Weak Contraction Condition Involving Cubic Terms of 𝑑(𝑥,𝑦) under the Fixed Point Consideration

Abstract

A fixed point theorem is presented for single-valued map with using generalized 𝜑-weak contractive condition involving various combinations of 𝑑(𝑥,𝑦) on a complete metric space. Our result is an extension as well as a generalization of Alber and Guerre-Delabriere (1997) in particular. It also generalizes the results of Rhoades (2001), Choudhury and Dutta, (2000), and Dutta and Choudhury, (2008).

1. Introduction

Let (𝑋,𝑑) be a metric space. A map 𝑇∶𝑋→𝑋 is a contraction if, for each 𝑥,𝑦∈𝑋, there exists a constant 𝑘∈(0,1) such that 𝑑(𝑇𝑥,𝑇𝑦)≤𝑘𝑑(𝑥,𝑦).

A map 𝑇∶𝑋→𝑋 is a 𝜑-weak contraction if, for each 𝑥,𝑦∈𝐸, there exists a function 𝜑∶[0,∞)→[0,∞), 𝜑(𝑡)>0 for all 𝑡>0 and 𝜑(0)=0 such that 𝑑(𝑇𝑥,𝑇𝑦)≤𝑑(𝑥,𝑦)−𝜑(𝑑(𝑥,𝑦)).

In [1] Alber and Guerre-Delabriere introduced the concept of weak contraction in Hilbert spaces. Rhoades [2] has shown that the result which Alber and Guerre-Delabriere had proved in [2] is also valid in complete metric spaces.

In this paper we introduced generalized 𝜑-weak contractive condition involving various combinations of 𝑑(𝑥,𝑦). Our result is an extension as well as a generalization of Alber and Guerre-Delabriere [1] and Rhoades [2] in particular. It also generalizes the results of [3, 4].

Now, We state the result of Rhoades as follows.

Theorem 1 (see [2, Theorem 2]). Let (𝑋,𝑑) be a complete metric space, and let 𝑇 be a 𝜑-weak contraction on 𝑋𝜑∶[0,+∞)→[0,+∞) which is a continuous and nondecreasing function with 𝜑(𝑡)>0 for all 𝑡∈(0,∞) and 𝜑(0)=0 then 𝑇 has a unique fixed point in 𝑋.
If one takes 𝜑(𝑡)=(1−𝑘)𝑡, where 0<𝑘<1, then weak contraction reduces to contraction mapping.
In this paper, a new type of inequality is introduced with cubic terms involving 𝑝(≥0) called a “generalized 𝜑-weak contractive condition with cubic terms involving 𝑝”.
Let (𝑋,𝑑) be a metric space and 𝑇 a self-map of 𝑋 satisfying the following condition: 1+𝑝𝑑𝑥,𝑦𝑑²1𝑇𝑥,𝑇𝑦≤𝑝max2𝑑²(𝑥,𝑇𝑥)𝑑𝑦,𝑇𝑦+𝑑(𝑥,𝑇𝑥)𝑑²,𝑑()𝑑𝑑,𝑑𝑑𝑑𝑚,𝑦,𝑇𝑦𝑥,𝑇𝑥𝑥,𝑇𝑦𝑦,𝑇𝑥𝑥,𝑇𝑦𝑦,𝑇𝑥𝑦,𝑇𝑦+𝑚𝑥,𝑦−𝜑𝑥,𝑦 (1) where 𝑚𝑑𝑥,𝑦=max²()𝑑𝑑,1𝑥,𝑦,𝑑𝑥,𝑇𝑥𝑦,𝑇𝑦,𝑑𝑥,𝑇𝑦𝑦,𝑇𝑥2𝑑(𝑥,𝑇𝑥)𝑑𝑥,𝑇𝑦+𝑑𝑦,𝑇𝑥𝑑𝑦,𝑇𝑦 (2)𝑝≥0 is a real number is and 𝜑∶[0,∞)→[0,∞) is a continuous function with 𝜑(𝑡)=0⇔𝑡=0 and 𝜑(𝑡)>0 for each 𝑡>0.

2. Main Result

Lemma 2. Let 𝑇 be a self map of a metric space 𝑋 satisfying (1). For any sequence {𝑥_𝑛} in 𝑋 defined by 𝑥_𝑛+1=𝑇𝑥_𝑛, 𝑛≥0. Then the sequence {𝑥_𝑛} is Cauchy in 𝑋.

Proof. Let 𝑥₀∈𝑋 be an arbitrary point. Constructing the sequence {𝑥_𝑛} follows 𝑥_𝑛+1=𝑇𝑥_𝑛,𝑛∈𝑁₀. (3) If 𝑥_𝑛=𝑥_𝑛+1 for some 𝑛, then trivially 𝑇 has a fixed point. We assume 𝑥_𝑛+1≠𝑥_𝑛, for all 𝑛∈𝑁₀. We write 𝛼_𝑛=𝑑(𝑥_𝑛,𝑥_𝑛+1).
First, we prove that {𝛼_𝑛} is a nonincreasing sequence and converges to 0.
Case 1. If 𝑛 is even, taking 𝑥=𝑥_2𝑛 and 𝑦=𝑥_2𝑛+1 in (1), we get 1+𝑝𝑑𝑥_2𝑛,𝑥_2𝑛+1𝑑²𝑇𝑥_2𝑛,𝑇𝑥_2𝑛+11≤𝑝max2𝑑²𝑥_2𝑛,𝑇𝑥_2𝑛𝑑𝑥_2𝑛+1,𝑇𝑥_2𝑛+1+𝑑𝑥_2𝑛,𝑇𝑥_2𝑛𝑑²𝑥_2𝑛+1,𝑇𝑥_2𝑛+1,𝑑𝑥_2𝑛,𝑇𝑥_2𝑛𝑑𝑥_2𝑛,𝑇𝑥_2𝑛+1×𝑑𝑥_2𝑛+1,𝑇𝑥_2𝑛,𝑑𝑥_2𝑛,𝑇𝑥_2𝑛+1×𝑑𝑥_2𝑛+1,𝑇𝑥_2𝑛𝑑𝑥_2𝑛+1,𝑇𝑥_2𝑛+1+𝑚𝑥_2𝑛,𝑥_2𝑛+1−𝜑𝑚𝑥_2𝑛,𝑥_2𝑛+1, (4) where 𝑚𝑥_2𝑛,𝑥_2𝑛+1=max𝑑²𝑥_2𝑛,𝑥_2𝑛+1,𝑑𝑥_2𝑛,𝑇𝑥_2𝑛𝑑𝑥_2𝑛+1,𝑇𝑥_2𝑛+1,𝑑𝑥_2𝑛,𝑇𝑥_2𝑛+1𝑑𝑥_2𝑛+1,𝑇𝑥_2𝑛,12𝑑𝑥_2𝑛,𝑇𝑥_2𝑛𝑑𝑥_2𝑛,𝑇𝑥_2𝑛+1+𝑑𝑥_2𝑛+1,𝑇𝑥_2𝑛𝑑𝑥_2𝑛+1,𝑇𝑥_2𝑛+1. (5)
By using (3), we get 1+𝑝𝑑𝑥_2𝑛,𝑥_2𝑛+1𝑑²𝑥_2𝑛+1,𝑥_2𝑛+21≤𝑝max2𝑑²𝑥_2𝑛,𝑥_2𝑛+1𝑑𝑥_2𝑛+1,𝑥_2𝑛+2+𝑑𝑥_2𝑛,𝑥_2𝑛+1𝑑²𝑥_2𝑛+1,𝑥_2𝑛+2,𝑑𝑥_2𝑛,𝑥_2𝑛+1𝑑𝑥_2𝑛,𝑥_2𝑛+2×𝑑𝑥_2𝑛+1,𝑥_2𝑛+1,𝑑𝑥_2𝑛,𝑥_2𝑛+2×𝑑𝑥_2𝑛+1,𝑥_2𝑛+1𝑑𝑥_2𝑛+1,𝑥_2𝑛+2+𝑚𝑥_2𝑛,𝑥_2𝑛+1−𝜑𝑚𝑥_2𝑛,𝑥_2𝑛+1, (6) where 𝑚𝑥_2𝑛,𝑥_2𝑛+1=max𝑑²𝑥_2𝑛,𝑥_2𝑛+1,𝑑𝑥_2𝑛,𝑥_2𝑛+1𝑑𝑥_2𝑛+1,𝑥_2𝑛+2,𝑑𝑥_2𝑛,𝑥_2𝑛+2𝑑𝑥_2𝑛+1,𝑥_2𝑛+1,12𝑑𝑥_2𝑛,𝑥_2𝑛+1𝑑𝑥_2𝑛,𝑥_2𝑛+2+𝑑𝑥_2𝑛+1,𝑥_2𝑛+1𝑑𝑥_2𝑛+1,𝑥_2𝑛+2. (7) Now consider 𝛼_2𝑛=𝑑(𝑥_2𝑛,𝑥_2𝑛+1); then we have 1+𝑝𝛼_2𝑛𝛼²_2𝑛+11≤𝑝max2𝛼²_2𝑛𝛼_2𝑛+1+𝛼_2𝑛𝛼²_2𝑛+1𝑥,0,0+𝑚_2𝑛,𝑥_2𝑛+1−𝜑𝑚𝑥_2𝑛,𝑥_2𝑛+1, (8) where 𝑚(𝑥_2𝑛,𝑥_2𝑛+1)=max{𝛼²_2𝑛,𝛼_2𝑛𝛼_2𝑛+1,0,  (1/2)[𝛼_2𝑛𝑑(𝑥_2𝑛,𝑥_2𝑛+2)+0]}.
By triangular inequality and using property of 𝜑, we get 𝑑𝑥_2𝑛,𝑥_2𝑛+2≤𝑑𝑥_2𝑛,𝑥_2𝑛+1+𝑑𝑥_2𝑛+1,𝑥_2𝑛+2=𝛼_2𝑛+𝛼_2𝑛+1. (9) Then 𝑚𝑥_2𝑛,𝑥_2𝑛+1≤𝑚^′𝛼𝑥,𝑦=max²_2𝑛,𝛼_2𝑛+1𝛼_2𝑛1,0,2𝛼_2𝑛𝛼_2𝑛+𝛼_2𝑛+1,0. (10)
If 𝛼_2𝑛<𝛼_2𝑛+1, then (8) reduces to 𝑝𝛼²_2𝑛+1≤𝑝𝛼²_2𝑛+1−𝜑(𝛼²_2𝑛+1), a contradiction. Thus 𝛼²_2𝑛+1≤𝛼²_2𝑛 implies that 𝛼_2𝑛+1≤𝛼_2𝑛.
In a similar way, if 𝑛 is odd, we can obtain 𝛼_2𝑛+2<𝛼_2𝑛+1.
It follows that the sequence {𝛼_𝑛} is decreasing.
Let lim_𝑛→∞𝛼_𝑛=𝑟, for some 𝑟≥0.
Suppose 𝑟>0; then from inequality (1), we have 1+𝑝𝑑𝑥_𝑛,𝑥_𝑛+1𝑑²𝑇𝑥_𝑛,𝑇𝑥_𝑛+11≤𝑝max2𝑑²𝑥_𝑛,𝑇𝑥_𝑛𝑑𝑥_𝑛+1,𝑇𝑥_𝑛+1+𝑑𝑥_𝑛,𝑇𝑥_𝑛𝑑²𝑥_𝑛+1,𝑇𝑥_𝑛+1,𝑑𝑥_𝑛,𝑇𝑥_𝑛𝑑𝑥_𝑛,𝑇𝑥_𝑛+1×𝑑𝑥_𝑛+1,𝑇𝑥_𝑛,𝑑𝑥_𝑛,𝑇𝑥_𝑛+1×𝑑𝑥_𝑛+1,𝑇𝑥_𝑛𝑑𝑥_𝑛+1,𝑇𝑥_𝑛+1+𝑚𝑥_𝑛,𝑥_𝑛+1−𝜑𝑚𝑥_𝑛,𝑥_𝑛+1, (11) where 𝑚𝑥_𝑛,𝑥_𝑛+1=max𝑑²𝑥_𝑛,𝑥_𝑛+1,𝑑𝑥_𝑛,𝑇𝑥_𝑛𝑑𝑥_𝑛+1,𝑇𝑥_𝑛+1,𝑑𝑥_𝑛,𝑇𝑥_𝑛+1𝑑𝑥_𝑛+1,𝑇𝑥_𝑛,12𝑑𝑥_𝑛,𝑇𝑥_𝑛𝑑𝑥_𝑛,𝑇𝑥_𝑛+1+𝑑𝑥_𝑛+1,𝑇𝑥_𝑛𝑑𝑥_𝑛+1,𝑇𝑥_𝑛+1. (12) By using (3), we get 1+𝑝𝑑𝑥_𝑛,𝑥_𝑛+1𝑑²𝑥_𝑛+1,𝑥_𝑛+21≤𝑝max2𝑑²𝑥_𝑛,𝑥_𝑛+1𝑑𝑥_𝑛+1,𝑥_𝑛+2+𝑑𝑥_𝑛,𝑥_𝑛+1𝑑²𝑥_𝑛+1,𝑥_𝑛+2,𝑑𝑥_𝑛,𝑥_𝑛+1𝑑𝑥_𝑛,𝑥_𝑛+2𝑑𝑥_𝑛+1,𝑥_𝑛+1,𝑑𝑥_𝑛,𝑥_𝑛+2𝑑𝑥_𝑛+1,𝑥_𝑛+1𝑑𝑥_𝑛+1,𝑥_𝑛+2+𝑚𝑥_𝑛,𝑥_𝑛+1−𝜑𝑚𝑥_𝑛,𝑥_𝑛+1, (13) where 𝑚𝑥_𝑛,𝑥_𝑛+1=max𝑑²𝑥_𝑛,𝑥_𝑛+1,𝑑𝑥_𝑛,𝑥_𝑛+1𝑑𝑥_𝑛+1,𝑥_𝑛+2,𝑑𝑥_𝑛,𝑥_𝑛+2𝑑𝑥_𝑛+1,𝑥_𝑛+1,12𝑑𝑥_𝑛,𝑥_𝑛+1𝑑𝑥_𝑛,𝑥_𝑛+2+𝑑𝑥_𝑛+1,𝑥_𝑛+1𝑑𝑥_𝑛+1,𝑥_𝑛+2. (14)
Using triangular inequality and property of 𝜑 and taking limits 𝑛→∞, we get 1+𝑝𝑟𝑟²≤𝑝𝑟³+𝑟²−𝜑𝑟². (15) Then 𝜑(𝑟²)≤0, since 𝑟 is positive; then by the property of 𝜑, we get 𝑟=0. We conclude that lim_𝑛→∞𝛼_𝑛=lim_𝑛→∞𝑑𝑥_𝑛,𝑥_𝑛+1=𝑟=0. (16)
Now we show that {𝑥_𝑛} is a Cauchy sequence. Suppose, we assume that {𝑥_𝑛} is not a Cauchy sequence; then there is 𝜀>0 for which we can find two sequences of positive integers {𝑚(𝑘)} and {𝑛(𝑘)} such that for all positive integers 𝑘, 𝑛(𝑘)>𝑚(𝑘)>𝑘𝑑𝑥_𝑚(𝑘),𝑥_𝑛(𝑘)≥𝜖,𝑑𝑥_𝑚(𝑘),𝑥_{𝑛(𝑘)−1}<𝜖. (17)
Now, 𝜖≤𝑑𝑥_𝑚(𝑘),𝑥_𝑛(𝑘)≤𝑑𝑥_𝑚(𝑘),𝑥_{𝑛(𝑘)−1}+𝑑𝑥_{𝑛(𝑘)−1},𝑥_𝑛(𝑘). (18) Letting 𝑘→∞, we get lim_𝑘→∞𝑑𝑥_𝑚(𝑘),𝑥_𝑛(𝑘)=𝜖. (19) Now, from the triangular inequality, we have 𝑑𝑥_𝑛(𝑘),𝑥_𝑚(𝑘)+1−𝑑𝑥_𝑚(𝑘),𝑥_𝑛(𝑘)≤𝑑𝑥_𝑚(𝑘),𝑥_𝑚(𝑘)+1. (20)
Taking limits as 𝑘→∞ and using (16) and (19), we have lim_𝑘→∞𝑑𝑥_𝑛(𝑘),𝑥_𝑚(𝑘)+1=𝜖. (21)
Again from the triangular inequality, we have 𝑑𝑥_𝑚(𝑘),𝑥_𝑛(𝑘)+1−𝑑𝑥_𝑚(𝑘),𝑥_𝑛(𝑘)≤𝑑𝑥_𝑛(𝑘),𝑥_𝑛(𝑘)+1. (22)
Taking limits as 𝑘→∞ and using (16) and (19), we have lim_𝑘→∞𝑑𝑥_𝑚(𝑘),𝑥_𝑛(𝑘)+1=𝜖. (23) Again by using triangular inequality, we have 𝑑𝑥_𝑚(𝑘)+1,𝑥_𝑛(𝑘)+1−𝑑𝑥_𝑚(𝑘),𝑥_𝑛(𝑘)≤𝑑𝑥_𝑚(𝑘),𝑥_𝑚(𝑘)+1+𝑑𝑥_𝑛(𝑘),𝑥_𝑛(𝑘)+1. (24) Taking limit 𝑘→∞ in the above inequality and using (16) and (19), we have lim_𝑘→∞𝑑𝑥_𝑛(𝑘)+1,𝑥_𝑚(𝑘)+1=𝜖. (25)
Again putting 𝑥=𝑥_𝑚(𝑘) and 𝑦=𝑥_𝑛(𝑘) in (1), we get 1+𝑝𝑑𝑥_𝑚(𝑘),𝑥_𝑛(𝑘)𝑑²𝑇𝑥_𝑚(𝑘),𝑇𝑥_𝑛(𝑘)1≤𝑝2𝑑²𝑥_𝑚(𝑘),𝑇𝑥_𝑚(𝑘)𝑑𝑥_𝑛(𝑘),𝑇𝑥_𝑛(𝑘)+𝑑𝑥_𝑚(𝑘),𝑇𝑥_𝑚(𝑘)𝑑²𝑥_𝑛(𝑘),𝑇𝑥_𝑛(𝑘),𝑑𝑥_𝑚(𝑘),𝑇𝑥_𝑚(𝑘)𝑑𝑥_𝑚(𝑘),𝑇𝑥_𝑛(𝑘)𝑑𝑥_𝑛(𝑘),𝑇𝑥_𝑚(𝑘),𝑑𝑥_𝑚(𝑘),𝑇𝑥_𝑛(𝑘)𝑑𝑥_𝑛(𝑘),𝑇𝑥_𝑚(𝑘)𝑑𝑥_𝑛(𝑘),𝑇𝑥_𝑛(𝑘)+𝑚𝑥_𝑚(𝑘),𝑥_𝑛(𝑘)−𝜑𝑚𝑥_𝑚(𝑘),𝑥_𝑛(𝑘), (26) where 𝑚𝑥_𝑚(𝑘),𝑥_𝑛(𝑘)=max𝑑²𝑥_𝑚(𝑘),𝑥_𝑛(𝑘),𝑑𝑥_𝑚(𝑘),𝑇𝑥_𝑚(𝑘)𝑑𝑥_𝑛(𝑘),𝑇𝑥_𝑛(𝑘),𝑑𝑥_𝑚(𝑘),𝑇𝑥_𝑛(𝑘)𝑑𝑥_𝑛(𝑘),𝑇𝑥_𝑚(𝑘),12𝑑𝑥_𝑚(𝑘),𝑇𝑥_𝑚(𝑘)𝑑𝑥_𝑚(𝑘),𝑇𝑥_𝑛(𝑘)+𝑑𝑥_𝑛(𝑘),𝑇𝑥_𝑚(𝑘)𝑑𝑥_𝑛(𝑘),𝑇𝑥_𝑛(𝑘). (27) Using (3), then we obtain 1+𝑝𝑑𝑥_𝑚(𝑘),𝑥_𝑛(𝑘)𝑑²𝑥_𝑚(𝑘)+1,𝑥_𝑛(𝑘)+11≤𝑝2𝑑²𝑥_𝑚(𝑘),𝑥_𝑚(𝑘)+1𝑑𝑥_𝑛(𝑘),𝑥_𝑛(𝑘)+1+𝑑𝑥_𝑚(𝑘),𝑥_𝑚(𝑘)+1𝑑²𝑥_𝑛(𝑘),𝑥_𝑛(𝑘)+1,𝑑𝑥_𝑚(𝑘),𝑥_𝑚(𝑘)+1𝑑𝑥_𝑚(𝑘),𝑥_𝑛(𝑘)+1×𝑑𝑥_𝑛(𝑘),𝑥_𝑚(𝑘)+1,𝑑𝑥_𝑚(𝑘),𝑥_𝑛(𝑘)+1×𝑑𝑥_𝑛(𝑘),𝑥_𝑚(𝑘)+1𝑑𝑥_𝑛(𝑘),𝑥_𝑛(𝑘)+1+𝑚𝑥_𝑚(𝑘),𝑥_𝑛(𝑘)−𝜑𝑚𝑥_𝑚(𝑘),𝑥_𝑛(𝑘), (28) where 𝑚𝑥_𝑚(𝑘),𝑥_𝑛(𝑘)=max𝑑²𝑥_𝑚(𝑘),𝑥_𝑛(𝑘),𝑑𝑥_𝑚(𝑘),𝑥_𝑚(𝑘)+1𝑑𝑥_𝑛(𝑘),𝑥_𝑛(𝑘)+1,𝑑𝑥_𝑚(𝑘),𝑥_𝑛(𝑘)+1𝑑𝑥_𝑛(𝑘),𝑥_𝑚(𝑘)+1,12𝑑𝑥_𝑚(𝑘),𝑥_𝑚(𝑘)+1𝑑𝑥_𝑚(𝑘),𝑥_𝑛(𝑘)+1+𝑑𝑥_𝑛(𝑘),𝑥_𝑚(𝑘)+1𝑑𝑥_𝑛(𝑘),𝑥_𝑛(𝑘)+1. (29) Letting 𝑘→∞ and using (16)–(25), we get 1+𝑝𝜖𝜖²1≤𝑝max2[0+0],0,0+𝜖²−𝜑𝜖²=𝜖²−𝜑𝜖², (30) a contradiction. Thus {𝑥_𝑛} is a Cauchy in 𝑋.

Theorem 3. Let 𝑇 be a self-map of a complete metric space 𝑋 satisfying (1). Then 𝑇 has a unique fixed point in 𝑋.

Proof. From Lemma 2 the sequence {𝑥_𝑛} is a Cauchy in 𝑋. Since (𝑋,𝑑) is a complete metric space, then there exists a point 𝑧∈𝑋 such that lim_𝑛→∞𝑥_𝑛=𝑧. (31)
Now we prove that 𝑧 is a fixed point of 𝑇.
Taking 𝑥=𝑥_𝑛 and 𝑦=𝑧 in (1), we have 1+𝑝𝑑𝑥_𝑛,𝑧𝑑²𝑇𝑥_𝑛1,𝑇𝑧≤𝑝max2𝑑²𝑥_𝑛,𝑇𝑥_𝑛𝑑()𝑥𝑧,𝑇𝑧+𝑑_𝑛,𝑇𝑥_𝑛𝑑²(,𝑑𝑥𝑧,𝑇𝑧)_𝑛,𝑇𝑥_𝑛𝑑𝑥_𝑛,𝑇𝑧𝑑𝑧,𝑇𝑥_𝑛,𝑑𝑥_𝑛,𝑇𝑧𝑑𝑧,𝑇𝑥_𝑛𝑑(𝑥𝑧,𝑇𝑧)+𝑚_𝑛,𝑧−𝜑𝑚𝑥_𝑛,𝑧, (32) where 𝑚𝑥_𝑛𝑑,𝑧=max²𝑥_𝑛,𝑑𝑥,𝑧_𝑛,𝑇𝑥_𝑛𝑑(𝑧,𝑇𝑧),𝑑𝑥_𝑛,𝑇𝑧𝑑𝑧,𝑇𝑥_𝑛,12𝑑𝑥_𝑛,𝑇𝑥_𝑛𝑑𝑥_𝑛,𝑇𝑥_𝑛+𝑑𝑧,𝑇𝑥_𝑛𝑑(𝑧,𝑇𝑧). (33)
Using (31) and (3), we get 1+𝑝𝑑(𝑧,𝑧)𝑑²()1𝑧,𝑇𝑧≤𝑝max2𝑑²(𝑧,𝑧)𝑑(𝑧,𝑇𝑧)+𝑑(𝑧,𝑧)𝑑²(,𝑑()𝑑()𝑑(),𝑑()𝑑()𝑑(()(𝑚(.𝑧,𝑇𝑧)𝑧,𝑧𝑧,𝑇𝑧𝑧,𝑧𝑧,𝑇𝑧𝑧,𝑧𝑧,𝑇𝑧)+𝑚𝑧,𝑧−𝜑𝑧,𝑧)) (34)
Hence 𝑑²(𝑧,𝑇𝑧)≤0⇒𝑇𝑧=𝑧.
Then 𝑇 has a fixed point in 𝑋.
To prove the uniqueness of the fixed point, we assume that 𝑧₁ and 𝑧₂ are two fixed points of 𝑇. Taking 𝑥=𝑧₁ and 𝑦=𝑧₂ in (1), we easily get 𝑑(𝑧₁,𝑧₂)=0 wich implies that 𝑧₁=𝑧₂. Therefore 𝑇 has a unique fixed point in 𝑋.

Corollary 4. Let 𝑇 be a self-map of a complete metric space 𝑋 satisfying the condition 𝑑²𝑇𝑥,𝑇𝑦≤𝑚𝑥,𝑦−𝜑𝑚𝑥,𝑦, (35) where 𝑚𝑑𝑥,𝑦=max²()𝑑,𝑑𝑑,1𝑥,𝑦,𝑑𝑥,𝑇𝑥𝑦,𝑇𝑦𝑥,𝑇𝑦𝑦,𝑇𝑥2𝑑(𝑥,𝑇𝑥)𝑑𝑥,𝑇𝑦+𝑑𝑦,𝑇𝑥𝑑𝑦,𝑇𝑦 (36) for all 𝑥,𝑦∈𝑋 and 𝜑∶[0,∞)→[0,∞) is a continuous function with 𝜑(𝑡)=0⇔𝑡=0 and 𝜑(𝑡)>0 for each 𝑡>0. Then 𝑇 has a unique fixed point in 𝑋.

Proof. 𝑝=0 in Theorem 3; we have the result.

Now we give an example to support our result.

Example 5. Let 𝑋={0,1,2}, and let 𝑑 be the usual metric on 𝑋. Let 𝑇∶𝑋→𝑋 be defined by 𝑇0=𝑇1=0 and 𝑇2=1. And define 𝜑∶[0,∞)→[0,∞) by 𝜑(𝑡)=𝑡/2. For any value of 𝑝>0 and 𝑥,𝑦∈𝑋, then it is easy to verify that inequality (1) holds. Hence Theorem 3 holds well.

Acknowledgment

Penumurthy Parvateesam Murthy is thankful to the University Grants Commission, New Delhi, India, for financial assistance through Major Reserch Project File no. 42-32/2013 (SR).