New ergodic convergence theorems for non-expansive mappings and m-accretive mappings

Abstract

Two new ergodic convergence theorems for approximating the common element of the set of zero points of an m-accretive mapping and the set of fixed points of an infinite family of non-expansive mappings in a real smooth and uniformly convex Banach space are obtained, which improves some of the previous work. The computational experiments to demonstrate the effectiveness of the proposed iterative algorithms in this paper are conducted.

1 Introduction and preliminaries

Let E be a real Banach space with norm $\|\cdot\|$ and let $E^{*}$ denote the dual space of E. We use ‘→’ and ‘⇀’ (or ‘$w\mbox{-}\!\lim$’) to denote strong and weak convergence either in E or in $E^{*}$, respectively. We denote the value of $f \in E^{*}$ at $x \in E$ by $\langle x,f \rangle$.

A Banach space E is said to be uniformly convex if, for each $\varepsilon\in(0,2]$, there exists $\delta> 0$ such that

$$\|x\| = \|y\| = 1,\quad \|x - y\| \geq \varepsilon\quad\Rightarrow\quad\bigl\| \frac{x+y}{2}\bigr\| \leq1-\delta. $$

A Banach space E is said to be smooth if

$$\lim_{t \rightarrow0}\frac{\|x+ty\|-\|x\|}{t} $$

exists for each $x , y \in\{z \in E : \|z\| = 1\}$.

The normalized duality mapping $J: E \rightarrow2^{E^{*}}$ is defined by

$$Jx : = \bigl\{ f \in E^{*}: \langle x, f\rangle= \|x\|^{2}= \|f \|^{2}\bigr\} ,\quad x \in E. $$

If E is reduced to the Hilbert space H, then $J \equiv I$ is the identity mapping. If E is smooth, then $J: E \rightarrow E^{*}$ is norm-to-norm continuous. The normalized duality mapping J is said to be weakly sequentially continuous at zero if $\{x_{n}\}$ is a sequence in E which converges weakly to 0; it follows that $\{Jx_{n}\}$ converges in weak^∗ to 0.

For a mapping $T: D(T) \sqsubseteq E \rightarrow E$, we use $\operatorname{Fix}(T)$ to denote the fixed point set of it; that is, $\operatorname{Fix}(T) : = \{x\in D(T): Tx = x\}$.

Let $T : D(T) \sqsubseteq E \rightarrow E$ be a mapping. Then T is said to be:

• (1) non-expansive if

  $$\|Tx - Ty\| \leq\|x-y\| \quad \mbox{for } \forall x,y \in D(T); $$

• (2) contraction if there exists $0 < k < 1$ such that

  $$\|Tx - Ty\| \leq k \|x - y\| \quad \mbox{for } \forall x,y \in D(T); $$

• (3) accretive if, for all $x, y \in D(T)$, there exists $j(x-y) \in J(x-y)$ such that

  $$\bigl\langle Tx - Ty, j(x-y)\bigr\rangle \geq0; $$

• (4) m-accretive if T is accretive and $R(I+\lambda T) = E$ for $\forall\lambda> 0$;

• (5) strongly positive (see [1]) if $D(T) = E$ where E is a real smooth Banach space and there exists $\overline{\gamma} > 0$ such that

  $$\langle Tx,Jx\rangle\geq\overline{\gamma} \|x\|^{2}\quad \mbox{for } \forall x \in E; $$

  in this case,

  $$\|aI-bT\| = \sup_{\|x\| \leq1}\bigl|\bigl\langle (aI - bT)x, J(x)\bigr\rangle \bigr|, $$

  where I is the identity mapping and $a \in [0,1]$, $b \in[-1,1]$;

• (6) demiclosed at p if whenever $\{x_{n}\}$ is a sequence in $D(T)$ such that $x_{n} \rightharpoonup x \in D(T)$ and $Tx_{n} \rightarrow p$ then $Tx =p$.

For the accretive mapping A, we use $A^{-1}0$ to denote the set of zero points of it; that is, $A^{-1}0: = \{x \in D(A) : Ax = 0\}$. If A is accretive, then we can define, for each $r>0$, a non-expansive single-valued mapping $J_{r}^{A} : R(I+rA)\rightarrow D(A)$ by $J_{r}^{A} : = (I+rA)^{-1}$, which is called the resolvent of A [2]. It is well known that $J^{A}_{r}$ is non-expansive and $A^{-1}0 = \operatorname{Fix}(J_{r}^{A})$.

Let C be a nonempty, closed, and convex subset of E and Q be a mapping of E onto C. Then Q is said to be sunny [3] if $Q(Q(x)+t(x-Q(x))) = Q(x)$, for all $x \in E$ and $t \geq 0$.

A mapping Q of E into E is said to be a retraction [3] if $Q^{2} = Q$. If a mapping Q is a retraction, then $Q(z) = z$ for every $z \in R(Q)$, where $R(Q)$ is the range of Q.

A subset C of E is said to be a sunny non-expansive retract of E [3] if there exists a sunny non-expansive retraction of E onto C and it is called a non-expansive retract of E if there exists a non-expansive retraction of E onto C.

The first mean ergodic theorem for non-expansive mappings was proved by Baillon in Hilbert space [4]. That is, for each $x \in C$, the Cesaro means

$$ S_{n}(x):= \frac{1}{n}\sum_{k = 0}^{n-1}T^{k} x,\quad n \geq1, $$

(1.1)

converge weakly to a fixed point of a non-expansive mapping T.

The implicit midpoint rule (IMR) for non-expansive mappings in a Hilbert space was introduced by [5]. This rule generates a sequence $\{x_{n}\}$ via the semi-implicit procedure:

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

(1.2)

$\{x_{n}\}$ is proved to be weakly convergent to a fixed point of the non-expansive mapping T.

The ergodic convergence of the sequence $\{x_{n}\}$ generated by (1.2) is considered in a Hilbert space in [6]. That is, the convergence of the means

$$ z_{n} : = \frac{1}{\sum_{k = 1}^{n} a_{k}}\sum_{k = 1}^{n} a_{k}x_{k},\quad n = 1,2, \ldots, $$

(1.3)

where $\{x_{k}\}$ satisfies (1.2), to a fixed point of the non-expansive mapping T is obtained.

In a Hilbert space, Marino et al. presented the following iterative algorithm in [7]:

$$ x_{0} \in C,\quad x_{n+1}= \alpha_{n} \gamma f(x_{n}) + (I-\alpha_{n} A)Tx_{n},\quad n \geq0, $$

(1.4)

where f is a contraction, A is a strongly positive linear bounded operator, and T is non-expansive. If $\operatorname{Fix}(T)\neq\emptyset$, they proved that $\{x_{n}\}$ converges strongly to $p \in \operatorname{Fix}(T)$, which solves the variational inequality $\langle(\gamma f - A)p, z - p\rangle\leq0$, for $\forall z \in \operatorname{Fix}(T)$ under some conditions.

Motivated by the previous work, we shall present two iterative algorithms and the ergodic convergence theorems are obtained. The connection between the strongly convergent point and the solution of one kind variational inequalities is being set up. Some examples are exemplified to illustrate the effectiveness of the proposed algorithms. The computational experiments are conducted and the codes are written in Visual Basic Six.

We need the following preliminaries.

Lemma 1.1

(see [8])

Let E be a Banach space and C be a nonempty closed and convex subset of E. Let $f: C \rightarrow C$ be a contraction. Then f has a unique fixed point $u \in C$.

Lemma 1.2

(see [9])

Let E be a real uniformly convex Banach space, C be a nonempty, closed, and convex subset of E and $T: C \rightarrow E$ be a non-expansive mapping such that $\operatorname{Fix}(T) \neq \emptyset$, then $I-T$ is demiclosed at zero.

Lemma 1.3

(see [10])

In a real Banach space E, the following inequality holds:

$$\|x+y\|^{2} \leq\|x\|^{2} + 2\bigl\langle y, j(x+y)\bigr\rangle , \quad \forall x,y \in E, $$

where $j(x+y) \in J(x+y)$.

Lemma 1.4

(see [11])

Let $\{a_{n}\}$ and $\{c_{n}\}$ be two sequences of nonnegative real numbers satisfying

$$a_{n+1}\leq(1-t_{n})a_{n} + b_{n}+c_{n}, \quad\forall n \geq0, $$

where $\{t_{n}\}\subset(0,1)$ and $\{b_{n}\}$ is a number sequence. Assume that $\sum_{n=0}^{\infty}t_{n} = +\infty$, $\limsup_{n \rightarrow\infty} \frac{b_{n}}{t_{n}} \leq0$, and $\sum_{n=0}^{\infty}c_{n} < +\infty$. Then $\lim_{n \rightarrow\infty }a_{n} = 0$.

Lemma 1.5

(see [12])

Let E be a Banach space and let A be an m-accretive mapping. For $\lambda>0$, $\mu>0$, and $x \in E$, one has

$$J_{\lambda}^{A}x = J_{\mu}^{A}\biggl( \frac{\mu}{\lambda}x+\biggl(1-\frac{\mu}{\lambda}\biggr)J_{\lambda}^{A} x\biggr), $$

where $J_{\lambda}^{A} = (I+\lambda A)^{-1}$ and $J_{\mu}^{A} = (I+\mu A)^{-1}$.

Lemma 1.6

(see [13])

Assume T is a strongly positive bounded operator with coefficient $\overline{\gamma} > 0$ on a real smooth Banach space E and $0 < \rho\leq \|T\|^{-1}$. Then $\|I-\rho T \| \leq1- \rho\overline{\gamma}$.

Lemma 1.7

(see [14])

Let E be a real smooth and uniformly convex Banach space and C be a nonempty, closed, and convex sunny non-expansive retract of E, and let $Q_{C}$ be the sunny non-expansive retraction of E onto C. Let $f : E \rightarrow E$ be a fixed contractive mapping with coefficient $k \in(0,1)$, $T: E \rightarrow E$ be a strongly positive linear bounded operator with coefficient γ̅ and $U : C \rightarrow C$ be a non-expansive mapping. Suppose that the duality mapping $J : E \rightarrow E^{*}$ is weakly sequentially continuous at zero, $0 < \eta< \frac{\overline{\gamma}}{2k}$ and $\operatorname{Fix}(U) \neq\emptyset$. If for each $t \in(0,1)$, define $T_{t} : E \rightarrow E$ by

$$T_{t} x : = t \eta f(x) + (I-t T)UQ_{C}x, $$

then $T_{t}$ has a fixed point $x_{t}$, for each $0 < t \leq \|T\|^{-1}$, which is convergent strongly to the fixed point of U, as $t \rightarrow0$. That is, $\lim_{t\rightarrow0}x_{t} = p_{0} \in \operatorname{Fix}(U)$. Moreover, $p_{0}$ satisfies the following variational inequality: for $\forall z \in \operatorname{Fix}(U)$,

$$\bigl\langle (T-\eta f)p_{0}, J(p_{0}-z)\bigr\rangle \leq0. $$

Lemma 1.8

(see [15])

Let E be a real strictly convex Banach space and let C be a nonempty closed and convex subset of E. Let $T_{m}: C \rightarrow C$ be a non-expansive mapping for each $m \geq1$. Let $\{a_{m}\}$ be a real number sequence in $(0,1)$ such that $\sum_{m = 1}^{\infty}a_{m} = 1$. Suppose that $\bigcap_{m=1}^{\infty}\operatorname{Fix}(T_{m}) \neq\emptyset$. Then the mapping $\sum_{m = 1}^{\infty}a_{m} T_{m}$ is non-expansive with $\operatorname{Fix}(\sum_{m = 1}^{\infty}a_{m}T_{m}) = \bigcap_{m = 1}^{\infty}\operatorname{Fix}(T_{m})$.

2 Strong convergence theorems

Lemma 2.1

Suppose $A : D(A) \subset E \rightarrow E$ is an m-accretive mapping, where E is a Banach space, and $\{r_{n}\}_{n=0}^{\infty} \subset(0,+\infty)$ is any real number sequence. Then, for $n \geq0$, $\forall x,y\in E$, if $r_{n} \leq r_{n+1}$,

$$ \bigl\| J_{r_{n+1}}^{A}x-J_{r_{n}}^{A}y\bigr\| \leq\|x-y \|+ \frac{r_{n+1}-r_{n}}{r_{n+1}}\bigl\| J_{r_{n+1}}^{A}x-y\bigr\| ; $$

(2.1)

if $r_{n+1}\leq r_{n}$,

$$ \bigl\| J_{r_{n+1}}^{A}x-J_{r_{n}}^{A}y\bigr\| \leq\|x-y \|+ \frac{r_{n}-r_{n+1}}{r_{n}}\bigl\| x-J_{r_{n}}^{A}y\bigr\| . $$

(2.2)

Proof

In fact, if $r_{n}\leq r_{n+1}$, then, using Lemma 1.5, we have

$$\begin{aligned} \bigl\| J_{r_{n+1}}^{A}x-J_{r_{n}}^{A}y\bigr\| &= \biggl\| J_{r_{n}}^{A}\biggl[\frac {r_{n}}{r_{n+1}}x+\biggl(1- \frac{r_{n}}{r_{n+1}}\biggr) J^{A}_{r_{n+1}}x\biggr]-J_{r_{n}}^{A}y \biggr\| \\ & \leq \frac{r_{n}}{r_{n+1}}\|x-y\|+\biggl(1-\frac{r_{n}}{r_{n+1}}\biggr) \bigl\| J^{A}_{r_{n+1}}x-y\bigr\| \\ &\leq\|x-y\|+\frac{r_{n+1}-r_{n}}{r_{n+1}}\bigl\| J_{r_{n+1}}^{A}x-y\bigr\| , \end{aligned}$$

which implies that (2.1) is true.

If $r_{n+1}\leq r_{n}$, then imitating the proof of (2.1), we have (2.2).

This completes the proof. □

2.1 Ergodic convergence of the first iterative algorithm

Theorem 2.2

Let E be a real smooth and uniformly convex Banach space. Let C be a nonempty, closed, and convex sunny non-expansive retract of E, and $Q_{C}$ be the sunny non-expansive retraction of E onto C. Let $f : C \rightarrow C$ be a contraction with contractive constant $k \in(0,1)$, $T: C \rightarrow C$ be a strongly positive linear bounded operator with coefficient γ̅. Let $A: C \rightarrow E$ be an m-accretive mapping, and $S_{i}: C \rightarrow C$ be non-expansive mappings, for $i = 1,2,\ldots$ . Let $\Omega: = \bigcap_{i = 1}^{\infty}\operatorname{Fix}(S_{i})\cap A^{-1}0\neq\emptyset$. Suppose the duality mapping $J: E \rightarrow E^{*}$ is weakly sequentially continuous at zero, and $0 < \eta< \frac{\overline{\gamma}}{2k}$. Suppose $\{\alpha_{n,i}\}$, $\{\delta_{n}\}$, $\{\beta_{n}\}$, $\{b_{n,i}\}$, $\{\gamma_{n}\} \subset(0,1)$ and $\{r_{n}\}\subset (0,+\infty)$, where $n \geq0$ and $i = 1,2,\ldots$ . Let $\{x_{n}\}$ be generated by the following iterative algorithm:

$$ \left \{ \textstyle\begin{array}{@{}l} x_{0} \in C,\\ y_{n} = \beta_{n}x_{n}+(1-\beta_{n})\sum_{i = 1}^{\infty}b_{n,i}[(1-\alpha _{n,i})J^{A}_{r_{n}}+\alpha_{n,i}S_{i}]Q_{C}x_{n},\\ u_{n}= (1-\delta_{n})y_{n} + \delta_{n}J^{A}_{r_{n}}(\frac{u_{n}+y_{n}}{2}),\\ x_{n+1}= \gamma_{n} \eta f(x_{n})+(I-\gamma_{n}T)u_{n}, \quad n \geq0, \end{array}\displaystyle \right . $$

(A)

where $J^{A}_{r_{n}} = (I+r_{n}A)^{-1}$, for $n \geq0$. Further suppose that the following conditions are satisfied:

• (i) $\sum_{n=0}^{\infty}\sum_{i = 1}^{\infty} |\alpha_{n+1,i}-\alpha_{n,i}| <+\infty$;

• (ii) $\sum_{n=0}^{\infty} \gamma_{n} = \infty$, $\gamma_{n} \rightarrow0$, as $n \rightarrow\infty$ and $\sum_{n=1}^{\infty} |\gamma_{n} -\gamma_{n-1}|<+\infty$;

• (iii) $\sum_{n=0}^{\infty}|r_{n+1} - r_{n}| < +\infty$ and $0 < \varepsilon\leq r_{n}$, for $n \geq0$;

• (iv) $\sum_{n=1}^{\infty} |\delta_{n} -\delta_{n-1}|<+\infty$, $\sum_{n=1}^{\infty}|\beta_{n}-\beta_{n-1}|<+\infty$, $\beta_{n} \rightarrow0$, $\delta_{n} \rightarrow0$, as $n \rightarrow \infty$;

• (v) $\sum_{n=0}^{\infty}\sum_{i = 1}^{\infty} |b_{n+1,i}-b_{n,i}|<+\infty$ and $\sum_{i = 1}^{\infty} b_{n,i} = 1$ for $n \geq0$.

Then the three sequences $\{y_{n}\}$, $\{u_{n}\}$, and $\{x_{n}\}$ converge strongly to the unique element $q_{0} \in\Omega$ which satisfies the following variational inequality: for $\forall y \in\Omega$,

$$ \bigl\langle (T - \eta f)q_{0}, J(q_{0} - y)\bigr\rangle \leq0. $$

(2.3)

Moreover, the ergodic convergence is obtained in the sense that

$$z_{n} : = \frac{1}{\sum_{k=1}^{n} a_{k}}\sum_{k=1}^{n} a_{k} x_{k}, \quad n \geq1, $$

converges strongly to the above $q_{0}$, under the assumption that $\{a_{n}\}$ is a sequence of positive numbers such that $\sum_{k=1}^{n} a_{k} \rightarrow\infty$, as $n \rightarrow \infty$.

To prove Theorem 2.2, we need the following lemmas.

Lemma 2.3

In Theorem  2.2, set $W_{n,i} = (1-\alpha_{n,i})J_{r_{n}}^{A} + \alpha_{n,i}S_{i}$. Then $W_{n,i}: C \rightarrow C$ is non-expansive for $n \geq0$ and $i = 1,2,\ldots$ . Moreover, $\Omega= \bigcap_{i = 1}^{\infty}\operatorname{Fix}(W_{n,i})$.

Proof

It can be easily obtained that $W_{n,i}: C \rightarrow C$ is non-expansive since both $J^{A}_{r_{n}}$ and $S_{i}$ are non-expansive, for $n \geq0$ and $i =1,2,\ldots$ .

The fact that $\Omega\subset\bigcap_{i = 1}^{\infty}\operatorname{Fix}(W_{n,i})$ is trivial. We are left to show that $\bigcap_{i = 1}^{\infty}\operatorname{Fix}(W_{n,i})\subset\Omega$.

For $p \in\bigcap_{i = 1}^{\infty}\operatorname{Fix}(W_{n,i})$, then $p = (1-\alpha_{n,i})J_{r_{n}}^{A} p+\alpha_{n,i}S_{i}p$. For $\forall q \in \Omega\subset\bigcap_{i = 1}^{\infty}\operatorname{Fix}(W_{n,i})$, we have

$$\begin{aligned} \|p - q\|&= \bigl\| (1-\alpha_{n,i}) \bigl(J_{r_{n}}^{A} p-J_{r_{n}}^{A} q\bigr)+\alpha_{n,i}(S_{i}p-S_{i}q) \bigr\| \\ & \leq(1-\alpha_{n,i})\bigl\| J_{r_{n}}^{A} p- q\bigr\| + \alpha_{n,i}\|S_{i}p-q\| \\ & \leq(1-\alpha_{n,i})\bigl\| J_{r_{n}}^{A}p- q\bigr\| + \alpha_{n,i}\|p-q\|, \end{aligned}$$

which implies that $\|J_{r_{n}}^{A} p- q\|=\|p-q\|$. Similarly, $\|p-q\|=\|S_{i}p-q\|$. Since E is uniformly convex, we have $p - q = J_{r_{n}}^{A} p - q= S_{i} p - q$, which implies that $p\in \bigcap_{i=1}^{\infty} \operatorname{Fix}(S_{i})\cap A^{-1}0$.

This completes the proof. □

Lemma 2.4

In Theorem  2.2, $\{z_{n}\}$ is well-defined.

Proof

In fact, it suffices to show that $\{u_{n}\}$ is well-defined.

For $t\in(0,1)$, define $U_{t}: C \rightarrow C$ by $U_{t} x: = (1-t)y + tU(\frac{x+y}{2})$, where $U: C \rightarrow C$ is non-expansive for $x,y\in C$. Then for $\forall x,y \in C$,

$$\|U_{t} x - U_{t}z\|\leq t\biggl\| \frac{x+y}{2}- \frac{y+z}{2}\biggr\| \leq\frac{t}{2} \|x-z\|. $$

Thus $U_{t}$ is a contraction, which ensures from Lemma 1.1 that there exists $x_{t}\in C$ such that $U_{t} x_{t} = x_{t}$. That is, $x_{t} = (1-t)y + tU(\frac{y+x_{t}}{2})$.

Note that $J_{r_{n}}^{A}$ is non-expansive, then $\{u_{n}\}$ is well-defined, and then $\{x_{n}\}$ and $\{z_{n}\}$ are all well-defined.

This completes the proof. □

Lemma 2.5

The variational inequality (2.3) in Theorem  2.2 has a unique solution in Ω.

Proof

Using Lemmas 1.7, 1.8 and 2.3, we know that there exists $v_{t}$ such that $v_{t} = t\eta f(v_{t})+(I-tT)\sum_{i = 1 }^{\infty}b_{n,i}W_{n,i}Q_{C}v_{t}$, for $t \in(0,1)$, where $W_{n,i}$ is the same as that in Lemma 2.3, for $i = 1,2,\ldots$ . Moreover, $v_{t} \rightarrow q_{0} \in\Omega$, as $t \rightarrow0$, which is the unique solution of the variational inequality (2.3).

This completes the proof. □

Proof of Theorem 2.2

Step 1. $\{u_{n}\}$, $\{y_{n}\}$, and $\{x_{n}\}$ are all bounded.

For $\forall p \in\Omega$, noticing Lemma 2.3, we see that for $n \geq0$,

$$ \|y_{n} - p\|\leq\|x_{n}-p\|. $$

(2.4)

Also,

$$\begin{aligned} \|u_{n} - p\|\leq(1-\delta_{n})\|y_{n}-p\|+ \delta_{n}\biggl\| \frac {y_{n}+u_{n}}{2}-p\biggr\| \leq \biggl(1-\frac{\delta_{n}}{2}\biggr)\|y_{n}-p\|+\frac{\delta_{n}}{2} \|u_{n}-p\|, \end{aligned}$$

which implies that

$$ \|u_{n} - p\| \leq\|y_{n}-p\|\leq\|x_{n} - p\|. $$

(2.5)

Using Lemma 1.6 and (2.5), we have, for $n \geq0$,

$$ \begin{aligned}[b] \|x_{n+1} - p\| &\leq \gamma_{n}\eta k \|x_{n}- p\| + \gamma _{n}\bigl\| \eta f(p) - Tp\bigr\| + (1- \gamma_{n}\overline{\gamma})\|u_{n} - p\| \\ & \leq\bigl[1-\gamma_{n} (\overline{\gamma}- k \eta)\bigr] \|x_{n} -p\| + \gamma_{n} \bigl\| \eta f(p) - Tp\bigr\| . \end{aligned} $$

(2.6)

By using the inductive method, we can easily get the following result from (2.6):

$$\|x_{n+1}-p\| \leq \max\biggl\{ \|x_{0} - p\|, \frac{\|\eta f(p) -Tp\|}{\overline{\gamma}-k \eta}\biggr\} . $$

Therefore, $\{x_{n}\}$ is bounded. Then both $\{y_{n}\}$ and $\{u_{n}\}$ are bounded in view of (2.5).

Moreover, we can easily know that $\{J_{r_{n}}^{A}Q_{C}x_{n}\}$, $\{S_{i}Q_{C}x_{n} \}$, $\{J_{r_{n}}^{A}(\frac{u_{n}+y_{n}}{2})\}$, $\{Q_{C}x_{n}\}$, $\{f(x_{n})\}$, and $\{W_{n,i}Q_{C}x_{n}\}$ are all bounded, for $n \geq0$ and $i = 1,2,\ldots$ .

Set

$$\begin{aligned} M' ={}& \sup\biggl\{ \|u_{n}\|, \|x_{n}\|, \|y_{n}\|, \|S_{i}Q_{C}x_{n} \|, \biggl\| J_{r_{n}}^{A}\biggl(\frac{u_{n}+y_{n}}{2}\biggr)\biggr\| , \bigl\| J_{r_{n}}^{A}Q_{C}x_{n}\bigr\| , \|Q_{C}x_{n}\|, \bigl\| f(x_{n})\bigr\| ,\\ &{} \|W_{n,i}Q_{C}x_{n}\|: n \geq0, i\geq1\biggr\} . \end{aligned}$$

Then $M'$ is a positive constant.

Step 2. $\lim_{n \rightarrow\infty} \|x_{n+1} - x_{n} \| = 0$.

In fact, using Lemma 2.1, we have, for $n \geq0$,

$$\begin{aligned} &\biggl\| J_{r_{n+1}}^{A}\biggl(\frac {u_{n+1}+y_{n+1}}{2} \biggr)-J_{r_{n}}^{A}\biggl(\frac{u_{n}+y_{n}}{2}\biggr)\biggr\| \\ &\quad \leq \frac{\|u_{n+1}-u_{n}\|}{2}+\frac{\|y_{n+1}-y_{n}\|}{2}+2\frac {|r_{n}-r_{n+1}|}{\varepsilon}M'. \end{aligned}$$

(2.7)

In view of (2.7) and Lemma 2.1, we have, for $n \geq0$,

$$\begin{aligned} &\|u_{n+1}-u_{n}\| \\ &\quad \leq(1- \delta_{n+1})\|y_{n+1}-y_{n}\|+| \delta_{n+1}-\delta_{n}|\|y_{n}\|\\ &\qquad{}+ \delta_{n+1} \biggl\| J_{r_{n+1}}^{A}\biggl(\frac{u_{n+1}+y_{n+1}}{2} \biggr)-J_{r_{n}}^{A}\biggl(\frac {u_{n}+y_{n}}{2}\biggr)\biggr\| +|\delta_{n+1}-\delta_{n}|\biggl\| J_{r_{n}}^{A} \biggl(\frac{u_{n}+y_{n}}{2}\biggr)\biggr\| \\ &\quad \leq (1-\delta_{n+1})\|y_{n+1}-y_{n}\|+2| \delta_{n+1}-\delta_{n}|M' \\ &\qquad{}+\frac{\delta_{n+1}}{2} \bigl(\|u_{n+1} - u_{n}\|+ \|y_{n+1} - y_{n}\|\bigr)+\frac{2}{\varepsilon}\delta_{n+1} |r_{n}-r_{n+1}|M', \end{aligned}$$

which implies that

$$\begin{aligned} \|u_{n+1}-u_{n}\| &\leq \|y_{n+1}-y_{n} \|+ \frac{4|\delta_{n+1}-\delta_{n}|M'}{2-\delta_{n+1}} +\frac{4\delta_{n+1} |r_{n}-r_{n+1}|M'}{(2-\delta_{n+1})\varepsilon} \\ & \leq\|y_{n+1}-y_{n}\|+ 4|\delta_{n+1}- \delta_{n}|M'+\frac{4M'}{\varepsilon}|r_{n+1}-r_{n}|. \end{aligned}$$

(2.8)

Now, in view of Lemma 2.1, computing the following:

$$\begin{aligned} &\|W_{n+1,i}Q_{C}x_{n+1}-W_{n,i}Q_{C}x_{n} \| \\ &\quad=\bigl\| \bigl[(1-\alpha_{n+1,i})J^{A}_{r_{n+1}}+ \alpha_{n+1,i}S_{i}\bigr]Q_{C}x_{n+1} - \bigl[(1-\alpha_{n,i})J^{A}_{r_{n}}+ \alpha_{n,i}S_{i}\bigr]Q_{C}x_{n}\bigr\| \\ &\quad \leq (1-\alpha_{n+1,i})\bigl\| J^{A}_{r_{n+1}}Q_{C}x_{n+1}-J^{A}_{r_{n}}Q_{C}x_{n} \bigr\| \\ &\qquad{}+\alpha_{n+1,i}\|S_{i}Q_{C}x_{n+1}-S_{i}Q_{C}x_{n} \| \\ &\qquad{}+|\alpha_{n+1,i}-\alpha_{n,i}|\bigl( \|S_{i}Q_{C}x_{n}\|+\bigl\| J^{A}_{r_{n}}Q_{C}x_{n} \bigr\| \bigr) \\ &\quad\leq (1-\alpha_{n+1,i})\|x_{n+1}-x_{n} \|+2M'\frac{|r_{n+1}-r_{n}|}{\varepsilon} \\ &\qquad{} + \alpha_{n+1,i}\|x_{n+1}-x_{n}\|+ 2M'|\alpha_{n+1,i}-\alpha_{n,i}| \\ &\quad = \|x_{n+1}-x_{n}\|+2M' \frac{|r_{n+1}-r_{n}|}{\varepsilon} + 2M'|\alpha_{n+1,i}-\alpha_{n,i}|. \end{aligned}$$

(2.9)

Using (2.9), we have, for $n\geq0$,

$$\begin{aligned} &\Biggl\| \sum_{i = 1}^{\infty}b_{n+1,i}W_{n+1,i}Q_{C}x_{n+1}- \sum_{i = 1}^{\infty }b_{n,i}W_{n,i}Q_{C}x_{n} \Biggr\| \\ &\quad \leq\Biggl\| \sum_{i = 1}^{\infty}b_{n+1,i}W_{n+1,i}Q_{C}x_{n+1}- \sum_{i = 1}^{\infty}b_{n+1,i}W_{n,i}Q_{C}x_{n} \Biggr\| \\ &\qquad{} + \Biggl\| \sum_{i = 1}^{\infty}b_{n+1,i}W_{n,i}Q_{C}x_{n}- \sum_{i = 1}^{\infty }b_{n,i}W_{n,i}Q_{C}x_{n} \Biggr\| \\ &\quad\leq\sum_{i = 1}^{\infty}b_{n+1,i}\| W_{n+1,i}Q_{C}x_{n+1}-W_{n,i}Q_{C}x_{n} \| \\ &\qquad{} + \sum_{i = 1}^{\infty}|b_{n+1,i}-b_{n,i}| \|W_{n,i}Q_{C}x_{n}\| \\ &\quad \leq\|x_{n+1}-x_{n}\|+2M' \frac{|r_{n+1}-r_{n}|}{\varepsilon} + 2M'\sum_{i = 1}^{\infty}| \alpha_{n+1,i}-\alpha_{n,i}| \\ &\qquad{}+M'\sum_{i = 1}^{\infty}|b_{n+1,i}-b_{n,i}|. \end{aligned}$$

(2.10)

Using (2.10), we know that

$$\begin{aligned} & \|y_{n+1}-y_{n}\| \\ &\quad\leq\beta_{n+1}\|x_{n+1}-x_{n}\|+|\beta _{n+1}-\beta_{n}|\|x_{n}\| \\ &\qquad{} + (1-\beta_{n+1}) \Biggl\| \sum_{i = 1}^{\infty }b_{n+1,i}W_{n+1,i}Q_{C}x_{n+1}- \sum_{i = 1}^{\infty }b_{n,i}W_{n,i}Q_{C}x_{n} \Biggr\| \\ &\qquad{} + |\beta_{n+1}-\beta_{n}|\sum _{i = 1}^{\infty}b_{n,i}\|W_{n,i}Q_{C}x_{n} \| \\ &\quad\leq \|x_{n+1}-x_{n}\|+2|\beta_{n+1}- \beta_{n}|M'+2M'\frac {|r_{n+1}-r_{n}|}{\varepsilon} \\ &\qquad{}+2M'\sum_{i = 1}^{\infty}| \alpha_{n+1,i}-\alpha_{n,i}| +M'\sum _{i = 1}^{\infty}|b_{n+1,i}-b_{n,i}|. \end{aligned}$$

(2.11)

Thus in view of (2.8) and (2.11), we have, for $n \geq0$,

$$\begin{aligned} &\|x_{n+2}-x_{n+1}\| \\ &\quad \leq\gamma_{n+1} \eta k\|x_{n+1}-x_{n}\|+ (1+ \eta)|\gamma_{n+1}-\gamma_{n}|M' \\ &\qquad{}+(1-\gamma_{n+1}\overline{\gamma})\Biggl[\|x_{n} - x_{n+1}\|+2M'|\beta_{n+1}-\beta_{n}|+2M' \frac{|r_{n+1}-r_{n}|}{\varepsilon} \\ &\qquad{}+M'\sum_{i = 1}^{\infty}|b_{n+1,i}-b_{n,i}|+2M' \sum_{i=1}^{\infty}|\alpha_{n+1,i}- \alpha_{n,i}| +4M'|\delta_{n+1}- \delta_{n}| +\frac{4M'}{\varepsilon}|r_{n+1}-r_{n}|\Biggr] \\ &\quad = \bigl[1-\gamma_{n+1}(\overline{\gamma}- \eta k)\bigr] \|x_{n+1}-x_{n}\|+(1+\eta) M'| \gamma_{n+1}-\gamma_{n}| \\ &\qquad{}+(1-\gamma_{n+1}\overline{\gamma})\Biggl[2M'\sum _{i = 1}^{\infty}|\alpha_{n+1,i}- \alpha_{n,i}|+ 2M'|\beta_{n+1}- \beta_{n}| \\ &\qquad{}+4M'|\delta_{n+1}-\delta_{n}|+M' \sum_{i = 1}^{\infty}|b_{n+1,i}-b_{n,i}|+ \frac{6M'}{\varepsilon}|r_{n+1}-r_{n}|\Biggr]. \end{aligned}$$

(2.12)

Using Lemma 1.4, we have from (2.12) $\lim_{n \rightarrow \infty}\|x_{n+1}-x_{n}\| = 0$. Since $\gamma_{n} \rightarrow0$, we have $x_{n+1}-u_{n} = \gamma_{n}(\eta f(x_{n})-Tu_{n})\rightarrow0$, as $n \rightarrow\infty$, which implies that $x_{n} -u_{n} \rightarrow0$, as $n \rightarrow\infty$.

Since $\delta_{n} \rightarrow0$, we have $u_{n} -y_{n} = \delta_{n}[J_{r_{n}}^{A}(\frac{u_{n}+y_{n}}{2})-y_{n}]\rightarrow0$, as $n \rightarrow\infty$. Therefore, $x_{n} - y_{n} \rightarrow0$, as $n \rightarrow\infty$.

Then

$$\begin{aligned} &\Biggl\| y_{n}- \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}Q_{C}y_{n} \Biggr\| \\ & \quad\leq\Biggl\| y_{n}- \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}Q_{C}x_{n} \Biggr\| +\Biggl\| \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}(Q_{C}x_{n}-Q_{C}y_{n}) \Biggr\| \\ & \quad\leq\beta_{n} \Biggl\| x_{n}- \sum _{i = 1}^{\infty}b_{n,i}W_{n,i}Q_{C}x_{n} \Biggr\| +\|x_{n}-y_{n}\|\rightarrow0, \end{aligned}$$

since $\beta_{n} \rightarrow0$. Moreover, $x_{n+1}- \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}Q_{C}y_{n} \rightarrow0$, as $n \rightarrow \infty$.

Step 3. $\limsup_{n\rightarrow+\infty}\langle\eta f(q_{0})-Tq_{0}, J(x_{n+1}-q_{0})\rangle\leq0$.

From Lemma 2.5, we know that $v_{t} = t\eta f(v_{t})+(I-tT)\sum_{i = 1}^{\infty}b_{n,i}W_{n,i}Q_{C}v_{t}$ for $t \in(0,1)$. Moreover, $v_{t} \rightarrow q_{0} \in\Omega$, as $t \rightarrow0$. $q_{0}$ is the unique solution of the variational inequality (2.3).

Since $\|v_{t}\|\leq\|v_{t} - q_{0} \|+\|q_{0}\|$, we see that $\{v_{t}\}$ is bounded, as $t \rightarrow0$. Using Lemma 1.3, we have

$$\begin{aligned} &\|v_{t} - y_{n}\|^{2} \\ &\quad= \Biggl\| v_{t} - \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}Q_{C}y_{n}+ \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}Q_{C}y_{n} - y_{n}\Biggr\| ^{2} \\ &\quad \leq\Biggl\| v_{t} - \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}Q_{C}y_{n} \Biggr\| ^{2} + 2\Biggl\langle \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}Q_{C}y_{n} - y_{n}, J(v_{t} - y_{n}) \Biggr\rangle \\ &\quad = \Biggl\| t\eta f(v_{t}) + (I-tT)\sum_{i = 1}^{\infty}b_{n,i}W_{n,i}Q_{C}v_{t} - \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}Q_{C}y_{n} \Biggr\| ^{2} \\ &\qquad{}+ 2\Biggl\langle \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}Q_{C}y_{n} - y_{n}, J(v_{t} - y_{n}) \Biggr\rangle \\ &\quad\leq\Biggl\| \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}Q_{C}v_{t} - \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}Q_{C}y_{n} \Biggr\| ^{2} \\ &\qquad{}+ 2t\Biggl\langle \eta f(v_{t}) - T\sum _{i = 1}^{\infty}b_{n,i}W_{n,i}Q_{C}v_{t}, J\Biggl(v_{t} - \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}Q_{C}y_{n} \Biggr) \Biggr\rangle \\ &\qquad{}+ 2\Biggl\langle \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}Q_{C}y_{n}- y_{n}, J(v_{t} - y_{n}) \Biggr\rangle \\ & \quad\leq\|v_{t} - y_{n}\|^{2} + 2t\Biggl\langle \eta f(v_{t}) - T\sum_{i = 1}^{\infty}b_{n,i}W_{n,i}Q_{C}v_{t}, J\Biggl(v_{t} - \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}Q_{C}y_{n} \Biggr) \Biggr\rangle \\ &\qquad{} + 2 \Biggl\| \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}Q_{C}y_{n} - y_{n}\Biggr\| \|v_{t} - y_{n}\|, \end{aligned}$$

which implies that

$$\begin{aligned} & t\Biggl\langle T\sum_{i = 1}^{\infty}b_{n,i}W_{n,i}Q_{C}v_{t}- \eta f(v_{t}), J\Biggl(v_{t} - \sum _{i = 1}^{\infty}b_{n,i}W_{n,i}Q_{C}y_{n} \Biggr) \Biggr\rangle \\ &\quad \leq\Biggl\| \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}Q_{C}y_{n} - y_{n}\Biggr\| \|v_{t}-y_{n}\|. \end{aligned}$$

So, $\lim_{t \rightarrow0}\limsup_{n\rightarrow+\infty}\langle T\sum_{i = 1}^{\infty}b_{n,i}W_{n,i}Q_{C}v_{t}-\eta f(v_{t}), J(v_{t} - \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}Q_{C}y_{n}) \rangle\leq0$ in view of Step 2.

Since $v_{t} \rightarrow q_{0}$, we have $\sum_{i = 1}^{\infty}b_{n,i}W_{n,i}Q_{C}v_{t} \rightarrow\sum_{i = 1}^{\infty}b_{n,i}W_{n,i}Q_{C}q_{0} = Q_{C}q_{0}=q_{0}$, as $t \rightarrow 0$. Noticing the fact that

$$\begin{aligned} & \Biggl\langle Tq_{0}-\eta f(q_{0}), J \Biggl(q_{0} - \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}Q_{C}y_{n} \Biggr)\Biggr\rangle \\ &\quad= \Biggl\langle Tq_{0}-\eta f(q_{0}), J \Biggl(q_{0} - \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}Q_{C}y_{n} \Biggr) - J\Biggl(v_{t} -\sum_{i = 1}^{\infty}b_{n,i}W_{n,i}Q_{C}y_{n} \Biggr)\Biggr\rangle \\ &\qquad{}+ \Biggl\langle Tq_{0}-\eta f(q_{0}), J \Biggl(v_{t} -\sum_{i = 1}^{\infty}b_{n,i}W_{n,i}Q_{C}y_{n} \Biggr)\Biggr\rangle \\ &\quad= \Biggl\langle Tq_{0}-\eta f(q_{0}), J \Biggl(q_{0} - \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}Q_{C}y_{n} \Biggr) - J\Biggl(v_{t} -\sum_{i = 1}^{\infty}b_{n,i}W_{n,i}Q_{C}y_{n} \Biggr)\Biggr\rangle \\ &\qquad{} + \Biggl\langle Tq_{0}-\eta f(q_{0})-T\sum _{i = 1}^{\infty}b_{n,i}W_{n,i}Q_{C}v_{t}+ \eta f(v_{t}), J\Biggl(v_{t} -\sum _{i = 1}^{\infty}b_{n,i}W_{n,i}Q_{C}y_{n} \Biggr)\Biggr\rangle \\ &\qquad{} + \Biggl\langle T\sum_{i = 1}^{\infty}b_{n,i}W_{n,i}Q_{C}v_{t}- \eta f(v_{t}), J\Biggl(v_{t} -\sum _{i = 1}^{\infty}b_{n,i}W_{n,i}Q_{C}y_{n} \Biggr)\Biggr\rangle , \end{aligned}$$

we have $\limsup_{n\rightarrow+\infty}\langle Tq_{0}-\eta f(q_{0}), J(q_{0} - \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}Q_{C}y_{n})\rangle\leq0$.

Since $\langle Tq_{0}-\eta f(q_{0}), J(q_{0} - x_{n+1})\rangle= \langle Tq_{0}-\eta f(q_{0}), J(q_{0} - x_{n+1})-J(q_{0} - \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}Q_{C}y_{n})\rangle+ \langle Tq_{0}-\eta f(q_{0}), J(q_{0} - \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}Q_{C}y_{n})\rangle$ and $x_{n+1}-\sum_{i = 1}^{\infty}b_{n,i}W_{n,i}Q_{C}y_{n} \rightarrow0$ in view of Step 2, we have $\limsup_{n \rightarrow \infty} \langle\eta f(q_{0})-Tq_{0}, J(x_{n+1}-q_{0}) \rangle\leq0$.

Step 4. $x_{n} \rightarrow q_{0}$, as $n \rightarrow+\infty$, where $q_{0} \in\Omega$ is the same as that in Step 3.

Since

$$\|u_{n} - q_{0}\| \leq (1-\delta_{n}) \|y_{n}-q_{0}\|+\delta_{n}\biggl\| \frac{y_{n}+u_{n}}{2}-q_{0} \biggr\| , $$

we have

$$\|u_{n} - q_{0}\| \leq\|y_{n}-q_{0}\| \leq\beta_{n} \|x_{n}-q_{0}\|+(1- \beta_{n})\|x_{n}-q_{0}\|=\|x_{n}-q_{0} \|. $$

Using Lemma 1.3, we have, for $n \geq0$,

$$\begin{aligned} &\|x_{n+1} - q_{0}\|^{2} \\ &\quad = \bigl\| \gamma_{n}\bigl(\eta f(x_{n})-Tq_{0} \bigr)+(I- \gamma_{n} T) (u_{n} - q_{0}) \bigr\| ^{2} \\ &\quad \leq(1-\gamma_{n} \overline{\gamma})^{2} \|u_{n}-q_{0}\|^{2}+2\gamma_{n} \bigl\langle \eta f(x_{n})-Tq_{0}, J(x_{n+1}-q_{0}) \bigr\rangle \\ &\quad \leq (1-\gamma_{n} \overline{\gamma})^{2} \|x_{n}-q_{0}\|^{2} \\ &\qquad{}+2\gamma_{n}\eta\bigl\langle f(x_{n})-f(q_{0}), J(x_{n+1}-q_{0})-J(x_{n}-q_{0})\bigr\rangle \\ &\qquad{} +2\gamma_{n} \eta\bigl\langle f(x_{n})- f(q_{0}), J(x_{n}-q_{0})\bigr\rangle +2 \gamma_{n}\bigl\langle \eta f(q_{0})-Tq_{0}, J(x_{n+1}-q_{0})\bigr\rangle \\ &\quad \leq\bigl[1-\gamma_{n} (\overline{\gamma}-2\eta k )\bigr] \|x_{n}-q_{0}\|^{2} \\ &\qquad{} +2\gamma_{n} \bigl[\bigl\langle \eta f(q_{0})-Tq_{0}, J(x_{n+1}-q_{0})\bigr\rangle +\eta \|x_{n}-q_{0} \|\|x_{n+1}-x_{n}\|\bigr]. \end{aligned}$$

(2.13)

Let $\delta_{n}^{(1)} = \gamma_{n}(\overline{\gamma}-2\eta k)$, $\delta_{n}^{(2)} = 2\gamma_{n}[\langle\eta f(q_{0})-Tq_{0}, J(x_{n+1}-q_{0})\rangle+\eta\|x_{n}-q_{0}\|\|x_{n+1}-x_{n}\|]$. Then (2.13) can be simplified as $\|x_{n+1}-p_{0}\|^{2} \leq (1-\delta_{n}^{(1)})\|x_{n}-q_{0}\|^{2} + \delta_{n}^{(2)}$.

Using the assumption (ii), the result of Steps 2 and 3 and Lemma 1.4, we know that $x_{n} \rightarrow q_{0}$, as $n \rightarrow +\infty$.

Combining with the result of Step 2, $y_{n} \rightarrow q_{0}$ and $u_{n} \rightarrow q_{0}$, as $n \rightarrow\infty$.

Step 5. $z_{n} \rightarrow q_{0}$, as $n \rightarrow\infty$.

Since $\sum_{k=1}^{n} a_{k} \rightarrow\infty$ and $\|x_{n} - q_{0}\| \rightarrow0$, as $n \rightarrow\infty$, we have, for $\forall \varepsilon> 0$, there exists $N^{*}$, such that, for all $n \geq N^{*}$, $\frac{1}{\sum_{k=1}^{n} a_{k}}\sum_{k=1}^{N^{*}}a_{k}M''< \frac{\varepsilon}{2}$ and $\|x_{n} - q_{0}\|< \frac{\varepsilon}{2}$, where $M''=\max\{\|x_{k}-q_{0}\|: k = 1,2,\ldots,N^{*}\}$. Then, for all $n > N^{*}$,

$$\begin{aligned} \|z_{n} -q_{0}\| &= \Biggl\| \frac{1}{\sum_{k=1}^{n} a_{k}}\sum _{k=1}^{n} a_{k} (x_{k}-q_{0}) \Biggr\| \\ &= \Biggl\| \frac{1}{\sum_{k=1}^{n} a_{k}}\Biggl[\sum_{k=1}^{N^{*}} a_{k}(x_{k}-q_{0})+\sum _{k=N^{*}+1}^{n} a_{k} (x_{k}-q_{0}) \Biggr]\Biggr\| \\ & \leq\frac{1}{\sum_{k=1}^{n} a_{k}}\sum_{k=1}^{N^{*}} a_{k}\|x_{k} -q_{0}\|+\frac{1}{\sum_{k=1}^{n} a_{k}}\sum _{k=N^{*}+1}^{n} a_{k} \|x_{k} -q_{0}\| \\ & \leq\frac{1}{\sum_{k=1}^{n} a_{k}}\sum_{k=1}^{N^{*}} a_{k}M''+\frac{\sum_{k=N^{*}+1}^{n} a_{k}}{\sum_{k=1}^{n} a_{k}}\frac{\varepsilon}{2} \\ & < \frac{\varepsilon}{2}+\frac{\varepsilon}{2} = \varepsilon, \end{aligned}$$

which implies that $z_{n} \rightarrow q_{0}$, as $n \rightarrow \infty$.

This completes the proof. □

Remark 2.6

The assumptions imposed on the real number sequences in Theorem 2.2 are reasonable if we take $\alpha_{n,i} = \frac{1}{(n+1)i^{2}}$, $\gamma_{n} = \delta_{n} = \beta_{n} = \frac{1}{1+n}$, $r_{n} = \varepsilon+\frac{1}{n+1}$, and $b_{n,i} = \frac{n+1}{(n+2)^{i}}$, for $n \geq0$ and $i = 1,2,\ldots$ .

2.2 Ergodic convergence of the second iterative algorithm

Theorem 2.7

Let E, Ω, f, A, $S_{i}$, $\{r_{n}\}$, $\{\delta_{n}\}$, $\{b_{n,i}\}$, and $\{\alpha_{n,i}\}$ be the same as those in Theorem  2.2. Let C be a nonempty, closed, and convex subset of E. Suppose $\Omega\neq\emptyset$, $0 < k < \frac{1}{2}$, the duality mapping $J: E \rightarrow E^{*}$ is weakly sequentially continuous at zero and $\{\zeta_{n}\}\subset(0,1)$. Let $\{x_{n}\}$ be generated by the following iterative algorithm:

$$ \left \{ \textstyle\begin{array}{@{}l} x_{0} \in C,\\ u_{n}= (1-\delta_{n})x_{n} + \delta_{n}J^{A}_{r_{n}}(\frac{u_{n}+x_{n}}{2}),\\ x_{n+1}=[\zeta_{n} f+(1-\zeta_{n})I]\sum_{i = 1}^{\infty }b_{n,i}[(1-\alpha_{n,i})J^{A}_{r_{n}}+\alpha_{n,i}S_{i}]u_{n}, \quad n \geq0. \end{array}\displaystyle \right . $$

(B)

Further suppose that the following conditions are satisfied:

• (vi) $\sum_{n=0}^{\infty} \zeta_{n} = \infty$, $\sum_{n = 0}^{\infty}|\zeta_{n+1}-\zeta_{n}|<+\infty$, and $\zeta_{n} \rightarrow0$, as $n \rightarrow\infty$.

Then both $\{u_{n}\}$ and $\{x_{n}\}$ converge strongly to the unique element $p_{0} \in\Omega$, which satisfies the following variational inequality: for $\forall y \in\Omega$,

$$ \bigl\langle p_{0} - f(p_{0}), J(p_{0} - y) \bigr\rangle \leq0. $$

(2.14)

Moreover, the ergodic convergence is obtained in the sense that

$$z_{n} : = \frac{1}{\sum_{k=1}^{n} a_{k}}\sum_{k=1}^{n} a_{k} x_{k}, \quad n \geq1, $$

converges strongly to the above $p_{0}$ under the assumption that $\{a_{n}\}$ is a sequence of positive numbers such that $\sum_{k=1}^{n} a_{k} \rightarrow\infty$, as $n \rightarrow \infty$.

Proof

Using the same method as that in Theorem 2.2, we know that $\{u_{n}\}$ is well-defined. Let $V_{n} = \zeta_{n} f+(1-\zeta_{n})I$.

Step 1. $V_{n} : C \rightarrow C$ is a contraction.

For $\forall x,y \in C$,

$$\begin{aligned} \|V_{n} x-V_{n} y\| &= \bigl\| \zeta_{n} \bigl(f(x)-f(y)\bigr)+(1-\zeta_{n}) (x-y)\bigr\| \\ &\leq \zeta_{n} k \|x-y\|+(1-\zeta_{n})\|x-y\|= \bigl[1-(1-k)\zeta_{n}\bigr]\|x-y\|. \end{aligned}$$

(2.15)

Step 2. $\{x_{n}\}$ is bounded.

Let $W_{n,i}$ be the same as that in Lemma 2.3 and Theorem 2.2, then for $p \in\Omega= \bigcap_{i = 1}^{\infty}\operatorname{Fix}(W_{n,i})$, we can easily know that $\|u_{n}-p\|\leq\|x_{n} - p\|$. Using (2.15), we have

$$\begin{aligned} \|x_{n+1}-p\| &= \Biggl\| V_{n} \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}u_{n} - V_{n} \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}p+V_{n} \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}p - p\Biggr\| \\ &\leq\bigl[1-(1-k)\zeta_{n}\bigr]\|u_{n}-p\|+ \zeta_{n}\bigl\| f(p)-p\bigr\| \\ &\leq\bigl[1-(1-k)\zeta_{n}\bigr]\|x_{n}-p\|+ \zeta_{n}\bigl\| f(p)-p\bigr\| \\ & \leq\max\biggl\{ \| x_{0}-p\|, \frac{\|f(p)-p\|}{1-k}\biggr\} . \end{aligned}$$

Therefore, $\{x_{n}\}$ is bounded. Thus $\{J^{A}_{r_{n}}(\frac{u_{n}+x_{n}}{2})\}$, $\{u_{n}\}$, $\{W_{n,i}u_{n}\}$ and $\{f(\sum_{i = 1}^{\infty}b_{n,i}W_{n,i}u_{n})\}$ are all bounded for $n \geq0$ and $i = 1,2,\ldots$ . Let

$$\begin{aligned} M'''={}& \sup\Biggl\{ M', \|x_{n}\|, \|u_{n}\|, \biggl\| J_{r_{n}}^{A}\biggl(\frac{u_{n}+x_{n}}{2}\biggr)\biggr\| , \|W_{n,i}u_{n}\|,\\ &{} \Biggl\| f\Biggl(\sum_{i = 1}^{\infty}b_{n,i}W_{n,i}u_{n}\Biggr)\Biggr\| : n \geq0, i = 1,2, \ldots\Biggr\} , \end{aligned}$$

where $M'$ is the same as that in Theorem 2.2, then $M'''$ is a positive constant.

Step 3. $\lim_{n \rightarrow\infty} \|x_{n+1} - x_{n} \| = 0$.

In fact, using Lemma 2.1, we know that

$$\begin{aligned} &\|u_{n+1}-u_{n}\| \\ &\quad \leq (1-\delta_{n+1})\|x_{n+1}-x_{n}\|+| \delta_{n+1}-\delta_{n}|\|x_{n}\|+ \delta_{n+1} \biggl\| J_{r_{n+1}}^{A}\biggl(\frac{u_{n+1}+x_{n+1}}{2} \biggr)-J_{r_{n}}^{A}\biggl(\frac {u_{n}+x_{n}}{2}\biggr)\biggr\| \\ &\qquad{}+ |\delta_{n+1}-\delta_{n}|\biggl\| J_{r_{n}}^{A} \biggl(\frac{u_{n}+x_{n}}{2}\biggr)\biggr\| \\ &\quad \leq (1-\delta_{n+1})\|x_{n+1}-x_{n} \|+2M'''|\delta_{n+1}- \delta_{n}|+\delta _{n+1}\biggl(\frac{\|u_{n+1}-u_{n}\|}{2}+ \frac{\|x_{n+1}-x_{n}\|}{2}\biggr) \\ &\qquad{}+\frac{2M'''}{\varepsilon} |r_{n+1}-r_{n}|, \end{aligned}$$

which implies that $\|u_{n+1}-u_{n}\|\leq\|x_{n+1}-x_{n}\|+4|\delta_{n+1}-\delta _{n}|M'''+4M'''\frac{|r_{n+1}-r_{n}|}{\varepsilon}$.

Thus using (2.15) and noticing the result of (2.9), we have, for $n \geq0$,

$$\begin{aligned} &\|x_{n+2}-x_{n+1}\| \\ &\quad= \Biggl\| V_{n+1} \sum_{i = 1}^{\infty }b_{n+1,i}W_{n+1,i}u_{n+1} -V_{n} \sum_{i = 1}^{\infty }b_{n,i}W_{n,i}u_{n} \Biggr\| \\ & \quad\leq\Biggl\| V_{n+1} \sum_{i = 1}^{\infty}b_{n+1,i}W_{n+1,i}u_{n+1} - V_{n+1} \sum_{i = 1}^{\infty}b_{n+1,i}W_{n,i}u_{n} \Biggr\| \\ &\qquad{}+ \Biggl\| V_{n+1} \sum_{i = 1}^{\infty}b_{n+1,i}W_{n,i}u_{n} -V_{n} \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}u_{n} \Biggr\| \\ &\quad \leq\bigl[1-(1-k)\zeta_{n+1}\bigr]\sum _{i = 1}^{\infty}b_{n+1,i}\| W_{n+1,i}u_{n+1} - W_{n,i}u_{n}\| \\ &\qquad{}+ \Biggl\| V_{n+1} \sum_{i = 1}^{\infty}b_{n+1,i}W_{n,i}u_{n} -V_{n+1} \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}u_{n} \Biggr\| +\Biggl\| (V_{n+1}-V_{n})\sum_{i = 1}^{\infty}b_{n,i}W_{n,i}u_{n} \Biggr\| \\ &\quad\leq \bigl[1-(1-k)\zeta_{n+1}\bigr]\sum _{i = 1}^{\infty}b_{n+1,i}\biggl[ \|u_{n+1}-u_{n}\|+ \frac{2M'''}{\varepsilon}|r_{n+1}-r_{n}|+2M'''| \alpha_{n+1,i}-\alpha _{n,i}|\biggr] \\ &\qquad{}+ \Biggl\| \zeta_{n+1} f\Biggl(\sum_{i = 1}^{\infty}b_{n+1,i}W_{n,i}u_{n} \Biggr)+(1-\zeta_{n+1})\sum_{i = 1}^{\infty}b_{n+1,i}W_{n,i}u_{n} \\ &\qquad{}-\zeta_{n+1} f\Biggl(\sum_{i = 1}^{\infty}b_{n,i}W_{n,i}u_{n} \Biggr)- (1-\zeta_{n+1})\sum_{i = 1}^{\infty}b_{n,i}W_{n,i}u_{n} \Biggr\| \\ &\qquad{}+ |\zeta_{n+1}-\zeta_{n}|\Biggl\| f\Biggl(\sum _{i = 1}^{\infty}b_{n,i}W_{n,i}u_{n} \Biggr)\Biggr\| +|\zeta_{n+1}-\zeta_{n}|\Biggl\| \sum _{i = 1}^{\infty}b_{n,i}W_{n,i}u_{n} \Biggr\| \\ &\quad \leq\bigl[1-(1-k)\zeta_{n+1}\bigr]\|x_{n+1}-x_{n} \|+2M'''\sum_{i = 1}^{\infty}| \alpha_{n+1,i}-\alpha_{n,i}|+\frac{6M'''}{\varepsilon }|r_{n+1}-r_{n}| \\ &\qquad{}+4M'''|\delta_{n+1}- \delta_{n}|+2M'''| \zeta_{n+1}-\zeta_{n}|+M''' \sum_{i =1}^{\infty}|b_{n+1,i}-b_{n,i}|. \end{aligned}$$

Then Lemma 1.4 implies that $\lim_{n \rightarrow \infty}\|x_{n+1}-x_{n}\| = 0$.

Since $\zeta_{n} \rightarrow0$, we have $x_{n+1}-\sum_{i = 1}^{\infty}b_{n,i}W_{n,i}u_{n}\rightarrow0$, as $n \rightarrow \infty$. Since $\delta_{n} \rightarrow0$, we have $x_{n} - u_{n} \rightarrow0$, which implies that $\sum_{i = 1}^{\infty}b_{n,i}W_{n,i}u_{n} - u_{n} \rightarrow0$, as $n \rightarrow\infty$.

Step 4. $\limsup_{n\rightarrow+\infty}\langle f(p_{0})-p_{0}, J(x_{n+1}-p_{0})\rangle\leq0$.

Noticing Lemmas 1.7 and 2.3, we know that there exists $z_{t}$ such that $z_{t} = t f(z_{t})+(1-t)\sum_{i = 1}^{\infty}b_{n,i}W_{n,i}z_{t}$ for $t \in(0,1)$. Moreover, $z_{t} \rightarrow p_{0} \in\Omega$, as $t \rightarrow0$. $p_{0}$ is the unique solution of the variational inequality (2.14).

Since $\|z_{t}\|\leq\|z_{t} - p_{0} \|+\|p_{0}\|$, $\{z_{t}\}$ is bounded, as $t \rightarrow0$. Using Lemma 1.3, we have

$$\begin{aligned} &\|z_{t} - u_{n}\|^{2} \\ &\quad= \Biggl\| z_{t} - \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}u_{n} + \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}u_{n} - u_{n}\Biggr\| ^{2} \\ &\quad \leq\Biggl\| z_{t} - \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}u_{n} \Biggr\| ^{2} + 2\Biggl\langle \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}u_{n} - u_{n}, J(z_{t} - u_{n}) \Biggr\rangle \\ &\quad = \Biggl\| tf(z_{t})+(1-t)\sum_{i = 1}^{\infty}b_{n,i}W_{n,i}z_{t} - \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}u_{n} \Biggr\| ^{2} \\ &\qquad{}+ 2\Biggl\langle \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}u_{n} - u_{n}, J(z_{t} - u_{n}) \Biggr\rangle \\ & \quad\leq\|z_{t}-u_{n}\|^{2}+2t\Biggl\langle f(z_{t})-\sum_{i = 1}^{\infty}b_{n,i}W_{n,i}z_{t}, J\Biggl(z_{t} - \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}u_{n} \Biggr) \Biggr\rangle \\ &\qquad{}+2\Biggl\langle \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}u_{n} - u_{n}, J(z_{t} - u_{n}) \Biggr\rangle \\ & \quad\leq\|z_{t} - u_{n}\|^{2} + 2t\Biggl\langle f(z_{t}) - \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}z_{t}, J\Biggl(z_{t} - \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}u_{n} \Biggr) \Biggr\rangle \\ &\qquad{} + 2 \Biggl\| \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}u_{n} - u_{n}\Biggr\| \|z_{t} - u_{n}\| , \end{aligned}$$

which implies that

$$t\Biggl\langle \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}z_{t}- f(z_{t}), J\Biggl(z_{t} - \sum _{i = 1}^{\infty}b_{n,i}W_{n,i}u_{n} \Biggr) \Biggr\rangle \leq\Biggl\| \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}u_{n} - u_{n}\Biggr\| \|z_{t} - u_{n}\|. $$

So, $\lim_{t \rightarrow0}\limsup_{n\rightarrow+\infty}\langle \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}z_{t} - f(z_{t}), J(z_{t} - \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}u_{n}) \rangle\leq0$ in view of Step 3.

Since $z_{t} \rightarrow p_{0}$, we have $\sum_{i = 1}^{\infty}b_{n,i}W_{n,i}z_{t} \rightarrow\sum_{i = 1}^{\infty}b_{n,i}W_{n,i}p_{0} = p_{0}$, as $t \rightarrow0$. Noticing the fact that

$$\begin{aligned} & \Biggl\langle p_{0}-f(p_{0}), J\Biggl(p_{0} - \sum_{i = 1}^{\infty }b_{n,i}W_{n,i}u_{n} \Biggr)\Biggr\rangle \\ &\quad= \Biggl\langle p_{0}-f(p_{0}), J \Biggl(p_{0} - \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}u_{n} \Biggr) - J\Biggl(z_{t} -\sum_{i = 1}^{\infty}b_{n,i}W_{n,i}u_{n} \Biggr)\Biggr\rangle \\ &\qquad{}+ \Biggl\langle p_{0}-f(p_{0}), J \Biggl(z_{t} -\sum_{i = 1}^{\infty }b_{n,i}W_{n,i}u_{n} \Biggr)\Biggr\rangle \\ &\quad= \Biggl\langle p_{0}-f(p_{0}), J \Biggl(p_{0} - \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}u_{n} \Biggr) - J\Biggl(z_{t} -\sum_{i = 1}^{\infty}b_{n,i}W_{n,i}u_{n} \Biggr)\Biggr\rangle \\ &\qquad{} + \Biggl\langle p_{0}-f(p_{0})-\sum _{i = 1}^{\infty}b_{n,i}W_{n,i}z_{t}+f(z_{t}), J\Biggl(z_{t} -\sum_{i = 1}^{\infty}b_{n,i}W_{n,i}u_{n} \Biggr)\Biggr\rangle \\ &\qquad{} + \Biggl\langle \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}z_{t}-f(z_{t}), J\Biggl(z_{t} -\sum_{i = 1}^{\infty}b_{n,i}W_{n,i}u_{n} \Biggr)\Biggr\rangle , \end{aligned}$$

we have $\limsup_{n\rightarrow+\infty}\langle p_{0}-f(p_{0}), J(p_{0} - \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}u_{n})\rangle\leq0$.

Since $\langle p_{0}-f(p_{0}), J(p_{0} - x_{n+1})\rangle= \langle p_{0}-f(p_{0}), J(p_{0} - x_{n+1})-J(p_{0} - \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}u_{n})\rangle+ \langle p_{0}- f(p_{0}), J(p_{0} - \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}u_{n})\rangle$ and $x_{n+1}-\sum_{i = 1}^{\infty}b_{n,i}W_{n,i}u_{n} \rightarrow0$ in view of Step 3, we have $\limsup_{n \rightarrow\infty} \langle f(p_{0})-p_{0}, J(x_{n+1}-p_{0}) \rangle\leq0$.

Step 5. $x_{n} \rightarrow p_{0}$, as $n \rightarrow+\infty$, where $p_{0} \in\Omega$ is the same as in Step 4.

Since

$$\|u_{n} - p_{0}\| \leq (1-\delta_{n}) \|x_{n}-p_{0}\|+\delta_{n}\|\frac{x_{n}+u_{n}}{2}-p_{0} \|, $$

we have

$$\|u_{n} - p_{0}\| \leq\|x_{n}-p_{0} \|. $$

Using Lemma 1.3, we have, for $n \geq0$,

$$\begin{aligned} &\|x_{n+1} - p_{0}\|^{2} \\ &\quad= \Biggl\| V_{n} \sum_{i = 1}^{\infty }b_{n,i}W_{n,i}u_{n} - p_{0}\Biggr\| ^{2} \\ &\quad = \Biggl\| V_{n} \sum_{i = 1}^{\infty }b_{n,i}W_{n,i}u_{n} - V_{n} \sum_{i = 1}^{\infty}b_{n,i}W_{n,i}p_{0}+V_{n}p_{0}- p_{0}\Biggr\| ^{2} \\ &\quad \leq \bigl[1-(1-k)\zeta_{n}\bigr]\|u_{n}-p_{0} \|^{2}+2\bigl\langle V_{n}p_{0}-p_{0}, J(x_{n+1}-p_{0})\bigr\rangle \\ &\quad \leq \bigl[1-(1-k)\zeta_{n}\bigr]\|x_{n}-p_{0} \|^{2}+2\zeta_{n} \bigl\langle f(p_{0})-p_{0}, J(x_{n+1}-p_{0})\bigr\rangle . \end{aligned}$$

Using Lemma 1.4, the assumptions and the result of Step 4, we know that $x_{n} \rightarrow p_{0}$, as $n \rightarrow+\infty$. Combining with the result of Step 3, $u_{n} \rightarrow p_{0}$, as $n \rightarrow\infty$.

Copy Step 5 in Theorem 2.2, $z_{n} \rightarrow p_{0}$, as $n \rightarrow\infty$.

This completes the proof. □

Remark 2.8

The assumptions imposed on the real number sequences in Theorem 2.7 are reasonable if we take $\alpha_{n,i} = \frac{1}{(n+1)i^{2}}$, $\delta_{n} = \zeta_{n} = \frac{1}{n+1}$, $b_{n,i} = \frac{n+1}{(n+2)^{i}}$, and $r_{n} = \varepsilon+\frac{1}{n+1}$, for $n \geq0$ and $i = 1,2, \ldots$ .

Remark 2.9

The four sequences $\{x_{n}\}$, $\{u_{n}\}$, $\{y_{n}\}$, and $\{z_{n}\}$ in Theorem 2.2 and the three sequences $\{u_{n}\}$, $\{x_{n}\}$, and $\{z_{n}\}$ in Theorem 2.7 are proved to be strongly convergent to the zero point of an m-accretive mapping and the fixed point of an infinite family of non-expansive mappings. The strongly convergent point is proved to be the unique solution of one kind variational inequalities.

Remark 2.10

In Theorem 2.7, $V_{n}$ can be regarded as an averaged mapping, whose definition can be seen in [16].

Remark 2.11

The discussions on Theorems 2.2 and 2.7 are undertaken in the frame of a real smooth and uniformly convex Banach space, which is more general than that in Hilbert space.

3 Examples and numerical experiments

In this section, we provide some numerical experiments to show that both algorithms (A) and (B) are effective. In our experiments, we consider the following examples.

Example 3.1

In algorithm (A), suppose $E = C = (-\infty, +\infty )$. Let $a_{i} = 1$, $\alpha_{n,i}=\frac{1}{2^{n+i}}$, $b_{n,i}=\frac {n+1}{(n+2)^{i}}$, $\gamma_{n} = \delta_{n} = \beta_{n} = \frac{1}{n+1}$, $r_{n} = \frac{n+2}{n+1}$, $\varepsilon= 1$, $f(x) = \frac{x}{14}$, $k = \frac{1}{7}$, $Ax = \frac{x}{2}$, $Tx = \frac{2x}{7}$, $\overline{\gamma} = \frac{2}{7}$, $\eta= \frac{1}{2}$, and $S_{i} x = \frac{x}{2^{i}}$. Then all of the assumptions in Theorem 2.2 are satisfied. And $\Omega= \bigcap_{i=1}^{\infty}\operatorname{Fix}(S_{i})\cap A^{-1}0 = \{0\}$.

Remark 3.1

All codes were written in Visual Basic Six. For the initial value $x_{0} = -4$, the values of $\{y_{n}\}$, $\{u_{n}\}$, $\{x_{n}\}$, and $\{z_{n}\}$ with different n are reported in Table 1.

Table 1 The values of $\pmb{\{y_{n}\}}$ , $\pmb{\{u_{n}\}}$ , $\pmb{\{x_{n}\}}$ , and $\pmb{\{ z_{n}\}}$ with initial value $\pmb{x_{0} = -4}$

Remark 3.2

In Figure 1, the abscissa denotes the iterative step and the ordinate denotes the values of $\{y_{n}\}$, $\{u_{n}\}$, $\{x_{n}\}$, and $\{z_{n}\}$ with different iterative step n.

Figure 1

The convergence of $\pmb{\{y_{n}\}}$ , $\pmb{\{u_{n}\}}$ , $\pmb{\{x_{n}\}}$ , and $\pmb{\{z_{n}\}}$ .

Remark 3.3

Table 1 and Figure 1 show that the sequences $\{y_{n}\}$, $\{u_{n}\}$, $\{x_{n}\}$, and $\{z_{n}\}$ converge to 0. Also, $\{0\} = \Omega$.

Remark 3.4

Figure 2 is a screen shot of Figure 1, whose ordinates are enlarged. Our purpose to draw Figure 2 is to let Figure 1 be clearer.

Figure 2

Screenshot of Figure 1 .

Remark 3.5

Figure 3 shows the values of ergodic sequence $\{z_{n}\}$ with the initial value $x_{0}$ being chosen arbitrarily in $[-2,10]$.

Figure 3

The convergence of $\pmb{\{z_{n}\}}$ under different initial values.

Example 3.2

In algorithm (B), suppose $E = (-\infty, +\infty)$ and $C = [-2,10]$. Let $a_{i} = 1$, $\alpha_{n,i}=\frac{1}{2^{n+i}}$, $b_{n,i}=\frac{n+1}{(n+2)^{i}}$, $\delta_{n} = \zeta_{n} = \frac{1}{n+1}$, $r_{n} = \frac{n+2}{n+1}$, $\varepsilon= 1$, $f(x) = \frac{x}{14}$, $k = \frac{1}{7}$, $Ax = \frac{x}{2}$, and $S_{i} x = \frac{x}{2^{i}}$. Then all of the assumptions in Theorem 2.7 are satisfied. Also $\Omega= \bigcap_{i=1}^{\infty}\operatorname{Fix}(S_{i})\cap A^{-1}0 = \{0\}$.

All codes were written in Visual Basic Six, the values of $\{u_{n}\}$, $\{x_{n}\}$, and $\{z_{n}\}$ with different n are reported in Table 2.

Table 2 The values of $\pmb{\{u_{n}\}}$ , $\pmb{\{x_{n}\}}$ , and $\pmb{\{z_{n}\}}$ with initial value $\pmb{x_{0} = 8}$

Remark 3.6

Table 2 and Figure 4 show that the sequences $\{u_{n}\}$, $\{x_{n}\}$, and $\{z_{n}\}$ converge to 0. Also, $\{0\} = \Omega$.

Figure 4

The convergence of $\pmb{\{u_{n}\}}$ , $\pmb{\{x_{n}\}}$ , and $\pmb{\{z_{n}\}}$ .

Remark 3.7

Figure 5 shows the values of ergodic sequence $\{z_{n}\}$ with the initial value $x_{0}$ being chosen arbitrarily in $[-2,10]$.

Figure 5

The convergence of $\pmb{\{z_{n}\}}$ for different initial values.