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# Strong convergent result for quasi-nonexpansive mappings in Hilbert spaces

Fixed Point Theory and Applications20112011:88

https://doi.org/10.1186/1687-1812-2011-88

• Accepted: 25 November 2011
• Published:

## Abstract

In this article, we study an iterative method over the class of quasi-nonexpansive mappings which are more general than nonexpansive mappings in Hilbert spaces. Our strong convergent theorems include several corresponding authors' results.

2000 MSC: 58E35; 47H09; 65J15.

## Keywords

• quasi-nonexpansive mapping
• Lipschitzian continuous
• strongly monotone
• nonlinear operator
• fixed point
• viscosity method

## 1. Introduction

Let H be a real Hilbert space with inner product 〈·,·〉, and induced norm ||·||. A mapping T: HH is called nonexpansive if ||Tx - Ty|| ≤ ||x - y|| for all x,y H. The set of the fixed points of T is denoted by Fix(T) := {x H: Tx = x}.

The viscosity approximation method was first introduced by Moudafi [1] in 2000. Starting with an arbitrary initial x0 H, define a sequence {x n } generated by
$\begin{array}{r}{x}_{n+1}=\frac{{\epsilon }_{n}}{1+{\epsilon }_{n}}f\left({x}_{n}\right)+\frac{1}{1+{\epsilon }_{n}}T{x}_{n},\phantom{\rule{1em}{0ex}}\forall \phantom{\rule{1em}{0ex}}n\ge 0,\end{array}$
(1.1)

where f is a contraction with a coefficient α [0,1) on H, i.e., ||f(x) - f(y)|| ≤ α||x - y|| for all x,y H, T is nonexpansive, and {ε n } is a sequence in (0,1) satisfying the following given conditions:

(i1) limn→∞ε n = 0;

(i2) ${\sum }_{n=0}^{\mathrm{\infty }}{\epsilon }_{n}=\mathrm{\infty }$;

(i3) $\underset{n\to \mathrm{\infty }}{lim}\left(\frac{1}{{\epsilon }_{n}}-\frac{1}{{\epsilon }_{n+1}}\right)=0$.

It is proved that the sequence {x n } generated by (1.1) converges strongly to the unique solution x* C(C := Fix(T)) of the variational inequality:
$\begin{array}{r}〈\left(I-f\right){x}^{*},x-{x}^{*}〉\ge 0,\phantom{\rule{1em}{0ex}}\forall \phantom{\rule{1em}{0ex}}x\in Fix\left(T\right).\end{array}$
In 2003, Xu [2] proved that the sequence {x n } defined by the below process where T is also nonexpansive, started with an arbitrary initial x0 H:
$\begin{array}{r}{x}_{n+1}={\alpha }_{n}b+\left(I-{\alpha }_{n}A\right)T{x}_{n},\phantom{\rule{1em}{0ex}}\forall \phantom{\rule{1em}{0ex}}n\ge 0,\end{array}$
(1.2)
converges strongly to the unique solution of the minimization problem (1.3) when the sequence {α n } satisfies certain conditions:
$\begin{array}{r}\underset{x\in C}{min}\frac{1}{2}〈Ax,x〉-〈x,b〉,\end{array}$
(1.3)

where C is the set of fixed points set of T on H and b is a given point in H.

In 2006, Marino and Xu [3] combined the iterative method (1.2) with the viscosity approximation method (1.1) and considered the following general iterative method:
$\begin{array}{r}{x}_{n+1}={\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(I-{\alpha }_{n}A\right)T{x}_{n},\phantom{\rule{1em}{0ex}}\forall \phantom{\rule{1em}{0ex}}n\ge 0.\end{array}$
(1.4)
It is proved that if the sequence {α n } satisfies appropriate conditions, the sequence {x n } generated by (1.4) converges strongly to the unique solution of the variational inequality:
$\begin{array}{r}〈\left(\gamma f-A\right)\stackrel{̃}{x},x-\stackrel{̃}{x}〉\le 0,\phantom{\rule{1em}{0ex}}\forall \phantom{\rule{1em}{0ex}}x\in C,\end{array}$
(1.5)

or equivalently $\stackrel{̃}{x}={P}_{Fix\left(T\right)}\left(I-A+\gamma f\right)\stackrel{̃}{x}$, where C is the fixed point set of a nonexpansive mapping T.

In 2009, Maingè [4] considered the viscosity approximation method (1.1), and expanded the strong convergence to quasi-nonexpansive mappings in Hilbert space.

In 2010, Tian [5] considered the following general iterative method under the frame of nonexpansive mappings:
$\begin{array}{r}{x}_{n+1}={\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(I-\mu {\alpha }_{n}F\right)T{x}_{n},\phantom{\rule{1em}{0ex}}\forall \phantom{\rule{1em}{0ex}}n\ge 0,\end{array}$
(1.6)

and gave some strong convergent theorems.

Very recently, Tian [6] extended (1.6) to a more general scheme, that is: the mapping f: H → H is no longer a contraction but a L-Lipschitzian continuous operator with coefficient L > 0, and proved that if the sequence {α n } satisfies appropriate conditions, the sequence {x n } generated by xn+1= α n γf(x n ) + (I - μα n F)Tx n converges strongly to the unique solution $\stackrel{̃}{x}\in Fix\left(T\right)$ of the variational inequality where T is still nonexpansive:
$\begin{array}{r}〈\left(\gamma f-\mu F\right)\stackrel{̃}{x},x-\stackrel{̃}{x}〉\le 0,\phantom{\rule{1em}{0ex}}\forall \phantom{\rule{1em}{0ex}}x\in Fix\left(T\right).\end{array}$
(1.7)
Motivated by Maingè [4] and Tian [6], we consider the following iterative process:
$\begin{array}{r}\left\{\begin{array}{c}{x}_{0}=x\in H\phantom{\rule{1em}{0ex}}\text{arbitrarily}\phantom{\rule{1em}{0ex}}\text{chosen},\hfill \\ {x}_{n+1}={\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(I-{\alpha }_{n}\mu F\right){T}_{\omega }{x}_{n},\phantom{\rule{1em}{0ex}}\forall \phantom{\rule{1em}{0ex}}n\ge 0,\hfill \end{array}\right\\end{array}$
(1.8)

where f is L-Lipschitzian, T ω = (1 - ω)I + ωT, and T is a quasi-nonexpansive mapping. Under some appropriate conditions on ω and {α n }, we obtain strong convergence over the class of quasi-nonexpansive mappings in Hilbert spaces. Our result is more general than Maingè's [4] conclusion.

## 2. Preliminaries

Throughout this article, we write x n x to indicate that the sequence {x n } converges weakly to x. x n → x implies that the sequence {x n } converges strongly to x. The following lemmas are useful for our article.

The following statements are valid in a Hilbert space H: for each x,y H, t [0,1]
1. (i)

||x + y|| ≤ ||x||2 + 2〈y, x + y〉;

2. (ii)

||(1 - t)x + ty||2 = (1 - t)||x||2 + t||y||2 - (1 - t)t||x - y||2;

3. (iii)

$〈x,y〉=-\frac{1}{2}\parallel x-y{\parallel }^{2}+\frac{1}{2}\parallel x{\parallel }^{2}+\frac{1}{2}\parallel y{\parallel }^{2}$.

Lemma 2.1. Let f: HH be a L-Lipschitzian continuous operator with coefficient L > 0. F: HH is a κ-Lipschitzian continuous and η-strongly monotone operator with κ > 0 and η > 0. Then, for 0 < γμη/L,
$\begin{array}{r}〈x-y,\left(\mu F-\gamma f\right)x-\left(\mu F-\gamma f\right)y〉\ge \left(\mu \eta -\gamma L\right)\parallel x-y{\parallel }^{2}.\end{array}$
(2.1)

That is, μF - γf is strongly monotone with coefficient$\mu \eta -\gamma L$.

Lemma 2.2.[4]Let T ω := (1 - ω)I + ωT, with T quasi-nonexpansive on H, Fix(T) ≠ , and ω (0,1]. Then, the following statements are reached:

(a1) Fix(T) = Fix(T ω );

(a2) T ω is quasi-nonexpansive;

(a3) ||T ω x - q||2 ≤ ||x - q||2 - ω(1 - ω)||Tx - x||2for all x H and q Fix(T);

(a4) $〈x-{T}_{\omega }x,x-q〉\ge \frac{\omega }{2}\parallel x-Tx{\parallel }^{2}$for all x H and q Fix(T).

Proposition 2.3. From the equality (iii) and the fact that T is quasi-nonexpansive, we have
$\begin{array}{r}〈x-Tx,x-q〉=-\frac{1}{2}\parallel Tx-q{\parallel }^{2}+\frac{1}{2}\parallel x-Tx{\parallel }^{2}+\frac{1}{2}\parallel x-q{\parallel }^{2}\ge \frac{1}{2}\parallel x-Tx{\parallel }^{2}.\end{array}$

(a4) is easily deduced by I-T ω = ω(I-T) and the previous inequality.

Lemma 2.4.[7]Let n } be a sequence of real numbers that does not decrease at infinity, in the sense that there exist a subsequence${\left\{{\mathrm{\Gamma }}_{{n}_{j}}\right\}}_{j\ge 0}$of n } which satisfies${\mathrm{\Gamma }}_{{n}_{j}}<{\mathrm{\Gamma }}_{{n}_{j}+1}$for all j ≥ 0. Also, consider the sequence of integers${\left\{\tau \left(n\right)\right\}}_{n\ge {n}_{0}}$defined by
$\begin{array}{r}\tau \left(n\right)=max\left\{k\le n\mid {\mathrm{\Gamma }}_{k}<{\mathrm{\Gamma }}_{k+1}\right\}.\end{array}$
Then, ${\left\{\tau \left(n\right)\right\}}_{n\ge {n}_{0}}$is a nondecreasing sequence verifying limn→∞τ(n) = ∞ and for all nn0, it holds that Γτ(n)< Γτ(n)+1and we have
$\begin{array}{r}{\mathrm{\Gamma }}_{n}\le {\mathrm{\Gamma }}_{\tau \left(n\right)+1}.\end{array}$
Recall the metric projection P K from a Hilbert space H to a closed convex subset K of H is defined: for each x H the unique element P K x K such that
$\begin{array}{r}\parallel x-{P}_{K}x\parallel :=inf\left\{\parallel x-y\parallel :y\in K\right\}.\end{array}$
Lemma 2.5. Let K be a closed convex subset of H. Given x H, and z K, z = P K x, if and only if there holds the inequality:
$\begin{array}{r}〈x-z,y-z〉\le 0,\phantom{\rule{1em}{0ex}}\forall \phantom{\rule{1em}{0ex}}y\in K.\end{array}$
Lemma 2.6. If x* is the solution of the variational inequality (1.7) with T: HH demi-closed and {y n } H is a bounded sequence such that ||Ty n - y n || → 0, then
$\begin{array}{r}{lim inf}_{n\to \mathrm{\infty }}〈\left(\mu F-\gamma f\right){x}^{*},{y}_{n}-{x}^{*}〉\ge 0.\end{array}$
(2.2)
Proof. We assume that there exists a subsequence $\left\{{y}_{{n}_{j}}\right\}$ of {y n } such that ${y}_{{n}_{j}}⇀ỹ$. From the given conditions $\parallel T{y}_{n}-{y}_{n}\parallel \to 0$ and T: HH demi-closed, we have that any weak cluster point of {y n } belongs to the fixed point set Fix(T). Hence, we conclude that $ỹ\in Fix\left(T\right)$, and also have that
$\begin{array}{r}{lim inf}_{n\to \mathrm{\infty }}〈\left(\mu F-\gamma f\right){x}^{*},{y}_{n}-{x}^{*}〉=\underset{j\to \mathrm{\infty }}{lim}〈\left(\mu F-\gamma f\right){x}^{*},{y}_{{n}_{j}}-{x}^{*}〉.\end{array}$
Recalling (1.7), we immediately obtain
$\begin{array}{r}{lim inf}_{n\to \mathrm{\infty }}〈\left(\mu F-\gamma f\right){x}^{*},{y}_{n}-{x}^{*}〉=〈\left(\mu F-\gamma f\right){x}^{*},ỹ-{x}^{*}〉\ge 0.\end{array}$

This completes the proof.   □

## 3. Main results

Let H be a real Hilbert space, let F be a κ-Lipschitzian and η-strongly monotone operator on H with k > 0, η > 0, and let T be a quasi-nonexpansive mapping on H, and f is a L-Lipschitzian mapping with coefficient L > 0 for all x,y H. Assume the set Fix(T) of fixed points of T is nonempty and we note that Fix(T) is closed and convex.

Theorem 3.1. Let$0<\mu <2\eta ∕{\kappa }^{2},0<\gamma <\mu \left(\eta -\frac{\mu {\kappa }^{2}}{2}\right)∕L=\tau ∕L$, and start with an arbitrary chosen x0 H, let the sequence {x n } be generated by
$\begin{array}{r}{x}_{n+1}={\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(I-{\alpha }_{n}\mu F\right){T}_{\omega }{x}_{n},\end{array}$
(3.1)

where the sequence {α n } (0,1) satisfies limn→∞α n = 0, and${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_{n}=\mathrm{\infty }$. Also$\omega \in \left(0,\frac{1}{2}\right)$, T ω := (1 - ω)I + ωI with two conditions on T:

(C1) ||Tx - q|| ≤ ||x - q|| for any x H, and q Fix(T); this means that T is a quasi-nonexpansive mapping;

(C2) T is demi-closed on H; that is: if {y k } H, y k z, and (I - T)y k → 0, then z Fix(T).

Then, {x n } converges strongly to the x* Fix(T) which is the unique solution of the VIP:
$\begin{array}{r}〈\left(\mu F-\gamma f\right){x}^{*},x-{x}^{*}〉\ge 0,\phantom{\rule{1em}{0ex}}\forall \phantom{\rule{1em}{0ex}}x\in Fix\left(T\right).\end{array}$
(3.2)

Proof. First, we show that {x n } is bounded.

Take any p Fix(T), by Lemma 2.2 (a3), we have
$\begin{array}{l}\parallel {x}_{n+1}-p\parallel \\ \phantom{\rule{1em}{0ex}}=\parallel {\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(I-{\alpha }_{n}\mu F\right){T}_{\omega }{x}_{n}-p\parallel \\ \phantom{\rule{1em}{0ex}}=\parallel {\alpha }_{n}\gamma \left(f\left({x}_{n}\right)-f\left(p\right)\right)+{\alpha }_{n}\left(\gamma f\left(p\right)-\mu Fp\right)+\left(I-{\alpha }_{n}\mu F\right){T}_{\omega }{x}_{n}-\left(I-{\alpha }_{n}\mu F\right)p\parallel \\ \phantom{\rule{1em}{0ex}}\le {\alpha }_{n}\gamma L\parallel {x}_{n}-p\parallel +{\alpha }_{n}\parallel \gamma f\left(p\right)-\mu Fp\parallel +\left(1-{\alpha }_{n}\tau \right)\parallel {x}_{n}-p\parallel \\ \phantom{\rule{1em}{0ex}}\le \left(1-{\alpha }_{n}\left(\tau -\gamma L\right)\right)\parallel {x}_{n}-p\parallel +{\alpha }_{n}\parallel \gamma f\left(p\right)-\mu Fp\parallel .\end{array}$
(3.3)
By induction, we have
$\begin{array}{r}\parallel {x}_{n}-p\parallel \le max\left\{\parallel {x}_{0}-p\parallel ,\frac{\parallel \gamma f\left(p\right)-\mu Fp\parallel }{\tau -\gamma L}\right\},\phantom{\rule{1em}{0ex}}\forall \phantom{\rule{1em}{0ex}}n\ge 0.\end{array}$

Hence, {x n } is bounded, so are the {f(x n )} and {F(x n )}.

From (3.1), we have
$\begin{array}{r}{x}_{n+1}-{x}_{n}+{\alpha }_{n}\left(\mu F{x}_{n}-\gamma f\left({x}_{n}\right)\right)=\left(I-{\alpha }_{n}\mu F\right){T}_{\omega }{x}_{n}-\left(I-{\alpha }_{n}\mu F\right){x}_{n}.\end{array}$
(3.4)
Since x* Fix(T), from Lemma 2.2 (a4), and together with (3.4), we obtain
$\begin{array}{l}〈{x}_{n+1}-{x}_{n}+{\alpha }_{n}\left(\mu F\left({x}_{n}\right)-\gamma f\left({x}_{n}\right)\right),{x}_{n}-{x}^{*}〉\\ \phantom{\rule{1em}{0ex}}=〈\left(I-{\alpha }_{n}\mu F\right){T}_{\omega }{x}_{n}-\left(I-{\alpha }_{n}\mu F\right){x}_{n},{x}_{n}-{x}^{*}〉\\ \phantom{\rule{1em}{0ex}}=\left(1-{\alpha }_{n}\right)〈{T}_{\omega }{x}_{n}-{x}_{n},{x}_{n}-{x}^{*}〉+{\alpha }_{n}〈\left(I-\mu F\right){T}_{\omega }{x}_{n}-\left(I-\mu F\right){x}_{n},{x}_{n}-{x}^{*}〉\\ \phantom{\rule{1em}{0ex}}\le -\frac{\omega }{2}\left(1-{\alpha }_{n}\right)\parallel {x}_{n}-T{x}_{n}{\parallel }^{2}+{\alpha }_{n}\parallel \left(I-\mu F\right){T}_{\omega }{x}_{n}-\left(I-\mu F\right){x}_{n}\parallel \parallel {x}_{n}-{x}^{*}\parallel \\ \phantom{\rule{1em}{0ex}}\le -\frac{\omega }{2}\left(1-{\alpha }_{n}\right)\parallel {x}_{n}-T{x}_{n}{\parallel }^{2}+{\alpha }_{n}\left(1-\tau \right)\parallel {T}_{\omega }{x}_{n}-{x}_{n}\parallel \parallel {x}_{n}-{x}^{*}\parallel \\ \phantom{\rule{1em}{0ex}}=-\frac{\omega }{2}\left(1-{\alpha }_{n}\right)\parallel {x}_{n}-T{x}_{n}{\parallel }^{2}+\omega {\alpha }_{n}\left(1-\tau \right)\parallel T{x}_{n}-{x}_{n}\parallel \parallel {x}_{n}-{x}^{*}\parallel ,\end{array}$
it follows from the previous inequality that
$\begin{array}{r}\begin{array}{c}\begin{array}{cc}\hfill -〈{x}_{n}-{x}_{n+1},{x}_{n}-{x}^{*}〉& \le -{\alpha }_{n}〈\left(\mu F-\gamma f\right){x}_{n},{x}_{n}-{x}^{*}〉-\frac{\omega }{2}\left(1-{\alpha }_{n}\right)\parallel {x}_{n}-T{x}_{n}{\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}+\omega {\alpha }_{n}\left(1-\tau \right)\parallel T{x}_{n}-{x}_{n}\parallel \parallel {x}_{n}-{x}^{*}\parallel .\hfill \end{array}\hfill \end{array}\end{array}$
(3.5)
From (iii), we obviously have
$\begin{array}{r}〈{x}_{n}-{x}_{n+1},{x}_{n}-{x}^{*}〉=-\frac{1}{2}\parallel {x}_{n+1}-{x}^{*}{\parallel }^{2}+\frac{1}{2}\parallel {x}_{n}-{x}^{*}{\parallel }^{2}+\frac{1}{2}\parallel {x}_{n+1}-{x}_{n}{\parallel }^{2}.\end{array}$
(3.6)
Set ${\mathrm{\Gamma }}_{n}:=\frac{1}{2}\parallel {x}_{n}-{x}^{*}{\parallel }^{2}$, and combine (3.5) with (3.6), it follows that
$\begin{array}{r}\begin{array}{c}\begin{array}{cc}\hfill {\mathrm{\Gamma }}_{n+1}-{\mathrm{\Gamma }}_{n}-\frac{1}{2}\parallel {x}_{n+1}-{x}_{n}{\parallel }^{2}& \le -{\alpha }_{n}〈\left(\mu F-\gamma f\right){x}_{n},{x}_{n}-{x}^{*}〉-\frac{\omega }{2}\left(1-{\alpha }_{n}\right)\parallel {x}_{n}-T{x}_{n}{\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}+\omega {\alpha }_{n}\left(1-\tau \right)\parallel T{x}_{n}-{x}_{n}\parallel \parallel {x}_{n}-{x}^{*}\parallel .\hfill \end{array}\hfill \end{array}\end{array}$
(3.7)

Now, we calculate ||x n +1 - x n ||.

From the given condition: T ω := (1 - ω)I + ωT, it is easy to deduce that ||T ω x n - x n || = ω||x n - Tx n ||. Thus, it follows from (3.4) that
$\begin{array}{r}\begin{array}{c}\begin{array}{cc}\hfill \parallel {x}_{n+1}-{x}_{n}{\parallel }^{2}& =\parallel {\alpha }_{n}\left(\gamma f\left({x}_{n}\right)-\mu F{x}_{n}\right)+\left(I-{\alpha }_{n}\mu F\right){T}_{\omega }{x}_{n}-\left(I-{\alpha }_{n}\mu F\right){x}_{n}{\parallel }^{2}\hfill \\ \le 2{\alpha }_{n}^{2}\parallel \gamma f\left({x}_{n}\right)-\mu F{x}_{n}{\parallel }^{2}+2{\left(1-{\alpha }_{n}\tau \right)}^{2}\parallel {T}_{\omega }{x}_{n}-{x}_{n}{\parallel }^{2}\hfill \\ =2{\alpha }_{n}^{2}\parallel \gamma f\left({x}_{n}\right)-\mu F{x}_{n}{\parallel }^{2}+2{\omega }^{2}{\left(1-{\alpha }_{n}\tau \right)}^{2}\parallel T{x}_{n}-{x}_{n}{\parallel }^{2}.\hfill \end{array}\hfill \end{array}\end{array}$
(3.8)
Then, from (3.7) and (3.8), we have
$\begin{array}{l}{\mathrm{\Gamma }}_{n+1}-{\mathrm{\Gamma }}_{n}+\left[\frac{\omega }{2}\left(1-{\alpha }_{n}\right)-{\omega }^{2}{\left(1-{\alpha }_{n}\tau \right)}^{2}\right]\parallel {x}_{n}-T{x}_{n}{\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le {\alpha }_{n}\left[{\alpha }_{n}\parallel \gamma f\left({x}_{n}\right)-\mu F{x}_{n}{\parallel }^{2}-〈\left(\mu F-\gamma f\right){x}_{n},{x}_{n}-{x}^{*}〉\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+\omega \left(1-\tau \right)\parallel T{x}_{n}-{x}_{n}\parallel \parallel {x}_{n}-{x}^{*}\parallel \right].\end{array}$
(3.9)

Finally, we prove x n x*. To this end, we consider two cases.

Case 1: Suppose that there exists n 0 such that ${\left\{{\mathrm{\Gamma }}_{n}\right\}}_{n\ge {n}_{0}}$ is nonincreasing, it is equal to Γn+1≤ Γ n for all nn0. It follows that limn→∞Γ n exists, so we conclude that
$\begin{array}{r}\underset{n\to \mathrm{\infty }}{lim}\left({\mathrm{\Gamma }}_{n+1}-{\mathrm{\Gamma }}_{n}\right)=0.\end{array}$
(3.10)
It follows from (3.9),(3.10) and combine with the fact that limn→∞α n = 0, we have limn→∞||x n - Tx n || = 0. Considering (3.9) again, from (3.10), we have
$\begin{array}{l}-{\alpha }_{n}\left[{\alpha }_{n}\parallel \gamma f\left({x}_{n}\right)-\mu F{x}_{n}{\parallel }^{2}-〈\left(\mu F-\gamma f\right){x}_{n},{x}_{n}-{x}^{*}〉\\ \phantom{\rule{1em}{0ex}}+\omega \left(1-\tau \right)\parallel T{x}_{n}-{x}_{n}\parallel \parallel {x}_{n}-{x}^{*}\parallel \right]\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\le {\mathrm{\Gamma }}_{n}-{\mathrm{\Gamma }}_{n+1}.\end{array}$
(3.11)
Then, by ${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_{n}=\mathrm{\infty }$, we conclude that
$\begin{array}{l}\underset{n\to \mathrm{\infty }}{lim inf}-\left[{\alpha }_{n}\parallel \gamma f\left({x}_{n}\right)-\mu F{x}_{n}{\parallel }^{2}-〈\left(\mu F-\gamma f\right){x}_{n},{x}_{n}-{x}^{*}〉\\ \phantom{\rule{1em}{0ex}}+\omega \left(1-\tau \right)\parallel T{x}_{n}-{x}_{n}\parallel \parallel {x}_{n}-{x}^{*}\parallel \right]\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\le 0.\end{array}$
(3.12)
Since {f(x n )} and {x n } are both bounded, as well as α n → 0, and limn→∞||x n - Tx n || = 0, it follows from (3.12) that
$\begin{array}{r}{lim inf}_{n\to \mathrm{\infty }}〈\left(\mu F-\gamma f\right){x}_{n},{x}_{n}-{x}^{*}〉\le 0.\end{array}$
(3.13)
From Lemma 2.1, it is obvious that
$\begin{array}{r}〈\left(\mu F-\gamma f\right){x}_{n},{x}_{n}-{x}^{*}〉\ge 〈\left(\mu F-\gamma f\right){x}^{*},{x}_{n}-{x}^{*}〉+2\left(\mu \eta -\gamma L\right){\mathrm{\Gamma }}_{n}.\end{array}$
(3.14)
Thus, from (3.14), and the fact that limn→∞Γ n exists, we immediately obtain*******
$\begin{array}{l}\underset{n\to \mathrm{\infty }}{lim inf}〈\left(\mu F-\gamma f\right){x}^{*},{x}_{n}-{x}^{*}〉+2\left(\mu \eta -\gamma L\right){\mathrm{\Gamma }}_{n}\\ \phantom{\rule{1em}{0ex}}=2\left(\mu \eta -\gamma L\right)\underset{n\to \mathrm{\infty }}{lim}{\mathrm{\Gamma }}_{n}+\underset{n\to \mathrm{\infty }}{lim inf}〈\left(\mu F-\gamma f\right){x}^{*},{x}_{n}-{x}^{*}〉\le 0,\end{array}$
(3.15)
or equivalently
$\begin{array}{r}\begin{array}{c}\begin{array}{c}\hfill 2\left(\mu \eta -\gamma L\right)\underset{n\to \mathrm{\infty }}{lim}{\mathrm{\Gamma }}_{n}\le -{lim inf}_{n\to \mathrm{\infty }}〈\left(\mu F-\gamma f\right){x}^{*},{x}_{n}-{x}^{*}〉.\end{array}\hfill \end{array}\end{array}$
(3.16)
Finally, by Lemma 2.6, we have
$\begin{array}{r}2\left(\mu \eta -\gamma L\right)\underset{n\to \mathrm{\infty }}{lim}{\mathrm{\Gamma }}_{n}\le 0,\end{array}$
(3.17)

so we conclude that limn→∞Γ n = 0, which equivalently means that {x n } converges strongly to x*.

Case 2: Assume that there exists a subsequence ${\left\{{\mathrm{\Gamma }}_{{n}_{j}}\right\}}_{j\ge 0}$ of {Γ n }n ≥ 0such that ${\mathrm{\Gamma }}_{{n}_{j}}<{\mathrm{\Gamma }}_{{n}_{j}+1}$ for all j . In this case, it follows from Lemma 2.4 that there exists a subsequence {Γτ(n)} of {Γ n } such that Γτ(n)+1> Γτ(n), and {τ(n)} is defined as in Lemma 2.4.

Invoking (3.9) again, it follows that
$\begin{array}{l}{\mathrm{\Gamma }}_{\tau \left(n\right)+1}-{\mathrm{\Gamma }}_{\tau \left(n\right)}+\left[\frac{\omega }{2}\left(1-{\alpha }_{\tau \left(n\right)}\right)-{\omega }^{2}{\left(1-{\alpha }_{\tau \left(n\right)}\tau \right)}^{2}\right]\parallel {x}_{\tau \left(n\right)}-T{x}_{\tau \left(n\right)}{\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le {\alpha }_{\tau \left(n\right)}\left[{\alpha }_{\tau \left(n\right)}\parallel \gamma f\left({x}_{\tau \left(n\right)}\right)-\mu F{x}_{\tau \left(n\right)}{\parallel }^{2}-〈\left(\mu F-\gamma f\right){x}_{\tau \left(n\right)},{x}_{\tau \left(n\right)}-{x}^{*}〉\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+\omega \left(1-\tau \right)\parallel T{x}_{\tau \left(n\right)}-{x}_{\tau \left(n\right)}\parallel \parallel {x}_{\tau \left(n\right)}-{x}^{*}\parallel \right].\end{array}$
Recalling the fact that Γτ(n)+1> Γτ(n), we have
$\begin{array}{l}\left[\frac{\omega }{2}\left(1-{\alpha }_{\tau \left(n\right)}\right)-{\omega }^{2}{\left(1-{\alpha }_{\tau \left(n\right)}\tau \right)}^{2}\right]\parallel {x}_{\tau \left(n\right)}-T{x}_{\tau \left(n\right)}{\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le {\alpha }_{\tau \left(n\right)}\left[{\alpha }_{\tau \left(n\right)}\parallel \gamma f\left({x}_{\tau \left(n\right)}\right)-\mu F{x}_{\tau \left(n\right)}{\parallel }^{2}-〈\left(\mu F-\gamma f\right){x}_{\tau \left(n\right)},{x}_{\tau \left(n\right)}-{x}^{*}〉\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+\omega \left(1-\tau \right)\parallel T{x}_{\tau \left(n\right)}-{x}_{\tau \left(n\right)}\parallel \parallel {x}_{\tau \left(n\right)}-{x}^{*}\parallel \right].\end{array}$
(3.18)
From the preceding results, we get the boundedness of {x n } and α n → 0 which obviously lead to
$\begin{array}{r}\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{\tau \left(n\right)}-T{x}_{\tau \left(n\right)}\parallel =0.\end{array}$
(3.19)
Hence, combining (3.18) with (3.19), we immediately deduce that
$\begin{array}{r}\begin{array}{c}\begin{array}{cc}\hfill 〈\left(\mu F-\gamma f\right){x}_{\tau \left(n\right)},{x}_{\tau \left(n\right)}-{x}^{*}〉& \le {\alpha }_{\tau \left(n\right)}\parallel \gamma f\left({x}_{\tau \left(n\right)}\right)-\mu F{x}_{\tau \left(n\right)}{\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}+\omega \left(1-\tau \right)\parallel T{x}_{\tau \left(n\right)}-{x}_{\tau \left(n\right)}\parallel \parallel {x}_{\tau \left(n\right)}-{x}^{*}\parallel .\hfill \end{array}\hfill \end{array}\end{array}$
(3.20)
Again, (3.14) and (3.20) yield
$\begin{array}{r}\begin{array}{c}\begin{array}{cc}\hfill 〈\left(\mu F-\gamma f\right){x}^{*},{x}_{\tau \left(n\right)}-{x}^{*}〉+2\left(\mu \eta -\gamma L\right){\mathrm{\Gamma }}_{\tau \left(n\right)}& \le {\alpha }_{\tau \left(n\right)}\parallel \gamma f\left({x}_{\tau \left(n\right)}\right)-\mu F{x}_{\tau \left(n\right)}{\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}+\omega \left(1-\tau \right)\parallel T{x}_{\tau \left(n\right)}-{x}_{\tau \left(n\right)}\parallel \parallel {x}_{\tau \left(n\right)}-{x}^{*}\parallel .\hfill \end{array}\hfill \end{array}\end{array}$
(3.21)
Recall that limn→∞ατ(n)= 0, from (3.19) and (3.21), we immediately have
$\begin{array}{r}2\left(\mu \eta -\gamma L\right){lim sup}_{n\to \mathrm{\infty }}{\mathrm{\Gamma }}_{\tau \left(n\right)}\le -{lim inf}_{n\to \mathrm{\infty }}〈\left(\mu F-\gamma f\right){x}^{*},{x}_{\tau \left(n\right)}-{x}^{*}〉.\end{array}$
(3.22)
By Lemma 2.6, we have
$\begin{array}{r}{lim inf}_{n\to \mathrm{\infty }}〈\left(\mu F-\gamma f\right){x}^{*},{x}_{\tau \left(n\right)}-{x}^{*}〉\ge 0.\end{array}$
(3.23)
Consider (3.22) again, we conclude that
$\begin{array}{r}{lim sup}_{n\to \mathrm{\infty }}{\mathrm{\Gamma }}_{\tau \left(n\right)}=0,\end{array}$
(3.24)

which means that limn→∞Γτ(n)= 0. By Lemma 2.4, it follows that Γ n ≤ Γτ(n), thus, we get limn→∞Γ n = 0, which is equivalent to x n x*.   □

Remark 3.2. Corollary 3.3 is only valid for $\omega \in \left(0,\frac{1}{2}\right)$. This is revised by Wongchan and Saejung [8].

corollary 3.3.[4]Let the sequence {x n } be generated by
$\begin{array}{r}{x}_{n+1}={\alpha }_{n}f\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right){T}_{\omega }{x}_{n},\end{array}$
(3.25)

where the sequence {α n } (0,1) satisfies limn→∞α n = 0, and${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_{n}=\mathrm{\infty }$. Also$\omega \in \left(0,\frac{1}{2}\right)$, and T ω := (1 - ω)I + ωT with two conditions on T:

(C1) ||Tx - q|| ≤ ||x - q|| for any x H, and q Fix(T); this means that T is a quasi-nonexpansive mapping;

(C2) T is demi-closed on H; that is: if {y k } H, y k z, and (I - T)y k → 0, z Fix(T).

Then, {x n } converges strongly to the x* Fix(T) which is the unique solution of the VIP(3.26):
$\begin{array}{r}〈\left(I-f\right){x}^{*},x-{x}^{*}〉\ge 0,\phantom{\rule{1em}{0ex}}\forall \phantom{\rule{1em}{0ex}}x\in Fix\left(T\right).\end{array}$
(3.26)

## Declarations

### Acknowledgements

The first author was supported by the Fundamental Research Funds for the Central Universities (No. ZXH2011C002).

## Authors’ Affiliations

(1)
College of Science, Civil Aviation University of China, Tianjin, 300300, China

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