Some Common Fixed Point Theorems in Generalized Metric Spaces

Abstract

First we prove common fixed point theorems for weakly compatible maps which generalize the results of Chen (2012). Secondly, we prove common fixed point theorems using property E.A. along with weakly compatible maps. At the end, we prove common fixed point theorems using common limit range property (CLR property) along with weakly compatible maps.

1. Introduction

In 2000, Branciari [1] introduced the notion of generalized metric space as follows.

Definition 1. Let 𝑋 be a nonempty set and let 𝑑∶𝑋×𝑋→[0,∞) be a mapping such that for all 𝑥,𝑦∈𝑋 and for all distinct points  𝑢,𝑣∈𝑋  each of them different from 𝑥 and 𝑦, one has the following:(i)𝑑(𝑥,𝑦)=0 if and only if 𝑥=𝑦,(ii)𝑑(𝑥,𝑦)=𝑑(𝑦,𝑥), (iii)𝑑(𝑥,𝑦)≤𝑑(𝑥,𝑢)+𝑑(𝑢,𝑣)+𝑑(𝑣,𝑦) (rectangular inequality).
Then (𝑋,𝑑) is called a generalized metric space (or shortly g.m.s.).

Definition 2. Let (𝑋,𝑑) be a g.m.s. Let {𝑥_𝑛} be a sequence in 𝑋 and  𝑥∈𝑋. We say that {𝑥_𝑛} is (i) g.m.s. convergent to 𝑥 if and only if  𝑑(𝑥_𝑛,𝑥)→0 as  𝑛→∞,(ii) g.m.s. Cauchy sequence if and only if, for each  𝜀>0, there exists 𝑛₀∈ℕ such that, 𝑑(𝑥_𝑚,𝑥_𝑛)<𝜀 for all 𝑛>𝑚>𝑛₀,(iii) complete g.m.s. if every g.m.s. Cauchy sequence is g.m.s. convergent in 𝑋.
In 1986, Jungck [2] introduced the notion of compatible maps as follows.
Let (𝑋,𝑑) be a metric space. Let 𝑆,𝐹∶𝑋→𝑋 be two single-valued functions. We say that 𝑆 and 𝐹 are compatible if  lim_𝑛→∞𝑑(𝑆𝐹𝑥_𝑛,𝐹𝑆𝑥_𝑛)=0, whenever {𝑥_𝑛} is a sequence in 𝑋 such that lim_𝑛→∞𝑑(𝐹𝑥_𝑛,𝑆𝑥_𝑛)=0.
In 1998, Jungck and Rhoades [3] introduced the notion of weakly compatible maps as follows.
Two maps are said to be weakly compatible if they commute at the coincidence point. Note that compatible maps are weakly compatible but the converse is not true in general.
In 2002, Aamri and El Moutawakil [4] introduced the notion of property E.A. as follows.
Let 𝐴 and 𝐵 be two self-mappings of a metric space (𝑋,𝑑). We say that 𝐴 and 𝐵 satisfy the property E.A. if there exists a sequence {𝑥_𝑛} such that lim_𝑛→∞𝐴𝑥_𝑛=lim_𝑛→∞𝐵𝑥_𝑛=𝑡 for some 𝑡 in 𝑋.
In 2011, Sintunavarat and Kumam [5] introduced the notion of (CLRg) property as follows.
Let (𝑋,𝑑) be a metric space. Two mappings 𝑓∶𝑋→𝑋 and 𝑔∶𝑋→𝑋 are said to satisfy (CLRg) property if there exists sequences {𝑥_𝑛} in 𝑋 such that lim_𝑛→∞𝑓𝑥_𝑛=lim_𝑛→∞𝑔𝑥_𝑛=𝑔(𝑡)forsome𝑡in𝑋. (1) Similarly, we can state (CLR_𝐹) and (CLR_𝐺) properties if the mapping  𝑔∶𝑋→𝑋  has been replaced by the mapping 𝐹∶𝑋→𝑋 and  𝐺∶𝑋→𝑋, respectively.
In 2012, Chen [6] introduced the notion of 𝒲 function and 𝒮 function as follows.

Definition 3. A function  𝜑∶ℝ⁺→ℝ⁺ is said to be a  𝒲 function if the function 𝜙 satisfies the following conditions:  (𝜙₁) 𝜙(𝑡)<𝑡 for all 𝑡>0 and 𝜙(0)=0,  (𝜙₂) lim_{𝑡𝑛→𝑡}inf𝜙(𝑡_𝑛)<𝑡 for all 𝑡>0.

Definition 4. A function  𝜑∶ℝ⁺→ℝ⁺ is said to be a 𝒮 function if the function 𝜙 satisfies the following conditions: (𝜙₁) 𝜙 is a strictly increasing, continuous function in each coordinate,  (𝜙₂) for all 𝑡>0, 𝜙(𝑡,𝑡,𝑡)<𝑡,  𝜙(𝑡,0,0)<𝑡,  𝜙(0,𝑡,0)<𝑡, and 𝜙(0,0,𝑡)<𝑡.
Before proving our results, we need the following lemma.

Lemma 5 (see [6]). Let 𝜙∶ℝ⁺→ℝ⁺ be a 𝒲 function. Then lim_𝑛→∞𝜙^𝑛(𝑡)=0for all 𝑡>0, where 𝜙^𝑛(𝑡) denotes the 𝑛th iteration of 𝜙.

2. Weakly Compatible Maps

In 2012, Chen [6] proved the following fixed point theorems for compatible maps.

Theorem 6. Let (𝑋,𝑑) be Hausdorff and complete g.m.s. and let 𝜙∶ℝ⁺→ℝ⁺ be a 𝒲 function. Let 𝑆,  𝑇,  𝐹,  and  𝐺  be self-maps on 𝑋 satisfying the following: 𝑇(𝑋)(𝑋)(𝑋)(𝑋),𝑑𝑑(){}{}⊆𝐹,𝑆⊆𝐺𝑆𝑥,𝑇𝑦≤𝜙max𝐹𝑥,𝐺𝑦,𝑑𝐹𝑥,𝑆𝑥,𝑑𝐺𝑦,𝑇𝑦∀𝑥,𝑦∈𝑋,𝑝𝑎𝑖𝑟𝑠𝑆,𝐹𝑎𝑛𝑑𝑇,𝐺𝑎𝑟𝑒𝑐𝑜𝑚𝑝𝑎𝑡𝑖𝑏𝑙𝑒. (2)(3)(4) If 𝐹 or 𝐺 is continuous, then 𝑆, 𝑇, 𝐹, and 𝐺 have a unique common fixed point in 𝑋.

Theorem 7. Let (𝑋,𝑑) be Hausdorff and complete g.m.s. and let 𝜙∶ℝ⁺³→ℝ⁺ be a 𝒮 function. Let  𝑆,  𝑇, 𝐹, and 𝐺 be self-maps on 𝑋 satisfying (2), (4), and the following: 𝑑𝑑()𝑆𝑥,𝑇𝑦≤𝜙𝐹𝑥,𝐺𝑦,𝑑𝐹𝑥,𝑆𝑥,𝑑𝐺𝑦,𝑇𝑦∀𝑥,𝑦∈𝑋. (5) If one of 𝑆(𝑋), 𝑇(𝑋), 𝐹(𝑋), and 𝐺(𝑋), is a complete subset of 𝑋 then 𝑆, 𝑇, 𝐹, and 𝐺 have a unique common fixed point in 𝑋.

Now, we prove our results for weakly compatible maps.

Theorem 8. Let (𝑋,𝑑) be Hausdorff and complete g.m.s. and let 𝜙∶ℝ⁺→ℝ⁺ be a 𝒲 function. Let 𝑆, 𝑇, 𝐹, and 𝐺 be self-maps on 𝑋 satisfying (2), (3) and the following: 𝑝𝑎𝑖𝑟𝑠{𝑆,𝐹}𝑎𝑛𝑑{𝑇,𝐺}𝑎𝑟𝑒𝑤𝑒𝑎𝑘𝑙𝑦𝑐𝑜𝑚𝑝𝑎𝑡𝑖𝑏𝑙𝑒. (6) If one of 𝑆(𝑋), 𝑇(𝑋), 𝐹(𝑋), and 𝐺(𝑋) is a complete subset of 𝑋, then 𝑆,  𝑇,  𝐹, and 𝐺 have a unique common fixed point in 𝑋.

Proof. Let 𝑥₀∈𝑋. Define the sequence {𝑥_𝑛} recursively as follows: 𝐺𝑥_2𝑛+1=𝑆𝑥_2𝑛=𝑧_2𝑛,𝐹𝑥_2𝑛+2=𝑇𝑥_2𝑛+1=𝑧_2𝑛+1. (7) By Theorem 6 [6], {𝑧_𝑛} is g.m.s. Cauchy sequence in 𝑋. Now suppose that 𝐹(𝑋) is complete subset of 𝑋, then the subsequence 𝑧_2𝑛+1=𝐹𝑥_2𝑛+2 must get a limit in 𝐹(𝑋). Call it to be 𝑢 and 𝑣∈𝐹⁻¹(𝑢). Then 𝐹(𝑣)=𝑢. As {𝑧_𝑛} is a Cauchy sequence containing a convergent subsequence {𝑧_2𝑛+1}, therefore the sequence {𝑧_𝑛} also converges implying, thereby, the convergence of {𝑧_2𝑛} being a subsequence of the convergent sequence {𝑧_𝑛}.
Taking 𝑥=𝑣 and 𝑦=𝑥_2𝑛+1 in (3), we get 𝑑𝑆𝑣,𝑇𝑥_2𝑛+1≤𝜙max𝑑𝐹𝑣,𝐺𝑥_2𝑛+1(),𝑑,𝑑𝐹𝑣,𝑆𝑣𝐺𝑥_2𝑛+1,𝑇𝑥_2𝑛+1. (8) Letting 𝑛→∞, we have 𝑑()({𝑑()()((𝑑(()𝑆𝑣,𝑢≤𝜙max𝑢,𝑢,𝑑𝑢,𝑆𝑣,𝑑𝑢,𝑢)})=𝜙𝑢,𝑆𝑣))<𝑑𝑆𝑣,𝑢,acontradiction. (9) Therefore, 𝑆𝑣=𝑢=𝐹𝑣 implies that the pair (𝑆,𝐹) has a point of coincidence.
As 𝑆(𝑋)⊆𝐺(𝑋), 𝑆𝑣=𝑢 implies that 𝑢∈𝐺(𝑋). Letting 𝑤∈𝐺⁻¹(𝑢), then 𝐺𝑤=𝑢.
Now, from (3), we have 𝑑𝑧_2𝑛,𝑇𝑤=𝑑𝑆𝑥_2𝑛𝑑,𝑇𝑤≤𝜙max𝐹𝑥_2𝑛,𝐺𝑤,𝑑𝐹𝑥_2𝑛,𝑆𝑥_2𝑛,𝑑(𝐺𝑤,𝑇𝑤). (10) Letting 𝑛→∞, we have 𝑑()({𝑑()()((𝑑(()𝑢,𝑇𝑤≤𝜙max𝑢,𝑢,𝑑𝑢,𝑢,𝑑𝑢,𝑇𝑤)})=𝜙𝑢,𝑇𝑤))<𝑑𝑢,𝑇𝑤,acontradiction. (11) Therefore, 𝑢=𝑇𝑤.
Thus, we have shown that 𝑢=𝑆𝑣=𝐹𝑣=𝑇𝑤=𝐺𝑤, which amounts to say that both pairs have point of coincidence. If one assumes 𝐺(𝑋) to be complete, then an analogous argument establishes this claim.
The remaining two cases pertain essentially to the previous cases. Indeed, if 𝑆(𝑋) is complete, then 𝑢∈𝑆(𝑋)⊆𝐺(𝑋), and if 𝑇(𝑋) is complete, then 𝑢∈𝑇(𝑋)⊆𝐹(𝑋). Thus 𝑆 and 𝐹,  𝑇, and 𝐺 have a point of coincidence.
Since the pairs (𝑆,𝐹) and (𝑇,𝐺) are weakly compatible at 𝑣 and 𝑤, respectively, then ()()()()𝑆𝑢=𝑆𝐹𝑣=𝐹𝑆𝑣=𝐹𝑢,𝑇𝑢=𝑇𝐺𝑤=𝐺𝑇𝑤=𝐺𝑢. (12) If 𝑆𝑢≠𝑢, then 𝑑()()({𝑑()()(({𝑑()(𝑑(()𝑆𝑢,𝑢=𝑑𝑆𝑢,𝑇𝑤≤𝜙max𝐹𝑢,𝐺𝑤,𝑑𝐹𝑢,𝑆𝑢,𝑑𝐺𝑤,𝑇𝑤)})=𝜙max𝑆𝑢,𝑢,0,0})=𝜙𝑆𝑢,𝑢))<𝑑𝑆𝑢,𝑢,acontradiction. (13) Therefore, 𝑆𝑢=𝑢.
Similarly, one can show that 𝑇𝑢=𝑢.
Thus 𝑢 is the common fixed point of 𝑆, 𝑇, 𝐹, and 𝐺.
Finally, we prove that 𝑆, 𝑇, 𝐹, and 𝐺 have a unique common fixed point. Let 𝑡 be another common fixed point of 𝑆, 𝑇, 𝐹, and 𝐺.
From (3), we have 𝑑()()({𝑑()()(({𝑑()()(()()𝑡,𝑢=𝑑𝑆𝑡,𝑇𝑢≤𝜙max𝐹𝑡,𝐺𝑢,𝑑𝐹𝑡,𝑆𝑡,𝑑𝐺𝑢,𝑇𝑢)})=𝜙max𝑡,𝑢,𝑑𝑡,𝑡,𝑑𝑢,𝑢)})<𝑑𝑡,𝑢,acontradictionunless𝑑𝑡,𝑢=0,thatis,𝑡=𝑢. (14) Hence 𝑢 is the unique common fixed point of 𝑆, 𝑇, 𝐹, and 𝐺 in 𝑋.
We give the following example to illustrate Theorem 8.

Example 9. Let 𝑋={𝑎₁,𝑎₂,𝑎₃,𝑎₄,𝑎₅}, where 𝑎₁, 𝑎₂, 𝑎₃, 𝑎₄, and 𝑎₅ are positive constants. We define 𝑑∶𝑋×𝑋→[0,∞) by(1)𝑑(𝑥,𝑥)=0, for all 𝑥∈𝑋,(2)𝑑(𝑥,𝑦)=𝑑(𝑦,𝑥), for all 𝑥,𝑦∈𝑋,(3)𝑑(𝑎₁,𝑎₂)=3𝑘, (4)𝑑(𝑎₁,𝑎₃)=𝑑(𝑎₂,𝑎₃)=𝑘, (5)𝑑(𝑎₁,𝑎₄)=𝑑(𝑎₂,𝑎₄)=𝑑(𝑎₃,𝑎₄)=2𝑘, (6)𝑑(𝑎₁,𝑎₅)=𝑑(𝑎₂,𝑎₅)=𝑑(𝑎₃,𝑎₅)=𝑑(𝑎₄,𝑎₅)=2𝑘,
where 𝑘>0 is a constant.
If 𝜙∶ℝ⁺→ℝ⁺, 𝜙(𝑡)=(2/3)𝑡, then 𝜙 is a 𝒲 function. We next define 𝑆,𝑇,𝐹,𝐺∶𝑋→𝑋 by 𝑆(𝑥)=𝑎₃,if𝑥≠𝑎₄,𝑎₅,if𝑥=𝑎₄,𝑇(𝑥)=𝑎₃,if𝑥≠𝑎₄,𝑎₁,if𝑥=𝑎₄, (15)𝐺(𝑥)=𝐼(𝑥)= the identity mapping, 𝐹(𝑥)=𝑎₃,if𝑥=𝑎₃,𝑎₁,if𝑥=𝑎₁,𝑎₂,𝑎₅,𝑎₂,if𝑥=𝑎₄. (16) Then all the conditions of Theorem 8 are satisfied and 𝑎₃ is a unique common fixed point of 𝑆, 𝑇, 𝐹, and 𝐺.

Theorem 10. Let (𝑋,𝑑) be Hausdorff and complete g.m.s. and let 𝜙∶ℝ⁺³→ℝ⁺ be a 𝒮 function. Let 𝑆, 𝑇, 𝐹, and 𝐺 be self-maps on 𝑋 satisfying (2), (5) and (6). If one of 𝑆(𝑋), 𝑇(𝑋), 𝐹(𝑋), and 𝐺(𝑋) is a complete subset of 𝑋, then 𝑆, 𝑇, 𝐹, and 𝐺 have a unique common fixed point in 𝑋.

Proof. Given that 𝑥₀∈𝑋, define the sequence {𝑥_𝑛} recursively as follows: 𝐺𝑥_2𝑛+1=𝑆𝑥_2𝑛=𝑧_2𝑛,𝐹𝑥_2𝑛+2=𝑇𝑥_2𝑛+1=𝑧_2𝑛+1. (17) Then due to Theorem 7 [6], {𝑧_𝑛} is g.m.s. Cauchy sequence in 𝑋. Now, suppose that 𝐹(𝑋) is complete subset of 𝑋; then the subsequence 𝑧_2𝑛+1=𝐹𝑥_2𝑛+2 must get a limit in 𝐹(𝑋). Call it to be 𝑢 and 𝑣∈𝐹⁻¹(𝑢). Then 𝐹(𝑣)=𝑢. As {𝑧_𝑛} is a Cauchy sequence containing a convergent subsequence {𝑧_2𝑛+1}, therefore the sequence {𝑧_𝑛} also converges implying thereby the convergence of {𝑧_2𝑛} being a subsequence of the convergent sequence {𝑧_𝑛}. On taking 𝑥=𝑣 and 𝑦=𝑥_2𝑛+1 in (5) one gets 𝑑𝑆𝑣,𝑇𝑥_2𝑛+1≤𝜙𝑑𝐹𝑣,𝐺𝑥_2𝑛+1,𝑑(𝐹𝑣,𝑆𝑣),𝑑𝐺𝑥_2𝑛+1,𝑇𝑥_2𝑛+1. (18) Letting 𝑛→∞, we have 𝑑()𝑑()()((𝑑(()𝑆𝑣,𝑢≤𝜙({𝑢,𝑢,𝑑𝑢,𝑆𝑣,𝑑𝑢,𝑢)})=𝜙𝑢,𝑆𝑣))<𝑑𝑆𝑣,𝑢,acontradiction. (19) Therefore, 𝑆𝑣=𝑢=𝐹𝑣, which shows that the pair (𝑆,𝐹) has a point of coincidence.
As 𝑆(𝑋)⊆𝐺(𝑋), 𝑆𝑣=𝑢 implies that 𝑢∈𝐺(𝑋). Letting 𝑤∈𝐺⁻¹(𝑢), then 𝐺𝑤=𝑢.
From (5), we have 𝑑𝑧_2𝑛,𝑇𝑤=𝑑𝑆𝑥_2𝑛𝑑,𝑇𝑤≤𝜙𝐹𝑥_2𝑛,𝐺𝑤,𝑑𝐹𝑥_2𝑛,𝑆𝑥_2𝑛,𝑑(𝐺𝑤,𝑇𝑤). (20) Letting 𝑛→∞, we have 𝑑()𝑑()()((𝑑(()𝑢,𝑇𝑤≤𝜙({𝑢,𝑢,𝑑𝑢,𝑢,𝑑𝑢,𝑇𝑤)})=𝜙𝑢,𝑇𝑤))<𝑑𝑢,𝑇𝑤,acontradiction. (21) Therefore, 𝑢=𝑇𝑤. Thus, we have shown that 𝑢=𝑆𝑣=𝐹𝑣=𝑇𝑤=𝐺𝑤, which amounts to say that both pairs have point of coincidence. If one assumes 𝐺(𝑋) to be complete, then an analogous argument establishes this claim.
The remaining two cases pertain essentially to the previous cases. Indeed if 𝑆(𝑋) is complete, then 𝑢∈𝑆(𝑋)⊆𝐺(𝑋), and if 𝑇(𝑋) is complete, then 𝑢∈𝑇(𝑋)⊆𝐹(𝑋).
Thus 𝑆 and 𝐹, 𝑇, and 𝐺 have a point of coincidence.
Since the pairs (𝑆,𝐹) and (𝑇,𝐺) are weakly compatible at 𝑣 and 𝑤, respectively, then ()()()()𝑆𝑢=𝑆𝐹𝑣=𝐹𝑆𝑣=𝐹𝑢,𝑇𝑢=𝑇𝐺𝑤=𝐺𝑇𝑤=𝐺𝑢. (22) If 𝑆𝑢≠𝑢, then 𝑑()()𝑑()()(𝑑()(𝑑(()𝑆𝑢,𝑢=𝑑𝑆𝑢,𝑇𝑤≤𝜙({𝐹𝑢,𝐺𝑤,𝑑𝐹𝑢,𝑆𝑢,𝑑𝐺𝑤,𝑇𝑤)})=𝜙({𝑆𝑢,𝑢,0,0})=𝜙𝑆𝑢,𝑢))<𝑑𝑆𝑢,𝑢,acontradiction. (23) Therefore, 𝑆𝑢=𝑢.
Similarly, one can show that 𝑇𝑢=𝑢.
Thus 𝑢 is the common fixed point of 𝑆, 𝑇, 𝐹, and 𝐺.
Finally, we prove that 𝑆, 𝑇, 𝐹, and  𝐺  have a unique common fixed point. Let 𝑡 be another common fixed point of 𝑆, 𝑇, 𝐹, and 𝐺. Then using (5), we have 𝑑()()𝑑()()(𝑑()()(()()𝑡,𝑢=𝑑𝑆𝑡,𝑇𝑢≤𝜙({𝐹𝑡,𝐺𝑢,𝑑𝐹𝑡,𝑆𝑡,𝑑𝐺𝑢,𝑇𝑢)})=𝜙({𝑡,𝑢,𝑑𝑡,𝑡,𝑑𝑢,𝑢)})<𝑑𝑡,𝑢,acontradictionunless𝑑𝑡,𝑢=0,thatis,𝑡=𝑢. (24) Hence 𝑢 is the unique common fixed point of 𝑆, 𝑇, 𝐹, and 𝐺 in 𝑋.

We give the following example to illustrate Theorem 10.

Example 11. Let 𝑋={𝑎₁,𝑎₂,𝑎₃,𝑎₄,𝑎₅}, where 𝑎₁, 𝑎₂, 𝑎₃, 𝑎₄, and 𝑎₅ are positive constants. We define 𝑑∶𝑋×𝑋→[0,∞) by(1)𝑑(𝑥,𝑥)=0, for all 𝑥∈𝑋,(2)𝑑(𝑥,𝑦)=𝑑(𝑦,𝑥), for all 𝑥,𝑦∈𝑋,(3)𝑑(𝑎₁,𝑎₂)=3𝑘,(4)𝑑(𝑎₁,𝑎₃)=𝑑(𝑎₂,𝑎₃)=𝑘,(5)𝑑(𝑎₁,𝑎₄)=𝑑(𝑎₂,𝑎₄)=𝑑(𝑎₃,𝑎₄)=2𝑘, (6)𝑑(𝑎₁,𝑎₅)=𝑑(𝑎₂,𝑎₅)=𝑑(𝑎₃,𝑎₅)=𝑑(𝑎₄,𝑎₅)=2𝑘,
where 𝑘>0 is a constant.
If 𝜙∶ℝ⁺³→ℝ⁺,   𝜙(𝑡)=(3/5)⋅max{𝑡₁,𝑡₂,𝑡₃}, then 𝜙 is a 𝒮 function. We next define 𝑆, 𝑇, 𝐹,  𝐺∶𝑋→𝑋 by 𝑆(𝑥)=𝑎₃,if𝑥≠𝑎₄,𝑎₅,if𝑥=𝑎₄,𝑇(𝑥)=𝑎₃,if𝑥≠𝑎₄,𝑎₁,if𝑥=𝑎₄,𝐺(𝑥)(𝑥)𝐹(𝑥)=𝑎=𝐼=theidentitymapping,₃,if𝑥=𝑎₃,𝑎₁,if𝑥=𝑎₁,𝑎₂,𝑎₅,𝑎₂,if𝑥=𝑎₄. (25) Then all the conditions of Theorem 10 are satisfied and 𝑎₃ is a unique common fixed point of 𝑆, 𝑇, 𝐹, and 𝐺.

3. Property E.A.

Theorem 12. Let (𝑋,𝑑) be Hausdorff and complete g.m.s. and let 𝜙∶ℝ⁺→ℝ⁺ be a 𝒲 function. Let 𝑆, 𝑇, 𝐹, and 𝐺 be self-maps on 𝑋 satisfying (2), (3), (6), and the following: {𝑆,𝐹}𝑜𝑟{𝑇,𝐺}𝑠𝑎𝑡𝑖𝑠𝑓𝑦𝑡ℎ𝑒𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑦𝐸.𝐴. (26) If the range of one of the maps 𝑆, 𝑇, 𝐹, or 𝐺 is a complete subspace of 𝑋, then 𝑆, 𝑇, 𝐹, and 𝐺 have a unique common fixed point in 𝑋.

Proof. If the pair {𝑇,𝐺} satisfies the property E.A., then there exists a sequence {𝑥_𝑛} in 𝑋 such that 𝑇𝑥_𝑛→𝑧 and 𝐺𝑥_𝑛→𝑧 for some 𝑧∈𝑋 as 𝑛→∞. Since 𝑇(𝑋)⊆𝐹(𝑋), there exists a sequence {𝑦_𝑛} in 𝑋 such that 𝑇𝑥_𝑛=𝐹𝑦_𝑛. Hence 𝐹𝑦_𝑛→𝑧 as 𝑛→∞. Also, since 𝑆(𝑋)⊆𝐺(𝑋), there exists a sequence {𝑦^′_𝑛} in 𝑋 such that 𝑆𝑦^′_𝑛=𝐺𝑦^′_𝑛. Hence 𝑆𝑦^′_𝑛→𝑧 as 𝑛→∞.
Suppose that 𝐹(𝑋) is a complete subspace of 𝑋. Then, 𝑧=𝐹𝑢 for some 𝑢∈𝑋.
Subsequently, we have 𝑆𝑦^′_𝑛→𝐹𝑢, 𝑇𝑥_𝑛→𝐹𝑢, 𝐹𝑦_𝑛→𝐹𝑢 as 𝑛→∞.
From (3), we have 𝑑𝑆𝑢,𝑇𝑥_𝑛≤𝜙max𝑑𝐹𝑢,𝐺𝑥_𝑛,𝑑(𝐹𝑢,𝑆𝑢),𝑑𝐺𝑥_𝑛,𝑇𝑥_𝑛. (27) Letting 𝑛→∞, we have 𝑑()({𝑑()()((𝑑((),𝑆𝑢,𝐹𝑢≤𝜙max𝐹𝑢,𝐹𝑢,𝑑𝐹𝑢,𝑆𝑢,𝑑𝐹𝑢,𝐹𝑢)})=𝜙𝐹𝑢,𝑆𝑢))<𝑑𝐹𝑢,𝑆𝑢 (28) which implies that 𝐹𝑢=𝑆𝑢.
The weak compatibility of 𝐹 and 𝑆 implies that 𝐹𝑆𝑢=𝑆𝐹𝑢 and then 𝑆𝑆𝑢=𝑆𝐹𝑢=𝐹𝑆𝑢=𝐹𝐹𝑢.
On the other hand, since 𝑆(𝑋)⊆𝐺(𝑋), there exists a 𝑣∈𝑋 such that 𝑆𝑢=𝐺𝑣.
Now, we show that 𝐺𝑣=𝑇𝑣.
From (3), we have 𝑑()({𝑑()()((𝑑((),𝑆𝑢,𝑇𝑣≤𝜙max𝐹𝑢,𝐺𝑣,𝑑𝐹𝑢,𝑆𝑢,𝑑𝐺𝑣,𝑇𝑣)})=𝜙𝐺𝑣,𝑇𝑣))<𝑑𝐺𝑣,𝑇𝑣impliesthat𝐺𝑣=𝑇𝑣,thatis,𝑆𝑢=𝐹𝑢=𝐺𝑣=𝑇𝑣. (29) The weak compatibility of 𝑇 and 𝐺 implies that 𝑇𝐺𝑣=𝐺𝑇𝑣 and 𝐺𝐺𝑣=𝐺𝑇𝑣=𝑇𝐺𝑣=𝑇𝑇𝑣.
Let us show that 𝑆𝑢 is the common fixed point of 𝑆,  𝑇,  𝐹,  and  𝐺.
From (3), we have 𝑑()({𝑑()()((𝑑((),𝑆𝑆𝑢,𝑇𝑣≤𝜙max𝐹𝑆𝑢,𝐺𝑣,𝑑𝐹𝑆𝑢,𝑆𝑆𝑢,𝑑𝐺𝑣,𝑇𝑣)})=𝜙𝑆𝑆𝑢,𝑇𝑣))<𝑑𝐺𝑣,𝑇𝑣impliesthat𝑆𝑆𝑢=𝑇𝑣=𝑆𝑢. (30) Therefore, 𝑆𝑢=𝑆𝑆𝑢=𝐹𝑆𝑢 and 𝑆𝑢 is the common fixed point of 𝑆 and 𝐹.
Similarly, we can prove that 𝑇𝑣 is the common fixed point of 𝑇 and 𝐺.
Since 𝑆𝑢=𝑇𝑣, we conclude that 𝑆𝑢 is the common fixed point of 𝑆,  𝑇,  𝐹,  and 𝐺.
The proof is similar when 𝐺(𝑋) is assumed to be complete subspace of 𝑋. The cases in which 𝑆(𝑋) or 𝑇(𝑋) is a complete subspace of 𝑋 are similar to the cases in which 𝐺(𝑋) or 𝐹(𝑋), respectively, is complete since 𝑆(𝑋)⊆𝐺(𝑋),𝑇(𝑋)⊆𝐹(𝑋). (31) If 𝑆𝑢=𝑇𝑢=𝐹𝑢=𝐺𝑢=𝑢 and 𝑆𝑣=𝑇𝑣=𝐹𝑣=𝐺𝑣=𝑣, then (3) gives 𝑑()({𝑑()()(()𝑆𝑢,𝑇𝑣≤𝜙max𝐹𝑢,𝐺𝑣,𝑑𝐹𝑢,𝑆𝑢,𝑑𝐺𝑣,𝑇𝑣)})<𝑑𝐹𝑢,𝐺𝑣,impliesthat𝑢=𝑣. (32) Therefore, 𝑢=𝑣 and the common fixed point is unique.

Theorem 13. Let (𝑋,𝑑) be Hausdorff and complete g.m.s. and let  𝜙∶ℝ⁺³→ℝ⁺  be a 𝒮 function. Let 𝑆,  𝑇, 𝐹, and 𝐺 be self-maps on 𝑋 satisfying (2), (5), (6), and (26). If the range of one of the maps 𝑆,  𝑇, 𝐹, or 𝐺 is a complete subspace of 𝑋, then 𝑆, 𝑇, 𝐹, and 𝐺 have a unique common fixed point in 𝑋.

Proof. If the pair {𝑇,𝐺} satisfies the property E.A., then there exists a sequence {𝑥_𝑛} in 𝑋 such that 𝑇𝑥_𝑛→𝑧 and 𝐺𝑥_𝑛→𝑧 for some 𝑧∈𝑋 as 𝑛→∞.
Since 𝑇(𝑋)⊆𝐹(𝑋), there exists a sequence {𝑦_𝑛} in 𝑋 such that𝑇𝑥_𝑛=𝐹𝑦_𝑛. Hence 𝐹𝑦_𝑛→𝑧 as 𝑛→∞. Also, since 𝑆(𝑋)⊆𝐺(𝑋), there exists a sequence {𝑦^′_𝑛} in 𝑋 such that 𝑆𝑦^′_𝑛=𝐺𝑦^′_𝑛. Hence 𝑆𝑦^′_𝑛→𝑧 as 𝑛→∞.
Suppose that 𝐹(𝑋) is a complete subspace of 𝑋. Then, 𝑧=𝐹𝑢 for some 𝑢∈𝑋.
Subsequently, we have 𝑆𝑦^′_𝑛→𝐹𝑢, 𝑇𝑥_𝑛→𝐹𝑢, 𝐹𝑦_𝑛→𝐹𝑢 as 𝑛→∞.
From (5), we have 𝑑𝑆𝑢,𝑇𝑥_𝑛≤𝜙𝑑𝐹𝑢,𝐺𝑥_𝑛,𝑑(𝐹𝑢,𝑆𝑢),𝑑𝐺𝑥_𝑛,𝑇𝑥_𝑛. (33) Letting 𝑛→∞, we have 𝑑()(𝑑()()((())(),𝑆𝑢,𝐹𝑢≤𝜙𝐹𝑢,𝐹𝑢,𝑑𝐹𝑢,𝑆𝑢,𝑑𝐹𝑢,𝐹𝑢))=𝜙0,𝑑𝐹𝑢,𝑆𝑢,0<𝑑𝐹𝑢,𝑆𝑢impliesthat,𝐹𝑢=𝑆𝑢. (34)
The weak compatibility of 𝐹 and 𝑆 implies that 𝐹𝑆𝑢=𝑆𝐹𝑢 and then 𝑆𝑆𝑢=𝑆𝐹𝑢=𝐹𝑆𝑢=𝐹𝐹𝑢.
On the other hand, since 𝑆(𝑋)⊆𝐺(𝑋), there exists a 𝑣∈𝑋 such that 𝑆𝑢=𝐺𝑣.
Now, we show that 𝐺𝑣=𝑇𝑣.
From (5), we have 𝑑()(𝑑()()((((),𝑆𝑢,𝑇𝑣≤𝜙𝐹𝑢,𝐺𝑣,𝑑𝐹𝑢,𝑆𝑢,𝑑𝐺𝑣,𝑇𝑣))=𝜙0,0,𝑑𝐺𝑣,𝑇𝑣))<𝑑𝐺𝑣,𝑇𝑣impliesthat𝐺𝑣=𝑇𝑣,thatis,𝑆𝑢=𝐹𝑢=𝐺𝑣=𝑇𝑣. (35) The weak compatibility of 𝑇 and 𝐺 implies that 𝑇𝐺𝑣=𝐺𝑇𝑣 and 𝐺𝐺𝑣=𝐺𝑇𝑣=𝑇𝐺𝑣=𝑇𝑇𝑣.
Let us show that 𝑆𝑢 is the common fixed point of 𝑆, 𝑇, 𝐹 and 𝐺.
From (5), we have 𝑑()(𝑑()()((𝑑())(),𝑆𝑆𝑢,𝑇𝑣≤𝜙𝐹𝑆𝑢,𝐺𝑣,𝑑𝐹𝑆𝑢,𝑆𝑆𝑢,𝑑𝐺𝑣,𝑇𝑣))=𝜙𝑆𝑆𝑢,𝑇𝑣,0,0<𝑑𝑆𝑆𝑢,𝑇𝑣impliesthat𝑆𝑆𝑢=𝑇𝑣=𝑆𝑢,thatis,𝑆𝑢=𝑆𝑆𝑢=𝐹𝑆𝑢. (36) Therefore, 𝑆𝑢 is the common fixed point of 𝑆 and 𝐹.
Similarly, one can prove that 𝑇𝑣 is the common fixed point of 𝑇 and 𝐺.
Since 𝑆𝑢=𝑇𝑣, we conclude that 𝑆𝑢 is the common fixed point of 𝑆, 𝑇, 𝐹, and 𝐺.
The proof is similar when 𝐺(𝑋) is assumed to be complete subspace of 𝑋. The cases in which 𝑆(𝑋) or 𝑇(𝑋) is a complete subspace of 𝑋 are similar to the cases in which 𝐺(𝑋) or 𝐹(𝑋), respectively, is complete since 𝑆(𝑋)⊆𝐺(𝑋) and 𝑇(𝑋)⊆𝐹(𝑋).
If 𝑆𝑢=𝑇𝑢=𝐹𝑢=𝐺𝑢=𝑢 and 𝑆𝑣=𝑇𝑣=𝐹𝑣=𝐺𝑣=𝑣, then (5) gives 𝑑()(𝑑()()(()𝑆𝑢,𝑇𝑣≤𝜙𝐹𝑢,𝐺𝑣,𝑑𝐹𝑢,𝑆𝑢,𝑑𝐺𝑣,𝑇𝑣))<𝑑𝐹𝑢,𝐺𝑣,impliesthat𝑢=𝑣. (37)
Therefore, 𝑢=𝑣 and the common fixed point is unique.

4. (CLR) Property

Theorem 14. Let (𝑋,𝑑) be Hausdorff and complete g.m.s. and let 𝜙∶ℝ⁺→ℝ⁺ be a 𝒲 function. Let 𝑆,  𝑇,  𝐹, and  𝐺 be self-maps on 𝑋 satisfying (3), (6), and the following: 𝑇(𝑋)⊆𝐹(𝑋)𝑎𝑛𝑑𝑡ℎ𝑒𝑝𝑎𝑖𝑟{𝑇,𝐺}𝑠𝑎𝑡𝑖𝑠𝑓𝑖𝑒𝑠CLR_𝐺𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑦, (38) or 𝑆(𝑋)⊆𝐺(𝑋)𝑎𝑛𝑑𝑡ℎ𝑒𝑝𝑎𝑖𝑟{𝑆,𝐹}𝑠𝑎𝑡𝑖𝑠𝑓𝑖𝑒𝑠CLR_𝐹𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑦. (39) Then 𝑆, 𝑇, 𝐹, and 𝐺 have a unique common fixed point.

Proof. Without loss of generality, we assume that 𝑇(𝑋)⊆𝐹(𝑋) and the pair {𝑇,𝐺} satisfies (CLR_𝐺) property; then there exists some sequence {𝑥_𝑛} in 𝑋 such that 𝑇𝑥_𝑛 and 𝐺𝑥_𝑛 converge to 𝐺𝑥, for some 𝑥 in 𝑋 as 𝑛→∞.
Since 𝑇(𝑋)⊆𝐹(𝑋) there exists a sequence {𝑦_𝑛} in 𝑋 such that 𝑇𝑥_𝑛=𝐹𝑦_𝑛; hence 𝐹𝑦_𝑛→𝐺𝑥 as 𝑛→∞.
Now, we claim that lim_𝑛→∞𝑆𝑦_𝑛=𝐺𝑥.
Let lim_𝑛→∞𝑆𝑦_𝑛=𝑧.
Taking 𝑥=𝑦_𝑛 and 𝑦=𝑥_𝑛 in (3), we have 𝑑𝑆𝑦_𝑛,𝑇𝑥_𝑛≤𝜙max𝑑𝐹𝑦_𝑛,𝐺𝑥_𝑛,𝑑𝐹𝑦_𝑛,𝑆𝑦_𝑛,𝑑𝐺𝑥_𝑛,𝑇𝑥_𝑛. (40) Letting 𝑛→∞, we have 𝑑()({𝑑()()((𝑑((),𝑧,𝐺𝑥≤𝜙max𝐺𝑥,𝐺𝑥,𝑑𝐺𝑥,𝑧,𝑑𝐺𝑥,𝐺𝑥)})=𝜙𝐺𝑥,𝑧))<𝑑𝑧,𝐺𝑥impliesthat𝑧=𝐺𝑥. (41) Subsequently, we have 𝑇𝑥_𝑛, 𝐺𝑥_𝑛, 𝐹𝑦_𝑛, and 𝑆𝑦_𝑛 that converge to 𝑧.
Now, we claim that 𝑇𝑥=𝑧.
Taking 𝑥=𝑦_𝑛 and 𝑦=𝑥 in (3), we have 𝑑𝑆𝑦_𝑛𝑑,𝑇𝑥≤𝜙max𝐹𝑦_𝑛,𝐺𝑥,𝑑𝐹𝑦_𝑛,𝑆𝑦_𝑛,𝑑(𝐺𝑥,𝑇𝑥). (42) Letting 𝑛→∞, we have 𝑑()({𝑑()()((𝑑((),𝑧,𝑇𝑥≤𝜙max𝐺𝑥,𝐺𝑥,𝑑𝑧,𝑧,𝑑𝑧,𝑇𝑥)})=𝜙𝑧,𝑇𝑥))<𝑑𝑧,𝑇𝑥impliesthat𝑧=𝑇𝑥=𝐺𝑥. (43) Since the pair (𝑇,𝐺) is weakly compatible, it follows that 𝑇𝑧=𝐺𝑧.
Also, since 𝑇(𝑋)⊆𝐹(𝑋), there exists some 𝑦 in 𝑋 such that 𝑇𝑥=𝐹𝑦(=𝑧).
We next show that 𝐹𝑦=𝑆𝑦(=𝑧).
Taking 𝑦=𝑥_𝑛, 𝑥=𝑦 in (3), we have 𝑑𝑆𝑦,𝑇𝑥_𝑛≤𝜙max𝑑𝐹𝑦,𝐺𝑥_𝑛,𝑑𝐹𝑦,𝑆𝑦,𝑑𝐺𝑥_𝑛,𝑇𝑥_𝑛. (44) Letting 𝑛→∞, we have 𝑑𝑑()(𝑑,𝑆𝑦,𝑧≤𝜙max𝑧,𝑧,𝑑𝑧,𝑆𝑦,𝑑𝑧,𝑧)=𝜙𝑧,𝑆𝑦<𝑑𝑆𝑦,𝑧impliesthat𝑆𝑦=𝑧=𝐹𝑦. (45) But the pair {𝑆,𝐹} is weakly compatible; it follows that 𝑆𝑧=𝐹𝑧.
Next, we claim that 𝑆𝑧=𝑇𝑧.
Taking 𝑥=𝑧,  𝑦=𝑧 in (3), we have 𝑑()({𝑑()()((𝑑(()𝑆𝑧,𝑇𝑧≤𝜙max𝐹𝑧,𝐺𝑧,𝑑𝐹𝑧,𝑆𝑧,𝑑𝐺𝑧,𝑇𝑧)})=𝜙𝑆𝑧,𝑇𝑧))<𝑑𝑆𝑧,𝑇𝑧,impliesthat𝑆𝑧=𝑇𝑧,thatis,𝑆𝑧=𝑇𝑧=𝐹𝑧=𝐺𝑧. (46)
Now, we show that 𝑧=𝑆𝑧.
Taking 𝑥=𝑧, 𝑦=𝑥 in (3), we have 𝑑()({𝑑()()((𝑑((),𝑆𝑧,𝑇𝑥≤𝜙max𝐹𝑧,𝐺𝑥,𝑑𝐹𝑧,𝑆𝑧,𝑑𝐺𝑥,𝑇𝑥)})=𝜙𝑆𝑧,𝑧))<𝑑𝑆𝑧,𝑧impliesthat𝑧=𝑆𝑧=𝑇𝑧=𝐹𝑧=𝐺𝑧. (47) Hence 𝑧 is the common fixed point of 𝑆, 𝑇, 𝐹, and 𝐺.
If 𝑆𝑧=𝑇𝑧=𝐹𝑧=𝐺𝑧=𝑧 and 𝑆𝑡=𝑇𝑡=𝐹𝑡=𝐺𝑡=𝑡, then (3) gives 𝑑()({𝑑()()((𝑑(()𝑆𝑧,𝑇𝑡≤𝜙max𝐹𝑧,𝐺𝑡,𝑑𝐹𝑧,𝑆𝑧,𝑑𝐺𝑡,𝑇𝑡)})=𝜙𝑆𝑧,𝑇𝑡))<𝑑𝑆𝑧,𝑇𝑡,impliesthat𝑆𝑧=𝑇𝑡,thatis,𝑧=𝑡. (48)
Therefore, uniqueness follows.

Theorem 15. Let (𝑋,𝑑) be Hausdorff and complete g.m.s. and let 𝜙∶ℝ⁺³→ℝ⁺ be a 𝒮 function. Let 𝑆, 𝑇, 𝐹, and 𝐺 be self-maps on 𝑋 satisfying (5), (6), and (38). Then 𝑆, 𝑇, 𝐹, and 𝐺 have a unique common fixed point.

Proof. Without loss of generality, we assume that 𝑇(𝑋)⊆𝐹(𝑋) and the pair {𝑇,𝐺} satisfies (CLR_𝐺) property; then there exists some sequence {𝑥_𝑛} in 𝑋 such that 𝑇𝑥_𝑛 and 𝐺𝑥_𝑛 converge to 𝐺𝑥, for some 𝑥 in 𝑋 as 𝑛→∞.
Since 𝑇(𝑋)⊆𝐹(𝑋) there exists a sequence {𝑦_𝑛} in 𝑋 such that 𝑇𝑥_𝑛=𝐹𝑦_𝑛; hence 𝐹𝑦_𝑛→𝐺𝑥 as 𝑛→∞.
Now, we claim that lim_𝑛→∞𝑆𝑦_𝑛=𝐺𝑥.
Let lim_𝑛→∞𝑆𝑦_𝑛=𝑧.
Taking 𝑥=𝑦_𝑛 and 𝑦=𝑥_𝑛 in (5), we have 𝑑𝑆𝑦_𝑛,𝑇𝑥_𝑛≤𝜙𝑑𝐹𝑦_𝑛,𝐺𝑥_𝑛,𝑑𝐹𝑦_𝑛,𝑆𝑦_𝑛,𝑑𝐺𝑥_𝑛,𝑇𝑥_𝑛. (49) Letting  𝑛→∞, we have 𝑑()(𝑑()()((())()𝑧,𝐺𝑥≤𝜙𝐺𝑥,𝐺𝑥,𝑑𝐺𝑥,𝑧,𝑑𝐺𝑥,𝐺𝑥))=𝜙0,𝑑𝐺𝑥,𝑧,0<𝑑𝑧,𝐺𝑥,impliesthat𝑧=𝐺𝑥. (50) Subsequently, we have 𝑇𝑥_𝑛, 𝐺𝑥_𝑛, 𝐹𝑦_𝑛, and 𝑆𝑦_𝑛 that converge to 𝑧.
Now, we claim that 𝑇𝑥=𝑧.
Taking 𝑥=𝑦_𝑛 and 𝑦=𝑥 in (5), we have 𝑑𝑆𝑦_𝑛,𝑇𝑥≤𝜙𝑑𝐹𝑦_𝑛,𝐺𝑥,𝑑𝐹𝑦_𝑛,𝑆𝑦_𝑛,𝑑(𝐺𝑥,𝑇𝑥). (51) Letting 𝑛→∞, we have 𝑑()(𝑑()()(((()𝑧,𝑇𝑥≤𝜙𝐺𝑥,𝐺𝑥,𝑑𝑧,𝑧,𝑑𝑧,𝑇𝑥))=𝜙0,0,𝑑𝑧,𝑇𝑥))<𝑑𝑧,𝑇𝑥,impliesthat𝑧=𝑇𝑥=𝐺𝑥. (52) Since the pair (𝑇,𝐺) is weakly compatible, it follows that 𝑇𝑧=𝐺𝑧.
Also, since 𝑇(𝑋)⊆𝐹(𝑋), there exists some 𝑦 in 𝑋 such that 𝑇𝑥=𝐹𝑦(=𝑧).
We next show that 𝐹𝑦=𝑆𝑦(=𝑧).
Taking 𝑦=𝑥_𝑛, 𝑥=𝑦 in (5), we have 𝑑𝑆𝑦,𝑇𝑥_𝑛≤𝜙𝑑𝐹𝑦,𝐺𝑥_𝑛,𝑑𝐹𝑦,𝑆𝑦,𝑑𝐺𝑥_𝑛,𝑇𝑥_𝑛. (53) Letting 𝑛→∞, we have 𝑑𝑑()(),𝑆𝑦,𝑧≤𝜙𝑧,𝑧,𝑑𝑧,𝑆𝑦,𝑑𝑧,𝑧=𝜙0,𝑑𝑧,𝑆𝑦,0<𝑑𝑆𝑦,𝑧impliesthat𝑆𝑦=𝑧=𝐹𝑦. (54) But the pair {𝑆,𝐹} is weakly compatible; it follows that 𝑆𝑧=𝐹𝑧.
Next, we claim that 𝑆𝑧=𝑇𝑧.
Taking 𝑥=𝑧, 𝑦=𝑧 in (5), we have 𝑑()(𝑑()()((𝑑())(),𝑆𝑧,𝑇𝑧≤𝜙𝐹𝑧,𝐺𝑧,𝑑𝐹𝑧,𝑆𝑧,𝑑𝐺𝑧,𝑇𝑧))=𝜙𝑆𝑧,𝑇𝑧,0,0<𝑑𝑆𝑧,𝑇𝑧impliesthat𝑆𝑧=𝑇𝑧. (55) Hence, 𝑆𝑧=𝑇𝑧=𝐹𝑧=𝐺𝑧.
Now, we show that 𝑧=𝑆𝑧.
Taking 𝑥=𝑧, 𝑦=𝑥 in (5), we have 𝑑()(𝑑()()((𝑑())(),𝑆𝑧,𝑇𝑥≤𝜙𝐹𝑧,𝐺𝑥,𝑑𝐹𝑧,𝑆𝑧,𝑑𝐺𝑥,𝑇𝑥))=𝜙𝑆𝑧,𝑧,0,0<𝑑𝑆𝑧,𝑧impliesthat𝑧=𝑆𝑧=𝑇𝑧=𝐹𝑧=𝐺𝑧. (56) Hence 𝑧 is the common fixed point of 𝑆,  𝑇, 𝐹 and 𝐺.
If 𝑆𝑧=𝑇𝑧=𝐹𝑧=𝐺𝑧=𝑧 and 𝑆𝑡=𝑇𝑡=𝐹𝑡=𝐺𝑡=𝑡, then (5) gives 𝑑()(𝑑()()((𝑑())(),𝑆𝑧,𝑇𝑡≤𝜙𝐹𝑧,𝐺𝑡,𝑑𝐹𝑧,𝑆𝑧,𝑑𝐺𝑡,𝑇𝑡))=𝜙𝑆𝑧,𝑇𝑡,0,0<𝑑𝑆𝑧,𝑇𝑡impliesthat𝑆𝑧=𝑇𝑡or𝑧=𝑡. (57) Therefore, 𝑧=𝑡 and the common fixed point is unique.

Acknowledgment

One of the authors (S. Kumar) is very grateful to UGC for providing MRP under Ref. F. no. 39-41/2010(SR).