An Iterative Approach to the Solutions of Proximal Split Feasibility Problems

Abstract

The proximal split feasibility problem is investigated in Hilbert spaces. An iterative procedure is introduced for finding the solution of the proximal split feasibility problem. Strong convergence analysis of the presented algorithm is proved.

1. Introduction

Recall that the split feasibility problem (SFP) seeks a point $u^{†}$ such that

$$\begin{array}{c}\hfill u^{†}\in \mathcal{C}and\mathcal{A}{u}^{†}\in \mathcal{Q},\end{array}$$

(1)

where $\varnothing \ne \mathcal{C}$ and $\varnothing \ne \mathcal{Q}$ are two closed convex subsets of two real Hilbert spaces ${\mathcal{H}}_1$ and ${\mathcal{H}}_2$, respectively and $\mathcal{A}:{\mathcal{H}}_1\to {\mathcal{H}}_2$ is a bounded linear operator.

In 1994, Censor and Elfving [1] refined the above mathematical model from the medical image reconstruction and phase retrievals. This provides us a useful tool to research inverse problems arising in science and engineering. One effective method for solving SFP (1) is algorithmic iteration. In the literature, there are several effective iterative algorithms presented by some authors (see, for instance [2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28].)

In this paper, our goal is to focus a general case of proximal split feasibility problems and to investigate the convergence analysis. To begin with, we first give several related concepts.

Let $\varphi :{\mathcal{H}}_2\to \mathcal{R}\cup \{+\infty \}$ be a lower semi-continuous, proper and convex function. Let $ϵ>0$ be a constant. Recall that the Moreau [29]-Yosida [30] regularization is defined by

$$\begin{array}{c}\hfill {\varphi }_{ϵ}\left(x\right)=\underset{u\in {\mathcal{H}}_2}{min}\left\{\varphi \left(u\right)+\frac{1}{2ϵ}{\parallel u-x\parallel }^2\right\}.\end{array}$$

Consequently, we can define the proximity operator of $\varphi$ by the form

$$\begin{array}{c}\hfill pro{x}_{ϵ\varphi }\left(x\right)=arg\underset{u\in {\mathcal{H}}_2}{min}\left\{\varphi \left(u\right)+\frac{1}{2ϵ}{\parallel u-x\parallel }^2\right\}.\end{array}$$

The subdifferential of $\varphi$ at $x^{†}$ denoted by $\partial \varphi \left(x^{†}\right)$ is defined as follows

$$\partial \varphi \left(x^{†}\right)=\{x^{*}\in {\mathcal{H}}_2:\varphi \left(x^{†}\right)+\langle x^{*},x^{‡}-x^{†}\rangle \le \varphi \left(x^{‡}\right),\forall x^{‡}\in {\mathcal{H}}_2\}.$$

It is easy to validate that $0\in \partial \varphi \left(x^{†}\right)\iff x^{†}=pro{x}_{ϵ\varphi }\left(x^{†}\right)$. This means that the minimizer of $\varphi$ is the fixed point of its proximity operator.

Let $\phi :{\mathcal{H}}_1\to \mathcal{R}\cup \{+\infty \}$ be a lower semi-continuous, proper and convex function. Recall that the proximal split feasibility problem seeks a point $x^{†}\in {\mathcal{H}}_1$ such that $x^{†}$ solves the following minimization problem

$$\begin{array}{c}\hfill \underset{x^{†}\in {\mathcal{H}}_1}{min}\{\phi \left(x^{†}\right)+{\varphi }_{ϵ}\left(\mathcal{A}{x}^{†}\right)\}.\end{array}$$

(2)

In what follows, we use $\mathsf{\Gamma }$ to denote the solution set of the problem (2).

The above problem (2) has been studied extensively in the literature, see for instance [31,32,33,34,35]. In order to solve problem (2), in [36], Moudafi and Thakur presented the following split proximal algorithm.

• Fixed an initialization $u_0\in {\mathcal{H}}_1$.
• Assume that $u_n\in {\mathcal{H}}_1$ has been obtained. Calculate $\nu \left(u_n\right)=\sqrt{\parallel \nabla g\left(u_n\right){\parallel }^2+{\parallel \nabla h\left(u_n\right)\parallel }^2},$ where ${\displaystyle g\left(u_n\right)=\frac{1}{2}{\parallel (I-pro{x}_{ϵ\varphi })\mathcal{A}{u}_n\parallel }^2}$ and ${\displaystyle h\left(u_n\right)=\frac{1}{2}{\parallel (I-pro{x}_{{\mu }_nϵ\phi })u_n\parallel }^2.}$
• If $\nu \left(u_n\right)=0,$ then the iterative procedure stops, otherwise continue to compute the next iterate

  $$\begin{array}{c}\hfill u_{n+1}=pro{x}_{{\mu }_nϵ\phi }(u_n-{\mu }_n{\mathcal{A}}^{*}(I-pro{x}_{ϵ\varphi })\mathcal{A}{u}_n),n\ge 0,\end{array}$$

  where ${\displaystyle {\mu }_n={\tau }_n\frac{g\left(u_n\right)+h\left(u_n\right)}{{\nu }^2\left(u_n\right)}}$.

Remark 1.

Note that the stepsize sequence $\left\{{\mu }_n\right\}$ is implicit because of the terms $g\left(u_n\right)$ and $\nu \left(u_n\right)$. This indicates that the computation of $u_{n+1}$ is complicated.

To overcome this difficulty, Shehu and Iyiola [37] suggested the following explicit algorithm to solve problem (2).

• Fixed $u\in {\mathcal{H}}_1$ and $u_1\in {\mathcal{H}}_1$.
• Set $n=1$ and calculate

  $$\begin{array}{c}\hfill \left\{\begin{array}{c}{y}_n={\zeta }_{n}u+(1-{\zeta }_n)u_n,\hfill \\ \nu \left(y_n\right)=\parallel {\mathcal{A}}^{*}(I-pro{x}_{ϵ\varphi })\mathcal{A}{y}_n+(I-pro{x}_{ϵ\phi })y_n\parallel ,\hfill \\ {\displaystyle z_n=y_n-{\tau }_n\frac{h\left(y_n\right)+l\left(y_n\right)}{{\nu }^2\left(y_n\right)}({\mathcal{A}}^{*}(I-pro{x}_{ϵ\varphi })\mathcal{A}{y}_n+(I-pro{x}_{ϵ\phi })y_n),}\hfill \\ u_{n+1}=(1-{\vartheta }_n)y_n+{\vartheta }_n{z}_n.\hfill \end{array}\right.\end{array}$$
• If ${\mathcal{A}}^{*}(I-pro{x}_{ϵ\varphi })\mathcal{A}{y}_n=0=(I-pro{x}_{ϵ\phi })y_n$ and $u_{n+1}=u_n$, then the iterative process stops, otherwise continue to the next step.
• Set $n\leftarrow n+1$ and repeat steps 2-3.

Remark 2.

In Step 3, we note that ${\mathcal{A}}^{*}(I-pro{x}_{ϵ\varphi })\mathcal{A}{y}_n=0=(I-pro{x}_{ϵ\phi })y_n$ implies $\nu \left(y_n\right)=0$. In this case, the iterates $z_n$ and $u_{n+1}$ have no meanings.

In the present paper, our goal is to mend the above gap and to suggest a modified proximal split feasibility algorithm for solving the proximal split feasibility problem (2). We prove that the presented sequence converges strongly to a solution of the proximal split feasibility problem (2).

2. Preliminaries

Let $\mathcal{H}$ be a real Hilbert space. Use $\langle ·,·\rangle$ and $\parallel ·\parallel$ to denote its inner product and norm, respectively. Let $\mathcal{C}$ be a nonempty closed convex subset of $\mathcal{H}$. Recall that a mapping $S:\mathcal{C}\to \mathcal{C}$ is said to be firmly nonexpansive [38] if

$$\begin{array}{c}\hfill {\parallel Su-Sv\parallel }^2\le \langle Su-Sv,u-v\rangle \end{array}$$

for all $u,v\in \mathcal{C}$.

Note that the proximal operators $I-pro{x}_{ϵ\phi }$ and $I-pro{x}_{ϵ\varphi }$ are firmly nonexpansive, namely,

$$\begin{array}{c}\hfill \parallel (I-pro{x}_{ϵ\phi })u-(I-pro{x}_{ϵ\phi }){v\parallel }^2\le \langle (I-pro{x}_{ϵ\phi })u-(I-pro{x}_{ϵ\phi })v,u-v\rangle \end{array}$$

(3)

for all $u,v\in {\mathcal{H}}_1$ and

$$\begin{array}{c}\hfill \parallel (I-pro{x}_{ϵ\varphi })u-(I-pro{x}_{ϵ\varphi }){v\parallel }^2\le \langle (I-pro{x}_{ϵ\varphi })u-(I-pro{x}_{ϵ\varphi })v,u-v\rangle \end{array}$$

(4)

for all $u,v\in {\mathcal{H}}_2$.

For $\forall u\in \mathcal{H}$, there exists a unique point in $\mathcal{C}$, denoted by $pro{j}_{\mathcal{C}}\left(u\right)$, such that

$$\parallel u-pro{j}_{\mathcal{C}}\left(u\right)\parallel \le \parallel u-u^{†}\parallel$$

for all $u^{†}\in \mathcal{C}$.

It is known that $pro{j}_{\mathcal{C}}$ is firmly-nonexpansive and has the following characteristic [39]

$$\langle u-pro{j}_{\mathcal{C}}\left(u\right),u^{†}-pro{j}_{\mathcal{C}}\left(u\right)\rangle \le 0$$

for all $u\in \mathcal{H}$ and $u^{†}\in \mathcal{C}$.

An operator F is called strongly positive if there exists a constant $\delta >0$ such that $\langle Fu,u\rangle \ge {\delta \parallel u\parallel }^2$ for all $u\in \mathcal{H}$.

The following expressions will be used in the sequel.

• $u_n\rightharpoonup u$ denotes the weak convergence of $\left\{u_n\right\}$ to u;
• $u_n\to u$ denotes the strong convergence of $\left\{u_n\right\}$ to u;
• $Fix\left(S\right)$ means the set of fixed points of S.

Lemma 1.

[40] In a real Hilbert space $\mathcal{H}$, the following identity holds

$$\begin{array}{c}\hfill \parallel \lambda x^{†}+(1-\lambda )\tilde{x}{\parallel }^2=\lambda \parallel x^{†}{\parallel }^2+(1-\lambda )\parallel \tilde{x}{\parallel }^2-\lambda (1-\lambda ){\parallel x^{†}-\tilde{x}\parallel }^2,\end{array}$$

for all $\lambda \in [0,1],\forall x^{†},\tilde{x}\in \mathcal{H}$.

Lemma 2.

[41] Suppose that three sequences $\left\{u_n\right\}$, $\left\{v_n\right\}$ and $\left\{{\theta }_n\right\}$ satisfy the following conditions

(i) 

$u_n\ge 0$ for all $n\ge 0$;

(ii) 

there exists a constant M such that $v_n\le M$ for all $n\ge 0$;

(iii) 

${\theta }_n\in [0,1]$ and ${\sum }_{n=0}^{\infty }{\theta }_n=\infty$;

(iv) 

$u_{n+1}\le (1-{\theta }_n)u_n+{\theta }_n{v}_n$ for all $n\ge 0$.

Then, we have ${\lim \sup }_{n\to \infty }{u}_n\le {\lim \sup }_{n\to \infty }{v}_n$.

Lemma 3.

[42] Suppose that $\mathcal{H}$ is a real Hilbert space and $\mathcal{C}\subset \mathcal{H}$ is a nonempty closed convex set. If T is a nonexpansive self-mapping of $\mathcal{C}$, then the operator $I-T$ is demi-closed at 0, i.e., $x_n\rightharpoonup x\in \mathcal{C}$ and $x_n-T{x}_n\to 0$ imply $x=Tx$.

Lemma 4.

[43] Assume that three sequences $\left\{{\rho }_n\right\}$, $\left\{{\eta }_n\right\}$ and $\left\{{\zeta }_n\right\}$ satisfy the following assumptions

(i) 

${\rho }_n\ge 0$ for all $n\ge 0$;

(ii) 

${\left\{{\zeta }_n\right\}}_{n\in \mathbb{N}}\subset [0,1]$ and ${\sum }_{n=1}^{\infty }{\zeta }_n=\infty ;$

(iii) 

${\lim \sup }_{n\to \infty }{\eta }_n\le 0$;

(iv) 

${\rho }_{n+1}\le (1-{\zeta }_n){\rho }_n+{\zeta }_n{\eta }_n$ for all $n\ge 0$.

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

3. Main Results

Suppose that

(i)

${\mathcal{H}}_1$ and ${\mathcal{H}}_2$ are two real Hilbert spaces and $\varnothing \ne C\subset {\mathcal{H}}_1$ and $\varnothing \ne Q\subset H_2$ are two closed convex sets;

(ii)

$\mathcal{A}:{\mathcal{H}}_1\to {\mathcal{H}}_2$ is a bounded linear operator, $\phi :{\mathcal{H}}_1\to \mathcal{R}\cup \{+\infty \}$ and $\varphi :{\mathcal{H}}_2\to \mathcal{R}\cup \{+\infty \}$ are two proper, convex and lower semi-continuous functions.

In what follows, assume $\mathsf{\Gamma }\ne \varnothing$. The following lemma plays a key role for constructing our algorithm and proving our main result.

Lemma 5.

[34] $z^{†}\in \mathsf{\Gamma }$ iff ${\mathcal{A}}^{*}(I-pro{x}_{ϵ\varphi })\mathcal{A}{z}^{†}+(I-pro{x}_{ϵ\phi })z^{†}=0$.

Next, we suggest the following algorithm by applying Lemma 5.

Let $f:{\mathcal{H}}_1\to {\mathcal{H}}_1$ be a $\mu$-contraction. Let $F:{\mathcal{H}}_1\to {\mathcal{H}}_1$ be a strongly positive linear bounded operator with coefficient $\delta >0$. Let $\left\{{\zeta }_n\right\}\subset (0,1)$, $\left\{{\vartheta }_n\right\}\subset (0,1)$ and $\left\{{\tau }_n\right\}\subset (0,+\infty )$ be three real number sequences. Let $\gamma$ be a constant such that $\delta /\mu >\gamma >0$.

• Given fixed point $x_0\in {\mathcal{H}}_1$. Set $n=0$.
• Calculate $y_n$ and $\nu \left(y_n\right)$ via the iterative procedures

  $$\begin{array}{c}\hfill y_n={\zeta }_n\gamma f\left(x_n\right)+(I-{\zeta }_{n}F)x_n,\end{array}$$

  (5)

  and

  $$\begin{array}{c}\hfill \nu \left(y_n\right)={\mathcal{A}}^{*}(I-pro{x}_{ϵ\varphi })\mathcal{A}{y}_n+(I-pro{x}_{ϵ\phi })y_n.\end{array}$$

  (6)
• If $\nu \left(y_n\right)=0$, then the iterative process stops (in this case, $y_n$ is a solution of (2) by Lemma 5), otherwise continuous to the next step.
• Compute

  $$\begin{array}{c}\hfill x_{n+1}=(1-{\vartheta }_n)y_n+{\vartheta }_n{z}_n,\end{array}$$

  (7)

  where

  $$\begin{array}{c}\hfill {\displaystyle z_n=y_n-{\tau }_n\frac{g\left(y_n\right)+h\left(y_n\right)}{\parallel \nu \left(y_n\right){\parallel }^2}\nu \left(y_n\right),}\end{array}$$

  (8)

  in which $g\left(y_n\right)=\frac{1}{2}{\parallel (I-pro{x}_{ϵ\varphi })\mathcal{A}{y}_n\parallel }^2$ and $h\left(y_n\right)=\frac{1}{2}{\parallel (I-pro{x}_{ϵ\phi })y_n\parallel }^2.$
• Set $n\leftarrow n+1$ and repeat steps 2–4.

Assume that the above iterates (5)–(8) do not terminate, that is, the sequence $\left\{x_n\right\}$ generated by (7) is very large. In this case, we demonstrate the convergence analysis of the sequence $\left\{x_n\right\}$.

Theorem 1.

Suppose that the control parameters $\left\{{\zeta }_n\right\}$, $\left\{{\vartheta }_n\right\}$ and $\left\{{\tau }_n\right\}$ satisfy the following restrictions

(C1): 

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

(C2): 

$0<{\lim \inf }_{n\to \infty }{\vartheta }_n\le {\lim \sup }_{n\to \infty }{\vartheta }_n<1$;

(C3): 

${\lim ;\inf }_{n\to \infty }{\tau }_n(4-{\tau }_n)>0$.

Then sequence $\left\{x_n\right\}$ generated by (7) strongly converges to z, where $z=pro{j}_{\mathsf{\Gamma }}(I-F+\gamma f)z$.

Proof. 

Firstly, it is easy to check that operator $pro{j}_{\mathsf{\Gamma }}(I-F+\gamma f)$ is a contraction under the restriction $\delta /\mu >\gamma >0$. Denote its unique fixed point by z, that is, $z=pro{j}_{\mathsf{\Gamma }}(I-F+\gamma f)z$. Next, we show the boundedness of the sequence $\left\{x_n\right\}$. In terms of the nonexpansivity of the operators $I-pro{x}_{ϵ\varphi }$ and $I-pro{x}_{ϵ\phi }$, from (3) and (4), we have

$$2h\left(y_n\right)={\parallel (I-pro{x}_{ϵ\phi })y_n\parallel }^2\le \langle (I-pro{x}_{ϵ\phi })y_n,y_n-z\rangle ,$$

(9)

and

$$\begin{array}{c}\hfill 2g\left(y_n\right)={\parallel (I-pro{x}_{ϵ\varphi })\mathcal{A}{y}_n\parallel }^2\le \langle (I-pro{x}_{ϵ\varphi })\mathcal{A}{y}_n,\mathcal{A}{y}_n-\mathcal{A}z\rangle .\end{array}$$

(10)

By (6), (9) and (10), we obtain

$$\begin{array}{cc}\hfill \langle \nu \left(y_n\right),y_n-z\rangle & =\langle {\mathcal{A}}^{*}(I-pro{x}_{ϵ\varphi })\mathcal{A}{y}_n+(I-pro{x}_{ϵ\phi })y_n,y_n-z\rangle \hfill \\ & =\langle (I-pro{x}_{ϵ\varphi })\mathcal{A}{y}_n,\mathcal{A}{y}_n-\mathcal{A}z\rangle +\langle (I-pro{x}_{ϵ\phi })y_n,y_n-z\rangle \hfill \\ & \ge 2g\left(y_n\right)+2h\left(y_n\right).\hfill \end{array}$$

(11)

This together with (8) implies that

$$\begin{array}{cc}\hfill \parallel z_n{-z\parallel }^2& =\parallel y_n-{\tau }_n\frac{g\left(y_n\right)+h\left(y_n\right)}{\parallel \nu \left(y_n\right){\parallel }^2}\nu \left(y_n\right)-z\parallel \hfill \\ & =\parallel y_n{-z\parallel }^2-2{\tau }_n\frac{g\left(y_n\right)+h\left(y_n\right)}{\parallel \nu \left(y_n\right){\parallel }^2}\langle \nu \left(y_n\right),y_n-z\rangle +{\tau }_n^2\frac{{(g\left(y_n\right)+h\left(y_n\right))}^2}{\parallel \nu \left(y_n\right){\parallel }^2}\hfill \\ & \le \parallel y_n{-z\parallel }^2-{\tau }_n(4-{\tau }_n)\frac{{(g\left(y_n\right)+h\left(y_n\right))}^2}{\parallel \nu \left(y_n\right){\parallel }^2}.\hfill \end{array}$$

(12)

By condition (C3), without loss of generality, we assume $0<a<{\tau }_n<b<4$ for all $n\ge 0$. In the light of (7) and (12), we have

$$\begin{array}{cc}\hfill \parallel x_{n+1}-z\parallel & \le (1-{\vartheta }_n)\parallel y_n-z\parallel +{\vartheta }_n\parallel z_n-z\parallel \hfill \\ & \le \parallel y_n-z\parallel \hfill \\ & =\parallel {\zeta }_n\gamma (f\left(x_n\right)-f\left(z\right))+{\zeta }_n(\gamma f\left(z\right)-F\left(z\right))+(I-{\zeta }_{n}F)(x_n-z)\parallel \hfill \\ & \le {\zeta }_n\gamma \mu \parallel x_n-z\parallel +{\zeta }_n\parallel \gamma f\left(z\right)-F\left(z\right)\parallel +(1-\delta {\zeta }_n)\parallel x_n-z\parallel \hfill \\ & =[1-(\delta -\gamma \mu ){\zeta }_n]\parallel x_n-z\parallel +{\zeta }_n\parallel \gamma f\left(z\right)-F\left(z\right)\parallel \hfill \\ & \le max\{\parallel x_n-z\parallel ,\frac{\parallel \gamma f\left(z\right)-F\left(z\right)\parallel }{\delta -\gamma \mu }\}\hfill \\ & \le \cdots \hfill \\ & \le max\{\parallel x_0-z\parallel ,\frac{\parallel \gamma f\left(z\right)-F\left(z\right)\parallel }{\delta -\gamma \mu }\}.\hfill \end{array}$$

(13)

Hence, the sequence $\left\{x_n\right\}$ is bounded. Consequently, we can check easily that the sequences $\left\{y_n\right\}$ and $\left\{z_n\right\}$ are bounded.

By virtue of (5), we have

$$\begin{array}{cc}\hfill \parallel y_n{-z\parallel }^2& =\parallel {\zeta }_n(\gamma f\left(x_n\right)-F\left(z\right))+(I-{\zeta }_{n}F)(x_n-z){\parallel }^2\hfill \\ & \le \parallel I-{\zeta }_n{F\parallel }^2\parallel x_n{-z\parallel }^2+{\zeta }_n^2{\parallel \gamma f\left(x_n\right)-F\left(z\right)\parallel }^2+2{\zeta }_n\langle \gamma f\left(x_n\right)-F\left(z\right),(I-{\zeta }_{n}F)(x_n-z)\rangle \hfill \\ & \le {(1-\delta {\zeta }_n)}^2\parallel x_n{-z\parallel }^2+{\zeta }_n^2{\parallel \gamma f\left(x_n\right)-F\left(z\right)\parallel }^2+2\gamma {\zeta }_n\langle f\left(x_n\right)-f\left(z\right),(I-{\zeta }_{n}F)(x_n-z)\rangle \hfill \\ & +2{\zeta }_n\langle \gamma f\left(z\right)-F\left(z\right),x_n-z\rangle -2{\zeta }_n^2\langle \gamma f\left(z\right)-F\left(z\right),F\left(x_n\right)-F\left(z\right)\rangle \hfill \\ & \le {(1-\delta {\zeta }_n)}^2\parallel x_n{-z\parallel }^2+{\zeta }_n^2\parallel \gamma f\left(x_n\right)-F\left(z\right){\parallel }^2+2\gamma \mu (1-\delta {\zeta }_n){\zeta }_n{\parallel x_n-z\parallel }^2\hfill \\ & +2\delta {\zeta }_n^2\parallel \gamma f\left(z\right)-F\left(z\right)\parallel \parallel x_n-z\parallel +2{\zeta }_n\langle \gamma f\left(z\right)-F\left(z\right),x_n-z\rangle \hfill \\ & \le [1-2(\delta -\gamma \mu ){\zeta }_n]{\parallel x_n-z\parallel }^2+2{\zeta }_n\langle \gamma f\left(z\right)-F\left(z\right),x_n-z\rangle +{\zeta }_n^2{M}_1,\hfill \end{array}$$

(14)

where $M_1\ge {\sup }_{n\ge 0}\{\parallel \gamma f\left(x_n\right)-F\left(z\right)\parallel ,2\delta \parallel \gamma f\left(x_n\right)-F\left(z\right)\parallel \parallel x_n-z\parallel ,{\delta }^2\parallel x_n-z{\parallel }^2\}$.

On the basis of (7) and Lemma 1, we get

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \parallel x_{n+1}{-z\parallel }^2& =\parallel (1-{\vartheta }_n)(y_n-z)+{\vartheta }_n(z_n-z){\parallel }^2\hfill \\ & =(1-{\vartheta }_n)\parallel y_n{-z\parallel }^2+{\vartheta }_n\parallel z_n{-z\parallel }^2-(1-{\vartheta }_n){\vartheta }_n{\parallel y_n-z_n\parallel }^2.\hfill \end{array}\end{array}$$

(15)

From (12) and (15), we deduce

$$\begin{array}{c}\hfill \parallel x_{n+1}{-z\parallel }^2\le \parallel y_n{-z\parallel }^2-(1-{\vartheta }_n){\vartheta }_n{\parallel y_n-z_n\parallel }^2.\end{array}$$

(16)

By (7), we note that

$$y_n-z_n=\frac{1}{{\vartheta }_n}(y_n-x_{n+1}).$$

(17)

Thus, combining (14), (16) with (25), we get

$$\begin{array}{cc}\hfill \parallel x_{n+1}{-z\parallel }^2& \le \parallel y_n{-z\parallel }^2-\frac{1-{\vartheta }_n}{{\vartheta }_n}{\parallel x_{n+1}-y_n\parallel }^2\hfill \\ & \le [1-2(\delta -\gamma \mu ){\zeta }_n]{\parallel x_n-z\parallel }^2+2{\zeta }_n\langle \gamma f\left(z\right)-F\left(z\right),x_n-z\rangle \hfill \\ & +{\zeta }_n^2{M}_1-\frac{1-{\vartheta }_n}{{\vartheta }_n}{\parallel x_{n+1}-y_n\parallel }^2\hfill \\ & =[1-2(\delta -\gamma \mu ){\zeta }_n]{\parallel x_n-z\parallel }^2+{\zeta }_n{\sigma }_n,\hfill \end{array}$$

(18)

where

$$\begin{array}{c}\hfill {\sigma }_n=2\langle \gamma f\left(z\right)-F\left(z\right),x_n-z\rangle +M_1{\zeta }_n-\frac{1-{\vartheta }_n}{{\zeta }_n{\vartheta }_n}{\parallel x_{n+1}-y_n\parallel }^2.\end{array}$$

(19)

According to the boundedness of the sequence $\left\{x_n\right\}$, from (27), we obtain ${\sigma }_n\le M_2(\forall n\ge 0)$ for some $M_2$. Applying Lemma 2 to (26), we get $0\le 2(\delta -\gamma \mu ){\lim \sup }_{n\to \infty }{\parallel x_n-z\parallel }^2\le {\lim \sup }_{n\to \infty }{\sigma }_n\le M_2$. Therefore, ${\lim \sup }_{n\to \infty }{\sigma }_n$ exists and there exists a subsequence $\left\{x_{n_i}\right\}$ of $\left\{x_n\right\}$ such that $x_{n_i}\rightharpoonup \tilde{x}$ and

$$\begin{array}{cc}\hfill \underset{n\to \infty }{\lim \sup }{\sigma }_n& =\underset{n\to \infty }{\lim \sup }(2\langle \gamma f\left(z\right)-F\left(z\right),x_n-z\rangle +M_1{\zeta }_n-\frac{1-{\vartheta }_n}{{\zeta }_n{\vartheta }_n}\parallel x_{n+1}-y_n{\parallel }^2)\hfill \\ & =\underset{i\to \infty }{\lim }(2\langle \gamma f\left(z\right)-F\left(z\right),\tilde{x}-z\rangle -\frac{1-{\vartheta }_{n_i}}{{\zeta }_{n_i}{\vartheta }_{n_i}}\parallel x_{n_i+1}-y_{n_i}{\parallel }^2).\hfill \end{array}$$

(20)

This indicates that ${\displaystyle \underset{i\to \infty }{\lim }\frac{1-{\vartheta }_{n_i}}{{\zeta }_{n_i}{\vartheta }_{n_i}}{\parallel x_{n_i+1}-y_{n_i}\parallel }^2}$ exists and by conditions (C1) and (C2), we deduce

$$\begin{array}{c}\hfill \underset{i\to \infty }{\lim }\parallel x_{n_i+1}-y_{n_i}\parallel =0.\end{array}$$

(21)

This together with (25) implies that

$$\begin{array}{c}\hfill \underset{i\to \infty }{\lim }\parallel y_{n_i}-z_{n_i}\parallel =0.\end{array}$$

Combining (5) with (21), we obtain

$$\begin{array}{c}\hfill \underset{i\to \infty }{\lim }\parallel x_{n_i+1}-x_{n_i}\parallel =\underset{i\to \infty }{\lim }\parallel y_{n_i}-x_{n_i}\parallel =0.\end{array}$$

By (12), we have

$$\begin{array}{cc}\hfill 0\le {\tau }_{n_i}(4-{\tau }_{n_i})\frac{{(g\left(y_{n_i}\right)+h\left(y_{n_i}\right))}^2}{\parallel \nu \left(y_{n_i}\right){\parallel }^2}& \le \parallel y_{n_i}{-z\parallel }^2-{\parallel z_{n_i}-z\parallel }^2\hfill \\ & \le \parallel y_{n_i}-z_{n_i}\parallel (\parallel y_{n_i}-z\parallel +\parallel z_{n_i}-z\parallel )\to 0(\mathrm{as}i\to \infty ).\hfill \end{array}$$

(22)

It follows that

$$\begin{array}{c}\hfill \underset{i\to \infty }{\lim }\frac{g\left(y_{n_i}\right)+h\left(y_{n_i}\right)}{\parallel \nu \left(y_{n_i}\right)\parallel }=0.\end{array}$$

(23)

Noting that $\nu \left(y_{n_i}\right)$ is bounded, from (23), we deduce ${\lim }_{i\to \infty }(g\left(y_{n_i}\right)+h\left(y_{n_i}\right))=0$. Thus, ${\lim }_{i\to \infty }g\left(y_{n_i}\right)={\lim }_{i\to \infty }h\left(y_{n_i}\right)=0$. That is,

$$\begin{array}{c}\hfill \underset{i\to \infty }{\lim }\parallel (I-pro{x}_{ϵ\phi })y_{n_i}\parallel =\underset{i\to \infty }{\lim }\parallel (I-pro{x}_{ϵ\varphi })\mathcal{A}{y}_{n_i}\parallel =0.\end{array}$$

This together with Lemma 3 implies that $\tilde{x}\in Fix\left(pro{x}_{ϵ\phi }\right)$ and $A\tilde{x}\in Fix\left(pro{x}_{ϵ\varphi }\right)$. Hence $\tilde{x}\in \mathsf{\Gamma }$. Therefore,

$$\begin{array}{c}\hfill \underset{n\to \infty }{\lim \sup }\langle \gamma f\left(z\right)-F\left(z\right),x_n-z\rangle =\langle \gamma f\left(z\right)-F\left(z\right),\tilde{x}-z\rangle \le 0.\end{array}$$

By (26), we have

$$\begin{array}{c}\hfill \parallel x_{n+1}{-z\parallel }^2\le [1-2(\delta -\gamma \mu ){\zeta }_n]{\parallel x_n-z\parallel }^2+2{\zeta }_n\langle \gamma f\left(z\right)-F\left(z\right),x_n-z\rangle +{\zeta }_n^2{M}_1.\end{array}$$

(24)

According to Lemma 4 and (24), we deduce that $x_n\to z$. This completes the proof. □

• Given fixed point $x_0\in {\mathcal{H}}_1$. Set $n=0$.
• Calculate $y_n$ and $\nu \left(y_n\right)$ via the iterative procedures

  $$\begin{array}{c}\hfill y_n=(1-{\zeta }_n)x_n,\end{array}$$

  (25)

  and

  $$\begin{array}{c}\hfill \nu \left(y_n\right)={\mathcal{A}}^{*}(I-pro{x}_{ϵ\varphi })\mathcal{A}{y}_n+(I-pro{x}_{ϵ\phi })y_n.\end{array}$$

  (26)
• If $\nu \left(y_n\right)=0$, then the iterative process stops (in this case, $y_n\in \mathsf{\Gamma }$ by Lemma 5), otherwise continuous to the next step.
• Compute

  $$\begin{array}{c}\hfill x_{n+1}=(1-{\vartheta }_n)y_n+{\vartheta }_n{z}_n,\end{array}$$

  (27)

  where

  $$\begin{array}{c}\hfill {\displaystyle z_n=y_n-{\tau }_n\frac{g\left(y_n\right)+h\left(y_n\right)}{\parallel \nu \left(y_n\right){\parallel }^2}\nu \left(y_n\right),}\end{array}$$

  (28)

  in which $g\left(y_n\right)=\frac{1}{2}{\parallel (I-pro{x}_{ϵ\varphi })\mathcal{A}{y}_n\parallel }^2$ and $h\left(y_n\right)=\frac{1}{2}{\parallel (I-pro{x}_{ϵ\phi })y_n\parallel }^2.$
• Set $n\leftarrow n+1$ and repeat steps 2–4.

Assume that the above iterates (25)–(28) do not terminate, that is, the sequence $\left\{x_n\right\}$ generated by (27) is very large.

Corollary 1.

Suppose that the control parameters $\left\{{\zeta }_n\right\}$, $\left\{{\vartheta }_n\right\}$ and $\left\{{\tau }_n\right\}$ satisfy the restrictions (C1)–(C3). Then sequence $\left\{x_n\right\}$ generated by (27) strongly converges to $z=pro{j}_{\mathsf{\Gamma }}\left(0\right)$, the minimum norm element in Γ.