Received 03rd June 2020, Revised 08th December 2020, Accepted 15th December 2020 DOI: 10.14456/past.2021.2
Almost Type of Simulation Functions Results
Paiwan Wongsasinchai
Department of Mathematics, Faculty of Science and Technology, Rambhai Barni Rajabhat University, Chanthaburi, Thailand.
E-mail: [email protected]
Abstract
In this paper, we introduce common point theorems for almost type simulation functions and a complete metric framework. We investigated both the existence and uniqueness of common fixed points of such mappings.
We used an example to illustrate the main result observed. Our main results cover several existing results in the corresponding literature.
Keywords: almost type, common point theorems, simulation functions
1. Introduction
The fixed problem can be considered a simple equation Οπ = π. In almost all scientific disciplines, most of the issues can be converted into fixed point equations. The first fixed point theorem was announced in Banachβs thesis in 1922 (1), in the setting of complete normed space, which can be described as the abstraction of the successive approximation method.
Theorem 1.1 (1) Let (π¬, π) be a complete metric space and Ο be a self-mapping on the set π¬ such that there exists π β [0,1) ,
π(Οπ, Οπ) β€ ππ(π, π) for all π, π β Ξ. (1.1) Then, Ο has a unique fixed point in π¬.
Theorem 1.2 (3) Let (π¬, π) be a complete metric space and a self-mapping Ο on the set π¬ be an almost contraction, that is, a mapping for which there exist πΏ β [0,1) and there exist πΏ β₯ 0such that π(Οπ, Οπ) β€ πΏπ(π, π) + πΏπ(π, Οπ), βπ, π β Ξ. (1.2) Then,
(i) πΉππ₯(Ο) β β , where πΉππ₯(Ο) = {π β π¬:Οπ = π};
(ii) For any π0β π¬, the Picard iteration {ππ} given by ππ+1=Οππ for each π β₯ 0 converges to some π ββ πΉππ₯(Ο);
(iii) The following estimate holds π(ππ+πβ1, π β) β€ πΏπ
1βπΏπ(ππ, ππβ1), βπ β₯ 0, π β₯ 1.
Babu et al. [5] defined the class of mappings satisfying condition (B) as follows:
Definition 1.3 (5) Let (Ξ, π) be a metric space and a self-mapping Ο on Ξ is said to satisfy condition (B) if there exist a constant πΏ β (0,1) and there exists πΏ β₯ 0 such that
π(Οπ, Οπ) β€ πΏπ(π, π) + πΏπΎ(π, π),βπ, π β Ξ, (1.3) where
πΎ(π, π) = πππ{ π(π, Οπ), π(π, Οπ), π(π, Οπ), π(π, Οπ)}.
They proved a fixed point theorem for such mappings in complete metric spaces. They also discussed quasi- contraction, almost contraction, and mappings class that satisfy condition (B) in detail.
Khojasteh et al. (6) originated the notion of π -contractions using a specific family of functions called simulation functions. Subsequently, many researchers generalized this idea in many ways (8-24) and proved many interesting results in the arena of fixed point theory. Recently, Heidary et al. (7) proposed a new notion, the π -simulation function.
The notion of the ππ-contraction covers several distinct types of contraction, including the π - contraction that was defined in (6). We denote Ξ¨: = {π: β0+β β0+|π is continuous and nondecreasing, and π(π) = 0 β π = 0}.
Definition 1.4 (7) We say that π: β0+Γ β0+β β is a π -simulation function, if there exists π β Ξ¨ such that:
(π1)π(π, π) < π(π) β π(π) for all π, π > 0;
(π2) if {ππ}, {ππ} are sequences in (0, β) such that
πββπππππ= πππ
πββππ> 0 implies πππ π π’π
πββ π(ππ, ππ) < 0.
Let ππ be the set of all π -simulation functions. Note that if we take π as an identity
mapping, then π -simulation function becomes a simulation function in the sense of (6).
Example 1.5. (7) Let π β Ξ¨
(i) π1(π, π) = ππ(π) β π(π) for all π, π β [0, β), where π β [0,1).
(ii) π2(π, π) = π(π(π)) β π(π) for all
π, π β [0, β), where π is a self-mapping from [0, β) to [0, β) so that π(0) = 0 and for each π > 0, π(π) < π,
πππ π π’π
πβπ π(π) < π.
(iii) π3(π, π) = π(π) β π(π) β π(π) for all π, π β [0, β) where π: [0, β) β [0, β) is a mapping such that, for each π > 0,
πππ πππ
πβπ π(π) < 0.
It is clear that π1, π2, π3β ππ.
Motivated and inspired by Babu et al.(5), Khojasteh et al. (6) and Heidary et al. (7), we define an Almost type simulation functions in metric spaces.
2. Main Results
Firstly, we present the following definition, which will be used in our main results.
Definition 2.1. Let (π¬, π) be a metric space.
We say that a pair of mappings Ο, π·: π¬ β π¬ is an almost type simulation functions whenever there is a constant πΏ β₯ 0, for all π, π β π¬, such that
1
2πππ{ π(π, Οπ), π(π, Ξ¦π)} β€ π(π, π) implies π(π(Οπ, Ξ¦π), π·(π, π) + πΏπΎ(π, π)) β₯0, (2.1) where
π·(π, π) = πππ₯ {π(π, π),[1+π(π,Οπ)]π(π,π·π) 1+π(π,π) } and πΎ(π, π) = πππ{π(π, Οπ), π(π, π·π), π(π, π·π), π(π, Οπ)}.
Lemma 2.2. Let (π¬, π) be a metric space and Ο, π·: π¬ β π¬ be two selfmaps. Assume that the pair (Ο, π·) is an almost type of simulation functions.
Then π’ is a fixed point of Ο if and only if π’ is a fixed point of π·. In that case, π’ is a common fixed point of Ο and π·, and π’ is unique.
Proof. Let π’ be a fixed point of Ο, i.e., Οπ’ = π’. Now, we prove that π’ is a fixed point of π·. Assume that π(π’, π·π’) > 0 Thus from (2.1) we have
0 =12πππ{ π(π’, Οπ’), π(π’, Ξ¦π’)} β€ π(π’, π’). (2.2) This implies
π(π(Οπ’, Ξ¦π’), π·(π’, π’) + πΏπΎ(π’, π’)) β₯ 0. (2.3) By (π1), we have
π(π(Οπ’, π·π’)) < π(π·(π’, π’) + πΏπΎ(π’, π’)).
Therefore,
π(Οπ’, Ξ¦π’) < π·(π’, π’) + πΏπΎ(π’, π’), (2.4)
where
π·(π’, π’) = πππ₯ {π(π’, π’),[1+ π(π’, Οπ’)]π(π’, π·π’) 1+ π(π’, π’) }
= π(π’, π·π’) and
πΎ(π’, π’) = πππ{π(π’, Οπ’), π(π’, π·π’), π(π’, π·π’), π(π’, Οπ’)} =0.
Hence, using the values of π·(π’, π’) and πΎ(π’, π’) in (2.4), we obtain π(π’, π·π’) < π(π’, π·π’) + πΏ0, which is a contradiction. Therefore π’ = π·π’. So, π’ is a fixed point of π·. Same the way, it is easy to see that if π’ is a fixed point of π· then π’ is a fixed point of Ο also.
Next, we prove that π’ is a unique common fixed point of Ο and π·. Let π’ and π£ be two common fixed points of Ο and π· such that π(π’, π£) > 0. Thus from (2.1), we have
0=12πππ{ π(π’, Οπ’), π(π£, Ξ¦π£)} β€ π(π’, π£). (2.5) This implies
π(π(Οπ’, Ξ¦π£), π·(π’, π£) + πΏπΎ(π’, π£)) β₯ 0. (2.6) By (π1) , we have
π(π(Οπ’, π·π£)) < π(π·(π’, π£) + πΏπΎ(π’, π£)).
Therefore,
π(Οπ’, Ξ¦π£) < π·(π’, π£) + πΏπΎ(π’, π£), (2.7) where
π·(π’, π£) = πππ₯ {π(π’, π£),[1+ π(π’, Οπ’)]π(π£, π·π£) 1+ π(π’, π£) }
= π(π’, π£) and
πΎ(π’, π£) = πππ{π(π’, Οπ’), π(π£, π·π£), π(π’, π·π£), π(π£, Οπ’)} =0.
Hence, using the values of π·(π’, π£) and πΎ(π’, π£) in (2.7), we obtain
( , ) ( , ) 0
d ο‘u v οΌd u v +L ,
which is a contradiction. We conclude that π(π’, π£) =0, i.e., π’ = π£ and the theorem is proven.
Theorem 2.3. Let (π¬, π) be a complete metric space and Ο, π·: π¬ β π¬ be two selfmaps.
Assume that the pair (Ο, π·) is an almost type of simulation functions. Then π’ is a unique common fixed point of Ο and π·.
Proof. Let π’0β π¬ be an arbitrary point. We define a sequence {π’π} β π¬ by π’2π+1= Οπ’2π and π’2π+2= π·π’2π+1 for π =0,1,2, . ..
We note that 1
2πππ{ π(π, Οπ), π(π, π·π)} β€ π(π, π) if and only if either
1
2π(π, Οπ) β€ π(π, π)} or 12π(π, π·π) β€ π(π, π)}.
Assume that π’2π= π’2π+1for some π, then π’2π= Οπ’2π. So, π’2π is a fixed point of Ο. Thus, by Lemma 2.2, we have π’2π is a fixed point of π· also π’2π is a common fixed point of Ο and π·. Same the way, if π’2π+1= π’2π+2 then π’2π+1 is a fixed point of π·. Using Lemma 2.2, we have π’2π+1 is a common fixed point of Ο and π·. Without loss
of generality, we can assume that π(π’π, π’π+1) >0 for π =0,1,2, . .. From (2.1), we get
1
2πππ{ π(π’2π, Οπ’2π), π(π’2π+1, π·π’2π+1)} β€ π(π’2π, π’2π+1).
This implies
π(π(Οπ’2π, π·π’2π+1), π·(π’2π, π’2π+1) + πΏπΎ(π’2π, π’2π+1)) β₯0.
By (π1), we have
π(π(Οπ’2π, π·π’2π+1)) < π(π·(π’2π, π’2π+1) +
πΏπΎ(π’2π, π’2π+1)). (2.8)
Therefore,
π(Οπ’2π, Ξ¦π’2π+1) < π·(π’2π, π’2π+1) +
πΏπΎ(π’2π, π’2π+1), (2.9)
where π·(π’2π, π’2π+1)
= πππ₯ {π(π’2π, π’2π+1),[1+π(π’2π1+π(π’,Οπ’2π)]π(π’2π+1,π·π’2π+1)
2π,π’2π+1) } = πππ₯ {π(π’2π, π’2π+1),[1+π(π’2π,π’2π+1)]π(π’2π+1,π’2π+2)
1+π(π’2π,π’2π+1) } = πππ₯{π(π’2π, π’2π+1), π(π’2π+1, π’2π+2)}, so that
π·(π’2π, π’2π+1)
= πππ₯{π(π’2π, π’2π+1), π(π’2π+1, π’2π+2)}
and
πΎ(π’2π, π’2π+1)
= πππ {π(π’2π, Οπ’2π), π(π’2π+1, π·π’2π+1), π(π’2π, π·π’2π+1), π(π’2π+1, Οπ’2π)} = πππ {π(π’2π, π’2π+1), π(π’2π+1, π’2π+2),
π(π’2π, π’2π+2), π(π’2π+1, π’2π+1)} =0.
Hence, using the values of π·(π’2π, π’2π+1) andπΎ(π’2π, π’2π+1) in (2.9), we obtain
2 1 2 2 2 2 1 2 1 2 2
( n , n ) max{ ( n, n ), ( n , n )} 0.
d u + u + οΌ d u u + d u + u + +L (2.10) If π(π’2π, π’2π+1) < π(π’2π+1, π’2π+2) for some π, then from (2.10), we have
π(π’2π+1, π’2π+2) < π(π’2π+1, π’2π+2) which is a contradiction. Therefore,
π(π’2π+1, π’2π+2) < π(π’2π, π’2π+1) and
π·(π’2π, π’2π+1) = π(π’2π, π’2π+1) Hence, from (2.8), we have
π(π(π’2π+1, π’2π+2)) < π(π(π’2π, π’2π+1)). (2.11) Same the way, we have
π(π(π’2π+2, π’2π+3)) < π(π(π’2π+1, π’2π+2)). (2.12) Therefore from (2.11) and (2.12), we have
π(π(π’π+1, π’π+2)) < π(π(π’π, π’π+1)) for π =0,1,2, . ..
which implies that
π(π’π+1, π’π+2) β€ π(π’π, π’π+1) for π =0,1,2, . . . , π.
(2.13)
Hence, the sequence {π(π’π, π’π+1)} is a non- increasing and bounded below. Then it is convergent and there exists a real number πΌ β₯0 such that πΌ = πππ
πββπ(π’π, π’π+1). (2.14)
To prove that πΌ =0, assume πΌ >0. For π β₯0, we consider
1
2πππ{π(π’2π, Οπ’2π), π(π’2π+1, π·π’2π+1)}
=1
2πππ{π(π’2π, π’2π+1), π(π’2π+1, π’2π+2)}
β€ π(π’2π, π’2π+1).
This implies
π(π(π’2π+1, π’2π+2), π(π’2π, π’2π+1) β₯0.
Hence, πππ π π’π
πββ
π(π(π’2π+1, π’2π+2), π(π’2π, π’2π+1) β₯0.
Using (2.14), we obtain
πββππππ(π’2π+1, π’2π+2) = πΌ >0. (2.15) Using (π2) with ππ= π(π’2π+1, π’2π+2) and ππ= π(π’2π, π’2π+1), we have
πππ π π’π
πββ π(π(π’2π+1, π’2π+2), π(π’2π, π’2π+1) <0, which is a contradiction. Therefore,
πββππππ(π’π, π’π+1) =0. (2.16) We know prove that {π’π}is a Cauchy sequence. On the contrary suppose that {π’π} is not Cauchy. Then there is an π > 0 and sequences of integers {2ππ}and {2ππ} with ππ> ππ> π such that
π(π’2ππ, π’2ππ) β₯ π and π(π’2ππβ2, π’2ππ) < π. (2.17) Case(i): We will proveπππ
πββπ(π’2ππ, π’2ππ) = π. From (2.17), we have π β€ π(π’2ππ, π’2ππ). Taking limit infimum as π ββ, we obtain
π β€ lim πππ
πββ π(π’2ππ, π’2ππ). (2.18) We consider
π(π’2ππ, π’2ππ) β€ π(π’2ππ, π’2ππβ2) + π(π’2ππβ2, π’2ππ) < π(π’2ππ, π’2ππβ2) + π.
Taking limit superior as π ββ, we obtain lim π π’π
πββ π(π’2ππ, π’2ππ) β€ π. (2.19) Using (2.18) and (2.19), we obtain
πββππππ(π’2ππ, π’2ππ) = π.
Same the way we prove the following:
Case(ii): πππ
πββπ(π’2ππ, π’2ππ+1) = π.
Case(iii): πππ
πββπ(π’2ππ+1, π’2ππ) = π.
Case(iv): πππ
πββπ(π’2ππ+1, π’2ππ+1) = π.
Case(v): πππ
πββπ(π’2ππ+1, π’2ππ+2) = π.
Now, from the definition of π·(π, π), we have
πββππππ·(π’2ππ, π’2ππβ1)
= πππ
πββπππ₯ {
π(π’2ππ, π’2ππβ1),
[1+ π(π’2ππ, Οπ’2ππ)]π(π’2ππβ1, π·π’2ππβ1) 1+ π(π’2ππ, π’2ππβ1)
}
= πππ
πββπππ₯ {
π(π’2ππ, π’2ππβ1),
[1+ π(π’2ππ, π’2ππ+1)]π(π’2ππβ1, π’2ππβ1) 1+ π(π’2ππ, π’2ππβ1)
}
= π and
πββππππΎ(π’2ππ, π’2ππβ1)
= πππ
πββπππ {π(π’2ππ, Οπ’2ππ), π(π’2ππβ1, π·π’2ππβ1), π(π’2ππ, π·π’2ππβ1), π(π’2ππβ1, Οπ’2ππ)}
= πππ
πββπππ {π(π’2ππ, π’2ππ+1), π(π’2ππβ1, π’2ππ), π(π’2ππ, π’2ππ), π(π’2ππβ1, π’2ππ+1)}
=0.
Taking π sufficiently large with ππ> ππ> π and since {π(π’π, π’π+1)}is a non-increasing,
π(π’2ππ, Οπ’2ππ) = π(π’2ππ, Οπ’2ππ+1) β€ π(π’2ππ, π’2ππ+1)
β€ π(π’2ππβ1, π’2ππ) β€ π(π’2ππβ1, π·π’2ππβ1).
So, 1
2πππ{π(π’2ππ, Οπ’2ππ), π(π’2ππβ1, Ξ¦π’2ππβ1)}
=1
2π(π’2ππ, Οπ’2ππ)
β€ π(π’2ππ, π’2ππ+1). (2.20) Using (2.16), there exists π1β β such that for anyπ >
π1,
π(π’2ππ, π’2ππ+1) <π 2.
Also, there exists π2β βsuch that for any π > π2, π(π’2ππβ1, π’2ππ) <π
2.
Thus, for any π > πππ₯{π1, π2}and ππ> ππ> π, we have
π β€ π(π’2ππ, π’2ππ)
β€ π(π’2ππ, π’2ππβ1) + π(π’2ππβ1, π’2ππ) β€π2+ π(π’2ππβ1, π’2ππ),
which implies that π
2β€ π(π’2ππβ1, π’2ππ).
For any π > πππ₯{π1, π2} π and ππ> ππ> π, π(π’2ππ, π’2ππ+1) <π
β€ π(π’2 2ππβ1, π’2ππ).
Therefore, from (2.20), we have 1
2πππ{π(π’2ππ, Οπ’2ππ), π(π’2ππβ1, π·π’2ππβ1)}
=1
2π(π’2ππ, π’2ππ+1) β€ π(π’2ππβ1, π’2ππ).
This implies
π ( π(Οπ’2ππ, π·π’2ππβ1),
π·(π’2ππ, π’2ππβ1) + πΏπΎ(π’2ππ, π’2ππβ1)) β₯0.
By (π1) , we have
π(π(Οπ’2ππ, π·π’2ππβ1)) < π(π·(π’2ππ, π’2ππβ1) + πΏπΎ(π’2ππ, π’2ππβ1)).
Since π is nondecreasing, we get π(Οπ’2ππ, π·π’2ππβ1) < π·(π’2ππ, π’2ππβ1) + πΏπΎ(π’2ππ, π’2ππβ1)).
Thus, πππ π π’π
πββ π(π’2ππ+1, π’2ππ)
< πππ π π’π
πββ
π·(π’2ππ, π’2ππβ1) + πΏπππ π π’π
πββ
πΎ(π’2ππ, π’2ππβ1) and then π < π, which is a contradiction. Therefore, {π’π} is a Cauchy sequence in π¬. Since (π¬, π) is a complete metric space, we have {π’π} is convergent to some point π’ in π¬. Therefore
π’ = πππ
πββπ’2π+1= πππ
πββΟπ’2π and π’ = πππ
πββπ’2π+2= πππ
πββπ·π’2π+1. Hence,
πββπππΟπ’2π= π’ = πππ
πββπ·π’2π+1.
We assume that Ο is continuous. Since π’2πβ π’ as π ββ , we have Οπ’2πβΟπ’ as π ββ. Hence 0β€ π(π’, Οπ’) β€ (π(π’, Οπ’2π) + π(Οπ’2π, Οπ’)) β0 as π ββ, so that π(π’, Οπ’) =0. Thus, π’ is a fixed point of Ο. Now by Lemma 2.2, we have π’ is a unique common fixed point of Ο and π·. Same the way, we can prove that π’ is a unique common fixed point of Ο and π· whenever π· is continuous.
Example Let π = [0,2] be endowed with the usual metric. Define a mapping Ο, π·: π β π as Ο, π· =2β π for all π β π. Then, Ο and π· are not a π - contraction with respect to π where for all π‘, π β [0,β)
π(π‘, π ) = πΌπ β π‘, πΌ β [0,1).
For all π β π, we get 1
2πππ{ π(π, Οπ), π(π, π·π)}
=1
2πππ{|π β (2β π)|, |π β (2β π)|}
=1
2πππ{|π β2+ π|, |π β2+ π|}
=1
2πππ{|2π β2|, |2π β2|}
β€ |π β π|
= π(π, π).
And
π(π(Οπ, π·π), π(π, π))
= πΌ|π β π| β |2β π β (2β π)|
= πΌ|π β π| β |π β π|
< |π β π| β |π β π|
=0.
Now, we show that Ο and Ξ¦ are a modified almost π -contraction with respect to π.
π(π(Οπ, Ξ¦π), π·(π, π) + πΏπΎ(π, π))
= πΌ[|π·(π, π) + πΏπΎ(π, π)|] β |2β π β (2β π)|
= πΌ[|π·(π, π) + πΏπΎ(π, π)|] β |π β π|,
where π·(π, π)
= πππ₯ {|π β π|,[1+ |π β (2β π)|]|π β (2β π)|
1+ |π β π| }
= πππ₯ {|π β π|,[1+|2πβ2|]|2πβ2|
1+|πβπ| } and
π·(π, π)
= πππ {|π β (2β π)|, |π β (2β π)|,
|π β (2β π)|, |π β (2β π)|}
= πππ{|2π β2|, |2π β2|, |2π β2|, |π + π β2|}
= πππ{|2π β2|, |2π β2|, |π + π β2|}.
We deduce that
π(π(Οπ, Ξ¦π), π·(π, π) + πΏπΎ(π, π)) = πΌ [πππ₯ {|π β π|,[1+|2πβ2|]|2πβ2|
1+|πβπ| } + πΎ πππ {|2π β2|, |2π β2|,
|π + π β2| }] β |π β π|.
Thus, we get two cases : Cases(i): If π = π, then
π(π(Οπ, Ξ¦π), π·(π, π) + πΏπΎ(π, π)) = πΌ [ [1+ |2π β2|]
|2π β2| + πΏ|2π β2|] β₯0.
Cases(ii): Suppose that π > π.
Then
π(π(Οπ, Ξ¦π), π·(π, π) + πΏπΎ(π, π)) = πΌ[1+ |2π β2|]|2π β2|
1+ |π β π| + πΌπΏ|2π β2| β |π β π|.
We choose πΌ =12 and πΏ =6, then we get π(π(Οπ, Ξ¦π), π·(π, π) + πΏπΎ(π, π)) =1
2
[1+ |2π β2|]|2π β2|
1+ |π β π| +3|2π β2| β |π β π|.
Thus, all of the conditions of Theorem 2.3 are satisfied. Thus, Ο and Ξ¦ have a unique common fixed point π’ =1.
Acknowledgements
The author thanks Rambhai Barni Rajabhat University for the support.
Declaration of conflicting interests
The author declared that they have no conflicts of interest in the research, authorship, and this article's publication.
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