
Using this notation, for any mappingT:E→BC(X), we have the following statements: • Thas an end point if and only if there exists an elementx∈Esuch thatD(x,Tx) = . • Thas the approximate endpoint property if and only if there exists a sequence{xn}in
Esuch thatD(xn,Txn)→.
Recall that a multivalued mappingT:E→BC(X) isnonexpansiveif it does not increase the distances, that is,
H(Tx,Ty)≤d(x,y) for allx,y∈E.
Let us first start with the following easy observation.
Proposition Let E be a nonempty subset of a metric space X,and T :E→BC(X)be given.Let x,y∈E.Then the following statements hold.
• Ify∈Ty,thend(x,y)≤D(x,Tx) +H(Tx,Ty).
• Ify∈TyandT is nonexpansive,thend(x,y)≤D(x,Tx) +d(x,y).
Proof We prove only the first assertion. Letx,y∈E, andx∈Txandy∈Ty. It follows that
Sincex∈Txis arbitrary, we have
dx,y≤D(x,Tx) +dy,Tx≤D(x,Tx) +H(Ty,Tx).
This completes the proof.
The following result follows easily from the preceding proposition.
Lemma Let E be a nonempty subset of a metric space X,and T:E→BC(X)be a contin-uous mapping.If{un}is a sequence in E such thatlimnD(un,Tun) = and{un}converges
to some element u∈E,then u∈End(T).
Proof Letu∈Tu. It follows from Proposition thatd(un,u)≤D(un,Tun) +H(Tu,Tun).
SinceTis continuous, we havelimnH(Tu,Tun) = . Then the sequence{un}converges to
u, and henceu=u∈Tu. This finishes the proof.
2.1 Endpoint results in strictly convex spaces and uniformly convex spaces A Banach spaceXisstrictly convexif the following implication holds:
u ≤, v ≤, u–v> ⇒
u+v< .
The uniform version of this property is as follows:Xisuniformly convex[] if for each
ε> , there existsδ> such that
u ≤, v ≤, u–v ≥ε ⇒
u+v ≤ –δ.
Every uniformly convex space is strictly convex. The converse is not true (see [], Exam-ple ..).
First, we give an affirmative answer to Question P. Moreover, we show that the uniform convexity can be weaken to the strict convexity. The following lemma seems to be known, but we give a proof for completeness.
Lemma If u and v are two elements in a strictly convex space such thatu+v=u+ v,thenvu=uv.
Proof If eitheru= orv= , then the conclusion holds. We now assume thatu= and v= . Lett=u/(u+v). Thent∈(, ). Sinceu+v=u+v, we have
t u
u+ ( –t) v v
= .
Without loss of generality, we may assume thatt≤/. (Otherwise, lett=v/(u+v).) It follows then that
=t u
u+ ( –t) v v
=t
u u+
v v
+ ( – t) v v
≤t
u u+
v v
In particular,
u
u+ v v
= .
Since the space is strictly convex, we have
u u=
v v.
This completes the proof.
Theorem Let E be a nonempty closed convex subset of a strictly convex Banach space X. If T:E→BC(X)is a nonexpansive mapping,thenEnd(T)is convex.
Proof Letu,v∈End(T) andz= ( –t)u+tvwheret∈(, ). We may assume thatu=v. To see thatz∈End(T), letz∈Tz. It follows from Proposition that
u–z≤ u–z and v–z≤ v–z.
We consider the following:
u–v=u–z+z–v ≤u–z+z–v ≤ u–z+z–v
=tu–v+ ( –t)u–v
=u–v.
It follows from Lemma that
z–vu–z=u–zz–v.
Moreover,
z–v=z–v= ( –t)u–v,
u–z=u–z=tu–v.
Hence, ( –t)(u–z) =t(z–v), that is,z= ( –t)u+tv=z∈Tz. This completes the proof.
We simultaneously extend Theorem P and give an affirmative answer to Question P. We also introduce the following concept, which is a multivalued version of [].
Definition LetE be a nonempty convex subset of a Banach spaceX. A multivalued mappingT:E→BC(X) is ofconvex typeiflimnD(zn,Tzn) = whenever{un}and{vn}are
sequences inEsuch thatlimnD(un,Tun) =limnD(vn,Tvn) = andzn=un+vnfor all
In the presence of uniform convexity, this class of mappings includes all nonexpansive ones. Note thatXis uniformly convex if and only iflimnxn–yn= whenever{xn}and
{yn}are sequences in Xsatisfyinglimnxn=limnyn=limnxn+yn= (see [],
Proposition ..).
Lemma Let E be a nonempty bounded closed convex subset of a uniformly convex Banach space X.If T:E→BC(X)is a nonexpansive mapping,then it is of convex type.
Proof Let{un}and{vn}be sequences inEsuch thatlimnD(un,Tun) =limnD(vn,Tvn) =
andzn=un+vnfor alln≥. For eachn, letzn∈Tznbe such that
D(zn,Tzn) –
n≤zn–z
n≤D(zn,Tzn).
It follows from Proposition that
lim sup n
un–zn≤limn D(un,Tun) +lim sup
n
un–zn
=
lim supn
un–vn.
Similarly, we have
lim sup n
vn–zn≤
lim supn
un–vn.
Without loss of generality, we may assume thatα:=limnun–zn,β:=limnvn–zn, and γ :=limnun–vndo exist. Ifγ = , then we are done. We now assume thatγ > . Let us
consider the following:
γ =lim
n un–vn
=lim n un–z
n
+zn–vn
≤lim n un–z
n+limn zn–vn
=α+β≤
γ + γ=γ.
It then follows thatα=β=γ> . In particular,
lim n
un–zn
α
=lim
n
zn–vn
β =lim n un–zn
α +
zn–vn β
= .
By the uniform convexity ofXwe have
lim n
un–zn
γ/ – zn –vn
γ/
=lim
n
un–zn
α –
zn–vn β
Hence,limnzn–zn=limn(un+vn) –zn= . This implies that
lim
n D(zn,Tzn)≤limn zn–z
n+
n
= .
Consequently,Tis of convex type.
Theorem Suppose that E is a nonempty bounded closed convex subset of a uniformly convex Banach space X.If T:E→BC(X)is a continuous multivalued mapping of convex type,then(A)holds.
Proof For eachr> , letB(r) denote the closed ball centered at zero and radiusr. Set
R:=r> :E∩B(r)=∅and inf
x∈E∩B(r)D(x,Tx) =
.
It follows from the boundedness ofEthatα:=infR<∞. Ifα= , then it follows that ∈E and ∈End(T). We now consider the caseα> . For eachn≥, letxn∈E∩B(α+ /n) and
D(xn,Txn)≤/n. In particular,lim supnxn ≤αandlimnD(xn,Txn) = . If{xn}contains a
convergent subsequence, then its limit point is an endpoint ofT. Suppose that there are a constantε> and a subsequence{xnk}such thatxnk–xnk+ ≥εfor allk≥. Setuk≡xnk,
vk≡xnk+, andzk≡ uk+
vk. It follows thatlimkD(uk,Tuk) =limkD(vk,Tvk) = . Note
thatlim supkuk ≤αandlim supkvk ≤α. By the uniform convexity ofXthere exists δ> such thatlim supkzk ≤( –δ)α<α. Moreover, it follows from the convexity type
ofTthatlimnD(zn,Tzn) = . We now obtain a contradiction. The proof is finished.
2.2 Endpoint results in reflexive spaces with Opial property
A Banach spaceXis said to have theOpial property[] if whenever{xn}is weakly
con-vergent toxandy=x, it follows that
lim sup
n
xn–x<lim sup
n
xn–y.
The following result is related to the demiclosedness property.
Proposition Let E be a nonempty subset of a Banach space X with the Opial prop-erty,and T:E→BC(X)be a nonexpansive mapping.If{un}is a sequence in E such that
D(un,Tun)→and{un}converges weakly to u∈E,then u∈End(T).
Proof Letu∈Tu. It follows from Proposition that
un–u≤D(un,Tun) +un–u.
In particular,
lim sup n
un–u≤lim sup
n
un–u.
It follows from the Opial property thatu=u∈Tu. This completes the proof.
Theorem Suppose that E is a nonempty bounded closed convex subset of a reflexive Banach space X with the Opial property.If T:E→BC(X)is a nonexpansive mapping, then(A)holds.
Proof Assume thatThas an approximate endpoint property. Let{un}be a sequence inE
such thatD(un,Tun)→. Since{un}is bounded, there exists a subsequence{unk}of{un}
such that{unk} converges weakly to some elementu∈E. It follows from the preceding
proposition thatu∈End(T).
2.3 Endpoint results in geodesic spaces whose curvature is bounded above In this section, we extend both Theorems P and P in a more general setting, that is, we consider geodesic spaces whose curvature is bounded above. Let us recall relevant definitions and a concept as follows. For more details on the subject, we refer to [].
Forκ∈R, letMκ := (Mκ,dκ) denote the unique simply connected surface (real
two-dimensional Riemannian manifold) with constant curvatureκ. That is,
Mκ:=
⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
(/√κ)S ifκ> ,
R ifκ= ,
(/√–κ)H ifκ< .
HereS,R, andH represent the two-dimensional sphere, Euclidean space, and
hyper-bolic space, respectively. In particular,
Dκ:=diamMκ=
⎧ ⎨ ⎩
π/√κ ifκ> ,
∞ ifκ≤.
A geodesic pathjoining two elementsx, yin a metric space (X,d) is an isometryc: [,l]→X, whered(x,y) =l, such thatc() =xandc(l) =y. The image of a geodesic path is called ageodesic segment. A metric space for which every two points can be joined by a geodesic segment is called a geodesic space. For any three elements x,y,z∈X, a geodesic triangle (x,y,z) is the union of three geodesic segments joining each two of them. Since there may be more than one geodesic segment joining each two points, the triangle(x,y,z) depends on the geodesic paths we choose. We say that a geodesic tri-angle(x,y,z), wherex,y,z∈Xandd(x,y) +d(y,z) +d(z,x) < Dκ, satisfiestheCAT(κ)
inequalityif there exists corresponding three pointsx,y,z∈Mκsuch that
• d(x,y) =dMκ(x,y),d(y,z) =dMκ(y,z), andd(x,z) =dMκ(x,z);
• d(p,q)≤dMκ(p,q)for allp,q∈ (x,y,z)andp,q∈ (x,y,z).
A geodesic metric space (X,d) is aCAT(κ)spaceif every geodesic triangleinXwith perimeter less than Dksatisfies theCAT(κ) inequality. IfXis aCAT(κ) space, then it is
Dκ-uniquely geodesic, that is, there exists a unique geodesic path joiningxandyfor all
x,y∈Xwithd(x,y) <Dκ. In this case, we denote the unique geodesic segment joiningx
andyby [x,y], and the elementz∈[x,y] satisfyingd(z,x) = ( –t)d(x,y) andd(z,y) =td(x,y) for somet∈[, ] is also denoted bytx⊕( –t)y. A subsetYof aCAT(κ) space isDκ-convex
Note that in this study we can consider onlyCAT() spaces because all the results can be easily extended to CAT(κ) spaces with κ> by resizing the space. Moreover, every
CAT(κ) space is aCAT(κ) space wheneverκ≤κ.
Lemma ([], Lemma .) Let(x,y,z)be a geodesic triangle in aCAT()space such that d(x,y) +d(x,z) +d(y,z) < π,and let w∈[x,y].Then
cosd(w,z)sind(x,y)≥cosd(x,z)sind(y,w) +cosd(y,z)sind(x,w).
Theorem Let E be a nonempty convex subset of a completeCAT()space X.If T:E→ BC(X)is a nonexpansive mapping,thenEnd(T)isπ-convex.
Proof Letu,v∈End(T) be such that <d(u,v) <π. Letw∈[u,v]. Note thatw∈E. We show thatw∈End(T). To see this, letw∈Tw. It follows from Proposition that
du,w≤d(u,w) and dv,w≤d(v,w). Moreover, we have the following:
sind(u,v)≥cosdw,wsind(u,v)
≥cosdu,wsind(v,w) +cosdv,wsind(u,w) ≥cosd(u,w)sind(v,w) +cosd(v,w)sind(u,w)
=sind(u,w) +d(v,w) =sind(u,v).
In particular,cosd(w,w) = . This implies thatd(w,w) = , that is,w=w∈Tw.
LetEbe a nonempty subset of aCAT() spaceX such thatd(u,v) <π for allu,v∈E. Analogously to Definition , we introduce the following concept: T :E→BC(X) is of convex typeif limnD(zn,Tzn) = whenever {un}and{vn} are sequences inE such that limnD(un,Tun) =limnD(vn,Tvn) = andzn=un⊕vnfor alln≥. It is shown in the
following theorem that this class of mappings includes all nonexpansive ones.
Theorem Let E be a nonempty convex subset of a completeCAT()space X such that
diam(E) <π.If T:E→BC(X)is nonexpansive,then it is of convex type.
Proof Let{un}and{vn}be sequences inEsuch thatlimnD(un,Tun) =limnD(vn,Tvn) =
andzn=un⊕vnfor alln≥. We show thatlimnD(zn,Tzn) = . To see this, letzn∈Tzn
be such that
D(zn,Tzn) –
n≤d
zn,zn
.
It follows from Proposition that
lim sup n
dun,zn
≤lim
n D(un,Tun) +lim supn d(un,zn) =
lim supn
Similarly, we have
lim sup n
dvn,zn
≤ lim supn
d(un,vn).
Passing to subsequences, we may assume thatα:=limnd(un,zn),β:=limnd(vn,zn), and γ :=limnd(un,vn) exist. Note thatα≤ γ andβ ≤ γ. If γ = , thenlimnd(zn,zn)≤
limn(d(zn,un) +d(un,zn))≤limnd(zn,un) +limnD(un,Tun) =limnd(un,vn) +limnD(un,
Tun) = . This implies thatlimnD(zn,Tzn)≤limn(d(zn,zn) +n) = . We now assume that γ > . Now let us consider the following:
sinγ ≥lim sup n
cosdzn,zn
sinγ
=lim sup n
cosdzn,zn
sind(un,vn)
≥lim sup n
cosdun,zn
sind(zn,vn) +cosd
vn,zn
sind(zn,un)
=cosαsin
γ+cosβsin γ
≥cos
γsin γ+cos
γsin
γ =sinγ.
In particular, sinceγ<π, we havelim supncosd(zn,zn) = . This implies thatlimnd(zn,zn) =
, that is,limnD(zn,Tzn)≤limn(d(zn,zn) +n) = . The proof is finished.
The proof of the following result is very similar to that of Theorem .
Theorem Let E be a nonempty closed convex subset of a completeCAT()space X such thatdiam(E) <πand d(x,y) +d(y,z) +d(z,x) < πfor all x,y,z∈E.If T:E→BC(X)is of convex type and continuous,then(A)holds.
Proof Assume thatThas the approximate endpoint property. Letx∈Ebe fixed andB(r)
denote the closed ball centered atxand radiusr> . Set
R:=
r> :E∩B(r)=∅and inf
x∈E∩B(r)D(x,Tx) =
.
Note that α:=infR<∞. If α= , then it follows thatx∈End(T). We now consider
the case α> . For eachn≥, letxn∈E∩B(α+ /n) andD(xn,Txn)≤/n. In
partic-ular, lim supnd(xn,x)≤α andlimnD(xn,Txn) = . If{xn} contains a convergent
subse-quence{xnk}such that{xnk}converges to some elementx∈E, thenx∈End(T). On the
other hand, we suppose that there are a constantε> and a subsequence{xnk}such that
d(xnk,xnk+)≥εfor allk≥. Setuk≡xnk,vk≡xnk+, andzk≡ uk⊕
vk. It follows that limkD(uk,Tuk) =limkD(vk,Tvk) = . SinceTis of convex type, we havelimkD(zk,Tzk) = .
allk≥. Sinced(x,uk) +d(uk,vk) +d(vk,x) < π, we have
cosd(zk,x)sind(uk,vk)
≥cosd(uk,x)sind(vk,zk) +cosd(vk,x)sind(uk,zk)
=cosd(uk,x) +cosd(vk,x)
sin
d(uk,vk).
In particular,
cosd(zk,x)≥
cosd(uk,x) +cosd(vk,x)
sind(uk,vk) sind(uk,vk)
=cosd(uk,x) +cosd(vk,x) cosd(uk,vk)
.
This implies that
lim inf
k cosd(zk,x)≥ cosα cos
ε
>cosα.
Sincet→costis strictly decreasing on (,π), there existsδ> such that
lim sup k
d(zk,x)≤α–δ<α,
which is a contradiction. The proof is finished.
The following result improves that of Espínola and Fernández-León []. It is clear that the conditiondiam(E) <π/ impliesdiam(E) <πandd(x,y) +d(y,z) +d(z,x) < πfor all x,y,z∈E. In the proof above, we do not use the modulus of convexity ofSendowed with
the spherical distance as was the case in [].
Corollary Let E be a nonempty closed convex subset of a completeCAT()space X such thatdiam(E) <πand d(x,y) +d(y,z) +d(z,x) < πfor all x,y,z∈E.Suppose that T:E→X is of convex type and is continuous.ThenFix(T)=∅if and only ifinf{d(x,Tx) :x∈E}= .
Competing interests
The author declares that he has no competing interests.
Acknowledgements
The author would like to thank the two referees for their comments and suggestions. This work is partially supported by the Research Center for Environmental and Hazardous Substance Management, Khon Kaen University.
Received: 29 September 2015 Accepted: 8 April 2016 References
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