Fixed Point Results for Various 𝛼-Admissible Contractive Mappings on Metric-Like Spaces

Abstract

We establish some fixed point theorems for 𝛼-admissible mappings in the context of metric-like space via various auxiliary functions. In particular, we prove the existence of a fixed point of the generalized Meir-Keeler type 𝛼−𝜙-contractive self-mapping 𝑓 defined on a metric-like space 𝑋. The given results generalize, improve, and unify several fixed point theorems for the generalized cyclic contractive mappings that have appeared recently in the literature.

1. Introduction and Preliminaries

Nonlinear functional analysis is one of the most dynamic research fields in mathematics. In particular, fixed point theory that has a wide application potential to several quantitative sciences has attracted a number of authors. In the recent decades, several new abstract spaces and new contractive type mappings have been considered to develop the fixed point theory and to increase application potential to existing open problems. Among them, Samet et al. [1] proved very interesting fixed point theorem by introducing the 𝛼-𝜓-contractive self-mapping 𝑓 in the setting of complete metric space (𝑋,𝑑). In this notion, 𝜓 is a 𝑐-distance function (see, e.g., [2–5]) and self-mapping is 𝛼-admissible. The notion of mapping 𝛼-𝜓-contractive mappings has charmed a number of authors (see, e.g., [1, 6–14]).

In this paper, we combine some of the notions to get more general results in the research field of fixed point theory. In particular, we investigate the existence of a fixed point of 𝛼-admissible mapping in the context of metric-like space via implicit functions.

Throughout this paper, by ℝ⁺, we denote the set of all nonnegative numbers, while ℕ is the set of all natural numbers. In 1994, Matthews [15] introduced the following notion of partial metric spaces.

Definition 1 (see [15]). A partial metric on a nonempty set 𝑋 is a function 𝑝∶𝑋×𝑋→[0,∞) such that for all 𝑥,𝑦,𝑧∈𝑋(𝑝₁)𝑥=𝑦 if and only if 𝑝(𝑥,𝑥)=𝑝(𝑥,𝑦)=𝑝(𝑦,𝑦);(𝑝₂)𝑝(𝑥,𝑥)≤𝑝(𝑥,𝑦);(𝑝₃)𝑝(𝑥,𝑦)=𝑝(𝑦,𝑥);(𝑝₄)𝑝(𝑥,𝑦)≤𝑝(𝑥,𝑧)+𝑝(𝑧,𝑦)−𝑝(𝑧,𝑧).A partial metric space is a pair (𝑋,𝑝) such that 𝑋 is a nonempty set and 𝑝 is a partial metric on 𝑋.

Remark 2. It is clear that if 𝑝(𝑥,𝑦)=0, then, from (𝑝₁) and (𝑝₂), 𝑥=𝑦. But if 𝑥=𝑦, 𝑝(𝑥,𝑦) may not be 0.

Later, fixed point theory has developed rapidly on partial metric spaces; see [16–23]. Further, in 2012, Amini-Harandi [24] introduced the concept of a metric-like space.

Definition 3 (see [24]). A function 𝜎∶𝑋×𝑋→[0,∞), where 𝑋 is a nonempty set, is said to be metric-like on 𝑋 if the following conditions are satisfied, for all 𝑥,𝑦,𝑧∈𝑋: (𝜎₁)if 𝜎(𝑥,𝑦)=0, then 𝑥=𝑦;(𝜎₂)𝜎(𝑥,𝑦)=𝜎(𝑦,𝑥);(𝜎₃)𝜎(𝑥,𝑦)≤𝜎(𝑥,𝑧)+𝜎(𝑧,𝑦).Then the pair (𝑋,𝜎) is called a metric-like space.

Remark 4 (see [24]). (1) A metric-like on 𝑋 satisfies all of the conditions of a metric except that 𝜎(𝑥,𝑥) may be positive for 𝑥∈𝑋.
(2) Every partial metric space is a metric-like space. But the converse is not true.

Each metric-like 𝜎 on 𝑋 generates a topology 𝜏_𝜎 on 𝑋 whose base is the family of open 𝜎-balls {𝐵_𝜎(𝑥,𝛾)∶𝑥∈𝑋,𝛾>0}, where 𝐵_𝜎(𝑥,𝛾)={𝑦∈𝑋∶|𝜎(𝑥,𝑦)−𝜎(𝑥,𝑥)|<𝛾} for all 𝑥∈𝑋 and 𝛾>0. We recall some definitions on a metric-like space as follows.

Definition 5 (see [24]). Let (𝑋,𝜎) be a metric-like space. Then (1)a sequence {𝑥_𝑛} in a metric-like space (𝑋,𝜎) converges to 𝑥∈𝑋 if and only if 𝜎(𝑥,𝑥)=lim_𝑛→∞𝜎(𝑥_𝑛,𝑥);(2)a sequence {𝑥_𝑛} in a metric-like space (𝑋,𝜎) is called a 𝜎-Cauchy sequence if and only if lim_{𝑚,𝑛→∞}𝜎(𝑥_𝑚,𝑥_𝑛) exists (and is finite);(3)a metric-like space (𝑋,𝜎) is said to be complete if every 𝜎-Cauchy sequence {𝑥_𝑛} in 𝑋 converges, with respect to 𝜏_𝜎, to a point 𝑥∈𝑋 such that 𝜎(𝑥,𝑥)=lim_𝑛→∞𝜎𝑥_𝑛,𝑥=lim_{𝑚,𝑛→∞}𝜎𝑥_𝑚,𝑥_𝑛, (1)(4)a mapping 𝑇∶𝑋→𝑋 is continuous, if the following limits exist (finite) and lim_𝑛→∞𝜎𝑥_𝑛,𝑥=lim_{𝑚,𝑛→∞}𝜎(𝑇𝑥,𝑥). (2)

Definition 6 (see [24]). Let (𝑋,𝜎) be a metric-like space and 𝑈 be a subset of 𝑋. Then 𝑈 is a 𝜎-open subset of 𝑋 if, for all 𝑥∈𝑋, there exists 𝛾>0 such that 𝐵_𝜎(𝑥,𝛾)⊂𝑈. Also, 𝑉⊂𝑋 is a 𝜎-closed subset of 𝑋 if 𝑋⧵𝑉 is a 𝜎-open subset of 𝑋.

Further, Karapınar and Salimi [25] proved the following crucial properties in the setting of metric-like space (𝑋,𝜎).

Lemma 7 (see [25]). Let (𝑋,𝜎) be a metric-like space. Then (A)if 𝜎(𝑥,𝑦)=0, 𝜎(𝑥,𝑥)=𝜎(𝑦,𝑦)=0;(B)if {𝑥_𝑛} is a sequence such that lim_𝑛→∞𝜎(𝑥_𝑛,𝑥_𝑛+1)=0, lim_𝑛→∞𝜎𝑥_𝑛,𝑥_𝑛=lim_𝑛→∞𝜎𝑥_𝑛+1,𝑥_𝑛+1=0; (3)(C)if 𝑥≠𝑦, 𝜎(𝑥,𝑦)>0;(D)𝜎(𝑥,𝑥)≤(2/𝑛)∑^𝑛_𝑖=1𝜎(𝑥,𝑥_𝑖) holds for all 𝑥,𝑥_𝑖∈𝑋, where 1≤𝑖≤𝑛.

Lemma 8. Let (𝑋,𝜎) be a metric-like space and {𝑥_𝑛} be a sequence in 𝑋 such that 𝑥_𝑛→𝑥 as 𝑛→∞ and 𝜎(𝑥,𝑥)=0. Then lim_𝑛→∞𝜎(𝑥_𝑛,𝑦)=𝜎(𝑥,𝑦) for all 𝑦∈𝑋.

We recall the notion of cyclic map which was introduced by Kirk et al. [26]. A mapping 𝑓∶𝐴∪𝐵→𝐴∪𝐵 is called cyclic if 𝑓(𝐴)⊂𝐵 and 𝑓(𝐵)⊂𝐴. Kirk et al. [26] proved the analog of the Banach contraction mapping principle via cyclic mappings.

Theorem 9 (see [26]). Let 𝐴 and 𝐵 be two nonempty closed subsets of a complete metric space (𝑋,𝑑), and suppose 𝑓∶𝐴∪𝐵→𝐴∪𝐵 satisfies the following:(i)𝑓 is a cyclic map,(ii)𝑑(𝑓𝑥,𝑓𝑦)≤𝑘⋅𝑑(𝑥,𝑦) for all 𝑥∈𝐴, 𝑦∈𝐵, and 𝑘∈(0,1).Then 𝐴∩𝐵 is nonempty and 𝑓 has a unique fixed point in 𝐴∩𝐵.

Furthermore, Kirk et al. [26] also introduced the following notion of the cyclic representation.

Definition 10 (see [26]). Let 𝑋 be a nonempty set, 𝑚∈ℕ, and 𝑓∶𝑋→𝑋 an operator. Then 𝑋=∪^𝑚_𝑖=1𝐴_𝑖 is called a cyclic representation of 𝑋 with respect to 𝑓 if(1)𝐴_𝑖, 𝑖=1,2,…,𝑚, are nonempty subsets of 𝑋;(2)𝑓(𝐴₁)⊂𝐴₂, 𝑓(𝐴₂)⊂𝐴₃,…,𝑓(𝐴_𝑚−1)⊂𝐴_𝑚, and 𝑓(𝐴_𝑚)⊂𝐴₁.

By using the notion in the definition above, Kirk et al. [26] proved the following theorem.

Theorem 11 (see [26]). Let (𝑋,𝑑) be a complete metric space, 𝑚∈ℕ, 𝐴₁,𝐴₂,…,𝐴_𝑚 be closed nonempty subsets of 𝑋, and 𝑋=∪^𝑚_𝑖=1𝐴_𝑖. Suppose that 𝑓 satisfies the following condition: 𝑑𝑓𝑥,𝑓𝑦≤𝜓𝑑𝑥,𝑦,∀𝑥∈𝐴_𝑖,𝑦∈𝐴_𝑖+1,𝑖∈{1,2,…,𝑚}, (4) where 𝜓∶ℝ⁺→ℝ⁺ is upper semicontinuous from the right and 0≤𝜓(𝑡)<𝑡 for 𝑡>0. Then 𝑓 has a fixed point 𝑧∈∩^𝑛_𝑖=1𝐴_𝑖.

In 2012, Karapınar et al. [22] investigated the existence and uniqueness of a fixed point for cyclic generalized 𝜙-𝜓-contractive type mappings 𝑓∶𝑋→𝑋 in the context of partial metric space. Very recently, Karapınar and Salimi [25] improved the results in [22] by introducing the notion of cyclic generalized 𝜙-𝜓-contractive mapping 𝑓∶𝑋→𝑋. In this paper [25], the authors proved fixed theorems for such a mapping in the setting of a metric-like space 𝑋 with a cyclic representation of 𝑋 with respect to 𝑓.

Definition 12 (see [25]). Let (𝑋,𝜎) be a metric-like space, 𝐴₁,𝐴₂,…,𝐴_𝑚 be 𝜎-closed nonempty subsets of 𝑋, and 𝑌=∪^𝑚_𝑖=1𝐴_𝑖. One says that 𝑇∶𝑌→𝑌 is called a generalized cyclic 𝜙-𝜓-contractive mapping if(1)𝑌=∪^𝑚_𝑖=1𝐴_𝑖 is a cyclic representation of 𝑌 with respect to 𝑇;(2)One considers 𝜓(𝑡)−𝜓(𝑠)+𝜙(𝑡)>0∀𝑡>0,𝑠=𝑡,or𝜓𝜎𝑀𝑠=0,𝑇𝑥,𝑇𝑦≤𝜓_𝜎𝑥,𝑦−𝜙𝑀_𝜎,𝑀𝑥,𝑦_𝜎𝜎(),𝜎𝑥,𝑦=max𝑥,𝑦,𝜎𝑥,𝑇𝑥,𝜎𝑦,𝑇𝑦𝑥,𝑇𝑦+𝜎𝑦,𝑇𝑥4, (5)for all 𝑥∈𝐴_𝑖 and 𝑦∈𝐴_𝑖+1, 𝑖=1,2,…,𝑚, where 𝜓∶ℝ⁺→ℝ⁺ is nondecreasing and continuous and 𝜙∶ℝ⁺→ℝ⁺ is lower semicontinuous.

Theorem 13 (see [25]). Let (𝑋,𝜎) be a metric-like space, 𝐴₁,𝐴₂,…,𝐴_𝑚 be 𝜎-closed nonempty subsets of 𝑋, and 𝑌=∪^𝑚_𝑖=1𝐴_𝑖. If 𝑇∶𝑌→𝑌 is a generalized cyclic 𝜙-𝜓-contractive mapping, then 𝑇 has a fixed point 𝜈∈∩^𝑛_𝑖=1𝐴_𝑖.

In this study, we also discuss the notion of 𝛼-admissible mappings. The following definition was introduced in [1].

Definition 14 (see [1]). For a nonempty set 𝑋, let 𝑇∶𝑋→𝑋 and 𝛼∶𝑋×𝑋→[0,∞) be mappings. One says that 𝑓 is 𝛼-admissible, if, for all 𝑥,𝑦∈𝑋, one has 𝛼𝑥,𝑦≥1⟹𝛼𝑇𝑥,𝑇𝑦≥1. (6)

Recall that Samet et al. [1] introduced the following concepts.

Definition 15 (see [1]). Let (𝑋,𝑑) be a metric space and let 𝑇∶𝑋→𝑋 be a given mapping. One says that 𝑇 is an 𝛼-𝜓 contractive mapping if there exist two functions 𝛼∶𝑋×𝑋→[0,∞) and a certain 𝜓 such that 𝛼𝑥,𝑦𝑑𝑇𝑥,𝑇𝑦≤𝜓𝑑𝑥,𝑦,∀𝑥,𝑦∈𝑋. (7)

It is evident that a mapping satisfying the Banach contraction is a 𝛼-𝜓 contractive mapping equipped with 𝛼(𝑥,𝑦)=1 for all 𝑥,𝑦∈𝑋 and 𝜓(𝑡)=𝑘𝑡, 𝑘∈(0,1).

The notion of transitivity of mapping 𝛼∶𝑋×𝑋→[0,+∞) was introduced in [13, 14] as follows.

Definition 16 (see [13, 14]). Let 𝑁∈ℕ. One says that 𝛼 is 𝑁-transitive (on 𝑋) if 𝑥₀,𝑥₁,…,𝑥_𝑁+1∈𝑋∶𝛼𝑥_𝑖,𝑥_𝑖+1≥1, (8) for all 𝑖∈{0,1,…,𝑁}⇒𝛼(𝑥₀,𝑥_𝑁+1)≥1.
In particular, one says that 𝛼 is transitive if it is 1-transitive; that is, 𝑥,𝑦,𝑧∈𝑋∶𝛼𝑥,𝑦≥1,𝛼𝑦,𝑧≥1⟹𝛼(𝑥,𝑧)≥1. (9)

As consequences of Definition 16, one obtains the following remarks.

Remark 17 (see [13, 14]). (1) Any function 𝛼∶𝑋×𝑋→[0,+∞) is 0-transitive.
(2) If 𝛼 is 𝑁-transitive, then it is 𝑘𝑁-transitive for all 𝑘∈ℕ.
(3) If 𝛼 is transitive, then it is 𝑁-transitive for all 𝑁∈ℕ.
(4) If 𝛼 is 𝑁-transitive, then it is not necessarily transitive for all 𝑁∈ℕ.

In this paper, we investigate the existence and uniqueness of a fixed point of several 𝛼-admissible mappings in the context of metric-like space. In particular, we establish fixed point theorem for the generalized cyclic Meir-Keeler type 𝜙-𝛼-contractive mappings, the generalized (𝜑,𝜙,𝜓,𝜉)-𝛼-contractive mappings, and the generalized weaker Meir-Keeler type (𝜙,𝜑)-𝛼-contractive mappings. Our results generalize or improve many recent fixed point theorems for the generalized cyclic contractive mappings in the literature.

2. Fixed Point Theorem via the 𝛼-Admissible Meir-Keeler Type Mappings

In this section, first of all, we will introduce the notion of the generalized Meir-Keeler type 𝛼−𝜙-contractive mappings. Later, we investigate the existence and uniqueness of such mappings in the context of metric-like spaces. We start with recalling the notion of the Meir-Keeler type mappings.

A function 𝛾∶[0,∞)→[0,∞) is said to be a Meir-Keeler type mapping (see [27]), if, for each 𝜂∈[0,∞), there exists 𝛿>0 such that, for 𝑡∈[0,∞) with 𝜂≤𝑡<𝜂+𝛿, we have 𝛾(𝑡)<𝜂.

Let Φ be the class of all function 𝜙∶ℝ⁺⁵→ℝ⁺ satisfying the following conditions:(𝜙₁)𝜙 is an increasing and continuous function in each coordinate;(𝜙₂)for 𝑡>0, 𝜙(𝑡,𝑡,𝑡,2𝑡,2𝑡)<𝑡, 𝜙(𝑡,0,0,𝑡,𝑡)<𝑡, and 𝜙(0,0,𝑡,𝑡,0)<𝑡;(𝜙₃)𝜙(𝑡₁,𝑡₂,𝑡₃,𝑡₄,𝑡₅)=0 if and only if 𝑡₁=𝑡₂=𝑡₃=𝑡₄=𝑡₅=0.

We will introduce the notion of the generalized Meir-Keeler type 𝛼−𝜙-contractive mappings in metric-like spaces as follows.

Definition 18. Let (𝑋,𝜎) be a metric-like space and let 𝛼∶𝑋×𝑋→[0,∞). One says that 𝑇∶𝑋→𝑋 is called a generalized Meir-Keeler type 𝛼−𝜙-contractive mapping if for each 𝜂>0 there exists 𝛿>0 such that 𝜎()𝜎𝜂≤𝜙𝑥,𝑦,𝜎𝑥,𝑇𝑥,𝜎𝑦,𝑇𝑦,𝜎𝑥,𝑇𝑦,𝜎𝑦,𝑇𝑥<𝜂+𝛿⟶𝛼𝑥,𝑦𝑇𝑥,𝑇𝑦<𝜂, (10) for all 𝑥,𝑦∈𝑋 and 𝜙∈Φ.

Remark 19. Note that if 𝑇 is a generalized Meir-Keeler type 𝛼−𝜙-contractive mapping, then we have, for all 𝑥,𝑦∈𝑋 and 𝜙∈Φ, 𝛼𝜎𝜎(),𝜎.𝑥,𝑦𝑇𝑥,𝑇𝑦≤𝜙𝑥,𝑦,𝜎𝑥,𝑇𝑥,𝜎𝑦,𝑇𝑦𝑥,𝑇𝑦,𝜎𝑦,𝑇𝑥 (11)

In what follows, we state the main fixed point theorem for a generalized Meir-Keeler type 𝛼−𝜙-contractive mapping in the setting of complete metric-like space.

Theorem 20. Let (𝑋,𝜎) be a complete metric-like space and let 𝑇∶𝑋→𝑋 be a generalized Meir-Keeler type 𝛼−𝜙-contractive mapping where 𝛼 is transitive. Suppose that(i)𝑇 is 𝛼-admissible;(ii)there exists 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1;(iii)𝑇 is continuous.Then there exists 𝑢∈𝑋 such that 𝑇𝑢=𝑢.

Proof. Our proof consists of four steps. In the first step, we prove that 𝛼(𝑥_𝑛,𝑥_𝑛+1)≥1, for all 𝑛=0,1,…. Due to assumption (ii) of the theorem, there exists 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1. We will construct an iterative sequence {𝑥_𝑛} in 𝑋 as follows: 𝑥_𝑛+1=𝑇𝑥_𝑛=𝑇^𝑛+1𝑥₀∀𝑛≥0. (12) If we have 𝑥_𝑛0=𝑥_𝑛0+1, for some 𝑛₀, then the proof is completed. Indeed, 𝑢=𝑥_𝑛0 is a fixed point of 𝑇. Hence, throughout the proof, we presume that 𝑥_𝑛≠𝑥_𝑛+1∀𝑛. (13) Since 𝑇 is 𝛼-admissible, we have 𝛼𝑥₀,𝑥₁=𝛼𝑥₀,𝑇𝑥₀≥1⟹𝛼𝑇𝑥₀,𝑇𝑥₁=𝛼𝑥₁,𝑥₂≥1. (14) By elementary calculations, we derive that 𝛼𝑥_𝑛,𝑥_𝑛+1≥1,∀𝑛=0,1,…. (15) In the second step, we will prove that lim_𝑛→∞𝜎(𝑥_𝑛,𝑥_𝑛+1)=0. Notice that we have 𝜎(𝑥_𝑛,𝑥_𝑛+1)>0 for all 𝑛=0,1,2,… by (13) and Lemma 7(C). Since 𝑇 is a generalized Meir-Keeler type 𝛼−𝜙-contractive mapping, by taking 𝑥=𝑥_𝑛−1 and 𝑦=𝑥_𝑛 in (11), we have 𝜎𝑥_𝑛,𝑥_𝑛+1=𝜎𝑇𝑥_𝑛−1,𝑇𝑥_𝑛≤𝛼𝑥_𝑛−1,𝑥_𝑛𝜎𝑇𝑥_𝑛−1,𝑇𝑥_𝑛≤𝜙𝜎𝑥_𝑛−1,𝑥_𝑛,𝜎𝑥_𝑛−1,𝑇𝑥_𝑛−1,𝜎𝑥_𝑛,𝑇𝑥_𝑛,𝜎𝑥_𝑛−1,𝑇𝑥_𝑛,𝜎𝑥_𝑛,𝑇𝑥_𝑛−1≤𝜙𝜎𝑥_𝑛−1,𝑥_𝑛,𝜎𝑥_𝑛−1,𝑥_𝑛,𝜎𝑥_𝑛,𝑥_𝑛+1,𝜎𝑥_𝑛−1,𝑥_𝑛+1,𝜎𝑥_𝑛,𝑥_𝑛≤𝜙𝜎𝑥_𝑛−1,𝑥_𝑛,𝜎𝑥_𝑛−1,𝑥_𝑛,𝜎𝑥_𝑛,𝑥_𝑛+1,𝜎𝑥_𝑛−1,𝑥_𝑛+𝜎𝑥_𝑛,𝑥_𝑛+1,𝜎𝑥_𝑛−1,𝑥_𝑛+𝜎𝑥_𝑛,𝑥_𝑛+1. (16) We assert that {𝜎(𝑥_𝑛,𝑥_𝑛+1)} is decreasing; that is, 𝜎(𝑥_𝑛,𝑥_𝑛+1)<𝜎(𝑥_𝑛−1,𝑥_𝑛) for all 𝑛∈ℕ. Suppose, on the contrary, that 𝜎(𝑥_𝑛0,𝑥_𝑛0+1)≥𝜎(𝑥_𝑛0−1,𝑥_𝑛0) for some 𝑛₀∈ℕ. By taking 𝑥=𝑥_𝑛0−1 and 𝑦=𝑥_𝑛0 in (11) and (16), we have 𝜎𝑥_𝑛0,𝑥_𝑛0+1=𝜎𝑇𝑥_𝑛0−1,𝑇𝑥_𝑛0≤𝛼𝑥_𝑛0−1,𝑥_𝑛0𝜎𝑇𝑥_𝑛0−1,𝑇𝑥_𝑛0≤𝜙𝜎𝑥_𝑛0,𝑥_𝑛0+1,𝜎𝑥_𝑛0,𝑥_𝑛0+1,𝜎𝑥_𝑛0,𝑥_𝑛0+1,2𝜎𝑥_𝑛0,𝑥_𝑛0+1,2𝜎𝑥_𝑛0,𝑥_𝑛0+1<𝜎𝑥_𝑛0,𝑥_𝑛0+1, (17) which is a contradiction. So the {𝜎(𝑥_𝑛,𝑥_𝑛+1)} is decreasing, and it must converge to some 𝛾≥0; that is, lim_𝑛→∞𝜎𝑥_𝑛,𝑥_𝑛+1=𝛾. (18) By condition (𝜙₁), inequality (16) becomes 𝜎𝑥_𝑛,𝑥_𝑛+1=𝜎𝑇𝑥_𝑛−1,𝑇𝑥_𝑛≤𝜙𝜎𝑥_𝑛−1,𝑥_𝑛,𝜎𝑥_𝑛−1,𝑥_𝑛,𝜎𝑥_𝑛−1,𝑥_𝑛,2𝜎𝑥_𝑛−1,𝑥_𝑛,2𝜎𝑥_𝑛−1,𝑥_𝑛. (19) We next claim that 𝛾=0. If not, we assume that 𝛾>0. By taking limit as 𝑛→∞ in (19), we have 𝛾=lim_𝑛→∞𝜎𝑥_𝑛,𝑥_𝑛+1=lim_𝑛→∞𝜎𝑇𝑥_𝑛−1,𝑇𝑥_𝑛≤𝜙𝛾,𝛾,𝛾,2𝛾,2𝛾<𝛾, (20) which is a contradiction. Hence, we have 𝛾=0.
In the third step, we will prove that {𝑥_𝑛} is a 𝜎-Cauchy sequence. We will use the method of reductio ad absurdum. Suppose, on the contrary, that {𝑥_𝑛} is not a 𝜎-Cauchy sequence. Hence, there exists 𝜖>0 and subsequences {𝑥_𝑛𝑘} and {𝑥_𝑚𝑘} of {𝑥_𝑛} with 𝑚_𝑘>𝑛_𝑘≥𝑘 satisfying 𝜎𝑥_𝑚𝑘,𝑥_𝑛𝑘≥𝜖,𝜎𝑥_𝑚𝑘−1,𝑥_𝑛𝑘<𝜖. (21) Since 𝛼 is transitive, from (15), we have 𝛼(𝑥_𝑛,𝑥_𝑛+𝑘)≥1 and hence 𝛼(𝑥_𝑛,𝑥_𝑚)≥1. Consider the following: 𝜖≤𝜎𝑥_𝑚𝑘,𝑥_𝑛𝑘≤𝜎𝑥_𝑚𝑘,𝑥_𝑚𝑘−1+𝜎𝑥_𝑚𝑘−1,𝑥_𝑛𝑘<𝜖+𝜎𝑥_𝑚𝑘,𝑥_𝑚𝑘−1. (22) Letting 𝑘→∞, we obtain that lim_𝑛→∞𝜎𝑥_𝑚𝑘,𝑥_𝑛𝑘=𝜖. (23) Also we have 𝜖≤𝜎𝑥_𝑚𝑘,𝑥_𝑛𝑘≤𝜎𝑥_𝑚𝑘,𝑥_𝑚𝑘+1+𝜎𝑥_𝑚𝑘+1,𝑥_𝑛𝑘𝜎𝑥_𝑚𝑘+1,𝑥_𝑛𝑘≤𝜎𝑥_𝑚𝑘+1,𝑥_𝑚𝑘+𝜎𝑥_𝑚𝑘,𝑥_𝑛𝑘. (24) We get 𝜖≤𝜎𝑥_𝑚𝑘,𝑥_𝑛𝑘≤𝜎𝑥_𝑚𝑘,𝑥_𝑚𝑘+1+𝜎𝑥_𝑚𝑘+1,𝑥_𝑚𝑘+𝜎𝑥_𝑚𝑘,𝑥_𝑛𝑘. (25) Letting 𝑘→∞ in the inequality above, we find that lim_𝑛→∞𝜎𝑥_𝑚𝑘+1,𝑥_𝑛𝑘=𝜖. (26) Analogously, we derive that lim_𝑛→∞𝜎𝑥_𝑚𝑘,𝑥_𝑛𝑘+1=𝜖. (27) Further, we have 𝜖≤𝜎𝑥_𝑚𝑘,𝑥_𝑛𝑘≤𝜎𝑥_𝑚𝑘,𝑥_𝑚𝑘+1+𝜎𝑥_𝑚𝑘+1,𝑥_𝑛𝑘+1+𝜎𝑥_𝑛𝑘+1,𝑥_𝑛𝑘≤𝜎𝑥_𝑚𝑘,𝑥_𝑚𝑘+1+𝜎𝑥_𝑚𝑘+1,𝑥_𝑛𝑘+𝜎𝑥_𝑛𝑘,𝑥_𝑛𝑘+1+𝜎𝑥_𝑛𝑘+1,𝑥_𝑛𝑘. (28) Letting 𝑘→∞ in the above inequality, we get that lim_𝑛→∞𝜎𝑥_𝑚𝑘+1,𝑥_𝑛𝑘+1=𝜖. (29)
Notice also that 𝜎𝑥_𝑚𝑘+1,𝑥_𝑛𝑘+1=𝜎𝑇𝑥_𝑚𝑘,𝑇𝑥_𝑛𝑘≤𝛼𝑥_𝑚𝑘,𝑥_𝑛𝑘𝜎𝑇𝑥_𝑚𝑘,𝑇𝑥_𝑛𝑘≤𝜙𝜎𝑥_𝑚𝑘,𝑥_𝑛𝑘,𝜎𝑥_𝑚𝑘,𝑥_𝑚𝑘+1,𝜎𝑥_𝑛𝑘,𝑥_𝑛𝑘+1,𝜎𝑥_𝑚𝑘,𝑥_𝑛𝑘+1,𝜎𝑥_𝑛𝑘,𝑥_𝑚𝑘+1. (30) Letting 𝑘→∞ in the above inequality and taking the property (𝜙₂) into account, we get that 𝜖=lim_𝑛→∞𝜎𝑥_𝑞𝑛+1,𝑥_𝑝𝑛+1≤𝜙(𝜖,0,0,𝜖,𝜖)<𝜖, (31) which is a contradiction. Thus, {𝑥_𝑛} is a 𝜎-Cauchy sequence.
In the fourth and last step, we will prove that 𝑇 has a fixed point 𝑢∈𝑋. Owing to the fact that (𝑋,𝜎) is complete, there exists 𝑢∈𝑋 such that lim_𝑛→∞𝑥_𝑛=𝑢; equivalently, 𝜎(𝑢,𝑢)=lim_𝑛→∞𝜎𝑥_𝑛,𝑢=lim_{𝑚,𝑛→∞}𝜎𝑥_𝑛,𝑥_𝑚=0. (32) Since 𝑇 is continuous, we obtain from (32) that lim_𝑛→∞𝜎𝑇𝑥_𝑛,𝑇𝑢=lim_𝑛→∞𝜎𝑥_𝑛,𝑢=𝜎(𝑢,𝑢)=0. (33) Due to Lemma 8, we also have lim_𝑛→∞𝜎𝑥_𝑛,𝑇𝑢=𝜎(𝑢,𝑇𝑢). (34) Combining (32)–(34) and Lemma 7(A), we get immediately that 𝑢 is a fixed point of 𝑇; that is, 𝑇𝑢=𝑢.

In the next theorem the continuity of 𝑇 is not required.

Theorem 21. Let (𝑋,𝜎) be a complete metric-like space and let 𝑇∶𝑋→𝑋 be a generalized Meir-Keeler type 𝛼−𝜙-contractive mapping, mapping, where 𝛼 is transitive. Suppose that(i)𝑇 is 𝛼-admissible;(ii)there exists 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1;(iii)if {𝑥_𝑛} is a sequence in 𝑋 such that 𝛼(𝑥_𝑛,𝑥_𝑛+1)≥1 for all 𝑛 and 𝑥_𝑛→𝑥∈𝑋 as 𝑛→∞, then 𝛼(𝑥_𝑛,𝑥)≥1 for all 𝑛.Then there exists 𝑢∈𝑋 such that 𝑇𝑢=𝑢.

Proof. Following the proof of Theorem 20, we know that the sequence {𝑥_𝑛} defined by 𝑥_𝑛+1=𝑇𝑥_𝑛, for all 𝑛≥0, converges to 𝑢 where 𝑢∈𝑋. It is enough to show that 𝑢∈𝑋 is the fixed point of 𝑇. Suppose, on the contrary, that 𝜎(𝑇𝑢,𝑢)=𝑡>0. From (15) and condition (iii), there exists a subsequence {𝑥_𝑛(𝑘)} of {𝑥_𝑛} such that 𝛼(𝑥_𝑛(𝑘),𝑢)≥1 for all 𝑘. Applying (11), for all 𝑘, we get that 𝜎𝑥_𝑛(𝑘)+1,𝑇𝑢=𝜎𝑇𝑥_𝑛(𝑘),𝑇𝑢≤𝛼𝑥_𝑛(𝑘),𝑢𝜎𝑇𝑥_𝑛(𝑘)𝜎𝑥,𝑇𝑢≤𝜙_𝑛(𝑘),𝑢,𝜎𝑥_𝑛(𝑘),𝑇𝑥_𝑛(𝑘)(),𝜎𝑥,𝜎𝑢,𝑇𝑢_𝑛(𝑘),𝑇𝑢,𝜎𝑢,𝑇𝑥_𝑛(𝑘)=𝜙𝜎𝑥_𝑛(𝑘),𝑢,𝜎𝑥_𝑛(𝑘),𝑥_𝑛(𝑘)+1(),𝜎𝑥,𝜎𝑢,𝑇𝑢_𝑛(𝑘),𝑇𝑢,𝜎𝑢,𝑥_𝑛(𝑘)+1. (35) Letting 𝑘→∞ in the above equality and taking (34) into account, we get that 𝑡≤𝜙(0,0,𝑡,𝑡,0). (36) By (𝜙₂) we get that 𝑡≤𝜙(0,0,𝑡,𝑡,0)<𝑡, (37) which is a contradiction. Thus we get 𝜎(𝑢,𝑇𝑢)=0, and, by Lemma 7(A), we have 𝑢=𝑇𝑢.

For the uniqueness, we need an additional condition.(U) For all 𝑥,𝑦∈Fix(𝑇), we have 𝛼(𝑥,𝑦)≥1, where Fix(𝑇) denotes the set of fixed points of 𝑇.In what follows we will show that 𝑢 is a unique fixed point of 𝑇.

Theorem 22. Adding condition (𝑈) to the hypotheses of Theorem 20 (resp., Theorem 21), one obtains that 𝑢 is the unique fixed point of 𝑇.

Proof. We will use the reductio ad absurdum. Let 𝑣 be another fixed point of 𝑇 with 𝑣≠𝑢 and hence 𝜎(𝑢,𝑣)=𝑡>0. By hypothesis (𝑈), 1≤𝛼(𝑢,𝑣)=𝛼(𝑇𝑢,𝑇𝑣). (38)
Due to inequality (11) we have𝛼()𝜎(𝜎()()(),𝜎()(.𝑢,𝑣𝑇𝑢,𝑇𝑦≤𝜙𝑢,𝑣,𝜎𝑢,𝑇𝑢,𝜎𝑣,𝑇𝑣𝑢,𝑇𝑣,𝜎𝑣,𝑇𝑢)) (39) Taking property (𝜙₂) into account, we get that 𝑡≤(𝑡,0,0,𝑡,𝑡)<𝑡, (40) which is a contradiction. Hence, 𝜎(𝑢,𝑣)=0. It follows from Lemma 7(A) that 𝑢=𝑣. Thus we proved that 𝑢 is the unique fixed point of 𝑇.

3. Fixed Point Theorem via Auxiliary Functions

In the section, we will discuss the notion of generalized (𝜑,𝜙,𝜓,𝜉)-𝛼-contractive mappings and prove fixed point theorems for these mappings in complete metric-like spaces. We denote by Ψ the class of functions 𝜓∶ℝ⁺→ℝ⁺ satisfying the following conditions:(𝜓₁)𝜓 is continuous and nondecreasing;(𝜓₂)𝜓(𝑡)>0 for all 𝑡>0 and 𝜓(0)=0.

Let Φ_𝜓 be the class of all function 𝜙∶ℝ⁺⁵→ℝ⁺ satisfying the following conditions:(ϕ₁)𝜙 is an increasing and continuous function in each coordinate;(ϕ₂)for 𝑡>0, 𝜙(𝜓(𝑡),𝜓(𝑡),𝜓(𝑡),𝜓(2𝑡),𝜓(2𝑡))≤𝜓(𝑡), 𝜙(𝑡,0,0,𝑡,𝑡)≤𝑡, and 𝜙(0,0,𝑡,𝑡,0)≤𝑡, where 𝜓∈Ψ;(ϕ₃)𝜙(𝑡₁,𝑡₂,𝑡₃,𝑡₄,𝑡₅)=0 if and only if 𝑡₁=𝑡₂=𝑡₃=𝑡₄=𝑡₅=0.

We use the following notations to specify the collection of the given functions: Ξ=𝜉∶ℝ⁺⟶ℝ⁺∶𝜉islowercontinuous,Θ=𝜑∶ℝ⁺⟶ℝ⁺∶𝜑iscontinuous. (41)

We now state the new notion of generalized (𝜑,𝜙,𝜓,𝜉)-𝛼-contractive mappings in metric-like spaces is as follows.

Definition 23. Let (𝑋,𝜎) be a metric-like space and let 𝛼∶𝑋×𝑋→ℝ⁺. One says that 𝑇 is called a generalized (𝜑,𝜙,𝜓,𝜉)-𝛼-contractive mapping if 𝑇 is 𝛼-admissible and satisfies the following inequality: 𝛼𝜑𝜎𝜓𝜎(𝜎(𝜎,𝜓𝜎𝜎𝜎(),𝜎𝑥,𝑦𝑇𝑥,𝑇𝑦≤𝜙𝑥,𝑦,𝜓𝑥,𝑇𝑥)),𝜓𝑦,𝑇𝑦𝑥,𝑇𝑦,𝜓𝑦,𝑇𝑥−𝜉max𝑥,𝑦,𝜎𝑥,𝑇𝑥,𝜎𝑦,𝑇𝑦𝑥,𝑇𝑦+𝜎𝑦,𝑇𝑥4, (42) for all 𝑥,𝑦∈𝑋, where 𝜓∈Ψ, 𝜙∈Φ_𝜓, 𝜑∈Θ, and 𝜉∈Ξ.

One now states the main fixed point of this section as follows.

Theorem 24. Let (𝑋,𝜎) be a complete metric-like space and let 𝑇∶𝑋→𝑋 be a (𝜑,𝜙,𝜓,𝜉)-𝛼-contractive mapping where 𝛼 is transitive. Suppose that(i)𝑇 is 𝛼-admissible;(ii)there exists 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1;(iii)𝑇 is continuous.Then there exists 𝑢∈𝑋 such that 𝑇𝑢=𝑢.

Proof. As in the proof of Theorem 20, we construct an iterative sequence {𝑥_𝑛} in 𝑋 as follows: 𝑥_𝑛+1=𝑇𝑥_𝑛=𝑇^𝑛+1𝑥₀∀𝑛≥0. (43) If we have 𝑥_𝑛0=𝑥_𝑛0+1, for some 𝑛₀, then the proof is completed. Indeed, 𝑢=𝑥_𝑛0 is a fixed point of 𝑇. Hence, from now on, we assume that 𝑥_𝑛≠𝑥_𝑛+1∀𝑛. (44) Moreover, due to Lemmas 7(C) and (D), we have 𝜎𝑥_𝑛,𝑥_𝑛+1>0. (45) Again, as in the proof of Theorem 20, since 𝑇 is 𝛼-admissible, we deduce that 𝛼𝑥_𝑛,𝑥_𝑛+1≥1,∀𝑛=0,1,…. (46)
Owing to the fact that 𝑇 is a generalized (𝜑,𝜙,𝜓,𝜉)-𝛼-contraction, by taking 𝑥=𝑥_𝑛−1 and 𝑦=𝑥_𝑛 in (42), we have 𝜑𝜎𝑥_𝑛,𝑥_𝑛+1=𝜑𝜎𝑇𝑥_𝑛−1,𝑇𝑥_𝑛≤𝛼𝑥_𝑛−1,𝑥_𝑛𝜑𝜎𝑇𝑥_𝑛−1,𝑇𝑥_𝑛≤𝜙𝜓𝜎𝑥_𝑛−1,𝑥_𝑛,𝜓𝜎𝑥_𝑛−1,𝑇𝑥_𝑛−1,𝜓𝜎𝑥_𝑛,𝑇𝑥_𝑛,𝜓𝜎𝑥_𝑛−1,𝑇𝑥_𝑛,𝜓𝜎𝑥_𝑛,𝑇𝑥_𝑛−1−𝜉max𝜎𝑥_𝑛−1,𝑥_𝑛,𝜎𝑥_𝑛−1,𝑇𝑥_𝑛−1,𝜎𝑥_𝑛,𝑇𝑥_𝑛,𝜎𝑥_𝑛−1,𝑇𝑥_𝑛+𝜎𝑥_𝑛,𝑇𝑥_𝑛−14≤𝜙𝜓𝜎𝑥_𝑛−1,𝑥_𝑛,𝜓𝜎𝑥_𝑛−1,𝑥_𝑛,𝜓𝜎𝑥_𝑛,𝑥_𝑛+1,𝜓𝜎𝑥_𝑛−1,𝑥_𝑛+1,𝜓𝜎𝑥_𝑛,𝑥_𝑛−𝜉max𝜎𝑥_𝑛−1,𝑥_𝑛,𝜎𝑥_𝑛−1,𝑥_𝑛,𝜎𝑥_𝑛,𝑥_𝑛+1,𝜎𝑥_𝑛−1,𝑥_𝑛+1+𝜎𝑥_𝑛,𝑥_𝑛4≤𝜙𝜓𝜎𝑥_𝑛−1,𝑥_𝑛,𝜓𝜎𝑥_𝑛,𝑥_𝑛+1,𝜓𝜎𝑥_𝑛−1,𝑥_𝑛+𝜎𝑥_𝑛,𝑥_𝑛+1,𝜓𝜎𝑥_𝑛−1,𝑥_𝑛+𝜎𝑥_𝑛,𝑥_𝑛+1−𝜉max𝜎𝑥_𝑛−1,𝑥_𝑛,𝜎𝑥_𝑛−1,𝑥_𝑛,𝜎𝑥_𝑛,𝑥_𝑛+1,𝜎𝑥_𝑛−1,𝑥_𝑛+𝜎𝑥_𝑛,𝑥_𝑛+12. (47) As a first step, we prove that lim_𝑛→∞𝜎𝑥_𝑛,𝑥_𝑛+1=0. (48) For this goal, we show that {𝜎(𝑥_𝑛,𝑥_𝑛+1)} is decreasing; that is, 𝜎(𝑥_𝑛,𝑥_𝑛+1)<𝜎(𝑥_𝑛−1,𝑥_𝑛) for all 𝑛∈ℕ. Suppose, on the contrary, that 𝜎(𝑥_𝑛0,𝑥_𝑛0+1)≥𝜎(𝑥_𝑛0−1,𝑥_𝑛0) for some 𝑛₀∈ℕ. By substituting 𝑥=𝑥_𝑛0−1 and 𝑦=𝑥_𝑛0 in (42) and (47), we have 𝜑𝜎𝑥_𝑛0,𝑥_𝑛0+1=𝜑𝜎𝑇𝑥_𝑛0−1,𝑇𝑥_𝑛0≤𝛼𝑥_𝑛0−1,𝑥_𝑛0𝜑𝜎𝑇𝑥_𝑛0−1,𝑇𝑥_𝑛0≤𝜙𝜓𝜎𝑥_𝑛0,𝑥_𝑛0+1,𝜓𝜎𝑥_𝑛0,𝑥_𝑛0+1,𝜓𝜎𝑥_𝑛0,𝑥_𝑛0+1,𝜓2𝜎𝑥_𝑛0,𝑥_𝑛0+1,𝜓2𝜎𝑥_𝑛0,𝑥_𝑛0+1−𝜉𝜎𝑥_𝑛0,𝑥_𝑛0+1≤𝜓𝜎𝑥_𝑛0,𝑥_𝑛0+1−𝜉𝜎𝑥_𝑛0,𝑥_𝑛0+1. (49) Regarding the condition 𝜑(𝑡)−𝜓(𝑡)+𝜉(𝑡)>0 for all 𝑡>0 and by using inequality (49), we derive that 𝜎(𝑥_𝑛0,𝑥_𝑛0+1)=0, which contradicts to (45). Hence, we deduce that 𝜎𝑥_𝑛,𝑥_𝑛+1<𝜎𝑥_𝑛−1,𝑥_𝑛∀𝑛∈ℕ. (50) From the arguments above, we also have, for each 𝑛∈ℕ, 𝜑𝜎𝑥_𝑛,𝑥_𝑛+1=𝜑𝜎𝑇𝑥_𝑛−1,𝑇𝑥_𝑛≤𝜓𝜎𝑥_𝑛−1,𝑥_𝑛−𝜉𝜎𝑥_𝑛−1,𝑥_𝑛. (51) It follows from (50) that the sequence {𝜎(𝑥_𝑛,𝑥_𝑛+1)} is monotone decreasing. Hence, it should be convergent to some 𝜂≥0; that is, lim_𝑛→∞𝜎𝑥_𝑛,𝑥_𝑛+1=𝜂. (52) Letting 𝑛→∞ in (51) and by using the continuities of 𝜓 and 𝜑 and the lower semicontinuity of 𝜉, we have 𝜑𝜂≤𝜓𝜂−𝜉𝜂, (53) which implies that 𝜂=0.
As in the proof of Theorem 20, we will use the same techniques, method of reductio ad absurdum, to prove that {𝑥_𝑛} is a 𝜎-Cauchy sequence. Suppose, on the contrary, that {𝑥_𝑛} is not a 𝜎-Cauchy sequence. Hence, there exists 𝜖>0 and subsequences {𝑥_𝑛𝑘} and {𝑥_𝑚𝑘} of {𝑥_𝑛} with 𝑚_𝑘>𝑛_𝑘≥𝑘 satisfying 𝜎𝑥_𝑚𝑘,𝑥_𝑛𝑘≥𝜖,𝜎𝑥_𝑚𝑘−1,𝑥_𝑛𝑘<𝜖. (54) By repeating the related lines in the proof of Theorem 20, we find the following limits: lim_𝑛→∞𝜎𝑥_𝑚𝑘,𝑥_𝑛𝑘=lim_𝑛→∞𝜎𝑥_𝑚𝑘+1,𝑥_𝑛𝑘=lim_𝑛→∞𝜎𝑥_𝑚𝑘,𝑥_𝑛𝑘+1=lim_𝑛→∞𝜎𝑥_𝑚𝑘+1,𝑥_𝑛𝑘+1=𝜖. (55) By assumption of the theorem, we have 𝜎𝑥_𝑚𝑘+1,𝑥_𝑛𝑘+1=𝜎𝑇𝑥_𝑚𝑘,𝑇𝑥_𝑛𝑘≤𝛼𝑥_𝑚𝑘,𝑥_𝑛𝑘𝜎𝑇𝑥_𝑚𝑘,𝑇𝑥_𝑛𝑘≤𝜙𝜓𝜎𝑥_𝑛𝑘,𝑥_𝑚𝑘,𝜓𝜎𝑥_𝑛𝑘,𝑇𝑥_𝑛𝑘,𝜓𝜎𝑥_𝑚𝑘,𝑇𝑥_𝑚𝑘,𝜓𝜎𝑥_𝑛𝑘,𝑇𝑥_𝑚𝑘,𝜓𝜎𝑥_𝑚𝑘,𝑇𝑥_𝑛𝑘−𝜉max𝜎𝑥_𝑛𝑘,𝑥_𝑚𝑘,𝜎𝑥_𝑛𝑘,𝑇𝑥_𝑛𝑘,𝜎𝑥_𝑚𝑘,𝑇𝑥_𝑚𝑘,𝜎𝑥_𝑛𝑘,𝑥_𝑚𝑘+1+𝜎𝑥_𝑚𝑘,𝑇𝑥_𝑛𝑘4≤𝜙𝜓𝜎𝑥_𝑛𝑘,𝑥_𝑚𝑘,𝜓𝜎𝑥_𝑛𝑘,𝑥_𝑛𝑘+1,𝜓𝜎𝑥_𝑚𝑘,𝑇𝑥_𝑚𝑘,𝜓𝜎𝑥_𝑛𝑘,𝑥_𝑚𝑘+1,𝜓𝜎𝑥_𝑚𝑘,𝑥_𝑛𝑘+1−𝜉max𝜎𝑥_𝑛𝑘,𝑥_𝑚𝑘,𝜎𝑥_𝑛𝑘,𝑥_𝑛𝑘+1,𝜎𝑥_𝑚𝑘,𝑥_𝑚𝑘+1,𝜎𝑥_𝑛𝑘,𝑥_𝑚𝑘+1+𝜎𝑥_𝑚𝑘,𝑥_𝑛𝑘+14. (56) Letting 𝑛→∞ in (56), we find that 𝜑(𝜖)≤𝜙𝜓(𝜖),0,0,𝜓(𝜖),𝜓(𝜖)−𝜉(𝜖)≤𝜓(𝜖)−𝜉(𝜖), (57) which implies that 𝜖=0. This is a contradiction. Therefore, the sequence {𝑥_𝑛} is a 𝜎-Cauchy sequence.
As a last step, we will prove that 𝑇 has a fixed point 𝑢∈𝑋. Owing to the fact that (𝑋,𝜎) is complete, there exists 𝑢∈𝑋 such that lim_𝑛→∞𝑥_𝑛=𝑢, equivalently, 𝜎(𝑢,𝑢)=lim_𝑛→∞𝜎𝑥_𝑛,𝑢=lim_{𝑚,𝑛→∞}𝜎𝑥_𝑛,𝑥_𝑚=0. (58) Since 𝑇 is continuous, we obtain from (58) that lim_𝑛→∞𝜎𝑇𝑥_𝑛,𝑇𝑢=lim_𝑛→∞=𝜎𝑥_𝑛,𝑢=𝜎(𝑢,𝑢)=0. (59) Due to Lemma 8, we also have lim_𝑛→∞𝜎𝑥_𝑛,𝑇𝑢=𝜎(𝑢,𝑇𝑢). (60) On account of (58)–(60) and Lemma 7(A), we derive that 𝑢 is a fixed point of 𝑇; that is, 𝑇𝑢=𝑢.

Theorem 25. Let (𝑋,𝜎) be a complete metric-like space and let 𝑇∶𝑋→𝑋 be a (𝜑,𝜙,𝜓,𝜉)-𝛼-contractive mapping where 𝛼 is transitive. Suppose that(i)𝑇 is 𝛼-admissible;(ii)there exists 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1;(iii)if {𝑥_𝑛} is a sequence in 𝑋 such that 𝛼(𝑥_𝑛,𝑥_𝑛+1)≥1 for all 𝑛 and 𝑥_𝑛→𝑥∈𝑋 as 𝑛→∞, then 𝛼(𝑥_𝑛,𝑥)≥1 for all 𝑛.Then there exists 𝑢∈𝑋 such that 𝑇𝑢=𝑢.

Proof. Following the proof of Theorem 24, we know that the sequence {𝑥_𝑛} defined by 𝑥_𝑛+1=𝑇𝑥_𝑛, for all 𝑛≥0, converges to 𝑢 where 𝑢∈𝑋. It is enough to show that 𝑢∈𝑋 is the fixed point of 𝑇. Suppose, on the contrary, that 𝜎(𝑇𝑢,𝑢)=𝑡>0. From (46) and condition (iii), there exists a subsequence {𝑥_𝑛(𝑘)} of {𝑥_𝑛} such that 𝛼(𝑥_𝑛(𝑘),𝑢)≥1 for all 𝑘. Applying (42), for all 𝑘, we get that 𝜎𝑥_𝑛(𝑘)+1,𝑇𝑢=𝜎𝑇𝑥_𝑛(𝑘),𝑇𝑢≤𝛼𝑥_𝑛(𝑘),𝑢𝜎𝑇𝑥_𝑛(𝑘)𝜓𝜎𝑥,𝑇𝑢≤𝜙_𝑛(𝑘),𝑢,𝜓𝜎𝑥_𝑛(𝑘),𝑇𝑥_𝑛(𝑘)(𝜎(,𝜓𝜎𝑥,𝜓𝑢,𝑇𝑢))_𝑛(𝑘),𝑇𝑢,𝜓𝜎𝑢,𝑇𝑥_𝑛(𝑘)−𝜉max𝜎𝑥_𝑛(𝑘),𝑢,𝜎𝑥_𝑛(𝑘),𝑇𝑥_𝑛(𝑘)(),𝜎𝑥,𝜎𝑢,𝑇𝑢_𝑛(𝑘),𝑇𝑢+𝜎𝑢,𝑇𝑥_𝑛(𝑘)4=𝜙𝜓𝜎𝑥_𝑛(𝑘),𝑢,𝜓𝜎𝑥_𝑛(𝑘),𝑥_𝑛(𝑘)+1(𝜎(,𝜓𝜎𝑥,𝜓𝑢,𝑇𝑢))_𝑛(𝑘),𝑇𝑢,𝜓𝜎𝑢,𝑥_𝑛(𝑘)+1−𝜉max𝜎𝑥_𝑛(𝑘),𝑢,𝜎𝑥_𝑛(𝑘),𝑥_𝑛(𝑘)+1(),𝜎𝑥,𝜎𝑢,𝑇𝑢_𝑛(𝑘),𝑇𝑢+𝜎𝑢,𝑥_𝑛(𝑘)+14. (61) Letting 𝑘→∞ in the above equality and taking (60) into account, we get that 𝑡≤𝜙(0,0,𝑡,𝑡,0)−𝜉(𝑡)≤𝜙(0,0,𝑡,𝑡,0). (62) By (𝜓₂) we get that 𝑡≤𝜙(0,0,𝑡,𝑡,0)<𝑡, (63) which is a contradiction. Thus we get 𝜎(𝑢,𝑇𝑢)=0, and, by Lemma 7(A), we have 𝑢=𝑇𝑢.

In the next theorem we will show that 𝑢 is a unique fixed point of 𝑇.

Theorem 26. Adding condition (𝑈) to the hypotheses of Theorem 24 (resp., Theorem 25), one obtains that 𝑢 is the unique fixed point of 𝑇.

Proof. We will use the reductio ad absurdum. Let 𝑣 be another fixed point of 𝑇 with 𝑣≠𝑢 and hence 𝜎(𝑢,𝑣)=𝑡>0. By hypothesis (𝑈), 1≤𝛼(𝑢,𝑣)=𝛼(𝑇𝑢,𝑇𝑣). (64) Due to inequality (42) we have 𝛼()𝜎()𝜓(𝜎((𝜎((𝜎(,𝜓(𝜎((𝜎(𝜎()()(),𝜎()()𝑢,𝑣𝑇𝑢,𝑇𝑣≤𝜙𝑢,𝑣)),𝜓𝑢,𝑇𝑢)),𝜓𝑣,𝑇𝑣))𝑢,𝑇𝑣)),𝜓𝑣,𝑇𝑢))−𝜉max𝑢,𝑣,𝜎𝑢,𝑇𝑢,𝜎𝑣,𝑇𝑣𝑢,𝑇𝑣+𝜎𝑣,𝑇𝑢4. (65) Taking property (𝜙₂) into account, we get that 𝑡≤(𝑡,0,0,𝑡,𝑡)−𝜉(𝑡)<𝑡−𝜉(𝑡) (66) which is a contradiction. Hence, 𝜎(𝑢,𝑣)=0. By Lemma 7(A) we get that 𝑢=𝑣. Thus we proved that 𝑢 is the unique fixed point of 𝑇.

4. Fixed Point Theorems via the Weaker Meir-Keeler Function 𝜇

In the section, we will investigate the existence and uniqueness of a fixed point of certain mappings by using the Meir-Keeler function. Now, we recall the notion of the weaker Meir-Keeler function 𝜇∶[0,∞)→[0,∞).

Definition 27 (see [28]). One calls 𝜇∶[0,∞)→[0,∞) a weaker Meir-Keeler function if, for each 𝜂>0, there exists 𝛿>0 such that, for 𝑡∈[0,∞) with 𝜂≤𝑡<𝜂+𝛿, there exists 𝑛₀∈ℕ such that 𝜇^𝑛0(𝑡)<𝜂.

One denotes by ℳ the class of nondecreasing functions 𝜇∶[0,∞)→[0,∞) satisfying the following conditions:(μ₁)𝜇∶[0,∞)→[0,∞) is a weaker Meir-Keeler function;(μ₂)𝜇(𝑡)>0 for 𝑡>0 and 𝜇(0)=0;(μ₃)for all 𝑡>0, {𝜇^𝑛(𝑡)}_𝑛∈ℕ is decreasing;(μ₄)if lim_𝑛→∞𝑡_𝑛=𝛾, then lim_𝑛→∞𝜇(𝑡_𝑛)≤𝛾.

And one denotes by Θ the class of functions 𝜑∶[0,∞)→[0,∞) satisfying the following conditions: (𝜑₁) 𝜑 is continuous; (𝜑₂) 𝜑(𝑡)>0 for 𝑡>0 and 𝜑(0)=0.

We state the notion of the generalized weaker Meir-Keeler type (𝜇,𝜑)-𝛼-contractive mappings in metric-like spaces as follows.

Definition 28. Let (𝑋,𝜎) be a metric-like space, and let 𝛼∶𝑋×𝑋→ℝ⁺. One says that 𝑇∶𝑌→𝑌 is called a generalized weaker Meir-Keeler type 𝛼-(𝜇,𝜑)-contractive mapping if 𝑇 is 𝛼-admissible and satisfies 𝛼𝑥,𝑦𝜎𝑇𝑥,𝑇𝑦≤𝜇𝑀𝑥,𝑦−𝜑𝑀𝑥,𝑦, (67) for all 𝑥,𝑦∈𝑋, where 𝜇∈ℳ, 𝜑∈Θ, and 𝑀𝜎,𝜎𝑥,𝑦=max𝑥,𝑦,𝜎𝑥,𝑓𝑥,𝜎𝑦,𝑓𝑦𝑥,𝑓𝑦+𝜎𝑦,𝑓𝑥4. (68)

The main result of this section is the following.

Theorem 29. Let (𝑋,𝜎) be a complete metric-like space and let 𝑇∶𝑋→𝑋 be a generalized weaker Meir-Keeler type 𝛼-(𝜇,𝜑)-contractive mapping where 𝛼 is transitive. Suppose that(i)𝑇 is 𝛼-admissible;(ii)there exists 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1;(iii)𝑇 is continuous.Then there exists 𝑢∈𝑋 such that 𝑇𝑢=𝑢.

Proof. Following the lines in the proof of Theorem 20, we construct an iterative sequence {𝑥_𝑛} in 𝑋 as follows: 𝑥_𝑛+1=𝑇𝑥_𝑛=𝑇^𝑛+1𝑥₀∀𝑛≥0. (69) If we have 𝑥_𝑛0=𝑥_𝑛0+1, for some 𝑛₀, then the proof is completed. Indeed, 𝑢=𝑥_𝑛0 is a fixed point of 𝑇. Hence, from now on, we assume that 𝑥_𝑛≠𝑥_𝑛+1∀𝑛. (70) Moreover, due to Lemmas 7(C) and (D), we have 𝜎𝑥_𝑛,𝑥_𝑛+1>0. (71) Again, by following the lines in the proof of Theorem 20, we get that 𝛼𝑥_𝑛,𝑥_𝑛+1≥1,∀𝑛=0,1,…. (72) We divide the proof into three steps.
Step 1. We will prove that lim_𝑛→∞𝜎(𝑥_𝑛,𝑥_𝑛+1)=0. Since 𝑇 is a generalized weaker Meir-Keeler type 𝛼-(𝜇,𝜑)-contractive mapping, by taking 𝑥=𝑥_𝑛−1 and 𝑦=𝑥_𝑛 in (67), we have 𝜎𝑥_𝑛,𝑥_𝑛+1=𝜎𝑇𝑥_𝑛−1,𝑇𝑥_𝑛≤𝛼𝑥_𝑛−1,𝑥_𝑛𝜎𝑇𝑥_𝑛−1,𝑇𝑥_𝑛≤𝜇𝑀𝑥_𝑛−1,𝑥_𝑛−𝜑𝑀𝑥_𝑛−1,𝑥_𝑛, (73) where 𝑀𝑥_𝑛−1,𝑥_𝑛=max𝜎𝑥_𝑛−1,𝑥_𝑛,𝜎𝑥_𝑛−1,𝑇𝑥_𝑛−1,𝜎𝑥_𝑛,𝑇𝑥_𝑛,𝜎𝑥_𝑛−1,𝑇𝑥_𝑛+𝜎𝑥_𝑛,𝑇𝑥_𝑛−14=max𝜎𝑥_𝑛−1,𝑥_𝑛,𝜎𝑥_𝑛−1,𝑥_𝑛,𝜎𝑥_𝑛,𝑥_𝑛+1,𝜎𝑥_𝑛−1,𝑥_𝑛+1+𝜎𝑥_𝑛,𝑥_𝑛4=max𝜎𝑥_𝑛−1,𝑥_𝑛,𝜎𝑥_𝑛,𝑥_𝑛+1,𝜎𝑥_𝑛−1,𝑥_𝑛+𝜎𝑥_𝑛,𝑥_𝑛+12=max𝜎𝑥_𝑛−1,𝑥_𝑛,𝜎𝑥_𝑛,𝑥_𝑛+1. (74) If 𝑀(𝑥_𝑛−1,𝑥_𝑛)=𝜎(𝑥_𝑛,𝑥_𝑛+1), then, by (73) and the properties of the functions 𝜇 and 𝜑, we have 𝜎𝑥_𝑛,𝑥_𝑛+1=𝜎𝑇𝑥_𝑛−1,𝑇𝑥_𝑛≤𝛼𝑥_𝑛−1,𝑥_𝑛𝜎𝑇𝑥_𝑛−1,𝑇𝑥_𝑛≤𝜇𝜎𝑥_𝑛,𝑥_𝑛+1−𝜑𝜎𝑥_𝑛,𝑥_𝑛+1≤𝜇𝜎𝑥_𝑛,𝑥_𝑛+1. (75) Since {𝜇^𝑛(𝑡)}_𝑛∈ℕ is decreasing, the inequality above yields a contradiction. Thus, we conclude that 𝑀(𝑥_𝑛−1,𝑥_𝑛)=𝜎(𝑥_𝑛−1,𝑥_𝑛) and inequality (71) becomes 𝜎𝑥_𝑛,𝑥_𝑛+1≤𝜇𝜎𝑥_𝑛−1,𝑥_𝑛, (76) for all 𝑛∈ℕ. Recursively, we conclude that 𝜎𝑥_𝑛,𝑥_𝑛+1≤𝜇^𝑛𝜎𝑥₀,𝑥₁, (77) for all 𝑛∈ℕ.
Since {𝜇^𝑛(𝜎(𝑥₀,𝑥₁))}_𝑛∈ℕ is decreasing, it must converge to some 𝜂≥0. We claim that 𝜂=0. On the contrary, assume that 𝜂>0. Then, by the definition of the weaker Meir-Keeler function 𝜇, there exists 𝛿>0 such that, for 𝑥₀,𝑥₁∈𝑋 with 𝜂≤𝜎(𝑥₀,𝑥₁)<𝛿+𝜂, there exists 𝑛₀∈ℕ such that 𝜇^𝑛0(𝜎(𝑥₀,𝑥₁))<𝜂. Since lim_𝑛→∞𝜇^𝑛(𝜎(𝑥₀,𝑥₁))=𝜂, there exists 𝑝₀∈ℕ such that 𝜂≤𝜇^𝑝(𝜎(𝑥₀,𝑥₁))<𝛿+𝜂, for all 𝑝≥𝑝₀. Thus, we conclude that 𝜇^𝑝0+𝑛0(𝜎(𝑥₀,𝑥₁))<𝜂. So we get a contradiction. Therefore lim_𝑛→∞𝜇^𝑛(𝜎(𝑥₀,𝑥₁))=0; that is, lim_𝑛→∞𝜎𝑥_𝑛,𝑥_𝑛+1=0. (78)
Step 2. We prove that {𝑥_𝑛} is a 𝜎-Cauchy sequence.
We will use the method of reductio ad absurdum, as in the proof of Theorem 20. Suppose, on the contrary, that {𝑥_𝑛} is not a 𝜎-Cauchy sequence. Hence, there exists 𝜖>0 and subsequences {𝑥_𝑛𝑘} and {𝑥_𝑚𝑘} of {𝑥_𝑛} with 𝑚_𝑘>𝑛_𝑘≥𝑘 satisfying 𝜎𝑥_𝑚𝑘,𝑥_𝑛𝑘≥𝜖,𝜎𝑥_𝑚𝑘−1,𝑥_𝑛𝑘<𝜖. (79) By repeating the related lines in the proof of Theorem 20, we find the following limits: lim_𝑛→∞𝜎𝑥_𝑚𝑘,𝑥_𝑛𝑘=lim_𝑛→∞𝜎𝑥_𝑚𝑘+1,𝑥_𝑛𝑘=lim_𝑛→∞𝜎𝑥_𝑚𝑘,𝑥_𝑛𝑘+1=lim_𝑛→∞𝜎𝑥_𝑚𝑘+1,𝑥_𝑛𝑘+1=𝜖. (80) By the assumption of the theorem, we have 𝜎𝑥_𝑛𝑘+1,𝑥_𝑚𝑘+1=𝛼𝑥_𝑛𝑘,𝑥_𝑚𝑘𝜎𝑇𝑥_𝑛𝑘,𝑇𝑥_𝑚𝑘≤𝜇𝑀𝑥_𝑛𝑘,𝑥_𝑚𝑘−𝜑𝑀𝑥_𝑛𝑘,𝑥_𝑚𝑘, (81) where 𝑀𝑥_𝑛𝑘,𝑥_𝑚𝑘=max𝜎𝑥_𝑛𝑘,𝑥_𝑚𝑘,𝜎𝑥_𝑛𝑘,𝑇𝑥_𝑛𝑘,𝜎𝑥_𝑚𝑘,𝑇𝑥_𝑚𝑘,𝜎𝑥_𝑛𝑘,𝑇𝑥_𝑚𝑘+𝜎𝑥_𝑚𝑘,𝑇𝑥_𝑛𝑘4=max𝜎𝑥_𝑛𝑘,𝑥_𝑚𝑘,𝜎𝑥_𝑛𝑘,𝑥_𝑛𝑘+1,𝜎𝑥_𝑚𝑘,𝑥_𝑚𝑘+1,𝜎𝑥_𝑛𝑘,𝑥_𝑚𝑘+1+𝜎𝑥_𝑚𝑘,𝑥_𝑛𝑘+14. (82)
Case 1. If 𝑀(𝑥_𝑛𝑘,𝑥_𝑚𝑘)=𝜎(𝑥_𝑛𝑘,𝑥_𝑚𝑘), letting 𝑛→∞, then (81) becomes 𝜖≤𝜖−𝜑(𝜖), (83) which yields that 𝜑(𝜖)=0, and so we conclude that 𝜖=0. Therefore, we get a contradiction.
Case 2. If 𝑀(𝑥_𝑛𝑘,𝑥_𝑚𝑘)=𝜎(𝑥_𝑛𝑘,𝑥_𝑛𝑘+1) or 𝑀(𝑥_𝑛𝑘,𝑥_𝑚𝑘)=𝜎(𝑥_𝑚𝑘,𝑥_𝑚𝑘+1), letting 𝑛→∞, then (81) turns into 𝜖≤𝜇(0)−𝜑(0)=0, (84) which yields that 𝜖=0. It is a contradiction.
Case 3. If 𝑀(𝑥_𝑛𝑘,𝑥_𝑚𝑘)=(𝜎(𝑥_𝑛𝑘,𝑥_𝑚𝑘+1)+𝜎(𝑥_𝑚𝑘,𝑥_𝑛𝑘+1))/4, letting 𝑛→∞, then (81) becomes 𝜖𝜖≤𝜖−𝜑2, (85) which yields that 𝜑(𝜖/2)=0, and hence 𝜖=0. So, we get a contradiction.
Following the arguments above, we show also that {𝑥_𝑛} is a 𝜎-Cauchy sequence.
Step 3. In this step, we prove that 𝑇 has a fixed point 𝑢∈𝑋. Since (𝑋,𝜎) is complete, there exists 𝑢∈𝑋 such that lim_𝑛→∞𝑥_𝑛=𝑢; equivalently, 𝜎(𝑢,𝑢)=lim_𝑛→∞𝜎𝑥_𝑛,𝑢=lim_{𝑚,𝑛→∞}𝜎𝑥_𝑛,𝑥_𝑚=0. (86) Since 𝑇 is continuous, we obtain from (86) that lim_𝑛→∞𝜎𝑇𝑥_𝑛,𝑇𝑢=lim_𝑛→∞=𝜎𝑥_𝑛,𝑢=𝜎(𝑢,𝑢)=0. (87) Due to Lemma 8, we also have lim_𝑛→∞𝜎𝑥_𝑛,𝑇𝑢=𝜎(𝑢,𝑇𝑢). (88) On account of (58)–(88) and Lemma 7(A), we derive that 𝑢 is a fixed point of 𝑇; that is, 𝑇𝑢=𝑢.

Theorem 30. Let (𝑋,𝜎) be a complete metric-like space and let 𝑇∶𝑋→𝑋 be a generalized weaker Meir-Keeler type 𝛼-(𝜇,𝜑)-contractive mapping where 𝛼 is transitive. Suppose that(i)𝑇 is 𝛼-admissible;(ii)there exists 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1;(iii)if {𝑥_𝑛} is a sequence in 𝑋 such that 𝛼(𝑥_𝑛,𝑥_𝑛+1)≥1 for all 𝑛 and 𝑥_𝑛→𝑥∈𝑋 as 𝑛→∞, then 𝛼(𝑥_𝑛,𝑥)≥1 for all 𝑛.Then there exists 𝑢∈𝑋 such that 𝑇𝑢=𝑢.

Proof. Following the proof of Theorem 29, we know that the sequence {𝑥_𝑛} defined by 𝑥_𝑛+1=𝑇𝑥_𝑛, for all 𝑛≥0, converges to 𝑢 where 𝑢∈𝑋. It is enough to show that 𝑢∈𝑋 is the fixed point of 𝑇. Suppose, on the contrary, that 𝜎(𝑇𝑢,𝑢)=𝑡>0. From (71) and condition (iii), there exists a subsequence {𝑥_𝑛(𝑘)} of {𝑥_𝑛} such that 𝛼(𝑥_𝑛(𝑘),𝑢)≥1 for all 𝑘. Applying (67), for all 𝑘, we get that 𝜎𝑥_𝑛(𝑘)+1,𝑇𝑢=𝜎𝑇𝑥_𝑛(𝑘)𝑥,𝑇𝑢≤𝛼_𝑛(𝑘),𝑢𝜎𝑇𝑥_𝑛(𝑘)𝜇𝑀𝑥,𝑇𝑢_𝑛(𝑘),𝑢−𝜑𝑀𝑥_𝑛(𝑘),𝑢, (89) where 𝑀_𝑛(𝑘)𝜎𝑥,𝑢=max_𝑛(𝑘),𝑢,𝜎𝑥_𝑛(𝑘),𝑇𝑥_𝑛(𝑘)(),𝜎𝑥,𝜎𝑢,𝑇𝑢_𝑛(𝑘),𝑇𝑢+𝜎𝑢,𝑇𝑥_𝑛(𝑘)4=max𝜎𝑥_𝑛(𝑘),𝑢,𝜎𝑥_𝑛(𝑘),𝑥_𝑛(𝑘)+1(),𝜎𝑥,𝜎𝑢,𝑇𝑢_𝑛(𝑘),𝑇𝑢+𝜎𝑢,𝑥_𝑛(𝑘)+14. (90) Letting 𝑘→∞ in equality (89) and taking (88) into account, we get that 𝑡≤𝜇(𝑡)−𝜑(𝑡). (91) Since 𝜇(𝑡)−𝜑(𝑡)<𝑡, for all 𝑡>0, we conclude that 𝜎(𝑢,𝑇𝑢)=0; that is, 𝑇𝑢=𝑢.

In what follows we will show that 𝑢 is a unique fixed point of 𝑇.

Theorem 31. Adding condition (𝑈) to the hypotheses of Theorem 29 (resp., Theorem 30), one obtains that 𝑢 is the unique fixed point of 𝑇.

Proof. We will use the reductio ad absurdum. Let 𝑣 be another fixed point of 𝑇 with 𝑣≠𝑢 and hence 𝜎(𝑢,𝑣)=𝑡>0. By hypothesis (𝑈), 1≤𝛼(𝑢,𝑣)=𝛼(𝑇𝑢,𝑇𝑣). (92)
Due to inequality (67), we have 𝜎()()𝜎()𝑀𝑀,𝑢,𝑣=𝜎𝑇𝑢,𝑇𝑣≤𝛼𝑢,𝑦𝑇𝑢,𝑇𝑣≤𝜇𝑢,𝑦−𝜑𝑢,𝑦 (93) where 𝑀()𝜎()()(),𝜎()()𝑢,𝑣=max𝑢,𝑣,𝜎𝑢,𝑇𝑢,𝜎𝑣,𝑇𝑣𝑢,𝑇𝑣+𝜎𝑣,𝑇𝑢4𝜎()()(),𝜎()()=max𝑢,𝑣,𝜎𝑢,𝑢,𝜎𝑣,𝑣𝑢,𝑣+𝜎𝑣,𝑢4=𝜎(𝑢,𝑣). (94) Hence, we have 𝜎𝑢,𝑦≤𝜇𝜎𝑢,𝑦−𝜑𝜎𝑢,𝑦, (95) since 𝜇(𝑡)−𝜑(𝑡)<𝑡, for all 𝑡>0, which is a contradiction. Thus we proved that 𝑢 is the unique fixed point of 𝑇.

5. Consequences

In this section, we will demonstrate that several existing fixed point results in the literature can be deduced easily from our main results: Theorem 22, Theorem 26, and Theorem 31.

5.1. Standard Fixed Point Theorems

If we substitute 𝛼(𝑥,𝑦)=1 for all 𝑥,𝑦∈𝑋 in Theorem 22, we derive immediately the following fixed point theorem.

Theorem 32. Let (𝑋,𝜎) be a complete metric-like space and let 𝑇∶𝑋→𝑋 be a mapping. Suppose that for each 𝜂>0 there exists 𝛿>0 such that 𝜂≤𝜙𝜎𝑥,𝑦,𝜎(𝑥,𝑇𝑥),𝜎𝑦,𝑇𝑦,𝜎𝑥,𝑇𝑦,𝜎𝑦,𝑇𝑥<𝜂+𝛿⟶𝜎𝑇𝑥,𝑇𝑦<𝜂, (96) for all 𝑥,𝑦∈𝑋 and 𝜙∈Φ. Then there exists a unique fixed point 𝑢∈𝑋 such that 𝑇𝑢=𝑢.

If we take 𝛼(𝑥,𝑦)=1 for all 𝑥,𝑦∈𝑋 in Theorem 26, we get the following fixed point theorem.

Theorem 33. Let (𝑋,𝜎) be a metric-like space and let 𝑇∶𝑋→𝑋 be self-mapping. Suppose that 𝑇 satisfies the following inequality: 𝜑𝜎𝜓𝜎(𝜎(𝜎,𝜓𝜎𝜎𝜎(),𝜎𝑇𝑥,𝑇𝑦≤𝜙𝑥,𝑦,𝜓𝑥,𝑇𝑥)),𝜓𝑦,𝑇𝑦𝑥,𝑇𝑦,𝜓𝑦,𝑇𝑥−𝜉max𝑥,𝑦,𝜎𝑥,𝑇𝑥,𝜎𝑦,𝑇𝑦𝑥,𝑇𝑦+𝜎𝑦,𝑇𝑥4, (97) for all 𝑥,𝑦∈𝑋, where 𝜓∈Ψ, 𝜙∈Φ_𝜓, 𝜑∈Θ, and 𝜉∈Ξ. Then there exists a unique fixed point 𝑢∈𝑋 such that 𝑇𝑢=𝑢.

If we take 𝛼(𝑥,𝑦)=1 for all 𝑥,𝑦∈𝑋 in Theorem 31, we get the following fixed point theorem.

Theorem 34. Let (𝑋,𝜎) be a complete metric-like space and let 𝑇∶𝑋→𝑋 be mapping. Suppose that 𝑇 satisfies 𝜎𝑇𝑥,𝑇𝑦≤𝜇𝑀𝑥,𝑦−𝜑𝑀𝑥,𝑦, (98) for all 𝑥,𝑦∈𝑋, where 𝜇∈ℳ, 𝜑∈Θ, and 𝑀𝜎,𝜎𝑥,𝑦=max𝑥,𝑦,𝜎𝑥,𝑓𝑥,𝜎𝑦,𝑓𝑦𝑥,𝑓𝑦+𝜎𝑦,𝑓𝑥4. (99) Then there exists a unique fixed point 𝑢∈𝑋 such that 𝑇𝑢=𝑢.

5.2. Fixed Point Theorems on Metric Spaces Endowed with a Partial Order

In the last decade, the investigation of the existence of fixed point on metric spaces endowed with partial orders has been appreciated by several authors. The initial results in this direction were reported by Turinici [29], Ran and Reurings in [30]. Now, we consider the partially ordered versions of our theorems. For this purpose, we need to recall some concepts.

Definition 35. Let (𝑋,≼) be a partially ordered set and let 𝑇∶𝑋→𝑋 be a given mapping. One says that 𝑇 is nondecreasing with respect to ≼ if 𝑥,𝑦∈𝑋,𝑥≼𝑦⟹𝑇𝑥≼𝑇𝑦. (100)

Definition 36. Let (𝑋,≼) be a partially ordered set. A sequence {𝑥_𝑛}⊂𝑋 is said to be nondecreasing with respect to ≼ if 𝑥_𝑛≼𝑥_𝑛+1 for all 𝑛.

Definition 37. Let (𝑋,≼) be a partially ordered set and let 𝑑 be a metric on 𝑋. One says that (𝑋,≼,𝑑) is regular if, for every nondecreasing sequence {𝑥_𝑛}⊂𝑋 such that 𝑥_𝑛→𝑥∈𝑋 as 𝑛→∞, there exists a subsequence {𝑥_𝑛(𝑘)} of {𝑥_𝑛} such that 𝑥_𝑛(𝑘)≼𝑥 for all 𝑘.

Theorem 38. Let (𝑋,𝜎) be a complete metric-like space and let 𝑇∶𝑋→𝑋 be a mapping. Suppose that for each 𝜂>0 there exists 𝛿>0 such that 𝜂≤𝜙𝜎𝑥,𝑦,𝜎(𝑥,𝑇𝑥),𝜎𝑦,𝑇𝑦,𝜎𝑥,𝑇𝑦,𝜎𝑦,𝑇𝑥<𝜂+𝛿⟶𝜎𝑇𝑥,𝑇𝑦<𝜂, (101) for all 𝑥,𝑦∈𝑋 and 𝜙∈Φ. Then there exists a unique fixed point 𝑢∈𝑋 such that 𝑇𝑢=𝑢.

We have the following result.

Corollary 39. Let (𝑋,≼) be a partially ordered set and let 𝜎 be a metric-like mapping on 𝑋 such that (𝑋,𝜎) is complete metric-like space. Let 𝑇∶𝑋→𝑋 be a nondecreasing mapping with respect to ≼. Suppose that for each 𝜂>0 there exists 𝛿>0 such that 𝜂≤𝜙𝜎𝑥,𝑦,𝜎(𝑥,𝑇𝑥),𝜎𝑦,𝑇𝑦,𝜎𝑥,𝑇𝑦,𝜎𝑦,𝑇𝑥<𝜂+𝛿⟶𝜎𝑇𝑥,𝑇𝑦<𝜂, (102) for all 𝑥,𝑦∈𝑋 with 𝑥≽𝑦 and 𝜙∈Φ. Suppose also that the following conditions hold:(i)there exists 𝑥₀∈𝑋 such that 𝑥₀≼𝑇𝑥₀;(ii)𝑇 is continuous or (𝑋,≼,𝜎) is regular.Then 𝑇 has a fixed point. Moreover, if for all 𝑥,𝑦∈𝑋 there exists 𝑧∈𝑋 such that 𝑥≼𝑧 and 𝑦≼𝑧, one has uniqueness of the fixed point.

Proof. Define the mapping 𝛼∶𝑋×𝑋→[0,∞) by 𝛼=1𝑥,𝑦if𝑥≼𝑦or0𝑥≽𝑦,otherwise. (103) Clearly, for each 𝜂>0, there exists 𝛿>0 such that 𝜎()𝜎𝜂≤𝜙𝑥,𝑦,𝜎𝑥,𝑇𝑥,𝜎𝑦,𝑇𝑦,𝜎𝑥,𝑇𝑦,𝜎𝑦,𝑇𝑥<𝜂+𝛿⟶𝛼𝑥,𝑦𝑇𝑥,𝑇𝑦<𝜂, (104) for all 𝑥,𝑦∈𝑋 and 𝜙∈Φ. From condition (i), we have 𝛼(𝑥₀,𝑇𝑥₀)≥1. Moreover, for all 𝑥,𝑦∈𝑋, from the monotone property of 𝑇, we have 𝛼𝑥,𝑦≥1⟹𝑥≽𝑦or𝑥≼𝑦⟹𝑇𝑥≽𝑇𝑦or𝑇𝑥≼𝑇𝑦⟹𝛼𝑇𝑥,𝑇𝑦≥1. (105) Thus 𝑇 is 𝛼-admissible. Now, if 𝑇 is continuous, the existence of a fixed point follows from Theorem 20. Suppose now that (𝑋,≼,𝜎) is regular. Let {𝑥_𝑛} be a sequence in 𝑋 such that 𝛼(𝑥_𝑛,𝑥_𝑛+1)≥1 for all 𝑛 and 𝑥_𝑛→𝑥∈𝑋 as 𝑛→∞. From the regularity hypothesis, there exists a subsequence {𝑥_𝑛(𝑘)} of {𝑥_𝑛} such that 𝑥_𝑛(𝑘)≼𝑥 for all 𝑘. This implies from the definition of 𝛼 that 𝛼(𝑥_𝑛(𝑘),𝑥)≥1 for all 𝑘. In this case, the existence of a fixed point follows from Theorem 21. To show the uniqueness, let 𝑥,𝑦∈𝑋. By the hypothesis, there exists 𝑧∈𝑋 such that 𝑥≼𝑧 and 𝑦≼𝑧, which implies from the definition of 𝛼 that 𝛼(𝑥,𝑧)≥1 and 𝛼(𝑦,𝑧)≥1. Thus we deduce the uniqueness of the fixed point by Theorem 22.

By using the same argument in the proof of Corollary 39, we can conclude the following two corollaries. We omit the proofs of these corollaries to avoid the repetition.

Corollary 40. Let (𝑋,≼) be a partially ordered set and let 𝜎 be a metric-like mapping on 𝑋 such that (𝑋,𝜎) is complete metric-like space. Let 𝑇∶𝑋→𝑋 be a nondecreasing mapping with respect to ≼. Suppose that 𝑇 satisfies the following inequality: 𝜑𝜎𝜓𝜎(𝜎(𝜎,𝜓𝜎𝜎𝜎(),𝜎𝑇𝑥,𝑇𝑦≤𝜙𝑥,𝑦,𝜓𝑥,𝑇𝑥)),𝜓𝑦,𝑇𝑦𝑥,𝑇𝑦,𝜓𝑦,𝑇𝑥−𝜉max𝑥,𝑦,𝜎𝑥,𝑇𝑥,𝜎𝑦,𝑇𝑦𝑥,𝑇𝑦+𝜎𝑦,𝑇𝑥4, (106) for all 𝑥,𝑦∈𝑋 with 𝑥≽𝑦, where 𝜓∈Ψ, 𝜙∈Φ_𝜓, 𝜑∈Θ, and 𝜉∈Ξ.
Suppose also that the following conditions hold:(i)there exists 𝑥₀∈𝑋 such that 𝑥₀≼𝑇𝑥₀;(ii)𝑇 is continuous or (𝑋,≼,𝜎) is regular.Then 𝑇 has a fixed point. Moreover, if for all 𝑥,𝑦∈𝑋 there exists 𝑧∈𝑋 such that 𝑥≼𝑧 and 𝑦≼𝑧, one has uniqueness of the fixed point.

Corollary 41. Let (𝑋,≼) be a partially ordered set and 𝜎 be a metric-like mapping on 𝑋 such that (𝑋,𝜎) is complete metric-like space. Let 𝑇∶𝑋→𝑋 be a nondecreasing mapping with respect to ≼. Suppose that 𝑇 satisfies 𝜎𝑇𝑥,𝑇𝑦≤𝜇𝑀𝑥,𝑦−𝜑𝑀𝑥,𝑦, (107) for all 𝑥,𝑦∈𝑋 with 𝑥≽𝑦, where 𝜇∈ℳ, 𝜑∈Θ, and 𝑀𝜎,𝜎𝑥,𝑦=max𝑥,𝑦,𝜎𝑥,𝑓𝑥,𝜎𝑦,𝑓𝑦𝑥,𝑓𝑦+𝜎𝑦,𝑓𝑥4. (108)
Suppose also that the following conditions hold:(i)there exists 𝑥₀∈𝑋 such that 𝑥₀≼𝑇𝑥₀;(ii)𝑇 is continuous or (𝑋,≼,𝜎) is regular.Then 𝑇 has a fixed point. Moreover, if for all 𝑥,𝑦∈𝑋 there exists 𝑧∈𝑋 such that 𝑥≼𝑧 and 𝑦≼𝑧, one has uniqueness of the fixed point.

5.3. Fixed Point Theorems for Cyclic Contractive Mappings

In this subsection, we consider the cyclic contraction and related fixed point as a consequence of our main results. Notice that this trend was initiated by Kirk et al. [31]. Following this paper [31], a number of fixed point theorems for cyclic contractive mappings have been reported (see, e.g., [32–37]).

We have the following result.

Corollary 42. Let {𝐴_𝑖}²_𝑖=1 be nonempty closed subsets of a complete metric-like space (𝑋,𝜎) and let 𝑇∶𝑌→𝑌 be a given mapping such that(I)𝑇(𝐴₁)⊆𝐴₂ and 𝑇(𝐴₂)⊆𝐴₁,where 𝑌=𝐴₁∪𝐴₂. Suppose that for each 𝜂>0 there exists 𝛿>0 such that 𝜂≤𝜙𝜎𝑥,𝑦,𝜎(𝑥,𝑇𝑥),𝜎𝑦,𝑇𝑦,𝜎𝑥,𝑇𝑦,𝜎𝑦,𝑇𝑥<𝜂+𝛿⟶𝜎𝑇𝑥,𝑇𝑦<𝜂, (109) for all (𝑥,𝑦)∈𝐴₁×𝐴₂ and 𝜙∈Φ. Then 𝑇 has a unique fixed point that belongs to 𝐴₁∩𝐴₂.

Proof. Since 𝐴₁ and 𝐴₂ are closed subsets of the complete metric space (𝑋,𝑑), then (𝑌,𝑑) is complete. Define the mapping 𝛼∶𝑌×𝑌→[0,∞) by 𝛼=1𝑥,𝑦if𝑥,𝑦∈𝐴₁×𝐴₂∪𝐴₂×𝐴₁,0otherwise. (110) From (111) and the definition of 𝛼, for each 𝜂>0, there exists 𝛿>0 such that 𝜎()𝜎𝜂≤𝜙𝑥,𝑦,𝜎𝑥,𝑇𝑥,𝜎𝑦,𝑇𝑦,𝜎𝑥,𝑇𝑦,𝜎𝑦,𝑇𝑥<𝜂+𝛿⟶𝛼𝑥,𝑦𝑇𝑥,𝑇𝑦<𝜂, (111) for all (𝑥,𝑦)∈𝐴₁×𝐴₂ and 𝜙∈Φ. Thus 𝑇 satisfies the contractive condition (104).
Let (𝑥,𝑦)∈𝑌×𝑌 such that 𝛼(𝑥,𝑦)≥1. If (𝑥,𝑦)∈𝐴₁×𝐴₂, from (I), (𝑇𝑥,𝑇𝑦)∈𝐴₂×𝐴₁, which implies that 𝛼(𝑇𝑥,𝑇𝑦)≥1. If (𝑥,𝑦)∈𝐴₂×𝐴₁, from (I), (𝑇𝑥,𝑇𝑦)∈𝐴₁×𝐴₂, which implies that 𝛼(𝑇𝑥,𝑇𝑦)≥1. Thus, in all cases, we have 𝛼(𝑇𝑥,𝑇𝑦)≥1. This implies that 𝑇 is 𝛼-admissible.
Also, from (I), for any 𝑎∈𝐴₁, we have (𝑎,𝑇𝑎)∈𝐴₁×𝐴₂, which implies that 𝛼(𝑎,𝑇𝑎)≥1.
Now, let {𝑥_𝑛} be a sequence in 𝑋 such that 𝛼(𝑥_𝑛,𝑥_𝑛+1)≥1 for all 𝑛 and 𝑥_𝑛→𝑥∈𝑋 as 𝑛→∞. This implies from the definition of 𝛼 that 𝑥_𝑛,𝑥_𝑛+1∈𝐴₁×𝐴₂∪𝐴₂×𝐴₁,∀𝑛. (112) Since (𝐴₁×𝐴₂)∪(𝐴₂×𝐴₁) is a closed set with respect to the Euclidean metric, we get that (𝑥,𝑥)∈𝐴₁×𝐴₂∪𝐴₂×𝐴₁, (113) which implies that 𝑥∈𝐴₁∩𝐴₂. Thus we get immediately from the definition of 𝛼 that 𝛼(𝑥_𝑛,𝑥)≥1 for all 𝑛.
Let 𝑥,𝑦∈𝑋 be distinct fixed points of 𝑇 from (I); this implies that 𝑥,𝑦∈𝐴₁∩𝐴₂. So, for any 𝑧∈𝑌, we have 𝛼(𝑥,𝑧)≥1 and 𝛼(𝑦,𝑧)≥1. Thus condition (𝑈) is satisfied.
Now, all the hypotheses of Theorem 22 are satisfied; we deduce that 𝑇 has a unique fixed point that belongs to 𝐴₁∩𝐴₂ (from (I)).

As in the previous section, we can conclude the following two corollaries by using the same argument in the proof of Corollary 42. We omit the proofs of the following corollaries to avoid the repetition.

Corollary 43. Let {𝐴_𝑖}²_𝑖=1 be nonempty closed subsets of a complete metric-like space (𝑋,𝜎) and let 𝑇∶𝑌→𝑌 be a given mapping such that(I)𝑇(𝐴₁)⊆𝐴₂ and 𝑇(𝐴₂)⊆𝐴₁, where 𝑌=𝐴₁∪𝐴₂. Suppose that 𝑇 satisfies the following inequality: 𝜑𝜎𝜓𝜎(𝜎(𝜎,𝜓𝜎𝜎𝜎(),𝜎𝑇𝑥,𝑇𝑦≤𝜙𝑥,𝑦,𝜓𝑥,𝑇𝑥)),𝜓𝑦,𝑇𝑦𝑥,𝑇𝑦,𝜓𝑦,𝑇𝑥−𝜉max𝑥,𝑦,𝜎𝑥,𝑇𝑥,𝜎𝑦,𝑇𝑦𝑥,𝑇𝑦+𝜎𝑦,𝑇𝑥4, (114)for all (𝑥,𝑦)∈𝐴₁×𝐴₂, where 𝜓∈Ψ, 𝜙∈Φ_𝜓, 𝜑∈Θ, and 𝜉∈Ξ. Then 𝑇 has a unique fixed point that belongs to 𝐴₁∩𝐴₂.

Corollary 44. Let {𝐴_𝑖}²_𝑖=1 be nonempty closed subsets of a complete metric-like space (𝑋,𝜎) and let 𝑇∶𝑌→𝑌 be a given mapping such that(I)𝑇(𝐴₁)⊆𝐴₂ and 𝑇(𝐴₂)⊆𝐴₁, where 𝑌=𝐴₁∪𝐴₂. Suppose that 𝑇 satisfies 𝜎𝑇𝑥,𝑇𝑦≤𝜇𝑀𝑥,𝑦−𝜑𝑀𝑥,𝑦, (115) for all (𝑥,𝑦)∈𝐴₁×𝐴₂, where 𝜇∈ℳ, 𝜑∈Θ, and 𝑀𝜎,𝜎𝑥,𝑦=max𝑥,𝑦,𝜎𝑥,𝑓𝑥,𝜎𝑦,𝑓𝑦𝑥,𝑓𝑦+𝜎𝑦,𝑓𝑥4. (116)Then 𝑇 has a unique fixed point that belongs to 𝐴₁∩𝐴₂.

Conflict of Interests

The authors declare that they have no conflict of interests.

Authors’ Contribution

All authors contributed equally and significantly to writing this paper. All authors read and approved the final paper.

Acknowledgments

This research was supported by Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah, Saudi Arabia. The authors give thanks to anonymous referees for their remarkable comments, suggestion, and ideas that help to improve this paper.