Some Common Fixed Point Results in Rectangular Metric Spaces

Abstract

We obtain sufficient conditions for the existence of unique common fixed point of (𝜓−𝜙)-weakly contractive mappings on complete rectangular metric spaces. In the process, we generalize several fixed point results from the literature. We also give an example to illustrate our work.

1. Introduction

Fixed point theorems are very important tools in nonlinear functional analysis. Banach Contraction Mapping Principle is the most frequently cited fixed point theorem in the literature. It asserts that if 𝑋 is a complete metric space and 𝑇∶𝑋→𝑋 is a contraction, that is, there exists 𝜆∈[0,1) such that for all 𝑥,𝑦∈𝑋, 𝑑𝑇𝑥,𝑇𝑦≤𝜆𝑑𝑥,𝑦,(1) then 𝑇 has a unique fixed point. The contraction definition (1) implies that 𝑇 is uniformly continuous, which is a very strong condition. It is quite natural to ask whether the inequality (1) can be replaced with another inequality which does not force 𝑇 to be continuous. This question was answered affirmatively by Kannan [1]. A self-mapping 𝑇∶𝑋→𝑋 has a unique fixed point in a complete metric space (𝑋,𝑑) if there are nonnegative real numbers 𝛼,𝛽 with 𝛼+𝛽<1 such that the following inequality is satisfied for all 𝑥,𝑦∈𝑋: 𝑑𝑇𝑥,𝑇𝑦≤𝛼𝑑(𝑥,𝑇𝑥)+𝛽𝑑𝑦,𝑇𝑦.(2)

In 2000, Branciari [2] introduced the notion of a generalized (rectangular) metric space where the triangle inequality of a metric space was replaced by another inequality, the so-called rectangular inequality. In this paper [2], the author also extended the celebrated Banach Contraction Mapping Principle in the context of generalized metric spaces. Later, Azam and Arshad [3] obtained sufficient conditions for existence of a unique fixed point of Kannan type mappings in the framework of generalized/rectangular metric spaces. Subsequently, Azam et al. [4] proved an analog of Banach Contraction Principle in the setting of rectangular cone metric spaces. Following this trend, a number of authors focused on rectangular metric spaces and proved the existence and uniqueness of a fixed point for certain type of mappings (see e.g., [5–14]).

Recently, Di Bari and Vetro [15] obtained some common fixed point theorems for mappings satisfying a (𝜓−𝜙)-weakly contractive condition in rectangular metric spaces. In this paper, we prove several fixed point results in rectangular metric spaces that can be considered as a continuation of [15].

We recall some basic definitions and necessary results on the topic in the literature.

Definition 1. Let 𝑋 be a nonempty set and let 𝑑∶𝑋×𝑋→[0,+∞) be a mapping such that for all 𝑥,𝑦∈𝑋 and for all distinct points 𝑢,𝑣∈𝑋, each of them different from 𝑥 and 𝑦, one has 𝑑()𝑑,()𝑑()()()𝑥,𝑦=0iff𝑥=𝑦,RM1𝑥,𝑦=𝑑𝑦,𝑥RM2𝑥,𝑦≤𝑑𝑥,𝑢+𝑑𝑢,𝑣+𝑑𝑣,𝑦(therectangularinequality).RM3
Then, the map 𝑑 is called rectangular (generalized) metric. The pair (𝑋,𝑑) is called a rectangular (generalized) metric space.

To avoid confusion, we prefer to use the term “rectangular metric space” for the spaces under consideration in this paper, because there are some other spaces that are also called generalized metric spaces. We abbreviate a rectangular metric spaces with RMS.

Definition 2 (see [2]). Let (𝑋,𝑑) be a RMS.(i)A sequence {𝑥_𝑛} is called RMS convergent to 𝑥∈𝑋 if and only if 𝑑(𝑥_𝑛,𝑥)→0 as 𝑛→+∞. In this case, we use the notation 𝑥_𝑛→𝑥.(ii)A sequence {𝑥_𝑛} in 𝑋 is called a RMS Cauchy if and only if for each 𝜖>0, there exists a natural number 𝑁(𝜖) such that 𝑑(𝑥_𝑛,𝑥_𝑚)<𝜖 for all 𝑛>𝑚>𝑁(𝜖).(iii)A RMS (𝑋,𝑑) is called a RMS complete if every RMS Cauchy sequence is RMS convergent in 𝑋.

Definition 3. Let 𝐹 and 𝑔 be self-mappings of a nonempty set 𝑋.(i)A point 𝑥∈𝑋 is said to be a common fixed point of 𝐹 and 𝑔 if 𝑥=𝐹𝑥=𝑔𝑥.(ii)A point 𝑥∈𝑋 is called a coincidence point of 𝐹 and 𝑔 if 𝐹𝑥=𝑔𝑥. And if 𝑤=𝐹𝑥=𝑔𝑥, then 𝑤 is said to be a point of coincidence of 𝐹 and 𝑔.(iii)The mappings 𝐹,𝑔∶𝑋→𝑋 are said to be weakly compatible if they commute at their coincidence point that is, 𝐹𝑔𝑥=𝑔𝐹𝑥 whenever 𝑔𝑥=𝐹𝑥.

Lemma 4. Let 𝑋 be a nonempty set. Suppose that the mappings 𝐹,𝑔∶𝑋→𝑋 have a unique coincidence point 𝑧 in 𝑋. If 𝐹 and 𝑔 are weakly compatible, then 𝐹 and 𝑔 have a unique common fixed point.

Proof. Let 𝑧∈𝑋 be the coincidence point of 𝐹,𝑔∶𝑋→𝑋, that is, 𝐹𝑧=𝑔𝑧=𝑡.(3) Since 𝐹 and 𝑔 are weakly compatible, we observe that 𝐹𝑧=𝑔𝑧⇒𝐹𝑔𝑧=𝑔𝐹𝑧⟺𝐹𝑡=𝑔𝑡.(4)𝐹,𝑔∶𝑋→𝑋 have a unique coincidence point, then 𝑧=𝑡. Hence, we have 𝐹𝑧=𝑔𝑧=𝑧 by (3).

Let Ψ denote all functions 𝜙∶[0,∞)→[0,∞) such that(a)𝜙 is continuous,(b)𝜙(𝑡)=0 if and only if 𝑡=0.

2. Main Result

We start this section with the following theorem.

Theorem 5. Let (𝑋,𝑑) be a Hausdorff RMS and let 𝐹,𝑔∶𝑋→𝑋 be self-mappings such that 𝐹𝑋⊂𝑔𝑋. Assume that (𝑔𝑋,𝑑) is a complete RMS. Suppose that the following condition holds: 𝜓𝑑𝐹𝑥,𝐹𝑦≤𝜓𝑀𝑔𝑥,𝑔𝑦−𝜙𝑀𝑔𝑥,𝑔𝑦,(5) for all 𝑥,𝑦∈𝑋 and 𝜓,𝜙∈Ψ, where 𝜓 is nondecreasing and 𝑀𝑔𝑥,𝑔𝑦=max𝑑𝑔𝑥,𝑔𝑦,𝑑𝑔𝑥,𝐹𝑥,𝑑𝑔𝑦,𝐹𝑦.(6) Then 𝐹 and 𝑔 have a unique coincidence point in 𝑋. Moreover, if 𝐹 and 𝑔 are weakly compatible, then 𝐹 and 𝑔 have a unique common fixed point.

Proof. We first prove that the coincidence point of 𝑔 and 𝐹 is unique if it exists. Let 𝑧 and 𝑤 be coincidence points of 𝑔 and 𝐹. Thus, there exists some 𝑥,𝑦∈𝑋 such that 𝑤=𝐹𝑥=𝑔𝑥 and 𝑧=𝐹𝑦=𝑔𝑦. By (5), we derive that 𝜓(𝑑(𝑑𝑀𝑀,𝑤,𝑧))=𝜓𝐹𝑥,𝐹𝑦≤𝜓𝑔𝑥,𝑔𝑦−𝜙𝑔𝑥,𝑔𝑦(7) where 𝑀𝑑().𝑔𝑥,𝑔𝑦=max𝑔𝑥,𝑔𝑦,𝑑𝑔𝑥,𝐹𝑥,𝑑𝑔𝑦,𝐹𝑦=𝑑𝑔𝑥,𝑔𝑦=𝑑𝑤,𝑧(8) Thus, we conclude that 𝑧=𝑤 by (7).
Remember that 𝑔 and 𝐹 are weakly compatible. Since 𝑧 is the unique coincidence point of 𝑔 and 𝐹, the point 𝑧 is the unique common fixed point of 𝑔 and 𝐹 by Lemma 4.
Now, we will prove the existence of a coincidence point of 𝑔 and 𝐹. Let 𝑥₀ be an arbitrary point. Since 𝐹𝑋⊂𝑔𝑋, we define two iterative sequences {𝑥_𝑛} and {𝑦_𝑛} in 𝑋 as follows: 𝑦_𝑛=𝑔𝑥_𝑛+1=𝐹𝑥_𝑛(9) for all 𝑛=0,1,2,…. If 𝑦_𝑛=𝑦_𝑛+1 then clearly 𝐹 and 𝑔 have a coincidence point in 𝑋. Indeed, 𝑦_𝑛=𝑔𝑥_𝑛+1=𝐹𝑥_𝑛=𝑔𝑥_𝑛+2=𝐹𝑥_𝑛+1=𝑦_𝑛+1 and 𝑥_𝑛+1 is the desired point. Thus, we assume that 𝑦_𝑛≠𝑦_𝑛+1, that is, 𝑑(𝑦_𝑛,𝑦_𝑛+1)>0 for all 𝑛=0,1,…. Moreover, if 𝑦_𝑛=𝐹𝑥_𝑛=𝐹𝑥_𝑛+𝑝=𝑦_𝑛+𝑝, then we choose 𝑥_𝑛+𝑝+1=𝑥_𝑛+1, for all 𝑛≥0.
We assert that lim_𝑛→∞𝑑𝑦_𝑛,𝑦_𝑛+1=0,lim_𝑛→∞𝑑𝑦_𝑛,𝑦_𝑛+2=0.(10) Now from (5), we have 𝜓𝑑𝑦_𝑛,𝑦_𝑛+1=𝜓𝑑𝐹𝑥_𝑛,𝐹𝑥_𝑛+1≤𝜓𝑀𝑔𝑥_𝑛,𝑔𝑥_𝑛+1−𝜙𝑀𝑔𝑥_𝑛,𝑔𝑥_𝑛+1,(11) where 𝑀𝑔𝑥_𝑛,𝑔𝑥_𝑛+1=max𝑑𝑔𝑥_𝑛,𝑔𝑥_𝑛+1,𝑑𝑔𝑥_𝑛,𝐹𝑥_𝑛,𝑑𝑔𝑥_𝑛+1,𝐹𝑥_𝑛+1=max𝑑𝑔𝑥_𝑛,𝑔𝑥_𝑛+1,𝑑𝑔𝑥_𝑛,𝑔𝑥_𝑛+1,𝑑𝑔𝑥_𝑛+1,𝑔𝑥_𝑛+2=max𝑑𝑦_𝑛−1,𝑦_𝑛,𝑑𝑦_𝑛−1,𝑦_𝑛,𝑑𝑦_𝑛,𝑦_𝑛+1=max𝑑𝑦_𝑛−1,𝑦_𝑛,𝑑𝑦_𝑛,𝑦_𝑛+1.(12) If 𝑀(𝑔𝑥_𝑛,𝑔𝑥_𝑛+1)= 𝑑(𝑦_𝑛,𝑦_𝑛+1), then we have 𝜓𝑑𝑦_𝑛,𝑦_𝑛+1≤𝜓𝑑𝑦_𝑛,𝑦_𝑛+1−𝜙𝑑𝑦_𝑛,𝑦_𝑛+1,(13) which implies that 𝜙(𝑑(𝑦_𝑛,𝑦_𝑛+1))=0, and hence 𝑑(𝑦_𝑛,𝑦_𝑛+1)=0. Then 𝑦_𝑛=𝑦_𝑛+1, which contradicts with the initial assumption. Thus, we have 𝑀(𝑔𝑥_𝑛,𝑔𝑥_𝑛+1)=𝑑(𝑦_𝑛−1,𝑦_𝑛), and hence 𝜓𝑑𝑦_𝑛,𝑦_𝑛+1≤𝜓𝑑𝑦_𝑛−1,𝑦_𝑛−𝜙𝑑𝑦_𝑛−1,𝑦_𝑛≤𝜓𝑑𝑦_𝑛−1,𝑦_𝑛.(14) Since 𝜓 is nondecreasing, then 𝑑(𝑦_𝑛,𝑦_𝑛+1)≤𝑑(𝑦_𝑛−1,𝑦_𝑛) for all 𝑛≥0, that is, the sequence {𝑑(𝑦_𝑛,𝑦_𝑛+1)} is nonincreasing and bounded below. Hence, it converges to a positive number, say 𝑟>0. Taking the limit as 𝑛→∞ in (14), we get 𝜓(𝑟)≤𝜓(𝑟)−𝜙(𝑟),(15) which leads to 𝜙(𝑟)=0, and hence 𝑟=0. Thus, lim_𝑛→∞𝑑𝑦_𝑛,𝑦_𝑛+1=0.(16) In this step, we show that the second limit in (5) is also 0. To prove this claim, we set 𝑥=𝑥_𝑛−1 and 𝑦=𝑥_𝑛+1 in (5) and get that 𝜓𝑑𝑦_𝑛,𝑦_𝑛+2=𝜓𝑑𝐹𝑥_𝑛,𝐹𝑥_𝑛+2≤𝜓𝑀𝑔𝑥_𝑛,𝑔𝑥_𝑛+2−𝜙𝑀𝑔𝑥_𝑛,𝑔𝑥_𝑛+2,(17) where 𝑀𝑔𝑥_𝑛,𝑔𝑥_𝑛+2=max𝑑𝑔𝑥_𝑛,𝑔𝑥_𝑛+2,𝑑𝑔𝑥_𝑛,𝐹𝑥_𝑛,𝑑𝑔𝑥_𝑛+2,𝐹𝑥_𝑛+2=max𝑑𝑔𝑥_𝑛,𝑔𝑥_𝑛+2,𝑑𝑔𝑥_𝑛,𝑔𝑥_𝑛+1,𝑑𝑔𝑥_𝑛+2,𝑔𝑥_𝑛+3=max𝑑𝑦_𝑛−1,𝑦_𝑛+1,𝑑𝑦_𝑛−1,𝑦_𝑛,𝑑𝑦_𝑛+1,𝑦_𝑛+2.(18) We consider all possible cases for 𝑀(𝑔𝑥_𝑛,𝑔𝑥_𝑛+2). If 𝑀(𝑔𝑥_𝑛,𝑔𝑥_𝑛+2)=𝑑(𝑦_𝑛+1,𝑦_𝑛+2), then we have 𝜓𝑑𝑦_𝑛,𝑦_𝑛+2≤𝜓𝑑𝑦_𝑛+1,𝑦_𝑛+2−𝜙𝑑𝑦_𝑛+1,𝑦_𝑛+2.(19) Letting 𝑛→∞ in (19), the right hand side of (19) tends to 0. Hence, lim_𝑛→∞𝜓(𝑑(𝑦_𝑛,𝑦_𝑛+2))=0. Since, 𝜓 is continuous, we find that lim_𝑛→∞𝑑(𝑦_𝑛,𝑦_𝑛+2)=0. For the case 𝑀(𝑔𝑥_𝑛,𝑔𝑥_𝑛+2)=𝑑(𝑦_𝑛−1,𝑦_𝑛), we get analogously lim_𝑛→∞𝑑(𝑦_𝑛,𝑦_𝑛+2)=0.
Let us consider the last case, that is, 𝑀(𝑔𝑥_𝑛,𝑔𝑥_𝑛+2)=𝑑(𝑦_𝑛−1,𝑦_𝑛+1). The inequality (17) turns into 0≤𝜓𝑑𝑦_𝑛,𝑦_𝑛+2≤𝜓𝑑𝑦_𝑛−1,𝑦_𝑛+1−𝜙𝑑𝑦_𝑛−1,𝑦_𝑛+1≤𝜓𝑑𝑦_𝑛−1,𝑦_𝑛+1.(20) Therefore, the sequence {𝑑(𝑦_𝑛,𝑦_𝑛+2)} is non-increasing and bounded below. Hence, the sequence {𝑑(𝑦_𝑛,𝑦_𝑛+2)} converges to a number, 𝑠≥0. Taking limit as 𝑛→∞ in (20), we get 0≤𝜓(𝑑(𝑠))≤𝜓(𝑠)−𝜙(𝑠),(21) which implies that 𝜙(𝑠)=0, and hence 𝑠=0. In other words, lim_𝑛→∞𝑑𝑦_𝑛,𝑦_𝑛+2=0.(22) Suppose that 𝑦_𝑛≠𝑦_𝑚 for all 𝑚≠𝑛 and prove that {𝑦_𝑛} is a RMS Cauchy sequence. If possible, let {𝑦_𝑛} be not a Cauchy sequence. Then there exists 𝜖>0 for which we can find subsequences {𝑦_𝑚𝑘} and {𝑦_𝑛𝑘} of {𝑦_𝑛} with 𝑛_𝑘>𝑚_𝑘≥𝑘 such that 𝑑𝑦_𝑚𝑘,𝑦_𝑛𝑘≥𝜖.(23) Furthermore, corresponding to 𝑚_𝑘, we can choose 𝑛_𝑘 in such a way that it is the smallest integer with 𝑛_𝑘>𝑚_𝑘 and satisfying (23). Then, 𝑑𝑦_𝑚𝑘,𝑦_𝑛𝑘−1<𝜖.(24) Using (23), (24), and the rectangular inequality (RM3), we have 𝜖≤𝑑𝑦_𝑚𝑘,𝑦_𝑛𝑘≤𝑑𝑦_𝑛𝑘,𝑦_𝑛𝑘−2+𝑑𝑦_𝑛𝑘−2,𝑦_𝑛𝑘−1+𝑑𝑦_𝑛𝑘−1,𝑦_𝑚𝑘≤𝑑𝑦_𝑛𝑘,𝑦_𝑛𝑘−2+𝑑𝑦_𝑛𝑘−2,𝑦_𝑛𝑘−1+𝜖.(25) Taking limit as 𝑘→∞ in (23) and using (16), (22) we get lim_𝑘→∞𝑑𝑦_𝑚𝑘,𝑦_𝑛𝑘=𝜖.(26) Again, using the rectangular inequality (RM3), we obtain 𝑑𝑦_𝑛𝑘,𝑦_𝑚𝑘−𝑑𝑦_𝑚𝑘,𝑦_𝑚𝑘−1−𝑑𝑦_𝑛𝑘−1,𝑦_𝑛𝑘≤𝑑𝑦_𝑛𝑘−1,𝑦_𝑚𝑘−1≤𝑑𝑦_𝑛𝑘−1,𝑦_𝑛𝑘+𝑑𝑦_𝑛𝑘,𝑦_𝑚𝑘+𝑑𝑦_𝑚𝑘,𝑦_𝑚𝑘−1.(27) Letting 𝑘→∞ in (27), and by using (16) and (22) we get lim_𝑘→∞𝑑𝑦_𝑛𝑘−1,𝑦_𝑚𝑘−1=𝜖.(28) Now, we substitute 𝑥=𝑥_𝑛𝑘 and 𝑦=𝑥_𝑚𝑘 in (5). Consider 𝜓𝑑𝑦_𝑛𝑘,𝑦_𝑚𝑘=𝜓𝑑𝐹𝑥_𝑛𝑘,𝐹𝑥_𝑚𝑘≤𝜓𝑀𝑔𝑥_𝑛𝑘,𝑔𝑥_𝑚𝑘−𝜙𝑀𝑔𝑥_𝑛𝑘,𝑔𝑥_𝑚𝑘(29) where 𝑀𝑔𝑥_𝑛𝑘,𝑔𝑥_𝑚𝑘=max𝑑𝑔𝑥_𝑛𝑘,𝑔𝑥_𝑚𝑘,𝑑𝑔𝑥_𝑛𝑘,𝐹𝑥_𝑛𝑘,𝑑𝑔𝑥_𝑚𝑘,𝐹𝑥_𝑚𝑘=max𝑑𝑦_𝑛𝑘−1,𝑦_𝑚𝑘−1,𝑑𝑦_𝑛𝑘−1,𝑦_𝑛𝑘,𝑑𝑦_𝑚𝑘−1,𝑦_𝑚𝑘.(30) Clearly, as 𝑘→∞, we have 𝑀𝑔𝑥_𝑛𝑘,𝑔𝑥_𝑚𝑘→max{𝜖,0,0}=𝜖.(31) Then letting 𝑘→∞ in (29), we have 0≤𝜓(𝜖)≤𝜓(𝜖)−𝜙(𝜖).(32) This implies that 𝜙(𝜖)=0,hence𝜖=0,(33) which contradicts the fact that 𝜖>0. Thus, {𝑦_𝑛} is a RMS Cauchy sequence. Since (𝑔𝑋,𝑑) is RMS complete, there exists 𝑧∈𝑔𝑋 such that 𝑦_𝑛→𝑧 as 𝑛→∞. Let 𝑦∈𝑋 such that 𝑔𝑦=𝑧. Applying the inequality (5), with 𝑥=𝑥_𝑛, we obtain 𝜓𝑑𝐹𝑥_𝑛,𝐹𝑦≤𝜓𝑀𝑔𝑥_𝑛,𝑔𝑦−𝜙𝑀𝑔𝑥_𝑛,𝑔𝑦,(34) where 𝑀𝑔𝑥_𝑛,𝑔𝑦=max𝑑𝑔𝑥_𝑛,𝑔𝑦,𝑑𝑔𝑥_𝑛,𝐹𝑥_𝑛𝑑,𝑑𝑔𝑦,𝐹𝑦=max𝑔𝑥_𝑛,𝑔𝑦,𝑑𝑔𝑥_𝑛,𝑔𝑥_𝑛+1,𝑑𝑔𝑦,𝐹𝑦.(35) Now, if 𝑀(𝑔𝑥_𝑛,𝑔𝑦)=𝑑(𝑔𝑥_𝑛,𝑔𝑦) or 𝑀(𝑔𝑥_𝑛,𝑔𝑦)=𝑑(𝑔𝑥_𝑛,𝑔𝑥_𝑛+1), we have 𝑑𝐹𝑥_𝑛,𝐹𝑦≤𝑑𝑔𝑥_𝑛𝑑,𝑔𝑦or𝐹𝑥_𝑛,𝐹𝑦≤𝑑𝑔𝑥_𝑛,𝑔𝑥_𝑛+1,(36) since 𝜓 is nondecreasing. In either case, letting 𝑛→∞, we get 𝑔𝑥_𝑛+1= 𝐹𝑥_𝑛→𝐹𝑦. Since 𝑋 is Hausdorff, we deduce that 𝑔𝑦=𝐹𝑦. If, on the other hand, 𝑀(𝑔𝑥_𝑛,𝑔𝑦)=𝑑(𝑔𝑦,𝐹𝑦), then taking limit as 𝑛→∞ in 𝜓𝑑𝐹𝑥_𝑛,𝐹𝑦≤𝜓𝑑𝑔𝑦,𝐹𝑦−𝜙𝑑𝑔𝑦,𝐹𝑦,(37) we get 𝜙(𝑑(𝑔𝑦,𝐹𝑦))=0, hence 𝑑(𝑔𝑦,𝐹𝑦)=0, that is, 𝑔𝑦=𝐹𝑦. Let 𝑧=𝑔𝑦=𝐹𝑦. Then 𝑧 is a point of coincidence of 𝐹 and 𝑔. Suppose that there exists 𝑛,𝑝∈ℕ such that 𝑦_𝑛=𝑦_𝑛+𝑝, we can choose in such a way that it is the smallest positive integer satisfying 𝑦_𝑛=𝑦_𝑛+𝑝. We aim to prove that 𝑝=1, then 𝑔𝑥_𝑛+1=𝐹𝑥_𝑛=𝐹𝑥_𝑛+1=𝑦_𝑛+1,(38) and so 𝑦_𝑛+1 is a point of coincidence of 𝑔 and 𝐹. Assume that 𝑝>1. This implies that 𝑑(𝑦_{𝑛+𝑝−1},𝑦_𝑛+𝑝)>0. Now we have 𝜓𝑑𝑦_𝑛,𝑦_𝑛+1=𝜓𝑑𝑦_𝑛+𝑝,𝑦_𝑛+𝑝+1=𝜓𝑑𝐹𝑥_𝑛+𝑝,𝐹𝑥_𝑛+𝑝+1≤𝜓𝑀𝑔𝑥_𝑛+𝑝,𝑔𝑥_𝑛+𝑝+1−𝜙𝑀𝑔𝑥_𝑛+𝑝,𝑔𝑥_𝑛+𝑝+1,(39) where 𝑀𝑔𝑥_𝑛+𝑝,𝑔𝑥_𝑛+𝑝+1=max𝑑𝑔𝑥_𝑛+𝑝,𝑔𝑥_𝑛+𝑝+1,𝑑𝑔𝑥_𝑛+𝑝,𝐹𝑥_𝑛+𝑝,𝑑𝑔𝑥_𝑛+𝑝+1,𝐹𝑥_𝑛+𝑝+1=max𝑑𝑔𝑥_𝑛+𝑝,𝑔𝑥_𝑛+𝑝+1,𝑑𝑔𝑥_𝑛+𝑝,𝑔𝑥_𝑛+𝑝+1,𝑑𝑔𝑥_𝑛+𝑝+1,𝑔𝑥_𝑛+𝑝+2=max𝑑𝑔𝑥_𝑛+𝑝,𝑔𝑥_𝑛+𝑝+1,𝑑𝑔𝑥_𝑛+𝑝+1,𝑔𝑥_𝑛+𝑝+2=max𝑑𝑦_{𝑛+𝑝−1},𝑦_𝑛+𝑝,𝑑𝑦_𝑛+𝑝,𝑦_𝑛+𝑝+1=max𝑑𝑦_𝑛−1,𝑦_𝑛,𝑑𝑦_𝑛,𝑦_𝑛+1.(40) If 𝑀(𝑔𝑥_𝑛+𝑝,𝑔𝑥_𝑛+𝑝+1)=𝑑(𝑦_𝑛,𝑦_𝑛+1), then we have 𝜓𝑑𝑦_𝑛,𝑦_𝑛+1=𝜓𝑑𝑦_𝑛+𝑝,𝑦_𝑛+𝑝+1<𝜓𝑑𝑦_𝑛,𝑦_𝑛+1,(41) which is a contradiction. Thus 𝑝=1.

Corollary 6. Let (𝑋,𝑑) be a Hausdorff and complete RMS and let 𝐹,𝑔∶𝑋→𝑋 be self-mappings such that 𝐹𝑋⊆𝑔𝑋 satisfying 𝑑𝐹𝑥,𝐹𝑦≤𝑘max𝑑𝑔𝑥,𝑔𝑦,𝑑𝑔𝑥,𝐹𝑥,𝑑𝑔𝑦,𝐹𝑦(42) for all 𝑥,𝑦∈𝑋 and 0≤𝑘<1. Then 𝐹 and 𝑔 have a unique common fixed point in 𝑋.

Proof. Let 𝜓(𝑡)=𝑡 and 𝜙(𝑡)=(1−𝑘)𝑡. Then by Theorem 5,  𝐹 and 𝑔 have a unique common fixed point.

Corollary 7. Let (𝑋,𝑑) be a Hausdorff and complete RMS and let 𝐹,𝑔∶𝑋→𝑋 be self-mappings such that 𝐹𝑋⊆𝑔𝑋 satisfying 𝑑𝐹𝑥,𝐹𝑦≤𝑘𝑑𝑔𝑥,𝑔𝑦+𝑑𝑔𝑥,𝐹𝑥+𝑑𝑔𝑦,𝐹𝑦,(43) for all 𝑥,𝑦∈𝑋 and 0≤𝑘<1. Then 𝐹 and 𝑔 have a unique common fixed point in 𝑋.

Proof. It is obvious that 𝑑𝑑.𝑔𝑥,𝑔𝑦+𝑑𝑔𝑥,𝐹𝑥+𝑑𝑔𝑦,𝐹𝑦≤3max𝑔𝑥,𝑔𝑦,𝑑𝑔𝑥,𝐹𝑥,𝑑𝑔𝑦,𝐹𝑦(44) Let 𝜓(𝑡)=𝑡 and 𝜙(𝑡)=(1−3𝑘)𝑡. Then by Theorem 5, 𝐹 and 𝑔 have a unique common fixed point.

Corollary 8. Let (𝑋,𝑑) be a Hausdorff and complete RMS and let 𝐹,𝑔∶𝑋→𝑋 be self-mappings such that 𝐹𝑋⊆𝑔𝑋 satisfying 𝑑𝐹𝑥,𝐹𝑦≤𝑀𝑔𝑥,𝑔𝑦−𝜙𝑀𝑔𝑥,𝑔𝑦,(45) for all 𝑥,𝑦∈𝑋 where 𝑀𝑑.𝑔𝑥,𝑔𝑦=max𝑔𝑥,𝑔𝑦,𝑑𝑔𝑥,𝐹𝑥,𝑑𝑔𝑦,𝐹𝑦(46) Then 𝐹 and 𝑔 have a unique common fixed point.

Proof. Let 𝜓(𝑡)=𝑡. Then by Theorem 5, 𝐹 and 𝑔 have a unique common fixed point.

Theorem 9. Let (𝑋,𝑑) be a Hausdorff and complete RMS and let 𝐹,𝑔∶𝑋→𝑋 be self-mappings such that 𝐹𝑋⊆𝑔𝑋 satisfying 𝜓𝑑𝐹𝑥,𝐹𝑦≤𝜓𝑀𝑔𝑥,𝑔𝑦−𝜙𝑀𝑔𝑥,𝑔𝑦,(47) for all 𝑥,𝑦∈𝑋 and 𝜓,𝜙∈Ψ, where 𝜓 is nondecreasing and 𝑀𝑑𝑔𝑥,𝑔𝑦=max𝑔𝑥,𝑔𝑦,𝑑𝑔𝑦,𝐹𝑦1+𝑑𝑔𝑥,𝐹𝑥.1+𝑑𝑔𝑦,𝑔𝑦(48) Then 𝐹 and 𝑔 have a unique common fixed point.

Proof. Let 𝑥₀∈𝑋 be an arbitrary point. Since 𝐹𝑋⊆𝑔𝑋, we define the sequence {𝑥_𝑛}⊂𝑋 as follows 𝑔𝑥_𝑛=𝐹𝑥_𝑛−1 for all 𝑛≥1. Assume that 𝑔𝑥_𝑛≠𝑔𝑥_𝑛+1=𝐹𝑥_𝑛 for all 𝑛≥0. Now from (5),we get 𝜓𝑑𝑔𝑥_𝑛,𝑔𝑥_𝑛+1=𝜓𝑑𝐹𝑥_𝑛−1,𝐹𝑥_𝑛≤𝜓𝑀𝑔𝑥_𝑛−1,𝑔𝑥_𝑛−𝜙𝑀𝑔𝑥_𝑛−1,𝑔𝑥_𝑛,(49) where 𝑀𝑔𝑥_𝑛−1,𝑔𝑥_𝑛=max𝑑𝑔𝑥_𝑛−1,𝑔𝑥_𝑛,𝑑𝑔𝑥_𝑛,𝐹𝑥_𝑛1+𝑑𝑔𝑥_𝑛−1,𝐹𝑥_𝑛−11+𝑑𝑔𝑥_𝑛−1,𝑔𝑥_𝑛𝑀𝑔𝑥_𝑛−1,𝑔𝑥_𝑛=max𝑑𝑔𝑥_𝑛−1,𝑔𝑥_𝑛,𝑑𝑔𝑥_𝑛,𝑔𝑥_𝑛+1.(50) The rest of the proof is the same as the proof of Theorem 5.

By Λ, we denote the class of functions 𝑓∶[0,∞)→[0,∞) satisfying the following.(a)𝑓 is Lebesgue integrable function on each compact subset of [0,∞),(b)∫^𝜀₀𝑓(𝑠)𝑑𝑠>0 for any 𝜀>0.

Theorem 10. Let (𝑋,𝑑) be a Hausdorff RMS and let 𝐹,𝑔∶𝑋→𝑋 be self-mappings such that 𝐹𝑋⊂𝑔𝑋. Assume that (𝑔𝑋,𝑑) is a complete RMS and that the following condition holds: ₀^{𝑑(𝐹𝑥,𝐹𝑦)}𝑓(𝑟)≤𝑑𝑟_{0𝑀(𝑔𝑥,𝑔𝑦)}𝑓(𝑟)𝑑𝑟−₀^{𝑀(𝑔𝑥,𝑔𝑦)}ℎ(𝑟)𝑑𝑟(51) for all 𝑥,𝑦∈𝑋 and 𝑓,ℎ∈Λ where 𝑀𝑔𝑥,𝑔𝑦=max𝑑𝑔𝑥,𝑔𝑦,𝑑𝑔𝑥,𝐹𝑥,𝑑𝑔𝑦,𝐹𝑦.(52) Then 𝐹 and 𝑔 have a unique common fixed point.

Proof. Let 𝜓(𝑟)=∫^𝑟₀𝑓(𝑣)𝑑𝑣 and 𝜙(𝑟)=∫^𝑟₀ℎ(𝑣)𝑑𝑣. Then 𝜓 and 𝜙 are function in Ψ. By Theorem 5,  𝐹 and 𝑔 have a unique common fixed point.

Theorem 11. Let (𝑋,𝑑) be a Hausdorff RMS and let 𝐹,𝑔∶𝑋→𝑋 be self-mappings such that 𝐹𝑋⊂𝑔𝑋. Assume that (𝑔𝑋,𝑑) is a complete RMS and that the following condition holds: ₀^{𝑑(𝐹𝑥,𝐹𝑦)}𝑓(𝑠)𝑑𝑠≤𝜆₀^{𝑀(𝑔𝑥,𝑔𝑦)}𝑓(𝑠)𝑑𝑠(53) for all 𝑥,𝑦∈𝑋 and 𝑓∈Λ and some 0≤𝜆<1, where 𝑀𝑔𝑥,𝑔𝑦=max𝑑𝑔𝑥,𝑔𝑦,𝑑𝑔𝑥,𝐹𝑥,𝑑𝑔𝑦,𝐹𝑦.(54) Then 𝐹 and 𝑔 have a unique common fixed point.

Proof. Let ℎ(𝑟)=(1−𝜆)𝑓(𝑟). Then by Theorem 10, 𝐹 and 𝑔 have a unique common fixed point.

Theorem 12. Let (𝑋,𝑑) be a Hausdorff RMS and let 𝐹,𝑔∶𝑋→𝑋 be a self-mappings such that 𝐹𝑋⊂𝑔𝑋. Assume that (𝑔𝑋,𝑑) is a complete RMS and that the following condition holds: ₀^{𝑑(𝐹𝑥,𝐹𝑦)}𝑓(𝑟)≤𝑑𝑟_{0𝑀(𝑔𝑥,𝑔𝑦)}𝑓(𝑟)𝑑𝑟−₀^{𝑀(𝑔𝑥,𝑔𝑦)}ℎ(𝑟)𝑑𝑟(55) for all 𝑥,𝑦∈𝑋 and 𝑓,ℎ∈Λ where 𝑀𝑑𝑔𝑥,𝑔𝑦=max𝑔𝑥,𝑔𝑦,𝑑𝑔𝑦,𝐹𝑦1+𝑑𝑔𝑥,𝐹𝑥.1+𝑑𝑔𝑦,𝑔𝑦(56) Then 𝐹 and 𝑔 have a unique common fixed point.

Proof. Let 𝜓(𝑟)=∫^𝑟₀𝑓(𝑣)𝑑𝑣 and 𝜙(𝑟)=∫^𝑟₀ℎ(𝑣)𝑑𝑣. Then 𝜓 and 𝜙 are function in Ψ. By Theorem 9,  𝐹 and 𝑔 have a unique common fixed point.

Example 13. Let 𝑋=𝐴∪𝐵, where 𝐴={1/2,2/3,3/4,4/5} and 𝐵=[1,3]. Define the generalized metric 𝑑 on 𝑋 as follows: 𝑑12,233=𝑑4,45𝑑1=0.2,(57)2,452=𝑑3,34𝑑1=0.3,(58)2,342=𝑑3,451=0.6,(59)2,122=𝑑3,233=𝑑4,344=𝑑5,45𝑑==0,(60)𝑥,𝑦𝑥−𝑦if𝑥,𝑦∈𝐵or𝑥∈𝐴,𝑦∈𝐵or𝑥∈𝐵,𝑦∈𝐴.(61)
It is easy to show that 𝑑 does not satisfy the triangle inequality on 𝐴. Indeed, 10.6=𝑑2,341≥𝑑2,232+𝑑3,34=0.2+0.3=0.5.(62) Thus (RM3) holds, so 𝑑 is a rectangular metric. Notice that (𝑋|_𝐵,𝑑) is usual metric space, and hence it is Hausdorff. On the other hand, each singleton is closed and open in (𝑋|_𝐴,𝑑), and hence (𝑋,𝑑) is Hausdorff rectangular metric space.
Let 𝐹,𝑔∶𝑋→𝑋 be defined as follows: 4𝐹𝑥=5[],3if𝑥∈1,341if𝑥∈2,23,34,234if𝑥=5,2𝑔𝑥=3[],3if𝑥∈1,341if𝑥∈2,34,452if𝑥=3,124if𝑥=5.(63) Define 𝜓(𝑡)=𝑡 and 𝜙(𝑡)=𝑡/3. Then 𝐹 and 𝑔 satisfy the condition of Theorem 5 and have a unique common fixed point of 𝑋, that is, 𝑥=3/4.

Acknowledgment

The authors thank the referees for their appreciation, valuable comments, and suggestions.