Volume 2012, Article ID 409872,17pages doi:10.1155/2012/409872
Research Article
New Meir-Keeler Type Tripled Fixed-Point Theorems on Ordered Partial Metric Spaces
Hassen Aydi
1and Erdal Karapınar
21Institut Sup´erieur d’Informatique et des Technologies de Communication de Hammam Sousse, Universit´e de Sousse, Route GP1 4011, Tunisia
2Department of Mathematics, Atilim University, Incek, 06836 Ankara, Turkey
Correspondence should be addressed to Erdal Karapınar,[email protected]
Received 3 February 2012; Accepted 27 March 2012 Academic Editor: Zheng-Guang Wu
Copyrightq2012 H. Aydi and E. Karapınar. 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.
In this paper, we prove some new Meir-Keeler type tripled fixed-point theorems on a partially ordered complete partial metric space. Also, as application, some results of integral type are given.
1. Introduction and Preliminaries
In the last century, the theory of fixed points has appeared as a crucial technique in the study of nonlinear phenomena. Particularly, the tools in fixed-point theory have an application in such diverse fields as biology, chemistry, physics, economics, computer sciences, and engineering.
Recently, fixed-point theorems are considered on partial metric spaces on which self- distance of some points may not be zero. This phenomenon was discovered by Matthews1 when he considered the tools of metric spaces in the field of semantics and domain theory in computer sciencesee, e.g.,2,3. After the initial results of Mathews, other papers have been released on partial metric spacessee e.g.,4–20.
Another important development is reported in fixed-point theory via ordered metric spaces. Fixed-point theorems in ordered sets were discussed by Ran and Reurings 21.
Subsequently, many results in this direction were givensee, e.g.,22–31.
In this paper, we combine two recent trends, partial metric spaces and ordered sets, and discuss the existence and uniqueness of some new Meir-Keeler type tripled fixed-point theorems in the context of partially ordered partial metric spaces.
LetXbe a nonempty set. A partial metric is a functionp:X×X → 0,∞satisfying the following conditions:
P1ifpx, x px, y py, y, thenxy,
P2px, y py, x,
P3px, x≤px, y,
P4px, z py, y≤px, y py, z,
for allx, y, z∈X. Then,X, pis called a partial metric space.
Ifpis a partial metricponX, then the functiondp:X×X → 0,∞given by dp
x, y 2p
x, y
−px, x−p y, y
1.1 is a metric onX. Each partial metricponXgenerates aT0topologyτponXwith a base of the family of openp-balls{Bpx, ε:x∈X, ε >0}, whereBpx, ε {y∈X:px, y< px, x ε}
for allx ∈ X andε > 0. Similarly, closedp-ball is defined asBpx, ε {y ∈ X : px, y ≤ px, x ε}. For more details, see1,5.
Definition 1.1see, e.g.,1,5,15. LetX, pbe a partial metric space.
iA sequence{xn}inXconverges tox∈Xwhenever limn→ ∞px, xn px, x.
iiA sequence{xn} inX is called Cauchy whenever limn,m→ ∞pxn, xmexistsand finite.
iii X, pis said to be complete if every Cauchy sequence{xn}inX converges, with respect toτp, to a pointx∈X, that is, limn,m→∞pxn, xm px, x.
Lemma 1.2see, e.g.,1,5,15. LetX, pbe a partial metric space.
aA sequence{xn}is Cauchy if and only if{xn}is a Cauchy sequence in the metric space X, dp.
b X, pis complete if and only if the metric spaceX, dpis complete. Moreover,
nlim→ ∞dpx, xn 0⇐⇒ lim
n→ ∞px, xn lim
n,m→ ∞pxn, xm px, x. 1.2
Lemma 1.3see, e.g.,4,15,16. LetX, pbe a partial metric space. Then, Aifpx, y 0, thenxy,
Bifx /y, thenpx, y>0.
Remark 1.4. Ifxy,px, ymay not be 0.
Lemma 1.5see, e.g.,4,15,16. Letxn → zasn → ∞in a partial metric spaceX, pwhere pz, z 0. Then, limn→ ∞pxn, y pz, yfor everyy∈X.
X, p,≤is called a partially ordered partial metric space ifX,≤is a partially ordered set andX, pis a partial metric space. Further, ifX, pis a complete partial metric space, then X, p,≤is called a partially ordered complete partial metric space. Hereafter, we assume that X /∅and we use the notation
XkX ×X× · · · ×X
k-many
. 1.3
Also, take the mappingP:X3×X3 → 0,∞such that Px,y:max
p x1, y1
, p x2, y2
, p
x3, y3 , 1.4
wherex x1, x2, x3andy y1, y2, y3∈X3.
Let X,≤ be a partially ordered set. We consider the following partial orderalso denoted by≤on the product spaceX3:
u, v, w≤ x, y, z
iffx≥u, y≤v, z≥w, 1.5
whereu, v, w,x, y, z∈X3. Moreover, we say thatx, y, zis equal tou, v, rif and only if xu,v, andzr. In the sequel, we need the following definitions.
Definition 1.6 see 32. Let X,≤ be a partially ordered set and F : X3 → X a given mapping. We say that F has the mixed monotone property if Fx, y, z is monotone nondecreasing inxandz, and it is monotone nonincreasing iny, that is, for anyx, y, z∈X,
x1, x2∈X, x1≤x2⇒F
x1, y, z
≤F
x2, y, z , y1, y2∈X, y1≤y2 ⇒F
x, y1, z
≥F
x, y2, z , z1, z2 ∈X, z1≤z2⇒F
x, y, z1
≤F x, y, z2
.
1.6
Definition 1.7see32. An elementx, y, z∈X3is called a tripled fixed point ofF:X3 → X if
F x, y, z
x, F y, x, y
y, F z, y, x
z. 1.7
Berinde and Borcut32proved the following theorem.
Theorem 1.8. LetX,≤be a partially ordered set andX, da complete metric space. LetF:X3 → Xbe a mapping having the mixed monotone property onX. Assume that there exist constantsa, b, c∈ 0,1such thatabc <1 for which
d F
x, y, z
, Fu, v, w
≤adx, u bd y, v
cdz, w 1.8
for allx≥u,y≤vandz≥w. Assume thatXhas the following properties:
iif a nondecreasing sequencexn → x, thenxn≤xfor alln, iiif a nonincreasing sequenceyn → y, thenyn≥yfor alln. If there existx0, y0, z0∈Xsuch that
x0 ≤F
x0, y0, z0
, y0≥F
y0, x0, y0
, z0≤F
z0, y0, x0
, 1.9
then there existx, y, z∈Xsuch that F
x, y, z
x, F y, x, y
y, F z, y, x
z. 1.10
Recently,Theorem 1.8is extended to cone metric spaces by Rao and Kishore33. On the other hand, very recently, Aydi et al.34introduced the following concepts.
Definition 1.9see34. LetX,≤be a partially ordered set andF :X3 → X. We say thatF has the mixed strict monotone property if, for anyx, y, z∈X,
x1, x2∈X, x1< x2⇒F
x1, y, z
< F
x2, y, z , y1, y2∈X, y1< y2 ⇒F
x, y1, z
> F
x, y2, z , z1, z2 ∈X, z1< z2⇒F
x, y, z1
< F x, y, z2
.
1.11
Definition 1.10 see34. LetX, d,≤be a partially ordered metric space. A mapping F : X3 → Xis said to be a generalized Meir-Keeler type contraction if, for anyε >0, there exists aδε>0 such that
ε≤max
dx, u, d y, v
, dz, r < εδε ⇒d F
x, y, z
, Fu, v, r
< ε, 1.12
for allx, y, z, u, v, r∈Xwithx≤u,y≥vandz≤r.
In the following, we consider the partial case ofDefinition 1.10and we introduce the following.
Definition 1.11. LetX, p,≤be a partially ordered partial metric space. A mappingF:X3 → X is said to be a generalizedp-Meir-Keeler type contraction if, for anyε > 0, there exists a δε>0 such that
ε≤max
px, u, p y, v
, pz, r < εδε ⇒p F
x, y, z
, Fu, v, r
< ε, 1.13
for allx, y, z, u, v, r∈Xwithx≤u,y≥vandz≤r.
Remark 1.12. It is immediate to show that ifF :X3 → Xis a generalizedp-Meir-Keeler type contraction, then
p F
x, y, z
, Fu, v, r
<max
px, u, p y, v
, pz, r 1.14
for allx, u, y, v, z, r,∈Xwithx < u, y≥v, z < rorx≤u, y > v, z≤r.
Proposition 1.13. LetX, p,≤be a partially ordered partial metric space andF :X3 → Xa given mapping. If1.8is satisfied, thenFis a generalizedp-Meir-Keeler type function.
Proof. Assume that1.8is satisfied. For allε >0, one can check that1.13is satisfied with
δε 1/abc−1ε.
In the sequel, we use the following notations given in34. LetF:X3 → X3be such that, fora, b, c∈X,
Fa, b, c Fa, b, c, Fb, a, b, Fc, b, a. 1.15
Letx0, y0, z0∈Xbe such that x0< F
x0, y0, z0
, y0≥F
y0, x0, y0
, z0< F
z0, y0, x0
. 1.16
We consider sequences{xn},{yn}, and{zn}such that
⎡
⎣xn
yn
zn
⎤
⎦
An
⎡
⎣F
xn−1, yn−1, zn−1 F
yn−1, xn−1, yn−1 F
zn−1, yn−1, xn−1
⎤
⎦
FA n−1
⎡
⎣Fnx0, y0, z0 Fny0, x0, y0 Fnz0, y0, x0
⎤
⎦
FnA0
, 1.17
forn1,2,3, . . ..
Our first auxiliary result is as follows.
Proposition 1.14. LetX, p,≤be a partially ordered partial metric space, and letF :X3 → Xbe a given mapping such that the following hypotheses hold:
iFhas the mixed strict monotone property, iiFis a generalizedp-Meir-Keeler type function,
iii∃x, y, z,u, v, r∈X3such thatx < u,y≥vandz < r.
Then,
P Fn
x, y, z
,Fnu, v, r
−→0, asn−→∞. 1.18
Proof. Letx, y, z x0, y0, z0andu, v, r u0, v0, r0. We show that xnFn
x0, y0, z0,
< Fnu0, v0, r0 un, ynFn
y0, x0, y0
> Fnv0, u0, v0 vn, znFn
z0, y0, x0
< Fnr0, v0, u0 rn,
∀n1,2, . . . , 1.19
withFF1.
Due to the fact that F has the mixed strict monotone property, together with the assumption thatx < u,y≥vandz < r, we obtain
x1F x, y, z
F
x0, y0, z0
< F
u0, y0, z0
⇒F
x0, y0, z0
< Fu0, v0, z0 ⇒F
x0, y0, z0
< Fu0, v0, r0 u1.
1.20
Analogously, we have y1F
y0, x0, y0
> Fv0, u0, v0 v1, z1F
z0, y0, x0
< Fr0, v0, u0 r1. 1.21
Thus,1.19holds forn1. By using the same arguments, we show that1.19holds also for n2. In fact,
x2F2
x0, y0, z0
F
x1, y1, z1
F
F
x0, y0, z0
, F
y0, x0, y0
, F
z0, y0, x0
< F
Fu0, v0, r0, F
y0, x0, y0
, F
z0, y0, x0
< F
Fu0, v0, r0, Fv0, u0, v0, F
z0, y0, x0
< FFu0, v0, r0, Fv0, u0, v0, Fr0, v0, u0 F2u0, v0, r0 Fu1, v1, r1 u2.
1.22
Similarly, we find
y2F2
y0, x0, y0
≥F2v0, u0, v0 v2, z2F2
z0, y0, x0
< F2r0, v0, u0 r2. 1.23
Inductively, we get that1.19holds.
ByRemark 1.12, together with1.19, we have
pxn2, un2 p Fn2
x0, y0, z0
, Fn2u0, v0, r0 p
F
xn1, yn1, zn1
, Fun1, vn1, rn1
<max
pxn1, un1, p
yn1, vn1
, pzn1, rn1 ,
1.24
pzn2, rn2 p Fn2
z0, y0, x0
, Fn2r0, v0, u0 p
F
zn1, yn1, xn1
, Frn1, vn1, un1
<max
pzn1, rn1, p
yn1, vn1
, pxn1, un1 ,
1.25
p
yn2, vn2 p
Fn2
y0, x0, y0
, Fn2v0, u0, v0 p
F
yn1, xn1, yn1
, Fvn1, un1, vn1
<max p
yn1, vn1
, pxn1, un1, p
yn1, vn1
≤max
pzn1, rn1, p
yn1, vn1
, pxn1, un1 .
1.26
LetΔn1:max{pxn1, un1, pyn1, vn1, pzn1, rn1}. Combining1.24–1.26, we get
Δn2<Δn1, ∀n1,2. . . . 1.27
If we denoteBn un, vn, rn, then, by definition of the partial metricPand1.27, we have
PAn2, Bn2< PAn1, Bn1. 1.28
Consequently, the sequence{tn} {PAn, Bn}is decreasing. Hence,{tn}converges, say to ε ≥0. Clearly, ifε 0, we have finished. Suppose, on the contrary,ε > 0. Thus, there exists k∈ {1,2, . . .}such that
ε≤tnPAn, Bn< εδε for any n≥k. 1.29
In particular, fornk, we have
ε≤tkPAk, Bk< εδε, 1.30
that is equal to
ε≤Δk< εδε. 1.31
It follows from1.19and the hypothesisiithat p
F
xk, yk, zk
, Fuk, vk, rk
< ε 1.32
which is equivalent to
pxk1, uk1< ε. 1.33
Moreover, we have
p
yk1, vk1
< ε, pzk1, rk1< ε. 1.34
Combining1.33and1.34, we have
Δk1< ε. 1.35
Thus,tk1PAk1, Bk1< εwhich is a contradiction with respect to1.29, and soε0.
We conclude that
PAn, Bn P Fn
x, y, z
,Fnu, v, r
−→0, asn−→∞. 1.36
Remark 1.15. The previous proposition remains true if, iniii, we change the assumption
∃ x, y, z
,u, v, r∈X3 such thatx < u, y≥v, z < r 1.37
with the following
∃ x, y, z
,u, v, r∈X3 such thatx≤u, y > v, z≤r. 1.38
2. Existence of Tripled Fixed Point
The following theorem is our first main result.
Theorem 2.1. LetX, p,≤be a partially ordered complete partial metric space. Suppose thatX has the following properties:
aif{xn}is a sequence such thatxn1 > xnfor eachn1,2, . . .andxn → x, thenxn < x for eachn1,2, . . .,
bif{yn}is a sequence such thatyn1 < yn for eachn1,2, . . .andyn → y, thenyn > y for eachn1,2, . . ..
Assume thatF:X3 → Xsatisfies the following hypotheses:
iFhas the mixed strict monotone property, iiFis a generalizedp-Meir-Keeler type function, iiithere existx0, y0, z0∈Xsuch that
x0< F
x0, y0, z0
, y0≥F
y0, x0, y0
, z0< F
z0, y0, x0,
. 2.1
Then,Fhas a tripled fixed point, that is, there existx, y, z∈Xsuch that F
x, y, z
x, F y, x, y
y, F z, y, x
z. 2.2
Also,px, x py, y pz, z 0.
Proof. Letx0, y0, z0 ∈Xbe as iniii. We construct sequences{xn},{yn}, and{zn}according to1.17.
We claim that, for alln≥2, we have
· · ·> xn> xn−1>· · ·> x1> x0,
· · ·< yn< yn−1<· · ·< y1≤y0,
· · ·> zn> zn−1>· · ·> z1> z0.
2.3
Indeed, we will use a mathematical induction to prove2.3. Clearly, we have x0< F
x0, y0, z0
x1, y0≥F
y0, x0, y0
y1, z0< F
z0, y0, x0
z1. 2.4
Suppose now that the inequalities in2.3hold for somen≥2. By the mixed strict monotone property ofF, together with1.17, we have
xnF
xn−1, yn−1, zn−1
< F
xn, yn, zn
xn1, ynF
yn−1, xn−1, yn−1
> F
yn, xn, yn
yn1, znF
zn−1, yn−1, xn−1
< F
zn, yn, xn
zn1.
2.5
Thus,2.3holds for alln≥2.
Puttingx, y, z A0andu, v, r A1and byProposition 1.14, we get
P
FnA0,FnA1
−→0, asn−→∞, 2.6
which is equivalent to
PAn, An1−→0, asn−→∞. 2.7
Take an arbitraryε >0. It follows from2.7that there existsk∈Nsuch that
PAk, Ak1< δε. 2.8
Without loss of the generality, assume thatδε≤εand define the following set:
Π:
A
x, y, z
∈X3:P
FA k,FA
< εδε, x > xk, y≤yk, z > zk
. 2.9
We claim that
FA ∈Π ∀A∈Π. 2.10
TakeA∈Π. Then, by2.8and the triangle inequalitywhich still holds for partial metrics, we have
P
Ak,FA
max p
xk, F
x, y, z , p
yk, F
y, x, y , p
zk, F z, y, x
≤max
pxk, xk1 p
xk1, F
x, y, z , p
yk, yk1 p
yk1, F
y, x, y
, pzk, zk1 p zk1, F
z, y, x max
pxk, xk1 p F
xk, yk, zk
, F
x, y, z , p
yk, yk1 p
F
yk, xk, yk
, F
y, x, y
, pzk, zk1 p F
zk, yk, xk
, F z, y, x
≤PAk, Ak1 P
FA k,FA
< δε P
FA k,FA .
2.11
We consider the following two cases.
Case 1PAk, A≤ε. ByRemark 1.12and the definition ofΠ, the inequality2.11turns into
P
Ak,FA
< δε P
FA k,FA
< δε PAk, A
< δε ε.
2.12
Case 2ε < PAk, A< δε ε. That is,
ε <max
px, xk, p y, yk
, pz, zk < δε ε. 2.13
Sincex > xk, z > zk, y≤yk, then, byii, we have
p F
x, y, z , F
xk, yk, zk
< ε, p
F y, x, y
, F
yk, xk, yk
< ε, p
F z, y, x
, F
zk, yk, xk
< ε.
2.14
Hence, combining2.14and2.11, we get
P
Ak,FA
< δε ε. 2.15
On the other hand, usingi, one can easily check that
F x, y, z
> xk, F y, x, y
≤yk, F z, y, x
> zk. 2.16
Hence, we conclude that2.10holds. By2.8, we have thatAk1 ∈Π, and so, by2.10we get
Ak1∈Π ⇒FA k1 Ak2∈Π ⇒FA k2 Ak3∈Π
· · ·
⇒An∈Π ∀n > k.
2.17
Then, for alln, m > k, we have
PAn, Am≤PAn, Ak PAk, Am<2εδε≤4ε. 2.18
By definition ofP, we have
n,mlim→ ∞pxn, xm lim
n,m→ ∞p yn, ym
lim
n,m→ ∞pzn, zm 0. 2.19
Consequently, by definition of the metricdp,dpx, y≤2px, y, so we get
n,m→ ∞lim dpxn, xm lim
n,m→ ∞dp
yn, ym
lim
n,m→ ∞dpzn, zm 0. 2.20
Therefore,{xn},{yn}, and{zn}are Cauchy sequences in the metric spaceX, dp. SinceX, p is a complete partial metric space, then, byLemma 1.2,X, dpis also a complete metric space.
Hence, there exists a pointx, y, z∈X3such that
dpxn, x, dp
yn, y
, dpzn, z−→0 asn−→∞. 2.21
Again, byLemma 1.2and2.19, we obtain px, x lim
n→ ∞pxn, x lim
n,m→ ∞pxn, xm 0, p
y, y lim
n→ ∞p yn, y
lim
n,m→ ∞p yn, ym
0, pz, z lim
n→ ∞pzn, z lim
n,m→ ∞pzn, zm 0.
2.22
We will prove that
F x, y, z
x, F y, x, y
y, F z, y, x
z. 2.23
To this aim, take an arbitraryε >0. Since xnFn
x0, y0, z0
−→x, ynFn
y0, x0, y0
−→y, znFn
z0, y0, x0
−→z, 2.24
then there existn1, n2, n3 ∈Nsuch that by2.22 pxl, x p
Fl
x0, y0, z0
, x
< px, x εε, p
yq, y p
Fq
y0, x0, y0
, y
< p y, y
εε, pzr, z p
Fr
z0, y0, x0
, z
< pz, z εε,
2.25
for alll≥n1,q≥n2,r ≥n3. Now, takingnmax{n1, n2, n3}and usingRemark 1.12with the assumption
xnFn
x0, y0, z0
< x, ynFn
y0, x0, y0
> y, znFn
z0, y0, x0
< z, 2.26
by2.25, we get p
x, F
x, y, z
≤px, xn1 p
xn1, F
x, y, z
px, xn1 p
Fn1
x0, y0, z0
, F
x, y, z px, xn1 p
F
xn, yn, zn
, F
x, y, z
< px, xn1 max
pxn, x, p yn, y
, pzn, z
<2ε.
2.27
Analogously, we get that
p y, F
y, x, y
<2ε, p z, F
z, y, x
<2ε, 2.28
which yield that
F x, y, z
x, F y, x, y
y, F z, y, x
z. 2.29
Remark 2.2. Theorem 2.1remains true if we replaceivwith one of the following statements.
There existx0, y0, z0∈Xsuch that
1
⎧⎪
⎪⎨
⎪⎪
⎩ x0≤F
x0, y0, z0
, y0 > F
y0, x0, y0
, z0< F
z0, y0, x0
,
2
⎧⎪
⎪⎨
⎪⎪
⎩ x0≤F
x0, y0, z0
, y0 > F
y0, x0, y0
, z0≤F
z0, y0, x0
,
3
⎧⎪
⎪⎨
⎪⎪
⎩ x0≤F
x0, y0, z0
, y0 ≥F
y0, x0, y0
, z0< F
z0, y0, x0
,
4
⎧⎪
⎪⎨
⎪⎪
⎩ x0< F
x0, y0, z0
, y0 ≥F
y0, x0, y0
, z0≤F
z0, y0, x0
,
5
⎧⎪
⎪⎨
⎪⎪
⎩ x0< F
x0, y0, z0
, y0 > F
y0, x0, y0
, z0≤F
z0, y0, x0
.
2.30
3. Uniqueness of Tripled Fixed Point
In this section, we will prove the uniqueness of the tripled fixed point.
Theorem 3.1. In addition to hypotheses ofTheorem 2.1, assume that, for allx, y, z,u, v, r∈X3, there existsa, b, c∈ X3that is comparable tox, y, zandu, v, r. Then,F has a unique tripled fixed point.
Proof. The set of tripled fixed points ofFis not empty due toTheorem 2.1. We suppose that A x, y, z, A∗ x∗, y∗, z∗ ∈ X3 are two tripled fixed points of F. We distinguish the following two cases.
Case 1. x, y, zis comparable tox∗, y∗, z∗with respect to the ordering inX3, where
nlim→∞p Fn
x0, y0, z0
, x
px, x 0,
nlim→∞P Fn
y0, x0, y0
, y p
y, y 0,
n→lim∞p Fn
z0, y0, x0
, z
pz, z 0.
3.1
Without loss of the generality, we may assume that
xF x, y, z
< F
x∗, y∗, z∗ x∗, yF
y, x, y
≥F
y∗, x∗, y∗ y∗, zF
z, y, x
< F
z∗, y∗, x∗ z∗.
3.2
By this, definition ofP,Lemma 1.3, andRemark 1.12, we have
0< PA, A∗ P x, y, z
,
x∗, y∗, z∗ max
px, x∗, p y, y∗
, pz, z∗ max
p F
x, y, z , F
x∗, y∗, z∗ , p
F y, x, y
, F
y∗, x∗, y∗ , p
F z, y, x
, F
z∗, y∗, x∗
<max
px, x∗, p y, y∗
, pz, z∗ PA, A∗,
3.3
which is a contradiction and therefore must beAA∗.
Case 2. x, y, zis not comparable tox∗, y∗, z∗. By assumption, there existsB a, b, c∈X3 which is comparable to bothAandA∗. Without loss of the generality, we may assume that
xF x, y, z
< a, F
x∗, y∗, z∗
x∗< a, yF
y, x, y
≥b, F
y∗, x∗, y∗
y∗≥b, zF
z, y, x
< c, F
z∗, y∗, x∗
z∗< c.
3.4
FromProposition 1.14and3.4, we have
nlim→∞P
FnA,FnB 0,
nlim→∞P
FnA∗,FnB
0. 3.5
By triangle inequality, we derive
PA, A∗ lim
n→∞P
FnA,FnA∗
≤ lim
n→∞P
FnA,FnB lim
n→∞P
FnB,FnA∗
0. 3.6
ByLemma 1.3, we getAA∗.
4. Results of Integral Type
Motivated by Suzuki 35and on the same lines of 31, Theorem 3.1, one can prove the following result.
Theorem 4.1. LetX, p,≤be a partially ordered complete partial metric space, and letF:X3 → X be a given mapping. Assume that there exists a function θ from 0,∞into itself satisfying the following:
Iθ0 0 andθt>0 for everyt >0, IIθis nondecreasing and right continuous, IIIfor everyε >0, there existsδε>0 such that ε≤θ
max
px, u, p y, v
, pz, r < εδε ⇒θ p
F x, y, z
, Fu, v, r
< ε, 4.1
for allx≥u, y≤vandz≥r.
Then,Fis a generalizedp-Meir-Keeler type function.
The following result is an immediate consequence of Theorems2.1and4.1.
Corollary 4.2. LetX, p,≤be a partially ordered complete partial metric spaceF :X3 → X be a mapping satisfying the following hypotheses:
iFhas the mixed strict monotone property, iifor everyε >0, there existsδε>0 such that
ε≤
max{px,u,py,v,pz,r}
0
φtdt < εδε ⇒
pFx,y,z,Fu,v,r
0
φtdt < ε, 4.2
for allx ≥ u, y ≤ v andz ≥ r, whereφ : 0,∞ → 0,∞is a locally integrable function satisfyings
0φtdt >0 for alls >0, iiithere existx0, y0, z0∈Xsuch that
x0< F
x0, y0, z0
, y0≥F
y0, x0, y0
, z0< F
z0, y0, x0
. 4.3
Assume that the hypotheses (a) and (b) given inTheorem 2.1 hold. Then,F has a tripled fixed point.
To end this paper, we give the following corollary.
Corollary 4.3. LetX, d,≤be a partially ordered complete partial metric spaceF : X3 → X be a mapping satisfying the following hypotheses:
iFhas the mixed strict monotone property, iifor all,x≥u, y≤vandz≥r,
pFx,y,z,Fu,v,r
0
φtdt≤k
max{px,u,py,v,pz,r}
0
φtdt, 4.4
wherek ∈0,1andφis a locally integrable function from0,∞into itself satisfying s
0φtdt >0 for alls >0,
iiithere existx0, y0, z0∈Xsuch that
x0 < F
x0, y0, z0
, y0≥F
y0, x0, y0
, z0< F
z0, y0, x0
. 4.5
Assume that the hypotheses (a) and (b) of Theorem 2.1 hold. Then,F has a tripled fixed point.
Proof. For allε >0, we takeδε 1/k−1εand we applyCorollary 4.2.
Remark 4.4. By taking φt 1, we retrieve the analogous of Theorem 1.8of Berinde and Borcut on ordered partial metric spaceswitha b c k/3. In fact, assume that1.8 holds forabck/3, that is,
p F
x, y, z
, Fu, v, w
≤ k 3
px, u p y, v
pz, w
4.6
for allx≥u, y≤v, z≥w. From this inequality, we get that p
F x, y, z
, Fu, v, w
≤kmax
px, u, p y, v
, pz, w , 4.7
which corresponds to4.4withφt 1. Then, we may applyCorollary 4.3.
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