Some results on Rockafellar-type iterative algorithms for zeros of accretive operators

Abstract

We provide a new proof technique to obtain strong convergence of the sequences generated by viscosity iterative methods for a Rockafellar-type iterative algorithm and a Halpern-type iterative algorithm to a zero of an accretive operator in Banach spaces. By using a new method different from previous ones, the main results improve and develop the recent well-known results in this area.

MSC:47H06, 47H09, 47H10, 47J25, 49M05, 65J15.

1 Introduction

Let E be a real Banach space with the norm $\parallel \cdot \parallel$ and the dual space $E^{\ast }$. The value of $x^{\ast }\in E^{\ast }$ at $y\in E$ is denoted by $〈y,x^{\ast }〉$ and the normalized duality mapping $\mathcal{J}$ from E into $2^{E^{\ast }}$ is defined by

$$\mathcal{J}(x)=\{x^{\ast }\in E^{\ast }:〈x,x^{\ast }〉=\parallel x\parallel \parallel x^{\ast }\parallel ,\parallel x\parallel =\parallel x^{\ast }\parallel \},\mathrm{\forall }x\in E.$$

Recall that a (possibly multivalued) operator $A:D(A)\subset E\to 2^E$ with the domain $D(A)$ and the range $R(A)$ in E is accretive if, for each $x_i\in D(A)$ and $y_i\in A{x}_i$ ($i=1,2$), there exists a $j\in \mathcal{J}(x_1-x_2)$ such that $〈y_1-y_2,j〉\ge 0$. (Here $\mathcal{J}$ is the normalized duality mapping.) In a Hilbert space, an accretive operator is also called a monotone operator. The set of zeros of A is denoted by $A^{-1}0$, that is,

$$A^{-1}0:=\{z\in D(A):0\in Az\}.$$

If $A^{-1}0\ne \mathrm{\varnothing }$, then the inclusion $0\in Ax$ is solvable.

Iterative methods have extensively been studied over the last forty years for constructions of zeros of accretive operators (see, e.g., [1–7]). In particular, in order to find a zero of a monotone operator, Rockafellar [7] introduced a powerful and successful algorithm in a Hilbert space H, which is recognized as the Rockafellar proximal point algorithm: For any initial point $x_0\in H$, a sequence $\{x_n\}$ is generated by

$$x_{n+1}=J_{r_n}(x_n+e_n),\mathrm{\forall }n\ge 0,$$

where $J_r={(I+rA)}^{-1}$ for all $r>0$ is the resolvent of A and $\{e_n\}$ is an error sequence in H. Bruck [2] proposed the following in a Hilbert space H: For any fixed point $u\in H$,

$$x_{n+1}=J_{r_n}(u),\mathrm{\forall }n\ge 0.$$

In 1991, Güler [8] gave an example showing that Rockafellar’s proximal algorithm does not converge strongly. Solodov and Svaitor [9] in 2000 proposed a modified proximal point algorithm which converges strongly to a solution of the equation $0\in Ax$ by using the projection method. In 2000, Kamimura and Takahashi [10–12] introduced the following iterative algorithms of Halpern type [13] and Mann type [14] in Hilbert spaces and Banach spaces: For any initial point $x_0$,

$$x_{n+1}={\alpha }_n{x}_0+(1-{\alpha }_n)J_{r_n}{x}_n,x_{n+1}={\alpha }_n{x}_0+(1-{\alpha }_n)J_{r_n}{x}_n+e_n,\mathrm{\forall }n\ge 0$$

and

$$x_{n+1}={\alpha }_n{x}_n+(1-{\alpha }_n)J_{r_n}{x}_n,x_{n+1}={\alpha }_n{x}_n+(1-{\alpha }_n)J_{r_n}{x}_n+e_n,\mathrm{\forall }n\ge 0,$$

where $\{{\alpha }_n\}\subset (0,1)$, $\{r_n\}\subset (0,\mathrm{\infty })$, and $\{e_n\}$ is an error sequence, and obtained strong and weak convergence of sequences generated by these algorithms.

Xu [15] in 2006 and Song and Yang [16] in 2009 obtained the strong convergence of the regularization method for Rockafellar’s proximal point algorithm in a Hilbert space H: For any initial point $x_0\in H$

$$x_{n+1}=J_{r_n}({\alpha }_{n}u+(1-{\alpha }_n)x_n+e_n),\mathrm{\forall }n\ge 0,$$

where $\{{\alpha }_n\}\subset (0,1)$, $\{e_n\}\subset H$ and $\{r_n\}\subset (0,\mathrm{\infty })$.

In 2012, as in [17], Zhang and Song [18] considered the following Rockafellar-type iterative algorithm (1.1) and Halpern-type iterative algorithm (1.2) for finding a zero of an accretive operator A in a uniformly convex Banach space E with a weakly continuous duality mapping ${\mathcal{J}}_{\phi }$ with gauge function φ or with a uniformly Gâteaux differentiable norm: For any initial point $x_0\in E$ and fixed point $u\in E$,

$$x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)J_{r_n}({\alpha }_{n}u+(1-{\alpha }_n)x_n),\mathrm{\forall }n\ge 0$$

(1.1)

and

$$x_{n+1}={\alpha }_{n}u+{\beta }_n{x}_n+(1-{\alpha }_n-{\beta }_n)J_{r_n}{x}_n,\mathrm{\forall }n\ge 0,$$

(1.2)

where the sequences $\{{\alpha }_n\},\{{\beta }_n\}\subset (0,1)$ and $\{r_n\}\subset (0,\mathrm{\infty })$ satisfy the conditions: (i) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$; (ii) ${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$; (iii) ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1$; and (iv) ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{r}_n>0$. In particular, in order to obtain strong convergence of the sequence generated by (1.2) to a zero of an accretive operator A, they utilized the well-known inequality in uniformly convex Banach spaces (see Xu [19]). The results of Zhang and Song [18] in a Banach space with a uniformly Gâteaux differentiable norm and the corresponding results of Song [17] are mutually complementary since Zhang and Song [18] assumed uniform convexity on the space instead of reflexivity on the space in Song [17] and relaxed the conditions $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1$ and ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{r}_n>0$, ${lim}_{n\to \mathrm{\infty }}\frac{r_n}{r_{n+1}}=1$ on sequences $\{{\beta }_n\}$ and $\{r_n\}$ in Song [17]. Yu [20] filled the gaps in the result of Zhang and Song [18] for the Halpern-type iterative algorithm (1.2) by utilizing the result on the sequence of real numbers in [21], which is of fundamental importance for the techniques of analysis. Also, Zhang and Song [18] studied the Rockafellar-type iterative algorithm (1.1) in a uniformly convex Banach space with a weakly continuous normalized duality mapping $\mathcal{J}$ or with a uniformly Gâteaux differentiable norm.

In this paper, motivated by the above mentioned results, we consider viscosity iterative methods for the Rockafellar-type iterative algorithm (1.1) and Halpern-type iterative algorithm (1.2). By using a new method different from ones in [18, 20] which recover the gaps in [18] as in [20], we establish results on strong convergence of the sequences generated by the proposed iterative methods to a zero of an accretive operator A, which solves a certain variational inequality in a uniformly convex Banach space having a weakly continuous duality mapping ${\mathcal{J}}_{\phi }$ with gauge function φ or having a uniformly Gâteaux differentiable norm. Our results improve, develop and complement the corresponding results of Song [17], Zhang and Song [18], Yu [20] and Song et al. [22] as well as many existing ones.

2 Preliminaries and lemmas

Let E be a real Banach space with the norm $\parallel \cdot \parallel$ and let $E^{\ast }$ be its dual. When $\{x_n\}$ is a sequence in E, then $x_n\to x$ (resp., $x_n\rightharpoonup x$, ) will denote strong (resp., weak, weak^∗) convergence of the sequence $\{x_n\}$ to x.

Recall that a mapping $f:E\to E$ is said to be contractive mapping on E if there exists a constant $k\in (0,1)$ such that $\parallel f(x)-f(y)\parallel \le k\parallel x-y\parallel$, $\mathrm{\forall }x,y\in E$. An accretive operator A is said to satisfy the range condition if $\overline{D(A)}\subset R(I+rA)$ for all $r>0$, where I is an identity operator of E and $\overline{D(A)}$ denotes the closure of the domain $D(A)$ of A. An accretive operator A is called m-accretive if $R(I+rA)=E$ for each $r>0$. If A is an accretive operator which satisfies the range condition, then we can define, for each $r>0$, a mapping $J_r:R(I+rA)\to D(A)$ defined by $J_r={(I+rA)}^{-1}$, which is called the resolvent of A. We know that $J_r$ is nonexpansive (i.e., $\parallel J_{r}x-J_{r}y\parallel \le \parallel x-y\parallel$, $\mathrm{\forall }x,y\in R(I+rA)$) and $A^{-1}0=F(J_r)=\{x\in D(J_r):J_{r}x=x\}$ for all $r>0$. Moreover, for $r>0$, $t>0$ and $x\in E$,

$$J_{r}x=J_t(\frac{t}{r}x+(1-\frac{t}{r})J_{r}x),$$

(2.1)

which is referred to as the resolvent identity (see [23, 24] where more details on accretive operators can be found).

The norm of E is said to be Gâteaux differentiable if

$$\underset{t\to 0}{lim}\frac{\parallel x+ty\parallel -\parallel x\parallel }{t}$$

exists for each x, y in its unit sphere $S=\{x\in E:\parallel x\parallel =1\}$. Such an E is called a smooth Banach space. The norm is said to be uniformly Gâteaux differentiable if for $y\in S$, the limit is attained uniformly for $x\in S$.

A Banach space E is said to be uniformly convex if for all $\epsilon \in [0,2]$, there exists ${\delta }_{\epsilon }>0$ such that

$$\parallel x\parallel =\parallel y\parallel =1\text{implies}\frac{\parallel x+y\parallel }{2}<1-{\delta }_{\epsilon }\text{whenever }\parallel x-y\parallel \ge \epsilon .$$

Let $l>1$ and $M>0$ be two fixed real numbers. Then a Banach space is uniformly convex if and only if there exists a continuous strictly increasing convex function $g:[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ with $g(0)=0$ such that

$${\parallel \lambda x+(1-\lambda )y\parallel }^l\le \lambda {\parallel x\parallel }^l+(1-\lambda ){\parallel y\parallel }^l-{\omega }_l(\lambda )g(\parallel x-y\parallel )$$

(2.2)

for all $x,y\in B_M(0)=\{x\in E:\parallel x\parallel \le M\}$, where ${\omega }_l(\lambda )={\lambda }^l(1-\lambda )+\lambda {(1-\lambda )}^l$. For more detail, see Xu [19].

By a gauge function we mean a continuous strictly increasing function φ defined on ${\mathbb{R}}^{+}:=[0,\mathrm{\infty })$ such that $\phi (0)=0$ and ${lim}_{r\to \mathrm{\infty }}\phi (r)=\mathrm{\infty }$. The mapping ${\mathcal{J}}_{\phi }:E\to 2^{E^{\ast }}$ defined by

$${\mathcal{J}}_{\phi }(x)=\{f\in E^{\ast }:〈x,f〉=\parallel x\parallel \parallel f\parallel ,\parallel f\parallel =\phi (\parallel x\parallel )\},\mathrm{\forall }x\in E$$

is called the duality mapping with gauge function φ. In particular, the duality mapping with gauge function $\phi (t)=t$ denoted by $\mathcal{J}$, is referred to as the normalized duality mapping. The following property of duality mapping is well known [23]:

$${\mathcal{J}}_{\phi }(\lambda x)=sign\lambda \left(\frac{\phi (|\lambda |\cdot \parallel x\parallel )}{\parallel x\parallel }\right)\mathcal{J}(x),\mathrm{\forall }x\in E\setminus 0,\lambda \in \mathbb{R},$$

(2.3)

where ℝ is the set of all real numbers; in particular, $\mathcal{J}(-x)=-\mathcal{J}(x)$, $\mathrm{\forall }x\in E$. It is well known that E is smooth if and only if the normalized duality mapping $\mathcal{J}$ is single-valued, and that in a Hilbert space H, the normalized duality mapping $\mathcal{J}$ is the identity.

We say that a Banach space E has a weakly continuous duality mapping if there exists a gauge function φ such that the duality mapping ${\mathcal{J}}_{\phi }$ is single-valued and continuous from the weak topology to the weak^∗ topology, that is, for any $\{x_n\}\in E$ with $x_n\rightharpoonup x$, . For example, every $l^p$ space ($1<p<\mathrm{\infty }$) has a weakly continuous duality mapping with gauge function $\phi (t)=t^{p-1}$ [23].

Let LIM be a continuous linear functional on $l^{\mathrm{\infty }}$ and $(a_1,a_2,\dots )\in l^{\mathrm{\infty }}$. We write ${LIM}_n(a_n)$ instead of $LIM((a_1,a_2,\dots ))$. LIM is said to be a Banach limit if LIM satisfies $\parallel LIM\parallel ={LIM}_n(1)=1$ and ${LIM}_n(a_{n+1})={LIM}_n(a_n)$ for all $(a_1,a_2,\dots )\in l^{\mathrm{\infty }}$. If LIM is a Banach limit, the following are well known [25]:

• (i) for all $n\ge 1$, $a_n\le c_n$ implies ${LIM}_n(a_n)\le {LIM}_n(c_n)$,

• (ii) ${LIM}_n(a_{n+N})={LIM}_n(a_n)$ for any fixed positive integer N,

• (iii) ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{a}_n\le {LIM}_n(a_n)\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{a}_n$ for all $(a_1,a_2,\dots )\in l^{\mathrm{\infty }}$.

We need the following lemmas for the proofs of our main results.

Lemma 2.1 [23, 25]

Let E be a real Banach space and φ be a continuous strictly increasing function on ${\mathbb{R}}^{+}$ such that $\phi (0)=0$ and ${lim}_{r\to \mathrm{\infty }}\phi (r)=\mathrm{\infty }$. Define

$$\mathrm{\Phi }(t)={\int }_0^t\phi (\tau )d\tau ,\mathrm{\forall }t\in {\mathbb{R}}^{+}.$$

Then the following inequalities hold:

$$\begin{array}{c}\mathrm{\Phi }(kt)\le k\mathrm{\Phi }(t),0<k<1,\hfill \\ \mathrm{\Phi }(\parallel x+y\parallel )\le \mathrm{\Phi }(\parallel x\parallel )+〈y,j_{\phi }(x+y)〉,\mathrm{\forall }x,y\in E,\hfill \end{array}$$

where $j_{\phi }(x+y)\in {\mathcal{J}}_{\phi }(x+y)$. In particular, if E is smooth, then one has

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

Lemma 2.2 [26]

Let $a\in \mathbb{R}$ be a real number and let a sequence $\{a_n\}\in l^{\mathrm{\infty }}$ satisfy the condition ${LIM}_n(a_n)\le a$ for all Banach limit LIM. If ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}(a_{n+1}-a_n)\le 0$, then ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{a}_n\le a$.

Lemma 2.3 [27]

Let $\{s_n\}$ be a sequence of non-negative real numbers satisfying

$$s_{n+1}\le (1-{\lambda }_n)s_n+{\lambda }_n{\delta }_n,\mathrm{\forall }n\ge 0,$$

where $\{{\lambda }_n\}$ and $\{{\delta }_n\}$ satisfy the following conditions:

• (i) $\{{\lambda }_n\}\subset [0,1]$ and ${\sum }_{n=0}^{\mathrm{\infty }}{\lambda }_n=\mathrm{\infty }$;

• (ii) ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\delta }_n\le 0$ or ${\sum }_{n=0}^{\mathrm{\infty }}{\lambda }_n{\delta }_n<\mathrm{\infty }$.

Then ${lim}_{n\to \mathrm{\infty }}{s}_n=0$.

Also, we will use the next lemma which is of fundamental importance for our proof.

Lemma 2.4 [21]

Let $\{s_n\}$ be a sequence of real numbers that does not decrease at infinity, in the sense that there exists a subsequence $\{s_{n_i}\}$ of $\{s_n\}$ such that $s_{n_i}<s_{n_i+1}$ for all $i\ge 0$. For every $n\ge n_0$, define the sequence of integers $\{\tau (n)\}$ by

$$\tau (n):=max\{k\le n:s_k<s_{k+1}\}.$$

Then ${\{\tau (n)\}}_{n\ge n_0}$ is a nondecreasing sequence verifying

$$\underset{n\to \mathrm{\infty }}{lim}\tau (n)=\mathrm{\infty }$$

and, for all $n\ge n_0$, the following two estimates hold:

$$s_{\tau (n)}\le s_{\tau (n)+1},s_n\le s_{\tau (n)+1}.$$

3 Main results

In this section, we study the convergence of the following two iterative algorithms: For an initial value $x_0\in C$,

$$x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)J_{r_n}({\alpha }_{n}f(x_n)+(1-{\alpha }_n)x_n),\mathrm{\forall }n\ge 0$$

(3.1)

and

$$x_{n+1}={\alpha }_{n}f(x_n)+{\beta }_n{x}_n+(1-{\alpha }_n-{\beta }_n)J_{r_n}{x}_n,\mathrm{\forall }n\ge 0.$$

(3.2)

Throughout this section, it is assumed that $A:D(A)\subset E\to 2^E$ is an accretive operator satisfying the range condition with $A^{-1}0\ne \mathrm{\varnothing }$; C is a nonempty closed convex subset of E such that $\overline{D(A)}\subset C\subset {\bigcap }_{r>0}R(I+rA)$; $f:C\to C$ is a contractive mapping with a constant $k\in (0,1)$; and $\{{\alpha }_n\},\{{\beta }_n\}\subset (0,1)$ and $\{r_n\}\subset (0,\mathrm{\infty })$ are sequences satisfying the conditions:

(C1) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$;

(C2) ${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

(C3) $0\le {\beta }_n\le a<1$ for some a;

(C4) ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{r}_n>0$.

We need the following result for the existence of solutions of a certain variational inequality.

Theorem J [28, 29]

Let E be a reflexive Banach space with a weakly continuous duality mapping ${\mathcal{J}}_{\phi }$ with gauge function φ. Let C be a nonempty closed convex subset of E, let $T:C\to C$ be a nonexpansive mapping with $F(T)\ne \mathrm{\varnothing }$ and $f:C\to C$ be a contractive mapping with a constant $k\in (0,1)$. For $t\in (0,1)$, let $\{x_t\}$ be the unique solution in C to the equation $x_t=tf(x_t)+(1-t)T{x}_t$. Then $\{x_t\}$ converges as $t\to 0^{+}$ strongly to a point q in $F(T)$, which solves the variational inequality

$$〈(I-f)(q),{\mathcal{J}}_{\phi }(q-p)〉\le 0,\mathrm{\forall }p\in F(T).$$

Using Theorem J, we have the following result.

Theorem 3.1 Let E be a reflexive Banach space having a weakly continuous duality mapping ${\mathcal{J}}_{\phi }$ with gauge function φ. Let $\{x_n\}$ be a sequence generated by (3.1) and $y_n={\alpha }_{n}f(x_n)+(1-{\alpha }_n)x_n$ for all $n\ge 0$. Let LIM be a Banach limit. If ${lim}_{n\to \mathrm{\infty }}\parallel y_n-J_{r_n}{y}_n\parallel =0$, then

$${LIM}_n\left(〈(I-f)(q),{\mathcal{J}}_{\phi }(q-x_n)〉\right)\le 0,$$

where $q:={lim}_{t\to 0^{+}}{x}_t$ with $x_t$ being defined by $x_t=tf(x_t)+(1-t)J_r{x}_t$ for each $r>0$.

Proof Let $x_t$ be defined by $x_t=tf(x_t)+(1-t)J_r{x}_t$ for $0<t<1$ and $r>0$. Then, since $A^{-1}0\ne \mathrm{\varnothing }$, $\{x_t\}$ is bounded. In fact, for $p\in A^{-1}0=F(J_r)$ for $r>0$, we have

$$\begin{array}{rcl}\parallel x_t-p\parallel & \le & t\parallel f(x_t)-p\parallel +(1-t)\parallel J_r{x}_t-J_{r_n}p\parallel \\ \le & t\parallel f(x_t)-p\parallel +(1-t)\parallel x_t-p\parallel .\end{array}$$

This gives that

$$\begin{array}{rcl}\parallel x_t-p\parallel & \le & \parallel f(x_t)-p\parallel \le \parallel f(x_t)-f(p)\parallel +\parallel f(p)-P\parallel \\ \le & k\parallel x_t-p\parallel +\parallel f(p)-p\parallel .\end{array}$$

Thus

$$\parallel x_t-p\parallel \le \frac{1}{1-k}\parallel f(p)-p\parallel ,t\in (0,1),$$

and hence $\{x_t\}$ is bounded. Also, by Theorem J, $\{x_t\}$ converges as $t\to 0^{+}$ strongly to a point in $F(J_r)=A^{-1}0$, which is denoted by $q:={lim}_{t\to 0^{+}}{x}_t$.

First we show that $\{x_n\}$ and $\{y_n\}$ are bounded. Since $A^{-1}0\ne \mathrm{\varnothing }$, we take $p\in A^{-1}0=F(J_r)$ for all $r>0$. From (3.1) and the nonexpansivity of $J_{r_n}$ for all n, we have

$$\begin{array}{c}\parallel x_{n+1}-p\parallel \hfill \\ \le {\beta }_n\parallel x_n-p\parallel +(1-{\beta }_n)\parallel J_{r_n}{y}_n-p\parallel \hfill \\ \le {\beta }_n\parallel x_n-p\parallel +(1-{\beta }_n)\parallel {\alpha }_{n}f(x_n)+(1-{\alpha }_n)x_n-p\parallel \hfill \\ \le {\beta }_n\parallel x_n-p\parallel +(1-{\beta }_n)[{\alpha }_n\parallel f(x_n)-f(p)\parallel +(1-{\alpha }_n)\parallel x_n-p\parallel +{\alpha }_n\parallel f(p)-p\parallel ]\hfill \\ \le {\beta }_n\parallel x_n-p\parallel +(1-{\beta }_n)[{\alpha }_{n}k\parallel x_n-p\parallel +(1-{\alpha }_n)\parallel x_n-p\parallel +{\alpha }_n\parallel f(p)-p\parallel ]\hfill \\ =(1-(1-{\beta }_n)(1-k){\alpha }_n)\parallel x_n-p\parallel +(1-{\beta }_n)(1-k){\alpha }_n\frac{\parallel f(p)-p\parallel }{1-k}\hfill \\ \le max\{\parallel x_n-p\parallel ,\frac{\parallel f(p)-p\parallel }{1-k}\}\hfill \\ \cdots \hfill \\ \le max\{\parallel x_0-p\parallel ,\frac{\parallel f(p)-p\parallel }{1-k}\}.\hfill \end{array}$$

Hence $\{x_n\}$ is bounded. Also, for $p\in A^{-1}0$, we get

$$\begin{array}{rcl}\parallel y_n-p\parallel & \le & {\alpha }_n\parallel f(x_n)-f(p)\parallel +(1-{\alpha }_n)\parallel x_n-p\parallel +{\alpha }_n\parallel f(p)-p\parallel \\ \le & {\alpha }_{n}k\parallel x_n-p\parallel +(1-{\alpha }_n)\parallel x_n-p\parallel +{\alpha }_n\parallel f(p)-p\parallel \\ =& (1-(1-k){\alpha }_n)\parallel x_n-p\parallel +(1-k){\alpha }_n\frac{\parallel f(p)-p\parallel }{1-k}\\ \le & max\{\parallel x_n-p\parallel ,\frac{\parallel f(p)-p\parallel }{1-k}\}\end{array}$$

and so $\{y_n\}$ is bounded. Moreover, since $\parallel J_{r_n}{y}_n-p\parallel \le \parallel y_n-p\parallel$, it follows that $\{J_{r_n}{y}_n\}$ is bounded. Also, $\{f(x_n)\}$ is bounded. As a consequence, with the control condition (C1), we get

$$\parallel y_n-x_n\parallel ={\alpha }_n\parallel f(x_n)-x_n\parallel \le {\alpha }_n(\parallel f(x_n)\parallel +\parallel x_n\parallel )\to 0(n\to \mathrm{\infty }).$$

(3.3)

Since ${lim}_{n\to \mathrm{\infty }}\parallel y_n-J_{r_n}{y}_n\parallel =0$, by (3.1) and (3.3), we obtain

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-J_{r_n}{y}_n\parallel \le \underset{n\to \mathrm{\infty }}{lim}(\parallel x_n-y_n\parallel +\parallel y_n-J_{r_n}{y}_n\parallel )=0$$

(3.4)

and

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_{n+1}-J_{r_n}{y}_n\parallel =\underset{n\to \mathrm{\infty }}{lim}{\beta }_n\parallel x_n-J_{r_n}{y}_n\parallel \le \underset{n\to \mathrm{\infty }}{lim}a\parallel x_n-J_{r_n}{y}_n\parallel =0.$$

(3.5)

Now, we show that ${LIM}_n(〈(I-f)(q),{\mathcal{J}}_{\phi }(q-x_n)〉)\le 0$, where $q={lim}_{t\to 0^{+}}{x}_t$ with $x_t$ being defined by $x_t=tf(x_t)+(1-t)J_r{x}_t$ for each $r>0$. Indeed, it follows that

$$x_t-x_{n+1}=(1-t)(J_r{x}_t-x_{n+1})+t(f(x_t)-x_{n+1}).$$

Applying Lemma 2.1, we have

$$\mathrm{\Phi }(\parallel x_t-x_{n+1}\parallel )\le \mathrm{\Phi }((1-t)\parallel J_r{x}_t-x_{n+1}\parallel )+t〈f(x_t)-x_{n+1},{\mathcal{J}}_{\phi }(x_t-x_{n+1})〉.$$

(3.6)

Along with using the resolvent identity (2.1), noting

$$\begin{array}{rcl}\parallel J_r{y}_n-J_{r_n}{y}_n\parallel & =& \parallel J_r(\frac{r}{r_n}{y}_n+(1-\frac{r}{r_n})J_{r_n}{y}_n)-J_r{y}_n\parallel \\ \le & |1-\frac{r}{r_n}|\parallel y_n-J_{r_n}{y}_n\parallel \\ \le & |1-\frac{r}{r_n}|(\parallel y_n-x_n\parallel +\parallel x_n-J_{r_n}{y}_n\parallel ),\end{array}$$

we observe also that

$$\begin{array}{rcl}\parallel J_r{x}_t-x_{n+1}\parallel & \le & \parallel J_r{x}_t-J_r{x}_n\parallel +\parallel J_r{x}_n-J_r{y}_n\parallel +\parallel J_r{y}_n-J_{r_n}{y}_n\parallel +\parallel J_{r_n}{y}_n-x_{n+1}\parallel \\ \le & \parallel x_t-x_n\parallel +\parallel x_n-y_n\parallel \\ +|1-\frac{r}{r_n}|(\parallel y_n-x_n\parallel +\parallel x_n-J_{r_n}{y}_n\parallel )+\parallel J_{r_n}{y}_n-x_{n+1}\parallel \\ =& \parallel x_t-x_n\parallel +{\epsilon }_n,\end{array}$$

where ${\epsilon }_n=(1+|1-\frac{r}{r_n}|)\parallel x_n-y_n\parallel +|1-\frac{r}{r_n}|\parallel x_n-J_{r_n}{y}_n\parallel +\parallel x_{n+1}-J_{r_n}{y}_n\parallel \to 0$ as $n\to \mathrm{\infty }$ (by (3.3), (3.4) and (3.5)), and

$$〈f(x_t)-x_{n+1},{\mathcal{J}}_{\phi }(x_t-x_{n+1})〉=〈f(x_t)-x_t,{\mathcal{J}}_{\phi }(x_t-x_{n+1})〉+\parallel x_t-x_{n+1}\parallel \phi (\parallel x_t-x_{n+1}\parallel ).$$

Thus it follows from (3.6) that

$$\begin{array}{rcl}\mathrm{\Phi }(\parallel x_t-x_{n+1}\parallel )& \le & \mathrm{\Phi }((1-t)(\parallel x_t-x_n\parallel ))+(1-t)\parallel {\epsilon }_n\parallel \phi (\parallel x_t-x_{n+1}\parallel )\\ +t(〈f(x_t)-x_t,{\mathcal{J}}_{\phi }(x_t-x_{n+1})〉+\parallel x_t-x_{n+1}\parallel \phi (\parallel x_t-x_{n+1}\parallel )).\end{array}$$

(3.7)

Applying the Banach limit LIM to (3.7), we have

$$\begin{array}{rcl}{LIM}_n\left(\mathrm{\Phi }(\parallel x_t-x_{n+1}\parallel )\right)& \le & {LIM}_n\left(\mathrm{\Phi }((1-t)(\parallel x_t-x_n\parallel ))\right)\\ +(1-t){LIM}_n(\parallel {\epsilon }_n\parallel \phi (\parallel x_t-x_{n+1}\parallel ))\\ +t{LIM}_n\left(〈f(x_t)-x_t,{\mathcal{J}}_{\phi }(x_t-x_{n+1})〉\right)\\ +t{LIM}_n(\parallel x_t-x_{n+1}\parallel \phi (\parallel x_t-x_{n+1}\parallel )).\end{array}$$

(3.8)

Hence, noting ${lim}_{n\to \mathrm{\infty }}{\epsilon }_n=0$ and applying the property ${LIM}_n(a_{n+1})={LIM}_n(a_n)$ of the Banach limit LIM to (3.8), we obtain

$$\begin{array}{c}{LIM}_n\left(〈x_t-f(x_t),{\mathcal{J}}_{\phi }(x_t-x_n)〉\right)\hfill \\ \le \frac{1}{t}{LIM}_n(\mathrm{\Phi }((1-t)\parallel x_t-x_n\parallel )-\mathrm{\Phi }(\parallel x_t-x_n\parallel ))+{LIM}_n(\parallel x_t-x_n\parallel \phi (\parallel x_t-x_n\parallel ))\hfill \\ =-\frac{1}{t}{LIM}_n({\int }_{(1-t)\parallel x_t-x_n\parallel }^{\parallel x_t-x_n\parallel }\phi (\tau )d\tau )+{LIM}_n(\parallel x_t-x_n\parallel \phi (\parallel x_t-x_n\parallel ))\hfill \\ ={LIM}_n(\parallel x_t-x_n\parallel (\phi (\parallel x_t-x_n\parallel )-\phi ({\theta }_n)))\hfill \end{array}$$

(3.9)

for some ${\tau }_n$ satisfying $(1-t)\parallel x_t-x_n\parallel \le {\theta }_n\le \parallel x_t-x_n\parallel$. Since φ is uniformly continuous on compact intervals on ${\mathbb{R}}^{+}$ and

$$\begin{array}{rcl}\parallel x_t-x_n\parallel -{\theta }_n& \le & t\parallel x_t-x_n\parallel \\ \le & t(\frac{2}{1-k}\parallel f(p)-p\parallel +\parallel x_0-p\parallel )\to 0(t\to 0),\end{array}$$

we conclude from (3.9) and $q={lim}_{t\to 0^{+}}{x}_t$ that

$$\begin{array}{rcl}{LIM}_n\left(〈(I-f)(q),{\mathcal{J}}_{\phi }(q-x_n)〉\right)& \le & \underset{t\to 0}{lim\hspace{0.17em}sup}{LIM}_n\left(〈x_t-f(x_t),{\mathcal{J}}_{\phi }(x_t-x_n)〉\right)\\ \le & \underset{t\to 0}{lim\hspace{0.17em}sup}{LIM}_n(\parallel x_t-x_n\parallel (\phi (\parallel x_t-x_n\parallel )-\phi ({\theta }_n)))\\ \le & 0.\end{array}$$

This completes the proof. □

By using Theorem 3.1, we establish the strong convergence of the Rockafellar-type iterative algorithm (3.1).

Theorem 3.2 Let E be a uniformly convex Banach space having a weakly continuous duality mapping ${\mathcal{J}}_{\phi }$ with gauge function φ. Then the sequence $\{x_n\}$ generated by (3.1) converges strongly to $q\in A^{-1}0$, where q is the unique solution of the variational inequality

$$〈(I-f)(q),{\mathcal{J}}_{\phi }(q-p)〉\le 0,\mathrm{\forall }p\in A^{-1}0.$$

(3.10)

Proof First, we note that by Theorem J, there exists a solution q of a variational inequality

$$〈(I-f)(q),{\mathcal{J}}_{\phi }(q-p)〉\le 0,\mathrm{\forall }p\in A^{-1}0,$$

where $q={lim}_{t\to 0^{+}}{x}_t\in A^{-1}0$ with $x_t$ being defined by $x_t=tf(x_t)+(1-t)J_r{x}_t$ for each $r>0$ and $0<t<1$. From now, we put $y_n={\alpha }_{n}f(x_n)+(1-{\alpha }_n)x_n$ for all $n\ge 0$.

We know that $\parallel x_n-p\parallel \le max\{\parallel x_0-p\parallel ,\frac{1}{1-k}\parallel f(p)-p\parallel \}$ for all $n\ge 0$ and all $p\in A^{-1}0$ and $\{x_n\}$, $\{y_n\}$, $\{f(x_n)\}$ and $\{J_{r_n}{y}_n\}$ are bounded by the proof of Theorem 3.1.

First, by using arguments similar to those of [18] with $u=f(x_n)$ and the inequality (2.2) ($l=2$, $\lambda =\frac{1}{2}$), we have

$${\parallel x_{n+1}-q\parallel }^2\le {\parallel x_n-q\parallel }^2+{\alpha }_n{\parallel f(x_n)-q\parallel }^2-(1-{\beta }_n)\frac{1}{4}g(\parallel y_n-J_{r_n}{y}_n\parallel ),$$

where $g:[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ is a continuous strictly increasing convex function in (2.2). From the condition (C3), it follows that $(1-{\beta }_n)\ge (1-a)>0$ for all $n\ge 0$ and

$$(1-a)\frac{1}{4}g(\parallel y_n-J_{r_n}{y}_n\parallel )-{\alpha }_n{\parallel f(x_n)-q\parallel }^2\le {\parallel x_n-q\parallel }^2-{\parallel x_{n+1}-q\parallel }^2.$$

(3.11)

In order to prove that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-q\parallel =0$, we consider two possible cases as in the proof of Yu [20].

Case 1. Assume that $\{\parallel x_n-q\parallel \}$ is a monotone sequence. In other words, for $n_0$ large enough, $\{\parallel x_n-q\parallel \}$ is either nondecreasing or nonincreasing. Hence $\{\parallel x_n-q\parallel \}$ converges (since $\{\parallel x_n-q\parallel \}$ is bounded). Thus, by (3.11) we obtain

$$\underset{n\to \mathrm{\infty }}{lim}g(\parallel y_n-J_{r_n}{y}_n\parallel )=0.$$

Thus, from the property of the function g in (2.2), it follows that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel y_n-J_{r_n}{y}_n\parallel =0.$$

Now, we proceed with the following steps.

Step 1. We know from (3.5) that ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel =0$.

Step 2. We show that ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}〈(I-f)(q),{\mathcal{J}}_{\phi }(q-x_n)〉\le 0$. To this end, put

$$a_n:=〈(I-f)(q),{\mathcal{J}}_{\phi }(q-x_n)〉,n\ge 0.$$

Then Theorem 3.1 implies that ${LIM}_n(a_n)\le 0$ for any Banach limit LIM. Since $\{x_n\}$ is bounded, there exists a subsequence $\{x_{n_j}\}$ of $\{x_n\}$ such that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}(a_{n+1}-a_n)=\underset{j\to \mathrm{\infty }}{lim}(a_{n_j+1}-a_{n_j})$$

and $x_{n_j}\rightharpoonup z\in E$. This implies that $x_{n_j+1}\rightharpoonup z$ since $\{x_n\}$ is weakly asymptotically regular by Step 1. From the weak continuity of a duality mapping ${\mathcal{J}}_{\phi }$, we have

$$w\text{-}\underset{j\to \mathrm{\infty }}{lim}{\mathcal{J}}_{\phi }(q-x_{n_j+1})=w\text{-}\underset{j\to \mathrm{\infty }}{lim}{\mathcal{J}}_{\phi }(q-x_{n_j})={\mathcal{J}}_{\phi }(q-z)$$

and so

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}(a_{n+1}-a_n)=\underset{j\to \mathrm{\infty }}{lim}〈(I-f)(q),{\mathcal{J}}_{\phi }(q-x_{n_j+1})-{\mathcal{J}}_{\phi }(q-x_{n_j})〉=0.$$

Then Lemma 2.2 implies that ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{a}_n\le 0$, that is,

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈(I-f)(q),{\mathcal{J}}_{\phi }(q-x_n)〉\le 0.$$

Step 3. We show that ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}〈(I-f)(q),{\mathcal{J}}_{\phi }(q-y_n)〉\le 0$. In fact, let $\{y_{n_i}\}$ be a subsequence of $\{y_n\}$ such that $y_{n_i}\rightharpoonup v\in E$ and

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈(I-f)(q),{\mathcal{J}}_{\phi }(q-y_n)〉=\underset{i\to \mathrm{\infty }}{lim}〈(I-f)(q),\mathcal{J}(q-y_{n_i})〉.$$

Since ${lim}_{n\to \mathrm{\infty }}\parallel x_n-y_n\parallel =0$ by (3.3) in the proof of Theorem 3.1, we have also $x_{n_i}\rightharpoonup v$. From the weak continuity of $J_{\phi }$, it follows that

$$w\text{-}\underset{i\to \mathrm{\infty }}{lim}{\mathcal{J}}_{\phi }(q-y_{n_i})=w\text{-}\underset{i\to \mathrm{\infty }}{lim}{\mathcal{J}}_{\phi }(q-x_{n_i})=J_{\phi }(q-v).$$

Hence, by Step 2, we have

$$\begin{array}{c}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈(I-f)(q),{\mathcal{J}}_{\phi }(q-y_n)〉\hfill \\ =\underset{i\to \mathrm{\infty }}{lim}〈(I-f)(q),{\mathcal{J}}_{\phi }(q-y_{n_i})-{\mathcal{J}}_{\phi }(q-x_{n_i})〉+\underset{i\to \mathrm{\infty }}{lim}〈(I-f)(q),{\mathcal{J}}_{\phi }(q-x_{n_i})〉\hfill \\ =\underset{i\to \mathrm{\infty }}{lim}〈(I-f)(q),{\mathcal{J}}_{\phi }(q-x_{n_i})〉\hfill \\ \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈(I-f)(q),{\mathcal{J}}_{\phi }(q-x_n)〉\le 0.\hfill \end{array}$$

Step 4. We show that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-q\parallel =0$. Indeed, by using (3.1), we obtain

$$\begin{array}{rcl}\parallel x_{n+1}-q\parallel \phi (\parallel x_{n+1}-q\parallel )& =& 〈{\beta }_n(x_n-q)+(1-{\beta }_n)(J_{r_n}{y}_n-q),{\mathcal{J}}_{\phi }(x_{n+1}-q)〉\\ \le & {\beta }_n\parallel x_n-q\parallel \phi (\parallel x_{n+1}-q\parallel )+(1-{\beta }_n)\parallel y_n-q\parallel \phi (\parallel x_{n+1}-q\parallel )\end{array}$$

and so

$$\parallel x_{n+1}-q\parallel \le {\beta }_n\parallel x_n-q\parallel +(1-{\beta }_n)\parallel y_n-q\parallel .$$

(3.12)

Since

$$y_n-q={\alpha }_n(f(x_n)-f(q))+{\alpha }_n(f(q)-q)+(1-{\alpha }_n)(x_n-q),$$

by Lemma 2.1, we also get

$$\begin{array}{rcl}\mathrm{\Phi }(\parallel y_n-q\parallel )& \le & \mathrm{\Phi }({\alpha }_n\parallel f(x_n)-f(q)\parallel +(1-{\alpha }_n)\parallel x_n-q\parallel )+{\alpha }_n〈f(q)-q,{\mathcal{J}}_{\phi }(y_n-q)〉\\ \le & \mathrm{\Phi }({\alpha }_{n}k\parallel x_n-q\parallel +(1-{\alpha }_n)\parallel x_n-q\parallel )+{\alpha }_n〈f(q)-q,{\mathcal{J}}_{\phi }(y_n-q)〉\\ \le & (1-(1-k){\alpha }_n)\mathrm{\Phi }(\parallel x_n-q\parallel )+{\alpha }_n〈f(q)-q,{\mathcal{J}}_{\phi }(y_n-q)〉.\end{array}$$

(3.13)

As a consequence, since Φ in Lemma 2.1 is an increasing convex function with $\mathrm{\Phi }(0)=0$, by (3.12) and (3.13), we have

$$\begin{array}{rcl}\mathrm{\Phi }(\parallel x_{n+1}-q\parallel )& \le & \mathrm{\Phi }({\beta }_n\parallel x_n-q\parallel +(1-{\beta }_n)\parallel y_n-q\parallel )\\ \le & {\beta }_n\mathrm{\Phi }(\parallel x_n-q\parallel )+(1-{\beta }_n)\mathrm{\Phi }(\parallel y_n-q\parallel )\\ \le & {\beta }_n\mathrm{\Phi }(\parallel x_n-q\parallel )+(1-{\beta }_n)(1-(1-k){\alpha }_n)\mathrm{\Phi }(\parallel x_n-q\parallel )\\ +(1-{\beta }_n){\alpha }_n〈f(q)-q,{\mathcal{J}}_{\phi }(y_n-q)〉\\ =& (1-(1-{\beta }_n)(1-k){\alpha }_n)\mathrm{\Phi }(\parallel x_n-q\parallel )\\ +(1-{\beta }_n){\alpha }_n〈f(q)-q,{\mathcal{J}}_{\phi }(y_n-q)〉.\end{array}$$

(3.14)

Put

$${\lambda }_n=(1-{\beta }_n)(1-k){\alpha }_n\text{and}{\delta }_n=\frac{1}{1-k}〈(I-f)(q),{\mathcal{J}}_{\phi }(q-y_n)〉.$$

From the conditions (C1)-(C3) and Step 3, it follows that ${\lambda }_n\to 0$, ${\sum }_{n=0}^{\mathrm{\infty }}{\lambda }_n=\mathrm{\infty }$ and ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\delta }_n\le 0$. Since (3.14) reduces to

$$\mathrm{\Phi }(\parallel x_{n+1}-q\parallel )\le (1-{\lambda }_n)\mathrm{\Phi }(\parallel x_n-q\parallel )+{\lambda }_n{\delta }_n,$$

from Lemma 2.3, we conclude that ${lim}_{n\to \mathrm{\infty }}\mathrm{\Phi }(\parallel x_n-q\parallel )=0$ and ${lim}_{n\to \mathrm{\infty }}\parallel x_n-q\parallel =0$.

Case 2. Assume that $\{\parallel x_n-q\parallel \}$ is not a monotone sequence. Then, we can define a sequence of integers $\{\tau (n)\}$ for all $n\ge n_0$ (for some $n_0$ large enough) by

$$\tau (n):=max\{k\in \mathbb{N}:k\le n,\parallel x_k-q\parallel <\parallel x_{k+1}-q\parallel \}.$$

Clearly, $\{\tau (n)\}$ is a nondecreasing sequence such that $\tau (n)\to \mathrm{\infty }$ as $n\to \mathrm{\infty }$ and

$$\parallel x_{\tau (n)}-q\parallel \le \parallel x_{\tau (n)+1}-q\parallel$$

for all $n\ge n_0$. In this case, we derive from (3.11) that

$$\underset{n\to \mathrm{\infty }}{lim}g(\parallel y_{\tau (n)}-J_{r_{\tau (n)}}{y}_{\tau (n)}\parallel )=0.$$

So, by the property of the function g in (2.2), we have

$$\parallel y_{\tau (n)}-J_{r_{\tau (n)}}{y}_{\tau (n)}\parallel =0.$$

From (3.3), (3.4) and (3.5), we also have

$$\begin{array}{c}\underset{n\to \mathrm{\infty }}{lim}\parallel x_{\tau (n)}-y_{\tau (n)}\parallel =0,\hfill \\ \underset{n\to \mathrm{\infty }}{lim}\parallel x_{\tau (n)}-J_{r_{\tau (n)}}{y}_{\tau (n)}\parallel =0\hfill \end{array}$$

and

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_{\tau (n)+1}-J_{r_{\tau (n)}}{y}_{\tau (n)}\parallel =0.$$

By using the same argument as in Theorem 3.1 with $\{x_{\tau (n)}\}$, $\{y_{\tau (n)}\}$ and $\{J_{r_{\tau (n)}}{y}_{\tau (n)}\}$, we obtain

$${LIM}_n\left(〈(I-f)(q),{\mathcal{J}}_{\phi }(q-x_{\tau (n)})〉\right)\le 0.$$

Moreover, by using the same argument as in Step 1-Step 4 of Case 1 with $\{x_{\tau (n)}\}$, $\{y_{\tau (n)}\}$ and $\{J_{r_{\tau (n)}}{y}_{\tau (n)}\}$, we obtain the following:

Step 1′ ${lim}_{n\to \mathrm{\infty }}\parallel x_{\tau (n)+1}-x_{\tau (n)}\parallel =0$;

Step 2′ ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}〈(I-f)(q),{\mathcal{J}}_{\phi }(q-x_{\tau (n)})〉\le 0$;

Step 3′ ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}〈(I-f)(q),{\mathcal{J}}_{\phi }(q-y_{\tau (n)})〉\le 0$;

Step 4′ ${lim}_{n\to \mathrm{\infty }}\mathrm{\Phi }(\parallel x_{\tau (n)}-q\parallel )=0$ and ${lim}_{n\to \mathrm{\infty }}\mathrm{\Phi }(\parallel x_{\tau (n)+1}-q\parallel )=0$. Hence

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_{\tau (n)}-q\parallel =0\text{and}\underset{n\to \mathrm{\infty }}{lim}\parallel x_{\tau (n)+1}-q\parallel =0.$$

From Lemma 2.4, we have

$$\parallel x_n-q\parallel \le \parallel x_{\tau (n)+1}-q\parallel .$$

Therefore, ${lim}_{n\to \mathrm{\infty }}\parallel x_n-q\parallel =0$. This completes the proof. □

By taking ${\beta }_n=0$ in Theorem 3.2, we obtain the following result, which is an extension of Corollary 3.4 of Zhang and Song [18] to the viscosity iteration method.

Corollary 3.1 Let E be a uniformly convex Banach space having a weakly continuous duality mapping ${\mathcal{J}}_{\phi }$ with gauge function φ. Let $\{x_n\}$ be a sequence generated by

$$x_{n+1}=J_{r_n}({\alpha }_{n}f(x_n)+(1-{\alpha }_n)x_n),\mathrm{\forall }n\ge 0.$$

Then $\{x_n\}$ converges strongly to $q\in A^{-1}0$, where q is the unique solution of the variational inequality (3.10).

Theorem 3.3 Let E be a uniformly convex Banach space having a uniformly Gâteaux differentiable norm. Then the sequence $\{x_n\}$ generated by (3.1) converges strongly to $q\in A^{-1}0$, where q is the unique solution of the variational inequality

$$〈(I-f)(q),\mathcal{J}(q-p)〉\le 0,\mathrm{\forall }p\in A^{-1}0.$$

(3.15)

Proof We also note that by [30], there exists a solution q of the variational inequality

$$〈(I-f)(q),\mathcal{J}(q-p)〉\le 0,\mathrm{\forall }p\in A^{-1}0,$$

where $q={lim}_{t\to 0^{+}}{x}_t\in A^{-1}0$ with $x_t$ being defined by $x_t=tf(x_t)+(1-t)J_r{x}_t$ for each $r>0$ and $0<t<1$. From now, we put $y_n={\alpha }_{n}f(x_n)+(1-{\alpha }_n)x_n$ for $n\ge 0$.

We also know that $\{x_n\}$, $\{y_n\}$, $\{J_{r_n}{x}_n\}$ and $\{f(x_n)\}$ are bounded by the proof of Theorem 3.1.

As in the proof of Theorem 3.2, we divide the proof into several steps. We only include the differences.

Step 1. By considering two cases as in the proof of Theorem 3.2, we have that ${lim}_{n\to \mathrm{\infty }}\parallel y_n-J_{r_n}{y}_n\parallel =0$ and ${lim}_{n\to \mathrm{\infty }}\parallel y_{\tau (n)}-J_{r_{\tau (n)}}{y}_{\tau (n)}\parallel =0$, where $\tau (n)$ is as in Case 2 in the proof of Theorem 3.2.

Step 2. (1) In the case when ${lim}_{n\to \mathrm{\infty }}\parallel y_n-J_{r_n}{y}_n\parallel =0$, we show that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈(I-f)(q),\mathcal{J}(q-y_n)〉\le 0.$$

To prove this, let a subsequence $\{y_{n_j}\}$ of $\{y_n\}$ be such that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈(I-f)(q),\mathcal{J}(q-y_n)〉=\underset{j\to \mathrm{\infty }}{lim}〈(I-f)(q),\mathcal{J}(q-y_{n_j})〉$$

and $y_{n_j}\rightharpoonup z$ for some $z\in E$. Since

$$x_t-y_n=(1-t)(J_r{x}_t-y_n)+t(f(x_t)-y_n),$$

by Lemma 2.1, we have

$${\parallel x_t-y_n\parallel }^2\le {(1-t)}^2{\parallel J_r{x}_t-y_n\parallel }^2+2t〈f(x_t)-y_n,\mathcal{J}(x_t-y_n)〉.$$

Along with using the resolvent identity (2.1), noting

$$\begin{array}{rcl}\parallel J_r{y}_n-J_{r_n}{y}_n\parallel & =& \parallel J_r(\frac{r}{r_n}{y}_n+(1-\frac{r}{r_n})J_{r_n}{y}_n)-J_r{y}_n\parallel \\ \le & |1-\frac{r}{r_n}|\parallel y_n-J_{r_n}{y}_n\parallel ,\end{array}$$

we observe also that

$$\begin{array}{rcl}\parallel J_r{x}_t-y_n\parallel & \le & \parallel J_r{x}_t-J_r{y}_n\parallel +\parallel J_r{y}_n-J_{r_n}{y}_n\parallel +\parallel J_{r_n}{y}_n-y_n\parallel \\ \le & \parallel x_t-y_n\parallel +(1+|1-\frac{r}{r_n}\left|\right)\parallel y_n-J_{r_n}{y}_n\parallel \\ =& \parallel x_t-y_n\parallel +{\epsilon }_n,\end{array}$$

where ${\epsilon }_n=(1+|1-\frac{r}{r_n}|)(\parallel y_n-J_{r_n}{y}_n\parallel )\to 0$ as $n\to \mathrm{\infty }$ (by Step 1 and condition (C4)). Putting

$$a_j(t)={(1-t)}^2{\epsilon }_{n_j}(2\parallel x_t-y_{n_j}\parallel +{\epsilon }_{n_j})\to 0(j\to \mathrm{\infty })$$

and using Lemma 2.1, we obtain

$$\begin{array}{rcl}{\parallel x_t-y_{n_j}\parallel }^2& \le & {(1-t)}^2{\parallel J_r{x}_t-y_{n_j}\parallel }^2+2t〈f(x_t)-y_{n_j},\mathcal{J}(x_t-y_{n_j})〉\\ \le & {(1-t)}^2{(\parallel J_r{x}_t-J_r{y}_{n_j}\parallel +\parallel J_r{y}_{n_j}-y_{n_j}\parallel )}^2\\ +2t〈f(x_t)-x_t,\mathcal{J}(x_t-y_{n_j})〉+2t{\parallel x_t-y_{n_j}\parallel }^2\\ \le & {(1-t)}^2{\parallel x_t-y_{n_j}\parallel }^2+a_j(t)\\ +2t〈f(x_t)-x_t,\mathcal{J}(x_t-y_{n_j})〉+2t{\parallel x_t-y_{n_j}\parallel }^2.\end{array}$$

The last inequality implies

$$〈x_t-f(x_t),\mathcal{J}(x_t-y_{n_j})〉\le \frac{t}{2}{\parallel x_t-y_{n_j}\parallel }^2+\frac{1}{2t}{a}_j(t).$$

It follows that

$$\underset{j\to \mathrm{\infty }}{lim}〈x_t-f(x_t),\mathcal{J}(x_t-y_{n_j})〉\le \frac{t}{2}M,$$

(3.16)

where $M>0$ is a constant such that $M\ge {\parallel x_t-y_n\parallel }^2$ for all $n\ge 0$ and $t\in (0,1)$. Taking the lim sup as $t\to 0$ in (3.16) and noticing the fact that the two limits are interchangeable due to the fact that J is uniformly continuous on bounded subsets of E from the strong topology of E to the weak^∗ topology of $E^{\ast }$, we have

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈(I-f)(q),\mathcal{J}(q-y_n)〉=\underset{j\to \mathrm{\infty }}{lim}〈(I-f)(q),\mathcal{J}(q-y_{n_j})〉\le 0.$$

• (2) In the case when ${lim}_{n\to \mathrm{\infty }}\parallel y_{\tau (n)}-J_{r_{\tau (n)}}{y}_{\tau (n)}\parallel =0$, by using the same argument with $\{y_{\tau (n)}\}$ and $\{J_{r_{\tau (n)}}{y}_{\tau (n)}\}$, we also have ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}〈(I-f)(q),\mathcal{J}(q-y_{\tau (n)})〉\le 0$.

Step 3. (1) In the case when ${lim}_{n\to \mathrm{\infty }}\parallel y_n-J_{r_n}{y}_n\parallel =0$, we conclude ${lim}_{n\to \mathrm{\infty }}\parallel x_n-q\parallel =0$. Indeed, by using (3.1) and applying Lemma 2.1, we obtain

$$\begin{array}{c}{\parallel x_{n+1}-q\parallel }^2\hfill \\ =〈{\beta }_n(x_n-q)+(1-{\beta }_n)(J_{r_n}{y}_n-q),\mathcal{J}(x_{n+1}-q)〉\hfill \\ \le {\beta }_n\parallel x_n-q\parallel \parallel x_{n+1}-q\parallel +(1-{\beta }_n)\parallel y_n-q\parallel \parallel x_{n+1}-q\parallel \hfill \\ \le {\beta }_n\frac{{\parallel x_n-q\parallel }^2+{\parallel x_{n+1}-q\parallel }^2}{2}+(1-{\beta }_n)\frac{{\parallel y_n-q\parallel }^2+{\parallel x_{n+1}-q\parallel }^2}{2}\hfill \end{array}$$

and

$$\begin{array}{c}{\parallel y_n-q\parallel }^2\hfill \\ =〈{\alpha }_{n}f(x_n)+(1-{\alpha }_n)x_n-p,\mathcal{J}(y_n-q)〉\hfill \\ =〈{\alpha }_n(f(x_n)-f(q))+(1-{\alpha }_n)(x_n-q)+{\alpha }_n(f(q)-q),\mathcal{J}(y_n-q)〉\hfill \\ \le ({\alpha }_{n}k\parallel x_n-q\parallel +(1-{\alpha }_n)\parallel x_n-q\parallel )\parallel y_n-q\parallel +{\alpha }_n〈f(q)-q,\mathcal{J}(y_n-q)〉\hfill \\ =(1-(1-k){\alpha }_n)\frac{{\parallel x_n-q\parallel }^2+{\parallel y_n-q\parallel }^2}{2}+{\alpha }_n〈f(q)-q,\mathcal{J}(y_n-q)〉.\hfill \end{array}$$

Thus

$${\parallel x_{n+1}-q\parallel }^2\le {\beta }_n{\parallel x_n-q\parallel }^2+(1-{\beta }_n){\parallel y_n-q\parallel }^2$$

(3.17)

and

$$\begin{array}{rcl}{\parallel y_n-q\parallel }^2& \le & (1-\frac{2(1-k){\alpha }_n}{1+(1-k){\alpha }_n}){\parallel x_n-q\parallel }^2\\ +\frac{2{\alpha }_n}{1+(1-k){\alpha }_n}〈f(q)-q,\mathcal{J}(y_n-q)〉.\end{array}$$

(3.18)

Combining (3.17) and (3.18) yields

$$\begin{array}{rcl}{\parallel x_{n+1}-q\parallel }^2& \le & {\beta }_n{\parallel x_n-q\parallel }^2+(1-{\beta }_n)(1-\frac{2(1-k){\alpha }_n}{1+(1-k){\alpha }_n}){\parallel x_n-q\parallel }^2\\ +\frac{2(1-{\beta }_n){\alpha }_n}{1+(1-k){\alpha }_n}〈f(q)-q,\mathcal{J}(y_n-q)〉\\ =& (1-(1-{\beta }_n)\frac{2(1-k){\alpha }_n}{1+(1-k){\alpha }_n}){\parallel x_n-q\parallel }^2\\ +(1-{\beta }_n)\frac{2(1-k){\alpha }_n}{(1+(1-k){\alpha }_n)(1-k)}〈f(q)-q,\mathcal{J}(y_n-q)〉.\end{array}$$

(3.19)

Put

$${\lambda }_n=(1-{\beta }_n)\frac{2(1-k){\alpha }_n}{1+(1-k){\alpha }_n}\text{and}{\delta }_n=\frac{1}{1-k}〈(I-f)(q),\mathcal{J}(q-y_n)〉.$$

From the conditions (C1)-(C3) and (1) of Step 2, it follows that ${\lambda }_n\to 0$, ${\sum }_{n=0}^{\mathrm{\infty }}{\lambda }_n=\mathrm{\infty }$ and ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\delta }_n\le 0$. Since (3.19) reduces to

$${\parallel x_{n+1}-q\parallel }^2\le (1-{\lambda }_n){\parallel x_n-q\parallel }^2+{\lambda }_n{\delta }_n,$$

from Lemma 2.3, we conclude that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-q\parallel =0$.

• (2) In the case when ${lim}_{n\to \mathrm{\infty }}\parallel y_{\tau (n)}-J_{r_{\tau (n)}}{y}_{\tau (n)}\parallel =0$, by using the same argument with $\{x_{\tau (n)}\}$, $\{y_{\tau (n)}\}$ and $\{J_{r_{\tau (n)}}{y}_{\tau (n)}\}$ and (2) of Step 2, we can obtain

  $$\underset{n\to \mathrm{\infty }}{lim}\parallel x_{\tau (n)}-q\parallel =0\text{and}\underset{n\to \mathrm{\infty }}{lim}\parallel x_{\tau (n)+1}-q\parallel =0.$$

From Lemma 2.4, we have

$$\parallel x_n-q\parallel \le \parallel x_{\tau (n)+1}-q\parallel .$$

Therefore, ${lim}_{n\to \mathrm{\infty }}\parallel x_n-q\parallel =0$. This completes the proof. □

By taking ${\beta }_n=0$, we also have the following.

Corollary 3.2 Let E be a uniformly convex Banach space having a uniformly Gâteaux differentiable norm. Let $\{x_n\}$ be a sequence generated by

$$x_{n+1}=J_{r_n}({\alpha }_{n}f(x_n)+(1-{\alpha }_n)x_n),\mathrm{\forall }n\ge 0.$$

Then $\{x_n\}$ converges strongly to $q\in A^{-1}0$, where q is the unique solution of the variational inequality (3.15).

Corollary 3.3 Let H be a Hilbert space. Assume that $A:D(A)\subset H\to 2^H$ is a monotone operator satisfying the range condition with $A^{-1}0\ne \mathrm{\varnothing }$ and that C is a nonempty closed convex subset of H such that $\overline{D(A)}\subset C\subset {\bigcap }_{r>0}R(I+rA)$. Let $\{x_n\}$ be a sequence generated by (3.1). Then $\{x_n\}$ converges strongly to $q\in A^{-1}0$, where q is the unique solution of the variational inequality

$$〈(I-f)(q),q-p〉\le 0,\mathrm{\forall }p\in A^{-1}0.$$

(3.20)

By taking ${\beta }_n=0$ in Corollary 3.3, we also have the following.

Corollary 3.4 Let H be a Hilbert space. Assume that $A:D(A)\subset H\to 2^H$ is a maximal monotone operator with $A^{-1}0\ne \mathrm{\varnothing }$. Let $\{x_n\}$ be a sequence generated by

$$x_{n+1}=J_{r_n}({\alpha }_{n}f(x_n)+(1-{\alpha }_n)x_n),\mathrm{\forall }n\ge 0.$$

Then $\{x_n\}$ converges strongly to $q\in A^{-1}0$, where q is the unique solution of the variational inequality (3.20).

Proof Since A is maximal monotone, A is monotone and satisfies the range condition $\overline{D(A)}\subset H=R(I+rA)$ for all $r>0$. Putting $C=H$ in Corollary 3.3, we can obtain the desired result. □

By using arguments similar to those in the proofs of Theorems 3.1, 3.2 and 3.3 and [20], we can obtain the following theorems for the Halpern-type iterative algorithm (3.2).

Theorem 3.4 Let E be a reflexive Banach space having a weakly continuous duality mapping ${\mathcal{J}}_{\phi }$ with gauge function φ. Let $\{x_n\}$ be a sequence generated by (3.2) and LIM be a Banach limit. If ${lim}_{n\to \mathrm{\infty }}\parallel x_n-J_{r_n}{x}_n\parallel =0$, then

$${LIM}_n\left(〈(I-f)(q),{\mathcal{J}}_{\phi }(q-x_n)〉\right)\le 0,$$

where $q:={lim}_{t\to 0^{+}}{x}_t$ with $x_t$ being defined by $x_t=tf(x_t)+(1-t)J_r{x}_t$ for each $r>0$.

Theorem 3.5 Let E be a uniformly convex Banach space having a weakly continuous duality mapping ${\mathcal{J}}_{\phi }$ with gauge function φ. Then the sequence $\{x_n\}$ generated by (3.2) converges strongly to $q\in A^{-1}0$, where q is the unique solution of the variational inequality (3.10).

Theorem 3.6 Let E be a uniformly convex Banach space having a uniformly Gâteaux differentiable norm. Then the sequence $\{x_n\}$ generated by (3.2) converges strongly to $q\in A^{-1}0$, where q is the unique solution of the variational inequality (3.15).

Corollary 3.5 Let H be a Hilbert space. Assume that $A:D(A)\subset H\to 2^H$ is a maximal monotone operator with $A^{-1}0\ne \mathrm{\varnothing }$. Let $\{x_n\}$ be a sequence generated by (3.2). Then $\{x_n\}$ converges strongly to $q\in A^{-1}0$, where q is the unique solution of the variational inequality (3.20).

Remark 3.1

(1) Theorem 3.2 improves and develops Theorem 3.7 of Zhang and Song [18] in the following aspects.

• (a) The following gaps, which authors in [18] overlooked, are corrected: there exist two subsequences $\{z_{n_i}\}$ and $\{z_{n_j}\}$ of $\{z_n\}$ satisfying

  $$\frac{1}{4}(1-{\beta }_{n_i})g(\parallel z_{n_i}-J_{r_n}{z}_{n_i}\parallel )\le {\alpha }_{n_i}{\parallel u-p\parallel }^2,\mathrm{\forall }i\ge 0$$

and

$$\frac{1}{4}(1-{\beta }_{n_j})g(\parallel z_{n_j}-J_{r_{n_j}}{z}_{n_j}\parallel )>{\alpha }_{n_j}{\parallel u-p\parallel }^2,\mathrm{\forall }j\ge 0,$$

where $x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)J_{r_n}{z}_n$, $z_n={\alpha }_{n}u+(1-{\alpha }_n)x_n$ and $p\in A^{-1}0$.

• (b) The case of an iterative scheme $z_n={\alpha }_{n}u+(1-{\alpha }_n)x_n$ in [[11], Theorem 3.7] is extended to the case of a viscosity iterative scheme $y_n={\alpha }_{n}f(x_n)+(1-{\alpha }_n)x_n$, where $f:C\to C$ is a contractive mapping with a constant $k\in (0,1)$.

• (c) We utilize the weakly continuous duality mapping ${\mathcal{J}}_{\phi }$ with gauge function φ instead of the weakly continuous normalized duality mapping $\mathcal{J}$ in [[11], Theorem 3.7].

• (2) Theorem 3.3 extends Theorem 3.8 of Zhang and Song [18] to the viscosity iterative method together with our proof, which corrects the gap in the proof of [18].

• (3) Theorem 3.2 and Theorem 3.3 improve Theorem 3.3 and Theorem 3.4 of Yu [20], which were given without proofs, to the case of the viscosity iterative method together with our proofs. Theorem 3.3 also develops and complements Theorem 4.2 of Song [17]. In particular, the limit point $q\in A^{-1}0$ of the sequence $\{x_n\}$ in Theorem 3.3 is the unique solution of the variational inequality (3.15) in comparison with [[10], Theorem 4.2].

• (4) Theorem 3.5 and Theorem 3.6 extend Theorems 3.1 and 3.2 of Zhang and Song [18] and Theorem 3.1 and Theorem 3.2 of Yu [20] to the viscosity iterative method.

• (5) Corollaries 3.1 and 3.2 improve the corresponding results of Zhang and Song [18] and Song et al. [22]. Corollary 3.4 also develops the corresponding results of Xu [15] and Song and Yang [16].

• (6) As in [22, 31, 32], we can replace the contractive mapping f in our algorithms by the weakly contractive mapping g (recall that a mapping $g:C\to C$ is said to be weakly contractive [33] if $\parallel g(x)-g(y)\parallel \le \parallel x-y\parallel -\psi (\parallel x-y\parallel )$, $\mathrm{\forall }x,y\in C$, where $\psi :[0,+\mathrm{\infty })\to [0,+\mathrm{\infty })$ is a continuous and strictly increasing function such that ψ is positive on $(0,\mathrm{\infty })$ and $\psi (0)=0$).