E
ex ac a ma hema icae Vol. 17, N´um. 3, 331 – 349 (2002)
IV Cu so Espacios de Banach y Ope ado es.
La edo, Agos o de 2001.
Geome ic P ope ies o Banach Spaces and
Me ic Fixed Poin Theo y
Tom´
as Dom´
ınguez Bena ides
Facul ad de Ma em´a icas, P.O. Box 1160, 41080 Se illa, Spain
e-mail: [email p o ec ed]
AMS Subjec Class. (2000): 47H09, 47H10
1. In oduc ion
Le Ta mapping om a se Xin o i sel . The mapping Thas a ixed
poin i he e exis s x0∈Xsuch ha Tx0=x0. The mos known ixed poin
heo em is he Con ac ion Mapping P inciple, due o S. Banach [3]. The
classical s a emen is he ollowing:
Theo em. (Banach Fixed Poin Theo em [3]) Le Xbe a comple e me ic
space and T:X→Xa con ac i e mapping, i.e. he e exis s k∈[0,1) such
ha
d(Tx, Ty)≤kd(x, y)
o e e y x, y ∈X. Then Thas a (unique) ixed poin xω. Fu he mo e
xω= limnTnx0 o any x0∈X.
The p oo o he Banach Theo em is a ely simple (p obably because he
wide di usion o Banach echniques). Indeed, le x0be an a bi a y poin and
se xn=Tnx0. We ha e
d(xn+1, xn)≤kd(xn, xn−1)≤... ≤knd(x1, x0).
Thus, {xn}is a Cauchy sequence, because
d(xn+j, xn)≤d(xn+j, xn+j−1) + ... +d(xn+1, xn)
≤(kn+j−1+... +kn)d(x1, x0)≤kn+j−1
1−k< ²
331
332 . dom´
ınguez
i nis la ge enough. Ob iously, he ixed poin is unique, because i x=Tx,
y=Ty we ha e d(x, y) = d(Tx, Ty)≤kd(x, y) which implies d(x, y) = 0.
Banach Theo em has p o ed o be e y use ul o sol e di e en heo e ical
and p ac ical p oblems. Fo ins ance, i is used o p o e he Pica d-Lindel¨o
Theo em abou exis ence o solu ion o di e en ial equa ions and o p o e
he In e se (o Implici ) Func ion Theo em in in ini e dimensional spaces.
A big numbe o gene aliza ions o he Banach Theo em ha e appea ed in
he li e a u e, ying o weak some assump ions (see, o ins ance, [11, Cap.
3]). Howe e , he mos na u al imp o emen would be o le he cons an
ka ain he alue 1, bu in his case he esul is no mo e ue. Indeed, a
ansla ion in Rsa is ies he assump ions and i is ixed poin ee. E en he
mild assump ion d(Tx, Ty)< d(x, y) does no assu e he exis ence o a ixed
poin . Indeed, conside X= [1,∞) and Tx =x+ 1/x. We ha e
d(Tx, Ty) = ¯¯¯¯
x+1
x−y−1
y¯¯¯¯
=|y−x|−¯¯¯¯
1
x−1
y¯¯¯¯
<|y−x|.
Howe e Tis ixed poin ee. The ail o he exis ence o ixed poin in his
example occu s because, in addi ion, he in e al [1,∞) is unbounded. Indeed,
o he wise we could apply B ouwe ’s Theo em:
Theo em. (B ouwe ’s Theo em (1912) [5]) Le Ma bounded con ex
closed subse o Rnand T:M→Ma con inuous mapping. Then Thas
a ixed poin .
Ha ing in mind bo h Banach and B ouwe Theo ems, a ques ion seems
o be na u al: Assume ha Mis a con ex, closed bounded subse o an
a bi a y Banach space and T:M→Mis a nonexpansi e mapping, i.e.,
kTx −Tyk ≤ kx−yk o e e y x, y ∈M. Does Thas a ixed poin ? We mus
ealize ha he assump ion on he space has been weakened ( he dimension
is no necessa ily ini e) bu s onge condi ions a e assumed on he mapping
(nonexpansi eness ins ead o con inui y). The answe in again nega i e:
Example. (Kaku ani (1943) [9]). Le B he uni ball in c0and T:B→B
de ined by T(x1, x2, ...) = (1 − kxk, x1, x2, ...). I is easy o check ha Tis
nonexpansi e and ixed poin ee.
As a consequence o hese ac s, he p oblems abou exis ence o ixed poin
o nonexpansi e mappings we e elega ed. Howe e in 1965, wo su p ising
heo ems appea ed:
geome ic p ope ies o banach spaces 333
Theo em. (B owde ’s Theo em [6]) Le Cbe a con ex bounded closed
subse o a uni o mly con ex Banach space (a p elimina y e sion was gi en
o Hilbe spaces) and T:C→Ca nonexpansi e mapping. Then Thas a
ixed poin .
Theo em. (Ki k’s Theo em [10]) Le Cbe a con ex bounded closed
subse o a e lexi e Banach space wi h no mal s uc u e. I T:C→C
is nonexpansi e, hen Thas a ixed poin .
In he second sec ion we will include a p oo o he la e heo em (which
includes B owde ’s Theo em). I is no ewo hy ha hese esul s s a e a b idge
be ween no ions which had usually been conside ed in Linea Func ional
Analysis (uni o m con exi y, e lexi i y, no mal s uc u e, e c) and p oblems
abou exis ence o ixed poin o nonlinea ope a o s. F om his s a ing poin
a big numbe o ixed poin esul s ha e been ob ained o di e en classes
o mappings using geome ic p ope ies o Banach spaces. In o hcoming
sec ions we will e iew some o hese esul s.
2. No mal s uc u e. Exis ence o ixed poin s o
nonexpansi e ope a o s
We ecall some de ini ions yielding o he no ion o no mal s uc u e.
De ini ion 1. Le Xbe a Banach space, Aa bounded subse o Xand
Ban a bi a y subse o X. The Chebyshe adius o Awi h espec o Bis
de ined by
(A, B) = in {sup{kx−yk:x∈A}:y∈B}
whe e we w i e (A) ins ead o (A, co (A)). The Chebyshe cen e o Awi h
espec o Bis de ined by
Z(A, B) = {y∈B: sup{kx−yk:x∈A}= (A, B)}
whe e we w i e Z(A) ins ead o Z(A, co (A)).
Rema k 1. Roughly speaking, we can say ha he Chebyshe adius
(A, B) is he adius o he smalles ball cen e ed a a poin in Band co e ing
he se A, he Chebyshe cen e Z(A, B) being he se o med by all cen e s o
hese smalles balls. Howe e , since he in imum appea ing in he de ini ion
334 . dom´
ınguez
is no , necessa ily a ained, he se Z(A, B) can be emp y. In opposi ion, i
o e e y ε > 0 we conside he se
Zε(A, B) = {y∈B: (A, y)≤ (A, B) + ε},
hen Zε(A, B) is a nonemp y, con ex, bounded and closed se i Bsa is ies he
same p ope ies. Thus, Zε(A, B) is con ex, nonemp y and weakly compac i
so is B. Since
ε>0
Zε(A, B) = Z(A, B),
he ini e in e sec ion p ope y implies ha Z(A, B) is nonemp y when Bis
a con ex and weakly compac se .
De ini ion 2. A bounded con ex closed subse Ao a Banach space X
is said o be diame al i diam (A) = (A). Equi alen ly, i Z(A) = A.
We say ha a Banach space Xhas no mal s uc u e ( espec i ely weak
no mal s uc u e) i e e y con ex closed nonemp y ( espec i ely con ex weakly
compac ) diame al subse o Xis a single on.
Rema k 2. Acco ding o he abo e de ini ion, a Banach space has no mal
s uc u e i e e y con ex se Awhich is no a single on can be co e ed by a
ball whose adius is less han he diame e o Aand cen e ed a a poin in A.
We could hink ha his is he case o e e y Banach space, and in ac , his
occu s o e e y uni o mly con ex space (we will see he de ini ion in he nex
sec ion), o ins ance, `pand Lp(Ω), 1 < p < +∞. Howe e he sequence space
c0 ails o ha e bo h no mal s uc u e and weak no mal s uc u e. Indeed,
conside he se A= co ({en:n∈N}) whe e{en}is he s anda d basis. We
ha e diam (A) = 1 and (A) = 1 because limn→∞ kx−enk= 1 o e e y
x∈c0. Fu he mo e, since he sequence {en}is weakly null, Ais a weakly
compac se . The same se can be conside ed in he sequence space `1, gi ing
us ha `1 ails o ha e no mal s uc u e ei he . Howe e , we will show in
he nex sec ion ha `1(and any Banach space wi h he Schu p ope y) has
weak no mal s uc u e.
Theo em 1. Le Xbe a Banach space wi h weak no mal s uc u e, Ca
con ex weakly compac subse o Xand T:C→Ca nonexpansi e mapping.
Then Thas a ixed poin .
P oo . Le Bbe he collec ion o all con ex weakly compac subse s o
Cwhich a e T-in a ian . I is easy o check ha B, o de ed by inclusion is
geome ic p ope ies o banach spaces 335
an induc i e amily. Zo n’s Lemma assu es he exis ence o a minimal se
K. Since T(K)⊂Kwe ha e co (T(K)) ⊂K. Thus, co (T(K)) is a con ex
weakly compac subse o Kwhich is T−in a ian .The minimali y o K
implies K= co (T(K)). Since Kis a weakly compac con ex se , we know
om Rema k 1 ha Z(K) is a nonemp y con ex weakly compac se . We will
p o e ha Z(K) is T-in a ian . Indeed, ake x∈Z(K), i.e. (K, x) = (K).
Fo e e y y∈Kwe ha e kTy−Txk ≤ ky−xk ≤ (K). Hence T(K) is co e ed
by he closed ball B(Tx, (K)) which implies co (T(K)) = K⊂B(Tx, (K)).
The e o e (K, Tx)≤ (K) which implies Tx ∈Z(K). The minimali y o
Kimplies Z(K) = Kand hus diam (K) = 0 because Xhas weak no mal
s uc u e. Hence Kis a single on and con ains a ixed poin o T.
3. Geome ic p ope ies which imply no mal s uc u e
In o de o s udy some geome ic p ope ies implying no mal s uc u e we
ecall a geome ic coe icien de ined by Bynum [7] in 1980 (a p elimina y o m
had been s udied by po J¨ung in 1901).
De ini ion 3. Le Xbe a Banach space. The no mal s uc u e coe icien
o Xis de ined by
N(X) = in ndiam (A)
(A):A⊂Xcon ex closed and
bounded wi h diam (A)>0o.
I is clea ha Xhas no mal s uc u e i N(X)>1. Howe e spaces
wi h no mal s uc u e exis which sa is y N(X) = 1. This coe icien can
be conside ed as a measu e o he “wo s ” possible ela ionship be ween he
diame e and he Chebyshe adius o a subse o X. Fo ins ance, in he
euclidean plane `2
2 his “wo s ” ela ionship is a ained a he equila e al
iangle and i s alue is √3. In he idimensional euclidean space `3
2i is
a ained a he e ahed on wi h a alue equal o 2√2/√3 and, in gene al, o
`n
2 he wo s alue co esponds o he “hipe e ahed on” wi h a alue equal
o √2p(n+ 1)/n. I is no easy o e alua e N(X) o a de e mined space X,
and, in ac , i s alue is unknown in many cases. The ollowing connec ion
be ween he alue o N(X) and he e lexi i y o he space is impo an :
Theo em 2. ([12]) Le Xbe a Banach space such ha N(X)>1. Then
Xis e lexi e.

336 . dom´
ınguez
P oo . I Xis no e lexi e, o e e y ε > 0 he e exis s a sequence {xn}
(see [13]) such ha 1−ε≤ ku1,n−un,ωk ≤ 1+ε o e e y u1,n ∈co ({xj}1≤j≤n),
un,ω ∈co ({xj}j>n), and o each n. Thus diam ({xn})≤1 + ε. Fu he mo e,
i belongs o co ({xn}) and nis la ge enough we ha e kxn− k ≥ 1−ε/2.
Since εis a bi a y, we ob ain N(X) = 1.
The no mal s uc u e coe icien is use ul o s udy he s abili y o he
no mal s uc u e unde eno ming. Recall ha i Xand Ya e isomo phic
Banach spaces, he Banach-Mazu dis ance be ween Xand Yis de ined by
d(X, Y ) = in ©kTkkT−1k:T∈Isom (X, Y )ª.
Clea ly, d(X, Y ) = 1 when Xand Ya e isome ic spaces.
Theo em 3. Le Xand Ybe isomo phic Banach spaces. Then
N(X)≤d(X, Y )N(Y).
P oo . Le Cbe a bounded con ex closed subse o Y. I U:Y→Xis
an isomo phism we ha e
(C)≤ kU−1k (U(C)) ≤ kU−1kdiam(U(C))/N(X)
≤ kU−1kkUkdiam(C)/N(X).
Thus (C)≤d(X, Y ) diam(C)/N(X) and his inequali y implies he esul .
The i s geome ic p ope y, ela ed o no mal s uc u e, which will be
conside ed is he uni o m con exi y. Le us ecall ha a Banach space is said o
be s ic ly con ex i he uni sphe e does no con ain a segmen . Equi alen ly:
De ini ion 4. We say ha a Banach space Xis s ic ly con ex i o
e e y ec o s xand yin Xwhich a e no collinea , we ha e
kx+yk<kxk+kyk.
A s onge no ion appea s i we assume his p ope y in a uni o m
sense, ha is, oughly speaking, assuming ha he e is no segmen wi h a
p ede e mined leng h as close o he uni sphe e as wan ed.
geome ic p ope ies o banach spaces 337
De ini ion 5. We say ha a Banach space Xis uni o mly con ex, i o
e e y ε∈(0,2] he e exis s δ > 0 such ha o x,y∈Xwi h
kx−yk ≥ ε
x, y ∈B(0,1))⇒1−°
°
°
°
x+y
2°
°
°
°
> δ.
Example 1. Hilbe spaces a e uni o mly con ex as a consequence o he
pa allelog am iden i y. Indeed, i x, y ∈B(0,1) and kx−yk ≥ ε, we ha e
°
°
°
°
x+y
2°
°
°
°≤ 1−³ε
2´2.
Conside ing δ=q1−(ε/2)2we ob ain he uni o m con exi y o he space. To
p o e ha `pspaces a e uni o mly con ex is mo e echnical (see, o ins ance,
[4]). On he o he hand, ecall ha a Banach space Xis ini ely ep esen able
in ano he Banach space Yi o e e y ini e dimensional subspace Eo X
and e e y ε > 0 he e exis s a subspace Fo Ysuch ha d(E, F)<1 + ε. I
is no di icul o p o e ha Lp(Ω) is ini ely ep esen able in `p. Indeed, i
Eis an n-dimensional subspace o Lp(Ω) and { 1, 2, ..., n}is a no malized
basis o E, wi h basic cons an c, o e e y ε > 0 we can ind simple
unc ions {s1, s2, ..., sn}such ha k k−skk< ε/nc(2 + ε) o k= 1, ..., n.
I =Pn
k=1 ak k, we de ine T =Pn
k=1 aksk. Then Tis an isomo phism
om Eon o span{s1, ..., sn}and kTkkT−1k<1 + ε. Since span{s1, ..., sn}
can be isome ically embedded in `p(by disc e iza ion o he measu e), he e
exis s a subspace Fo `psuch ha d(E, F)<1 + ε. Since he de ini ion o
he uni o m con exi y only depends on 2-dimensional subspaces and Lp([0,1])
con ains isome ically o `p, we can assu e ha Lp([0,1]) spaces a e uni o mly
con ex o he same choice o δas in `p.
We will need a measu e o he uni o m con exi y o he space:
De ini ion 6. Le Xbe a Banach space. The modulus o con exi y o
X,δX(ε), is de ined by
δX(ε) = in ½1−°
°
°
°
x+y
2°
°
°
°
:x, y ∈B(0,1),kx−yk ≥ ε¾.
Theo em 4. Le Xbe a Banach space wi h modulus o con exi y δX.
Then N(X)≥(1 −δX(1))−1.
338 . dom´
ınguez
P oo . Le Abe a closed con ex bounded subse o Xwhich is no a
single on and choose ε > 0. Deno e d= diam (A) and = (A). Choose x
and yin Asuch ha kx−yk ≥ d−ε. W i e w= (x+y)/2, and choose zin
Asuch ha kz−wk ≥ −ε. Since k(z−x)/dk ≤ 1,k(z−y)/dk ≤ 1 and
k(z−x)/d −(z−y)/dk>(d−ε)/d om he de ini ion o δXwe ob ain
kz−wk ≤ dµ1−δXµd−ε
d¶¶.
Thus
≤ε+dµ1−δXµd−ε
d¶¶
which implies he esul using he con inui y o he no m.
Rema k 3. No ice ha uni o m con exi y implies no mal s uc u e, bu
his p ope y is also sha ed by e e y space such ha segmen s wi h leng h
equal o 1 a e sepa a ed om he uni sphe e. On he o he hand, i is well
known ha uni o mly con ex spaces a e e lexi e. F om heo ems 2 and 4 a
s onge esul is ob ained: he condi ion δX(1) >0 implies e lexi i y.
Nex we will e iew ano he coe icien which implies weak no mal
s uc u e. We will use he no ion o asymp o ically equila e al sequence
De ini ion 7. Le Xbe a me ic space. A sequence {xn}in Xis said o
be asymp o ically equila e al i limn,m ;n6=md(xn, xm) exis s, i.e., a numbe d
exis s such ha o e e y ε > 0 he e is a nonnega i e in ege n0such ha
d−ε < d(xn, xm)< d +εi n, m > n0and n6=m.
To s udy he exis ence o asymp o ically equila e al sequence we will use
a simple e sion o Ramsey Lemma. Le us ix he no a ion: By Nwe deno e
he se o nonnega i e in ege s, [N] he collec ion o i s in ini e subse s and o
e e y se Cin [N], [C]2will deno e he se o med by all o de ed pai o med
wi h numbe s in C.
Lemma. (Ramsey Lemma) Le : [N]2→ {1,2}be a unc ion. Then,
he e exis s C∈[N]such ha he es ic ion o o [C]2is a cons an .
Theo em 5. Le {xn}be a bounded sequence in a me ic space. Then
{xn}con ains an asymp o ically equila e al subsequence.
geome ic p ope ies o banach spaces 339
P oo . Fo e e y subsequence {yn}o {xn}we deno e φ({yn}) = in {ε >
0 : {yn}can be co e ed by ini ely many se s wi h diame e ≤ε}.
Claim.“The e exis s a subsequence {yn}o {xn}such ha φ({zn}) = φ({yn})
o e e y subsequence {zn}o {yn}.” To p o e he claim, de ine by induc ion
{z0
n}={xn}and
φm+1 = in {φ({zn}) : {zn}subsequence o {zm
n}}.
Le {zm+1
n}be a subsequence o {zm
n}such ha
φ({zm+1
n)< φm+1 +1
m+ 1.
Conside he diagonal subsequence {zn
n}. We will show ha his sequence
sa is ies he equi ed condi ion. Since {zn
n}is a subsequence o {zm
n} o n > m
we ha e φ({zn
n})≤φ({zm
n}) o each m. Assume ha {zn}is a subsequence
o {zn
n}. Hence {zn}is a subsequence o {zm
n} o n > m. Thus
φ({zn
n})≤φ({zm
n})< φm+1
m≤φ({zn}) + 1
m.
Since mis a bi a y we ob ain
φ({zn
n})≤φ({zn})≤φ({zn
n})
and he claim is p o ed. Choose now an a bi a y ε > 0 and a subsequence
{yn}o {xn}sa is ying he p ope y in he claim. Taking a subsequence (which
“a o io i” sa is ies he same p ope y) we can assume φ({yn})+ε≥ kyn−ymk
o e e y n, m. De ine he ollowing unc ion om [N]2in o {1,2}: (n, m) = 1
i kyn−ymk> φ({yn})−εand (n, m) = 2 i kyn−ymk ≤ φ({yn})−ε. By
Ramsey’s Lemma, he e exis s a subsequence {zn}o {yn}sa is ying ei he
kzn−zmk> φ({yn})−ε o e e y n, m ;n6=mo kzn−zmk ≤ φ({yn})−ε
o e e y n, m. Since he second possibili y is a con adic ion acco ding o he
p ope y sa is ied by {yn}, we deduce ha he i s possibili y always holds
and we ha e
φ({yn})−ε≤ kzn−zmk< φ({yn}) + ε
o e e y n, m ;n6=m. Choosing ε= 1/n we can conclude he p oo by a
diagonal a gumen .
To in oduce a coe icien o weak no mal s uc u e we need some p e ious
de ini ions.
346 . dom´
ınguez
whe e x+and x−a e ec o s whose coo dina es a e
(x+)i= max{xi,0}=|xi|+xi
2
(x−)i= max{−xi,0}=|xi|−xi
2.
I is no di icul o p o e ha WCS(`2,k·k2,1) = √2 (see [2, Theo em
VI.3.11]) and so his space has no mal s uc u e. Fu he mo e, i s conjuga e
space is (`2,k·k2,∞). We will show ha his space ails o ha e no mal
s uc u e. Indeed, he s anda d basic sequence is a diame al sequence because
ken−emkp,∞= 1 i n6=mand o e e y poin u=Pn
i=1 αiei, αi≥
0,Pn
i=1 αi= 1 we ha e ken+1 −ukp
p,∞= sup{1,Pαp
i}= 1.
Rema k 6. Many o he geome ic p ope ies o Banach spaces implying
no mal s uc u e can be ound in he books [8] and [2, Chap e VI].
4. Exis ence o ixed poin in absence o no mal s uc u e
Un il now, we ha e s udied he exis ence o ixed poin o nonexpansi e
mappings as a consequence o he no mal s uc u e. In his sec ion we will
show some geome ic p ope ies which imply he exis ence o ixed poin s in
absence o no mal s uc u e. The mos impo an case co esponds o he
space c0which ails o ha e weak no mal s uc u e. Howe e , we will see ha
Theo em 1 s ill holds o his space. We i s ecall a “classical” esul in
Me ic Fixed Poin Theo y.
Lemma. (Goebel-Ka lo i z Lemma) Le Kbe a con ex weakly compac
subse o a Banach space X, and T:K→Ka nonexpansi e mapping.
Assume ha Kis minimal wi h hese p ope ies and le {xn}be an
app oxima ed ixed poin sequence o Tin K, i.e. limnkxn−Txnk= 0.
Then
lim
n→∞ ky−xnk= diam (K)
o e e y y∈K.
P oo . F om he p oo o Theo em 1 we know ha K=Z(K) which
implies ha Kis a diame al se . We will p o e ha Za({xn}, K) = K. Le
Za,ε({xn}, K) = {y∈K: lim sup
n→∞ kxn−yk ≤ a({xn}, K) + ε}.

geome ic p ope ies o banach spaces 347
I is easy o check ha Za,ε({xn}, K) is a nonemp y closed con ex
and T-in a ian se . Thus Za,ε({xn}, K) = Kand Za({xn}, K) =
Tε>0Za,ε({xn}, K) = K. We claim ha lim supn→∞ ky−xnk= diam (K)
o e e y y∈K. Indeed, assume ha he e exis s y∈Ksuch ha
lim supn→∞ ky−xnk<diam (K). Deno e = lim supn→∞ ky−xnk,d=
diam (K) and conside he collec ion {B(z, ( +d)/2) ∩K:z∈K}. Choose
an a bi a y posi i e numbe εsuch ha ε < (d− )/2. F om he i s pa
in he p oo we know ha lim supn→∞ kxn−zk= o e e y z∈K. Thus,
o e e y ini e subse {z1, ..., zk}o K he e exis s a nonnega i e in ege N
such ha kxN−zik ≤ +ε= ( +d)/2 o i= 1, ..., k. Hence xNbelongs
o Tk
i=1 B(zi,( +d)/2). The weak compac ness o Kimplies he exis ence o
x0∈Tz∈KB(z, ( +d)/2) ∩Kand his poin is no diame al because
sup
z∈Kkz−x0k< +d
2< d = diam (K).
This con adic ion p o es he claim. I lim in n→∞ ky−xnk<diam (K)
o some y∈K he e exis s a subsequence {yn}o {xn}such ha
lim supn→∞ kyn−yk= lim in n→∞ kxn−yk<diam (K), which is a
con adic ion acco ding o he claim applied o he sequence {yn}which is
again an app oxima ed ixed poin sequence.
Theo em 12. Le Kbe a con ex weakly compac subse o a Banach
space X, and T:K→Ka nonexpansi e mapping. Assume ha Kis
minimal o hese condi ions, diam(K)=1and {xn}is an app oxima ed
ixed poin sequence which is weakly null. Then, o e e y ε > 0and ∈[0,1],
he e exis s a sequence {zn}in Ksuch ha : (i) {zn}is weakly con e gen o
a poin z∈K.(ii) kznk>1−ε o e e y n∈N.(iii) kzn−zmk ≤ o e e y
n, m ∈N.(i ) lim supnkzn−xnk ≤ 1− .
P oo . Since {wn}is an app oxima ed ixed poin sequence in K,
diam(K) = 1 and 0 lies in K, om Goebel-Ka lo i z Lemma we deduce
limnkwnk= 1. Hence, o e e y ε > 0 he e exis s δ(ε)>0 such ha
kxk>1−εi x∈Kand kTx −xk< δ(ε). Indeed, o he wise, he e
exis s ε0>0 such ha we can ind wn∈Ksa is ying kTwn−wnk<1/n
and kwnk ≤ 1−ε0 o e e y n∈N. The e o e, he sequence {wn}is an
app oxima ed ixed poin sequence in Ksa is ying lim supnkwnk ≤ 1−ε0.
Le ε > 0 and ∈[0,1]. Choose γ < min{1, δ(ε)}and o any n∈Nde ine
he con ac ion Sn:K→Kby
Sn(x) = (1 −γ)T(x) + γ xn.
348 . dom´
ınguez
The Con ac i e Mapping P inciple assu es ha he e exis s a (unique) ixed
poin zno Sn. Since Kis a weakly compac se , we can assume, aking a
subsequence i necessa y, ha {zn}sa is ies (i). Since kzn−Tznk< γ we
know ha {zn}sa is ies (ii). Condi ion (iii) is easily ob ained and (i ) is a
consequence o he inequali ies
kzn−xnk ≤ k(1 −γ)Tzn+γ xn−xnk
≤(1 −γ)kTzn−Txnk+ (1 −γ)kTxn−xnk+γ(1 − )kxnk.
Thus,
kzn−xnk ≤ 1− +1−γ
γkTxn−xnk.
Taking limsup as n ends o in ini y, we ob ain (i ).
De ini ion 12. Le Xbe a Banach space. We de ine he coe icien
R(X) = sup{lim in
n→∞ kxn+xk}
whe e he sup emum is aken o e all x∈Xwi h kxk ≤ 1 and o e all weakly
null sequences in B(0,1).
Theo em 13. Le Xbe a Banach space wi h R(X)<2. I Cis a con ex
weakly compac subse o Xand T:C→Cis a nonexpansi e mapping, hen
Thas a ixed poin .
P oo . O he wise, we can ind a con ex weakly compac T-in a ian subse
o Xwhich is no a single on and which is minimal o hese condi ions.
By mul iplica ion we assume ha i s diame e is 1. Fu he mo e, om he
Con ac i e Mapping P inciple i is easy o cons uc an app oxima ed ixed
poin sequence {xn}in K. We can assume ha {xn}is weakly con e gen and,
by ansla ion, ha {xn}is weakly null. Conside a sequence {zn}sa is ying
(i)-(i ) in Theo em 12 o = 1/2. Taking again a subsequence, i necessa y,
we can assume ha limnkzn−zkexis s. Fu he mo e, limnkzn−zk ≤
limnlimmkzn−zmk ≤ 1/2. We can choose η > 0 such ha ηR(X)<
1−R(X)/2. Fo nla ge enough, we ha e kzn−zk ≤ 1/2 + η. Fu he mo e
kzk ≤ lim in n→∞ kzn−xnk ≤ 1/2. Thus
°
°
°
°
zn
1/2 + η°
°
°
°
=°
°
°
°
zn−z
1/2 + η+z
1/2 + η°
°
°
°≤R(X).
The e o e lim supn→∞ kznk ≤ R(X)(1/2 + η)<1 which is a con adic ion
because 0 ∈K.
geome ic p ope ies o banach spaces 349
Rema k 7. Fo X=c0i is easy o check ha R(c0) = 1. Thus Theo em
1 holds o c0e en hough his space ails o ha e weak no mal s uc u e.
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ınguez, T., L´
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