# A Unification ofG-Metric, Partial Metric, and b-Metric Spaces

Authors: Nawab Hussain; Jamal Rezaei Roshan; Vahid Parvaneh; Abdul Latif
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

Using the concepts ofG-metric, partial metric, and b-metric spaces, we define a new concept of generalized partial b-metric space. Topological and structural properties of the new space are investigated and certain fixed point theorems for contractive mappings in such spaces are obtained. Some examples are provided here to illustrate the usability of the obtained results.


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## Body

## 1. Introduction and Mathematical Preliminaries

The concept of ab-metric space was introduced by Czerwik in [1, 2]. After that, several interesting results about the existence of fixed point for single-valued and multivalued operators in (ordered) b-metric spaces have been obtained (see, e.g., [3–13]).Definition 1 (see [1]).
LetX be a (nonempty) set and s
≥
1 a given real number. A function d
:
X
×
X
→
ℝ
+ is a b-metric on X if, for all x
,
y
,
z
∈
X, the following conditions hold:(
b
1)
d
(
x
,
y
)
=
0 if and only if x
=
y,
(b
2)
d
(
x
,
y
)
=
d
(
y
,
x
),
(b
3)
d
(
x
,
z
)
≤
s
[
d
(
x
,
y
)
+
d
(
y
,
z
)
].
In this case, the pair(
X
,
d
) is called a b-metric space.The concept of a generalized metric space, or aG-metric space, was introduced by Mustafa and Sims [14].Definition 2 (see [14]).
LetX be a nonempty set and G
:
X
×
X
×
X
→
ℝ
+ a function satisfying the following properties:(G
1)
G
(
x
,
y
,
z
)
=
0 if x
=
y
=
z;
(G
2)
0
<
G
(
x
,
x
,
y
), for all x
,
y
∈
X with x
≠
y;
(G
3)
G
(
x
,
x
,
y
)
≤
G
(
x
,
y
,
z
), for all x
,
y
,
z
∈
X with z
≠
y;
(G
4)
G
(
x
,
y
,
z
)
=
G
(
p
{
x
,
y
,
z
}
), where p is any permutation of x
,
y
,
z (symmetry in all the three variables);
(G
5)
G
(
x
,
y
,
z
)
≤
G
(
x
,
a
,
a
)
+
G
(
a
,
y
,
z
), for all x
,
y
,
z
,
a
∈
X (rectangle inequality).
Then, the functionG is called a G-metric on X and the pair (
X
,
G
) is called a G-metric space.Aghajani et al. in [15] introduced the class of generalized b-metric spaces (G
b-metric spaces) and then they presented some basic properties of G
b-metric spaces.The following is their definition ofG
b-metric spaces.Definition 3 (see [15]).
LetX be a nonempty set and s
≥
1 a given real number. Suppose that a mapping G
:
X
×
X
×
X
→
ℝ
+ satisfies(G
b
1)
G
(
x
,
y
,
z
)
=
0 if x
=
y
=
z,
(G
b
2)
0
<
G
(
x
,
x
,
y
) for all x
,
y
∈
X with x
≠
y,
(G
b
3)
G
(
x
,
x
,
y
)
≤
G
(
x
,
y
,
z
) for all x
,
y
,
z
∈
X with y
≠
z,
(G
b
4)
G
(
x
,
y
,
z
)
=
G
(
p
{
x
,
y
,
z
}
), where p is a permutation of x
,
y
,
z (symmetry),
(G
b
5)
G
(
x
,
y
,
z
)
≤
s
[
G
(
x
,
a
,
a
)
+
G
(
a
,
y
,
z
)
] for all x
,
y
,
z
,
a
∈
X (rectangle inequality).
ThenG is called a generalized b-metric and the pair (
X
,
G
) is called a generalized b-metric space or a G
b-metric space.On the other hand, Matthews [16] has introduced the notion of a partial metric space as a part of the study of denotational semantics of dataflow networks. In partial metric spaces, self-distance of an arbitrary point need not to be equal to zero.Definition 4 (see [16]).
A partial metric on a nonempty setX is a mapping p
:
X
×
X
→
ℝ
+ such that, for all x
,
y
,
z
∈
X:(p
1)
x
=
y if and only if 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
),
(p
4)
p
(
x
,
y
)
≤
p
(
x
,
z
)
+
p
(
z
,
y
)
-
p
(
z
,
z
).
In this case,(
X
,
p
) is called a partial metric space.For a survey of fixed point theory, its applications, and comparison of different contractive conditions and related results in both partial metric spaces andG-metric spaces we refer the reader to, for example, [17–27] and references mentioned therein.Recently, Zand and Nezhad [28] have introduced a new generalized metric space (G
p-metric spaces) as a generalization of both partial metric spaces and G-metric spaces.We will use the following definition of aG
p-metric space.Definition 5 (see [29]).
LetX be a nonempty set. Suppose that a mapping G
p
:
X
×
X
×
X
→
ℝ
+ satisfies(G
p
1)
x
=
y
=
z if G
p
(
x
,
y
,
z
)
=
G
p
(
z
,
z
,
z
)
=
G
p
(
y
,
y
,
y
)
=
G
p
(
x
,
x
,
x
);
(G
p
2)
G
p
(
x
,
x
,
x
)
≤
G
p
(
x
,
x
,
y
)
≤
G
p
(
x
,
y
,
z
) for all x
,
y
,
z
∈
X with z
≠
y;
(G
p
3)
G
p
(
x
,
y
,
z
)
=
G
p
(
p
{
x
,
y
,
z
}
), where p is any permutation of x
,
y, and z (symmetry in all the three variables);
(G
p
4)
G
p
(
x
,
y
,
z
)
≤
G
p
(
x
,
a
,
a
)
+
G
p
(
a
,
y
,
z
)
-
G
p
(
a
,
a
,
a
) for all x
,
y
,
z
,
a
∈
X (rectangle inequality).
ThenG
p is called a G
p-metric and (
X
,
G
p
) is called a G
p-metric space.As a generalization and unification of partial metric andb-metric spaces, Shukla [30] presented the concept of a partial b-metric space as follows.Definition 6 (see [30]).
A partialb-metric on a nonempty set X is a mapping p
b
:
X
×
X
→
ℝ
+ such that, for all x
,
y
,
z
∈
X:(p
b
1)
x
=
y if and only if p
b
(
x
,
x
)
=
p
b
(
x
,
y
)
=
p
b
(
y
,
y
),
(p
b
2)
p
b
(
x
,
x
)
≤
p
b
(
x
,
y
),
(p
b
3)
p
b
(
x
,
y
)
=
p
b
(
y
,
x
),
(p
b
4)
p
b
(
x
,
y
)
≤
s
[
p
b
(
x
,
z
)
+
p
b
(
z
,
y
)
]
-
p
b
(
z
,
z
).
A partialb-metric space is a pair (
X
,
p
b
) such that X is a nonempty set and p
b is a partial b-metric on X. The number s
≥
1 is called the coefficient of (
X
,
p
b
).In a partialb-metric space (
X
,
p
b
), if x
,
y
∈
X and p
b
(
x
,
y
)
=
0, then x
=
y, but the converse may not be true. It is clear that every partial metric space is a partial b-metric space with coefficient s
=
1 and every b-metric space is a partial b-metric space with the same coefficient and zero self-distance. However, the converse of these facts needs not to be hold.Example 7 (see [30]).
LetX
=
ℝ
+, q
>
1 a constant, and p
b
:
X
×
X
→
ℝ
+ defined by
(1)
p
b
(
x
,
y
)
=
[
max
⁡
{
x
,
y
}
]
q
+
|
x
-
y
|
q
∀
x
,
y
∈
X
.
Then (
X
,
p
b
) is a partial b-metric space with the coefficient s
=
2
q
-
1
>
1, but it is neither a b-metric nor a partial metric space.Note that in a partialb-metric space the limit of a convergent sequence may not be unique (see [30, Example 2]).In [31], Mustafa et al. introduced a modified version of ordered partial b-metric spaces in order to obtain that each partial b-metric p
b generates a b-metric d
p
b.Definition 8 (see [31]).
LetX be a (nonempty) set and s
≥
1 a given real number. A function p
b
:
X
×
X
→
ℝ
+ is a partial b-metric if, for all x
,
y
,
z
∈
X, the following conditions are satisfied:(p
b
1)
x
=
y
⇔
p
b
(
x
,
x
)
=
p
b
(
x
,
y
)
=
p
b
(
y
,
y
),
(p
b
2)
p
b
(
x
,
x
)
≤
p
b
(
x
,
y
),
(p
b
3)
p
b
(
x
,
y
)
=
p
b
(
y
,
x
),
(p
b
4
′)
p
b
(
x
,
y
)
≤
s
(
p
b
(
x
,
z
)
+
p
b
(
z
,
y
)
-
p
b
(
z
,
z
)
)
+
(
(
1
-
s
)
/
2
)
(
p
b
(
x
,
x
)
+
p
b
(
y
,
y
)
).
The pair(
X
,
p
b
) is called a partial b-metric space.Sinces
≥
1, from (
p
b
4
′
), we have
(2)
p
b
(
x
,
y
)
≤
s
(
p
b
(
x
,
z
)
+
p
b
(
z
,
y
)
-
p
b
(
z
,
z
)
)
≤
s
(
p
b
(
x
,
z
)
+
p
b
(
z
,
y
)
)
-
p
b
(
z
,
z
)
.
Hence, a partial b-metric in the sense of Definition 8 is also a partial b-metric in the sense of Definition 6.The following example shows that a partialb-metric on X (in the sense of Definition 8) is neither a partial metric nor a b-metric on X.Example 9 (see [31]).
Let(
X
,
d
) be a metric space and p
b
(
x
,
y
)
=
d
(
x
,
y
)
q
+
a, where q
>
1 and a
≥
0 are real numbers. Then p
b is a partial b-metric with s
=
2
q
-
1.Proposition 10 (see [31]).
Every partialb-metric p
b defines a b-metric d
p
b, where
(3)
d
p
b
(
x
,
y
)
=
2
p
b
(
x
,
y
)
-
p
b
(
x
,
x
)
-
p
b
(
y
,
y
)
for all x
,
y
∈
X.Hence, the importance of our definition of partialb-metric is that by using it we can define a dependent b-metric which we call the b-metric associated with p
b.Now, we present some definitions and propositions in a partialb-metric space.Definition 11 (see [31]).
Let(
X
,
p
b
) be a partial b-metric space. Then, for an x
∈
X and an ϵ
>
0, the p
b-ball with center x and radius ϵ
>
0 is
(4)
B
p
b
(
x
,
ϵ
)
=
{
y
∈
X
∣
p
b
(
x
,
y
)
<
p
b
(
x
,
x
)
+
ϵ
}
.Lemma 12 (see [31]).
Let(
X
,
p
b
) be a partial b-metric space. Then,(A)
ifp
b
(
x
,
y
)
=
0, then x
=
y;
(B)
ifx
≠
y, then p
b
(
x
,
y
)
>
0.Proposition 13 (see [31]).
Let(
X
,
p
b
) be a partial b-metric space, x
∈
X, and ϵ
>
0. If y
∈
B
p
b
(
x
,
ϵ
) then there exists a δ
>
0 such that B
p
b
(
y
,
δ
)
⊆
B
p
b
(
x
,
ɛ
).Thus, from the above proposition the family of all openp
b-balls
(5)
Δ
=
{
B
p
b
(
x
,
r
)
∣
x
∈
X
,
r
>
0
}
is a base of a T
0-topology τ
p
b on X which we call the p
b-metric topology.The topological space(
X
,
p
b
) is T
0 but need not be T
1.The following lemma shows the relationship between the concepts ofp
b-convergence, p
b-Cauchyness, and p
b-completeness in two spaces (
X
,
p
b
) and (
X
,
d
p
b
).Lemma 14 (see [31]).
(1) A sequence{
x
n
} is a p
b-Cauchy sequence in a partial b-metric space (
X
,
p
b
) if and only if it is a b-Cauchy sequence in the b-metric space (
X
,
d
p
b
).
(2) A partialb-metric space (
X
,
p
b
) is p
b-complete if and only if the b-metric space (
X
,
d
p
b
) is b-complete. Moreover, lim
⁡
n
→
∞
⁡
d
p
b
(
x
,
x
n
)
=
0 if and only if
(6)
lim
⁡
n
→
∞
p
b
(
x
,
x
n
)
=
lim
⁡
n
,
m
→
∞
p
b
(
x
n
,
x
m
)
=
p
b
(
x
,
x
)
.Now, we introduce the concept of generalized partialb-metric space, a G
p
b-metric space, as a generalization of both partial b-metric space and G-metric space.Definition 15.
LetX be a nonempty set. Suppose that the mapping G
p
b
:
X
×
X
×
X
→
ℝ
+ satisfies the following conditions:(G
p
b
1)
x
=
y
=
z if G
p
b
(
x
,
y
,
z
)
=
G
p
b
(
z
,
z
,
z
)
=
G
p
b
(
y
,
y
,
y
)
=
G
p
b
(
x
,
x
,
x
);
(G
p
b
2)
G
p
b
(
x
,
x
,
x
)
≤
G
p
b
(
x
,
x
,
y
)
≤
G
p
b
(
x
,
y
,
z
) for all x
,
y
,
z
∈
X with z
≠
y;
(G
p
b
3)
G
p
b
(
x
,
y
,
z
)
=
G
p
b
(
p
{
x
,
y
,
z
}
), where p is any permutation of x
,
y, or z (symmetry in all three variables);
(G
p
b
4)
G
p
b
(
x
,
y
,
z
)
≤
s
[
G
p
b
(
x
,
a
,
a
)  +  G
p
b
(
a
,
y
,
z
)
-
G
p
b
(
a
,
a
,
a
)
]  +  (
(
1
-
s
)
/
3
)
[
G
p
b
(
x
,
x
,
x
)  +   G
p
b
(
y
,
y
,
y
)  +   G
p
b
(
z
,
z
,
z
)
] for all x
,
y
,
z
,
a
∈
X (rectangle inequality).
ThenG
p
b is called a G
p
b-metric and (
X
,
G
p
b
) is called a G
p
b-metric space.Sinces
≥
1, so from G
p
b
4 we have the following inequality:
(7)
G
p
b
(
x
,
y
,
z
)
≤
s
[
G
p
b
(
x
,
a
,
a
)
+
G
p
b
(
a
,
y
,
z
)
-
G
p
b
(
a
,
a
,
a
)
]
.
The G
p
b-metric space G
p
b is called symmetric if
(8)
G
p
b
(
x
,
x
,
y
)
=
G
p
b
(
x
,
y
,
y
)
holds for all x
,
y
∈
X. Otherwise, G
p
b is an asymmetric G
p
b-metric.Now we present some examples ofG
p
b-metric space.Example 16.
LetX
=
[
0
,
+
∞
) and let G
p
b
:
X
3
→
ℝ
+ be given by G
p
b
(
x
,
y
,
z
)
=
[
max
⁡
{
x
,
y
,
z
}
]
p, where p
>
1. Obviously, (
X
,
G
p
b
) is a symmetric G
p
b-metric space which is not a G-metric space. Indeed, if x
=
y
=
z
>
0, then G
p
b
(
x
,
y
,
z
)
=
x
p
>
0. It is easy to see that G
p
b
1
–
G
p
b
3 are satisfied. Now we show that G
p
b
4 holds. For each x
,
y
,
z
,
a
∈
X, we have
(9)
x
p
+
y
p
+
z
p
3
≤
[
max
⁡
{
x
,
y
,
z
}
]
p
,
so
(10)
[
max
⁡
{
x
,
y
,
z
}
]
p
+
s
-
1
3
(
x
p
+
y
p
+
z
p
)
+
s
a
p
≤
[
max
⁡
{
x
,
y
,
z
}
]
p
+
(
s
-
1
)
[
max
⁡
{
x
,
y
,
z
}
]
p
+
s
a
p
=
s
[
max
⁡
{
x
,
y
,
z
}
]
p
+
s
a
p
≤
s
[
max
⁡
{
x
,
a
}
]
p
+
s
[
max
⁡
{
a
,
y
,
z
}
]
p
.
Thus,
(11)
[
max
⁡
{
x
,
y
,
z
}
]
p
≤
s
(
[
max
⁡
{
x
,
a
}
]
p
+
[
max
⁡
{
a
,
y
,
z
}
]
p
-
a
p
)
+
1
-
s
3
(
x
p
+
y
p
+
z
p
)
which implies the required inequality
(12)
G
p
b
(
x
,
y
,
z
)
≤
s
[
G
p
b
(
x
,
a
,
a
)
+
G
p
b
(
a
,
y
,
z
)
-
G
p
b
(
a
,
a
,
a
)
]
+
1
-
s
3
[
G
p
b
(
x
,
x
,
x
)
+
G
p
b
(
y
,
y
,
y
)
+
G
p
b
(
z
,
z
,
z
)
]
.Example 17.
LetX
=
{
0,1
,
2,3
}. Let
(13)
A
=
{
(
1,0
,
0
)
,
(
0,1
,
0
)
,
(
0,0
,
1
)
,
(
2,0
,
0
)
,
(
0,2
,
0
)
,
hhl
(
0,0
,
2
)
,
(
3,0
,
0
)
,
(
0,3
,
0
)
,
(
0,0
,
3
)
,
(
1,2
,
2
)
,
hhl
(
2,1
,
2
)
,
(
2,2
,
1
)
,
(
2,3
,
3
)
,
(
3,2
,
3
)
,
(
3,3
,
2
)
}
,
B
=
{
(
0,1
,
1
)
,
(
1,0
,
1
)
,
(
1,1
,
0
)
,
(
0,2
,
2
)
,
(
2,0
,
2
)
,
hhl
(
2,2
,
0
)
,
(
0,3
,
3
)
,
(
3,0
,
3
)
,

(
3,3
,
0
)
,

(
2,1
,
1
)
,
hhl
(
1,2
,
1
)
,
(
1,1
,
2
)
,
(
3,2
,
2
)
,
(
2,3
,
2
)
,
(
2,2
,
3
)
}
.
Define G
p
b
:
X
3
→
ℝ
+ by
(14)
G
p
b
(
x
,
y
,
z
)
=
{
3
2
,
if
x
=
y
=
z
≠
2
,
0
,
if
x
=
y
=
z
=
2
,
2
,
if
(
x
,
y
,
z
)
∈
A
,
5
2
,
if
(
x
,
y
,
z
)
∈
B
,
3
,
otherwise
.
It is easy to see that (
X
,
G
p
b
) is an asymmetric G
p
b-metric space with coefficient s
≥
6
/
5.Proposition 18.
EveryG
p
b-metric space (
X
,
G
p
b
) defines a b-metric space (
X
,
d
G
p
b
), where
(15)
d
G
p
b
(
x
,
y
)
=
G
p
b
(
x
,
y
,
y
)
+
G
p
b
(
y
,
x
,
x
)
-
G
p
b
(
x
,
x
,
x
)
-
G
p
b
(
y
,
y
,
y
)
,
for all x
,
y
∈
X.Proof.
Letx
,
y
,
z
∈
X. Then we have
(16)
d
G
p
b
(
x
,
y
)
=
G
p
b
(
x
,
y
,
y
)
+
G
p
b
(
y
,
x
,
x
)
-
G
p
b
(
x
,
x
,
x
)
-
G
p
b
(
y
,
y
,
y
)
≤
s
(
G
p
b
(
x
,
z
,
z
)
+
G
p
b
(
z
,
y
,
y
)
-
G
p
b
(
z
,
z
,
z
)
)
h
h
h
h
h
h

+
(
1
-
s
3
)
(
G
p
b
(
x
,
x
,
x
)
+
2
G
p
b
(
y
,
y
,
y
)
)

+
s
(
G
p
b
(
y
,
z
,
z
)
+
G
p
b
(
z
,
x
,
x
)
-
G
p
b
(
z
,
z
,
z
)
)
h
h
h
h
h
h

+
(
1
-
s
3
)
(
2
G
p
b
(
x
,
x
,
x
)
+
G
p
b
(
y
,
y
,
y
)
)

-
G
p
b
(
x
,
x
,
x
)
-
G
p
b
(
y
,
y
,
y
)
=
s
[
G
p
b
(
x
,
z
,
z
)
+
G
p
b
(
z
,
x
,
x
)
-
G
p
b
(
x
,
x
,
x
)
h
h
h
h
h
h
-
G
p
b
(
z
,
z
,
z
)
+
G
p
b
(
z
,
y
,
y
)
h
h
h
h
h
h
+
G
p
b
(
y
,
z
,
z
)
-
G
p
b
(
z
,
z
,
z
)
-
G
p
b
(
y
,
y
,
y
)
]
=
s
[
d
G
p
b
(
x
,
z
)
+
d
G
p
b
(
z
,
y
)
]
.With straightforward calculations, we have the following proposition.Proposition 19.
LetX be a G
p
b-metric space. Then for each x
,
y
,
z
,
a
∈
X it follows that(1)
G
p
b
(
x
,
y
,
z
)
≤
s
G
p
b
(
x
,
a
,
a
)
+
s
2
G
p
b
(
y
,
a
,
a
)
+
s
2
G
p
b
(
z
,
a
,
a
)
-
(
s
+
s
2
)
G
p
b
(
a
,
a
,
a
);
(2)
G
p
b
(
x
,
y
,
z
)
≤
s
[
G
p
b
(
x
,
x
,
y
)  +  G
p
b
(
x
,
x
,
z
)  −  G
p
b
(
x
,
x
,
x
)
]  +  (
(
1
-
s
)
/
3
)
[
G
p
b
(
x
,
x
,
x
)
+
G
p
b
(
y
,
y
,
y
)
+
G
p
b
(
z
,
z
,
z
)
];
(3)
G
p
b
(
x
,
y
,
y
)
≤
2
s
G
p
b
(
x
,
x
,
y
)  +  (
(
1
-
4
s
)
/
3
)
G
p
b
(
x
,
x
,
x
)  + (
2
/
3
)
(
1
-
s
)
G
p
b
(
y
,
y
,
y
);
(4)
G
p
b
(
x
,
y
,
z
)
≤
s
[
G
p
b
(
x
,
a
,
z
)  +  G
p
b
(
a
,
y
,
z
)  −  G
p
b
(
a
,
a
,
a
)
]  +  (
(
1
-
s
)
/
3
)
[
G
p
b
(
x
,
x
,
x
)
+
G
p
b
(
y
,
y
,
y
)
+
G
p
b
(
z
,
z
,
z
)
], a
≠
z.Lemma 20.
Let(
X
,
G
p
b
) be a G
p
b-metric space. Then,(A)
ifG
p
b
(
x
,
y
,
z
)
=
0, then x
=
y
=
z;
(B)
ifx
≠
y, then G
p
b
(
x
,
y
,
y
)
>
0.Proof.
IfG
p
b
(
x
,
y
,
z
)
=
0, then from G
p
b
2 we have G
p
b
(
x
,
x
,
x
)
=
G
p
b
(
y
,
y
,
y
)
=
G
p
b
(
z
,
z
,
z
)
=
0, so from G
p
b
1 we obtain (A). To prove (B), on the contrary, if G
p
b
(
x
,
y
,
y
)
=
0, then from (A) we have x
=
y, a contradiction.Definition 21.
Let(
X
,
G
p
b
) be a G
p
b-metric space. Then for an x
∈
X and an ϵ
>
0, the G
p
b-ball with center x and radius ϵ
>
0 is
(17)
B
G
p
b
(
x
,
ϵ
)
=
{
y
∈
X
∣
G
p
b
(
x
,
x
,
y
)
<
G
p
b
(
x
,
x
,
x
)
+
ϵ
}
.By motivation of Proposition 4 in [31], we provide the following proposition.Proposition 22.
Let(
X
,
G
p
b
) be a G
p
b-metric space, x
∈
X, and ϵ
>
0. If y
∈
B
G
p
b
(
x
,
ϵ
), then there exists a δ
>
0 such that B
G
p
b
(
y
,
δ
)
⊆
B
G
p
b
(
x
,
ɛ
).Proof.
Lety
∈
B
G
p
b
(
x
,
ϵ
); if y
=
x, then we choose δ
=
ϵ. Suppose that y
≠
x; then, by Lemma 20, we have G
p
b
(
x
,
x
,
y
)
≠
0. Now, we consider two cases.
Case 1. IfG
p
b
(
x
,
x
,
y
)
=
G
p
b
(
x
,
x
,
x
), then for s
=
1 we choose δ
=
ϵ. If s
>
1, then we consider the set
(18)
A
=
{
n
∈
ℕ
∣
ϵ
2
s
n
+
1
(
s
-
1
)
<
G
p
b
(
x
,
x
,
x
)
}
.
By Archmedean property,A is a nonempty set; then by the well ordering principle, A has a least element m. Since m
-
1
∉
A, we have G
p
b
(
x
,
x
,
x
)
≤
(
ϵ
/
(
2
s
m
(
s
-
1
)
)
) and we choose δ
=
ϵ
/
2
s
m
+
1. Let z
∈
B
G
p
b
(
y
,
δ
); by property G
p
b
4 we have
(19)
G
p
b
(
x
,
x
,
z
)
≤
s
(
G
p
b
(
x
,
x
,
y
)
+
G
p
b
(
y
,
y
,
z
)
-
G
p
b
(
y
,
y
,
y
)
)
≤
s
(
G
p
b
(
x
,
x
,
x
)
+
δ
)
≤
G
p
b
(
x
,
x
,
x
)
+
ϵ
2
s
m
+
ϵ
2
s
m
=
G
p
b
(
x
,
x
,
x
)
+
ϵ
s
m
<
G
p
b
(
x
,
x
,
x
)
+
ϵ
.
Hence, B
G
p
b
(
y
,
δ
)
⊆
B
G
p
b
(
x
,
ϵ
).
Case 2. IfG
p
b
(
x
,
x
,
y
)
≠
G
p
b
(
x
,
x
,
x
), then, from property G
p
b
2, we have G
p
b
(
x
,
x
,
x
)
<
G
p
b
(
x
,
x
,
y
), and. for s
=
1, we consider the set
(20)
B
=
{
n
∈
ℕ
∣
ϵ
2
n
+
3
<
G
p
b
(
x
,
x
,
y
)
-
G
p
b
(
x
,
x
,
x
)
}
.
Similarly, by the well ordering principle there exists an elementm such that G
p
b
(
x
,
x
,
y
)
-
G
p
b
(
x
,
x
,
x
)
≤
ϵ
/
2
m
+
2, and we choose δ
=
ϵ
/
2
m
+
2. One can easily obtain that B
G
p
b
(
y
,
δ
)
⊆
B
G
p
b
(
x
,
ϵ
).
Fors
>
1, we consider the set
(21)
C
=
{
n
∈
ℕ
∣
ϵ
2
s
n
+
2
<
G
p
b
(
x
,
x
,
y
)
-
1
s
G
p
b
(
x
,
x
,
x
)
}
and by the well ordering principle there exists an element m such that G
p
b
(
x
,
x
,
y
)
-
(
1
/
s
)
G
p
b
(
x
,
x
,
x
)
≤
ϵ
/
2
s
m
+
1 and we choose δ
=
ϵ
/
2
s
m
+
1. Let z
∈
B
G
p
b
(
y
,
δ
). By property G
p
b
4 we have
(22)
G
p
b
(
x
,
x
,
z
)
≤
s
(
G
p
b
(
x
,
x
,
y
)
+
G
p
b
(
y
,
y
,
z
)
-
G
p
b
(
y
,
y
,
y
)
)
≤
s
(
G
p
b
(
x
,
x
,
y
)
+
δ
)
≤
G
p
b
(
x
,
x
,
x
)
+
ϵ
2
s
m
+
ϵ
2
s
m
=
G
p
b
(
x
,
x
,
x
)
+
ϵ
s
m
<
G
p
b
(
x
,
x
,
x
)
+
ϵ
.
Hence, B
G
p
b
(
y
,
δ
)
⊆
B
G
p
b
(
x
,
ϵ
).Thus, from the above proposition the family of all openG
p
b-balls
(23)
ϝ
=
{
B
G
p
b
(
x
,
ϵ
)
∣
x
∈
X
,
ϵ
>
0
}
is a base of a T
0-topology τ
G
p
b on X which we call the G
p
b-metric topology.The topological space(
X
,
G
p
b
) is T
0, but need not be T
1.Definition 23.
Let(
X
,
G
p
b
) be a G
p
b-metric space. Let {
x
n
} be a sequence in X.(1)
A pointx
∈
X is said to be a limit of the sequence {
x
n
}, denoted by x
n
→
x, if lim
⁡
n
,
m
→
∞
G
p
b
(
x
,
x
n
,
x
m
)
=
G
p
b
(
x
,
x
,
x
).
(2)
{
x
n
} is said to be a G
p
b-Cauchy sequence, if lim
⁡
n
,
m
→
∞
G
p
b
(
x
n
,
x
m
,
x
m
) exists (and is finite).
(3)
(
X
,
G
p
b
) is said to be G
p
b-complete if every G
p
b-Cauchy sequence in X is G
p
b-convergent to an x
∈
X.Using the above definitions, one can easily prove the following proposition.Proposition 24.
Let(
X
,
G
p
b
) be a G
p
b-metric space. Then, for any sequence {
x
n
} in X and a point x
∈
X, the following statements are equivalent:(1)
{
x
n
}  is  G
p
b
-convergent to  x.
(2)
G
p
b
(
x
n
,
x
n
,
x
)
→
G
p
b
(
x
,
x
,
x
)
, as  n
→
∞.
(3)
G
p
b
(
x
n
,
x
,
x
)
→
G
p
b
(
x
,
x
,
x
)
, as  n
→
∞.Based on Lemma 2.2 of [27], we prove the following essential lemma.Lemma 25.
(1) A sequence{
x
n
} is a G
p
b-Cauchy sequence in a G
p
b-metric space (
X
,
G
p
b
) if and only if it is a b-Cauchy sequence in the b-metric space (
X
,
d
G
p
b
).
(2) AG
p
b-metric space (
X
,
G
p
b
) is G
p
b-complete if and only if the b-metric space (
X
,
d
G
p
b
) is b-complete. Moreover, lim
⁡
n
→
∞
d
G
p
b
(
x
,
x
n
)
=
0 if and only if
(24)
lim
⁡
n
→
∞
G
p
b
(
x
,
x
n
,
x
n
)
=
lim
⁡
n
→
∞
G
p
b
(
x
n
,
x
,
x
)
=
lim
⁡
n
,
m
→
∞
G
p
b
(
x
n
,
x
n
,
x
m
)
=
G
p
b
(
x
,
x
,
x
)
.Proof.
First, we show that everyG
p
b-Cauchy sequence in (
X
,
G
p
b
) is a b-Cauchy sequence in (
X
,
d
G
p
b
). Let {
x
n
} be a G
p
b-Cauchy sequence in (
X
,
G
p
b
). Then, there exists α
∈
ℝ such that, for arbitrary ɛ
>
0, there is n
ɛ
∈
ℕ with
(25)
|
G
p
b
(
x
n
,
x
m
,
x
m
)
-
α
|
<
ɛ
4
,
for all n
,
m
≥
n
ɛ. Hence,
(26)
|
d
G
p
b
(
x
n
,
x
m
)
|
=
G
p
b
(
x
n
,
x
m
,
x
m
)
+
G
p
b
(
x
m
,
x
n
,
x
n
)
-
G
p
b
(
x
n
,
x
n
,
x
n
)
-
G
p
b
(
x
m
,
x
m
,
x
m
)
=
|
G
p
b
(
x
n
,
x
m
,
x
m
)
-
α
+
α
-
G
p
b
(
x
n
,
x
n
,
x
n
)
hhhhh
+
G
p
b
(
x
m
,
x
n
,
x
n
)
-
α
+
α
-
G
p
b
(
x
m
,
x
m
,
x
m
)
|
≤
|
G
p
b
(
x
n
,
x
m
,
x
m
)
-
α
|
+
|
α
-
G
p
b
(
x
n
,
x
n
,
x
n
)
|
+
|
G
p
b
(
x
m
,
x
n
,
x
n
)
-
α
|
+
|
α
-
G
p
b
(
x
m
,
x
m
,
x
m
)
|
<
ɛ
,
for all n
,
m
≥
n
ɛ. Hence, we conclude that {
x
n
} is a b-Cauchy sequence in (
X
,
d
G
p
b
).
Next, we prove thatb-completeness of (
X
,
d
G
p
b
) implies G
p
b-completeness of (
X
,
G
p
b
). Indeed, if {
x
n
} is a G
p
b-Cauchy sequence in (
X
,
G
p
b
), then it is also a b-Cauchy sequence in (
X
,
d
G
p
b
). Since the b-metric space (
X
,
d
G
p
b
) is b-complete we deduce that there exists y
∈
X such that lim
⁡
n
→
∞
d
G
p
b
(
y
,
x
n
)
=
0. Hence,
(27)
lim
⁡
n
→
∞
[
G
p
b
(
x
n
,
y
,
y
)
-
G
p
b
(
y
,
y
,
y
)
+
G
p
b
(
y
,
x
n
,
x
n
)
hhhhh
-
G
p
b
(
x
n
,
x
n
,
x
n
)
]
=
0
,
therefore; lim
⁡
n
→
∞
[
G
p
b
(
x
n
,
y
,
y
)
-
G
p
b
(
y
,
y
,
y
)
]
=
0.
On the other hand,(28)
lim
⁡
n
,
m
→
∞
G
p
b
(
x
n
,
x
m
,
y
)
≤
lim
⁡
n
,
m
→
∞
s
G
p
b
(
x
n
,
y
,
y
)
+
lim
⁡
n
,
m
→
∞
s
G
p
b
(
x
m
,
y
,
y
)
-
s
G
p
b
(
y
,
y
,
y
)
+
1
-
s
3
[
lim
⁡
n
,
m
→
∞
G
p
b
(
x
n
,
x
n
,
x
n
)
h
h
h
h
h
h
h
h
h
h
+
lim
⁡
n
,
m
→
∞
G
p
b
(
x
m
,
x
m
,
x
m
)
h
h
h
h
h
h
h
h
h
h
lim
⁡
n
,
m
→
∞
+
G
p
b
(
y
,
y
,
y
)
]
=
G
p
b
(
y
,
y
,
y
)
.
Also, from (
G
p
b
2
),
(29)
G
p
b
(
y
,
y
,
y
)
≤
lim
⁡
n
,
m
→
∞
G
p
b
(
x
n
,
x
m
,
y
)
.
Hence, we obtain that {
x
n
} is a G
p
b-convergent sequence in (
X
,
G
p
b
).
Now, we prove that everyb-Cauchy sequence {
x
n
} in (
X
,
d
G
p
b
) is a G
p
b-Cauchy sequence in (
X
,
G
p
b
). Let ɛ
=
1
/
2. Then, there exists n
0
∈
ℕ such that d
G
p
b
(
x
n
,
x
m
)
<
1
/
2 for all n
,
m
≥
n
0. Since
(30)
G
p
b
(
x
n
,
x
n
0
,
x
n
0
)
-
G
p
b
(
x
n
0
,
x
n
0
,
x
n
0
)
≤
d
G
p
b
(
x
n
,
x
n
0
)
<
1
2
,
then
(31)
G
p
b
(
x
n
,
x
n
,
x
n
)
≤
d
G
p
b
(
x
n
,
x
n
0
)
+
G
p
b
(
x
n
,
x
n
0
,
x
n
0
)
<
1
2
+
G
p
b
(
x
n
0
,
x
n
0
,
x
n
0
)
.
Consequently, the sequence {
G
p
b
(
x
n
,
x
n
,
x
n
)
} is bounded in ℝ and so there exists a
∈
ℝ such that a subsequence {
G
p
b
(
x
n
k
,
x
n
k
,
x
n
k
)
} is convergent to a; that is,
(32)
lim
⁡
k
→
∞
G
p
b
(
x
n
k
,
x
n
k
,
x
n
k
)
=
a
.
Now, we prove that {
G
p
b
(
x
n
,
x
n
,
x
n
)
} is a Cauchy sequence in ℝ. Since {
x
n
} is a b-Cauchy sequence in (
X
,
d
G
p
b
), for given ɛ
>
0, there exists n
ɛ
∈
ℕ such that d
G
p
b
(
x
n
,
x
m
)
<
ɛ, for all n
,
m
≥
n
ɛ. Thus, for all n
,
m
≥
n
ɛ,
(33)
G
p
b
(
x
n
,
x
n
,
x
n
)
-
G
p
b
(
x
m
,
x
m
,
x
m
)
≤
G
p
b
(
x
n
,
x
m
,
x
m
)
-
G
p
b
(
x
m
,
x
m
,
x
m
)
≤
d
G
p
b
(
x
m
,
x
n
)
<
ɛ
.
Therefore, lim
⁡
n
→
∞
G
p
b
(
x
n
,
x
n
,
x
n
)
=
a.
On the other hand,(34)
|
G
p
b
(
x
n
,
x
m
,
x
m
)
-
a
|
=
|
G
p
b
(
x
n
,
x
m
,
x
m
)
-
G
p
b
(
x
n
,
x
n
,
x
n
)
hh
+
G
p
b
(
x
n
,
x
n
,
x
n
)
-
a
|
≤
d
G
p
b
(
x
m
,
x
n
)
+
|
G
p
b
(
x
n
,
x
n
,
x
n
)
-
a
|
,
for all n
,
m
≥
n
ɛ. Hence, lim
⁡
n
,
m
→
∞
G
p
b
(
x
n
,
x
m
,
x
m
)
=
a, and consequently {
x
n
} is a G
p
b-Cauchy sequence in (
X
,
G
p
b
).
Conversely, let{
x
n
} be a b-Cauchy sequence in (
X
,
d
G
p
b
). Then, {
x
n
} is a G
p
b-Cauchy sequence in (
X
,
G
p
b
) and so it is G
p
b-convergent to a point x
∈
X with
(35)
lim
⁡
n
→
∞
G
p
b
(
x
,
x
n
,
x
n
)
=
lim
⁡
n
→
∞
G
p
b
(
x
n
,
x
,
x
)
=
lim
⁡
n
,
m
→
∞
G
p
b
(
x
,
x
m
,
x
n
)
=
G
p
b
(
x
,
x
,
x
)
.
Then, for given ɛ
>
0, there exists n
ɛ
∈
ℕ such that
(36)
G
p
b
(
x
,
x
n
,
x
n
)
-
G
p
b
(
x
,
x
,
x
)
<
ɛ
4
,
G
p
b
(
x
n
,
x
,
x
)
-
G
p
b
(
x
,
x
,
x
)
<
ɛ
4
.
Therefore,
(37)
|
d
G
p
b
(
x
n
,
x
)
|
=
|
G
p
b
(
x
n
,
x
,
x
)
-
G
p
b
(
x
n
,
x
n
,
x
n
)
hhhhl
+
G
p
b
(
x
n
,
x
n
,
x
)
-
G
p
b
(
x
,
x
,
x
)
|
≤
|
G
p
b
(
x
n
,
x
,
x
)
-
G
p
b
(
x
,
x
,
x
)
|
+
|
G
p
b
(
x
,
x
,
x
)
-
G
p
b
(
x
n
,
x
n
,
x
n
)
|
+
|
G
p
b
(
x
n
,
x
n
,
x
)
-
G
p
b
(
x
,
x
,
x
)
|
<
ɛ
,
whenever n
≥
n
ɛ. Therefore, (
X
,
d
G
p
b
) is b-complete.
Finally, letlim
⁡
n
→
∞
d
G
p
b
(
x
n
,
x
)
=
0. So
(38)
lim
⁡
n
→
∞
[
G
p
b
(
x
n
,
x
,
x
)
-
G
p
b
(
x
n
,
x
n
,
x
n
)
]
+
lim
⁡
n
→
∞
[
G
p
b
(
x
n
,
x
n
,
x
)
-
G
p
b
(
x
,
x
,
x
)
]
=
0
,
lim
⁡
n
→
∞
[
G
p
b
(
x
n
,
x
,
x
)
-
G
p
b
(
x
,
x
,
x
)
]
+
lim
⁡
n
→
∞
[
G
p
b
(
x
n
,
x
n
,
x
)
-
G
p
b
(
x
n
,
x
n
,
x
n
)
]
=
0
.
On the other hand,
(39)
lim
⁡
n
,
m
→
∞
G
p
b
(
x
n
,
x
m
,
x
m
)
-
G
p
b
(
x
,
x
,
x
)
≤
s
[
lim
⁡
n
→
∞
G
p
b
(
x
n
,
x
,
x
)
+
lim
⁡
m
→
∞
G
p
b
(
x
,
x
m
,
x
m
)
h
h
h
h
h
h
lim
⁡
n
→
∞
G
p
b
(
x
n
,
x
,
x
)
-
G
p
b
(
x
,
x
,
x
)
]
+
1
-
s
3
×
[
lim
⁡
n
→
∞
G
p
b
(
x
n
,
x
n
,
x
n
)
+
2
lim
⁡
m
→
∞
G
p
b
(
x
m
,
x
m
,
x
m
)
]
-
G
p
b
(
x
,
x
,
x
)
=
0
.Definition 26.
Let(
X
,
G
p
b
) and (
X
′
,
G
p
b
′
) be two generalized partial b-metric spaces and let f
:
(
X
,
G
p
b
)
→
(
X
′
,
G
p
b
′
) be a mapping. Then f is said to be G
p
b-continuous at a point a
∈
X if, for a given ɛ
>
0, there exists δ
>
0 such that x
∈
X and G
p
b
(
a
,
a
,
x
)
<
δ
+
G
p
b
(
a
,
a
,
a
) imply that G
p
b
′
(
f
(
a
)
,
f
(
a
)
,
f
(
x
)
)
<
ɛ
+
G
p
b
′
(
f
(
a
)
,
f
(
a
)
,
f
(
a
)
). The mapping f is G
p
b-continuous on X if it is G
p
b-continuous at all a
∈
X. For simplicity, we say that f is continuous.From the above definition, with straightforward calculations, we have the following proposition.Proposition 27.
Let(
X
,
G
p
b
) and (
X
′
,
G
p
b
′
) be two generalized partial b-metric spaces. Then a mapping f
:
X
→
X
′ is G
p
b-continuous at a point x
∈
X if and only if it is G
p
bsequentially continuous at x; that is, whenever {
x
n
} is G
p
b-convergent to x, {
f
(
x
n
)
} is G
p
b
′-convergent to f
(
x
).Definition 28.
A triple(
X
,
⪯
,
G
p
b
) is called an ordered generalized partial b-metric space if (
X
,
⪯
) is a partially ordered set and G
p
b is a generalized partial b-metric on X.We will need the following simple lemma of theG
p
b-convergent sequences in the proof of our main results.Lemma 29.
Let(
X
,
G
p
b
) be a G
p
b-metric space and suppose that {
x
n
}, {
y
n
} and {
z
n
} are G
p
b-convergent to x, y, and z, respectively. Then we have
(40)
1
s
3
G
p
b
(
x
,
y
,
z
)
-
1
s
2
G
p
b
(
x
,
x
,
x
)
-
1
s
G
p
b
(
y
,
y
,
y
)
-
G
p
b
(
z
,
z
,
z
)
≤
liminf
⁡
n
→
∞
G
p
b
(
x
n
,
y
n
,
z
n
)
≤
limsup
⁡
n
→
∞
G
p
b
(
x
n
,
y
n
,
z
n
)
≤
s
3
G
p
b
(
x
,
y
,
z
)
+
s
G
p
b
(
x
,
x
,
x
)
+
s
2
G
p
b
(
y
,
y
,
y
)
+
s
3
G
p
b
(
z
,
z
,
z
)
.
In particular, if {
y
n
}
=
{
z
n
}
=
a are constant, then
(41)
1
s
G
p
b
(
x
,
a
,
a
)
-
G
p
b
(
x
,
x
,
x
)
≤
liminf
⁡
n
→
∞
G
p
b
(
x
n
,
a
,
a
)
≤
limsup
⁡
n
→
∞
G
p
b
(
x
n
,
a
,
a
)
≤
s
G
p
b
(
x
,
a
,
a
)
+
s
G
p
b
(
x
,
x
,
x
)
.Proof.
Using the rectangle inequality, we obtain(42)
G
p
b
(
x
,
y
,
z
)
≤
s
G
p
b
(
x
,
x
n
,
x
n
)
+
s
2
G
p
b
(
y
,
y
n
,
y
n
)
+
s
3
G
p
b
(
z
,
z
n
,
z
n
)
+
s
3
G
p
b
(
x
n
,
y
n
,
z
n
)
,
G
p
b
(
x
n
,
y
n
,
z
n
)
≤
s
G
p
b
(
x
n
,
x
,
x
)
+
s
2
G
p
b
(
y
n
,
y
,
y
)
+
s
3
G
p
b
(
z
n
,
z
,
z
)
+
s
3
G
p
b
(
x
,
y
,
z
)
.
Taking the lower limit as n
→
∞ in the first inequality and the upper limit as n
→
∞ in the second inequality we obtain the desired result.
If{
y
n
}
=
{
z
n
}
=
a, then
(43)
G
p
b
(
x
,
a
,
a
)
≤
s
G
p
b
(
x
,
x
n
,
x
n
)
+
s
G
p
b
(
x
n
,
a
,
a
)
,
G
p
b
(
x
n
,
a
,
a
)
≤
s
G
p
b
(
x
n
,
x
,
x
)
+
s
G
p
b
(
x
,
a
,
a
)
.Let𝔖 denote the class of all real functions β
:
[
0
,
+
∞
)
→
[
0,1
) satisfying the condition
(44)
β
(
t
n
)
⟶
1
implies
t
n
⟶
0
,
as
n
⟶
∞
.In order to generalize the Banach contraction principle, Geraghty proved the following result.Theorem 30 (see [32]).
Let(
X
,
d
) be a complete metric space and let f
:
X
→
X be a self-map. Suppose that there exists β
∈
𝔖 such that
(45)
d
(
f
x
,
f
y
)
≤
β
(
d
(
x
,
y
)
)
d
(
x
,
y
)
holds for all x
,
y
∈
X. Then f has a unique fixed point z
∈
X and for each x
∈
X the Picard sequence {
f
n
x
} converges to z.In [33], some fixed point theorems for mappings satisfying Geraghty-type contractive conditions are proved in various generalized metric spaces.As in [33], we will consider the class of functions ℱ, where β
∈
ℱ if β
:
[
0
,
∞
)
→
[
0,1
/
s
) and has the property
(46)
β
(
t
n
)
⟶
1
s
implies
t
n
⟶
0
,
as
n
⟶
∞
.Theorem 31 (see [33]).
Lets
>
1 and let (
X
,
D
,
s
) be a complete metric type space. Suppose that a mapping f
:
X
→
X satisfies the condition
(47)
D
(
f
x
,
f
y
)
≤
β
(
D
(
x
,
y
)
)
D
(
x
,
y
)
for all x
,
y
∈
X and some β
∈
ℱ. Then f has a unique fixed point z
∈
X and for each x
∈
X
{
f
n
x
} converges to z in (
X
,
D
,
s
).The aim of this paper is to present certain new fixed point theorems for hybrid rational Geraghty-type andψ-contractive mappings in partially ordered G
p
b-metric spaces. Our results improve and generalize many comparable results in literature. Some examples are established to prove the generality of our results.


## 2. Main Results

Recall thatℱ denotes the class of all functions β
:
[
0
,
∞
)
→
[
0,1
/
s
) satisfying the following condition:
(48)
β
(
t
n
)
⟶
1
s
implies
t
n
⟶
0
,
as
n
⟶
∞
.Theorem 32.
Let(
X
,
⪯
) be a partially ordered set and suppose that there exists a generalized partial b-metric G
p
b on X such that (
X
,
G
p
b
) is a G
p
b-complete G
p
b-metric space and let f
:
X
→
X be an increasing mapping with respect to ⪯ with x
0
⪯
f
(
x
0
) for some x
0
∈
X. Suppose that
(49)
s
G
p
b
(
f
x
,
f
y
,
f
z
)
≤
β
(
G
p
b
(
x
,
y
,
z
)
)
M
(
x
,
y
,
z
)
for all comparable elements x
,
y
,
z
∈
X, where
(50)
M
(
x
,
y
,
z
)
=
max
⁡
{
G
p
b
(
x
,
y
,
z
)
,
G
p
b
(
x
,
x
,
f
x
)
G
p
b
(
y
,
y
,
f
y
)
G
p
b
(
z
,
z
,
f
z
)
1
+
[
s
G
p
b
(
f
x
,
f
y
,
f
z
)
]
2
h
h
h
h
h
h
h
h
h
G
p
b
(
x
,
x
,
f
x
)
G
p
b
(
y
,
y
,
f
y
)
G
p
b
(
z
,
z
,
f
z
)
1
+
[
s
G
p
b
(
f
x
,
f
y
,
f
z
)
]
2
}
.
If f is continuous, then f has a fixed point.Proof.
Putx
n
=
f
n
(
x
0
). Since x
0
⪯
f
(
x
0
) and f is an increasing function we obtain by induction that
(51)
x
0
⪯
f
(
x
0
)
⪯
f
2
(
x
0
)
⪯
⋯
⪯
f
n
(
x
0
)
⪯
f
n
+
1
(
x
0
)
⪯
⋯
.
Step
1. We will show that lim
⁡
n
→
∞
G
p
b
(
x
n
,
x
n
+
1
,
x
n
+
2
)
=
0. Since x
n
⪯
x
n
+
1 for each n
∈
ℕ, then by (49) we have
(52)
s
G
p
b
(
x
n
,
x
n
+
1
,
x
n
+
2
)
=
s
G
p
b
(
f
x
n
-
1
,
f
x
n
,
f
x
n
+
1
)
≤
β
(
G
p
b
(
x
n
-
1
,
x
n
,
x
n
+
1
)
)
M
(
x
n
-
1
,
x
n
,
x
n
+
1
)
≤
1
s
G
p
b
(
x
n
-
1
,
x
n
,
x
n
+
1
)
≤
G
p
b
(
x
n
-
1
,
x
n
,
x
n
+
1
)
,
because
(53)
M
(
x
n
-
1
,
x
n
,
x
n
+
1
)
=
max
⁡
{
G
p
b
(
x
n
-
1
,
x
n
,
x
n
+
1
)
,
(
1
+
[
s
G
p
b
(
f
x
n
-
1
,
f
x
n
,
f
x
n
+
1
)
]
2
)
-
1
h
h
h
h
h
h
h
h
l
h
(
G
p
b
(
x
n
-
1
,
x
n
-
1
,
f
x
n
-
1
)
G
p
b
(
x
n
,
x
n
,
f
x
n
)
h
h
h
h
h
h
h
h
h
l
h
×
G
p
b
(
x
n
+
1
,
x
n
+
1
,
f
x
n
+
1
)
)
h
h
h
h
h
h
h
h
l
h
×
(
1
+
[
s
G
p
b
(
f
x
n
-
1
,
f
x
n
,
f
x
n
+
1
)
]
2
)
-
1
}
=
max
⁡
⁡
{
G
p
b
(
x
n
-
1
,
x
n
,
x
n
+
1
)
,
(
1
+
[
s
G
p
b
(
f
x
n
-
1
,
f
x
n
,
f
x
n
+
1
)
]
2
)
-
1
h
h
h
h
h
h
h
h
l
h
(
G
p
b
(
x
n
-
1
,
x
n
-
1
,
x
n
)
G
p
b
(
x
n
,
x
n
,
x
n
+
1
)
h
h
h
h
h
h
h
h
h
l
h
×
G
p
b
(
x
n
+
1
,
x
n
+
1
,
x
n
+
2
)
)
h
h
h
h
h
h
h
h
l
h
×
(
1
+
[
s
G
p
b
(
x
n
,
x
n
+
1
,
x
n
+
2
)
]
2
)
-
1
}
=
G
p
b
(
x
n
-
1
,
x
n
,
x
n
+
1
)
.
Therefore, {
G
p
b
(
x
n
,
x
n
+
1
,
x
n
+
2
)
} is decreasing. Then there exists r
≥
0 such that lim
⁡
n
→
∞
G
p
b
(
x
n
,
x
n
+
1
,
x
n
+
2
)
=
r. Letting n
→
∞ in (52) we have
(54)
s
r
≤
r
.
Since s
>
1, we deduce that r
=
0; that is,
(55)
lim
⁡
n
→
∞
G
p
b
(
x
n
,
x
n
+
1
,
x
n
+
2
)
=
0
.
Step
2. Now, we prove that the sequence {
x
n
} is a G
p
b-Cauchy sequence. By rectangular inequality and (49), we have
(56)
G
p
b
(
x
n
,
x
m
,
x
m
)
≤
s
G
p
b
(
x
n
,
x
n
+
1
,
x
n
+
1
)
+
s
2
G
p
b
(
x
n
+
1
,
x
m
+
1
,
x
m
+
1
)
+
s
2
G
p
b
(
x
m
+
1
,
x
m
,
x
m
)
≤
s
G
p
b
(
x
n
,
x
n
+
1
,
x
n
+
1
)
+
s
β
(
G
p
b
(
x
n
,
x
m
,
x
m
)
)
×
M
(
x
n
,
x
m
,
x
m
)
+
s
2
G
p
b
(
x
m
+
1
,
x
m
,
x
m
)
.
Lettingm
,
n
→
∞ in the above inequality and applying (55) we have
(57)
lim
⁡
m
,
n
→
∞
G
p
b
(
x
n
,
x
m
,
x
m
)
≤
s
lim
⁡
n
,
m
→
∞
β
(
G
p
b
(
x
n
,
x
m
,
x
m
)
)
lim
⁡
n
,
m
→
∞
M
(
x
n
,
x
m
,
x
m
)
.
Here,
(58)
G
p
b
(
x
n
,
x
m
,
x
m
)
≤
M
(
x
n
,
x
m
,
x
m
)
=
max
⁡
{
G
p
b
(
x
n
,
x
m
,
x
m
)
,
G
p
b
(
x
,
x
,
f
x
)
G
p
b
(
y
,
y
,
f
y
)
G
p
b
(
z
,
z
,
f
z
)
1
+
[
s
G
p
b
(
f
x
,
f
y
,
f
z
)
]
2
hhhhlh
G
p
b
(
x
n
,
x
n
,
f
(
x
n
)
)
[
G
p
b
(
x
m
,
x
m
,
f
(
x
m
)
)
]
2
1
+
[
s
G
p
b
(
f
x
n
,
f
x
m
,
f
x
m
)
]
2
}
=
max
⁡
{
G
p
b
(
x
n
,
x
m
,
x
m
)
,
G
p
b
(
x
,
x
,
f
x
)
G
p
b
(
y
,
y
,
f
y
)
G
p
b
(
z
,
z
,
f
z
)
1
+
[
s
G
p
b
(
f
x
,
f
y
,
f
z
)
]
2
hhhhlh
G
p
b
(
x
n
,
x
n
,
x
n
+
1
)
[
G
p
b
(
x
m
,
x
m
,
x
m
+
1
)
]
2
1
+
[
s
G
p
b
(
x
n
+
1
,
x
m
+
1
,
x
m
+
1
)
]
2
}
.
Letting m
,
n
→
∞ in the above inequality we get
(59)
lim
⁡
m
,
n
→
∞
M
(
x
n
,
x
m
,
x
m
)
=
lim
⁡
m
,
n
→
∞
G
p
b
(
x
n
,
x
m
,
x
m
)
.
Hence, from (57) and (59), we obtain
(60)
lim
⁡
m
,
n
→
∞
G
p
b
(
x
n
,
x
m
,
x
m
)
≤
s
lim
⁡
m
,
n
→
∞
β
(
G
p
b
(
x
n
,
x
m
,
x
m
)
)
lim
⁡
m
,
n
→
∞
G
p
b
(
x
n
,
x
m
,
x
m
)
and so we get
(61)
1
s
≤
lim
⁡
m
,
n
→
∞
β
(
G
p
b
(
x
n
,
x
m
,
x
m
)
)
.
Since β
∈
ℱ we deduce that
(62)
lim
⁡
m
,
n
→
∞
G
p
b
(
x
n
,
x
m
,
x
m
)
=
0
.
Consequently, {
x
n
} is a G
p
b-Cauchy sequence in X. Thus, from Lemma 25, {
x
n
} is a b-Cauchy sequence in the b-metric space (
X
,
d
G
p
b
). Since (
X
,
G
p
b
) is G
p
b-complete, then, from Lemma 25, (
X
,
d
G
p
b
) is a b-complete b-metric space. Therefore, the sequence {
x
n
}
b-converges to some u
∈
X; that is, lim
⁡
n
→
∞
d
G
p
b
(
x
n
,
u
)
=
0. Again, from Lemma 25 and (62),
(63)
lim
⁡
n
→
∞
G
p
b
(
u
,
x
n
,
x
n
)
=
lim
⁡
m
,
n
→
∞
G
p
b
(
x
n
,
x
m
,
x
m
)
=
G
p
b
(
u
,
u
,
u
)
=
0
.
Step
3. Now, we show that u is a fixed point of f. Suppose to the contrary; that is, f
u
≠
u; then, from Lemma 20, we have G
p
b
(
u
,
u
,
f
u
)
>
0.
Using the rectangular inequality, we get(64)
G
p
b
(
u
,
u
,
f
u
)
≤
s
G
p
b
(
f
u
,
f
x
n
,
f
x
n
)
+
s
G
p
b
(
f
x
n
,
u
,
u
)
.
Letting n
→
∞ and using the continuity of f and (63), we get
(65)
G
p
b
(
u
,
u
,
f
u
)
≤
s
lim
⁡
n
→
∞
G
p
b
(
f
u
,
f
x
n
,
f
x
n
)
+
s
lim
⁡
n
→
∞
G
p
b
(
f
x
n
,
u
,
u
)
=
s
G
p
b
(
f
u
,
f
u
,
f
u
)
.
Note that, from (49), we have
(66)
s
G
p
b
(
f
u
,
f
u
,
f
u
)
≤
β
(
G
p
b
(
u
,
u
,
u
)
)
M
(
u
,
u
,
u
)
,
where by (65)
(67)
M
(
u
,
u
,
u
)
=
max
⁡
{
G
p
b
(
u
,
u
,
u
)
,
G
p
b
(
u
,
u
,
f
u
)
G
p
b
(
u
,
u
,
f
u
)
G
p
b
(
u
,
u
,
f
u
)
1
+
[
s
G
p
b
(
f
u
,
f
u
,
f
u
)
]
2
h
h
h
h
h
h
h
l
h
h
G
p
b
(
u
,
u
,
f
u
)
G
p
b
(
u
,
u
,
f
u
)
G
p
b
(
u
,
u
,
f
u
)
1
+
[
s
G
p
b
(
f
u
,
f
u
,
f
u
)
]
2
}
≤
G
p
b
(
u
,
u
,
f
u
)
.
Hence, as β
(
t
)
≤
1 for all t
∈
[
0
,
∞
), we have s
G
p
b
(
f
u
,
f
u
,
f
u
)
≤
G
p
b
(
u
,
u
,
f
u
). Thus, by (65) we obtain that s
G
p
b
(
f
u
,
f
u
,
f
u
)
=
G
p
b
(
u
,
u
,
f
u
). But then, using (66), we get that G
p
b
(
u
,
u
,
f
u
)
=
s
G
p
b
(
f
u
,
f
u
,
f
u
)
≤
β
(
G
p
b
(
u
,
u
,
u
)
)
M
(
u
,
u
,
u
)
<
G
p
b
(
u
,
u
,
f
u
), which is a contradiction. Hence, we have f
u
=
u. Thus, u is a fixed point of f.Now we replace the continuity off in Theorem 32 by the regularity of the space to get the required conclusion.Theorem 33.
Under the same hypotheses of Theorem32, instead of the continuity assumption of f, assume that, whenever {
x
n
} is a nondecreasing sequence in X such that x
n
→
u, one has x
n
⪯
u for all n
∈
ℕ. Then f has a fixed point.Proof.
Repeating the proof of Theorem32, we construct an increasing sequence {
x
n
} in X such that x
n
→
u
∈
X. Using the assumption on X we have x
n
⪯
u. Now, we show that u
=
f
u. By Lemma 29 and (63)
(68)
s
[
1
s
G
p
b
(
u
,
f
u
,
f
u
)
-
G
p
b
(
u
,
u
,
u
)
]
≤
s
limsup
⁡
n
→
∞
G
p
b
(
x
n
+
1
,
f
u
,
f
u
)
≤
limsup
⁡
n
→
∞
(
β
(
G
p
b
(
x
n
,
u
,
u
)
)
M
(
x
n
,
u
,
u
)
)
≤
1
s
limsup
⁡
n
→
∞
M
(
x
n
,
u
,
u
)
,
where
(69)
lim
⁡
n
→
∞
M
(
x
n
,
u
,
u
)
=
lim
⁡
n
→
∞
max
⁡
{
G
p
b
(
x
n
,
u
,
u
)
,
[
G
p
b
(
x
n
,
x
n
,
f
x
n
)
,
G
p
b
(
u
,
u
,
f
u
)
]
2
1
+
[
s
G
p
b
(
f
x
n
,
f
u
,
f
u
)
]
2
h
h
h
h
h
h
h
h
h
h
h
h
h
h
[
G
p
b
(
x
n
,
x
n
,
f
x
n
)
,
G
p
b
(
u
,
u
,
f
u
)
]
2
1
+
[
s
G
p
b
(
f
x
n
,
f
u
,
f
u
)
]
2
}
=
lim
⁡
n
→
∞
max
⁡
{
G
p
b
(
x
n
,
x
n
,
u
)
,
[
G
p
b
(
x
n
,
x
n
,
f
x
n
)
,
G
p
b
(
u
,
u
,
f
u
)
]
2
1
+
[
s
G
p
b
(
f
x
n
,
f
u
,
f
u
)
]
2
h
h
h
h
h
h
h
h
h
h
h
h
h
h
[
G
p
b
(
x
n
,
x
n
,
x
n
+
1
)
,
G
p
b
(
u
,
u
,
f
u
)
]
2
1
+
[
s
G
p
b
(
x
n
+
1
,
f
u
,
f
u
)
]
2
}
=
max
⁡
{
G
p
b
(
u
,
u
,
u
)
,
0
}
=
0
(
see
(
55
)
and
(
63
)
)
.
Therefore, we deduce that G
p
b
(
u
,
f
u
,
f
u
)
≤
s
G
p
b
(
u
,
u
,
u
)
=
0. Hence, we have u
=
f
u.If in the above theorems we assumeβ
(
t
)
=
r, where 0
≤
r
≤
1
/
s, we obtain the following corollary.Corollary 34.
Let(
X
,
⪯
) be a partially ordered set and suppose that there exists a G
p
b-metric on X such that (
X
,
G
p
b
) is a G
p
b-complete G
p
b-metric space, and let f
:
X
→
X be an increasing mapping with x
0
⪯
f
(
x
0
) for some x
0
∈
X. Suppose that
(70)
s
G
p
b
(
f
x
,
f
y
,
f
z
)
≤
r
M
(
x
,
y
,
z
)
for all comparable elements x
,
y
,
z
∈
X, where 0
≤
r
<
1
/
s and
(71)
M
(
x
,
y
,
z
)
=
max
⁡
⁡
{
G
p
b
(
x
,
y
,
z
)
,
G
p
b
(
x
,
x
,
f
x
)
G
p
b
(
y
,
y
,
f
y
)
G
p
b
(
z
,
z
,
f
z
)
1
+
[
s
G
p
b
(
f
x
,
f
y
,
f
z
)
]
2
h
h
h
h
h
h
h
h
h
G
p
b
(
x
,
x
,
f
x
)
G
p
b
(
y
,
y
,
f
y
)
G
p
b
(
z
,
z
,
f
z
)
1
+
[
s
G
p
b
(
f
x
,
f
y
,
f
z
)
]
2
}
.
If f is continuous or for any nondecreasing sequence {
x
n
} in X such that x
n
→
u
∈
X one has x
n
⪯
u for all n
∈
ℕ, then f has a fixed point.Corollary 35.
Let(
X
,
⪯
) be a partially ordered set and suppose that there exists a G
p
b-metric space G
p
b on X such that (
X
,
G
p
b
) is a G
p
b-complete G
p
b-metric space, and let f
:
X
→
X be an increasing mapping with respect to ⪯ such that there exists an element x
0
∈
X with x
0
⪯
f
(
x
0
). Suppose that
(72)
s
G
p
b
(
f
x
,
f
y
,
f
z
)
≤
a
G
p
b
(
x
,
y
,
z
)
+
b
G
p
b
(
x
,
x
,
f
x
)
G
p
b
(
y
,
y
,
f
y
)
G
p
b
(
z
,
z
,
f
z
)
1
+
[
s
G
p
b
(
f
x
,
f
y
,
f
z
)
]
2
for all comparable elements x
,
y
,
z
∈
X, where a
,
b
≥
0 and a
+
b
≤
1
/
s.
Iff is continuous or for any nondecreasing sequence {
x
n
} in X such that x
n
→
u
∈
X one has x
n
⪯
u for all n
∈
ℕ, then f has a fixed point.Proof.
Since(73)
a
G
p
b
(
x
,
y
,
z
)
+
b
G
p
b
(
x
,
x
,
f
x
)
G
p
b
(
y
,
y
,
f
y
)
G
p
b
(
z
,
z
,
f
z
)
1
+
[
s
G
p
b
(
f
x
,
f
y
,
f
z
)
]
2
≤
(
a
+
b
)
×
max
⁡
⁡
{
G
p
b
(
x
,
y
,
z
)
,
G
p
b
(
x
,
x
,
f
x
)
G
p
b
(
y
,
y
,
f
y
)
G
p
b
(
z
,
z
,
f
z
)
1
+
[
s
G
p
b
(
f
x
,
f
y
,
f
z
)
]
2
h
h
h
h
h
h
h
h
h
G
p
b
(
x
,
x
,
f
x
)
G
p
b
(
y
,
y
,
f
y
)
G
p
b
(
z
,
z
,
f
z
)
1
+
[
s
G
p
b
(
f
x
,
f
y
,
f
z
)
]
2
}
,
taking r
=
a
+
b, all the conditions of Corollary 34 hold and hence f has a fixed point.LetΨ be the family of all continuous and nondecreasing functions ψ
:
[
0
,
∞
)
→
[
0
,
∞
) such that
(74)
lim
⁡
n
→
∞
ψ
n
(
t
)
=
0
for all t
>
0. It is known that, if ψ
∈
Ψ, then ψ
(
0
)
=
0 and ψ
(
t
)
<
t for all t
>
0.Theorem 36.
Let(
X
,
⪯
) be a partially ordered set and suppose that there exists a generalized partial b-metric G
p
b on X such that (
X
,
G
p
b
) is a G
p
b-complete G
p
b-metric space, and let f
:
X
→
X be an increasing mapping with x
0
⪯
f
(
x
0
) for some x
0
∈
X. Suppose that
(75)
s
G
p
b
(
f
x
,
f
y
,
f
z
)
≤
ψ
(
M
(
x
,
y
,
z
)
)
,
where
(76)
M
(
x
,
y
,
z
)
=
max
⁡
{
G
p
b
(
x
,
y
,
z
)
,
G
p
b
(
x
,
x
,
f
x
)
G
p
b
(
y
,
y
,
f
y
)
G
p
b
(
z
,
z
,
f
z
)
1
+
[
s
G
p
b
(
f
x
,
f
y
,
f
z
)
]
2
h
h
h
h
h
h
h
h
h
h
G
p
b
(
x
,
x
,
f
x
)
G
p
b
(
y
,
y
,
f
y
)
G
p
b
(
z
,
z
,
f
z
)
1
+
[
s
G
p
b
(
f
x
,
f
y
,
f
z
)
]
2
}
for all comparable elements x
,
y
,
z
∈
X. If f is continuous, then f has a fixed point.Proof.
Sincex
0
⪯
f
(
x
0
) and f is an increasing function we obtain by induction that
(77)
x
0
⪯
f
(
x
0
)
⪯
f
2
(
x
0
)
⪯
⋯
⪯
f
n
(
x
0
)
⪯
f
n
+
1
(
x
0
)
⪯
⋯
.
Putting x
n
=
f
n
(
x
0
), we have
(78)
x
0
⪯
x
1
⪯
x
2
⪯
⋯
⪯
x
n
⪯
x
n
+
1
⪯
⋯
.
If there exists n
0
∈
ℕ such that x
n
0
=
x
n
0
+
1 then x
n
0
=
f
x
n
0 and so we have nothing to prove. Hence, for all n
∈
ℕ, we assume that x
n
≠
x
n
+
1.
Step
1. We will prove that
(79)
lim
⁡
n
→
∞
G
p
b
(
x
n
,
x
n
+
1
,
x
n
+
2
)
=
0
.
Using condition (75), we obtain
(80)
G
p
b
(
x
n
,
x
n
+
1
,
x
n
+
2
)
≤
s
G
p
b
(
x
n
,
x
n
+
1
,
x
n
+
2
)
=
s
G
p
b
(
f
x
n
-
1
,
f
x
n
,
f
x
n
+
1
)
≤
ψ
(
M
(
x
n
-
1
,
x
n
,
x
n
+
1
)
)
.
Here,
(81)
M
(
x
n
-
1
,
x
n
,
x
n
+
1
)
=
max
⁡
{
G
p
b
(
x
n
-
1
,
x
n
,
x
n
+
1
)
,
(
1
+
[
G
p
b
(
f
x
n
-
1
,
f
x
n
,
f
x
n
+
1
)
]
)
-
1
h
h
h
h
h
h
h
h
h
×
(
G
p
b
(
x
n
-
1
,
x
n
-
1
,
f
x
n
-
1
)
G
p
b
(
x
n
,
x
n
,
f
x
n
)
h
h
h
h
h
h
h
h
h
h
h
l
h
×
G
p
b
(
x
n
+
1
,
x
n
+
1
,
f
x
n
+
1
)
)
h
h
h
h
h
h
h
h
l
×
(
1
+
[
G
p
b
(
f
x
n
-
1
,
f
x
n
,
f
x
n
+
1
)
]
)
-
1
}
=
G
p
b
(
x
n
-
1
,
x
n
,
x
n
+
1
)
.
Hence,
(82)
G
p
b
(
x
n
,
x
n
+
1
,
x
n
+
2
)
≤
s
G
p
b
(
x
n
,
x
n
+
1
,
x
n
+
2
)
≤
ψ
(
G
p
b
(
x
n
-
1
,
x
n
,
x
n
+
1
)
)
.
By induction, we get that
(83)
G
p
b
(
x
n
+
2
,
x
n
+
1
,
x
n
)
≤
ψ
(
G
p
b
(
x
n
+
1
,
x
n
,
x
n
-
1
)
)
≤
ψ
2
(
G
p
b
(
x
n
,
x
n
-
1
,
x
n
-
2
)
)
≤
⋯
≤
ψ
n
(
G
p
b
(
x
2
,
x
1
,
x
0
)
)
.
As ψ
∈
Ψ, we conclude that
(84)
lim
⁡
n
→
∞
G
p
b
(
x
n
,
x
n
+
1
,
x
n
+
2
)
=
0
.
Step
2. We will prove that {
x
n
} is a G
p
b-Cauchy sequence.
Suppose to the contrary that{
x
n
} is not a G
p
b-Cauchy sequence. Then there exists ɛ
>
0 for which we can find two subsequences {
x
m
i
} and {
x
n
i
} of {
x
n
} such that n
i is the smallest index for which
(85)
n
i
>
m
i
>
i
,
G
p
b
(
x
m
i
,
x
n
i
,
x
n
i
)
≥
ɛ
.
This means that
(86)
G
p
b
(
x
m
i
,
x
n
i
-
1
,
x
n
i
-
1
)
<
ɛ
.
From (85) and using the rectangle inequality, we get
(87)
ɛ
≤
G
p
b
(
x
m
i
,
x
n
i
,
x
n
i
)
≤
s
G
p
b
(
x
m
i
,
x
m
i
+
1
,
x
m
i
+
1
)
+
s
G
p
b
(
x
m
i
+
1
,
x
n
i
,
x
n
i
)
.
Taking the upper limit as i
→
∞, we get
(88)
ɛ
s
≤
limsup
⁡
i
→
∞
G
p
b
(
x
m
i
+
1
,
x
n
i
,
x
n
i
)
.
From the definition of M
(
x
,
y
) we have
(89)
M
(
x
m
i
,
x
n
i
-
1
,
x
n
i
-
1
)
=
max
⁡
⁡
{
G
p
b
(
x
m
i
,
x
n
i
-
1
,
x
n
i
-
1
)
,
(
1
+
[
s
G
p
b
(
f
x
m
i
,
f
x
n
i
-
1
,
f
x
n
i
-
1
)
]
2
)
-
1
h
h
h
h
h
h
h
h
l
h
(
G
p
b
(
x
m
i
,
x
m
i
,
f
x
m
i
)
[
G
p
b
(
x
n
i
-
1
,
x
n
i
-
1
,
f
x
n
i
-
1
)
]
2
h
h
h
h
h
h
h
h
h
h
l
h
×
[
G
p
b
(
x
n
i
-
1
,
x
n
i
-
1
,
f
x
n
i
-
1
)
]
2
)
h
h
h
h
h
h
h
h
l
h
×
(
1
+
[
s
G
p
b
(
f
x
m
i
,
f
x
n
i
-
1
,
f
x
n
i
-
1
)
]
2
)
-
1
}
=
max
⁡
⁡
{
G
p
b
(
x
m
i
,
x
n
i
-
1
,
x
n
i
-
1
)
,
(
1
+
[
s
G
p
b
(
f
x
m
i
,
f
x
n
i
-
1
,
f
x
n
i
-
1
)
]
2
)
-
1
h
h
h
h
h
h
h
h
l
h
(
G
p
b
(
x
m
i
,
x
m
i
,
x
m
i
+
1
)
[
G
p
b
(
x
n
i
-
1
,
x
n
i
-
1
,
x
n
i
)
]
2
h
h
h
h
h
h
h
h
h
h
l
h
×
[
G
p
b
(
x
n
i
-
1
,
x
n
i
-
1
,
x
n
i
)
]
2
)
h
h
h
h
h
h
h
h
l
h
×
(
1
+
[
s
G
p
b
(
x
m
i
+
1
,
x
n
i
,
x
n
i
)
]
2
)
-
1
}
and if i
→
∞, by (84) and (86), we have
(90)
limsup
⁡
i
→
∞
M
(
x
m
i
,
x
n
i
-
1
,
x
n
i
-
1
)
≤
ɛ
.
Now, from (75) we have
(91)
s
G
p
b
(
x
m
i
+
1
,
x
n
i
,
x
n
i
)
=
s
G
p
b
(
f
x
m
i
,
f
x
n
i
-
1
,
f
x
n
i
-
1
)
≤
ψ
(
M
(
x
m
i
,
x
n
i
-
1
,
x
n
i
-
1
)
)
.
Again, if i
→
∞ by (88) we obtain
(92)
ɛ
=
s
(
ɛ
s
)
≤
(
s
limsup
⁡
i
→
∞
G
p
b
(
x
m
i
+
1
,
x
n
i
,
x
n
i
)
)
≤
ψ
(
ɛ
)
<
ɛ
,
which is a contradiction. Consequently, {
x
n
} is a G
p
b-Cauchy sequence in X. Thus, from Lemma 25 we have proved that {
x
n
} is a b-Cauchy sequence in the b-metric space (
X
,
d
G
p
b
). Since (
X
,
G
p
b
) is G
p
b-complete, then, from Lemma 25, (
X
,
d
G
p
b
) is a b-complete b-metric space. Therefore, the sequence {
x
n
}
b-converges to some u
∈
X; that is, lim
⁡
n
→
∞
d
G
p
b
(
x
n
,
u
)
=
0. Again, from Lemma 25 and (62),
(93)
lim
⁡
n
→
∞
G
p
b
(
u
,
x
n
,
x
n
)
=
lim
⁡
m
,
n
→
∞
G
p
b
(
x
n
,
x
m
,
x
m
)
=
G
p
b
(
u
,
u
,
u
)
=
0
.
Step
3. Now we show that u is a fixed point of f. Suppose to the contrary, that f
u
≠
u; then, from Lemma 20, we have G
p
b
(
u
,
u
,
f
u
)
>
0.
Using the rectangle inequality, we get(94)
G
p
b
(
u
,
u
,
f
u
)
≤
s
G
p
b
(
f
u
,
f
x
n
,
f
x
n
)
+
s
G
p
b
(
f
x
n
,
u
,
u
)
.
Letting n
→
∞ and using the continuity of f, we get
(95)
G
p
b
(
u
,
u
,
f
u
)
≤
s
G
p
b
(
f
u
,
f
u
,
f
u
)
.
Note that, from (75), we have
(96)
s
G
p
b
(
f
u
,
f
u
,
f
u
)
≤
ψ
(
M
(
u
,
u
,
u
)
)
,
where
(97)
M
(
u
,
u
,
u
)
=
max
⁡
⁡
{
G
p
b
(
u
,
u
,
u
)
,
G
p
b
(
u
,
u
,
f
u
)
G
p
b
(
u
,
u
,
f
u
)
G
p
b
(
u
,
u
,
f
u
)
1
+
[
s
G
p
b
(
f
u
,
f
u
,
f
u
)
]
2
h
h
h
h
h
h
h
h
l
h
G
p
b
(
u
,
u
,
f
u
)
G
p
b
(
u
,
u
,
f
u
)
G
p
b
(
u
,
u
,
f
u
)
1
+
[
s
G
p
b
(
f
u
,
f
u
,
f
u
)
]
2
}

≤
G
p
b
(
u
,
u
,
f
u
)
.
Hence, as ψ is nondecreasing, we have s
G
p
b
(
f
u
,
f
u
,
f
u
)
≤
G
p
b
(
u
,
u
,
f
u
). Thus, by (95) we obtain that
(98)
G
p
b
(
u
,
u
,
f
u
)
=
s
G
p
b
(
f
u
,
f
u
,
f
u
)
.
Equation (96) yields that G
p
b
(
u
,
u
,
f
u
)
≤
ψ
(
M
(
u
,
u
,
u
)
)
≤
ψ
(
G
p
b
(
u
,
u
,
f
u
)
). This is impossible, according to our assumptions on ψ. Hence, we have f
u
=
u. Thus, u is a fixed point of f.Theorem 37.
Under the hypotheses of Theorem36, instead of the continuity assumption of f, assume that, whenever {
x
n
} is a nondecreasing sequence in X such that x
n
→
u
∈
X, one has x
n
⪯
u for all n
∈
ℕ. Then f has a fixed point.Proof.
Following the proof of Theorem36, we construct an increasing sequence {
x
n
} in X such that x
n
→
u
∈
X. Using the given assumption on X we have x
n
⪯
u. Now, we show that u
=
f
u. By (75) we have
(99)
s
G
p
b
(
f
u
,
f
u
,
x
n
)
=
s
G
p
b
(
f
u
,
f
u
,
f
x
n
-
1
)
≤
ψ
(
M
(
u
,
u
,
x
n
-
1
)
)
,
where
(100)
M
(
u
,
u
,
x
n
-
1
)
=
max
⁡
⁡
{
G
p
b
(
u
,
u
,
x
n
-
1
)
,
[
G
p
b
(
u
,
u
,
f
u
)
]
2
G
p
b
(
x
n
-
1
,
x
n
-
1
,
f
x
n
-
1
)
1
+
[
s
G
p
b
(
f
u
,
f
u
,
f
x
n
-
1
)
]
2
h
h
h
h
h
h
h
h
l
h
[
G
p
b
(
u
,
u
,
f
u
)
]
2
G
p
b
(
x
n
-
1
,
x
n
-
1
,
f
x
n
-
1
)
1
+
[
s
G
p
b
(
f
u
,
f
u
,
f
x
n
-
1
)
]
2
}
.
Letting n
→
∞ in the above, from (93), we get
(101)
lim
⁡
n
→
∞
M
(
u
,
u
,
x
n
-
1
)
=
0
.
Again, taking the upper limit as n
→
∞ in (99) and using Lemma 29 and (101) we get
(102)
s
[
1
s
G
p
b
(
u
,
f
u
,
f
u
)
-
G
p
b
(
u
,
u
,
u
)
]
≤
s
limsup
⁡
n
→
∞
G
p
b
(
x
n
,
f
u
,
f
u
)
≤
limsup
⁡
n
→
∞
ψ
(
M
(
u
,
u
,
x
n
-
1
)
)
=
0
.
So we get G
p
b
(
u
,
f
u
,
f
u
)
=
0. That is, f
u
=
u.Corollary 38.
Let(
X
,
⪯
) be a partially ordered set and suppose that there exists a generalized partial b-metric G
p
b on X such that (
X
,
G
p
b
) is a G
p
b-complete G
p
b-metric space, and let f
:
X
→
X be an increasing mapping with x
0
⪯
f
(
x
0
) for some x
0
∈
X. Suppose that
(103)
s
G
p
b
(
f
x
,
f
y
,
f
z
)
≤
k
M
(
x
,
y
,
z
)
,
where 0
≤
k
<
1
/
s and
(104)
M
(
x
,
y
,
z
)
=
max
⁡
{
G
p
b
(
x
,
y
,
z
)
,
G
p
b
(
x
,
x
,
f
x
)
G
p
b
(
y
,
y
,
f
y
)
G
p
b
(
z
,
z
,
f
z
)
1
+
[
s
G
p
b
(
f
x
,
f
y
,
f
z
)
]
2
h
h
h
h
h
h
h
h
h
G
p
b
(
x
,
x
,
f
x
)
G
p
b
(
y
,
y
,
f
y
)
G
p
b
(
z
,
z
,
f
z
)
1
+
[
s
G
p
b
(
f
x
,
f
y
,
f
z
)
]
2
}
for all comparable elements x
,
y
,
z
∈
X. If f is continuous or, for any nondecreasing sequence {
x
n
} in X such that x
n
→
u
∈
X, we have x
n
⪯
u for all n
∈
ℕ, then f has a fixed point.We conclude this section by presenting some examples that illustrate our results.Example 39.
LetX
=
[
0,1
] be equipped with the usual order and G
p
b-metric function G
p
b given by G
p
b
(
x
,
y
,
z
)
=
[
max
⁡
{
x
,
y
,
z
}
]
2
=
max
⁡
{
x
2
,
y
2
,
z
2
} with s
=
2. Consider the mapping f
:
X
→
X defined by f
(
x
)
=
(
1
/
4
)
x
(
e
-
x
2
)
1
/
2 and the function β
∈
ℱ given by β
(
t
)
=
(
1
/
2
)
e
-
t
,
t
>
0, and β
(
0
)
∈
[
0,1
/
2
). It is easy to see that f is an increasing function on X and 0
≤
f
(
0
)
=
0. We show that f is G
p
b-continuous on X. By Proposition 27 it is sufficient to show that f is G
p
bsequentially continuous on X. Let {
x
n
} be a sequence in X such that lim
⁡
n
→
∞
G
p
b
(
x
n
,
x
,
x
)
=
G
p
b
(
x
,
x
,
x
), so we have lim
⁡
n
→
∞
max
⁡
{
x
n
2
,
x
2
}
=
x
2, equally max
⁡
{
lim
⁡
n
→
∞
x
n
2
,
x
2
}
=
x
2, and hence lim
⁡
n
→
∞
x
n
2
=
α
≤
x
2. On the other hand we have
(105)
lim
⁡
n
→
∞
G
p
b
(
f
x
n
,
f
x
,
f
x
)
=
lim
⁡
n
→
∞
max
⁡
{
(
f
x
n
)
2
,
(
f
x
)
2
}
=
lim
⁡
n
→
∞
max
⁡
{
1
16
x
n
2
e
-
x
n
2
,
1
16
x
2
e
-
x
2
}
=
max
⁡
{
1
16
lim
⁡
n
→
∞
x
n
2
e
-
x
n
2
,
1
16
x
2
e
-
x
2
}
=
max
⁡
{
1
16
α
e
-
α
,
1
16
x
2
e
-
x
2
}
=
1
16
x
2
e
-
x
2
=
max
⁡
{
1
16
x
2
e
-
x
2
,
1
16
x
2
e
-
x
2
,
1
16
x
2
e
-
x
2
}
=
G
p
b
(
f
x
,
f
x
,
f
x
)
.
So f is G
p
bsequentially continuous on X.
For all comparable elementsx
,
y
,
z
∈
X and the fact that g
(
x
)
=
x
2
e
-
x
2 is an increasing function on X we have
(106)
s
G
p
b
(
f
x
,
f
y
,
f
z
)
=
2
max
⁡
{
1
16
x
2
e
-
x
2
,
1
16
y
2
e
-
y
2
,
1
16
z
2
e
-
z
2
}
=
1
8
max
⁡
{
x
2
e
-
x
2
,
y
2
e
-
y
2
,
z
2
e
-
z
2
}
=
1
8
e
-
max
⁡
{
x
2
,
y
2
,
z
2
}
max
⁡
{
x
2
,
y
2
,
z
2
}
≤
1
2
e
-
max
⁡
{
x
2
,
y
2
,
z
2
}
max
⁡
{
x
2
,
y
2
,
z
2
}
=
β
(
G
p
b
(
x
,
y
,
z
)
)
G
p
b
(
x
,
y
,
z
)
≤
β
(
G
p
b
(
x
,
y
,
z
)
)
M
(
x
,
y
,
z
)
.
Hence, f satisfies all the assumptions of Theorem 32 and thus it has a fixed point (which is u
=
0
).Example 40.
LetX
=
[
0,1
] be equipped with the usual order and G
p
b-metric function G
p
b given by G
p
b
(
x
,
y
,
z
)
=
[
max
⁡
{
x
,
y
,
z
}
]
2
=
max
⁡
{
x
2
,
y
2
,
z
2
} with s
=
2. Consider the mapping f
:
X
→
X defined by f
(
x
)
=
(
1
/
4
)
ln
⁡
(
x
2
+
1
) and the function ψ
∈
Ψ given by ψ
(
t
)
=
(
1
/
8
)
t
,
t
≥
0. It is easy to see that f is increasing function and 0
≤
f
(
0
)
=
0. Now we show that f is G
p
b-continuous function on X.
Let{
x
n
} be a sequence in X such that lim
⁡
n
→
∞
G
p
b
(
x
n
,
x
,
x
)
=
G
p
b
(
x
,
x
,
x
), so we have lim
⁡
n
→
∞
max
⁡
{
x
n
2
,
x
2
}
=
x
2, equally max
⁡
{
lim
⁡
n
→
∞
x
n
2
,
x
2
}
=
x
2, and hence lim
⁡
n
→
∞
x
n
2
=
α
≤
x
2. On the other hand we have
(107)
lim
⁡
n
→
∞
G
p
b
(
f
x
n
,
f
x
,
f
x
)
=
lim
⁡
n
→
∞
max
⁡
{
(
f
x
n
)
2
,
(
f
x
)
2
}
=
lim
⁡
n
→
∞
max
⁡
{
1
16
ln
⁡
(
x
n
2
+
1
)
,
1
16
ln
⁡
(
x
2
+
1
)
}
=
max
⁡
{
1
16
ln
⁡
(
lim
⁡
n
→
∞
x
n
2
+
1
)
,
1
16
ln
⁡
(
x
2
+
1
)
}
=
max
⁡
{
1
16
ln
⁡
⁡
(
α
+
1
)
,
1
16
ln
⁡
⁡
(
x
2
+
1
)
}
=
1
16
ln
⁡
(
x
2
+
1
)
=
max
⁡
{
1
16
ln
⁡
(
x
2
+
1
)
,
1
16
ln
⁡
(
x
2
+
1
)
,
1
16
ln
⁡
(
x
2
+
1
)
}
=
G
p
b
(
f
x
,
f
x
,
f
x
)
.
So f is G
p
b-sequentially continuous on X.
For all comparable elementsx
,
y
,
z
∈
X, we have
(108)
s
G
p
b
(
f
x
,
f
y
,
f
z
)
=
2
max
⁡
{
(
1
4
ln
⁡
(
x
2
+
1
)
)
2
,
(
1
4
ln
⁡
(
y
2
+
1
)
)
2
,
h
h
h
h
h
h
h
l
h
h
h
(
1
4
ln
⁡
(
z
2
+
1
)
)
2
}
=
2
max
⁡
{
1
16
ln
⁡
(
x
2
+
1
)
,
1
16
ln
⁡
(
y
2
+
1
)
,
1
16
ln
⁡
(
z
2
+
1
)
}
=
1
8
max
⁡
{
ln
⁡
⁡
(
x
2
+
1
)
,
ln
⁡
⁡
(
y
2
+
1
)
,
ln
⁡
⁡
(
z
2
+
1
)
}
≤
1
8
max
⁡
{
x
2
,
y
2
,
z
2
}
=
ψ
(
G
p
b
(
x
,
y
,
z
)
)
≤
ψ
(
M
(
x
,
y
,
z
)
)
.
Hence, f satisfies all the assumptions of Theorem 36 and thus it has a fixed point (which is u
=
0
).



---

*Source: 180698-2014-03-05.xml*