# Fixed Points of GeneralizedF-Suzuki Type Contraction in Complete b-Metric Spaces

Authors: Hamed Hamdan Alsulami; Erdal Karapınar; Hossein Piri
Journal: Discrete Dynamics in Nature and Society
 (2015)

---

## Abstract

We introduce the notion of generalizedF-Suzuki type contraction in b-metric spaces and investigate the existence of fixed points of such mappings. The presented results generalize and improve several results of the topics in the literature.


---

## Body

## 1. Introduction and Preliminaries

The concept ofb-metric was introduced by Czerwik [1] as a generalization of metric (see also Bakhtin [2, 3]) to extend celebrated Banach contraction mapping principle. Following this initial paper of Czerwik [1], a number of researchers in nonlinear analysis investigate the topology of the paper and proved several fixed point theorems in the context of complete b-metric spaces (see, e.g., [4–8] and the related references therein).Definition 1 (see [1]).
LetX be a nonempty set and let s
≥
1 be a given real number. A mapping d
:
X
×
X
→
[
0
,
∞
) is said to be a b-metric if for all x
,
y
,
z
∈
X the following conditions are satisfied:(
b
M
1
)
d
(
x
,
y
)
=
0 if and only if x
=
y;
(
b
M
2
)
d
(
x
,
y
)
=
d
(
y
,
x
);
(
b
M
3
)
d
(
x
,
z
)
≤
s
[
d
(
x
,
y
)
+
d
(
y
,
z
)
].In this case, the pair (
X
,
d
) is called a b-metric space (with constant s).Remark 2.
It is clear that the notion ofb-metric is a real generalization of usual metric since a b-metric space is a metric space when s
=
1. For more details and examples on b-metric, see, for example, [1, 3, 9, 10].Example 3.
LetX
=
{
0,1
,
2,3
} and let a mapping d
:
X
×
X
→
[
0
,
∞
) be defined as follows:
(1)
d
x
,
y
=
0
,
if
x
=
y
,
x
-
y
-
1
,
if
x
≠
y
.
Then, (
X
,
d
) is a b-metric space with coefficient s
=
7
/
12. But it is not a metric space since the triangle inequality is not satisfied. Indeed,
(2)
d
0,1
>
d
0,3
+
d
3,1
,
d
0,1
>
d
0,4
+
d
4,1
,
d
3,4
>
d
3,2
+
d
2,4
.Definition 4 (see [11]).
Let(
X
,
d
) be a b-metric space. Then, a sequence {
x
n
}
n
=
1
∞ in X is called(A)
convergent if and only if there existsx
∈
X such that d
(
x
n
,
x
)
→
0 as n
→
∞ and in this case we write li
m
n
→
x
n
=
x;
(B)
Cauchy if and only ifd
(
x
n
,
x
m
)
→
0 as m
,
n
→
∞.Remark 5 (see [11]).
Notice that in ab-metric space (
X
,
d
) the following assertions hold:(A)
a convergent sequence has a unique limit;
(B)
each convergent sequence is Cauchy;
(C)
in general, ab-metric is not continuous;
(D)
in general, ab-metric does not induce a topology on X.Definition 6 (see [11]).
Theb-metric space (
X
,
d
) is complete if every Cauchy sequence in X converges in X.Definition 7.
Let(
X
,
d
X
) and (
Y
,
d
Y
) be b-metric spaces; a mapping f
:
X
→
Y is called (A)
continuous at a pointx
∈
X, if for every sequence {
x
n
}
n
=
1
∞ in X such that x
n
→
x, then f
(
x
n
)
→
f
(
x
);
(B)
continuous onX, if it is continuous at each point x
∈
X.


## 2. Main Result

In this section we state and prove our main results. Throughout the paper, we assume thatb-metric is continuous.Inspired by the notion ofF-contraction, defined by Wardowski [12], we introduce the notion of generalized F-Suzuki type contraction as follows. N, R, R
+ denote the set of natural numbers, set of real numbers, and the set of non-negative real numbers.Definition 8.
Let(
X
,
d
) be a b-metric space with constant s
≥
1. A mapping T
:
X
→
X is called a generalized F-Suzuki type contraction if there exists τ
>
0 such that for all x
,
y
∈
X with x
≠
y
(3)
1
2
s
d
x
,
T
x
<
d
x
,
y
⟹
τ
+
F
d
T
x
,
T
y
≤
α
F
d
x
,
y
+
β
F
d
x
,
T
x
+
γ
F
d
y
,
T
y
,
where γ
∈
[
0,1
) and α
,
β
∈
[
0,1
] are real numbers with α
+
β
+
γ
=
1 and F
:
R
+
→
R is a mapping satisfying the following conditions:(
F
1
)
F is strictly increasing; that is, for all α
,
β
∈
R
+ such that α
<
β,  F
(
α
)
<
F
(
β
);
(
F
2
)
for every sequence{
α
n
}
n
=
1
∞ of positive real numbers lim
n
→
∞
α
n
=
0 if and only if lim
n
→
∞
F
(
α
n
)
=
-
∞.The following is the main result of this paper.Theorem 9.
Let(
X
,
d
) be a complete b-metric space with constant s
≥
1. If T is a generalized F-Suzuki type contraction, then T has a fixed point v
∈
X.Proof.
Fixx
∈
X. We establish an iterative sequence {
x
n
} in the following way:
(4)
x
1
=
T
x
,
x
2
=
T
x
1
=
T
2
x
,
…
,
x
n
=
T
x
n
-
1
=
T
n
x
,
∀
n
∈
N
.
Throughout the proof, we assume that
(5)
d
x
n
,
T
x
n
>
0
,
for
every
n
∈
N
.
Indeed, if there exists n
0
∈
N such that d
(
x
n
0
,
T
x
n
0
)
=
0, then the proof is completed trivially.
Due to assumption (5), we have
(6)
1
2
s
d
x
n
,
T
x
n
<
d
x
n
,
T
x
n
,
∀
n
∈
N
.
Thus, by hypotheses of theorem, we have
(7)
τ
+
F
d
T
x
n
,
T
2
x
n
≤
α
F
d
x
n
,
T
x
n
+
β
F
d
x
n
,
T
x
n
+
γ
F
d
T
x
n
,
T
2
x
n
,
which is equivalent to
(8)
τ
+
1
-
γ
F
d
T
x
n
,
T
2
x
n
≤
α
+
β
F
d
x
n
,
T
x
n
.
Since α
+
β
+
γ
=
1, inequality (8) turns into
(9)
F
d
T
x
n
,
T
2
x
n
≤
F
d
x
n
,
T
x
n
-
τ
α
+
β
<
F
d
x
n
,
T
x
n
.
From (
F
1
), we conclude that
(10)
d
x
n
+
1
,
T
x
n
+
1
=
d
T
x
n
,
T
2
x
n
<
d
x
n
,
T
x
n
,
∀
n
∈
N
.
Therefore, {
d
(
x
n
,
T
x
n
)
}
n
=
1
∞ is a decreasing sequence of real numbers which is bounded from below. Therefore, {
d
(
x
n
,
T
x
n
)
}
n
=
1
∞ converges and
(11)
lim
n
→
∞
d
x
n
,
T
x
n
=
β
=
inf
⁡
d
x
n
,
T
x
n
:
n
∈
N
.
We will show that β
=
0. Suppose, on the contrary, that β
>
0. In other words, for every ϵ
>
0 there exists m
∈
N, such that
(12)
d
x
m
,
T
x
m
<
β
+
ϵ
.
From (
F
1
), we find that
(13)
F
d
x
m
,
T
x
m
<
F
β
+
ϵ
.
On the other hand, we have
(14)
1
2
s
d
x
m
,
T
x
m
<
d
x
m
,
T
x
m
,
due to (6). Since T is generalized F-Suzuki type contraction, we derive
(15)
τ
+
F
d
T
x
m
,
T
2
x
m
≤
α
F
d
x
m
,
T
x
m
+
β
F
d
x
m
,
T
x
m
+
γ
F
d
T
x
m
,
T
2
x
m
,
which yields that
(16)
τ
+
1
-
γ
F
d
T
x
m
,
T
2
x
m
≤
α
+
β
F
d
x
m
,
T
x
m
.
Taking the fact that α
+
β
+
γ
=
1 into account, we find
(17)
F
d
T
x
m
,
T
2
x
m
≤
F
d
x
m
,
T
x
m
-
τ
α
+
β
.
Analogously, again by (6), we have (
1
/
2
s
)
d
(
T
x
m
,
T
2
x
m
)
<
d
(
T
x
m
,
T
2
x
m
). Owing to the fact that T is generalized F-Suzuki type contraction, we conclude that
(18)
τ
+
F
d
T
2
x
m
,
T
3
x
m
≤
α
F
d
T
x
m
,
T
2
x
m
+
β
F
d
T
x
m
,
T
2
x
m
+
γ
F
d
T
2
x
m
,
T
3
x
m
.
It implies that
(19)
F
d
T
2
x
m
,
T
3
x
m
≤
F
d
T
x
m
,
T
2
x
m
-
τ
α
+
β
,
since α
+
β
+
γ
=
1. Furthermore, by combining (17) and (20), we get
(20)
F
d
T
2
x
m
,
T
3
x
m
≤
F
(
d
(
T
x
m
,
T
2
x
m
)
)
-
τ
α
+
β
≤
F
(
d
(
x
m
,
T
x
m
)
)
-
2
τ
α
+
β
.
Iteratively, we obtain that(21)
F
d
T
n
x
m
,
T
n
+
1
x
m
≤
F
d
T
n
x
m
,
T
n
-
1
x
m
-
τ
α
+
β
≤
F
d
T
n
-
1
x
m
,
T
n
-
2
x
m
-
2
τ
α
+
β
⋮
≤
F
d
T
x
m
,
x
m
-
n
τ
α
+
β
<
F
β
+
ϵ
-
n
τ
α
+
β
.
By lettingn
→
∞, we find that
(22)
lim
n
→
∞
F
d
T
n
x
m
,
T
n
+
1
x
m
=
-
∞
.
Consequently, from (F
2), we derive that lim
n
→
∞
d
(
T
n
x
m
,
T
n
+
1
x
m
)
=
0. Thus, there exists N
1
∈
N such that
(23)
d
T
n
x
m
,
T
n
+
1
x
m
<
β
,
∀
n
≥
N
1
,
and from (4) we get
(24)
d
x
m
+
n
,
T
x
m
+
n
<
β
,
∀
n
≥
N
1
.
This is a contradiction with the definition of β. Hence, we have
(25)
lim
n
→
∞
d
(
x
n
,
T
x
n
)
=
0
.
In what follows, we will prove that(26)
lim
n
,
m
→
∞
d
(
x
n
,
x
m
)
=
0
.
Suppose, on the contrary, that there exist ϵ
>
0 and sequences {
p
(
n
)
}
n
=
1
∞ and {
q
(
n
)
}
n
=
1
∞ of natural numbers such that
(27)
p
(
n
)
>
q
(
n
)
>
n
,
d
x
p
n
,
x
q
n
≥
ϵ
,
d
x
p
n
-
1
,
x
q
n
<
ϵ
,
∀
n
∈
N
.
From the triangle inequality, we have
(28)
d
x
p
n
,
x
q
n
≤
s
d
x
p
n
,
x
p
n
-
1
+
d
x
p
n
-
1
,
x
q
n
≤
s
d
x
p
n
,
x
p
n
-
1
+
s
ϵ
=
s
d
x
p
n
-
1
,
T
x
p
n
-
1
+
s
ϵ
,
∀
n
∈
N
.
Owing to (25), there exists N
2
∈
N such that
(29)
d
x
p
n
,
T
x
p
n
<
ϵ
,
∀
n
>
N
2
.
Taking (29) into account, (28) yields that
(30)
d
x
p
n
,
x
q
n
<
2
s
ϵ
,
∀
n
>
N
2
.
So from (
F
2
), we obtain
(31)
F
d
x
p
n
,
x
q
n
<
F
2
s
ϵ
,
∀
n
>
N
2
.
On the other hand, we can easily get that(32)
1
2
s
d
x
p
n
,
T
x
p
n
<
ϵ
2
s
<
ϵ
≤
d
x
p
n
,
x
q
n
,
∀
n
>
N
2
,
from (27) and (29). Since T is generalized F-Suzuki type contraction, for all n
>
N
2 we find that
(33)
τ
+
F
d
T
x
p
n
,
T
x
q
n
≤
α
F
d
x
p
n
,
x
q
n
+
β
F
d
x
p
n
,
T
x
p
n
+
γ
F
d
x
q
n
,
T
x
q
n
.
Taking (31) into account, (33) turns into
(34)
τ
+
F
d
T
x
p
n
,
T
x
q
n
<
α
F
2
s
ϵ
+
β
F
d
x
p
n
,
T
x
p
n
+
γ
F
d
x
q
n
,
T
x
q
n
,
∀
n
>
N
2
.
Regarding (25) and (
F
2
), we obtain that
(35)
lim
n
→
∞
F
d
T
x
p
n
,
T
x
q
n
=
-
∞
.
From (
F
2
), we get that
(36)
lim
n
→
∞
d
T
x
p
n
,
T
x
q
n
=
0
⟺
lim
n
→
∞
d
x
p
n
+
1
,
x
q
n
+
1
=
0
.
This is a contradiction with relations in (27). Hence, lim
m
,
n
→
∞
d
(
x
n
,
x
m
)
=
0; that is, {
x
n
}
n
=
1
∞ is a Cauchy sequence in X. On account of the completeness of (
X
,
d
), there exists v
∈
X such that
(37)
lim
n
→
∞
d
(
x
n
,
v
)
=
0
.
We claim that, for every n
∈
N,
(38)
1
2
s
d
x
n
,
T
x
n
<
d
x
n
,
v
or
1
2
s
d
T
x
n
,
T
2
x
n
<
d
T
x
n
,
v
,
∀
n
∈
N
.
We will prove the claim above by the method ofreductio ad absurdum. Suppose, on the contrary, that there exists m
∈
N such that
(39)
1
2
s
d
x
m
,
T
x
m
≥
d
x
m
,
v
,
1
2
s
d
T
x
m
,
T
2
x
m
≥
d
T
x
m
,
v
.
From (10) and (
F
1
), we have
(40)
d
T
x
m
,
T
2
x
m
<
d
x
m
,
T
x
m
.
It follows from (39) and (40) that
(41)
d
x
m
,
T
x
m
≤
s
d
x
m
,
v
+
s
d
v
,
T
x
m
≤
1
2
d
x
m
,
T
x
m
+
1
2
d
T
x
m
,
T
2
x
m
<
1
2
d
x
m
,
T
x
m
+
1
2
d
x
m
,
T
x
m
=
d
x
m
,
T
x
m
.
This is a contradiction. Hence, (38) holds. Since T is generalized F-Suzuki type contraction, (38) yields that, for every n
∈
N, either
(42)
τ
+
F
d
T
x
n
,
T
v
≤
α
F
d
x
n
,
v
+
β
F
d
x
n
,
T
x
n
+
γ
F
d
v
,
T
v
or
(43)
τ
+
F
d
T
2
x
n
,
T
v
≤
α
F
d
T
x
n
,
v
+
β
F
d
T
x
n
,
T
2
x
n
+
γ
F
d
v
,
T
v
holds. On account of (
F
2
), the limits in (25) and (37) imply that
(44)
lim
n
→
∞
F
σ
x
n
,
v
=
-
∞
,
lim
n
→
∞
F
σ
x
n
,
T
x
n
=
-
∞
.
Thus, letting n
→
∞ in (42), we conclude that
(45)
lim
n
→
∞
F
d
T
x
n
,
T
v
=
-
∞
.
Again by using (
F
2
), we observe that
(46)
lim
n
→
∞
d
T
x
n
,
T
v
=
0
.
By regarding the triangle inequality with (4), we derive that
(47)
d
v
,
T
v
≤
s
d
v
,
T
x
n
+
d
T
x
n
,
T
v
=
s
d
v
,
x
n
+
1
+
s
d
T
x
n
,
T
v
.
By letting n
→
∞ in the inequality above together with the limits in (37) and (46), we conclude that d
(
v
,
T
v
)
=
0. Thus, v is a fixed point of T; that is, v
=
T
v.
Let us analyze the second case (42). Regarding (4), we have
(48)
F
d
T
2
x
n
,
T
v
<
τ
+
F
d
T
2
x
n
,
T
v
≤
α
F
d
T
x
n
,
v
+
β
F
d
T
x
n
,
T
2
x
n
+
γ
F
(
d
(
v
,
T
v
)
)
=
α
F
d
x
n
+
1
,
v
+
β
F
d
x
n
+
1
,
T
x
n
+
1
+
γ
F
d
v
,
T
v
.
As it was discussed above, from (25), (37), and (
F
2
), we conclude that
(49)
lim
n
→
∞
F
d
T
2
x
n
,
T
v
=
-
∞
.
From (
F
2
) equivalently, we get
(50)
lim
n
→
∞
d
T
2
x
n
,
T
v
=
0
.
Again by the triangle inequality together with (4), we find that
(51)
d
v
,
T
v
≤
s
d
v
,
T
2
x
n
+
d
T
2
x
n
,
T
v
=
s
d
v
,
x
n
+
2
+
d
T
2
x
n
,
T
v
.
By letting n
→
∞ in the inequality above together with the limits in (37) and (50), we obtain d
(
v
,
T
v
)
=
0. Thus, v is a fixed point of T and that completes the proof.Definition 10.
Let(
X
,
d
) be a b-metric space with constant s
≥
1. A mapping T
:
X
→
X is said to be a F-Suzuki type contraction if there exists τ
>
0 such that for all x
,
y
∈
X with x
≠
y
(52)
1
2
s
d
x
,
T
x

<

d
x
,
y
⟹
τ
+
F
d
T
x
,
T
y

≤

F
d
x
,
y
,
where F
:
R
+
→
R is a mapping satisfying the following conditions.(F
1)
F is strictly increasing; that is, for all α
,
β
∈
R
+ such that α
<
β,  F
(
α
)
<
F
(
β
).
(F
2)
For every sequence{
α
n
}
n
=
1
∞ of positive real numbers lim
n
→
∞
α
n
=
0 if and only if lim
n
→
∞
F
(
α
n
)
=
-
∞.Theorem 11.
Let(
X
,
d
) be complete b-metric space with constant s
≥
1 and let T
:
X
→
X be a F-Suzuki type contraction mapping. Then, T has a fixed point.Proof.
By takingα
=
1 and β
=
γ
=
0 in Theorem 9, the proof is complete.Corollary 12.
LetT be a self-mapping on a complete metric space (
X
,
d
). Let γ
∈
[
0,1
) and α
,
β
∈
[
0,1
] be real numbers such that α
+
β
+
γ
=
1. Assume that there exists τ
>
0 such that, for all x
,
y
∈
X with x
≠
y,
(53)
1
2
d
x
,
T
x
<
d
x
,
y
⟹
τ
+
F
d
T
x
,
T
y
≤
α
F
d
x
,
y
+
β
F
d
x
,
T
x
+
γ
F
d
y
,
T
y
,
where F
:
R
+
→
R is satisfied in conditions (
F
1
) and (
F
2
). Then, T has a fixed point v
∈
X; that is, T
v
=
v.Proof.
Since any metric space is ab-metric space with constant s
=
1, so from Theorem 9 the proof is complete.Corollary 13.
LetT be a self-mapping on a complete metric space (
X
,
d
). Assume that there exists τ
>
0 such that, for all x
,
y
∈
X with x
≠
y,
(54)
1
2
d
x
,
T
x
<
d
x
,
y
⟹
τ
+
F
d
T
x
,
T
y
≤
F
d
x
,
y
,
where F
:
R
+
→
R is satisfied in conditions (
F
1
) and (
F
2
). Then, T has a fixed point v
∈
X; that is, T
v
=
v.Proof.
Since any metric space is ab-metric space with constant s
=
1, so by taking α
=
1 and β
=
γ
=
0 in Theorem 9 the proof is complete.Theorem 14.
Let(
X
,
d
) be complete b-metric space with constant s
≥
1 and let T be a continuous self-mapping on X. Let γ
∈
[
0,1
) and α
,
β
∈
[
0,1
] be real numbers such that α
+
β
+
γ
=
1. Assume that there exists τ
>
0 such that for all x
,
y
∈
X
(55)
d
T
x
,
T
y
>
0
⟹
τ
+
F
d
T
x
,
T
y
<
α
F
d
x
,
y
+
β
F
d
x
,
T
x
+
γ
F
d
y
,
T
y
,
where F
:
R
+
→
R is satisfied in conditions (
F
1
) and (
F
2
). Then, T has a fixed point v
∈
X; that is, T
v
=
v.Proof.
Choosex
∈
X. Set
(56)
x
1
=
T
x
,
x
2
=
T
x
1
=
T
2
x
,
…
,
x
n
=
T
x
n
-
1
=
T
n
x
0
,
∀
n
∈
N
.
If there exists n
∈
N such that d
(
x
n
,
T
x
n
)
=
0, the proof is complete. So we assume that
(57)
0
<
d
x
n
,
T
x
n
=
d
T
x
n
-
1
,
T
x
n
,
∀
n
∈
N
.
So from the assumption of theorem, we have
(58)
τ
+
F
d
T
x
n
-
1
,
T
x
n
≤
α
F
d
x
n
-
1
,
x
n
+
β
F
d
x
n
-
1
,
T
x
n
-
1
+
γ
F
d
x
n
,
T
x
n
=
α
F
d
x
n
-
1
,
x
n
+
β
F
d
x
n
-
1
,
x
n
+
γ
F
d
T
x
n
-
1
,
T
x
n
and hence
(59)
τ
+
1
-
γ
F
d
T
x
n
-
1
,
T
x
n
≤
α
+
β
F
d
x
n
-
1
,
x
n
.
Since α
+
β
+
γ
=
1, we get
(60)
F
d
T
x
n
-
1
,
T
x
n
≤
F
d
x
n
-
1
,
x
n
-
τ
α
+
β
<
F
d
x
n
-
1
,
x
n
.
So from(
F
1
), we conclude that
(61)
d
x
n
,
T
x
n

=

d
T
x
n
-
1
,
T
x
n

<

d
x
n
-
1
,
x
n

=

d
x
n
-
1
,
T
x
n
-
1
,
∀
n
∈
N
.
Therefore, {
d
(
x
n
,
T
x
n
)
}
n
=
1
∞ is a decreasing sequence of real numbers which is bounded from below. Therefore, {
d
(
x
n
,
T
x
n
)
}
n
=
1
∞ converges and
(62)
lim
n
→
∞
d
x
n
,
T
x
n
=
β
=
inf
⁡
d
x
n
,
T
x
n
:
n
∈
N
.
We will show that β
=
0. Arguing by contradiction, we assume β
>
0. For every ϵ
>
0, there exists m
∈
N, such that
(63)
d
(
x
m
,
T
x
m
)
<
β
+
ϵ
.
Hence, from (
F
1
), we get
(64)
F
d
x
m
,
T
x
m
<
F
β
+
ϵ
.
On the other hand (57), we have
(65)
0
<
d
x
m
,
T
x
m
=
d
T
x
m
-
1
,
T
x
m
.
So from assumption of theorem, we obtain
(66)
τ
+
F
d
T
x
m
-
1
,
T
x
m
≤
α
F
d
x
m
-
1
,
x
m
+
β
F
d
x
m
-
1
,
T
x
m
-
1
+
γ
F
d
x
m
,
T
x
m
=
α
F
d
x
m
-
1
,
x
m
+
β
F
d
x
m
-
1
,
x
m
+
γ
F
d
T
x
m
-
1
,
T
x
m
,
and hence
(67)
τ
+
1
-
γ
F
d
T
x
m
-
1
,
T
x
m
<
α
+
β
F
d
x
m
-
1
,
x
m
.
Sinceα
+
β
+
γ
=
1, we get
(68)
F
d
T
x
m
-
1
,
T
x
m
≤
F
d
x
m
-
1
,
x
m
-
τ
α
+
β
.
Also from (57), we have 0
<
d
(
T
x
m
,
T
x
m
+
1
), and thus, by assumption of theorem, we have
(69)
τ
+
F
d
T
x
m
,
T
x
m
+
1
≤
α
F
d
x
m
,
x
m
+
1
+
β
F
d
x
m
,
T
x
m
+
γ
F
d
x
m
+
1
,
T
x
m
+
1
=
α
F
d
x
m
,
x
m
+
1
+
β
F
d
x
m
,
x
m
+
1
+
γ
F
d
T
x
m
,
T
x
m
+
1
,
and therefore,
(70)
τ
+
1
-
γ
F
d
T
x
m
,
T
x
m
+
1
≤
α
+
β
F
d
x
m
,
x
m
+
1
.
Since α
+
β
+
γ
=
1, we get
(71)
F
d
T
x
m
,
T
x
m
+
1
≤
F
d
x
m
,
x
m
+
1
-
τ
α
+
β
.
Now by using (64) and continuing similar method as used in (68) and (71), we obtain
(72)
F
d
x
m
+
n
,
x
m
+
n
+
1
=
F
d
T
x
m
+
n
-
1
,
T
x
m
+
n
≤
F
d
x
m
+
n
-
1
,
x
m
+
n
-
τ
α
+
β
=
F
d
T
x
m
+
n
-
2
,
T
x
m
+
n
-
1
-
τ
α
+
β
≤
F
d
x
m
+
n
-
2
,
x
m
+
n
-
1
-
2
τ
α
+
β
=
F
d
T
x
m
+
n
-
3
,
T
x
m
+
n
-
2
-
2
τ
α
+
β
≤
F
d
x
m
+
n
-
3
,
x
m
+
n
-
2
-
3
τ
α
+
β
⋮
≤
F
d
x
m
+
1
,
x
m
+
2
-
n
-
1
τ
α
+
β
=
F
d
T
x
m
,
T
x
m
+
1
-
n
-
1
τ
α
+
β
≤
F
β
+
ϵ
-
n
τ
α
+
β
.
This implies that(73)
lim
n
→
∞
F
d
x
m
+
n
,
x
m
+
n
+
1
=
-
∞
.
So from (F
2), we have lim
n
→
∞
d
(
x
m
+
n
,
x
m
+
n
+
1
)
=
0, so that there exists N
1
∈
N such that
(74)
d
x
m
+
n
,
x
m
+
n
+
1
<
β
,
∀
n
≥
N
1
,
and so from (56) we get
(75)
d
x
m
+
n
,
T
x
m
+
n
<
β
,
∀
n
≥
N
1
.
This is a contradiction with definition ofβ. So, β
=
0 and from (62) we have
(76)
lim
n
→
∞
d
x
n
,
T
x
n
=
0
.
Now, we claim that(77)
lim
n
,
m
→
∞
d
x
n
,
x
m
=
0
.
Arguing by contradiction, we assume that there exist ϵ
>
0 and sequences {
p
(
n
)
}
n
=
1
∞ and {
q
(
n
)
}
n
=
1
∞ of natural numbers such that
(78)
p
n
>
q
n
>
n
,
d
x
p
n
,
x
q
n
≥
ϵ
,
d
x
p
n
-
1
,
x
q
n
<
ϵ
,
∀
n
∈
N
.
So
(79)
d
x
p
n
,
x
q
n
≤
s
d
x
p
n
,
x
p
n
-
1
+
d
x
p
n
-
1
,
x
q
n
≤
s
d
x
p
n
,
x
p
n
-
1
+
s
ϵ
=
s
d
x
p
n
-
1
,
T
x
p
n
-
1
+
s
ϵ
,
∀
n
∈
N
.
From (76), there exists N
2
∈
N such that
(80)
d
(
x
p
(
n
)
,
T
x
p
(
n
)
)
<
ϵ
,
∀
n
>
N
2
.
It follows from (79) and (80) that
(81)
d
(
x
p
(
n
)
,
x
q
(
n
)
)
<
2
s
ϵ
,
∀
n
>
N
2
.
So, from (
F
2
), we obtain
(82)
F
d
x
p
n
,
x
q
n
<
F
2
s
ϵ
,
∀
n
>
N
2
.
On the other hand from (78), we have
(83)
ϵ
≤
d
x
p
n
+
1
,
x
q
n
+
1
=
d
T
x
p
n
,
T
x
q
n
,
∀
n
>
N
2
.
Hence, from (82) and assumption of theorem, we have
(84)
τ
+
F
d
T
x
p
n
,
T
x
q
n
≤
α
F
d
x
p
n
,
x
q
n
+
β
F
d
x
p
n
,
T
x
p
n
+
γ
F
d
x
q
n
,
T
x
q
n
<
α
F
2
s
ϵ
+
β
F
d
x
p
n
,
T
x
p
n
+
γ
F
d
x
q
n
,
T
x
q
n
,
∀
n
>
N
2
.
Using (79) and (
F
2
), we obtain
(85)
lim
n
→
∞
F
d
T
x
p
n
,
T
x
q
n
=
-
∞
.
Thus, from this and(
F
2
), we get
(86)
lim
n
→
∞
d
T
x
p
n
,
T
x
q
n
=
0
⟺
lim
n
→
∞
d
x
p
n
+
1
,
x
q
n
+
1
=
0
.
This is contradiction with relation (78). Hence, lim
m
,
n
→
∞
d
(
x
n
,
x
m
)
=
0. By completeness of (
X
,
d
), there exists v
∈
X such that
(87)
lim
n
→
∞
d
x
n
,
v
=
0
.
Since T is continuous, we get
(88)
lim
n
→
∞
d
(
T
x
n
,
T
v
)
=
0
.
Since d
(
v
,
T
v
)
≤
s
[
d
(
v
,
x
n
)
+
d
(
x
n
,
T
v
)
], therefore d
(
v
,
T
v
)
=
0 and hence v
=
T
v.Theorem 15.
Let(
X
,
d
) be complete b-metric space with constant s
≥
1 and let T be a continuous self-mapping on X. Assume that there exists τ
>
0 such that for all x
,
y
∈
X
(89)
d
T
x
,
T
y
>
0
⟹
τ
+
F
d
T
x
,
T
y
<
F
d
x
,
y
,
where F
:
R
+
→
R is satisfied in conditions (
F
1
) and (
F
2
). Then, T has a fixed point v
∈
X; that is, T
v
=
v.Proof.
By takingα
=
1 and β
=
γ
=
0 in Theorem 14, the proof is complete.Theorem 16.
Let(
X
,
d
) be complete metric space and let T be a continuous self-mapping on X. Let γ
∈
[
0,1
) and α
,
β
∈
[
0,1
] be real numbers such that α
+
β
+
γ
=
1. Assume that there exists τ
>
0 such that for all x
,
y
∈
X
(90)
d
T
x
,
T
y
>
0
⟹
τ
+
F
d
T
x
,
T
y
<
α
F
d
x
,
y
+
β
F
d
x
,
T
x
+
γ
F
d
y
,
T
y
,
where F
:
R
+
→
R is satisfied in conditions (
F
1
) and (
F
2
). Then, T has a fixed point v
∈
X; that is, T
v
=
v.Proof.
It is sufficient to takes
=
1 in Theorem 14.Theorem 17.
Let(
X
,
d
) be complete metric space and let T be a continuous self-mapping on X. Assume that there exists τ
>
0 such that for all x
,
y
∈
X
(91)
d
T
x
,
T
y
>
0
⟹
τ
+
F
d
T
x
,
T
y
<
F
d
x
,
y
,
where F
:
R
+
→
R is satisfied in conditions (
F
1
) and (
F
2
). Then, T has a unique fixed point v
∈
X; that is, T
v
=
v.Proof.
Since every metric space is ab-metric space with constant s
=
1, from Theorem 15  T has a fixed point v
∈
X. Indeed, if there is another fixed point w
∈
X of T, such that v
≠
w, therefore d
(
T
v
,
T
w
)
>
0. Since τ
>
0, from assumption of theorem we obtain
(92)
F
d
v
,
w
=
F
d
T
v
,
T
w
<
τ
+
F
d
T
v
,
T
w
≤
α
F
d
v
,
w
.
This is contradiction. So T has a unique fixed point.Remark 18.
Theorem17 gives all consequence of Theorem 2.1 of [12] without assumption (
F
3
) used by [12]. Notice also that the results in [12] can be also concluded from the main theorem in [13] that is also a proper extension of the results in [12] in different aspect, more precisely, in complete metric-like spaces.



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*Source: 969726-2015-02-22.xml*