Contractive Mapping in Generalized, Ordered Metric Spaces with Application in Integral Equations

Abstract

We consider the concept of Ω-distance on a complete, partially ordered 𝐺-metric space and prove some fixed point theorems. Then, we present some applications in integral equations of our obtained results.

1. Introduction

The Banach fixed point theorem for contraction mapping has been generalized and extended in many directions [1–11]. Nieto and Rodríguez-López [10], Ran and Reurings [12], and Petrusel and Rus [13] presented some new results for contractions in partially ordered metric spaces. The main idea in [10, 12, 14] involves combining the ideas of an iterative technique in the contraction mapping principle with those in the monotone technique. Also, Mustafa and Sims [15] introduced the concept of 𝐺-metric. Some authors [16, 17] have proved some fixed point theorems in these spaces. Recently, Saadati et al. [18], using the concept of 𝐺-metric, defined an Ω-distance on complete 𝐺-metric space and generalized the concept of 𝑤-distance due to Kada et al. [19].

In this paper, we extend some recent fixed point theorems by using this concept and prove various fixed point theorems in generalized partially ordered 𝐺-metric spaces.

At first we recall some definitions and lemmas. For more information see [15–18, 20–23].

Definition 1 (see [15]). Let 𝑋 be a nonempty set. A function 𝐺∶𝑋×𝑋×𝑋→[0,∞) is called a 𝐺-metric if the following conditions are satisfied:(i)𝐺(𝑥,𝑦,𝑧)=0 if 𝑥=𝑦=𝑧 (coincidence), (ii)𝐺(𝑥,𝑥,𝑦)>0 for all 𝑥,𝑦∈𝑋, where 𝑥≠𝑦, (iii)𝐺(𝑥,𝑥,𝑧)≤𝐺(𝑥,𝑦,𝑧) for all 𝑥,𝑦,𝑧∈𝑋, with 𝑧≠𝑦, (iv)𝐺(𝑥,𝑦,𝑧)=𝐺(𝑝{𝑥,𝑦,𝑧}), where 𝑝 is a permutation of 𝑥,𝑦,𝑧 (symmetry), (v)𝐺(𝑥,𝑦,𝑧)≤𝐺(𝑥,𝑎,𝑎)+𝐺(𝑎,𝑦,𝑧) for all 𝑥,𝑦,𝑧,𝑎∈𝑋 (rectangle inequality). A 𝐺-metric is said to be symmetric if 𝐺(𝑥,𝑦,𝑦)=𝐺(𝑦,𝑥,𝑥) for all 𝑥,𝑦∈𝑋.

Definition 2. Let (𝑋,𝐺) be a 𝐺-metric space,(1)a sequence {𝑥_𝑛} in 𝑋 is said to be 𝐺-Cauchy sequence if, for each 𝜀>0, there exists a positive integer 𝑛₀ such that for all 𝑚,𝑛,𝑙≥𝑛₀, 𝐺(𝑥_𝑛,𝑥_𝑚,𝑥_𝑙)<𝜀; (2)a sequence {𝑥_𝑛} in 𝑋 is said to be 𝐺-convergent to a point 𝑥∈𝑋 if, for each 𝜀>0, there exists a positive integer 𝑛₀ such that for all 𝑚,𝑛≥𝑛₀, 𝐺(𝑥_𝑚,𝑥_𝑛,𝑥)<𝜀.

Definition 3 (see [15]). Let (𝑋,𝐺) be a 𝐺-metric space. Then a function Ω∶𝑋×𝑋×𝑋→[0,∞) is called an Ω-distance on 𝑋 if the following conditions are satisfied: (a)Ω(𝑥,𝑦,𝑧)≤Ω(𝑥,𝑎,𝑎)+Ω(𝑎,𝑦,𝑧) for all 𝑥,𝑦,𝑧,𝑎∈𝑋, (b)for any 𝑥,𝑦∈𝑋,Ω(𝑥,𝑦,⋅),Ω(𝑥,⋅,𝑦)∶𝑋→[0,∞) are lower semicontinuous, (c)for each 𝜀>0, there exists a 𝛿>0 such that Ω(𝑥,𝑎,𝑎)≤𝛿 and Ω(𝑎,𝑦,𝑧)≤𝛿 imply 𝐺(𝑥,𝑦,𝑧)≤𝜀.

Example 1 (see [18]). Let (𝑋,𝑑) be a metric space and 𝐺∶𝑋³→[0,∞) defined by 𝐺𝑥,𝑦,𝑧=max𝑑𝑥,𝑦,𝑑𝑦,𝑧,𝑑(𝑥,𝑧), (1.1) for all 𝑥,𝑦,𝑧∈𝑋. Then Ω=𝐺 is an Ω-distance on 𝑋.

Example 2 (see [18]). In 𝑋=ℝ we consider the 𝐺-metric 𝐺 defined by 𝐺=1𝑥,𝑦,𝑧3𝑥−𝑦+𝑦−𝑧+|𝑥−𝑧|, (1.2) for all 𝑥,𝑦,𝑧∈ℝ. Then Ω∶ℝ³→[0,∞) defined by Ω=1𝑥,𝑦,𝑧3|𝑧−𝑥|+𝑥−𝑦, (1.3) for all 𝑥,𝑦,𝑧∈ℝ is an Ω-distance on ℝ.

For more example see [18].

Lemma 1.1 (see [18]). Let 𝑋 be a metric space with metric 𝐺 and Ω be an Ω-distance on 𝑋. Let 𝑥_𝑛,𝑦_𝑛 be sequences in 𝑋,𝛼_𝑛,𝛽_𝑛 be sequences in [0,∞) converging to zero and let 𝑥,𝑦,𝑧,𝑎∈𝑋. Then one has the following.(1)If Ω(𝑦,𝑥_𝑛,𝑥_𝑛)≤𝛼_𝑛 and Ω(𝑥_𝑛,𝑦,𝑧)≤𝛽_𝑛 for 𝑛∈ℕ, then 𝐺(𝑦,𝑦,𝑧)<𝜀 and hence 𝑦=𝑧. (2)If Ω(𝑦_𝑛,𝑥_𝑛,𝑥_𝑛)≤𝛼_𝑛 and Ω(𝑥_𝑛,𝑦_𝑚,𝑧)≤𝛽_𝑛 for 𝑚>𝑛 then 𝐺(𝑦_𝑛,𝑦_𝑚,𝑧)→0 and hence 𝑦_𝑛→𝑧. (3)If Ω(𝑥_𝑛,𝑥_𝑚,𝑥_𝑙)≤𝛼_𝑛 for any 𝑙,𝑚,𝑛∈ℕ with 𝑛≤𝑚≤𝑙, then 𝑥_𝑛 is a G-Cauchy sequence. (4)If Ω(𝑥_𝑛,𝑎,𝑎)≤𝛼_𝑛 for any 𝑛∈ℕ then 𝑥_𝑛 is a G-Cauchy sequence.

Definition 4 (see [18]). 𝐺-metric space 𝑋 is said to be Ω-bounded if there is a constant 𝑀>0 such that Ω(𝑥,𝑦,𝑧)≤𝑀 for all 𝑥,𝑦,𝑧∈𝑋.

2. Fixed Point Theorems on Partially Ordered 𝐺-Metric Spaces

Definition 5. Suppose (𝑋,≤) is a partially ordered space and 𝑇∶𝑋→𝑋 is a mapping of 𝑋 into itself. We say that 𝑇 is nondecreasing if for 𝑥,𝑦∈𝑋, 𝑥≤𝑦⟹𝑇(𝑥)≤𝑇𝑦. (2.1)

Theorem 2.1. Let (𝑋,≤) be a partially ordered space. Suppose that there exists a 𝐺-metric on 𝑋 such that (𝑋,𝐺) is a complete 𝐺-metric space and Ω is an Ω-distance on 𝑋 such that 𝑋 is Ω-bounded. Let 𝑓∶𝑋→𝑋 and 𝑔∶𝑋→𝑋 weakly compatible and 𝑓,𝑔 be non-decreasing mapping such that (a)𝑔(𝑋)⊆𝑓(𝑋); (b)Ω(𝑔𝑥,𝑔𝑦,𝑔𝑧)≤𝑘max{Ω(𝑓𝑥,𝑓𝑦,𝑓𝑧), Ω(𝑓𝑥,𝑔𝑥,𝑓𝑧), Ω(𝑓𝑦,𝑔𝑦,𝑓𝑧),Ω(𝑓𝑥,𝑔𝑦,𝑓𝑧),Ω(𝑓𝑦,𝑔𝑥,𝑓𝑧)};for all 𝑥,𝑦,𝑧∈𝑋and0≤𝑘<1,(c)for every 𝑥∈𝑋 and 𝑦∈𝑋 with 𝑓(𝑦)≠𝑔(𝑦), inf{Ω(𝑓𝑥,𝑦,𝑓𝑥)+Ω(𝑓𝑥,𝑦,𝑔𝑥)+Ω(𝑓𝑥,𝑔𝑥,𝑦)∶𝑓(𝑥)≤𝑔(𝑥)}>0;(d)there exist 𝑥₀∈𝑋 that 𝑓(𝑥₀)≤𝑔(𝑥₀); then 𝑓 and 𝑔 have a unique common fixed point 𝑢 in 𝑋 and Ω(𝑢,𝑢,𝑢)=0.

Proof. Let 𝑥₀∈𝑋 that 𝑓(𝑥₀)≤𝑔(𝑥₀). By part (a), we can choose 𝑥₁∈𝑋 such that 𝑓(𝑥₁)=𝑔(𝑥₀). Again from part (a), we can choose 𝑥₂∈𝑋 such that 𝑓(𝑥₂)=𝑔(𝑥₁). Continuing this process we can construct sequences {𝑥_𝑛} in 𝑋 such that, 𝑦_𝑛=𝑔𝑥_𝑛=𝑓𝑥_𝑛+1𝑥,∀𝑛≥0,_𝑛≤𝑥_𝑛+1. (2.2) Now, since 𝑔 is non-decreasing mapping then, 𝑔𝑥_𝑛≤𝑔𝑥_𝑛+1,∀𝑛≥0, (2.3) so, for all 𝑠≥0, Ω𝑦_𝑛,𝑦_𝑛+1,𝑦_𝑛+𝑠=Ω𝑔𝑥_𝑛,𝑔𝑥_𝑛+1,𝑔𝑥_𝑛+𝑠≤𝑘maxΩ𝑓𝑥_𝑛,𝑓𝑥_𝑛+1,𝑓𝑥_𝑛+𝑠,Ω𝑓𝑥_𝑛,𝑔𝑥_𝑛,𝑓𝑥_𝑛+𝑠,Ω𝑓𝑥_𝑛+1,𝑔𝑥_𝑛+1,𝑓𝑥_𝑛+𝑠,Ω𝑓𝑥_𝑛,𝑔𝑥_𝑛+1,𝑓𝑥_𝑛+𝑠,Ω𝑓𝑥_𝑛+1,𝑔𝑥_𝑛,𝑓𝑥_𝑛+𝑠=𝑘maxΩ𝑦_𝑛−1,𝑦_𝑛,𝑦_{𝑛+𝑠−1},Ω𝑦_𝑛−1,𝑦_𝑛,𝑦_{𝑛+𝑠−1},Ω𝑦_𝑛,𝑦_𝑛+1,𝑦_{𝑛+𝑠−1},Ω𝑦_𝑛−1,𝑦_𝑛+1,𝑦_{𝑛+𝑠−1},Ω𝑦_𝑛,𝑦_𝑛,𝑦_{𝑛+𝑠−1}. (2.4) Then, Ω𝑦_𝑛,𝑦_𝑛+1,𝑦_𝑛+𝑠≤𝑘maxΩ𝑦_𝑛−1,𝑦_𝑛,𝑦_{𝑛+𝑠−1},Ω𝑦_𝑛,𝑦_𝑛+1,𝑦_{𝑛+𝑠−1},Ω𝑦_𝑛−1,𝑦_𝑛+1,𝑦_{𝑛+𝑠−1},Ω𝑦_𝑛,𝑦_𝑛,𝑦_{𝑛+𝑠−1}. (2.5) Now since, Ω𝑦_𝑛−1,𝑦_𝑛+1,𝑦_{𝑛+𝑠−1}≤𝑘maxΩ𝑦_𝑛−2,𝑦_𝑛,𝑦_{𝑛+𝑠−2},Ω𝑦_𝑛−2,𝑦_𝑛−1,𝑦_{𝑛+𝑠−2},Ω𝑦_𝑛,𝑦_𝑛+1,𝑦_{𝑛+𝑠−2},Ω𝑦_𝑛−2,𝑦_𝑛+1,𝑦_{𝑛+𝑠−2},Ω𝑦_𝑛,𝑦_𝑛−1,𝑦_{𝑛+𝑠−2}Ω𝑦_𝑛,𝑦_𝑛,𝑦_{𝑛+𝑠−1}≤𝑘maxΩ𝑦_𝑛−1,𝑦_𝑛−1,𝑦_{𝑛+𝑠−2},Ω𝑦_𝑛−1,𝑦_𝑛,𝑦_{𝑛+𝑠−2},Ω𝑦_𝑛−1,𝑦_𝑛,𝑦_{𝑛+𝑠−2},Ω𝑦_𝑛−1,𝑦_𝑛,𝑦_{𝑛+𝑠−2},Ω𝑦_𝑛−1,𝑦_𝑛,𝑦_{𝑛+𝑠−2}, (2.6) thus, Ω𝑦_𝑛,𝑦_𝑛+1,𝑦_𝑛+𝑠≤𝑘²maxΩ𝑦_𝑖,𝑦_𝑗,𝑦_𝑡⋮,𝑛−2≤𝑖≤𝑛,𝑛−1≤𝑗≤𝑛+1,𝑛+𝑠−2≤𝑡≤𝑛+𝑠−1≤𝑘^𝑛−1maxΩ𝑦_𝑖,𝑦_𝑗,𝑦_𝑡;1≤𝑖≤𝑛,2≤𝑗≤𝑛+1,𝑠+1≤𝑡≤𝑛+𝑠−1. (2.7) So Ω(𝑦_𝑛,𝑦_𝑛+1,𝑦_𝑛+𝑠)≤𝑘^𝑛−1𝑀_𝑛,𝑠 where 𝑀_𝑛,𝑠∶=maxΩ𝑦_𝑖,𝑦_𝑗,𝑦_𝑡,1≤𝑖≤𝑛,2≤𝑗≤𝑛+1,𝑠+1≤𝑡≤𝑛+𝑠−1. (2.8) Now, for any 𝑙>𝑚>𝑛 with 𝑚=𝑛+𝑘 and 𝑙=𝑚+𝑡(𝑘,𝑡∈ℕ), we have, lim_{𝑚,𝑛,𝑙→∞}Ω𝑦_𝑛,𝑦_𝑚,𝑦_𝑙=0. (2.9) Since 𝑋 is Ω-bounded and Ω𝑦_𝑛,𝑦_𝑚,𝑦_𝑙≤Ω𝑦_𝑛,𝑦_𝑛+1,𝑦_𝑛+1+Ω𝑦_𝑛+1,𝑦_𝑚,𝑦_𝑙≤Ω𝑦_𝑛,𝑦_𝑛+1,𝑦_𝑛+1+Ω𝑦_𝑛+1,𝑦_𝑛+2,𝑦_𝑛+2+⋯+Ω𝑦_𝑚−1,𝑦_𝑚,𝑦_𝑙≤𝑘^𝑛−1𝑀_𝑛,1+𝑘^𝑛𝑀_𝑛+1,2+⋯+𝑘^𝑚−2𝑀_{𝑚−1,𝑡+1}≤^{𝑛−𝑚+2}_𝑗=1𝑘^𝑛−𝑗𝑘𝑀≤^𝑛−11−𝑘𝑀, (2.10) so, by Part (3) of Lemma 1.1, {𝑦_𝑛} is a 𝐺-Cauchy sequence. Since 𝑋 is 𝐺-complete, {𝑦_𝑛} converges to a point 𝑦∈𝑋. Thus, for 𝜀>0 and by the lower semicontinuity of Ω, we have Ω𝑦_𝑛,𝑦_𝑚,𝑦≤liminf_𝑝→∞Ω𝑦_𝑛,𝑦_𝑚,𝑦_𝑝Ω𝑦≤𝜀,𝑚≥𝑛_𝑛,𝑦,𝑦_𝑙≤liminf_𝑝→∞Ω𝑦_𝑛,𝑦_𝑝,𝑦_𝑙≤𝜀,𝑙≥𝑛. (2.11) Assume that 𝑓𝑦≠𝑔𝑦. Since, 𝑦_𝑛=𝑓𝑥_𝑛+1=𝑔𝑥_𝑛≤𝑔𝑥_𝑛+1=𝑓𝑥_𝑛+2=𝑦_𝑛+1, (2.12) so, 𝑦_𝑛≤𝑦_𝑛+1, and, 0<infΩ𝑦_𝑛,𝑦,𝑦_𝑛+Ω𝑦_𝑛,𝑦_𝑛+1,𝑦+Ω𝑦_𝑛,𝑦,𝑦_𝑛+1≤3𝜀, (2.13) for every 𝜀>0, that is a contraction. So, we have 𝑓𝑦=𝑔𝑦. Then, by (b), Ω𝑔𝑦,𝑔𝑦,𝑔𝑦≤𝑘Ω𝑔𝑦,𝑔𝑦,𝑔𝑦, (2.14) so, Ω(𝑔𝑦,𝑔𝑦,𝑔𝑦)=0. Similarly, Ω(𝑔²𝑦,𝑔²𝑦,𝑔𝑦)=0.
Now, Ω𝑔𝑦,𝑔²𝑦,𝑔𝑦≤𝑘maxΩ𝑔𝑦,𝑔²𝑦,𝑔𝑦,Ω𝑔²,Ω𝑔𝑦,𝑔𝑦,𝑔𝑦₂𝑦,𝑔²Ω𝑦,𝑔𝑦,Ω𝑔𝑦,𝑔𝑦,𝑔𝑦=𝑘max𝑔𝑦,𝑔²𝑦,𝑔𝑦,Ω𝑔²Ω𝑔𝑦,𝑔𝑦,𝑔𝑦₂𝑦,𝑔𝑦,𝑔𝑦≤𝑘maxΩ𝑔𝑦,𝑔²𝑦,𝑔𝑦,Ω𝑔²𝑦,𝑔𝑦,𝑔𝑦. (2.15) Thus, Ω𝑔𝑦,𝑔²𝑦,𝑔𝑦=0,Ω𝑔²𝑦,𝑔𝑦,𝑔𝑦=0. (2.16) By Part (c) of Definition 3, 𝐺(𝑔²𝑦,𝑔²𝑦,𝑔𝑦)=0 and consequently 𝑔²𝑦=𝑔𝑦 which implies that 𝑔𝑦 is a fixed point for 𝑔. Now, 𝑓𝑔𝑦=𝑔𝑓𝑦=𝑔²𝑦=𝑔𝑦. (2.17) So, it is enough to put 𝑔𝑦=𝑢, then 𝑢 is a common fixed point of 𝑓 and 𝑔.
Uniqueness: Assume that there exist 𝑣∈𝑋 such that 𝑓𝑣=𝑔𝑣=𝑣. Hence, we have, Ω(𝑣,𝑣,𝑣)≤𝑘Ω(𝑣,𝑣,𝑣), (2.18) and so Ω(𝑣,𝑣,𝑣)=0. Also, Ω(𝑣,𝑣,𝑢)=0. On the other hand, Ω(){Ω()(,Ω(){Ω()(,𝑣,𝑢,𝑢≤𝑘max𝑣,𝑢,𝑢,Ω𝑢,𝑣,𝑢)}𝑢,𝑣,𝑢≤𝑘max𝑢,𝑣,𝑢,Ω𝑣,𝑢,𝑢)} (2.19) which follows that, Ω(𝑣,𝑢,𝑢)=Ω(𝑢,𝑣,𝑢)=0. Then by Part (c) of Definition 3, 𝑢=𝑣 and Ω(𝑢,𝑢,𝑢)=0.

The following corollary is a generalization of [24, Theorem 2.1].

Corollary 2.2. Let (𝑋,≤) be a partially ordered space. Suppose that there exists a 𝐺-metric on 𝑋 such that (𝑋,𝐺) is a 𝐺-metric space and Ω is an Ω-distance on 𝑋 such that 𝑋 be Ω-bounded. Let 𝑓∶𝑋→𝑋 and 𝑔∶𝑋→𝑋 be weakly compatible and 𝑓,𝑔 be a non-decreasing mapping such that (a)𝑔(𝑋)⊆𝑓(𝑋) and either 𝑓(𝑋) or 𝑔(𝑋) is complete;(b)for all 𝑥,𝑦,𝑧∈𝑋 and 0≤𝑘<1, Ω(𝑔𝑥,𝑔𝑦,𝑔𝑧)≤𝑘Ω(𝑓𝑥,𝑓𝑦,𝑓𝑧);(c)for every 𝑥∈𝑋 and 𝑦∈𝑋 with 𝑓(𝑦)≠𝑔(𝑦),  inf{Ω(𝑓𝑥,𝑦,𝑓𝑥)+Ω(𝑓𝑥,𝑦,𝑔𝑥)+Ω(𝑓𝑥,𝑔𝑥,𝑦)∶𝑓(𝑥)≤𝑔(𝑥)}>0;(d)there exist 𝑥₀∈𝑋 that 𝑓(𝑥₀)≤𝑔(𝑥₀);then 𝑓 and 𝑔 have a unique common fixed point 𝑦 in 𝑋 and Ω(𝑦,𝑦,𝑦)=0.

Definition 6 (see [25]). Let Φ be the set of all functions 𝜑 such that 𝜑∶[0,∞)→[0,∞) is a continuous and nondecreasing function with 𝜑(𝑡)<𝑡for all𝑡∈ℝ⁺ and ∑^∞_𝑛=1𝜑^𝑛(𝑡)<∞for each𝑡∈ℝ⁺. The function 𝜑 is called a growth or control function of 𝑇∶𝑋→𝑋.
It is clear that lim_𝑛→∞𝜑^𝑛(𝑡)=0,∀𝑡∈ℝ⁺,𝜑^𝑛(0)=0. (2.20)

Theorem 2.3. Let (𝑋,≤) be a partially ordered space. Suppose that there exists a 𝐺-metric on 𝑋 such that (𝑋,𝐺) is a complete 𝐺-metric space and Ω is an Ω-distance on 𝑋 and 𝑇 is a non-decreasing mapping from 𝑋 into itself. Let 𝑋 be Ω-bounded. Suppose that 𝜑∈Φ and Ω𝑇𝑥,𝑇²𝑥,𝑇𝑤≤𝜑(Ω(𝑥,𝑇𝑥,𝑤))∀𝑥≤𝑇𝑥,𝑤∈𝑋. (2.21) Also, for every 𝑥∈𝑋infΩ𝑥,𝑦,𝑥+Ω𝑥,𝑦,𝑇𝑥+Ω𝑥,𝑇²𝑥,𝑦∶𝑥≤𝑇𝑥>0, (2.22) for every 𝑦∈𝑋 with 𝑦≠𝑇𝑦. If there exists an 𝑥₀∈𝑋 with 𝑥₀≤𝑇𝑥₀, then 𝑇 has a unique fixed point. Moreover, if 𝑣=𝑇𝑣, then Ω(𝑣,𝑣,𝑣)=0.

Proof. If 𝑥₀=𝑇𝑥₀, then the proof is finished. Suppose that 𝑇𝑥₀≠𝑥₀. since 𝑥₀≤𝑇𝑥₀ and 𝑇 is non-decreasing, we obtain 𝑥₀≤𝑇𝑥₀≤𝑇²𝑥₀≤⋯≤𝑇^𝑛+1𝑥₀≤⋯ (2.23) For all 𝑛∈ℕ and 𝑡≥0, Ω𝑇^𝑛𝑥₀,𝑇^𝑛+1𝑥₀,𝑇^𝑛+𝑡𝑥₀≤𝜑Ω𝑇^𝑛−1𝑥₀,𝑇^𝑛𝑥₀,𝑇^{𝑛+𝑡−1}𝑥₀≤𝜑²Ω𝑇^𝑛−2𝑥₀,𝑇^𝑛−1𝑥₀,𝑇^{𝑛+𝑡−2}𝑥₀⋮≤𝜑^𝑛Ω𝑥₀,𝑇𝑥₀,𝑇^𝑡𝑥₀. (2.24) We claim that for 𝑚=𝑛+𝑘 and 𝑙=𝑚+𝑡(𝑘,𝑡∈ℕ) with 𝑙>𝑚>𝑛, lim_{𝑚,𝑛,𝑙→∞}Ω𝑇^𝑛𝑥₀,𝑇^𝑚𝑥₀,𝑇^𝑙𝑥₀=0. (2.25) We prove by, Ω𝑇^𝑛𝑥₀,𝑇^𝑚𝑥₀,𝑇^𝑙𝑥₀≤Ω𝑇^𝑛𝑥₀,𝑇^𝑛+1𝑥₀,𝑇^𝑛+1𝑥₀+Ω𝑇^𝑛+1𝑥₀,𝑇^𝑚𝑥₀,𝑇^𝑙𝑥₀≤Ω𝑇^𝑛𝑥₀,𝑇^𝑛+1𝑥₀,𝑇^𝑛+1𝑥₀+Ω𝑇^𝑛+1𝑥₀,𝑇^𝑛+2𝑥₀,𝑇^𝑛+2𝑥₀+⋯+Ω𝑇^𝑚−1𝑥₀,𝑇^𝑚𝑥₀,𝑇^𝑙𝑥₀≤𝜑^𝑛Ω𝑥₀,𝑇𝑥₀,𝑇𝑥₀+𝜑^𝑛+1Ω𝑥₀,𝑇𝑥₀,𝑇𝑥₀+⋯+𝜑^𝑚−2Ω𝑥₀,𝑇𝑥₀,𝑇𝑥₀+𝜑^𝑚−1Ω𝑥₀,𝑇𝑥₀,𝑇^𝑡+1𝑥₀≤𝜑^𝑛−1(𝑀)^∞_𝑛=1𝜑^𝑛(𝑀). (2.26) Since ∑^∞_𝑛=1𝜑^𝑛(𝑀)<∞, so, lim_{𝑚,𝑛→∞}Ω𝑇^𝑛𝑥₀,𝑇^𝑚𝑥₀,𝑇^𝑙𝑥₀=0. (2.27) By Part (c) of Lemma 1.1{𝑇^𝑛𝑥₀} is a 𝐺-Cauchy sequence. Since 𝑋 is 𝐺-complete, {𝑇^𝑛𝑥₀} converges to a point 𝑢∈𝑋. Let 𝑛∈ℕ be fixed. By lower semicontinuity of Ω, Ω𝑇^𝑛𝑥₀,𝑇^𝑚𝑥₀,𝑢≤liminf_𝑝→∞Ω𝑇^𝑛𝑥₀,𝑇^𝑚𝑥₀,𝑇^𝑝𝑥₀Ω𝑇≤𝜀,𝑚>𝑛,_𝑛𝑥₀,𝑢,𝑇^𝑙𝑥₀≤liminf_𝑝→∞Ω𝑇^𝑛𝑥₀,𝑇^𝑝𝑥₀,𝑇^𝑚𝑥₀≤𝜀,𝑙≥𝑛. (2.28) Assume that 𝑢≠𝑇𝑢. Since 𝑇^𝑛𝑥₀≤𝑇^𝑛+1𝑥₀, 0<infΩ𝑇^𝑛𝑥₀,𝑢,𝑇^𝑛𝑥₀+Ω𝑇^𝑛𝑥₀,𝑢,𝑇^𝑛+1𝑥₀+Ω𝑇^𝑛𝑥₀,𝑇^𝑛+2𝑥₀,𝑢∶𝑛∈ℕ≤3𝜀, (2.29) for every 𝜀>0, which is a contraction. Therefore, we have 𝑢=𝑇𝑢.
Uniqueness: let 𝑣 be another fixed point of 𝑇, then Ω(𝑢,𝑢,𝑣)=Ω𝑇𝑢,𝑇²𝑢,𝑇𝑣≤𝜑(Ω(𝑢,𝑇𝑢,𝑣))<Ω(𝑢,𝑢,𝑣), (2.30) which is a contraction. Therefore, fixed point 𝑢 is unique. Now, if 𝑣=𝑇𝑣, we have, Ω(𝑣,𝑣,𝑣)=Ω𝑇𝑣,𝑇²𝑣,𝑇³𝑣≤𝜑Ω𝑣,𝑇𝑣,𝑇²𝑣=𝜑(Ω(𝑣,𝑣,𝑣)). (2.31) So Ω(𝑣,𝑣,𝑣)=0.

Corollary 2.4. Let the assumptions of Theorem 2.3 hold and Ω𝑇^𝑚𝑥,𝑇^𝑚+1𝑥,𝑇^𝑚𝑤≤𝜑(Ω(𝑥,𝑇𝑥,𝑤))∀𝑚∈ℕ,𝑥≤𝑇𝑥,𝑤∈𝑋, (2.32) then 𝑇 has a unique fixed point.

Proof. From Theorem 2.3, 𝑇^𝑚 has a unique fixed point 𝑢. However, 𝑇𝑢=𝑇𝑇^𝑚𝑢=𝑇^𝑚+1𝑢=𝑇^𝑚𝑇𝑢, (2.33) so 𝑇𝑢 is also a fixed point of 𝑇^𝑚. Since the fixed point of 𝑇^𝑚 is unique, it must be the case that 𝑇𝑢=𝑢.

Corollary 2.5. Let the assumptions of Theorem 2.3 hold and 𝑇∶𝑋→𝑋 satisfies, Ω𝑇𝑥,𝑇²𝑥,𝑇𝑥≤𝜑(Ω(𝑥,𝑇𝑥,𝑥))∀𝑥≤𝑇𝑥. (2.34) Then 𝑇 has a unique fixed point.

Proof. Take 𝑤=𝑥, and apply Theorem 2.3.

Theorem 2.6. Let (𝑋,≤) be a partially ordered space. Suppose that there exists a 𝐺-metric on 𝑋 such that (𝑋,𝐺) is a complete 𝐺-metric space, Ω is an Ω-distance on 𝑋, and 𝑇 is a non-decreasing mapping from 𝑋 into itself. Let 𝑋 be Ω-bounded. Suppose that Ω𝑇𝑥,𝑇²𝑥,𝑇𝑤≤𝑘Ω𝑥,𝑇²𝑥,𝑇𝑤+Ω(𝑥,𝑇𝑥,𝑇𝑥), (2.35) where 𝑥≤𝑇𝑥,𝑤∈𝑋,𝑘∈[0,1/3). Also for every 𝑥∈𝑋, infΩ𝑥,𝑦,𝑥+Ω𝑥,𝑦,𝑇𝑥+Ω𝑥,𝑇²𝑥,𝑦∶𝑥≤𝑇𝑥>0, (2.36) for every 𝑦∈𝑋 with 𝑦≠𝑇𝑦. If there exists an 𝑥₀∈𝑋 with 𝑥₀≤𝑇𝑥₀, then 𝑇 has a unique fixed point say 𝑢 and Ω(𝑢,𝑢,𝑢)=0.

Proof. Let 𝑥₀∈𝑋 be an arbitrary point, and define the sequence 𝑥_𝑛 by 𝑥_𝑛=𝑇^𝑛𝑥₀. By (2.35) and for all 𝑡≥0, Ω𝑥_𝑛,𝑥_𝑛+1,𝑥_𝑛+𝑡≤𝑘Ω𝑥_𝑛−1,𝑥_𝑛+1,𝑥_𝑛+𝑡+Ω𝑥_𝑛−1,𝑥_𝑛,𝑥_𝑛. (2.37) But by Part (a) of Definition 3, Ω𝑥_𝑛−1,𝑥_𝑛+1,𝑥_𝑛+𝑡≤Ω𝑥_𝑛−1,𝑥_𝑛,𝑥_𝑛+Ω𝑥_𝑛,𝑥_𝑛+1,𝑥_𝑛+𝑡. (2.38) Hence, Ω𝑥_𝑛,𝑥_𝑛+1,𝑥_𝑛+𝑡≤𝑘2Ω𝑥_𝑛−1,𝑥_𝑛,𝑥_𝑛+Ω𝑥_𝑛,𝑥_𝑛+1,𝑥_𝑛+𝑡, (2.39) which implies, Ω𝑥_𝑛,𝑥_𝑛+1,𝑥_𝑛+𝑡≤2𝑘Ω𝑥1−𝑘^𝑛−1,𝑥_𝑛,𝑥_𝑛. (2.40) Let 𝑟=2𝑘/(1−𝑘), then 𝑟<1 and by repeated application of (2.40), we have Ω𝑥_𝑛,𝑥_𝑛+1,𝑥_𝑛+𝑡≤𝑟^𝑛Ω𝑥₀,𝑥₁,𝑥₁. (2.41) Now, for any 𝑙>𝑚>𝑛 with 𝑚=𝑛+𝑘 and 𝑙=𝑚+𝑡(𝑘,𝑡∈ℕ), we have, Ω𝑥_𝑛,𝑥_𝑚,𝑥_𝑙≤Ω𝑥_𝑛,𝑥_𝑛+1,𝑥_𝑛+1+Ω𝑥_𝑛+1,𝑥_𝑛+2,𝑥_𝑛+2+Ω𝑥_𝑛+2,𝑥_𝑛+3,𝑥_𝑛+3+⋯+Ω𝑥_𝑚−1,𝑥_𝑚,𝑥_𝑙≤𝑟^𝑛+𝑟^𝑛+1+⋯+𝑟^𝑚−1Ω𝑥₀,𝑥₁,𝑥₁≤𝑟^𝑛Ω𝑥1−𝑟⁰,𝑥₁,𝑥₁. (2.42) So, lim_{𝑚,𝑛,𝑙→∞}Ω𝑥_𝑛,𝑥_𝑚,𝑥_𝑙=0. (2.43) By Part (3) of Lemma 1.1, 𝑥_𝑛 is a 𝐺-Cauchy sequence. Since 𝑋 is 𝐺-complete, 𝑥_𝑛 converges to a point 𝑢∈𝑋. Now, similar to proving Theorem 2.1, 𝑇 has a unique fixed point and Ω(𝑢,𝑢,𝑢)=0.

Corollary 2.7. Let the assumptions of Theorem 2.6 hold and Ω𝑇^𝑚𝑥,𝑇^𝑚+2𝑥,𝑇^𝑚𝑤≤𝑘Ω𝑥,𝑇^𝑚+2𝑥,𝑇^𝑚𝑤+Ω𝑥,𝑇^𝑚𝑥,𝑇^𝑚𝑥 (2.44) where 𝑘∈[0,1/3), then 𝑇 has a unique fixed point.

Proof. The argument is similar to that used in the proof of Corollary 2.4.

3. Applications

In this section, we give an existence theorem for a solution of a class of integral equations. Denote by Λ the set of all functions 𝜆∶[0,+∞)→[0,+∞) satisfying the following hypotheses: (i)𝜆 is a Lebesgue-integrable mapping on each compact of [0,+∞), (ii)for every 𝜖>0, we have ∫^𝜖₀𝜆(𝑠)𝑑𝑠>0,(iii)‖𝜆‖<1, where ‖𝜆‖ denotes to the norm of 𝜆.

Now, we have the following results.

Theorem 3.1. Let (𝑋,≤) be a partially ordered space. Suppose that there exists a 𝐺-metric on 𝑋 such that (𝑋,𝐺) is a complete 𝐺-metric space and Ω is an Ω-distance on 𝑋 and 𝑇 is a non-decreasing mapping from 𝑋 into itself. Let 𝑋 be Ω-bounded. Suppose that Ω𝑇𝑥,𝑇²𝑥,𝑇𝑤≤₀^{Ω(𝑥,𝑇𝑥,𝑤)}𝛼(𝑠)𝑑𝑠, (3.1) where 𝛼∈Λ. Also, suppose that for every 𝑥∈𝑋infΩ𝑥,𝑦,𝑥+Ω𝑥,𝑦,𝑇𝑥+Ω𝑥,𝑇²𝑥,𝑦∶𝑥≤𝑇𝑥>0, (3.2) for every 𝑦∈𝑋 with 𝑦≠𝑇𝑦. If there exists an 𝑥₀∈𝑋 with 𝑥₀≤𝑇𝑥₀, then 𝑇 has a unique fixed point.

Proof. Define 𝜙∶[0,+∞)→[0,+∞) by 𝜙(𝑡)=∫^𝑡₀𝛼(𝑠)𝑑𝑠. It is clear that 𝜙 is nondecreasing and continuous. From (iii), we have 𝜙(𝑡)=𝜙(𝑡)=^𝑡₀𝜆(𝑠)𝑑𝑠≤^𝑡₀|𝜆(𝑠)|𝑑𝑠≤‖𝜆‖𝑡<𝑡. (3.3) Also, note that 𝜙²(𝑡)=𝜙𝜙(𝑡)≤‖𝜆‖𝜙(𝑡)≤‖𝜆‖²𝑡. (3.4) In general, we have 𝜙^𝑛(𝑡)≤‖𝜆‖^𝑛𝑡. Thus, we have ^∞_𝑛=1𝜙^𝑛(𝑡)≤^∞_𝑛=1‖𝜆‖^𝑛‖𝜆‖𝑡𝑡=1−‖𝜆‖<+∞. (3.5) Therefore 𝜙 satisfies all the hypotheses of Definition 6. By inequality (3.1), we have Ω(𝑇𝑥,𝑇²𝑥,𝑇𝑤)≤𝜙(Ω(𝑥,𝑇𝑥,𝑤). Therefore by Theorem 2.3, 𝑇 has a unique fixed point.

Now, our aim is to give an existence theorem for a solution of the following integral equation:𝑢(𝑡)=¹₀𝐾(𝑡,𝑠,𝑢(𝑠))𝑑𝑠+𝑔(𝑡),𝑡∈[0,1]. (3.6) Let 𝑋=𝐶([0,1]) be the set of all continuous functions defined on [0,1]. Define 𝐺∶𝑋×𝑋×𝑋⟶ℝ⁺ (3.7) by 𝐺𝑥,𝑦,𝑧=max𝑥−𝑦,‖𝑥−𝑧‖,𝑦−𝑧, (3.8) where ‖𝑥‖=sup{|𝑥(𝑡)|∶𝑡∈[0,1]}. Then (𝑋,𝐺) is a complete 𝐺-metric space. Let Ω=𝐺. Then Ω is an Ω-distance on 𝑋.

Define an ordered relation ≤ on 𝑋 by 𝑥≤𝑦iff𝑥(𝑡)≤𝑦(𝑡),∀𝑡∈[0,1]. (3.9) Then (𝑋,≤) is a partially ordered set. Now, we prove the following result.

Theorem 3.2. Suppose the following hypotheses hold. (a)𝐾∶[0,1]×[0,1]×ℝ⁺→ℝ⁺ and 𝑔∶ℝ⁺→ℝ⁺ are continuous. (b)𝐾 is nondecreasing in its first coordinate and 𝑔 is nondecreasing. (c)There exist a continuous function 𝐺∶[0,1]×[0,1]→[0,+∞] such that |𝐾(𝑡,𝑠,𝑢)−𝐾(𝑡,𝑠,𝑣)|≤𝐺(𝑡,𝑠)|𝑢−𝑣|, (3.10)for each comparable 𝑢,𝑣∈ℝ⁺ and each 𝑡,𝑠∈[0,1]. (d)sup_𝑡∈[0,1]∫¹₀𝐺(𝑡,𝑠)𝑑𝑠≤𝑟 for some 𝑟<1. Then the integral equation (3.6) has a solution 𝑢∈𝐶([0,1]).

Proof. Define 𝑇∶𝐶([0,1])→𝐶([0,1]) by 𝑇𝑥(𝑡)=¹₀𝐾(𝑡,𝑠,𝑥(𝑠))𝑑𝑠+𝑔(𝑡),𝑡∈[0,1]. (3.11) By hypothesis (b), we have that 𝑇 is nondecreasing.
Now, if infΩ𝑥,𝑦,𝑥+Ω𝑥,𝑦,𝑇𝑥+Ω𝑥,𝑇²𝑥,𝑦∶𝑥≤𝑇𝑥=0, (3.12) for 𝑦∈𝐶([0,1]) with 𝑦≠𝑇𝑦, then for each 𝑛∈ℕ there exists 𝑥_𝑛∈𝐶([0,1]) with 𝑥_𝑛≤𝑇𝑥_𝑛 such that Ω𝑥_𝑛,𝑦,𝑥_𝑛+Ω𝑥_𝑛,𝑦,𝑇𝑥_𝑛+Ω𝑥_𝑛,𝑇²𝑥_𝑛≤1,𝑦𝑛. (3.13) So, we have Ω𝑥_𝑛,𝑦,𝑇𝑥_𝑛=max𝑥_𝑛−𝑦,𝑥_𝑛−𝑇𝑥_𝑛,𝑦−𝑇𝑥_𝑛≤1𝑛. (3.14) Therefore, for each 𝑡∈[0,1], we have lim_𝑛→+∞𝑥_𝑛(𝑡)=𝑦(𝑡),lim_𝑛→+∞𝑇𝑥_𝑛(𝑡)=𝑦(𝑡). (3.15) By the continuity of 𝐾, we have 𝑦(𝑡)=lim_𝑛→+∞𝑇𝑥_𝑛(𝑡)=¹₀𝐾𝑡,𝑠,lim_𝑛→+∞𝑥_𝑛(𝑠)𝑑𝑠+𝑔(𝑡)=¹₀𝐾𝑡,𝑠,𝑦(𝑠)𝑑𝑠+𝑔(𝑡)=𝑇𝑦(𝑡). (3.16) Thus, we have 𝑦=𝑇𝑦, a contradiction. Thus, infΩ𝑥,𝑦,𝑥+Ω𝑥,𝑦,𝑇𝑥+Ω𝑥,𝑇²𝑥,𝑦∶𝑥≤𝑇𝑥>0. (3.17) Define 𝜙∶[0,+∞)→[0,+∞) by 𝜙(𝑡)=𝑟𝑡. For 𝑥∈𝐶([0,𝑇]) with 𝑥≤𝑇𝑥, we have Ω𝑇𝑥,𝑇²𝑥,𝑇𝑥=sup_𝑡∈[0,1]𝑇𝑥(𝑡)−𝑇²𝑥(𝑡)=sup_𝑡∈[0,1]¹₀𝐾(𝑡,𝑠,𝑥(𝑠))−𝐾(𝑡,𝑠,𝑇𝑥(𝑠))𝑑𝑠≤sup_𝑡∈[0,1]¹₀|𝐾(𝑡,𝑠,𝑥(𝑠))−𝐾(𝑡,𝑠,𝑇𝑥(𝑠))|𝑑𝑠≤sup_[]𝑡∈0,1¹₀𝐺(𝑡,𝑠)|𝑥(𝑠)−𝑇𝑥(𝑠)|𝑑𝑠≤sup_[]𝑡∈0,1|𝑥(𝑡)−𝑇𝑥(𝑡)|sup_[]𝑡∈0,1^𝑇₀𝐺()()𝑡,𝑠𝑑𝑠=Ω𝑥,𝑇𝑥,𝑥sup_[]𝑡∈0,1¹₀𝐺()()(Ω(.𝑡,𝑠𝑑𝑠≤𝑟Ω𝑥,𝑇𝑥,𝑥=𝜙𝑥,𝑇𝑥,𝑥)) (3.18) Moreover, take 𝑥₀=0, then 𝑥₀≤𝑇𝑥₀. Thus all the required hypotheses of Corollary 2.5 are satisfied. Thus there exists a solution 𝑢∈𝐶([0,𝑇]) of the integral equation (3.6).

Acknowledgment

The authors would like to thank the referee and area editor Professor Cristian Toma for providing useful suggestions and comments for the improvement of this paper.