Some new iterative algorithms with errors for common solutions of two finite families of accretive mappings in a Banach space

Abstract

The purpose of this paper is to prove some new theorems of strong convergence to common solutions for two finite families of accretive mappings in a real uniformly smooth and uniformly convex Banach space by means of some new iterative algorithms with errors, which extend the corresponding works by some authors. As applications, the theorems of strong convergence to common fixed points of two finite families of pseudo-contractive mappings are presented.

MSC:47H05, 47H09, 47H10.

1 Introduction

Let E be a real Banach space with norm $\parallel \cdot \parallel$ and let $E^{\ast }$ denote the dual space of E. We use → and ⇀ to denote strong and weak convergence, respectively. We denote the value of $f\in E^{\ast }$ at $x\in E$ by $〈x,f〉$.

We use J to denote the normalized duality mapping from E to $2^{E^{\ast }}$, which is defined by

$$Jx:=\{f\in E^{\ast }:〈x,f〉={\parallel x\parallel }^2={\parallel f\parallel }^2\},x\in E.$$

It is well known that J is single-valued if $E^{\ast }$ is strictly convex. Moreover, $J(cx)=cJx$, for $\mathrm{\forall }x\in E$ and $c\in R^1$. We call J weakly sequentially continuous if each $\{x_n\}\subset E$ which converges weakly to x implies that $\{J{x}_n\}$ converges in the sense of weak^∗ to Jx.

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 [1] if $Q(Q(x)+t(x-Q(x)))=Q(x)$, for all $x\in E$ and $t\ge 0$.

A mapping Q of E into E is said to be a retraction [1] 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 mapping $T:C\to C$ is said to be nonexpansive if $\parallel Tx-Ty\parallel \le \parallel x-y\parallel$, for $\mathrm{\forall }x,y\in C$. We use $F(T)$ to denote the fixed point set of T, that is, $F(T):=\{x\in C:Tx=x\}$. A mapping $T:E\supset D(T)\to R(T)\subset E$ is said to be demiclosed at p if whenever $\{x_n\}$ is a sequence in $D(T)$ such that $x_n\rightharpoonup x\in D(T)$ and $T{x}_n\to p$ then $Tx=p$.

A subset C of E is said to be a sunny nonexpansive retract of E [2] if there exists a sunny nonexpansive retraction of E onto C and it is called a nonexpansive retract of E if there exists a nonexpansive retraction of E onto C. If E is reduced to a Hilbert space H, then the metric projection $P_C$ is a sunny nonexpansive retraction from H to any closed and convex subset C of H. But this is not true in a general Banach space. We note that if E is smooth and Q is a retraction of C onto $F(T)$, then Q is sunny and nonexpansive if and only if for $\mathrm{\forall }x\in C$, $z\in F(T)$, $〈Qx-x,J(Qx-z)〉\le 0$ [3].

A mapping $T:C\to C$ is called pseudo-contractive [2] if there exists $j(x-y)\in J(x-y)$ such that $〈Tx-Ty,j(x-y)〉\le {\parallel x-y\parallel }^2$ holds for all $x,y\in C$.

Interest in pseudo-contractive mappings stems mainly from their firm connection with the important class of nonlinear accretive mappings. A mapping $A:D(A)\subset E\to E$ is said to be accretive if $\parallel x_1-x_2\parallel \le \parallel x_1-x_2+r(y_1-y_2)\parallel$, for $\mathrm{\forall }{x}_i\in D(A)$, $y_i\in A{x}_i$, $i=1,2$, and $r>0$. If A is accretive, then we can define, for each $r>0$, a nonexpansive single-valued mapping $J_r^A:R(I+rA)\to D(A)$ by $J_r^A:={(I+rA)}^{-1}$, which is called the resolvent of A. We also know that for an accretive mapping A, $N(A)=F(J_r^A)$, where $N(A)=\{x\in D(A):Ax=0\}$. An accretive mapping A is said to be m-accretive if $R(I+\lambda A)=E$, for $\mathrm{\forall }\lambda >0$.

It is well known that if A is an accretive mapping, then the solutions of the problem $0\in Ax$ correspond to the equilibrium points of some evolution equations. Hence, the problem of finding a solution $x\in E$ with $0\in Ax$ has been studied by many researchers (see [4–12] and the references contained therein).

One classical method for studying the problem $0\in Ax$, where A is an m-accretive mapping, is the following so-called proximal method (cf. [4]), presented in a Hilbert space:

$$x_0\in H,x_{n+1}\approx J_{r_n}^A{x}_n,n\ge 0,$$

(1.1)

where $J_{r_n}^A:={(I+r_{n}A)}^{-1}$. It was shown that the sequence generated by (1.1) converges weakly or strongly to a zero point of A under some conditions.

On the other hand, one explicit iterative process was first introduced, in 1967, by Halpern [13] in the frame of Hilbert spaces:

$$u\in C,x_0\in C,x_{n+1}={\alpha }_{n}u+(1-{\alpha }_n)T{x}_n,n\ge 0,$$

(1.2)

where $\{{\alpha }_n\}\subset [0,1]$ and $T:C\to C$ is a nonexpansive mapping. It was proved that under some conditions, the sequence $\{x_n\}$ produced by (1.2) converges strongly to a point in $F(T)$.

In 2007, Qin and Su [6] presented the following iterative algorithm:

$$\begin{array}{r}{x}_1\in C,\\ y_n={\beta }_n{x}_n+(1-{\beta }_n)J_{r_n}^A{x}_n,\\ x_{n+1}={\alpha }_{n}u+(1-{\alpha }_n)y_n.\end{array}$$

(1.3)

They showed that $\{x_n\}$ generated by (1.3) converges strongly to a point in $N(A)$.

Motivated by iterative algorithms (1.1) and (1.2), Zegeye and Shahzad extended their discussion to the case of finite m-accretive mappings. They presented in [14] the following iterative algorithm:

$$x_0\in C,x_{n+1}={\alpha }_{n}u+(1-{\alpha }_n)S_r{x}_n,n\ge 0,$$

(1.4)

where $S_r=a_{0}I+a_1{J}_{A_1}+a_2{J}_{A_2}+\cdots +a_l{J}_{A_l}$ with $J_{A_i}={(I+A_i)}^{-1}$ and ${\sum }_{i=0}^l{a}_i=1$. If ${\bigcap }_{i=1}^{l}N(A_i)\ne \mathrm{\varnothing }$, they proved that $\{x_n\}$ generated by (1.4) converges strongly to the common point in $N(A_i)$ ($i=1,2,\dots ,l$) under some conditions.

The work in [14] was then extended to the following one presented by Hu and Liu in [15]:

$$x_0\in C,x_{n+1}={\alpha }_{n}u+{\beta }_n{x}_n+{\vartheta }_n{S}_{r_n}{x}_n,n\ge 0,$$

(1.5)

where $S_{r_n}=a_{0}I+a_1{J}_{r_n}^{A_1}+a_2{J}_{r_n}^{A_2}+\cdots +a_l{J}_{r_n}^{A_l}$ with $J_{r_n}^{A_i}={(I+r_n{A}_i)}^{-1}$ and ${\sum }_{i=0}^l{a}_i=1$. We have $\{{\alpha }_n\},\{{\beta }_n\},\{{\vartheta }_n\}\subset (0,1)$ and ${\alpha }_n+{\beta }_n+{\vartheta }_n=1$. If ${\bigcap }_{i=1}^{l}N(A_i)\ne \mathrm{\varnothing }$, they proved that $\{x_n\}$ converges strongly to the common point in $N(A_i)$ ($i=1,2,\dots ,l$) under some conditions.

In 2009, Yao et al. presented the following iterative algorithm in the frame of Hilbert space in [16]:

$$\begin{array}{r}{x}_1\in C,\\ y_n=P_C[(1-{\alpha }_n)x_n],\\ x_{n+1}=(1-{\beta }_n)x_n+{\beta }_{n}T{y}_n,n\ge 1.\end{array}$$

(1.6)

Here $T:C\to C$ is a nonexpansive mapping with $F(T)\ne \mathrm{\varnothing }$. Suppose $\{{\alpha }_n\}$ and $\{{\beta }_n\}$ are two real sequences in $(0,1)$ satisfying

• (a) ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=+\mathrm{\infty }$ and ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$;

• (b) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1$.

Then $\{x_n\}$ constructed by (1.6) converges strongly to a point in $F(T)$.

The following lemma is commonly used in proving the convergence of the iterative algorithms in a Banach space.

Lemma 1.1 ([17])

Let E be a real uniformly smooth Banach space, then there exists a nondecreasing continuous function $\beta :[0,+\mathrm{\infty })\to [0,+\mathrm{\infty })$ with ${lim}_{t\to 0^{+}}\beta (t)=0$ and $\beta (ct)\le c\beta (t)$ for $c\ge 1$, such that for all $x,y\in E$, the following inequality holds:

$${\parallel x+y\parallel }^2\le {\parallel x\parallel }^2+2〈y,Jx〉+max\{\parallel x\parallel ,1\}\parallel y\parallel \beta (\parallel y\parallel ).$$

Motivated by the work in [14] and [16], and after imposing an additional condition on the function β in Lemma 1.1 that

$$\beta (t)\le \frac{t}{max\{1,2r_1\}},$$

(1.7)

where $r_1>0$ is a constant satisfying some conditions, Shehu and Ezeora presented the following result.

Theorem 1.1 ([2])

Let E be a real uniformly smooth and uniformly convex Banach space, and let C be a nonempty, closed, and convex sunny nonexpansive retract of E, where $Q_C$ is the sunny nonexpansive retraction of E onto C. Supposed the duality mapping $J:E\to E^{\ast }$ is weakly sequentially continuous. For each $i=1,2,\dots ,N$, let $A_i:C\to E$ be an m-accretive mapping such that ${\bigcap }_{i=1}^{N}N(A_i)\ne \mathrm{\varnothing }$. Let $\{{\alpha }_n\},\{{\beta }_n\}\subset (0,1)$ satisfy (a) and (b). Let $\{x_n\}$ be generated iteratively by

$$\begin{array}{r}{x}_1\in C,\\ y_n=Q_C[(1-{\alpha }_n)x_n],\\ x_{n+1}=(1-{\beta }_n)x_n+{\beta }_n{S}_N{y}_n,n\ge 1.\end{array}$$

(1.8)

Here $S_N:=a_{0}I+a_1{J}_{A_1}+a_2{J}_{A_2}+\cdots +a_N{J}_{A_N}$ with $J_{A_i}={(I+A_i)}^{-1}$, for $i=1,2,\dots ,N$. $0<a_k<1$, for $k=0,1,2,\dots ,N$, and ${\sum }_{k=0}^N{a}_k=1$. Then $\{x_n\}$ converges strongly to the common point in $N(A_i)$, where $i=1,2,\dots ,N$.

How do we show the convergence of the iterative sequence $\{x_n\}$ in (1.8) if β loses the additional condition (1.7)? How about the convergence of $\{x_n\}$ if different $A_i$ has different coefficient in (1.8)?

To answer these questions, Wei and Tan presented the following iterative scheme in [18]:

$$\begin{array}{r}{x}_1\in C,\\ u_n=Q_C[(1-{\alpha }_n)(x_n+e_n)],\\ v_n=(1-{\beta }_n)x_n+{\beta }_n{S}_n{u}_n,\\ x_{n+1}={\gamma }_n{x}_n+(1-{\gamma }_n)S_n{v}_n,n\ge 1,\end{array}$$

(1.9)

where $\{e_n\}\subset E$ is the error sequence and ${\{A_i\}}_{i=1}^N$ is a finite family of m-accretive mappings. $S_n:=a_{0}I+a_1{J}_{r_{n,1}}^{A_1}+a_2{J}_{r_{n,2}}^{A_2}+\cdots +a_N{J}_{r_{n,N}}^{A_N}$, $J_{r_{n,i}}^{A_i}={(I+r_{n,i}{A}_i)}^{-1}$, for $i=1,2,\dots ,N$, ${\sum }_{k=0}^N{a}_k=1$, $0<a_k<1$, for $k=0,1,2,\dots ,N$. Some strong convergence theorems are obtained.

In this paper, our main purpose is to extend the discussion of (1.9) from one family of m-accretive mappings ${\{A_i\}}_{i=1}^N$ to that of two families of m-accretive mappings ${\{A_i\}}_{i=1}^N$ and ${\{B_j\}}_{j=1}^M$. We shall first present and study the following three-step iterative algorithm (A) with errors $\{e_n\}\subset E$:

$$\begin{array}{r}{x}_1\in C,\\ u_n=Q_C[(1-{\alpha }_n)(x_n+e_n)],\\ v_n=(1-{\beta }_n)x_n+{\beta }_n{S}_n{u}_n,\\ x_{n+1}={\gamma }_n{x}_n+(1-{\gamma }_n)W_n{S}_n{v}_n,n\ge 1,\end{array}$$

(A)

where $S_n:=a_{0}I+a_1{J}_{r_{n,1}}^{A_1}+a_2{J}_{r_{n,2}}^{A_2}+\cdots +a_N{J}_{r_{n,N}}^{A_N}$, and $W_n:=b_{0}I+b_1{J}_{s_{n,1}}^{B_1}+b_2{J}_{s_{n,2}}^{B_2}+\cdots +b_M{J}_{s_{n,M}}^{B_M}$. For $i=1,2,\dots ,N$, $J_{r_{n,i}}^{A_i}={(I+r_{n,i}{A}_i)}^{-1}$. For $j=1,2,\dots ,M$, $J_{s_{n,j}}^{B_j}={(I+s_{n,j}{B}_j)}^{-1}$. $a_0,a_1,\dots ,a_N$ and $b_0,b_1,\dots ,b_M$ are real numbers in $(0,1)$ and ${\sum }_{i=0}^N{a}_i=1$, ${\sum }_{j=0}^M{b}_j=1$. $r_{n,i}>0$, for $i=1,2,\dots ,N$, and $s_{n,j}>0$, for $j=1,2,\dots ,M$ and $n\ge 1$.

Later, we introduce and study the following one:

$$\begin{array}{r}{x}_1\in C,\\ u_n=Q_C[(1-{\alpha }_n)(x_n+e_n)],\\ v_n=(1-{\beta }_n)x_n+{\beta }_n{S}_n{u}_n,\\ x_{n+1}={\gamma }_n{x}_n+(1-{\gamma }_n)U_n{S}_n{v}_n,n\ge 1,\end{array}$$

(B)

where $U_n:=c_{0}I+c_1{J}_{t_{n,1}}^{B_1}+c_2{J}_{t_{n,2}}^{B_2}{J}_{t_{n,1}}^{B_1}+\cdots +c_M{J}_{t_{n,M}}^{B_M}{J}_{t_{n,M-1}}^{B_{M-1}}\cdots J_{t_{n,1}}^{B_1}$, $c_0,c_1,\dots ,c_M$ are real numbers in $(0,1)$, ${\sum }_{j=0}^M{c}_j=1$, and $J_{t_{n,j}}^{B_j}={(I+t_{n,j}{B}_j)}^{-1}$ and $t_{n,j}>0$, for $j=1,2,\dots ,M$ and $n\ge 1$.

More details will be presented in Section 3. Some strong convergence theorems are obtained, which can be regarded as the extension of the work done in [2, 6, 14, 15, 18], etc. As a consequence, some new iterative algorithms are constructed to converge strongly to the common fixed point of two finite families of pseudo-contractive mappings from C to E.

2 Preliminaries

Now, we list some results we need in sequel.

Lemma 2.1 ([19])

Let E be a real uniformly convex Banach space and let C be a nonempty, closed, and convex subset of E and $T:C\to C$ is a nonexpansive mapping such that $F(T)\ne \mathrm{\varnothing }$, then $I-T$ is demiclosed at zero.

Lemma 2.2 ([15])

Let E be a strictly convex Banach space which has a uniformly Gâteaux differential norm, and let C be a nonempty, closed, and convex subset of E. Let ${\{A_i\}}_{i=1}^N$ be a finite family of accretive mappings with ${\bigcap }_{i=1}^{N}N(A_i)\ne \mathrm{\varnothing }$, satisfying the following range conditions:

$$\overline{D(A_i)}\subseteq C\subset \bigcap _{r>0}R(I+r{A}_i),i=1,2,\dots ,N.$$

Let $a_0,a_1,\dots ,a_N$ be real numbers in $(0,1)$ such that ${\sum }_{i=0}^N{a}_i=1$ and $S_{r_n}=a_{0}I+a_1{J}_{r_n}^{A_1}+a_2{J}_{r_n}^{A_2}+\cdots +a_N{J}_{r_n}^{A_N}$, where $J_{r_n}^{A_i}={(I+r_n{A}_i)}^{-1}$ and $r_n>0$, then $S_{r_n}$ is nonexpansive and $F(S_{r_n})={\bigcap }_{i=1}^{N}N(A_i)$.

Lemma 2.3 ([12])

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

$${\parallel x+y\parallel }^2\le {\parallel x\parallel }^2+2〈y,j(x+y)〉,\mathrm{\forall }x,y\in E,$$

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

Lemma 2.4 ([20])

Let $\{a_n\}$, $\{b_n\}$, and $\{c_n\}$ be three sequences of nonnegative real numbers satisfying

$$a_{n+1}\le (1-c_n)a_n+b_n{c}_n,\mathrm{\forall }n\ge 1,$$

where $\{c_n\}\subset (0,1)$ such that (i) $c_n\to 0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{c}_n=+\mathrm{\infty }$, (ii) either ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{b}_n\le 0$ or ${\sum }_{n=1}^{\mathrm{\infty }}|b_n{c}_n|<+\mathrm{\infty }$. Then ${lim}_{n\to \mathrm{\infty }}{a}_n=0$.

Lemma 2.5 ([21])

Let $\{x_n\}$ and $\{y_n\}$ be two bounded sequences in a Banach space E such that $x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)y_n$, for $n\ge 1$. Suppose $\{{\beta }_n\}\subset (0,1)$ satisfying $0<{lim\hspace{0.17em}inf}_{n\to +\mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to +\mathrm{\infty }}{\beta }_n<1$. If ${lim\hspace{0.17em}sup}_{n\to +\mathrm{\infty }}(\parallel y_{n+1}-y_n\parallel -\parallel x_{n+1}-x_n\parallel )\le 0$, then ${lim}_{n\to +\mathrm{\infty }}\parallel y_n-x_n\parallel =0$.

Lemma 2.6 ([22])

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

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

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

3 Main results

Lemma 3.1 ([2])

Let E be a real uniformly smooth and uniformly convex Banach space. Let C be a nonempty, closed, and convex sunny nonexpansive retract of E, and $Q_C$ be the sunny nonexpansive retraction of E onto C. Let $T:C\to C$ be nonexpansive with $F(T)\ne \mathrm{\varnothing }$. Suppose that the duality mapping $J:E\to E^{\ast }$ is weakly sequentially continuous. If for each $t\in (0,1)$, define $T_t:C\to C$ by

$$T_{t}x:=T{Q}_C[(1-t)x].$$

(3.1)

Then $T_t$ is a contraction and has a fixed point $z_t$, which satisfies $\parallel z_t-T{z}_t\parallel \to 0$, as $t\to 0$.

Lemma 3.2 ([2])

Under the assumptions of Lemma  3.1, suppose further that β in Lemma  1.1 satisfies (1.7), where $r_1>0$ is a sufficiently large constant such that $z_t\in C\bigcap \{z\in E:\parallel z-x^{\ast }\parallel \le r_1\}$, $x^{\ast }$ is in $F(T)$ and $t\in (0,1)$, then ${lim}_{t\to 0}{z}_t=z_0\in F(T)$.

Remark 3.1 Lemma 1.1 with additional condition (1.7) is employed as a key tool to prove Lemma 3.2. In the following lemma, we shall show that Lemma 2.3 can be used instead of Lemma 1.1, which simplifies the proof and weakens the assumption.

Lemma 3.3 Only under the assumptions of Lemma  3.1, the result of Lemma  3.2 is true, which ensures that the assumption is weaker than that in Lemma  3.2.

Proof To show that ${lim}_{t\to 0}{z}_t=z_0\in F(T)$, it suffices to show that for any sequence $\{t_n\}$ such that $t_n\to 0$, we have ${lim}_{n\to \mathrm{\infty }}{z}_{t_n}=z_0\in F(T)$.

In fact, Lemma 3.1 implies that $z_t\in F(T)$ such that $z_t=T{Q}_C[(1-t)z_t]$, $t\in (0,1)$. By using Lemma 2.3, we have for $\mathrm{\forall }p\in F(T)$,

$$\begin{array}{rl}{\parallel z_t-p\parallel }^2& ={\parallel T{Q}_C[(1-t)z_t]-T{Q}_{C}p\parallel }^2\\ \le {\parallel z_t-p-t{z}_t\parallel }^2\\ \le {\parallel z_t-p\parallel }^2-2t{\parallel z_t-p-t{z}_t\parallel }^2-2t〈p+t{z}_t,J(z_t-p-t{z}_t)〉.\end{array}$$

This implies that

$${\parallel z_t-p-t{z}_t\parallel }^2\le 〈p,J(p+t{z}_t-z_t)〉+t〈z_t,J(p+t{z}_t-z_t)〉.$$

(3.2)

In particular,

$${\parallel z_{t_n}-p-t_n{z}_{t_n}\parallel }^2\le 〈p,J(p+t_n{z}_{t_n}-z_{t_n})〉+t_n〈z_{t_n},J(p+t_n{z}_{t_n}-z_{t_n})〉.$$

(3.3)

Since $\mathrm{\forall }p\in F(T)$,

$$\begin{array}{rl}\parallel z_t-p\parallel & =\parallel T{Q}_C[(1-t)z_t]-T{Q}_{C}p\parallel \\ \le \parallel Q_C[(1-t)z_t]-Q_{C}p\parallel \\ \le \parallel (1-t)z_t-p\parallel =\parallel (1-t)(z_t-p)-tp\parallel \\ \le (1-t)\parallel z_t-p\parallel +t\parallel p\parallel ,\end{array}$$

$\{z_t\}$ is bounded.

Without loss of generality, we can assume that $\{z_{t_n}\}$ converges weakly to $z_0$. Using Lemma 3.1 and Lemma 2.1, we have $z_0\in F(T)$.

Substituting $z_0$ for p in (3.3), we obtain

$${\parallel z_{t_n}-z_0-t_n{z}_{t_n}\parallel }^2\le 〈z_0,J(z_0+t_n{z}_{t_n}-z_{t_n})〉+t_n〈z_{t_n},J(z_0+t_n{z}_{t_n}-z_{t_n})〉.$$

(3.4)

Then from (3.4) and the weak convergence of J, we have $z_{t_n}-z_0-t_n{z}_{t_n}\to 0$, as $n\to \mathrm{\infty }$.

Then from $\parallel z_{t_n}-z_0\parallel \le \parallel z_{t_n}-z_0-t_n{z}_{t_n}\parallel +t_n\parallel z_{t_n}\parallel$, we see that $z_{t_n}\to z_0$, as $n\to \mathrm{\infty }$.

Suppose there exists another sequence $z_{t_m}\rightharpoonup x_0$, as $t_m\to 0$ and $m\to \mathrm{\infty }$. Then from Lemma 3.1 that $\parallel z_{t_m}-T{z}_{t_m}\parallel \to 0$ and $I-T$ is demi-closed at zero, we have $x_0\in F(T)$. Moreover, repeating the above proof, we have $z_{t_m}\to x_0$, as $m\to \mathrm{\infty }$. Next, we want to show that $z_0=x_0$.

Using (3.2), we have

$${\parallel z_{t_m}-z_0-t_m{z}_{t_m}\parallel }^2\le 〈z_0,J(z_0+t_m{z}_{t_m}-z_{t_m})〉+t_m〈z_{t_m},J(z_0+t_m{z}_{t_m}-z_{t_m})〉.$$

(3.5)

By letting $m\to \mathrm{\infty }$, (3.5) implies that

$${\parallel x_0-z_0\parallel }^2\le 〈z_0,J(z_0-x_0)〉.$$

(3.6)

Interchanging $x_0$ and $z_0$ in (3.6), we obtain

$${\parallel z_0-x_0\parallel }^2\le 〈x_0,J(x_0-z_0)〉.$$

(3.7)

Then (3.6) and (3.7) ensure

$$2{\parallel x_0-z_0\parallel }^2\le {\parallel x_0-z_0\parallel }^2,$$

(3.8)

which implies that $x_0=z_0$.

Therefore, ${lim}_{t\to 0}{z}_t=z_0\in F(T)$.

This completes the proof. □

Lemma 3.4 Let E be a strictly convex Banach space and let C be a nonempty, closed, and convex subset of E. Let $A_i:C\to E$ ($i=1,2,\dots ,N$) be a finite family of m-accretive mappings such that ${\bigcap }_{i=1}^{N}N(A_i)\ne \mathrm{\varnothing }$.

Let $a_0,a_1,\dots ,a_N$ be real numbers in $(0,1)$ such that ${\sum }_{i=0}^N{a}_i=1$ and $S_n=a_{0}I+a_1{J}_{r_{n,1}}^{A_1}+a_2{J}_{r_{n,2}}^{A_2}+\cdots +a_N{J}_{r_{n,N}}^{A_N}$, where $J_{r_{n,i}}^{A_i}={(I+r_{n,i}{A}_i)}^{-1}$ and $r_{n,i}>0$, for $i=1,2,\dots ,N$, and $n\ge 1$, then $S_n:C\to C$ is nonexpansive and $F(S_n)={\bigcap }_{i=1}^{N}N(A_i)$, for $n\ge 1$.

Proof The proof is from [18]. For later use, we present the proof in the following.

It is easy to check that $S_n:C\to C$ is nonexpansive and ${\bigcap }_{i=1}^{N}N(A_i)\subset F(S_n)$.

On the other hand, for $\mathrm{\forall }p\in F(S_n)$, then $p=S_{n}p=a_{0}p+a_1{J}_{r_{n,1}}^{A_1}p+a_2{J}_{r_{n,2}}^{A_2}p+\cdots +a_N{J}_{r_{n,N}}^{A_N}p$.

For $\mathrm{\forall }q\in {\bigcap }_{i=1}^{N}N(A_i)\subset F(S_n)$, we have

$$\begin{array}{rl}\parallel p-q\parallel & \le a_0\parallel p-q\parallel +a_1\parallel J_{r_{n,1}}^{A_1}p-q\parallel +\cdots +a_N\parallel J_{r_{n,N}}^{A_N}p-q\parallel \\ \le (a_0+a_1+\cdots +a_{N-1})\parallel p-q\parallel +a_N\parallel J_{r_{n,N}}^{A_N}p-q\parallel \\ =(1-a_N)\parallel p-q\parallel +a_N\parallel J_{r_{n,N}}^{A_N}p-q\parallel \\ \le \parallel p-q\parallel .\end{array}$$

Therefore, $\parallel p-q\parallel =(1-a_N)\parallel p-q\parallel +a_N\parallel J_{r_{n,N}}^{A_N}p-q\parallel$, which implies that $\parallel p-q\parallel =\parallel J_{r_{n,N}}^{A_N}p-q\parallel$. Similarly, $\parallel p-q\parallel =\parallel J_{r_{n,1}}^{A_1}p-q\parallel =\cdots =\parallel J_{r_{n,N}}^{A_N}p-q\parallel$.

Then $\parallel p-q\parallel =\parallel \frac{a_1}{{\sum }_{i=1}^N{a}_i}(J_{r_{n,1}}^{A_1}p-q)+\frac{a_2}{{\sum }_{i=1}^N{a}_i}(J_{r_{n,2}}^{A_2}p-q)+\cdots +\frac{a_N}{{\sum }_{i=1}^N{a}_i}(J_{r_{n,N}}^{A_N}p-q)\parallel$, which implies from the strict convexity of E that $p-q=J_{r_{n,1}}^{A_1}p-q=J_{r_{n,2}}^{A_2}p-q=\cdots =J_{r_{n,N}}^{A_N}p-q$.

Therefore, $J_{r_{n,i}}^{A_i}p=p$, for $i=1,2,\dots ,N$. We have $p\in {\bigcap }_{i=1}^{N}N(A_i)$, which completes the proof. □

Similar to Lemma 3.4, we have the following lemma.

Lemma 3.5 Let E and C be the same as those in Lemma  3.4. Let ${\{B_j\}}_{j=1}^M$ be a finite family of m-accretive mappings such that ${\bigcap }_{j=1}^{M}N(B_j)\ne \mathrm{\varnothing }$.

Let $b_0,b_1,\dots ,b_M$ be real numbers in $(0,1)$ such that ${\sum }_{j=0}^M{b}_j=1$ and $W_n=b_{0}I+b_1{J}_{s_{n,1}}^{B_1}+b_2{J}_{s_{n,2}}^{B_2}+\cdots +b_M{J}_{s_{n,M}}^{B_M}$, where $J_{s,j}^{B_j}={(I+s_{n,j}{B}_j)}^{-1}$ and $s_{n,j}>0$, for $j=1,2,\dots ,M$, then $W_n:C\to C$ is nonexpansive and $F(W_n)={\bigcap }_{j=1}^{M}N(B_j)$, for $n\ge 1$.

Lemma 3.6 Let E, C, $S_n$, and $W_n$ be the same as those in Lemmas 3.4 and 3.5. Suppose $D:=({\bigcap }_{i=1}^{N}N(A_i))\cap ({\bigcap }_{j=1}^{M}N(B_j))\ne \mathrm{\varnothing }$. Then $W_n{S}_n,S_n{W}_n:C\to C$ are nonexpansive and $F(W_n{S}_n)=F(S_n{W}_n)=D$.

Proof From Lemmas 3.4 and 3.5, we can easily check that $W_n{S}_n,S_n{W}_n:C\to C$ are nonexpansive and $F(S_n)\cap F(W_n)=D$. So, it suffices to show that $F(S_n)\cap F(W_n)\supset F(W_n{S}_n)$ since $F(S_n)\cap F(W_n)\subset F(W_n{S}_n)$ is trivial.

For $\mathrm{\forall }p\in F(W_n{S}_n)$, then $p=W_n{S}_{n}p$.

For $\mathrm{\forall }q\in F(S_n)\cap F(W_n)\subset F(W_n{S}_n)$, then $q=W_n{S}_{n}q$. Now,

$$\parallel p-q\parallel \le \parallel S_{n}p-S_{n}q\parallel \le a_0\parallel p-q\parallel +a_1\parallel J_{r_{n,1}}^{A_1}p-q\parallel +\cdots +a_N\parallel J_{r_{n,N}}^{A_N}p-q\parallel .$$

Then repeating the discussion in Lemma 3.4, we know that $p\in F(S_n)$. Then $p=W_n{S}_{n}p=W_{n}p$, thus $p\in F(W_n)$, which completes the proof. □

Theorem 3.1 Let E be a real uniformly smooth and uniformly convex Banach space. Let C be a nonempty, closed, and convex sunny nonexpansive retract of E, where $Q_C$ is the sunny nonexpansive retraction of E onto C. Let $A_i,B_j:C\to E$ be m-accretive mappings, where $i=1,2,\dots N$, $j=1,2,\dots ,M$. Suppose that the duality mapping $J:E\to E^{\ast }$ is weakly sequentially continuous and $D:=({\bigcap }_{i=1}^{N}N(A_i))\cap ({\bigcap }_{j=1}^{M}N(B_j))\ne \mathrm{\varnothing }$. Let $\{x_n\}$ be generated by the iterative algorithm (A), where $S_n:=a_{0}I+a_1{J}_{r_{n,1}}^{A_1}+a_2{J}_{r_{n,2}}^{A_2}+\cdots +a_N{J}_{r_{n,N}}^{A_N}$, and $J_{r_{n,i}}^{A_i}={(I+r_{n,i}{A}_i)}^{-1}$, for $i=1,2,\dots ,N$, $0<a_k<1$, for $k=0,1,2,\dots ,N$, ${\sum }_{k=0}^N{a}_k=1$. $W_n=b_{0}I+b_1{J}_{s_{n,1}}^{B_1}+b_2{J}_{s_{n,2}}^{B_2}+\cdots +b_M{J}_{s_{n,M}}^{B_M}$, where $J_{s_{n,j}}^{B_j}={(I+s_{n,j}{B}_j)}^{-1}$, for $j=1,2,\dots ,M$, $0<b_k<1$, for $k=0,1,2,\dots ,M$, ${\sum }_{k=0}^M{b}_k=1$. Suppose $\{e_n\}\subset E$, $\{{\alpha }_n\}$, $\{{\beta }_n\}$, and $\{{\gamma }_n\}$ are three sequences in $(0,1)$, and $\{r_{n,i}\},\{s_{n,j}\}\subset (0,+\mathrm{\infty })$ satisfy the following conditions:

• (i) ${\alpha }_n\to 0$, ${\beta }_n\to 0$, as $n\to \mathrm{\infty }$;

• (ii) ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n{\beta }_n=+\mathrm{\infty }$;

• (iii) $0<{lim\hspace{0.17em}inf}_{n\to +\mathrm{\infty }}{\gamma }_n\le {lim\hspace{0.17em}sup}_{n\to +\mathrm{\infty }}{\gamma }_n<1$;

• (iv) ${\sum }_{n=1}^{\mathrm{\infty }}|r_{n+1,i}-r_{n,i}|<+\mathrm{\infty }$ and $r_{n,i}\ge \epsilon >0$, for $n\ge 1$ and $i=1,2,\dots ,N$;

• (v) ${\sum }_{n=1}^{\mathrm{\infty }}|s_{n+1,j}-s_{n,j}|<+\mathrm{\infty }$ and $s_{n,j}\ge \epsilon >0$, for $n\ge 1$ and $j=1,2,\dots ,M$;

• (vi) $\frac{\parallel e_n\parallel }{{\alpha }_n}\to 0$, as $n\to +\mathrm{\infty }$, and ${\sum }_{n=1}^{\mathrm{\infty }}\parallel e_n\parallel <+\mathrm{\infty }$.

Then $\{x_n\}$ converges strongly to a point $p_0\in D$.

Proof We shall split the proof into five steps:

Step 1. $\{x_n\}$, $\{u_n\}$, $\{S_n{u}_n\}$, $\{v_n\}$, and $\{S_n{x}_n\}$ are all bounded.

We shall first show that $\mathrm{\forall }p\in D$,

$$\parallel x_{n+1}-p\parallel \le M_1+\sum _{i=1}^n\parallel e_i\parallel ,$$

(3.9)

where $M_1=max\{\parallel x_1-p\parallel ,\parallel p\parallel \}$.

By using the induction method, we see that for $n=1$, $\mathrm{\forall }p\in D$,

$$\begin{array}{rl}\parallel x_2-p\parallel & \le {\gamma }_1\parallel x_1-p\parallel +(1-{\gamma }_1)\parallel W_1{S}_1{v}_1-p\parallel \\ \le {\gamma }_1\parallel x_1-p\parallel +(1-{\gamma }_1)\parallel v_1-p\parallel \\ \le {\gamma }_1\parallel x_1-p\parallel +(1-{\gamma }_1)(1-{\beta }_1)\parallel x_1-p\parallel +{\beta }_1(1-{\gamma }_1)\parallel u_1-p\parallel \\ \le {\gamma }_1\parallel x_1-p\parallel +(1-{\gamma }_1)(1-{\beta }_1)\parallel x_1-p\parallel +{\beta }_1(1-{\gamma }_1)\parallel (1-{\alpha }_1)(x_1+e_1)-p\parallel \\ \le [1-{\alpha }_1{\beta }_1(1-{\gamma }_1)]\parallel x_1-p\parallel +{\alpha }_1{\beta }_1(1-{\gamma }_1)\parallel p\parallel +(1-{\alpha }_1){\beta }_1(1-{\gamma }_1)\parallel e_1\parallel \\ \le M_1+\parallel e_1\parallel .\end{array}$$

Suppose that (3.9) is true for $n=k$. Then, for $n=k+1$,

$$\begin{array}{rl}\parallel x_{k+2}-p\parallel \le & {\gamma }_{k+1}\parallel x_{k+1}-p\parallel +(1-{\gamma }_{k+1})\parallel v_{k+1}-p\parallel \\ \le & {\gamma }_{k+1}\parallel x_{k+1}-p\parallel +(1-{\gamma }_{k+1})[(1-{\beta }_{k+1})\parallel x_{k+1}-p\parallel +{\beta }_{k+1}\parallel u_{k+1}-p\parallel ]\\ \le & {\gamma }_{k+1}\parallel x_{k+1}-p\parallel +(1-{\gamma }_{k+1})[(1-{\beta }_{k+1})\parallel x_{k+1}-p\parallel \\ +{\beta }_{k+1}\parallel (1-{\alpha }_{k+1})(x_{k+1}+e_{k+1})-p\parallel ]\\ \le & [1-{\alpha }_{k+1}{\beta }_{k+1}(1-{\gamma }_{k+1})]\parallel x_{k+1}-p\parallel +{\alpha }_{k+1}{\beta }_{k+1}(1-{\gamma }_{k+1})\parallel p\parallel \\ +{\beta }_{k+1}(1-{\alpha }_{k+1})(1-{\gamma }_{k+1})\parallel e_{k+1}\parallel \\ \le & M_1+[1-{\alpha }_{k+1}{\beta }_{k+1}(1-{\gamma }_{k+1})]\sum _{i=1}^k\parallel e_i\parallel +(1-{\alpha }_{k+1}){\beta }_{k+1}(1-{\gamma }_{k+1})\parallel e_{k+1}\parallel \\ \le & M_1+\sum _{i=1}^{k+1}\parallel e_i\parallel .\end{array}$$

Thus (3.9) is true for all $n\in N$. Since ${\sum }_{n=1}^{\mathrm{\infty }}\parallel e_n\parallel <+\mathrm{\infty }$, (3.9) ensures that $\{x_n\}$ is bounded.

For $\mathrm{\forall }p\in D$, from $\parallel u_n-p\parallel \le \parallel (1-{\alpha }_n)(x_n+e_n)-p\parallel \le \parallel x_n\parallel +\parallel e_n\parallel +\parallel p\parallel$, we see that $\{u_n\}$ is bounded.

Since $\parallel S_n{u}_n\parallel \le \parallel S_n{u}_n-S_{n}p\parallel +\parallel p\parallel \le \parallel u_n-p\parallel +\parallel p\parallel$, $\{S_n{u}_n\}$ is bounded. Since both $\{S_n{u}_n\}$ and $\{x_n\}$ are bounded, $\{v_n\}$ is bounded. Similarly, $\{S_n{x}_n\}$, $\{S_n{v}_n\}$, $\{J_{r_{n,i}}^{A_i}{u}_n\}$, $\{J_{r_{n,i}}^{A_i}{v}_n\}$, and $\{J_{s_{n,j}}^{B_j}{S}_n{v}_n\}$ are all bounded, for $i=1,2,\dots ,N$; $j=1,2,\dots ,M$.

Then we set $M_2=sup\{\parallel u_n\parallel ,\parallel J_{r_{n,i}}^{A_i}{u}_n\parallel ,\parallel v_n\parallel ,\parallel J_{r_{n,i}}^{A_i}{v}_n\parallel ,\parallel S_n{u}_n\parallel ,\parallel S_n{v}_n\parallel ,\parallel x_n\parallel ,\parallel J_{s_{n,j}}^{B_j}{S}_n{v}_n\parallel :n\ge 1,i=1,2,\dots ,N;j=1,2,\dots ,M\}$.

Step 2. ${lim}_{n\to \mathrm{\infty }}\parallel x_n-W_n{S}_n{v}_n\parallel =0$ and ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel =0$.

In fact,

$$\begin{array}{r}\parallel W_{n+1}{S}_{n+1}{v}_{n+1}-W_n{S}_n{v}_n\parallel \\ \le b_0\parallel S_{n+1}{v}_{n+1}-S_n{v}_n\parallel +\sum _{j=1}^M{b}_j\parallel J_{s_{n+1,j}}^{B_j}{S}_{n+1}{v}_{n+1}-J_{s_{n,j}}^{B_j}{S}_n{v}_n\parallel .\end{array}$$

(3.10)

Next, we discuss $\parallel J_{s_{n+1,j}}^{B_j}{S}_{n+1}{v}_{n+1}-J_{s_{n,j}}^{B_j}{S}_n{v}_n\parallel$.

If $s_{n,j}\le s_{n+1,j}$, then, using Lemma 2.6,

$$\begin{array}{r}\parallel J_{s_{n+1,j}}^{B_j}{S}_{n+1}{v}_{n+1}-J_{s_{n,j}}^{B_j}{S}_n{v}_n\parallel \\ =\parallel J_{s_{n,j}}^{B_j}(\frac{s_{n,j}}{s_{n+1,j}}{S}_{n+1}{v}_{n+1}+(1-\frac{s_{n,j}}{s_{n+1,j}})J_{s_{n+1,j}}^{B_j}{S}_{n+1}{v}_{n+1})-J_{s_{n,j}}^{B_j}{S}_n{v}_n\parallel \\ \le \parallel \frac{s_{n,j}}{s_{n+1,j}}{S}_{n+1}{v}_{n+1}+(1-\frac{s_{n,j}}{s_{n+1,j}})J_{s_{n+1,j}}^{B_j}{S}_{n+1}{v}_{n+1}-S_n{v}_n\parallel \\ \le \frac{s_{n,j}}{s_{n+1,j}}\parallel S_{n+1}{v}_{n+1}-S_n{v}_n\parallel +(1-\frac{s_{n,j}}{s_{n+1,j}})\parallel J_{s_{n+1,j}}^{B_j}{S}_{n+1}{v}_{n+1}-S_n{v}_n\parallel \\ \le \parallel S_{n+1}{v}_{n+1}-S_n{v}_n\parallel +2M_2\frac{s_{n+1,j}-s_{n,j}}{\epsilon }.\end{array}$$

(3.11)

If $s_{n+1,j}\le s_{n,j}$, then imitating the proof of (3.11), we have

$$\parallel J_{s_{n+1,j}}^{B_j}{S}_{n+1}{v}_{n+1}-J_{s_{n,j}}^{B_j}{S}_n{v}_n\parallel \le \parallel S_{n+1}{v}_{n+1}-S_n{v}_n\parallel +2M_2\frac{s_{n,j}-s_{n+1,j}}{\epsilon }.$$

(3.12)

Combining (3.11) and (3.12), we have

$$\begin{array}{r}\parallel J_{s_{n+1,j}}^{B_j}{S}_{n+1}{v}_{n+1}-J_{s_{n,j}}^{B_j}{S}_n{v}_n\parallel \\ \le \parallel S_{n+1}{v}_{n+1}-S_n{v}_n\parallel +2M_2\frac{|s_{n,j}-s_{n+1,j}|}{\epsilon }.\end{array}$$

(3.13)

Putting (3.13) into (3.10), we have

$$\parallel W_{n+1}{S}_{n+1}{v}_{n+1}-W_n{S}_n{v}_n\parallel \le \parallel S_{n+1}{v}_{n+1}-S_n{v}_n\parallel +\frac{2M_2}{\epsilon }\sum _{j=1}^M|s_{n,j}-s_{n+1,j}|.$$

(3.14)

Similarly, we have

$$\parallel S_{n+1}{u}_{n+1}-S_n{u}_n\parallel \le \parallel u_{n+1}-u_n\parallel +\frac{2M_2}{\epsilon }\sum _{i=1}^N|r_{n,i}-r_{n+1,i}|$$

(3.15)

and

$$\parallel S_{n+1}{v}_{n+1}-S_n{v}_n\parallel \le \parallel v_{n+1}-v_n\parallel +\frac{2M_2}{\epsilon }\sum _{i=1}^N|r_{n,i}-r_{n+1,i}|.$$

(3.16)

Therefore,

$$\begin{array}{r}\parallel W_{n+1}{S}_{n+1}{v}_{n+1}-W_n{S}_n{v}_n\parallel \\ \le \parallel v_{n+1}-v_n\parallel +\frac{2M_2}{\epsilon }\sum _{j=1}^M|s_{n,j}-s_{n+1,j}|+\frac{2M_2}{\epsilon }\sum _{i=1}^N|r_{n,i}-r_{n+1,i}|\\ \le \parallel x_{n+1}-x_n\parallel +{\beta }_n\parallel x_n\parallel +{\beta }_{n+1}\parallel x_{n+1}\parallel +|{\beta }_{n+1}-{\beta }_n|\parallel S_{n+1}{u}_{n+1}\parallel +{\beta }_n\parallel S_{n+1}{u}_{n+1}-S_n{u}_n\parallel \\ +\frac{2M_2}{\epsilon }\sum _{j=1}^M|s_{n,j}-s_{n+1,j}|+\frac{2M_2}{\epsilon }\sum _{i=1}^N|r_{n,i}-r_{n+1,i}|\\ \le \parallel x_{n+1}-x_n\parallel +{\beta }_n\parallel x_n\parallel +{\beta }_{n+1}\parallel x_{n+1}\parallel +|{\beta }_{n+1}-{\beta }_n|\parallel S_{n+1}{u}_{n+1}\parallel +{\beta }_n\parallel u_{n+1}-u_n\parallel \\ +\frac{4M_2}{\epsilon }\sum _{i=1}^N|r_{n,i}-r_{n+1,i}|+\frac{2M_2}{\epsilon }\sum _{j=1}^M|s_{n,j}-s_{n+1,j}|\\ \le \parallel x_{n+1}-x_n\parallel +{\beta }_n\parallel x_n\parallel +{\beta }_{n+1}\parallel x_{n+1}\parallel +|{\beta }_{n+1}-{\beta }_n|\parallel S_{n+1}{u}_{n+1}\parallel \\ +{\beta }_n\parallel (1-{\alpha }_{n+1})(x_{n+1}+e_{n+1})-(1-{\alpha }_n)(x_n+e_n)\parallel \\ +\frac{4M_2}{\epsilon }\sum _{i=1}^N|r_{n,i}-r_{n+1,i}|+\frac{2M_2}{\epsilon }\sum _{j=1}^M|s_{n,j}-s_{n+1,j}|\\ \le (1+{\beta }_n)\parallel x_{n+1}-x_n\parallel +({\beta }_n+{\alpha }_n{\beta }_n)\parallel x_n\parallel +({\beta }_{n+1}+{\alpha }_{n+1}{\beta }_n)\parallel x_{n+1}\parallel \\ +|{\beta }_{n+1}-{\beta }_n|\parallel S_{n+1}{u}_{n+1}\parallel +{\beta }_n\parallel e_{n+1}-e_n\parallel +{\beta }_n\parallel {\alpha }_{n+1}{e}_{n+1}-{\alpha }_n{e}_n\parallel \\ +\frac{4M_2}{\epsilon }\sum _{i=1}^N|r_{n,i}-r_{n+1,i}|+\frac{2M_2}{\epsilon }\sum _{j=1}^M|s_{n,j}-s_{n+1,j}|.\end{array}$$

(3.17)

Thus ${lim\hspace{0.17em}sup}_{n\to +\mathrm{\infty }}(\parallel W_{n+1}{S}_{n+1}{v}_{n+1}-W_n{S}_n{v}_n\parallel -\parallel x_{n+1}-x_n\parallel )\le 0$. Using Lemma 2.5, we have from (3.17) ${lim}_{n\to \mathrm{\infty }}\parallel x_n-W_n{S}_n{v}_n\parallel =0$ and then ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel ={lim}_{n\to \mathrm{\infty }}(1-{\gamma }_n)\parallel W_n{S}_n{v}_n-x_n\parallel =0$.

Step 3. ${lim}_{n\to \mathrm{\infty }}\parallel x_n-W_n{S}_n{x}_n\parallel =0$.

In fact,

$$\begin{array}{rl}\parallel x_n-W_n{S}_n{x}_n\parallel & \le \parallel x_{n+1}-x_n\parallel +\parallel x_{n+1}-W_n{S}_n{x}_n\parallel \\ \le \parallel x_{n+1}-x_n\parallel +\parallel {\gamma }_n{x}_n+(1-{\gamma }_n)W_n{S}_n{v}_n-W_n{S}_n{x}_n\parallel \\ \le \parallel x_{n+1}-x_n\parallel +{\gamma }_n\parallel x_n-W_n{S}_n{v}_n\parallel +\parallel W_n{S}_n{v}_n-W_n{S}_n{x}_n\parallel \\ \le \parallel x_{n+1}-x_n\parallel +{\gamma }_n\parallel x_n-W_n{S}_n{v}_n\parallel +{\beta }_n\parallel S_n{u}_n-x_n\parallel \\ \le \parallel x_{n+1}-x_n\parallel +{\gamma }_n\parallel x_n-W_n{S}_n{v}_n\parallel +2{\beta }_n{M}_2.\end{array}$$

(3.18)

Then (3.18) and step 2 imply that $\parallel x_n-W_n{S}_n{x}_n\parallel \to 0$, as $n\to +\mathrm{\infty }$, since ${\beta }_n\to 0$.

Step 4. ${lim\hspace{0.17em}sup}_{n\to +\mathrm{\infty }}〈p_0,J(p_0-x_n)〉\le 0$, where $p_0$ is an element in D.

From Lemma 3.6, we know that $W_n{S}_n:C\to C$ is nonexpansive and $F(W_n{S}_n)=D$. Then Lemma 3.1 and Lemma 3.3 imply that there exists $z_t\in C$ such that $z_t=W_n{S}_n{Q}_C[(1-t)z_t]$ for $t\in (0,1)$. Moreover, $z_t\to p_0\in D$, as $t\to 0$.

Since $\parallel z_t-p_0\parallel \le \parallel (1-t)z_t-p_0\parallel \le (1-t)\parallel z_t-p_0\parallel +\parallel p_0\parallel$, $\{z_t\}$ is bounded. Let $M_3=sup\{\parallel z_t-x_n\parallel :n\ge 1,t>0\}$. Then from step 1, we know that $M_3$ is a positive constant. Using Lemma 2.3, we have

$$\begin{array}{rl}{\parallel z_t-x_n\parallel }^2& ={\parallel z_t-W_n{S}_n{x}_n+W_n{S}_n{x}_n-x_n\parallel }^2\\ \le {\parallel z_t-W_n{S}_n{x}_n\parallel }^2+2〈W_n{S}_n{x}_n-x_n,J(z_t-x_n)〉\\ \le {\parallel z_t-W_n{S}_n{x}_n\parallel }^2+2\parallel W_n{S}_n{x}_n-x_n\parallel \parallel z_t-x_n\parallel \\ \le {\parallel (1-t)z_t-x_n\parallel }^2+2\parallel W_n{S}_n{x}_n-x_n\parallel \parallel z_t-x_n\parallel \\ \le {\parallel z_t-x_n\parallel }^2-2t〈z_t,J[(1-t)z_t-x_n]〉+2M_3\parallel W_n{S}_n{x}_n-x_n\parallel .\end{array}$$

So $〈z_t,J[(1-t)z_t-x_n]〉\le \frac{M_3}{t}\parallel W_n{S}_n{x}_n-x_n\parallel$, which implies that ${lim}_{t\to 0}{lim\hspace{0.17em}sup}_{n\to +\mathrm{\infty }}〈z_t,J[(1-t)z_t-x_n]〉\le 0$ in view of step 3.

Since $\{x_n\}$ is bounded and J is uniformly continuous on each bounded subset of E, $〈p_0,J(p_0-x_n)-J[(1-t)z_t-x_n]〉\to 0$, as $t\to 0$.

Moreover, noticing the fact that

$$\begin{array}{rl}〈p_0,J(p_0-x_n)〉=& 〈p_0,J(p_0-x_n)-J[(1-t)z_t-x_n]〉\\ +〈p_0-z_t,J[(1-t)z_t-x_n]〉+〈z_t,J[(1-t)z_t-x_n]〉,\end{array}$$

we have ${lim\hspace{0.17em}sup}_{n\to +\mathrm{\infty }}〈p_0,J(p_0-x_n)〉\le 0$.

Since $〈p_0,J[p_0-x_n-(1-{\alpha }_n)e_n+{\alpha }_n{x}_n]〉=〈p_0,J[p_0-x_n-(1-{\alpha }_n)e_n+{\alpha }_n{x}_n]-J(p_0-x_n)〉+〈p_0,J(p_0-x_n)〉$, and J is uniformly continuous on each bounded subset of E,

$$\underset{n\to +\mathrm{\infty }}{lim\hspace{0.17em}sup}〈p_0,J[p_0-x_n-(1-{\alpha }_n)e_n+{\alpha }_n{x}_n]〉\le 0.$$

(3.19)

Step 5. $x_n\to p_0$, as $n\to +\mathrm{\infty }$, where $p_0\in D$ is the same as in step 4.

Let $M_4=sup\{\parallel (1-{\alpha }_n)(x_n+e_n)-p_0\parallel :n\ge 1\}$. By using Lemma 2.3 again, we have

$$\begin{array}{r}{\parallel x_{n+1}-p_0\parallel }^2\\ \le {\gamma }_n{\parallel x_n-p_0\parallel }^2+(1-{\gamma }_n){\parallel v_n-p_0\parallel }^2\\ \le {\gamma }_n{\parallel x_n-p_0\parallel }^2+(1-{\gamma }_n)(1-{\beta }_n){\parallel x_n-p_0\parallel }^2+(1-{\gamma }_n){\beta }_n{\parallel u_n-p_0\parallel }^2\\ =(1-{\beta }_n+{\beta }_n{\gamma }_n){\parallel x_n-p_0\parallel }^2+(1-{\gamma }_n){\beta }_n{\parallel u_n-p_0\parallel }^2\\ \le (1-{\beta }_n+{\beta }_n{\gamma }_n){\parallel x_n-p_0\parallel }^2+(1-{\gamma }_n){\beta }_n{\parallel (1-{\alpha }_n)(x_n+e_n)-p_0\parallel }^2\\ \le (1-{\beta }_n+{\beta }_n{\gamma }_n){\parallel x_n-p_0\parallel }^2+(1-{\gamma }_n){\beta }_n(1-{\alpha }_n){\parallel x_n-p_0\parallel }^2\\ +2(1-{\gamma }_n){\beta }_n(1-{\alpha }_n)〈e_n,J[(1-{\alpha }_n)(x_n+e_n)-p_0]〉\\ +2{\alpha }_n{\beta }_n(1-{\gamma }_n)〈p_0,J[p_0-x_n-(1-{\alpha }_n)e_n+{\alpha }_n{x}_n]〉\\ \le [1-{\alpha }_n{\beta }_n(1-{\gamma }_n)]{\parallel x_n-p_0\parallel }^2+2(1-{\gamma }_n)(1-{\alpha }_n){\beta }_n{M}_4\parallel e_n\parallel \\ +2{\alpha }_n{\beta }_n(1-{\gamma }_n)〈p_0,J[p_0-x_n-(1-{\alpha }_n)e_n+{\alpha }_n{x}_n]〉.\end{array}$$

(3.20)

Let $c_n=(1-{\gamma }_n){\alpha }_n{\beta }_n$, then (3.20) reduces to ${\parallel x_{n+1}-p_0\parallel }^2\le (1-c_n){\parallel x_n-p_0\parallel }^2+2c_n\{〈p_0,J[p_0-x_n-(1-{\alpha }_n)e_n+{\alpha }_n{x}_n]〉+(1-{\alpha }_n)M_4\frac{\parallel e_n\parallel }{{\alpha }_n}\}$.

From (3.19), (3.20), and the assumptions, by using Lemma 2.4, we know that $x_n\to p_0$, as $n\to +\mathrm{\infty }$.

This completes the proof. □

If in Theorem 3.1, $C=E$, then we have the following theorem.

Theorem 3.2 Let E and D be the same as those in Theorem  3.1. Suppose that the duality mapping $J:E\to E^{\ast }$ is weakly sequentially continuous. Let $A_i:E\to E$ ($i=1,2,\dots ,N$) and $B_j:E\to E$ ($j=1,2,\dots ,M$) be two finite families of m-accretive mappings. Let $\{e_n\}\subset E$, $\{{\alpha }_n\},\{{\beta }_n\},\{{\gamma }_n\}\subset (0,1)$, and $\{r_{n,i}\},\{s_{n,j}\}\subset (0,+\mathrm{\infty })$ satisfy the some conditions presented in Theorem  3.1.

Let $\{x_n\}$ be generated by the following scheme:

$$\begin{array}{r}{x}_1\in E,\\ u_n=(1-{\alpha }_n)(x_n+e_n),\\ v_n=(1-{\beta }_n)x_n+{\beta }_n{S}_n{u}_n,\\ x_{n+1}={\gamma }_n{x}_n+(1-{\gamma }_n)W_n{S}_n{v}_n,n\ge 1.\end{array}$$

(C)

Then $\{x_n\}$ converges strongly to a point $p_0\in D$, where $S_n$ and $W_n$ are the same as those in Theorem  3.1.

Lemma 3.7 Let E, C and ${\{B_j\}}_{j=1}^M$ be the same as those in Lemma  3.5. ${\bigcap }_{j=1}^{M}N(B_j)\ne \mathrm{\varnothing }$.

Let $c_0,c_1,\dots ,c_M$ be real numbers in $(0,1)$ such that ${\sum }_{j=0}^M{c}_j=1$ and $U_n=c_{0}I+c_1{J}_{t_{n,1}}^{B_1}+c_2{J}_{t_{n,2}}^{B_2}{J}_{t_{n,1}}^{B_1}+\cdots +c_M{J}_{t_{n,M}}^{B_M}{J}_{t_{n,M-1}}^{B_{M-1}}\cdots J_{t_{n,1}}^{B_1}$, where $J_{t_{n,j}}^{B_j}={(I+t_{n,j}{B}_j)}^{-1}$ and $t_{n,j}>0$, for $j=1,2,\dots ,M$, and $n\ge 1$, then $U_n:C\to C$ is nonexpansive and $F(U_n)={\bigcap }_{j=1}^{M}N(B_j)$, for $n\ge 1$.

Proof It is easy to check that $U_n:C\to C$ is nonexpansive and ${\bigcap }_{j=1}^{M}N(B_j)\subset F(U_n)$.

On the other hand, for $\mathrm{\forall }p\in F(U_n)$, then $p=U_{n}p=c_{0}p+c_1{J}_{t_{n,1}}^{B_1}p+c_2{J}_{t_{n,2}}^{B_2}{J}_{t_{n,1}}^{B_1}p+\cdots +c_M{J}_{t_{n,M}}^{B_M}{J}_{t_{n,M-1}}^{B_{M-1}}\cdots J_{t_{n,1}}^{B_1}p$.

For $\mathrm{\forall }q\in {\bigcap }_{j=1}^{M}N(B_j)\subset F(U_n)$, then

$$\begin{array}{rl}\parallel p-q\parallel & \le c_0\parallel p-q\parallel +c_1\parallel J_{t_{n,1}}^{B_1}p-q\parallel +\cdots +c_M\parallel J_{t_{n,M}}^{B_M}{J}_{t_{n,M-1}}^{B_{M-1}}\cdots J_{t_{n,1}}^{B_1}p-q\parallel \\ \le (c_0+c_2+\cdots +c_M)\parallel p-q\parallel +c_1\parallel J_{t_{n,1}}^{B_1}p-q\parallel \\ =(1-c_1)\parallel p-q\parallel +c_1\parallel J_{t_{n,1}}^{B_1}p-q\parallel \\ \le \parallel p-q\parallel .\end{array}$$

Therefore, $\parallel p-q\parallel =(1-c_1)\parallel p-q\parallel +c_1\parallel J_{t_{n,1}}^{B_1}p-q\parallel$, which implies that $\parallel p-q\parallel =\parallel J_{t_{n,1}}^{B_1}p-q\parallel$. Similarly, $\parallel p-q\parallel =\parallel J_{t_{n,1}}^{B_1}p-q\parallel =\parallel J_{t_{n,2}}^{B_2}{J}_{t_{n,1}}^{B_1}p-q\parallel =\cdots =\parallel J_{t_{n,M}}^{B_M}{J}_{t_{n,M-1}}^{B_{M-1}}\cdots J_{t_{n,1}}^{B_1}p-q\parallel$.

Then $\parallel p-q\parallel =\parallel \frac{c_1}{{\sum }_{j=1}^M{c}_j}(J_{t_{n,1}}^{B_1}p-q)+\frac{c_2}{{\sum }_{j=1}^M{c}_j}(J_{t_{n,2}}^{B_2}{J}_{t_{n,1}}^{B_1}p-q)+\cdots +\frac{c_M}{{\sum }_{j=1}^M{c}_j}(J_{t_{n,M}}^{B_M}{J}_{t_{n,M-1}}^{B_{M-1}}\cdots J_{t_{n,1}}^{B_1}p-q)\parallel$, which implies from the strict convexity of E that $p-q=J_{t_{n,1}}^{B_1}p-q=J_{t_{n,2}}^{B_2}{J}_{t_{n,1}}^{B_1}p-q=\cdots =J_{t_{n,M}}^{B_M}{J}_{t_{n,M-1}}^{B_{M-1}}\cdots J_{t_{n,1}}^{B_1}p-q$.

Therefore, $J_{t_{n,1}}^{B_1}p=p$, and then we can easily see that $J_{t_{n,j}}^{B_j}p=p$, for $j=2,\dots ,M$. Thus $p\in {\bigcap }_{j=1}^{M}N(B_j)$, which completes the proof. □

Lemma 3.8 Let E and C be the same as those in Lemma  3.4. Let $S_n$ and $U_n$ be the same as those in Lemmas 3.4 and 3.7, respectively. Suppose $D:=({\bigcap }_{i=1}^{N}N(A_i))\cap ({\bigcap }_{j=1}^{M}N(B_j))\ne \mathrm{\varnothing }$. Then $S_n{U}_n,U_n{S}_n:C\to C$ are nonexpansive and $F(U_n{S}_n)=F(S_n{U}_n)=D$.

Proof From Lemmas 3.4 and 3.7, we can easily check that $U_n{S}_n,S_n{U}_n:C\to C$ are nonexpansive and $F(S_n)\cap F(U_n)=D$. So, it suffices to show that $F(S_n)\cap F(U_n)\supset F(U_n{S}_n)$ since $F(S_n)\cap F(U_n)\subset F(U_n{S}_n)$ is trivial.

For $\mathrm{\forall }p\in F(U_n{S}_n)$, then $p=U_n{S}_{n}p$.

For $\mathrm{\forall }q\in F(S_n)\cap F(U_n)\subset F(U_n{S}_n)$, then $q=U_n{S}_{n}q$. Now,

$$\parallel p-q\parallel =\parallel U_n{S}_{n}p-q\parallel \le \parallel S_{n}p-S_{n}q\parallel \le \parallel p-q\parallel .$$

Then repeating the discussion in Lemma 3.4, we know that $p\in F(S_n)$. Then $p=U_n{S}_{n}p=U_{n}p$, thus $p\in F(U_n)$, which completes the proof. □

Theorem 3.3 Let E, C, $Q_C$, $S_n$, and D be the same as those in Theorem  3.1. Let $A_i,B_j:C\to E$ be m-accretive mappings, for $i=1,2,\dots N$, and $j=1,2,\dots ,M$. Suppose that the duality mapping $J:E\to E^{\ast }$ is weakly sequentially continuous and $D\ne \mathrm{\varnothing }$. Let $\{x_n\}$ be generated by the iterative algorithm (B), where $U_n:=c_{0}I+c_1{J}_{t_{n,1}}^{B_1}+c_2{J}_{t_{n,2}}^{B_2}{J}_{t_{n,1}}^{B_1}+\cdots +c_M{J}_{t_{n,M}}^{B_M}{J}_{t_{n,M-1}}^{B_{M-1}}\cdots J_{t_{n,1}}^{B_1}$, and $J_{t_{n,j}}^{B_j}={(I+t_{n,j}{B}_j)}^{-1}$, for $j=1,2,\dots ,M$, $0<c_k<1$, for $k=0,1,2,\dots ,M$, and ${\sum }_{k=0}^M{c}_k=1$. Suppose $\{e_n\}\subset E$, $\{{\alpha }_n\}$, $\{{\beta }_n\}$, and $\{{\gamma }_n\}$ are three sequences in $(0,1)$ and $\{r_{n,i}\},\{t_{n,j}\}\subset (0,+\mathrm{\infty })$ satisfy the following conditions:

• (i) ${\alpha }_n\to 0$, ${\beta }_n\to 0$, as $n\to \mathrm{\infty }$;

• (ii) ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n{\beta }_n=+\mathrm{\infty }$;

• (iii) $0<{lim\hspace{0.17em}inf}_{n\to +\mathrm{\infty }}{\gamma }_n\le {lim\hspace{0.17em}sup}_{n\to +\mathrm{\infty }}{\gamma }_n<1$;

• (iv) ${\sum }_{n=1}^{\mathrm{\infty }}|r_{n+1,i}-r_{n,i}|<+\mathrm{\infty }$ and $r_{n,i}\ge \epsilon >0$, for $n\ge 1$ and $i=1,2,\dots ,N$;

• (v) ${\sum }_{n=1}^{\mathrm{\infty }}|t_{n+1,j}-t_{n,j}|<+\mathrm{\infty }$ and $t_{n,j}\ge \epsilon >0$, for $n\ge 1$ and $j=1,2,\dots ,M$;

• (vi) $\frac{\parallel e_n\parallel }{{\alpha }_n}\to 0$, as $n\to +\mathrm{\infty }$, and ${\sum }_{n=1}^{\mathrm{\infty }}\parallel e_n\parallel <+\mathrm{\infty }$.

Then $\{x_n\}$ converges strongly to a point $p_0\in D$.

Proof We shall split the proof into five steps:

Step 1. $\{x_n\}$, $\{u_n\}$, $\{S_n{u}_n\}$, $\{v_n\}$, $\{S_n{v}_n\}$ and $\{S_n{x}_n\}$ are all bounded.

Similar to the proof of step 1 in Theorem 3.1, we can get the result of step 1.

Then $\{J_{t_{n,1}}^{B_1}{S}_n{v}_n\},\{J_{t_{n,2}}^{B_2}{J}_{t_{n,1}}^{B_1}{S}_n{v}_n\},\dots ,\{J_{t_{n,M}}^{B_M}{J}_{t_{n,M-1}}^{B_{M-1}}\cdots J_{t_{n,1}}^{B_1}{S}_n{v}_n\}$ are all bounded.

Step 2. ${lim}_{n\to \mathrm{\infty }}\parallel x_n-U_n{S}_n{v}_n\parallel =0$ and ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel =0$.

In fact,

$$\begin{array}{r}\parallel U_{n+1}{S}_{n+1}{v}_{n+1}-U_n{S}_n{v}_n\parallel \\ \le c_0\parallel S_{n+1}{v}_{n+1}-S_n{v}_n\parallel +c_1\parallel J_{t_{n+1,1}}^{B_1}{S}_{n+1}{v}_{n+1}-J_{t_{n,1}}^{B_1}{S}_n{v}_n\parallel \\ +c_2\parallel J_{t_{n+1,2}}^{B_2}{J}_{t_{n+1,1}}^{B_1}{S}_{n+1}{v}_{n+1}-J_{t_{n,2}}^{B_2}{J}_{t_{n,1}}^{B_1}{S}_n{v}_n\parallel +\cdots \\ +c_M\parallel J_{t_{n+1,M}}^{B_M}{J}_{t_{n+1,M-1}}^{B_{M-1}}\cdots J_{t_{n+1,1}}^{B_1}{S}_{n+1}{v}_{n+1}-J_{t_{n,M}}^{B_M}{J}_{t_{n,M-1}}^{B_{M-1}}\cdots J_{t_{n,1}}^{B_1}{S}_n{v}_n\parallel .\end{array}$$

(3.21)

Similar to (3.13), we know that

$$\parallel J_{t_{n+1,1}}^{B_1}{S}_{n+1}{v}_{n+1}-J_{t_{n,1}}^{B_1}{S}_n{v}_n\parallel \le \parallel S_{n+1}{v}_{n+1}-S_n{v}_n\parallel +2M_5\frac{|t_{n+1,1}-t_{n,1}|}{\epsilon },$$

(3.22)

where $M_5=sup\{\parallel S_n{v}_n\parallel ,\parallel J_{t_{n,1}}^{B_1}{S}_n{v}_n\parallel ,\parallel J_{t_{n,2}}^{B_2}{J}_{t_{n,1}}^{B_1}{S}_n{v}_n\parallel ,\dots ,\parallel J_{t_{n,M}}^{B_M}{J}_{t_{n,M-1}}^{B_{M-1}}\cdots J_{t_{n,1}}^{B_1}{S}_n{v}_n\parallel :n\ge 1\}$.

Repeating (3.22), we have

$$\begin{array}{r}\parallel J_{t_{n+1,2}}^{B_2}{J}_{t_{n+1,1}}^{B_1}{S}_{n+1}{v}_{n+1}-J_{t_{n,2}}^{B_2}{J}_{t_{n,1}}^{B_1}{S}_n{v}_n\parallel \\ \le \parallel J_{t_{n+1,1}}^{B_1}{S}_{n+1}{v}_{n+1}-J_{t_{n,1}}^{B_1}{S}_n{v}_n\parallel +\frac{2M_5}{\epsilon }|t_{n+1,2}-t_{n,2}|.\end{array}$$

(3.23)

Then (3.22) and (3.23) imply that

$$\begin{array}{r}\parallel J_{t_{n+1,2}}^{B_2}{J}_{t_{n+1,1}}^{B_1}{S}_{n+1}{v}_{n+1}-J_{t_{n,2}}^{B_2}{J}_{t_{n,1}}^{B_1}{S}_n{v}_n\parallel \\ \le \parallel S_{n+1}{v}_{n+1}-S_n{v}_n\parallel +\frac{2M_5}{\epsilon }(|t_{n+1,2}-t_{n,2}|+|t_{n+1,1}-t_{n,1}|).\end{array}$$

(3.24)

By induction, we have

$$\begin{array}{r}\parallel J_{t_{n+1,M}}^{B_M}{J}_{t_{n+1,M-1}}^{B_{M-1}}\cdots J_{t_{n,1}}^{B_1}{S}_{n+1}{v}_{n+1}-J_{t_{n,M}}^{B_M}{J}_{t_{n,M-1}}^{B_{M-1}}\cdots J_{t_{n,1}}^{B_1}{S}_n{v}_n\parallel \\ \le \parallel S_{n+1}{v}_{n+1}-S_n{v}_n\parallel \\ +\frac{2M_5}{\epsilon }(|t_{n+1,M}-t_{n,M}|+\cdots +|t_{n+1,2}-t_{n,2}|+|t_{n+1,1}-t_{n,1}|).\end{array}$$

(3.25)

Going back to (3.21), we have

$$\begin{array}{r}\parallel U_{n+1}{S}_{n+1}{v}_{n+1}-U_n{S}_n{v}_n\parallel \\ \le \parallel S_{n+1}{v}_{n+1}-S_n{v}_n\parallel \\ +\frac{2M_5}{\epsilon }(\sum _{j=1}^M{c}_j|t_{n,1}-t_{n+1,1}|+\sum _{j=2}^M{c}_j|t_{n,2}-t_{n+1,2}|+\cdots +c_M|t_{n,M}-t_{n+1,M}|).\end{array}$$

(3.26)

Therefore, similar to (3.17), we have

$$\begin{array}{r}\parallel U_{n+1}{S}_{n+1}{v}_{n+1}-U_n{S}_n{v}_n\parallel \\ \le \parallel v_{n+1}-v_n\parallel +\frac{2M_2}{\epsilon }\sum _{i=1}^N|r_{n,i}-r_{n+1,i}|\\ +\frac{2M_5}{\epsilon }(\sum _{j=1}^M{c}_j|t_{n,1}-t_{n+1,1}|+\sum _{j=2}^M{c}_j|t_{n,2}-t_{n+1,2}|+\cdots +c_M|t_{n,M}-t_{n+1,M}|)\\ \le (1+{\beta }_n)\parallel x_{n+1}-x_n\parallel +({\beta }_n+{\alpha }_n{\beta }_n)\parallel x_n\parallel +({\beta }_{n+1}+{\alpha }_{n+1}{\beta }_n)\parallel x_{n+1}\parallel \\ +|{\beta }_{n+1}-{\beta }_n|\parallel S_{n+1}{u}_{n+1}\parallel +{\beta }_n\parallel e_{n+1}-e_n\parallel +{\beta }_n\parallel {\alpha }_{n+1}{e}_{n+1}-{\alpha }_n{e}_n\parallel \\ +\frac{4M_2}{\epsilon }\sum _{i=1}^N|r_{n,i}-r_{n+1,i}|\\ +\frac{2M_5}{\epsilon }(\sum _{j=1}^M{c}_j|t_{n,1}-t_{n+1,1}|+\sum _{j=2}^M{c}_j|t_{n,2}-t_{n+1,2}|+\cdots +c_M|t_{n,M}-t_{n+1,M}|).\end{array}$$

(3.27)

Thus ${lim\hspace{0.17em}sup}_{n\to +\mathrm{\infty }}(\parallel U_{n+1}{S}_{n+1}{v}_{n+1}-U_n{S}_n{v}_n\parallel -\parallel x_{n+1}-x_n\parallel )\le 0$. Using Lemma 2.5, we have from (3.27) ${lim}_{n\to \mathrm{\infty }}\parallel x_n-U_n{S}_n{v}_n\parallel =0$ and then ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel ={lim}_{n\to \mathrm{\infty }}(1-{\gamma }_n)\parallel U_n{S}_n{v}_n-x_n\parallel =0$.

Similar to Theorem 3.1, we have

Step 3. ${lim}_{n\to \mathrm{\infty }}\parallel x_n-U_n{S}_n{x}_n\parallel =0$.

Step 4. ${lim\hspace{0.17em}sup}_{n\to +\mathrm{\infty }}〈p_0,J(p_0-x_n)〉\le 0$, where $p_0$ is an element in D.

From Lemma 3.8, we know that $U_n{S}_n:C\to C$ is nonexpansive and $F(U_n{S}_n)=D$. Then Lemma 3.1 and Lemma 3.3 imply that there exists $z_t\in C$ such that $z_t=U_n{S}_n{Q}_C[(1-t)z_t]$ for $t\in (0,1)$. Moreover, $z_t\to p_0\in D$, as $t\to 0$. Then copy step 4 in Theorem 3.1, the result follows.

Step 5. $x_n\to p_0\in D$, which is the same as that in step 4.

Copy step 5 in Theorem 3.1, the result follows.

This completes the proof. □

If in Theorem 3.3, $C=E$, then we have the following theorem.

Theorem 3.4 Let E and D be the same as those in Theorem  3.3. Suppose that the duality mapping $J:E\to E^{\ast }$ is weakly sequentially continuous. Let $A_i:E\to E$ ($i=1,2,\dots ,N$) and $B_j:E\to E$ ($j=1,2,\dots ,M$) be two finite families of m-accretive mappings. Let $\{e_n\}\subset E$, $\{{\alpha }_n\},\{{\beta }_n\},\{{\gamma }_n\}\subset (0,1)$ and $\{r_{n,i}\},\{t_{n,j}\}\subset (0,+\mathrm{\infty })$ satisfy the some conditions presented in Theorem  3.3.

Let $\{x_n\}$ be generated by the following scheme:

$$\begin{array}{r}{x}_1\in E,\\ u_n=(1-{\alpha }_n)(x_n+e_n),\\ v_n=(1-{\beta }_n)x_n+{\beta }_n{S}_n{u}_n,\\ x_{n+1}={\gamma }_n{x}_n+(1-{\gamma }_n)U_n{S}_n{v}_n,n\ge 1.\end{array}$$

(D)

Then $\{x_n\}$ converges strongly to a point $p_0\in D$, where $S_n$ and $U_n$ are the same as those in Theorem  3.3.

Next, we apply Theorems 3.1 and 3.3 to the cases of finite pseudo-contractive mappings.

Theorem 3.5 Let E be a real uniformly smooth and uniformly convex Banach space. Let C be a nonempty, closed, and convex sunny nonexpansive retract of E, where $Q_C$ is the sunny nonexpansive retraction of E onto C. Let $T_i^{(1)},T_j^{(2)}:C\to E$ be pseudo-contractive mappings such that $(I-T_i^{(1)})$ and $(I-T_j^{(2)})$ are m-accretive, where $i=1,2,\dots N$, $j=1,2,\dots ,M$. Suppose that the duality mapping $J:E\to E^{\ast }$ is weakly sequentially continuous and $D:=({\bigcap }_{i=1}^{N}F(T_i^{(1)}))\cap ({\bigcap }_{j=1}^{M}F(T_j^{(2)}))\ne \mathrm{\varnothing }$. Let $\{x_n\}$ be generated by the iterative algorithm (A), where $S_n:=a_{0}I+a_1{J}_{r_{n,1}}^{I-T_1^{(1)}}+a_2{J}_{r_{n,2}}^{I-T_2^{(1)}}+\cdots +a_N{J}_{r_{n,N}}^{I-T_N^{(1)}}$, and $J_{r_{n,i}}^{I-T_i^{(1)}}={[I+r_{n,i}(I-T_i^{(1)})]}^{-1}$, for $i=1,2,\dots ,N$, $0<a_k<1$, for $k=0,1,2,\dots ,N$, ${\sum }_{k=0}^N{a}_k=1$. $W_n=b_{0}I+b_1{J}_{s_{n,1}}^{I-T_1^{(2)}}+b_2{J}_{s_{n,2}}^{I-T_2^{(2)}}+\cdots +b_M{J}_{s_{n,M}}^{I-T_M^{(2)}}$, where $J_{s_{n,j}}^{I-T_j^{(2)}}={[I+s_{n,j}(I-T_j^{(2)})]}^{-1}$, for $j=1,2,\dots ,M$, $0<b_k<1$, for $k=0,1,2,\dots ,M$, ${\sum }_{k=0}^M{b}_k=1$. Suppose $\{e_n\}\subset E$, $\{{\alpha }_n\}$, $\{{\beta }_n\}$, and $\{{\gamma }_n\}$ are three sequences in $(0,1)$ and $\{r_{n,i}\},\{s_{n,j}\}\subset (0,+\mathrm{\infty })$ satisfying the following conditions:

• (i) ${\alpha }_n\to 0$, ${\beta }_n\to 0$, as $n\to \mathrm{\infty }$;

• (ii) ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n{\beta }_n=+\mathrm{\infty }$;

• (iii) $0<{lim\hspace{0.17em}inf}_{n\to +\mathrm{\infty }}{\gamma }_n\le {lim\hspace{0.17em}sup}_{n\to +\mathrm{\infty }}{\gamma }_n<1$;

• (iv) ${\sum }_{n=1}^{\mathrm{\infty }}|r_{n+1,i}-r_{n,i}|<+\mathrm{\infty }$ and $r_{n,i}\ge \epsilon >0$, for $n\ge 1$ and $i=1,2,\dots ,N$;

• (v) ${\sum }_{n=1}^{\mathrm{\infty }}|s_{n+1,j}-s_{n,j}|<+\mathrm{\infty }$ and $s_{n,j}\ge \epsilon >0$, for $n\ge 1$ and $j=1,2,\dots ,M$;

• (vi) $\frac{\parallel e_n\parallel }{{\alpha }_n}\to 0$, as $n\to +\mathrm{\infty }$, and ${\sum }_{n=1}^{\mathrm{\infty }}\parallel e_n\parallel <+\mathrm{\infty }$.

Then $\{x_n\}$ converges strongly to a point $p_0\in D$.

Proof Let $A_i=(I-T_i^{(1)})$ and $B_j=(I-T_j^{(2)})$, for $i=1,2,\dots ,N$ and $j=1,2,\dots ,M$. Then the result follows from Theorem 3.1. □

Similarly, from Theorem 3.3, we have the following result.

Theorem 3.6 Let E, C, $Q_C$ and D be the same as those in Theorem  3.5. Let $T_i^{(1)},T_j^{(2)}:C\to E$ be pseudo-contractive mappings such that $(I-T_i^{(1)})$ and $(I-T_j^{(2)})$ are m-accretive mappings, where $i=1,2,\dots N$, $j=1,2,\dots ,M$. Suppose that the duality mapping $J:E\to E^{\ast }$ is weakly sequentially continuous and $D\ne \mathrm{\varnothing }$. Let $\{x_n\}$ be generated by the iterative algorithm (B), where $S_n$ is the same as that in Theorem  3.5 and $U_n=c_{0}I+c_1{J}_{t_{n,1}}^{I-T_1^{(2)}}+c_2{J}_{t_{n,2}}^{I-T_2^{(2)}}{J}_{t_{n,1}}^{I-T_1^{(2)}}+\cdots +c_M{J}_{t_{n,M}}^{I-T_M^{(2)}}{J}_{t_{n,M-1}}^{I-T_{M-1}^{(2)}}\cdots J_{t_{n,1}}^{I-T_1^{(2)}}$, where $J_{t_{n,j}}^{I-T_j^{(2)}}={[I+t_{n,j}(I-T_j^{(2)})]}^{-1}$, for $j=1,2,\dots ,M$, $0<c_k<1$, for $k=0,1,2,\dots ,M$, ${\sum }_{k=0}^M{c}_k=1$. Suppose $\{e_n\}\subset E$, $\{{\alpha }_n\}$, $\{{\beta }_n\}$, and $\{{\gamma }_n\}$ are three sequences in $(0,1)$ and $\{r_{n,i}\},\{t_{n,j}\}\subset (0,+\mathrm{\infty })$ satisfying the following conditions:

• (i) ${\alpha }_n\to 0$, ${\beta }_n\to 0$, as $n\to \mathrm{\infty }$;

• (ii) ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n{\beta }_n=+\mathrm{\infty }$;

• (iii) $0<{lim\hspace{0.17em}inf}_{n\to +\mathrm{\infty }}{\gamma }_n\le {lim\hspace{0.17em}sup}_{n\to +\mathrm{\infty }}{\gamma }_n<1$;

• (iv) ${\sum }_{n=1}^{\mathrm{\infty }}|r_{n+1,i}-r_{n,i}|<+\mathrm{\infty }$ and $r_{n,i}\ge \epsilon >0$, for $n\ge 1$ and $i=1,2,\dots ,N$;

• (v) ${\sum }_{n=1}^{\mathrm{\infty }}|t_{n+1,j}-t_{n,j}|<+\mathrm{\infty }$ and $s_{n,j}\ge \epsilon >0$, for $n\ge 1$ and $j=1,2,\dots ,M$;

• (vi) $\frac{\parallel e_n\parallel }{{\alpha }_n}\to 0$, as $n\to +\mathrm{\infty }$, and ${\sum }_{n=1}^{\mathrm{\infty }}\parallel e_n\parallel <+\mathrm{\infty }$.

Then $\{x_n\}$ converges strongly to a point $p_0\in D$.