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# Iterative algorithms for quasi-variational inclusions and fixed point problems of pseudocontractions

Fixed Point Theory and Applications20142014:82

https://doi.org/10.1186/1687-1812-2014-82

• Received: 27 February 2014
• Accepted: 17 March 2014
• Published:

## Abstract

In this paper, quasi-variational inclusions and fixed point problems of pseudocontractions are considered. An iterative algorithm is presented. A strong convergence theorem is demonstrated.

MSC:49J40, 47J20, 47H09, 65J15.

## Keywords

• quasi-variational inclusions
• fixed point problem
• pseudocontractions
• maximal monotone
• firmly nonexpansive mappings

## 1 Introduction

Let C be a nonempty closed convex subset of a real Hilbert space H. Let $A:C\to H$ be a single-valued nonlinear mapping and $B:H\to {2}^{H}$ be a multi-valued mapping. The ‘so called’ quasi-variational inclusion problem is to find an $u\in {2}^{H}$ such that
$0\in Au+Bu.$
(1.1)
The set of solutions of (1.1) is denoted by ${\left(A+B\right)}^{-1}\left(0\right)$. A number of problems arising in structural analysis, mechanics, and economics can be studied in the framework of this kind of variational inclusions; see for instance . For related work, see . The problem (1.1) includes many problems as special cases.
1. (1)
If $B=\partial \varphi :H\to {2}^{H}$, where $\varphi :H\to R\cup +\mathrm{\infty }$ is a proper convex lower semi-continuous function and ∂ϕ is the subdifferential of ϕ, then the variational inclusion problem (1.1) is equivalent to finding $u\in H$ such that
$〈Au,y-u〉+\varphi \left(y\right)-\varphi \left(u\right)\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }y\in H,$

which is called the mixed quasi-variational inequality (see ).

2. (2)
If $B=\partial {\delta }_{C}$, where C is a nonempty closed convex subset of H and ${\delta }_{C}:H\to \left[0,\mathrm{\infty }\right]$ is the indicator function of C, i.e.,
${\delta }_{C}=\left\{\begin{array}{cc}0,\hfill & x\in C,\hfill \\ +\mathrm{\infty },\hfill & x\notin C,\hfill \end{array}$

then the variational inclusion problem (1.1) is equivalent to finding $u\in C$ such that
$〈Au,v-u〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }v\in C.$

This problem is called the Hartman-Stampacchia variational inequality (see ).

Let $T:C\to C$ be a nonlinear mapping. The iterative scheme of Mann’s type for approximating fixed points of T is the following: ${x}_{0}\in C$ and
${x}_{n+1}={\alpha }_{n}{x}_{n}+\left(1-{\alpha }_{n}\right)T{x}_{n},$
for all $n\ge 1$, where $\left\{{\alpha }_{n}\right\}$ is a sequence in $\left[0,1\right]$; see . For two nonlinear mappings S and T, Takahashi and Tamura  considered the following iteration procedure: ${x}_{0}\in C$ and
${x}_{n+1}={\alpha }_{n}{x}_{n}+\left(1-{\alpha }_{n}\right)S\left({\beta }_{n}{x}_{n}+\left(1-{\beta }_{n}\right)T{x}_{n}\right),$

for all $n\ge 1$, where $\left\{{\alpha }_{n}\right\}$ and $\left\{{\beta }_{n}\right\}$ are two sequences in $\left[0,1\right]$. Algorithms for finding the fixed points of nonlinear mappings or for finding the zero points of maximal monotone operators have been studied by many authors. The reader can refer to . Especially, Takahashi et al.  recently gave the following convergence result.

Theorem 1.1 Let C be a closed and convex subset of a real Hilbert space H. Let A be an α-inverse strongly monotone mapping of C into H and let B be a maximal monotone operator on H, such that the domain of B is included in C. Let ${J}_{\lambda }^{B}={\left(I+\lambda B\right)}^{-1}$ be the resolvent of B for $\lambda >0$ and let T be a nonexpansive mapping of C into itself, such that $F\left(T\right)\cap {\left(A+B\right)}^{-1}0\ne \mathrm{\varnothing }$. Let ${x}_{1}=x\in C$ and let $\left\{{x}_{n}\right\}\subset C$ be a sequence generated by
${x}_{n+1}={\beta }_{n}{x}_{n}+\left(1-{\beta }_{n}\right)T\left({\alpha }_{n}x+\left(1-{\alpha }_{n}\right){J}_{{\lambda }_{n}}^{B}\left({x}_{n}-{\lambda }_{n}A{x}_{n}\right)\right),$
for all $n\ge 0$, where $\left\{{\lambda }_{n}\right\}\subset \left(0,2\alpha \right)$, $\left\{{\alpha }_{n}\right\}\subset \left(0,1\right)$ and $\left\{{\beta }_{n}\right\}\subset \left(0,1\right)$ satisfy
$\begin{array}{c}0

Then $\left\{{x}_{n}\right\}$ converges strongly to a point of $F\left(T\right)\cap {\left(A+B\right)}^{-1}0$.

Recently, Zhang et al.  introduced a new iterative scheme for finding a common element of the set of solutions to the inclusion problem and the set of fixed points of nonexpansive mappings in Hilbert spaces. Peng et al.  introduced another iterative scheme by the viscosity approximate method for finding a common element of the set of solutions of a variational inclusion with set-valued maximal monotone mapping and inverse strongly monotone mappings, the set of solutions of an equilibrium problem, and the set of fixed points of a nonexpansive mapping.

Motivated and inspired by the works in this field, the purpose of this paper is to consider the quasi-variational inclusions and fixed point problems of pseudocontractions. An iterative algorithm is presented. A strong convergence theorem is demonstrated.

## 2 Notations and lemmas

Let H be a real Hilbert space with inner product $〈\cdot \phantom{\rule{0.2em}{0ex}},\cdot 〉$ and norm $\parallel \cdot \parallel$, respectively. Let C be a nonempty closed convex subset of H. It is well known that in a real Hilbert space H, the following equality holds:
${\parallel tx+\left(1-t\right)y\parallel }^{2}=t{\parallel x\parallel }^{2}+\left(1-t\right){\parallel y\parallel }^{2}-t\left(1-t\right){\parallel x-y\parallel }^{2}$
(2.1)

for all $x,y\in H$ and $t\in \left[0,1\right]$.

Recall that a mapping $T:C\to C$ is called

(D1) L-Lipschitzian there exists $L>0$ such that $\parallel Tx-Ty\parallel \le L\parallel x-y\parallel$ for all $x,y\in C$; in the case of $L=1$, T is said to be nonexpansive;

(D2) Firmly nonexpansive ${\parallel Tx-Ty\parallel }^{2}\le {\parallel x-y\parallel }^{2}-{\parallel \left(I-T\right)x-\left(I-T\right)y\parallel }^{2}⟺{\parallel Tx-Ty\parallel }^{2}\le 〈Tx-Ty,x-y〉$ for all $x,y\in C$;

(D3) Pseudocontractive $〈Tx-Ty,x-y〉\le {\parallel x-y\parallel }^{2}⟺{\parallel Tx-Ty\parallel }^{2}\le {\parallel x-y\parallel }^{2}+{\parallel \left(I-T\right)x-\left(I-T\right)y\parallel }^{2}$ for all $x,y\in C$;

(D4) Strongly monotone there exists a positive constant $\stackrel{˜}{\gamma }$ such that $〈Tx-Ty,x-y〉\ge \stackrel{˜}{\gamma }\parallel x-y\parallel$ for all $x,y\in C$;

(D5) Inverse strongly monotone $⟹〈Tx-Ty,x-y〉\ge \alpha {\parallel Tx-Ty\parallel }^{2}$ for some $\alpha >0$ and for all $x,y\in C$.

Let B be a mapping of H into ${2}^{H}$. The effective domain of B is denoted by $dom\left(B\right)$, that is, $dom\left(B\right)=\left\{x\in H:Bx\ne \mathrm{\varnothing }\right\}$. A multi-valued mapping B is said to be a monotone operator on H iff
$〈x-y,u-v〉\ge 0$

for all $x,y\in dom\left(B\right)$, $u\in Bx$, and $v\in By$. A monotone operator B on H is said to be maximal iff its graph is not strictly contained in the graph of any other monotone operator on H. Let B be a maximal monotone operator on H and let ${B}^{-1}0=\left\{x\in H:0\in Bx\right\}$.

For a maximal monotone operator B on H and $\lambda >0$, we may define a single-valued operator ${J}_{\lambda }^{B}={\left(I+\lambda B\right)}^{-1}:H\to dom\left(B\right)$, which is called the resolvent of B for λ. It is known that the resolvent ${J}_{\lambda }^{B}$ is firmly nonexpansive, i.e.,
${\parallel {J}_{\lambda }^{B}x-{J}_{\lambda }^{B}y\parallel }^{2}\le 〈{J}_{\lambda }^{B}x-{J}_{\lambda }^{B}y,x-y〉$

for all $x,y\in C$ and ${B}^{-1}0=Fix\left({J}_{\lambda }^{B}\right)$ for all $\lambda >0$.

Usually, the convergence of fixed point algorithms requires some additional smoothness properties of the mapping T such as demi-closedness.

Recall that a mapping T is said to be demiclosed if, for any sequence $\left\{{x}_{n}\right\}$ which weakly converges to $\stackrel{˜}{x}$, and if the sequence $\left\{T\left({x}_{n}\right)\right\}$ strongly converges to z, then $T\left(\stackrel{˜}{x}\right)=z$. For the pseudocontractions, the following demiclosed principle is well known.

Lemma 2.1 ()

Let H be a real Hilbert space, C a closed convex subset of H. Let $U:C\to C$ be a continuous pseudo-contractive mapping. Then
1. (i)

$Fix\left(U\right)$ is a closed convex subset of C,

2. (ii)

$\left(I-U\right)$ is demiclosed at zero.

Lemma 2.2 ()

Let $\left\{{r}_{n}\right\}$ be a sequence of real numbers. Assume $\left\{{r}_{n}\right\}$ does not decrease at infinity, that is, there exists at least a subsequence $\left\{{r}_{{n}_{k}}\right\}$ of $\left\{{r}_{n}\right\}$ such that ${r}_{{n}_{k}}\le {r}_{{n}_{k}+1}$ for all $k\ge 0$. For every $n\ge N$, define an integer sequence $\left\{\tau \left(n\right)\right\}$ as
$\tau \left(n\right)=max\left\{i\le n:{r}_{{n}_{i}}<{r}_{{n}_{i}+1}\right\}.$
Then $\tau \left(n\right)\to \mathrm{\infty }$ as $n\to \mathrm{\infty }$, and for all $n\ge N$
$max\left\{{r}_{\tau \left(n\right)},{r}_{n}\right\}\le {r}_{\tau \left(n\right)+1}.$

Lemma 2.3 ()

Assume $\left\{{a}_{n}\right\}$ is a sequence of nonnegative real numbers such that
${a}_{n+1}\le \left(1-{\gamma }_{n}\right){a}_{n}+{\delta }_{n}{\gamma }_{n},$
where $\left\{{\gamma }_{n}\right\}$ is a sequence in $\left(0,1\right)$ and $\left\{{\delta }_{n}\right\}$ is a sequence such that
1. (1)

${\sum }_{n=1}^{\mathrm{\infty }}{\gamma }_{n}=\mathrm{\infty }$;

2. (2)

${lim sup}_{n\to \mathrm{\infty }}{\delta }_{n}\le 0$ or ${\sum }_{n=1}^{\mathrm{\infty }}|{\delta }_{n}{\gamma }_{n}|<\mathrm{\infty }$.

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

In the sequel we shall use the following notations:
1. 1.

denote the weak ω-limit set of $\left\{{u}_{n}\right\}$;

2. 2.

${u}_{n}⇀x$ stands for the weak convergence of $\left\{{u}_{n}\right\}$ to x;

3. 3.

${u}_{n}\to x$ stands for the strong convergence of $\left\{{u}_{n}\right\}$ to x;

4. 4.

$Fix\left(T\right)$ stands for the set of fixed points of T.

## 3 Main results

In this section, we consider a strong convergence theorem for quasi-variational inclusions and fixed point problems of pseudocontractive mappings in a Hilbert space.

Algorithm 3.1 Let C be a nonempty closed and convex subset of a real Hilbert space H. Let A be an α-inverse strongly monotone mapping of C into H and let B be a maximal monotone operator on H, such that the domain of B is included in C. Let ${J}_{\lambda }^{B}={\left(I+\lambda B\right)}^{-1}$ be the resolvent of B for λ. Let $F:C\to H$ be an ${L}_{1}$-Lipschitzian and ς strongly monotone mapping and $f:C\to C$ be a ρ-contraction such that $\rho . Let $T:C\to C$ be an ${L}_{2}\left(>1\right)$-Lipschitzian pseudocontraction. For ${x}_{0}\in C$, define a sequence $\left\{{x}_{n}\right\}$ as follows:
$\left\{\begin{array}{c}{z}_{n}={J}_{\lambda }^{B}\left(I-\lambda A\right){x}_{n},\hfill \\ {y}_{n}=\nu {z}_{n}+\left(1-\nu \right)T\left(\left(1-\zeta \right){z}_{n}+\zeta T{z}_{n}\right),\hfill \\ {x}_{n+1}={\alpha }_{n}{x}_{n}+\left(1-{\alpha }_{n}\right)\left({\beta }_{n}f\left({x}_{n}\right)+\left(I-{\beta }_{n}F\right){y}_{n}\right),\hfill \end{array}$
(3.1)

for all $n\in \mathbb{N}$, where λ, ν and ζ are three constants, $\left\{{\alpha }_{n}\right\}$ and $\left\{{\beta }_{n}\right\}$ are two sequences in $\left[0,1\right]$.

Now, we demonstrate the convergence analysis of the algorithm (3.1).

Theorem 3.2 Suppose $\mathrm{\Gamma }:=Fix\left(T\right)\cap {\left(A+B\right)}^{-1}\left(0\right)\ne \mathrm{\varnothing }$. Assume the following conditions are satisfied:

(C1) ${\alpha }_{n}\in \left[a,b\right]\subset \left(0,1\right)$;

(C2) ${lim}_{n\to \mathrm{\infty }}{\beta }_{n}=0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\beta }_{n}=\mathrm{\infty }$;

(C3) $\lambda \in \left(0,2\alpha \right)$ and $0<1-\nu \le \zeta <\frac{1}{\sqrt{1+{L}_{2}^{2}}+1}$.

Then the sequence $\left\{{x}_{n}\right\}$ defined by (3.1) converges strongly to $u={P}_{\mathrm{\Gamma }}\left(I-F+f\right)u$.

Proof Let ${x}^{\ast }\in Fix\left(T\right)\cap {\left(A+B\right)}^{-1}\left(0\right)$. Then, we get ${x}^{\ast }={J}_{\lambda }^{B}\left(I-\lambda A\right){x}^{\ast }=T{x}^{\ast }$. From (3.1), we have
$\begin{array}{rl}{\parallel {z}_{n}-{x}^{\ast }\parallel }^{2}& ={\parallel {J}_{\lambda }^{B}\left(I-\lambda A\right){x}_{n}-{J}_{\lambda }^{B}\left(I-\lambda A\right){x}^{\ast }\parallel }^{2}\\ \le {\parallel {x}_{n}-{x}^{\ast }-\lambda \left(A{x}_{n}-A{x}^{\ast }\right)\parallel }^{2}\\ ={\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}-2\lambda 〈A{x}_{n}-A{x}^{\ast },{x}_{n}-{x}^{\ast }〉+{\lambda }^{2}{\parallel A{x}_{n}-A{x}^{\ast }\parallel }^{2}\\ \le {\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}-2\lambda \alpha {\parallel A{x}_{n}-A{x}^{\ast }\parallel }^{2}+{\lambda }^{2}{\parallel A{x}_{n}-A{x}^{\ast }\parallel }^{2}\\ ={\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}-\lambda \left(2\alpha -\lambda \right){\parallel A{x}_{n}-A{x}^{\ast }\parallel }^{2}\\ \le {\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}.\end{array}$
(3.2)
It follows that
$\parallel {z}_{n}-{x}^{\ast }\parallel \le \parallel {x}_{n}-{x}^{\ast }\parallel .$
(3.3)
Since ${x}^{\ast }\in Fix\left(T\right)$, we have from (D3) that
${\parallel Tx-{x}^{\ast }\parallel }^{2}\le {\parallel x-{x}^{\ast }\parallel }^{2}+{\parallel Tx-x\parallel }^{2},$
(3.4)

for all $x\in C$.

Thus,
$\begin{array}{rl}{\parallel T\left(\left(1-\zeta \right)I+\zeta T\right){z}_{n}-{x}^{\ast }\parallel }^{2}\le & {\parallel \left(1-\zeta \right)\left({z}_{n}-{x}^{\ast }\right)+\zeta \left(T{z}_{n}-{x}^{\ast }\right)\parallel }^{2}\\ +{\parallel \left(\left(1-\zeta \right)I+\zeta T\right){z}_{n}-T\left(\left(1-\zeta \right)I+\zeta T\right){z}_{n}\parallel }^{2}.\end{array}$
(3.5)
By (3.4), (3.5), and (2.1), we obtain
$\begin{array}{r}{\parallel T\left(\left(1-\zeta \right)I+\zeta T\right){z}_{n}-{x}^{\ast }\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le {\parallel \left(1-\zeta \right)\left({z}_{n}-{x}^{\ast }\right)+\zeta \left(T{z}_{n}-{x}^{\ast }\right)\parallel }^{2}\\ \phantom{\rule{2em}{0ex}}+{\parallel \left(\left(1-\zeta \right)I+\zeta T\right){z}_{n}-T\left(\left(1-\zeta \right)I+\zeta T\right){z}_{n}\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}={\parallel \left(1-\zeta \right)\left({z}_{n}-T\left(\left(1-\zeta \right)I+\zeta T\right){z}_{n}\right)+\zeta \left(T{z}_{n}-T\left(\left(1-\zeta \right)I+\zeta T\right){z}_{n}\right)\parallel }^{2}\\ \phantom{\rule{2em}{0ex}}+{\parallel \left(1-\zeta \right)\left({z}_{n}-{x}^{\ast }\right)+\zeta \left(T{z}_{n}-{x}^{\ast }\right)\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}=\left(1-\zeta \right){\parallel {z}_{n}-T\left(\left(1-\zeta \right)I+\zeta T\right){z}_{n}\parallel }^{2}+\zeta {\parallel T{z}_{n}-T\left(\left(1-\zeta \right)I+\zeta T\right){z}_{n}\parallel }^{2}\\ \phantom{\rule{2em}{0ex}}-\zeta \left(1-\zeta \right){\parallel {z}_{n}-T{z}_{n}\parallel }^{2}+\left(1-\zeta \right){\parallel {z}_{n}-{x}^{\ast }\parallel }^{2}+\zeta {\parallel T{z}_{n}-{x}^{\ast }\parallel }^{2}-\zeta \left(1-\zeta \right){\parallel {z}_{n}-T{z}_{n}\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le \left(1-\zeta \right){\parallel {z}_{n}-{x}^{\ast }\parallel }^{2}+\zeta \left({\parallel {z}_{n}-{x}^{\ast }\parallel }^{2}+{\parallel {z}_{n}-T{z}_{n}\parallel }^{2}\right)\\ \phantom{\rule{2em}{0ex}}-2\zeta \left(1-\zeta \right){\parallel {z}_{n}-T{z}_{n}\parallel }^{2}+\left(1-\zeta \right){\parallel {z}_{n}-T\left(\left(1-\zeta \right)I+\zeta T\right){z}_{n}\parallel }^{2}\\ \phantom{\rule{2em}{0ex}}+\zeta {\parallel T{z}_{n}-T\left(\left(1-\zeta \right)I+\zeta T\right){z}_{n}\parallel }^{2}.\end{array}$
Noting that T is ${L}_{2}$-Lipschitzian and ${z}_{n}-\left(\left(1-\zeta \right)I+\zeta T\right){z}_{n}=\zeta \left({z}_{n}-T{z}_{n}\right)$, we have
$\begin{array}{r}{\parallel T\left(\left(1-\zeta \right)I+\zeta T\right){z}_{n}-{x}^{\ast }\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le \left(1-\zeta \right){\parallel {z}_{n}-{x}^{\ast }\parallel }^{2}+\zeta \left({\parallel {z}_{n}-{x}^{\ast }\parallel }^{2}+{\parallel {z}_{n}-T{z}_{n}\parallel }^{2}\right)\\ \phantom{\rule{2em}{0ex}}-2\zeta \left(1-\zeta \right){\parallel {z}_{n}-T{z}_{n}\parallel }^{2}+\left(1-\zeta \right){\parallel {z}_{n}-T\left(\left(1-\zeta \right)I+\zeta T\right){z}_{n}\parallel }^{2}+{\zeta }^{3}{L}_{2}^{2}{\parallel {z}_{n}-T{z}_{n}\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}={\parallel {z}_{n}-{x}^{\ast }\parallel }^{2}+\left(1-\zeta \right){\parallel {z}_{n}-T\left(\left(1-\zeta \right)I+\zeta T\right){z}_{n}\parallel }^{2}\\ \phantom{\rule{2em}{0ex}}-\zeta \left(1-2\zeta -{\zeta }^{2}{L}_{2}^{2}\right){\parallel {z}_{n}-T{z}_{n}\parallel }^{2}.\end{array}$
(3.6)
Since $\zeta <\frac{1}{\sqrt{1+{L}_{2}^{2}}+1}$, we have $1-2\zeta -{\zeta }^{2}{L}_{2}^{2}>0$. From (3.6), we can deduce
${\parallel T\left(\left(1-\zeta \right)I+\zeta T\right){z}_{n}-{x}^{\ast }\parallel }^{2}\le {\parallel {z}_{n}-{x}^{\ast }\parallel }^{2}+\left(1-\zeta \right){\parallel {z}_{n}-T\left(\left(1-\zeta \right)I+\zeta T\right){z}_{n}\parallel }^{2}.$
(3.7)
Hence,
$\begin{array}{rl}{\parallel {y}_{n}-{x}^{\ast }\parallel }^{2}=& {\parallel \nu {z}_{n}+\left(1-\nu \right)T\left(\left(1-\zeta \right)I+\zeta T\right){z}_{n}-{x}^{\ast }\parallel }^{2}\\ =& {\parallel \nu \left({z}_{n}-{x}^{\ast }\right)+\left(1-\nu \right)\left(T\left(\left(1-\zeta \right)I+\zeta T\right){z}_{n}-{x}^{\ast }\right)\parallel }^{2}\\ =& \nu {\parallel {z}_{n}-{x}^{\ast }\parallel }^{2}+\left(1-\nu \right){\parallel T\left(\left(1-\zeta \right)I+\zeta T\right){z}_{n}-{x}^{\ast }\parallel }^{2}\\ -\nu \left(1-\nu \right){\parallel T\left(\left(1-\zeta \right)I+\zeta T\right){z}_{n}-{z}_{n}\parallel }^{2}\\ \le & \nu {\parallel {z}_{n}-{x}^{\ast }\parallel }^{2}+\left(1-\nu \right)\left[{\parallel {z}_{n}-{x}^{\ast }\parallel }^{2}+\left(1-\zeta \right){\parallel {z}_{n}-T\left(\left(1-\zeta \right)I+\zeta T\right){z}_{n}\parallel }^{2}\right]\\ -\nu \left(1-\nu \right){\parallel T\left(\left(1-\zeta \right)I+\zeta T\right){z}_{n}-{z}_{n}\parallel }^{2}\\ =& {\parallel {z}_{n}-{x}^{\ast }\parallel }^{2}+\left(1-\nu \right)\left(1-\zeta -\nu \right){\parallel T\left(\left(1-\zeta \right)I+\zeta T\right){z}_{n}-{z}_{n}\parallel }^{2}.\end{array}$
(3.8)
By (C3) and (3.8), we obtain
$\parallel {y}_{n}-{x}^{\ast }\parallel \le \parallel {z}_{n}-{x}^{\ast }\parallel .$
(3.9)
Let ${u}_{n}={\beta }_{n}f\left({x}_{n}\right)+\left(I-{\beta }_{n}F\right){y}_{n}$ for all $n\ge 0$. Then, we have
$\begin{array}{rl}\parallel {u}_{n}-{x}^{\ast }\parallel =& \parallel {\beta }_{n}f\left({x}_{n}\right)+\left(I-{\beta }_{n}F\right){y}_{n}-{x}^{\ast }\parallel \\ \le & {\beta }_{n}\parallel f\left({x}_{n}\right)-F{x}^{\ast }\parallel +\parallel \left(I-{\beta }_{n}F\right){y}_{n}-\left(I-{\beta }_{n}F\right){x}^{\ast }\parallel \\ \le & {\beta }_{n}\parallel f\left({x}_{n}\right)-f\left({x}^{\ast }\right)\parallel +{\beta }_{n}\parallel f\left({x}^{\ast }\right)-F{x}^{\ast }\parallel \\ +\parallel \left(I-{\beta }_{n}F\right){y}_{n}-\left(I-{\beta }_{n}F\right){x}^{\ast }\parallel \\ \le & {\beta }_{n}\rho \parallel {x}_{n}-{x}^{\ast }\parallel +{\beta }_{n}\parallel f\left({x}^{\ast }\right)-F{x}^{\ast }\parallel \\ +\parallel \left(I-{\beta }_{n}F\right){y}_{n}-\left(I-{\beta }_{n}F\right){x}^{\ast }\parallel .\end{array}$
(3.10)
Since F is ${L}_{1}$-Lipschitzian and ς strongly monotone, we have
$\begin{array}{r}{\parallel \left(I-{\beta }_{n}F\right){y}_{n}-\left(I-{\beta }_{n}F\right){x}^{\ast }\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}={\parallel \left({y}_{n}-{x}^{\ast }\right)-{\beta }_{n}\left(F{y}_{n}-F{x}^{\ast }\right)\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}={\parallel {y}_{n}-{x}^{\ast }\parallel }^{2}-2{\beta }_{n}〈F{y}_{n}-F{x}^{\ast },{y}_{n}-{x}^{\ast }〉+{\beta }_{n}^{2}{\parallel F{y}_{n}-F{x}^{\ast }\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le {\parallel {y}_{n}-{x}^{\ast }\parallel }^{2}-2{\beta }_{n}\varsigma {\parallel {y}_{n}-{x}^{\ast }\parallel }^{2}+{\beta }_{n}^{2}{L}_{1}^{2}{\parallel {y}_{n}-{x}^{\ast }\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}=\left(1-2{\beta }_{n}\varsigma +{\beta }_{n}^{2}{L}_{1}^{2}\right){\parallel {y}_{n}-{x}^{\ast }\parallel }^{2}.\end{array}$
(3.11)
Noting that ${L}_{1}\ge \varsigma$ and ${lim}_{n\to \mathrm{\infty }}{\beta }_{n}=0$, without loss of generality, we assume that ${\beta }_{n}<\frac{\varsigma }{{L}_{1}^{2}-\varsigma /4}$ for all $n\ge 0$. Thus, $1-2{\beta }_{n}\varsigma +{\beta }_{n}^{2}{L}_{1}^{2}\le {\left(1-{\beta }_{n}\frac{\varsigma }{2}\right)}^{2}$. So,
$\parallel \left(I-{\beta }_{n}F\right){y}_{n}-\left(I-{\beta }_{n}F\right){x}^{\ast }\parallel \le \left(1-{\beta }_{n}\frac{\varsigma }{2}\right)\parallel {y}_{n}-{x}^{\ast }\parallel .$
(3.12)
We have from (3.9), (3.10), and (3.12)
$\begin{array}{rl}\parallel {u}_{n}-{x}^{\ast }\parallel & \le {\beta }_{n}\rho \parallel {x}_{n}-{x}^{\ast }\parallel +{\beta }_{n}\parallel f\left({x}^{\ast }\right)-F{x}^{\ast }\parallel +\left(1-{\beta }_{n}\frac{\varsigma }{2}\right)\parallel {x}_{n}-{x}^{\ast }\parallel \\ =\left[1-\left(\frac{\varsigma }{2}-\rho \right){\beta }_{n}\right]\parallel {x}_{n}-{x}^{\ast }\parallel +{\beta }_{n}\parallel f\left({x}^{\ast }\right)-F{x}^{\ast }\parallel .\end{array}$
(3.13)
From (3.1) and (3.13), we have
$\begin{array}{rl}\parallel {x}_{n+1}-{x}^{\ast }\parallel =& \parallel {\alpha }_{n}\left({x}_{n}-{x}^{\ast }\right)+\left(1-{\alpha }_{n}\right)\left({u}_{n}-{x}^{\ast }\right)\parallel \\ \le & \left(1-{\alpha }_{n}\right)\left(\left[1-\left(\frac{\varsigma }{2}-\rho \right){\beta }_{n}\right]\parallel {x}_{n}-{x}^{\ast }\parallel +{\beta }_{n}\parallel f\left({x}^{\ast }\right)-F{x}^{\ast }\parallel \right)\\ +{\alpha }_{n}\parallel {x}_{n}-{x}^{\ast }\parallel \\ =& \left[1-\left(\frac{\varsigma }{2}-\rho \right)\left(1-{\alpha }_{n}\right){\beta }_{n}\right]\parallel {x}_{n}-{x}^{\ast }\parallel +\left(1-{\alpha }_{n}\right){\beta }_{n}\parallel f\left({x}^{\ast }\right)-F{x}^{\ast }\parallel .\end{array}$
(3.14)
By the definition of ${x}_{n}$, we have
$\begin{array}{rl}{x}_{n+1}-{x}_{n}& ={\alpha }_{n}{x}_{n}+\left(1-{\alpha }_{n}\right)\left({\beta }_{n}f\left({x}_{n}\right)+\left(I-{\beta }_{n}F\right){y}_{n}\right)-{x}_{n}\\ =\left(1-{\alpha }_{n}\right)\left[{\beta }_{n}f\left({x}_{n}\right)-{\beta }_{n}F{y}_{n}+{y}_{n}-{x}_{n}\right].\end{array}$
(3.15)
Hence,
$\begin{array}{rl}〈{x}_{n+1}-{x}_{n},{x}_{n}-{x}^{\ast }〉=& 〈\left(1-{\alpha }_{n}\right)\left[{\beta }_{n}f\left({x}_{n}\right)-{\beta }_{n}F{y}_{n}+{y}_{n}-{x}_{n}\right],{x}_{n}-{x}^{\ast }〉\\ =& \left(1-{\alpha }_{n}\right){\beta }_{n}〈f\left({x}_{n}\right),{x}_{n}-{x}^{\ast }〉-\left(1-{\alpha }_{n}\right){\beta }_{n}〈F{y}_{n},{x}_{n}-{x}^{\ast }〉\\ +\left(1-{\alpha }_{n}\right)〈{y}_{n}-{x}_{n},{x}_{n}-{x}^{\ast }〉.\end{array}$
(3.16)
Since $2〈{x}_{n+1}-{x}_{n},{x}_{n}-{x}^{\ast }〉={\parallel {x}_{n+1}-{x}^{\ast }\parallel }^{2}-{\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}-{\parallel {x}_{n+1}-{x}_{n}\parallel }^{2}$ and $2〈{y}_{n}-{x}_{n},{x}_{n}-{x}^{\ast }〉={\parallel {y}_{n}-{x}^{\ast }\parallel }^{2}-{\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}-{\parallel {y}_{n}-{x}_{n}\parallel }^{2}$, it follows from (3.16), (3.3), and (3.9) that
$\begin{array}{r}{\parallel {x}_{n+1}-{x}^{\ast }\parallel }^{2}-{\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}-{\parallel {x}_{n+1}-{x}_{n}\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}=2\left(1-{\alpha }_{n}\right){\beta }_{n}〈f\left({x}_{n}\right),{x}_{n}-{x}^{\ast }〉-2\left(1-{\alpha }_{n}\right){\beta }_{n}〈F{y}_{n},{x}_{n}-{x}^{\ast }〉\\ \phantom{\rule{2em}{0ex}}+\left(1-{\alpha }_{n}\right)\left[{\parallel {y}_{n}-{x}^{\ast }\parallel }^{2}-{\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}-{\parallel {y}_{n}-{x}_{n}\parallel }^{2}\right]\\ \phantom{\rule{1em}{0ex}}\le 2\left(1-{\alpha }_{n}\right){\beta }_{n}〈f\left({x}_{n}\right),{x}_{n}-{x}^{\ast }〉-2\left(1-{\alpha }_{n}\right){\beta }_{n}〈F{y}_{n},{x}_{n}-{x}^{\ast }〉\\ \phantom{\rule{2em}{0ex}}-\left(1-{\alpha }_{n}\right){\parallel {y}_{n}-{x}_{n}\parallel }^{2}.\end{array}$
(3.17)
By (3.15), we obtain
$\begin{array}{rl}{\parallel {x}_{n+1}-{x}_{n}\parallel }^{2}\le & {\left(1-{\alpha }_{n}\right)}^{2}{\left[{\beta }_{n}\parallel f\left({x}_{n}\right)-F{y}_{n}\parallel +\parallel {y}_{n}-{x}_{n}\parallel \right]}^{2}\\ =& {\left(1-{\alpha }_{n}\right)}^{2}\left[{\beta }_{n}^{2}{\parallel f\left({x}_{n}\right)-F{y}_{n}\parallel }^{2}+{\parallel {y}_{n}-{x}_{n}\parallel }^{2}\\ +2{\beta }_{n}\parallel f\left({x}_{n}\right)-F{y}_{n}\parallel \parallel {y}_{n}-{x}_{n}\parallel \right].\end{array}$
(3.18)
Combining (3.17) and (3.18) to deduce
$\begin{array}{r}{\parallel {x}_{n+1}-{x}^{\ast }\parallel }^{2}-{\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le 2\left(1-{\alpha }_{n}\right){\beta }_{n}〈f\left({x}_{n}\right),{x}_{n}-{x}^{\ast }〉-2\left(1-{\alpha }_{n}\right){\beta }_{n}〈F{y}_{n},{x}_{n}-{x}^{\ast }〉\\ \phantom{\rule{2em}{0ex}}-\left(1-{\alpha }_{n}\right){\parallel {y}_{n}-{x}_{n}\parallel }^{2}+{\left(1-{\alpha }_{n}\right)}^{2}\left[{\beta }_{n}^{2}{\parallel f\left({x}_{n}\right)-F{y}_{n}\parallel }^{2}\\ \phantom{\rule{2em}{0ex}}+{\parallel {y}_{n}-{x}_{n}\parallel }^{2}+2{\beta }_{n}\parallel f\left({x}_{n}\right)-F{y}_{n}\parallel \parallel {y}_{n}-{x}_{n}\parallel \right]\\ \phantom{\rule{1em}{0ex}}\le 2\left(1-{\alpha }_{n}\right){\beta }_{n}〈f\left({x}_{n}\right),{x}_{n}-{x}^{\ast }〉-2\left(1-{\alpha }_{n}\right){\beta }_{n}〈F{y}_{n},{x}_{n}-{x}^{\ast }〉-\left(1-{\alpha }_{n}\right){\alpha }_{n}{\parallel {y}_{n}-{x}_{n}\parallel }^{2}\\ \phantom{\rule{2em}{0ex}}+{\left(1-{\alpha }_{n}\right)}^{2}\left[{\beta }_{n}^{2}{\parallel f\left({x}_{n}\right)-F{y}_{n}\parallel }^{2}+2{\beta }_{n}\parallel f\left({x}_{n}\right)-F{y}_{n}\parallel \parallel {y}_{n}-{x}_{n}\parallel \right].\end{array}$
Hence, we obtain
$\begin{array}{r}{\parallel {x}_{n+1}-{x}^{\ast }\parallel }^{2}-{\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}+\left(1-{\alpha }_{n}\right){\alpha }_{n}{\parallel {y}_{n}-{x}_{n}\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le 2\left(1-{\alpha }_{n}\right){\beta }_{n}〈f\left({x}_{n}\right),{x}_{n}-{x}^{\ast }〉-2\left(1-{\alpha }_{n}\right){\beta }_{n}〈F{y}_{n},{x}_{n}-{x}^{\ast }〉\\ \phantom{\rule{2em}{0ex}}+{\left(1-{\alpha }_{n}\right)}^{2}\left[{\beta }_{n}^{2}{\parallel f\left({x}_{n}\right)-F{y}_{n}\parallel }^{2}+2{\beta }_{n}\parallel f\left({x}_{n}\right)-F{y}_{n}\parallel \parallel {y}_{n}-{x}_{n}\parallel \right].\end{array}$
It follows that, hence, we obtain
$\begin{array}{r}\left(1-{\alpha }_{n}\right){\alpha }_{n}{\parallel {y}_{n}-{x}_{n}\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le {\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}-{\parallel {x}_{n+1}-{x}^{\ast }\parallel }^{2}+2\left(1-{\alpha }_{n}\right){\beta }_{n}〈f\left({x}_{n}\right),{x}_{n}-{x}^{\ast }〉\\ \phantom{\rule{2em}{0ex}}-2\left(1-{\alpha }_{n}\right){\beta }_{n}〈F{y}_{n},{x}_{n}-{x}^{\ast }〉\\ \phantom{\rule{2em}{0ex}}+{\left(1-{\alpha }_{n}\right)}^{2}\left[{\beta }_{n}^{2}{\parallel f\left({x}_{n}\right)-F{y}_{n}\parallel }^{2}+2{\beta }_{n}\parallel f\left({x}_{n}\right)-F{y}_{n}\parallel \parallel {y}_{n}-{x}_{n}\parallel \right].\end{array}$
(3.19)

Next we divide our proof into two possible cases.

Case 1. There exists an integer number m such that $\parallel {x}_{n+1}-{x}^{\ast }\parallel \le \parallel {x}_{n}-{x}^{\ast }$ for all $n\ge m$. In this case, we have ${lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-{x}^{\ast }\parallel$ exists. Since ${\alpha }_{n}\in \left[a,b\right]\subset \left(0,1\right)$ and ${lim}_{n\to \mathrm{\infty }}{\beta }_{n}=0$, by (3.19), we derive
$\underset{n\to \mathrm{\infty }}{lim}\parallel {y}_{n}-{x}_{n}\parallel =0.$
(3.20)
This together with (3.18) implies that
$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n+1}-{x}_{n}\parallel =0.$
(3.21)
Note that
$\begin{array}{rl}\parallel {u}_{n}-{y}_{n}\parallel & =\parallel {\beta }_{n}f\left({x}_{n}\right)+\left(I-{\beta }_{n}F\right){y}_{n}-{y}_{n}\parallel \\ \le {\beta }_{n}\parallel f\left({x}_{n}\right)-F{y}_{n}\parallel .\end{array}$
So,
$\underset{n\to \mathrm{\infty }}{lim}\parallel {u}_{n}-{y}_{n}\parallel =0.$
(3.22)
By (3.20) and (3.22), we obtain
$\underset{n\to \mathrm{\infty }}{lim}\parallel {u}_{n}-{x}_{n}\parallel =0.$
(3.23)
From (3.2) and (3.9), we have
${\parallel {y}_{n}-{x}^{\ast }\parallel }^{2}\le {\parallel {z}_{n}-{x}^{\ast }\parallel }^{2}\le {\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}-\lambda \left(2\alpha -\lambda \right){\parallel A{x}_{n}-A{x}^{\ast }\parallel }^{2}.$
Hence,
$\begin{array}{rl}\lambda \left(2\alpha -\lambda \right){\parallel A{x}_{n}-A{x}^{\ast }\parallel }^{2}& \le {\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}-{\parallel {y}_{n}-{x}^{\ast }\parallel }^{2}\\ \le \parallel {x}_{n}-{y}_{n}\parallel \left(\parallel {x}_{n}-{x}^{\ast }\parallel +\parallel {y}_{n}-{x}^{\ast }\parallel \right).\end{array}$
Therefore,
$\underset{n\to \mathrm{\infty }}{lim}\parallel A{x}_{n}-A{x}^{\ast }\parallel =0.$
(3.24)
Since ${J}_{\lambda }^{B}$ is firmly nonexpansive and A is monotone, we have
$\begin{array}{rcl}{\parallel {z}_{n}-{x}^{\ast }\parallel }^{2}& =& {\parallel {J}_{\lambda }^{B}\left(I-\lambda A\right){x}_{n}-{J}_{\lambda }^{B}\left(I-\lambda A\right){x}^{\ast }\parallel }^{2}\\ \le & 〈\left(I-\lambda A\right){x}_{n}-\left(I-\lambda A\right){x}^{\ast },{z}_{n}-{x}^{\ast }〉\\ =& 〈{z}_{n}-{x}^{\ast },{x}_{n}-{x}^{\ast }〉-\lambda 〈{z}_{n}-{x}^{\ast },A{x}_{n}-A{x}^{\ast }〉\\ =& \frac{1}{2}\left({\parallel {z}_{n}-{x}^{\ast }\parallel }^{2}+{\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}-{\parallel {z}_{n}-{x}_{n}\parallel }^{2}\right)\\ -\lambda 〈{x}_{n}-{x}^{\ast },A{x}_{n}-A{x}^{\ast }〉-\lambda 〈{z}_{n}-{x}_{n},A{x}_{n}-A{x}^{\ast }〉\\ \le & \frac{1}{2}\left({\parallel {z}_{n}-{x}^{\ast }\parallel }^{2}+{\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}-{\parallel {z}_{n}-{x}_{n}\parallel }^{2}\right)\\ +\lambda \parallel {z}_{n}-{x}_{n}\parallel \parallel A{x}_{n}-A{x}^{\ast }\parallel .\end{array}$
It follows that
${\parallel {z}_{n}-{x}^{\ast }\parallel }^{2}\le {\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}-{\parallel {z}_{n}-{x}_{n}\parallel }^{2}+2\lambda \parallel {z}_{n}-{x}_{n}\parallel \parallel A{x}_{n}-A{x}^{\ast }\parallel .$
(3.25)
By (3.25) and (3.9), we deduce
${\parallel {y}_{n}-{x}^{\ast }\parallel }^{2}\le {\parallel {z}_{n}-{x}^{\ast }\parallel }^{2}\le {\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}-{\parallel {z}_{n}-{x}_{n}\parallel }^{2}+2\lambda \parallel {z}_{n}-{x}_{n}\parallel \parallel A{x}_{n}-A{x}^{\ast }\parallel .$
Therefore,
$\begin{array}{rcl}{\parallel {z}_{n}-{x}_{n}\parallel }^{2}& \le & {\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}-{\parallel {y}_{n}-{x}^{\ast }\parallel }^{2}+2\lambda \parallel {z}_{n}-{x}_{n}\parallel \parallel A{x}_{n}-A{x}^{\ast }\parallel \\ \le & \parallel {x}_{n}-{y}_{n}\parallel \left(\parallel {x}_{n}-{x}^{\ast }\parallel +\parallel {y}_{n}-{x}^{\ast }\parallel \right)+2\lambda \parallel {z}_{n}-{x}_{n}\parallel \parallel A{x}_{n}-A{x}^{\ast }\parallel .\end{array}$
(3.26)
Equations (3.20), (3.24), and (3.26) imply that
$\underset{n\to \mathrm{\infty }}{lim}\parallel {z}_{n}-{x}_{n}\parallel =0.$
(3.27)
Notice that $F-f$ is $\left(\varsigma -\rho \right)$ strongly monotone. Thus, the variational inequality of finding $y\in \mathrm{\Gamma }$ such that $〈\left(F-f\right)y,x-y〉\ge 0$ for all $x\in \mathrm{\Gamma }$ has a unique solution, denoted by ${x}^{\ast }$, that is, ${x}^{\ast }={P}_{\mathrm{\Gamma }}\left(I-V+F\right)\left({x}^{\ast }\right)$. Next, we prove that
$\underset{n\to \mathrm{\infty }}{lim sup}〈\left(f-F\right){x}^{\ast },{u}_{n}-{x}^{\ast }〉\le 0.$
Since ${u}_{n}$ is bounded, without loss of generality, we assume that there exists a subsequence $\left\{{z}_{{n}_{i}}\right\}$ of $\left\{{u}_{n}\right\}$ such that ${u}_{{n}_{i}}⇀\stackrel{˜}{x}$ for some $\stackrel{˜}{x}\in H$ and
$\underset{n\to \mathrm{\infty }}{lim sup}〈\left(f-F\right){x}^{\ast },{u}_{n}-{x}^{\ast }〉=\underset{i\to \mathrm{\infty }}{lim sup}〈\left(f-F\right){x}^{\ast },{u}_{{n}_{i}}-{x}^{\ast }〉.$
Thus, we have that ${x}_{{n}_{i}}⇀\stackrel{˜}{x}$ and
$\underset{i\to \mathrm{\infty }}{lim}\parallel {J}_{\lambda }^{B}\left(I-\lambda A\right){x}_{{n}_{i}}-{x}_{{n}_{i}}\parallel =0.$

Therefore, $\stackrel{˜}{x}\in Fix\left({J}_{\lambda }^{B}\left(I-\lambda A\right)\right)={\left(A+B\right)}^{-1}\left(0\right)$.

Next we show that $\stackrel{˜}{x}\in Fix\left(T\right)$. First, we show that $Fix\left(T\right)=Fix\left(T\left(\left(1-\zeta \right)I+\zeta T\right)\right)$. As a matter of fact, $Fix\left(T\right)\subset Fix\left(T\left(\left(1-\zeta \right)I+\zeta T\right)\right)$ is obvious. Next, we show that $Fix\left(T\left(\left(1-\zeta \right)I+\zeta T\right)\right)\subset Fix\left(T\right)$.

Take any ${x}^{\ast }\in Fix\left(T\left(\left(1-\zeta \right)I+\zeta T\right)\right)$. We have $T\left(\left(1-\zeta \right)I+\zeta T\right){x}^{\ast }={x}^{\ast }$. Set $S=\left(1-\zeta \right)I+\zeta T$. We have $TS{x}^{\ast }={x}^{\ast }$. Write $S{x}^{\ast }={y}^{\ast }$. Then, $T{y}^{\ast }={x}^{\ast }$. Now we show ${x}^{\ast }={y}^{\ast }$. In fact,
$\begin{array}{rl}\parallel {x}^{\ast }-{y}^{\ast }\parallel & =\parallel T{y}^{\ast }-S{x}^{\ast }\parallel =\parallel T{y}^{\ast }-\left(1-\zeta \right){x}^{\ast }-\zeta T{x}^{\ast }\parallel \\ =\zeta \parallel T{y}^{\ast }-T{x}^{\ast }\parallel \le \zeta {L}_{2}\parallel {y}^{\ast }-{x}^{\ast }\parallel .\end{array}$

Since, $\zeta <\frac{1}{\sqrt{1+{L}_{2}^{2}}+1}<\frac{1}{{L}_{2}}$, we deduce ${y}^{\ast }={x}^{\ast }\in Fix\left(S\right)=Fix\left(T\right)$. Thus, ${x}^{\ast }\in Fix\left(T\right)$. Hence, $Fix\left(T\left(\left(1-\zeta \right)I+\zeta T\right)\right)\subset Fix\left(T\right)$. Therefore, $Fix\left(T\left(\left(1-\zeta \right)I+\zeta T\right)\right)=Fix\left(T\right)$.

By (3.1), (3.20), and (3.27), we deduce
$\underset{n\to \mathrm{\infty }}{lim}\parallel T\left(\left(1-\zeta \right)I+\zeta T\right){x}_{n}-{x}_{n}\parallel =0.$
(3.28)

Next we prove that $T\left(\left(1-\zeta \right)I+\zeta T\right)-I$ is demiclosed at 0. Let the sequence $\left\{{w}_{n}\right\}\subset {H}_{2}$ satisfying ${w}_{n}⇀{x}^{†}$ and ${w}_{n}-T\left(\left(1-\zeta \right)I+\zeta T\right){w}_{n}\to 0$. Next, we will show that ${x}^{†}\in Fix\left(T\left(\left(1-\zeta \right)I+\zeta T\right)\right)=Fix\left(T\right)$.

Since T is ${L}_{2}$-Lipschizian, we have
$\begin{array}{rl}\parallel {w}_{n}-T{w}_{n}\parallel & \le \parallel {w}_{n}-T\left(\left(1-\zeta \right)I+\zeta T\right){w}_{n}\parallel +\parallel T\left(\left(1-\zeta \right)I+\zeta T\right){w}_{n}-T{w}_{n}\parallel \\ \le \parallel {w}_{n}-T\left(\left(1-\zeta \right)I+\zeta T\right){w}_{n}\parallel +\zeta L\parallel {w}_{n}-T{w}_{n}\parallel .\end{array}$
It follows that
$\parallel {w}_{n}-T{w}_{n}\parallel \le \frac{1}{1-\zeta L}\parallel {w}_{n}-T\left(\left(1-\zeta \right)I+\zeta T\right){w}_{n}\parallel .$
Hence,
$\underset{n\to \mathrm{\infty }}{lim}\parallel {w}_{n}-T{w}_{n}\parallel =0.$
Since $T-I$ is demiclosed at 0 by Lemma 2.1, we immediately deduce ${x}^{†}\in Fix\left(T\right)=Fix\left(T\left(\left(1-\zeta \right)I+\zeta T\right)\right)$. Therefore, $T\left(\left(1-\zeta \right)I+\zeta T\right)-I$ is demiclosed at 0. By (3.28), we deduce $\stackrel{˜}{x}\in Fix\left(T\right)$. Hence, $\stackrel{˜}{x}\in \mathrm{\Gamma }$. So,
$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim sup}〈\left(f-F\right){x}^{\ast },{u}_{n}-{x}^{\ast }〉& =& \underset{i\to \mathrm{\infty }}{lim sup}〈\left(f-F\right){x}^{\ast },{u}_{{n}_{i}}-{x}^{\ast }〉\\ =& 〈\left(f-F\right){x}^{\ast },\stackrel{˜}{x}-{x}^{\ast }〉\\ \le & 0.\end{array}$
(3.29)
Note that
$\begin{array}{rcl}{\parallel {u}_{n}-{x}^{\ast }\parallel }^{2}& =& {\parallel {\beta }_{n}\left(f\left({x}_{n}\right)-f\left({x}^{\ast }\right)\right)+{\beta }_{n}\left(f\left({x}^{\ast }\right)-F{x}^{\ast }\right)+\left(I-{\beta }_{n}F\right)\left({y}_{n}-{x}^{\ast }\right)\parallel }^{2}\\ \le & {\parallel \left(I-{\beta }_{n}F\right)\left({y}_{n}-{x}^{\ast }\right)\parallel }^{2}+2{\beta }_{n}〈f\left({x}_{n}\right)-f\left({x}^{\ast }\right),{u}_{n}-{x}^{\ast }〉\\ +2{\beta }_{n}〈f\left({x}^{\ast }\right)-F{x}^{\ast },{u}_{n}-{x}^{\ast }〉\\ \le & {\left(1-{\beta }_{n}\frac{\varsigma }{2}\right)}^{2}{\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}+2{\beta }_{n}\rho \parallel {x}_{n}-{x}^{\ast }\parallel \parallel {u}_{n}-{x}^{\ast }\parallel \\ +2{\beta }_{n}〈f\left({x}^{\ast }\right)-F{x}^{\ast },{u}_{n}-{x}^{\ast }〉\\ \le & {\left(1-{\beta }_{n}\frac{\varsigma }{2}\right)}^{2}{\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}+2{\beta }_{n}\rho {\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}+\frac{1}{2}{\parallel {u}_{n}-{x}^{\ast }\parallel }^{2}\\ +2{\beta }_{n}〈f\left({x}^{\ast }\right)-F{x}^{\ast },{u}_{n}-{x}^{\ast }〉.\end{array}$
It follows that
$\begin{array}{rcl}{\parallel {u}_{n}-{x}^{\ast }\parallel }^{2}& \le & \left[1-2\left(\frac{\varsigma }{2}-\rho \right){\beta }_{n}\right]{\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}+{\beta }_{n}^{2}{\frac{\varsigma }{4}}^{2}{\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}\\ +4{\beta }_{n}〈f\left({x}^{\ast }\right)-F{x}^{\ast },{u}_{n}-{x}^{\ast }〉.\end{array}$
So,
$\begin{array}{rcl}{\parallel {x}_{n+1}-{x}^{\ast }\parallel }^{2}& =& {\parallel {\alpha }_{n}\left({x}_{n}-{x}^{\ast }\right)+\left(1-{\alpha }_{n}\right)\left({u}_{n}-{x}^{\ast }\right)\parallel }^{2}\\ \le & {\alpha }_{n}{\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}+\left(1-{\alpha }_{n}\right){\parallel {u}_{n}-{x}^{\ast }\parallel }^{2}\\ \le & \left[1-2\left(\frac{\varsigma }{2}-\rho \right)\left(1-{\alpha }_{n}\right){\beta }_{n}\right]{\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}+\left(1-{\alpha }_{n}\right){\beta }_{n}^{2}\frac{{\varsigma }^{2}}{4}{\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}\\ +4\left(1-{\alpha }_{n}\right){\beta }_{n}〈f\left({x}^{\ast }\right)-F{x}^{\ast },{u}_{n}-{x}^{\ast }〉\\ =& \left[1-\left(\varsigma -2\rho \right)\left(1-{\alpha }_{n}\right){\beta }_{n}\right]{\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}\\ +\left(\varsigma -2\rho \right)\left(1-{\alpha }_{n}\right){\beta }_{n}\left\{{\beta }_{n}\frac{{\varsigma }^{2}}{4\left(\varsigma -2\rho \right)}{\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}\\ +\frac{4}{\varsigma -2\rho }〈f\left({x}^{\ast }\right)-F{x}^{\ast },{u}_{n}-{x}^{\ast }〉\right\}.\end{array}$
(3.30)

Applying Lemma 2.3 to (3.30) we deduce ${x}_{n}\to {x}^{\ast }$.

Case 2. Assume there exists an integer ${n}_{0}$ such that $\parallel {x}_{{n}_{0}}-{x}^{\ast }\parallel \le \parallel {x}_{{n}_{0}+1}-{x}^{\ast }\parallel$. In this case, we set ${\omega }_{n}=\left\{\parallel {x}_{n}-{x}^{\ast }\parallel \right\}$. Then, we have ${\omega }_{{n}_{0}}\le {\omega }_{{n}_{0}+1}$. Define an integer sequence $\left\{{\tau }_{n}\right\}$ for all $n\ge {n}_{0}$ as follows:
$\tau \left(n\right)=max\left\{l\in \mathbb{N}|{n}_{0}\le l\le n,{\omega }_{l}\le {\omega }_{l+1}\right\}.$
It is clear that $\tau \left(n\right)$ is a non-decreasing sequence satisfying
$\underset{n\to \mathrm{\infty }}{lim}\tau \left(n\right)=\mathrm{\infty }$
and
${\omega }_{\tau \left(n\right)}\le {\omega }_{\tau \left(n\right)+1},$
for all $n\ge {n}_{0}$. From (3.19), we get
$\begin{array}{r}\left(1-{\alpha }_{\tau \left(n\right)}\right){\alpha }_{\tau \left(n\right)}{\parallel {y}_{\tau \left(n\right)}-{x}_{\tau \left(n\right)}\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le {\parallel {x}_{\tau \left(n\right)}-{x}^{\ast }\parallel }^{2}-{\parallel {x}_{\tau \left(n\right)+1}-{x}^{\ast }\parallel }^{2}+2\left(1-{\alpha }_{\tau \left(n\right)}\right){\beta }_{\tau \left(n\right)}〈f\left({x}_{\tau \left(n\right)}\right),{x}_{\tau \left(n\right)}-{x}^{\ast }〉\\ \phantom{\rule{2em}{0ex}}-2\left(1-{\alpha }_{\tau \left(n\right)}\right){\beta }_{\tau \left(n\right)}〈F{y}_{\tau \left(n\right)},{x}_{\tau \left(n\right)}-{x}^{\ast }〉\\ \phantom{\rule{2em}{0ex}}+{\left(1-{\alpha }_{\tau \left(n\right)}\right)}^{2}\left[{\beta }_{\tau \left(n\right)}^{2}{\parallel f\left({x}_{\tau \left(n\right)}\right)-F{y}_{\tau \left(n\right)}\parallel }^{2}\\ \phantom{\rule{2em}{0ex}}+2{\beta }_{\tau \left(n\right)}\parallel f\left({x}_{\tau \left(n\right)}\right)-F{y}_{\tau \left(n\right)}\parallel \parallel {y}_{\tau \left(n\right)}-{x}_{\tau \left(n\right)}\parallel \right].\end{array}$
(3.31)
It follows that
$\underset{n\to \mathrm{\infty }}{lim}\parallel {y}_{\tau \left(n\right)}-{x}_{\tau \left(n\right)}\parallel =0.$
By a similar argument to that of (3.29) and (3.30), we can prove that
$\underset{n\to \mathrm{\infty }}{lim sup}〈\left(f-F\right){x}^{\ast },{u}_{\tau \left(n\right)}-{x}^{\ast }〉\le 0,$
(3.32)
and
$\begin{array}{rcl}{\omega }_{\tau \left(n\right)+1}^{2}& \le & \left[1-2\left(\frac{\varsigma }{2}-\rho \right)\left(1-{\alpha }_{\tau \left(n\right)}\right){\beta }_{\tau \left(n\right)}\right]{\omega }_{\tau \left(n\right)}^{2}\\ +\left(1-{\alpha }_{\tau \left(n\right)}\right){\beta }_{\tau \left(n\right)}^{2}\frac{{\varsigma }^{2}}{4}{\omega }_{\tau \left(n\right)}^{2}\\ +4\left(1-{\alpha }_{\tau \left(n\right)}\right){\beta }_{\tau \left(n\right)}〈f\left({x}^{\ast }\right)-F{x}^{\ast },{u}_{\tau \left(n\right)}-{x}^{\ast }〉.\end{array}$
(3.33)
Since ${\omega }_{\tau \left(n\right)}\le {\omega }_{\tau \left(n\right)+1}$, we have from (3.33)
${\omega }_{\tau \left(n\right)}^{2}\le \frac{16}{4\left({\varsigma }^{2}-2\rho \right)-{\varsigma }^{2}{\beta }_{\tau \left(n\right)}}〈f\left({x}^{\ast }\right)-F{x}^{\ast },{u}_{\tau }\left(n\right)-{x}^{\ast }〉.$
(3.34)
Combining (3.33) and (3.34), we have
$\underset{n\to \mathrm{\infty }}{lim sup}{\omega }_{\tau \left(n\right)}\le 0,$
and hence
$\underset{n\to \mathrm{\infty }}{lim}{\omega }_{\tau \left(n\right)}=0.$
(3.35)
From (3.33), we also obtain
$\underset{n\to \mathrm{\infty }}{lim sup}{\omega }_{\tau \left(n\right)+1}\le \underset{n\to \mathrm{\infty }}{lim sup}{\omega }_{\tau \left(n\right)}.$
This together with (3.35) imply that
$\underset{n\to \mathrm{\infty }}{lim}{\omega }_{\tau \left(n\right)+1}=0.$
Applying Lemma 2.2 to get
$0\le {\omega }_{n}\le max\left\{{\omega }_{\tau \left(n\right)},{\omega }_{\tau \left(n\right)+1}\right\}.$

Therefore, ${\omega }_{n}\to 0$. That is, ${x}_{n}\to {x}^{\ast }$. This completes the proof. □

## Declarations

### Acknowledgements

Yonghong Yao was supported in part by NSFC 71161001-G0105. Yeong-Cheng Liou was supported in part by NSC 101-2628-E-230-001-MY3 and NSC 101-2622-E-230-005-CC3.

## Authors’ Affiliations

(1)
Department of Mathematics, Tianjin Polytechnic University, Tianjin, 300387, China
(2)
Department of Mathematics, Texas A&M University, Kingsville, USA
(3)
Nonlinear Analysis and Applied Mathematics Research Group (NAAM), King Abdulaziz University, P.O. Box 80203, Jeddah, 21589, Saudi Arabia
(4)
Department of Information Management, Cheng Shiu University, Kaohsiung, 833, Taiwan
(5)
Center for General Education, Kaohsiung Medical University, Kaohsiung, 807, Taiwan

## References

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