Fixed Point Theorems for Various Classes of Cyclic Mappings

Abstract

We introduce new classes of cyclic mappings and we study the existence and uniqueness of fixed points for such mappings. The presented theorems generalize and improve several existing results in the literature.

1. Introduction

The Banach contraction principle is one of the most important results on fixed point theory. Several extensions and generalizations of this result have appeared in the literature and references therein, see [1–11]. One of the most interesting extensions was given by Kirk et al. in [12].

Theorem 1.1 (see [12]). Let 𝐴 and 𝐵 be two nonempty closed subsets of a complete metric space (𝑋,𝑑). Suppose that 𝑇∶𝐴∪𝐵→𝐴∪𝐵 satisfies (i)𝑇(𝐴)⊆𝐵, 𝑇(𝐵)⊆𝐴; (ii)there exists a constant 𝑘∈(0,1) such that 𝑑𝑇𝑥,𝑇𝑦≤𝑘𝑑𝑥,𝑦,∀𝑥∈𝐴,𝑦∈𝐵. (1.1) Then 𝑇 has a unique fixed point that belongs to 𝐴∩𝐵.

A mapping satisfying (i) is called cyclic. Recently, many results dealing with cyclic contractions have appeared in several works (see, e.g., [13–30]). Karpagam and Agrawal [31, 32] introduced the notion of a cyclic orbital contraction, and obtained a unique fixed point theorem for such class of mappings (see also [33, 34]).

In what follows, ℕ is the set of positive integers.

Definition 1.2 (see [31]). Let 𝐴 and 𝐵 be nonempty subsets of a metric space (𝑋,𝑑), let 𝑇∶𝐴∪𝐵→𝐴∪𝐵 be a cyclic mapping such that for some 𝑥∈𝐴, there exists a 𝑘_𝑥∈(0,1) such that 𝑑𝑇^2𝑛𝑥,𝑇𝑦≤𝑘_𝑥𝑑𝑇^2𝑛−1𝑥,𝑦, (1.2) for all 𝑛∈ℕ and 𝑦∈𝐴. Then 𝑇 is called a cyclic orbital contraction.

Theorem 1.3 (see [31]). Let 𝐴 and 𝐵 be two nonempty closed subsets of a complete metric space (𝑋,𝑑), and let 𝑇∶𝐴∪𝐵→𝐴∪𝐵 be a cyclic orbital contraction. Then 𝑇 has a unique fixed point that belongs to 𝐴∩𝐵.

Very recently, Chen [19] introduced the class of cyclic orbital stronger Meir-Keeler contraction.

Definition 1.4 (see [19]). Let (𝑋,𝑑) be a metric space. A mapping 𝜓∶[0,∞)→[0,1) is called a stronger Meir-Keeler type mapping in 𝑋 if the following condition holds for all 𝜀≥0, there exist 𝛿_𝜀>0 and 𝛾_𝜀∈(0,1) such that for all 𝑥,𝑦∈𝑋, we have 𝜀≤𝑑𝑥,𝑦<𝜀+𝛿_𝜀⇒𝜓𝑑𝑥,𝑦<𝛾_𝜀. (1.3)

Definition 1.5 (see [19]). Let 𝐴 and 𝐵 be nonempty subsets of a metric space (𝑋,𝑑). Suppose that 𝑇∶𝐴∪𝐵→𝐴∪𝐵 is a cyclic map such that for some 𝑥∈𝐴, there exists a stronger Meir-Keeler type mapping 𝜓_𝑥∶[0,∞)→[0,1) such that 𝑑𝑇^2𝑛𝑥,𝑇𝑦≤𝜓_𝑥𝑑𝑇^2𝑛−1𝑥,𝑦𝑑𝑇^2𝑛−1𝑥,𝑦, (1.4) for all 𝑛∈ℕ and 𝑦∈𝐴. Then 𝑇 is called a a cyclic orbital stronger Meir-Keeler 𝜓_𝑥-contraction.

Clearly, if 𝑇∶𝐴∪𝐵→𝐴∪𝐵 is a cyclic orbital contraction, then 𝑇 is a cyclic orbital stronger Meir-Keeler 𝜓_𝑥-contraction, where 𝜓_𝑥(𝑡)=𝑘_𝑥 for all 𝑡≥0.

Theorem 1.6 (see [19]). Let 𝐴 and 𝐵 be two nonempty closed subsets of a complete metric space (𝑋,𝑑), and let 𝑇∶𝐴∪𝐵→𝐴∪𝐵 be a cyclic orbital stronger Meir-Keeler 𝜓_𝑥-contraction, for some 𝑥∈𝐴. Then 𝑇 has a unique fixed point that belongs to 𝐴∩𝐵.

In this work, we introduce new classes of cyclic contractive mappings and we study the existence and uniqueness of fixed points for such mappings. Our obtained results extend and generalize several existing fixed point theorems in the literature, including Theorems 1.3 and 1.6.

2. Main Results

We denote by Ψ the class of stronger Meir-Keeler type mappings (see Definition 1.4).

The following lemma will be useful later.

Lemma 2.1. Let (𝑋,𝑑) be a metric space, 𝜓∈Ψ, {𝑥_𝑛},{𝑦_𝑛}⊂𝑋. If 𝑑(𝑥_𝑛,𝑦_𝑛)→𝑟 as 𝑛→∞, then there exist 𝛾=𝛾_𝑟∈(0,1) and 𝑁∈ℕ such that 𝜓𝑑𝑥_𝑛,𝑦_𝑛<𝛾,∀𝑛≥𝑁. (2.1)

Proof. Since 𝜓∈Ψ, there exist 𝛿_𝑟>0 and 𝛾=𝛾_𝑟∈(0,1) such that for all 𝑥,𝑦∈𝑋, we have 𝑟≤𝑑𝑥,𝑦<𝑟+𝛿_𝑟⇒𝜓𝑑𝑥,𝑦<𝛾_𝑟. (2.2) Since 𝑑(𝑥_𝑛,𝑦_𝑛)→𝑟⁺ as 𝑛→∞, there exists 𝑁∈ℕ such that 𝑟≤𝑑𝑥_𝑛,𝑦_𝑛<𝑟+𝛿_𝑟,∀𝑛≥𝑁. (2.3) It follows from (2.2) that 𝜓𝑑𝑥_𝑛,𝑦_𝑛<𝛾_𝑟,∀𝑛≥𝑁. (2.4) This makes end to the proof.

Our first result is the following.

Theorem 2.2. Let 𝐴 and 𝐵 be two nonempty closed subsets of a complete metric space (𝑋,𝑑). Let 𝑇∶𝐴∪𝐵→𝐴∪𝐵 be a cyclic mapping. Suppose that for some 𝑥∈𝐴 there exists 𝜓_𝑥∈Ψ such that 𝑑𝑇^2𝑛𝑥,𝑇𝑦≤𝜓_𝑥𝑑𝑇^2𝑛−1𝑑𝑇𝑥,𝑦^2𝑛−1𝑥,𝑇^2𝑛𝑥+𝑑𝑇𝑦,𝑦2, (2.5) for all 𝑛∈ℕ and 𝑦∈𝐴. Then 𝑇 has a unique fixed point that belongs to 𝐴∩𝐵.

Proof. At first, since 𝑇 is a cyclic mapping and 𝑥∈𝐴, we have {𝑇^2𝑛𝑥∶𝑛∈ℕ}⊂𝐴. So, taking 𝑦=𝑇^2𝑛𝑥 in (2.5), for all 𝑛∈ℕ, we get 𝑑𝑇^2𝑛𝑥,𝑇^2𝑛+1𝑥≤𝜓_𝑥𝑑𝑇^2𝑛−1𝑥,𝑇^2𝑛𝑥𝑑𝑇^2𝑛−1𝑥,𝑇^2𝑛𝑥+𝑑𝑇^2𝑛𝑥,𝑇^2𝑛+1𝑥2≤𝑑𝑇^2𝑛−1𝑥,𝑇^2𝑛𝑥+𝑑𝑇^2𝑛𝑥,𝑇^2𝑛+1𝑥2. (2.6) Thus for all 𝑛∈ℕ, we have 𝑑𝑇^2𝑛𝑥,𝑇^2𝑛+1𝑥≤𝑑𝑇^2𝑛−1𝑥,𝑇^2𝑛𝑥. (2.7) Again, taking 𝑦=𝑇^2𝑛𝑥 in (2.5), for all 𝑛∈ℕ, we get 𝑑𝑇^2𝑛+1𝑥,𝑇^2𝑛+2𝑥=𝑑𝑇^2𝑛+2𝑥,𝑇^2𝑛+1𝑥≤𝜓_𝑥𝑑𝑇^2𝑛+1𝑥,𝑇^2𝑛𝑥𝑑𝑇^2𝑛+1𝑥,𝑇^2𝑛+2𝑥+𝑑𝑇^2𝑛+1𝑥,𝑇^2𝑛𝑥2≤𝑑𝑇^2𝑛+1𝑥,𝑇^2𝑛+2𝑥+𝑑𝑇^2𝑛+1𝑥,𝑇^2𝑛𝑥2. (2.8) Thus for all 𝑛∈ℕ, we have 𝑑𝑇^2𝑛+1𝑥,𝑇^2𝑛+2𝑥≤𝑑𝑇^2𝑛𝑥,𝑇^2𝑛+1𝑥. (2.9) It follows from (2.7) and (2.9) that {𝑑(𝑇^𝑛𝑥,𝑇^𝑛+1𝑥)} is a decreasing sequence. Then there exists 𝜀≥0 such that lim_𝑛→∞𝑑𝑇^𝑛𝑥,𝑇^𝑛+1𝑥=𝜀. (2.10) Applying Lemma 2.1 with 𝑟=𝜀, we obtain that there exist 𝑛₀∈ℕ and 𝛾_𝜀∈(0,1) such that 𝜓_𝑥𝑑𝑇^𝑛𝑥,𝑇^𝑛+1𝑥<𝛾_𝜀,∀𝑛≥𝑛₀. (2.11) Denote 𝑘₀=[(𝑛₀+1)/2]+1, where [(𝑛₀+1)/2] is the integer part of (𝑛₀+1)/2. By (2.11) and (2.5), for all 𝑘≥𝑘₀, we have 𝑑𝑇^2𝑘𝑥,𝑇^2𝑘+1𝑥≤𝜓_𝑥𝑑𝑇^2𝑘−1𝑥,𝑇^2𝑘𝑥𝑑𝑇^2𝑘−1𝑥,𝑇^2𝑘𝑥+𝑑𝑇^2𝑘+1𝑥,𝑇^2𝑘𝑥2≤𝛾_𝜀𝑑𝑇^2𝑘−1𝑥,𝑇^2𝑘𝑥+𝑑𝑇^2𝑘+1𝑥,𝑇^2𝑘𝑥2. (2.12) This implies that 𝑑𝑇^2𝑘𝑥,𝑇^2𝑘+1𝑥≤𝛾_𝜀2−𝛾_𝜀𝑑𝑇^2𝑘−1𝑥,𝑇^2𝑘𝑥,∀𝑘≥𝑘₀. (2.13) Note that 𝛾_𝜀/(2−𝛾_𝜀)∈(0,1). Similarly, from (2.11) and (2.5), for all 𝑘≥𝑘₀, we have 𝑑𝑇^2𝑘+2𝑥,𝑇^2𝑘+1𝑥≤𝜓_𝑥𝑑𝑇^2𝑘+1𝑥,𝑇^2𝑘𝑥𝑑𝑇^2𝑘+1𝑥,𝑇^2𝑘+2𝑥+𝑑𝑇^2𝑘𝑥,𝑇^2𝑘+1𝑥2≤𝛾_𝜀𝑑𝑇^2𝑘+1𝑥,𝑇^2𝑘+2𝑥+𝑑𝑇^2𝑘𝑥,𝑇^2𝑘+1𝑥2. (2.14) This implies that 𝑑𝑇^2𝑘+1𝑥,𝑇^2𝑘+2𝑥≤𝛾_𝜀2−𝛾_𝜀𝑑𝑇^2𝑘𝑥,𝑇^2𝑘+1𝑥,∀𝑘≥𝑘₀. (2.15) Now, it follows from (2.13) and (2.15) that 𝑑𝑇^2𝑘0+𝑛𝑥,𝑇^{2𝑘0+𝑛+1}𝑥≤𝛾_𝜀2−𝛾_𝜀^𝑛𝑑𝑇^2𝑘0−1𝑥,𝑇^2𝑘0𝑥,∀𝑛∈ℕ∪{0}. (2.16) Denote 𝑞_𝜀=𝛾_𝜀/(2−𝛾_𝜀). From (2.16), for all 𝑝∈ℕ, we have 𝑑𝑇^2𝑘0+𝑛𝑥,𝑇^{2𝑘0+𝑛+𝑝}𝑥≤^{𝑛+𝑝−1}_𝑖=𝑛𝑑𝑇^2𝑘0+𝑖𝑥,𝑇^{2𝑘0+𝑖+1}𝑥≤𝑞^𝑛_𝜀1−𝑞_𝜀𝑑𝑇^2𝑘0−1𝑥,𝑇^2𝑘0𝑥→0as𝑛→∞. (2.17) This implies that {𝑇^𝑛𝑥} is a Cauchy sequence. Since (𝑋,𝑑) is complete, 𝐴 and 𝐵 are closed, and {𝑇^𝑛𝑥}⊂𝐴∪𝐵, there exists a 𝑧∈𝐴∪𝐵 such that 𝑇^𝑛𝑥→𝑧 as 𝑛→∞. On the other hand, since 𝑇 is a cyclic mapping and 𝑥∈𝐴, we have {𝑇^2𝑛𝑥}⊂𝐴 and {𝑇^2𝑛+1𝑥}⊂𝐵. Since 𝐴 and 𝐵 are closed, this implies that 𝑧∈𝐴∩𝐵. On the other hand, since 𝑑(𝑇^2𝑛−1𝑥,𝑧)→0 as 𝑛→∞, by Lemma 2.1, there exist 𝑁∈ℕ and 𝛾₀∈(0,1) such that 𝜓𝑑𝑇^2𝑛−1𝑥,𝑧<𝛾₀,∀𝑛≥𝑁. (2.18) From (2.5) and (2.18), for all 𝑛≥𝑁, we have 𝑑𝑇^2𝑛𝑥,𝑇𝑧≤𝜓_𝑥𝑑𝑇^2𝑛−1𝑑𝑇𝑥,𝑧^2𝑛−1𝑥,𝑇^2𝑛𝑥+𝑑(𝑧,𝑇𝑧)2≤𝛾₀𝑑𝑇^2𝑛−1𝑥,𝑇^2𝑛𝑥+𝑑(𝑧,𝑇𝑧)2. (2.19) Letting 𝑛→∞ in the above inequality, we get 𝑑(𝑧,𝑇𝑧)≤𝛾₀𝑑(𝑧,𝑇𝑧)2, (2.20) which implies (since 𝛾₀∈(0,1)) that 𝑑(𝑧,𝑇𝑧)=0, that is, 𝑧 is a fixed point of 𝑇. To show the uniqueness of the fixed point, suppose that 𝑤∈𝐴∪𝐵 is also a fixed point of 𝑇. Clearly, since 𝑇 is a cyclic mapping, we have 𝑤=𝑇𝑤∈𝐴∩𝐵. Now, from (2.5), for all 𝑛∈ℕ, we have 𝑑𝑇^2𝑛𝑥,𝑤=𝑑𝑇^2𝑛𝑥,𝑇𝑤≤𝜓_𝑥𝑑𝑇^2𝑛−1𝑑𝑇𝑥,𝑤^2𝑛−1𝑥,𝑇^2𝑛𝑥+𝑑(𝑇𝑤,𝑤)2≤𝑑𝑇^2𝑛−1𝑥,𝑇^2𝑛𝑥2→0as𝑛→∞. (2.21) Thus we have 𝑇^2𝑛𝑥→𝑤 as 𝑛→∞. From the uniqueness of the limit, we have 𝑧=𝑤. Thus 𝑧 is the unique fixed point of 𝑇.

Example 2.3. Let 𝑋=[0,∞) with the usual metric 𝑑(𝑥,𝑦)=|𝑥−𝑦| for all 𝑥,𝑦∈𝑋. Consider 𝐴=[0,1/2] and 𝐵=[1/2,1]. We have 𝐴∩𝐵=1/2. Define 𝑇∶𝐴∪𝐵→𝐴∪𝐵 by 1𝑇𝑥=2if𝑥∈[0,1),0if𝑥=1. (2.22) Fix any 𝑥 in 𝐴. For all 𝑛∈ℕ and 𝑦∈𝐴, we have 𝑑𝑇^2𝑛𝑥,𝑇𝑦=0, (2.23) that is, (2.49) holds. Applying Theorem 2.2, the mapping 𝑇 has a unique fixed point which is 𝑧=1/2.

Our second main result is the following.

Theorem 2.4. Let 𝐴 and 𝐵 be two nonempty closed subsets of a complete metric space (𝑋,𝑑). Let 𝑇∶𝐴∪𝐵→𝐴∪𝐵 be a cyclic mapping. Suppose that for some 𝑥∈𝐴 there exists 𝜓_𝑥∈Ψ such that 𝑑𝑇^2𝑛𝑥,𝑇𝑦≤𝜓_𝑥𝑑𝑇^2𝑛−1𝑑𝑇𝑥,𝑦^2𝑛−1𝑥,𝑇𝑦+𝑑𝑇^2𝑛𝑥,𝑦2, (2.24) for all 𝑛∈ℕ and 𝑦∈𝐴. Then 𝑇 has a unique fixed point that belongs to 𝐴∩𝐵.

Proof. Taking 𝑦=𝑇^2𝑛𝑥 in (2.24), for all 𝑛∈ℕ, we get 𝑑𝑇^2𝑛𝑥,𝑇^2𝑛+1𝑥≤𝜓_𝑥𝑑𝑇^2𝑛−1𝑥,𝑇^2𝑛𝑥𝑑𝑇^2𝑛−1𝑥,𝑇^2𝑛+1𝑥+𝑑𝑇^2𝑛𝑥,𝑇^2𝑛𝑥2≤𝑑𝑇^2𝑛−1𝑥,𝑇^2𝑛+1𝑥2≤𝑑𝑇^2𝑛−1𝑥,𝑇^2𝑛𝑥+𝑑𝑇^2𝑛𝑥,𝑇^2𝑛+1𝑥2. (2.25) Thus for all 𝑛∈ℕ, we have 𝑑𝑇^2𝑛𝑥,𝑇^2𝑛+1𝑥≤𝑑𝑇^2𝑛−1𝑥,𝑇^2𝑛𝑥. (2.26) Again, taking 𝑦=𝑇^2𝑛𝑥 in (2.24), for all 𝑛∈ℕ, we get 𝑑𝑇^2𝑛+1𝑥,𝑇^2𝑛+2𝑥=𝑑𝑇^2𝑛+2𝑥,𝑇^2𝑛+1𝑥≤𝜓_𝑥𝑑𝑇^2𝑛+1𝑥,𝑇^2𝑛𝑥𝑑𝑇^2𝑛+1𝑥,𝑇^2𝑛+1𝑥+𝑑𝑇^2𝑛+2𝑥,𝑇^2𝑛𝑥2≤𝑑𝑇^2𝑛+2𝑥,𝑇^2𝑛𝑥2≤𝑑𝑇^2𝑛+2𝑥,𝑇^2𝑛+1𝑥+𝑑𝑇^2𝑛+1𝑥,𝑇^2𝑛𝑥2. (2.27) Thus for all 𝑛∈ℕ, we have 𝑑𝑇^2𝑛+1𝑥,𝑇^2𝑛+2𝑥≤𝑑𝑇^2𝑛𝑥,𝑇^2𝑛+1𝑥. (2.28) It follows from (2.26) and (2.28) that {𝑑(𝑇^𝑛𝑥,𝑇^𝑛+1𝑥)} is a decreasing sequence. Then there exists 𝜀≥0 such that lim_𝑛→∞𝑑𝑇^𝑛𝑥,𝑇^𝑛+1𝑥=𝜀. (2.29) Applying Lemma 2.1 with 𝑟=𝜀, we obtain that there exist 𝑛₀∈ℕ and 𝛾_𝜀∈(0,1) such that 𝜓_𝑥𝑑𝑇^𝑛𝑥,𝑇^𝑛+1𝑥<𝛾_𝜀,∀𝑛≥𝑛₀. (2.30) Denote 𝑘₀=[(𝑛₀+1)/2]+1. By (2.30) and (2.24), for all 𝑘≥𝑘₀, we have 𝑑𝑇^2𝑘𝑥,𝑇^2𝑘+1𝑥≤𝜓_𝑥𝑑𝑇^2𝑘−1𝑥,𝑇^2𝑘𝑥𝑑𝑇^2𝑘−1𝑥,𝑇^2𝑘+1𝑥+𝑑𝑇^2𝑘𝑥,𝑇^2𝑘𝑥2≤𝛾_𝜀𝑑𝑇^2𝑘−1𝑥,𝑇^2𝑘+1𝑥2≤𝛾_𝜀𝑑𝑇^2𝑘−1𝑥,𝑇^2𝑘𝑥+𝑑𝑇^2𝑘𝑥,𝑇^2𝑘+1𝑥2. (2.31) This implies that 𝑑𝑇^2𝑘𝑥,𝑇^2𝑘+1𝑥≤𝑞_𝜀𝑑𝑇^2𝑘−1𝑥,𝑇^2𝑘𝑥,∀𝑘≥𝑘₀, (2.32) where 𝑞_𝜀=𝛾_𝜀/(2−𝛾_𝜀)∈(0,1). Similarly, we can show that 𝑑𝑇^2𝑘+1𝑥,𝑇^2𝑘+2𝑥≤𝑞_𝜀𝑑𝑇^2𝑘𝑥,𝑇^2𝑘+1𝑥,∀𝑘≥𝑘₀. (2.33) Now, it follows from (2.32) and (2.33) that 𝑑𝑇^2𝑘0+𝑛x,𝑇^{2𝑘0+𝑛+1}𝑥≤𝑞^𝑛_𝜀𝑑𝑇^2𝑘0−1𝑥,𝑇^2𝑘0𝑥,∀𝑛∈ℕ∪{0}. (2.34) As in the proof of Theorem 2.2, we obtain from the above inequality that {𝑇^𝑛𝑥} is a Cauchy sequence in (𝑋,𝑑), which implies that there exists a 𝑧∈𝐴∩𝐵 such that 𝑇^𝑛𝑥→𝑧 as 𝑛→∞. On the other hand, by Lemma 2.1, there exist 𝑁∈ℕ and 𝛾₀∈(0,1) such that 𝜓𝑑𝑇^2𝑛−1𝑥,𝑧<𝛾₀,∀𝑛≥𝑁. (2.35) From (2.24) and (2.35), for all 𝑛≥𝑁, we have 𝑑𝑇^2𝑛𝑥,𝑇𝑧≤𝜓_𝑥𝑑𝑇^2𝑛−1𝑑𝑇𝑥,𝑧^2𝑛−1𝑥,𝑇𝑧+𝑑𝑇^2𝑛𝑥,𝑧2≤𝛾₀𝑑𝑇^2𝑛−1𝑥,𝑇𝑧+𝑑𝑇^2𝑛𝑥,𝑧2. (2.36) Letting 𝑛→∞ in the above inequality, we get 𝑑(𝑧,𝑇𝑧)≤𝛾₀𝑑(𝑧,𝑇𝑧)2, (2.37) which implies (since 𝛾₀∈(0,1)) that 𝑑(𝑧,𝑇𝑧)=0, that is, 𝑧 is a fixed point of 𝑇. Suppose now that 𝑤∈𝐴∩𝐵 is also a fixed point of 𝑇. From (2.24), for all 𝑛∈ℕ, we have 𝑑𝑇^2𝑛𝑥,𝑤=𝑑𝑇^2𝑛𝑥,𝑇𝑤≤𝜓_𝑥𝑑𝑇^2𝑛−1𝑑𝑇𝑥,𝑤^2𝑛−1𝑥,𝑇𝑤+𝑑𝑇^2𝑛𝑥,𝑤2≤𝑑𝑇^2𝑛−1𝑥,𝑤+𝑑𝑇^2𝑛𝑥,𝑤2. (2.38) This implies that 𝑑𝑇^2𝑛𝑥,𝑤≤𝑑𝑇^2𝑛−1𝑥,𝑤,∀𝑛∈ℕ. (2.39) Then there exists 𝑎≥0 such that 𝑑(𝑇^2𝑛𝑥,𝑤)→𝑎 as 𝑛→∞. On the other hand, we have 𝑑𝑇^2𝑛𝑥,𝑤−𝑑𝑇^2𝑛−1𝑥,𝑤=𝑑𝑇^2𝑛−1𝑥,𝑤−𝑑𝑇^2𝑛𝑇𝑥,𝑤≤𝑑^2𝑛𝑥,𝑇^2𝑛−1𝑥→0as𝑛→∞. (2.40) Thus we have 𝑑(𝑇^2𝑛−1𝑥,𝑤)→𝑎 as 𝑛→∞. By Lemma 2.1, there exist 𝑞∈ℕ and 𝛾_𝑎∈(0,1) such that 𝜓𝑑𝑇^2𝑛−1𝑥,𝑤<𝛾_𝑎,∀𝑛≥𝑞. (2.41) From (2.41) and (2.24), for all 𝑛≥𝑞, we have 𝑑𝑇^2𝑛𝑥,𝑤=𝑑𝑇^2𝑛𝑥,𝑇𝑤≤𝜓_𝑥𝑑𝑇^2𝑛−1𝑑𝑇𝑥,𝑤^2𝑛−1𝑥,𝑇𝑤+𝑑𝑇^2𝑛𝑥,𝑤2≤𝛾_𝑎𝑑𝑇^2𝑛−1𝑥,𝑤+𝑑𝑇^2𝑛𝑥,𝑤2. (2.42) Letting 𝑛→∞, we get 𝑎≤𝛾_𝑎𝑎, (2.43) which implies (since 𝛾_𝑎∈(0,1)) that 𝑎=0. Thus we have 𝑑(𝑇^2𝑛𝑥,𝑤)→0 as 𝑛→∞. By the uniqueness of the limit, we get that 𝑤=𝑧. Thus 𝑧 is the unique fixed point of 𝑇.

The following result is an immediate consequence of Theorem 2.2.

Theorem 2.5. Let 𝐴 and 𝐵 be two nonempty closed subsets of a complete metric space (𝑋,𝑑). Let 𝑇∶𝐴∪𝐵→𝐴∪𝐵 be a cyclic mapping. Suppose that for some 𝑥∈𝐴 there exists 𝑘_𝑥∈(0,1) such that 𝑑𝑇^2𝑛𝑥,𝑇𝑦≤𝑘_𝑥𝑑𝑇^2𝑛−1𝑥,𝑇^2𝑛𝑥+𝑑𝑇𝑦,𝑦2, (2.44) for all 𝑛∈ℕ and 𝑦∈𝐴. Then 𝑇 has a unique fixed point that belongs to 𝐴∩𝐵.

The following result is an immediate consequence of Theorem 2.4.

Theorem 2.6. Let 𝐴 and 𝐵 be two nonempty closed subsets of a complete metric space (𝑋,𝑑). Let 𝑇∶𝐴∪𝐵→𝐴∪𝐵 be a cyclic mapping. Suppose that for some 𝑥∈𝐴 there exists 𝑘_𝑥∈(0,1) such that 𝑑𝑇^2𝑛𝑥,𝑇𝑦≤𝑘_𝑥𝑑𝑇^2𝑛−1𝑥,𝑇𝑦+𝑑𝑇^2𝑛𝑥,𝑦2, (2.45) for all 𝑛∈ℕ and 𝑦∈𝐴. Then 𝑇 has a unique fixed point that belongs to 𝐴∩𝐵.

Now, we present a data dependence result.

Theorem 2.7. Let 𝐴 and 𝐵 be nonempty closed subsets of a complete metric space (X,𝑑) and 𝑇,𝑆∶𝐴∪𝐵→𝐴∪𝐵 are two mappings satisfying (i)𝑇 is a cyclic mapping; (ii)𝑇 satisfies (2.5); (iii)𝑆 has a fixed point 𝑥_𝑆∈𝐴; (iv)there exists 𝜂>0 such that 𝑑(𝑇𝑢,𝑆𝑢)≤𝜂 for all 𝑢∈𝐴. Then 𝑑(𝑥_𝑇,𝑥_𝑆)≤3/2𝜂, where 𝑥_𝑇 is the unique fixed point of 𝑇.

Proof. From Theorem 2.2, conditions (i) and (ii) imply that 𝑇 has a unique fixed point 𝑥_𝑇∈𝐴∩𝐵. Using (2.5), we have 𝑑𝑥_𝑇,𝑥_𝑆=𝑑𝑇𝑥_𝑇,𝑆𝑥_𝑆≤𝑑𝑇^2𝑛𝑥,𝑇𝑥_𝑇+𝑑𝑇^2𝑛𝑥,𝑇𝑥_𝑆+𝑑𝑇𝑥_𝑆,𝑆𝑥_𝑆≤𝜓_𝑥𝑑𝑇^2𝑛−1𝑥,𝑥_𝑇𝑑𝑇^2𝑛−1𝑥,𝑇^2𝑛𝑥+𝑑𝑇𝑥_𝑇,𝑥_𝑇2+𝜓_𝑥𝑑𝑇^2𝑛−1𝑥,𝑥_𝑆𝑑𝑇^2𝑛−1𝑥,𝑇^2𝑛𝑥+𝑑𝑇𝑥_𝑆,𝑥_𝑆2+𝑑𝑇𝑥_𝑆,𝑆𝑥_𝑆≤𝑑𝑇^2𝑛−1𝑥,𝑇^2𝑛𝑥+𝑑𝑇𝑥_𝑇,𝑥_𝑇2+𝑑𝑇^2𝑛−1𝑥,𝑇^2𝑛𝑥+𝑑𝑇𝑥_𝑆,𝑥_𝑆2+𝑑𝑇𝑥_𝑆,𝑆𝑥_𝑆. (2.46) Letting 𝑛→∞ and using (iv), we get that 𝑑𝑥_𝑇,𝑥_𝑆≤32𝑑𝑇𝑥_𝑆,𝑆𝑥_𝑆≤32𝜂. (2.47)

We introduce now the following class of cyclic contractive mappings. At first define the following set Let Φ be the set of functions 𝜑∶[0,∞)→[0,∞) satisfying the following conditions (Φ₁)𝜑 is lower semicontinuous; (Φ₂)𝜑(𝑡)=0 if and only if 𝑡=0.

Definition 2.8. Let 𝐴 and 𝐵 be nonempty subsets of a metric space (𝑋,𝑑). A cyclic map 𝑇∶𝐴∪𝐵→𝐴∪𝐵 is said to be a cyclic weakly orbital contraction if for some 𝑥∈𝐴, there exists 𝜑_𝑥∈Φ such that 𝑑𝑇^2𝑛𝑥,𝑇𝑦≤𝑑𝑇^2𝑛−1𝑥,𝑦−𝜑_𝑥𝑑𝑇^2𝑛−1𝑥,𝑦,∀𝑦∈𝐴,𝑛∈ℕ. (2.48)

We have the following result.

Theorem 2.9. Let 𝐴 and 𝐵 be two nonempty closed subsets of a complete metric space (𝑋,𝑑). Let 𝑇∶𝐴∪𝐵→𝐴∪𝐵 be a cyclic weakly orbital contraction. Then 𝑇 has a unique fixed point that belongs to 𝐴∩𝐵.

Proof. Since 𝑇 is a cyclic weakly orbital contraction, there exist 𝑥∈𝐴 and 𝜑_𝑥∈Φ such that (2.48) is satisfied. Taking 𝑦=𝑇^2𝑛𝑥 in (2.48), for all 𝑛∈ℕ, we get 𝑑𝑇^2𝑛𝑥,𝑇^2𝑛+1𝑥≤𝑑𝑇^2𝑛−1𝑥,𝑇^2𝑛𝑥−𝜑_𝑥𝑑𝑇^2𝑛−1𝑥,𝑇^2𝑛𝑥≤𝑑𝑇^2𝑛−1𝑥,𝑇^2𝑛𝑥. (2.49) Thus for all 𝑛∈ℕ, we have 𝑑𝑇^2𝑛𝑥,𝑇^2𝑛+1𝑥≤𝑑𝑇^2𝑛−1𝑥,𝑇^2𝑛𝑥. (2.50) Again, taking 𝑦=𝑇^2𝑛𝑥 in (2.48), for all 𝑛∈ℕ, we get 𝑑𝑇^2𝑛+1𝑥,𝑇^2𝑛+2𝑥=𝑑𝑇^2𝑛+2𝑥,𝑇^2𝑛+1𝑥≤𝑑𝑇^2𝑛+1𝑥,𝑇^2𝑛𝑥−𝜑_𝑥𝑑𝑇^2𝑛+1𝑥,𝑇^2𝑛𝑥≤𝑑𝑇^2𝑛𝑥,𝑇^2𝑛+1𝑥. (2.51) Thus for all 𝑛∈ℕ, we have 𝑑𝑇^2𝑛+1𝑥,𝑇^2𝑛+2𝑥≤𝑑𝑇^2𝑛𝑥,𝑇^2𝑛+1𝑥. (2.52) Then {𝑑(𝑇^𝑛𝑥,𝑇^𝑛+1𝑥)} is a decreasing sequence, and there exists 𝑟≥0 such that lim_𝑛→∞𝑑𝑇^𝑛𝑥,𝑇^𝑛+1𝑥=𝑟. (2.53) Letting 𝑛→∞ in (2.49), using (2.53) and the lower semicontinuity of 𝜑_𝑥, we obtain 𝑟≤𝑟−𝜑_𝑥(𝑟), (2.54) which implies from condition (Φ₂) that 𝑟=0. Then we have lim_𝑛→∞𝑑𝑇^𝑛𝑥,𝑇^𝑛+1𝑥=0. (2.55) Now, we shall prove that {𝑇^𝑛𝑥} is Cauchy. From (2.55), it is sufficient to show that {𝑇^2𝑛𝑥} is Cauchy. We argue by contradiction, suppose that {𝑇^2𝑛𝑥} is not a Cauchy sequence. Then there exists 𝜀>0 and two subsequences {𝑇^2𝑛(𝑘)𝑥} and {𝑇^2𝑚(𝑘)𝑥} of {𝑇^2𝑛𝑥} such that for all 𝑘, 𝑛(𝑘)>𝑚(𝑘)>𝑘 and 𝑑𝑇^2𝑛(𝑘)𝑥,𝑇^2𝑚(𝑘)𝑥≥𝜀. (2.56) We can take 𝑛(𝑘) the smallest integer such that the above inequality is satisfied. So, for all 𝑘, we have 𝑑𝑇^{2𝑛(𝑘)−2}𝑥,𝑇^2𝑚(𝑘)𝑥<𝜀. (2.57) Using (2.56) and (2.57), we obtain 𝜀≤𝑑𝑇^2𝑛(𝑘)𝑥,𝑇^2𝑚(𝑘)𝑥≤𝑑𝑇^2𝑛(𝑘)𝑥,𝑇^{2𝑛(𝑘)−1}𝑥+𝑑𝑇^{2𝑛(𝑘)−1}𝑥,𝑇^{2𝑛(𝑘)−2}𝑥+𝑑𝑇^{2𝑛(𝑘)−2}𝑥,𝑇^2𝑚(𝑘)𝑥<𝑑𝑇^2𝑛(𝑘)𝑥,𝑇^{2𝑛(𝑘)−1}𝑥+𝑑𝑇^{2𝑛(𝑘)−1}𝑥,𝑇^{2𝑛(𝑘)−2}𝑥+𝜀. (2.58) Letting 𝑘→∞ and using (2.55), we have lim_𝑘→∞𝑑𝑇^2𝑛(𝑘)𝑥,𝑇^2𝑚(𝑘)𝑥=𝜀. (2.59) On the other hand, for all 𝑘, we have 𝑑𝑇^2𝑛(𝑘)𝑥,𝑇^{2𝑚(𝑘)+1}𝑥−𝑑𝑇^2𝑛(𝑘)𝑥,𝑇^2𝑚(𝑘)𝑥≤𝑑𝑇^{2𝑚(𝑘)+1}𝑥,𝑇^2𝑚(𝑘)𝑥. (2.60) Letting 𝑘→∞, using (2.55) and (2.59), we have lim_𝑘→∞𝑑𝑇^2𝑛(𝑘)𝑥,𝑇^{2𝑚(𝑘)+1}𝑥=𝜀. (2.61) Similarly, for all 𝑘, we have 𝑑𝑇^{2𝑛(𝑘)−1}𝑥,𝑇^2𝑚(𝑘)𝑥−𝑑𝑇^2𝑛(𝑘)𝑥,𝑇^2𝑚(𝑘)𝑥≤𝑑𝑇^{2𝑛(𝑘)−1}𝑥,𝑇^2𝑛(𝑘)𝑥. (2.62) Letting 𝑘→∞, using (2.55) and (2.59), we have lim_𝑘→∞𝑑𝑇^{2𝑛(𝑘)−1}𝑥,𝑇^2𝑚(𝑘)𝑥=𝜀. (2.63) Now, applying (2.48) with 𝑦=𝑇^2𝑚(𝑘)𝑥, for all 𝑘, we have 𝑑𝑇^2𝑛(𝑘)𝑥,𝑇^{2𝑚(𝑘)+1}𝑥≤𝑑𝑇^{2𝑛(𝑘)−1}𝑥,𝑇^2𝑚(𝑘)𝑥−𝜑_𝑥𝑑𝑇^{2𝑛(𝑘)−1}𝑥,𝑇^2𝑚(𝑘)𝑥. (2.64) Letting 𝑘→∞ in the above inequality, using (2.59), (2.63), and the lower semi-continuity of 𝜑_𝑥, we obtain 𝜀≤𝜀−𝜑_𝑥(𝜀), (2.65) which implies from (Φ₂) that 𝜀=0, a contradiction. Thus we proved that {𝑇^𝑛𝑥} is a Cauchy sequence. Since (𝑋,𝑑) is complete, 𝐴 and 𝐵 are closed, and {𝑇^𝑛𝑥}⊂𝐴∪𝐵, there exists a 𝑧∈𝐴∪𝐵 such that 𝑇^𝑛𝑥→𝑧 as 𝑛→∞. On the other hand, since 𝑇 is a cyclic mapping and 𝑥∈𝐴, we have {𝑇^2𝑛𝑥}⊂𝐴 and {𝑇^2𝑛+1𝑥}⊂𝐵. Since 𝐴 and 𝐵 are closed, this implies that 𝑧∈𝐴∩𝐵. Now, we will prove that 𝑧 is a fixed point of 𝑇. From (2.48), for all 𝑛∈ℕ, we have 𝑑𝑇^2𝑛𝑥,𝑇𝑧≤𝑑𝑇^2𝑛−1𝑥,𝑧−𝜑_𝑥𝑑𝑇^2𝑛−1𝑥,𝑧. (2.66) Letting 𝑛→∞ and using the properties of 𝜑_𝑥, we get 𝑑(𝑧,𝑇𝑧)≤𝑑(𝑧,𝑧)−𝜑_𝑥(𝑑(𝑧,𝑧))=0, (2.67) which implies that 𝑑(𝑧,𝑇𝑧)=0, that is, 𝑧=𝑇𝑧. Then 𝑧 is a fixed point of 𝑇. Suppose now that 𝑤∈𝐴∩𝐵 is a fixed point of 𝑇. From (2.48), for all 𝑛∈ℕ, we have 𝑑(𝑧,𝑤)=𝑑𝑇^2𝑛𝑧,𝑇𝑤≤𝑑𝑇^2𝑛−1𝑥,𝑤−𝜑_𝑥𝑑𝑇^2𝑛−1𝑥,𝑤. (2.68) Letting 𝑛→∞, we get 𝑑(𝑧,𝑤)≤𝑑(𝑧,𝑤)−𝜑_𝑥(𝑑(𝑧,𝑤)), (2.69) which implies that 𝑑(𝑧,𝑤)=0, that is, 𝑧=𝑤. Then 𝑇 has a unique fixed point.

Denote now by Λ the set of functions 𝛼∶[0,∞)→[0,∞) satisfying (Λ₁)𝛼 is locally integrable on [0,∞); (Λ₂) for all 𝜀>0, we have ∫^𝜀₀𝛼(𝑠)𝑑𝑠>0. The following result is an immediate consequence of Theorem 2.9.

Theorem 2.10. Let 𝐴 and 𝐵 be two nonempty closed subsets of a complete metric space (𝑋,𝑑). Let 𝑇∶𝐴∪𝐵→𝐴∪𝐵 be a cyclic mapping. Suppose that for some 𝑥∈𝐴 there exists 𝛼_𝑥∈Λ such that 𝑑𝑇^2𝑛𝑥,𝑇𝑦≤𝑑𝑇^2𝑛−1𝑥,𝑦−^{𝑑(𝑇2𝑛−1}₀^𝑥,𝑦)𝛼_𝑥(𝑠)𝑑𝑠, (2.70) for all 𝑛∈ℕ and 𝑦∈𝐴. Then 𝑇 has a unique fixed point that belongs to 𝐴∩𝐵.

Remark 2.11. Clearly, any cyclic orbital contraction is a cyclic weakly orbital contraction. So Theorem 1.3 is a particular case of our Theorem 2.9.