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 OALibJ  Vol.6 No.12 , December 2019
Suzuki-Type Fixed Point Theorem in b2-Metric Spaces
Abstract: In this paper, we establish a fixed point theorem for two mappings under a contraction condition in b2-metric space, and this theorem is related to a Suzuki-type of contraction.

1. Introduction

Banach [1] proved a principle, and this famous Banach contraction principle has many generalizations, see [2] - [7], and in 2008, Suzuki [8] established one of those generalizations, and this generalization is called Suzuki principle.

The aim of this paper is to prove a fixed point result generalized from the above mentioned principle in b2-metric space [9].

2. Preliminaries

Before giving our results, these definitions and results as follows will be needed to present.

Definition 2.1 [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 .

1) 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 2.2 [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 2.3 [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 2.4 [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 ) .

Lemma 2.5 [9] Let ( X , d ) be a b2-metric space and suppose that { x n } and { y n } are b2-convergent to x and y, respectively. Then we have

1 s 2 d ( x , y , a ) lim n inf d ( x n , y n , a ) lim n sup d ( x n , y n , a ) s 2 d ( x , y , a ) , for all a in X. In particular, if y n = y is a constant, then

1 s d ( x , y , a ) lim n inf d ( x n , y , a ) lim n sup d ( x n , y , a ) s d ( x , y , a ) , for all a in X.

Lemma 2.6 [10] Let ( X , d ) be a b2 metric space with s 1 and let { x n } n = 0 be a sequence in X such that

d ( x n , x n + 1 , a ) λ d ( x n 1 , x n , a ) , (2.1)

for all n N and all a X , where λ [ 0 , 1 s ) . Then { x n } is a b2-Cauchy sequence in ( X , d ) .

3. Main Results

Theorem 3.1. Let ( X , d ) be a complete b2-metric space. Let f , g : X X be two self-maps and ϕ : [ 0 , 1 ) ( 1 2 , 1 ] be defined as follows

ϕ ( ρ ) = { 1 , 0 ρ 5 1 2 1 ρ ρ 2 , 5 1 2 ρ 1 2 1 1 + ρ , 1 2 ρ < 1 (3.1)

Assume there exists ρ [ 0 , 1 ) such that for every x , y X , the following condition is satisfied

1 s ϕ ( ρ ) min { d ( x , f x , a ) , d ( f x , f y , a ) } d ( x , y , a ) max { d ( g x , g y , a ) , d ( g x , f y , a ) , d ( f x , f y , a ) , d ( g y , f x , a ) } ρ s 2 d ( x , y , a ) . (3.2)

Then f , g have a unique common fixed point z X .

Proof in (3.2), we take y = f x

1 s ϕ ( ρ ) min { d ( x , f x , a ) , d ( x , g x , a ) } d ( x , g x , a ) max { d ( g x , g 2 x , a ) , d ( g x , f g x , a ) , d ( f x , f g x , a ) , d ( g 2 x , f x , a ) } ρ s 2 d ( x , g x , a ) , for x X .(3.3)

therefore,

d ( g x , f g x , a ) ρ s 2 d ( x , g x , a ) . (3.4)

Now we take y = f x in (3.2)

1 s ϕ ( ρ ) min { d ( x , f x , a ) , d ( x , g y , a ) } d ( x , f y , a ) max { d ( g x , g f y , a ) , d ( g x , f 2 y , a ) , d ( f x , f 2 x , a ) , d ( g f x , f x , a ) } ρ s 2 d ( x , f x , a ) , for all x X .(3.5)

therefore,

d ( f x , f 2 x , a ) ρ s 2 d ( x , f x , a ) , (3.6)

and

d ( g f x , f x , a ) ρ s 2 d ( x , f x , a ) . (3.7)

Given an arbitrary point x 0 in X thenby x 2 n + 1 = g x 2 n and x 2 n + 1 = f x 2 n + 1 we construct a sequence { x n } , for n N .

From (3.4), we get

d ( x 2 n + 1 , x 2 n + 2 , a ) = d ( g x 2 n , f g x 2 n , a ) ρ s 2 d ( x 2 n , g x 2 n , a ) = ρ s 2 d ( x 2 n , x 2 n + 1 , a ) . (3.8)

From (3.7) and (3.8) we get

d ( x 2 n + 1 , x 2 n , a ) = d ( g f x 2 n 1 , f x 2 n 1 , a ) ρ s 2 d ( x 2 n , f x 2 n 1 , a ) = ρ s 2 d ( x 2 n 1 , x 2 n , a ) ,

that is,

d ( x n + 1 , x n , a ) ρ s 2 d ( x n , x n 1 , a ) , since ρ s 2 [ 0 , 1 ) , by Lemma 2.6, we get { x n } is a Cauchy sequence.

Since X is complete, there exists z in X, such that lim n x n = z , that is lim n g x 2 n = lim n x 2 n + 1 = z , and lim n f x 2 n + 1 = lim n x 2 n + 2 = z .

Now let us give that

d ( f x , z , a ) ρ d ( x , z , a ) , for every x z . For { d ( x 2 n , g x 2 n , a ) } is convergent to 0, and by Lemma 2.5, we get

1 s d ( x , z , a ) lim n sup d ( x 2 n , x , a ) , thus we have lim n sup d ( x 2 n , x , a ) > 0 , thus from the above relation, there exists a point x 2 n k in X such that

1 s ϕ ( ρ ) min { d ( x 2 n k , g x 2 n k , a ) , d ( x 2 n k , f x 2 n k , a ) } d ( x 2 n k , x , a ) .

For such x 2 n k , (3.2) implies that

d ( g x 2 n k , f x , a ) max { d ( g x 2 n k , g x , a ) , d ( g x 2 n k , f x , a ) , d ( f x 2 n k , f x , a ) , d ( g x , f x 2 n k , a ) } ρ s 2 d ( x 2 n k , x , a ) ,

therefore by Lemma 3.5,

1 s d ( f x , z , a ) lim sup n d ( g x 2 n k , f x , a ) ρ s 2 lim sup n d ( x 2 n k , x , a ) ρ s d ( x , z , a ) ,

therefore we get

d ( f x , z , a ) ρ d ( x , z , a ) , for each x z . (3.9)

Now we show that for each n N ,

d ( f n z , z , a ) d ( f z , z , a ) , (3.10)

It is obvious that the above inequality is true for n = 1 , assume that the relation holds for some m N . We get (3.10) is true when we have f m z = f z if f m z = z , then if f m z z , we get the following relation from (3.9) and induction hypothesis, and that is

d ( z , f m + 1 z , a ) ρ d ( z , f m z , a ) ρ 2 d ( z , f m 1 z , a ) ρ m + 1 d ( z , f z , a ) ρ d ( f z , z , a ) d ( f z , z , a ) ,

then (3.10) is proved.

Now we consider the following two possible cases in order to prove that f has a fixed point z in X, and that is f z = z .

Case 1 0 ρ < 1 2 , therefore, ϕ ( ρ ) 1 ρ ρ 2 . First, we prove the following relation

d ( f n z , f z , a ) ρ s d ( f z , z , a ) , for n N . (3.11)

When n = 1 it is obvious, and it follows from (3.6) when n = 2 , from (3.10) and take a = f z we have

d ( f n z , z , f z ) d ( f z , z , f z ) = 0 , then we get d ( f n z , f z , z ) = 0 .

Now suppose that (3.11) holds for some n > 2 ,

d ( f z , z , a ) s ( d ( z , f n z , a ) + d ( f n z , f z , a ) + d ( f n z , f z , z ) ) s d ( z , f n z , a ) + s d ( z , f z , a ) ,

Therefore, we get

( 1 ρ ) d ( z , f z , a ) s d ( z , f n z , a ) , that is d ( z , f z , a ) s 1 ρ s d ( z , f n z , a ) , (3.11.1)

then by taking x = f n 1 z in (3.6)

d ( f n z , f n + 1 z , a ) ρ s 2 d ( f n 1 z , f n z , a ) ρ n s 2 n d ( z , f z , a ) , (3.11.2)

using the above two relations, (3.11.1) and (3.11.2) we have

1 s ϕ ( ρ ) min { d ( g f n z , f n z , a ) , d ( f n z , f n + 1 z , a ) } 1 ρ s ρ 2 d ( f n z , f n + 1 z , a ) 1 ρ s ρ n d ( f n z , f n + 1 z , a ) 1 ρ s ρ n ρ n s 2 n d ( z , f z , a ) = 1 ρ s 2 n + 1 d ( z , f z , a ) 1 ρ s 2 n + 1 s 1 ρ d ( z , f n z , a ) 1 s 2 n d ( z , f n z , a ) d ( z , f n z , a ) .

From (3.2) and (3.10) with x = f n z and y = z , we have

max { d ( g f n z , g z , a ) , d ( g f n z , f z , a ) , d ( f n + 1 z , f z , a ) , d ( g z , f n + 1 z , a ) } ρ s 2 d ( z , f n z , a ) ρ s 2 d ( z , f z , a ) ρ s d ( z , f z , a ) .

Therefore,

d ( f n + 1 z , f z , a ) ρ s d ( f z , z , a ) . (3.12)

So by induction we prove the relation of (3.11).

Now (3.11) and f z z show that for every n N f n z z , thus, (3.9) shows that

d ( z , f n + 1 z , a ) ρ d ( z , f n z , a ) ρ 2 d ( z , f n 1 z , a ) ρ n d ( z , f z , a ) .

Therefore lim n d ( z , f n + 1 z , a ) = 0 . Furthermore by using Lemma 2.5, we get

1 s d ( z , lim inf n f n + 1 z , a ) lim inf n d ( z , f n + 1 z , a ) = 0 ,

so

d ( z , lim inf n f n + 1 z , a ) = 0.

In the same way,

d ( z , lim sup n f n + 1 z , a ) = 0 , thus we have d ( z , lim n f n + 1 z , a ) = 0 , that is f n + 1 z z , and by using Lemma 2.5 in (3.12), we get

1 s d ( z , f z , a ) lim sup n d ( f n + 1 z , f z , a ) ρ s d ( z , f z , a ) , which claims that d ( z , f z , a ) = 0 , and that is a contraction.

Case 2. 1 2 ρ < 1 , and that is when ϕ ( ρ ) = 1 1 + ρ . We now prove that we can find a subsequence { x n k } of { x n } such that

1 s ( 1 + ρ ) min { d ( x n k , g x n k , a ) , d ( x n k , f x n k , a ) } d ( x n k , z , a ) , for k N . (3.13)

The contraries of the above relation are as follows

1 s ( 1 + ρ ) d ( x n , f x n , a ) 1 s ( 1 + ρ ) min { d ( x n , g x n , a ) , d ( x n , f x n , a ) } > d ( x n , z , a ) ,

and

1 s ( 1 + ρ ) d ( x n , f x n , a ) 1 s ( 1 + ρ ) min { d ( x n , g x n , a ) , d ( x n , f x n , a ) } > d ( x n , z , a ) ,

for n N . If n is even we have

1 s ( 1 + ρ ) d ( x 2 n , g x 2 n , a ) 1 s ( 1 + ρ ) min { d ( x 2 n , g x 2 n , a ) , d ( x 2 n , f x 2 n , a ) } > d ( x 2 n , z , a ) ,

if n is odd then we get

1 s ( 1 + ρ ) d ( x 2 n + 1 , f x 2 n + 1 , a ) 1 s ( 1 + ρ ) min { d ( x 2 n + 1 , g x 2 n + 1 , a ) , d ( x 2 n + 1 , f x 2 n + 1 , a ) } > d ( x 2 n + 1 , z , a ) ,

for n N . By (3.8) we have

d ( x 2 n , x 2 n + 1 , a ) s ( d ( x 2 n , z , a ) + d ( x 2 n + 1 , z , a ) + d ( x 2 n , x 2 n + 1 , z ) ) < s s ( 1 + ρ ) d ( x 2 n , g x 2 n , a ) + s s ( 1 + ρ ) d ( x 2 n + 1 , f x 2 n + 1 , a ) + s s ( 1 + ρ ) d ( x 2 n , g x 2 n , x 2 n + 1 ) = 1 1 + ρ ( d ( x 2 n , x 2 n + 1 , a ) + d ( x 2 n + 1 , x 2 n + 2 , a ) + d ( x 2 n , x 2 n + 1 , x 2 n + 1 ) )

1 1 + ρ d ( x 2 n , x 2 n + 1 , a ) + ρ s 2 ( 1 + ρ ) d ( x 2 n + 1 , x 2 n , a ) 1 1 + ρ d ( x 2 n , x 2 n + 1 , a ) + ρ 1 + ρ d ( x 2 n + 1 , x 2 n , a ) = d ( x 2 n , x 2 n + 1 , a ) ,

this is impossible. Therefore, one of the following relations is true for every n N ,

1 s ϕ ( ρ ) min { d ( x 2 n , g x 2 n , a ) , d ( x 2 n , f x 2 n , a ) } d ( x 2 n , z , a ) ,

or

1 s ϕ ( ρ ) min { d ( x 2 n + 1 , g x 2 n + 1 , a ) , d ( x 2 n + 1 , f x 2 n + 1 , a ) } d ( x 2 n + 1 , z , a ) .

That means there exists a subsequence { x n k } of { x n } such that (3.13) is true for every k N . Thus (3.2) shows that

d ( g x 2 n , f z , a ) max { d ( f x 2 n , g z , a ) , d ( f z , g x 2 n , a ) , d ( f x 2 n , f z , a ) , d ( g z , f x 2 n , a ) } ρ s 2 d ( x 2 n , z , a ) .

or

d ( f x 2 n + 1 , f z , a ) max { d ( g x 2 n + 1 , g z , a ) , d ( f z , g x 2 n + 1 , a ) , d ( f x 2 n + 1 , f z , a ) , d ( g z , f x 2 n + 1 , a ) } ρ s 2 d ( x 2 n + ! , z , a ) .

From Lemma 2.5, we have

1 s d ( z , f z , a ) lim sup n d ( g x 2 n , f z , a ) ρ s 2 lim sup n d ( x 2 n , z , a ) ρ s d ( z , z , a ) = 0 ,

or

1 s d ( z , f z , a ) lim sup n d ( f x 2 n + 1 , f z , a ) ρ s 2 lim sup n d ( x 2 n + 1 , z , a ) ρ s d ( z , z , a ) = 0 ,

Therefore d ( z , f z , a ) 0 , which is impossible unless f z = z . hence z in X is a fixed point of f. From the process of the above proof, we know f z = z , then by

0 = 1 s ϕ ( ρ ) min { d ( z , f z , a ) , d ( z , g z , a ) } d ( z , f z , a ) ,

it implies

d ( g z , z , a ) max { d ( g z , g f z , a ) , d ( g z , f 2 z , a ) , d ( f z , f 2 z , a ) , d ( g f z , f z , a ) } ρ s 2 d ( f z , z , a ) = 0 ,

this proves that g z = z . By (3.2) we can prove the uniqueness of the common fixed point z,

1 s ϕ ( ρ ) min { d ( z , f z , a ) , d ( z , g z , a ) } d ( z , z , a ) , so (3.2) shows that

d ( z , z , a ) = max { d ( g z , g z , a ) , d ( f z , f z , a ) , d ( g z , f z , a ) , d ( g z , f z , a ) } ρ s 2 d ( z , z , a ) ,

which is impossible unless z = z . □

NOTES

*Corresponding author.

Cite this paper: Wu, C. , Cui, J. and Zhong, L. (2019) Suzuki-Type Fixed Point Theorem in b2-Metric Spaces. Open Access Library Journal, 6, 1-9. doi: 10.4236/oalib.1105974.
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