A Generalization of Ćirić Quasicontractions

Abstract

We proved a fixed point theorem for a class of maps that satisfy Ćirić's contractive condition dependent on another function. We presented an example to show that our result is a real generalization.

1. Introduction and Preliminaries

The fixed point theorems have extensive applications in many disciplines such as in mathematics, economics, and computer science (see, e.g., [1–5]). The Banach contraction mapping principle [6] is the main motivation of this theory. A self-mapping 𝑇 on a metric space 𝑋 is called a contraction if, for each 𝑥,𝑦∈𝑋, there exists a constant 𝑘∈[0,1) such that 𝑑(𝑇𝑥,𝑇𝑦)≤𝑘𝑑(𝑥,𝑦). Banach [6] proved that every contraction on a complete metric space has a unique fixed point.

Many authors have studied on various generalizations of Banach contraction mapping principle. For important and famous examples of such generalizations, we may cite Kannan [7], Reich [8–10], Chatterjea [11], Hardy and Rogers [12], and Ćirić [13, 14]. In this manuscript, we focused on Ćirić type contractions and generalized some of his results [13, 14].

In [14], Ćirić introduced a class of self-maps on a metric space (𝑋,𝑑) which satisfy the following condition:𝑑𝑆𝑥,𝑆𝑦≤𝑞max𝑑𝑥,𝑦,𝑑(𝑥,𝑆𝑥),𝑑𝑦,𝑆𝑦,𝑑𝑥,𝑆𝑦,𝑑𝑦,𝑆𝑥, (1.1) for every 𝑥,𝑦∈𝑋 and 0≤𝑞<1. The maps satisfying Condition (1.1) are said to be quasicontractions.

Let 𝑇 be a self-mapping on a metric space (𝑋,𝑑). For 𝑌⊂𝑋 and for each 𝑥∈𝑋, we set (cf. [14, 15]) (1)𝛿(𝑌)=sup{𝑑(𝑥,𝑦)∶𝑥,𝑦∈𝑌}, (2)𝑂(𝑥,𝑛)={𝑥,𝑇𝑥,𝑇²𝑥,…,𝑇^𝑛𝑥}for 𝑛∈ℕ, (3)𝑂(𝑥,∞)={𝑥,𝑇𝑥,𝑇²𝑥,…,}.

A metric space (𝑋,𝑑) is called 𝑇-orbitally complete if and only if every Cauchy sequence of 𝑂(𝑥,∞) (for some 𝑥∈𝑋) converges to a point in 𝑋. It is clear that if 𝑋 is complete, then 𝑋 is 𝑆-orbitally complete for every 𝑇∶𝑋→𝑋. We now state the theorem of Ćirić [14].

Theorem 1.1. Let 𝑇 be a quasicontraction on 𝑇-orbitally complete metric space 𝑋. Then (1)𝑇 has a unique fixed point 𝑢 in 𝑋, (2)lim_𝑛→∞𝑇^𝑛𝑥=𝑢, for 𝑥∈𝑋.

Other generalizations of Banach's contraction principle and analogous results for metric spaces are abundantly present in the literature (see [7–19] and the references given therein). Rhoades [19] considered many types of contractive definitions and analyzed the relationships among them. Recently, some authors have provided a result on the existence of fixed points for a new class of contractive mappings.

Definition 1.2 (see, e.g., [20, 21]). Let (𝑋,𝑑) be a metric and 𝑇,𝑆∶𝑋→𝑋 be two functions. A mapping 𝑆 is called to be a 𝑇contraction if there exists 𝑞∈(0,1) such that 𝑑𝑇𝑆𝑥,𝑇𝑆𝑦≤𝑞𝑑𝑇𝑥,𝑇𝑦,∀𝑥,𝑦∈𝑋. (1.2)

In the definition above, if we choose 𝑇𝑥=𝑥 for all 𝑥∈𝑋, then the 𝑇contraction becomes a contraction. In [20, 22], some fixed point theorems are proved for 𝑇contractions in metric spaces and in 2-metric spaces. In this paper, by the using the same techniques, we shall extend the Ćirić's theorem (Theorem 1.1). For this purpose, we recall the following elementary definition.

Definition 1.3 (see, e.g., [21, 22]). Let (𝑋,𝑑) be a metric space. A mapping 𝑇∶𝑋→𝑋 is called sequentially convergent if the statement {𝑇𝑦_𝑛} is convergent and implies that {𝑦_𝑛} is a convergent sequence for every sequence {𝑦_𝑛}.

2. Main Results

The aim of this section is to prove the following result.

Theorem 2.1. Let (𝑋,𝑑) be a metric space. Let 𝑇∶𝑋→𝑋 be a one-to-one, continuous, and sequentially convergent mapping and 𝑆 be a self-mapping of 𝑋. Suppose that there exists 𝑞∈[0,1), 𝑑𝑇𝑆𝑥,𝑇𝑆𝑦≤𝑞max𝑑𝑇𝑥,𝑇𝑦,𝑑(𝑇𝑥,𝑇𝑆𝑥),𝑑𝑇𝑦,𝑇𝑆𝑦,𝑑𝑇𝑥,𝑇𝑆𝑦,𝑑𝑇𝑦,𝑇𝑆𝑥, (2.1) for every 𝑥,𝑦∈𝑋 and every Cauchy sequence of the form {𝑇𝑆^𝑛𝑥} for 𝑥∈𝑋 converges in 𝑋. Then (i)𝑑(𝑇𝑆^𝑖,𝑇𝑆^𝑗)≤𝑞𝛿[𝑂(𝑇𝑥,𝑛)], for all 𝑖,𝑗∈{1,2,…,𝑛}, for all 𝑥∈𝑋 and 𝑛∈ℕ,(ii)𝛿[𝑂(𝑇𝑥,∞)]≤1/(1−𝑞)𝑑(𝑇𝑥,𝑇𝑆𝑥), for all 𝑥∈𝑋,(iii)𝑆 has a unique fixed point 𝑏∈𝑋, (iv)lim𝑇𝑆^𝑛𝑥=𝑇𝑏.

Proof. We will mainly follow the arguments in the proof of the Ćirić's theorem. Let 𝑇,𝑆 be defined as in Theorem 2.1. We start with the proof of (i). Let 𝑥∈𝑋 be arbitrary element. Let 𝑛≥1 be integer and 𝑖,𝑗∈{1,2,…,𝑛}. Then 𝑇𝑆^𝑖−1𝑥,𝑇𝑆^𝑖,𝑇𝑆^𝑗−1𝑥,𝑇𝑆^𝑗𝑥∈𝑂(𝑇𝑥,𝑛). (2.2) Note that we use 𝑆⁰ as the identity self-mapping. Due to (2.1), we have 𝑑𝑇𝑆^𝑖𝑥,𝑇𝑆^𝑗𝑥=𝑑𝑇𝑆𝑆^𝑖−1𝑥,𝑇𝑆𝑆^𝑗−1𝑥≤𝑞max𝑑𝑇𝑆^𝑖−1𝑥,𝑇𝑆^𝑗−1𝑥;𝑑𝑇𝑆^𝑖−1𝑥,𝑇𝑆^𝑖𝑥,𝑑𝑇𝑆^𝑗−1𝑥,𝑇𝑆^𝑗𝑥,𝑑𝑇𝑆^𝑖−1𝑥,𝑇𝑆^𝑗𝑥,𝑑𝑇𝑆^𝑖𝑥,𝑇𝑆^𝑗−1𝑥≤𝑞𝛿[𝑂(𝑇𝑥,𝑛)]. (2.3) Thus, (i) is proved. Let us show (ii). Observe that 𝛿[𝑂(𝑇𝑥,1)]≤𝛿[𝑂(𝑇𝑥,2)]≤⋯≤𝛿[𝑂(𝑇𝑥,𝑛)]≤𝛿[𝑂(𝑇𝑥,𝑛+1)]≤⋯, (2.4) which implies that 𝛿[𝑂(𝑇𝑥,∞)]=sup{𝛿[𝑂(𝑇𝑥,𝑛)]∶𝑛∈ℕ}. (2.5) So we only need to show that 𝛿[𝑂(𝑇𝑥,𝑛)]≤1/(1−𝑞)𝑑(𝑇𝑥,𝑇𝑆𝑥) for all 𝑛∈ℕ. Indeed, since 𝑂(𝑇𝑥,𝑛)={𝑇𝑥,𝑇𝑆𝑥,…,𝑇𝑆^𝑛𝑥} is a finite set for each 𝑛 and 𝑇𝑆^𝑖,𝑇𝑆^𝑗≤𝑞𝛿[𝑂(𝑇𝑥,𝑛)],∀𝑖,𝑗∈{1,2,…,𝑛} (2.6) with 0≤𝑞<1, we infer that there exists 𝑘∈{1,2,…,𝑛} such that 𝑑𝑇𝑥,𝑇𝑆^𝑘𝑥=𝛿[𝑂(𝑇𝑥,𝑛)]. (2.7) Applying the triangle inequality and by (i), we get that 𝑑𝑇𝑥,𝑇𝑆^𝑘𝑥≤𝑑(𝑇𝑥,𝑇𝑆𝑥)+𝑑𝑇𝑆𝑥,𝑇𝑆^𝑘𝑥()[𝑂(()≤𝑑𝑇𝑥,𝑇𝑆𝑥+𝑞𝛿𝑇𝑥,𝑛)]=𝑑𝑇𝑥,𝑇𝑆𝑥+𝑞𝑑𝑇𝑥,𝑇𝑆^𝑘𝑥. (2.8) Hence, 𝛿[𝑂(𝑇𝑥,𝑛)]=𝑑(𝑇𝑥,𝑇𝑆^𝑘𝑥)≤1/(1−𝑞)𝑑(𝑇𝑥,𝑇𝑆𝑥). Thus, (ii) is proved.

To prove (iii), choose 𝑥₀∈𝑋 and define the iterative sequence {𝑥_𝑛} and {𝑦_𝑛} as follows:𝑥_𝑛+1=𝑆𝑥_𝑛=𝑆^𝑛+1𝑥₀,𝑦_𝑛=𝑇𝑥_𝑛=𝑇𝑆^𝑛𝑥₀,𝑛=0,1,2,… (2.9) We assert that {𝑦_𝑛} is a Cauchy sequence. Without loss of generality, we may assume that 𝑛<𝑚, where 𝑛,𝑚∈ℕ. By (i), we obtain 𝑑𝑦_𝑛,𝑦_𝑚=𝑑𝑇𝑆^𝑛𝑥₀,𝑇𝑆^𝑚𝑥₀=𝑑𝑇𝑆𝑆^𝑛−1𝑥₀,𝑇𝑆^{𝑚−𝑛+1}𝑆^𝑛−1𝑥₀≤𝑞𝛿𝑂𝑇𝑆^𝑛−1𝑥₀,𝑚−𝑛+1. (2.10) According to the assumption 𝑛<𝑚, there exists 𝑙∈{1,𝑚−𝑛+1} such that 𝛿𝑂𝑇𝑆^𝑛−1𝑥₀,𝑚−𝑛+1=𝑑𝑇𝑆^𝑛−1𝑥₀,𝑇𝑆^𝑙𝑆^𝑛−1𝑥₀. (2.11) Again, by the mentioned assumption, we have 𝑑𝑇𝑆^𝑛−1𝑥₀,𝑇𝑆^𝑙𝑆^𝑛−1𝑥₀=𝑑𝑇𝑆𝑆^𝑛−2𝑥₀,𝑇𝑆^𝑙+1𝑆^𝑛−2𝑥₀≤𝑞𝛿𝑂𝑇𝑆^𝑛−2𝑥₀,𝑙+1. (2.12) So we can get the following system of inequalities: 𝑑𝑦_𝑛,𝑦_𝑚=𝑑𝑇𝑆^𝑛𝑥₀,𝑇𝑆^𝑚𝑥₀≤𝑞𝛿𝑂𝑇𝑆^𝑛−1𝑥₀,𝑚−𝑛+1≤𝑞²𝛿𝑂𝑇𝑆^𝑛−2𝑥₀,𝑚−𝑛+2. (2.13) By the same argument, we obtain 𝑑𝑦_𝑛,𝑦_𝑚=𝑑𝑇𝑆^𝑛𝑥₀,𝑇𝑆^𝑚𝑥₀≤⋯≤𝑞^𝑛𝛿𝑂𝑇𝑆𝑥₀,𝑚. (2.14) Since 𝛿[𝑂(𝑇𝑆𝑥₀,𝑚)]≤𝛿[𝑂(𝑇𝑆𝑥₀,∞)]≤1/(1−𝑞)𝑑(𝑇𝑥₀,𝑇𝑆𝑥₀), we have 𝑑𝑦_𝑛,𝑦_𝑚=𝑑𝑇𝑆^𝑛𝑥₀,𝑇𝑆^𝑚𝑥₀≤𝑞^𝑛𝑑1−𝑞𝑇𝑥₀,𝑇𝑆𝑥₀. (2.15) This implies that {𝑦_𝑛}={𝑇𝑆^𝑛𝑥₀} is a Cauchy sequence. Since every Cauchy sequence of the form {𝑇𝑆^𝑛𝑥₀} converges in 𝑋, we deduce that lim𝑇𝑆^𝑛𝑥₀=𝑎∈𝑋. (2.16) Since 𝑇 is sequentially convergent, it follows that lim𝑆^𝑛𝑥₀=𝑏∈𝑋. By the hypothesis that 𝑇 is continuous, we getlim𝑇𝑆^𝑛𝑥₀=𝑇𝑏. (2.17) Therefore, 𝑇𝑏=𝑎. We shall show that 𝑆𝑏=𝑏. To achieve this, we consider the following inequalities: 𝑑(𝑇𝑆𝑏,𝑇𝑏)≤𝑑𝑇𝑏,𝑇𝑆^𝑛+1𝑥₀+𝑑𝑇𝑆^𝑛+1𝑥₀,𝑇𝑆𝑏≤𝑑𝑇𝑏,𝑇𝑆^𝑛+1𝑥₀+𝑞max𝑑𝑇𝑏,𝑇𝑆^𝑛𝑥₀(),𝑑,𝑑𝑇𝑏,𝑇𝑆𝑏𝑇𝑆^𝑛𝑥₀,𝑇𝑆^𝑛+1𝑥₀,𝑑𝑇𝑏,𝑇𝑆^𝑛+1𝑥₀,𝑑𝑇𝑆^𝑛𝑥₀,𝑇𝑆𝑏≤𝑑𝑇𝑏,𝑇𝑆^𝑛+1𝑥₀+𝑞max𝑑𝑇𝑏,𝑇S^𝑛𝑥₀,𝑑(𝑇𝑏,𝑇𝑆𝑏),𝑑𝑇𝑆^𝑛𝑥₀,𝑇𝑆^𝑛+1𝑥₀,𝑑𝑇𝑏,𝑇𝑆^𝑛+1𝑥₀,𝑑𝑇𝑆^𝑛𝑥₀,𝑇𝑏+𝑑(𝑇𝑏,𝑇𝑆𝑏)≤𝑑𝑇𝑏,𝑇𝑆^𝑛+1𝑥₀+𝑞𝑑𝑇𝑏,𝑇𝑆^𝑛+1𝑥₀+𝑑𝑇𝑆^𝑛𝑥₀,𝑇S^𝑛+1𝑥₀+𝑑𝑇𝑆^𝑛𝑥₀,𝑇𝑏+𝑑(𝑇𝑏,𝑇𝑆𝑏). (2.18) Hence, 𝑑()≤1𝑇𝑏,𝑇𝑆𝑏𝑑1−𝑞1+𝑞𝑇𝑏,𝑇𝑆^𝑛+1𝑥₀+𝑞𝑑𝑇𝑏,𝑇𝑆^𝑛𝑥₀+𝑞𝑑𝑇𝑆^𝑛𝑥₀,𝑇𝑆^𝑛+1𝑥₀. (2.19) Letting 𝑛→∞ and combining with the fact that lim𝑇𝑆^𝑛𝑥₀=𝑇𝑏, we can conclude that 𝑑(𝑇𝑆𝑏,𝑇𝑏)=0. (2.20)

Therefore, 𝑇𝑆𝑏=𝑇𝑏. Since 𝑇 is one to one, we obtain that 𝑆𝑏=𝑏. Now we will show that 𝑏 is the unique fixed point of 𝑆. Suppose that 𝑏 and 𝑏^′ are fixed points of 𝑆. Then 𝑆𝑏=𝑏, 𝑆𝑏^′=𝑏^′, and applying (2.1), 𝑑𝑇𝑏,𝑇𝑏^′=𝑑𝑇𝑆𝑏,𝑇𝑆𝑏^′≤𝑞max𝑑(𝑇𝑏,𝑇𝑆𝑏),𝑑𝑇𝑏^′,𝑇𝑆𝑏^′,𝑑𝑇𝑏,𝑇𝑆𝑏^′,𝑑𝑇𝑏^′,𝑇𝑆𝑏,𝑑𝑇𝑏,𝑇𝑏^′=𝑞𝑑𝑇𝑏,𝑇𝑏^′. (2.21) Since 0≤𝑞<1, we infer that 𝑑(𝑇𝑏,𝑇𝑏^′)=0. Hence, 𝑇𝑏=𝑇𝑏^′. Since 𝑇 is one-to-one, we may conclude that 𝑏=𝑏^′. This proves (iii) of Theorem 2.1. Proof of (iv) is directly obtained from (2.17). The theorem is proved.

Remark 2.2. Theorem 1.1 is obtained from Theorem 2.1 if we choose 𝑇𝑥=𝑥 for every 𝑥∈𝑋.

The following example shows that Theorem 2.1 is indeed an extension of Theorem 1.1.

Example 2.3. Let 𝑋=[0,+∞) with metric induced by 𝑑(𝑥,𝑦)=|𝑥−𝑦|. Then 𝑋 is a complete metric space. Consider the function 𝑆𝑥=𝑥²/(𝑥+1) for all 𝑥∈𝑋. It is easy to compute that 𝑑(𝑆𝑥,𝑆2𝑥)=4𝑥²−𝑥2𝑥+1²=𝑥𝑥+1²(2𝑥+3)(),𝑑()𝑥2𝑥+1)(𝑥+1𝑥,𝑆𝑥=𝑥−²=𝑥𝑥+1,𝑑()𝑥+12𝑥,𝑆2𝑥=2𝑥−4𝑥²=2𝑥+12𝑥,𝑑()=2𝑥+1𝑥,𝑆2𝑥4𝑥²=2𝑥+1−𝑥2𝑥²−𝑥,𝑑()𝑥2𝑥+12𝑥,𝑆𝑥=2𝑥−²=𝑥𝑥+1²+2𝑥,𝑑()𝑥+1𝑥,2𝑥=𝑥. (2.22) Therefore, if 𝑥 is large enough, we have {𝑑()()()()(=𝑥max𝑥,2𝑥,𝑑𝑥,𝑆𝑥,𝑑2𝑥,𝑆2𝑥,𝑑𝑥,𝑆2𝑥,𝑑2𝑥,𝑆𝑥)}²+2𝑥.𝑥+1 (2.23) It follows that the quasicontractive condition 𝑑(𝑆𝑥,𝑆2𝑥)≤𝑞max{𝑑(𝑥,2𝑥),𝑑(𝑥,𝑆𝑥),𝑑(2𝑥,𝑆2𝑥),𝑑(𝑥,𝑆2𝑥),𝑑(2𝑥,𝑆𝑥)} (2.24) is equivalent to 𝑥(2𝑥+3)(2𝑥+1)(𝑥+2)≤𝑞. (2.25) Letting 𝑥→∞, we get that 𝑞≥1 which contradicts the fact that 𝑞∈(0,1). Hence 𝑆 is not a quasicontraction. Therefore, we can not apply Theorem 1.1 for 𝑆. However, it is easy to see that 0 is the unique fixedpoint of 𝑆.
Now we shall show that 𝑆 satisfies Theorem 2.1 with 𝑇𝑥=𝑒^𝑥−1 and 𝑞=1/2. Indeed, 𝑇 is one-to-one, continuous and sequentially convergent on [0,+∞). We have 𝑑𝑇𝑆𝑥,𝑇𝑆𝑦=𝑒^{𝑥2/(𝑥+1)}−𝑒^{𝑦2/(𝑦+1)},𝑑𝑇𝑥,𝑇𝑦=𝑒^𝑥−𝑒^𝑦. (2.26) Now, we will show that 𝑑≤1𝑇𝑆𝑥,𝑇𝑆𝑦2𝑑𝑇𝑥,𝑇𝑦. (2.27) If 𝑥=𝑦, then the inequality above holds. So we can assume that 𝑥>𝑦. Then (2.27) is equivalent to 𝑒^{𝑥2/(𝑥+1)}−𝑒^{𝑦2/(𝑦+1)}≤12𝑒^𝑥−𝑒^𝑦,∀𝑥>𝑦. (2.28) It follows from the fact that 𝜙 and 𝜓 are increasing, (2.28) reduces to 𝑒^{𝑥2/(𝑥+1)}−𝑒^𝑥2≤𝑒^{𝑦2/(𝑦+1)}−𝑒^𝑦2,∀𝑥>𝑦. (2.29) Consider the function 𝜑(𝑡)=𝑒^{𝑡2/(𝑡+1)}−𝑒^𝑡/2, (𝑡≥0). It is easy to check that 𝜑 is decreasing on [0,+∞). So we can deduce that 𝑑≤1𝑇𝑆𝑥,𝑇𝑆𝑦2𝑑≤1𝑇𝑥,𝑇𝑦2𝑑();𝑑max𝑇𝑥,𝑇𝑦;𝑑𝑇𝑥,𝑇𝑆𝑥𝑇𝑦,𝑇𝑆𝑦,𝑑𝑇𝑥,𝑇𝑆𝑦,𝑑𝑇𝑦,𝑇𝑆𝑥 (2.30) for every 𝑥,𝑦∈𝑋. Therefore, we can apply Theorem 2.1 for 𝑆.

As an immediate consequence of Theorem 2.1, we have the following corollary.

Corollary 2.4. Let (𝑋,𝑑) be a metric space. Let 𝑇∶𝑋→𝑋 be a one-to-one, continuous, and sequentially convergent mapping, and let 𝑆 be a self-mapping of 𝑋. Suppose that there exists 𝑞∈[0,1), 𝑑𝑑(),𝑑𝑇𝑆𝑥,𝑇𝑆𝑦≤𝑞max𝑇𝑥,𝑇𝑦,𝑑𝑇𝑥,𝑇𝑆𝑥,𝑑𝑇𝑦,𝑇𝑆𝑦𝑇𝑥,𝑇𝑆𝑦+𝑑𝑇𝑦,𝑇𝑆𝑥2 (2.31) for every x,𝑦∈𝑋 and every Cauchy sequence of the form {𝑇𝑆^𝑛𝑥} for 𝑥∈𝑋 converges in 𝑋. Then,(a)𝑆 has a unique fixed point 𝑏∈𝑋, (b)lim𝑇𝑆^𝑛𝑥=𝑇𝑏 for 𝑥∈𝑋.

This corollary is also a consequence of the main result of [21], which is established by a contractive condition of integral type.

We also get the following corollary.

Corollary 2.5. Let (𝑋,𝑑) be a metric space. Let 𝑇∶𝑋→𝑋 be a one-to-one, continuous, and sequentially convergent mapping and 𝑆 be a self-mapping of 𝑋. Suppose that there exists 𝑞∈[0,1), 𝑑𝑇𝑆^𝑘𝑥,𝑇𝑆^𝑘𝑦≤𝑞max𝑑𝑇𝑥,𝑇𝑦,𝑑𝑇𝑥,𝑇𝑆^𝑘𝑥,𝑑𝑇𝑦,𝑇𝑆^𝑘𝑦,𝑑𝑇𝑥,𝑇𝑆^𝑘𝑦,𝑑𝑇𝑦,𝑇𝑆^𝑘𝑥 (2.32) for every 𝑥,𝑦∈𝑋, for some positive integer 𝑘 and every Cauchy sequence of the form {𝑇𝑆^𝑛𝑥} for 𝑥∈𝑋 converges in 𝑋. Then, 𝑆 has a unique fixed point 𝑏∈𝑋.

Proof. Applying Theorem 2.1 for the map 𝑆^𝑘, we can deduce that 𝑆^𝑘 has a unique fixed point 𝑏∈𝑋. This yields that 𝑆^𝑘𝑏=𝑏 and 𝑆^𝑘(𝑆𝑏)=𝑆^𝑘+1𝑏=𝑆(𝑆^𝑘𝑏)=𝑆𝑏. It follows that 𝑆𝑏 is a fixed point of 𝑆^𝑘. By the uniqueness of fixed point of 𝑆^𝑘, we infer that 𝑆𝑏=𝑏. Therefore, 𝑏 is a fixed point of 𝑆. Now, if 𝑎 is a fixed point of 𝑆, then 𝑆^𝑘𝑎=𝑆^𝑘−1𝑎=⋯=𝑆𝑎=𝑎. Hence, 𝑎 is a fixed point of 𝑆^𝑘, so that 𝑎=𝑏. This proves the uniqueness of fixed point of 𝑆.

The following corollary is an extension of the main result of [12].

Corollary 2.6. Let (𝑋,𝑑) be a metric space. Let 𝑇∶𝑋→𝑋 be a one-to-one, continuous, and sequentially convergent mapping, and let 𝑆 be a self-mapping of 𝑋. Suppose that there exists 𝑎_𝑖≥0, 𝑖=1,2,…,5 satisfying that ∑^𝑛_𝑖=1𝑎_𝑖<1 such that 𝑑𝑇𝑆𝑥,𝑇𝑆𝑦≤𝑎₁𝑑𝑇𝑥,𝑇𝑦+𝑎₂𝑑(𝑇𝑥,𝑇𝑆𝑥)+𝑎₃𝑑𝑇𝑦,𝑇𝑆𝑦+𝑎₄𝑑𝑇𝑥,𝑇𝑆𝑦+𝑎₅𝑑𝑇𝑦,𝑇𝑆𝑥 (2.33) for every 𝑥,𝑦∈𝑋 and every Cauchy sequence of the form {𝑇𝑆^𝑛𝑥} for 𝑥∈𝑋 converges in 𝑋. Then (a)𝑆 has a unique fixed point 𝑏∈𝑋. (b)lim𝑇𝑆^𝑛𝑥=𝑇𝑏.

Proof. It is easy to see that the condition (2.1) is a consequence of the condition (2.33). This is enough to prove the corollary.

Remark 2.7. In the same way in Corollary 2.6, we can get the extensions of Kannan's contractions (see [7]), and Reich's contractions (see [8–10]).