Systems of Variational Inequalities with Nonlinear Operators

Abstract

In this work, we concern ourselves with the problem of solving a general system of variational inequalities whose solutions also solve a common fixed-point problem of a family of countably many nonlinear operators via a hybrid viscosity implicit iteration method in 2 uniformly smooth and uniformly convex Banach spaces. An application to common fixed-point problems of asymptotically nonexpansive and pseudocontractive mappings and variational inequality problems for strict pseudocontractive mappings is also given in Banach spaces.

1. Introduction

Let E be a real Banach space whose topological dual space is denoted by $E^{*}$. Recall that the normalized duality mapping $J:E\to 2^{E^{*}}$ is defined by

$$J\left(x\right)=\{\phi \in E^{*}:\parallel \phi \parallel =\parallel x\parallel ,\langle x,\phi \rangle =\parallel x{\parallel }^2\},\forall x\in E,$$

where $\langle ·,·\rangle$ is the duality pair on E and $E^{*}$. J is single-valued in a smooth Banach space. In the sequel, we shall denote by j the single-valued duality mapping, that is, $j\left(x\right)\in J\left(x\right)$. Let C be a convex closed set in E. A mapping $f:C\to C$ is said to be $\delta$-Lipschitzian on C if $\delta \in (0,+\infty )$ and $\parallel f\left(x\right)-f\left(y\right)\parallel \le \delta \parallel x-y\parallel$ for all $x,y\in C$. If $\delta <1$, then f is called a $\delta$-contraction mapping or a contraction mapping with coefficient $\delta$. Each contraction $f:C\to C$ has a unique fixed point from the well known the Banach contractive principal. A mapping $f:C\to C$ is said to be nonexpansive if it is Lipschitzian with $\delta =1$. We use $\mathrm{Fix}\left(f\right)$ to denote the set of fixed points of f, i.e., $\mathrm{Fix}\left(f\right)=\{x\in C:f(x)=x\}$. Moreover, a mapping $T:C\to C$ is said to be asymptotically nonexpansive [1] if there exists a sequence $\left\{{\theta }_n\right\}\subset [0,+\infty )$ with ${lim}_{n\to \infty }{\theta }_n=0$ such that

$$\parallel T^{n}x-T^{n}y\parallel \le \parallel x-y\parallel +{\theta }_n\parallel x-y\parallel ,\forall x,y\in C,\forall n\ge 0.$$

If

$$\underset{n\to \infty }{lim\; sup}(\underset{x,y\in C}{sup}(\parallel T^{n}x-T^{n}y\parallel -\parallel x-y\parallel ))\le 0,$$

(1)

and T enjoys the continuity, then T is called an asymptotically nonexpansive mapping in the intermediate sense; see [2]. Throughout this paper, we assume

$$c_n:=max\{0,\underset{x,y\in C}{sup}(\parallel T^{n}x-T^{n}y\parallel -\parallel x-y\parallel \left)\right\}.$$

(2)

Hence, $c_n\ge 0\forall n\ge 0,c_n\to 0(n\to \infty )$, and the definition is reduced to

$$\parallel T^{n}x-T^{n}y\parallel \le \parallel x-y\parallel +c_n,\forall x,y\in C,\forall n\ge 0.$$

Recall that a mapping T with domain $D\left(T\right)$ and range $R\left(T\right)$ in E is called pseudocontractive if the inequality holds

$$\parallel x-y\parallel \le \parallel r\left(\right(I-T)x-(I-T\left)y\right)+(x-y)\parallel ,\forall x,y\in D\left(T\right),\forall r>0.$$

From a result of Kato [3], we know that the notion of pseudocontraction is equivalent to the one that for each $x,y\in D\left(T\right)$, there exists $j(x-y)\in J(x-y)$ such that

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

It is well known that the class of pseudocontractive mappings is a crucial generation of nonexpansive mappings. Moreover, focus on pseudocontractive mappings are also from their relation with the class of accretive mappings in Banach spaces (monotone in Hilbert spaces). A mapping A with domain $D\left(A\right)$ and range $R\left(A\right)$ in E is called accretive if for each $x,y\in D\left(T\right)$, there exists $j(x-y)\in J(x-y)$ such that

$$\langle Ax-Ay,j(x-y)\rangle \ge 0.$$

It will be called a monotone mapping if the space is Hilbert. If for each $x,y\in D\left(A\right)$, there exists $j(x-y)\in J(x-y)$ such that

$${\alpha \parallel Ax-Ay\parallel }^2\le \langle Ax-Ay,j(x-y)\rangle \mathrm{for}\mathrm{some}\alpha >0,$$

then A is called $\alpha$-inverse-strongly accretive.

Recently, fixed/zero points of pseudocontraction/accretive operators were investigated by many authors for solving various convex optimization problems; see [4,5,6,7,8,9,10,11,12,13] and the references therein.

Let E be a smooth Banach space. Let $B_1,B_2$ be two non-self-mappings from C to E. The general system of variational inequalities (GSVI) is to find $(x^{*},y^{*})\in C\times C$ such that

$$\left\{\begin{array}{c}\langle {\mu }_1{B}_1{y}^{*}-y^{*}+x^{*},j(x-x^{*})\rangle \ge 0,\forall x\in C,\hfill \\ \langle {\mu }_2{B}_2{x}^{*}-x^{*}+y^{*},j(x-y^{*})\rangle \ge 0,\forall x\in C,\hfill \end{array}\right.$$

(3)

where ${\mu }_1$ and ${\mu }_2$ are two positive coefficients.

In particular, if $B_1=B_2=B$, then problem (3) reduces to the following system of variational inequalities (SVI) in Banach spaces:

Find $(x^{*},y^{*})\in C\times C$ such that

$$\left\{\begin{array}{c}\langle {\mu }_{1}B{y}^{*}-y^{*}+x^{*},j(x-x^{*})\rangle \ge 0,\forall x\in C,\hfill \\ \langle {\mu }_{2}B{x}^{*}-x^{*}+y^{*},j(x-y^{*})\rangle \ge 0,\forall x\in C.\hfill \end{array}\right.$$

(4)

Furthermore, if $x^{*}=y^{*}$, then we obtain the following variational inequality (VI) in Banach spaces: Find $x^{*}\in C$ such that

$$\langle \mu B{x}^{*},j(x-x^{*})\rangle \ge 0,\forall x\in C.$$

(5)

Let $\mathrm{VI}(C,B)$ denote the set of solutions to problem (5). Whenever $E=H$ a real Hilbert space, it is easy to see that the GSVI (3) reduces to the following problem of finding $(x^{*},y^{*})\in C\times C$ such that

$$\left\{\begin{array}{c}\langle {\mu }_1{B}_1{y}^{*}-y^{*}+x^{*},x-x^{*}\rangle \ge 0,\forall x\in C,\hfill \\ \langle {\mu }_2{B}_2{x}^{*}-x^{*}+y^{*},x-y^{*}\rangle \ge 0,\forall x\in C,\hfill \end{array}\right.$$

(6)

which is called a GSVI in Hilbert spaces. In [11], the GSVI (6) was transformed into a fixed-point problem by Ceng, Wang and Yao in the following way.

Lemma 1.

[14] For chosen $x^{*},y^{*}\in C$, $x^{*}$-$y^{*}$ is a solution of GSVI (1.6) if and only if $x^{*}\in \mathrm{GSVI}(C,B_1,B_2)$, where $\mathrm{GSVI}(C,B_1,B_2)$ is the fixed-point set of the mapping $G:=P_C(P_C(I-\eta B_2)-\rho B_1{P}_C(I-\eta B_2))$, and $y^{*}=P_C(I-\eta B_2)x^{*}$.

Recently, many authors studied problems (3)–(6) via projection-based methods in Hilbert or Banach spaces; see [15,16,17,18,19,20,21,22] and the references therein. In this paper, we introduce a hybrid viscosity implicit iteration method that is based on Korpelevich’s extragradient method, the viscosity approximation method and the Mann iteration method for finding a common solution of the GSVI (3) for two inverse-strongly accretive mappings, a common fixed-point problem (CFPP) of a countable family of uniformly Lipschitzian pseudocontractive mappings and an asymptotically nonexpansive mapping in the intermediate sense. We prove the strong convergence of the proposed method to a common solution of the GSVI (3) and the CFPP, which solves a certain variational inequality on their common solution set in 2-uniformly smooth and uniformly convex Banach spaces. Additionally, we give an application to solve CFPPs of asymptotically nonexpansive and pseudocontractive mappings, and variational inequality problems for strict pseudocontractive mappings in Banach spaces.

2. Preliminaries

Throughout this paper we write $x_n\rightharpoonup x$ (respectively, $x_n\to x$) to indicate that the sequence $\left\{x_n\right\}$ converges weakly (respectively, strongly) to x. Without loss of generality, we assume that E is a real Banach space and the dual will be presented by $E^{*}$ in this paper.

Definition 1.

Let ${\left\{S_n\right\}}_{n=0}^{\infty }$ be a vector sequence of pseudocontractive continuous self-mappings on C, a convex closed convex subset of Banach space E. Recall that ${\left\{S_n\right\}}_{n=0}^{\infty }$ is said to be a countable family of ℓ-uniformly Lipschitzian pseudocontractive self-mappings provided that there exists a constant $\ell >0$ such that each $S_n$ is a ℓ-Lipschitz continuous mapping.

In a smooth Banach space E, an operator A is said to be strongly positive if there exists a constant $\overline{\gamma }>0$ with the property

$$\parallel aI-bA\parallel =\underset{\parallel x\parallel \le 1}{sup}\left|\langle (aI-bA)x,j\left(x\right)\rangle \right|,\langle Ax,j\left(x\right)\rangle \ge \overline{\gamma }{\parallel x\parallel }^{2}a\in [0,1],b\in [-1,1],$$

where I is the identity mapping and $j(·)$ is the single-valued normalized duality mapping.

Recall that a Banach space E is said to be strictly convex if for any $x,y\in \{x\in E:\parallel x\parallel =1\}$, $x\ne y\Rightarrow \parallel {\displaystyle \frac{x+y}{2}}\parallel <1$. It is also said to be uniformly convex if for each $ϵ\in (0,2]$, there exists $\delta >0$ such that for any $x,y\in \{x\in E:\parallel x\parallel =1\}$, $\parallel x-y\parallel \ge ϵ\Rightarrow \parallel x+y\parallel \le 2-2\delta$. Clearly, if E is uniformly convex, then it is strictly convex. A Banach space E is said to have a Gâteaux differentiable norm if the limit

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

exists for each $x,y\in \{x\in E:\parallel x\parallel =1\}$ and in this case we call E smooth; E is said to have a uniformly Gâteaux differentiable norm if for each $y\in \{x\in E:\parallel x\parallel =1\}$, the above limit is attained uniformly for $x\in \{x\in E:\parallel x\parallel =1\}$. Moreover, it is said to have a uniformly Fréchet differentiable norm if the above limit is attained uniformly for $x,y\in \{x\in E:\parallel x\parallel =1\}$ and in this case we call E uniformly smooth. The norm of E is said to be the Fréchet differentiable if for each $x\in \{x\in E:\parallel x\parallel =1\}$, the above limit is attained uniformly for $y\in \{x\in E:\parallel x\parallel =1\}$. The modulus of smoothness of E is defined by

$$\varrho \left(\tau \right)=sup\{{\displaystyle \frac{(\parallel x+y\parallel +\parallel x-y\parallel )-2}{2}}:x,y\in E,\parallel x\parallel =1,\parallel y\parallel =\tau \},$$

where $\varrho :[0,\infty )\to [0,\infty )$ is a function. It is known that E is uniformly smooth if and only if ${lim}_{\tau \to 0}{\displaystyle \frac{\varrho \left(\tau \right)}{\tau }}=0$. Let q be a fixed real number with $1<q\le 2$. A Banach space E is said to be q-uniformly smooth if there exists a constant $\kappa >0$ such that ${\displaystyle \frac{\varrho \left(\tau \right)}{\kappa }}\le {\tau }^q$ for all $\tau >0$. From [23], we know the following relation. Let q be a fixed number with $1<q\le 2$ and E be a Banach space. Then E is q-uniformly smooth if and only if there exists a constant $c>0$ such that

$${\parallel x\parallel }^q+{\parallel \kappa y\parallel }^q\ge {\displaystyle \frac{{\parallel x+y\parallel }^q+{\parallel x-y\parallel }^q}{2}},\forall x,y\in E.$$

The best constant $\kappa$ in the above inequality is called the q-uniformly smooth constant of E; see [23] for more details. In addition, no Banach space is q-uniformly smooth for $q>2$; see [24] for more details. If E be a 2-uniformly smooth Banach space. Then

$${\parallel x+y\parallel }^2-{\parallel x\parallel }^2\le 2\langle y,j\left(x\right)\rangle +2{\parallel \kappa y\parallel }^2,\forall x,y\in E,$$

where $\kappa$ is the 2-uniformly smooth constant of E.

In particular, if E is a Hilbert space, then the duality pairing $\langle ·,·\rangle$ reduces to the inner product, $j=I$ the identity mapping of E, and $\kappa =\sqrt{2}/2$.

For $q>1$, the generalized duality mapping $J_q:E\to 2^{E^{*}}$ is defined by

$$J_q\left(x\right)=\{\phi \in E^{*}:\parallel \phi \parallel ={\parallel x\parallel }^{q-1},\langle x,\phi \rangle ={\parallel x\parallel }^q\},\forall x\in E.$$

In particular, $J=J_2$ is called the normalized duality mapping. It is known that $J\left(x\right)={\displaystyle \frac{J_q\left(x\right)}{{\parallel x\parallel }^{q-2}}}$ for all $x\in E$. If E is a Hilbert space, then $J=I$ (the identity mapping). Recall that the following statements hold:

(1)

if E is smooth, then J is norm-to-weak* continuous single-valued on E;

(2)

if E is uniformly smooth, then J is norm-to-norm uniformly continuous single-valued on bounded subsets of E;

(3)

if E has a uniformly Gáteaux differentiable norm, then J is norm-to-weak* uniformly continuous single-valued on bounded subsets of E;

Proposition 1.

(see [25]). Let C be a convex nonempty closed set in a Banach space E. Let $S_0,S_1,\dots$ be a sequence of mappings of C into itself. Suppose ${\sum }_{n=1}^{\infty }sup\{\parallel S_{n}x-S_{n-1}x\parallel :x\in C\}<\infty$. For each $y\in C,$$\left\{S_{n}y\right\}$ converges in norm to some point of C. Moreover, let S be a mapping defined by $Sy={lim}_{n\to \infty }{S}_{n}y$ for all $y\in C$. $\{\parallel Sx-S_{n}x\parallel :x\in C\}\to 0$ as $n\to \infty$.

Proposition 2.

(see [26]). Let C be a convex closed set in a Banach space E and $T:C\to C$ be a strong continuous pseudocontraction mapping. Then, T has a fixed point. Indeed, it is the unique fixed point in C for T.

Let D be a nonempty set in C and let $\mathrm{\Pi }$ be a mapping from C to D. Then $\mathrm{\Pi }$ is said to be a sunny if $\mathrm{\Pi }\left[\right(1-t\left)\mathrm{\Pi }\right(x)+tx]=\mathrm{\Pi }\left(x\right)$, when $(1-t)\mathrm{\Pi }\left(x\right)+tx\in C$ for all $x\in C$ and $t\ge 0$. A mapping $\mathrm{\Pi }$ of C into itself is called a retraction if ${\mathrm{\Pi }}^2=\mathrm{\Pi }$. If a mapping $\mathrm{\Pi }$ of C into itself is a retraction, then $\mathrm{\Pi }\left(z\right)=z$ for each $z\in R\left(\mathrm{\Pi }\right)$, where $R\left(\mathrm{\Pi }\right)$ is the range of $\mathrm{\Pi }$. A subset D of C is called a sunny nonexpansive retract of C if there exists a sunny nonexpansive retraction from C onto D.

In a smooth Banach space E, a duality mapping J is said to be weakly sequentially continuous [27], if for each $\left\{x_n\right\}\subset E$ with $x_n\rightharpoonup x$, then $\left\{j\left(x_n\right)\right\}$ converges weakly* to $j\left(x\right)$. In [27], Gossez and Lami Dozo showed that a space with a weakly continuous duality mapping satisfies Opial’s condition. Conversely, we know from [28] that if a space satisfies Opial’s condition and has a uniformly Gáteaux differentiable norm, then it has a weakly continuous duality mapping.

Proposition 3.

(see [29]). Let C be a nonempty closed convex subset of a smooth Banach space E, D be a nonempty subset of C and Π be a retraction of C onto D. Then the following are equivalent:

(i) 

Π is sunny and nonexpansive;

(ii) 

$\langle x-y,j(\mathrm{\Pi }\left(x\right)-\mathrm{\Pi }\left(y\right))\rangle \ge {\parallel \mathrm{\Pi }\left(x\right)-\mathrm{\Pi }\left(y\right)\parallel }^2,\forall x,y\in C$;

(iii) 

$\langle x-\mathrm{\Pi }\left(x\right),j(y-\mathrm{\Pi }(x\left)\right)\rangle \le 0,\forall x\in C,y\in D$.

If E is a Hilbert space, then a sunny nonexpansive retraction ${\mathrm{\Pi }}_C$ of E onto C coincides with the nearest projection of E onto C and it is well known that if C is a convex closed set in a reflexive Banach space E with a uniformly Gáteaux differentiable norm and D is a nonexpansive retract of C, then it is a sunny nonexpansive retract of C; see, e.g., [30,31] and the references therein.

To prove our main results, we need to use some lemmas in the sequel. The following Lemma is an immediate consequence of the subdifferential inequality of the function ${\displaystyle \frac{1}{2}}{\parallel ·\parallel }^2$.

Lemma 2.

[32] Let E be a real Banach space and J be the normalized duality mapping on E. Then for any given $x,y\in E$, the following inequality holds:

$${\parallel x+y\parallel }^2-{\parallel x\parallel }^2\le 2\langle y,j(x+y)\rangle ,\forall j(x+y)\in J(x+y).$$

If C is a convex closed set in a smooth Banach space E and ${\mathrm{\Pi }}_C$ a sunny nonexpansive retraction from E onto C, we have

$$\mathrm{VI}(C,B)=\mathrm{Fix}\left({\mathrm{\Pi }}_C(I-\lambda B)\right).$$

where $B:C\to E$ be an accretive mapping and $\lambda >0$,

Using Proposition 3, we immediately obtain the following lemmas.

Lemma 3.

Let C be a nonempty closed convex subset of a smooth Banach space E and $B_1,B_2:C\to E$ be two nonlinear mappings. Let ${\mathrm{\Pi }}_C$ be a sunny nonexpansive retraction from E onto C. For given $x^{*},y^{*}\in C,(x^{*},y^{*})$ is a solution of the GSVI (3) if and only if $x^{*}\in \mathrm{GSVI}(C,B_1,B_2)$ where $\mathrm{GSVI}(C,B_1,B_2)$ is the set of fixed points of the mapping

$$G:={\mathrm{\Pi }}_C({\mathrm{\Pi }}_C(I-{\mu }_2{B}_2)-{\mu }_1{B}_1{\mathrm{\Pi }}_C(I-{\mu }_2{B}_2))$$

and $y^{*}={\mathrm{\Pi }}_C(I-{\mu }_2{B}_2)x^{*}$.

Lemma 4.

Let C be a nonempty closed convex subset of a 2-uniformly smooth Banach space E. Let the mapping $A:C\to E$ be α-inverse-strongly accretive. Then, for any given $\lambda \ge 0$,

$${\parallel (I-\lambda A)x-(I-\lambda A)y\parallel }^2{-\parallel x-y\parallel }^2\le 2\lambda ({\kappa }^2\lambda -\alpha ){\parallel Ax-Ay\parallel }^2.$$

In particular, if $0\le \lambda {\kappa }^2\le \alpha$, then $I-\lambda A$ is a nonexpansive operator. Let ${\mathrm{\Pi }}_C$ be a sunny nonexpansive retraction from E onto C. Let the mappings $B_1,B_2:C\to E$ be α-inverse-strongly accretive and β-inverse-strongly accretive, respectively. Let the mapping $G:C\to C$ be defined as $G:={\mathrm{\Pi }}_C(I-{\mu }_1{B}_1){\mathrm{\Pi }}_C(I-{\mu }_2{B}_2)$. If $0\le {\mu }_1{\kappa }^2\le \alpha$ and $0\le {\mu }_2{\kappa }^2\le \beta$, then $G:C\to C$ is nonexpansive.

Let C be a nonempty closed convex subset of a uniformly convex Banach space E and $T:C\to C$ be an asymptotically nonexpansive mapping in the intermediate sense. Given any bounded subset $\mathcal{K}\subset C$. For every $\epsilon >0$ and every integer $n\ge 2$ there exist an integer $N_{\epsilon }\ge 1$ and ${\delta }_{\epsilon }>0$, where both $N_{\epsilon }$ and ${\delta }_{\epsilon }$ are independent of n, such that if $k\ge N_{\epsilon },z_1,z_2,\dots ,z_n\in \mathcal{K}$ and if

$$\parallel z_i-z_j\parallel -\parallel T^k{z}_i-T^k{z}_j\parallel \le {\delta }_{\epsilon }$$

for $1\le i,j\le n$, then

$$\parallel T^k(\sum _{i=1}^n{\lambda }_i{z}_i)-\sum _{i=1}^n{\lambda }_i{T}^k{z}_i\parallel <\epsilon$$

for all $\lambda =({\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n)$ such that ${\lambda }_i\ge 0$ for $i=1,2,\dots ,n$ and ${\sum }_{i=1}^n{\lambda }_i=1$; see ([33], Lemma 4) for details.

From the above results, we know that if ${\left\{x_m\right\}}_{m=0}^{\infty }$ is a sequence in C converging weakly to x and if ${lim}_{m\to \infty }\parallel x_m-T{x}_m\parallel =0$, then $Tx=x$, where $T:C\to C$ is a uniformly continuous self-mapping on C, which is asymptotically nonexpansive in the intermediate sense.

Lemma 5.

(see [34]). Let E be a smooth and uniformly convex Banach space, and let $r>0$. Then there exists a strictly increasing, continuous, and convex function $g:[0,2r]\to \mathbf{R},g\left(0\right)=0$ such that

$$g(\parallel x-y\parallel )+2\langle x,j\left(y\right)\rangle \le {\parallel x\parallel }^2+{\parallel y\parallel }^2,\forall x,y\in \{x\in E:\parallel x\parallel \le r\}.$$

Lemma 6.

(see [35]). Let E be a reflexive Banach space, C be a convex nonempty, closed subset of E, and $T:C\to E$ be a nonexpansive mapping. Suppose that E admits a weakly sequentially continuous duality mapping. Then the mapping $I-T$ is demiclosed on C, where I is the identity mapping.

Lemma 7.

(see [36]). Let $\left\{a_n\right\}$ be a sequence of nonnegative real numbers satisfying

$$a_{n+1}\le a_n+s_n{t}_n+{\nu }_n-s_n{a}_n,\forall n\ge 0,$$

where $\left\{s_n\right\},\left\{t_n\right\}$ and $\left\{{\nu }_n\right\}$ satisfy the conditions:

(i) 

${lim\; sup}_{n\to \infty }{t}_n\le 0$;

(ii) 

$\left\{s_n\right\}\subset [0,1]$ and ${\sum }_{n=0}^{\infty }{s}_n=\infty$;

(iii) 

${\nu }_n\ge 0,\forall n\ge 0$, and ${\sum }_{n=0}^{\infty }{\nu }_n<\infty$.

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

3. Main Results

In this section, we suggest and analyze a hybrid viscosity implicit iteration method for solving the GSVI (3) with the hierarchical variational inequality (HVI) constraint for countably many uniformly Lipschitzian pseudocontractive mappings and an asymptotically nonexpansive mapping in the intermediate sense in a 2-uniformly smooth and uniformly convex Banach space.

Theorem 1.

Let C be a convex closed set in a 2-uniformly smooth and uniformly convex Banach space E which admits a weakly sequentially continuous duality mapping. Let ${\mathrm{\Pi }}_C$ be a sunny nonexpansive retraction from E onto C. Let the mappings $B_1,B_2:C\to E$ be α-inverse-strongly accretive and β-inverse-strongly accretive, respectively. Let $f:C\to C$ be a contraction mapping with coefficient $\gamma \in [0,1)$ and $F:E\to E$ be a strongly positive linear bounded operator with the coefficient $\overline{\gamma }$ such that $0<\gamma <\overline{\gamma }\theta$ and $0<\theta \le {\parallel F\parallel }^{-1}$. Let $T:C\to C$ be uniformly continuous and asymptotically nonexpansive mapping in the intermediate sense, and ${\left\{S_n\right\}}_{n=0}^{\infty }$ be a countable family of ℓ-uniformly Lipschitzian pseudocontractive self-mappings on C such that $\mathrm{\Omega }:={\bigcap }_{n=0}^{\infty }\mathrm{Fix}\left(S_n\right)\cap \mathrm{GSVI}(C,B_1,B_2)\cap \mathrm{Fix}\left(T\right)\ne \varnothing$ where $\mathrm{GSVI}(C,B_1,B_2)$ is the fixed-point set of the mapping $G:={\mathrm{\Pi }}_C({\mathrm{\Pi }}_C(I-{\mu }_2{B}_2)-{\mu }_1{B}_1{\mathrm{\Pi }}_C(I-{\mu }_2{B}_2))$ with $0<{\mu }_1{\kappa }^2<\alpha$ and $0<{\mu }_2{\kappa }^2<\beta$ for κ the 2-uniformly smooth constant of E. Assume that ${\sum }_{n=0}^{\infty }{c}_n<\infty$, where $c_n$ is defined by (2). For arbitrarily given $x_0\in C$, let $\left\{x_n\right\}$ be the sequence generated by

$$\left\{\begin{array}{c}{u}_n={\beta }_n(x_n-S_n{u}_n)+S_n{u}_n,\hfill \\ z_n={\mathrm{\Pi }}_C(u_n-{\mu }_2{B}_2{u}_n),\hfill \\ y_n={\mathrm{\Pi }}_C(z_n-{\mu }_1{B}_1{z}_n),\hfill \\ x_{n+1}={\mathrm{\Pi }}_C[(T^n{y}_n-{\alpha }_n\theta F{T}^n{y}_n)+{\alpha }_{n}f\left(x_n\right)],\forall n\ge 0,\hfill \end{array}\right.$$

(7)

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

(i) 

${lim}_{n\to \infty }{\alpha }_n=0$, ${\sum }_{n=1}^{\infty }|{\alpha }_n-{\alpha }_{n-1}|<\infty$ and ${\sum }_{n=0}^{\infty }{\alpha }_n=\infty$,

(ii) 

$0<{lim\; inf}_{n\to \infty }{\beta }_n\le {lim\; sup}_{n\to \infty }{\beta }_n<1$ and ${\sum }_{n=1}^{\infty }|{\beta }_n-{\beta }_{n-1}|<\infty$.

Assume that ${\sum }_{n=1}^{\infty }{sup}_{x\in D}\parallel S_{n}x-S_{n-1}x\parallel <\infty$ for any bounded subset D of C, and let S be a mapping of C into itself defined by $Sx={lim}_{n\to \infty }{S}_{n}x$ for all $x\in C$, and suppose that $\mathrm{Fix}\left(S\right)={\bigcap }_{n=0}^{\infty }\mathrm{Fix}\left(S_n\right)$. If ${\sum }_{n=0}^{\infty }\parallel T^{n+1}{y}_n-T^n{y}_n\parallel <\infty$, then $\left\{x_n\right\}$ converges strongly to $x^{*}\in \mathrm{\Omega }$. In this case,

(a) 

$x^{*}\in \mathrm{\Omega }$ solves the VI: $\langle f\left(x^{*}\right)-\theta F\left(x^{*}\right),j(x-x^{*})\rangle \le 0,\forall x\in \mathrm{\Omega }$;

(b) 

$(x^{*},y^{*})$ is a solution of GSVI (1.3) with $y^{*}={\mathrm{\Pi }}_C(I-{\mu }_2{B}_2)x^{*}$.

Proof. 

First of all, from ${\alpha }_n\to 0(n\to \infty )$, we may assume, without loss of generality, that ${\alpha }_n\le {\displaystyle \frac{{\parallel F\parallel }^{-1}}{\theta }}$ for all $n\ge 0$. Since F is strongly positive linear bounded, it follows that $1-{\alpha }_n\overline{\gamma }\theta \ge \parallel I-{\alpha }_n\theta F\parallel .$ Taking into account that $\left\{{\beta }_n\right\}$ is bounded away from 0 and 1, we may assume that $\left\{{\beta }_n\right\}\subset [a,b]\subset (0,1)$ for some $a,b\in (0,1)$. Please note that the mapping $G:C\to C$ is defined as $G:={\mathrm{\Pi }}_C({\mathrm{\Pi }}_C(I-{\mu }_2{B}_2)-{\mu }_1{B}_1{\mathrm{\Pi }}_C(I-{\mu }_2{B}_2))$, where $0<{\mu }_1{\kappa }^2\le \alpha$ and $0<{\mu }_2{\kappa }^2\le \beta$ for $\kappa$ the 2-uniformly smooth constant of E. Therefore, by Lemma 3, we obtain that G is nonexpansive. It is easy to see that for each $n\ge 0$ there exists a unique element $u_n\in C$ such that

$$u_n={\beta }_n(x_n-S_n{u}_n)+S_n{u}_n.$$

(8)

As a matter of fact, consider the mapping

$$F_{n}x={\beta }_n(x_n-S_{n}x)+S_{n}x,\forall x\in C.$$

Since $S_n:C\to C$ is a continuous pseudocontraction mapping, we deduce that

$$(1-{\beta }_n){\parallel x-y\parallel }^2\ge (1-{\beta }_n)\langle S_{n}x-S_{n}y,j(x-y)\rangle =\langle F_{n}x-F_{n}y,j(x-y)\rangle ,\forall x,y\in C.$$

Also, from $\left\{{\beta }_n\right\}\subset [a,b]\subset (0,1)$ we get $0<1-{\beta }_n<1$ for all $n\ge 0$. Thus, $F_n$ is a continuous and strong pseudocontraction mapping of C into itself. By Proposition 2, we know that for each $n\ge 0$ there exists a unique element $u_n\in C$, satisfying (8). Therefore, it can be readily seen that the hybrid viscosity implicit iterative scheme (7) can be rewritten as

$$\left\{\begin{array}{c}{u}_n={\beta }_n(x_n-S_n{u}_n)+S_n{u}_n,\hfill \\ y_n=G{u}_n,x_{n+1}={\mathrm{\Pi }}_C[(T^n{y}_n-{\alpha }_n\theta F{T}^n{y}_n)+{\alpha }_{n}f\left(x_n\right)],\forall n\ge 0,\hfill \end{array}\right.$$

(9)

Next, we divide the rest of the proof into several steps.

Step 1. We claim that $\left\{x_n\right\},\left\{y_n\right\},\left\{z_n\right\},\left\{u_n\right\},\left\{f\left(x_n\right)\right\},\left\{T^n{y}_n\right\}$ and $\left\{F\left(T^n{y}_n\right)\right\}$ are bounded vector sequences. Indeed, take an element $p\in \mathrm{\Omega }={\bigcap }_{n=0}^{\infty }\mathrm{Fix}\left(S_n\right)\cap \mathrm{GSVI}(C,B_1,B_2)\cap \mathrm{Fix}\left(T\right)$ arbitrarily. Then we have $S_{n}p=p$, $Gp=p$ and $Tp=p$. Since each $S_n:C\to C$ is a pseudocontraction mapping, it follows that

$$\begin{array}{cc}\parallel p-u_n{\parallel }^2\hfill & =\langle {\beta }_n(p-x_n)+(1-{\beta }_n)(p-S_n{u}_n),j(p-u_n)\rangle \hfill \\ & =(1-{\beta }_n)\langle p-S_n{u}_n,j(p-u_n)\rangle +{\beta }_n\langle p-x_n,j(p-u_n)\rangle \hfill \\ & \le (1-{\beta }_n)\parallel p-u_n{\parallel }^2+{\beta }_n\parallel x_n-p\parallel \parallel p-u_n\parallel ,\hfill \end{array}$$

which hence yields

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

(10)

Then we observe

$$\parallel y_n-p\parallel =\parallel G{u}_n-p\parallel \le \parallel u_n-p\parallel \le \parallel x_n-p\parallel .$$

(11)

Combining (9) and (11), we have

$$\begin{array}{cc}\parallel x_{n+1}-p\parallel \hfill & \le \parallel {\alpha }_n(f\left(x_n\right)-\theta F\left(p\right))+(I-{\alpha }_n\theta F)T^n{y}_n-(I-{\alpha }_n\theta F)p\parallel \hfill \\ & \le {\alpha }_n\gamma \parallel x_n-p\parallel +{\alpha }_n\parallel f\left(p\right)-\theta F\left(p\right)\parallel +(1-{\alpha }_n\overline{\gamma }\theta )\parallel T^n{y}_n-p\parallel \hfill \\ & \le {\alpha }_n\gamma \parallel x_n-p\parallel +{\alpha }_n\parallel f\left(p\right)-\theta F\left(p\right)\parallel +(1-{\alpha }_n\overline{\gamma }\theta )(\parallel y_n-p\parallel +c_n)\hfill \\ & \le c_n+{\alpha }_n\parallel f\left(p\right)-\theta F\left(p\right)\parallel +{\alpha }_n\gamma \parallel x_n-p\parallel +(1-{\alpha }_n\overline{\gamma }\theta )\parallel x_n-p\parallel \hfill \\ & \le c_n+max\{\parallel x_n-p\parallel ,{\displaystyle \frac{\parallel f\left(p\right)-\theta F\left(p\right)\parallel }{\overline{\gamma }\theta -\gamma }}\}.\hfill \end{array}$$

By induction, we get

$$\parallel x_n-p\parallel \le \sum _{n=0}^{\infty }{c}_n+max\{\parallel x_0-p\parallel ,{\displaystyle \frac{\parallel f\left(p\right)-\theta F\left(p\right)\parallel }{\overline{\gamma }\theta -\gamma }}\},\forall n\ge 0.$$

It immediately follows that $\left\{x_n\right\}$ is bounded, and so are the sequences $\left\{y_n\right\},\left\{u_n\right\},\left\{f\left(x_n\right)\right\},\left\{T^n{y}_n\right\}$ and $\left\{F\left(T^n{y}_n\right)\right\}$ (due to (10), (11) and the Lipschitz continuity of f). Taking into account that $\left\{S_n\right\}$ is ℓ-uniformly Lipschitzian on C, we know that

$$\ell \parallel u_n-p\parallel +\parallel p\parallel \ge |S_n{u}_n-p\parallel +\parallel p\parallel \ge \parallel S_n{u}_n\parallel ,$$

which implies that $\left\{S_n{u}_n\right\}$ is bounded. In addition, from Lemma 3 and $p\in \mathrm{\Omega }\subset \mathrm{GSVI}(C,B_1,B_2)$, it also follows that $(p,q)$ is a solution of GSVI (3) where $q={\mathrm{\Pi }}_C(p-{\mu }_2{B}_{2}p)$. Please note that $z_n={\mathrm{\Pi }}_C(I-{\mu }_2{B}_2)u_n$ for all $n\ge 0$. Then by Lemma 4 we get

$$\begin{array}{cc}\parallel z_n\parallel \hfill & \le \parallel {\mathrm{\Pi }}_C(u_n-{\mu }_2{B}_2{u}_n)-{\mathrm{\Pi }}_C(p-{\mu }_2{B}_{2}p)\parallel +\parallel q\parallel \hfill \\ & \le \parallel (u_n-{\mu }_2{B}_2{u}_n)-(p-{\mu }_2{B}_{2}p)\parallel +\parallel q\parallel \hfill \\ & \le \parallel q\parallel +\parallel u_n-p\parallel .\hfill \end{array}$$

This shows that $\left\{z_n\right\}$ is bounded.

Step 2. We claim that $\parallel x_{n+1}-x_n\parallel \to 0$ and $\parallel y_{n+1}-y_n\parallel \to 0$ as $n\to \infty$. Indeed, from (9) we have

$$\begin{array}{c}\parallel x_{n+1}-x_n\parallel \hfill \\ \le \parallel {\alpha }_{n}f\left(x_n\right)+(I-{\alpha }_n\theta F)T^n{y}_n-{\alpha }_{n-1}f\left(x_{n-1}\right)-(I-{\alpha }_{n-1}\theta F)T^{n-1}{y}_{n-1}\parallel \hfill \\ \le {\alpha }_n\gamma \parallel x_n-x_{n-1}\parallel +|{\alpha }_n-{\alpha }_{n-1}|\parallel f\left(x_{n-1}\right)-\theta F{T}^n{y}_{n-1}\parallel +(1-{\alpha }_n\overline{\gamma }\theta )(\parallel y_n-y_{n-1}\parallel +c_n)\hfill \\ +(1-{\alpha }_{n-1}\overline{\gamma }\theta )\parallel T^n{y}_{n-1}-T^{n-1}{y}_{n-1}\parallel \hfill \\ \le {\alpha }_n\gamma \parallel x_n-x_{n-1}\parallel +|{\alpha }_n-{\alpha }_{n-1}|M_1+(1-{\alpha }_n\overline{\gamma }\theta )\parallel u_n-u_{n-1}\parallel \hfill \\ +\parallel T^n{y}_{n-1}-T^{n-1}{y}_{n-1}\parallel +c_n,\hfill \end{array}$$

(12)

where

$$M_1=\underset{n\ge 1}{sup}\parallel \theta F{T}^n{y}_{n-1}-f\left(x_{n-1}\right)\parallel .$$

Also, simple calculations show that

$$\begin{array}{c}\parallel u_n-u_{n-1}{\parallel }^2\hfill \\ \le {\beta }_n\parallel x_n-x_{n-1}\parallel \parallel u_n-u_{n-1}\parallel +(1-{\beta }_n)[\parallel S_n{u}_n-S_{n-1}{u}_n\parallel \parallel u_n-u_{n-1}\parallel \hfill \\ +\parallel u_n-u_{n-1}{\parallel }^2]+|{\beta }_n-{\beta }_{n-1}|\parallel x_{n-1}-S_{n-1}{u}_{n-1}\parallel \parallel u_n-u_{n-1}\parallel .\hfill \end{array}$$

(13)

So, it follows from (13) that

$$\begin{array}{cc}\parallel u_n-u_{n-1}\parallel \hfill & \le {\beta }_n\parallel x_n-x_{n-1}\parallel +(1-{\beta }_n)[\parallel S_n{u}_n-S_{n-1}{u}_n\parallel \hfill \\ & +\parallel u_n-u_{n-1}\parallel ]+|{\beta }_n-{\beta }_{n-1}|\parallel x_{n-1}-S_{n-1}{u}_{n-1}\parallel ,\hfill \end{array}$$

which immediately leads to

$$\begin{array}{cc}\parallel u_n-u_{n-1}\parallel -\parallel x_n-x_{n-1}\parallel \hfill & \le {\displaystyle \frac{1-{\beta }_n}{{\beta }_n}}\parallel S_n{u}_n-S_{n-1}{u}_n\parallel +|{\beta }_n-{\beta }_{n-1}|{\displaystyle \frac{\parallel x_{n-1}-S_{n-1}{u}_{n-1}\parallel }{{\beta }_n}}\hfill \\ & \le {\displaystyle \frac{1}{a}}\parallel S_n{u}_n-S_{n-1}{u}_n\parallel +|{\beta }_n-{\beta }_{n-1}|{\displaystyle \frac{\parallel x_{n-1}-S_{n-1}{u}_{n-1}\parallel }{a}}.\hfill \end{array}$$

(14)

Putting $D=\{u_n:n\ge 0\}$, we know that D is a bounded subset of C. Then by the assumption we get ${\sum }_{n=1}^{\infty }{sup}_{x\in D}\parallel S_{n}x-S_{n-1}x\parallel <\infty$. Noticing that

$$\underset{x\in D}{sup}\parallel S_{n}x-S_{n-1}x\parallel \ge \parallel S_n{u}_n-S_{n-1}{u}_n\parallel ,\forall n\ge 1,$$

we have

$$\sum _{n=1}^{\infty }\parallel S_n{u}_n-S_{n-1}{u}_n\parallel <\infty .$$

(15)

Therefore, from (12) and (14) we deduce that

$$\begin{array}{cc}\parallel x_{n+1}-x_n\parallel \hfill & \le {\alpha }_n\gamma \parallel x_n-x_{n-1}\parallel +|{\alpha }_n-{\alpha }_{n-1}|M_1+(1-{\alpha }_n\overline{\gamma }\theta )\{\parallel x_n-x_{n-1}\parallel \hfill \\ & +{\displaystyle \frac{1}{a}}\parallel S_n{u}_n-S_{n-1}{u}_n\parallel +|{\beta }_n-{\beta }_{n-1}|{\displaystyle \frac{\parallel x_{n-1}-S_{n-1}{u}_{n-1}\parallel }{a}}\}\hfill \\ & +c_n+\parallel T^n{y}_{n-1}-T^{n-1}{y}_{n-1}\parallel \hfill \\ & \le |{\alpha }_n-{\alpha }_{n-1}|M_1+{\displaystyle \frac{1}{a}}\parallel S_n{u}_n-S_{n-1}{u}_n\parallel +[1-{\alpha }_n(\overline{\gamma }\theta -\gamma )]\parallel x_n-x_{n-1}\parallel \hfill \\ & +|{\beta }_n-{\beta }_{n-1}|{\displaystyle \frac{\parallel x_{n-1}-S_{n-1}{u}_{n-1}\parallel }{a}}+c_n+\parallel T^n{y}_{n-1}-T^{n-1}{y}_{n-1}\parallel \hfill \\ & \le M_2(|{\alpha }_n-{\alpha }_{n-1}|+|{\beta }_n-{\beta }_{n-1}|+[1-{\alpha }_n(\overline{\gamma }\theta -\gamma )]\parallel x_n-x_{n-1}\parallel \hfill \\ & +\parallel S_n{u}_n-S_{n-1}{u}_n\parallel +c_n+\parallel T^n{y}_{n-1}-T^{n-1}{y}_{n-1}\parallel ),\hfill \end{array}$$

(16)

where

$$M_2=\underset{n\ge 1}{sup}\{{\displaystyle \frac{1}{a}}+{\displaystyle \frac{\parallel x_{n-1}-S_{n-1}{u}_{n-1}\parallel }{a}}+M_1+1\}.$$

From (15), the assumption ${\sum }_{n=0}^{\infty }{c}_n<\infty$ and conditions (i) and (ii), it can be readily seen that ${\sum }_{n=0}^{\infty }{\alpha }_n(\overline{\gamma }\theta -\gamma )=\infty$ and

$$\sum _{n=1}^{\infty }{M}_2(\parallel S_n{u}_n-S_{n-1}{u}_n\parallel +c_n+|{\alpha }_n-{\alpha }_{n-1}|+|{\beta }_n-{\beta }_{n-1}|+\parallel T^n{y}_{n-1}-T^{n-1}{y}_{n-1}\parallel )<\infty .$$

So, it follows from Lemma 7 and (16) that

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

(17)

Again from (9) and (14) we conclude that

$$\begin{array}{c}a\parallel y_n-y_{n-1}\parallel =a\parallel G{u}_n-G{u}_{n-1}\parallel \le a\parallel u_n-u_{n-1}\parallel \hfill \\ \le a\parallel x_n-x_{n-1}\parallel +\parallel S_n{u}_n-S_{n-1}{u}_n\parallel +|{\beta }_n-{\beta }_{n-1}|\parallel x_{n-1}-S_{n-1}{u}_{n-1}\parallel \to 0(n\to \infty ).\hfill \end{array}$$

That is,

$$\underset{n\to \infty }{lim}\parallel y_{n+1}-y_n\parallel =0.$$

(18)

Step 3. We claim that $\parallel x_n-G{x}_n\parallel \to 0$ as $n\to \infty$. Indeed, note that $q={\mathrm{\Pi }}_C(p-{\mu }_2{B}_{2}p),z_n={\mathrm{\Pi }}_C(u_n-{\mu }_2{B}_2{u}_n)$ and $y_n={\mathrm{\Pi }}_C(z_n-{\mu }_1{B}_1{z}_n)$. Then $y_n=G{u}_n$. By Lemma 4 we have

$$\begin{array}{cc}\parallel z_n{-q\parallel }^2\hfill & \le \parallel u_n-p-{\mu }_2(B_2{u}_n-B_{2}p){\parallel }^2\hfill \\ & \le 2{\mu }_2({\kappa }^2{\mu }_2-\beta )\parallel B_2{u}_n-B_2{p\parallel }^2+{\parallel u_n-p\parallel }^2\hfill \end{array}$$

(19)

and

$$\begin{array}{cc}\parallel y_n{-p\parallel }^2\hfill & \le \parallel z_n-q-{\mu }_1(B_1{z}_n-B_{1}q){\parallel }^2\hfill \\ & \le 2{\mu }_1({\kappa }^2{\mu }_1-\alpha )\parallel B_1{z}_n-B_1{q\parallel }^2+{\parallel z_n-q\parallel }^2.\hfill \end{array}$$

(20)

Substituting (19) for (20), we obtain

$$\parallel y_n{-p\parallel }^2+2{\mu }_2(\beta -{\kappa }^2{\mu }_2)\parallel B_2{u}_n-B_2{p\parallel }^2+2{\mu }_1(\alpha -{\kappa }^2{\mu }_1)\parallel B_1{z}_n-B_1{q\parallel }^2\le {\parallel u_n-p\parallel }^2.$$

(21)

Let $v_n:=(T^n{y}_n-{\alpha }_n\theta F{T}^n{y}_n)+{\alpha }_{n}f\left(x_n\right)$. Then, from (7) and Lemma 2 we obtain

$$\begin{array}{cc}\parallel x_{n+1}{-p\parallel }^2\hfill & \le \parallel v_n{-p\parallel }^2\hfill \\ & \le 2{\alpha }_n\langle f\left(x_n\right)-\theta F{T}^n{y}_n,j(v_n-p)\rangle +{\parallel T^n{y}_n-p\parallel }^2\hfill \\ & \le 2{\alpha }_n\parallel f\left(x_n\right)-\theta F{T}^n{y}_n\parallel \parallel v_n-p\parallel +(\parallel y_n-p{\parallel +c_n)}^2\hfill \\ & \le \parallel y_n{-p\parallel }^2+c_n{M}_2+{\alpha }_n{M}_3,\hfill \end{array}$$

(22)

where $M_2={sup}_{n\ge 0}\{2\parallel y_n-p\parallel +c_n\}$ and

$$M_3=\underset{n\ge 0}{sup}\{2\parallel f\left(x_n\right)-\theta F{T}^n{y}_n\parallel \parallel v_n-p\parallel \}.$$

Substituting (21) to (22), we deduce from (10) that

$$\begin{array}{c}\parallel x_{n+1}{-p\parallel }^2+2{\mu }_2(\beta -{\kappa }^2{\mu }_2)\parallel B_2{u}_n-B_2{p\parallel }^2+2{\mu }_1(\alpha -{\kappa }^2{\mu }_1){\parallel B_1{z}_n-B_{1}q\parallel }^2\hfill \\ \le \parallel x_n{-p\parallel }^2+c_n{M}_2+{\alpha }_n{M}_3,\hfill \end{array}$$

which immediately yields

$$\begin{array}{c}2{\mu }_2(\beta -{\kappa }^2{\mu }_2)\parallel B_2{u}_n-B_2{p\parallel }^2+2{\mu }_1(\alpha -{\kappa }^2{\mu }_1){\parallel B_1{z}_n-B_{1}q\parallel }^2\hfill \\ \le {\alpha }_n{M}_3-\parallel x_{n+1}{-p\parallel }^2+c_n{M}_2+{\parallel x_n-p\parallel }^2\hfill \\ \le c_n{M}_2+{\alpha }_n{M}_3+(\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel )\parallel x_n-x_{n+1}\parallel .\hfill \end{array}$$

Since ${\mu }_1{\kappa }^2\in (0,\alpha ),{\mu }_2{\kappa }^2\in (0,\beta ),{lim}_{n\to \infty }{c}_n=0$ and ${lim}_{n\to \infty }{\alpha }_n=0$, we obtain from (17) that

$$\underset{n\to \infty }{lim}\parallel B_2{u}_n-B_{2}p\parallel =0\mathrm{and}\underset{n\to \infty }{lim}\parallel B_1{z}_n-B_{1}q\parallel =0.$$

(23)

On the other hand, from Proposition 3 and Lemma 5 we have

$$\begin{array}{cc}2\parallel z_n{-q\parallel }^2\hfill & \le 2\langle u_n-{\mu }_2{B}_2{u}_n-(p-{\mu }_2{B}_{2}p),j(z_n-q)\rangle \hfill \\ & \le [\parallel u_n{-p\parallel }^2+\parallel z_n{-q\parallel }^2-g_1(\parallel u_n-z_n-(p-q)\parallel \left)\right]+2{\mu }_2\parallel B_{2}p-B_2{u}_n\parallel \parallel z_n-q\parallel ,\hfill \end{array}$$

which implies that

$$\parallel z_n{-q\parallel }^2+g_1(\parallel u_n-z_n-(p-q)\parallel )\le \parallel u_n{-p\parallel }^2+2{\mu }_2\parallel B_{2}p-B_2{u}_n\parallel \parallel z_n-q\parallel .$$

(24)

In the same way, we derive

$$\begin{array}{cc}2\parallel y_n{-p\parallel }^2\hfill & \le 2\langle z_n-{\mu }_1{B}_1{z}_n-(q-{\mu }_1{B}_{1}q),j(y_n-p)\rangle \hfill \\ & \le [\parallel z_n{-q\parallel }^2+\parallel y_n{-p\parallel }^2-g_2(\parallel z_n-y_n+(p-q)\parallel \left)\right]+2{\mu }_1\parallel B_{1}q-B_1{z}_n\parallel \parallel y_n-p\parallel ,\hfill \end{array}$$

which implies that

$$\parallel y_n{-p\parallel }^2+g_2(\parallel z_n-y_n+(p-q)\parallel )\le \parallel z_n{-q\parallel }^2+2{\mu }_1\parallel B_{1}q-B_1{z}_n\parallel \parallel y_n-p\parallel .$$

(25)

Substituting (24) for (25), we deduce from (10) that

$$\begin{array}{cc}\parallel y_n{-p\parallel }^2\hfill & \le \parallel u_n{-p\parallel }^2-g_1(\parallel u_n-z_n-(p-q)\parallel )-g_2(\parallel z_n-y_n+(p-q)\parallel )\hfill \\ & +2{\mu }_2\parallel B_{2}p-B_2{u}_n\parallel \parallel z_n-q\parallel +2{\mu }_1\parallel B_{1}q-B_1{z}_n\parallel \parallel y_n-p\parallel \hfill \\ & \le \parallel x_n{-p\parallel }^2-g_1(\parallel u_n-z_n-(p-q)\parallel )-g_2(\parallel z_n-y_n+(p-q)\parallel )\hfill \\ & +2{\mu }_1\parallel B_{1}q-B_1{z}_n\parallel \parallel y_n-p\parallel +2{\mu }_2\parallel B_{2}p-B_2{u}_n\parallel \parallel z_n-q\parallel .\hfill \end{array}$$

(26)

Substituting (26) for (22), we have

$$\begin{array}{c}\parallel x_{n+1}{-p\parallel }^2+g_1(\parallel u_n-z_n-(p-q)\parallel )+g_2(\parallel z_n-y_n+(p-q)\parallel )\hfill \\ \le \parallel x_n{-p\parallel }^2+2{\mu }_2\parallel B_{2}p-B_2{u}_n\parallel \parallel z_n-q\parallel +2{\mu }_1\parallel B_{1}q-B_1{z}_n\parallel \parallel y_n-p\parallel +c_n{M}_2+{\alpha }_n{M}_3,\hfill \end{array}$$

which hence yields

$$\begin{array}{c}{g}_1(\parallel u_n-z_n-(p-q)\parallel )+g_2(\parallel z_n-y_n+(p-q)\parallel )\hfill \\ \le (\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel )\parallel x_n-x_{n+1}\parallel +2{\mu }_2\parallel B_{2}p-B_2{u}_n\parallel \parallel z_n-q\parallel \hfill \\ +2{\mu }_1\parallel B_{1}q-B_1{z}_n\parallel \parallel y_n-p\parallel +{\alpha }_n{M}_3+c_n{M}_2.\hfill \end{array}$$

Since ${lim}_{n\to \infty }{c}_n=0$ and ${lim}_{n\to \infty }{\alpha }_n=0$, we conclude from (17) and (23) that

$$\underset{n\to \infty }{lim}{g}_1(\parallel u_n-z_n-(p-q)\parallel )=0\mathrm{and}\underset{n\to \infty }{lim}{g}_2(\parallel z_n-y_n+(p-q)\parallel )=0.$$

Using the properties of $g_1$ and $g_2$, we obtain

$$\underset{n\to \infty }{lim}\parallel u_n-z_n-(p-q)\parallel =0\mathrm{and}\underset{n\to \infty }{lim}\parallel z_n-y_n+(p-q)\parallel =0.$$

(27)

It follows that

$$\parallel u_n-y_n\parallel \le \parallel u_n-z_n-(p-q)\parallel +\parallel z_n-y_n+(p-q)\parallel \to 0(n\to \infty ).$$

That is,

$$\underset{n\to \infty }{lim}\parallel u_n-G{u}_n\parallel =\underset{n\to \infty }{lim}\parallel u_n-y_n\parallel =0.$$

(28)

Also, according to (8) we have

$$\begin{array}{c}\parallel u_n{-p\parallel }^2\le {\beta }_n\langle x_n-p,j(u_n-p)\rangle +(1-{\beta }_n){\parallel u_n-p\parallel }^2,\hfill \end{array}$$

which together with Lemma 5, yields

$$\begin{array}{cc}2\parallel u_n{-p\parallel }^2\hfill & \le 2\langle x_n-p,j(u_n-p)\rangle \hfill \\ & \le [\parallel x_n{-p\parallel }^2+\parallel u_n{-p\parallel }^2-g(\parallel x_n-u_n\parallel \left)\right].\hfill \end{array}$$

This immediately implies that

$$\parallel u_n{-p\parallel }^2+g(\parallel x_n-u_n\parallel )\le \parallel x_n{-p\parallel }^2,$$

which together with (22), yields

$$\begin{array}{cc}\parallel x_{n+1}{-p\parallel }^2\hfill & \le \parallel u_n{-p\parallel }^2+c_n{M}_2+{\alpha }_n{M}_3\hfill \\ & \le \parallel x_n{-p\parallel }^2-g(\parallel x_n-u_n\parallel )+c_n{M}_2+{\alpha }_n{M}_3.\hfill \end{array}$$

Hence we have

$$\begin{array}{cc}g(\parallel u_n-x_n\parallel )\hfill & \le \parallel x_n{-p\parallel }^2+c_n{M}_2-{\parallel x_{n+1}-p\parallel }^2+{\alpha }_n{M}_3\hfill \\ & \le (\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel )\parallel x_n-x_{n+1}\parallel +{\alpha }_n{M}_3+c_n{M}_2.\hfill \end{array}$$

Since ${lim}_{n\to \infty }{c}_n=0$ and ${lim}_{n\to \infty }{\alpha }_n=0$, we obtain from (17) that ${lim}_{n\to \infty }g(\parallel x_n-u_n\parallel )=0$. Using the properties of g, we have

$$\underset{n\to \infty }{lim}\parallel x_n-u_n\parallel =0.$$

(29)

Also, observe that

$$\parallel x_n-u_n\parallel +\parallel u_n-G{u}_n\parallel \ge \parallel x_n-y_n\parallel ,$$

and

$$\parallel x_n-y_n\parallel +\parallel u_n-x_n\parallel \ge \parallel x_n-y_n\parallel +\parallel G{u}_n-G{x}_n\parallel \ge \parallel x_n-G{x}_n\parallel .$$

Then from (28) and (29) it follows that

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

(30)

Step 4. We claim that $\parallel T{x}_n-x_n\parallel \to 0$ and $\parallel S_n{x}_n-x_n\parallel \to 0$ as $n\to \infty$. Indeed, combining (8) with (29), we obtain that

$$\parallel S_n{u}_n-u_n\parallel ={\displaystyle \frac{{\beta }_n}{1-{\beta }_n}}\parallel x_n-u_n\parallel \le {\displaystyle \frac{b}{1-b}}\parallel x_n-u_n\parallel \to 0(n\to \infty ).$$

That is,

$$\underset{n\to \infty }{lim}\parallel S_n{u}_n-u_n\parallel =0.$$

(31)

Since ${\left\{S_n\right\}}_{n=0}^{\infty }$ is ℓ-uniformly Lipschitzian on C, we deduce from (29) and (31) that

$$\begin{array}{cc}\parallel S_n{x}_n-x_n\parallel \hfill & \le \parallel S_n{x}_n-S_n{u}_n\parallel +\parallel S_n{u}_n-u_n\parallel +\parallel u_n-x_n\parallel \hfill \\ & \le (\ell +1)\parallel x_n-u_n\parallel +\parallel S_n{u}_n-u_n\parallel \to 0(n\to \infty ).\hfill \end{array}$$

That is,

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

(32)

We note that

$$\begin{array}{cc}\parallel x_n-T^n{y}_n\parallel \hfill & \le \parallel x_n-x_{n+1}\parallel +\parallel x_{n+1}-T^n{y}_n\parallel \hfill \\ & =\le \parallel x_n-x_{n+1}\parallel +{\alpha }_n\parallel f\left(x_n\right)-\theta F{T}^n{y}_n\parallel .\hfill \end{array}$$

Then we have

$$\begin{array}{cc}\parallel y_n-T^n{y}_n\parallel \hfill & \le \parallel y_n-x_n\parallel +\parallel x_n-T^n{y}_n\parallel \hfill \\ & \le \parallel y_n-x_n\parallel +\parallel x_n-x_{n+1}\parallel +{\alpha }_n\parallel f\left(x_n\right)-\theta F{T}^n{y}_n\parallel .\hfill \end{array}$$

Consequently, from (17), (30) and ${lim}_{n\to \infty }{\alpha }_n=0$, it follows that

$$\underset{n\to \infty }{lim}\parallel y_n-T^n{y}_n\parallel =0.$$

(33)

We also note that

$$\parallel y_n-T{y}_n\parallel \le \parallel y_n-T^n{y}_n\parallel +\parallel T^n{y}_n-T^{n+1}{y}_n\parallel +\parallel T^{n+1}{y}_n-T{y}_n\parallel .$$

By the assumption ${\sum }_{n=0}^{\infty }\parallel T^{n+1}{y}_n-T^n{y}_n\parallel <\infty$, (33) and the condition that $T:C\to C$ is uniformly continuous, we get

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

(34)

In addition, noticing that

$$\parallel x_n-T{x}_n\parallel \le \parallel x_n-y_n\parallel +\parallel y_n-T{y}_n\parallel +\parallel T{y}_n-T{x}_n\parallel ,$$

we deduce from (30), (34) and the uniform continuity of T that

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

(35)

Step 5. We claim that $\parallel x_n-\overline{S}{x}_n\parallel \to 0$ as $n\to \infty$ where $\overline{S}:={(2I-S)}^{-1}$. Indeed, first, let us show that $S:C\to C$ is pseudocontractive and ℓ-Lipschitzian such that ${lim}_{n\to \infty }\parallel S{x}_n-x_n\parallel =0$ where $Sx={lim}_{n\to \infty }{S}_{n}x\forall x\in C$. Observe that for all $x,y\in C$, ${lim}_{n\to \infty }\parallel S_{n}x-Sx\parallel =0$ and ${lim}_{n\to \infty }\parallel S_{n}y-Sy\parallel =0$. Since $S_n$ is pseudocontractive operator, we get

$$\langle Sx-Sy,j(x-y)\rangle \le {\parallel x-y\parallel }^2.$$

This means that S is pseudocontractive. Noting that ${\left\{S_n\right\}}_{n=0}^{\infty }$ is ℓ-uniformly Lipschitzian on C, we have

$$\parallel Sx-Sy\parallel =\underset{n\to \infty }{lim}\parallel S_{n}x-S_{n}y\parallel \le \ell \parallel x-y\parallel ,\forall x,y\in C.$$

This means that S is ℓ-Lipschitzian. Taking into account the boundedness of $\left\{x_n\right\}$ and putting $D=\overline{\mathrm{conv}}\{x_n:n\ge 0\}$ (the closed convex hull of the set $\{x_n:n\ge 0\}$), by the assumption we have ${\sum }_{n=1}^{\infty }{sup}_{x\in D}\parallel S_{n-1}x-S_{n}x\parallel <\infty$. Hence, by Proposition 1 we get

$$\underset{n\to \infty }{lim}\underset{x\in D}{sup}\parallel S_{n}x-Sx\parallel =0,$$

which immediately yields

$$\underset{n\to \infty }{lim}\parallel S_n{x}_n-S{x}_n\parallel =0.$$

(36)

Thus, combining (32) with (36) we have

$$\parallel x_n-S{x}_n\parallel \le \parallel x_n-S_n{x}_n\parallel +\parallel S_n{x}_n-S{x}_n\parallel \to 0(n\to \infty ).$$

That is,

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

(37)

Now, let us show that if we define $\overline{S}:={(2I-S)}^{-1}$, then $\overline{S}:C\to C$ is nonexpansive, $\mathrm{Fix}\left(\overline{S}\right)=\mathrm{Fix}\left(S\right)={\bigcap }_{n=0}^{\infty }\mathrm{Fix}\left(S_n\right)$ and ${lim}_{n\to \infty }\parallel x_n-\overline{S}{x}_n\parallel =0$. Indeed, put $\overline{S}:={(2I-S)}^{-1}$, where I is the identity mapping on E. Then it is known that $\overline{S}$ is nonexpansive and the fixed-point set $\mathrm{Fix}\left(S\right)=\mathrm{Fix}\left(\overline{S}\right)={\bigcap }_{n=0}^{\infty }\mathrm{Fix}\left(S_n\right)$. From (37) it follows that

$$\begin{array}{cc}\parallel x_n-\overline{S}{x}_n\parallel \hfill & =\parallel \overline{S}{\overline{S}}^{-1}{x}_n-\overline{S}{x}_n\parallel \hfill \\ & \le \parallel {\overline{S}}^{-1}{x}_n-x_n\parallel \hfill \\ & =\parallel (2I-S)x_n-x_n\parallel =\parallel x_n-S{x}_n\parallel \to 0(n\to \infty ).\hfill \end{array}$$

That is,

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

(38)

Step 6. We claim that

$$\underset{n\to \infty }{lim\; sup}\langle f\left(x^{*}\right)-\theta F\left(x^{*}\right),j(x_n-x^{*})\rangle \le 0,$$

(39)

where $x^{*}={\mathrm{\Pi }}_{\mathrm{\Omega }}(f+I-\theta F)\left(x^{*}\right)$. Indeed, there exists a subsequence $\left\{x_{n_i}\right\}$ of $\left\{x_n\right\}$ such that

$$\underset{n\to \infty }{lim\; sup}\langle f\left(x^{*}\right)-\theta F\left(x^{*}\right),j(x_n-x^{*})\rangle =\underset{i\to \infty }{lim}\langle f\left(x^{*}\right)-\theta F\left(x^{*}\right),j(x_{n_i}-x^{*})\rangle .$$

Now we show that ${\mathrm{\Pi }}_{\mathrm{\Omega }}(f+I-\theta F)$ is a contraction mapping. Since F is bounded linear strongly positive, for all $x,y\in C$, we have

$$\begin{array}{c}\parallel {\mathrm{\Pi }}_{\mathrm{\Omega }}(f+I-\theta F)\left(x\right)-{\mathrm{\Pi }}_{\mathrm{\Omega }}(f+I-\theta F)\left(y\right)\parallel \hfill \\ \le \parallel f\left(x\right)-f\left(y\right)\parallel +\parallel (I-\theta F)\left(x\right)-(I-\theta F)\left(y\right)\parallel \hfill \\ \le [1+(\gamma -\overline{\gamma }\theta )]\parallel x-y\parallel ,\hfill \end{array}$$

which implies that ${\mathrm{\Pi }}_{\mathrm{\Omega }}(f+I-\theta F)$ is a contraction mapping. Banach’s contraction mapping principle guarantees that ${\mathrm{\Pi }}_{\mathrm{\Omega }}(f+I-\theta F)$ has a unique fixed point. Say $x^{*}\in C$, that is, $x^{*}={\mathrm{\Pi }}_{\mathrm{\Omega }}(f+I-\theta F)\left(x^{*}\right)$. Since $\left\{x_n\right\}$ is a bounded sequence in C, we may assume that $x_{n_i}\rightharpoonup \overline{x}\in C$. Please note that G and $\overline{S}$ are nonexpansive and that T is asymptotically nonexpansive in the intermediate sense. Since $(I-G)x_n\to 0$ and $(I-\overline{S})x_n\to 0$ (due to (30) and (37)), by Lemma 6 we have that $\overline{x}\in \mathrm{Fix}\left(G\right)=\mathrm{GSVI}(C,B_1,B_2)$ and $\overline{x}\in \mathrm{Fix}\left(\overline{S}\right)=\mathrm{Fix}\left(S\right)={\bigcap }_{n=0}^{\infty }\mathrm{Fix}\left(S_n\right)$. From (35), we have that ${lim}_{i\to \infty }\parallel x_{n_i}-T{x}_{n_i}\parallel =0$ for the subsequence $\left\{x_{n_i}\right\}$ of $\left\{x_n\right\}$. It follows that $\overline{x}\in \mathrm{Fix}\left(T\right)$. Then, $\overline{x}\in \mathrm{\Omega }={\bigcap }_{n=0}^{\infty }\mathrm{Fix}\left(S_n\right)\cap \mathrm{GSVI}(C,B_1,B_2)\cap \mathrm{Fix}\left(T\right)$. Since E admits a weakly sequentially continuous duality mapping $j(·)$ and $x_{n_i}\rightharpoonup \overline{x}$, we obtain

$$\begin{array}{cc}{\displaystyle \underset{n\to \infty }{lim\; sup}}\langle f\left(x^{*}\right)-\theta F\left(x^{*}\right),j(x_n-x^{*})\rangle \hfill & ={\displaystyle \underset{i\to \infty }{lim}}\langle f\left(x^{*}\right)-\theta F\left(x^{*}\right),j(x_{n_i}-x^{*})\rangle \hfill \\ & =\langle f\left(x^{*}\right)-\theta F\left(x^{*}\right),j(\overline{x}-x^{*})\rangle \le 0,\hfill \end{array}$$

which implies that (39) holds. Noticing that $j(·)$ is also norm-to-norm uniformly continuous on bounded subsets of E, we obtain from (17) that

$$\underset{n\to \infty }{lim\; sup}\langle f\left(x^{*}\right)-\theta F\left(x^{*}\right),j(x_{n+1}-x^{*})\rangle \le 0.$$

(40)

Step 7. We claim that $x_n\to x^{*}$ as $n\to \infty$. Indeed, it follows from $x_{n+1}={\mathrm{\Pi }}_C{v}_n$ and Proposition 3 (iii) that

$$\langle {\mathrm{\Pi }}_C{v}_n-v_n,j({\mathrm{\Pi }}_C{v}_n-x^{*})\rangle \le 0,$$

which leads to

$$\langle x_{n+1}-v_n,j(x_{n+1}-x^{*})\rangle \le 0.$$

Then, from (11) we have

$$\begin{array}{c}2\parallel x_{n+1}-x^{*}{\parallel }^2\hfill \\ \le 2\langle x^{*}-v_n,j(x^{*}-x_{n+1})\rangle \hfill \\ \le 2{\alpha }_n\gamma \parallel x_n-x^{*}\parallel \parallel x_{n+1}-x^{*}\parallel +2(1-{\alpha }_n\overline{\gamma }\theta )(\parallel y_n-x^{*}\parallel +c_n)\parallel x_{n+1}-x^{*}\parallel \hfill \\ +2{\alpha }_n\langle f\left(x^{*}\right)-\theta F\left(x^{*}\right),j(x_{n+1}-x^{*})\rangle \hfill \\ \le 2[1-{\alpha }_n(\overline{\gamma }\theta -\gamma )]\parallel x_n-x^{*}\parallel \parallel x_{n+1}-x^{*}\parallel +2c_n{M}_4+2{\alpha }_n\langle f\left(x^{*}\right)-\theta F\left(x^{*}\right),j(x_{n+1}-x^{*})\rangle \hfill \\ \le \parallel x_{n+1}-x^{*}{\parallel }^2+(1-{\alpha }_n(\overline{\gamma }\theta -\gamma )){\parallel x_n-x^{*}\parallel }^2+2c_n{M}_4+2{\alpha }_n\langle f\left(x^{*}\right)-\theta F\left(x^{*}\right),j(x_{n+1}-x^{*})\rangle ,\hfill \end{array}$$

where $M_4={sup}_{n\ge 0}\parallel x_{n+1}-x^{*}\parallel$. This immediately implies that

$$\begin{array}{cc}\parallel x_{n+1}-x^{*}{\parallel }^2\hfill & \le [1-{\alpha }_n(\overline{\gamma }\theta -\gamma )]{\parallel x_n-x^{*}\parallel }^2+{\alpha }_n(\overline{\gamma }\theta -\gamma ){\displaystyle \frac{2\langle f\left(x^{*}\right)-\theta F\left(x^{*}\right),j(x_{n+1}-x^{*})\rangle }{\overline{\gamma }\theta -\gamma }}+2c_n{M}_4.\hfill \end{array}$$

(41)

Applying Lemma 7 to (41), we infer that $x_n\to x^{*}$ as $n\to \infty$. The proof is completed.  □

It is remarkable that according to the proof of Theorem 1, we know that $\left\{y_n\right\}$ is bounded. We now give two examples to illustrate partial conditions of Theorem 1 to be satisfied.

Example 1.

Let $T:C\to C$ be a contraction mapping with a constant $\beta \in (0,1)$. We take $S_n:=T^n$ and obtain

$$\underset{x\in D}{sup}\parallel S_{n}x-S_{n-1}x\parallel =\underset{x\in D}{sup}\parallel T^{n}x-T^{n-1}x\parallel \le {\beta }^{n-1}·\underset{x\in D}{sup}\parallel Tx-x\parallel ,\forall n\ge 1,$$

for any bounded subset D of C. Therefore, it follows that

$$+\infty >\sum _{n=1}^{\infty }{\beta }^{n-1}·\underset{x\in D}{sup}\parallel Tx-x\parallel \ge \sum _{n=1}^{\infty }\underset{x\in D}{sup}\parallel S_{n}x-S_{n-1}x\parallel .$$

In particular, whenever D is a bounded sequence ${\left\{x_n\right\}}_{n=0}^{\infty }$ in C, we have

$$+\infty >\sum _{n=0}^{\infty }\underset{x\in D}{sup}\parallel S_{n+1}x-S_{n}x\parallel \ge \sum _{n=0}^{\infty }\parallel S_{n+1}{x}_n-S_n{x}_n\parallel =\sum _{n=0}^{\infty }\parallel T^{n+1}{x}_n-T^n{x}_n\parallel .$$

Since T is a contraction mapping, Banach’s Contraction Mapping Principle guarantees that T has a unique fixed point. Say $p\in C$. We define $Sx:=p$ for all $x\in C$. It is easy to see that $Sx={lim}_{n\to \infty }{S}_{n}x$ for all $x\in C$ and $\mathrm{Fix}\left(S\right)={\bigcap }_{n=0}^{\infty }\mathrm{Fix}\left(S_n\right)$.

Example 2.

Let $E=\mathbf{R}$ and $C=[-1,1]$, and let $T:C\to C$ be an identity mapping, i.e., $Tx=x$ for all $x\in C$. Moreover, we define

$$S_{n}x:={\displaystyle \frac{sinx+x}{2+n}},\forall x\in C,\forall n\ge 0.$$

Then we obtain

$$\begin{array}{c}|S_{n}x-S_{n}y|=|{\displaystyle \frac{sinx+}{n+2}}-{\displaystyle \frac{siny+y}{n+2}}|\hfill \\ \le {\displaystyle \frac{|y-x|+|siny-sinx|}{n+2}}\le {\displaystyle \frac{2|x-y|}{n+2}}\le |x-y|,\forall x,y\in C,\forall n\ge 0,\hfill \end{array}$$

and

$$\begin{array}{cc}{\displaystyle \underset{x\in C}{sup}}|S_{n}x-S_{n-1}x|\hfill & ={\displaystyle \underset{x\in C}{sup}}|{\displaystyle \frac{x+sinx}{n+2}}-{\displaystyle \frac{x+sinx}{n+1}}|={\displaystyle \underset{x\in C}{sup}}\left|{\displaystyle \frac{x+sinx}{(n+2)(n+1)}}\right|\hfill \\ & \le {\displaystyle \frac{2}{(n+2)(n+1)}}\le {\displaystyle \frac{2}{{(n+1)}^2}},\forall n\ge 0.\hfill \end{array}$$

Therefore, it follows that for any bounded subset D of C,

$$\sum _{n=1}^{\infty }\underset{x\in D}{sup}|S_{n}x-S_{n-1}x|<\infty .$$

In addition, whenever ${\left\{x_n\right\}}_{n=0}^{\infty }$ is a bounded sequence in C, it is clear that

$$\sum _{n=0}^{\infty }|T^n{x}_n-T^{n+1}{x}_n|<\infty .$$

Also, we define $Sx:=0$ for all $x\in C$. Then, it is clear that $Sx={lim}_{n\to \infty }{S}_{n}x\forall x\in C$, and $\mathrm{Fix}\left(S\right)={\bigcap }_{n=0}^{\infty }\mathrm{Fix}\left(S_n\right)$.

4. Applications

Now, we give an application to solve CFPPs of asymptotically nonexpansive and pseudocontractive mappings, and variational inequality problems for strict pseudocontractive mappings in Banach spaces.

Let C be a nonempty, closed, and convex subset of a real Banach space E. A mapping $T:C\to C$ is said to be $\lambda$-strictly pseudocontractive if for every $x,y\in C$ there exists $j(x-y)\in J(x-y)$ such that

$${\langle Tx-Ty,j(x-y)\rangle +\lambda \parallel (I-T)x-(I-T)y\parallel }^2\le {\parallel x-y\parallel }^2,\mathrm{for}\mathrm{some}\lambda \in (0,1).$$

(42)

A simple computation shows that (42) is equivalent to the following inequality:

$$\langle (I-T)x-(I-T)y,j(x-y)\rangle \ge {\lambda \parallel (I-T)x-(I-T)y\parallel }^2,$$

(43)

for every $x,y\in C$ and for some $j(x-y)\in J(x-y)$. Therefore, $I-T$ is $\lambda$-inverse-strongly accretive.

By Theorem 1, we can obtain the following results easily.

Theorem 2.

Let C be a convex closed set in a 2-uniformly smooth and uniformly convex Banach space E which admits a weakly sequentially continuous duality mapping. Let ${\mathrm{\Pi }}_C$ be a sunny nonexpansive retraction from E onto C. Let the mappings $B_1,B_2:C\to C$ be α-strictly pseudocontractive and β-strictly pseudocontractive, respectively. Let $f:C\to C$ be a contraction mapping with coefficient $\gamma \in [0,1)$ and $F:E\to E$ be a strongly positive linear bounded operator with the coefficient $\overline{\gamma }$ such that $0<\gamma <\overline{\gamma }\theta$ and $0<\theta \le {\parallel F\parallel }^{-1}$. Let $T:C\to C$ be uniformly continuous and asymptotically nonexpansive mapping in the intermediate sense, and ${\left\{S_n\right\}}_{n=0}^{\infty }$ be a countable family of ℓ-uniformly Lipschitzian pseudocontractive self-mappings on C such that $\mathrm{\Omega }:={\bigcap }_{n=0}^{\infty }\mathrm{Fix}\left(S_n\right)\cap \mathrm{GSVI}(C,I-B_1,I-B_2)\cap \mathrm{Fix}\left(T\right)\ne \varnothing$ where $\mathrm{GSVI}(C,I-B_1,I-B_2)$ is the fixed-point set of the mapping $G:=[(1-{\mu }_1)I+{\mu }_1{B}_1][(1-{\mu }_2)I+{\mu }_2{B}_2]$ with $0<{\mu }_1{\kappa }^2<min\{1,\alpha \}$ and $0<{\mu }_2{\kappa }^2<min\{1,\beta \}$ for κ the 2-uniformly smooth constant of E. Assume that ${\sum }_{n=0}^{\infty }{c}_n<\infty$, where $c_n$ is defined by (2). For arbitrarily given $x_0\in C$, let $\left\{x_n\right\}$ be the sequence generated by

$$\left\{\begin{array}{c}{y}_n=(1-{\mu }_1)z_n+{\mu }_1{B}_1{z}_n,\hfill \\ z_n=(1-{\mu }_2)u_n+{\mu }_2{B}_2{u}_n,\hfill \\ u_n={\beta }_n(x_n-S_n{u}_n)+S_n{u}_n,\hfill \\ x_{n+1}={\mathrm{\Pi }}_C[(T^n{y}_n-{\alpha }_n\theta F{T}^n{y}_n)+{\alpha }_{n}f\left(x_n\right)],\forall n\ge 0,\hfill \end{array}\right.$$

(44)

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

(i) 

${\sum }_{n=0}^{\infty }{\alpha }_n=\infty$, ${\sum }_{n=1}^{\infty }|{\alpha }_n-{\alpha }_{n-1}|<\infty$ and ${lim}_{n\to \infty }{\alpha }_n=0$;

(ii) 

$0<{lim\; inf}_{n\to \infty }{\beta }_n\le {lim\; sup}_{n\to \infty }{\beta }_n<1$ and ${\sum }_{n=1}^{\infty }|{\beta }_n-{\beta }_{n-1}|<\infty$.

Assume that ${\sum }_{n=1}^{\infty }{sup}_{x\in D}\parallel S_{n}x-S_{n-1}x\parallel <\infty$ for any bounded subset D of C, and let S be a mapping of C into itself defined by $Sx={lim}_{n\to \infty }{S}_{n}x$ for all $x\in C$, and suppose that $\mathrm{Fix}\left(S\right)={\bigcap }_{n=0}^{\infty }\mathrm{Fix}\left(S_n\right)$. If ${\sum }_{n=0}^{\infty }\parallel T^{n+1}{y}_n-T^n{y}_n\parallel <\infty$, then $\left\{x_n\right\}$ converges strongly to $x^{*}\in \mathrm{\Omega }$. In this case,

(a) 

$x^{*}\in \mathrm{\Omega }$ solves the VI: $\langle f\left(x^{*}\right)-\theta F\left(x^{*}\right),j(x-x^{*})\rangle \le 0,\forall x\in \mathrm{\Omega }$;

(b) 

$(x^{*},y^{*})$ is a solution of GSVI (1.3) for two inverse-strongly accretive mappings $I-B_i,i=1,2$, where $y^{*}=(1-{\mu }_2)x^{*}+{\mu }_2{B}_2{x}^{*}$.

Proof. 

Since the mappings $B_1,B_2:C\to C$ are $\alpha$-strictly pseudocontractive and $\beta$-strictly pseudocontractive, respectively, it can be seen readily that $I-B_1,I-B_2:C\to E$ are $\alpha$-inverse-strongly accretive and $\beta$-inverse-strongly accretive, respectively. Please note that $0<{\mu }_1{\kappa }^2<min\{1,\alpha \}$ and $0<{\mu }_2{\kappa }^2<min\{1,\beta \}$ for $\kappa$ the 2-uniformly smooth constant of E. Then, $\mathrm{GSVI}(C,I-B_1,I-B_2)$ is the fixed-point set of the following mapping

$$\begin{array}{cc}Gx:\hfill & ={\mathrm{\Pi }}_C[(1-{\mu }_1)I+{\mu }_1{B}_1]{\mathrm{\Pi }}_C[(1-{\mu }_2)x+{\mu }_2{B}_{2}x]\hfill \\ & ={\mathrm{\Pi }}_C\{(1-{\mu }_1)[(1-{\mu }_2)x+{\mu }_2{B}_{2}x]+{\mu }_1{B}_1[(1-{\mu }_2)x+{\mu }_2{B}_{2}x]\}\hfill \\ & =[(1-{\mu }_1)I+{\mu }_1{B}_1][(1-{\mu }_2)I+{\mu }_2{B}_2]x,\forall x\in C.\hfill \end{array}$$

In this case, it is easy to see that the iterative scheme (7) reduces to (44). Therefore, by Theorem 1 we obtain the desired result.  □

Theorem 3.

Let C be a bounded, convex and closed set in a 2-uniformly smooth and uniformly convex Banach space E which admits a weakly sequentially continuous duality mapping. Let ${\mathrm{\Pi }}_C$ be a sunny nonexpansive retraction from E onto C. Let the mappings $B_1,B_2:C\to C$ be α-strictly pseudocontractive and β-strictly pseudocontractive, respectively. Let $f:C\to C$ be a contraction mapping with coefficient $\gamma \in [0,1)$ and $F:E\to E$ be a strongly positive linear bounded operator with the coefficient $\overline{\gamma }$ such that $0<\gamma <\overline{\gamma }\theta$ and $0<\theta \le {\parallel F\parallel }^{-1}$. Let $T:C\to C$ be an asymptotically nonexpansive mapping with a sequence $\left\{{\theta }_n\right\}\subset [0,\infty )$ satisfying ${\sum }_{n=0}^{\infty }{\theta }_n<\infty$, and ${\left\{S_n\right\}}_{n=0}^{\infty }$ be a countable family of ℓ-uniformly Lipschitzian pseudocontractive self-mappings on C such that $\mathrm{\Omega }:={\bigcap }_{n=0}^{\infty }\mathrm{Fix}\left(S_n\right)\cap \mathrm{GSVI}(C,I-B_1,I-B_2)\cap \mathrm{Fix}\left(T\right)\ne \varnothing$ where $\mathrm{GSVI}(C,I-B_1,I-B_2)$ is the fixed-point set of the mapping $G:=[(1-{\mu }_1)I+{\mu }_1{B}_1][(1-{\mu }_2)I+{\mu }_2{B}_2]$ with $0<{\mu }_1{\kappa }^2<min\{1,\alpha \}$ and $0<{\mu }_2{\kappa }^2<min\{1,\beta \}$ for κ the 2-uniformly smooth constant of E. For arbitrarily given $x_0\in C$, let $\left\{x_n\right\}$ be the sequence generated by

$$\left\{\begin{array}{c}{z}_n=(1-{\mu }_2)u_n+{\mu }_2{B}_2{u}_n,\hfill \\ u_n={\beta }_n(x_n-S_n{u}_n)+S_n{u}_n,\hfill \\ y_n=(1-{\mu }_1)z_n+{\mu }_1{B}_1{z}_n,\hfill \\ x_{n+1}={\mathrm{\Pi }}_C[{\alpha }_{n}f\left(x_n\right)+(I-{\alpha }_n\theta F)T^n{y}_n],\forall n\ge 0,\hfill \end{array}\right.$$

(45)

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

(i) 

${\sum }_{n=0}^{\infty }{\alpha }_n=\infty$, ${\sum }_{n=1}^{\infty }|{\alpha }_n-{\alpha }_{n-1}|<\infty$ and ${lim}_{n\to \infty }{\alpha }_n=0$;

(ii) 

$0<{lim\; inf}_{n\to \infty }{\beta }_n\le {lim\; sup}_{n\to \infty }{\beta }_n<1$ and ${\sum }_{n=1}^{\infty }|{\beta }_n-{\beta }_{n-1}|<\infty$.

Assume that ${\sum }_{n=1}^{\infty }{sup}_{x\in D}\parallel S_{n}x-S_{n-1}x\parallel <\infty$ for any bounded subset D of C, and let S be a mapping of C into itself defined by $Sx={lim}_{n\to \infty }{S}_{n}x$ for all $x\in C$, and suppose that $\mathrm{Fix}\left(S\right)={\bigcap }_{n=0}^{\infty }\mathrm{Fix}\left(S_n\right)$. If ${\sum }_{n=0}^{\infty }\parallel T^{n+1}{y}_n-T^n{y}_n\parallel <\infty$, then $\left\{x_n\right\}$ converges strongly to $x^{*}\in \mathrm{\Omega }$. In this case,

(a) 

$x^{*}\in \mathrm{\Omega }$ solves the VI: $\langle f\left(x^{*}\right)-\theta F\left(x^{*}\right),j(x-x^{*})\rangle \le 0,\forall x\in \mathrm{\Omega }$;

(b) 

$(x^{*},y^{*})$ is a solution of GSVI (1.3) for two inverse-strongly accretive mappings $I-B_i,i=1,2$, where $y^{*}=(1-{\mu }_2)x^{*}+{\mu }_2{B}_2{x}^{*}$.

Proof. 

Since set C is a bounded set, we know that $\mathrm{diam}\left(C\right)={sup}_{x,y\in C}\parallel x-y\parallel <\infty$. We have that $T:C\to C$ is an asymptotically nonexpansive mapping with a sequence $\left\{{\theta }_n\right\}\subset [0,\infty )$ satisfying ${\sum }_{n=0}^{\infty }{\theta }_n<\infty$. Then, we deduce that for all $x,y\in C$,

$$\parallel T^{n}x-T^{n}y\parallel \le (1+{\theta }_n)\parallel x-y\parallel =\parallel x-y\parallel +{\theta }_n\parallel x-y\parallel \le \parallel x-y\parallel +{\theta }_n\mathrm{diam}\left(C\right).$$

Hence, we get

$$c_n:=max\{0,\underset{x,y\in C}{sup}(\parallel T^{n}x-T^{n}y\parallel -\parallel x-y\parallel \left)\right\}\le {\theta }_n\mathrm{diam}\left(C\right),$$

which immediately attains ${\sum }_{n=0}^{\infty }{c}_n<\infty$. Therefore, by Theorem 1 we derive the desired result.  □

5. Conclusions

In this work, we studied problem of solving a general system of monotone variational inequalities whose solutions are also the solutions of the CFPP of countably many nonlinear operators via a hybrid viscosity implicit iteration method. Strong convergence theorems were established in 2-uniformly smooth and uniformly convex Banach spaces. An application to CFPPs of asymptotically nonexpansive and pseudocontractive mappings, and variational inequality problems for strict pseudocontractive mappings was also given in Banach spaces. We also provided two examples to support the main results of this paper.