Fixed point and coupled fixed point theorems on b-metric-like spaces

There exist many generalizations of the concept of metric spaces in the literature. In particular, Matthews [1] introduced the notion of partial metric space and proved that the Banach contraction mapping theorem can be generalized to the partial metric context for applications in program verification. After that, fixed point results in partial metric spaces have been studied by many authors [1, 2]. The concept of b-metric space was introduced and studied by Bakhtin [3] and Czerwik [4]. Since then several papers have dealt with fixed point theory for single-valued and multi-valued operators in b-metric spaces (see [5‐8] and references therein). Recently, Amini-Harandi [9, 10] introduced the notion of metric-like space, which is an interesting generalization of partial metric space and dislocated metric space [11‐13]. In this paper, we first introduce a new generalization of metric-like space and partial metric space which is called a b-metric-like space. Then, we give some fixed point results in such spaces. Our fixed point theorems, even in the case of metric-like spaces and partial metric spaces, generalize and improve some well-known results in the literature. Moreover, some examples and an application to integral equations are provided to illustrate the usability of the obtained results.

Matthews [1] introduced the concept of a partial metric space as follows.

Definition 2.1 A mapping p : X × X → R +, where X is a nonempty set, is said to be a partial metric on X if for any x , y , z ∈ X the following four conditions hold true:

(P1) x = y if and only if p ( x , x ) = p ( y , y ) = p ( x , y );

(P2) p ( x , x ) ≤ p ( x , y );

(P3) p ( x , y ) = p ( y , x );

(P4) p ( x , z ) ≤ p ( x , y ) + p ( y , z ) − p ( y , y ).

The pair ( X , p ) is then called a partial metric space.

Definition 2.2 [9]

A mapping σ : X × X → R +, where X is a nonempty set, is said to be a metric-like on X if for any x , y , z ∈ X the following three conditions hold true:

(σ 1) σ ( x , y ) = 0 ⇒ x = y;

(σ 2) σ ( x , y ) = σ ( y , x );

(σ 3) σ ( x , z ) ≤ σ ( x , y ) + σ ( y , z ).

The pair ( X , σ ) is then called a metric-like space. A metric-like on X satisfies all of the conditions of a metric except that σ ( x , x ) may be positive for x ∈ X.

Every partial metric space is a metric-like space but not conversely in general (see [9, 10]).

The concept of b-metric space was introduced by Czerwik in [4]. Since then, several papers have been published on the fixed point theory of various classes of single-valued and multi-valued operators in b-metric spaces (see, e.g., [6‐8]).

Definition 2.3 A b-metric on a nonempty set X is a function D : X × X → [ 0 , + ∞ ) such that for all x , y , z ∈ X and a constant K ≥ 1 the following three conditions hold true:

( 1) if D ( x , y ) = 0 ⇔ x = y;

( 2) D ( x , y ) = D ( y , x );

( 3) D ( x , y ) ≤ K ( D ( x , z ) + D ( z , y ) ).

The pair ( X , D ) is called a b-metric space.

Definition 2.4 A b-metric-like on a nonempty set X is a function D : X × X → [ 0 , + ∞ ) such that for all x , y , z ∈ X and a constant K ≥ 1 the following three conditions hold true:

( 1) if D ( x , y ) = 0 ⇒ x = y;

( 2) D ( x , y ) = D ( y , x );

( 3) D ( x , y ) ≤ K ( D ( x , z ) + D ( z , y ) ).

The pair ( X , D ) is called a b-metric-like space.

Example 2.5 Let X = [ 0 , ∞ ). Define the function D : X 2 → [ 0 , ∞ ) by D ( x , y ) = ( x + y ) 2. Then ( X , D ) is a b-metric-like space with constant K = 2. Clearly, ( X , D ) is not a b-metric or metric-like space. Indeed, for all x , y , z ∈ X,

and so ( 3) holds. Clearly, ( 1) and ( 2) hold.

Similarly, we have the following example.

Example 2.6 Let X = [ 0 , ∞ ). Define the function D : X 2 → [ 0 , ∞ ) by D ( x , y ) = ( max { x , y } ) 2. Then ( X , D ) is a b-metric-like space with constant K = 2. Clearly, ( X , D ) is not a b-metric or metric-like space.

Example 2.7 Let C b ( X ) = { f : X → R : sup x ∈ X | f ( x ) | < + ∞ }. The function D : X × X → R +, defined by

is a b-metric-like with constant K = 4 3, and so ( X , D , 4 3 ) is a b-metric-like space.

For this, note that if a, b are two nonnegative real numbers, then

This implies that

Let ( X , D ) be a b-metric-like space. Let x ∈ X and r > 0, then the set

is called an open ball with center at x and radius r > 0.

Now we have the following definitions.

Definition 2.8 Let ( X , D ) be a b-metric-like space, and let { x n } be a sequence of points of X. A point x ∈ X is said to be the limit of the sequence { x n } if lim n → + ∞ D ( x , x n ) = D ( x , x ), and we say that the sequence { x n } is convergent to x and denote it by x n ⟶ x as n → ∞.

Definition 2.9 Let ( X , D ) be a b-metric-like space.

(S1) A sequence { x n } is called Cauchy if and only if lim m , n → ∞ D ( x n , x m ) exists and is finite.

(S2) A b-metric-like space ( X , D ) is said to be complete if and only if every Cauchy sequence { x n } in X converges to x ∈ X so that

Proposition 2.10Let( X , D , K )be ab-metric-like space, and let{ x n }be a sequence inXsuch thatlim n → ∞ D ( x n , x ) = 0. Then

Assume that there exists a y ∈ X such that lim n → ∞ D ( x n , y ) = 0, then

Hence, from ( 1) we have y = x.

(B)

From (D3) we have

and so

 □

Definition 2.11 Let ( X , D ) be a b-metric-like space, and let U be a subset of X. We say U is an open subset of X if for all x ∈ U there exists r > 0 such that B ( x , r ) ⊆ U. Also, V ⊆ X is a closed subset of X if X ∖ V is an open subset of X.

Proposition 2.12Let( X , D , K )be ab-metric-like space, and letVbe a subset ofX. ThenVis closed if and only if for any sequence{ x n }inV, which converges tox, we havex ∈ V.

Proof At first, we suppose that V is closed. Let x ∉ V. By the above definition, X ∖ V is an open set. Then there is an r > 0 such that B ( x , r ) ⊆ X ∖ V. On the other hand, since x n → x as n → ∞, then

Hence, there exists n 0 ∈ N such that for all n ≥ n 0 we have

That is, for all n ≥ n 0, { x n } ⊆ B ( x , r ) ⊆ X ∖ V, which is a contradiction. Since for all n ∈ N, { x n } ⊆ V. Conversely, suppose that for any sequence { x n } in V which converges to x, we have x ∈ V. Let y ∉ V. Let us prove that there exists r 0 > 0 such that B ( y , r 0 ) ∩ V = ∅. Assume to the contrary that for all r > 0, we have B ( y , r ) ∩ V ≠ ∅. Then, for all n ∈ N, chose x n ∈ B ( y , 1 / n ) ∩ V ≠ ∅. Therefore, | D ( x n , y ) − D ( y , y ) | < 1 / n for all n ∈ N. Hence, x n → y as n → ∞. Our assumption on V implies y ∈ V, which is a contradiction. Then, for all y ∉ V, there exists r 0 > 0 such that B ( y , r 0 ) ∩ V = ∅. That is, V is closed. □

Lemma 2.13Let( X , D , K )be ab-metric-like space, and let{ x k } k = 0 n ⊂ X. Then

From Lemma 2.13, we deduce the following result.

Lemma 2.14Let{ y n }be a sequence in ab-metric-like space( X , D , K )such that

for someλ, 0 < λ < 1 / K, and eachn ∈ N. Thenlim m , n → ∞ D ( y m , y n ) = 0.

Let ( X , D , K ) be a b-metric-like space. Define D s : X 2 → [ 0 , ∞ ) by

Clearly, D s ( x , x ) = 0 for all x ∈ X.

The study of expansive mappings is a very interesting research area in fixed point theory (see, e.g., [14‐21]). In this section we prove some new fixed point results on expansive mappings in the setting of a b-metric-like space. Our results generalize and extend some old and recent fixed point results in the literature.

Theorem 3.1Let( X , D , K )be a completeb-metric-like space. Assume that the mappingT : X → Xis onto and satisfies

for allx , y ∈ X, whereR > K, L ≥ 0. ThenThas a fixed point.

Proof Let x 0 ∈ X, since T is onto, then there exists x 1 ∈ X such that x 0 = T x 1. By continuing this process, we get x n = T x n + 1 for all n ∈ N ∪ { 0 }. In case x n 0 = x n 0 + 1 for some n 0 ∈ N ∪ { 0 }, then it is clear that x n 0 is a fixed point of T. Now assume that x n ≠ x n + 1 for all n. From (3.1) with x = x n and y = x n + 1 we get

which implies

and so

Then by Lemma 2.14 we have lim m , n → ∞ D ( x n , x m ) = 0. Now, since lim m , n → ∞ D ( x n , x m ) = 0 exists (and is finite), so { x n } is a Cauchy sequence. Since ( X , D , K ) is a complete b-metric-like space, the sequence { x n } in X converges to z ∈ X so that

Since T is onto, there exists w ∈ X such that z = T w. From (3.1) we have

Taking limit as n → ∞ in the above inequality, we get

which implies lim n → ∞ D ( x n + 1 , w ) = 0. By Proposition 2.10 (A), we get z = w. That is, z = T z. □

If in Theorem 3.1 we take L = 0, then we deduce the following corollary.

Corollary 3.2Let( X , D , K )be a completeb-metric-like space. Assume that the mappingT : X → Xis onto and satisfies

for allx , y ∈ X, whereR > K. ThenThas a fixed point.

Example 3.3 Let X = [ 0 , ∞ ) and let a b-metric-like D : X × X → R + be defined by

Clearly, ( X , D , 2 ) is a complete b-metric-like space. Let T : X → X be defined by

★ Let x , y ∈ [ 0 , 1 ), then

★ Let x , y ∈ [ 1 , 2 ), then

★ Let x , y ∈ [ 2 , ∞ ), then

★ Let x ∈ [ 0 , 1 ) and y ∈ [ 1 , 2 ), then

★ Let x ∈ [ 0 , 1 ) and y ∈ [ 2 , ∞ ), then

★ Let x ∈ [ 1 , 2 ) and y ∈ [ 2 , ∞ ), then

That is, D ( T x , T y ) ≥ R D ( x , y ) for all x , y ∈ X, where R = 3 > 2 = K. The conditions of Corollary 3.2 are satisfied and T has a fixed point x = 0.

Let Ψ B L denote the class of those functions B : ( 0 , ∞ ) → ( L 2 , ∞ ) which satisfy the condition B ( t n ) → ( L 2 ) + ⇒ t n → 0, where L > 0.

Theorem 3.4Let( X , D , K )be a completeb-metric-like space. Assume that the mappingT : X → Xis onto and satisfies

for allx , y ∈ X, whereB ∈ Ψ B K. ThenThas a fixed point.

Proof Let x 0 ∈ X, since T is onto, so there exists x 1 ∈ X such that x 0 = T x 1. By continuing this process, we get x n = T x n + 1 for all n ∈ N ∪ { 0 }. In case x n 0 = x n 0 + 1 for some n 0 ∈ N ∪ { 0 }, then it is clear that x n 0 is a fixed point of T. Now assume that x n ≠ x n + 1 for all n. From (3.2) with x = x n and y = x n + 1, we get

Then the sequence { D ( x n , x n + 1 ) } is a decreasing sequence in R + and so there exists s ≥ 0 such that lim n → ∞ D ( x n , x n + 1 ) = s. Let us prove that s = 0. Suppose to the contrary that s > 0. By (3.3) we can deduce

By taking limit as n → ∞ in the above inequality, we have lim n → ∞ B ( D ( x n , x n + 1 ) ) = K 2. Hence,

which is a contradiction. That is, s = 0. We shall show that lim sup m , n → ∞ D ( x n , x m ) = 0. Suppose to the contrary that

By (3.2) we have

That is,

Then by ( 3) we get

Therefore,

By taking limit as m , n → ∞ in the above inequality, since lim sup m , n → ∞ D ( x n , x m ) > 0 and s = lim n → ∞ D ( x n , x n + 1 ) = 0, then we obtain

which implies

and so

which is a contradiction. Hence, lim sup m , n → ∞ D ( x n , x m ) = 0. Now, since lim m , n → ∞ D ( x n , x m ) = 0 exists (and is finite), so { x n } is a Cauchy sequence. Since ( X , D , K ) is a complete b-metric-like space, the sequence { x n } in X converges to z ∈ X so that

As T is onto, so there exists w ∈ X such that z = T w. Let us prove that w = z. Suppose to the contrary that z ≠ w. Then by (3.2) we have

By taking limit as n → ∞ in the above inequality and applying Proposition 2.10(B), we have

and hence

which is a contradiction. Indeed, lim n → ∞ B ( D ( x n + 1 , z ) ) ≥ K 2. Since B ( t ) > K 2 for all t ∈ [ 0 , ∞ ), therefore z = w. That is, z = T w = T z. □

Example 3.5 Let X = [ 0 , ∞ ) and D : X × X → R + be defined by

Clearly, ( X , D , 2 ) is a complete b-metric-like space. Let T : X → X be defined by

Also define B : ( 0 , ∞ ) → ( 4 , ∞ ) by B ( t ) = 4 ( 1 + t ). At first we show that T is an onto mapping. For a given a ∈ X, we choose x 0 = 1 2 4 + a 2 − 2. Then

So, T is an onto mapping. Without loss of generality, we assume that x ≤ y. Now, since

so

equivalently,

and hence

That is,

The conditions of Theorem 3.4 hold and T has a fixed point (here, x = 0 is a fixed point of T).

Note that b-metric-like spaces are a proper extension of partial metric, metric-like and b-metric spaces. Hence, we can deduce the following corollaries in the settings of partial metric, metric-like and b-metric spaces, respectively.

Corollary 3.6Let( X , p )be a complete partial metric space. Assume that the mappingT : X → Xis onto and satisfies

for allx , y ∈ X, whereB ∈ Ψ B 1. ThenThas a fixed point.

Corollary 3.7Let( X , σ )be a complete metric-like space. Assume that the mappingT : X → Xis onto and satisfies

for allx , y ∈ X, whereB ∈ Ψ B 1. ThenThas a fixed point.

Corollary 3.8Let( X , d , K )be a completeb-metric space. Assume that the mappingT : X → Xis onto and satisfies

for allx , y ∈ X, whereB ∈ Ψ B K. ThenThas a fixed point.

In this section we prove certain new fixed point theorems in partially ordered b-metric-like spaces which generalize and extend corresponding results of Amini-Harandi [9, 10] and many others (see [22]).

Let Ψ L L denote the class of those functions L : ( 0 , ∞ ) → ( 0 , 1 L 2 ) which satisfy the condition L ( t n ) → ( 1 L 2 ) + ⇒ t n → 0, where L > 0.

Theorem 4.1Let( X , D , K , ⪯ )be a partially ordered completeb-metric-like space, and letT : X → Xbe a non-decreasing mapping such that

for allx , y ∈ Xwithx ⪯ y, whereL ∈ Ψ L K, J : [ 0 , ∞ ) → [ 0 , ∞ )is a bounded function and

and

Also, suppose that the following assertions hold:

thenThas a fixed point.

Proof Let x 0 ⪯ T x 0. If x 0 = T x 0, then the result is proved. Hence we suppose that x 0 ≺ f x 0. Define a sequence { x n } by x n = T n x 0 = T x n − 1 for all n ∈ N. Since T is non-decreasing and x 0 ≺ T x 0, then

and hence { x n } is a non-decreasing sequence. If x n = x n + 1 = T x n for some n ∈ N, then the result is proved as x n is a fixed point of T. In what follows we will suppose that x n ≠ x n + 1 for all n ∈ N. From (4.1) and (4.2) we have

where

Then

On the other hand, from ( 3) we have

and

Then

and hence

That is,

Now by (4.3) we get

If max { D ( x n − 1 , x n ) , D ( x n , x n + 1 ) } = D ( x n , x n + 1 ), then

which is a contradiction. Hence,

and so the sequence { D ( x n , x n + 1 ) } is a decreasing sequence in R +. Then there exists s ≥ 0 such that lim n → ∞ D ( x n , x n + 1 ) = s. By (4.4) we can write

Taking limit as n → ∞ in the above inequality, we get

and so s = lim n → ∞ D ( x n − 1 , x n ) = 0. Now we want to show that lim sup m , n → ∞ D ( x n , x m ) = 0. Suppose to the contrary that

At first,