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].

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 ) .

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 ′ . □

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

Wu, C., Cui, J.X. and Zhong, L.N. (2019) Suzuki-Type Fixed Point Theorem in b₂-Metric Spaces. Open Access Library Journal, 6: e5974. https://doi.org/10.4236/oalib.1105974