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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

1

and Erdal Karapınar

2

1Institut 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:

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P1ifpx, x px, y py, y, thenxy,

P2px, y py, x,

P3px, xpx, y,

P4px, z py, ypx, y py, z,

for allx, y, zX. 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, xp y, y

1.1 is a metric onX. Each partial metricponXgenerates aT0topologyτponXwith a base of the family of openp-balls{Bpx, ε:xX, ε >0}, whereBpx, ε {y∈X:px, y< px, x ε}

for allxX andε > 0. Similarly, closedp-ball is defined asBpx, ε {y ∈ X : px, ypx, 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 toxXwhenever 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 pointxX, that is, limn,mpxn, 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. Letxnzasn → ∞in a partial metric spaceX, pwhere pz, z 0. Then, limn→ ∞pxn, y pz, yfor everyyX.

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

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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, y3X3.

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

iffxu, yv, zw, 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 : X3X 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, zX,

x1, x2X, x1x2F

x1, y, z

F

x2, y, z , y1, y2X, y1y2F

x, y1, z

F

x, y2, z , z1, z2X, z1z2F

x, y, z1

F x, y, z2

.

1.6

Definition 1.7see32. An elementx, y, z∈X3is called a tripled fixed point ofF:X3X 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:X3Xbe 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 allxu,yvandzw. Assume thatXhas the following properties:

iif a nondecreasing sequencexnx, thenxnxfor alln, iiif a nonincreasing sequenceyny, thenynyfor alln. If there existx0, y0, z0Xsuch that

x0F

x0, y0, z0

, y0F

y0, x0, y0

, z0F

z0, y0, x0

, 1.9

then there existx, y, zXsuch that F

x, y, z

x, F y, x, y

y, F z, y, x

z. 1.10

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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 :X3X. We say thatF has the mixed strict monotone property if, for anyx, y, zX,

x1, x2X, x1< x2F

x1, y, z

< F

x2, y, z , y1, y2X, y1< y2F

x, y1, z

> F

x, y2, z , z1, z2X, z1< z2F

x, y, z1

< F x, y, z2

.

1.11

Definition 1.10 see34. LetX, d,≤be a partially ordered metric space. A mapping F : X3Xis 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, rXwithxu,yvandzr.

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:X3X 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, rXwithxu,yvandzr.

Remark 1.12. It is immediate to show that ifF :X3Xis 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, yv, z < rorxu, y > v, zr.

Proposition 1.13. LetX, p,≤be a partially ordered partial metric space andF :X3Xa 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:X3X3be such that, fora, b, cX,

Fa, b, c Fa, b, c, Fb, a, b, Fc, b, a. 1.15

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Letx0, y0, z0Xbe such that x0< F

x0, y0, z0

, y0F

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 :X3Xbe 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,yvandz < 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,yvandz < r, we obtain

x1F x, y, z

F

x0, y0, z0

< F

u0, y0, z0

F

x0, y0, z0

< Fu0, v0, z0F

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

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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

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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

εtnPAn, Bn< εδε for any nk. 1.29

In particular, fornk, we have

εtkPAk, 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, yv, z < r 1.37

with the following

x, y, z

,u, v, r∈X3 such thatxu, y > v, zr. 1.38

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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, . . .andxnx, thenxn < x for eachn1,2, . . .,

bif{yn}is a sequence such thatyn1 < yn for eachn1,2, . . .andyny, thenyn > y for eachn1,2, . . ..

Assume thatF:X3Xsatisfies the following hypotheses:

iFhas the mixed strict monotone property, iiFis a generalizedp-Meir-Keeler type function, iiithere existx0, y0, z0Xsuch that

x0< F

x0, y0, z0

, y0F

y0, x0, y0

, z0< F

z0, y0, x0,

. 2.1

Then,Fhas a tripled fixed point, that is, there existx, y, zXsuch 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, z0Xbe 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<· · ·< y1y0,

· · ·> 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, y0F

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.

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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 thatδεεand define the following set:

Π:

A

x, y, z

X3:P

FA k,FA

< εδε, x > xk, yyk, 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, Aε. 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

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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, yyk, 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, AmPAn, Ak PAk, Am<δε≤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

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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 allln1,qn2,rn3. 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

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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, z0Xsuch that

1

⎧⎪

⎪⎨

⎪⎪

x0F

x0, y0, z0

, y0 > F

y0, x0, y0

, z0< F

z0, y0, x0

,

2

⎧⎪

⎪⎨

⎪⎪

x0F

x0, y0, z0

, y0 > F

y0, x0, y0

, z0F

z0, y0, x0

,

3

⎧⎪

⎪⎨

⎪⎪

x0F

x0, y0, z0

, y0F

y0, x0, y0

, z0< F

z0, y0, x0

,

4

⎧⎪

⎪⎨

⎪⎪

x0< F

x0, y0, z0

, y0F

y0, x0, y0

, z0F

z0, y0, x0

,

5

⎧⎪

⎪⎨

⎪⎪

x0< F

x0, y0, z0

, y0 > F

y0, x0, y0

, z0F

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, zX3 are two tripled fixed points of F. We distinguish the following two cases.

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Case 1. x, y, zis comparable tox, y, zwith respect to the ordering inX3, where

nlim→∞p Fn

x0, y0, z0

, x

px, x 0,

nlimP Fn

y0, x0, y0

, y p

y, y 0,

n→limp 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

yb, 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

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By triangle inequality, we derive

PA, A lim

n→∞P

FnA,FnA

≤ lim

n→∞P

FnA,FnB lim

nP

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:X3X 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 εθ

max

px, u, p y, v

, pz, r < εδε θ p

F x, y, z

, Fu, v, r

< ε, 4.1

for allxu, yvandzr.

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 :X3X 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 allxu, yv andzr, whereφ : 0,∞ → 0,∞is a locally integrable function satisfyings

0φtdt >0 for alls >0, iiithere existx0, y0, z0Xsuch that

x0< F

x0, y0, z0

, y0F

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.

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To end this paper, we give the following corollary.

Corollary 4.3. LetX, d,≤be a partially ordered complete partial metric spaceF : X3X be a mapping satisfying the following hypotheses:

iFhas the mixed strict monotone property, iifor all,xu, yvandzr,

pFx,y,z,Fu,v,r

0

φtdtk

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, z0Xsuch that

x0 < F

x0, y0, z0

, y0F

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 allxu, yv, zw. 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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