Iterative algorithms for infinite accretive mappings and applications to p-Laplacian-like differential systems

Abstract

Some new iterative algorithms with errors for approximating common zero point of an infinite family of m-accretive mappings in a real Banach space are presented. A path convergence theorem and some new weak and strong convergence theorems are proved by means of some new techniques, which extend the corresponding works by some authors. As applications, an infinite p-Laplacian-like differential system is investigated, from which we construct an infinite family of m-accretive mappings and discuss the connections between the equilibrium solution of the differential systems and the zero point of the m-accretive mappings.

1 Introduction

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, respectively. We denote the value of $f \in E^{*}$ at $x \in E$ by $\langle x,f \rangle$.

Define a function $\rho_{E} : [0, +\infty) \rightarrow[0, +\infty)$ called the modulus of smoothness of E as follows:

$$\rho_{E}(t) = \sup \biggl\{ \frac{\|x+y\|+\|x-y\|}{2}-1: x,y\in E, \|x\| = 1, \| y\| \leq t\biggr\} . $$

A Banach space E is said to be uniformly smooth if $\frac{\rho _{E}(t)}{t}\rightarrow 0$, as $t \rightarrow0$.

A Banach space E is said to be strictly convex if and only if $\|x\|= \|y\|= \|(1-\lambda)x+\lambda y\|$ for $x, y \in E$ and $0 < \lambda<1$ implies that $x = y$. A Banach space E is said to be uniformly convex if for any $\varepsilon\in(0,2]$ there exists $\delta>0$ such that

$$\|x\| = \|y\| = 1, \qquad \|x-y\| \geq\varepsilon\quad \Rightarrow\quad \biggl\Vert \frac {x+y}{2}\biggr\Vert \leq 1-\delta. $$

It is well known that a uniformly convex Banach space is reflexive and strictly convex.

An operator $B: E\rightarrow E^{*}$ is said to be monotone if $\langle u - v, Bu - Bv\rangle\geq0$, for all $u,v \in D(B)$. The monotone operator B is said to be maximal monotone if the graph of B, $G(B)$, is not contained properly in any other monotone subset of $E \times E^{*}$.

A single-valued mapping $F:D(F) = E \rightarrow E^{*}$ is said to be hemi-continuous [1] if $w\mbox{-}\!\lim_{t\rightarrow0}F(x+ty) = Fx$, for any $x,y\in E$. A single-valued mapping $F:D(F) = E \rightarrow E^{*}$ is said to be demi-continuous [1] if $w\mbox{-}\!\lim_{n\rightarrow\infty}Fx_{n} = Fx$, for any sequence $\{x_{n}\}$ strongly convergent to x in E.

Following from [1] or [2], the function h is said to be a proper convex function on E if h is defined from E onto $(-\infty, +\infty]$, h is not identically +∞ such that $h((1-\lambda)x+\lambda y)\leq(1-\lambda)h(x)+\lambda h(y)$, whenever $x,y \in E$ and $0 \leq\lambda\leq1$. h is said to be strictly convex if $h((1-\lambda)x+\lambda y)< (1-\lambda)h(x)+\lambda h(y)$, for all $0 < \lambda< 1$ and $x,y \in E$ with $x \neq y$, $h(x) <+\infty$ and $h(y) < +\infty$. The function $h: E \rightarrow(-\infty, +\infty]$ is said to be lower-semi-continuous on E if $\liminf_{y \rightarrow x}h(y) \geq h(x)$, for any $x \in E$.

A continuous strictly increasing function $\varphi: [0,+\infty) \rightarrow[0,+\infty)$ is called a gauge function [2] if $\varphi(0) = 0$ and $\varphi(t) \rightarrow\infty$, as $t \rightarrow\infty$. The duality mapping $J_{\varphi}: E \rightarrow2^{E^{*}}$ associated with the gauge function φ is defined by [2]

$$J_{\varphi}(x) = \bigl\{ f \in E^{*}: \langle x, f \rangle= \|x\|\varphi\bigl( \Vert x\Vert \bigr), \|f\|= \varphi\bigl(\Vert x\Vert \bigr)\bigr\} ,\quad x \in E. $$

It can be seen from [2] that the duality mapping $J_{\varphi}$ has the following properties:

• (i) $J_{\varphi}(-x) = - J_{\varphi}(x)$ and $J_{\varphi}(kx)= \frac{\varphi(\|kx\|)}{\varphi(\|x\|)} J_{\varphi}(x)$, for $\forall x \in E$ and $k >0$;

• (ii) if $E^{*}$ is uniformly convex, then $J_{\varphi}$ is uniformly continuous on each bounded subset in E;

• (iii) the reflexivity of E and strict convexity of $E^{*}$ imply that $J_{\varphi}$ is single-valued, monotone and demi-continuous.

In the case $\varphi(t)\equiv t$, we call $J_{\varphi}$ the normalized duality mapping, which is usually denoted by J.

For the gauge function φ, the function $\Phi: [0,+\infty) \rightarrow[0,+\infty)$ defined by

$$ \Phi(t)= \int_{0}^{t} \varphi(s)\, ds $$

(1.1)

is a continuous convex strictly increasing function on $[0,+\infty)$.

Following the result in [3], a Banach space E is said to have a weakly continuous duality mapping if there is a gauge φ for which the duality mapping $J_{\varphi}(x)$ is single-valued and weak-to-weak^∗ sequentially continuous (i.e., if $\{ x_{n}\}$ is a sequence in E weakly convergent to a point x, then the sequence $J_{\varphi}(x_{n})$ converges weakly^∗ to $J_{\varphi}(x)$). It is well known that $l^{p}$ has a weakly continuous duality mapping with a gauge function $\varphi(t) = t^{p-1}$ for all $1 < p < +\infty$.

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 [4] 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 [4] 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 $f : C \rightarrow C$ is called a contraction with contractive constant $k \in(0,1)$ if $\|f(x) - f(y)\| \leq k \|x - y\|$, for $\forall x,y \in C$.

A mapping $T : C \rightarrow C$ is said to be nonexpansive if $\|Tx - Ty\| \leq\|x-y\|$, for $\forall x,y \in C$. We use $\operatorname{Fix}(T)$ to denote the fixed point set of T. That is, $\operatorname{Fix}(T) : = \{x\in C: Tx = x\}$. A mapping $T : E \supset D(T) \rightarrow R(T) \subset E$ is said to be demi-closed 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 $Tx =p$.

A subset C of E is said to be a sunny nonexpansive retract of E [5, 6] 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.

A mapping $A : D(A) \subset E \rightarrow E$ is said to be accretive if $\|x_{1}-x_{2}\| \leq\|x_{1}-x_{2}+r(y_{1}-y_{2})\|$, for $\forall x_{i} \in D(A)$, $y_{i} \in Ax_{i}$, $i = 1,2$, and $r>0$. For the accretive mapping A, we use $N(A)$ to denote the set of zero points of it; that is, $N(A): = \{x \in D(A) : Ax = 0\}$. If A is accretive, then we can define, for each $r>0$, a nonexpansive 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 [1]. We also know that, for an accretive mapping A, $N(A) = \operatorname{Fix}(J_{r}^{A})$. An accretive mapping A is said to be m-accretive if $R(I+\lambda A) = E$, for $\forall\lambda > 0$.

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 [7–15] and the references 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. [7]), presented in a Hilbert space:

$$ x_{0} \in H,\quad x_{n+1} \approx J_{r_{n}}^{A}x_{n}, \quad n \geq0, $$

(1.2)

where $J_{r_{n}}^{A}:= (I+r_{n} A)^{-1}$. It was shown that the sequence generated by (1.2) converges weakly or strongly to a zero point of A under some conditions.

An explicit iterative process to approximate fixed point of a nonexpansive mapping $T : C \rightarrow C$ was introduced in 1967 by Halpern [16] in the frame of Hilbert spaces:

$$ u\in C, x_{0} \in C, \quad x_{n+1}= \alpha_{n} u +(1-\alpha_{n})Tx_{n}, \quad n \geq 0, $$

(1.3)

where $\{\alpha_{n}\} \subset[0,1]$.

In 2007, based on (1.2) and (1.3), Qin and Su [9] presented the following iterative algorithm:

$$ \left \{ \textstyle\begin{array}{l} x_{1}\in C\quad \text{chosen arbitrarily}, \\ y_{n} = \beta_{n} x_{n} + (1-\beta_{n})J_{r_{n}}^{A}x_{n}, \\ x_{n+1}= \alpha_{n} u+(1-\alpha_{n})y_{n}. \end{array}\displaystyle \right . $$

(1.4)

They showed that $\{x_{n}\}$ generated by (1.4) converges strongly to a zero point of an m-accretive mapping A.

Motivated by iterative algorithms (1.2) and (1.3), Zegeye and Shahzad extended their discussion to the case of finite m-accretive mappings $\{A_{i}\}_{i = 1}^{l}$. They presented in [17] the following iterative algorithm:

$$ x_{0} \in C, \quad x_{n+1}= \alpha_{n} u + (1- \alpha_{n}) S_{r}x_{n},\quad n \geq0, $$

(1.5)

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 $\bigcap_{i=1}^{l} N(A_{i}) \neq\emptyset$, they proved that $\{x_{n}\}$ generated by (1.5) converges strongly to the common zero point of $A_{i}$ ($i = 1,2,\ldots,l$) under some conditions.

The work in [17] was then extended to the following one presented by Hu and Liu in [18]:

$$ x_{0} \in C,\quad x_{n+1}= \alpha_{n} u + \beta_{n} x_{n} + \vartheta_{n} S_{r_{n}}x_{n}, \quad n \geq0, $$

(1.6)

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 $\sum_{i = 0}^{l} a_{i} = 1$. $\{\alpha_{n}\}, \{\beta_{n}\}, \{\vartheta_{n}\} \subset(0,1)$ and $\alpha_{n} + \beta_{n} + \vartheta_{n} = 1$. If $\bigcap_{i=1}^{l} N(A_{i}) \neq\emptyset$, they proved that $\{x_{n}\}$ converges strongly to the common point in $N(A_{i})$ ($i = 1,2,\ldots,l$) under some conditions.

In 2009, Yao et al. presented the following iterative algorithm in the frame of Hilbert space in [19]:

$$ \left \{ \textstyle\begin{array}{l} x_{1}\in C, \\ y_{n} = P_{C}[(1-\alpha_{n})x_{n}], \\ x_{n+1}= (1- \beta_{n})x_{n} + \beta_{n} Ty_{n},\quad n \geq1, \end{array}\displaystyle \right . $$

(1.7)

where $T : C \rightarrow C$ is nonexpansive with $\operatorname{Fix}(T)\neq\emptyset$. Suppose $\{\alpha_{n}\}$ and $\{\beta_{n}\}$ are two real sequences in $(0,1)$ satisfying

• (a) $\sum_{n=1}^{\infty} \alpha_{n} = +\infty$ and $\lim_{n \rightarrow\infty} \alpha_{n} = 0$;

• (b) $0 < \liminf_{n \rightarrow\infty}\beta_{n} \leq \limsup_{n \rightarrow\infty}\beta_{n} < 1$.

Then $\{x_{n}\}$ constructed by (1.7) converges strongly to a point in $\operatorname{Fix}(T)$.

Motivated by the work in [17] and [19], Shehu and Ezeora [5] presented the following iterative algorithm and the discussion is undertaken in the frame of a real uniformly smooth and uniformly convex Banach space:

$$ \left \{ \textstyle\begin{array}{l} x_{1}\in C, \\ y_{n} = Q_{C}[(1-\alpha_{n})x_{n}], \\ x_{n+1}= (1- \beta_{n})x_{n} + \beta_{n} S_{N}y_{n},\quad n \geq1. \end{array}\displaystyle \right . $$

(1.8)

Here $Q_{C}$ is the sunny nonexpansive retraction of E onto C. $A_{i} : C \rightarrow E$ is m-accretive mapping with $\bigcap_{i=1}^{N} N(A_{i}) \neq\emptyset$. $S_{N} : = a_{0} I + a_{1}J_{A_{1}}+a_{2}J_{A_{2}}+ \cdots+a_{N}J_{A_{N}}$ with $J_{A_{i}} = (I+A_{i})^{-1}$, for $i = 1,2,\ldots, N$. $0 < a_{k} <1$, for $k = 0,1,2,\ldots, N$, and $\sum_{k=0}^{N} a_{k} = 1$. $\{\alpha_{n}\}, \{\beta_{n}\}\subset(0,1)$. Then $\{x_{n}\}$ converges strongly to the common zero point of $A_{i}$, where $i = 1,2,\ldots,N$.

In 2014, by modifying iterative algorithm (1.8) and employing new techniques, Wei and Tan [20] presented and studied the following three-step iterative algorithm:

$$ \left \{ \textstyle\begin{array}{l} x_{1}\in C, \\ u_{n} = Q_{C}[(1-\alpha_{n})(x_{n}+e_{n})], \\ v_{n}= (1- \beta_{n})x_{n} + \beta_{n} S_{n} u_{n}, \\ x_{n+1}=\gamma_{n} x_{n}+(1-\gamma_{n})S_{n}v_{n}, \quad n \geq1, \end{array}\displaystyle \right . $$

(1.9)

where $Q_{C}$ is the sunny nonexpansive retraction of E onto C, $\{e_{n}\} \subset E$ 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 $i = 1,2,\ldots,N$, $\sum_{k = 0}^{N}a_{k} = 1$, $0 < a_{k} < 1$, for $k = 0 , 1, 2, \ldots, N$. And, some strong convergence theorems to approximate common zero point of $A_{i}$ ($i = 1,2, \ldots, N$) are obtained.

In 2015, Wang and Zhang [21] extended the discussion of the finite family of m-accretive mappings $\{A_{i}\}_{i = 1}^{N}$ to that of infinite family of m-accretive mappings $\{A_{i}\}_{i = 1}^{\infty}$. They presented the following two-step iterative algorithms with errors $\{e_{n}\} \subset E$:

$$ \left\{ \textstyle\begin{array}{l} x_{1} \in\bigcap_{i = 1}^{\infty} \overline{D(A_{i})}\quad \mbox{chosen arbitrarily}, \\ y_{n} = \alpha_{n}'x_{n}+ \beta_{n}'\sum_{i = 1}^{\infty} \delta _{n,i}J_{r_{i}}x_{n} + \gamma'_{n} e_{n}, \\ x_{n+1} = \alpha_{n}f(x_{n})+\beta_{n} x_{n} + \gamma_{n} y_{n},\quad n \geq1, \end{array}\displaystyle \right. $$

(1.10)

where f is a contraction on $\bigcap_{i = 1}^{\infty}\overline{D(A_{i})}$. For $i = 1,2,\ldots$ , $J_{r_{i}} = (I + r_{i} A_{i} )^{-1}$. $\{\alpha_{n}\}$, $\{\beta_{n}\}$, $\{\gamma_{n}\}$, $\{\alpha_{n}'\}$, $\{\beta_{n}'\}$, and $\{\gamma_{n}'\}$ are real number sequences in $(0,1)$. Some strong convergence theorems to approximate common zero point of $A_{i}$ ($i = 1,2, \ldots$) are obtained under some conditions.

Inspired by the work in [5, 9, 17–21], we shall design a four-step iterative algorithm with errors in a Banach space in Section 3. Some weak and strong convergence theorems for approximating common zero point of an infinite family of m-accretive mappings are obtained. Some new proof techniques can be found. In Section 4, we shall present an example of infinite p-Laplacian-like differential systems, to highlight the significance of the studies on iterative construction for zero points of accretive mappings in applied mathematics and engineering. We demonstrate the applications of the main results in Section 3.

Our main contributions are:

• (i) a new four-step iterative algorithm is designed by combining the ideas of famous iterative algorithms such as proximal methods, Halpern methods, convex combination methods, and viscosity methods;

• (ii) three sequences constructed in the new iterative algorithm are proved to be weakly or strongly to the common zero point of an infinite family of m-accretive mappings;

• (iii) the characteristic of the weakly convergent point of the new iterative algorithm is pointed out;

• (iv) under the new assumptions, a path convergence theorem for nonexpansive mappings is proved;

• (v) some new techniques are employed, for example, the tool $\|\cdot\|^{2}$ for estimating the convergence of the iterative sequence $\{x_{n}\}$ in most of the existing related work is partly replaced by function Φ defined by (1.1);

• (vi) the discussion is undertaken in the frame of a Banach space, which is more general than that in Hilbert space; the assumption that ‘the normalized duality mapping J is weakly sequentially continuous’ in most of the existing related work is weakened to ‘$J_{\varphi}$ is weakly sequentially continuous for a given gauge function φ’;

• (vii) compared to (1.10), $J_{r_{i}}$ is replaced by $J_{r_{n,i}}$; compared to (1.9), a contraction f is considered; compared to the work in [5, 9, 17–20], an infinite family of m-accretive mappings is discussed;

• (viii) in Section 4, the applications of the main results in Section 3 on approximating the equilibrium solution of the nonlinear p-Laplacian-like differential systems are demonstrated.

2 Preliminaries

Now, we list some results we need in the sequel.

Lemma 2.1

(see [22])

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 nonexpansive 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 nonexpansive with $\operatorname{Fix}(\sum_{m = 1}^{\infty}a_{m}T_{m}) = \bigcap_{m = 1}^{\infty}\operatorname{Fix}(T_{m})$.

Lemma 2.2

(see [23])

Assume that a real Banach space E has a weakly continuous duality mapping $J_{\varphi}$ with a gauge φ. Then Φ defined by (1.1) has the following properties.

• (i) $\Phi(\|x+y\|) \leq\Phi(\|x\|) + \langle y, J_{\varphi}(x+y) \rangle$, $\forall x,y \in E$.

• (ii) Assume that a sequence $\{x_{n}\}$ in E converges weakly to a point $x \in E$. Then

  $$\limsup_{n \rightarrow\infty} \Phi\bigl(\Vert x_{n} - y\Vert \bigr) = \limsup_{n \rightarrow \infty} \Phi\bigl(\Vert x_{n} - x \Vert \bigr)+\Phi\bigl(\Vert y-x\Vert \bigr), \quad \forall y \in E. $$

Lemma 2.3

(see [24])

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

$$a_{n+1}\leq(1-t_{n})a_{n} + b_{n}+c_{n}, \quad \forall n \geq1, $$

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

Lemma 2.4

(see [20])

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 $n \geq1$. Suppose $\{\beta_{n}\}\subset(0,1)$ satisfying $0 < \liminf_{n\rightarrow +\infty}\beta_{n} \leq \limsup_{n\rightarrow+\infty}\beta_{n} < 1$. If $\limsup_{n\rightarrow+\infty}(\|y_{n+1}-y_{n}\|-\|x_{n+1}-x_{n}\|) \leq0$, then $\lim_{n\rightarrow+\infty}\|y_{n}-x_{n}\| = 0$.

Lemma 2.5

(see [25])

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

Lemma 2.6

(see [1])

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

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

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

Lemma 2.7

(see [20])

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 let $Q_{C}$ be the sunny nonexpansive retraction of E onto C. Let $T : C \rightarrow C$ be a nonexpansive mapping with $\operatorname{Fix}(T) \neq \emptyset$. If for each $t \in(0,1)$, define $T_{t} : C \rightarrow C$ by

$$T_{t} x : = TQ_{C}\bigl[(1-t)x\bigr]. $$

Then $T_{t}$ is a contraction and has a fixed point $z_{t}$, which satisfies $\|z_{t} - Tz_{t}\|\rightarrow0$, as $t \rightarrow 0$.

3 Weak and strong convergence theorems

Lemma 3.1

Let E be a real strictly convex Banach space and C be a nonempty, closed, and convex subset of E. Let $A_{i}: C \rightarrow E$ be m-accretive mappings, where $i = 1,2,\ldots$ . Suppose $D: = \bigcap_{i = 1}^{\infty} N(A_{i}) \neq\emptyset$ and $\{r_{n,i}\}\subset(0,+\infty)$ for $i = 1,2, \ldots$ . If $\{a_{i}\}_{i = 0}^{\infty}\subset(0,1)$ satisfies $\sum_{i = 0}^{\infty}a_{i} = 1$. Then $(a_{0} I + \sum_{i =1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}}) : E \rightarrow E$ is nonexpansive and

$$\operatorname{Fix}\Biggl(a_{0} I + \sum_{i =1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}} \Biggr) = D, $$

where $J_{r_{n,i}}^{A_{i}} = (I+r_{n,i}A_{i})^{-1}$ for $n \geq1$ and $i = 1,2,\ldots$ .

Proof

If we set $T_{1} = I$ and $T_{i+1} = J_{r_{n,i}}^{A_{i}}$, then $(a_{0} I + \sum_{i =1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}}) = \sum_{i = 0 }^{\infty}a_{i}T_{i+1}$. Since both I and $J_{r_{n,i}}^{A_{i}}$ are nonexpansive, Lemma 2.1 implies that $(a_{0} I + \sum_{i =1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}})$ is nonexpansive and

$$\operatorname{Fix}\Biggl(a_{0} I + \sum_{i =1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}} \Biggr) = \bigcap_{i =1}^{\infty} \operatorname{Fix}\bigl(J_{r_{n,i}}^{A_{i}}\bigr) = \bigcap _{i =1}^{\infty} N(A_{i}) = D. $$

This completes the proof. □

Theorem 3.1

Let E be a real strictly convex Banach space which has a weakly continuous duality mapping $J_{\varphi}$. 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 $f : C \rightarrow C$ be a contraction with contractive constant $k \in (0,1)$. Let $A_{i}: C \rightarrow E$ be m-accretive mappings, for $i = 1,2,\ldots$ . Let $D : = \bigcap_{i = 1}^{\infty}N(A_{i})\neq \emptyset$. Suppose $\{\alpha_{n}\}, \{\beta_{n}\}, \{\mu_{n}\}, \{\gamma_{n}\}, \{\delta_{n}\}, \{\zeta_{n}\} \subset(0,1)$, and $\{r_{n,i}\}\subset(0,+\infty)$ for $i = 1,2, \ldots$ . Suppose $\{a_{i}\}_{i = 0}^{\infty}\subset(0,1)$ with $\sum_{i = 0}^{\infty}a_{i} = 1$ and $\{e_{n}\} \subset E$ is the error sequence. Let $\{x_{n}\}$ be generated by the following iterative algorithm:

$$ \left \{ \textstyle\begin{array}{l} x_{1}\in C, \\ u_{n} = Q_{C}[(1-\alpha_{n})(x_{n}+e_{n})], \\ v_{n}= (1- \beta_{n})x_{n} + \beta_{n} (a_{0}I+\sum_{i = 1}^{\infty }a_{i}J_{r_{n,i}}^{A_{i}})u_{n}, \\ w_{n} = \mu_{n} f(x_{n}) + \gamma_{n} x_{n} + \delta_{n} v_{n}, \\ x_{n+1}= (1-\zeta_{n}) w_{n} + \zeta_{n} x_{n}, \quad n \geq 1. \end{array}\displaystyle \right . $$

(3.1)

Further suppose that the following conditions are satisfied:

• (i) $\alpha_{n} \rightarrow0$, $\delta_{n} \rightarrow0$, $\mu_{n} \rightarrow0$, as $n \rightarrow\infty$;

• (ii) $\mu_{n} + \gamma_{n}+\delta_{n} \equiv1$, $n \geq1$;

• (iii) $0 < \liminf_{n \rightarrow+\infty}\beta_{n}\leq \limsup_{n \rightarrow+\infty} \beta_{n} < 1$ and $0 < \liminf_{n \rightarrow+\infty}\zeta_{n}\leq \limsup_{n \rightarrow+\infty} \zeta_{n} < 1$;

• (iv) $\sum_{n=1}^{\infty}|r_{n+1,i} - r_{n,i}| < +\infty$ and $r_{n,i}\geq\varepsilon> 0$, for $n \geq1$ and $i = 1,2,\ldots$ ;

• (v) $\gamma_{n+1}-\gamma_{n} \rightarrow0$, $\beta_{n+1}-\beta_{n} \rightarrow0$, as $n \rightarrow\infty$;

• (vi) $\sum_{n=1}^{\infty}\|e_{n}\| < +\infty$.

Then the three sequences $\{x_{n}\}$, $\{u_{n}\}$, and $\{w_{n}\}$ converge weakly to the unique element $q_{0} \in D$, which satisfies, for $\forall y \in D$,

$$ \limsup_{n \rightarrow\infty}\Phi\bigl(\Vert x_{n} - q_{0}\Vert \bigr)= \min_{y \in D}\limsup _{n \rightarrow\infty}\Phi\bigl(\Vert x_{n} - y\Vert \bigr). $$

(3.2)

Proof

We shall split the proof into five steps.

Step 1. $\{x_{n}\}$, $\{u_{n}\}$, $\{\sum_{i = 1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}}u_{n}\}$, $\{v_{n}\}$, and $\{f(x_{n})\}$ are all bounded.

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

$$ \|x_{n+1}-p\| \leq M_{1} + \sum_{i = 1}^{n} \|e_{i}\|, $$

(3.3)

where $M_{1} = \max \{\|x_{1} -p\|, \frac{\|f(p)-p\|}{1-k}, \|p\|\}$.

By using Lemma 3.1 and the induction method, we see that, for $n=1$, $\forall p \in D$,

$$\begin{aligned} \Vert x_{2} - p\Vert \leq&(1-\zeta_{1})\Vert w_{1}-p\Vert +\zeta_{1}\Vert x_{1}-p\Vert \\ \leq&(1-\zeta_{1})\bigl[\mu_{1} \bigl\Vert f(x_{1}) - p\bigr\Vert + \gamma_{1} \Vert x_{1} - p\Vert +\delta_{1} \Vert v_{1}-p\Vert \bigr]+ \zeta_{1}\Vert x_{1}-p\Vert \\ \leq&(1-\zeta_{1})\Biggl[k \mu_{1} \Vert x_{1} - p\Vert + \mu_{1} \bigl\Vert f(p)-p\bigr\Vert + \gamma_{1}\Vert x_{1}-p\Vert \\ &{}+\delta_{1}(1-\beta_{1})\Vert x_{1}-p\Vert +\delta_{1}\beta_{1} \Biggl\Vert \Biggl(a_{0}I+ \sum_{i = 1}^{\infty}a_{i}J_{r_{1,i}}^{A_{i}} \Biggr)u_{1} - p\Biggr\Vert \Biggr]+\zeta_{1}\Vert x_{1}-p\Vert \\ \leq&(1-\zeta_{1})\bigl[k\mu_{1}+\gamma_{1}+ \delta_{1}(1-\beta_{1})\bigr]\Vert x_{1}-p\Vert + (1-\zeta_{1})\mu_{1}\bigl\Vert f(p)-p\bigr\Vert + \zeta_{1}\Vert x_{1}-p\Vert \\ &{}+ (1-\zeta_{1})\delta_{1}\beta_{1}\Vert u_{1}-p\Vert \\ \leq& (1-\zeta_{1})\bigl[k\mu_{1}+\gamma_{1}+ \delta_{1}(1-\beta_{1})\bigr]\Vert x_{1}-p\Vert + (1-\zeta_{1})\mu_{1}\bigl\Vert f(p)-p\bigr\Vert + \zeta_{1}\Vert x_{1}-p\Vert \\ &{}+ (1-\zeta_{1})\delta _{1}\beta_{1}\bigl\Vert (1-\alpha_{1}) (x_{1}+e_{1})-p\bigr\Vert \\ \leq& \bigl[1-(1-k)\mu_{1}(1-\zeta_{1})- \alpha_{1}\beta_{1}\delta_{1}(1-\zeta_{1}) \bigr] \Vert x_{1} - p\Vert + (1-\zeta_{1}) (1- \alpha_{1})\beta_{1}\delta_{1}\Vert e_{1}\Vert \\ &{}+(1-\zeta_{1})\alpha_{1}\beta_{1} \delta_{1}\Vert p\Vert +(1-\zeta_{1})\mu_{1}(1-k) \frac{\Vert f(p)-p\Vert }{1-k} \\ \leq& M_{1} + \Vert e_{1}\Vert . \end{aligned}$$

Suppose that (3.3) is true for $n = k$. Then, for $n = k+1$,

$$\begin{aligned} \Vert x_{k+2} - p\Vert \leq&(1-\zeta_{k+1})\Vert w_{k+1}-p\Vert +\zeta _{k+1}\Vert x_{k+1}-p\Vert \\ \leq&(1-\zeta_{k+1})\bigl[\mu_{k+1} \bigl\Vert f(x_{k+1}) - p\bigr\Vert + \gamma_{k+1} \Vert x_{k+1} - p\Vert +\delta_{k+1} \Vert v_{k+1}-p \Vert \bigr] \\ &{}+\zeta_{k+1}\Vert x_{k+1}-p\Vert \\ \leq&(1-\zeta_{k+1})\Biggl[k \mu_{k+1} \Vert x_{k+1} - p\Vert + \mu_{k+1} \bigl\Vert f(p)-p\bigr\Vert +\gamma_{k+1}\Vert x_{k+1}-p\Vert \\ &{}+\delta_{k+1}(1-\beta_{k+1})\Vert x_{k+1}-p \Vert +\delta_{k+1}\beta_{k+1} \Biggl\Vert \Biggl(a_{0}I+\sum_{i = 1}^{\infty}a_{i}J_{r_{{k+1},i}}^{A_{i}} \Biggr)u_{k+1} - p\Biggr\Vert \Biggr] \\ &{}+\zeta_{k+1}\Vert x_{k+1}-p\Vert \\ \leq& (1-\zeta_{k+1})\bigl[k\mu_{k+1}+\gamma_{k+1}+ \delta_{k+1}(1-\beta_{k+1})\bigr]\Vert x_{k+1}-p \Vert \\ &{}+ (1-\zeta_{k+1})\mu_{k+1}\bigl\Vert f(p)-p\bigr\Vert + \zeta_{k+1}\Vert x_{k+1}-p\Vert + (1- \zeta_{k+1})\delta_{k+1}\beta_{k+1}\Vert u_{k+1}-p\Vert \\ \leq& \bigl\{ (1-\zeta_{k+1})\bigl[k\mu_{k+1}+ \gamma_{k+1}+\delta_{k+1}(1-\beta_{k+1})\bigr]+ \zeta_{k+1}\bigr\} \Vert x_{k+1}-p\Vert \\ &{}+ (1- \zeta_{k+1})\mu_{k+1}\bigl\Vert f(p)-p\bigr\Vert \\ &{}+ (1-\zeta_{k+1})\delta_{k+1}\beta_{k+1}\bigl\Vert (1-\alpha _{k+1}) (x_{k+1}+e_{k+1})-p\bigr\Vert \\ \leq& \bigl[1-(1-k)\mu_{k+1}(1-\zeta_{k+1})- \alpha_{k+1}\beta_{k+1}\delta _{k+1}(1- \zeta_{k+1})\bigr] \Vert x_{k+1} - p\Vert + \Vert e_{k+1}\Vert \\ &{}+(1-\zeta_{k+1})\alpha_{k+1}\beta_{k+1} \delta_{k+1}\Vert p\Vert +(1-\zeta_{k+1}) \mu_{k+1}(1-k)\frac{\Vert f(p)-p\Vert }{1-k} \\ \leq& M_{1} + \sum_{i = 1}^{k+1} \Vert e_{i}\Vert . \end{aligned}$$

Thus (3.3) is true for all $n \in N^{+}$. Since $\sum_{n=1}^{\infty}\|e_{n}\| < +\infty$, (3.3) ensures that $\{x_{n}\}$ is bounded.

For $\forall p \in D$, from $\|u_{n} - p\| \leq \|(1-\alpha_{n})(x_{n}+e_{n})-p\|\leq\|x_{n}\|+\|e_{n}\|+\|p\|$, we see that $\{u_{n}\}$ is bounded.

Since $\|\sum_{i = 1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}}u_{n}\|\leq \|\sum_{i = 1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}}u_{n} - \sum_{i = 1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}}p\|+(1-a_{0})\|p\|\leq\|u_{n} -p\|+\|p\|$, $\{\sum_{i = 1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}}u_{n}\}$ is bounded. Since both $\{\sum_{i = 1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}}u_{n}\}$ and $\{x_{n}\}$ are bounded, $\{v_{n}\}$ is bounded. From the definition of a contraction, we can easily see that $\{f(x_{n})\}$ is bounded.

Set $M_{2} = \sup\{ \|u_{n}\|, \|J_{r_{n,i}}^{A_{i}}u_{n}\|, \|\sum_{i = 1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}}u_{n}\|, \|f(x_{n})\|, \|v_{n}\|, \|x_{n}\|, : n \geq1, i\geq1\}$.

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

In fact, if $r_{n,i}\leq r_{n+1,i}$, then, using Lemma 2.6,

$$\begin{aligned}& \Biggl\Vert \sum_{i = 1}^{\infty}a_{i}J_{r_{n+1,i}}^{A_{i}}u_{n+1}- \sum_{i = 1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}}u_{n} \Biggr\Vert \\& \quad \leq\sum_{i = 1}^{\infty}a_{i} \bigl\Vert J_{r_{n+1,i}}^{A_{i}}u_{n+1}-J_{r_{n,i}}^{A_{i}}u_{n} \bigr\Vert \\& \quad = \sum_{i = 1}^{\infty}a_{i} \biggl\Vert J_{r_{n,i}}^{A_{i}}\biggl(\frac{r_{n,i}}{r_{n+1,i}}u_{n+1}+ \biggl(1-\frac {r_{n,i}}{r_{n+1,i}}\biggr)J_{r_{n+1,i}}^{A_{i}}u_{n+1} \biggr) - J_{r_{n,i}}^{A_{i}}u_{n}\biggr\Vert \\& \quad \leq\sum_{i = 1}^{\infty}a_{i} \biggl\Vert \frac{r_{n,i}}{r_{n+1,i}}u_{n+1}+\biggl(1-\frac {r_{n,i}}{r_{n+1,i}} \biggr)J_{r_{n+1,i}}^{A_{i}}u_{n+1}-u_{n}\biggr\Vert \\& \quad \leq(1-a_{0})\Vert u_{n+1}-u_{n}\Vert + \sum_{i = 1}^{\infty}a_{i}\biggl(1- \frac{r_{n,i}}{r_{n+1,i}}\biggr)\bigl\Vert J_{r_{n+1,i}} ^{A_{i}}u_{n+1}-u_{n} \bigr\Vert \\& \quad \leq(1-a_{0})\Vert u_{n+1}-u_{n}\Vert + \frac{2M_{2}}{\varepsilon}\sum_{i = 1}^{\infty}(r_{n+1,i}-r_{n,i}). \end{aligned}$$

(3.4)

If $r_{n+1,i}\leq r_{n,i}$, then imitating the proof of (3.4), we have

$$\begin{aligned}& \Biggl\Vert \sum_{i = 1}^{\infty}a_{i}J_{r_{n+1,i}}^{A_{i}}u_{n+1}- \sum_{i = 1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}}u_{n} \Biggr\Vert \\& \quad \leq(1-a_{0})\Vert u_{n+1}-u_{n}\Vert + \frac{2M_{2}}{\varepsilon}\sum_{i = 1}^{\infty}(r_{n,i}-r_{n+1,i}). \end{aligned}$$

(3.5)

Combining (3.4) and (3.5), we have

$$\begin{aligned} \begin{aligned}[b] &\Biggl\Vert \sum_{i = 1}^{\infty}a_{i}J_{r_{n+1,i}}^{A_{i}}u_{n+1}- \sum_{i = 1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}}u_{n} \Biggr\Vert \\ &\quad \leq(1-a_{0})\Vert u_{n+1}-u_{n}\Vert + \frac{2M_{2}}{\varepsilon}\sum_{i = 1}^{\infty}|r_{n,i}-r_{n+1,i}|. \end{aligned} \end{aligned}$$

(3.6)

On the other hand,

$$\begin{aligned} \|u_{n+1}-u_{n}\| \leq& \|x_{n+1}-x_{n} \|+\alpha_{n+1}\|x_{n+1}\|+\alpha_{n} \|x_{n}\| \\ &{}+\|\alpha _{n+1}e_{n+1}-\alpha_{n}e_{n} \|+\|e_{n+1}-e_{n}\|. \end{aligned}$$

(3.7)

In view of (3.6) and (3.7), we have

$$\begin{aligned} \Vert v_{n+1}-v_{n}\Vert \leq& (1-\beta_{n+1}) \Vert x_{n+1}-x_{n}\Vert +a_{0} \beta_{n+1}\Vert u_{n+1}-u_{n}\Vert \\ &{}+\beta_{n+1} \Biggl\Vert \sum_{i = 1}^{\infty}a_{i}J_{r_{n+1,i}}^{A_{i}}u_{n+1} - \sum_{i = 1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}}u_{n} \Biggr\Vert \\ &{} +|\beta_{n+1}-\beta_{n}|\Vert x_{n}\Vert + a_{0}|\beta_{n+1}-\beta_{n}|\Vert u_{n} \Vert +|\beta_{n+1}-\beta_{n}|\Biggl\Vert \sum _{i = 1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}}u_{n} \Biggr\Vert \\ \leq& (1-\beta_{n+1})\Vert x_{n+1}-x_{n}\Vert +a_{0}\beta_{n+1}\bigl[\Vert x_{n+1}-x_{n} \Vert +\alpha _{n+1}\Vert x_{n+1}\Vert \\ &{}+\alpha_{n}\Vert x_{n}\Vert + \Vert e_{n+1}-e_{n}\Vert +\Vert \alpha_{n+1}e_{n+1}- \alpha_{n}e_{n}\Vert \bigr] \\ &{} + \beta_{n+1}\Biggl[ (1-a_{0})\Vert u_{n+1}-u_{n}\Vert +\frac{2M_{2}}{\varepsilon}\sum _{i = 1}^{\infty} |r_{n,i}-r_{n+1,i}|\Biggr] \\ &{} + |\beta_{n+1}-\beta_{n}|\Vert x_{n}\Vert + a_{0}|\beta_{n+1}-\beta_{n}|\Vert u_{n}\Vert +|\beta_{n+1}-\beta_{n}|\Biggl\Vert \sum_{i = 1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}}u_{n} \Biggr\Vert \\ \leq&\Vert x_{n+1}-x_{n}\Vert + \Vert x_{n} \Vert +\Vert x_{n+1}\Vert +\Vert \alpha_{n+1}e_{n+1}- \alpha_{n}e_{n}\Vert +\Vert e_{n+1}-e_{n} \Vert \\ &{}+|\beta_{n+1}-\beta_{n}|\Biggl\Vert \sum _{i = 1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}}u_{n} \Biggr\Vert +|\beta_{n+1}-\beta_{n}|\Vert x_{n} \Vert +a_{0}|\beta_{n+1}-\beta_{n}|\Vert u_{n}\Vert \\ &{} +\frac{2M_{2}}{\varepsilon}\beta_{n+1}\sum_{i = 1}^{\infty} |r_{n,i}-r_{n+1,i}|. \end{aligned}$$

(3.8)

Thus in view of (3.8), we have

$$\begin{aligned} \Vert w_{n+1}-w_{n}\Vert \leq& (k\mu_{n+1}+ \gamma_{n+1}+\delta_{n+1})\Vert x_{n+1}-x_{n} \Vert +|\mu_{n+1}-\mu _{n}|\bigl\Vert f(x_{n}) \bigr\Vert \\ &{}+|\gamma_{n+1}-\gamma_{n}|\Vert x_{n}\Vert +\delta_{n+1}\Vert x_{n}\Vert +\delta_{n+1} \Vert x_{n+1}\Vert +|\delta_{n+1}-\delta_{n}| \Vert v_{n}\Vert \\ &{} + \delta_{n+1}\Vert e_{n+1}-e_{n}\Vert + \delta_{n+1}\Vert \alpha_{n+1}e_{n+1}-\alpha _{n}e_{n}\Vert +\delta_{n+1}|\beta_{n+1}- \beta_{n}|\Vert x_{n}\Vert \\ &{}+\delta_{n+1}|\beta_{n+1}-\beta_{n}|\Biggl\Vert \sum_{i = 1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}}u_{n} \Biggr\Vert +\delta_{n+1}a_{0}|\beta_{n+1}-\beta _{n}|\Vert u_{n}\Vert \\ &{}+\delta_{n+1}\beta_{n+1}\frac{2M_{2}}{\varepsilon}\sum _{i = 1}^{\infty }|r_{n,i}-r_{n+1,i}| \\ \leq&\bigl(1-(1-k)\mu_{n+1}\bigr)\Vert x_{n+1}-x_{n} \Vert \\ &{}+2M_{2}\delta_{n+1} + M_{2}\bigl(| \mu_{n+1}-\mu_{n}|+|\gamma_{n+1}-\gamma_{n}|+| \delta_{n+1}-\delta _{n}|+3|\beta_{n+1}- \beta_{n}|\bigr) \\ &{}+\Vert e_{n+1}-e_{n}\Vert + \Vert \alpha_{n+1}e_{n+1}-\alpha_{n}e_{n}\Vert \\ &{}+ \delta_{n+1}\beta_{n+1}\frac {2M_{2}}{\varepsilon}\sum _{i = 1}^{\infty}|r_{n,i}-r_{n+1,i}|. \end{aligned}$$

(3.9)

Using Lemma 2.4, we have from (3.9) $\lim_{n \rightarrow \infty} \|x_{n} - w_{n}\| = 0$ and then $\lim_{n \rightarrow\infty} \|x_{n+1} - x_{n} \| = 0$.

Step 3. $\lim_{n \rightarrow\infty} \|x_{n} - u_{n}\| = 0$ and $\lim_{n \rightarrow\infty} \|u_{n} - (a_{0}I+\sum_{i = 1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}})u_{n}\| = 0$.

We compute the following:

$$\begin{aligned} \Vert v_{n+1}-v_{n}\Vert \leq&(1-\beta_{n+1}) \Vert x_{n+1}-x_{n}\Vert + |\beta_{n+1}- \beta_{n}|\Vert x_{n}\Vert \\ &{} +\beta_{n+1}\Biggl\Vert \Biggl(a_{0}I+\sum _{i = 1}^{\infty}a_{i}J_{r_{n+1,i}}^{A_{i}} \Biggr)u_{n+1} - \Biggl(a_{0}I+\sum _{i = 1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}} \Biggr)u_{n}\Biggr\Vert \\ &{}+ |\beta_{n+1}-\beta_{n}| \Biggl\Vert \Biggl(a_{0}I+\sum_{i = 1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}} \Biggr)u_{n}\Biggr\Vert \\ \leq&(1-\beta_{n+1})\Vert x_{n+1}-x_{n}\Vert + 2|\beta_{n+1}-\beta_{n}|M_{2} \\ &{}+\Biggl\Vert \Biggl(a_{0}I+\sum_{i = 1}^{\infty}a_{i}J_{r_{n+1,i}}^{A_{i}} \Biggr)u_{n+1} - \Biggl(a_{0}I+\sum _{i = 1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}} \Biggr)u_{n}\Biggr\Vert . \end{aligned}$$

Using the result of Step 2 and Lemma 2.4, we have

$$ \lim_{n \rightarrow\infty} \Biggl\Vert \Biggl(a_{0}I+\sum _{i = 1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}} \Biggr)u_{n}-v_{n}\Biggr\Vert = 0. $$

(3.10)

Since

$$\Biggl\Vert \Biggl(a_{0}I+\sum_{i = 1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}} \Biggr)u_{n}-v_{n}\Biggr\Vert = (1-\beta_{n}) \Biggl\Vert x_{n} - \Biggl(a_{0}I+\sum _{i = 1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}} \Biggr)u_{n}\Biggr\Vert $$

and $0 < \liminf_{n \rightarrow\infty}\beta_{n} \leq \limsup_{n \rightarrow\infty}\beta_{n} <1$, (3.10) implies that

$$ \lim_{n \rightarrow\infty}\Biggl\Vert x_{n} - \Biggl(a_{0}I+\sum_{i = 1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}} \Biggr)u_{n}\Biggr\Vert = 0. $$

(3.11)

Moreover,

$$\|u_{n} - x_{n}\| = \bigl\Vert Q_{C}\bigl[(1- \alpha_{n}) (x_{n}+e_{n})\bigr]-Q_{C}x_{n} \bigr\Vert \leq \alpha_{n}\|x_{n}\|+ (1- \alpha_{n})\|e_{n}\|, $$

then

$$ \lim_{n\rightarrow\infty}\|x_{n} - u_{n}\| = 0. $$

(3.12)

Noticing (3.11) and (3.12), we have

$$ \lim_{n \rightarrow \infty}\Biggl\Vert u_{n} - \Biggl(a_{0}I+\sum_{i = 1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}} \Biggr)u_{n} \Biggr\Vert = 0. $$

(3.13)

Step 4. $\omega(x_{n})\subset D$, where $\omega(x_{n})$ is the set of all of the weak limit points of $\{x_{n}\}$.

Since $\{x_{n}\}$ is bounded, there exists a subsequence of $\{x_{n}\}$, which is denoted by $\{x_{n_{k}}\}$, such that $x_{n_{k}} \rightharpoonup q_{0}$, as $k \rightarrow\infty$. From (3.12), we have $u_{n_{k}} \rightharpoonup q_{0}$, as $k \rightarrow\infty$. Then Lemma 2.2(ii) implies that

$$ \limsup_{k \rightarrow\infty} \Phi\bigl(\Vert u_{n_{k}}-x\Vert \bigr) = \limsup_{k \rightarrow\infty} \Phi\bigl(\Vert u_{n_{k}}- q_{0}\Vert \bigr)+ \Phi\bigl(\Vert q_{0}-x\Vert \bigr),\quad \forall x \in E. $$

(3.14)

Lemma 3.1, Lemma 2.2(i), and (3.13) imply that

$$\begin{aligned}& \limsup_{k \rightarrow\infty} \Phi\Biggl(\Biggl\Vert u_{n_{k}}- \Biggl(a_{0}I+\sum_{i = 1}^{\infty}a_{i}J_{r_{n_{k},i}}^{A_{i}} \Biggr)q_{0}\Biggr\Vert \Biggr) \\& \quad \leq \limsup_{k \rightarrow\infty} \Phi\Biggl(\Biggl\Vert \sum _{i = 1}^{\infty}a_{i}J_{r_{n_{k},i}}^{A_{i}}u_{n_{k}}- \Biggl(a_{0}I+\sum_{i = 1}^{\infty}a_{i}J_{r_{n_{k},i}}^{A_{i}} \Biggr)q_{0}+a_{0}u_{n_{k}}\Biggr\Vert \Biggr) \\& \qquad {}+ \limsup_{k \rightarrow\infty} \Biggl\langle (1-a_{0})u_{n_{k}}- \sum_{i = 1}^{\infty}a_{i}J_{r_{n_{k},i}}^{A_{i}}u_{n_{k}}, J_{\varphi}\Biggl(u_{n_{k}}-\Biggl(a_{0}I+\sum _{i = 1}^{\infty}a_{i}J_{r_{n_{k},i}}^{A_{i}} \Biggr)q_{0}\Biggr)\Biggr\rangle \\& \quad \leq \limsup_{k \rightarrow\infty} \Phi\bigl(\Vert u_{n_{k}}-q_{0}\Vert \bigr). \end{aligned}$$

(3.15)

From (3.14) and (3.15), we know that $\Phi(\|(a_{0}I+\sum_{i = 1}^{\infty}a_{i}J_{r_{n_{k},i}}^{A_{i}})q_{0}-q_{0}\|)\leq0$, which implies that

$$a_{0}q_{0}+\sum_{i = 1}^{\infty}a_{i}J_{r_{n_{k},i}}^{A_{i}}q_{0} = q_{0}. $$

Then Lemma 3.1 ensures that $q_{0} \in D$.

Step 5. $x_{n} \rightharpoonup q_{0}$, as $n \rightarrow\infty$, where $q_{0}\in D$ is the unique element which satisfies (3.2).

Now, we define $h(y) = \limsup_{n \rightarrow\infty}\Phi(\|x_{n} - y\|)$, for $y \in D$. Then $h(y) : D \rightarrow R^{+}$ is proper, strictly convex, lower-semi-continuous, and $h(y) \rightarrow +\infty$, as $\|y\| \rightarrow+\infty$. Therefore, there exists a unique element $q_{0} \in D$ such that $h(q_{0}) = \min_{y \in D}h(y)$. That is, $q_{0}$ satisfies (3.2).

Next, we shall show that $x_{n} \rightharpoonup q_{0}$, as $n \rightarrow\infty$.

In fact, suppose there exists a subsequence $\{x_{n_{m}}\}$ of $\{x_{n}\}$ (for simplicity, we still denote it by $\{x_{n}\}$) such that $x_{n} \rightharpoonup v_{0}$, as $n \rightarrow\infty$, then $v_{0} \in D$ in view of Step 4. Using Lemma 2.2(ii), $h(q_{0}) = h(v_{0}) + \Phi(\|v_{0} - q_{0}\|)$, which implies that $\Phi(\|v_{0} - q_{0}\|)= 0$, and then $v_{0} = q_{0}$.

If there exists another subsequence $\{x_{n_{k}}\}$ of $\{x_{n}\}$ such that $x_{n_{k}} \rightharpoonup p_{0}$, as $k \rightarrow \infty$. Then repeating the above process, we have $p_{0} = q_{0}$.

Since all of the weakly convergent subsequences of $\{x_{n}\}$ converge to the same element $q_{0}$, the whole sequence $\{x_{n}\}$ converges weakly to $q_{0}$. Combining the results of Steps 2 and 3, $u_{n} \rightharpoonup q_{0}$, $w_{n} \rightharpoonup q_{0}$, as $n \rightarrow\infty$.

This completes the proof. □

Remark 3.1

Compared to the work in [20], the smoothness of E is not needed. The iterative algorithm (3.1) is more general than those discussed in [5, 9, 17–21].

Remark 3.2

The properties of the function Φ defined by (1.1) are widely used in the proof of Steps 4 and 5, which can be regarded as a new proof technique compared to the existing work.

Remark 3.3

Three sequences are proved to be weakly convergent to the common zero point of an infinite family of m-accretive mappings. The characteristic of the weakly convergent point $q_{0}$ of $\{x_{n}\}$ is presented in Theorem 3.1.

Remark 3.4

The assumptions imposed on the real number sequences in Theorem 3.1 are reasonable if we take $\alpha_{n} = \frac{1}{n}$, $\mu_{n} = \frac{n}{n^{2}}$, $\gamma_{n} = \frac{n^{2} - n -1}{n^{2}}$, $\delta_{n} = \frac{1}{n^{2}}$, and $\beta_{n} = \zeta_{n} = \frac{n+1}{2n}$, for $n \geq1$.

Lemma 3.2

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 $T : C \rightarrow C$ be a nonexpansive mapping with $\operatorname{Fix}(T) \neq\emptyset$. Let $T_{t} : C \rightarrow C$ be defined by $T_{t} x:= TQ_{C}[(1-t)x]$, $x \in C$. Then:

• (i) $T_{t}$ is a contraction and has a fixed point $z_{t}$, which satisfies $\|z_{t} - Tz_{t}\| \rightarrow0$, as $t\rightarrow0$;

• (ii) further suppose that E has a weakly continuous duality mapping $J_{\varphi}$, then $\lim_{t\rightarrow0}z_{t} = z_{0} \in \operatorname{Fix}(T)$.

Proof

Lemma 2.7 ensures the result of (i).

To show that (ii) holds, it suffices to show that, for any sequence $\{t_{n}\}$ such that $t_{n} \rightarrow0$, we have $\lim_{n\rightarrow\infty}z_{t_{n}} = z_{0} \in \operatorname{Fix}(T)$.

In fact, the result of (i) implies that there exists $z_{t} \in \operatorname{Fix}(T)$ such that $z_{t} = TQ_{C}[(1-t)z_{t}]$, $t \in(0,1)$. For $\forall p \in \operatorname{Fix}(T)$, since

$$\|z_{t} -p\| = \bigl\Vert TQ_{C}\bigl[(1-t)z_{t} \bigr] - TQ_{C} p\bigr\Vert \leq\|z_{t} - p -tz_{t}\| \leq(1-t)\|z_{t}-p\|+t\|p\|, $$

$\{z_{t}\}$ is bounded. Without loss of generality, we may assume that $z_{t_{n}}\rightharpoonup z_{0}$. Using (i) and Lemma 2.5, we have $z_{0} \in \operatorname{Fix}(T)$.

Using Lemma 2.2, we have, for $\forall p \in \operatorname{Fix}(T)$,

$$\begin{aligned} \Phi\bigl(\Vert z_{t} -p\Vert \bigr) =& \Phi\bigl(\bigl\Vert TQ_{C}\bigl[(1-t)z_{t}\bigr] - TQ_{C} p\bigr\Vert \bigr) \\ \leq&\Phi\bigl(\bigl\Vert (1-t)z_{t}-p\bigr\Vert \bigr) \\ \leq&\Phi\bigl(\Vert z_{t} -p\Vert \bigr)-t\bigl\langle z_{t}, J_{\varphi}(z_{t}-p-tz_{t})\bigr\rangle \\ = &\Phi\bigl(\Vert z_{t} -p\Vert \bigr)-t \bigl\langle z_{t}-p-tz_{t}, J_{\varphi}(z_{t}-p-tz_{t}) \bigr\rangle - t\bigl\langle p+tz_{t}, J_{\varphi}(z_{t}-p-tz_{t}) \bigr\rangle , \end{aligned}$$

which implies that

$$\|z_{t} -p -tz_{t}\| \varphi\bigl(\Vert z_{t} -p -tz_{t}\Vert \bigr) \leq\bigl\langle p,J_{\varphi} (p+tz_{t}-z_{t})\bigr\rangle +t \bigl\langle z_{t}, J_{\varphi}(p+tz_{t}-z_{t})\bigr\rangle . $$

In particular,

$$\begin{aligned}& \|z_{t_{n}} -p -{t_{n}}z_{t_{n}}\|\varphi\bigl(\Vert z_{t_{n}} -p -{t_{n}}z_{t_{n}}\Vert \bigr) \\& \quad \leq\bigl\langle p,J_{\varphi }(p+{t_{n}}z_{t_{n}}-z_{t_{n}}) \bigr\rangle + {t_{n}} \bigl\langle z_{t_{n}}, J_{\varphi}(p+{t_{n}}z_{t_{n}}-z_{t_{n}})\bigr\rangle . \end{aligned}$$

Thus

$$\begin{aligned}& \|z_{t_{n}} -z_{0} -{t_{n}}z_{t_{n}}\| \varphi\bigl(\Vert z_{t_{n}} -z_{0} -{t_{n}}z_{t_{n}} \Vert \bigr) \\& \quad \leq\bigl\langle z_{0},J_{\varphi}(z_{0}+{t_{n}}z_{t_{n}}-z_{t_{n}}) \bigr\rangle + {t_{n}} \bigl\langle z_{t_{n}}, J_{\varphi}(z_{0}+{t_{n}}z_{t_{n}}-z_{t_{n}}) \bigr\rangle . \end{aligned}$$

(3.16)

Since E has a weakly continuous duality mapping $J_{\varphi}$, (3.16) implies that $z_{t_{n}} -z_{0} -{t_{n}}z_{t_{n}}\rightarrow 0$, as $n \rightarrow\infty$.

From $\|z_{t_{n}}-z_{0}\|\leq \|z_{t_{n}}-z_{0}-t_{n}z_{t_{n}}\|+t_{n}\|z_{t_{n}}\|$, we see that $z_{t_{n}}\rightarrow z_{0}$, as $n \rightarrow\infty$.

Suppose there exists another sequence $z_{t_{m}} \rightharpoonup x_{0}$, as $t_{m} \rightarrow0$ and $m \rightarrow\infty$. Then from (i) $\|z_{t_{m}}-Tz_{t_{m}}\| \rightarrow0$ and $I-T$ being demi-closed at zero, we have $x_{0} \in \operatorname{Fix}(T)$. Moreover, repeating the above proof, we have $z_{t_{m}}\rightarrow x_{0}$, as $m \rightarrow\infty$. Next, we want to show that $z_{0} = x_{0}$.

Similar to (3.16), we have

$$\begin{aligned}& \|z_{t_{m}} - z_{0} -{t_{m}}z_{t_{m}}\| \varphi\bigl(\Vert z_{t_{m}} - z_{0} -{t_{m}}z_{t_{m}} \Vert \bigr) \\& \quad \leq\bigl\langle z_{0},J_{\varphi}(z_{0}+{t_{m}}z_{t_{m}}-z_{t_{m}}) \bigr\rangle + {t_{m}}\bigl\langle z_{t_{m}}, J_{\varphi}(z_{0}+{t_{m}}z_{t_{m}}-z_{t_{m}}) \bigr\rangle . \end{aligned}$$

Letting $m \rightarrow\infty$,

$$ \|x_{0}-z_{0}\|\varphi\bigl(\Vert x_{0}-z_{0} \Vert \bigr) \leq\bigl\langle z_{0}, J_{\varphi}(z_{0} -x_{0})\bigr\rangle . $$

(3.17)

Interchanging $x_{0}$ and $z_{0}$ in (3.17), we obtain

$$ \|z_{0}-x_{0}\|\varphi\bigl(\Vert z_{0}-x_{0} \Vert \bigr) \leq\bigl\langle x_{0}, J_{\varphi}(x_{0} -z_{0})\bigr\rangle . $$

(3.18)

Then (3.17) and (3.18) ensure

$$2\|x_{0}-z_{0}\|\varphi\bigl(\Vert x_{0}-z_{0} \Vert \bigr) \leq\bigl\langle x_{0} - z_{0}, J_{\varphi}(x_{0}-z_{0})\bigr\rangle = \|x_{0} - z_{0} \|\varphi\bigl(\Vert x_{0}-z_{0} \Vert \bigr), $$

which implies that $x_{0} = z_{0}$.

Therefore, $\lim_{t\rightarrow0}z_{t} = z_{0} \in \operatorname{Fix}(T)$.

This completes the proof. □

Remark 3.5

Compared to the proof of Lemma 2.7 in [20], a different method is used in the proof of the result (ii) in Lemma 3.2.

Theorem 3.2

Further suppose that E is real uniformly convex and uniformly smooth and (vii) $\sum_{n=1}^{\infty}(1-\zeta_{n})\alpha_{n}\beta_{n}\delta_{n} = +\infty$; (viii) $\mu_{n} = o(\alpha_{n}\beta_{n}\delta_{n})$. The other restrictions are the same as those in Theorem  3.1, then the iterative sequence $\{x_{n}\}$ generated by (3.1) converges strongly to $p_{0} \in D$.

Proof

We shall split the proof into six steps. The proofs of Steps 1, 2, and 3 are the same as those in Theorem 3.1.

Step 4. $\limsup_{n\rightarrow+\infty}\langle p_{0}, J_{\varphi}(p_{0}-x_{n})\rangle\leq0$, where $p_{0}$ is an element in D.

From Lemma 3.2, we know that there exists $y_{t} \in C$ such that

$$y_{t} = a_{0}Q_{C}\bigl[(1-t)y_{t} \bigr]+\sum_{i = 1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}}Q_{C} \bigl[(1-t)y_{t}\bigr] $$

for $t \in (0,1)$. Moreover, $y_{t} \rightarrow p_{0} \in D$, as $t \rightarrow 0$.

Since $\|y_{t}\|\leq\|y_{t} - p_{0} \|+\|p_{0}\|$, $\{y_{t}\}$ is bounded. Using Lemma 2.2, we have

$$\begin{aligned} \Phi\bigl(\Vert y_{t} - x_{n}\Vert \bigr) \leq&\Phi \bigl(\bigl\Vert (1-t)y_{t}-x_{n}\bigr\Vert \bigr) \\ &{}+ \Biggl\langle \Biggl(a_{0}I+\sum_{i = 1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}} \Biggr)x_{n}-x_{n}, J_{\varphi}(y_{t}-x_{n}) \Biggr\rangle \\ \leq&\Phi\bigl(\Vert y_{t}-x_{n}\Vert \bigr) - t \bigl\langle y_{t}, J_{\varphi}\bigl[(1-t)y_{t} -x_{n}\bigr]\bigr\rangle \\ &{}+ K_{1}\Biggl\Vert \Biggl(a_{0}I+\sum_{i = 1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}} \Biggr)x_{n}-x_{n}\Biggr\Vert , \end{aligned}$$

where $K_{1} = \sup \{\varphi(\|y_{t} - x_{n}\|): n \geq1, t > 0\}$ and from the result of Step 1 we know that $K_{1}$ is a positive constant.

Thus $\langle y_{t}, J_{\varphi}[(1-t)y_{t} -x_{n}]\rangle\leq \frac{K_{1}}{t}\|(a_{0}I+\sum_{i = 1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}})x_{n}-x_{n}\|$. Noticing (3.12) and (3.13), we have

$$\begin{aligned}& \Biggl\Vert \Biggl(a_{0}I+\sum_{i = 1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}} \Biggr)x_{n}-x_{n}\Biggr\Vert \\& \quad \leq\Biggl\Vert \Biggl(a_{0}I+\sum _{i = 1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}} \Biggr)x_{n}-\Biggl(a_{0}I+\sum_{i = 1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}} \Biggr)u_{n}\Biggr\Vert \\& \qquad {}+\Biggl\Vert \Biggl(a_{0}I+\sum _{i = 1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}} \Biggr)u_{n}-u_{n}\Biggr\Vert +\Vert u_{n}-x_{n} \Vert \\& \quad \leq2\Vert x_{n} - u_{n}\Vert +\Biggl\Vert \Biggl(a_{0}I+\sum_{i = 1}^{\infty}a_{i}J_{r_{n,i}}^{A_{i}} \Biggr)u_{n}-u_{n}\Biggr\Vert \rightarrow0, \end{aligned}$$

as $n \rightarrow\infty$. Therefore,

$$\lim_{t \rightarrow0}\limsup_{n\rightarrow+\infty}\bigl\langle y_{t}, J_{\varphi}\bigl[(1-t)y_{t} -x_{n} \bigr]\bigr\rangle \leq0. $$

From the assumption on $J_{\varphi}$ and the following fact:

$$\begin{aligned} \bigl\langle p_{0}, J_{\varphi}(p_{0} - x_{n})\bigr\rangle =& \bigl\langle p_{0}, J_{\varphi}(p_{0} - x_{n}) - J_{\varphi} \bigl[(1-t)y_{t} -x_{n}\bigr]\bigr\rangle \\ &{} + \bigl\langle p_{0} - y_{t}, J_{\varphi} \bigl[(1-t)y_{t} -x_{n}\bigr]\bigr\rangle + \bigl\langle y_{t}, J_{\varphi}\bigl[(1-t)y_{t} -x_{n} \bigr]\bigr\rangle , \end{aligned}$$

we have $\limsup_{n\rightarrow+\infty}\langle p_{0}, J_{\varphi}(p_{0} - x_{n})\rangle\leq0$.

Then

$$\begin{aligned}& \limsup_{n \rightarrow\infty}\bigl\langle p_{0}, J_{\varphi } \bigl[p_{0}-x_{n}-(1-\alpha_{n})e_{n} + \alpha_{n} x_{n}\bigr]\bigr\rangle \\& \quad \leq \limsup_{n \rightarrow\infty}\bigl\langle p_{0}, J_{\varphi}\bigl[p_{0}-x_{n}-(1-\alpha_{n})e_{n} + \alpha_{n} x_{n}\bigr]-J_{\varphi}(p_{0}-x_{n}) \bigr\rangle + \limsup_{n \rightarrow\infty}\bigl\langle p_{0}, J_{\varphi}(p_{0}-x_{n})\bigr\rangle \\& \quad = \limsup_{n \rightarrow\infty}\bigl\langle p_{0}, J_{\varphi}(p_{0}-x_{n})\bigr\rangle \leq0. \end{aligned}$$

(3.19)

Step 5. Φ defined by (1.1) satisfies $\Phi(kt) \leq k \Phi(t)$, for $t \in[0, +\infty)$, where $k \in[0,1]$.

In fact, let $F(t) = \int_{0}^{kt} \varphi(s) \,ds - k \int_{0}^{t}\varphi(s)\,ds$, for $t \in[0,+\infty)$. Then $F'(t) = k \varphi(kt) - k\varphi(t) \leq0$, since $k \in(0,1)$ and the gauge function φ is increasing. Thus F is decreasing on $[0,+\infty)$. That is, $F(t) \leq F(0)$, for $t \in[0,+\infty)$. Then $\int_{0}^{kt}\varphi(s) \,ds \leq k \int_{0}^{t} \varphi(s) \,ds$, for $t \in[0,+\infty)$, which implies that $\Phi(kt) \leq k\Phi(t)$, for $t \in[0,+\infty)$.

Step 6. $x_{n} \rightarrow p_{0}$, as $n \rightarrow+\infty$, where $p_{0} \in D$ is the same as that in Step 4.

Since Φ is convex, we have

$$ \Phi\bigl(\Vert x_{n+1} - p_{0}\Vert \bigr)\leq(1- \zeta_{n}) \Phi\bigl(\Vert w_{n}-p_{0}\Vert \bigr)+\zeta_{n} \Phi\bigl(\Vert x_{n}-p_{0} \Vert \bigr). $$

(3.20)

Using Lemma 2.2 and the result of Step 5, we have

$$\begin{aligned} \Phi\bigl(\Vert w_{n} - p_{0}\Vert \bigr) \leq& \mu_{n} \Phi\bigl(\bigl\Vert f(x_{n})-p_{0}\bigr\Vert \bigr)+\gamma_{n} \Phi\bigl(\Vert x_{n}-p_{0} \Vert \bigr)+\delta_{n} \Phi\bigl(\Vert v_{n} - p_{0}\Vert \bigr) \\ \leq&(\mu_{n} k +\gamma_{n})\Phi\bigl(\Vert x_{n}-p_{0}\Vert \bigr)+ \mu_{n} \bigl\langle f(p_{0}) - p_{0}, J_{\varphi} \bigl(f(x_{n}) - p_{0}\bigr) \bigr\rangle \\ &{}+\delta_{n} \Phi\bigl(\Vert v_{n} - p_{0} \Vert \bigr). \end{aligned}$$

(3.21)

Similarly, we have

$$\begin{aligned} \Phi\bigl(\Vert v_{n} - p_{0}\Vert \bigr) \leq&(1- \beta_{n}) \Phi\bigl(\Vert x_{n}-p_{0}\Vert \bigr)+\beta_{n} \Phi\bigl(\Vert u_{n} - p_{0} \Vert \bigr) \\ \leq&(1-\beta_{n}) \Phi\bigl(\Vert x_{n}-p_{0} \Vert \bigr)+\beta_{n} (1-\alpha_{n})\Phi\bigl(\Vert x_{n}-p_{0}\Vert \bigr) \\ &{}+ \beta_{n} \bigl\langle (1-\alpha_{n})e_{n} - \alpha_{n} p_{0}, J_{\varphi} \bigl((1- \alpha_{n}) (x_{n}+e_{n})-p_{0}\bigr) \bigr\rangle \\ =&(1-\beta_{n} \alpha_{n})\Phi\bigl(\Vert x_{n}-p_{0}\Vert \bigr) \\ &{}+ \beta_{n} (1-\alpha_{n})\bigl\langle e_{n}, J_{\varphi} \bigl((1-\alpha_{n}) (x_{n}+e_{n})-p_{0} \bigr)\bigr\rangle \\ &{}+\beta_{n} \alpha_{n} \bigl\langle p_{0}, J_{\varphi} \bigl(p_{0}-x_{n}-(1-\alpha_{n})e_{n}+ \alpha_{n}x_{n}\bigr)\bigr\rangle . \end{aligned}$$

(3.22)

Let $K_{2} = \sup \{\varphi(\|(1-\alpha_{n})(x_{n}+e_{n})-p_{0}\|), \varphi(\|f(x_{n})-p_{0}\|): n \geq1\}$. Then $K_{2}$ is a positive constant. Using (3.20)-(3.22), we have

$$\begin{aligned} \Phi\bigl(\Vert x_{n+1}-p_{0}\Vert \bigr) \leq&\bigl\{ 1-(1-\zeta_{n})\bigl[\mu_{n}(1-k)+\alpha_{n} \beta_{n} \delta_{n}\bigr]\bigr\} \Phi\bigl(\Vert x_{n}-p_{0}\Vert \bigr) \\ &{} +(1-\zeta_{n})\beta_{n}(1-\alpha_{n}) \delta_{n}\bigl\langle e_{n}, J_{\varphi}\bigl[(1- \alpha_{n}) (x_{n}+e_{n})-p_{0}\bigr] \bigr\rangle \\ &{}+\alpha_{n}\beta_{n}\delta_{n}(1- \zeta_{n})\bigl\langle p_{0},J_{\varphi } \bigl[p_{0}-x_{n}-(1-\alpha_{n})e_{n} + \alpha_{n} x_{n}\bigr]\bigr\rangle \\ &{} + (1-\zeta_{n})\mu_{n} \bigl\langle f(p_{0})-p_{0}, J_{\varphi}\bigl(f(x_{n}) - p_{0} \bigr)\bigr\rangle \\ \leq& \bigl[1-\alpha_{n}\beta_{n}\delta_{n}(1- \zeta_{n})\bigr]\Phi\bigl(\Vert x_{n}-p_{0} \Vert \bigr) \\ &{}+(1-\zeta_{n}) (1-\alpha_{n}) \beta_{n} \delta_{n}K_{2} \Vert e_{n} \Vert \\ &{}+\alpha_{n}\beta_{n}\delta_{n}(1- \zeta_{n})\bigl\langle p_{0},J_{\varphi} \bigl[p_{0}-x_{n}-(1-\alpha_{n})e_{n} + \alpha_{n} x_{n}\bigr]\bigr\rangle \\ &{} +(1-\zeta_{n})\mu_{n} \bigl\langle f(p_{0})-p_{0},J_{\varphi} \bigl(f(x_{n})-p_{0}\bigr) \bigr\rangle . \end{aligned}$$

(3.23)

Let $c_{n} = \alpha_{n}\beta_{n}\delta_{n}(1-\zeta_{n})$, then (3.23) reduces to $\Phi(\|x_{n+1}-p_{0}\|) \leq(1-c_{n})\Phi(\|x_{n}-p_{0}\|) + c_{n} \{\langle p_{0}, J_{\varphi}[p_{0}-x_{n}-(1-\alpha_{n})e_{n} + \alpha_{n} x_{n}]\rangle+\frac{\mu_{n}}{\alpha_{n}\beta_{n}\delta_{n}}K_{2}\| f(p_{0})-p_{0}\|\} + K_{2}\|e_{n}\|$.

From Lemma 2.3, the assumptions (vii) and (viii), (3.19), and (3.23), we know that $\Phi(\|x_{n}-p_{0}\|) \rightarrow0$, which implies that $x_{n} \rightarrow p_{0}$, as $n \rightarrow+\infty$. Combining the results in Steps 2 and 3, we can also know that $w_{n} \rightarrow p_{0}$, $u_{n} \rightarrow p_{0}$, as $n \rightarrow +\infty$.

This completes the proof. □

Remark 3.6

Actually, the three sequences $\{x_{n}\}$, $\{w_{n}\}$, and $\{u_{n}\}$ are proved to be strongly convergent to the common zero point $p_{0}$ of an infinite family of m-accretive mappings. The $p_{0}$ in Theorem 3.2 also satisfies (3.2).

Remark 3.7

The assumptions imposed on the real sequences in Theorem 3.2 are reasonable if we take $\alpha_{n} = \delta_{n} = \frac{1}{n^{\frac{1}{3}}}$, $\mu_{n} = \frac{1}{n^{2}}$, $\gamma_{n} = 1-\frac{1}{n^{2}}-\frac{1}{n^{\frac{1}{3}}}$, $\zeta_{n} = \frac{n+1}{2n}$, and $\beta_{n} = \frac{1+n^{\frac{1}{3}}}{2n^{\frac{1}{3}}}$, for $n \geq1$.

4 Example: infinite p-Laplacian-like differential systems

Remark 4.1

In the next of this paper, we shall present an example of infinite p-Laplacian-like differential systems. Based on the example, we shall construct an infinite family of m-accretive mappings, present characteristic of the common zero point of theirs, and demonstrate the applications of Theorems 3.1 and 3.2.

Now, we investigate the following p-Laplacian-like differential systems:

$$ \left \{ \textstyle\begin{array}{l} -\operatorname{div}[(C(x)+|\nabla u|^{2})^{\frac{s_{i}}{2}}|\nabla u|^{m_{i}-1}\nabla u] = f(x),\quad \text{a.e. in } \Omega, \\ -\langle\vartheta,(C(x)+|\nabla u|^{2})^{\frac{s_{i}}{2}}|\nabla u|^{m_{i}-1}\nabla u\rangle = 0,\quad \text{a.e. on } \Gamma, \\ \quad i = 1,2,\ldots, \end{array}\displaystyle \right . $$

(4.1)

where Ω is a bounded conical domain of the Euclidean space $R^{N}$ ($N \geq1$) with its boundary $\Gamma\in C^{1}$ [26].

$m_{i} + s_{i} + 1 = p_{i}$, $m_{i} \geq0$ and $\frac{2N}{N+1} < p_{i} < +\infty$, for $i = 1,2,\ldots$ . $|\cdot|$ is the Euclidean norm in $R^{N}$ and $\langle \cdot,\cdot \rangle$ is the Euclidean inner-product. ϑ is the exterior normal derivative of Γ. $C(x)\geq0$ and $C(x)\in L^{p_{i}}(\Omega)$, $i \in N^{+}$.

We use $\|\cdot\|_{p_{i}}$ and $\|\cdot\|_{1,p_{i},\Omega}$ to denote the norms in $L^{p_{i}}(\Omega)$ and $W^{1,p_{i}}(\Omega)$, respectively. Let $\frac{1}{p_{i}} + \frac{1}{p'_{i}} = 1$.

Remark 4.2

If $s_{i} = 0$ and $m_{i} = p_{i} - 1$, $i \in N^{+}$, then (4.1) is reduced to the case of infinite p-Laplacian differential systems.

Remark 4.3

([27])

The mapping $J_{p_{i}}:L^{p_{i}}(\Omega)\rightarrow L^{p'_{i}}(\Omega)$ defined by $J_{p_{i}}u = |u|^{p_{i}-1}\operatorname{sgn} u$, for $u\in L^{p_{i}}(\Omega)$, is the duality mapping with the gauge function $\varphi(r) = r^{p_{i}-1}$ for $i\in N^{+}$. This presents a vivid example of a duality mapping in $L^{p_{i}}(\Omega)$.

Lemma 4.1

([28])

Let E be a real Banach space and $E^{*}$ be its duality space. If $B: E \rightarrow E^{*}$ is maximal monotone and coercive, then B is a surjection.

Lemma 4.2

([29])

For each $i\in N^{+}$, define the mapping $B_{i}:W^{1,p_{i}}(\Omega)\rightarrow(W^{1,p_{i}}(\Omega))^{*}$ by

$$\langle v,B_{i}u\rangle= \int_{\Omega}\bigl\langle \bigl(C(x)+|\nabla u|^{2} \bigr)^{\frac{s_{i}}{2}}|\nabla u |^{m_{i}-1}\nabla u , \nabla v \bigr\rangle \, dx $$

for any $u,v \in W^{1,p_{i}}(\Omega)$. Then $B_{i}$ is everywhere defined, monotone, and hemi-continuous.

Remark 4.4

Based on $B_{i}$, we shall construct two groups of mappings $\widetilde{A}_{i}:L^{2}(\Omega)\rightarrow L^{2}(\Omega)$ and $A_{i}:L^{p_{i}}(\Omega)\rightarrow L^{p_{i}} (\Omega)$ in the following. Since $B_{i}$ may not be coercive, different proof methods are employed while showing that both $\widetilde{A}_{i}$ and $A_{i}$ are m-accretive mappings, compared to the work done in [29].

Definition 4.1

For each $i\in N^{+}$, define the mapping $\widetilde{A}_{i}:L^{2}(\Omega)\rightarrow L^{2}(\Omega)$ in the following way:

• $D(\widetilde{A}_{i})= \{u(x) \in L^{2}(\Omega): \mbox{there exists } h_{i}(x) \in L^{2}(\Omega)\mbox{ such that }h_{i}(x) = B_{i}u \}$, for any $u \in D(\widetilde{A}_{i})$, $\widetilde{A}_{i}u = h_{i}(x)$.

Proposition 4.1

The mapping $\widetilde{A}_{i}$ is m-accretive.

Proof

First, for every $\lambda> 0$, the mapping $T_{i,\lambda} : H^{1}(\Omega)\rightarrow2^{(H^{1}(\Omega))^{*}}$ defined by $T_{i,\lambda} = u + \lambda\widetilde{A}_{i}u$ is maximal monotone and coercive. It follows from the fact that $L^{2}(\Omega)\subset(H^{1}(\Omega))^{*}$ and Lemma 4.1 that $R(I+\lambda\widetilde{A}_{i}) = L^{2}(\Omega)$ for every $\lambda > 0$.

Secondly, for any $u_{i} \in D(\widetilde{A}_{i})$, $\langle u_{1}-u_{2}, \widetilde{A}_{i}u_{1}-\widetilde{A}_{i}u_{2}\rangle= \langle u_{1}-u_{2}, B_{i}u_{1}-B_{i}u_{2}\rangle\geq0$ since $B_{i}$ is monotone.

This complete the proof. □

Lemma 4.3

If $f, g \in L^{2}(\Omega)$ and there exist $u , v \in L^{2}(\Omega)$ such that $f = u + \lambda\widetilde{A}_{i}u$, $g = v + \lambda\widetilde{A}_{i}v$. Then

$$\int_{\Omega}|u-v|^{p_{i}}\, dx \leq \int_{\Omega}|f-g|^{p_{i}}\, dx, $$

where $p_{i} \geq2$. (Functions $u(x)$ and $v(x)$ exist from Proposition  4.1.)

Proof

Define $\psi: R\rightarrow R$ by $\psi(t) = \frac{1}{p_{i}}|t|^{p_{i}}$, let $\partial\psi: R\rightarrow R$ denote its subdifferential and for $\mu> 0$, let $(\partial\psi)_{\mu} : R\rightarrow R$ denote the Yosida approximation of ∂ψ. Let $\psi_{\mu}$ be an indefinite integral of $(\partial \psi)_{\mu}$ so that $(\partial\psi)_{\mu}= \partial\psi_{\mu}$. We write $\partial\psi_{\mu}= \chi_{\mu}: R\rightarrow R$ and observe that $\chi_{\mu}$ is monotone, Lipschitz with constant $\frac{1}{\mu}$ and differential everywhere except at $t = 0$.

Since

$$\psi_{\mu}\bigl(f(x)-g(x)\bigr)-\psi_{\mu}\bigl(u(x)-v(x) \bigr)\geq \bigl[\chi_{\mu}\bigl(u(x)-v(x)\bigr)\bigr] \bigl[ \bigl(f(x)-g(x)\bigr)-\bigl(u(x)-v(x)\bigr)\bigr] $$

for $x \in\Omega$, we have

$$\int_{\Omega}\psi_{\mu}(f-g)\, dx \geq \int_{\Omega}\psi _{\mu}(u-v)\, dx + \int_{\Omega}\bigl[\chi_{\mu}(u-v)\bigr] \bigl[(f-g)-(u-v)\bigr]\, dx. $$

So, to prove the lemma, it suffices to prove that

$$\int_{\Omega}\chi_{\mu}(u-v) (\widetilde{A}_{i}u- \widetilde{A}_{i}v)\, dx \geq0 $$

for every $\mu> 0$ in view of the fact that $\psi _{\mu}(t)\uparrow\psi(t)$ for every $t\in R$ and the monotone convergence theorem.

Now, since

$$\begin{aligned}& \int_{\Omega}\chi_{\mu}(u-v) (\widetilde{A}_{i}u- \widetilde{A}_{i}v)\,dx \\& \quad = \bigl\langle \chi_{\mu}(u-v),B_{i}u-B_{i}v \bigr\rangle \\& \quad \geq \int_{\Omega} \bigl[\bigl(C(x)+|\nabla u|^{2} \bigr)^{\frac{s_{i}}{2}}|\nabla u|^{m_{i}+1} - \bigl(C(x)+|\nabla u|^{2}\bigr)^{\frac{s_{i}}{2}}|\nabla u|^{m_{i}}|\nabla v| \\& \qquad {}-\bigl(C(x)+|\nabla v|^{2}\bigr)^{\frac{s_{i}}{2}}|\nabla v|^{m_{i}}|\nabla u|+ \bigl(C(x)+|\nabla v|^{2} \bigr)^{\frac{s_{i}}{2}}|\nabla v|^{m_{i}+1}\bigr] \chi'_{\mu }(u-v) \,dx \\& \quad = \int_{\Omega} \bigl[\bigl(C(x)+|\nabla u|^{2} \bigr)^{\frac{s_{i}}{2}}|\nabla u|^{m_{i}} - \bigl(C(x)+|\nabla v|^{2}\bigr)^{\frac{s_{i}}{2}}|\nabla v|^{m_{i}}\bigr] \bigl( \vert \nabla u\vert - |\nabla v| \bigr)\chi'_{\mu}(u-v) \,dx \\& \quad \geq0, \end{aligned}$$

the last inequality is available since $\chi_{\mu}$ is monotone and $\chi_{\mu}(0)=0$.

This completes the proof. □

Definition 4.2

For $i\in N^{+}$, define the mapping $A_{i}:L^{p_{i}}(\Omega)\rightarrow L^{p_{i}} (\Omega)$ in the following way:

• (i) if $p_{i} \geq2$, $D(A_{i})= \{u(x) \in L^{p_{i}}(\Omega): \mbox{there exists }h_{i}(x) \in L^{p_{i}}(\Omega)\mbox{ such that }h_{i}(x) = B_{i}u \}$, for any $u \in D(A_{i})$, $A_{i} u = h_{i}(x)$;

• (ii) if $\frac{2N}{N+1} < p_{i} < 2$, we define $A_{i}: L^{p_{i}}(\Omega) \rightarrow L^{p_{i}}(\Omega)$ as the $L^{p_{i}}$-closure of $\widetilde{A_{i}}: L^{2}(\Omega)\rightarrow L^{2}(\Omega)$ defined in Definition 4.1.

Proposition 4.2

For $2 \leq p_{i} < + \infty$, the mapping $A_{i}:L^{p_{i}}(\Omega)\rightarrow L^{p_{i}}(\Omega)$ is m-accretive, where $i\in N^{+}$.

Proof

First, we show that

$$R(I+\lambda A_{i}) = L^{p_{i}}(\Omega) $$

for every $\lambda> 0$.

In fact, since $p_{i} \geq2$, then Proposition 4.1 implies that

$$R(I+\lambda\widetilde{A}_{i}) = L^{2}(\Omega)\supseteq L^{p_{i}}(\Omega). $$

Then for any $f(x)\in L^{p_{i}}(\Omega)$, there is a $u\in D(\widetilde{A}_{i})$ such that $f = u+\lambda\widetilde{A}_{i}u$. Since $0 = 0 + \lambda\widetilde{A}_{i}0$, by Lemma 4.3, we have

$$\int_{\Omega}|u|^{p_{i}}\, dx \leq \int_{\Omega}|f|^{p_{i}}\, dx < +\infty. $$

That is, $u\in L^{p_{i}}(\Omega)$ and so $R(I+\lambda A_{i}) = L^{p_{i}}(\Omega)$ in view of the definition of $A_{i}$.

Secondly, we shall show that $A_{i}$ is accretive.

For any $u_{j} \in D(A_{i})$, $j = 1,2$, we are left to show that

$$\bigl\langle |u_{1}-u_{2}|^{p_{i}-1} \operatorname{sgn}(u_{1}-u_{2}), A_{i}u_{1}-A_{i}u_{2} \bigr\rangle \geq0 . $$

It suffices for us to show that

$$\bigl\langle |u_{1}-u_{2}|^{p_{i}-1} \operatorname{sgn}(u_{1}-u_{2}) , B_{i}u_{1}-B_{i}u_{2} \bigr\rangle \geq0. $$

To this aim, take for a constant $k > 0$, define $\chi_{k}: R\rightarrow R$ by

$$\chi_{k}(t) = \bigl\vert (t\wedge k)\vee(-k)\bigr\vert ^{p_{i}-1}\operatorname{sgn}t. $$

Then $\chi_{k}$ is monotone, Lipschitz with $\chi_{k}(0) = 0$, and $\chi' _{k}$ is continuous except at finitely many points on R. This shows that

$$\begin{aligned}& \bigl\langle |u_{1}-u_{2}|^{p_{i}-1} \operatorname{sgn}(u_{1}-u_{2}),B_{i}u_{1}-B_{i}u_{2} \bigr\rangle \\& \quad = \lim_{k\rightarrow+\infty} \int_{\Omega}\bigl\langle \bigl(C(x)+|\nabla u|^{2} \bigr)^{\frac{s_{i}}{2}}|\nabla u|^{m_{i}-1}\nabla u - \bigl(C(x)+|\nabla v|^{2}\bigr)^{\frac{s_{i}}{2}}|\nabla v|^{m_{i}-1}\nabla v,\nabla u - \nabla v\bigr\rangle \\& \qquad {}\times\chi'_{k}(u_{1}-u_{2}) \,dx \\& \quad \geq \int_{\Omega} \bigl[\bigl(C(x)+|\nabla u|^{2} \bigr)^{\frac{s_{i}}{2}}|\nabla u|^{m_{i}} - \bigl(C(x)+|\nabla v|^{2}\bigr)^{\frac{s_{i}}{2}}|\nabla v|^{m_{i}}\bigr] \\& \qquad {}\times\bigl( \vert \nabla u\vert - |\nabla v| \bigr)\chi'_{k}(u_{1}-u_{2}) \,dx \\& \quad \geq0. \end{aligned}$$

This completes the proof. □

Proposition 4.3

For $\frac{2N}{N+1} < p_{i} < 2$, the mapping $A_{i}:L^{p_{i}}(\Omega)\rightarrow L^{p_{i}}(\Omega)$ is m-accretive, where $i \in N^{+}$.

Proof

For $f(x) \in L^{p_{i}}(\Omega)$, we may choose a sequence $\{f_{n}\}\subset L^{2}(\Omega)$ such that $f_{n}(x) \rightarrow f(x)$ in $L^{p_{i}}(\Omega)$, as $n \rightarrow\infty$.

Proposition 4.1 implies that there is a $u_{n} \in L^{2}(\Omega)$ such that $u_{n} + \lambda\widetilde{A}_{i}u_{n} = f_{n}$, for $n \geq1$.

Using Lemma 4.3, we have

$$\int_{\Omega}|u_{n}-u_{m}|^{p_{i}}\, dx \leq \int_{\Omega}|f_{n}-f_{m}|^{p_{i}}\, dx. $$

This implies that there is a $u(x) \in L^{p_{i}}(\Omega)$ such that $u_{n} \rightarrow u$ in $L^{p_{i}}(\Omega)$ and then Definition 4.2 ensures that $R(I+\lambda A_{i})= L^{p_{i}}(\Omega)$.

The nonexpansive property of $(I+\lambda A_{i})^{-1} : L^{p_{i}}(\Omega)\rightarrow L^{p_{i}}(\Omega)$ follows from Lemma 4.3, which implies that $A_{i}$ is accretive.

This completes the proof. □

Lemma 4.4

([30])

Let Ω be a bounded conical domain in $R^{N}$. If $mp>N$, then $W^{m,p}(\Omega) \hookrightarrow\hookrightarrow C_{B}(\Omega)$; if $0< mp\leq N$ and $q_{0}=\frac{Np}{N-mp}$, then $W^{m,p}(\Omega)\hookrightarrow\hookrightarrow L^{q}(\Omega)$, where $1\leq q< q_{0}$ and ‘↪↪’ means compact embedding.

Lemma 4.5

([31])

Let $X_{0}$ denote the closed subspace of all constant functions in $W^{1,p}(\Omega)$. Let X be the quotient space $W^{1,p}(\Omega)/X_{0}$. For $u\in W^{1,p}(\Omega)$, define the mapping $P:W^{1,p}(\Omega) \rightarrow X_{0}$ by $Pu=\frac{1}{\operatorname{meas}(\Omega)}\int_{\Omega}u \, dx$. Then there is a constant $C>0$, such that $\forall u\in W^{1,p}(\Omega)$,

$$\|u-Pu\|_{p}\leq C\|\nabla u\|_{(L^{p}(\Omega))^{N}}. $$

Theorem 4.1

For $i\in R^{+}$, $N(A_{i}) = \{u \in L^{p_{i}}(\Omega): u(x) \equiv \mathit{constant}\textit{ on }\Omega\}$.

Proof

(i) $p_{i} \geq2$.

Let $u(x) \in N(A_{i})$, then $0 = \langle u,B_{i}u \rangle= \int_{\Omega}(C(x)+|\nabla u|^{2})^{\frac{s_{i}}{2}}|\nabla u |^{m_{i}+1}\, dx \geq\int_{\Omega}|\nabla u|^{p_{i}}\, dx \geq0$, which implies that $u(x) \equiv \mathrm{constant}$. That is, $N(A_{i}) \subset \{u \in L^{p_{i}}(\Omega): u(x) \equiv \mathrm{constant}\mbox{ on }\Omega\}$.

On the other hand, suppose $u(x) \equiv \mathrm{constant}$. Then $0 = \langle v, B_{i}u\rangle$, for $\forall v \in W^{1,p_{i}}(\Omega)$. Then $u \in N(A_{i})$. The result follows.

(ii) $\frac{2N}{N+1} < p_{i} < 2$.

Suppose $u \in L^{p_{i}}(\Omega)$ and $u(x) \equiv \mathrm{constant}$. Let $u_{n} \equiv u$, then $\widetilde{A}_{i}u_{n} = 0$ in view of (i). Thus $\{u \in L^{p_{i}}(\Omega): u(x) \equiv \mathrm{constant}\mbox{ on }\Omega\} \subset N(A_{i})$ in view of the definition of $A_{i}$.

On the other hand, let $u \in N(A_{i})$. Then there exist $\{u_{n}\}$ and $\{f_{n}\}$ in $L^{2}(\Omega)$ such that $u_{n} \rightarrow u$, $f_{n} \rightarrow0$ in $L^{p_{i}}(\Omega)$ and $\widetilde{A}_{i} u_{n} = f_{n}$. Now, define the following functions:

$$ \eta(t) = \left \{ \textstyle\begin{array}{l@{\quad}l} |t|^{p_{i}-1}\operatorname{sgn}t, &\mbox{if }|t|\geq1, \\ t , &\mbox{if }|t| < 1 \end{array}\displaystyle \right . $$

and

$$ \xi(t) = \left \{ \textstyle\begin{array}{l@{\quad}l} |t|^{2-\frac{2}{p_{i}}}\operatorname{sgn}t,& \mbox{if }|t|\geq1, \\ t ,& \mbox{if }|t| < 1. \end{array}\displaystyle \right . $$

Then for $u \in L^{2}(\Omega)$, the function $t \in R \rightarrow \int_{\Omega}\xi(u+t) \, dx \in R$ is continuous and $\lim_{t \rightarrow\pm\infty}\int_{\Omega}\xi(u+t) \, dx = \pm\infty$. Therefore, there exists $t_{u} \in R$ such that $\int_{\Omega}\xi(u+t_{u})\, dx = 0$. So, for $u_{n} \in L^{2}(\Omega)$, we may assume that there exists $t_{n} \in R$ such that $\int_{\Omega}\xi(u_{n}+t_{n}) \, dx = 0$ and $\widetilde{A}_{i} u_{n} = f_{n}$, for $n\geq1$. Let $v_{n} = u_{n} + t_{n}$, then $\widetilde{A}_{i} v_{n} = \widetilde{A}_{i} u_{n} = f_{n}$, for $n\geq1$.

Now, we compute the following:

$$\begin{aligned}& \Vert f_{n}\Vert _{p_{i}}\biggl( \int_{|v_{n}|\leq1} |v_{n}|^{p_{i}'}\,dx + \int _{|v_{n}|\geq1} |v_{n}|^{2} \,dx \biggr)^{\frac{p_{i} -1}{p_{i}}} \\& \quad \geq \Vert f_{n}\Vert _{p_{i}}\biggl( \int_{|v_{n}|\leq1} |v_{n}|^{p_{i}'}\,dx + \int _{|v_{n}|\geq1} |v_{n}|^{p_{i}} \,dx \biggr)^{\frac{p_{i} -1}{p_{i}}} \\& \quad = \Vert f_{n}\Vert _{p_{i}}\bigl\Vert \eta(v_{n})\bigr\Vert _{p_{i}'}\geq\bigl\langle \eta(v_{n}),f_{n} \bigr\rangle \\& \quad = \bigl\langle \eta(v_{n}),\widetilde{A}_{i}v_{n} \bigr\rangle \geq \int_{\Omega}|\nabla v_{n}|^{p_{i}} \eta'(v_{n})\,dx \\& \quad \geq \mathrm{const.} \int_{\Omega}\bigl\vert \nabla\bigl(\xi(v_{n})\bigr) \bigr\vert ^{p_{i}}\,dx . \end{aligned}$$

(4.2)

Using Lemma 4.5,

$$ \int_{\Omega}\bigl\vert \nabla\bigl(\xi(v_{n})\bigr) \bigr\vert ^{p_{i}}\, dx \geq \mathrm{const.}\bigl\Vert \xi(v_{n})\bigr\Vert ^{p_{i}}_{1,p_{i},\Omega}. $$

(4.3)

Then Lemma 4.4 implies that

$$\begin{aligned} \bigl\Vert \xi(v_{n})\bigr\Vert ^{p_{i}}_{1,p_{i},\Omega} \geq& \mathrm{const.} \bigl\Vert \xi(v_{n})\bigr\Vert ^{p_{i}}_{p_{i}'} \\ =& \mathrm{const.}\biggl( \int_{|v_{n}|\leq1} |v_{n}|^{p_{i}'}\, dx + \int_{|v_{n}|\geq1} |v_{n}|^{(2-\frac{2}{p_{i}})p_{i}'} \, dx \biggr)^{\frac{p_{i}}{p_{i}'}} \\ =& \mathrm{const.}\biggl( \int_{|v_{n}|\leq1} |v_{n}|^{p_{i}'}\, dx + \int_{|v_{n}|\geq1} |v_{n}|^{2}\, dx \biggr)^{\frac{p_{i}}{p_{i}'}}. \end{aligned}$$

(4.4)

From (4.2)-(4.4), we know that $\|\xi(v_{n})\|^{p_{i}}_{1,p_{i},\Omega}\leq \mathrm{const.}\|f_{n}\|_{p_{i}} \rightarrow0$, as $n \rightarrow\infty$. Then $\xi(v_{n}) \rightarrow0$ in $L^{p_{i}'}(\Omega)$. Since the Nemytskyi mapping $u \in L^{p_{i}'}(\Omega) \rightarrow\xi^{-1}(u) \in L^{p_{i}}(\Omega)$ is continuous, $v_{n} \rightarrow0$ in $L^{p_{i}}(\Omega)$. Then $u(x) \equiv \mathrm{constant}$. Thus $N(A_{i})\subset\{u \in L^{p_{i}}(\Omega): u(x) \equiv \mathrm{constant}\mbox{ on }\Omega\}$.

This completes the proof. □

Remark 4.5

Two infinite families of m-accretive mappings related to nonlinear p-Laplacian-like differential systems are constructed, which emphasizes the importance of the study on approximating common zero points of infinite nonlinear m-accretive mappings.

Remark 4.6

Theorem 4.1 helps us to see the assumption that $\bigcap_{i = 1}^{\infty}N(A_{i})\neq\emptyset$ in Theorems 3.1 and 3.2 are valid.

Definition 4.3

If $f(x) \equiv0$ in (4.1), then the solution $u(x)$ of (4.1) is called the equilibrium solution to the p-Laplacian-like differential systems (4.1).

Theorem 4.2

For $i \in N^{+}$, $u(x) \in N(A_{i})$ if and only if $u(x)$ is the equilibrium solution of (4.1).

Proof

It is easy to see that if $u(x) \in N(A_{i})$, then $u(x)$ is the equilibrium solution of (4.1). On the other hand, if $u(x)$ is the equilibrium solution of (4.1), then

$$-\operatorname{div}\bigl[\bigl(C(x)+|\nabla u|^{2} \bigr)^{\frac{s_{i}}{2}}|\nabla u|^{m_{i}-1}\nabla u\bigr] = 0,\quad \mbox{a.e. in }\Omega. $$

Thus for $\forall\varphi\in C_{0}^{\infty}(\Omega)$, by using the property of generalized function, we have

$$\begin{aligned} 0 &= \bigl\langle \varphi, -\operatorname{div}\bigl[\bigl(C(x)+|\nabla u|^{2}\bigr)^{\frac {s_{i}}{2}}|\nabla u|^{m_{i}-1}\nabla u\bigr]\bigr\rangle \\ & = \int_{\Omega}-\operatorname{div}\bigl[\bigl(C(x)+|\nabla u|^{2}\bigr)^{\frac{s_{i}}{2}}|\nabla u|^{m_{i}-1}\nabla u\bigr] \varphi \, dx \\ & = \int_{\Omega}\bigl\langle \bigl(C(x)+|\nabla u|^{2} \bigr)^{\frac{s_{i}}{2}}|\nabla u|^{m_{i}-1}\nabla u, \nabla\varphi\bigr\rangle \, dx = \langle\varphi,B_{i}u \rangle, \end{aligned}$$

which implies that $u(x) \in N(A_{i})$.

This completes the proof. □

Remark 4.7

Based on Theorem 4.2, Theorems 3.1 and 3.2 can be applied to approximate the equilibrium solution of (4.1).