**2. Preliminaries**

Basic notions and notations for a good understanding of this manuscript are given in [5]. Nevertheless, we recall here the essential facts. Throughout this manuscript, *X* always stands for a nonempty set. A *binary relation on X* is a nonempty subset S of the product space *X* × *X*. If (*<sup>x</sup>*, *y*) ∈ S, we denote it by *<sup>x</sup>*S*y*. We write *x*S∗*y* when *x*S*y* and *x* = *y*. Notice that S∗, if it is nonempty, is another binary relation on *X*. Two points *x* and *y* are S*-comparable* if *x*S*y* or *y*S*<sup>x</sup>*. A binary relation S is:


Reflexive and transitive binary relations are called *preorders* (or *quasiorders*), and, if they are also antisymmetric, then they are *partial orders*. The trivial partial order S*X* is defined by *x*S*Xy* for each *x*, *y* ∈ *X*.

From now on, N = {0, 1, 2, 3, ...} stands for the set of all nonnegative integers and N∗ = N{0}. Henceforth, let *T* : *X* → *X* be a map from *X* into itself, let (*<sup>X</sup>*, *d*) be a metric space and let *A* ⊆ R be a nonempty subset of the set of all real numbers. The range (or image) of *d* is ran(*d*) = {*d*(*<sup>x</sup>*, *y*) : *x*, *y* ∈ *X*} ⊆ [0, <sup>∞</sup>).

If *Tx* = *x*, then *x* is a *fixed point of T*. The maps {*Tn* : *X* → *<sup>X</sup>*}*n*∈<sup>N</sup> defined by *T*<sup>0</sup> =identity, *T*<sup>1</sup> = *T* and *Tn*+<sup>1</sup> = *T* ◦ *Tn* for all *n* ≥ 2 are known as the *iterates* of *T*. The *Picard sequence of T based on x*0 ∈ *X* is the sequence {*xn*}*n*∈<sup>N</sup> given by *xn*+1 = *Txn* for all *n* ∈ N (hence, *xn* = *Tnx*0 for each *n* ∈ N). When any Picard sequence of *T* converges to a fixed point of *T*, we say that *T* is a *weakly Picard operator*, and if it has a unique fixed point, then *T* is known as *Picard operator*.

In [5], the authors used the following terminology. Let S be a binary relation on a metric space (*<sup>X</sup>*, *d*), let *Y* ⊆ *X* be a nonempty subset, let {*xn*} be a sequence in *X* and let *T* : *X* → *X* be a self-mapping. We say that:


We follow the notation given in [12,13]. Next, we list a collection of properties that can be satisfied by a function *φ* : [0, ∞) → [0, <sup>∞</sup>).

(*s*).

(P1) *φ* is non-decreasing, that is, if 0 ≤ *t* ≤ *s*, then *φ* (*t*) ≤ *φ* (P10) The series ∑ *n*≥1 *φn* (*t*) converges for all *t* > 0. (P11) lim*n*→∞ *φn* (*t*) = 0 for all *t* > 0. (P12) *φ* (*t*) < *t* for all *t* > 0. (P13) lim*t*→0<sup>+</sup> *φ* (*t*) = 0. (P)*φ*(0)=0.

It is clear that (P10) ⇒ (P11) and, on the other hand, (P12) ⇒ (P13).

**Proposition 1** ([12,13])**.** *If* (P1) *holds, then* (P10) ⇒ (P11) ⇒ (P12) ⇒ (P13) ⇒ (P)*.*

Given a function *α* : *X* × *X* → [0, <sup>∞</sup>), it is possible to redefine the previous notions in terms of *α* (transitivity, *α*-admissibility, *α*-nondecreasing character, *α*-nondecreasing-continuity, *α*-strictly-increasing-regularity, (*<sup>α</sup>*, *d*)-strictly-increasing-completeness, (*<sup>α</sup>*, *d*)-strictly-increasingprecompleteness, etc.). For details, see [5]. Such properties can be translated to the previous setting by using the binary relation S*α* on *X* given, for *x*, *y* ∈ *X*, by

$$\mathbf{x} \mathbf{S}\_{\mathbf{a}} \mathbf{y} \quad \text{if} \quad \mathbf{a}(\mathbf{x}, \mathbf{y}) \succeq \mathbf{1}. \tag{1}$$

**Lemma 1.** *Let* (*<sup>X</sup>*, *d*) *be a metric space, let T* : *X* → *X be a self-mapping and let α* : *X* × *X* → [0, ∞) *be a function. Then, the following properties hold.*


In [5], Shahzad et al. introduced the following notions.

**Definition 1.** *Let* {*an*} *and* {*bn*} *be two sequences of real numbers. We say that* {(*an*, *bn*)} *is a* (*<sup>T</sup>*, S)-sequence *if there exist two sequences* {*xn*}, {*yn*} ⊆ *X such that*

$$\mathbf{x}\_n \mathbf{S} y\_n \quad a\_n = d(T \mathbf{x}\_n, T y\_n) > 0 \quad \text{and} \quad b\_n = d(\mathbf{x}\_n, y\_n) > 0 \quad \text{for all } n \in \mathbb{N}.$$

*If* S *is the trivial binary relation* S*X, then* {(*an*, *bn*)} *is called a T*-sequence*.*

**Remark 1.** *Notice that* {(*an* = *d*(*Txn*, *Tyn*), *bn* = *d*(*xn*, *yn*))} *is a* (*<sup>T</sup>*, S)*-sequence if, and only if,*

$$x\_n \mathcal{S}^\* y\_n \quad \text{and} \quad a\_n > 0 \quad \text{for all } n \in \mathbb{N}.$$

**Definition 2.** *We say that T* : *X* → *X is an* (A, S)-contraction *if there exists a function* : *A* × *A* → R *such that T and satisfy the following four conditions:*

(A1) ran(*d*) ⊆ *A.* (A2) *If* {*xn*} ⊆ *X is a Picard* S*-nondecreasing sequence of T such that*

> *xn* = *xn*+1 *and* (*d* (*xn*+1, *xn*+<sup>2</sup>), *d* (*xn*, *xn*+<sup>1</sup>)) > 0 *for all n* ∈ N,

*then* {*xn*} *is asymptotically regular on* (*<sup>X</sup>*, *d*) *(that is,* {*d* (*xn*, *xn*+<sup>1</sup>)} → 0*).*


*In such a case, we say that T is an* (A, S)-contraction with respect to *. We denote the family of all* (A, S)*-contractions from* (*<sup>X</sup>*, *d*) *into itself with respect to by* <sup>A</sup>*<sup>X</sup>*,*d*,S,,*<sup>A</sup> or, for simplicity, by* A *when no confusion is possible.*

*If* S *is the trivial binary relation* S*X, then T is called an* A*-contraction (with respect to ).*

Condition (A1) implies that *A* is a nonempty set. In some cases, we also consider the following properties.

(A2) If *x*1, *x*2 ∈ *X* are two points such that

$$T^n \mathbf{x}\_1 \mathcal{S}^\* T^n \mathbf{x}\_2 \quad \text{and} \quad \varrho(d\left(T^{n+1} \mathbf{x}\_1, T^{n+1} \mathbf{x}\_2\right), d\left(T^n \mathbf{x}\_1, T^n \mathbf{x}\_2\right)) > 0 \quad \text{for all } n \in \mathbb{N}\_+$$

then {*d* (*Tnx*1, *Tnx*2)} → 0. (A5) If {(*an*, *bn*)} is a (*<sup>T</sup>*, S)-sequence such that {*bn*} → 0 and (*an*, *bn*) > 0 for all *n* ∈ N, then {*an*} → 0.

#### **3. Ample Spectrum Contractions**

In this section, we slightly modify the axioms given in [5] in a subtle way in order to consider a wider class of contractions. In what follows, let (*<sup>X</sup>*, *d*) be a metric space, let S be a binary relation on *X* and let *T* : *X* → *X* be a self-mapping.

**Definition 3.** *Let* {*an*} *and* {*bn*} *be two sequences of real numbers. We say that* {(*an*, *bn*)} *is a* (*<sup>T</sup>*, S∗)-sequence *if there exist two sequences* {*xn*}, {*yn*} ⊆ *X such that*

*xn*S∗*yn*, *Txn*S∗*Tyn*, *an* = *d*(*Txn*, *Tyn*) > 0 *and bn* = *d*(*xn*, *yn*) > 0 *for all n* ∈ N.

**Proposition 2.** *Every* (*<sup>T</sup>*, S∗)*-sequence is a* (*<sup>T</sup>*, S)*-sequence.*

**Definition 4.** *We say that T* : *X* → *X is a* ample spectrum contraction *if there exists a function* : *A* × *A* → R *such that T and satisfy the following four conditions:*


*xn* = *xn*+1 *and* (*d* (*xn*+1, *xn*+<sup>2</sup>), *d* (*xn*, *xn*+<sup>1</sup>)) ≥ 0 *for all n* ∈ N,

*then* {*d* (*xn*, *xn*+<sup>1</sup>)} → 0*.*


*In such a case, we say that T is a* ample spectrum contraction with respect to S and *. We denote the family of all ample spectrum contractions from* (*<sup>X</sup>*, *d*) *into itself with respect to* S *and by* <sup>B</sup>*<sup>X</sup>*,*d*,S,,*A.*

In some cases, we also consider the following properties:

(B2) If *x*1, *x*2 ∈ *X* are two points such that

> *Tnx*1S∗*Tnx*2 and (*d <sup>T</sup><sup>n</sup>*+1*x*1, *<sup>T</sup><sup>n</sup>*+1*x*<sup>2</sup> , *d* (*Tnx*1, *<sup>T</sup>nx*2)) ≥ 0 for all *n* ∈ N,

then {*d* (*Tnx*1, *Tnx*2)} → 0. (B5) If {(*an*, *bn*)} is a (*<sup>T</sup>*, S∗)-sequence such that {*bn*} → 0 and (*an*, *bn*) ≥ 0 for all *n* ∈ N, then {*an*} → 0.

**Remark 2.** *The reader can observe the following facts about the previous assumptions:*


**Proposition 3.** *If* (*<sup>t</sup>*,*<sup>s</sup>*) ≤ *s* − *t for all t*,*<sup>s</sup>* ∈ *A* ∩ (0, <sup>∞</sup>)*, then* (B5) *holds.*

**Proof.** Assume that {*an*}, {*bn*} ⊂ (0, ∞) ∩ *A* are two sequences such that {*bn*} → 0 and (*an*, *bn*) ≥ 0 for all *n* ∈ N. Since *an*, *bn* ∈ (0, ∞) ∩ *A*, then 0 < (*an*, *bn*) ≤ *bn* − *an* for all *n* ∈ N. As a consequence, 0 < *an* ≤ *bn* for all *n* ∈ N, which means that {*an*} → 0.

The previous definition generalizes the notion of (A, S)-contraction, as we prove in the following result:

**Theorem 1.** *Every* (A, S)*-contraction is an ample spectrum contraction (with respect to the same function ). Furthermore, if it satisfies* (A2) *(respectively,* (A5)*), then it also verifies* (B2) *(respectively,* (B5)*).*

In particular, we prove the following implications:

$$\begin{aligned} (\mathcal{A}\_1) &\Rightarrow (\mathcal{B}\_1), \\ (\mathcal{A}\_4) &\Rightarrow (\mathcal{B}\_4), \\ (\mathcal{A}\_2) + (\mathcal{A}\_4) &\Rightarrow (\mathcal{B}\_2), \\ (\mathcal{A}\_3) + (\mathcal{A}\_4) &\Rightarrow (\mathcal{B}\_3), \\ (\mathcal{A}\_4) + (\mathcal{A}\_5) &\Rightarrow (\mathcal{B}\_5), \\ (\mathcal{A}\_2') + (\mathcal{A}\_4) &\Rightarrow (\mathcal{B}\_2'). \end{aligned}$$

**Proof.** Let (*<sup>X</sup>*, *d*) be a metric space, let *T* : *X* → *X* be a mapping and let : *A* × *A* → R be a function. Clearly, (A1) ⇒ (B1) and (A4) ⇒ (B4). Next, we prove the rest of conditions.

[ (A2) + (A4) ⇒ (B2) ] Let *x*1, *x*2 ∈ *X* be two points such that

$$d\left(T^{n}\mathbf{x}\_{1}\mathbf{S}^{\*}T^{n}\mathbf{x}\_{2}\quad\text{and}\quad\varrho\left(d\left(T^{n+1}\mathbf{x}\_{1},T^{n+1}\mathbf{x}\_{2}\right),d\left(T^{n}\mathbf{x}\_{1},T^{n}\mathbf{x}\_{2}\right)\right)\geq 0\quad\text{for all }n\in\mathbb{N}.\tag{10.4}$$

Let us denote

$$\mathbf{x}\_n^1 = T^\eta \mathbf{x}\_1 \quad \text{and} \quad \mathbf{x}\_n^2 = T^\eta \mathbf{x}\_2 \quad \text{for all } n \in \mathbb{N}.$$

Hence, by hypothesis, *x*1*n* = *Tnx*1S∗*Tnx*2 = *x*2*n* and *Tx*1*n* = *Tn*+1*x*1S∗*Tn*+1*x*2 = *Tx*2*n*. Applying Condition (A4), for all *n* ∈ N,

$$\varrho(d\left(T^{n+1}\mathbf{x}\_1, T^{n+1}\mathbf{x}\_2\right), d\left(T^n\mathbf{x}\_1, T^n\mathbf{x}\_2\right)) = \varrho(d\left(T\mathbf{x}\_{n'}^1, T\mathbf{x}\_n^2\right), d\left(\mathbf{x}\_{n'}^1, \mathbf{x}\_n^2\right)) > 0.$$

Therefore, Condition (A2) implies that {*d*(*Tnx*1, *Tnx*2)} → 0.

[ (A2) + (A4) ⇒ (B2) ] It follows as in the previous implication by using *x*1 = *x*0 and *x*2 = *Tx*0. [ (A3) + (A4) ⇒ (B3) ] Let {(*an*, *bn*)} ⊆ *A* × *A* be a (*<sup>T</sup>*, S∗)-sequence such that {*an*} and {*bn*} converge to the same limit *L* ≥ 0 and verifying that *L* < *an* and (*an*, *bn*) ≥ 0 for all *n* ∈ N. By definition, there are two sequences {*xn*}, {*yn*} ⊆ *X* such that

*xn*S∗*yn*, *Txn*S∗*Tyn*, *an* = *d*(*Txn*, *Tyn*) > 0 and *bn* = *d*(*xn*, *yn*) > 0 for all *n* ∈ N.

As *xn*S∗*yn* and *Txn*S∗*Tyn*, then it follows from (A4) that

$$\varrho(a\_{n\prime}b\_n) = \varrho(d(T\mathbf{x}\_n, T\mathbf{y}\_n), d(\mathbf{x}\_n\mathbf{y}\_n)) > 0 \quad \text{for all } n \in \mathbb{N}.$$

Therefore, applying (A3), we conclude that *L* = 0.

[ (A4) + (A5) ⇒ (B5) ] Let {(*an*, *bn*)} be a (*<sup>T</sup>*, S∗)-sequence such that {*bn*} → 0 and (*an*, *bn*) ≥ 0 for all *n* ∈ N. By definition, there exist two sequences {*xn*}, {*yn*} ⊆ *X* such that

$$\text{tr}\_n \mathbb{S}^\* y\_n \quad \text{Tx}\_n \mathbb{S}^\* Ty\_n \quad a\_n = d(\text{Tx}\_n, Ty\_n) > 0 \quad \text{and} \quad b\_n = d(\text{x}\_n, y\_n) > 0 \quad \text{for all } n \in \mathbb{N}.$$

As *xn*S∗*yn* and *Txn*S∗*Tyn*, then it follows from (A4) that

$$d\varrho(a\_{n\_{\ell}}b\_{n}) = \varrho(d(T\mathbf{x}\_{n\_{\ell}}Ty\_{n}), d(\mathbf{x}\_{n\_{\ell}}y\_{n})) > 0 \quad \text{for all } n \in \mathbb{N}.$$

Therefore, applying (A5), we conclude that {*an*} → 0.

The previous theorem provides us a large list of ample spectrum contractions because every (A, S)-contraction is an ample spectrum contraction. In particular, as the authors proved in [3,5], the following ones are examples of ample spectrum contractions:


The converse of Theorem 1 is false, as we show in the following example:

**Example 1.** *Let X* = {0, 1, 3} *be endowed with the Euclidean metric dE*(*<sup>x</sup>*, *y*) = | *x* − *y* | *and the usual order* ≤*. Hence,* (*<sup>X</sup>*, *dE*) *is a complete metric space. Let A* = ran(*dE*) = {0, 1, 2, 3} *and let T* : *X* → *X and* : *A* × *A* → R *be defined by*

$$T\mathbf{x} = \begin{cases} \ 0, & \text{if } \mathbf{x} \in \{0, 1\}, \\\ 1, & \text{if } \mathbf{x} = 3; \end{cases} \qquad \varrho(t, \mathbf{s}) = 0 \text{ for all } t, \mathbf{s} \in A.$$

*Then, T is not an* (A, ≤)*-contraction with respect to because, if x* = 1 *and y* = 3*, then x* < *y and Tx* < *Ty, but* (*d*(*Tx*, *Ty*), *d*(*<sup>x</sup>*, *y*)) = 0*. Let us show that T is an ample spectrum contraction with respect to and* ≤*. Condition* (B4) *is obvious. Properties* (B2) *and* (B2) *follows from the fact that any Picard sequence* {*xn*} *of T must verify xn* = 0 *for all n* ≥ 3*. Taking into account that any convergent sequence on A is almost constant (because it is discrete), Axioms* (B3) *and* (B5) *are satisfied because such kind of sequences do not exist. Hence, T is an ample spectrum contraction with respect to and* ≤*.*

The notion of (*<sup>T</sup>*, S∗)-sequence plays a key role in the definition of ample spectrum contraction. In fact, if we had not changed the notion of (*<sup>T</sup>*, S)-sequence by the concept of (*<sup>T</sup>*, S∗)-sequence in Definition 4, then there would have not been any relationship between (A, S)-contractions and ample spectrum contractions. We illustrate this affirmation with the following example.

**Example 2.** *Let X* = [0, 2] ∪ *C* ∪ *D, where C* = { 10*m* ∈ N : *m* ∈ N∗ } *and D* = { 10*m* + 4 ∈ N : *m* ∈ N∗ }*. Assume that X is endowed with the Euclidean metric dE*(*<sup>x</sup>*, *y*) = | *x* − *y* | *and the usual order* ≤*. Hence,* (*<sup>X</sup>*, *dE*) *is a complete metric space. The range of dE can be expressed as*

$$\text{ran}(d\_E) = [0,2] \cup \{4\} \cup B \quad \text{where } B \subsetneq [6, \infty) \dots$$

*Let A* = ran(*dE*) *and let T* : *X* → *X and* : *A* × *A* → R *be defined by*

$$T\mathbf{x} = \begin{cases} \frac{\mathbb{X}}{\mathbf{4}}, & \text{if } \mathbf{x} \in [0, 2] \\ 1 + \frac{1}{m}, & \text{if } \mathbf{x} = 10m \in \mathbb{C} \quad \text{(for some } m \in \mathbb{N}^\*), \\ 0, & \text{if } \mathbf{x} = 10m + 4 \in D \quad \text{(for some } m \in \mathbb{N}^\*); \end{cases}$$

$$\varrho(t, \mathbf{s}) = \begin{cases} 0, & \text{if } t > 1 \text{ and } \mathbf{s} \ge 1, \\ \frac{\mathbf{s}}{2} - t, & \text{otherwise.} \end{cases}$$

*Notice that T satisfies the following properties.*

(*p*1) *T*(*X*) ⊂ [0, 2]*. In particular,* | *Tx* − *Ty* | ≤ 2 *for all x*, *y* ∈ *X.*


*Let us show that T is an ample spectrum contraction with respect to and* ≤*.* (B2) *Let* {*xn*} ⊆ *X be a Picard* S*-nondecreasing sequence of T such that*

> *xn* = *xn*+1 *and* (*dE* (*xn*+1, *xn*+<sup>2</sup>), *dE* (*xn*, *xn*+<sup>1</sup>)) ≥ 0 *for all n* ∈ N.

*Since* {*xn*} → 0*,* {*dE* (*xn*, *xn*+<sup>1</sup>)} → 0*.*

(B3) *Let* {(*an*, *bn*)} ⊆ *A* × *A be a* (*<sup>T</sup>*, <sup>&</sup>lt;)*-sequence such that* {*an*} *and* {*bn*} *converge to the same limit L* ≥ 0 *and verifying that L* < *an and* (*an*, *bn*) ≥ 0 *for all n* ∈ N*. By definition, there are two sequences* {*xn*}, {*yn*} ⊆ *X such that*

$$\text{tr}\_n < y\_n, \quad \text{Tx}\_n < \text{Ty}\_n, \quad a\_n = d\_E(\text{Tx}\_n, \text{Ty}\_n) > 0 \quad \text{and} \quad b\_n = d(\text{x}\_n, y\_n) > 0 \quad \text{for all } n \in \mathbb{N}.$$

*As an* = *dE*(*Txn*, *Tyn*) ∈ [0, 2]*, then L* ≤ 2*. Since* {*bn* = *dE*(*xn*, *yn*)} → *L* ≤ 2*, there exists n*0 ∈ N *such that dE*(*xn*, *yn*) < 4 *for all n* ≥ *n*0*. By* (*p*2)*, we have that xn*, *yn* ∈ [0, 2] *for all n* ≥ *n*0*. Therefore, for all n* ≥ *n*0*,*

$$d\_n = d\_E(T \mathbf{x}\_{n\prime} T \mathbf{y}\_n) = \left\lfloor \frac{\mathbf{x}\_n}{4} - \frac{\mathbf{y}\_n}{4} \right\rfloor = \frac{\left\lfloor \mathbf{x}\_n - \mathbf{y}\_n \right\rfloor}{4} = \frac{b\_n}{4}.$$

.

*Letting n* → <sup>∞</sup>*, we deduce that L* = *L*/4*, so L* = 0*.*

(B4) *Let x*, *y* ∈ *X be two points such that x* < *y and Tx* < *Ty. To prove that* (*d*(*Tx*, *Ty*), *d*(*<sup>x</sup>*, *y*)) ≥ 0*, we observe three cases.*


$$\log\left(d(T\mathbf{x}, T\mathbf{y}), d(\mathbf{x}, \mathbf{y})\right) = \frac{|\mathbf{x} - \mathbf{y}|}{2} - |T\mathbf{x} - T\mathbf{y}| = \frac{\mathbf{y} - \mathbf{x}}{2} - (T\mathbf{y} - T\mathbf{x}) \text{ (by }$$

*which corresponds to the case in which* | *Tx* − *Ty* | ≤ 1 *or* | *x* − *y* | < 1*.*


$$\varrho\left(d(Tx, Ty), d(x, y)\right) = \frac{\left\lfloor x - y \right\rfloor}{2} - \left\lfloor Tx - Ty \right\rfloor \ge \frac{4}{2} - 2 = 0.$$


$$\log\left(d(\mathbf{T}\mathbf{x}, \mathbf{T}y), d(\mathbf{x}, y)\right) = \frac{\|\mathbf{x} - \mathbf{y}\|}{2} - \left|\mathbf{T}\mathbf{x} - \mathbf{T}y\right| = \frac{\|\mathbf{x} - \mathbf{y}\|}{2} - \left|\frac{\mathbf{x}}{4} - \frac{\mathbf{y}}{4}\right| = \frac{\|\mathbf{x} - \mathbf{y}\|}{4} > 0,$$

*which means that* (B4) *holds.*

*In any case,*(B4) *holds.*

The following result is useful in order to study when an ample spectrum contraction can have multiple fixed points.

**Proposition 4.** *Let* (*<sup>X</sup>*, *d*) *be a metric space endowed with a binary relation* S *and let T* : *X* → *X and* : *A* × *A* → R *be two maps such that* (B1)*,* (B 2) *and* (B4) *holds. If ω*, *ω* ∈ *X are two* S*-comparable fixed points of T, then ω* = *ω .*

**Proof.** Reasoning by contradiction, assume that *ω* and *ω* are two distinct fixed points of *T*. As *ω* and *ω* are S-comparable, we can suppose, without loss of generality, that *ω* S *<sup>ω</sup>*. Hence, *ω* S∗*ω* and also *T ω* S∗*T <sup>ω</sup>*. Let *an* = *d* (*<sup>ω</sup>*, *ω* ) > 0 for all *n* ∈ N. By using (B4), for all *n* ∈ N,

$$\varrho\left(a\_{n+1}, a\_n\right) = \varrho\left(d\left(\omega, \omega'\right), d\left(\omega, \omega'\right)\right) \\ = \varrho\left(d\left(T\omega, T\omega'\right), d\left(\omega, \omega'\right)\right) \ge 0.$$

Therefore, it follows from (B 2) that {*an* = *d* (*<sup>ω</sup>*, *ω* )} → 0, which is a contradiction. Thus, *ω* = *ω* .

#### **4. Fixed Point Theorems Involving Ample Spectrum Contractions**

Once we have changed the notions of (*<sup>T</sup>*, S)-sequence and (A, S)-contraction by the concepts of (*<sup>T</sup>*, S∗)-sequence and ample spectrum contraction, we are ready to introduce the main results of the manuscript, which is the aim of the current section. Concretely, as we show below, the following one is the most general theorem of this manuscript.

**Theorem 2.** *Let* (*<sup>X</sup>*, *d*) *be a metric space endowed with a transitive binary relation* S *and let T* : *X* → *X be an* S*-nondecreasing ample spectrum contraction with respect to* : *A* × *A* → R*. Suppose that T*(*X*) *is* (S, *d*)*-strictly-increasing-precomplete and there exists a point x*0 ∈ *X such that <sup>x</sup>*0S*Tx*0*. Assume that at least one of the following conditions is fulfilled:*


*Then, the Picard sequence of T based on x*0 *converges to a fixed point of T. In particular, T has at least a fixed point.*

Notice that the metric space (*<sup>X</sup>*, *d*) needs not to be complete.

**Proof.** Let *x*0 ∈ *X* be a point such that *<sup>x</sup>*0S*Tx*0 and let {*xn*+<sup>1</sup> = *Txn*}*n*≥<sup>0</sup> be the Picard sequence of *T* based on *x*0. If there exists some *n*0 ∈ N such that *xn*0+<sup>1</sup> = *xn*0, then *xn*0is a fixed point of *T*,

and {*xn*} converges to such point. On the contrary case, assume that *xn* = *xn*+1 for all *n* ∈ N. As *T* is S-nondecreasing and *<sup>x</sup>*0S*Tx*0 = *x*1, then *xn*S*xn*+<sup>1</sup> for all *n* ∈ N, and, as S is transitive,

$$\mathbf{x}\_n \mathbf{S} \mathbf{x}\_m \quad \text{for all } n, m \in \mathbb{N} \text{ such that } n < m. \tag{2}$$

In fact, as *xn* = *xn*+1 for all *n* ∈ N, then

$$\text{tr}\_{\mathfrak{U}} \mathbb{S}^\* \mathbf{x}\_{n+1} \quad \text{and} \quad T \mathbf{x}\_{\mathfrak{U}} \mathbb{S}^\* T \mathbf{x}\_{n+1} \quad \text{for all } n \in \mathbb{N}. \tag{3}$$

Let consider the sequence {*d*(*xn*, *xn*+<sup>1</sup>)} ⊆ *A*. Taking into account Equation (3) and the fact that *T* is an ample spectrum contraction, Condition (B4) implies that, for all *n* ∈ N,

$$\varrho\left(d\left(T^{n+1}\mathbf{x}\_0, T^{n+2}\mathbf{x}\_0\right), d\left(T^n\mathbf{x}\_0, T^{n+1}\mathbf{x}\_0\right)\right) = \varrho\left(d(T\mathbf{x}\_n, T\mathbf{x}\_{n+1}), d(\mathbf{x}\_n, \mathbf{x}\_{n+1})\right) \ge 0.$$

Applying (B2) ,we deduce that {*xn* = *Tnx*0} is an asymptotically regular sequence on (*<sup>X</sup>*, *d*), that is, {*d*(*xn*, *xn*+<sup>1</sup>)} → 0.

Let us show that {*xn*} is an S-strictly-increasing sequence. Indeed, in view of Equation (2), assume that there exists *n*0, *m*0 ∈ N such that *n*0 < *m*0 and *xn*0 = *xm*0 . If *p*0 = *m*0 − *n*0 ∈ N{0}, then *xn*0 = *xn*0+*<sup>k</sup> p*0 for all *k* ∈ N. In particular, the sequence {*d*(*xn*, *xn*+<sup>1</sup>)} contains the constant subsequence

$$d\left\{d(\mathbf{x}\_{n\_0+k}\mathbf{z}\_{p\_0\prime}\mathbf{x}\_{n\_0+k}\mathbf{p}\_0+1) = d(\mathbf{x}\_{n\_0\prime}\mathbf{x}\_{n\_0+1}) > 0\right\}\_{k\in\mathbb{N}}\prime$$

which contradicts the fact that {*d*(*xn*, *xn*+<sup>1</sup>)} → 0. This contradiction guarantees that *xn* = *xm* for all *n* = *m*, thus *xn*S∗*xm* for all *n*, *m* ∈ N such that *n* < *m*, that is, {*xn*} is an S-strictly-increasing sequence.

Next, we show that {*xn*} is a Cauchy sequence reasoning by contradiction. If {*xn*} is not a Cauchysequence, then there exist *ε*0 > 0 and two subsequences {*xn*(*k*)} and {*xm*(*k*)} of {*xn*} such that

$$\begin{aligned} k \le n(k) &< m(k), \quad d(\mathbf{x}\_{n(k)}, \mathbf{x}\_{m(k)-1}) \le \varepsilon\_0 < d(\mathbf{x}\_{n(k)}, \mathbf{x}\_{m(k)}) \quad \text{for all } k \in \mathbb{N}, \\\lim\_{k \to \infty} d(\mathbf{x}\_{n(k)}, \mathbf{x}\_{m(k)}) &= \lim\_{k \to \infty} d(\mathbf{x}\_{n(k)-1}, \mathbf{x}\_{m(k)-1}) = \varepsilon\_0. \end{aligned}$$

Let *L* = *ε*0 > 0, {*ak* = *<sup>d</sup>*(*xn*(*k*), *xm*(*k*))} → *L* and {*bk* = *<sup>d</sup>*(*xn*(*k*)−1, *xm*(*k*)−<sup>1</sup>)} → *L*. As *n*(*k*) < *m*(*k*) (and *n*(*k*) − 1 < *m*(*k*) − 1), then *xn*(*k*)S∗*xm*(*k*) and *xn*(*k*)−<sup>1</sup>S∗*xm*(*k*)−1. Thus, {(*ak*, *bk*)} is a (*<sup>T</sup>*, S∗)-sequence. Since *L* = *ε*0 < *<sup>d</sup>*(*xn*(*k*), *xm*(*k*)) = *ak* and

$$\begin{aligned} \varrho\left(a\_k, b\_k\right) &= \varrho\left(d(\mathfrak{x}\_{\mathfrak{n}(k)}, \mathfrak{x}\_{\mathfrak{m}(k)}), d(\mathfrak{x}\_{\mathfrak{n}(k)-1}, \mathfrak{x}\_{\mathfrak{m}(k)-1})\right) \\ &= \varrho\left(d(T\mathfrak{x}\_{\mathfrak{n}(k)-1}, T\mathfrak{x}\_{\mathfrak{m}(k)-1}), d(\mathfrak{x}\_{\mathfrak{n}(k)-1}, \mathfrak{x}\_{\mathfrak{m}(k)-1})\right) \ge 0 \end{aligned}$$

for all *k* ∈ N, Condition (B3) guarantees that *ε*0 = *L* = 0, which is a contradiction. As a consequence, {*xn*} is a Cauchy sequence. Since {*xn*}*n*≥<sup>1</sup> ⊆ *T*(*X*) and *T*(*X*) is (S, *d*)-strictly-increasing-precomplete, there is a subset *Z* ⊆ *X* such that *T*(*X*) ⊆ *Z* ⊆ *X* and *Z* is (S, *d*)-strictly-increasing-complete. In particular, as {*xn*} is an S-strictly-increasing and Cauchy sequence, there exists *z* ∈ *Z* ⊆ *X* such that {*xn*} → *z*. Let us show that *z* is a fixed point of *T* considering three cases.

*Case 1. Assume that T is* S*-strictly-increasing-continuous.* In this case, {*xn*+<sup>1</sup> = *Txn*} → *Tz*, so *Tz* = *z*.

*Case 2. Assume that* (*<sup>X</sup>*, *d*) *is* S*-strictly-increasing-regular and condition* (B5) *holds.* In this case, as {*xn*} is an S-strictly-increasing sequence such that {*xn*} → *z*, it follows that

$$\text{tr}\_{\mathfrak{n}} \mathbb{S}z \text{ for all } n \in \mathbb{N}. \tag{4}$$

Since *T* is *S*-nondecreasing,

$$T\mathbf{x}\_{\mathbb{N}}\mathbf{S}T\mathbf{z}\text{ for all }\mathfrak{n}\in\mathbb{N}.\tag{5}$$

Let *an* = *d*(*xn*+1, *Tz*) = *d*(*Txn*, *Tz*) and *bn* = *d*(*xn*, *z*) for all *n* ∈ N. Clearly, {*bn*} → 0. Notice that

$$b\_n = 0 \quad \Rightarrow \quad a\_n = 0 \tag{6}$$

because

$$b\_{\mathbb{N}} = 0 \quad \Leftrightarrow \quad \mathbf{x}\_{\mathbb{N}} = z \quad \Rightarrow \quad \mathbf{x}\_{\mathbb{n}+1} = T\mathbf{x}\_{\mathbb{n}} = Tz \quad \Leftrightarrow \quad a\_{\mathbb{n}} = 0.$$

Let consider the set

$$\Omega = \{ n \in \mathbb{N} : a\_n = 0 \} = \{ n \in \mathbb{N} : d(\mathfrak{x}\_{n+1}, Tz) = 0 \}.$$

*Subcase 2.1. Assume that* Ω *is finite.* In this case, there exists *n*0 ∈ N such that *d*(*xn*+1, *Tz*) = *an* > 0 for all *n* ≥ *n*0. By (6), *d*(*xn*, *z*) = *bn* > 0 for all *n* ≥ *n*0. In this case, {(*an*, *bn*)}*n*≥*<sup>n</sup>*0 is a (*<sup>T</sup>*, S)-sequence (because *an* = *d*(*Txn*, *Tz*) > 0 and *bn* = *d*(*xn*, *z*) > 0 for all *n* ≥ *n*0). In particular, *xn* = *z* and *Txn* = *Tz* for all *n* ≥ *n*0. By Equations (4) and (5), we deduce that *xn*S∗*z* and *Txn*S∗*Tz* for all *n* ≥ *n*0. It follows from (B4) that

$$
\varrho(a\_n, b\_n) = \varrho\left(d(T\mathbf{x}\_n, Tz), d(\mathbf{x}\_n, z)\right) \ge 0 \quad \text{for all } n \ge n\_0.
$$

As a consequence, as (B5) holds, we conclude that {*an* = *d*(*xn*+1, *Tz*)} → 0, that is, {*xn*+<sup>1</sup>} → *Tz*, which guarantees that *Tz* = *z*.

*Subcase 2.2. Assume that* Ω *is not finite.* In this case, there exists a subsequence {*xn*(*k*)} of {*xn*} such that

$$d(\mathfrak{x}\_{\mathfrak{n}(k)+1'}Tz) = 0 \quad \text{for all } k \in \mathbb{N}.$$

Hence, *xn*(*k*)+1 = *Tz* for all *k* ∈ N. Since {*xn*} → *z* and *xn*(*k*)+1 → *Tz*, *Tz* = *z*.

*Case 3. Assume that* (*<sup>X</sup>*, *d*) *is* S*-strictly-increasing-regular and* (*<sup>t</sup>*,*<sup>s</sup>*) ≤ *s* − *t for all t*,*<sup>s</sup>* ∈ *A* ∩ (0, <sup>∞</sup>)*.* Proposition 3 guarantees that Item (*b*) is applicable.

In any case, we conclude that *z* is a fixed point of *T*.

In the following result, we describe sufficient conditions in order to guarantee uniqueness of the fixed point.

**Theorem 3.** *Under the hypotheses of Theorem 2, assume that the following properties are fulfilled:*


*Then, T has a unique fixed point.*

**Proof.** Let *x*, *y* ∈ Fix(*T*) be two fixed points of *T*. By hypothesis, there exists *z*0 ∈ *X* such that *z*0 is, at the same time, S-comparable to *x* and S-comparable to *y*. Let {*zn*} be the Picard sequence of *T* based on *z*0, that is, *zn*+1 = *Tzn* for all *n* ∈ N. We prove that *x* = *y* by showing that {*zn*} → *x* and {*zn*} → *y*. We first use *x*, but the same argumen<sup>t</sup> is valid for *y*.

Since *z*0 is S-comparable to *x*, assume that *<sup>z</sup>*0S*<sup>x</sup>* (the case *<sup>x</sup>*S*<sup>z</sup>*0 is similar). As *T* is S-nondecreasing, *zn*S*x* for all *n* ∈ N. If there exists *n*0 ∈ N such that *zn*0 = *x*, then *zn* = *x* for all *n* ≥ *n*0. In particular, {*zn*} → *x* and the proof is finished. On the contrary case, assume that *zn* = *x* for all *n* ∈ N. Therefore *zn*S∗*x* and *Tzn*S∗*Tx* for all *n* ∈ N. Using the contractivity Condition (B4), for all *n* ∈ N,

$$0 \le \varrho(d(Tz\_n, Tx), d(z\_n, x)) = \varrho(d(T^{n+1}z\_{0\prime}T^{n+1}x), d(T^n z\_{0\prime}T^n x)).$$

It follows from (B2) that {*d*(*Tnz*0, *Tnx*)} → 0, that is, {*zn*} → *x*.

## **5. Consequences**

In this section, we illustrate how many well known theorems in fixed point theory (that involve only *d*(*<sup>x</sup>*, *y*) and *d*(*Tx*, *Ty*) in their contractivity conditions) can be deduced from our main results.

#### *5.1. Meir–Keeler Contractions*

Meir and Keeler generalized the Banach theorem in a way that have attracted much attention in the last 40 years.

**Definition 5** (Meir and Keeler [15])**.** *A* Meir–Keeler contraction *is a mapping T* : *X* → *X from a metric space* (*<sup>X</sup>*, *d*) *into itself such that for all ε* > 0*, there exists δ* > 0 *verifying that if x*, *y* ∈ *X and ε* ≤ *d*(*<sup>x</sup>*, *y*) < *ε* + *δ, then d*(*Tx*, *Ty*) < *ε.*

Lim characterized this kind of mappings in terms of a contractivity condition using the following class of auxiliary functions.

**Definition 6** (Lim [14])**.** *A function φ* : [0, ∞) → [0, ∞) *is called an* L-function *if*

*(a) φ*(0) = 0*;*


Each *L*-function must satisfy:

$$
\phi(t) \le t \quad \text{for all } t \in [0, \infty) \,. \tag{7}
$$

**Theorem 4** (Lim [14], Theorem 1)**.** *Let* (*<sup>X</sup>*, *d*) *be a metric space and let T* : *X* → *X be a self-mapping. Then, T is a Meir–Keeler mapping if, and only if, there exists an (non-decreasing, right-continuous) L-map φ such that*

$$d(Tx, Ty) < \phi (d(x, y)) \quad \text{for all } x, y \in X \text{ verify} \\ \text{big } d(x, y) > 0. \tag{8}$$

Meir and Keeler [15] demonstrated the following fixed point theorem by using a result of Chu and Diaz [18].

**Theorem 5** (Meir and Keeler [15])**.** *Every Meir–Keeler contraction from a complete metric space into itself has a unique fixed point.*

We prove that this result can be immediately deduced from our main statements.

**Theorem 6.** *Every Meir–Keeler contraction is an ample spectrum contraction that also verifies* (B 2) *and* (B5)*.*

**Proof.** Let (*<sup>X</sup>*, *d*) be a metric space and let *T* : *X* → *X* be a Meir–Keeler contraction. By Theorem 4, there exists an *L*-map *φ* : [0, ∞) → [0, ∞) verifying Equation (8). Let *A* = ran(*d*) and let define *φ* : *A* × *A* → R by *φ*(*<sup>t</sup>*,*<sup>s</sup>*) = *φ* (*s*) − *t* for all *t*,*<sup>s</sup>* ∈ *A*. Let us show that *T* is an ample spectrum contraction with respect to *φ*.

(B 2) Let *x*1, *x*2 ∈ *X* be two points such that

$$T^{\mathfrak{n}}\mathbf{x}\_1 \neq T^{\mathfrak{n}}\mathbf{x}\_2 \quad \text{and} \quad \varrho\_{\varPhi}(d\left(T^{\mathfrak{n}+1}\mathbf{x}\_1, T^{\mathfrak{n}+1}\mathbf{x}\_2\right), d\left(T^{\mathfrak{n}}\mathbf{x}\_1, T^{\mathfrak{n}}\mathbf{x}\_2\right)) \geq 0 \quad \text{for all } \mathfrak{n} \in \mathbb{N}.$$

As *d* (*Tnx*1, *Tnx*2) > 0, it follows from Equations (7) and (8) that, for all *n* ∈ N,

$$d\left(T^{n+1}\mathbf{x}\_1, T^{n+1}\mathbf{x}\_2\right) = d\left(TT^{\mathfrak{u}}\mathbf{x}\_1, TT^{\mathfrak{u}}\mathbf{x}\_2\right) \\ < \phi\left(d\left(T^{\mathfrak{u}}\mathbf{x}\_1, T^{\mathfrak{u}}\mathbf{x}\_2\right)\right) \\ \le d\left(T^{\mathfrak{u}}\mathbf{x}\_1, T^{\mathfrak{u}}\mathbf{x}\_2\right).$$

As {*d*(*Tnx*1, *Tnx*2)} is a bounded-below decreasing sequence of real numbers, it is convergent. Let *L* ≥ 0 be its limit. To prove that *L* = 0, we reason by contradiction. Assume that *L* > 0. Hence,

$$0 < L \le d(T^{n+1}\mathbf{x}\_1, T^{n+1}\mathbf{x}\_2) < \phi(d(T^n\mathbf{x}\_1, T^n\mathbf{x}\_2)) \le d(T^n\mathbf{x}\_1, T^n\mathbf{x}\_2) \quad \text{for all } n \in \mathbb{N}.$$

Letting *ε* = *L* > 0 in Condition (*c*) of Definition 6, there exists *δ* > 0 such that *φ*(*t*) ≤ *ε* = *L* for all *t* ∈ [*<sup>ε</sup>*,*<sup>ε</sup>* + *δ*]. As {*d*(*Tnx*1, *Tnx*2)} *L*+, there exists *n*0 ∈ N such that *L* < *d*(*Tn*0 *x*1, *Tn*0 *<sup>x</sup>*2) < *L* + *δ* for all *n* ≥ *n*0. Therefore,

$$\phi(d(T^{n\_0}\mathbf{x}\_1, T^{n\_0}\mathbf{x}\_2)) \le \varepsilon = L < \phi(d(T^{n\_0}\mathbf{x}\_1, T^{n\_0}\mathbf{x}\_2)),$$

which is a contradiction. Thus, *L* = 0 and {*d* (*Tnx*1, *Tnx*2)} → 0.

(B2) It follows from (B2).

(B3) Let {(*an*, *bn*)} ⊆ *A* × *A* be a *T*-sequence such that {*an*} and {*bn*} converge to the same limit *L* ≥ 0 and verifying that *L* < *an* and *φ*(*an*, *bn*) ≥ 0 for all *n* ∈ N. By definition, there exist two sequences {*xn*}, {*yn*} ⊆ *X* such that

$$a\_{\mathbb{N}} = d(T\mathbf{x}\_{\mathbb{N}}, T\mathbf{y}\_{\mathbb{N}}) > 0 \quad \text{and} \quad b\_{\mathbb{N}} = d(\mathbf{x}\_{\mathbb{N}}, \mathbf{y}\_{\mathbb{N}}) > 0 \quad \text{for all } \mathbb{N} \in \mathbb{N}.$$

Notice that, from Equation (8), for all *n* ∈ N,

$$L < a\_n = d(T\mathfrak{x}\_n, Ty\_n) < \phi(d(\mathfrak{x}\_n, y\_n)) = \phi(b\_n) \le b\_n.$$

To prove that *L* = 0, assume that *L* > 0. Letting *ε* = *L* > 0 in Condition (*c*) of Definition 6, there exists *δ* > 0 such that

$$
\phi(t) \le \varepsilon = L \quad \text{for all } t \in \left[\varepsilon, \varepsilon + \delta\right].
$$

As {*d* (*xn*, *yn*)} *L*+, there exists *n*0 ∈ N such that *L* < *d* (*xn*, *yn*) < *L* + *δ* for all *n* ≥ *n*0. Therefore,

$$\phi(d\left(\mathfrak{x}\_{n\_{0}\prime}y\_{n\_{0}}\right)) \le \varepsilon = L < \phi(d\left(\mathfrak{x}\_{n\_{0}\prime}y\_{n\_{0}}\right)),$$

which is a contradiction. Thus, *L* = 0.

(B4) It is clear that, for all *x*, *y* ∈ *X* such that *d*(*<sup>x</sup>*, *y*) > 0 and *d*(*Tx*, *Ty*) > 0, Theorem 4 guarantees that

$$\varrho\_{\Phi} \left( d(Tx, Ty), d(x, y) \right) = \phi \left( d(x, y) \right) - d(Tx, Ty) > 0.$$

(B5) Let {(*an*, *bn*)} be a *T*-sequence such that {*bn*} → 0 and *φ*(*an*, *bn*) ≥ 0 for all *n* ∈ N. Then, for all *n* ∈ N,

$$0 \le \varrho\_{\Phi}(a\_{\mathcal{U}} b\_{\mathcal{U}}) = \phi\left(b\_{\mathcal{U}}\right) - a\_{\mathcal{U}}.$$

which means that 0 ≤ *an* ≤ *φ* (*bn*) ≤ *bn*. Therefore, {*bn*} → 0 implies {*an*} → 0.

**Theorem 7.** *Theorem 5 follows from Theorems 2 and 3.*

**Proof.** From Theorem 6, every Meir–Keeler contraction is an ample spectrum contraction that also verifies (B2) and (B5), thus Theorems 2 and 3 are applicable in order to conclude that every Meir–Keeler contraction has a unique fixed point.

#### *5.2. Samet et al.'s Contractions*

In [9], Samet et al. introduced the following kind of contractions and proved the following results. Let us denote by Ψ the family of nondecreasing functions *ψ* : [0, ∞) → [0, ∞) such that <sup>Σ</sup>*n*∈<sup>N</sup>*ψ<sup>n</sup>*(*t*) < ∞ for each *t* > 0, where *ψn* is the *n*th iterate of *ψ*.

**Definition 7.** *Let* (*<sup>X</sup>*, *d*) *be a metric space and T* : *X* → *X be a given mapping. We say that T is an α - ψ - contractive mapping if there exist two functions α* : *X* × *X* → [0, ∞) *and ψ* ∈ Ψ *such that*

$$d(\mathbf{x}, \mathbf{y}) \, d(T\mathbf{x}, T\mathbf{y}) \le \psi(d(\mathbf{x}, \mathbf{y})) \quad \text{for all } \mathbf{x}, \mathbf{y} \in X. \tag{9}$$

The main results in [9] can be summarized as follows.

**Theorem 8** (Samet, Vetro and Vetro [9], Theorems 2.1, 2.2 and 2.3)**.** *Let* (*<sup>X</sup>*, *d*) *be a complete metric space and T* : *X* → *X be an α - ψ - contractive mapping satisfying the following conditions:*

(*i*) *T is α - admissible (that is, if <sup>α</sup>*(*<sup>x</sup>*, *y*) ≥ 1*, then <sup>α</sup>*(*Tx*, *Ty*) ≥ 1*);*

(*ii*) *there exists x*0 ∈ *X such that <sup>α</sup>*(*<sup>x</sup>*0, *Tx*0) ≥ 1*; and*

(*iii*) *at least, one of the following conditions holds:*

> (*iii*.1) *T is continuous; or* (*iii*.2) *if* {*xn*} *is a sequence in X such that <sup>α</sup>*(*xn*, *xn*+<sup>1</sup>) ≥ 1 *for all n and* {*xn*} → *x* ∈ *X as n* → <sup>∞</sup>*, then <sup>α</sup>*(*xn*, *x*) ≥ 1 *foralln.*

*Then, T has a fixed point, that is, there exists x*<sup>∗</sup> ∈ *X such that Tx*∗ = *<sup>x</sup>*<sup>∗</sup>*. Furthermore, adding the condition:*

(*H*) *for all x*, *y* ∈ *X, there exists z* ∈ *X such that <sup>α</sup>*(*<sup>x</sup>*, *z*) ≥ 1 *and <sup>α</sup>*(*y*, *z*) ≥ 1*,*

*we obtain uniqueness of the fixed point of T .*

To show that the previous theorem can be seen as a consequence of our main results, we present the following statement in which we use a more general class of auxiliary functions.

**Theorem 9.** *Let* (*<sup>X</sup>*, *d*) *be a metric space and T* : *X* → *X be a given mapping. Assume that there exist two functions α* : *X* × *X* → [0, ∞) *and ψ* : [0, ∞) → [0, ∞) *such that ψ is nondecreasing,* lim*n*→∞ *ψn* (*t*) = 0 *for all t* > 0*, and also*

$$d(\mathbf{x}, \mathbf{y}) \, d(T\mathbf{x}, T\mathbf{y}) \le \psi(d(\mathbf{x}, \mathbf{y})) \quad \text{for all } \mathbf{x}, \mathbf{y} \in X. \tag{10}$$

*Then, T is an ample spectrum contraction with respect to* S*α that also verifies* (B2) *and* (B5)*.*

**Proof.** Let S*α* be the binary relation on *X* given in (1). Let *A* = ran(*d*) and let define *γ* : *A* → R and *γ* : *A* × *A* → R by, for all *t*,*<sup>s</sup>* ∈ *A*,

$$\begin{aligned} \gamma\left(s\right) &= \inf\left(\left\{a\left(x,y\right) : d\left(x,y\right) = s\right\}\right), \\ \varrho\_{\vargamma}\left(t,s\right) &= \psi\left(s\right) - t\,\gamma\left(s\right). \end{aligned}$$

Notice that *γ* is well defined because if *s* ∈ *A* = ran(*d*), then there exist *xs*, *ys* ∈ *X* such that *d*(*xs*, *ys*) = *s*, and we can take infimum in a nonempty, subset of non-negative real numbers. Furthermore, as *γ* (*d*(*<sup>x</sup>*, *y*)) ≤ *<sup>α</sup>*(*<sup>x</sup>*, *y*) for all *x*, *y* ∈ *X*, then, by (10),

$$\begin{aligned} \varrho\_{\gamma} \left( d(Tx, Ty), d(x, y) \right) &= \psi \left( d(x, y) \right) - d(Tx, Ty) \,\gamma \left( d(x, y) \right) \\ &\ge \psi \left( d(x, y) \right) - d(Tx, Ty) \,\mathfrak{a} \left( x, y \right) \ge 0. \end{aligned}$$

Hence, (B4) holds. Let us prove the rest of properties.

(B2) Let *x*1, *x*2 ∈ *X* be two points such that

> *<sup>T</sup>nx*1S∗*α Tnx*2 and *φ*(*<sup>d</sup> <sup>T</sup><sup>n</sup>*+1*x*1, *<sup>T</sup><sup>n</sup>*+1*x*<sup>2</sup> , *d* (*Tnx*1, *<sup>T</sup>nx*2)) ≥ 0 for all *n* ∈ N.

Since *Tnx*1S∗*α Tnx*1, then *<sup>α</sup>*(*Tnx*1, *Tnx*2) ≥ 1 and *Tnx*1 = *Tnx*2 for all *n* ∈ N. By using Equation (10) and Proposition 1, for all *n* ∈ N,

$$\begin{aligned} d\left(T^{n+1}\mathbf{x}\_{1\prime}T^{n+1}\mathbf{x}\_2\right) &\leq \mathfrak{a}\left(T^{n}\mathbf{x}\_{1\prime}T^{n}\mathbf{x}\_2\right)d\left(TT^{n}\mathbf{x}\_{1\prime}TT^{n}\mathbf{x}\_2\right) \\ &\leq \psi\left(d\left(T^{n}\mathbf{x}\_{1\prime}T^{n}\mathbf{x}\_2\right)\right) \leq d\left(T^{n}\mathbf{x}\_{1\prime}T^{n}\mathbf{x}\_2\right). \end{aligned}$$

As {*d*(*Tnx*1, *Tnx*2)} is a bounded-below non-increasing sequence of real numbers, it is convergent. Let *L* ≥ 0 be its limit. Hence,

$$0 \le L \le d(T^{n+1}\mathbf{x}\_1, T^{n+1}\mathbf{x}\_2) \le \psi(d(T^n\mathbf{x}\_1, T^n\mathbf{x}\_2)) \le d(T^n\mathbf{x}\_1, T^n\mathbf{x}\_2) \quad \text{for all } n \in \mathbb{N}.$$

As *ψ* is nondecreasing, for all *n* ∈ N,

$$d\left(T^{\mathfrak{n}}\mathbf{x}\_1, T^{\mathfrak{n}}\mathbf{x}\_2\right) \le \psi\left(d\left(T^{\mathfrak{n}-1}\mathbf{x}\_1, T^{\mathfrak{n}-1}\mathbf{x}\_2\right)\right) \le \psi^2\left(d\left(T^{\mathfrak{n}-2}\mathbf{x}\_1, T^{\mathfrak{n}-2}\mathbf{x}\_2\right)\right) \le \dots \le \psi^n\left(d\left(\mathbf{x}\_1, \mathbf{x}\_2\right)\right).$$

Taking into account that *d* (*<sup>x</sup>*1, *<sup>x</sup>*2) > 0, then lim*n*→∞ *ψn* (*d* (*<sup>x</sup>*1, *<sup>x</sup>*2)) = 0, and letting *n* → ∞ in

$$0 \le L \le d(T^n \mathfrak{x}\_1, T^n \mathfrak{x}\_2) \le \psi^n(d\left(\mathfrak{x}\_1, \mathfrak{x}\_2\right))\_n$$

we conclude that *L* = lim*n*→∞ *d*(*Tnx*1, *Tnx*2) = 0.

(B2) It follows from (B2).

(B3) Let {(*an*, *bn*)} ⊆ *A* × *A* be a (*<sup>T</sup>*, S*α*)-sequence such that {*an*} and {*bn*} converge to the same limit *L* ≥ 0 and verifying that *L* < *an* and *γ*(*an*, *bn*) ≥ 0 for all *n* ∈ N. By definition, there are two sequences {*xn*}, {*yn*} ⊆ *X* such that

$$\mathbf{x}\_{\mathsf{n}} \mathsf{S}\_{\mathsf{n}} y\_{\mathsf{n}}, \quad a\_{\mathsf{n}} = d(T \mathbf{x}\_{\mathsf{n}}, T y\_{\mathsf{n}}) > 0 \quad \text{and} \quad b\_{\mathsf{n}} = d(\mathbf{x}\_{\mathsf{n}}, y\_{\mathsf{n}}) > 0 \quad \text{for all } \mathsf{n} \in \mathbb{N}.$$

Hence, *α* (*xn*, *yn*) ≥ 1 for all *n* ∈ N. To prove that *L* = 0, we reason by contradiction. Assume that *L* > 0. By Property (P12) of Proposition 1, *ψ*(*L*) < *L*. It follows from Equation (10) that

$$\Psi(L) < L < a\_n = d(T\mathbf{x}\_n, \mathbf{T}y\_n) \le a\left(\mathbf{x}\_n, y\_n\right) \, d(T\mathbf{x}\_n, \mathbf{T}y\_n) \le \psi\left(d\left(\mathbf{x}\_n, y\_n\right)\right) \le d\left(\mathbf{x}\_n, y\_n\right) = b\_n. \tag{11}$$

Since {*bn*} → *L*, then lim*n*→∞ *ψ* (*d* (*xn*, *yn*)) = *L*. As *ψ* is nondecreasing, the following limit exists and takes the value

$$\lim\_{s \to L^{+}} \psi \left( s \right) = \lim\_{u \to \infty} \psi \left( d \left( \mathfrak{x}\_{n}, y\_{n} \right) \right) = L.$$

As *ψ* is nondecreasing, *ψ*(*L*) ≤ *ψ*(*s*) ≤ *ψ*(*t*) for all *L* ≤ *s* ≤ *t*, so

$$
\psi(L) < L = \lim\_{s \to L^{+}} \psi \left( s \right) \le \psi \left( t \right) \quad \text{for all } t \in \left( L, \infty \right) \dots
$$

Taking in mind that *L* ≤ *ψ* (*t*) for all *t* ∈ (*<sup>L</sup>*, <sup>∞</sup>), next, we distinguish two cases.

*(Case 1) Assume that ψ*(*t*) > *L for all t* ∈ (*<sup>L</sup>*, <sup>∞</sup>)*.* In this case, let *t*0 ∈ (*<sup>L</sup>*, ∞) be arbitrary. Then, *ψ*(*<sup>t</sup>*0) > *L*. Therefore, *ψ*<sup>2</sup>(*<sup>t</sup>*0) = *ψ*(*ψ*(*<sup>t</sup>*0)) > *L*. Repeating this argument, *ψ*<sup>3</sup>(*<sup>t</sup>*0) = *ψ*(*ψ*<sup>2</sup>(*<sup>t</sup>*0)) > *L*. Similarly, by induction, *ψ<sup>n</sup>*(*<sup>t</sup>*0) > *L* for all *n* ∈ N, which contradicts the fact that lim*n*→∞ *ψn* (*<sup>t</sup>*0) = 0.

*(Case 2) Assume that there exists L* > *L such that ψ*(*L*) = *L.* In this case, as *ψ* is nondecreasing, for all *t* ∈ (*<sup>L</sup>*, *<sup>L</sup>*], we have that *L* ≤ *ψ* (*t*) ≤ *ψ*(*L*) = *L*, so *ψ*(*t*) = *L* for all *t* ∈ (*<sup>L</sup>*, *<sup>L</sup>*]. Since {*bn* = *d* (*xn*, *yn*)} *L*+, there exists *n*0 ∈ N such that *d* (*xn*0 , *yn*0 ) ∈ (*<sup>L</sup>*, *<sup>L</sup>*]. Hence, *ψ* (*d* (*xn*0 , *yn*0 )) = *L*, which contradicts the strict inequality in Equation (11) because

$$L < a\_{n\_0} \le \psi\left(d\left(\mathfrak{x}\_{n\_0\prime} y\_{n\_0}\right)\right).$$

In any case, we ge<sup>t</sup> a contradiction, so *L* = 0.

(B5) Let {(*an*, *bn*)} be a (*<sup>T</sup>*, S*α*)-sequence such that {*bn*} → 0 and *γ*(*an*, *bn*) ≥ 0 for all *n* ∈ N. By definition, there exist two sequences {*xn*}, {*yn*} ⊆ *X* such that

$$d\_n x\_n \mathbb{S}\_n y\_n, \quad a\_n = d(T x\_n, Ty\_n) > 0 \quad \text{and} \quad b\_n = d(x\_n, y\_n) > 0 \quad \text{for all } n \in \mathbb{N}.$$

In particular, *α* (*xn*, *yn*) ≥ 1 for all *n* ∈ N. It follows from Equation (10) that

$$0 < a\_{\mathbb{N}} = d(T\mathbf{x}\_{\mathbb{N}}, T\mathbf{y}\_{\mathbb{N}}) \le a\left(\mathbf{x}\_{\mathbb{N}}, \mathbf{y}\_{\mathbb{N}}\right)d(T\mathbf{x}\_{\mathbb{N}}, T\mathbf{y}\_{\mathbb{N}}) \le \psi\left(d\left(\mathbf{x}\_{\mathbb{N}}, \mathbf{y}\_{\mathbb{N}}\right)\right) \le d\left(\mathbf{x}\_{\mathbb{N}}, \mathbf{y}\_{\mathbb{N}}\right) = b\_{\mathbb{N}}.$$

Since {*bn*} → 0, then {*an*} → 0.

**Corollary 1.** *Every Samet et al.'s α - ψ - contraction (in the sense of Definition 7) is an ample spectrum contraction with respect to* S*α that also verifies* (B2) *and* (B5)*.*

**Proof.** It follows from the fact that, if *ψ* ∈ Ψ, then Theorem 9 is applicable because *ψ* is nondecreasing and lim*n*→∞ *ψn* (*t*) = 0 for all *t* > 0 (recall Proposition 1).

**Theorem 10.** *Theorem 8 immediately follows from Theorems 2 and 3.*

**Proof.** By Corollary 1, every Samet et al.'s *α* - *ψ* - contraction is an ample spectrum contraction with respect to S*α* that also verifies (B2) and (B5), thus Theorems 2 and 3 are applicable.

*5.3. Some Meditations about a Nonsymmetric Condition*

In [1], Khojasteh et al. introduced the notion of simulation function as a mapping *ζ* : [0, ∞) × [0, ∞) → R satisfying the following conditions:

(*ζ*1) *ζ*(0, 0) = 0; (*ζ*2) *ζ*(*<sup>t</sup>*,*<sup>s</sup>*) < *s* − *t* for all *t*,*<sup>s</sup>* > 0; and (*ζ*3) if {*tn*}, {*sn*} are sequences in (0, ∞) such that lim*n*→∞ *tn* = lim*n*→∞ *sn* > 0, then

$$\limsup\_{n \to \infty} \zeta(t\_{n\prime}s\_n) < 0.$$

Shortly after, Roldán López de Hierro et al. [2] pointed out that Condition (*ζ*3) is symmetric in both arguments of *ζ*, which is not necessary. Hence, these authors introduced the following variation in Axiom (*ζ*3):

(*ζ*3) if {*tn*}, {*sn*} are sequences in (0, ∞) such that lim*n*→∞*tn* = lim*n*→∞*sn* > 0 and *tn* < *sn* for all *n* ∈ N, then

$$\cdots \longrightarrow \longrightarrow$$

$$\limsup\_{n \to \infty} \zeta(t\_{n\prime}s\_n) < 0.$$

In this way, they removed the symmetry of a key function involved in the contractivity condition. After that, Roldán López de Hierro and Shahzad [3] presented the concept of *R*-contraction, which is intimately associated to an *R*-function : *A* × *A* → R. Such kind of functions must satisfy the following conditions (see [3], Definition 12):


Questions immediately arise: Why did the authors impose

$$L < a\_n \quad \text{for all } n \in \mathbb{N} \tag{12}$$

in Assumption (2)? Why did they not consider

$$L < b\_n \quad \text{for all } n \in \mathbb{N} \tag{13}$$

rather than Equation (12)? A first response we can give is that both assumptions are interesting in order to remove the symmetry in the variables of in Assumption (2) because the role of the sequence {*an*} is different from the role of {*bn*}. However, are Equations (12) and (13) equivalent? The response is no: we do believe that the condition in Equation (12) is better than the one in Equation (13). We justify it by the following fact: using the hypothesis in Equation (12), it is easy to check that every Meir–Keeler condition is an *R*-condition (see Theorem 25 in [3]). However, if we have only assumed that Equation (13) holds, then some Meir–Keeler contractions would not have been *R*-contractions. To illustrate it, we modify Example 2 in the following way.

**Example 3.** *Let X* = [0, 1] ∪ *C* ∪ *D, where C* = { 10*m* ∈ N : *m* ∈ N∗ } *and D* = { 10*m* + 1 + 1*m* ∈ N : *m* ∈ N∗ }*. If X is furnished with the Euclidean metric dE*(*<sup>x</sup>*, *y*) = | *x* − *y* | *for all x*, *y* ∈ *X, then* (*<sup>X</sup>*, *dE*) *is a complete metric space. Let T* : *X* → *X be the self-mapping defined by*

$$T\mathbf{x} = \begin{cases} \frac{\mathbb{X}}{4}, & \text{if } \mathbf{x} \in [0, 1], \\ 0, & \text{if } \mathbf{x} = 10m \in \mathbb{C} \quad \text{(for some } m \in \mathbb{N}^\*), \\ 1 - \frac{1}{2m}, & \text{if } \mathbf{x} = 10m + 1 + \frac{1}{m} \in D \quad \text{(for some } m \in \mathbb{N}^\*); \end{cases}$$

*Notice that Tx* ∈ [0, 1) *for all x* ∈ *X. Therefore,*

$$d\_E(Tx, Ty) < 1 \quad \text{for all } x, y \in X. \tag{14}$$

*Let us show that T is a Meir–Keeler contraction in* (*<sup>X</sup>*, *dE*)*. Indeed, let φ* : [0, ∞) → [0, ∞) *be the function given by*

$$\phi(t) = \begin{cases} -\frac{t}{2}, & \text{if } t \in [0, 1], \\\ 1, & \text{if } t > 1. \end{cases}$$

*Clearly, φ is an L-function, and we claim that Equation (8) holds. Let x*, *y* ∈ *X be such that d* (*<sup>x</sup>*, *y*) > 0*. Suppose, without loss of generality, that x* < *y.*

• *If x*, *y* ∈ [0, 1]*, then dE*(*<sup>x</sup>*, *y*) ≤ 1 *and*

$$d\_E(Tx, Ty) = d\_E\left(\frac{x}{4}, \frac{y}{4}\right) = \left\lfloor \frac{x}{4} - \frac{y}{4} \right\rfloor = \frac{\left\lfloor x - y \right\rfloor}{4} < \frac{\left\lfloor x - y \right\rfloor}{2} = \phi(d\_E(x, y)).$$

• *If x* ∈ [0, 1] *and y* ∈ *C* ∪ *D, then dE*(*<sup>x</sup>*, *y*) > 1*, and it follows from Equation (14) that*

$$d\_E(T\mathbf{x}, T\mathbf{y}) < 1 = \phi(d\_E(\mathbf{x}, \mathbf{y})) .$$

• *If x*, *y* ∈ *C* ∪ *D, then dE*(*<sup>x</sup>*, *y*) > 1 *and, similarly, dE*(*Tx*, *Ty*) < 1 = *φ*(*dE*(*<sup>x</sup>*, *y*))*.*

*In any case, Equation (8) holds and Theorem 4 ensures us that T is a Meir–Keeler contraction in* (*<sup>X</sup>*, *dE*)*. In fact, Theorem 21 in [3] guarantees that the function φ* : [0, ∞) × [0, ∞) → R *given by*

$$
\varrho\_{\phi}(t, s) = \phi(s) - t \quad \text{for all } t, s \in [0, \infty) \text{ }.
$$

*is an R-function on* [0, ∞) *verifying* (3)*. In particular, it satisfies Axiom* (2)*. Let us show that φ would not satisfy* (2) *if we replace Equation (12) with Equation (13). Indeed, let* {*xn*}*n*∈N<sup>∗</sup> *and* {*yn*}*n*∈N<sup>∗</sup> *be the sequences in X given by*

$$x\_{\mathbb{N}} = 10n \quad \text{and} \quad y\_{\mathbb{N}} = 10n + 1 + \frac{1}{n} \quad \text{for all } n \in \mathbb{N}.$$

> *Therefore, for all n* ∈ N*,*

$$\begin{aligned} a\_{n} &= d\_{\mathbb{E}}(T \mathbf{x}\_{n}, T \mathbf{y}\_{n}) = d\_{\mathbb{E}}\left(0, 1 - \frac{1}{2n}\right) = 1 - \frac{1}{2n} > 0 \quad \text{and} \\ b\_{n} &= d\_{\mathbb{E}}(\mathbf{x}\_{n}, \mathbf{y}\_{n}) = d\_{\mathbb{E}}\left(10n, 10n + 1 + \frac{1}{n}\right) = 1 + \frac{1}{n} > 1. \end{aligned}$$

*Hence, for all n* ∈ N*,*

$$\begin{aligned} \varrho\_{\Phi}(a\_n, b\_n) &= \varrho\_{\Phi}\left(1 - \frac{1}{2n}, 1 + \frac{1}{n}\right) = \phi\left(1 + \frac{1}{n}\right) - \left(1 - \frac{1}{2n}\right) \\ &= 1 - \left(1 - \frac{1}{2n}\right) = \frac{1}{2n} > 0 \end{aligned}$$

*However, L* = 1 *is not zero. Therefore, φ does not satisfy* (2) *if we replace Equation (12) with Equation (13). Thus, in this case, there would be Meir–Keeler contractions that are not R-contractions.*

As it can be easily checked, Property (2) that *R*-functions must satisfy leads to Condition (A3) for (A, S)-contractions and Condition (B3) for ample spectrum contractions.

(B3) If {(*an*, *bn*)} ⊆ *A* × *A* is a (*<sup>T</sup>*, S∗)-sequence such that {*an*} and {*bn*} converge to the same limit *L* ≥ 0 and verifying that *L* < *an* and (*an*, *bn*) ≥ 0 for all *n* ∈ N, then *L* = 0.

If we have assumed the condition in Equation (13) rather than the condition in Equation (12) in (B3), then the same arguments given in Example 3 prove that there would be Meir–Keeler contractions that are not ample spectrum contractions. As a consequence, we conclude that the assumption in Equation (12) is more appropriate than the one in Equation (13) in the context of fixed point theory.

Nevertheless, in the next subsection, we are going to show that, under some very recent contractivity conditions, they would be equivalent.

#### *5.4. Shahzad et al.'s Contractions*

In [10], Shahzad et al. presented some coincidence point results for a new class of contractive mappings that they called (*<sup>α</sup>*, *ψ*, *φ*)*-contractions*. They used the following kind of auxiliary functions.

**Definition 8** (Roldán López de Hierro [10], Definition 3.5)**.** *Let* FA *be the family of all pairs* (*ψ*, *φ*) *where ψ*, *φ* : [0, ∞) → [0, ∞) *are two functions verifying the following two conditions:*

<sup>F</sup>1A *If* {*an*} ⊂ (0, ∞) *is a sequence such that ψ* (*an*+<sup>1</sup>) ≤ *φ*(*an*) *for all n* ∈ N*, then* {*an*} → 0*.* <sup>F</sup><sup>2</sup>A *If* {*an*}, {*bn*} ⊂ [0, ∞) *are two sequences converging to the same limit L and such that L* < *an and ψ* (*bn*) ≤ *φ*(*an*) *for all n* ∈ N*, then L* = 0*.*

As a consequence of their main coincidence results, they presented the following statement (see the necessary preliminaries in [10]).

**Theorem 11** (Shahzad, Karapınar and Roldán López de Hierro [10], Theorem 6.1)**.** *Let* (*<sup>X</sup>*, *d*) *be a metric space, let α* : *X* × *X* → [0, ∞) *be a function and let T* : *X* → *X be a mapping such that the following conditions are fulfilled:*


$$
\mu(\mathbf{x}, \boldsymbol{y}) \,\,\psi\left(d(T\mathbf{x}, T\boldsymbol{y})\right) \le \phi\left(d(\mathbf{x}, \boldsymbol{y})\right) \quad \text{for all } \mathbf{x}, \boldsymbol{y} \in \mathcal{X}; \tag{15}
$$

*and*

	- *(a) there exists x*0 ∈ *X such that <sup>α</sup>*(*<sup>x</sup>*0, *Tx*0) ≥ 1 *and T is* (*d*, *α*)*-right-continuous; or*
	- *(b) there exists x*0 ∈ *X such that <sup>α</sup>*(*Tx*0, *<sup>x</sup>*0) ≥ 1 *and T is* (*d*, *α*)*-left-continuous.*

*Then, T has, at least, a fixed point. Additionally, assume that φ*(0) = 0*, ψ*−<sup>1</sup>({0}) = {0}*, and the following property holds:*

(*U*) *for all fixed points x and y of T, there exists z* ∈ *X such that z is, at the same time, α-comparable to x and to y.*

*Then, T has a unique fixed point.*

In the following definition, we modify the second condition.

**Definition 9.** *Let* G A *be the family of all pairs* (*ψ*, *φ*) *where ψ*, *φ* : [0, ∞) → [0, ∞) *are two functions verifying the following two conditions:*

 F<sup>1</sup> A *If* {*an*} ⊂ (0, ∞) *is a sequence such that ψ* (*an*+<sup>1</sup>) ≤ *φ*(*an*) *for all n* ∈ N*, then* {*an*} → 0*.* G2 A *If* {*an*}, {*bn*} ⊂ [0, ∞) *are two sequences converging to the same limit L and such that L* < *bn and ψ* (*bn*) ≤ *φ*(*an*) *for all n* ∈ N*, then L* = 0*.*

The same theorem can be proved in this case.

**Theorem 12.** *Let* (*<sup>X</sup>*, *d*) *be a metric space, let α* : *X* × *X* → [0, ∞) *be a function and let T* : *X* → *X be a mapping such that the following conditions are fulfilled:*


$$d(\mathbf{x}, y)\,\psi\left(d(T\mathbf{x}, Ty)\right) \le \phi\left(d(\mathbf{x}, y)\right) \quad \text{for all } \mathbf{x}, y \in \mathcal{X}.\tag{16}$$

	- *(a) there exists x*0 ∈ *X such that <sup>α</sup>*(*<sup>x</sup>*0, *Tx*0) ≥ 1 *and T is* (*d*, *α*)*-right-continuous; or*
	- *(b) there exists x*0 ∈ *X such that <sup>α</sup>*(*Tx*0, *<sup>x</sup>*0) ≥ 1 *and T is* (*d*, *α*)*-left-continuous.*

*Then, T has, at least, a fixed point. Additionally, assume that φ*(0) = 0*, ψ*−<sup>1</sup>({0}) = {0}*, and the following property holds:*

(*U*) *For all fixed points x and y of T, there exists z* ∈ *X such that z is, at the same time, α-comparable to x and to y.*

*Then, T has a unique fixed point.*

Let us show how this last result can be deduced from Theorems 2 and 3. The key is the following result.

**Lemma 2.** *Let* (*<sup>X</sup>*, *d*) *be a metric space, let α* : *X* × *X* → [0, ∞) *be a function and let T* : *X* → *X be a mapping such that the following conditions are fulfilled:*

*1. There exists* (*ψ*, *φ*) ∈ G A *such that*

$$d(\mathbf{x}, \mathbf{y}) \,\,\psi\left(d(T\mathbf{x}, T\mathbf{y})\right) \le \phi\left(d(\mathbf{x}, \mathbf{y})\right) \quad \text{for all } \mathbf{x}, \mathbf{y} \in \mathcal{X}.\tag{17}$$

*2. There exist two distinct points x*0, *x*1 ∈ *X such that α* (*<sup>x</sup>*0, *<sup>x</sup>*1) ≥ 1*.*

*Then, T is an ample spectrum contraction with respect to a function and* S*α that also verifies* (B2)*.*

**Proof.** Let us consider

$$\begin{aligned} A &= \{ \, d\left(\mathbf{x}, y\right) \in \left[0, \infty\right) : \mathbf{x}, y \in X, \, \mathbf{x} \mathcal{S}\_a^\* y \} \\ &= \{ \, d\left(\mathbf{x}, y\right) \in \left[0, \infty\right) : \mathbf{x}, y \in X, \, \mathbf{x} \neq y, \, \mathbf{a}\left(\mathbf{x}, y\right) \ge 1 \} \dots \end{aligned}$$

As *d* (*<sup>x</sup>*0, *<sup>x</sup>*1) ∈ *A*, then *A* is nonempty. Let us define the function *γ* : *A* → R, for all *t* ∈ *A*, by

$$\gamma\left(t\right) = \inf\left(\left\{a\left(\mathbf{x},\boldsymbol{y}\right) : \mathbf{x}, \boldsymbol{y} \in X, \,\mathbf{x}\big\mathcal{S}\_a^\*\boldsymbol{y} \text{ and } d\left(\mathbf{x},\boldsymbol{y}\right) = t\right\}\right).$$

To prove that *γ* is well defined, let *t* ∈ *A* be arbitrary and let

$$\Omega\_t = \left\{ a(\mathbf{x}, \boldsymbol{y}) : \mathbf{x}, \boldsymbol{y} \in X, \, \mathbf{x} \mathbf{S}\_a^\* \boldsymbol{y} \text{ and } d(\mathbf{x}, \boldsymbol{y}) = t \right\}.$$

By definition, as *t* ∈ *A*, there exist *xt*, *yt* ∈ *X* such that *xt*S∗*α yt* and *t* = *d* (*xt*, *yt*). Therefore, *α* (*xt*, *yt*) ∈ Ω*<sup>t</sup>*, so this set is nonempty. Moreover, let *x*, *y* ∈ *X* be arbitrary points such that *<sup>x</sup>*S∗*α y* and *d*(*<sup>x</sup>*, *y*) = *t*. Hence, *<sup>α</sup>*(*<sup>x</sup>*, *y*) ≥ 1. This proves that *<sup>α</sup>*(*<sup>x</sup>*, *y*) ≥ 1 for all number *<sup>α</sup>*(*<sup>x</sup>*, *y*) ∈ Ω*<sup>t</sup>*. Taking into account that Ω*t* is nonempty and bounded below by 1, we can take infimum, which means that *γ*(*t*) is well defined. In particular, we have proved the following facts:

$$
\gamma(t) = \inf \Omega\_t \ge 1 \quad \text{for all } t \in A; \tag{18}
$$

$$
\gamma\left(d\left(\mathbf{x},\boldsymbol{y}\right)\right) \le a\left(\mathbf{x},\boldsymbol{y}\right) \quad \text{for all } \mathbf{x},\boldsymbol{y} \in X \text{ such that } \mathbf{x}\mathbf{S}\_{\mathbf{a}}^{\*}\boldsymbol{y}.\tag{19}
$$

Considering the pair (*ψ*, *φ*) ∈ GA, let : *A* × *A* → R be defined, for all *t*,*<sup>s</sup>* ∈ *A*, by

$$
\boldsymbol{\varrho}\left(t, \mathbf{s}\right) = \boldsymbol{\Phi}\left(\mathbf{s}\right) - \boldsymbol{\gamma}\left(\mathbf{s}\right)\,\boldsymbol{\Psi}\left(t\right) \quad \text{for all } t, \mathbf{s} \in A.
$$

We claim that *T* is an ample spectrum contraction with respect to and S*α* that also verifies (B2). We demonstrate each condition. (B1) is obvious.

(B4) Let *x*, *y* ∈ *X* be arbitrary points such that *<sup>x</sup>*S∗*α y* and *Tx*S∗*α Ty*, that is, *α* (*<sup>x</sup>*, *y*) ≥ 1, *α* (*Tx*, *Ty*) ≥ 1, *x* = *y* and *Tx* = *Ty*. Therefore, applying Equation (17),

$$d\left(\mathbf{x}, y\right)\psi\left(d(T\mathbf{x}, Ty)\right) \le \phi\left(d(\mathbf{x}, y)\right). \tag{20}$$

In particular, it follows from Equations (19) and (20) that

$$\begin{aligned} \left(\varrho\left(d(T\mathbf{x},T\mathbf{y}),d(\mathbf{x},\mathbf{y})\right)\right) &= \phi\left(d(\mathbf{x},\mathbf{y})\right) - \gamma\left(d(\mathbf{x},\mathbf{y})\right)\left\|\psi\left(d(T\mathbf{x},T\mathbf{y})\right)\right\|\\ &\geq \phi\left(d(\mathbf{x},\mathbf{y})\right) - a(\mathbf{x},\mathbf{y})\,\psi\left(d(T\mathbf{x},T\mathbf{y})\right) \geq 0,\end{aligned}$$

so (B4) holds.

> (B2) Let *x*1, *x*2 ∈ *X* be two points such that

$$d\left(T^{\mathfrak{n}}\mathbf{x}\_{1}\mathcal{S}\_{\mathfrak{n}}^{\*}T^{\mathfrak{n}}\mathbf{x}\_{2} \quad \text{and} \quad \varrho\big(d\left(T^{\mathfrak{n}+1}\mathbf{x}\_{1}, T^{\mathfrak{n}+1}\mathbf{x}\_{2}\right), d\left(T^{\mathfrak{n}}\mathbf{x}\_{1}, T^{\mathfrak{n}}\mathbf{x}\_{2}\right)\big) \geq 0 \quad \text{for all } \mathfrak{n} \in \mathbb{N} .$$

Notice that *Tnx*1S∗*α Tnx*2 and *Tn*+1*x*1S∗*α Tn*+1*x*2 imply that *d* (*Tnx*1, *Tnx*2) and *d Tn*+1*x*1, *Tn*+1*x*2 belong to *A*. Let

$$a\_n = d\left(T^n \mathbf{x}\_1, T^n \mathbf{x}\_2\right) > 0 \quad \text{for all } n \in \mathbb{N}.$$

In particular, as *γ* ≥ 1, then

$$\begin{aligned} 0 \le & \varrho(d\left(T^{n+1}\mathbf{x}\_1, T^{n+1}\mathbf{x}\_2\right), d\left(T^{\mathbf{u}}\mathbf{x}\_1, T^{\mathbf{u}}\mathbf{x}\_2\right)) = \varrho\left(a\_{n+1}, a\_n\right) \\ = & \phi\left(a\_n\right) - \gamma\left(a\_n\right)\,\psi\left(a\_{n+1}\right) \le \phi\left(a\_n\right) - \psi\left(a\_{n+1}\right) \end{aligned}$$

that is, *ψ* (*an*+<sup>1</sup>) ≤ *φ* (*an*), for all *n* ∈ N. Since (*φ*, *ψ*) ∈ GA, Condition <sup>F</sup>1A implies that {*an*} → 0, that is, {*d* (*Tnx*1, *Tnx*2)} → 0, which means that (B2) holds.

(B2) It immediately follows from (B2).

(B3) Let {(*an*, *<sup>b</sup>n*)} ⊆ *A* × *A* be a (*<sup>T</sup>*, S∗*α* )-sequence such that {*an*} and {*bn*} converge to the same limit *L* ≥ 0 and verifying that *L* < *an* and (*an*, *<sup>b</sup>n*) ≥ 0 for all *n* ∈ N. By definition, there exist two sequences {*xn*}, {*yn*} ⊆ *X* such that

$$\text{tr}\_n \mathbb{S}^\* y\_n \quad \text{Tx}\_n \mathbb{S}^\* Ty\_n \quad a'\_n = d(\text{Tx}\_n, Ty\_n) > 0 \quad \text{and} \quad b'\_n = d(\text{x}\_n, y\_n) > 0 \quad \text{for all } n \in \mathbb{N}.$$

As *γ* ≥ 1, then

$$0 \le \varrho(a'\_{n'}b'\_n) = \phi\left(b'\_n\right) - \gamma\left(b'\_n\right)\ \psi\left(a'\_n\right) \le \phi\left(b'\_n\right) - \psi\left(a'\_n\right),$$

that is, *ψ* (*an*) ≤ *φ* (*bn*), for all *n* ∈ N. Since (*φ*, *ψ*) ∈ GA, Condition <sup>G</sup><sup>2</sup>A (applied to {*an*} = {*bn*} and {*bn*} = {*an*}) implies that *L* = 0, which means that (B3) holds.

As a consequence, we conclude that *T* is an ample spectrum contraction with respect to and S*α* that also verifies (B2).

Lemma 2 permits us to show that Theorem 12 is a particular case of the above-presented main statements.

#### **Theorem 13.** *Theorem 12 follows from Theorems 2 and 3.*

**Proof.** Assume that all the hypotheses of Theorem 12 hold. For instance, assume that there exists *x*0 ∈ *X* such that *<sup>α</sup>*(*<sup>x</sup>*0, *Tx*0) ≥ 1 and *T* is (*d*, *α*)-right-continuous (notice that Condition (4.*b*) requires a version of Theorems 2 and 3 in which *T* is non-increasing). Let {*xn*+<sup>1</sup> = *Txn*}*n*≥<sup>0</sup> be the Picard sequence of *T* based on *x*0. If there exists some *n*0 ∈ N such that *xn*0+<sup>1</sup> = *xn*0 , then *xn*0 is a fixed point of *T*, and {*xn*} converges to such point. In this case, the part about existence of a fixed point of *T* is finished. On the contrary case, assume that *xn* = *xn*+1 for all *n* ∈ N. Let S*α* be the binary relation on *X* given, for *x*, *y* ∈ *X*, by

$$\mathbf{x} \mathbf{S}\_{\mathbf{a}} y \quad \text{if} \quad \mathfrak{a}(\mathbf{x}, y) \ge 1. \tag{21}$$

By Lemma 1:


By Hypothesis 1 of Theorem 12, there exists a subset *A* ⊆ *X* such that *T*(*X*) ⊆ *A* and (*<sup>A</sup>*, *d*) is complete. In particular, *T*(*X*) is (S*<sup>α</sup>*, *d*)-strictly-increasing-precomplete. Finally, Lemma 2 guarantees that *T* is a an ample spectrum contraction with respect to and S*α* that also verifies (B2). As all hypotheses of Theorem 2 are satisfied, *T* has at least a fixed point.

Following the statement of Theorem 12, additionally, assume that *φ*(0) = 0, *ψ*−<sup>1</sup>({0}) = {0}, and the following property holds:

(*U*) For all fixed points *x* and *y* of *T*, there exists *z* ∈ *X* such that *z* is, at the same time, *α*-comparable to *x* and to *y*.

Then, Theorem 3 is applicable, thus *T* has a unique fixed point.

**Remark 3.** *Notice that, in fact, we have proved that every Shahzad et al.'s contraction in the sense of Theorem 11 is an ample spectrum contraction with respect to an appropriate function .*

*5.5. Wardowski's F-Contractions*

**Definition 10** (Wardowski [11], Definition 2.1)**.** *Given a function F* : (0, ∞) → R *, let consider the following properties:*

(*<sup>F</sup>*1) *F is strictly increasing, that is, F*(*t*) < *<sup>F</sup>*(*s*) *for all t*,*<sup>s</sup>* ∈ (0, ∞) *such that t* < *s.*


*If* (*<sup>X</sup>*, *d*) *is a metric space, a mapping T* : *X* → *X is an* F-contraction *if there exist a positive number τ* > 0 *and a function F* : (0, ∞) → R *satisfying properties* (*<sup>F</sup>*1)*-*(*<sup>F</sup>*3) *such that*

$$\forall \tau \in F \left( d \left( Tx, Ty \right) \right) \le F \left( d \left( x, y \right) \right) \qquad \text{for all } x, y \in X \text{ such that } d \left( Tx, Ty \right) > 0.$$

**Theorem 14** (Wardowski [11], Theorem 2.1)**.** *Let* (*<sup>X</sup>*, *d*) *be a complete metric space and let T* : *X* → *X be an F-contraction. Then, T has a unique fixed point x*<sup>∗</sup> ∈ *X, and for every x*0 ∈ *X a sequence* {*Tnx*0}*n*∈<sup>N</sup> *is convergent to <sup>x</sup>*<sup>∗</sup>*.*

**Lemma 3.** *Every F-contraction is an ample spectrum contraction.*

Notice that in the following proof we do not use Property (*<sup>F</sup>*3).

**Proof.** Let (*<sup>X</sup>*, *d*) be a metric space and let *T* : *X* → *X* be an *F*-contraction with respect to a constant *τ* > 0 and a function *F* : (0, ∞) → R. Let *λ* = e<sup>−</sup>*<sup>τ</sup>* ∈ (0, <sup>1</sup>), let *A* = [0, ∞) and let *φ* : (0, ∞) → (0, ∞) and : *A* × *A* → R be the functions:

$$\begin{aligned} \boldsymbol{\phi}\left(t\right) &= \begin{cases} \mathbf{e}^{\boldsymbol{F}\left(t\right)}, & \text{if } t > 0, \\ \boldsymbol{0}, & \text{if } t = 0; \end{cases} \\ \boldsymbol{\varrho}\left(t, \mathbf{s}\right) &= \lambda \,\boldsymbol{\phi}\left(\mathbf{s}\right) - \boldsymbol{\phi}\left(t\right) \quad \text{for all } t, \mathbf{s} \in \left[0, \infty\right) \end{aligned}$$

Property (*<sup>F</sup>*1) implies that *φ* is strictly increasing on (0, ∞) and Property (*<sup>F</sup>*2) guarantees that for each sequence {*tn*}*n*∈<sup>N</sup> of positive real numbers we have that

$$\{t\_n\} \to 0 \text{ if, and only if, } \{\phi(t\_n)\} \to 0. \tag{22}$$

We claim that *T* is an ample spectrum contraction with respect to and the trivial preorder S*<sup>X</sup>*. Property (B1) is obvious.

(B2) Let {*xn*} ⊆ *X* be a Picard sequence of *T* such that

$$d(\mathbf{x}\_n \neq \mathbf{x}\_{n+1} \quad \text{and} \quad \varrho\left(d\left(\mathbf{x}\_{n+1}, \mathbf{x}\_{n+2}\right), d\left(\mathbf{x}\_n, \mathbf{x}\_{n+1}\right)\right) \geq 0 \quad \text{for all } n \in \mathbb{N}.$$

Therefore, for all *n* ∈ N, *d* (*xn*, *xn*+<sup>1</sup>) > 0 and

$$0 \le \varrho\left(d\left(\mathbf{x}\_{n+1}, \mathbf{x}\_{n+2}\right), d\left(\mathbf{x}\_n, \mathbf{x}\_{n+1}\right)\right) = \lambda\,\phi\left(d\left(\mathbf{x}\_n, \mathbf{x}\_{n+1}\right)\right) - \phi\left(d\left(\mathbf{x}\_{n+1}, \mathbf{x}\_{n+2}\right)\right),$$

so

$$0 \le \phi\left(d\left(\mathfrak{x}\_{n+1}, \mathfrak{x}\_{n+2}\right)\right) \le \lambda \phi\left(d\left(\mathfrak{x}\_{n\prime}\mathfrak{x}\_{n+1}\right)\right)\dots$$

In particular, {*φ* (*d* (*xn*, *xn*+<sup>1</sup>))} → 0, and the property in Equation (22) guarantees that {*d* (*xn*, *xn*+<sup>1</sup>)} → 0.

(B3) Let {(*an*, *bn*)} ⊆ *A* × *A* be a (*<sup>T</sup>*, S∗*X*)-sequence such that {*an*} and {*bn*} converge to the same limit *L* ≥ 0 and verifying that *L* < *an* and (*an*, *bn*) ≥ 0 for all *n* ∈ N. By Definition 3, *an* > 0 and *bn* > 0 for all *n* ∈ *N*. To prove that *L* = 0, assume, by contradiction, that *L* > 0. Notice that for all *n* ∈ N,

$$0 \le \varrho(a\_n, b\_n) = \lambda \, \phi\left(b\_n\right) - \phi\left(a\_n\right).$$

As *φ* is strictly increasing,

$$0 < \phi\left(L\right) < \phi\left(a\_n\right) \le \lambda\left\phi\left(b\_n\right) < \phi\left(b\_n\right).$$

This means that *L* < *an* < *bn*. Since *φ* is strictly increasing, the following limit exists:

$$L' = \lim\_{s \to L^{+}} \phi\left(s\right).$$

Furthermore, 0 < *φ* (*L*) ≤ *L*. As {*an*} → *L*, {*bn*} → *L* and *L* < *an* < *bn* for all *n* ∈ N, then

$$L' = \lim\_{s \to L^{+}} \phi \left( s \right) = \lim\_{n \to \infty} \phi \left( a\_n \right) = \lim\_{n \to \infty} \phi \left( b\_n \right).$$

Taking limit as *n* → ∞ in *φ* (*an*) ≤ *λ φ* (*bn*), we deduce that *L* ≤ *λ L*, which contradicts the fact that *L* > 0. Therefore, *L* = 0.

(B4) Let *x*, *y* ∈ *X* be two points such that *Tx* = *Ty*. In particular, *d* (*Tx*, *Ty*) > 0. Hence,

$$\begin{split} & \tau + F\left(d\left(T\mathbf{x}, T\mathbf{y}\right)\right) \le F\left(d\left(\mathbf{x}, \mathbf{y}\right)\right) \quad \Leftrightarrow \quad \mathbf{e}^{\tau + F\left(d\left(T\mathbf{x}, T\mathbf{y}\right)\right)} \le \mathbf{e}^{F\left(d\left(\mathbf{x}, \mathbf{y}\right)\right)} \\ & \Leftrightarrow \quad \mathbf{e}^{F\left(d\left(T\mathbf{x}, T\mathbf{y}\right)\right)} \le \mathbf{e}^{-\tau}\mathbf{e}^{F\left(d\left(\mathbf{x}, \mathbf{y}\right)\right)} \quad \Leftrightarrow \quad \boldsymbol{\phi}\left(d\left(T\mathbf{x}, T\mathbf{y}\right)\right) \le \boldsymbol{\lambda}\,\boldsymbol{\phi}\left(d\left(\mathbf{x}, \mathbf{y}\right)\right) \\ & \Leftrightarrow \quad \boldsymbol{\lambda}\,\boldsymbol{\phi}\left(d\left(\mathbf{x}, \mathbf{y}\right)\right) - \boldsymbol{\phi}\left(d\left(T\mathbf{x}, T\mathbf{y}\right)\right) \ge 0 \quad \Leftrightarrow \quad \boldsymbol{\phi}\left(d\left(T\mathbf{x}, T\mathbf{y}\right), d\left(\mathbf{x}, \mathbf{y}\right)\right) \ge 0. \end{split}$$

Therefore, *T* is an ample spectrum contraction with respect to and S*<sup>X</sup>*.

As a consequence, Theorem 14 is a simple consequence of Theorems 2 and 3.

Finally, we point out that the present techniques can be easily generalized to guarantee existence and uniqueness of multidimensional coincidence/fixed points following the techniques described in [19–25].

**Author Contributions:** Conceptualization, A.F.R.L.d.H. and N.S.; Methodology, A.F.R.L.d.H. and N.S.; Writing-Original Draft Preparation, A.F.R.L.d.H. and N.S.; Writing-Review & Editing, A.F.R.L.d.H. and N.S.

**Funding:** This article was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah.

**Acknowledgments:** The authors acknowledge with thanks DSR for financial support. A.F. Roldán López de Hierro is grateful to Junta de Andalucía,y project FQM-268 of the Andalusian CICYE and Project TIN2017-89517-P of the Ministerio de Economía, Industria y Competitividad.

**Conflicts of Interest:** The authors declare no conflict of interest.
