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Common Fixed Point Results in Metric-Type Spaces


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Volume 2010, Article ID 978121,15pages doi:10.1155/2010/978121

Research Article

Common Fixed Point Results in

Metric-Type Spaces

Mirko Jovanovi ´c,


Zoran Kadelburg,


and Stojan Radenovi ´c


1Faculty of Electrical Engineering, University of Belgrade, Bulevar kralja Aleksandra 73,

11000 Beograd, Serbia

2Faculty of Mathematics, University of Belgrade, Studentski Trg 16, 11000 Beograd, Serbia 3Faculty of Mechanical Engineering, University of Belgrade, Kraljice Marije 16,

11120 Beograd, Serbia

Correspondence should be addressed to Stojan Radenovi´c,sradenovic@mas.bg.ac.rs

Received 16 October 2010; Accepted 8 December 2010

Academic Editor: Tomonari Suzuki

Copyrightq2010 Mirko Jovanovi´c 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.

Several fixed point and common fixed point theorems are obtained in the setting of metric-type spaces introduced by M. A. Khamsi in 2010.

1. Introduction

Symmetric spaces were introduced in 1931 by Wilson1, as metric-like spaces lacking the triangle inequality. Several fixed point results in such spaces were obtained, for example, in

2–4. A new impulse to the theory of such spaces was given by Huang and Zhang5when they reintroduced cone metric spaces replacing the set of real numbers by a cone in a Banach space, as the codomain of a metricsuch spaces were known earlier under the name of K-metric spaces, see6. Namely, it was observed in7that ifdx, yis a cone metric on the setXin the sense of5, thenDx, y dx, yis symmetric with some special properties, particularly in the case when the underlying cone is normal. The spaceX, Dwas then called the symmetric space associated with cone metric spaceX, d.

The last observation also led Khamsi8to introduce a new type of spaces which he called metric-type spaces, satisfying basic properties of the associated spaceX, D,Dd. Some fixed point results were obtained in metric-type spaces in the papers7–10.


2. Preliminaries

Let X be a nonempty set. Suppose that a mapping D : X ×X → 0,∞ satisfies the following:

s1Dx, y 0 if and only ifxy;

s2Dx, y Dy, x, for allx, yX.

ThenDis called asymmetriconX, andX, Dis called asymmetric space1.

LetEbe a real Banach space. A nonempty subsetP /{0}ofEis called aconeifP is closed, ifa, b∈R,a, b≥0, andx, yPimplyaxbyP, and ifP∩−P {0}. Given a cone

PE, we define the partial ordering with respect toP byx yif and only ifyxP. Let X be a nonempty set. Suppose that a mapping d : X × XE satisfies the following:

co 10 dx, yfor allx, yXanddx, y 0 if and only ifxy;

co 2dx, y dy, xfor allx, yX;

co 3dx, y dx, z dz, yfor allx, y, zX.

Thendis called acone metriconXandX, dis called acone metric space5.

IfX, dis a cone metric space, the functionDx, y dx, yis easily seen to be a symmetric onX7,8. Following7, the spaceX, Dwill then be calledassociated symmetric spacewith the cone metric spaceX, d. If the underlying conePofX, dis normali.e., if, for somek≥1, 0 x yalways impliesxky, the symmetricDsatisfies some additional properties. This led M.A. Khamsi to introduce a new type of spaces which he called metric type spaces. We will use the following version of his definition.

Definition 2.1 see8. LetX be a nonempty set, letK ≥ 1 be a real number, and let the functionD:X×X → Rsatisfy the following properties:

aDx, y 0 if and only ifxy;

bDx, y Dy, xfor allx, yX;

cDx, zKDx, y Dy, zfor allx, y, zX.

ThenX, D, Kis called ametric-type space.

Obviously, forK1, metric-type space is simply a metric space.

A metric type space may satisfy some of the following additional properties:

dDx, zKDx, y1 Dy1, y2 · · · Dyn, z for arbitrary points x, y1,

y2, . . . , yn, zX;

efunctionDis continuous in two variables; that is,

xn−→x, yn−→yinX, D, KimplyD

xn, yn

−→Dx, y. 2.1


Conditiondwas used instead ofcin the original definition of a metric-type space by Khamsi8. Both conditionsdandeare satisfied by the symmetricDx, y dx, y

which is associated with a cone metricdwith a normal cone see7–9. Note that the weaker version of propertye:

exnxandynxinX, D, Kimply thatDxn, yn → 0

is satisfied in an arbitrary metric type space. It can also be proved easily that the limit of a sequence in a metric type space is unique. Indeed, ifxnxandxnyinX, D, Kand

Dx, y ε >0, then

0≤Dx, yKDx, xn D

xn, y

< K ε

2K ε


ε 2.2

for sufficiently largen, which is impossible.

The notions such asconvergent sequence,Cauchy sequence,andcomplete spaceare defined in an obvious way.

We prove in this paper several versions of fixed point and common fixed point results in metric type spaces. We start with versions of classical Banach, Kannan and Zamfirescu results then proceed with Hardy-Rogers-type theorems, and with quasicontractions of ´Ciri´c and Das-Naik, and results for four mappings of Fisher and finally conclude with a result for strict contractions.

Recall also that a mappingf : XX is said to have property P11if Fixfn

Fixffor eachn∈N, where Fixfstands for the set of fixed points off.

A pointwX is called apoint of coincidenceof a pair of self-mapsf, g:XXand

uXis itscoincidence pointiffuguw. Mappingsfandgareweakly compatibleiffgu gfufor each of their coincidence pointsu12,13. The notion ofoccasionally weak compatibility

is also used in some papers, but it was shown in14that it is actually superfluous.

3. Results

We begin with a simple, but useful lemma.

Lemma 3.1. Let{yn}be a sequence in a metric type spaceX, D, Ksuch that

Dyn, yn1

λDyn−1, yn


for someλ,0< λ <1/K, and eachn1,2, . . . .Then{yn}is a Cauchy sequence inX, D, K.

Proof. Let m, n ∈ N and m < n. Applying the triangle-type inequality c to triples


Dym, yn

KDym, ym1

Dym1, yn

KDym, ym1

K2Dym1, ym2

Dym2, yn

≤ · · · ≤KDym, ym1

K2Dym1, ym2

· · ·

Knm−1Dyn−2, yn−1

Dyn−1, yn

KDym, ym1

K2Dym1, ym2

· · ·

Knm−1Dyn−2, yn−1

KnmDyn−1, yn



Now3.1andKλ <1 imply that

Dym, yn

KλmK2λm1· · ·Knmλn−1Dy0, y1

Kλm1 · · · Kλnm−1Dy0, y1



y0, y1

−→0 whenm−→ ∞.


It follows that{yn}is a Cauchy sequence.

Remark 3.2. If, instead of triangle-type inequality c, we use stronger conditiond, then a weaker condition 0< λ <1 can be used in the previous lemma to obtain the same conclusion. The proof is similar.

Next is the simplest: Banach-type version of a fixed point result for contractive mappings in a metric type space.

Theorem 3.3. LetX, D, Kbe a complete metric type space, and letf:XXbe a map such that for someλ,0< λ <1/K,

Dfx, fyλDx, y 3.4

holds for allx, yX. Thenfhas a unique fixed pointz, and for everyx0∈X, the sequence{fnx0}

converges toz.

Proof. Take an arbitraryx0∈Xand denoteynfnx0. Then

Dyn, yn1

Dfyn−1, fyn

λDyn−1, yn


for eachn1,2. . . .Lemma3.1implies that{yn}is a Cauchy sequence, and sinceX, D, K

is complete, there existszXsuch thatynzwhenn → ∞. Then

Dfz, zKDfz, fyn

Dyn1, z

Dz, yn

Dyn1, z

−→0, 3.6


If z1 is another fixed point of f, then Dz, z1 Dfz, fz1 ≤ λDz, z1 which is

possible only ifzz1.

Remark 3.4. In a standard way we prove that the following estimate holds for the sequence

{fnx 0}:

Dfmx0, z



x0, fx0


for eachm∈N. Indeed, form < n,

Dfmx0, z

KDfmx0, fnx0

Dfnx0, z



x0, fx0

KDfnx0, z



and passing to the limit whenn → ∞, we obtain estimate3.7. Note that continuity of functionDpropertyewas not used.

The first part of the following result was obtained, under the additional assumption of boundedness of the orbit, in8, Theorem 3.3.

Theorem 3.5. LetX, D, Kbe a complete metric type space. Letf : XX be a map such that for everyn ∈ Nthere isλn ∈ 0,1such thatDfnx, fnyλnDx, yfor all x, yX and let

limn→ ∞λn0. Thenfhas a unique fixed pointz. Moreover,fhas the property P.

Proof. Takeλsuch that 0 < λ < 1/K. Sinceλn → 0,n → ∞, there existsn0 ∈ Nsuch that

λn < λfor each nn0. Then Dfnx, fnyλDx, y for allx, yX whenever nn0.

In other words, for any mn0, g fm satisfies Dgx, gyλDx, y for allx, yX.

Theorem 3.3implies thatg has a unique fixed point, say z. Thenfmz z, implying that

fm1zfmfz fzandfzis a fixed point ofgfm. Since the fixed point ofgis unique, it follows thatfzzandzis also a fixed point off.

From the given condition we get thatDfx, f2x Dfx, ffxλ

1Dx, fxfor some

λ1 <1 and eachxX. This property, together with Fixf/∅, implies, in the same way as in

11, Theorem 1.1, thatfhas the property P.

Remark 3.6. If, in addition to the assumptions of previous theorem, we suppose that the series

n1λnconverges and thatDsatisfies propertyd, we can prove that, for eachxX, the

respective Picard sequence{fnx}converges to the fixed pointz.

Indeed, letm, n∈Nandn > m. Then

Dfmx, fnxKDfmx, fm1x· · ·Dfn−1xfnx

KDfmx, fmfx· · ·Dfn−1x, fn−1fx

Kλm· · ·λn−1D

x, fx−→0,


whenm → ∞due to the convergence of the given series. So,{fnx}is a Cauchy sequence


andgnx z whenn → ∞, but {fmnx}is a subsequence of {fnx}which is convergent;

hence, the latter converges toz.

The next is a common fixed point theorem of Hardy-Rogers typesee, e.g., 15 in metric type spaces.

Theorem 3.7. LetX, D, Kbe a metric type space, and letf, g:XXbe two mappings such that fXgXand one of these subsets ofXis complete. Suppose that there exist nonnegative coefficients ai,i1, . . . ,5such that

2Ka1 K1a2a3

K2Ka4a5<2 3.10

and that for allx, yX

Dfx, fya1D

gx, gya2D

gx, fxa3D

gy, fya4D

gx, fya5D

gy, fx


holds. Then f and g have a unique point of coincidence. If, moreover, the pair f, g is weakly compatible, thenfandghave a unique common fixed point.

Note that condition3.10is satisfied, for example, when5i1ai < 1/K2. Note also

that whenK1 it reduces to the standard Hardy-Rogers condition in metric spaces.

Proof. Suppose, for example, thatgXis complete. Take an arbitrary x0 ∈X and, using that

fXgX, construct a Jungck sequence{yn}defined byyn fxn gxn1,n0,1,2, . . . .Let

us prove that this is a Cauchy sequence. Indeed, using3.11, we get that

Dyn, yn1

Dfxn, fxn1


gxn, gxn1


gxn, fxn


gxn1, fxn1


gxn, fxn1


gxn1, fxn


yn−1, yn


yn−1, yn


yn, yn1


yn−1, yn1

a5·0 ≤a1a2D

yn−1, yn


yn, yn1


Dyn−1, yn

Dyn, yn1


yn−1, yn


yn, yn1



Similarly, we conclude that


Dfxn1, fxn


yn−1, yn


yn, yn1

. 3.13

Adding the last two inequalities, we get that

2Dyn, yn1


yn−1, yn


yn, yn1



that is,

Dyn, yn1

≤ 2a1a2a3Ka4Ka5


Dyn−1, yn

λDyn−1, yn

. 3.15

The assumption3.10implies that


λ 2a1a2a3Ka4Ka5

2−a2−a3−Ka4−Ka5 <




Lemma3.1implies that{yn} is a Cauchy sequence ingX and so there iszX such that

fxngxn1 → gzwhenn → ∞. We will prove thatfzgz.

Using3.11we conclude that

Dfxn, fz


gxn, gz


gxn, fxn


gz, fz


gxn, fz


gz, fxn


gxn, gz


gxn, fxn


Dgz, fxn

Dfxn, fz


Dgxn, fxn

Dfxn, fz


gz, fxn


gxn, gz


gxn, fxn


gz, fxn


fxn, fz




Dfz, fxn


gxn, gz


gz, fxn


fxn, fz


fxn, gxn

. 3.18

Adding up, one concludes that




gxn, gz


fxn, gxn

Ka2a3 a4a5D

fxn, gz



The right-hand side of the last inequality tends to 0 whenn → ∞. SinceKa2a3a4a5<

2Ka1K1a2a3K2Ka4a5<2because of3.10, it is 2−Ka2a3a4a5>0,

and so also the left-hand side tends to 0, andfxnfz. Since the limit of a sequence is


Suppose thatw1 fz1 gz1is another point of coincidence forfandg. Then3.11

implies that

Dw, w1 D

fz, fz1


gz, gz1


gz, fza3D

gz1, fz1


gz, fz1


gz1, fz

a1Dw, w1 a2·0a3·0a4Dw, w1 a5Dw1, w

a1a4a5Dw, w1.


Sincea1a4a5 <1because of3.10, the last relation is possible only ifw w1. So, the

point of coincidence is unique.

Iff, gis weakly compatible, then13, Proposition 1.12implies thatfandghave a unique common fixed point.

Taking special values for constantsai, we obtain as special cases Theorem3.3as well

as metric type versions of some other well-known theoremsKannan, Zamfirescu, see, e.g.,


Corollary 3.8. LetX, D, Kbe a metric type space, and letf, g:XXbe two mappings such that fXgXand one of these subsets ofXis complete. Suppose that one of the following three conditions holds:

1◦Dfx, fya1Dgx, gyfor somea1<1/Kand allx, yX;

2◦Dfx, fya2Dgx, fx Dgy, fyfor somea2<1/K1and allx, yX;

3◦Dfx, fya4Dgx, fy Dgy, fxfor somea4<1/K2Kand allx, yX.

Then f and g have a unique point of coincidence. If, moreover, the pair f, g is weakly compatible, thenfandghave a unique common fixed point.

Puttingg iX in Theorem3.7, we get metric type version of Hardy-Rogers theorem

which is obviously a special case forK1.

Corollary 3.9. LetX, D, Kbe a complete metric type space, and letf:XXsatisfy

Dfx, fya1D

x, ya2D

x, fxa3D

y, fya4D

x, fya5D

y, fx 3.21

for someai,i 1, . . . ,5 satisfying 3.10and for allx, yX. Thenf has a unique fixed point.


Proof. We have only to prove the last assertion. For arbitraryxX, we have that

Dfx, f2xDfx, ffx


x, fxa2D

x, fxa3D

fx, f2xa4D

x, f2xa5D

fx, fx


x, fx a3Ka4D

fx, f2x,


and similarly

Df2x, fxDffx, fxa1a3Ka5D

x, fx a2Ka5D

fx, f2x. 3.23

Adding the last two inequalities, we obtain

Dfx, f2x≤ 2a1a2a3Ka4a5


Dx, fxλDx, fx. 3.24

Similarly as in the proof of Theorem3.7, we get thatλ < 1/K < 1. Now11, Theorem 1.1 implies thatfhas property P.

Remark 3.10. If the metric-type function D satisfies both properties d and e, then it is easy to see that condition 3.10in Theorem3.7and the last corollary can be weakened to

a1a2a3Ka4a5<1. In particular, this is the case whenDx, y dx, yfor a cone

metricdonXover a normal cone, see7.

The next is a possible metric-type variant of a common fixed point result for ´Ciri´c and Das-Naik quasicontractions16,17.

Theorem 3.11. LetX, D, Kbe a metric type space, and letf, g :XX be two mappings such thatfXgXand one of these subsets ofX is complete. Suppose that there existsλ,0 < λ <1/K such that for allx, yX

Dfx, fyλmaxMf, g;x, y, 3.25


Mf, g;x, y

Dgx, gy, Dgx, fx, Dgy, fy,D

gx, fy

2K ,

Dgy, fx


. 3.26


Proof. Letx0 ∈ Xbe arbitrary and, using conditionfXgX, construct a Jungck sequence {yn}satisfying yn fxn gxn1,n 0,1,2, . . . .Suppose thatDyn, yn1 > 0 for each n

otherwise the conclusion follows easily. Using3.25we conclude that

Dyn1, yn

Dfxn1, fxn


Dgxn1, gxn

, Dgxn1, fxn1

, Dgxn, fxn


gxn1, fxn

2K ,

Dgxn, fxn1



Dyn, yn−1

, Dyn, yn1

, Dyn−1, yn


yn−1, yn1


λmax Dyn, yn−1



Dyn−1, yn

Dyn, yn1



If Dyn, yn−1 < Dyn1, yn, then Dyn, yn−1 < 1/2Dyn−1, yn Dyn, yn1 <

Dyn, yn1, and it would follow from 3.27 that Dyn1, ynλDyn1, yn which is

impossible sinceλ < 1. For the same reason the termDyn, yn1 was omitted in the last

row of the previous series of inequalities. Hence, Dyn, yn−1 > Dyn1, yn and 3.27

becomesDyn1, ynλDyn, yn−1. Using Lemma3.1, we conclude that{yn}is a Cauchy

sequence ingX. Supposing that, for example, the last subset ofX is complete, we conclude thatyn fxngxn1 → gzwhenn → ∞for somezX.

To prove thatfzgz, putxxnandyzin3.25to get

Dfxn, fz


Dgxn, gz

, Dgxn, fxn

, Dgz, fz,D

gxn, fz

2K ,

Dgz, fxn




Note that fxngz and gxngz when n → ∞, implying that Dgxn, fxn

KDgxn, gz Dgz, fxn → 0 whenn → ∞. It follows that the only possibilities are

the following:

1◦Dfxn, fzλDgz, fzλKDgz, fxn Dfxn, fz; in this case 1 −

λKDfxn, fzλKDgz, fxn → 0, and since 1− λK > 0, it follows that


2◦Dfxn, fzλ1/2KDgxn, fzλ/2Dgxn, fxn Dfxn, fz; in this case,

1−λ/2Dfxn, fzλ/2Dgxn, fxn → 0, so againfxnfz,n → ∞.


PuttinggiX, we obtain the first part of the following corollary.

Corollary 3.12. LetX, D, Kbe a complete metric type space, and letf :XXbe such that for someλ,0< λ <1/K, and for allx, yX,

Dfx, fyλmax

Dx, y, Dx, fx, Dy, fy,D

x, fy

2K ,

Dy, fx



holds. Thenfhas a unique fixed point, sayz. Moreover, the functionfis continuous at pointzand it has the property P.

Proof. Letxnzwhenn → ∞. Then

Dfxn, fz


Dxn, z, D

xn, fxn

, Dz, fz,D

xn, fz

2K ,

Dfxn, z



Dxn, z, D

xn, fxn


fxn, z




SinceDxn, z → 0 andDxn, fxnKDxn, z Dfz, fxn, the only possibility is that

Dfx,fzλKDxn, zDfz, fxn, implying that1−λKDfxn, fzλKDxn, z → 0,

n → ∞. Since 0< λK <1, it follows thatfxnfzz,n → ∞, andfis continuous at the


We will prove thatfsatisfies

Dfx, f2xhDx, fx 3.31

for someh, 0< h <1 and eachxX.

Applying3.29to the pointsxandfxfor anyxX, we conclude that

Dfx, f2xλmax

Dx, fx, Dx, fx, Dfx, f2x,D

x, f2x

2K ,

Dfx, fx



Dx, fx, Dfx, f2x,D

x, f2x




The following cases are possible:

1◦Dfx, f2xλDx, fx, and3.31holds withhλ;

2◦Dfx, f2xλDfx, f2x, which is only possible ifDfx, f2x 0 and then3.31

obviously holds.

3◦Dfx, f2xλ/2KDx, f2xλ/2KKDx, fx Dfx, f2x, implying that

1−λ/2Dfx, f2xλ/2Dx, fxandDfx, f2xhDx, fx, where 0< h


So, relation3.31holds for someh, 0 < h <1 and eachxX. Using the mentioned analogue of11, Theorem 1.1, one obtains thatfsatisfies property P.

We will now prove a generalization and an extension of Fisher’s theorem on four mappings from 18 to metric type spaces. Note that, unlike in 18, we will not use the case whenfandS, as well asgandT, commute, neither whenSandTare continuous. Also, functionDneed not be continuousi.e., we do not use propertye.

Theorem 3.13. LetX, D, Kbe a metric type space, and letf, g, S, T :XXbe four mappings such thatfXTXandgXSX, and suppose that at least one of these four subsets ofXis complete. Let

Dfx, gyλDSx, Ty 3.33

holds for someλ,0< λ <1/Kand allx, yX. Then pairsf, Sandg, Thave a unique common point of coincidence. If, moreover, pairsf, Sandg, Tare weakly compatible, thenf,g,S, andT have a unique common fixed point.

Proof. Letx0∈Xbe arbitrary and construct sequences{xn}and{yn}such that

fx2n−2Tx2n−1y2n−1, gx2n−1Sx2ny2n 3.34

forn1,2, . . . .We will prove that condition3.1holds forn1,2, . . . .Indeed,

Dy2n1, y2n2

Dfx2n, gx2n1

λDSx2n, Tx2n1 λD

y2n, y2n1


Dy2n3, y2n2

Dfx2n2, gx2n1

λDSx2n2, Tx2n1 λD

y2n2, y2n1



Using Lemma3.1, we conclude that{yn}is a Cauchy sequence. Suppose, for example, that

SX is a complete subset ofX. Then ynu Sv,n → ∞, for somev ∈ X. Of course,

subsequences{y2n−1}and{y2n}also converge tou. Let us prove thatfv u. Using3.33,

we get that

Dfv, uKDfv, gx2n−1

Dgx2n−1, u

KλDSv, Tx2n−1 D

gx2n−1, u

−→·00 0. 3.36

hencefvuSv. SinceufXTX, we get that there existswXsuch thatTwu. Let us prove that alsogwu. Using3.33, again we conclude that

Dgw, uKDgw, fx2n

Dfx2n, u

KλDSx2n, Tw D

fx2n, u

−→·00 0,



If now these pairs are weakly compatible, then for example,fufSv SfvSu z1 and gu gTw Tgw Tu z2for example, . Moreover, Dz1, z2 Dfu,gu

λDSu, Tu λDz1, z2and 0 < λ < 1 implies thatz1 z2. So, we have thatfu gu

Su Tu. It remains to prove that, for example, u gu. Indeed,Du, gu Dfv, guλDSv, Tu λDu, gu, implying thatu gu. The proof that this common fixed point of

f, g, S, andT is unique is straightforward.

We conclude with a metric type version of a fixed point theorem for strict contractions. The proof is similar to the respective proof, for example, for cone metric spaces in5. An example follows showing that additional condition ofsequential compactness cannot be omitted.

Theorem 3.14. Let a metric type spaceX, D, Kbe sequentially compact, and letDbe a continuous function (satisfying property (e)). Iff :XXis a mapping such that

Dfx, fy< Dx, y, forx, yX, x /y, 3.38

thenfhas a unique fixed point.

Proof. According to 9, Theorem 3.1, sequential compactness and compactness are equivalent in metric type spaces, and also continuity is a sequential property. The given condition3.38of strict continuity implies that a fixed point off is uniqueif it existsand that both mappingsfandf2are continuous. Letx0∈Xbe an arbitrary point, and let{xn}be

the respective Picard sequencei.e.,xnfnx0. Ifxnxn1for somen, thenxnis aunique

fixed point. Ifxn/xn1for eachn0,1,2, . . ., then

Dn:Dxn1, xn D

fn1x 0, fnx0

< Dfnx

0, fn−1x0

Dn−1. 3.39

Hence, there existsD∗, such that 0 ≤ D∗ ≤ Dn for each nandDnD∗,n → ∞. Using

sequential compactness of X, choose a subsequence{xni} of {xn} that converges to some

x∗∈Xwheni → ∞. The continuity offandf2implies that

fxni −→fx, f 2x


2xwheni−→ ∞, 3.40

and the continuity of the symmetricDimplies that

Dfxni, xni

−→Dfx, x, Df2xni, fxni

−→Df2x, fx∗ wheni−→ ∞. 3.41

It follows thatDfxni, xni DniDDfx, x∗. It remains to prove that fxx∗. If

fx/x∗, thenD>0 and3.41implies that

D∗ lim

i→ ∞Dni1ilim→ ∞D

f2xni, fxni

Df2x, fx< Dfx, xD. 3.42


Example 3.15. LetE R2,P {x, y E : x, y 0},X 1,, andd : X×X Ebe

defined bydx, y |xy|,|xy|. ThenX, dis a cone metric space over a normal cone with the normal constantK1see, e.g.,5. The associated symmetric is in this case simply the metricDx, y dx, y|xy|√2.

Letf:XXbe defined byfxx1/x. Then

Dfx, fyxy

1− 1


2<xy√2Dx, y 3.43

for all x, yX. Hence, f satisfies condition 3.38 but it has no fixed points. Obviously,

X, D, Kis notsequentiallycompact.


The authors are thankful to the Ministry of Science and Technological Development of Serbia.


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