Fixed Point Results of Locally Contractive Mappings in Ordered Quasi-Partial Metric Spaces

Abstract

Fixed point results for a self-map satisfying locally contractive conditions on a closed ball in an ordered 0-complete quasi-partial metric space have been established. Instead of monotone mapping, the notion of dominated mappings is applied. We have used weaker metric, weaker contractive conditions, and weaker restrictions to obtain unique fixed points. An example is given which shows that how this result can be used when the corresponding results cannot. Our results generalize, extend, and improve several well-known conventional results.

1. Introduction and Preliminaries

Fixed points results of mappings satisfying certain contractive conditions on the entire domain have been at the centre of vigorous research activity, (see [1–3]) and it has a wide range of applications in different areas such as nonlinear and adaptive control systems, parameterize estimation problems, fractal image decoding, computing magnetostatic fields in a nonlinear medium, and convergence of recurrent networks (see [4–6]).

Recently, many results appeared related to fixed point theorem in complete metric spaces endowed with a partial ordering in the literature. Ran and Reurings [7] proved an analogue of Banach's fixed point theorem in metric space endowed with a partial order and gave applications to matrix equations. In this way, they weakened the usual contraction condition. Subsequently, Nieto and Rodríguez-López [8] extended the result in [7] for nondecreasing mappings and applied it to obtain a unique solution for a 1st-order ordinary differential equation with periodic boundary conditions. Thereafter, many works related to fixed point problems have also been considered in partially ordered metric spaces (see [7–11]).

On the other hand notion of a partial metric space was introduced by Matthews in [12]. In partial metric spaces, the distance of a point from itself may not be zero. Partial metric spaces have applications in theoretical computer science (see [13]). Altun and Erduran [14] and Paesano and Vetro [15] used the idea of partial metric space and partial order and gave some fixed point theorems for contractive condition on ordered partial metric spaces. Recently, Karapinar et al. [16] introduced the concept of quasi-partial metric space. Romaguera [17] has given the idea of 0-complete partial metric space. Nashine et al. [18] used this concept and proved some classical results.

From the application point of view the situation is not yet completely satisfactory because it frequently happens that a mapping 𝑇 is a contraction not on the entire space 𝑋 but merely on a subset 𝑌 of 𝑋. However, if 𝑌 is closed and a sequence {𝑥_𝑛} in 𝑋 converges to some 𝑥 in 𝑋, then by imposing a subtle restriction on the choice of 𝑥₀, one may force the sequence to stay eventually in 𝑌. In this case, one can establish the existence of a fixed point of 𝑇. Arshad et al. [19] proved a significant result concerning the existence of fixed points of a mapping satisfying a contractive conditions on closed ball in a complete dislocated metric space. Other results on closed ball can be seen in [20, 21]. In this paper we have obtained fixed point theorems for dominated self-mappings in a 0-complete ordered quasi-partial metric space on closed ball under several contractive conditions to generalize, extend and improve some classical fixed point results. We have used weaker contractive condition and weaker restrictions to obtain unique fixed point. Our results do not exist even yet in metric spaces. An example shows how this result can be used when the corresponding results cannot.

Consistent with [16, 22, 23] the following definitions and results will be needed in the sequel.

Definition 1 (see [16]). A quasi-partial metric is a function 𝑞∶𝑋×𝑋→𝑅⁺ satisfying the following:(i)if 0≤𝑞(𝑥,𝑥)=𝑞(𝑥,𝑦)=𝑞(𝑦,𝑦), then 𝑥=𝑦 (equality);(ii)𝑞(𝑥,𝑥)≤𝑞(𝑦,𝑥) (small self-distances);(iii)𝑞(𝑥,𝑥)≤𝑞(𝑥,𝑦) (small self-distances);(iv)𝑞(𝑥,𝑧)+𝑞(𝑦,𝑦)≤𝑞(𝑥,𝑦)+𝑞(𝑦,𝑧) (triangle inequality), for all 𝑥,𝑦,𝑧∈𝑋.The pair (𝑋,𝑞) is called a quasi-partial metric space.
Note that if 𝑞(𝑥,𝑦)=𝑞(𝑦,𝑥) for all 𝑥,𝑦∈𝑋, then (𝑋,𝑞) becomes a partial metric space (𝑋,𝑝). Moreover if 𝑞(𝑥,𝑥)=0 for all 𝑥∈𝑋, then (𝑋,𝑞) and (𝑋,𝑝) become a quasimetric space and a metric space, respectively. Also 𝑝_𝑞(𝑥,𝑦)=(1/2)[𝑞(𝑥,𝑦)+𝑞(𝑦,𝑥)],𝑥,𝑦∈𝑋 is a partial metric on 𝑋. The function 𝑑_𝑝𝑞∶𝑋×𝑋→𝑅⁺ defined by 𝑑_𝑝𝑞(𝑥,𝑦)=𝑞(𝑥,𝑦)+𝑞(𝑦,𝑥)−𝑞(𝑥,𝑥)−𝑞(𝑦,𝑦) is a (usual) metric on 𝑋. The ball 𝐵(𝑥,𝜀), where 𝐵(𝑥,𝜀)={𝑦∈𝑋∶𝑞(𝑥,𝑦)≤𝜀+𝑞(𝑥,𝑥)}, is a closed ball in quasi-partial metric space, for some 𝑥∈𝑋 and 𝜀>0.

Definition 2 (see [16]). Let (𝑋,𝑞) be a quasi-partial metric. Then, we have the following; (a)A sequence {𝑥_𝑛} in (𝑋,𝑞) converges to a point 𝑥∈𝑋 if and only if lim_𝑛→∞𝑞(𝑥,𝑥_𝑛)=lim_𝑛→∞𝑞(𝑥_𝑛,𝑥)=𝑞(𝑥,𝑥).(b)A sequence {𝑥_𝑛} in (𝑋,𝑞) is called a Cauchy sequence if the limits lim_{𝑛,𝑚→∞}𝑞(𝑥_𝑛,𝑥_𝑚) and lim_{𝑛,𝑚→∞}𝑞(𝑥_𝑚,𝑥_𝑛) exist (and are finite).(c)The space (𝑋,𝑞) is said to be complete if every Cauchy sequence {𝑥_𝑛} in (𝑋,𝑞) converges to a point 𝑥∈𝑋 such that (𝑥,𝑥)=lim_{𝑛,𝑚→∞}𝑞(𝑥_𝑛,𝑥_𝑚)=lim_{𝑛,𝑚→∞}𝑞(𝑥_𝑚,𝑥_𝑛).

Lemma 3 (see [16]). Let (𝑋,𝑞) be a quasi-partial metric space, let (𝑋,𝑝_𝑞) be the corresponding partial metric space, and let (𝑋,𝑑_𝑝𝑞) be the corresponding metric space. These statements are equivalent. (i) The sequence {𝑥_𝑛} is Cauchy in (𝑋,𝑞). (ii) The sequence {𝑥_𝑛} is Cauchy in (𝑋,𝑝_𝑞). (iii) The sequence {𝑥_𝑛} is Cauchy in (𝑋,𝑑_𝑝𝑞). These statements are also equivalent. (i) (𝑋,𝑞) is complete. (ii) (𝑋,𝑝_𝑞) is complete. (iii) (𝑋,𝑑_𝑝𝑞) is complete.

Definition 4. Let (𝑋,⪯) be a partial ordered set. Then 𝑥,𝑦∈𝑋 are called comparable if 𝑥⪯𝑦 or 𝑦⪯𝑥 holds.

Definition 5 (see [22]). Let (𝑋,⪯) be a partially ordered set. A self-mapping 𝑓 on 𝑋 is called dominated if 𝑓𝑥⪯𝑥 for each 𝑥 in 𝑋.

Example 6 (see [22]). Let 𝑋=[0,1] be endowed with the usual ordering and 𝑓∶𝑋→𝑋 defined by 𝑓𝑥=𝑥^𝑛 for some 𝑛∈ℕ. Since 𝑓𝑥=𝑥^𝑛≤𝑥 for all 𝑥∈𝑋, therefore 𝑓 is a dominated map.

Theorem 7 (see [23]). Let (𝑋,𝑑) be a complete metric space, 𝑆∶𝑋→𝑋 a mapping, 𝑟>0,and 𝑥₀ an arbitrary point in 𝑋. Suppose there exists 𝑘∈[0,1) with 𝑑𝑆𝑥,𝑆𝑦≤𝑘𝑑𝑥,𝑦,𝑓𝑜𝑟𝑎𝑙𝑙𝑥,𝑦∈𝑌=𝐵𝑥₀,𝑟 (1) and 𝑑(𝑥₀,𝑆𝑥₀)<(1−𝑘)𝑟. Then there exists a unique point 𝑥^∗ in 𝐵(𝑥₀,𝑟) such that 𝑥^∗=𝑆𝑥^∗.

2. Fixed Points of Dominated Mapping

In this section we introduce the notion of a 0-complete quasi-partial metric spaces. We also prove some results in these spaces.

Definition 8. Let (𝑋,𝑞) be a quasi-partial metric space.(a)A sequence {𝑥_𝑛} in (𝑋,𝑞) is called 0-Cauchy if lim_{𝑛,𝑚→∞}𝑞(𝑥_𝑛,𝑥_𝑚)=0 or lim_{𝑛,𝑚→∞}𝑞(𝑥_𝑚,𝑥_𝑛)=0.(b)The space (𝑋,𝑞) is called 0-complete if every 0-Cauchy sequence in 𝑋 converges to a point 𝑥∈𝑋 such that 𝑞(𝑥,𝑥)=0.
It is easy to see that every 0-Cauchy sequence in (𝑋,𝑞) is Cauchy in (𝑋,𝑑_𝑝𝑞) and if (𝑋,𝑞) is complete, then it is 0-complete but the converse assertions do not hold. For example, the space 𝑋=[0,+∞)∩𝑄 with 𝑞(𝑥,𝑦)=|𝑥−𝑦|+|𝑥| is a 0-complete quasi-partial metric space but it is not complete (since 𝑑_𝑝𝑞(𝑥,𝑦)=2|𝑥−𝑦| and (𝑋,𝑑_𝑝𝑞) is not complete).
Moreover, if (𝑋,𝑑) is complete quasimetric space which is also quasi-partial, then it is 0-complete, quasi-partial metric space.

Definition 9. Let 𝑋 be a nonempty set. Then (𝑋,⪯,𝑞) is called an ordered quasi-partial metric space if (i) 𝑞 is a quasi-partial metric on 𝑋 and (ii) ⪯ is a partial order on 𝑋.

Theorem 10. Let (𝑋,⪯,𝑞) be a 0-complete ordered quasi-partial metric space, 𝑆∶𝑋→𝑋 a dominated map, and 𝑥₀ an arbitrary point in 𝑋. Suppose that there exists 𝑘∈[0,1) such that 𝑞𝑆𝑥,𝑆𝑦≤𝑘𝑞𝑥,𝑦𝑓𝑜𝑟𝑎𝑙𝑙𝑐𝑜𝑚𝑝𝑎𝑟𝑎𝑏𝑙𝑒𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠𝑥,𝑦𝑖𝑛𝐵𝑥₀,𝑟. (2) And 𝑞𝑥₀,𝑆𝑥₀≤(1−𝑘)𝑟+𝑞𝑥₀,𝑥₀. (3) If, for a nonincreasing sequence {𝑥_𝑛} in 𝐵(𝑥₀,𝑟),{𝑥_𝑛}→𝑢 implies that 𝑢⪯𝑥_𝑛, then there exists a point 𝑥^∗ in 𝐵(𝑥₀,𝑟) such that 𝑥^∗=𝑆𝑥^∗ and 𝑞(𝑥^∗,𝑥^∗)=0. Moreover, 𝑥^∗ is unique, if for every pair of elements 𝑥, 𝑦 in 𝑋there exists a point 𝑧∈𝑋 such that 𝑧⪯𝑥 and 𝑧⪯𝑦.

Proof. Consider a Picard sequence 𝑥_𝑛+1=𝑆𝑥_𝑛 with initial guess 𝑥₀, as 𝑥_𝑛+1=𝑆𝑥_𝑛⪯𝑥_𝑛 for all 𝑛∈{0}∪𝑁. We will prove that 𝑥_𝑛∈𝐵(𝑥₀,𝑟) for all 𝑛∈𝑁 by mathematical induction. By using inequality (3), we have 𝑞𝑥₀,𝑥₁≤(1−𝑘)𝑟+𝑞𝑥₀,𝑥₀≤𝑟+𝑞𝑥₀,𝑥₀. (4) Therefore 𝑥₁∈𝐵(𝑥₀,𝑟). Now let 𝑥₂,…,𝑥_𝑗∈𝐵(𝑥₀,𝑟) for some 𝑗∈𝑁. As 𝑥_𝑛+1⪯𝑥_𝑛, so using inequality (2), we obtain 𝑞𝑥_𝑗,𝑥_𝑗+1=𝑞𝑆𝑥_𝑗−1,𝑆𝑥_𝑗≤𝑘𝑞𝑥_𝑗−1,𝑥_𝑗, (5) which implies that, 𝑞𝑥_𝑗,𝑥_𝑗+1≤𝑘²𝑞𝑥_𝑗−2,𝑥_𝑗−1≤⋯≤𝑘^𝑗𝑞𝑥₀,𝑥₁. (6) Now 𝑞𝑥₀,𝑥_𝑗+1≤𝑞𝑥₀,𝑥₁+⋯+𝑞𝑥_𝑗,𝑥_𝑗+1−𝑞𝑥₁,𝑥₁+⋯+𝑞𝑥_𝑗,𝑥_𝑗≤𝑞𝑥₀,𝑥₁1+⋯+𝑘^𝑗−1+𝑘^𝑗,𝑞𝑥byinequality(6)₀,𝑥_𝑗+1≤(1−𝑘)𝑟+𝑞𝑥₀,𝑥₀1−𝑘^𝑗+1.1−𝑘byinequality(3) (7) Thus 𝑥_𝑗+1∈𝐵(𝑥₀,𝑟). Hence 𝑥_𝑛∈𝐵(𝑥₀,𝑟) for all 𝑛∈𝑁. Also 𝑥_𝑛+1⪯𝑥_𝑛 for all 𝑛∈𝑁. It implies that 𝑞𝑥_𝑛,𝑥_𝑛+1≤𝑘^𝑛𝑞𝑥₀,𝑥₁forall𝑛∈𝑁. (8) It follows that 𝑞𝑥_𝑛,𝑥_𝑛+𝑖≤𝑞𝑥_𝑛,𝑥_𝑛+1+⋯+𝑞𝑥_{𝑛+𝑖−1},𝑥_𝑛+𝑖≤𝑘^𝑛𝑞𝑥₀,𝑥₁1+⋯+𝑘^𝑖−2+𝑘^𝑖−1⟶0as𝑛⟶∞. (9) Notice that the sequence {𝑥_𝑛} is a 0-Cauchy sequence in (𝐵(𝑥₀,𝑟),𝑞). As 𝐵(𝑥₀,𝑟) is closed so is 0-complete. Therefore there exists a point 𝑥^∗∈𝐵(𝑥₀,𝑟) with 𝑞𝑥^∗,𝑥^∗=lim_𝑛→∞𝑞𝑥_𝑛,𝑥^∗=lim_𝑛→∞𝑞𝑥^∗,𝑥_𝑛=0. (10) Now, 𝑞𝑥^∗,𝑆𝑥^∗≤𝑞𝑥^∗,𝑥_𝑛+𝑞𝑆𝑥_𝑛−1,𝑆𝑥^∗−𝑞𝑥_𝑛,𝑥_𝑛. (11) On taking limit as 𝑛→∞ and using the fact that 𝑥^∗⪯𝑥_𝑛⪯𝑥_𝑛−1, when 𝑥_𝑛→𝑥^∗, we have, 𝑞𝑥^∗,𝑆𝑥^∗≤lim_𝑛→∞𝑞𝑥^∗,𝑥_𝑛+𝑘𝑞𝑥_𝑛−1,𝑥^∗. (12) Then by inequality (10), we have 𝑞𝑥^∗,𝑆𝑥^∗≤0. (13) Similarly, 𝑞𝑆𝑥^∗,𝑥^∗≤0, (14) and hence 𝑥^∗=𝑆𝑥^∗.

Uniqueness. Let 𝑦 be another point in 𝐵(𝑥₀,𝑟) such that 𝑦=𝑆𝑦. If 𝑥^∗ and 𝑦 are comparable, then 𝑞𝑥^∗,𝑦=𝑞𝑆𝑥^∗,𝑆𝑦≤𝑘𝑞𝑥^∗,𝑦. (15) This shows that 𝑥^∗=𝑦. Now if 𝑥^∗ and 𝑦 are not comparable then there exists a point 𝑧∈𝐵(𝑥₀,𝑟) which is lower bound of both 𝑥^∗ and 𝑦, that is, 𝑧⪯𝑥^∗ and 𝑧⪯𝑦. Moreover by assumption 𝑧⪯𝑥^∗⪯𝑥_𝑛⋯⪯𝑥₀. Now we will prove that 𝑆^𝑛𝑧∈𝐵(𝑥₀,𝑟): 𝑞𝑥₀,𝑆𝑧≤𝑞𝑥₀,𝑥₁+𝑞𝑥₁,𝑆𝑧−𝑞𝑥₁,𝑥₁≤(1−𝑘)𝑟+𝑞𝑥₀,𝑥₀+𝑘𝑞𝑥₀,𝑞𝑥,𝑧byinequality(2)andinequality(3)₀,𝑆𝑧≤(1−𝑘)𝑟+(1−𝑘)𝑞𝑥₀,𝑥₀+𝑘𝑟+𝑞𝑥₀,𝑥₀=𝑟−𝑟𝑘+𝑞𝑥₀,𝑥₀−𝑘𝑞𝑥₀,𝑥₀𝑥+𝑟𝑘+𝑘𝑞₀,𝑥₀=𝑟+𝑞𝑥₀,𝑥₀. (16) It follows that 𝑆𝑧∈𝐵(𝑥₀,𝑟). Now let 𝑆^𝑗𝑧∈𝐵(𝑥₀,𝑟)𝑞𝑥₀,𝑆^𝑗+1𝑧≤𝑞𝑥₀,𝑥₁+𝑞𝑥₁,𝑆^𝑗+1𝑧−𝑞𝑥₁,𝑥₁≤(1−𝑘)𝑟+𝑞𝑥₀,𝑥₀+𝑘𝑞𝑥₀,𝑆^𝑗𝑧,𝑞𝑥byinequality(2)andinequality(3)₀,𝑆^𝑗+1𝑧≤(1−𝑘)𝑟+(1−𝑘)𝑞𝑥₀,𝑥₀+𝑘𝑟+𝑞𝑥₀,𝑥₀≤𝑟−𝑟𝑘+𝑞𝑥₀,𝑥₀−𝑘𝑞𝑥₀,𝑥₀+𝑟𝑘+𝑘𝑞𝑥₀,𝑥₀≤𝑟+𝑞𝑥₀,𝑥₀. (17) It follows that 𝑆^𝑗+1𝑧∈𝐵(𝑥₀,𝑟). Hence 𝑆^𝑛𝑧∈𝐵(𝑥₀,𝑟) for all 𝑛∈𝑁. Nowas 𝑆 is dominated, it follows that 𝑆^𝑛−1𝑧⪯𝑆^𝑛−2𝑧⪯⋯⪯𝑧⪯𝑥^∗ and 𝑆^𝑛−1𝑧⪯𝑦 for all 𝑛∈𝑁,which further implies 𝑆^𝑛−1𝑧⪯𝑆^𝑛𝑥^∗ and 𝑆^𝑛−1𝑧⪯𝑆^𝑛𝑦 for all 𝑛∈𝑁 as 𝑆^𝑛𝑥^∗=𝑥^∗ and 𝑆^𝑛𝑦=𝑦 for all 𝑛∈𝑁: 𝑞𝑥^∗,𝑦=𝑞𝑆^𝑛𝑥^∗,𝑆^𝑛𝑦≤𝑞𝑆^𝑛𝑥^∗,𝑆^𝑛−1𝑧+𝑞𝑆^𝑛−1𝑧,𝑆^𝑛𝑦−𝑞𝑆^𝑛−1𝑧,𝑆^𝑛−1𝑧≤𝑘𝑞𝑆^𝑛−1𝑥^∗,𝑆^𝑛−2𝑧+𝑘𝑞𝑆^𝑛−2𝑧,𝑆^𝑛−1𝑦⋮∵byinequality(2)≤𝑘^𝑛−2𝑞𝑥^∗,𝑆𝑧+𝑘^𝑛−2𝑞𝑆𝑧,𝑦⟶0as𝑛⟶∞. (18) Hence 𝑥^∗=𝑦.

Example 11. Let 𝑋=[0,+∞)∩𝑄 be endowed with order 𝑥⪯𝑦 if 𝑞(𝑥,𝑥)≤𝑞(𝑦,𝑦) and let 𝑞∶𝑋×𝑋→𝑅⁺ be the 0-complete ordered quasi-partial metric on 𝑋 defined by 𝑞(𝑥,𝑦)=max{𝑦−𝑥,0}+𝑥. Define 1𝑆𝑥=[]410𝑥if𝑥∈0,1∩𝑋𝑥−9if𝑥∈(1,∞)∩𝑋. (19) Clearly, 𝑆 is dominated mapping. Take 𝑥₀=1/2,𝑟=1/2; then 𝐵(𝑥₀,𝑟)=[0,1]∩𝑋; we have 𝑞(𝑥₀,𝑥₀)=1/2,𝑘=4/9 with (1−𝑘)𝑟+𝑞𝑥₀,𝑥₀=41−912+12=59,𝑞𝑥₀,𝑆𝑥₀1=𝑞21,𝑆21=𝑞2,1=1202<59. (20) Also if 𝑥,𝑦∈(1,∞)∩𝑋, then, 𝑞4𝑆𝑥,𝑆𝑦=max𝑦−94−𝑥+94,0+𝑥−94=max𝑦−𝑥,0+𝑥−9≥49max𝑦−𝑥,0+𝑥=𝑘𝑞𝑥,𝑦. (21) So the contractive condition does not hold on 𝑋 in each case.

Now if 𝑥,𝑦∈𝐵(𝑥₀,𝑟)∩𝑋, (22) then, 𝑞1𝑆𝑥,𝑆𝑦=max110𝑦−+110𝑥,0𝑥=110𝑞<410𝑥,𝑦9𝑞𝑥,𝑦. (23) Therefore, all the conditions of Theorem 13 are satisfied. Moreover, 0 is the unique fixed point of 𝑆 and 𝑞(0,0)=0.

In Theorem 10, the condition (3) is imposed to restrict the condition (2) only for 𝑥, 𝑦 in 𝐵(𝑥₀,𝑟) and Example 11 explains the utility of this restriction. However, the following result relaxes the condition (3) but imposes the condition (2) for all comparable elements in the whole space 𝑋.

Theorem 12. Let (𝑋,⪯,𝑞) be a 0-complete ordered quasi-partial metric space, 𝑆∶𝑋→𝑋 a dominated map, and 𝑥₀ an arbitrary point in 𝑋. Suppose there exists 𝑘∈[0,1) with 𝑞𝑆𝑥,𝑆𝑦≤𝑘𝑞𝑥,𝑦,𝑓𝑜𝑟𝑎𝑙𝑙𝑐𝑜𝑚𝑝𝑎𝑟𝑎𝑏𝑙𝑒𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠𝑥,𝑦𝑖𝑛𝑋. (24) If, for a nonincreasing sequence {𝑥_𝑛} in 𝑋,{𝑥_𝑛}→𝑢 implies that 𝑢⪯𝑥_𝑛, then there exists a point 𝑥^∗ in 𝑋 such that 𝑥^∗=𝑆𝑥^∗ and 𝑞(𝑥^∗,𝑥^∗)=0. Moreover, 𝑥^∗ is unique, if for every pair of elements 𝑥, 𝑦 in 𝑋, there exists a point 𝑧∈𝑋 such that 𝑧⪯𝑥 and 𝑧⪯𝑦.

In Theorem 10, the existence of a lower bound and for a nonincreasing sequence {𝑥_𝑛} in 𝑋,{𝑥_𝑛}→𝑢 implies that 𝑢⪯𝑥_𝑛 are imposed to restrict the condition (2) only for comparable elements. However, the following result relaxes these conditions but imposes the condition (2) for all elements in 𝐵(𝑥₀,𝑟).

Theorem 13. Let (𝑋,𝑞) be a 0-complete partial metric space, 𝑆∶𝑋→𝑋 a map, and 𝑥₀ an arbitrary point in 𝑋. Suppose there exists 𝑘∈[0,1) with 𝑞𝑆𝑥,𝑆𝑦≤𝑘𝑞𝑥,𝑦,𝑓𝑜𝑟𝑎𝑙𝑙𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠𝑥,𝑦𝑖𝑛𝐵𝑥₀,𝑞𝑥,𝑟₀,𝑆𝑥₀≤(1−𝑘)𝑟+𝑞𝑥₀,𝑥₀. (25) Then there exists a unique point 𝑥^∗ in 𝐵(𝑥₀,𝑟) such that 𝑥^∗=𝑆𝑥^∗. Further 𝑞(𝑥^∗,𝑥^∗)=0.

Metric version of Theorem 10 is given below.

Theorem 14. Let (𝑋,⪯,𝑑) be a complete ordered metric space, 𝑆∶𝑋→𝑋 a dominated map, and 𝑥₀ an arbitrary point in 𝑋. Suppose that there exists 𝑘∈[0,1) such that 𝑑𝑆𝑥,𝑆𝑦≤𝑘𝑑𝑥,𝑦𝑓𝑜𝑟𝑎𝑙𝑙𝑐𝑜𝑚𝑝𝑎𝑟𝑎𝑏𝑙𝑒𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠𝑥,𝑦𝑖𝑛𝐵𝑥₀,𝑑𝑥,𝑟₀,𝑆𝑥₀≤(1−𝑘)𝑟. (26) If, for a nonincreasing sequence {𝑥_𝑛} in 𝐵(𝑥₀,𝑟),{𝑥_𝑛}→𝑢 implies that 𝑢⪯𝑥_𝑛, then there exists a point 𝑥^∗ in 𝐵(𝑥₀,𝑟) such that 𝑥^∗=𝑆𝑥^∗ and 𝑞(𝑥^∗,𝑥^∗)=0. Moreover, 𝑥^∗ is unique, if for every pair of elements 𝑥, 𝑦 in 𝑋 there exists a point 𝑧∈𝑋 such that 𝑧⪯𝑥 and 𝑧⪯𝑦.

Theorem 15. Let (𝑋,⪯,𝑞) be a 0-complete ordered quasi-partial metric space, 𝑆∶𝑋→𝑋 a dominated map, and 𝑥₀ an arbitrary point in 𝑋. Suppose that there exists 𝑏∈[0,1/2) such that 𝑞𝑆𝑥,𝑆𝑦≤𝑏𝑞(𝑥,𝑆𝑥)+𝑞𝑦,𝑆𝑦 (27) for all comparable elements 𝑥, 𝑦 in 𝐵(𝑥₀,𝑟).
And 𝑞𝑥₀,𝑆𝑥₀≤(1−𝑘)𝑟+𝑞𝑥₀,𝑥₀, (28) where 𝑘=𝑏/(1−𝑏). If, for a nonincreasing sequence {𝑥_𝑛} in 𝐵(𝑥₀,𝑟),{𝑥_𝑛}→𝑢 implies that 𝑢⪯𝑥_𝑛, then there exists a point 𝑥^∗ in 𝐵(𝑥₀,𝑟) such that 𝑥^∗=𝑆𝑥^∗ and 𝑞(𝑥^∗,𝑥^∗)=0. Moreover, 𝑥^∗ is unique, if for any two points 𝑥,𝑦 in 𝐵(𝑥₀,𝑟) there exists a point 𝑧∈𝐵(𝑥₀,𝑟) such that 𝑧⪯𝑥 and 𝑧⪯𝑦, and 𝑞𝑥₀,𝑆𝑥₀+𝑞(𝑧,𝑆𝑧)≤𝑞𝑥₀,𝑧+𝑞𝑆𝑥₀,𝑆𝑧𝑓𝑜𝑟𝑎𝑙𝑙𝑧⪯𝑆𝑥₀. (29)

Proof. Consider a Picard sequence 𝑥_𝑛+1=𝑆𝑥_𝑛 with initial guess 𝑥₀, as 𝑥_𝑛+1=𝑆𝑥_𝑛⪯𝑥_𝑛 for all 𝑛∈{0}∪𝑁. We will prove that 𝑥_𝑛∈𝐵(𝑥₀,𝑟) for all 𝑛∈𝑁 by mathematical induction. By using inequality (28), we have 𝑞𝑥₀,𝑥₁≤(1−𝑘)𝑟+𝑞𝑥₀,𝑥₀≤𝑟+𝑞𝑥₀,𝑥₀. (30) Therefore 𝑥₁∈𝐵(𝑥₀,𝑟). Now let 𝑥₂,…,𝑥_𝑗∈𝐵(𝑥₀,𝑟) for some 𝑗∈𝑁. As 𝑥_𝑗+1⪯𝑥_𝑗, so using inequality (27), we obtain 𝑞𝑥_𝑗,𝑥_𝑗+1=𝑞𝑆𝑥_𝑗−1,𝑆𝑥_𝑗≤𝑏𝑞𝑥_𝑗−1,𝑥_𝑗+𝑞𝑥_𝑗,𝑥_𝑗+1≤𝑘𝑞𝑥_𝑗−1,𝑥_𝑗, (31) which implies that 𝑞𝑥_𝑗,𝑥_𝑗+1≤𝑘²𝑞𝑥_𝑗−2,𝑥_𝑗−1≤⋯≤𝑘^𝑗𝑞𝑥₀,𝑥₁. (32) Now, 𝑞𝑥₀,𝑥_𝑗+1≤𝑞𝑥₀,𝑥₁+⋯+𝑞𝑥_𝑗,𝑥_𝑗+1−𝑞𝑥₁,𝑥₁+⋯+𝑞𝑥_𝑗,𝑥_𝑗≤𝑞𝑥₀,𝑥₁1+⋯+𝑘^𝑗−1+𝑘^𝑗,𝑞𝑥byinequality(32)₀,𝑥_𝑗+1≤(1−𝑘)𝑟+𝑞𝑥₀,𝑥₀1−𝑘^𝑗+1.1−𝑘byinequality(28) (33) Thus 𝑥_𝑗+1∈𝐵(𝑥₀,𝑟). Hence 𝑥_𝑛∈𝐵(𝑥₀,𝑟) for all 𝑛∈𝑁. Also 𝑥_𝑛+1⪯𝑥_𝑛 for all 𝑛∈𝑁. It implies that 𝑞𝑥_𝑛,𝑥_𝑛+1≤𝑘^𝑛𝑞𝑥₀,𝑥₁forall𝑛∈𝑁. (34) It follows that 𝑞𝑥_𝑛,𝑥_𝑛+𝑖≤𝑞𝑥_𝑛,𝑥_𝑛+1+⋯+𝑞𝑥_{𝑛+𝑖−1},𝑥_𝑛+𝑖𝑞𝑥_𝑛,𝑥_𝑛+𝑖≤𝑘^𝑛𝑞𝑥₀,𝑥₁1+⋯+𝑘^𝑖−2+𝑘^𝑖−1→0as𝑛→∞. (35) Notice that the sequence {𝑥_𝑛} is a 0-Cauchy sequence in (𝐵(𝑥₀,𝑟),𝑞). As 𝐵(𝑥₀,𝑟) is closed, so is 0-complete. Therefore there exists a point 𝑥^∗∈𝐵(𝑥₀,𝑟) with 𝑞𝑥^∗,𝑥^∗=lim_𝑛→∞𝑞𝑥_𝑛,𝑥^∗=lim_𝑛→∞𝑞𝑥^∗,𝑥_𝑛=0. (36) Now, 𝑞𝑥^∗,𝑆𝑥^∗≤𝑞𝑥^∗,𝑥_𝑛+𝑞𝑆𝑥_𝑛−1,𝑆𝑥^∗−𝑞𝑥_𝑛,𝑥_𝑛. (37) On taking limit as 𝑛→∞ and using the fact that 𝑥^∗⪯𝑥_𝑛⪯𝑥_𝑛−1, when 𝑥_𝑛→𝑥^∗, we have, 𝑞𝑥^∗,𝑆𝑥^∗≤lim_𝑛→∞𝑞𝑥^∗,𝑥_𝑛+𝑏𝑞𝑥_𝑛−1,𝑆𝑥_𝑛−1+𝑞𝑥^∗,𝑆𝑥^∗≤lim_𝑛→∞𝑏𝑘^𝑛−1𝑞𝑥₀,𝑥₁+𝑞𝑥^∗,𝑆𝑥^∗≤lim_𝑛→∞𝑘^𝑛𝑞𝑥₀,𝑥₁≤0. (38) Similarly, 𝑞𝑆𝑥^∗,𝑥^∗≤0, (39) and hence 𝑥^∗=𝑆𝑥^∗. Now, 𝑞𝑥^∗,𝑥^∗=𝑞𝑆𝑥^∗,𝑆𝑥^∗≤𝑏𝑞𝑥^∗,𝑆𝑥^∗+𝑞𝑥^∗,𝑆𝑥^∗(1−2𝑏)𝑞𝑥^∗,𝑥^∗≤0. (40) This implies that 𝑞𝑥^∗,𝑥^∗=0. (41)

Uniqueness. Let 𝑦 be another point in 𝐵(𝑥₀,𝑟) such that 𝑦=𝑆𝑦. If 𝑥^∗ and 𝑦 are comparable, then 𝑞𝑥^∗,𝑦=𝑞𝑆𝑥^∗𝑞𝑥,𝑆𝑦≤𝑏^∗,𝑆𝑥^∗𝑞𝑥+𝑞𝑦,𝑆𝑦=𝑏^∗,𝑥^∗.+𝑞𝑦,𝑦=𝑏𝑞𝑦,𝑦≤𝑞𝑦,𝑦 (42) Using the fact that 𝑞(𝑦,𝑦)≤𝑞(𝑥^∗,𝑦),we have 𝑞(𝑥^∗,𝑦)=0. Similarly, 𝑞(𝑦,𝑥^∗)=0. Now if 𝑥^∗ and 𝑦 are not comparable to then there exists a point 𝑧∈𝐵(𝑥₀,𝑟) which is a lower bound of both 𝑥^∗ and 𝑦. Moreover by assumption 𝑆𝑧⪯𝑧⪯𝑥^∗⪯𝑥_𝑛⋯⪯𝑥₀. Now by using inequality (27), we have 𝑞𝑆𝑥₀,𝑆𝑧≤𝑏𝑞𝑥₀,𝑥₁(𝑞𝑥+𝑞𝑧,𝑆𝑧)≤𝑏₀,𝑧+𝑞𝑥₁,𝑞𝑥,𝑆𝑧byinequality(29)₁,𝑆𝑧≤𝑘𝑞𝑥₀.𝑞𝑥,𝑧₀,𝑆𝑧≤𝑞𝑥₀,𝑥₁+𝑞𝑥₁,𝑆𝑧−𝑞𝑥₁,𝑥₁≤𝑞𝑥₀,𝑥₁+𝑘𝑞𝑥₀,𝑞𝑥,𝑧byinequality(43)₀,𝑆𝑧≤(1−𝑘)𝑟+𝑞𝑥₀,𝑥₀+𝑘𝑟+𝑞𝑥₀,𝑥₀=𝑟+𝑞𝑥₀,𝑥₀. (43)(44) It follows that 𝑆𝑧∈𝐵(𝑥₀,𝑟). Now, we will prove that 𝑆^𝑛𝑧∈𝐵(𝑥₀,𝑟), by using mathematical induction to apply inequality (27). Let 𝑆²𝑧,…,𝑆^𝑗𝑧∈𝐵(𝑥₀,𝑟) for some 𝑗∈𝑁. As 𝑆^𝑗𝑧⪯𝑆^𝑗−1𝑧⪯⋯⪯𝑧⪯𝑥^∗⪯𝑥_𝑛⋯⪯𝑥₀; then, 𝑞𝑆^𝑗𝑧,𝑆^𝑗+1𝑧=𝑞𝑆𝑆^𝑗−1𝑧,𝑆𝑆^𝑗𝑧≤𝑏𝑞𝑆^𝑗−1𝑧,𝑆^𝑗𝑧+𝑞𝑆^𝑗𝑧,𝑆^𝑗+1𝑧, (45) which implies that 𝑞𝑆^𝑗𝑧,𝑆^𝑗+1𝑧≤𝑘𝑞𝑆^𝑗−1𝑧,𝑆^𝑗𝑧≤𝑘²𝑞𝑆^𝑗−2𝑧,𝑆^𝑗−1𝑧≤⋯≤𝑘^𝑗𝑞(𝑧,𝑆𝑧). (46) Now, 𝑞𝑥_𝑗+1,𝑆^𝑗+1𝑧=𝑞𝑆𝑥_𝑗,𝑆𝑆^𝑗𝑧≤𝑏𝑞𝑥_𝑗,𝑆𝑥_𝑗+𝑞𝑆^𝑗𝑧,𝑆^𝑗+1𝑧≤𝑏𝑘^𝑗𝑞𝑥₀,𝑥₁+𝑘^𝑗𝑞(𝑧,𝑆𝑧)(byinequality(32)andinequality(46))≤𝑏𝑘^𝑗𝑞𝑥₀,𝑧+𝑞𝑥₁,𝑆𝑧(byinequality(29))≤𝑏𝑘^𝑗𝑞𝑥₀,𝑧+𝑘𝑞𝑥₀,𝑧=𝑘^𝑗+1𝑞𝑥₀,𝑧₀. (47) Now, 𝑞𝑥₀,𝑆^𝑗+1𝑧≤𝑞𝑥₀,𝑥₁+𝑞𝑥₁,𝑥₂+⋯+𝑞𝑥_𝑗,𝑥_𝑗+1+𝑞𝑥_𝑗+1,𝑆^𝑗+1𝑧≤𝑞𝑥₀,𝑥₁+𝑘𝑞𝑥₀,𝑥₁+⋯+𝑘^𝑗+1𝑞𝑥₀,𝑞𝑥,𝑧byinequality(32)andinequality(47)₀,𝑆^𝑗+1𝑧≤𝑞𝑥₀,𝑥₁1+𝑘+⋯𝑘^𝑗+𝑘^𝑗+1𝑟,as𝑧∈𝐵𝑥₀𝑞𝑥,𝑟₀,𝑆^𝑗+1𝑧≤(1−𝑘)𝑟1−𝑘^𝑗+11−𝑘+𝑘^𝑗+1𝑟=𝑟. (48) It follows that 𝑆^𝑗+1𝑧∈𝐵(𝑥₀,𝑟) and hence 𝑆^𝑛𝑧∈𝐵(𝑥₀,𝑟). Now inequality (46) can be written as 𝑞𝑆^𝑛𝑧,𝑆^𝑛+1𝑧≤𝑘^𝑛𝑞(𝑧,𝑆𝑧)→0as𝑛→∞. (49) Now, 𝑞𝑥^∗,𝑦=𝑞𝑆𝑥^∗,𝑆𝑦≤𝑞𝑆𝑥^∗,𝑆^𝑛+1𝑧+𝑞𝑆^𝑛+1𝑞𝑥𝑧,𝑆𝑦≤𝑏^∗,𝑆𝑥^∗+𝑞𝑆^𝑛𝑧,𝑆^𝑛+1𝑧+𝑏𝑞𝑆^𝑛𝑧,𝑆^𝑛+1𝑧𝑥+𝑞𝑦,𝑆𝑦≤𝑏𝑞^∗,𝑥^∗+2𝑏𝑞𝑆^𝑛𝑧,𝑆^𝑛+1𝑧.+𝑏𝑞𝑦,𝑦≤𝑏𝑞𝑦,𝑦≤𝑞𝑦,𝑦byinequality(41)andinequality(49) (50) Using the fact that 𝑞(𝑦,𝑦)≤𝑞(𝑥^∗,𝑦),we have 𝑞(𝑥^∗,𝑦)=0. Similarly, 𝑞(𝑦,𝑥^∗)=0. Hence 𝑥^∗=𝑦.

In Theorem 15, the conditions (29), the existence of a lower bound and for a nonincreasing sequence {𝑥_𝑛} in 𝑋,{𝑥_𝑛}→𝑢 implies that 𝑢⪯𝑥_𝑛 are imposed to restrict the condition (27) only for comparable elements. However, the following result relaxes these conditions but imposes the condition (27) for all elements in 𝐵(𝑥₀,𝑟).

Theorem 16. Let (𝑋,𝑑_𝑞) be a complete dislocated quasimetric space, 𝑆∶𝑋→𝑋 a map, and 𝑥₀ an arbitrary point in 𝑋. Suppose there exists 𝑘∈[0,1/2) with 𝑑_𝑞𝑆𝑥,𝑆𝑦≤𝑘𝑑_𝑞(𝑥,𝑆𝑥)+𝑑_𝑞𝑦,𝑆𝑦, (51) for all elements 𝑥,𝑦 in 𝐵(𝑥₀,𝑟)and 𝑑_𝑞𝑥₀,𝑆𝑥₀≤(1−𝜃)𝑟, (52) where 𝜃=𝑘/(1−𝑘). Then there exists a unique point 𝑥^∗ in 𝐵(𝑥₀,𝑟) such that 𝑥^∗=𝑆𝑥^∗ and 𝑑_𝑞(𝑥^∗,𝑥^∗)=0.

In Theorem 10, the conditions (28) and (29) are imposed to restrict the condition (27) only for comparable elements. However, the following result relaxes these conditions but imposes the condition (27) for all elements in 𝑋.

Theorem 17. Let (𝑋,⪯,𝑞) be a 0-complete ordered quasi-partial metric space and let 𝑆∶𝑋→𝑋 be a dominated map. Suppose that there exists 𝑏∈[0,1/2) such that 𝑞𝑆𝑥,𝑆𝑦≤𝑏𝑞(𝑥,𝑆𝑥)+𝑞𝑦,S𝑦 (53) for all comparable elements 𝑥,𝑦 in 𝑋. If, for a nonincreasing sequence {𝑥_𝑛} in 𝑋,{𝑥_𝑛}→𝑢 implies that 𝑢⪯𝑥_𝑛, then there exists a point 𝑥^∗ in 𝑋 such that 𝑥^∗=𝑆𝑥^∗ and 𝑞(𝑥^∗,𝑥^∗)=0. Moreover, 𝑥^∗ is unique, if for any two points 𝑥,𝑦 in 𝐵(𝑥₀,𝑟) there exists a point 𝑧∈𝐵(𝑥₀,𝑟) such that 𝑧⪯𝑥 and 𝑧⪯𝑦.

Theorem 18. Let (𝑋,⪯,𝑞) be a 0-complete ordered quasi-partial metric space, 𝑆∶𝑋→𝑋 a dominated map, and 𝑥₀ an arbitrary point in 𝑋. Suppose that there exists 𝑐∈[0,1/2) such that 𝑞𝑆𝑥,𝑆𝑦≤𝑐𝑞𝑥,𝑆𝑦+𝑞𝑆𝑥,𝑦 (54) for all comparable elements 𝑥,𝑦 in 𝐵(𝑥₀,𝑟).
And 𝑞𝑥₀,𝑆𝑥₀≤(1−𝑘)𝑟+𝑞𝑥₀,𝑥₀, (55) where 𝑘=𝑐/(1−𝑐). If, for a nonincreasing sequence {𝑥_𝑛} in 𝐵(𝑥₀,𝑟),{𝑥_𝑛}→𝑢 implies that 𝑢⪯𝑥_𝑛, then there exists a point 𝑥^∗ in 𝐵(𝑥₀,𝑟) such that 𝑥^∗=𝑆𝑥^∗. Further 𝑞(𝑥^∗,𝑥^∗)=0.

Proof. Consider a Picard sequence 𝑥_𝑛+1=𝑆𝑥_𝑛 with initial guess 𝑥₀. As 𝑥_𝑛+1=𝑆𝑥_𝑛⪯𝑥_𝑛 for all 𝑛∈{0}∪𝑁. We will prove that 𝑥_𝑛∈𝐵(𝑥₀,𝑟) for all 𝑛∈𝑁 by mathematical induction. By using inequality (55), we have 𝑞𝑥₀,𝑥₁≤𝑟+𝑞𝑥₀,𝑥₀. (56) Therefore 𝑥₁∈𝐵(𝑥₀,𝑟). Now let 𝑥₂,…,𝑥_𝑗∈𝐵(𝑥₀,𝑟) for some 𝑗∈𝑁. As 𝑥_𝑛+1⪯𝑥_𝑛, so using inequality (54), we obtain 𝑞𝑥_𝑗,𝑥_𝑗+1≤𝑐𝑞𝑥_𝑗−1,𝑥_𝑗+𝑞𝑥_𝑗,𝑥_𝑗+1−𝑞𝑥_𝑗,𝑥_𝑗+𝑞𝑥_𝑗,𝑥_𝑗≤𝑘𝑞𝑥_𝑗−1,𝑥_𝑗, (57) which implies that 𝑞𝑥_𝑗,𝑥_𝑗+1≤𝑘²𝑞𝑥_𝑗−2,𝑥_𝑗−1≤⋯≤𝑘^𝑗𝑞𝑥₀,𝑥₁. (58) Now 𝑞𝑥₀,𝑥_𝑗+1≤𝑞𝑥₀,𝑥₁+⋯+𝑞𝑥_𝑗,𝑥_𝑗+1−𝑞𝑥₁,𝑥₁+⋯+𝑞𝑥_𝑗,𝑥_𝑗𝑞𝑥₀,𝑥_𝑗+1≤(1−𝑘)𝑟+𝑞𝑥₀,𝑥₀1−𝑘^𝑗+1.1−𝑘byinequality(55) (59) Thus 𝑥_𝑗+1∈𝐵(𝑥₀,𝑟). Hence 𝑥_𝑛∈𝐵(𝑥₀,𝑟) for all 𝑛∈𝑁. Also 𝑥_𝑛+1⪯𝑥_𝑛 for all 𝑛∈𝑁. It implies that 𝑞𝑥_𝑛,𝑥_𝑛+1≤𝑘^𝑛𝑞𝑥₀,𝑥₁forall𝑛∈𝑁. (60) It follows that 𝑞𝑥_𝑛,x_𝑛+𝑖≤𝑞𝑥_𝑛,𝑥_𝑛+1+⋯+𝑞𝑥_{𝑛+𝑖−1},𝑥_𝑛+𝑖𝑞𝑥_𝑛,𝑥_𝑛+𝑖≤𝑘^𝑛𝑞𝑥₀,𝑥₁1+⋯+𝑘^𝑖−2+𝑘^𝑖−1→0as𝑛→∞. (61) Notice that the sequence {𝑥_𝑛} is a 0-Cauchy sequence in (𝐵(𝑥₀,𝑟),𝑞). As 𝐵(𝑥₀,𝑟) is closed so is 0-complete. Therefore there exists a point 𝑥^∗∈𝐵(𝑥₀,𝑟) with 𝑞𝑥^∗,𝑥^∗=lim_𝑛→∞𝑞𝑥_𝑛,𝑥^∗=lim_𝑛→∞𝑞𝑥^∗,𝑥_𝑛=0. (62) Now, 𝑞𝑥^∗,𝑆𝑥^∗≤𝑞𝑥^∗,𝑥_𝑛+𝑞𝑆𝑥_𝑛−1,𝑆𝑥^∗−𝑞𝑥_𝑛,𝑥_𝑛. (63) On taking limit as 𝑛→∞ and using the fact that 𝑥^∗⪯𝑥_𝑛⪯𝑥_𝑛−1, when 𝑥_𝑛→𝑥^∗, we have 𝑞𝑥^∗,𝑆𝑥^∗≤lim_𝑛→∞𝑞𝑥^∗,𝑥_𝑛+𝑐𝑞𝑥_𝑛−1,𝑆𝑥^∗+𝑞𝑥_𝑛,𝑥^∗≤lim_𝑛→∞𝑐𝑞𝑥_𝑛−1,𝑥^∗+𝑞𝑥^∗,𝑆𝑥^∗−𝑞𝑥^∗,𝑥^∗+𝑞𝑥_𝑛,𝑥^∗(1−𝑐)𝑞𝑥^∗,𝑆𝑥^∗≤0byinequality(62). (64) Similarly, 𝑞𝑆𝑥^∗,𝑥^∗≤0 (65) and hence 𝑥^∗=𝑆𝑥^∗.

Remark 19. We can obtain the partial metric, quasi-metric, and metric version of all theorems which are still not present in the literature.

Conflict of Interests

The authors declare that they have no competing interests.