Strong convergence of an iterative process for a family of strictly pseudocontractive mappings

Abstract

In this article, fixed point problems of a family of strictly pseudocontractive mappings are investigated based on an iterative process. Strong convergence of the iterative process is obtained in a real 2-uniformly Banach space.

MSC:47H09, 47J05, 47J25.

1 Introduction and preliminaries

Throughout this paper, we always assume that E is a real Banach space. Let $E^{\ast }$ be the dual space of E. Let $J_q$ ($q>1$) denote the generalized duality mapping from E into $2^{E^{\ast }}$ given by

$$J_q(x)=\{f^{\ast }\in E^{\ast }:〈x,f^{\ast }〉={\parallel x\parallel }^q,\parallel f^{\ast }\parallel ={\parallel x\parallel }^{q-1}\},\mathrm{\forall }x\in E,$$

where $〈\cdot ,\cdot 〉$ denotes the generalized duality pairing. In particular, $J_2$ is called the normalized duality mapping, which is usually denoted by J. In this paper, we use j to denote the single-valued normalized duality mapping. It is known that $J_q(x)={\parallel x\parallel }^{q-2}J(x)$ if $x\ne 0$. If E is a Hilbert space, then $J=I$, the identity mapping. Further, we have the following properties of the generalized duality mapping $J_q$:

• (1) $J_q(tx)=t^{q-1}{J}_q(x)$ for all $x\in E$ and $t\in [0,\mathrm{\infty })$;

• (2) $J_q(-x)=-J_q(x)$ for all $x\in E$.

A Banach space E is said to be smooth if the limit

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

exists for all $x,y\in U_E$. It is also said to be uniformly smooth if the limit is attained uniformly for all $x,y\in U_E$. The norm of E is said to be Fréchet differentiable if, for any $x\in U_E$, the above limit is attained uniformly for all $y\in U_E$. The modulus of smoothness of E is the function ${\rho }_E:[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ defined by

$${\rho }_E(\tau )=sup\{\frac{1}{2}(\parallel x+y\parallel +\parallel x-y\parallel )-1:\parallel x\parallel \le 1,\parallel y\parallel \le \tau \},\mathrm{\forall }\tau \ge 0.$$

The Banach space E is uniformly smooth if and only if ${lim}_{\tau \to \mathrm{\infty }}\frac{{\rho }_E(\tau )}{\tau }=0$. Let $q>1$. The Banach space E is said to be q-uniformly smooth if there exists a constant $c>0$ such that ${\rho }_E(\tau )\le c{\tau }^q$. It is shown in [1] that there is no Banach space which is q-uniformly smooth with $q>2$. Hilbert spaces, $L^p$ (or $l^p$) spaces and Sobolev space $W_m^p$, where $p\ge 2$, are 2-uniformly smooth.

Let C be a nonempty closed convex subset of E and let $T:C\to C$ be a mapping. In this paper, we use $F(T)$ to denote the fixed point set of T. A mapping T is said to be κ-contractive iff there exists a constant $\kappa \in (0,1)$ such that

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

A mapping T is said to be nonexpansive iff

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

A mapping T is said to be κ-strictly pseudocontractive iff there exist a constant $\kappa \in (0,1)$ and $j(x-y)\in J(x-y)$ such that

$$〈Tx-Ty,j(x-y)〉\le {\parallel x-y\parallel }^2-\kappa {\parallel (I-T)x-(I-T)y\parallel }^2,\mathrm{\forall }x,y\in C.$$

(1.1)

It is clear that (1.1) is equivalent to the following:

$$〈(I-T)x-(I-T)y,j(x-y)〉\ge \kappa {\parallel (I-T)x-(I-T)y\parallel }^2,\mathrm{\forall }x,y\in C.$$

(1.2)

The class of κ-strictly pseudocontractive mappings was first introduced by Browder and Petryshyn [2] in Hilbert spaces. A mapping T is said to be pseudocontractive iff there exists $j(x-y)\in J(x-y)$ such that

$$〈Tx-Ty,j(x-y)〉\le {\parallel x-y\parallel }^2,\mathrm{\forall }x,y\in C.$$

(1.3)

A mapping T is said to be κ-strongly pseudocontractive iff there exist a constant $\kappa \in (0,1)$ and $j(x-y)\in J(x-y)$ such that

$$〈Tx-Ty,j(x-y)〉\le \kappa {\parallel x-y\parallel }^2,\mathrm{\forall }x,y\in C.$$

(1.4)

In 1974, Deimling [3] proved the existence of fixed points of continuous κ-strongly pseudocontractive mappings in Banach spaces; see [3] for more details. We remark that the class of κ-strongly pseudocontractive mappings is independent of the class of κ-strictly pseudocontractive mappings. This can be seen from Zhou [4]. Lipschitz pseudocontractive mappings may not be κ-strictly pseudocontractive mappings, which can be seen from Chidume and Mutangadura [5].

One classical way to study nonexpansive mappings is to use contractions to approximate a nonexpansive mapping; for more details, see [6–12] and the references therein. More precisely, take $t\in (0,1)$ and define a contraction $T_t:C\to C$ by

$$T_{t}x=tu+(1-t)Tx,\mathrm{\forall }x\in C,$$

(1.5)

where $u\in C$ is a fixed point. Banach’s contraction mapping principle guarantees that $T_t$ has a unique fixed point $x_t$ in C. In the case of T having a fixed point, Browder [7] proved that $x_t$ converges strongly to a fixed point of T in the framework of Hilbert spaces. Reich [10] extended Browder’s result to the setting of Banach spaces and proved that if E is a uniformly smooth Banach space, then $x_t$ converges strongly to a fixed point of T and the limit defines the (unique) sunny nonexpansive retraction from C onto $F(T)$; see [10] for more details.

Recall that the normal Mann iterative process was introduced by Mann [13] in 1953. Recently, the construction of fixed points for nonexpansive mappings via the normal Mann iterative process has been extensively investigated by many authors. The normal Mann iterative process generates a sequence $\{x_n\}$ in the following manner:

$$\{\begin{array}{c}{x}_1\in C\text{chosen arbitrarily},\hfill \\ x_{n+1}=(1-{\alpha }_n)x_n+{\alpha }_{n}T{x}_n,\mathrm{\forall }n\ge 1,\hfill \end{array}$$

(1.6)

where the sequence $\{{\alpha }_n\}$ is in the interval $(0,1)$.

In an infinite-dimensional Hilbert space, the normal Mann iteration algorithm has only weak convergence; see [14] for more details. In many disciplines, including economics [15], image recovery [16] and control theory [17], problems arise in infinite dimension spaces. In such problems, strong convergence is often much more desirable than weak convergence for it translates the physically tangible property so that the energy $\parallel x_n-x\parallel$ of the error between the iterate $x_n$ and the solution x eventually becomes arbitrarily small.

Recently, many authors have tried to modify the normal Mann iteration process to have strong convergence for nonexpansive mappings and κ-strictly pseudocontractive mappings; see [18–36] and the references therein.

Let D be a nonempty subset of C. Let $Q:C\to D$. Q is said to be a contraction iff $Q^2=Q$; sunny iff for each $x\in C$ and $t\in (0,1)$, we have $Q(tx+(1-t)Qx)=Qx$; sunny nonexpansive retraction iff Q 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 [37] and [38], describes a characterization of sunny nonexpansive retractions on a smooth Banach space.

Let $Q:E\to C$ be a retraction, and let 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) $〈x-Qx,j(y-Qx)〉\le 0$, $\mathrm{\forall }x\in E$, $y\in C$.

In this paper, we investigate the problem of modifying the normal Mann iteration process for a family of κ-strictly pseudocontractive mappings. Strong convergence of the purposed iterative process is obtained in a real 2-uniformly Banach space. In order to prove our main results, we need the following tools.

Lemma 1.1 [1]

Let E be a real 2-uniformly smooth Banach space with the best smooth constant K. Then the following inequality holds:

$${\parallel x+y\parallel }^2\le {\parallel x\parallel }^2+2〈y,Jx〉+2{\parallel Ky\parallel }^2,\mathrm{\forall }x,y\in E.$$

Lemma 1.2 [31]

Let C be a nonempty subset of a real 2-uniformly smooth Banach space E and let $T:C\to C$ be a κ-strict pseudocontraction. For $\alpha \in (0,1)$, we define $T_{\alpha }x=(1-\alpha )x+\alpha Tx$ for every $x\in C$ . Then, as $\alpha \in (0,\frac{\kappa }{K^2}]$, $T_{\alpha }$ is nonexpansive such that $F(T_{\alpha })=F(T)$.

Lemma 1.3 [39]

Assume that $\{{\alpha }_n\}$ is a sequence of nonnegative real numbers such that

$${\alpha }_{n+1}\le (1-{\gamma }_n){\alpha }_n+{\delta }_n,$$

where $\{{\gamma }_n\}$ is a sequence in $(0,1)$ and $\{{\delta }_n\}$ is a sequence such that

• (i) ${\sum }_{n=1}^{\mathrm{\infty }}{\gamma }_n=\mathrm{\infty }$;

• (ii) ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\delta }_n/{\gamma }_n\le 0$ or ${\sum }_{n=1}^{\mathrm{\infty }}|{\delta }_n|<\mathrm{\infty }$.

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

Lemma 1.4 [40]

Let E be a real smooth Banach space. Then the following inequality holds:

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

Lemma 1.5 [29]

Let E be a smooth Banach space and let C be a nonempty convex subset of E. Given an integer $N\ge 1$, assume that ${\{T_i\}}_{i=1}^N:C\to C$ is a finite family of ${\kappa }_i$-strict pseudocontractions such that ${\bigcap }_{i=1}^{N}F(T_i)\ne \mathrm{\varnothing }$. Assume that ${\{{\lambda }_i\}}_{i=1}^r$ is a positive sequence such that ${\sum }_{i=1}^N{\lambda }_i=1$. Then $F({\sum }_{i=1}^{N}F(T_i))={\bigcap }_{i=1}^{N}F(T_i)$.

Lemma 1.6 [10]

Let E be a real uniformly smooth Banach space and let C be a nonempty closed convex subset of E. Let $T:C\to C$ be a nonexpansive mapping with a fixed point and let $f:C\to C$ be a contraction. For each $t\in (0,1)$, let $z_t$ be the unique solution of the equation $x=tf(x)+(1-t)Tx$. Then $\{z_t\}$ converges to a fixed point of T as $t\to 0$ and $Q(f)=s-{lim}_{t\to 0}{z}_t$ defines the unique sunny nonexpansive retraction from C onto $F(T)$.

2 Main results

Theorem 2.1 Let C be a nonempty closed convex subset of a real 2-uniformly smooth Banach space E with the best smooth constant K and let N be some positive integer. Let $T_i:C\to C$ be a ${\kappa }_i$-strictly pseudocontractive mapping for each $1\le i\le N$. Assume that ${\bigcap }_{i=1}^{N}F(T_i)\ne \mathrm{\varnothing }$. Let f be an α-contractive mapping. Let $\{x_n\}$ be a sequence generated in the following process:

$$\{\begin{array}{c}{x}_0\in C\mathit{\text{arbitrarily chosen}},\hfill \\ y_n={\beta }_n{x}_n+(1-{\beta }_n){\sum }_{i=1}^N{\lambda }_i{T}_i{x}_n,\hfill \\ x_{n+1}={\alpha }_{n}f(y_n)+(1-{\alpha }_n)y_n,n\ge 0,\hfill \end{array}$$

where $\{{\alpha }_n\}$, $\{{\beta }_n\}$ and $\{{\lambda }_i\}$ are real number sequences in $[0,1]$ satisfying the following restrictions:

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

• (b) $1-\frac{\kappa }{K^2}\le {\beta }_n\le \beta <1$, ${\sum }_{n=1}^{\mathrm{\infty }}|{\beta }_n-{\beta }_{n-1}|<\mathrm{\infty }$,

• (c) ${\sum }_{n=1}^N{\lambda }_i=1$,

where β is some real number, and $\kappa :=min\{{\kappa }_i:1\le i\le N\}$. Then $\{x_n\}$ converges strongly as $n\to \mathrm{\infty }$ to some point in ${\bigcap }_{i=1}^{N}F(T_i)$, which is the unique solution in ${\bigcap }_{i=1}^{N}F(T_i)$ to the following variational inequality:

$$〈f\left(x^{\ast }\right)-x^{\ast },j(x^{\ast }-p)〉\ge 0,\mathrm{\forall }p\in \bigcap _{i=1}^{N}F(T_i).$$

Proof The proof is split into four steps.

Step 1. Show that $\{x_n\}$ and $\{y_n\}$ are bounded.

Putting $T:={\sum }_{i=1}^N{\lambda }_i{T}_i$, we see that T is a κ-strictly pseudocontractive mapping. Indeed, we have the following:

$$\begin{array}{r}〈Tx-Ty,j(x-y)〉\\ ={\lambda }_1〈T_{1}x-T_{1}y,j(x-y)〉+{\lambda }_2〈T_{2}x-T_{2}y,j(x-y)〉+\cdots \\ +{\lambda }_N〈T_{N}x-T_{N}y,j(x-y)〉\\ \le {\lambda }_1({\parallel x-y\parallel }^2-{\kappa }_1{\parallel (I-T_1)x-(I-T_1)y\parallel }^2)\\ +{\lambda }_2({\parallel x-y\parallel }^2-{\kappa }_2{\parallel (I-T_2)x-(I-T_2)y\parallel }^2)+\cdots \\ +{\lambda }_N({\parallel x-y\parallel }^2-{\kappa }_N{\parallel (I-T_N)x-(I-T_N)y\parallel }^2)\\ \le {\parallel x-y\parallel }^2-\kappa ({\lambda }_1{\parallel (I-T_1)x-(I-T_1)y\parallel }^2\\ +{\lambda }_2{\parallel (I-T_2)x-(I-T_2)y\parallel }^2+\cdots +{\lambda }_N{\parallel (I-T_N)x-(I-T_N)y\parallel }^2)\\ \le {\parallel x-y\parallel }^2-\kappa {\parallel (I-T)x-(I-T)y\parallel }^2.\end{array}$$

This proves that T is a κ-strictly pseudocontractive mapping. Fix $p\in {\bigcap }_{i=1}^{N}F(T_i)$. It follows from Lemma 1.1 that

$$\begin{array}{rcl}{\parallel y_n-p\parallel }^2& =& {\parallel (x_n-p)+(1-{\beta }_n)(T{x}_n-x_n)\parallel }^2\\ \le & {\parallel x_n-p\parallel }^2+2(1-{\beta }_n)〈T{x}_n-x_n,j(x_n-p)〉\\ +2K^2{(1-{\beta }_n)}^2{\parallel (T{x}_n-x_n)\parallel }^2\\ =& {\parallel x_n-p\parallel }^2+2(1-{\beta }_n)〈T{x}_n-p,j(x_n-p)〉-2(1-{\beta }_n){\parallel x_n-p\parallel }^2\\ +2K^2{(1-{\beta }_n)}^2{\parallel (T{x}_n-x_n)\parallel }^2\\ \le & {\parallel x_n-p\parallel }^2+2(1-{\beta }_n)({\parallel x_n-p\parallel }^2-\kappa {\parallel T{x}_n-x_n\parallel }^2)\\ -2(1-{\beta }_n){\parallel x_n-p\parallel }^2+2K^2{(1-{\beta }_n)}^2{\parallel (T{x}_n-x_n)\parallel }^2\\ =& {\parallel x_n-p\parallel }^2-2(1-{\beta }_n)(\kappa -K^2(1-{\beta }_n)){\parallel T{x}_n-x_n\parallel }^2\\ \le & {\parallel x_n-p\parallel }^2.\end{array}$$

(2.1)

This implies that

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

This in turn implies that

$$\parallel x_n-p\parallel \le max\{\parallel x_0-p\parallel ,\frac{\parallel p-f(p)\parallel }{1-\alpha }\},$$

which gives that the sequence $\{x_n\}$ is bounded, so is $\{y_n\}$. This completes step 1.

Step 2. Show that $\parallel T_{\frac{\kappa }{K^2}}{x}_n-x_n\parallel \to 0$ as $n\to \mathrm{\infty }$.

Put $T_{{\beta }_n}x={\beta }_{n}x+(1-{\beta }_n)Tx$, $\mathrm{\forall }x\in C$. It follows from Lemma 1.2 that

$$\begin{array}{rl}\parallel y_n-y_{n-1}\parallel & =\parallel T_{{\beta }_n}{x}_n-T_{{\beta }_{n-1}}{x}_{n-1}\parallel \\ \le \parallel T_{{\beta }_n}{x}_n-T_{{\beta }_n}{x}_{n-1}\parallel +\parallel T_{{\beta }_n}{x}_{n-1}-T_{{\beta }_{n-1}}{x}_{n-1}\parallel \\ \le \parallel x_n-x_{n-1}\parallel +\parallel {\beta }_n{x}_{n-1}+(1-{\beta }_n)T{x}_{n-1}-{\beta }_{n-1}{x}_{n-1}-(1-{\beta }_{n-1})T{x}_{n-1}\parallel \\ \le \parallel x_n-x_{n-1}\parallel +|{\beta }_n-{\beta }_{n-1}|\parallel x_{n-1}-T{x}_{n-1}\parallel .\end{array}$$

(2.2)

Notice that

$$x_{n+1}-x_n={\alpha }_n(f(y_n)-f(y_{n-1}))+(1-{\alpha }_n)(y_n-y_{n-1})+({\alpha }_n-{\alpha }_{n-1})(f(y_{n-1})-y_{n-1}).$$

It follows from (2.2) that

$$\begin{array}{r}\parallel x_{n+1}-x_n\parallel \\ \le {\alpha }_n\parallel f(y_n)-f(y_{n-1})\parallel +(1-{\alpha }_n)\parallel y_n-y_{n-1}\parallel +|{\alpha }_n-{\alpha }_{n-1}|\parallel f(y_{n-1})-y_{n-1}\parallel \\ \le (1-{\alpha }_n(1-\alpha ))\parallel y_n-y_{n-1}\parallel +|{\alpha }_n-{\alpha }_{n-1}|\parallel f(y_{n-1})-y_{n-1}\parallel \\ \le (1-{\alpha }_n(1-\alpha ))\parallel x_n-x_{n-1}\parallel +|{\beta }_n-{\beta }_{n-1}|\parallel x_{n-1}-T{x}_{n-1}\parallel \\ +|{\alpha }_n-{\alpha }_{n-1}|\parallel f(y_{n-1})-y_{n-1}\parallel .\end{array}$$

In view of Lemma 1.3, we obtain from the restrictions (a) and (b) that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_{n+1}-x_n\parallel =0.$$

(2.3)

Notice that

$$x_{n+1}-x_n={\alpha }_n(f(x_n)-x_n)+(1-{\alpha }_n)(y_n-x_n).$$

In view of the restriction (a), we obtain that ${lim}_{n\to \mathrm{\infty }}\parallel y_n-x_n\parallel =0$. On the other hand, we have $y_n-x_n=(1-{\beta }_n)(T{x}_n-x_n)$. This in turn implies that ${lim}_{n\to \mathrm{\infty }}\parallel T{x}_n-x_n\parallel =0$. It follows from the restriction (b) that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel T_{\frac{\kappa }{K^2}}{x}_n-x_n\parallel =0.$$

(2.4)

This completes step 2.

Step 3. Show that

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

(2.5)

where $z=Qf(z)$, where Q is a sunny nonexpansive retraction from C onto ${\bigcap }_{i=1}^{N}F(T_i)$, is the strong limit of the sequence $z_t$ defined by

$$z_t=tf(z_t)+(1-t)T_{\frac{\kappa }{K^2}}{z}_t,\mathrm{\forall }t\in (0,1).$$

It follows that

$$z_t-x_n=(1-t)(T_{\frac{\kappa }{K^2}}{z}_t-x_n)+t(f(z_t)-x_n),\mathrm{\forall }t\in (0,1).$$

For any $t\in (0,1)$, we see from Lemma 1.4 that

$$\begin{array}{rcl}{\parallel z_t-x_n\parallel }^2& \le & {(1-t)}^2{\parallel T_{\frac{\kappa }{K^2}}{z}_t-x_n\parallel }^2+2t〈f(z_t)-x_n,j(z_t-x_n)〉\\ \le & {(1-t)}^2({\parallel T_{\frac{\kappa }{K^2}}{z}_t-T_{\frac{\kappa }{K^2}}{x}_n\parallel }^2+{\parallel T_{\frac{\kappa }{K^2}}{x}_n-x_n\parallel }^2\\ +2\parallel T_{\frac{\kappa }{K^2}}{z}_t-T_{\frac{\kappa }{K^2}}{x}_n\parallel \parallel T_{\frac{\kappa }{K^2}}{x}_n-x_n\parallel )+2t〈f(z_t)-z_t,j(z_t-x_n)〉\\ +2t〈z_t-x_n,j(z_t-x_n)〉\\ \le & {(1-t)}^2{\parallel z_t-x_n\parallel }^2+{\lambda }_n(t)+2t〈f(z_t)-z_t,j(z_t-x_n)〉+2t{\parallel z_t-x_n\parallel }^2,\end{array}$$

(2.6)

where

$${\lambda }_n(t)={\parallel T_{\frac{\kappa }{K^2}}{x}_n-x_n\parallel }^2+2\parallel z_t-x_n\parallel \parallel T_{\frac{\kappa }{K^2}}{x}_n-x_n\parallel .$$

It follows from (2.6) that

$$〈z_t-f(z_t),j(z_t-x_n)〉\le \frac{t}{2}{\parallel z_t-x_n\parallel }^2+\frac{1}{2t}{\lambda }_n(t).$$

This implies that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈z_t-f(z_t),j(z_t-x_n)〉\le \frac{t}{2}{\parallel z_t-x_n\parallel }^2.$$

Since E is 2-uniformly smooth, $J:E\to E^{\ast }$ is uniformly continuous on any bounded sets of E, which ensures that the ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}$ and ${lim\hspace{0.17em}sup}_{t\to 0}$ are interchangeable, and hence

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

This shows that (2.5) holds. This completes the proof of step 3.

Step 4. Show that $x_n\to z$ as $n\to \mathrm{\infty }$.

It follows from (2.1) that $\parallel y_n-z\parallel \le \parallel x_n-z\parallel$. In view of Lemma 1.4, we see that

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

It then follows that

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

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

$$\underset{n\to \mathrm{\infty }}{lim}\frac{2(1-\alpha ){\alpha }_n}{1-{\alpha }_n\alpha }=0,\sum _{n=1}^{\mathrm{\infty }}\frac{2(1-\alpha ){\alpha }_n}{1-{\alpha }_n\alpha }=\mathrm{\infty }$$

and

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}(\frac{1}{1-\alpha }〈f(z)-z,J(x_{n+1}-z)〉+\frac{{\alpha }_n}{2(1-\alpha )}{\parallel x_n-z\parallel }^2)\le 0.$$

This implies from Lemma 1.3 that $x_n\to z$ as $n\to \mathrm{\infty }$. This completes the proof. □

For a single mapping, we have the following.

Corollary 2.2 Let C be a nonempty closed convex subset of a real 2-uniformly smooth Banach space E with the best smooth constant K. Let $T:C\to C$ be a κ-strictly pseudocontractive mapping such that $F(T)\ne \mathrm{\varnothing }$. Let f be an α-contractive mapping. Let $\{x_n\}$ be a sequence generated in the following process:

$$\{\begin{array}{c}{x}_0\in C\mathit{\text{arbitrarily chosen}},\hfill \\ y_n={\beta }_n{x}_n+(1-{\beta }_n)T{x}_n,\hfill \\ x_{n+1}={\alpha }_{n}f(y_n)+(1-{\alpha }_n)y_n,n\ge 0,\hfill \end{array}$$

where $\{{\alpha }_n\}$, $\{{\beta }_n\}$ and $\{{\lambda }_i\}$ are real number sequences in $[0,1]$ satisfying the following restrictions:

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

• (b) $1-\frac{\kappa }{K^2}\le {\beta }_n\le \beta <1$, ${\sum }_{n=1}^{\mathrm{\infty }}|{\beta }_n-{\beta }_{n-1}|<\mathrm{\infty }$,

where β is some real number, and $\kappa :=min\{{\kappa }_i:1\le i\le N\}$. Then $\{x_n\}$ converges strongly as $n\to \mathrm{\infty }$ to some point in $F(T)$, which is the unique solution in $F(T)$, to the following variational inequality:

$$〈f\left(x^{\ast }\right)-x^{\ast },j(x^{\ast }-p)〉\ge 0,\mathrm{\forall }p\in F(T).$$

If E is a Hilbert space, then the best smooth constant $K=\frac{\sqrt{2}}{2}$. The following result can be deduced from Theorem 2.1 immediately.

Corollary 2.3 Let C be a nonempty closed convex subset of a real Hilbert space E and let N be some positive integer. Let $T_i:C\to C$ be a ${\kappa }_i$-strictly pseudocontractive mapping for each $1\le i\le N$. Assume that ${\bigcap }_{i=1}^{N}F(T_i)\ne \mathrm{\varnothing }$. Let f be an α-contractive mapping. Let $\{x_n\}$ be a sequence generated in the following process:

$$\{\begin{array}{c}{x}_0\in C\mathit{\text{arbitrarily chosen}},\hfill \\ y_n={\beta }_n{x}_n+(1-{\beta }_n){\sum }_{i=1}^N{\lambda }_i{T}_i{x}_n,\hfill \\ x_{n+1}={\alpha }_{n}f(y_n)+(1-{\alpha }_n)y_n,n\ge 0,\hfill \end{array}$$

where $\{{\alpha }_n\}$, $\{{\beta }_n\}$ and $\{{\lambda }_i\}$ are real number sequences in $[0,1]$ satisfying the following restrictions:

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

• (b) $1-2\kappa \le {\beta }_n\le \beta <1$, ${\sum }_{n=1}^{\mathrm{\infty }}|{\beta }_n-{\beta }_{n-1}|<\mathrm{\infty }$;

• (c) ${\sum }_{n=1}^N{\lambda }_i=1$,

where β is some real number, and $\kappa :=min\{{\kappa }_i:1\le i\le N\}$. Then $\{x_n\}$ converges strongly as $n\to \mathrm{\infty }$ to some point in ${\bigcap }_{i=1}^{N}F(T_i)$, which is the unique solution in ${\bigcap }_{i=1}^{N}F(T_i)$, to the following variational inequality:

$$〈f\left(x^{\ast }\right)-x^{\ast },j(x^{\ast }-p)〉\ge 0,\mathrm{\forall }p\in \bigcap _{i=1}^{N}F(T_i).$$