Volume 2010, Article ID 978121,15pages doi:10.1155/2010/978121
Research Article
Common Fixed Point Results in
Metric-Type Spaces
Mirko Jovanovi ´c,
1Zoran Kadelburg,
2and Stojan Radenovi ´c
31Faculty 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, y∈X.
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, 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 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 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 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, z ≤ KDx, y1 Dy1, y2 · · · Dyn, z for arbitrary points x, y1,
y2, . . . , yn, z∈X;
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:
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≤Dx, y≤KDx, xn D
xn, y
< K ε
2K ε
2K
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 : 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 apoint of coincidenceof a pair of self-mapsf, g:X → Xand
u∈Xis 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
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
Dym, yn
≤KDym, ym1
Dym1, yn
≤KDym, ym1
K2Dym1, ym2
Dym2, yn
≤ · · · ≤KDym, ym1
K2Dym1, ym2
· · ·
Kn−m−1Dyn−2, yn−1
Dyn−1, yn
≤KDym, ym1
K2Dym1, ym2
· · ·
Kn−m−1Dyn−2, yn−1
Kn−mDyn−1, yn
.
3.2
Now3.1andKλ <1 imply that
Dym, yn
≤KλmK2λm1· · ·Kn−mλn−1Dy0, y1
Kλm1 Kλ · · · Kλn−m−1Dy0, y1
≤ Kλm
1−KλD
y0, y1
−→0 whenm−→ ∞.
3.3
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:X → Xbe a map such that for someλ,0< λ <1/K,
Dfx, fy≤λDx, 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
Dyn, yn1
Dfyn−1, fyn
≤λDyn−1, yn
3.5
for eachn1,2. . . .Lemma3.1implies that{yn}is a Cauchy sequence, and sinceX, D, K
is complete, there existsz∈Xsuch thatyn → zwhenn → ∞. Then
Dfz, z≤KDfz, fyn
Dyn1, z
≤KλDz, 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
≤ K2λm
1−KλD
x0, fx0
3.7
for eachm∈N. Indeed, form < n,
Dfmx0, z
≤KDfmx0, fnx0
Dfnx0, z
≤ K2λm
1−KλD
x0, fx0
KDfnx0, 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
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 n ≥ n0. Then Dfnx, fny ≤ λDx, y for allx, y ∈ X whenever n ≥ n0.
In other words, for any m ≥ n0, g fm satisfies Dgx, gy ≤ λDx, y for allx, y ∈ X.
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, ffx≤λ
1Dx, 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
Dfmx, fnx≤KDfmx, fm1x· · ·Dfn−1xfnx
KDfmx, fmfx· · ·Dfn−1x, fn−1fx
≤Kλm· · ·λn−1D
x, fx−→0,
3.9
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:X → Xbe two mappings such that fX⊂gXand 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, y∈X
Dfx, fy≤a1D
gx, gya2D
gx, fxa3D
gy, fya4D
gx, fya5D
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, 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
fX⊂gX, 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
≤a1D
gxn, gxn1
a2D
gxn, fxn
a3D
gxn1, fxn1
a4D
gxn, fxn1
a5D
gxn1, fxn
a1D
yn−1, yn
a2D
yn−1, yn
a3D
yn, yn1
a4D
yn−1, yn1
a5·0 ≤a1a2D
yn−1, yn
a3D
yn, yn1
a4K
Dyn−1, yn
Dyn, yn1
a1a2Ka4D
yn−1, yn
a3Ka4D
yn, yn1
.
3.12
Similarly, we conclude that
Dyn1,yn
Dfxn1, fxn
≤a1a3Ka5D
yn−1, yn
a2Ka5D
yn, yn1
. 3.13
Adding the last two inequalities, we get that
2Dyn, yn1
≤2a1a2a3Ka4Ka5D
yn−1, yn
a2a3Ka4Ka5D
yn, yn1
,
that is,
Dyn, yn1
≤ 2a1a2a3Ka4Ka5
2−a2−a3−Ka4−Ka5
Dyn−1, yn
λDyn−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
fxngxn1 → gzwhenn → ∞. We will prove thatfzgz.
Using3.11we conclude that
Dfxn, fz
≤a1D
gxn, gz
a2D
gxn, fxn
a3D
gz, fz
a4D
gxn, fz
a5D
gz, fxn
≤a1D
gxn, gz
a2D
gxn, fxn
a3K
Dgz, fxn
Dfxn, fz
a4K
Dgxn, fxn
Dfxn, fz
a5D
gz, fxn
a1D
gxn, gz
a2Ka4D
gxn, fxn
Ka3a5D
gz, fxn
Ka3a4D
fxn, fz
.
3.17
Similarly,
Dfz, fxn
≤a1D
gxn, gz
Ka2a4D
gz, fxn
Ka2a5D
fxn, fz
a3Ka5D
fxn, gxn
. 3.18
Adding up, one concludes that
2−Ka2a3a4a5D
fxn,fz
≤2a1D
gxn, gz
a2a3Ka4a5D
fxn, gxn
Ka2a3 a4a5D
fxn, 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
Suppose thatw1 fz1 gz1is another point of coincidence forfandg. Then3.11
implies that
Dw, w1 D
fz, fz1
≤a1D
gz, gz1
a2D
gz, fza3D
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 constantsai, 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
Dfx, fy≤a1D
x, ya2D
x, fxa3D
y, fya4D
x, fya5D
y, fx 3.21
for someai,i 1, . . . ,5 satisfying 3.10and for allx, y ∈ X. Thenf has a unique fixed point.
Proof. We have only to prove the last assertion. For arbitraryx∈X, we have that
Dfx, f2xDfx, ffx
≤a1D
x, fxa2D
x, fxa3D
fx, f2xa4D
x, f2xa5D
fx, fx
≤a1a2Ka4D
x, fx a3Ka4D
fx, f2x,
3.22
and similarly
Df2x, fxDffx, fx≤a1a3Ka5D
x, fx a2Ka5D
fx, f2x. 3.23
Adding the last two inequalities, we obtain
Dfx, f2x≤ 2a1a2a3Ka4a5
2−a2−a3−Ka4a5
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 :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
Dfx, fy≤λmaxMf, g;x, y, 3.25
where
Mf, g;x, y
Dgx, gy, Dgx, fx, Dgy, fy,D
gx, fy
2K ,
Dgy, fx
2K
. 3.26
Proof. Letx0 ∈ Xbe arbitrary and, using conditionfX ⊂ gX, 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
≤λmax
Dgxn1, gxn
, Dgxn1, fxn1
, Dgxn, fxn
,D
gxn1, fxn
2K ,
Dgxn, fxn1
2K
λmax
Dyn, yn−1
, Dyn, yn1
, Dyn−1, yn
,0,D
yn−1, yn1
2K
≤λmax Dyn, yn−1
,1
2
Dyn−1, yn
Dyn, yn1
.
3.27
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, yn ≤ λDyn1, 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, yn ≤ λDyn, 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 somez∈X.
To prove thatfzgz, putxxnandyzin3.25to get
Dfxn, fz
≤λmax
Dgxn, gz
, Dgxn, fxn
, Dgz, fz,D
gxn, fz
2K ,
Dgz, fxn
2K
.
3.28
Note that fxn → gz and gxn → gz when n → ∞, implying that Dgxn, fxn ≤
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
fxn → 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 → ∞.
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,
Dfx, fy≤λmax
Dx, y, Dx, fx, Dy, fy,D
x, fy
2K ,
Dy, 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
Dfxn, fz
≤λmax
Dxn, z, D
xn, fxn
, Dz, fz,D
xn, fz
2K ,
Dfxn, z
2K
λmax
Dxn, z, D
xn, fxn
,D
fxn, z
2K
.
3.30
SinceDxn, z → 0 andDxn, fxn≤ 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 thatfxn → fzz,n → ∞, andfis continuous at the
pointz.
We will prove thatfsatisfies
Dfx, f2x≤hDx, fx 3.31
for someh, 0< h <1 and eachx∈X.
Applying3.29to the pointsxandfxfor anyx∈X, we conclude that
Dfx, f2x≤λmax
Dx, fx, Dx, fx, Dfx, f2x,D
x, f2x
2K ,
Dfx, fx
2K
λmax
Dx, fx, Dfx, 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
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
Dfx, gy≤λDSx, 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{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
.
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{y2n−1}and{y2n}also converge tou. Let us prove thatfv u. Using3.33,
we get that
Dfv, u≤KDfv, gx2n−1
Dgx2n−1, u
≤KλDSv, Tx2n−1 D
gx2n−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
Dgw, u≤KDgw, fx2n
Dfx2n, u
≤KλDSx2n, Tw D
fx2n, u
−→Kλ·00 0,
3.37
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
Dfx, fy< Dx, 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.,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 nandDn → D∗,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
ni → f
2x∗ wheni−→ ∞, 3.40
and the continuity of the symmetricDimplies that
Dfxni, xni
−→Dfx∗, x∗, Df2xni, fxni
−→Df2x∗, fx∗ wheni−→ ∞. 3.41
It follows thatDfxni, xni Dni → D∗ Dfx∗, x∗. It remains to prove that fx∗ x∗. If
fx∗/x∗, thenD∗>0 and3.41implies that
D∗ lim
i→ ∞Dni1ilim→ ∞D
f2xni, fxni
Df2x∗, fx∗< Dfx∗, x∗D∗. 3.42
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
Dfx, fyx−y
1− 1
xy
√
2<x−y√2Dx, 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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