Fixed Point Theorems for a Class of Weakly C-Contractive Mappings in a Setting of 2-Banach Space

Abstract

In this paper, the authors have proved some existence theorems of fixed points for a class of weakly C-contractive mappings in a setting of 2-Banach space. The authors have supported the results with the help of suitable examples.

1. Introduction

The theory of 2-Banach spaces was investigated by Gähler in [1] and Iseki in [2] who proved some fixed point theorems in such spaces and these new spaces have subsequently been studied by many authors. For example, we refer to [3, 4] where authors have dealt with mappings that are of contractive nature. Our findings as presented here have added further findings in fixed point theory in a 2-Banach space by working with a class of mappings so general that take care of those as found in [5, 6].

We now state some definitions before presenting our main results.

2. Preliminaries

Definition 1. Let 𝑋 be a real linear space and ‖⋅,⋅‖ a non-negative real valued function defined on 𝑋 satisfying the following conditions:(i)‖𝑥,𝑦‖=0 if and only if 𝑥 and 𝑦 are linearly dependent,(ii)‖𝑥,𝑦‖=‖𝑦,𝑥‖ for all 𝑥,𝑦∈𝑋,(iii)‖𝑥,𝛼𝑦‖=|𝛼|‖𝑥,𝑦‖, 𝛼 being real for all 𝑥,𝑦∈𝑋,(iv)‖𝑥,𝑦+𝑧‖≤‖𝑥,𝑦‖+‖𝑥,𝑧‖ for all 𝑥,𝑦,𝑧∈𝑋.Then ‖⋅,⋅‖ is called a 2-norm and the pair (𝑋,‖⋅,⋅‖) is called a linear 2-normed space. So a 2-norm ‖𝑥,𝑦‖ always satisfies ‖𝑥,𝑦+𝛼𝑥‖=‖𝑥,𝑦‖ for all 𝑥,𝑦∈𝑋 and all scalars 𝛼.

Definition 2. A sequence {𝑥_𝑛} in a linear 2-normed space (𝑋,‖⋅,⋅‖) is said to be a Cauchy sequence if lim_{𝑛,𝑚→∞}‖𝑥_𝑛−𝑥_𝑚,𝑎‖=0 for all 𝑎 in 𝑋.

Definition 3. A sequence {𝑥_𝑛} in a linear 2-normed space (𝑋,‖⋅,⋅‖) is said to be convergent in 𝑋 if there exists a point 𝑥 in 𝑋 such that lim_𝑛→∞‖𝑥_𝑛−𝑥,𝑎‖=0 for all 𝑎 in 𝑋. If {𝑥_𝑛} converges to 𝑥, we write 𝑥_𝑛→𝑥 as 𝑛→∞.

Definition 4. A linear 2-normed space 𝑋 is said to be complete if every Cauchy sequence is convergent to an element of 𝑋. A complete linear 2-normed space 𝑋 is called a 2-Banach space.
The main purpose of our paper is to find out the existence of fixed point for a class of weakly C-contractive results due to Choudhury [7] in a setting of 2-Banach space. We have also established a common fixed point theorem for such type of mappings with a supporting example. Some allied theorems have also been presented here.

3. Main Results

Definition 5. A mapping 𝑇∶(𝑋,‖⋅,⋅‖)→(𝑋,‖⋅,⋅‖) is said to be weakly 𝐶-contractive or a weak 𝐶-contraction if for all 𝑥,𝑦,𝑎 in 𝑋, 𝑇(𝑥)𝑦≤1−𝑇,𝑎2𝑦+(𝑥)𝑦,(𝑥),𝑥−𝑇,𝑎𝑦−𝑇,𝑎−𝜙𝑥−𝑇,𝑎𝑦−𝑇,𝑎 (1) where 𝜙∶[0,∞)²→[0,∞) is a continuous mapping such that 𝜙(𝑥,𝑦)=0 if and only if 𝑥=𝑦=0.

Theorem 6. Let (𝑋,‖⋅,⋅‖) be a 2-Banach space and let 𝑇∶𝑋→𝑋 be a weak C-contraction. Then 𝑇 has a unique fixed point in 𝑋.

Proof. Let 𝑥₀∈𝑋. Let 𝑥_𝑛+1=𝑇(𝑥_𝑛), 𝑛=1,2,….
If 𝑥_𝑛=𝑥_𝑛+1=𝑇(𝑥_𝑛), then 𝑇 has a fixed point 𝑥_𝑛.
Assume that 𝑥_𝑛≠𝑥_𝑛+1. Now for any 𝑎∈𝑋, 𝑥_𝑛−𝑥_𝑛+1,𝑎=𝑇𝑥_𝑛−1−𝑇𝑥_𝑛≤1,𝑎2𝑥_𝑛−1−𝑇𝑥_𝑛,𝑎+𝑥_𝑛−𝑇𝑥_𝑛−1𝑥,𝑎−𝜙_𝑛−1−𝑇𝑥_𝑛,𝑎,𝑥_𝑛−𝑇𝑥_𝑛−1≤1,𝑎2𝑥_𝑛−1−𝑥_𝑛,𝑎+𝑥_𝑛−𝑥_𝑛+1𝑥,𝑎−𝜙_𝑛−1−𝑥_𝑛+1,𝑎,0. (2) Therefore, 𝑥_𝑛−𝑥_𝑛+1,𝑎≤𝑥_𝑛−1−𝑥_𝑛,𝑎. (3) So {‖𝑥_𝑛−𝑥_𝑛+1,𝑎‖} is a monotone decreasing sequence of real numbers and hence {‖𝑥_𝑛−𝑥_𝑛+1,𝑎‖} is convergent and let ‖𝑥_𝑛−𝑥_𝑛+1,𝑎‖→𝑟 (say) as 𝑛→∞
We now show that 𝑟=0. If not, then 𝑥_𝑛−𝑥_𝑛+1,𝑎=𝑇𝑥_𝑛−1−𝑇𝑥_𝑛≤1,𝑎2𝑥_𝑛−1−𝑥_𝑛+1≤1,𝑎2𝑥_𝑛−1−𝑥_𝑛,𝑎+𝑥_𝑛−𝑥_𝑛+1,𝑎. (4) Taking 𝑛→∞, we get 1𝑟≤2lim_𝑛→∞𝑥_𝑛−1−𝑥_𝑛+1+1,𝑎2lim_𝑛→∞𝑥_𝑛−𝑥_𝑛+1≤1,𝑎21𝑟+2𝑟, (5) implying that lim_𝑛→∞𝑥_𝑛−1−𝑥_𝑛+1,𝑎=2𝑟. (6)
Also from (2) and using continuity of 𝜙, we get 𝑟≤𝑟−𝜙[2𝑟,0] showing that 𝜙[2𝑟,0]≤0, a contradiction, unless 𝑟=0. So, ‖𝑥_𝑛−𝑥_𝑛+1,𝑎‖→0 as 𝑛→∞, that is lim_𝑛→∞𝑥_𝑛−𝑥_𝑛+1,𝑎=0. (7) We shall now show that {𝑥_𝑛} is Cauchy sequence in 𝑋.
If not, then there exists a 𝜀>0 and for each positive integer 𝑘, there exist integers 𝑛(𝑘) and 𝑚(𝑘) with 𝑘<𝑚(𝑘)<𝑛(𝑘) such that ‖𝑥_𝑚(𝑘)−𝑥_𝑛(𝑘),𝑎‖≥𝜀 and ‖𝑥_𝑚(𝑘)−𝑥_{𝑛(𝑘)−1},𝑎‖<𝜀. Then 𝜀≤𝑥_𝑚(𝑘)−𝑥_𝑛(𝑘),𝑎=𝑇𝑥_{𝑚(𝑘)−1}−𝑇𝑥_{𝑛(𝑘)−1}≤1,𝑎2𝑥_{𝑚(𝑘)−1}−𝑥_𝑛(𝑘),𝑎+𝑥_{𝑛(𝑘)−1}−𝑥_𝑚(𝑘)𝑥,𝑎−𝜙_{𝑚(𝑘)−1}−𝑥_𝑛(𝑘),𝑎,𝑥_{𝑛(𝑘)−1}−𝑥_𝑚(𝑘),𝑎. (8) Also, 𝜀≤𝑥_𝑚(𝑘)−𝑥_𝑛(𝑘),𝑎≤𝑥_𝑚(𝑘)−𝑥_{𝑛(𝑘)−1}+𝑥,𝑎_{𝑛(𝑘)−1}−𝑥_𝑛(𝑘)𝑥,𝑎<𝜀+_{𝑛(𝑘)−1}−𝑥_𝑛(𝑘),𝑎. (9) So by using (7), we obtain lim_𝑛→∞𝑥_𝑚(𝑘)−𝑥_𝑛(𝑘),𝑎=𝜀. (10) Now 𝑥_𝑚(𝑘)−𝑥_{𝑛(𝑘)−1},𝑎≤𝑥_𝑚(𝑘)−𝑥_{𝑚(𝑘)−1}+𝑥,𝑎_{𝑚(𝑘)−1}−𝑥_𝑛(𝑘)+𝑥,𝑎_𝑛(𝑘)−𝑥_{𝑛(𝑘)−1},𝑎. (11) Also 𝑥_{𝑚(𝑘)−1}−𝑥_𝑛(𝑘),𝑎≤𝑥_{𝑚(𝑘)−1}−𝑥_𝑚(𝑘)+𝑥,𝑎_𝑚(𝑘)−𝑥_𝑛(𝑘),𝑎. (12) Therefore, lim_𝑘→∞𝑥_{𝑚(𝑘)−1}−𝑥_𝑛(𝑘),𝑎=𝜀. (13)
Similarly, it can be shown that lim_𝑘→∞𝑥_{𝑛(𝑘)−1}−𝑥_𝑚(𝑘),𝑎=𝜀. (14) So from (8), we get 1𝜀≤2(𝜀+𝜀)−𝜙[𝜀,𝜀] (15) implying that 𝜙[𝜀,𝜀]≤0, a contradiction.
Hence {𝑥_𝑛} is a Cauchy sequence and therefore {𝑥_𝑛} is a convergent in 𝑋 and let 𝑥_𝑛→𝑧∈𝑋 (say) as 𝑛→∞. Now for any 𝑎∈𝑋, ‖𝑧−𝑇(𝑧),𝑎‖≤𝑧−𝑥_𝑛+1,𝑎+𝑥_𝑛+1(𝑧)≤−𝑇,𝑎𝑧−𝑥_𝑛+1,𝑎+𝑇𝑥_𝑛(𝑧)≤−𝑇,𝑎𝑧−𝑥_𝑛+1+1,𝑎2𝑧−𝑇𝑥_𝑛,𝑎+𝑥_𝑛(𝑧)𝑥−𝑇,𝑎−𝜙𝑧−𝑇_𝑛,𝑎,𝑥_𝑛−𝑇(𝑧),𝑎. (16) So by the continuity of 𝜙 and taking 𝑛→∞, we get ‖(𝑧)‖≤1𝑧−𝑇,𝑎2(𝑧)‖[‖(𝑧)≤1‖𝑧−𝑇,𝑎−𝜙0,𝑧−𝑇,𝑎‖]2‖𝑧−𝑇(𝑧),𝑎‖ (17) implying that 𝑧=𝑇(𝑧).
We shall now show that the fixed point 𝑧 is unique. If possible let 𝑧₁∈𝑋 be another fixed point of 𝑇, then for any 𝑎∈𝑋𝑧−𝑧₁,𝑎=𝑇(𝑧)−𝑇𝑧₁≤1,𝑎2𝑧−𝑇𝑧₁,𝑎+𝑧₁(𝑧)𝑧−𝑇,𝑎−𝜙𝑧−𝑇₁,𝑎,𝑧₁(𝑧)≤−𝑇,𝑎𝑧−𝑧₁,𝑎−𝜙𝑧−𝑧₁,𝑎,𝑧−𝑧₁,𝑎, (18) which by property of 𝜙 is a contradiction unless ‖𝑧−𝑧₁,𝑎‖=0 for all 𝑎∈𝑋 that is 𝑧=𝑧₁.
This completes the proof.

Example 7. Let 𝑋 be ℝ³ and consider the following 2-norm on 𝑋 as 𝑥,𝑦=det𝑥𝑖𝑗𝑘₁𝑥₂𝑥₃𝑦₁𝑦₂𝑦₃, (19) where 𝑥=(𝑥₁,𝑥₂,𝑥₃), 𝑦=(𝑦₁,𝑦₂,𝑦₃)∈𝑋. Then (𝑋,‖⋅,⋅‖) is a 2-Banach space (see [8, 9]).
𝑇∶(𝑋,‖⋅,⋅‖)→(𝑋,‖⋅,⋅‖) is defined by 𝑇(𝑥,𝑦,𝑧)=(0,0,1) for all (𝑥,𝑦,𝑧)∈𝑋. Also let 𝑇(0,0,1)=(0,0,1).
Now setting 𝜙𝑡,𝑡^′=12𝑡+𝑡^′;𝑡,𝑡^′∈[0,∞)×[0,∞). (20)𝜙∶[0,∞)²→[0,∞) is a continuous mapping such that 𝜙((𝑡,𝑡^′))=0 if and only if 𝑡=𝑡^′=0. Now 𝑇(𝑥)=𝑇(𝑥₁,𝑥₂,𝑥₃)=(0,0,1) and 𝑇(𝑦)=𝑇(𝑦₁,𝑦₂,𝑦₃)=(0,0,1). Take 𝑎=(𝑎₁,𝑎₂,𝑎₃)∈𝑋.
Now 𝑇(𝑥)𝑦=𝑎−𝑇,𝑎det𝑖𝑗𝑘000₁𝑎₂𝑎₃=0. (21) Also, 𝑇(𝑥)𝑦≤1−𝑇,𝑎2𝑦+(𝑥)𝑦,(𝑥).𝑥−𝑇,𝑎𝑦−𝑇,𝑎−𝜙𝑥−𝑇,𝑎𝑦−𝑇,𝑎 (22) All the conditions of Theorem 6 are satisfied.
Hence 𝑇 has a unique fixed point (0,0,1) in ℝ³.

Theorem 8. Let 𝐹 and 𝐺 be two self-mappings of a 2-Banach space 𝑋. Suppose that for any 𝑎∈𝑋, 𝐹(𝑥)𝑦≤1−𝐺,𝑎2𝑦+(𝑥)𝑦,(𝑥),𝑥−𝐺,𝑎𝑦−𝐹,𝑎−𝜙𝑥−𝐺,𝑎𝑦−𝐹,𝑎 (23) where 𝜙∶[0,∞)²→[0,∞) is a continuous mapping such that 𝜙(𝑥,𝑦)=0 if and only if 𝑥=𝑦=0. Then 𝐹 and 𝐺 have a unique common fixed point in 𝑋.

Proof. For 𝑥₀∈𝑋, define {𝑥_𝑛}_𝑛≥1 by 𝑥_2𝑛+1=𝐹(𝑥_2𝑛),𝑥_2𝑛+2=𝐺(𝑥_2𝑛+1). So for any 𝑎∈𝑋, 𝑥_2𝑛+1−𝑥_2𝑛+2,𝑎=𝐹𝑥_2𝑛−𝐺𝑥_2𝑛+1≤1,𝑎2𝑥_2𝑛−𝐺𝑥_2𝑛+1+𝑥,𝑎_2𝑛+1−𝐹𝑥_2𝑛𝑥,𝑎−𝜙_2𝑛−𝐺𝑥_2𝑛+1,𝑥,𝑎_2𝑛+1−𝐹𝑥_2𝑛≤1,𝑎2𝑥_2𝑛−𝑥_2𝑛+2≤1,𝑎2𝑥_2𝑛−𝑥_2𝑛+1+𝑥,𝑎_2𝑛+1−𝑥_2𝑛+2,𝑎. (24)Hence 𝑥_2𝑛+1−𝑥_2𝑛+2,𝑎≤𝑥_2𝑛−𝑥_2𝑛+1,𝑎. (25)Similarly 𝑥_2𝑛−𝑥_2𝑛+1,𝑎≤𝑥_2𝑛−1−𝑥_2𝑛,𝑎. (26) Thus {‖𝑥_𝑛−𝑥_𝑛+1,𝑎‖} is a monotone decreasing sequence of real numbers and hence is convergent and
Let lim_𝑛→∞𝑥_𝑛−𝑥_𝑛+1,𝑎=𝑟≥0say (27)
We now claim that 𝑟=0.
If not, 𝑥_2𝑛+1−𝑥_2𝑛+2≤1,𝑎2𝑥_2𝑛−𝑥_2𝑛+2≤1,𝑎2𝑥_2𝑛−𝑥_2𝑛+1+𝑥,𝑎_2𝑛+1−𝑥_2𝑛+2,𝑎. (28) So, lim_𝑛→∞𝑥_2𝑛−𝑥_2𝑛+2,𝑎=2𝑟 (29) from (24) taking 𝑛→∞𝑟≤𝑟−𝜙[2𝑟,0] (30) implying that 𝜙[2𝑟,0]≤0, a contradiction.
Therefore, 𝑟=0.
Now for any integer 𝑝>0𝑥_𝑛−𝑥_𝑛+𝑝,𝑎≤𝑥_𝑛−𝑥_𝑛+1,𝑎+𝑥_𝑛+1−𝑥_𝑛+2𝑥,𝑎+⋯+_{𝑛+𝑝−1}−𝑥_𝑛+𝑝,𝑎→0as𝑛→∞. (31)
So {𝑥_𝑛} is Cauchy in 𝑋 and hence by completeness of 𝑋, let lim_𝑛→∞𝑥_𝑛=𝑧∈𝑋. (32) Now for all 𝑎∈𝑋, ‖𝑧−𝐹(𝑧),𝑎‖≤𝑧−𝑥_𝑛+2,𝑎+𝑥_2𝑛+2(𝑧)≤−𝐹,𝑎𝑧−𝑥_2𝑛+2,𝑎+𝐹(𝑧)−𝐺𝑥_2𝑛+1,𝑎. (33) By routine check up, we can see that for sufficiently large values of 𝑛, Hence, 𝑧=𝐹𝑧.
Similarly we can show that 𝑧=𝐺(𝑧).
We shall now show that 𝑧 is unique.
If not let 𝑧₁∈𝑋 be another fixed point of 𝐹 and 𝐺. Then for any 𝑎∈𝑋𝑧−𝑧₁,𝑎=𝐹(𝑧)−𝐺𝑧₁≤1,𝑎2𝑧−𝑧₁,𝑎+𝑧₁−𝑧,𝑎−𝜙𝑧−𝑧₁,𝑎,𝑧₁−𝑧,𝑎. (34)
So ‖𝑧−𝑧₁,𝑎‖≤‖𝑧−𝑧₁,𝑎‖−𝜙[‖𝑧−𝑧₁,𝑎‖,‖𝑧₁−𝑧,𝑎‖], showing that 𝜙[‖𝑧−𝑧₁,𝑎‖,‖𝑧₁−𝑧,𝑎‖]≤0. So 𝑧−𝑧₁=0 implies 𝑧=𝑧₁.

Example 9. Let 𝑋 be ℝ³ and consider the following 2-norm on 𝑋 as 𝑥,𝑦=det𝑥𝑖𝑗𝑘₁𝑥₂𝑥₃𝑦₁𝑦₂𝑦₃, (35) where 𝑥=(𝑥₁,𝑥₂,𝑥₃), 𝑦=(𝑦₁,𝑦₂,𝑦₃)∈𝑋. Then (𝑋,‖⋅,⋅‖) is a 2-Banach space (see [8, 9]).
Define 𝐹,𝐺∶𝑋→𝑋 by 𝐹(𝑥,𝑦,𝑧)=((1/4)𝑥,(1/4)𝑦,(1/4)𝑧) for all (𝑥,𝑦,𝑧)∈𝑋 and 𝐺(𝑥,𝑦,𝑧)=((1/3)𝑥,(1/3)𝑦,(1/3)𝑧) for all (𝑥,𝑦,𝑧)∈𝑋.
Now setting 𝜙𝑡,𝑡^′=12max𝑡,𝑡^′;𝑡,𝑡^′∈[0,∞)×[0,∞). (36)𝜙∶[0,∞)²→[0,∞) is a continuous mapping such that 𝜙((𝑡,𝑡^′))=0 if and only if 𝑡=𝑡^′=0. Take 𝑎=(𝑎₁,𝑎₂,𝑎₃)∈𝑋.
Now 𝐹(𝑥)𝑦=1−𝐺,𝑎det𝑖𝑗𝑘4𝑥₁−13𝑦₁14𝑥₂−13𝑦₂14𝑥₃−13𝑦₃𝑎₁𝑎₂𝑎₃𝑦=𝑥𝑥−𝐺,𝑎det𝑖𝑗𝑘₁−13𝑦₁𝑥₂−13𝑦₂𝑥₃−13𝑦₃𝑎₁𝑎₂𝑎₃,(𝑥)=𝑦𝑦−𝐹,𝑎det𝑖𝑗𝑘₁−13𝑥₁𝑦₂−13𝑥₂𝑦₃−13𝑥₃𝑎₁𝑎₂𝑎₃. (37) By routine calculation, it can be seen that the condition in Theorem 8𝐹(𝑥)𝑦≤1−𝐺,𝑎2𝑦+(𝑥)𝑦,(𝑥)𝑥−𝐺,𝑎𝑦−𝐹,𝑎−𝜙𝑥−𝐺,𝑎𝑦−𝐹,𝑎 (38) is satisfied. Also 𝐹 and 𝐺 have a unique common fixed point (0,0,0) in 𝑋.

Theorem 10. Let (𝑋,‖⋅,⋅‖) be a 2-Banach space with 𝑇∶𝑋→𝑋 and let {𝑇_𝑛∶𝑋→𝑋} be a sequence of mappings such that(i) for all 𝑢∈𝑋 and 𝑛=1,2,…𝑇_𝑛(𝑥)−𝑇_𝑛𝑦≤1,𝑢2𝑥−𝑇_𝑛𝑦+,𝑢𝑦−𝑇_𝑛(𝑥),𝑢−𝜙𝑥−𝑇_𝑛𝑦,𝑢,𝑦−𝑇_𝑛(𝑥),𝑢, (39)where 𝜙∶[0,∞)²→[0,∞) is a continuous mapping such that 𝜙(𝑥,𝑦)=0 if and only if 𝑥=𝑦=0,(ii)lim_𝑛→∞𝑇_𝑛(𝑥)=𝑇(𝑥) for each 𝑥∈𝑋. Then 𝑇 has a unique fixed point 𝑧 in 𝑋 such that 𝑧=lim_𝑛→∞𝑧_𝑛 where 𝑧_𝑛 is the fixed point of 𝑇_𝑛(𝑛=1,2,…).

Proof. 𝑇(𝑥)𝑦≤𝑇(𝑥)−𝑇,𝑢−𝑇_𝑛(𝑥)+1,𝑢2𝑥−𝑇_𝑛𝑦+,𝑢𝑦−𝑇_𝑛(𝑥),𝑢−𝜙𝑥−𝑇_𝑛𝑦,𝑢,𝑦−𝑇_𝑛(𝑥)+𝑇𝑦,𝑢−𝑇_𝑛𝑦,𝑢. (40) Now 𝑥−𝑇_𝑛𝑦≤𝑦+𝑇,𝑢𝑥−𝑇,𝑢_𝑛𝑦𝑦→𝑦−𝑇,𝑢𝑥−𝑇,𝑢as𝑛→∞. (41) Similarly 𝑦−𝑇_𝑛(𝑥),𝑢→𝑦−𝑇(𝑥),𝑢as𝑛→∞. (42) As 𝜙 is continuous, 𝜙𝑥−𝑇_𝑛𝑦,𝑢,𝑦−𝑇_𝑛(𝑥)𝑦,(𝑥).,𝑢→𝜙𝑥−𝑇,𝑢𝑦−𝑇,𝑢 (43) So from (40) taking limit as 𝑛→∞, we obtain 𝑇(𝑥)𝑦≤1−𝑇,𝑢2𝑦+(𝑥)𝑦,(𝑥).𝑥−𝑇,𝑢𝑦−𝑇,𝑢−𝜙𝑥−𝑇,𝑢𝑦−𝑇,𝑢 (44) Hence by Theorem 6, 𝑇 has a unique fixed point 𝑧 in 𝑋.
If possible, let lim_𝑛→∞𝑧_𝑛≠𝑧∈𝑋. (45) Now for 𝑢∈𝑋 let lim_𝑛→∞𝑧_𝑛−𝑧,𝑢=𝛼>0𝑧−𝑧_𝑛,𝑢=𝑇(𝑧)−𝑇_𝑛𝑧_𝑛≤𝑇(𝑧),𝑢−𝑇_𝑛(𝑧),𝑢+𝑇_𝑛(𝑧)−𝑇_𝑛𝑧_𝑛,𝑢. (46)(47) Now 𝑇_𝑛(𝑧)−𝑇_𝑛𝑧_𝑛≤1,𝑢2𝑧−𝑇_𝑛𝑧_𝑛+𝑧,𝑢_𝑛−𝑇_𝑛(𝑧),𝑢−𝜙𝑧−𝑇_𝑛𝑧_𝑛,𝑧,𝑢_𝑛−𝑇_𝑛(𝑧),𝑢. (48) Now 𝑧_𝑛−𝑇_𝑛(𝑧),𝑢≤𝑧_𝑛+‖(𝑧)‖+𝑇(𝑧)−𝑧,𝑢𝑧−𝑇,𝑢−𝑇_𝑛(𝑧)=𝑧,𝑢_𝑛−𝑧,𝑢+𝑇(𝑧)−𝑇_𝑛(𝑧),𝑢. (49) So from (46), (47), (48), and (49), we get lim_𝑛→∞𝑧−𝑧_𝑛,𝑢≤lim_𝑛→∞𝑇_𝑛(𝑧)−𝑇_𝑛𝑧_𝑛,𝑢≤lim_𝑛→∞12𝑧−𝑧_𝑛,𝑢+𝑧_𝑛−𝑧,𝑢−lim_𝑛→∞𝜙𝑧−𝑧_𝑛,𝑢,𝑧_𝑛−𝑧,𝑢 (50)
Hence, 𝛼≤(1/2)(2𝛼)−𝜙[𝛼,𝛼] showing that 𝜙[𝛼,𝛼]≤0 which is possible only when 𝛼=0
Hence lim_𝑛→∞‖𝑧−𝑧_𝑛,𝑢‖=0, implying that lim_𝑛→∞𝑧_𝑛=𝑧.

Theorem 11. Let 𝑋 be a 2-Banach space, {𝑇_𝑛} a sequence of mappings of 𝑋 onto itself with fixed points 𝑧_𝑛. Let 𝑇 be a mapping of 𝑋 into itself satisfying the condition of Theorem 6 with fixed point 𝑧, such that 𝑇_𝑛→𝑇 uniformly on {𝑧_𝑛∶𝑛=1,2,…}. Then lim_𝑛→∞𝑧_𝑛=𝑧∈𝑋.

Proof. Fix 𝜀>0. From the uniform convergence of {𝑇_𝑛} on {𝑧_𝑛∶𝑛=1,2,…}, there exists an integer 𝑁 such that for all 𝑛≥𝑁, and for all 𝑢∈𝑋, ‖𝑇(𝑧_𝑛)−𝑇_𝑛(𝑧_𝑛),𝑢‖<𝜀 for all 𝑧_𝑛.
Now 𝑧−𝑧_𝑛,𝑢=𝑇(𝑧)−𝑇_𝑛𝑧_𝑛≤𝑇(𝑧)𝑧,𝑢−𝑇_𝑛,𝑢+𝑇𝑧_𝑛−𝑇_𝑛𝑧_𝑛,𝑢. (51) If possible let lim_𝑛→∞𝑧_𝑛≠𝑧∈𝑋 that is there exists 𝑢∈𝑋, lim_𝑛→∞𝑧_𝑛−𝑧,𝑢=𝛼>0. (52) Now for all 𝑢∈𝑋, 𝑇(𝑧)−𝑇𝑧_𝑛≤1,𝑢2𝑧−𝑇𝑧_𝑛,𝑢𝑧_𝑛(𝑧)𝑧−𝑇,𝑢−𝜙𝑧−𝑇_𝑛,𝑢,𝑧_𝑛−𝑇(𝑧),𝑢. (53) So for 𝑢∈𝑋 satisfying (52) 𝑧−𝑇𝑧_𝑛,𝑢≤𝑧−𝑧_𝑛,𝑢+𝑇_𝑛𝑧_𝑛−𝑇𝑧_𝑛,𝑢. (54) Therefore lim_𝑛→∞𝑧−𝑇𝑧_𝑛,𝑢≤𝜀+lim_𝑛→∞𝑧−𝑧_𝑛,𝑢=𝜀+𝛼. (55) As 𝜀>0 is arbitrary, lim_𝑛→∞𝑧−𝑇𝑧_𝑛,𝑢≤𝛼. (56) Again 𝑧−𝑧_𝑛,𝑢≤𝑧−𝑇𝑧_𝑛,𝑢+𝑇𝑧_𝑛−𝑇_𝑛𝑧_𝑛,𝑢. (57) Hence lim_𝑛→∞𝑧−𝑧_𝑛,𝑢≤lim_𝑛→∞𝑧−𝑇𝑧_𝑛,𝑢+𝜀. (58) As 𝜀>0 is arbitrary, 𝛼≤lim_𝑛→∞𝑧−𝑇𝑧_𝑛,𝑢. (59) Therefore lim_𝑛→∞𝑧−𝑇𝑧_𝑛,𝑢=𝛼. (60) Hence lim_𝑛→∞𝜙𝑧−𝑇𝑧_𝑛,𝑢,𝑧_𝑛−𝑇(𝑧),𝑢→𝜙[𝛼,𝛼]. (61) Taking 𝑛→∞ in (53), we get 𝛼≤(1/2)(2𝛼)−𝜙[𝛼,𝛼], showing that 𝜙[𝛼,𝛼]≤0 which possible only when 𝛼=0, a contradiction to (52).
Hence lim_𝑛→∞‖𝑧_𝑛−𝑧,𝑢‖=0, implying that lim_𝑛→∞𝑧_𝑛=𝑧.

Theorem 12. Let (𝑋,‖⋅,⋅‖) be a 2-Banach space and for any 𝑥∈𝑋, {𝑇_𝑛(𝑥)}_𝑛=1,2,… a sequence of mappings of 𝑋 onto itself. Suppose there exists a sequence of nonnegative integers {𝑚_𝑛} such that for all 𝑥,𝑦,𝑢∈𝑋, and every pair 𝑖,𝑗 with 𝑖≠𝑗 satisfying 𝑇^𝑚𝑖_𝑖(𝑥)−𝑇^𝑚𝑗_𝑗𝑦≤1,𝑢2𝑥−𝑇^𝑚𝑗_𝑗𝑦+,𝑢𝑦−𝑇^𝑚𝑖_𝑖(𝑥),𝑢−𝜙𝑥−𝑇^𝑚𝑗_𝑗𝑦,𝑢,𝑦−𝑇^𝑚𝑖_𝑖(𝑥),𝑢, (62) where 𝜙∶[0,∞)²→[0,∞) is a continuous mappings such that 𝜙(𝑥,𝑦)=0 if and only if 𝑥=𝑦=0. Then {𝑇_𝑛} have a unique common fixed point in 𝑋.

Proof. Define 𝑆_𝑖=𝑇^𝑚𝑖_𝑖,𝑖=1,2,….
Then 𝑆_𝑖(𝑥)−𝑆_𝑗𝑦≤1,𝑢2𝑥−𝑆_𝑗𝑦+,𝑢𝑦−𝑆_𝑖(𝑥),𝑢−𝜙𝑥−𝑆_𝑗𝑦,𝑢,𝑦−𝑆_𝑖(𝑥),𝑢. (63)
Pick 𝑥₀∈𝑋 and define 𝑥_𝑛=𝑆_𝑛(𝑥_𝑛−1),𝑛=1,2,….
Now 𝑥_𝑛−𝑥_𝑛+1,𝑢=𝑆_𝑛𝑥_𝑛−1−𝑆_𝑛+1𝑥_𝑛≤1,𝑢2𝑥_𝑛−1−𝑆_𝑛+1𝑥_𝑛+𝑥,𝑢_𝑛−𝑆_𝑛𝑥_𝑛−1𝑥,𝑢−𝜙_𝑛−1−𝑆_𝑛+1𝑥_𝑛,𝑥,𝑢_𝑛−𝑆_𝑛𝑥_𝑛−1,𝑢. (64) By routine calculation, we can see that 𝑥_𝑛−𝑥_𝑛+1,𝑢≤𝑥_𝑛−1−𝑥_𝑛,𝑢. (65) Implying that {‖𝑥_𝑛−𝑥_𝑛+1,𝑢‖} is a monotone decreasing sequence of real numbers and hence is convergent and let lim_𝑛→∞𝑥_𝑛−𝑥_𝑛+1,𝑢=𝑟>0. (66) We shall now show that 𝑟=0.
If not, 𝑥_𝑛−𝑥_𝑛+1≤1,𝑢2𝑥_𝑛−1−𝑥_𝑛+1≤1,𝑢2𝑥_𝑛−1−𝑥_𝑛,𝑢+𝑥_𝑛−𝑥_𝑛−1,𝑢. (67) Taking 𝑛→∞, 1𝑟≤2lim_𝑛→∞𝑥_𝑛−1−𝑥_𝑛+1≤1,𝑢2lim_𝑛→∞𝑥_𝑛−1−𝑥_𝑛,𝑢+lim_𝑛→∞𝑥_𝑛−𝑥_𝑛+1,𝑢=𝑟. (68) Hence lim_𝑛→∞𝑥_𝑛−1−𝑥_𝑛+1,𝑢=2𝑟. (69) Taking 𝑛→∞ in (64), we get 𝑟≤𝑟−𝜙[2𝑟,0], showing that 𝜙[2𝑟,0]≤0 which is possible only when 𝑟=0.
Now for any integer 𝑝>0𝑥_𝑛−𝑥_𝑛+𝑝,𝑢≤𝑥_𝑛−𝑥_𝑛+1,𝑢+𝑥_𝑛+1−𝑥_𝑛+2𝑥,𝑢+⋯+_{𝑛+𝑝−1}−𝑥_𝑛+𝑝,𝑢→0as𝑛→∞. (70) Therefore {𝑥_𝑛} is Cauchy in 𝑋 and hence by completeness of 𝑋, let lim_𝑛→∞𝑥_𝑛=𝑧∈𝑋. (71)
Now for any integer 𝑚, 𝑧−𝑆_𝑚(𝑧),𝑢≤𝑧−𝑥_𝑛,𝑢+𝑥_𝑛−𝑆_𝑚(𝑧)=,𝑢𝑧−𝑥_𝑛,𝑢+𝑆_𝑛𝑥_𝑛−1−𝑆_𝑚(𝑧)≤,𝑢𝑧−𝑥_𝑛+1,𝑢2𝑥_𝑛−1−𝑆_𝑚(𝑧)+,𝑢𝑧−𝑆_𝑛𝑥_𝑛−1≤,𝑢𝑧−𝑥_𝑛+1,𝑢2𝑥_𝑛−1+1−𝑧,𝑢2𝑧−𝑆_𝑚(𝑧),𝑢+𝑧−𝑥_𝑛,𝑢. (72) Hence 12𝑧−𝑆_𝑚(𝑧),𝑢≤2𝑧−𝑥_𝑛+1,𝑢2𝑥_𝑛−1−𝑧,𝑢→0as𝑛→∞ (73) Therefore ‖𝑧−𝑆_𝑚(𝑧),𝑢‖=0 implying that 𝑆_𝑚(𝑧)=𝑧 for all 𝑚.
Hence {𝑆_𝑚} has a common fixed point 𝑧.
If 𝑧₁∈𝑋 is another fixed point of {𝑆_𝑛}, then 𝑧−𝑧₁,𝑢=𝑆_𝑛(𝑧)−𝑆_𝑛𝑧₁≤1,𝑢2𝑧−𝑧₁,𝑢+𝑧−𝑧₁,𝑢−𝜙𝑧−𝑧₁,𝑢,𝑧−𝑧₁,𝑢. (74) Implying that 𝜙[‖𝑧−𝑧₁,𝑢‖,‖𝑧−𝑧₁,𝑢‖]≤0. Hence 𝑧−𝑧₁=0 implying that 𝑧=𝑧₁.
Now for each 𝑚, 𝑇_𝑛(𝑧)=𝑇_𝑛𝑆_𝑛(𝑧)=𝑇_𝑛𝑇^𝑚𝑛_𝑛(𝑧)=𝑇^𝑚𝑛_𝑛𝑇_𝑛(𝑧), (75) showing that 𝑇_𝑛(𝑧) is the fixed point of 𝑆_𝑛.
Also 𝑆_𝑛(𝑧)=𝑧 and 𝑆_𝑛(𝑇_𝑛(𝑧))=𝑇_𝑛(𝑧). Hence 𝑧=𝑇_𝑛(𝑧) for all 𝑛.