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# Some results on zero points of m-accretive operators in reflexive Banach spaces

Fixed Point Theory and Applications20142014:118

https://doi.org/10.1186/1687-1812-2014-118

• Accepted: 30 April 2014
• Published:

## Abstract

A modified proximal point algorithm is proposed for treating common zero points of a finite family of m-accretive operators. A strong convergence theorem is established in a reflexive, strictly convex Banach space with the uniformly Gâteaux differentiable norm.

## Keywords

• accretive operator
• nonexpansive mapping
• resolvent
• fixed point
• zero point

## 1 Introduction and preliminaries

Let E be a Banach space and let ${E}^{\ast }$ be the dual of E. Let $〈\cdot ,\cdot 〉$ denote the pairing between E and ${E}^{\ast }$. The normalized duality mapping $J:E\to {2}^{{E}^{\ast }}$ is defined by
$J\left(x\right)=\left\{f\in {E}^{\ast }:〈x,f〉={\parallel x\parallel }^{2}={\parallel f\parallel }^{2}\right\},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in E.$

A Banach space E is said to strictly convex if and only if $\parallel x\parallel =\parallel y\parallel =\parallel \left(1-\lambda \right)x+\lambda y\parallel$ for $x,y\in E$ and $0<\lambda <1$ implies that $x=y$. Let ${U}_{E}=\left\{x\in E:\parallel x\parallel =1\right\}$. The norm of E is said to be Gâteaux differentiable if the limit ${lim}_{t\to 0}\frac{\parallel x+ty\parallel -\parallel x\parallel }{t}$ exists for each $x,y\in {U}_{E}$. In this case, E is said to be smooth. The norm of E is said to be uniformly Gâteaux differentiable if for each $y\in {U}_{E}$, the limit is attained uniformly for all $x\in {U}_{E}$. The norm of E is said to be Fréchet differentiable if for each $x\in {U}_{E}$, the limit is attained uniformly for all $y\in {U}_{E}$. The norm of E is said to be uniformly Fréchet differentiable if the limit is attained uniformly for all $x,y\in {U}_{E}$. It is well known that (uniform) Fréchet differentiability of the norm of E implies (uniform) Gâteaux differentiability of the norm of E.

Let ${\rho }_{E}:\left[0,\mathrm{\infty }\right)\to \left[0,\mathrm{\infty }\right)$ be the modulus of smoothness of E by
${\rho }_{E}\left(t\right)=sup\left\{\frac{\parallel x+y\parallel -\parallel x-y\parallel }{2}-1:x\in {U}_{E},\parallel y\parallel \le t\right\}.$

A Banach space E is said to be uniformly smooth if $\frac{{\rho }_{E}\left(t\right)}{t}\to 0$ as $t\to 0$. It is well known that if the norm of E is uniformly Gâteaux differentiable, then the duality mapping J is single valued and uniformly norm to weak continuous on each bounded subset of E.

Recall that a closed convex subset C of a Banach space E is said to have a normal structure if for each bounded closed convex subset K of C which contains at least two points, there exists an element x of K which is not a diametral point of K, i.e., $sup\left\{\parallel x-y\parallel :y\in K\right\}, where $d\left(K\right)$ is the diameter of K.

Let D be a nonempty subset of a set C. Let ${\mathit{Proj}}_{D}:C\to D$. Q is said to be
1. (1)

sunny if for each $x\in C$ and $t\in \left(0,1\right)$, we have ${\mathit{Proj}}_{D}\left(tx+\left(1-t\right){\mathit{Proj}}_{D}x\right)={\mathit{Proj}}_{D}x$;

2. (2)

a contraction if ${\mathit{Proj}}_{D}^{2}={\mathit{Proj}}_{D}$;

3. (3)

a sunny nonexpansive retraction if ${\mathit{Proj}}_{D}$ is sunny, nonexpansive, and a contraction.

D is said to be a nonexpansive retract of C if there exists a nonexpansive retraction from C onto D. The following result, which was established in , describes a characterization of sunny nonexpansive retractions on a smooth Banach space.

Let E be a smooth Banach space and let C be a nonempty subset of E. Let ${\mathit{Proj}}_{C}:E\to C$ be a retraction and ${J}_{\phi }$ be the duality mapping on E. Then the following are equivalent:
1. (1)

${\mathit{Proj}}_{C}$ is sunny and nonexpansive;

2. (2)

$〈x-{\mathit{Proj}}_{C}x,{J}_{\phi }\left(y-{\mathit{Proj}}_{C}x\right)〉\le 0$, $\mathrm{\forall }x\in E$, $y\in C$;

3. (3)

${\parallel {\mathit{Proj}}_{C}x-{\mathit{Proj}}_{C}y\parallel }^{2}\le 〈x-y,{J}_{\phi }\left({\mathit{Proj}}_{C}x-{\mathit{Proj}}_{C}y\right)〉$, $\mathrm{\forall }x,y\in E$.

It is well known that if E is a Hilbert space, then a sunny nonexpansive retraction ${\mathit{Proj}}_{C}$ is coincident with the metric projection from E onto C. Let C be a nonempty closed convex subset of a smooth Banach space E, let $x\in E$, and let ${x}_{0}\in C$. Then we have from the above that ${x}_{0}={\mathit{Proj}}_{C}x$ if and only if $〈x-{x}_{0},{J}_{\phi }\left(y-{x}_{0}\right)〉\le 0$ for all $y\in C$, where ${\mathit{Proj}}_{C}$ is a sunny nonexpansive retraction from E onto C. For more additional information on nonexpansive retracts, see  and the references therein.

Let C be a nonempty closed convex subset of E. Let $T:C\to C$ be a mapping. In this paper, we use $F\left(T\right)$ to denote the set of fixed points of T. Recall that T is said to be an α-contractive mapping iff there exists a constant $\alpha \in \left[0,1\right)$ such that $\parallel Tx-Ty\parallel \le \alpha \parallel x-y\parallel$, $\mathrm{\forall }x,y\in C$. The Picard iterative process is an efficient method to study fixed points of α-contractive mappings. It is well known that α-contractive mappings have a unique fixed point. T is said to be nonexpansive iff $\parallel Tx-Ty\parallel \le \parallel x-y\parallel$, $\mathrm{\forall }x,y\in C$. It is well known that nonexpansive mappings have fixed points if the set C is closed and convex, and the space E is uniformly convex. The Krasnoselski-Mann iterative process is an efficient method for studying fixed points of nonexpansive mappings. The Krasnoselski-Mann iterative process generates a sequence $\left\{{x}_{n}\right\}$ in the following manner:
${x}_{1}\in C,\phantom{\rule{1em}{0ex}}{x}_{n+1}={\alpha }_{n}T{x}_{n}+\left(1-{\alpha }_{n}\right){x}_{n},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1.$
It is well known that the Krasnoselski-Mann iterative process only has weak convergence for nonexpansive mappings in infinite-dimensional Hilbert spaces; see  for more details and the references therein. In many disciplines, including economics, image recovery, quantum physics, and control theory, problems arise in infinite-dimensional spaces. In such problems, strong convergence (norm convergence) is often much more desirable than weak convergence, for it translates the physically tangible property that the energy $\parallel {x}_{n}-x\parallel$ of the error between the iterate ${x}_{n}$ and the solution x eventually becomes arbitrarily small. To improve the weak convergence of a Krasnoselski-Mann iterative process, so-called hybrid projections have been considered; see  for more details and the references therein. The Halpern iterative process was initially introduced in ; see  for more details and the references therein. The Halpern iterative process generates a sequence $\left\{{x}_{n}\right\}$ in the following manner:
${x}_{1}\in C,\phantom{\rule{1em}{0ex}}{x}_{n+1}={\alpha }_{n}u+\left(1-{\alpha }_{n}\right)T{x}_{n},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1,$

where ${x}_{1}$ is an initial and u is a fixed element in C. Strong convergence of Halpern iterative process does not depend on metric projections. The Halpern iterative process has recently been extensively studied for treating accretive operators; see  and the references therein.

Let I denote the identity operator on E. An operator $A\subset E×E$ with domain $D\left(A\right)=\left\{z\in E:Az\ne \mathrm{\varnothing }\right\}$ and range $R\left(A\right)=\bigcup \left\{Az:z\in D\left(A\right)\right\}$ is said to be accretive if for each ${x}_{i}\in D\left(A\right)$ and ${y}_{i}\in A{x}_{i}$, $i=1,2$, there exists $j\left({x}_{1}-{x}_{2}\right)\in J\left({x}_{1}-{x}_{2}\right)$ such that $〈{y}_{1}-{y}_{2},j\left({x}_{1}-{x}_{2}\right)〉\ge 0$. An accretive operator A is said to be m-accretive if $R\left(I+rA\right)=E$ for all $r>0$. In this paper, we use ${A}^{-1}\left(0\right)$ to denote the set of zero points of A. For an accretive operator A, we can define a nonexpansive single valued mapping ${J}_{r}:R\left(I+rA\right)\to D\left(A\right)$ by ${J}_{r}={\left(I+rA\right)}^{-1}$ for each $r>0$, which is called the resolvent of A.

Now, we are in a position to give the lemmas to prove main results.

Lemma 1.1 

Let $\left\{{a}_{n}\right\}$, $\left\{{b}_{n}\right\}$, $\left\{{c}_{n}\right\}$, and $\left\{{d}_{n}\right\}$ be four nonnegative real sequences satisfying ${a}_{n+1}\le \left(1-{b}_{n}\right){a}_{n}+{b}_{n}{c}_{n}+{d}_{n}$, $\mathrm{\forall }n\ge {n}_{0}$, where ${n}_{0}$ is some positive integer, $\left\{{b}_{n}\right\}$ is a number sequence in $\left(0,1\right)$ such that ${\sum }_{n={n}_{0}}^{\mathrm{\infty }}{b}_{n}=\mathrm{\infty }$, $\left\{{c}_{n}\right\}$ is a number sequence such that ${lim sup}_{n\to \mathrm{\infty }}{c}_{n}\le 0$, and $\left\{{d}_{n}\right\}$ is a positive number sequence such that ${\sum }_{n={n}_{0}}^{\mathrm{\infty }}{d}_{n}<\mathrm{\infty }$. Then ${lim}_{n\to \mathrm{\infty }}{a}_{n}=0$.

Lemma 1.2 

Let C be a closed convex subset of a strictly convex Banach space E. Let $N\ge 1$ be some positive integer and let ${T}_{i}:C\to C$ be a nonexpansive mapping for each $i\in \left\{1,2,\dots ,N\right\}$. Let $\left\{{\delta }_{i}\right\}$ be a real number sequence in $\left(0,1\right)$ with ${\sum }_{i=1}^{N}{\delta }_{i}=1$. Suppose that ${\bigcap }_{i=1}^{N}F\left({T}_{i}\right)$ is nonempty. Then the mapping ${\bigcap }_{i=1}^{N}{T}_{i}$ is defined to be nonexpansive with $F\left({\bigcap }_{i=1}^{N}{T}_{i}\right)={\bigcap }_{i=1}^{N}F\left({T}_{i}\right)$.

Lemma 1.3 

Let $\left\{{x}_{n}\right\}$ and $\left\{{y}_{n}\right\}$ be bounded sequences in a Banach space E and let ${\beta }_{n}$ be a sequence in $\left[0,1\right]$ with $0<{lim inf}_{n\to \mathrm{\infty }}{\beta }_{n}\le {lim sup}_{n\to \mathrm{\infty }}{\beta }_{n}<1$. Suppose that ${x}_{n+1}=\left(1-{\beta }_{n}\right){y}_{n}+{\beta }_{n}{x}_{n}$ for all $n\ge 0$ and
$\underset{n\to \mathrm{\infty }}{lim sup}\left(\parallel {y}_{n+1}-{y}_{n}\parallel -\parallel {x}_{n+1}-{x}_{n}\parallel \right)\le 0.$

Then ${lim}_{n\to \mathrm{\infty }}\parallel {y}_{n}-{x}_{n}\parallel =0$.

Lemma 1.4 

Let E be a real reflexive Banach space with the uniformly Gâteaux differentiable norm and the normal structure, and let C be a nonempty closed convex subset of E. Let $f:C\to C$ be α-contractive mapping and let $T:C\to C$ be a nonexpansive mapping with a fixed point. Let $\left\{{x}_{t}\right\}$ be a sequence generated by the following: ${x}_{t}=tf\left({x}_{t}\right)+\left(1-t\right)T{x}_{t}$, where $t\in \left(0,1\right)$. Then $\left\{{x}_{t}\right\}$ converges strongly as $t\to 0$ to a fixed point ${x}^{\ast }$ of T, which is the unique solution in $F\left(T\right)$ to the following variational inequality: $〈f\left({x}^{\ast }\right)-{x}^{\ast },j\left({x}^{\ast }-p\right)〉\ge 0$, $\mathrm{\forall }p\in F\left(T\right)$.

## 2 Main results

Theorem 2.1 Let E be a real reflexive, strictly convex Banach space with the uniformly Gâteaux differentiable norm. Let $N\ge 1$ be some positive integer. Let ${A}_{m}$ be an m-accretive operator in E for each $m\in \left\{1,2,\dots ,N\right\}$. Assume that $C:={\bigcap }_{m=1}^{N}\overline{D\left({A}_{m}\right)}$ is convex and has the normal structure. Let $f:C\to C$ be an α-contractive mapping. Let $\left\{{\alpha }_{n}\right\}$, $\left\{{\beta }_{n}\right\}$, and $\left\{{\gamma }_{n}\right\}$ be real number sequences in $\left(0,1\right)$ with the restriction ${\alpha }_{n}+{\beta }_{n}+{\gamma }_{n}=1$. Let $\left\{{\delta }_{n,i}\right\}$ be a real number sequence in $\left(0,1\right)$ with the restriction ${\delta }_{n,1}+{\delta }_{n,2}+\cdots +{\delta }_{n,N}=1$. Let $\left\{{r}_{m}\right\}$ be a positive real numbers sequence and $\left\{{e}_{n,i}\right\}$ a sequence in E for each $i\in \left\{1,2,\dots ,N\right\}$. Assume that ${\bigcap }_{i=1}^{N}{A}_{i}^{-1}\left(0\right)$ is not empty. Let $\left\{{x}_{n}\right\}$ be a sequence generated in the following manner:
${x}_{1}\in C,\phantom{\rule{1em}{0ex}}{x}_{n+1}={\alpha }_{n}f\left({x}_{n}\right)+{\beta }_{n}{x}_{n}+{\gamma }_{n}\sum _{i=1}^{N}{\delta }_{n,i}{J}_{{r}_{i}}\left({x}_{n}+{e}_{n,i}\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1,$
where ${J}_{{r}_{i}}={\left(I+{r}_{i}{A}_{i}\right)}^{-1}$. Assume that the control sequences $\left\{{\alpha }_{n}\right\}$, $\left\{{\beta }_{n}\right\}$, $\left\{{\gamma }_{n}\right\}$, and $\left\{{\delta }_{n,i}\right\}$ satisfy the following restrictions:
1. (a)

${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$, ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_{n}=\mathrm{\infty }$;

2. (b)

$0<{lim inf}_{n\to \mathrm{\infty }}{\beta }_{n}\le {lim sup}_{n\to \mathrm{\infty }}{\beta }_{n}<1$;

3. (c)

${\sum }_{n=1}^{\mathrm{\infty }}\parallel {e}_{n,m}\parallel <\mathrm{\infty }$;

4. (d)

${lim}_{n\to \mathrm{\infty }}{\delta }_{n,i}={\delta }_{i}\in \left(0,1\right)$.

Then the sequence $\left\{{x}_{n}\right\}$ converges strongly to $\overline{x}$, which is the unique solution to the following variational inequality: $〈f\left(\overline{x}\right)-\overline{x},J\left(p-\overline{x}\right)〉\le 0$, $\mathrm{\forall }p\in {\bigcap }_{i=1}^{N}{A}_{i}^{-1}\left(0\right)$.

Proof Put ${y}_{n}={\sum }_{i=1}^{N}{\delta }_{n,i}{J}_{{r}_{i}}\left({x}_{n}+{e}_{n,i}\right)$. Fixing $p\in {\bigcap }_{i=1}^{N}{A}_{i}^{-1}\left(0\right)$, we have
$\begin{array}{rl}\parallel {y}_{n}-p\parallel & \le \sum _{i=1}^{N}{\delta }_{n,i}\parallel {J}_{{r}_{i}}\left({x}_{n}+{e}_{n,i}\right)-p\parallel \\ \le \sum _{i=1}^{N}{\delta }_{n,i}\parallel \left({x}_{n}+{e}_{n,i}\right)-p\parallel \\ \le \parallel {x}_{n}-p\parallel +\sum _{i=1}^{N}\parallel {e}_{n,i}\parallel .\end{array}$
Hence, we have
$\begin{array}{rcl}\parallel {x}_{n+1}-p\parallel & \le & {\alpha }_{n}\parallel f\left({x}_{n}\right)-p\parallel +{\beta }_{n}\parallel {x}_{n}-p\parallel +{\gamma }_{n}\parallel {y}_{n}-p\parallel \\ \le & {\alpha }_{n}\alpha \parallel {x}_{n}-p\parallel +{\alpha }_{n}\parallel f\left(p\right)-p\parallel +{\beta }_{n}\parallel {x}_{n}-p\parallel +{\gamma }_{n}\parallel {x}_{n}-p\parallel +{\gamma }_{n}\sum _{i=1}^{N}\parallel {e}_{n,i}\parallel \\ \le & \left(1-{\alpha }_{n}\left(1-\alpha \right)\right)\parallel {x}_{n}-p\parallel +{\alpha }_{n}\left(1-\alpha \right)\frac{\parallel f\left(p\right)-p\parallel }{1-\alpha }+\sum _{i=1}^{N}\parallel {e}_{n,i}\parallel \\ \le & max\left\{\parallel {x}_{n}-p\parallel ,\parallel f\left(p\right)-p\parallel \right\}+\sum _{i=1}^{N}\parallel {e}_{n,i}\parallel \\ ⋮\\ \le & max\left\{\parallel {x}_{1}-p\parallel ,\parallel f\left(p\right)-p\parallel \right\}+\sum _{j=1}^{\mathrm{\infty }}\sum _{i=1}^{N}\parallel {e}_{j,i}\parallel .\end{array}$
This proves that the sequence $\left\{{x}_{n}\right\}$ is bounded, and so is $\left\{{y}_{n}\right\}$. Since
$\begin{array}{rl}{y}_{n}-{y}_{n-1}=& \sum _{i=1}^{N}{\delta }_{n,i}\left({J}_{{r}_{m}}\left({x}_{n}+{e}_{n,i}\right)-{J}_{{r}_{i}}\left({x}_{n-1}+{e}_{n-1,i}\right)\right)\\ +\sum _{i=1}^{N}\left({\delta }_{n,i}-{\delta }_{n-1,i}\right){J}_{{r}_{i}}\left({x}_{n-1}+{e}_{n-1,i}\right),\end{array}$
we have
$\begin{array}{rcl}\parallel {y}_{n}-{y}_{n-1}\parallel & \le & \sum _{i=1}^{N}{\delta }_{n,i}\parallel {J}_{{r}_{i}}\left({x}_{n}+{e}_{n,i}\right)-{J}_{{r}_{i}}\left({x}_{n-1}+{e}_{n-1,i}\right)\parallel \\ +\sum _{i=1}^{N}|{\delta }_{n,i}-{\delta }_{n-1,i}|\parallel {J}_{{r}_{i}}\left({x}_{n-1}+{e}_{n-1,i}\right)\parallel \\ \le & \parallel {x}_{n}-{x}_{n-1}\parallel +\sum _{i=1}^{N}\parallel {e}_{n,i}\parallel +\sum _{i=1}^{N}\parallel {e}_{n-1,i}\parallel \\ +\sum _{i=1}^{N}|{\delta }_{n,i}-{\delta }_{n-1,i}|\parallel {J}_{{r}_{i}}\left({x}_{n-1}+{e}_{n-1,i}\right)\parallel \\ \le & \parallel {x}_{n}-{x}_{n-1}\parallel +\sum _{i=1}^{N}\parallel {e}_{n,i}\parallel +\sum _{i=1}^{N}\parallel {e}_{n-1,i}\parallel +{M}_{1}\sum _{i=1}^{N}|{\delta }_{n,i}-{\delta }_{n-1,i}|,\end{array}$
where ${M}_{1}$ is an appropriate constant such that
${M}_{1}=max\left\{\underset{n\ge 1}{sup}\parallel {J}_{{r}_{1}}\left({x}_{n}+{e}_{n,1}\right)\parallel ,\underset{n\ge 1}{sup}\parallel {J}_{{r}_{2}}\left({x}_{n}+{e}_{n,2}\right)\parallel ,\dots ,\underset{n\ge 1}{sup}\parallel {J}_{{r}_{N}}\left({x}_{n}+{e}_{n,N}\right)\parallel \right\}.$
Define a sequence $\left\{{z}_{n}\right\}$ by ${z}_{n}:=\frac{{x}_{n+1}-{\beta }_{n}{x}_{n}}{1-{\beta }_{n}}$, that is, ${x}_{n+1}={\beta }_{n}{x}_{n}+\left(1-{\beta }_{n}\right){z}_{n}$. It follows that
$\begin{array}{rcl}\parallel y{z}_{n}-{z}_{n-1}\parallel & \le & \frac{{\alpha }_{n}}{1-{\beta }_{n}}\parallel f\left({x}_{n}\right)-{y}_{n}\parallel +\frac{{\alpha }_{n-1}}{1-{\beta }_{n-1}}\parallel f\left({x}_{n-1}\right)-{y}_{n-1}\parallel +\parallel {y}_{n}-{y}_{n-1}\parallel \\ \le & \frac{{\alpha }_{n}}{1-{\beta }_{n}}\parallel f\left({x}_{n}\right)-{y}_{n}\parallel +\frac{{\alpha }_{n-1}}{1-{\beta }_{n-1}}\parallel f\left({x}_{n-1}\right)-{y}_{n-1}\parallel +\parallel {x}_{n}-{x}_{n-1}\parallel \\ +\sum _{i=1}^{N}|{\delta }_{n,i}-{\delta }_{n-1,i}|\parallel {J}_{{r}_{i}}{x}_{n-1}\parallel \\ \le & \frac{{\alpha }_{n}}{1-{\beta }_{n}}\parallel f\left({x}_{n}\right)-{y}_{n}\parallel +\frac{{\alpha }_{n-1}}{1-{\beta }_{n-1}}\parallel f\left({x}_{n-1}\right)-{y}_{n-1}\parallel +\parallel {x}_{n}-{x}_{n-1}\parallel \\ +{M}_{2}\left(\sum _{i=1}^{N}|{\delta }_{n,i}-{\delta }_{i}|+\sum _{i=1}^{N}|{\delta }_{i}-{\delta }_{n-1,i}|\right),\end{array}$
where ${M}_{2}$ is an appropriate constant such that
${M}_{2}=max\left\{\underset{n\ge 1}{sup}\parallel {J}_{{r}_{1}}{x}_{n}\parallel ,\underset{n\ge 1}{sup}\parallel {J}_{{r}_{2}}{x}_{n}\parallel ,\dots ,\underset{n\ge 1}{sup}\parallel {J}_{{r}_{N}}{x}_{n}\parallel \right\}.$
This implies that
$\begin{array}{c}\parallel {z}_{n}-{z}_{n-1}\parallel -\parallel {x}_{n}-{x}_{n-1}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{{\alpha }_{n}}{1-{\beta }_{n}}\parallel f\left({x}_{n}\right)-{y}_{n}\parallel +\frac{{\alpha }_{n-1}}{1-{\beta }_{n-1}}\parallel f\left({x}_{n-1}\right)-{y}_{n-1}\parallel \hfill \\ \phantom{\rule{2em}{0ex}}+{M}_{2}\left(\sum _{i=1}^{N}|{\delta }_{n,i}-{\delta }_{i}|+\sum _{i=1}^{N}|{\delta }_{i}-{\delta }_{n-1,i}|\right).\hfill \end{array}$
From the restrictions (a), (b), (c), and (d), we find that
$\underset{n\to \mathrm{\infty }}{lim sup}\left(\parallel {z}_{n}-{z}_{n-1}\parallel -\parallel {x}_{n}-{x}_{n-1}\parallel \right)\le 0.$
Using Lemma 1.4, we find that ${lim}_{n\to \mathrm{\infty }}\parallel {z}_{n}-{x}_{n}\parallel =0$. This further shows that ${lim sup}_{n\to \mathrm{\infty }}\parallel {x}_{n+1}-{x}_{n}\parallel =0$. Put $T={\sum }_{i=1}^{N}{\delta }_{i}{J}_{{r}_{i}}$. It follows from Lemma 1.3 that T is nonexpansive with $F\left(T\right)={\bigcap }_{i=1}^{N}F\left({J}_{{r}_{i}}\right)={\bigcap }_{i=1}^{N}{A}_{i}^{-1}\left(0\right)$. Note that
$\begin{array}{c}\parallel {x}_{n}-T{x}_{n}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}\le \parallel {x}_{n}-{x}_{n+1}\parallel +\parallel {x}_{n+1}-T{x}_{n}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}\le \parallel {x}_{n}-{x}_{n+1}\parallel +{\alpha }_{n}\parallel f\left({x}_{n}\right)-T{x}_{n}\parallel +{\beta }_{n}\parallel {x}_{n}-T{x}_{n}\parallel +{\gamma }_{n}\parallel {y}_{n}-T{x}_{n}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}\le \parallel {x}_{n}-{x}_{n+1}\parallel +{\alpha }_{n}\parallel f\left({x}_{n}\right)-T{x}_{n}\parallel +{\beta }_{n}\parallel {x}_{n}-T{x}_{n}\parallel +{M}_{2}\sum _{i=1}^{N}|{\delta }_{n,i}-{\delta }_{i}|.\hfill \end{array}$
This implies that
$\left(1-{\beta }_{n}\right)\parallel {x}_{n}-T{x}_{n}\parallel \le \parallel {x}_{n}-{x}_{n+1}\parallel +{\alpha }_{n}\parallel f\left({x}_{n}\right)-T{x}_{n}\parallel +{M}_{2}\sum _{i=1}^{N}|{\delta }_{n,i}-{\delta }_{i}|.$
It follows from the restrictions (a), (b), and (d) that
$\underset{n\to \mathrm{\infty }}{lim}\parallel T{x}_{n}-{x}_{n}\parallel =0.$
Now, we are in a position to prove that ${lim sup}_{n\to \mathrm{\infty }}〈f\left(\overline{x}\right)-\overline{x},J\left({x}_{n}-\overline{x}\right)〉\le 0$, where $\overline{x}={lim}_{t\to 0}{x}_{t}$, and ${x}_{t}$ solves the fixed point equation
${x}_{t}=tf\left({x}_{t}\right)+\left(1-t\right)T{x}_{t},\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in \left(0,1\right).$
It follows that
$\begin{array}{rcl}{\parallel {x}_{t}-{x}_{n}\parallel }^{2}& =& t〈f\left({x}_{t}\right)-{x}_{n},J\left({x}_{t}-{x}_{n}\right)〉+\left(1-t\right)〈T{x}_{t}-{x}_{n},j\left({x}_{t}-{x}_{n}\right)〉\\ =& t〈f\left({x}_{t}\right)-{x}_{t},J\left({x}_{t}-{x}_{n}\right)〉+t〈{x}_{t}-{x}_{n},J\left({x}_{t}-{x}_{n}\right)〉\\ +\left(1-t\right)〈T{x}_{t}-T{x}_{n},J\left({x}_{t}-{x}_{n}\right)〉+\left(1-t\right)〈T{x}_{n}-{x}_{n},J\left({x}_{t}-{x}_{n}\right)〉\\ \le & t〈f\left({x}_{t}\right)-{x}_{t},J\left({x}_{t}-{x}_{n}\right)〉+{\parallel {x}_{t}-{x}_{n}\parallel }^{2}+\parallel T{x}_{n}-{x}_{n}\parallel \parallel {x}_{t}-{x}_{n}\parallel ,\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in \left(0,1\right).\end{array}$
This implies that
$〈{x}_{t}-f\left({x}_{t}\right),J\left({x}_{t}-{x}_{n}\right)〉\le \frac{1}{t}\parallel T{x}_{n}-{x}_{n}\parallel \parallel {x}_{t}-{x}_{n}\parallel ,\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in \left(0,1\right).$
Since ${lim}_{n\to \mathrm{\infty }}\parallel T{x}_{n}-{x}_{n}\parallel =0$, we find that ${lim sup}_{n\to \mathrm{\infty }}〈{x}_{t}-f\left({x}_{t}\right),J\left({x}_{t}-{x}_{n}\right)〉\le 0$. Since J is strong to weak uniformly continuous on bounded subsets of E, we find that
$\begin{array}{c}|〈f\left(\overline{x}\right)-\overline{x},J\left({x}_{n}-\overline{x}\right)〉-〈{x}_{t}-f\left({x}_{t}\right),J\left({x}_{t}-{x}_{n}\right)〉|\hfill \\ \phantom{\rule{1em}{0ex}}\le |〈f\left(\overline{x}\right)-\overline{x},J\left({x}_{n}-\overline{x}\right)〉-〈f\left(\overline{x}\right)-\overline{x},J\left({x}_{n}-{x}_{t}\right)〉|\hfill \\ \phantom{\rule{2em}{0ex}}+|〈f\left(\overline{x}\right)-\overline{x},J\left({x}_{n}-{x}_{t}\right)〉-〈{x}_{t}-f\left({x}_{t}\right),J\left({x}_{t}-{x}_{n}\right)〉|\hfill \\ \phantom{\rule{1em}{0ex}}\le |〈f\left(\overline{x}\right)-\overline{x},J\left({x}_{n}-\overline{x}\right)-J\left({x}_{n}-{x}_{t}\right)〉|+|〈f\left(\overline{x}\right)-\overline{x}+{x}_{t}-f\left({x}_{t}\right),J\left({x}_{n}-{x}_{t}\right)〉|\hfill \\ \phantom{\rule{1em}{0ex}}\le \parallel f\left({x}_{t}\right)-\overline{x}\parallel \parallel J\left({x}_{n}-\overline{x}\right)-J\left({x}_{n}-{x}_{t}\right)\parallel +\left(1+\alpha \right)\parallel \overline{x}-{x}_{t}\parallel \parallel {x}_{n}-{x}_{t}\parallel .\hfill \end{array}$
Since ${x}_{t}\to \overline{x}$, as $t\to 0$, we have
$\underset{t\to 0}{lim}|〈f\left(\overline{x}\right)-\overline{x},J\left({x}_{n}-\overline{x}\right)〉-〈f\left({x}_{t}\right)-{x}_{t},J\left({x}_{n}-{x}_{t}\right)〉|=0.$
For $ϵ>0$, there exists $\delta >0$ such that $\mathrm{\forall }t\in \left(0,\delta \right)$, we have
$〈f\left(\overline{x}\right)-\overline{x},J\left({x}_{n}-\overline{x}\right)〉\le 〈f\left({x}_{t}\right)-{x}_{t},J\left({x}_{n}-{x}_{t}\right)〉+ϵ.$

This implies that ${lim sup}_{n\to \mathrm{\infty }}〈f\left(\overline{x}\right)-\overline{x},J\left({x}_{n}-\overline{x}\right)〉\le 0$.

Finally, we show that ${x}_{n}\to \overline{x}$ as $n\to \mathrm{\infty }$. Since ${\parallel \cdot \parallel }^{2}$ is convex, we see that
$\begin{array}{rcl}{\parallel {y}_{n}-\overline{x}\parallel }^{2}& =& {\parallel \sum _{i=1}^{N}{\delta }_{n,i}{J}_{{r}_{i}}\left({x}_{n}+{e}_{n,i}\right)-\overline{x}\parallel }^{2}\\ \le & \sum _{i=1}^{N}{\delta }_{n,i}{\parallel {J}_{{r}_{i}}\left({x}_{n}+{e}_{n,i}\right)-\overline{x}\parallel }^{2}\\ \le & {\parallel {x}_{n}-\overline{x}\parallel }^{2}+\sum _{i=1}^{N}\parallel {e}_{n,i}\parallel \left(\parallel {e}_{n,i}\parallel +2\parallel {x}_{n}-\overline{x}\parallel \right).\end{array}$
It follows that
$\begin{array}{rcl}{\parallel {x}_{n+1}-\overline{x}\parallel }^{2}& =& {\alpha }_{n}〈f\left({x}_{n}\right)-\overline{x},J\left({x}_{n+1}-\overline{x}\right)〉+{\beta }_{n}〈{x}_{n}-\overline{x},J\left({x}_{n+1}-\overline{x}\right)〉\\ +{\gamma }_{n}〈{y}_{n}-\overline{x},J\left({x}_{n+1}-\overline{x}\right)〉\\ \le & {\alpha }_{n}\alpha \parallel {x}_{n}-\overline{x}\parallel \parallel {x}_{n+1}-\overline{x}\parallel +{\alpha }_{n}〈f\left(\overline{x}\right)-\overline{x},J\left({x}_{n+1}-\overline{x}\right)〉\\ +{\beta }_{n}\parallel {x}_{n}-\overline{x}\parallel \parallel {x}_{n+1}-\overline{x}\parallel +{\gamma }_{n}\parallel {y}_{n}-\overline{x}\parallel \parallel {x}_{n+1}-\overline{x}\parallel \\ \le & \frac{{\alpha }_{n}\alpha }{2}\left({\parallel {x}_{n}-\overline{x}\parallel }^{2}+{\parallel {x}_{n+1}-\overline{x}\parallel }^{2}\right)+{\alpha }_{n}〈f\left(\overline{x}\right)-\overline{x},J\left({x}_{n+1}-\overline{x}\right)〉\\ +\frac{{\beta }_{n}}{2}\left({\parallel {x}_{n}-\overline{x}\parallel }^{2}+{\parallel {x}_{n+1}-\overline{x}\parallel }^{2}\right)+\frac{{\gamma }_{n}}{2}{\parallel {x}_{n}-\overline{x}\parallel }^{2}\\ +\sum _{i=1}^{N}\parallel {e}_{n,i}\parallel \left(\parallel {e}_{n,i}\parallel +2\parallel {x}_{n}-\overline{x}\parallel \right)+\frac{{\gamma }_{n}}{2}{\parallel {x}_{n+1}-\overline{x}\parallel }^{2}.\end{array}$
Hence, we have
$\begin{array}{rcl}{\parallel {x}_{n+1}-\overline{x}\parallel }^{2}& \le & \left(1-{\alpha }_{n}\left(1-\alpha \right)\right){\parallel {x}_{n}-\overline{x}\parallel }^{2}+2{\alpha }_{n}〈f\left(\overline{x}\right)-\overline{x},J\left({x}_{n+1}-\overline{x}\right)〉\\ +\sum _{i=1}^{N}\parallel {e}_{n,i}\parallel \left(\parallel {e}_{n,i}\parallel +2\parallel {x}_{n}-\overline{x}\parallel \right).\end{array}$

Using Lemma 1.1, we find ${x}_{n}\to \overline{x}$ as $n\to \mathrm{\infty }$. This completes the proof. □

Remark 2.2 There are many spaces satisfying the restriction in Theorem 2.1, for example ${L}^{p}$, where $p>1$.

Corollary 2.3 Let E be a Hilbert space and let $N\ge 1$ be some positive integer. Let ${A}_{m}$ be a maximal monotone operator in E for each $m\in \left\{1,2,\dots ,N\right\}$. Assume that $C:={\bigcap }_{m=1}^{N}\overline{D\left({A}_{m}\right)}$ is convex and has the normal structure. Let $f:C\to C$ be an α-contractive mapping. Let $\left\{{\alpha }_{n}\right\}$, $\left\{{\beta }_{n}\right\}$, and $\left\{{\gamma }_{n}\right\}$ be real number sequences in $\left(0,1\right)$ with the restriction ${\alpha }_{n}+{\beta }_{n}+{\gamma }_{n}=1$. Let $\left\{{\delta }_{n,i}\right\}$ be a real number sequence in $\left(0,1\right)$ with the restriction ${\delta }_{n,1}+{\delta }_{n,2}+\cdots +{\delta }_{n,N}=1$. Let $\left\{{r}_{m}\right\}$ be a positive real numbers sequence and $\left\{{e}_{n,i}\right\}$ a sequence in E for each $i\in \left\{1,2,\dots ,N\right\}$. Assume that ${\bigcap }_{i=1}^{N}{A}_{i}^{-1}\left(0\right)$ is not empty. Let $\left\{{x}_{n}\right\}$ be a sequence generated in the following manner:
${x}_{1}\in C,\phantom{\rule{1em}{0ex}}{x}_{n+1}={\alpha }_{n}f\left({x}_{n}\right)+{\beta }_{n}{x}_{n}+{\gamma }_{n}\sum _{i=1}^{N}{\delta }_{n,i}{J}_{{r}_{i}}\left({x}_{n}+{e}_{n,i}\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1,$
where ${J}_{{r}_{i}}={\left(I+{r}_{i}{A}_{i}\right)}^{-1}$. Assume that the control sequences $\left\{{\alpha }_{n}\right\}$, $\left\{{\beta }_{n}\right\}$, $\left\{{\gamma }_{n}\right\}$, and $\left\{{\delta }_{n,i}\right\}$ satisfy the following restrictions:
1. (a)

${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$, ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_{n}=\mathrm{\infty }$;

2. (b)

$0<{lim inf}_{n\to \mathrm{\infty }}{\beta }_{n}\le {lim sup}_{n\to \mathrm{\infty }}{\beta }_{n}<1$;

3. (c)

${\sum }_{n=1}^{\mathrm{\infty }}\parallel {e}_{n,m}\parallel <\mathrm{\infty }$;

4. (d)

${lim}_{n\to \mathrm{\infty }}{\delta }_{n,i}={\delta }_{i}\in \left(0,1\right)$.

Then the sequence $\left\{{x}_{n}\right\}$ converges strongly to $\overline{x}$, which is the unique solution to the following variational inequality: $〈f\left(\overline{x}\right)-\overline{x},p-\overline{x}〉\le 0$, $\mathrm{\forall }p\in {\bigcap }_{i=1}^{N}{A}_{i}^{-1}\left(0\right)$.

## 3 Applications

In this section, we consider a variational inequality problem. Let $A:C\to {E}^{\ast }$ be a single valued monotone operator which is hemicontinuous; that is, continuous along each line segment in C with respect to the weak topology of ${E}^{\ast }$. Consider the following variational inequality:
The solution set of the variational inequality is denoted by $VI\left(C,A\right)$. Recall that the normal cone ${N}_{C}\left(x\right)$ for C at a point $x\in C$ is defined by
${N}_{C}\left(x\right)=\left\{{x}^{\ast }\in {E}^{\ast }:〈y-x,{x}^{\ast }〉\le 0,\mathrm{\forall }y\in C\right\}.$

Now, we are in a position to give the convergence theorem.

Theorem 3.1 Let E be a real reflexive, strictly convex Banach space with the uniformly Gâteaux differentiable norm. Let $N\ge 1$ be some positive integer and let C be nonempty closed and convex subset of E. Let ${A}_{i}:C\to {E}^{\ast }$ a single valued, monotone and hemicontinuous operator. Assume that ${\bigcap }_{i=1}^{N}VI\left(C,{A}_{i}\right)$ is not empty and C has the normal structure. Let $f:C\to C$ be an α-contractive mapping. Let $\left\{{\alpha }_{n}\right\}$, $\left\{{\beta }_{n}\right\}$, and $\left\{{\gamma }_{n}\right\}$ be real number sequences in $\left(0,1\right)$ with the restriction ${\alpha }_{n}+{\beta }_{n}+{\gamma }_{n}=1$. Let $\left\{{\delta }_{n,i}\right\}$ be a real number sequence in $\left(0,1\right)$ with the restriction ${\delta }_{n,1}+{\delta }_{n,2}+\cdots +{\delta }_{n,N}=1$. Let $\left\{{r}_{m}\right\}$ be a positive real numbers sequence and $\left\{{e}_{n,i}\right\}$ a sequence in E for each $i\in \left\{1,2,\dots ,N\right\}$. Let $\left\{{x}_{n}\right\}$ be a sequence generated in the following manner:
${x}_{1}\in C,\phantom{\rule{1em}{0ex}}{x}_{n+1}={\alpha }_{n}f\left({x}_{n}\right)+{\beta }_{n}{x}_{n}+{\gamma }_{n}\sum _{i=1}^{N}{\delta }_{n,i}VI\left(C,{A}_{i}+\frac{1}{{r}_{i}}\left(I-{x}_{n}\right)\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1.$
Assume that the control sequences $\left\{{\alpha }_{n}\right\}$, $\left\{{\beta }_{n}\right\}$, $\left\{{\gamma }_{n}\right\}$, and $\left\{{\delta }_{n,i}\right\}$ satisfy the following restrictions:
1. (a)

${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$, ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_{n}=\mathrm{\infty }$;

2. (b)

$0<{lim inf}_{n\to \mathrm{\infty }}{\beta }_{n}\le {lim sup}_{n\to \mathrm{\infty }}{\beta }_{n}<1$;

3. (c)

${\sum }_{n=1}^{\mathrm{\infty }}\parallel {e}_{n,m}\parallel <\mathrm{\infty }$;

4. (d)

${lim}_{n\to \mathrm{\infty }}{\delta }_{n,i}={\delta }_{i}\in \left(0,1\right)$.

Then the sequence $\left\{{x}_{n}\right\}$ converges strongly to $\overline{x}$, which is the unique solution to the following variational inequality: $〈f\left(\overline{x}\right)-\overline{x},J\left(p-\overline{x}\right)〉\le 0$, $\mathrm{\forall }p\in {\bigcap }_{i=1}^{N}VI\left(C,{A}_{i}\right)$.

Proof Define a mapping ${T}_{i}\subset E×{E}^{\ast }$ by
${T}_{i}x:=\left\{\begin{array}{cc}{A}_{i}x+{N}_{C}x,\hfill & x\in C,\hfill \\ \mathrm{\varnothing },\hfill & x\notin C.\hfill \end{array}$
From Rockafellar , we find that ${T}_{i}$ is maximal monotone with ${T}_{i}^{-1}\left(0\right)=VI\left(C,{A}_{i}\right)$. For each ${r}_{i}>0$, and ${x}_{n}\in E$, we see that there exists a unique ${x}_{{r}_{i}}\in D\left({T}_{i}\right)$ such that ${x}_{n}\in {x}_{{r}_{i}}+{r}_{i}{T}_{i}\left({x}_{{r}_{i}}\right)$, where ${x}_{{r}_{i}}={\left(I+{r}_{i}{T}_{i}\right)}^{-1}{x}_{n}$. Notice that
${y}_{n,i}=VI\left(C,{A}_{i}+\frac{1}{{r}_{i}}\left(I-{x}_{n}\right)\right),$
which is equivalent to
$〈y-{y}_{n,i},{A}_{i}{y}_{n,i}+\frac{1}{{r}_{i}}\left({y}_{n,i}-{x}_{n}\right)〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }y\in C,$

that is, $-{A}_{i}{y}_{n,i}+\frac{1}{{r}_{i}}\left({x}_{n}-{y}_{n,i}\right)\in {N}_{C}\left({y}_{n,i}\right)$. This implies that ${y}_{n,i}={\left(I+{r}_{i}{T}_{i}\right)}^{-1}{x}_{n}$. Using Theorem 2.1, we find the desired conclusion immediately. □

From Theorem 3.1, the following result is not hard to derive.

Corollary 3.2 Let E be a real reflexive, strictly convex Banach space with the uniformly Gâteaux differentiable norm. Let C be nonempty closed and convex subset of E. Let $A:C\to {E}^{\ast }$ a single valued, monotone and hemicontinuous operator with $VI\left(C,A\right)$. Assume that C has the normal structure. Let $f:C\to C$ be an α-contractive mapping. Let $\left\{{\alpha }_{n}\right\}$, $\left\{{\beta }_{n}\right\}$, and $\left\{{\gamma }_{n}\right\}$ be real number sequences in $\left(0,1\right)$ with the restriction ${\alpha }_{n}+{\beta }_{n}+{\gamma }_{n}=1$. Let $\left\{{x}_{n}\right\}$ be a sequence generated in the following manner:
${x}_{1}\in C,\phantom{\rule{1em}{0ex}}{x}_{n+1}={\alpha }_{n}f\left({x}_{n}\right)+{\beta }_{n}{x}_{n}+{\gamma }_{n}VI\left(C,A+\frac{1}{r}\left(I-{x}_{n}\right)\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1.$
Assume that the control sequences $\left\{{\alpha }_{n}\right\}$, $\left\{{\beta }_{n}\right\}$, and $\left\{{\gamma }_{n}\right\}$ satisfy the following restrictions:
1. (a)

${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$, ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_{n}=\mathrm{\infty }$;

2. (b)

$0<{lim inf}_{n\to \mathrm{\infty }}{\beta }_{n}\le {lim sup}_{n\to \mathrm{\infty }}{\beta }_{n}<1$.

Then the sequence $\left\{{x}_{n}\right\}$ converges strongly to $\overline{x}$, which is the unique solution to the following variational inequality: $〈f\left(\overline{x}\right)-\overline{x},J\left(p-\overline{x}\right)〉\le 0$, $\mathrm{\forall }p\in VI\left(C,{A}_{i}\right)$.

## Declarations

### Acknowledgements

The authors are grateful to the editor and the reviewers for useful suggestions which improved the contents of the article.

## Authors’ Affiliations

(1) 