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.
In 2008 Suzuki5introduced a new type of mappings and presented a generalization of the Banach contraction principle in which the completeness can also be characterized by the existence of fixed points of these mappings. Define a nonincreasing functionθfrom0,1
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≤d
x, y impliesH
Tx, Ty ≤rd
x, y , 1.2
for allx, y∈X, then there existsz∈Xsuch thatz∈Tz.
Obviously, the converse ofTheorem 1.3is valid since 1/1r≤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 impliesH
Tx, Ty ≤ad
x, y bdx, Tx cd
y, 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 Section 3, the Suzuki’s result Theorem 1.2 to more general classes of mappings. We then consider multivalued mappings inSection 4. We also show in this section that the converse ofTheorem 1.4is true.
2. Preliminaries
LetX, dbe a complete metric space and letf : X → X be a mapping. We say that f is a Caristi mapping if there exists a lower semicontinuous functionϕ : X → R such thatϕis bounded below and
d
x, fx ≤ϕx−ϕ
fx , forx∈X. 2.1
Recall that a mappingf is lower semicontinuous if 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∈Ada, 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. If T is a mapping of a complete metric space X into the family of all nonempty closed subsets of X andϕ:X → R∪ {∞}is a lower semicontinuous function such that the following condition holds:
inf d
x, y ϕ
y :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≤d
x, y impliesd
Tx, Ty ≤αdx, Tx αd
y, Ty , 3.2
for allx, y∈X, thenT has a unique fixed pointzand limnTnxzholds for everyx∈X.
Theorem 3.2see9. LetX, dbe a complete metric space and letTbe a mapping onX. Suppose that there existsr∈0,1such that
θrdx, Tx≤d
x, y impliesd
Tx, Ty ≤r
dx, Tx∨d
y, Ty , 3.3
for allx, y∈X.ThenThas a unique fixed pointzand limnTnxzholds 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
d
Tx, T2x
≤rdx, Tx∨rd
Tx, T2x
, ∀x∈X. 3.4
IfdTx, T2x≤rdTx, T2xfor somex∈X, thenTxTTx,and we get a fixed pointTxof T.
Suppose now that
d
Tx, 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 n → ∞we getdz, Tx≤rdz, x∨rdx, Tx,for allx /zand we obtain3.6.
As in the proof of5, Theorem 1.2, we show thatTkz zfor somek from which it is proved thatzis a fixed point ofT.For this purpose, we assumeTkz /zfor allkand find a contradiction. We show, by induction, that
d
Tk1z, z
≤rkdz, Tz, hold ∀k. 3.8
From3.6we have
d z, T2z
≤rdz, Tz∨rd
Tz, T2z
≤rdz, Tz∨r2dz, Tz rdz, Tz.
3.9
Supposedz, Tk1z≤rkdz, Tz. Thus
d
z, Tk2z
≤rd
Tk1z, Tk2z
∨rd
z, Tk1z
by3.6,
≤r·rk1dz, Tz∨r·rkdz, 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 haver2r−1<0.
AssumedT2z, z< dT2z, T3zthen dz, Tz≤d
z, T2z d
T2z, Tz
< d
T2z, T3z d
T2z, Tz
≤r2dz, Tz rdz, Tz< dz, Tz, 3.11
which is a contradiction. So
d
T3z, Tz
≤r d
z, T2z
∨dz, Tz∨d
T2z, T3z
≤r
rdz, Tz∨dz, Tz∨r2dz, Tz
rdz, 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 2r2<1.
We show, by induction, that
θrd
Tkz, Tk1z
≤d z, Tkz
, ∀k≥2 ∗
Ifdz, T2z< θrdT2z, T3z,then
dz, Tz≤d z, T2z
d
T2z, Tz
< θrd
T2z, T3z
rdz, Tz
≤ 1−r
r2
r2dz, Tz rdz, Tz dz, Tz, 3.13
which is a contradiction. ThereforeθrdT2z, T3z≤dz, T2z.
SupposeθrdTkz, Tk1z≤dz, Tkz. Thus
d
Tz, Tk1z
≤r d
z, Tkz
∨dz, Tz∨d
Tkz, Tk1z
≤r
rk−1dz, Tz∨dz, Tz∨rkdz, Tz
rdz, Tz.
3.14
Ifdz, Tk1z< θrdTk1z, Tk2z,then
dz, Tz≤d
z, Tk1z d
Tk1z, Tz
< θrd
Tk1z, Tk2z
rdz, Tz
≤ 1−r
r2
rk1dz, Tz rdz, Tz<1−rrdz, Tz dz, Tz,
3.15
which is a contradiction. HenceθrdTk1z, Tk2z≤dz, Tk1z,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 thatθrdx2n, 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≤lim
k rdxnk, z∨rdxnk, Txnk∨rdz, Tz
rdz, Tz. 3.17
ThusTzz, which is a contradiction.
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 impliesd
Tx, Ty ≤rd
x, y , 3.18
for allx, y∈Xhas a fixed point;
iiifor eachr :2α∈0,1, every mappingT onXsuch that θrdx, Tx≤d
x, y impliesd
Tx, Ty ≤αdx, Tx αd
y, Ty , 3.19
for allx, y∈Xhas a fixed point;
ivfor eachr ∈0,1, every mappingT onXsuch that θrdx, Tx≤d
x, y impliesd
Tx, Ty ≤rdx, Tx∨rd
y, 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 impliesd Tx, Ty
≤ad
x, y bdx, Tx cd y, Ty ,
3.21
for allx, y∈Xhas a fixed point;
vifor eachr :2βγ∈0,1, every mappingTonXsuch that θrdx, Tx≤d
x, y impliesd Tx, Ty
≤γd
x, y β
dx, Tx d
y, Ty , 3.22
for allx, y∈Xhas a fixed point;
viifor eachr ∈0,1, every mappingT onXsuch that θrdx, Tx≤d
x, y impliesd Tx, Ty
≤rd
x, y ∨rdx, Tx∨rd
y, Ty , 3.23
for allx, y∈Xhas a fixed point.
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
d
fx, Tfx ≤H
Tx, Tfx
≤rd
x, fx ∨rd
fx, Tfx ∨rdx, Tx. 4.1
LetK:rdx, fx∨rdfx, Tfx∨rdx, Tx.
CaseKrdx, fx:dfx, Tfx≤rdx, fx.
CaseKrdx, Tx:dfx, Tfx≤rdx, Tx≤rdx, fx.
CaseKrdfx, Tfx:dfx, Tfx≤rdfx, Tfxwhich is impossible.
Hence
d
x, fx 1 ε
εrd
x, fx −rd
x, fx
≤ 1 ε
εr· 1
εrdx, Tx−d
fx, Tfx ϕx−ϕ fx .
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−f y ≤d
x, y ≤fx f
y , ∀y∈Tx. 4.3
We have
fx−gx≤dx, Tx≤fx gx, 4.4
H
Tx, Ty ≤gx g
y . 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≤d
x, y impliesH
Tx, Ty ≤rd
x, y ∨rdx, Tx∨rd
y, Ty . 4.6
Observe that
d
x, y ≥θrdx, Tx
≥θr
fx−gx
≥θr
1− θrr
3rθrr
fx
θr3r 3rθrr
fx.
4.7
Case 1fy≥fx.
H
Tx, Ty ≤gx g
y 3r 3
gx g y
−r 3
gx g y
by4.5
≤ 3r
3 · θrr
3rθrr
fx f y
−r 3
gx g y
r 3
f
y −fx
≤ 3r 3 · r
3r
fx f y
−r 3
gx g y
r 3
f
y −fx
≤ r
3dx, Tx r 3d
y, Ty r 3d
x, y by4.4
≤rdx, Tx∨rd
y, Ty ∨d x, y .
4.8
Case 2fydfx, dx, y< θrdy, Ty.
H
Tx, Ty ≤gx g
y ≤ θrr
3rθrr
fx f y
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∨d
x, y ∨rd y, Ty .
4.9
Therefore4.6holds.
It remains to show that the mappingx→dx, Txis lower semicontinuous, that is,
∀ε >0,∃δ >0 such thatd
y, 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 infukm∈Tykdyk, ukm,∃{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, ukm−dx, yk−1/k≤dyk, ukm−1/k≤dyk, Tyk< dx, Tx−ε, for all largem.So for thosem, dx, ukm−2/k ε < dx, Tx. Consequently,
fx− 2
kεlim
m
d
x, ukm
− 2 k ε
≤dx, Tx, ∀k, 4.11
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
H
Tx, Ty ≤ad
x, y bdx, Tx cd
y, Ty . 4.13
Observe that
d
x, 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.
H
Tx, Ty ≤fTx f Ty
1β fTx f Ty
−β
fTx f Ty
≤
1β · β 1β
fx f y
−β
fTx f Ty
≤βdx, Tx βd
y, Ty ≤ad
x, y bdx, Tx cd y, Ty .
4.15
Case 2fy≤fx, dx, y<1−b−c/1ady, Ty.
H
Tx, Ty ≤fTx f
Ty ≤ β 1β
fx f y
≤βdx, Tx β
1βfx≤βdx, Tx β 1β 1β
1a 1−b−cd
x, y
≤βdx, Tx β1a 1−b−c
1−b−c 1a d
y, Ty βdx, Tx βd y, Ty
≤ad
x, y bdx, Tx cd y, 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 ∈Rwith 2βγ ∈0,1and for eachx, y∈X,
1
2βγ1dx, Tx≤d
x, y impliesH
Tx, Ty ≤βdx, Tx βd
y, Ty γd x, 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
1xn1 ∈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, for alln ∈N.Hence, asn → ∞we get1−βdx, Tx≤ 1γdx, zand soηdx, Tx ≤ dx, zimplying thatHTx, Tz≤βdx, Tx βdz, Tz γdx, zforx∈X.
Finally, we obtain
dz, Tz lim
n dxn1, Tz≤lim
n 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:
bd
x, y ≤dx, Tx, d
y, Ty ≤cd
x, 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
rlim→tsupϕr<1, for each t∈0,∞, 5.2
and such that for anyx∈Xthere isy∈Txsatisfying the following two conditions:
d
x, y ≤ 2−ϕ
d x, y
dx, Tx, d
y, Ty ≤ϕ d
x, y d
x, 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
bd
x, fx ≤dx, Tx, d
fx, Tfx ≤cd
x, fx . 5.4
We show thatdx, fx≤ϕx−ϕfx, x∈X. Letx∈X. Clearly,
d
x, fx 1 b−c
bd
x, fx −cd
x, fx
≤ 1 b−c
bd
x, fx −d
fx, Tfx
≤ 1 b−c
dx, Tx−d
fx, Tfx
ϕx−ϕ fx .
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
d
x, fx ≤ 2−ϕ
d
x, fx
dx, Tx, d
fx, Tfx ≤ϕ d
x, fx d
x, fx . 5.6
Furthermore,dx, fx≤ψx−ψfx, x∈X. Indeed,
d
x, fx 1 k
ϕ d
x, fx k d
x, fx −ϕ d
x, fx d
x, fx
≤ 1 k
ϕ
d
x, fx
ϕ d
x, fx
−1 2 2−ϕ
d
x, fx
d x, fx
−ϕ d
x, fx d
x, fx
≤ 1 k
1
2−ϕ d
x, fx
d
x, fx −ϕ d
x, fx d
x, fx
≤ 1 k
dx, Tx−d
fx, Tfx
ψx−ψ fx .
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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