On a Theorem of Khan in a Generalized Metric Space

Abstract

Existence and uniqueness of fixed points are established for a mapping satisfying a contractive condition involving a rational expression on a generalized metric space. Several particular cases and applications as well as some illustrative examples are given.

1. Introduction and Preliminaries

In the last decades, several attempts have been made in order to generalize the concept of metric space, and many important results have been reported. For instance, see quasi metric spaces [1], generalized quasi metric spaces [2], pseudometric spaces ([3], Chapter 2), approach spaces [4], 𝑚-spaces [5], inframetric spaces [6], and 𝐺-metric spaces [7]. Sometimes, as in [8, 9], even the very notion of generalized metric spaces (or even gms [10]) is used, but it has a different meaning. In 2000, Branciari [11] introduced the notion of generalized metric space where the triangle inequality of a metric space is replaced by a rectangular inequality involving four terms instead of three. He also extended the Banach’s contraction principle in such spaces. In 2008, Azam and Arshad [12] obtained sufficient conditions for existence of unique fixed point of Kannan type mappings defined on generalized metric spaces. Recently, Samet [13] and Sarma et al. [14] showed that some propositions in [11] are not true. Moreover, in [14], a rigorous and nice proof of the Banach’s contraction principle is presented, by assuming that the generalized metric space is Hausdorff. Afterwards, many authors studied various existence theorems of fixed points in such spaces. For more details about fixed point theory in generalized metric spaces, we refer the reader to [13, 15–24].

On the other hand, in [25] Khan proved the following fixed point theorem.

Theorem 1. Let (𝑋,𝑑) be a complete metric space and let 𝑇 be a self-mapping on 𝑋 that satisfies the following contractive condition: 𝑑𝑑()𝑑𝑑𝑇𝑥,𝑇𝑦≤𝜇𝑥,𝑇𝑥𝑥,𝑇𝑦+𝑑𝑦,𝑇𝑥𝑦,𝑇𝑦𝑑𝑥,𝑇𝑦+𝑑𝑦,𝑇𝑥 (1) for all 𝑥,𝑦∈𝑋 and for some 𝜇∈[0,1). Then 𝑇 has a unique fixed point in 𝑋.

Remark 2. In (1) if the denominator vanishes, then 𝑥=𝑇𝑦 and 𝑦=𝑇𝑥 and consequently also the numerator vanishes. Moreover, we have 𝑑(𝑇𝑥,𝑇𝑦)=𝑑(𝑦,𝑥), and so the contractive condition is not well defined.

The aim of this paper is to give a version of Theorem 1 in the setting of generalized metric spaces.

The following definitions will be needed in the sequel.

Definition 3 (see [11]). 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 (gm1)𝑑(𝑥,𝑦)=0 if and only if 𝑥=𝑦,(gm2)𝑑(𝑥,𝑦)=𝑑(𝑦,𝑥), (gm3)𝑑(𝑥,𝑦)≤𝑑(𝑥,𝑢)+𝑑(𝑢,𝑣)+𝑑(𝑣,𝑦) (the rectangular inequality).

Then 𝑑 is called generalized metric and the pair (𝑋,𝑑) is called generalized metric space (gms, for short). Note that the rectangular inequality and simple induction show that this inequality holds with 2𝑙+1 terms, where 𝑙≥1.

The next example gives a method of construction a new generalized metric space from a family of given generalized metric spaces.

Example 4. Let {(𝑋_𝑛,𝑑_𝑛)∶𝑛∈𝐽⊂ℕ} be a family of disjoint generalized metric spaces and let 𝑋=⋃{𝑋_𝑛∶𝑛∈𝐽}. Define 𝑑∶𝑋×𝑋→[0,+∞) by 𝑑(𝑥,𝑦)=𝑑(𝑦,𝑥) with 𝑑=𝑑𝑥,𝑦_𝑛𝑥,𝑦,if𝑥,𝑦∈𝑋_𝑛,𝑛∈𝐽,1,if𝑥∈𝑋_𝑛,𝑦∈𝑋_𝑚,𝑚,𝑛∈𝐽,𝑚≠𝑛. (2) Then, (𝑋,𝑑) is a gms.

By referring to the Minkowski’s inequality for sums, we give an example of a gms that is not a metric space.

Example 5. Let 𝑋=ℝ², 𝑝∈(0,1) and define the generalized metric 𝑑∶𝑋×𝑋→[0,+∞) by 𝑑𝑥,𝑦=²_𝑖∑₌₁|𝑥_𝑖−𝑦_𝑖|^𝑝1/𝑝,if𝑥_𝑖,𝑦_𝑖∈{0,1₂},_𝑖∑₌₁𝑥_𝑖−𝑦_𝑖,otherwise. (3) Clearly, (gm1) and (gm2) are obvious. Also (gm3) is immediate if at least one between 𝑥_𝑖 and 𝑦_𝑖 is not in {0,1}. Then, consider 𝐴=[0,0]^𝑇, 𝐵=[0,1]^𝑇, 𝐶=[1,0]^𝑇, and 𝐷=[1,1]^𝑇 so that we have 𝑑(𝐴,𝐵)=𝑑(𝐴,𝐶)=𝑑(𝐶,𝐷)=𝑑(𝐵,𝐷)=1 and 𝑑(𝐴,𝐷)=𝑑(𝐶,𝐵)=2^1/𝑝. Note that 𝑑(𝐶,𝐵)=2^1/𝑝<2+2^1/𝑝=𝑑(𝐶,𝐷)+𝑑(𝐴,𝐷)+𝑑(𝐴,𝐵) (4) and so (gm3) holds true. Then (𝑋,𝑑) is a gms. On the other hand, 𝑑(𝐴,𝐵)+𝑑(𝐴,𝐶)=2<2^1/𝑝=𝑑(𝐶,𝐵) since 𝑝∈(0,1). Here, the triangle inequality is violated; that is, (𝑋,𝑑) is not a metric space.

Definition 6 (see [11]). Let (𝑋,𝑑) be a gms, and let {𝑥_𝑛} be a sequence in 𝑋 and 𝑥∈𝑋. We say that {𝑥_𝑛} is gms convergent to 𝑥 if and only if 𝑑(𝑥_𝑛,𝑥)→0 as 𝑛→+∞. We denote this by 𝑥_𝑛→𝑥.

Definition 7 (see [11]). Let (𝑋,𝑑) be a gms and let {𝑥_𝑛} be a sequence in 𝑋. We say that {𝑥_𝑛} is a gms Cauchy sequence 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).

Definition 8 (see [11]). Let (𝑋,𝑑) be a gms. (𝑋,𝑑) is called a complete gms if every gms Cauchy sequence is gms convergent in 𝑋.

Definition 9. Let 𝑋 be a set and let 𝜏 be a topology on 𝑋. (𝑋,𝜏) is a Hausdorff topological space, if for every 𝑥,𝑦∈𝑋 with 𝑥≠𝑦 there are open subsets 𝑈,𝑉 of 𝑋 such that 𝑥∈𝑈, 𝑦∈𝑉, and 𝑈,𝑉 are disjoint. Equivalently, a topological space (𝑋,𝜏) is Hausdorff if and only if every convergent net in 𝑋 has a unique limit, and in particular the limit of a convergent sequence is unique.

Now, we recall an example of a complete gms that is not Hausdorff.

Example 10 (see [14]). Let 𝑋={0,2}∪{1/𝑛∶𝑛∈ℕ} and define 𝑑∶𝑋×𝑋→[0,+∞) by 𝑑(𝑥,𝑦)=𝑑(𝑦,𝑥) with 𝑑={}{}{}{}.𝑥,𝑦0,if𝑥=𝑦,1,if𝑥,𝑦∈0,2or𝑥,𝑦∈1/𝑛∶𝑛∈ℕ,𝑥≠𝑦,𝑦,if𝑥∈0,2,𝑦∈1/𝑛∶𝑛∈ℕ (5) Then (𝑋,𝑑) is a complete gms.

2. Main Results

Our main theorem is essentially inspired by Khan [25]. More precisely, we state and prove the following result.

Theorem 11. Let (𝑋,𝑑) be a complete gms and let 𝑇∶𝑋→𝑋 be a self-mapping such that 𝑑≤𝑑()𝑑𝑑𝑇𝑥,𝑇𝑦𝜆𝑑𝑥,𝑦+𝜇𝑥,𝑇𝑥𝑥,𝑇𝑦+𝑑𝑦,𝑇𝑥𝑦,𝑇𝑦𝑑,𝑥,𝑇𝑦+𝑑𝑦,𝑇𝑥𝑖𝑓𝑑𝑥,𝑇𝑦+𝑑𝑦,𝑇𝑥≠0,0,𝑖𝑓𝑑𝑥,𝑇𝑦+𝑑𝑦,𝑇𝑥=0, (6) for all 𝑥,𝑦∈𝑋 and 𝑥≠𝑦, and for some 𝜆,𝜇∈[0,1) with 𝜆+𝜇<1. Then 𝑇 has a unique fixed point in 𝑋.

Proof. Let 𝑥₀∈𝑋 be an arbitrary point. Define 𝑥_𝑛+1=𝑇𝑥_𝑛=𝑇^𝑛+1𝑥₀, for 𝑛=0,1,2,…. If 𝑥_𝑛=𝑥_𝑛−1 for some 𝑛≥1, then 𝑥_𝑛−1=𝑇𝑥_𝑛−1 and hence 𝑥_𝑛−1 is a fixed point of 𝑇. This completes the proof. Therefore, we suppose 𝑥_𝑛≠𝑥_𝑛−1 for all 𝑛≥1 and distinguish two cases.
Case??1. Assume that 𝑑(𝑥_𝑛−1,𝑇𝑥_𝑛)+𝑑(𝑥_𝑛,𝑇𝑥_𝑛−1)≠0, for 𝑛=0,1,2,…. Then 𝑑𝑥_𝑛,𝑥_𝑛+1=𝑑𝑇𝑥_𝑛−1,𝑇𝑥_𝑛≤𝜆𝑑𝑥_𝑛−1,𝑥_𝑛+𝜇𝑑𝑥_𝑛−1,𝑇𝑥_𝑛−1𝑑𝑥_𝑛−1,𝑇𝑥_𝑛+𝑑𝑥_𝑛,𝑇𝑥_𝑛−1𝑑𝑥_𝑛,𝑇𝑥_𝑛×𝑑𝑥_𝑛−1,𝑇𝑥_𝑛+𝑑𝑥_𝑛,𝑇𝑥_𝑛−1⁻¹=𝜆𝑑𝑥_𝑛−1,𝑥_𝑛+𝜇𝑑𝑥_𝑛−1,𝑥_𝑛𝑑𝑥_𝑛−1,𝑥_𝑛+1+𝑑𝑥_𝑛,𝑥_𝑛𝑑𝑥_𝑛,𝑥_𝑛+1×𝑑𝑥_𝑛−1,𝑥_𝑛+1+𝑑𝑥_𝑛,𝑥_𝑛⁻¹, (7) that is, 𝑑𝑥_𝑛,𝑥_𝑛+1≤𝜆𝑑𝑥_𝑛−1,𝑥_𝑛+𝜇𝑑𝑥_𝑛−1,𝑥_𝑛, (8) or equivalently 𝑑𝑇^𝑛𝑥₀,𝑇^𝑛+1𝑥₀≤𝜆+𝜇𝑑𝑇^𝑛−1𝑥₀,𝑇^𝑛𝑥₀. (9) Thus in general, for 𝑛≥1, we have 𝑑𝑇^𝑛𝑥₀,𝑇^𝑛+1𝑥₀≤𝜆+𝜇𝑑𝑇^𝑛−1𝑥₀,𝑇^𝑛𝑥₀≤(𝜆+𝜇)²𝑑𝑇^𝑛−2𝑥₀,𝑇^𝑛−1𝑥₀≤⋯≤(𝜆+𝜇)^𝑛𝑑𝑥₀,𝑇𝑥₀, (10) and so, since 𝜆+𝜇<1, 𝑑(𝑇^𝑛𝑥₀,𝑇^𝑛+1𝑥₀)→0 as 𝑛→+∞. Now, we consider the following two subcases.
Subcase??1.1. We assume that 𝑇^𝑚𝑥₀=𝑇^𝑛𝑥₀ for some 𝑚,𝑛∈ℕ with 𝑚≠𝑛. Letting 𝑚>𝑛, then 𝑇^𝑚−𝑛(𝑇^𝑛𝑥₀)=𝑇^𝑛𝑥₀, that is, 𝑇^𝑘𝑦=𝑦 where 𝑘=𝑚−𝑛, 𝑦=𝑇^𝑛𝑥₀. Now, if 𝑘>1, from (6) we have 𝑑𝑦,𝑇𝑦=𝑑𝑇^𝑘𝑦,𝑇^𝑘+1𝑦≤⋯≤(𝜆+𝜇)^𝑘𝑑𝑦,𝑇𝑦, (11) that implies 𝑦=𝑇𝑦, since 𝜆+𝜇<1.
Subcase??1.2. We assume that 𝑇^𝑚𝑥₀≠𝑇^𝑛𝑥₀ for all 𝑚,𝑛∈ℕ with 𝑚≠𝑛. Clearly we have 𝑑𝑇^𝑛𝑥₀,𝑇^𝑛+1𝑥₀≤(𝜆+𝜇)^𝑛𝑑𝑥₀,𝑇𝑥₀ (12) and, after routine calculations, it is also easy to obtain 𝑑𝑇^𝑛𝑥₀,𝑇^𝑛+2𝑥₀≤𝜆𝑑𝑇^𝑛−1𝑥₀,𝑇^𝑛+1𝑥₀+𝜇𝑑𝑇^𝑛−1𝑥₀,𝑇^𝑛𝑥₀𝑑𝑇^𝑛−1𝑥₀,𝑇^𝑛+2𝑥₀+𝑑𝑇^𝑛+1𝑥₀,𝑇^𝑛𝑥₀𝑑𝑇^𝑛+1𝑥₀,𝑇^𝑛+2𝑥₀×𝑑𝑇^𝑛−1𝑥₀,𝑇^𝑛+2𝑥₀+𝑑𝑇^𝑛+1𝑥₀,𝑇^𝑛𝑥₀⁻¹≤𝜆𝑑𝑇^𝑛−1𝑥₀,𝑇^𝑛+1𝑥₀+𝜇𝑑𝑇^𝑛−1𝑥₀,𝑇^𝑛𝑥₀+𝜇𝑑𝑇^𝑛+1𝑥₀,𝑇^𝑛+2𝑥₀≤𝜆𝑑𝑇^𝑛−1𝑥₀,𝑇^𝑛+1𝑥₀+𝜇𝜆+𝜇^𝑛−1𝑑𝑥₀,𝑇𝑥₀+𝜇(𝜆+𝜇)^𝑛+1𝑑𝑥₀,𝑇𝑥₀≤𝜆𝑑𝑇^𝑛−1𝑥₀,𝑇^𝑛+1𝑥₀+2𝜆+𝜇^𝑛−1𝑑𝑥₀,𝑇𝑥₀. (13) Using the previous inequality, we get 𝑑𝑇^𝑛𝑥₀,𝑇^𝑛+2𝑥₀≤𝜆²𝑑𝑇^𝑛−2𝑥₀,𝑇^𝑛𝑥₀+2𝜆(𝜆+𝜇)^𝑛−2×𝑑𝑥₀,𝑇𝑥₀+2𝜆+𝜇^𝑛−1𝑑𝑥₀,𝑇𝑥₀≤𝜆²𝑑𝑇^𝑛−2𝑥₀,𝑇^𝑛𝑥₀+4(𝜆+𝜇)^𝑛−1𝑑𝑥₀,𝑇𝑥₀ (14) and hence 𝑑𝑇^𝑛𝑥₀,𝑇^𝑛+2𝑥₀≤𝜆^𝑛𝑑𝑥₀,𝑇²𝑥₀+2𝑛𝜆+𝜇^𝑛−1𝑑𝑥₀,𝑇𝑥₀. (15) Now, if 𝑚>2 is odd, then writing 𝑚=2𝑙+1, 𝑙≥1 and using the fact that 𝑇^𝑚𝑥₀≠𝑇^𝑛𝑥₀ for 𝑚,𝑛∈ℕ with 𝑚≠𝑛, we can easily show that 𝑑𝑇^𝑛𝑥₀,𝑇^𝑛+𝑚𝑥₀≤𝑑𝑇^𝑛𝑥₀,𝑇^𝑛+1𝑥₀+𝑑𝑇^𝑛+1𝑥₀,𝑇^𝑛+2𝑥₀+⋯+𝑑𝑇^𝑛+2𝑙𝑥₀,𝑇^𝑛+2𝑙+1𝑥₀≤𝜆+𝜇^𝑛𝑑𝑥₀,𝑇𝑥₀+(𝜆+𝜇)^𝑛+1𝑑𝑥₀,𝑇𝑥₀+⋯+(𝜆+𝜇)^𝑛+2𝑙𝑑𝑥₀,𝑇𝑥₀≤(𝜆+𝜇)^𝑛𝑑𝑥1−𝜆−𝜇⁰,𝑇𝑥₀. (16) On the other hand, if 𝑚>2 is even, then writing 𝑚=2𝑙, 𝑙≥2 and using the fact that 𝑇^𝑚𝑥₀≠𝑇^𝑛𝑥₀ for 𝑚,𝑛∈ℕ with 𝑚≠𝑛, we can easily show that 𝑑𝑇^𝑛𝑥₀,𝑇^𝑛+𝑚𝑥₀≤𝑑𝑇^𝑛𝑥₀,𝑇^𝑛+2𝑥₀+𝑑𝑇^𝑛+2𝑥₀,𝑇^𝑛+3𝑥₀+⋯+𝑑𝑇^{𝑛+2𝑙−1}𝑥₀,𝑇^𝑛+2𝑙𝑥₀≤𝜆^𝑛𝑑𝑥₀,𝑇²𝑥₀+2𝑛(𝜆+𝜇)^𝑛−1𝑑𝑥₀,𝑇𝑥₀+(𝜆+𝜇)^𝑛+2𝑑𝑥₀,𝑇𝑥₀+⋯+(𝜆+𝜇)^{𝑛+2𝑙−1}𝑑𝑥₀,𝑇𝑥₀≤𝜆^𝑛𝑑𝑥₀,𝑇²𝑥₀+2𝑛(𝜆+𝜇)^𝑛−1𝑑𝑥₀,𝑇𝑥₀+(𝜆+𝜇)^𝑛+2𝑑𝑥1−𝜆−𝜇⁰,𝑇𝑥₀. (17) Thus, combining all the cases, for all 𝑚,𝑛∈ℕ we have 𝑑𝑇^𝑛𝑥₀,𝑇^𝑛+𝑚𝑥₀≤(𝜆+𝜇)^𝑛𝑑𝑥1−𝜆−𝜇⁰,𝑇𝑥₀+𝜆^𝑛𝑑𝑥₀,𝑇²𝑥₀+2𝑛(𝜆+𝜇)^𝑛−1𝑑𝑥₀,𝑇𝑥₀. (18)
Therefore, 𝑑(𝑇^𝑛𝑥₀,𝑇^𝑛+𝑚𝑥₀)→0 as 𝑛→+∞. This implies that {𝑥_𝑛} is a Cauchy sequence in 𝑋. Since 𝑋 is complete, there exists a 𝑢∈𝑋 such that 𝑥_𝑛→𝑢. If 𝑑(𝑢,𝑇^𝑛𝑥₀)+𝑑(𝑇^𝑛−1𝑥₀,𝑇𝑢)=0, for some 𝑛≥1, then 𝑢=𝑇^𝑛𝑥₀, and, by (6), we get 𝑑(𝑇𝑢,𝑇^𝑛𝑥₀)=0. It follows that 𝑢=𝑇^𝑛𝑥₀=𝑇𝑢; that is, 𝑢 is a fixed point of 𝑇. Then, without loss of generality we can assume that 𝑑(𝑢,𝑇^𝑛𝑥₀)+𝑑(𝑇^𝑛−1𝑥₀,𝑇𝑢)>0 for all 𝑛≥1. Now, if 𝑢≠𝑇𝑢, 𝑇𝑢≠𝑇^𝑛𝑥₀ and 𝑇^𝑛+1𝑥₀≠𝑢, by the rectangular inequality we have 𝑑(𝑇𝑢,𝑢)≤𝑑𝑇𝑢,𝑇^𝑛𝑥₀+𝑑𝑇^𝑛𝑥₀,𝑇^𝑛+1𝑥₀+𝑑𝑇^𝑛+1𝑥₀,𝑢≤𝜆𝑑𝑢,𝑇^𝑛−1𝑥₀+𝜇𝑑(𝑢,𝑇𝑢)𝑑𝑢,𝑇^𝑛𝑥₀+𝑑𝑇^𝑛−1𝑥₀𝑇,𝑇𝑢×𝑑^𝑛−1𝑥₀,𝑇^𝑛𝑥₀×𝑑𝑢,𝑇^𝑛𝑥₀+𝑑𝑇^𝑛−1𝑥₀,𝑇𝑢⁻¹+𝜆+𝜇^𝑛𝑑𝑥₀,𝑇𝑥₀+𝑑𝑇^𝑛+1𝑥₀,𝑢≤𝜆𝑑𝑢,𝑇^𝑛−1𝑥₀+𝜇𝑑(𝑢,𝑇𝑢)+𝑑𝑇^𝑛−1𝑥₀,𝑇^𝑛𝑥₀+𝜆+𝜇^𝑛𝑑𝑥₀,𝑇𝑥₀+𝑑𝑇^𝑛+1𝑥₀,𝑢≤𝜆𝑑𝑢,𝑇^𝑛−1𝑥₀+𝜇𝑑(𝑢,𝑇𝑢)+𝜆+𝜇^𝑛−1𝑑𝑥₀,𝑇𝑥₀+𝜆+𝜇^𝑛𝑑𝑥₀,𝑇𝑥₀+𝑑𝑇^𝑛+1𝑥₀,𝑢. (19) Letting 𝑛→+∞, since 𝑇^𝑛𝑥₀→𝑢, we obtain easily (1−𝜇)𝑑(𝑢,𝑇𝑢)≤0 that leads to a contradiction since 𝜇<1 and so must be 𝑢=𝑇𝑢.
On the other hand, if 𝑇𝑢=𝑇^𝑛𝑥₀ for every 𝑛∈𝐽⊂ℕ and 𝐽 is infinite, then lim_{𝑛→+∞,𝑛∈𝐽}𝑑(𝑇𝑢,𝑢)=lim_{𝑛→+∞,𝑛∈𝐽}𝑑𝑇^𝑛𝑥₀,𝑢=0, (20) that implies 𝑑(𝑇𝑢,𝑢)=0 and so 𝑇𝑢=𝑢. If 𝐽 is finite, then 𝑇^𝑛−1𝑥₀=𝑢 (or generally 𝑇^𝑛+1𝑥₀=𝑢) only for a finite set of values of 𝑛, and therefore (19) holds for all 𝑛 large enough. Now, we show that 𝑇 has a unique fixed point. For this, we assume that 𝑢^∗ is another fixed point of 𝑇 in 𝑋 such that 𝑑(𝑢,𝑢^∗)+𝑑(𝑢^∗,𝑢)>0. By (6) we have 𝑑𝑢,𝑢^∗=𝑑𝑇𝑢,𝑇𝑢^∗≤𝜆𝑑𝑢,𝑢^∗𝑑()𝑑+𝜇𝑢,𝑇𝑢𝑢,𝑇𝑢^∗+𝑑𝑢^∗,𝑇𝑢𝑑𝑢^∗,𝑇𝑢^∗𝑑(𝑢,𝑇𝑢^∗)+𝑑(𝑢^∗,𝑇𝑢)=𝜆𝑑𝑢,𝑢^∗𝑑()𝑑+𝜇𝑢,𝑢𝑢,𝑢^∗+𝑑𝑢^∗,𝑢𝑑𝑢^∗,𝑢^∗𝑑(𝑢,𝑢^∗)+𝑑(𝑢^∗),,𝑢 (21) or, equivalently, (1−𝜆)𝑑(𝑢,𝑢^∗)≤0 that leads to a contradiction and hence 𝑢=𝑢^∗.
Case??2. Assume that 𝑑(𝑥_𝑛−1,𝑇𝑥_𝑛)+𝑑(𝑥_𝑛,𝑇𝑥_𝑛−1)=0, for some 𝑛. By condition (6), it follows that 𝑑(𝑇𝑥_𝑛−1,𝑇𝑥_𝑛)=0 and hence 𝑥_𝑛=𝑇𝑥_𝑛. This completes the proof of the existence of a fixed point of 𝑇, say 𝑢.

The uniqueness follows as in Case 1.

As consequence of Theorem 11, we give the following corollary.

Corollary 12. Let (𝑋,𝑑) be a complete gms and let 𝑇∶𝑋→𝑋 be a self-mapping such that 𝑑𝑇^𝑛𝑥,𝑇^𝑛𝑦≤𝑑𝜆𝑑𝑥,𝑦+𝜇𝑥,𝑇^𝑛𝑥𝑑𝑥,𝑇^𝑛𝑦+𝑑𝑦,𝑇^𝑛𝑥𝑑𝑦,𝑇^𝑛𝑦𝑑𝑥,𝑇^𝑛𝑦+𝑑𝑦,𝑇^𝑛𝑥,𝑖𝑓𝑑𝑥,𝑇^𝑛𝑦+𝑑𝑦,𝑇^𝑛𝑥≠0,0,𝑖𝑓𝑑𝑥,𝑇^𝑛𝑦+𝑑𝑦,𝑇^𝑛𝑥=0, (22) for all 𝑥,𝑦∈𝑋 and 𝑥≠𝑦, for some 𝑛≥1 and for some 𝜆,𝜇∈[0,1) with 𝜆+𝜇<1. Then 𝑇 has a unique fixed point in 𝑋.

Proof. By Theorem 11, let 𝑢 be the unique fixed point of the mapping 𝑇^𝑛. Then, we have 𝑇^𝑛(𝑇𝑢)=𝑇(𝑇^𝑛𝑢)=𝑇𝑢 and so 𝑇𝑢 is a fixed point of 𝑇^𝑛. Therefore, by uniqueness of the fixed point it must be that 𝑇𝑢=𝑢.

By setting 𝜇=0 in Corollary 12, we draw the following result which can be viewed as an extension of Bryant’s theorem [26] to generalized metric spaces.

Corollary 13. Let (𝑋,𝑑) be a complete gms and let 𝑇∶𝑋→𝑋 be a self-mapping such that 𝑑𝑇^𝑛𝑥,𝑇^𝑛𝑦≤𝜆𝑑𝑥,𝑦 (23) for all 𝑥,𝑦∈𝑋 and 𝑥≠𝑦, for some 𝑛≥1 and for some 𝜆∈[0,1). Then 𝑇 has a unique fixed point in 𝑋.

Proof. Note that if 𝑑(𝑥,𝑇^𝑛𝑦)+𝑑(𝑦,𝑇^𝑛𝑥)=0, then 𝑥=𝑇^𝑛𝑦 and 𝑦=𝑇^𝑛𝑥 and so from (23) we have 𝑑𝑇^𝑛𝑥,𝑇^𝑛𝑦=𝑑𝑥,𝑦≤𝜆𝑑𝑥,𝑦. (24) Since 𝜆<1, this implies 𝑥=𝑦, but by hypothesis, condition (23) must hold true only for all 𝑥,𝑦∈𝑋 with 𝑥≠𝑦. Then, without loss of generality we can assume that 𝑑(𝑥,𝑇^𝑛𝑦)+𝑑(𝑦,𝑇^𝑛𝑥)≠0. The rest of the proof is a direct consequence of Corollary 12 and Theorem 11.

Our first example illustrates Theorem 11.

Example 14. Let 𝑋={1,−1,𝑖,−𝑖} and define the generalized metric 𝑑∶𝑋×𝑋→[0,+∞) by 𝑑(𝑥,𝑦)=𝑑(𝑦,𝑥) with 𝑑={}{}𝑥,𝑦0,if𝑥=𝑦,1,if𝑥∈1,−1,𝑦=𝑖,3,if𝑥,𝑦∈1,−1,𝑥≠𝑦,5,otherwise. (25) By routine calculations, it is easy to check that (gm3) holds and so 𝑑 is a generalized metric; obviously, (𝑋,𝑑) is complete. On the other hand, 𝑑 does not satisfy the triangle inequality. Indeed, we get 3=𝑑(1,−1)>𝑑(1,𝑖)+𝑑(𝑖,−1)=2. (26) Now, define 𝑇∶𝑋→𝑋 by 𝑇𝑥=𝑖,if𝑥≠−𝑖,1,if𝑥=−𝑖. (27) The reader can easily show that the mapping 𝑇 satisfies the contractive condition of Theorem 11 with 𝜆=1/3 and 𝜇=1/4, for example. Then, 𝑥=𝑖 is the unique fixed point of 𝑇.

The following example illustrates Corollary 12 but essentially shows the usefulness of Bryant’s theorem in respect of Banach’s contraction principle.

Example 15. Let 𝑋=[0,2] and 𝐴={1/2,2/3,3/4,4/5}. Define the generalized metric 𝑑∶𝑋×𝑋→[0,+∞) by 𝑑(𝑥,𝑦)=𝑑(𝑦,𝑥) with 𝑑12,233=𝑑4,45𝑑1=0.2,2,452=𝑑3,34𝑑1=0.3,2,342=𝑑3,45𝑑1=0.6,2,122=𝑑3,233=𝑑4,344=𝑑5,45𝑑==0,𝑥,𝑦𝑥−𝑦if𝑥,𝑦∈𝑋⧵𝐴or𝑥∈𝐴,𝑦∈𝑋⧵𝐴. (28)
By routine calculations, it is easy to check that (gm3) holds and so 𝑑 is a generalized metric. On the other hand, 𝑑 does not satisfy the triangle inequality on 𝐴. Indeed, we get 10.6=𝑑2,341≥𝑑2,232+𝑑3,34=0.5. (29) Next, we show that (𝑋,𝑑) is complete. Indeed, if there exists 𝑚∈ℕ such that 𝑥_𝑚=𝑥_𝑛 for all 𝑛≥𝑚, then the sequence {𝑥_𝑛} is Cauchy. Now, we assume that for all 𝑚∈ℕ there exists 𝑛≥𝑚 such that 𝑥_𝑚≠𝑥_𝑛. Let 𝜀<0.2 be fixed and assume that there exists 𝑛(𝜀) such that 𝑑(𝑥_𝑛,𝑥_𝑚)<𝜀. This implies that {𝑥_𝑛∶𝑛≥𝑛(𝜀)}∩𝐴 is at most singleton (each two distinct points in 𝐴 have distance greater than 𝜀, by definition). Also by definition of 𝑑, it follows that {𝑥_𝑛} is Cauchy in 𝑋, endowed with the usual metric of real numbers, and so convergent in [0,2]. Consequently, the same holds in the gms (𝑋,𝑑).
Now, define 𝑇∶𝑋→𝑋 by 1𝑇𝑥=0,if𝑥∈2,23,34,42,if𝑥=5,1,otherwise. (30) Then, for 𝑥=1/2 and 𝑦=4/5, we get 𝑑𝑇124,𝑇5()1=𝑑0,2=2≤𝜆𝑑2,45=𝜆(0.3), (31) which leads to a contradiction since 𝜆∈[0,1). Therefore, the Banach’s contraction principle is not applicable in this case. On the other hand, since 𝑇²𝑥=1, we have 0=𝑑𝑇²𝑥,𝑇²𝑦≤𝜆𝑑𝑥,𝑦, (32) which shows that 𝑇 satisfies the contractive condition of Corollary 12 and 𝑥=1 is the unique fixed point of 𝑇.

Finally, as an application of Theorem 11 we give the following result.

Corollary 16. Let (𝑋,𝑑) be a complete gms and let 𝑇∶𝑋→𝑋 be a self-mapping satisfying (6). If 𝑆∶𝑋→𝑋 is a self-mapping such that 𝑆𝑇=𝑇𝑆, then 𝑇 and 𝑆 have a unique common fixed point in 𝑋; that, is there exists 𝑢∈𝑋 such that 𝑢=𝑇𝑢=𝑆𝑢.

Proof. By Theorem 11, let 𝑢 be the unique fixed point of the mapping 𝑇. Then, we have 𝑇(𝑆𝑢)=𝑆(𝑇𝑢)=𝑆𝑢 and so 𝑆𝑢 is a fixed point of 𝑇. Therefore, by uniqueness of the fixed point it must be that 𝑆𝑢=𝑢=𝑇𝑢.

Acknowledgments

The authors would like to thank the editor and reviewers for their helpful comments. The third author is supported by Università degli Studi di Palermo, Local University Project R.S. ex 60%.