# Fixed Point Results for Single-Valued Mappings on a Set with Two Metrics using a Jaggi-Type Bilateral Contraction

DOI : 10.17577/IJERTV11IS110074

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#### Fixed Point Results for Single-Valued Mappings on a Set with Two Metrics using a Jaggi-Type Bilateral Contraction

Muhammad Umair, Asad Hassan, Ali Hassan, Adan Qureshi, Arfatul Mamun, Ajabul Islam, Usman Ahmed, Nabeel Iftikhar, Aneela Bashir

Jiangsu Normal University

Abstract:- This is a research paper exploring some new fixed point results using a bilateral contraction. First, we recall the work on fixed-point results in different research papers. Then we made the new results by combining the results of two papers, one is of Rus, in which different fixed point results on a set with two metrics were discussed, and the second is of Chen, in which different fixed point results were proved using bilateral contractions. The principal aim of this paper is to make new results for single-valued mappings on a single set with two different metrics. All this has been done using the idea of a Jaggi-type bilateral contraction.

Keywords: Bilateral contractions, Jaggi-type bilateral contraction, fixed point results on a set with two metrics

INTRODUCTION AND PRELIMINARIES

Fixed point theory gives important aspects to solve the problems of different branches of mathematics. In the last five decades, fixed point theory has been a growing area of mathematics [1]. A metric space is a non-empty set with metric (or distance function) defined on it. There is much use of metric spaces in different fields and applications, so it is expanded in many ways [2] [3] [5] [9] [16]. In [6] Zhang and Huang explained cone metric spaces. They briefly explained Banach's fixed point theorem for such spaces. Banach's fixed point theorem explains the conditions for the uniqueness of fixed points.

In 1968, Maia investigated the famous result of the Banach contraction principle using two metrics on a non-empty set [10]. In 1975, Iseki described a fixed point theorem in a metric space [7]. In 1977, Rus proved a fixed point theorem in a set containing two metrics [12]. In 1981, Sigh and Pant proved a fixed point theorem in two metrics [14]. In 1989, Kaneko and Sessa established an idea about a fixed point theorem for contractive single and multivalued mappings [8]. In 1996, Takahashi introduced a fixed point of the multivalued mappings in convex metric spaces [13]. In 2007, Muresan gave some results about the fixed point theorem of Maia and expressed how to use these results in the sets with two metrics [11]. In 2019, Joonaghany and Karapinar enhanced the composition by combining the execution of results of two bilateral contractions; Jaggi-type bilateral contraction and Dass Gupta-type bilateral contraction [4]. In 2020, Stinson, Almuthaybiri and Tisdell described a notation about the development of fixed point theorems in a set containing two metrics with the help of iterated method [15].

MAIN BODY

At the start of this section, we define a Jaggi-type bilateral contraction, which is cited in a well-known paper on bilateral contractions by Chen [4].

Definition 1. Let ( , ) be a non-empty set. The function is called Jaggi-type bilateral contraction, if there is a

[0 , ) such that for all distinct ,

implies

( , ) > 0

( , ) [() ()] max{( , ) , ( , )}

Firstly, suppose that max{( , ) , ( , )} = ( , ) then take a set with two metrics, and we make the new result, which

is:

Theorem 1. Let be a non-empty set. Suppose 1 and 2 be two metrics on and ( , 1) ( , 1) be a function. If there is a [0 , ) and for all ,

(a) 1( , ) [() ()] 2( , )

(b) ( , 1) is a complete metric space (c) ( , 1) ( , 1) is continuous

(d) (0 ,1) we have 2( , ) 2( , )

Then has a unique fixed point.

Proof. We prove the theorem by the iterative method. For any , let

0 =

1 = 0

where .

2 = 1

= 1

This implies that {} converges in .

If = then our theorem has been proved.

Suppose . Then for any distinct 1 , , let = 1(1 , ) then by the given condition

+1 = 1( , +1)

= 1(1 , )

[(1) (1)] 2(1 , )

= [( 1) ()] 2(1 , )

It follows

1(1 , +1)

2(1 , )

(1) ()

1( , +1)

0 <

2(1 , )

(1) ()

0 < (1) ()

(1) > ()

We conclude that the sequence {()} is not only strictly decreasing but also necessarily positive. So {()} converges to some limit 0.

Now for each we have

=1

1( , +1)

2(1 , )

[(1) ()]

=1

[(0) (1)] + [(1) (2)] + + [(1) ()]

(0) (1) + (1) (2) + (2) + (1) + (1) ()

(0) ()

If then ()

=1

1( , +1)

2(1 , )

(0) <

1 ( , +1 )

2 1

In other words, we can say =1 ( , ) is a finite positive number.

By induction,

1 ( , +1 )

2(1 ,)

is bounded in (0 , 1), then there exists some (0 , 1) we have

1( , +1)

2(1 , )

1( , +1) 2(1 , )

2 2(2 , 1)

3 2(3 , 2)

2(0 , 1)

Now, for each , with < such that

1( , ) [() ()] 2(1 , 1)

[() (+1)] 2(2 , 2)

[() (+1)]2 2(3 , 3)

[() (+1)]1 2(0 , )

Since, () is strictly decreasing, then [() (+1)] is very small and (0 , 1) then we can conclude that

[() (+1)]1 < then

1( , ) < 2(0 , )

<

This implies that {} is the Cauchy sequence.

Since is complete. By the continuity of ( , 1) ( , 1) , for any 0

0 = lim [(0)]

= lim [. 1(0)]

= ( lim [1(0)])

= (0)

Thus, 0 is a fixed point of .

Suppose 0 is another fixed point of , then

2(0 , 0) = (0 , 0)

2(0 , 0) (1 ) 2(0 , 0) 0

2(0 , 0) = 0

0 = 0

Hence, 0 is a unique fixed point of .

By applying some more conditions to the above theorem, we make a new result. More conditions were taken from the paper by

Rus [12].

Theorem 2. Let be a non-empty set. Suppose 1 and 2 be two metrics on and , be the functions. If for all

, such that

1. ( , 1) , ( , 2) and satisfy the hypothesis of Theorem 1

2. The sequence uniformly converges on ( , 1) to

(c) > 0 we have 2( , ) 1( , )

Then for every , sequence {} converges to a unique fixed point 0 of .

Proof. We prove that every sequence {} converges to a unique fixed point 0 . Since for some

() =

Now,

2 2

1( , 0) = 1 ( () , (0))

2 2 2 2

1 ( () , ()) + 1 ( () , (0))

2 2 2 2

1 ( () , ()) + [() ()] 2 ( () , (0))

2 2

1 ( () , ()) + [() ()] 2( , 0)

2 2

Since

Then

1 ( ( , ()) + [() ()] 1( , 0)

[() ()] < 1

=> [() ()] 0

=> [() ()] <

2 2

1( , 0) 1 ( () , ()) + 1( , 0)

2 2

(1 ) 1( , 0) 1 ( () , ())

2 2

1( , 0) (1 )1 1 ( () , ())

2 2

(1 )1 [1 ( () , . ()) + 1 (. () , ())]

1 1

(1 )1 [1 (2( ) , . ( )) + ( ( ) , ( ))]

1

It is given that uniformly converges to in metric 1 , then 1 (2( ) , ( )) 0 and ( ( ) , ( )) 0 as

2

. Thus

1( , 0) (1 )1 [1 ( () , ()) + 1 1 (() , ())] 0

It means as .

1( , 0) 0

Hence, {} converges in ( , 1) to a unique fixed point 0 of .

Now, suppose in the definition-1, if max{( , ) , ( , )} = ( , ) then one more new result is generated.

Theorem 3 Let be a non-empty set. Suppose 1 and 2 be two metrics on and ( , 1) ( , 1) be a function. If there is a [0 , ) and for all , such that

(a) 1( , ) [() ()] 2( , )

(b) ( , 1) is a complete metric space (c) ( , 1) ( , 1) is continuous

(d) (0 , 1) we have 2( , ) 2( , )

Then has a unique fixed point.

Proof. We prove the theorem by the iterative method. For any , let

0 =

1 = 0

2 = 1

= 1

where .

This implies that {} converges in .

If = then our theorem has been proved.

Suppose . Then for any distinct 1 , , let = 1(1 , ) then by the given condition

+1 = 1( , +1)

= 1(1 , )

[(1 (1)]

2(1 , 1) 2( , )

2(1 , )

[(1) ()]

2(1 , ) 2( , +1)

2(1 , )

It follows

[(1) ()] 2( , +1)

1( , +1)

(1) ()

2( , +1)

1( , +1)

0 <

2( , +1)

(1) ()

0 < (1) ()

(1) > ()

We conclude that the sequence {()} is not only strictly decreasing but also necessarily positive. So {()} converges to some limit 0.

Now for each we have

=1

1( , +1)

2( , +1)

[(1) ()]

=1

[(0) (1)] + [(1) (2)] + + [(1) ()]

(0) (1) + (1) (2) + (2) + (1) + (1) ()

(0) ()

If then ()

=1

1( , +1)

2( , +1)

(0) <

1 ( , +1 )

In other words, we can say is a finite positive number.

=1

2( ,+1)

By induction,

1( ,+1)

2( ,+1)

is bounded in (0 , 1), then there exists some (0 , 1) we have

1( , +1)

2( , +1)

1( , +1) 2( , +1)

2 2(1 , )

3 2(2 , 1)

+1 2(0 , 1)

Now, for each , with < such that

1( , ) [() ()] 2( , )

[() (+1)] 2(1 , 1)

[() (+1)]2 2(2 , 2)

[() (+1)] 2(0 , )

Since, () is strictly decreasing, then [() (+1)] is very small and (0 , 1) then we can conclude that

[() (+1)] < then

1( , ) < 2(0 , )

<

This implies that {} is the Cauchy sequence.

Since is complete. By the continuity of ( , 1) ( , 1) , for any 0

0 = lim [(0)]

= lim [. 1(0)]

= ( lim [1(0)])

= (0)

Thus, 0 is a fixed point of .

Suppose 0 is another fixed point of , then

2(0 , 0) = (0 , 0)

2(0 , 0) (1 ) 2(0 , 0) 0

2(0 , 0) = 0

0 = 0

Hence, 0 is a unique fixed point of .

Similarly, by applying some more conditions to the above result, we make a new result.

Theorem 4. Let be a non-empty set. Suppose 1 and 2 be two metrics on and , be the functions. If for all

, such that

1. ( , 1) , ( , 2) and satisfy the hypothesis of Theorem 3

2. The sequence uniformly converges on ( , 1) to

(c) > 0 we have 2( , ) 1( , )

Then for every , sequence {} converges to a unique fixed point 0 of .

Proof. We prove that every sequence {} converges to a unique fixed point 0 . Since for some

() =

Now,

2 2

1( , 0) = 1 ( () , (0))

2 2 2 2

1 ( () , ()) + 1 ( () , (0))

2 2 2 2

1 ( () , ()) + [() (+1)] 2 ( () , (0))

2 2

1 ( () , ()) + [() (+1)] 2( , 0)

2 2

1 ( () , ()) + [() (+1)] 1( , 0)

Since () is strictly decreasing then [() (+1)] is very small and (0 , 1) then we conclude that

[() (+1)] < then

2 2

1( , 0) 1 ( () , ()) + 1( , 0)

2 2

(1 ) 1( , 0) 1 ( () , ())

2 2

1( , 0) (1 )1 1 ( () , ())

2 2

(1 )1 [1 ( () , . ()) + 1 (. () , ())]

1 1

(1 )1 [1 (2( ) , . ( )) + ( ( ) , ( ))]

1

It is given that uniformly converges to in metric 1 , then 1 (2( ) , ( )) 0 and ( ( ) , ( )) 0 as

. This implies that

It means as .

1 1

1( , 0) (1 )1 [1 (2( ) , ( )) + ( ( ) , ( ))] 0

1( , 0) 0

Hence, {} converges in ( , 1) to a unique fixed point 0 of .

By the above theorems, we proved the new fixed point results on a set with two metrics using the idea of a bilateral contraction.

Now, we will take an example, which helps us to prove the inequalities, which we used in the above results and disprove the other contraction inequalities.

Example 1. Let = {0 , 1 , 2} endowed with the metric 1 and 2 defined for all ,

( , ) = { 0 = and ( , ) = | |

1 1 2

Let defined by

Define [0 , ) as

Prove that for all , , satisfies

(0) = 0 , (1) = 2 (2) = 0

(0) = 0 , (1) = 4 (2) = 2

1( , ) [() ()] 2( , )

We prove for all ,

(i) For ( , ) = (0 , 0) :

1(0 , 0) [(0) (0)] 2(0 , 0)

1(0 , 0) [(0) 1(0)] 2(0 , 0) 0 [4 2] |1 0|

1 2

(iii) For ( , ) = (1 , 1) :

1(1 , 1) [(1) (1)] 2(1 , 1)

1(2 , 2) [(1) (2)] 2(1 , 1) 0 [4 2] |1 1|

0 0

(iv) For ( , ) = (1 , 2) :

1(1 , 2) [(1) (1)] 2(1 , 2)

1(2 , 0) [(1) (2)] 2(1 , 2) 1 [4 2] |1 2|

1 2

(v) For ( , ) = (2 , 0) :

1(2 , 0) [(2) (2)] 2(2 , 0)

1(0 , 0) [(2) (0)] 2(2 , 0) 0 [2 0] |2 0|

0 4

(vi) For ( , ) = (2 , 2) :

1(2 , 2) [(2) (2)] 2(2 , 2)

1(0 , 0) [(2) (0)] 2(2 , 2) 0 [2 0] |2 2|

0 0

Hence, for all , , satisfied the given inequality.

Now, we check doesnt satisfy other contraction inequalities. Suppose the contraction inequality on two metrics:

1( , ) 2( , )

for some > 0.

For ( , ) = (1 , 2) :

1(1 , 2) 2(1 , 2) 1 |1 2|

1 |1|

1

This is false for 0 < < 1. So

1

Hence, doesnt satisfy 1( , ) 2( , ).

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