IHJPAS. 36(2)2023 

367 
This work is licensed under a Creative Commons Attribution 4.0 International License. 

 

   

 

 

Zena Hussein Maibed  
mrs_ zena.hussein@yahoo.com 

Department of Mathematics, College of Education for Pure Science Ibn Al-Haitham 

University of Baghdad, Baghdad, Iraq. 

 

 

 
 

 

 

Abstract 

The goal of this article  is to study a new explicit iterative processes method  approach to zeroes 

for solving maximal monotone(M.M ) multivalued operators in Hilbert spaces, utilizing a finite 

family of different types of  mappings as nonexpansive map,contraction map, resolvent operator 

and nearest point metric projection map. On the other hand, The results in this paper  are develop 
and extend the main and important findings of previous studies. Then, utilizing various structural 

conditions in Hilbert space and variational inequality problems, also,we examine the strong 

convergence to nearest projection map for these explicit iterative processes methods under the 

presence of two important conditions for convergence, namely closure and convexity. The findings 

reported in this research strengthen, improve  and extend key previous findings from the literature. 

 

Keywords: Projection Mapping, Iterative Method, Nonexpamsive Mapping, Monotone Operators  

                   ,s-convergence, Fixed Point. 

 

1.Introduction  

Assume𝛨 is a true Hilbert space and M is the biggest monotone mapping. Many scholars have 

explored the challenge of finding zeroes, including [1] and [2]. The Proximal method Algorithm 

is a popular approach for solving 0 ∈ Mx. Rockafellar demonstrated w-convergence of the peri-

point technique in 1976, but it did not s-converge [3]. The human-like iterative approach is 

properly arranged in the f- point process to solve numerous nonlinear issues [4]. However, Mann-

likealgorithm processes in Hilbert space are only w- smoothed. [5] presented a viscosity,for 

resolving f- point of a nonlinear maps . The contour of Hilbert space identifies common solution 

s- convergence theorems. The  f- point theory solutions have been demonstrated to be a successful 

and influential way for addressing a wide range of real-world problems that can be broken down 

into identical f- point problems.In order to get a rough resolution of f-point difficulties, many 

algorithm procedures must be devised [6- 18]. The purpose of this paper is to expand and improve 
the proximal methods of multivalued  operators. One of the most important examples of f- point 

theory is the challenge of solving zero of (V.I) problem as                                                                                                                                                                     

doi.org/10.30526/36.2.3023 

Article history: Received 18 September 2022, Accepted 12 December 2022, Published in April 2023. 

 

 

 

Ibn Al-Haitham Journal for Pure and Applied Sciences 

Journal homepage: jih.uobaghdad.edu.iq 

 

Convergence To Approximate Solutions of Multivalued Operators 

 

https://creativecommons.org/licenses/by/4.0/
about:blank
http://en.uobaghdad.edu.iq/?page_id=15060


IHJPAS. 36(2)2023 

368 
 

{
𝑎2𝑛+1 = 𝑎𝑛𝑏 + 𝛿𝑛𝑎2𝑛 + 𝛾𝑛𝑅𝑒𝑠𝛽𝑛

𝑁 (𝑎2𝑛) + ℯ𝑛            𝑛 ≥ 0

𝑎2𝑛 = 𝜆𝑛𝑏 + 𝜌𝑛𝑎2𝑛−1 + 𝛿𝑛𝑅𝑒𝑠𝜇𝑛
𝑀 (𝑎2𝑛−1) + ℯ𝑛 

′         𝑛 ≥ 0
 

For given  u , a0 ∈ 𝛨, where (ℯ n) and (ℯ𝑛 
′ )   are sequences of calculation mistakes , N and M are 

M.M operators and 𝑎𝑛 , 𝛿𝑛 , 𝛾𝑛are seq lies in (0,1)   and 𝛽𝑛,  𝜇𝑛 ∈ (0, ∞)  . Here 𝑅𝑒𝑠𝛽
𝑁 =

(𝐼 + 𝛽𝑁 )−1 , 𝛽 > 0  ( the Resovent operator of N) .  
 

Now, we will review some concepts and lemmas known . Let ∅ ≠C be a closed,convex in H and 
ProC denote the nearest mapping  from H onto C,A mapping 𝐾is called nonexpansive if 
‖𝐾𝑢 − 𝐾𝑣 ‖ ≤ ‖𝑢 − 𝑣‖, ∀𝑢, 𝑣 ∈ 𝐻 and 𝛼 −inverse strongly monotone if there exist α > 0 
such that  
〈𝑆𝑢 − 𝑆𝑣 , 𝑢 − 𝑣〉 ≥ 𝛼‖𝐾𝑢 − 𝐾𝑣 ‖

2, ∀𝑢, 𝑣 ∈H . 
Lemma 1 . [16] Let {𝑎𝑛}𝑛=1

∞  be a sequence in  𝑅+ satisfying the following relation : 
                                                𝑎𝑛+1 ≤ (1 − 𝛼𝑛)𝑎𝑛 + 𝛼𝑛𝛿𝑛  , 𝑛 ≥ 𝑛𝑜  
Where{𝛼𝑛}𝑛=1

∞  ⊏ (0,1) and{𝛿𝑛}𝑛=1
∞  ⊂ 𝑅 satisfy the following condition : ∑ 𝛼𝑛

∞
𝑛=1 = ∞ and 

lim
𝑛→∞

𝑠𝑢𝑝𝛿𝑛 ≤ 0  or    ∑ 𝛼𝑛
∞
𝑛=1 𝛿𝑛 = ∞ . Then  lim

𝑛→∞
𝛼𝑛 = 0  . 

lemma 2 [18]  Let 𝑎 ∈ 𝐻   be given . The nearest mapping  is characterized by:  
(i) 𝑃𝑟𝑜𝐶 𝑎 ∈ 𝐶, 〈𝑎 − 𝑃𝑟𝑜𝐶 𝑎, 𝑃𝑟𝑜𝐶 𝑎 − 𝑐〉 ≥ 0    , for all 𝑐 ∈ 𝐶 .  
(ii) 𝑃𝑟𝑜𝐶   is firmly nonexpansive. 
Lemma 3 [19] Let{𝑎𝑛}𝑛=1

∞   and {𝑏𝑛}𝑛=1
∞  be bounded sequence in H and let{𝛽𝑛}𝑛=1

∞  be a sequence 

in (0,1) with  

0 ≤ lim
𝑛→∞

𝑖𝑛𝑓𝛽𝑛 ≤ lim
𝑛→∞

𝑠𝑢𝑝𝛽𝑛 ≤ 1.   

   Suppose 𝑎𝑛+1 = (1 − 𝛽𝑛)𝑏𝑛 + 𝛽𝑛𝑎𝑛 for all integers 𝑛 ≥ 0 and   lim
𝑛→∞

sup (‖𝑏𝑛+1 − 𝑏𝑛‖ −

‖𝑎𝑛+1 − 𝑎𝑛‖) ≤ 0 .Then lim
𝑛→∞

‖𝑏𝑛 − 𝑎𝑛‖ = 0     . 

 

2 .Main Results 

In this part, we introduced a new methods and we study the s- convergence, 

 

Theorem : 2.1 Let  ℳ𝑖 : 𝛨 ⟶ 2
𝐻 be a M.M. operators such that                                         

𝐹(𝑃𝑟𝑜𝑠) ⋂ ℳ𝑖
−10 ≠ ∅𝑙𝑖=1 .If the sequences {𝜉𝑛

𝑖 }
𝑛=2

∞
, {𝘙𝑛𝑖 }𝑛=1

∞
, {𝘚𝑛}𝑛=1

∞  𝑎𝑛𝑑{𝘘𝑛}𝑛=1
∞ are  seqs in 

(0,1)satisfy the following conditions :  

       (i)  ∑ |𝜉𝑛+1
𝑖 − 𝜉𝑛

𝑖 | < ∞ 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑖 = 1,2, … 𝑙∞𝑛=1  . 

        (ii)   ∑ |𝘙𝑛+1
𝑖 − 𝘙𝑛

𝑖 | < ∞ for all 𝑖 = 1,2, … 𝑙∞𝑛=1  .  

       (iii)   ∑ |𝘘𝑛+1 − 𝘘𝑛|
∞
𝑛=1 < ∞ . 

‖𝘚(𝑛+1)𝑖 𝑓(𝑥𝑛+1) − 𝘚𝑛𝑖 𝑓(𝑥𝑛)‖ ≤  𝘚𝑛𝑖 ‖𝑓(𝑥𝑛+1) − 𝑓(𝑥𝑛)‖and 

‖(1 − 𝘚(𝑛+1)𝑖 ) 𝑔(𝑥𝑛+1) − (1 − 𝘚𝑛𝑖 ) 𝑔(𝑥𝑛)‖ ≤ (1 − 𝘚𝑛𝑖 ) ‖ 𝑔(𝑥𝑛+1) −  𝑔(𝑥𝑛)‖.Then the 

sequences {𝑥𝑛 } defined by  𝑥 ∈ 𝐶 and  

  

      𝑦𝑛
𝑖−1 = 𝘚𝑛𝑖 𝑓(𝑥𝑛) + (1 − 𝘚𝑛𝑖 ) 𝑔(𝑥𝑛)                                                                                

𝑦𝑛
𝑚 = 𝘙𝑛1 𝑦𝑛

0 + ∑ 𝘙𝑛𝑖  𝑅𝑒𝑠𝑖,𝑛𝑦𝑛
𝑖−1𝑙

𝑖=2     where  𝑓 , 𝑔: 𝐻 → 𝑆 is nonexpansive

𝑥𝑛+1 = 𝘘𝑛𝑃𝑟𝑜𝛹𝑥 + (1 − 𝘘𝑛)𝑦𝑛
𝑚, 𝑛, 𝑚 ≥ 0                                               

  

 Is s-converges to 𝑃𝑟𝑜𝛹 𝑥 . 
Proof : Let 𝑒 ∈ 𝑆. Then we have 
               ‖𝑥𝑛+1 − 𝑒‖ = ‖𝘘𝑛𝑃𝑟𝑜𝛹𝑥 + (1 − 𝘘𝑛)𝑦𝑛

𝑚 − 𝑒‖  
= ‖𝘘𝑛𝑃𝑟𝑜𝛹 𝑥 + (1 − 𝘘𝑛)𝑦𝑛

𝑚 − [(1 − 𝘘𝑛)𝑒 + 𝘘𝑛𝑒]‖ 
 ≤  𝘘𝑛‖𝑃𝑟𝑜𝛹𝑥 − 𝑒‖ + (1 − 𝘘𝑛)‖𝑦𝑛

𝑚 − 𝑒‖ ≤ 𝘘𝑛‖𝑥 − 𝑒‖ + (1 − 𝘘𝑛)‖𝑦𝑛
𝑚 − 𝑒‖           (2.1)                    

   

‖𝑦𝑛
𝑚 − 𝑒‖ = ‖𝘙𝑛1 𝑦𝑛

0 + ∑ 𝘙𝑛𝑖  𝑅𝑒𝑠𝑖,𝑛𝑦𝑛
𝑖−1

𝑙

𝑖=2

− 𝑒‖ 



IHJPAS. 36(2)2023 

369 
 

 =  ‖𝘙𝑛1 𝑦𝑛
0 + ∑ 𝘙𝑛𝑖 𝑅𝑒𝑠𝑖,𝑛𝑦𝑛

𝑖−1𝑙
𝑖=2 − [𝘙𝑛1 𝑒 + ∑ 𝘙𝑛𝑖  𝑒

𝑚
𝑖=1 ]‖ 

    = ‖(𝘙𝑛1 )𝑦𝑛
0 + ∑ 𝘙𝑛𝑖  𝑅𝑒𝑠𝑖,𝑛𝑦𝑛

𝑖−1

𝑙

𝑖=2

− [𝘙𝑛1 𝑒 + ∑ 𝘙𝑛𝑖  𝑒

𝑚

𝑖=2

] ‖ 

≤   (𝘙𝑛1 )‖𝑦𝑛
𝑜 − 𝑒‖ + ∑ 𝘙𝑛𝑖

𝑙
𝑖=2 ‖ 𝑦𝑛

𝑖−1 − 𝑒‖                             

   ≤   ∑ 𝘙𝑛𝑖

𝑙

𝑖=1

‖ 𝑦𝑛
𝑖−1 − 𝑒‖ ≤   ‖ 𝑦𝑛

𝑖−1 − 𝑒‖ = ‖  𝘚𝑛𝑖 𝑓(𝑥𝑛) + (1 − 𝘚𝑛𝑖 ) 𝑔(𝑥𝑛)   − 𝑒‖ 

     

≤  ‖𝘚𝑛𝑖 𝑓(𝑥𝑛) − 𝑒‖ + 𝑘(1 − 𝘚𝑛𝑖 )‖  𝑔(𝑥𝑛 ) − 𝑒‖ 
≤ 𝘚𝑛𝑖  ‖𝑥𝑛 − 𝑒‖ + (1 − 𝘚𝑛𝑖 )‖ 𝑥𝑛 − 𝑒‖  ≤ ‖  𝑥𝑛 − 𝑝‖         (2.2)    
                                                                                                       

From (2.1) and (2.2) . we get  

               ‖𝑥𝑛+1 − 𝑒‖ ≤ 𝘘𝑛‖𝑃𝑟𝑜𝛹𝑥 − 𝑒‖ + (1 − 𝘘𝑛)  ‖ 𝑥𝑛 − 𝑒‖ 
                                   ≤ max{‖𝑥 − 𝑒‖, ‖ 𝑥𝑛 − 𝑒‖ }        
                                    

                                   . 

                                   .  

                                   ≤ max{‖𝑥 − 𝑒‖, ‖𝑥0 − 𝑒‖} 
 

Hence ,{𝑥𝑛 }𝑛=1
∞  is  bounded . So {𝑦𝑛

𝑚}𝑛=1
∞  and {𝑅𝑒𝑠𝑖,𝑛𝑦𝑛

𝑖 }
𝑛=1

∞
  is bounded . 

Now , 

  ‖𝑦𝑛+1
𝑖 − 𝑦𝑛

𝑖 ‖  ∀ 𝑖 = 1,2, … 𝑙 .   

‖𝑦𝑛+1
𝑚 − 𝑦𝑛

𝑚‖  = ‖𝘙𝑛1+1 𝑦𝑛+1
0 + ∑ 𝘙𝑛𝑖+1 𝑅𝑒𝑠𝑖,𝑛+1𝑦𝑛+1

𝑖−1𝑙
𝑖=2 − 𝘙𝑛1 𝑦𝑛

𝑜 − ∑ 𝘙𝑛𝑖 𝑅𝑒𝑠𝑖,𝑛𝑦𝑛
𝑖−1𝑙

𝑖=2   ‖ 

=
‖

‖
𝘙𝑛1+1𝑦𝑛+1

0 − 𝘙𝑛1+1𝑦𝑛
0 + 𝘙𝑛1+1𝑦𝑛

0 + ∑ 𝘙𝑛𝑖+1 𝑅𝑒𝑠𝑖,𝑛+1𝑦𝑛+1
𝑖−1

𝑙

𝑖=2

−

∑ 𝘙𝑛𝑖+1 𝑅𝑒𝑠𝑖,𝑛𝑦𝑛
𝑖−1

𝑙

𝑖=2

+ ∑ 𝘙𝑛𝑖  𝑅𝑒𝑠𝑖,𝑛𝑦𝑛
𝑖−1

𝑙

𝑖=2

  − 𝘙𝑛1 𝑦𝑛
0 − ∑ 𝘙𝑛𝑖+1 𝑅𝑒𝑠𝑖,𝑛𝑦𝑛

𝑖−1

𝑙

𝑖=2

 
‖

‖
 

≤ 𝘙𝑛1+1‖𝑦𝑛+1
0 − 𝑦𝑛

0‖ + |𝘙𝑛1+1 − 𝘙𝑛1 |(‖𝑦𝑛
0‖) + ∑ 𝘙𝑛𝑖+1 

𝑙

𝑖=2

‖𝑅𝑒𝑠𝑖,𝑛+1𝑦𝑛+1
𝑖−1 − 𝑅𝑒𝑠𝑖,𝑛𝑦𝑛

𝑖−1‖ 

+  ∑  

𝑙

𝑖=2

|( 𝘙𝑛𝑖 −  𝘙 𝑛𝑖+1)|‖𝑅𝑒𝑠𝑖,𝑛𝑦𝑛
𝑖−1‖ 

    

≤ 𝘙𝑛1+1‖𝑦𝑛+1
0 − 𝑦𝑛

0‖ +  (∑  

𝑙

𝑖=2

|( 𝘙𝑛𝑖 −   𝘙𝑛𝑖+1)|‖𝑅𝑒𝑠𝑖,𝑛𝑦𝑛
𝑖−1‖ + |𝘙𝑛1+1 − 𝘙𝑛1 |(‖𝑦𝑛

0‖))

+ ∑ 𝘙𝑛𝑖+1 

𝑙

𝑖=2

‖𝑅𝑒𝑠𝑖,𝑛+1𝑦𝑛+1
𝑖−1 − 𝑅𝑒𝑠𝑖,𝑛𝑦𝑛

𝑖−1‖ 

≤ 𝘙𝑛1+1‖𝑦𝑛+1
0 − 𝑦𝑛

0‖ +  ((∑  

𝑙

𝑖=2

|( 𝘙𝑛𝑖 −   𝘙𝑛𝑖+1)| + |𝘙𝑛0+1 − 𝘙𝑛0 |) 𝐾)

+ ∑ 𝘙𝑛𝑖+1 

𝑙

𝑖=2

‖𝑅𝑒𝑠𝑖,𝑛+1𝑦𝑛+1
𝑖−1 − 𝑅𝑒𝑠𝑖,𝑛𝑦𝑛

𝑖−1‖ 

 



IHJPAS. 36(2)2023 

370 
 

≤ 𝘙𝑛1+1K +  ((∑  

𝑙

𝑖=1

|( 𝘙𝑛𝑖 −   𝘙𝑛𝑖+1)| + |𝘙𝑛1+1 − 𝘙𝑛1 |) 𝐾)

+ ∑ 𝘙𝑛𝑖+1 

𝑙

𝑖=2

‖𝑅𝑒𝑠𝑖,𝑛+1𝑦𝑛+1
𝑖−1 − 𝑅𝑒𝑠𝑖,𝑛𝑦𝑛

𝑖−1‖ 

𝐾 = 𝑚𝑎𝑥{‖𝑅𝑒𝑠𝑖,𝑛𝑦𝑛
𝑖−1‖, ‖𝑦𝑛

0‖, ‖𝑦𝑛+1
0 − 𝑦𝑛

0‖} 

Next , to find, ‖𝑅𝑒𝑠𝑖,𝑛+1𝑦𝑛+1
𝑖−1 − 𝑅𝑒𝑠𝑖,𝑛𝑦𝑛

𝑖−1‖  

For  𝜉𝑛+1
𝑖 ≥ 𝜉𝑛

𝑖     , we have  
 

  ‖𝑅𝑒𝑠𝑖,𝑛+1𝑦𝑛+1
𝑖−1 − 𝑅𝑒𝑠𝑖,𝑛𝑦𝑛

𝑖−1‖ 
    

= ‖𝑅𝑒𝑠𝑖,𝑛 (
𝜉𝑛

𝑖

𝜉𝑛+1
𝑖 𝑦𝑛+1

𝑖−1 + (1 −
𝜉𝑛

𝑖

𝜉𝑛+1
𝑖 ) 𝑅𝑒𝑠𝑖,𝑛+1𝑦𝑛+1

𝑖−1 ) − 𝑅𝑒𝑠𝑖,𝑛𝑦𝑛
𝑖−1‖ 

  

   ≤  ‖
𝜉𝑛

𝑖

𝜉𝑛+1
𝑖

𝑦𝑛+1
𝑖−1 + (1 −

𝜉𝑛
𝑖

𝜉𝑛+1
𝑖

) 𝑅𝑒𝑠𝑖,𝑛+1𝑦𝑛+1
𝑖−1 − 𝑦𝑛

𝑖−1‖ 

=  ‖
𝜉𝑛

𝑖

𝜉𝑛+1
𝑖 𝑦𝑛+1

𝑖−1 + (1 −
𝜉𝑛

𝑖

𝜉𝑛+1
𝑖 ) 𝑅𝑒𝑠𝑖,𝑛+1𝑦𝑛+1

𝑖−1 −
𝜉𝑛

𝑖

𝜉𝑛+1
𝑖 𝑦𝑛

𝑖−1 + (1 −
𝜉𝑛

𝑖

𝜉𝑛+1
𝑖 )  𝑦𝑛

𝑖−1  ‖ 

 

= ‖
𝜉𝑛

𝑖

𝜉𝑛+1
𝑖 𝑦𝑛+1

𝑖−1 −
𝜉𝑛

𝑖

𝜉𝑛+1
𝑖 𝑦𝑛

𝑖−1 + (1 −
𝜉𝑛

𝑖

𝜉𝑛+1
𝑖 ) 𝑅𝑒𝑠𝑖,𝑛+1𝑦𝑛+1

𝑖−1 + (1 −
𝜉𝑛

𝑖

𝜉𝑛+1
𝑖 ) 𝑦𝑛

𝑖−1‖ 

 

≤
𝜉𝑛

𝑖

𝜉𝑛+1
𝑖 ‖𝑦𝑛+1

𝑖−1 − 𝑦𝑛
𝑖−1‖ +|1 −

𝜉𝑛
𝑖

𝜉𝑛+1
𝑖 | (‖𝑅𝑒𝑠𝑖,𝑛+1𝑦𝑛+1

𝑖−1 + 𝑦𝑛+1
𝑖−1 )‖) 

≤ ‖𝑦𝑛+1
𝑖−1 − 𝑦𝑛

𝑖−1‖ +|1 −
𝜉𝑛

𝑖

𝜉𝑛+1
𝑖 | (‖𝑅𝑒𝑠𝑖,𝑛+1𝑦𝑛+1

𝑖−1 + 𝑦𝑛+1
𝑖−1 ‖) 

≤ ‖𝑦𝑛+1
𝑖−1 − 𝑦𝑛

𝑖−1‖ +|1 −
𝜉𝑛

𝑖

𝜉𝑛+1
𝑖 | 𝑆               where 𝑆 = sup {‖𝑅𝑒𝑠𝑖,𝑛+1𝑦𝑛+1

𝑖−1 + 𝑦𝑛+1
𝑖−1 ‖, 𝑛 ≥ 0} 

≤ ‖𝑦𝑛+1
𝑖−1 − 𝑦𝑛

𝑖−1‖ +|
𝜉𝑛+1

𝑖 −𝜉𝑛
𝑖

𝜉𝑛+1
𝑖 | 𝑆 

≤ ‖𝑦𝑛+1
𝑖−1 − 𝑦𝑛

𝑖−1‖  +  
 𝑆

𝜉𝑛+1
𝑖 |𝜉𝑛+1

𝑖 − 𝜉𝑛
𝑖  | 

≤ ‖𝑦𝑛+1
𝑖−1 − 𝑦𝑛

𝑖−1‖  +  
𝑠

𝜉
|𝜉𝑛+1

𝑖 − 𝑟𝑛
𝑖  | 

≤ ‖𝘚(𝑛+1)𝑖 𝑓(𝑥𝑛+1) + (1 − 𝘚(𝑛+1)𝑖 ) 𝑔(𝑥𝑛+1) − 𝘚𝑛𝑖 𝑓(𝑥𝑛) + (1 − 𝘚𝑛𝑖 ) 𝑔(𝑥𝑛)‖  +  
𝑆

𝜉
|𝜉𝑛+1

𝑖 − 𝜉𝑛
𝑖  | 

≤ ‖𝘚(𝑛+1)𝑖 𝑓(𝑥𝑛+1) − 𝘚𝑛𝑖 𝑓(𝑥𝑛)‖ + ‖(1 − 𝘚(𝑛+1)𝑖 ) 𝑔(𝑥𝑛+1) − (1 − 𝘚𝑛𝑖 ) 𝑔(𝑥𝑛)‖ +
𝑆

𝜉
|𝜉𝑛+1

𝑖 − 𝜉𝑛
𝑖  | 

≤ 𝘚𝑛𝑖 ‖(𝑥𝑛+1) − (𝑥𝑛)‖ + (1 − 𝘚𝑛𝑖 )‖ (𝑥𝑛+1) −  (𝑥𝑛)‖ +
𝑆

𝜉
|𝜉𝑛+1

𝑖 − 𝜉𝑛
𝑖  | 

≤ ‖𝑥𝑛+1 − 𝑥𝑛‖ +
𝑆

𝜉
|𝜉𝑛+1

𝑖 − 𝜉𝑛
𝑖  | 

‖𝑦𝑛+1
𝑚 − 𝑦𝑛

𝑚‖   

≤ 𝘙𝑛0+1K +  ((∑ ( 
𝑙
𝑖=1 |( 𝘙𝑛𝑖 −   𝘙𝑛𝑖+1)| + |𝘙𝑛0+1 − 𝘙𝑛0 |)𝐾) + ∑ 𝘙𝑛𝑖+1 (

𝑙
𝑖=1 ‖𝑥𝑛+1 − 𝑥𝑛‖  +  

𝑆

𝜉
|𝜉𝑛+1

𝑖 − 𝜉𝑛
𝑖  |) 

 

‖𝑦𝑛+1
𝑚 − 𝑦𝑛

𝑚‖   ≤ ‖𝑥𝑛+1 − 𝑥𝑛‖ +  𝘙𝑛0+1K +  ((∑ ( 
𝑙
𝑖=1 |( 𝘙𝑛𝑖 −   𝘙𝑛𝑖+1)| + |𝘙𝑛0+1 − 𝘙𝑛0 |)𝐾) +

𝑆

𝜉
∑ 𝘙𝑛𝑖+1    |𝜉𝑛+1

𝑖 − 𝜉𝑛
𝑖  |𝑙𝑖=1   (2.3) 

 



IHJPAS. 36(2)2023 

371 
 

Now,we have  

 

‖𝑦𝑛+1
𝑖 − 𝑦𝑛

𝑖 ‖  ≤ ‖𝑥𝑛+1 − 𝑥𝑛 ‖ +
𝑆

𝜉
∑ (|𝜉𝑛+1

𝑖 − 𝜉𝑛
𝑖 |)𝑙𝑖=1  + 𝘙𝑛0+1K +  ((∑ ( 

𝑙
𝑖=0 |( 𝘙𝑛𝑖 −   𝘙𝑛𝑖+1)|)𝐾) 

Now, we estimate ‖𝑥𝑛+2 − 𝑥𝑛+1‖ . since 𝑥𝑛+1 = 𝛼𝑛 𝑥 + (1 − 𝛼𝑛)𝑦𝑛
𝑚 

= ‖𝑥𝑛+2 − 𝑥𝑛+1‖ = ‖𝘘𝑛+1𝑃𝑟𝑜𝛹𝑥 + (1 − 𝘘𝑛+1)𝑦𝑛+1
𝑚 − 𝘘𝑛𝑥 + (1 − 𝘘𝑛)𝑦𝑛

𝑚‖             
≤ 1 − 𝘘𝑛 )‖𝑦𝑛+1

𝑚 − 𝑦𝑛
𝑚‖ + |𝘘𝑛+1 − 𝘘𝑛|(‖𝑃𝑟𝑜𝛹𝑥‖ +  ‖𝑦𝑛+1

𝑚 ‖) 
                                 

≤ (1 − 𝘘𝑛)‖𝑦𝑛+1
𝑚 − 𝑦𝑛

𝑚‖ + L|𝘘𝑛+1 − 𝘘𝑛+1| 
 

Where L =‖𝑃𝑟𝑜𝛹𝑥‖ +  ‖𝑦𝑛+1
𝑚 ‖ .So, we  get    

‖𝑥𝑛+2 − 𝑥𝑛+1‖ ≤ (1 − 𝘘𝑛)‖𝑥𝑛+1 − 𝑥𝑛‖ + 𝘙𝑛0+1K +  ((∑ ( 
𝑙
𝑖=0 |( 𝘙𝑛𝑖 −   𝘙𝑛𝑖+1)|)𝐾)  +

𝑆

𝜉
∑ 𝘙𝑛𝑖+1    |𝜉𝑛+1

𝑖 − 𝜉𝑛
𝑖  |𝑙𝑖=1 + L|𝘘𝑛+1 − 𝘘𝑛+1| 

So, ‖𝑥𝑛+1 − 𝑥𝑛‖ → 0 𝑎𝑠 𝑛 → ∞     
      ‖𝑥𝑛 − 𝑦𝑛

𝑚‖ ≤ ‖𝑥𝑛+1 − 𝑥𝑛‖ + ‖𝑥𝑛+1 − 𝑦𝑛
𝑚‖        

                           ≤ ‖𝑥𝑛+1 − 𝑥𝑛‖ + ‖𝘘𝑛 𝑥 + (1 − 𝘘𝑛)𝑦𝑛
𝑚 − 𝑦𝑛

𝑚‖         
                           = ‖𝑥𝑛+1 − 𝑥𝑛‖ + ‖𝘘𝑛𝑥 + 𝑦𝑛

𝑚 − 𝘘𝑛𝑦𝑛
𝑚 − 𝑦𝑛

𝑚‖   
                           = ‖𝑥𝑛+1 − 𝑥𝑛‖ + ‖𝘘𝑛𝑥 − 𝘘𝑛𝑦𝑛

𝑚‖    
                            ≤ ‖𝑥𝑛+1 − 𝑥𝑛‖ + 𝘘𝑛 ‖𝑥 − 𝑦𝑛

𝑚‖      
 

We have      ‖𝑥𝑛 − 𝑦𝑛
𝑚 ‖ → 0    𝑎𝑠  𝑛 → ∞  .  

Now , we show that ‖𝑦𝑛
𝑖−1 − 𝑅𝑒𝑠𝑖,𝑛𝑦𝑛

𝑖−1‖ → 0 for all 𝑖 = 1,2, . . 𝑀 .  In fact ,  

Let {𝑥𝑛𝑘 } be a subsequence of {𝑥𝑛} such that 

 

lim
𝑛→∞

𝑠𝑢𝑝 ‖𝑦𝑛
𝑖−1 − 𝑅𝑒𝑠𝑖,𝑛𝑦𝑛

𝑖−1‖ =  lim
𝑘→∞

‖𝑦𝑛𝑘
𝑖−1 − 𝑅𝑒𝑠𝑖,𝑛𝑘 𝑦𝑛𝑘

𝑖−1‖ , 

 

And  let {𝑥𝑛𝑘𝑙
} be a subsequence of {𝑥𝑛𝑘 } such that 

 

lim
𝑘→∞

𝑠𝑢𝑝 ‖𝑥𝑛𝑘 − 𝑒‖ =  lim𝑙→∞
‖𝑥𝑛𝑘𝑙

− 𝑒‖  . 

 

 Then we have  

‖𝑥𝑛𝑘𝑙
− 𝑒‖ = ‖𝑥𝑛𝑘𝑙

− 𝑦𝑛𝑘𝑙
𝑚 + 𝑦𝑛𝑘𝑙

𝑚 − 𝑒‖ 

≤ ‖𝑥𝑛𝑘𝑙
− 𝑦𝑛𝑘𝑙

𝑚 ‖ + ‖𝑦𝑛𝑘𝑙
𝑚 − 𝑒‖ 

≤ ‖𝑥𝑛𝑘𝑙
− 𝑦𝑛𝑘𝑙

𝑚 ‖ + ‖𝘙𝑛0 𝑦𝑛
𝑜 + ∑ 𝘙𝑛𝑖  𝑅𝑒𝑠𝑖,𝑛𝑦𝑛

𝑖

𝑚

𝑖=2

− 𝑒‖ 

                                   

≤ ‖𝑥𝑛𝑘𝑙
− 𝑦𝑛𝑘𝑙

𝑚 ‖ + ‖𝘙𝑛0 𝑦𝑛𝑘𝑙
𝑜 − 𝑒‖ +    ‖∑ 𝘙𝑛𝑖  𝑅𝑒𝑠𝑖,𝑛𝑦𝑛𝑘𝑙

𝑖

𝑙

𝑖=2

− 𝑒‖                                

≤ ‖𝑥𝑛𝑘𝑙
− 𝑦𝑛𝑘𝑙

𝑚 ‖ + 𝘙𝑛0 ‖𝑦𝑛𝑘𝑙
𝑜 − 𝑒‖ +   ∑ 𝘙𝑛𝑖  

𝑙

𝑖=2

 ‖𝑅𝑒𝑠𝑖,𝑛𝑦𝑛𝑘𝑙
𝑖 − 𝑒‖ 

≤ ‖𝑥𝑛𝑘𝑙
− 𝑦𝑛𝑘𝑙

𝑚 ‖ + 𝘙𝑛0 ‖𝑦𝑛𝑘𝑙
𝑜 − 𝑒‖ +    ∑ 𝘙𝑛𝑖  

𝑙

𝑖=2

‖𝑦𝑛𝑘𝑙
𝑖 − 𝑒‖ 

 



IHJPAS. 36(2)2023 

372 
 

≤ ‖𝑥𝑛𝑘𝑙
− 𝑦𝑛𝑘𝑙

𝑚 ‖ + 𝘙𝑛0 ‖𝑦𝑛𝑘𝑙
1 − 𝑒‖ +    ∑ 𝘙𝑛𝑖  

𝑙

𝑖=1

‖𝑦𝑛𝑘𝑙
𝑖 − 𝑒‖ 

≤ ‖𝑥𝑛𝑘𝑙
− 𝑦𝑛𝑘𝑙

𝑚 ‖ + 𝘙𝑛0 ‖𝑦𝑛𝑘𝑙
1 − 𝑒‖ +    ∑ 𝘙𝑛𝑖  

𝑙

𝑖=2

‖𝑦𝑛𝑘𝑙
𝑖 − 𝑒‖ 

 

≤ ‖𝑥𝑛𝑘𝑙
− 𝑦𝑛𝑘𝑙

𝑚 ‖ + ∑ 𝘙𝑛𝑖  

𝑙

𝑖=0

‖𝑦𝑛𝑘𝑙 − 𝑒‖ 

‖𝑥𝑛𝑘𝑙
− 𝑒‖=‖𝑥𝑛𝑘𝑙

− 𝑦𝑛𝑘𝑙
𝑀 ‖ + ‖𝑦𝑛𝑘𝑙 − 𝑒‖                               

Therefore  

 

lim
𝑙→∞

‖𝑥𝑛𝑘𝑙
− 𝑒‖  ≤ lim

𝑙→∞
‖𝑦𝑛𝑘𝑙

𝑚 − 𝑒‖  

‖𝑦𝑛𝑘𝑙
𝑚 − 𝑒‖

2

= ‖𝘙𝑛0 𝑦𝑛𝑘𝑙
1 + ∑ 𝘙𝑛𝑖𝑘𝑙  𝑅𝑒𝑠𝑖,𝑛𝑦𝑛𝑘𝑙

𝑖

𝑙

𝑖=2

− 𝑒 ‖

2

= 

(1 − ∑ 𝛽𝑛𝑖𝑘𝑙

𝑚𝑙

𝑖=1

) ‖𝑦𝑛𝑘𝑙
1 − 𝑒‖

2

+ ∑ 𝘙𝑛𝑖𝑘𝑙

𝑙

𝑖=1

‖𝑅𝑒𝑠𝑖,𝑛𝑘𝑙
𝑦𝑛𝑘𝑙

𝑖 − 𝑒‖
2

− (1 − ∑ 𝘙𝑛𝑖𝑘𝑙

𝑙

𝑖=1

) ∑ 𝘙𝑛𝑖𝑘𝑙

𝑙

𝑖=1

‖𝑦𝑛𝑘𝑙
𝑖 − 𝑅𝑒𝑠𝑖,𝑛𝑘𝑙

𝑦𝑛𝑘𝑙
𝑖 ‖

2

 

     

                      ≤  (1 − ∑ 𝘙𝑛𝑖𝑘𝑙
𝑙
𝑖=1 + ∑ 𝘙𝑛𝑖𝑘𝑙

𝑙
𝑖=1 ) ‖𝑦𝑛𝑘𝑙

𝑖 − 𝑒‖
2

− (1 −

∑ 𝘙𝑛𝑖𝑘𝑙
𝑙
𝑖=1 ) ∑ 𝘙𝑛𝑖𝑘𝑙

𝑙
𝑖=1 ‖𝑦𝑛𝑘𝑙

𝑖 − 𝑅𝑒𝑠𝑖,𝑛𝑘𝑙
𝑦𝑛𝑘𝑙

1 ‖
2

 

 

≤  ‖𝑦𝑛𝑘𝑙
𝑖−1 − 𝑒‖

2

− (1 − ∑ 𝘙𝑛𝑖𝑘𝑙

𝑙

𝑖=1

) ∑ 𝘙𝑛𝑖𝑘𝑙

𝑙

𝑖=1

‖𝑦𝑛𝑘𝑙
𝑖 − 𝑅𝑒𝑠𝑖,𝑛𝑘𝑙

𝑦𝑛𝑘𝑙
𝑖 ‖

2

 

                 

≤  ‖𝑥𝑛𝑘𝑙
− 𝑒‖

2

− (1 − ∑ 𝘙𝑛𝑖𝑘𝑙

𝑙

𝑖=1

) ∑ 𝘙𝑛𝑖𝑘𝑙

𝑙

𝑖=1

‖𝑦𝑛𝑘𝑙
1 − 𝑅𝑒𝑠𝑖,𝑛𝑘𝑙

𝑦𝑛𝑘𝑙
𝑖 ‖

2

 

 

Hence,(1 − ∑ 𝘙𝑛𝑖𝑘𝑙
𝑙
𝑖=1 ) ∑ 𝘙𝑛𝑖𝑘𝑙

𝑙
𝑖=1 ‖𝑦𝑛𝑘𝑙

𝑖 − 𝑅𝑒𝑠𝑖,𝑛𝑘𝑙
𝑦𝑛𝑘𝑙

𝑖 ‖
2

 

≤  ‖𝑥𝑛𝑘𝑙
− 𝑒‖

2

− ‖𝑦𝑛𝑘𝑙
𝑖 − 𝑒‖

2

→ 0   ,  as 𝑙 → ∞ 

 

Therefore, we have    ‖𝑦𝑛𝑘𝑙
𝑖 − 𝑅𝑒𝑠𝑖,𝑛𝑘𝑙

𝑦𝑛𝑘𝑙
𝑖 ‖  → 0 .   which implies that 

 

lim
𝑛→∞

𝑠𝑢𝑝‖𝑦𝑛
𝑖 − 𝑅𝑒𝑠𝑖,𝑛𝑦𝑛

𝑖 ‖ = 0. Hence   ‖𝑦𝑛
𝑖 − 𝑅𝑒𝑠𝑖,𝑛𝑦𝑛

𝑖 ‖ → 0   ,  for all   i .       

 

As the same way, we obtain ‖𝑥𝑛 − 𝑅𝑒𝑠𝑖,𝑛𝑥𝑛‖ → 0.Next we show that lim
𝑛→∞

𝑠𝑢𝑝 < 𝑥 −

𝑃𝑟𝑜𝛹(𝑥) , 𝑥𝑛 − 𝑃𝑟𝑜𝛹(𝑥) >  ≤ 0. Let {𝑥𝑛𝑘 } be a subsequence of {𝑥𝑛} ⇀  𝑥
∗ and 𝑥 ∗ ∈ 𝑆 

    lim
𝑛→∞

𝑠𝑢𝑝 < 𝑣 − 𝑃𝑟𝑜𝛹(𝑣) , 𝑥𝑛 − 𝑃𝑟𝑜𝛹(𝑣) > = lim
𝑘→∞

< 𝑣 − 𝑃𝑟𝑜𝛹(𝑣) , 𝑥𝑛𝑘 − 𝑃𝑟𝑜𝛹(𝑣) >  . 

= < 𝑣 − 𝑃𝑟𝑜𝛹(𝑣), 𝑥
∗ − 𝑃𝑟𝑜𝛹(𝑣) ≤ 0. 



IHJPAS. 36(2)2023 

373 
 

Finally , we show that ‖𝑥𝑛 − 𝑃𝑟𝑜𝛹(𝑣)‖ → 0 . From lemma, We have  
‖𝑥𝑛+1 − 𝑃𝑟𝑜𝑠(𝑣)‖

2 = ‖𝘘𝑛𝑃𝑟𝑜𝛹𝑥 + (1 − 𝘘𝑛)𝑦𝑛
𝑚 − 𝑃𝑟𝑜𝛹 (𝑣)‖

2 

                            = ‖𝘘𝑛𝑃𝑟𝑜𝛹 𝑥 + (1 − 𝘘𝑛)𝑦𝑛
𝑚   − [(1 − 𝘘𝑛)𝑃𝑟𝑜𝛹(𝑣) − 𝘘𝑛 𝑃𝑟𝑜𝛹(𝑣)]‖

2 

                            ≤ ‖𝘘𝑛(𝑃𝑟𝑜𝑠𝑥 − 𝑃𝑟𝑜𝑠(𝑣) + (1 − 𝘘𝑛)(𝑦𝑛
𝑚 −  𝑃𝑟𝑜𝑠(𝑣)) ‖

2 

                            ≤  (1 − 𝘘𝑛)‖𝑦𝑛
𝑚 − 𝑃𝑟𝑜𝛹(𝑣)‖

2 + 𝘘𝑛 ‖ 𝑥 − 𝑃𝑟𝑜𝛹(𝑣)‖
2    

≤  (1 − 𝘘𝑛 )‖𝑥𝑛 − 𝑃𝑟𝑜𝛹(𝑣)‖
2 + 𝘘𝑛‖ 𝑥 − 𝑃𝑟𝑜𝛹(𝑣)‖

2           
That ‖𝑥𝑛+1 − 𝑃𝑟𝑜𝛹 (𝑣)‖ → 0 𝑎𝑠 𝑛 → ∞.    This completes the proof . 
Corollary : 2.2 .If  ℳ𝑖  are M.M. operators such that  

                                                𝐹(𝑃𝑟𝑜𝑠) ⋂ ℳ𝑖
−1

0 ≠ ∅𝑙𝑖=1  
Under (i-ii) in theorem (3.1). Then {𝑥𝑛} defined by  𝑥 ∈ 𝐶 and  

   {

      𝑦𝑛
𝑖−1 = 𝘚𝑛𝑖 (𝑥𝑛 ) + (1 − 𝘚𝑛𝑖 ) (𝑥𝑛)                                                                                

𝑦𝑛
𝑚 = 𝘙𝑛1 𝑦𝑛

0 + ∑ 𝘙𝑛𝑖  𝑅𝑒𝑠𝑖,𝑛𝑦𝑛
𝑖−1𝑙

𝑖=2                                                                   

𝑥𝑛+1 = 𝘘𝑛𝑃𝑟𝑜𝛹𝑥 + (1 − 𝘘𝑛)𝑦𝑛
𝑚, 𝑛, 𝑚 ≥ 0                                               

 

     Converges- s to 𝑃𝑟𝑜𝛹𝑥 . 

Corollary : 2.3 .If  ℳ𝑖  are M.M. operators such that ℳ𝑖
−1

0 ≠ ∅.Under (i-ii) in theorem (3.1).    
Then the sequences {𝑥𝑛} defined by  𝑥 ∈ 𝐶 and  

   {

      𝑦𝑛
𝑖−1 = 𝘚𝑛𝑖 (𝑥𝑛 ) + (1 − 𝘚𝑛𝑖 ) (𝑥𝑛)                                                                                

𝑦𝑛
𝑚 = 𝘙𝑛1 𝑦𝑛

0 + ∑ 𝘙 𝑅𝑒𝑠𝑖,𝑛𝑦𝑛
𝑖−1𝑙

𝑖=2    

𝑥𝑛+1 = 𝘘𝑛 𝑥 + (1 − 𝘘𝑛)𝑦𝑛
𝑚, 𝑛, 𝑚 ≥ 0                                               

  

   s-converges  to 𝑥 . 
 

3. Conclusion  

In this article, a new iterative methods of  approximation fixed point  are presented. On the other 

hand, the s-convergence by using multivalued operators  also proven.  

 

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