# New Meir-Keeler Type Tripled Fixed-Point Theorems on Ordered Partial Metric Spaces

Authors: Hassen Aydi; Erdal Karapınar
Journal: Mathematical Problems in Engineering
 (2012)

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## Abstract

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.


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## Body

## 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 Matthews [1] when he considered the tools of metric spaces in the field of semantics and domain theory in computer science (see, e.g., [2, 3]). After the initial results of Mathews, other papers have been released on partial metric spaces (see 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 given (see, 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.LetX be a nonempty set. A partial metric is a function p:X×X→[0,∞) satisfying the following conditions:(P1)
ifp(x,x)=p(x,y)=p(y,y), then x=y,(P2)
p(x,y)=p(y,x),(P3)
p(x,x)≤p(x,y),(P4)
p(x,z)+p(y,y)≤p(x,y)+p(y,z),for allx,y,z∈X. Then, (X,p) is called a partial metric space.Ifp is a partial metric p on X, then the function dp:X×X→[0,∞) given by
(1.1)dp(x,y)=2p(x,y)-p(x,x)-p(y,y)
is a metric on X. Each partial metric p on X generates a T0 topology τp on X with a base of the family of open p-balls {Bp(x,ε):x∈X,ε>0}, where Bp(x,ε)={y∈X:p(x,y)<p(x,x)+ε} for all x∈X and ε>0. Similarly, closed p-ball is defined as Bp[x,ε]={y∈X:p(x,y)≤p(x,x)+ε}. For more details, see [1, 5].Definition 1.1 (see, e.g., [1, 5, 15]).
Let(X,p) be a partial metric space.
(i)
A sequence{xn} in X converges to x∈X whenever limn→∞p(x,xn)=p(x,x).(ii)
A sequence{xn} in X is called Cauchy whenever limn,m→∞p(xn,xm) exists (and finite).(iii)
(X,p) is said to be complete if every Cauchy sequence {xn} in X converges, with respect to τp, to a point x∈X, that is, limn,m→∞p(xn,xm)=p(x,x).Lemma 1.2 (see, e.g., [1, 5, 15]).
Let(X,p) be a partial metric space.
(a)
A sequence{xn} is Cauchy if and only if {xn} is a Cauchy sequence in the metric space (X,dp).(b)
(X,p) is complete if and only if the metric space (X,dp) is complete. Moreover,
(1.2)lim⁡n→∞dp(x,xn)=0⟺lim⁡n→∞p(x,xn)=lim⁡n,m→∞p(xn,xm)=p(x,x).Lemma 1.3 (see, e.g., [4, 15, 16]).
Let(X,p) be a partial metric space. Then, (A)
ifp(x,y)=0, then x=y,(B)
ifx≠y, then p(x,y)>0.Remark 1.4.
Ifx=y,p(x,y) may not be 0.Lemma 1.5 (see, e.g., [4, 15, 16]).
Letxn→z as  n→∞ in a partial metric space (X,p) where p(z,z)=0.  Then, limn→∞p(xn,y)=p(z,y) for every y∈X.(X,p,≤) is called a partially ordered partial metric space if (X,≤) is a partially ordered set and (X,p) is a partial metric space. Further, if (X,p) is 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
(1.3)Xk=X×X×⋯×X︸k-many.
Also, take the mapping P:X3×X3→[0,+∞) such that
(1.4)P(x,y):=max⁡{p(x1,y1),p(x2,y2),p(x3,y3)},
where x=(x1,x2,x3) and y=(y1,y2,y3)∈X3.Let(X,≤) be a partially ordered set. We consider the following partial order (also denoted by ≤) on the product space X3:
(1.5)(u,v,w)≤(x,y,z)iffx≥u,y≤v,z≥w,
where (u,v,w),(x,y,z)∈X3. Moreover, we say that (x,y,z) is equal to (u,v,r) if and only if x=u,=v, and z=r. 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 F(x,y,z) is monotone nondecreasing in x and z,and it is monotone nonincreasing in y, that is, for anyx,y,z∈X,
(1.6)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).Definition 1.7 (see [32]).
An element(x,y,z)∈X3 is called a tripled fixed point of F:X3→X if
(1.7)F(x,y,z)=x,F(y,x,y)=y,F(z,y,x)=z.Berinde and Borcut [32] proved the following theorem.Theorem 1.8.
Let(X,≤) be a partially ordered set and  (X,d) a complete metric space. Let  F:X3→X be a mapping having the mixed monotone property on  X. Assume that there exist constants  a,b,c∈[0,1) such that  a+b+c<1 for which
(1.8)d(F(x,y,z),F(u,v,w))≤ad(x,u)+bd(y,v)+cd(z,w)
for all x≥u,y≤v and z≥w. Assume that X has the following properties:(i)
if a nondecreasing sequencexn→x, then xn≤x for all n,(ii)
if a nonincreasing sequenceyn→y, then yn≥y for all n.If there exist x0,y0,z0∈X such that
(1.9)x0≤F(x0,y0,z0),y0≥F(y0,x0,y0),z0≤F(z0,y0,x0),
then there exist x,y,z∈X such that
(1.10)F(x,y,z)=x,F(y,x,y)=y,F(z,y,x)=z.Recently, Theorem1.8 is extended to cone metric spaces by Rao and Kishore [33]. On the other hand, very recently, Aydi et al. [34] introduced the following concepts.Definition 1.9 (see [34]).
Let(X,≤) be a partially ordered set and F:X3→X.We say that F has the mixed strict monotone property if, for anyx,y,z∈X,
(1.11)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).Definition 1.10 (see [34]).
Let(X,d,≤) be a partially ordered metric space. A mapping F:X3→X is said to be a generalized Meir-Keeler type contraction if, for any ε>0, there exists a δ(ε)>0 such that
(1.12)ε≤max⁡{d(x,u),d(y,v),d(z,r)}<ε+δ(ε)⟹d(F(x,y,z),F(u,v,r))<ε,
for all x,y,z,u,v,r∈X with x≤u, y≥v and z≤r.In the following, we consider the partial case of Definition1.10 and we introduce the following.Definition 1.11.
Let(X,p,≤) be a partially ordered partial metric space. A mapping F:X3→X is said to be a generalized p-Meir-Keeler type contraction if, for any ε>0, there exists a δ(ε)>0 such that
(1.13)ε≤max⁡{p(x,u),p(y,v),p(z,r)}<ε+δ(ε)⟹p(F(x,y,z),F(u,v,r))<ε,
for all x,y,z,u,v,r∈X with x≤u, y≥v and z≤r.Remark 1.12.
It is immediate to show that ifF:X3→X is a generalized p-Meir-Keeler type contraction, then
(1.14)p(F(x,y,z),F(u,v,r))<max⁡{p(x,u),p(y,v),p(z,r)}
for all x,u,y,v,z,r,∈X with x<u,y≥v,z<r or x≤u,y>v,z≤r.Proposition 1.13.
Let(X,p,≤) be a partially ordered partial metric space and F:X3→X a given mapping. If (1.8) is satisfied, then F is a generalized p-Meir-Keeler type function.Proof.
Assume that (1.8) is satisfied. For all ε>0, one can check that (1.13) is satisfied with δ(ε)=(1/(a+b+c)-1)ε.In the sequel, we use the following notations given in [34]. Let F̃:X3→X3 be such that, fora,b,c∈X,
(1.15)F̃(a,b,c)=(F(a,b,c),F(b,a,b),F(c,b,a)).
Let x0,y0,z0∈X be such that
(1.16)x0<F(x0,y0,z0),y0≥F(y0,x0,y0),z0<F(z0,y0,x0).
We consider sequences {xn}, {yn}, and {zn} such that
(1.17)[xnynzn]︸An=[F(xn-1,yn-1,zn-1)F(yn-1,xn-1,yn-1)F(zn-1,yn-1,xn-1)]︸F̃(An-1)=[Fn(x0,y0,z0)Fn(y0,x0,y0)Fn(z0,y0,x0)]︸Fñ(A0),
for n=1,2,3,….Our first auxiliary result is as follows.Proposition 1.14.
Let(X,p,≤)be a partially ordered partial metric space, and letF:X3→Xbe a given mapping such that the following hypotheses hold:(i)
F has the mixed strict monotone property,(ii)
F is a generalized p-Meir-Keeler type function,(iii)
∃(x,y,z),(u,v,r)∈X3 such that x<u, y≥v and z<r.Then,(1.18)P(Fñ(x,y,z),Fñ(u,v,r))⟶0,asn⟶+∞.Proof.
Let(x,y,z)=(x0,y0,z0) and (u,v,r)=(u0,v0,r0). We show that
(1.19)xn=Fn(x0,y0,z0,)<Fn(u0,v0,r0)=un,yn=Fn(y0,x0,y0)>Fn(v0,u0,v0)=vn,zn=Fn(z0,y0,x0)<Fn(r0,v0,u0)=rn,∀n=1,2,…,
with F=F1.
Due to the fact thatF has the mixed strict monotone property, together with the assumption that x<u, y≥v and z<r, we obtain
(1.20)x1=F(x,y,z)=F(x0,y0,z0)<F(u0,y0,z0)⟹F(x0,y0,z0)<F(u0,v0,z0)⟹F(x0,y0,z0)<F(u0,v0,r0)=u1.
Analogously, we have
(1.21)y1=F(y0,x0,y0)>F(v0,u0,v0)=v1,z1=F(z0,y0,x0)<F(r0,v0,u0)=r1.
Thus, (1.19) holds for n=1. By using the same arguments, we show that (1.19) holds also for n=2. In fact,
(1.22)x2=F2(x0,y0,z0)=F(x1,y1,z1)=F(F(x0,y0,z0),F(y0,x0,y0),F(z0,y0,x0))<F(F(u0,v0,r0),F(y0,x0,y0),F(z0,y0,x0))<F(F(u0,v0,r0),F(v0,u0,v0),F(z0,y0,x0))<F(F(u0,v0,r0),F(v0,u0,v0),F(r0,v0,u0))=F2(u0,v0,r0)=F(u1,v1,r1)=u2.
Similarly, we find
(1.23)y2=F2(y0,x0,y0)≥F2(v0,u0,v0)=v2,z2=F2(z0,y0,x0)<F2(r0,v0,u0)=r2.
Inductively, we get that (1.19) holds.
By Remark1.12, together with (1.19), we have
(1.24)p(xn+2,un+2)=p(Fn+2(x0,y0,z0),Fn+2(u0,v0,r0))=p(F(xn+1,yn+1,zn+1),F(un+1,vn+1,rn+1))<max⁡{p(xn+1,un+1),p(yn+1,vn+1),p(zn+1,rn+1)},(1.25)p(zn+2,rn+2)=p(Fn+2(z0,y0,x0),Fn+2(r0,v0,u0))=p(F(zn+1,yn+1,xn+1),F(rn+1,vn+1,un+1))<max⁡{p(zn+1,rn+1),p(yn+1,vn+1),p(xn+1,un+1)},(1.26)p(yn+2,vn+2)=p(Fn+2(y0,x0,y0),Fn+2(v0,u0,v0))=p(F(yn+1,xn+1,yn+1),F(vn+1,un+1,vn+1))<max⁡{p(yn+1,vn+1),p(xn+1,un+1),p(yn+1,vn+1)}≤max⁡{p(zn+1,rn+1),p(yn+1,vn+1),p(xn+1,un+1)}.
Let Δn+1:=max{p(xn+1,un+1),p(yn+1,vn+1),p(zn+1,rn+1)}. Combining (1.24)–(1.26), we get
(1.27)Δn+2<Δn+1,∀n=1,2….
If we denote Bn=(un,vn,rn), then, by definition of the partial metric P and (1.27), we have
(1.28)P(An+2,Bn+2)<P(An+1,Bn+1).
Consequently, the sequence {tn}={P(An,Bn)} is decreasing. Hence, {tn} converges, say to ε≥0. Clearly, if ε=0, we have finished. Suppose, on the contrary, ε>0. Thus, there exists k∈{1,2,…} such that
(1.29)ε≤tn=P(An,Bn)<ε+δ(ε)foranyn≥k.
In particular, for n=k, we have
(1.30)ε≤tk=P(Ak,Bk)<ε+δ(ε),
that is equal to
(1.31)ε≤Δk<ε+δ(ε).
It follows from (1.19) and the hypothesis (ii) that
(1.32)p(F(xk,yk,zk),F(uk,vk,rk))<ε
which is equivalent to
(1.33)p(xk+1,uk+1)<ε.
Moreover, we have
(1.34)p(yk+1,vk+1)<ε,p(zk+1,rk+1)<ε.
Combining (1.33) and (1.34), we have
(1.35)Δk+1<ε.
Thus, tk+1=P(Ak+1,Bk+1)<ε which is a contradiction with respect to (1.29), and so ε=0. We conclude that
(1.36)P(An,Bn)=P(Fñ(x,y,z),Fñ(u,v,r))⟶0,asn⟶+∞.Remark 1.15.
The previous proposition remains true if, in (iii), we change the assumption(1.37)∃(x,y,z),(u,v,r)∈X3suchthatx<u,y≥v,z<r
with the following
(1.38)∃(x,y,z),(u,v,r)∈X3suchthatx≤u,y>v,z≤r.


## 2. Existence of Tripled Fixed Point

The following theorem is our first main result.Theorem 2.1.
Let(X,p,≤) be a partially ordered complete partial metric space.  Suppose that X has the following properties:
(a)
if{xn} is a sequence such that xn+1>xn for each n=1,2,… and xn→x, then xn<x for each n=1,2,…,(b)
if{yn} is a sequence such that yn+1<yn for each n=1,2,… and yn→y, then yn>y for each n=1,2,….
Assume thatF:X3→X satisfies the following hypotheses:(i)
F has the mixed strict monotone property,(ii)
F is a generalized p-Meir-Keeler type function,(iii)
there existx0,y0,z0∈X such that
(2.1)x0<F(x0,y0,z0),y0≥F(y0,x0,y0),z0<F(z0,y0,x0,).
Then, F has a tripled fixed point, that is, there exist x,y,z∈X such that
(2.2)F(x,y,z)=x,F(y,x,y)=y,F(z,y,x)=z.
Also, p(x,x)=p(y,y)=p(z,z)=0.Proof.
Letx0,y0,z0∈X be as in (iii). We construct sequences {xn}, {yn}, and {zn} according to (1.17).
We claim that, for alln≥2, we have
(2.3)⋯>xn>xn-1>⋯>x1>x0,⋯<yn<yn-1<⋯<y1≤y0,⋯>zn>zn-1>⋯>z1>z0.
Indeed, we will use a mathematical induction to prove (2.3). Clearly, we have
(2.4)x0<F(x0,y0,z0)=x1,y0≥F(y0,x0,y0)=y1,z0<F(z0,y0,x0)=z1.
Suppose now that the inequalities in (2.3) hold for some n≥2. By the mixed strict monotone property of F, together with (1.17), we have
(2.5)xn=F(xn-1,yn-1,zn-1)<F(xn,yn,zn)=xn+1,yn=F(yn-1,xn-1,yn-1)>F(yn,xn,yn)=yn+1,zn=F(zn-1,yn-1,xn-1)<F(zn,yn,xn)=zn+1.
Thus, (2.3) holds for all n≥2.
Putting(x,y,z)=A0 and (u,v,r)=A1 and by Proposition 1.14, we get
(2.6)P(Fñ(A0),Fñ(A1))⟶0,asn⟶+∞,
which is equivalent to
(2.7)P(An,An+1)⟶0,asn⟶+∞.
Take an arbitrary ε>0. It follows from (2.7) that there exists k∈ℕ such that
(2.8)P(Ak,Ak+1)<δ(ε).
Without loss of the generality, assume that δ(ε)≤ε and define the following set:
(2.9)Π:={A=(x,y,z)∈X3:P(F̃(Ak),F̃(A))<ε+δ(ε),x>xk,y≤yk,z>zk}.
We claim that
(2.10)F̃(A)∈Π∀A∈Π.
Take A∈Π. Then, by (2.8) and the triangle inequality (which still holds for partial metrics), we have
(2.11)P(Ak,F̃(A))=max⁡{p(xk,F(x,y,z)),p(yk,F(y,x,y)),p(zk,F(z,y,x))}≤max⁡{p(xk,xk+1)+p(xk+1,F(x,y,z)),p(yk,yk+1)+p(yk+1,F(y,x,y)),p(zk,zk+1)+p(zk+1,F(z,y,x))}=max⁡{p(xk,xk+1)+p(F(xk,yk,zk),F(x,y,z)),p(yk,yk+1)+p(F(yk,xk,yk),F(y,x,y)),p(zk,zk+1)+p(F(zk,yk,xk),F(z,y,x))}≤P(Ak,Ak+1)+P(F̃(Ak),F̃(A))<δ(ε)+P(F̃(Ak),F̃(A)).
We consider the following two cases.Case 1(P(Ak,A)≤ε).
By Remark1.12 and the definition of Π, the inequality (2.11) turns into
(2.12)P(Ak,F̃(A))<δ(ε)+P(F̃(Ak),F̃(A))<δ(ε)+P(Ak,A)<δ(ε)+ε.Case 2 (ε<P(Ak,A)<δ(ε)+ε).
That is,(2.13)ε<max{p(x,xk),p(y,yk),p(z,zk)}<δ(ε)+ε.
Since x>xk,z>zk,y≤yk, then, by (ii), we have
(2.14)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.
Hence, combining (2.14) and (2.11), we get
(2.15)P(Ak,F̃(A))<δ(ε)+ε.
On the other hand, using (i), one can easily check that
(2.16)F(x,y,z)>xk,F(y,x,y)≤yk,F(z,y,x)>zk.
Hence, we conclude that (2.10) holds. By (2.8), we have that Ak+1∈Π, and so, by (2.10) we get
(2.17)Ak+1∈Π⟹F̃(Ak+1)=Ak+2∈Π⟹F̃(Ak+2)=Ak+3∈Π⋯⟹An∈Π∀n>k.
Then, for all n,m>k, we have
(2.18)P(An,Am)≤P(An,Ak)+P(Ak,Am)<2(ε+δ(ε))≤4ε.
By definition of P, we have
(2.19)lim⁡n,m→∞p(xn,xm)=lim⁡n,m→∞p(yn,ym)=lim⁡n,m→∞p(zn,zm)=0.
Consequently, by definition of the metric dp, dp(x,y)≤2p(x,y), so we get
(2.20)lim⁡n,m→∞dp(xn,xm)=lim⁡n,m→∞dp(yn,ym)=lim⁡n,m→∞dp(zn,zm)=0.
Therefore, {xn}, {yn},  and {zn} are Cauchy sequences in the metric space (X,dp). Since (X,p) is a complete partial metric space, then, by Lemma 1.2, (X,dp) is also a complete metric space. Hence, there exists a point (x,y,z)∈X3 such that
(2.21)dp(xn,x),dp(yn,y),dp(zn,z)⟶0asn⟶+∞.
Again, by Lemma 1.2 and (2.19), we obtain
(2.22)p(x,x)=lim⁡n→∞p(xn,x)=lim⁡n,m→∞p(xn,xm)=0,p(y,y)=lim⁡n→∞p(yn,y)=lim⁡n,m→∞p(yn,ym)=0,p(z,z)=lim⁡n→∞p(zn,z)=lim⁡n,m→∞p(zn,zm)=0.
We will prove that
(2.23)F(x,y,z)=x,F(y,x,y)=y,F(z,y,x)=z.
To this aim, take an arbitrary ε>0. Since
(2.24)xn=Fn(x0,y0,z0)⟶x,yn=Fn(y0,x0,y0)⟶y,zn=Fn(z0,y0,x0)⟶z,
then there exist n1,n2,n3∈ℕ such that by (2.22)
(2.25)p(xl,x)=p(Fl(x0,y0,z0),x)<p(x,x)+ε=ε,p(yq,y)=p(Fq(y0,x0,y0),y)<p(y,y)+ε=ε,p(zr,z)=p(Fr(z0,y0,x0),z)<p(z,z)+ε=ε,
for all l≥n1, q≥n2, r≥n3. Now, taking n=max{n1,n2,n3} and using Remark 1.12 with the assumption
(2.26)xn=Fn(x0,y0,z0)<x,yn=Fn(y0,x0,y0)>y,zn=Fn(z0,y0,x0)<z,
by (2.25), we get
(2.27)p(x,F(x,y,z))≤p(x,xn+1)+p(xn+1,F(x,y,z))=p(x,xn+1)+p(Fn+1(x0,y0,z0),F(x,y,z))=p(x,xn+1)+p(F(xn,yn,zn),F(x,y,z))<p(x,xn+1)+max⁡{p(xn,x),p(yn,y),p(zn,z)}<2ε.
Analogously, we get that
(2.28)p(y,F(y,x,y))<2ε,p(z,F(z,y,x))<2ε,
which yield that
(2.29)F(x,y,z)=x,F(y,x,y)=y,F(z,y,x)=z.Remark 2.2.
Theorem2.1 remains true if we replace (iv) with one of the following statements.
There existx0,y0,z0∈X such that
(2.30)(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).


## 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 of Theorem2.1, assume that, for all (x,y,z), (u,v,r)∈X3, there exists (a,b,c)∈X3 that is comparable to (x,y,z) and (u,v,r). Then, F has a unique tripled fixed point.Proof.
The set of tripled fixed points ofF is not empty due to Theorem 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,z) is comparable to (x*,y*,z*) with respect to the ordering in X3, where
(3.1)lim⁡n→+∞p(Fn(x0,y0,z0),x)=p(x,x)=0,lim⁡n→+∞P(Fn(y0,x0,y0),y)=p(y,y)=0,lim⁡n→+∞p(Fn(z0,y0,x0),z)=p(z,z)=0.
Without loss of the generality, we may assume that
(3.2)x=F(x,y,z)<F(x*,y*,z*)=x*,y=F(y,x,y)≥F(y*,x*,y*)=y*,z=F(z,y,x)<F(z*,y*,x*)=z*.
By this, definition of P, Lemma 1.3, and Remark 1.12, we have
(3.3)0<P(A,A*)=P((x,y,z),(x*,y*,z*))=max⁡{p(x,x*),p(y,y*),p(z,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⁡{p(x,x*),p(y,y*),p(z,z*)}=P(A,A*),
which is a contradiction and therefore must be A=A*.Case 2.
(x,y,z) is not comparable to (x*,y*,z*). By assumption, there exists B=(a,b,c)∈X3 which is comparable to both A and A*. Without loss of the generality, we may assume that
(3.4)x=F(x,y,z)<a,F(x*,y*,z*)=x*<a,y=F(y,x,y)≥b,F(y*,x*,y*)=y*≥b,z=F(z,y,x)<c,F(z*,y*,x*)=z*<c.
From Proposition 1.14 and (3.4), we have
(3.5)lim⁡n→+∞P(Fñ(A),Fñ(B))=0,lim⁡n→+∞P(Fñ(A*),Fñ(B))=0.
By triangle inequality, we derive
(3.6)P(A,A*)=lim⁡n→+∞P(Fñ(A),Fñ(A*))≤lim⁡n→+∞P(Fñ(A),Fñ(B))+lim⁡n→+∞P(Fñ(B),Fñ(A*))=0.
By Lemma 1.3, we get A=A*.


## 4. Results of Integral Type

Motivated by Suzuki [35] and on the same lines of  [31, Theorem  3.1], one can prove the following result.Theorem 4.1.
Let(X,p,≤) be a partially ordered complete partial metric space, and let F:X3→X be  a given mapping. Assume that there exists a function θ from [0,+∞) into itself satisfying the following:(I)
θ(0)=0 and θ(t)>0 for every t>0,(II)
θ is nondecreasing and right continuous,(III)
for everyε>0, there exists δ(ε)>0 such that
(4.1)ε≤θ(max⁡⁡{p(x,u),p(y,v),p(z,r)})<ε+δ(ε)⟹θ(p(F(x,y,z),F(u,v,r)))<ε,
for all x≥u,y≤v and z≥r.
Then,F is a generalized p-Meir-Keeler type function.The following result is an immediate consequence of Theorems2.1 and 4.1.Corollary 4.2.
Let(X,p,≤) be a partially ordered complete partial metric space F:X3→X be a mapping satisfying the following hypotheses:(i)
F has the mixed strict monotone property,(ii)
for everyε>0, there exists δ(ε)>0 such that
(4.2)ε≤∫0max⁡{p(x,u),p(y,v),p(z,r)}ϕ(t)dt<ε+δ(ε)⟹∫0p(F(x,y,z),F(u,v,r))ϕ(t)dt<ε,
for all x≥u,y≤v and z≥r, where ϕ:[0,+∞)→[0,+∞) is a locally integrable function satisfying ∫0sϕ(t)dt>0 for all s>0,(iii)
there existx0,y0,z0∈X such that
(4.3)x0<F(x0,y0,z0),y0≥F(y0,x0,y0),z0<F(z0,y0,x0).
Assume that the hypotheses (a) and (b) given in Theorem 2.1 hold. Then, F has a tripled fixed point.To end this paper, we give the following corollary.Corollary 4.3.
Let(X,d,≤) be a partially ordered complete partial metric space F:X3→X be a mapping satisfying the following hypotheses:
(i)
F has the mixed strict monotone property,(ii)
for all,x≥u,y≤v andz≥r,
(4.4)∫0p(F(x,y,z),F(u,v,r))ϕ(t)dt≤k∫0max⁡{p(x,u),p(y,v),p(z,r)}ϕ(t)dt,
where k∈(0,1) and ϕ is a locally integrable function from [0,+∞) into itself satisfying ∫0sϕ(t)dt>0 for all s>0,(iii)
there existx0,y0,z0∈X such that
(4.5)x0<F(x0,y0,z0),y0≥F(y0,x0,y0),z0<F(z0,y0,x0).
Assume that the hypotheses (a) and (b) of Theorem 2.1 hold. Then, F has a tripled fixed point.Proof.
For allε>0, we take δ(ε)=(1/k-1)ε and we apply Corollary 4.2.Remark 4.4.
By takingϕ(t)=1,we retrieve the analogous of Theorem 1.8 of Berinde and Borcut on ordered partial metric spaces (with a=b=c=k/3). In fact, assume that (1.8) holds for  a=b=c=k/3, that is,
(4.6)p(F(x,y,z),F(u,v,w))≤k3(p(x,u)+p(y,v)+p(z,w))
for all x≥u,y≤v,z≥w. From this inequality, we get that
(4.7)p(F(x,y,z),F(u,v,w))≤kmax⁡{p(x,u),p(y,v),p(z,w)},
which corresponds to (4.4) with ϕ(t)=1. Then, we may apply Corollary 4.3.



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*Source: 409872-2012-05-29.xml*