Fixed points and approximate fixed points in product spaces
Abstract
The paper deals with the general theme of what is known about the existence of fixed points and approximate fixed points for mappings which satisfy geometric conditions in product spaces. In particular it is shown that if X and Y are metric spaces each of which has the fixed point property for nonexpansive mappings, then the product space (X ×Y )∞ has the fixed point property for nonexpansive mappings satisfying various contractive conditions. It is also shown that the product space H = (M × K)∞ has the approximate fixed point property for nonexpansive mappings whenever M is a metric space which has the approximate fixed point property for such mappings and K is a bounded convex subset of a Banach space.
Full text
TAIWANESE JOURNAL OF MATHEMATICS
Vol. 5, No. 2, pp. 405-416, June 2001
This pape is a ailable online a h p://www.ma h.n hu.edu. w/ jm/
FIXED POINTS AND APPROXIMATE FIXED POINTS IN
PRODUCT SPACES
R. Espínola and W. A. Ki k
Abs ac . The pape deals wi h he gene al heme o wha is known abou
he exis ence o ixed poin s and app oxima e ixed poin s o mappings which
sa is y geome ic condi ions in p oduc spaces. In pa icula i is shown ha
i Xand Ya e me ic spaces each o which has he ixed poin p ope y o
nonexpansi e mappings, hen he p oduc space (X×Y)∞has he ixed poin
p ope y o nonexpansi e mappings sa is ying a ious con ac i e condi ions.
I is also shown ha he p oduc space H=(M×K)∞has he app oxima e
ixed poin p ope y o nonexpansi e mappings whene e Mis a me ic space
which has he app oxima e ixed poin p ope y o such mappings and Kis
a bounded con ex subse o a Banach space.
1. INTRODUCTION
The s udy o ixed poin heo y o nonexpansi e mappings in p oduc spaces
is an ou g ow h o i s analog o con inuous mappings. A opological space is said
o ha e he ixed poin p ope y i e e y con inuous sel -map o he space has a
ixed poin . I has been known o some ime ha i bo h Xand Yha e he ixed
poin p ope y o con inouus mappings, hen i need no be he case ha X×Yhas
he ixed poin p ope y o mappings :X×Y→X×Ywhich a e con inuous
ela i e o he p oduc opology. Indeed, an example is gi en in [4] o a me ic
space Xwhich has he ixed poin p ope y, ye he space X×X ails o ha e he
ixed poin p ope y. See, o example, [6] (speci ically, Theo em 4.9) o a mo e
ex ensi e discussion.
Recei ed Feb ua y 9, 2000; e ised Decembe 15, 2000.
Communica ed by M.-H. Shih.
2001 Ma hema ics Subjec Classi ica ion: 54H25, 47H09; Seconda y 47H10.
Key wo ds and ph ases: Nonexpansi e mapping, p oduc space, ixed poin , app oxima e ixed poin
p ope y.
This esea ch was conduc ed while he i s au ho was isi ing he Uni e si y o Iowa. He acknowl-
edges he kind hospi ali y o he Uni e si y o Iowa and also he suppo o DGICYT esea ch p ojec
PB96-1338-C02-01.
405
406 R. Espíinola and W. A. Ki k
In 1968, Nadle [17] ini ia ed a s udy o ixed poin p ope ies o mappings
T:X×Y→X×Y, whe e Xis a opological space wi h he ixed poin p ope y,
Yis a me ic space, and Tis a con inuous mapping which is also a local con ac ion
in i s second coo dina e. Fo a con inues his app oach in [7].
The abo e esul s lead na u ally o he ques ion o wha happens i bo h Xand
Ya e me ic spaces wi h he con ac i e condi ions placed di ec ly on T. Fo his
discussion we need o ix some e minology. A mapping o a me ic space (M,d)
in o a me ic space (N, )is said o be nonexpansi e i ( (x), (y)) ≤d(x, y) o
all x, y ∈M. I ( (x), (y)) <d(x, y) o all x, y ∈Mwi h x6=y, hen is
said o be s ic ly con ac i e. A mapping is said o be a gene alized con ac ion
i o each x∈M he e exis s α(x)∈(0,1) such ha o each y∈M, ( (x),
(y)) ≤α(x)d(x, y).I αis a cons an map, hen o cou se is a con ac ion
mapping in he sense o Banach.
We shall use ix ( ) o deno e he se o ixed poin s o a mapping :M→M.
I (X, ρ)and (Y,d)a e me ic spaces, hen he me ic d∞on X×Yis de ined
in he usual way:
d∞((x, u),(y, )) = max{ρ(x, y),d(u, )}
o (x, u),(y, )∈X×Y. We shall con ine ou sel es o he me ic d∞in his
pape , al hough all o he esul s, indeed in some ins ances e en s onge ones, seem
o hold o he me ics dp,p∈[1,∞),
dp((x, u),(y, )) = [(ρ(x, y))p+(d(u, ))p]1/p.
A basic ques ion now becomes: I (X, ρ)and (Y,d)ha e he ixed poin p ope y
o nonexpansi e mappings and i T:X×Y→X×Yis nonexpansi e ela i e o
he me ic d∞, hen does Tnecessa ily ha e a ixed poin ? Al hough sha p esul s
ha e been ob ained, he ull answe o his ques ion emains open.
In he nex sec ion we summa ize wha is known abou me ic ixed poin heo y
in p oduc spaces. In Sec ion 3 we p o e some new esul s o mappings sa is ying
`con ac i e' condi ions. In Sec ion 4 we p o e a new esul abou he exis ence
o `app oxima e ixed poin s' o nonexpansi e mappings in p oduc spaces by ap-
plying a well-known esul abou asymp o ic egula i y o `a e aged' nonexpansi e
mappings.
2. OVERVIEW
We begin by summa izing he esul s o Nadle and Fo a. He e and h oughou
we use P1( esp., P2) o deno e he na u al coo dina e p ojec ion o X×Yon o X
Fixed Poin s and App oxima e Fixed Poin s in P oduc Space 407
( esp., on o Y). Ve sion (C) o his esul is due o Nadle ; e sion (C0) o Fo a.
Al e na e p oo s o hese esul s a e gi en in [11].
Theo em 2.1. Suppose Xis a opological space which has he ixed poin
p ope y wi h espec o con inuous mappings,suppose Yis a comple e me ic
space,and suppose T:X×Y→X×Yis a con inuous mapping which sa is ies
(C) o each x∈X, he e exis s a numbe λ(x)∈(0,1) such ha o all u, ∈Y,
d(P2◦T(x, u),P
2◦T(x, )) ≤λ(x)d(u, ).
Then Thas a ixed poin i ei he
(a) Tis uni o mly con inuous;o
(b) Yis locally compac .
Assump ions (a) and (b) can be d opped i condi ion (C) is s eng hened o
(C0) o each x∈X, he e exis s a numbe λ(x)∈(0,1) and a neighbo hood Vx
such ha o each w∈Vxand all u, ∈Y,
d(P2◦T(w, u),P
2◦T(w, )) ≤λ(x)d(u, ).
We now u n o nonexpansi e mappings in p oduc spaces. In [15], i was
shown ha i a bounded closed con ex subse Ho a Banach space has he ixed
poin p ope y o nonexpansi e mappings, and i Kis a bounded closed con ex
subse o ei he a uni o mly con ex o uni o mly smoo h Banach space, hen e e y
nonexpansi e T:H×K→H×Khas a ixed poin . This esul led o a sequence
o gene aliza ions, culmina ing in a ema kable esul o T. Kuczumow [16].
In o de o desc ibe Kuczumow's esul , we need some addi ional ac s. I is
known ha , in gene al, a weakly compac con ex subse o a Banach space need no
ha e he ixed poin p ope y o nonexpansi e mappings (Alspach [1]), bu a he
same ime weak compac ness (o e lexi i y o he unde lying space) in conjunc ion
wi h a a ie y o o he geome ic condi ions (e.g., see [9]) does in ac assu e ha
any closed con ex se has he ixed poin p ope y o nonexpansi e mappings. In
iew o his, he ollowing de ini ion is qui e na u al.
De ini ion 2.1. A closed con ex subse Kis said o ha e he gene ic ixed
poin p ope y ( o nonexpansi e mappings)i o e e y nonexpansi e T:K→K
and e e y T-in a ian nonemp y closed con ex H⊆K, ix (T)∩H6=∅.
Kuczumow used a e ac ion app oach based on a me hod o B uck [3] o p o e
he ollowing.
Theo em 2.2. Le Xbe a Banach space. Suppose K⊆Xis weakly compac
con ex and has he gene ic ixed poin p ope y,and suppose (Y,d)is a me ic
408 R. Espíinola and W. A. Ki k
space which has he ixed poin p ope y o nonexpansi e mappings. Then e e y
nonexpansi e T:(K×Y)∞→(K×Y)∞has a ixed poin .
Kuczumow obse ed ha i Xis a conjuga e space, hen he weak opology in
he abo e esul can be eplaced by he weak∗ opology.
B uck's pape [3] is ema kably ich in ideas, and in ac a di e en app oach
ound in he same pape can be modi ied o p o e he ollowing esul . The de ails
a e ound in [12].
Theo em 2.3. Le Ebe a Banach space. Suppose X⊆Eis a sepa able closed
con ex subse o Ewhich has he gene ic ixed poin p ope y,and suppose (Y,d)
is a sepa able me ic space which has he ixed poin p ope y o nonexpansi e
mappings. Then e e y nonexpansi e T:(X×Y)∞→(X×Y)∞has a ixed
poin .
While i s me hod o p oo is di e en , i is no clea o wha ex en , i any,
Theo em 2.3 is ac ually quali a i ely mo e gene al han Theo em 2.2. This is because
he e is no known example o a closed con ex subse o a Banach space which has
he gene ic ixed poin p ope y ye ails o be weakly compac .
The e a e pe haps wo addi ional esul s which should be men ioned. While
we a e basically in e es ed he e in he case p=∞,i is qui e easy o p o e he
ollowing o 1≤p<∞.
Theo em 2.4. Le Eand Fbe Banach spaces. Suppose X⊆Eand Y⊆
Fbo h ha e he ixed poin p ope y o nonexpansi e mappings. Then e e y
nonexpansi e T:(X×Y)p→(X×Y)phas a ixed poin o 1≤p<∞.
A p oo o he abo e esul is gi en in [14], based on an a gumen gi en o he
ollowing esul in [11].
Theo em 2.5. Le Eand Fbe Banach spaces. Suppose X⊆Eand Y⊆
Fbo h ha e he ixed poin p ope y o gene alized con ac ions. Then e e y
gene alized con ac ion T:(X×Y)p→(X×Y)phas a ixed poin o 1≤p≤∞.
This comple es an o e iew o wha appea o be he mos impo an known
esul s. We now u n o some new obse a ions.
3. CONTRACTIVE MAPPINGS IN PRODUCT SPACES
I he assump ion o nonexpansi eness is s eng hened, hen i is possible o
p o e addi ional esul s in a ai ly di ec manne .
Fixed Poin s and App oxima e Fixed Poin s in P oduc Space 409
Theo em 3.1. Le (X, ρ)and (Y,d)be me ic spaces. Suppose Yhas he
ixed poin p ope y o nonexpansi e mappings and suppose Xhas he ixed poin
p ope y o s ic ly con ac i e mappings,and suppose T:(X×Y)∞→(X×Y)∞
is a nonexpansi e mapping which sa is ies he addi ional condi ion
ρ(P1◦T(x, u),P
1◦T(y, )) <d
∞((x, u),(y, ))
o all (x, u),(y, )∈X×Ysa is ying ρ(x, y)6=d(u, ).Then Thas a ixed
poin .
Theo em 3.2. Le (X, ρ)and (Y,d)be me ic spaces,each o which has he
ixed poin p ope y o s ic ly con ac i e mappings. Then e e y s ic ly con ac-
i e mapping T:(X×Y)∞→(X×Y)∞has a ixed poin .
P oo o Theo em 3.1. Fix u∈Yand de ine Tu:X→Xby se ing
Tu(x)=P1◦T(x, u),x∈X.
Then i x6=y, i ollows ha ρ(x, y)6=d(u, u)=0,and we ha e
ρ(Tu(x),T
u(y))= ρ(P1◦T(x, u)),ρ(P1◦T(y,u)) <d
∞((x, u),(y,u))
=ρ(x, y).
Thus Tuis s ic ly con ac i e and by assump ion has a unique ixed g(u)∈X.
Now de ine
ϕ(u)=P2◦T(g(u),u).
We show ha ϕis nonexpansi e. No e ha since Tu(g(u)) = g(u)and T (g( )) =
g( ),we ha e
g(u)=P1◦T(g(u),u); g( )=P1◦T(g( ), ),
and, mo eo e , i ρ(g(u),g( )) 6=d(u, ), hen
ρ(g(u),g( ))= ρ(P1◦T(g(u),u),P
1◦T(g( ), ))
<d
∞((g(u),u),(g( ), ))
= max{ρ(g(u),g( )),d(u, )}
=d(u, ).
The e o e, ρ(g(u),g( )) ≤d(u, ) o all u, ∈Y. I ollows ha
d(ϕ(u),ϕ( )) = d(P2◦T(g(u),u),P
2◦T(g( ), ))
≤max{ρ(P1◦T(g(u),u),P
1◦T(g( ), )),d(P2◦T(g(u),u),P
2◦T(g( ), ))}
=d∞(T(g(u),u),T(g( ), )) ≤d∞((g(u),u),(g( ), ))
= max{ρ(g(u),g( )),d(u, )}=d(u, ).
410 R. Espíinola and W. A. Ki k
The e o e, ϕ:Y→Yis nonexpansi e. Since Yhas he ixed poin p ope y o
nonexpansi e mappings, he e exis s u∈Ysuch ha ϕ(u)=u;whence
u=ϕ(u)=P2◦T(g(u),u).
Since by assump ion g(u)∈ ix (Tu),we ha e Tu(g(u)) = P1◦T(g(u),u).
P oo o Theo em 3.2. The a gumen ollows he p e ious one, excep in his
case we mus show ha ϕis s ic ly con ac i e. The ac ha Tis s ic ly con ac i e
assu es ha
max{ρ(P1◦T(x, u),P
1◦T(y, )),d(P2◦T(x, u),P
2◦T(y, ))}
<max{ρ(x, y),d(u, )}
i x6=yo u6= . Following he p e ious a gumen s ep-by-s ep, we conclude ha
o u∈Yand x6=y, he mapping Tuis s ic ly con ac i e and has a unique ixed
poin g(u). Also, i u6= we ha e
d(ϕ(u),ϕ( ))= d(P2◦T(g(u),u),P
2◦T(g( ), ))
≤d∞(T(g(u),u),T(g( ), ))
<d
∞((g(u),u),(g( ), ))
=d(u, ).
The conclusion now ollows as in Theo em 3.1.
The ollowing is a a ian o Theo em 3.1. The assump ions on he mapping T
do no seem o be compa able.
Theo em 3.3. Le (X, ρ)and (Y,d)be me ic spaces,each o which has he
ixed poin p ope y o nonexpansi e mappings,and suppose T:(X×Y)∞→
(X×Y)∞is a nonexpansi e mapping which sa is ies he addi ional condi ion
ρ(P1◦T(x, u, P1◦T(y, )) <d
∞((x, u),(y, ))
o all (x, u),(y, )∈X×Ysa is ying u6= and x6=y. Then Thas a ixed
poin .
Theo em 3.3 has he ollowing immedia e co olla y.
Co olla y 3.1. Le (X, ρ)and (Y,d)be me ic spaces,each o which has he
ixed poin p ope y o nonexpansi e mappings,and suppose T:(X×Y)∞→
(X×Y)∞is a nonexpansi e mapping which is quasi-con ac i e in he sense ha
d∞(T(x, u),T(y, )) <d
∞((x, u),(y, ))
Fixed Poin s and App oxima e Fixed Poin s in P oduc Space 411
o all (x, u),(y, )∈X×Ysa is ying u6= and x6=y. Then Thas a (unique)
ixed poin .
No e ha he condi ion o he co olla y is weake han he con ac i e condi ion
o Theo em 3.2. In exchange, a li le mo e is assumed abou he spaces; speci ically
ha Xhas he ixed poin p ope y o nonexpansi e mappings.
P oo o Theo em 3.3. Fix u∈Yand as be o e de ine Tu:X→Xby se ing
Tu(x)=P1◦T(x, u),x∈X.
Then
ρ(Tu(x),T
u(y))
≤max{ρ(P1◦T(x, u),P
1◦T(y,u)),d(P2◦T(x, u),P
2◦T(y,u))}
=d∞(T(x, u),T(y,u))
≤d∞((x, u),(y,u))
=ρ(x, y).
Thus Tuis nonexpansi e and by assump ion has a nonemp y ixed poin se
ix (Tu)⊆X. Le gbe any selec ion o he mapping
u7→ ix (Tu)
and de ine ϕas in Theo em 3.1. We show ha ϕis nonexpansi e. Since g(u)∈
ix (Tu)and g( )∈ ix (T ), we ha e
g(u)=P1◦T(g(u),u); g( )=P1◦T(g( ), ).
Now le u, ∈Y. The e a e wo cases.
1. I g(u)=g( ), hen ob iously ρ(g(u),g( )) ≤d(u, )and we ha e
d(ϕ(u),ϕ( )) = d(P2◦T(g(u),u),P
2◦T(g( ), ))
≤max{ρ(P1◦T(g(u),u),P
1◦T(g( ), )),d(P2◦T(g(u),u),P
2◦T(g( ), ))}
=d∞(T(g(u),u),T(g( ), )) ≤d∞((g(u),u),(g( ), )) ≤d(u, ).
2. On he o he hand, i g(u)6=g( ), hen i mus also be he case ha u6= .
The e o e,
ρ(g(u),g( )) = ρ(P1◦T(g(u),u),P
1◦T(g( ), ))
<d
∞((g(u),u),(g( ), ))
= max{ρ(g(u),g( )),d(u, )}
=d(u, )
412 R. Espíinola and W. A. Ki k
and i ollows ha
d(ϕ(u),ϕ( )) = d(P2◦T(g(u),u),P
2◦T(g( ), ))
≤max{ρ(P1◦T(g(u),u),P
1◦T(g( ), )),d(P2◦T(g(u),u),P
2◦T(g( ), ))}
=d∞(T(g(u),u),T(g( ), ))
≤d∞((g(u),u),(g( ), ))
= max{ρ(g(u),g( )),d(u, )}=d(u, ).
The e o e, in ei he case, d(ϕ(u),ϕ( )) ≤d(u, ),and ϕ:Y→Yis nonex-
pansi e. The conclusion again ollows as in Theo em 3.1.
In he p eceding p oo , he ques ion migh a ise as o whe he ix (Tu)is a
single on. Suppose o he wise, and le g1(u)and g2(u)be dis inc choices o he
selec ion u7→ ix (Tu).Then acco ding o case 2, o any ∈Y,
d(g1(u),g
2(u)) ≤d(g1(u), )+d(g2(u), )<2d(u, ).
Ob iously, his can happen only i uis an isola ed poin o Y.
To acili a e compa ison, we summa ize he o egoing esul s as ollows:
Theo em 3.4. Le (X, ρ)and (Y,d)be me ic spaces,and suppose T:(X×
Y)∞→(X×Y)∞is a nonexpansi e mapping. Then Thas a ixed poin i any
one o he ollowing condi ions holds.
(a) Xand Yha e he ixed poin p ope y o nonexpansi e mappings and T
sa is ies
d∞(T(x, u),T(y, )) <d
∞((x, u),(y, ))
o all (x, u),(y, )∈X×Ysa is ying u6= and x6=y.
(b) Yhas he ixed poin p ope y o nonexpansi e mappings, Xhas he ixed
poin p ope y o s ic ly con ac i e mappings, and T:(X×Y)∞→
(X×Y)∞sa is ies
ρ(P1◦T(x, u),P
1◦T(y, )) <d
∞((x, u),(y, ))
o all (x, u),(y, )∈X×Ysa is ying ρ(x, y)6=d(u, ).
(c) Xand Yha e he ixed poin p ope y o s ic ly con ac i e mappings and
Tis s ic ly con ac i e.
Fixed Poin s and App oxima e Fixed Poin s in P oduc Space 413
4. APPROXIMATE FIXED POINTS IN PRODUCT SPACES
In his sec ion we p o e an app oxima e ixed poin heo em o nonexpansi e
mappings in ce ain p oduc spaces. A me ic space (M,d)is said o ha e he
app oxima e ixed poin p ope y i any nonexpansi e mapping T:M→Mhas
an app oxima e ixed poin sequence, ha is, a sequence {un}in M o which
limnd(un,T(un)) = 0.This o cou se is equi alen o saying
in {d(x, T(x)) : x∈M}=0.
Ou heo em is based on he ollowing esul , which was p o ed o a single
mapping (and o a mo e gene al con e gence p ocess) by Ishikawa [10]. Edels ein
and O'B ien [5] showed ha he con e gence is uni o m o e K, and subsequen ly
Goebel and Ki k [8] showed ha in ac he con e gence is uni o m o e x0in K
and o e he class o all nonexpansi e mappings T:K→K. Ano he p oo o his
ac is gi en in [13]. Fo a echnical s udy o he a e o uni o m con e gence and
a comp ehensi e e iew o he li e a u e, see [2].
Theo em 4.1. Le Kbe a bounded con ex subse o a Banach space and le
ε>0.Then he e exis s N∈Nsuch ha i n≥N, i x0∈K, and i T:K→K
is nonexpansi e, hen
k n(x0)− n+1(x0)k≤ε,
whe e =(1/2)(I+T).
An in e es ing ea u e o he p oo gi en below is he ac ha he penul ima e
s ep o he p oo equi es he uni o mi y o he con e gence o { n(x0)}in he
abo e esul o e he class o all nonexpansi e T:K→K.
Theo em 4.2. Suppose Mis a me ic space which has he app oxima e ixed
poin p ope y o nonexpansi e mappings and suppose Kis a bounded closed
con ex subse o a Banach space X. Le
H=(K×M)∞.
Then Hhas he app oxima e ixed poin p ope y o nonexpansi e mappings.
P oo . Le T:H→Hbe nonexpansi e and le P1and P2deno e he espec i e
coo dina e p ojec ions o Hon o Kand M. Fix y∈Mand de ine Ty:K→K
by se ing
Ty(x)=P1◦T(x, y),x∈K.
Now ix x0∈Kand se y=(I+Ty)/2.