# Fixed Point Theorems for Geraghty Type Contractive Mappings and Coupled Fixed Point Results in 0-Complete Ordered Partial Metric Spaces

Authors: Esra Yolacan
Journal: International Journal of Analysis
 (2016)

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## Abstract

We establish new fixed point theorems in 0-complete ordered partial metric spaces. Also, we give remark on coupled generalized Banach contraction. Some examples illustrate the usability of our results. The theorems presented in this paper are generalizations and improvements of the several well known results in the literature.


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## Body

## 1. Introduction and Preliminaries

Henceforward, the lettersR, R
+, and N will indicate the set of real numbers, the set of nonnegative real numbers, and the set of positive integer numbers, respectively.Definition 1 (see [1]).
A partial metric on a nonempty setX is a function P
:
X
2
→
R
+ such that, for all x
,
y
,
z
∈
X, (P
1) x
=
y
⇔
P
x
,
x
=
P
x
,
y
=
P
y
,
y, (P
2) P
x
,
x
≤
P
x
,
y, (P
3) P
x
,
y
=
P
y
,
x, and (P
4) P
x
,
y
≤
P
x
,
z
+
P
z
,
y
-
P
z
,
z. The pair X
,
P is called a partial metric space.IfP is a partial metric on X, then the function P
s
:
X
2
→
R
+ given by P
s
x
,
y
=
-
P
y
,
y
-
P
x
,
x
+
2
P
x
,
y is a metric on X. Each partial metric P on X introduces a T
0 topology τ
P on X which has as a base the family of open balls B
P
x
,
ε
=
p
∈
X
:
P
x
,
p
<
P
x
,
x
+
ε for all x
∈
X and ε
>
0
.LetX
,
P be a partial metric space, and let x
n be any sequence in X and x
∈
X. Then (i) a sequence x
n is convergent to x with respect to τ
P, if P
x
n
,
x
→
P
x
,
x as n
→
∞, (ii) a sequence x
n is a Cauchy sequence in X
,
P if lim
n
,
m
→
∞
⁡
P
x
n
,
x
m exists and is finite; (iii) X
,
P is called complete if for every Cauchy sequence x
n in X there exists x
∈
X such that P
x
n
,
x
m
→
P
x
,
x as n
,
m
→
∞
.Romaguera [2] introduced the notion of 0-Cauchy sequence, 0-complete partial metric spaces and proved some characterizations of partial metric spaces in terms of completeness and 0-completeness. After that many authors extended the results of [2] and studied fixed point theorems in 0-complete partial metric space (see [2–10]).Definition 2 (see [2]).
LetX
,
P be a partial metric space. A sequence x
n in X is called a 0-Cauchy sequence if P
x
n
,
x
m
→
0 as n
,
m
→
∞. The space X
,
P is said to be 0-complete if every 0-Cauchy sequence in X converges with respect to τ
P to a point x
∈
X such that P
x
,
x
=
0
.Remark 3 (see [11, 12]).
(1) LetX
,
P be a partial metric space. If P
x
n
,
ρ
→
P
ρ
,
ρ
=
0 as n
→
∞, then P
x
n
,
y
→
P
ρ
,
y as n
→
∞ for all y
∈
X.
(2) Iff
:
X
→
X is a continuous at ρ, then for each sequence x
n in X, we have P
x
n
,
ρ
→
P
ρ
,
ρ
=
0 as n
→
∞
⇒
P
f
x
n
,
f
ρ
→
P
f
ρ
,
f
ρ
=
0 as n
→
∞ (see [5]).Let₣ be the class of functions θ
:
0
,
∞
→
0,1 with θ
t
n
→
1 implying t
n
→
0
. Amini-Harandi and Emami [13] presented the following results.Theorem 4 (see [13]).
LetX
,
≼ be an ordered set endowed with a metric d and let f
:
X
→
X be a given mapping. Suppose that the following conditions hold:(i)
X
,
d is complete.
(ii)
(1)
f is continuous or
(2)
if a nondecreasing sequencex
n in X converges to some point x
∈
X, then x
n
≼
x for all n
∈
N.
(iii)
f is nondecreasing.
(iv)
There existsx
0
∈
X such that x
0
≼
f
x
0.
(v)
There existsθ
∈
₣ such that for all x
,
y
∈
X with x
≽
y,(1)dfx,fy≤θdx,ydx,y.
Thenf has a fixed point. Moreover, if for all x
,
y
∈
X
2 there exists a z
∈
X such that x
≼
z and y
≼
z, we obtain uniqueness of the fixed point.In this paper, we establish new fixed point theorems in 0-complete ordered partial metric spaces (briefly 0-COPMS). Also, we give remark on coupled generalized Banach contraction. Some examples illustrate the usability of our results. The theorems presented in this paper are generalizations and improvements of the several well known results in the literature.


## 2. Main Results

Theorem 5.
LetX
,
P be a 0-COPMS. Let f
:
X
→
X be a nondecreasing mapping such that(2)Pfx,fy≤θPx,yPx,yfor all x
,
y
∈
X with x
≽
y
o
r
y
≽
x and θ
∈
₣. Also suppose that there exists x
0
∈
X such that x
0
≼
f
x
0
o
r
x
0
≽
f
x
0. One assumes (1)
f is continuous or
(2)
if a nondecreasing sequencex
n in X converges to some point x
∈
X, then x
n
≼
x
o
r
x
n
≽
x for all n
∈
N.
Thenf has a fixed point ρ.Proof.
By assumption there existsx
0
∈
X such that x
0
≼
f
x
0. Define x
1
∈
X as x
1
=
f
x
0. Then we have x
1
≽
x
0. In a similar manner, we get x
2
∈
X as x
2
=
f
x
1. In that case, x
2
=
f
x
1
≽
f
x
0
=
x
1. Continuing this procedure we have x
n in X such that(3)fxn=xn+1,n∈N.If x
n
+
1
=
x
n for some n
∈
N, then the proof is completed. Suppose farther that x
n
+
1
≠
x
n for each n
∈
N. Consider, as f is nondecreasing, we obtain that(4)x0≼fx0=x1≼fx1=x2≼⋯≼xn=fxn-1≼xn+1=fxn≼⋯.From (2), (3), and (4), for all n
≥
1, we get that(5)Pxn+1,xn+2=Pfxn,fxn+1≤θPxn,xn+1Pxn,xn+1≤Pxn,xn+1.Then P
x
n
,
x
n
+
1 is a monotone decreasing. Hence P
x
n
,
x
n
+
1
→
c
≥
0 as n
→
∞. Assume c
>
0. Then by (2) we have(6)Pxn+1,xn+2Pxn,xn+1≤θPxn,xn+1.Equation (6) yields θ
P
x
n
,
x
n
+
1
→
1 as n
→
∞. By virtue of θ
∈
₣, this implies that(7)Pxn,xn+1⟶0asn⟶∞.Now we claim that x
n is a 0-Cauchy sequence. Conversely, suppose that(8)lim⁡supn,m→∞⁡Pxn,xm>0.By (
P
4
) and (2), we have, for n
>
m,(9)Pxn,xm≤Pxn,xn+1+Pxn+1,xm+1+Pxm+1,xm-Pxm+1,xm+1-Pxn+1,xn+1≤Pxn,xn+1+Pxm+1,xm+θPxn,xmPxn,xm≤Pxn,xn+1+Pxm+1,xm1-θPxn,xm.Owing to (7) and (8), we get that(10)limn,m→∞⁡11-θPxn,xm=∞,from which we have lim
n
,
m
→
∞
⁡
θ
P
x
n
,
x
m
≥
1 which implies lim
n
,
m
→
∞
⁡
θ
P
x
n
,
x
m
=
1. Since θ
∈
₣, we obtain lim
n
,
m
→
∞
⁡
P
x
n
,
x
m
=
0. It is a contradiction. Thus x
n is a 0-Cauchy sequence. As X
,
P is 0-complete, it follows that there exists ρ
∈
X such that x
n
→
ρ in X
,
P and P
ρ
,
ρ
=
0. Furthermore,(11)limn→∞⁡Pxn,ρ=Pρ,ρ=0.We will show that f
ρ
=
ρ. Consider two cases.
Case 1. If f is continuous, then(12)fρ=limn→∞⁡fxn=limn→∞⁡xn+1=ρ;
hencef
ρ
=
ρ.
Case 2. If (2) holds, then,(13)Pρ,fρ≤Pρ,xn+1+Pxn+1,fρ-Pxn+1,xn+1≤Pρ,xn+1+Pxn+1,fρ≤Pρ,xn+1+θPxn,ρPxn,ρ≤Pρ,xn+1+Pxn,ρ.In view of P
x
n
,
ρ
→
0 as n
→
∞, then we have f
ρ
=
ρ
.The following is example which illustrate Theorem5 and that the generalizations are proper.Example 6.
LetX
=
0
,
+
∞
∩
Q, and let P
:
X
×
X
→
R
+ be defined by P
x
,
y
=
m
a
x
x
,
y for all x
,
y
∈
X. Then X
,
P is a 0-COPMS. Yet it is not complete partial metric space. We endow X with the partial order(14)x≼y⟺x=yorx,y∈0,1withx≤y.Let θ
t
=
t
+
1
-
1 for all t
≥
0. Then it is clear that θ
∈
₣. Define f
:
X
→
X as(15)
f
x
=
x
3
1
+
x
3
x
∈
0,1
,
1
x
2
x
>
1
.Assume that y
≼
x. Then we have two cases.
Case 1. If x
∈
0,1
y
∈
0,1, then(16)Pfx,fy=max⁡x31+x3,y31+y3=x31+x3,Px,y=max⁡x,y=x.Therefore, we have(17)θPx,yPx,y-Pfx,fy=x1+x-x31+x3=x-x21+x3≥0.Hence, for x
∈
0,1, P
f
x
,
f
y
≤
θ
P
x
,
y
P
x
,
y.
Case 2. If x
>
1
y
=
x, then(18)Pfx,fy=max⁡1x2,1y2=1x2,Px,y=max⁡x,y=x.Hence, we get(19)θPx,yPx,y-Pfx,fy=x1+x-1x2=x3-x-1x3+x2=1-k≥0,where(20)0<k=x2+x+1x3+x2<1.
Thus, forx
>
1, P
f
x
,
f
y
≤
θ
P
x
,
y
P
x
,
y.
Moreover, by Cases 1 and 2, it is clear that both assumptions (1) and (2) of Theorem5 are satisfied, and for x
0
=
0, we have x
0
≼
f
x
0. Hence, all assumptions of Theorem 5 are satisfied, and f has a fixed point ρ
=
0
.
On the contrary, consider Example6 in the standard metric d
x
,
y. If x
=
1
/
3 and y
=
1, then(21)df13,f1=1/331+1/33-12=0.46,d13,1=13-1=0.66and so(22)df13,f1=0.46>0,4=θd13,1d13,1.Thus, d
f
x
,
f
y
≤
θ
d
x
,
y
d
x
,
y is not satisfied.


## 3. Remark on Coupled Generalized Banach Contraction

The following result generalizes and extends Theorem 2.1. in [14]. When making the proof of the theorem, Radenović’s technique [15] is used.Theorem 7.
LetX
,
p be a 0-COPMS and let f
:
X
2
→
X be a mapping. Suppose that, for all x
,
y
,
a
,
b
∈
X and θ
∈
₣, the following condition(23)pFx,y,Fa,b≤θpx,a+py,b×2-1·px,a+py,b×2-1holds. Then F has a fixed point.Proof.
Consider the metricsP
:
X
2
×
X
2
→
0
,
∞ defined by(24)PA,B=px,a+py,b,∀A=x,y,B=a,b∈X2.If X
2
,
p is complete, then X
2
,
P is complete (resp. 0-complete), too. Now, define the operator T
:
X
2
→
X
2 by(25)TA=Fx,y,Fy,x,∀A=x,y∈X2.Let P
′
:
X
2
×
X
2
→
0
,
∞ be a metric on X
2 defined by P
′
A
,
B
=
P
A
,
B
/
2 for all A
=
x
,
y
,
B
=
a
,
b
∈
X
2
.
From (23), for all x
,
y, a
,
b
∈
X
2 with x
≼
a and y
≽
b, we get(26)pFx,y,Fa,b≤θpx,a+py,b×2-1·px,a+py,b×2-1,pFb,a,Fy,x≤θpx,a+py,b×2-1·px,a+py,b×2-1.This implies that for all x
,
y, a
,
b
∈
X
2 with x
≼
a and y
≽
b,(27)pFx,y,Fa,b+pFb,a,Fy,x×2-1≤θpx,a+py,b×2-1·px,a+py,b×2-1;that is,(28)P′TA,TB≤θP′A,BP′A,Bwhich is in fact condition (2), for all A
=
x
,
y, B
=
a
,
b
∈
X
2 with A
=
x
,
y
≽
a
,
b
=
B. Hence, all conditions of Theorem 5 are satisfied. In this case, applying Theorem 5, we have that T has a fixed point. From the definition of T, we have x
=
F
x
,
y and y
=
F
y
,
x; that is to say, x
,
y is a coupled fixed point of F
.Remark 8.
Ifθ
t
=
k,  0
≤
k
<
1 in the inequality (23), then we obtain results of Bhaskar and Lakshmikantham [16] in 0-COPMS.The following example illustrates the case when Theorem7 is applicable, while Theorem 2.1. in [14] is not.Example 9.
LetX
=
0,1, and let p
:
X
2
→
R
+ be defined by p
x
,
y
=
m
a
x
x
,
y
+
x
-
y for all x
,
y
∈
X. Then X
,
P is a 0-COPMS. Yet it is not complete partial metric space. We consider the following order relation on X: (29)x,y∈X,x≼y⟺x=yorx,y∈0,0,0,1,1,1.Let θ
t
=
t
+
1
-
1 for all t
≥
0. Then it is clear thatθ
∈
₣. Define F
:
X
2
→
X as F
x
,
y
=
y
-
x
/
5 for all x
,
y
∈
X
.
We have the following cases.
Case 1. For x
,
y
=
a
,
b
=
0,0 or x
,
y
=
a
,
b
=
1,1 or x
,
y
=
0,0 and a
,
b
=
1,1, we have p
F
x
,
y
,
F
a
,
b
=
0. Thus, (23) holds.
Case 2. For x
,
y
=
0,0 and a
,
b
=
0,1, we have(30)pF0,0,F0,1=25<12=θp0,0+p0,1×2-1·p0,0+p0,1×2-1.Thus, (23) holds.
Case 3. For x
,
y
=
0,1 and a
,
b
=
1,1, we have(31)pF0,1,F1,1=25<35=θp0,1+p1,1×2-1·p0,1+p1,1×2-1.Thus, (23) holds.
Therefore, all the conditions of Theorem7 are satisfied and 0,0 is a coupled fixed point of F.



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*Source: 8947020-2016-08-03.xml*