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FOR INFINITE FAMILIES OF NONEXPANSIVE MAPPINGSIN BANACH SPACES TOMONARI SUZUKI Received 19 August 2005; Revised 24 February 2006; Accepted 26 February 2006 We prove Browder’s type strong

FOR INFINITE FAMILIES OF NONEXPANSIVE MAPPINGS

IN BANACH SPACES

TOMONARI SUZUKI

Received 19 August 2005; Revised 24 February 2006; Accepted 26 February 2006

We prove Browder’s type strong convergence theorems for infinite families of nonexpan-sive mappings One of our main results is the following: letC be a bounded closed convex

subset of a uniformly smooth Banach spaceE Let {T n:n ∈ N}be an infinite family of commuting nonexpansive mappings on C Let {α n }and {t n } be sequences in (0, 1/2)

satisfying limn t n =limn α n /t 

n =0 for ∈ N Fixu ∈ C and define a sequence {u n }inC

byu n =(1− α n)((1−n k=1t k

n)T1u n+n

k=1t k

n T k+1 u n) +α n u for n ∈ N Then {u n } con-verges strongly toPu, where P is the unique sunny nonexpansive retraction from C onto

∞

n=1F(T n)

Copyright © 2006 Tomonari Suzuki This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited

1 Introduction

LetC be a closed convex subset of a Banach space E A mapping T on C is called a non-expansive mapping if Tx − T y ≤ x − yfor allx, y ∈ C We denote by F(T) the set of

fixed points ofT We know that F(T) is nonempty in the case that E is uniformly smooth

andC is bounded; see Baillon [1] WhenE has the Opial property and C is weakly

com-pact,F(T) is also nonempty; see [11,13] See also [4,5,10] and others Fixu ∈ C Then

for eachα ∈(0, 1), there exists a unique pointx αinC satisfying x α =(1− α)Tx α+αu

be-cause the mappingx →(1− α)Tx + αu is contractive; see [2] In 1967 Browder [6] proved the following strong convergence theorem

Theorem 1.1 (Browder [6]) LetC be a bounded closed convex subset of a Hilbert space E and let T be a nonexpansive mapping on C Let {α n } be a sequence in (0, 1) converging to 0 Fix u ∈ C and define a sequence {u n } in C by

u n =1− α n

for n ∈ N Then {u n } converges strongly to the element of F(T) nearest to u.

Hindawi Publishing Corporation

Fixed Point Theory and Applications

Volume 2006, Article ID 59692, Pages 1 16

DOI 10.1155/FPTA/2006/59692

Reich extended this theorem to uniformly smooth Banach spaces in [17] Using the notion of Bochner integral and (invariant) mean, Shioji and Takahashi in [18] proved Browder’s type strong convergence theorems for families of nonexpansive mappings Very recently, the author proved the following Browder’s type strong convergence the-orem for one-parameter nonexpansive semigroups This is a generalization of the results

in [19,25] We remark that we do not use the notion of Bochner integral

Theorem 1.2 [24] LetC be a weakly compact convex subset of a Banach space E Assume that either of the following holds:

(i)E is uniformly convex with uniformly Gˆateaux differentiable norm;

(ii)E is uniformly smooth; or

(iii)E is a smooth Banach space with the Opial property and the duality mapping J of E

is weakly sequentially continuous at zero.

Let {T(t) : t ≥0} be a one-parameter nonexpansive semigroup on C Let τ be a nonnegative real number Let {α n } and {t n } be sequences of real numbers satisfying 0 < α n < 1, 0 < τ + t n

and t n = 0 for n ∈ N, and lim n t n =limn α n /t n = 0 Fix u ∈ C and define a sequence {u n } in

C by

u n =1− α nTτ + t nu n+α n u (1.2)

for n ∈ N Then {u n } converges strongly to Pu, where P is the unique sunny nonexpansive retraction from C ontot≥0F(T(t)).

Also, very recently, the author proved Krasnoselskii and Mann’s type convergence the-orems for infinite families of nonexpansive mappings in [21] See also [20] In this paper, using the idea in [21], we prove Browder’s type strong convergence theorems for infinite families of nonexpansive mappings without assuming the strict convexity of the Banach space We remark that if we assume the strict convexity, its proof is very easy because the set of common fixed points of countable families of nonexpansive mappings is the set of fixed points of some single nonexpansive mapping; see Bruck [8] We also remark that

we do not use the notion of (invariant) mean

2 Preliminaries

Throughout this paper, we denote byN, Z, Q, and Rthe set of all positive integers, all integers, all rational numbers, and all real numbers, respectively

Let{x n }be a sequence in a topological spaceX By the axiom of choice, there exist a

directed set (D,≤ ) and a universal subnet {x f (ν):ν ∈ D}of{x n }, that is,

(i) f is a mapping from D intoNsuch that for eachn ∈ Nthere existsν0∈ D such

thatν ≥ ν0implies f (ν) ≥ n;

(ii) for each subsetA of X, there exists ν0∈ D such that either {x f (ν):ν ≥ ν0} ⊂ A or {x f (ν):ν ≥ ν0} ⊂ X \ A holds.

In this paper, we often use{x ν:ν ∈ D}instead of{x f (ν):ν ∈ D}, for short We know that

if a net{x ν }is universal andg is a mapping from X into an arbitrary set Y, then {g(x ν)}

is also universal We also know that ifX is compact, then a universal net {x ν } always converges See [12] for details

LetE be a real Banach space We denote by E ∗the dual ofE E is called uniformly convex if for each ε > 0, there exists δ > 0 such that x + y/2 < 1 − δ for all x, y ∈ E

withx = y =1 andx − y ≥ ε E is said to be smooth or said to have a Gˆateaux

di fferentiable norm if the limit

limt→

0

x + ty − x

exists for eachx, y ∈ E with x = y =1.E is said to have a uniformly Gˆateaux differ-entiable norm if for each y ∈ E with y =1, the limit is attained uniformly inx ∈ E with

x =1.E is said to be uniformly smooth or said to have a uniformly Fr´echet differentiable norm if the limit is attained uniformly in x, y ∈ E with x =  y =1.E is said to have the Opial property [14] if for each weakly convergent sequence {x n }inE with weak limit

x0,

lim infn→∞ x n − x0< liminf

holds for allx ∈ E with x = x0 We remark that we may replace “lim inf” by “lim sup.” That is,E has the Opial property if and only if for each weakly convergent sequence {x n }

inE with weak limit x0,

lim sup

n→∞

x n − x0< limsup

n→∞

holds for allx ∈ E with x = x0

LetE be a smooth Banach space The duality mapping J from E into E ∗is defined by



x,J(x)= x2=J(x) 2

(2.4) for allx ∈ E J is said to be weakly sequentially continuous at zero if for every sequence {x n }inE which converges weakly to 0 ∈ E, {J(x n)}converges weakly∗to 0∈ E ∗

A convex subsetC of a Banach space E is said to have normal structure [3] if for every bounded convex subsetK of C which contains more than one point, there exists z ∈ K

such that

sup

x∈K x − z < sup

We know that compact convex subsets of any Banach spaces and closed convex subsets

of uniformly convex Banach spaces have normal structure Turett [27] proved that uni-formly smooth Banach spaces have normal structure Also, Gossez and Lami Dozo [11] proved that every weakly compact convex subset of a Banach space with the Opial prop-erty has normal structure We recall that a closed convex subsetC of a Banach space E

is said to have the fixed point property for nonexpansive mappings (FPP, for short) if for

every bounded closed convex subsetK of C, every nonexpansive mapping on K has a

fixed point So, by Kirk’s fixed point theorem [13], every weakly compact convex subset with normal structure has FPP

LetC and K be subsets of a Banach space E A mapping P from C into K is called sunny

[7] if

forx ∈ C with Px + t (x − Px) ∈ C and t ≥0 The following is proved in [15]

Lemma 2.1 (Reich [15]) LetE be a smooth Banach space and let C be a convex subset of

E Let K be a subset of C and let P be a retraction from C onto K Then the following are equivalent:

(i)x − Px, J(Px − y) 0 for all x ∈ C and y ∈ K;

(ii)P is both sunny and nonexpansive.

Hence, there is at most one sunny nonexpansive retraction from C onto K.

The following lemma is proved in [24] However, it is essentially proved in [16] See also [26]

Lemma 2.2 (Reich [16]) LetC be a nonempty closed convex subset of a Banach space E with a uniformly Gˆateaux differentiable norm Let {x α:α ∈ D} be a net in E and let z ∈ C Suppose that the limits of {x α − y} exist for all y ∈ C Then the following are equivalent:

(i) limα∈ x α − z =miny∈Climα∈ x α − y;

(ii) lim supα∈  y − z, J(x α − z) 0 for all y ∈ C;

(iii) lim infα∈ y − z, J(x α − z) 0 for all y ∈ C.

The following lemma is well known

Lemma 2.3 Let {u n } be a sequence in a Banach space E and let z belong to E Assume that every subsequence {u n i } of {u n } has a subsequence converging to z Then {u n } itself converges to z.

FromLemma 2.3, we obtain the following

Lemma 2.4 Let {u n } be a sequence in a Banach space E Assume that {u n } has at most one cluster point, and every subsequence of {u n } has a cluster point Then {u n } converges Proof Since {u n }is a subsequence of{u n },{u n }has a cluster pointz ∈ E Let {u n i }be

an arbitrary subsequence of{u n } Then by assumption{u n i }has a cluster pointw ∈ E.

Sincew is also a cluster point of {u n }, we havew = z Hence, {u n i }has a cluster point

z ∈ E That is, there exists a subsequence of {u n i } converging toz So, byLemma 2.3,

3 Fixed point theorem

The following theorem is one of the most famous fixed point theorems for families of nonexpansive mappings

Theorem 3.1 (Bruck [9]) Suppose a closed convex subsetC of a Banach space E has the fixed point property for nonexpansive mappings, and C is either weakly compact, or bounded and separable Then for any commuting family S of nonexpansive mappings on C, the set of common fixed points of S is a nonempty nonexpansive retract of C.

UsingTheorem 3.1, we prove the following fixed point theorem.

Theorem 3.2 Let C be a closed convex subset of a Banach space E Let A be a weakly compact convex subset of C Assume that A has the fixed point property for nonexpansive mappings Let {T n:n ∈ N} be an infinite family of commuting nonexpansive mappings on

C such that

T1(A) ⊂ A, T +1

A ∩



k=1

FT k

for all  ∈ N Then there exists a common fixed point z0∈ A of {T n:n ∈ N}.

Proof We put B := A ∩(

k =1F(T k)) for ∈ N We first showB is nonempty and there exists a nonexpansive retractionP fromA onto B for all ∈ N From the assumption of

T1(A) ⊂ A, there exists a fixed point z1∈ A of T1, that is,B1 =∅ ByTheorem 3.1, there exists a nonexpansive retractionP1fromA onto B1 We assumeB is nonempty and there exists a nonexpansive retractionP fromA onto B for some ∈ N From the assumption

ofT +1(B )⊂ A, we have that T +1 ◦ P  is a nonexpansive mapping onA We note that

B +1 = F(T +1 ◦ P ) Indeed,B +1 ⊂ F(T +1 ◦ P ) is obvious Conversely, we assumez2∈ A

satisfiesT +1 ◦ P  z2= z2 Fork ∈ Nwithk ≤ , we have

T k z2= T k ◦ T +1 ◦ P  z2= T +1 ◦ T k ◦ P  z2= T +1 ◦ P  z2= z2, (3.2) that is,z2∈ B and henceP  z2= z2 Thus, we also have

T +1 z2= T +1 ◦ P  z2= z2. (3.3) Thereforez2∈ B +1 and henceB +1 ⊃ F(T +1 ◦ P ) We have shownB +1 = F(T +1 ◦ P ) SinceA has the fixed point property, we have

B +1 = FT +1 ◦ P 

ByTheorem 3.1again, there exists a nonexpansive retractionP +1fromA onto B +1 So,

by induction, we have shown thatB is nonempty and there exists a nonexpansive retrac-tionP  fromA onto B  for all ∈ N Define a sequence{Q n:n ∈ N}of nonexpansive mappings onA by

Q n:= P n ◦ P n −1◦ ··· ◦ P2◦ P1 (3.5) forn ∈ N SinceP m x = P n ◦ P m x for x ∈ A, m,n ∈ Nwithm ≥ n, we have

Q m ◦ Q n = Pmax{m,n} ◦ Pmax{m,n}−1◦ ··· ◦ P2◦ P1 (3.6) for allm,n ∈ Nand henceQ m ◦ Q n = Q n ◦ Q mfor allm,n ∈ N So, byTheorem 3.1, there exists a common fixed pointz0∈ A of {Q n:n ∈ N} Let us prove thatz0is also a common fixed point of{T n:n ∈ N} Since

we havez0∈ B1, that is,T1z0= z0 We assume

T1z0= T2z0= ··· = T  z0= z0 (3.8) for some ∈ N Then

z0= Q +1 z0= P +1 ◦ P  ◦ ··· ◦ P2◦ P1z0

= P +1 ◦ P  ◦ ··· ◦ P2z0= ··· = P +1 ◦ P  z0= P +1 z0

(3.9)

and hencez0∈ B +1, that is,T +1 z0= z0 So, by induction,z0is a common fixed point of

4 Lemmas

In this section, we prove some lemmas which are used in the proofs of our main results

Lemma 4.1 Let C be a closed convex subset of a Banach space E Let {T n:n ∈ N} be an infinite family of commuting nonexpansive mappings on C with a common fixed point Let {α n } and {t n } be sequences in (0, 1/2) satisfying lim n t n =limn α n /t 

n = 0 for  ∈ N Let {I n }

be a sequence of nonempty subsets ofNsuch that I n ⊂ I n+1 for n ∈ N, and∞ n=1I n = N For

I ⊂ N and t ∈(0, 1/2) with I =∅, define nonexpansive mappingsS(I,t) on C by

S(I,t)x :=

1−

k∈I

t k

T1x +

k∈I

t k T k+1 x

(4.1)

for x ∈ C Fix u ∈ C and define a sequence {u n } in C by

u n =(1− α n)S(I n,t n)u n+α n u (4.2)

for n ∈ N Let {u n β:β ∈ D} be a subnet of {u n } Then the following hold.

(i) lim supβ u n β − T1x ≤lim supβ u n β − x for x ∈ C.

(ii) If x ∈ C satisfies T1x = x, then limsup β u n β − T2x ≤lim supβ u n β − x.

(iii) If x ∈ C satisfies T1x = T2x = ··· = T −1x = x for some  ∈ N with  ≥ 3, then

lim supβ u n β − T  x ≤lim supβ u n β − x.

Proof Let v be a common fixed point of {T n:n ∈ N} It is obvious thatS(I,t)v = v for

allI ⊂ Nandt ∈(0, 1/2) with I =∅ Forx ∈ C and k ∈ N, we have

T k x  ≤  T k x − v+v =T k x − T k v+v ≤ x − v+v. (4.3) Hence,{T k x : k ∈ N}is bounded for everyx ∈ C Therefore S(I,t) is well defined for every

I ⊂ Nandt ∈(0, 1/2) with I =∅ It is obvious thatS(I,t) is a nonexpansive mapping on

C for every I and t Since

u n − v  = 1 − α n

SI n,t n

u n+α n u − v

≤1− α nS

I n,t n

u n − v+α n u − v

≤1− α nu n − v+α n u − v, (4.4)

we haveu n − v ≤ u − vforn ∈ N Therefore{u n }is bounded Since

T k u n  ≤  T k u n − v+v ≤u n − v+v ≤u n+ 2v (4.5) for alln,k ∈ N,{T k u n:n,k ∈ N}is also bounded We fixx ∈ C and we put

M :=max



u,v, sup

n∈N

u n, sup

n,k∈N

T k u n,x, sup

k∈N

T k x< ∞. (4.6)

It is obvious thatS(I,t)u n  ≤ M and S(I,t)x ≤ M for all n ∈ N,I ⊂ Nandt ∈(0, 1/2)

withI =∅ From the assumption, we have

SI n,t n

u n − u n = α n

SI n,t n

forn ∈ N We have

u n β − T1x  ≤  u n β − SI n β,t n β



u n β+S

I n β,t n β



u n β − SI n β,t n β



x+S

I n β,t n β



x − T1x

≤ α n βS

I n β, t n β



u n β − u+u n β − x+

 − k∈I nβ

t k

n β T1x +

k∈I nβ

t k

n β T k+1 x





≤2Mα n β+u n

β − x+ 2M

k∈I nβ

t k

n β ≤2Mα n β+u n

β − x+ 2M t n β

1− t n β

(4.8) forβ ∈ D and hence

lim sup

β∈

u n

β − T1x  ≤lim sup

β∈

u n

This is (i) We next show (ii) We assume thatT1x = x Then T1◦ T2x = T2◦ T1x = T2x.

Forβ ∈ D with 1,2 ∈ I n β, we have

u n β − T2x  ≤  u n β − SI n β,t n β



u n β+S

I n β, t n β



u n β − T2x

≤ α n βS

I n β, t n β



u n β − u+ 1−

k∈I nβ

t k

n β

T1u n β − T2x

+t n βT2u n β − T2x+

k∈I nβ \{1}

t k

n βT k+1 u n β − T2x

≤2Mα n β+

1− t n βT1u n β − T2x+t n βu n β − x+ 2M

k∈I nβ \{1}

t k

n β

≤2Mα n β+

1− t n βT1u n β − T1◦ T2x+t n βu n β − x+ 2M t2

n β

1− t n β

≤2Mα n β+

1− t n βu n β − T2x+t n βu n β − x+ 2M t2

n β

1− t n β

(4.10)

and hence

u n β − T2x  ≤2M α n β

t n β

+u n β − x+ 2M t n β

1− t n β

Therefore we obtain

lim sup

β∈

u n β − T2x  ≤lim sup

β∈

u n β − x. (4.12)

Let us prove (iii) We assumeT1x = T2x = ··· = T −1x = x for some  ∈ Nwith  ≥

3 Then T m ◦ T  x = T  ◦ T m x = T  x for every m ∈ N with 1≤ m <  For β ∈ D with

1, 2, , −1∈ I n β, we have

u n

β − T  x  ≤  u n β − SI n β,t n β



u n β+S

I n β, t n β



u n β − T  x

≤ α n βS

I n β, t n β



u n β − u+ 1−

k ∈ I nβ

t k

n β

T1u n

β − T  x

+

−2

m 1

t m

n βT m+1 u n β − T  x+t  −1

n β T  u n β − T  x

+

k∈I nβ \{1,2, ,−1}

t k

n βT k+1 u n β − T  x

≤2Mα n β+

1−

−1

m 1

t m

n β

T1u n β − T  x

+

−2

m 1

t m

n βT m+1 u n β − T  x+t −1

n β u n β − x+ 2M

k∈I nβ \{1,2, ,−1}

t k

n β

≤2Mα n β+

1−

−1

m 1

t m

n β

T1u n β − T1◦ T  x

+

−2

m 1

t m

n βT m+1 u n β − T m+1 ◦ T  x+t −1

n β u n β − x+ 2M t n  β

1− t n β

≤2Mα n β+

1−

−1

m 1

t m

n β

u n

β − T  x

+

−2

m 1

t m

n βu n β − T  x+t −1

n β u n β − x+ 2M t n  β

1− t n β

=2Mα n β+

1− t −1

n β

u n β − T  x+t −1

n β u n β − x+ 2M t  n β

1− t n β

(4.13)

and hence

u n β − T  x  ≤2M α n β

t −1

n β

+u n β − x+ 2M t n β

1− t n β

Therefore we obtain

lim sup

β∈

u n β − T  x  ≤lim sup

β∈

u n β − x. (4.15)

Remark 4.2 Let g be a strictly increasing mapping onN Then it is obvious that limn t g(n)

=limn α g(n) /t 

g(n) =0 for all ∈ N,I g(n) ⊂ I g(n+1)forn ∈ N, and∞

n=1I g(n) = N Thus, the same conclusions of Lemmas4.3–4.6also hold for{u g(n) }

Lemma 4.3 Let E, C, {T n }, {α n }, {t n }, {I n }, u, and {u n } be as in Lemma 4.1 Assume that {u n } converges strongly to some point x ∈ C Then x is a common fixed point of {T n:n ∈ N}. Proof FromLemma 4.1(i), we have

lim sup

n→∞

u n − T1x  ≤ n→∞limu n − x  =0. (4.16) This means {u n } converges to T1x and hence T1x = x We assume that T1x = ··· =

T −1x = x for some  ∈ Nwith ≥2 Then fromLemma 4.1(ii) and (iii), we have

lim sup

n→∞

u n − T  x  ≤ n→∞limu n − x  =0. (4.17) This means{u n }converges toT  x and hence T  x = x So, by induction, we obtain T n x = x

Lemma 4.4 Let E, C, {T n }, {α n }, {t n }, {I n }, u, and {u n } be as in Lemma 4.1 Assume that

E is smooth and z ∈ C is a common fixed point of {T n:n ∈ N} Then



for all n ∈ N.

Proof Since α n(u n − u) =(1− α n)(S(I n,t n)u n − u n), we have

α n

1− α n



u n − u, Ju n − z=SI n,t n

u n − u n,Ju n − z

=SI n,t nu n − z,Ju n − z+z − u n,Ju n − z

=SI n,t n

u n − SI n,t n

z,Ju n − z−u n − z 2

≤S

I n,t n

u n − SI n,t n

z u n − z − u n − z 2

≤u n − z 2

−u n − z 2

=0.

(4.19)

Thus we obtain



Lemma 4.5 Let E, C, {T n }, {α n }, {t n }, {I n }, u, and {u n } be as in Lemma 4.1 Assume that

E is smooth Then {u n } has at most one cluster point.

Proof We assume that a subsequence {u n i } of{u n } converges strongly tox, and that

another subsequence{u n j }of{u n }converges strongly to y ApplyingLemma 4.3to the subsequences{u n i }and{u n j }, we have thatx and y are common fixed points of {T n:n ∈ N} So, byLemma 4.4, we have

u n i − u, Ju n i − y≤0 (4.21)

for alli ∈ N Therefore we obtain



Similarly we can prove



So we obtain

x − y2=x − y, J(x − y)

=x − u, J(x − y)+

u − y, J(x − y)

=x − u, J(x − y)+

y − u, J(y − x)≤0.

(4.24)

Lemma 4.6 Let E be a reflexive Banach space with uniformly Gˆateaux differentiable norm and let C be a closed convex subset of E with the fixed point property for nonexpansive map-pings Let {T n }, {α n }, {t n }, {I n }, u, and {u n } be as in Lemma 4.1 Then {u n } has a cluster point which is a common fixed point of {T n:n ∈ N}.

Proof From the proof ofLemma 4.1, we have that{u n }is bounded Take a universal subnet{u ν:ν ∈ D}of{u n } Define a continuous convex function f from C into [0,∞) by

f (x) :=lim

for allx ∈ C We note that f is well defined because {u ν − x}is a universal net in some compact subset ofRfor eachx ∈ C From the reflexivity of E and lim x→∞ f (x) = ∞, we can putr :=minx∈C f (x) and define a nonempty weakly compact convex subset A of C

by

The following theorem is one of the most famous fixed point theorems for families of nonexpansive mappings

Theorem 3.1 (Bruck [9]) Suppose a closed convex subsetC of a Banach space... points of S is a nonempty nonexpansive retract of C.

UsingTheorem 3.1, we prove the following