New Meir-Keeler Type Tripled Fixed-Point Theorems on Ordered Partial Metric Spaces

Abstract

In this paper, we prove some new Meir-Keeler type tripled fixed-point theorems on a partially ordered complete partial metric space. Also, as application, some results of integral type are given.

1. Introduction and Preliminaries

In the last century, the theory of fixed points has appeared as a crucial technique in the study of nonlinear phenomena. Particularly, the tools in fixed-point theory have an application in such diverse fields as biology, chemistry, physics, economics, computer sciences, and engineering.

Recently, fixed-point theorems are considered on partial metric spaces on which self-distance of some points may not be zero. This phenomenon was discovered by Matthews [1] when he considered the tools of metric spaces in the field of semantics and domain theory in computer science (see, e.g., [2, 3]). After the initial results of Mathews, other papers have been released on partial metric spaces (see e.g., [4–20]).

Another important development is reported in fixed-point theory via ordered metric spaces. Fixed-point theorems in ordered sets were discussed by Ran and Reurings [21]. Subsequently, many results in this direction were given (see, e.g., [22–31]).

In this paper, we combine two recent trends, partial metric spaces and ordered sets, and discuss the existence and uniqueness of some new Meir-Keeler type tripled fixed-point theorems in the context of partially ordered partial metric spaces.

Let 𝑋 be a nonempty set. A partial metric is a function 𝑝∶𝑋×𝑋→[0,∞) satisfying the following conditions:(P1)if 𝑝(𝑥,𝑥)=𝑝(𝑥,𝑦)=𝑝(𝑦,𝑦), then 𝑥=𝑦,(P2)𝑝(𝑥,𝑦)=𝑝(𝑦,𝑥),(P3)𝑝(𝑥,𝑥)≤𝑝(𝑥,𝑦),(P4)𝑝(𝑥,𝑧)+𝑝(𝑦,𝑦)≤𝑝(𝑥,𝑦)+𝑝(𝑦,𝑧),

for all 𝑥,𝑦,𝑧∈𝑋. Then, (𝑋,𝑝) is called a partial metric space.

If 𝑝 is a partial metric 𝑝 on 𝑋, then the function 𝑑_𝑝∶𝑋×𝑋→[0,∞) given by 𝑑_𝑝𝑥,𝑦=2𝑝𝑥,𝑦−𝑝(𝑥,𝑥)−𝑝𝑦,𝑦 (1.1) is a metric on 𝑋. Each partial metric 𝑝 on 𝑋 generates a 𝑇₀ topology 𝜏_𝑝 on 𝑋 with a base of the family of open 𝑝-balls {𝐵_𝑝(𝑥,𝜀)∶𝑥∈𝑋,𝜀>0}, where 𝐵_𝑝(𝑥,𝜀)={𝑦∈𝑋∶𝑝(𝑥,𝑦)<𝑝(𝑥,𝑥)+𝜀} for all 𝑥∈𝑋 and 𝜀>0. Similarly, closed 𝑝-ball is defined as 𝐵_𝑝[𝑥,𝜀]={𝑦∈𝑋∶𝑝(𝑥,𝑦)≤𝑝(𝑥,𝑥)+𝜀}. For more details, see [1, 5].

Definition 1.1 (see, e.g., [1, 5, 15]). Let (𝑋,𝑝) be a partial metric space.
(i)A sequence {𝑥_𝑛} in 𝑋 converges to 𝑥∈𝑋 whenever lim_𝑛→∞𝑝(𝑥,𝑥_𝑛)=𝑝(𝑥,𝑥).(ii) A sequence {𝑥_𝑛} in 𝑋 is called Cauchy whenever lim_{𝑛,𝑚→∞}𝑝(𝑥_𝑛,𝑥_𝑚) exists (and finite).(iii)(𝑋,𝑝) is said to be complete if every Cauchy sequence {𝑥_𝑛} in 𝑋 converges, with respect to 𝜏_𝑝, to a point 𝑥∈𝑋, that is, lim_{𝑛,𝑚→∞}𝑝(𝑥_𝑛,𝑥_𝑚)=𝑝(𝑥,𝑥).

Lemma 1.2 (see, e.g., [1, 5, 15]). Let (𝑋,𝑝) be a partial metric space.
(a)A sequence {𝑥_𝑛} is Cauchy if and only if {𝑥_𝑛} is a Cauchy sequence in the metric space (𝑋,𝑑_𝑝).(b)(𝑋,𝑝) is complete if and only if the metric space (𝑋,𝑑_𝑝) is complete. Moreover, lim_𝑛→∞𝑑_𝑝𝑥,𝑥_𝑛=0⟺lim_𝑛→∞𝑝𝑥,𝑥_𝑛=lim_{𝑛,𝑚→∞}𝑝𝑥_𝑛,𝑥_𝑚=𝑝(𝑥,𝑥). (1.2)

Lemma 1.3 (see, e.g., [4, 15, 16]). Let (𝑋,𝑝) be a partial metric space. Then, (A) if 𝑝(𝑥,𝑦)=0, then 𝑥=𝑦,(B) if 𝑥≠𝑦, then 𝑝(𝑥,𝑦)>0.

Remark 1.4. If 𝑥=𝑦, 𝑝(𝑥,𝑦) may not be 0.

Lemma 1.5 (see, e.g., [4, 15, 16]). Let 𝑥_𝑛→𝑧 as 𝑛→∞ in a partial metric space (𝑋,𝑝) where 𝑝(𝑧,𝑧)=0. Then, lim_𝑛→∞𝑝(𝑥_𝑛,𝑦)=𝑝(𝑧,𝑦) for every 𝑦∈𝑋.

(𝑋,𝑝,≤) is called a partially ordered partial metric space if (𝑋,≤) is a partially ordered set and (𝑋,𝑝) is a partial metric space. Further, if (𝑋,𝑝) is a complete partial metric space, then (𝑋,𝑝,≤) is called a partially ordered complete partial metric space. Hereafter, we assume that 𝑋≠∅ and we use the notation 𝑋^𝑘=𝑋×𝑋×⋯×𝑋_𝑘-many. (1.3) Also, take the mapping 𝑃∶𝑋³×𝑋³→[0,+∞) such that 𝑃(𝐱,𝐲)∶=max𝑝𝑥₁,𝑦₁,𝑝𝑥₂,𝑦₂,𝑝𝑥₃,𝑦₃, (1.4) where 𝐱=(𝑥₁,𝑥₂,𝑥₃) and 𝐲=(𝑦₁,𝑦₂,𝑦₃)∈𝑋³.

Let (𝑋,≤) be a partially ordered set. We consider the following partial order (also denoted by ≤) on the product space 𝑋³: (𝑢,𝑣,𝑤)≤𝑥,𝑦,𝑧iff𝑥≥𝑢,𝑦≤𝑣,𝑧≥𝑤, (1.5) where (𝑢,𝑣,𝑤),(𝑥,𝑦,𝑧)∈𝑋³. Moreover, we say that (𝑥,𝑦,𝑧) is equal to (𝑢,𝑣,𝑟) if and only if 𝑥=𝑢,=𝑣, and 𝑧=𝑟. In the sequel, we need the following definitions.

Definition 1.6 (see [32]). Let (𝑋,≤) be a partially ordered set and 𝐹∶𝑋³→𝑋 a given mapping. We say that 𝐹 has the mixed monotone property if 𝐹(𝑥,𝑦,𝑧) is monotone nondecreasing in 𝑥 and 𝑧, and it is monotone nonincreasing in 𝑦, that is, for any𝑥,𝑦,𝑧∈𝑋, 𝑥₁,𝑥₂∈𝑋,𝑥₁≤𝑥₂⟹𝐹𝑥₁,𝑦,𝑧≤𝐹𝑥₂,𝑦,𝑦,𝑧₁,𝑦₂∈𝑋,𝑦₁≤𝑦₂⟹𝐹𝑥,𝑦₁,𝑧≥𝐹𝑥,𝑦₂,𝑧,𝑧₁,𝑧₂∈𝑋,𝑧₁≤𝑧₂⟹𝐹𝑥,𝑦,𝑧₁≤𝐹𝑥,𝑦,𝑧₂. (1.6)

Definition 1.7 (see [32]). An element (𝑥,𝑦,𝑧)∈𝑋³ is called a tripled fixed point of 𝐹∶𝑋³→𝑋 if 𝐹𝑥,𝑦,𝑧=𝑥,𝐹𝑦,𝑥,𝑦=𝑦,𝐹𝑧,𝑦,𝑥=𝑧. (1.7)

Berinde and Borcut [32] proved the following theorem.

Theorem 1.8. Let (𝑋,≤) be a partially ordered set and (𝑋,𝑑) a complete metric space. Let 𝐹∶𝑋³→𝑋 be a mapping having the mixed monotone property on 𝑋. Assume that there exist constants 𝑎,𝑏,𝑐∈[0,1) such that 𝑎+𝑏+𝑐<1 for which 𝑑𝐹𝑥,𝑦,𝑧,𝐹(𝑢,𝑣,𝑤)≤𝑎𝑑(𝑥,𝑢)+𝑏𝑑𝑦,𝑣+𝑐𝑑(𝑧,𝑤) (1.8) for all 𝑥≥𝑢,𝑦≤𝑣 and 𝑧≥𝑤. Assume that 𝑋 has the following properties:(i)if a nondecreasing sequence 𝑥_𝑛→𝑥, then 𝑥_𝑛≤𝑥 for all 𝑛,(ii)if a nonincreasing sequence 𝑦_𝑛→𝑦, then 𝑦_𝑛≥𝑦 for all 𝑛.If there exist 𝑥₀,𝑦₀,𝑧₀∈𝑋 such that 𝑥₀≤𝐹𝑥₀,𝑦₀,𝑧₀,𝑦₀≥𝐹𝑦₀,𝑥₀,𝑦₀,𝑧₀≤𝐹𝑧₀,𝑦₀,𝑥₀, (1.9) then there exist 𝑥,𝑦,𝑧∈𝑋 such that 𝐹𝑥,𝑦,𝑧=𝑥,𝐹𝑦,𝑥,𝑦=𝑦,𝐹𝑧,𝑦,𝑥=𝑧. (1.10)

Recently, Theorem 1.8 is extended to cone metric spaces by Rao and Kishore [33]. On the other hand, very recently, Aydi et al. [34] introduced the following concepts.

Definition 1.9 (see [34]). Let (𝑋,≤) be a partially ordered set and 𝐹∶𝑋³→𝑋. We say that 𝐹 has the mixed strict monotone property if, for any𝑥,𝑦,𝑧∈𝑋, 𝑥₁,𝑥₂∈𝑋,𝑥₁<𝑥₂⟹𝐹𝑥₁,𝑦,𝑧<𝐹𝑥₂,𝑦,𝑦,𝑧₁,𝑦₂∈𝑋,𝑦₁<𝑦₂⟹𝐹𝑥,𝑦₁,𝑧>𝐹𝑥,𝑦₂,𝑧,𝑧₁,𝑧₂∈𝑋,𝑧₁<𝑧₂⟹𝐹𝑥,𝑦,𝑧₁<𝐹𝑥,𝑦,𝑧₂. (1.11)

Definition 1.10 (see [34]). Let (𝑋,𝑑,≤) be a partially ordered metric space. A mapping 𝐹∶𝑋³→𝑋 is said to be a generalized Meir-Keeler type contraction if, for any 𝜀>0, there exists a 𝛿(𝜀)>0 such that 𝜀≤max𝑑(𝑥,𝑢),𝑑𝑦,𝑣,𝑑(𝑧,𝑟)<𝜀+𝛿(𝜀)⟹𝑑𝐹𝑥,𝑦,𝑧,𝐹(𝑢,𝑣,𝑟)<𝜀, (1.12) for all 𝑥,𝑦,𝑧,𝑢,𝑣,𝑟∈𝑋 with 𝑥≤𝑢, 𝑦≥𝑣 and 𝑧≤𝑟.

In the following, we consider the partial case of Definition 1.10 and we introduce the following.

Definition 1.11. Let (𝑋,𝑝,≤) be a partially ordered partial metric space. A mapping 𝐹∶𝑋³→𝑋 is said to be a generalized 𝑝-Meir-Keeler type contraction if, for any 𝜀>0, there exists a 𝛿(𝜀)>0 such that 𝜀≤max𝑝(𝑥,𝑢),𝑝𝑦,𝑣,𝑝(𝑧,𝑟)<𝜀+𝛿(𝜀)⟹𝑝𝐹𝑥,𝑦,𝑧,𝐹(𝑢,𝑣,𝑟)<𝜀, (1.13) for all 𝑥,𝑦,𝑧,𝑢,𝑣,𝑟∈𝑋 with 𝑥≤𝑢, 𝑦≥𝑣 and 𝑧≤𝑟.

Remark 1.12. It is immediate to show that if 𝐹∶𝑋³→𝑋 is a generalized 𝑝-Meir-Keeler type contraction, then 𝑝𝐹𝑥,𝑦,𝑧,𝐹(𝑢,𝑣,𝑟)<max𝑝(𝑥,𝑢),𝑝𝑦,𝑣,𝑝(𝑧,𝑟) (1.14) for all 𝑥,𝑢,𝑦,𝑣,𝑧,𝑟,∈𝑋 with 𝑥<𝑢,𝑦≥𝑣,𝑧<𝑟 or 𝑥≤𝑢,𝑦>𝑣,𝑧≤𝑟.

Proposition 1.13. Let (𝑋,𝑝,≤) be a partially ordered partial metric space and 𝐹∶𝑋³→𝑋 a given mapping. If (1.8) is satisfied, then 𝐹 is a generalized 𝑝-Meir-Keeler type function.

Proof. Assume that (1.8) is satisfied. For all 𝜀>0, one can check that (1.13) is satisfied with 𝛿(𝜀)=(1/(𝑎+𝑏+𝑐)−1)𝜀.

In the sequel, we use the following notations given in [34]. Let 𝐹∶𝑋³→𝑋³ be such that, for𝑎,𝑏,𝑐∈𝑋, 𝐹(𝑎,𝑏,𝑐)=(𝐹(𝑎,𝑏,𝑐),𝐹(𝑏,𝑎,𝑏),𝐹(𝑐,𝑏,𝑎)). (1.15) Let 𝑥₀,𝑦₀,𝑧₀∈𝑋 be such that 𝑥₀<𝐹𝑥₀,𝑦₀,𝑧₀,𝑦₀≥𝐹𝑦₀,𝑥₀,𝑦₀,𝑧₀<𝐹𝑧₀,𝑦₀,𝑥₀. (1.16) We consider sequences {𝑥_𝑛}, {𝑦_𝑛}, and {𝑧_𝑛} such that 𝑥_𝑛𝑦_𝑛𝑧_𝑛𝐴𝑛=𝐹𝑥_𝑛−1,𝑦_𝑛−1,𝑧_𝑛−1𝐹𝑦_𝑛−1,𝑥_𝑛−1,𝑦_𝑛−1𝐹𝑧_𝑛−1,𝑦_𝑛−1,𝑥_𝑛−1𝐹_{(𝐴𝑛−1)}=𝐹^𝑛(𝑥₀,𝑦₀,𝑧₀)𝐹^𝑛(𝑦₀,𝑥₀,𝑦₀)𝐹^𝑛(𝑧₀,𝑦₀,𝑥₀)𝐹^𝑛_(𝐴0), (1.17) for 𝑛=1,2,3,….

Our first auxiliary result is as follows.

Proposition 1.14. Let (𝑋,𝑝,≤)be a partially ordered partial metric space, and let𝐹∶𝑋³→𝑋be a given mapping such that the following hypotheses hold:(i)𝐹 has the mixed strict monotone property,(ii)𝐹 is a generalized 𝑝-Meir-Keeler type function,(iii)∃(𝑥,𝑦,𝑧),(𝑢,𝑣,𝑟)∈𝑋³ such that 𝑥<𝑢, 𝑦≥𝑣 and 𝑧<𝑟.

Then, 𝑃𝐹^𝑛𝑥,𝑦,𝑧,𝐹^𝑛(𝑢,𝑣,𝑟)⟶0,as𝑛⟶+∞. (1.18)

Proof. Let (𝑥,𝑦,𝑧)=(𝑥₀,𝑦₀,𝑧₀) and (𝑢,𝑣,𝑟)=(𝑢₀,𝑣₀,𝑟₀). We show that 𝑥_𝑛=𝐹^𝑛𝑥₀,𝑦₀,𝑧₀,<𝐹^𝑛𝑢₀,𝑣₀,𝑟₀=𝑢_𝑛,𝑦_𝑛=𝐹^𝑛𝑦₀,𝑥₀,𝑦₀>𝐹^𝑛𝑣₀,𝑢₀,𝑣₀=𝑣_𝑛,𝑧_𝑛=𝐹^𝑛𝑧₀,𝑦₀,𝑥₀<𝐹^𝑛𝑟₀,𝑣₀,𝑢₀=𝑟_𝑛,∀𝑛=1,2,…, (1.19) with 𝐹=𝐹¹.
Due to the fact that 𝐹 has the mixed strict monotone property, together with the assumption that 𝑥<𝑢, 𝑦≥𝑣 and 𝑧<𝑟, we obtain 𝑥₁=𝐹𝑥,𝑦,𝑧=𝐹𝑥₀,𝑦₀,𝑧₀<𝐹𝑢₀,𝑦₀,𝑧₀⟹𝐹𝑥₀,𝑦₀,𝑧₀<𝐹𝑢₀,𝑣₀,𝑧₀⟹𝐹𝑥₀,𝑦₀,𝑧₀<𝐹𝑢₀,𝑣₀,𝑟₀=𝑢₁. (1.20) Analogously, we have 𝑦₁=𝐹𝑦₀,𝑥₀,𝑦₀>𝐹𝑣₀,𝑢₀,𝑣₀=𝑣₁,𝑧₁=𝐹𝑧₀,𝑦₀,𝑥₀<𝐹𝑟₀,𝑣₀,𝑢₀=𝑟₁. (1.21) Thus, (1.19) holds for 𝑛=1. By using the same arguments, we show that (1.19) holds also for 𝑛=2. In fact, 𝑥₂=𝐹²𝑥₀,𝑦₀,𝑧₀=𝐹𝑥₁,𝑦₁,𝑧₁=𝐹𝐹𝑥₀,𝑦₀,𝑧₀,𝐹𝑦₀,𝑥₀,𝑦₀,𝐹𝑧₀,𝑦₀,𝑥₀<𝐹𝐹𝑢₀,𝑣₀,𝑟₀,𝐹𝑦₀,𝑥₀,𝑦₀,𝐹𝑧₀,𝑦₀,𝑥₀<𝐹𝐹𝑢₀,𝑣₀,𝑟₀,𝐹𝑣₀,𝑢₀,𝑣₀,𝐹𝑧₀,𝑦₀,𝑥₀<𝐹𝐹𝑢₀,𝑣₀,𝑟₀,𝐹𝑣₀,𝑢₀,𝑣₀,𝐹𝑟₀,𝑣₀,𝑢₀=𝐹²𝑢₀,𝑣₀,𝑟₀=𝐹𝑢₁,𝑣₁,𝑟₁=𝑢₂. (1.22) Similarly, we find 𝑦₂=𝐹²𝑦₀,𝑥₀,𝑦₀≥𝐹²𝑣₀,𝑢₀,𝑣₀=𝑣₂,𝑧₂=𝐹²𝑧₀,𝑦₀,𝑥₀<𝐹²𝑟₀,𝑣₀,𝑢₀=𝑟₂. (1.23) Inductively, we get that (1.19) holds.
By Remark 1.12, together with (1.19), we have 𝑝𝑥_𝑛+2,𝑢_𝑛+2=𝑝𝐹^𝑛+2𝑥₀,𝑦₀,𝑧₀,𝐹^𝑛+2𝑢₀,𝑣₀,𝑟₀=𝑝𝐹𝑥_𝑛+1,𝑦_𝑛+1,𝑧_𝑛+1,𝐹𝑢_𝑛+1,𝑣_𝑛+1,𝑟_𝑛+1<max𝑝𝑥_𝑛+1,𝑢_𝑛+1,𝑝𝑦_𝑛+1,𝑣_𝑛+1,𝑝𝑧_𝑛+1,𝑟_𝑛+1,𝑝𝑧_𝑛+2,𝑟_𝑛+2=𝑝𝐹^𝑛+2𝑧₀,𝑦₀,𝑥₀,𝐹^𝑛+2𝑟₀,𝑣₀,𝑢₀=𝑝𝐹𝑧_𝑛+1,𝑦_𝑛+1,𝑥_𝑛+1,𝐹𝑟_𝑛+1,𝑣_𝑛+1,𝑢_𝑛+1<max𝑝𝑧_𝑛+1,𝑟_𝑛+1,𝑝𝑦_𝑛+1,𝑣_𝑛+1,𝑝𝑥_𝑛+1,𝑢_𝑛+1,𝑝𝑦_𝑛+2,𝑣_𝑛+2=𝑝𝐹^𝑛+2𝑦₀,𝑥₀,𝑦₀,𝐹^𝑛+2𝑣₀,𝑢₀,𝑣₀=𝑝𝐹𝑦_𝑛+1,𝑥_𝑛+1,𝑦_𝑛+1,𝐹𝑣_𝑛+1,𝑢_𝑛+1,𝑣_𝑛+1<max𝑝𝑦_𝑛+1,𝑣_𝑛+1,𝑝𝑥_𝑛+1,𝑢_𝑛+1,𝑝𝑦_𝑛+1,𝑣_𝑛+1≤max𝑝𝑧_𝑛+1,𝑟_𝑛+1,𝑝𝑦_𝑛+1,𝑣_𝑛+1,𝑝𝑥_𝑛+1,𝑢_𝑛+1. (1.24)(1.25)(1.26) Let Δ_𝑛+1∶=max{𝑝(𝑥_𝑛+1,𝑢_𝑛+1),𝑝(𝑦_𝑛+1,𝑣_𝑛+1),𝑝(𝑧_𝑛+1,𝑟_𝑛+1)}. Combining (1.24)–(1.26), we get Δ_𝑛+2<Δ_𝑛+1,∀𝑛=1,2…. (1.27) If we denote 𝐵_𝑛=(𝑢_𝑛,𝑣_𝑛,𝑟_𝑛), then, by definition of the partial metric 𝑃 and (1.27), we have 𝑃𝐴_𝑛+2,𝐵_𝑛+2<𝑃𝐴_𝑛+1,𝐵_𝑛+1. (1.28) Consequently, the sequence {𝑡_𝑛}={𝑃(𝐴_𝑛,𝐵_𝑛)} is decreasing. Hence, {𝑡_𝑛} converges, say to 𝜀≥0. Clearly, if 𝜀=0, we have finished. Suppose, on the contrary, 𝜀>0. Thus, there exists 𝑘∈{1,2,…} such that 𝜀≤𝑡_𝑛=𝑃𝐴_𝑛,𝐵_𝑛<𝜀+𝛿(𝜀)forany𝑛≥𝑘. (1.29) In particular, for 𝑛=𝑘, we have 𝜀≤𝑡_𝑘=𝑃𝐴_𝑘,𝐵_𝑘<𝜀+𝛿(𝜀), (1.30) that is equal to 𝜀≤Δ_𝑘<𝜀+𝛿(𝜀). (1.31) It follows from (1.19) and the hypothesis (ii) that 𝑝𝐹𝑥_𝑘,𝑦_𝑘,𝑧_𝑘,𝐹𝑢_𝑘,𝑣_𝑘,𝑟_𝑘<𝜀 (1.32) which is equivalent to 𝑝𝑥_𝑘+1,𝑢_𝑘+1<𝜀. (1.33) Moreover, we have 𝑝𝑦_𝑘+1,𝑣_𝑘+1<𝜀,𝑝𝑧_𝑘+1,𝑟_𝑘+1<𝜀. (1.34) Combining (1.33) and (1.34), we have Δ_𝑘+1<𝜀. (1.35) Thus, 𝑡_𝑘+1=𝑃(𝐴_𝑘+1,𝐵_𝑘+1)<𝜀 which is a contradiction with respect to (1.29), and so 𝜀=0. We conclude that 𝑃𝐴_𝑛,𝐵_𝑛=𝑃𝐹^𝑛𝑥,𝑦,𝑧,𝐹^𝑛(𝑢,𝑣,𝑟)⟶0,as𝑛⟶+∞. (1.36)

Remark 1.15. The previous proposition remains true if, in (iii), we change the assumption ∃𝑥,𝑦,𝑧,(𝑢,𝑣,𝑟)∈𝑋³suchthat𝑥<𝑢,𝑦≥𝑣,𝑧<𝑟 (1.37) with the following ∃𝑥,𝑦,𝑧,(𝑢,𝑣,𝑟)∈𝑋³suchthat𝑥≤𝑢,𝑦>𝑣,𝑧≤𝑟. (1.38)

2. Existence of Tripled Fixed Point

The following theorem is our first main result.

Theorem 2.1. Let (𝑋,𝑝,≤) be a partially ordered complete partial metric space. Suppose that 𝑋 has the following properties:
(a)if {𝑥_𝑛} is a sequence such that 𝑥_𝑛+1>𝑥_𝑛 for each 𝑛=1,2,… and 𝑥_𝑛→𝑥, then 𝑥_𝑛<𝑥 for each 𝑛=1,2,…,(b)if {𝑦_𝑛} is a sequence such that 𝑦_𝑛+1<𝑦_𝑛 for each 𝑛=1,2,… and 𝑦_𝑛→𝑦, then 𝑦_𝑛>𝑦 for each 𝑛=1,2,….
Assume that 𝐹∶𝑋³→𝑋 satisfies the following hypotheses:(i)𝐹 has the mixed strict monotone property,(ii)𝐹 is a generalized 𝑝-Meir-Keeler type function,(iii)there exist 𝑥₀,𝑦₀,𝑧₀∈𝑋 such that 𝑥₀<𝐹𝑥₀,𝑦₀,𝑧₀,𝑦₀≥𝐹𝑦₀,𝑥₀,𝑦₀,𝑧₀<𝐹𝑧₀,𝑦₀,𝑥₀,. (2.1) Then, 𝐹 has a tripled fixed point, that is, there exist 𝑥,𝑦,𝑧∈𝑋 such that 𝐹𝑥,𝑦,𝑧=𝑥,𝐹𝑦,𝑥,𝑦=𝑦,𝐹𝑧,𝑦,𝑥=𝑧. (2.2) Also, 𝑝(𝑥,𝑥)=𝑝(𝑦,𝑦)=𝑝(𝑧,𝑧)=0.

Proof. Let 𝑥₀,𝑦₀,𝑧₀∈𝑋 be as in (iii). We construct sequences {𝑥_𝑛}, {𝑦_𝑛}, and {𝑧_𝑛} according to (1.17).
We claim that, for all 𝑛≥2, we have ⋯>𝑥_𝑛>𝑥_𝑛−1>⋯>𝑥₁>𝑥₀,⋯<𝑦_𝑛<𝑦_𝑛−1<⋯<𝑦₁≤𝑦₀,⋯>𝑧_𝑛>𝑧_𝑛−1>⋯>𝑧₁>𝑧₀. (2.3) Indeed, we will use a mathematical induction to prove (2.3). Clearly, we have 𝑥₀<𝐹𝑥₀,𝑦₀,𝑧₀=𝑥₁,𝑦₀≥𝐹𝑦₀,𝑥₀,𝑦₀=𝑦₁,𝑧₀<𝐹𝑧₀,𝑦₀,𝑥₀=𝑧₁. (2.4) Suppose now that the inequalities in (2.3) hold for some 𝑛≥2. By the mixed strict monotone property of 𝐹, together with (1.17), we have 𝑥_𝑛=𝐹𝑥_𝑛−1,𝑦_𝑛−1,𝑧_𝑛−1<𝐹𝑥_𝑛,𝑦_𝑛,𝑧_𝑛=𝑥_𝑛+1,𝑦_𝑛=𝐹𝑦_𝑛−1,𝑥_𝑛−1,𝑦_𝑛−1>𝐹𝑦_𝑛,𝑥_𝑛,𝑦_𝑛=𝑦_𝑛+1,𝑧_𝑛=𝐹𝑧_𝑛−1,𝑦_𝑛−1,𝑥_𝑛−1<𝐹𝑧_𝑛,𝑦_𝑛,𝑥_𝑛=𝑧_𝑛+1. (2.5) Thus, (2.3) holds for all 𝑛≥2.
Putting (𝑥,𝑦,𝑧)=𝐴₀ and (𝑢,𝑣,𝑟)=𝐴₁ and by Proposition 1.14, we get 𝑃𝐹^𝑛𝐴₀,𝐹^𝑛𝐴₁⟶0,as𝑛⟶+∞, (2.6) which is equivalent to 𝑃𝐴_𝑛,𝐴_𝑛+1⟶0,as𝑛⟶+∞. (2.7) Take an arbitrary 𝜀>0. It follows from (2.7) that there exists 𝑘∈ℕ such that 𝑃𝐴_𝑘,𝐴_𝑘+1<𝛿(𝜀). (2.8) Without loss of the generality, assume that 𝛿(𝜀)≤𝜀 and define the following set: Π∶=𝐴=𝑥,𝑦,𝑧∈𝑋³∶𝑃𝐹𝐴_𝑘,𝐹(𝐴)<𝜀+𝛿(𝜀),𝑥>𝑥_𝑘,𝑦≤𝑦_𝑘,𝑧>𝑧_𝑘. (2.9) We claim that 𝐹(𝐴)∈Π∀𝐴∈Π. (2.10) Take 𝐴∈Π. Then, by (2.8) and the triangle inequality (which still holds for partial metrics), we have 𝑃𝐴_𝑘,𝐹(𝐴)=max𝑝𝑥_𝑘,𝐹𝑥,𝑦,𝑧,𝑝𝑦_𝑘,𝐹𝑦,𝑥,𝑦,𝑝𝑧_𝑘𝑝𝑥,𝐹𝑧,𝑦,𝑥≤max_𝑘,𝑥_𝑘+1+𝑝𝑥_𝑘+1,𝐹𝑥,𝑦,𝑧,𝑝𝑦_𝑘,𝑦_𝑘+1+𝑝𝑦_𝑘+1,𝐹𝑦,𝑥,𝑦,𝑝𝑧_𝑘,𝑧_𝑘+1+𝑝𝑧_𝑘+1𝑝𝑥,𝐹𝑧,𝑦,𝑥=max_𝑘,𝑥_𝑘+1+𝑝𝐹𝑥_𝑘,𝑦_𝑘,𝑧_𝑘,𝐹𝑥,𝑦,𝑧,𝑝𝑦_𝑘,𝑦_𝑘+1+𝑝𝐹𝑦_𝑘,𝑥_𝑘,𝑦_𝑘,𝐹𝑦,𝑥,𝑦,𝑝𝑧_𝑘,𝑧_𝑘+1+𝑝𝐹𝑧_𝑘,𝑦_𝑘,𝑥_𝑘𝐴,𝐹𝑧,𝑦,𝑥≤𝑃_𝑘,𝐴_𝑘+1+𝑃𝐹𝐴_𝑘,𝐹(𝐴)<𝛿(𝜀)+𝑃𝐹𝐴_𝑘,𝐹(𝐴). (2.11) We consider the following two cases.Case 1 (𝑃(𝐴_𝑘,𝐴)≤𝜀). By Remark 1.12 and the definition of Π, the inequality (2.11) turns into 𝑃𝐴_𝑘,𝐹(𝐴)<𝛿(𝜀)+𝑃𝐹𝐴_𝑘,𝐹(𝐴)<𝛿(𝜀)+𝑃𝐴_𝑘(𝜀),𝐴<𝛿+𝜀. (2.12)Case 2 (𝜀<𝑃(𝐴_𝑘,𝐴)<𝛿(𝜀)+𝜀). That is, 𝜀<𝑚𝑎𝑥𝑝𝑥,𝑥_𝑘,𝑝𝑦,𝑦_𝑘,𝑝𝑧,𝑧_𝑘<𝛿(𝜀)+𝜀. (2.13) Since 𝑥>𝑥_𝑘,𝑧>𝑧_𝑘,𝑦≤𝑦_𝑘, then, by (ii), we have 𝑝𝐹𝑥,𝑦,𝑧,𝐹𝑥_𝑘,𝑦_𝑘,𝑧_𝑘𝑝𝐹𝑦<𝜀,𝑦,𝑥,𝑦,𝐹_𝑘,𝑥_𝑘,𝑦_𝑘𝑝𝐹𝑧<𝜀,𝑧,𝑦,𝑥,𝐹_𝑘,𝑦_𝑘,𝑥_𝑘<𝜀. (2.14) Hence, combining (2.14) and (2.11), we get 𝑃𝐴_𝑘,𝐹(𝐴)<𝛿(𝜀)+𝜀. (2.15) On the other hand, using (i), one can easily check that 𝐹𝑥,𝑦,𝑧>𝑥_𝑘,𝐹𝑦,𝑥,𝑦≤𝑦_𝑘,𝐹𝑧,𝑦,𝑥>𝑧_𝑘. (2.16) Hence, we conclude that (2.10) holds. By (2.8), we have that 𝐴_𝑘+1∈Π, and so, by (2.10) we get 𝐴_𝑘+1∈Π⟹𝐹𝐴_𝑘+1=𝐴_𝑘+2⟹𝐹𝐴∈Π_𝑘+2=𝐴_𝑘+3⋯∈Π⟹𝐴_𝑛∈Π∀𝑛>𝑘. (2.17) Then, for all 𝑛,𝑚>𝑘, we have 𝑃𝐴_𝑛,𝐴_𝑚≤𝑃𝐴_𝑛,𝐴_𝑘+𝑃𝐴_𝑘,𝐴_𝑚<2(𝜀+𝛿(𝜀))≤4𝜀. (2.18) By definition of 𝑃, we have lim_{𝑛,𝑚→∞}𝑝𝑥_𝑛,𝑥_𝑚=lim_{𝑛,𝑚→∞}𝑝𝑦_𝑛,𝑦_𝑚=lim_{𝑛,𝑚→∞}𝑝𝑧_𝑛,𝑧_𝑚=0. (2.19) Consequently, by definition of the metric 𝑑_𝑝, 𝑑_𝑝(𝑥,𝑦)≤2𝑝(𝑥,𝑦), so we get lim_{𝑛,𝑚→∞}𝑑_𝑝𝑥_𝑛,𝑥_𝑚=lim_{𝑛,𝑚→∞}𝑑_𝑝𝑦_𝑛,𝑦_𝑚=lim_{𝑛,𝑚→∞}𝑑_𝑝𝑧_𝑛,𝑧_𝑚=0. (2.20) Therefore, {𝑥_𝑛}, {𝑦_𝑛}, and {𝑧_𝑛} are Cauchy sequences in the metric space (𝑋,𝑑_𝑝). Since (𝑋,𝑝) is a complete partial metric space, then, by Lemma 1.2, (𝑋,𝑑_𝑝) is also a complete metric space. Hence, there exists a point (𝑥,𝑦,𝑧)∈𝑋³ such that 𝑑_𝑝𝑥_𝑛,𝑥,𝑑_𝑝𝑦_𝑛,𝑦,𝑑_𝑝𝑧_𝑛,𝑧⟶0as𝑛⟶+∞. (2.21) Again, by Lemma 1.2 and (2.19), we obtain 𝑝(𝑥,𝑥)=lim_𝑛→∞𝑝𝑥_𝑛,𝑥=lim_{𝑛,𝑚→∞}𝑝𝑥_𝑛,𝑥_𝑚𝑝=0,𝑦,𝑦=lim_𝑛→∞𝑝𝑦_𝑛,𝑦=lim_{𝑛,𝑚→∞}𝑝𝑦_𝑛,𝑦_𝑚𝑝()=0,𝑧,𝑧=lim_𝑛→∞𝑝𝑧_𝑛,𝑧=lim_{𝑛,𝑚→∞}𝑝𝑧_𝑛,𝑧_𝑚=0. (2.22) We will prove that 𝐹𝑥,𝑦,𝑧=𝑥,𝐹𝑦,𝑥,𝑦=𝑦,𝐹𝑧,𝑦,𝑥=𝑧. (2.23) To this aim, take an arbitrary 𝜀>0. Since 𝑥_𝑛=𝐹^𝑛𝑥₀,𝑦₀,𝑧₀⟶𝑥,𝑦_𝑛=𝐹^𝑛𝑦₀,𝑥₀,𝑦₀⟶𝑦,𝑧_𝑛=𝐹^𝑛𝑧₀,𝑦₀,𝑥₀⟶𝑧, (2.24) then there exist 𝑛₁,𝑛₂,𝑛₃∈ℕ such that by (2.22) 𝑝𝑥_𝑙,𝑥=𝑝𝐹^𝑙𝑥₀,𝑦₀,𝑧₀()𝑝𝑦,𝑥<𝑝𝑥,𝑥+𝜀=𝜀,_𝑞,𝑦=𝑝𝐹^𝑞𝑦₀,𝑥₀,𝑦₀𝑝𝑧,𝑦<𝑝𝑦,𝑦+𝜀=𝜀,_𝑟,𝑧=𝑝𝐹^𝑟𝑧₀,𝑦₀,𝑥₀,𝑧<𝑝(𝑧,𝑧)+𝜀=𝜀, (2.25) for all 𝑙≥𝑛₁, 𝑞≥𝑛₂, 𝑟≥𝑛₃. Now, taking 𝑛=max{𝑛₁,𝑛₂,𝑛₃} and using Remark 1.12 with the assumption 𝑥_𝑛=𝐹^𝑛𝑥₀,𝑦₀,𝑧₀<𝑥,𝑦_𝑛=𝐹^𝑛𝑦₀,𝑥₀,𝑦₀>𝑦,𝑧_𝑛=𝐹^𝑛𝑧₀,𝑦₀,𝑥₀<𝑧, (2.26) by (2.25), we get 𝑝𝑥,𝐹𝑥,𝑦,𝑧≤𝑝𝑥,𝑥_𝑛+1+𝑝𝑥_𝑛+1,𝐹𝑥,𝑦,𝑧=𝑝𝑥,𝑥_𝑛+1+𝑝𝐹^𝑛+1𝑥₀,𝑦₀,𝑧₀,𝐹𝑥,𝑦,𝑧=𝑝𝑥,𝑥_𝑛+1+𝑝𝐹𝑥_𝑛,𝑦_𝑛,𝑧_𝑛,𝐹𝑥,𝑦,𝑧<𝑝𝑥,𝑥_𝑛+1+max𝑝𝑥_𝑛,𝑥,𝑝𝑦_𝑛,𝑦,𝑝𝑧_𝑛,𝑧<2𝜀. (2.27) Analogously, we get that 𝑝𝑦,𝐹𝑦,𝑥,𝑦<2𝜀,𝑝𝑧,𝐹𝑧,𝑦,𝑥<2𝜀, (2.28) which yield that 𝐹𝑥,𝑦,𝑧=𝑥,𝐹𝑦,𝑥,𝑦=𝑦,𝐹𝑧,𝑦,𝑥=𝑧. (2.29)

Remark 2.2. Theorem 2.1 remains true if we replace (iv) with one of the following statements.
There exist 𝑥₀,𝑦₀,𝑧₀∈𝑋 such that (1)𝑥₀≤𝐹𝑥₀,𝑦₀,𝑧₀,𝑦₀>𝐹𝑦₀,𝑥₀,𝑦₀,𝑧₀<𝐹𝑧₀,𝑦₀,𝑥₀,(2)𝑥₀≤𝐹𝑥₀,𝑦₀,𝑧₀,𝑦₀>𝐹𝑦₀,𝑥₀,𝑦₀,𝑧₀≤𝐹𝑧₀,𝑦₀,𝑥₀,(3)𝑥₀≤𝐹𝑥₀,𝑦₀,𝑧₀,𝑦₀≥𝐹𝑦₀,𝑥₀,𝑦₀,𝑧₀<𝐹𝑧₀,𝑦₀,𝑥₀,(4)𝑥₀<𝐹𝑥₀,𝑦₀,𝑧₀,𝑦₀≥𝐹𝑦₀,𝑥₀,𝑦₀,𝑧₀≤𝐹𝑧₀,𝑦₀,𝑥₀,(5)𝑥₀<𝐹𝑥₀,𝑦₀,𝑧₀,𝑦₀>𝐹𝑦₀,𝑥₀,𝑦₀,𝑧₀≤𝐹𝑧₀,𝑦₀,𝑥₀. (2.30)

3. Uniqueness of Tripled Fixed Point

In this section, we will prove the uniqueness of the tripled fixed point.

Theorem 3.1. In addition to hypotheses of Theorem 2.1, assume that, for all (𝑥,𝑦,𝑧), (𝑢,𝑣,𝑟)∈𝑋³, there exists (𝑎,𝑏,𝑐)∈𝑋³ that is comparable to (𝑥,𝑦,𝑧) and (𝑢,𝑣,𝑟). Then, 𝐹 has a unique tripled fixed point.

Proof. The set of tripled fixed points of 𝐹 is not empty due to Theorem 2.1. We suppose that 𝐴=(𝑥,𝑦,𝑧),𝐴^∗=(𝑥^∗,𝑦^∗,𝑧^∗)∈𝑋³ are two tripled fixed points of 𝐹. We distinguish the following two cases.Case 1. (𝑥,𝑦,𝑧) is comparable to (𝑥^∗,𝑦^∗,𝑧^∗) with respect to the ordering in 𝑋³, where lim_𝑛→+∞𝑝𝐹^𝑛𝑥₀,𝑦₀,𝑧₀,𝑥=𝑝(𝑥,𝑥)=0,lim_𝑛→+∞𝑃𝐹^𝑛𝑦₀,𝑥₀,𝑦₀,𝑦=𝑝𝑦,𝑦=0,lim_𝑛→+∞𝑝𝐹^𝑛𝑧₀,𝑦₀,𝑥₀,𝑧=𝑝(𝑧,𝑧)=0. (3.1) Without loss of the generality, we may assume that 𝑥=𝐹𝑥,𝑦,𝑧<𝐹𝑥^∗,𝑦^∗,𝑧^∗=𝑥^∗,𝑦=𝐹𝑦,𝑥,𝑦≥𝐹𝑦^∗,𝑥^∗,𝑦^∗=𝑦^∗,𝑧=𝐹𝑧,𝑦,𝑥<𝐹𝑧^∗,𝑦^∗,𝑥^∗=𝑧^∗. (3.2) By this, definition of 𝑃, Lemma 1.3, and Remark 1.12, we have 0<𝑃𝐴,𝐴^∗=𝑃𝑥,𝑦,𝑧,𝑥^∗,𝑦^∗,𝑧^∗=max𝑝𝑥,𝑥^∗,𝑝𝑦,𝑦^∗,𝑝𝑧,𝑧^∗=max𝑝𝐹𝑥,𝑦,𝑧,𝐹𝑥^∗,𝑦^∗,𝑧^∗,𝑝𝐹𝑦,𝑥,𝑦,𝐹𝑦^∗,𝑥^∗,𝑦^∗,𝑝𝐹𝑧,𝑦,𝑥,𝐹𝑧^∗,𝑦^∗,𝑥^∗<max𝑝𝑥,𝑥^∗,𝑝𝑦,𝑦^∗,𝑝𝑧,𝑧^∗=𝑃𝐴,𝐴^∗, (3.3) which is a contradiction and therefore must be 𝐴=𝐴^∗.Case 2. (𝑥,𝑦,𝑧) is not comparable to (𝑥^∗,𝑦^∗,𝑧^∗). By assumption, there exists 𝐵=(𝑎,𝑏,𝑐)∈𝑋³ which is comparable to both 𝐴 and 𝐴^∗. Without loss of the generality, we may assume that 𝑥=𝐹𝑥,𝑦,𝑧<𝑎,𝐹𝑥^∗,𝑦^∗,𝑧^∗=𝑥^∗𝑦<𝑎,𝑦=𝐹𝑦,𝑥,𝑦≥𝑏,𝐹^∗,𝑥^∗,𝑦^∗=𝑦^∗𝑧≥𝑏,𝑧=𝐹𝑧,𝑦,𝑥<𝑐,𝐹^∗,𝑦^∗,𝑥^∗=𝑧^∗<𝑐. (3.4) From Proposition 1.14 and (3.4), we have lim_𝑛→+∞𝑃𝐹^𝑛(𝐴),𝐹^𝑛(𝐵)=0,lim_𝑛→+∞𝑃𝐹^𝑛𝐴^∗,𝐹^𝑛(𝐵)=0. (3.5) By triangle inequality, we derive 𝑃𝐴,𝐴^∗=lim_𝑛→+∞𝑃𝐹^𝑛(𝐴),𝐹^𝑛𝐴^∗≤lim_𝑛→+∞𝑃𝐹^𝑛(𝐴),𝐹^𝑛(𝐵)+lim_𝑛→+∞𝑃𝐹^𝑛(𝐵),𝐹^𝑛𝐴^∗=0. (3.6) By Lemma 1.3, we get 𝐴=𝐴^∗.

4. Results of Integral Type

Motivated by Suzuki [35] and on the same lines of [31, Theorem  3.1], one can prove the following result.

Theorem 4.1. Let (𝑋,𝑝,≤) be a partially ordered complete partial metric space, and let 𝐹∶𝑋³→𝑋 be a given mapping. Assume that there exists a function 𝜃 from [0,+∞) into itself satisfying the following:(I)𝜃(0)=0 and 𝜃(𝑡)>0 for every 𝑡>0,(II)𝜃 is nondecreasing and right continuous,(III)for every 𝜀>0, there exists 𝛿(𝜀)>0 such that 𝜀≤𝜃max𝑝(𝑥,𝑢),𝑝𝑦,𝑣,𝑝(𝑧,𝑟)<𝜀+𝛿(𝜀)⟹𝜃𝑝𝐹𝑥,𝑦,𝑧,𝐹(𝑢,𝑣,𝑟)<𝜀, (4.1) for all 𝑥≥𝑢,𝑦≤𝑣 and 𝑧≥𝑟.
Then, 𝐹 is a generalized 𝑝-Meir-Keeler type function.

The following result is an immediate consequence of Theorems 2.1 and 4.1.

Corollary 4.2. Let (𝑋,𝑝,≤) be a partially ordered complete partial metric space 𝐹∶𝑋³→𝑋 be a mapping satisfying the following hypotheses:(i)𝐹 has the mixed strict monotone property,(ii)for every 𝜀>0, there exists 𝛿(𝜀)>0 such that 𝜀≤₀^{max{𝑝(𝑥,𝑢),𝑝(𝑦,𝑣),𝑝(𝑧,𝑟)}}𝜙(𝑡)𝑑𝑡<𝜀+𝛿(𝜀)⟹₀^{𝑝(𝐹(𝑥,𝑦,𝑧),𝐹(𝑢,𝑣,𝑟))}𝜙(𝑡)𝑑𝑡<𝜀, (4.2) for all 𝑥≥𝑢,𝑦≤𝑣 and 𝑧≥𝑟, where 𝜙∶[0,+∞)→[0,+∞) is a locally integrable function satisfying ∫^𝑠₀𝜙(𝑡)𝑑𝑡>0 for all 𝑠>0,(iii)there exist 𝑥₀,𝑦₀,𝑧₀∈𝑋 such that 𝑥₀<𝐹𝑥₀,𝑦₀,𝑧₀,𝑦₀≥𝐹𝑦₀,𝑥₀,𝑦₀,𝑧₀<𝐹𝑧₀,𝑦₀,𝑥₀. (4.3) Assume that the hypotheses (a) and (b) given in Theorem 2.1 hold. Then, 𝐹 has a tripled fixed point.

To end this paper, we give the following corollary.

Corollary 4.3. Let (𝑋,𝑑,≤) be a partially ordered complete partial metric space 𝐹∶𝑋³→𝑋 be a mapping satisfying the following hypotheses:
(i)𝐹 has the mixed strict monotone property,(ii)for all, 𝑥≥𝑢,𝑦≤𝑣 and𝑧≥𝑟, ₀^{𝑝(𝐹(𝑥,𝑦,𝑧),𝐹(𝑢,𝑣,𝑟))}𝜙(𝑡)𝑑𝑡≤𝑘₀^{max{𝑝(𝑥,𝑢),𝑝(𝑦,𝑣),𝑝(𝑧,𝑟)}}𝜙(𝑡)𝑑𝑡, (4.4) where 𝑘∈(0,1) and 𝜙 is a locally integrable function from [0,+∞) into itself satisfying ∫^𝑠₀𝜙(𝑡)𝑑𝑡>0 for all 𝑠>0,(iii)there exist 𝑥₀,𝑦₀,𝑧₀∈𝑋 such that 𝑥₀<𝐹𝑥₀,𝑦₀,𝑧₀,𝑦₀≥𝐹𝑦₀,𝑥₀,𝑦₀,𝑧₀<𝐹𝑧₀,𝑦₀,𝑥₀. (4.5) Assume that the hypotheses (a) and (b) of Theorem 2.1 hold. Then, 𝐹 has a tripled fixed point.

Proof. For all 𝜀>0, we take 𝛿(𝜀)=(1/𝑘−1)𝜀 and we apply Corollary 4.2.

Remark 4.4. By taking 𝜙(𝑡)=1, we retrieve the analogous of Theorem 1.8 of Berinde and Borcut on ordered partial metric spaces (with 𝑎=𝑏=𝑐=𝑘/3). In fact, assume that (1.8) holds for 𝑎=𝑏=𝑐=𝑘/3, that is, 𝑝𝐹()≤𝑘𝑥,𝑦,𝑧,𝐹𝑢,𝑣,𝑤3𝑝(𝑥,𝑢)+𝑝𝑦,𝑣+𝑝(𝑧,𝑤) (4.6) for all 𝑥≥𝑢,𝑦≤𝑣,𝑧≥𝑤. From this inequality, we get that 𝑝𝐹𝑥,𝑦,𝑧,𝐹(𝑢,𝑣,𝑤)≤𝑘max𝑝(𝑥,𝑢),𝑝𝑦,𝑣,𝑝(𝑧,𝑤), (4.7) which corresponds to (4.4) with 𝜙(𝑡)=1. Then, we may apply Corollary 4.3.