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

Abstract

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

1 Introduction and preliminaries

Let E be a Banach space and let $E^{\ast }$ be the dual of E. Let $〈\cdot ,\cdot 〉$ denote the pairing between E and $E^{\ast }$. The normalized duality mapping $J:E\to 2^{E^{\ast }}$ is defined by

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

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

Let ${\rho }_E:[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ be the modulus of smoothness of E by

$${\rho }_E(t)=sup\{\frac{\parallel x+y\parallel -\parallel x-y\parallel }{2}-1:x\in U_E,\parallel y\parallel \le t\}.$$

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

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

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

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

• (2) a contraction if ${\mathit{Proj}}_D^2={\mathit{Proj}}_D$;

• (3) a sunny nonexpansive retraction if ${\mathit{Proj}}_D$ is sunny, nonexpansive, and a contraction.

D is said to be a nonexpansive retract of C if there exists a nonexpansive retraction from C onto D. The following result, which was established in [1–3], describes a characterization of sunny nonexpansive retractions on a smooth Banach space.

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

• (1) ${\mathit{Proj}}_C$ is sunny and nonexpansive;

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

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

It is well known that if E is a Hilbert space, then a sunny nonexpansive retraction ${\mathit{Proj}}_C$ is coincident with the metric projection from E onto C. Let C be a nonempty closed convex subset of a smooth Banach space E, let $x\in E$, and let $x_0\in C$. Then we have from the above that $x_0={\mathit{Proj}}_{C}x$ if and only if $〈x-x_0,J_{\phi }(y-x_0)〉\le 0$ for all $y\in C$, where ${\mathit{Proj}}_C$ is a sunny nonexpansive retraction from E onto C. For more additional information on nonexpansive retracts, see [4] and the references therein.

Let C be a nonempty closed convex subset of E. Let $T:C\to C$ be a mapping. In this paper, we use $F(T)$ to denote the set of fixed points of T. Recall that T is said to be an α-contractive mapping iff there exists a constant $\alpha \in [0,1)$ such that $\parallel Tx-Ty\parallel \le \alpha \parallel x-y\parallel$, $\mathrm{\forall }x,y\in C$. The Picard iterative process is an efficient method to study fixed points of α-contractive mappings. It is well known that α-contractive mappings have a unique fixed point. T is said to be nonexpansive iff $\parallel Tx-Ty\parallel \le \parallel x-y\parallel$, $\mathrm{\forall }x,y\in C$. It is well known that nonexpansive mappings have fixed points if the set C is closed and convex, and the space E is uniformly convex. The Krasnoselski-Mann iterative process is an efficient method for studying fixed points of nonexpansive mappings. The Krasnoselski-Mann iterative process generates a sequence $\{x_n\}$ in the following manner:

$$x_1\in C,x_{n+1}={\alpha }_{n}T{x}_n+(1-{\alpha }_n)x_n,\mathrm{\forall }n\ge 1.$$

It is well known that the Krasnoselski-Mann iterative process only has weak convergence for nonexpansive mappings in infinite-dimensional Hilbert spaces; see [5–7] for more details and the references therein. In many disciplines, including economics, image recovery, quantum physics, and control theory, problems arise in infinite-dimensional spaces. In such problems, strong convergence (norm convergence) is often much more desirable than weak convergence, for it translates the physically tangible property that the energy $\parallel x_n-x\parallel$ of the error between the iterate $x_n$ and the solution x eventually becomes arbitrarily small. To improve the weak convergence of a Krasnoselski-Mann iterative process, so-called hybrid projections have been considered; see [8–22] for more details and the references therein. The Halpern iterative process was initially introduced in [23]; see [23] for more details and the references therein. The Halpern iterative process generates a sequence $\{x_n\}$ in the following manner:

$$x_1\in C,x_{n+1}={\alpha }_{n}u+(1-{\alpha }_n)T{x}_n,\mathrm{\forall }n\ge 1,$$

where $x_1$ is an initial and u is a fixed element in C. Strong convergence of Halpern iterative process does not depend on metric projections. The Halpern iterative process has recently been extensively studied for treating accretive operators; see [24–31] and the references therein.

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 this paper, we use $A^{-1}(0)$ to denote the set of zero points of A. For an accretive operator A, we can define a nonexpansive single valued mapping $J_r:R(I+rA)\to D(A)$ by $J_r={(I+rA)}^{-1}$ for each $r>0$, which is called the resolvent of A.

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

Lemma 1.1 [32]

Let $\{a_n\}$, $\{b_n\}$, $\{c_n\}$, and $\{d_n\}$ be four nonnegative real sequences satisfying $a_{n+1}\le (1-b_n)a_n+b_n{c}_n+d_n$, $\mathrm{\forall }n\ge n_0$, where $n_0$ is some positive integer, $\{b_n\}$ is a number sequence in $(0,1)$ such that ${\sum }_{n=n_0}^{\mathrm{\infty }}{b}_n=\mathrm{\infty }$, $\{c_n\}$ is a number sequence such that ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{c}_n\le 0$, and $\{d_n\}$ is a positive number sequence such that ${\sum }_{n=n_0}^{\mathrm{\infty }}{d}_n<\mathrm{\infty }$. Then ${lim}_{n\to \mathrm{\infty }}{a}_n=0$.

Lemma 1.2 [33]

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

Lemma 1.3 [34]

Let $\{x_n\}$ and $\{y_n\}$ be bounded sequences in a Banach space E and let ${\beta }_n$ be a sequence in $[0,1]$ with $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1$. Suppose that $x_{n+1}=(1-{\beta }_n)y_n+{\beta }_n{x}_n$ for all $n\ge 0$ 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 1.4 [35]

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

2 Main results

Theorem 2.1 Let E be a real reflexive, strictly convex Banach space with the uniformly Gâteaux differentiable norm. Let $N\ge 1$ be some positive integer. Let $A_m$ be an m-accretive operator in E for each $m\in \{1,2,\dots ,N\}$. Assume that $C:={\bigcap }_{m=1}^N\overline{D(A_m)}$ is convex and has the normal structure. Let $f:C\to C$ be an α-contractive mapping. Let $\{{\alpha }_n\}$, $\{{\beta }_n\}$, and $\{{\gamma }_n\}$ be real number sequences in $(0,1)$ with the restriction ${\alpha }_n+{\beta }_n+{\gamma }_n=1$. Let $\{{\delta }_{n,i}\}$ be a real number sequence in $(0,1)$ with the restriction ${\delta }_{n,1}+{\delta }_{n,2}+\cdots +{\delta }_{n,N}=1$. Let $\{r_m\}$ be a positive real numbers sequence and $\{e_{n,i}\}$ a sequence in E for each $i\in \{1,2,\dots ,N\}$. Assume that ${\bigcap }_{i=1}^N{A}_i^{-1}(0)$ is not empty. Let $\{x_n\}$ be a sequence generated in the following manner:

$$x_1\in C,x_{n+1}={\alpha }_{n}f(x_n)+{\beta }_n{x}_n+{\gamma }_n\sum _{i=1}^N{\delta }_{n,i}{J}_{r_i}(x_n+e_{n,i}),\mathrm{\forall }n\ge 1,$$

where $J_{r_i}={(I+r_i{A}_i)}^{-1}$. Assume that the control sequences $\{{\alpha }_n\}$, $\{{\beta }_n\}$, $\{{\gamma }_n\}$, and $\{{\delta }_{n,i}\}$ satisfy the following restrictions:

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

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

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

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

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

Proof Put $y_n={\sum }_{i=1}^N{\delta }_{n,i}{J}_{r_i}(x_n+e_{n,i})$. Fixing $p\in {\bigcap }_{i=1}^N{A}_i^{-1}(0)$, we have

$$\begin{array}{rl}\parallel y_n-p\parallel & \le \sum _{i=1}^N{\delta }_{n,i}\parallel J_{r_i}(x_n+e_{n,i})-p\parallel \\ \le \sum _{i=1}^N{\delta }_{n,i}\parallel (x_n+e_{n,i})-p\parallel \\ \le \parallel x_n-p\parallel +\sum _{i=1}^N\parallel e_{n,i}\parallel .\end{array}$$

Hence, we have

$$\begin{array}{rcl}\parallel x_{n+1}-p\parallel & \le & {\alpha }_n\parallel f(x_n)-p\parallel +{\beta }_n\parallel x_n-p\parallel +{\gamma }_n\parallel y_n-p\parallel \\ \le & {\alpha }_n\alpha \parallel x_n-p\parallel +{\alpha }_n\parallel f(p)-p\parallel +{\beta }_n\parallel x_n-p\parallel +{\gamma }_n\parallel x_n-p\parallel +{\gamma }_n\sum _{i=1}^N\parallel e_{n,i}\parallel \\ \le & (1-{\alpha }_n(1-\alpha ))\parallel x_n-p\parallel +{\alpha }_n(1-\alpha )\frac{\parallel f(p)-p\parallel }{1-\alpha }+\sum _{i=1}^N\parallel e_{n,i}\parallel \\ \le & max\{\parallel x_n-p\parallel ,\parallel f(p)-p\parallel \}+\sum _{i=1}^N\parallel e_{n,i}\parallel \\ ⋮\\ \le & max\{\parallel x_1-p\parallel ,\parallel f(p)-p\parallel \}+\sum _{j=1}^{\mathrm{\infty }}\sum _{i=1}^N\parallel e_{j,i}\parallel .\end{array}$$

This proves that the sequence $\{x_n\}$ is bounded, and so is $\{y_n\}$. Since

$$\begin{array}{rl}{y}_n-y_{n-1}=& \sum _{i=1}^N{\delta }_{n,i}(J_{r_m}(x_n+e_{n,i})-J_{r_i}(x_{n-1}+e_{n-1,i}))\\ +\sum _{i=1}^N({\delta }_{n,i}-{\delta }_{n-1,i})J_{r_i}(x_{n-1}+e_{n-1,i}),\end{array}$$

we have

$$\begin{array}{rcl}\parallel y_n-y_{n-1}\parallel & \le & \sum _{i=1}^N{\delta }_{n,i}\parallel J_{r_i}(x_n+e_{n,i})-J_{r_i}(x_{n-1}+e_{n-1,i})\parallel \\ +\sum _{i=1}^N|{\delta }_{n,i}-{\delta }_{n-1,i}|\parallel J_{r_i}(x_{n-1}+e_{n-1,i})\parallel \\ \le & \parallel x_n-x_{n-1}\parallel +\sum _{i=1}^N\parallel e_{n,i}\parallel +\sum _{i=1}^N\parallel e_{n-1,i}\parallel \\ +\sum _{i=1}^N|{\delta }_{n,i}-{\delta }_{n-1,i}|\parallel J_{r_i}(x_{n-1}+e_{n-1,i})\parallel \\ \le & \parallel x_n-x_{n-1}\parallel +\sum _{i=1}^N\parallel e_{n,i}\parallel +\sum _{i=1}^N\parallel e_{n-1,i}\parallel +M_1\sum _{i=1}^N|{\delta }_{n,i}-{\delta }_{n-1,i}|,\end{array}$$

where $M_1$ is an appropriate constant such that

$$M_1=max\{\underset{n\ge 1}{sup}\parallel J_{r_1}(x_n+e_{n,1})\parallel ,\underset{n\ge 1}{sup}\parallel J_{r_2}(x_n+e_{n,2})\parallel ,\dots ,\underset{n\ge 1}{sup}\parallel J_{r_N}(x_n+e_{n,N})\parallel \}.$$

Define a sequence $\{z_n\}$ by $z_n:=\frac{x_{n+1}-{\beta }_n{x}_n}{1-{\beta }_n}$, that is, $x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)z_n$. It follows that

$$\begin{array}{rcl}\parallel y{z}_n-z_{n-1}\parallel & \le & \frac{{\alpha }_n}{1-{\beta }_n}\parallel f(x_n)-y_n\parallel +\frac{{\alpha }_{n-1}}{1-{\beta }_{n-1}}\parallel f(x_{n-1})-y_{n-1}\parallel +\parallel y_n-y_{n-1}\parallel \\ \le & \frac{{\alpha }_n}{1-{\beta }_n}\parallel f(x_n)-y_n\parallel +\frac{{\alpha }_{n-1}}{1-{\beta }_{n-1}}\parallel f(x_{n-1})-y_{n-1}\parallel +\parallel x_n-x_{n-1}\parallel \\ +\sum _{i=1}^N|{\delta }_{n,i}-{\delta }_{n-1,i}|\parallel J_{r_i}{x}_{n-1}\parallel \\ \le & \frac{{\alpha }_n}{1-{\beta }_n}\parallel f(x_n)-y_n\parallel +\frac{{\alpha }_{n-1}}{1-{\beta }_{n-1}}\parallel f(x_{n-1})-y_{n-1}\parallel +\parallel x_n-x_{n-1}\parallel \\ +M_2(\sum _{i=1}^N|{\delta }_{n,i}-{\delta }_i|+\sum _{i=1}^N|{\delta }_i-{\delta }_{n-1,i}|),\end{array}$$

where $M_2$ is an appropriate constant such that

$$M_2=max\{\underset{n\ge 1}{sup}\parallel J_{r_1}{x}_n\parallel ,\underset{n\ge 1}{sup}\parallel J_{r_2}{x}_n\parallel ,\dots ,\underset{n\ge 1}{sup}\parallel J_{r_N}{x}_n\parallel \}.$$

This implies that

$$\begin{array}{c}\parallel z_n-z_{n-1}\parallel -\parallel x_n-x_{n-1}\parallel \hfill \\ \le \frac{{\alpha }_n}{1-{\beta }_n}\parallel f(x_n)-y_n\parallel +\frac{{\alpha }_{n-1}}{1-{\beta }_{n-1}}\parallel f(x_{n-1})-y_{n-1}\parallel \hfill \\ +M_2(\sum _{i=1}^N|{\delta }_{n,i}-{\delta }_i|+\sum _{i=1}^N|{\delta }_i-{\delta }_{n-1,i}|).\hfill \end{array}$$

From the restrictions (a), (b), (c), and (d), we find that

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

Using Lemma 1.4, we find that ${lim}_{n\to \mathrm{\infty }}\parallel z_n-x_n\parallel =0$. This further shows that ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel =0$. Put $T={\sum }_{i=1}^N{\delta }_i{J}_{r_i}$. It follows from Lemma 1.3 that T is nonexpansive with $F(T)={\bigcap }_{i=1}^{N}F(J_{r_i})={\bigcap }_{i=1}^N{A}_i^{-1}(0)$. Note that

$$\begin{array}{c}\parallel x_n-T{x}_n\parallel \hfill \\ \le \parallel x_n-x_{n+1}\parallel +\parallel x_{n+1}-T{x}_n\parallel \hfill \\ \le \parallel x_n-x_{n+1}\parallel +{\alpha }_n\parallel f(x_n)-T{x}_n\parallel +{\beta }_n\parallel x_n-T{x}_n\parallel +{\gamma }_n\parallel y_n-T{x}_n\parallel \hfill \\ \le \parallel x_n-x_{n+1}\parallel +{\alpha }_n\parallel f(x_n)-T{x}_n\parallel +{\beta }_n\parallel x_n-T{x}_n\parallel +M_2\sum _{i=1}^N|{\delta }_{n,i}-{\delta }_i|.\hfill \end{array}$$

This implies that

$$(1-{\beta }_n)\parallel x_n-T{x}_n\parallel \le \parallel x_n-x_{n+1}\parallel +{\alpha }_n\parallel f(x_n)-T{x}_n\parallel +M_2\sum _{i=1}^N|{\delta }_{n,i}-{\delta }_i|.$$

It follows from the restrictions (a), (b), and (d) that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel T{x}_n-x_n\parallel =0.$$

Now, we are in a position to prove that ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}〈f(\overline{x})-\overline{x},J(x_n-\overline{x})〉\le 0$, where $\overline{x}={lim}_{t\to 0}{x}_t$, and $x_t$ solves the fixed point equation

$$x_t=tf(x_t)+(1-t)T{x}_t,\mathrm{\forall }t\in (0,1).$$

It follows that

$$\begin{array}{rcl}{\parallel x_t-x_n\parallel }^2& =& t〈f(x_t)-x_n,J(x_t-x_n)〉+(1-t)〈T{x}_t-x_n,j(x_t-x_n)〉\\ =& t〈f(x_t)-x_t,J(x_t-x_n)〉+t〈x_t-x_n,J(x_t-x_n)〉\\ +(1-t)〈T{x}_t-T{x}_n,J(x_t-x_n)〉+(1-t)〈T{x}_n-x_n,J(x_t-x_n)〉\\ \le & t〈f(x_t)-x_t,J(x_t-x_n)〉+{\parallel x_t-x_n\parallel }^2+\parallel T{x}_n-x_n\parallel \parallel x_t-x_n\parallel ,\mathrm{\forall }t\in (0,1).\end{array}$$

This implies that

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

Since ${lim}_{n\to \mathrm{\infty }}\parallel T{x}_n-x_n\parallel =0$, we find that ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}〈x_t-f(x_t),J(x_t-x_n)〉\le 0$. Since J is strong to weak^∗ uniformly continuous on bounded subsets of E, we find that

$$\begin{array}{c}|〈f(\overline{x})-\overline{x},J(x_n-\overline{x})〉-〈x_t-f(x_t),J(x_t-x_n)〉|\hfill \\ \le |〈f(\overline{x})-\overline{x},J(x_n-\overline{x})〉-〈f(\overline{x})-\overline{x},J(x_n-x_t)〉|\hfill \\ +|〈f(\overline{x})-\overline{x},J(x_n-x_t)〉-〈x_t-f(x_t),J(x_t-x_n)〉|\hfill \\ \le \left|〈f(\overline{x})-\overline{x},J(x_n-\overline{x})-J(x_n-x_t)〉\right|+\left|〈f(\overline{x})-\overline{x}+x_t-f(x_t),J(x_n-x_t)〉\right|\hfill \\ \le \parallel f(x_t)-\overline{x}\parallel \parallel J(x_n-\overline{x})-J(x_n-x_t)\parallel +(1+\alpha )\parallel \overline{x}-x_t\parallel \parallel x_n-x_t\parallel .\hfill \end{array}$$

Since $x_t\to \overline{x}$, as $t\to 0$, we have

$$\underset{t\to 0}{lim}|〈f(\overline{x})-\overline{x},J(x_n-\overline{x})〉-〈f(x_t)-x_t,J(x_n-x_t)〉|=0.$$

For $ϵ>0$, there exists $\delta >0$ such that $\mathrm{\forall }t\in (0,\delta )$, we have

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

This implies that ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}〈f(\overline{x})-\overline{x},J(x_n-\overline{x})〉\le 0$.

Finally, we show that $x_n\to \overline{x}$ as $n\to \mathrm{\infty }$. Since ${\parallel \cdot \parallel }^2$ is convex, we see that

$$\begin{array}{rcl}{\parallel y_n-\overline{x}\parallel }^2& =& {\parallel \sum _{i=1}^N{\delta }_{n,i}{J}_{r_i}(x_n+e_{n,i})-\overline{x}\parallel }^2\\ \le & \sum _{i=1}^N{\delta }_{n,i}{\parallel J_{r_i}(x_n+e_{n,i})-\overline{x}\parallel }^2\\ \le & {\parallel x_n-\overline{x}\parallel }^2+\sum _{i=1}^N\parallel e_{n,i}\parallel (\parallel e_{n,i}\parallel +2\parallel x_n-\overline{x}\parallel ).\end{array}$$

It follows that

$$\begin{array}{rcl}{\parallel x_{n+1}-\overline{x}\parallel }^2& =& {\alpha }_n〈f(x_n)-\overline{x},J(x_{n+1}-\overline{x})〉+{\beta }_n〈x_n-\overline{x},J(x_{n+1}-\overline{x})〉\\ +{\gamma }_n〈y_n-\overline{x},J(x_{n+1}-\overline{x})〉\\ \le & {\alpha }_n\alpha \parallel x_n-\overline{x}\parallel \parallel x_{n+1}-\overline{x}\parallel +{\alpha }_n〈f(\overline{x})-\overline{x},J(x_{n+1}-\overline{x})〉\\ +{\beta }_n\parallel x_n-\overline{x}\parallel \parallel x_{n+1}-\overline{x}\parallel +{\gamma }_n\parallel y_n-\overline{x}\parallel \parallel x_{n+1}-\overline{x}\parallel \\ \le & \frac{{\alpha }_n\alpha }{2}({\parallel x_n-\overline{x}\parallel }^2+{\parallel x_{n+1}-\overline{x}\parallel }^2)+{\alpha }_n〈f(\overline{x})-\overline{x},J(x_{n+1}-\overline{x})〉\\ +\frac{{\beta }_n}{2}({\parallel x_n-\overline{x}\parallel }^2+{\parallel x_{n+1}-\overline{x}\parallel }^2)+\frac{{\gamma }_n}{2}{\parallel x_n-\overline{x}\parallel }^2\\ +\sum _{i=1}^N\parallel e_{n,i}\parallel (\parallel e_{n,i}\parallel +2\parallel x_n-\overline{x}\parallel )+\frac{{\gamma }_n}{2}{\parallel x_{n+1}-\overline{x}\parallel }^2.\end{array}$$

Hence, we have

$$\begin{array}{rcl}{\parallel x_{n+1}-\overline{x}\parallel }^2& \le & (1-{\alpha }_n(1-\alpha )){\parallel x_n-\overline{x}\parallel }^2+2{\alpha }_n〈f(\overline{x})-\overline{x},J(x_{n+1}-\overline{x})〉\\ +\sum _{i=1}^N\parallel e_{n,i}\parallel (\parallel e_{n,i}\parallel +2\parallel x_n-\overline{x}\parallel ).\end{array}$$

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

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

Corollary 2.3 Let E be a Hilbert space and let $N\ge 1$ be some positive integer. Let $A_m$ be a maximal monotone operator in E for each $m\in \{1,2,\dots ,N\}$. Assume that $C:={\bigcap }_{m=1}^N\overline{D(A_m)}$ is convex and has the normal structure. Let $f:C\to C$ be an α-contractive mapping. Let $\{{\alpha }_n\}$, $\{{\beta }_n\}$, and $\{{\gamma }_n\}$ be real number sequences in $(0,1)$ with the restriction ${\alpha }_n+{\beta }_n+{\gamma }_n=1$. Let $\{{\delta }_{n,i}\}$ be a real number sequence in $(0,1)$ with the restriction ${\delta }_{n,1}+{\delta }_{n,2}+\cdots +{\delta }_{n,N}=1$. Let $\{r_m\}$ be a positive real numbers sequence and $\{e_{n,i}\}$ a sequence in E for each $i\in \{1,2,\dots ,N\}$. Assume that ${\bigcap }_{i=1}^N{A}_i^{-1}(0)$ is not empty. Let $\{x_n\}$ be a sequence generated in the following manner:

$$x_1\in C,x_{n+1}={\alpha }_{n}f(x_n)+{\beta }_n{x}_n+{\gamma }_n\sum _{i=1}^N{\delta }_{n,i}{J}_{r_i}(x_n+e_{n,i}),\mathrm{\forall }n\ge 1,$$

where $J_{r_i}={(I+r_i{A}_i)}^{-1}$. Assume that the control sequences $\{{\alpha }_n\}$, $\{{\beta }_n\}$, $\{{\gamma }_n\}$, and $\{{\delta }_{n,i}\}$ satisfy the following restrictions:

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

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

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

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

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

3 Applications

In this section, we consider a variational inequality problem. Let $A:C\to E^{\ast }$ be a single valued monotone operator which is hemicontinuous; that is, continuous along each line segment in C with respect to the weak^∗ topology of $E^{\ast }$. Consider the following variational inequality:

$$\text{find }x\in C\text{ such that }〈y-x,Ax〉\ge 0,\mathrm{\forall }y\in C.$$

The solution set of the variational inequality is denoted by $VI(C,A)$. Recall that the normal cone $N_C(x)$ for C at a point $x\in C$ is defined by

$$N_C(x)=\{x^{\ast }\in E^{\ast }:〈y-x,x^{\ast }〉\le 0,\mathrm{\forall }y\in C\}.$$

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

Theorem 3.1 Let E be a real reflexive, strictly convex Banach space with the uniformly Gâteaux differentiable norm. Let $N\ge 1$ be some positive integer and let C be nonempty closed and convex subset of E. Let $A_i:C\to E^{\ast }$ a single valued, monotone and hemicontinuous operator. Assume that ${\bigcap }_{i=1}^{N}VI(C,A_i)$ is not empty and C has the normal structure. Let $f:C\to C$ be an α-contractive mapping. Let $\{{\alpha }_n\}$, $\{{\beta }_n\}$, and $\{{\gamma }_n\}$ be real number sequences in $(0,1)$ with the restriction ${\alpha }_n+{\beta }_n+{\gamma }_n=1$. Let $\{{\delta }_{n,i}\}$ be a real number sequence in $(0,1)$ with the restriction ${\delta }_{n,1}+{\delta }_{n,2}+\cdots +{\delta }_{n,N}=1$. Let $\{r_m\}$ be a positive real numbers sequence and $\{e_{n,i}\}$ a sequence in E for each $i\in \{1,2,\dots ,N\}$. Let $\{x_n\}$ be a sequence generated in the following manner:

$$x_1\in C,x_{n+1}={\alpha }_{n}f(x_n)+{\beta }_n{x}_n+{\gamma }_n\sum _{i=1}^N{\delta }_{n,i}VI(C,A_i+\frac{1}{r_i}(I-x_n)),\mathrm{\forall }n\ge 1.$$

Assume that the control sequences $\{{\alpha }_n\}$, $\{{\beta }_n\}$, $\{{\gamma }_n\}$, and $\{{\delta }_{n,i}\}$ satisfy the following restrictions:

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

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

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

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

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

Proof Define a mapping $T_i\subset E\times E^{\ast }$ by

$$T_{i}x:=\{\begin{array}{cc}{A}_{i}x+N_{C}x,\hfill & x\in C,\hfill \\ \mathrm{\varnothing },\hfill & x\notin C.\hfill \end{array}$$

From Rockafellar [36], we find that $T_i$ is maximal monotone with $T_i^{-1}(0)=VI(C,A_i)$. For each $r_i>0$, and $x_n\in E$, we see that there exists a unique $x_{r_i}\in D(T_i)$ such that $x_n\in x_{r_i}+r_i{T}_i(x_{r_i})$, where $x_{r_i}={(I+r_i{T}_i)}^{-1}{x}_n$. Notice that

$$y_{n,i}=VI(C,A_i+\frac{1}{r_i}(I-x_n)),$$

which is equivalent to

$$〈y-y_{n,i},A_i{y}_{n,i}+\frac{1}{r_i}(y_{n,i}-x_n)〉\ge 0,\mathrm{\forall }y\in C,$$

that is, $-A_i{y}_{n,i}+\frac{1}{r_i}(x_n-y_{n,i})\in N_C(y_{n,i})$. This implies that $y_{n,i}={(I+r_i{T}_i)}^{-1}{x}_n$. Using Theorem 2.1, we find the desired conclusion immediately. □

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

Corollary 3.2 Let E be a real reflexive, strictly convex Banach space with the uniformly Gâteaux differentiable norm. Let C be nonempty closed and convex subset of E. Let $A:C\to E^{\ast }$ a single valued, monotone and hemicontinuous operator with $VI(C,A)$. Assume that C has the normal structure. Let $f:C\to C$ be an α-contractive mapping. Let $\{{\alpha }_n\}$, $\{{\beta }_n\}$, and $\{{\gamma }_n\}$ be real number sequences in $(0,1)$ with the restriction ${\alpha }_n+{\beta }_n+{\gamma }_n=1$. Let $\{x_n\}$ be a sequence generated in the following manner:

$$x_1\in C,x_{n+1}={\alpha }_{n}f(x_n)+{\beta }_n{x}_n+{\gamma }_{n}VI(C,A+\frac{1}{r}(I-x_n)),\mathrm{\forall }n\ge 1.$$

Assume that the control sequences $\{{\alpha }_n\}$, $\{{\beta }_n\}$, and $\{{\gamma }_n\}$ satisfy the following restrictions:

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

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

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