A damped algorithm for the split feasibility and fixed point problems

Abstract

The purpose of this paper is to study the split feasibility problem and the fixed point problem. We suggest a damped algorithm. Convergence theorem is proven.

MSC:47J25, 47H09, 65J15, 90C25.

1 Introduction

Let C and Q be two closed convex subsets of two Hilbert spaces $H_1$ and $H_2$, respectively, and let $A:H_1\to H_2$ be a bounded linear operator. Finding a point $x^{\ast }$ satisfies

$$x^{\ast }\in C\text{and}A{x}^{\ast }\in Q.$$

(1.1)

This problem, referred to as the split problem, has been studied by some authors. See, e.g., [1–8] and [9]. Some algorithms for solving (1.1) have been presented. One is Byrne’s CQ algorithm [1]

$$x_{n+1}=P_C(x_n-\tau A^{\ast }(I-P_Q)A{x}_n),n\in \mathbb{N},$$

where $\tau \in (0,\frac{2}{L})$ with L being the largest eigenvalue of the matrix $A^{\ast }A$, I is the unit matrix or operator, and $P_C$ and $P_Q$ denote the orthogonal projections onto C and Q, respectively. Motivated by Byrne’s CQ algorithm, Xu [6] suggested a single step regularized method

$$x_{n+1}=P_C((1-{\alpha }_n{\gamma }_n)x_n-{\gamma }_n{A}^{\ast }(I-P_Q)A{x}_n),n\in \mathbb{N}.$$

(1.2)

Very recently, Dang and Gao [5] introduced the following damped projection algorithm

$$x_{n+1}=(1-{\beta }_n)x_n+{\beta }_n{P}_C((1-{\alpha }_n)(x_n-\tau A^{\ast }(I-P_Q)A{x}_n)),n\in \mathbb{N}.$$

If every closed convex subset of a Hilbert space is the fixed point set of its associating projection, then the split feasibility problem becomes a special case of the split common fixed point problem of finding a point $x^{\ast }$ with the property

$$x^{\ast }\in Fix(U)\text{and}A{x}^{\ast }\in Fix(T).$$

This problem was first introduced by Censor and Segal [10], who invented an algorithm, which generates a sequence $\{x_n\}$ according to the iterative procedure

$$x_{n+1}=U(x_n-\gamma A^{\ast }(I-T)A{x}_n),n\in \mathbb{N}.$$

Recently, Cui, Su and Wang [11] extended the damped projection algorithm to the split common fixed point problems. For some related work, please refer to [12] and [13, 14].

Motivated by these results, the purpose of this paper is to study the following split feasibility problem and fixed point problem

$$\text{Find }{x}^{\ast }\in C\cap Fix(T)\text{ such that }A{x}^{\ast }\in Q\cap Fix(S),$$

(1.3)

where $S:Q\to Q$ and $T:C\to C$ are two nonexpansive mappings. We suggest a damped algorithm for solving (1.3). Convergence theorem is proven.

2 Preliminaries

Let H be a real Hilbert space with the inner product $〈\cdot ,\cdot 〉$ and the norm $\parallel \cdot \parallel$, respectively. Let C be a nonempty closed convex subset of H.

Definition 2.1 A mapping $T:C\to C$ is called nonexpansive if

$$\parallel Tx-Ty\parallel \le \parallel x-y\parallel$$

for all $x,y\in C$.

We will use $Fix(T)$ to denote the set of fixed points of T, that is, $Fix(T)=\{x\in C:x=Tx\}$.

Definition 2.2 We call $P_C:H\to C$ the metric projection if for each $x\in H$

$$\parallel x-P_C(x)\parallel =inf\{\parallel x-y\parallel :y\in C\}.$$

It is well known that the metric projection $P_C:H\to C$ is characterized by

$$〈x-P_C(x),y-P_C(x)〉\le 0$$

for all $x\in H$, $y\in C$. From this, we can deduce that $P_C$ is firmly-nonexpansive, that is,

$${\parallel P_C(x)-P_C(y)\parallel }^2\le 〈x-y,P_C(x)-P_C(y)〉$$

(2.1)

for all $x,y\in H$. Hence $P_C$ is also nonexpansive.

It is well known that in a real Hilbert space H, the following two equalities hold

$${\parallel tx+(1-t)y\parallel }^2=t{\parallel x\parallel }^2+(1-t){\parallel y\parallel }^2-t(1-t){\parallel x-y\parallel }^2$$

(2.2)

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

$${\parallel x+y\parallel }^2={\parallel x\parallel }^2+2〈x,y〉+{\parallel y\parallel }^2$$

(2.3)

for all $x,y\in H$. It follows that

$${\parallel x+y\parallel }^2\le {\parallel x\parallel }^2+2〈y,x+y〉$$

(2.4)

for all $x,y\in H$.

Lemma 2.3 [15]

Let C be a closed convex subset of a real Hilbert space H, and let $S:C\to C$ be a nonexpansive mapping. Then, the mapping $I-S$ is demiclosed. That is, if $\{x_n\}$ is a sequence in C such that $x_n\to x^{\ast }$ weakly and $(I-S)x_n\to y$ strongly, then $(I-S)x^{\ast }=y$.

Lemma 2.4 [16]

Assume that $\{a_n\}$ is a sequence of nonnegative real numbers such that

$$a_{n+1}\le (1-{\gamma }_n)a_n+{\delta }_n,n\in \mathbb{N},$$

where $\{{\gamma }_n\}$ is a sequence in $(0,1)$, and $\{{\delta }_n\}$ is a sequence such that

• (1) ${\sum }_{n=1}^{\mathrm{\infty }}{\gamma }_n=\mathrm{\infty }$;

• (2) ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}\frac{{\delta }_n}{{\gamma }_n}\le 0$ or ${\sum }_{n=1}^{\mathrm{\infty }}|{\delta }_n|<\mathrm{\infty }$.

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

3 Main results

Let C and Q be two nonempty closed convex subsets of real Hilbert spaces $H_1$ and $H_2$, respectively. Let $A:H_1\to H_2$ be a bounded linear operator with its adjoint $A^{\ast }$. Let $S:Q\to Q$ and $T:C\to C$ be two nonexpansive mappings. We use Γ to denote the set of solutions of (1.3), that is, $\mathrm{\Gamma }=\{x^{\ast }|x^{\ast }\in C\cap Fix(T),A{x}^{\ast }\in Q\cap Fix(S)\}$. Now, we present our algorithm.

Algorithm 3.1 For $x_0\in H_1$ arbitrarily, let $\{x_n\}$ be a sequence defined by

$$x_{n+1}=T{P}_C((1-{\alpha }_n)(x_n-\delta A^{\ast }(I-S{P}_Q)A{x}_n))\text{for all }n\in \mathbb{N},$$

(3.1)

where ${\{{\alpha }_n\}}_{n\in \mathbb{N}}$ and ${\{{\beta }_n\}}_{n\in \mathbb{N}}$ are two real number sequences in $(0,1)$ and $\delta \in (0,\frac{1}{{\parallel A\parallel }^2})$.

Theorem 3.2 Suppose $\mathrm{\Gamma }\ne \mathrm{\varnothing }$. Assume the sequence ${\{{\alpha }_n\}}_{n\in \mathbb{N}}$ satisfies three conditions

(C1) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$;

(C2) ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

(C3) ${lim}_{n\to \mathrm{\infty }}\frac{{\alpha }_{n+1}}{{\alpha }_n}=1$.

Then the sequence $\{x_n\}$, generated by algorithm (3.1), converges strongly to $x^{\ast }=P_{\mathrm{\Gamma }}(0)$.

Proof For the convenience, we write $z_n=P_{Q}A{x}_n$, $y_n=(1-{\alpha }_n)(x_n-\delta A^{\ast }(I-S{P}_Q)A{x}_n)$ and $u_n=P_C((1-{\alpha }_n)(x_n-\delta A^{\ast }(I-S{P}_Q)A{x}_n))$ for all $n\in \mathbb{N}$. Thus $u_n=P_C{y}_n$ for all $n\in \mathbb{N}$.

Let $x^{\ast }=P_{\mathrm{\Gamma }}(0)$. Hence, $x^{\ast }\in C\cap Fix(T)$ and $A{x}^{\ast }\in Q\cap Fix(S)$. By the firmly-nonexpansivity of $P_C$ and $P_Q$, we can deduce the following conclusions

$$\parallel z_n-A{x}^{\ast }\parallel =\parallel P_{Q}A{x}_n-P_{Q}A{x}^{\ast }\parallel \le \parallel A{x}_n-A{x}^{\ast }\parallel ,$$

(3.2)

$$\parallel u_n-x^{\ast }\parallel =\parallel P_C{y}_n-P_C{x}^{\ast }\parallel \le \parallel y_n-x^{\ast }\parallel ,$$

(3.3)

$${\parallel S{z}_n-A{x}^{\ast }\parallel }^2\le {\parallel z_n-A{x}^{\ast }\parallel }^2\le {\parallel A{x}_n-A{x}^{\ast }\parallel }^2-{\parallel z_n-A{x}_n\parallel }^2,$$

(3.4)

$$\parallel u_{n+1}-u_n\parallel =\parallel P_C{y}_{n+1}-P_C{y}_n\parallel \le \parallel y_{n+1}-y_n\parallel$$

(3.5)

and

$$\parallel z_{n+1}-z_n\parallel =\parallel P_{Q}A{x}_{n+1}-P_{Q}A{x}_n\parallel \le \parallel A{x}_{n+1}-A{x}_n\parallel .$$

(3.6)

From (3.1) and (3.3), we have

$$\parallel x_{n+1}-x^{\ast }\parallel =\parallel T{u}_n-x^{\ast }\parallel \le \parallel u_n-x^{\ast }\parallel \le \parallel y_n-x^{\ast }\parallel .$$

(3.7)

Using (2.3), we get

$$\begin{array}{rcl}{\parallel y_n-x^{\ast }\parallel }^2& =& {\parallel (1-{\alpha }_n)(x_n-x^{\ast }+\delta A^{\ast }(S{z}_n-A{x}_n))-{\alpha }_n{x}^{\ast }\parallel }^2\\ \le & (1-{\alpha }_n){\parallel (x_n-x^{\ast }+\delta A^{\ast }(S{z}_n-A{x}_n)\parallel }^2+{\alpha }_n{\parallel x^{\ast }\parallel }^2\\ =& (1-{\alpha }_n)[\parallel x_n-x^{\ast }\parallel +{\delta }^2{\parallel A^{\ast }(S{z}_n-A{x}_n)\parallel }^2\\ +2\delta 〈x_n-x^{\ast },A^{\ast }(S{z}_n-A{x}_n)〉]+{\alpha }_n{\parallel x^{\ast }\parallel }^2.\end{array}$$

(3.8)

Since A is a linear operator with its adjoint $A^{\ast }$, we have

$$\begin{array}{c}〈x_n-x^{\ast },A^{\ast }(S{z}_n-A{x}_n)〉\hfill \\ =〈A(x_n-x^{\ast }),S{z}_n-A{x}_n〉\hfill \\ =〈A{x}_n-A{x}^{\ast }+S{z}_n-A{x}_n-(S{z}_n-A{x}_n),S{z}_n-A{x}_n〉\hfill \\ =〈S{z}_n-A{x}^{\ast },S{z}_n-A{x}_n〉-{\parallel S{z}_n-A{x}_n\parallel }^2.\hfill \end{array}$$

(3.9)

Again using (2.3), we obtain

$$〈S{z}_n-A{x}^{\ast },S{z}_n-A{x}_n〉=\frac{1}{2}({\parallel S{z}_n-A{x}^{\ast }\parallel }^2+{\parallel S{z}_n-A{x}_n\parallel }^2-{\parallel A{x}_n-A{x}^{\ast }\parallel }^2).$$

(3.10)

By (3.4), (3.9) and (3.10), we get

$$\begin{array}{rcl}〈x_n-x^{\ast },A^{\ast }(S{z}_n-A{x}_n)〉& =& \frac{1}{2}({\parallel S{z}_n-A{x}^{\ast }\parallel }^2+{\parallel S{z}_n-A{x}_n\parallel }^2-{\parallel A{x}_n-A{x}^{\ast }\parallel }^2)\\ -{\parallel S{z}_n-A{x}_n\parallel }^2\\ \le & \frac{1}{2}({\parallel A{x}_n-A{x}^{\ast }\parallel }^2-{\parallel z_n-A{x}_n\parallel }^2+{\parallel S{z}_n-A{x}_n\parallel }^2\\ -{\parallel A{x}_n-A{x}^{\ast }\parallel }^2)-{\parallel S{z}_n-A{x}_n\parallel }^2\\ =& -\frac{1}{2}{\parallel z_n-A{x}_n\parallel }^2-\frac{1}{2}{\parallel S{z}_n-A{x}_n\parallel }^2.\end{array}$$

(3.11)

Substituting (3.11) into (3.8), we deduce

$$\begin{array}{rcl}{\parallel y_n-x^{\ast }\parallel }^2& \le & (1-{\alpha }_n)[{\parallel x_n-x^{\ast }\parallel }^2+{\delta }^2{\parallel A\parallel }^2{\parallel S{z}_n-A{x}_n\parallel }^2\\ +2\delta (-\frac{1}{2}{\parallel z_n-A{x}_n\parallel }^2-\frac{1}{2}{\parallel S{z}_n-A{x}_n\parallel }^2)]+{\alpha }_n{\parallel x^{\ast }\parallel }^2\\ =& (1-{\alpha }_n)[{\parallel x_n-x^{\ast }\parallel }^2+({\delta }^2{\parallel A\parallel }^2-\delta ){\parallel S{z}_n-A{x}_n\parallel }^2\\ -\delta {\parallel z_n-A{x}_n\parallel }^2]+{\alpha }_n{\parallel x^{\ast }\parallel }^2\\ \le & (1-{\alpha }_n){\parallel x_n-x^{\ast }\parallel }^2+{\alpha }_n{\parallel x^{\ast }\parallel }^2.\end{array}$$

(3.12)

It follows from (3.7) that

$$\begin{array}{rcl}{\parallel x_{n+1}-x^{\ast }\parallel }^2& \le & {\parallel y_n-x^{\ast }\parallel }^2\\ \le & (1-{\alpha }_n){\parallel x_n-x^{\ast }\parallel }^2+{\alpha }_n{\parallel x^{\ast }\parallel }^2\\ \le & max\{{\parallel x_n-x^{\ast }\parallel }^2,{\parallel x^{\ast }\parallel }^2\}.\end{array}$$

The boundedness of the sequence $\{x_n\}$ yields.

Next, we estimate $\parallel x_{n+1}-x_n\parallel$. Set $v_n=x_n-\delta A^{\ast }(I-S{P}_Q)A{x}_n$. According to (2.3) and (3.5), we have

$$\begin{array}{rcl}{\parallel v_{n+1}-v_n\parallel }^2& =& {\parallel x_{n+1}-x_n+\delta [A^{\ast }(S{P}_Q-I)A{x}_{n+1}-A^{\ast }(S{P}_Q-I)A{x}_n]\parallel }^2\\ =& {\parallel x_{n+1}-x_n\parallel }^2+{\delta }^2{\parallel A^{\ast }[(S{P}_Q-I)A{x}_{n+1}-(S{P}_Q-I)A{x}_n]\parallel }^2\\ +2\delta 〈x_{n+1}-x_n,A^{\ast }[(S{P}_Q-I)A{x}_{n+1}-(S{P}_Q-I)A{x}_n]〉\\ \le & {\parallel x_{n+1}-x_n\parallel }^2+{\delta }^2{\parallel A\parallel }^2{\parallel S{z}_{n+1}-S{z}_n-(A{x}_{n+1}-A{x}_n)\parallel }^2\\ +2\delta 〈A{x}_{n+1}-A{x}_n,S{z}_{n+1}-S{z}_n-(A{x}_{n+1}-A{x}_n)〉\\ =& {\parallel x_{n+1}-x_n\parallel }^2+{\delta }^2{\parallel A\parallel }^2{\parallel S{z}_{n+1}-S{z}_n-(A{x}_{n+1}-A{x}_n)\parallel }^2\\ +2\delta 〈S{z}_{n+1}-S{z}_n,S{z}_{n+1}-S{z}_n-(A{x}_{n+1}-A{x}_n)〉\\ -2\delta {\parallel S{z}_{n+1}-S{z}_n-(A{x}_{n+1}-A{x}_n)\parallel }^2\\ =& {\parallel x_{n+1}-x_n\parallel }^2+{\delta }^2{\parallel A\parallel }^2{\parallel S{z}_{n+1}-S{z}_n-(A{x}_{n+1}-A{x}_n)\parallel }^2\\ +\delta ({\parallel S{z}_{n+1}-S{z}_n\parallel }^2+{\parallel S{z}_{n+1}-S{z}_n-(A{x}_{n+1}-A{x}_n)\parallel }^2\\ -{\parallel A{x}_{n+1}-A{x}_n\parallel }^2)-2\delta {\parallel S{z}_{n+1}-S{z}_n-(A{x}_{n+1}-A{x}_n)\parallel }^2\\ =& {\parallel x_{n+1}-x_n\parallel }^2+({\delta }^2{\parallel A\parallel }^2-\delta ){\parallel S{z}_{n+1}-S{z}_n-(A{x}_{n+1}-A{x}_n)\parallel }^2\\ +\delta ({\parallel S{z}_{n+1}-S{z}_n\parallel }^2-{\parallel A{x}_{n+1}-A{x}_n\parallel }^2)\\ \le & {\parallel x_{n+1}-x_n\parallel }^2+({\delta }^2{\parallel A\parallel }^2-\delta ){\parallel S{z}_{n+1}-S{z}_n-(A{x}_{n+1}-A{x}_n)\parallel }^2\\ +\delta ({\parallel z_{n+1}-z_n\parallel }^2-{\parallel A{x}_{n+1}-A{x}_n\parallel }^2).\end{array}$$

(3.13)

Since $\delta \in (0,\frac{1}{{\parallel A\parallel }^2})$, we derive by virtue of (3.6) and (3.13) that

$$\parallel v_{n+1}-v_n\parallel \le \parallel x_{n+1}-x_n\parallel .$$

(3.14)

From (3.5) and (3.14), we have

$$\begin{array}{rcl}\parallel x_{n+1}-x_n\parallel & \le & \parallel y_{n+1}-y_n\parallel \\ =& \parallel (1-{\alpha }_{n+1})v_{n+1}-(1-{\alpha }_n)v_n\parallel \\ =& \parallel (1-{\alpha }_{n+1})(v_{n+1}-v_n)+({\alpha }_n-{\alpha }_{n+1})v_n\parallel \\ \le & (1-{\alpha }_{n+1})\parallel v_{n+1}-v_n\parallel +|{\alpha }_{n+1}-{\alpha }_n|\parallel v_n\parallel \\ \le & (1-{\alpha }_{n+1})\parallel x_{n+1}-x_n\parallel +|{\alpha }_{n+1}-{\alpha }_n|\parallel v_n\parallel .\end{array}$$

It follows that

$$\parallel x_{n+1}-x_n\parallel \le \frac{|{\alpha }_{n+1}-{\alpha }_n|}{{\alpha }_{n+1}}\parallel v_n\parallel .$$

This, together with condition (C3), implies that

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

(3.15)

That is,

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

(3.16)

Using the firmly-nonexpansiveness of $P_C$, we have

$$\begin{array}{rcl}{\parallel u_n-x^{\ast }\parallel }^2& =& {\parallel P_C{y}_n-x^{\ast }\parallel }^2\\ \le & {\parallel y_n-x^{\ast }\parallel }^2-{\parallel P_C{y}_n-y_n\parallel }^2\\ =& {\parallel y_n-x^{\ast }\parallel }^2-{\parallel u_n-y_n\parallel }^2.\end{array}$$

(3.17)

Thus,

$$\begin{array}{rcl}{\parallel x_{n+1}-x^{\ast }\parallel }^2& \le & {\parallel u_n-x^{\ast }\parallel }^2\\ \le & {\parallel y_n-x^{\ast }\parallel }^2-{\parallel u_n-y_n\parallel }^2\\ \le & (1-{\alpha }_n){\parallel x_n-x^{\ast }\parallel }^2+{\alpha }_n{\parallel x^{\ast }\parallel }^2-{\parallel u_n-y_n\parallel }^2.\end{array}$$

(3.18)

It follows that

$$\begin{array}{rcl}{\parallel u_n-y_n\parallel }^2& \le & {\parallel x_n-x^{\ast }\parallel }^2-{\parallel x_{n+1}-x^{\ast }\parallel }^2+{\alpha }_n{\parallel x^{\ast }\parallel }^2\\ \le & (\parallel x_n-x^{\ast }\parallel +\parallel x_{n+1}-x^{\ast }\parallel )\parallel x_{n+1}-x_n\parallel +{\alpha }_n{\parallel x^{\ast }\parallel }^2.\end{array}$$

This, together with (3.15) and (C1), implies that

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

(3.19)

Returning to (3.18) and using (3.12), we have

$$\begin{array}{rcl}{\parallel x_{n+1}-x^{\ast }\parallel }^2& \le & {\parallel y_n-x^{\ast }\parallel }^2\\ \le & (1-{\alpha }_n){\parallel x_n-x^{\ast }\parallel }^2+(1-{\alpha }_n)({\delta }^2{\parallel A\parallel }^2-\delta ){\parallel S{z}_n-A{x}_n\parallel }^2\\ -(1-{\alpha }_n)\delta {\parallel z_n-A{x}_n\parallel }^2+{\alpha }_n{\parallel x^{\ast }\parallel }^2.\end{array}$$

Hence,

$$\begin{array}{c}(1-{\alpha }_n)(\delta -{\delta }^2{\parallel A\parallel }^2){\parallel S{z}_n-A{x}_n\parallel }^2+(1-{\alpha }_n)\delta {\parallel z_n-A{x}_n\parallel }^2\hfill \\ \le {\parallel x_n-x^{\ast }\parallel }^2-{\parallel x_{n+1}-x^{\ast }\parallel }^2+{\alpha }_n{\parallel x^{\ast }\parallel }^2\hfill \\ \le (\parallel x_n-x^{\ast }\parallel +\parallel x_{n+1}-x^{\ast }\parallel )\parallel x_{n+1}-x_n\parallel +{\alpha }_n{\parallel x^{\ast }\parallel }^2,\hfill \end{array}$$

which implies that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel S{z}_n-A{x}_n\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel z_n-A{x}_n\parallel =0.$$

(3.20)

So,

$$\underset{n\to \mathrm{\infty }}{lim}\parallel S{z}_n-z_n\parallel =0.$$

(3.21)

Note that

$$\begin{array}{rcl}\parallel y_n-x_n\parallel & =& \parallel \delta A^{\ast }(S{P}_Q-I)A{x}_n+{\alpha }_n{v}_n\parallel \\ \le & \delta \parallel A\parallel \parallel S{z}_n-A{x}_n\parallel +{\alpha }_n\parallel v_n\parallel .\end{array}$$

It follows from (3.20) that

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

(3.22)

From (3.16), (3.19) and (3.22), we get

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

(3.23)

Now, we show that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈x^{\ast },y_n-x^{\ast }〉\ge 0.$$

Choose a subsequence $\{y_{n_i}\}$ of $\{y_n\}$ such that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈x^{\ast },y_n-x^{\ast }〉=\underset{i\to \mathrm{\infty }}{lim}〈x^{\ast },y_{n_i}-x^{\ast }〉.$$

(3.24)

Since the sequence $\{y_{n_i}\}$ is bounded, we can choose a subsequence $\{y_{n_{i_j}}\}$ of $\{y_{n_i}\}$ such that $y_{n_{i_j}}\rightharpoonup z$. For the sake of convenience, we assume (without loss of generality) that $y_{n_i}\rightharpoonup z$. Consequently, we derive from the above conclusions that

$$x_{n_i}\rightharpoonup z,u_{n_i}\rightharpoonup z,A{x}_{n_i}\rightharpoonup Az\text{and}{z}_{n_i}\rightharpoonup Az.$$

(3.25)

By the demiclosed principle of the nonexpansive mappings S and T (see Lemma 2.3), we deduce that $z\in Fix(T)$ and $Az\in Fix(S)$ (according to (3.23) and (3.21), respectively). Note that $u_{n_i}=P_C{y}_{n_i}\in C$ and $z_{n_i}=P_{Q}A{x}_{n_i}\in Q$. From (3.25), we deduce $z\in C$ and $Az\in Q$. To this end, we deduce that $z\in C\cap Fix(T)$ and $Az\in Q\cap Fix(S)$. That is to say, $z\in \mathrm{\Gamma }$. Therefore,

$$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈x^{\ast },y_n-x^{\ast }〉& =& \underset{i\to \mathrm{\infty }}{lim}〈x^{\ast },y_{n_i}-x^{\ast }〉\\ =& \underset{i\to \mathrm{\infty }}{lim}〈x^{\ast },z-x^{\ast }〉\\ \ge & 0.\end{array}$$

(3.26)

Finally, we prove that $x_n\to x^{\ast }$. From (3.1), we have

$$\begin{array}{rcl}{\parallel x_{n+1}-x^{\ast }\parallel }^2& \le & {\parallel y_n-x^{\ast }\parallel }^2\\ =& {\parallel (1-{\alpha }_n)(v_n-x^{\ast })-{\alpha }_n{x}^{\ast }\parallel }^2\\ \le & (1-{\alpha }_n){\parallel v_n-x^{\ast }\parallel }^2-2{\alpha }_n〈x^{\ast },y_n-x^{\ast }〉\\ \le & (1-{\alpha }_n){\parallel x_n-x^{\ast }\parallel }^2-2{\alpha }_n〈x^{\ast },y_n-x^{\ast }〉.\end{array}$$

(3.27)

Applying Lemma 2.4 and (3.26) to (3.27), we deduce that $x_n\to x^{\ast }$. The proof is completed. □