OF NONEXPANSIVE MAPPINGS

OF NONEXPANSIVE MAPPINGS

TOMONARI SUZUKI

Received 14 September 2004 and in revised form 24 March 2005

Using the notion of Banach limits, we discuss the characterization of fixed points of non- expansive mappings in Banach spaces. Indeed, we prove that the two sets of fixed points of a nonexpansive mapping and some mapping generated by a Banach limit coincide. In our discussion, we may not assume the strict convexity of the Banach space.

1. Introduction

LetCbe a closed convex subset of a Banach spaceE. A mappingT onCis called anon- expansive mappingifTx−T y ≤ x−yfor allx,y∈C. We denote byF(T) the set of fixed points ofT. Kirk [17] proved thatF(T) is nonempty in the case thatCis weakly compact and has normal structure. See also [2,3,5,6,11] and others.

Convergence theorems to fixed points are also proved by many authors; see [1,7,8, 9,10,13,15,18,23,30] and others. Very recently, the author proved the convergence theorems for two nonexpansive mappings without the assumption of the strict convexity of the Banach space. To prove this, the author proved the following theorem, which plays an extremely important role in [26].

Theorem1.1 (see [26]). LetCbe a compact convex subset of a Banach spaceEand letT be a nonexpansive mapping onC. Thenz∈Cis a fixed point ofTif and only if

lim inf

n→∞

1

n n i=1

Tⁱz−z=0 (1.1)

holds.

The author also proved the following theorem. Using it, we give one nonexpansive retraction onto the set of all fixed points.

Theorem1.2 (see [27]). LetEbe a Banach space with the Opial property and letCbe a weakly compact convex subset ofE. LetTbe a nonexpansive mapping onC. Put

M(n,x)=1 n

n i=1

Tⁱx (1.2)

Copyright©2005 Hindawi Publishing Corporation

International Journal of Mathematics and Mathematical Sciences 2005:11 (2005) 1723–1735 DOI:10.1155/IJMMS.2005.1723

forn∈Nandx∈C. Then forz∈C, the following are equivalent:

(i)zis a fixed point ofT;

(ii){M(n,z)}converges weakly toz;

(iii)there exists a subnet{M(νβ,z) :β∈D}of a sequence{M(n,z)}inCconverging weakly toz.

In this paper, using the notion of Banach limits, we generalize Theorems1.1and1.2.

We remark that the proofs of our results are simpler than the proofs of Theorems1.1and 1.2. In our discussion, we may not assume the strict convexity of the Banach space.

2. Preliminaries

Throughout this paper, we denote byNthe set of all positive integers and byRthe set of all real numbers. For a subsetAofN, we define a functionIAfromNintoRby

IA(n)=





1 ifn∈A,

0 ifn∈A. (2.1)

LetEbe a Banach space. We denote byE^∗ the dual ofE. We recall that Eis said to have theOpial property[21] if for each weakly convergent sequence{xn}inEwith weak limitx0, lim inf_nxn−x0<lim inf_nxn−xfor allx∈Ewithx=x0. All Hilbert spaces, all finite-dimensional Banach spaces, and^p(1≤p <∞) have the Opial property. A Ba- nach space with a duality mapping which is weakly sequentially continuous also has the Opial property; see [12]. We know that every separable Banach space can be equivalently renormed so that it has the Opial property; see [31]. Gossez and Lami Dozo [12] prove that every weakly compact convex subset of a Banach space with the Opial property has normal structure. See also [19,20,22,25] and others.

We denote by^∞the Banach space consisting of all bounded functions fromNintoR (i.e., all bounded real sequences) with supremum norm. We recall thatµ∈(^∞)^∗is called amean ifµ =µ(IN)=1. It is equivalent to

ninf∈Na(n)≤µ(a)≤sup

n∈Na(n) (2.2)

for alla∈^∞. We also know that ifa(n)≤b(n) for alln∈N, thenµ(a)≤µ(b) holds.

Sometimes, we denote byµn(a(n)) the valueµ(a).µ∈(^∞)^∗is called aBanach limitif the following hold:

(i)µis a mean;

(ii)µ(a)=µn(a(n+ 1)) for alla∈^∞. That is, puttingb(n)=a(n+ 1) forn∈N, we haveµ(a)=µ(b).

It is obvious that

µ(a)=µn

a(n+k) (2.3)

for a Banach limitµ,a∈^∞, andk∈N. We know that Banach limits exist; see [4]. We also know that

lim inf

n→∞ a(n)≤µ(a)≤lim sup

n→∞ a(n) (2.4)

for alla∈^∞.

LetT be a nonexpansive mapping on a weakly compact convex subsetCof a Banach spaceE. Letµbe a Banach limit. Then we know that for x∈C, there exists a unique elementx0ofCsatisfying

µn

fTⁿx=fx0

(2.5)

for allf ∈E^∗; see [14,19]. Following Rod´e [24], we denote suchx0byTµx. We also know thatTµis a nonexpansive mapping onC.

We now prove the following lemmas, which are used inSection 3.

Lemma2.1. Leta,b∈^∞and letµbe a Banach limit. Then the following hold.

(i)If there existsn0∈Nsuch thata(n)≤b(n)for alln≥n0, thenµ(a)≤µ(b)holds.

(ii)If there existsn0∈Nsuch thata(n)=b(n)for alln≥n0, thenµ(a)=µ(b)holds.

Proof. We first show (i). We note thata(n0+n)≤b(n0+n) for alln∈N. Sinceµis a Banach limit, we have

µn

a(n)=µn

an0+n≤µn

bn0+n=µn

b(n). (2.6) It is obvious that (ii) follows from (i). This completes the proof.

Lemma2.2. LetA1,A2,A3,...,Akbe subsets ofNand letµbe a Banach limit. Put A=

k j=1

Aj, α= k j=1

µIAj

−k+ 1. (2.7)

Suppose thatα >0. Then,

µIA

≥α (2.8)

holds and

n∈N:n≥n0

∩A=∅ (2.9)

holds for alln0∈N.

Proof. It is obvious thatn∈Aif and only if k j=1

IAj(n)=k, (2.10)

andn∈N\Aif and only if

k j=1

IAj(n)≤k−1. (2.11)

Therefore we obtain

IA(n)≥ k j=1

IAj(n)−k+ 1 (2.12)

for alln∈N. Hence,

µIA

≥µn



^k

j=1

IAj(n)−k+ 1





=^k

j=1

µIAj

−k+ 1

=α >0.

(2.13)

We suppose that{n∈N:n≥n0} ∩A=∅for somen0∈N. ThenIA(n)=0 forn≥n0. So, fromLemma 2.1, we obtain

0< α≤µIA

=µ(0)=0. (2.14)

This is a contradiction. This completes the proof.

3. Main results

In this section, we prove our main results.

We first prove the following theorem, which plays an important role in this paper.

Theorem3.1. LetEbe a Banach space and letCbe a weakly compact convex subset ofE.

LetTbe a nonexpansive mapping onC. Letµbe a Banach limit. Suppose thatTµz=zfor somez∈C. Then there exist sequences{pn}inNand{fn}inE^∗such that

pn+1> pn, T^pnz−z≥λ− 1

3ⁿ⁺¹, f

T^pnz−z≤2⁺¹

3⁺¹ for=1, 2,...,n−1, fn=1, fn

T^pnz−z=T^pnz−z

(3.1)

for alln∈N, where

λ=lim sup

n→∞

Tⁿz−z. (3.2)

Before proving this theorem, we need some preliminaries. In the following lemmas and the proof ofTheorem 3.1, we put

A(f,ε)= n∈N:fTⁿz−z≤ε (3.3) for f ∈E^∗andε >0, and

B(ε)= n∈N:Tⁿz−z≥λ−ε (3.4) forε >0.

Lemma3.2. For everyn∈N,

Tⁿz−z≤λ (3.5)

holds.

Proof. Sinceµis a Banach limit, we haveµn(Tⁿz−z)≤λ. Fixm∈N. By the Hahn- Banach theorem, there exists f ∈E^∗such that

f =1, fT^mz−z=T^mz−z. (3.6) Forn∈N, we have

T^mz−z=fT^mz−z

=fT^mz−T^m+nz+fT^m+nz−z

≤ fT^mz−T^m+nz+fT^m+nz−z

=T^mz−T^m+nz+fT^m+nz−z

≤Tⁿz−z+fT^m+nz−z.

(3.7)

Hence

T^mz−z=µnT^mz−z

≤µnTⁿz−z+fT^m+nz−z

=µnTⁿz−z+µn

fT^m+nz−f(z)

=µnTⁿz−z+µn

fTⁿz−f(z)

=µnTⁿz−z+fTµz−f(z)

=µnTⁿz−z

≤λ.

(3.8)

This completes the proof.

Lemma3.3. Suppose thatm∈N,f ∈E^∗, andδ >0satisfy that

f =1, fT^mz−z=T^mz−z≥λ−δ. (3.9) Then

µIA(f,ε)

≥ ε

ε+δ (3.10)

holds for allε >0.

Proof. Forn > m, byLemma 3.2, we have

fTⁿz−z= fTⁿz−T^mz+fT^mz−z

≥ −fTⁿz−T^mz+ fT^mz−z

= −Tⁿz−T^mz+T^mz−z

≥ −T^n−mz−z+T^mz−z

≥ −λ+λ−δ= −δ.

(3.11)

On the other hand, by the definition ofA(f,ε),

fTⁿz−z> ε (3.12)

for alln∈N\A(f,ε). Therefore, forn∈Nwithn > m, we have fTⁿz−z≥ −δIA(f,ε)(n) +εIN\A(f,ε)(n)

= −δIA(f,ε)(n) +εIN\A(f,ε)(n) + (ε−ε)IA(f,ε)(n)

= −(δ+ε)IA(f,ε)(n) +εIN(n)

= −(δ+ε)IA(f,ε)(n) +ε.

(3.13)

ByLemma 2.1, we have

0= fTµz−f(z)

=µn

fTⁿz−f(z)

=µn

fTⁿz−z

≥µn

−(δ+ε)IA(f,ε)(n) +ε

= −(δ+ε)µIA(f,ε) +ε.

(3.14)

Hence, we obtain

µIA(f,ε)

≥ ε

ε+δ . (3.15)

This completes the proof.

Lemma3.4. µ(IB(ε))=1holds for allε >0.

Proof. We fixε >0 andη∈Rwith 1/2< η <1 and put δ=ε(1−η)

2η . (3.16)

We note that 0< δ < ε/2. By the definition ofλ, there existsm∈Nsuch that

T^mz−z≥λ−δ. (3.17)

Fix f ∈E^∗with

f =1, fT^mz−z=T^mz−z. (3.18) So, usingLemma 3.3, we have

µIA(f,ε/2)

≥ ε/2

ε/2 +δ ⁼η. (3.19)

Forn∈Nwithm+n∈A(f,ε/2), we have

Tⁿz−z≥T^mz−T^m+nz

≥fT^mz−T^m+nz

=fT^mz−z+fz−T^m+nz

=T^mz−z+fz−T^m+nz

≥λ−δ−ε 2

≥λ−ε,

(3.20)

and hencen∈B(ε). Therefore

IB(ε)(n)≥IA(f,ε/2)(m+n) (3.21)

for alln∈N. So we obtain

µIB(ε)

≥µn

IA(f,ε/2)(m+n)

=µn

IA(f,ε/2)(n)

≥η.

(3.22)

Sinceηis arbitrary, we obtain the desired result.

Proof ofTheorem 3.1. By the definition ofλ, there existsp1∈Nsuch that T^p1z−z≥λ− 1

3². (3.23)

Fix f1∈E^∗with

f1=1, f1

T^p1z−z=T^p1z−z. (3.24)

ByLemma 3.3, we have

µIA(f1,(2/3)²)

≥ 2²

2²+ 1. (3.25)

We now define inductively sequences{pn}inNand{fn}inE^∗. Suppose thatpk∈Nand fk∈E^∗are known. Since

µIB(1/3^k+2)

+ k =1

µIA(f,(2/3)⁺¹)

−k

≥1 + k =1

2⁺¹ 2⁺¹+ 1⁻k

≥1 + k =1

2⁺¹−1

2^{+1 −}k=1 + k =1

−1 2⁺¹

>1 2>0,

(3.26)

we have

m∈N:m≥pk+ 1∩B 1 3^k+2

∩^k

=1

A

f, 2

3 +1

=∅ (3.27)

byLemma 2.2. So we can choosepk+1∈Nsuch thatpk+1> pk, T^pk+1z−z≥λ− 1

3^k+2, f

T^pk+1z−z≤2⁺¹

3⁺¹ (3.28)

for=1, 2,...,k. Fix fk+1∈E^∗with fk+1=1, fk+1

T^pk+1z−z=T^pk+1z−z. (3.29) Note that

µIA(fk+1,(2/3)^k+2)

≥ 2^k+2

2^k+2+ 1 (3.30)

byLemma 3.3. Hence we have defined{pn}and{fn}. Now, we prove our main results.

Theorem3.5. LetCbe a weakly compact convex subset of a Banach spaceEwith the Opial property. LetTbe a nonexpansive mapping onC. Letµbe a Banach limit. Thenz∈Cis a fixed point ofTif and only ifTµz=z.

Proof. We first assume thatz∈Cis a fixed point ofT. Then, we have fTµz=µn

fTⁿz=µn

f(z)=f(z) (3.31)

for all f ∈E^∗, and henceTµz=z. Conversely, we assume thatTµz=z. ByTheorem 3.1, there exist sequences{pn}inNand{fn}inE^∗satisfying the conclusion ofTheorem 3.1.

We putλ=lim sup_nTⁿz−z. Since Cis weakly compact, there exists a subsequence {pnk}of{pn}such that{T^pnkz}converges weakly to some pointu∈C. Ifnk> , then

f

T^pnkz−z≤2⁺¹

3⁺¹. (3.32)

So we obtain

f(u−z)≤2⁺¹

3⁺¹ (3.33)

for all∈N. Since

T^pz−u=fT^pz−u

≥f

T^pz−u

=f

T^pz−z+f(z−u)

=T^pz−z+f(z−u)

≥λ− 1 3⁺¹⁻

2⁺¹ 3⁺¹

(3.34)

for∈N, we have

lim inf

→∞ T^pz−u≥λ, (3.35)

and hence

lim inf

k→∞

T^pnkz−z≤λ≤lim inf

→∞

T^pz−u

≤lim inf

k→∞ T^pnkz−u. (3.36) By the Opial property ofE, we obtainz=u. We also have

lim inf

k→∞ T^pnkz−Tz≤lim inf

k→∞ T^pnk−1z−z≤λ, (3.37) and henceTz=u. ThereforeTz=z. This completes the proof.

Theorem3.6. LetCbe a compact convex subset of a Banach spaceE. LetTbe a nonexpan- sive mapping onC. Letµbe a Banach limit. Thenz∈Cis a fixed point ofTif and only if Tµz=z.

Proof. From the proof ofTheorem 3.5, we know thatTz=zimplies thatTµz=z. Con- versely, we assume thatTµz=z. ByTheorem 3.1, there exist sequences{pn}inNand{fn} inE^∗satisfying the conclusion ofTheorem 3.1. We putλ=lim sup_nTⁿz−z. SinceCis compact, there exists a subsequence{pnk}of{pn}such that{T^pnkz}converges strongly to some pointu∈C. As in the proof ofTheorem 3.5, we obtain lim infT^pz−u ≥λ.

This implies thatλ=0, and hence{Tⁿz}converges toz. So we have Tz=Tlim

n→∞Tⁿz=lim

n→∞Tⁿ⁺¹z=z. (3.38)

This completes the proof.

Appendix

In some sense, Theorems3.5and3.6are generalizations of Theorems1.2and1.1, respec- tively. To show this, we use the notion of universal nets. We recall that a net{yβ:β∈D} in a topological spaceYis universal if for each subsetAofY, there existsβ0∈Dsatisfying either of the following:

(i) yβ∈Afor allβ∈Dwithβ≥β0; or (ii) yβ∈Y\Afor allβ∈Dwithβ≥β0.

For every net{yβ:β∈D}, by the axiom of choice, there exists a universal subnet{yβγ: γ∈D}of{yβ:β∈D}. If f is a mapping fromY into a topological spaceZ and{yβ: β∈D}is a universal net inY, then{f(yβ) :β∈D}is also a universal net inZ. IfY is compact, then a universal net{yβ:β∈D}inYalways converges. See [16] and others for details.

PropositionA.1. Let{νβ:β∈D}be a universal subnet of a sequence{n:n∈N}inN. Define a functionµfrom^∞intoRby

µ(a)=lim

β∈D

1 νβ

νβ

i=1

a(i) (A.1)

for alla∈^∞. Thenµis a Banach limit.

Proof. We note thatµis well defined because{νβ:β∈D}is universal. It is obvious that µis linear. Fora∈^∞, we have

µ(a)=lim

β∈D

1 νβ

νβ

i=1

a(i)≤lim

β∈D

1 νβ

νβ

i=1

a =lim

β∈Da = a. (A.2)

Similarly, we obtainµ(a)≥ −a. Henceµ ≤1. Sinceµ(IN)=1, we haveµ =µ(IN)= 1, that is,µis a mean on^∞. We also have

µn

a(n+ 1)=lim

β∈D

1 νβ

νβ

i=1

a(i+ 1)

=lim

β∈D



1 νβ

νβ

i=1

a(i)−a(1)

νβ +aνβ+ 1 νβ





=µ(a)

(A.3)

for alla∈^∞. This completes the proof.

PropositionA.2. LetCbe a weakly compact convex subset of a Banach spaceEand letT be a nonexpansive mapping onC. DefineM(·,·)as inTheorem 1.2. Take a universal subnet {νβ:β∈D}of a sequence{n:n∈N}inNand define a mappingUonCby

Ux=weak-lim

β∈D Mνβ,x (A.4)

for allx∈C. Then there exists a Banach limitµsatisfyingTµx=Uxfor allx∈C.

Proof. Define a Banach limitµas inProposition A.1. Then forx∈Candf ∈E^∗, we have f(Ux)=lim

β∈Df 1

νβ νβ

i=1

Tⁱx

=lim

β∈D

1 νβ

νβ

i=1

fTⁱx

=µn

f(Tⁿx)

=fTµx.

(A.5)

Since f is arbitrary, we haveUx=Tµxfor allx∈C. This completes the proof.

UsingProposition A.2, we obtain the following proposition.

PropositionA.3. LetCbe a weakly compact convex subset of a Banach spaceEand letT be a nonexpansive mapping onC. Then ifz∈CsatisfiesTheorem 1.2(iii), then there exists a Banach limitµsatisfyingTµz=z.

Proof. DefineM(·,·) as inTheorem 1.2. Take a universal subnet{νβγ:γ∈D}of{νβ: β∈D}. We note that{νβγ:γ∈D}is also a universal subnet of a sequence{n:n∈N}in N. We also note that{M(νβγ,z) :γ∈D}converges weakly tozbecause{M(νβ,z) :β∈D} does. Define a mappingUonCby

Ux=weak-lim

γ∈D Mνβγ,x (A.6)

for allx∈C. It is obvious thatUz=z. ByProposition A.2, there exists a Banach limitµ satisfyingU=Tµ. Then we haveTµz=z. This completes the proof.

In the case thatEis a Hilbert space, orEis a uniformly convex Banach space with a Fr´echet differentiable norm,Tµitself is a nonexpansive retraction fromContoF(T); see Baillon [1] and Bruck [8]. In general, this does not hold. We finally give an example.

Example A.4(see [27,28]). Define a compact convex subsetCof (R², · ∞) by C= x1,x2

: 0≤x2≤1,−x2≤x1≤x2

. (A.7)

Define a nonexpansive mappingTonCby Tx1,x2

=

−x1,x1 (A.8)

for all (x1,x2)∈C. Then,F(T)= {(0, 0)}and Tµ

x1,x2

=

0,x1 (A.9)

for (x1,x2)∈Cand a Banach limitµ. That is,Tµis not a nonexpansive retraction fromC ontoF(T).

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Tomonari Suzuki: Department of Mathematics, Kyushu Institute of Technology, Sensui-cho, To- bata-ku, Kitakyushu 804-8550, Japan

E-mail address:[email protected]