# Integral Type F-Contractions in Partial Metric Spaces

Authors: Vildan Ozturk
Journal: Journal of Function Spaces
 (2019)

---

## Abstract

Partial metric spaces were introduced as a generalization of usual metric spaces where the self-distance for any point need not be equal to zero. In this work, we defined generalized integral type F-contractions and proved common fixed point theorems for four mappings satisfying this type (Branciari type) of contractions in partial metric spaces.


---

## Body

## 1. Introduction and Preliminaries

LetX be a nonempty set and p
:
X
×
X
→
0
,
∞ satisfy(
P
M
1
):  x
=
y
⇔
p
x
,
x
=
p
y
,
y
=
p
x
,
y,
(
P
M
2
):  p
x
,
x
≤
p
x
,
y,
(
P
M
3
):  p
x
,
y
=
p
y
,
x,
(
P
M
4
):  p
x
,
y
≤
p
x
,
z
+
p
z
,
y
-
p
z
,
zfor allx
,
y and z
∈
X. Then the pair X
,
p is called a partial metric space (in short PMS) and p is called a partial metric on X ([1]).LetX
,
p be a PMS. Then, the functions d
p
,
d
m
:
X
×
X
→
0
,
∞ given by(1)
d
p
x
,
y
=
2
p
x
,
y
-
p
x
,
x
-
p
y
,
y
,
(2)
d
m
x
,
y
=
max
⁡
p
x
,
y
-
p
x
,
x
,
p
x
,
y
-
p
y
,
yare (usual) metrics on X. It is clear that d
p and d
m are equivalent ([1]).Definition 1 (see [1]).
(i)
A sequencex
n in a PMS X
,
p converges to x
∈
X if and only if p
x
,
x
=
l
i
m
n
→
∞
p
x
,
x
n
.
(ii)
A sequencex
n in a PMS X
,
p is called a Cauchy sequence if and only if l
i
m
n
,
m
→
∞
p
x
n
,
x
m exists (and finite).
(iii)
A PMSX
,
p is said to be complete if every Cauchy sequence x
n in X converges, with respect to τ
p, to a point x
∈
X such that p
x
,
x
=
l
i
m
n
,
m
→
∞
p
x
n
,
x
m
.
(iv)
A mappingf
:
X
→
X is said to be continuous at x
0
∈
X if for every ε
>
0, there exists δ
>
0 such that f
B
x
0
,
δ
⊂
B
f
x
0
,
ε
.Lemma 2 (see [1]).
(i)
A sequencex
n is Cauchy in a PMS X
,
p if and only if x
n is Cauchy in a metric space X
,
d
p.
(ii)
A PMSX
,
p is complete if and only if the metric space X
,
d
p is complete. Moreover, (3)
lim
n
→
∞
⁡
d
p
x
,
x
n
=
0
⇔
p
x
,
x
=
lim
n
→
∞
⁡
p
x
,
x
n
=
lim
n
,
m
→
∞
⁡
p
x
n
,
x
m
wherex is a limit of x
n in X
,
d
p.Remark 3 (see [2]).
LetX
,
p be a PMS. Therefore,(i)
ifp
x
,
y
=
0, then x
=
y;
(ii)
ifx
≠
y, then p
x
,
y
>
0.Lemma 4 (see [3]).
Assumex
n
→
z as n
→
∞ in a PMS X
,
p such that p
z
,
z
=
0. Then l
i
m
n
→
∞
p
x
n
,
y
=
p
z
,
y for every y
∈
X.In literature, there are many generalizations of Banach contraction principle in metric and generalized metric spaces. One of them is integral type contraction which was defined by Brianciari ([4]). On the other hand, Wardowski [5] introduced F
-contraction in metric spaces as a generalization Banach contraction principle. For more details, you can see [5–9]. In this work, we will introduce generalized integral type F
-contraction in partial metric spaces and prove common fixed point theorems.Definition 5 (see [5]).
Let a mappingF
:
0
,
∞
→
R satisfy the following:(F1)
F is strictly increasing, i.e., for all α
,
β
∈
0
,
∞ such that α
<
β, F
α
<
F
β;
(F2)
for each sequenceα
n
n
∈
N, l
i
m
n
→
∞
α
n
=
0
⇔
l
i
m
n
→
∞
F
(
α
n
)
=
-
∞;
(F3)
there existsk
∈
0,1 such that l
i
m
α
→
0
+
α
k
F
a
=
0.Definition 6 (see [5]).
A mappingT
:
X
→
X is said to be F
-contraction if there exists τ
>
0 such that(4)
∀
x
,
y
∈
X
,
d
T
x
,
T
y
>
0
⇒
τ
+
F
d
T
x
,
T
y
≤
F
d
x
,
y
.Theorem 7 (see [5]).
Let(
X
,
d
) be a complete metric space and let T
:
X
→
X be an F
-contraction. Then T has a unique fixed point in X.Example 8 (see [5]).
LetF
:
R
+
→
R be given by F
α
=
ln
⁡
α. F satisfies (F1), (F2), and (F3). Each mapping T
:
X
→
X is an F-contraction such that, for all x
,
y in X and T
x
≠
T
y,(5)
d
T
x
,
T
y
≤
e
-
τ
d
x
,
yIt is clear that for x
,
y
∈
X such that T
x
=
T
y the inequality d
T
x
,
T
y
≤
e
-
τ
d
x
,
y also holds; i.e., T is a Banach contraction.Definition 9 (see [10]).
The mappingsf
,
g
:
X
→
X are said to be weakly compatible if f and g commute at each coincidence point; i.e., f
x
=
g
x for some x
∈
X.


## 2. Main Results

Theorem 10.
LetX
,
p be a complete partial metric space and f
,
g
,
S, T
:
X
→
X are mappings satisfying f
X
⊆
T
X and g
X
⊆
S
X. Suppose there exist F
∈
F and τ
>
0 such that for all x
,
y
∈
X satisfying p
f
x
,
g
y
>
0
(6)
τ
+
F
∫
0
p
f
x
,
g
y
φ
t
d
t
≤
F
∫
0
M
x
,
y
φ
t
d
twhere(7)
M
x
,
y
=
max
⁡
p
S
x
,
T
y
,
p
f
x
,
S
x
,
p
g
y
,
T
y
,
p
S
x
,
g
y
+
p
f
x
,
T
y
2and φ
:
0
,
∞
→
0
,
∞ is a Lebesgue integrable mapping which is summable, nonnegative and for each μ
>
0
(8)
∫
0
μ
φ
t
d
t
>
0
.
If(i)
f
X, g
X
,
S
X, or T
X is closed,
(ii)
F is continuous,
(iii)
f
,
S and g
,
T are weakly compatible,
then the pairsf
,
g
,
S, and T have a unique common fixed point.Proof.
Letx
0
∈
X be arbitrary. Define a sequence x
n for n
≥
0 by(9)
y
2
n
+
1
=
f
x
2
n
=
T
x
2
n
+
1
and
y
2
n
+
2
=
g
x
2
n
+
1
=
S
x
2
n
+
2
.
Step I. Prove that p
y
n
,
y
n
+
1
→
0 as n
→
∞.
By (6),(10)
τ
+
F
∫
0
p
y
2
n
+
1
,
y
2
n
+
2
φ
t
d
t
=
τ
+
F
∫
0
p
f
x
2
n
,
g
x
2
n
+
1
φ
t
d
t
≤
F
∫
0
M
x
2
n
,
x
2
n
+
1
φ
t
d
twhere(11)
M
x
2
n
,
x
2
n
+
1
=
max
⁡
p
S
x
2
n
,
f
x
2
n
,
p
T
x
2
n
+
1
,
g
x
2
n
+
1
,
ρ
S
x
2
n
,
T
x
2
n
+
1
,
p
T
x
2
n
+
1
,
f
x
2
n
+
p
S
x
2
n
,
g
x
2
n
+
1
2
=
max
⁡
p
y
2
n
,
y
2
n
+
1
,
p
y
2
n
+
1
,
y
2
n
+
2
,
p
y
2
n
,
y
2
n
+
1
,
p
y
2
n
+
1
,
y
2
n
+
1
+
p
y
2
n
,
y
2
n
+
2
2
≤
max
⁡
p
y
2
n
,
y
2
n
+
1
,
p
y
2
n
+
1
,
y
2
n
+
2
,
p
y
2
n
+
1
,
y
2
n
+
1
+
p
y
2
n
,
y
2
n
+
1
+
p
y
2
n
+
1
,
y
2
n
+
2
-
p
y
2
n
+
1
,
y
2
n
+
1
2
=
max
⁡
p
y
2
n
,
y
2
n
+
1
,
p
y
2
n
+
1
,
y
2
n
+
2
.If m
a
x
p
y
2
n
,
y
2
n
+
1
,
p
y
2
n
+
1
,
y
2
n
+
2
=
p
y
2
n
+
1
,
y
2
n
+
2, then it follows from (10) (12)
τ
+
F
∫
0
p
y
2
n
+
1
,
y
2
n
+
2
φ
t
d
t
≤
F
∫
0
p
y
2
n
+
1
,
y
2
n
+
2
φ
t
d
twhich is a contradiction (as τ
>
0). Thus (13)
max
⁡
p
y
2
n
,
y
2
n
+
1
,
p
y
2
n
+
1
,
y
2
n
+
2
=
p
y
2
n
,
y
2
n
+
1
.From (10),(14)
F
∫
0
p
y
2
n
+
1
,
y
2
n
+
2
φ
t
d
t
≤
F
∫
0
p
y
2
n
,
y
2
n
+
1
φ
t
d
t
-
τ
.Continuing this way, we have(15)
F
∫
0
p
y
2
n
,
y
2
n
+
1
φ
t
d
t
≤
F
∫
0
p
y
2
n
-
1
,
y
2
n
φ
t
d
t
-
τ
.Using (14) and (15),(16)
F
∫
0
p
y
2
n
+
1
,
y
2
n
+
2
φ
t
d
t
≤
F
∫
0
p
y
2
n
,
y
2
n
+
1
φ
t
d
t
-
τ
≤
F
∫
0
p
y
2
n
-
1
,
y
2
n
φ
t
d
t
-
2
τ
≤
⋯
≤
F
∫
0
p
y
0
,
y
1
φ
t
d
t
-
2
n
+
1
τ
.And(17)
F
∫
0
p
y
2
n
,
y
2
n
+
1
φ
t
d
t
≤
F
∫
0
p
y
2
n
-
1
,
y
2
n
φ
t
d
t
-
τ
≤
F
∫
0
p
y
2
n
-
2
,
y
2
n
-
1
φ
t
d
t
-
2
τ
≤
⋯
≤
F
∫
0
p
y
0
,
y
1
φ
t
d
t
-
2
n
τThen, it follows l
i
m
n
→
∞
F
∫
0
p
y
n
,
y
n
+
1
φ
t
d
t
=
-
∞. By F
∈
F and (F2), we have(18)
lim
n
→
∞
⁡
p
y
n
,
y
n
+
1
=
0
.
Step II. Now, we prove that y
n is p
-Cauchy sequence. By F
∈
F and (F3), there exits k
∈
0,1 such that(19)
lim
n
→
∞
⁡
p
y
n
,
y
n
+
1
k
F
p
y
n
,
y
n
+
1
=
0
.By (16) and (17),(20)
p
y
2
n
+
1
,
y
2
n
+
2
k
F
∫
0
p
y
2
n
+
1
,
y
2
n
+
2
φ
t
d
t
-
F
∫
0
p
y
0
,
y
1
φ
t
d
t
≤
-
2
n
+
1
p
y
2
n
+
1
,
y
2
n
+
2
k
τ
≤
0and (21)
p
y
2
n
,
y
2
n
+
1
k
F
∫
0
p
y
2
n
,
y
2
n
+
1
φ
t
d
t
-
F
∫
0
p
y
0
,
y
1
φ
t
d
t
≤
-
2
n
p
y
2
n
,
y
2
n
+
1
k
τ
≤
0Using the above inequalities and (19),(22)
lim
n
→
∞
⁡
n
p
y
n
,
y
n
+
1
k
=
0
.Therefore, there exists n
1
∈
N such that n
p
y
n
,
y
n
+
1
k
<
1 for all n
>
n
1, or (23)
p
y
n
,
y
n
+
1
<
1
n
1
/
k
.Let m
,
n
∈
N with m
>
n
>
n
1; using triangular inequality, we have (24)
p
y
n
,
y
m
=
p
y
n
,
y
n
+
1
+
p
y
n
+
1
,
y
n
+
2
+
⋯
+
p
y
m
-
1
,
y
m
-
p
y
n
+
1
,
y
n
+
1
+
p
y
n
+
2
,
y
n
+
2
+
⋯
+
p
y
m
-
1
,
y
m
-
1
≤
p
y
n
,
y
n
+
1
+
p
y
n
+
1
,
y
n
+
2
+
⋯
+
p
y
m
-
1
,
y
m
=
∑
i
=
n
m
-
1
p
y
i
,
y
i
+
1
≤
∑
i
=
n
∞
p
y
i
,
y
i
+
1
≤
∑
i
=
n
∞
1
i
1
/
k
.As k
∈
0,1, the series ∑
i
=
n
∞
‍
1
/
i
1
/
k converges, so(25)
lim
n
,
m
→
∞
⁡
p
y
n
,
y
m
=
0
.Thus y
n is a Cauchy sequence in X
,
p. Therefore, y
n is a Cauchy sequence in X
,
d
p. Since X
,
p is complete partial metric space, then X
,
d
p is complete metric space. Then, there exists a u
∈
X such that l
i
m
n
→
∞
d
p
y
n
,
u
=
0. Moreover (26)
p
u
,
u
=
lim
n
→
∞
⁡
p
y
n
,
u
=
lim
n
,
m
→
∞
⁡
p
y
n
,
y
m
=
0
.Since y
n
→
u, then f
x
2
n
,
T
x
2
n
+
1,g
x
2
n
+
1
,
S
x
2
n
+
2 converge to u.
Step III. We will prove that f
,
g
,
S, and T have a coincidence point.
SupposeT
X is closed, there exists v
∈
X such that T
v
=
u
. We shall show that g
v
=
u
. Then from (6),(27)
τ
+
F
∫
0
p
f
x
2
n
,
g
v
φ
t
d
t
≤
F
∫
0
M
x
2
n
,
v
φ
t
d
twhere(28)
M
x
2
n
,
v
=
max
⁡
p
S
x
2
n
,
f
x
2
n
,
p
T
v
,
g
v
,
p
S
x
2
n
,
T
v
,
p
T
v
,
f
x
2
n
+
p
S
x
2
n
,
g
v
2
=
max
⁡
p
S
x
2
n
,
f
x
2
n
,
p
u
,
g
v
,
p
S
x
2
n
,
u
,
p
u
,
f
x
2
n
+
p
S
x
2
n
,
g
v
2
.
Passing to limit asn
→
∞, (29)
τ
+
F
∫
0
p
u
,
g
v
φ
t
d
t
≤
F
∫
0
p
u
,
g
v
φ
t
d
t
.This is a contradiction with τ
>
0. Thus we have g
v
=
u
. Therefore T
v
=
g
v
=
u. Since g and T are weakly compatible g
u
=
g
T
v
=
T
g
v
=
T
u.
Now we show thatg
u
=
u.(30)
τ
+
F
∫
0
p
f
x
2
n
,
g
u
φ
t
d
t
≤
F
∫
0
M
x
2
n
,
u
φ
t
d
twhere(31)
M
x
2
n
,
u
=
max
⁡
p
S
x
2
n
,
f
x
2
n
,
p
T
u
,
g
u
,
p
S
x
2
n
,
T
u
,
p
T
u
,
f
x
2
n
+
p
S
x
2
n
,
g
u
2
=
max
⁡
p
S
x
2
n
,
f
x
2
n
,
p
g
u
,
g
u
,
p
S
x
2
n
,
g
u
,
p
g
u
,
f
x
2
n
+
ρ
S
x
2
n
,
g
u
2
.Passing to the limit as n
→
∞ and using continuity of F, we have (32)
τ
+
F
∫
0
p
u
,
g
u
φ
t
d
t
≤
F
∫
0
p
u
,
g
u
φ
t
d
t
,which is a contradiction. Therefore p
u
,
g
u
=
0; that is, u is a fixed point of g and T.
Now we show thatu is a fixed point of f and S
. Since g
X
⊆
S
X, there exists a point z
∈
X such that g
u
=
S
z. Suppose that f
z
≠
S
z, then (33)
τ
+
F
∫
0
p
f
z
,
g
u
φ
t
d
t
≤
F
∫
0
M
z
,
u
φ
t
d
twhere(34)
M
z
,
u
=
max
⁡
p
S
z
,
f
z
,
p
T
u
,
g
u
,
p
S
z
,
T
u
,
p
T
u
,
f
z
+
p
S
z
,
g
u
2
=
max
⁡
p
S
z
,
f
z
,
p
T
u
,
f
z
+
p
S
z
,
g
u
2
=
p
S
z
,
f
z
=
p
g
u
,
f
z
.Thus(35)
τ
+
F
∫
0
p
f
z
,
g
u
φ
t
d
t
≤
F
∫
0
p
f
z
,
g
u
φ
t
d
t
,which is a contradiction. Thus f
z
=
g
u
=
S
z. By weak compatibility of f and S, S
u
=
S
f
z
=
f
S
z
=
f
u. Finally we show that f
u
=
u. From (6), (36)
τ
+
F
∫
0
p
f
u
,
u
φ
t
d
t
=
τ
+
F
∫
0
p
f
u
,
g
u
φ
t
d
t
≤
F
∫
0
M
u
,
u
φ
t
d
twhere(37)
M
u
,
u
=
max
⁡
p
S
u
,
f
u
,
p
T
u
,
g
u
,
p
S
u
,
T
u
,
p
T
u
,
f
u
+
p
S
u
,
g
u
2
=
max
⁡
p
f
u
,
f
u
,
p
u
,
u
,
p
f
u
,
u
,
p
u
,
f
u
+
p
f
u
,
u
2
=
p
f
u
,
u
.Thus,(38)
τ
+
F
∫
0
p
f
u
,
u
φ
t
d
t
≤
F
∫
0
p
f
u
,
u
φ
t
d
tand we have f
u
=
S
u
=
u.
Sou is a common fixed point of f, g
,
S, and T.
Step IV. We show uniqueness of common fixed point. Let w be another common fixed point of f and g and u
≠
w.
From (6), we have (39)
τ
+
F
∫
0
p
u
,
w
φ
t
d
t
=
τ
+
F
∫
0
p
f
u
,
f
w
φ
t
d
t
≤
F
∫
0
M
u
,
w
φ
t
d
twhere(40)
M
u
,
w
=
max
⁡
p
S
u
,
f
u
,
p
T
w
,
g
w
,
p
S
u
,
T
w
,
p
T
w
,
f
u
+
p
S
u
,
g
w
2
,
=
max
⁡
p
u
,
u
,
p
w
,
w
,
p
u
,
w
,
p
w
,
u
+
p
u
,
w
2
.Hence,(41)
τ
+
F
∫
0
p
u
,
w
φ
t
d
t
≤
F
∫
0
p
u
,
w
φ
t
d
twhich is a contradiction. So u
=
w
.Corollary 11.
LetX
,
p be a complete partial metric space and f, g
:
X
→
X are two mappings. Suppose there exist F
∈
F and τ
>
0 such that for all x
,
y
∈
X satisfying p
f
x
,
g
y
>
0
(42)
τ
+
F
∫
0
p
f
x
,
g
y
φ
t
d
t
≤
F
∫
0
M
x
,
y
φ
t
d
t
where(43)
M
x
,
y
=
max
⁡
p
x
,
y
,
p
x
,
f
x
,
p
y
,
g
y
,
p
y
,
f
x
+
ρ
x
,
g
y
2
.and φ
:
0
,
∞
→
0
,
∞ is a Lebesgue integrable mapping which is summable, nonnegative and for each μ
>
0
(44)
∫
0
μ
φ
t
d
t
>
0
.
If(i)
f
X or g
X is closed,
(ii)
F is continuous,
then the pairsf and g have a unique common fixed point.Corollary 12.
LetX
,
p be a complete partial metric space and f
,
g
:
X
→
X are two mappings. Suppose there exist F
∈
F and τ
>
0 such that for all x
,
y
∈
X satisfying p
f
x
,
g
y
>
0
(45)
τ
+
F
∫
0
p
f
x
,
g
y
φ
t
d
t
≤
F
∫
0
p
x
,
y
φ
t
d
t
.
Andφ
:
0
,
∞
→
0
,
∞ is a Lebesgue integrable mapping which is summable, nonnegative and for each μ
>
0
(46)
∫
0
μ
φ
t
d
t
>
0
.If(i)
f
X or g
X is closed,
(ii)
F is continuous,
then the pairsf and g have a unique common fixed point.Theorem 13.
LetX
,
p be a complete partial metric space and f
,
g
:
X
→
X be mappings. Suppose there exist F
∈
F and τ
>
0 such that for all x
,
y
∈
X satisfying p
f
x
,
g
y
>
0
(47)
τ
+
F
∫
0
p
f
x
,
g
y
φ
t
d
t
≤
F
∫
0
M
x
,
y
φ
t
d
twhere (48)
M
x
,
y
=
max
⁡
p
x
,
y
,
p
x
,
f
x
,
p
y
,
g
y
,
p
y
,
f
x
+
p
x
,
g
y
2
.And φ
:
0
,
∞
→
0
,
∞ is a Lebesgue integrable mapping which is summable, nonnegative and for each μ
>
0
(49)
∫
0
μ
φ
t
d
t
>
0
.If(i)
f or g is continuous, or
(ii)
F is continuous,
then the pairsf and g have a unique common fixed point.Proof.
Letx
0
∈
X be arbitrary. Define a sequence x
n for n
≥
0 by(50)
x
2
n
+
1
=
f
x
2
n
and
x
2
n
+
2
=
g
x
2
n
+
1
.
Step I. Prove that p
x
n
,
x
n
+
1
→
0 as n
→
∞.
By (47),(51)
τ
+
F
∫
0
p
x
2
n
+
1
,
x
2
n
+
2
φ
t
d
t
=
τ
+
F
∫
0
p
f
x
2
n
,
g
x
2
n
+
1
φ
t
d
t
≤
F
∫
0
M
x
2
n
,
x
2
n
+
1
φ
t
d
t
where(52)
M
x
2
n
,
x
2
n
+
1
=
max
⁡
p
x
2
n
,
x
2
n
+
1
,
p
x
2
n
,
f
x
2
n
,
p
x
2
n
+
1
,
g
x
2
n
+
1
,
p
x
2
n
+
1
,
f
x
2
n
+
p
x
2
n
g
x
2
n
+
1
2
=
max
⁡
p
x
2
n
,
x
2
n
+
1
,
p
x
2
n
,
x
2
n
+
1
,
p
x
2
n
+
1
,
x
2
n
+
2
,
p
x
2
n
+
1
,
x
2
n
+
1
+
p
x
2
n
,
x
2
n
+
2
2
≤
max
⁡
p
x
2
n
,
x
2
n
+
1
,
p
x
2
n
+
1
,
x
2
n
+
2
,
p
x
2
n
,
x
2
n
+
1
+
p
x
2
n
+
1
,
x
2
n
+
2
2
=
max
⁡
p
x
2
n
,
x
2
n
+
1
,
p
x
2
n
+
1
,
x
2
n
+
2
.Then the proof is similar proof of Theorem 10.
We will prove thatf and g have common fixed point. Since X
,
p is complete partial metric space, then X
,
d
p is complete metric space. Then, there exists u
∈
X such that l
i
m
n
→
∞
d
p
y
n
,
u
=
0. Moreover (53)
p
u
,
u
=
lim
n
→
∞
⁡
p
y
n
,
u
=
lim
n
,
m
→
∞
⁡
p
y
n
,
y
m
=
0
.We consider two cases.
Case 1. Suppose f is continuous. Then, u
=
l
i
m
n
→
∞
x
n
=
l
i
m
n
→
∞
x
2
n
=
l
i
m
n
→
∞
x
2
n
+
1
=
l
i
m
n
→
∞
f
x
2
n
=
f
u. Thus u is a fixed point of f.
Now we proveu is a fixed point of g. On the contrary, we assume g
u
≠
u. From (47),(54)
τ
+
F
∫
0
p
f
x
2
n
,
g
u
φ
t
d
t
≤
F
∫
0
M
x
2
n
,
u
φ
t
d
t where (55)
M
x
2
n
,
u
=
max
⁡
p
x
2
n
,
u
,
p
x
2
n
,
f
x
2
n
,
p
u
,
g
u
,
p
u
,
f
x
2
n
+
p
x
2
n
,
g
u
2
=
max
⁡
p
x
2
n
,
u
,
p
x
2
n
,
x
2
n
+
1
,
p
u
,
g
u
,
p
u
,
x
2
n
+
1
+
p
x
2
n
,
g
u
2
.
Lettingn
→
∞, (56)
lim
n
→
∞
⁡
M
x
2
n
,
u
=
p
u
,
g
u
.
τ
+
F
∫
0
p
u
,
g
u
φ
t
d
t
≤
F
∫
0
p
u
,
g
u
φ
t
d
t
.This is a contradiction with τ
>
0. Thus we have g
u
=
u.
Similarly, we have the same results wheng is continuous.
Case 2. Now, we suppose that F is continuous. We can assume there exists n
1
∈
N such that f
x
n
+
1
≠
u (i.e., p
x
n
,
u
>
0) for all n
≥
n
1. Then from (47) we have(57)
τ
+
F
∫
0
p
f
u
,
g
x
2
n
+
1
φ
t
d
t
≤
F
∫
0
M
u
,
x
2
n
+
1
φ
t
d
twhere(58)
M
u
,
x
2
n
+
1
=
max
⁡
p
u
,
x
2
n
+
1
,
p
u
,
f
u
,
p
x
2
n
+
1
,
g
x
2
n
+
1
,
p
x
2
n
+
1
,
f
u
+
p
u
,
g
x
2
n
+
1
2
=
max
⁡
p
u
,
x
2
n
+
1
,
p
u
,
f
u
,
p
x
2
n
+
1
,
x
2
n
+
2
,
p
x
2
n
+
1
,
f
u
+
p
u
,
x
2
n
+
2
2
.Then there exists n
2
∈
N such that, for all n
≥
n
2, we have(59)
max
⁡
p
u
,
x
2
n
+
1
,
p
u
,
f
u
,
p
x
2
n
+
1
,
x
2
n
+
2
,
p
x
2
n
+
1
,
u
+
p
u
,
f
u
+
p
u
,
x
2
n
+
2
-
p
u
,
u
2
=
p
u
,
f
u
.Thus, we have (60)
τ
+
F
∫
0
p
f
u
,
g
x
2
n
+
1
φ
t
d
t
≤
F
∫
0
p
u
,
f
u
φ
t
d
t
,for all n
≥
m
a
x
n
1
,
n
2
. Since F is continuous, taking the limit as n
→
∞, we get (61)
τ
+
F
∫
0
p
f
u
,
u
φ
t
d
t
≤
F
∫
0
p
u
,
f
u
φ
t
d
twhich is a contradiction. Therefore p
f
u
,
u
=
0 and u is a fixed point of f
.
Now we show thatu is a fixed point of g
.
(62)
τ
+
F
∫
0
p
u
,
g
u
φ
t
d
t
=
τ
+
F
∫
0
p
f
u
,
g
u
φ
t
d
t
≤
F
∫
0
M
u
,
u
φ
t
d
twhere(63)
M
u
,
u
=
max
⁡
p
u
,
u
,
p
u
,
f
u
,
p
u
,
g
u
,
p
u
,
f
u
+
p
u
,
g
u
2
=
max
⁡
p
u
,
u
,
p
u
,
u
,
p
u
,
g
u
,
p
u
,
u
+
p
u
,
g
u
2
=
p
u
,
g
u
.Thus (64)
τ
+
F
∫
0
p
u
,
g
u
φ
t
d
t
≤
F
∫
0
p
u
,
g
u
φ
t
d
t
.Hence, u
=
g
u.
Sou is a common fixed point of f and g.
Now we prove uniqueness of common fixed point. We assume thatv is another common fixed point of f and g and u
≠
v.
From (47), we have (65)
τ
+
F
∫
0
p
u
,
v
φ
t
d
t
=
τ
+
F
∫
0
p
f
u
,
g
v
φ
t
d
t
≤
F
∫
0
M
u
,
v
φ
t
d
twhere (66)
M
u
,
v
=
max
⁡
p
u
,
v
,
p
u
,
f
u
,
p
v
,
g
v
,
p
v
,
f
u
+
p
u
,
g
v
2
,
=
max
⁡
p
u
,
v
,
p
u
,
u
,
p
v
,
v
,
p
v
,
u
+
p
u
,
v
2
.Hence, (67)
τ
+
F
∫
0
p
u
,
v
φ
t
d
t
≤
F
∫
0
p
u
,
v
φ
t
d
t
,which is a contradiction with τ
>
0. So u
=
v.Corollary 14.
LetX
,
p be a complete partial metric space and f
,
g
:
X
→
X two mappings. Suppose there exist F
∈
F and τ
>
0 such that for all x
,
y
∈
X satisfying p
f
x
,
g
y
>
0
(68)
τ
+
F
∫
0
p
f
x
,
g
v
φ
t
d
t
≤
F
∫
0
p
x
,
v
φ
t
d
tand φ
:
0
,
∞
→
0
,
∞ is a Lebesgue integrable mapping which is summable, nonnegative and for each μ
>
0
(69)
∫
0
μ
φ
t
d
t
>
0
.If(i)
f or g is continuous, or
(ii)
F is continuous, then the pairs f and g have a unique common fixed point.Example 15.
LetX
=
0,1, and p
x
,
y
=
max
⁡
x
,
y for all x
,
y
∈
X. Then X
,
p is complete partial metric space. Let f
,
g
,
S
,
T
:
X
→
X and φ
:
0
,
∞
→
0
,
∞
(70)
f
x
=
x
8
,
g
x
=
0
,
S
x
=
3
x
4
a
n
d
T
x
=
x
,
φ
t
=
2
t
.Consider F in Example 8. Then all conditions of Theorem 10 and the contractive condition (6) are satisfied for some τ
>
0 and for p
x
,
y
>
0.
If3
x
/
4
>
y,(71)
τ
+
F
∫
0
p
f
x
,
g
y
φ
t
d
t
=
τ
+
ln
⁡
x
2
128
≤
ln
⁡
9
x
2
16
=
F
∫
0
M
x
,
y
φ
t
d
t
.If 3
x
/
4
<
y,(72)
τ
+
F
∫
0
p
f
x
,
g
y
φ
t
d
t
=
τ
+
ln
⁡
x
2
128
<
τ
+
ln
⁡
16
y
2
3.128
≤
ln
⁡
y
2
2
=
F
∫
0
M
x
,
y
φ
t
d
t
.Therefore 0 is a fixed point of f
,
g
,
S, and T.



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*Source: 5193862-2019-03-25.xml*