Mann iteration for monotone nonexpansive mappings in ordered CAT(0) space with an application to integral equations

Uddin e al. Jou nal o Inequali ies and Applica ions (2018) 2018:339
h ps://doi.o g/10.1186/s13660-018-1925-2
RESEARCH Open Access
Mann i e a ion o mono one
nonexpansi e mappings in o de ed CAT(0)
space wi h an applica ion o in eg al
equa ions
Izha Uddin1, Chanchal Ga odia1* and Juan Jose Nie o2
*Co espondence:
c.ga [email p o ec ed]
1Depa men o Ma hema ics,
Facul y o Na u al Sciences, Jamia
Millia Islamia, New Delhi, India
Full lis o au ho in o ma ion is
a ailable a he end o he a icle
Abs ac
In his pape , we es ablish some con e gence esul s o a mono one nonexpansi e
mapping in a CAT(0) space. We p o e he - and s ong con e gence o he Mann
i e a ion scheme. Fu he , we p o ide a nume ical example o illus a e he
con e gence o ou i e a ion scheme, and also, as an applica ion, we discuss he
solu ion o in eg al equa ion. Ou esul s ex end some o he ele an esul s.
MSC: 47H09; 47H10
Keywo ds: CAT(0) space; Fixed poin ; -con e gence; Mono one nonexpansi e
mapping
1 In oduc ion
TheBanachcon ac ionp inciple[1]isoneo hemos undamen al esul sinfixedpoin
heo yandhasbeenu ilizedwidely o p o ing heexis enceo solu ionso diffe en non-
linea unc ional equa ions. In he las ew yea s, many effo s ha e been made o ob ain
fixedpoin sinpa iallyo de edse s.In2004,RanandReu ings[2]gene alized heBanach
con ac ion p inciple o o de ed me ic spaces. La e on, in 2005, Nie o and Rod iguez
[3] used he same app oach o u he ex end some mo e esul s o fixed poin heo y in
pa ially o de ed me ic spaces and u ilized hem o s udy he exis ence o solu ions o
diffe en ial equa ions.
No e ha he Banach con ac ion p inciple is no longe ue o nonexpansi e map-
pings, ha is, a nonexpansi emappingneedno admi afixedpoin onacomple eme ic
space. Also, Pica d i e a ion need no con e ge o a nonexpansi e map in a comple e
me ic space. This led o he beginning o a new e a o fixed poin heo y o nonexpan-
si e mappings by using geome ic p ope ies. In 1965, B owde [4], Göhde [5], and Ki k
[6] ga e h ee basic exis ence esul s o nonexpansi e mappings. Wi h a iew o loca ing
fixed poin s o nonexpansi e mappings, Mann [7]andIshikawa[8]in oduced wobasic
i e a ion schemes.
Now,fixedpoin heo yo mono onenonexpansi emappingsisgainingmucha en ion
among he esea che s.Recen ly,Bacha andKhamsi[9],Abdulla i e al.[10],andSonge
©The Au ho (s) 2018. This a icle is dis ibu ed unde he e ms o he C ea i e Commons A ibu ion 4.0 In e na ional License
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ided you gi e app op ia e c edi o he o iginal au ho (s) and he sou ce, p o ide a link o he C ea i e Commons license, and
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Uddin e al. Jou nal o Inequali ies and Applica ions (2018) 2018:339 Page 2 o 13
al.[11]p o ed some exis enceand con e gence esul s o mono onenonexpansi emap-
pings. Dehaish and Khamsi [12] p o ed he weak con e gence o he Mann i e a ion o
a mono one nonexpansi e mapping. In 2016, Song e al. [11] conside ed he weak con-
e gence o he Mann i e a ion scheme o a mono one nonexpansi e mapping Tunde
some mild diffe en condi ions in a Banach space.
Theaimo hispape is os udy hecon e gencebeha io o hewell-knownManni e -
a ion [7]inaCAT(0) space o a mono one nonexpansi e mapping. Fu he , we p o ide a
nume icalexampleandapplica ion ela ed osolu iono anin eg alequa ion.Ou esul s
gene alize and imp o e se e al exis ing esul s in he li e a u e.
2P elimina ies
To make ou pape sel -con ained, we ecall some basic defini ions and ele an esul s.
Ame icspaceXis a CAT(0) space i i is geodesically connec ed and i e e y geodesic
iangleinXisa leas as hinasi scompa ison ianglein heEuclideanplane.Fo u he
in o ma ionabou hesespaces and he undamen al ole heyplayin a ious b ancheso
ma hema ics, we e e o B idson and Haeflige [13] and Bu ago e al. [14]. E e y con ex
subse o EuclideanspaceRnendowedwi h heinducedme icisaCAT(0)space.Fu he ,
he class o Hilbe spaces a e examples o CAT(0) spaces.
The fixed poin heo y in CAT(0) spaces is gaining a en ion o esea che s, and many
esul s ha e been ob ained o single- and mul i alued mappings in a CAT(0) space. Fo
diffe en aspec so fixedpoin heo yinCAT(0)spaces,we e e o[15–24].The ollowing
ew esul s a e necessa y o ou subsequen discussion.
Lemma 2.1 ([21]) Le (X,d)be a CAT(0) space.Fo e, ∈Xandz∈[0,1], he e exis s a
unique h∈[e, ]such ha
d(e,h)=zd(e, )and d( ,h)=(1–z)d(e, ).
We use he no a ion (1–z)e⊕z o he unique poin ho he lemma.
Lemma2.2([21]) Le (X,d)be a CAT(0) space.Fo e, ,h∈Xandz∈[0,1], we ha e
d(1–z)e⊕z ,h≤(1–z)d(e,h)+zd( ,h).
Lemma2.3([21]) Le X be a CAT(0) space.Then
d(1–z)e⊕z ,h2≤(1–z)d(e,h)2+zd( ,h)2–z(1–z)d(e, )2
o all e, ,h∈Xandz∈[0,1].
Le {un}beaboundedsequenceinacomple eCAT(0) space X.Fo u∈X,wedeno e
u,{un}=limsup
n→∞ d(u,un).
The asymp o ic adius ({un})isgi enby
{un}=in  (u,un):u∈X,
Uddin e al. Jou nal o Inequali ies and Applica ions (2018) 2018:339 Page 3 o 13
and he asymp o ic cen e A({un})o {un}is defined as
A{un}=u∈X: (u,un)= {un}.
I is known ha in a CAT(0) space, A({un}) consis s o exac ly one poin [25,P oposi-
ion 5].
In 1976, Lim [26] in oduced he concep o -con e gence in a me ic space. La e
on, Ki k and Panyanak [22]p o ed ha CAT(0) spaces p esen ed a na u al amewo k o
Lim’s concep and p o ided p ecise analogs o se e al esul s in Banach spaces in ol ing
weak con e gence in CAT(0) space se ing.
Defini ion 2.4 Asequence{un}in Xis said o be -con e gen o u∈Xi uis he
unique asymp o ic cen e o { n} o e e y subsequence { n}o {un}.In hiscase,wew i e
-limnun=uand say ha uis he -limi o {un}.
Defini ion 2.5 A Banach space Xis said o sa is y Opial’s condi ion i o any sequence
{un}in Xwi h unu(deno es weak con e gence), we ha e limsupn→∞ un–u<
limsupn→∞un–  o all ∈Xwi h =u.
Examples o Banach spaces sa is ying his condi ion a e Hilbe spaces and all lpspaces
(1<p<∞). On he o he hand, Lp[0,2π]wi h1<p=2 ail o sa is y Opial’s condi ion.
No ice ha i gi en a sequence {un}in Xsuch ha {un}-con e ge o u, hen o ∈X
wi h =u,weha e
limsup
n→∞ un–u<limsup
n→∞ un– .
So,e e y CAT(0) space sa isfies Opial’s p ope y.
Lemma 2.6 ([22]) E e y bounded sequence in a comple e CAT(0) space admi s a -
con e gen subsequence.
Lemma 2.7 ([21]) I G is a closed con ex subse o a comple e CAT(0) space X and i {un}
is a bounded sequence in G, hen he asymp o ic cen e o {un}is in G.
Nex , we in oduce he concep o pa ial o de in he se ing o CAT(0) spaces.
Le Xbe a comple e CAT(0) space endowed wi h pa ial o de “”. An o de in e al is
any o he subse s
[a,→)={u∈X;au}o (←,a]={u∈X:ua}
o any a∈X. So, an o de in e al [u, ] o all u, ∈Xis gi en by
[u, ]={w∈X:uw }.
Th oughou we will assume ha he o de in e als a e closed and con ex subse s o an
o de ed CAT(0) space (X,).
Uddin e al. Jou nal o Inequali ies and Applica ions (2018) 2018:339 Page 4 o 13
Defini ion 2.8 Le Gbe a nonemp y subse o an o de ed me ic space X. A mapping
P:G→Gis said o be:
(i) mono one i PuP o all u, ∈Gwi h u ,
(ii) mono one nonexpansi e i P is mono one and
d(Pu,P )≤d(u, )
o all u, ∈Gwi h u .
Now we p esen he Mann i e a ion scheme in he se ing o o de ed CAT(0) spaces
(X,). Le Gbe anonemp ycon exsubse o a CAT(0) space X. Then he Mann i e a ion
is as ollows:
u1∈G,
un+1 =(1–κn)un⊕κnPun,n∈N,(2.1)
whe e {κn}⊂[0,1]. In his pape , we p o e some -con e gence and s ong con e gence
esul s in CAT(0) spaces.
3Some-con e gence and s ong con e gence heo ems
We begin wi h he ollowing impo an lemma.
Lemma3.1 Le Gbeanonemp yclosedcon exsubse o acomple eo de ed CAT(0)space
(X,), and le P :G→G be a mono one nonexpansi e mapping.Fix u1∈Gsuch ha
u1Pu1.I {un}is defined by (2.1)wi h condi ion ∞
n=1 κn(1–κn)=∞, hen we ha e:
(i) unun+1 Pun o any n≥1,
(ii) unu,p o ided ha {un}-con e ges o a poin u∈G.
P oo (i) We will p o e he esul by induc ion on n.No e ha i q1,q2∈Ga e such ha
q1q2, henq1λq1+(1–λ)q2q2 o any λ∈[0,1]. This is ue because we ha e
assumed ha o de in e alsa econ ex.Thusweonlyneed oshow ha unPun o any
n≥1. We ha e al eady assumed ha u1Pu1, and hence he inequali y holds o n=1.
Assume ha unPun o n≥2. Since κn∈[0,1] o all n,weha e
un(1–κn)un⊕κnPunPun,
ha is, unun+1 Pun.SincePis mono one, we ha e PunPun+1. By using he an-
si i i y o he o de we ge un+1 Pun+1. Thus by induc ion he inequali y is ue o any
n≥1.
(ii) Le ube he -limi o {un}.F ompa (i)weha eunun+1 o all n≥1since{un}
is inc easing and he o de in e al [um,→) is closed and con ex. The e o e u∈[um,→)
o a fixed m∈N;o he wise,i u/∈[um,→), hen we could cons uc a subsequence {u }
o {un}by lea ing he fi s m–1 e mso hesequence{un}, and hen he asymp o ic
cen e o {u }would no be u, which con adic s he assump ion ha uis he -limi o
he sequence {un}. This comple es he p oo o pa (ii). 
Uddin e al. Jou nal o Inequali ies and Applica ions (2018) 2018:339 Page 5 o 13
Lemma3.2 Le G be a nonemp y closed con ex subse o a comple e CAT(0) space (X,),
and le P :G→G be amono one nonexpansi emapping.Fix u1∈Gsuch ha u
1Pu1.I
{un}is a sequence desc ibed as in (2.1)and F(P)=∅wi h ∈F(P)such ha u1, hen:
(i) limn→∞d(un, )exis s,and
(ii) limn→∞d(Pun,un)=0.
P oo (i)Since u1,usingpa (i)o Lemma3.1,weha eunun+1 Pun.Inpa icula ,
o n=1,weha eu1u2Pu1. Using he ansi i i y o he o de , we ge u2.By
ma hema ical induc ion we ha e un o all n≥1. Now we ha e
d(un+1, )=d(1–κn)un⊕κnPun, 
≤(1–κn)d(un, )+κnd(Pun, )
=(1–κn)d(un, )+κnd(Pun,P ).
Since Pis a mono one map and un o all n≥1, we ha e
d(un+1, )≤(1–κn)d(un, )+κnd(un, )
=d(un, ).
Thus we ha e d(un+1, )≤d(un, ) o all n≥1. So {d(un, )}is a dec easing eal sequence
bounded below by ze o. Hence limn→∞ d(un, )exis s.
(ii) Fi s , conside
d(Pun+1,un+1)=dPun+1,(1–κn)un⊕κnPun
≤(1–κn)d(Pun+1,un)+κnd(Pun+1,Pun)
≤(1–κn)d(Pun+1,un)+κnd(un+1,un)
≤(1–κn)d(Pun+1,Pun)+d(Pun,un)+κnd(un+1,un)
≤(1–κn)d(un+1,un)+d(Pun,un)+κnd(un+1,un)
=(1–κn)d(Pun,un)+d(un+1,un)
=(1–κn)d(Pun,un)+d(1–κn)un⊕κnPun,un
≤(1–κn)d(Pun,un)+(1–κn)d(un,un)+κnd(Pun,un)
=d(Pun,un).
So limn→∞d(Pun,un)exis s.
Since u1, using he Lemma 3.1,weha e u1un o all n≥1. Then, since Pis a
nonexpansi e map and is a fixed poin o P,weha e
d(un+1, )2=d(1–κn)un⊕κnPun, 2
≤(1–κn)d(un, )2+κnd(Pun, )2–(1–κn)κnd(un,Pun)2
=(1–κn)d(un, )2+κnd(Pun,P )2–(1–κn)κnd(un,Pun)2

Uddin e al. Jou nal o Inequali ies and Applica ions (2018) 2018:339 Page 6 o 13
≤(1–κn)d(un, )2+κnd(un, )2–(1–κn)κnd(un,Pun)2
=d(un, )2–(1–κn)κnd(un,Pun)2.
F om his we ge
∞

n=1(1–κn)κnd(un,Pun)2≤d(u1, )2<∞. (3.1)
Since ∞
n=1(1–κn)κn=∞, he eexis sasubsequence{unk}o {un}such ha
lim
n→∞d(Punk,unk)=0.
Since limn→∞d(Pun,un) exis s, i ollows ha limn→∞d(Pun,un) = 0, and his p o es he
esul . 
The ollowing lemma is an analogue o Theo em 3.7 o [22].
Lemma3.3 Le G be a nonemp y closed con ex subse o a comple e CAT(0) space (X,),
and le P :G→G be a mono one nonexpansi e mapping.Fix u1∈Gsuch ha u
1
Pu1.I {un}is a sequence desc ibed as in (2.1), hen he condi ions -limnun=uand
limn→∞d(Pun,un)=0imply ha u is a fixed poin o P.
P oo Since -limnun=u, by Lemma 3.1 we ge unu o all n≥1. Then om he non-
expansi eness o Pand limn→∞ d(Pun,un)=0 i ollows ha
d(Pu,un)≤d(Pu,Pun)+d(Pun,un),
limsup
n→∞ d(Pu,un)≤limsup
n→∞ d(Pu,Pun)+d(Pun,un)
=limsup
n→∞ d(Pu,Pun)
≤limsup
n→∞ d(u,un).
Thus by he uniqueness o asymp o ic cen e we ge Pu =u, which p o es he desi ed
esul . 
Theo em3.4 Le Gbeanonemp yclosedcon exsubse o acomple eCAT(0)space(X,),
and le P :G→G be a mono one nonexpansi e mapping wi h F(P)=∅.Fix u1∈Gsuch
ha u1Pu1.I {un}is a sequence desc ibed as in (2.1), hen {un}-con e ges o a fixed
poin o P.
P oo F om Lemma 3.2 we ha e ha limn→∞ d(un, ) exis s o each ∈F(P), so he se-
quence {un}is bounded, and limn→∞ d(un,Pun)=0.
Le Wω({un})=:X({ n}), whe e he union is aken o e all subsequences { n}o e
{un}. To show he -con e gence o {un} o a fixed poin o P,wewillfi s p o e ha
Wω({un})⊂F(P) and he ea e a gue ha Wω({un}) is a single on se . To show ha
Wω({un})⊂F(P), le y∈Wω({un}). Then he e exis s a subsequence {yn}o {un}such
Uddin e al. Jou nal o Inequali ies and Applica ions (2018) 2018:339 Page 7 o 13
ha X({yn})=y. By Lemmas 2.6 and 2.7 he e exis s a subsequence {zn}o {yn}such ha
-limnzn=zandz∈G.Sincelimn→∞d(Pun,un)=0and{zn}isasubsequenceo {un},we
ha e ha limn→∞d(zn,Pzn)= 0.In iewo Lemma3.3,weha ez=Pz,andhencez∈F(P).
Nowwewish oshow ha z=y. I , on he con a y, z=y, hen we would ha e
limsup
n→∞ d(zn,z)<limsup
n→∞ d(zn,y)
≤limsup
n→∞ d(yn,y)
<limsup
n→∞ d(yn,z)
=limsup
n→∞ d(un,z)
=limsup
n→∞ d(zn,z),
whichisacon adic ionsince Xsa isfies heOpial condi ion andhence z=y∈F(P).Now
i emains o show ha Wω({un}) consis s o a single elemen only. Fo his, le {yn}be a
subsequence o {un}. Again, using Lemmas 2.6 and 2.7,wecanfindasubsequence{zn}o
{yn}such ha -limnzn=z.Le X({yn})=yand X({un})=u. P e iously, we ha e al eady
p o ed ha y=z; he e o e, i suffices o show ha z=u.I z=u, hensincez∈F(P),
{d(un,z)}is con e gen by Lemma 3.2, By he uniqueness o asymp o ic cen e we ha e
limsup
n→∞ d(zn,z)<limsup
n→∞ d(zn,u)
≤limsup
n→∞ d(un,u)
<limsup
n→∞ d(un,z)
=limsup
n→∞ d(zn,z),
which gi es a con adic ion. The e o e we mus ha e z=u,whichp o es ha Wω({un})
is a single on se and ha a pa icula elemen is a fixed poin o P.Hence heconclusion
ollows. 
Theo em 3.5 Le X be a comple e CAT(0) space endowed wi h pa ial o de ing ,and
le G be a nonemp y closed con ex subse o X.Le P :G→G beamono one nonexpansi e
mapping such ha F(P)=∅.Fix u1∈Gsuch ha andu
1Pu1.I {un}is a sequence de-
sc ibed as in (2.1)such ha ∞
n=1 κn(1–κn)=∞, hen {un}con e ges o a fixed poin o P
i and only i limin n→∞d(un,F(P))=0.
P oo I he sequence {un}con e ges o a poin u∈F(P), hen i is ob ious ha
limin n→∞d(un,F(P))=0.
Fo he con e se pa , assume ha limin n→∞ d(un,F(P)) = 0. F om Lemma 3.2(i) we
ha e
d(un+1, )≤d(un, ) o any ∈F(P),
so ha
dun+1,F(P)≤dun,F(P).
Uddin e al. Jou nal o Inequali ies and Applica ions (2018) 2018:339 Page 8 o 13
Thus {d(un,F(P))} o ms a dec easing sequence ha is bounded below by ze o, so
limn→∞d(un,F(P)) exis s. As limin n→∞ d(un,F(P))=0, we ha e limn→∞d(un,F(P))=0.
Now we p o e ha {un}is a Cauchy sequence in G.Le >0 be a bi a y. Since
limin n→∞d(un,F(P))= 0, he e exis s n0such ha , o all n≥n0,weha e
dun,F(P)<
4.
In pa icula ,
in d(un0, ): ∈F(P)<
4,
so he e mus exis ∈F(P)such ha
d(un0, )<
2.
Thus, o m,n≥n0,weha e
d(un+m,un)≤d(un+m, )+d(un, )<2d(un0, )<2
2=,
which shows ha {un}is a Cauchy sequence. Since Gis a closed subse o a comple e
me ic space X,soGi sel is a comple e me ic space, and he e o e {un}mus con e ge
in G.Le limin n→∞ un=q.
Now Pis a mono one nonexpansi e mapping, and om Lemma 3.3(i) we ha e
limn→∞d(Pun,un)=0.Also, om hep oo o Lemma3.1in[12] we can easily deduce
ha unq o any n≥1. The e o e we ha e
d(q,Pq)≤d(q,un)+d(un,Pun)+d(Pun,Pq)
≤d(q,un)+dun,P(un)+d(un,q)
→0asn→∞,
and hence q=Pq.Thusq∈F(P). 
4 Nume ical example
In his sec ion, we p esen a nume ical example o illus a e he con e gence beha io o
ou i e a ion scheme (2.1).
Le X=[0,+∞) be a comple e me ic space wi h he me ic
d(u, )=|u– |,u, ∈X.
Now, conside he o de ela ion u as
u, ∈[0,1] and u≤ o
u, ∈(n,n+1] o somen=1,2,... and u≤ .
Uddin e al. Jou nal o Inequali ies and Applica ions (2018) 2018:339 Page 9 o 13
Le Pbe defined by
P(0)= 0, P(u)=n
2+u
2,u∈(n,n+1],n=0,1,2,....
Then, clea ly, Pis no con inuous a =n+1 o n=0,1,2,...,since
Pn+1–=n+1
2=n+1=Pn+1+.
Also, i u , henu, ∈[0,1] o u, ∈(n,n+1] o somen=1,2,...,and
dP(u),P( )=dn
2+u
2,n
2+
2=1
2d(u, ).
So, Pis a mono one nonexpansi e map bu no a nonexpansi e map, and 0 is he unique
fixed poin o P.
Now, we show he con e gence o (2.1) using wo diffe en se s o alues.
I is e iden om he ables (Table 1and Table 2)andg aphs(Fig.1and Fig. 2) ha ou
sequence (2.1) con e ges o 0, which is a fixed poin o P.
Table 1 (κn=2n
5n+2 o all n∈N)
S ep When u1=0.25 u1=0.45 u1=0.65
1 0.25 0.45 0.65
2 0.1607142857142857 0.2892857142857142 0.4178571428571429
3 0.1071428571428571 0.1928571428571428 0.2785714285714286
4 0.07247899159663865 0.1304621848739496 0.1884453781512605
5 0.04941749427043545 0.0889514896867838 0.1284854851031322
6 0.03386013496307614 0.06094824293353705 0.088036350903998
7 0.02327884278711485 0.04190191701680672 0.0605249912464986
8 0.01604352678571429 0.02887834821428571 0.04171316964285715
9 0.01107767325680272 0.0199398118622449 0.02880195046768708
10 0.007660093209491246 0.01378816777708424 0.01991624234467724
11 0.005303141452724708 0.00954565461490447 0.01378816777708424
12 0.003674983989168876 0.006614971180503976 0.00955495837183908
13 0.002548779218294543 0.004587802592930178 0.006626825967565813
14 0.001768928860458153 0.003184071948824675 0.004599215037191199
15 0.001228422819762606 0.002211161075572691 0.003193899331382777
16 0.000853514556588304 0.001536326201858948 0.002219137847129592
17 0.0005932967039699188 0.001067934067145854 0.00154257143032179
18 0.0004125798918411505 0.0007426438053140709 0.001072707718786992
19 0.0002870120986721047 0.0005166217776097884 0.0007462314565474726
20 0.0001997249140244027 0.0003595048452439249 0.0005192847764634474
21 0.0001390242048601234 0.0002502435687482223 0.0003614629326363212
22 0.0000967972267484037 0.0001742350081471267 0.0002516727895458498
23 0.00006741235434263828 0.0001213422378167489 0.0001752721212908597
24 0.0000469581784523506 0.0000845247212142311 0.0001220912639761116
25 0.00003271676367581804 0.0000588901746164725 0.000085063585557127
26 0.00002279868964810942 0.00004103764136659699 0.00005927659308508453
27 0.000015889995815349 0.00002860199246762819 0.00004131398911990741
28 0.00001107660292237831 0.00001993788526028096 0.00002879916759818363
29 7.722420347291922 ×10–6 0.00001390035662512546 0.00002007829290295901
30 5.384680854404231 ×10–6 9.69242553792 ×10–6 0.00001400017022145
31 3.755106385308214 ×10–6 6.759191493554787 ×10–6 9.76327660180 ×10–6