Fixed Point Theorems for Expansive Mappings in js-Metric β¦54 Manoj Kumar, Vishnu Narayan Mishra,...
Transcript of Fixed Point Theorems for Expansive Mappings in js-Metric β¦54 Manoj Kumar, Vishnu Narayan Mishra,...
Advances in Dynamical Systems and Applications.
ISSN 0973-5321, Volume 12, Number 1, (2017) pp. 49-63
Β© Research India Publications
http://www.ripublication.com
Fixed Point Theorems for Expansive Mappings
in js-Metric Space
Manoj Kumar1, Vishnu Narayan Mishra2,3*, Asha Rani4,
Asha Rani5, Kumari Jyoti6 1Department of Mathematics, Lovely Professional University, Punjab, India.
2Applied Mathematics and Humanities Dept., S.V. National Institute of Technology, Surat 395 007, Gujarat, India.
3Department of Mathematics, Indira Gandhi National Tribal University, Lalpur, Amarkantak, Anuppur, Madhya Pradesh 484 887, India.
4,5,6Department of Mathematics, SRM University, Haryana, India.
Abstract
In this paper, we introduce Kannan, Chatterjea, Zamfirscu and Rhodes type
expansive mappings in the setting of generalized metric spaces (js-metric
spaces). The results proved in the setting of generalized metric spaces
generalizes the results in metric spaces, dislocated metric spaces, b-metric
spaces, and modular spaces. We also illustrate our results with the help of
certain examples.
Keywords: Fixed point, js-metric space, Kannan expansion, Zamfirscu
expansion, Rhodes expansion.
1. INTRODUCTION
In the last few decades, fixed point theory is being one of the most interesting
research subject in non linear analysis. In 1922, Banach [1] gave a new direction in
research by introducing Banach Contraction Principle. After that Kannan [2],
Corresponding Author
50 Manoj Kumar, Vishnu Narayan Mishra, Asha Rani, Asha Rani & Kumari Jyoti
Chatterjea [3], Zamfirscu [4], Rhodes[5] gave generalisation of this result. In 1984,
Wang [6] introduced the expansive mappings and proved fixed point results for them.
Now a days, the concept of standard metric spaces plays a role of fundamental tool in
fixed point theory and also attract many researchers because of development of fixed
point theory in standard metric spaces. In last few years, several generalization of
standard metric spaces came into existence like b-metric spaces[7], dislocated metric
spaces[8], modular spaces[9]. Recently in 2015, Jleli and Samet [10] introduced a
new generalization of metric spaces (js-metric space).
In this paper, we establish some new results of fixed points by defining Kannan type,
Zamfirscu, Rhodes fixed point theorem in generalized metric spaces (js-metric
spaces) which recovers several topological spaces including dislocated metric spaces,
b-metric spaces, modular spaces.
2. PRELIMINARIES
Definition 2.1[10] Let π be a nonempty and π:π Γ π β [0,β] be a given mapping.
For every π β π, let us define the set
πΆ(π, π, π) = {{ππ} β π: limπββ
π(ππ, π) = 0}.
Definition 2.2 [10] Let π be a non empty set and π:π Γ π β [0,β]be a mapping.
Then (π,π) is a generalized metric on π if it satisfies the following conditions:
(π1) for every (π, π) β π Γ π, we have π(π, π) = 0 βΉ π = π;
(π2) for every (π, π) β π Γ π, we have π(π, π) = π(π, π);
(π3) there exists πΆ > 0 such that if (π, π) β π Γ π, {ππ} β πΆ(π, π, π),
then π(π, π) β€ πΆππππ π’ππββπ(ππ, π).
In this case, the pair (π,π) is a generalized metric space.
Remark 2.3 [10] If the set πΆ(π, π, π) is empty for every π β π, then (π, π) is a
generalized metric space if and only if (π1) and (π2) are satisfied.
Definition 2.4 [10] Let (π,π) be a generalized metric space and {ππ} be a sequence
in π and π β π. We say that {ππ} π-converges to π if π β πΆ(π, π, π).
Fixed Point Theorems for Expansive Mappings in js-Metric Space 51
Proposition 2.5 [10] Let (π,π) be a generalized metric space and {ππ} be a sequence
in π and (π, π) β π Γ π. If {ππ} π-converges to π and {ππ} π-converges to π, then
π = π.
Definition 2.6 [10] Let (π,π)be a generalized metric space and {ππ} be a sequence in
π. We say that {ππ} is a π-cauchy sequence if limπββ
π(ππ, ππ+π) = 0.
Definition 2.7 [10] Let (π,π)be a generalized metric space. It is said to be π -
complete if every Cauchy sequence in π is convergent to some element in π.
Definition 2.8 [10] Let (π,π)be a generalized metric space and π: π β π be a
mapping. Let π β (0,1) then π is said to be π -contraction if π(π(π), π(π)) β€
ππ(π, π), for every (π, π) β π Γ π.
Definition 2.9 [10] Suppose that π is a π-contraction for some π β (0,1). Then any
fixed point π’ β π of π satisfies π(π’, π’) < ββΉ π·(π’, π’) = 0.
Definition 2.10 [10] For every π β π, let πΏ(π, π, π) = π π’π {π (ππ(π), ππ(π)) : π, π β β}.
Definition 2.11 [10] Suppose that the following conditions hold:
(i) (π, π)is complete;
(ii) π is a π-contraction for some π β (0,1);
(iii) there exists π0 β π such that πΏ(π, π, π0) < β.
Then {ππ(π0)} converges to π’ β π, a fixed point of π. Moreover, if π£ β π is another
fixed point of π such that π(π’, π£) < β, then π’ = π£.
Definition 2.12[10] Let π β (0,1), it is said to be π-quasicontraction if
π(π(π), π(π)) β€ π πππ₯{π(π, π), π(π, ππ),π(π, ππ),π(π, ππ),π(π, ππ)},
for every (π, π) β π Γ π.
Theorem 2.13[10] Suppose that the following conditions hold:
(i) (π, π)is complete;
(ii) π is a π-quasicontraction for some π β (0,1);
(iii) there exists π0 β π such that πΏ(π, π, π0) < β.
Then {ππ(π0)} converges to π’ β π, a fixed point of π. Moreover, if π£ β π is another
fixed point of π such that π(π’, π£) < β, then π’ = π£.
52 Manoj Kumar, Vishnu Narayan Mishra, Asha Rani, Asha Rani & Kumari Jyoti
3. MAIN RESULT
Definition 3.1 Let (π, π)be a generalized metric space and π: π β π be a onto
mapping. Let π > 1, then π is said to be π-expansion if
π(π(π₯), π(π¦)) β₯ ππ(π₯, π¦), for every (π₯, π¦) β π Γ π.
Definition 3.2 Let (π, π)be a generalized metric space and π: π β π be a onto
mapping Let π > 1, it is said to be π-quasiexpansion if
π(π(π₯), π(π¦)) β₯
π πππ₯{π(π₯, π¦),π(π₯, ππ₯),π(π¦, ππ¦),π(π₯, ππ¦),π(π¦, ππ₯)}, for every
(π₯, π¦) β π Γ π.
Definition 3.3 Let (π, π)be a generalized metric space and π: π β π be a onto
mapping. Let π β₯1
2 then π is said to be k-Kannan expansion if
π(π(π₯), π(π¦)) β₯ π[π(π₯, ππ₯) + π(π¦, ππ¦)], for every (π₯, π¦) β π Γ π.
Definition 3.4 Let (π, π)be a generalized metric space and π: π β π be a mapping.
Let π β₯1
2 then π is said to be k- Chatterjea expansion if
π(π(π₯), π(π¦)) β₯ π[π(π₯ , ππ¦) + π(π¦, ππ₯)], for every (π₯, π¦) β π Γ π.
Definition 3.5 Let (π, π)be a generalized metric space and π: π β π be a mapping.
Let π β (0,1) then π is said to be k-Zamfirscu expansion if
π(π(π₯), π(π¦)) β₯ πmax {π(π₯, π¦),π(π₯,ππ₯)+π(π¦,ππ¦)
2,π(π₯,ππ¦)+π(π¦,ππ₯)
2}, for every (π₯, π¦) β
π Γ π.
Definition 3.6 Let (π, π)be a generalized metric space and π: π β π be a mapping.
Let π β (0,1) then π is said to be k-Rhodes expansion if
π(π(π₯), π(π¦)) β₯ πmax {π(π₯, π¦),π(π₯,ππ₯)+π(π¦,ππ¦)
2, π(π₯. ππ¦),π(π¦, ππ₯)}, for every
(π₯, π¦) β π Γ π.
Theorem 3.7 Let (π, π) is complete generalized metric space and π: π β π be a
onto mapping which satisfies the following conditions :
(i) π is a π-expansion for some π β (0,1);
Fixed Point Theorems for Expansive Mappings in js-Metric Space 53
(ii) there exists π₯0 β π such that limπββ
πΏ(π,π,π₯0)
ππ< β.
Then {π(π₯0)} converges to π β π, a fixed point of π. Moreover, if πβ² β π is another
fixed point of π such that π(π,πβ²) < β, then π = πβ².
Proof Let π β β (π β₯ 1). Since π is a π-expansion, for all π, π β β,we have
π (ππβ1+π(π₯0), ππβ1+π(π₯0)) β₯ ππ (ππ+π(π₯0), π
π+π(π₯0)),
which implies that
πΏ(π, π, ππ(π₯0)) β€1
ππΏ(π, π, ππβ1(π₯0)).
Then, for every π β β, we have
πΏ(π, π, ππ(π₯0)) β€1
πππΏ(π, π, π₯0).
Using the above inequality, for every π,π β β, we have
π(ππ(π₯0), ππ+π(π₯0)) β€ πΏ(π, π, π
π(π₯0)) β€1
πππΏ(π, π, π₯0).
Since limπββ
πΏ(π,π,π₯0)
ππ< β and π > 1, we get
limπ,πββ
π(ππ(π₯0), ππ+π(π₯0)) = 0,
which implies that {ππ(π₯0)} is a π-cauchy sequence.
But (π,π) is π-complete so there exists some π’ β π such that {ππ(π₯0)} is a π-
convergent to π’.
Since π is onto, so there exist π β π, such that π β πβ1(π’) βΉ π(π) = π’
But by condition (i), π is π-expansion, for all π β β, we have
π(ππ(π₯0), π’) = π(ππ+1(π₯0), π(π)) β₯ ππ(ππ(π₯0),π).
Letting π β β in the above inequality, we get
limπββ
π(ππ(π₯0),π) = 0.
54 Manoj Kumar, Vishnu Narayan Mishra, Asha Rani, Asha Rani & Kumari Jyoti
Then {ππ(π₯0)} is π-convergent to π. By the proposition 2.5 the uniqueness of the
limit we get, π’ = π, Hence, π’ is a fixed point of π
Now, suppose that π£ β π is a fixed point of π such that π(π’, π£) < β. Since π is a π-
expansion, we have
π(π’, π£) = π(π(π’), π(π£)) β₯ ππ(π’, π£),
By property (π1), we get π’ = π£.
Observe that we can replace condition (ii) in Theorem 3.2 by
(H) there exists π₯0 β π such that π π’π{π(π₯0, ππ(π₯0))} < β.
Example 3.8 Let π = [0,1] be a complete generalized metric space with π =
πππ{π₯, π¦} with π =3
2. Define the function π: π β π such that π(π₯) = 2π₯
If π₯, π¦ β [0,1], without lose of generality π₯ < π¦ then
π(ππ₯, ππ¦) = πππ{ππ₯, ππ¦} = πππ{2π₯, 2π¦} = 2πππ{π₯, π¦} =
2π(π₯, π¦).
Clearly, π is an expansive mapping. Now, for every π¦ β π, there exists an π₯ =π¦
2β π,
such that, π¦ = π(π₯). So, π is onto. Clearly for all π,0 β€ ππ(π₯) β€ 2π, which implies
that limπββ
πΏ(π,π,π₯0)
ππ< β. So the condition (ii) of the Theorem 3.7 is also satisfied. So
all the conditions of theorem 3.7 is satisfied with unique fixed point is π₯ = 0.
Corollary 3.9 Let (π,π)be a complete b-metric space and π: π β π be a mapping.
Suppose that for some π > 1, we have
π(π(π₯), π(π¦)) β₯ ππ(π₯, π¦), for every (π₯, π¦) β π Γ π.
If there exists π₯0 β π such that π π’π {π (ππ(π₯0), ππ(π₯0)) : π, π β β} < β.
Then the sequence {ππ(π₯0)} converges to a fixed point of π. Moreover, π has one and
only one fixed point.
Corollary 3.10 Let (π,π)be a complete dislocated metric space and π: π β π be a
mapping. Suppose that for some π > 1, we have
π(π(π₯), π(π¦)) β₯ ππ(π₯, π¦), for every (π₯, π¦) β π Γ π.
Fixed Point Theorems for Expansive Mappings in js-Metric Space 55
If there exists π₯0 β π such that π π’π {π (ππ(π₯0), ππ(π₯0)) : π, π β β} < β.
Then the sequence {ππ(π₯0)} converges to a fixed point of π. Moreover, π has one and
only one fixed point.
Theorem 3.11 Let (π, π) is complete generalized metric space and π: π β π be an
onto mapping which satisfies following conditions:
(i) π is a π-quasi expansion for some π β₯ 1;
(ii) there exists π₯0 β π such that limπββ
πΏ(π,π,π₯0)
ππ< β.
Then {ππ(π₯0)} converges to π’ β π, a fixed point of π. Moreover, if π£ β π is another
fixed point of π such that π(π’, π£) < β, then π’ = π£.
Proof Let π β β (π β₯ 1). Since π is a π-quasi expansion, for all π, π β β,we have
π (ππβ1+π(π₯0), ππβ1+π(π₯0)) β₯ π πππ₯
{
π (ππ+π(π₯0), π
π+π(π₯0)) ,
π (ππ+π(π₯0), ππβ1+π(π₯0)) ,
π (ππβ1+π(π₯0), ππ+π(π₯0)) ,
π (ππ+π(π₯0), ππβ1+π(π₯0)) ,
π (ππ+π(π₯0), ππβ1+π(π₯0)) }
.
which implies that
πΏ(π, π, ππ(π₯0)) β₯ π[πΏ(π, π, ππβ1(π₯0))].
Then, for every π β₯ 1, we have
πΏ(π, π, ππ(π₯0)) β€1
πππΏ(π, π, π₯0).
Using the above inequality, for every π,π β β, we have
π(ππ(π₯0), ππ+π(π₯0)) β€ πΏ(π, π, π
π(π₯0)) β€1
πππΏ(π, π, π₯0).
Since limπββ
πΏ(π,π,π₯0)
ππ< β.and π β₯ 1, we get
limπ,πββ
π(ππ(π₯0), ππ+π(π₯0)) = 0,
56 Manoj Kumar, Vishnu Narayan Mishra, Asha Rani, Asha Rani & Kumari Jyoti
which implies that {ππ(π₯0)} is a π-cauchy sequence.
By condition (i), (π,π) is π-complete, there exists some π’ β π such that {ππ(π₯0)} is
a π-convergent to π’.
But (π,π) is π-complete so there exists some π’ β π such that {ππ(π₯0)} is a π-
convergent to π’.
Since π is onto, so there exist π β π, such that π β πβ1(π’) βΉ π(π) = π’
But by condition (i), π is π-quasi expansion, for all π β β, we have
π(ππ(π₯0), π’) = π(ππ+1(π₯0), π(π)) β₯
ππππ₯ {
π(ππ(π₯0),π),π(ππ(π₯0), π
π+1(π₯0)),
π(π, π(π)),π(ππ(π₯0), π(π)),
π(π, ππ+1(π₯0))
}.
Letting π β β in the above inequality, we get
limπββ
π(ππ(π₯0),π) β₯ ππ(π, π’),
π(π, π’) β₯ ππ(π, π’).
Implies that π’ = π, Hence, π’ is a fixed point of π.
Remark 3.12 Using theorem 3.11 we can prove fixed point results for k-
quasiexpansion in b-metric spaces, dislocated spaces and modular spaces.
Theorem 3.13 Let (π, π) is complete generalized metric space and π: π β π be an
onto mapping which satisfies following conditions:
(i) π is a π-Kannan expansion for some π β₯1`
2;
(ii) there exists π₯0 β π such that limπββ
πΏ(π,π,π₯0)
ππ< β.
Then {ππ(π₯0)} converges to π’ β π, a fixed point of π Moreover, if π£ β π is another
fixed point of π such that π(π’, π£) < β, then π’ = π£.
Proof Let π β β (π β₯ 1). Since π is a π-Kannan expansion, for all π, π β β,we have
π (ππβ1+π(π₯0), ππβ1+π(π₯0)) β₯ π [π (ππ+π(π₯0), π
πβ1+π(π₯0)) +
π (ππ+π(π₯0), ππβ1+π(π₯0))].
Fixed Point Theorems for Expansive Mappings in js-Metric Space 57
which implies that
πΏ(π, π, ππ(π₯0)) β₯ π[πΏ(π, π, ππβ1(π₯0)) + πΏ(π·, π, π
πβ1(π₯0))] β₯ π[2πΏ(π, π, π₯0)].
Then, for every π β₯ 1, we have
πΏ(π, π, ππ(π₯0)) β€1
(2π)ππΏ(π, π, π₯0).
Using the above inequality, for every π,π β β, we have
π(ππ(π₯0), ππ+π(π₯0)) β€ πΏ(π, π, π
π(π₯0)) β€1
(2π)ππΏ(π, π, π₯0).
Since limπββ
πΏ(π,π,π₯0)
ππ< β.and π β₯
1
2, we get
limπ,πββ
π(ππ(π₯0), ππ+π(π₯0)) = 0,
which implies that {ππ(π₯0)} is a π-cauchy sequence.
By condition (i), (π,π) is π-complete, there exists some π’ β π such that {ππ(π₯0)} is
a π-convergent to π’.
But (π,π) is π-complete so there exists some π’ β π such that {ππ(π₯0)} is a π-
convergent to π’.
Since π is onto, so there exist π β π, such that π β πβ1(π’) βΉ π(π) = π’
But by condition (i), π is π-Kannan expansion, for all π β β, we have
π(ππ(π₯0), π’) = π(ππ+1(π₯0), π(π)) β₯ π[π(ππ(π₯0), π
π+1(π₯0)) + π(π, π(π))].
Letting π β β in the above inequality, we get
limπββ
π(ππ(π₯0),π) β₯ ππ(π, π’),
π(π, π’) β₯ ππ(π, π’).
Implies that π’ = π, Hence, π’ is a fixed point of π.
Remark 3.14 Using theorem 3.13 we can prove fixed point results for k-
quasiexpansion in b-metric spaces, dislocated spaces and modular spaces.
58 Manoj Kumar, Vishnu Narayan Mishra, Asha Rani, Asha Rani & Kumari Jyoti
Example 3.15 Let π = [0,2] be a complete generalized metric space with π =
|π₯ β π¦| with π =1
2. Define the function π: π β π such that π(π₯) = {
2π₯ π₯ < 12π₯ β 1 π₯ β₯ 1
Clearly π satisfies the condition
π(π(π₯), π(π¦)) β₯ π[π(π₯, ππ₯) + π(π¦, ππ¦)].
Now, for every π¦ β π, there exists an π₯ = {
π¦
2 πππ π¦ < 2
π¦+1
2πππ π¦ β₯ 2
β π, such that, π¦ = π(π₯).
So, π is onto. Clearly for all π when π₯ < 1, 0 β€ ππ(π₯) β€ 2πand 0 β€ ππ(π₯) β€ 2π β 1
when π₯ β₯ 1which implies that limπββ
πΏ(π,π,π₯0)
ππ< β. So the condition (ii) of the
Theorem 3.7 is also satisfied. So all the conditions of theorem 3.13 is satisfied with
two fixed points π₯ = 0 and π₯ = 1.
Theorem 3.16 Let (π, π) is complete generalized metric space and π: π β π be an
onto mapping which satisfies following conditions:
(i) π is a π- Chatterjea expansion for some π β₯1
2;
(ii) there exists π₯0 β π such that limπββ
πΏ(π,π,π₯0)
ππ< β.
Then {ππ(π₯0)} converges to π’ β π, a fixed point of π. Moreover, if π£ β π is another
fixed point of π such that π(π’, π£) < β, then π’ = π£.
Proof Let π β β (π β₯ 1). Since π is a π- Chatterjea expansion, for all π, π β β,we
have
π (ππβ1+π(π₯0), ππβ1+π(π₯0)) β₯ π [π (ππ+π(π₯0), π
πβ1+π(π₯0)) +
π (ππ+π(π₯0), ππβ1+π(π₯0))].
which implies that
πΏ(π, π, ππ(π₯0)) β₯ π[πΏ(π, π, ππβ1(π₯0)) + πΏ(π·, π, π
πβ1(π₯0))] β₯ π[2πΏ(π, π, π₯0)].
Then, for every π β₯ 1, we have
πΏ(π, π, ππ(π₯0)) β€1
(2π)ππΏ(π, π, π₯0).
Fixed Point Theorems for Expansive Mappings in js-Metric Space 59
Using the above inequality, for every π,π β β, we have
π(ππ(π₯0), ππ+π(π₯0)) β€ πΏ(π, π, π
π(π₯0)) β€1
(2π)ππΏ(π, π, π₯0).
Since limπββ
πΏ(π,π,π₯0)
ππ< β.and π β₯
1
2, we get
limπ,πββ
π(ππ(π₯0), ππ+π(π₯0)) = 0,
which implies that {ππ(π₯0)} is a π-cauchy sequence.
By condition (i), (π,π) is π-complete, there exists some π’ β π such that {ππ(π₯0)} is
a π-convergent to π’.
But (π,π) is π-complete so there exists some π’ β π such that {ππ(π₯0)} is a π-
convergent to π’.
Since π is onto, so there exist π β π, such that π β πβ1(π’) βΉ π(π) = π’
But by condition (i), π is π- Chatterjea expansion, for all π β β, we have
π(ππ(π₯0), π’) = π(ππ+1(π₯0), π(π)) β₯ π[π(ππ(π₯0), π(π)) + π(π, π
π+1(π₯0))].
Letting π β β in the above inequality, we get
limπββ
π(ππ(π₯0),π) β₯ π[π(π’, π) + π(π, π’)],
π(π, π’) β₯ 2ππ(π, π’). Since π β₯1
2
Which implies that π’ = π, Hence, π’ is a fixed point of π.
Theorem 3.17 Let (π, π) is complete generalized metric space and π: π β π be an
onto mapping which satisfies following conditions:
(i) π is a π- zamfirscu expansion for some π β₯ 1;
(ii) there exists π₯0 β π such that limπββ
πΏ(π,π,π₯0)
ππ< β.
Then {ππ(π₯0)} converges to π’ β π, a fixed point of π. Moreover, if π£ β π is another
fixed point of π such that π(π’, π£) < β, then π’ = π£.
Proof Let π β β (π β₯ 1). Since π is a π - zamfirscu expansion, for all π, π β β,we
have
60 Manoj Kumar, Vishnu Narayan Mishra, Asha Rani, Asha Rani & Kumari Jyoti
π (ππβ1+π(π₯0), ππβ1+π(π₯0)) β₯
π πππ₯
{
π (ππ+π(π₯0), π
π+π(π₯0)) ,
π(ππ+π(π₯0),ππβ1+π(π₯0))+π(π
π+π(π₯0),ππβ1+π(π₯0))
2,
π(ππ+π(π₯0),ππβ1+π(π₯0))+π(π
π+π(π₯0),ππβ1+π(π₯0))
2 }
;
which implies that
πΏ(π, π, ππ(π₯0)) β₯ π[πΏ(π, π, ππβ1(π₯0))].
Then, for every π β₯ 1, we have
πΏ(π, π, ππ(π₯0)) β€1
πππΏ(π, π, π₯0).
Using the above inequality, for every π,π β β, we have
π(ππ(π₯0), ππ+π(π₯0)) β€ πΏ(π, π, π
π(π₯0)) β€1
πππΏ(π, π, π₯0).
Since limπββ
πΏ(π,π,π₯0)
ππ< β and π β₯ 1, we get
limπ,πββ
π(ππ(π₯0), ππ+π(π₯0)) = 0,
which implies that {ππ(π₯0)} is a π-cauchy sequence.
By condition (i), (π,π) is π-complete, there exists some π’ β π such that {ππ(π₯0)} is
a π-convergent to π’.
But (π,π) is π-complete so there exists some π’ β π such that {ππ(π₯0)} is a π-
convergent to π’.
Since π is onto, so there exist π β π, such that π β πβ1(π’) βΉ π(π) = π’
But by condition (ii), π is π- zamfirscu expansion, for all π β β, we have
π(ππ(π₯0), π’) = π(ππ+1(π₯0), π(π)) β₯ ππππ₯
{
π(ππ(π₯0), π),
π(ππ(π₯0),ππ+1(π₯0))+π(π,π(π))
2,
π(ππ(π₯0),π(π))+π(π,ππ+1(π₯0))
2 }
.
Fixed Point Theorems for Expansive Mappings in js-Metric Space 61
Letting π β β in the above inequality, we get
limπββ
π(ππ(π₯0),π) β₯ ππ(π, π’),
π(π, π’) β₯ ππ(π, π’).
Implies that π’ = π, Hence, π’ is a fixed point of π.
Remark 3.18 Using theorem 3.17, we can prove fixed point results for k- zamfirscu
expansion in b-metric spaces, dislocated spaces and modular spaces.
Theorem 3.19 Let (π, π) is complete generalized metric space and π: π β π be an
onto mapping which satisfies following conditions:
(i) π is a π-Rhodes expansion for some π β₯ 1;
(ii) there exists π₯0 β π such that limπββ
πΏ(π,π,π₯0)
ππ< β.
Then {ππ(π₯0)} converges to π’ β π, a fixed point of π. Moreover, if π£ β π is another
fixed point of π such that π(π’, π£) < β, then π’ = π£.
Proof Let π β β (π β₯ 1). Since π is a π- Rhodes expansion, for all π, π β β,we have
π (ππβ1+π(π₯0), ππβ1+π(π₯0)) β₯
π πππ₯
{
π (ππ+π(π₯0), π
π+π(π₯0)) ,
π(ππ+π(π₯0),ππβ1+π(π₯0))+π(π
π+π(π₯0),ππβ1+π(π₯0))
2,
π (ππ+π(π₯0), ππβ1+π(π₯0)) ,
π (ππ+π(π₯0), ππβ1+π(π₯0)) }
.
which implies that
πΏ(π, π, ππ(π₯0)) β₯ π[πΏ(π, π, ππβ1(π₯0))].
Then, for every π β₯ 1, we have
πΏ(π, π, ππ(π₯0)) β€1
πππΏ(π, π, π₯0).
Using the above inequality, for every π,π β β, we have
π(ππ(π₯0), ππ+π(π₯0)) β€ πΏ(π, π, π
π(π₯0)) β€1
πππΏ(π, π, π₯0).
62 Manoj Kumar, Vishnu Narayan Mishra, Asha Rani, Asha Rani & Kumari Jyoti
Since limπββ
πΏ(π,π,π₯0)
ππ< β.and π β₯ 1, we get
limπ,πββ
π(ππ(π₯0), ππ+π(π₯0)) = 0,
which implies that {ππ(π₯0)} is a π-cauchy sequence.
By condition (i), (π,π) is π-complete, there exists some π’ β π such that {ππ(π₯0)} is
a π-convergent to π’.
But (π,π) is π-complete so there exists some π’ β π such that {ππ(π₯0)} is a π-
convergent to π’.
Since π is onto, so there exist π β π, such that π β πβ1(π’) βΉ π(π) = π’
But by condition (ii), π is π- Rhodes expansion, for all π β β, we have
π(ππ(π₯0), π’) = π(ππ+1(π₯0), π(π)) β₯
ππππ₯ {π(ππ(π₯0),π),
π(ππ(π₯0),ππ+1(π₯0))+π(π,π(π))
2,
π(ππ(π₯0), π(π)),π(π, ππ+1(π₯0))
}.
Letting π β β in the above inequality, we get
limπββ
π(ππ(π₯0),π) β₯ ππ(π, π’),
π(π, π’) β₯ ππ(π, π’).
Implies that π’ = π, Hence, π’ is a fixed point of π.
Remark 3.20 Using theorem 3.19 we can prove fixed point results for k- Rhodes
expansion in b-metric spaces, dislocated spaces and modular spaces.
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64 Manoj Kumar, Vishnu Narayan Mishra, Asha Rani, Asha Rani & Kumari Jyoti