Modified Mann-Type Algorithm for Two Countable Families of Nonexpansive Mappings and Application to Monotone Inclusion and Image Restoration Problems

Abstract

In this paper, we introduce and study a modified Mann-type algorithm that combines inertial terms for solving common fixed point problems of two countable families of nonexpansive mappings in Hilbert spaces. Under appropriate assumptions on the sequences of parameters, we establish a strong convergence result for the sequence generated by the proposed method in finding a common fixed point of two countable families of nonexpansive mappings. This method can be applied to solve the monotone inclusion problem. Additionally, we employ a modified Mann-type iterative algorithm to address image restoration problems. Furthermore, we present numerical results across different scenarios to demonstrate the superior efficiency of our algorithm compared to existing algorithms.

1. Introduction

Let H be a real Hilbert space with inner product $\langle ·,·\rangle$ and associated norm $\parallel ·\parallel =\sqrt{\langle ·,·\rangle }$. Recall that a mapping $T:H\to H$ is said to be a strict pseudo-contraction if there exists $k\in (-\infty ,1)$ such that

$${∥Tx-Ty∥}^2\le {∥x-y∥}^2+k{∥(I-T)x-(I-T)y∥}^2,\forall x,y\in H.$$

(1)

In particular, it is well known that if $k=-1$ in (1), then the mapping T in (1) is said to be a firmly nonexpansive mapping and the equivalent form of this mapping can be written as ${∥Tx-Ty∥}^2\le 〈x-y,Tx-Ty〉,\forall x,y\in H.$ Furthermore, if $k=0$ in (1), then the mapping T in (1) is called a nonexpansive mapping which we can view (1) in this particular case as $∥Tx-Ty∥\le ∥x-y∥,\forall x,y\in H$, see [1,2] for more details.

We denote by $Fix\left(T\right):=\{x\in H:Tx=x\}$ the set of all fixed points of T. Let C be a nonempty closed convex subset of H. Then for every $z\in H$ there exists a unique $x^{\ast }\in C$ such that $∥z-x^{\ast }∥=\underset{x\in C}{inf}∥z-x∥$ and if we define $P_C:H\to C$ by $P_C\left(z\right)=x^{\ast }$ for all $z\in H$ then $P_C$ is called the metric projection (or nearest point projection) of H onto C, see [3,4,5] more details.

Fixed Point Problem: The following is the fixed point problem for the mapping T usually represented by:

$$\mathrm{find}x\in H\mathrm{such}\mathrm{that}x=Tx.$$

(2)

Fixed point problems have found wide-ranging applications across various fields of mathematics, engineering, computer science, and other disciplines. These problems involve finding points in a function’s domain that remain unchanged after applying the function, making them essential in solving equations, optimization, economic equilibrium analysis, signal processing, control systems, machine learning, and fractal geometry. By providing a framework for iterative methods and stability analysis, fixed point problems offer powerful tools to tackle complex problems and uncover solutions in diverse areas of study.

The well known and important iterative algorithm for finding a solution of (2) was proposed by Mann [6] as follows:

$$\left\{\begin{array}{c}{x}_1\in C,\hfill \\ x_{n+1}=(1-{\alpha }_n)x_n+{\alpha }_{n}T{x}_n,\forall n\ge 1,\hfill \end{array}\right.$$

(3)

where ${\left({\alpha }_n\right)}_{n\ge 1}$ is chosen from $[0,1].$ Reich [7] showed that if T in (3) is a nonexpansive with a fixed point and ${\left({\alpha }_n\right)}_{n\ge 1}$ satisfies some appropriate conditions, then (3) converges weakly to a fixed point of T. Later, Ishikawa [8] relied on Mann’s ideas and proposed the following iterative method for a Lipschitzian pseudo-contraction in Hilbert spaces as follows:

$$\left\{\begin{array}{c}{x}_1\in C,\hfill \\ y_n=(1-{\alpha }_n)x_n+{\alpha }_{n}T{x}_n,\hfill \\ x_{n+1}=(1-{\beta }_n)x_n+{\beta }_{n}T{y}_n,\forall n\ge 1.\hfill \end{array}\right.$$

(4)

Under certain suitable conditions of C, ${\left({\alpha }_n\right)}_{n\ge 1}$, and ${\left({\beta }_n\right)}_{n\ge 1}$, he was able to prove that (4) converges strongly to a fixed point of T. On the other hand, Halpern [9] presented a new approach for finding a fixed point of a nonexpansive mapping T which differs from Mann’s method by forcing one point of the vector to rest as follows:

$$\left\{\begin{array}{c}u,x_1\in C,\hfill \\ x_{n+1}=(1-{\alpha }_n)u+{\alpha }_{n}T{x}_n,\forall n\ge 1,\hfill \end{array}\right.$$

(5)

where ${\left({\alpha }_n\right)}_{n\ge 1}$ is chosen from $[0,1].$ It can be proved under some appropriate conditions that (5) converges strongly to a fixed point of T. After that, Moudafi [10] developed a new iterative method which guarantees strong convergence that later came to be known as the viscosity approximation method by combining the Halpern iterative method with the contraction mapping. Many researchers studied the concept of viscosity method and then developed this method in many directions, see [11,12,13,14,15] for more details. In 2019, Bot et al. [16] modified and developed (3) to induce strong convergence to a fixed point of a nonexpansive mapping without relying on the concept of viscosity approximation method. Their method is as follows:

$$\left\{\begin{array}{c}{x}_1\in H,\hfill \\ x_{n+1}=(1-{\alpha }_n){\delta }_n{x}_n+{\alpha }_{n}T{\delta }_n{x}_n,\forall n\ge 1,\hfill \end{array}\right.$$

(6)

when ${\left({\alpha }_n\right)}_{n\ge 1}$ and ${\left({\delta }_n\right)}_{n\ge 1}$ are sequences chosen from $(0,1]$ with suitable assumptions.

On the other hand, it is well known that the fixed point problem (2) has the strong relationship with the monotone inclusion problem which it can be written as follows:

$$\mathrm{find}x\in H\mathrm{such}\mathrm{that}0\in Vx,$$

(7)

where $V:H\to 2^H$ is a multi-value operator. Indeed, for example, if V is a multi-value maximal monotone operator, then resolvent operator $J_{\gamma V}:H\to H$ is a single-value and firmly nonexpansive mapping where $\gamma >0$. Furthermore, for solving all solutions of (7) is equivalent to solve all solutions of the fixed point problem of $J_{\gamma V}$ or equivalently to say that $zer\left(V\right)=Fix\left(J_{\gamma V}\right)$, where $zer\left(V\right)=\left\{\left.x\in H\right|0\in Vx\right\}$ is the set of all zero points of V (see Section 4 and [3,4] for more details).

In 1964, Polyak [17] proposed several strategies to improve the convergence rate of iteration methods. These techniques include modifications to the classical iterative schemes, such as the use of variable relaxation parameters and acceleration techniques by using the inertial extrapolation term, that is, ${\theta }_n(x_n-x_{n-1})$ where $\left({\theta }_n\right)$ is a sequence that satisfies appropriate conditions. Consequently, the inertial extrapolation has received attention and has been studied by many authors, see [18,19,20,21,22,23,24,25] more details. In 2019, Shehu [26] proposed an algorithm that combines the inertial method, the Halpern method, and error terms to solve a fixed point of a nonexpansive mapping. Later, Kitkuan et al. [27] employed the idea of inertial viscosity to find some solutions of monotone inclusion problems and apply to solve image restoration problems. Recently, Akutsah et al. [28] introduced a new algorithm for finding a common fixed point of generalized nonexpansive mappings. Moreover, the fixed point problems have been studied in many aspects, see for instance [29,30,31,32,33]. Inspired by Bot et al. [16], Artsawang and Ungchittrakool [25] introduced the inertial Mann-type algorithm for finding a fixed point of a nonexpansive mapping and applying to monotone inclusion and image restoration problems as follows:

$$\left\{\begin{array}{c}{x}_0,x_1\in H,\hfill \\ y_n=x_n+{\theta }_n(x_n-x_{n-1}),\hfill \\ x_{n+1}={\delta }_n{y}_n+{\alpha }_n(T{\delta }_n{y}_n-{\delta }_n{y}_n)+{\epsilon }_n,\forall n\ge 1,\hfill \end{array}\right.$$

where ${\left({\theta }_n\right)}_{n\ge 1},{\left({\alpha }_n\right)}_{n\ge 1},{\left({\delta }_n\right)}_{n\ge 1}$ are some appropriate sequences chosen from $[0,1]$.

Common Fixed Point Problem: the common fixed point problem for the mapping S and T is generally denoted by the following:

$$\mathrm{find}x\in H\mathrm{such}\mathrm{that}x=Sx=Tx.$$

(8)

The common fixed point problem (8) has many applications in various disciplines such as numerical analysis, optimization, dynamical systems, game theory, control theory, topology, computer science, and economics, etc. To approximate the common fixed points of two mappings, Das and Debata [34] and Takahashi and Tamura [35] generalized the Ishikawa iterative algorithm for mappings S and T as follows:

$$\left\{\begin{array}{c}{x}_1\in C,\hfill \\ y_n=(1-{\alpha }_n)x_n+{\alpha }_{n}S{x}_n,\hfill \\ x_{n+1}=(1-{\beta }_n)x_n+{\beta }_{n}T{y}_n,\forall n\ge 1,\hfill \end{array}\right.$$

where ${\left({\alpha }_n\right)}_{n\ge 1},{\left({\beta }_n\right)}_{n\ge 1}$ are some sequences chosen in $[0,1]$. Later, Khan and Fukhar-ud-din [36] introduced and studied a scheme with errors for two nonexpansive mappings.

In 2007, Aoyama et al. [37] introduced and studied Halpern iterative algorithm for a countable family of nonexpansive mappings and provided some important conditions on a countable family of nonexpansive mappings which later known as AKTT condition to guarantee the convergence result to a common fixed point solution.

As mentioned above, all of these researches inspired and drove us to develop a new iterative algorithm, which is as follows:

Let ${\left(S_n\right)}_{n\ge 0},{\left(T_n\right)}_{n\ge 0}:H\to H$ be any two countable families of nonexpansive mappings such that ${\displaystyle \bigcap _{n=0}^{\infty }}Fix\left(S_n\right)\cap {\displaystyle \bigcap _{n=0}^{\infty }}Fix\left(T_n\right)\ne \varnothing$. We propose the following algorithm:

$$\left\{\begin{array}{c}{x}_0,x_1\in H,\hfill \\ y_n=x_n+{\theta }_n(x_n-x_{n-1}),\hfill \\ z_n=(1-{\alpha }_n){\delta }_n{y}_n+{\alpha }_n{S}_n{\delta }_n{y}_n,\hfill \\ x_{n+1}=z_n+{\beta }_n\left(T_n{\delta }_n{y}_n-z_n\right)+{\epsilon }_n,\hfill \end{array}\right.$$

(9)

for all $n\ge 1$, where ${\left({\theta }_n\right)}_{n\ge 0}\subseteq [0,\theta ]$ with $\theta \in [0,1)$, ${\left({\alpha }_n\right)}_{n\ge 0},{\left({\beta }_n\right)}_{n\ge 0},$ and ${\left({\delta }_n\right)}_{n\ge 0}$ are sequences in $(0,1]$ and ${\left({\epsilon }_n\right)}_{n\ge 0}$ is a sequence in H.

Moreover, we can apply Equation (9) as a tool to find a zero point solution of the sum of three monotone operators $U,V,W$ as follows:

$$\mathrm{find}x\in \mathcal{H}\mathrm{such}\mathrm{that}0\in \left({\mu }_{1}Ux+\eta Vx+{\nu }_{1}Wx\right)\cap \left({\mu }_{2}Ux+\eta Vx+{\nu }_{2}Wx\right),$$

(10)

where $U,V$ are some monotone operators on a Hilbert space H, W is $\kappa$-cocoercive with parameter $\kappa$, and ${\mu }_1,{\mu }_2,\eta >0$ and ${\nu }_1,{\nu }_2\in (0,2\kappa )$ (see Section 4 for more details). The problem (10) can be viewed as a generalization of Artsawang and Ungchittrakool [25] and Davis and Yin [38]. It is useful to note the importance and association that (10) can be rewritten to be some common fixed point problems of nonexpansive mappings (see Section 4 for more details). For this reason, finding a solution of the monotone inclusion problem (10), it can be applied the ideas of common fixed point problem and fixed point iterative algorithms to solve (10). For practical applications, we can use by-products obtained from Equation (9) to solve the image restoration problems. Furthermore, the effectiveness of our novel algorithm is substantiated by presenting numerical outcomes in different scenarios. These results conclusively establish that our algorithm outperforms its predecessor, as evidenced by the higher performance exhibited in the numerical analysis.

2. Preliminaries

In this section, we present the collected results from real Hilbert spaces, which are relevant and beneficial for the conducted convergence analysis in this study.

Lemma 1 

([3,4]). Let H be a real Hilbert space. The following conditions are hold.

1. 

${\parallel x+y\parallel }^2\le {\parallel x\parallel }^2+2\langle x+y,y\rangle$ for all $x,y\in H$,

2. 

${\parallel tx+(1-t)y\parallel }^2={t\parallel x\parallel }^2+{(1-t)\parallel y\parallel }^2-t(1-t){\parallel x-y\parallel }^2$ for all $t\in \mathbb{R}$ and $x,y\in H.$

Lemma 2 

([39,40]). Let ${\left(s_n\right)}_{n\ge 0},{\left({ϵ}_n\right)}_{n\ge 0}\subseteq [0,+\infty )$, ${\left({\sigma }_n\right)}_{n\ge 0}\subseteq [0,1]$ and ${\left({\nu }_n\right)}_{n\ge 0}\subseteq \mathbb{R}$ be sequences such that

$$s_{n+1}\le (1-{\sigma }_n)s_n+{\sigma }_n{\nu }_n+{ϵ}_n,\forall n\ge 0.$$

Assume that ${\displaystyle \sum _{n=0}^{\infty }}{ϵ}_n<+\infty .$ Then, the statements below are true.

1. 

If ${\sigma }_n{\nu }_n\le r{\sigma }_n$ (where $r\ge 0$), then ${\left(s_n\right)}_{n\ge 0}$ is bounded.

2. 

If ${\displaystyle \sum _{n=0}^{\infty }}{\sigma }_n=+\infty$ and $\underset{n\to \infty }{limsup}{\nu }_n\le 0$, then $\underset{n\to \infty }{lim}{s}_n=0$.

Lemma 3 

([5]). Let $T:H\to H$ be a nonexpansive operator, ${\left(x_n\right)}_{n\ge 0}\subseteq H$, and $x\in H$ be such that $x_n\rightharpoonup x$ as $n\to \infty$ and $∥x_n-T{x}_n∥\to 0$ as $n\to \infty$. Then $x\in Fix\left(T\right)$.

The following lemma is one important characterization of the metric projection.

Lemma 4 

([3,4,5]). Let C be a nonempty closed convex subset of H. Then for every $z\in H$ and $x^{\ast }\in C$,

$x^{\ast }=P_C\left(z\right)$ if and only if $〈z-x^{\ast },y-x^{\ast }〉\le 0,\forall y\in C.$

Aoyama, Kimura, Takahashi, and Toyoda [37] (see also Plubtieng and Ungchittrakool [41]) observed some behaviors of a countable family of mappings and provided a nice condition as the following definition.

Definition 1

(AKTT condition, Aoyama et al. [37]). Let D be a nonempty subset of a Hilbert space H and let ${\left(T_n\right)}_{n\ge 0}$ be a countable family of mappings from D to H. For a subset Q of D, we say that $\left({\left(T_n\right)}_{n\ge 0},Q\right)$ satisfies condition AKTT if ${\displaystyle \sum _{n=1}^{\infty }}sup\left\{\left.∥T_{n}z-T_{n-1}z∥\right|z\in Q\right\}<+\infty .$

Additionally, they verified the following result, which is important to our main task.

Lemma 5 

([37]). Let D be a nonempty subset of a Hilbert space H and let ${\left(T_n\right)}_{n\ge 0}$ be a countable family of mappings from D to H. Let Q be a subset of D with $\left({\left(T_n\right)}_{n\ge 0},Q\right)$ satisfying condition AKTT, then there exists a mapping $T:Q\to H$ such that

$$Tx=\underset{n\to \infty }{lim}{T}_{n}x,\forall x\in Q,$$

and $\underset{n\to \infty }{lim}sup\left\{\left.∥Tz-T_{n}z∥\right|z\in Q\right\}=0.$

3. Main Results

In this section, we analyze the convergence of Equation (9), starting with the boundedness property of the algorithm, as stated in the following lemma.

Assumption 1.

Let ${\left({\alpha }_n\right)}_{n\ge 0}\subseteq (0,1],{\left({\beta }_n\right)}_{n\ge 0},{\left({\delta }_n\right)}_{n\ge 0}\subseteq [0,1]$, and ${\left({\epsilon }_n\right)}_{n\ge 0}\subseteq H$ be consistent with the following conditions:

1. 

 $\underset{n\to \infty }{liminf}{\beta }_n(1-{\beta }_n){\alpha }_n>0.$

2. 

 ${\displaystyle \sum _{n=1}^{\infty }}\left|{\alpha }_n-{\alpha }_{n-1}\right|<+\infty .$

3. 

 ${\displaystyle \sum _{n=1}^{\infty }}|{\beta }_n-{\beta }_{n-1}|<+\infty .$

4. 

(a) 

$\underset{n\to \infty }{lim}{\delta }_n=1.$

(b) 

${\displaystyle \sum _{n=0}^{\infty }}(1-{\delta }_n)=+\infty ,$ and ${\displaystyle \sum _{n=1}^{\infty }}|{\delta }_n-{\delta }_{n-1}|<+\infty .$

5. 

 ${\displaystyle \sum _{n=0}^{\infty }}∥{\epsilon }_n∥<+\infty .$

We have confirmed the validity of Assumption 1, as demonstrated in the following remark.

Remark 1.

Let $z\in H$. We set ${\delta }_n=1-\frac{1}{n+2}$, ${\alpha }_n={\beta }_n=\frac{1}{4}-\frac{1}{{(n+3)}^2}$ and ${\epsilon }_n=\frac{z}{{(n+1)}^3}$ for all $n\ge 0$. It is easy to see that the Assumption 1 is satisfied.

Lemma 6. 

Let ${\left(S_n\right)}_{n\ge 0},{\left(T_n\right)}_{n\ge 0}:H\to H$ be two countable families of nonexpansive mappings such that ${\displaystyle \bigcap _{n=0}^{\infty }}Fix\left(S_n\right)\cap {\displaystyle \bigcap _{n=0}^{\infty }}Fix\left(T_n\right)\ne \varnothing$ and let ${\left(x_n\right)}_{n\ge 0}$ be generated by Equation (9). Let ${\left({\theta }_n\right)}_{n\ge 0}$ be a sequence in $[0,\theta ]$ with $\theta \in [0,1)$ such that ${\displaystyle \sum _{n=1}^{\infty }}{\theta }_n∥x_n-x_{n-1}∥<+\infty .$ Suppose 5. in Assumption 1 holds. Then ${\left(x_n\right)}_{n\ge 0}$ is bounded.

Proof. 

Let $p\in {\displaystyle \bigcap _{n=0}^{\infty }}Fix\left(S_n\right)\cap {\displaystyle \bigcap _{n=0}^{\infty }}Fix\left(T_n\right)$. Then, let us consider

$$\begin{array}{cc}\hfill ∥y_n-p∥& =∥x_n+{\theta }_n(x_n-x_{n-1})-p∥=∥x_n-p+{\theta }_n(x_n-x_{n-1})∥\hfill \\ \hfill & \le ∥x_n-p∥+{\theta }_n∥x_n-x_{n-1}∥.\hfill \end{array}$$

(11)

By using (11), we obtain

$$\begin{array}{cc}\hfill ∥{\delta }_n{y}_n-p∥& =∥{\delta }_n{y}_n-{\delta }_{n}p+{\delta }_{n}p-p∥\le {\delta }_n∥y_n-p∥+\left(1-{\delta }_n\right)∥p∥\hfill \\ \hfill & \le {\delta }_n∥x_n-p∥+{\theta }_n∥x_n-x_{n-1}∥+\left(1-{\delta }_n\right)∥p∥.\hfill \end{array}$$

(12)

Further, we observe that

$$\begin{array}{cc}\hfill ∥z_n-p∥& =∥(1-{\alpha }_n){\delta }_n{y}_n+{\alpha }_n{S}_n{\delta }_n{y}_n-p∥\hfill \\ \hfill & \le (1-{\alpha }_n)∥{\delta }_n{y}_n-p∥+{\alpha }_n∥S_n{\delta }_n{y}_n-p∥\hfill \\ \hfill & \le ∥{\delta }_n{y}_n-p∥.\hfill \end{array}$$

(13)

Using (12) and (13), we will have

$$\begin{array}{cc}\hfill ∥x_{n+1}-p∥& =∥(1-{\beta }_n)(z_n-p)+{\beta }_n(T_n{\delta }_n{y}_n-p)+{\epsilon }_n∥\hfill \\ \hfill & \le (1-{\beta }_n)∥z_n-p∥+{\beta }_n∥T_n{\delta }_n{y}_n-p∥+∥{\epsilon }_n∥\hfill \\ \hfill & \le (1-{\beta }_n)∥{\delta }_n{y}_n-p∥+{\beta }_n∥{\delta }_n{y}_n-p∥+∥{\epsilon }_n∥\hfill \\ \hfill & =∥{\delta }_n{y}_n-p∥+∥{\epsilon }_n∥\hfill \\ \hfill & \le {\delta }_n∥x_n-p∥+{\theta }_n∥x_n-x_{n-1}∥+\left(1-{\delta }_n\right)∥p∥+∥{\epsilon }_n∥\hfill \\ \hfill & =\left(1-\left(1-{\delta }_n\right)\right)∥x_n-p∥+\left(1-{\delta }_n\right)∥p∥+{\theta }_n∥x_n-x_{n-1}∥+∥{\epsilon }_n∥.\hfill \end{array}$$

(14)

Notice that ${\displaystyle \sum _{n=1}^{\infty }}\left({\theta }_n∥x_n-x_{n-1}∥+∥{\epsilon }_n∥\right)<+\infty$, we can apply Lemma 2 (1) to (14) and then it shows that ${\left(x_n\right)}_{n\ge 0}$ is bounded. □

Lemma 7. 

Let ${\left(S_n\right)}_{n\ge 0},{\left(T_n\right)}_{n\ge 0}:H\to H$ be two countable families of nonexpansive mappings such that ${\displaystyle \bigcap _{n=0}^{\infty }}Fix\left(S_n\right)\cap {\displaystyle \bigcap _{n=0}^{\infty }}Fix\left(T_n\right)\ne \varnothing$ and let ${\left(x_n\right)}_{n\ge 0}$ be generated by Equation (9). Let ${\left({\theta }_n\right)}_{n\ge 0}$ be a sequence in $[0,\theta ]$ with $\theta \in [0,1)$ such that ${\displaystyle \sum _{n=1}^{\infty }}{\theta }_n∥x_n-x_{n-1}∥<+\infty .$ Suppose 2., 3., 4b., and 5. in Assumption 1 hold and for any bounded subset Q of H, $\left({\left(S_n\right)}_{n\ge 0},Q\right)$ and $\left({\left(T_n\right)}_{n\ge 0},Q\right)$ satisfy AKTT condition. Then $\parallel x_{n+1}-x_n\parallel \to 0$ as $n\to \infty$.

Proof. 

Let us consider

$$\begin{array}{cc}\hfill ∥y_n-y_{n-1}∥& =∥x_n+{\theta }_n(x_n-x_{n-1})-\left(x_{n-1}+{\theta }_{n-1}(x_{n-1}-x_{n-2})\right)∥\hfill \\ \hfill & =∥x_n-x_{n-1}+{\theta }_n(x_n-x_{n-1})-{\theta }_{n-1}(x_{n-1}-x_{n-2})∥\hfill \\ \hfill & \le ∥x_n-x_{n-1}∥+{\theta }_n∥x_n-x_{n-1}∥+{\theta }_{n-1}∥x_{n-1}-x_{n-2}∥.\hfill \end{array}$$

(15)

By using (15), it can be observed that

$$\begin{array}{cc}\hfill & ∥{\delta }_n{y}_n-{\delta }_{n-1}{y}_{n-1}∥\hfill \\ \hfill & =∥{\delta }_n(y_n-y_{n-1})+({\delta }_n-{\delta }_{n-1})y_{n-1}∥\hfill \\ \hfill & \le {\delta }_n∥y_n-y_{n-1}∥+|{\delta }_n-{\delta }_{n-1}|∥y_{n-1}∥\hfill \\ \hfill & \le {\delta }_n\left(∥x_n-x_{n-1}∥+\parallel {\theta }_n(x_n-x_{n-1})-{\theta }_{n-1}(x_{n-1}-x_{n-2})\parallel \right)+|{\delta }_n-{\delta }_{n-1}|M_1\hfill \\ \hfill & \le {\delta }_n∥x_n-x_{n-1}∥+{\theta }_n∥x_n-x_{n-1}∥+{\theta }_{n-1}∥x_{n-1}-x_{n-2}∥+|{\delta }_n-{\delta }_{n-1}|M_1\hfill \\ \hfill & ={\delta }_n∥x_n-x_{n-1}∥+{\xi }_n^{\left(1\right)},\hfill \end{array}$$

(16)

where ${\xi }_n^{\left(1\right)}:={\theta }_n∥x_n-x_{n-1}∥+{\theta }_{n-1}∥x_{n-1}-x_{n-2}∥+|{\delta }_n-{\delta }_{n-1}|M_1$

and $M_1:=sup\left\{\left.∥y_{n-1}∥\right|n\in \mathbb{N}\right\}$.

Moreover, we can find that

$$\begin{array}{cc}\hfill & ∥z_n-z_{n-1}∥\hfill \\ \hfill & =∥(1-{\alpha }_n){\delta }_n{y}_n+{\alpha }_n{S}_n{\delta }_n{y}_n-((1-{\alpha }_{n-1}){\delta }_{n-1}{y}_{n-1}+{\alpha }_{n-1}{S}_{n-1}{\delta }_{n-1}{y}_{n-1})∥\hfill \\ \hfill & =∥(1-{\alpha }_n){\delta }_n{y}_n+{\alpha }_n\left(S_n{\delta }_n{y}_n-S_n{\delta }_{n-1}{y}_{n-1}+S_n{\delta }_{n-1}{y}_{n-1}\right)\hfill \\ \hfill & +\left(-{\alpha }_{n-1}+{\alpha }_{n-1}\right)S_n{\delta }_{n-1}{y}_{n-1}-((1-{\alpha }_n+{\alpha }_n-{\alpha }_{n-1}){\delta }_{n-1}{y}_{n-1}\hfill \\ \hfill & +{\alpha }_{n-1}{S}_{n-1}{\delta }_{n-1}{y}_{n-1})∥\hfill \\ \hfill & =∥(1-{\alpha }_n)({\delta }_n{y}_n-{\delta }_{n-1}{y}_{n-1})+{\alpha }_n(S_n{\delta }_n{y}_n-S_n{\delta }_{n-1}{y}_{n-1})\hfill \\ \hfill & +({\alpha }_n-{\alpha }_{n-1})S_n{\delta }_{n-1}{y}_{n-1}+{\alpha }_{n-1}(S_n{\delta }_{n-1}{y}_{n-1}-S_{n-1}{\delta }_{n-1}{y}_{n-1})\hfill \\ \hfill & +({\alpha }_{n-1}-{\alpha }_n){\delta }_{n-1}{y}_{n-1}∥\hfill \\ \hfill & \le (1-{\alpha }_n)∥{\delta }_n{y}_n-{\delta }_{n-1}{y}_{n-1}∥+{\alpha }_n∥S_n{\delta }_n{y}_n-S_n{\delta }_{n-1}{y}_{n-1}∥\hfill \\ \hfill & +\left|{\alpha }_n-{\alpha }_{n-1}\right|\left(∥S_n{\delta }_{n-1}{y}_{n-1}∥+∥{\delta }_{n-1}{y}_{n-1}∥\right)\hfill \\ \hfill & +{\alpha }_{n-1}∥S_n{\delta }_{n-1}{y}_{n-1}-S_{n-1}{\delta }_{n-1}{y}_{n-1}∥\hfill \\ \hfill & \le ∥{\delta }_n{y}_n-{\delta }_{n-1}{y}_{n-1}∥+\left|{\alpha }_n-{\alpha }_{n-1}\right|M_2+sup\left\{\left.∥S_{n}z-S_{n-1}z∥\right|z\in Q\right\}\hfill \\ \hfill & =∥{\delta }_n{y}_n-{\delta }_{n-1}{y}_{n-1}∥+{\xi }_n^{\left(2\right)},\hfill \end{array}$$

(17)

where $M_2:=sup\left\{\left.∥S_n{\delta }_{n-1}{y}_{n-1}∥+∥{\delta }_{n-1}{y}_{n-1}∥\right|n\in \mathbb{N}\right\}$, $Q\subseteq H$ is some bounded set such that ${\left({\delta }_{n-1}{y}_{n-1}\right)}_{n\ge 1}\subseteq Q$, and ${\xi }_n^{\left(2\right)}:=\left|{\alpha }_n-{\alpha }_{n-1}\right|M_2+sup\left\{\left.∥S_{n}z-S_{n-1}z∥\right|z\in Q\right\}.$ In the final step, by employing (16) and (17), it will obtain the following result

$$\begin{array}{cc}\hfill ∥x_{n+1}-x_n∥& =∥z_n+{\beta }_n(T_n{\delta }_n{y}_n-z_n)+{\epsilon }_n-(z_{n-1}+{\beta }_{n-1}(T_{n-1}{\delta }_{n-1}{y}_{n-1}-z_{n-1})+{\epsilon }_{n-1})∥\hfill \\ \hfill & =∥(1-{\beta }_n)z_n+{\beta }_n{T}_n{\delta }_n{y}_n+{\epsilon }_n-((1-{\beta }_{n-1})z_{n-1}+{\beta }_{n-1}{T}_{n-1}{\delta }_{n-1}{y}_{n-1}+{\epsilon }_{n-1})∥\hfill \\ \hfill & =∥(1-{\beta }_n)z_n+{\beta }_n\left(T_n{\delta }_n{y}_n-T_n{\delta }_{n-1}{y}_{n-1}+T_n{\delta }_{n-1}{y}_{n-1}\right)+\left(-{\beta }_{n-1}+{\beta }_{n-1}\right)T_n{\delta }_{n-1}{y}_{n-1}\hfill \\ \hfill & -(1-{\beta }_n+{\beta }_n-{\beta }_{n-1})z_{n-1}-{\beta }_{n-1}{T}_{n-1}{\delta }_{n-1}{y}_{n-1}+({\epsilon }_n-{\epsilon }_{n-1})∥\hfill \\ \hfill & =∥(1-{\beta }_n)(z_n-z_{n-1})+{\beta }_n(T_n{\delta }_n{y}_n-T_n{\delta }_{n-1}{y}_{n-1})+({\beta }_n-{\beta }_{n-1})T_n{\delta }_{n-1}{y}_{n-1}\hfill \\ \hfill & +{\beta }_{n-1}(T_n{\delta }_{n-1}{y}_{n-1}-T_{n-1}{\delta }_{n-1}{y}_{n-1})+({\beta }_{n-1}-{\beta }_n)z_{n-1}+({\epsilon }_n-{\epsilon }_{n-1})∥\hfill \\ \hfill & \le (1-{\beta }_n)∥z_n-z_{n-1}∥+{\beta }_n∥T_n{\delta }_n{y}_n-T_n{\delta }_{n-1}{y}_{n-1}∥+\left|{\beta }_n-{\beta }_{n-1}\right|\left(∥T_n{\delta }_{n-1}{y}_{n-1}∥+∥z_{n-1}∥\right)\hfill \\ \hfill & +{\beta }_{n-1}∥T_n{\delta }_{n-1}{y}_{n-1}-T_{n-1}{\delta }_{n-1}{y}_{n-1}∥+∥{\epsilon }_n-{\epsilon }_{n-1}∥\hfill \\ \hfill & \le (1-{\beta }_n)\left(∥{\delta }_n{y}_n-{\delta }_{n-1}{y}_{n-1}∥+{\xi }_n^{\left(2\right)}\right)+{\beta }_n∥{\delta }_n{y}_n-{\delta }_{n-1}{y}_{n-1}∥+|{\beta }_n-{\beta }_{n-1}|M_3\hfill \\ \hfill & +sup\left\{\left.∥T_{n}z-T_{n-1}z∥\right|z\in Q\right\}+∥{\epsilon }_n-{\epsilon }_{n-1}∥\hfill \\ \hfill & \le ∥{\delta }_n{y}_n-{\delta }_{n-1}{y}_{n-1}∥+{\xi }_n^{\left(2\right)}+{\xi }_n^{\left(3\right)}\hfill \\ \hfill & \le {\delta }_n∥x_n-x_{n-1}∥+{\xi }_n^{\left(1\right)}+{\xi }_n^{\left(2\right)}+{\xi }_n^{\left(3\right)}\hfill \\ \hfill & =(1-(1-{\delta }_n))∥x_n-x_{n-1}∥+(1-{\delta }_n)0+{\xi }_n^{\left(1\right)}+{\xi }_n^{\left(2\right)}+{\xi }_n^{\left(3\right)},\hfill \end{array}$$

(18)

where $M_3=sup\left\{\left.∥T_n{\delta }_{n-1}{y}_{n-1}∥+∥z_{n-1}∥\right|n\in \mathbb{N}\right\}$ and

${\xi }_n^{\left(3\right)}:=|{\beta }_n-{\beta }_{n-1}|M_3+sup\left\{\left.∥T_{n}z-T_{n-1}z∥\right|z\in Q\right\}+∥{\epsilon }_n-{\epsilon }_{n-1}∥.$ By using 2., 3., 4b., and 5. in Assumption 1 and AKTT conditon, we obtain that ${\displaystyle \sum _{n=1}^{\infty }}\left({\xi }_n^{\left(1\right)}+{\xi }_n^{\left(2\right)}+{\xi }_n^{\left(3\right)}\right)<+\infty$. By applying Lemma 2 (2) to (18), it yields that $∥x_{n+1}-x_n∥\to 0$  as  $n\to +\infty .$  □

Theorem 1.

Let ${\left(S_n\right)}_{n\ge 0},{\left(T_n\right)}_{n\ge 0}:H\to H$ be two countable families of nonexpansive mappings such that ${\displaystyle \bigcap _{n=0}^{\infty }}Fix\left(S_n\right)\cap {\displaystyle \bigcap _{n=0}^{\infty }}Fix\left(T_n\right)\ne \varnothing$ and let ${\left(x_n\right)}_{n\ge 0}$ be generated by Equation (9). Let ${\left({\theta }_n\right)}_{n\ge 0}$ be a sequence in $[0,\theta ]$ with $\theta \in [0,1)$ such that ${\displaystyle \sum _{n=1}^{\infty }}{\theta }_n∥x_n-x_{n-1}∥<+\infty .$ Suppose Assumption 1 holds and for any bounded subset Q of H, $\left({\left(S_n\right)}_{n\ge 0},Q\right)$ and $\left({\left(T_n\right)}_{n\ge 0},Q\right)$ satisfy AKTT condition. Let $S,T:H\to H$ be defined by $Sz=\underset{n\to \infty }{lim}{S}_{n}z$ and $Tz=\underset{n\to \infty }{lim}{T}_{n}z$ for all $z\in H$ and suppose that $Fix\left(S\right)={\displaystyle \bigcap _{n=0}^{\infty }}Fix\left(S_n\right)$ and $Fix\left(T\right)={\displaystyle \bigcap _{n=0}^{\infty }}Fix\left(T_n\right)$ and let $\mathsf{\Omega }:={\displaystyle \bigcap _{n=0}^{\infty }}Fix\left(S_n\right)\cap {\displaystyle \bigcap _{n=0}^{\infty }}Fix\left(T_n\right)=Fix\left(S\right)\cap Fix\left(T\right)$. Then, the sequence ${\left(x_n\right)}_{n\ge 0}$ strongly converges to $x^{\ast }:=P_{\mathsf{\Omega }}\left(0\right)$.

Proof. 

From Lemma 6, we have that ${\left(x_n\right)}_{n\ge 0}$ is bounded. By (11) and (13), it is not hard to see that ${\left(y_n\right)}_{n\ge 0}$ and ${\left(z_n\right)}_{n\ge 0}$ are also bounded. Let $x^{\ast }=P_{\mathsf{\Omega }}\left(0\right)$. The first step let us consider the following inequality and equality.

$$\begin{array}{cc}\hfill {∥z_n-x^{\ast }∥}^2& ={∥(1-{\alpha }_n){\delta }_n{y}_n+{\alpha }_n{S}_n{\delta }_n{y}_n-x^{\ast }∥}^2\hfill \\ \hfill & ={∥(1-{\alpha }_n)({\delta }_n{y}_n-x^{\ast })+{\alpha }_n(S_n{\delta }_n{y}_n-x^{\ast })∥}^2\hfill \\ \hfill & =(1-{\alpha }_n){∥{\delta }_n{y}_n-x^{\ast }∥}^2+{\alpha }_n{∥S_n{\delta }_n{y}_n-x^{\ast }∥}^2\hfill \\ \hfill & -{\alpha }_n(1-{\alpha }_n){∥{\delta }_n{y}_n-S_n{\delta }_n{y}_n∥}^2\hfill \\ \hfill & \le {∥{\delta }_n{y}_n-x^{\ast }∥}^2-{\alpha }_n(1-{\alpha }_n){∥{\delta }_n{y}_n-S_n{\delta }_n{y}_n∥}^2,\hfill \end{array}$$

(19)

and

$$\begin{array}{cc}\hfill {∥z_n-T_n{\delta }_n{y}_n∥}^2& ={∥(1-{\alpha }_n){\delta }_n{y}_n+{\alpha }_n{S}_n{\delta }_n{y}_n-T_n{\delta }_n{y}_n∥}^2\hfill \\ \hfill & ={∥(1-{\alpha }_n)\left({\delta }_n{y}_n-T_n{\delta }_n{y}_n\right)+{\alpha }_n\left(S_n{\delta }_n{y}_n-T_n{\delta }_n{y}_n\right)∥}^2\hfill \\ \hfill & =(1-{\alpha }_n){∥{\delta }_n{y}_n-T_n{\delta }_n{y}_n∥}^2+{\alpha }_n{∥S_n{\delta }_n{y}_n-T_n{\delta }_n{y}_n∥}^2\hfill \\ \hfill & -{\alpha }_n(1-{\alpha }_n){∥{\delta }_n{y}_n-S_n{\delta }_n{y}_n∥}^2.\hfill \end{array}$$

(20)

By using Lemma 1 (1), we obtain the following inequality.

$$\begin{array}{cc}\hfill {∥{\delta }_n{y}_n-x^{\ast }∥}^2& ={∥{\delta }_n(y_n-x^{\ast })+({\delta }_n-1)x^{\ast }∥}^2\hfill \\ \hfill & ={\delta }_n^2{∥y_n-x^{\ast }∥}^2+2{\delta }_n(1-{\delta }_n)\langle -x^{\ast },y_n-x^{\ast }\rangle +{(1-{\delta }_n)}^2{∥x^{\ast }∥}^2\hfill \\ \hfill & \le {\delta }_n{∥y_n-x^{\ast }∥}^2+(1-{\delta }_n)\left(2{\delta }_n\langle -x^{\ast },y_n-x^{\ast }\rangle +(1-{\delta }_n){∥x^{\ast }∥}^2\right)\hfill \\ \hfill & ={\delta }_n{∥x_n+{\theta }_n(x_n-x_{n-1})-x^{\ast }∥}^2+(1-{\delta }_n)\left(2{\delta }_n\langle -x^{\ast },y_n-x^{\ast }\rangle +(1-{\delta }_n){∥x^{\ast }∥}^2\right)\hfill \\ \hfill & ={\delta }_n{∥x_n-x^{\ast }+{\theta }_n(x_n-x_{n-1})∥}^2+(1-{\delta }_n)\left(2{\delta }_n\langle -x^{\ast },y_n-x^{\ast }\rangle +(1-{\delta }_n){∥x^{\ast }∥}^2\right)\hfill \\ \hfill & \le {\delta }_n{∥x_n-x^{\ast }∥}^2+2{\delta }_n{\theta }_n〈y_n-x^{\ast },x_n-x_{n-1}〉+(1-{\delta }_n)\left(2{\delta }_n\langle -x^{\ast },y_n-x^{\ast }\rangle +(1-{\delta }_n){∥x^{\ast }∥}^2\right)\hfill \\ \hfill & \le {\delta }_n{∥x_n-x^{\ast }∥}^2+\left(2{\delta }_n∥y_n-x^{\ast }∥\right){\theta }_n∥x_n-x_{n-1}∥+(1-{\delta }_n)\left(2{\delta }_n\langle -x^{\ast },y_n-x^{\ast }\rangle +(1-{\delta }_n){∥x^{\ast }∥}^2\right)\hfill \\ \hfill & \le {\delta }_n{∥x_n-x^{\ast }∥}^2+K_1{\theta }_n∥x_n-x_{n-1}∥+(1-{\delta }_n)\left(2{\delta }_n\langle -x^{\ast },y_n-x^{\ast }\rangle +(1-{\delta }_n){∥x^{\ast }∥}^2\right),\hfill \end{array}$$

(21)

where $K_1:=sup\left\{\left.2{\delta }_n∥y_n-x^{\ast }∥\right|n\in \mathbb{N}\right\}$. Applying Lemma 1 (1), (19) and (20), we can see that

$$\begin{array}{cc}\hfill {∥x_{n+1}-x^{\ast }∥}^2& ={∥z_n+{\beta }_n(T_n{\delta }_n{y}_n-z_n)+{\epsilon }_n-x^{\ast }∥}^2\hfill \\ \hfill & ={∥(1-{\beta }_n)(z_n-x^{\ast })+{\beta }_n(T_n{\delta }_n{y}_n-x^{\ast })+{\epsilon }_n∥}^2\hfill \\ \hfill & \le {∥(1-{\beta }_n)(z_n-x^{\ast })+{\beta }_n(T_n{\delta }_n{y}_n-x^{\ast })∥}^2+2\langle {\epsilon }_n,x_{n+1}-x^{\ast }\rangle \hfill \\ \hfill & =(1-{\beta }_n){∥z_n-x^{\ast }∥}^2+{\beta }_n{∥T_n{\delta }_n{y}_n-x^{\ast }∥}^2-{\beta }_n(1-{\beta }_n){∥z_n-T_n{\delta }_n{y}_n∥}^2+2\langle {\epsilon }_n,x_{n+1}-x^{\ast }\rangle \hfill \\ \hfill & \le (1-{\beta }_n)\left({∥{\delta }_n{y}_n-x^{\ast }∥}^2-{\alpha }_n(1-{\alpha }_n){∥{\delta }_n{y}_n-S_n{\delta }_n{y}_n∥}^2\right)+{\beta }_n{∥{\delta }_n{y}_n-x^{\ast }∥}^2+2∥x_{n+1}-x^{\ast }∥∥{\epsilon }_n∥\hfill \\ \hfill & -{\beta }_n(1-{\beta }_n)\left((1-{\alpha }_n){∥{\delta }_n{y}_n-T_n{\delta }_n{y}_n∥}^2+{\alpha }_n{∥S_n{\delta }_n{y}_n-T_n{\delta }_n{y}_n∥}^2-{\alpha }_n(1-{\alpha }_n){∥{\delta }_n{y}_n-S_n{\delta }_n{y}_n∥}^2\right)\hfill \\ \hfill & ={∥{\delta }_n{y}_n-x^{\ast }∥}^2-(1-{\beta }_n){\alpha }_n(1-{\alpha }_n){∥{\delta }_n{y}_n-S_n{\delta }_n{y}_n∥}^2+2∥x_{n+1}-x^{\ast }∥∥{\epsilon }_n∥\hfill \\ \hfill & -{\beta }_n(1-{\beta }_n)(1-{\alpha }_n){∥{\delta }_n{y}_n-T_n{\delta }_n{y}_n∥}^2-{\beta }_n(1-{\beta }_n){\alpha }_n{∥S_n{\delta }_n{y}_n-T_n{\delta }_n{y}_n∥}^2\hfill \\ \hfill & +{\beta }_n(1-{\beta }_n){\alpha }_n(1-{\alpha }_n){∥{\delta }_n{y}_n-S_n{\delta }_n{y}_n∥}^2\hfill \\ \hfill & \le {\delta }_n{∥x_n-x^{\ast }∥}^2+K_1{\theta }_n∥x_n-x_{n-1}∥+(1-{\delta }_n)\left(2{\delta }_n〈-x^{\ast },y_n-x^{\ast }〉+(1-{\delta }_n){∥x^{\ast }∥}^2\right)\hfill \\ \hfill & +K_2∥{\epsilon }_n∥-{(1-{\beta }_n)}^2{\alpha }_n(1-{\alpha }_n){∥{\delta }_n{y}_n-S_n{\delta }_n{y}_n∥}^2-{\beta }_n(1-{\beta }_n)(1-{\alpha }_n){∥{\delta }_n{y}_n-T_n{\delta }_n{y}_n∥}^2\hfill \\ \hfill & -{\beta }_n(1-{\beta }_n){\alpha }_n{∥S_n{\delta }_n{y}_n-T_n{\delta }_n{y}_n∥}^2\hfill \\ \hfill & \le {\delta }_n{∥x_n-x^{\ast }∥}^2+(1-{\delta }_n)\left(2{\delta }_n〈-x^{\ast },y_n-x^{\ast }〉+(1-{\delta }_n){∥x^{\ast }∥}^2\right)+K_1{\theta }_n∥x_n-x_{n-1}∥\hfill \\ \hfill & +K_2∥{\epsilon }_n∥-{\beta }_n(1-{\beta }_n){\alpha }_n{∥S_n{\delta }_n{y}_n-T_n{\delta }_n{y}_n∥}^2\hfill \\ \hfill & \le {∥x_n-x^{\ast }∥}^2+(1-{\delta }_n)\left(2{\delta }_n〈-x^{\ast },y_n-x^{\ast }〉+(1-{\delta }_n){∥x^{\ast }∥}^2\right)+K_1{\theta }_n∥x_n-x_{n-1}∥+K_2∥{\epsilon }_n∥\hfill \\ \hfill & -{\beta }_n(1-{\beta }_n){\alpha }_n{∥S_n{\delta }_n{y}_n-T_n{\delta }_n{y}_n∥}^2,\hfill \end{array}$$

(22)

where $K_2:=sup\left\{\left.2∥x_{n+1}-x^{\ast }∥\right|n\in \mathbb{N}\right\}.$

$$\begin{array}{cc}\hfill & {\beta }_n(1-{\beta }_n){\alpha }_n{∥S_n{\delta }_n{y}_n-T_n{\delta }_n{y}_n∥}^2\hfill \\ \hfill & \le {∥x_n-x^{\ast }∥}^2-{∥x_{n+1}-x^{\ast }∥}^2+(1-{\delta }_n)\left(2{\delta }_n〈-x^{\ast },y_n-x^{\ast }〉+(1-{\delta }_n){∥x^{\ast }∥}^2\right)\hfill \\ \hfill & +K_1{\theta }_n∥x_n-x_{n-1}∥+K_2∥{\epsilon }_n∥\hfill \\ \hfill & \le ∥x_n-x_{n+1}∥\left(∥x_n-x^{\ast }∥+∥x_{n+1}-x^{\ast }∥\right)\hfill \\ \hfill & +(1-{\delta }_n)\left(2{\delta }_n〈-x^{\ast },y_n-x^{\ast }〉+(1-{\delta }_n){∥x^{\ast }∥}^2\right)+K_1{\theta }_n∥x_n-x_{n-1}∥+K_2∥{\epsilon }_n∥\hfill \end{array}$$

(23)

Since $\underset{n\to \infty }{liminf}{\beta }_n(1-{\beta }_n){\alpha }_n>0$, so (23) yields that

$$∥S_n{\delta }_n{y}_n-T_n{\delta }_n{y}_n∥\to 0\mathrm{as}n\to \infty .$$

(24)

On the other hand, let us consider

$$\begin{array}{cc}\hfill ∥T_n{\delta }_n{y}_n-{\delta }_n{y}_n∥& =∥T_n{\delta }_n{y}_n-x_{n+1}+x_{n+1}-{\delta }_n{y}_n∥\hfill \\ \hfill & \le ∥T_n{\delta }_n{y}_n-x_{n+1}∥+∥x_{n+1}-{\delta }_n{y}_n∥\hfill \\ \hfill & =∥T_n{\delta }_n{y}_n-\left(z_n+{\beta }_n\left(T_n{\delta }_n{y}_n-z_n\right)+{\epsilon }_n\right)∥+∥(1-{\delta }_n)x_{n+1}+{\delta }_n{x}_{n+1}-{\delta }_n{y}_n∥\hfill \\ \hfill & \le ∥(1-{\beta }_n)(T_n{\delta }_n{y}_n-z_n)-{\epsilon }_n∥+(1-{\delta }_n)∥x_{n+1}∥+{\delta }_n∥x_{n+1}-y_n∥\hfill \\ \hfill & \le (1-{\beta }_n)∥z_n-T_n{\delta }_n{y}_n∥+∥{\epsilon }_n∥+(1-{\delta }_n)∥x_{n+1}∥+{\delta }_n(∥x_{n+1}-x_n∥+{\theta }_n∥x_n-x_{n-1}∥)\hfill \\ \hfill & =(1-{\beta }_n)∥(1-{\alpha }_n)({\delta }_n{y}_n-T_n{\delta }_n{y}_n)+{\alpha }_n(S_n{\delta }_n{y}_n-T_n{\delta }_n{y}_n)∥+∥{\epsilon }_n∥+(1-{\delta }_n)∥x_{n+1}∥\hfill \\ \hfill & +{\delta }_n(∥x_{n+1}-x_n∥+{\theta }_n∥x_n-x_{n-1}∥)\hfill \\ \hfill & \le (1-{\beta }_n)(1-{\alpha }_n)∥{\delta }_n{y}_n-T_n{\delta }_n{y}_n∥+(1-{\beta }_n){\alpha }_n∥S_n{\delta }_n{y}_n-T_n{\delta }_n{y}_n∥+∥{\epsilon }_n∥+(1-{\delta }_n)∥x_{n+1}∥\hfill \\ \hfill & +{\delta }_n(∥x_{n+1}-x_n∥+{\theta }_n∥x_n-x_{n-1}∥)\hfill \\ \hfill & \le (1-{\alpha }_n)∥{\delta }_n{y}_n-T_n{\delta }_n{y}_n∥+(1-{\beta }_n){\alpha }_n∥S_n{\delta }_n{y}_n-T_n{\delta }_n{y}_n∥+∥{\epsilon }_n∥+(1-{\delta }_n)∥x_{n+1}∥\hfill \\ \hfill & +{\delta }_n(∥x_{n+1}-x_n∥+{\theta }_n∥x_n-x_{n-1}∥)\hfill \end{array}$$

(25)

By using (25) and (24), it yields the following result.

$$\begin{array}{cc}\hfill & ∥T_n{\delta }_n{y}_n-{\delta }_n{y}_n∥\hfill \\ \hfill & \le (1-{\beta }_n)∥S_n{\delta }_n{y}_n-T_n{\delta }_n{y}_n∥\hfill \\ \hfill & +\frac{1}{{\alpha }_n}\left(∥{\epsilon }_n∥+(1-{\delta }_n)∥x_{n+1}∥+{\delta }_n(∥x_{n+1}-x_n∥+{\theta }_n∥x_n-x_{n-1}∥)\right)\hfill \\ \hfill & \to 0\mathrm{as}n\to \infty .\hfill \end{array}$$

(26)

Next, by employing (24) and (26), we obtain the following result.

$$\begin{array}{cc}\hfill ∥{\delta }_n{y}_n-S_n{\delta }_n{y}_n∥& \le ∥{\delta }_n{y}_n-T_n{\delta }_n{y}_n+T_n{\delta }_n{y}_n-S_n{\delta }_n{y}_n∥\hfill \\ \hfill & \le ∥{\delta }_n{y}_n-T_n{\delta }_n{y}_n∥+∥T_n{\delta }_n{y}_n-S_n{\delta }_n{y}_n∥\hfill \\ \hfill & \to 0\mathrm{as}n\to \infty .\hfill \end{array}$$

(27)

By the virtue of (27), it allows us to obtain

$$\begin{array}{cc}\hfill ∥S{\delta }_n{y}_n-{\delta }_n{y}_n∥& \le ∥S{\delta }_n{y}_n-S_n{\delta }_n{y}_n∥+∥S_n{\delta }_n{y}_n-{\delta }_n{y}_n∥\hfill \\ \hfill & \le sup\left\{\left.∥Sz-S_{n}z∥\right|z\in Q\right\}+∥S_n{\delta }_n{y}_n-{\delta }_n{y}_n∥\hfill \\ \hfill & \to 0\mathrm{as}n\to \infty .\hfill \end{array}$$

(28)

Similarly, the inequality (26) yields the result below

$$\begin{array}{cc}\hfill ∥T{\delta }_n{y}_n-{\delta }_n{y}_n∥& \le ∥T{\delta }_n{y}_n-T_n{\delta }_n{y}_n∥+∥T_n{\delta }_n{y}_n-{\delta }_n{y}_n∥\hfill \\ \hfill & \le sup\left\{\left.∥Tz-T_{n}z∥\right|z\in Q\right\}+∥T_n{\delta }_n{y}_n-{\delta }_n{y}_n∥\hfill \\ \hfill & \to 0\mathrm{as}n\to \infty .\hfill \end{array}$$

(29)

It is not hard to see that (22) can be rewritten to be the following

$$\begin{array}{cc}\hfill & {∥x_{n+1}-x^{\ast }∥}^2\hfill \\ \hfill & \le (1-(1-{\delta }_n)){∥x_n-x^{\ast }∥}^2+(1-{\delta }_n)\left(2{\delta }_n〈-x^{\ast },y_n-x^{\ast }〉+(1-{\delta }_n){∥x^{\ast }∥}^2\right)\hfill \\ \hfill & +K_1{\theta }_n∥x_n-x_{n-1}∥+K_2∥{\epsilon }_n∥.\hfill \end{array}$$

(30)

In order to prove that ${\left(x_n\right)}_{n\ge 0}$ converges strongly to $x^{\ast }$, it is enough to show that

$${\displaystyle \underset{n\to \infty }{lim\; sup}\langle -x^{\ast },y_n-x^{\ast }\rangle \le 0.}$$

(31)

Assume to a contrary that (31) is not true, then there exists a real number $k>0$ and a subsequence ${\left(y_{n_i}\right)}_{i\ge 0}$ of ${\left(y_n\right)}_{n\ge 0}$ such that

$$\langle -x^{\ast },y_{n_i}-x^{\ast }\rangle \ge k>0,\forall i\ge 0.$$

Since ${\left(y_{n_i}\right)}_{i\ge 0}$ is bounded on a Hilbert space H, we obtain that there exists a subsequence ${(y_{n_{i_j}})}_{j\ge 0}$ of ${(y_{n_i})}_{i\ge 0}$ such that $y_{n_{i_j}}\rightharpoonup y\in H$ as $j\to \infty$. Therefore,

$$0<k\le \underset{j\to \infty }{lim}〈-x^{\ast },y_{n_{i_j}}-x^{\ast }〉=〈-x^{\ast },y-x^{\ast }〉.$$

(32)

Notice that ${\displaystyle \underset{n\to \infty }{lim}{\delta }_n=1}$, we obtain ${\delta }_{n_{i_j}}{y}_{n_{i_j}}\rightharpoonup y$ as $j\to \infty .$ Applying (28), (29) and Lemma 3, we obtain that $y\in Fix\left(S\right)\cap Fix\left(T\right).$ The characterization of metric projection via Lemma 4 yields that $〈-x^{\ast },y-x^{\ast }〉=〈0-x^{\ast },y-x^{\ast }〉\le 0$ which lead to a contradiction. For this reason, the inequality (31) holds. So, it is unlocked and allows us to obtain

$$\underset{n\to \infty }{limsup}\left(2{\delta }_n〈-x^{\ast },y_n-x^{\ast }〉+(1-{\delta }_n){∥x^{\ast }∥}^2\right)\le 0.$$

Applying Lemma 2 (2) to (30), it can be concluded that ${\displaystyle \underset{n\to \infty }{lim}{x}_n=x^{\ast }.}$ This complete the proof. □

Remark 2.

The assumption of the sequence ${\left({\theta }_n\right)}_{n\ge 0}$ in Theorem 1 is verified, if we choose ${\theta }_n$ such that $0\le {\theta }_n\le {\overline{\theta }}_n$, where

$${\overline{\theta }}_n=\left\{\begin{array}{cccc}& min\left\{\theta ,\frac{c_n}{∥x_n-x_{n-1}∥}\right\},\hfill & & \mathit{if}{x}_n\ne x_{n-1},\hfill \\ & \theta ,\hfill & & \mathit{otherwise},\hfill \end{array}\right.$$

and ${\sum }_{n\ge 0}{c}_n<+\infty .$

It is worth to mention that The implication of Theorem 1 can be applied for a wide range of the previous existing results as follows:

The first one, in the sense of constant sequences of nonexpansive mappings as in the followings:

Corollary 1.

Let $S,T:H\to H$ be two nonexpansive mappings such that $Fix\left(S\right)\cap Fix\left(T\right)\ne \varnothing$ and let ${\left(x_n\right)}_{n\ge 0}$ be generated by

$$\left\{\begin{array}{cc}\hfill & x_0,x_1\in H,\hfill \\ \hfill & y_n=x_n+{\theta }_n(x_n-x_{n-1}),\hfill \\ \hfill & z_n=(1-{\alpha }_n){\delta }_n{y}_n+{\alpha }_{n}S{\delta }_n{y}_n,\hfill \\ \hfill & x_{n+1}=z_n+{\beta }_n\left(T{\delta }_n{y}_n-z_n\right)+{\epsilon }_n,\hfill \end{array}\right.$$

(33)

where ${\left({\theta }_n\right)}_{n\ge 0}$ is a sequence in $[0,\theta ]$ with $\theta \in [0,1)$ such that ${\displaystyle \sum _{n=1}^{\infty }}{\theta }_n∥x_n-x_{n-1}∥<+\infty .$ Suppose Assumption 1 holds. Then, the sequence ${\left(x_n\right)}_{n\ge 0}$ strongly converges to $x^{\ast }:=P_{Fix\left(S\right)\cap Fix\left(T\right)}\left(0\right)$.

Proof. 

If $S_n=S$ and $T_n=T$ for all $n\in \mathbb{N}\cup \left\{0\right\}$ where $S,T:H\to H$ are nonexpansive mappings, then it is obvious that $\left({\left(S_n\right)}_{n\ge 0},Q\right)$ and $\left({\left(T_n\right)}_{n\ge 0},Q\right)$ satisfy AKTT condition for any nonempty subset $Q\subseteq H$. Moreover, $Fix\left(S\right)={\displaystyle \bigcap _{n=0}^{\infty }}Fix\left(S_n\right)$ and $Fix\left(T\right)={\displaystyle \bigcap _{n=0}^{\infty }}Fix\left(T_n\right)$. Therefore, the desired result is obtained by the implication of Theorem 1. □

Lemma 8. 

Let $S:H\to H$ be a nonexpansive mapping with $Fix\left(S\right)\ne ⌀$. Define $T_n:=(1-{\alpha }_n)I+{\alpha }_{n}S$ for all $n\in \mathbb{N}\cup \left\{0\right\}$. Then the following statements hold:

(i) 

$T_n:H\to H$ are nonexpansive mappings for all $n\in \mathbb{N}\cup \left\{0\right\}$.

(ii) 

If ${\left({\alpha }_n\right)}_{n\ge 0}\subseteq [0,1]$ with the property that $\underset{n\to \infty }{liminf}{\alpha }_n>0$, then $Fix\left(S\right)={\displaystyle \bigcap _{n=0}^{\infty }}Fix\left(T_n\right)$.

(iii) 

If ${\displaystyle \sum _{n=1}^{\infty }}\left|{\alpha }_n-{\alpha }_{n-1}\right|<+\infty$, then $\left({\left(T_n\right)}_{n\ge 0},Q\right)$ satisfies AKTT condition for any nonempty bounded subset Q of H.

Proof. 

It is not hard to verify the proof of (i). For the proof of (ii), let us consider

$$\begin{array}{cc}\hfill x\in Fix\left(S\right)& \Rightarrow x=Sx\Rightarrow 0={\alpha }_n(Sx-x)\hfill \\ \hfill & \Rightarrow x=x+{\alpha }_n(Sx-x)=(1-{\alpha }_n)x+{\alpha }_{n}Sx=T_{n}x,\forall n\in \mathbb{N}\hfill \\ \hfill & \Rightarrow x\in {\displaystyle \bigcap _{n=0}^{\infty }}Fix\left(T_n\right).\hfill \end{array}$$

On the other hand, it is found that

$$\begin{array}{cc}\hfill x\in {\displaystyle \bigcap _{n=0}^{\infty }}Fix\left(T_n\right)& \Rightarrow x=T_{n}x=(1-{\alpha }_n)x+{\alpha }_{n}Sx=x+{\alpha }_n(Sx-x),\forall n\in \mathbb{N}\hfill \\ \hfill & \Rightarrow 0={\alpha }_n(Sx-x),\forall n\in \mathbb{N}\Rightarrow \left(\underset{n\to \infty }{liminf}{\alpha }_n\right)∥Sx-x∥=0\hfill \\ \hfill & \Rightarrow ∥Sx-x∥=\frac{0}{\underset{n\to \infty }{liminf}{\alpha }_n}=0\hfill \\ \hfill & \Rightarrow x\in Fix\left(S\right).\hfill \end{array}$$

Therefore, Lemma 8(ii) holds. For the proof of Lemma 8(iii), let $Q\subseteq H$ be a nonempty and bounded subset. Then, since $Fix\left(S\right)\ne ⌀$, it is not hard to see that the set $\left\{\left.∥z∥+∥Sz∥\right|z\in Q\right\}$ is also bounded. And then it can be observed that for any $n\in \mathbb{N}$,

$$\begin{array}{cc}\hfill ∥T_{n}z-T_{n-1}z∥& =∥({\alpha }_n-{\alpha }_{n-1})z+({\alpha }_n-{\alpha }_{n-1})Sz∥\hfill \\ \hfill & \le \left|{\alpha }_n-{\alpha }_{n-1}\right|∥z∥+\left|{\alpha }_n-{\alpha }_{n-1}\right|∥Sz∥\hfill \\ \hfill & =\left|{\alpha }_n-{\alpha }_{n-1}\right|\left(∥z∥+∥Sz∥\right),\forall z\in Q.\hfill \end{array}$$

This implies that $sup\left\{\left.∥T_{n}z-T_{n-1}z∥\right|z\in Q\right\}\le L\left|{\alpha }_n-{\alpha }_{n-1}\right|,$

where $L:=sup\left\{\left.∥z∥+∥Sz∥\right|z\in Q\right\}$. Therefore,

$${\displaystyle \sum _{n=1}^{\infty }}sup\left\{\left.∥T_{n}z-T_{n-1}z∥\right|z\in Q\right\}\le L{\displaystyle \sum _{n=1}^{\infty }}\left|{\alpha }_n-{\alpha }_{n-1}\right|<+\infty .$$

This shows that Lemma 8(iii) is true. □

Corollary 2 

([25]). Let $S:H\to H$ be a nonexpansive mapping such that $Fix\left(S\right)\ne ⌀$ and let ${\left(x_n\right)}_{n\ge 0}$ be generated by

$$\left\{\begin{array}{c}{x}_0,x_1\in H,\hfill \\ y_n=x_n+{\theta }_n(x_n-x_{n-1}),\hfill \\ x_{n+1}={\delta }_n{y}_n+{\alpha }_n\left(S{\delta }_n{y}_n-{\delta }_n{y}_n\right)+{\epsilon }_n,\hfill \end{array}\right.$$

(34)

where ${\left({\theta }_n\right)}_{n\ge 0}$ is a sequence in $[0,\theta ]$ with $\theta \in [0,1)$ such that ${\displaystyle \sum _{n=1}^{\infty }}{\theta }_n∥x_n-x_{n-1}∥<+\infty .$ Suppose that ${\left({\alpha }_n\right)}_{n\ge 0},{\left({\delta }_n\right)}_{n\ge 0}\subseteq (0,1]$ and ${\left({\epsilon }_n\right)}_{n\ge 0}\subseteq H$ are consistent with the following conditions:

(a) 

$\underset{n\to \infty }{liminf}{\alpha }_n>0$ and ${\displaystyle \sum _{n=1}^{\infty }}\left|{\alpha }_n-{\alpha }_{n-1}\right|<+\infty ,$

(b) 

$\underset{n\to \infty }{lim}{\delta }_n=1,{\displaystyle \sum _{n=0}^{\infty }}(1-{\delta }_n)=+\infty ,$ and ${\displaystyle \sum _{n=1}^{\infty }}\left|{\delta }_n-{\delta }_{n-1}\right|<+\infty ,$

(c) 

 ${\displaystyle \sum _{n=0}^{\infty }}∥{\epsilon }_n∥<+\infty .$

Then, the sequence ${\left(x_n\right)}_{n\ge 0}$ strongly converges to $x^{\ast }:=P_{Fix\left(S\right)}\left(0\right).$

Proof. 

First, it is not hard to verify that

$$x_{n+1}={\delta }_n{y}_n+{\alpha }_n(S{\delta }_n{y}_n-{\delta }_n{y}_n)+{\epsilon }_n\Rightarrow \left\{\begin{array}{cc}& z_n=(1-{\alpha }_n){\delta }_n{y}_n+{\alpha }_{n}S{\delta }_n{y}_n,\hfill \\ & x_{n+1}=z_n+\frac{1}{2}\left(T_n{\delta }_n{y}_n-z_n\right)+{\epsilon }_n,\hfill \end{array}\right.$$

where $z_n=(1-{\alpha }_n){\delta }_n{y}_n+{\alpha }_{n}S{\delta }_n{y}_n$ and $T_n=(1-{\alpha }_n)I+{\alpha }_{n}S$. Indeed, from the third line of (34), it is observed that

$$\begin{array}{cc}\hfill x_{n+1}& ={\delta }_n{y}_n+{\alpha }_n(S{\delta }_n{y}_n-{\delta }_n{y}_n)+{\epsilon }_n=(1-{\alpha }_n){\delta }_n{y}_n+{\alpha }_{n}S{\delta }_n{y}_n+{\epsilon }_n=z_n+0+{\epsilon }_n\hfill \\ \hfill & =z_n+\frac{1}{2}\left(z_n-z_n\right)+{\epsilon }_n=z_n+\frac{1}{2}\left((1-{\alpha }_n){\delta }_n{y}_n+{\alpha }_{n}S{\delta }_n{y}_n-z_n\right)+{\epsilon }_n\hfill \\ \hfill & =z_n+\frac{1}{2}\left(((1-{\alpha }_n)I+{\alpha }_{n}S){\delta }_n{y}_n-z_n\right)+{\epsilon }_n\hfill \\ \hfill & =z_n+\frac{1}{2}\left(T_n{\delta }_n{y}_n-z_n\right)+{\epsilon }_n.\hfill \end{array}$$

Setting ${\beta }_n=\frac{1}{2}$ for all $n\in \mathbb{N}\cup \left\{0\right\}$ and since $\underset{n\to \infty }{liminf}{\alpha }_n>0$, so we have $\underset{n\to \infty }{liminf}{\beta }_n(1-{\beta }_n){\alpha }_n=\underset{n\to \infty }{liminf}\left(\frac{1}{4}{\alpha }_n\right)=\frac{1}{4}\underset{n\to \infty }{liminf}{\alpha }_n>0.$ On the other hand, the definition of $\left\{T_n\right\}$ and the condition (a) imply that ${\left(T_n\right)}_{n\ge 0}$ satisfies (i)–(iii) of Lemma 8. Then, by applying Theorem 1, we have the desired result. □

Corollary 3

([16] Theorem 3). Let $S:H\to H$ be a nonexpansive mapping such that $Fix\left(S\right)\ne ⌀$ and let ${\left(x_n\right)}_{n\ge 0}$ be generated by

$$\left\{\begin{array}{c}{x}_0\in H,\hfill \\ x_{n+1}={\delta }_n{x}_n+{\alpha }_n\left(S{\delta }_n{x}_n-{\delta }_n{x}_n\right).\hfill \end{array}\right.$$

Suppose that ${\left({\alpha }_n\right)}_{n\ge 0},{\left({\delta }_n\right)}_{n\ge 0}\subseteq (0,1]$ are consistent with the following conditions:

1. 

$\underset{n\to \infty }{liminf}{\alpha }_n>0$ and ${\displaystyle \sum _{n=1}^{\infty }}\left|{\alpha }_n-{\alpha }_{n-1}\right|<+\infty ,$

2. 

$\underset{n\to \infty }{lim}{\delta }_n=1,{\displaystyle \sum _{n=0}^{\infty }}(1-{\delta }_n)=+\infty ,$ and ${\displaystyle \sum _{n=1}^{\infty }}\left|{\delta }_n-{\delta }_{n-1}\right|<+\infty .$

Then, the sequence ${\left(x_n\right)}_{n\ge 0}$ strongly converges to $x^{\ast }:=P_{Fix\left(S\right)}\left(0\right).$

Proof. 

By letting ${\theta }_n=0$ and ${\epsilon }_n=0$ for all $n\in \mathbb{N}\cup \left\{0\right\}$ in Corollary 2. Then, we obtain the desired result. □

4. Applications

This section focuses on exploring the potential uses of Equation (9) and its by-products in specific cases from Equation (9) to solve monotone inclusion problems.

Recall that the operator $W:H\to H$ is said to be:

• monotone if it satisfies $\langle Wx-Wy,x-y\rangle \ge 0$ for all $x,y\in H$.
• κ-cocoercive if there exists $\kappa >0$ such that $\langle Wx-Wy,x-y\rangle \ge {\kappa \parallel Wx-Wy\parallel }^2$ for all $x,y\in H.$

The set of all zeros of the operator W is denoted by $\mathrm{zer}\left(W\right):=\{z\in H:0=Wz\}.$

Let $\mathrm{G}\left(V\right):=\left\{\left.(x,y)\in H\times H\right|y\in Vx\right\}$ be the graph of a multivalued operator $V:H\to 2^H$. Then, V is called:

• monotone if for all $x,y\in H$, $u\in Vx$ and $v\in Vy$ implies that $\langle u-v,x-y\rangle \ge 0$.
• maximal monotone if $\mathrm{G}\left(V\right)$ is not properly contained in any graph of other multivalued monotone operator, that is, if $\hat{V}:H\to 2^H$ is a multivalued monotone operator such that $G\left(V\right)\subseteq G\left(\hat{V}\right)$, then $G\left(V\right)=G\left(\hat{V}\right)$.
• λ-strongly monotone if there exists $\lambda >0$ such that $\langle x-y,u-v\rangle \ge {\lambda \parallel x-y\parallel }^2$ for all $(x,u),(y,v)\in \mathrm{G}\left(V\right)$.

The multivalued operator $J_V:H\to 2^H$ which is defined by

$$J_V\left(x\right)={(I+V)}^{-1}\left(x\right),\forall x\in H$$

is called the resolvent operator associated with V. It is well known that if $V:H\to 2^H$ is maximal monotone and $\gamma >0$, then $J_{\gamma V}$ is single-valued and firmly nonexpansive. By employing $J_{\gamma V}$, the Yosida approximation$V_{\gamma }$ of V is defined by $V_{\gamma }:=\frac{1}{\gamma }\left(I-J_{\gamma V}\right)$.

Let us consider the following monotone inclusion problem:

$$\mathrm{find}x\in H\mathrm{such}\mathrm{that}0\in \left({\mu }_{1}Ux+\eta Vx+{\nu }_{1}Wx\right)\cap \left({\mu }_{2}Ux+\eta Vx+{\nu }_{2}Wx\right)$$

(35)

where ${\mu }_1,{\mu }_2,\eta ,{\nu }_1,{\nu }_2>0$, $U,V:H\to 2^H$ are maximal monotone operators and $W:H\to H$ is a $\kappa$-cocoercive operator with $\kappa >0$. For applying Equation (9) to solve (35), the important tool is as follows:

Proposition 1

([38], Proposition 2.1). Let $T_1,T_2:H\to H$ be two firmly nonexpansive operators and W be a κ-cocoercive operator with $\kappa >0$. Let $\nu \in (0,2\kappa )$. Then operator

$$T:=T_1\circ (2T_2-I-\nu W\circ T_2)+I-T_2$$

is α-averaged with coefficient $\alpha :=\frac{2\kappa }{4\kappa -\nu }<1$. In particular, the following inequality holds for all $x,y\in H$

$${\parallel Tx-Tx\parallel }^2\le {\parallel x-y\parallel }^2-\frac{(1-\alpha )}{\alpha }{\parallel (I-T)x-(I-T)y\parallel }^2.$$

The following lemma provides a characterization of $\mathrm{zer}\left(\mu U+\eta V+\nu W\right)$.

Lemma 9 

(Fixed point encoding). Let $\mu ,\eta ,\nu >0$, $U,V:H\to 2^H$ be maximal monotone operators and $W:H\to H$ be an operator. Suppose that $\mathrm{zer}\left(\mu U+\eta V+\nu W\right)\ne \varnothing .$ Then,

$$\mathrm{zer}\left(\mu U+\eta V+\nu W\right)=J_{\eta V}\left(Fix\left(T\right)\right),$$

where $T=J_{\mu U}\circ \left(2J_{\eta V}-I-\nu W\circ J_{\eta V}\right)+\left(I-J_{\eta V}\right).$

Proof. 

Let $x\in \mathrm{zer}\left(\mu U+\eta V+\nu W\right)$. Then, $0\in \mu Ux+\eta Vx+\nu Wx$. This implies that $\exists y_U\in Ux$ and $\exists y_V\in Vx$ such that $\mu y_U+\eta y_V+\nu Wx=0.$ Next, we expect that $x=J_{\eta V}\left(z\right)$ for some $z\in Fix\left(T\right)$. Let $z=x+\eta y_V$ and since $z=x+\eta y_V\in x+\eta Vx=\left(I+\eta V\right)x$. This means that $x=J_{\eta V}\left(z\right)$. Moreover, it can be observed that $2J_{\eta V}\left(z\right)-z-\nu W{J}_{\eta V}\left(z\right)=2x-z-\nu Wx=x-\nu Wx-\eta y_V=x+\mu y_U$. It remains to show that $z\in Fix\left(T\right)$. Let us consider

$$\begin{array}{cc}\hfill Tz& =J_{\mu U}\left(2J_{\eta V}\left(z\right)-z-\nu W{J}_{\eta V}\left(z\right)\right)+\left(z-J_{\eta V}\left(z\right)\right)\hfill \\ \hfill & =J_{\mu U}\left(x+\mu y_U\right)+\left(z-x\right)=x+z-x=z.\hfill \end{array}$$

This shows that $z\in Fix\left(T\right)$ and then $x=J_{\eta V}\left(z\right)\in J_{\eta V}\left(Fix\left(T\right)\right)$.

On the other hand, let $x\in J_{\eta V}\left(Fix\left(T\right)\right).$ Then, there exists $z\in Fix\left(T\right)$ such that $x=J_{\eta V}\left(z\right).$ Moreover, it can be observed that

$$z=Tz=J_{\mu U}\left(2J_{\eta V}\left(z\right)-z-\nu W{J}_{\eta V}\left(z\right)\right)+\left(z-J_{\eta V}\left(z\right)\right).$$

This implies that $x=J_{\eta V}\left(z\right)=J_{\mu U}\left(2J_{\eta V}\left(z\right)-z-\nu W{J}_{\eta V}\left(z\right)\right)$ and by using some properties of Yosida approximation $\frac{1}{\mu }\left(I-J_{\mu U}\right)$ of U and $\frac{1}{\eta }\left(I-J_{\eta V}\right)$ of V, it yields that

$$\begin{array}{cc}\hfill z& =Tz\hfill \\ \hfill & =J_{\mu U}\circ \left(2J_{\eta V}-I-\nu W\circ J_{\eta V}\right)+\left(I-J_{\eta V}\right)z\hfill \\ \hfill & =z+J_{\mu U}\left(2J_{\eta V}\left(z\right)-z-\nu W{J}_{\eta V}\left(z\right)\right)-J_{\eta V}\left(z\right)\hfill \\ \hfill & =z+J_{\mu U}\left(2J_{\eta V}\left(z\right)-z-\nu W{J}_{\eta V}\left(z\right)\right)-\left(2J_{\eta V}\left(z\right)-z-\nu W{J}_{\eta V}\left(z\right)\right)+J_{\eta V}\left(z\right)\hfill \\ \hfill & -z-\nu W{J}_{\eta V}\left(z\right)\hfill \\ \hfill & =z-\mu \left(\frac{1}{\mu }\left(I-J_{\mu U}\right)\left(2J_{\eta V}\left(z\right)-z-\nu W{J}_{\eta V}\left(z\right)\right)\right)-\eta \left(\frac{1}{\eta }\left(I-J_{\eta V}\right)\left(z\right)\right)\hfill \\ \hfill & -\nu W{J}_{\eta V}\left(z\right)\hfill \\ \hfill & \in z-\mu U{J}_{\mu U}\left(2J_{\eta V}\left(z\right)-z-\nu W{J}_{\eta V}\left(z\right)\right)-\eta V{J}_{\eta V}\left(z\right)-\nu W{J}_{\eta V}\left(z\right)\hfill \\ \hfill & =z-\mu Ux-\eta Vx-\nu Wx.\hfill \end{array}$$

By simple calculation, we have that $0\in \mu Ux+\eta Vx+\nu Wx$ and then

$x\in zer\left(\mu U+\eta V+\nu W\right).$□

Setting

$$S:=J_{{\mu }_{1}U}\circ \left(2J_{\eta V}-I-{\nu }_{1}W\circ J_{\eta V}\right)+\left(I-J_{\eta V}\right)$$

and

$$T:=J_{{\mu }_{2}U}\circ \left(2J_{\eta V}-I-{\nu }_{2}W\circ J_{\eta V}\right)+\left(I-J_{\eta V}\right).$$

Assume that $Fix\left(S\right)\cap Fix\left(T\right)\ne \varnothing$ and by employing (33), then the following algorithm can be constructed for solving (35) as follows:

$$\left\{\begin{array}{c}{x}_0,x_1\in H,\hfill \\ y_n=x_n+{\theta }_n(x_n-x_{n-1}),\hfill \\ j_n=J_{\eta V}\left({\delta }_n{y}_n\right),\hfill \\ z_n={\delta }_n{y}_n+{\alpha }_n\left(J_{{\mu }_{1}U}\left(2j_n-{\delta }_n{y}_n-{\nu }_{1}W\left(j_n\right)\right)-j_n\right),\hfill \\ x_{n+1}=(1-{\beta }_n)z_n+{\beta }_n\left(J_{{\mu }_{2}U}\left(2j_n-{\delta }_n{y}_n-{\nu }_{2}W\left(j_n\right)\right)+\left({\delta }_n{y}_n-j_n\right)\right)+{\epsilon }_n,\hfill \end{array}\right.$$

(36)

for all $n\ge 1$, where ${\mu }_1,{\mu }_2,\eta >0$ and ${\nu }_1,{\nu }_2\in (0,2\kappa )$.

Theorem 2.

Let $U,V:H\to 2^H$ be two maximal monotone operators and $W:H\to H$ be κ-cocoercive with $\kappa >0$. Suppose that ${\mu }_1,{\mu }_2,\eta ,{\nu }_1,{\nu }_2>0$, $S=J_{{\mu }_{1}U}\circ \left(2J_{\eta V}-I-{\nu }_{1}W\circ J_{\eta V}\right)+\left(I-J_{\eta V}\right)$, $T=J_{{\mu }_{2}U}\circ \left(2J_{\eta V}-I-{\nu }_{2}W\circ J_{\eta V}\right)+\left(I-J_{\eta V}\right)$ and $Fix\left(S\right)\cap Fix\left(T\right)\ne \varnothing$, that is, $\varnothing \ne J_{\eta V}\left(Fix\left(S\right)\cap Fix\left(T\right)\right)\subseteq \mathrm{zer}\left({\mu }_{1}U+\eta V+{\nu }_{1}W\right)\cap \mathrm{zer}\left({\mu }_{2}U+\eta V+{\nu }_{2}W\right)$. Let ${\left({\theta }_n\right)}_{n\ge 1}$ be a sequence in $[0,\theta ]$ with $\theta \in [0,1)$ and ${\mu }_1,{\mu }_2,\eta >0$ and ${\nu }_1,{\nu }_2\in (0,2\kappa )$. Let ${\left(x_n\right)}_{n\ge 0}$ and ${\left(j_n\right)}_{n\ge 1}$ be generated by Equation (36). Assume that ${\displaystyle \sum _{n=1}^{\infty }}{\theta }_n∥x_n-x_{n-1}∥<+\infty$ and the Assumption 1 holds. Then the following statements are true:

(A) 

${\left(x_n\right)}_{n\ge 0}$ strongly converges to $x^{\ast }:=P_{Fix\left(S\right)\cap Fix\left(T\right)}\left(0\right)$.

(B) 

${\left(j_n\right)}_{n\ge 1}$ strongly converges to $J_{\eta V}\left(x^{\ast }\right)\in \mathrm{zer}\left({\mu }_{1}U+\eta V+{\nu }_{1}W\right)\cap \mathrm{zer}\left({\mu }_{2}U+\eta V+{\nu }_{2}W\right)$.

Proof. 

(A): Let ${\left(x_n\right)}_{n\ge 0}$ be generated by Equation (36). By Proposition 1, we obtain S and T are nonexpansive. By applying Corollary 1, we have the sequence ${\left(x_n\right)}_{n\ge 0}$ strongly converges to $x^{\ast }:=P_{Fix\left(S\right)\cap Fix\left(T\right)}\left(0\right)$ as $n\to +\infty$.

(B): Notice that $y_n\to x^{\ast }$ and ${\delta }_n\to 1$ as $n\to +\infty$, thus ${\delta }_n{y}_n\to x^{\ast }$ as $n\to +\infty$. Since $J_{\eta V}$ is continuous, so $j_n=J_{\eta V}\left({\delta }_n{y}_n\right)\to J_{\eta V}\left(x^{\ast }\right)\in J_{\eta V}(Fix\left(S\right)\cap Fix\left(T\right))$. By applying Lemma 9, we have that $zer\left({\mu }_{1}U+\eta V+{\nu }_{1}W\right)=J_{\eta V}\left(Fix\left(S\right)\right)$ and $\mathrm{zer}\left({\mu }_{2}U+\eta V+{\nu }_{2}W\right)=J_{\eta V}\left(Fix\left(T\right)\right)$. Then,

$$\begin{array}{cc}\hfill J_{\eta V}\left(Fix\left(S\right)\cap Fix\left(T\right)\right)& \subseteq J_{\eta V}\left(Fix\left(S\right)\right)\cap J_{\eta V}\left(Fix\left(T\right)\right)\hfill \\ & =zer\left({\mu }_{1}U+\eta V+{\nu }_{1}W\right)\cap zer\left({\mu }_{2}U+\eta V+{\nu }_{2}W\right).\hfill \end{array}$$

This shows that $j_n\to J_{\eta V}\left(x^{\ast }\right)\in \mathrm{zer}\left({\mu }_{1}U+\eta V+{\nu }_{1}W\right)\cap \mathrm{zer}\left({\mu }_{2}U+\eta V+{\nu }_{2}W\right).$  □

In Theorem 2, if we set $Vx=0$, then $J_{\eta V}\left(x\right)=I\left(x\right)$ for all $x\in H$. By applying Lemma 9, it yields $\mathrm{zer}\left({\mu }_{1}U+{\nu }_{1}W\right)\cap \mathrm{zer}\left({\mu }_{2}U+{\nu }_{2}W\right)=Fix\left(S\right)\cap Fix\left(T\right)$ where $S:=J_{{\mu }_{1}U}\circ \left(I-{\nu }_{1}W\right)$ and $T:=J_{{\mu }_{2}U}\circ \left(I-{\nu }_{2}W\right)$. All of which leads to the following corollary.

Corollary 4.

Let $U:H\to 2^H$ be a maximal monotone operator and $W:H\to H$ a κ-cocoercive operator with $\kappa >0$ and $\mathrm{zer}\left({\mu }_{1}U+{\nu }_{1}W\right)\cap \mathrm{zer}\left({\mu }_{2}U+{\nu }_{2}W\right)=\varnothing .$ Let ${\left(x_n\right)}_{n\ge 0}$ be generated by the following

$$\left\{\begin{array}{c}{x}_0,x_1\in H,\hfill \\ y_n=x_n+{\theta }_n(x_n-x_{n-1}),\hfill \\ z_n=(1-{\alpha }_n){\delta }_n{y}_n+{\alpha }_n{J}_{{\mu }_{1}U}({\delta }_n{y}_n-{\nu }_{1}W\left({\delta }_n{y}_n\right)),\hfill \\ x_{n+1}=(1-{\beta }_n)z_n+{\beta }_n{J}_{{\mu }_{2}U}({\delta }_n{y}_n-{\nu }_{2}W\left({\delta }_n{y}_n\right))+{\epsilon }_n,\hfill \end{array}\right.$$

(37)

for all $n\ge 1$, where ${\mu }_1,{\mu }_2>0$, ${\nu }_1,{\nu }_2\in (0,2\kappa )$, and ${\left({\theta }_n\right)}_{n\ge 1}\subseteq [0,\theta ]$ with $\theta \in [0,1)$. Assume that the Assumption 1 holds and ${\displaystyle \sum _{n=1}^{\infty }}{\theta }_n∥x_n-x_{n-1}∥<+\infty$. Then, the sequence ${\left(x_n\right)}_{n\ge 0}$ strongly converges to $P_{\mathrm{zer}\left({\mu }_{1}U+{\nu }_{1}W\right)\cap \mathrm{zer}\left({\mu }_{2}U+{\nu }_{2}W\right)}\left(0\right)$.

5. Numerical Experiments

To examine the performance of the iterative method proposed in this study, we conducted a numerical example focusing on image restoration problems. The entire experimentation was conducted using MATLAB R2016b on a MacBook Air 13-inch, early 2017 model, which is equipped with a 1.8 GHz Intel Core i5 processor and 8 GB 1600 MHz DDR3 memory.

Image Restoration Problems

We utilize the suggested algorithm in this section to address image restoration issues, encompassing the deblurring and denoising of images. Our focus is on the degradation model that accurately represents real-world image restoration problems, or at least the most relevant mathematical approximations of them.

$$y=\Pi x+w,$$

(38)

where $y,\Pi ,x$ and w refer to the corrupted image, degradation operator (or blurring operator), pristine image, and noise operator, respectively.

To obtain the reconstructed image, we solve the subsequent regularized least-squares problem

$$\underset{x}{min}\left\{\frac{1}{2}{\parallel \Pi x-y\parallel }_2^2+\tau \varphi \left(x\right)\right\},$$

(39)

where the regularization parameter is denoted by $\tau >0$, while $\varphi (·)$ is the regularization function. An established regularization functional utilized to eliminate noise in restoration problems is the $l_1$ norm, which is commonly referred to as Tikhonov regularization [42]. The problem (39) can be expressed as follows:

$$\mathrm{find}x\in \underset{x\in {\mathbb{R}}^k}{argmin}\left\{\frac{1}{2}{\parallel \Pi x-y\parallel }_2^2+\tau {\parallel x\parallel }_1\right\},$$

(40)

where y refers to the corrupted image, and $\Pi$ represents a bounded linear operator. It is worth noting that problem (40) is a specific instance of problem (10) when configured by setting $U=\partial f(·)$, $V=0$, and $W=\nabla L(·)$ where $f\left(x\right)={\parallel x\parallel }_1,\tau =0.001,$ and $L\left(x\right)=\frac{1}{2}{\parallel \Pi x-y\parallel }_2^2$. With this configuration, it follows that $W\left(x\right)=\nabla L\left(x\right)={\Pi }^{\ast }(\Pi x-y)$, where ${\Pi }^{\ast }$ is a transpose of $\Pi$. To commence the problem, we first select images and corrupt them through various blurring techniques. By utilizing the configuration outlined in Corollary 4, our algorithm is utilized to solve the problem (40) with the following settings ${\alpha }_n=0.99+\frac{1}{100n^2},{\beta }_n=0.99+\frac{1}{100n^2},{\delta }_n=1-\frac{1}{100n^2},{\mu }_1={\mu }_2=0.000099,{\nu }_1=0.7,{\nu }_2=0.99,{\epsilon }_n=0$ and ${\theta }_n$ is defined by

$$\begin{array}{c}\hfill {\theta }_n=\left\{\begin{array}{cccc}& min\left\{\frac{70n-9}{100n},\frac{1}{{(n+1)}^2∥x_n-x_{n-1}∥}\right\},\hfill & & \mathrm{if}{x}_n\ne x_{n-1},\hfill \\ & \frac{70n-9}{100n},\hfill & & \mathrm{otherwise}.\hfill \end{array}\right.\end{array}$$

(41)

We compare our proposed algorithm with the algorithm presented in [28] Algorithm (4.1), the algorithm was presented in [25], Equaltion (34), and the inertial Mann-type algorithm introduced by Kitkuan et al. [27].

In the case of the algorithm presented in [28] Algorithm (4.1), we set ${\alpha }_n={\beta }_n={\gamma }_n=0.99+\frac{1}{100n^2}$. For the Equaltion (34), we select ${\alpha }_n=0.99+\frac{1}{100n^2},{\delta }_n=1-\frac{1}{100n+1},{\lambda }_n=0.7$. Regarding the algorithm discussed in ([27], Algorithm in Theorem 3.1), we opt for ${\varsigma }_n={\theta }_n,{\alpha }_n=\frac{1}{{(n+10)}^2},{\lambda }_n=0.7$ and $h\left(x\right)=\frac{x^2}{12}$. We assess the quality of the reconstructed image by using the signal to noise ratio (SNR) for images which is defined by

$$\mathbf{SNR}\left(n\right)=20{log}_{10}\frac{{\parallel x\parallel }_2^2}{\parallel x-x_n{\parallel }_2^2},$$

where x and $x_n$ denote the original and the restored image at iteration n, respectively.

The corresponding numerical results for the aforementioned selections are presented in Figure 1, Figure 2, Figure 3 and Figure 4.

Our algorithm has demonstrated superior performance in image restoration compared to other algorithms, as shown in

Table 1. The performance of the signal to noise ratio (SNR) is assessed for two images.

n	Car				Peppers
	Kitkuan Alg.	Artsawang Alg.	Akutsah Alg.	Our Alg.	Kitkuan Alg.	Artsawang Alg.	Akutsah Alg.	Our Alg.
1	29.1979	31.2609	30.8224	31.2717	20.0085	21.7953	21.4784	21.8055
50	34.1247	34.2846	33.8063	34.4055	28.6085	28.9211	26.1883	29.1169
100	34.5512	34.6842	34.3091	34.8931	31.3053	31.6493	28.4961	31.8981
200	34.9508	35.0692	34.7918	35.4585	33.9631	34.2778	31.2576	34.5329
300	35.1721	35.2809	35.0944	35.8563	35.3848	35.6774	32.8735	35.9750
500	35.3848	35.4698	35.5312	36.4348	37.0055	37.2620	34.7713	37.7622
1000	35.2761	35.2815	36.2586	37.3087	38.6425	38.7880	37.2158	40.1373

.

6. Conclusions

This paper presents and studies the modified Mann-type algorithm, referred to as Equation (9), for two countable families of nonexpansive mappings, incorporating an inertial extrapolation term. Under some mild conditions that are available on nonexpansive mappings together with AKTT condition (see Definition 1) and appropriate control conditions of scalar parameters, this new theoretical tool that we have devised strongly converges to a common fixed point of two countable families of nonexpansive mappings. In terms of of applications, we applied Equation (36), a by-product of Equation (9), to solve the monotone inclusion problem (35). Furthermore, for solving the image restoration problem (40), we can apply the new tool that is Equation (37), a by-product of Equation (36), to find a solution of (40). Numerical results in various instances were provided to showcase the superior efficiency of our newly developed algorithm. These numerical findings unambiguously demonstrate that our algorithm performs significantly better than the previous results.