A general iterative method for quasi-nonexpansive mappings in Hilbert space

Abstract

Iterative algorithms have been extensively studied over the class of nonexpansive mappings in Hilbert spaces. Recall that nonexpansive mappings belong to quasi-nonexpansive mappings. The aim of this article is expanding the general approximation method proposed by Marino and Xu to quasi-nonexpansive mappings in Hilbert spaces.

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}.

Iterative theory and methods for nonlinear mappings and variational inequalities have recently been applied to solve convex minimization problems, zero point problems and many others; see, e.g., [1–9] and references therein.

The viscosity approximation method was first introduced by Moudafi [10]. Starting with an arbitrary initial x₀ ∈ H, define a sequence {x _n } generated by:

$$x_{n+1}=\frac{{\epsilon }_n}{1+{\epsilon }_n}f\left(x_n\right)+\frac{1}{1+{\epsilon }_n}T{x}_n,\forall n\ge 0,$$

(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, and {ε _n } is a sequence in (0,1) satisfying the following given conditions:

• (1) lim_n→∞ε _n = 0;

• (2) ${\sum }_{n=0}^{\infty }{\epsilon }_n=\infty$;

• (3) ${\lim }_{n\to \infty }(\frac{1}{{\epsilon }_n}-\frac{1}{{\epsilon }_{n+1}})=0$.

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:

$$⟨\left(I-f\right)x^{*},x-x^{*}⟩\ge 0,\forall x\in Fix\left(T\right).$$

In [1], Xu proved that the sequence {x _n } defined by the below process started with an arbitrary initial x₀ ∈ H:

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

(1.2)

converges strongly to the unique solution of the minimization problem (1.3) provided the the sequence {α _n } satisfies certain conditions:

$$\underset{x\in C}{\text{min}}\frac{1}{2}⟨Ax,x⟩-⟨x,b⟩,$$

(1.3)

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

In [2], Marino and Xu combined the iterative method (1.2) with the viscosity approximation method (1.1) and considered the following general iterative method:

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

(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:

$$⟨\left(\gamma f-A\right)\tilde{x},x-\tilde{x}⟩\le 0,\forall x\in C,$$

(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 [11], Maingé considered the viscosity approximation method (1.1), and expanded the strong convergence to quasi-nonexpansive mappings in Hilbert space. Motivated by Marino and Xu [2] and Maingé [11], we consider the following iterative process:

$$\left\{\begin{array}{c}{x}_0=x\in Harbitrarilychosen,\hfill \\ x_{n+1}={\alpha }_n\gamma f\left(x_n\right)+\left(I-{\alpha }_{n}A\right)T_{\omega }{x}_n,\forall n\ge 0,\hfill \end{array}\right.$$

(1.6)

where 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 Maingé's [11] conclusion, and also extends the iterative method (1.4) to quasi-nonexpansive mappings.

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 identities are valid in a Hilbert space H: for each x,y ∈ H, t ∈ [0, 1]

• (i) ||x + y||² ≤ ||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) $⟨x,y⟩=-\frac{1}{2}{∥x-y∥}^2+\frac{1}{2}{∥x∥}^2+\frac{1}{2}{∥y∥}^2$.

Lemma 2.1. [2] Let H be a Hilbert space H. Given x ∈ H, C is a closed convex subset of H, f : H → H is a contraction with coefficient 0 < α < 1, and A is a strongly positive linear bounded operator with coefficient $\bar{\gamma }$. Then for $0<\gamma <\bar{\gamma }/\alpha$,

$$⟨x-y,\left(A-\gamma f\right)x-\left(A-\gamma f\right)y⟩\ge \left(\bar{\gamma }-\gamma \alpha \right){∥x-y∥}^2,\forall x,y\in H.$$

That is, A - γ f is strongly monotone with coefficient $\bar{\gamma }-\gamma \alpha$.

Lemma 2.2. [2] Assume A is a strongly positive linear bounded operator on a Hilbert space H with coefficient $\bar{\gamma }>0$ and 0 < ρ ≤ ||A||^-1. Then $∥I-\rho A∥\le 1-\rho \bar{\gamma }$.

Lemma 2.3. [11] Let T _ω : = (1 - ω)I + ωT, with T being a quasi-nonexpansive mapping on H, $Fix\left(T\right)\ne \mathrm{0̸}$, 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||² for all x ∈ H and q ∈ Fix(T);

(a4) $⟨x-T_{\omega }x,x-q⟩\ge \frac{\omega }{2}{∥x-Tx∥}^2$ for all x ∈ H and q ∈ Fix(T).

Remark 2.4. (a4) was revised by Wongchan and Saejung [12] (Proposition 2).

Lemma 2.5. [13] Let {Γ _n } be a sequence of real numbers that does not decrease at infinity, in the sense that there exist a subsequence ${\left\{{\Gamma }_{n_j}\right\}}_{j\ge 0}$ of {Γ _n } which satisfies ${\Gamma }_{n_j}<{\Gamma }_{n_j+1}$ for all j ≥ 0. Also consider the sequence of integers ${\left\{\tau \left(n\right)\right\}}_{n\ge n_0}$ defined by

$$\tau \left(n\right)=\text{max}\left\{k\le n|{\Gamma }_k<{\Gamma }_{k+1}\right\}.$$

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

$${\Gamma }_n\le {\Gamma }_{\tau \left(n\right)+1}.$$

Recall the metric projection P _K form a Hilbert space H to a closed convex subset K of H is defined: for each x ∈ H, there exists a unique element P _K x ∈ K such that

$$∥x-P_{K}x∥:=\text{inf}\left\{∥x-y∥:y\in K\right\}.$$

Lemma 2.6. 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:

$$⟨x-z,y-z⟩\le 0,\forall y\in K.$$

Lemma 2.7. If x* is the solution of the variational inequality (1.5) with demi-closedness of T and {y _n } ∈ H is a bounded sequence such that ||Ty _n - y _n || → 0, then

$$\underset{n\to \infty }{\text{lim inf}}〈(A-\gamma f)x^{*},y_n-x^{*}〉\ge 0.$$

(2.1)

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 ||Ty _n - y _n || → 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

$$\underset{n\to \infty }{\text{lim inf}}〈(A-\gamma f)x^{*},y_n-x^{*}〉=\underset{j\to \infty }{\text{lim}}〈(A-\gamma f)x^{*},y_{n_j}-x^{*}〉.$$

Recalling the (1.5), we immediately obtain

$$\underset{n\to \infty }{\text{lim inf}}〈(A-\gamma f)x^{*},y_n-x^{*}〉=〈(A-\gamma f)x^{*},\tilde{y}-x^{*}〉\ge 0.$$

This completes the proof.

3. Main results

Let H be a real Hilbert space, let A be a bounded linear operator on H, and let T be a quasi-nonexpansive mapping on H, and f is a contraction with coefficient α; that is ||f (x) - f(y)|| ≤ α||x - y|| 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 (see [14] for more general results).

Throughout this article, we assume that A is strongly positive; that is, there exist a constant $\bar{\gamma }>0$ such that $⟨Ax,x⟩\ge \bar{\gamma }{∥x∥}^2$, for all x ∈ H. Let $0<\gamma <\bar{\gamma }/\alpha$.

Theorem 3.1. Starting with an arbitrary chosen x₀ ∈ H, let the sequence {x _n } be generated by

$$x_{n+1}={\alpha }_n\gamma f\left(x_n\right)+\left(I-{\alpha }_{n}A\right)T_{\omega }{x}_n,$$

(3.1)

where the sequence {α _n } ⊂ (0,1) satisfies lim_n→∞α _n = 0, and ${\sum }_{n=0}^{\infty }{\alpha }_n=\infty$. Also $\omega \in \left(0,\frac{1}{2}\right)$, 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 demiclosed 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:

$$⟨\left(\gamma f-A\right)x^{*},x-x^{*}⟩\le 0,\forall x\in Fix\left(T\right).$$

(3.2)

Remark 3.2. Equivalently, from the VIP (3.2), we have

$$x^{*}=P_{Fix\left(T\right)}\circ \left(I-A+\gamma f\right)x^{*}.$$

(3.3)

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

Take any p ∈ Fix(T), from Lemma 2.3 (a3), we have

$$\begin{array}{ll}\hfill ∥x_{n+1}-p∥& =∥{\alpha }_n\gamma f\left(x_n\right)+\left(I-{\alpha }_{n}A\right)T_{\omega }{x}_n-p∥\\ =∥{\alpha }_n\gamma \left(f\left(x_n\right)-f\left(p\right)\right)+{\alpha }_n\left(\gamma f\left(p\right)-Ap\right)+\left(I-{\alpha }_{n}A\right)\left(T_{\omega }{x}_n-p\right)∥\\ \le {\alpha }_n\gamma \alpha ∥f\left(x_n\right)-f\left(p\right)∥+{\alpha }_n∥\gamma f\left(p\right)-Ap∥+\left(1-{\alpha }_n\bar{\gamma }\right)∥x_n-p∥\\ \le \left(1-{\alpha }_n\left(\bar{\gamma }-\gamma \alpha \right)\right)∥x_n-p∥+{\alpha }_n∥\gamma f\left(p\right)-Ap∥.\end{array}$$

(3.4)

By induction

$$∥x_n-p∥\le \text{max}\left\{∥x_0-p∥,\frac{∥\gamma f\left(p\right)-Ap∥}{\bar{\gamma }-\gamma \alpha }\right\},\forall n\ge 0.$$

Hence {x _n } is bounded, so are the {f(x _n )} and {A(x _n )}.

Let x* = P_Fix(T)o(I - A + γf)x* From (3.1), we have

$$x_{n+1}-x_n+{\alpha }_n\left(A{x}_n-\gamma f\left(x_n\right)\right)=\left(I-{\alpha }_{n}A\right)\left(T_{\omega }{x}_n-x_n\right).$$

(3.5)

Since x* ∈ Fix(T), from (a4), and together with (3.5), we obtain

$$\begin{array}{c}⟨x_{n+1}-x_n+{\alpha }_n\left(A{x}_n-\gamma f\left(x_n\right)\right),x_n-x^{*}⟩\\ =⟨\left(I-{\alpha }_{n}A\right)\left(T_{\omega }{x}_n-x_n\right),x_n-x^{*}⟩\\ =\left(1-{\alpha }_n\right)⟨T_{\omega }{x}_n-x_n,x_n-x^{*}⟩+{\alpha }_n⟨\left(I-A\right)\left(T_{\omega }{x}_n-x_n\right),x_n-x^{*}⟩\\ \le -\frac{\omega }{2}\left(1-{\alpha }_n\right){∥x_n-T{x}_n∥}^2+\omega {\alpha }_n⟨\left(I-A\right)\left(T-I\right)x_n,x_n-x^{*}⟩,\end{array}$$

it follows from the previous inequality that

$$\begin{array}{ll}\hfill -⟨x_n-x_{n+1},x_n-x^{*}⟩& \le -{\alpha }_n⟨\left(A-\gamma f\right)x_n,x_n-x^{*}⟩-\frac{\omega }{2}\left(1-{\alpha }_n\right){∥x_n-T{x}_n∥}^2\\ +\omega {\alpha }_n⟨\left(I-A\right)\left(T-I\right)x_n,x_n-x^{*}⟩.\end{array}$$

(3.6)

From (iii), we obviously have

$$⟨x_n-x_{n+1},x_n-x^{*}⟩=-\frac{1}{2}{∥x_{n+1}-x^{*}∥}^2+\frac{1}{2}{∥x_n-x^{*}∥}^2+\frac{1}{2}{∥x_{n+1}-x_n∥}^2.$$

(3.7)

Set ${\Gamma }_n:=\frac{1}{2}{∥x_n-x^{*}∥}^2$, and combine with (3.6), it follows that

$$\begin{array}{ll}\hfill {\Gamma }_{n+1}-{\Gamma }_n-\frac{1}{2}{∥x_{n+1}-x_n∥}^2& \le -{\alpha }_n⟨\left(A-\gamma f\right)x_n,x_n-x^{*}⟩-\frac{\omega }{2}\left(1-{\alpha }_n\right){∥x_n-T{x}_n∥}^2\\ +\omega {\alpha }_n⟨\left(I-A\right)\left(T-I\right)x_n,x_n-x^{*}⟩.\end{array}$$

(3.8)

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 (3.5) that

$$\begin{array}{ll}\hfill {∥x_{n+1}-x_n∥}^2& ={∥{\alpha }_n\left(\gamma f\left(x_n\right)-A{x}_n\right)+\left(I-{\alpha }_{n}A\right)\left(T_{\omega }{x}_n-x_n\right)∥}^2\\ \le 2{\alpha }_n^2{∥\gamma f\left(x_n\right)-A{x}_n∥}^2+2{\left(1-{\alpha }_n\bar{\gamma }\right)}^2{∥T_{\omega }{x}_n-x_n∥}^2\\ \le 2{\alpha }_n^2{∥\gamma f\left(x_n\right)-A{x}_n∥}^2+2\left(1-{\alpha }_n\bar{\gamma }\right){∥T_{\omega }{x}_n-x_n∥}^2\\ \le 2{\alpha }_n^2{∥\gamma f\left(x_n\right)-A{x}_n∥}^2+2{\omega }^2\left(1-{\alpha }_n\bar{\gamma }\right){∥T{x}_n-x_n∥}^2.\end{array}$$

(3.9)

Then from (3.8) and (3.9), we have

$$\begin{array}{c}{\Gamma }_{n+1}-{\Gamma }_n+\left[\frac{\omega }{2}\left(1-{\alpha }_n\right)-{\omega }^2\left(1-{\alpha }_n\bar{\gamma }\right)\right]{∥x_n-T{x}_n∥}^2\\ \le {\alpha }_n\left[{\alpha }_n{∥\gamma f\left(x_n\right)-A{x}_n∥}^2-⟨\left(A-\gamma f\right)x_n,x_n-x^{*}⟩\right.\\ \left.+\omega ⟨\left(I-A\right)\left(T-I\right)x_n,x_n-x^{*}⟩\right].\end{array}$$

(3.10)

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

Case 1: Suppose that there exists n₀ such that ${\left\{{\Gamma }_n\right\}}_{n\ge n_0}$ 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

$$\underset{n\to \infty }{\text{lim}}\left({\Gamma }_{n+1}-{\Gamma }_n\right)=0.$$

(3.11)

It follows from (3.10), (3.11) and the fact that lim_n→∞α _n = 0, we have lim_n→∞||x _n -Tx _n || = 0. Again, from (3.10), we have

$$\begin{array}{c}-{\alpha }_n\left[{\alpha }_n{∥\gamma f\left(x_n\right)-A{x}_n∥}^2-⟨\left(A-\gamma f\right)x_n,x_n-x^{*}⟩+\omega ⟨\left(I-A\right)\left(T-I\right)x_n,x_n-x^{*}⟩\right]\\ \le {\Gamma }_n-{\Gamma }_{n+1}.\end{array}$$

(3.12)

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

$$\underset{n\to \infty }{\text{lim inf}}-[{\alpha }_n{\Vert (\gamma f-A)x_n\Vert }^2-〈(A-\gamma f)x_n,x_n-x^{*}〉+\omega 〈(I-A)(T-I)x_n,x_n-x^{*}〉]\le 0.$$

(3.13)

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 (3.13) that

$$\underset{n\to \infty }{\text{lim inf}}〈(A-\gamma f)x_n,x_n-x^{*}〉\le 0.$$

(3.14)

From Lemma 2.1, it is obvious that

$$⟨\left(A-\gamma f\right)x_n,x_n-x^{*}⟩\ge ⟨\left(A-\gamma f\right)x^{*},x_n-x^{*}⟩+2\left(\bar{\gamma }-\gamma \alpha \right){\Gamma }_n.$$

(3.15)

Thus, from (3.14), (3.15) and the fact that lim_n→∞Γ _n exists, we immediately obtain

$$\begin{array}{l}\underset{n\to \infty }{\text{lim inf}}[〈(A-\gamma f)x^{*},x_n-x^{*}〉+2(\overline{\gamma }-\gamma \alpha ){\Gamma }_n]\\ =2(\overline{\gamma }-\gamma \alpha )\underset{n\to \infty }{\text{lim}}{\Gamma }_n+\underset{n\to \infty }{\text{lim inf}}〈(A-\gamma f)x^{*},x_n-x^{*}〉\le 0,\end{array}$$

(3.16)

or equivalently

$$2(\overline{\gamma }-\gamma \alpha )\underset{n\to \infty }{\text{lim}}{\Gamma }_n\le -\underset{n\to \infty }{\text{lim inf}}〈(A-\gamma f)x^{*},x_n-x^{*}〉.$$

(3.17)

Finally, by Lemma 2.7, we have

$$2\left(\bar{\gamma }-\gamma \alpha \right)\underset{n\to \infty }{\text{lim}}{\Gamma }_n\le 0,$$

(3.18)

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${\left\{{\Gamma }_{n_j}\right\}}_{j\ge 0}$ of {Γ _n }_n≥0such that ${\Gamma }_{n_j}<{\Gamma }_{n_j+1}$ for all j ∈ ℕ. In this case, it follows from Lemma 2.5 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 2.5.

Invoking the (3.10) again, it follows that

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

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

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

(3.19)

From the preceding results, we get the boundedness of {x _n } and α _n → 0, which obviously lead to

$$\underset{n\to \infty }{\text{lim}}∥x_{\tau \left(n\right)}-T{x}_{\tau \left(n\right)}∥=0.$$

(3.20)

Hence, combining (3.19) with (3.20), we immediately deduce that

$$\begin{array}{ll}\hfill ⟨\left(A-\gamma f\right)x_{\tau \left(n\right)},x_{\tau \left(n\right)}-x^{*}⟩& \le {\alpha }_{\tau \left(n\right)}{∥\gamma f\left(x_{\tau \left(n\right)}\right)-A{x}_{\tau \left(n\right)}∥}^2\\ +\omega ⟨\left(I-A\right)\left(T-I\right)x_{\tau \left(n\right)},x_{\tau \left(n\right)}-x^{*}⟩.\end{array}$$

(3.21)

Again, (3.15) and (3.21) yield

$$\begin{array}{ll}\hfill ⟨\left(A-\gamma f\right)x^{*},x_{\tau \left(n\right)}-x^{*}⟩+2\left(\bar{\gamma }-\gamma \alpha \right){\Gamma }_{\tau \left(n\right)}& \le {\alpha }_{\tau \left(n\right)}{∥\gamma f\left(x_{\tau \left(n\right)}\right)-A{x}_{\tau \left(n\right)}∥}^2\\ +\omega ⟨\left(I-A\right)\left(T-I\right)x_{\tau \left(n\right)},x_{\tau \left(n\right)}-x^{*}⟩.\end{array}$$

(3.22)

Recall that lim_n→∞α _τ(n)= 0 and (3.20), we immediately have

$$2\left(\bar{\gamma }-\gamma \alpha \right)\underset{n\to \infty }{\text{lim sup}}{\Gamma }_{\tau \left(n\right)}\le -\underset{n\to \infty }{\text{lim inf}}⟨\left(A-\gamma f\right)x^{*},x_{\tau \left(n\right)}-x^{*}⟩$$

(3.23)

By Lemma 2.7, we have

$$\underset{n\to \infty }{\text{lim inf}}⟨\left(A-\gamma f\right)x^{*},x_{\tau \left(n\right)}-x^{*}⟩\ge 0.$$

(3.24)

Consider (3.23) again, we conclude that

$$\underset{n\to \infty }{\text{lim sup}}{\Gamma }_{\tau \left(n\right)}=0,$$

(3.25)

which means that lim_n→∞Γ_τ(n)= 0. By Lemma 2.5, it follows that Γ _n ≤ Γ_τ(n), thus, we get lim_n→∞Γ _n = 0, which is equivalent to x _n → x*.

Corollary 3.3. [11] Let the sequence {x _n } be generated by

$$x_{n+1}={\alpha }_{n}f\left(x_n\right)+\left(1-{\alpha }_n\right)T_{\omega }{x}_n,$$

(3.26)

where the sequence {α _n } ⊂ (0,1) satisfies lim_n→∞α _n = 0, and ${\sum }_{n=0}^{\infty }{\alpha }_n=\infty$. 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 demiclosed 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 (3.27):

$$⟨\left(I-f\right)x^{*},x-x^{*}⟩\ge 0,\forall x\in Fix\left(T\right).$$

(3.27)