Strong Convergence Theorems of Viscosity Iterative Algorithms for Split Common Fixed Point Problems

Abstract

In this paper, we propose a viscosity approximation method to solve the split common fixed point problem and consider the bounded perturbation resilience of the proposed method in general Hilbert spaces. Under some mild conditions, we prove that our algorithms strongly converge to a solution of the split common fixed point problem, which is also the unique solution of the variational inequality problem. Finally, we show the convergence and effectiveness of the algorithms by two numerical examples.

1. Introduction

Let $H_1$ and $H_2$ be two real Hilbert spaces with the inner product $\langle ·,·\rangle$ and the induced norm $\parallel ·\parallel$.

The split feasibility problem (SFP for short) is as follows:

$$\mathrm{Find}\mathrm{a}\mathrm{point}{x}^{*}\in C\mathrm{such}\mathrm{that}A{x}^{*}\in Q,$$

(1)

where C and Q are nonempty closed convex subsets of $H_1$ and $H_2$, respectively, and A is a bounded linear operator of $H_1$ into $H_2$.

If the set of solutions of the problem (1) is nonempty, then $x^{*}$ solving problem (1) is equivalent to

$$x^{*}=P_C(I-\tau A^{*}(I-P_Q)A)x^{*},$$

(2)

where $\tau >0$ and $P_C$ denotes the metric projection of $H_1$ onto C and $A^{*}$ is the corresponding adjoint operator of A.

Recently, many problems in engineering and technology can be modeled by problem (1) and many authors have shown that the SFP has many applications in our real life such as image reconstruction, signal processing and intensity-modulated radiation therapy (see [1,2,3]).

In 1994, Censor and Elfving [4] used their algorithm to solve the SFP in the finite-dimensional Euclidean space. In 2002, Byrne [5] improved the algorithm of Censor and Elfving and presented a new method called the $CQ$ algorithm for solving the SFP (1) as follows:

$$x_{n+1}=P_C(I-\tau A^{*}(I-P_Q)A)x_n,\forall n\in \mathbb{N}.$$

(3)

The split common fixed point problem (shortly, SCFPP) is formulated as follows:

$$\mathrm{Find}\mathrm{a}\mathrm{point}{x}^{*}\in Fix\left(U\right)\mathrm{such}\mathrm{that}A{x}^{*}\in Fix\left(T\right),$$

(4)

where $U:H_1\to H_1$ and $T:H_2\to H_2$ are nonlinear mappings; here, $Fix\left(U\right)$ denotes the set of fixed points of the mapping U. We use S to denote the solution set of problem (4).

Note that, since every closed convex subset of a Hilbert space is the fixed point set of its associating projection if $U:=P_C$ and $T:=P_Q$, the SFP becomes a special case of the SCFPP.

In 2007, Censor and Segal [6] first studied the SCFPP and, to solve the SCFPP, they proposed the following iterative algorithm:

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

(5)

where $\tau$ is a properly chosen stepsize. Algorithm (5) was originally designed to solve the problem (4) for directed mappings.

In 2010, Moudafi [7] proposed an iterative method to solve the SCFPP for quasi-nonexpansive mappings. In 2014, combining the Moudafi method with the Halpern iterative method, Kraikaew and Saejung [8] proposed a new iterative algorithm which does not involve the projection operator to solve the split common fixed point problem. More specifically, their algorithm generates a sequence $\left\{x_n\right\}$ via the recursions:

$$x_{n+1}={\alpha }_n{x}_0+(1-{\alpha }_n)U(x_n-\tau A^{*}(I-T)A{x}_n),\forall n\in \mathbb{N},$$

where $x_0\in H$ is a fixed element, U and T are quasi-nonexpansive operators.

Recently, many authors have studied the SCFPP, the generalized SCFPP and some relative problems (see, for instance, refs. [3,4,5,9,10,11,12,13] and they have also proposed a lot of algorithms to solve the SCFPP (see [14,15,16,17] and the references therein).

On the other hand, the bounded perturbation resilience and superiorization of iterative methods have been studied by some authors (see [18,19,20,21,22,23]). These problems have received much attention because of their applications in convex feasibility problems [24], image reconstruction [25] and inverse problems of radiation therapy [26] and so on.

Let $\mathbf{P}$ denote an algorithm operator. If the iteration $x_{n+1}=\mathbf{P}{x}_n$ is replaced by

$$x_{n+1}=\mathbf{P}(x_n+{\beta }_n{\nu }_n),$$

where ${\beta }_n$ is a sequence of nonnegative real numbers and $\left\{{\nu }_n\right\}$ is a sequence in H such that

$$\sum _{n=0}^{\infty }{\beta }_n<\infty \mathrm{and}\parallel {\nu }_n\parallel \le M.$$

(6)

Then, the algorithm is still convergent and so the algorithm $\mathbf{P}$ is the bounded perturbation resilient [19].

In 2016, Jin, Censor and Jiang [21] introduced the projected scaled gradient method (PSG for short) with bounded perturbations for solving the following minimization problem:

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

(7)

where f is a continuous differentiable, convex function. The method PSG generates a sequence $\left\{x_n\right\}$ defined by

$$x_{n+1}=P_C(x_n-{\gamma }_{n}D\left(x_n\right)\nabla f\left(x_n\right)+e\left(x_n\right)),\forall n\ge 0,$$

(8)

where $D\left(x_n\right)$ is a diagonal scaling matrix and $e\left(x_n\right)$ denotes the sequence of outer perturbations satisfying ${\sum }_{n=0}^{\infty }\parallel e\left(x_n\right)\parallel <\infty$.

Recently, Xu [22] projected the superiorization techniques for the relaxed PSG as follows:

$$x_{n+1}=(1-{\tau }_n)x_n+{\tau }_n{P}_C(x_n-{\gamma }_{n}D\left(x_n\right)\nabla f\left(x_n\right)+e\left(x_n\right)),\forall n\ge 0,$$

(9)

where ${\tau }_n$ is a sequence in $[0,1].$

Recently, for solving minimization problem of the combination of two convex functions ${min}_{x\in H}f\left(x\right)+g\left(x\right)$, Guo and Cui [20] considered the modified proximal gradient method:

$$x_{n+1}={\alpha }_{n}h\left(x_n\right)+(1-{\alpha }_n)pro{x}_{{\lambda }_{n}g}(I-{\lambda }_n\nabla f)\left(x_n\right)+e\left(x_n\right),\forall n\ge 0,$$

(10)

and, under suitable conditions, they proved some strong convergence theorems of the method. The definition of proximal operator $pro{x}_{\lambda \phi }$ is as follows.

Definition 1 

(see [27]). Let ${\mathsf{\Gamma }}_0\left(H\right)$ be the space of functions on a real Hilbert space H that are proper, lower semicontinuous and convex. The proximal operator of $\phi \in {\mathsf{\Gamma }}_0\left(H\right)$ is defined by

$$pro{x}_{\phi }\left(x\right)=arg\underset{\nu \in H}{min}\{\phi \left(\nu \right)+\frac{1}{2}{\parallel \nu -x\parallel }^2\},x\in H.$$

The proximal operator of φ of order $\lambda >0$ is defined as the proximal operator of $\lambda \phi$, that is,

$$pro{x}_{\lambda \phi }\left(x\right)=arg\underset{\nu \in H}{min}\{\phi \left(\nu \right)+\frac{1}{2\lambda }{\parallel \nu -x\parallel }^2\},x\in H.$$

Now, we propose a viscosity method for the problem (4) as follows:

$$x_{n+1}={\alpha }_{n}h(x_n+e\left(x_n\right))+(1-{\alpha }_n)U(x_n-{\tau }_n{A}^{*}(I-T)A{x}_n+e\left(x_n\right)),\forall n\ge 0.$$

(11)

If we treat the above algorithm as the basic algorithm $\mathbf{P}$, the bounded perturbation of it is a sequence $\left\{x_n\right\}$ generated by the iterative process:

$$\left\{\begin{array}{c}{y}_n=x_n+{\beta }_n{\nu }_n,\hfill \\ x_{n+1}={\alpha }_{n}h(y_n+e\left(y_n\right))\hfill \\ +(1-{\alpha }_n)U(y_n-{\tau }_n{A}^{*}(I-T)A{y}_n+e\left(y_n\right)),\forall n\ge 0.\hfill \end{array}\right.$$

(12)

In this paper, mainly based on the above works [6,20,22], we prove that our main iterative method (11) is the bounded perturbation resilient and, under some mild conditions, our algorithms strongly converge to a solution of the split common fixed point problem, which is also the unique solution of the variational inequality problem (13). Finally, we give two numerical examples to demonstrate the effectiveness of our iterative schemes.

2. Preliminaries

Let $\left\{x_n\right\}$ be a sequence in the real Hilbert space H. We adopt the following notations:

(1)

Denote $\left\{x_n\right\}$ converging weakly to x by $x_n\rightharpoonup x$ and $\left\{x_n\right\}$ converging strongly to x by $x_n\to x$.

(2)

Denote the weak $\omega$-limit set of $\left\{x_n\right\}$ by ${\omega }_w\left(x_n\right):=\{x:\exists x_{n_j}\rightharpoonup x\}$.

Definition 2.

A mapping $F:H\to H$ is said to be:

(i) 

Lipschitz if there exists a positive constant L such that

$$\parallel Fx-Fy\parallel \le L\parallel x-y\parallel ,\forall x,y\in H.$$

In particular, if $L=1$, then we say that F is nonexpansive, namely,

$$\parallel Fx-Fy\parallel \le \parallel x-y\parallel ,\forall x,y\in H.$$

If $L\in [0,1)$, then we say F is contractive.

(ii) 

α-averaged mapping (shortly, α-av) if

$$F=(1-\alpha )I+\alpha T,$$

where $\alpha \in [0,1)$ and $T:H\to H$ is nonexpansive.

Definition 3.

A mapping $B:H\to H$ is said to be:

(i) 

monotone if

$$\langle Bx-By,x-y\rangle \ge 0,\forall x,y\in H.$$

(ii) 

η-strongly monotone if there exists a positive constant η such that

$$\langle Bx-By,x-y\rangle \ge {\eta \parallel x-y\parallel }^2,\forall x,y\in H.$$

(iii) 

α-inverse strongly monotone (shortly, α-ism) if there exists a positive constant α such that

$$\langle Bx-By,x-y\rangle \ge {\alpha \parallel Bx-By\parallel }^2,\forall x,y\in H.$$

In particular, if $\alpha =1$, then we say B is firmly nonexpansive, namely,

$$\langle Bx-By,x-y\rangle \ge {\parallel Bx-By\parallel }^2,\forall x,y\in H.$$

Using the Cauchy–Schwartz inequality, it is easy to deduce that B is $\frac{1}{\alpha }-$Lipschitz if it is $\alpha$-ism.

Now, we give the following lemmas and propositions needed in the proof of the main results.

Lemma 1 

([28]). Let H be a real Hilbert space. Then, the following inequality holds:

$${\parallel x+y\parallel }^2\le {\parallel x\parallel }^2+2\langle x+y,y\rangle ,\forall x,y\in H.$$

Lemma 2 

([29]). Let $h:H\to H$ be a ρ-contraction with $\rho \in (0,1)$ and $T:H\to H$ be a nonexpansive mapping. Then,

(i) 

$I-h$ is $(1-\rho )$-strongly monotone, that is,

$$\langle (I-h)x-(I-h)y,x-y\rangle \ge {(1-\rho )\parallel x-y\parallel }^2,\forall x,y\in H.$$

(ii) 

$I-T$ is monotone, that is,

$$\langle (I-T)x-(I-T)y,x-y\rangle \ge 0,\forall x,y\in H.$$

Proposition 1 

([30]).

(i) 

If $T_1,T_2,\cdots ,T_n$ are averaged mappings, then we can get that $T_n{T}_{n-1}\cdots T_1$ is averaged. In particular, if $T_i$ is ${\alpha }_i$-av for each $i=1,2$, where ${\alpha }_i\in (0,1)$, then $T_2{T}_1$ is $({\alpha }_2+{\alpha }_1-{\alpha }_2{\alpha }_1)$-av.

(ii) 

If the mappings ${\left\{T_i\right\}}_{i=1}^N$ are averaged and have a common fixed point, then we have

$$\bigcap _{i=1}^{N}Fix\left(T_i\right)=Fix(T_1\cdots T_N).$$

(iii) 

A mapping T is nonexpansive if and only if $I-T$ is $\frac{1}{2}-$ism.

(iv) 

If T is ν-ism, then, for any $\tau >0$, $\tau T$ is $\frac{\nu }{\tau }$-ism.

(v) 

T is averaged if and only if $I-T$ is ν-ism for some $\nu >\frac{1}{2}$. Indeed, for any $0<\alpha <1$, T is $\alpha -$averaged if and only if $I-T$ is $\frac{1}{2\alpha }$-ism.

Lemma 3 

([31]). Let H be a real Hilbert space and $T:H\to H$ be a nonexpansive mapping with $Fix\left(T\right)\ne \varnothing$. If $\left\{x_n\right\}$ is a sequence in H weakly converging to x and $\left\{(I-T)x_n\right\}$ converges strongly to y, then $(I-T)x=y$. In particular, if $y=0$, then $x\in Fix\left(T\right).$

Lemma 4 

([32] or [33]). Assume that $\left\{s_n\right\}$ is a sequence of nonnegative real numbers such that

$$s_{n+1}\le (1-{\gamma }_n)s_n+{\gamma }_n{\delta }_n,s_{n+1}\le s_n-{\eta }_n+{\phi }_n,\forall n\ge 0,$$

where $\left\{{\gamma }_n\right\}$ is a sequence in $(0,1)$, $\left\{{\eta }_n\right\}$ is a sequence of nonnegative real numbers, $\left\{{\delta }_n\right\}$ and $\left\{{\phi }_n\right\}$ are two sequences in $\mathbb{R}$ such that

(i) 

${\sum }_{n=0}^{\infty }{\gamma }_n=\infty$;

(ii) 

${lim}_{n\to \infty }{\phi }_n=0$;

(iii) 

${lim}_{k\to \infty }{\eta }_{n_k}=0$ implies ${lim\; sup}_{k\to \infty }{\delta }_{n_k}\le 0$ for any subsequence $\left\{n_k\right\}\subset \left\{n\right\}$.

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

Lemma 5.

Assume that $A:H_1\to H_2$ is a bounded linear operator and $A^{*}$ is the corresponding adjoint operator of A. Let $T:H_2\to H_2$ be a nonexpansive mapping. If there exists a point $z\in H_1$ such that $Az\in Fix\left(T\right)$, then

$$(I-T)Ax=0⟺A^{*}(I-T)Ax=0,\forall x\in H_1.$$

Proof. 

It is clear that $(I-T)Ax=0$ implies $A^{*}(I-T)Ax=0$ for all $x\in H_1$.

To see the converse, let $x\in H$ such that $A^{*}(I-T)Ax=0$. Take $Az\in Fix\left(T\right)$. Since T is nonexpansive, we have

$$\begin{array}{c}\hfill {\parallel TAx-Az\parallel }^2={\parallel TAx-TAz\parallel }^2\le {\parallel Ax-Az\parallel }^2\end{array}$$

and

$${\parallel TAx-Az\parallel }^2={\parallel Ax-TAx-(Ax-Az)\parallel }^2={\parallel Ax-TAx\parallel }^2-2\langle Ax-TAx,Ax-Az\rangle +{\parallel Ax-Az\parallel }^2.$$

Combine the above two formulas, we have

$${\parallel Ax-TAx\parallel }^2\le 2\langle Ax-TAx,Ax-Az\rangle =2\langle A^{*}(I-T)Ax,x-z\rangle =0.$$

This completes the proof. □

3. The Main Results

In 2000, Moudafi [34] proposed the viscosity approximation method:

$$x_{n+1}={\alpha }_{n}h\left(x_n\right)+(1-{\alpha }_n)N{x}_n,\forall n\ge 0,$$

which converges strongly to a fixed point $x^{*}$ of the nonexpansive mapping N (see [35,36]). In 2004, Xu [29] further proved that $x^{*}\in Fix\left(N\right)$ is also the unique solution of the following variational inequality problem:

$$\langle (I-h)x^{*},\tilde{x}-x^{*}\rangle \ge 0,\forall \tilde{x}\in Fix\left(N\right),$$

(13)

where $h:H\to H$ is a $\rho$-contraction. By Lemma 2, we get $I-h$ is strongly monotone, hence the solution of problem (13) is unique.

In this section, we present a viscosity iterative method for solving problem (4). Meanwhile, the algorithm approximates the unique fixed point of variational inequality problem (13).

Putting $e_n:=e\left(x_n\right),$ we can rewrite the iteration (11) as follows:

$$\begin{array}{cc}\hfill x_{n+1}& ={\alpha }_{n}h(x_n+e_n)+(1-{\alpha }_n)U(x_n-{\tau }_n{A}^{*}(I-T)A{x}_n+e_n)\hfill \\ & ={\alpha }_{n}h\left(x_n\right)+(1-{\alpha }_n)U(x_n-{\tau }_n{A}^{*}(I-T)A{x}_n)+{\tilde{e}}_n,\forall n\ge 0,\hfill \end{array}$$

(14)

where

$$\begin{array}{cc}\hfill {\tilde{e}}_n& ={\alpha }_n(h(x_n+e_n)-h\left(x_n\right))\hfill \\ & +(1-{\alpha }_n)(U(x_n-{\tau }_n{A}^{*}(I-T)A{x}_n+e_n)-U(x_n-{\tau }_n{A}^{*}(I-T)A{x}_n)).\hfill \end{array}$$

Since U is nonexpansive and h is contractive, it is easy to get

$$\begin{array}{cc}\hfill \parallel {\tilde{e}}_n\parallel & \le {\alpha }_n\parallel h(x_n+e_n)-h\left(x_n\right)\parallel \hfill \\ & +(1-{\alpha }_n)\parallel U(x_n-{\tau }_n{A}^{*}(I-T)A{x}_n+e_n)-U(x_n-{\tau }_n{A}^{*}(I-T)A{x}_n)\parallel \hfill \\ & \le ({\alpha }_n\rho +1-{\alpha }_n)\parallel e_n\parallel \hfill \\ & \le \parallel e_n\parallel .\hfill \end{array}$$

(15)

Theorem 1.

Let $H_1$,$H_2$ be two real Hilbert spaces and $A:H_1\to H_2$ be a bounded linear operator with $L=\parallel A^{*}A\parallel$, where $A^{*}$ is the adjoint of A. Suppose that $U:H_1\to H_1$ and $T:H_2\to H_2$ are two averaged mappings with the coefficients ${\gamma }_1$ and ${\gamma }_2$, respectively. Assume that the problem $\left(4\right)$ is consistent (i.e., $S\ne \varnothing$). Let $h:H_1\to H_1$ be a ρ-contraction with $0\le \rho <1$. For any $x_0\in H_1$, define the sequence $\left\{x_n\right\}$ by $\left(14\right)$. If the following conditions are satisfied:

(i) 

${lim}_{n\to \infty }{\alpha }_n=0$ and ${\sum }_{n=0}^{\infty }{\alpha }_n=\infty$;

(ii) 

$0<{lim\; inf}_{n\to \infty }{\tau }_n\le {lim\; sup}_{n\to \infty }{\tau }_n<\frac{1}{{\gamma }_{2}L}$;

(iii) 

${\sum }_{n=0}^{\infty }\parallel e_n\parallel <\infty$.

Then, the sequence $\left\{x_n\right\}$ converges strongly to a point $x^{*}\in S$, which is also the unique solution of the variational inequality problem $\left(13\right)$.

Proof. 

Set $V_{{\tau }_n}:=U(I-{\tau }_n{A}^{*}(I-T)A)$. Then, by Proposition 1, it follows that $U(I-{\tau }_n{A}^{*}(I-T)A)$ is $({\gamma }_1+(1-{\gamma }_1){\gamma }_2{\tau }_{n}L)$-$av$ as $0<{\tau }_n<\frac{1}{{\gamma }_{2}L}$.

Step 1. Show that $\left\{x_n\right\}$ is bounded. For any $z\in S$, we have

$$\begin{array}{c}\parallel x_{n+1}-z\parallel \hfill \\ =\parallel {\alpha }_{n}h\left(x_n\right)+(1-{\alpha }_n)V_{{\tau }_n}{x}_n+{\tilde{e}}_n-z\parallel \hfill \\ =\parallel {\alpha }_n(h\left(x_n\right)-z)+(1-{\alpha }_n)(V_{{\tau }_n}{x}_n-z)+{\tilde{e}}_n\parallel \hfill \\ \le {\alpha }_n\parallel h\left(x_n\right)-h\left(z\right)\parallel +{\alpha }_n\parallel h\left(z\right)-z\parallel +(1-{\alpha }_n)\parallel V_{{\tau }_n}{x}_n-z\parallel +\parallel {\tilde{e}}_n\parallel \hfill \\ \le {\alpha }_n\rho \parallel x_n-z\parallel +{\alpha }_n\parallel h\left(z\right)-z\parallel +(1-{\alpha }_n)\parallel x_n-z\parallel +\parallel {\tilde{e}}_n\parallel \hfill \\ =(1-{\alpha }_n(1-\rho ))\parallel x_n-z\parallel +{\alpha }_n(1-\rho )\frac{\parallel h\left(z\right)-z\parallel +\parallel {\tilde{e}}_n\parallel /{\alpha }_n}{1-\rho }.\hfill \end{array}$$

(16)

Note that the condition (iii) and (15) imply that ${\sum }_{n=0}^{\infty }\parallel {\tilde{e}}_n\parallel <\infty$ and, from the conditions (i), (iii) and ${\alpha }_n>0$, it is easy to show that $\{\parallel {\tilde{e}}_n\parallel /{\alpha }_n\}$ is bounded. Therefore, there exists $M_1>0$, such that $M_1:={sup}_{n\in \mathbb{N}}\{\parallel h\left(z\right)-z\parallel +\parallel {\tilde{e}}_n\parallel /{\alpha }_n\}$. Thus, since the induction argument shows that

$$\parallel x_n-z\parallel \le max\left\{\parallel x_0-z\parallel ,\frac{M_1}{1-\rho }\right\},$$

it turns out that the sequence $\left\{x_n\right\}$ is bounded and so are $\left\{h\left(x_n\right)\right\}$, $\left\{V_{{\tau }_n}{x}_n\right\}$ and $\left\{A^{*}(I-T)A{x}_n\right\}$.

Step 2. Show that, for any sequence $\left\{n_k\right\}\subset \left\{n\right\},$ if ${\eta }_{n_k}\to 0,$ then ${lim}_{k\to \infty }\parallel x_{n_k}-V_{{\tau }_{n_k}}{x}_{n_k}\parallel =0.$ First, if $z\in S$, then we have

$$\begin{array}{c}\parallel x_{n+1}{-z\parallel }^2\hfill \\ =\parallel {\alpha }_{n}h\left(x_n\right)+(1-{\alpha }_n)V_{{\tau }_n}{x}_n+{\tilde{e}}_n{-z\parallel }^2\hfill \\ =\parallel {\alpha }_{n}h\left(x_n\right)+(1-{\alpha }_n)V_{{\tau }_n}{x}_n{-z\parallel }^2+2\langle {\alpha }_{n}h\left(x_n\right)+(1-{\alpha }_n)V_{{\tau }_n}{x}_n-z,{\tilde{e}}_n\rangle +{\parallel {\tilde{e}}_n\parallel }^2\hfill \\ \le {\alpha }_n^2\parallel h\left(x_n\right){-z\parallel }^2+{(1-{\alpha }_n)}^2{\parallel V_{{\tau }_n}{x}_n-z\parallel }^2+2{\alpha }_n(1-{\alpha }_n)\langle h\left(x_n\right)-z,V_{{\tau }_n}{x}_n-z\rangle \hfill \\ +(2{\alpha }_n\parallel h\left(x_n\right)-z\parallel +2(1-{\alpha }_n)\parallel x_n-z\parallel +\parallel {\tilde{e}}_n\parallel )\parallel {\tilde{e}}_n\parallel \hfill \\ \le 2{\alpha }_n^2(\parallel h\left(x_n\right)-h\left(z\right){\parallel }^2+\parallel h\left(z\right){-z\parallel }^2)+{(1-{\alpha }_n)}^2{\parallel V_{{\tau }_n}{x}_n-z\parallel }^2\hfill \\ +2{\alpha }_n(1-{\alpha }_n)\langle h\left(x_n\right)-z,V_{{\tau }_n}{x}_n-z\rangle +M_2\parallel {\tilde{e}}_n\parallel \hfill \\ \le 2{\alpha }_n^2(\parallel h\left(x_n\right)-h\left(z\right){\parallel }^2+\parallel h\left(z\right){-z\parallel }^2)+{(1-{\alpha }_n)}^2{\parallel V_{{\tau }_n}{x}_n-z\parallel }^2\hfill \\ +2{\alpha }_n(1-{\alpha }_n)(\parallel h\left(x_n\right)-h\left(z\right)\parallel \parallel x_n-z\parallel +\langle h\left(z\right)-z,V_{{\tau }_n}{x}_n-z\rangle )+M_2\parallel {\tilde{e}}_n\parallel \hfill \\ \le (1-{\alpha }_n(2-{\alpha }_n(1+2{\rho }^2)-2(1-{\alpha }_n)\rho )){\parallel x_n-z\parallel }^2\hfill \\ +2{\alpha }_n(1-{\alpha }_n)\langle h\left(z\right)-z,V_{{\tau }_n}{x}_n-z\rangle +2{\alpha }_n^2\parallel h\left(z\right){-z\parallel }^2+M_2\parallel {\tilde{e}}_n\parallel ,\hfill \end{array}$$

(17)

where $M_2:={sup}_{n\in \mathbb{N}}\{2{\alpha }_n\parallel h\left(x_n\right)-z\parallel +2(1-{\alpha }_n)\parallel x_n-z\parallel +\parallel {\tilde{e}}_n\parallel \}.$

Second, we can rewrite $V_{{\tau }_n}$ as

$$\begin{array}{c}\hfill V_{{\tau }_n}=U(I-{\tau }_n{A}^{*}(I-T)A)=(1-w_n)I+w_n{W}_n,\end{array}$$

(18)

where $w_n={\gamma }_1+(1-{\gamma }_1){\gamma }_2{\tau }_{n}L$ and $W_n$ is nonexpansive. By the condition (ii), we get ${\gamma }_1<{lim\; inf}_{n\to \infty }{w}_n\le {lim\; sup}_{n\to \infty }{w}_n<1$. Thus, it follows from (14), (17) and (18) that

$$\begin{array}{c}\parallel x_{n+1}{-z\parallel }^2\hfill \\ =\parallel {\alpha }_{n}h\left(x_n\right)+(1-{\alpha }_n)V_{{\tau }_n}{x}_n+{\tilde{e}}_n{-z\parallel }^2\hfill \\ \le \parallel {\alpha }_{n}h\left(x_n\right)+(1-{\alpha }_n)V_{{\tau }_n}{x}_n{-z\parallel }^2+M_2\parallel {\tilde{e}}_n\parallel \hfill \\ =\parallel V_{{\tau }_n}{x}_n-z+{\alpha }_n(h\left(x_n\right)-V_{{\tau }_n}{x}_n){\parallel }^2+M_2\parallel {\tilde{e}}_n\parallel \hfill \\ =\parallel V_{{\tau }_n}{x}_n{-z\parallel }^2+{{\alpha }_n}^2{\parallel h\left(x_n\right)-V_{{\tau }_n}{x}_n\parallel }^2\hfill \\ +2{\alpha }_n\langle V_{{\tau }_n}{x}_n-z,h\left(x_n\right)-V_{{\tau }_n}{x}_n\rangle +M_2\parallel {\tilde{e}}_n\parallel \hfill \\ =\parallel (1-w_n)x_n+w_n{W}_n{x}_n{-z\parallel }^2+{{\alpha }_n}^2{\parallel h\left(x_n\right)-V_{{\tau }_n}{x}_n\parallel }^2\hfill \\ +2{\alpha }_n\langle V_{{\tau }_n}{x}_n-z,h\left(x_n\right)-V_{{\tau }_n}{x}_n\rangle +M_2\parallel {\tilde{e}}_n\parallel \hfill \\ =(1-w_n)\parallel x_n{-z\parallel }^2+w_n\parallel W_n{x}_n-W_n{z\parallel }^2-w_n(1-w_n){\parallel W_n{x}_n-x_n\parallel }^2\hfill \\ +{{\alpha }_n}^2\parallel h\left(x_n\right)-V_{{\tau }_n}{x}_n{\parallel }^2+2{\alpha }_n\langle V_{{\tau }_n}{x}_n-z,h\left(x_n\right)-V_{{\tau }_n}{x}_n\rangle +M_2\parallel {\tilde{e}}_n\parallel \hfill \\ \le \parallel x_n{-z\parallel }^2-w_n(1-w_n)\parallel W_n{x}_n-x_n{\parallel }^2+{{\alpha }_n}^2{\parallel h\left(x_n\right)-V_{{\tau }_n}{x}_n\parallel }^2\hfill \\ +2{\alpha }_n\langle V_{{\tau }_n}{x}_n-z,h\left(x_n\right)V_{{\tau }_n}{x}_n\rangle +M_2\parallel {\tilde{e}}_n\parallel .\hfill \end{array}$$

(19)

Furthermore, set

$$s_n={\parallel x_n-z\parallel }^2,{\gamma }_n={\alpha }_n(2-{\alpha }_n(1+2{\rho }^2)-2(1-{\alpha }_n)\rho ),$$

$$\begin{array}{l}{\delta }_n=\frac{1}{2-{\alpha }_n(1+2{\rho }^2)-2(1-{\alpha }_n)\rho }[2{\alpha }_n{\parallel h\left(z\right)-z\parallel }^2\\ +M_2\frac{\parallel {\tilde{e}}_n\parallel }{{\alpha }_n}+2(1-{\alpha }_n)\langle h\left(z\right)-z,V_{{\tau }_n}{x}_n-z\rangle ],\end{array}$$

$${\eta }_n=w_n(1-w_n){\parallel W_n{x}_n-x_n\parallel }^2,$$

$${\phi }_n={{\alpha }_n}^2\parallel h\left(x_n\right)-V_{{\tau }_n}{x}_n{\parallel }^2+2{\alpha }_n\langle V_{{\tau }_n}{x}_n-z,h\left(x_n\right)-V_{{\tau }_n}{x}_n\rangle +M_2\parallel {\tilde{e}}_n\parallel .$$

Using the condition (i), it is easy to get ${\gamma }_n\to 0,{\Sigma }_{n=0}^{\infty }{\gamma }_n=\infty$ and ${\phi }_n\to 0$. In order to complete the proof, from Lemma 4, it suffices to verify that ${\eta }_{n_k}\to 0$ as $k\to \infty$, which implies that

$$\underset{k\to \infty }{lim\; sup}{\delta }_{n_k}\le 0$$

for any subsequence $\left\{n_k\right\}\subset \left\{n\right\}$. Indeed, ${\eta }_{n_k}\to 0$ as $k\to \infty$ implies that $\parallel W_{n_k}{x}_{n_k}-x_{n_k}\parallel \to 0$ as $k\to \infty$ from the condition (iii). Thus, from (18), it follows that

$$\begin{array}{c}\hfill \parallel x_{n_k}-V_{{\tau }_{n_k}}{x}_{n_k}\parallel =w_{n_k}\parallel x_{n_k}-W_{n_k}{x}_{n_k}\parallel \to 0.\end{array}$$

(20)

Step 3. Show that

$$\begin{array}{c}\hfill {\omega }_w\left(x_{n_k}\right)\subset S,\end{array}$$

(21)

where ${\omega }_w\left(x_{n_k}\right)$ is the set of all weak cluster points of $\left\{x_{n_k}\right\}$. To see (21), we prove the following:

Take $\tilde{x}\in {\omega }_w\left\{x_{n_k}\right\}$ and assume that $\left\{x_{n_{k_j}}\right\}$ is a subsequence of $\left\{x_{n_k}\right\}$ weakly converging to $\tilde{x}$. Without loss of generality, we still use $\left\{x_{n_k}\right\}$ to denote $\left\{x_{n_{k_j}}\right\}$. Assume ${\tau }_{n_k}\to \tau$. Then, we have $0<\tau <\frac{1}{{\gamma }_{2}L}$. Setting $V=U(I-\tau A^{*}(I-T)A)$, we deduce that

$$\begin{array}{c}\parallel V_{{\tau }_{n_k}}{x}_{n_k}-V{x}_{n_k}\parallel \hfill \\ =\parallel U(x_{n_k}-{\tau }_{n_k}{A}^{*}(I-T)A{x}_{n_k})-U(x_{n_k}-\tau A^{*}(I-T)A{x}_{n_k})\parallel \hfill \\ \le |{\tau }_{n_k}-\tau |\parallel A^{*}(I-T)A{x}_{n_k}\parallel .\hfill \end{array}$$

(22)

Since ${\tau }_{n_k}\to \tau$ as $k\to \infty$, it follows immediately from (22) that

$$\parallel V_{{\tau }_{n_k}}{x}_{n_k}-V{x}_{n_k}\parallel \to 0$$

as $k\to \infty$. Thus, we have

$$\parallel x_{n_k}-V{x}_{n_k}\parallel \le \parallel x_{n_k}-V_{{\tau }_{n_k}}{x}_{n_k}\parallel +\parallel V_{{\tau }_{n_k}}{x}_{n_k}-V{x}_{n_k}\parallel \to 0.$$

(23)

Using Lemma 3, we get ${\omega }_w\left(x_{n_k}\right)\subset Fix\left(V\right)$. Since both U and T are averaged, it follows from Proposition 1 (ii) that

$${\omega }_w\left(x_{n_k}\right)\subset Fix\left(U\right),{\omega }_w\left(x_{n_k}\right)\subset Fix(I-\tau A^{*}(I-T)A).$$

Then, by Lemma 5, we obtain ${\omega }_w\left(x_{n_k}\right)\subset S$ immediately. Meanwhile, we have

$$\begin{array}{c}\hfill \underset{k\to \infty }{lim\; sup}\langle h\left(x^{*}\right)-x^{*},x_{n_k}-x^{*}\rangle =\langle h\left(x^{*}\right)-x^{*},\tilde{x}-x^{*}\rangle ,\forall \tilde{x}\in S.\end{array}$$

(24)

In addition, since $x^{*}$ is the unique solution of the variational inequality problem (13), we have

$$\underset{k\to \infty }{lim\; sup}\langle h\left(x^{*}\right)-x^{*},x_{n_k}-x^{*}\rangle \le 0$$

together with (20) and hence ${lim\; sup}_{k\to \infty }{\delta }_{n_k}\le 0$. This completes the proof. □

Next, we consider the bounded perturbation of (14) generated by the following iterative process:

$$\left\{\begin{array}{c}{y}_n=x_n+{\beta }_n{\nu }_n,\hfill \\ x_{n+1}={\alpha }_{n}h(y_n+e\left(y_n\right))+(1-{\alpha }_n)U(I-{\tau }_n{A}^{*}(I-T)A{y}_n+e\left(y_n\right)).\hfill \end{array}\right.$$

(25)

Theorem 2.

Assume that the sequences $\left\{{\beta }_n\right\}$ and $\left\{{\nu }_n\right\}$ satisfy the condition $\left(6\right)$. Let $H_1$,$H_2$ be two real Hilbert spaces and $A:H_1\to H_2$ be a bounded linear operator with $L=\parallel A^{*}A\parallel$, where $A^{*}$ is the adjoint of A. Suppose that $U:H_1\to H_1$ and $T:H_2\to H_2$ are two averaged mappings with the coefficients ${\gamma }_1$ and ${\gamma }_2$, respectively. Assume that problem $\left(4\right)$ is consistent (i.e., $S\ne \varnothing$). Let $h:H_1\to H_1$ be a ρ-contraction with $0\le \rho <1$. For any $x_0\in H_1$, define the sequence $\left\{x_n\right\}$ by $\left(25\right)$. If the following conditions are satisfied:

(i) 

${lim}_{n\to \infty }{\alpha }_n=0$ and ${\sum }_{n=0}^{\infty }{\alpha }_n=\infty$;

(ii) 

$0<{lim\; inf}_{n\to \infty }{\tau }_n\le {lim\; sup}_{n\to \infty }{\tau }_n<\frac{1}{{\gamma }_{2}L}$;

(iii) 

${\sum }_{n=0}^{\infty }\parallel e\left(y_n\right)\parallel <\infty$.

Then, the sequence $\left\{x_n\right\}$ converges strongly to $x^{*}$, where $x^{*}$ is a solution of the problem $\left(4\right)$, which is also the unique solution of the variational inequality problem $\left(13\right)$.

Proof. 

Now, put

$$\begin{array}{cc}\hfill {\tilde{e}}_n=& {\alpha }_n(h(y_n+e\left(y_n\right))-h\left(x_n\right))\hfill \\ & +(1-{\alpha }_n)(U(y_n-{\tau }_n{A}^{*}(I-T)A{y}_n+e\left(y_n\right))-U(x_n-{\tau }_n{A}^{*}(I-T)A{x}_n)).\hfill \end{array}$$

Then, Equation (25) can be rewritten as follows:

$$\begin{array}{c}\hfill x_{n+1}={\alpha }_{n}h\left(x_n\right)+(1-{\alpha }_n)U(I-{\tau }_n{A}^{*}(I-T)A)\left(x_n\right)+{\tilde{e}}_n,\end{array}$$

(26)

In fact, by Proposition 1 (iii) and the nonexpansiveness of T, it is not hard to show that $A^{*}(I-T)A$ is $2L-$Lipschitz. Thus, we have

$$\begin{array}{cc}\hfill \parallel {\tilde{e}}_n\parallel & \le {\alpha }_n\parallel h(y_n+e\left(y_n\right))-h\left(x_n\right)\parallel \hfill \\ & +(1-{\alpha }_n)\parallel y_n-x_n-{\tau }_n(A^{*}(I-T)A{y}_n+e\left(y_n\right)-A^{*}(I-T)A{x}_n)\parallel \hfill \\ & \le {\alpha }_n\rho \parallel y_n-x_n+e\left(y_n\right)\parallel \hfill \\ & +(1-{\alpha }_n)(\parallel y_n-x_n\parallel +2{\tau }_{n}L\parallel y_n-x_n\parallel +\parallel e\left(y_n\right)\parallel )\hfill \\ & \le ({\alpha }_n\rho +(1-{\alpha }_n)(1+2{\tau }_{n}L)){\beta }_n\parallel {\nu }_n\parallel +({\alpha }_n\rho +(1-{\alpha }_n))\parallel e\left(y_n\right)\parallel .\hfill \end{array}$$

(27)

From the condition (iii) and condition (6), we have ${\sum }_{n=0}^{\infty }\parallel {\tilde{e}}_n\parallel <\infty$. Consequently, using Theorem 1, it follows that the algorithm (14) is bounded perturbation resilient. This completes the proof. □

4. Numerical Results

In this section, we consider the following numerical examples to present the effectiveness, realization and convergence of Theorems 1 and 2:

Example 1.

Let $H_1=H_2={\mathbb{R}}^2$. Suppose $h\left(x\right)=\frac{1}{10}x$ and

$$A=\left(\begin{array}{cc}2& 0\\ 0& 3\end{array}\right).$$

Take $U=P_C$ and $T=P_Q$, where C and Q are defined as follows:

$$C=\{x\in {\mathbb{R}}^2:\parallel x{\parallel }_2\le 4\}$$

and

$$Q=\{y\in {\mathbb{R}}^2:6\le y\left(i\right)\le 12,i=1,2\},$$

where $y\left(i\right)$ denotes the $ith$ element of y.

We can compute the solution set $S=\{x:3\le x\left(1\right)\le 6,2\le x\left(2\right)\le 4,\parallel x{\parallel }_2\le 4\}.$

Take the experiment parameters ${\tau }_n=\frac{1}{L}$ and ${\alpha }_n=\frac{1}{n+1}$ in the following iterative algorithms and the stopping criteria is $\parallel x_{n+1}-x_n\parallel <error.$ According to the iterative process of Theorem 1, the sequence $\left\{x_n\right\}$ is generated by

$$\begin{array}{c}\hfill x_{n+1}=\frac{1}{n+1}\ast \frac{1}{10}{x}_n+(1-\frac{1}{n+1})U(x_n-{\tau }_n{A}^T(I-T)A{x}_n).\end{array}$$

(28)

As $n\to \infty$, we have $x_n\to x^{*}$. Then, taking the random initial guess $x_0$ and using MATLAB software (MATLAB R2012a, MathWorks, Natick, MA, USA), we obtain the numerical experiment results in

Table 1. $x_0=rand(2,1)$, results without bounded perturbation.

$\mathit{\tau }$	n	Time (s)	${\mathit{x}}_{\mathit{n}}$	Error
$0.1111$	52	$0.02657$	${\left[2.99951.9998\right]}^T$	${10}^{-5}$
$0.1111$	135	$0.07538$	${\left[2.99982.0000\right]}^T$	${10}^{-6}$
$0.1111$	216	$0.15873$	${\left[3.00002.0000\right]}^T$	${10}^{-7}$

.

Next, we consider the algorithm with bounded perturbation resilience. Choose the the bounded sequence $\left\{{\nu }_n\right\}$ and the summarable nonnegative real sequence $\left\{{\beta }_n\right\}$ as follows:

$$v_n=\left\{\begin{array}{cc}\hfill -\frac{d_n}{\parallel d_n\parallel },& \mathrm{if}0\ne d_n\in \partial I_C\left(x_n\right),\hfill \\ \hfill 0,& \mathrm{if}0=d_n\in \partial I_C\left(x_n\right),\hfill \end{array}\right.$$

(29)

and

$${\beta }_n=c^n$$

for some $c\in (0,1)$, where the indicator function

$$I_C\left(x\right)=\left\{\begin{array}{c}0,\mathrm{if}x\in C,\hfill \\ \infty ,\mathrm{if}x\notin C,\hfill \end{array}\right.$$

and

$$\partial I_C\left(x\right)=N_C\left(x\right)=\left\{\begin{array}{cc}\{u\in H:\langle u,x-y\rangle \ge 0,\forall y\in C\},\hfill & \hfill \mathrm{if}x\in C,\\ \varnothing ,\hfill & \hfill \mathrm{if}x\notin C,\end{array}\right.$$

is the normal cone to C. The point $d_n$ is taken from $N_C\left(x_n\right)$. Setting $c=0.5,$ the numerical results can be seen in

Table 2. $x_0=rand(2,1)$, results with bounded perturbation.

$\mathit{\tau }$	n	Time (s)	${\mathit{x}}_{\mathit{n}}$	Error
$0.1111$	45	$0.02342$	${\left[2.99921.9998\right]}^T$	${10}^{-5}$
$0.1111$	98	$0.05317$	${\left[299982.0000\right]}^T$	${10}^{-6}$
$0.1111$	153	$0.10256$	${\left[3.00002.0000\right]}^T$	${10}^{-7}$

.

As we have seen above, the accuracy of the solution is improved with the decrease of the stop criteria. In addition, the sequence $\left\{x_n\right\}$ converges to the point $(3,2)$, which is a solution of the numerical example. Of course, it is also the unique solution of the variational inequality $\langle (I-h)x^{*},x-x^{*}\rangle \ge 0.$

In addition, we contrast the approximate value of solution $x^{*}$ of Example 1 under the same parameter conditions, the same iterative numbers and the same initial value. The numerical results are reported in

Table 3. $x_0=2\ast rand(2,1)$, results without bounded perturbation.

$\mathit{\tau }$	n	${\mathit{x}}_{\mathit{n}}^{\left(1\right)}$	${\mathit{x}}_{\mathit{n}}^{\left(2\right)}$	Error
$0.2179$	52	${\left[2.98841.9993\right]}^T$	${\left[2.98501.9940\right]}^T$	${10}^{-5}$
$0.2179$	132	${\left[3.00001.9995\right]}^T$	${\left[2.99421.9988\right]}^T$	${10}^{-6}$
$0.2179$	208	${\left[3.00001.9999\right]}^T$	${\left[2.99951.9996\right]}^T$	${10}^{-7}$

and

Table 4. $x_0=2\ast rand(2,1)$, results with bounded perturbation.

$\mathit{\tau }$	n	${\mathit{x}}_{\mathit{n}}^{\left(1\right)}$	${\mathit{x}}_{\mathit{n}}^{\left(2\right)}$	Error
$0.2179$	32	${\left[2.99901.9997\right]}^T$	${\left[2.99201.9987\right]}^T$	${10}^{-5}$
$0.2179$	56	${\left[2.99971.9999\right]}^T$	${\left[2.99421.9988\right]}^T$	${10}^{-6}$
$0.2179$	115	${\left[3.00002.0000\right]}^T$	${\left[2.99951.9998\right]}^T$	${10}^{-7}$

, where $\left\{x_n^{\left(1\right)}\right\}$ and $\left\{x_n^{\left(2\right)}\right\}$ denote the iterative sequences generated by the algorithm (14) in this paper and Theorem 3.2 in Ref. [8], respectively.

Example 2.

Let $H_1=H_2={\mathbb{R}}^3$. Suppose $h\left(x\right)=\frac{1}{3}x$ and

$$A=\left(\begin{array}{ccc}1& 0& -8\\ 0& 2& 0\\ 0& 0& 5\end{array}\right).$$

Define $T:{\mathbb{R}}^3\to {\mathbb{R}}^3$ by

$$T:y={(y\left(1\right),y\left(2\right),y\left(3\right))}^T\mapsto {(y\left(1\right),y\left(2\right),\frac{y\left(3\right)+siny\left(3\right)}{2})}^T.$$

(30)

It is obvious that T is $\frac{1}{2}-av$ and the set of fixed points $Fix\left(T\right)=\left\{y\right|{(y\left(1\right),y\left(2\right),0)}^T\}$ is nonempty. Let $U=P_C$ and $C=\{x\in {\mathbb{R}}^3|\parallel x\parallel \le 1\}.$ Then, we use the iterative algorithm of Theorem 1 to approximate a point $x^{*}\in C$ such that $A{x}^{*}\in Fix\left(T\right)$.

Take the experiment parameters ${\tau }_n=\frac{1.9\ast n}{(n+1)L}$ and ${\alpha }_n=\frac{1}{n+1}$ in the following iterative algorithms. Let $F\left(x\right)=\frac{1}{2}{\parallel (I-T)Ax\parallel }^2+I_C\left(x\right)$ and the stopping criteria is $F\left(x\right)<error.$

Then, taking the random initial guess $x_0$ and using MATLAB software, we obtain the numerical experiment results in

Table 5. $x_0=10\ast rand(3,1)$, results without bounded perturbation.

$\mathit{\tau }$	n	Time (s)	${\mathit{x}}_{\mathit{n}}$	Error
$0.0255$	120	$0.0155$	${\left[0.00850.01670.0037\right]}^T\ast {10}^{-2}$	${10}^{-5}$
$0.0256$	235	$0.0461$	${\left[0.00780.00130.0013\right]}^T\ast {10}^{-3}$	${10}^{-6}$
$0.0257$	518	$0.1881$	${\left[0.00290.00150.0002\right]}^T\ast {10}^{-4}$	${10}^{-7}$

.

Next, we consider the bounded perturbation. The definitions of $v_n$ and ${\beta }_n$ are similar to the Example 1. Setting $c=0.8,$ the numerical results can be seen in Table 4.

As we have seen in Table 5 and

Table 6. $x_0=10\ast rand(3,1)$, results with bounded perturbation.

$\mathit{\tau }$	n	Time (s)	${\mathit{x}}_{\mathit{n}}$	Error
$0.0246$	22	$0.0036$	${\left[0.01080.01390.0015\right]}^T\ast {10}^{-2}$	${10}^{-5}$
$0.0249$	32	$0.0040$	${\left[0.00850.00140.0014\right]}^T\ast {10}^{-3}$	${10}^{-6}$
$0.0250$	36	$0.0058$	${\left[0.00280.00250.0034\right]}^T\ast {10}^{-4}$	${10}^{-7}$

, the sequence $\left\{x_n\right\}$ approximates to the point ${(0,0,0)}^T$, which is a solution of the numerical example. Of course, it is also the unique solution of the variational inequality $\langle (I-h)x^{*},x-x^{*}\rangle \ge 0.$

5. Conclusions

The SCFPP is an inverse problem that consists in finding a point in a fixed point set such that its image under a bounded linear operator belongs to another fixed point set. Many iterative algorithms have been developed to solve these kinds of problems. In this paper, we have introduced a viscosity iterative sequence and obtained the strong convergence. We prove the main result using the weaker conditions than many existing similar methods—for example, Xu’s algorithm [37] for the SFP. More specifically, his algorithm generates a sequence $\left\{x_n\right\}$ via the following recursions:

$$x_{n+1}={\alpha }_{n}u+(1-{\alpha }_n)P_C(x_n-{\tau }_n{A}^{*}(I-P_Q)A{x}_n),$$

where u is a a fixed element and $\left\{{\alpha }_n\right\}\subset [0,1]$ satisfies the assumptions:

(i)

${lim}_{n\to \infty }{\alpha }_n=0$ and ${\sum }_{n=0}^{\infty }{\alpha }_n=\infty$,

(ii)

either ${\sum }_{n=0}^{\infty }\parallel {\alpha }_{n+1}-{\alpha }_n\parallel <\infty$ or ${lim}_{n\to \infty }({\alpha }_{n+1}/{\alpha }_n)=1.$

The second condition is not necessary in our theorems. We also consider the bounded perturbation resilience of the proposed method and get theoretical convergence results. Finally, numerical experiments have been presented to illustrate the effectiveness of the proposed algorithms.