Proximal point algorithms for zero points of nonlinear operators

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.

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 $\{x\in E:Ax\ne 0\}$. 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 [1–20] and the references therein. One of the most popular techniques for solving the inclusion problem goes back to the work of Browder [21]. 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 $R_A={(I+A)}^{-1}$, 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 [22] and the references therein. Regularization methods recently have been investigated for treating zero points of monotone operators; for [23–33] 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., $sup\{\parallel x-y\parallel :y\in K\}<d(K)$, where $d(K)$ 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 [34] and the references therein.

Let $U_E=\{x\in E:\parallel x\parallel =1\}$. 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

$$J(x)=\{f\in E^{\ast }:〈x,f〉={\parallel x\parallel }^2={\parallel f\parallel }^2\}$$

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 $F(T)$ to denote the set of fixed points of T. Recall that T is said to be contractive if there exists a constant $\alpha \in (0,1)$ such that

$$\parallel Tx-Ty\parallel \le \alpha \parallel x-y\parallel ,\mathrm{\forall }x,y\in C.$$

For such a case, we also call T an α-contraction. T is said to be nonexpansive if

$$\parallel Tx-Ty\parallel \le \parallel x-y\parallel ,\mathrm{\forall }x,y\in C.$$

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 $t\in (0,1)$, we have $Q(tx+(1-t)Qx)=Qx$; 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 [34], 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) Q is sunny and nonexpansive;

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

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

Let I denote the identity operator on E. An operator $A\subset E\times E$ with domain $D(A)=\{z\in E:Az\ne \mathrm{\varnothing }\}$ and range $R(A)=\bigcup \{Az:z\in D(A)\}$ is said to be accretive if for each $x_i\in D(A)$ and $y_i\in A{x}_i$, $i=1,2$, there exists $j(x_1-x_2)\in J(x_1-x_2)$ such that $〈y_1-y_2,j(x_1-x_2)〉\ge 0$. An accretive operator A is said to be m-accretive if $R(I+rA)=E$ 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 $A^{-1}(0)$ to denote the set of zeros of A. For an accretive operator A, we can define a nonexpansive single-valued mapping $J_r:R(I+rA)\to D(A)$ by $J_r={(I+rA)}^{-1}$ 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 [35]

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

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

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

Lemma 2.2 [36]

Let $\{a_n\}$ be a sequence of nonnegative numbers satisfying the condition $a_{n+1}\le (1-t_n)a_n+t_n{b}_n+c_n$, $\mathrm{\forall }n\ge 0$, where $\{t_n\}$ is a number sequence in $(0,1)$ such that ${lim}_{n\to \mathrm{\infty }}{t}_n=0$ and ${\sum }_{n=0}^{\mathrm{\infty }}{t}_n=\mathrm{\infty }$, $\{b_n\}$ is a number sequence such that ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{b}_n\le 0$, and $\{c_n\}$ 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 [37]

Let $\{x_n\}$ and $\{y_n\}$ be bounded sequences in a Banach space E, and $\{{\beta }_n\}$ be a sequence in $(0,1)$ with

$$0<\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{\beta }_n\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\beta }_n<1.$$

Suppose that $x_{n+1}=(1-{\beta }_n)y_n+{\beta }_n{x}_n$, $\mathrm{\forall }n\ge 1$ and

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}(\parallel y_{n+1}-y_n\parallel -\parallel x_{n+1}-x_n\parallel )\le 0.$$

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

Lemma 2.4 [31]

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 $\alpha \in (0,1)$. Let $\{x_t\}$ be a sequence generated by the following $x_t=tf{x}_t+(1-t)S{x}_t$, where $t\in (0,1)$. Then $\{x_t\}$ converges strongly as $t\to 0$ to a fixed point $x^{\ast }$ of S, which is the unique solution in $F(S)$ to the following variational inequality $〈f(x^{\ast }-x^{\ast }),j(x^{\ast }-p)〉\ge 0$, $\mathrm{\forall }p\in F(S)$.

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 $C:=\overline{D(A)}$ is convex and has the normal structure. Let $f:C\to C$ be a fixed α-contraction. Let $\{{\alpha }_n\}$, $\{{\beta }_n\}$, $\{{\gamma }_n\}$, and $\{{\delta }_n\}$ be real number sequences in $(0,1)$ such that ${\alpha }_n+{\beta }_n+{\gamma }_n+{\delta }_n=1$. Let $Q_C$ be the sunny nonexpansive retraction from E onto C and $\{x_n\}$ be a sequence generated in the following manner:

$$x_0\in C,x_{n+1}={\alpha }_{n}f(x_n)+{\beta }_n{J}_{r_n}(x_n+e_{n+1})+{\gamma }_n{x}_n+{\delta }_n{Q}_C(g_n),\mathrm{\forall }n\ge 0,$$

(ϒ)

where $\{e_n\}$ is a sequence in E, $\{g_n\}$ is a bounded sequence in E, $\{r_n\}$ is a positive real numbers sequence, and $J_{r_n}={(I+r_{n}A)}^{-1}$. Assume that $A^{-1}(0)$ is not empty and the above control sequences satisfy the following restrictions:

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

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

• (c) ${\sum }_{n=1}^{\mathrm{\infty }}\parallel e_n\parallel <\mathrm{\infty }$ and ${\sum }_{n=0}^{\mathrm{\infty }}{\delta }_n<\mathrm{\infty }$;

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

Proof Fixing $p\in A^{-1}(0)$, we find that

$$\begin{array}{rl}\parallel x_1-p\parallel \le & {\alpha }_0\parallel f(x_0)-p\parallel +{\beta }_0\parallel J_{r_0}(x_0+e_1)-p\parallel +{\gamma }_0\parallel x_0-p\parallel +{\delta }_0\parallel Q_C(g_0)-p\parallel \\ \le & {\alpha }_0\alpha \parallel x_0-p\parallel +{\alpha }_0\parallel f(p)-p\parallel +{\beta }_0\parallel x_0-p\parallel +{\beta }_0\parallel e_1\parallel \\ +{\gamma }_0\parallel x_0-p\parallel +{\delta }_0\parallel g_0-p\parallel \\ \le & (1-{\alpha }_0(1-\alpha ))\parallel x_0-p\parallel +{\alpha }_0\parallel f(p)-p\parallel +\parallel e_1\parallel +{\delta }_0\parallel g_0-p\parallel .\end{array}$$

(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 $M_1=max\{\parallel x_0-p\parallel ,\frac{\parallel f(p)-p\parallel }{1-\alpha }\}<\mathrm{\infty }$. 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

$$\begin{array}{rl}\parallel x_{m+1}-p\parallel \le & {\alpha }_m\parallel f(x_m)-p\parallel +{\beta }_n\parallel J_{r_m}(x_m+e_{m+1})-p\parallel +{\gamma }_m\parallel x_m-p\parallel \\ +{\delta }_m\parallel Q_C(g_m)-p\parallel \\ \le & {\alpha }_m\alpha \parallel x_m-p\parallel +{\alpha }_m\parallel f(p)-p\parallel +{\beta }_m\parallel x_m-p\parallel +{\beta }_m\parallel e_{m+1}\parallel \\ +{\gamma }_m\parallel x_m-p\parallel +{\delta }_m\parallel Q_C(g_m)-p\parallel \\ \le & (1-{\alpha }_m(1-\alpha ))\parallel x_m-p\parallel +{\alpha }_m(1-\alpha )\frac{\parallel f(p)-p\parallel }{1-\alpha }\\ +\parallel e_{m+1}\parallel +{\delta }_m\parallel g_m-p\parallel \\ \le & M_1+\sum _{i=1}^{m+1}\parallel e_i\parallel +\sum _{i=1}^m{\delta }_i\parallel g_i\parallel .\end{array}$$

This shows that (3.2) holds. In view of the restriction (c), we find that the sequence $\{x_n\}$ is bounded. Put $y_n=J_{r_n}(x_n+e_{n+1})$ 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

$$\begin{array}{rl}{z}_{n+1}-z_n=& \frac{{\alpha }_{n+1}}{1-{\gamma }_{n+1}}f(x_{n+1})+\frac{{\beta }_{n+1}}{1-{\gamma }_{n+1}}{y}_{n+1}+\frac{{\delta }_{n+1}}{1-{\gamma }_{n+1}}{Q}_C(g_{n+1})\\ -\frac{{\alpha }_n}{1-{\gamma }_n}f(x_n)-\frac{{\beta }_n}{1-{\gamma }_n}{y}_n-\frac{{\delta }_n}{1-{\gamma }_n}{Q}_C(g_n)\\ =& \frac{{\alpha }_{n+1}}{1-{\gamma }_{n+1}}(f(x_{n+1})-y_{n+1})+y_{n+1}+\frac{{\delta }_{n+1}}{1-{\gamma }_{n+1}}(Q_C(g_{n+1})-y_{n+1})\\ -\frac{{\alpha }_n}{1-{\gamma }_n}(f(x_n)-y_n)-y_n-\frac{{\delta }_n}{1-{\gamma }_n}(Q_C(g_n)-y_n).\end{array}$$

This yields

$$\begin{array}{rl}\parallel z_{n+1}-z_n\parallel \le & \frac{{\alpha }_{n+1}}{1-{\gamma }_{n+1}}\parallel f(x_{n+1})-y_{n+1}\parallel +\parallel y_{n+1}-y_n\parallel +\frac{{\alpha }_n}{1-{\gamma }_n}\parallel f(x_n)-y_n\parallel \\ +\frac{{\delta }_{n+1}}{1-{\gamma }_{n+1}}\parallel Q_C(g_{n+1})-y_{n+1}\parallel +\frac{{\delta }_n}{1-{\gamma }_n}\parallel Q_C(g_n)-y_n\parallel .\end{array}$$

(3.3)

Next, we estimate $\parallel y_{n+1}-y_n\parallel$. In view of Lemma 2.1, we find that

$$\begin{array}{r}\parallel y_n-y_{n+1}\parallel \\ \le \parallel \frac{r_n}{r_{n+1}}(x_n+e_{n+1})+(1-\frac{r_n}{r_{n+1}})J_{r_{n+1}}(x_n+e_{n+1})-(x_{n+1}+e_{n+2})\parallel \\ =\parallel \frac{r_n}{r_{n+1}}((x_n+e_{n+1})-(x_{n+1}+e_{n+2}))+\frac{r_{n+1}-r_n}{r_{n+1}}(J_{r_{n+1}}(x_n+e_{n+1})-(x_{n+1}+e_{n+2}))\parallel \\ \le \parallel x_n-x_{n+1}\parallel +\parallel e_{n+1}\parallel +\parallel e_{n+2}\parallel +\frac{M_2}{\mu }(r_{n+1}-r_n),\end{array}$$

(3.4)

where $M_2$ is an appropriate constant such that

$$M_2\ge \underset{n\ge 1}{sup}\left\{\parallel J_{r_{n+1}}(x_n+e_n)-(x_{n+1}+e_{n+1})\parallel \right\}.$$

Substituting (3.4) into (3.3), we arrive at

$$\begin{array}{r}\parallel z_{n+1}-z_n\parallel -\parallel x_n-x_{n+1}\parallel \\ \le \frac{{\alpha }_{n+1}}{1-{\gamma }_{n+1}}\parallel f(x_{n+1})-y_{n+1}\parallel +\parallel e_{n+1}\parallel +\parallel e_{n+2}\parallel +\frac{M_2}{\mu }(r_{n+1}-r_n)+\frac{{\alpha }_n}{1-{\gamma }_n}\parallel f(x_n)-y_n\parallel \\ +\frac{{\delta }_{n+1}}{1-{\gamma }_{n+1}}\parallel Q_C(g_{n+1})-y_{n+1}\parallel +\frac{{\delta }_n}{1-{\gamma }_n}\parallel Q_C(g_n)-y_n\parallel .\end{array}$$

In view of the restrictions (a), (b), (c), and (d), we find that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}(\parallel z_{n+1}-z_n\parallel -\parallel x_n-x_{n+1}\parallel )\le 0.$$

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

$$\begin{array}{r}\parallel x_n-J_{r_n}(x_n+e_{n+1})\parallel \\ \le \parallel x_n-x_{n+1}\parallel +\parallel x_{n+1}-J_{r_n}(x_n+e_{n+1})\parallel \\ \le \parallel x_n-x_{n+1}\parallel +{\alpha }_n\parallel f(x_n)-J_{r_n}(x_n+e_{n+1})\parallel +{\gamma }_n\parallel x_n-J_{r_n}(x_n+e_{n+1})\parallel \\ +{\delta }_n\parallel Q_C(g_n)-J_{r_n}(x_n+e_{n+1})\parallel .\end{array}$$

It follows that

$$\begin{array}{r}(1-{\gamma }_n)\parallel x_n-J_{r_n}(x_n+e_{n+1})\parallel \\ \le \parallel x_n-x_{n+1}\parallel +{\alpha }_n\parallel f(x_n)-J_{r_n}(x_n+e_{n+1})\parallel +{\delta }_n\parallel Q_C(g_n)-J_{r_n}(x_n+e_{n+1})\parallel .\end{array}$$

In view of the restrictions (a), (b), and (c), we find from (3.5) that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-J_{r_n}(x_n+e_{n+1})\parallel =0.$$

(3.6)

Notice that

$$\begin{array}{rl}\parallel x_n-J_{r_n}{x}_n\parallel & \le \parallel x_n-J_{r_n}(x_n+e_{n+1})\parallel +\parallel J_{r_n}(x_n+e_{n+1})-J_{r_n}{x}_n\parallel \\ \le \parallel x_n-J_{r_n}(x_n+e_{n+1})\parallel +\parallel e_{n+1}\parallel .\end{array}$$

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

$$\begin{array}{rl}\parallel J_{r_n}{x}_n-J_r{x}_n\parallel & =\parallel J_r(\frac{r}{r_n}{x}_n+(1-\frac{r}{r_n})J_{r_n}{x}_n)-J_r{x}_n\parallel \\ \le \parallel (1-\frac{r}{r_n})(J_{r_n}{x}_n-x_n)\parallel \\ \le \parallel J_{r_n}{x}_n-x_n\parallel .\end{array}$$

(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 ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}〈f(\overline{x})-\overline{x},j(x_n-\overline{x})〉\le 0$, where $\overline{x}={lim}_{t\to 0}{z}_t$, and $z_t$ solves the fixed point equation $z_t=tf(z_t)+(1-t)J_r{z}_t$, $\mathrm{\forall }t\in (0,1)$, from which it follows that

$$\parallel z_t-x_n\parallel =\parallel (1-t)(J_r{z}_t-x_n)+t(f(z_t)-x_n)\parallel .$$

For any $t\in (0,1)$, we see that

$$\begin{array}{rl}{\parallel z_t-x_n\parallel }^2=& (1-t)〈J_r{z}_t-x_n,j(z_t-x_n)〉+t〈f(z_t)-x_n,j(z_t-x_n)〉\\ =& (1-t)(〈J_r{z}_t-J_r{x}_n,j(z_t-x_n)〉+〈J_r{x}_n-x_n,j(z_t-x_n)〉)\\ +t〈f(z_t)-z_t,j(z_t-x_n)〉+t〈z_t-x_n,j(z_t-x_n)〉\\ \le & (1-t)({\parallel z_t-x_n\parallel }^2+\parallel J_r{x}_n-x_n\parallel \parallel z_t-x_n\parallel )\\ +t〈f(z_t)-z_t,j(z_t-x_n)〉+t{\parallel z_t-x_n\parallel }^2\\ \le & {\parallel z_t-x_n\parallel }^2+\parallel J_r{x}_n-x_n\parallel \parallel z_t-x_n\parallel +t〈f(z_t)-z_t,j(z_t-x_n)〉.\end{array}$$

It follows that

$$〈z_t-f(z_t),j(z_t-x_n)〉\le \frac{1}{t}\parallel J_r{x}_n-x_n\parallel \parallel z_t-x_n\parallel ,\mathrm{\forall }t\in (0,1).$$

By virtue of (3.8), we find that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈z_t-f(z_t),j(z_t-x_n)〉\le 0.$$

(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

$$\begin{array}{r}|〈f(\overline{x})-\overline{x},j(x_n-\overline{x})〉-〈z_t-f(z_t),j(z_t-x_n)〉|\\ \le |〈f(\overline{x})-\overline{x},j(x_n-\overline{x})〉-〈f(\overline{x})-\overline{x},j(x_n-z_t)〉|\\ +|〈f(\overline{x})-\overline{x},j(x_n-z_t)〉-〈z_t-f(z_t),j(z_t-x_n)〉|\\ \le \left|〈f(\overline{x})-\overline{x},j(x_n-\overline{x})-j(x_n-z_t)〉\right|+\left|〈f(\overline{x})-\overline{x}+z_t-f(z_t),J(x_n-z_t)〉\right|\\ \le \parallel f(\overline{x})-\overline{x}\parallel \parallel j(x_n-\overline{x})-j(x_n-z_t)\parallel \\ +\parallel f(\overline{x})-\overline{x}+z_t-f(z_t)\parallel \parallel x_n-z_t\parallel \to 0,\text{as }t\to 0.\end{array}$$

Hence, for any $ϵ>0$, there exists $\lambda >0$ such that $\mathrm{\forall }t\in (0,\lambda )$ the following inequality holds:

$$〈f(\overline{x})-\overline{x},j(x_n-\overline{x})〉\le 〈z_t-f(z_t),j(z_t-x_n)〉+ϵ.$$

This implies that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈f(\overline{x})-\overline{x},j(x_n-\overline{x})〉\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈z_t-f(z_t),j(z_t-x_n)〉+ϵ.$$

Since ϵ is arbitrary and (3.9), one finds that ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}〈f(\overline{x})-\overline{x},j(x_n-\overline{x})〉\le 0$. This implies that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈f(\overline{x})-\overline{x},j(x_{n+1}-\overline{x})〉\le 0.$$

(3.10)

Finally, we prove that $x_n\to \overline{x}$ as $n\to \mathrm{\infty }$. Note that

$$\begin{array}{rl}{\parallel x_{n+1}-\overline{x}\parallel }^2\le & {\alpha }_n〈f(x_n)-\overline{x},j(x_{n+1}-\overline{x})〉+{\beta }_n\parallel y_n-\overline{x}\parallel \parallel \parallel x_{n+1}-\overline{x}\parallel \\ +{\gamma }_n\parallel x_n-\overline{x}\parallel \parallel x_{n+1}-\overline{x}\parallel +{\delta }_n\parallel Q_C(g_n)-\overline{x}\parallel \parallel x_{n+1}-\overline{x}\parallel \\ \le & {\alpha }_n〈f(x_n)-\overline{x},j(x_{n+1}-\overline{x})〉+\frac{{\beta }_n}{2}({\parallel y_n-\overline{x}\parallel }^2+{\parallel x_{n+1}-\overline{x}\parallel }^2)\\ +\frac{{\gamma }_n}{2}({\parallel x_n-\overline{x}\parallel }^2+{\parallel x_{n+1}-\overline{x}\parallel }^2)+\frac{{\delta }_n}{2}({\parallel Q_C(g_n)-\overline{x}\parallel }^2+{\parallel x_{n+1}-\overline{x}\parallel }^2).\end{array}$$

Note that $\parallel y_n-\overline{x}\parallel \le \parallel x_n-\overline{x}\parallel +\parallel e_{n+1}\parallel$. It follows that

$$\begin{array}{rl}{\parallel x_{n+1}-\overline{x}\parallel }^2\le & 2{\alpha }_n〈f(x_n)-\overline{x},j(x_{n+1}-\overline{x})〉+{\beta }_n{\parallel y_n-\overline{x}\parallel }^2\\ +{\gamma }_n{\parallel x_n-\overline{x}\parallel }^2+{\delta }_n{\parallel Q_C(g_n)-\overline{x}\parallel }^2\\ \le & 2{\alpha }_n〈f(x_n)-\overline{x},j(x_{n+1}-\overline{x})〉+(1-{\alpha }_n){\parallel x_n-\overline{x}\parallel }^2+{\nu }_n,\end{array}$$

(3.11)

where ${\nu }_n=\parallel e_{n+1}\parallel (\parallel e_{n+1}\parallel +2\parallel x_n-\overline{x}\parallel )+{\delta }_n{\parallel Q_C(g_n)-\overline{x}\parallel }^2$. In view of the restriction (c), we find that ${\sum }_{n=1}^{\mathrm{\infty }}{\nu }_n<\mathrm{\infty }$. Let ${\rho }_n=max\{〈f(x_n)-\overline{x},j(x_{n+1}-\overline{x})〉,0\}$. 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

$$〈f(x_n)-\overline{x},j(x_{n+1}-\overline{x})〉<ϵ,\mathrm{\forall }n\ge n_1.$$

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

$${\parallel x_{n+1}-\overline{x}\parallel }^2\le (1-{\alpha }_n){\parallel x_n-\overline{x}\parallel }^2+2{\alpha }_n{\rho }_n+{\nu }_n.$$

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 $C:=\overline{D(A)}$ is convex and has the normal structure. Let $\{{\alpha }_n\}$, $\{{\beta }_n\}$, and $\{{\gamma }_n\}$ be real number sequences in $(0,1)$ such that ${\alpha }_n+{\beta }_n+{\gamma }_n=1$. Let $\{x_n\}$ be a sequence generated in the following manner:

$$x_0\in C,x_{n+1}={\alpha }_{n}u+{\beta }_n{J}_{r_n}(x_n+e_{n+1})+{\gamma }_n{x}_n,\mathrm{\forall }n\ge 0,$$

where u is a fixed element in C, $\{e_n\}$ is a sequence in E, $\{g_n\}$ is a bounded sequence in E, $\{r_n\}$ is a positive real numbers sequence, and $J_{r_n}={(I+r_{n}A)}^{-1}$. Assume that $A^{-1}(0)$ is not empty and the above control sequences satisfy the following restrictions:

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

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

• (c) ${\sum }_{n=1}^{\mathrm{\infty }}\parallel e_n\parallel <\mathrm{\infty }$;

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

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 $w:H\to (-\mathrm{\infty },\mathrm{\infty }]$, the subdifferential mapping ∂w of w is defined by

$$\partial w(x)=\{x^{\ast }\in H:w(x)+〈y-x,x^{\ast }〉\le w(y),\mathrm{\forall }y\in H\},\mathrm{\forall }x\in H.$$

Rockafellar [38] proved that ∂w is a maximal monotone operator. It is easy to verify that $0\in \partial w(v)$ if and only if $w(v)={min}_{x\in H}g(x)$.

Theorem 4.1 Let $w:H\to (-\mathrm{\infty },+\mathrm{\infty }]$ be a proper convex lower semicontinuous function such that ${(\partial w)}^{-1}(0)$ is not empty. Let $f:H\to H$ be a κ-contraction and let $\{x_n\}$ be a sequence in H in the following process: $x_0\in H$ and

$$\{\begin{array}{c}{y}_n={arg\hspace{0.17em}min}_{z\in H}\{w(z)+\frac{{\parallel z-x_n-e_{n+1}\parallel }^2}{2r_n}\},\hfill \\ x_{n+1}={\alpha }_{n}f(x_n)+{\beta }_n{y}_n+{\gamma }_n{x}_n,\mathrm{\forall }n\ge 0,\hfill \end{array}$$

where $\{e_n\}$ is a sequence in H, $\{g_n\}$ is a bounded sequence in H, and $\{r_n\}$ 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 $\{x_n\}$ converges strongly to $\overline{x}$, which is the unique solution to the following variational inequality: $〈f(\overline{x})-\overline{x},j(p-\overline{x})〉\le 0$, $\mathrm{\forall }p\in {(\partial w)}^{-1}(0)$.

Proof Since $w:H\to (-\mathrm{\infty },\mathrm{\infty }]$ is a proper convex and lower semicontinuous function, we see that subdifferential ∂w of w is maximal monotone. We note that

$$y_n=\underset{z\in H}{arg\hspace{0.17em}min}\{w(z)+\frac{{\parallel z-x_n-e_{n+1}\parallel }^2}{2r_n}\}$$

is equivalent to $0\in \partial w(y_n)+\frac{1}{r_n}(y_n-x_n-e_{n+1})$. It follows that

$$x_n+e_{n+1}\in y_n+r_n\partial w(y_n).$$

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