Fixed Point Theorem for Meir-Keeler Type Function in b2-Metric Spaces

DOI: 10.4236/oalib.1105973    96 Downloads   213 Views  

ABSTRACT

In this paper, we prove fixed point theorems of a generalization which is related to the concept of Meir-Keeler function in a complete b2-metric space. And we know it extends and generalizes some known results in metric space to b2-metric space.

Cite this paper

Tian, Z. , Cui, J. and Zhong, L. (2019) Fixed Point Theorem for Meir-Keeler Type Function in b2-Metric Spaces. Open Access Library Journal, 6, 1-8. doi: 10.4236/oalib.1105973.

1. Introduction

Many mathematicians have studied fixed point theory over the last several decades since Banach contraction principle [1] was introduced in 1992. The notion of Meir-Keeler function [2] was introduced in 1969. Then the concept of weaker Meir-Keeler function [3] was introduced by Chi-Ming Chen in 2012. And in this paper, we establish fixed point for Meir-Keeler function and weaker Meir-Keeler function in a complete new type of generalized matric space, which is called by b2-metric space, and this space was generalized from both 2-metric space [4] [5] [6] and b-metric space [7] [8].

2. Preliminaries

Throughout this paper N will denote the set of all positive integers and R will denote the set of all real numbers.

Before stating our main results, some necessary definitions might be introduced as follows.

Definition 2.1 [2] Let X be a nonempty subsets, m N and f : X X an operator. Then X = i = 1 m A i is called a cyclic representation of X with respect to f if

1) A i , i = 1 , 2 , , m are empty subsets of X,

2) f ( A i ) A 2 , f ( A 2 ) A 3 , , f ( A m 1 ) A m , f ( A m ) A 1 .

Definition 2.2 [2] A function ϕ : ( 0 ] ( 0 ] is said to be a Meir-Keeler function if for each η > 0 , there exists δ > 0 such that for each t ( 0 ] with η t η + δ , we have ϕ ( t ) < η .

Definition 2.3 [3] We call ϕ : ( 0 ] ( 0 ] a weak Meir-Keeler function if for each η > 0 such that for each t ( 0 ] with η t η + δ , there exists n 0 N such that ϕ n 0 ( t ) < η .

Definition 2.4 [4] [5] [6] Let X be an nonempty set and let d : X × X × X R be a map satisfying the following conditions:

1) For every pair of distinct points x , y X , there exists a point z X such that d ( x , y , z ) 0 .

2) If at least two of three points x , y , z are the same, then d ( x , y , z ) = 0 ,

3) The symmetry:

d ( x , y , z ) = d ( x , z , y ) = d ( y , x , z ) = d ( y , z , x ) = d ( z , x , y ) = d ( z , x , y ) for all x , y , z X .

4)The rectangle inequality:

d ( x , y , z ) d ( x , y , a ) + d ( y , z , a ) + d ( z , x , a ) for all x , y , z , a X .

Then d is called a 2 metric on X and ( X , d ) is called a 2 metric space.

Definition 2.5 [7] [8] Let X be a nonempty set and s 1 be 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:

1) d ( x , y ) = 0 if and only if x = y .

2) d ( x , y ) = d ( y , x ) .

3) d ( x , y ) s [ d ( x , y ) + d ( y , z ) ] .

In this case, the pair ( X , d ) is called a b metric space.

Definition 2.6 [9] Let X be a nonempty set, s 1 be a real number and let d : X × X × X R be a map satisfying the following conditions:

1) For every pair of distinct points x , y X , there exists a point z X such that d ( x , y , z ) 0 .

2) If at least two of three points x , y , z are the same, then d ( x , y , z ) = 0 ,

3) The symmetry:

d ( x , y , z ) = d ( x , z , y ) = d ( y , x , z ) = d ( y , z , x ) = d ( z , x , y ) = d ( z , x , y ) for all x , y , z X .

4) The rectangle inequality:

d ( x , y , z ) s [ d ( x , y , a ) + d ( y , z , a ) + d ( z , x , a ) ] , for all x , y , z , a X .

Then d is called a b2 metric on X and ( X , d ) is called a b2 metric space with parameter s. Obviously, for s = 1 , b2 metric reduces to 2-metric.

Definition 2.7 [9] Let { x n } be a sequence in a b2 metric space ( X , d ) .

1) A sequence { x n } is said to be b2-convergent to x X , written as lim n x n = x , if all a X lim n d ( x n , x , a ) = 0 .

2) { x n } is Cauchy sequence if and only if d ( x n , x m , a ) 0 , when n , m . for all a X .

3) ( X , d ) is said to be complete if every b2-Cauchy sequence is a b2-convergent sequence.

Definition 2.8 [9] Let ( X , d ) and ( X , d ) be two b2-metric spaces and let f : X X be a mapping. Then f is said to be b2-continuous,at a point z X if for a given ε > 0 , there exists δ > 0 such that x X and d ( z , x , a ) < δ for all a X imply that d ( f z , f x , a ) < ε . The mapping f is b2-continuous on X if it is b2-continuous at all z X .

Definition 2.9 [9] Let ( X , d ) and ( X , d ) be two b2-metric spaces. Then a mapping f : X X is b2-continuous at a point x X if and only if it is b2-sequentially continuous at x; that is, whenever { x n } is b2-convergent to x, { f x n } is b2-convergent to f ( x ) .

3. Main Results

In this section, we give and prove a generalization of the Meir-Keeler fixed point theorem [2].

Theorem 3.1. Let ( X , d ) be a complete b2-metric space and let f be a mapping on X, for each ε > 0 , there exists δ ( s ε , ( 2 s 1 ) ε ) such that

(a) 1 2 s d ( x , f x , a ) < d ( x , y , a ) and d ( x , y , a ) < ε + δ imply d ( f x , f y , a ) ε

(b) 1 2 s d ( x , f x , a ) < d ( x , y , a ) implies d ( f x , f y , a ) < d ( x , y , a ) for all x , y X . Then there exists a unique fixed point z of f. Moreover lim n f n x = z for all x X .

Proof If f x x , then we can easily get that d ( x , f x , a ) < 2 s d ( x , f x , a ) . So, by hypothesis, d ( f x , f 2 x , a ) < d ( x , f x , a ) holds for all x X with f x x . We also get

d ( f x , f 2 x , a ) d ( x , f x , a ) for all x X (3.1)

Fix point x 0 in X and define a sequence { x n } in X by x n + 1 = f x n = f n x 0 for n N . From the above (3.1) we get d ( x n , x n + 1 , a ) d ( x n 1 , x n , a ) , so we know that { d ( x n , x n + 1 , a ) } is a decreasing sequence, and the sequence { d ( x n , x n + 1 , a ) } converges to some β 0 . We assume that β > 0 , then we know that d ( x n , x n + 1 , a ) > β for every n N , then there exists δ such that (a) is true with ε = β , for the definition of β , there exists i N such that d ( x i , x i + 1 , a ) < β + δ , so we have d ( x i + 1 , x i + 2 , a ) β , which is a contraction. Therefore β = 0 , and that is:

lim n d ( x n , x n + 1 , a ) = 0 .

Now we show that d ( x i , x j , x k ) = 0 .

From part 2 of Definition 2.6, the equation d ( x m , x m , x m 1 ) = 0 is obtained. Since { d ( x n , x n + 1 , a ) } is decreasing, if d ( x n 1 , x n , a ) = 0 , then d ( x n , x n + 1 , a ) = 0 , then it is easy to get

d ( x n , x n + 1 , x m ) = 0 , for all n + 1 m . (3.2)

For 0 n + 1 < m , we get m 1 n + 1 and that is m 2 n , from (3.2)

d ( x m 1 , x m , x n + 1 ) = d ( x m 1 , x m , x n ) = 0 , (3.3)

From (3.2) and triangular inequality,

And since, and from the inequality above,

, for all. (3.4)

Now for all, the condition of is considered here, from the above equation

(3.5)

From (3.5) and triangular inequality, therefore

In conclusion, the result below is true

, for all. (3.6)

Now we fix, then there exists such that (a) is true. Let such that

, for all with.(3.7)

Now we will show that

for (3.8)

By induction, when, it is true for (3.8). We assume that (3.8) holds for some.

In one case, we have

From (3.6) and (3.7) we have

(3.9)

In other case, where, since

We get and then we have

(3.10)

So for (3.9) and (3.10), (3.8) is true for every. Therefore we have

, for all. This shows that is a Cauchy sequence.

Since X is complete, there exists a point such that sequence converges to it. From the following two respectively cases, we will show that this point is a fixed point for f.

Case one: There exists such that.

Case two:, for all.

In the first case, we know that for. Since as, then we get for. This prove that.

In the second case, we know that, for all, so we get sequence is strictly decreasing. If we assume that

and

for some. For the first inequality of the above assumption, we choose, then we have

(3.11)

Then we have

This is a contraction. So we get either

or for all. Since as, the above inequality prove that there exists a sub sequence of sequence, which converges to fz. This shows that z is a fixed point of f. Next we prove that z is the unique fixed point of f. Suppose that z and y are two different fixed point of f, from the assumption of this theorem, we get

from the above inequality we have

This is a contraction. Hence z is a unique fixed point of f. £

In this section, we prove a fixed point theory for the cyclic weaker Meir-Keeler function in b2-metric space. Now we give some comments as follows:

is a set, where is a weaker Meir-Keeler function and satisfying the following conditions:

()for, and;

() For all, is decreasing;

() For, if, then.

, where is a non-increasing and continuous function with for all and.

We now introduce the following definition of cyclic weaker -contraction mapping in b2-metric space:

Definition 3.2 Let be a b2-metric space, are all nonempty subsets of. A mapping is said to be cyclic weaker -contraction in b2-metric space if satisfying the following condition:

1) with respect of f, it is a cyclic representation of X.

2), for any, such that

, where, and.

Theorem 3.3 Let be a b2-metric space, are all nonempty subsets of. Let be cyclic weaker -contraction in b2-metric space, then f has a unique fixed point in.

Proof Let be an arbitrary point in X and we define a sequence by, for all, if there exists some such that then. Thus is a fixed point of f. Suppose that for all, we know that there exists such that and for any. Since be cyclic weaker -contraction, we get

Since sequence is decreasing for all, and this sequence must converge to some. We get by the following assumption.

First we assume that, since is defined as a weaker Meir-Keeler function, there exists such that for, there exists such that, from , we know that there exists such that, for all. Thus we get a conclusion, which is a contraction. Thus, and that is,.

Now we prove that is a Cauchy sequence.

Suppose to the contrary, that is, is not a Cauchy sequence. Then there exists for which we can find two sub sequences and such that and

and (3.12)

From the part 4 of Definition 3.6 and (3.6), we get

Taking, from (3.6) and (3.12) we have

(3.13)

Now by using the condition that f is a cyclic weaker -contraction, we get

Letting and using the condition of, we get

(3.14)

From (3.13) and (3.14), which is a contraction. Therefore is a Cauchy sequence in X.

Since X is a complete set, there exists a point such that,. For is a cyclic representation of X respect to f, thus in each for, the sequence has infinite term. A sub sequence of, we take this sub sequence and it also all converge to z, for all. Since

From the above inequality, letting, we get, so.

Now we prove the fixed point is unique for f. Suppose there exists another fixed point y, since f gets the cyclic character, we have. Since f is a cyclic weaker -contraction, we get

then we get

, that is, we get the result of the uniqueness of point z. £

NOTES

*Corresponding author.

Conflicts of Interest

The authors declare no conflicts of interest regarding the publication of this paper.

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