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# Proximal point algorithms for zero points of nonlinear operators

Fixed Point Theory and Applications20142014:42

https://doi.org/10.1186/1687-1812-2014-42

• Received: 26 November 2013
• Accepted: 4 February 2014
• Published:

## Abstract

A proximal point algorithm with double computational errors for treating zero points of accretive operators is investigated. Strong convergence theorems of zero points are established in a Banach space.

MSC:47H05, 47H09, 47H10, 65J15.

## Keywords

• accretive operator
• fixed point
• nonexpansive mapping
• proximal point algorithm
• zero point

## 1 Introduction

In this paper, we are concerned with the problem of finding zero points of an operator $A:E\to {2}^{{E}^{\ast }}$; that is, finding $x\in domA$ such that $0\in Ax$. The domain domA of A is defined by the set $\left\{x\in E:Ax\ne 0\right\}$. Many important problems have reformulations which require finding zero points, for instance, evolution equations, complementarity problems, mini-max problems, variational inequalities and optimization problems; see  and the references therein. One of the most popular techniques for solving the inclusion problem goes back to the work of Browder . One of the basic ideas in the case of a Hilbert space H is reducing the above inclusion problem to a fixed point problem of the operator ${R}_{A}:H\to {2}^{H}$ defined by ${R}_{A}={\left(I+A\right)}^{-1}$, which is called the classical resolvent of A. If A has some monotonicity conditions, the classical resolvent of A is with full domain and firmly nonexpansive. Rockafellar introduced the algorithm ${x}_{n+1}={R}_{A}{x}_{n}$ and call it the proximal point algorithm; for more detail, see  and the references therein. Regularization methods recently have been investigated for treating zero points of monotone operators; for  and the references therein. Methods for finding zero points of monotone mappings in the framework of Hilbert spaces are based on the good properties of the resolvent ${R}_{A}$, but these properties are not available in the framework of Banach spaces.

In this paper, we investigate a proximal point algorithm with double computational errors based on regularization ideas in the framework of Banach spaces. The organization of this paper is as follows. In Section 2, we provide some necessary preliminaries. In Section 3, strong convergence of the algorithm is obtained in a general Banach space. In Section 4, an application is provided to support the main results.

## 2 Preliminaries

In what follows, we always assume that E is Banach space with the dual ${E}^{\ast }$. Recall that a closed convex subset C of E is said to have 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. It is well known that a closed convex subset of uniformly convex Banach space has the normal structure and a compact convex subset of a Banach space has the normal structure; for more details, see  and the references therein.

Let ${U}_{E}=\left\{x\in E:\parallel x\parallel =1\right\}$. E is said to be smooth or said to be have a Gâteaux differentiable norm if the limit ${lim}_{t\to 0}\frac{\parallel x+ty\parallel -\parallel x\parallel }{t}$ exists for each $x,y\in {U}_{E}$. E is said to have a uniformly Gâteaux differentiable norm if for each $y\in {U}_{E}$, the limit is attained uniformly for all $x\in {U}_{E}$. E is said to be uniformly smooth or said to have a uniformly Fréchet differentiable norm if the limit is attained uniformly for $x,y\in {U}_{E}$. Let $〈\cdot ,\cdot 〉$ denote the pairing between E and ${E}^{\ast }$. The normalized duality mapping $J:E\to {2}^{{E}^{\ast }}$ is defined by
$J\left(x\right)=\left\{f\in {E}^{\ast }:〈x,f〉={\parallel x\parallel }^{2}={\parallel f\parallel }^{2}\right\}$

for all $x\in E$. In the sequel, we use j to denote the single-valued normalized duality mapping. It is 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.

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 contractive if there exists a constant $\alpha \in \left(0,1\right)$ such that
$\parallel Tx-Ty\parallel \le \alpha \parallel x-y\parallel ,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in C.$
For such a case, we also call T an α-contraction. T is said to be nonexpansive if
$\parallel Tx-Ty\parallel \le \parallel x-y\parallel ,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in C.$

Let D be a nonempty subset of C. Let $Q:C\to D$. Q is said to be a contraction if ${Q}^{2}=Q$; sunny if for each $x\in C$ and $t\in \left(0,1\right)$, we have $Q\left(tx+\left(1-t\right)Qx\right)=Qx$; sunny nonexpansive retraction if Q is sunny, nonexpansive, and contraction. K 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 C be a nonempty subset of E. Let $Q:E\to C$ be a retraction and j be the normalized duality mapping on E. Then the following are equivalent:
1. (1)

Q is sunny and nonexpansive;

2. (2)

${\parallel Qx-Qy\parallel }^{2}\le 〈x-y,j\left(Qx-Qy\right)〉$, $\mathrm{\forall }x,y\in E$;

3. (3)

$〈x-Qx,j\left(y-Qx\right)〉\le 0$, $\mathrm{\forall }x\in E$, $y\in C$.

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 a real Hilbert space, an operator A is m-accretive if and only if A is maximal monotone. In this paper, we use ${A}^{-1}\left(0\right)$ to denote the set of zeros 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.

In order to prove our main results, we also need the following lemmas.

Lemma 2.1 

Let E be a Banach space, and A an m-accretive operator. For $\lambda >0$, $\mu >0$, and $x\in E$, we have
${J}_{\lambda }x={J}_{\mu }\left(\frac{\mu }{\lambda }x+\left(1-\frac{\mu }{\lambda }\right){J}_{\lambda }x\right),$

where ${J}_{\lambda }={\left(I+\lambda A\right)}^{-1}$ and ${J}_{\mu }={\left(I+\mu A\right)}^{-1}$.

Lemma 2.2 

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

Lemma 2.3 

Let $\left\{{x}_{n}\right\}$ and $\left\{{y}_{n}\right\}$ be bounded sequences in a Banach space E, and $\left\{{\beta }_{n}\right\}$ be a sequence in $\left(0,1\right)$ with
$0<\underset{n\to \mathrm{\infty }}{lim inf}{\beta }_{n}\le \underset{n\to \mathrm{\infty }}{lim sup}{\beta }_{n}<1.$
Suppose that ${x}_{n+1}=\left(1-{\beta }_{n}\right){y}_{n}+{\beta }_{n}{x}_{n}$, $\mathrm{\forall }n\ge 1$ 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 2.4 

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

## 3 Main results

Theorem 3.1 Let E be a real reflexive Banach space with the uniformly Gâteaux differentiable norm and A be an m-accretive operators in E. Assume that $C:=\overline{D\left(A\right)}$ is convex and has the normal structure. Let $f:C\to C$ be a fixed α-contraction. Let $\left\{{\alpha }_{n}\right\}$, $\left\{{\beta }_{n}\right\}$, $\left\{{\gamma }_{n}\right\}$, and $\left\{{\delta }_{n}\right\}$ be real number sequences in $\left(0,1\right)$ such that ${\alpha }_{n}+{\beta }_{n}+{\gamma }_{n}+{\delta }_{n}=1$. Let ${Q}_{C}$ be the sunny nonexpansive retraction from E onto C and $\left\{{x}_{n}\right\}$ be a sequence generated in the following manner:
${x}_{0}\in C,\phantom{\rule{1em}{0ex}}{x}_{n+1}={\alpha }_{n}f\left({x}_{n}\right)+{\beta }_{n}{J}_{{r}_{n}}\left({x}_{n}+{e}_{n+1}\right)+{\gamma }_{n}{x}_{n}+{\delta }_{n}{Q}_{C}\left({g}_{n}\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 0,$
(ϒ)
where $\left\{{e}_{n}\right\}$ is a sequence in E, $\left\{{g}_{n}\right\}$ is a bounded sequence in E, $\left\{{r}_{n}\right\}$ is a positive real numbers sequence, and ${J}_{{r}_{n}}={\left(I+{r}_{n}A\right)}^{-1}$. Assume that ${A}^{-1}\left(0\right)$ is not empty and the above control sequences satisfy the following restrictions:
1. (a)

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

2. (b)

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

3. (c)

${\sum }_{n=1}^{\mathrm{\infty }}\parallel {e}_{n}\parallel <\mathrm{\infty }$ and ${\sum }_{n=0}^{\mathrm{\infty }}{\delta }_{n}<\mathrm{\infty }$;

4. (d)

${r}_{n}\ge \mu$ for each $n\ge 1$ and ${lim}_{n\to \mathrm{\infty }}|{r}_{n}-{r}_{n+1}|=0$.

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

Proof Fixing $p\in {A}^{-1}\left(0\right)$, we find that
$\begin{array}{rl}\parallel {x}_{1}-p\parallel \le & {\alpha }_{0}\parallel f\left({x}_{0}\right)-p\parallel +{\beta }_{0}\parallel {J}_{{r}_{0}}\left({x}_{0}+{e}_{1}\right)-p\parallel +{\gamma }_{0}\parallel {x}_{0}-p\parallel +{\delta }_{0}\parallel {Q}_{C}\left({g}_{0}\right)-p\parallel \\ \le & {\alpha }_{0}\alpha \parallel {x}_{0}-p\parallel +{\alpha }_{0}\parallel f\left(p\right)-p\parallel +{\beta }_{0}\parallel {x}_{0}-p\parallel +{\beta }_{0}\parallel {e}_{1}\parallel \\ +{\gamma }_{0}\parallel {x}_{0}-p\parallel +{\delta }_{0}\parallel {g}_{0}-p\parallel \\ \le & \left(1-{\alpha }_{0}\left(1-\alpha \right)\right)\parallel {x}_{0}-p\parallel +{\alpha }_{0}\parallel f\left(p\right)-p\parallel +\parallel {e}_{1}\parallel +{\delta }_{0}\parallel {g}_{0}-p\parallel .\end{array}$
(3.1)
Next, we prove that
$\parallel {x}_{n}-p\parallel \le {M}_{1}+\sum _{i=1}^{n}\parallel {e}_{i}\parallel +\sum _{i=1}^{n-1}{\delta }_{i}\parallel {g}_{i}\parallel ,$
(3.2)
where ${M}_{1}=max\left\{\parallel {x}_{0}-p\parallel ,\frac{\parallel f\left(p\right)-p\parallel }{1-\alpha }\right\}<\mathrm{\infty }$. In view of (3.1), we find that (3.2) holds for $n=1$. We assume that the result holds for some m. Notice that
$\begin{array}{rl}\parallel {x}_{m+1}-p\parallel \le & {\alpha }_{m}\parallel f\left({x}_{m}\right)-p\parallel +{\beta }_{n}\parallel {J}_{{r}_{m}}\left({x}_{m}+{e}_{m+1}\right)-p\parallel +{\gamma }_{m}\parallel {x}_{m}-p\parallel \\ +{\delta }_{m}\parallel {Q}_{C}\left({g}_{m}\right)-p\parallel \\ \le & {\alpha }_{m}\alpha \parallel {x}_{m}-p\parallel +{\alpha }_{m}\parallel f\left(p\right)-p\parallel +{\beta }_{m}\parallel {x}_{m}-p\parallel +{\beta }_{m}\parallel {e}_{m+1}\parallel \\ +{\gamma }_{m}\parallel {x}_{m}-p\parallel +{\delta }_{m}\parallel {Q}_{C}\left({g}_{m}\right)-p\parallel \\ \le & \left(1-{\alpha }_{m}\left(1-\alpha \right)\right)\parallel {x}_{m}-p\parallel +{\alpha }_{m}\left(1-\alpha \right)\frac{\parallel f\left(p\right)-p\parallel }{1-\alpha }\\ +\parallel {e}_{m+1}\parallel +{\delta }_{m}\parallel {g}_{m}-p\parallel \\ \le & {M}_{1}+\sum _{i=1}^{m+1}\parallel {e}_{i}\parallel +\sum _{i=1}^{m}{\delta }_{i}\parallel {g}_{i}\parallel .\end{array}$
This shows that (3.2) holds. In view of the restriction (c), we find that the sequence $\left\{{x}_{n}\right\}$ is bounded. Put ${y}_{n}={J}_{{r}_{n}}\left({x}_{n}+{e}_{n+1}\right)$ and ${z}_{n}=\frac{{x}_{n+1}-{\gamma }_{n}{x}_{n}}{1-{\gamma }_{n}}$. Now, we compute $\parallel {z}_{n+1}-{z}_{n}\parallel$. Note that
$\begin{array}{rl}{z}_{n+1}-{z}_{n}=& \frac{{\alpha }_{n+1}}{1-{\gamma }_{n+1}}f\left({x}_{n+1}\right)+\frac{{\beta }_{n+1}}{1-{\gamma }_{n+1}}{y}_{n+1}+\frac{{\delta }_{n+1}}{1-{\gamma }_{n+1}}{Q}_{C}\left({g}_{n+1}\right)\\ -\frac{{\alpha }_{n}}{1-{\gamma }_{n}}f\left({x}_{n}\right)-\frac{{\beta }_{n}}{1-{\gamma }_{n}}{y}_{n}-\frac{{\delta }_{n}}{1-{\gamma }_{n}}{Q}_{C}\left({g}_{n}\right)\\ =& \frac{{\alpha }_{n+1}}{1-{\gamma }_{n+1}}\left(f\left({x}_{n+1}\right)-{y}_{n+1}\right)+{y}_{n+1}+\frac{{\delta }_{n+1}}{1-{\gamma }_{n+1}}\left({Q}_{C}\left({g}_{n+1}\right)-{y}_{n+1}\right)\\ -\frac{{\alpha }_{n}}{1-{\gamma }_{n}}\left(f\left({x}_{n}\right)-{y}_{n}\right)-{y}_{n}-\frac{{\delta }_{n}}{1-{\gamma }_{n}}\left({Q}_{C}\left({g}_{n}\right)-{y}_{n}\right).\end{array}$
This yields
$\begin{array}{rl}\parallel {z}_{n+1}-{z}_{n}\parallel \le & \frac{{\alpha }_{n+1}}{1-{\gamma }_{n+1}}\parallel f\left({x}_{n+1}\right)-{y}_{n+1}\parallel +\parallel {y}_{n+1}-{y}_{n}\parallel +\frac{{\alpha }_{n}}{1-{\gamma }_{n}}\parallel f\left({x}_{n}\right)-{y}_{n}\parallel \\ +\frac{{\delta }_{n+1}}{1-{\gamma }_{n+1}}\parallel {Q}_{C}\left({g}_{n+1}\right)-{y}_{n+1}\parallel +\frac{{\delta }_{n}}{1-{\gamma }_{n}}\parallel {Q}_{C}\left({g}_{n}\right)-{y}_{n}\parallel .\end{array}$
(3.3)
Next, we estimate $\parallel {y}_{n+1}-{y}_{n}\parallel$. In view of Lemma 2.1, we find that
$\begin{array}{r}\parallel {y}_{n}-{y}_{n+1}\parallel \\ \phantom{\rule{1em}{0ex}}\le \parallel \frac{{r}_{n}}{{r}_{n+1}}\left({x}_{n}+{e}_{n+1}\right)+\left(1-\frac{{r}_{n}}{{r}_{n+1}}\right){J}_{{r}_{n+1}}\left({x}_{n}+{e}_{n+1}\right)-\left({x}_{n+1}+{e}_{n+2}\right)\parallel \\ \phantom{\rule{1em}{0ex}}=\parallel \frac{{r}_{n}}{{r}_{n+1}}\left(\left({x}_{n}+{e}_{n+1}\right)-\left({x}_{n+1}+{e}_{n+2}\right)\right)+\frac{{r}_{n+1}-{r}_{n}}{{r}_{n+1}}\left({J}_{{r}_{n+1}}\left({x}_{n}+{e}_{n+1}\right)-\left({x}_{n+1}+{e}_{n+2}\right)\right)\parallel \\ \phantom{\rule{1em}{0ex}}\le \parallel {x}_{n}-{x}_{n+1}\parallel +\parallel {e}_{n+1}\parallel +\parallel {e}_{n+2}\parallel +\frac{{M}_{2}}{\mu }\left({r}_{n+1}-{r}_{n}\right),\end{array}$
(3.4)
where ${M}_{2}$ is an appropriate constant such that
${M}_{2}\ge \underset{n\ge 1}{sup}\left\{\parallel {J}_{{r}_{n+1}}\left({x}_{n}+{e}_{n}\right)-\left({x}_{n+1}+{e}_{n+1}\right)\parallel \right\}.$
Substituting (3.4) into (3.3), we arrive at
$\begin{array}{r}\parallel {z}_{n+1}-{z}_{n}\parallel -\parallel {x}_{n}-{x}_{n+1}\parallel \\ \phantom{\rule{1em}{0ex}}\le \frac{{\alpha }_{n+1}}{1-{\gamma }_{n+1}}\parallel f\left({x}_{n+1}\right)-{y}_{n+1}\parallel +\parallel {e}_{n+1}\parallel +\parallel {e}_{n+2}\parallel +\frac{{M}_{2}}{\mu }\left({r}_{n+1}-{r}_{n}\right)+\frac{{\alpha }_{n}}{1-{\gamma }_{n}}\parallel f\left({x}_{n}\right)-{y}_{n}\parallel \\ \phantom{\rule{2em}{0ex}}+\frac{{\delta }_{n+1}}{1-{\gamma }_{n+1}}\parallel {Q}_{C}\left({g}_{n+1}\right)-{y}_{n+1}\parallel +\frac{{\delta }_{n}}{1-{\gamma }_{n}}\parallel {Q}_{C}\left({g}_{n}\right)-{y}_{n}\parallel .\end{array}$
In view of the restrictions (a), (b), (c), and (d), we find that
$\underset{n\to \mathrm{\infty }}{lim sup}\left(\parallel {z}_{n+1}-{z}_{n}\parallel -\parallel {x}_{n}-{x}_{n+1}\parallel \right)\le 0.$
It follows from Lemma 2.3 that ${lim}_{n\to \mathrm{\infty }}\parallel {z}_{n}-{x}_{n}\parallel =0$. It follows from the restriction (b) that
$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n+1}-{x}_{n}\parallel =0.$
(3.5)
Notice that
$\begin{array}{r}\parallel {x}_{n}-{J}_{{r}_{n}}\left({x}_{n}+{e}_{n+1}\right)\parallel \\ \phantom{\rule{1em}{0ex}}\le \parallel {x}_{n}-{x}_{n+1}\parallel +\parallel {x}_{n+1}-{J}_{{r}_{n}}\left({x}_{n}+{e}_{n+1}\right)\parallel \\ \phantom{\rule{1em}{0ex}}\le \parallel {x}_{n}-{x}_{n+1}\parallel +{\alpha }_{n}\parallel f\left({x}_{n}\right)-{J}_{{r}_{n}}\left({x}_{n}+{e}_{n+1}\right)\parallel +{\gamma }_{n}\parallel {x}_{n}-{J}_{{r}_{n}}\left({x}_{n}+{e}_{n+1}\right)\parallel \\ \phantom{\rule{2em}{0ex}}+{\delta }_{n}\parallel {Q}_{C}\left({g}_{n}\right)-{J}_{{r}_{n}}\left({x}_{n}+{e}_{n+1}\right)\parallel .\end{array}$
It follows that
$\begin{array}{r}\left(1-{\gamma }_{n}\right)\parallel {x}_{n}-{J}_{{r}_{n}}\left({x}_{n}+{e}_{n+1}\right)\parallel \\ \phantom{\rule{1em}{0ex}}\le \parallel {x}_{n}-{x}_{n+1}\parallel +{\alpha }_{n}\parallel f\left({x}_{n}\right)-{J}_{{r}_{n}}\left({x}_{n}+{e}_{n+1}\right)\parallel +{\delta }_{n}\parallel {Q}_{C}\left({g}_{n}\right)-{J}_{{r}_{n}}\left({x}_{n}+{e}_{n+1}\right)\parallel .\end{array}$
In view of the restrictions (a), (b), and (c), we find from (3.5) that
$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n}-{J}_{{r}_{n}}\left({x}_{n}+{e}_{n+1}\right)\parallel =0.$
(3.6)
Notice that
$\begin{array}{rl}\parallel {x}_{n}-{J}_{{r}_{n}}{x}_{n}\parallel & \le \parallel {x}_{n}-{J}_{{r}_{n}}\left({x}_{n}+{e}_{n+1}\right)\parallel +\parallel {J}_{{r}_{n}}\left({x}_{n}+{e}_{n+1}\right)-{J}_{{r}_{n}}{x}_{n}\parallel \\ \le \parallel {x}_{n}-{J}_{{r}_{n}}\left({x}_{n}+{e}_{n+1}\right)\parallel +\parallel {e}_{n+1}\parallel .\end{array}$
Since ${\sum }_{n=1}^{\mathrm{\infty }}\parallel {e}_{n}\parallel <\mathrm{\infty }$, we see from (3.6) that
$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n}-{J}_{{r}_{n}}{x}_{n}\parallel =0.$
Take a fixed number r such that $ϵ>r>0$. In view of Lemma 2.1, we obtain
$\begin{array}{rl}\parallel {J}_{{r}_{n}}{x}_{n}-{J}_{r}{x}_{n}\parallel & =\parallel {J}_{r}\left(\frac{r}{{r}_{n}}{x}_{n}+\left(1-\frac{r}{{r}_{n}}\right){J}_{{r}_{n}}{x}_{n}\right)-{J}_{r}{x}_{n}\parallel \\ \le \parallel \left(1-\frac{r}{{r}_{n}}\right)\left({J}_{{r}_{n}}{x}_{n}-{x}_{n}\right)\parallel \\ \le \parallel {J}_{{r}_{n}}{x}_{n}-{x}_{n}\parallel .\end{array}$
(3.7)
Note that
$\parallel {x}_{n}-{J}_{r}{x}_{n}\parallel \le \parallel {x}_{n}-{J}_{{r}_{n}}{x}_{n}\parallel +\parallel {J}_{{r}_{n}}{x}_{n}-{J}_{r}{x}_{n}\parallel \le 2\parallel {x}_{n}-{J}_{{r}_{n}}{x}_{n}\parallel .$
This combines with (3.7), yielding
$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n}-{J}_{r}{x}_{n}\parallel =0.$
(3.8)
Next, we claim that ${lim sup}_{n\to \mathrm{\infty }}〈f\left(\overline{x}\right)-\overline{x},j\left({x}_{n}-\overline{x}\right)〉\le 0$, where $\overline{x}={lim}_{t\to 0}{z}_{t}$, and ${z}_{t}$ solves the fixed point equation ${z}_{t}=tf\left({z}_{t}\right)+\left(1-t\right){J}_{r}{z}_{t}$, $\mathrm{\forall }t\in \left(0,1\right)$, from which it follows that
$\parallel {z}_{t}-{x}_{n}\parallel =\parallel \left(1-t\right)\left({J}_{r}{z}_{t}-{x}_{n}\right)+t\left(f\left({z}_{t}\right)-{x}_{n}\right)\parallel .$
For any $t\in \left(0,1\right)$, we see that
$\begin{array}{rl}{\parallel {z}_{t}-{x}_{n}\parallel }^{2}=& \left(1-t\right)〈{J}_{r}{z}_{t}-{x}_{n},j\left({z}_{t}-{x}_{n}\right)〉+t〈f\left({z}_{t}\right)-{x}_{n},j\left({z}_{t}-{x}_{n}\right)〉\\ =& \left(1-t\right)\left(〈{J}_{r}{z}_{t}-{J}_{r}{x}_{n},j\left({z}_{t}-{x}_{n}\right)〉+〈{J}_{r}{x}_{n}-{x}_{n},j\left({z}_{t}-{x}_{n}\right)〉\right)\\ +t〈f\left({z}_{t}\right)-{z}_{t},j\left({z}_{t}-{x}_{n}\right)〉+t〈{z}_{t}-{x}_{n},j\left({z}_{t}-{x}_{n}\right)〉\\ \le & \left(1-t\right)\left({\parallel {z}_{t}-{x}_{n}\parallel }^{2}+\parallel {J}_{r}{x}_{n}-{x}_{n}\parallel \parallel {z}_{t}-{x}_{n}\parallel \right)\\ +t〈f\left({z}_{t}\right)-{z}_{t},j\left({z}_{t}-{x}_{n}\right)〉+t{\parallel {z}_{t}-{x}_{n}\parallel }^{2}\\ \le & {\parallel {z}_{t}-{x}_{n}\parallel }^{2}+\parallel {J}_{r}{x}_{n}-{x}_{n}\parallel \parallel {z}_{t}-{x}_{n}\parallel +t〈f\left({z}_{t}\right)-{z}_{t},j\left({z}_{t}-{x}_{n}\right)〉.\end{array}$
It follows that
$〈{z}_{t}-f\left({z}_{t}\right),j\left({z}_{t}-{x}_{n}\right)〉\le \frac{1}{t}\parallel {J}_{r}{x}_{n}-{x}_{n}\parallel \parallel {z}_{t}-{x}_{n}\parallel ,\phantom{\rule{1em}{0ex}}\mathrm{\forall }t\in \left(0,1\right).$
By virtue of (3.8), we find that
$\underset{n\to \mathrm{\infty }}{lim sup}〈{z}_{t}-f\left({z}_{t}\right),j\left({z}_{t}-{x}_{n}\right)〉\le 0.$
(3.9)
Since ${z}_{t}\to \overline{x}$, as $t\to 0$ and the fact that j is strong to weak uniformly continuous on bounded subsets of E, we see that
Hence, for any $ϵ>0$, there exists $\lambda >0$ such that $\mathrm{\forall }t\in \left(0,\lambda \right)$ the following inequality holds:
$〈f\left(\overline{x}\right)-\overline{x},j\left({x}_{n}-\overline{x}\right)〉\le 〈{z}_{t}-f\left({z}_{t}\right),j\left({z}_{t}-{x}_{n}\right)〉+ϵ.$
This implies that
$\underset{n\to \mathrm{\infty }}{lim sup}〈f\left(\overline{x}\right)-\overline{x},j\left({x}_{n}-\overline{x}\right)〉\le \underset{n\to \mathrm{\infty }}{lim sup}〈{z}_{t}-f\left({z}_{t}\right),j\left({z}_{t}-{x}_{n}\right)〉+ϵ.$
Since ϵ is arbitrary and (3.9), one finds that ${lim sup}_{n\to \mathrm{\infty }}〈f\left(\overline{x}\right)-\overline{x},j\left({x}_{n}-\overline{x}\right)〉\le 0$. This implies that
$\underset{n\to \mathrm{\infty }}{lim sup}〈f\left(\overline{x}\right)-\overline{x},j\left({x}_{n+1}-\overline{x}\right)〉\le 0.$
(3.10)
Finally, we prove that ${x}_{n}\to \overline{x}$ as $n\to \mathrm{\infty }$. Note that
$\begin{array}{rl}{\parallel {x}_{n+1}-\overline{x}\parallel }^{2}\le & {\alpha }_{n}〈f\left({x}_{n}\right)-\overline{x},j\left({x}_{n+1}-\overline{x}\right)〉+{\beta }_{n}\parallel {y}_{n}-\overline{x}\parallel \parallel \parallel {x}_{n+1}-\overline{x}\parallel \\ +{\gamma }_{n}\parallel {x}_{n}-\overline{x}\parallel \parallel {x}_{n+1}-\overline{x}\parallel +{\delta }_{n}\parallel {Q}_{C}\left({g}_{n}\right)-\overline{x}\parallel \parallel {x}_{n+1}-\overline{x}\parallel \\ \le & {\alpha }_{n}〈f\left({x}_{n}\right)-\overline{x},j\left({x}_{n+1}-\overline{x}\right)〉+\frac{{\beta }_{n}}{2}\left({\parallel {y}_{n}-\overline{x}\parallel }^{2}+{\parallel {x}_{n+1}-\overline{x}\parallel }^{2}\right)\\ +\frac{{\gamma }_{n}}{2}\left({\parallel {x}_{n}-\overline{x}\parallel }^{2}+{\parallel {x}_{n+1}-\overline{x}\parallel }^{2}\right)+\frac{{\delta }_{n}}{2}\left({\parallel {Q}_{C}\left({g}_{n}\right)-\overline{x}\parallel }^{2}+{\parallel {x}_{n+1}-\overline{x}\parallel }^{2}\right).\end{array}$
Note that $\parallel {y}_{n}-\overline{x}\parallel \le \parallel {x}_{n}-\overline{x}\parallel +\parallel {e}_{n+1}\parallel$. It follows that
$\begin{array}{rl}{\parallel {x}_{n+1}-\overline{x}\parallel }^{2}\le & 2{\alpha }_{n}〈f\left({x}_{n}\right)-\overline{x},j\left({x}_{n+1}-\overline{x}\right)〉+{\beta }_{n}{\parallel {y}_{n}-\overline{x}\parallel }^{2}\\ +{\gamma }_{n}{\parallel {x}_{n}-\overline{x}\parallel }^{2}+{\delta }_{n}{\parallel {Q}_{C}\left({g}_{n}\right)-\overline{x}\parallel }^{2}\\ \le & 2{\alpha }_{n}〈f\left({x}_{n}\right)-\overline{x},j\left({x}_{n+1}-\overline{x}\right)〉+\left(1-{\alpha }_{n}\right){\parallel {x}_{n}-\overline{x}\parallel }^{2}+{\nu }_{n},\end{array}$
(3.11)
where ${\nu }_{n}=\parallel {e}_{n+1}\parallel \left(\parallel {e}_{n+1}\parallel +2\parallel {x}_{n}-\overline{x}\parallel \right)+{\delta }_{n}{\parallel {Q}_{C}\left({g}_{n}\right)-\overline{x}\parallel }^{2}$. In view of the restriction (c), we find that ${\sum }_{n=1}^{\mathrm{\infty }}{\nu }_{n}<\mathrm{\infty }$. Let ${\rho }_{n}=max\left\{〈f\left({x}_{n}\right)-\overline{x},j\left({x}_{n+1}-\overline{x}\right)〉,0\right\}$. Next, we show that ${lim}_{n\to \mathrm{\infty }}{\rho }_{n}=0$. Indeed, from (3.10), for any give $ϵ>0$, there exists a positive integer ${n}_{1}$ such that
$〈f\left({x}_{n}\right)-\overline{x},j\left({x}_{n+1}-\overline{x}\right)〉<ϵ,\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge {n}_{1}.$
This implies that $0\le {\rho }_{n}<ϵ$, $\mathrm{\forall }n\ge {n}_{1}$. Since $ϵ>0$ is arbitrary, we see that ${lim}_{n\to \mathrm{\infty }}{\rho }_{n}=0$. In view of (3.11), we find that
${\parallel {x}_{n+1}-\overline{x}\parallel }^{2}\le \left(1-{\alpha }_{n}\right){\parallel {x}_{n}-\overline{x}\parallel }^{2}+2{\alpha }_{n}{\rho }_{n}+{\nu }_{n}.$

In view of Lemma 2.2, we find the desired conclusion immediately. □

If the mapping f maps any element in C into a fixed element u and ${\delta }_{n}=0$, then we have the following result.

Corollary 3.2 Let E be a real reflexive Banach space with the uniformly Gâteaux differentiable norm and A be an m-accretive operators in E. Assume that $C:=\overline{D\left(A\right)}$ is convex and has the normal structure. Let $\left\{{\alpha }_{n}\right\}$, $\left\{{\beta }_{n}\right\}$, and $\left\{{\gamma }_{n}\right\}$ be real number sequences in $\left(0,1\right)$ such that ${\alpha }_{n}+{\beta }_{n}+{\gamma }_{n}=1$. Let $\left\{{x}_{n}\right\}$ be a sequence generated in the following manner:
${x}_{0}\in C,\phantom{\rule{1em}{0ex}}{x}_{n+1}={\alpha }_{n}u+{\beta }_{n}{J}_{{r}_{n}}\left({x}_{n}+{e}_{n+1}\right)+{\gamma }_{n}{x}_{n},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 0,$
where u is a fixed element in C, $\left\{{e}_{n}\right\}$ is a sequence in E, $\left\{{g}_{n}\right\}$ is a bounded sequence in E, $\left\{{r}_{n}\right\}$ is a positive real numbers sequence, and ${J}_{{r}_{n}}={\left(I+{r}_{n}A\right)}^{-1}$. Assume that ${A}^{-1}\left(0\right)$ is not empty and the above control sequences satisfy the following restrictions:
1. (a)

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

2. (b)

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

3. (c)

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

4. (d)

${r}_{n}\ge \mu$ for each $n\ge 1$ and ${lim}_{n\to \mathrm{\infty }}|{r}_{n}-{r}_{n+1}|=0$.

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

Remark 3.3 We remark here that the algorithm (ϒ) is convergence under mild restrictions. However, it does not include the Halpern iterative algorithm as a special case because of the restriction (b). It is of interest to develop a different analysis technique for the algorithm without the restriction or under mild restrictions.

## 4 Applications

In this section, we give an application of Theorem 3.1 in the framework of Hilbert spaces.

For a proper lower semicontinuous convex function $w:H\to \left(-\mathrm{\infty },\mathrm{\infty }\right]$, the subdifferential mapping ∂w of w is defined by
$\partial w\left(x\right)=\left\{{x}^{\ast }\in H:w\left(x\right)+〈y-x,{x}^{\ast }〉\le w\left(y\right),\mathrm{\forall }y\in H\right\},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in H.$

Rockafellar  proved that ∂w is a maximal monotone operator. It is easy to verify that $0\in \partial w\left(v\right)$ if and only if $w\left(v\right)={min}_{x\in H}g\left(x\right)$.

Theorem 4.1 Let $w:H\to \left(-\mathrm{\infty },+\mathrm{\infty }\right]$ be a proper convex lower semicontinuous function such that ${\left(\partial w\right)}^{-1}\left(0\right)$ is not empty. Let $f:H\to H$ be a κ-contraction and let $\left\{{x}_{n}\right\}$ be a sequence in H in the following process: ${x}_{0}\in H$ and
$\left\{\begin{array}{c}{y}_{n}={arg min}_{z\in H}\left\{w\left(z\right)+\frac{{\parallel z-{x}_{n}-{e}_{n+1}\parallel }^{2}}{2{r}_{n}}\right\},\hfill \\ {x}_{n+1}={\alpha }_{n}f\left({x}_{n}\right)+{\beta }_{n}{y}_{n}+{\gamma }_{n}{x}_{n},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 0,\hfill \end{array}$

where $\left\{{e}_{n}\right\}$ is a sequence in H, $\left\{{g}_{n}\right\}$ is a bounded sequence in H, and $\left\{{r}_{n}\right\}$ is a positive real numbers sequence. Assume that the above control sequences satisfy the restrictions (a), (b), (d), and ${\sum }_{n=1}^{\mathrm{\infty }}\parallel {e}_{n}\parallel <\mathrm{\infty }$. Then the sequence $\left\{{x}_{n}\right\}$ converges strongly to $\overline{x}$, which is the unique solution to the following variational inequality: $〈f\left(\overline{x}\right)-\overline{x},j\left(p-\overline{x}\right)〉\le 0$, $\mathrm{\forall }p\in {\left(\partial w\right)}^{-1}\left(0\right)$.

Proof Since $w:H\to \left(-\mathrm{\infty },\mathrm{\infty }\right]$ is a proper convex and lower semicontinuous function, we see that subdifferential ∂w of w is maximal monotone. We note that
${y}_{n}=\underset{z\in H}{arg min}\left\{w\left(z\right)+\frac{{\parallel z-{x}_{n}-{e}_{n+1}\parallel }^{2}}{2{r}_{n}}\right\}$
is equivalent to $0\in \partial w\left({y}_{n}\right)+\frac{1}{{r}_{n}}\left({y}_{n}-{x}_{n}-{e}_{n+1}\right)$. It follows that
${x}_{n}+{e}_{n+1}\in {y}_{n}+{r}_{n}\partial w\left({y}_{n}\right).$

Putting ${\delta }_{n}=0$ in Theorem 3.1, we draw the desired conclusion from Theorem 3.1 immediately. □

## Declarations

### Acknowledgements

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

## Authors’ Affiliations

(1)
Department of Mathematics, Hangzhou Normal University, Hangzhou, 310036, China
(2)
Department of Mathematics, Gyeongsang National University, Jinju, 660-701, Korea

## References

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