Common fixed points of G-nonexpansive mappings on Banach spaces with a graph

Abstract

In this paper, we prove the weak and strong convergence of a sequence $\{x_{n}\}$ generated by the Ishikawa iteration to some common fixed points of two G-nonexpansive mappings defined on a Banach space endowed with a graph.

1 Introduction

In 1922, Banach proved a remarkable and powerful result called the Banach contraction principle. Because of its fruitful applications, the principle has been generalized in many directions. The recent version of the theorem was given in Banach spaces endowed with a graph. In 2008, Jachymski [1] gave a generalization of the Banach contraction principle to mappings on a metric space endowed with a graph. In 2012, Aleomraninejad et al. [2] presented some iterative scheme results for G-contractive and G-nonexpansive mappings on graphs. In 2015, Alfuraidan and Khamsi [3] defined the concept of G-monotone nonexpansive multivalued mappings defined on a metric space with a graph. In the same year, Alfuraidan [4] gave a new definition of the G-contraction and obtained sufficient conditions for the existence of fixed points for multivalued mappings on a metric space with a graph, and also in [5], he proved the existence of a fixed point of monotone nonexpansive mapping defined in a Banach space endowed with a graph. Recently, Tiammee et al. [6] proved Browder’s convergence theorem for G-nonexpansive mapping in a Banach space with a directed graph. They also proved the strong convergence of the Halpern iteration for a G-nonexpansive mapping.

Inspired by all aforementioned references, the author proves strong and weak convergence theorems for G-nonexpansive mappings using the Ishikawa iteration generated from arbitrary $x_{0}$ in a closed convex subset C of a uniformly convex Banach space X endowed with a graph.

2 Preliminaries

In this section, we recall some standard graph notations and terminology and also some needed results.

Let $(X, d)$ be a metric space, and $\triangle= \{(x, x) | x \in X\}$. Consider a directed graph G for which the set $V(G)$ of its vertices coincides with X and the set $E(G)$ of its edges contains all loops. Assume that G has no parallel edges. Then $G = (V(G), E(G))$, and by assigning to each edge the distance between its vertices, G may be treated as a weighted graph.

Definition 2.1

The conversion of a graph G is the graph obtained from G by reversing the direction of edges denoted by $G^{-1}$, and

$$ E\bigl(G^{-1}\bigr) = \bigl\{ (x,y) \in X \times X | (y,x) \in E(G)\bigr\} . $$

Definition 2.2

Let x and y be vertices of a graph G. A path in G from x to y of length N ($N \in\mathbb{N} \cup\{0\}$) is a sequence $\{ x_{i}\}_{i=0}^{N}$ of $N+1$ vertices for which

$$x_{0} = x,\qquad x_{N} =y,\quad \text{and} \quad (x_{i}, x_{i+1}) \in E(G)\quad \text{for } i = 0,1, \ldots,N-1. $$

Definition 2.3

A graph G is said to be connected if there is a path between any two vertices of the graph G.

Definition 2.4

A directed graph $G=(V(G), E(G))$ is said to be transitive if, for any $x,y,z \in V(G)$ such that $(x,y)$ and $(y,z)$ are in $E(G)$, we have $(x,z) \in E(G)$.

The definition of a G-nonexpansive mapping is given as follows.

Definition 2.5

Let C be a nonempty convex subset of a Banach space X, and $G =(V(G), E(G))$ a directed graph such that $V(G) =C$. Then a mapping $T: C \to C$ is G-nonexpansive (see [3], Definition 2.3(iii)) if it satisfies the following conditions.

• (i) T is edge-preserving.

• (ii) $\|Tx -Ty\| \leq\|x-y\|$ whenever $(x,y) \in E(G)$ for any $x,y \in C$.

Definition 2.6

([7])

Let C be a nonempty closed convex subset of a real uniformly convex Banach space X. The mappings $T_{i}$ ($i = 1,2$) on C are said to satisfy Condition B if there exists a nondecreasing function $f : [0, \infty) \rightarrow [0, \infty)$ with $f(0) = 0$ and $f(r) > 0$ for all $r > 0$ such that, for all $x \in C$,

$$\max\bigl\{ \Vert x -T_{1}x\Vert , \Vert x-T_{2}x \Vert \bigr\} \geq f\bigl(d(x,F)\bigr), $$

where $F = F(T_{1}) \cap F(T_{2})$ and $F(T_{i})$ ($i=1,2$) are the sets of fixed points of $T_{i}$.

Definition 2.7

([7])

Let C be a subset of a metric space $(X, d)$. A mapping T is semicompact if for a sequence $\{x_{n}\}$ in C with $\lim_{n \rightarrow\infty} d(x_{n}, Tx_{n}) = 0$, there exists a subsequence $\{x_{n_{i}}\}$ of $\{x_{n}\}$ such that $x_{n_{i}} \rightarrow p \in C$.

Definition 2.8

A Banach space X is said to satisfy Opial’s property if the following inequality holds for any distinct elements x and y in X and for each sequence $\{x_{n}\}$ weakly convergent to x:

$$\liminf_{n \to\infty} \|x_{n} -x\| < \liminf _{n \to\infty} \|x_{n} -y\|. $$

Definition 2.9

Let X be a Banach space. A mapping T with domain D and range R in X is demiclosed at 0 if, for any sequence $\{x_{n}\}$ in D such that $\{x_{n}\}$ converges weakly to $x \in D$ and $\{Tx_{n}\}$ converges strongly to 0, we have $Tx = 0$.

Lemma 2.10

([8])

Let X be a uniformly convex Banach space, and $\{\alpha_{n}\}$ a sequence in $[\delta, 1 - \delta]$ for some $\delta\in(0,1)$. Suppose that sequences $\{x_{n}\}$ and $\{y_{n}\}$ in X are such that $\limsup_{n \to\infty} \|x_{n}\| \leq c$, $\limsup_{n \to\infty} \| y_{n}\| \leq c$ and $\limsup_{n \to\infty} \|\alpha x_{n} + (1-\alpha _{n})y_{n}\| = c$ for some $c \geq0$. Then $\lim_{n \to\infty} \| x_{n} -y_{n}\| = 0$.

Lemma 2.11

([9])

Let X be a Banach space, and $R>1$ be a fixed number. Then X is uniformly convex if and only if there exists a continuous, strictly increasing, and convex function $g : [0,\infty) \to[0,\infty)$ with $g(0) =0$ such that

$$\bigl\Vert \lambda x + (1-\lambda)y\bigr\Vert ^{2} \leq\lambda \Vert x\Vert ^{2} + (1-\lambda)\Vert y\Vert ^{2} - \lambda(1-\lambda)g\bigl(\Vert x-y\Vert \bigr) $$

for all $x,y \in B_{R}(0) = \{x \in X| \|x\| \leq R\}$ and $\lambda\in[0,1]$.

Lemma 2.12

([10])

Let X be a Banach space that satisfies Opial’s property, and let $\{ x_{n}\}$ be a sequence in X. Let x, y in X be such that $\lim_{n \to\infty} \|x_{n} - x\|$ and $\lim_{n \to\infty} \|x_{n} - y\|$ exist. If $\{x_{n_{j}}\}$ and $\{x_{n_{k}}\}$ are subsequences of $\{ x_{n}\}$ that converge weakly to x and y, respectively, then $x=y$.

3 Main results

Throughout the section, we let C be a nonempty closed convex subset of a Banach space X endowed with a directed graph G such that $V(G) = C$ and $E(G)$ is convex. We also suppose that the graph G is transitive. The mappings $T_{i}$ ($i =1,2$) are G-nonexpansive from C to C with $F = F(T_{1}) \cap F(T_{2})$ nonempty. Let $\{x_{n}\}$ be a sequence generated from arbitrary $x_{0} \in C$,

$$\begin{aligned}& x_{n+1} = (1-\alpha_{n}) x_{n} + \alpha_{n} T_{1} y_{n}, \\& y_{n} = (1-\beta_{n}) x_{n} + \beta_{n} T_{2} x_{n}, \end{aligned}$$

where $\{\alpha_{n}\}$ and $\{\beta_{n}\}$ are real sequences in $[0,1]$.

We first begin by proving the following useful results.

Proposition 3.1

Let $z_{0} \in F$ be such that $(x_{0}, z_{0})$, $(y_{0}, z_{0})$, $(z_{0}, x_{0})$, and $(z_{0}, y_{0})$ are in $E(G)$. Then $(x_{n}, z_{0})$, $(y_{n}, z_{0})$, $(z_{0}, x_{n})$, $(z_{0}, y_{n})$, and $(x_{n}, y_{n})$ are in $E(G)$.

Proof

We divide the proof into three parts. In the first part, with the assumption $(x_{0}, z_{0}), (y_{0}, z_{0}) \in E(G)$, we will show by induction that $(x_{n}, z_{0}), (y_{n}, z_{0}) \in E(G)$. Then, with the assumption $(z_{0}, x_{0}), (z_{0}, y_{0})\in E(G)$, we will again prove by induction that $(z_{0}, x_{n}), (z_{0}, y_{n}) \in E(G)$. In the third part, we combine these two results using transitivity of G to get the statement in the proposition. Let $(x_{0},z_{0})$ and $(y_{0},z_{0}) \in E(G)$. Then $(T_{1}y_{0}, z_{0}), (T_{2}x_{0}, z_{0}) \in E(G)$ since $T_{i}$ ($i = 1,2$) are edge-preserving. By the convexity of $E(G)$ and $(T_{1}y_{0}, z_{0}), (x_{0},z_{0}) \in E(G)$, we have $(x_{1},z_{0}) \in E(G)$. Then, by edge-preserving of $T_{2}$, $(T_{2}x_{1}, z_{0}) \in E(G)$. Again, by the convexity of $E(G)$ and $(T_{2}x_{1}, z_{0}), (x_{1},z_{0}) \in E(G)$, we get $(y_{1},z_{0}) \in E(G)$ and then $(T_{1}y_{1},z_{0}) \in E(G)$. Next, we assume that $(x_{k}, z_{0}), (y_{k},z_{0}) \in E(G)$. Then $(T_{2}x_{k}, z_{0}), (T_{1}y_{k}, z_{0}) \in E(G)$ since $T_{i}$ ($i=1,2$) are edge-preserving. Since $E(G)$ is convex, $(x_{k+1},z_{0}) \in E(G)$. Indeed,

$$\alpha(T_{1}y_{k}, z_{0}) +(1-\alpha) (x_{k}, z_{0}) = \bigl(\alpha T_{1}y_{k} + (1-\alpha)x_{k}, z_{0}\bigr)=(x_{k+1},z_{0}) \in E(G). $$

Since $T_{2}$ is edge-preserving, $(T_{2}x_{k+1}, z_{0}) \in E(G)$. Using the convexity of $E(G)$, we get $(y_{k+1}, z_{0}) \in E(G)$. To be explicit,

$$\beta(T_{2}x_{k+1},z_{0}) + (1- \beta) (x_{k+1}, z_{0}) = \bigl(\beta T_{2}x_{k+1} + (1-\beta)x_{k+1}, z_{0}\bigr) = (y_{k+1},z_{0}) \in E(G). $$

Hence, by induction, $(x_{n},z_{0}), (y_{n},z_{0}) \in E(G)$ for all $n \geq1$. Using a similar argument, we can show that $(z_{0},x_{n}), (z_{0},y_{n}) \in E(G)$ under the assumption that $(z_{0},x_{0}), (z_{0},y_{0}) \in E(G)$. Therefore, $(x_{n},y_{n}) \in E(G)$ by the transitivity of G. □

Lemma 3.2

Let $z_{0} \in F$. Suppose that $(x_{0}, z_{0}), (y_{0}, z_{0}), (z_{0}, x_{0}), (z_{0}, y_{0})\in E(G)$ for arbitrary $x_{0}$ in C. Then $\lim_{n \to\infty} \|x_{n} -z_{0}\|$ exists.

Proof

Notice that

$$\begin{aligned} \Vert x_{n+1} -z_{0}\Vert &= \bigl\Vert (1- \alpha_{n})x_{n} + \alpha_{n}T_{1}y_{n} -z_{0}\bigr\Vert \\ &\leq(1-\alpha_{n}) \Vert x_{n} - z_{0} \Vert + \alpha_{n}\Vert T_{1}y_{n} -z_{0}\Vert \\ &\leq(1-\alpha_{n}) \Vert x_{n} - z_{0} \Vert + \alpha_{n}\Vert y_{n} -z_{0}\Vert \\ &= (1-\alpha_{n}) \Vert x_{n} - z_{0}\Vert + \alpha_{n}\bigl\Vert (1-\beta_{n})x_{n} - (1-\beta_{n})z_{0} +\beta_{n}(T_{2}x_{n} - z_{0})\bigr\Vert \\ &\leq (1- \alpha_{n}) \Vert x_{n} - z_{0} \Vert + \alpha_{n}(1-\beta_{n})\Vert x_{n} -z_{0} \Vert + \alpha_{n} \beta_{n}\Vert x_{n} -z_{0} \Vert \\ & = (1- \alpha_{n}) \Vert x_{n} - z_{0} \Vert + \alpha_{n} \Vert x_{n} -z_{0} \Vert \\ &=\Vert x_{n} - z_{0}\Vert . \end{aligned}$$

Thus, $\lim_{n \to\infty} \|x_{n}-z_{0}\|$ exists. In particular, the sequence $\{x_{n}\}$ is bounded. □

Lemma 3.3

If X is uniformly convex, $\{\alpha_{n}\}, \{\beta_{n}\} \subset [\delta, 1-\delta]$ for some $\delta\in(0,\frac{1}{2})$, and $(x_{0}, z_{0}), (y_{0}, z_{0}), (z_{0}, x_{0}), (z_{0}, y_{0})\in E(G)$ for arbitrary $x_{0}$ in C and $z_{0} \in F$, then

$$\lim_{n \to\infty}\|x_{n} - T_{1}x_{n} \| = 0= \lim_{n \to\infty} \| {x_{n} - T_{2}x_{n}} \|. $$

Proof

Let $z_{0} \in F$. Then, by the boundedness of $\{x_{n}\}$ and $\{T_{2}x_{n}\}$ there exists $r > 0$ such that $x_{n} - z_{0}, y_{n} -z_{0} \in B_{r}(0)$ for all $n \geq1$. Put $c = \lim_{n \to\infty} \|x_{n} - z_{0}\|$. If $c = 0$, then by the G-nonexpansiveness of $T_{i}$ ($i=1,2$) we have

$$\|x_{n} - T_{i}x_{n}\| \leq\|x_{n} -z_{0}\| + \|z_{0} -T_{i}x_{n}\| \leq \|x_{n} -z_{0}\| + \|z_{0} -x_{n}\|. $$

Therefore, the result follows. Suppose that $c > 0$. Hence, by Lemma 2.11 together with the G-nonexpansiveness of $T_{2}$, we have

$$\begin{aligned} \Vert y_{n} -z_{0}\Vert ^{2} &= \bigl\Vert (1-\beta_{n})x_{n} + \beta_{n} T_{2}x_{n} -z_{0}\bigr\Vert ^{2} \\ &= \bigl\Vert \beta_{n}(T_{2}x_{n} - z_{0}) + (1 - \beta_{n}) (x_{n} -z_{0})\bigr\Vert ^{2} \\ &\leq\beta_{n} \Vert T_{2}x_{n} -z_{0}\Vert ^{2} + (1-\beta_{n})\Vert x_{n} -z_{0}\Vert ^{2} - \beta_{n}(1- \beta_{n}) g\bigl(\Vert T_{2}x_{n} -x_{n}\Vert \bigr) \\ &\leq\beta_{n}\Vert x_{n}-z_{0}\Vert ^{2} + (1-\beta_{n}) \Vert x_{n} - z_{0}\Vert ^{2} \\ &=\Vert x_{n} -z_{0}\Vert ^{2}. \end{aligned}$$

Thus,

$$\limsup_{n \to\infty} \|y_{n} -z_{0}\| \leq \limsup_{n \to\infty} \| x_{n} -z_{0}\| \leq c. $$

Notice also that

$$\begin{aligned} \Vert x_{n+1} -z_{0}\Vert ^{2} &= \bigl\Vert (1-\alpha_{n})x_{n} + \alpha_{n}T_{1}y_{n} -z_{0}\bigr\Vert ^{2} \\ & \leq\alpha_{n}\Vert y_{n} -z_{0}\Vert ^{2} +(1 - \alpha_{n}) \Vert x_{n} -z_{0}\Vert ^{2} - \alpha_{n}(1- \alpha_{n}) g\bigl(\Vert T_{1}y_{n} -x_{n}\Vert \bigr) \\ & \leq \Vert x_{n} -z_{0}\Vert ^{2} - \alpha_{n}(1-\alpha_{n})g\bigl(\Vert T_{1}y_{n} -x_{n}\Vert \bigr) \\ &\leq \Vert x_{n} -z_{0}\Vert ^{2} - \delta^{2} g\bigl(\Vert T_{1}y_{n} -x_{n}\Vert \bigr). \end{aligned}$$

Thus,

$$ \delta^{2} g\bigl(\Vert T_{1}y_{n} -x_{n}\Vert \bigr) \leq \Vert x_{n} -z_{0} \Vert ^{2} - \Vert x_{n+1} -z_{0}\Vert ^{2}. $$

This implies that $\lim_{n \to\infty} g (\|T_{1} y_{n} -x_{n}\|) =0$, and since g is strictly increasing and continuous at 0,

$$ \lim_{n \to\infty} \|T_{1} y_{n} - x_{n}\| =0. $$

(1)

Since $T_{1}$ is G-nonexpansive, we have

$$ \|x_{n} -z_{0}\| \leq\|x_{n} -T_{1}y_{n}\| + \|T_{1}y_{n} -T_{1}z_{0}\| \leq \|x_{n} - T_{1}y_{n}\| + \|y_{n} -z_{0}\|. $$

Taking lim inf yields

$$ c \leq\liminf_{n \to\infty} \|y_{n} -z_{0}\|. $$

Hence, we have

$$\lim_{n \to\infty} \|y_{n} -z_{0}\| = c. $$

Since

$$\lim_{n \to\infty} \bigl\Vert \beta_{n}(T_{2}x_{n} -z_{0}) + (1-\beta _{n}) (x_{n} -z_{0}) \bigr\Vert = \lim_{n \to\infty} \Vert y_{n} -z_{0}\Vert = c $$

and

$$\limsup_{n \to\infty} \|T_{2} x_{n} - z_{0}\| \leq c, $$

by Lemma 2.10 we have

$$ \lim_{n \to\infty} \|T_{2}x_{n} -x_{n}\| =0. $$

(2)

By the G-nonexpansiveness of $T_{1}$ together with $\|x_{n} -y_{n}\| \leq\|T_{2}x_{n} -x_{n}\|$ we have

$$\begin{aligned} \|T_{1} x_{n} -x_{n}\| & \leq \|T_{1}x_{n} -T_{1}y_{n}\|+\| T_{1}y_{n}-x_{n}\| \\ & \leq\|x_{n} - y_{n}\| + \|T_{1}y_{n} -x_{n}\| \\ & \leq \|T_{2}x_{n} -x_{n}\| + \|T_{1}y_{n} -x_{n}\|. \end{aligned}$$

Using (1) and (2), $\lim_{n \to\infty} \|T_{1}x_{n} - x_{n}\| =0$. Hence, the lemma is proved. □

Lemma 3.4

Suppose that X satisfies the Opial’s property and that $(x_{0},z_{0})$, $(y_{0},z_{0})$ are in $E(G)$ for $z_{0} \in F$ and arbitrary $x_{0} \in C$. Then $I -T_{i}$ ($i=1,2$) are demiclosed.

Proof

Suppose that $\{x_{n}\}$ is a sequence in C that converges weakly to q. From Lemma 3.3 we have $\lim_{n \to \infty}\|x_{n} -T_{i}x_{n}\| =0$. Suppose for contradiction that $q \neq T_{i}q$. Then, by Opial’s property we have

$$\begin{aligned} \limsup_{n \to\infty} \Vert x_{n}-q\Vert &< \limsup _{n \to\infty} \Vert x_{n} - T_{i}q\Vert \\ & \leq\limsup_{n \to\infty} \bigl( \Vert x_{n}-T_{i}x_{n} \Vert + \Vert T_{i}x_{n} -T_{i}q\Vert \bigr) \\ &\leq\limsup_{n \to\infty} \Vert x_{n}-q\Vert , \end{aligned}$$

a contradiction. Hence, $T_{i}q = q$, so the conclusion holds. □

Theorem 3.5

Suppose X is uniformly convex, $\{\alpha_{n}\} , \{\beta_{n}\} \subset[\delta, 1-\delta]$ for some $\delta\in(0,\frac{1}{2})$, $T_{i}$ ($i =1,2$) satisfy Condition B, F is dominated by $x_{0}$, F dominates $x_{0}$, and $(x_{0}, z), (y_{0}, z), (z, x_{0}), (z, y_{0})\in E(G)$ for each $z \in F$ and arbitrary $x_{0}\in C$. Then $\{ x_{n}\}$ converges strongly to a common fixed point of $T_{i}$.

Proof

Let $z\in F$. Recall the following facts from Lemma 3.2:

• (i) $\{x_{n}\}$ is bounded;

• (ii) $\lim_{n \to\infty} \|x_{n} - z\|$ exists;

• (iii) $\|x_{n+1} -z\| \leq\|x_{n} -z\|$ for all $n \geq1$.

They imply that

$$d(x_{n +1}, F) \leq d(x_{n}, F). $$

Thus $\lim_{n \to\infty} d(x_{n}, F)$ exists. Since each $T_{i}$ ($i = 1,2$) satisfies Condition B and $\lim_{n \to\infty}\|x_{n} - T_{i}x_{n}\| =0$, we have

$$\lim_{n \to\infty} f\bigl(d(x_{n}, F)\bigr) = 0 $$

and then

$$\lim_{n \to\infty} d(x_{n}, F) = 0. $$

Hence, there are a subsequence $\{x_{n_{j}}\}$ of $\{x_{n}\}$ and a sequence $\{z_{j}\} \subset F$ satisfying

$$\|x_{n_{j}} - z_{j}\| \leq\frac{1}{2^{j}}. $$

Put $n_{j+1} = n_{j} + k$ for some $k \geq1$. Then

$$\|x_{n_{j+1}} - z_{j}\| \leq\|x_{n_{j} + k-1} - z_{j} \| \leq\| x_{n_{j}} -z_{j}\| \leq\frac{1}{2^{j}}. $$

Hence,

$$\|z_{j+1} - z_{j} \| \leq\frac{3}{2^{j+1}}, $$

so that $\{z_{j}\}$ is a Cauchy sequence. We assume that $z_{j} \to q \in C$ as $n \to\infty$. Since F is closed, $q \in F$. Hence, we have $x_{n_{j}} \to q$ as $j \to\infty$, and since $\lim_{n \to\infty } \|x_{n} -q\|$ exists, the conclusion follows. □

Theorem 3.6

Suppose that X is uniformly convex, $\{\alpha_{n}\}, \{\beta_{n}\} \subset[\delta, 1-\delta]$ for some $\delta\in(0,\frac{1}{2})$, one of $T_{i}$ ($i =1,2$) is semicompact, F is dominated by $x_{0}$, F dominates $x_{0}$, and $(x_{0}, z_{0}), (y_{0}, z_{0}), (z_{0}, x_{0}), (z_{0}, y_{0})\in E(G)$ for $z_{0} \in F$ and arbitrary $x_{0} \in C$. Then $\{x_{n}\}$ converges strongly to a common fixed point of $T_{i}$.

Proof

Suppose that $T_{2}$ is semicompact; by Lemma 3.2 and Lemma 3.3 we have a bounded sequence $\{x_{n}\}$, and $\lim_{n \to\infty} \|x_{n} -T_{i}x_{n}\| = 0$. Hence, by the semicompactness of $T_{2}$ there exist $q \in C$ and a subsequence $\{x_{n_{j}}\}$ of $\{x_{n}\}$ such that $x_{n_{j}} \to q$ as $j \to\infty$ and $\lim_{n \to\infty} \|x_{n_{j}} -T_{i}x_{n_{j}}\| = 0$. Notice that

$$\begin{aligned} \|q - T_{i}q\| & \leq\|q-x_{n_{j}}\| + \|x_{n_{j}} - T_{i}x_{n_{j}}\| + \|T_{i}x_{n_{j}} - T_{i}q\| \\ & \leq\|q-x_{n_{j}}\| + \|x_{n_{j}} - T_{i}x_{n_{j}} \| + \|x_{n_{j}} -q\| \\ & \to0 \quad \text{as }n \to\infty. \end{aligned}$$

Hence, $q \in F$. Since $\lim_{n \to\infty} d(x_{n}, F)=0$, it follows, by repeating the same argument as in the proof of Theorem 3.5, that $\{x_{n}\}$ converges strongly to a common fixed point of $T_{i}$ ($i = 1,2$), and the proof is complete. □

Theorem 3.7

Suppose that X is uniformly convex, $\{\alpha_{n}\}, \{\beta_{n}\} \subset[\delta, 1-\delta]$ for some $\delta\in(0,\frac{1}{2})$. If X satisfies Opial’s property, $I-T_{i}$ is demiclosed at zero for each i, F is dominated by $x_{0}$, F dominates $x_{0}$, and $(x_{0}, z_{0}), (y_{0}, z_{0}), (z_{0}, x_{0}), (z_{0}, y_{0})\in E(G)$ for $z_{0} \in F$ and arbitrary $x_{0} \in C$, then $\{x_{n}\}$ converges weakly to a common fixed point of $T_{i}$.

Proof

Note that by Lemma 3.2, for each $q \in F$,

$$ \lim_{n \to\infty} \|x_{n} -q\|\quad \text{exists}. $$

(3)

Let $\{x_{n_{k}}\}$ and $\{x_{n_{j}}\}$ be subsequences of the sequence $\{x_{n}\}$ with two weak limits $q_{1}$ and $q_{2}$, respectively. Notice that, by Lemma 3.3,

$$\begin{aligned}& \|x_{n_{j}} - T_{i}x_{n_{j}}\| \to0 \quad \text{as }n \to\infty\quad \text{and} \\& \|x_{n_{k}} - T_{i}x_{n_{k}}\| \to0 \quad \text{as }n \to\infty. \end{aligned}$$

Hence, $T_{i}q_{1} = q_{1}$ and $T_{i}q_{2} = q_{2}$. By Lemma 3.4 we have $q_{1}, q_{2} \in F$. In particular, $q_{1} =q_{2}$ by Lemma 2.12. Therefore, $\{x_{n}\}$ converges weakly to a common fixed point in F. □