Fixed point results in generalized suprametric spaces

Definition 2.2

Let ( X , D ) be a generalized semimetric space. Then, D is called generalized suprametric on X if the following additional condition holds:

1. there exist C 0 ≥ 0 and C 1 ≥ 1 such that if x ∈ X and { x n } ∈ C ( D , X , x ) , then

   D ( x , y ) ≤ C 1 limsup n → ∞ D ( x n , y ) + C 0 limsup n → ∞ D ( x , x n ) D ( x n , y ) .

Example 2.3

Let d : R × R → [ 0 , ∞ ] be a function given by:

where a n = 1 + 1 n + 1 and n ∈ N . Observe that d ( x , y ) = 0 ⇒ x = y ; however, the converse is not true since d ( 2 , 2 ) = 1 . It is not difficult to see that ( D 1 )–( D 3 ) are satisfied for C 0 = C 1 = 1 , which means that d is a generalized suprametric.

Definition 2.4

Let ( X , D ) be a generalized suprametric space. Let { x n } be a sequence in X and x ∈ X. We say that { x n } D -converges to x if

Definition 2.5

Let ( X , D ) be a generalized suprametric space and let { x n } be a sequence in X . We say that { x n } is a D -Cauchy sequence if

Definition 2.6

Let ( X , D ) be a generalized suprametric space. We say that ( X , D ) is complete or X is D -complete if every D -Cauchy sequence, D -converges in X .

Definition 2.7

A triple ( X , D , ≼ ) is said to be partially ordered generalized suprametric space if ( X , D ) is a generalized suprametric space and ≼ is a partial order on X . We denote by E ≼ the subset of X × X defined by:

Definition 2.8

Let ( X , D ) be a generalized suprametric space and f : X → X be a given mapping. We say that f is weak continuous if { x n } ∈ C ( D , X , x ) , then there exists a subsequence { x n ( k ) } such that { f x n ( k ) } ∈ C ( D , X , f x ) .

Definition 2.9

Let ( X , D , ≼ ) be a partially ordered generalized suprametric space and f : X → X be a given mapping. We say that f is nondecreasing if

Definition 2.10

Let ( X , D , ≼ ) be a partially ordered generalized suprametric space. We say that the pair ( X , ≼ ) is D -regular if for every sequence { x n } ⊂ X satisfying ( x n , x n + 1 ) ∈ E ≼ , for every n large enough, if { x n } ∈ C ( D , X , x ) , then there exists a subsequence { x n ( k ) } such that ( x n ( k ) , x ) ∈ E ≼ , for every k large enough.

Proposition 2.11

Let ( X , D ) be a generalized suprametric space. Let { x n } be a sequence in X and x , y ∈ X . If { x n }   D -converges simultaneously to x and y , then x = y .

Proof

From ( D 3 ), follows that

then by ( D 1 ), we deduce that x = y .□

Definition 2.12

Let ( X , D ) be a generalized semimetric space. Then, D is called JS-metric on X if the following additional condition holds:

1. there exists C ≥ 0 such that if x ∈ X and { x n } ∈ C ( D , X , x ) , then

   D ( x , y ) ≤ C limsup n → ∞ D ( x n , y ) .

Proposition 2.13

[12] Every standard metric space, respectively, b-metric space, dislocated metric space, and modular space with the Fatou property, is a JS-metric space.

Remark 2.14

Note that every JS-metric space with constant C is a JS-metric space with constant C ′ > C . Clearly, if ( X , D ) is a nontrivial JS-metric space, then the set of constants for which (D 3 ′ ) holds is nonempty. The infimum of this set is greater than or equal to 1 when C ( D , X , x ) is nonempty for every x ∈ X . This can be seen by taking the constant sequence x n = x for all n ∈ N in (D 3 ′ ) (see [10, Remark 2.14]).

Proposition 2.15

Any JS-metric space is a generalized suprametric space.

Definition 2.16

Let X be a nonempty set, b ≥ 1 and ρ ≥ 0 . A function d : X × X → [ 0 , ∞ ) is called b -suprametric if for all x , y , z ∈ X the following properties hold:

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

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

3. d ( x , y ) ≤ b ( d ( x , z ) + d ( z , y ) ) + ρ d ( x , z ) d ( z , y ) .

Proposition 2.17

Every b-metric space, respectively, suprametric space, is a b-suprametric space.

Remark 2.18

Note that every b -suprametric space is a b ′ -suprametric space with constant b ′ > b .

Definition 2.19

A dislocated b -suprametric space is a pair ( X , d ) , where X is a nonempty set and d : X × X → [ 0 , ∞ ) is a function satisfying (i’): d ( x , y ) = 0 implies x = y ; and the conditions (ii) and (iii) of Definition 2.16.

Proposition 2.20

Any dislocated b-suprametric space is a generalized suprametric space.

Proof

Let ( X , d ) be a dislocated b -suprametric space. The conditions ( D 1 ) and ( D 2 ) are clearly satisfied by (i’) and (ii), respectively. It suffices to prove that ( D 3 ) is satisfied. Let x ∈ X and { x n } ∈ C ( d , X , x ) . For every y ∈ X , by the property (iii) of Definition 2.16, we have

for every n ∈ N . Thus, we have

Then, ( D 3 ) holds for ( C 0 , C 1 ) = ( ρ , b ) .□

Definition 2.21

[15] Let X be a nonempty set. A function m : X × X → [ 0 , ∞ ) is said to be a supermetric if the following conditions hold:

1. m ( x , y ) = 0 , then x = y ,

2. m ( x , y ) = m ( y , x ) for all x , y ∈ X ,

3. there exists s ≥ 1 such that for all y ∈ X , there exist distinct sequences { x n } , { y n } ⊂ X with lim n → ∞ m ( x n , y n ) = 0 such that

   limsup n → ∞ m ( y n , y ) ≤ s limsup n → ∞ m ( x n , y ) .

Remark 2.22

It is worth to note that the super metric is different from the generalized suprametric. In fact, the same arguments as in [15, Example 2.2] show that a super metric is not a generalized suprametric. Besides that the supermetric does not reach infinity, the converse is generally not true. To see this, let n ∈ N and d : R × R → [ 0 , ∞ ) be a function given by:

where b n = n − 1 n , n > 1 . Let s ≥ 1 and take y = 1 , and observe that there are no sequences { x n } and { y n } with lim n → ∞ d ( x n , y n ) = 0 such that

Note that d is a 3-suprametric but not a metric, since d ( 1 , b 2 ) + d ( b 2 , 2 ) < d ( 1 , 2 ) .

Theorem 3.5

Let ( X , D ) be a complete generalized suprametric space and f : X → X be a given mapping. Assume that there exists ϕ ∈ Φ such that

and there exists an x 0 ∈ X such that δ 0 < ∞ . Then, the sequence { f n x 0 } n ∈ N D -converges to a fixed point x ∗ of f. In addition, if y ∗ is a fixed point of f such that D ( x ∗ , y ∗ ) < ∞ , then x ∗ = y ∗ .

Proof

First, observing that if p , q ∈ N with p ≥ q , we deduce from (2) that δ p ≤ δ q ≤ δ 0 < ∞ . In particular, δ k ≤ δ k − 1 ≤ δ 0 < ∞ for all k ≥ 1 . Now, for every p , q ∈ N such that p ≥ q ≥ k ≥ 1 , D ( f p − 1 x 0 , f q − 1 x 0 ) ≤ δ k − 1 , thus it follows from (2), (3), and ( i ϕ ) that

Since for k ≥ 1 , the previous inequality holds for all p ≥ q ≥ k , we conclude that δ k ≤ ϕ ( δ k − 1 ) . By induction, the monotonicity of ϕ and Remark 3.1, we deduce that δ k ≤ ϕ k ( δ 0 ) for all k ≥ 1 . Thus, using (1), we obtain

for all m , n ∈ N . Using ( i i ϕ ) and the fact that δ 0 < ∞ , we deduce that

Consequently, the sequence { f n x 0 } n ∈ N is D -Cauchy. As ( X , D ) is D -complete, then { f n x 0 } n ∈ N   D -converges to a point x ∗ ∈ X , say. Now, we shall show that x ∗ is a fixed point of f . Using (3) and Remark 3.1, we obtain

so taking n → ∞ in the previous inequality, we deduce that

Thus, the sequence { f n x 0 } n ∈ N   D -converges simultaneously to x ∗ and f x ∗ , which implies by Proposition 2.11 that x ∗ = f x ∗ , i.e., x ∗ is a fixed point of f . Assume now that y ∗ is a fixed point of f such that D ( x ∗ , y ∗ ) < ∞ . Then, we obtain

which implies a contradiction if D ( x ∗ , y ∗ ) ≠ 0 , since if D ( x ∗ , y ∗ ) ≠ 0 , we have ϕ ( D ( x ∗ , y ∗ ) ) < D ( x ∗ , y ∗ ) . We conclude by ( D 1 ) that x ∗ = y ∗ .□

Proposition 3.6

Theorem 3.5  generalizes [12, Theorem 3.3].

Theorem 3.7

[5, Theorem 3.1] Let ( X , d ) be a complete b-suprametric space and f : X → X be a mapping. Assume that there exists a comparison function ϕ such that

Then, f has a unique fixed point x ∗ and { f n x } n ∈ N converges to x ∗ for every x ∈ X .

Proposition 3.8

Theorem 3.5 extends Theorem 3.7.

Theorem 3.9

Let ( X , D ) be a complete generalized suprametric space, ρ : ( 0 , 1 ) → R + be a given function, and f : X → X be a given mapping. Assume that there exists ψ ∈ Ψ ρ such that

and there exists an x 0 ∈ X such that

where λ 0 ≔ δ 0 − 1 ψ ( δ 0 ) . Then, the sequence { f n x 0 } n ∈ N D -converges to a fixed point x ∗ of f. In addition, if

and y ∗ is a fixed point of f such that D ( x ∗ , y ∗ ) < ∞ , then x ∗ = y ∗ .

Proof

Observe first that by (4), if ψ ( δ 0 ) = 0 , δ 1 = 0 , which implies that f x 0 is a fixed point of f , and we are done. Assume next that ψ ( δ 0 ) ≠ 0 , so λ 0 ∈ ( 0 , 1 ) , and from (4), we obtain δ 1 ≤ ψ ( δ 0 ) < δ 0 . Using (4) and the properties of ψ , we obtain

By induction, we easily obtain δ n + 1 ≤ ρ ( λ 0 ) n ψ ( δ 0 ) , for all n ∈ N . Hence, it follows from (1) that

Consequently, { f n x 0 } n ∈ N is D -Cauchy sequence. As ( X , D ) is D -complete, the sequence { f n x 0 } n ∈ N   D -converges to a point x ∗ ∈ X , say. Now, we shall show that x ∗ is a fixed point of f . Using (3), we obtain

Taking n → ∞ in the previous inequality, we deduce by Remark 3.2 that

Thus, the sequence { f n x 0 } n ∈ N   D -converges simultaneously to x ∗ and f x ∗ , so by Proposition 2.11, x ∗ = f x ∗ , i.e., x ∗ is a fixed point of f . Assume now that (6) holds and y ∗ is a fixed point of f such that D ( x ∗ , y ∗ ) < ∞ . Then, we obtain

which implies a contradiction by (6) if D ( x ∗ , y ∗ ) ≠ 0 . Hence, D ( x ∗ , y ∗ ) = 0 , and from ( D 1 ), we deduce that x ∗ = y ∗ .□

Proposition 3.10

Theorem 3.9  extends [7, Theorem 2.1].

Corollary 3.11

Let ( X , D ) be a complete JS-metric space, ρ : ( 0 , 1 ) → R + be a given function, and f : X → X be a given mapping. Assume that there exists ψ ∈ Ψ ρ such that

and there exists an x 0 ∈ X such that ψ ( δ 0 ) < δ 0 < ∞ and ρ ( λ 0 ) < 1 , where λ 0 ≔ δ 0 − 1 ψ ( δ 0 ) . Then, the sequence { f n x 0 } n ∈ N   D -converges to a fixed point x ∗ of f. In addition, if ψ ( t ) < t for all t ∈ ( 0 , ∞ ) and y ∗ is a fixed point of f such that D ( x ∗ , y ∗ ) < ∞ , then x ∗ = y ∗ .

Theorem 4.1

Let ( X , D , ≼ ) be a partially ordered generalized suprametric space such that ( X , D ) is complete. Let f : X → X be a weak continuous and nondecreasing mapping. Assume that there exists ϕ ∈ Φ such that

and there exists an x 0 ∈ X such that ( x 0 , f x 0 ) ∈ E ≼ and δ 0 < ∞ . Then, the sequence { f n x 0 } n ∈ N   D -converges to a fixed point x ∗ of f . If y ∗ is a fixed point of f such that x ∗ ≤ y ∗ and D ( x ∗ , y ∗ ) < ∞ , then x ∗ = y ∗ .

Proof

Let x 0 ∈ X such that ( x 0 , f x 0 ) ∈ E ≼ , so using the monotonicity of f , we deduce by induction that ( f n x 0 , f n + 1 x 0 ) ∈ E ≼ for all n ∈ N . Due to the transitivity of the partial order, observe that if for p , q ∈ N , p ≼ q , then f p x 0 ≼ f q x 0 . Hence, it follows by induction and the symmetry of D that

for all integer n ≥ 1 and i , j ∈ N , which implies that δ n ≤ ϕ ( δ n − 1 ) , and by induction, we deduce that δ n ≤ ϕ n ( δ 0 ) for all n ∈ N . Hence, using that δ 0 < ∞ and ( i i ϕ ), we obtain

Consequently, { f n x 0 } n ∈ N is D -Cauchy sequence. As ( X , D ) is D -complete, the sequence { f n x 0 } n ∈ N   D -converges to a point x ∗ ∈ X . Now, we shall show that x ∗ is a fixed point of f . The mapping f is weak continuous, so there exists a subsequence { f n ( k ) x 0 } such that { f n ( k ) + 1 x 0 } is D -convergent to f x ∗ . By Proposition 2.11, it follows that x ∗ is a fixed point of f . Assume now that y ∗ is a fixed point of f such that x ∗ ≤ y ∗ and D ( x ∗ , y ∗ ) < ∞ , then by (7), we have

which is a contradiction by Remark 3.1 if D ( x ∗ , y ∗ ) ≠ 0 . So D ( x ∗ , y ∗ ) = 0 and by ( D 1 ) follows that x ∗ = y ∗ .□

Theorem 4.2

Let ( X , D , ≼ ) be a partially ordered generalized suprametric space such that ( X , D ) is complete and ( X , ≼ ) is D -regular. Let f : X → X be a nondecreasing mapping. Assume that there exists ϕ ∈ Φ such that

and there exists an x 0 ∈ X such that ( x 0 , f x 0 ) ∈ E ≼ and δ 0 < ∞ . Then, the sequence { f n x 0 } n ∈ N   D -converges to a fixed point x ∗ of f . If y ∗ is a fixed point of f such that x ∗ ≤ y ∗ and D ( x ∗ , y ∗ ) < ∞ , then x ∗ = y ∗ .

Proof

As in the proof of Theorem 4.1, we obtain that { f n x 0 } n ∈ N   D -converges to a point x ∗ ∈ X . Now, since ( X , ≼ ) is D -regular, then there exists a subsequence { f n ( k ) x 0 } such that ( f n ( k ) x 0 , x ∗ ) ∈ E ≼ for all k large enough. Moreover, from (8), we have

for all k large enough. So letting k → ∞ , then by Remark 3.1, D ( f n ( k ) + 1 x 0 , f x ∗ ) → 0 , and this implies that { f n ( k ) x 0 } D -converges simultaneously to x ∗ and f x ∗ . Hence, by Proposition 2.11, we deduce that x ∗ = f x ∗ . Finally, if y ∗ is a fixed point of f such that x ∗ ≤ y ∗ and D ( x ∗ , y ∗ ) < ∞ , we deduce as above that x ∗ = y ∗ .□