Some Common Fixed-Point Theorems for Generalized-Contractive-Type Mappings on Complex-Valued Metric Spaces

Abstract

Fixed-point theory in complex valued metric spaces has greatly developed in recent times. In this paper, we prove certain common fixed-point theorems for two single-valued mappings in such spaces. The mappings we consider here are assumed to satisfy certain metric inequalities with generalized fixed-point theorems due to Rouzkard and Imdad (2012). This extends and subsumes many results of other authors which were obtained for mappings on complex-valued metric spaces.

1. Introduction

The existence and uniqueness of fixed-point theorems of operators or mappings has been a subject of great interest since the work of Banach in 1992 [1]. The Banach contraction mapping principle is widely recognized as the source of metric fixed-point theory. A mapping 𝑇∶𝑋→𝑋, where (𝑋,𝑑) is a metric space, is said to be a contraction mapping if for all 𝑥,𝑦∈𝑋, 𝑑𝑇𝑥,𝑇𝑦≤𝜆𝑑𝑥,𝑦,where0<𝜆<1. (1) According to the Banach contraction mapping principle, any mapping 𝑇 satisfying (1) in a complete metric space will have a unique fixed point. This principle includes different directions in different spaces adopted by mathematicians; for example, metric spaces, 𝐺-metric spaces, partial metric spaces, cone metric spaces, quasimetric spaces have already been obtained.

A new space called the complex-valued metric space which is more general than well-known metric spaces has been introduced by Azam et al. [2]. Azam proved some fixed-point theorems for mappings satisfying a rational inequality. Naturally, this new idea can be utilized to define complex-valued normed spaces and complex-valued inner product spaces which, in turn, offer a wide scope for further investigation. Several authors studied many common fixed point results on complex-valued metric spaces (see [3–5]).

In 2012, Rouzkard and Imdad [6] extended and improved the common fixed-point theorems which are more general than the result of Azam et al. [2].

Theorem 1 (see [6, Theorem 1]). If 𝑆 and 𝑇 are self-mappings defined on a complete complex-valued metric space (𝑋,𝑑) satisfying the condition 𝑑+()𝑑𝑆𝑥,𝑇𝑦⪷𝜆𝑑𝑥,𝑦𝜇𝑑𝑥,𝑆𝑥𝑦,𝑇𝑦+𝑑1+𝑑𝑥,𝑦𝛾𝑑𝑦,𝑆𝑥𝑥,𝑇𝑦1+𝑑𝑥,𝑦 (2) for all 𝑥,𝑦∈𝑋 where 𝜆, 𝜇, and 𝛾 are nonnegative with 𝜆+𝜇+𝛾<1, then 𝑆 and 𝑇 have a unique common fixed point.

Though complex-valued metric spaces from a spacial class of cone metric spaces, yet this idea is intended to define rational expressions which are not meaningful in cone metric spaces, and thus many results of analysis cannot be generalized to cone metric spaces. The aim of this paper is to establish some common fixed-point theorems for two nonlinear general contraction mappings in complex-valued metric spaces. Our results generalized Theorem 1.

2. Preliminaries

Let ℂ be the set of complex numbers and 𝑧₁,𝑧₂∈ℂ, we define a partial order ≺ and ⪷ on ℂ as follows:(i)𝑧₁≺𝑧₂ if and only if Re(𝑧₁)<Re(𝑧₂) and Im(𝑧₁)<Im(𝑧₂);(ii)𝑧₁⪷𝑧₂ if and only if Re(𝑧₁)≤Re(𝑧₂) and Im(𝑧₁)≤Im(𝑧₂).

Now, we briefly review the notation about complex valued metric space and some lemma for prove our main results.

Definition 2. Let 𝑋 be a nonempty set. Suppose that the mapping 𝑑∶𝑋×𝑋→ℂ satisfies the following conditions:(d₁)0⪷𝑑(𝑥,𝑦) for all 𝑥,𝑦∈𝑋;(d₂)𝑑(𝑥,𝑦)=0 if and only if 𝑥=𝑦 for all 𝑥,𝑦∈𝑋;(d₃)𝑑(𝑥,𝑦)=𝑑(𝑦,𝑥) for all 𝑥,𝑦∈𝑋;(d₄)𝑑(𝑥,𝑦)⪷𝑑(𝑥,𝑧)+𝑑(𝑧,𝑦) for all 𝑥,𝑦,𝑧∈𝑋.
Then, 𝑑 is called a complex-valued metric on 𝑋, and (𝑋,𝑑) is called a complex valued metric space.

Definition 3. Let (𝑋,𝑑) be a complex-valued metric space.(i)A point 𝑥∈𝑋 is called interior point of a set 𝐵⊆𝑋 whenever there exists 0≺𝑟∈ℂ such that 𝑁(𝑥,𝑟)∶={𝑦∈𝑋∶𝑑(𝑥,𝑦)≺𝑟}⊆𝐵.(ii)A point 𝑥∈𝑋 is called limit point of a set 𝐵⊆𝑋 whenever for every 0≺𝑟∈ℂ such that 𝑁(𝑥,𝑦)∩(𝑋−𝐵)≠∅.(iii)A subset 𝐵⊆𝑋 is called open whenever each element of 𝐵 is an interior point of 𝐵.(iv)A subset 𝐵⊆𝑋 is called closed whenever each limit point of 𝐵 belongs to 𝐵. (v)The family 𝐹={𝑁(𝑥,𝑟)∶𝑥∈𝑋,0≺𝑟} is a subbasis for a topology on 𝑋. We denote this complex topology by 𝜏_𝑐. Indeed, the topology 𝜏_𝑐 is Hausdorff.

Definition 4 (see [2]). Let (𝑋,𝑑) be a complex-valued metric space, and let {𝑥_𝑛} be a sequence in 𝑋 and 𝑥∈𝑋.(i)If for every 𝑐∈ℂ, with 0≺𝑐 there is 𝑁∈ℕ such that for all 𝑛>𝑁, 𝑑(𝑥_𝑛,𝑥)≺𝑐, then {𝑥_𝑛} is said to be convergent, {𝑥_𝑛} converges to 𝑥 and 𝑥 is limit point of {𝑥_𝑛}. We denote this by 𝑥_𝑛→𝑥 as 𝑛→∞ or lim_𝑛→∞𝑥_𝑛=𝑥.(ii)If for every 𝑐∈ℂ, with 0≺𝑐 there is 𝑁∈ℕ such that for all 𝑛>𝑁, 𝑑(𝑥_𝑛,𝑥_𝑛+𝑚)≺𝑐, where 𝑚∈ℕ, then {𝑥_𝑛} is said to be Cauchy sequence. (iii)If every Cauchy sequence in 𝑋 is convergent, then (𝑋,𝑑) is said to be a complete complex-valued metric space.

Lemma 5 (see [2]). Let (𝑋,𝑑) be a complex-valued metric space, and let {𝑥_𝑛} be a sequence in 𝑋. Then, {𝑥_𝑛} converges to 𝑥 if and only if |𝑑(𝑥_𝑛,𝑥)|→0 as 𝑛→∞.

Lemma 6 (see [2]). Let (𝑋,𝑑) be a complex-valued metric space, and let {𝑥_𝑛} be a sequence in 𝑋. Then, {𝑥_𝑛} is a Cauchy sequence if and only if |𝑑(𝑥_𝑛,𝑥_𝑛+𝑚)|→0 as 𝑛→∞, where 𝑚∈ℕ.

Definition 7. Two families of self-mappings {𝑇_𝑖}^𝑚_𝑖=1 and {𝑆_𝑖}^𝑛_𝑖=1 are said to be pairwise commuting if:(i)𝑇_𝑖𝑇_𝑗=𝑇_𝑗𝑇_𝑖, 𝑖,𝑗∈{1,2,…,𝑚}.(ii)𝑆_𝑖𝑆_𝑗=𝑆_𝑗𝑆_𝑖, 𝑖,𝑗∈{1,2,…,𝑛}.(iii)𝑇_𝑖𝑆_𝑗=𝑆_𝑗𝑇_𝑖, 𝑖∈{1,2,…,𝑚}, 𝑗∈{1,2,…,𝑛}.

Definition 8. Let 𝑆 and 𝑇 be self-mappings of a nonempty set 𝑋.(i)A point 𝑥∈𝑋 is said to be a fixed point of 𝑇 if 𝑇𝑥=𝑥.(ii)A point 𝑥∈𝑋 is said to be a common fixed point of 𝑇 and 𝑆 if 𝑇𝑥=𝑆𝑥=𝑥.

Remark 9. We obtain that the following statements hold:(i)If 𝑧₁⪷𝑧₂ and 𝑧₂⪷𝑧₃, then 𝑧₁⪷𝑧₃.(ii)If 𝑧∈ℂ, 𝑎,𝑏∈ℝ, and 𝑎≤𝑏, then 𝑎𝑧⪷𝑏𝑧.(iii)If 0⪷𝑧₁⪷𝑧₂, then |𝑧₁|≤|𝑧₂|.

3. Main Results

In this section, we will prove some common fixed-point theorems for the generalized contractive mappings in complex-valued metric space.

Theorem 10. If 𝑆 and 𝑇 are self-mappings defined on a complete complex valued metric space (𝑋,𝑑) satisfying the condition 𝑑+()𝑑𝑆𝑥,𝑇𝑦⪷𝐴𝑑𝑥,𝑦𝐵𝑑𝑥,𝑆𝑥𝑦,𝑇𝑦+𝑑1+𝑑𝑥,𝑦𝐶𝑑𝑦,𝑆𝑥𝑥,𝑇𝑦+()𝑑1+𝑑𝑥,𝑦𝐷𝑑𝑥,𝑆𝑥𝑥,𝑇𝑦+𝑑1+𝑑𝑥,𝑦𝐸𝑑𝑦,𝑆𝑥𝑦,𝑇𝑦1+𝑑𝑥,𝑦 (3) for all 𝑥,𝑦∈𝑋, where 𝐴, 𝐵, 𝐶, 𝐷, and 𝐸 are nonnegative with 𝐴+𝐵+𝐶+2𝐷+2𝐸<1, then 𝑆 and 𝑇 have a unique common fixed point.

Proof. Let 𝑥₀ be an arbitrary in 𝑋. Since 𝑆(𝑋)⊆𝑋 and 𝑇(𝑋)⊆𝑋, we construct the sequence {𝑥_𝑘} in 𝑋 such that 𝑥_2𝑘+1=𝑆𝑥_2𝑘 and 𝑥_2𝑘+2=𝑇𝑥_2𝑘+1 for all 𝑘≥0. From the definition of {𝑥_𝑘} and (3), we obtain that 𝑑𝑥_2𝑘+1,𝑥_2𝑘+2=𝑑𝑆𝑥_2𝑘,𝑇𝑥_2𝑘+1⪷𝐴𝑑𝑥_2𝑘,𝑥_2𝑘+1+𝐵𝑑𝑥_2𝑘,𝑆𝑥_2𝑘𝑑𝑥_2𝑘+1,𝑇𝑥_2𝑘+11+𝑑𝑥_2𝑘,𝑥_2𝑘+1+𝐶𝑑𝑥_2𝑘+1,𝑆𝑥_2𝑘𝑑𝑥_2𝑘,𝑇𝑥_2𝑘+11+𝑑𝑥_2𝑘,𝑥_2𝑘+1+𝐷𝑑𝑥_2𝑘,𝑆𝑥_2𝑘𝑑𝑥_2𝑘,𝑇𝑥_2𝑘+11+𝑑𝑥_2𝑘,𝑥_2𝑘+1+𝐸𝑑𝑥_2𝑘+1,𝑆𝑥_2𝑘𝑑𝑥_2𝑘+1,𝑇𝑥_2𝑘+11+𝑑𝑥_2𝑘,𝑥_2𝑘+1. (4) Since 𝑥_2𝑘+1=𝑆𝑥_2𝑘 implies that 𝑑(𝑥_2𝑘+1,𝑆𝑥_2𝑘)=0; therefore, 𝑑𝑥_2𝑘+1,𝑥_2𝑘+2⪷𝐴𝑑𝑥_2𝑘,𝑥_2𝑘+1+𝐵𝑑𝑥_2𝑘,𝑆𝑥_2𝑘𝑑𝑥_2𝑘+1,𝑇𝑥_2𝑘+11+𝑑𝑥_2𝑘,𝑥_2𝑘+1+𝐷𝑑𝑥_2𝑘,𝑆𝑥_2𝑘𝑑𝑥_2𝑘,𝑇𝑥_2𝑘+11+𝑑𝑥_2𝑘,𝑥_2𝑘+1, (5) by Remark 9 and |1+𝑑(𝑥_2𝑘,𝑥_2𝑘+1)|>|𝑑(𝑥_2𝑘,𝑥_2𝑘+1)|, we have 𝑑𝑥_2𝑘+1,𝑥_2𝑘+2≤𝐴𝑑𝑥_2𝑘,𝑥_2𝑘+1+𝐵𝑑𝑥_2𝑘,𝑆𝑥_2𝑘𝑑𝑥_2𝑘+1,𝑇𝑥_2𝑘+11+𝑑𝑥_2𝑘,𝑥_2𝑘+1+𝐷𝑑𝑥_2𝑘,𝑆𝑥_2𝑘𝑑𝑥_2𝑘,𝑇𝑥_2𝑘+11+𝑑𝑥_2𝑘,𝑥_2𝑘+1=𝐴𝑑𝑥_2𝑘,𝑥_2𝑘+1+𝐵𝑑𝑥_2𝑘+1,𝑥_2𝑘+2𝑑𝑥_2𝑘,𝑥_2𝑘+11+𝑑𝑥_2𝑘,𝑥_2𝑘+1+𝐷𝑑𝑥_2𝑘,𝑥_2𝑘+2𝑑𝑥_2𝑘,𝑥_2𝑘+11+𝑑𝑥_2𝑘,𝑥_2𝑘+1<𝐴𝑑𝑥_2𝑘,𝑥_2𝑘+1+𝐵𝑑𝑥_2𝑘+1,𝑥_2𝑘+2+𝐷𝑑𝑥_2𝑘,𝑥_2𝑘+2. (6) From (6) and Definition 2, we have 𝑑𝑥_2𝑘+1,𝑥_2𝑘+2<𝐴𝑑𝑥_2𝑘,𝑥_2𝑘+1+𝐵𝑑𝑥_2𝑘+1,𝑥_2𝑘+2+𝐷𝑑𝑥_2𝑘,𝑥_2𝑘+1+𝐷𝑑𝑥_2𝑘+1,𝑥_2𝑘+2 (7) it follows that |𝑑(𝑥_2𝑘+1,𝑥_2𝑘+2)|<((𝐴+𝐷)/(1−𝐵−𝐷))|𝑑(𝑥_2𝑘,𝑥_2𝑘+1)|.
Similarly, we get 𝑑𝑥_2𝑘+2,𝑥_2𝑘+3=𝑑𝑥_2𝑘+3,𝑥_2𝑘+2=𝑑𝑆𝑥_2𝑘+2,𝑇𝑥_2𝑘+1⪷𝐴𝑑𝑥_2𝑘+2,𝑥_2𝑘+1+𝐵𝑑𝑥_2𝑘+2,𝑆𝑥_2𝑘+2𝑑𝑥_2𝑘+1,𝑇𝑥_2𝑘+11+𝑑𝑥_2𝑘+2,𝑥_2𝑘+1+𝐶𝑑𝑥_2𝑘+1,𝑆𝑥_2𝑘+2𝑑𝑥_2𝑘+2,𝑇𝑥_2𝑘+11+𝑑𝑥_2𝑘+2,𝑥_2𝑘+1+𝐷𝑑𝑥_2𝑘+2,𝑆𝑥_2𝑘+2𝑑𝑥_2𝑘+2,𝑇𝑥_2𝑘+11+𝑑𝑥_2𝑘+2,𝑥_2𝑘+1+𝐸𝑑𝑥_2𝑘+1,𝑆𝑥_2𝑘+2𝑑𝑥_2𝑘+1,𝑇𝑥_2𝑘+11+𝑑𝑥_2𝑘+2,𝑥_2𝑘+1. (8) Since 𝑥_2𝑘+2=𝑇𝑥_2𝑘+1 implies that 𝑑(𝑥_2𝑘+2𝑇𝑥_2𝑘+1)=0; therefore, 𝑑𝑥_2𝑘+2,𝑥_2𝑘+3⪷𝐴𝑑𝑥_2𝑘+2,𝑥_2𝑘+1+𝐵𝑑𝑥_2𝑘+2,𝑆𝑥_2𝑘+2𝑑𝑥_2𝑘+1,𝑇𝑥_2𝑘+11+𝑑𝑥_2𝑘+2,𝑥_2𝑘+1+𝐸𝑑𝑥_2𝑘+1,𝑆𝑥_2𝑘+2𝑑𝑥_2𝑘+1,𝑇𝑥_2𝑘+11+𝑑𝑥_2𝑘+2,𝑥_2𝑘+1; (9) by Remark 9 and |1+𝑑(𝑥_2𝑘+2,𝑥_2𝑘+1)|>|𝑑(𝑥_2𝑘+2,𝑥_2𝑘+1)|, we have 𝑑𝑥_2𝑘+2,𝑥_2𝑘+3≤𝐴𝑑𝑥_2𝑘+2,𝑥_2𝑘+1+𝐵𝑑𝑥_2𝑘+2,𝑆𝑥_2𝑘+2𝑑𝑥_2𝑘+1,𝑇𝑥_2𝑘+11+𝑑𝑥_2𝑘+2,𝑥_2𝑘+1+𝐸𝑑𝑥_2𝑘+1,𝑆𝑥_2𝑘+2𝑑𝑥_2𝑘+1,𝑇𝑥_2𝑘+11+𝑑𝑥_2𝑘+2,𝑥_2𝑘+1=𝐴𝑑𝑥_2𝑘+2,𝑥_2𝑘+1+𝐵𝑑𝑥_2𝑘+2,𝑥_2𝑘+3𝑑𝑥_2𝑘+1,𝑥_2𝑘+21+𝑑𝑥_2𝑘+2,𝑥_2𝑘+1+𝐸𝑑𝑥_2𝑘+1,𝑥_2𝑘+3𝑑𝑥_2𝑘+1,𝑥_2𝑘+21+𝑑𝑥_2𝑘+2,𝑥_2𝑘+1<𝐴𝑑𝑥_2𝑘+2,𝑥_2𝑘+1+𝐵𝑑𝑥_2𝑘+2,𝑥_2𝑘+3+𝐸𝑑𝑥_2𝑘+1,𝑥_2𝑘+3. (10) From (10) and Definition 2, we have 𝑑𝑥_2𝑘+1,𝑥_2𝑘+2<𝐴𝑑𝑥_2𝑘+2,𝑥_2𝑘+1+𝐵𝑑𝑥_2𝑘+2,𝑥_2𝑘+3+𝐸𝑑𝑥_2𝑘+1,𝑥_2𝑘+2+𝐷𝑑𝑥_2𝑘+2,𝑥_2𝑘+3; (11) it follows that |𝑑(𝑥_2𝑘+2,𝑥_2𝑘+3)|<((𝐴+𝐸)/(1−𝐵−𝐸))|𝑑(𝑥_2𝑘+1,𝑥_2𝑘+2)|.
Putting 𝑘=max{((𝐴+𝐷)/(1−𝐵−𝐷)),((𝐴+𝐸)/(1−𝐵−𝐸))}, we obtain that 𝑑𝑥_𝑛,𝑥_𝑛+1≤𝑘𝑑𝑥_𝑛−1,𝑥_𝑛≤𝑘²𝑑𝑥_𝑛−2,𝑥_𝑛−1≤⋯≤𝑘^𝑛𝑑𝑥₀,𝑥₁∀𝑛. (12) Thus, for any 𝑛∈ℕ, we have 𝑑𝑥_𝑛,𝑥_𝑚≤𝑑𝑥_𝑛,𝑥_𝑛+1≤𝑑𝑥_𝑛+1,𝑥_𝑛+2≤⋯≤𝑑𝑥_𝑚−1,𝑥_𝑚≤𝑘^𝑛+𝑘^𝑛+1+𝑘^𝑛+2+⋯+𝑘^𝑚−1𝑑𝑥₀,𝑥₁≤𝑘^𝑛𝑑𝑥1−𝑘⁰,𝑥₁; (13) it follows that |𝑑(𝑥_𝑛,𝑥_𝑚)|≤(𝑘^𝑛/(1−𝑘))|𝑑(𝑥₀,𝑥₁)|→0 as 𝑛→∞.
By Lemma 6, the sequence {𝑥_𝑛} is a Cauchy. Since 𝑋 is compete, there exists a point 𝑧∈𝑋 such that 𝑥_𝑛→𝑧 as 𝑛→∞.
Next, we will show that 𝑆𝑧=𝑧. By the notion of a complete complex-valued metric 𝑑, we have 𝑑(𝑧,𝑆𝑧)⪷𝑑𝑧,𝑥_2𝑘+2+𝑑𝑥_2𝑘+2,𝑆𝑧=𝑑𝑧,𝑥_2𝑘+2+𝑑𝑆𝑧,𝑇𝑥_2𝑘+1⪷𝑑𝑧,𝑥_2𝑘+2+𝐴𝑑𝑧,𝑥_2𝑘+1+𝐵𝑑(𝑧,𝑆𝑧)𝑑𝑥_2𝑘+1,𝑇𝑥_2𝑘+11+𝑑𝑧,𝑥_2𝑘+1+𝐶𝑑𝑥_2𝑘+1,𝑆𝑧𝑑𝑧,𝑇𝑥_2𝑘+11+𝑑𝑧,𝑥_2𝑘+1+𝐷𝑑(𝑧,𝑆𝑧)𝑑𝑧,𝑇𝑥_2𝑘+11+𝑑𝑧,𝑥_2𝑘+1+𝐸𝑑𝑥_2𝑘+1,𝑆𝑧𝑑𝑥_2𝑘+1,𝑇𝑥_2𝑘+11+𝑑𝑧,𝑥_2𝑘+1, (14) which implies that |𝑑(𝑧,𝑆𝑧)|≤𝑑𝑧,𝑥_2𝑘+2+𝐴𝑑𝑧,𝑥_2𝑘+1+𝐵|𝑑(𝑧,𝑆𝑧)|𝑑𝑥_2𝑘+1,𝑥_2𝑘+21+𝑑𝑧,𝑥_2𝑘+1+𝐶𝑑𝑥_2𝑘+1,𝑆𝑧𝑑𝑧,𝑥_2𝑘+21+𝑑𝑧,𝑥_2𝑘+1+𝐷|𝑑(𝑧,𝑆𝑧)|𝑑𝑧,𝑥_2𝑘+21+𝑑𝑧,𝑥_2𝑘+1+𝐸𝑑𝑥_2𝑘+1,𝑆𝑧𝑑𝑥_2𝑘+1,𝑥_2𝑘+21+𝑑𝑧,𝑥_2𝑘+1. (15) Taking 𝑘→∞, we have |𝑑(𝑧,𝑆𝑧)|=0; it is obtained that 𝑑(𝑧,𝑆𝑧)=0. Thus, 𝑆𝑧=𝑧. It follows that similarly 𝑇𝑧=𝑧. Therefore, 𝑧 is common fixed point of 𝑆 and 𝑇.
Finally, to prove the uniqueness of common fixed point, let 𝑧^∗∈𝑋 be another common fixed point of 𝑆 and 𝑇 such that 𝑆𝑧^∗=𝑇𝑧^∗=𝑧^∗. Consider 𝑑𝑧,𝑧^∗=𝑑𝑆𝑧,𝑇𝑧^∗⪷𝐴𝑑𝑧,𝑧^∗+𝐵𝑑(𝑧,𝑆𝑧)𝑑𝑧^∗,𝑇𝑧^∗1+𝑑(𝑧,𝑧^∗)+𝐶𝑑𝑧^∗,𝑆𝑧𝑑𝑧,𝑇𝑧^∗1+𝑑(𝑧,𝑧^∗)+𝐷𝑑(𝑧,𝑆𝑧)𝑑𝑧,𝑇𝑧^∗1+𝑑(𝑧,𝑧^∗)+𝐸𝑑𝑧^∗,𝑆𝑧𝑑𝑧^∗,𝑇𝑧^∗1+𝑑(𝑧,𝑧^∗) (16) so that 𝑑𝑧,𝑧^∗≤𝐴𝑑𝑧,𝑧^∗+𝐶𝑑𝑧^∗,𝑆𝑧𝑑𝑧,𝑇𝑧^∗|1+𝑑(𝑧,𝑧^∗𝑑)|=𝐴𝑧,𝑧^∗+𝐶𝑑𝑧^∗𝑑,𝑧𝑧,𝑧^∗|1+𝑑(𝑧,𝑧^∗.)| (17) Since |1+𝑑(𝑧,𝑧^∗)|>|𝑑(𝑧,𝑧^∗)|, therefore |𝑑(𝑧,𝑧^∗)|<𝐴|𝑑(𝑧,𝑧^∗)|+𝐶|𝑑(𝑧^∗,𝑧)|=(𝐴+𝐶)|𝑑(𝑧,𝑧^∗)|.
This is contradiction to 𝐴+𝐶<1. Hence, 𝑧=𝑧^∗. Therefore, 𝑧 is a unique common fixed point of 𝑆 and 𝑇.

Corollary 11. If 𝑇 is a self-mapping defined on a complete complex-valued metric space (𝑋,𝑑) satisfying the condition 𝑑+()𝑑𝑇𝑥,𝑇𝑦⪷𝐴𝑑𝑥,𝑦𝐵𝑑𝑥,𝑇𝑥𝑦,𝑇𝑦+𝑑1+𝑑𝑥,𝑦𝐶𝑑𝑦,𝑇𝑥𝑥,𝑇𝑦+()𝑑1+𝑑𝑥,𝑦𝐷𝑑𝑥,𝑇𝑥𝑥,𝑇𝑦+𝑑1+𝑑𝑥,𝑦𝐸𝑑𝑦,𝑇𝑥𝑦,𝑇𝑦1+𝑑𝑥,𝑦 (18) for all 𝑥,𝑦∈𝑋, where 𝐴, 𝐵, 𝐶, 𝐷, and 𝐸 are nonnegative with 𝐴+𝐵+𝐶+2𝐷+2𝐸<1, then 𝑇 has a unique fixed point.

Proof. We can prove this result by applying Theorem 10 by setting 𝑇=𝑆.

Corollary 12. If 𝑆 and 𝑇 are self-mappings defined on a complete complex valued metric space (𝑋,𝑑) satisfying the condition 𝑑+()𝑑𝑆𝑥,𝑇𝑦⪷𝐴𝑑𝑥,𝑦𝐵𝑑𝑥,𝑆𝑥𝑦,𝑇𝑦+𝑑1+𝑑𝑥,𝑦𝐶𝑑𝑦,𝑆𝑥𝑥,𝑇𝑦+()𝑑1+𝑑𝑥,𝑦𝐷𝑑𝑥,𝑆𝑥𝑥,𝑇𝑦1+𝑑𝑥,𝑦 (19) for all 𝑥,𝑦∈𝑋, where 𝐴, 𝐵, 𝐶, and 𝐷 are nonnegative with 𝐴+𝐵+𝐶+2𝐷<1, then 𝑆 and 𝑇 have a unique common fixed point.

Proof. We can prove this result by applying Theorem 10 by setting 𝐸=0.

Corollary 13. If 𝑇 is a self-mapping defined on a complete complex valued metric space (𝑋,𝑑) satisfying the condition 𝑑+()𝑑𝑇𝑥,𝑇𝑦⪷𝐴𝑑𝑥,𝑦𝐵𝑑𝑥,𝑇𝑥𝑦,𝑇𝑦+𝑑1+𝑑𝑥,𝑦𝐶𝑑𝑦,𝑇𝑥𝑥,𝑇𝑦+()𝑑1+𝑑𝑥,𝑦𝐷𝑑𝑥,𝑇𝑥𝑥,𝑇𝑦1+𝑑𝑥,𝑦 (20) for all 𝑥,𝑦∈𝑋, where 𝐴, 𝐵, 𝐶, and 𝐷 are nonnegative with 𝐴+𝐵+𝐶+2𝐷<1, then 𝑇 has a unique fixed point.

Proof. We can prove this result by applying Corollary 12 by setting 𝑇=𝑆 and 𝐸=0.

Corollary 14. If 𝑆 and 𝑇 are self-mappings defined on a complete complex valued metric space (𝑋,𝑑) satisfying the condition 𝑑+()𝑑𝑆𝑥,𝑇𝑦⪷𝐴𝑑𝑥,𝑦𝐵𝑑𝑥,𝑆𝑥𝑦,𝑇𝑦+𝑑1+𝑑𝑥,𝑦𝐶𝑑𝑦,𝑆𝑥𝑥,𝑇𝑦+𝑑1+𝑑𝑥,𝑦𝐸𝑑𝑦,𝑆𝑥𝑦,𝑇𝑦1+𝑑𝑥,𝑦 (21) for all 𝑥,𝑦∈𝑋 where 𝐴, 𝐵, 𝐶, and 𝐸 are nonnegative with 𝐴+𝐵+𝐶+2𝐸<1, then 𝑆 and 𝑇 have a unique common fixed point.

Proof. We can prove this result by applying Theorem 10 by setting 𝐷=0.

Corollary 15. If 𝑇 is a self-mapping defined on a complete complex valued metric space (𝑋,𝑑) satisfying the condition 𝑑+()𝑑𝑇𝑥,𝑇𝑦⪷𝐴𝑑𝑥,𝑦𝐵𝑑𝑥,𝑇𝑥𝑦,𝑇𝑦+𝑑1+𝑑𝑥,𝑦𝐶𝑑𝑦,𝑇𝑥𝑥,𝑇𝑦+()𝑑1+𝑑𝑥,𝑦𝐷𝑑𝑥,𝑇𝑥𝑥,𝑇𝑦+𝑑1+𝑑𝑥,𝑦𝐸𝑑𝑦,𝑇𝑥𝑦,𝑇𝑦1+𝑑𝑥,𝑦 (22) for all 𝑥,𝑦∈𝑋, where 𝐴, 𝐵, 𝐶, and 𝐸 are nonnegative with 𝐴+𝐵+𝐶+2𝐸<1, then 𝑇 has a unique fixed point.

Proof. We can prove this result by applying Corollary 14 by setting 𝑇=𝑆.

Remark 16. (i) By choosing 𝐷=𝐸=0 in Theorem 10, we get Theorem 1 of [6].
(ii) By choosing 𝐷=𝐸=0 and 𝑆=𝑇 in Theorem 10, we get Corollary 3 of [6].
(iii) By choosing 𝐶=𝐷=𝐸=0 in Theorem 10, we get Theorem 4 of Azam et al. [2].
(iv) By choosing 𝐶=𝐷=𝐸=0 and 𝑆=𝑇 in Theorem 10, we get Corollary 5 of Azam et al. [2].

Theorem 17. If {𝑇_𝑖}^𝑚₁ and {𝑆_𝑖}^𝑛₁ are two finite pairwise commuting finite families of self-mapping defined on complete complex-valued metric space (𝑋,𝑑) such that the mappings 𝑇 and 𝑆 with 𝑇=𝑇₁𝑇₂⋯𝑇_𝑚 and 𝑆=𝑆₁𝑆₂⋯𝑆_𝑛 satisfy condition (3), then the component maps of the two families {𝑇_𝑖}^𝑚₁ and {𝑆_𝑖}^𝑛₁ have a unique common fixed point.

Proof. By Theorem 10, one can infer that 𝑇 and 𝑆 have a unique common fixed point 𝑍(i.e.,𝑇𝑧=𝑆𝑧=𝑧). Now, we will show that 𝑧 is a common fixed point of all the component maps of both families. In view of pairwise commutativity of the families {𝑇_𝑖}^𝑚₁ and {𝑆_𝑖}^𝑛₁, for every 1≤𝑘≤𝑚, we can write 𝑇_𝑘𝑧=𝑇_𝑘𝑆𝑧=𝑆𝑇_𝑘𝑇𝑧,_𝑘𝑧=𝑇_𝑘𝑇𝑧=𝑇𝑇_𝑘𝑧. (23) It implies that 𝑇_𝑘𝑧(∀𝑘) is also a common fixed point of 𝑇 and 𝑆. By using the uniqueness of common fixed point, we have 𝑇_𝑘𝑧=𝑧(∀𝑘). Hence, 𝑧 is a common fixed point of the family {𝑇_𝑖}^𝑚₁. Similarly, we can show that 𝑧 is a common fixed point of the family {𝑆_𝑖}^𝑛₁. This completes the proof of the theorem.

Corollary 18. If 𝐹 and 𝐺 are self-mappings defined on a complete complex-valued metric space (𝑋,𝑑) satisfying the condition 𝑑𝐹^𝑚𝑥,𝐺^𝑛𝑦⪷𝐴𝑑𝑥,𝑦+𝐵𝑑𝑥,𝐹^𝑚𝑥𝑑𝑦𝐺^𝑛𝑦+1+𝑑𝑥,𝑦𝐶𝑑𝑦,𝐹^𝑚𝑥𝑑𝑥,𝐺^𝑛𝑦+1+𝑑𝑥,𝑦𝐷𝑑𝑥,𝐹^𝑚𝑥𝑑𝑥,𝐺^𝑛𝑦+1+𝑑𝑥,𝑦𝐸𝑑𝑦,𝐹^𝑚𝑥𝑑𝑦,𝐺^𝑛𝑦1+𝑑𝑥,𝑦 (24) for all 𝑥,𝑦∈𝑋, where 𝐴, 𝐵, 𝐶, 𝐷, and 𝐸 are nonnegative with 𝐴+𝐵+𝐶+2𝐷+2𝐸<1, then 𝐹 and 𝐺 have a unique common fixed point.

Proof. We can prove this result by applying Theorem 17 by setting 𝑇₁=𝑇₂=⋯=𝑇_𝑚=𝐹 and 𝑆₁=𝑆₂=⋯=𝑆_𝑛=𝐺.

Corollary 19. If 𝑇 is a self-mapping defined on a complete complex valued metric space (𝑋,𝑑) satisfying the condition 𝑑𝑇^𝑛𝑥,𝑇^𝑛𝑦⪷𝐴𝑑𝑥,𝑦+𝐵𝑑𝑥,𝑇^𝑛𝑥𝑑𝑦,𝑇^𝑛𝑦+1+𝑑𝑥,𝑦𝐶𝑑𝑦,𝑇^𝑛𝑥𝑑𝑥,𝑇^𝑛𝑦+1+𝑑𝑥,𝑦𝐷𝑑𝑥,𝑇^𝑛𝑥𝑑𝑥,𝑇^𝑛𝑦+1+𝑑𝑥,𝑦𝐸𝑑𝑦,𝑇^𝑛𝑥𝑑𝑦,𝑇^𝑛𝑦1+𝑑𝑥,𝑦 (25) for all 𝑥,𝑦∈𝑋, where 𝐴, 𝐵, 𝐶, 𝐷, and 𝐸 are nonnegative with 𝐴+𝐵+𝐶+2𝐷+2𝐸<1, then 𝑇 has a unique fixed point.

Proof. We can prove this result by applying Corollary 18 by setting 𝐹=𝐺=𝑇.

Remark 20. (i) By choosing 𝐷=𝐸=0 in Theorem 17, we get Theorem 1 of [6].
(ii) By choosing 𝐷=𝐸=0 in Corollary 18, we get Corollary 6 of [6].
(iii) By choosing 𝐷=𝐸=0 in Corollary 19, we get Corollary 7 of [6].
(iv) By choosing 𝐶=𝐷=𝐸=0 in Corollary 19, we get Corollary 6 of Azam et al. [2].

Corollary 21 (see [5]). If 𝑇∶𝑋→𝑋 is a mapping defined on a complete complex-valued metric space (𝑋,𝑑) satisfying the condition 𝑑𝑇^𝑛𝑥,𝑇^𝑛𝑦⪷𝜆𝑑𝑥,𝑦 (26) for all 𝑥,𝑦∈𝑋, where 𝜆 is nonnegative reals 𝜆<1, then 𝑇 has a unique fixed point.

The following example demonstrates the superiority of Bryant theorem [5] over Banach contraction theorem.

Example 22. Let 𝑋=ℂ be the set of complex numbers. Define 𝑑∶ℂ×ℂ→ℂ as 𝑑𝑧₁,𝑧₂=𝑥₁−𝑥₂+𝑖𝑦₁−𝑦₂, (27) where 𝑧₁=𝑥₁+𝑖𝑦₁ and 𝑧₂=𝑥₂+𝑖𝑦₂. Then, (ℂ,𝑑) is a complete complex-valued metric space. Define 𝑇∶ℂ→ℂ as 𝑇𝑥+𝑖𝑦=0;𝑥,𝑦∈𝑄,2+2𝑖;𝑥,𝑦∈𝑄^𝑐,2;𝑥∈𝑄^𝑐,𝑦∈𝑄,2𝑖;𝑥∈𝑄,𝑦∈𝑄^𝑐. (28) Now, for 𝑥=1/√3 and 𝑦=0, we get 𝑑𝑇1√3(0)()1,𝑇=𝑑2,0=2⪷𝜆𝑑√31,0=𝜆√3. (29) Thus, 𝜆≥√3, which is a contradiction as 0≤𝜆<1. However, notice that 𝑇²𝑧=0, so that 0=𝑑(𝑇²𝑧₁,𝑇²𝑧₂)⪷𝜆𝑑(𝑧₁,𝑧₂), which shows that 𝑇² satisfies the requirement of Bryant theorem and 𝑧=0 is the unique fixed point of 𝑇.

Acknowledgment

The authors would like to thank the Faculty of Science, Naresuan University, Phitsanulok, Thailand, for the financial support.