General modified viscosity implicit rules for generalized asymptotically nonexpansive mappings in complete $\operatorname{CAT}(0)$ spaces

Abstract

It is well known that the concept (generalized) asymptotically nonexpansive is closely related to the theory of fixed points in Banach spaces, and the implicit midpoint rule is one of the powerful numerical methods for solving differential equations. The purpose of this paper is to introduce a class of new general modified viscosity implicit rules of generalized asymptotically nonexpansive mappings in complete $\operatorname{CAT}(0)$ spaces, and to prove some strong convergence theorems of the procedure generated by the new general modified viscosity implicit rules under some suitable conditions. The results presented in this paper improve and extend varieties of results in the recent literature.

1 Introduction

In 1972, Goebel and Kirk [1] introduced the concept of asymptotically nonexpansive mapping, which is closely related to the theory of fixed points in Banach spaces. Whereafter, Zhou et al. [2] discussed convergence of modified Ishikawa and Mann iterative sequences for approximating the fixed points of a class of generalized asymptotically nonexpansive mappings. Recently, the iterative approximation problems of fixed points for nonexpansive mapping, asymptotically nonexpansive mapping, and asymptotically nonexpansive type mapping in Hilbert space or Banach spaces have been studied by many authors. See, for example, [3,4,5,6,7] and the references therein. It is well known that Mann’s and Ishikawa’s iterations have the only weak convergence theorem even in a Hilbert space. As a counterexample, Bauschke et al. [8] also showed that the algorithm only converges weakly but not strongly. In order to obtain strong convergence theorems, Moudafi [9] introduced the viscosity approximation method for nonexpansive mappings in Hilbert spaces according to the ideas of Attouch [10]. It is an extension of Halpern iteration method, the refinements of which in Hilbert spaces and extensions to Banach spaces were obtained by Xu [11].

Furthermore, Shi and Chen [12] first studied the convergence theorems of the following Moudafi viscosity iteration method for a nonexpansive mapping $T: C\to C$ with $F(T)=\{x\in C| x=T(x) \}\neq \emptyset$ in a $\operatorname{CAT}(0)$ space X: For any $x_{0}\in C$,

$$ x_{n+1}=\alpha _{n}f(x_{n}) \oplus (1-\alpha _{n})Tx_{n}, \quad n\geq 0, $$

(1.1)

where $C\subseteq X$ is a nonempty closed convex subset, $f: C\to C$ is a contraction mapping, $\{\alpha _{n}\}\subseteq [0,1]$ is a sequence satisfying certain conditions. In connection with work in the Hilbert ball studied by Reich and Shemen [13] and Kopecká and Reich [14], very recently, we considered strong convergence of viscosity iterative approximation methods for set-valued nonexpansive mappings in [15, 16].

On the other hand, for initial value problem of the ordinary differential equation

$$ x'(t)=f\bigl(x(t)\bigr), \qquad x(0)=x_{0}, $$

(1.2)

the implicit midpoint rule, which is one of the powerful numerical methods for solving ordinary differential equations and differential algebraic equations, generates a sequence $\{x_{n}\}$ by the recursion procedure

$$\begin{aligned} x_{n+1}=x_{n}+hf \biggl( \frac{x_{n}+x_{n+1}}{2} \biggr), \quad n\geq 0, \end{aligned}$$

(1.3)

where $h>0$ is a step size. It is known that if $f: {\mathbb{R}}^{m} \to {\mathbb{R}}^{m}$ is Lipschitz continuous and sufficiently smooth, then the sequence $\{x_{n}\}$ generated by (1.3) converges to the exact solution of (1.2) as $h\to 0$ uniformly over $t\in [0,u]$ for any fixed $u>0$. Based on the above fact, Alghamdi et al. [17] proposed the following implicit midpoint rule for nonexpansive mappings in Hilbert space H:

$$ x_{n+1}=\alpha _{n}x_{n}+(1-\alpha _{n})T \biggl(\frac{x_{n}+x_{n+1}}{2} \biggr), \quad n\geq 0, $$

(1.4)

where $\alpha _{n}\in [0,1]$ and $T:H\to H$ is a nonexpansive mapping. The authors proved the weak convergence of (1.4) under some additional conditions on $\{\alpha _{n}\}$. In 2015, using the viscosity approximation method associated with implicit midpoint rule, Xu et al. [18] presented the following viscosity implicit midpoint rule for nonexpansive mappings in a Hilbert space:

$$ x_{n+1}=\alpha _{n}f(x_{n})+(1-\alpha _{n})T \biggl(\frac{x_{n}+x_{n+1}}{2} \biggr), \quad n\geq 0, $$

(1.5)

under some suitable conditions and proved that the sequence $\{x_{n}\}$ generated by (1.5) converges strongly to a point $q\in F(T)$, which is also the unique solution of the variational inequality

$$ \bigl\langle (I-f)q,x-q \bigr\rangle \geq 0, \quad \forall x \in F(T), $$

where I is an identity mapping. Since then, the iterative method (1.5) was generalized to Banach spaces, $\operatorname{CAT}(0)$ space, and geodesic spaces, which were studied by Luo et al. [19], Zao et al. [20], and Preechasilp [21], respectively. In connection with variational inequalities, asymptotically nonexpansive mappings, equilibrium problems, algorithm of (modified) viscosity implicit rules, and so on have been studied by many authors. See, for example, [21,22,23,24,25,26,27,28,29] and the references therein.

Recently, Li and Liu [30] extended viscosity implicit midpoint rule to the following asymptotically nonexpansive mapping in a $\operatorname{CAT}(0)$ space X: For any $x_{0}\in C\subseteq X$,

$$ x_{n+1}=\alpha _{n}f(x_{n})\oplus (1- \alpha _{n})T^{n} \biggl(\frac{x _{n}\oplus x_{n+1}}{2} \biggr), \quad n \geq 0, $$

(1.6)

where $T: C\to C$ is an asymptotically nonexpansive mapping and $f: C\to C$ is a contractive mapping and $\alpha _{n}\in [0,1]$. Moreover, under some suitable conditions, the authors proved that the sequence $\{x_{n}\}$ generated by (1.6) converges strongly to a fixed point $q\in F(T)$ such that $q=P_{F(T)}f(q)$, which is also the unique solution of the variational inequality

$$\begin{aligned} \bigl\langle \overrightarrow{qf(q)}, \overrightarrow{x q} \bigr\rangle \geq 0, \quad x\in F(T). \end{aligned}$$

(1.7)

Similar research was accomplished by Zao et al. [31] in Hilbert spaces. Further, Ke and Ma [32] and Yan and Hu [33] established strong convergence theorems for the generalized viscosity implicit rules of (asymptotically) nonexpansive mappings in Hilbert spaces to a generalized viscosity rule of nonexpansive mappings in Hilbert spaces. Here, the rule is an extension of the implicit midpoint rule presented in [30, 31, 34].

Motivated and inspired by the above work, in this paper, we consider the following new general modified viscosity implicit rules of generalized asymptotically nonexpansive mappings in a complete $\operatorname{CAT}(0)$ space $(X,d)$:

$$\begin{aligned} x_{n+1}=\alpha _{n}f(x_{n})\oplus (1- \alpha _{n})T^{n} \bigl(s_{n}x_{n} \oplus (1-s_{n})x_{n+1} \bigr), \quad \forall n\geq 0, \end{aligned}$$

(1.8)

where $x_{0}\in C\subseteq X$ is an arbitrary fixed element, $\{\alpha _{n}\},\{s_{n}\}\subseteq [0,1]$, $T: C\to C$ is a generalized asymptotically nonexpansive mapping, and $f: C\to C$ is a contractive mapping. We shall prove that the sequence $\{x_{n}\}$ defined by (1.8) converges strongly to $q\in F(T)$ such that $q=P_{F(T)}f(q)$ is the unique solution of the variational inequality

$$ \bigl\langle \overrightarrow{qf(q)}, \overrightarrow{x q} \bigr\rangle \geq 0, \quad x\in F(T). $$

2 Preliminaries

In this section, we will give some definitions and lemmas for proving our main results.

Throughout this paper, let $(X,d)$ be a metric space, $C\subseteq X$ be a nonempty subset.

Definition 2.1

A nonlinear mapping $T: C\to C$ is said to be

• (i) contraction if there exists a constant $\kappa \in [0,1)$ such that

  $$ d\bigl(T(x),T(y)\bigr)\leq \kappa d(x,y), \quad \forall x, y\in C; $$

  when $\kappa =1$, then T is called nonexpansive;

• (ii) asymptotically nonexpansive if there exists a real number sequence $\{k_{n}\}\subseteq [1,+\infty )$ with $\lim_{n\to \infty }k_{n}=1$ such that

  $$ d\bigl(T^{n}x,T^{n}y\bigr)\leq k_{n}d(x,y),\quad \forall x,y\in C, n\geq 1; $$

• (iii) asymptotically nonexpansive mapping in the intermediate sense if T is uniformly continuous and

  $$ \limsup_{n\to \infty }\sup_{x,y\in C} \bigl\{ d \bigl(T^{n}x,T^{n}y\bigr)-d(x,y) \bigr\} \leq 0; $$

• (iv) generalized asymptotically nonexpansive if there exist two real number sequences $\{k_{n}\}\subseteq [1,+\infty )$ and $\{\xi _{n}\}\subseteq [0,+\infty )$ with $\lim_{n\to \infty }k_{n}=1$ and $\lim_{n\to \infty }\xi _{n}=0$ such that

  $$ d\bigl(T^{n}x,T^{n}y\bigr)\leq k_{n}d(x,y)+\xi _{n},\quad \forall x,y\in C, n \ge 1; $$

• (v) $(\{\mu _{n}\},\{\xi _{n}\},\zeta )$-total asymptotically nonexpansive if there exist two nonnegative real number sequences $\{\mu _{n}\}$ and $\{\xi _{n}\}$ with $\lim_{n\to \infty }\mu _{n}=0$, $\lim_{n\to \infty }\xi _{n}=0$ and a strictly increasing continuous function $\zeta : [0,+\infty )\to [0,+\infty )$ with $\zeta (0)=0$ such that

  $$ d \bigl(T^{n}x,T^{n}y \bigr)\leq d(x,y)+\mu _{n} \xi \bigl(d(x,y) \bigr)+ \xi _{n},\quad \forall x,y\in C, n\ge 1; $$

• (vi) uniformly L-Lipschitzian if there exists a constant $L>0$ such that

  $$ d\bigl(T^{n}x,T^{n}y\bigr)\leq Ld(x,y),\quad \forall x,y\in C, n\ge 1. $$

Remark 2.1

(i) For an asymptotically nonexpansive mapping in the intermediate sense, if we let $\xi _{n}=\max \{0,\sup_{x,y\in C}\{d(T ^{n}x,T^{n}y)-d(x,y)\}\}$, then $\lim_{n\to \infty }\xi _{n}=0$. It follows that relational expression of (iii) is reduced to $d(T^{n}x,T ^{n}y)\leq d(x,y)+\xi _{n}$ for any $x,y\in C$ and $n\ge 1$.

(ii) By the above definitions, one can know that nonexpansive mapping, asymptotically nonexpansive mapping, asymptotically nonexpansive mapping in the intermediate sense is generalized asymptotically nonexpansive, a generalized asymptotically nonexpansive mapping is totally asymptotically nonexpansive. But the converse does not hold. For other relevant details of asymptotically nonexpansive mapping in the intermediate sense, totally asymptotically nonexpansive mappings, and so on, see, for example, [1, 7, 35,36,37] and the references therein.

A geodesic path joining $x\in X$ to $y\in X$ (or, more briefly, a geodesic from x to y) is a map ϕ from a closed interval $[0,l]\subseteq \mathbb{R}$ to X such that $\phi (0)=x$, $\phi (l)=y$, and $d(\phi (s),\phi (t))=|s-t|$ for all $s, t\in [0,l]$. In particular, ϕ is an isometry and $d(x,y)=l$. The image of ϕ is called a geodesic segment (or metric) joining x and y when unique is denoted by $[x,y]$. The space $(X,d)$ is said to be a geodesic space if every two points in X are joined by a geodesic, and X is said to be uniquely geodesic if there is exactly one geodesic joining x and y for each $x,y\in X$. A subset C of X is said to be convex if every pair of points $x,y\in C$ can be joined by a geodesic in X and the image of every such geodesic is contained in C.

A geodesic triangle $\triangle (p, q, r)$ in a geodesic space $(X,d)$ consists of three points p, q, r in X (the vertices of △) and a choice of three geodesic segments $[p, q]$, $[q, r]$, $[r, p]$ (the edge of △) joining them. A comparison triangle for geodesic triangle $\triangle (p, q, r)$ in X is a triangle $\overline{\triangle }(\bar{p}, \bar{q}, \bar{r})$ in the Euclidean plane $\mathbb{R}^{2}$ such that

$$ d_{\mathbb{R}^{2}}( \bar{p},\bar{q})=d(p, q),\qquad d_{\mathbb{R}^{2}} (\bar{q}, \bar{r})=d(q, r),\quad \text{and}\quad d_{\mathbb{R}^{2}}( \bar{r},\bar{p}) =d(r, p). $$

A point $\bar{u}\in [\bar{p}, \bar{q}]$ is called a comparison point for $u\in [p, q]$ if $d(p, u)=d_{\mathbb{R}^{2}}(\bar{p},\bar{u})$. Similarly, the comparison points on $[\bar{q}, \bar{r}]$ and $[\bar{r}, \bar{p}]$ can be defined in the same way.

A geodesic space is said to be a $\operatorname{CAT}(0)$ space if all geodesic triangles of appropriate size satisfy the following comparison axiom. Let △ be a geodesic triangle in $(X,d)$, and let △̅ be a comparison triangle for △, then △ is said to satisfy the $\operatorname{CAT}(0)$ inequality if, for any $u,v\in \triangle$ and for their comparison points $\bar{u}, \bar{v}\in \overline{\triangle }$, one has

$$ d(u,v)\leq d_{\mathbb{R}^{2}}(\bar{u}, \bar{v}). $$

Complete $\operatorname{CAT}(0)$ space is often called Hadamard space (see [38]). For other equivalent definitions and basic properties of $\operatorname{CAT}(0)$ spaces, we refer the reader to standard texts, such as [39]. It is well known that every $\operatorname{CAT}(0)$ space is uniquely geodesic and any complete, simply connected Riemannian manifold having non-positive sectional curvature is a $\operatorname{CAT}(0)$ space. Other examples include pre-Hilbert spaces [39], $\mathbb{R}$-trees [40], Euclidean buildings [41], the complex Hilbert ball with a hyperbolic metric [42], and many others.

Let C be a nonempty closed convex subset of a complete $\operatorname{CAT}(0)$ space $(X,d)$. It follows from Proposition 2.4 of [39] that, for every $x\in X$, there exists a unique point $x_{0}\in C$ such that

$$ d(x, x_{0})=\inf \bigl\{ d(x,y) : y\in C\bigr\} . $$

In this case, $x_{0}$ is called the unique nearest point of x in C. The metric projection of X onto C is the mapping $P_{C}: X\to C$ defined by

$$ P_{C}(x):= \text{the unique nearest point of } x \text{ in } C. $$

Let $(X,d)$ be a $\operatorname{CAT}(0)$ space. For each $x,y\in X$ and $t\in [0,1]$, it follows from Lemma 2.1 in [43] that there exists a unique point $z\in [x,y]$ such that

$$ d(x,z)=(1-t)d(x,y) \quad \text{and} \quad d(y,z)=t d(x,y). $$

(2.1)

We shall denote by $tx\oplus (1-t)y$ the unique point z satisfying (2.1). Now, we collect some elementary facts about $\operatorname{CAT}(0)$ spaces which will be used in the proof of our main theorem.

Lemma 2.1

([43])

Assume that $(X,d)$ is a $\operatorname{CAT}(0)$ space. Then, for any $x,y,z\in X$ and $t\in [0,1]$,

• (i) $d(tx\oplus (1-t)y,z)\leq td(x,z)+(1-t)d(y,z)$,

• (ii) $d^{2}(t x\oplus (1-t)y,z)\leq td^{2}(x,z)+(1-t)d^{2}(y,z)-t(1-t)d ^{2}(x,y)$.

Lemma 2.2

([11])

Let $\{s_{n}\}$ be a non-negative real number sequence satisfying

$$ s_{n+1}\leq (1-\alpha _{n})s_{n}+\alpha _{n}\beta _{n}, \quad \forall n\geq 1, $$

where $\{\alpha _{n}\}\subset [0,1]$ and $\{\beta _{n}\}\subset \mathbb{R}$ such that

• (i) $\sum_{n=1}^{\infty } \alpha _{n}=\infty$;

• (ii) $\limsup_{n\to \infty }\beta _{n} \leq 0$ or $\sum_{n=1} ^{\infty } |\alpha _{n}\beta _{n}| < \infty$.

Then $\{s_{n}\}$ converges to zero as $n\to \infty$.

We note that the concept of Δ-convergence due to Lim [44] was shown by Kirk and Panyanak [45] in $\operatorname{CAT}(0)$ spaces to be very similar to the weak convergence in Banach space setting. Next, we give the concept of Δ-convergence and collect some basic properties.

Let $\{x_{n}\}$ be a bounded sequence in a $\operatorname{CAT}(0)$ space $(X,d)$. For $x\in X$, we define a function

$$ r\bigl(x, \{x_{n}\}\bigr)=\limsup_{n\to \infty }d(x,x_{n}). $$

The asymptotic radius $r(\{x_{n}\})$ of $\{x_{n}\}$ is given by

$$ r\bigl(\{x_{n}\}\bigr)=\inf \bigl\{ r\bigl(x, \{x_{n}\} \bigr):x\in X\bigr\} , $$

and the asymptotic center $A(\{x_{n}\})$ of $\{x_{n}\}$ is the set

$$ A\bigl(\{x_{n}\}\bigr)=\bigl\{ x\in X: r\bigl(x, \{x_{n}\} \bigr)=r\bigl(\{x_{n}\}\bigr)\bigr\} . $$

It is well known from [46, Proposition 7] that in $\operatorname{CAT}(0)$ spaces, $A(\{x_{n}\})$ consists of exactly one point. A sequence $\{x_{n}\}\subseteq X$ is called Δ-convergence to $q\in X$ if $A(\{x_{n_{k}}\})=\{q\}$ for any subsequence $\{x_{n_{k}} \}$ of $\{x_{n}\}$. In this case, we write Δ-$\lim_{n\to \infty }x_{n}=q$ and call q the Δ-limit of $\{x_{n}\}$. For more details, see, for example, [47] and the references therein.

The uniqueness of an asymptotic center implies that a complete $\operatorname{CAT}(0)$ space $(X,d)$ satisfies Opial’s property [48] if, whenever $\{x_{n}\}\subseteq X$ Δ-converges to $q\in X$ and for given $y\in X$ with $y\neq q$,

$$ \limsup_{n\to \infty }d(x_{n},q)< \limsup_{n\to \infty }d(x_{n},y). $$

It is well known that every Hilbert space satisfies Opial’s property (see [48]).

Since it is not possible to formulate the concept of demiclosedness in a $\operatorname{CAT}(0)$ setting, as stated in linear spaces, let us formally say that “$I-T$ is demiclosed at zero” if the conditions $\{x_{n}\} \subseteq C$ Δ-converges to $q\in X$ and $d(x_{n},Tx_{n}) \to 0$ as $n\to \infty$ imply $q\in F(T)$, where C is a nonempty convex closed subset of X.

Lemma 2.3

([45])

Each bounded sequence in a complete $\operatorname{CAT}(0)$ space always has a Δ-convergent subsequence.

Lemma 2.4

([43])

If C is a closed convex subset of a complete $\operatorname{CAT}(0)$ space $(X,d)$ and $\{x_{n}\}$ is a bounded sequence in C, then the asymptotic center $A(\{x_{n}\})$ of $\{x_{n}\}$ is in C.

Lemma 2.5

([49])

Suppose that C is a closed convex subset of a complete $\operatorname{CAT}(0)$ space $(X,d)$ and $T: C\to C$ is a totally asymptotically nonexpansive (specially, T is a generalized asymptotically nonexpansive) and uniformly Lipschitzian mapping, and $\{x_{n}\}$ is a bounded sequence in C such that $\{x_{n}\}$ Δ-converges to q and $\lim_{n\to \infty }(x_{n},Tx_{n})=0$. Then $q\in F(T)$.

The concept of quasi-linearization was introduced by Berg and Nikolaev [50]. Let us denote a pair $(a, b)$ in $X\times X$ by $\overrightarrow{ab}$ and call it a vector. The quasi-linearization is a map $\langle \cdot , \cdot \rangle$: $(X \times X)\times (X\times X)\to \mathbb{R}$ defined by

$$ \langle \overrightarrow{ab}, \overrightarrow{cd} \rangle = \frac{1}{2} \bigl(d^{2}(a, d) + d^{2}(b, c) -d^{2}(a, c)- d^{2}(b, d) \bigr), \quad \forall a, b, c, d\in X. $$

It is easy to see that $\langle \overrightarrow{ab}, \overrightarrow{cd} \rangle = \langle \overrightarrow{cd}, \overrightarrow{ab} \rangle$, $\langle \overrightarrow{ab}, \overrightarrow{cd} \rangle =- \langle \overrightarrow{ba}, \overrightarrow{cd} \rangle$, and $\langle \overrightarrow{ax}, \overrightarrow{cd} \rangle + \langle \overrightarrow{xb}, \overrightarrow{cd} \rangle = \langle \overrightarrow{ab}, \overrightarrow{cd} \rangle$ for all $a, b, c, d, x \in X$. A geodesic metric space $(X,d)$ is said to satisfy the Cauchy–Schwarz inequality if

$$ \bigl\vert \langle \overrightarrow{ab}, \overrightarrow{cd} \rangle \bigr\vert \leq d(a, b)d(c, d),\quad \forall a, b, c, d\in X. $$

From [50, Corollary 3], it follows that a geodesic space X is a $\operatorname{CAT}(0)$ space if and only if X satisfies the Cauchy–Schwarz inequality. Some other properties of quasi-linearization are included as follows.

Lemma 2.6

([51])

Let C be a nonempty closed convex subset of a complete $\operatorname{CAT}(0)$ space $(X,d)$, $u\in X$, and $x\in C$. Then

$$ x=P_{C}u \quad \textit{if and only if} \quad \langle \overrightarrow{xu}, \overrightarrow{yx} \rangle \ge 0, \quad \forall y\in C. $$

Lemma 2.7

([52])

Let $(X,d)$ be a $\operatorname{CAT}(0)$ space. Then

$$ d^{2}(x,u) \leq d^{2}(y,u) + 2 \langle \overrightarrow{xy}, \overrightarrow{xu} \rangle , \quad \forall u, x, y\in X. $$

Lemma 2.8

([53])

Assume that $(X,d)$ is a complete $\operatorname{CAT}(0)$ space, $\{x_{n}\}$ is a sequence in X, and $q\in X$. Then $\{x_{n}\}$ Δ-converges to q if and only if $\limsup_{n\to \infty } \langle \overrightarrow{x _{n}q}, \overrightarrow{yq} \rangle \leq 0$ for any $y\in X$.

Lemma 2.9

([30])

Let $(X,d)$ be a complete $\operatorname{CAT}(0)$ space, and $z_{1}=\sigma x\oplus (1-\sigma )u$ and $z_{2}=\sigma y\oplus (1-\sigma )u$ for each $u,x,y\in X$ and all $\sigma \in [0,1]$. Then the following inequality holds:

$$ \langle \overrightarrow{z_{1}z_{2}}, \overrightarrow{xz_{2}} \rangle \le \sigma \langle \overrightarrow{xy}, \overrightarrow{xz_{2}} \rangle . $$

3 Main results

In this section, by using the pre-requisite and elementary facts presented in Sect. 2, we will prove strong convergence of the sequences generated by the new general modified viscosity implicit rules under some suitable conditions.

Theorem 3.1

Let C be a nonempty closed convex subset of a complete $\operatorname{CAT}(0)$ space $(X,d)$, $T: C\to C$ be a uniformly Lipschitzian and generalized asymptotically nonexpansive mapping with sequences $\{k_{n}\}\subset [1,+\infty )$ and $\{\xi _{n}\}\subset [0,+\infty )$, and $f: C\to C$ be a contraction with coefficient $\kappa \in [0,1 )$. If $\lim_{n\to \infty }d(T^{n}x_{n},x _{n})=0$ and $F(T)\neq \emptyset$, and the following conditions hold:

($C_{1}$):

$\{\alpha _{n}\}\subseteq (0,1)$ and $\lim_{n\to \infty } \alpha _{n} =0$,

($C_{2}$):

$\sum^{+\infty }_{n=1} \alpha _{n} = +\infty$,

($C_{3}$):

$\lim_{n\to \infty }\frac{k^{2}_{n}-1}{\alpha _{n}}=0$,

($C_{4}$):

$s_{n}\in (0,1]$ for all $n\ge 0$ and $\lim_{n\to \infty } s _{n} =s\in (0,1]$,

($C_{5}$):

$\sum^{+\infty }_{n=1}\xi _{n}<+\infty$,

then the sequence $\{x_{n}\}$ generated by (1.8) converges strongly as $n\to \infty$ to $q=P_{F(T)}f(q)$, which solves the following variational inequality:

$$ \bigl\langle \overrightarrow{qf(q)}, \overrightarrow{x q} \bigr\rangle \ge 0, \quad x\in F(T). $$

Proof

We divide the proof into five steps as follows.

Step (I) We first prove that $\{x_{n}\}$ is a bounded sequence.

In fact, by (1.8) and Lemma 2.1, for any $p\in F(T)$, we have

$$\begin{aligned} d(x_{n+1}, p) \le& \alpha _{n} d\bigl(f(x_{n}), p \bigr)+(1-\alpha _{n})d\bigl(T^{n} \bigl(s_{n}x_{n} \oplus (1-s_{n}) x_{n+1} \bigr),p\bigr) \\ \le& \alpha _{n} \bigl(d\bigl(f(x_{n}), f(p)\bigr)+d \bigl(f(p),p\bigr) \bigr) \\ &{} +(1-\alpha _{n}) \bigl[k_{n}d\bigl(s_{n}x_{n} \oplus (1-s_{n}) x_{n+1},p\bigr)+ \xi _{n} \bigr] \\ \le& \alpha _{n} \bigl(\kappa d(x_{n}, p)+d\bigl(f(p),p \bigr) \bigr) \\ &{} +(1-\alpha _{n}) \bigl[k_{n}s_{n}d(x_{n},p)+k_{n}(1-s_{n})d( x _{n+1},p)+\xi _{n} \bigr], \end{aligned}$$

that is,

$$\begin{aligned}& \bigl(1-k_{n}(1-s_{n}) (1-\alpha _{n}) \bigr)d(x_{n+1},p) \\& \quad \leq \bigl(k_{n}s_{n}(1-\alpha _{n})+\kappa \alpha _{n} \bigr)d(x_{n},p)+ \alpha _{n}d \bigl(f(p),p\bigr)+(1-\alpha _{n})\xi _{n}. \end{aligned}$$

(3.1)

By conditions $(C_{1})$, $(C_{3})$, and $(C_{4})$, for any given positive number ϵ ($0<\epsilon <1-\kappa$), there exists a sufficiently large positive integer N such that, for any $n>N$,

$$\begin{aligned} k_{n}-1\leq \frac{1}{2}\bigl(k^{2}_{n}-1 \bigr)\leq \epsilon \alpha _{n} \end{aligned}$$

(3.2)

and

$$\begin{aligned} \frac{1-\alpha _{n}}{1-k_{n}(1-s_{n})(1-\alpha _{n})}\leq \frac{2}{s}, \end{aligned}$$

(3.3)

where $\lim_{n\to \infty }\frac{1-\alpha _{n}}{1-k_{n}(1-s_{n})(1- \alpha _{n})}=\frac{1}{s}$.

By (3.1)–(3.3), after simplifying, for any $n>N$, we have

$$\begin{aligned} d(x_{n+1},p) \le& \frac{k_{n}s_{n}(1-\alpha _{n}) +\kappa \alpha _{n}}{1-k _{n}(1-s_{n})(1-\alpha _{n})}d(x_{n},p) \\ &{} +\frac{\alpha _{n} d(f(p),p)}{1-k_{n}(1-s_{n})(1-\alpha _{n})}+\frac{(1- \alpha _{n})\xi _{n}}{1-k_{n}(1-s_{n})(1-\alpha _{n})} \\ \le& \biggl[1+\frac{k_{n}-1-k_{n}\alpha _{n}+\kappa \alpha _{n}}{1-k _{n}(1-s_{n})(1-\alpha _{n})} \biggr]d(x_{n},p) \\ &{} +\frac{\alpha _{n} d(f(p),p)}{1-k_{n}(1-s_{n})(1-\alpha _{n})}+\frac{(1- \alpha _{n})\xi _{n}}{1-k_{n}(1-s_{n})(1-\alpha _{n})} \\ \le& \biggl[1-\frac{(k_{n}-\epsilon -\kappa )\alpha _{n}}{1-k_{n}(1-s _{n})(1-\alpha _{n})} \biggr]d(x_{n},p) \\ &{} +\frac{\alpha _{n} d(f(p),p)}{1-k_{n}(1-s_{n})(1-\alpha _{n})}+\frac{(1- \alpha _{n})\xi _{n}}{1-k_{n}(1-s_{n})(1-\alpha _{n})} \\ \le& \biggl[1-\frac{(1-\epsilon -\kappa )\alpha _{n}}{1-k_{n}(1-s_{n})(1- \alpha _{n})} \biggr]d(x_{n},p) \\ &{} +\frac{(1-\epsilon -\kappa )\alpha _{n}}{1-k_{n}(1-s_{n})(1- \alpha _{n})}\frac{d(f(p),p)}{1-\epsilon -\kappa }+\frac{2\xi _{n}}{s} \\ \le& \max \biggl\{ d(x_{n},p),\frac{d(f(p),p)}{1-\kappa -\epsilon } \biggr\} + \frac{2 \xi _{n}}{s}. \end{aligned}$$

By induction, we have

$$\begin{aligned} d(x_{n},p)\le \max \biggl\{ d(x_{N+1},p), \frac{d(f(p),p)}{1-\kappa - \epsilon } \biggr\} +\frac{2}{s}\sum^{n-1}_{k=N+1} \xi _{k}. \end{aligned}$$

(3.4)

Since $\sum^{+\infty }_{n=1}\xi _{n}<+\infty$, there exists a positive constant $M_{0}$ such that

$$\begin{aligned} \sum^{n-1}_{k=N+1}\xi _{k}\leq M_{0}. \end{aligned}$$

(3.5)

By (3.4) and (3.5), we know that $\{x_{n}\}$ is bounded, and so are $\{f(x_{n})\}$, $\{T^{n}(x_{n}) \}$, $\{T^{n}(x_{n+1})\}$, and $\{T^{n}(s_{n}x_{n}\oplus (1-s_{n}) x_{n+1}) \}$.

Step (II) We show that $\lim_{n\to \infty }d(x_{n},x_{n+1})=0$.

Indeed, it follows from (1.8) that

$$\begin{aligned}& d(x_{n}, x_{n+1}) \\& \quad \le d\bigl(x_{n+1},T^{n}x_{n}\bigr)+d \bigl(T^{n}x_{n},x_{n}\bigr) \\& \quad \le d\bigl(\alpha _{n}f( x_{n})\oplus (1-\alpha _{n})T^{n} \bigl(s_{n}x _{n} \oplus (1-s_{n}) x_{n+1} \bigr),T^{n}x_{n}\bigr) \\& \qquad {}+d\bigl(T^{n}x_{n},x_{n}\bigr) \\& \quad \le \alpha _{n}d\bigl(f(x_{n}),T^{n}x_{n} \bigr)+(1-\alpha _{n})d\bigl(T^{n} \bigl(s _{n}x_{n} \oplus (1-s_{n}) x_{n+1} \bigr),T^{n}x_{n}\bigr) \\& \qquad {}+d\bigl(T^{n}x_{n},x_{n}\bigr) \\& \quad \le \alpha _{n}d\bigl(f(x_{n}),T^{n}x_{n} \bigr)+d\bigl(T^{n}x_{n},x_{n}\bigr) \\& \quad \quad {}+(1-\alpha _{n}) \bigl(k_{n} d\bigl(s_{n}x_{n} \oplus (1-s_{n}) x_{n+1},x _{n}\bigr)+\xi _{n} \bigr) \\& \quad \le \alpha _{n}d\bigl(f(x_{n}),T^{n}x_{n} \bigr)+d\bigl(T^{n}x_{n},x_{n}\bigr) \\& \quad \quad {}+k_{n}(1-s_{n}) (1-\alpha _{n}) d( x_{n},x_{n+1})+(1-\alpha _{n}) \xi _{n}. \end{aligned}$$

Since $\{f(x_{n})\}$ and $\{T^{n}x_{n}\}$ are bounded, there exists $M_{1}>0$ such that $M_{1}\geq \sup_{n\geq 1}d(f(x_{n}), T^{n}x_{n})$. Thus,

$$\begin{aligned} d(x_{n}, x_{n+1})\le \alpha _{n}M_{1}+d \bigl(T^{n}x_{n},x_{n}\bigr)+k_{n}(1-s _{n}) (1-\alpha _{n}) d( x_{n},x_{n+1})+\xi _{n}, \end{aligned}$$

and so, for all $n>N$,

$$\begin{aligned} d(x_{n}, x_{n+1}) \le& \frac{\alpha _{n} M_{1}}{1-k_{n}(1-s_{n})(1- \alpha _{n})}+ \frac{d(T^{n}x_{n}, x_{n})}{1-k_{n}(1-s_{n})(1-\alpha _{n})} \\ &{} +\frac{\xi _{n} }{1-k_{n}(1-s_{n})(1-\alpha _{n})}. \end{aligned}$$

(3.6)

From $\sum^{+\infty }_{n=1}\xi _{n}<+\infty$, one can clearly see

$$\begin{aligned} \lim_{n\to \infty }\xi _{n}=0. \end{aligned}$$

(3.7)

By virtue of conditions $(C_{1})$, $(C_{3})$, and $(C_{4})$, $\lim_{n\to \infty }d(T^{n}x_{n},x_{n})=0$ and (3.6)–(3.7), we now know

$$\begin{aligned} \lim_{n\to \infty }d(x_{n},x_{n+1})=0. \end{aligned}$$

(3.8)

Step (III) $\lim_{n\to \infty }d(Tx_{n},x_{n})=0$ will be displayed.

From (1.8), we have

$$\begin{aligned} d\bigl(x_{n+1},T^{n} x_{n+1}\bigr) =&d\bigl(\alpha _{n}f( x_{n})\oplus (1-\alpha _{n})T ^{n} \bigl(s_{n}x_{n} \oplus (1-s_{n}) x_{n+1} \bigr),T^{n}x_{n+1}\bigr) \\ \le& \alpha _{n}d\bigl(f(x_{n}),T^{n}x_{n+1} \bigr) \\ &{} +(1-\alpha _{n})d\bigl(T^{n}\bigl(s_{n}x_{n} \oplus (1-s_{n}) x_{n+1}\bigr),T ^{n}x_{n+1} \bigr) \\ \le& \alpha _{n}d\bigl(f(x_{n}),T^{n}x_{n+1} \bigr) \\ &{} +(1-\alpha _{n}) \bigl(k_{n}d \bigl(s_{n}x_{n} \oplus (1-s_{n}) x _{n+1},x_{n+1} \bigr)+\xi _{n} \bigr) \\ \le& \alpha _{n}d\bigl(f(x_{n}),T^{n}x_{n+1} \bigr) \\ &{} +k_{n}s_{n}(1-\alpha _{n})d(x_{n}, x_{n+1})+(1-\alpha _{n})\xi _{n} \\ \le& \alpha _{n}M_{1}+k_{n}s_{n}(1- \alpha _{n})d(x_{n}, x_{n+1})+\xi _{n}. \end{aligned}$$

It follows from conditions $(C_{1})$, $(C_{3})$, $(C_{4})$ and (3.7), (3.8) that

$$\begin{aligned} \lim_{n\to \infty }d\bigl(x_{n+1},T^{n}x_{n+1} \bigr)=0. \end{aligned}$$

(3.9)

Further, by the continuity of the generalized asymptotically nonexpansive mapping T, we obtain

$$\begin{aligned} \lim_{n\to \infty }d\bigl(Tx_{n+1},T^{n+1}x_{n+1} \bigr)=0. \end{aligned}$$

(3.10)

Moreover,

$$\begin{aligned} d(Tx_{n+1},x_{n+1})\le d\bigl(Tx_{n+1},T^{n+1}x_{n+1} \bigr)+d\bigl(T^{n+1}x_{n+1},x _{n+1}\bigr). \end{aligned}$$

(3.11)

It follows from (3.9)–(3.11) that $\lim_{n\to \infty }d(Tx _{n+1},x_{n+1})=0$, which implies that

$$\begin{aligned} \lim_{n\to \infty }d(Tx_{n},x_{n})=0. \end{aligned}$$

(3.12)

Step (IV) We prove that

$$\begin{aligned} w_{\Delta }\{x_{n}\}:=\bigcup _{\{u_{n}\}\subset \{x_{n}\}}\bigl\{ A\bigl(\{u_{n} \}\bigr)\bigr\} \subset F(T), \end{aligned}$$

(3.13)

where $A(\{u_{n}\})$ is the asymptotic center of $\{u_{n}\}$. Let $u\in w_{\Delta }\{x_{n}\}$. Then there exists a subsequence $\{u_{n}\}$ of $\{x_{n}\}$ such that $A(\{u_{n}\})=\{u\}$. From Lemma 2.3, it follows that there exists a subsequence $\{v_{n}\}$ of $\{u_{n}\}$ such that Δ-$\lim_{n\to \infty }v_{n}=u$. In view of (3.12), we have

$$ \lim_{n\to \infty }d(Tv_{n},v_{n})=0. $$

By Lemma 2.5 and demi-closedness of $I-T$ at zero, also by Lemma 2.4, we know that $u\in C$ and $Tu=u$. Hence, $u\in F(T)$, i.e., $w_{\Delta }\{x_{n}\}\subset F(T)$.

Step (V) We show $x_{n}\to q$ as $n\to \infty$, where $q\in F(T)$ is the unique fixed point of contraction $P_{F(T)}f$, that is, $q=P_{F(T)}f(q)$, which solves the following variational inequality:

$$ \bigl\langle \overrightarrow{qf(q)}, \overrightarrow{x q} \bigr\rangle \geq 0, \quad x\in F(T). $$

Firstly, mapping $P_{F(T)}f$ is a contraction, thus, obviously, there exists a unique $q\in F(T)$ such that $q=P_{F(T)}f(q)$. It follows from Lemma 2.6 that, for any $x\in F(T)$,

$$ \bigl\langle \overrightarrow{qf(q)}, \overrightarrow{x q} \bigr\rangle \geq 0. $$

Next, we reveal

$$ \limsup_{n\to \infty } \bigl\langle \overrightarrow{qf(q)}, \overrightarrow{q x_{n}} \bigr\rangle \leq 0. $$

As a matter of fact, since $\{x_{n}\}$ is bounded, it follows from Lemma 2.3 that there exists a subsequence $\{x_{n_{k}}\}\subset \{x_{n}\}$, which Δ-converges to a point p. By Lemma 2.8 and (3.13), we get

$$\begin{aligned} \limsup_{n\to \infty } \bigl\langle \overrightarrow{qf(q)}, \overrightarrow{q x_{n}} \bigr\rangle =\lim _{k\to \infty } \bigl\langle \overrightarrow{qf(q)}, \overrightarrow{q x_{n_{k}}} \bigr\rangle = \bigl\langle \overrightarrow{qf(q)}, \overrightarrow{q p} \bigr\rangle \leq 0. \end{aligned}$$

(3.14)

Finally, for each $n\geq 0$, let $z_{n}=\alpha _{n}q\oplus (1-\alpha _{n})T^{n} (s_{n}x_{n} \oplus (1-s_{n}) x_{n+1} )$. By Lemma 2.9, we know

$$ \langle \overrightarrow{z_{n}x_{n+1}}, \overrightarrow{q x_{n+1}} \rangle \leq \alpha _{n} \bigl\langle \overrightarrow{qf(x _{n})}, \overrightarrow{q x_{n+1}} \bigr\rangle . $$

Thus, it follows from Lemmas 2.7 and 2.1, the definition of quasi-linearization, and the Cauchy–Schwarz inequality that

$$\begin{aligned}& d^{2}(x_{n+1},q) \\& \quad \le d^{2}(z_{n},q)+2 \langle \overrightarrow{z_{n}x_{n+1}}, \overrightarrow{q x_{n+1}} \rangle \\& \quad \le (1-\alpha _{n})^{2}d^{2} \bigl(T^{n} \bigl(s_{n}x_{n} \oplus (1-s _{n}) x_{n+1} \bigr),q \bigr)+2 \langle \overrightarrow{z_{n}x _{n+1}}, \overrightarrow{q x_{n+1}} \rangle \\& \quad \le (1-\alpha _{n})^{2} \bigl(k_{n}d \bigl(s_{n}x_{n} \oplus (1-s_{n}) x_{n+1},q \bigr)+ \xi _{n} \bigr)^{2}+2\alpha _{n} \bigl\langle \overrightarrow{qf(x_{n})}, \overrightarrow{q x_{n+1}} \bigr\rangle \\& \quad \le (1-\alpha _{n})^{2} \bigl(k^{2}_{n}d^{2} \bigl(s_{n}x_{n} \oplus (1-s _{n}) x_{n+1},q \bigr)+M\xi _{n} \bigr) \\& \quad \quad{} +2\alpha _{n}d\bigl(f(q),f(x_{n})\bigr)d( x_{n+1},q)+2\alpha _{n} \bigl\langle \overrightarrow{qf(q)}, \overrightarrow{q x_{n+1}} \bigr\rangle \\& \quad \le (1-\alpha _{n})^{2}k^{2}_{n}d^{2} \bigl(s_{n}x_{n} \oplus (1-s_{n}) x _{n+1},q \bigr) \\& \quad \quad {}+M(1-\alpha _{n})^{2}\xi _{n}+2\kappa \alpha _{n}d(x_{n},q)d( x _{n+1},q)+2\alpha _{n} \bigl\langle \overrightarrow{qf(q)}, \overrightarrow{q x_{n+1}} \bigr\rangle \\& \quad \le (1-\alpha _{n})^{2}k^{2}_{n} \bigl(s_{n}d^{2}(x_{n},q)+(1-s_{n})d ^{2}(x_{n+1},q)-s_{n}(1-s_{n})d^{2}(x_{n+1},x_{n}) \bigr) \\& \quad \quad {}+M\xi _{n}+\kappa \alpha _{n} \bigl(d^{2}(x_{n},q)+d^{2}(x_{n+1},q) \bigr)+2\alpha _{n} \bigl\langle \overrightarrow{qf(q)}, \overrightarrow{q x_{n+1}} \bigr\rangle \\& \quad \le \bigl[k^{2}_{n}s_{n}(1-2\alpha _{n})+\kappa \alpha _{n} \bigr]d^{2}(x _{n},q)+ \bigl[k^{2}_{n}(1-s_{n}) (1-2 \alpha _{n})+\kappa \alpha _{n} \bigr]d^{2}(x_{n+1},q) \\& \quad \quad {}+M\xi _{n}+k^{2}_{n}\alpha ^{2}_{n} \bigl(s_{n}d^{2}(x_{n},q)+(1-s _{n})d^{2}(x_{n+1},q) \bigr)+2\alpha _{n} \bigl\langle \overrightarrow{qf(q)}, \overrightarrow{q x_{n+1}} \bigr\rangle . \end{aligned}$$

Since $\{x_{n}\}$ is bounded, there exists $M>0$ such that

$$ \sup_{n\ge 1} \bigl\{ k_{n}^{2}d^{2}(x_{n},q),2k_{n}d \bigl(s_{n}x_{n} \oplus (1-s_{n}) x_{n+1},q \bigr)+\xi _{n} \bigr\} \leq M, $$

and so

$$\begin{aligned} d^{2}(x_{n+1},q) \le& \bigl[k^{2}_{n}s_{n}(1-2 \alpha _{n})+\kappa \alpha _{n} \bigr]d^{2}(x_{n},q) \\ &{} + \bigl[k^{2}_{n}(1-s_{n}) (1-2\alpha _{n})+\kappa \alpha _{n} \bigr]d^{2}(x_{n+1},q) \\ &{} +M\xi _{n}+\alpha ^{2}_{n}M+2\alpha _{n} \bigl\langle \overrightarrow{qf(q)}, \overrightarrow{q x_{n+1}} \bigr\rangle . \end{aligned}$$

(3.15)

From conditions $(C_{1})$, $(C_{3})$, and $(C_{4})$, it follows that $\lim_{n\to \infty } (1-k_{n}^{2}(1-s_{n})(1-2\alpha _{n})-\kappa \alpha _{n} )=s>0$, for sufficiently large $n>N$, we have

$$ 1-k_{n}^{2}(1-s_{n}) (1-2\alpha _{n})- \kappa \alpha _{n}>0, $$

and so

$$\begin{aligned}& \frac{k^{2}_{n}s_{n}(1-2\alpha _{n})+\kappa \alpha _{n}}{1-k_{n}^{2}(1-s _{n})(1-2\alpha _{n})-\kappa \alpha _{n}} \\& \quad = \biggl[1+\frac{(k_{n}^{2}-1)-2k_{n}^{2}\alpha _{n}+2\kappa \alpha _{n}}{1-k_{n}^{2}(1-s_{n})(1-2\alpha _{n})-\kappa \alpha _{n}} \biggr] \\& \quad \le \biggl[1-\frac{2(1-\epsilon -\kappa )\alpha _{n}}{1-k_{n}^{2}(1-s _{n})(1-2\alpha _{n})-\kappa \alpha _{n}} \biggr]. \end{aligned}$$

(3.16)

It follows from (3.15) and (3.16) that

$$\begin{aligned} d^{2}(x_{n+1},q) \le& \frac{ [k^{2}_{n}s_{n}(1-2\alpha _{n}) + \kappa \alpha _{n} ]d^{2}(x_{n},q)}{1-k_{n}^{2}(1-s_{n})(1-2 \alpha _{n})-\kappa \alpha _{n}} \\ &{} +\frac{M\xi _{n}+M\alpha ^{2}_{n}+2\alpha _{n} \langle \overrightarrow{qf(q)}, \overrightarrow{q x_{n+1}} \rangle }{1-k_{n}^{2}(1-s_{n})(1-2 \alpha _{n})-\kappa \alpha _{n}} \\ \le &\biggl[1-\frac{2(1-\epsilon -\kappa )\alpha _{n}}{1-k_{n}^{2}(1-s _{n})(1-2\alpha _{n})-\kappa \alpha _{n}} \biggr]d^{2}(x_{n},q) \\ &{} +\frac{\alpha _{n} (M\frac{\xi _{n}}{\alpha _{n}}+M\alpha _{n}+2 \langle \overrightarrow{qf(q)}, \overrightarrow{q x_{n+1}} \rangle )}{1-k_{n}^{2}(1-s _{n})(1-2\alpha _{n})-\kappa \alpha _{n}} \\ =&(1-\gamma _{n})d^{2}(x_{n},q)+\delta _{n}, \end{aligned}$$

(3.17)

where

$$\begin{aligned}& \gamma _{n}=\frac{2(1-\epsilon -\kappa )\alpha _{n}}{1-k_{n}^{2}(1-s _{n})(1-2\alpha _{n})-\kappa \alpha _{n}}, \\& \delta _{n}=\frac{\alpha _{n} (M\frac{\xi _{n}}{\alpha _{n}}+M\alpha _{n}+2 \langle \overrightarrow{qf(q)}, \overrightarrow{q x_{n+1}} \rangle )}{1-k_{n}^{2}(1-s _{n})(1-2\alpha _{n})-\kappa \alpha _{n}}. \end{aligned}$$

By conditions $(C_{1})$, $(C_{2})$, and $(C_{5})$,

$$\begin{aligned} \lim_{n\to \infty }\frac{\xi _{n}}{\alpha _{n}}=0. \end{aligned}$$

(3.18)

It follows from conditions $(C_{1})$–$(C_{4})$, (3.14), and (3.18) that $\gamma _{n}\subset (0,1)$ and $\sum^{\infty }_{n=1} \gamma _{n}=\infty$,

$$\begin{aligned} \limsup_{n\to \infty }\frac{\delta _{n}}{\gamma _{n}}=\limsup_{n\to \infty } \frac{\frac{\xi _{n}}{\alpha _{n}}M+\alpha _{n}M+2 \langle \overrightarrow{qf(q)},\overrightarrow{qx _{n+1}} \rangle }{2(1-\epsilon -\kappa )}\leq 0. \end{aligned}$$

By (3.17) and Lemma 2.2, we know that $x_{n}\to q= P_{F(T)}f(q)$, which solves the following variational inequality:

$$ \bigl\langle \overrightarrow{f(q)q},\overrightarrow{qx} \bigr\rangle \geq 0, \quad \forall x \in F(T). $$

This completes the proof of Theorem 3.1. □

If $k_{n}\equiv 1$, for any $n\geq 1$, that is, T is an asymptotically nonexpansive mapping in the intermediate sense, then from Theorem 3.1 we have the following result.

Corollary 3.1

Assume that f, C, and $(X,d)$ are the same as in Theorem 3.1. Let $T: C\to C$ be an asymptotically nonexpansive mapping in the intermediate sense with sequence $\{\xi _{n}\}\subset [0,+\infty )$. If $\lim_{n\to \infty }d(T^{n}x_{n},x_{n})=0$, $F(T)\neq \emptyset$, and conditions ($C_{1}$), ($C_{2}$), ($C_{4}$), and ($C_{5}$) in Theorem 3.1 hold, then, for an arbitrary initial point $x_{0}\in C$, the sequence $\{x_{n}\}$ defined by

$$\begin{aligned} x_{n+1}=\alpha _{n}f( x_{n})\oplus (1-\alpha _{n})T^{n} \bigl(s_{n}x _{n} \oplus (1-s_{n}) x_{n+1} \bigr), \quad \forall n\geq 0, \end{aligned}$$

converges strongly to $q=P_{F(T)}f(q)$, which is the unique solution of the following variational inequality:

$$ \bigl\langle \overrightarrow{qf(q)}, \overrightarrow{x q} \bigr\rangle \geq 0, \quad x\in F(T). $$

Remark 3.1

Corollary 3.1 still is a new consequence.

If $T: C\to C$ is an asymptotically nonexpansive mapping, then Theorem 3.1 should be rewritten as the following two corollaries.

Corollary 3.2

Let f, C, and $(X,d)$ be the same as in Theorem 3.1, $T: C\to C$ be an asymptotically nonexpansive mapping with sequence $\{k_{n}\}\subset [1,+\infty )$. If $\lim_{n\to \infty }d(T^{n}x_{n},x_{n})=0$, $F(T)\neq \emptyset$, and conditions ($C_{1}$)–($C_{4}$) in Theorem 3.1 hold, then for an arbitrary initial point $x_{0}\in C$, the following sequence $\{x_{n}\}$

$$ x_{n+1}=\alpha _{n}f( x_{n})\oplus (1-\alpha _{n})T^{n} \bigl(s_{n}x _{n} \oplus (1-s_{n}) x_{n+1} \bigr), \quad \forall n\geq 0, $$

converges strongly to $q=P_{F(T)}f(q)$, which is the unique solution of the variational inequality

$$ \bigl\langle \overrightarrow{qf(q)}, \overrightarrow{x q} \bigr\rangle \geq 0, \quad x\in F(T). $$

Proof

Take $\xi _{n}\equiv 0$ for any $n\geq 1$ in Theorem 3.1 and note that condition ($C_{5}$) in Theorem 3.1 is satisfied automatically. Hence the conclusion of Corollary 3.2 can be obtained from Theorem 3.1 immediately. □

Remark 3.2

Corollary 3.2 improves and extends the main results of [34] in regard to parameter $\alpha _{n}$.

Corollary 3.3

Suppose that T, f, C, and $(X,d)$ are the same as in Corollary 3.2. If $\lim_{n\to \infty }d(T^{n}x_{n},x_{n})=0$, $F(T)\neq \emptyset$, and conditions ($C_{1}$)–($C_{3}$) in Theorem 3.1 hold, then for any given initial point $x_{0}\in C$, the sequence $\{x_{n}\}$ generated by

$$\begin{aligned} x_{n+1}=\alpha _{n}f( x_{n})\oplus (1-\alpha _{n})T^{n} \biggl(\frac{x _{n} \oplus x_{n+1}}{2} \biggr), \quad \forall n \geq 0, \end{aligned}$$

converges strongly as $n\to \infty$ to $q=P_{F(T)}f(q)$, which solves the following variational inequality:

$$ \bigl\langle \overrightarrow{qf(q)}, \overrightarrow{x q} \bigr\rangle \geq 0, \quad x\in F(T). $$

Proof

Take $s_{n}=\frac{1}{2}$ for any $n\geq 1$ in Corollary 3.2. Then condition ($C_{4}$) in Corollary 3.2 holds. From Corollary 3.2, this completes the proof. □

Remark 3.3

Corollary 3.3 is the main result of Li and Liu [30]. It also extends and improves the main results of [31] from a Hilbert space to a $\operatorname{CAT}(0)$ space.

Corollary 3.4

Let f, C, and $(X,d)$ be the same as in Theorem 3.1, $T: C\to C$ be a nonexpansive mapping with $F(T)\neq \emptyset$. If conditions ($C _{1}$), ($C_{2}$), and ($C_{4}$) in Theorem 3.1 hold, then for any initial point $x_{0}\in C$, the sequence $\{x_{n}\}$ defined by

$$\begin{aligned} x_{n+1}=\alpha _{n}f( x_{n})\oplus (1-\alpha _{n})T \bigl(s_{n}x_{n} \oplus (1-s_{n}) x_{n+1} \bigr), \quad \forall n\geq 0, \end{aligned}$$

converges strongly to $q=P_{F(T)}f(q)$, which is the unique solution of the following variational inequality:

$$ \bigl\langle \overrightarrow{qf(q)}, \overrightarrow{x q} \bigr\rangle \geq 0, \quad x\in F(T). $$

Proof

For each $n\geq 1$, let $\xi _{n}\equiv 0$ and $k_{n}\equiv 1$ in Theorem 3.1. Then conditions ($C_{3}$) and ($C_{5}$) in Theorem 3.1 are satisfied. The condition of $\lim_{n\to \infty }d(T ^{n}x_{n},x_{n})=0$ is not needed, because it can be proved very similarly to [31]. Thus, the conclusion of Corollary 3.4 can be obtained from Theorem 3.1 immediately. □

Remark 3.4

Corollary 3.4 improves and extends the corresponding results of Theorem 2.1 in [32] from a Hilbert space to a $\operatorname{CAT}(0)$ space, monotonic increase of sequence $\{s_{n}\}$ and condition $\sum^{+\infty }_{n=1}|\alpha _{n+1}- \alpha _{n}|<+\infty$ are not also needed.