Algorithms for Solving the Resolvent of the Sum of Two Maximal Monotone Operators with a Finite Family of Nonexpansive Operators

Abstract

In this paper, we address a variational problem involving the sum of two maximal monotone operators combined with a finite family of nonexpansive operators. To solve this problem, we propose iterative algorithms based on single-valued mappings. First, we examine cases involving two or three maximal monotone operators, introducing novel algorithms to obtain their solutions. Secondly, we extend our analysis by applying the Ishikawa iterative scheme within the framework of fixed-point theory. This allows us to establish strong convergence results. Finally, we provide an illustrative example to demonstrate the effectiveness and applicability of the proposed methods.

1. Introduction

Convex analysis, monotone operator theory, and the theory of nonexpansive mappings constitute the foundational pillars of nonlinear analysis; they are intricately connected through their shared mathematical structures. A wide range of practical minimization and variational problems can be elegantly reformulated as monotone inclusion problems, underscoring the unifying role these theories play in addressing complex mathematical challenges (see [1,2,3,4,5]).

One of the most classical problems in this framework is given as follows:

$$\mathrm{Find}x\in \mathbb{H}\mathrm{such}\mathrm{that}0\in A\left(x\right),$$

where $\mathbb{H}$ is a real Hilbert space endowed with an inner product $\langle ·,·\rangle$ and associated norm $\parallel ·\parallel$, and A is a maximal monotone operator. This problem has attracted the attention of numerous researchers, notably Rockafellar (1976) (see [6]), who developed several iterative algorithms to approximate a solution set.

The set of zeros for operator A can be equivalently characterized as follows:

$$S=\{x\in \mathbb{H}\mid J_{\lambda }^A\left(x\right)=x\},$$

where for every $\lambda >0$, the operator

$$J_{\lambda }^A:={(I+\lambda A)}^{-1}$$

denotes the resolvent of A. Among the various algorithms proposed to approximate points in S, the well-known Mann iterative process (see [7,8]) is defined by the following:

$$\left\{\begin{array}{c}{x}_0\in \mathbb{H},\hfill \\ x_{k+1}=(1-a_k)x_k+a_k{J}_{\lambda }^A\left(x_k\right),k\ge 0,\hfill \end{array}\right.$$

and, under suitable assumptions of the sequence $\left\{a_k\right\}$, it converges strongly to a fixed point of $J_{\lambda }^A$, which corresponds to a solution of the first inclusion problem defined at the beginning of this paper.

In this work, we extend this line of research by studying a more general problem formulated as follows:

$$0\in A\left(x\right)+B\left(x\right)+\sum _{i=1}^n{C}_i\left(x\right),$$

where $A:\mathbb{H}\rightrightarrows \mathbb{H}$ is an $\alpha$-inverse strongly monotone operator; $B:\mathbb{H}\rightrightarrows \mathbb{H}$ is a $\beta$-strongly monotone operator; and ${\left\{C_i\right\}}_{1\le i\le n}$ represents nonexpansive mappings on $\mathbb{H}$.

If $C_i=0$ for all $1\le i\le n$, and both A and B are maximal monotone operators defined on a Hilbert space, the problem reduces to the well-studied case addressed by many authors (see [1,9,10,11,12]) using the Douglas–Rachford iterative algorithm (DRIA), defined by the following:

$$Z_{n+1}=J_{\lambda }^B\left(2J_{\lambda }^A-I\right)\left(Z_n\right)+\left(I-J_{\lambda }^B\right)\left(Z_n\right).$$

A. Beddani further introduced an alternative approach for finding ${(A+B)}^{-1}\left(0\right)$ (see [13,14]), through the following function:

$$f_{\lambda }:{\mathbb{R}}^2\to {\mathbb{R}}^2,f_{\lambda }(x,y)=\left(\begin{array}{c}{J}_{\lambda }^A\left(x\right)-{\displaystyle \frac{x+y}{2}}\\ J_{\lambda }^B\left(y\right)-{\displaystyle \frac{x+y}{2}}\end{array}\right).$$

If $(x,y)\in {\mathbb{R}}^2$ exists such that $\parallel f_{\lambda }(x,y)\parallel =0$, then $0\in A\left(J_{\lambda }^{A}x\right)+B\left(J_{\lambda }^{A}x\right)$.

In our study, we adopt and generalize the Douglas–Rachford Splitting Algorithm (DRSA) to address problems involving both monotone and nonexpansive operators (see [15,16]). We introduce a new operator defined below,

$${\Psi }_{\lambda }\left(x\right)=J_{\lambda }^B\left(-\lambda \sum _{i=1}^n{C}_i\left(J_{\lambda }^{A}x\right)-x+2J_{\lambda }^{A}x\right)+\lambda A_{\lambda }\left(x\right),$$

and based on this formulation, we propose several iterative schemes. One of them is the Ishikawa-type iterative sequence (see [17]), which ensures strong convergence under appropriate conditions.

To establish these results, we first recall some preliminary concepts from convex analysis and monotone operator theory, which serve as the foundation for the proposed algorithms.

2. Preliminaries

2.1. Operators and Monotonicity

Let $\mathbb{H}$ be a real Hilbert space, and let $A:\mathbb{H}\to 2^{\mathbb{H}}$ be a set-valued operator. We denote the domain of A by $\mathrm{dom}\left(A\right)$ by the following:

$$\mathrm{dom}\left(A\right)=\{x\in \mathbb{H}:A(x)\ne \varnothing \}.$$

We state that A has the full domain if $\mathrm{dom}\left(A\right)=\mathbb{H}$. The range of A is defined as follows:

$$\mathrm{Im}\left(A\right)=\{y\in \mathbb{H}:\exists x\in \mathbb{H},y\in A(x\left)\right\}.$$

The graph of A is given by the following:

$$\mathrm{gph}\left(A\right):=\left\{\right(x,y)\in \mathbb{H}\times \mathbb{H}:x\in \mathrm{dom}(A),y\in A(x\left)\right\}.$$

Let ${\left\{A_i\right\}}_{i=1}^n$ be a finite family of operators. Then the sum operator is defined as follows:

$$\left(\sum _{i=1}^n{A}_i\right)\left(x\right):=\left\{\sum _{i=1}^n{y}_i:y_i\in A_i\left(x\right),i=1,\dots ,n\right\}.$$

Definition 1

([18]). The operator A is said to be monotone if

$$〈y_1-y_2,x_1-x_2〉\ge 0forall(x_i,y_i)\in \mathrm{gph}\left(A\right),i=1,2.$$

Definition 2

([18]). The operator A is said to be $\alpha$-strongly monotone with $\alpha >0$ if 

$$\langle y_1-y_2,x_1-x_2\rangle \ge \alpha {\parallel x_1-x_2\parallel }^{2}forall(x_i,y_i)\in \mathrm{gph}\left(A\right),i=1,2.$$

An operator A is said to be nonexpansive if

$$\parallel y_1-y_2\parallel \le \parallel x_1-x_2\parallel forall(x_i,y_i)\in \mathrm{gph}\left(A\right),i=1,2.$$

Proposition 1

([1]). For all $\alpha >0$ and $\lambda >0$, the following is true:

$\left(1\right)$

If A is α-strongly monotone, then $J_{\lambda }^A$ is ${\displaystyle \frac{1}{\alpha \lambda +1}}$-Lipschitz continuous.

$\left(2\right)$

If A is α-inverse strongly monotone, then the Yosida approximation $A_{\lambda }$ is ${\displaystyle \frac{1}{\alpha +\lambda }}$-Lipschitz continuous.

Proposition 2

([19]). $A_{\lambda }\left(y\right)\in A\left(J_{\lambda }^A\left(y\right)\right),\forall y\in \mathbb{H}.$

Proposition 3

([19]). For any $\lambda >0$, $\delta >0$, we have

$$J_{\lambda }^A\left(x\right)=J_{\delta }^A\left({\displaystyle \frac{\delta }{\lambda }}x+\left(1-{\displaystyle \frac{\delta }{\lambda }}\right)J_{\lambda }^A\left(x\right)\right).$$

2.2. Maximal Monotone Operators and Convex Functions

Operator A is said to be maximal monotone if it satisfies the following:

• A is monotone.
• If B is another monotone operator such that the graph of A (i.e., the set of all pairs $(x,A(x\left)\right)$) is contained in the graph of B, then $B=A$.

Let $X,Y\subseteq \mathbb{H}$ be convex subsets of a Hilbert space $\mathbb{H}$, and let $f:X\to \mathbb{R}$ be a function.

Definition 3

([15]). A function f is said to be convex if

$$\forall x,y\in Xandforall\lambda \in [0,1],f\left(\right(1-\lambda )x+\lambda y)\le (1-\lambda )f\left(x\right)+\lambda f\left(y\right).$$

If the inequality is strict for all $x\ne y$, then f is called strictly convex. Moreover, f is said to be α-strongly convex if

$$f((1-\lambda )x+\lambda y)\le (1-\lambda )f\left(x\right)+\lambda f\left(y\right)-{\displaystyle \frac{\lambda (1-\lambda )}{2}}{\parallel x-y\parallel }^2.$$

Definition 4

([15]). Let $f:\mathbb{H}\to \mathbb{R}$ be a convex function. The set

$$\mathsf{\partial }f\left(x\right):=\left\{g\in \mathbb{H}:\langle g,y-x\rangle \le f\left(y\right)-f\left(x\right)forally\in \mathbb{H}\right\}$$

is called the subdifferential of f at x. The function f is said to be subdifferentiable at x if $\mathsf{\partial }f\left(x\right)\ne \varnothing$. An element of the subdifferential is called a subgradient.

A classical example of a maximal monotone operator is given by the subdifferential of the function $f\left(x\right)=\left|x\right|$ defined on $\mathbb{R}$. The subdifferential of f is expressed as follows:

$$\mathsf{\partial }f\left(x\right)=\left\{\begin{array}{cc}-1,\hfill & \mathrm{if}x<0,\hfill \\ [-1,1],\hfill & \mathrm{if}x=0,\hfill \\ 1,\hfill & \mathrm{if}x>0.\hfill \end{array}\right.$$

It can be verified that no monotone operator on $\mathbb{R}$ properly contains $\mathsf{\partial }f$. Consequently, $\mathsf{\partial }f$ is a maximal monotone operator.

Lemma 1

([20]). Given any maximal monotone operator A, a real number $\lambda >0$, and $x\in \mathbb{H}$, we have $0\in A\left(x\right)$ if and only if $J_{\lambda }^A\left(x\right)=x$.

Lemma 2

([17]). Let a real sequence ${\left\{x_k\right\}}_{k=1}^{\infty }$ satisfy the following condition:

$$x_{k+1}\le \sigma x_k+{\rho }_k$$

where $x_k\ge 0$, ${\rho }_k\ge 0$, and ${\displaystyle \underset{k\to \infty }{lim}{\rho }_k=0}$, $0\le \sigma <1$. Then, ${\displaystyle \underset{k\to \infty }{lim}{x}_k=0}$.

3. Main Result

In this section, we present our main results related to the problem under consideration. Our objective is to address and solve various cases of the following monotone inclusion problem:

$$0\in A\left(x\right)+B\left(x\right)+\sum _{i=1}^n{C}_i\left(x\right),$$

(1)

where $A:\mathbb{H}\to 2^{\mathbb{H}}$ is $\alpha$-inverse strongly monotone operator; $B:\mathbb{H}\to 2^{\mathbb{H}}$ is $\beta$-strongly monotone; and ${\left\{C_i\right\}}_{1\le i\le n}$ represents a finite family of nonexpansive operators $C_i:\mathbb{H}\to \mathbb{H}$.

Let us define the operator $F\left({\Psi }_{\lambda }\right)=\{x^{*}\in \mathbb{H}\mid {\Psi }_{\lambda }\left(x^{*}\right)=x^{*}\}$, and S as follows:

$$S\left(x\right)=\left\{x\in \mathbb{H}\mid 0\in A\left(x\right)+B\left(x\right)+\sum _{i=1}^n{C}_i\left(x\right)\right\}.$$

First, we aim to study the problem defined by the sum of two maximally monotone opertaors, A and B, both defined on a Hilbert space $\mathbb{H}$. This problem can be formulated as follows. Find element $x\in \mathbb{H}$ such that

$$0\in A\left(x\right)+B\left(x\right).$$

(2)

So, let us propose a simple algorithm using the Yosida approximation, which can be used to solve (2).

Proposition 4.

For any $\delta >0$, $\lambda >0$, we have $A_{\lambda }\left(x\right)=A_{\delta }(x+\delta A_{\lambda }\left(x\right)).$

Proof. 

Let $\delta >0$, $\lambda >0$, and we have

$$J_{\lambda }^A\left(x\right)=J_{\delta }^A\left({\displaystyle \frac{\delta }{\lambda }}x+\left(1-{\displaystyle \frac{\delta }{\lambda }}\right)J_{\lambda }^A\left(x\right)\right),$$

which is equivalent to

$$x-\lambda A_{\lambda }\left(x\right)={\displaystyle \frac{\delta }{\lambda }}x+\left(1-{\displaystyle \frac{\delta }{\lambda }}\right)(x-\lambda A_{\lambda }\left(x\right))-\delta A_{\delta }\left({\displaystyle \frac{\delta }{\lambda }}x+\left(1-{\displaystyle \frac{\delta }{\lambda }}\right)(x-\lambda A_{\lambda }\left(x\right))\right),$$

after simplifying

$$x-\lambda A_{\lambda }\left(x\right)={\displaystyle \frac{\delta }{\lambda }}x+x-{\displaystyle \frac{\delta }{\lambda }}x-\lambda A_{\lambda }\left(x\right)+\delta A_{\lambda }\left(x\right)-\delta A_{\delta }\left({\displaystyle \frac{\delta }{\lambda }}x+\left(1-{\displaystyle \frac{\delta }{\lambda }}\right)(x-\lambda A_{\lambda }\left(x\right))\right).$$

Therefore,

$$-\lambda A_{\lambda }\left(x\right)=-\lambda A_{\lambda }\left(x\right)+\delta A_{\lambda }\left(x\right)-\delta A_{\delta }\left({\displaystyle \frac{\delta }{\lambda }}x+\left(1-{\displaystyle \frac{\delta }{\lambda }}\right)(x-\lambda A_{\lambda }\left(x\right))\right),$$

so we conclude that

$$A_{\lambda }\left(x\right)=A_{\delta }\left((x+\delta A_{\lambda }\left(x\right))\right).$$

This completes the proof. □

Proposition 5.

Let $\lambda >0$ and define the operator ${\theta }_{\lambda }:\mathbb{H}\to \mathbb{H}$ by

$${\theta }_{\lambda }\left(x\right)=x+A_{\lambda }\left(x\right)+B_{\lambda }\left(x-2\lambda A_{\lambda }\left(x\right)\right).$$

If $x^{★}$ is a fixed point of ${\theta }_{\lambda }$, then $x^{★}-\lambda A_{\lambda }\left(x^{★}\right)$ is a solution of the monotone inclusion problem (2).

Proof. 

Assume that $x^{*}$ is a fixed point of ${\theta }_{\lambda }$. By definition, this implies the following:

$$x^{*}+A_{\lambda }\left(x^{*}\right)+B_{\lambda }(x^{*}-2\lambda A_{\lambda }\left(x^{*}\right))=x^{*}.$$

Subtracting $x^{*}$ from both sides, we obtain

$$A_{\lambda }\left(x^{*}\right)+B_{\lambda }(x^{*}-2\lambda A_{\lambda }\left(x^{*}\right))=0.$$

Let us define $z=x^{*}-\lambda A_{\lambda }\left(x^{*}\right)$. Then,

$$x^{*}=z+\lambda A_{\lambda }\left(x^{*}\right)\mathrm{and}{x}^{*}-2\lambda A_{\lambda }\left(x^{*}\right)=z-\lambda A_{\lambda }\left(x^{*}\right).$$

Substituting these into the previous equation gives the following:

$$A_{\lambda }\left(x^{*}\right)+B_{\lambda }(z-\lambda A_{\lambda }\left(x^{*}\right))=0.$$

Thus, we have

$$0\in A\left(z\right)+B\left(z\right),$$

which means that z is a solution of the monotone inclusion (2), as claimed. □

Theorem 1.

For all $\lambda >0$, the sequence $\left\{x_k\right\}$ is defined as

$$\left\{\begin{array}{c}{x}_0,x_1\in \mathbb{H},\hfill \\ x_{k+1}=(1-\alpha )x_k+\alpha {\theta }_{\lambda }\left(x_k\right)+{ϵ}_k(x_k-x_{k-1}),k\ge 0.\hfill \end{array}\right.$$

where $0<\alpha <1$, ${ϵ}_k\in ]0,1[$ and ${\sum }_{i=1}^{\infty }{ϵ}_k<\infty$. If $\left\{x_k\right\}$ converge to l then $l-\lambda A_{\lambda }\left(l\right)$ solve (2).

We now turn to the study of the principal problem (1).

Theorem 2.

For all $\lambda >0$, if ${\Psi }_{\lambda }\left(x\right)=J_{\lambda }^B(-\lambda {\sum }_{i=1}^n{C}_i\left(J_{\lambda }^A\left(x\right)\right)-x+2J_{\lambda }^A\left(x\right))+\lambda A_{\lambda }\left(x\right)$ if $S\ne \varnothing$, then $\left\{J_{\lambda }^A\left(F\left({\Psi }_{\lambda }\right)\right)\right\}\subseteq S$.

Proof. 

Let $x^{*}$ be a fixed point of ${\Psi }_{\lambda }$, so

$$\begin{array}{cc}\hfill x^{*}\in F({\Psi }_{\lambda })& \Rightarrow {\Psi }_{\lambda }(x^{*})=x^{*}\hfill \\ & \Rightarrow J_{\lambda }^B(-\lambda \sum _{i=1}^n{C}_i(J_{\lambda }^A(x^{*}))-x^{*}+2J_{\lambda }^A(x^{*}))+\lambda A_{\lambda }(x^{*})=x^{*}\hfill \\ \hfill & \Rightarrow x^{*}-\lambda A_{\lambda }(x^{*})=J_{\lambda }^B(-\lambda \sum _{i=1}^n{C}_i(J_{\lambda }^A(x^{*}))-x^{*}+2J_{\lambda }^A(x^{*}))\hfill \\ \hfill & \Rightarrow J_{\lambda }^A(x^{*})=J_{\lambda }^B(-\lambda \sum _{i=1}^n{C}_i(J_{\lambda }^A(x^{*}))-x^{*}+2J_{\lambda }^A(x^{*}))\hfill \\ \hfill & \Rightarrow -\lambda \sum _{i=1}^n{C}_i(J_{\lambda }^A(x^{*}))-x^{*}+2J_{\lambda }^A(x^{*})\in \lambda B(J_{\lambda }^A(x^{*}))+J_{\lambda }^A(x^{*})\hfill \\ \hfill & \Rightarrow -x^{*}+J_{\lambda }^A(x^{*})\in \lambda \sum _{i=1}^n{C}_i(J_{\lambda }^A(x^{*}))+\lambda B(J_{\lambda }^A(x^{*}))\hfill \\ \hfill & \mathrm{and}{x}^{*}-J_{\lambda }^A(x^{*})\in \lambda A(J_{\lambda }^A(x^{*}))\hfill \\ \hfill & \Rightarrow 0\in \lambda \sum _{i=1}^n{C}_i(J_{\lambda }^A(x^{*}))+\lambda B(J_{\lambda }^A(x^{*}))+\lambda A(J_{\lambda }^A(x^{*}))\hfill \\ \hfill & \Rightarrow J_{\lambda }^A(x^{*})\in \mathrm{zer}(\sum _{i=1}^n{C}_i+A+B).\hfill \end{array}$$

This completes the proof. □

3.1. Algorithm 1

In this algorithm, we impose an additional condition on the family ${\left\{C_i\right\}}_{1\le i\le n}$, which is that the operators $I+\lambda C_i$ must be bijective.

Proposition 6.

For all $\lambda >0$, if $F\left({\Psi }_{\lambda }\right)\ne \varnothing$, then the system defined as follows:

$$\left\{\begin{array}{c}{J}_{\lambda }^A\left(x\right)={\displaystyle \frac{x+y+{\sum }_{i=1}^{i=n}{z}_i}{n+2}}\hfill \\ J_{\lambda }^B\left(y\right)={\displaystyle \frac{x+y+{\sum }_{i=1}^{i=n}{z}_i}{n+2}}\hfill \\ J_{\lambda }^{C_1}\left(z_1\right)={\displaystyle \frac{x+y+{\sum }_{i=1}^{i=n}{z}_i}{n+2}}\hfill \\ .\hfill \\ .\hfill \\ .\hfill \\ J_{\lambda }^{C_n}\left(z_n\right)={\displaystyle \frac{x+y+{\sum }_{i=1}^{i=n}{z}_i}{n+2}}.\hfill \end{array}\right.$$

(3)

which has the solution $(x,y,z_1,\dots ,z_n)\in {\mathbb{H}}^{n+2}$.

Proof. 

Let $F\left({\Psi }_{\lambda }\right)\ne \varnothing$, then exist $x^{★}\in \mathbb{H}$ such that ${\Psi }_{\lambda }\left(x^{★}\right)=x^{★}$; this implies

$$J_{\lambda }^A\left(x^{★}\right)=J_{\lambda }^B(-\lambda \sum _{i=1}^n{C}_i\left(J_{\lambda }^A\left(x^{★}\right)\right)-x^{★}+2J_{\lambda }^A\left(x^{★}\right)).$$

Let us propose,

$$\left\{\begin{array}{c}{x}^{★}=x,\hfill \\ -\lambda \sum _{i=1}^n{C}_i\left(J_{\lambda }^A\left(x^{★}\right)\right)-x^{★}+2J_{\lambda }^A\left(x^{★}\right))=y,\hfill \\ J_{\lambda }^A\left(x^{★}\right)+\lambda C_i\left(J_{\lambda }^A\left(x^{★}\right)\right)=z_i.\hfill \end{array}\right.$$

Therefore,

$$\left\{\begin{array}{c}{J}_{\lambda }^A\left(x\right)=J_{\lambda }^B\left(y\right),\hfill \\ J_{\lambda }^A\left(x\right)=J_{\lambda }^{C_i}\left(z_i\right),\hfill \\ x+y+\sum _{i=1}^{i=n}{z}_i=(n+2)J_{\lambda }^A\left(x\right).\hfill \end{array}\right.$$

Consequently,

$$\left\{\begin{array}{c}{J}_{\lambda }^A\left(x\right)={\displaystyle \frac{x+y+{\sum }_{i=1}^{i=n}{z}_i}{n+2}},\hfill \\ J_{\lambda }^B\left(y\right)={\displaystyle \frac{x+y+{\sum }_{i=1}^{i=n}{z}_i}{n+2}},\hfill \\ J_{\lambda }^{C_1}\left(z_1\right)={\displaystyle \frac{x+y+{\sum }_{i=1}^{i=n}{z}_i}{n+2}},\hfill \\ .\hfill \\ .\hfill \\ .\hfill \\ J_{\lambda }^{C_n}\left(z_n\right)={\displaystyle \frac{x+y+{\sum }_{i=1}^{i=n}{z}_i}{n+2}}.\hfill \end{array}\right.$$

This implies that $(x,y,z_1,\dots ,z_n)$ is the solution of (3). □

Theorem 3.

For all $\lambda >0$, the system of sequences is defined as follows:

$$\left\{\begin{array}{c}(x_0,y_0,{z_1}_0,\dots ,{z_n}_0)\in {\mathbb{H}}^{n+2},\hfill \\ x_{k+1}=(n+2)J_{\lambda }^A\left(x_k\right)-y_k-\sum _{i=1}^{i=n}{z_i}_k,\hfill \\ y_{k+1}=(n+2)J_{\lambda }^A\left(y_k\right)-x_k-\sum _{i=1}^{i=n}{z_i}_k,\hfill \\ {z_i}_{k+1}=J_{\lambda }^A\left(x_k\right)+\lambda C_i\left(J_{\lambda }^A\left(x_k\right)\right),\hfill \end{array}\right.k\ge 0.$$

Converge to $(x^{★},y^{★},z_1^{★},\dots ,z_n^{★})$ then $J_{\lambda }^A\left(x^{★}\right)$ is the solution of (1).

Proof. 

Let us assume that the last system converges to $(x^{★},y^{★},z_1^{★},\dots ,z_n^{★})$ in ${\mathbb{H}}^{n+2}$ so we have

$$\left\{\begin{array}{c}{x}^{★}=(n+2)J_{\lambda }^A\left(x^{★}\right)-y_k-\sum _{i=1}^{i=n}{z_i}^{★},\hfill \\ y^{★}=(n+2)J_{\lambda }^A\left(y^{★}\right)-x^{★}-\sum _{i=1}^{i=n}{z_i}^{★},\hfill \\ {z_i}^{★}=J_{\lambda }^A\left(x^{★}\right)+\lambda C_i\left(J_{\lambda }^A\left(x^{★}\right)\right).\hfill \end{array}\right.$$

Therefore,

$$\left\{\begin{array}{c}{x}^{★}+y^{★}+\sum _{i=1}^{i=n}{z_i}^{★}=(n+2)J_{\lambda }^A\left(x^{★}\right),\hfill \\ y^{★}+x^{★}+\sum _{i=1}^{i=n}{z_i}^{★}=(n+2)J_{\lambda }^A\left(y^{★}\right),\hfill \\ {z_i}^{★}=J_{\lambda }^A{x}^{★}+\lambda C_i\left(J_{\lambda }^A\left(x^{★}\right)\right).\hfill \end{array}\right.$$

Then,

$$x^{★}+y^{★}+n{J}_{\lambda }^A{x}^{★}+\lambda \sum _{i=1}^{i=n}{C}_i\left(J_{\lambda }^A{x}^{★}\right)=(n+2)J_{\lambda }^A\left(x^{★}\right),$$

After simplifying,

$$y^{★}=2J_{\lambda }^A{x}^{★}-\lambda \sum _{i=1}^{i=n}{C}_i\left(J_{\lambda }^A{x}^{★}\right)-x^{★}$$

We have also, $J_{\lambda }^A\left(x^{★}\right)=J_{\lambda }^B\left(y^{★}\right)$.

Consequently,

$$J_{\lambda }^A\left(x^{★}\right)=J_{\lambda }^B(-\lambda \sum _{i=1}^n{C}_i\left(J_{\lambda }^A\left(x^{★}\right)\right)-x^{★}+2J_{\lambda }^A{x}^{★}).$$

So we conclude that $x^{★}$ is the fixed point of ${\Psi }_{\lambda }$, which proves that $J_{\lambda }^A\left(x^{★}\right)$ is the solution of (1). □

Below we will examine the case when $n=1$, so the problem (1) is defined as follows. Find an element x in the Hilbert space $\mathbb{H}$ such that

$$0\in A\left(x\right)+B\left(x\right)+C\left(x\right)$$

(4)

where A and B are two maximal monotone operators defined on Hilbert space $\mathbb{H}$ and C is a nonexpansive single valued mapping also defined on $\mathbb{H}$. Then the algorithm is defined as follows:

$$\left\{\begin{array}{c}(x_0,y_0,z_0)\in {\mathbb{H}}^3,\hfill \\ x_{k+1}=3J_{\lambda }^A\left(x_k\right)-y_k-z_k,\hfill \\ y_{k+1}=3J_{\lambda }^A\left(y_k\right)-x_k-z_k,\hfill \\ z_{k+1}=J_{\lambda }^A\left(x_k\right)+\lambda C\left(J_{\lambda }^A\left(x_k\right)\right),\hfill \end{array}\right.k\ge 0.$$

3.2. Algorithm 2

Proposition 7.

Let ${\left\{C_i\right\}}_{1\le i\le n}$ be a finite family of nonexpansive operators defined on H, where A is α-inverse strongly monotone operator and B is a β-strongly monotone operator defined on a real Hilbert space. For all $\alpha >0$, $\beta >0$ and $\lambda >0$, ${\Psi }_{\lambda }$ is a L-lipschitzian operator where

$$L={\displaystyle \frac{n{\lambda }^2+(\alpha n+2)\lambda +\alpha }{(\beta \lambda +1)(\lambda +\alpha )}}+{\displaystyle \frac{\lambda }{\lambda +\alpha }}.$$

Proof. 

$$\begin{array}{cc}\hfill \parallel {\Psi }_{\lambda }\left(x\right)-{\Psi }_{\lambda }\left(y\right)\parallel & =∥J_{\lambda }^B\left(-\lambda \sum _{i=1}^n{C}_i\left(J_{\lambda }^{A}x\right)-x+2J_{\lambda }^{A}x\right)\hfill \\ \hfill & -J_{\lambda }^B\left(-\lambda \sum _{i=1}^n{C}_i\left(J_{\lambda }^{A}y\right)-y+2J_{\lambda }^{A}y\right)-\lambda A_{\lambda }\left(x\right)+\lambda A_{\lambda }\left(y\right)∥\hfill \\ \hfill & <∥J_{\lambda }^B\left(-\lambda \sum _{i=1}^n{C}_i\left(J_{\lambda }^{A}x\right)-(x-J_{\lambda }^{A}x)\right)\hfill \\ \hfill & -J_{\lambda }^B\left(-\lambda \sum _{i=1}^n{C}_i\left(J_{\lambda }^{A}y\right)-(y-J_{\lambda }^{A}y)\right)∥+{\displaystyle \frac{\lambda }{\lambda +\alpha }}\parallel x-y\parallel \hfill \\ \hfill & <{\displaystyle \frac{1}{\beta \lambda +1}}∥-\lambda \sum _{i=1}^n{C}_i\left(J_{\lambda }^{A}x\right)-\lambda A_{\lambda }\left(x\right)-J_{\lambda }^{A}x\hfill \\ \hfill & +\lambda \sum _{i=1}^n{C}_i\left(J_{\lambda }^{A}y\right)+\lambda A_{\lambda }\left(y\right)+J_{\lambda }^{A}y∥+{\displaystyle \frac{\lambda }{\lambda +\alpha }}\parallel x-y\parallel \hfill \\ \hfill & <{\displaystyle \frac{1}{\beta \lambda +1}}\left[\lambda \sum _{i=1}^n\parallel C_i\left(J_{\lambda }^{A}x\right)-C_i\left(J_{\lambda }^{A}y\right)\parallel +\left({\displaystyle \frac{\lambda }{\lambda +\alpha }}+1\right)\parallel x-y\parallel \right]\hfill \\ \hfill & +{\displaystyle \frac{\lambda }{\lambda +\alpha }}\parallel x-y\parallel \hfill \\ \hfill & <{\displaystyle \frac{1}{\beta \lambda +1}}\left[\lambda \sum _{i=1}^n\parallel J_{\lambda }^{A}x-J_{\lambda }^{A}y\parallel +\left({\displaystyle \frac{\lambda }{\lambda +\alpha }}+1\right)\parallel x-y\parallel \right]\hfill \\ \hfill & +{\displaystyle \frac{\lambda }{\lambda +\alpha }}\parallel x-y\parallel \hfill \\ \hfill & <{\displaystyle \frac{1}{\beta \lambda +1}}\left(\lambda n+{\displaystyle \frac{2\lambda +\alpha }{\lambda +\alpha }}\right)\parallel x-y\parallel +{\displaystyle \frac{\lambda }{\lambda +\alpha }}\parallel x-y\parallel \hfill \\ \hfill & <\left[{\displaystyle \frac{n{\lambda }^2+(\alpha n+2)\lambda +\alpha }{(\beta \lambda +1)(\lambda +\alpha )}}+{\displaystyle \frac{\lambda }{\lambda +\alpha }}\right]\parallel x-y\parallel .\hfill \end{array}$$

□

Theorem 4.

For all λ, α and β are positive real numbers if $\beta >n$, $\alpha >{\displaystyle \frac{2}{\beta -n}}$ and $0<\lambda <{\displaystyle \frac{(\beta -n)\alpha -2}{n}}$, where $n\in {\mathbb{N}}^{*}$. Then, ${\Psi }_{\lambda }$ is contractive mapping.

Proof. 

If ${\Psi }_{\lambda }$ is contractive, then the inequality satisfies the following:

$$\begin{array}{cc}\hfill {\displaystyle \frac{n{\lambda }^2+(\alpha n+2)\lambda +\alpha }{(\beta \lambda +1)(\lambda +\alpha )}}+{\displaystyle \frac{\lambda }{\lambda +\alpha }}<1& \iff {\displaystyle \frac{(n+\beta ){\lambda }^2+(3+\alpha n)\lambda +\alpha }{(\beta \lambda +1)(\alpha +\lambda )}}-1<0\hfill \\ \hfill & \iff {\displaystyle \frac{n{\lambda }^2+(2+\alpha n-\alpha \beta )\lambda }{(\beta \lambda +1)(\alpha +\lambda )}}<0.\hfill \end{array}$$

After simplification, the next step is to solve the resulting polynomial inequality involving parameters $\alpha$, n, and $\beta$:

$$n{\lambda }^2+(2+\alpha n-\alpha \beta )\lambda <0,$$

where we have

$$n{\lambda }^2+(2+\alpha n-\alpha \beta )\lambda <0\iff \lambda (n\lambda +2+n\alpha -\alpha \beta )<0.$$

This implies

$$\left\{\begin{array}{c}\lambda >0,\hfill \\ n\lambda +2+n\alpha -\alpha \beta <0,n>0.\hfill \end{array}\right.$$

Hence, we deduce the following conditions:

$$\left\{\begin{array}{c}\beta >n,\hfill \\ \alpha >{\displaystyle \frac{2}{\beta -n}},\hfill \\ 0<\lambda <{\displaystyle \frac{\alpha \beta -2-n\alpha }{n}},\hfill \end{array}\right.n>0.$$

This completes the proof. □

In this part, we modify the Ishikawa algorithm to achieve faster convergence of our sequence $\left\{{\Psi }_{\lambda }{x}_k\right\}$. Accordingly, we present the following theorem that defines the modified algorithm:

Theorem 5.

Let $\mathbb{H}$ be a real Hilbert space and $\mathbb{C}$ be a closed convex subspace of $\mathbb{H}$. Let ${\Psi }_{\lambda }:\mathbb{C}\to \mathbb{C}$ be contractive mapping. Let $\left\{x_k\right\}$ be a sequence defined iteratively for each integer $k\ge 0$ by

$$\left\{\begin{array}{c}{x}_0\in \mathbb{H},\hfill \\ x_{k+1}=a_k{x}_k+b_k{\Psi }_{\lambda }{y}_k,\hfill \\ y_k=c_k{x}_k+d_k{\Psi }_{\lambda }{x}_k,\hfill \end{array}\right.k\ge 0,$$

where $\left\{a_k\right\}$ and $\left\{b_k\right\}$ are sequences of positive numbers satisfying the following conditions:

• $0\le d_k\le b_k<1$,
• $b_k+a_k=1$,
• $c_k+d_k=1$.

If $\left\{x_k\right\}$ converges, then it converges to a unique fixed point of ${\Psi }_{\lambda }$.

3.3. Convergence Analysis

Ishikawa has shown that for any points x, y, z in a Hilbert space and any real number $\lambda$, the following is true:

$$\left|\right|\lambda x+(1-\lambda )y-{z\left|\right|}^2=\lambda \left|\right|x-{z\left|\right|}^2+(1-\lambda )\left|\right|y-{z\left|\right|}^2-\lambda (1-\lambda )\left|\right|x-{y\left|\right|}^2.$$

Let $x^{*}$ be a fixed point of ${\Psi }_{\lambda }$, then we have

$$\begin{array}{cc}\hfill \left|\right|x_{k+1}-x^{*}{\left|\right|}^2& =\left|\right|a_k{x}_k+b_k{\Psi }_{\lambda }{y}_k-x^{*}{\left|\right|}^2\hfill \\ \hfill & =b_k\left|\right|{\Psi }_{\lambda }{y}_k-x^{*}{\left|\right|}^2+a_k\left|\right|x_k-x^{*}{\left|\right|}^2-b_k{a}_k\left|\right|x_k-{\Psi }_{\lambda }{y}_k{\left|\right|}^2\hfill \end{array}$$

(5)

From the contraction condition we have

$$\parallel {\Psi }_{\lambda }{y}_k-x^{*}{\parallel }^2=\left|\right|{\Psi }_{\lambda }{y}_k-{\Psi }_{\lambda }{x}^{*}{\left|\right|}^2\le \parallel y_k-x^{*}{\parallel }^2+h{\parallel y_k-{\Psi }_{\lambda }{y}_k\parallel }^2,\mathrm{where}h=L^2.$$

(6)

On the other hand,

$$\parallel y_k-x^{*}{\parallel }^2={\parallel c_k{x}_k+d_k{\Psi }_{\lambda }{x}_k-x^{*}\parallel }^2,$$

which expands to

$$\parallel y_k-x^{*}{\parallel }^2=d_k\parallel {\Psi }_{\lambda }{x}_k-x^{*}{\parallel }^2+c_k\parallel x_k-x^{*}{\parallel }^2-d_k{c}_k{\parallel x_k-{\Psi }_{\lambda }{x}_k\parallel }^2.$$

(7)

Similarly, we can express the following:

$$\parallel y_k-{\Psi }_{\lambda }{y}_k{\parallel }^2=d_k\parallel {\Psi }_{\lambda }{x}_k-{\Psi }_{\lambda }{y}_k{\parallel }^2+c_k\parallel x_k-{\Psi }_{\lambda }{y}_k{\parallel }^2-d_k{c}_k{\parallel x_k-{\Psi }_{\lambda }{x}_k\parallel }^2.$$

(8)

Moreover, we have the following inequality:

$$\parallel {\Psi }_{\lambda }{x}_k-x^{*}{\parallel }^2\le \parallel x_k-x^{*}{\parallel }^2+h{\parallel x_k-{\Psi }_{\lambda }{x}_k\parallel }^2.$$

(9)

By introducing Equations (7)–(9) into (6), we obtain

$$\parallel {\Psi }_{\lambda }{y}_k-x^{*}{\parallel }^2\le c_k\parallel x_k-x^{*}{\parallel }^2+h{c}_k\parallel x_k-{\Psi }_{\lambda }{x}_k{\parallel }^2+d_k{\parallel x_k-x^{*}\parallel }^2$$

$$-d_k{c}_k\parallel x_k-{\Psi }_{\lambda }{x}_k{\parallel }^2+h{d}_k\parallel {\Psi }_{\lambda }{x}_k-{\Psi }_{\lambda }{y}_k{\parallel }^2+h{c}_k{\parallel x_k-{\Psi }_{\lambda }{y}_k\parallel }^2.$$

Thus,

$$\parallel {\Psi }_{\lambda }{y}_k-x^{*}{\parallel }^2\le \parallel x_k-x^{*}{\parallel }^2-d_k(c_k-h{d}_k){\parallel x_k-{\Psi }_{\lambda }{x}_k\parallel }^2$$

$$+h{d}_k\parallel {\Psi }_{\lambda }{x}_k-{\Psi }_{\lambda }{y}_k{\parallel }^2+h{c}_k{\parallel x_k-{\Psi }_{\lambda }{y}_k\parallel }^2.$$

(10)

Substituting Equation (10) into Equation (5), we obtain the following:

$$\parallel x_{k+1}-x^{*}{\parallel }^2\le \parallel x_k-x^{*}{\parallel }^2+h{b}_k{\parallel {\Psi }_{\lambda }{x}_k-{\Psi }_{\lambda }{y}_k\parallel }^2$$

$$-b_k{d}_k(c_k-h{d}_k)\parallel x_k-{\Psi }_{\lambda }{x}_k{\parallel }^2-b_k(a_k-h+h{d}_k){\parallel x_k-{\Psi }_{\lambda }{y}_k\parallel }^2.$$

This shows that $\{\parallel x_k-x^{*}{\parallel }^2\}$ is decreasing for all sufficiently large k. There exists a subsequence $\left\{x_{k_m}\right\}$ of $\left\{x_k\right\}$ such that

$$\underset{m\to \infty }{lim}\parallel x_{k_m}-{\Psi }_{\lambda }{x}_{k_m}\parallel =0.$$

Now, we show that $\left\{{\Psi }_{\lambda }{x}_k\right\}$ is a Cauchy sequence. Indeed,

$$\parallel {\Psi }_{\lambda }{x}_{k_m}-{\Psi }_{\lambda }{x}_{k_{\ell }}\parallel \le \parallel x_{k_{\ell }}-{\Psi }_{\lambda }{x}_{k_m}\parallel +\parallel x_{k_{\ell }}-{\Psi }_{\lambda }{x}_{k_{\ell }}\parallel$$

Taking the limit as $m,\ell \to \infty$, we have

$$\parallel {\Psi }_{\lambda }{x}_{k_m}-{\Psi }_{\lambda }{x}_{k_{\ell }}\parallel \to 0.$$

Thus, $\left\{{\Psi }_{\lambda }{x}_k\right\}$ is a Cauchy sequence, and hence, convergent.

Call the limit $x^{*}$. Then,

$$\underset{m\to \infty }{lim}{\Psi }_{\lambda }{x}_{k_m}=\underset{m\to \infty }{lim}{x}_{k_m}=x^{*}.$$

Using the contraction of ${\Psi }_{\lambda }$, we have

$$\parallel {\Psi }_{\lambda }{x}^{*}-{\Psi }_{\lambda }{x}_{k_m}\parallel \le \parallel x^{*}-x_{k_m}\parallel +\parallel x_{k_m}-{\Psi }_{\lambda }{x}_{k_m}\parallel +\parallel {\Psi }_{\lambda }{x}^{*}-{\Psi }_{\lambda }{x}_{k_m}\parallel .$$

Taking the limit as $m\to \infty$, we obtain the following:

$$\underset{m\to \infty }{lim}\parallel {\Psi }_{\lambda }{x}^{*}-{\Psi }_{\lambda }{x}_{k_m}\parallel =0.$$

Hence, we conclude that

$$\parallel x^{*}-{\Psi }_{\lambda }{x}^{*}\parallel \le \parallel x^{*}-x_{k_m}\parallel +\parallel x_{k_m}-{\Psi }_{\lambda }{x}_{k_m}\parallel +\parallel {\Psi }_{\lambda }{x}_{k_m}-{\Psi }_{\lambda }{x}^{*}\parallel .$$

Taking the limit as $m\to \infty$, we deduce that $\parallel x^{*}-{\Psi }_{\lambda }{x}^{*}\parallel =0$, i.e., $x^{*}={\Psi }_{\lambda }{x}^{*}$. Now, we aim to prove that the sequence $\left\{x_k\right\}$ converges to the unique fixed point of ${\Psi }_{\lambda }$.

$$\begin{array}{cc}\hfill \parallel x_{k+1}-x^{*}{\parallel }^2& =\parallel a_k{x}_k+b_k{\Psi }_{\lambda }{y}_k-x^{*}{\parallel }^2\hfill \\ \hfill & =b_k\parallel {\Psi }_{\lambda }{y}_k-x^{*}{\parallel }^2+a_k\parallel x_k-x^{*}{\parallel }^2-b_k{a}_k{\parallel {\Psi }_{\lambda }{y}_k-x^{*}\parallel }^2.\hfill \end{array}$$

(11)

We know that

$$\parallel {\Psi }_{\lambda }{y}_k-x^{*}{\parallel }^2\le L^2\parallel y_k-x^{*}{\parallel }^2+L^2{\parallel y_k-{\Psi }_{\lambda }{y}_k\parallel }^2.$$

Suppose that $L^2=h$; then,

$$\parallel {\Psi }_{\lambda }{y}_k-x^{*}{\parallel }^2\le h\parallel y_k-x^{*}{\parallel }^2+h{\parallel y_k-{\Psi }_{\lambda }{y}_k\parallel }^2.$$

On the other hand,

$$\begin{array}{cc}\hfill \parallel y_k-x^{*}{\parallel }^2& =\parallel d_k{\Psi }_{\lambda }{x}_k+c_k{x}_k-x^{*}{\parallel }^2\hfill \\ \hfill & =d_k\parallel {\Psi }_{\lambda }{x}_k-x^{*}{\parallel }^2+c_k\parallel x_k-x^{*}{\parallel }^2-d_k{c}_k{\parallel {\Psi }_{\lambda }{x}_k-x_k\parallel }^2.\hfill \end{array}$$

(12)

And similarly,

$$\begin{array}{cc}\hfill \parallel y_k-{\Psi }_{\lambda }{y}_k{\parallel }^2& =\parallel d_k{\Psi }_{\lambda }{x}_k+c_k{x}_k-{\Psi }_{\lambda }{y}_k{\parallel }^2\hfill \\ \hfill & =c_k\parallel {\Psi }_{\lambda }{x}_k-{\Psi }_{\lambda }{y}_k{\parallel }^2+c_k\parallel x_k-{\Psi }_{\lambda }{y}_k{\parallel }^2-d_k{c}_k{\parallel {\Psi }_{\lambda }{x}_k-x_k\parallel }^2.\hfill \end{array}$$

(13)

Hence (13) can be rewritten as follows:

$$\begin{array}{cc}\hfill \parallel {\Psi }_{\lambda }{y}_k-x^{*}{\parallel }^2& \le h{d}_k\parallel {\Psi }_{\lambda }{x}_k-x^{*}{\parallel }^2+h{c}_k{\parallel x_k-x^{*}\parallel }^2\hfill \\ \hfill & -h{d}_k{c}_k\parallel {\Psi }_{\lambda }{x}_k-x_k{\parallel }^2+h{d}_k{\parallel {\Psi }_{\lambda }{x}_k-{\Psi }_{\lambda }{y}_k\parallel }^2\hfill \\ \hfill & +h{c}_k\parallel x_k-{\Psi }_{\lambda }{y}_k{\parallel }^2-h{d}_k{c}_k{\parallel {\Psi }_{\lambda }{x}_k-x_k\parallel }^2.\hfill \end{array}$$

(14)

However, we also have

$$\parallel {\Psi }_{\lambda }{x}_k-x^{*}{\parallel }^2\le h\parallel x_k-x^{*}{\parallel }^2+h{\parallel {\Psi }_{\lambda }{x}_k-x_k\parallel }^2.$$

(15)

By substituting (15) into (14), we obtain the following:

$$\begin{array}{cc}\hfill \parallel {\Psi }_{\lambda }{y}_k-x^{*}{\parallel }^2& \le h^2{d}_k\parallel x_k-x^{*}{\parallel }^2+h^2\parallel {\Psi }_{\lambda }{x}_k-x_k{\parallel }^2+h{c}_k{\parallel x_k-x^{*}\parallel }^2\hfill \\ \hfill & -h{d}_k{c}_k\parallel {\Psi }_{\lambda }{x}_k-x_k{\parallel }^2+h{d}_k{c}_k{\parallel {\Psi }_{\lambda }{x}_k-{\Psi }_{\lambda }{y}_k\parallel }^2\hfill \\ \hfill & +h{c}_k\parallel x_k-{\Psi }_{\lambda }{y}_k{\parallel }^2-h{d}_k{c}_k{\parallel {\Psi }_{\lambda }{x}_k-x_k\parallel }^2\hfill \\ \hfill & \le (h{c}_k+h{d}_k)\parallel x_k-x^{*}{\parallel }^2-h{d}_k(2-2d_k){\parallel {\Psi }_{\lambda }{x}_k-x_k\parallel }^2\hfill \\ \hfill & +h{c}_k\parallel x_k-{\Psi }_{\lambda }{y}_k{\parallel }^2+h{d}_k{\parallel {\Psi }_{\lambda }{x}_k-{\Psi }_{\lambda }{y}_k\parallel }^2.\hfill \end{array}$$

(16)

Incorporating (16) into (11) yields the following:

$$\begin{array}{cc}\hfill \parallel x_{k+1}-x^{*}{\parallel }^2& \le b_{k}h\parallel x_k-x^{*}{\parallel }^2-b_k{d}_{k}h(2-h-2d_k){\parallel {\Psi }_{\lambda }{x}_k-x_k\parallel }^2\hfill \\ \hfill & +b_k{c}_{k}h\parallel x_k-{\Psi }_{\lambda }{y}_k{\parallel }^2+b_k{d}_{k}h{\parallel {\Psi }_{\lambda }{x}_k-{\Psi }_{\lambda }{y}_k\parallel }^2\hfill \\ \hfill & +a_k\parallel x_k-x^{*}{\parallel }^2-b_k{a}_{k}h{\parallel x_k-{\Psi }_{\lambda }{y}_k\parallel }^2\hfill \\ \hfill & \le a_k(1-h)\parallel x_k-x^{*}{\parallel }^2-b_k{d}_{k}h(2-h-2d_k){\parallel {\Psi }_{\lambda }{x}_k-x_k\parallel }^2\hfill \\ \hfill & +b_k{d}_{k}h\parallel {\Psi }_{\lambda }{x}_k-{\Psi }_{\lambda }{y}_k{\parallel }^2-b_k{a}_k(1-h+h{d}_k))\parallel x_k-{\Psi }_{\lambda }{y}_k{\parallel }^2.\hfill \end{array}$$

(17)

Given that ${\displaystyle \frac{1-h}{2}}\le b_k\le 1-h$, $0<h<1$, $d_k\ge 0$, and $lim{d}_k=0$, there exists a natural number N such that for $k>N$,

$$2-h-2d_k\ge 0\mathrm{and}{a}_k-h+h{d}_k\ge 0.$$

Thus, for $k\ge N$, we have

$$\parallel x_{k+1}-x^{*}{\parallel }^2\le \tilde{h}\parallel x_k-x^{*}{\parallel }^2+b_k{d}_{k}h{\parallel {\Psi }_{\lambda }{x}_k-{\Psi }_{\lambda }{y}_k\parallel }^2,$$

where $0<\tilde{h}=1-{\displaystyle \frac{{(1-h)}^2}{2}}$.

From the boundedness of C, it follows that $\parallel {\Psi }_{\lambda }{x}_k-{\Psi }_{\lambda }{y}_k{\parallel }^2$ is bounded. Therefore, we conclude that

$$\underset{k\to \infty }{lim}{b}_k{d}_{k}h{\parallel {\Psi }_{\lambda }{x}_k-{\Psi }_{\lambda }{y}_k\parallel }^2=0.$$

From Lemma 2, we conclude that ${\displaystyle \underset{k\to \infty }{lim}{x}_k=x^{*}}$. This completes the proof.

3.4. Maximal Monotone Operators and Minimization Problem

We consider the following composite convex optimization problem:

$$\underset{x\in {\mathbb{R}}^n}{min}\left[f\left(x\right)+G\left(x\right)+H\left(x\right)\right],$$

(18)

where

• $f:{\mathbb{R}}^n\to \mathbb{R}$ is a continuously differentiable function with a Lipschitz continuous gradient, i.e., $\nabla f$ is 1-Lipschitz;
• G and H are convex, closed, and proper functions.

Proposition 8.

Let $A=\mathsf{\partial }G$, $B=\mathsf{\partial }H$ and $C=\nabla f$. Then, the minimization problem (18) is equivalent to finding a zero of the sum of maximal monotone operators; that is,

$$\mathit{Find}x\in {\mathbb{R}}^n\mathit{such}\mathit{that}0\in A\left(x\right)+B\left(x\right)+C\left(x\right).$$

(19)

3.5. Example

Let f, G, and H be three real-valued functions defined on $\mathbb{R}$ as follows:

$$f\left(x\right)={\displaystyle \frac{1}{2}}{x}^2+2,G\left(x\right)={\displaystyle \frac{1}{10}}{x}^2,H\left(x\right)=2x^2.$$

We consider the following minimization problem:

$$\underset{x\in \mathbb{R}}{min}\left(f\left(x\right)+G\left(x\right)+H\left(x\right)\right).$$

Let us define the following monotone operators corresponding to the gradients of G, H, and f:

$$A\left(x\right)=\{\nabla G\left(x\right)\}=\left\{{\displaystyle \frac{1}{5}}x\right\},B\left(x\right)=\{\nabla H\left(x\right)\}=\left\{4x\right\},C\left(x\right)=\nabla f\left(x\right)=x.$$

Then, the minimization problem above is equivalent to the inclusion problem:

$$\mathrm{Find}x\in \mathbb{R}\mathrm{such}\mathrm{that}0\in A\left(x\right)+B\left(x\right)+C\left(x\right).$$

We know the resolvents of the operators are given by the following:

$$J_{\lambda A}\left(x\right)={\displaystyle \frac{5}{5+\lambda }}x,J_{\lambda B}\left(x\right)={\displaystyle \frac{1}{1+4\lambda }}x,$$

and hence,

$${\Psi }_{\lambda }\left(x\right)={\displaystyle \frac{4{\lambda }^2-5\lambda +10}{(4\lambda +1)(\lambda +5)}}x.$$

According to Theorem 5, we proceed by choosing the parameters:

$$\lambda =2,b_n={\displaystyle \frac{1}{{(n+1)}^2}},d_n={\displaystyle \frac{1}{n+1}}.$$

Application of the Algorithm 2

We demonstrate single-valued mapping:

$${\Psi }_{\lambda }\left(x\right)={\displaystyle \frac{4{\lambda }^2-5\lambda +10}{(4\lambda +1)(\lambda +5)}}x.$$

For $\lambda =2$, this becomes

$${\Psi }_2\left(x\right)={\displaystyle \frac{16}{63}}x.$$

We initialize the process as follows:

$$x_0=y_0=1,$$

and define

$$d_n={\displaystyle \frac{1}{{(n+1)}^2}},b_n={\displaystyle \frac{1}{n+1}},a_n=1-b_n,c_n=1-d_n.$$

Iteration Steps

At each iteration $n\ge 0$, we update the following:

$$\begin{array}{cc}\hfill y_n& =c_n{x}_n+d_n{\Psi }_2\left(x_n\right)=\left(1-{\displaystyle \frac{1}{{(n+1)}^2}}\right)x_n+{\displaystyle \frac{16}{63{(n+1)}^2}}{x}_n,\hfill \\ \hfill x_{n+1}& =a_n{x}_n+b_n{\Psi }_2\left(y_n\right)=\left(1-{\displaystyle \frac{1}{n+1}}\right)x_n+{\displaystyle \frac{16}{63(n+1)}}{y}_n.\hfill \end{array}$$

These choices still satisfy the assumptions of Theorem 5 and ensure that the sequence $\left\{x_n\right\}$ converges to the unique fixed point of ${\Psi }_2$, which is

$$x^{*}={\Psi }_2\left(x^{*}\right)\Rightarrow x^{*}=0.$$

4. Conclusions

In conclusion, problem (1) has been extensively studied by numerous authors under various settings. The research began with the case of finding the zeros of a single maximal monotone operator and was later extended to the case involving the sum of two maximal monotone operators, where several authors proposed generalized algorithms.

In our work, we studied problem (1) under different scenarios and proposed several algorithms that contributed to its solution. We also support our theoretical results with a set of simple numerical examples, illustrating the convergence of the sequences proposed in this paper to the same solution of the problem.

Nevertheless, the development of alternative algorithms under suitable conditions that can efficiently handle this class of problems remains an open area of research, providing valuable opportunities for further exploration and advancement.