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RETRACTIONS IN BANACH SPACESARKADY ALEYNER AND SIMEON REICH Received 22 March 2005 An explicit algorithmic scheme for constructing the unique sunny nonexpansive retrac-tion onto the comm

RETRACTIONS IN BANACH SPACES

ARKADY ALEYNER AND SIMEON REICH

Received 22 March 2005

An explicit algorithmic scheme for constructing the unique sunny nonexpansive retrac-tion onto the common fixed point set of a nonlinear semigroup of nonexpansive map-pings in a Banach space is analyzed and a proof of convergence is given

1 Introduction

Throughout this paper all vector spaces are real and we denote by N and R + the set

of nonnegative integers and nonnegative real numbers, respectively Let (X,
·
) be a Banach space and letX ∗be its dual The value ofy ∈ X ∗atx ∈ X will be denoted by

 x, y  We also denote byJ : X →2X ∗the normalized duality map fromX into the family

of nonempty (by the Hahn-Banach theorem) weak-star compact convex subsets ofX ∗, which is defined byJx = { x ∗ ∈ X ∗: x,x ∗  =
x
2=
x ∗
2}for allx ∈ X The Banach

spaceX is said to be smooth or to have a Gˆateaux differentiable norm if the limit

lim

t →0

x + ty
−
x

exists for eachx, y ∈ X with
x
=
y
=1 The spaceX is said to have a uniformly

Gˆateaux differentiable norm if, for each y∈ X with
y
=1, the limit (1.1) is attained uniformly inx ∈ X with
x
=1 It is known [12, Lemma 2.2] that if the norm ofX

is uniformly Gˆateaux differentiable, then the duality map is single-valued and norm to weak star uniformly continuous on each bounded subset ofX Let C be a nonempty,

closed and convex subset ofX and E be a nonempty subset of C A mapping Q : C → X

is nonexpansive if
Qx − Qy
≤
x − y
for allx, y ∈ C A mapping Q : C → E is called a retraction from C onto E if Qx = x for all x ∈ E A retraction Q from C onto E is called sunny if Q has the following property: Q(Qx + t(x − Qx)) = Qx for all x ∈ C and t ≥0 withQx + t(x − Qx) ∈ C It is known [6, Lemma 13.1] that in a smooth Banach spaceX,

a retractionQ from C onto E is both sunny and nonexpansive if and only if

for allx ∈ C and y ∈ E Hence, there is at most one sunny nonexpansive retraction from

C onto E For example, if E is a nonempty, closed and convex subset of a Hilbert space

Copyright©2005 Hindawi Publishing Corporation

Fixed Point Theory and Applications 2005:3 (2005) 295–305

DOI: 10.1155/FPTA.2005.295

H, then the nearest point projection P EfromH onto E is the unique sunny

nonexpan-sive retraction ofH onto E This is not true for all Banach spaces, since outside Hilbert

space, nearest point projections, although sunny, are no longer nonexpansive On the other hand, sunny nonexpansive retractions do sometimes play a similar role in Banach spaces to that of nearest point projections in a Hilbert space So an interesting problem arises: for which subsets of a Banach space does a sunny nonexpansive retraction exist?

If it does exist, how can one find it? It is known [6, Theorem 13.2] that ifC is a closed

convex subset of a uniformly smooth Banach space andT : C → C is nonexpansive, then

the fixed point set ofT is a sunny nonexpansive retract of C More generally, Bruck [3, Theorem 2] proves that ifC is a closed convex subset of a reflexive Banach space every

bounded, closed and convex subset of which has the fixed point property for nonexpan-sive mappings andT : C → C is nonexpansive, then its fixed point set is a nonexpansive

retract ofC (It is still an open question whether all bounded, closed and convex subsets

of reflexive Banach spaces have this fixed point property.) For a weak sufficient condi-tion on the underlying space which guarantees that nonexpansive retracts are, in fact, sunny nonexpansive retracts see [10, Theorem 4.1] In the present paper we show that if

F is the nonempty common fixed point set of a commuting family of nonexpansive

self-mappings of closed convex subsetsC of certain Banach spaces X, satisfying an asymptotic

regularity condition, then it is possible to construct the sunny nonexpansive retractionQ

ofC onto F in an explicit iterative way The origin of our work lies in a recent publication

by Dom´ınguez Benavides, L ´opez Acedo and Xu [5] who attempted to construct sunny nonexpansive retractions using both implicit and explicit iterative schemes (cf the dis-cussion in [1]) Our work improves, corrects and generalizes some of the results obtained

in the above paper It is also related to a result of Reich [11], where the case of a single mapping is dealt with In this connection we would also like to refer the interested reader

to the results obtained by Suzuki [14] who deals with an implicit scheme for construct-ing the sunny nonexpansive retraction onto the common fixed point set of some one-parameter semigroups of nonexpansive mappings For related results in Hilbert space see Aleyner and Censor [1], Bauschke [2], Deutsch and Yamada [4], Halpern [7], Lions [8], and Wittmann [15]

2 Preliminaries and notations

Letl ∞ denote the real Banach space of all bounded sequencesa =(a1,a2, ) with the

norm defined by
a
=supn | a n | A continuous linear functional LIM onl ∞is called a

Banach limit when LIM satisfies LIM(a) ≥0 if a n ≥0, n =1, 2, , LIM( { a n })=

LIM({ a n+1 }) and
LIM
=LIM(1)=1 To prove our theorem, we need the following two propositions [13, Propositions 1 and 2], which can be deduced from the arguments

in the proof of [9, Theorem 1] We sketch their proofs for the sake of completeness

Proposition 2.1 Let α be a real number and let a =(a1,a2, ) ∈ l ∞ Then LIM(a) ≤ α for all Banach limits LIM if and only if for each ε > 0, there exists n0∈ N such that

a k+a k+1+···+a k+n −1

for all n ≥ n and k ∈ N

Proof First we prove the necessity of (2.1) Assume LIM(a) ≤ α for all Banach limits LIM.

Define a sublinear functionalβ from l ∞into the real lineRby

βb1,b2, =lim sup

n →∞ sup

k ∈N

1

n

k+n−1

i = k

where (b1,b2, ) ∈ l ∞ By the Hahn-Banach theorem, there exists a linear functional µ

froml ∞intoRsuch thatµ ≤ β and µ(a) = β(a) It is not difficult to see that µ is a Banach

limit Sinceµ(a) ≤ α, there exists, for each ε > 0, a natural number n0∈ Nwhich satisfies (2.1) Next we prove that (2.1) is sufficient Let µ be a Banach limit and let ε > 0 By the hypothesis, there existsn0such that (2.1) is satisfied Hence we have

µ(a) = µa k+a k+1+···+a k+n0 −1

n0



Sinceε is an arbitrary positive number, we see that µ(a) ≤ α.

Proposition 2.2 Let α be a real number and let a =(a1,a2, ) ∈ l ∞ be such that

for all Banach limits LIM, and

lim sup

Then

lim sup

Proof Let ε > 0 ByProposition 2.1, there existsn ≥2 such that

a k+a k+1+···+a k+n −1

for allk ∈ N Choosek0∈ Nsuch thata k+1 − a k < ε/(n −1) for allk ≥ k0 Letk ≥ k0+n.

Then we have

a k = a k − i+

a k − i+1 − a k − i

+

a k − i+2 − a k − i+1

+···+

a k − a k −1



≤ a k − i+ iε

n −1 (2.8) for eachi =0, 1, 2, ,n −1 So we obtain

a k ≤ a k+a k −1+···+a k − n+1

1

n ·

n(n −1)

n −1≤ α + ε. (2.9) Hence we have

lim sup

Sinceε is an arbitrary positive number, the proposition is proved.

3 Convergence theorem

LetX be a Banach space, C a nonempty, closed and convex subset of X, G an unbounded

subset ofR +such that

t + h ∈ G ∀ t,h ∈ G,

andΓ= { T t:t ∈ G }a family of nonexpansive self-mappings ofC such that the set F of

the common fixed points ofΓ is nonempty We make the following assumptions

Assumptions on the space X is a reflexive Banach space with a uniformly Gˆateaux

dif-ferentiable norm such that each nonempty, bounded, closed and convex subsetK of X

has the common fixed point property for nonexpansive mappings; that is, any family of commuting nonexpansive self-mappings ofK has a common fixed point Note that all

these assumptions are fulfilled wheneverX is uniformly smooth.

Assumptions on the mappings. Γ is a uniformly asymptotically regular semigroup on bounded subsets ofC, that is,

for allt,s ∈ G, x ∈ C, and for all bounded subsets K of C there holds

lim

r →∞sup

K

T s T r x − T r x  =0, (3.3) uniformly for alls ∈ G Note that both these assumptions hold when the trajectories of

the semigroupΓ converge uniformly on bounded subsets of X.

Assumptions on the parameters { λ n }is a sequence of numbers in [0, 1) with the following properties:

∞

n =0



1− λ n

=0; equivalently,

∞



n =0

∞



n =0

Observe that given points f ∈ F, u,x0∈ C, and the bounded subset D = { x ∈ C :
x −

f
≤max(
x0− f
,
u − f
)}, there exists a sequence{ r n } ⊆ G such that

r0< r1< r2< ··· < r n < ···, nlim

∞



n =0

sup

D

T s T r n x − T r n x< ∞, (3.8)

uniformly for alls ∈ G We now define the sequence { x n }by

x n+1 = λ n u +1− λ nT r n x n, (3.9) wheren ≥0; we say that{ x n }has anchoru and initial point x0.

Theorem 3.1 If the above assumptions on the space, mappings and parameters hold, then the sequence generated by ( 3.9 ) converges in norm to Qu, where Q is the unique sunny non-expansive retraction from C onto F.

Proof We first prove the result for the special case x0= u and then extend it to the general

case We divide our proof into a sequence of separate claims

Claim 3.2 For all n ≥0 and every f ∈ F,

We proceed by induction onn Fix f ∈ F Clearly, (3.10) holds forn =0 If
x n − f
≤

u − f
, then

x n+1 − f  ≤ λ n
u − f
+

1− λ nT r n x n − f

≤ λ n
u − f
+

1− λ nx n − f

≤
u − f
,

(3.11)

as required

Claim 3.3 The following strong convergence holds:

This is true because (3.10) guarantees that{ x n }is bounded, which, in turn, implies that

{ T r n x n }is also bounded The boundedness of{ T r n x n }together with (3.4) imply, in view

of (3.9), our assertion

Claim 3.4 The differences of consecutive iterates strongly converge to zero, namely,

Indeed, it follows from (3.10) thatx n ∈ D for all n ≥0 By the boundedness of{ x n }and

{ T r n x n }there exists some constantL ≥0 such that
x n+1 − x n
≤ L and
u − T r n x n
≤ L

for alln ≥0 Therefore, for alln ≥1 we get

x n+1 − x n  =  λ n − λ n −1



u − T r n −1x n −1



+

1− λ n

T r n x n − T r n −1x n −1 

≤λ n − λ n −1u − T r n −1x n −1 +1− λ nT r n x n − T r n x n −1 

+1− λ n

T r n x n −1− T r n −1x n −1 

≤ λ n − λ n −1 u − T r n −1x n −1 +

1− λ nx n − x n −1

+T r n x n −1− T r n

−1x n −1 

≤ λ n − λ n −1 L +1− λ nx n − x n −1

+T r x n −1− T r x n −1.

(3.14)

SinceΓ is a semigroup, we are able to rewrite the last term as follows:

T r

n x n −1− T r n −1x n −1 =  T r n − r n −1T r n −1x n −1− T r n −1x n −1 . (3.15)

Thus

x n+1 − x n λ n − λ n −1 L +

1− λ nx n − x n −1

+T r n − r n

−1T r n −1x n −1− T r n −1x n −1  (3.16) for alln ≥1 Hence, inductively,

x n+1 − x n  ≤ Ln

k =

λ k − λ k −1 +x m − x m 1 n

k =



1− λ k

+

n



k =

T r k − r k

−1T r k −1x k −1− T r k −1x k −1 , (3.17)

for alln ≥ m ≥1 Taking now the limit asn tends to + ∞, we obtain

lim sup

n →∞

x n+1 − x n

≤ L∞

k =

λ k − λ k −1 +L ∞

k =



1− λ k

+

∞



k =

sup

D

T r k − r k

−1T r k −1x − T r k −1x

= L∞

k =

λ k − λ k −1 +

∞



k =

sup

D

T r k − r k −

1T r k −1x − T r k −1x

(3.18)

by (3.5) On the other hand, conditions (3.6) and (3.8) imply that

lim

m →∞

∞



k =

λ k − λ k −1 0,

lim

m →∞

∞



k =

sup

D

T r k − r k

−1T r k −1x − T r k −1x  =0.

(3.19)

Altogether, by lettingm tend to ∞, we conclude thatx n+1 − x n →0, as claimed

Claim 3.5 For each fixed s ∈ G,

Indeed, lets ∈ G Then

T s x n − x n  ≤  T s x n − T s T r n x n+T s T r n x n − T r n x n+T r n x n − x n

≤2x n − T r n x n+ sup

D

T s T r n x − T r n x

≤2x n − x n+1+x n+1 − T r n x n+ sup

D

T s T r n x − T r n x. (3.21)

Combining (3.12), (3.13), and (3.8), we see thatT s x n − x n →0, as asserted

Let LIM be a Banach limit and let { α s } s ∈ G be a net in the interval (0, 1) such that lims →∞ α s =0 By Banach’s fixed point theorem, for eachs ∈ G, there exists a unique point

z s ∈ C satisfying the equation z s = α s u + (1 − α s)T s z s Since the following claim is essen-tially proved in [5], we include only a sketch of its proof

Claim 3.6.

whereQ : C → F is the unique sunny nonexpansive retraction from C onto F.

Indeed, let{ s n }be a subsequence ofG such that lim n →∞ s n = ∞ Since{ z s n }is bounded,

we can define a functionalg on C by

g(x) =LIM 

z s n − x 2

We have for eachr ∈ G,

gT r x=LIM 

z s n − T r x 2

=LIM 

T r T s n z s n − T r x 2

≤LIM 

T s n z s n − x 2

=LIM 

z s n − x 2

,

(3.24)

by (3.3) In other words,

for allr ∈ G and x ∈ C Let K = { x ∈ C : g(x) =minC g } Sinceg is convex and

continu-ous, lim
x
→∞ g(x) = ∞andX is reflexive, K is a nonempty, closed, bounded and convex

subset ofC From (3.25) we see thatK is invariant under each T r; that is,T r(K) ⊂ K,

r ∈ G Hence K contains a common fixed point of Γ Let q ∈ KF be such a common

fixed point Sinceq is a minimizer of g over C, it follows that for each x ∈ C,

0≤lim

λ →0 +

1

λ



gq + λ(x − q)− g(q)

=LIM



lim

λ →0 +

1

λ



z s n − q+λ(q − x) 2

−z s n − q 2 

=LIM

2

q − x,Jz s n − q.

(3.26)

Thus,

LIM

x − q,Jz s n − q≤0 (3.27) for allx ∈ C On the other hand, for any f ∈ F,

z s − f =1− α s

T s z s − f+α s(u − f ). (3.28)

It follows that

z s n − f 2

=1− α s n



T s n z s n − f ,Jz s n − f+α s n

u − f ,Jz s n − f

≤1− α s nz s n − f 2

+α s n

u − f ,Jz s n − f. (3.29)

Hence

z s

n − f 2

≤
u − f ,Jz s n − f. (3.30) Combining (3.27) and (3.30), we get

LIM 

z s n − q 2

Hence there is a subsequence{ z r j }of{ z s n }such that limj →∞
z r j − q
=0 Assume that there exists another subsequence{ z p k } of { z s n } such that limk →∞
z p k − q˜
=0, where

˜

q ∈ KF Then (3.30) implies that

q − q˜ 2

≤
u − q,J˜ q − q˜. (3.32) Similarly we have

q˜− q 2

≤
u − q,Jq˜− q. (3.33) Adding up (3.32) and (3.33) we obtainq = q Therefore˜ { z s }converges in norm to a point

inF Now we define Q : C → F by Qu =lims →∞ z s ThenQ is a retraction from C onto F.

Moreover, by (3.30) we get for all f ∈ F,

Qu − f 2

That is,

for all f ∈ F Therefore Q is the unique sunny nonexpansive retraction from C onto F

(see (1.2))

Claim 3.7.

lim sup

n →∞

u − Qu,Jx n − Qu≤0. (3.36)

SinceT sis nonexpansive, (3.20) implies that

LIM 

x n − T s z s 2

=LIM 

T s x n − T s z s 2

≤LIM 

x n − z s 2

. (3.37)

Since (1− α s)(x n − T s z s)=(x n − z s)− α s(x n − u), we have



1− α s 2 x n − T s z s 2

≥x n − z s 2

−2α s
x n − u,Jx n − z s

=x n − z s 2

−2α s

x n − z s+z s − u,Jx n − z s

=x n − z s 2

−2α s

x n − z s,Jx n − z s

−2α s

z s − u,Jx n − z s

=1−2α sx n − z s 2

+ 2α s
u − z s,Jx n − z s.

(3.38) Therefore



1− α s 2

LIM 

x n − z s 2

≥1−2α s

LIM 

x n − z s 2

+ 2α sLIM

u − z s,Jx n − z s (3.39)

for eachn ≥0 These inequalities yield

α s

2 LIM



x n − z s 2

≥LIM

u − z s,Jx n − z s

Since

u − z s,Jx n − z s

−
u − Qu,Jx n − Qu

=
u − z s −(u − Qu),Jx n − z s

+

u − Qu,Jx n − z s

− Jx n − Qu, (3.41)

we obtain by lettings tend to ∞that

0≥LIM

u − Qu,Jx n − Qu (3.42) becauseX has a uniformly Gˆateaux differentiable norm and (3.22) holds On the other hand, we have

lim

n →∞ u − Qu,Jx n+1 − Qu−
u − Qu,Jx n − Qu 0 (3.43)

by (3.13) Hence we obtain byProposition 2.2,

lim sup

n →∞

u − Qu,Jx n − Qu≤0, (3.44)

as claimed

Now we can conclude the proof for the special casex0= u.

Claim 3.8.

Indeed, since



1− λ n

T r x n − Qu=x n+1 − Qu− λ n(u − Qu), (3.46)

we have

1− λ n

T r n x n − Qu 2

≥x n+1 − Qu 2

−2λ n

u − Qu,Jx n+1 − Qu. (3.47) Hence

x n+1 − Qu 2

≤1− λ nx n − Qu 2

+ 2

1−1− λ n

u − Qu,Jx n+1 − Qu (3.48) for eachn ≥0 Letε > 0 be given By (3.36), there existsm ≥0 such that

u − Qu,Jx n − Qu≤ ε

for alln ≥ m Therefore

x n+m − Qu 2

≤

n+m −1

k =



1− λ kx m − Qu 2

+



1− n+m −1

k =



1− λ k

ε (3.50)

for alln ≥1 Hence by (3.5) we get

lim sup

n →∞

x n − Qu 2

=lim sup

n →∞

x n+m − Qu 2

Sinceε is an arbitrary positive real number, we conclude that { x n }converges strongly to

Qu; that is, the special case is verified.

Finally, we extend the proof to the general case Let{ x n }be the sequence generated by (3.9) with an initial pointx0(possibly different from u) and let{ y n }be another sequence generated by (3.9) with an initial pointy0= u On the one hand, by the special case,

On the other hand, it is easily checked that

x n − y n  ≤  x0− y0 n −1

k =0



1− λ k

(3.53)

for alln ≥1 Thus,x n − y n →0 and, altogether,x n → Qu.

Note added in proof We are now able to proveTheorem 3.1under much weaker assump-tions on the mappings and the parameters We expect the details to be part of a forthcom-ing paper

Acknowledgments

The research of the second author was partially supported by the Israel Science Founda-tion founded by the Israel Academy of Sciences and Humanities (Grant 592/00), by the Fund for the Promotion of Research at the Technion, and by the Technion VPR Fund

. (3.37)

Since (1− α s)(x... s(u − f ). (3.28)

It follows that

z... n(u − Qu), (3.46)

we have

1−