Strong convergent result for quasi-nonexpansive mappings in Hilbert spaces

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.

1. Introduction

Let H be a real Hilbert space with inner product 〈·,·〉, and induced norm ||·||. A mapping T: H → H 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

$$\begin{array}{r}{x}_{n+1}=\frac{{\epsilon }_n}{1+{\epsilon }_n}f\left(x_n\right)+\frac{1}{1+{\epsilon }_n}T{x}_n,\forall n\ge 0,\end{array}$$

(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) lim_n→∞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:

$$\begin{array}{r}〈\left(I-f\right)x^{*},x-x^{*}〉\ge 0,\forall x\in Fix\left(T\right).\end{array}$$

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:

$$\begin{array}{r}{x}_{n+1}={\alpha }_{n}b+\left(I-{\alpha }_{n}A\right)T{x}_n,\forall n\ge 0,\end{array}$$

(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:

$$\begin{array}{r}{x}_{n+1}={\alpha }_n\gamma f\left(x_n\right)+\left(I-{\alpha }_{n}A\right)T{x}_n,\forall n\ge 0.\end{array}$$

(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:

$$\begin{array}{r}〈\left(\gamma f-A\right)\tilde{x},x-\tilde{x}〉\le 0,\forall x\in C,\end{array}$$

(1.5)

or equivalently $\tilde{x}=P_{Fix\left(T\right)}\left(I-A+\gamma f\right)\tilde{x}$, 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:

$$\begin{array}{r}{x}_{n+1}={\alpha }_n\gamma f\left(x_n\right)+\left(I-\mu {\alpha }_{n}F\right)T{x}_n,\forall n\ge 0,\end{array}$$

(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 x_n+1= α _n γf(x _n ) + (I - μα _n F)Tx _n converges strongly to the unique solution $\tilde{x}\in Fix\left(T\right)$ of the variational inequality where T is still nonexpansive:

$$\begin{array}{r}〈\left(\gamma f-\mu F\right)\tilde{x},x-\tilde{x}〉\le 0,\forall x\in Fix\left(T\right).\end{array}$$

(1.7)

Motivated by Maingè [4] and Tian [6], we consider the following iterative process:

$$\begin{array}{r}\left\{\begin{array}{c}{x}_0=x\in H\text{arbitrarily}\text{chosen},\hfill \\ x_{n+1}={\alpha }_n\gamma f\left(x_n\right)+\left(I-{\alpha }_n\mu F\right)T_{\omega }{x}_n,\forall n\ge 0,\hfill \end{array}\right.\end{array}$$

(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]

• (i) ||x + y|| ≤ ||x||² + 2〈y, x + y〉;

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

• (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: H →H be a L-Lipschitzian continuous operator with coefficient L > 0. F: H → H is a κ-Lipschitzian continuous and η-strongly monotone operator with κ > 0 and η > 0. Then, for 0 < γ ≤ μη/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 {Γ _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 {Γ _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 lim_n→∞τ(n) = ∞ and for all n ≥ n₀, it holds that Γ_τ(n)< Γ_τ(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:

$$\begin{array}{r}〈x-z,y-z〉\le 0,\forall y\in K.\end{array}$$

Lemma 2.6. If x* is the solution of the variational inequality (1.7) with T: H → H 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 $y_{n_j}\rightharpoonup ỹ$. From the given conditions $\parallel T{y}_n-y_n\parallel \to 0$ and T: H → H 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 lim_n→∞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:

$$\begin{array}{r}〈\left(\mu F-\gamma f\right)x^{*},x-x^{*}〉\ge 0,\forall x\in Fix\left(T\right).\end{array}$$

(3.2)

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

Take any p ∈ Fix(T), by Lemma 2.2 (a3), we have

$$\begin{array}{l}\parallel x_{n+1}-p\parallel \\ =\parallel {\alpha }_n\gamma f\left(x_n\right)+\left(I-{\alpha }_n\mu F\right)T_{\omega }{x}_n-p\parallel \\ =\parallel {\alpha }_n\gamma \left(f\left(x_n\right)-f\left(p\right)\right)+{\alpha }_n\left(\gamma f\left(p\right)-\mu Fp\right)+\left(I-{\alpha }_n\mu F\right)T_{\omega }{x}_n-\left(I-{\alpha }_n\mu F\right)p\parallel \\ \le {\alpha }_n\gamma L\parallel x_n-p\parallel +{\alpha }_n\parallel \gamma f\left(p\right)-\mu Fp\parallel +\left(1-{\alpha }_n\tau \right)\parallel x_n-p\parallel \\ \le \left(1-{\alpha }_n\left(\tau -\gamma L\right)\right)\parallel x_n-p\parallel +{\alpha }_n\parallel \gamma f\left(p\right)-\mu Fp\parallel .\end{array}$$

(3.3)

By induction, we have

$$\begin{array}{r}\parallel x_n-p\parallel \le max\left\{\parallel x_0-p\parallel ,\frac{\parallel \gamma f\left(p\right)-\mu Fp\parallel }{\tau -\gamma L}\right\},\forall n\ge 0.\end{array}$$

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

$$\begin{array}{l}〈x_{n+1}-x_n+{\alpha }_n\left(\mu F\left(x_n\right)-\gamma f\left(x_n\right)\right),x_n-x^{*}〉\\ =〈\left(I-{\alpha }_n\mu F\right)T_{\omega }{x}_n-\left(I-{\alpha }_n\mu F\right)x_n,x_n-x^{*}〉\\ =\left(1-{\alpha }_n\right)〈T_{\omega }{x}_n-x_n,x_n-x^{*}〉+{\alpha }_n〈\left(I-\mu F\right)T_{\omega }{x}_n-\left(I-\mu F\right)x_n,x_n-x^{*}〉\\ \le -\frac{\omega }{2}\left(1-{\alpha }_n\right)\parallel x_n-T{x}_n{\parallel }^2+{\alpha }_n\parallel \left(I-\mu F\right)T_{\omega }{x}_n-\left(I-\mu F\right)x_n\parallel \parallel x_n-x^{*}\parallel \\ \le -\frac{\omega }{2}\left(1-{\alpha }_n\right)\parallel x_n-T{x}_n{\parallel }^2+{\alpha }_n\left(1-\tau \right)\parallel T_{\omega }{x}_n-x_n\parallel \parallel x_n-x^{*}\parallel \\ =-\frac{\omega }{2}\left(1-{\alpha }_n\right)\parallel x_n-T{x}_n{\parallel }^2+\omega {\alpha }_n\left(1-\tau \right)\parallel T{x}_n-x_n\parallel \parallel x_n-x^{*}\parallel ,\end{array}$$

it follows from the previous inequality that

$$\begin{array}{r}\begin{array}{c}\begin{array}{cc}\hfill -〈x_n-x_{n+1},x_n-x^{*}〉& \le -{\alpha }_n〈\left(\mu F-\gamma f\right)x_n,x_n-x^{*}〉-\frac{\omega }{2}\left(1-{\alpha }_n\right)\parallel x_n-T{x}_n{\parallel }^2\hfill \\ +\omega {\alpha }_n\left(1-\tau \right)\parallel T{x}_n-x_n\parallel \parallel x_n-x^{*}\parallel .\hfill \end{array}\hfill \end{array}\end{array}$$

(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

$$\begin{array}{r}\begin{array}{c}\begin{array}{cc}\hfill {\mathrm{\Gamma }}_{n+1}-{\mathrm{\Gamma }}_n-\frac{1}{2}\parallel x_{n+1}-x_n{\parallel }^2& \le -{\alpha }_n〈\left(\mu F-\gamma f\right)x_n,x_n-x^{*}〉-\frac{\omega }{2}\left(1-{\alpha }_n\right)\parallel x_n-T{x}_n{\parallel }^2\hfill \\ +\omega {\alpha }_n\left(1-\tau \right)\parallel T{x}_n-x_n\parallel \parallel x_n-x^{*}\parallel .\hfill \end{array}\hfill \end{array}\end{array}$$

(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

$$\begin{array}{l}{\mathrm{\Gamma }}_{n+1}-{\mathrm{\Gamma }}_n+\left[\frac{\omega }{2}\left(1-{\alpha }_n\right)-{\omega }^2{\left(1-{\alpha }_n\tau \right)}^2\right]\parallel x_n-T{x}_n{\parallel }^2\\ \le {\alpha }_n[{\alpha }_n\parallel \gamma f\left(x_n\right)-\mu F{x}_n{\parallel }^2-〈\left(\mu F-\gamma f\right)x_n,x_n-x^{*}〉\\ +\omega \left(1-\tau \right)\parallel T{x}_n-x_n\parallel \parallel x_n-x^{*}\parallel ].\end{array}$$

(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 Γ_n+1≤ Γ _n for all n ≥ n₀. It follows that lim_n→∞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 lim_n→∞α _n = 0, we have lim_n→∞||x _n - Tx _n || = 0. Considering (3.9) again, from (3.10), we have

$$\begin{array}{l}-{\alpha }_n[{\alpha }_n\parallel \gamma f\left(x_n\right)-\mu F{x}_n{\parallel }^2-〈\left(\mu F-\gamma f\right)x_n,x_n-x^{*}〉\\ +\omega \left(1-\tau \right)\parallel T{x}_n-x_n\parallel \parallel x_n-x^{*}\parallel ]\\ \le {\mathrm{\Gamma }}_n-{\mathrm{\Gamma }}_{n+1}.\end{array}$$

(3.11)

Then, by ${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$, we conclude that

$$\begin{array}{l}\underset{n\to \mathrm{\infty }}{lim\; inf}-[{\alpha }_n\parallel \gamma f\left(x_n\right)-\mu F{x}_n{\parallel }^2-〈\left(\mu F-\gamma f\right)x_n,x_n-x^{*}〉\\ +\omega \left(1-\tau \right)\parallel T{x}_n-x_n\parallel \parallel x_n-x^{*}\parallel ]\\ \le 0.\end{array}$$

(3.12)

Since {f(x _n )} and {x _n } are both bounded, as well as α _n → 0, and lim_n→∞||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 lim_n→∞Γ _n exists, we immediately obtain*******

$$\begin{array}{l}\underset{n\to \mathrm{\infty }}{lim\; inf}〈\left(\mu F-\gamma f\right)x^{*},x_n-x^{*}〉+2\left(\mu \eta -\gamma L\right){\mathrm{\Gamma }}_n\\ =2\left(\mu \eta -\gamma L\right)\underset{n\to \mathrm{\infty }}{lim}{\mathrm{\Gamma }}_n+\underset{n\to \mathrm{\infty }}{lim\; inf}〈\left(\mu F-\gamma f\right)x^{*},x_n-x^{*}〉\le 0,\end{array}$$

(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 lim_n→∞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 ≥ 0}such that ${\mathrm{\Gamma }}_{n_j}<{\mathrm{\Gamma }}_{n_j+1}$ for all j ∈ ℕ. In this case, it follows from Lemma 2.4 that there exists a subsequence {Γ_τ(n)} of {Γ _n } such that Γ_τ(n)+1> Γ_τ(n), and {τ(n)} is defined as in Lemma 2.4.

Invoking (3.9) again, it follows that

$$\begin{array}{l}{\mathrm{\Gamma }}_{\tau \left(n\right)+1}-{\mathrm{\Gamma }}_{\tau \left(n\right)}+\left[\frac{\omega }{2}\left(1-{\alpha }_{\tau \left(n\right)}\right)-{\omega }^2{\left(1-{\alpha }_{\tau \left(n\right)}\tau \right)}^2\right]\parallel x_{\tau \left(n\right)}-T{x}_{\tau \left(n\right)}{\parallel }^2\\ \le {\alpha }_{\tau \left(n\right)}[{\alpha }_{\tau \left(n\right)}\parallel \gamma f\left(x_{\tau \left(n\right)}\right)-\mu F{x}_{\tau \left(n\right)}{\parallel }^2-〈\left(\mu F-\gamma f\right)x_{\tau \left(n\right)},x_{\tau \left(n\right)}-x^{*}〉\\ +\omega \left(1-\tau \right)\parallel T{x}_{\tau \left(n\right)}-x_{\tau \left(n\right)}\parallel \parallel x_{\tau \left(n\right)}-x^{*}\parallel ].\end{array}$$

Recalling the fact that Γ_τ(n)+1> Γ_τ(n), we have

$$\begin{array}{l}\left[\frac{\omega }{2}\left(1-{\alpha }_{\tau \left(n\right)}\right)-{\omega }^2{\left(1-{\alpha }_{\tau \left(n\right)}\tau \right)}^2\right]\parallel x_{\tau \left(n\right)}-T{x}_{\tau \left(n\right)}{\parallel }^2\\ \le {\alpha }_{\tau \left(n\right)}[{\alpha }_{\tau \left(n\right)}\parallel \gamma f\left(x_{\tau \left(n\right)}\right)-\mu F{x}_{\tau \left(n\right)}{\parallel }^2-〈\left(\mu F-\gamma f\right)x_{\tau \left(n\right)},x_{\tau \left(n\right)}-x^{*}〉\\ +\omega \left(1-\tau \right)\parallel T{x}_{\tau \left(n\right)}-x_{\tau \left(n\right)}\parallel \parallel x_{\tau \left(n\right)}-x^{*}\parallel ].\end{array}$$

(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

$$\begin{array}{r}\begin{array}{c}\begin{array}{cc}\hfill 〈\left(\mu F-\gamma f\right)x_{\tau \left(n\right)},x_{\tau \left(n\right)}-x^{*}〉& \le {\alpha }_{\tau \left(n\right)}\parallel \gamma f\left(x_{\tau \left(n\right)}\right)-\mu F{x}_{\tau \left(n\right)}{\parallel }^2\hfill \\ +\omega \left(1-\tau \right)\parallel T{x}_{\tau \left(n\right)}-x_{\tau \left(n\right)}\parallel \parallel x_{\tau \left(n\right)}-x^{*}\parallel .\hfill \end{array}\hfill \end{array}\end{array}$$

(3.20)

Again, (3.14) and (3.20) yield

$$\begin{array}{r}\begin{array}{c}\begin{array}{cc}\hfill 〈\left(\mu F-\gamma f\right)x^{*},x_{\tau \left(n\right)}-x^{*}〉+2\left(\mu \eta -\gamma L\right){\mathrm{\Gamma }}_{\tau \left(n\right)}& \le {\alpha }_{\tau \left(n\right)}\parallel \gamma f\left(x_{\tau \left(n\right)}\right)-\mu F{x}_{\tau \left(n\right)}{\parallel }^2\hfill \\ +\omega \left(1-\tau \right)\parallel T{x}_{\tau \left(n\right)}-x_{\tau \left(n\right)}\parallel \parallel x_{\tau \left(n\right)}-x^{*}\parallel .\hfill \end{array}\hfill \end{array}\end{array}$$

(3.21)

Recall that lim_n→∞α_τ(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 lim_n→∞Γ_τ(n)= 0. By Lemma 2.4, it follows that Γ _n ≤ Γ_τ(n), thus, we get lim_n→∞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 lim_n→∞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):

$$\begin{array}{r}〈\left(I-f\right)x^{*},x-x^{*}〉\ge 0,\forall x\in Fix\left(T\right).\end{array}$$

(3.26)