Iterative algorithms for finding the zeroes of sums of operators

Abstract

Let $H_1$, $H_2$ be real Hilbert spaces, $C\subseteq H_1$ be a nonempty closed convex set, and $0\notin C$. Let $A:H_1\to H_2$, $B:H_1\to H_2$ be two bounded linear operators. We consider the problem to find $x\in C$ such that $Ax=-Bx$ ($0=Ax+Bx$). Recently, Eckstein and Svaiter presented some splitting methods for finding a zero of the sum of monotone operator A and B. However, the algorithms are largely dependent on the maximal monotonicity of A and B. In this paper, we describe some algorithms for finding a zero of the sum of A and B which ignore the conditions of the maximal monotonicity of A and B.

1 Introduction and preliminaries

Let $H_1$, $H_2$, $H_3$ be real Hilbert spaces, $C\subseteq H_1$ be a nonempty closed convex set and $0\notin C$. Let $A:H_1\to H_2$, $B:H_1\to H_2$ be two bounded linear operators. We consider the interesting problem of finding $x\in C$ such that

$$Ax=-Bx(\text{or }0=Ax+Bx).$$

(1.1)

For convenience, we denote the problem by $\mathcal{P}$.

For $\mathcal{P}$ it is generally difficult to find zeroes of A and B separately. To overcome this difficulty, Eckstein and Svaiter [1] present the splitting methods for finding a zero of the sum of monotone operator A and B. Three basic families of splitting methods for this problem were identified in [1]:

• (i) The Douglas/Peaceman-Rachford family, whose iteration is given by

  $$\begin{array}{c}{y}_k=[2{(I+\xi B)}^{-1}+I]x_k,\hfill \\ z_k=[2{(I+\xi A)}^{-1}+I]y_k,\hfill \\ x_{k+1}=(1-{\rho }_k)x_k+{\rho }_k{z}_k,\hfill \end{array}$$

where $\xi >0$ is a fixed scalar, and $\{{\rho }_k\}\subseteq (0,1]$ is a sequence of relaxation parameters.

• (ii) The double backward splitting method, with iteration given by

  $$\begin{array}{c}{y}_k={(I+{\lambda }_{k}B)}^{-1}{x}_k,\hfill \\ x_{k+1}={(I+{\lambda }_{k}A)}^{-1}{y}_k,\hfill \end{array}$$

where $\{{\lambda }_k\}\subseteq (0,\mathrm{\infty })$ a sequence of regularization parameters.

• (iii) The forward-backward splitting method, with iteration given by

  $$\begin{array}{c}{y}_k\in {(I-{\lambda }_{k}A)}^{-1}{x}_k,\hfill \\ x_{k+1}={(I+{\lambda }_{k}B)}^{-1}{y}_k,\hfill \end{array}$$

where $\{{\lambda }_k\}\subseteq (0,\mathrm{\infty })$ a sequence of regularization parameters.

Convergence results for the scheme (i), in the case in which $\{{\rho }_k\}$ is contained in a compact subset of $(0,1)$, can be found in [2]; the convergence analysis of the double backward scheme given by (ii), which can be found in [3] and [4]; the standard convergence analysis for (iii) one can see [5]. However, the convergence results are largely dependent on the maximal monotonicity of A and B. It is therefore the aim of this paper to construct new algorithms for problem $\mathcal{P}$ which ignore the conditions of the maximal monotonicity of A and B.

The paper is organized as follows. In Section 2, we define the concept of the minimal norm solution of the problem $\mathcal{P}$ (1.1). Using Tychonov regularization, we obtain a net of solutions for some minimization problem approximating such minimal norm solution (see Theorem 2.4). In Section 3, we introduce an algorithm and prove the strong convergence of the algorithm, more importantly, its limit is the minimum-norm solution of the problem $\mathcal{P}$ (1.1) (see Theorem 3.2). In Section 4, we introduce KM-CQ-like iterative algorithm which converge strongly to a solution of the problem $\mathcal{P}$ (1.1) (see Theorem 4.3).

Throughout the rest of this paper, I denotes the identity operator on Hilbert space H, $Fix(T)$ the set of the fixed points of an operator T and ∇f the gradient of the functional $f:H\to R$. An operator T on a Hilbert space H is nonexpansive if, for each x and y in H, $\parallel Tx-Ty\parallel \le \parallel x-y\parallel$. T is said to be averaged, if there exist $0<\alpha <1$ and a nonexpansive operator N such that $T=(1-\alpha )I+\alpha N$.

We know that the projection $P_C$ from H onto a nonempty closed convex subset C of H is a typical example of a nonexpansive and averaged mapping, which is defined by

$$P_C(w)=arg\underset{x\in C}{min}\parallel x-w\parallel .$$

It is well known that $P_C(w)$ is characterized by the inequality

$$〈w-P_C(w),x-P_C(w〉\le 0,\mathrm{\forall }x\in C.$$

We now collect some elementary facts which will be used in the proofs of our main results.

Lemma 1.1 [6, 7]

Let X be a Banach space, C a closed convex subset of X, and $T:C\to C$ a nonexpansive mapping with $Fix(T)\ne \mathrm{\varnothing }$. If $\{x_n\}$ is a sequence in C weakly converging to x and if $\{(I-T)x_n\}$ converges strongly to y, then $(I-T)x=y$.

Lemma 1.2 [8]

Let $\{s_n\}$ be a sequence of nonnegative real numbers, $\{{\alpha }_n\}$ a sequence of real numbers in $[0,1]$ with ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$, $\{u_n\}$ a sequence of nonnegative real numbers with ${\sum }_{n=1}^{\mathrm{\infty }}{u}_n<\mathrm{\infty }$, and $\{t_n\}$ a sequence of real numbers with ${lim\hspace{0.17em}sup}_n{t}_n\le 0$. Suppose that

$$s_{n+1}=(1-{\alpha }_n)s_n+{\alpha }_n{t}_n+u_n,\mathrm{\forall }n\in N.$$

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

Lemma 1.3 [9]

Let $\{w_n\}$, $\{z_n\}$ be bounded sequences in a Banach space and let $\{{\beta }_n\}$ be a sequence in $[0,1]$ which satisfies the following condition: $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1$. Suppose that $w_{n+1}=(1-{\beta }_n)w_n+{\beta }_n{z}_n$ and ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}\parallel z_{n+1}-z_n\parallel -\parallel w_{n+1}-w_n\parallel \le 0$, then ${lim}_{n\to \mathrm{\infty }}\parallel z_n-w_n\parallel =0$.

Lemma 1.4 [10]

Let f be a convex and differentiable functional and let C be a closed convex subset of H. Then $x\in C$ is a solution of the problem

$$\underset{x\in C}{min}f(x)$$

if and only if $x\in C$ satisfies the following optimality condition:

$$〈\mathrm{\nabla }f(x),v-x〉\ge 0,\mathrm{\forall }v\in C.$$

Moreover, if f is, in addition, strictly convex and coercive, then the minimization problem has a unique solution.

Lemma 1.5 [11]

Let A and B be averaged operators and suppose that $Fix(A)\cap Fix(B)$ is nonempty. Then $Fix(A)\cap Fix(B)=Fix(AB)=Fix(BA)$.

2 The minimum-norm solution of the problem $\mathcal{P}$

In this section, we propose the concept of the minimal norm solution of $\mathcal{P}$ (1.1). Then, using Tychonov regularization, we obtain the minimal norm solution by a net of solution for some minimization problem.

We use Γ to denote the solution set of $\mathcal{P}$, i.e.,

$$\mathrm{\Gamma }=\{x\in H_1,Ax=-Bx,x\in C\}$$

and assume consistency of $\mathcal{P}$. Hence Γ is closed, convex, and nonempty.

Let $H=H_1\times H_1$, $M=\{(x,x),x\in H_1\}\subseteq H$, P be the linear operator from $H_1$ onto M, and P has the matrix form

$$P=\left[\begin{array}{c}I\\ I\end{array}\right],$$

that is to say, $P(x)=(x,x)$, $\mathrm{\forall }x\in H_1$.

Define $G:H\to H_2$ by $G((x,y))=Ax+By$, $\mathrm{\forall }(x,y)\in H_2$. Then G has the matrix form $G=[A,B]$, and $GP=A+B$, $P{G}^{\ast }GP=A^{\ast }A+A^{\ast }B+B^{\ast }A+B^{\ast }B$.

The problem can now be reformulated as finding $x\in C$ with $GPx=0$, or solving the following minimization problem:

$$\underset{x\in C}{min}f(x)=\frac{1}{2}{\parallel GPx\parallel }^2,$$

(2.1)

which is ill-posed. A classical way is the well-known Tychonov regularization, which approximates a solution of problem (2.1) by the unique minimizer of the regularized problem:

$$\underset{x\in C}{min}{f}_{\alpha }(x)=\frac{1}{2}{\parallel GPx\parallel }^2+\frac{1}{2}\alpha {\parallel x\parallel }^2,$$

(2.2)

where $\alpha >0$ is the regularization parameter. Denote by $x_{\alpha }$ the unique solution of (2.2).

Proposition 2.1 For $\alpha >0$, the solution $x_{\alpha }$ of (2.2) is uniquely defined. $x_{\alpha }$ is characterized by the inequality

$$〈P^{\ast }{G}^{\ast }GP{x}_{\alpha }+\alpha x_{\alpha },x-x_{\alpha }〉\ge 0,\mathrm{\forall }x\in C.$$

Proof Obviously, $f(x)=\frac{1}{2}{\parallel GPx\parallel }^2$ is convex and differentiable with gradient $\mathrm{\nabla }f(x)=P^{\ast }{G}^{\ast }GPx$. Recall that $f_{\alpha }(x)=f(x)+\frac{1}{2}\alpha {\parallel x\parallel }^2$, we see that $f_{\alpha }$ is strictly convex and differentiable with gradient

$$\mathrm{\nabla }{f}_{\alpha }(x)=P^{\ast }{G}^{\ast }GPx+\alpha x.$$

According Lemma 1.4, $x_{\alpha }$ is characterized by the inequality

$$〈P^{\ast }{G}^{\ast }GP{x}_{\alpha }+\alpha x_{\alpha },x-x_{\alpha }〉\ge 0,\mathrm{\forall }x\in C.$$

(2.3)

 □

Definition 2.2 An element $\tilde{x}\in \mathrm{\Gamma }$ is said to be the minimal norm solution of SEP (1.1) if $\parallel \tilde{x}\parallel ={inf}_{x\in \mathrm{\Gamma }}\parallel x\parallel$.

The following proposition collects some useful properties of $\{x_{\alpha }\}$ the unique solution of (2.2).

Proposition 2.3 Let $x_{\alpha }$ be given as the unique solution of (2.2). Then we have:

• (i) $\parallel x_{\alpha }\parallel$ is decreasing for $\alpha \in (0,\mathrm{\infty })$.

• (ii) $\alpha \mapsto x_{\alpha }$ defines a continuous curve from $(0,\mathrm{\infty })$ to $H_1$.

Proof Let $\alpha >\beta >0$, since $x_{\alpha }$ and $x_{\beta }$ are the unique minimizers of $f_{\alpha }$ and $f_{\beta }$, respectively, we get

$$\begin{array}{c}\frac{1}{2}{\parallel GP{x}_{\alpha }\parallel }^2+\frac{1}{2}\alpha {\parallel x_{\alpha }\parallel }^2\le \frac{1}{2}{\parallel GP{x}_{\beta }\parallel }^2+\frac{1}{2}\alpha {\parallel x_{\beta }\parallel }^2,\hfill \\ \frac{1}{2}{\parallel GP{x}_{\beta }\parallel }^2+\frac{1}{2}\beta {\parallel x_{\beta }\parallel }^2\le \frac{1}{2}{\parallel GP{x}_{\alpha }\parallel }^2+\frac{1}{2}\beta {\parallel x_{\alpha }\parallel }^2.\hfill \end{array}$$

It follows that $\parallel x_{\alpha }\parallel \le \parallel x_{\beta }\parallel$. Thus $\parallel x_{\alpha }\parallel$ is decreasing for $\alpha \in (0,\mathrm{\infty })$.

According to Proposition 2.1, we get

$$〈P^{\ast }{G}^{\ast }GP{x}_{\alpha }+\alpha x_{\alpha },x_{\beta }-x_{\alpha }〉\ge 0,$$

and

$$〈P^{\ast }{G}^{\ast }GP{x}_{\beta }+\beta x_{\beta },x_{\alpha }-x_{\beta }〉\ge 0.$$

It follows that

$$〈x_{\alpha }-x_{\beta },\alpha x_{\alpha }-\beta x_{\beta }〉\le 〈x_{\alpha }-x_{\beta },P^{\ast }{G}^{\ast }GP(x_{\beta }-x_{\alpha })〉\le 0.$$

Thus

$$\alpha \parallel x_{\alpha }-x_{\beta }\parallel \le (\alpha -\beta )〈x_{\alpha }-x_{\beta },x_{\beta }〉.$$

It turns out that

$${\parallel x_{\alpha }-x_{\beta }\parallel }^2\le \frac{|\alpha -\beta |}{\alpha }\parallel x_{\beta }\parallel .$$

Hence, $\alpha \mapsto x_{\alpha }$ is a continuous curve from $(0,\mathrm{\infty })$ to $H_1$. □

Theorem 2.4 Let $x_{\alpha }$ be the unique solution of (2.2). Then $x_{\alpha }$ converges strongly to the minimum-norm solution $\tilde{x}$ of $\mathcal{P}$ (1.1) with $\alpha \to 0$.

Proof For any $0<\alpha <\mathrm{\infty }$, $x_{\alpha }$ is given as (2.2), we get

$$\frac{1}{2}{\parallel GP{x}_{\alpha }\parallel }^2+\frac{1}{2}\alpha {\parallel x_{\alpha }\parallel }^2\le \frac{1}{2}{\parallel GP\tilde{x}\parallel }^2+\frac{1}{2}\alpha {\parallel \tilde{x}\parallel }^2.$$

Since $\tilde{x}\in \mathrm{\Gamma }$ is a solution for $\mathcal{P}$,

$$\frac{1}{2}{\parallel GP{x}_{\alpha }\parallel }^2+\frac{1}{2}\alpha {\parallel x_{\alpha }\parallel }^2\le \frac{1}{2}\alpha {\parallel \tilde{x}\parallel }^2.$$

It follows that $\parallel x_{\alpha }\parallel \le \parallel \tilde{x}\parallel$ for all $\alpha >0$. Thus $\{x_{\alpha }\}$ is a bounded net in $H_1$.

All we need to prove is that for any sequence $\{{\alpha }_n\}$ such that ${\alpha }_n\to 0$, $\{x_{{\alpha }_n}\}$ contains a subsequence converging strongly to $\tilde{x}$. For convenience, we set $x_n=x_{{\alpha }_n}$.

In fact $\{x_n\}$ is bounded, by passing to a subsequence if necessary, we may assume that $\{x_n\}$ converges weakly to a point $\hat{x}\in S$. Due to Proposition 2.1, we get

$$〈P^{\ast }{G}^{\ast }GP{x}_n+{\alpha }_n{x}_n,\tilde{x}-x_n〉\ge 0.$$

It turns out that

$$〈GP{x}_n,GP\tilde{x}-GP{x}_n〉\ge {\alpha }_n〈x_n,x_n-\tilde{x}〉.$$

Since $\tilde{x}\in \mathrm{\Gamma }$, it follows that

$$〈GP{x}_n,-GP{x}_n〉\ge {\alpha }_n〈x_n,x_n-\tilde{x}〉.$$

Noting that $\parallel x_n\parallel \le \parallel \tilde{x}\parallel$, we have

$$\parallel GP{x}_n\parallel \le 2{\alpha }_n\parallel \tilde{x}\parallel \to 0.$$

Moreover, note that $\{x_n\}$ converges weakly to a point $\hat{x}\in C$, thus $\{GP{x}_n\}$ converges weakly to $GP\hat{x}$. It follows that $GP\hat{x}=0$, i.e. $\hat{x}\in \mathrm{\Gamma }$.

Finally, we prove that $\hat{x}=\tilde{x}$ and this finishes the proof.

Recall that $\{x_n\}$ converges weakly to $\hat{x}$ and $\parallel x_n\parallel \le \parallel \tilde{x}\parallel$, one can deduce that

$$\parallel \hat{x}\parallel \le \underset{n}{lim\hspace{0.17em}inf}\parallel x_n\parallel \le \parallel \tilde{x}\parallel =min\{\parallel x\parallel :x\in \mathrm{\Gamma }\}.$$

This shows that $\hat{x}$ is also a point in Γ with minimum-norm. By the uniqueness of minimum-norm element, we get $\hat{x}=\tilde{x}$. □

Finally, we will introduce another method to get the minimum-norm solution of the problem $\mathcal{P}$.

Lemma 2.5 Let $T=I-\gamma P^{\ast }{G}^{\ast }GP$, where $0<\gamma <2/\rho (P^{\ast }{G}^{\ast }GP)$ with $\rho (P^{\ast }{G}^{\ast }GP)$ being the spectral radius of the self-adjoint operator $P^{\ast }{G}^{\ast }GP$ on $H_1$. Then we have the following:

• (1) $\parallel T\parallel \le 1$ (i.e. T is nonexpansive) and averaged;

• (2) $Fix(T)=\{x\in H_1,Ax=-Bx\}$, $Fix(P_{C}T)=Fix(P_C)\cap Fix(T)=\mathrm{\Gamma }$;

• (3) $x\in Fix(P_{C}T)$ if and only if x is a solution of the variational inequality $〈P^{\ast }{G}^{\ast }GPx,v-x〉\ge 0$, $\mathrm{\forall }v\in C$.

Proof (1) It is easily proved that $\parallel T\parallel \le 1$, we only need to prove that $T=I-\gamma P^{\ast }{G}^{\ast }GP$ is averaged. Indeed, choose $0<\beta <1$, such that $\gamma /(1-\beta )<2/\rho (P^{\ast }{G}^{\ast }GP)$, then $T=I-\gamma P^{\ast }{G}^{\ast }GP=\beta I+(1-\beta )V$, where $V=I-\gamma /(1-\beta )P^{\ast }{G}^{\ast }GP$ is a nonexpansive mapping. That is to say T is averaged.

• (2) If $x\in \{x\in H_1,Ax=-Bx\}$, it is obviously that $x\in Fix(T)$. Conversely, assume that $x\in Fix(T)$, we have $x=x-\gamma P^{\ast }{G}^{\ast }GPx$, hence $\gamma P^{\ast }{G}^{\ast }GPx=0$ then ${\parallel GPx\parallel }^2=〈P^{\ast }{G}^{\ast }GPx,x〉=0$, we get $x\in \{x\in H_1,Ax=-Bx\}$. We have $Fix(T)=\{x\in H_1,Ax=-Bx\}$.

Now we prove $Fix(P_{C}T)=Fix(P_C)\cap Fix(T)=\mathrm{\Gamma }$. By $Fix(T)=\{x\in H_1,Ax=-Bx\}$, $Fix(P_C)\cap Fix(T)=\mathrm{\Gamma }$ is obviously. On the other hand, since $Fix(P_C)\cap Fix(T)=\mathrm{\Gamma }\ne \mathrm{\varnothing }$, and both $P_C$ and T are averaged, from Lemma 1.5, we have $Fix(P_{C}T)=Fix(P_C)\cap Fix(T)$.

• (3)
  $$\begin{array}{rcl}〈P^{\ast }{G}^{\ast }GPx,v-x〉\ge 0,\mathrm{\forall }v\in C& \iff & 〈x-(x-\gamma P^{\ast }{G}^{\ast }GPx),v-x〉\ge 0,\mathrm{\forall }v\in S\\ \iff & w=P_C(w-\gamma P^{\ast }{G}^{\ast }GPx)\\ \iff & w\in Fix(P_{C}T).\end{array}$$

 □

Remark 2.6 Choose a constant γ satisfying that $0<\gamma <2/\rho (P^{\ast }{G}^{\ast }GP)$. For $\alpha \in (0,\frac{2-\gamma \parallel P^{\ast }{G}^{\ast }GP\parallel }{2\gamma })$, we define a mapping

$$W_{\alpha }(x):=P_C[(1-\alpha \gamma )I-\gamma P^{\ast }{G}^{\ast }GP]x.$$

It is clear that $W_{\alpha }$ is a contractive. Hence, $W_{\alpha }$ has a unique fixed point $x_{\alpha }$, we have

$$x_{\alpha }=P_C[(1-\alpha \gamma )I-\gamma P^{\ast }{G}^{\ast }GP]x_{\alpha }.$$

(2.4)

Theorem 2.7 Let $x_{\alpha }$ be given as (2.4). Then $x_{\alpha }$ converges strongly to the minimum-norm solution $\tilde{x}$ of the problem $\mathcal{P}$ (1.1) when $\alpha \to 0$.

Proof Choose $\check{x}\in \mathrm{\Gamma }$, noting that $\alpha \in (0,\frac{2-\gamma \parallel P^{\ast }{G}^{\ast }GP\parallel }{2\gamma })$, $I-\frac{\gamma }{(1-\alpha \gamma )}{P}^{\ast }{G}^{\ast }GP$ is nonexpansive, it turns out that

$$\begin{array}{rcl}\parallel x_{\alpha }-\check{x}\parallel & =& \parallel P_C[(1-\alpha \gamma )I-\gamma P^{\ast }{G}^{\ast }GP]x_{\alpha }-P_C[\check{x}-\gamma P^{\ast }{G}^{\ast }GP\check{x}]\parallel \\ \le & \parallel [(1-\alpha \gamma )I-\gamma P^{\ast }{G}^{\ast }GP]x_{\alpha }-[\check{x}-\gamma P^{\ast }{G}^{\ast }GP\check{x}]\parallel \\ =& \parallel (1-\alpha \gamma )[x_{\alpha }-\frac{\gamma }{1-\alpha \gamma }{P}^{\ast }{G}^{\ast }GP{x}_{\alpha }]\\ -(1-\alpha \gamma )[\check{x}-\frac{\gamma }{1-\alpha \gamma }{P}^{\ast }{G}^{\ast }GP\check{x}]-\alpha \gamma \check{x}\parallel \\ \le & (1-\alpha \gamma )\parallel (x_{\alpha }-\frac{\gamma }{1-\alpha \gamma }{P}^{\ast }{G}^{\ast }GP{x}_{\alpha })-(\check{x}-\frac{\gamma }{1-\alpha \gamma }{P}^{\ast }{G}^{\ast }GP\check{x})\parallel +\alpha \gamma \parallel \check{x}\parallel \\ \le & (1-\alpha \gamma )\parallel x_{\alpha }-\check{x}\parallel +\alpha \gamma \parallel \check{x}\parallel .\end{array}$$

That is,

$$\parallel x_{\alpha }-\check{x}\parallel \le \parallel \check{x}\parallel .$$

Hence $\{x_{\alpha }\}$ is bounded.

Taking into account of (2.4), we have

$$\parallel x_{\alpha }-P_C[I-\gamma P^{\ast }{G}^{\ast }GP]x_{\alpha }\parallel \le \alpha \parallel \gamma x_{\alpha }\parallel \to 0.$$

We assert that $\{x_{\alpha }\}$ is relatively norm compact as $\alpha \to 0^{+}$. In fact, assume that $\{{\alpha }_n\}\subseteq (0,\frac{2-\gamma \parallel P^{\ast }{G}^{\ast }GP\parallel }{2\gamma })$ and ${\alpha }_n\to 0^{+}$ as $n\to \mathrm{\infty }$. For convenience, we put $x_n:=x_{{\alpha }_n}$, we get

$$\parallel x_n-P_C[I-\gamma P^{\ast }{G}^{\ast }GP]x_n\parallel \le {\alpha }_n\parallel \gamma x_n\parallel \to 0.$$

Since $P_C$ is nonexpansive, one concludes that

$$\begin{array}{rcl}{\parallel x_{\alpha }-\check{x}\parallel }^2& =& {\parallel P_C[(1-\alpha \gamma )I-\gamma P^{\ast }{G}^{\ast }GP]x_{\alpha }-P_C[\check{x}-\gamma P^{\ast }{G}^{\ast }GP\check{x}]\parallel }^2\\ \le & 〈[(1-\alpha \gamma )I-\gamma P^{\ast }{G}^{\ast }GP]x_{\alpha }-[\check{x}-\gamma P^{\ast }{G}^{\ast }GP\check{x}],x_{\alpha }-\check{x}〉\\ =& 〈(1-\alpha \gamma )[x_{\alpha }-\frac{\gamma }{1-\alpha \gamma }{P}^{\ast }{G}^{\ast }GP{x}_{\alpha }]\\ -(1-\alpha \gamma )[\check{x}-\frac{\gamma }{1-\alpha \gamma }{P}^{\ast }{G}^{\ast }GP\check{x}],x_{\alpha }-\check{x}〉-\alpha \gamma 〈\check{x},x_{\alpha }-\check{x}〉\\ \le & (1-\alpha \gamma ){\parallel x_{\alpha }-\check{x}\parallel }^2-\alpha \gamma 〈\check{x},x_{\alpha }-\check{x}〉.\end{array}$$

That is ,

$${\parallel x_{\alpha }-\check{x}\parallel }^2\le 〈-\check{x},x_{\alpha }-\check{x}〉.$$

Thus,

$${\parallel x_n-\check{x}\parallel }^2\le 〈-\check{x},x_n-\check{x}〉,\mathrm{\forall }\check{x}\in \mathrm{\Gamma }.$$

Due to $\{x_n\}$ is bounded, there exists a subsequence of $\{x_n\}$ which converges weakly to a point $\tilde{x}$. Without loss of generality, we may assume that $\{x_n\}$ converges weakly to $\tilde{x}$. Noting that

$$\parallel x_n-P_C[I-\gamma P^{\ast }{G}^{\ast }GP]x_n\parallel \le {\alpha }_n\parallel \gamma x_n\parallel \to 0,$$

and applying Lemma 1.1, we obtain $\tilde{x}\in Fix(P_C[I-\gamma P^{\ast }{G}^{\ast }GP])=\mathrm{\Gamma }$.

Since

$${\parallel x_n-\check{x}\parallel }^2\le 〈-\check{x},x_n-\check{x}〉,\mathrm{\forall }\check{x}\in \mathrm{\Gamma },$$

it concludes that

$${\parallel x_n-\tilde{x}\parallel }^2\le 〈-\tilde{x},x_n-\tilde{x}〉.$$

Hence, if $\{x_n\}$ converges weakly to $\tilde{x}$, then $\{x_n\}$ converges strongly to $\tilde{x}$. That is to say $\{x_{\alpha }\}$ is relatively norm compact as $\alpha \to 0^{+}$.

Moreover, again using

$${\parallel x_n-\check{x}\parallel }^2\le 〈-\check{x},x_n-\check{x}〉,\mathrm{\forall }\check{x}\in \mathrm{\Gamma },$$

let $n\to \mathrm{\infty }$, we have

$${\parallel \tilde{x}-\check{x}\parallel }^2\le 〈-\check{x},\tilde{x}-\check{x}〉,\mathrm{\forall }\check{x}\in \mathrm{\Gamma }.$$

This implies that

$$〈-\check{x},\check{x}-\tilde{x}〉\le 0,\mathrm{\forall }\check{x}\in \mathrm{\Gamma }.$$

This is equivalent to

$$〈-\tilde{x},\check{x}-\tilde{x}〉\le 0,\mathrm{\forall }\check{x}\in \mathrm{\Gamma }.$$

It turns out that $\tilde{x}\in P_C(0)$. Consequently, each cluster point of $x_{\alpha }$ is equals $\tilde{x}$. Thus $x_{\alpha }\to \tilde{x}(\alpha \to 0)$ the minimum-norm solution of the problem $\mathcal{P}$. □

3 Iterative algorithm for the minimum-norm solution of the problem $\mathcal{P}$

In this section, we introduce the following algorithm and prove the strong convergence of the algorithm, more importantly, its limit is the minimum-norm solution of the problem $\mathcal{P}$.

Algorithm 3.1 For an arbitrary point $x_0\in H_1$ the sequence $\{x_n\}$ is generated by the iterative algorithm

$$x_{n+1}=P_C\{(1-{\alpha }_n)[I-\gamma P^{\ast }{G}^{\ast }GP]x_n\},$$

(3.1)

where ${\alpha }_n>0$ is a sequence in $(0,1)$ such that

• (i) ${lim}_n{\alpha }_n=0$;

• (ii) ${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (iii) ${\sum }_{n=0}^{\mathrm{\infty }}|{\alpha }_{n+1}-{\alpha }_n|<\mathrm{\infty }$ or ${lim}_n|{\alpha }_{n+1}-{\alpha }_n|/{\alpha }_n=0$.

Now, we prove the strong convergence of the iterative algorithm.

Theorem 3.2 The sequence $\{x_n\}$ generated by algorithm (3.1) converges strongly to the minimum-norm solution $\tilde{x}$ of the problem $\mathcal{P}$ (1.1).

Proof Let $R_n$ and R be defined by

$$\begin{array}{c}{R}_{n}x:=P_C\{(1-{\alpha }_n)[I-\gamma P^{\ast }{G}^{\ast }GP]\}x=P_C[(1-{\alpha }_n)Tx],\hfill \\ Rx:=P_C(I-\gamma P^{\ast }{G}^{\ast }GP)x=P_C(Tx),\hfill \end{array}$$

where $T=I-\gamma P^{\ast }{G}^{\ast }GP$, by Lemma 2.5 , it is easy to see that $R_n$ is a contraction with contractive constant $1-{\alpha }_n$. Algorithm (3.1) can be written as $x_{n+1}=R_n{x}_n$.

For any $\hat{x}\in \mathrm{\Gamma }$, we have

$$\begin{array}{rcl}\parallel R_n\hat{x}-\hat{x}\parallel & =& \parallel P_C[(1-{\alpha }_n)T\hat{x}]-\hat{x}\parallel \\ =& \parallel P_C[(1-{\alpha }_n)T\hat{x}]-P_S(T\hat{x})\parallel \\ \le & \parallel (1-{\alpha }_n)T\hat{x}-T\hat{x}\parallel \\ =& {\alpha }_n\parallel T\hat{x}\parallel \le {\alpha }_n\parallel \hat{x}\parallel .\end{array}$$

Hence,

$$\begin{array}{rcl}\parallel x_{n+1}-\hat{x}\parallel & =& \parallel R_n{x}_n-\hat{x}\parallel \le \parallel R_n{x}_n-R_n\hat{x}\parallel +\parallel R_n\hat{x}-\hat{x}\parallel \\ \le & \parallel P_C[(1-{\alpha }_n)T\hat{x}]-P_S(T\hat{x})\parallel \\ \le & (1-{\alpha }_n)\parallel x_n-\hat{x}\parallel +{\alpha }_n\parallel \hat{x}\parallel \\ \le & max\{\parallel x_n-\hat{x}\parallel ,\parallel \hat{x}\parallel \}.\end{array}$$

It follows that $\parallel x_n-\hat{x}\parallel \le max\{\parallel x_0-\hat{x}\parallel ,\parallel \hat{x}\parallel \}$. So $\{x_n\}$ is bounded.

Next we prove that ${lim}_n\parallel x_{n+1}-x_n\parallel =0$.

Indeed,

$$\begin{array}{rcl}\parallel x_{n+1}-x_n\parallel & =& \parallel R_n{x}_n-R_{n-1}{x}_{n-1}\parallel \\ \le & \parallel R_n{x}_n-R_n{x}_{n-1}\parallel +\parallel R_n{x}_{n-1}-R_{n-1}{x}_{n-1}\parallel \\ \le & (1-{\alpha }_n)\parallel x_n-x_{n-1}\parallel +\parallel R_n{x}_{n-1}-R_{n-1}{x}_{n-1}\parallel .\end{array}$$

Notice that

$$\begin{array}{rcl}\parallel R_n{x}_{n-1}-R_{n-1}{x}_{n-1}\parallel & =& \parallel P_C[(1-{\alpha }_n)T{x}_{n-1}]-P_C[(1-{\alpha }_{n-1})T{x}_{n-1}]\parallel \\ \le & \parallel (1-{\alpha }_n)T{x}_{n-1}-(1-{\alpha }_{n-1})T{x}_{n-1}\parallel \\ =& |{\alpha }_n-{\alpha }_{n-1}|\parallel T{x}_{n-1}\parallel \\ \le & |{\alpha }_n-{\alpha }_{n-1}|\parallel x_{n-1}\parallel .\end{array}$$

Hence

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

By virtue of the assumptions (1)-(3) and Lemma 1.2, we have

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

Therefore,

$$\begin{array}{rcl}\parallel x_n-R{x}_n\parallel & \le & \parallel x_{n+1}-x_n\parallel +\parallel R_n{x}_n-R{x}_n\parallel \\ \le & \parallel x_{n+1}-x_n\parallel +\parallel (1-{\alpha }_n)T{x}_n-T{x}_n\parallel \\ \le & \parallel x_{n+1}-x_n\parallel +{\alpha }_n\parallel x_n\parallel \to 0.\end{array}$$

By the demiclosedness principle ensures that each weak limit point of $\{x_n\}$ is a fixed point of the nonexpansive mapping $R=P_{C}T$, that is, a point of the solution set Γ of SEP (1.1).

Finally, we will prove that ${lim}_n\parallel x_{n+1}-\tilde{x}\parallel =0$.

Choose $0<\beta <1$, such that $\gamma /(1-\beta )<2/\rho (P^{\ast }{G}^{\ast }GP)$, then $T=I-\gamma P^{\ast }{G}^{\ast }GP=\beta I+(1-\beta )V$, where $V=I-\gamma /(1-\beta )P^{\ast }{G}^{\ast }GP$ is a nonexpansive mapping. Taking $z\in \mathrm{\Gamma }$, we deduce that

$$\begin{array}{rcl}{\parallel x_{n+1}-z\parallel }^2& =& {\parallel P_C[(1-{\alpha }_n)T{x}_n]-z\parallel }^2\\ \le & {\parallel (1-{\alpha }_n)T{x}_n-z\parallel }^2\\ \le & (1-{\alpha }_n){\parallel T{x}_n-z\parallel }^2+{\alpha }_n{\parallel z\parallel }^2\\ \le & {\parallel \beta (x_n-z)+(1-\beta )(V{x}_n-z)\parallel }^2+{\alpha }_n{\parallel z\parallel }^2\\ \le & \beta {\parallel (x_n-z)\parallel }^2+(1-\beta ){\parallel (V{x}_n-z)\parallel }^2-\beta (1-\beta ){\parallel x_n-V{x}_n\parallel }^2+{\alpha }_n{\parallel z\parallel }^2\\ \le & {\parallel (x_n-z)\parallel }^2-\beta (1-\beta ){\parallel x_n-V{x}_n\parallel }^2+{\alpha }_n{\parallel z\parallel }^2.\end{array}$$

Then

$$\begin{array}{rcl}\beta (1-\beta )\parallel x_n-V{x}_n\parallel & \le & {\parallel x_n-z\parallel }^2-{\parallel x_{n+1}-z\parallel }^2+{\alpha }_n{\parallel z\parallel }^2\\ \le & (\parallel x_n-z\parallel +\parallel x_{n+1}-z\parallel )(\parallel x_n-z\parallel -\parallel x_{n+1}-z\parallel ){\alpha }_n{\parallel z\parallel }^2\\ \le & (\parallel x_n-z\parallel +\parallel x_{n+1}-z\parallel )(\parallel x_n-x_{n+1}\parallel ){\alpha }_n{\parallel z\parallel }^2\to 0.\end{array}$$

Note that $T=I-\gamma P^{\ast }{G}^{\ast }GP=\beta I+(1-\beta )V$, it follows that ${lim}_n\parallel T{x}_n-x_n\parallel =0$.

Take a subsequence $\{x_{n_k}\}$ of $\{x_n\}$ such that ${lim\hspace{0.17em}sup}_n〈x_n-\tilde{x},-\tilde{x}〉={lim}_k〈x_{n_k}-\tilde{x},-\tilde{x}〉$.

By virtue of the boundedness of $x_n$, we may further assume with no loss of generality that $x_{n_k}$ converges weakly to a point $\check{x}$. Since $\parallel R{x}_n-x_n\parallel \to 0$, using the demiclosedness principle, $\check{x}\in Fix(R)=Fix(P_{C}T)=\mathrm{\Gamma }$. Noticing that $\tilde{x}$ is the projection of the origin onto Γ, we get

$$\underset{n}{lim\hspace{0.17em}sup}〈x_n-\tilde{x},-\tilde{x}〉=\underset{k}{lim}〈x_{n_k}-\tilde{x},-\tilde{x}〉=〈\check{x}-\tilde{x},-\tilde{x}〉\le 0.$$

Finally, we compute

$$\begin{array}{rcl}{\parallel x_{n+1}-\tilde{x}\parallel }^2& =& {\parallel P_C[(1-{\alpha }_n)T{x}_n]-\tilde{x}\parallel }^2\\ =& {\parallel P_C[(1-{\alpha }_n)T{x}_n]-P_{C}T\tilde{x}\parallel }^2\\ \le & {\parallel (1-{\alpha }_n)T{x}_n-T\tilde{x}\parallel }^2\\ =& {\parallel (1-{\alpha }_n)T{x}_n-\tilde{x}\parallel }^2\\ =& {\parallel (1-{\alpha }_n)(T{x}_n-\tilde{x})+{\alpha }_n(-\tilde{x})\parallel }^2\\ =& {(1-{\alpha }_n)}^2{\parallel (T{x}_n-\tilde{x})\parallel }^2+{\alpha }_n^2{\parallel \tilde{x}\parallel }^2+2{\alpha }_n(1-{\alpha }_n)〈T{x}_n-\tilde{x},-\tilde{x}〉\\ \le & (1-{\alpha }_n){\parallel (T{x}_n-\tilde{x})\parallel }^2+{\alpha }_n[{\alpha }_n{\parallel \tilde{x}\parallel }^2+2(1-{\alpha }_n)〈T{x}_n-\tilde{x},-\tilde{x}〉].\end{array}$$

Since, ${lim\hspace{0.17em}sup}_n〈x_n-\tilde{x},-\tilde{x}〉\le 0$, $\parallel x_n-T{x}_n\parallel \to 0$, we know that ${lim\hspace{0.17em}sup}_n({\alpha }_n{\parallel \tilde{x}\parallel }^2+2(1-{\alpha }_n)〈T{x}_n-\tilde{x},-\tilde{x}〉)\le 0$, by Lemma 1.2, we conclude that ${lim}_n\parallel x_{n+1}-\tilde{x}\parallel =0$. This completes the proof. □

4 KM-CQ-like iterative algorithm for the problem $\mathcal{P}$

In this section, we establish a KM-CQ-like algorithm converges strongly to a solution of the problem $\mathcal{P}$.

Algorithm 4.1 For an arbitrary initial point $x_0$, sequence $\{x_n\}$ is generated by the iteration:

$$x_{n+1}=(1-{\beta }_n)x_n+{\beta }_n{P}_C[(1-{\alpha }_n)(I-\gamma P^{\ast }{G}^{\ast }GP)]x_n,$$

(4.1)

where ${\alpha }_n>0$ is a sequence in $(0,1)$ such that

• (i) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$, ${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (ii) ${lim}_{n\to \mathrm{\infty }}|{\alpha }_{n+1}-{\alpha }_n|=0$;

• (iii) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1$.

Lemma 4.2 If $z\in Fix(T)=Fix(I-\gamma P^{\ast }{G}^{\ast }GP)$, then for any x we have ${\parallel Tx-z\parallel }^2\le {\parallel x-z\parallel }^2-\beta (1-\beta ){\parallel Vx-x\parallel }^2$, where β and V are the same in Lemma 2.5(1).

Proof By Lemma 2.5(1), we know that $T=\beta I+(1-\beta )V$, where $0<\beta <1$ and V is a nonexpansive. It is clear that $z\in Fix(T)=Fix(V)$, and

$$\begin{array}{rcl}{\parallel Tx-z\parallel }^2& =& {\parallel \beta x+(1-\beta )Vx-z\parallel }^2\\ \le & \beta {\parallel x-z\parallel }^2+(1-\beta ){\parallel Vx-z\parallel }^2-\beta (1-\beta ){\parallel Vx-x\parallel }^2\\ \le & \beta {\parallel x-z\parallel }^2+(1-\beta ){\parallel x-z\parallel }^2-\beta (1-\beta ){\parallel Vx-x\parallel }^2\\ =& {\parallel x-z\parallel }^2-\beta (1-\beta ){\parallel Vx-x\parallel }^2.\end{array}$$

 □

Theorem 4.3 The sequence $\{x_n\}$ generated by algorithm (4.1) converges strongly to a solution of the problem $\mathcal{P}$.

Proof For any solution $\hat{x}$ of the problem $\mathcal{P}$, according to Lemma 2.5, $\hat{x}\in Fix(P_{C}T)=Fix(P_C)\cap Fix(T)$, where $T=I-\gamma P^{\ast }{G}^{\ast }GP$, and

$$\begin{array}{rcl}\parallel x_{n+1}-\hat{x}\parallel & =& \parallel (1-{\beta }_n)x_n+{\beta }_n{P}_C[(1-{\alpha }_n)T]x_n-\hat{x}\parallel \\ =& \parallel (1-{\beta }_n)(x_n-\hat{x})+{\beta }_n(P_C[(1-{\alpha }_n)T]x_n-\hat{x})\parallel \\ \le & (1-{\beta }_n)\parallel x_n-\hat{x}\parallel +{\beta }_n\parallel P_C[(1-{\alpha }_n)T]x_n-\hat{x}\parallel \\ \le & (1-{\beta }_n)\parallel x_n-\hat{x}\parallel \\ +{\beta }_n\parallel P_C[(1-{\alpha }_n)T]x_n-P_C[(1-{\alpha }_n)T]\hat{x}\parallel \\ +{\beta }_n\parallel P_C[(1-{\alpha }_n)T]\hat{x}-\hat{x}\parallel \\ \le & (1-{\beta }_n)\parallel x_n-\hat{x}\parallel +{\beta }_n(1-{\alpha }_n)\parallel x_n-\hat{x}\parallel +{\beta }_n{\alpha }_n\parallel \hat{x}\parallel \\ =& (1-{\beta }_n{\alpha }_n)\parallel x_n-\hat{x}\parallel +{\beta }_n{\alpha }_n\parallel \hat{x}\parallel \\ \le & max\{\parallel x_n-\hat{x}\parallel ,\parallel \hat{x}\parallel \}.\end{array}$$

One can deduce that

$$\parallel x_n-\hat{x}\parallel \le max\{\parallel x_0-\hat{x}\parallel ,\parallel \hat{x}\parallel \}.$$

Hence, $\{x_n\}$ is bounded and so is $\{T{x}_n\}$. Moreover,

$$\begin{array}{rcl}\parallel P_C[(1-{\alpha }_n)T]x_n-\hat{x}\parallel & \le & \parallel (1-{\alpha }_n)T{x}_n-\hat{x}\parallel \\ =& \parallel (1-{\alpha }_n)[T{x}_n-\hat{x}]-{\alpha }_n\hat{x}\parallel \\ \le & (1-{\alpha }_n)\parallel x_n-\hat{x}\parallel +{\alpha }_n\parallel \hat{x}\parallel \\ \le & max\{\parallel x_n-\hat{x}\parallel ,\parallel \hat{x}\parallel \}.\end{array}$$

Since $\{x_n\}$ is bounded, we see that $\{T{x}_n\}$, $(1-{\alpha }_n)T{x}_n$, and $\{P_C[(1-{\alpha }_n)T]x_n\}$ are also bounded.

Let $z_n=P_C[(1-{\alpha }_n)T]x_n$, and $M>0$ such that $M={sup}_{n\ge 1}\{T{x}_n\}$. Noting that

$$\begin{array}{rcl}\parallel P_C[(1-{\alpha }_{n+1})T]x_n-P_C[(1-{\alpha }_n)T]x_n\parallel & \le & \parallel (1-{\alpha }_{n+1})T{x}_n-(1-{\alpha }_n)T{x}_n\parallel \\ =& \parallel ({\alpha }_n-{\alpha }_{n+1})T{x}_n\parallel \\ \le & M|{\alpha }_n-{\alpha }_{n+1}|.\end{array}$$

One concludes that

$$\begin{array}{rcl}\parallel z_{n+1}-z_n\parallel & =& \parallel P_C[(1-{\alpha }_{n+1})T]x_{n+1}-P_C[(1-{\alpha }_n)T]x_n\parallel \\ \le & \parallel P_C[(1-{\alpha }_{n+1})T]x_{n+1}-P_C[(1-{\alpha }_{n+1})T]x_n\parallel \\ +\parallel P_C[(1-{\alpha }_{n+1})T]x_n-P_C[(1-{\alpha }_n)T]x_n\parallel \\ \le & (1-{\alpha }_{n+1})\parallel x_{n+1}-x_n\parallel +\parallel P_C[(1-{\alpha }_{n+1})T]x_n-P_C[(1-{\alpha }_n)T]x_n\parallel \\ \le & (1-{\alpha }_{n+1})\parallel x_{n+1}-x_n\parallel +M|{\alpha }_n-{\alpha }_{n+1}|.\end{array}$$

Since $0<{\alpha }_n<1$ and ${lim}_{n\to \mathrm{\infty }}|{\alpha }_{n+1}-{\alpha }_n|=0$, we have

$$\parallel z_{n+1}-z_n\parallel -\parallel x_{n+1}-x_n\parallel \le M|{\alpha }_n-{\alpha }_{n+1}|,$$

and

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel z_{n+1}-z_n\parallel -\parallel x_{n+1}-x_n\parallel \le 0.$$

Applying Lemma 1.3, we get

$$\underset{n\to \mathrm{\infty }}{lim}\parallel P_C[(1-{\alpha }_n)T]x_n-x_n\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel z_n-x_n\parallel =0.$$

Hence,

$$\begin{array}{rcl}\parallel x_{n+1}-x_n\parallel & =& \parallel (1-{\beta }_n)x_n+{\beta }_n{P}_C[(1-{\alpha }_n)T]x_n-x_n\parallel \\ =& {\beta }_n\parallel P_C[(1-{\alpha }_n)T]x_n-x_n\parallel \to 0.\end{array}$$

Let $R_n$ and R be defined by

$$\begin{array}{c}{R}_{n}x:=P_C\{(1-{\alpha }_n)[I-\gamma P^{\ast }{G}^{\ast }GP]\}x=P_C[(1-{\alpha }_n)Tx],\hfill \\ Rx:=P_C(I-\gamma P^{\ast }{G}^{\ast }GP)x=P_C(Tx).\hfill \end{array}$$

Noting that

$$\begin{array}{rcl}\parallel x_n-R{x}_n\parallel & \le & \parallel x_n-x_{n+1}\parallel +\parallel x_{n+1}-R{x}_n\parallel \\ =& \parallel x_n-x_{n+1}\parallel +\parallel (1-{\beta }_n)x_n+{\beta }_n{R}_n{x}_n-R{x}_n\parallel \\ \le & \parallel x_n-x_{n+1}\parallel +(1-{\beta }_n)\parallel x_n-R{x}_n\parallel +{\beta }_n\parallel R_n{x}_n-R{x}_n\parallel .\end{array}$$

So, we have

$$\begin{array}{rcl}\parallel x_n-R{x}_n\parallel & \le & \parallel x_n-x_{n+1}\parallel /{\beta }_n+\parallel R_n{x}_n-R{x}_n\parallel \\ =& \parallel x_n-x_{n+1}\parallel /{\beta }_n+\parallel P_C[(1-{\alpha }_n)T]x_n-P_{C}T{x}_n\parallel \\ \le & \parallel x_n-x_{n+1}\parallel /{\beta }_n+\parallel (1-{\alpha }_n)T{x}_n-T{x}_n\parallel \\ \le & \parallel x_n-x_{n+1}\parallel /{\beta }_n+M{\alpha }_n.\end{array}$$

By assumption, we have

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

Furthermore, $\{x_n\}$ is bounded, there exists a subsequence of $\{x_n\}$ which converges weakly to a point $\check{x}$. Without loss of generality, we may assume that $\{x_n\}$ converges weakly to $\check{x}$. Since $\parallel R{x}_n-x_n\parallel \to 0$, using the demiclosedness principle we know that $\check{x}\in Fix(R)=Fix(P_{C}T)=Fix(P_C)\cap Fix(T)=\mathrm{\Gamma }$.

Finally, we will prove that ${lim}_n\parallel x_{n+1}-\check{x}\parallel =0$. In fact,

$$\begin{array}{rcl}{\parallel x_{n+1}-\check{x}\parallel }^2& =& {\parallel (1-{\beta }_n)x_n+{\beta }_n{P}_C[(1-{\alpha }_n)T]x_n-P_{C}T\check{x}\parallel }^2\\ \le & (1-{\beta }_n){\parallel x_n-\check{x}\parallel }^2+{\beta }_n{\parallel P_C[(1-{\alpha }_n)T]x_n-P_{C}T\check{x}\parallel }^2\\ \le & (1-{\beta }_n){\parallel x_n-\check{x}\parallel }^2+{\beta }_n{\parallel (1-{\alpha }_n)T{x}_n-\check{x}\parallel }^2\\ =& (1-{\beta }_n){\parallel x_n-\check{x}\parallel }^2+{\beta }_n{\parallel (1-{\alpha }_n)(T{x}_n-\check{x})+{\alpha }_n\check{x}\parallel }^2\\ =& (1-{\beta }_n){\parallel x_n-\check{x}\parallel }^2+{\beta }_n[{(1-{\alpha }_n)}^2{\parallel T{x}_n-\check{x}\parallel }^2+{\alpha }_n^2{\parallel \check{x}\parallel }^2\\ +2{\alpha }_n(1-{\alpha }_n)〈T{x}_n-\check{x},-\check{x}〉]\\ \le & (1-{\beta }_n){\parallel x_n-\check{x}\parallel }^2+{\beta }_n[(1-{\alpha }_n){\parallel x_n-\check{x}\parallel }^2+{\alpha }_n^2{\parallel \check{x}\parallel }^2\\ +2{\alpha }_n(1-{\alpha }_n)〈T{x}_n-\check{x},-\check{x}〉]\\ =& (1-{\alpha }_n{\beta }_n){\parallel x_n-\check{x}\parallel }^2+{\alpha }_n{\beta }_n[2(1-{\alpha }_n)〈T{x}_n-\check{x},-\check{x}〉+{\alpha }_n{\parallel \check{x}\parallel }^2].\end{array}$$

Using Lemma 1.2, we only need to prove that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈T{x}_n-\check{x},-\check{x}〉\le 0.$$

Applying Lemma 2.5, T is averaged, that is $T=\beta I+(1-\beta )V$, where $0<\beta <1$ and V is nonexpansive. Hence, for $z\in Fix(P_{C}T)$, we have

$$\begin{array}{rcl}{\parallel x_{n+1}-z\parallel }^2& =& {\parallel (1-{\beta }_n)x_n+{\beta }_n{P}_C[(1-{\alpha }_n)T]x_n-z\parallel }^2\\ \le & (1-{\beta }_n){\parallel x_n-z\parallel }^2+{\beta }_n{\parallel (1-{\alpha }_n)T{x}_n-z\parallel }^2\\ =& (1-{\beta }_n){\parallel x_n-z\parallel }^2+{\beta }_n{\parallel (1-{\alpha }_n)(T{x}_n-z)-{\alpha }_{n}z\parallel }^2\\ \le & (1-{\beta }_n){\parallel x_n-z\parallel }^2+{\beta }_n[(1-{\alpha }_n){\parallel T{x}_n-z\parallel }^2+{\alpha }_n{\parallel z\parallel }^2]\\ \le & (1-{\beta }_n){\parallel x_n-z\parallel }^2+{\beta }_n[{\parallel T{x}_n-z\parallel }^2+{\alpha }_n{\parallel z\parallel }^2].\end{array}$$

By Lemma 4.2, we have

$$\begin{array}{rcl}{\parallel x_{n+1}-z\parallel }^2& \le & (1-{\beta }_n){\parallel x_n-z\parallel }^2\\ +{\beta }_n[{\parallel x_n-z\parallel }^2-\beta (1-\beta ){\parallel V{x}_n-x_n\parallel }^2+{\alpha }_n{\parallel z\parallel }^2]\\ \le & {\parallel x_n-z\parallel }^2-{\beta }_n\beta (1-\beta ){\parallel V{x}_n-x_n\parallel }^2+{\beta }_n{\alpha }_n{\parallel z\parallel }^2.\end{array}$$

Let $N>0$ such that $\parallel x_n-z\parallel \le N$ for all n, then it concludes that

$$\begin{array}{rcl}{\beta }_n\beta (1-\beta ){\parallel V{x}_n-x_n\parallel }^2& \le & {\parallel x_n-z\parallel }^2-{\parallel x_{n+1}-z\parallel }^2+{\beta }_n{\alpha }_n{\parallel z\parallel }^2\\ \le & 2N|\parallel x_n-z\parallel -\parallel x_{n+1}-z\parallel |+{\beta }_n{\alpha }_n{\parallel z\parallel }^2\\ \le & 2N\parallel x_n-x_{n+1}\parallel +{\beta }_n{\alpha }_n{\parallel z\parallel }^2.\end{array}$$

Hence,

$$\beta (1-\beta ){\parallel V{x}_n-x_n\parallel }^2\le \frac{2N\parallel x_n-x_{n+1}\parallel }{{\beta }_n}+{\alpha }_n{\parallel z\parallel }^2.$$

Since $\parallel x_n-x_{n+1}\parallel \to 0$, we get

$$\parallel V{x}_n-x_n\parallel \to 0.$$

Therefore,

$$\parallel T{x}_n-x_n\parallel \to 0.$$

It follows that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈T{x}_n-\check{x},-\check{x}〉=\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈x_n-\check{x},-\check{x}〉.$$

Since $\{x_n\}$ converges weakly to $\check{x}$, it follows that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈T{x}_n-\check{x},-\check{x}〉\le 0.$$

 □

Similar to the proof of Theorem 4.3, we can get the result that the following iterative algorithm converges strongly to a solution of the problem $\mathcal{P}$ also. Since the proof is similar to Theorem 4.3, we omit it.

Algorithm 4.4 For an arbitrary initial point $x_0$, sequence $\{x_n\}$ is generated by the iteration:

$$x_{n+1}=(1-{\beta }_n)(1-{\alpha }_n)(I-\gamma P^{\ast }{G}^{\ast }GP)x_n+{\beta }_n{P}_C[(1-{\alpha }_n)(I-\gamma P^{\ast }{G}^{\ast }GP)]x_n,$$

(4.2)

where ${\alpha }_n>0$ is a sequence in $(0,1)$ such that

• (i) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$, ${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (ii) ${lim}_{n\to \mathrm{\infty }}|{\alpha }_{n+1}-{\alpha }_n|=0$;

• (iii) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1$.