83 

 

Ibn Al-Haitham Jour. for Pure & Appl. Sci. 32 (3) 2019 

 

 

  

 

 

 
 

Zahra Mahmood Mohamed Hasan                                    Salwa Salman Abed
 
 

                    
 
 
 
 

 

 
 

Abstract 

     In this paper, we study some cases of a common fixed point theorem for classes of firmly 

nonexpansive and generalized nonexpansive maps. In addition, we establish that the Picard-

Mann iteration is faster than Noor iteration and we used Noor iteration to find the solution of 

delay differential equation. 

 

Keywords: Banach space, common fixed point, strong convergence, nonexpansive map, 

condition (A). 

MSC: 49J40; 47J20 
 

1. Introduction  

     Let B be a non-empty  subset of a Banach space M. A mapT on B is called nonexpansive 

[1]. if ‖𝑇𝑎−𝑇𝑏‖ ≤ ‖𝑎 −𝑏‖ for all 𝑎,𝑏 𝜖 𝐵  and 𝐹(𝑇) denoted the set of all fixed points of 

T. In 1973, Bruck [2]. introduced a map called firmly nonexpansive map in Banach space. 

Certainly, every firmly nonexpansive is nonexpansive. 

      To discuss the convergence theorem for a pair of nonexpansive maps S and T on B to 

itself, a generalization of Mann and Ishikawa  iterations was given by Das and Debata [3]. and 

Takahashi and Tamura [4]. This iteration dealt with two maps: 

𝑎1 ∈ 𝐵 

 𝑏𝑛 = 𝛽𝑛𝑎𝑛 +(1−𝛽𝑛)𝑇𝑎𝑛  
         𝑎𝑛+1 = 𝛼𝑛𝑎𝑛 +(1−𝛼𝑛)𝑆𝑏𝑛, ∀𝑛 ∈ 𝑁 

where (𝛼𝑛) and (𝛽𝑛) ∈ [0,1]. 
      The aim of this paper is to prove some strongly convergenve theorems for approximating 

common fixed points of firmly nonexpansive and generalized nonexpansive. 
 

2. Preliminaries  

      We will suppose that 𝑀 is a Banach space and B is a non-empty closed convex subset of 

M. 𝐹(𝑇,𝑆) denoted the set of all fixed points of S and T.  

A sequences (𝑎𝑛) in 𝐵 is called : 

      Picard-Mann hybrid [5].  

Ibn Al Haitham Journal for Pure and Applied Science 

Journal homepage: http://jih.uobaghdad.edu.iq/index.php/j/index 

 

Doi:10.30526/32.3.2285  

Department of Mathematics, College of Education for Pure Sciences,Ibn Al-Haitham, 
University of Baghdad, Baghdad, Iraq. 

 

Strongly Convergence of Two Iterations For a Common Fixed Point 

with an Application 
 

zahramoh1990@gmail.com salwaalbundi@yahoo.com 

Article history: Received 26 December 2018, Accepted 3 March 2019, Publish September 

2019 

 

mailto:zahramoh1990@gmail.com
mailto:salwaalbundi@yahoo.com


  

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Ibn Al-Haitham Jour. for Pure & Appl. Sci. 32 (3) 2019 

 

                                     𝑎𝑛+1 = 𝑆𝑏𝑛  
         𝑏𝑛 = (1−𝛼𝑛)𝑎𝑛 +𝛼𝑛𝑇𝑎𝑛  ,      ∀ 𝑛 ∈ 𝑁                                                                                (1) 
where (𝛼𝑛) is a sequence in (0,1). 

 And a sequence (𝑤𝑛) in 𝐵 is called: 

      Noor  iteration [6].  

                                      𝑤𝑛+1 = (1−𝛼𝑛)𝑤𝑛 +𝛼𝑛𝑆𝑢𝑛  
                                      𝑢𝑛 = (1−𝛽𝑛)𝑤𝑛 +𝛽𝑛𝑇𝑣𝑛   
                                      𝑣𝑛 = (1−𝛾𝑛)𝑤𝑛 +𝛾𝑛𝑇𝑤𝑛        ,∀ 𝑛 ∈ 𝑁                                                  (2)   
where (𝛼𝑛),(𝛽𝑛) and (𝛾𝑛) are sequences in [0,1]. 
Defintion(1)[2]: A map 𝑇:𝐵 → 𝑀 is called firmly nonexpansive map if ‖𝑇𝑎 −𝑇𝑏‖ ≤

‖(1−𝑡)(𝑇𝑎 −𝑇𝑏)+𝑡(𝑎 −𝑏)‖,∀ 𝑎,𝑏 ∈ 𝐵 and 𝑡 ≥ 0. 

Defintion(2)[7]: A map 𝑇:𝐵 → 𝑀 is said to be generalized nonexpansive map if there are 

nonnegative constants 𝛿,𝜇 and 𝜔 with 𝛿 +2𝜇 +2𝜔 ≤ 1 such that ∀ 𝑎,𝑏 ∈  𝐵 

 ‖𝑇𝑎−𝑇𝑏‖ ≤ 𝛿‖𝑎 −𝑏‖+𝜇{‖𝑎 −𝑇𝑎‖+‖𝑏 −𝑇𝑏‖}+𝜔{‖𝑎 −𝑇𝑏‖+‖𝑏 −𝑇𝑎‖} 

      Khan. And  Fukhar-ud-din [8]. 2005, introduced  the concept of condition (𝐴′) to prove the 

convergence of  two-step iterative scheme with errors to common fixed points of two 

nonexpansive mappings, see also [9,10]. and [11]. 

Definition(3)[9]: Two maps are called satisfing condition (A) if there is a nondecreasing 

function 𝑔:[0,∞) → [0,∞) with 𝑔(0) = 0,𝑔(𝑖) > 0,∀ 𝑖 𝜖 (0,∞) such that: Either ‖𝑎 −

𝑇𝑎‖ ≥ 𝑔(𝐷(𝑎,𝐹)) 𝑜𝑟 ‖𝑎 −𝑆𝑎‖ ≥ 𝑔(𝐷(𝑎,𝐹)),∀ 𝑎 ∈ 𝐵 where 𝐷(𝑎,𝐹) = 𝑖𝑛𝑓{‖𝑎 −

𝑎∗‖; 𝑎∗ 𝜖 𝐹} and 𝐹 = 𝐹(𝑇)∩𝐹(𝑆). 

Definition(4)[12]: A map 𝑇:𝐵 → 𝐵 is called affine if B is convex and  𝑇(𝑟𝑎 +(1−𝑟)𝑏) =

𝑟𝑇(𝑎)+(1−𝑟)𝑇𝑏,∀ 𝑎,𝑏 ∈ 𝐵 and 𝑟 ∈ [0,1]. 

Definition(5)[5]: Let (𝑓𝑛) and (𝑔𝑛) be two sequences of real numbers that converge to 

𝑓 and 𝑔, resepectively. Assume that there exists a real number 𝑙 such that: 

lim
𝑛→∞

‖𝑓𝑛 −𝑓‖

‖𝑔𝑛 −𝑔‖
= 𝑙. 

If 𝑙 = 0, then we say that (𝑓𝑛) converges faster to 𝑓 than (𝑔𝑛) to 𝑔. 

Lemma(6)[13]: Let (𝛶𝑛)
∞

n=0 and (𝜘𝑛)
∞

n=0 be nonnegative real sequences satisfying the 

inequality: 

𝛶𝑛+1 ≤ (1−𝜏𝑛)𝛶𝑛 +𝜘𝑛 

where 𝜏𝑛 ∈ (0,1),∀ 𝑛 ≥ 𝑛0,∑ 𝜏𝑛
∞
𝑛=1 = ∞ and 

𝜘𝑛

𝜏𝑛
→ 0 𝑎𝑠 𝑛 → ∞,then lim𝑛→∞ 𝛶𝑛 = 0. 

Lemma(7)[14]: Let M be a uniformly convex Banach space and 0 < 𝑙 ≤ 𝑡𝑛 ≤ 𝑘 < 1,∀ 𝑛 ∈

𝑁. Suppose that (𝑎𝑛) and (𝑏𝑛) are two sequences of M such that lim𝑛→∞‖𝑎𝑛‖ ≤

𝑚,lim𝑛→∞‖𝑏𝑛‖ ≤ 𝑚  and lim𝑛→∞‖𝑡𝑛𝑎𝑛 +(1−𝑡𝑛)𝑏𝑛‖ = 𝑚 hold for some 𝑚 ≥ 0. Then 

lim𝑛→∞‖𝑎𝑛 −𝑏𝑛‖ = 0. 
 

3. The Main Results 

Lemma (3.1): Let M be a Banach space,𝐵 ⊆ 𝑀, 𝑇:𝐵 → 𝐵 be a Lipschitzain and firmly 

nonexpansive map and  𝑆:𝐵 → 𝐵 be Lipschitzain and generalized nonexpansive map. Let 

1-(𝑎𝑛) defined in (1) where (𝛼𝑛) ∈ (0,1),𝑛 ∈ 𝑁.  

2-(𝑤𝑛) defined in (2) where (𝛼𝑛),(𝛽𝑛) and (𝛾𝑛) ∈ [0,1].  
If 𝐹(𝑆,𝑇) ≠ ∅, then lim𝑛→∞‖𝑎𝑛 −𝑎

∗‖ and lim𝑛→∞‖𝑤𝑛 −𝑎
∗‖ exist ∀ 𝑎∗ ∈ 𝐹(𝑆,𝑇).  

Proof: Let 𝑎
*
∈ 𝐹(𝑇,𝑆). 

1- Now, to proof lim𝑛→∞‖𝑎𝑛 −𝑎
∗‖ exists. Let the sequence (𝑎𝑛) be as shown in step (1), so  



  

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 ‖𝑎𝑛+1 −𝑎
∗‖ = ‖𝑆𝑏𝑛 −𝑎

∗‖ 

                            

≤  𝛿‖𝑏𝑛 −𝑎
∗‖+𝜇{‖𝑎∗ −𝑎∗‖+‖𝑏𝑛 −𝑆𝑏𝑛‖}+𝜔{‖𝑎

∗ −𝑆𝑏𝑛‖

+‖𝑏𝑛 −𝑎
∗‖}  

                         ≤ (𝛿 +2𝐾𝜇 +2𝐾𝜔)‖𝑏𝑛 −𝑎
∗‖ 

                            ≤ (𝛿 +2𝜇 +2𝜔)‖𝑏𝑛 −𝑎
∗‖    

                         = ‖(1−𝛼𝑛)𝑎𝑛 +𝛼𝑛𝑇𝑎𝑛 −𝑎
∗‖        

                         ≤ (1−𝛼𝑛)‖𝑎𝑛 −𝑎
∗‖+𝛼𝑛(1−𝑡)‖𝑇𝑎𝑛 −𝑎

∗‖+  𝛼𝑛𝑡‖𝑎𝑛 −𝑎
∗‖                                                                                    

                             ≤ (1−𝛼𝑛)‖𝑎𝑛 −𝑎
∗‖+𝛼𝑛[(1−𝑡)𝑘 +𝑡]‖𝑎𝑛 −𝑎

∗‖ 

                             ≤ ‖𝑎𝑛 −𝑎
∗‖ 

Then, lim𝑛→∞‖𝑎𝑛 −𝑎
∗‖ exists ∀𝑎∗ ∈  𝐹(𝑇,𝑆). 

2- Now, to proof lim𝑛→∞‖𝑤𝑛 −𝑎
∗‖ exists. Let the sequence (𝑤𝑛) be as shown in step (2), so 

 ‖𝑣𝑛 −𝑎
∗‖ ≤ (1−𝛾𝑛)‖𝑤𝑛 −𝑎

∗‖+𝛾𝑛‖𝑇𝑤𝑛 −𝑎
∗‖ 

                        ≤ (1−𝛾𝑛)‖𝑤𝑛 −𝑎
∗‖+𝛾𝑛[(1−𝑡)𝑘 +𝑡]‖𝑤𝑛 −𝑎

∗‖ 

                     ≤ (1−𝛾𝑛 +𝛾𝑛)‖𝑤𝑛 −𝑎
∗‖ 

                     ≤ ‖𝑤𝑛 −𝑎
∗‖ 

 ‖𝑢𝑛 −𝑎
∗‖ ≤ (1−𝛽𝑛)‖𝑤𝑛 −𝑎

∗‖+𝛽𝑛‖𝑇𝑣𝑛 −𝑎
∗‖ 

                      ≤ (1−𝛽𝑛)‖𝑤𝑛 −𝑎
∗‖+𝛽𝑛[(1−𝑡)𝑘 +𝑡]‖𝑣𝑛 −𝑎

∗‖ 

                    ≤ (1−𝛽𝑛 +𝛽𝑛)‖𝑤𝑛 −𝑎
∗‖ 

                    ≤ ‖𝑤𝑛 −𝑎
∗‖ 

Now, 

‖𝑤𝑛+1 −𝑎
∗‖ ≤ (1−𝛼𝑛)‖𝑤𝑛 −𝑎

∗‖+𝛼𝑛‖𝑆𝑢𝑛 −𝑎
∗‖ 

                          ≤ (1−𝛼𝑛)‖𝑤𝑛 −𝑎
∗‖+𝛼𝑛(𝛿 +2𝜇 +2𝜔)‖𝑢𝑛 −𝑎

∗‖ 

                       ≤ (1−𝛼𝑛 +𝛼𝑛)‖𝑤𝑛 −𝑎
∗‖ 

                       ≤ ‖𝑤𝑛 −𝑎
∗‖ 

Then, lim𝑛→∞‖𝑤𝑛 −𝑎
∗‖ exists ∀𝑎∗ ∈ 𝐹(𝑇,𝑆). 

Lemma(3.2): Let M be a uniformly convex Banach space and 𝐵 ⊆ 𝑀. Let 

1- 𝑇:𝐵 → 𝐵 be a Lipschitzain and firmly nonexpansive map, 𝑆:𝐵 → 𝐵 be affine, Lipschitzain 

and generalized nonexpansive map and (𝑎𝑛) defined in (1) . 

2- 𝑇:𝐵 → 𝐵 be a Lipschitzain and firmly nonexpansive map, 𝑆:𝐵 → 𝐵 be Lipschitzain and 

generalized nonexpansive map and (𝑤𝑛) defined in (2). Suppose that ‖𝑎 −𝑇𝑏‖ ≤

‖𝑆𝑎 −𝑇𝑏‖,∀ 𝑎,𝑏 ∈ 𝐵 holds. If 𝐹(𝑆,𝑇) ≠ ∅, then: 

lim
𝑛→∞

‖𝑇𝑎𝑛 −𝑎𝑛‖ = 0 = lim
𝑛→∞

‖𝑆𝑎𝑛 −𝑎𝑛‖  and lim
𝑛→∞

‖𝑇𝑤𝑛 −𝑤𝑛‖ = 0 = lim
𝑛→∞

‖𝑆𝑤𝑛 −𝑤𝑛‖. 

Proof: Let 𝑎
*
∈ 𝐹(𝑇,𝑆).  

1-By Lemma (3.1) lim𝑛→∞‖𝑎𝑛 −𝑎
∗‖ exists ∀𝑎∗ ∈ 𝐹(𝑇,𝑆). Suppose that lim𝑛→∞‖𝑎𝑛 −

𝑎∗‖ = 𝑐,∀ 𝑐 ≥ 0. 

If 𝑐 = 0, no prove is needed. 

Now suppose 𝑐 > 0, 

‖𝑎𝑛+1 −𝑎
∗‖ = ‖𝑆𝑏𝑛 −𝑎

∗‖ 

                      ≤ ‖𝑏𝑛 −𝑎
∗‖ 

By Lemma (3.1), we show that ‖𝑏𝑛 −𝑎
∗‖ ≤ ‖𝑎𝑛 −𝑎

∗‖ . 

This implies to:                                         



  

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lim
𝑛→∞

‖𝑏𝑛 −𝑎
∗‖ = 𝑐 

Next consider.  

𝑐 = ‖𝑏𝑛 −𝑎
∗‖ ≤ (1−𝛼𝑛)‖𝑎𝑛 −𝑎

∗‖+𝛼𝑛‖𝑇𝑎𝑛 −𝑎
∗‖ 

By applying Lemma (2.7), we get: 

lim
𝑛→∞

‖𝑎𝑛 −𝑇𝑎𝑛‖ = 0 

𝑐 = lim
𝑛→∞

‖𝑎𝑛+1 −𝑎
∗‖ = lim

𝑛→∞
‖𝑆𝑏𝑛 −𝑎

∗‖  

and, 

‖𝑆[(1−𝛼𝑛)𝑎𝑛 +𝛼𝑛𝑇𝑎𝑛]−𝑎
∗‖ ≤ (1−𝛼𝑛)‖𝑆𝑎𝑛 −𝑎

∗‖  +𝛼𝑛‖𝑆𝑇𝑎𝑛 −𝑎
∗‖ 

By applying Lemma (2.7), we get: 

lim
𝑛→∞

‖𝑆𝑎𝑛 −𝑆𝑇𝑎𝑛‖ = 0 

Now, 

‖𝑆𝑎𝑛 −𝑎𝑛‖ ≤ ‖𝑆𝑎𝑛 −𝑆𝑇𝑎𝑛‖+‖𝑆𝑇𝑎𝑛 −𝑎𝑛‖ 

By using the hypothesis condition, we obtain: 

‖𝑆𝑎𝑛 −𝑎𝑛‖ ≤ 2‖𝑆𝑎𝑛 −𝑆𝑇𝑎𝑛‖ → 0 𝑎𝑠 𝑛 → ∞. 

Thus, 

lim
𝑛→∞

‖𝑆𝑎𝑛 −𝑎𝑛‖ = 0. 

2-By Lemma (3.1) lim𝑛→∞‖𝑤𝑛 −𝑎
∗‖ exists ∀𝑎∗ ∈ 𝐹(𝑇,𝑆). Suppose that lim𝑛→∞‖𝑤𝑛 −

𝑎∗‖ = 𝑐,∀ 𝑐 ≥ 0. 

If 𝑐 = 0, no prove is needed. 

Now, suppose 𝑐 > 0, 

Since ‖𝑇𝑤𝑛 −𝑎
∗‖ ≤ ‖𝑤𝑛 −𝑎

∗‖,
 
and as proved by Lemma (3.1) 

‖𝑆𝑢𝑛 −𝑎
∗‖ ≤ ‖𝑢𝑛 −𝑎

∗‖ and ‖𝑇𝑣𝑛 −𝑎
∗‖ ≤ ‖𝑣𝑛 −𝑎

∗‖. 

Then, lim𝑛→∞‖𝑇𝑤𝑛 −𝑎
∗‖ ≤ 𝑐, lim𝑛→∞‖𝑆𝑢𝑛 −𝑎

∗‖ ≤ 𝑐 and lim𝑛→∞‖𝑇𝑣𝑛 −𝑎
∗‖ ≤ 𝑐  

Moreover,  

lim
𝑛→∞

‖𝑤𝑛+1 −𝑎
∗‖ = 𝑐

 

𝑐 = ‖𝑤𝑛+1 −𝑎
∗‖ ≤ (1−𝛼𝑛)‖𝑤𝑛 −𝑎

∗‖+𝛼𝑛‖𝑆𝑢𝑛 −𝑎
∗‖ 

By applying Lemma (2.7), we obtain:  

lim
𝑛→∞

‖𝑤𝑛 −𝑆𝑢𝑛‖ = 0 

Next, 

‖𝑤𝑛 −𝑎
∗‖ ≤ ‖𝑤𝑛 −𝑆𝑢𝑛‖+‖𝑆𝑢𝑛 −𝑎

∗‖
𝑦𝑖𝑒𝑙𝑑𝑠
→    𝑐 ≤ lim

𝑛→∞
𝑖𝑛𝑓‖𝑢𝑛 −𝑎

∗‖ 

Therefore, we get: 

lim
𝑛→∞

‖𝑢𝑛 −𝑎
∗‖ = 𝑐 



  

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Ibn Al-Haitham Jour. for Pure & Appl. Sci. 32 (3) 2019 

 

𝑐 = ‖𝑢𝑛 −𝑎
∗‖ ≤ (1−𝛽𝑛)‖𝑤𝑛 −𝑎

∗‖+𝛽𝑛‖𝑇𝑣𝑛 −𝑎
∗‖ 

                             ≤ (1−𝛽𝑛)‖𝑤𝑛 −𝑎
∗‖+𝛽𝑛[(1−𝑡)𝑘 +𝑡]‖𝑣𝑛 −𝑎

∗‖ 

                             ≤ (1−𝛽𝑛)‖𝑤𝑛 −𝑎
∗‖+𝛽𝑛(1−𝛾𝑛)‖𝑤𝑛 −𝑎

∗‖ 

                             +𝛽𝑛𝛾𝑛‖𝑇𝑤𝑛 −𝑎
∗‖ 

                             ≤ (1−𝛽𝑛𝛾𝑛)‖𝑤𝑛 −𝑎
∗‖+𝛽𝑛𝛾𝑛‖𝑇𝑤𝑛 −𝑎

∗‖ 

So, by applying Lemma (2.7), we obtain:  

lim
𝑛→∞

‖𝑤𝑛 −𝑇𝑤𝑛‖ = 0. 

Next, 

‖𝑤𝑛 −𝑆𝑤𝑛‖ ≤ ‖𝑤𝑛 −𝑆𝑢𝑛‖+‖𝑆𝑢𝑛 −𝑤𝑛‖+‖𝑤𝑛 −𝑆𝑤𝑛‖  

Letting n→∞, we obtain: 

‖𝑤𝑛 −𝑆𝑤𝑛‖ ≤ ‖𝑤𝑛 −𝑆𝑤𝑛‖ 

That means lim𝑛→∞‖𝑤𝑛 −𝑆𝑤𝑛‖ = 0. 
Theorem (3.3): Let 𝑇:𝐵 → 𝐵 be a Lipschitzain, firmly nonexpansive map, 𝑆:𝐵 → 𝐵 be a 

Lipschitzain and generalized nonexpansive map, with 𝐹(𝑆,𝑇) ≠ ∅ and, 

1- (𝑎𝑛) defined in (1) and (𝛼𝑛) ∈ (0,1) satisfying ∑ 𝛼𝑖
∞
𝑖=0 = ∞. 

2- (𝑤𝑛) defined in (2) and (𝛼𝑛),(𝛽𝑛) and (𝛾𝑛) ∈ [0,1] satisfying ∑ 𝛼𝑖
∞
𝑖=0 𝛽𝑖𝛾𝑖 = ∞. 

Then (𝑎𝑛) and (𝑤𝑛)  converge to a unique common fixed point 𝑎
∗ ∈ 𝐹(𝑆,𝑇). 

Proof: 

1-‖𝑏𝑛 −𝑎
∗‖ ≤ (1−𝛼𝑛)‖𝑎𝑛 −𝑎

∗‖+𝛼𝑛‖𝑇𝑎𝑛 −𝑎
∗‖ 

                        ≤ (1−𝛼𝑛)‖𝑎𝑛 −𝑎
∗‖+𝛼𝑛[(1−𝑡)𝑘 +𝑡]‖𝑎𝑛 −𝑎

∗‖ 

Suppose 𝜉 = (1−𝑡)𝑘 +𝑡 

                     ≤ (1−(1−𝜉)𝛼𝑛)‖𝑎𝑛 −𝑎
∗‖ 

‖𝑎𝑛+1 −𝑎
∗‖ = ‖𝑆𝑏𝑛 −𝑎

∗‖ 

                         ≤  𝛿‖𝑏𝑛 −𝑎
∗‖+𝜇{‖𝑎∗ −𝑎∗‖+‖𝑏𝑛 −𝑆𝑏𝑛‖}+𝜔{‖𝑎

∗ −𝑆𝑏𝑛‖+‖𝑏𝑛 −𝑎
∗‖}      

                         ≤ (𝛿 +2𝜇 +2𝜔)‖𝑏𝑛 −𝑎
∗‖ 

                         ≤ ‖𝑏𝑛 −𝑎
∗‖ 

                         ≤ (1−(1−𝜉)𝛼𝑛)‖𝑎𝑛 −𝑎
∗‖ 

By induction: 

 ‖𝑎𝑛+1 −𝑎
∗‖ ≤ ∏ (1−(1−𝜉)𝛼𝑖)‖𝑎0 −𝑎

∗‖𝑛𝑖=0  

                          ≤ ‖𝑎0 −𝑎
∗‖𝑒−(1−𝜉)∑ 𝛼𝑖

𝑛
𝑖=0                   

Since ∑ 𝛼𝑖
∞
𝑖=0 = ∞,𝑒

−(1−𝜉)∑ 𝛼𝑖
𝑛
𝑖=0 → 0 𝑎𝑠 𝑛 → ∞.

 

Thus, lim𝑛→∞‖𝑎𝑛 −𝑎
∗‖ = 0.

 

2-‖𝑣𝑛 −𝑎
∗‖ ≤ (1−𝛾𝑛)‖𝑤𝑛 −𝑎

∗‖+𝛾𝑛‖𝑇𝑤𝑛 −𝑎
∗‖ 

                        ≤ (1−𝛾𝑛)‖𝑤𝑛 −𝑎
∗‖+𝛾𝑛[(1−𝑡)𝑘 +𝑡]‖𝑤𝑛 −𝑎

∗‖ 

Setting 𝜉 = (1−𝑡)𝑘 +𝑡 

                        ≤ (1−𝛾𝑛 +𝛾𝑛𝜉)‖𝑤𝑛 −𝑎
∗‖ 

‖𝑢𝑛 −𝑎
∗‖ ≤ (1−𝛽𝑛)‖𝑤𝑛 −𝑎

∗‖+𝛽𝑛‖𝑇𝑣𝑛 −𝑎
∗‖ 

                     ≤ (1−𝛽𝑛)‖𝑤𝑛 −𝑎
∗‖+𝛽𝑛𝜉‖𝑣𝑛 −𝑎

∗‖ 

                     ≤ (1−𝛽𝑛)‖𝑤𝑛 −𝑎
∗‖+𝛽𝑛𝜉(1−𝛾𝑛 +𝛾𝑛𝜉)‖𝑤𝑛 −𝑎

∗‖ 

                     ≤ (1−𝛽𝑛)‖𝑤𝑛 −𝑎
∗‖+𝛽𝑛(1−𝛾𝑛 +𝛾𝑛𝜉)‖𝑤𝑛 −𝑎

∗‖ 



  

88 

 

Ibn Al-Haitham Jour. for Pure & Appl. Sci. 32 (3) 2019 

 

Now, 

‖𝑤𝑛+1 −𝑎
∗‖ ≤ (1−𝛼𝑛)‖𝑤𝑛 −𝑎

∗‖+𝛼𝑛‖𝑆𝑢𝑛 −𝑎
∗‖ 

                          ≤ (1−𝛼𝑛)‖𝑤𝑛 −𝑎
∗‖+𝛼𝑛[ 𝛿‖𝑢𝑛 −𝑎

∗‖+𝜇{‖𝑎∗ −𝑎∗‖+‖𝑢𝑛 −𝑆𝑢𝑛‖}

+𝜔{‖𝑎∗ −𝑆𝑢𝑛‖+‖𝑢𝑛 −𝑎
∗‖}]  

                          ≤ (1−𝛼𝑛)‖𝑤𝑛 −𝑎
∗‖+𝛼𝑛(𝛿 +2𝜇 +2𝜔)‖𝑢𝑛 −𝑎

∗‖ 

                          ≤ (1−𝛼𝑛)‖𝑤𝑛 −𝑎
∗‖+𝛼𝑛(1−𝛽𝑛)‖𝑤𝑛 −𝑎

∗‖ 

                           +𝛼𝑛𝛽𝑛(1−𝛾𝑛 +𝛾𝑛𝜉)‖𝑤𝑛 −𝑎
∗‖ 

                          ≤ [1−𝛼𝑛𝛽𝑛𝛾𝑛 +𝛼𝑛𝛽𝑛𝛾𝑛𝜉]‖𝑤𝑛 −𝑎
∗‖ 

                       ≤ [1−𝛼𝑛𝛽𝑛𝛾𝑛]‖𝑤𝑛 −𝑎
∗‖ 

By induction: 

 ‖𝑤𝑛+1 −𝑎
∗‖ ≤ ∏ [1−𝛼𝑖𝛽𝑖𝛾𝑖]‖𝑤0 −𝑎

∗‖𝑛𝑖=0  

                          ≤ ‖𝑤0 −𝑎
∗‖𝑒−∑ 𝛼𝑖𝛽𝑖𝛾𝑖

𝑛
𝑖=0  

Since ∑ 𝛼𝑖
∞
𝑖=0 𝛽𝑖𝛾𝑖 = ∞,𝑒

−∑ 𝛼𝑖𝛽𝑖𝛾𝑖
𝑛
𝑖=0 → 0 𝑎𝑠 𝑛 → ∞. Thus, lim𝑛→∞‖𝑤𝑛 −𝑎

∗‖ = 0. 

Theorem(3.4): Let 𝐵,𝑆,𝑇,(𝑎𝑛) and (𝑤𝑛) be as in Lemma (3.2) and S,T satisfying condition 

(A). If 𝐹(𝑆,𝑇) ≠ ∅, then (𝑎𝑛) and (𝑤𝑛) converge strongly to a common fixed point of S and 

T. 

Proof: Now, we will show that (𝑎𝑛) strong convergence. By Lemma (3.1), 

lim𝑛→∞ ║𝑎n−𝑎
*
║exists. Suppose that lim𝑛→∞ ║𝑎n−𝑎

*
║ = 𝑐,𝑐 ≥ 0. From Lemma (3.1), we 

have, 

║𝑎n+1−𝑎
*
║ ≤ ║𝑎n−𝑎

*
║

 

That gives: 

𝑖𝑛𝑓𝑎∗∈𝐹 ║𝑎n+1−𝑎
*
║ ≤ 𝑖𝑛𝑓𝑎∗∈𝐹 ║𝑎n−𝑎

*
║ 

Which means, 𝐷(𝑎n+1,𝐹) ≤ 𝐷(𝑎n,𝐹)
𝑦𝑖𝑒𝑙𝑑𝑠
→     lim𝑛→∞ 𝐷(𝑎n,𝐹) exists. 

By using condition (A), we have: 

lim𝑛→∞ 𝑔(𝐷(𝑎n,𝐹)) ≤ lim𝑛→∞ ║𝑎n−𝑇𝑎n║ = 0. 
Or, 

lim𝑛→∞ 𝑔(𝐷(𝑎n,𝐹)) ≤ lim𝑛→∞ ║𝑎n−𝑆𝑎n║ = 0. 
In both situation, we obtain 

lim𝑛→∞ 𝑔(𝐷(𝑎n,𝐹)) = 0. 
Since g is a non-decreasing function and 𝑔(0) = 0, it follows that: 

lim𝑛→∞ 𝐷(𝑎𝑛 ,𝐹) = 0. Now to show that (𝑎𝑛) is a Cauchy sequence in A. Let 𝜖 > 0, 

lim𝑛→∞ 𝐷(𝑎𝑛 ,𝐹) = 0,∃ a positive integer 𝑛0, such that:  

𝐷(𝑎𝑛,𝐹) <
𝜖

4
,           ∀𝑛 ≥ 𝑛0 

In particular, 

inf {‖𝑎𝑛 −𝑎
∗‖,𝑎∗ ∈ 𝐹} <

𝜖

2
 

Thus must exist  𝑎∗∗ ∈ 𝐹 such that ‖𝑎𝑛 −𝑎
∗∗‖ <

𝜖

2
 . 

Now, ∀ 𝑛,𝑤 ≥ 𝑛0, we obtain: 

‖𝑎𝑛+𝑤 −𝑎𝑛‖ ≤ ‖𝑎𝑛+𝑤 −𝑎
∗∗‖+‖𝑎𝑛 −𝑎

∗∗‖ <
𝜖

2
+
𝜖

2
= 𝜖

 

Hence, (𝑎𝑛) is Cauchy sequence in B of M. Then (𝑎𝑛) converges to a point 𝑝 ∈ 𝐵. 

Lim
𝑛→∞

𝐷(𝑎𝑛 ,𝐹) = 0 
𝑦𝑖𝑒𝑙𝑑𝑠
→     𝐷(𝑝,𝐹) = 0. 



  

89 

 

Ibn Al-Haitham Jour. for Pure & Appl. Sci. 32 (3) 2019 

 

Since F is closed, hence 𝑝 ∈ 𝐹. 

By utilizing the same procedure, we can prove (𝑤𝑛) convergence strongly. 

Theorem (3.5): Let 𝑀 be a Banach space, ∅ ≠ 𝐵 ⊆ 𝑀. Let  𝑇:𝐵 → 𝐵 be lipschitzain, firmly 

nonexpansive maps, 𝑆:𝐵 → 𝐵 be Lipschitzain, generalized nonexpansive map and 𝑎∗ ∈ 𝐵 be 

a common fixed point of S and T. Let (𝑎𝑛) and (𝑤𝑛) be the Picard-Mann and Noor iterations 

defined in (1) and (2). Suppose (𝛼𝑛),(𝛽𝑛) and (𝛾𝑛)satisfied the following conditions: 

1- (𝛼𝑛),(𝛽𝑛) and (𝛾𝑛) ∈ (0,1),∀ 𝑛 ≥ 0. 

2- ∑𝛼𝑛 = ∞. 

3- ∑𝛼𝑛𝛽𝑛 < ∞. 

If 𝑤0 = 𝑎0 and 𝑅(𝑇),𝑅(𝑆) are bounded, then the Picard-Mann iteration sequence 𝑎𝑛 → 𝑎
∗ 

and The Noor iteration sequence 𝑤𝑛 → 𝑎
∗. 

Proof: Since the range of T and S are bounded, let 

𝑀1 = 𝑠𝑢𝑝𝑎∈𝐵{‖𝑇𝑎‖}+‖𝑎0‖ < ∞  

And,                                                    

𝑀2 = 𝑠𝑢𝑝𝑎∈𝐵{‖𝑆𝑎‖}+‖𝑎0‖ < ∞ 

Let 𝑀 = 𝑚𝑎𝑥{𝑀1,𝑀2} 

Then,  

‖𝑎𝑛‖ ≤ 𝑀,‖𝑏𝑛‖ ≤ 𝑀,‖𝑤𝑛‖ ≤ 𝑀,‖𝑢𝑛‖ ≤ 𝑀,‖𝑣𝑛‖ ≤ 𝑀                 

Therefore,  

‖𝑇𝑎𝑛‖ ≤ 𝑀,‖𝑇𝑤𝑛‖ ≤ 𝑀 . 

‖𝑎𝑛+1 −𝑤𝑛+1‖ = ‖𝑆𝑏𝑛 −(1−𝛼𝑛)𝑤𝑛 −𝛼𝑛𝑆𝑢𝑛‖      

                              ≤ ‖𝑆𝑏𝑛 −𝑤𝑛‖+𝛼𝑛‖𝑆𝑢𝑛 −𝑤𝑛‖      

                              ≤ (𝛿 +2𝜇 +2𝜔)‖𝑏𝑛 −𝑎
∗‖+𝛼𝑛(𝛿 +2𝜇 +2𝜔)‖𝑢𝑛 −𝑎

∗‖+(1+𝛼𝑛) 

                              (𝛿 +2𝜇 +2𝜔)‖𝑤𝑛 −𝑎
∗‖ 

                             ≤ ‖𝑏𝑛 −𝑎
∗‖+𝛼𝑛‖𝑢𝑛 −𝑎

∗‖+(1+𝛼𝑛)‖𝑤𝑛 −𝑎
∗‖ 

‖𝑏𝑛 −𝑎
∗‖ ≤ (1−𝛼𝑛)‖𝑎𝑛 −𝑎

∗‖+𝛼𝑛(𝑀 +‖𝑎
∗‖)                        

‖𝑣𝑛 −𝑎
∗‖ ≤ (1−𝛾𝑛)‖𝑤𝑛 −𝑎

∗‖+𝛾𝑛‖𝑇𝑤𝑛 −𝑎
∗‖         

Since T is lipschitzain and firmly nonexpansive, setting 𝜉 = 𝑘 −𝑘𝑡 +𝑡 

                  ≤ (1−𝛾𝑛)‖𝑤𝑛 −𝑎
∗‖+𝛾𝑛𝜉‖𝑤𝑛 −𝑎

∗‖ 

                    ≤ ‖𝑤𝑛 −𝑎
∗‖                                                                           

‖𝑢𝑛 −𝑎
∗‖ ≤ (1−𝛽𝑛)‖𝑤𝑛 −𝑎

∗‖+𝛽𝑛‖𝑇𝑣𝑛 −𝑎
∗‖ 

                    ≤ (1−𝛽𝑛)‖𝑤𝑛 −𝑎
∗‖+𝛽𝑛𝜉‖𝑣𝑛 −𝑎

∗‖  

                    ≤ ‖𝑤𝑛 −𝑎
∗‖ 

                    ≤ 𝑀 +‖𝑎∗‖                                                                                            

Then, 

‖𝑎𝑛+1 −𝑤𝑛+1‖ ≤ ‖𝑏𝑛 −𝑎
∗‖+𝛼𝑛‖𝑢𝑛 −𝑎

∗‖+(1+𝛼𝑛)‖𝑤𝑛 −𝑎
∗‖   

                               ≤ (1−𝛼𝑛)‖𝑎𝑛 −𝑎
∗‖+𝛼𝑛(𝑀 +‖𝑎

∗‖)+ 

                                   𝛼𝑛( 𝑀 +‖𝑎
∗‖)+(1+𝛼𝑛)(𝑀+‖𝑎

∗‖) 

                               ≤ (1−𝛼𝑛)‖𝑎𝑛 −𝑤𝑛‖+(1−𝛼𝑛)(𝑀+‖𝑎
∗‖) 

                                   +2𝛼𝑛(𝑀 +‖𝑎
∗‖)+(1+𝛼𝑛)(𝑀+‖𝑎

∗‖)  

                               ≤  (1−𝛼𝑛)‖𝑎𝑛 −𝑤𝑛‖+2 (1+𝛼𝑛)(𝑀 +‖𝑎
∗‖)         

Let 𝛶𝑛 = ‖𝑎𝑛 −𝑤𝑛‖,𝜘𝑛 = (2+2𝛼𝑛)(𝑀 +‖𝑎
∗‖) and  

𝜘𝑛

𝜏𝑛
 → 0 𝑎𝑠 𝑛 → ∞. By applying 

Lemma (2.6), we get 

lim
𝑛→∞

‖𝑎𝑛 −𝑤𝑛‖ = 0 



  

90 

 

Ibn Al-Haitham Jour. for Pure & Appl. Sci. 32 (3) 2019 

 

If  𝑎𝑛 → 𝑎
∗ ∈ 𝐹(𝑆,𝑇), then 

‖𝑤𝑛 −𝑎
∗‖ ≤ ‖𝑤𝑛 −𝑎𝑛‖+‖𝑎𝑛 −𝑎

∗‖ → 0 𝑎𝑠 𝑛 → ∞  

If 𝑤𝑛 → 𝑎
∗ ∈ 𝐹(𝑆,𝑇),

 
then 

‖𝑎𝑛 −𝑎
∗‖ ≤ ‖𝑎𝑛 −𝑤𝑛‖+‖𝑤𝑛 −𝑎

∗‖ → 0 𝑎𝑠 𝑛 → ∞.  

Theorem (3.6): Let 𝑇:𝐵 → 𝐵 be a Lipschitzain, firmly nonexpansive map with 𝑘𝑡 < 1,  

𝑆:𝐵 → 𝐵 be a Lipschitzain and generalized nonexpansive map.Suppose that the Picard-Mann 

and Noor iterations converge to the same common fixed point 𝑎∗.Then picard-Mann iteration 

converges faster than Noor iteration. 

Proof: Let 𝑎∗ ∈ 𝐹(𝑇,𝑆). Then for Picard-Mann iteration. 

‖𝑏𝑛 −𝑎
∗‖ ≤ (1−𝛼𝑛)‖𝑎𝑛 −𝑎

∗‖+𝛼𝑛‖𝑇𝑎𝑛 −𝑎
∗‖ 

Setting 𝜉 = (1−𝑡)𝑘 +𝑡, then we have 

                                   ≤ (1−(1−𝜉)𝛼𝑛)‖𝑎𝑛 −𝑎
∗‖ 

Next, 

‖𝑎𝑛+1 −𝑎
∗‖ = ‖𝑆𝑏𝑛 −𝑎

∗‖ 

                         ≤  𝛿‖𝑏𝑛 −𝑎
∗‖+𝜇{‖𝑎∗ −𝑎∗‖+‖𝑏𝑛 −𝑆𝑏𝑛‖}+𝜔{‖𝑎

∗ −𝑆𝑏𝑛‖+‖𝑏𝑛 −𝑎
∗‖}  

                      ≤ (𝛿 +2𝑘𝜇 +2𝑘𝜔)‖𝑏𝑛 −𝑎
∗‖ 

                         ≤ (𝛿 +2𝜇 +2𝜔)‖𝑏𝑛 −𝑎
∗‖ 

                         ≤ ‖𝑏𝑛 −𝑎
∗‖ 

                         ≤ (1−(1−𝜉)𝛼𝑛)‖𝑎𝑛 −𝑎
∗‖ 

                           .  

                           .  

                      ≤ (1−(1−𝜉)𝛼)𝑛‖𝑎1 −𝑎
∗‖ 

Let 𝑓𝑛 = (1−(1−𝜉)𝛼)
𝑛‖𝑎1 −𝑎

∗‖ 

Now, Noor iteration, 

‖𝑣𝑛 −𝑎
∗‖ ≤ (1−𝛾𝑛)‖𝑤𝑛 −𝑎

∗‖+𝛾𝑛‖𝑇𝑤𝑛 −𝑎
∗‖  

                    ≤ (1−𝛾𝑛)‖𝑤𝑛 −𝑎
∗‖+𝛾𝑛𝜉‖𝑤𝑛 −𝑎

∗‖ 

                  = ‖𝑤𝑛 −𝑎
∗‖       

‖𝑢𝑛 −𝑎
∗‖ ≤ (1−𝛽𝑛)‖𝑤𝑛 −𝑎

∗‖+𝛽𝑛‖𝑇𝑣𝑛 −𝑎
∗‖ 

                      .  

                     ≤ (1−(1−𝜉)𝛽𝑛)‖𝑤𝑛 −𝑎
∗‖ 

Then, 

‖𝑤𝑛+1 −𝑎
∗‖ ≤ (1−𝛼𝑛)‖𝑤𝑛 −𝑎

∗‖+𝛼𝑛‖𝑆𝑢𝑛 −𝑎
∗‖ 

                          ≤ (1−𝛼𝑛)‖𝑤𝑛 −𝑎
∗‖+𝛼𝑛(𝛿 +2𝜇 +2𝜔)‖𝑢𝑛 −𝑎

∗‖ 

                          ≤ (1−𝛼𝑛 +𝛼𝑛(1−(1−𝜉)𝛽𝑛)‖𝑤𝑛 −𝑎
∗‖ 

Assume that 𝛼𝑛 ≤ (1−𝛼𝑛 +𝛼𝑛(1−(1−𝜉)𝛽𝑛) 

      ≤ 𝛼𝑛‖𝑤𝑛 −𝑎
∗‖ 

       . 

      . 

      ≤ 𝛼𝑛‖𝑤1 −𝑎
∗‖  

Let 𝑔𝑛 = 𝛼
𝑛‖𝑤1 −𝑎

∗‖ 

Now,  



  

91 

 

Ibn Al-Haitham Jour. for Pure & Appl. Sci. 32 (3) 2019 

 

𝑓𝑛

𝑔𝑛
=
(1−(1−𝜉)𝛼)𝑛‖𝑎1−𝑎

∗‖

𝛼𝑛‖𝑤1−𝑎
∗‖

≤ (1−(1−𝜉))
𝑛 ‖𝑎1−𝑎

∗‖

‖𝑤1−𝑎
∗‖
→ 0 𝑎𝑠 𝑛 → ∞. 

Then, (𝑎𝑛) converges faster than (𝑤𝑛) to 𝑎
∗. 

Example (3.7): Let 𝑇,𝑆:𝑅 → 𝑅 (where R is the set of all real numbers) be two maps defined 

by 𝑇𝑎 =
2𝑎

3
 and 𝑆𝑎 =

𝑎

2
  ∀𝑎 ∈ 𝑅. Choose 𝛼𝑛 =

3

7
,𝛽𝑛 =

1

7
,𝛾𝑛 =

3

7
 ,∀𝑛 with initial value 

𝑎1 = 20. The Picard-Mann iteration converges faster than Noor iteration, it is clear from 

Table 1. and Figure 1. 
 

Table 1. Numerical results corresponding to 𝑎1 = 20 for 50 steps. 

 

n Picard-Mann n Noor Iteration n Picard-Mann  n Noor iteration 

0 20.0000 0 20.0000 26 - 26 0.0244 

1 8.5714 1 15.4519 27 - 27 0.0189 

2 3.6735 2 11.9381 28 - 28 0.0146 

3 1.5747 3 9.2233 29 - 29 0.0113 

4 0.6747 4 7.1259 30 - 30 0.0087 

5 0.2892 5 5.5054 31 - 31 0.0067 

6 0.1239 6 4.2534 32 - 32 0.0052 

7 0.0531 7 3.2862 33 - 33 0.0040 

8 0.0228 8 2.5389 34 - 34 0.0031 

9 0.0098 9 1.9615 35 - 35 0.0024 

10 0.0042 10 1.5155 36 - 36 0.0019 

11 0.0018 11 1.1708 37 - 37 0.0014 

12 0.0008 12 0.9046 38 - 38 0.0011 

13 0.0003 13 0.6989 39 - 39 0.0009 

14 0.0001 14 0.5400 40 - 40 0.0007 

15 0.0001 15 0.4172 41 - 41 0.0005 

16 0.0000 16 0.3223 42 - 42 0.0004 

17 0.0000 17 0.2490 43 - 43 0.0003 

18 - 18 0.1924 44 - 44 0.0002 

19 - 19 0.1486 45 - 45 0.0002 

20 - 20 0.1148 46 - 46 0.0001 

21 - 21 0.0887 47 - 47 0.0001 

22 - 22 0.0685 48 - 48 0.0001 

23 - 23 0.0530 49 - 49 0.0001 

24 - 24 0.0409 50 - 50 0.0000 

25 - 25 0.0319     
 

 

Figure 1. Convergence behavior corresponding to 𝑎1 = 20 for 50 steps. 



  

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4. Applications  

      Let the space C([f, h]) of all continuous real valued functions on a closed interval [f, h] be 

endowed with the chebyshev norm ‖𝑎 −𝑏‖∞ = max𝑡∈[𝑓,ℎ]|𝑎(𝑡)−𝑏(𝑡)|. 

(𝐶[𝑓,ℎ],‖.‖∞) be a Banach space. The following delay differential equation: 

𝑤΄ (𝑡) = 𝑔(𝑡,𝑤(𝑡),𝑤(𝑡 −𝜏)),𝑡 ∈ [𝑡0,ℎ] with initial condition 𝑤(𝑡) = 𝜗(𝑡),𝑡 ∈

[𝑡0 −𝜏,𝑡0]                                                                                                                                               (3)   

Assume the conditions are satisfied: 

i- 𝑡0,ℎ ∈ 𝑅,𝜏 > 0. 

ii- 𝑔 ∈ 𝐶([𝑡0,ℎ]×𝑅
2,𝑅). 

iii- 𝜗 ∈ 𝐶([𝑡0 −𝜏,ℎ],𝑅). 

iv- There is 𝐿𝑔 > 0 such that 

|𝑔(𝑡,𝑥1,𝑥2)−𝑔(𝑡,𝑦1,𝑦2)| ≤ 𝐿𝑔 ∑ |𝑥𝑖 −𝑦𝑖| 
2

𝑖=1
∀ 𝑥𝑖,𝑦𝑖 ∈ 𝑅,𝑖 = 1,2,𝑡 ∈ [𝑡0,ℎ]. 

v- 2𝐿𝑔(ℎ− 𝑡0) < 1. 

Now, let us consider the following integral equation: 

𝑤(𝑡) = {

𝜗(𝑡)                                                             𝑡 ∈ [𝑡0 −𝜏,𝑡0]

𝜗(𝑡)+∫ 𝑔(𝑟,𝑤(𝑟),𝑤(𝑟−𝜏))𝑑𝑟             𝑡 ∈ [𝑡0,ℎ]    
𝑡

𝑡0

 

This is the solution of the above delay differential equation [15]. 

Theorem (4.1): Suppose the conditions (i-v) are accomplished the problem (3) has a unique 

solution 𝑎∗ 𝑖𝑛 𝐶([𝑡0 −𝜏,ℎ],𝑅)∩𝐶
−1([𝑡0,ℎ,𝑅]) and the Noor iteration converges to 𝑎

∗. 

Proof: Let (𝑤𝑛) be an iterative sequence generated by Noor for an map defined by 

    𝑇𝑤(𝑡) = {

𝜗(𝑡)                                                             𝑡 ∈ [𝑡0 −𝜏,𝑡0]

𝜗(𝑡)+∫ 𝑔(𝑟,𝑤(𝑟),𝑤(𝑟 −𝜏))𝑑𝑟               𝑡 ∈ [𝑡0,ℎ]    
𝑡

𝑡0

 

Let (𝑎∗) be a fixed point. Now, it is easy to see 𝑤𝑛 → 𝑎
∗ for each 𝑡 ∈ [𝑡0 −𝜏,𝑡0].  

Next, for 𝑡 ∈ [𝑡0,ℎ], we get: 

‖𝑣𝑛 −𝑎
∗‖∞ = ‖(1−𝛾𝑛)𝑤𝑛 +𝛾𝑛𝑇𝑤𝑛 −𝑎

∗‖∞ 

≤ (1−𝛾𝑛)‖𝑤𝑛 −𝑎
∗‖∞ +𝛾𝑛 max𝑡∈[𝑡0−𝜏,𝑡0]|𝑇𝑤𝑛(𝑡)−𝑇𝑎

∗(𝑡)|  

≤ (1−𝛾𝑛)‖𝑤𝑛 −𝑎
∗‖∞ 

+𝛾𝑛 max
𝑡∈[𝑡0−𝜏,𝑡0]

|𝜗(𝑡)+∫ 𝑔(𝑟,𝑤(𝑟),𝑤(𝑟−𝜏))𝑑𝑟 −𝜗(𝑡) −∫ 𝑔(𝑟,𝑎∗(𝑟),𝑎∗(𝑟−𝜏))𝑑𝑟
𝑡

𝑡0

𝑡

𝑡0

| 

≤ (1−𝛾𝑛)‖𝑤𝑛 −𝑎
∗‖∞

+𝛾𝑛 max
𝑡∈[𝑡0−𝜏,𝑡0]

∫ (|𝑔(𝑟,𝑤(𝑟),𝑤(𝑟−𝜏))|
𝑡

𝑡0

+|𝑔(𝑟,𝑎∗(𝑟),𝑎∗(𝑟−𝜏))|)𝑑𝑟 



  

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≤ (1−𝛾𝑛)‖𝑤𝑛 −𝑎
∗‖∞ +𝛾𝑛 ∫ (𝐿𝑔( max

𝑡∈[𝑡0−𝜏,𝑡0]
|𝑤𝑛(𝑡)−𝑎

∗(𝑡)|
𝑡

𝑡0

+ max
𝑡∈[𝑡0−𝜏,𝑡0]

|𝑤𝑛(𝑡 −𝜏)−𝑎
∗(𝑡 −𝜏)|)𝑑𝑟 

≤ (1−𝛾𝑛)‖𝑤𝑛 −𝑎
∗‖∞ +𝛾𝑛 ∫ 𝐿𝑔(‖𝑤𝑛 −𝑎

∗‖∞ +‖𝑤𝑛 −𝑎
∗‖∞)𝑑𝑟

𝑡

𝑡0

 

≤ (1−𝛾𝑛)‖𝑤𝑛 −𝑎
∗‖∞ +2𝛾𝑛𝐿𝑔(𝑡 −𝑡0)‖𝑤𝑛 −𝑎

∗‖∞ 

≤ [1−(1−2𝐿𝑔(ℎ −𝑡0)𝛾𝑛]‖𝑤𝑛 −𝑎
∗‖∞ 

Next, 

‖𝑢𝑛 −𝑎
∗‖∞ = ‖(1−𝛽𝑛)𝑤𝑛 +𝛽𝑛𝑇𝑣𝑛 −𝑎

∗‖∞ 

≤ (1−𝛽𝑛)‖𝑤𝑛 −𝑎
∗‖∞ +𝛽𝑛 max

𝑡∈ [𝑡0−𝜏,𝑡0]
|𝑇𝑣𝑛(𝑡)−𝑇𝑎

∗(𝑡)| 

≤ [1−(1−2𝐿𝑔(ℎ −𝑡0)𝛽𝑛]‖𝑣𝑛 −𝑎
∗‖∞ 

Therefore, 

‖𝑤𝑛+1 −𝑎
∗‖∞ = ‖(1−𝛼𝑛)𝑤𝑛 +𝛼𝑛𝑇𝑢𝑛 −𝑎

∗‖∞ 

≤ (1−𝛼𝑛)‖𝑤𝑛 −𝑎
∗‖∞ +𝛼𝑛 max

𝑡∈[𝑡0−𝜏,𝑡0]
|𝑇𝑢𝑛(𝑡)−𝑇𝑎

∗(𝑡)| 

≤ [1−(1−2𝐿𝑔(ℎ −𝑡0)𝛼𝑛]‖𝑢𝑛 −𝑎
∗‖∞ 

≤ [1−(1−2𝐿𝑔(ℎ −𝑡0)𝛼𝑛][1−(1−2𝐿𝑔(ℎ −𝑡0)𝛽𝑛][1−(1

−2𝐿𝑔(ℎ −𝑡0)𝛾𝑛]‖𝑤𝑛 −𝑎
∗‖∞ 

≤ [1−(𝛼𝑛 +𝛽𝑛 +𝛾𝑛)[1−(1−2𝐿𝑔(ℎ −𝑡0)]‖𝑤𝑛 −𝑎
∗‖∞ 

setting 𝜆𝑛 = 𝛼𝑛 +𝛽𝑛 +𝛾𝑛 and by condition (v) 2𝐿𝑔(ℎ −𝑡0) < 1. 

Now, under the conditions (i-v) and using theorem (3.3) , therefore the delay differential 

equation has a unique solution 𝑎∗ 𝑖𝑛 𝐶([𝑡0 −𝜏,ℎ],𝑅)∩𝐶
−1([𝑡0,ℎ,𝑅]) and the Noor iteration 

converges to 𝑎∗. 

    In support of this work, we would like to refer to our other results in this field in [16]. 

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