Fixed Points for Multivalued Mappings and the Metric Completeness

Volume 2009, Article ID 972395,15pages doi:10.1155/2009/972395

Research Article

Fixed Points for Multivalued Mappings and

the Metric Completeness

S. Dhompongsa and H. Yingtaweesittikul

Department of Mathematics, Faculty of Science, Chiang Mai University, Chiang Mai 50200, Thailand

Correspondence should be addressed to S. Dhompongsa,[email protected]

Received 24 December 2008; Accepted 6 May 2009

Recommended by Wataru Takahashi

We consider the equivalence of the existence of fixed points of single-valued mappings and multivalued mappings for some classes of mappings by proving some equivalence theorems for the completeness of metric spaces.

Copyrightq2009 S. Dhompongsa and H. Yingtaweesittikul. 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

The Banach contraction principle 1states that for a complete metric space X, d, every

contractionT onX, that is, for somer ∈ 0,1,dTx, Ty ≤ rdx, yfor allx, y ∈ X, has a uniquefixed point.

Connell 2 gave an example of a noncomplete metric space X on which every contraction on X has a fixed point. Thus contractions cannot characterize the metric completeness ofX.

Theorem 1.1see3, Kannan. LetX, dbe a complete metric space. LetTbe a Kannan mapping onX, that is, for someα ∈0,1/2,dTx, Ty ≤αdx, Tx αdy, Tyfor allx, y ∈X.ThenT

has a (unique) fixed point.

Subrahmanyam 4 proved that Kannan mappings can be used to characterize the completeness of the metric. That is, a metric spaceXis complete if and only if every Kannan mapping onXhas a fixed point.

onto1/2,1by

θr ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩

1 if 0≤r≤√5−1/2,

1−rr−2 _if√_{5−1/2≤r≤2}−1/2_,

1r−1 if 2−1/2_{≤r <1.}

1.1

Theorem 1.2see5. For a metric spaceX, d, the following are equivalent:

iXis complete;

iievery mappingT onX such that there existsr ∈ 0,1,θrdx, Tx≤ dx, yimplies

dTx, Ty≤rdx, yfor allx, y∈Xhas a fixed point.

In 2008, Kikkawa and Suzuki6partially extendedTheorem 1.2to multivalued mappings.

Theorem 1.3see6. Define a strictly decreasing functionηfrom0,1onto1/2,1byηr

1/1r. LetX, dbe a complete metric space and letT :X → 2X\ ∅be a multivalued mapping with bounded and closed values. Assume that there existsr∈0,1such that

ηrdx, Tx≤dx, y impliesHTx, Ty ≤rdx, y , 1.2

for allx, y∈X, then there existsz∈Xsuch thatz∈Tz.

Obviously, the converse ofTheorem 1.3is valid since1/1r≤θrfor allr ∈0,1.

Mot¸ and Petrus¸el 7 proved the following theorem which is a generalization of Kikkawa and Suzuki Theorem.

Theorem 1.4see7. LetX, dbe a complete metric space and letT :X → 2X_{\∅be a multivalued}

mapping with closed values and satisfies the following: if for nonnegative numbersa, b, cwithab c∈0,1and for eachx, y∈Y, one has

1−b−c

1a dx, Tx≤d

x, y impliesHTx, Ty ≤adx, y bdx, Tx cdy, Ty . 1.3

ThenT has a fixed point.

In this paper, we will characterize the completeness of a metric space by the existence of fixed points for both single-valued and multivalued mappings. We first aim to extend, in

2. Preliminaries

LetX, dbe a complete metric space and letf : X → X be a mapping. We say that f is a Caristi mappingif there exists a lower semicontinuous functionϕ : X → R such thatϕis bounded below and

dx, fx ≤ϕx−ϕfx , forx∈X. 2.1

Recall that a mappingf islower semicontinuousif for eachx0 ∈ Xand for everyε >0, there

exists a neighborhoodUofx0such thatfx≥fx0−εfor allx∈U.

For a metric spaceX, d, letClXandCBXdenote, respectively, a collection of all nonempty closed subsets ofXand a collection of all nonempty bounded closed subsets ofX. LetHbe the Hausdorffmetric onCBX. That is, forA, B∈CBX,

HA, B max

sup

a∈A

da, B,sup

b∈B

db, A

, 2.2

wheredx, D:inf{dx, y:y∈D}is the distance from a pointxinXto a subsetDofX. The next theorem plays important roles in this paper.

Theorem 2.1see cf. 8. IfT is a mapping of a complete metric spaceX into the family of all nonempty closed subsets ofXandϕ:X → R∪ {∞}is a lower semicontinuous function such that

the following condition holds:

infdx, y ϕy :y∈Tx≤ϕx, for eachx∈X, 2.3

thenT has at least one fixed point.

3. Completeness and Single-valued Mappings

In 2008, Kikkawa and Suzuki9proved fixed point theorems for some generalized Kannan mappings. Letϕbe a nonincreasing function defined from0,1onto1/2,1by

ϕr ⎧ ⎨ ⎩

1 if 0≤r≤2−1/2_,

1r−1 if 2−1/2_{≤r <1.} 3.1

Theorem 3.1see9. LetX, dbe a complete metric space and let T be a mapping onX. Let

α∈0,1/2and putr:α/1−α∈0,1.Assume that

ϕrdx, Tx≤dx, y impliesdTx, Ty ≤αdx, Tx αdy, Ty , 3.2

Theorem 3.2see9. LetX, dbe a complete metric space and letTbe a mapping onX. Suppose that there existsr∈0,1such that

θrdx, Tx≤dx, y impliesdTx, Ty ≤rdx, Tx∨dy, Ty , 3.3

for allx, y∈X.ThenThas a unique fixed pointzandlimnTnxzholds for everyx∈X.

The above theorems inspire us to present another version ofTheorem 1.2. Before doing that we present first the following theorem. The proof of which is a mild modification of the proofs in5,9.

Theorem 3.3. LetX, dbe a complete metric space and let T be a mapping onX such that there existsr∈0,1,θrdx, Tx≤dx, yimpliesdTx, Ty≤rdx, Tx∨rdy, Ty∨rdx, yfor allx, y∈X, thenThas a fixed point.

Proof. Sinceθr≤1,θrdx, Tx≤dx, Txholds for everyx∈X, and thus

dTx, T2_{x≤rdx, Tx∨rdTx, T}2_{x, ∀x∈X. 3.4}

IfdTx, T2_{x≤rdTx, T}2_{xfor somex∈X, thenTxTTx,and we get a fixed pointTxof}

T.

Suppose now that

dTx, T2x≤rdx, Tx, ∀x∈X. 3.5

We fixx0∈Xand define a sequence{xn}inXbyxnTnx0.

Thendxn, xn1 ≤ rndx0, Tx0, and so∞_n1dxn, xn1 < ∞. Thus{xn}is a Cauchy

sequence. SinceXis complete,{xn}converges to some pointz∈X.

We show that

dz, Tx≤rdz, x∨rdx, Tx, for eachx∈X\ {z}. 3.6

Supposex /z.Sincexn → zasn → ∞, there existsn0∈Nsuch thatdxn, z≤1/3dz, x

for eachn≥n0.Observe that

θrdxn, Txn≤dxn, Txn dxn, xn1≤dxn, z dz, xn1

≤ 2

3dx, z≤dx, z−dxn, z≤dxn, x.

3.7

Hencedxn1, Tx dTxn, Tx≤rdxn, Txn∨rdx, Tx∨rdxn, xfor eachn≥n0.Letting

As in the proof of5, Theorem 1.2, we show thatTk_{z zfor somek from which it}

is proved thatzis a fixed point ofT.For this purpose, we assumeTk_{z /zfor allkand find a}

contradiction. We show, by induction, that

dTk1z, z≤rkdz, Tz, hold ∀k. 3.8

From3.6we have

dz, T2_{z≤rdz, Tz∨rdTz, T}2_z

≤rdz, Tz∨r2dz, Tz rdz, Tz.

3.9

Supposedz, Tk1_z≤rk_{dz, Tz. Thus}

dz, Tk2z≤rdTk1z, Tk2z∨rdz, Tk1z by3.6,

≤r·rk1_{dz, Tz∨r·r}k_{dz, Tz,}

rk1dz, Tz by3.5.

3.10

Thus3.8holds and now we find a contradiction in each of the following cases.

Case 10≤r <√5−1/2. We haver2_r−1<0.

AssumedT2_{z, z< dT}2_{z, T}3_zthen

dz, Tz≤dz, T2_zdT2_{z, Tz< dT}2_{z, T}3_zdT2_{z, Tz}

≤r2dz, Tz rdz, Tz< dz, Tz,

3.11

which is a contradiction. So

dT3z, Tz≤rdz, T2z∨dz, Tz∨dT2z, T3z

≤rrdz, Tz∨dz, Tz∨r2_{dz, Tzrdz, Tz.}

3.12

Hence dz, Tz ≤ dz, T3z dT3z, Tz ≤ r2dz, Tz rdz, Tz < dz, Tz, which is a

contradiction.

Case 2√5−1/2≤r <2−1/2_{. We have 2r}2_<1.

We show, by induction, that

Ifdz, T2_{z< θrdT}2_{z, T}3_z,then

dz, Tz≤dz, T2zdT2z, Tz< θrdT2z, T3zrdz, Tz

≤

1−r r2

r2dz, Tz rdz, Tz dz, Tz,

3.13

which is a contradiction. ThereforeθrdT2_{z, T}3_{z≤dz, T}2_z.

SupposeθrdTk_{z, T}k1_{z≤dz, T}k_{z. Thus}

dTz, Tk1z≤rdz, Tkz∨dz, Tz∨dTkz, Tk1z

≤rrk−1dz, Tz∨dz, Tz∨rkdz, Tzrdz, Tz.

3.14

Ifdz, Tk1_{z< θrdT}k1_{z, T}k2_z,then

dz, Tz≤dz, Tk1zdTk1z, Tz< θrdTk1z, Tk2zrdz, Tz

≤1−r

r2

rk1dz, Tz rdz, Tz<1−rrdz, Tz dz, Tz,

3.15

which is a contradiction. HenceθrdTk1_{z, T}k2_{z≤dz, T}k1_{z,and thus∗holds.}

For k ≥ 2, dz, Tz ≤ dz, Tk1z dTk1z, Tz ≤ rkdz, Tz rdz, Tz. We have

dz, Tz≤rdz, Tz< dz, Tz, which is a contradiction.

Case 3 2−1/2 _{≤ r < 1. We claim thatθrdx}

2n, x2n1 ≤ dx2n, zor θrdx2n1, x2n2 ≤

dx2n1, z.Suppose not,

dx2n, x2n1≤dx2n, z dz, x2n1

< θrdx2n, x2n1 dx2n1, x2n2

≤θr1rdx2n, x2n1

dx2n, x2n1,

3.16

which is a contradiction. So there exists a subsequence{nk}of{n}such thatθrdxnk, xnk1≤

dxnk, z:

dz, Tz lim

k dxnk1, Tz≤limk rdxnk, z∨rdxnk, Txnk∨rdz, Tz

rdz, Tz.

3.17

In fact the following theorem shows that the converse ofTheorem 3.3is valid.

Theorem 3.4. LetX, dbe a metric space. Then the following are equivalent:

iXis complete;

iifor eachr ∈0,1, every mappingT onXsuch that

1

1rdx, Tx≤d

x, y impliesdTx, Ty ≤rdx, y , 3.18

for allx, y∈Xhas a fixed point;

iiifor eachr :2α∈0,1, every mappingT onXsuch that

θrdx, Tx≤dx, y impliesdTx, Ty ≤αdx, Tx αdy, Ty , 3.19

for allx, y∈Xhas a fixed point;

ivfor eachr ∈0,1, every mappingT onXsuch that

θrdx, Tx≤dx, y impliesdTx, Ty ≤rdx, Tx∨rdy, Ty 3.20

for allx, y∈Xhas a fixed point;

vFor nonnegative numbersa, b, cwithabc∈0,1, every mappingTonXsuch that

1−b−c

1a dx, Tx≤d

x, y impliesdTx, Ty

≤adx, y bdx, Tx cdy, Ty ,

3.21

for allx, y∈Xhas a fixed point;

vifor eachr :2βγ∈0,1, every mappingTonXsuch that

θrdx, Tx≤dx, y impliesdTx, Ty

≤γdx, y βdx, Tx dy, Ty , 3.22

for allx, y∈Xhas a fixed point;

viifor eachr ∈0,1, every mappingT onXsuch that

θrdx, Tx≤dx, y impliesdTx, Ty

≤rdx, y ∨rdx, Tx∨rdy, Ty , 3.23

Proof. The implicationi⇒viiis exactlyTheorem 3.3.

vii⇒vi. LetTsatisfy3.22. We show thatTsatisfies3.23to obtain a fixed point for

T. Letθrdx, Tx≤dx, y,r:2βγ. ThusdTx, Ty≤γdx, y βdx, Tx dy, Ty≤

2βγmax{dx, y, dx, Tx, dy, Ty}rdx, y∨rdx, Tx∨rdy, Ty, and3.23holds. vi⇒v. Let T satisfy3.21. To show T satisfies3.22, letθrdx, Tx ≤ dx, y, aγ,bcβandr2βγ. Notice thatθr≥1/1r 1/12βγ 1/1abc≥

1−b−c/1a.Thus1−b−c/1adx, Tx ≤ θrdx, Tx ≤ dx, y.So we get

dTx, Ty ≤adx, y bdx, Tx cdy, Ty γdx, y βdx, Tx βdy, Ty, and3.22

holds.

v⇒ii. LetTsatisfy3.18. To showTsatisfies3.21, let1−b−c/1adx, Tx≤ dx, y, ar, bc0.Thus1/1rdx, Tx 1−b−c/1adx, Tx≤dx, y, and

sodTx, Ty≤rdx, y adx, y bdx, Tx cdy, Tyand3.21holds.

ii⇒i. Follows the same proof ofTheorem 1.2. Notice that, for 0≤r <2−1/2_,1/1

r≤θr.

vii⇒iv. LetT satisfy3.20. To showT satisfies3.23, letθrdx, Tx ≤ dx, y.

ThusdTx, Ty≤rdx, Tx∨rdy, Ty≤rdx, y∨rdx, Tx∨rdy, Ty.

iv⇒iii. LetT satisfy3.19. We showT satisfies3.20. Letθrdx, Tx ≤ dx, y, r2α. ThusdTx, Ty≤αdx, Tx αdy, Ty≤2αmax{dx, Tx, dy, Ty}rdx, Tx∨ rdy, Ty.

iii⇒i. We know that every Kannan mapping belongs to the class of mappings in iii. ThusXis complete by Subrahmanyam4.

4. Completeness and Multivalued Mappings

Inspired byTheorem 1.2andTheorem 1.3, we prove the following theorem for a larger class of mappings under some certain assumptions.

Theorem 4.1. LetX, dbe a metric space. Then the following are equivalent: iXis complete;

iifor eachr ∈ 0,1, every mappingT : X → ClXsuch thatθrdx, Tx ≤ dx, y

impliesHTx, Ty≤rdx, y∨rdx, Tx∨rdy, Ty,x, y∈Xand the functionx→ dx, Txis lower semicontinuous has a fixed point.

Observe thatTheorem 4.1is not covered byTheorem 3.4when considering as single-valued mappings.

Proof ofTheorem 4.1. i⇒ii. Letε > 0 be small enough so thatεr < 1 and defineϕx

1/εdx, Tx. For anyx ∈ X, we can find somefx ∈ Txsatisfyingdx, fx ≤ 1/ε rdx, Tx. To applyTheorem 2.1, it remains to show thatdx, fx≤ϕx−ϕfx, x∈X. We have dx, Tx ≤ dx, fx ≤ 1/θrdx, fx.Thus HTx, Tfx ≤ rdx, fx∨ rdfx, Tfx∨rdx, Tx.Note that

dfx, Tfx ≤HTx, Tfx

≤rdx, fx ∨rdfx, Tfx ∨rdx, Tx. 4.1

CaseKrdx, fx:dfx, Tfx≤rdx, fx.

CaseKrdx, Tx:dfx, Tfx≤rdx, Tx≤rdx, fx.

CaseKrdfx, Tfx:dfx, Tfx≤rdfx, Tfxwhich is impossible. Hence

dx, fx 1 ε

εrdx, fx −rdx, fx

≤ 1

ε

εr· 1

εrdx, Tx−d

fx, Tfx ϕx−ϕfx .

4.2

ThusThas a fixed point byTheorem 2.1. ii⇒i. SupposeXis not complete.

Define a functionfas in the proof ofTheorem 1.2and a mappingTas follows: for each x ∈ X, since fx > 0 and limnfun 0, there exists υ ∈ N satisfying

fuυ≤θrr/3rθrrfx.

We putTx{un:fun≤θrr/3rθrrfx}and writegx supy∈Txfy.

It is obvious thatgx ≤θrr/3rθrrfxfor allx∈X.Sincefy < fx,

for ally∈Tx, for allx∈X, thusx /∈Tx.That is,T does not have a fixed point. Note that

fx−fy ≤dx, y ≤fx fy , ∀y∈Tx. 4.3

We have

fx−gx≤dx, Tx≤fx gx, 4.4

HTx, Ty ≤gx gy . 4.5

Fix x, y ∈ X withθrdx, Tx ≤ dx, y. To show that the mappingT satisfies the condition inii, that is, for allx, y∈X,

θrdx, Tx≤dx, y impliesHTx, Ty ≤rdx, y ∨rdx, Tx∨rdy, Ty . 4.6

Observe that

dx, y ≥θrdx, Tx

≥θrfx−gx

≥θr

1− θrr 3rθrr

fx

θr3r

3rθrr

fx.

Case 1fy≥fx.

HTx, Ty ≤gx gy 3r

3

gx gy −r

3

gx gy by4.5

≤ 3r 3 ·

θrr

3rθrr

fx fy −r

3

gx gy r

3

fy −fx

≤ 3r 3 ·

r

3r

fx fy −r

3

gx gy r

3

fy −fx

≤ r

3dx, Tx

r

3d

y, Ty r

3d

x, y by4.4

≤rdx, Tx∨rdy, Ty ∨dx, y .

4.8

Case 2fydfx, dx, y< θrdy, Ty.

HTx, Ty ≤gx gy ≤ θrr

3rθrr

fx fy by4.5

≤ θrr

3rdx, Tx

θrr

3rθrrfx

≤ θrr

3rdx, Tx

θrr

3rθrr

3rθrr θr3r d

x, y

≤ r

3dx, Tx

r

3d

x, y ≤rdx, Tx∨dx, y ∨rdy, Ty .

4.9

Therefore4.6holds.

It remains to show that the mappingx→dx, Txis lower semicontinuous, that is,

∀ε >0,∃δ >0 such thatdy, Ty ≥dx, Tx−ε, ∀y∈Bdx, δ. 4.10

Suppose not, then there exists ε > 0 such that dyk, Tyk < dx, Tx − ε, for all yk ∈

Bdx,1/k, for each k. Since dyk, Tyk infuk

m∈Tykdyk, u

k

m,∃{ukm} ⊆ Tyk such that

limmdyk, ukm dyk, Tyk.We havedyk, ukm−1/k≤dyk, Tyk< dx, Tx−ε, for all large

m. Thus for eachk,dx, uk

m−dx, yk−1/k≤dyk, ukm−1/k≤dyk, Tyk< dx, Tx−ε,

for all largem.So for thosem, dx, uk

m−2/k ε < dx, Tx. Consequently,

fx− 2

kεlimm

dx, uk_m− 2 k ε

which impliest that

fx εlim

k

fx− 2 k ε

≤dx, Tx≤dx, um, ∀um∈Tx

≤lim

m dx, um

lim

n dx, un fx,

4.12

a contradiction. Thus the mappingx→dx, Txis lower semicontinuous.

The converse ofTheorem 1.4is also valid by following the same proof ofTheorem 1.2. Assuming that X is not complete, we find a fixed point free mapping T satisfying the condition inTheorem 1.4. Following the same proof ofTheorem 1.2by replacingηr/3ηr

byβ/1βwhereβ b∧c, we obtainfTx ≤ β/1βfxfor allx ∈XandT is fixed point free. We now verify the condition inTheorem 1.4forT.

Fixx, y∈Xwith1−b−c/1adx, Tx≤dx, y.We show that

HTx, Ty ≤adx, y bdx, Tx cdy, Ty . 4.13

Observe that

dx, y ≥ 1−b−c

1a dx, Tx≥

1−b−c

1a

fx−fTx

≥ 1−b−c 1a

1− β

1β

fx 1−b−c

1a

1 1β

fx.

4.14

Case 1fy≥fx.

HTx, Ty ≤fTx fTy

1β fTx fTy −βfTx fTy

≤1β · β

1β

fx fy −βfTx fTy

≤βdx, Tx βdy, Ty ≤adx, y bdx, Tx cdy, Ty .

Case 2fy≤fx, dx, y<1−b−c/1ady, Ty.

HTx, Ty ≤fTx fTy ≤ β

1β

fx fy

≤βdx, Tx β

1βfx≤βdx, Tx

β1β

1β

1a

1−b−cd

x, y

≤βdx, Tx β1a

1−b−c

1−b−c

1a d

y, Ty βdx, Tx βdy, Ty

≤adx, y bdx, Tx cdy, Ty .

4.16

Therefore4.13holds, and the proof of the converse ofTheorem 1.4is complete.

Moreover, by following the proof ofTheorem 1.4, we can partially extend the class of mappings and still obtain their fixed points. Notice that1−2β/1γ≤1/2βγ1.

Theorem 4.2. LetX, dbe a metric space. Then the following are equivalent: iXis complete.

iievery mappingT :X → ClXsuch that for eachβ, γ ∈Rwith2βγ ∈0,1and for

eachx, y∈X,

1

2βγ1dx, Tx≤d

x, y impliesHTx, Ty ≤βdx, Tx βdy, Ty γdx, y ,

4.17

has a fixed point.

Proof. i⇒ii. Following the same proof ofTheorem 1.4by replacingη: 1−b−c/1a

in its proof byη:1/2βγ1.Thus we obtain a sequence{xn}such that

1x_n1 ∈Txn, for eachn∈Nand;

2dxn, xn1≤kβγ/1−kβndx0, x1,forn∈N.

Choosekso that 1< k <1/γ2βand therefore 0≤kβγ/1−kβ<1. We see that the sequence{xn}is Cauchy in X, and so{xn}converges to somez ∈ X.We showdz, Tx ≤

γdz, x βdx, Tx,for eachx∈X\ {z}.

Suppose x /z.Sincexn → zasn → ∞, there existsn0 ∈ N such thatdxn, z ≤

1/3dz, xfor each n ≥ n0.We haveηdxn, Txn ≤ dxn, Txn ≤ dxn, xn1 ≤ dxn, z

dz, xn1≤ 2/3dx, z≤ dx, z−dxn, z≤ dxn, x.Hencedxn1, Tx ≤ HTxn, Tx ≤

βdxn, Txn βdx, Tx γdxn, xfor each n ≥ n0.Letting n → ∞, we get dz, Tx ≤

γdz, x βdx, Tx,for allx /zas desired.

Next, we showHTx, Tz≤βdx, Txβdz, Tzγdx, z,for allx∈X.Forx /z,we obtain for eachn∈N, yn∈Txsuch thatdz, yn≤dz, Tx 1/ndx, z.Clearlydx, Tx≤

dx, yn≤dx, zdz, yn≤dx, zdz, Tx1/ndx, z≤1γ1/ndx, zβdx, Tx,

Finally, we obtain

dz, Tz lim

n dxn1, Tz≤limn HTxn, Tz,

≤lim

n

βdxn, Txn βdz, Tz γdxn, z 0.

4.18

ThuszTzandThas a fixed point.

ii⇒i. Let β 0, and γ r, we have 1/r 1dx, Tx ≤ dx, y implying

HTx, Ty≤rdx, y. HenceXis complete by the converse ofTheorem 1.4.

5. Caristi Set-Valued Mappings

In 2008, ´Ciri´c10proved the following fixed point theorems.

Theorem 5.110. LetX, dbe a complete metric space and letT : X → ClX.If there exist constantsb, c ∈0,1, c < b,such that for anyx ∈Xthere isy ∈Txsatisfying the following two conditions:

bdx, y ≤dx, Tx, dy, Ty ≤cdx, y . 5.1

ThenT has a fixed point inXprovided a functionfx dx, Txis lower semicontinuous.

Theorem 5.210. LetX, dbe a complete metric space andT : X → ClX.If there exists a functionϕ:0,∞ → 0,1satisfying

lim

r→tsupϕr<1, for each t∈0,∞, 5.2

and such that for anyx∈Xthere isy∈Txsatisfying the following two conditions:

dx, y ≤2−ϕdx, y dx, Tx, dy, Ty ≤ϕdx, y dx, y . 5.3

ThenT has a fixed point inXprovided a functionfx dx, Txis lower semicontinuous.

We give a simple proof of each of these theorems.

Proof ofTheorem 5.1. Define a lower semi-continuous functionϕbyϕx 1/b−cdx, Tx.

For anyx∈X,we can find somefx∈Txsatisfying

We show thatdx, fx≤ϕx−ϕfx, x∈X. Letx∈X. Clearly,

dx, fx 1 b−c

bdx, fx −cdx, fx ≤ 1 b−c

bdx, fx −dfx, Tfx

≤ 1

b−c

dx, Tx−dfx, Tfx ϕx−ϕfx .

5.5

HenceThas a fixed point byTheorem 2.1.

Proof ofTheorem 5.2. Letkinfϕr−12/2−ϕr>0 andψx 1/kdx, Tx. For each x∈X,there existsfx∈Txsuch that

dx, fx ≤2−ϕdx, fx dx, Tx, dfx, Tfx ≤ϕdx, fx dx, fx .

5.6

Furthermore,dx, fx≤ψx−ψfx, x∈X. Indeed,

dx, fx 1 k

ϕdx, fx k dx, fx −ϕdx, fx dx, fx

≤ 1

k

ϕdx, fx

ϕdx, fx −1 2 2−ϕdx, fx

dx, fx

−ϕdx, fx dx, fx

≤ 1

k

1

2−ϕdx, fx

dx, fx −ϕdx, fx dx, fx

≤ 1

k

dx, Tx−dfx, Tfx ψx−ψfx .

5.7

ThusThas a fixed point byTheorem 2.1.

Acknowledgment

The authors would like to thank the Thailand Research Fund grant BRG4780016 for its support.

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