A Pre-Conditioning CQ Algorithm with Double Inertia and Self-Updated Stepsizes for Split Feasibility Problems in Hilbert Spaces

Abstract

In this work, we propose a double inertial pre-conditioning CQ Algorithm for split feasibility problem in real Hilbert spaces, in which, we use the double inertial steps and new stepsizes to speed up the convergent rate. We also compute the only one projection onto the nonempty closed convex set C in our proposed method. These help improve the numerical results. Next, we establish the weak convergence of the sequence generated by our method. Finally, we use a numerical experiment to demonstrate our theoretical results.

1. Introduction

In this paper, let H be a real Hilbert space with inner product $\langle ·,·\rangle$ and the induced norm $\parallel ·\parallel .$ The split feasibility problem (SFP) is as follows.

$$\begin{array}{c}\begin{array}{c}\mathrm{Find}{g}^{\ast }\in C\mathrm{such} \mathrm{that}A{g}^{\ast }\in Q.\hfill \end{array}\hfill \end{array}$$

Its solution set is suggested as

$$\Omega =\{g^{\ast }\in C:A{g}^{\ast }\in Q\}.$$

Here, C and Q are nonempty closed convex subsets of real Hilbert spaces $H_1$ and $H_2$, respectively, and $A:H_1\to H_2$ is a bounded linear operator.

The split feasibility problem (SFP) proposed by Censor and Elfving [1] was used to study inverse problems, and then it was applied to signal recovery [2] and image deblurring problems [3]. From then, the SFP has attracted attention, and many related algorithms have been created to investigate it deeply. For instance, Byrne [2,3] stated a classical CQ algorithm as follows:

$$j_{n+1}=P_C\left(j_n-\Gamma A^{\ast }(I-P_Q)A{j}_n\right),\forall n\ge 1,$$

(1)

where $\Gamma \in \left(0,{\displaystyle \frac{2}{{\parallel A\parallel }^2}}\right)$, I represents an identity operator, $P_C$ and $P_Q$ are the projections onto C and Q, respectively, and $A^{\ast }$ means the adjoint of A. For the CQ algorithms, more details can be found in [4,5,6,7,8,9,10]. We know that the algorithm (1) begins the iteration; it first computes the operator norm $\parallel A\parallel$, which creates a small stepsize $\Gamma$, and thus, the convergent rate of the algorithm is slow. Meanwhile, it is not easy to calculate the operator norm $\parallel A\parallel$ in practice. From this consideration, Lopéz [11] suggested the following self-adaptive stepsize by modifying Yang [12].

$${\Gamma }_n={\displaystyle \frac{tU\left(j_n\right)}{\parallel \nabla U\left(j_n\right){\parallel }^2}}.$$

(2)

Here, $inft(4-t)>0$, and $U\left(j\right)={\displaystyle \frac{1}{2}}{\parallel (I-P_Q)Aj\parallel }^2.$ Under the stepsize (2), the weak convergence of the algorithm (1) is proved. Meanwhile, several CQ algorithms with self-updated stepsizes have been proposed to solve the SFP; (see Refs. [13,14,15,16,17] and the references therein).

To introduce an efficient stepsize, Altiparmak et al. [18] presented a pre-conditioning CQ algorithm for solving the SFP, as follows:

$$\left\{\begin{array}{c}{y}_n=P_C^B\left(j_n-{\Gamma }_n\nabla U_{\tilde{E}}\left(j_n\right)\right);j_{n+1}=P_C^B(y_n-{\gamma }_n\nabla U_{\tilde{E}}\left(y_n\right));\hfill \\ \mathrm{where} \mathrm{the} \mathrm{stepsizes} {\Gamma }_n\mathrm{and} {\gamma }_n\mathrm{is} \mathrm{updated} \mathrm{respectively} \mathrm{via} \hfill \\ {\Gamma }_n∥\nabla U\left(y_n\right)-\nabla U\left(j_n\right)∥\le \delta \parallel y_n-j_n\parallel ,0<\delta <1,\hfill \\ {\gamma }_n={\displaystyle \frac{tU\left(y_n\right)}{\parallel \nabla U_{\tilde{E}}\left(y_n\right){\parallel }^2}},0<t<4,\hfill \end{array}\right.$$

(3)

where $\nabla U_{\tilde{E}}\left(j\right)=E{A}^{\ast }\left(I-P_Q\right)Aj$,  $\nabla U\left(j\right)=A^{\ast }\left(I-P_Q\right)Aj,$ E is defined in Section 3, ${\Gamma }_n=\sigma {\rho }^{m_n}$, $\sigma >0$, $0<\rho <1$, and $m_n$ is the smallest nonnegative integer. The authors analyzed the weak convergence of the algorithm (3) under these stepsizes. Pre-conditioning methods are usually applied to speed up the convergence of the algorithms, for more details, see [19,20,21].

On the other hand, the inertial extrapolation technique can be used to accelerate the convergence rate of the algorithms; for this, Nesterov [22] proposed the following classical iterative scheme

$$\begin{array}{c}{v}_n=j_n+u_n(j_n-j_{n-1})\hfill \\ j_{n+1}={Ϝ}_n-{\Gamma }_{n}F\left(v_n\right),n\ge 1,\hfill \end{array}$$

(4)

where $u_n\in [0,1)$, and ${\Gamma }_n>0$. The notation $v_n$ is the inertial step. Inspired by the inertial idea [23,24,25,26,27], some inertial splitting algorithms have been proposed to solve the SFP [28,29,30,31]. Recently, Wang et al. [32] proposed the following double inertial splitting methods to solve monotone inclusion problems.

$$\left\{\begin{array}{c}{v}_n=j_n+u_n(j_n-j_{n-1}),\hfill \\ s_n=j_n+q_n(j_n-j_{n-1}),\hfill \\ y_n={(I+{\Gamma }_{n}Y)}^{-1}(I-{\Gamma }_{n}L)v_n;\hfill \\ j_{n+1}=(1-{\alpha }_n)s_n+{\alpha }_n(y_n-{\lambda }_n(L{y}_n-L{v}_n)),\hfill \end{array}\right.$$

(5)

where the stepsize ${\Gamma }_n$ is updated by

$${\Gamma }_{n+1}=\left\{\begin{array}{cc}\min \left\{{\displaystyle \frac{({\mu }_n+\delta )\parallel v_n-y_n\parallel }{\parallel L{v}_n-L{y}_n\parallel }},{\Gamma }_n+{\widehat{P}}_n\right\},\hfill & \mathrm{if}L{v}_n-L{y}_n\ne 0,\hfill \\ {\Gamma }_n+{\widehat{P}}_n,\hfill & \mathrm{otherwise},\hfill \end{array}\right.$$

Here, I is the identity operator on H. $Y:H\to 2^H$ is maximally monotone mapping, $L:H\to H$ is monotone and Lipschtz continuous mapping, $\delta \in (0,1)$, $u_n\in [0,1]$, $0\le q_n\le q_{n+1}<{\displaystyle \frac{3+2d-\sqrt{8d+17}}{2d}}, d\in (1,+\infty )$; $0<{\alpha }_n\le {\alpha }_{n+1}\le {\displaystyle \frac{1}{1+d}}, d\in (1,+\infty )$, and ${\displaystyle \sum _{n=1}^{\infty }}{\widehat{P}}_n<\infty$. Its convergence is established under suitable conditions in real Hilbert spaces.

It is observed that, in algorithm (3), the proper choice of the stepsize ${\Gamma }_n$ is required to satisfy the Armijo-like linesearch techniques; that is, the projections of the iterate point $A{y}_n$ onto Q and the projections onto C may be calculated many times at each iteration. On the other hand, the algorithm (3) needs to compute the double projections onto C at least for finishing an iteration. These considerations affect its effectiveness. Furthermore, their weak convergence is guaranteed by employing some strong conditions such as the firm nonexpansiveness of $I-P_Q$. However, the investigation of the algorithm (3) with double inertial effects (5) has yet to be conducted. The following question is asked.

Question: Can we develop new modifications of the algorithm (3) such that numerical improvement and new convergence results can be obtained under weaker conditions than the firm nonexpansiveness of $I-P_Q$?

To answer this question, our contributions in this article are as follows:

• To adopt double inertial steps to speed up the convergent rate of the algorithm (3) and create a new convergence result;
• To design a new stepsize not only to reduce the computational projections of $y_n$ onto Q but also to avoid the Armijo-like linesearch techniques [13,14,18]. Moreover, the projections $P_C^B$ for the computation of the next iteration $j_n$ can be removed, and thus, an accelerated convergent algorithm is generated;
• To provide some practical examples such as signal recovery problems for demonstration and illustration.

The paper is organized as follows. In Section 2, we list some useful results for the convergence analysis of the algorithm. In Section 3, we propose a double inertial pre-conditioning CQ algorithm and prove its weak convergence. Numerical experiments are described in Section 4.

2. Preliminaries

For the reader’s convenience, we first introduce two notations; that is, ⇀ stands for weak convergence, while → stands for strong convergence. Second, we let H be a real Hilbert space, and we recall some well-known concepts related to the SFP.

Definition 1.

Let C be a subset of H and $U:C\to H$ be an operator; then,

(i) 

U is called nonexpansive on C if

$$\parallel Uj-Uy\parallel \le \parallel j-y\parallel ,\forall j,y\in C.$$

(ii) 

F is said to be firmly nonexpansive on C if

$$\langle Uj-Uy,j-y\rangle \ge {\parallel Uj-Uy\parallel }^2,\forall j,y\in C.$$

Definition 2

([18]). Let H be a real Hilbert space; a differentiable function $F:H\to R$ is convex if and only if

$$U\left(y\right)-U\left(j\right)\ge \langle \nabla U\left(j\right),y-j\rangle ,\forall j,y\in H.$$

Lemma 1.

For all $j,y,z\in H$, then

(i) $\parallel \tilde{\alpha }j+(1-\tilde{\alpha }){y\parallel }^2=\tilde{\alpha }{\parallel j\parallel }^2+(1-\tilde{\alpha }){\parallel y\parallel }^2-\tilde{\alpha }(1-\tilde{\alpha }){\parallel j-y\parallel }^2$, $\tilde{\alpha }\in (0,1).$

(ii) $\langle j-y,j-z\rangle ={\displaystyle \frac{1}{2}}{\parallel j-y\parallel }^2+{\displaystyle \frac{1}{2}}{\parallel j-z\parallel }^2-{\displaystyle \frac{1}{2}}{\parallel y-z\parallel }^2.$

Next, we provide some concepts about a positive-definite operator and a self-adjoint operator.

Let $B:H\to H$ be a bounded linear operator on a real Hilbert space. B is named self adjoint if $B^{\ast }=B$, where $B^{\ast }$ is the adjoint operator. Moreover, a self adjoint operator is called positive definite if

$$\langle j,Bj\rangle >0,\forall j\in H,j\ne 0.$$

For more details, see [33,34].

Let B be a self-adjoint positive-definite operator, and $E=B^{-1}$. Thereby, we define the norm ${\parallel ·\parallel }_B$ as

$${\parallel j\parallel }_B^2=\langle j,Bj\rangle ={\langle j,j\rangle }_B,\forall j\in H,$$

as in [18]. The $B$—projection operator on C related to the norm ${\parallel ·\parallel }_B$ denotes $P_C^B$; that is,

$$P_C^B\left(j\right)=\arg \underset{z\in C}{\min }{\{\parallel j-z\parallel }_B\}.$$

Now, we give the following related properties.

Lemma 2

([18]). Let C be a nonempty closed convex subset of a real Hilbert space H; the $B$—projection operator on C fulfills the following relationships.

(i) $\langle B(j-P_C^{B}j),y-P_C^{B}j\rangle \le 0,\forall j\in Handy\in C.$

(ii) ${\parallel j\pm z\parallel }_B^2={\parallel j\parallel }_B^2\pm 2\langle j,Bz\rangle +{\parallel z\parallel }_B^2,\forall j,z\in H.$

(iii) $\parallel P_C^{B}j-P_C^B{y\parallel }_B^2\le {\parallel j-y\parallel }_B^2-{∥(P_C^{B}j-j)-(P_C^{B}y-y)∥}_B^2,\forall j,y\in H.$

(iv)$\parallel P_C^B{j-y\parallel }_B^2\le {\parallel j-y\parallel }_B^2-{\parallel j-P_C^{B}j\parallel }_B^2,\forall j,y\in H.$

Lemma 3

([29,35]). Let $\left\{{\phi }_n\right\}$ be a sequence of nonnegative numbers fulfilling

$${\phi }_{n+1}\le {\varphi }_n{\phi }_n+U_n,\forall n\in \mathbb{N},$$

where $\left\{{\varphi }_n\right\}$ and $\left\{U_n\right\}$ are sequences of nonnegative numbers such that $\left\{{\varphi }_n\right\}\subset [1,+\infty ),$ ${\displaystyle \sum _{n=1}^{\infty }}({\varphi }_n-1)<\infty$, and  ${\displaystyle \sum _{n=1}^{\infty }}{U}_n<\infty$. Then, ${\displaystyle \underset{n\to \infty }{\lim }}{\phi }_n$ exists.

Lemma 4

([31]). Let $C\subset H$ be a non-empty set, and let $\left\{j_n\right\}$ be a sequence in H that fulfills the following conditions:

(i) ${\displaystyle \underset{n\to \infty }{\lim }}\parallel j_n-j\parallel$ exists for all $j\in C$;

(ii) Every weak sequential cluster point of $\left\{j_n\right\}$ belongs to C.

Then, $\left\{j_n\right\}$ converges weakly to a point in C.

Lemma 5

(Lemma 2.2 [31]). Let $\left\{Q_n\right\}\subset [0,\infty )$ and $\left\{{\tilde{b}}_n\right\}\subset [0,\infty )$ be the sequences such that

(i) $Q_{n+1}-Q_n\le {\varrho }_n(Q_n-Q_{n-1})+b_n;$

(ii) ${\displaystyle \sum _{n=1}^{\infty }}{\tilde{b}}_n<\infty ;$

(iii) $\left\{{\varrho }_n\right\}\subset [0,\varrho ],$ where $\varrho \in [0,1).$

• Then, $\left\{Q_n\right\}$ is a converging sequence, and ${\displaystyle \sum _{n=1}^{\infty }}{[Q_n-Q_{n-1}]}_{+}<\infty ,$ where ${\left[\zeta \right]}_{+}=max\{\zeta ,0\}$ (for any $\zeta \in \mathbb{R}$).

3. Weak Convergence

In this section, we propose a double inertial pre-conditioning CQ algorithm for solving SFP and analyze a weak convergence result under mild conditions. Next, we present some assumptions.

$\left(P_1\right)$ The solution set of SFP is nonempty; that is, $\Omega \ne \varnothing .$

$\left(P_2\right)$ Let C and Q be nonempty closed convex subsets of real Hilbert spaces $H_1$ and $H_2$, respectively, and $A:H_1\to H_2$ be a bounded linear operator with its adjoint $A^{\ast }$. Let B and $\tilde{E}$ be self adjoint positive-definite operators such that $A^{\ast }\tilde{E}=E{A}^{\ast }$, with $E=B^{-1}.$

$\left(P_3\right)$ Let ${\kappa }_{\min }\left(B\right)$ and ${\kappa }_{max}\left(B\right)$ be the minimum and the largest eigenvalues of B, respectively. For the ${\parallel ·\parallel }_B^2$, the result is as follows:

$${\kappa }_{\min }{\left(B\right)\parallel z\parallel }^2\le {\parallel z\parallel }_B^2\le {\kappa }_{max}\left(B\right){\parallel z\parallel }^2,\forall z\in H,$$

which is suggested in Ref. [36].

$\left(P_4\right)$ The sequences $\left\{u_n\right\},\left\{q_n\right\},\left\{{\alpha }_n\right\}$, and $\left\{h_n\right\}$ have the following inequalities.

(i) $0\le u_n\le 1;$

(ii) $0\le q_n\le q_{n+1}\le q<{\displaystyle \frac{3+2d-\sqrt{8d+17}}{2d}},d\in (1,\infty );$

(iii) $0<\alpha <{\alpha }_n\le {\alpha }_{n+1}\le {\displaystyle \frac{1}{1+d}},d\in (1,\infty );$

(iv) $h_n=(1-{\alpha }_n)q_n+{\alpha }_n{u}_n$ is a non-decreasing sequence.

$\left(P_5\right)$ Let $P_Q$ be $\tilde{E}$-projection on Q related to the norm ${\parallel ·\parallel }_{\tilde{E}}.$ For the continuously differentiable convex functions $U,U_{\tilde{E}}:H_1\to R$, we define

$$U\left(j\right)={\displaystyle \frac{1}{2}}{\parallel (I-P_Q)Ax\parallel }^2,$$

and

$$U_{\tilde{E}}\left(j\right)={\displaystyle \frac{1}{2}}{\parallel (I-P_Q)Aj\parallel }_{\tilde{E}}^2={\displaystyle \frac{1}{2}}\langle \tilde{E}\left((I-P_Q)Aj\right),(I-P_Q)Aj\rangle ;$$

their corresponding gradients are the following, respectively.

$$\nabla U_{\tilde{E}}\left(j\right)=E{A}^{\ast }(I-P_Q)Aj,\nabla U\left(j\right)=A^{\ast }(I-P_Q)Aj.$$

Proposition 1

([18]). Assume that the condition $P_1$ holds. The results are equivalent below.

(i) $g^{\ast }\in \Omega ,$

(ii) $g^{\ast }\in C$ and $F\left(g^{\ast }\right)=0;$

(iii) $g^{\ast }\in C$ and $\nabla F\left(g^{\ast }\right)=0.$

In what follows, we illustrate that our designed stepsize scheme is well-defined.

Lemma 6.

Let $\left\{{\Gamma }_n\right\}$ be the sequence produced by Algorithm 1. Then, we need to prove that ${\displaystyle \underset{n\to \infty }{\lim }}{\Gamma }_n$ exists and denote ${\displaystyle \underset{n\to \infty }{\lim }}{\Gamma }_n=\tilde{\Gamma }$, where $\tilde{\Gamma }\ge \min \left\{{\displaystyle \frac{\delta }{{\parallel A\parallel }^2}},{\Gamma }_1\right\}$ and ${\Gamma }_1>0$, is an initial stepsize.

Proof. 

In the case of $M_n>0$, and since $P_Q$ is of firm nonexpansiveness, we derive

$$\begin{array}{c}\hfill \begin{array}{c}〈\nabla U\left(y_n\right)-\nabla U\left(v_n\right),y_n-v_n〉=〈A^{\ast }\left(I-P_Q\right)A{y}_n-A^{\ast }\left(I-P_Q\right)A{v}_n,y_n-v_n〉\hfill \\ =〈\left(I-P_Q\right)A{y}_n-\left(I-P_Q\right)A{v}_n,A{y}_n-A{v}_n〉\hfill \\ =\parallel A{y}_n-A{v}_n{\parallel }^2-〈P_Q\left(A{y}_n\right)-P_Q\left(A{v}_n\right),A{y}_n-A{v}_n〉\hfill \\ \le \parallel A{y}_n-A{v}_n{\parallel }^2-{∥P_Q\left(A{y}_n\right)-P_Q\left(A{v}_n\right)∥}^2\hfill \\ \le \parallel A{y}_n-A{v}_n{\parallel }^2\hfill \\ \le {\parallel A\parallel }^2{\parallel y_n-v_n\parallel }^2,\hfill \end{array}\end{array}$$

which shows that

$${\displaystyle \frac{\delta \parallel y_n-v_n{\parallel }^2}{〈\nabla F\left(y_n\right)-\nabla F\left(v_n\right),y_n-v_n〉}}\ge {\displaystyle \frac{\delta }{{\parallel A\parallel }^2}}.$$

This further discloses that

$$\begin{array}{c}{\Gamma }_{n+1}=\min \left\{{\displaystyle \frac{\delta \parallel y_n-v_n{\parallel }^2}{〈\nabla U\left(y_n\right)-\nabla U\left(v_n\right),y_n-v_n〉}},{\varphi }_n{\Gamma }_n\right\}\ge \min \left\{{\displaystyle \frac{\delta }{{\parallel A\parallel }^2}},{\Gamma }_n\right\},\hfill \end{array}$$

where ${\varphi }_n\ge 1$. By induction, the sequence $\left\{{\Gamma }_n\right\}$ has a lower bound $\min \left\{{\displaystyle \frac{\delta }{{\parallel A\parallel }^2}},{\Gamma }_1\right\}$. From (8), it verifies that

$${\Gamma }_{n+1}\le {\varphi }_n{\Gamma }_n.$$

Thanks to Lemma 3, we see that ${\displaystyle \underset{n\to \infty }{\lim }}{\Gamma }_n$ exists and denote ${\displaystyle \underset{n\to \infty }{\lim }}{\Gamma }_n=\tilde{\Gamma }$. Since the sequence $\left\{{\Gamma }_n\right\}$ has the lower bound $\min \left\{{\displaystyle \frac{\delta }{{\parallel A\parallel }^2}},{\Gamma }_1\right\}$, we disclose that $\tilde{\Gamma }>0$.    □

Now, our algorithm is described as follows.

Algorithm 1 A Pre-conditioning CQ Algorithm With Double Inertia and Self-updated Stepsizes For Solving SFP

Step 0. Given $j_0,j_1\in H_1$,${\Gamma }_1>0$, $\delta \in \left(0,{\displaystyle \frac{{\kappa }_{\min }\left(B\right)}{2}}\right)$. Take the sequence $\left\{{\varphi }_n\right\}$ fulfilling Lemma 2.3.  Step 1. Given $j_{n-1}$, $j_n(n\ge 1)$ and compute $$\begin{array}{c}{v}_n=j_n+u_n(j_n-j_{n-1}),s_n=j_n+q_n(j_n-j_{n-1}),\hfill \end{array}$$ (6) $$\begin{array}{c}{y}_n=P_C^B(v_n-{\Gamma }_n\nabla U_{\tilde{E}}\left(v_n\right)),j_{n+1}=(1-{\alpha }_n)s_n+{\alpha }_n(y_n-{\gamma }_n\nabla U_{\tilde{E}}\left(y_n\right)),\hfill \end{array}$$ (7) where the stepsize ${\Gamma }_n$ is computed by $${\Gamma }_{n+1}=\left\{\begin{array}{cc}\min \left\{{\displaystyle \frac{\delta \parallel y_n-v_n{\parallel }^2}{〈\nabla U\left(y_n\right)-\nabla U\left(v_n\right),y_n-v_n〉}},{\varphi }_n{\Gamma }_n\right\},\hfill & \mathrm{if}{M}_n>0,\hfill \\ {\varphi }_n{\Gamma }_n,\hfill & \mathrm{otherwise},\hfill \end{array}\right.$$ (8) where $M_n=〈\nabla U\left(y_n\right)-\nabla U\left(v_n\right),y_n-v_n〉.$ And where the stepsize ${\gamma }_n$ is computed by $${\gamma }_n={\displaystyle \frac{tU\left(y_n\right)}{\parallel \nabla U_{\tilde{E}}\left(y_n\right){\parallel }_B^2}},0<t<2.$$ (9) If $y_n=v_n$, then stop, $y_n$ is a solution of SFP. Otherwise, go to Step 1.

Remark 1.

From Algorithm 1, we observe that we can directly compute the iterative point $j_n$ by using the step (7), which is not involved in the projection $P_C^B$, compared to that in the original algorithm [18]. Thus, the numerical result is improved.

Theorem 1.

Assume that conditions $\left(P_1\right)–\left(P_5\right)$ hold. The sequence $\left\{j_n\right\}$ produced by Algorithm 1 converges weakly to a point $z\in \Omega$.

Proof. 

Let $z\in \Omega$; it yields that $\nabla F\left(z\right)=0$, $z=P_C^{B}z,$ and $P_{Q}Az=Az.$ Now, we have the following result by using the Pythagorean theorem.

$$\begin{array}{c}{∥y_n-{\gamma }_n\nabla U_{\tilde{E}}\left(y_n\right)-z∥}_B^2\hfill \\ =\parallel y_n{-z\parallel }_B^2+{\gamma }_n^2{\parallel \nabla U_{\tilde{E}}\left(y_n\right)\parallel }_B^2-2{\gamma }_n{〈y_n-z,\nabla U_{\tilde{E}}\left(y_n\right)〉}_B.\hfill \end{array}$$

(10)

By Lemma 1(ii), we show that

$$\begin{array}{c}{〈y_n-z,-\nabla U_{\tilde{E}}\left(y_n\right)〉}_B=〈y_n-z,-\nabla U\left(y_n\right)〉=〈A{y}_n-Az,\left(P_Q-I\right)A{y}_n〉\hfill \\ =〈A{y}_n-P_Q\left(A{y}_n\right)+P_Q\left(A{y}_n\right)-Az,\left(P_Q-I\right)A{y}_n〉\hfill \\ =〈P_Q\left(A{y}_n\right)-Az,P_Q\left(A{y}_n\right)-A{y}_n〉-{∥P_Q\left(A{y}_n\right)-A{y}_n∥}^2\hfill \\ ={\displaystyle \frac{1}{2}}\left[{∥P_Q\left(A{y}_n\right)-Az∥}^2+{∥P_Q\left(A{y}_n\right)-A{y}_n∥}^2-{\parallel A{y}_n-Az\parallel }^2\right]\hfill \\ -{∥P_Q\left(A{y}_n\right)-A{y}_n∥}^2\hfill \\ \le -{\displaystyle \frac{1}{2}}{∥P_Q\left(A{y}_n\right)-A{y}_n∥}^2;\hfill \end{array}$$

(11)

we also prove

$$\begin{array}{c}{〈v_n-z,-\nabla U_{\tilde{E}}\left(v_n\right)〉}_B\le -{\displaystyle \frac{1}{2}}{∥P_Q\left(A{v}_n\right)-A{v}_n∥}^2.\hfill \end{array}$$

(12)

By the iterate $y_n$, we have

$$\begin{array}{c}{∥y_n-z∥}_B^2={∥P_C^B\left(v_n-{\Gamma }_n\nabla U_{\tilde{E}}\left(v_n\right)\right)-z∥}_B^2\hfill \\ \le {∥v_n-{\Gamma }_n\nabla U_{\tilde{E}}\left(v_n\right)-z∥}_B^2-{∥y_n-v_n+{\Gamma }_n\nabla U_{\tilde{E}}\left(v_n\right)∥}_B^2\hfill \\ ={∥v_n-z∥}_B^2-2{\Gamma }_n{\langle v_n-z,\nabla U_{\tilde{E}}\left(v_n\right)\rangle }_B-{\parallel y_n-v_n\parallel }_B^2-2{\Gamma }_n{\langle y_n-v_n,\nabla U_{\tilde{E}}\left(v_n\right)\rangle }_B.\hfill \end{array}$$

(13)

Now, we need to deduce the above part ${\langle y_n-v_n,\nabla U_{\tilde{E}}\left(v_n\right)\rangle }_B$. By the differentiable convexity of U, we obtain

$$\begin{array}{c}{\langle y_n-v_n,\nabla U_{\tilde{E}}\left(v_n\right)\rangle }_B=\langle y_n-v_n,\nabla U\left(v_n\right)\rangle \hfill \\ =\langle y_n-v_n,\nabla U\left(v_n\right)-\nabla U\left(y_n\right)\rangle +\langle y_n-v_n,\nabla U\left(y_n\right)\rangle \hfill \\ =-\langle y_n-v_n,\nabla U\left(y_n\right)-\nabla U\left(v_n\right)\rangle +\langle y_n-v_n,\nabla U\left(y_n\right)\rangle \hfill \\ \ge -\langle y_n-v_n,\nabla U\left(y_n\right)-\nabla U\left(v_n\right)\rangle +U\left(y_n\right)-U\left(v_n\right)\hfill \\ \ge -\langle y_n-v_n,\nabla U\left(y_n\right)-\nabla U\left(v_n\right)\rangle -U\left(v_n\right).\hfill \end{array}$$

(14)

After arrangement by substituting (12) and (14) into (13), we have

$$\begin{array}{c}{∥y_n-z∥}_B^2\le {∥v_n-z∥}_B^2+2{\Gamma }_n\langle y_n-v_n,\nabla U\left(y_n\right)-\nabla U\left(v_n\right)\rangle -{\parallel y_n-v_n\parallel }_B^2\hfill \\ \le {∥v_n-z∥}_B^2+2{\displaystyle \frac{{\Gamma }_n}{{\Gamma }_{n+1}}}{\Gamma }_{n+1}\langle y_n-v_n,\nabla U\left(y_n\right)-\nabla U\left(v_n\right)\rangle -{\parallel y_n-v_n\parallel }_B^2\hfill \\ \le {∥v_n-z∥}_B^2+2{\displaystyle \frac{{\Gamma }_n}{{\Gamma }_{n+1}}}\delta \parallel y_n-v_n\parallel -\parallel y_n-v_n{\parallel }_B^2\hfill \\ ={∥v_n-z∥}_B^2-\left({\kappa }_{\min }\left(B\right)-2{\displaystyle \frac{{\Gamma }_n}{{\Gamma }_{n+1}}}\delta \right){\parallel y_n-v_n\parallel }^2.\hfill \end{array}$$

(15)

Since

$$\underset{n\to \infty }{\lim }\left({\kappa }_{\min }\left(B\right)-2{\displaystyle \frac{{\Gamma }_n}{{\Gamma }_{n+1}}}\delta \right)={\kappa }_{\min }\left(B\right)-2\delta >0,$$

where $\delta \in \left(0,{\displaystyle \frac{{\kappa }_{\min }\left(B\right)}{2}}\right)$, then $\exists N\ge 0$, $\forall n\ge N$, ${\kappa }_{\min }\left(B\right)-2{\displaystyle \frac{{\Gamma }_n}{{\Gamma }_{n+1}}}\delta >0$. So, for any $n\ge N$,

$${∥y_n-z∥}_B^2\le {∥v_n-z∥}_B^2-\left({\kappa }_{\min }\left(B\right)-2\delta \right){\parallel y_n-v_n\parallel }^2.$$

(16)

By Lemma 1(i), we deduce from the iterative step $j_{n+1}$ that

$$\begin{array}{c}\parallel j_{n+1}{-z\parallel }_B^2={\parallel (1-{\alpha }_n)s_n+{\alpha }_n(y_n-{\gamma }_n\nabla U_{\tilde{E}}\left(y_n\right))-z\parallel }_B^2\hfill \\ =(1-{\alpha }_n)\parallel s_n{-z\parallel }_B^2+{\alpha }_n{\parallel y_n-{\gamma }_n\nabla U_{\tilde{E}}\left(y_n\right)-z\parallel }_B^2\hfill \\ -(1-{\alpha }_n){\alpha }_n{\parallel s_n-y_n+{\gamma }_n\nabla U_{\tilde{E}}\left(y_n\right)\parallel }_B^2\hfill \\ =(1-{\alpha }_n)\parallel s_n{-z\parallel }_B^2+{\alpha }_n(\parallel y_n{-z\parallel }_B^2+{\gamma }_n^2\parallel \nabla U_{\tilde{E}}\left(y_n\right){\parallel }_B^2-2{\gamma }_n{\langle y_n-z,\nabla U_{\tilde{E}}\left(y_n\right)\rangle }_B)\hfill \\ -(1-{\alpha }_n){\alpha }_n{\parallel s_n-y_n+{\gamma }_n\nabla U_{\tilde{E}}\left(y_n\right)\parallel }_B^2\hfill \end{array}$$

(17)

Substituting (11) and (16) into (17), and for any $n\ge N$, one has

$$\begin{array}{c}{∥j_{n+1}-z∥}_B^2\le (1-{\alpha }_n)\parallel s_n{-z\parallel }_B^2+{\alpha }_n(\parallel v_n{-z\parallel }_B^2-\left({\kappa }_{\min }\left(B\right)-2\delta \right)\parallel y_n-v_n{\parallel }^2)\hfill \\ +{\alpha }_n({\gamma }_n^2\parallel \nabla U_{\tilde{E}}\left(y_n\right){\parallel }_B^2-2{\gamma }_{n}U\left(y_n\right))-(1-{\alpha }_n){\alpha }_n{\parallel s_n-y_n+{\gamma }_n\nabla U_{\tilde{E}}\left(y_n\right)\parallel }_B^2\hfill \\ =(1-{\alpha }_n)\parallel s_n{-z\parallel }_B^2+{\alpha }_n\parallel v_n{-z\parallel }_B^2-{\alpha }_n\left({\kappa }_{\min }\left(B\right)-2\delta \right){\parallel y_n-v_n\parallel }^2\hfill \\ +{\alpha }_n\left({\displaystyle \frac{t^2{U}^2\left(y_n\right)}{\parallel \nabla U_{\tilde{E}}\left(y_n\right){\parallel }_B^4}}{\parallel \nabla U_{\tilde{E}}\left(y_n\right)\parallel }_B^2-2{\displaystyle \frac{tU\left(y_n\right)}{\parallel \nabla U_{\tilde{E}}\left(y_n\right){\parallel }_B^2}}U\left(y_n\right)\right)-(1-{\alpha }_n){\alpha }_n{\parallel s_n-y_n+{\gamma }_n\nabla U_{\tilde{E}}\left(y_n\right)\parallel }_B^2\hfill \\ =(1-{\alpha }_n)\parallel s_n{-z\parallel }_B^2+{\alpha }_n\parallel v_n{-z\parallel }_B^2-{\alpha }_n\left({\kappa }_{\min }\left(B\right)-2\delta \right){\parallel y_n-v_n\parallel }^2\hfill \\ -{\alpha }_{n}t(2-t){\displaystyle \frac{U^2\left(y_n\right)}{\parallel \nabla U_{\tilde{E}}\left(y_n\right){\parallel }_B^2}}-(1-{\alpha }_n){\alpha }_n{\parallel s_n-y_n+{\gamma }_n\nabla U_{\tilde{E}}\left(y_n\right)\parallel }_B^2.\hfill \end{array}$$

(18)

Next, the inertial steps $s_n$ and $v_n$ respectively yield that

$$\begin{array}{c}\parallel s_n{-z\parallel }_B^2={\parallel j_n+q_n(j_n-j_{n-1})-z\parallel }_B^2\hfill \\ =\parallel (1+q_n)(j_n-z)-q_n(j_{n-1}-z){\parallel }_B^2\hfill \\ =(1+q_n)\parallel j_n{-z\parallel }_B^2-q_n\parallel j_{n-1}{-z\parallel }_B^2+q_n(1+q_n){\parallel j_n-j_{n-1}\parallel }_B^2,\hfill \end{array}$$

(19)

and

$$\parallel v_n{-z\parallel }_B^2=(1+u_n)\parallel j_n{-z\parallel }_B^2-u_n\parallel j_{n-1}{-z\parallel }_B^2+u_n(1+u_n){\parallel j_n-j_{n-1}\parallel }_B^2.$$

(20)

Substituting (19) and (20) into (18), and for any $n\ge N$, we directly deduce that

$$\begin{array}{c}{∥j_{n+1}-z∥}_B^2\le (1-{\alpha }_n)((1+q_n)\parallel j_n{-z\parallel }_B^2-q_n\parallel j_{n-1}{-z\parallel }_B^2+q_n(1+q_n)\parallel j_n-j_{n-1}{\parallel }^2)\hfill \\ +{\alpha }_n((1+u_n)\parallel j_n{-z\parallel }_B^2-u_n\parallel j_{n-1}{-z\parallel }_B^2+u_n(1+u_n)\parallel j_n-j_{n-1}{\parallel }_B^2)\hfill \\ -{\alpha }_n\left({\kappa }_{\min }\left(B\right)-2\delta \right){\parallel y_n-v_n\parallel }^2\hfill \\ -{\alpha }_{n}t(2-t){\displaystyle \frac{U^2\left(y_n\right)}{\parallel \nabla U_{\tilde{E}}\left(y_n\right){\parallel }_B^2}}-(1-{\alpha }_n){\alpha }_n{\parallel s_n-y_n+{\gamma }_n\nabla U_{\tilde{E}}\left(y_n\right)\parallel }_B^2.\hfill \end{array}$$

(21)

Since

$$\begin{array}{c}{\alpha }_n^2{\parallel s_n-y_n+{\gamma }_n\nabla U_{\tilde{E}}\left(y_n\right)\parallel }_B^2={\parallel j_{n+1}-s_n\parallel }_B^2\hfill \\ =\parallel j_{n+1}-j_n-q_n(j_n-j_{n-1}){\parallel }_B^2\hfill \\ \ge \parallel j_{n+1}-j_n{\parallel }_B^2+q_n^2\parallel j_n-j_{n-1}{\parallel }_B^2-2q_n\parallel j_{n+1}-j_n{\parallel }_B{\parallel j_n-j_{n-1}\parallel }_B\hfill \\ \ge (1-q_n)\parallel j_{n+1}-j_n{\parallel }_B^2+(q_n^2-q_n){\parallel j_n-j_{n-1}\parallel }_B^2,\hfill \end{array}$$

(22)

after arrangement by substituting (22) into (21), and for any $n\ge N$, the obtained result is the following:

$$\begin{array}{c}{∥j_{n+1}-z∥}_B^2\le (1-{\alpha }_n)((1+q_n)\parallel j_n{-z\parallel }_B^2-q_n\parallel j_{n-1}{-z\parallel }_B^2+q_n(1+q_n)\parallel j_n-j_{n-1}{\parallel }_B^2)\hfill \\ +{\alpha }_n((1+u_n)\parallel j_n{-z\parallel }_B^2-u_n\parallel j_{n-1}{-z\parallel }_B^2+u_n(1+u_n)\parallel j_n-j_{n-1}{\parallel }_B^2)\hfill \\ -{\alpha }_n\left({\kappa }_{\min }\left(B\right)-2\delta \right){\parallel y_n-v_n\parallel }^2\hfill \\ -{\alpha }_{n}t(2-t){\displaystyle \frac{U^2\left(y_n\right)}{\parallel \nabla U_{\tilde{E}}\left(y_n\right){\parallel }_B^2}}-{\displaystyle \frac{1-{\alpha }_n}{{\alpha }_n}}((1-q_n)\parallel j_{n+1}-j_n{\parallel }_B^2+(q_n^2-q_n)\parallel j_n-j_{n-1}{\parallel }_B^2)\hfill \\ \le (1+(1-{\alpha }_n)q_n+{\alpha }_n{u}_n)\parallel j_n{-z\parallel }_B^2-((1-{\alpha }_n)q_n+{\alpha }_n{u}_n){\parallel j_{n-1}-z\parallel }_B^2\hfill \\ +\left((1-{\alpha }_n)(1+q_n)q_n+{\alpha }_n(1+u_n)u_n-{\displaystyle \frac{1-{\alpha }_n}{{\alpha }_n}}(q_n^2-q_n)\right){\parallel j_n-j_{n-1}\parallel }_B^2\hfill \\ -{\displaystyle \frac{1-{\alpha }_n}{{\alpha }_n}}(1-q_n){\parallel j_{n+1}-j_n\parallel }_B^2\hfill \\ \le (1+h_n)\parallel j_n{-z\parallel }_B^2-h_n\parallel j_{n-1}{-z\parallel }_B^2+g_n\parallel j_n-j_{n-1}{\parallel }_B^2-c_n{\parallel j_{n+1}-j_n\parallel }_B^2,\hfill \end{array}$$

(23)

where

$$0<t<2,h_n=(1-{\alpha }_n)q_n+{\alpha }_n{u}_n,g_n=(1-{\alpha }_n)(1+q_n)q_n+{\alpha }_n(1+u_n)u_n-{\displaystyle \frac{1-{\alpha }_n}{{\alpha }_n}}(q_n^2-q_n),c_n={\displaystyle \frac{1-{\alpha }_n}{{\alpha }_n}}(1-q_n).$$

Define

$${\Lambda }_n=\parallel j_n{-z\parallel }_B^2-h_n\parallel j_{n-1}{-z\parallel }_B^2+g_n{\parallel j_n-j_{n-1}\parallel }_B^2.$$

Since the sequence $h_n$ is non-decreasing, by (23), one obtains

$$\begin{array}{c}{\Lambda }_n-{\Lambda }_{n-1}=\parallel j_{n+1}{-z\parallel }_B^2-(1+h_{n+1})\parallel j_n{-z\parallel }_B^2+h_n{\parallel j_{n-1}-z\parallel }_B^2\hfill \\ -g_n\parallel j_n-j_{n-1}{\parallel }_B^2+g_{n+1}{\parallel j_{n+1}-j_n\parallel }_B^2\hfill \\ \le \parallel j_{n+1}{-z\parallel }_B^2-(1+h_n)\parallel j_n{-z\parallel }_B^2+h_n{\parallel j_{n-1}-z\parallel }_B^2\hfill \\ -g_n\parallel j_n-j_{n-1}{\parallel }_B^2+g_{n+1}{\parallel j_{n+1}-j_n\parallel }_B^2\hfill \\ \le -(c_n-g_{n+1}){\parallel j_{n+1}-j_n\parallel }_B^2.\hfill \end{array}$$

(24)

According to $A_4$, we discuss that

$$\begin{array}{c}{c}_n-g_{n+1}={\displaystyle \frac{1-{\alpha }_n}{{\alpha }_n}}(1-q_n)-(1-q_{n+1})(1+q_{n+1})q_{n+1}-q_{n+1}(1+u_{n+1})u_{n+1}\hfill \\ +{\displaystyle \frac{1-q_{n+1}}{q_{n+1}}}(q_{n+1}^2-q_{n+1})\hfill \\ \ge {\displaystyle \frac{1-q_{n+1}}{q_{n+1}}}(1-2q_{n+1}+q_{n+1}^2)-(1-q_{n+1})(q_{n+1}+q_{n+1}^2)-2q_{n+1}\hfill \\ \ge d(1-2q+q^2)-{\displaystyle \frac{d}{1+d}}(q+q^2)-{\displaystyle \frac{2}{1+d}}\hfill \\ ={\displaystyle \frac{1}{1+d}}(d^2{q}^2-(3d+2d^2)q+(d^2+d-2)).\hfill \end{array}$$

(25)

The work (24) with (25) implies that

$${\Lambda }_n-{\Lambda }_{n-1}\le -\tilde{\kappa }{\parallel j_{n+1}-j_n\parallel }_B^2,$$

(26)

where $\tilde{\kappa }={\displaystyle \frac{d^2{q}^2-(3d+2d^2)q+(d^2+d-2)}{1+d}}.$ We find that $q<{\displaystyle \frac{3+2d-\sqrt{8d+17}}{2d}},d\in (1,\infty ),$ and one has $\tilde{\kappa }>0.$ With the help of (26), we see that

$${\Lambda }_n-{\Lambda }_{n-1}\le 0.$$

(27)

Thereby, the sequence ${\Lambda }_n$ is non-increasing. We know from $A_4$ that $g_n>0.$ Therefore,

$$\begin{array}{c}{\Lambda }_n=\parallel j_n{-z\parallel }_B^2-h_n\parallel j_{n-1}{-z\parallel }_B^2+g_n{\parallel j_n-j_{n-1}\parallel }_B^2\hfill \\ \ge \parallel j_n{-z\parallel }_B^2-h_n{\parallel j_{n-1}-z\parallel }_B^2.\hfill \end{array}$$

(28)

This is the equivalent to the following analysis:

$$\begin{array}{c}\parallel j_n{-z\parallel }_B^2\le h_n{\parallel j_{n-1}-z\parallel }_B^2+{\Lambda }_n\hfill \\ \le h\parallel j_{n-1}{-z\parallel }_B^2+{\Lambda }_n\hfill \\ \le \cdots \le h^n{\parallel j_0-z\parallel }_B^2+{\Lambda }_1(1+h+\cdots +h^{n-1})\hfill \\ \le h^n{\parallel j_0-z\parallel }_B^2+{\displaystyle \frac{{\Lambda }_1}{1-h}},\hfill \end{array}$$

(29)

where $h={\displaystyle \frac{5+2d-\sqrt{8d+17}}{2+2d}}<1.$

Using the definition of the sequence ${\Lambda }_n$, it shows that

$$\begin{array}{c}{\Lambda }_{n+1}=\parallel j_{n+1}{-z\parallel }_B^2-h_{n+1}\parallel j_{n-1}{-z\parallel }_B^2+b_{n+1}{\parallel j_n-j_{n-1}\parallel }_B^2\hfill \\ \ge -h_{n+1}{\parallel j_n-z\parallel }_B^2.\hfill \end{array}$$

(30)

Combining (29) together with (30), we deduce that

$$\begin{array}{c}-{\Lambda }_{n+1}\le h_{n+1}\parallel j_n{-z\parallel }_B^2\le h\parallel j_n{-z\parallel }_B^2\le h^{n+1}{\parallel j_0-z\parallel }_B^2+{\displaystyle \frac{h{\Lambda }_1}{1-h}}.\hfill \end{array}$$

(31)

Using (26) and (31) again, we see that

$$\begin{array}{c}\tilde{\kappa }{\displaystyle \sum _{k=1}^n}\parallel j_{k+1}-j_k{\parallel }_B^2\le {\Lambda }_1-{\Lambda }_{n+1}\le h^{n+1}{\parallel j_0-z\parallel }_B^2+{\displaystyle \frac{{\Lambda }_1}{1-h}}\hfill \\ \le \parallel j_0{-z\parallel }_B^2+{\displaystyle \frac{{\Lambda }_1}{1-h}}.\hfill \end{array}$$

(32)

So, we conclude that

$$\begin{array}{c}{\displaystyle \sum _{k=1}^n}\parallel j_{k+1}-j_k{\parallel }_B^2<\infty .\hfill \end{array}$$

(33)

Thereby,

$$\begin{array}{c}{\displaystyle \underset{n\to \infty }{\lim }}{\parallel j_{n+1}-j_n\parallel }_B=0,\hfill \end{array}$$

(34)

which combined with the boundedness of $u_n,$ directly shows that

$$\begin{array}{c}\parallel j_{n+1}-v_n{\parallel }_B^2=\parallel j_{n+1}-j_n{\parallel }_B^2+u_n^2{\parallel j_n-j_{n-1}\parallel }_B^2\hfill \\ -2u_n\langle j_{n+1}-j_n,j_n-j_{n-1}\rangle \to 0\mathrm{as}n\to \infty .\hfill \end{array}$$

(35)

We also have

$$\begin{array}{c}\parallel j_n-v_n{\parallel }_B\le \parallel j_n-j_{n+1}{\parallel }_B+{\parallel j_{n+1}-v_n\parallel }_B\to 0\mathrm{as}n\to \infty .\hfill \end{array}$$

(36)

From (23), we know that

$$\begin{array}{c}{∥j_{n+1}-z∥}_B\le (1+h_n)\parallel j_n{-z\parallel }_B^2-h_n\parallel j_{n-1}{-z\parallel }_B^2+g_n\parallel j_n-j_{n-1}{\parallel }_B^2-c_n{\parallel j_{n+1}-j_n\parallel }_B^2\hfill \\ \le (1+h_n)\parallel j_n{-z\parallel }_B^2-h_n\parallel j_{n-1}{-z\parallel }_B^2+g_n{\parallel j_n-j_{n-1}\parallel }_B^2\hfill \\ =\parallel j_n{-z\parallel }_B^2+h_n(\parallel j_n{-z\parallel }_B^2-\parallel j_{n-1}{-z\parallel }_B^2)+g_n{\parallel j_n-j_{n-1}\parallel }_B^2\hfill \end{array}$$

(37)

Owing to $0\le h_n\le h<1$ and the boundedness of $g_n$, and by Lemma 2.5 and (33), we seek a point $g^{\ast }\in [0,\infty )$ such that

$$\underset{n\to \infty }{\lim }{\parallel j_n-z\parallel }_B=g^{\ast }.$$

(38)

In view of (19) and the boundedness of $q_n,$ we arrive at

$$\begin{array}{c}\parallel s_n{-z\parallel }_B^2=(1+q_n)\parallel j_n{-z\parallel }_B^2-q_n\parallel j_{n-1}{-z\parallel }_B^2+q_n(1+q_n){\parallel j_n-j_{n-1}\parallel }_B^2\hfill \\ =\parallel j_n{-z\parallel }_B^2+q_n(\parallel j_n{-z\parallel }_B^2-\parallel j_{n-1}{-z\parallel }_B^2)+q_n(1+q_n){\parallel j_n-j_{n-1}\parallel }_B^2\to g^{\ast }\mathrm{as}n\to \infty .\hfill \end{array}$$

(39)

Similarly, and by the boundedness of $u_n$, one gets

$$\underset{n\to \infty }{\lim }{\parallel v_n-z\parallel }_B=g^{\ast }.$$

(40)

The inequality (18) can be rewritten as

$$\begin{array}{c}{\alpha }_n\left(\left({\kappa }_{\min }\left(B\right)-2\delta \right)\parallel y_n-v_n{\parallel }^2+t(2-t){\displaystyle \frac{U^2\left(y_n\right)}{\parallel \nabla U_{\tilde{E}}\left(y_n\right){\parallel }_B^2}}+(1-{\alpha }_n){\parallel s_n-y_n+{\Lambda }_n\nabla U_{\tilde{E}}\left(y_n\right)\parallel }_B^2\right)\hfill \\ \le (1-{\alpha }_n)\parallel s_n{-z\parallel }_B^2+{\alpha }_n{\parallel v_n-z\parallel }_B^2-{∥j_{n+1}-z∥}_B.\hfill \end{array}$$

(41)

This work, together with (38)–(40), and $0<{\alpha }_n<{\displaystyle \frac{1}{1+d}},$ $d\in (1,\infty )$, shows that

$$\begin{array}{c}{\displaystyle \underset{n\to \infty }{\lim }}{\parallel y_n-v_n\parallel }^2=0,\hfill \\ {\displaystyle \underset{n\to \infty }{\lim }}{\displaystyle \frac{U^2\left(y_n\right)}{\parallel \nabla U_{\tilde{E}}\left(y_n\right){\parallel }_B^2}}=0,\hfill \\ {\displaystyle \underset{n\to \infty }{\lim }}{\parallel s_n-y_n+{\Lambda }_n\nabla U_{\tilde{E}}\left(y_n\right)\parallel }_B^2=0.\hfill \end{array}$$

(42)

Since the gradient mapping $\nabla U$ is Lipschitzian, $\nabla U_{\tilde{E}}$ is Lipschitzian too. Thus, $\parallel \nabla U_{\tilde{E}}\left(y_n\right)\parallel$ is bounded. Therefore,

$$\begin{array}{c}{\displaystyle \underset{n\to \infty }{\lim }}U\left(y_n\right)=0.\hfill \end{array}$$

(43)

By ${\displaystyle \underset{n\to \infty }{\lim }}{\parallel y_n-v_n\parallel }^2=0$, we get

$$\begin{array}{c}{\displaystyle \underset{n\to \infty }{\lim }}{\parallel y_n-v_n\parallel }_B^2=0.\hfill \end{array}$$

(44)

Let $\overline{z}\in C$ and $\overline{z}\in w_w\left(j_n\right)\subset \Omega$, and thus, by the boundedness of the sequence $j_n$, there exists a subsequence $j_{n_i}$ of $j_n$ such that $j_{n_i}\to \overline{z}$ as $i\to \infty .$ From (36) and (44), we know that $y_{n_i}\to \overline{z}$ as $i\to \infty .$ Using (43), one sees that

$$\begin{array}{c}{\displaystyle \underset{i\to \infty }{\lim }}U\left(y_{n_i}\right)=U\left(\overline{z}\right)=\parallel A\overline{z}-P_{Q}A\overline{z}\parallel =0,\hfill \end{array}$$

(45)

which indicates $A\overline{z}\in Q$, and from Proposition 1(ii) and Lemma 4, it proves that $\overline{z}\in \Omega .$ □

4. Numerical Experiments

In this section, some numerical experiments are providing for comparing our proposed Algorithm 1 (abbreviated, Alg.1) with algorithm 3.2 [18] (PCQ), algorithm 3.1 [31] (Sahu 2021), and algorithm 1 [37] (Suantai 2022). All codes were written by MATLAB R2018a and run on a PC Desktop Intel(R) Core(TM) i7-6700 CPU @ 3.40 GHZ computer with RAM 8.00 GB.

Example 1

(LASSO problem [38]). The concrete form of the LASSO problem is as follows:

$$\min \left\{{\displaystyle \frac{1}{2}}{\parallel Aj-V\parallel }^2:j\in {\mathbb{R}}^N,{\parallel j\parallel }_1\le \kappa \right\},$$

(46)

where $A\in {\mathbb{R}}^{M\times N},M<N,$$V\in {\mathbb{R}}^M$, and $\kappa >0.$ This problem is devoted to finding a sparse solution to the SFP. The sampling matrice A is produced by a standard normal distribution with mean zero and unit variance. The truly sparse signal $j^{\ast }$ is built by distributing W nonzero elements uniformly at random in the interval $[-2,2]$, while the rest is kept zero. The sample data $V=A{j}^{\ast }$.

In consideration of SFP, we define $C=\left\{\tilde{j}\right|\parallel \tilde{j}{\parallel }_1\le \kappa \},\kappa =W,$ and $Q=\left\{V\right\}$. We suggest the subgradient projection, because of the fact that there is no closed-form expression for the projection of a closed convex set C. Let $c\left(\tilde{j}\right)={\parallel \tilde{j}\parallel }_1-\kappa$ and

$$C_n=\{\tilde{j}:c\left(v_n\right)+\langle {\omega }_n,\tilde{j}-v_n\rangle \le 0\},$$

where ${\omega }_n\in \partial c\left(v_n\right)$. The orthogonal projection of a point $\tilde{p}\in {\mathbb{R}}^N$ onto $C_n$ can be achieved by

$$P_{C_n}\left(\tilde{j}\right)=\left\{\begin{array}{c}j,ifc\left(v_n\right)+\langle {\omega }_n,\tilde{j}-v_n\rangle \le 0,\hfill \\ \tilde{j}-{\displaystyle \frac{c\left(v_n\right)+\langle {\omega }_n,\tilde{j}-y_n\rangle }{\parallel {\omega }_n{\parallel }^2}}{\omega }_n,otherwise,\hfill \end{array}\right.$$

The subdifferential $\partial c$ at $v_n$ is

$$\partial c\left(v_n\right)=\left\{\begin{array}{cc}1,\hfill & if{v}_n>0,\hfill \\ [-1,1],\hfill & if{v}_n=0,\hfill \\ -1,\hfill & if{v}_n<0.\hfill \end{array}\right.$$

We initialize the algorithms at the original and define

$$E_n={\displaystyle \frac{\parallel j_n-j^{\ast }\parallel }{max\{1,\parallel j_n\parallel \}}}.$$

To test all methods under the same iteration error $E_n$ for different $M,N,andW$, we limit the number of iterations to 7000 and a smaller value of $E_n$; the algorithms behave better. Meanwhile, the corresponding results are reported in

Table 1. Results for Example 1.

$(\mathit{M},\mathit{N},\mathit{W})$	Algorithm 1		Sahu 2021 [31]		Suantai 2022 [37]		PCQ
	${\mathit{E}}_{\mathit{n}}$	Time	${\mathit{E}}_{\mathit{n}}$	Time	${\mathit{E}}_{\mathit{n}}$	Time	${\mathit{E}}_{\mathit{n}}$	Time
(240, 1024, 30)	5.4772	1.7094	1.3328$\times {10}^{-9}$	2.0715	4.2953$\times {10}^{-5}$	2.9206	6.8805$\times {10}^{-16}$	3.9255
(480, 2048, 60)	7.7460	5.9345	1.7791$\times {10}^{-11}$	7.8245	2.5638$\times {10}^{-5}$	14.2064	7.3943$\times {10}^{-16}$	32.5748
(720, 3072, 90)	9.4868	51.0043	6.8132$\times {10}^{-12}$	70.7382	2.1654$\times {10}^{-5}$	117.7369	6.3745$\times {10}^{-16}$	134.6671
(960, 4096, 120)	10.9545	46.2216	3.8179$\times {10}^{-11}$	58.2518	1.9982$\times {10}^{-5}$	92.2356	6.6815$\times {10}^{-16}$	132.1087

.

The second problem is the recovery of the signal $j^{\ast }$ when $M=960,N=4096,W=120$, and $M\times N$ matrix A is randomly obtained with independent samples of standard Gaussian distribution. The original signal $j^{\ast }$ contains 120 randomly placed $\pm 1$ spikes. The initial point $j_0=0$, and we adopt the following mean square error for measuring the restoration accuracy:

$$MSE={\displaystyle \frac{1}{N}}{\parallel j_n-j^{\ast }\parallel }^2.$$

The smaller the MSE, the better the signal recovery.

The same parameters for all algorithms were chosen in the following manner:

For Algorithm 1 $q_n=0.1-{\displaystyle \frac{1}{1000+n}}$, ${\alpha }_n=0.46-{\displaystyle \frac{1}{1000+n}}$, and ${\varphi }_n={\displaystyle \frac{1}{{(n+1)}^{1.1}}}+1$;

For PCQ, $\sigma =1$, $\rho =0.5$, $t=1.2$, and $0<\delta <{\displaystyle \frac{{\kappa }_{\min }\left(B\right)}{2}}$, which is suggested in [18];

For the Sahu 2021, $\theta =0.3$, and $\delta =0.2$, which is suggested in [31];

For Suantai 2022, we set $\sigma =1$, $\rho =0.5$, $\delta =0.3$, $t=2\times {10}^{-4}$, and $\theta =0.3$ in [37].

We present the original signal and restored signal, as shown in Figure 1. Then, we compare the CPU time and $E_n$ for all the algorithms as shown in Table 1.

Remark 2.

From Figure 1, all algorithms finish the signal recovery successfully; we observe that Algorithm 1 uses less CPU time to obtain a smaller MSE than Sahu 2021 [31], Suantai 2022 [37], and PCQ. Thereby, our proposed algorithm runs faster than the others. According to Table 1, the $E_n$ we obtain is larger than the other compared algorithms under the same parameters. However, it is difficult for our proposed algorithm to recover the signal, if we choose a larger or smaller parameter than the fixed parameter that properly makes our algorithm achieve the signal recovery.

5. Conclusions

In this paper, we use the double inertia and new stepsizes to modify the pre-conditioning CQ algorithm for solving SFP. The weak convergence of the algorithm is obtained under mild conditions. Then, we use the LASSO problem to show that our algorithm behaves well.