Common Fixed Point Theorems for 𝛼-𝜓-Contractive Type Mappings

Abstract

Recently, Samet et al. (2012) introduced the notion of 𝛼-𝜓-contractive type mappings. They established some fixed point theorems for these mappings in complete metric spaces. In this paper, we introduce the notion of a coupled 𝛼-𝜓-contractive mapping and give a common fixed point result about the mapping. Also, we give a result of common fixed points of some coupled self-maps on complete metric spaces satisfying a contractive condition.

1. Introduction

We know fixed point theory has many applications and was extended by several authors from different views (see, e.g., [1–33]). Recently, Samet et al. introduced the notion of 𝛼-𝜓-contractive type mappings [3]. Denote with Ψ the family of nondecreasing functions 𝜓∶[0,∞)→[0,∞) such that ∑^∞_𝑛=1𝜓^𝑛(𝑡)<∞ for all 𝑡>0, where 𝜓^𝑛 is the 𝑛th iterate of 𝜓. It is known that 𝜓(𝑡)<𝑡 for all 𝑡>0 and 𝜓∈Ψ [3]. Let (𝑋,𝑑) be a metric space, 𝑇 a self map on 𝑋, 𝜓∈Ψ and 𝛼∶𝑋×𝑋→[0,∞). Then, 𝑇 is called a 𝛼-𝜓-contractive mapping whenever 𝛼(𝑥,𝑦)𝑑(𝑇𝑥,𝑇𝑦)≤𝜓(𝑑(𝑥,𝑦)) for all 𝑥,𝑦∈𝑋. Also, we say that 𝑇 is 𝛼-admissible whenever 𝛼(𝑥,𝑦)≥1 implies 𝛼(𝑇𝑥,𝑇𝑦)≥1 for all 𝑥,𝑦∈𝑋 [3]. Also, we say that 𝑋 has the property (𝐵_𝛼) if {𝑥_𝑛} is a sequence in 𝑋 such that 𝛼(𝑥_𝑛,𝑥_𝑛+1)≥1 for all 𝑛≥1 and 𝑥_𝑛→𝑥, then 𝛼(𝑥_𝑛,𝑥)≥1 for all 𝑛≥1. Let (𝑋,𝑑) be a complete metric space and let 𝑇 a 𝛼-admissible 𝛼-𝜓-contractive mapping on 𝑋. Suppose that there exists 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1. If 𝑇 is continuous or 𝑋 has the property (𝐵_𝛼), then 𝑇 has a fixed point (see [3]; Theorems 2.1 and 2.2). Finally, we say that 𝑋 has the property (𝐻_𝛼) whenever for each 𝑥,𝑦∈𝑋 there exists 𝑧∈𝑋 such that 𝛼(𝑥,𝑧)≥1 and 𝛼(𝑦,𝑧)≥1. If 𝑋 has the property (𝐻_𝛼) in the Theorems  2.1 and 2.2, then 𝑋 has a unique fixed point ([3]; Theorem 2.3). It is considerable that the results of Samet et al. generalize similar ordered results in the literature (see the results of the third section [3]). The aim of this paper is introducing the notion of generalized coupled 𝛼-𝜓-contractive mappings and give a common fixed point result about the mappings.

Definition 1. Let  ϝ  the family of functions 𝑓∶[0,∞)→ℝ satisfy:(i)𝑓(0)=0 and 𝑓(𝑡)>0 for all 𝑡∈(0,+∞);(ii)𝑓 is continuous;(iii)𝑓 is nondecreasing on [0,+∞);(iv)𝑓(𝑡₁+𝑡₂)≤𝑓(𝑡₁)+𝑓(𝑡₂) for all 𝑡₁,𝑡₂∈(0,+∞).

Definition 2. Let Ψ the family of functions 𝜓∶[0,+∞)→[0,+∞) satisfy(𝜓₁)𝜓 is nondecreasing;(𝜓₂)∑^+∞_𝑛=1𝜓^𝑛(𝑡)<∞, for all 𝑡∈(0,+∞).

These functions are known in the literature as (𝑐)-comparison functions. It is easily proved that if 𝜓 is a (𝑐)-comparison function, then 𝜓(𝑡)<𝑡 for all 𝑡>0.

Definition 3. Let (𝑋,𝑑) be a metric space, and let 𝑇,𝑆∶𝑋→𝑋 with given coupled mappings. Let 𝛼∶𝑋×𝑋→[0,+∞), 𝑓∈ϝ, 𝜓∈Ψ, and let 𝑚𝑑(),1𝐴𝑥,𝐵𝑦=max𝑥,𝑦,𝑑𝑥,𝐴𝑥,𝑑𝑦,𝐵𝑦2𝑑𝑥,𝐵𝑦+𝑑𝑦,𝐴𝑥, (1) for all coupled mappings 𝐴,𝐵∶𝑋→𝑋 and 𝑥,𝑦∈𝑋. One says that 𝑇,  𝑆 are generalized coupled 𝛼-𝜓-contractive mappings whenever 𝛼𝑓𝑑𝑓𝑚,𝛼𝑓𝑑𝑓𝑚,𝑥,𝑦𝑇𝑥,𝑆𝑦≤𝜓𝑇𝑥,𝑆𝑦𝑥,𝑦𝑆𝑥,𝑇𝑦≤𝜓𝑆𝑥,𝑇𝑦 (2) for all 𝑥,𝑦∈𝑋.

Definition 4. Let 𝑇,𝑆∶𝑋→𝑋, and let 𝛼∶𝑋×𝑋→[0,+∞). One says that 𝑇,  𝑆 are coupled 𝛼-admissible if 𝛼𝑥,𝑦≥1⟹𝛼𝑇𝑥,𝑆𝑦≥1or𝛼𝑆𝑥,𝑇𝑦≥1 (3) for all 𝑥,𝑦∈𝑋.

Definition 5. Let (𝑋,𝑑) be a compete metric space. For two subsets 𝐴,  𝐵 of 𝑋, one marks 𝐴⪯𝐵, if for all 𝑎∈𝐴, there exists 𝑏∈𝐵 such that 𝑎⪯𝑏.

Definition 6. A partial metric on a nonempty set 𝑋 is a function 𝜌∶𝑋×𝑋→ℝ⁺ such that for all 𝑥,𝑦,𝑧∈𝑋:(𝜌₁)𝑥=𝑦⇔𝜌(𝑥,𝑥)=𝜌(𝑥,𝑦)=𝜌(𝑦,𝑦);(𝜌₂)𝜌(𝑥,𝑥)≤𝜌(𝑥,𝑦);(𝜌₃)𝜌(𝑥,𝑦)=𝜌(𝑦,𝑥);(𝜌₄)𝜌(𝑥,𝑦)≤𝜌(𝑥,𝑧)+𝜌(𝑧,𝑦)−𝜌(𝑧,𝑧).

A partial metric space is a pair (𝑋,𝜌) such that 𝑋 is a nonempty set, and 𝜌 is a partial metric on 𝑋. It is clear that if 𝜌(𝑥,𝑦)=0, then from (𝜌₁) and (𝜌₂), 𝑥=𝑦. But if 𝑥=𝑦, 𝜌(𝑥,𝑦) may not be 0. A basic example of a partial metric is the pair (ℝ⁺,𝜌), where 𝜌(𝑥,𝑦)=max{𝑥,𝑦} for all 𝑥,𝑦∈ℝ⁺. If 𝜌 is a partial metric on 𝑋, then the function 𝜌^𝑠∶𝑋×𝑋→ℝ⁺ given by 𝜌^𝑠(𝑥,𝑦)=2𝜌(𝑥,𝑦)−𝜌(𝑥,𝑥)−𝜌(𝑦,𝑦) is a metric on 𝑋.

Example 7. Let 𝑋=ℝ endowed with the standard metric 𝑑(𝑥,𝑦)=|𝑥−𝑦| for all 𝑥,𝑦∈ℝ. Define the coupled mappings 𝑇,𝑆∶𝑋→𝑋 by 𝑇𝑥=𝑥+1,if𝑥𝑥>1,₂,if0≤𝑥≤1,0,if𝑥<0,𝑆𝑥=𝑥−1,if𝑥𝑥>1,₃,if0≤𝑥≤1,0,if𝑥<0. (4) We define the mapping 𝛼∶𝑋×𝑋→[0,+∞) by 𝛼𝑥,𝑦=1,if𝑥,𝑦∈[0,1],0,otherwise. (5) If 𝛼𝑥,𝑦≥1⟹𝑥,𝑦∈[0,1]⟹𝑥²,𝑦³∈[]𝑥0,1⟹𝛼𝑇𝑥,𝑆𝑦=𝛼²,𝑦³=1. (6) Similarly, 𝛼(𝑥,𝑦)≥1⇒𝛼(𝑆𝑥,𝑇𝑦)≥1. This shows that 𝑇,  𝑆 are coupled 𝛼-admissible.

Lemma 8. Let (𝑋,𝑑) be a metric space. Suppose that 𝑇,𝑆∶𝑋→𝑋 are generalized coupled 𝛼-𝜓-contractive mappings. Then, ℱ(𝑇)=ℱ(𝑆).

Proof. We first show that any fixed point of 𝑇 is also a fixed point of 𝑆 and conversely. Define 𝛼(𝑥,𝑦)=1 for all 𝑥,𝑦∈𝑋. Since ℱ(𝑇)≠ℱ(𝑆), we may assume there exists 𝑥^∗∈𝑋 such that 𝑥^∗∈ℱ(𝑇), but 𝑥^∗∉ℱ(𝑆). Since 𝑑(𝑥^∗,𝑆𝑥^∗)>0, we have 𝑚𝑇𝑥^∗,𝑆𝑥^∗=max𝑑𝑥^∗,𝑥^∗,𝑑𝑥^∗,𝑇𝑥^∗,𝑑𝑥^∗,𝑆𝑥^∗,12𝑑𝑥^∗,𝑆𝑥^∗+𝑑𝑥^∗,𝑇𝑥^∗=𝑑𝑥^∗,𝑆𝑥^∗,𝑓𝑑𝑥^∗,𝑆𝑥^∗=𝑓𝑑𝑇𝑥^∗,𝑆𝑥^∗≤𝛼𝑥^∗,𝑥^∗𝑓𝑑𝑇𝑥^∗,𝑆𝑥^∗≤𝜓𝑓𝑚𝑇𝑥^∗,𝑆𝑥^∗≤𝜓𝑓𝑑𝑥^∗,𝑆𝑥^∗<𝑓𝑑𝑥^∗,𝑆𝑥^∗. (7) This contradiction establishes that ℱ(𝑇)⊆ℱ(𝑆). A similar argument establishes the reverse containment, and therefore ℱ(𝑇)=ℱ(𝑆).

2. Main Results

Now, we are ready to state and prove our main results.

Theorem 9. Let (𝑋,𝑑) be a complete metric space. Suppose that 𝑇,𝑆∶𝑋→𝑋 are generalized coupled 𝛼-𝜓-contractive mappings and satisfy the following conditions:(i)𝑇,  𝑆 are coupled 𝛼-admissible;(ii) there exists 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1; or(ii)^∗ there exists 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1;(iii)𝑇 or 𝑆 is continuous.Then 𝑇,  𝑆 have common fixed point 𝑥^∗∈𝑋. Further, for each 𝑥₀∈𝑋, the iterated sequence {𝑥_𝑛} with 𝑥_2𝑛+1=𝑇𝑥_2𝑛 and 𝑥_2𝑛+2=𝑆𝑥_2𝑛+1 converges to the common fixed point of 𝑇,𝑆.

Proof. By Lemma 8, we have ℱ(𝑇)=ℱ(𝑆). Let 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1. Define the sequence {𝑥_𝑛} in 𝑋 by 𝑥_2𝑛+1=𝑇𝑥_2𝑛 and 𝑥_2𝑛+2=𝑆𝑥_2𝑛+1 for all 𝑛∈ℕ₀. If 𝑥_𝑛0=𝑥_𝑛0+1 for some 𝑛₀>1, then 𝑥^∗=𝑥_𝑛0 are a common fixed point for 𝑇,  𝑆. So, we can assume that 𝑥_2𝑛≠𝑇𝑥_2𝑛 and 𝑥_2𝑛+1≠𝑆𝑥_2𝑛+1 for all 𝑛∈ℕ₀. Since 𝑇,  𝑆 are coupled 𝛼-admissible, we have 𝛼𝑥₀,𝑥₁=𝛼𝑥₀,𝑇𝑥₀≥1⟹𝛼𝑇𝑥₀,𝑆𝑥₁=𝛼𝑥₁,𝑥₂≥1⟹𝛼𝑆𝑥₁,𝑇𝑥₂=𝛼𝑥₂,𝑥₃≥1. (8) Inductively, we have 𝛼𝑥_2𝑛,𝑥_2𝑛+1𝛼𝑥≥1,_2𝑛+1,𝑥_2𝑛+2≥1, (9) for all 𝑛∈ℕ₀. We obtain 𝑓𝑑𝑥_2𝑛+1,𝑥_2𝑛+2=𝑓𝑑𝑇𝑥_2𝑛,𝑆𝑥_2𝑛+1≤𝛼𝑥_2𝑛,𝑥_2𝑛+1𝑓𝑑𝑇𝑥_2𝑛,𝑆𝑥_2𝑛+1≤𝜓𝑓𝑚𝑇𝑥_2𝑛,𝑆𝑥_2𝑛+1. (10) Now, 𝑚𝑇𝑥_2𝑛,𝑆𝑥_2𝑛+1=max𝑑𝑥_2𝑛,𝑥_2𝑛+1,𝑑𝑥_2𝑛,𝑇𝑥_2𝑛,𝑑𝑥_2𝑛+1,𝑆𝑥_2𝑛+1,12𝑑𝑥_2𝑛,𝑆𝑥_2𝑛+1+𝑑𝑥_2𝑛+1,𝑇𝑥_2𝑛=max𝑑𝑥_2𝑛,𝑥_2𝑛+1,𝑑𝑥_2𝑛,𝑥_2𝑛+1,𝑑𝑥_2𝑛+1,𝑥_2𝑛+2,12𝑑𝑥_2𝑛,𝑥_2𝑛+2+𝑑𝑥_2𝑛+1,𝑥_2𝑛+1≤max𝑑𝑥_2𝑛,𝑥_2𝑛+1,𝑑𝑥_2𝑛+1,𝑥_2𝑛+2,12𝑑𝑥_2𝑛,𝑥_2𝑛+1+𝑑𝑥_2𝑛+1,𝑥_2𝑛+2𝑑𝑥+0=max_2𝑛,𝑥_2𝑛+1,𝑑𝑥_2𝑛+1,𝑥_2𝑛+2 (11) if max𝑑𝑥_2𝑛,𝑥_2𝑛+1,𝑑𝑥_2𝑛+1,𝑥_2𝑛+2=𝑑𝑥_2𝑛+1,𝑥_2𝑛+2. (12) So, in general, 𝑓𝑑𝑥_2𝑛+1,𝑥_2𝑛+2≤𝜓𝑓𝑑𝑥_2𝑛+1,𝑥_2𝑛+2<𝑓𝑑𝑥_2𝑛+1,𝑥_2𝑛+2, (13) which is a contradiction since 𝑑(𝑥_2𝑛+1,𝑥_2𝑛+2)>0. Thus, 𝑓𝑑𝑥_2𝑛+1,𝑥_2𝑛+2≤𝜓𝑓𝑑𝑥_2𝑛,𝑥_2𝑛+1. (14) Similarly, if 𝑓𝑑𝑥_2𝑛,𝑥_2𝑛+1≤𝜓𝑓𝑑𝑥_2𝑛−1,𝑥_2𝑛, (15) we have 𝑓𝑑𝑥_𝑛,𝑥_𝑛+1≤𝜓𝑓𝑑𝑥_𝑛−1,𝑥_𝑛, (16) for all 𝑛∈ℕ₀. By induction, we get 𝑓𝑑𝑥_𝑛,𝑥_𝑛+1≤𝜓𝜓𝑓𝑑𝑥_𝑛−2,𝑥_𝑛−1≤⋯≤𝜓^𝑛𝑓𝑑𝑥₀,𝑥₁, (17) for all 𝑛∈ℕ₀. Fix 𝜖>0, and let 𝑛(𝜖)∈ℕ₀ such that ^∞_{𝑛=𝑛(𝜖)}𝜓^𝑛𝑓𝑑𝑥₀,𝑥₁<𝜖. (18) Let 𝑛,𝑚∈ℕ₀ with 𝑚>𝑛>𝑛(𝜖). Using the triangle inequality, we obtain 𝑓𝑑𝑥_𝑛,𝑥_𝑚≤^𝑚−1_𝑘=𝑛𝑓𝑑𝑥_𝑘,𝑥_𝑘+1≤^𝑚−1_𝑘=𝑛𝜓^𝑘𝑓𝑑𝑥₀,𝑥₁≤^∞_{𝑛=𝑛(𝜖)}𝜓^𝑛𝑓𝑑𝑥₀,𝑥₁<𝜖. (19) Thus we proved that {𝑥_𝑛} is a Cauchy sequence in the metric space (𝑋,𝑑).
Since (𝑋,𝑑) is a complete metric space, there exists 𝑥^∗∈𝑋 such that 𝑥_𝑛→𝑥^∗ as 𝑛→+∞. From the continuity of 𝑇, it follows that 𝑥_2𝑛+1=𝑇𝑥_2𝑛→𝑇𝑥^∗ as 𝑛→+∞, then 𝑥^∗=𝑇𝑥^∗. Similarly if 𝑆 is continuous, we have 𝑥^∗=𝑆𝑥^∗.

Corollary 10. Let (𝑋,𝑑) be a complete metric space. Suppose that 𝑇∶𝑋→𝑋 is a generalized 𝛼-𝜓-contractive mapping and satisfies the following conditions:(i)𝑇 is 𝛼-admissible;(ii) there exists 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1; or(ii)^∗ there exists 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1;(iii)𝑇 is continuous.Then, 𝑇 has a fixed point 𝑥^∗∈𝑋. Further, for each 𝑥₀∈𝑋, the iterated sequence {𝑥_𝑛} with 𝑥_𝑛+1=𝑇𝑥_𝑛 converges to the fixed point of 𝑇.

Example 11. Let 𝑋=ℝ endowed with the standard metric 𝑑(𝑥,𝑦)=|𝑥−𝑦| for all 𝑥,𝑦∈ℝ. Define the coupled mappings 𝑇,𝑆∶𝑋→𝑋 by 3𝑇𝑥=2𝑥−2,if𝑥𝑥>1,2,if0≤𝑥≤1,0,if5𝑥<0,𝑆𝑥=2𝑥−3,if𝑥𝑥>1,2,if0≤𝑥≤1,0,if𝑥<0. (20) We define the mapping 𝛼∶𝑋×𝑋→[0,+∞) by 𝛼𝑥,𝑦=1,if𝑥,𝑦∈[0,1],0,otherwise. (21) If 𝑓(𝑡)=𝑡 and 𝜓(𝑡)=(1/2)𝑡 for all 𝑡≥0, we have 𝛼𝑓𝑑=𝑓𝑑,𝑥,𝑦𝑇𝑥,𝑆𝑦𝑇𝑥,𝑆𝑦if𝑥,𝑦∈[0,1],0,otherwise,𝛼𝑓𝑑𝑑=𝑥𝑥,𝑦𝑇𝑥,𝑆𝑦≤𝑓𝑇𝑥,𝑆𝑦2−𝑦2≤12𝑥−𝑦≤𝜓𝑓𝑚𝑇𝑥,𝑆𝑦, (22) for all 𝑥,𝑦∈𝑋. Thus, 𝑇,  𝑆 are generalized coupled 𝛼-𝜓-contractive mappings. Moreover, there exists 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1. In fact, for 𝑥₀=1, we have 𝛼(1,𝑇1)=1. Obviously, 𝑇 is continuous, and so it remains to show that 𝑇,  𝑆 are coupled 𝛼-admissible. To do so, let 𝑥,𝑦∈𝑋 such that 𝛼(𝑥,𝑦)≥1. This implies that 𝑥,𝑦∈[0,1] by the definition of 𝛼. We then have 𝑇𝑥∈[0,1], 𝑆𝑦∈[0,1] and 𝛼(𝑇𝑥,𝑆𝑦))=1. Then, 𝑇,  𝑆 are coupled 𝛼-admissible. Now, all the hypotheses of Theorem 9 are satisfied.
Consequently, 𝑇,  𝑆 have common fixed points. In this example, 0 is at least one common fixed point of 𝑇 and 𝑆.

Now, we omit the continuity hypothesis of 𝑇 and 𝑆.

Theorem 12. Let (𝑋,𝑑) be a complete metric space. Suppose that 𝑇,𝑆∶𝑋→𝑋 are generalized coupled 𝛼-𝜓-contractive mappings and satisfy the following conditions:(i)𝑇,  𝑆 are coupled 𝛼-admissible;(ii) there exists 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1; or(ii)^∗ there exists 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1;(iii) if {𝑥_𝑛} is a sequence in 𝑋 such that 𝛼(𝑥_𝑛,𝑥_𝑛+1)≥1 for all 𝑛 and 𝑥_𝑛→𝑥∈𝑋 as 𝑛→∞, then there exists a subsequence {𝑥_𝑛(𝑘)} of {𝑥_𝑛} such that 𝛼(𝑥_𝑛(𝑘),𝑥)≥1 for all 𝑘; or(iii)^∗ if {𝑥_𝑛} is a sequence in 𝑋 such that 𝛼(𝑥_𝑛+1,𝑥_𝑛)≥1 for all 𝑛 and 𝑥_𝑛→𝑥∈𝑋 as 𝑛→∞, then there exists a subsequence {𝑥_𝑛(𝑘)} of {𝑥_𝑛} such that 𝛼(𝑥,𝑥_𝑛(𝑘))≥1 for all 𝑘.Then, 𝑇,  𝑆 have common fixed point 𝑥^∗∈𝑋. Further, for each 𝑥₀∈𝑋, the iterated sequence {𝑥_𝑛} with 𝑥_2𝑛+1=𝑇𝑥_2𝑛 and 𝑥_2𝑛+2=𝑆𝑥_2𝑛+1 converges to the common fixed point of 𝑇,𝑆.

Proof. Following the proof of Theorem 9, we know that {𝑥_𝑛} is a Cauchy sequence in the complete metric space (𝑋,𝑑). Then, there exists 𝑥^∗∈𝑋 such that 𝑥_𝑛→𝑥^∗ as 𝑛→+∞. From Theorem 9 and condition (iii), there exists a subsequence {𝑥_𝑛(𝑘)} of {𝑥_𝑛} such that 𝛼(𝑥_2𝑛(𝑘),𝑥^∗)≥1 for all 𝑘.
Applying Theorem 9, for all 𝑘, we get that 𝑓𝑑𝑥_{2𝑛(𝑘)+1},𝑆𝑥^∗=𝑓𝑑𝑇𝑥_2𝑛(𝑘),𝑆𝑥^∗≤𝛼𝑥_2𝑛(𝑘),𝑥^∗𝑑𝑇𝑥_2𝑛(𝑘),𝑆𝑥^∗≤𝜓𝑓𝑚𝑇𝑥_2𝑛(𝑘),𝑆𝑥^∗. (23) On the other hand, we have 𝑚𝑇𝑥_2𝑛(𝑘),𝑆𝑥^∗=max𝑑𝑥_2𝑛(𝑘),𝑥^∗,𝑑𝑥_2𝑛(𝑘),𝑇𝑥_2𝑛(𝑘),𝑑𝑥^∗,𝑆𝑥^∗,12𝑑𝑥_2𝑛(𝑘),𝑆𝑥^∗+𝑑𝑥^∗,𝑇𝑥_2𝑛(𝑘). (24) Letting 𝑘→+∞, in the above equality, we get that lim_𝑘→∞𝑚𝑇𝑥_2𝑛(𝑘),𝑆𝑥^∗=𝑑𝑥^∗,𝑆𝑥^∗. (25) Suppose that 𝑑(𝑥^∗,𝑆𝑥^∗)>0. From (25), for 𝑘 large enough, we have 𝑚(𝑇𝑥_2𝑛(𝑘),𝑆𝑥^∗)>0, which implies that 𝜓𝑓𝑚𝑇𝑥_2𝑛(𝑘),𝑆𝑥^∗<𝑓𝑚𝑇𝑥_2𝑛(𝑘),𝑆𝑥^∗. (26) Thus, from (23), we have 𝑓(𝑑(𝑥_{2𝑛(𝑘)+1},𝑆𝑥^∗))<𝑚(𝑇𝑥_2𝑛(𝑘),𝑆𝑥^∗). Letting 𝑘→∞ in the above inequality, and using (25), we obtain that 𝑓(𝑑(𝑥^∗,𝑆𝑥^∗))<𝑓(𝑑(𝑥^∗,𝑆𝑥^∗)) which is a contradiction. Thus, we have 𝑑(𝑥^∗,𝑆𝑥^∗)=0; that is, 𝑥^∗=𝑆𝑥^∗. Similarly, it can be shown that 𝑥^∗=𝑇𝑥^∗.

Corollary 13. Let (𝑋,𝑑) be a complete metric space. Suppose that 𝑇∶𝑋→𝑋 is generalized 𝛼-𝜓-contractive mapping and satisfies thefollowing conditions:(i)𝑇 is 𝛼-admissible;(ii) there exists 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1; or(ii)^∗ there exists 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1;(iii) if {𝑥_𝑛} is a sequence in 𝑋 such that 𝛼(𝑥_𝑛,𝑥_𝑛+1)≥1 for all 𝑛 and 𝑥_𝑛→𝑥∈𝑋 as 𝑛→∞, then there exists a subsequence {𝑥_𝑛(𝑘)} of {𝑥_𝑛} such that 𝛼(𝑥_𝑛(𝑘),𝑥)≥1 for all 𝑘; or(iii)^∗ if {𝑥_𝑛} is a sequence in 𝑋 such that 𝛼(𝑥_𝑛+1,𝑥_𝑛)≥1 for all 𝑛 and 𝑥_𝑛→𝑥∈𝑋 as 𝑛→∞, then there exists a subsequence {𝑥_𝑛(𝑘)} of {𝑥_𝑛} such that 𝛼(𝑥,𝑥_𝑛(𝑘))≥1 for all 𝑘.Then, 𝑇 has a fixed point 𝑥^∗∈𝑋. Further, for each 𝑥₀∈𝑋, the iterated sequence {𝑥_𝑛} with 𝑥_𝑛+1=𝑇𝑥_𝑛 converges to the fixed point of 𝑇.

Example 14. Let 𝑋=ℝ endowed with the standard metric 𝑑(𝑥,𝑦)=|𝑥−𝑦| for all 𝑥,𝑦∈ℝ. Define the coupled mappings 𝑇,𝑆∶𝑋→𝑋 by 3𝑇𝑥=2𝑥−4,if𝑥𝑥>1,2,if0≤𝑥≤1,0,if5𝑥<0,𝑆𝑥=2𝑥−6,if𝑥𝑥>1,2,if0≤𝑥≤1,0,if𝑥<0. (27) We define the mapping 𝛼∶𝑋×𝑋→[0,+∞) by 𝛼𝑥,𝑦=1,if𝑥,𝑦∈[0,1],0,otherwise. (28) If 𝑓(𝑡)=𝑡 and 𝜓(𝑡)=(1/2)𝑡 for all 𝑡≥0, we have 𝛼𝑓𝑑=𝑓𝑑,𝑥,𝑦𝑇𝑥,𝑆𝑦𝑇𝑥,𝑆𝑦if𝑥,𝑦∈[0,1],0,otherwise,𝛼𝑓𝑑𝑑=𝑥𝑥,𝑦𝑇𝑥,𝑆𝑦≤𝑓𝑇𝑥,𝑆𝑦2−𝑦2≤12𝑥−𝑦≤𝜓𝑓𝑚𝑇𝑥,𝑆𝑦, (29) for all 𝑥,𝑦∈𝑋. Thus, 𝑇,  𝑆 are generalized coupled 𝛼-𝜓-contractive mappings. Moreover, there exists 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1. In fact, for 𝑥₀=1, we have 𝛼(1,𝑇1)=1. Let {𝑥_𝑛}∈𝑋,  𝑥_𝑛→𝑥 and 𝛼(𝑥_𝑛+1,𝑥_𝑛)≥1 for all 𝑛≥1. By the definition of 𝛼 we have, 𝑥_𝑛,𝑥_𝑛+1∈[0,1] for all 𝑛, so 𝑥∈[0,1] and 𝛼(𝑥_𝑛,𝑥)≥1. It remains to show that 𝑇,  𝑆 are coupled 𝛼-admissible. In doing so, let 𝑥,𝑦∈𝑋 such that 𝛼(𝑥,𝑦)≥1. This implies that 𝑥,𝑦∈[0,1] by the definition of 𝛼. We have 𝑇𝑥∈[0,1],  𝑆𝑦∈[0,1], and 𝛼(𝑇𝑥,𝑆𝑦))=1. Then 𝑇,  𝑆 are coupled 𝛼-admissible. Now, all the hypotheses of Theorem 12 are satisfied. Consequently, 𝑇,  𝑆 have common fixed points. In this example, 0 is at least one common fixed point of 𝑇 and 𝑆.

3. Fixed Point Theorems on Ordered Metric Space

Theorem 15. Let (𝑋,⪯,𝑑) be a complete ordered metric space, 𝜓∈Ψ, 𝑓∈ϝ, and 𝑇,  𝑆 be coupled mappings on 𝑋 such that 𝑓(𝑑(𝑇𝑥,𝑆𝑦))≰𝜓(𝑓(𝑚(𝑇𝑥,𝑆𝑦))) and 𝑓(𝑑(𝑆𝑥,𝑇𝑦))≤𝜓(𝑓(𝑚(𝑆𝑥,𝑇𝑦))) for all 𝑥,𝑦∈𝑋 with 𝑥⪯𝑦. Suppose that there exists 𝑥₀∈𝑋 such that 𝑥₀⪯𝑇𝑥₀ or (𝑇𝑥₀⪯𝑥₀), and if {𝑥_𝑛} is a sequence in 𝑋 such that 𝑥_𝑛⪯𝑥_𝑛+1 or (𝑥_𝑛+1⪯𝑥_𝑛) for all 𝑛≥1 and 𝑥_𝑛→𝑥, then 𝑥_𝑛⪯𝑥 or (𝑥⪯𝑥_𝑛) for all 𝑛≥1. If 𝑥⪯𝑦 implies 𝑇𝑥⪯𝑆𝑦  or  𝑆𝑥⪯𝑇𝑦 or (𝑇𝑦⪯𝑆𝑥  or  𝑆𝑦⪯𝑇𝑥), then 𝑇,𝑆 have common fixed points.

Proof. Define 𝛼∶𝑋×𝑋→[0,∞) by 𝛼(𝑥,𝑦)=1 whenever 𝑥⪯𝑦, and define 𝛼(𝑥,𝑦)=0 whenever 𝑥⋠𝑦. It is easy to check that 𝑇,  𝑆 are a coupled 𝛼-admissible and generalized coupled 𝛼-𝜓-contractive mappings on 𝑋. Now, by using Theorem 9, 𝑇,  𝑆 have common fixed points.

Corollary 16. Let (𝑋,⪯,𝑑) be a complete ordered metric space, 𝜓∈Ψ, 𝑓∈ϝ, and 𝑇 a mapping on 𝑋 such that 𝑓(𝑑(𝑇𝑥,𝑇𝑦))≤𝜓(𝑓(𝑚(𝑇𝑥,𝑇𝑦))) for all 𝑥,𝑦∈𝑋 with 𝑥⪯𝑦. Suppose that there exists 𝑥₀∈𝑋 such that 𝑥₀⪯𝑇𝑥₀ or (𝑇𝑥₀⪯𝑥₀) and if {𝑥_𝑛} is a sequence in 𝑋 such that 𝑥_𝑛⪯𝑥_𝑛+1 or (𝑥_𝑛+1⪯𝑥_𝑛) for all 𝑛≥1 and 𝑥_𝑛→𝑥, then 𝑥_𝑛⪯𝑥 or (𝑥⪯𝑥_𝑛) for all 𝑛≥1. If 𝑥⪯𝑦 implies 𝑇𝑥⪯𝑇𝑦 or (𝑇𝑦⪯𝑇𝑥), then 𝑇 has a fixed point.

4. Fixed Point Theorems on Metric Spaces Endowed with Partial Metric

If we substitute a partial metric 𝜌 instead the metric 𝑑 in Theorem 9, it is easy to check that a similar result holds for the partial metric space case as follows. We define 𝑚^∗𝜌(),𝜌,1𝐴𝑥,𝐵𝑦=max𝑥,𝑦,𝜌𝑥,𝐴𝑥𝑦,𝐵𝑦2𝜌𝑥,𝐵𝑦+𝜌𝑦,𝐴𝑥 (30) for all coupled mappings 𝐴,𝐵∶𝑋→𝑋 and 𝑥,𝑦∈𝑋.

Theorem 17. Let (𝑋,𝜌) be a complete partial metric space,  𝛼∶𝑋×𝑋→[0,∞) a function, 𝜓∈Ψ, 𝑓∈ϝ, and 𝑇,  𝑆 self maps on 𝑋 such that 𝛼𝑥,𝑦𝑓𝜌𝑇𝑥,𝑆𝑦≤𝜓𝑓𝑚^∗,𝛼𝑓𝜌𝑓𝑚𝑇𝑥,𝑆𝑦𝑥,𝑦𝑆𝑥,𝑇𝑦≤𝜓^∗𝑆𝑥,𝑇𝑦 (31) for all 𝑥,𝑦∈𝑋. Suppose that 𝑇,  𝑆 are coupled 𝛼-admissible and there exists 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1 or (𝛼(𝑇𝑥₀,𝑥₀)≥1). Assume that if {𝑥_𝑛} is a sequence in 𝑋 such that 𝛼(𝑥_𝑛,𝑥_𝑛+1)≥1 or (𝛼(𝑥_𝑛+1,𝑥_𝑛)≥1) for all 𝑛 and 𝑥_𝑛→𝑥 as 𝑛→+∞, then there exists a subsequence {𝑥_𝑛(𝑘)} of {𝑥_𝑛} such that 𝛼(𝑥_𝑛(𝑘),𝑥)≥1 or (𝛼(𝑥,𝑥_𝑛(𝑘))≥1) for all 𝑘. Then 𝑇,𝑆 have common fixed points.

Corollary 18. Let (𝑋,𝜌) be a complete partial metric space,  𝛼∶𝑋×𝑋→[0,∞) a function, 𝜓∈Ψ, 𝑓∈ϝ, and 𝑇 a self map on 𝑋 such that 𝛼𝑥,𝑦𝑓𝜌𝑇𝑥,𝑇𝑦≤𝜓𝑓𝑚^∗𝑇𝑥,𝑇𝑦 (32) for all 𝑥,𝑦∈𝑋. Suppose that 𝑇 is 𝛼-admissible and there exists 𝑥₀∈𝑋 such that 𝛼(𝑥₀,𝑇𝑥₀)≥1 or (𝛼(𝑇𝑥₀,𝑥₀)≥1). Assume that if {𝑥_𝑛} is a sequence in 𝑋 such that 𝛼(𝑥_𝑛,𝑥_𝑛+1)≥1 or (𝛼(𝑥_𝑛+1,𝑥_𝑛)≥1) for all 𝑛 and 𝑥_𝑛→𝑥 as 𝑛→+∞, then there exists a subsequence {𝑥_𝑛(𝑘)} of {𝑥_𝑛} such that 𝛼(𝑥_𝑛(𝑘),𝑥)≥1 or (𝛼(𝑥,𝑥_𝑛(𝑘))≥1) for all 𝑘. Then 𝑇 has a fixed point.

Example 19. Let 𝑋=[0,+∞) endowed with the partial metric 𝜌(𝑥,𝑦)=max{𝑥,𝑦}, for all 𝑥,𝑦∈𝑋. Define the mapping 𝑇∶𝑋→𝑋, by 3𝑇𝑥=2𝑥−2,if𝑥𝑥>1,₂,1+𝑥if0≤𝑥≤1,0,if𝑥<0. (33) We define the mapping 𝛼∶𝑋×𝑋→[0,+∞) by 𝛼𝑥,𝑦=1,if𝑥,𝑦∈[0,1],𝑥≤𝑦,0,otherwise,𝛼[],[],𝑥,𝑦≥1⟺𝑥,𝑦∈0,1𝑥≤𝑦⟺𝑇𝑥,𝑇𝑦∈0,1𝑇𝑥≤𝑇𝑦⟺𝛼𝑇𝑥,𝑇𝑦≥1 (34) since 𝑇 is 𝛼-admissible. If 𝜓(𝑡)=𝑡²/(1+𝑡), we have 𝛼𝜌𝑥𝑥,𝑦𝑇𝑥,𝑇𝑦≤𝜌²,𝑦1+𝑥²𝑥1+𝑦=max²,𝑦1+𝑥²=𝑦1+𝑦₂.1+𝑦≤𝑦=𝜌𝑥,𝑦 (35) Suppose that {𝑥_𝑛} is a sequence in 𝑋 such that 𝛼(𝑥_𝑛,𝑥_𝑛+1)≥1 for all 𝑛 and 𝑥_𝑛→𝑥 as 𝑛→+∞; by definition of 𝛼, we have 𝑥_𝑛,  𝑥_𝑛+1 and 𝑥∈[0,1], on the other hand 𝑥_𝑛≤𝑥_𝑛+1⇒𝑥_𝑛(𝑘)≤𝑥; then there exists a subsequence {𝑥_𝑛(𝑘)} of {𝑥_𝑛} such that 𝛼(𝑥_𝑛(𝑘),𝑥)≥1 for all 𝑘. There exists 𝑥₀=1 such that 𝛼(𝑥₀,𝑇𝑥₀)=1. This show that all conditions of Corollary 18 are satisfied, and so 𝑇 has a fixed point in 𝑋.

Acknowledgment

The author would like to thank Tehran Science and Research Branch, Islamic Azad University, for the financial support of this research, which is based on a research project contract.