A new approach in the context of ordered incomplete partial b-metric spaces

Definition 1.1

[2] Let X be a nonempty set and b : X × X → [ 0 , ∞ ) be a function verifying:

1. b ( ρ , ξ ) = 0 iff ρ = ξ ;

2. b ( ρ , ξ ) = b ( ξ , ρ ) ;

3. There exists s ≥ 1 such that b ( ρ , ξ ) ≤ s [ b ( ρ , τ ) + b ( τ , ξ ) ] ,

   for all ρ , ξ , τ ∈ X . Then, b is called a b-metric on X and ( X , b ) is called a b-metric space with a coefficient s ≥ 1 .

Definition 1.2

[3] Let X be a nonempty set and p : X × X → [ 0 , ∞ ) be a function such that:

1. p ( ρ , ρ ) = p ( ρ , ξ ) = p ( ξ , ξ ) iff ρ = ξ ;

2. p ( ρ , ρ ) ≤ p ( ρ , ξ ) ;

3. p ( ρ , ξ ) = p ( ξ , ρ ) ;

4. p ( ρ , ξ ) ≤ p ( ρ , τ ) + p ( τ , ξ ) − p ( τ , τ ) ,

   for all ρ , ξ , τ ∈ X . Then, p is called a partial metric on X and ( X , p ) is called a partial metric space.

Definition 1.3

[7,36] Let X be a nonempty set and d : X × X → [ 0 , ∞ ) be a function such that:

1. d ( ρ , ρ ) = d ( ρ , ξ ) = d ( ξ , ξ ) iff ρ = ξ ;

2. d ( ρ , ρ ) ≤ d ( ρ , ξ ) ;

3. d ( ρ , ξ ) = d ( ξ , ρ ) ;

4. There exists s ≥ 1 such that d ( ρ , ξ ) ≤ s [ d ( ρ , τ ) + d ( τ , ξ ) ] − d ( τ , τ ) , for all ρ , ξ , τ ∈ X . Then, d is called a partial b-metric on X and ( X , d ) is called a partial b-metric space with coefficient s ≥ 1 .

Example 1.1

Set p > 1 . The function d : [ 0 , ∞ ) 2 → [ 0 , ∞ ) 2 given as

for all ρ , ξ ∈ X , is a partial b-metric (here, s = 2 p − 1 > 1 ). It is neither a partial metric nor a b-metric.

Definition 1.4

[7] Given a sequence { ζ n } in a partial b-metric space ( X , d ) .

1. { ζ n } converges to ζ ∈ X (written as lim n → ∞ ζ n = ζ ) if

   lim n → ∞ d ( ζ n , ζ ) = d ( ζ , ζ ) .

2. { ζ n } is called Cauchy in X if lim n , m → ∞ d ( ζ n , ζ m ) exists and is finite.

3. ( X , d ) is said complete if for every Cauchy sequence { ζ n } , there is ζ ∈ X so that

Here, the limit of a convergent sequence may not be unique.

Example 1.2

[7] Take d ( ρ , ξ ) = max { ρ , ξ } + a for all ρ , ξ ∈ [ 0 , ∞ ) , where a > 0 . Then, ( X , d ) is a partial b-metric space with an arbitrary coefficient s ≥ 1 . If ζ n = 1 for each n ∈ ℕ , then lim n → ∞ ζ n = η , for each η ≥ 1 .

Definition 1.5

[36] Let ( X , ≼ ) be a partially ordered set and d be a partial b-metric on X. Then, the triplet ( X , d , ≼ ) is called an ordered partial b-metric space.

Definition 1.6

[37] Let f be a self-map on a metric space ( X , d ) . For A ⊂ X , define δ ( A ) = sup { d ( α , β ) : α , β ∈ A } . For each θ ∈ X and n ≥ 1 , take

and

The metric space ( X , d ) is called f-orbitally complete if each Cauchy sequence contained in O ( θ , ∞ ) converges in X.

Definition 2.1

Let f be a self-map on a metric space ( X , d ) . For a subset A ⊆ X with θ ∈ A , state

and

We say that ( X , d ) is f-orbitally complete with respect to A, if every Cauchy sequence contained in O A ( θ , ∞ ) converges in A.

Example 2.1

Let X = ( − 1 , 2 ) be endowed with the usual metric of ℝ . Define f : X → X by f ( ζ ) = ζ 2 , for all ζ ∈ X . Let A = [ 0 , 1 ) . Then, ( X , d ) is f-orbitally complete with respect to A.

Example 2.2

Let X = ℝ be endowed with the usual metric. Define f : X → X by f ( ζ ) = ζ 2 , for all ζ ∈ X . Let A = [ 0 , 1 ) . Again ( X , d ) is f-orbitally complete with respect to A.

Definition 2.2

Let ( X , d , ≼ ) be an ordered partial b-metric space. Then, for any subset A ⊆ X , ( X , d , ≼ ) is said to be f ¯ -orbitally complete with respect to A, if every strictly increasing Cauchy sequence contained in O A ( θ , ∞ ) for some θ ∈ A converges in A and the limit of the sequence is a strict upper bound of the sequence, i.e., if { ζ n } is a strictly increasing Cauchy sequence contained in O A ( ζ , ∞ ) , then there is ζ ∈ A such that ζ n → ζ and ζ n ≺ ζ , for all n ∈ ℕ .

Example 2.3

Let X = ( − 2 , 3 ) be endowed with usual metric and ≼ be defined as the natural ordering ≤ . Consider f ( ζ ) = ζ 3 for all ζ ∈ X . Take A = ( − 1 , 0 ] . Note that ( X , d , ≼ ) is f ¯ -orbitally complete with respect to A.

Example 2.4

Let X = ( − ∞ , 1 ) be endowed with the usual metric and ≼ be the natural ordering ≤ . Take f ( ζ ) = ζ 3 for all ζ ∈ X . Let A = ( − 1 , 0 ] . Then, ( X , d , ≼ ) is f ¯ -orbitally complete with respect to A.

Example 2.5

Let X = ( 0 , 2 ] be endowed with the usual metric and ≼ be the natural ordering ≤ . Define f : X → X by

Clearly, X is not f-orbitally complete. It can be verified by taking ζ = 2 3 and the sequence { a n = f n ( ζ ) , n even } in O ( ζ , ∞ ) . But, if we take A = [ 1 , 2 ] , then X is f-orbitally complete with respect to A.

Example 2.6

Let X = ( 0 , ∞ ) be endowed with the usual metric and ≼ be the natural ordering ≤ . Define f : X → X by

Clearly, X is not f-orbitally complete. If we take A = [ 1 , ∞ ) , then X is f-orbitally complete with respect to A.

Lemma 2.1

Let ( X , d , ≼ ) be a partial b-metric space with a coefficient s ≥ 1 . For each sequence { ζ n } in X, we have

for all n , m ∈ ℕ with n < m .

Proof

By using triangular inequality, one writes

Continuing in this way, we get

□

Theorem 2.1

Let ( X , d , ≼ ) be an ordered partial b-metric space with a coefficient s ≥ 1 . Suppose that the set A = { θ ∈ X : θ ≼ f ( θ ) } is nonempty and that

for all ζ , ξ ∈ A with ζ ≺ ξ , where λ ∈ [ 0 , 1 s ) . If further, f ( A ) ⊆ A and ( X , d , ≼ ) is f ¯ -orbitally complete with respect to A, then there is a fixed point of f in A.

Proof

Since A ≠ ∅ , let ζ 0 ∈ A . Then, f ( ζ 0 ) ∈ A . If f ( ζ 0 ) = ζ 0 , we are done. Otherwise, we choose ζ 1 = f ( ζ 0 ) . Now, f ( ζ 1 ) ∈ A , and if f ( ζ 1 ) = ζ 1 , then we are through. Otherwise, choose ζ 2 = f ( ζ 1 ) . Now, f ( ζ 2 ) ∈ A . As ζ 0 , ζ 1 ∈ A with ζ 0 ≺ ζ 1 , then by (1), we have

Again, as ζ 1 , ζ 2 ∈ A with ζ 1 ≺ ζ 2 , then by (1), we have

Using (2) in (3), we get

Continuing similarly, we obtain

Continuing the process, we get a strictly increasing sequence { ζ n } in A so that

Now, we show that { ζ n } is a Cauchy sequence in O A ( ζ 0 , ∞ ) . For this, take n , m ∈ ℕ with n < m . From Lemma 2.1, one writes

Now, by using (5) and (4), we get

That is,

Thus, the strictly increasing sequence { ζ n } is Cauchy in O A ( ζ 0 , ∞ ) . But since ( X , d , ≼ ) is f ¯ -orbitally complete with respect to A, there is u ∈ A so that ζ n → u and ζ n ≺ u for each n ∈ ℕ .

Since ζ n , u ∈ A with ζ n ≺ u for all n ∈ ℕ , by (1) and (4), we have

Making n → ∞ leads to d ( u , f ( u ) ) = 0 . Hence, u is a fixed point of f in A.□

Corollary 2.1

It is clear from Eq. (4) and (7) that Theorem 2.1 still holds if Eq. (1) is replaced by

for ρ , ξ ∈ A with ρ ≺ ξ , where δ > 0 .

Definition 3.1

Let { ζ n } be a sequence in an ordered set ( X , ≼ ) . An element ζ ∈ X is an upper bound of { ζ n } if ζ n ≼ ζ , for each n ∈ ℕ . Such ζ is said to be a strict upper bound of { ζ n } if ζ n ≺ ζ , for each n ∈ ℕ .

Definition 3.2

Let ( X , d , ≼ ) be an ordered partial b-metric space ( s ≥ 1 ) . For A ⊆ X , the triplet ( X , d , ≼ ) is said partially complete with respect to A, if every strictly increasing Cauchy sequence in A has a strict upper bound in A, i.e., if { ζ n } is a strictly increasing Cauchy sequence in A, then there exists ζ ∈ A such that ζ n ≺ ζ , for each n ∈ ℕ .

Example 3.1

Endow X = ℚ with the usual metric of ℝ and the natural order ≤ . Note that ( X , d , ≤ ) is an ordered partial b-metric space (with s = 1 ), but it is not complete. However, if we take A = ( 1 , 3 ] ∩ ℚ , then ( X , d , ≤ ) is partially complete with respect to A. Here, 3 is the strict upper bound for every strictly increasing Cauchy sequence in A.

Remark 3.1

Let ( X , ≼ ) be a totally ordered set. Then, every ordered complete metric space ( X , d , ≼ ) is partially complete with respect to A, where A is any closed subset of X. The converse of the above statement is not true in general. It can be seen by taking X = ( 0 , 5 ) and A = ( 1 , 3 ] . With the usual metric and the natural ordering ≤ , X is partially complete with respect to A, but it is not complete. Also, if we take A = [ 1 , 3 ] , then X is still not complete, but it is partially complete with respect to A.

Theorem 3.1

Let f be a self-map on an ordered partial b-metric space ( X , d , ≼ , s ≥ 1) such that the set A = { ζ ∈ X : ζ ≼ f ( ζ ) } is nonempty and

for all ζ , ξ ∈ A with ζ ≺ ξ , where λ ∈ [ 0 , 1 s ) . If further, f ( A ) ⊆ A and ( X , d , ≼ ) is partially complete with respect to A, then f has a fixed point in A.

Proof

Going through the same lines of proof of Theorem 2.1, we get a strictly increasing Cauchy sequence { ζ n } in A such that

Since ( X , d , ≼ ) is partially complete with respect to A, there is ζ ∈ A such that ζ n ≺ ζ , for each n ∈ ℕ . Thus, from (9) and (11), we get

Since s ≥ 1 , we have that d ( ζ , f ( ζ ) ) = 0 , i.e., ζ is the fixed point of f in A.□

Example 3.2

Let X = ℚ be endowed with d ( ζ , ξ ) = | ζ − ξ | − min { ζ , ξ } . Let ≼ be the natural order ≤ . Note that ( X , d , ≼ ) is an ordered partial b-metric space (with s = 2 ), which is not complete. Consider f : X → X by

Since X is not a complete partial b-metric space, we cannot apply the previous results. Take A = { 2 n − 1 2 n : n ≥ 1 } ∪ { 1 } . Here, f ( A ) ⊆ A and ζ ≤ f ( ζ ) for each ζ ∈ A . Also, ( X , d , ≼ ) is partially complete with respect to A. Therefore, it remains to prove that (9) is satisfied. Let ζ , ξ ∈ A be such that ζ < ξ . We have d ( ζ , f ( ζ ) ) = 1 − 3 ζ 2 , d ( ξ , f ( ξ ) ) = 1 − 3 ξ 2 and 2 ξ − ζ ≥ 1 3 . For λ = 1 2 , one writes

Hence, all the conditions of Theorem 3.1 hold and f has a fixed point in A, which is, u = 1 .

Example 3.3

Let X = ( − ∞ , 0 ] ∩ ℚ be endowed with d ( ζ , ξ ) = | ζ − ξ | − min { ζ , ξ } and ≼ be the natural order ≤ . Note that ( X , d , ≼ ) is an ordered partial b-metric space (with s = 2 ), which is not complete. Take

Take A = { a n : a n + 1 = a n 5 , n ≥ 0 with a 0 = − 1 2 } ∪ { 0 } . Then, A = { − 1 2 , − 1 10 , − 1 50 , … } ∪ { 0 } . Here, f ( A ) ⊆ A and ζ ≤ f ( ζ ) for each ζ ∈ A . Also, ( X , d , ≼ ) is partially complete with respect to A. Therefore, it remains to prove that (9) is satisfied. Let ζ , ξ ∈ A such that ζ < ξ . We have d ( ζ , f ( ζ ) ) = − 11 ζ 5 , d ( ξ , f ( ξ ) ) = − 11 ξ 5 and 4 ξ − ζ ≥ 0 . For λ = 1 4 , one writes

Thus, d ( ξ , f ( ξ ) ) ≤ λ d ( ζ , f ( ζ ) ) . Hence, all the conditions of Theorem 3.1 hold and f has a fixed point in A.

Corollary 3.1

Let f be a self-map on an ordered partial b-metric space ( X , d , ≼ , s ≥ 1) . Assume that f ( X ) is nonempty with ζ ≼ f ( ζ ) , for each ζ ∈ f ( X ) , and

for all ζ , ξ ∈ f ( X ) with ζ ≺ ξ , ζ ≠ f ( ζ ) and ξ ≠ f ( ξ ) , where λ ∈ [ 0 , 1 s ) . Furthermore, if ( X , d , ≼ ) is partially complete with respect to f ( X ) , then f has a fixed point in f ( X ) .

Proof

As f ( X ) ≠ ∅ , let ζ 0 ∈ f ( X ) . Note that ζ 0 ≼ f ( ζ 0 ) . If ζ 0 = f ( ζ 0 ) , then we have nothing to prove. Otherwise, choose ζ 1 = f ( ζ 0 ) ∈ f ( X ) . Then, again ζ 1 ≼ f ( ζ 1 ) . If ζ 1 = f ( ζ 1 ) , then we are through. Otherwise, choose ζ 2 = f ( ζ 1 ) ∈ f ( X ) . Continuing the same process, we get a strictly increasing sequence { ζ n } in f ( X ) such that

Now, repeating the process as in Theorem 2.1, we get for each n ∈ ℕ ,

Let n , m ∈ ℕ with n < m . Using Lemma 2.1, (14) and the fact that s ≥ 1 , we get

Thus, { ζ n } is a strictly increasing Cauchy sequence in f ( X ) . But, ( X , d , ≼ ) is partially complete with respect to f ( X ) , then { ζ n } has a strict upper bound in f ( X ) , say ζ .

Since ζ n , ζ ∈ f ( X ) , for each n with ζ n ≺ ζ , by (12) and (14), we have

As s ≥ 1 , so we have d ( ζ , f ( ζ ) ) = 0 , so ζ is a fixed point of f in f ( X ) .□

Example 3.4

Let A = { a n : a n + 1 = a n 3 , n ≥ 0 with a 0 = − 2 } and B = ( 0 , 3 ] ∩ ℚ . Then, A = { − 2 , − 2 3 , − 2 9 , − 2 27 , − 2 81 , … } . Consider X = A ∪ B . If we take ≼ as the natural ordering ≤ and d ( ζ , ξ ) = | ζ − ξ | , then ( X , d , ≼ ) is an ordered partial b-metric space with s = 1 . Choose

Here, f ( X ) = X . Also, every strictly increasing Cauchy sequence has a strict upper bound in f ( X ) . Thus, ( X , d , ≼ ) is partially complete with respect to f ( X ) . Note that ζ ≼ f ( ζ ) , for each ζ ∈ f ( X ) . Let ζ , ξ ∈ f ( X ) with ζ < ξ , ζ ≠ f ( ζ ) and ξ ≠ f ( ξ ) . We have ζ , ξ ∈ A , where ζ < ξ ⇒ 3 ξ − ζ ≥ 0 . Furthermore, for such ζ , ξ ∈ A , d ( ζ , f ( ζ ) ) = − 2 ζ 3 and d ( ξ , f ( ξ ) ) = − 2 ξ 3 . For λ = 1 3 , one writes

Hence, d ( ξ , f ( ξ ) ) ≤ λ d ( ζ , f ( ζ ) ) for all ζ , ξ ∈ f ( X ) with ζ < ξ , ζ ≠ f ( ζ ) and ξ ≠ f ( ξ ) . Thus, all the assumptions of Corollary 3.1 hold and there is at least one fixed point of f in f ( X ) .

Theorem 4.1

Let ( S 1 , p ) and ( S 2 , p ) be two partial b-metric spaces, A ⊆ S 1 and c ∈ A . Given a function f : A → S 2 . Then, if f is continuous at c, then for each convergent sequence { ζ n } to some c ∈ A , we have { f ( ζ n ) } is convergent to f ( c ) .

Theorem 4.2

Let ( X , d , ≼ ) be a complete ordered partial b-metric space ( s ≥ 1) and f be a continuous self-mapping on X. Let A = { ζ ∈ X : ζ ≼ f ( ζ ) } ≠ ∅ such that f ( A ) ⊆ A . Assume that

for all ζ , ξ ∈ A with ζ ≺ ξ , where λ ∈ [ 0 , 1 s ) . Suppose in addition that the partial b-metric is continuous, then f has a fixed point in X.

Proof

Since A ≠ ∅ , let ζ 0 ∈ A . By definition of A, we have ζ 0 ≼ f ( ζ 0 ) . As f ( A ) ⊆ A , we have ζ 0 , f ( ζ 0 ) ∈ A . If f ( ζ 0 ) = ζ 0 , then we are done. Otherwise, choose ζ 1 ∈ A such that ζ 0 ≺ f ( ζ 0 ) = ζ 1 . Again, ζ 1 ≼ f ( ζ 1 ) . If ζ 1 = f ( ζ 1 ) , then nothing to prove. Otherwise, we choose ζ 2 ∈ A such that ζ 1 ≺ f ( ζ 1 ) = ζ 2 . Continuing in this process, we get an increasing sequence { ζ n } in A such that

Also, { ζ n } is a Cauchy sequence in X (see the proof of Theorem 2.1). Hence, there is at least one ζ ∈ X such that ζ n → ζ . Since f is continuous, we get f ( ζ n ) → f ( ζ ) . By taking n → ∞ , we obtain that

As the partial b-metric is assumed to be continuous, we get that

so d ( ζ , f ( ζ ) ) = 0 . Hence, f ( ζ ) = ζ .□

Example 4.1

Let X = ℝ be endowed with d ( ζ , ξ ) = | ζ − ξ | . Let ≼ be the natural order ≤ . Note that ( X , d , ≼ ) is an ordered partial b-metric space (with s = 1 ). Define f : X → X by

Clearly, for ζ = 1 and ξ = 3 , we have d ( f ζ , f ξ ) > k d ( ζ , ξ ) for each k ∈ ( 0 , 1 ) . Consider A = { a n : a n + 1 = a n 3 , n ≥ 0 with a 0 = − 1 2 } . Note that A = { − 1 2 , − 1 6 , − 1 18 , … } . Here, f ( A ) ⊆ A and ζ ≤ f ( ζ ) for each ζ ∈ A . Then, it remains to prove that (16) is satisfied. Let ζ , ξ ∈ A be such that ζ < ξ . For λ = 1 2 ∈ [ 0 , 1 ) , we have

Thus, d ( ξ , f ( ξ ) ) ≤ λ d ( ζ , f ( ζ ) ) . Hence, all the conditions of Theorem 4.2 hold and f has a fixed point in X.