Convergence Theorems for Modified Implicit Iterative Methods with Perturbation for Pseudocontractive Mappings

Abstract

In this paper, first, we introduce a path for a convex combination of a pseudocontractive type of mappings with a perturbed mapping and prove strong convergence of the proposed path in a real reflexive Banach space having a weakly continuous duality mapping. Second, we propose two modified implicit iterative methods with a perturbed mapping for a continuous pseudocontractive mapping in the same Banach space. Strong convergence theorems for the proposed iterative methods are established. The results in this paper substantially develop and complement the previous well-known results in this area.

1. Introduction

Let E be a real Banach space, and let $E^{*}$ be the dual space of E. Let C be a nonempty closed convex subset of E. Recall that a mapping $f:C\to C$ is called contractive if there exists $k\in (0,1)$ such that $\parallel fx-fy\parallel \le k\parallel x-y\parallel ,\forall x,y\in C$ and that a mapping $S:C\to C$ is called nonexpansive if $\parallel Sx-Sy\parallel \le \parallel x-y\parallel ,\forall x,y\in C$.

Let J denote the normalized duality mapping from E into $2^{X^{*}}$ defined by

$$J\left(x\right)=\{f\in E^{*}:\langle x,f\rangle =\parallel x\parallel \parallel f\parallel ,\parallel f\parallel =\parallel x\parallel \},x\in E,$$

where $\langle ·,·\rangle$ denotes the generalized duality pair between E and $E^{*}$. The mapping $T:C\to C$ is called pseudocontractive (respectively, strong pseudocontractive), if there exists $j(x-y)\in J(x-y)$ such that

$$\langle Tx-Ty,j(x-y)\rangle \le {\parallel x-y\parallel }^2,\forall x,y\in C,$$

(respectively, $\langle Tx-Ty,j(x-y)\rangle \le {\beta \parallel x-y\parallel }^2$ for some $\beta \in (0,1)$).

The class of pseudocontractive mappings is one of the most important classes of mappings in nonlinear analysis, and it has been attracting mathematician’s interest. Apart from them being a generalization of nonexpansive mappings, interest in pseudocontractive mappings stems mainly from their firm connection with the class of accretive mappings, where a mapping A with domain $D\left(A\right)$ and range $R\left(A\right)$ in E is called accretive if the inequality

$$\parallel x-y\parallel \le \parallel x-y+s(Ax-Ay)\parallel ,$$

holds for every $x,y\in D\left(A\right)$ and for all $s>0$.

Within the past 50 years or so, many authors have been devoting their study to the existence of zeros of accretive mappings or fixed points of pseudocontractive mappings and several iterative methods for finding zeros of accretive mappings or fixed points of pseudocontractive mappings. We can refer to References [1,2,3,4,5,6,7,8,9,10,11,12,13,14] and the references in therein.

In 2007, Morales [15] introduced the following viscosity iterative method for pseudocontractive mapping:

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

(1)

where $T:C\to E$ is a continuous pseudocontractive mapping satisfying the weakly inward condition and $f:C\to C$ is a bounded continuous strongly pseudocontractive mapping. In a reflexive Banach space with a uniformly Gâteaux differentiable norm such that every closed convex bounded subset of C has the fixed point property for nonexpansive self-mappings, he proved the strong convergence of the sequences generated by the iterative method in Equation (1) to a point q in $Fix\left(T\right)$ (the set of fixed points of T), where q is the unique solution to the following variational inequality:

$$\langle fq-q,J(p-q)\rangle \le 0,\forall p\in Fix\left(T\right).$$

(2)

In 2009, using the method of Reference [16], Ceng et al. [17] introduced the following modified viscosity iterative method and modified implicit viscosity iterative method with a perturbed mapping for a pseudocontractive mapping:

$$x_t=tf{x}_t+r_{t}S{x}_t+(1-t-r_t)T{x}_t,t\in (0,1),$$

(3)

where $0<r_t<1-t$, $T:C\to C$ is a continuous pseudocontractive mapping, $S:C\to C$ is a nonexpansive mapping, and $f:C\to C$ is a Lipschitz strongly pseudocontractive mapping.

$$\left\{\begin{array}{c}{y}_n={\alpha }_n{x}_n+(1-{\alpha }_n)T{y}_n,\hfill \\ x_{n+1}={\beta }_{n}f{y}_n+{\gamma }_{n}S{y}_n+(1-{\beta }_n-{\gamma }_n)y_n,\hfill \end{array}\right.$$

(4)

and

$$\left\{\begin{array}{c}{x}_n={\alpha }_n{y}_n+(1-{\alpha }_n)T{y}_n,\hfill \\ y_n={\beta }_{n}f{x}_{n-1}+{\gamma }_{n}S{x}_{n-1}+(1-{\beta }_n-{\gamma }_n)x_{n-1},\hfill \end{array}\right.$$

(5)

where $f:C\to C$ is a contractive mapping, $x_0\in C$ is an arbitrary initial point, and $\left\{{\alpha }_n\right\}$, $\left\{{\beta }_n\right\}$, $\left\{{\gamma }_n\right\}$$\subset (0,1]$ such that ${lim}_{n\to \infty }({\gamma }_n/{\beta }_n)=0$ and ${\beta }_n+{\gamma }_n<1$. In a reflexive and strictly convex Banach space with a uniformly Gâteaux differentiable norm, they proved the strong convergence of the sequences generated by the iterative methods in Equations (3)–(5) to a point q in $Fix\left(T\right)$, where q is the unique solution to the variational inequality in Equation (2). Their results developed and improved the corresponding results of Song and Chen [11], Zeng and Yao [16], Xu [18], Xu and Ori [19], and Chen et al. [20].

In this paper, as a continuation of study in this direction, in a reflexive Banach space having a weakly sequentially continuous duality mapping $J_{\phi }$ with gauge function $\phi$, we consider the viscosity iterative methods in Equations (3)–(5) for a continuous pseudocontractive mapping T, a continuous bounded strongly pseudocontractive mapping f, and a nonexpansive mapping S. We establish strong convergence of the sequences generated by proposed iterative methods to a fixed point of the mapping T, which solves a variational inequality related to f. The main results develop and supplement the corresponding results of Song and Chen [11], Morales [15], Ceng et al. [17], and Xu [18] to different Banach space as well as Zeng and Yao [16], Xu and Ori [19], Chen et al. [20], and the references therein.

2. Preliminaries

Throughout the paper, we use the following notations: $“\rightharpoonup ”$ for weak convergence, $“\stackrel{*}{\rightharpoonup }”$ for weak${}^{*}$ convergence, and $“\to ”$ for strong convergence.

Let E be a real Banach space with the norm $\parallel ·\parallel$, and let $E^{*}$ be its dual. The value of $x^{*}\in E^{*}$ at $x\in E$ will be denoted by $\langle x,x^{*}\rangle$. Let C be a nonempty closed convex subset of E, and let $T:C\to C$ be a mapping. We denote the set of fixed points of the mapping T by $Fix\left(T\right)$. That is, $Fix\left(T\right):=\{x\in C:Tx=x\}$.

Recall that a Banach space E is said to be smooth if for each $x\in S_E=\{x\in E:\parallel x\parallel =1\}$, there exists a unique functional $j_x\in E^{*}$ such that $\langle x,j_x\rangle =\parallel x\parallel$ and $\parallel j_x\parallel =1$ and that a Banach space E is said to be strictly convex [21] if the following implication holds for $x,y\in E$:

$$\parallel x\parallel \le 1,\parallel y\parallel \le 1,\parallel x-y\parallel >0\Rightarrow ∥\frac{x+y}{2}∥<1.$$

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

$$J_{\phi }\left(x\right)=\{f\in E^{*}:\langle x,f\rangle =\parallel x\parallel \parallel f\parallel ,\parallel f\parallel =\phi (\parallel x\parallel )\}\mathrm{for}\mathrm{all}x\in E$$

is called the duality mapping with gauge function $\phi$. In particular, the duality mapping with gauge function $\phi \left(t\right)=t$ denoted by J is referred to as the normalized duality mapping. It is known that a Banach space E is smooth if and only if the normalized duality mapping J is single-valued. The following property of duality mapping is also well-known:

$$J_{\phi }\left(\lambda x\right)=\mathrm{sign}\lambda \left(\frac{\phi \left(\right|\lambda |·\parallel x\parallel )}{\parallel x\parallel }\right)J\left(x\right)\mathrm{for}\mathrm{all}x\in E\backslash \left\{0\right\},\lambda \in \mathbb{R},$$

(6)

where $\mathbb{R}$ is the set of all real numbers. The following are some elementary properties of the duality mapping J [21,22]:

(i)

For $x\in E$, $J\left(x\right)$ is nonempty, bounded, closed, and convex;

(ii)

$J\left(0\right)=0$;

(iii)

for $x\in E$ and a real $\alpha$, $J\left(\alpha x\right)=\alpha J\left(x\right)$;

(iv)

for $x,y\in E$, $f\in J\left(x\right)$ and $g\in J\left(y\right)$, $\langle x-y,f-g\rangle \ge 0$;

(v)

for $x,y\in E$, $f\in J\left(x\right)$, ${\parallel x\parallel }^2-{\parallel y\parallel }^2\ge 2\langle x-y,f\rangle$.

We say that a Banach space E has a weakly continuous duality mapping if there exists a gauge function $\phi$ such that the duality mapping $J_{\phi }$ is single-valued and continuous from the weak topology to the weak${}^{*}$ topology, that is, for any $\left\{x_n\right\}\in E$ with $x_n\rightharpoonup x$, $J_{\phi }\left(x_n\right)\stackrel{*}{\rightharpoonup }{J}_{\phi }\left(x\right)$. A duality mapping $J_{\phi }$ is weakly continuous at 0 if $J_{\phi }$ is single-valued and if $x_n\rightharpoonup 0$, $J_{\phi }\left(x_n\right)\stackrel{*}{\rightharpoonup }0$. For example, every $l^p$ space $(1<p<\infty )$ has a weakly continuous duality mapping with gauge function $\phi \left(t\right)=t^{p-1}$ [21,22,23]. Set

$$\mathsf{\Phi }\left(t\right)={\int }_0^t\phi \left(\tau \right)d\tau \mathrm{for}\mathrm{all}t\in {\mathbb{R}}^{+}.$$

Then it is known that $J_{\phi }\left(x\right)$ is the subdifferential of the convex functional $\mathsf{\Phi }(\parallel ·\parallel )$ at x. A Banach space E that has a weakly continuous duality mapping implies that E satisfies Opial’s property. This means that whenever $x_n\rightharpoonup x$ and $y\ne x$, we have ${lim\; sup}_{n\to \infty }\parallel x_n-x\parallel <{lim\; sup}_{n\to \infty }\parallel x_n-y\parallel$ [21,23].

The following lemma is Lemma 2.1 of Jung [24].

Lemma 1.

([24]) Let E be a reflexive Banach space having a weakly continuous duality mapping $J_{\phi }$ with gauge function φ. Let $\left\{x_n\right\}$ be a bounded sequence of E and $f:E\to E$ be a continuous mapping. Let $g:E\to \mathbb{R}$ be defined by

$$g\left(z\right)=\underset{n\to \infty }{lim\; sup}\langle z-fz,J_{\phi }(z-x_n)\rangle$$

for $z\in E$. Then, g is a real valued continuous function on E.

We need the following well-known lemma for the proof of our main result [21,22].

Lemma 2.

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

$$\mathsf{\Phi }\left(t\right)={\int }_0^t\phi \left(\tau \right)d\tau \mathrm{for}\mathrm{all}t\in {\mathbb{R}}^{+}.$$

Then, the following inequalities hold:

$$\mathsf{\Phi }\left(kt\right)\le k\mathsf{\Phi }\left(t\right),0<k<1,$$

$$\mathsf{\Phi }(\parallel x+y\parallel )\le \mathsf{\Phi }(\parallel x\parallel )+\langle y,j_{\phi }(x+y)\rangle \mathrm{for}\mathrm{all}x,y\in E,$$

where$j_{\phi }(x+y)\in J_{\phi }(x+y)$.

The following lemma can be found in Reference [18].

Lemma 3.

([18]) Let $\left\{s_n\right\}$ be a sequence of nonnegative real numbers satisfying

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

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

(i)

$\left\{{\lambda }_n\right\}\subset [0,1]$and${\sum }_{n=0}^{\infty }{\lambda }_n=\infty$or, equivalently,${\prod }_{n=0}^{\infty }(1-{\lambda }_n)=0,$

(ii)

${lim\; sup}_{n\to \infty }{\delta }_n\le 0$or${\sum }_{n=0}^{\infty }{\lambda }_n\left|{\delta }_n\right|<\infty ,$

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

Let C be a nonempty closed convex subset of a real Banach space E. Recall that $S:C\to C$ is called accretive if $I-S$ is pseudocontractive. If $T:C\to C$ is a pseudocontractive mapping, then $I-T$ is accretive. We denote $A=J_1={(2I-T)}^{-1}$. Then, $Fix\left(A\right)=Fix\left(T\right)$ and the operator $A:R(2I-T)\to C$ is nonexpansive and single-valued, where I denotes the identity mapping.

We also need the following result which can be found in Reference [11].

Lemma 4.

([11]) Let C be a nonempty closed convex subset of a real Banach space E, and let $T:C\to C$ be a continuous pseudocontractive mapping. We denote $A={(2I-T)}^{-1}$.

(i) 

The mapping A is nonexpansive self-mapping on C, i.e., for all$x,y\in nC$, there holds

$$\parallel Ax-Ay\parallel \le \parallel x-y\parallel ,andAx\in C.$$

(ii) 

If${lim}_{n\to \infty }\parallel x_n-T{x}_n\parallel =0$, then${lim}_{n\to \infty }\parallel x_n-A{x}_n\parallel =0$.

The following Lemmas, which are well-known, can be found in many books in the geometry of Banach spaces (see References [21,23]).

Lemma 5.

(Demiclosedness Principle) Let C be a nonempty closed convex subset of a Banach space E, and let $T:C\to C$ be a nonexpansive mapping. Then, $x_n\rightharpoonup x$ in C and $(I-T)x_n\to y$ imply that $(I-T)x=y$.

Lemma 6.

If E is a Banach space such that$E^{*}$is strictly convex, then E is smooth and any duality mapping is norm-to-weak${}^{*}$-continuous.

Finally, we need the following result which was given by Deimling [4].

Lemma 7.

([4]) Let C be a nonempty closed convex subset of a Banach space E, and let $T:C\to C$ be a continuous strong pseudocontractive mapping with a pseudocontractive coefficient $\beta \in (0,1)$. Then, T has a unique fixed point in C.

3. Convergence of Path with Perturbed Mapping

As we know, the path convergency plays an important role in proving the convergence of iterative methods to approximate fixed points. In this direction, we first prove the existence of a path for a convex combination of a pseudocontractive type of mappings with a perturbed mapping and boundedness of the path.

Proposition 1.

Let C be a nonempty closed convex subset of a real Banach space E. Let$T:C\to C$be a continuous pseudocontractive mapping, let$S:C\to C$be a nonexpansive mapping, and let$f:C\to C$be a continuous strongly pseudocontractive mapping with a pseudocontractive coefficient$\beta \in (0,1)$.

(i) 

There exists a unique path$t\mapsto x_t\in C$,$t\in (0,1)$, satisfying

$$x_t=tf{x}_t+r_{t}S{x}_t+(1-t-r_t)T{x}_t,$$

(7)

provided$r_t:(0,1)\to [0,1-t)$is continuous and${lim}_{t\to 0}(r_t/t)=0$.

(ii) 

In particular, if T has a fixed point in C, then the path$\left\{x_t\right\}$is bounded.

Proof. 

(i) For each $t\in (0,1)$, define the mapping $T_{(S,f)}:C\to C$ as follows:

$$T_{(S,f)}=tf+r_{t}S+(1-t-r_t)T,$$

where $0<r_t<1-t$ and ${lim}_{t\to 0}(r_t/t)=0$. Then, it is easy to show that the mapping $T_{(S,f)}$ is a continuous strongly pseudocontractive self-mapping of C. Therefore, by Lemma 7, $T_{(S,f)}$ has a unique fixed point in C, i.e., for each given $t\in (0,1)$, there exists $x_t\in C$ such that

$$x_t=tf{x}_t+r_{t}S{x}_t+(1-t-r_t)T{x}_t.$$

To show continuity, let $t,t_0\in (0,1)$. Then, there exists $j\in J(x_t-x_{t_0})$ such that

$$\begin{array}{cc}\hfill \langle x_t-x_{t_0},j\rangle & =\langle tf{x}_t+r_{t}S{x}_t+(1-t-r_t)T{x}_t-(t_{0}f{x}_{t_0}+r_{t}S{x}_t+(1-t_0-r_{t_0})T{x}_{t_0}),j\rangle \hfill \\ & =t\langle f{x}_t-f{x}_{t_0},j\rangle +(t-t_0)\langle f{x}_{t_0},j\rangle +r_t\langle S{x}_t-S{x}_{t_0},j\rangle +(r_t-r_{t_0})\langle S{x}_{t_0},j\rangle \hfill \\ & \hspace{1em}+(1-t-r_t)\langle T{x}_t-T{x}_{t_0},j\rangle +((t-t_0)+(r_t-r_{t_0}))\langle T{x}_{t_0},j\rangle ,\hfill \end{array}$$

and this implies that

$$\begin{array}{cc}\hfill \parallel x_t-x_{t_0}{\parallel }^2& \le t\beta \parallel x_t-x_{t_0}{\parallel }^2+|t-t_0|\parallel f{x}_{t_0}\parallel \parallel x_t-x_{t_0}\parallel \hfill \\ & \hspace{1em}+r_t\parallel x_t-x_{t_0}{\parallel }^2+|r_t-r_{t_0}|\parallel S{x}_{t_0}\parallel \parallel x_t-x_{t_0}\parallel \hfill \\ & \hspace{1em}+(1-t-r_t)\parallel x_t-x_{t_0}{\parallel }^2+|t-t_0|\parallel T{x}_{t_0}\parallel \parallel x_t-x_{t_0}\parallel +|r_t-r_{t_0}|\parallel T{x}_{t_0}\parallel \parallel x_t-x_{t_0}\parallel .\hfill \end{array}$$

and, hence,

$$\begin{array}{cc}\hfill \parallel x_t-x_{t_0}\parallel & \le t\beta \parallel x_t-x_{t_0}\parallel +|t-t_0|\parallel f{x}_{t_0}\parallel +|r_t-r_{t_0}|\parallel S{x}_{t_0}\parallel \hfill \\ & \hspace{1em}+(1-t-r_t)\parallel x_t-x_{t_0}\parallel +|t-t_0|\parallel T{x}_{t_0}\parallel +|r_t-r_{t_0}|\parallel T{x}_{t_0}\parallel \hfill \\ & =(1-(1-\beta )t)\parallel x_t-x_{t_0}\parallel +(\parallel f{x}_{t_0}\parallel +\parallel T{x}_{t_0}\parallel \left)\right|t-t_0|+(\parallel S{x}_{t_0}\parallel +\parallel T{x}_{t_0}\parallel \left)\right|r_t-r_{t_0}|.\hfill \end{array}$$

Therefore,

$$\parallel x_t-x_{t_0}\parallel \le \frac{\parallel f{x}_{t_0}\parallel +\parallel T{x}_{t_0}\parallel }{(1-\beta )t}|t-t_0|+\frac{\parallel S{x}_{t_0}\parallel +\parallel T{x}_{t_0}\parallel }{(1-\beta )t}|r_t-r_{t_0}|,$$

which guarantees continuity.

(ii) By the same argument as in the proof of Theorem 2.1 of Reference [17], we can prove that $\left\{x_t\right\}$ defined by Equation (7) is bounded for $t\in (0,t_0)$ for some $t_0\in (0,1)$, and so we omit its proof. □

The above path of Equation (7) is called the modified viscosity iterative method with perturbed mapping, where S is called the perturbed mapping.

The following result gives conditions for existence of a solution of a variational inequality:

$$\langle (I-f)q,J_{\phi }(q-p)\rangle \le 0,\forall p\in Fix\left(T\right).$$

(8)

Theorem 1.

Let E be a Banach space such that$E^{*}$is strictly convex. Let C be a nonempty closed convex subset of a real Banach space E. Let$T:C\to C$be a continuous pseudocontractive mapping with$Fix\left(T\right)\ne Ø$, let$S:C\to C$be a nonexpansive mapping, and let$f:C\to C$be a continuous strongly pseudocontractive mapping with a pseudocontractive coefficient$\beta \in (0,1)$. Suppose that$\left\{x_t\right\}$defined by Equation (7) converges strongly to a point in$Fix\left(T\right)$. If we define$q:={lim}_{t\to 0}{x}_t$, then q is a solution of the variational inequality in Equation (8).

Proof. 

First, from Lemma 6, we note that E is smooth and $J_{\phi }$ is norm-to-weak${}^{*}$-continuous.

Since

$$(I-f)x_t=-\frac{1-t-r_t}{t}(I-T)x_t-\frac{r_t}{t}(I-S)x_t,$$

we have for $p\in Fix\left(T\right)$

$$\begin{array}{cc}\hfill \langle (I-f)x_t,J_{\phi }(x_t-p)\rangle & =-\frac{1-t-r_t}{t}\langle (I-T)x_t-(I-T)p,J_{\phi }(x_t-p)\rangle \hfill \\ & \hspace{1em}+\frac{r_t}{t}\langle (S-I)x_t,J_{\phi }(x_t-p)\rangle .\hfill \end{array}$$

(9)

Since $I-T$ is accretive and $J(x_t-p)$ is a positive-scalar multiple of $J_{\phi }(x_t-p)$ (see Equation (6)), it follow from Equation (9) that

$$\begin{array}{cc}\hfill \langle (I-f)x_t,J_{\phi }(x_t-p)\rangle & \le \frac{r_t}{t}\langle (S-I)x_t,J_{\phi }(x_t-p)\rangle \hfill \\ & \le \frac{r_t}{t}\parallel (S-I)x_t\parallel \phi (\parallel x_t-p\parallel ).\hfill \end{array}$$

(10)

Taking the limit as $t\to 0$, by ${lim}_{t\to 0}\frac{r_t}{t}=0$, we obtain

$$\langle (I-f)q,J_{\phi }(q-p)\rangle \le 0,\forall p\in Fix\left(T\right).$$

This completes the proof. □

The following lemma provides conditions under which $\left\{x_t\right\}$ defined by Equation (7) converges strongly to a point in $Fix\left(T\right)$.

Lemma 8.

Let E be a reflexive smooth Banach space having Opial’s property and having some duality mapping$J_{\phi }$weakly continuous at 0. Let C be a nonempty closed convex subset of E. Let$T:C\to C$be a continuous pseudocontractive mapping with$Fix\left(T\right)\ne Ø$, let$S:C\to C$be a nonexpansive mapping, and let$f:C\to C$be a continuous bounded strongly pseudocontractive mapping with a pseudocontractive coefficient$\beta \in (0,1)$. Then,$\left\{x_t\right\}$defined by Equation (7) converges strongly to a point in$Fix\left(T\right)$as$t\to 0$.

Proof. 

First, from Proposition 1 (ii), we know that $\{x_t:t\in (0,t_0)\}$ is bounded for $t\in (0,t_0)$ for some $t_0\in (0,1)$.

Since f is a bounded mapping and S is a nonexpansive mapping, $\{f{x}_t:t\in (0,t_0)\}$ and $\{S{x}_t:t\in (0,t_0)\}$ are bounded. Moreover, noting that $x_t=tf{x}_t+r_{t}S{x}_t+(1-t-r_t)T{x}_t$, we have

$$T{x}_t=\frac{1}{1-t-r_t}{x}_t-\frac{t}{1-t-r_t}f{x}_t-\frac{r_t}{1-t-r_t}S{x}_t,$$

which implies that

$$\parallel T{x}_t\parallel \le \frac{1}{1-t-r_t}\parallel x_t\parallel +\frac{t}{1-t-r_t}\parallel f{x}_t\parallel +\frac{r_t}{1-t-r_t}\parallel S{x}_t\parallel .$$

Thus, we obtain

$$\parallel T{x}_t\parallel \le 2\parallel x_t\parallel +2t\parallel f{x}_t\parallel +2r_t\parallel S{x}_t\parallel ,\forall t\in (0,t_0)$$

and so $\{T{x}_t:t\in (0,t_0)\}$ is bounded. This implies that

$$\underset{t\to 0}{lim}\parallel x_t-T{x}_t\parallel \le \underset{t\to 0}{lim}t\parallel f{x}_t-T{x}_t\parallel +\underset{t\to 0}{lim}{r}_t\parallel S{x}_t-T{x}_t\parallel =0.$$

(11)

Now, let $t_m\in (0,t_0)$ for some $t_0\in (0,1)$ be such that $t_m\to 0$, and let $\left\{x_m\right\}:=\left\{x_{t_m}\right\}$ be a subsequence of $\left\{x_t\right\}$. Then,

$$x_m=t_{m}f{x}_m+r_m{S}_m+(1-t_m-r_m)T{x}_m.$$

Let $p\in Fix\left(T\right)$. Then, we have

$$x_m-p=t_m(f{x}_m-p)+r_m(S{x}_m-p)+(1-t_m-r_m)(T{x}_m-Tp)$$

and

$$\begin{array}{cc}\hfill \parallel x_m-p\parallel \phi (\parallel x_m-p\parallel )& =\langle x_m-p,J_{\phi }(x_m-p)\rangle \hfill \\ & \le t_m\langle f{x}_m-p,J_{\phi }(x_m-p)\rangle +r_m\langle S{x}_m-p,J_{\phi }(x_m-p)\rangle \hfill \\ & \hspace{1em}+(1-t_m-r_m)\parallel x_m-p\parallel \phi (\parallel x_m-p\parallel ).\hfill \end{array}$$

Thus, it follows that

$$\parallel x_m-p\parallel \phi (\parallel x_m-p\parallel )\le \frac{t_m}{t_m+r_m}\langle f{x}_m-p,J_{\phi }(x_m-p)\rangle +\frac{r_m}{t_m+r_m}\langle S{x}_m-p,J_{\phi }(x_m-p)\rangle .$$

(12)

Hence, we get

$$\langle p-f{x}_m,J_{\phi }(x_m-p)\rangle \le -\frac{t_m+r_m}{t_m}\parallel x_m-p\parallel \phi (\parallel x_m-p\parallel )+\frac{r_m}{t_m}\langle S{x}_m-p,J_{\phi }(x_m-p)\rangle ,$$

that is,

$$\langle p-f{x}_m,J_{\phi }(p-x_m)\rangle \ge \frac{t_m+r_m}{t_m}\parallel x_m-p\parallel \phi (\parallel x_m-p\parallel )+\frac{r_m}{t_m}\langle p-S{x}_m,J_{\phi }(x_m-p)\rangle .$$

Therefore, we have

$$\begin{array}{cc}\hfill \langle x_m-f{x}_m,J_{\phi }(p-x_m)\rangle & =\langle x_m-p,J_{\phi }(p-x_m)\rangle +\langle p-f{x}_m,J_{\phi }(p-x_m)\rangle \hfill \\ & \ge -\parallel x_m-p\parallel \phi (\parallel x_m-p\parallel )+\frac{t_m+r_m}{t_m}\parallel x_m-p\parallel \phi (\parallel x_m-p\parallel )\hfill \\ & \hspace{1em}+\frac{r_m}{t_m}\langle p-S{x}_m,J_{\phi }(x_m-p)\rangle \hfill \\ & =\frac{r_m}{t_m}\parallel x_m-p\parallel \phi (\parallel x_m-p\parallel )+\frac{r_m}{t_m}\langle p-S{x}_m,J_{\phi }(x_m-p)\rangle .\hfill \end{array}$$

On the other hand, since $\left\{x_m\right\}$ is bounded and E is reflexive, $\left\{x_m\right\}$ has a weakly convergent subsequence $\left\{x_{m_k}\right\}$, say, $x_{m_k}\rightharpoonup u\in E$. From Equation (11), it follows that

$$\parallel x_m-T{x}_m\parallel \le t_m\parallel f{x}_m-T{x}_m\parallel +r_m\parallel S{x}_m-T{x}_m\parallel \to 0.$$

From Lemma 4, we know that the mapping $A={(2I-T)}^{-1}:C\to C$ is nonexpansive, that $Fix\left(A\right)=Fix\left(T\right)$, and that $\parallel x_m-A{x}_m\parallel \to 0$. Thus, by Lemma 5, $u\in Fix\left(A\right)=Fix\left(T\right)$. Therefore, by Equation (12) and the assumption that $J_{\phi }$ is weakly continuous at 0, we obtain

$$\begin{array}{cc}\hfill \parallel x_{m_k}-u\parallel \phi (\parallel x_{m_k}-u\parallel )& \le \frac{t_{m_k}}{t_{m_k}+r_{m_k}}\langle f{x}_{m_k}-u,J_{\phi }(x_{m_k}-u)\rangle +\frac{r_{m_k}}{t_{m_k}+r_{m_k}}\langle S{x}_{m_k}-u,J_{\phi }(x_{m_k}-u)\rangle \hfill \\ & \le |\langle f{x}_{m_k}-u,J_{\phi }(x_{m_k}-u)\rangle |+\frac{r_{m_k}}{t_{m_k}}|\langle S{x}_{m_k}-u,J_{\phi }(x_{m_k}-u)\rangle |\to 0.\hfill \end{array}$$

Since $\phi$ is continuous and strictly increasing, we must have $x_{m_k}\to u$.

Now, we will show that every weakly convergent subsequence of $\left\{x_m\right\}$ has the same limit. Suppose that $x_{m_k}\rightharpoonup u$ and $x_{m_j}\rightharpoonup v$. Then, by the above proof, we have $u,v\in Fix\left(T\right)$ and $x_{m_k}\to u$ and $x_{m_j}\to v$. By Equation (12), we have the following for all $p\in Fix\left(T\right)$:

$$\begin{array}{cc}\hfill \parallel x_{m_k}-p\parallel \phi (\parallel x_{m_k}-p\parallel )& \le \frac{t_{m_k}}{t_{m_k}+r_{m_k}}\langle f{x}_{m_k}-p,J_{\phi }(x_{m_k}-p)\rangle +\frac{r_{m_k}}{t_{m_k}+r_{m_k}}\langle S{x}_{m_k}-p,J_{\phi }(x_{m_k}-p)\rangle \hfill \\ & \le \frac{t_{m_k}}{t_{m_k}+r_{m_k}}\langle f{x}_{m_k}-p,J_{\phi }(x_{m_k}-p)\rangle +\frac{r_{m_k}}{t_{m_k}}\left|\langle S{x}_{m_k}-p,J_{\phi }(x_{m_k}-p)\rangle \right|\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill \parallel x_{m_j}-p\parallel \phi (\parallel x_{m_j}-p\parallel )& \le \frac{t_{m_j}}{t_{m_j}+r_{m_j}}\langle f{x}_{m_j}-p,J_{\phi }(x_{m_j}-p)\rangle +\frac{r_{m_j}}{t_{m_j}+r_{m_j}}\langle S{x}_{m_j}-p,J_{\phi }(x_{m_j}-p)\rangle \hfill \\ & \le \frac{t_{m_j}}{t_{m_j}+r_{m_k}}\langle f{x}_{m_j}-p,J_{\phi }(x_{m_k}-p)\rangle +\frac{r_{m_k}}{t_{m_k}}\left|\langle S{x}_{m_k}-p,J_{\phi }(x_{m_k}-p)\rangle \right|.\hfill \end{array}$$

Taking limits, we get

$$\mathsf{\Phi }(\parallel u-v\parallel )=\parallel u-v\parallel \phi (\parallel u-v\parallel )\le \langle fu-v,J_{\phi }(u-v)\rangle$$

(13)

and

$$\mathsf{\Phi }(\parallel v-u\parallel )=\parallel v-u\parallel \phi (\parallel v-u\parallel )\le \langle fv-u,J_{\phi }(v-u)\rangle .$$

(14)

Adding up Equations (13) and (14) yields

$$\begin{array}{cc}\hfill 2\mathsf{\Phi }(\parallel u-v\parallel )=2\parallel u-v\parallel \phi (\parallel u-v\parallel )& \le \parallel u-v\parallel \phi (\parallel u-v\parallel )+\langle fu-fv,J_{\phi }(u-v)\rangle \hfill \\ & \le (1+\beta )\parallel u-v\parallel \phi (\parallel u-v\parallel )=(1+\beta )\mathsf{\Phi }(\parallel u-v\parallel ).\hfill \end{array}$$

Since $\beta \in (0,1)$, this implies $\mathsf{\Phi }(\parallel u-v\parallel )\le 0$, that is, $u=v$. Hence, $\left\{x_m\right\}$ is strongly convergent to a point in $Fix\left(T\right)$ as $t_m\to 0$.

The same argument shows that, if $t_l\to 0$, then the subsequence $\left\{x_l\right\}:=\left\{x_{t_l}\right\}$ of $\{x_t:t\in (0,t_0)\}$ for some $t_0\in (0,1)$ is strongly convergent to the same limit. Thus, as $t\to 0$, $\left\{x_t\right\}$ converges strongly to a point in $Fix\left(T\right)$. □

Using Theorem 1 and Lemma 8, we show the existence of a unique solution of the variational inequality in Equation (8) in a reflexive Banach space having a weakly continuous duality mapping.

Theorem 2.

Let E be a reflexive Banach space having a weakly continuous duality mapping$J_{\phi }$with gauge function φ, and let C be a nonempty closed convex subset of E. Let$T:C\to C$be a continuous pseudocontractive mapping such that$Fix\left(T\right)\ne Ø$, let$S:C\to C$be a nonexpansive mapping, and let$f:C\to C$be a continuous bounded strongly pseudocontractive mapping with a pseudocontractive coefficient$\beta \in (0,1)$. Then, there exists the unique solution in$q\in Fix\left(T\right)$of the variational inequality in Equation (8), where$q:={lim}_{t\to \infty }{x}_t$with$x_t$being defined by Equation (7).

Proof. 

We notice that the definition of the weak continuity of the duality mapping $J_{\phi }$ implies that E is smooth. Thus, $E^{*}$ is strictly convex for reflexivity of E. By Lemma 8, $\left\{x_t\right\}$ defined by Equation (7) converges strongly to a point q in $Fix\left(T\right)$ as $t\to 0$. Hence, by Theorem 1, q is the unique solution of the variational inequality in Equation (8). In fact, suppose that $q,p\in Fix\left(T\right)$ satisfy the variational inequality in Equation (8). Then, we have

$$\langle (I-f)q,J_{\phi }(q-p)\rangle \le 0\mathrm{and}\langle (I-f)p,J_{\phi }(p-q)\rangle \le 0.$$

Adding these two inequalities, we have

$$(1-\beta )\mathsf{\Phi }(\parallel q-p\parallel )=(1-\beta )\parallel q-p\parallel \phi (\parallel q-p\parallel )\le \langle (I-f)q-(I-f)p,J_{\phi }(q-p)\rangle \le 0,$$

and so $q=p$. □

As a direct consequence of Theorem 2, we have the following result.

Corollary 1.

([20, Theorem 3.2]) Let E be a reflexive Banach space having a weakly continuous duality mapping $J_{\phi }$ with gauge function φ, and let C be a nonempty closed convex subset of E. Let $T:C\to C$ be a continuous pseudocontractive mapping such that $Fix\left(T\right)\ne Ø$, and let $f:C\to C$ be a continuous bounded strongly pseudocontractive mapping with a pseudocontractive coefficient $\beta \in (0,1)$. Let $\left\{x_t\right\}$ be defined by

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

Then, as$t\to 0$,$x_t$converges strongly to a some point of T such that q is the unique solution of the variational inequality in Equation (8).

Proof. 

Put $S=I$ and $r_t=0$ for all $t\in (0,1)$. Then, the result follows immediately from Theorem 2. □

Remark 1.

(1) 

Theorem 2 develops and supplements Theorem 2.1 of Ceng et al. [17] in the following aspects:

(i) 

The space is replaced by the space having a weakly continuous duality mapping$J_{\phi }$with gauge function φ.

(ii) 

The Lipischiz strongly pseudocontractive mapping f in Theorem 2.1 in Reference [17] is replaced by a bounded continuous strongly pseudocontractive mapping f in Theorem 2.

(2) 

Corollary 1 complements Theorem 2.1 of Song and Chen [11] and Corollary 2.2 of Cent et al. [17] by replacing the Lipischiz strongly pseudocontractive mapping f in References [11,17] by the bounded continuous strongly pseudocontractive mapping f in Corollary 3.5 in a reflexive Banach space having a weakly continuous duality mapping$J_{\phi }$with gauge function φ.

(3) 

Corollary 1 also develops Theorem 2 of Morales [15] to a reflexive Banach space having a weakly continuous duality mapping$J_{\phi }$with gauge function φ.

4. Modified Implicit Iterative Methods with Perturbed Mapping

First, we prepare the following result.

Theorem 3.

Let E be a reflexive Banach space having a weakly continuous duality mapping$J_{\phi }$with gauge function φ, and let C be a nonempty closed convex subset of E. Let$T:C\to C$be a continuous pseudocontractive mapping such that$Fix\left(T\right)\ne Ø$, let$S:C\to C$be a nonexpansive mapping, and let$f:C\to C$be a continuous bounded strongly pseudocontractive mapping with a pseudocontractive coefficient$\beta \in (0,1)$. Let$\left\{x_t\right\}$be defined by Equation (7). If there exists a bounded sequence$\left\{x_n\right\}$such that${lim}_{n\to \infty }\parallel x_n-T{x}_n\parallel =0$and$q={lim}_{t\to 0}{x}_t$, then

$$\underset{n\to \infty }{lim\; sup}\langle fq-q,J_{\phi }(x_n-q)\rangle \le 0.$$

Proof. 

Using the equality

$$x_t-x_n=(1-t-r_t)(T{x}_t-x_n)+t(f{x}_t-x_n)+r_t(S{x}_t-x_n)$$

and the inequality

$$\langle Tx-Ty,J_{\phi }(x-y)\rangle \le \parallel x-y\parallel \phi (\parallel x-y\parallel ),\forall x,y\in C,$$

we derive

$$\begin{array}{cc}\hfill \parallel x_t-x_n\parallel \phi (\parallel x_t-x_n\parallel )& =(1-t-r_t)\langle T{x}_t-x_n,J_{\phi }(x_t-x_n)\rangle +t\langle f{x}_t-x_n,J_{\phi }(x_t-x_n)\rangle \hfill \\ & \hspace{1em} +r_t\langle S{x}_t-x_n,J_{\phi }(x_t-x_n)\rangle \hfill \\ & =(1-t-r_t)(\langle T{x}_t-T{x}_n,J_{\phi }(x_t-x_n)\rangle +\langle T{x}_n-x_n,J_{\phi }(x_t-x_n)\rangle \hfill \\ & \hspace{1em}t\langle f{x}_t-x_t,J_{\phi }(x_t-x_n)\rangle +t\parallel x_t-x_n\parallel \phi (\parallel x_t-x_n\parallel )\hfill \\ & \hspace{1em} +r_t\langle S{x}_t-x_t,J_{\phi }(x_t-x_n)\rangle +r_t\parallel x_t-x_n\parallel \phi (\parallel x_t-x_n\parallel )\hfill \\ & \le \parallel x_t-x_n\parallel \phi (\parallel x_t-x_n\parallel )+\parallel T{x}_n-x_n\parallel \phi (\parallel x_t-x_n\parallel )\hfill \\ & \hspace{1em}t\langle f{x}_t-x_t,J_{\phi }(x_t-x_n)\rangle +r_t\parallel S{x}_t-x_n\parallel \phi (\parallel x_t-x_n\parallel )\hfill \end{array}$$

and, hence,

$$\langle x_t-f{x}_t,J_{\phi }(x_t-x_n)\rangle \le \frac{\parallel T{x}_n-x_n\parallel }{t}\phi (\parallel x_t-x_n\parallel )+\frac{r_t}{t}\parallel S{x}_t-x_t\parallel \phi (\parallel x_t-x_n\parallel ).$$

Therefore, by ${lim\; sup}_{n\to \infty }\phi (\parallel x_t-x_n\parallel )<\infty$, we have

$$\begin{array}{cc}\hfill \underset{n\to \infty }{lim\; sup}\langle x_t-f{x}_t,J_{\phi }(x_t-x_n)\rangle & \le \underset{n\to \infty }{lim\; sup}\frac{\parallel T{x}_n-x_n\parallel }{t}\phi (\parallel x_t-x_n\parallel )\hfill \\ & \hspace{1em}+\underset{n\to \infty }{lim\; sup}\frac{r_t}{t}\parallel S{x}_t-x_t\parallel \phi (\parallel x_t-x_n\parallel )\hfill \\ & =\underset{n\to \infty }{lim\; sup}\frac{r_t}{t}\parallel S{x}_t-x_t\parallel \phi (\parallel x_t-x_n\parallel )\hfill \\ & =\frac{r_t}{t}\parallel S{x}_t-x_t\parallel \underset{n\to \infty }{lim\; sup}\phi (\parallel x_t-x_n\parallel ).\hfill \end{array}$$

Thus, noting that ${lim}_{t\to 0}{lim\; sup}_{n\to \infty }\phi (\parallel x_t-x_n\parallel )<\infty$, by Lemma 1, we conclude

$$\begin{array}{cc}\hfill \underset{n\to \infty }{lim\; sup}\langle fq-q,J_{\phi }(x_n-q)\rangle & =\underset{t\to 0}{lim}\underset{n\to \infty }{lim\; sup}\langle f{x}_t-x_t,J_{\phi }(x_n-x_t)\rangle \hfill \\ & \le \underset{t\to 0}{lim}\left[\frac{r_t}{t}\parallel S{x}_t-x_t\parallel \right]\underset{t\to 0}{lim}\underset{n\to \infty }{lim\; sup}\phi (\parallel x_t-x_n\parallel )\hfill \\ & =0\times \underset{t\to 0}{lim}\underset{n\to \infty }{lim\; sup}\phi (\parallel x_t-x_n\parallel )=0.\hfill \end{array}$$

This completes the proof. □

Theorem 4.

Let E be a reflexive Banach space having a weakly continuous duality mapping$J_{\phi }$with gauge function φ, and let C be a nonempty closed convex subset of E. Let$T:C\to C$be a continuous pseudocontractive mapping such that$Fix\left(T\right)\ne Ø$, let$S:C\to C$be a nonexpansive mapping, and let$f:C\to C$be a contractive mapping with a contractive coefficient$k\in (0,1)$. For$x_0\in C$, let$\left\{x_n\right\}$be defined by the following iterative scheme:

$$\left\{\begin{array}{c}{y}_n={\alpha }_n{x}_n+(1-{\alpha }_n)T{y}_n\hfill \\ x_{n+1}={\beta }_{n}f{y}_n+{\gamma }_{n}S{y}_n+(1-{\beta }_n-{\gamma }_n)y_n,\forall n\ge 0,\hfill \end{array}\right.$$

(15)

where$\left\{{\alpha }_n\right\}$,$\left\{{\beta }_n\right\}$, and$\left\{{\gamma }_n\right\}$are three sequences in$(0,1]$satisfying the following conditions:

(i) 

${lim}_{n\to \infty }{\alpha }_n=0$;

(ii) 

${lim}_{n\to \infty }{\beta }_n=0$,${\sum }_{n=0}^{\infty }{\beta }_n=\infty$;

(iii) 

${lim}_{n\to \infty }({\gamma }_n/{\beta }_n)=0$,${\beta }_n+{\gamma }_n\le 1$,$\forall n\ge 0$.

Then,$\left\{x_n\right\}$converges strongly to a fixed point$x^{*}$of T, which is the unique solution of the following variational inequality

$$\langle (I-f)x^{*},J_{\phi }(x^{*}-p)\rangle \le 0,\forall p\in Fix\left(T\right).$$

(16)

Proof. 

First, put $z_t=tf{z}_t+r_{t}S{z}_t+(1-t-r_t)T{z}_t$. Then, it follows from Theorem 2 that, as $t\to 0$, $z_t$ converges strongly to some fixed point $x^{*}$ of T such that $x^{*}$ is the unique solution in $Fix\left(T\right)$ to the variational inequality in Equation (16).

Now, we divide the proof into several steps.

Step 1. We show that $\left\{x_n\right\}$ is bounded. To this end, let $p\in Fix\left(T\right)$. Then, we have

$$\begin{array}{cc}\hfill \parallel y_n-p\parallel \phi (\parallel y_n-p\parallel )& =\langle {\alpha }_n{x}_n+(1-{\alpha }_n)T{y}_n-p,J_{\phi }(y_n-p)\rangle \hfill \\ & \le (1-{\alpha }_n)\langle T{y}_n-Tp,J_{\phi }(y_n-p)\rangle +{\alpha }_n\parallel x_n-p\parallel \phi (\parallel y_n-p\parallel )\hfill \\ & \le (1-{\alpha }_n)\parallel y_n-p\parallel \phi (\parallel y_n-p\parallel )+{\alpha }_n\parallel x_n-p\parallel \phi (\parallel y_n-p\parallel )\hfill \end{array}$$

and, hence,

$$\parallel y_n-p\parallel \le \parallel x_n-p\parallel ,\forall n\ge 0.$$

Thus, we obtain

$$\begin{array}{cc}\hfill \parallel x_{n+1}-p\parallel & \le {\beta }_n\parallel f{y}_n-p\parallel +{\gamma }_n\parallel S{y}_n-p\parallel +(1-{\beta }_n-{\gamma }_n)\parallel y_n-p\parallel \hfill \\ & \le {\beta }_n(\parallel f{y}_n-fp\parallel +\parallel fp-p\parallel )+{\gamma }_n(\parallel S{y}_n-Sp\parallel +\parallel Sp-p\parallel )\hfill \\ & \hspace{1em}+(1-{\beta }_n-{\gamma }_n)\parallel x_n-p\parallel \hfill \\ & \le {\beta }_{n}k\parallel y_n-p\parallel +{\beta }_n\parallel fp-p\parallel +{\gamma }_n\parallel y_n-p\parallel +{\gamma }_n\parallel Sp-p\parallel \hfill \\ & \hspace{1em}+(1-{\beta }_n-{\gamma }_n)\parallel x_n-p\parallel \le {\beta }_{n}k\parallel x_n-p\parallel +{\beta }_n\parallel fp-p\parallel +{\gamma }_n\parallel x_n-p\parallel +{\gamma }_n\parallel Sp-p\parallel +(1-{\beta }_n-{\gamma }_n)\parallel x_n-p\parallel \hfill \\ & =(1-(1-k){\beta }_n)\parallel x_n-p\parallel +{\beta }_n\parallel fp-p\parallel +{\gamma }_n\parallel Sp-p\parallel .\hfill \end{array}$$

(17)

Since ${lim}_{n\to \infty }({\gamma }_n/{\beta }_n)=0$, we may assume without loss of generality that ${\gamma }_n\le {\beta }_n$ for all $n>0$. Therefore, it follows from Equation (17) that

$$\begin{array}{cc}\hfill \parallel x_{n+1}-p\parallel & \le (1-(1-k){\beta }_n)\parallel x_n-p\parallel +(1-k){\beta }_n·\frac{1}{1-k}(\parallel fp-p\parallel +\parallel Sp-p\parallel )\hfill \\ & \le max\left\{\parallel x_n-p\parallel ,\frac{1}{1-k}(\parallel fp-p\parallel +\parallel Sp-p\parallel )\right\}.\hfill \end{array}$$

By induction, we derive

$$\parallel x_n-p\parallel \le max\left\{\parallel x_0-p\parallel ,\frac{1}{1-k}(\parallel fp-p\parallel +\parallel Sp-p\parallel )\right\},\forall n\ge 0.$$

This show that $\left\{x_n\right\}$ is bounded and so is $\left\{y_n\right\}$.

Step 2. We show that $\left\{f{y}_n\right\}$, $\left\{S{y}_n\right\}$, and $\left\{T{y}_n\right\}$ are bounded. Indeed, observe that

$$\parallel f{y}_n\parallel \le \parallel f{y}_n-fp\parallel +\parallel fp\parallel \le k\parallel y_n-p\parallel +\parallel fp\parallel$$

and

$$\parallel S{y}_n\parallel \le \parallel S{y}_n-Sp\parallel +\parallel Sp\parallel \le \parallel y_n-p\parallel +\parallel Sp\parallel .$$

Thus, $\left\{f{y}_n\right\}$ and $\left\{S{y}_n\right\}$ are bounded. Since ${lim}_{n\to \infty }{\alpha }_n=0$, there exist $n_0\ge 0$ and $a\in (0,1)$ such that ${\alpha }_n\le a$ for all $n\ge n_0$. Noting that $y_n={\alpha }_n{x}_n+(1-{\alpha }_n)T{y}_n$, we have

$$T{y}_n=\frac{1}{1-{\alpha }_n}{y}_n-\frac{{\alpha }_n}{1-{\alpha }_n}{x}_n$$

and so

$$\parallel T{y}_n\parallel \le \frac{1}{1-{\alpha }_n}\parallel y_n\parallel +\frac{{\alpha }_n}{1-{\alpha }_n}\parallel x_n\parallel \le \frac{1}{1-a}\parallel y_n\parallel +\frac{a}{1-a}\parallel x_n\parallel .$$

Consequently, the sequence $\left\{T{y}_n\right\}$ is also bounded.

Step 3. We show that ${lim\; sup}_{n\to \infty }\langle f{x}^{*}-x^{*},J_{\phi }(y_n-x^{*})\rangle \le 0.$ In fact, from condition (i) and boundedness of $\left\{x_n\right\}$ and $\left\{T{y}_n\right\}$, we get

$$\parallel y_n-T{y}_n\parallel ={\alpha }_n\parallel x_n-T{y}_n\parallel \to 0(n\to \infty ).$$

(18)

Thus, it follows from Equation (18) and Theorem 3 that ${lim\; sup}_{n\to \infty }\langle f{x}^{*}-x^{*},J_{\phi }(y_n-x^{*})\rangle \le 0.$

Step 4. We show that ${lim\; sup}_{n\to \infty }\langle f{x}^{*}-x^{*},J_{\phi }(x_{n+1}-x^{*})\rangle \le 0.$ Indeed, by Equations (15) and (18), we have

$$\begin{array}{cc}\hfill \parallel x_{n+1}-y_n\parallel & =\parallel {\beta }_{n}f{y}_n+{\gamma }_{n}S{y}_n+(1-{\beta }_n-{\gamma }_n)y_n-({\alpha }_n{x}_n+(1-{\alpha }_n)T{y}_n)\parallel \hfill \\ & \le {\alpha }_n\parallel x_n-T{y}_n\parallel +{\beta }_n\parallel f{y}_n-y_n\parallel +{\gamma }_n\parallel S{y}_n-y_n\parallel +\parallel y_n-T{y}_n\parallel \to 0(n\to \infty ).\hfill \end{array}$$

Since the duality mapping $J_{\phi }$ is single-valued and weakly continuous, we have

$$\underset{n\to \infty }{lim}\langle f{x}^{*}-x^{*},J_{\phi }(x_{n+1}-x^{*})-J_{\phi }(y_n-x^{*})\rangle =0.$$

Therefore, we obtain from step 3 that

$$\begin{array}{cc}\hfill \underset{n\to \infty }{lim\; sup}\langle f{x}^{*}-x^{*},J_{\phi }(x_{n+1}-x^{*})\rangle & \le \underset{n\to \infty }{lim\; sup}\langle f{x}^{*}-x^{*},J_{\phi }(y_n-x^{*})\rangle \hfill \\ & \hspace{1em}+\underset{n\to \infty }{lim\; sup}\langle f{x}^{*}-x^{*},J_{\phi }(x_{n+1}-x^{*})-J_{\phi }(y_n-x^{*})\rangle \hfill \\ & =\underset{n\to \infty }{lim\; sup}\langle f{x}^{*}-x^{*},J_{\phi }(y_n-x^{*})\rangle \le 0.\hfill \end{array}$$

Step 5. We show that ${lim}_{n\to \infty }\parallel x_n-x^{*}\parallel =0$. In fact, it follows from Equation (15) that

$$\begin{array}{cc}\hfill x_{n+1}-x^{*}& ={\beta }_n(f{y}_n-f{x}^{*})+{\gamma }_n(S{y}_n-S{x}^{*})+(1-{\beta }_n-{\gamma }_n)(y_n-x^{*})\hfill \\ & \hspace{1em}+{\beta }_n(f{x}^{*}-x^{*})+{\gamma }_n(S{x}^{*}-x^{*}).\hfill \end{array}$$

Therefore, using inequalities $\parallel y_n-x^{*}\parallel$ $\le \parallel x_n-x^{*}\parallel$, $\parallel fx-fy\parallel \le k\parallel x-y\parallel$, and $\parallel Sx-Sy\parallel \le \parallel x-y\parallel$ and using Lemma 2, we have

$$\begin{array}{cc}\hfill \mathsf{\Phi }(\parallel x_{n+1}-x^{*}\parallel )& \le \mathsf{\Phi }(\parallel {\beta }_n(f{y}_n-f{x}^{*})+{\gamma }_n(S{y}_n-S{x}^{*})+(1-{\beta }_n-{\gamma }_n)(y_n-x^{*})\parallel )\hfill \\ & \hspace{1em}+{\beta }_n\langle f{x}^{*}-x^{*},J_{\phi }(x_{n+1}-x^{*})\rangle +{\gamma }_n\langle S{x}^{*}-x^{*},J_{\phi }(x_{n+1}-x^{*})\rangle \hfill \\ & \le \mathsf{\Phi }({\beta }_{n}k\parallel y_n-x^{*}\parallel +{\gamma }_n\parallel y_n-x^{*}\parallel +(1-{\beta }_n-{\gamma }_n)\parallel y_n-x^{*}\parallel )\hfill \\ & \hspace{1em}+{\beta }_n\langle f{x}^{*}-x^{*},J_{\phi }(x_{n+1}-x^{*})\rangle +{\gamma }_n\langle S{x}^{*}-x^{*},J_{\phi }(x_{n+1}-x^{*})\rangle \hfill \\ & \le \mathsf{\Phi }((1-(1-k){\beta }_n)\parallel x_n-x^{*}\parallel )\hfill \\ & \hspace{1em}+{\beta }_n\langle f{x}^{*}-x^{*},J_{\phi }(x_{n+1}-x^{*})\rangle +{\gamma }_n\langle S{x}^{*}-x^{*},J_{\phi }(x_{n+1}-x^{*})\rangle \hfill \\ & \le (1-(1-k){\beta }_n)\mathsf{\Phi }(\parallel x_n-x^{*}\parallel )+{\beta }_n\langle f{x}^{*}-x^{*},J_{\phi }(x_{n+1}-x^{*})\rangle \hfill \\ & \hspace{1em}+{\gamma }_n\parallel S{x}^{*}-x^{*}\parallel \phi (\parallel x_{n+1}-x^{*}\parallel )\hfill \\ & \le (1-{\lambda }_n)\mathsf{\Phi }(\parallel x_n-x^{*}\parallel )+{\lambda }_n{\delta }_n,\hfill \end{array}$$

(19)

where ${\lambda }_n=(1-k){\beta }_n$ and

$${\delta }_n=\frac{1}{1-k}\left[\langle f{x}^{*}-x^{*},J_{\phi }(x_{n+1}-x^{*})\rangle +\frac{{\gamma }_n}{{\beta }_n}\parallel S{x}^{*}-x^{*}\parallel \phi (\parallel x_{n+1}-x^{*}\parallel )\right].$$

From conditions (ii) and (iii) and from step 4, it is easily seen that ${\sum }_{n=0}^{\infty }{\lambda }_n=\infty$ and ${lim\; sup}_{n\to \infty }{\delta }_n\le 0$. Thus, applying Lemma 3 to Equation (19), we conclude that ${lim}_{n\to \infty }\mathsf{\Phi }(\parallel x_n-x^{*}\parallel )=0$ and, hence, ${lim}_{n\to \infty }\parallel x_n-x^{*}\parallel =0$. This completes the proof. □

Theorem 5.

Let E be a reflexive Banach space having a weakly continuous duality mapping$J_{\phi }$with gauge function φ, and let C be a nonempty closed convex subset of E. Let$T:C\to C$be a continuous pseudocontractive mapping such that$Fix\left(T\right)\ne Ø$, let$S:C\to C$be a nonexpansive mapping, and let$f:C\to C$be a contractive mapping with a contractive coefficient$k\in (0,1)$. For$x_0\in C$, let$\left\{x_n\right\}$be defined by the following iterative scheme:

$$\left\{\begin{array}{c}{x}_n={\alpha }_n{y}_n+(1-{\alpha }_n)T{x}_n\hfill \\ y_n={\beta }_{n}f{x}_{n-1}+{\gamma }_{n}S{x}_{n-1}+(1-{\beta }_n-{\gamma }_n)x_{n-1},\forall n\ge 0,\hfill \end{array}\right.$$

(20)

where$\left\{{\alpha }_n\right\}$,$\left\{{\beta }_n\right\}$, and$\left\{{\gamma }_n\right\}$are three sequences in$(0,1]$satisfying the following conditions:

(i)

${lim}_{n\to \infty }{\alpha }_n=0$;

(ii)

${\sum }_{n=1}^{\infty }{\beta }_n=\infty$;

(iii)

${lim}_{n\to \infty }({\gamma }_n/{\beta }_n)=0$,${\beta }_n+{\gamma }_n\le 1$,$\forall n\ge 0$.

Then,$\left\{x_n\right\}$converges strongly to a fixed point$x^{*}$of T, which is the unique solution of the variational inequality in Equation (16).

Proof. 

First, as in Theorem 4, we put $z_t=tf{z}_t+r_{t}S{z}_t+(1-t-r_t)T{z}_t$. Then, from Theorem 2, it follows that, as $t\to 0$, $z_t$ converges strongly to some fixed point $x^{*}$ of T such that $x^{*}$ is the unique solution in $Fix\left(T\right)$ to the variational inequality in Equation (16).

Now, we divide the proof into several steps.

Step 1. We show that $\left\{x_n\right\}$ is bounded. To this end, let $p\in Fix\left(T\right)$. Then, by Equation (20), we have

$$\begin{array}{cc}\hfill \parallel x_n-p\parallel \phi (\parallel x_n-p\parallel )& =\langle {\alpha }_n{y}_n+(1-{\alpha }_n)T{x}_n-p,J_{\phi }(x_n-p)\rangle \hfill \\ & \le (1-{\alpha }_n)\langle T{x}_n-Tp,J_{\phi }(x_n-p)\rangle +{\alpha }_n\parallel y_n-p\parallel \phi (\parallel x_n-p\parallel )\hfill \\ & \le (1-{\alpha }_n)\parallel x_n-p\parallel \phi (\parallel x_n-p\parallel )+{\alpha }_n\parallel y_n-p\parallel \phi (\parallel y_n-p\parallel )\hfill \end{array}$$

and, hence,

$$\parallel x_n-p\parallel \le \parallel y_n-p\parallel ,\forall n\ge 0.$$

Thus, we obtain

$$\begin{array}{cc}\hfill \parallel x_n-p\parallel & \le \parallel y_n-p\parallel \hfill \\ & \le {\beta }_n\parallel f{x}_{n-1}-p\parallel +{\gamma }_n\parallel S{x}_{n-1}-p\parallel +(1-{\beta }_n-{\gamma }_n)\parallel x_{n-1}-p\parallel \hfill \\ & \le {\beta }_n(\parallel f{x}_{n-1}-fp\parallel +\parallel fp-p\parallel )+{\gamma }_n(\parallel S{x}_{n-1}-Sp\parallel +\parallel Sp-p\parallel )\hfill \\ & \hspace{1em}+(1-{\beta }_n-{\gamma }_n)\parallel x_{n-1}-p\parallel \hfill \\ & \le {\beta }_{n}k\parallel x_{n-1}-p\parallel +{\beta }_n\parallel fp-p\parallel +{\gamma }_n\parallel x_{n-1}-p\parallel +{\gamma }_n\parallel Sp-p\parallel \hfill \\ & \hspace{1em}+(1-{\beta }_n-{\gamma }_n)\parallel x_{n-1}-p\parallel \hfill \\ & =(1-(1-k){\beta }_n)\parallel x_{n-1}-p\parallel +{\beta }_n\parallel fp-p\parallel +{\gamma }_n\parallel Sp-p\parallel .\hfill \end{array}$$

(21)

Since ${lim}_{n\to \infty }({\gamma }_n/{\beta }_n)=0$, we may assume without loss of generality that ${\gamma }_n\le {\beta }_n$ for all $n>0$. Therefore, it follows from Equation (21) that

$$\begin{array}{cc}\hfill \parallel x_n-p\parallel & \le (1-(1-k){\beta }_n)\parallel x_{n-1}-p\parallel +(1-k){\beta }_n·\frac{1}{1-k}(\parallel fp-p\parallel +\parallel Sp-p\parallel )\hfill \\ & \le max\left\{\parallel x_{n-1}-p\parallel ,\frac{1}{1-k}(\parallel fp-p\parallel +\parallel Sp-p\parallel )\right\}.\hfill \end{array}$$

By induction, we derive

$$\parallel x_n-p\parallel \le max\left\{\parallel x_0-p\parallel ,\frac{1}{1-k}(\parallel fp-p\parallel +\parallel Sp-p\parallel )\right\},\forall n\ge 0.$$

This show that $\left\{x_n\right\}$ is bounded and so is $\left\{y_n\right\}$.

Step 2. We show that $\left\{f{x}_n\right\}$, $\left\{S{x}_n\right\}$, and $\left\{T{x}_n\right\}$ are bounded. Indeed, observe that

$$\parallel f{x}_n\parallel \le \parallel f{x}_n-fp\parallel +\parallel fp\parallel \le k\parallel x_n-p\parallel +\parallel fp\parallel$$

and

$$\parallel S{x}_n\parallel \le \parallel S{x}_n-Sp\parallel +\parallel Sp\parallel \le \parallel x_n-p\parallel +\parallel Sp\parallel .$$

Thus, $\left\{f{x}_n\right\}$ and $\left\{S{x}_n\right\}$ are bounded. Since ${lim}_{n\to \infty }{\alpha }_n=0$, there exist $n_0\ge 0$ and $a\in (0,1)$ such that ${\alpha }_n\le a$ for all $n\ge n_0$. Noting that $x_n={\alpha }_n{y}_n+(1-{\alpha }_n)T{x}_n$, we have

$$T{x}_n=\frac{1}{1-{\alpha }_n}{x}_n-\frac{{\alpha }_n}{1-{\alpha }_n}{y}_n$$

and so

$$\parallel T{x}_n\parallel \le \frac{1}{1-{\alpha }_n}\parallel x_n\parallel +\frac{{\alpha }_n}{1-{\alpha }_n}\parallel y_n\parallel \le \frac{1}{1-a}\parallel x_n\parallel +\frac{a}{1-a}\parallel y_n\parallel .$$

Consequently, the sequence $\left\{T{x}_n\right\}$ is also bounded.

Step 3. We show that ${lim\; sup}_{n\to \infty }\langle f{x}^{*}-x^{*},J_{\phi }(x_n-x^{*})\rangle \le 0.$ In fact, from condition (i) and boundedness of $\left\{x_n\right\}$ and $\left\{T{x}_n\right\}$, we get

$$\parallel x_n-T{x}_n\parallel ={\alpha }_n\parallel y_n-T{x}_n\parallel \to 0(n\to \infty ).$$

(22)

Thus, it follows from Equation (22) and Theorem 3 that ${lim\; sup}_{n\to \infty }\langle f{x}^{*}-x^{*},J_{\phi }(x_n-x^{*})\rangle \le 0.$

Step 4. We show that ${lim}_{n\to \infty }\parallel x_n-x^{*}\parallel =0$. In fact, using the equality

$$\begin{array}{cc}\hfill x_n-x^{*}& ={\alpha }_n[{\beta }_n(f{x}_{n-1}-f{x}^{*})+{\gamma }_n(S{x}_{n-1}-S{x}^{*})+(1-{\beta }_n-{\gamma }_n)(x_{n-1}-x^{*})]\hfill \\ & \hspace{1em}\hspace{1em}+{\alpha }_n[{\beta }_n(f{x}^{*}-x^{*})+{\gamma }_n(S{x}^{*}-x^{*})]+(1-{\alpha }_n)(T{x}_n-x^{*})\hfill \end{array}$$

by Equation (20) and the inequalities $\langle Tx-Ty,J_{\phi }(x-y)\rangle \le \parallel x-y\parallel \phi (\parallel x-y\parallel )$$=\mathsf{\Phi }(\parallel x-y\parallel )$, $\parallel fx-fy\parallel \le k\parallel x-y\parallel$, and $\parallel Sx-Sy\parallel \le \parallel x-y\parallel$, from Lemma 2, we derive

$$\begin{array}{cc}\hfill \mathsf{\Phi }(\parallel x_n-x^{*}\parallel )& =\mathsf{\Phi }({\alpha }_n\parallel {\beta }_n(f{x}_{n-1}-f{x}^{*})+{\gamma }_n(S{x}_{n-1}-S{x}^{*})+(1-{\beta }_n-{\gamma }_n)(x_{n-1}-x^{*})\parallel )\hfill \\ & \hspace{1em}+{\alpha }_n{\beta }_n\langle f{x}^{*}-x^{*},J_{\phi }(x_n-x^{*})\rangle +{\alpha }_n{\gamma }_n\langle S{x}^{*}-x^{*},J_{\phi }(x_n-x^{*})\rangle \hfill \\ & \hspace{1em}+(1-{\alpha }_n)\langle T{x}_n-x^{*},J_{\phi }(x_n-x^{*})\rangle \hfill \\ & \le {\alpha }_n\mathsf{\Phi }({\beta }_{n}k\parallel x_{n-1}-x^{*}\parallel +{\gamma }_n\parallel x_{n-1}-x^{*}\parallel +(1-{\beta }_n-{\gamma }_n)\parallel x_{n-1}-x^{*}\parallel )\hfill \\ & \hspace{1em}+{\alpha }_n{\beta }_n\langle f{x}^{*}-x^{*},J_{\phi }(x_n-x^{*})\rangle +{\alpha }_n{\gamma }_n\langle S{x}^{*}-x^{*},J_{\phi }(x_n-x^{*})\rangle \hfill \\ & \hspace{1em}+(1-{\alpha }_n)\parallel x_n-x^{*}\parallel \phi (\parallel x_n-x^{*}\parallel )\hfill \\ & \le {\alpha }_n(1-(1-k){\beta }_n)\mathsf{\Phi }(\parallel x_{n-1}-x^{*}\parallel )\hfill \\ & \hspace{1em}+{\alpha }_n{\beta }_n\langle f{x}^{*}-x^{*},J_{\phi }(x_n-x^{*})\rangle +{\alpha }_n{\gamma }_n\parallel S{x}^{*}-x^{*}\parallel \phi (\parallel x_n-x^{*}\parallel )\hfill \\ & \hspace{1em}+(1-{\alpha }_n)\mathsf{\Phi }(\parallel x_n-x^{*}\parallel ).\hfill \end{array}$$

(23)

By Equation (23), we obtain

$$\begin{array}{cc}\hfill \mathsf{\Phi }(\parallel x_n-x^{*}\parallel )& \le (1-(1-k){\beta }_n)\mathsf{\Phi }(\parallel x_{n-1}-x^{*}\parallel )+{\beta }_n\langle f{x}^{*}-x^{*},J_{\phi }(x_n-x^{*})\rangle \hfill \\ & \hspace{1em}+{\gamma }_n\parallel S{x}^{*}-x^{*}\parallel \phi (\parallel x_n-x^{*}\parallel )\hfill \\ & \le (1-(1-k){\beta }_n)\parallel x_{n-1}-x^{*}\parallel +{\beta }_n\langle f{x}^{*}-x^{*},J_{\phi }(x_n-x^{*})\rangle +{\gamma }_n\parallel S{x}^{*}-x^{*}\parallel M,\hfill \end{array}$$

(24)

where $M>0$ is a constant such that $\phi (\parallel x_n-x^{*}\parallel )\le M$ for all $n\ge 1$. Put ${\lambda }_n=(1-k){\beta }_n$ and

$${\delta }_n=\frac{1}{1-k}\left[\langle f{x}^{*}-x^{*},J_{\phi }(x_n-x^{*})\rangle +\frac{{\gamma }_n}{{\beta }_n}\parallel S{x}^{*}-x^{*}\parallel M\right].$$

From conditions (ii) and (iii) and from step 3, it easily seen that ${\sum }_{n=0}^{\infty }{\lambda }_n=\infty$ and ${lim\; sup}_{n\to \infty }{\delta }_n\le 0$. Since Equation (24) reduces to

$$\mathsf{\Phi }(\parallel x_n-x^{*}\parallel )\le (1-{\lambda }_n)\mathsf{\Phi }(\parallel x_{n-1}-x^{*}\parallel )+{\lambda }_n{\delta }_n,$$

(25)

applying Lemma 3 to Equation (25), we conclude that ${lim}_{n\to \infty }\mathsf{\Phi }(\parallel x_n-x^{*}\parallel )=0$ and, hence, ${lim}_{n\to \infty }\parallel x_n-x^{*}\parallel =0$. This completes the proof. □

Remark 2.

(1) 

Theorem 3 develops Theorem 2.3 of Ceng et al. [17] in the following aspects:

(i) 

The space is replaced by the space having a weakly continuous duality mapping$J_{\phi }$with gauge function φ.

(ii) 

The Lipischiz strongly pseudocontractive mapping f in Theorem 2.3 in Reference [17] is replaced by a bounded continuous strongly pseudocontractive mapping f in Theorem 3.

(2) 

Theorem 4 complements Theorem 3.1 as well as Theorem 3.4 of Ceng et al. [17] in a reflexive Banach space having a weakly continuous duality mapping$J_{\phi }$with gauge function φ.

(3) 

Theorem 5 also means that Theorem 3.2 as well as Theorem 3.5 of Ceng et al. [17] hold in a reflexive Banach space having a weakly continuous duality mapping$J_{\phi }$with gauge function φ.

(4) 

Whenever$S=I$and${\gamma }_n=0$for all$n\ge 0$in Theorem 5, it is easily seen that Theorem 3.1 Theorem 3.4 of Song and Chen [11] hold in a reflexive Banach space which has a weakly continuous duality mapping$J_{\phi }$with gauge function φ.