A Suzuki Type Fixed-Point Theorem

Abstract

We present a fixed-point theorem for a single-valued map in a complete metric space using implicit relation, which is a generalization of several previously stated results including that of Suziki (2008).

1. Introduction

There are a lot of generalizations of Banach fixed-point principle in the literature. See [1–5]. One of the most interesting generalizations is that given by Suzuki [6]. This interesting fixed-point result is as follows.

Theorem 1.1. Let (𝑋,𝑑) be a complete metric space, and let 𝑇 be a mapping on 𝑋. Define a nonincreasing function 𝜃 from [0,1) into (1/2,1] by 𝜃(𝑟)=√1,0≤𝑟≤5−12,1−𝑟𝑟²,√5−121≤𝑟≤√2,1,11+𝑟√2≤𝑟<1. (1.1) Assume that there exists 𝑟∈[0,1), such that 𝜃(𝑟)𝑑(𝑥,𝑇𝑥)≤𝑑𝑥,𝑦implies𝑑𝑇𝑥,𝑇𝑦≤𝑟𝑑𝑥,𝑦, (1.2) for all 𝑥,𝑦∈𝑋, then there exists a unique fixed-point 𝑧 of 𝑇. Moreover, lim_𝑛𝑇^𝑛𝑥=𝑧 for all 𝑥∈𝑋.

Like other generalizations mentioned above in this paper, the Banach contraction principle does not characterize the metric completeness of 𝑋. However, Theorem 1.1 does characterize the metric completeness as follows.

Theorem 1.2. Define a nonincreasing function 𝜃 as in Theorem 1.1, then for a metric space (𝑋,𝑑) the following are equivalent: (i)𝑋 is complete,(ii)Every mapping 𝑇 on 𝑋 satisfying (1.2) has a fixed point.

In addition to the above results, Kikkawa and Suzuki [7] provide a Kannan type version of the theorems mentioned before. In [8], it is provided a Chatterjea type version. Popescu [9] gives a Ciric type version. Recently, Kikkawa and Suzuki also provide multivalued versions which can be found in [10, 11]. Some fixed-point theorems related to Theorems 1.1 and 1.2 have also been proven in [12, 13].

The aim of this paper is to generalize the above results using the implicit relation technique in such a way that𝐹𝑑𝑇𝑥,𝑇𝑦,𝑑𝑥,𝑦,𝑑(𝑥,𝑇𝑥),𝑑𝑦,𝑇𝑦,𝑑𝑥,𝑇𝑦,𝑑𝑦,𝑇𝑥≤0, (1.3) for 𝑥,𝑦∈𝑋, where 𝐹∶[0,∞)⁶→ℝ is a function as given in Section 2.

2. Implicit Relation

Implicit relations on metric spaces have been used in many papers. See [1, 14–16].

Let ℝ₊ denote the nonnegative real numbers, and let Ψ be the set of all continuous functions 𝐹∶[0,∞)⁶→ℝ satisfying the following conditions:𝐹₁:𝐹(𝑡₁,…,𝑡₆) is nonincreasing in variables 𝑡₂,…,𝑡₆,𝐹₂: there exists 𝑟∈[0,1), such that 𝐹(𝑢,𝑣,𝑣,𝑢,𝑢+𝑣,0)≤0 (2.1) or 𝐹(𝑢,𝑣,0,𝑢+𝑣,𝑢,𝑣)≤0 (2.2) or 𝐹(𝑢,𝑣,𝑣,𝑣,𝑣,𝑣)≤0 (2.3) implies 𝑢≤𝑟𝑣,𝐹₃:𝐹(𝑢,0,0,𝑢,𝑢,0)>0, for all 𝑢>0.

Example 2.1. 𝐹(𝑡₁,…,𝑡₆)=𝑡₁−𝑟𝑡₂, where 𝑟∈[0,1). It is clear that 𝐹∈Ψ.

Example 2.2. 𝐹(𝑡₁,…,𝑡₆)=𝑡₁−𝛼[𝑡₃+𝑡₄], where 𝛼∈[0,1/2).
Let 𝐹(𝑢,𝑣,𝑣,𝑢,𝑢+𝑣,0)=𝑢−𝛼[𝑢+𝑣]≤0, then we have 𝑢≤(𝛼/(1−𝛼))𝑣. Similarly, let 𝐹(𝑢,𝑣,0,𝑢+𝑣,𝑢,𝑣)≤0, then we have 𝑢≤(𝛼/(1−𝛼))𝑣. Again, let 𝐹(𝑢,𝑣,𝑣,𝑣,𝑣,𝑣)≤0, then 𝑢≤2𝛼𝑣. Since 𝛼/(1−𝛼)≤2𝛼<1, 𝐹₂ is satisfied with 𝑟=2𝛼. Also 𝐹(𝑢,0,0,𝑢,𝑢,0)=(1−𝛼)𝑢>0, for all 𝑢>0. Therefore, 𝐹∈Ψ.

Example 2.3. 𝐹(𝑡₁,…,𝑡₆)=𝑡₁−𝛼max{𝑡₃,𝑡₄}, where 𝛼∈[0,1/2).
Let 𝐹(𝑢,𝑣,𝑣,𝑢,𝑢+𝑣,0)=𝑢−𝛼max{𝑢,𝑣}≤0, then we have 𝑢≤𝛼𝑣≤(𝛼/(1−𝛼))𝑣. Similarly, let 𝐹(𝑢,𝑣,0,𝑢+𝑣,𝑢,𝑣)≤0, then we have 𝑢≤(𝛼/(1−𝛼))𝑣. Again, let 𝐹(𝑢,𝑣,𝑣,𝑣,𝑣,𝑣)≤0, then 𝑢≤𝛼𝑣≤(𝛼/(1−𝛼))𝑣. Thus, 𝐹₂ is satisfied with 𝑟=𝛼/(1−𝛼). Also 𝐹(𝑢,0,0,𝑢,𝑢,0)=(1−𝛼)𝑢>0, for all 𝑢>0. Therefore, 𝐹∈Ψ.

Example 2.4. 𝐹(𝑡₁,…,𝑡₆)=𝑡₁−𝛼[𝑡₅+𝑡₆], where 𝛼∈[0,1/2).
Let 𝐹(𝑢,𝑣,𝑣,𝑢,𝑢+𝑣,0)=𝑢−𝛼[𝑢+𝑣]≤0, then we have 𝑢≤(𝛼/(1−𝛼))𝑣. Similarly, let 𝐹(𝑢,𝑣,0,𝑢+𝑣,𝑢,𝑣)≤0, then we have 𝑢≤(𝛼/(1−𝛼))𝑣. Again, let 𝐹(𝑢,𝑣,𝑣,𝑣,𝑣,𝑣)≤0, then 𝑢≤2𝛼𝑣. Since 𝛼/(1−𝛼)≤2𝛼<1, 𝐹₂ is satisfied with 𝑟=2𝛼. Also 𝐹(𝑢,0,0,𝑢,𝑢,0)=(1−𝛼)𝑢>0, for all 𝑢>0. Therefore, 𝐹∈Ψ.

Example 2.5. 𝐹(𝑡₁,…,𝑡₆)=𝑡₁−𝑎𝑡₃−𝑏𝑡₄, where 𝑎,𝑏∈[0,1/2).
Let 𝐹(𝑢,𝑣,𝑣,𝑢,𝑢+𝑣,0)=𝑢−𝑎𝑣−𝑏𝑢≤0, then we have 𝑢≤(𝑎/(1−𝑏))𝑣. Similarly, let 𝐹(𝑢,𝑣,0,𝑢+𝑣,𝑢,𝑣)≤0, then we have 𝑢≤(𝑏/(1−𝑏))𝑣. Again, let 𝐹(𝑢,𝑣,𝑣,𝑣,𝑣,𝑣)≤0, then 𝑢≤(𝑎+𝑏)𝑣. Thus, 𝐹₂ is satisfied with 𝑟=max{𝑎/(1−𝑏),𝑏/(1−𝑏),𝑎+𝑏}. Also 𝐹(𝑢,0,0,𝑢,𝑢,0)=(1−𝑏)𝑢>0, for all 𝑢>0. Therefore, 𝐹∈Ψ.

3. Main Result

Theorem 3.1. Let (𝑋,𝑑) be a complete metric space, and let 𝑇 be a mapping on 𝑋. Define a nonincreasing function 𝜃 from [0,1) into (1/2,1] as in Theorem 1.1. Assume that there exists 𝐹∈Ψ, such that 𝜃(𝑟)𝑑(𝑥,𝑇𝑥)≤𝑑(𝑥,𝑦) implies 𝐹𝑑𝑇𝑥,𝑇𝑦,𝑑𝑥,𝑦,𝑑(𝑥,𝑇𝑥),𝑑𝑦,𝑇𝑦,𝑑𝑥,𝑇𝑦,𝑑𝑦,𝑇𝑥≤0, (3.1) for all 𝑥,𝑦∈𝑋, then 𝑇 has a unique fixed-point 𝑧 and lim_𝑛𝑇^𝑛𝑥=𝑧 holds for every 𝑥∈𝑋.

Proof. Since 𝜃(𝑟)≤1, 𝜃(𝑟)𝑑(𝑥,𝑇𝑥)≤𝑑(𝑥,𝑇𝑥) holds for every 𝑥∈𝑋, by hypotheses, we have 𝐹𝑑𝑇𝑥,𝑇²𝑥,𝑑(𝑥,𝑇𝑥),𝑑(𝑥,𝑇𝑥),𝑑𝑇𝑥,𝑇²𝑥,𝑑𝑥,𝑇²𝑥,0≤0, (3.2) and so from (𝐹₁), 𝐹𝑑𝑇𝑥,𝑇²𝑥,𝑑(𝑥,𝑇𝑥),𝑑(𝑥,𝑇𝑥),𝑑𝑇𝑥,𝑇²𝑥,𝑑(𝑥,𝑇𝑥)+𝑑𝑇𝑥,𝑇²𝑥,0≤0. (3.3) By (𝐹₂), we have 𝑑𝑇𝑥,𝑇²𝑥≤𝑟𝑑(𝑥,𝑇𝑥), (3.4) for all 𝑥∈𝑋. Now fix 𝑢∈𝑋 and define a sequence {𝑢_𝑛} in 𝑋 by 𝑢_𝑛=𝑇^𝑛𝑢. Then from (3.4), we have 𝑑𝑢_𝑛,𝑢_𝑛+1=𝑑𝑇𝑢_𝑛−1,𝑇²𝑢_𝑛−1≤𝑟𝑑𝑢_𝑛−1,𝑇𝑢_𝑛−1≤⋯≤𝑟^𝑛𝑑(𝑢,𝑇𝑢). (3.5) This shows that ∑^∞_𝑛=1𝑑(𝑢_𝑛,𝑢_𝑛+1)<∞, that is, {𝑢_𝑛} is Cauchy sequence. Since 𝑋 is complete, {𝑢_𝑛} converges to some point 𝑧∈𝑋. Now, we show that 𝑑(𝑇𝑥,𝑧)≤𝑟𝑑(𝑥,𝑧)∀𝑥∈𝑋⧵{𝑧}. (3.6) For 𝑥∈𝑋⧵{𝑧}, there exists 𝑛₀∈ℕ, such that 𝑑(𝑢_𝑛,𝑧)≤𝑑(𝑥,𝑧)/3 for all 𝑛≥𝑛₀. Then, we have 𝜃(𝑟)𝑑𝑢_𝑛,𝑇𝑢_𝑛≤𝑑𝑢_𝑛,𝑇𝑢_𝑛=𝑑𝑢_𝑛,𝑢_𝑛+1≤𝑑𝑢_𝑛,𝑧+𝑑𝑧,𝑢_𝑛+1≤23𝑑()()−𝑑()𝑥,𝑧=𝑑𝑥,𝑧𝑥,𝑧3≤𝑑(𝑥,𝑧)−𝑑𝑢_𝑛,𝑧≤𝑑𝑢_𝑛,𝑥. (3.7) Hence, by hypotheses, we have 𝐹𝑑𝑇𝑢_𝑛,𝑇𝑥,𝑑𝑢_𝑛,𝑥,𝑑𝑢_𝑛,𝑇𝑢_𝑛,𝑑(𝑥,𝑇𝑥),𝑑𝑢_𝑛,𝑇𝑥,𝑑𝑥,𝑇𝑢_𝑛≤0, (3.8) and so 𝐹𝑑𝑢_𝑛+1,𝑇𝑥,𝑑𝑢_𝑛,𝑥,𝑑𝑢_𝑛,𝑢_𝑛+1,𝑑(𝑥,𝑇𝑥),𝑑𝑢_𝑛,𝑇𝑥,𝑑𝑥,𝑢_𝑛+1≤0. (3.9) Letting 𝑛→∞, we have 𝐹(𝑑(𝑧,𝑇𝑥),𝑑(𝑧,𝑥),0,𝑑(𝑥,𝑇𝑥),𝑑(𝑧,𝑇𝑥),𝑑(𝑥,𝑧))≤0, (3.10) and so 𝐹(𝑑(𝑧,𝑇𝑥),𝑑(𝑧,𝑥),0,𝑑(𝑥,𝑧)+𝑑(𝑧,𝑇𝑥),𝑑(𝑧,𝑇𝑥),𝑑(𝑥,𝑧))≤0. (3.11) By (𝐹₂), we have 𝑑(𝑧,𝑇𝑥)≤𝑟𝑑(𝑥,𝑧), (3.12) and this shows that (3.6) is true.
Now, we assume that 𝑇^𝑚𝑧≠𝑧 for all 𝑚∈ℕ, then from (3.6), we have 𝑑𝑇^𝑚+1𝑧,𝑧≤𝑟^𝑚𝑑(𝑇𝑧,𝑧), (3.13) for all 𝑚∈ℕ.
Case 1. Let 0≤𝑟≤(√5−1)/2. In this case, 𝜃(𝑟)=1. Now, we show by induction that 𝑑𝑇^𝑛𝑧,𝑇𝑧≤𝑟𝑑(𝑧,𝑇𝑧), (3.14) for 𝑛≥2. From (3.4), (3.14) holds for 𝑛=2. Assume that (3.14) holds for some 𝑛 with 𝑛≥2. Since 𝑑(𝑧,𝑇𝑧)≤𝑑𝑧,𝑇^𝑛𝑧+𝑑𝑇^𝑛𝑧,𝑇𝑧≤𝑑𝑧,𝑇^𝑛𝑧+𝑟𝑑(𝑧,𝑇𝑧), (3.15) we have 𝑑()≤1𝑧,𝑇𝑧𝑑1−𝑟𝑧,𝑇^𝑛𝑧, (3.16) and so 𝜃(𝑟)𝑑𝑇^𝑛𝑧,𝑇^𝑛+1𝑧=𝑑𝑇^𝑛𝑧,𝑇^𝑛+1𝑧≤𝑟^𝑛𝑑()≤𝑟𝑧,𝑇𝑧_𝑛𝑑1−𝑟𝑧,𝑇^𝑛𝑧≤𝑟²𝑑1−𝑟𝑧,𝑇^𝑛𝑧≤𝑑𝑧,𝑇^𝑛𝑧. (3.17) Therefore, by hypotheses, we have 𝐹𝑑𝑇^𝑛+1𝑧,𝑇𝑧,𝑑𝑇^𝑛𝑧,𝑧,𝑑𝑇^𝑛𝑧,𝑇^𝑛+1𝑧,𝑑(𝑧,𝑇𝑧),𝑑𝑇^𝑛𝑧,𝑇𝑧,𝑑𝑧,𝑇^𝑛+1𝑧≤0, (3.18) and so 𝐹𝑑𝑇^𝑛+1𝑧,𝑇𝑧,𝑟^𝑛−1𝑑(𝑇𝑧,𝑧),𝑟^𝑛𝑑(𝑧,𝑇𝑧),𝑑(𝑧,𝑇𝑧),𝑟𝑑(𝑧,𝑇𝑧),𝑟^𝑛𝑑(𝑧,𝑇𝑧)≤0, (3.19) then 𝐹𝑑𝑇^𝑛+1𝑧,𝑇𝑧,𝑑(𝑇𝑧,𝑧),𝑑(𝑧,𝑇𝑧),𝑑(𝑧,𝑇𝑧),𝑑(𝑧,𝑇𝑧),𝑑(𝑧,𝑇𝑧)≤0, (3.20) and by (𝐹₂), we have 𝑑𝑇^𝑛+1𝑧,𝑇𝑧≤𝑟𝑑(𝑇𝑧,𝑧). (3.21) Therefore, (3.14) holds.
Now, from (3.6), we have 𝑑𝑇^𝑛+1𝑧,𝑧≤𝑟𝑑𝑇^𝑛𝑧,𝑧≤𝑟^𝑛𝑑(𝑇𝑧,𝑧). (3.22) This shows that 𝑇^𝑛𝑧→𝑧, which contradicts (3.14).
Case 2. Let (√5−1)/2≤𝑟≤√2/2. In this case, 𝜃(𝑟)=(1−𝑟)/𝑟². Again we want to show that (3.14) is true for 𝑛≥2. From (3.4), (3.14) holds for 𝑛=2. Assume that (3.14) holds for some 𝑛 with 𝑛≥2. Since 𝑑(𝑧,𝑇𝑧)≤𝑑𝑧,𝑇^𝑛𝑧+𝑑𝑇^𝑛𝑧,𝑇𝑧≤𝑑𝑧,𝑇^𝑛𝑧+𝑟𝑑(𝑧,𝑇𝑧), (3.23) we have 𝑑()≤1𝑧,𝑇𝑧𝑑1−𝑟𝑧,𝑇^𝑛𝑧, (3.24) and so 𝜃(𝑟)𝑑𝑇^𝑛𝑧,𝑇^𝑛+1𝑧=1−𝑟𝑟²𝑑𝑇^𝑛𝑧,𝑇^𝑛+1𝑧≤1−𝑟𝑟^𝑛𝑑𝑇^𝑛𝑧,𝑇^𝑛+1𝑧≤(1−𝑟)𝑑(𝑧,𝑇𝑧)≤𝑑𝑧,𝑇^𝑛𝑧. (3.25) Therefore, as in the previous case, we can prove that (3.14) is true for 𝑛≥2. Again from (3.6), we have 𝑑𝑇^𝑛+1𝑧,𝑧≤𝑟𝑑𝑇^𝑛𝑧,𝑧≤𝑟^𝑛𝑑(𝑇𝑧,𝑧). (3.26) This shows that 𝑇^𝑛𝑧→𝑧, which contradicts (3.14).Case 3. Let √2/2≤𝑟<1. In this case, 𝜃(𝑟)=1/(1+𝑟). Note that for 𝑥,𝑦∈𝑋, either 𝜃(𝑟)𝑑(𝑥,𝑇𝑥)≤𝑑𝑥,𝑦 (3.27) or 𝜃(𝑟)𝑑𝑇𝑥,𝑇²𝑥≤𝑑𝑇𝑥,𝑦 (3.28) holds. Indeed, if 𝜃(𝑟)𝑑(),𝜃(𝑟)𝑑𝑥,𝑇𝑥>𝑑𝑥,𝑦𝑇𝑥,𝑇²𝑥>𝑑𝑇𝑥,𝑦, (3.29) then we have 𝑑(𝑥,𝑇𝑥)≤𝑑𝑥,𝑦+𝑑𝑇𝑥,𝑦<𝜃(𝑟)𝑑(𝑥,𝑇𝑥)+𝑑𝑇𝑥,𝑇²𝑥≤𝜃(𝑟)[𝑑(𝑥,𝑇𝑥)+𝑟𝑑(𝑥,𝑇𝑥)]=𝑑(𝑥,𝑇𝑥), (3.30) which is a contradiction. Therefore, either 𝜃(𝑟)𝑑𝑢_2𝑛,𝑇𝑢_2𝑛≤𝑑𝑢_2𝑛,𝑧 (3.31) or 𝜃(𝑟)𝑑𝑢_2𝑛+1,𝑇𝑢_2𝑛+1≤𝑑𝑢_2𝑛+1,𝑧 (3.32) holds for every 𝑛∈ℕ. If 𝜃(𝑟)𝑑𝑢_2𝑛,𝑇𝑢_2𝑛≤𝑑𝑢_2𝑛,𝑧 (3.33) holds, then by hypotheses we have 𝐹𝑑𝑇𝑢_2𝑛,𝑇𝑧,𝑑𝑢_2𝑛,𝑧,𝑑𝑢_2𝑛,𝑇𝑢_2𝑛,𝑑(𝑧,𝑇𝑧),𝑑𝑢_2𝑛,𝑇𝑧,𝑑𝑧,𝑇𝑢_2𝑛≤0, (3.34) and so 𝐹𝑑𝑢_2𝑛+1,𝑇𝑧,𝑑𝑢_2𝑛,𝑧,𝑑𝑢_2𝑛,𝑢_2𝑛+1,𝑑(𝑧,𝑇𝑧),𝑑𝑢_2𝑛,𝑇𝑧,𝑑𝑧,𝑢_2𝑛+1≤0. (3.35) Letting 𝑛→∞, we have 𝐹(𝑑(𝑧,𝑇𝑧),0,0,𝑑(𝑧,𝑇𝑧),𝑑(𝑧,𝑇𝑧),0)≤0, (3.36) which contradicts (𝐹₃). If 𝜃(𝑟)𝑑𝑢_2𝑛+1,𝑇𝑢_2𝑛+1≤𝑑𝑢_2𝑛+1,𝑧 (3.37) holds, then by hypotheses we have 𝐹𝑑𝑇𝑢_2𝑛+1,𝑇𝑧,𝑑𝑢_2𝑛+1,𝑧,𝑑𝑢_2𝑛+1,𝑇𝑢_2𝑛+1,𝑑(𝑧,𝑇𝑧),𝑑𝑢_2𝑛+1,𝑇𝑧,𝑑𝑧,𝑇𝑢_2𝑛+1≤0, (3.38) and so 𝐹𝑑𝑢_2𝑛+2,𝑇𝑧,𝑑𝑢_2𝑛+1,𝑧,𝑑𝑢_2𝑛+1,𝑢_2𝑛+2,𝑑(𝑧,𝑇𝑧),𝑑𝑢_2𝑛+1,𝑇𝑧,𝑑𝑧,𝑢_2𝑛+2≤0. (3.39) Letting 𝑛→∞, we have 𝐹(𝑑(𝑧,𝑇𝑧),0,0,𝑑(𝑧,𝑇𝑧),𝑑(𝑧,𝑇𝑧),0)≤0, (3.40) which contradicts (𝐹₃).
Therefore, in all the cases, there exists 𝑚∈ℕ, such that 𝑇^𝑚𝑧=𝑧. Since {𝑇^𝑛𝑧} is Cauchy sequence, we obtain 𝑇𝑧=𝑧. That is, 𝑧 is a fixed point of 𝑇. The uniqueness of fixed point follows easily from (3.6).

Remark 3.2. If we combine Theorem 3.1 with Examples 2.1, 2.2, 2.3, and 2.4, we have Theorem 2 of [6], Theorem 2.2 of [7], Theorem 3.1 of [7], and Theorem 4 of [8], respectively.

Using Example 2.5, we obtain the following result.

Corollary 3.3. Let (𝑋,𝑑) be a complete metric space, and let 𝑇 be a mapping on 𝑋. Define a nonincreasing function 𝜃 from [0,1) into (1/2,1] as in Theorem 1.1. Assume that 𝜃(𝑟)𝑑(𝑥,𝑇𝑥)≤𝑑𝑥,𝑦 (3.41) implies 𝑑𝑇𝑥,𝑇𝑦≤𝑎𝑑(𝑥,𝑇𝑥)+𝑏𝑑𝑦,𝑇𝑦, (3.42) for all 𝑥,𝑦∈𝑋, where 𝑎,𝑏∈[0,1/2), then there exists a unique fixed point of 𝑇.

Remark 3.4. We obtain some new results, if we combine Theorem 3.1 with some examples of 𝐹.