Common Attractive Point Results for Two Generalized Nonexpansive Mappings in Uniformly Convex Banach Spaces

Abstract

In this work, we study some basic properties of the set of common attractive points and prove strong convergence results for common attractive points of two generalized nonexpansive mappings in a uniformly convex Banach space. As a consequence, we obtain a common fixed point result of such mappings and apply it to solving the convex minimization problem. Finally, numerical experiments are given to support our results.

1. Introduction

Throughout this paper, let $\mathbb{N}$ and $\mathbb{R}$ be the sets of positive integers and real numbers, respectively. When C is a subset of a Banach space $X,$ a mapping $T:C\to X$ is called nonexpansive if $\parallel Tx-Tu\parallel \le \parallel x-u\parallel$ for every $x,u\in C.$ The set of fixed points of T is denoted by $F\left(T\right)$; that is, $F\left(T\right)=\{z\in C:Tz=z\}.$ A mapping $T:C\to X$ is said to be quasi-nonexpansive [1] if $F\left(T\right)\ne \varnothing$ and $\parallel Tx-z\parallel \le \parallel x-z\parallel$ for every $x\in C$ and for every $z\in F\left(T\right)$. It is easy to see that nonexpansive mappings with fixed points are included in the class of quasi-nonexpansive mappings.

Fixed point theory can solve not only problems in sciences and economics but also real-world problems (see [2,3,4,5,6,7] for examples). Specifically, the theory of nonexpansive mappings plays a crucial role, because it can be applied to plenty of problems, such as convex minimization problems, optimization problems, equilibrium problems and problems of proving the existence of solutions to integral and differential equations. Consequently, many generalized nonexpansive mappings have been studied in a variety of directions. In this work, we focus on a generalized mapping defined by Hardy and Rogers [8] as: a mapping $T:C\to X$ is called generalized nonexpansive if

$$\begin{array}{cc}\hfill \parallel Tx-Tu\parallel & \le a_1\parallel x-u\parallel +a_2\parallel Tx-x\parallel +a_3\parallel Tu-u\parallel \hfill \\ \hfill & \hspace{1em}+a_4\parallel Tu-x\parallel +a_5\parallel Tx-u\parallel ,\hfill \end{array}$$

(1)

for every $x,u\in C,$ where $a_1,a_2,a_3,a_4$ and $a_5$ are nonnegative constants such that $a_1+a_2+a_3+a_4+a_5\le 1$. It was also mentioned that condition (1) is equivalent with

$$\begin{array}{cc}\hfill \parallel Tx-Tu\parallel & \le a\parallel x-u\parallel +b(\parallel Tx-x\parallel +\parallel Tu-u\parallel )\hfill \\ \hfill & \hspace{1em}+c(\parallel Tu-x\parallel +\parallel Tx-u\parallel ),\hfill \end{array}$$

(2)

for every $x,u\in C,$ where $a,$ b and c are nonnegative constants such that $a+2b+2c\le 1.$ By letting $a=1$ and $b=c=0$, we can see that every nonexpansive mapping is a generalized nonexpansive mapping. It is well-known from [9] that every generalized nonexpansive mapping with a fixed point is a quasi-nonexpansive mapping.

In 1953, Mann [10] introduced the following iterative scheme to approximate fixed points of a nonexpansive mapping T:

$$\left\{\begin{array}{c}{x}_1\in C\hfill \\ x_{n+1}=(1-a_n)x_n+a_{n}T{x}_n, n\in \mathbb{N},\hfill \end{array}\right.$$

(3)

where $\left\{a_n\right\}$ is a sequence in $(0,1)$.

It is also known that, in general, Mann iteration does not necessarily converge to fixed points of nonexpansive mappings. Thus, in 1974, Ishikawa iteration [11] was introduced to approximate fixed points of such mappings as:

$$\left\{\begin{array}{c}{x}_1\in C\hfill \\ x_{n+1}=(1-a_n)x_n+a_{n}T{y}_n\hfill \\ y_n=(1-b_n)x_n+b_{n}T{x}_n, n\in \mathbb{N},\hfill \end{array}\right.$$

(4)

where $\left\{a_n\right\}$ and $\left\{b_n\right\}$ are sequences in $(0,1)$.

In 1998, Xu [12] introduced the following iterative scheme, called Mann iteration with error terms, for nonexpansive mappings:

$$\left\{\begin{array}{c}{x}_1\in C\hfill \\ x_{n+1}=a_n{x}_n+b_{n}T{x}_n+c_n{u}_n, n\in \mathbb{N},\hfill \end{array}\right.$$

(5)

where $\left\{a_n\right\}$, $\left\{b_n\right\}$ and $\left\{c_n\right\}$ are sequences in $(0,1)$ such that $a_n+b_n+c_n=1$ and $\left\{u_n\right\}$ is a bounded sequence in C. This scheme reduces to Mann iteration if $c_n=0$.

When studying two mappings $T_1,T_2:C\to X,$ we recall that $x\in C$ is a common fixed point of $T_1$ and $T_2$ if $T_{1}x=x=T_{2}x.$ The set of all common fixed points of $T_1$ and $T_2$ is denoted by $F(T_1,T_2)$. Moreover, $F(T_1,T_2)=F\left(T_1\right)\cap F\left(T_2\right).$

To study convergence theorems for common fixed points of two mappings, Das and Debata [13] and Takahashi and Tamura [14] constructed the following iterative scheme:

$$\left\{\begin{array}{c}{x}_1\in C\hfill \\ x_{n+1}=(1-a_n)x_n+a_n{T}_1{y}_n\hfill \\ y_n=(1-b_n)x_n+b_n{T}_2{x}_n, n\in \mathbb{N},\hfill \end{array}\right.$$

(6)

where $\left\{a_n\right\}$ and $\left\{b_n\right\}$ are sequences in $(0,1)$. If $T_1=T_2$, then the scheme reduces to an Ishikawa iterative scheme.

Lui et al. [15] introduced the following iterative process to prove weak and strong convergence theorems of common fixed points for a pair of nonexpansive and asymptytotically nonexpansive mappings:

$$\left\{\begin{array}{c}{x}_1\in C\hfill \\ x_{n+1}=(1-a_n)T_1{x}_n+a_n{T}_2{y}_n\hfill \\ y_n=(1-b_n)T_1{x}_n+b_n{T}_2{x}_n, n\in \mathbb{N},\hfill \end{array}\right.$$

(7)

where $\left\{a_n\right\}$ and $\left\{b_n\right\}$ are sequences in $(0,1)$.

Recently, Ali and Ali [16] proved a convergence theorem for common fixed points of the Mann iteration for two generalized nonexpansive mappings in uniformly convex Banach spaces. They defined a sequence $\left\{x_n\right\}$ as follows:

$$\left\{\begin{array}{c}{x}_1\in C\hfill \\ x_{n+1}=a_n{x}_n+b_n{T}_1{x}_n+c_n{T}_2{x}_n, n\in \mathbb{N},\hfill \end{array}\right.$$

(8)

where $\left\{a_n\right\}$, $\left\{b_n\right\}$ and $\left\{c_n\right\}$ are sequences in $(0,1)$ such that $a_n+b_n+c_n=1$.

Another concept that relates to fixed points is the concept of attractive points, which was first introduced in Hilbert spaces by Takahashi and Takeuchi [17]. Let H be a Hilbert space, and let $T:C\to H$ be a mapping, where C is a nonempty subset of H. The set of all attractive points of T is denoted by $A\left(T\right)$; that is,

$$A\left(T\right)=\{z\in H: \parallel z-Ty\parallel \le \parallel z-y\parallel , \forall y\in C\}.$$

(9)

The authors proved an ergodic convergence theorem to find an attractive point of some nonlinear mappings without assuming convexity of its domain. Moreover, the authors [17] proved that $F\left(T\right)$ relates to $A\left(T\right)$ as follows.

Lemma 1.

Let C be a nonempty closed convex subset of H, and let T be a mapping from C into itself. If $A\left(T\right)\ne \varnothing$, then $F\left(T\right)\ne \varnothing$.

Lemma 2.

Let C be a nonempty subset of H, and let T be a mapping from C into H. Then, $A\left(T\right)$ is a closed convex subset of H.

Furthermore, we also know the following lemma from Takahashi et al. [18] for quasi-nonexpansive mappings.

Lemma 3.

Let C be a nonempty subset of H, and let T be a quasi-nonexpansive mapping from C into $H.$ Then, $A\left(T\right)\cap C=F\left(T\right)$.

In 2018, Khan [19] extended the concept of attractive points to the case of two mappings in Hilbert spaces. Let $T_1,T_2:C\to H$, where C is a nonempty subset of $H.$ The set of all common attractive points for $T_1$ and $T_2$ is denoted by $A(T_1,T_2)$; that is,

$$A(T_1,T_2)=\{z\in H:max(\parallel T_{1}x-z\parallel ,\parallel T_{2}x-z\parallel )\le \parallel x-z\parallel , \forall x\in C\}.$$

(10)

Moreover, $A(T_1,T_2)=A\left(T_1\right)\cap A\left(T_2\right).$ The next properties were proven by Khan [19].

Lemma 4.

Let C be a nonempty closed convex subset of H, and let $T_1,T_2:C\to C$ be two mappings. If $A(T_1,T_2)\ne \varnothing$, then $F(T_1,T_2)\ne \varnothing$.

Lemma 5.

Let C be a nonempty subset of H, and let $T_1,T_2:C\to H$ be two mappings. Then, $A(T_1,T_2)$ is a closed convex subset of H.

Lemma 6.

Let C be a nonempty subset of H, and let $T_1,T_2:C\to H$ be two quasi-nonexpansive mappings. Then, $A(T_1,T_2)\cap C=F(T_1,T_2)$.

Furthermore, there are many results of common attractive point theorems in a Hilbert space (see [19,20,21], and references therein).

In 2013, Lin and Takahashi [22] introduced the concept of attractive points of a nonlinear mapping in the Banach spaces setting. In 2015, Zheng [23] proved convergence theorems for attractive points, defined in (9), of some generalized nonexpansive mappings in uniformly convex Banach spaces. Moreover, there are more studies about attractive points for nonlinear mappings in Banach spaces (see [22,23,24,25,26]).

In this paper, motivated by all results mentioned above, we study some basic properties of the set of common attractive points, defined in (10), for two nonlinear mappings in the setting of uniformly convex Banach spaces. Furthermore, we prove strong convergence theorems of common attractive points of the Mann iteration (8) for two generalized nonexpansive mappings in a uniformly convex Banach space without closedness of the domain of such mappings. Using this result, we obtain strong convergence theorems of common fixed points in a uniformly convex Banach space and solve a convex minimization problem in Hilbert spaces. Finally, to support our results, a numerical example is given.

2. Preliminaries

Let $\left\{x_n\right\}$ be a sequence in a Banach space X. We denote the strong convergence of $\left\{x_n\right\}$ to $x\in X$ by $x_n\to x$. A Banach space X is called uniformly convex if for each $ϵ\in [0,2],$ there is ${\delta }_{ϵ}>0$ such that

$$\parallel x\parallel =\parallel u\parallel =1 \mathrm{and} \parallel x-u\parallel \ge ϵ \mathrm{implies} \frac{\parallel x+u\parallel }{2}<1-{\delta }_{ϵ}.$$

Remark 1.

Hilbert spaces are uniformly convex (see [27]).

Definition 1

([28]). A subset C of a normed linear space X is said to be an existence subset of X if and only if for every element $x\in X$ there is an element $u\in C$ such that

$$\parallel x-u\parallel =d(x,C)=inf\{\parallel x-z\parallel :z\in C\},$$

where u is called the best approximation of x denoted by $\pi (x,C).$ It is well-known that if C is a closed and convex subset of a reflexive Banach space X, then the best approximation element $\pi (x,C)$ exists and is unique for every $x\in X$.

Every uniformly convex Banach space is reflexive (see [29]). We can see that if C is a closed and convex subset of a uniformly convex Banach space X, then for every $x\in X$ there exists a unique best approximation element $\pi (x,C).$

Definition 2

([28]). Let C be a nonempty closed and convex subset of a normed space X. If for every $x\in X$ there exists a unique $\pi (x,C)\in C,$ then the mapping $\pi (x,C)$ is called a metric projection onto C; that is,

$$\parallel x-\pi (x,C)\parallel =d(x,C), \forall x\in X.$$

It is well-known that if C is a closed and convex subset of a uniformly convex Banach space X, then there exists a metric projection from X onto $C.$

The following result is useful for our main theorem.

Lemma 7

([30]). Suppose X is a uniformly convex Banach space and $0<s\le r_n\le t<1$ for all $n\ge 1$. If $\left\{x_n\right\}$ and $\left\{u_n\right\}$ are two sequences in X such that ${lim\; sup}_{n\to \infty }\parallel x_n\parallel \le d,{lim\; sup}_{n\to \infty }\parallel u_n\parallel \le d$ and ${lim\; sup}_{n\to \infty }\parallel r_n{x}_n+(1-r_n)u_n\parallel =d$ hold for some $d\ge 0$. Then, ${lim}_{n\to \infty }\parallel x_n-u_n\parallel =0$.

3. Main Results

In this section, we begin with studying some basic properties of common attractive points for two mappings in the framework of uniformly convex Banach spaces as follows:

Lemma 8.

Let C be a nonempty, closed and convex subset of a uniformly convex Banach space X, and let $T_1,T_2:C\to C$ be two mappings. If $A(T_1,T_2)\ne \varnothing$, then $F(T_1,T_2)\ne \varnothing$. In particular, if $u\in A(T_1,T_2),$ then $\pi (u,C)\in F(T_1,T_2).$

Proof. 

Let $u\in A(T_1,T_2)$; then, $u\in A\left(T_1\right)$ and $u\in A\left(T_2\right)$; that is,

$$\parallel T_{1}y-u\parallel \le \parallel y-u\parallel \mathrm{and} \parallel T_{2}y-u\parallel \le \parallel y-u\parallel , \forall y\in C.$$

Since C is closed and convex, a unique element $x=\pi (u,C)$ exists in C and

$$\parallel x-u\parallel \le \parallel y-u\parallel , \forall y\in C.$$

We know that $T_{1}x\in C$, so $\parallel x-u\parallel \le \parallel T_{1}x-u\parallel$. Since $u\in A\left(T_1\right),$ we get

$$\parallel T_{1}x-u\parallel \le \parallel x-u\parallel \le \parallel T_{1}x-u\parallel .$$

Combing with the uniqueness of x, it follows that $T_{1}x=x.$ Thus $x\in F\left(T_1\right).$ Similarly, we also obtain $x\in F\left(T_2\right).$ Therefore, $x=\pi (u,C)\in F\left(T_1\right)\cap F\left(T_2\right)=F(T_1,T_2)$. □

Lemma 9.

Let C be a nonempty subset of a Banach space X, and let $T_1,T_2:C\to X$ be two mappings. Then, $A(T_1,T_2)$ is a closed subset of X.

Proof. 

Let $\left\{u_n\right\}\subset A(T_1,T_2)$ be a sequence converging strongly to some $u\in X.$ We shall prove that $u\in A(T_1,T_2).$ Indeed, for all $x\in C$ and $u_n\in A(T_1,T_2)$, the following results:

$$\parallel T_{1}x-u_n\parallel \le \parallel x-u_n\parallel \mathrm{and} \parallel T_{2}x-u_n\parallel \le \parallel x-u_n\parallel , \forall n\in \mathbb{N}.$$

By letting $n\to \infty ,$ we obtain

$$\parallel T_{1}x-u\parallel \le \parallel x-u\parallel \mathrm{and} \parallel T_{2}x-u\parallel \le \parallel x-u\parallel , \forall x\in C.$$

Thus, $u\in A(T_1,T_2)$, and hence $A(T_1,T_2)$ is closed. □

Lemma 10.

Let C be a nonempty subset of a Banach space X, and let $T_1,T_2:C\to X$ be two quasi-nonexpansive mappings. Then, $A(T_1,T_2)\cap C=F(T_1,T_2).$

Proof. 

Let $u\in A(T_1,T_2)\cap C$. Then, $u\in A\left(T_1\right)$ and $u\in A\left(T_2\right)$. That is

$$\parallel T_{1}y-u\parallel \le \parallel y-u\parallel \mathrm{and} \parallel T_{2}y-u\parallel \le \parallel y-u\parallel , \forall y\in C.$$

In particular, by choosing $y=u\in C$, we get

$$\parallel T_{1}u-u\parallel \le \parallel u-u\parallel =0 \mathrm{and} \parallel T_{2}u-u\parallel \le \parallel u-u\parallel =0.$$

It follows that $T_{1}u=u$ and $T_{2}u=u$. That is $u\in F\left(T_1\right)\cap F\left(T_2\right)=F(T_1,T_2)$.

Conversely, let $u\in F(T_1,T_2).$ Since $T_1$ and $T_2$ are quasi-nonexpansive mappings, we have

$$\parallel T_{1}y-u\parallel \le \parallel y-u\parallel \mathrm{and} \parallel T_{2}y-u\parallel \le \parallel y-u\parallel , \forall y\in C.$$

Then,

$$max(\parallel T_{1}y-u\parallel ,\parallel T_{2}y-u\parallel )\le \parallel y-u\parallel , \forall y\in C.$$

It follows that $u\in A(T_1,T_2).$ Since $u\in C$, we get $u\in A(T_1,T_2)\cap C$.

Thus, $A(T_1,T_2)\cap C=F(T_1,T_2).$ □

Next, we prove our main results, using the iterative scheme (8) for two generalized nonexpansive mappings in uniformly convex Banach spaces. Before proving the results, we need the following important tools.

Lemma 11.

Let C be a nonempty and convex subset of a Banach space X and $T_1,T_2:C\to C$ be two generalized nonexpansive mappings with $A(T_1,T_2)\ne \varnothing$. Let $\left\{x_n\right\}$ be a sequence generated by (8):

$$\left\{\begin{array}{c}{x}_1\in C\hfill \\ x_{n+1}=a_n{x}_n+b_n{T}_1{x}_n+c_n{T}_2{x}_n, \forall n\in \mathbb{N},\hfill \end{array}\right.$$

with $a_n,b_n,c_n\in (0,1)$ such that $a_n+b_n+c_n=1.$ Then, ${lim}_{n\to \infty }\parallel x_n-u\parallel$ exists for any $u\in A(T_1,T_2).$

Proof. 

Let $\left\{x_n\right\}$ be a sequence generated by (8) and $u\in A(T_1,T_2).$ Then,

$$max(\parallel T_{1}x-u\parallel ,\parallel T_{2}x-u\parallel )\le \parallel x-u\parallel , \forall x\in C.$$

It follows that

$$\parallel T_{1}x-u\parallel \le \parallel x-u\parallel \mathrm{and} \parallel T_{2}x-u\parallel \le \parallel x-u\parallel , \forall x\in C.$$

Consider

$$\begin{array}{cc}\hfill \parallel x_{n+1}-u\parallel & =\parallel a_n{x}_n+b_n{T}_1{x}_n+c_n{T}_2{x}_n-u\parallel \hfill \\ \hfill & =\parallel a_n(x_n-u)+b_n(T_1{x}_n-u)+c_n(T_2{x}_n-u)\parallel \hfill \\ \hfill & \le a_n\parallel x_n-u\parallel +b_n\parallel T_1{x}_n-u\parallel +c_n\parallel T_2{x}_n-u\parallel \hfill \\ \hfill & \le a_n\parallel x_n-u\parallel +b_n\parallel x_n-u\parallel +c_n\parallel x_n-u\parallel \hfill \\ \hfill & =(a_n+b_n+c_n)\parallel x_n-u\parallel \hfill \\ \hfill & =\parallel x_n-u\parallel , \forall n\in \mathbb{N}.\hfill \end{array}$$

This shows that the sequence $\{\parallel x_n-u\parallel \}$ is nonincreasing and bounded below for all $u\in A(T_1,T_2).$ Therefore, ${lim}_{n\to \infty }\parallel x_n-u\parallel$ exists. □

Lemma 12.

Let C be a nonempty and convex subset of a uniformly convex Banach space X and $T_1,T_2:C\to C$ be two generalized nonexpansive mappings with $A(T_1,T_2)\ne \varnothing .$ Let $\left\{x_n\right\}$ be a sequence generated by (8):

$$\left\{\begin{array}{c}{x}_1\in C\hfill \\ x_{n+1}=a_n{x}_n+b_n{T}_1{x}_n+c_n{T}_2{x}_n, \forall n\in \mathbb{N},\hfill \end{array}\right.$$

with $a_n,b_n,c_n\in (0,1)$ such that $a_n+b_n+c_n=1$ and $0<s\le a_n\le t<1.$ Then, $\left\{x_n\right\}$ is bounded, ${lim}_{n\to \infty }\parallel x_n-T_1{x}_n\parallel =0$ and ${lim}_{n\to \infty }\parallel x_n-T_2{x}_n\parallel =0.$

Proof. 

Suppose that $A(T_1,T_2)\ne \varnothing$ and $u\in A(T_1,T_2).$ By using Lemma 11, we have that ${lim}_{n\to \infty }\parallel x_n-u\parallel$ exists for all $u\in A(T_1,T_2).$ Therefore, $\{\parallel x_n-u\parallel \}$ is bounded, and so is $\left\{x_n\right\}$. Let

$$\underset{n\to \infty }{lim}\parallel x_n-u\parallel =\alpha .$$

(11)

We show that ${lim}_{n\to \infty }\parallel x_n-T_1{x}_n\parallel =0$ and ${lim}_{n\to \infty }\parallel x_n-T_2{x}_n\parallel =0.$

Since $u\in A(T_1,T_2)$, we obtain that

$$\parallel T_1{x}_n-u\parallel \le \parallel x_n-u\parallel \mathrm{and} \parallel T_2{x}_n-u\parallel \le \parallel x_n-u\parallel , \forall n\in \mathbb{N}.$$

Therefore

$$\underset{n\to \infty }{lim\; sup}\parallel T_1{x}_n-u\parallel \le \underset{n\to \infty }{lim\; sup}\parallel x_n-u\parallel =\alpha$$

(12)

and

$$\underset{n\to \infty }{lim\; sup}\parallel T_2{x}_n-u\parallel \le \underset{n\to \infty }{lim\; sup}\parallel x_n-u\parallel =\alpha .$$

(13)

According to (11), we derive

$$\begin{array}{cc}\hfill \alpha & =\underset{n\to \infty }{lim}\parallel x_{n+1}-u\parallel \hfill \\ \hfill & =\underset{n\to \infty }{lim}\parallel a_n{x}_n+b_n{T}_1{x}_n+c_n{T}_2{x}_n-u\parallel \hfill \\ \hfill & =\underset{n\to \infty }{lim}\parallel a_n(x_n-u)+b_n(T_1{x}_n-u)+c_n(T_2{x}_n-u)\parallel \hfill \\ \hfill & =\underset{n\to \infty }{lim}\parallel (1-c_n)\left(\frac{a_n}{1-c_n}(x_n-u)+\frac{b_n}{1-c_n}(T_1{x}_n-u)\right)+c_n(T_2{x}_n-u)\parallel .\hfill \end{array}$$

(14)

By using (12), we have

$$\begin{array}{cc}\hfill \underset{n\to \infty }{lim\; sup}\parallel \frac{a_n}{1-c_n}(x_n-u)+\frac{b_n}{1-c_n}(T_1{x}_n-u)\parallel & \le \underset{n\to \infty }{lim\; sup}\frac{a_n+b_n}{1-c_n}\parallel x_n-u\parallel \hfill \\ \hfill & =\underset{n\to \infty }{lim\; sup}\parallel x_n-u\parallel \hfill \\ \hfill & =\alpha .\hfill \end{array}$$

(15)

To apply Lemma 7, we choose

$${\mathbf{x}}_{\mathbf{n}}=T_2{x}_n-u,{\mathbf{u}}_{\mathbf{n}}=\frac{a_n}{1-c_n}(x_n-u)+\frac{b_n}{1-c_n}(T_1{x}_n-u), \mathrm{and} {\mathbf{r}}_{\mathbf{n}}=c_n, \forall n\in \mathbb{N}.$$

Since $0<s\le a_n\le t<1$, we get $0<s\le 1-(b_n+c_n)\le t<1.$ Hence, there exist $v,w\in \mathbb{R}$ such that $0<v\le c_n\le w<1$ for all $n\in \mathbb{N}$. We have from (13) and (15) that

$$\underset{n\to \infty }{lim\; sup}\parallel {\mathbf{x}}_{\mathbf{n}}\parallel =\underset{n\to \infty }{lim\; sup}\parallel T_2{x}_n-u\parallel \le \alpha$$

and

$$\underset{n\to \infty }{lim\; sup}\parallel {\mathbf{u}}_{\mathbf{n}}\parallel =\underset{n\to \infty }{lim\; sup}\parallel \frac{a_n}{1-c_n}(x_n-u)+\frac{b_n}{1-c_n}(T_1{x}_n-u)\parallel \le \alpha .$$

Furthermore, from (14), we have

$$\begin{array}{cc}\hfill \underset{n\to \infty }{lim\; sup}\parallel {\mathbf{r}}_{\mathbf{n}}{\mathbf{x}}_{\mathbf{n}}& +(1-{\mathbf{r}}_{\mathbf{n}}){\mathbf{u}}_{\mathbf{n}}\parallel \hfill \\ \hfill & =\underset{n\to \infty }{lim}\parallel (1-c_n)\left(\frac{a_n}{1-c_n}(x_n-u)+\frac{b_n}{1-c_n}(T_1{x}_n-u)\right)+c_n(T_2{x}_n-u)\parallel \hfill \\ \hfill & =\alpha .\hfill \end{array}$$

Therefore, the sequences $\left\{{\mathbf{x}}_{\mathbf{n}}\right\},\left\{{\mathbf{u}}_{\mathbf{n}}\right\},$ and $\left\{{\mathbf{r}}_{\mathbf{n}}\right\}$ satisfy the assumptions of Lemma 7, which imply that

$$\underset{n\to \infty }{lim}\parallel \frac{a_n}{1-c_n}(x_n-u)+\frac{b_n}{1-c_n}(T_1{x}_n-u)-(T_2{x}_n-u)\parallel =\underset{n\to \infty }{lim}\parallel {\mathbf{u}}_{\mathbf{n}}-{\mathbf{x}}_{\mathbf{n}}\parallel =0.$$

(16)

Consider

$$\begin{array}{cc}\hfill \underset{n\to \infty }{lim}& \frac{1}{1-c_n}\parallel x_{n+1}-T_2{x}_n\parallel \hfill \\ \hfill & =\underset{n\to \infty }{lim}\frac{1}{1-c_n}\parallel a_n{x}_n+b_n{T}_1{x}_n+c_n{T}_2{x}_n-T_2{x}_n+u-(a_n+b_n+c_n)u\parallel \hfill \\ \hfill & =\underset{n\to \infty }{lim}\frac{1}{1-c_n}\parallel a_n(x_n-u)+b_n(T_1{x}_n-u)-(1-c_n)(T_2{x}_n-u)\parallel \hfill \\ \hfill & =\underset{n\to \infty }{lim}\parallel \frac{a_n}{1-c_n}(x_n-u)+\frac{b_n}{1-c_n}(T_1{x}_n-u)-(T_2{x}_n-u)\parallel .\hfill \end{array}$$

We can conclude that

$$\underset{n\to \infty }{lim}\frac{1}{1-c_n}\parallel x_{n+1}-T_2{x}_n\parallel =0.$$

Since $0<v\le c_n\le w<1$ for all $n\in \mathbb{N}$, we have

$$\underset{n\to \infty }{lim}\parallel x_{n+1}-T_2{x}_n\parallel =0.$$

(17)

Next, we will show that ${lim}_{n\to \infty }\parallel x_{n+1}-T_1{x}_n\parallel =0.$ According to (11), we have

$$\begin{array}{cc}\hfill \alpha & =\underset{n\to \infty }{lim}\parallel x_{n+1}-u\parallel \hfill \\ \hfill & =\underset{n\to \infty }{lim}\parallel a_n{x}_n+b_n{T}_1{x}_n+c_n{T}_2{x}_n-u\parallel \hfill \\ \hfill & =\underset{n\to \infty }{lim}\parallel a_n(x_n-u)+b_n(T_1{x}_n-u)+c_n(T_2{x}_n-u)\parallel \hfill \\ \hfill & =\underset{n\to \infty }{lim}\parallel (1-b_n)\left(\frac{a_n}{1-b_n}(x_n-u)+\frac{c_n}{1-b_n}(T_2{x}_n-u)\right)+b_n(T_1{x}_n-u)\parallel .\hfill \end{array}$$

(18)

By using (13), we have

$$\begin{array}{cc}\hfill \underset{n\to \infty }{lim\; sup}\parallel \frac{a_n}{1-b_n}(x_n-u)+\frac{c_n}{1-b_n}(T_2{x}_n-u)\parallel & \le \underset{n\to \infty }{lim\; sup}\frac{a_n+c_n}{1-b_n}\parallel x_n-u\parallel \hfill \\ \hfill & =\underset{n\to \infty }{lim\; sup}\parallel x_n-u\parallel \hfill \\ \hfill & =\alpha .\hfill \end{array}$$

(19)

Applying Lemma 7 again, we choose

$${\mathbf{x}}_{\mathbf{n}}=T_1{x}_n-u,{\mathbf{u}}_{\mathbf{n}}=\frac{a_n}{1-b_n}(x_n-u)+\frac{c_n}{1-b_n}(T_2{x}_n-u), \mathrm{and} {\mathbf{r}}_{\mathbf{n}}=b_n, \forall n\in \mathbb{N}.$$

Since $0<s\le a_n\le t<1$, we get $0<s\le 1-(b_n+c_n)\le t<1.$ Hence, there exist $v^{\prime },w^{\prime }\in \mathbb{R}$ such that $0<v^{\prime }\le b_n\le w^{\prime }<1$ for all $n\in \mathbb{N}$. We have from (12) and (19) that

$$\underset{n\to \infty }{lim\; sup}\parallel {\mathbf{x}}_{\mathbf{n}}\parallel =\underset{n\to \infty }{lim\; sup}\parallel T_1{x}_n-u\parallel \le \alpha$$

and

$$\underset{n\to \infty }{lim\; sup}\parallel {\mathbf{u}}_{\mathbf{n}}\parallel =\underset{n\to \infty }{lim\; sup}\parallel \frac{a_n}{1-b_n}(x_n-u)+\frac{c_n}{1-b_n}(T_2{x}_n-u)\parallel \le \alpha .$$

Furthermore, from (18), we have

$$\begin{array}{cc}\hfill \underset{n\to \infty }{lim\; sup}\parallel {\mathbf{r}}_{\mathbf{n}}{\mathbf{x}}_{\mathbf{n}}& +(1-{\mathbf{r}}_{\mathbf{n}}){\mathbf{u}}_{\mathbf{n}}\parallel \hfill \\ \hfill & =\underset{n\to \infty }{lim}\parallel (1-b_n)\left(\frac{a_n}{1-b_n}(x_n-u)+\frac{c_n}{1-b_n}(T_2{x}_n-u)\right)+b_n(T_1{x}_n-u)\parallel \hfill \\ \hfill & =\alpha .\hfill \end{array}$$

Therefore, we can apply Lemma 7 to get

$$\underset{n\to \infty }{lim}\parallel \frac{a_n}{1-b_n}(x_n-u)+\frac{c_n}{1-b_n}(T_2{x}_n-u)-(T_1{x}_n-u)\parallel =\underset{n\to \infty }{lim}\parallel {\mathbf{u}}_{\mathbf{n}}-{\mathbf{x}}_{\mathbf{n}}\parallel =0.$$

(20)

Consider

$$\begin{array}{cc}\hfill \underset{n\to \infty }{lim}& \frac{1}{1-b_n}\parallel x_{n+1}-T_1{x}_n\parallel \hfill \\ \hfill & =\underset{n\to \infty }{lim}\frac{1}{1-b_n}\parallel a_n{x}_n+b_n{T}_1{x}_n+c_n{T}_2{x}_n-T_1{x}_n+u-(a_n+b_n+c_n)u\parallel \hfill \\ \hfill & =\underset{n\to \infty }{lim}\frac{1}{1-b_n}\parallel a_n(x_n-u)+c_n(T_2{x}_n-u)-(1-b_n)(T_1{x}_n-u)\parallel \hfill \\ \hfill & =\underset{n\to \infty }{lim}\parallel \frac{a_n}{1-b_n}(x_n-u)+\frac{c_n}{1-b_n}(T_2{x}_n-u)-(T_1{x}_n-u)\parallel ,\hfill \end{array}$$

we can conclude that

$$\underset{n\to \infty }{lim}\frac{1}{1-b_n}\parallel x_{n+1}-T_1{x}_n\parallel =0.$$

Since $0<v^{\prime }\le b_n\le w^{\prime }<1$ for all $n\in \mathbb{N}$, we get

$$\underset{n\to \infty }{lim}\parallel x_{n+1}-T_1{x}_n\parallel =0.$$

(21)

Finally, we will show that ${lim}_{n\to \infty }\parallel x_{n+1}-x_n\parallel =0.$ According to (11), we can derive

$$\begin{array}{cc}\hfill \alpha & =\underset{n\to \infty }{lim}\parallel x_{n+1}-u\parallel \hfill \\ \hfill & =\underset{n\to \infty }{lim}\parallel a_n{x}_n+b_n{T}_1{x}_n+c_n{T}_2{x}_n-u\parallel \hfill \\ \hfill & =\underset{n\to \infty }{lim}\parallel a_n(x_n-u)+b_n(T_1{x}_n-u)+c_n(T_2{x}_n-u)\parallel \hfill \\ \hfill & =\underset{n\to \infty }{lim}\parallel a_n(x_n-u)+(1-a_n)\left(\frac{b_n}{1-a_n}(T_1{x}_n-u)+\frac{c_n}{1-a_n}(T_2{x}_n-u)\right)\parallel .\hfill \end{array}$$

(22)

By using (12) and (13), we have

$$\begin{array}{cc}\hfill \underset{n\to \infty }{lim\; sup}\parallel \frac{b_n}{1-a_n}(T_1{x}_n-u)+\frac{c_n}{1-a_n}(T_2{x}_n-u)\parallel & \le \underset{n\to \infty }{lim\; sup}\frac{b_n+c_n}{1-a_n}\parallel x_n-u\parallel \hfill \\ \hfill & =\underset{n\to \infty }{lim\; sup}\parallel x_n-u\parallel \hfill \\ \hfill & =\alpha .\hfill \end{array}$$

(23)

To apply Lemma 7, we choose

$${\mathbf{x}}_{\mathbf{n}}=x_n-u,{\mathbf{u}}_{\mathbf{n}}=\frac{b_n}{1-a_n}(T_1{x}_n-u)+\frac{c_n}{1-a_n}(T_2{x}_n-u), \mathrm{and} {\mathbf{r}}_{\mathbf{n}}=a_n, \forall n\in \mathbb{N}.$$

We have from (11) and (23) that

$$\underset{n\to \infty }{lim\; sup}\parallel {\mathbf{x}}_{\mathbf{n}}\parallel =\underset{n\to \infty }{lim\; sup}\parallel x_n-u\parallel \le \alpha$$

and

$$\underset{n\to \infty }{lim\; sup}\parallel {\mathbf{u}}_{\mathbf{n}}\parallel =\underset{n\to \infty }{lim\; sup}\parallel \frac{b_n}{1-a_n}(T_1{x}_n-u)+\frac{c_n}{1-a_n}(T_2{x}_n-u)\parallel \le \alpha .$$

Furthermore, from (22), we have

$$\begin{array}{cc}\hfill \underset{n\to \infty }{lim\; sup}\parallel {\mathbf{r}}_{\mathbf{n}}{\mathbf{x}}_{\mathbf{n}}& +(1-{\mathbf{r}}_{\mathbf{n}}){\mathbf{u}}_{\mathbf{n}}\parallel \hfill \\ \hfill & =\underset{n\to \infty }{lim}\parallel (1-a_n)\left(\frac{b_n}{1-a_n}(T_1{x}_n-u)+\frac{c_n}{1-a_n}(T_2{x}_n-u)\right)+a_n(x_n-u)\parallel \hfill \\ \hfill & =\alpha .\hfill \end{array}$$

Therefore, we can apply Lemma 7 to get

$$\underset{n\to \infty }{lim}\parallel \frac{b_n}{1-a_n}(T_1{x}_n-u)+\frac{c_n}{1-a_n}(T_2{x}_n-u)-(x_n-u)\parallel =\underset{n\to \infty }{lim}\parallel {\mathbf{u}}_{\mathbf{n}}-{\mathbf{x}}_{\mathbf{n}}\parallel =0.$$

(24)

Consider

$$\begin{array}{cc}\hfill \underset{n\to \infty }{lim}& \frac{1}{1-a_n}\parallel x_{n+1}-x_n\parallel \hfill \\ \hfill & =\underset{n\to \infty }{lim}\frac{1}{1-a_n}\parallel a_n{x}_n+b_n{T}_1{x}_n+c_n{T}_2{x}_n-x_n+u-(a_n+b_n+c_n)u\parallel \hfill \\ \hfill & =\underset{n\to \infty }{lim}\frac{1}{1-a_n}\parallel b_n(T_1{x}_n-u)+c_n(T_2{x}_n-u)-(1-a_n)(x_n-u)\parallel \hfill \\ \hfill & =\underset{n\to \infty }{lim}\parallel \frac{b_n}{1-a_n}(T_1{x}_n-u)+\frac{c_n}{1-a_n}(T_2{x}_n-u)-(x_n-u)\parallel .\hfill \end{array}$$

We can conclude that

$$\underset{n\to \infty }{lim}\frac{1}{1-a_n}\parallel x_{n+1}-x_n\parallel =0.$$

Since $0<s\le a_n\le t<1$, we get

$$\underset{n\to \infty }{lim}\parallel x_{n+1}-x_n\parallel =0.$$

(25)

Note that

$$\parallel x_n-T_1{x}_n\parallel \le \parallel x_n-x_{n+1}\parallel +\parallel x_{n+1}-T_1{x}_n\parallel , \forall n\in \mathbb{N},$$

(26)

and

$$\parallel x_n-T_2{x}_n\parallel \le \parallel x_n-x_{n+1}\parallel +\parallel x_{n+1}-T_2{x}_n\parallel , \forall n\in \mathbb{N}.$$

(27)

By (17), (21), (25) and $n\to \infty$ in (26) and (27), we get

$$\underset{n\to \infty }{lim}\parallel x_n-T_1{x}_n\parallel =0 \mathrm{and} \underset{n\to \infty }{lim}\parallel x_n-T_2{x}_n\parallel =0$$

as desired. □

Next, we establish a strong convergence theorem of common attractive points for two generalized nonexpansive mappings, in the sense of Hardy and Roger [8] defined in (2), in a uniformly convex Banach space without assuming the closedness of the domain of such mappings.

Theorem 1.

Let C be a nonempty and convex subset of a uniformly convex Banach space X and $T_1,T_2:C\to C$ be two generalized nonexpansive mappings with $A(T_1,T_2)\ne \varnothing .$ Let $\left\{x_n\right\}$ be a sequence generated by (8):

$$\left\{\begin{array}{c}{x}_1\in C\hfill \\ x_{n+1}=a_n{x}_n+b_n{T}_1{x}_n+c_n{T}_2{x}_n, \forall n\in \mathbb{N},\hfill \end{array}\right.$$

with $a_n,b_n,c_n\in (0,1)$ such that $a_n+b_n+c_n=1$ and $0<s\le a_n\le t<1.$ Then, $\left\{x_n\right\}$ converges strongly to $z\in A(T_1,T_2)$ if and only if ${lim\; inf}_{n\to \infty }d(x_n,A(T_1,T_2))=0$ or ${lim\; sup}_{n\to \infty }d(x_n,A(T_1,T_2))=0.$

Proof. 

Suppose that $x_n\to u\in A(T_1,T_2).$ Then, for each $ϵ>0,$ there exists $m_0\in \mathbb{N}$ such that

$$\parallel x_n-u\parallel <ϵ$$

for all $n\ge m_0.$ Therefore, we obtain

$$d(x_n,A(T_1,T_2))=inf\{\parallel x_n-u\parallel :u\in A(T_1,T_2)\}\le \parallel x_n-u\parallel <ϵ, \forall n\ge m_0.$$

It follows that ${lim}_{n\to \infty }d(x_n,A(T_1,T_2))=0,$ and hence

$$\underset{n\to \infty }{lim\; inf}d(x_n,A(T_1,T_2))=0 \mathrm{and} \underset{n\to \infty }{lim\; sup}d(x_n,A(T_1,T_2))=0.$$

Conversely, if ${lim\; sup}_{n\to \infty }d(x_n,A(T_1,T_2))=0$, then ${lim}_{n\to \infty }d(x_n,A(T_1,T_2))=0.$

Assume that ${lim\; inf}_{n\to \infty }d(x_n,A(T_1,T_2))=0.$ This means that $\left\{d(x_n,A(T_1,T_2))\right\}$ contains a subsequence $\left\{d(x_{n_k},A(T_1,T_2))\right\}$ such that

$$\underset{k\to \infty }{lim}d(x_{n_k},A(T_1,T_2))=0.$$

By Lemma 11, we have that ${lim}_{n\to \infty }\parallel x_n-u\parallel$ exists for all $u\in A(T_1,T_2).$ Then,

$$\underset{n\to \infty }{lim}\parallel x_n-u\parallel =\underset{k\to \infty }{lim}\parallel x_{n_k}-u\parallel .$$

It follows that

$$\underset{n\to \infty }{lim}(inf\{\parallel x_n-u\parallel :u\in A(T_1,T_2)\left\}\right)=\underset{k\to \infty }{lim}(inf\{\parallel x_{n_k}-u\parallel :u\in A(T_1,T_2)\left\}\right)=0.$$

That is,

$$\underset{n\to \infty }{lim}d(x_n,A(T_1,T_2))=0.$$

Next, we shall show that $\left\{x_n\right\}$ is a Cauchy sequence. By Lemma 11, we have $\parallel x_{n+1}-u\parallel \le \parallel x_n-u\parallel$ for all $u\in A(T_1,T_2).$ In fact, for any $n,m\in \mathbb{N},$ without of generality, we may set $m>n$. Then,

$$\parallel x_m-u\parallel \le \parallel x_n-u\parallel , \forall u\in A(T_1,T_2).$$

Consider

$$\parallel x_n-x_m\parallel \le \parallel x_n-u\parallel +\parallel u-x_m\parallel \le 2\parallel x_n-u\parallel .$$

Since u is arbitrary, we may take infimum all over $u\in A(T_1,T_2)$ on both sides to get

$$\parallel x_n-x_m\parallel \le 2inf\{\parallel x_n-u\parallel :u\in A(T_1,T_2)\}=2d(x_n,A(T_1,T_2)).$$

From ${lim}_{n\to \infty }d(x_n,A(T_1,T_2))=0,$ we get ${lim}_{n\to \infty }\parallel x_n-x_m\parallel =0.$ This means that $\left\{x_n\right\}$ is a Cauchy sequence in a uniformly convex Banach space $X.$ Thus, there exists $z\in X$ such that

$$\underset{n\to \infty }{lim}\parallel x_n-z\parallel =0.$$

By Lemma 12, we have

$$\underset{n\to \infty }{lim}\parallel x_n-T_1{x}_n\parallel =0 \mathrm{and} \underset{n\to \infty }{lim}\parallel x_n-T_2{x}_n\parallel =0.$$

Thus,

$$\underset{n\to \infty }{lim}\parallel z-T_1{x}_n\parallel =0 \mathrm{and} \underset{n\to \infty }{lim}\parallel z-T_2{x}_n\parallel =0.$$

Next, we show that $z\in A(T_1,T_2).$ Since $T_1$ is generalized nonexpansive mapping, we can use (1) to get that there exist nonnegative constants $a,b,c$ with $a+2b+2c\le 1$ such that

$$\begin{array}{cc}\hfill \parallel T_{1}x-T_1{x}_n\parallel & \le a\parallel x-x_n\parallel +b(\parallel x-T_{1}x\parallel +\parallel x_n-T_1{x}_n\parallel )\hfill \\ \hfill & +c(\parallel x-T_1{x}_n\parallel +\parallel x_n-T_{1}x\parallel ), \forall x\in C, \forall n\in \mathbb{N}.\hfill \end{array}$$

Thus

$$\begin{array}{cc}\hfill \parallel T_{1}x-T_1{x}_n\parallel & \le a\parallel x-x_n\parallel +b(\parallel x-x_n\parallel +\parallel x_n-T_{1}x\parallel +\parallel x_n-T_1{x}_n\parallel )\hfill \\ \hfill & +c(\parallel x-T_1{x}_n\parallel +\parallel x_n-T_{1}x\parallel ), \forall x\in C, \forall n\in \mathbb{N}.\hfill \end{array}$$

Taking limit on both sides, we get

$$\begin{array}{cc}\hfill \parallel T_{1}x-z\parallel & \le a\parallel x-z\parallel +b(\parallel x-z\parallel +\parallel z-T_{1}x\parallel +\parallel z-z\parallel )+c(\parallel x-z\parallel +\parallel z-T_{1}x\parallel )\hfill \\ \hfill & =(a+b+c)\parallel x-z\parallel +(b+c)\parallel z-T_{1}x\parallel .\hfill \end{array}$$

Hence, $(1-b-c)\parallel T_{1}x-z\parallel \le (a+b+c)\parallel x-z\parallel , \forall x\in C.$

Since $a+2b+2c\le 1,$ we have

$$\frac{a+b+c}{1-b-c}\le \frac{1-b-c}{1-b-c}=1.$$

Thus, $\parallel T_{1}x-z\parallel \le \parallel x-z\parallel$ for all $x\in C.$ This means that $z\in A\left(T_1\right).$

Similarly, we have $z\in A\left(T_2\right).$ Therefore, $z\in A(T_1,T_2).$ □

By applying Theorem 1, we also obtain the following result.

Corollary 1.

Let C be a nonempty and convex subset of a uniformly convex Banach space X and $T_1,T_2:C\to C$ be two generalized nonexpansive mappings with $F(T_1,T_2)\ne \varnothing .$ Let $\left\{x_n\right\}$ be a sequence generated by (8) with $a_n,b_n,c_n\in (0,1)$ such that $a_n+b_n+c_n=1$ and $0<s\le a_n\le t<1$ for all $n\in \mathbb{N}.$

(1) Suppose that ${lim\; inf}_{n\to \infty }d(x_n,F(T_1,T_2))=0$ or ${lim\; sup}_{n\to \infty }d(x_n,F(T_1,T_2))=0.$ Then, $\left\{x_n\right\}$ converges strongly to $z\in A(T_1,T_2).$ If C is closed, then $\left\{x_n\right\}$ converges strongly to $z\in F(T_1,T_2).$

(2) Suppose that $\left\{x_n\right\}$ converges strongly to the common attractive point of $T_1$ and $T_2$; then,

${lim\; inf}_{n\to \infty }d(x_n,A(T_1,T_2))=0$ or ${lim\; sup}_{n\to \infty }d(x_n,A(T_1,T_2))=0.$ If C is closed, then ${lim\; inf}_{n\to \infty }d(x_n,F(T_1,T_2))=0$ or ${lim\; sup}_{n\to \infty }d(x_n,F(T_1,T_2))=0.$

Proof. 

Since $F(T_1,T_2)\ne \varnothing ,$ we have $T_1,T_2$ are quasi-nonexpansive mappings. By Lemma 10, we have $A(T_1,T_2)\cap C=F(T_1,T_2),$ which implies that, $A(T_1,T_2)\ne \varnothing$.

(1) Suppose that

$$\underset{n\to \infty }{lim\; inf}d(x_n,F(T_1,T_2))=0 \mathrm{or} \underset{n\to \infty }{lim\; sup}d(x_n,F(T_1,T_2))=0.$$

Since $F(T_1,T_2)\subset A(T_1,T_2),$ we have

$$d(x_n,A(T_1,T_2))\le d(x_n,F(T_1,T_2)), \forall n\in \mathbb{N}.$$

Then,

$$\underset{n\to \infty }{lim\; inf}d(x_n,A(T_1,T_2))\le \underset{n\to \infty }{lim\; inf}d(x_n,F(T_1,T_2))=0,$$

or

$$\underset{n\to \infty }{lim\; sup}d(x_n,A(T_1,T_2))\le \underset{n\to \infty }{lim\; sup}d(x_n,F(T_1,T_2))=0.$$

By Theorem 1, we get $x_n\to z\in A(T_1,T_2).$ If C is closed, then $z\in C.$ It follows that $x_n\to z\in F(T_1,T_2).$

(2) Assume that $x_n\to z\in A(T_1,T_2).$ By Theorem 1, we get

$$\underset{n\to \infty }{lim\; inf}d(x_n,A(T_1,T_2))=0 \mathrm{or} \underset{n\to \infty }{lim\; sup}d(x_n,A(T_1,T_2))=0.$$

If C is closed, then $z\in C.$ Since $A(T_1,T_2)\cap C=F(T_1,T_2),$ we have $x_n\to z\in F(T_1,T_2).$ It follows that ${lim\; inf}_{n\to \infty }d(x_n,F(T_1,T_2))=0$ or ${lim\; sup}_{n\to \infty }d(x_n,F(T_1,T_2))=0.$ □

Next, we aim to apply Corollary 1 to solve a convex minimization problem in Hilbert spaces. The general formulation of the convex minimization problem is defined by

$$\underset{x}{min}\left\{f\left(x\right)+g\left(x\right)\right\}.$$

(28)

In a Hilbert space H, the solution of problem (28) is usually considered under the following assumptions:

(i)

g is a lower semicontinuous function and properly convex from H into $\mathbb{R}\cup \{+\infty \}$;

(ii)

f is a convex differentiable function from H into $\mathbb{R}$, with $\nabla f$ being ℓ-Lipschitz constant for some $\ell >0$—that is, $\parallel \nabla f\left(x\right)-\nabla f\left(y\right)\parallel \le \ell \parallel x-y\parallel$ for all $x,y\in {\mathbb{R}}^n.$

We denote the set of all solutions of (28) by argmin$(f+g)$, and it is well-known that finding a solution of problem (28) is equivalent to finding a zero-solution $x^{🟉}$ such that:

$$0\in \partial g\left(x^{🟉}\right)+\nabla f\left(x^{🟉}\right),$$

(29)

where $\nabla f$ is the gradient operator of function f and $\partial g$ is the subdifferential of function g; see [31] for more details. Dealing with fixed point theory, Parikh and Boyd [32] solved the problem (29) by using the proximal gradient technique; that is, if $x^{🟉}$ solves (29), then $x^{🟉}$ is a fixed point of the proximal operator:

$$x^{🟉}={\mathrm{prox}}_{\kappa g}(I-\kappa \nabla f)\left(x^{🟉}\right),$$

where $\kappa$ is a positive parameter, ${\mathrm{prox}}_{\kappa g}={(I+\kappa \partial g)}^{-1}$ and I is the identity operator.

If we set $T_1:=$ prox${}_{{\kappa }_{1}g}(I-{\kappa }_1\nabla f)$ and $T_2:=$ prox${}_{{\kappa }_{2}g}(I-{\kappa }_2\nabla f)$ where ${\kappa }_1,{\kappa }_2\in \left(0,\frac{2}{\ell }\right)$, then $T_1$ and $T_2$ are nonexpansive mappings; see [33,34,35] for more details.

We denote $S^{🟉}=\{x^{🟉}:x^{🟉}\in$ argmin$(f+g)\}$. The following result is a consequence of Corollary 1.

Corollary 2.

Let g be a lower semicontinuous function and proper convex from a real Hilbert space H into $\mathbb{R}\cup \{+\infty \}$, and let f be a convex differentiable function from H into $\mathbb{R}$ with $\nabla f$ being ℓ-Lipschitz constant for some $\ell >0.$ Let $\left\{x_n\right\}$ be a sequence generated by (8) under the same conditions of parameters as in Corollary 1 with $S^{🟉}\ne \varnothing$. If ${lim\; inf}_{n\to \infty }d(x_n,S^{🟉})=0$ or ${lim\; sup}_{n\to \infty }d(x_n,S^{🟉})=0$, then $\left\{x_n\right\}$ converges strongly to an element in argmin$(f+g)$.

Proof. 

Let $T_1$ and $T_2$ be the forward–backward operators of f and g with respect to ${\kappa }_1$ and ${\kappa }_2$, respectively, where ${\kappa }_1,{\kappa }_2\in \left(0,\frac{2}{\ell }\right)$. Then, $T_1:=$ prox${}_{{\kappa }_{1}g}(I-{\kappa }_1\nabla f)$ and $T_2:=$ prox${}_{{\kappa }_{2}g}(I-{\kappa }_2\nabla f)$ are nonexpansive mappings which imply generalized nonexpansiveness. By Proposition 26.1 in [31], $F(T_1,T_2)=F\left(T_1\right)\cap F\left(T_2\right)=argmin(f+g)$. Using (1) of Corollary 1, we obtain that $\left\{x_n\right\}$ converges strongly to an element in argmin$(f+g)$. □

Remark 2.

A convex minimization problem (28) can be applied to solving many real world problems, such as image/signal processing, regression and data classification; see [36,37,38]. Hence, the convergence result of Corollary 2 can be applied for solving those important problems.

Another condition for strong convergence results was introduced by Senter and Dotson [39]. A mapping $T:C\to C$ satisfies condition $\left(A\right),$ if there exists a nondecreasing function $h:[0,\infty )\to [0,\infty )$ with $h\left(0\right)=0$ and $h\left(\gamma \right)>0$ for every $\gamma >0$ such that

$$h\left(d\right(u,A\left(T\right)\left)\right)\le \parallel u-Tu\parallel , \forall u\in C.$$

Chidume and Ali extended above condition to two mappings [40] as follows: Two mappings $T_1,T_2:C\to C$ are said to satisfy condition $\left(A^{\prime }\right)$ if there exists a nondecreasing function $h:[0,\infty )\to [0,\infty )$ with $h\left(0\right)=0$ and $h\left(\gamma \right)>0$ for every $\gamma >0$ such that

$$h\left(d(u,A(T_1,T_2))\right)\le \parallel u-T_{1}u\parallel \mathrm{or} h\left(d(u,A(T_1,T_2))\right)\le \parallel u-T_{2}u\parallel , \forall u\in C.$$

In the following theorem, we obtain a strong convergence theorem for common attractive points of two generalized nonexpansive mappings that satisfy condition $\left(A^{\prime }\right).$

Theorem 2.

Let C be a nonempty and convex subset of a uniformly convex Banach space X and $T_1,T_2:C\to C$ be two generalized nonexpansive mappings. Let $\left\{x_n\right\}$ be a sequence generated by (8):

$$\left\{\begin{array}{c}{x}_1\in C\hfill \\ x_{n+1}=a_n{x}_n+b_n{T}_1{x}_n+c_n{T}_2{x}_n, \forall n\in \mathbb{N},\hfill \end{array}\right.$$

with $a_n,b_n,c_n\in (0,1)$ such that $a_n+b_n+c_n=1$ and $0<s\le a_n\le t<1.$ Suppose that $A(T_1,T_2)\ne \varnothing$ and $T_1,T_2$ satisfy condition $\left(A^{\prime }\right);$ then, $\left\{x_n\right\}$ converges strongly to a common attractive point of $T_1$ and $T_2.$

Proof. 

By Lemma 12, we have

$$\underset{n\to \infty }{lim}\parallel x_n-T_1{x}_n\parallel =0=\underset{n\to \infty }{lim}\parallel x_n-T_2{x}_n\parallel .$$

Since $T_1$ and $T_2$ satisfy condition $\left(A^{\prime }\right)$, there exists a nondecreasing function $h:[0,\infty )\to [0,\infty )$ with $h\left(0\right)=0$ and $h\left(\gamma \right)>0$ for all $\gamma >0$ such that

$$h\left(d(x_n,A(T_1,T_2))\right)\le \parallel x_n-T_1{x}_n\parallel \mathrm{or} h\left(d(x_n,A(T_1,T_2))\right)\le \parallel x_n-T_2{x}_n\parallel .$$

It follows that

$$0\le \underset{n\to \infty }{lim}h\left(d(x_n,A(T_1,T_2))\right)\le \underset{n\to \infty }{lim}\parallel x_n-T_1{x}_n\parallel =0,$$

or

$$0\le \underset{n\to \infty }{lim}h\left(d(x_n,A(T_1,T_2))\right)\le \underset{n\to \infty }{lim}\parallel x_n-T_2{x}_n\parallel =0.$$

In both cases, we get

$$\underset{n\to \infty }{lim}h\left(d(x_n,A(T_1,T_2))\right)=0.$$

Since $h:[0,\infty )\to [0,\infty )$ is a nondecreasing function satisfying $h\left(0\right)=0$ and $h\left(\gamma \right)>0$ for all $\gamma >0$, we get

$$\underset{n\to \infty }{lim}d(x_n,A(T_1,T_2))=0.$$

It follows that

$$\underset{n\to \infty }{lim\; inf}d(x_n,A(T_1,T_2))=0 \mathrm{and} \underset{n\to \infty }{lim\; sup}d(x_n,A(T_1,T_2))=0.$$

By Theorem 1, we conclude that $\left\{x_n\right\}$ converges strongly to a common attractive point of $T_1$ and $T_2.$ □

Corollary 3.

Let C be a nonempty and convex subset of a uniformly convex Banach space X and $T_1,T_2:C\to C$ be two generalized nonexpansive mappings. Let $\left\{x_n\right\}$ be a sequence generated by (8) with $a_n,b_n,c_n\in (0,1)$ such that $a_n+b_n+c_n=1$ and $0<s\le a_n\le t<1$ for all $n\in \mathbb{N}.$ Suppose that $F(T_1,T_2)\ne \varnothing$ and $T_1,T_2$ satisfy condition $\left(A^{\prime }\right);$ then, $\left\{x_n\right\}$ converges strongly to a common attractive point of $T_1$ and $T_2.$ If C is closed, then $\left\{x_n\right\}$ converges strongly to a common fixed point of $T_1$ and $T_2.$

Proof. 

Since $F(T_1,T_2)\ne \varnothing ,$ we have $T_1,T_2$ are quasi-nonexpansive mappings. By Lemma 10, we have $A(T_1,T_2)\cap C=F(T_1,T_2).$ It follows that $A(T_1,T_2)\ne \varnothing .$ By Theorem 2, we have $x_n\to z\in A(T_1,T_2).$ If C is closed, then $z\in F(T_1,T_2).$ □

We end this section by providing some numerical experiments to illustrate the performance of iteration (8) for supporting our main results.

Example 1.

Let $X=\mathbb{R}$ with the usual norm and $C=(0,\infty )$. Suppose $T_1,T_2:C\to C$ are defined by

$$T_{1}x=\left\{\begin{array}{c}\frac{x}{4},x\in (0,1)\hfill \\ \frac{x}{16},x\in [1,\infty ),\hfill \end{array}\right.$$

and

$$T_{2}x=\left\{\begin{array}{c}\frac{x}{5},x\in (0,1)\hfill \\ \frac{x}{25},x\in [1,\infty ).\hfill \end{array}\right.$$

Then, $T_1$ and $T_2$ are generalized nonexpansive mappings such that $0\in A(T_1,T_2)$ but $F(T_1,T_2)=\varnothing .$ We chose the parameters $a_n=\frac{n}{50n-1},b_n=\frac{n}{50n-1},c_n=\frac{48n-1}{50n-1}$ and initial point $x_1=\frac{1}{2}.$ 

Table 1. Numerical experiment of the iteration process (8).

Iteration No.	${\mathit{x}}_{\mathit{n}}$	$|{\mathit{x}}_{\mathit{n}}-0|$	$|{\mathit{x}}_{\mathit{n}}-{\mathit{T}}_1{\mathit{x}}_{\mathit{n}}|$	$|{\mathit{x}}_{\mathit{n}}-{\mathit{T}}_2{\mathit{x}}_{\mathit{n}}|$
1	$0.5000$	$0.5000$	$0.4688$	$0.4800$
2	$0.0300$	$0.0300$	$0.0281$	$0.0288$
3	$0.0018$	$0.0018$	$0.0017$	$0.0017$
4	$1.0743\times {10}^{-4}$	$1.0743\times {10}^{-4}$	$1.0071\times {10}^{-4}$	$1.0313\times {10}^{-4}$
5	$6.4186\times {10}^{-6}$	$6.4186\times {10}^{-6}$	$6.0175\times {10}^{-6}$	$6.1619\times {10}^{-6}$
6	$3.8338\times {10}^{-7}$	$3.8338\times {10}^{-7}$	$3.5942\times {10}^{-7}$	$3.6804\times {10}^{-7}$
7	$2.2894\times {10}^{-8}$	$2.2894\times {10}^{-8}$	$2.1463\times {10}^{-8}$	$2.1978\times {10}^{-8}$
8	$1.3669\times {10}^{-9}$	$1.3669\times {10}^{-9}$	$1.2815\times {10}^{-9}$	$1.3122\times {10}^{-9}$
9	$8.1603\times {10}^{-11}$	$8.1603\times {10}^{-11}$	$7.6502\times {10}^{-11}$	$7.8339\times {10}^{-11}$
10	$4.8712\times {10}^{-12}$	$4.8712\times {10}^{-12}$	$4.5667\times {10}^{-12}$	$4.6763\times {10}^{-12}$

shows the values of $x_n$, $|x_n-0|$, $|x_n-T_1{x}_n|$ and $|x_n-T_2{x}_n|$ of iteration $n=1,2,3,\dots ,10.$

It is evident from Table 1 that $x_n\to 0\in A(T_1,T_2),$ with the errors $|x_n-0|\to 0,$$|x_n-T_1{x}_n|\to 0$ and $|x_n-T_2{x}_n|\to 0.$ Moreover, Figure 1 shows the convergence behavior of the iterative process (8).

Next, under control conditions from Theorem 1, we compared the rates of convergence for the sequences $\left\{x_n\right\}$ generated by (6)–(8), as shown in

Table 2. The values of $\left\{x_n\right\}$ for different iteration processes.

Iteration No.	${\mathit{x}}_{\mathit{n}}$ of Iteration (6)	${\mathit{x}}_{\mathit{n}}$ of Iteration (7)	${\mathit{x}}_{\mathit{n}}$ of Iteration (8)
1	$0.5000$	$0.5000$	$0.5000$
2	$0.4904$	$0.0306$	$0.0300$
3	$0.4811$	$0.0019$	$0.0018$
4	$0.4720$	$1.1509\times {10}^{-4}$	$1.0743\times {10}^{-4}$
5	$0.4631$	$7.0541\times {10}^{-6}$	$6.4186\times {10}^{-6}$
6	$0.4544$	$4.3238\times {10}^{-7}$	$3.8338\times {10}^{-7}$
7	$0.4458$	$2.6503\times {10}^{-8}$	$2.2894\times {10}^{-8}$
8	$0.4374$	$1.6245\times {10}^{-9}$	$1.3669\times {10}^{-9}$
9	$0.4292$	$9.9578\times {10}^{-11}$	$8.1603\times {10}^{-11}$
10	$0.4211$	$6.1039\times {10}^{-12}$	$4.8712\times {10}^{-12}$

.

From Table 2, we can see that iteration process (8) performs with a better rate of convergence than iteration processes (6) and (7).

4. Conclusions

In this paper, we studied some basic properties of the set of common attractive points, defined in (10), of two nonlinear mappings in the setting of uniformly convex Banach spaces. Furthermore, using the Mann iteration (8), we proved strong convergence theorems of common attractive points for two generalized nonexpansive mappings, in the sense of Hardy and Rogers [8], in a uniformly convex Banach space without the closedness assumption of the domain of such mappings. Using this result, we obtained strong convergence theorems of common fixed points in a uniformly convex Banach space and solved some convex minimize problems. Finally, we constructed a numerical example to support our main result.