Common Fixed Point Results in Metric-Type Spaces

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,[email protected] 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.

In this paper we prove several other fixed point and common fixed point results in metric-type spaces. In particular, metric-type versions of very well-known results of Hardy- Rogers, ´Ciri´c, Das-Naik, Fisher, and others are obtained.

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, y∈X.

ThenDis called a symmetric onX, andX, Dis called a symmetric space1.

LetEbe a real Banach space. A nonempty subsetP /{0}ofEis called a cone ifP is closed, ifa, b∈R,a, b≥0, andx, y∈Pimplyaxby∈P, and ifP∩−P {0}. Given a cone P ⊂E, we define the partial ordering with respect toP byx yif and only ify−x∈P.

Let X be a nonempty set. Suppose that a mapping d : X × X → E satisfies the following:

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

co 2dx, y dy, xfor allx, y∈X;

co 3dx, y dx, z dz, yfor allx, y, z∈X.

Thendis called a cone metric onXandX, dis called a cone 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 called associated symmetric space with the cone metric spaceX, d. If the underlying conePofX, dis normali.e., if, for somek≥1, 0 x yalways impliesx ≤ky, 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, y∈X;

cDx, z≤KDx, y Dy, zfor allx, y, z∈X.

ThenX, D, Kis called a metric-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, z ≤ KDx, y1 Dy1, y2 · · · Dyn, z for arbitrary points x, y1, y₂, . . . , y_n, z∈X;

efunctionDis continuous in two variables; that is,

x_n−→x, y_n−→yinX, D, KimplyD x_n, y_n

−→D x, y

. 2.1

The last condition is in the theory of symmetric spaces usually called “propertyHE”.

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:

exn → xandyn → xinX, 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, ifxn → xandxn → yinX, D, Kand Dx, y ε >0, then

0≤D x, y

≤K

Dx, xn D x_n, y

< K ε 2K ε

2K

ε 2.2

for sufficiently largen, which is impossible.

The notions such as convergent sequence, Cauchy sequence, and complete space are 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 : X → X is said to have property P11if Fixfⁿ Fixffor eachn∈N, where Fixfstands for the set of fixed points off.

A pointw ∈X is called a point of coincidence of a pair of self-mapsf, g:X → Xand u∈Xis its coincidence point iffuguw. Mappingsfandgare weakly compatible iffgu gfufor each of their coincidence pointsu12,13. The notion of occasionally 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

D

y_n, y_n1

≤λD

y_n−1, y_n

3.1

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 {ym, y_m1, y_n},{y_m1, y_m2, y_n}, . . . ,{y_n−2, y_n−1, y_n}we obtain

D ym, yn

≤K D

ym, y_m1 D

y_m1, yn

≤KD

y_m, y_m1 K²

D

y_m1, y_m2 D

y_m2, y_n

≤ · · · ≤KD

ym, y_m1

K²D

y_m1, y_m2 · · · K^n−m−1

D

y_n−2, y_n−1 D

y_n−1, y_n

≤KD

ym, y_m1

K²D

y_m1, y_m2 · · · K^n−m−1D

y_n−2, y_n−1

K^n−mD

y_n−1, y_n .

3.2

Now3.1andKλ <1 imply that D

y_m, y_n

≤

Kλ^mK²λ^m1· · ·K^n−mλⁿ⁻¹ D

y₀, y₁ Kλ^m

1 Kλ · · · Kλ^n−m−1 D

y₀, y₁

≤ Kλ^m 1−KλD

y0, y1

−→0 whenm−→ ∞.

3.3

It follows that{yn}is a Cauchy sequence.

Remark 3.2. If, instead of triangle-type inequalityc, 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:X → Xbe a map such that for someλ, 0< λ <1/K,

D fx, fy

≤λD x, y

3.4 holds for allx, y∈X. Thenfhas a unique fixed pointz, and for everyx₀∈X, the sequence{fⁿx₀} converges toz.

Proof. Take an arbitraryx0∈Xand denoteynfⁿx0. Then D

y_n, y_n1 D

fy_n−1, fy_n

≤λD

y_n−1, y_n

3.5 for eachn1,2. . . .Lemma3.1implies that{yn}is a Cauchy sequence, and sinceX, D, K is complete, there existsz∈Xsuch thaty_n → zwhenn → ∞. Then

D fz, z

≤K D

fz, fyn

D

y_n1, z

≤K λD

z, yn

D

y_n1, z

−→0, 3.6 whenn → ∞. Hence,Dfz, z 0 andzis a fixed point off.

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 {fⁿx0}:

D

f^mx0, z

≤ K²λ^m 1−KλD

x0, fx0

3.7

for eachm∈N. Indeed, form < n, D

f^mx₀, z

≤K D

f^mx₀, fⁿx₀ D

fⁿx₀, z

≤ K²λ^m 1−KλD

x₀, fx₀ KD

fⁿx₀, z ,

3.8 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 : X → X be a map such that for everyn ∈ Nthere isλn ∈ 0,1such thatDfⁿx, fⁿy ≤ λnDx, yfor all x, y ∈ X and let lim_{n→ ∞}λ_n0. 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 n ≥ n0. Then Dfⁿx, fⁿy ≤ λDx, y for allx, y ∈ X whenever n ≥ n0. In other words, for any m ≥ n₀, g f^m satisfies Dgx, gy ≤ λDx, y for allx, y ∈ X.

Theorem 3.3implies thatg has a unique fixed point, say z. Thenf^mz z, implying that f^m1zf^mfz fzandfzis a fixed point ofgf^m. Since the fixed point ofgis unique, it follows thatfzzandzis also a fixed point off.

From the given condition we get thatDfx, f²x Dfx, ffx≤λ1Dx, fxfor some λ1 <1 and eachx∈X. 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 eachx ∈ X, the respective Picard sequence{fⁿx}converges to the fixed pointz.

Indeed, letm, n∈Nandn > m. Then D

f^mx, fⁿx

≤K D

f^mx, f^m1x

· · ·D

fⁿ⁻¹xfⁿx K

D

f^mx, f^mfx

· · ·D

fⁿ⁻¹x, fⁿ⁻¹fx

≤Kλm· · ·λ_n−1D x, fx

−→0,

3.9

whenm → ∞due to the convergence of the given series. So,{fⁿx}is a Cauchy sequence and it is convergent. Formchosen in the proof of Theorem3.5such thatf^mg, it isgⁿf^mn

andgⁿx → z whenn → ∞, but {f^mnx}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:X → Xbe two mappings such that fX⊂gXand one of these subsets ofXis complete. Suppose that there exist nonnegative coefficients ai,i1, . . . ,5 such that

2Ka1 K1a2a3

K²K

a4a5<2 3.10

and that for allx, y∈X D

fx, fy

≤a1D gx, gy

a2D gx, fx

a3D gy, fy

a4D gx, fy

a5D gy, fx

3.11 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, when₅

i1ai < 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 fX⊂gX, construct a Jungck sequence{yn}defined byyn fxn gx_n1,n0,1,2, . . . .Let us prove that this is a Cauchy sequence. Indeed, using3.11, we get that

D

y_n, y_n1 D

fx_n, fx_n1

≤a₁D

gx_n, gx_n1 a₂D

gx_n, fx_n a3D

gx_n1, fx_n1 a4D

gxn, fx_n1

a5D

gx_n1, fxn

a₁D

y_n−1, y_n a₂D

y_n−1, y_n

a₃D

y_n, y_n1 a4D

y_n−1, y_n1 a5·0

≤a1a2D y_n−1, yn

a3D

yn, y_n1 a₄K

D

y_n−1, y_n D

y_n, y_n1 a1a2Ka4D

y_n−1, yn

a3Ka4D

yn, y_n1 .

3.12

Similarly, we conclude that D

y_n1,y_n D

fx_n1, fx_n

≤a1a₃Ka₅D

y_n−1, y_n

a₂Ka₅D

y_n, y_n1

. 3.13 Adding the last two inequalities, we get that

2D

y_n, y_n1

≤2a1a₂a₃Ka₄Ka₅D

y_n−1, y_n

a₂a₃Ka₄Ka₅D

y_n, y_n1 , 3.14

that is,

D

yn, y_n1

≤ 2a1a2a3Ka4Ka5

2−a₂−a₃−Ka₄−Ka₅ D y_n−1, yn

λD y_n−1, yn

. 3.15

The assumption3.10implies that

2Ka1Ka2Ka3K²a4a5<2−a2−a3−Ka4−Ka5, λ 2a₁a₂a₃Ka₄Ka₅

2−a2−a3−Ka4−Ka5 < 1 K.

3.16

Lemma3.1implies that{yn} is a Cauchy sequence ingX and so there isz ∈ X such that fxngx_n1 → gzwhenn → ∞. We will prove thatfzgz.

Using3.11we conclude that D

fxn, fz

≤a1D

gxn, gz a2D

gxn, fxn

a3D gz, fz a₄D

gx_n, fz a₅D

gz, fx_n

≤a₁D

gx_n, gz a₂D

gx_n, fx_n

a₃K D

gz, fx_n D

fx_n, fz a4K

D

gxn, fxn

D

fxn, fz a5D

gz, fxn

a₁D

gx_n, gz

a₂Ka₄D

gx_n, fx_n Ka3a5D

gz, fxn

Ka3a4D

fxn, fz .

3.17

Similarly,

D fz, fxn

≤a1D

gxn, gz

Ka2a4D gz, fxn

Ka2a₅D

fx_n, fz

a₃Ka₅D

fx_n, gx_n

. 3.18

Adding up, one concludes that

2−Ka2a₃a₄a₅D

fx_n,fz

≤2a1D

gxn, gz

a2a3Ka4a5D

fxn, gxn

Ka2a₃ a₄a₅D

fx_n, gz .

3.19

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, andfxn → fz. Since the limit of a sequence is unique, it follows thatfzgzwandfandghave a point of coincidencew.

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

Dw, w1 D fz, fz1

≤a₁D

gz, gz₁ a₂D

gz, fz

a₃D

gz₁, fz₁ a₄D

gz, fz₁ a₅D

gz₁, fz

a1Dw, w1 a2·0a3·0a4Dw, w1 a5Dw1, w a₁a₄a₅Dw, w1.

3.20

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 constantsa_i, we obtain as special cases Theorem3.3as well as metric type versions of some other well-known theoremsKannan, Zamfirescu, see, e.g., 15:

Corollary 3.8. LetX, D, Kbe a metric type space, and letf, g:X → Xbe two mappings such that fX⊂gXand one of these subsets ofXis complete. Suppose that one of the following three conditions holds:

1^◦Dfx, fy≤a₁Dgx, gyfor somea₁<1/Kand allx, y∈X;

2^◦Dfx, fy≤a₂Dgx, fx Dgy, fyfor somea₂<1/K1and allx, y∈X;

3^◦Dfx, fy≤a4Dgx, fy Dgy, fxfor somea4<1/K²Kand allx, y∈X.

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:X → Xsatisfy

D

fx, fy

≤a₁D x, y

a₂D x, fx

a₃D y, fy

a₄D x, fy

a₅D y, fx

3.21

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

Moreover,fhas property P.

Proof. We have only to prove the last assertion. For arbitraryx∈X, we have that D

fx, f²x D

fx, ffx

≤a1D x, fx

a2D x, fx

a3D

fx, f²x

a4D x, f²x

a5D fx, fx

≤a1a₂Ka₄D x, fx

a₃Ka₄D

fx, f²x ,

3.22

and similarly D

f²x, fx D

ffx, fx

≤a1a₃Ka₅D x, fx

a₂Ka₅D

fx, f²x

. 3.23

Adding the last two inequalities, we obtain

D

fx, f²x

≤ 2a1a2a3Ka4a5 2−a₂−a₃−Ka4a₅ D

x, fx λD

x, 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 :X → X be two mappings such thatfX⊂ gXand one of these subsets ofX is complete. Suppose that there existsλ, 0< λ <1/K such that for allx, y∈X

D fx, fy

≤λmaxM

f, g;x, y

, 3.25

where

M

f, g;x, y

D

gx, gy , D

gx, fx , D

gy, fy ,D

gx, fy

2K ,D

gy, fx 2K

. 3.26

Thenf andg have a unique point of coincidence. If, moreover, the pairf, gis weakly compatible, thenfandghave a unique common fixed point.

Proof. Letx0 ∈ Xbe arbitrary and, using conditionfX ⊂ gX, construct a Jungck sequence {yn}satisfying y_n fx_n gx_n1,n 0,1,2, . . . .Suppose thatDyn, y_n1 > 0 for each n otherwise the conclusion follows easily. Using3.25we conclude that

D

y_n1, y_n D

fx_n1, fxn

≤λmax

D

gx_n1, gxn

, D

gx_n1, fx_n1 , D

gxn, fxn

,D

gx_n1, fx_n

2K ,D

gx_n, fx_n1 2K

λmax

D

yn, y_n−1 , D

yn, y_n1 , D

y_n−1, yn

,0,D

y_n−1, y_n1 2K

≤λmax D

yn, y_n−1 ,1

2 D

y_n−1, yn

D

yn, y_n1 .

3.27

If Dyn, y_n−1 < Dyn1, y_n, then Dyn, y_n−1 < 1/2Dyn−1, y_n Dyn, y_n1 <

Dyn, y_n1, and it would follow from 3.27 that Dyn1, yn ≤ λDyn1, yn which is impossible sinceλ < 1. For the same reason the termDyn, y_n1 was omitted in the last row of the previous series of inequalities. Hence, Dyn, y_n−1 > Dyn1, y_n and 3.27 becomesDyn1, yn ≤ λDyn, y_n−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 thaty_n fx_ngx_n1 → gzwhenn → ∞for somez∈X.

To prove thatfzgz, putxxnandyzin3.25to get

D

fxn, fz

≤λmax

D

gxn, gz , D

gxn, fxn

, D gz, fz

,D

gx_n, fz

2K ,D

gz, fx_n 2K

. 3.28

Note that fx_n → gz and gx_n → gz when n → ∞, implying that Dgxn, fx_n ≤ KDgxn, gz Dgz, fxn → 0 whenn → ∞. It follows that the only possibilities are the following:

1^◦Dfxn, fz ≤ λDgz, fz ≤ λKDgz, fxn Dfxn, fz; in this case 1 − λKDfxn, fz ≤ λKDgz, fxn → 0, and since 1− λK > 0, it follows that fx_n → fz.

2^◦Dfxn, fz≤ λ1/2KDgxn, fz≤λ/2Dgxn, fxn Dfxn, fz; in this case, 1−λ/2Dfxn, fz≤λ/2Dgxn, fxn → 0, so againfxn → fz,n → ∞.

Since the limit of a sequence is unique, it follows thatfzgz.

The rest of conclusion follows as in the proof of Theorem3.7.

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

Corollary 3.12. LetX, D, Kbe a complete metric type space, and letf :X → Xbe such that for someλ, 0< λ <1/K, and for allx, y∈X,

D fx, fy

≤λmax

D x, y

, D x, fx

, D y, fy

,D x, fy 2K ,D

y, fx 2K

3.29

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

Proof. Letxn → zwhenn → ∞. Then

D

fxn, fz

≤λmax

Dxn, z, D xn, fxn

, D z, fz

,D x_n, fz

2K ,D fx_n, z

2K

λmax

Dxn, z, D

x_n, fx_n ,D

fxn, z 2K

.

3.30

SinceDxn, z → 0 andDxn, fx_n≤ KDxn, z Dfz, fxn, the only possibility is that Dfx,fz≤λKDxn, zDfz, fxn, implying that1−λKDfxn, fz≤λKDxn, z → 0, n → ∞. Since 0< λK <1, it follows thatfx_n → fzz,n → ∞, andfis continuous at the pointz.

We will prove thatfsatisfies D

fx, f²x

≤hD x, fx

3.31 for someh, 0< h <1 and eachx∈X.

Applying3.29to the pointsxandfxfor anyx∈X, we conclude that

D

fx, f²x

≤λmax

D x, fx

, D x, fx

, D

fx, f²x ,D

x, f²x 2K ,D

fx, fx 2K

λmax

D x, fx

, D

fx, f²x ,D

x, f²x 2K

.

3.32

The following cases are possible:

1^◦Dfx, f²x≤λDx, fx, and3.31holds withhλ;

2^◦Dfx, f²x≤λDfx, f²x, which is only possible ifDfx, f²x 0 and then3.31 obviously holds.

3^◦Dfx, f²x≤λ/2KDx, f²x≤λ/2KKDx, fx Dfx, f²x, implying that 1−λ/2Dfx, f²x≤λ/2Dx, fxandDfx, f²x≤hDx, fx, where 0< h λ/2−λ<1 since 0< λ <1.

So, relation3.31holds for someh, 0 < h <1 and eachx ∈X. 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 :X → Xbe four mappings such thatfX⊂TXandgX⊂SX, and suppose that at least one of these four subsets ofXis complete.

Let

D fx, gy

≤λD

Sx, Ty

3.33 holds for someλ, 0< λ <1/Kand allx, y∈X. 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{y_n}such that

fx_2n−2Tx_2n−1y_2n−1, gx_2n−1Sx_2ny_2n 3.34

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

y_2n1, y_2n2 D

fx2n, gx_2n1

≤λDSx2n, Tx_2n1 λD

y2n, y_2n1 , D

y_2n3, y_2n2 D

fx_2n2, gx_2n1

≤λDSx2n2, Tx_2n1 λD

y_2n2, y_2n1

. 3.35

Using Lemma3.1, we conclude that{yn}is a Cauchy sequence. Suppose, for example, that SX is a complete subset ofX. Then yn → u Sv,n → ∞, for somev ∈ X. Of course, subsequences{y_2n−1}and{y2n}also converge tou. Let us prove thatfv u. Using3.33, we get that

D fv, u

≤K D

fv, gx_2n−1 D

gx_2n−1, u

≤K

λDSv, Tx_2n−1 D

gx_2n−1, u

−→Kλ·00 0. 3.36

hencefvuSv. Sinceu∈fX⊂TX, we get that there existsw∈Xsuch thatTwu. Let us prove that alsogwu. Using3.33, again we conclude that

D gw, u

≤K D

gw, fx2n

D

fx2n, u

≤K

λDSx2n, Tw D

fx2n, u

−→Kλ·00 0, 3.37

implying thatgw u Tw. We have proved thatuis a common point of coincidence for pairsf, Sandg, T.

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, gu ≤ λDSv, 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 :X → Xis a mapping such that

D fx, fy

< D x, y

, forx, y∈X, 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.,x_nfⁿx₀. Ifx_nx_n1for somen, thenx_nis aunique fixed point. Ifxn/x_n1for eachn0,1,2, . . ., then

D_n:Dx_n1, x_n D

fⁿ¹x₀, fⁿx₀

< D

fⁿx₀, fⁿ⁻¹x₀

D_n−1. 3.39

Hence, there existsD^∗, such that 0 ≤ D^∗ ≤ Dn for each nandDn → D^∗,n → ∞. Using sequential compactness of X, choose a subsequence{xni} of {xn} that converges to some x^∗∈Xwheni → ∞. The continuity offandf²implies that

fx_ni −→fx^∗, f²x_ni → f²x^∗ wheni−→ ∞, 3.40 and the continuity of the symmetricDimplies that

D

fx_ni, x_ni

−→D

fx^∗, x^∗ , D

f²x_ni, fx_ni

−→D

f²x^∗, fx^∗

wheni−→ ∞. 3.41 It follows thatDfxni, x_ni D_ni → D^∗ Dfx^∗, x^∗. It remains to prove that fx^∗ x^∗. If fx^∗/x^∗, thenD^∗>0 and3.41implies that

D^∗ lim

i→ ∞Dni1 lim

i→ ∞D

f²xni, fxni

D

f²x^∗, fx^∗

< D

fx^∗, x^∗

D^∗. 3.42

This is a contradiction.

Example 3.15. LetE R²,P {x, y ∈E : x, y ≥ 0},X 1,∞, andd : X×X → Ebe defined bydx, y |x−y|,|x−y|. 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|x−y|√

2.

Letf:X → Xbe defined byfxx1/x. Then

D fx, fy

x−y 1− 1

xy √

2<x−y√ 2D

x, y

3.43

for all x, y ∈ X. Hence, f satisfies condition 3.38 but it has no fixed points. Obviously, X, D, Kis notsequentiallycompact.

Acknowledgment

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

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