Discussions on the almost 𝒵-contraction

Definition 1.1

(See [1]). A simulation function is a mapping ζ : [ 0 , ∞ ) × [ 0 , ∞ ) → ℝ satisfying the following conditions:

( ζ 1 ) ζ ( t , s ) < s − t for all t , s > 0 ;

( ζ 2 ) if { t n } , { s n } are sequences in ( 0 , ∞ ) such that lim n → ∞ t n = lim n → ∞ s n > 0 , then

Observe first that ( ζ 1 ) implies

Indeed, this simulation function is obtained by the abstraction of the Banach contraction mapping principle. We underline that in [1], there was an additional axiom ζ ( 0 , 0 ) = 0 . Since it is a superfluous condition, we omit it. Throughout the paper, the letter Z denotes the family of all functions ζ : [ 0 , ∞ ) × [ 0 , ∞ ) → ℝ . A function ζ ( t , s ) ≔ k s − t , where k ∈ [ 0 , 1 ) for all s , t ∈ [ 0 , ∞ ) , is an instantaneous example of a simulation function. For further and more interesting examples, we refer e.g. [1,2,3,4,5,6] and related references therein. In particular, in [7,8] the simulation functions work for establishing also common fixed points and coincidence points, both in a metric space and in a partial metric space.

We say that a self-mapping f, defined on a metric space ( X , d ) , is a Z -contraction with respect to ζ ∈ Z [1], if

Theorem 1.1

Every Z -contraction on a complete metric space has a unique fixed point.It is clear that the immediate example ζ ( t , s ) ≔ k s − t is obtained by abstraction of the Banach contraction mapping principle. In other words, with this function ζ ( t , s ) ≔ k s − t , Theorem 1.1 yields the Banach contraction mapping principle.

Lemma 1.1

[9] Let ( X , d ) be a metric space and let { p n } be a sequence in X such that d ( p n + 1 , p n ) is non-increasing and that

If { p 2 n } is not a Cauchy sequence then there exist a δ > 0 and two strictly increasing sequences { m k } and { n k } of positive integers such that the following sequences tend to δ when k → ∞ :

One of the interesting notions, α-admissibility was introduced by Samet-Vetro-Vetro [10], see also [11]. This study, which attracts the attention of many researchers, has been developed and generalized in many respects. In particular, [12,13], the author depicts applications of fixed point methodologies to the solution of a first-order periodic differential problem, converting such a problem into an integral equation. Moreover, in [14], the authors prove an existence theorem producing a periodic solution of some non-linear integral equations, using the Krasnoselskii-Schaefer-type method and technical assumptions.

Definition 1.2

Let f be a self-mapping on a non-empty set X, and α : X × X → [ 0 , ∞ ) be mapping. We say that f is extended-α-admissible if, for all x , y ∈ X , we have

Definition 2.1

Let f be a self-mapping, defined on a metric space ( X , d ) , and α : X × X → [ 0 , ∞ ) be a function. Here, f : X → X is called an almost- Z -contraction with respect to ζ ∈ Z if there exists ζ ∈ Z , and β ∈ G , and L ≥ 0 such that for all x , y ∈ X

where

with

and

Theorem 2.1

Suppose that a self-mapping f, defined on a complete metric space ( X , d ) , forms an almost- Z -contraction. Furthermore, we suppose, for all x , y ∈ X , that

1. f is an extended-α-admissible pair

2. there exists x 0 ∈ X such that α ( x 0 , f x 0 ) ≥ 1

3. either

   1. f is continuous,

or

1. if { x n } is a sequence in X such that α ( x n , x n + 1 ) ≥ 1 for all n and as n → ∞ , then there exists a subsequence { x n ( k ) } of x n such that α ( x n ( k ) , x ) ≥ 1 for all k.

Then, f has a fixed point.

Proof

On account of (ii), there is a point x 0 ∈ X such that α ( x 0 , f x 0 ) ≥ 1 . By starting this initial point, we shall build a sequence { x n } n ∈ ℕ 0 in X by x n + 1 = f x n for all n ≥ 0 . Throughout the proof, without loss of generality, we shall assume that

Indeed, in the opposite case, where x n 0 = x n 0 + 1 for some n 0 ∈ N , we conclude that x 0 is the desired fixed point, i.e., x n 0 + 1 = f x n 0 = x n 0 . This implies a trivial solution that is not interesting and that is why we exclude this case.

On the other hand, by taking both (i) and (ii) into account, we observe that

continuing in this way we get

Furthermore, by regarding (5), we derive that

As a first step, we want to conclude that { d ( x n , x n + 1 ) } is non-increasing. Suppose, in contrast, that d ( x n , x n + 1 ) < d ( x n + 1 , x n + 2 ) . Since α ( x n , x n + 1 ) ≥ 1 , we find that

which implies that

where

and

Hence, inequality (7) turns into

a contradiction. Consequently, we deduce that d ( x n + 1 , x n + 2 ) < d ( x n , x n + 1 ) , for each n. Since the sequence { d ( x n , x n + 1 ) } is non-increasing.

As a next step, lim n → ∞ d ( x n , x n + 1 ) = 0 . Indeed, since { d ( x n , x n + 1 ) } is non-increasing and bounded below, we conclude that it converges to some non-negative real numbers, say r,

It is evident that

We assert that r = 0 . Suppose, in contrast, that r ≠ 0 . Then, from Eq. (8) and ( ζ 2 ) , and taking limit as n → ∞ . Therefore,

Attendantly, r = 0 and also

In what follows, we claim that sequence { x n } is a Cauchy sequence. Assume that { x n } is not a Cauchy sequence, then there exists ε > 0 and sequences { x n k } , { x m k } ; n k > m k > k such that

Now take x = x m k − 1 and y = x n k − 1 in (6), we have

implies

where

and

Due to Lemma 1.1, we have

Since

using (13) and (9), we have

Let t n k = d ( x m k , x n k ) and s n k = K ( x m k − 1 , x n k − 1 ) we have lim k → ∞ s n k = lim k → ∞ t n k = ε and letting k → ∞ in (12)

Then, by (13), (14) and keeping ( ζ 2 ) in mind, we have

a contradiction. As a result, our claim is correct and { x n } is a Cauchy sequence.

Since ( X , d ) is a complete metric space, the sequence converges to some point u ∈ X as n → ∞

Now we shall show that f u = u .

Suppose we have (iiia). Since f is continuous, we derive the desired results obviously, that is,

Suppose we have (iiib). We shall use the method of reductio ad absurdum. Suppose, in contrast, that f u ≠ u , that is, d ( u , f u ) > 0 . By (iiib), there exists a subsequence { x n k } of { x n } such that α ( x n k , u ) ≥ 1 for all k.

It implies that

where

and

By letting k → ∞ in (16), together with the observation above, we have

is a contradiction. Hence, u is a fixed point of f.□

Theorem 2.2

In addition to the axioms of Theorem 2.1, we assume that

1. for all p , q ∈ S f ( X ) we have α ( p , q ) ≥ 1 , where S f ( X ) ⊂ X is the set of all fixed points of f. Then, f has a unique fixed point.

Proof

We shall use the method of reductio ad absurdum to reach our goal. Suppose that there are two distinct fixed points of f, that is, namely, p , q ∈ S f ( X ) with f p = p ≠ q = f q . On account of the additional assumption (iv), we have α ( p , q ) ≥ 1 , which implies

where

with

and

Hence, expression (17) turns into

a contradiction. This completes the proof.□

Example 2.1

Let X = [ 0 , 1 ] be endowed with metric d ( x , y ) = | x − y | for all x , y ∈ X . Let ζ ( t , s ) = s − t and considering β : [ 0 , ∞ ) → [ 0 , 1 ) , β ( t ) = 1 1 + t for all t ≥ 0 and L ≥ 0 . Let f : X → X be defined by f x = x 2 for all x ∈ [ 0 , 1 ] and α : X × X → [ 0 , ∞ ) be defined by

Since α ( x , y ) = 1 , x , y ∈ [ 0 , 1 ] implies

Therefore, f is almost- Z -contraction with respect to ζ ∈ Z . Hence, all the assumptions of Theorem 2.2 are satisfied, and hence f has a unique fixed point.

Theorem 3.1

Let f be a self-mapping, defined on a complete metric space ( X , d ) , and α : X × X → [ 0 , ∞ ) be a function. Suppose that there exists ζ ∈ Z , and β ∈ G , and L ≥ 0 such that for all x , y ∈ X

where

Furthermore, we suppose, for all x , y ∈ X , that

1. f is an extended-α-admissible pair;

2. there exists x 0 ∈ X such that α ( x 0 , f x 0 ) ≥ 1 ;

3. either

1. f is continuous,

or

1. if { x n } is a sequence in X such that α ( x n , x n + 1 ) ≥ 1 for all n and as n → ∞ , then there exists a subsequence { x n ( k ) } of x n such that α ( x n ( k ) , x ) ≥ 1 for all k;

1. for all p , q ∈ S f ( X ) we have α ( p , q ) ≥ 1 , where S f ( X ) ⊂ X is the set of all fixed points of f. Then, f has a fixed point.

Theorem 3.2

Let f be a self-mapping, defined on a complete metric space ( X , d ) . Suppose that there exists ζ ∈ Z , and β ∈ G , and L ≥ 0 such that for all x , y ∈ X

where K ( x , y ) , E ( x , y ) and N ( x , y ) are defined as in Theorem 2.1. Then, f has a unique fixed point.

Proof

It is sufficient to set α ( x , y ) = 1 for all x , y ∈ X , in Theorem 2.2 (and hence Theorem 2.1).□

Theorem 3.3

Let f be a self-mapping, defined on a complete metric space ( X , d ) endowed with a partial order ≼ on X. Suppose that there exists ζ ∈ Z , and β ∈ G , and L ≥ 0 such that for all x , y ∈ X with x ≼ y

where K ( x , y ) , E ( x , y ) and N ( x , y ) are defined as in Theorem 2.1. Suppose also that the following conditions hold:

1. there exists x 0 ∈ X such that x 0 ≼ f x 0 ;

2. f is continuous or ( X , ≼ , d ) is regular.

Then, f has a fixed point. Moreover, if for all x , y ∈ X there exists z ∈ X such that x ≼ z and y ≼ z , we have uniqueness of the fixed point.

Proof

It is sufficient to define the mapping α : X × X → [ 0 , ∞ ) by

Clearly, f is an almost- Z -contraction with respect to ζ ∈ Z . From condition (i), we have α ( x 0 , f x 0 ) ≥ 1 . Moreover, for all x , y ∈ X , from the monotone property of f, we have

The rest is satisfied in a straightway.□

Theorem 3.4

Let f be a self-mapping, defined on a complete metric space ( X , d ) , and α : X × X → [ 0 , ∞ ) be a function. Suppose that there exists ϕ 1 , ϕ 2 ∈ Φ with ϕ 1 ( t ) < t ≤ ϕ 2 ( t ) for all t > 0 , and β ∈ G , and L ≥ 0 such that for all x , y ∈ X

where K ( x , y ) , E ( x , y ) and N ( x , y ) are defined as in Theorem 2.1. Furthermore, we suppose, for all x , y ∈ X , that

1. f is an extended-α-admissible pair;

2. there exists x 0 ∈ X such that α ( x 0 , f x 0 ) ≥ 1 ;

3. either

1. f is continuous,

or

1. if { x n } is a sequence in X such that α ( x n , x n + 1 ) ≥ 1 for all n and as n → ∞ , then there exists a subsequence { x n ( k ) } of x n such that α ( x n ( k ) , x ) ≥ 1 for all k;

1. for all p , q ∈ S f ( X ) we have α ( p , q ) ≥ 1 , where S f ( X ) ⊂ X is the set of all fixed points of f. Then, f has a fixed point.

Proof

Let ζ ( t , s ) = ϕ 1 ( s ) − ϕ 2 ( t ) for all t , s ≥ 0 , where ϕ 1 ( t ) < t ≤ ϕ 2 ( t ) for all t > 0 . It is clear that ζ ∈ Z , see, e.g. [1,2]. Thus, the desired results follow from Theorem 2.2.□

Theorem 3.5

Let f be a self-mapping, defined on a complete metric space ( X , d ) , and α : X × X → [ 0 , ∞ ) be a function. Suppose that there exists ϕ Φ , and β ∈ G , and L ≥ 0 such that for all x , y ∈ X

where K ( x , y ) , E ( x , y ) and N ( x , y ) are defined as in Theorem 2.1. Furthermore, we suppose, for all x , y ∈ X , that

1. f is an extended-α-admissible pair;

2. there exists x 0 ∈ X such that α ( x 0 , f x 0 ) ≥ 1 ;

3. either

1. f is continuous,

or

1. if { x n } is a sequence in X such that α ( x n , x n + 1 ) ≥ 1 for all n and as n → ∞ , then there exists a subsequence { x n ( k ) } of x n such that α ( x n ( k ) , x ) ≥ 1 for all k;

1. for all p , q ∈ S f ( X ) we have α ( p , q ) ≥ 1 , where S f ( X ) ⊂ X is the set of all fixed points of f. Then, f has a fixed point.

Proof

Let ζ ( t , s ) = s − ϕ ( s ) − t for all t , s ≥ 0 . It is clear that ζ ∈ Z , see, e.g. [1,2,3]. Thus, the desired results follow from Theorem 2.2.□

Theorem 3.6

Let f be a self-mapping, defined on a complete metric space ( X , d ) , and α : X × X → [ 0 , ∞ ) be a function. Suppose that there exists ϕ 1 , ϕ 2 ∈ Φ with ϕ 1 ( t ) < t ≤ ϕ 2 ( t ) for all t > 0 , and β ∈ G , and L ≥ 0 such that for all x , y ∈ X

where K ( x , y ) , E ( x , y ) and N ( x , y ) are defined as in Theorem 2.1. Furthermore, we suppose, for all x , y ∈ X , that

1. f is an extended- α -admissible pair;

2. there exists x 0 ∈ X such that α ( x 0 , f x 0 ) ≥ 1 ;

3. either

1. f is continuous,

or

1. if { x n } is a sequence in X such that α ( x n , x n + 1 ) ≥ 1 for all n and as n → ∞ , then there exists a subsequence { x n ( k ) } of x n such that α ( x n ( k ) , x ) ≥ 1 for all k;

1. for all p , q ∈ S f ( X ) we have α ( p , q ) ≥ 1 ,

where S f ( X ) ⊂ X is the set of all fixed points of f. Then, f has a fixed point.

Proof

Let ζ ( t , s ) = s − ∫ 0 t μ ( u ) d u for all t , s ≥ 0 .

It is clear that ζ ∈ Z , see, e.g. [3,1,2]. Thus, the desired results follow from Theorem 2.2.□

Theorem 3.7

Let f be a self-mapping, defined on a complete metric space ( X , d ) , and α : X × X → [ 0 , ∞ ) be a function, and β ∈ G , and L ≥ 0 . Suppose that there exist φ : [ 0 , ∞ ) → [ 0 , ∞ ) which is upper semi-continuous and such that φ ( t ) < t for all t > 0 and φ ( 0 ) = 0 . Assume, for all x , y ∈ X , that