Browder-Krasnoselskii-Type Fixed Point Theorems in Banach Spaces

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Volume 2010, Article ID 243716,20pages doi:10.1155/2010/243716

Research Article

Browder-Krasnoselskii-Type Fixed Point Theorems

in Banach Spaces

Ravi P. Agarwal,

1, 2

_{Donal O’Regan,}

3

_{and Mohamed-Aziz Taoudi}

4

1_{Department of Mathematical Sciences, Florida Institute of Technology, 150 West University Boulevard,}

Melbourne, FL 32901, USA

2_{Mathematics and Statistics Department, King Fahd University of Petroleum and Minerals,}

Dhahran 31261, Saudi Arabia

3_{Department of Mathematics, National University of Ireland, Galway, Ireland}

4_{Laboratoire de Math´ematiques et de Dynamique de Populations, Universit´e Cadi Ayyad,}

Marrakech, Morocco

Correspondence should be addressed to Ravi P. Agarwal,[email protected]

Received 29 January 2010; Accepted 6 July 2010

Academic Editor: Hichem Ben-El-Mechaiekh

Copyrightq2010 Ravi P. Agarwal et al. 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.

We present some fixed point theorems for the sumABof a weakly-strongly continuous map and a nonexpansive map on a Banach spaceX. Our results cover several earlier works by Edmunds, Reinermann, Singh, and others.

1. Introduction

LetMbe a nonempty subset of a Banach spaceXandT :M → Xa mapping. We say that

T isweakly-strongly continuousif for each sequence{xn}inMwhich converges weakly tox

inM,the sequence{Txn}converges strongly toTx.The mappingT is called nonexpansiveif

Tx−Ty ≤ x−yfor allx, y∈M.

In1, Edmunds proved the following fixed point theorem

Theorem 1.1. LetMbe a nonempty bounded closed convex subset of a Hilbert spaceHandA, Btwo maps fromMintoXsuch that

iAis weakly-strongly continuous,

iiBis a nonexpansive mapping,

iiiAxBy∈Mfor allx, y∈M.

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It is apparent that Theorem 1.1is an important supplement to both Krasnoselskii’s fixed point2, Theorem 4.4.1and Browder’s fixed point theorems2, Theorem 5.1.3. The proof ofTheorem 1.1depends heavily upon the fact thatF I−AwhereIis the identity mapis monotone, that is,Fx−Fy, x−y ≥0 for allx, y,and uses the Krasnoselskii fixed point theorem for the sum of a completely continuous and a strict contraction mapping2,

3. In 4, Reinermann extended the above result to uniform Banach spaces. The methods used in the Hilbert space setting involving monotone operators do not apply in the more general context of uniform Banach spaces. The author follows another strategy of proof which is based on a demiclosedness principle for nonexpansive mapping defined on a uniformly convex Banach space and uses the fact that every uniformly convex space is reflexive. In5, Singh extendedTheorem 1.1 to reflexive Banach spaces by assuming further thatI −B is demiclosed. Notice that all the aforementioned extensions ofTheorem 1.1depend strongly upon the geometry of the ambient Banach space. In this paper we propose an extension of

Theorem 1.1to an arbitrary Banach space. Also, we discuss the existence of a fixed point for the sum of a compact mapping and a nonexpansive mapping for both the weak and the strong topology of a Banach space and under Krasnosel’skii-, Leray Schauder-, and Furi-Pera-type conditions. First we recall the following well-known result.

Theorem 1.2see2, Theorem 5.1.2. LetMbe a bounded closed convex subset of a Banach space

X andT a nonexpansive mapping ofMintoM.Then for eachε > 0,there is axε ∈ Msuch that

Txε−xε< ε.

Now, let us recall some definitions and results which will be needed in our further considerations. Let X be a Banach space, ΩX the collection of all nonempty bounded subsets ofX, andWXthe subset ofΩXconsisting of all weakly compact subsets ofX.

LetBr denote the closed ball inX centered at 0 with radiusr >0.In6De Blasi introduced

the following mapw:ΩX → 0,∞defined by

wM inf{r >0 : there exists a setN∈ WXsuch thatM⊆NBr}, 1.1

for allM∈ΩX. For completeness we recall some properties ofw·needed belowfor the proofs we refer the reader to6.

Lemma 1.3. LetM1, M2∈ΩX, then one has the following:

iifM1⊆M2, thenwM1≤wM2,

iiwM1 0if and only ifM1is relatively weakly compact,

iiiwMw₁ wM1, whereMw₁ is the weak closure ofM1,

ivwλM1 |λ|wM1for allλ∈R,

vwcoM1 wM1,

viwM1M2≤wM1 wM2,

viiifMn_n≥1is a decreasing sequence of nonempty, bounded, and weakly closed subsets ofX

withlimn→ ∞wMn 0,then∞n1Mn/∅andw∞n1Mn 0, that is,w∞n1Mn

is relatively weakly compact.

Throughout this paper, a measure of weak noncompactness will be a mappingψ :

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Definition 1.4. LetX be a Banach space, and let ψ be a measure of weak noncompactness onX.A mappingB : DB ⊆ X → X is said to beψ-contractive if it maps bounded sets into bounded sets and there is β ∈ 0,1such that ψBS ≤ βψS for all bounded sets

S ⊆DB.The mappingB:DB⊆ X → X is said to beψ-condensing if it maps bounded sets into bounded sets andψBS < ψSwheneverS is a bounded subset ofDBsuch thatψS>0.

LetJbe a nonlinear operator fromDJ⊆XintoX.In what follows, we will use the following two conditions.

H1Ifxn_n∈_Nis a weakly convergent sequence inDJ,then

Jxn_n∈_Nhas a strongly convergent subsequence inX.

H2Ifxnn∈Nis a weakly convergent sequence inDJ,then

Jxnn∈Nhas a weakly convergent subsequence inX.

Remark 1.5. 1Operators satisfyingH1orH2are not necessarily weakly continuoussee

7–9.

2Everyw-contractive map satisfiesH2.

3A mappingJsatisfiesH2if and only if it maps relatively weakly compact sets into relatively weakly compact onesuse the Eberlein-ˇSmulian theorem10, page 430.

4A mappingJsatisfiesH1if and only if it maps relatively weakly compact sets into relatively compact ones.

5ConditionH2holds true for every bounded linear operator.

6 ConditionH1holds true for the class of weakly compact operators acting on Banach spaces with the Dunford-Pettis property.

7 Continuous mappings satisfying H1 are sometimes called ws-compact operatorssee11, Definition 2.

The following fixed point theorems are crucial for our purposes.

Theorem 1.6 see7, Theorem 2.3. Let Mbe a nonempty closed bounded convex subset of a Banach spaceX.Suppose thatA:M → XandB:X → Xsuch that

iAis continuous,AMis relatively weakly compact, andAsatisfiesH1,

iiBis a strict contraction satisfyingH2,

Then there is ax∈Msuch thatAxBxx.

Theorem 1.7 see12, Theorem 2.1. LetMbe a nonempty closed bounded convex subset of a Banach spaceX.Suppose thatA:M → XandB:X → Xare sequentially weakly continuous such that

iAMis relatively weakly compact,

iiBis a strict contraction,

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Theorem 1.8see13,14. LetXbe a Banach space withC⊆Xclosed and convex. Assume thatU

is a relatively open subset ofCwith0∈U, FUbounded, andF :U → Ca condensing map. Then eitherFhas a fixed point inUor there is a pointu∈∂Uandλ∈0,1withuλFu, hereUand

∂Udenote the closure ofUinCand the boundary ofUinC,respectively.

Theorem 1.9see13,14. LetX be a Banach space andQa closed convex bounded subset ofX

with0∈Q.In addition, assume thatF:Q → Xis a condensing map with

ifxj, λj ∞

j1is a sequence in∂Q×0,1converging tox, λwith

xλFxand 0< λ <1, thenλjF

xj

∈Q forj suﬃciently large,

FP

holding. ThenFhas a fixed point.

2. Fixed Point Theorems

Now we are ready to state and prove the following result.

Theorem 2.1. LetM be a nonempty bounded closed convex subset of a Banach spaceX.Let A :

M → XandB:X → Xsatisfy the following:

iAis weakly-strongly continuous andAMis relatively weakly compact,

iiBis a nonexpansive mapping satisfyingH2,

iiiifxnis a sequence ofMsuch thatI−Bxnis weakly convergent, then the sequence

xnhas a weakly convergent subsequence,

ivI−Bis demiclosed,

vAxBy∈M,for allx, y∈M.

Then there is anx∈Msuch thatAxBxx.

Proof. Suppose first that 0∈M.By hypothesisvwe have for eachλ∈0,1andx, y∈M

λAxλBy∈M. 2.1

Thus the mappingsλAandλBsatisfy the conditions ofTheorem 1.6. Thus, for allλ ∈0,1 there is anxλ ∈Msuch thatλAxλλBxλ xλ.Now, choose a sequence{λn}in0,1such

thatλn → 1 and consider the corresponding sequence{xn}of elements ofMsatisfying

λnAxnλnBxnxn. 2.2

Using the fact thatAMis weakly compact and passing eventually to a subsequence, we may assume that{Axn}converges weakly to somey∈M.Hence

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Since{xn}is a sequence inM, then it is norm bounded and so is{Bxn}.Consequently

xn−Bxn−xn−λnBxn 1−λnBxn −→0. 2.4

As a result

xn−Bxn y. 2.5

By hypothesis iiithe sequence{xn}has a subsequence {xnk}which converges weakly to

somex∈M.SinceAis weakly-strongly continuous, then{Axnk}converges strongly toAx.

As a result

I−λnkBxnk −→Ax. 2.6

Arguing as above we get

xn−Bxn−→Ax. 2.7

The demiclosedness ofI−ByieldsAxBxx.

To complete the proof it remains to consider the case 0/∈M.In such a case let us fix any elementx0 ∈ M, and letM0 {x−x0, x ∈ M}. Define the maps A0 : M0 → Xand

B0:M0 → XbyA0x−x0 Ax−1/2x0andB0x−x0 Bx−1/2x0,forx∈M.Applying

the result of the first case toA0andB0we get anx∈Msuch thatA0x−x0B0x−x0 x−x0,

that is,AxBxx.

Remark 2.2. 1 The new feature about the result of Theorem 2.1 is that no additional assumption on the Banach spaceXis required.

2IfXis reflexive, then the strong continuity plainly implies compactness. Moreover, assumption iiiof Theorem 2.1is always verified. Also, every continuous mapping on X

satisfies conditionH2.If in addition we suppose thatXis a uniformly convex Banach space, thenBis nonexpansive implying thatI−Bis demiclosedsee4,15.

In the light of the aforementioned remarks we obtain the following consequences of

Theorem 2.1. The first is proved in4while the second in stated in5.

Corollary 2.3. LetMbe a nonempty bounded closed convex subset of a uniformly convex Banach spaceX.LetA:M → XandB:M → Xsatisfy the following:

iiBis nonexpansive,

iiiAxBy∈M,for allx, y∈M.

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Corollary 2.4. LetMbe a nonempty bounded closed convex subset of a reflexive Banach spaceX.Let

A:M → XandB:M → Xsatisfy the following:

iiBis nonexpansive andI−Bis demiclosed,

Our next result is the following.

Theorem 2.5. LetM be a nonempty bounded closed convex subset of a Banach spaceX.Let A :

M → XandB:M → Xsatisfy the following:

iAis sequentially weakly continuous, andAMis relatively weakly compact,

iiBis sequentially weakly continuous nonexpansive mapping,

xnhas a convergent subsequence,

ivAxBy∈M,for allx, y∈M.

Proof. Without loss of generality, we may assume that 0∈M.By hypothesisvwe have for eachλ∈0,1andx, y∈M

λAxλBy∈M. 2.8

Thus the mappingsλAandλBsatisfy the conditions ofTheorem 1.7. Thus, for allλ ∈0,1 there is anxλ ∈Msuch thatλAxλλBxλ xλ.Now choose a sequence{λn}in0,1such

thatλn → 1 and consider the corresponding sequence{xn}of elements ofMsatisfying

λnAxnλnBxnxn. 2.9

Using the fact thatAMis weakly compact and passing eventually to a subsequence, we may assume that{Axn}converges weakly to somey∈M.As a result

I−λnBxn y. 2.10

Since{xn}is a sequence inM, then it is norm bounded and so is{Bxn}.Consequently

xn−Bxn−xn−λnBxn 1−λnBxn −→0. 2.11

This amounts to

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some x ∈ M. Since Aand Bare weakly sequentially continuous, then {Axnk} converges

weakly toAxand{Bxnk}converges weakly toBx.Hence,xAxBx.

We next establish the following result which is a sharpening of16, Theorem 2.3. This result is of fundamental importance for our subsequent analysis.

Theorem 2.6. LetXbe a Banach space, and letψbe a measure of weak noncompactness onX.LetQ

andCbe closed, bounded, convex subsets ofXwithQ⊆C.In addition, letUbe a weakly open subset ofQwith0 ∈U,andF :Uw _→ _C_{a weakly sequentially continuous and}_ψ_{-condensing map. Then}

either

F has a fixed point, 2.13

or

there is a pointu∈∂QUandλ∈0,1 withuλFu, 2.14

here∂QUis the weak boundary ofUinQ.

Proof. Suppose that 2.14 does not occur and F does not have a fixed point on ∂QU

otherwise we are finished since2.13occurs. Let

Mx∈Uw_:_x_λFx _{for some}_λ_∈₀_,₁_. _2.15

The setMis nonempty since 0∈ U.Also Mis weakly sequentially closed. Indeed letxn

be sequence ofMwhich converges weakly to somex ∈ Uw_{, and let}_λ

nbe a sequence of

0,1satisfyingxn λnFxn.By passing to a subsequence if necessary, we may assume that

λnconverges to some λ ∈ 0,1.SinceF is weakly sequentially continuous, thenFxn

Fx.ConsequentlyλnFxn λFx.Hencex λFxand thereforex ∈ M.ThusMis weakly

sequentially closed. We now claim thatMis relatively weakly compact. Suppose thatψM>

0.SinceM⊆coFM∪ {0}, then

ψM≤ψcoFM∪ {0} ψFM< ψM, 2.16

which is a contradiction. HenceψM 0 and thereforeMw _{is compact. This proves our}

claim. Now letx ∈ Mw_._Since_Mw _{is weakly compact, then there is a sequence}_x_n_in_M

which converges weakly tox.SinceMis weakly sequentially closed we havex ∈M.Thus

Mw _M._Hence_M_{is weakly closed and therefore weakly compact. From our assumptions}

we haveM∩∂QU∅.SinceXendowed with the weak topology is a locally convex space,

then there exists a continuous mappingρ:Uw _→ ₀_,₁_with_ρ_M _{1 and}_ρ_∂

QU 0see

17. Let

Tx

⎧ ⎨ ⎩

ρxFx, x∈Uw_,

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Clearly T : C → C is weakly sequentially continuous since F is weakly sequentially continuous. Moreover, for anyS⊆Cwe have

TS⊆coFS∩U∪ {0}. 2.18

This implies that

ψTS≤ψcoFS∩U∪ {0} ψFS∩U≤ψFS< ψS 2.19

ifψS > 0.ThusT : C → Cis weakly sequentially continuous andψ-condensing. By18, Theorem 12there exists x ∈ Csuch thatTx x.Nowx ∈ Usince 0 ∈ U.Consequently

xρxFxand sox∈M.This implies thatρx 1 and soxFx.

Remark 2.7. In16, Theorem 2.3,Uw_{is assumed to be weakly compact.}

Lemma 2.8. LetXbe a Banach space andB:X → Xak-Lipschitzian map, that is,

∀x, y∈X, Bx−By≤kx−y. 2.20

In addition, suppose thatBverifiesH2.Then for each bounded subsetSofXone has

wBS≤kwS, 2.21

here,wis the De Blasi measure of weak noncompactness.

Proof. LetSbe a bounded subset ofX andr > wS.There exist 0 ≤ r0 < r and a weakly

compact subsetKofXsuch thatS⊆KBr0.Now we show that

BS⊆BKBkr0⊆BKwBkr0. 2.22

To see this letx ∈S.Then there is ay ∈ Ksuch thatx−y ≤r0.SinceBisk-Lipschizian,

thenBx−By ≤kx−y ≤kr0.This proves2.22. Further, sinceBsatisfiesH2, then the

Eberlein-ˇSmulian theorem10, page 430implies thatBKw_{is weakly compact. Consequently}

wBS≤kr0≤kr. 2.23

Lettingr → wSwe get

wBS≤kwS. 2.24

Now we are in a position to prove our next result.

Theorem 2.9. LetQandCbe closed, bounded, convex subsets of a Banach spaceX withQ⊆C.In addition, letUbe a weakly open subset ofQwith0∈U. Suppose thatA:Uw _→ _X_and_B_:_X _→ _X

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iAUw_{is relatively weakly compact,}

iiBis a nonexpansive map,

xnhas a convergent subsequence,

ivAxBx∈Cfor allx∈Uw_.

Then either

AB has a fixed point, 2.25

or

there is a pointu∈∂QUand λ∈0,1 with uλABu, 2.26

here∂QUis the weak boundary ofUinQ.

Proof. Letμ ∈ 0,1.We first show that the mappingFμ : μAμBisw-contractive with

constantμ. To see this let S be a bounded subset of Uw_. _{Using the homogeneity and the}

subadditivity of the De Blasi measure of weak noncompactness we obtain

wFμS

≤wμASμBS≤μwAS μwBS. 2.27

Keeping in mind thatAis weakly compact and usingLemma 2.8we deduce that

wFμS

≤μwS. 2.28

This proves thatFμisw-contractive with constantμ.Moreover, taking into account that 0∈U

and using assumptionivwe infer thatFμmapsUw intoC.Next suppose that2.26does

not occur andFμdoes not have a fixed point on∂QUotherwise we are finished since2.25

occurs. If there exists au∈∂QUandλ∈0,1withuλFμu,thenuλμAuλμBuwhich is

impossible sinceλμ∈0,1.ByTheorem 2.6there existsxμ∈Uwsuch thatxμFμxμ.Now

choose a sequence{μn}in0,1such thatμn → 1 and consider the corresponding sequence

{xn}of elements ofUwsatisfying

Fμnxn μnAxnμnBxnxn. 2.29

Using the fact thatAUw_{is weakly compact and passing eventually to a subsequence, we}

may assume that{Axn}converges weakly to somey∈Uw.Hence

I−μnB

xn y. 2.30

Since{xn}is a sequence inUw, then it is norm bounded and so is{Bxn}.Consequently _x_n₋_Bx_n₋

xn−μnBxn

1−μn

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As a result

xn−Bxn y. 2.32

somex∈Uw_._{The weak sequential continuity of}_A_and_B_{implies that}_x_Bx_Ax.

The following result is a sharpening of16, Theorem 2.4.

Theorem 2.10. LetX be a separable Banach space,Ca closed bounded convex subset ofX,andQ

a closed convex subset ofCwith 0 ∈ Q.Also, assume thatF : Q → Cis a weakly sequentially continuous and a weakly compact map. In addition, assume that the following conditions are satisfied:

ithere exists a weakly continuous retractionr:X → Q,

iithere exists aδ >0and a weakly compact setQδwithΩδ{x∈X : dx, Q≤δ} ⊆Qδ,

heredx, y x−y,

iiifor anyΩ{x∈X :dx, Q≤, 0< ≤δ},if{xj, λj}_j∞₁is a sequence inQ×0,1

with xj x ∈ ∂ΩQ, λj → λ, and x λFx,0 ≤ λ < 1,thenλjFxj ∈ Q forj

suﬃciently large, here∂ΩQis the weak boundary ofQrelative toΩ.

ThenFhas a fixed point inQ.

Proof. Consider

B{x∈X :xFrx}. 2.33

We first show that_{B /}∅.To see this, considerrF:Q → Q.ClearlyrFis weakly sequentially continuous, since F is weakly sequentially continuous and r is weakly continuous. Also

rFQis relatively weakly compact sinceFQis relatively weakly compact andris weakly continuous. Applying the Arino-Gautier Penot fixed point theorem19we infer that there existsy∈QwithrFy y.LetzFy,soFrz FrFy Fy z.Thusz ∈Band

B /∅.In additionBis weakly sequentially closed, sinceFris weakly sequentially continuous. Moreover, sinceB ⊆ FrB ⊆ FQ, thenBis relatively weakly compact. Now letx ∈ Bw_.

SinceBw_{is weakly compact, then there is a sequence}_x_n_{of elements of}_B_{which converges}

weakly to somex.SinceBis weakly sequentially closed, then x ∈ B.Thus,Bw _B._This

implies thatBis weakly compact. We now show thatB∩Q /∅.Suppose thatB∩Q∅.Then, sinceBis weakly compact andQis weakly closed, we have from20thatdB, Q>0.Thus there exists, 0< < δ,withΩ∩B∅,hereΩ{x∈X : dx, Q≤}.NowΩis closed

convex andΩ⊆Qδ.From our assumptions it follows thatΩis weakly compact. Also since

Xis separable, then the weak topology onΩis metrizable3,10; letd∗denote the metric.

Fori∈ {0,1, . . .}, let

Ui

x∈Ω:d∗x, Q<

i

. 2.34

For eachi∈ {0,1, . . .}fixed,Uiis open with respect todand soUiis weakly open inΩ.Also

Uw

i Udi

x∈Ω:d∗x, Q≤

i

, ∂ΩUi

x∈Ω:d∗x, Q

i

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Keeping in mind that Ω∩B ∅,Theorem 2.6guarantees that there existyi ∈ ∂ΩUi and

λi∈0,1withyiλiFryi.We now consider

D{x∈X:xλFrx, for someλ∈0,1}. 2.36

The same reasoning as above implies thatDis weakly compact. Then, up to a subsequence, we may assume thatλi → λ∗ ∈ 0,1and yi y∗ ∈ ∂ΩUi.HenceλiFryi λ∗Fry∗

and therefore y∗ λ∗Fry∗.Notice thatλ∗Fry∗/∈Qsince y∗ ∈ ∂ΩUi.Thusλ∗/1 since

B∩Q∅.From assumptioniiiit follows thatλiFryi∈Qforjsuﬃciently large, which is

a contradiction. ThusB∩Q /∅,so there existsx∈QwithxFrx,that is,xFx.

Remark 2.11. In16, Theorem 2.4,Qis assumed to be weakly compact.

Theorem 2.12. LetX be a separable Banach space,Ca closed bounded convex subset ofX,andQ

a closed convex subset ofCwith0 ∈ Q.Suppose thatA : Q → X andB : X → X are weakly sequentially continuous mappings satisfying the following:

iAQis relatively weakly compact,

iiBis a nonexpansive map andI−Bis injective,

iiiAQ⊆I−BC,

ivifxnis a sequence ofMsuch thatI−Bxnis weakly convergent, then the sequence

vthere exists a weakly continuous retractionr:X → Q,

vithere exists aδ >0and a weakly compact setQδwithΩδ{x∈X : dx, Q≤δ} ⊆Qδ,

heredx, y x−y,

viifor anyΩ{x∈X:dx, Q≤, 0< ≤δ},if{xj, λj}_j∞₁is a sequence inQ×0,1

withxj x∈∂ΩQ, λj → λandx∈λI−B −1

Ax,0≤λ <1(I−B−1Axis the inverse image ofAxunderI−B), then{λjI−B−1Axj} ⊆ Qforjsuﬃciently large,

here∂ΩQis the weak boundary ofQrelative toΩ.

ThenABhas a fixed point inQ.

Proof. Let us denote byFthe map which assigns to eachy∈Qthe pointFy∈Csuch that

I−BFy Ay.SinceI−Bis injective, thenF : Q → Cis well defined. Now we show thatF fulfills the conditions ofTheorem 2.10. We first claim thatFQis relatively weakly compact. Indeed let xn be a sequence of elements of Q.Since AQ is weakly compact,

then, by extracting a subsequence if necessary, we may assume thatAxnconverges weakly

to somex ∈ X.HenceI −BFxnconverges weakly tox.By assumptionivwe deduce

thatFxnhas a weakly convergent subsequence. This proves our claim. Now we show that

F:Q → Cis weakly sequentially continuous. To see this letxnnbe a sequence inQwhich

converges weakly tox.SinceFQis relatively weakly compact, there is a subsequencexnk

ofxnsuch that

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SinceI−BFxnk Axnk, then the weak sequential continuity ofAandBimpliesI −

Bz Ax.By the definition ofFwe haveI−BFx Ax.This giveszFxsinceI−B

is injective. Thus,

Fxnk Fx. 2.38

Now we show that

Fxn Fx. 2.39

Suppose the contrary, then there exists a weak neighborhoodNw_of_F_x_{and a subsequence}

xnj of xn such that Fxnj∈/Nw for all j ≥ 1. Sincexnj converges weakly to x,then

arguing as before we may extract a subsequence xn_jkof xnj such thatFxn_jk Fx.

This is not possible since Fxn_jk∈/Nw for allk ≥ 1. As a resultF is weakly sequentially

continuous. Now let Ω {x ∈ X : dx, Q ≤ , 0 < ≤ δ}, and let {xj, λj}∞_j₁ be a

sequence in Q×0,1 with xj x ∈ ∂ΩQ, λj → λ, and x λFx, 0 ≤ λ < 1.Then

I−Bx/λ I−BFx Ax.Hencex∈λI−B−1Ax.By assumptionviiwe infer that

{λjI−B−1Axj} ⊆Qforjsuﬃciently large. This implies thatλjFxj∈Qforjsuﬃciently

large. The result follows fromTheorem 2.10.

Theorem 2.13. LetMbe a nonempty bounded closed convex subset of a Banach spaceX.Suppose thatA, B:M → Xare two continuous mappings satisfying the following:

ithe set

F:x∈E:xBxAy for somey∈M 2.40

is relatively compact,

iiBis nonexpansive,

ivI−Bis injective and demiclosed,

ThenABhas at least one fixed point inM.

Proof. Let z ∈ AM. The map which assigns to each x ∈ M the valueBxz defines a nonexpansive mapping fromMintoM. In view ofTheorem 1.2, there exists a sequencexn

inMsuch that

I−Bxn−z−→0. 2.41

By assumptioniiiwe have thatxnhas a subsequence, sayxnk,which converges to some

x ∈ M.SinceI−Bis demiclosed, thenz I−Bx.Hencez ∈ I−BM.Consequently

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τy ∈ M such thatI−Bτy Ay.SinceI−Bis injective, thenτ : M → Mis well defined. Notice thatτM ⊆ F, then from assumptioniit follows thatτMis relatively compact. Now we show thatτ:M → Mis continuous. To see this letxnbe a sequence of

Mwhich converges to somex∈M.SinceτMis relatively compact, there is a subsequence

xnkofxnsuch that

τxnk−→u. 2.42

By definition ofτwe have

τxnk Axnk Bτxnk. 2.43

The continuity ofAandByieldsuBuAx.SinceI−Bτx AxandI−Bis injective, then we haveuτx.As a result

τxnk−→τx. 2.44

Now we show that

τxn−→τx. 2.45

Suppose the contrary, then there exists a > 0 and a subsequence xnj ofxn such that

τxnj−τx > for allj ≥ 1.Sincexnjconverges tox,then arguing as before we may

extract a subsequence xn_jk ofxnj such that τxn_jk → τx.This is not possible since

τxn_jk−τx > for allk ≥ 1. Consequently,τ : M → Mis continuous. Applying the

Schauder fixed point theorem we infer that there existsx∈Msuch that

xτx Bτx AxBxAx. 2.46

An easy consequence ofTheorem 2.13is the following.

Corollary 2.14. LetMbe a nonempty bounded closed convex subset of a reflexive Banach spaceX.

Suppose thatA, B:M → Xare two continuous mappings satisfying the following:

ithe set

F:x∈E :xBxAyfor some y∈M 2.47

iiBis nonexpansive,

iiiI−Bis injective and demi-closed,

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Proof. Keeping in mind that every bounded subset in a reflexive Banach space is relatively weakly compact, the result follows fromTheorem 2.13.

Corollary 2.15. LetMbe a nonempty bounded closed convex subset of a uniformly convex Banach spaceX.Suppose thatA, B:M → Xare two continuous mappings satisfying the following:

ithe set

F:x∈E :xBxAyfor some y∈M 2.48

iiBis nonexpansive andI−Bis injective,

Proof. Note that in a uniformly convex space we have thatBis nonexpansive implying that

I−Bis demiclosedsee4,15. Moreover, every uniformly convex Banach space is reflexive. The result follows fromCorollary 2.14.

Recall also the following definition.

Definition 2.16see2. LetXbe a Banach space,Ma nonempty subset ofXandT:M → X

be a mapping. We will callTa shrinking mapping if for allx, y∈Msuch that_{x /}ywe have

_Tx₋_Ty_<_x₋_y_. 2.49

Thus a shrinking mapping is nonexpansive but need not be a contraction mapping. IfT is a shrinking mapping, thenI−B−1 exists but need not be continuous. The following result is also an immediate consequence ofTheorem 2.13.

Corollary 2.17. LetMbe a nonempty bounded closed convex subset of a Banach spaceX.Suppose thatA, B:M → Xare two continuous mappings satisfying:

iThe set

F:x∈E:xBxAy for somey∈M 2.50

iiBis a shrinking map,

ivI−Bis demiclosed,

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The following example, which is taken from 21, shows that condition i in

Theorem 2.13andCorollary 2.15cannot be replaced by the compactness ofA.

Example 2.18. Let H be a separable Hilbert space andenn∈Z an orthonormal basis forH.

LetM be the closed unit ball ofH. DefineA andBas follows. For x n∞−∞xnen, Bx _∞

n−∞xnen1 and Ax 1− xe0.We have that Ais nonexpansive and maps bounded

sets into relatively compact sets. The mappingB is weakly continuous and nonexpansive. Moreover,I−Bis injective andABM⊆M.However,ABhas no fixed point inM.

In the case where A is compact and B is nonexpansive, we add an additional assumption onBto guarantee the existence of a fixed point for the sumABas follows.

Theorem 2.19. LetMbe a nonempty bounded closed convex subset of a Banach spaceX.Suppose thatA, B:M → Xare two continuous mappings satisfying the following:

iAis compact,

iiBis nonexpansive,

iiiifxnis a sequence ofMsuch thatI−Bxnis strongly convergent, then the sequence

xnhas a strongly convergent subsequence,

Proof. Arguing exactly in the same way as in the proof ofTheorem 2.5and using22, Theorem 2instead ofTheorem 1.7we get the desired result.

Now, we state the following fixed point theorem of Furi-Pera type.

Theorem 2.20. Let Q be a closed convex subset of a Banach space X and 0 ∈ Q. Suppose that

A:Q → XandB:X → Xare continuous mapping satisfyingthe following:

ithe set

F:x∈E:xBxAyfor some y∈Q 2.51

iiI−Bis injective,

iiiAQ⊆I−BX,

ivif{xj, λj}_j∞₁is a sequence of∂Q×0,1converging tox, λwithxλI−B−1Ax

and0≤λ <1,thenλjI−B−1Axj∈Qforjsuﬃciently large.

Proof. Lety ∈ Q be fixed. From assumptionsiiand iiiit follows that there is a unique

zy ∈X such thatAy I−Bzy.Let us denote byH:Q → Xthe map which assigns toy

the unique pointHy zy.Notice thatHQ⊆ F, then from assumptioniit follows that

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be a sequence ofQwhich converges to somex∈Q.SinceHQis relatively compact, then there is a subsequencexnkofxnsuch that

Hxnk−→u. 2.52

By definition ofHwe have

Hxnk Axnk BHxnk. 2.53

The continuity ofAandByieldsuBuAx.SinceI−BHx AxandI−Bis injective, then we haveuHx.As a result

Hxnk−→Hx. 2.54

The reasoning inTheorem 2.13shows that

Hxn−→Hx. 2.55

Now let{xj, λj}_j∞₁ be a sequence of∂Q×0,1converging to x, λwithx λHxand

0≤λ <1,thenxλI−B−1Ax.From our assumptions it follows thatλjI−B−1Axj∈Q

forj suﬃciently large. ThusλjHxj∈ Qforj suﬃciently large. NowTheorem 1.9implies

that there is anx∈QwithxHxand soxAxBHxAxBx.

Also, we give the following fixed point theorem of Leray-Schauder type.

Theorem 2.21. LetXbe a Banach space withC⊆Xclosed and convex. Assume thatUis a relatively open subset ofCwith0 ∈U.Suppose thatA:U → X andB :X → Xare continuous mapping satisfying the following:

ithe set

F:x∈E:xBxAyfor somey∈U 2.56

iiI−Bis injective,

iiiAU⊆I−BC.

Then either

ABhasafixed point inU, 2.57

or

there is a point u∈∂Uandλ∈0,1 withuλBu λ

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Proof. Lety ∈ Ube fixed. From assumptionsiiandiiiit follows that there is a unique

zy ∈ Csuch thatAy I−Bzy.Let us denote byH : U → Cthe map which assigns to

ythe unique pointHy zy.Notice thatHQ ⊆ F, then from assumptioniit follows

thatHQis relatively compact. The reasoning inTheorem 2.20shows thatHis continuous. NowTheorem 1.8implies that either there is au∈Usuch thatuHu,that is,uBuAu,

or there is a pointu ∈ ∂Uand λ ∈ 0,1withu λHu.Thusu λI−B−1Auwhich is equivalent touλBu

λ

λAu.

Theorem 2.22. LetUbe a bounded open convex set in a Banach spaceXwith0 ∈U.Suppose that

A:U → XandB:X → Xare continuous mapping satisfying the following:

iAUis compact andAis weakly-strongly continuous,

iiiifxnis a sequence ofUsuch thatI−Bxnis weakly convergent, then the sequence

xnhas a weakly convergent subsequence.

Then either

ABhas a fixed point inU, 2.59

or

there is a pointu∈∂Uandλ∈0,1withuλBuλAu. 2.60

Proof. Suppose that2.60does not occur, and letμ∈0,1.The mappingFμ:μAμBis the

sum of a compact map and a strict contraction. This implies thatFμis a condensing mapsee

23. ByTheorem 1.8we deduce that there is anxμ∈Usuch thatFμxμμAxμμBxμxμ.

Now, choose a sequence {μn} in0,1 such that μn → 1 and consider the corresponding

sequence{xn}of elements ofUsatisfying

μnAxnμnBxnxn. 2.61

Keeping in mind thatAUis weakly compact and passing eventually to a subsequence, we may assume that{Axn}converges weakly to somey∈U.Hence

I−μnB

xn y. 2.62

Since{xn}is a sequence inU, then it is norm bounded and so is{Bxn}.Consequently _x_n₋_Bx_n₋

xn−μnBxn

1−μn

Bxn −→0. 2.63

As a result

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somex∈U.SinceAis weakly-strongly continuous, then{Axnk}converges strongly toAx.

Consequently

Standard arguments yields

xnk−Bxnk −→Ax. 2.66

The demiclosedness ofI−BimpliesAxBxx.

Theorem 2.23. LetQbe a closed convex bounded set in a Banach spaceXwith0∈Q.Suppose that

A:Q → XandB:X → Xare continuous mapping satisfying the following:

iAQis compact andAis weakly-strongly continuous,

iiiifxnis a sequence ofQsuch thatI−Bxnis weakly convergent, then the sequence

ivif{xj, λj}_j∞₁is a sequence of∂Q×0,1converging tox, λwithxλAxλBxand

0≤λ <1,thenλjAxjλjBxj ∈Qforjsuﬃciently large.

Proof. Letμ∈ 0,1be fixed. SinceFμ : μAμBis the sum of a compact map and a strict

contraction, thenFμ is a condensing mapsee23. Now let{xj, λj}_j∞₁ be a sequence of

∂Q×0,1converging tox, λwithxλFμxand 0≤λ <1.ThenxμλAxμλBx.From

assumptionivit follows thatμλjAxj μλjBxj ∈ Qfor j suﬃciently large. Consequently

λjFμxj∈Qforjsuﬃciently large. ApplyingTheorem 1.9toFμwe deduce that there is an

xμ∈Qsuch thatFμxμ μAxμμBxμ xμ.Now, choose a sequence{μn}in0,1such that

μn → 1 and consider the corresponding sequence{xn}of elements ofQsatisfying

μnAxnμnBxnxn. 2.67

Keeping in mind thatAQis weakly compact and passing eventually to a subsequence, we may assume that{Axn}converges weakly to somey∈Q.Hence

I−μnB

xn y. 2.68

As inTheorem 2.22this implies that

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somex∈ Q.SinceAis weakly-strongly continuous, then{Axnk}converges strongly toAx.

Consequently

Standard arguments yield

xnk−Bxnk −→Ax. 2.71

The demiclosedness ofI−Bimplies thatAxBxx.

Definition 2.24see18, Definition 14. A mappingB : DB ⊆ X → X is said to be φ -expansive if there exists a functionφ:0,∞→ 0,∞satisfying the following:

iφ0 0,

iiφr>0 forr >0,

iiieither it is continuous or it is nondecreasing, such that, for everyx, y ∈DB,the inequalityBx−By ≥φx−yholds.

It was proved in18that ifMis a nonempty bounded closed and convex subset of a Banach spaceXandB:M → Xis a nonexpansive mapping such thatI−B:M → Xis

φ-expansive, thenI−B−1 :I−BM → Mexists and is continuous. In the light of this fact we obtain the following result which is an immediate consequence ofTheorem 2.19.

Corollary 2.25. LetMbe a nonempty bounded closed convex subset of a Banach spaceX.Suppose thatA, B:M → Xare two continuous mappings satisfying the following:

iAis compact,

iiBis nonexpansive andI−Bisφ-expansive,

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