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On the convergences and applications of the inertial-like proximal point methods for null point problems

Yan Tang1,2, Shiqing Zhang*1
1College of Mathematics,Sichuan University, Chengdu,China
2College of Mathematics and Statistics, Chongqing Technology
and Business University,Chongqing, China
tangyan@ctbu.edu.cn; zhangshiqing@scu.edu.cn
Abstract

Motivated and inspired by the discretization of the nonsmooth system of a nonlinear oscillator with damping, we propose what we call the inertial-like proximal point algorithms for finding the null point of the sum of two maximal operators, which has many applied backgrounds, such as, convex optimization and variational inequality problems, compressed sensing etc.. The common feature of the presented algorithms is using the new inertial-like proximal point method which does not involve the computation for the norm of the difference between two adjacent iterates xnx_{n} and xn1x_{n-1} in advance, and avoids complex inertial parameters satisfying the traditional and difficult checking conditions. Numerical experiments are presented to illustrate the performances of the algorithms.


2000 Mathematics Subject Classification: 65K10; 65K05; 47H10; 47L25.

Keywords: null point problems; inertial-like proximal methods; maximal monotone operators.

1 Introduction

Let A:H2HA:H\rightarrow 2^{H} be a set-valued operator in real Hilbert space HH.

(1)  The graph of AA is given by

gphA={(x,y)H×H:yAx}.gphA=\{(x,y)\in H\times H:y\in Ax\}.

(2)  The operator AA is said to be monotone if

xy,uv0,uAx,vAy,x,yH.\langle x-y,u-v\rangle\geq 0,u\in Ax,v\in Ay,\,\,\,\forall x,y\in H.

(3)  The operator AA is said to be maximal monotone if gphAgphA is not properly contained in the graph of any other monotone operator. Let A,BA,B be two maximal monotone operators in HH. We are concern with the well-studied null point problem that is formulated as follows:

0(A+B)x\displaystyle 0\in(A+B)x^{*} (1.1)

with solution set denoted by Ω\Omega. A special interesting case of (1.1) is a minimization of sum of two proper, lower semi-continuous and convex functions f,g:Hf,g:H\rightarrow\mathbb{R}, that is,

minxH{f(x)+g(x)}.\displaystyle\min_{x\in H}\{f(x)+g(x)\}. (1.2)

So this is equivalent to (1.1) with A=fA=\partial f and B=gB=\partial g being the subdifferentials of ff and gg, respectively. We recall the resolvent operator JrA=(I+rA)1,r>0,J_{r}^{A}=(I+rA)^{-1},r>0, which is called the backward operator and plays an significant role in the approximation theory for zero points of maximal monotone operators. Due to the work of Aoyama et al. [3], we have the following properties:

JrAxy,xJrAx0,yA1(0),\displaystyle\langle J_{r}^{A}x-y,x-J_{r}^{A}x\rangle\geq 0,y\in A^{-1}(0), (1.3)

where A1(0)={zH:0Az}A^{-1}(0)=\{z\in H:0\in Az\}. Moreover, the following key facts that will be needed in the sequel.

Fact 1: The resolvent is not only always single-valued, but also firmly monotone:

xy2(IJrA)x(IJrA)y2JrAxJrAy2.\displaystyle\|x-y\|^{2}-\|(I-J_{r}^{A})x-(I-J_{r}^{A})y\|^{2}\geq\|J_{r}^{A}x-J_{r}^{A}y\|^{2}. (1.4)

Fact 2: Using the resolvent operator, we can write down inclusion problem (1.1) as a fixed point problem. It is known that

x=JλA(IλB)x,λ>0.\displaystyle x^{*}=J_{\lambda}^{A}(I-\lambda{B})x^{*},\lambda>0.

Douglas-Rachford‘s splitting method (DRSM) (or forward-backward splitting method) was first introduced in [10] as an operator splitting technique to solve partial differential equations arising in heat conduction and soon later has been extended to find solutions for the sum of two maximal monotone operators by Lions and Mercier [13]. Douglas-Rachford‘s splitting method(DRSM) is formulated as

xn+1=JλA(IλB)xn,x_{n+1}=J^{A}_{\lambda}(I-\lambda B)x_{n}, (1.5)

where λ>0\lambda>0, and (IλB)(I-\lambda B) is called the forward operator. Basing on the method (1.5), many researchers improved and modified the algorithms for the inclusion problem (1.1) and obtained nice results, see e.g., Boikanyo [7], Dadashi and Postolache[8], Kazmi and Rizvi [12], Moudafi[20][19], Sitthithakerngkiet et al. [27]. On the other hand, one classical way of looking at the null point problem 0Ax0\in Ax is to consider the forward discretization for dxdtxn+1xnhn,hn>0\frac{dx}{dt}\approx\frac{x_{n+1}-x_{n}}{h_{n}},\forall h_{n}>0 in the evolution system:

{dxdt+Ax(t)0,x(0)=x0,\displaystyle\begin{cases}\frac{dx}{dt}+Ax(t)\ni 0,\\ x(0)=x_{0},\end{cases} (1.6)

and then the evolution system is discretized as

xn+1xnhn+Axn+10(I+hnF)xn+1=xn,\frac{x_{n+1}-x_{n}}{h_{n}}+Ax_{n+1}\ni 0\,\,\Longleftrightarrow\,\,(I+h_{n}F)x_{n+1}=x_{n},

which inspired Alvarez and Attouch [2] to introduce the inertial method for the following nonsmooth case of a nonlinear oscillator with damping

d2xdt2+γdxdt+A(x(t))=0,a.e.t0.\displaystyle\frac{d^{2}x}{dt^{2}}+\gamma\frac{dx}{dt}+A(x(t))=0,a.e.t\geq 0. (1.7)

By discretizing, Alvarez and Attouch [2] obtained the implicit iterative sequence

xn+1xnαn(xnxn1)+λnA(xn+1)0,\displaystyle x_{n+1}-x_{n}-\alpha_{n}(x_{n}-x_{n-1})+\lambda_{n}A(x_{n+1})\ni 0, (1.8)

where αn=1γhn\alpha_{n}=1-\gamma h_{n} and λn=hn2\lambda_{n}=h_{n}^{2}, which yielded the Inertial-Prox algorithm

xn+1=JλnA(xn+αn(xnxn1)),\displaystyle x_{n+1}=J_{\lambda_{n}}^{A}(x_{n}+\alpha_{n}(x_{n}-x_{n-1})), (1.9)

the extrapolation term αn(xnxn1)\alpha_{n}(x_{n}-x_{n-1}) is called inertial term. Note that when αn0\alpha_{n}\equiv 0, the recursion (1.8) corresponds to the standard proximal iteration

xn+1xn+λnA(xn+1)0,\displaystyle x_{n+1}-x_{n}+\lambda_{n}A(x_{n+1})\ni 0,

which has been well studied by Martinet [15]and Moreau[17] and other researchers, and the weak convergence of xnx_{n} to a solution of 0Ax0\in Ax has being well known since the classical work of Rockafellar[25]. For ensuring the convergence of the Inertial-Prox sequence, Alvarez and Attouch[2] pointed the following key assumption: there exists α(0,1)\alpha\in(0,1) such that αn[0,α]\alpha_{n}\in[0,\alpha] and

n=1αnxnxn12<.\displaystyle\sum_{n=1}^{\infty}\alpha_{n}\|x_{n}-x_{n-1}\|^{2}<\infty. (1.10)

Inertial method is shown to have nice convergence properties in the field of continuous optimization and is studied intensively in split inverse problem by many authors soon later (because which could be utilized in some situations to accelerate the convergence of the sequences). For some recent works applied to various fields, see Alvarez [1], Attouch et al. [6, 5, 4], Ochs et al. [23, 22]. Although Alvarez and Attouch [2] pointed that one can choose appropriate rule to enable the the assumption (1.10) applicable, the parameter αn\alpha_{n} involving the iterates xnx_{n} and xn1x_{n-1} should be computed in advance. Namely, to make the condition (1.10) hold, the researchers have to set constraints on the inertia coefficient αn\alpha_{n} and estimate the value of the xnxn1\|x_{n}-x_{n-1}\| before choosing αn\alpha_{n}. In recent works, Gibali et al. [11] improved the inertial control condition by constraining inertial coefficient αn\alpha_{n} such that 0<αn<α¯n0<\alpha_{n}<\bar{\alpha}_{n}, where α(0,1),\alpha\in(0,1), ϵn[0,)\epsilon_{n}\in[0,\infty) and n=0ϵn<\sum_{n=0}^{\infty}\epsilon_{n}<\infty,

α¯n={min{α,ϵnxnxn12},ifxnxn1,α,otherwise.\displaystyle\bar{\alpha}_{n}=\begin{cases}\min\{\alpha,\frac{\epsilon_{n}}{\|x_{n}-x_{n-1}\|^{2}}\},&\hbox{if}\,\,x_{n}\neq x_{n-1},\\ \alpha,&\text{otherwise}.\end{cases}

More works on inertial methods, one can refer Dang et al. [9], Moudafi et al.[18], Suantai et al. [26], Tang [28], and therein. Theoretically, the condition (1.10) on the parameter αn\alpha_{n} is too strict for the convergence of the inertial algorithm. Practically, estimating the value of the xnxn1\|x_{n}-x_{n-1}\| before choosing the inertial parameter αn\alpha_{n} may need large amount of computation. The two drawbacks may make the Inertial-Prox method inconvenient in the practical test in the sense. So it is natural to think about the following question: Question 1.1 Can we delete the condition (1.10) in inertial method? Namely, can we construct a new inertial algorithm for solving (1.1) without any constraint on the the inertial parameter or the computation of norm of the difference between xnx_{n} and xn1x_{n-1} before choosing the inertial parameter? The purpose of this paper is to present an affirmative answer to the above question. In this paper, we study the convergence problem of a new inertial-like technique for the solution of the null point problem (1.1) without the assumption (1.10) and the prior computation of the xnxn1\|x_{n}-x_{n-1}\| before choosing the inertial parameter θn\theta_{n}. The outline of the paper is as follows. In section 2, we collect definitions and results which are needed for our further analysis. In section 3, our novel approach for the null point problem is introduced and analyzed, the convergence theorems of the presented algorithms are obtained. Moreover, convex optimization and variational inequality problem are studied as the applications of the null point problem in section 4. Finally, in section 5,some numerical experiments, using the inertial-like method, are carried out in order to support our approach.

2 Preliminaries

Let ,\langle\cdot,\cdot\rangle and \|\cdot\| be the inner product and the induced norm in a Hilbert space HH, respectively. For a sequence {xn}\{x_{n}\} in HH, denote xnxx_{n}\rightarrow x and xnxx_{n}\rightharpoonup x by the strong and weak convergence to xx of {xn}\{x_{n}\}, respectively. Moreover, the symbol ωw(xn)\omega_{w}(x_{n}) represents the ω\omega-weak limit set of {xn}\{x_{n}\}, that is,

ωw(xn):={xH:xnjxforsomesubsequence{xnj}of{xn}}.\displaystyle\omega_{w}(x_{n}):=\{x\in H:x_{n_{j}}\rightharpoonup x\hskip 5.69046pt\mathrm{for}\hskip 5.69046pt\mathrm{some}\hskip 5.69046pt\mathrm{subsequence}\hskip 5.69046pt\{x_{n_{j}}\}\hskip 5.69046pt\mathrm{of}\hskip 5.69046pt\{x_{n}\}\}.

The identity below is useful:

αx+βy+γz2\displaystyle\|\alpha x+\beta y+\gamma z\|^{2} =αx2+βy2+γz2\displaystyle=\alpha\|x\|^{2}+\beta\|y\|^{2}+\gamma\|z\|^{2}
αβxy2βγyz2γαxz2\displaystyle\quad-\alpha\beta\|x-y\|^{2}-\beta\gamma\|y-z\|^{2}-\gamma\alpha\|x-z\|^{2} (2.1)

for all x,y,zRx,y,z\in R and α+β+γ=1\alpha+\beta+\gamma=1.

Definition 2.1.

Let HH be a real Hilbert space, DHD\subset H and T:DHT:D\rightarrow H some given operator.

(1)  The operator TT is said to be Lipschitz continuous with constant κ>0\kappa>0 on DD if

T(x)T(y)κxy,x,yD.\|T(x)-T(y)\|\leq\kappa\|x-y\|,\,\,\,\forall x,y\in D.

(2)  The operator TT is said to be γ\gamma-cocoercive if there exists γ>0\gamma>0 such that

TxTy,xyγT(x)T(y)2,x,yD.\langle Tx-Ty,x-y\rangle\geq\gamma\|T(x)-T(y)\|^{2},\,\,\,\forall x,y\in D.
Remark 2.2.

(1)  If TT is γ\gamma-cocoercive, then it is 1γ\frac{1}{\gamma}-Lipschitz continuous.

(2)  From Fact 1, we can conclude that JrAJ_{r}^{A} is a non-expansive operator if AA is a maximal monotone mapping.

Definition 2.3.

Let CC be a nonempty closed convex subset of HH. We use PCP_{C} to denote the projection from HH onto CC; namely,

PCx=argmin{xy:yC},xH.\displaystyle P_{C}x=\arg\min\{\|x-y\|:y\in C\},\quad x\in H.

The following significant characterization of the projection PCP_{C} should be recalled : Given xHx\in H and yCy\in C,

PCx=zxz,yz0,yC.\displaystyle P_{C}x=z\quad\Longleftrightarrow\quad\langle x-z,y-z\rangle\leq 0,\quad y\in C. (2.2)
Lemma 2.4.

(Xu [29]) Assume that {an}\{a_{n}\} is a sequence of nonnegative real numbers such that

an+1(1βn)an+βnbn+cn,n0,\displaystyle a_{n+1}\leq(1-\beta_{n})a_{n}+\beta_{n}b_{n}+c_{n},\quad n\geq 0,

where {βn}\{\beta_{n}\} is a sequence in (0,1)(0,1) and {cn}(0,)\{c_{n}\}\subset(0,\infty) and {bn}\{b_{n}\}\subset\mathbb{R} such that

  • (1)

    n=1βn=\sum_{n=1}^{\infty}\beta_{n}=\infty;

  • (2)

    lim supnbn0\limsup_{n\rightarrow\infty}b_{n}\leq 0 or n=1βn|bn|<\sum_{n=1}^{\infty}\beta_{n}|b_{n}|<\infty;

  • (3)

    n=1cn<\sum_{n=1}^{\infty}c_{n}<\infty.

Then limnan=0\lim_{n\rightarrow\infty}a_{n}=0.

Lemma 2.5.

(see e.g., Opial [21]) Let HH be a real Hilbert space and {xn}\{x_{n}\} be a bounded sequence in HH. Assume there exists a nonempty subset SHS\subset H satisfying the properties:

  • (i)

    limnxnz\lim_{n\rightarrow\infty}\|x_{n}-z\| exists for every zSz\in S,

  • (ii)

    ωw(xn)S\omega_{w}(x_{n})\subset S.

Then, there exists x¯S\bar{x}\in S such that {xn}\{x_{n}\} converges weakly to x¯\bar{x}.

Lemma 2.6.

(Maingé [16]) Let {Γn}\{\Gamma_{n}\} be a sequence of real numbers that does not decrease at the infinity in the sense that there exists a subsequence {Γnj}\{\Gamma_{n_{j}}\} of {Γn}\{\Gamma_{n}\} such that Γnj<Γnj+1\Gamma_{n_{j}}<\Gamma_{n_{j}+1} for all j0j\geq 0. Also consider the sequence of integers {σ(n)}nn0\{\sigma(n)\}_{n\geq n_{0}} defined by

σ(n)=max{kn:ΓkΓk+1}.\displaystyle\sigma(n)=\max\{k\leq n:\Gamma_{k}\leq\Gamma_{k+1}\}.

Then {σ(n)}nn0\{\sigma(n)\}_{n\geq n_{0}} is a nondecreasing sequence verifying limnσ(n)=\lim_{n\rightarrow\infty}\sigma(n)=\infty and, for all nn0n\geq n_{0},

max{Γσ(n),Γn}Γσ(n)+1.\displaystyle\max\{\Gamma_{\sigma(n)},\Gamma_{n}\}\leq\Gamma_{\sigma(n)+1}.

3 Main Results

3.1 Motivation of Inertial-Like Proximal Technique

Inspired and motivated by the discretization (1.8), we consider the following iterative sequence

xn+1xn1θn(xnxn1)+λnA(xn+1)0,\displaystyle x_{n+1}-x_{n-1}-\theta_{n}(x_{n}-x_{n-1})+\lambda_{n}A(x_{n+1})\ni 0,

where x0x_{0}, x1x_{1} are two arbitrary initial points, and {λn}\{\lambda_{n}\} is a real nonnegative number sequence. This recursion can be rewritten as

{wn=xn1+θn(xnxn1)xn+1=JλnAwn.\displaystyle\begin{cases}w_{n}=x_{n-1}+\theta_{n}(x_{n}-x_{n-1})\\ x_{n+1}=J_{\lambda_{n}}^{A}w_{n}.\end{cases} (3.1)

The discretization sequence {xn}\{x_{n}\} in (3.1) always exists because the sequence {xn}\{x_{n}\} satisfying (1.8) always exists for any choice of the sequence {αn}\{\alpha_{n}\} according to Alvarez and Attouch [2]. In addition, it can be deduced that if θn=1+αn\theta_{n}=1+\alpha_{n}, the formula (3.1) can cover Alvarez and Attouch‘s Inertial-Prox algorithm. More relevantly, the inertial coefficient θn\theta_{n} can be considered 1 in our new inertial proximal point algorithms. We thus obtain what we call the Inertial-Like Proximal point algorithm.

3.2 Some Conventions and Inertial-like Proximal Algorithms

C1:Throughout the rest of this paper, we always assume that HH is a Hilbert space. We rephrase the null point problem as follows:

0(A+B)x\displaystyle 0\in(A+B)x^{*} (3.2)

where A,B:H2HA,B:H\rightarrow 2^{H} are two maximal monotone set-valued operators with BB γ\gamma-cocoercive.
C2: Denote by Ω\Omega the solution set of the null point problem; namely,

Ω={xH:0(A+B)x}\displaystyle\Omega=\left\{x^{*}\in H:0\in(A+B)x^{*}\right\}

and we always assume Ω\Omega\not=\emptyset.

Now, combining the Fact 2 and the inertial-like technique (3.1), we introduce the following algorithms.

Algorithm 3.1

Initialization: Choose a positive sequence {θn}[0,1]\{\theta_{n}\}\subset[0,1]. Select arbitrary initial points x0x_{0}, x1x_{1}.

Iterative Step: After the nn-iterate xnx_{n} is constructed, compute

wn=xn1+θn(xnxn1),n1,\displaystyle w_{n}=x_{n-1}+\theta_{n}(x_{n}-x_{n-1}),\hskip 5.69046ptn\geq 1, (3.3)

and define the (n+1)(n+1)th iterate by

xn+1=JτnA(IτnB)wn.\displaystyle x_{n+1}=J_{\tau_{n}}^{A}(I-\tau_{n}B)w_{n}. (3.4)

Remark 3.1 It is not hard to find that if wn=xn+1w_{n}=x_{n+1}, for some n0n\geq 0, then xn+1x_{n+1} is a solution of the inclusion problem (3.2), and the iteration process is terminated in finite iterations. If θn=1\theta_{n}=1, Algorithm 3.1 reduces to the general forward-backward algorithm in Moudafi[19].

Algorithm 3.2

Initialization: Choose a sequence {θn}[0,1]\{\theta_{n}\}\subset[0,1] satisfying one of the three cases: (I.) θn(0,1)\theta_{n}\in(0,1) such that lim¯nθn(1θn)>0\underline{\lim}_{n\rightarrow\infty}\theta_{n}(1-\theta_{n})>0; (II.) θn0\theta_{n}\equiv 0; (III.) θn1\theta_{n}\equiv 1. Choose {αn}\{\alpha_{n}\} and {βn}\{\beta_{n}\} in (0,1)(0,1) such that

lim¯nαn>0;lim¯nαn<1;limnβn=0,n=0βn=.\displaystyle\underline{\lim}_{n\rightarrow\infty}\alpha_{n}>0;\overline{\lim}_{n\rightarrow\infty}\alpha_{n}<1;\hskip 5.69046pt\lim_{n\rightarrow\infty}\beta_{n}=0,\quad\sum_{n=0}^{\infty}\beta_{n}=\infty.

Select arbitrary initial points x0x_{0}, x1x_{1}.

Iterative Step: After the nn-iterate xnx_{n} is constructed, compute

wn=xn1+θn(xnxn1),\displaystyle w_{n}=x_{n-1}+\theta_{n}(x_{n}-x_{n-1}),\hskip 5.69046pt

and define the (n+1)(n+1)th iterate by

xn+1=(1αnβn)wn+αnJτnA(IτnB)wn.\displaystyle x_{n+1}=(1-\alpha_{n}-\beta_{n})w_{n}+\alpha_{n}J_{\tau_{n}}^{A}(I-\tau_{n}B)w_{n}. (3.5)

Remark 3.2. In the subsequent convergence analysis, we will always assume that the two algorithms generate an infinite sequence, namely, the algorithms are not terminated in finite iterations. In addition, in the simulation experiments, in order to be practical, we will give a stop criterion to end the iteration for practice. Otherwise, set n:=n+1n:=n+1 and return to Iterative Step.

3.3 Convergence Analysis of Algorithms

Theorem 3.1.

If the assumptions 𝐂𝟏𝐂𝟐{\bf C1}-{\bf C2} are satisfied and τn(ϵ,2γϵ)\tau_{n}\in(\epsilon,2\gamma-\epsilon) for some given ϵ>0\epsilon>0 small enough, we obtain the weak convergence result, namely the sequence {xn}\{x_{n}\} generated by Algorithm 3.1 converges weakly to a point x¯Ω\bar{x}\in\Omega.

Proof.

Without loss of generality, we take zΩz\in\Omega and then we have get z=JτnA(IτnB)zz=J_{\tau_{n}}^{A}(I-\tau_{n}B)z from Fact 2. It turns out from (2) and (3.3) that

wnz2\displaystyle\|w_{n}-z\|^{2} =\displaystyle= xn1+θn(xnxn1)z2\displaystyle\|x_{n-1}+\theta_{n}(x_{n}-x_{n-1})-z\|^{2} (3.6)
=\displaystyle= θnxnz2+(1θn)xn1z2θn(1θn)xnxn12.\displaystyle\theta_{n}\|x_{n}-z\|^{2}+(1-\theta_{n})\|x_{n-1}-z\|^{2}-\theta_{n}(1-\theta_{n})\|x_{n}-x_{n-1}\|^{2}.

Since JrAJ_{r}^{A} is firmly nonexpansive, it follows from (3.4) and Fact 1 that

xn+1z2\displaystyle\|x_{n+1}-z\|^{2} =\displaystyle= JτnA(IτnB)wnz2\displaystyle\|J_{\tau_{n}}^{A}(I-\tau_{n}B)w_{n}-z\|^{2} (3.7)
\displaystyle\leq (IτnB)wn(IτnB)z2(IJτnA)(IτnB)wn(IJτnA)(IτnB)z2\displaystyle\|(I-\tau_{n}B)w_{n}-(I-\tau_{n}B)z\|^{2}-\|(I-J_{\tau_{n}}^{A})(I-\tau_{n}B)w_{n}-(I-J_{\tau_{n}}^{A})(I-\tau_{n}B)z\|^{2}
=\displaystyle= (wnz)τn(BwnBz)2wnxn+1τn(BwnBz)2\displaystyle\|(w_{n}-z)-\tau_{n}(Bw_{n}-Bz)\|^{2}-\|w_{n}-x_{n+1}-\tau_{n}(Bw_{n}-Bz)\|^{2}
=\displaystyle= wnz22wnz,τn(BwnBz)+τn2BwnBz2\displaystyle\|w_{n}-z\|^{2}-2\langle w_{n}-z,\tau_{n}(Bw_{n}-Bz)\rangle+\tau_{n}^{2}\|Bw_{n}-Bz\|^{2}
wnxn+12+2wnxn+1,τn(BwnBz)τn2BwnBz2\displaystyle-\|w_{n}-x_{n+1}\|^{2}+2\langle w_{n}-x_{n+1},\tau_{n}(Bw_{n}-Bz)\rangle-\tau_{n}^{2}\|Bw_{n}-Bz\|^{2}
=\displaystyle= wnz2wnxn+122wnz,τn(BwnBz)\displaystyle\|w_{n}-z\|^{2}-\|w_{n}-x_{n+1}\|^{2}-2\langle w_{n}-z,\tau_{n}(Bw_{n}-Bz)\rangle
+2τnwnxn+1,BwnBz.\displaystyle+2\tau_{n}\langle w_{n}-x_{n+1},Bw_{n}-Bz\rangle.

It follows from the fact BB is γ\gamma-cocoercive that

BwnBz,wnzγBwnBz2,\displaystyle\langle Bw_{n}-Bz,w_{n}-z\rangle\geq\gamma\|Bw_{n}-Bz\|^{2},

so we have from (3.7) that

xn+1z2\displaystyle\|x_{n+1}-z\|^{2} \displaystyle\leq wnz2wnxn+122γτnBwnBz2\displaystyle\|w_{n}-z\|^{2}-\|w_{n}-x_{n+1}\|^{2}-2\gamma\tau_{n}\|Bw_{n}-Bz\|^{2}
+2τnwnxn+1,BwnBz.\displaystyle+2\tau_{n}\langle w_{n}-x_{n+1},Bw_{n}-Bz\rangle.

On the other hand, we have

2γτnBwnBzwnxn+12γ2\displaystyle 2\gamma\tau_{n}\Big{\|}Bw_{n}-Bz-\frac{w_{n}-x_{n+1}}{2\gamma}\Big{\|}^{2}
=\displaystyle= 2γτn(BwnBz2+wnxn+12γ22BwnBz,wnxn+12γ)\displaystyle 2\gamma\tau_{n}\Big{(}\|Bw_{n}-Bz\|^{2}+\Big{\|}\frac{w_{n}-x_{n+1}}{2\gamma}\Big{\|}^{2}-2\Big{\langle}Bw_{n}-Bz,\frac{w_{n}-x_{n+1}}{2\gamma}\Big{\rangle}\Big{)}
=\displaystyle= 2γτnBwnBz2+τn2γwnxn+122τnBwnBz,wnxn+1\displaystyle 2\gamma\tau_{n}\|Bw_{n}-Bz\|^{2}+\frac{\tau_{n}}{2\gamma}\|w_{n}-x_{n+1}\|^{2}-2\tau_{n}\langle Bw_{n}-Bz,w_{n}-x_{n+1}\rangle

and, furthermore,

2γτnBwnBz22τnBwnBz,wnxn+1\displaystyle 2\gamma\tau_{n}\|Bw_{n}-Bz\|^{2}-2\tau_{n}\langle Bw_{n}-Bz,w_{n}-x_{n+1}\rangle
=\displaystyle= 2γτnBwnBzwnxn+12γ2τn2γwnxn+12.\displaystyle 2\gamma\tau_{n}\Big{\|}Bw_{n}-Bz-\frac{w_{n}-x_{n+1}}{2\gamma}\Big{\|}^{2}-\frac{\tau_{n}}{2\gamma}\|w_{n}-x_{n+1}\|^{2}.

Hence we obtain from (3.6), (3.7) that

xn+1z2\displaystyle\|x_{n+1}-z\|^{2} \displaystyle\leq wnz2wnxn+122γτnBwnBz2\displaystyle\|w_{n}-z\|^{2}-\|w_{n}-x_{n+1}\|^{2}-2\gamma\tau_{n}\|Bw_{n}-Bz\|^{2} (3.8)
+2τnwnxn+1,BwnBz\displaystyle+2\tau_{n}\langle w_{n}-x_{n+1},Bw_{n}-Bz\rangle
=\displaystyle= wnz22γτnBwnBzwnxn+12γ2\displaystyle\|w_{n}-z\|^{2}-2\gamma\tau_{n}\Big{\|}Bw_{n}-Bz-\frac{w_{n}-x_{n+1}}{2\gamma}\Big{\|}^{2}
+(τn2γ1)wnxn+12\displaystyle+(\frac{\tau_{n}}{2\gamma}-1)\|w_{n}-x_{n+1}\|^{2}
=\displaystyle= θnxnz2+(1θn)xn1z2θn(1θn)xnxn12\displaystyle\theta_{n}\|x_{n}-z\|^{2}+(1-\theta_{n})\|x_{n-1}-z\|^{2}-\theta_{n}(1-\theta_{n})\|x_{n}-x_{n-1}\|^{2}
2γτnBwnBzwnxn+12γ2+(τn2γ1)wnxn+12.\displaystyle-2\gamma\tau_{n}\Big{\|}Bw_{n}-Bz-\frac{w_{n}-x_{n+1}}{2\gamma}\Big{\|}^{2}+(\frac{\tau_{n}}{2\gamma}-1)\|w_{n}-x_{n+1}\|^{2}.

Since τn(ϵ,2γϵ)\tau_{n}\leq(\epsilon,2\gamma-\epsilon), it follows from (3.8) that

xn+1z2\displaystyle\|x_{n+1}-z\|^{2} \displaystyle\leq θnxnz2+(1θn)xn1z2\displaystyle\theta_{n}\|x_{n}-z\|^{2}+(1-\theta_{n})\|x_{n-1}-z\|^{2}
\displaystyle\leq max{xnz2,xn1z2}\displaystyle\max\{\|x_{n}-z\|^{2},\|x_{n-1}-z\|^{2}\}
\displaystyle\leq \displaystyle\cdots
\displaystyle\leq max{x1z2,x0z2},\displaystyle\max\{\|x_{1}-z\|^{2},\|x_{0}-z\|^{2}\},

which means that the sequence {xnz}\{\|x_{n}-z\|\} is bounded, and so in turn {wn}\{w_{n}\} is. Now we claim that there exists the limit of the sequence {xnz}\{\|x_{n}-z\|\}. For this purpose, two situations are discussed as follows:. Case 1. There exists an integer N00N_{0}\geq 0 such that xn+1zxnz\|x_{n+1}-z\|\leq\|x_{n}-z\| for all nN0n\geq N_{0}. Then there exists the limit of the sequence {xnz}\{\|x_{n}-z\|\}, denoted by l=limnxnz2l=\lim_{n\rightarrow}\|x_{n}-z\|^{2}, and so

limn(xnz2xn+1z2)=0.\lim_{n\rightarrow\infty}(\|x_{n}-z\|^{2}-\|x_{n+1}-z\|^{2})=0.

In addition, we have

n=0(xn+1z2xnz2)=limn(xn+1zx0z)<\displaystyle\sum_{n=0}^{\infty}(\|x_{n+1}-z\|^{2}-\|x_{n}-z\|^{2})=\lim_{n\rightarrow\infty}(\|x_{n+1}-z\|-\|x_{0}-z\|)<\infty

and therefore from (3.8) we get

limnθn(1θn)xnxn12(xnz2xn+1z2)+(1θn)(xn1z2xnz2)\displaystyle\lim_{n\rightarrow}\theta_{n}(1-\theta_{n})\|x_{n}-x_{n-1}\|^{2}\leq(\|x_{n}-z\|^{2}-\|x_{n+1}-z\|^{2})+(1-\theta_{n})(\|x_{n-1}-z\|^{2}-\|x_{n}-z\|^{2})

and so

limnθn(1θn)xnxn12=0,n=0θn(1θn)xnxn12<.\displaystyle\lim_{n\rightarrow}\theta_{n}(1-\theta_{n})\|x_{n}-x_{n-1}\|^{2}=0,\quad\sum_{n=0}^{\infty}\theta_{n}(1-\theta_{n})\|x_{n}-x_{n-1}\|^{2}<\infty.

Now it remains to show that

ωw(xn)Ω.\omega_{w}(x_{n})\subset\Omega.

Since the sequence {xn}\{x_{n}\} is bounded, let x¯ωw(xn)\bar{x}\in\omega_{w}(x_{n}) and {xnk}\{x_{n_{k}}\} be a subsequence of {xn}\{x_{n}\} weakly converging to x¯\bar{x}. To this end, it remains to verify that x¯(A+B)1(0)\bar{x}\in(A+B)^{-1}(0). Notice again (3.8), we have

2γτnBwnBzwnxn+12γ2+(1τn2γ)wnxn+12+θn(1θn)xnxn12\displaystyle 2\gamma\tau_{n}\Big{\|}Bw_{n}-Bz-\frac{w_{n}-x_{n+1}}{2\gamma}\Big{\|}^{2}+(1-\frac{\tau_{n}}{2\gamma})\|w_{n}-x_{n+1}\|^{2}+\theta_{n}(1-\theta_{n})\|x_{n}-x_{n-1}\|^{2}
\displaystyle\leq θnxnz2+(1θn)xn1z2xn+1z2\displaystyle\theta_{n}\|x_{n}-z\|^{2}+(1-\theta_{n})\|x_{n-1}-z\|^{2}-\|x_{n+1}-z\|^{2}
=\displaystyle= (xnz2xn+1z2)+(1θn)(xn1z2xnz2),\displaystyle(\|x_{n}-z\|^{2}-\|x_{n+1}-z\|^{2})+(1-\theta_{n})(\|x_{n-1}-z\|^{2}-\|x_{n}-z\|^{2}),

which means that

BwnBzwnxn+12γ20;\Big{\|}Bw_{n}-Bz-\frac{w_{n}-x_{n+1}}{2\gamma}\Big{\|}^{2}\rightarrow 0;

and

wnxn+120.\|w_{n}-x_{n+1}\|^{2}\rightarrow 0.

At the same time, it follows from (3.4) that

wnxn+1τn(BwnBxn+1)(A+B)xn+1,\displaystyle\frac{w_{n}-x_{n+1}}{\tau_{n}}-(Bw_{n}-Bx_{n+1})\in(A+B)x_{n+1}, (3.9)

Since BB is γ\gamma-cocoercive, we have that BB is 1γ\frac{1}{\gamma}-Lipschitz continuous. Passing to the limit on the subsequence {xnk}\{x_{n_{k}}\} of {xn}\{x_{n}\} converging weakly to x¯\bar{x} in (3.9) and taking account that A+BA+B is maximal monotone and thus its graph is weakly-strongly closed, it follows that

0(A+B)x¯,0\in(A+B)\bar{x},

which means that x¯Ω\bar{x}\in\Omega. In view of the fact that the choice of x¯\bar{x} in ωw(xn)\omega_{w}(x_{n}) was arbitrary, namely ωw(xn)Ω\omega_{w}(x_{n})\subset\Omega, and we conclude from Lemma 2.5 that {xn}\{x_{n}\} converges weakly to x¯Ω\bar{x}\in\Omega. Case 2. If the sequence {xnkz}\{\|x_{n_{k}}-z\|\} does not decrease at infinity in the sense, then there exists a sub-sequence {nk}\{n_{k}\} of {n}\{n\} such that xnkzxnk+1z\|x_{n_{k}}-z\|\leq\|x_{{n_{k}}+1}-z\| for all k0.k\geq 0. Furthermore, by Lemma 2.6, there exists an integer, non-decreasing sequence σ(n)\sigma(n) for nN1n\geq N_{1} (for some N1N_{1} large enough) such that σ(n)\sigma(n)\rightarrow\infty as nn\rightarrow\infty,

xσ(n)zxσ(n)+1z,xnzxσ(n)+1z\|x_{\sigma(n)}-z\|\leq\|x_{{\sigma(n)}+1}-z\|,\hskip 2.84544pt\|x_{n}-z\|\leq\|x_{{\sigma(n)}+1}-z\|

for each n0n\geq 0. Notice the boundedness of the sequence {xnz}\{\|x_{n}-z\|\}, which implies that there exists the limit of the sequence {xσ(n)z}\{\|x_{\sigma(n)}-z\|\} and hence we conclude that

limn(xσ(n)+1z2xσ(n)z2)=0.\displaystyle\lim_{n\rightarrow\infty}(\|x_{\sigma(n)+1}-z\|^{2}-\|x_{\sigma(n)}-z\|^{2})=0.

From (3.8) with nn replaced by σ(n)\sigma(n), we have

xσ(n)+1z2xσ(n)z2\displaystyle\|x_{\sigma(n)+1}-z\|^{2}-\|x_{\sigma(n)}-z\|^{2}
\displaystyle\leq (θσ(n)1)(xσ(n)z2xσ(n)1z2)θσ(n)(1θσ(n))xσ(n)xσ(n)12\displaystyle(\theta_{\sigma(n)}-1)(\|x_{\sigma(n)}-z\|^{2}-\|x_{\sigma(n)-1}-z\|^{2})-\theta_{\sigma(n)}(1-\theta_{\sigma(n)})\|x_{\sigma(n)}-x_{\sigma(n)-1}\|^{2}
2γτσ(n)Bwσ(n)Bzwσ(n)xσ(n)+12γ2+(τσ(n)2γ1)wσ(n)xσ(n)+12.\displaystyle-2\gamma\tau_{\sigma(n)}\Big{\|}Bw_{\sigma(n)}-Bz-\frac{w_{\sigma(n)}-x_{\sigma(n)+1}}{2\gamma}\Big{\|}^{2}+(\frac{\tau_{\sigma(n)}}{2\gamma}-1)\|w_{\sigma(n)}-x_{\sigma(n)+1}\|^{2}.

By a similar argument to Case 1, we obtain

Bwσ(n)Bzwσ(n)xσ(n)+12γ20;\Big{\|}Bw_{\sigma(n)}-Bz-\frac{w_{\sigma(n)}-x_{\sigma(n)+1}}{2\gamma}\Big{\|}^{2}\rightarrow 0;

and

wσ(n)xσ(n)+120.\|w_{\sigma(n)}-x_{\sigma(n)+1}\|^{2}\rightarrow 0.

These together with (3.4) implies that

wσ(n)xσ(n)+1τσ(n)(Bwσ(n)Bxσ(n)+1)(A+B)xσ(n)+1.\displaystyle\frac{w_{\sigma(n)}-x_{{\sigma(n)}+1}}{\tau_{\sigma(n)}}-(Bw_{\sigma(n)}-Bx_{{\sigma(n)}+1})\in(A+B)x_{{\sigma(n)}+1}.

Since the sequence {xn}\{x_{n}\} is bounded, let x¯ωw(xn)\bar{x}\in\omega_{w}(x_{n}) and {xσ(n)}\{x_{\sigma(n)}\} be a subsequence of {xn}\{x_{n}\} weakly converging to x¯\bar{x}. Passing to the limit on the subsequence {xσ(n)}\{x_{\sigma(n)}\} of {xn}\{x_{n}\} converging weakly to x¯\bar{x} in the above inequality and taking account that A+BA+B is maximal monotone and thus its graph is weakly-strongly closed, it follows that

0(A+B)x¯,0\in(A+B)\bar{x},

which means that x¯Ω\bar{x}\in\Omega. In view of the fact that the choice of x¯\bar{x} in ωw(xn)\omega_{w}(x_{n}) was arbitrary, namely ωw(xn)Ω\omega_{w}(x_{n})\subset\Omega, and we conclude from Lemma 2.5 that {xn}\{x_{n}\} converges weakly to x¯Ω\bar{x}\in\Omega. This completes the proof. ∎

Next we prove the strong convergence of Algorithm 3.2.

Theorem 3.2.

If the assumptions 𝐂𝟏𝐂𝟐{\bf C1}-{\bf C2} are satisfied and τn(ϵ,2γϵ)\tau_{n}\in(\epsilon,2\gamma-\epsilon) for some given ϵ>0\epsilon>0 small enough, we obtain the strong convergence result, namely the sequence {xn}\{x_{n}\} generated by Algorithm 3.2 converges in norm to z=PΩ(0)z=P_{\Omega}(0) (i.e., the minimum-norm element of the solution set Ω\Omega).

Proof.

To illustrate the result clearly, the following three situations should be discussed: (I).  θn(0,1)\theta_{n}\in(0,1); (II). θn0\theta_{n}\equiv 0 and (III). θn1\theta_{n}\equiv 1, respectively. (I). First we consider the strong convergence under the assumption θn(0,1)\theta_{n}\in(0,1). Similar to the previous proof of weak convergence, we begin by showing the boundedness of the sequence {xn}\{x_{n}\}. To see this, we denote zn=JτnA(IτnB)wnz_{n}=J_{\tau_{n}}^{A}(I-\tau_{n}B)w_{n} and we use the projection z:=PΩ(0)z:=P_{\Omega}(0) in a similar way to the proof of (3.7) and (3.8) of Theorem 3.1 to get that

znz2\displaystyle\|z_{n}-z\|^{2} \displaystyle\leq wnz2wnzn22wnz,τn(BwnBz)\displaystyle\|w_{n}-z\|^{2}-\|w_{n}-z_{n}\|^{2}-2\langle w_{n}-z,\tau_{n}(Bw_{n}-Bz)\rangle (3.10)
+2τnwnzn,BwnBz\displaystyle+2\tau_{n}\langle w_{n}-z_{n},Bw_{n}-Bz\rangle
\displaystyle\leq wnz2+(τn2γ1)wnzn22γτnBwnBzwnzn2γ2,\displaystyle\|w_{n}-z\|^{2}+(\frac{\tau_{n}}{2\gamma}-1)\|w_{n}-z_{n}\|^{2}-2\gamma\tau_{n}\Big{\|}Bw_{n}-Bz-\frac{w_{n}-z_{n}}{2\gamma}\Big{\|}^{2},

hence one can see znzwnz\|z_{n}-z\|\leq\|w_{n}-z\|. It turns out from (3.3) and (3.5) that

xn+1z\displaystyle\|x_{n+1}-z\| =\displaystyle= (1αnβn)wn+αnznz\displaystyle\|(1-\alpha_{n}-\beta_{n})w_{n}+\alpha_{n}z_{n}-z\|
=\displaystyle= (1αnβn)(wnz)+αn(znz)+βn(z)\displaystyle\|(1-\alpha_{n}-\beta_{n})(w_{n}-z)+\alpha_{n}(z_{n}-z)+\beta_{n}(-z)\|
\displaystyle\leq (1αnβn)wnz+αnznz+βnz\displaystyle(1-\alpha_{n}-\beta_{n})\|w_{n}-z\|+\alpha_{n}\|z_{n}-z\|+\beta_{n}\|z\|
\displaystyle\leq (1βn)[θnxnz+(1θn)xn1z]+βnz\displaystyle(1-\beta_{n})[\theta_{n}\|x_{n}-z\|+(1-\theta_{n})\|x_{n-1}-z\|]+\beta_{n}\|z\|
\displaystyle\leq (1βn)(max{xnz,xn1z})+βnz\displaystyle(1-\beta_{n})(\max\{\|x_{n}-z\|,\|x_{n-1}-z\|\})+\beta_{n}\|z\|
\displaystyle\leq \displaystyle\cdots
\displaystyle\leq max{x0z,x1z,z},\displaystyle\max\{\|x_{0}-z\|,\|x_{1}-z\|,\|z\|\},

which implies that the sequence {xn}\{x_{n}\} is bounded, and so are the sequences {wn}\{w_{n}\}, {zn}\{z_{n}\}. Applying the identity (2) we deduce that

xn+1z2\displaystyle\|x_{n+1}-z\|^{2} =\displaystyle= (1αnβn)wn+αnznz2\displaystyle\|(1-\alpha_{n}-\beta_{n})w_{n}+\alpha_{n}z_{n}-z\|^{2} (3.11)
=\displaystyle= (1αnβn)(wnz)+αn(znz)+βn(z)2\displaystyle\|(1-\alpha_{n}-\beta_{n})(w_{n}-z)+\alpha_{n}(z_{n}-z)+\beta_{n}(-z)\|^{2}
\displaystyle\leq (1αnβn)wnz2+αnznz2+βnz2\displaystyle(1-\alpha_{n}-\beta_{n})\|w_{n}-z\|^{2}+\alpha_{n}\|z_{n}-z\|^{2}+\beta_{n}\|z\|^{2}
(1αnβn)αnznwn2.\displaystyle-(1-\alpha_{n}-\beta_{n})\alpha_{n}\|z_{n}-w_{n}\|^{2}.

Substituting (3.10) into (3.11) and after some manipulations, we obtain

xn+1z2\displaystyle\|x_{n+1}-z\|^{2} \displaystyle\leq (1αnβn)wnz2+αn[wnz2+(τn2γ1)wnzn2\displaystyle(1-\alpha_{n}-\beta_{n})\|w_{n}-z\|^{2}+\alpha_{n}[\|w_{n}-z\|^{2}+(\frac{\tau_{n}}{2\gamma}-1)\|w_{n}-z_{n}\|^{2}
2γτnBwnBzwnzn2γ2]+βnz2\displaystyle-2\gamma\tau_{n}\Big{\|}Bw_{n}-Bz-\frac{w_{n}-z_{n}}{2\gamma}\Big{\|}^{2}]+\beta_{n}\|z\|^{2}
(1αnβn)αnznwn2.\displaystyle-(1-\alpha_{n}-\beta_{n})\alpha_{n}\|z_{n}-w_{n}\|^{2}.

Combining (3.6) we have

xn+1z2\displaystyle\|x_{n+1}-z\|^{2} \displaystyle\leq (1βn)wnz2++βnz2(1αnβn)αnznwn2\displaystyle(1-\beta_{n})\|w_{n}-z\|^{2}++\beta_{n}\|z\|^{2}-(1-\alpha_{n}-\beta_{n})\alpha_{n}\|z_{n}-w_{n}\|^{2} (3.12)
+αn[(τn2γ1)wnzn22γτnBwnBzwnzn2γ2]\displaystyle+\alpha_{n}[(\frac{\tau_{n}}{2\gamma}-1)\|w_{n}-z_{n}\|^{2}-2\gamma\tau_{n}\Big{\|}Bw_{n}-Bz-\frac{w_{n}-z_{n}}{2\gamma}\Big{\|}^{2}]
=\displaystyle= (1βn)[θnxnz2+(1θn)xn1z2θn(1θn)xnxn12]\displaystyle(1-\beta_{n})[\theta_{n}\|x_{n}-z\|^{2}+(1-\theta_{n})\|x_{n-1}-z\|^{2}-\theta_{n}(1-\theta_{n})\|x_{n}-x_{n-1}\|^{2}]
+βnz2+αn(τn2γ2+αn+βn)wnzn2\displaystyle+\beta_{n}\|z\|^{2}+\alpha_{n}(\frac{\tau_{n}}{2\gamma}-2+\alpha_{n}+\beta_{n})\|w_{n}-z_{n}\|^{2}
2γτnαnBwnBzwnzn2γ2.\displaystyle-2\gamma\tau_{n}\alpha_{n}\Big{\|}Bw_{n}-Bz-\frac{w_{n}-z_{n}}{2\gamma}\Big{\|}^{2}.

We explain the strong convergence under two situations, respectively. Case 1. There exists an integer N00N_{0}\geq 0 such that xn+1zxnz\|x_{n+1}-z\|\leq\|x_{n}-z\| for all nN0n\geq N_{0}. Then there exists the limit of the sequence {xnz}\{\|x_{n}-z\|\}, denoted by l=limnxnz2l=\lim_{n\rightarrow}\|x_{n}-z\|^{2}, and so

limn(xnz2xn+1z2)=0.\lim_{n\rightarrow\infty}(\|x_{n}-z\|^{2}-\|x_{n+1}-z\|^{2})=0.

In addition, we have

n=0(xn+1z2xnz2)=limn(xn+1zx0z)<\displaystyle\sum_{n=0}^{\infty}(\|x_{n+1}-z\|^{2}-\|x_{n}-z\|^{2})=\lim_{n\rightarrow\infty}(\|x_{n+1}-z\|-\|x_{0}-z\|)<\infty

and therefore from (3.12) we obtain

αn(2τn2γαnβn)wnzn2+2γτnαnBwnBzwnzn2γ2\displaystyle\alpha_{n}(2-\frac{\tau_{n}}{2\gamma}-\alpha_{n}-\beta_{n})\|w_{n}-z_{n}\|^{2}+2\gamma\tau_{n}\alpha_{n}\Big{\|}Bw_{n}-Bz-\frac{w_{n}-z_{n}}{2\gamma}\Big{\|}^{2}
+θn(1θn)(1βn)xnxn12\displaystyle+\theta_{n}(1-\theta_{n})(1-\beta_{n})\|x_{n}-x_{n-1}\|^{2}
\displaystyle\leq (1βn)[θnxnz2+(1θn)xn1z2]+βnz2xn+1z2\displaystyle(1-\beta_{n})[\theta_{n}\|x_{n}-z\|^{2}+(1-\theta_{n})\|x_{n-1}-z\|^{2}]+\beta_{n}\|z\|^{2}-\|x_{n+1}-z\|^{2}
\displaystyle\leq θnxnz2+(1θn)xn1z2+βnz2xn+1z2\displaystyle\theta_{n}\|x_{n}-z\|^{2}+(1-\theta_{n})\|x_{n-1}-z\|^{2}+\beta_{n}\|z\|^{2}-\|x_{n+1}-z\|^{2}
=\displaystyle= xnz2xn+1z2+(1θn)(xn1z2xnz2)+βnz2,\displaystyle\|x_{n}-z\|^{2}-\|x_{n+1}-z\|^{2}+(1-\theta_{n})(\|x_{n-1}-z\|^{2}-\|x_{n}-z\|^{2})+\beta_{n}\|z\|^{2},

and then from the condition βn0\beta_{n}\rightarrow 0, we get

limnθn(1θn)(1βn)xnxn12=0;\displaystyle\lim_{n\rightarrow\infty}\theta_{n}(1-\theta_{n})(1-\beta_{n})\|x_{n}-x_{n-1}\|^{2}=0;
limnαn(2τn2γαnβn)wnzn2=0;\displaystyle\lim_{n\rightarrow\infty}\alpha_{n}(2-\frac{\tau_{n}}{2\gamma}-\alpha_{n}-\beta_{n})\|w_{n}-z_{n}\|^{2}=0;
limnγτnαnBwnBzwnzn2γ2=0.\displaystyle\lim_{n\rightarrow\infty}\gamma\tau_{n}\alpha_{n}\Big{\|}Bw_{n}-Bz-\frac{w_{n}-z_{n}}{2\gamma}\Big{\|}^{2}=0.

Notice αn(2τn2γαnβn)=αn(1αnβn)+αn(1τn2γ)\alpha_{n}(2-\frac{\tau_{n}}{2\gamma}-\alpha_{n}-\beta_{n})=\alpha_{n}(1-\alpha_{n}-\beta_{n})+\alpha_{n}(1-\frac{\tau_{n}}{2\gamma}) and notice the assumptions on the parameters θn,βn,τn\theta_{n},\beta_{n},\tau_{n} and αn\alpha_{n}, hence we have

limnxnxn12=0;limnwnzn2=0;\displaystyle\lim_{n\rightarrow\infty}\|x_{n}-x_{n-1}\|^{2}=0;\hskip 5.69046pt\lim_{n\rightarrow\infty}\|w_{n}-z_{n}\|^{2}=0;
limnBwnBzwnzn2γ2=0,\displaystyle\lim_{n\rightarrow\infty}\Big{\|}Bw_{n}-Bz-\frac{w_{n}-z_{n}}{2\gamma}\Big{\|}^{2}=0,

which imply that wnxn=|1θn|xnxn10\|w_{n}-x_{n}\|=|1-\theta_{n}|\cdot\|x_{n}-x_{n-1}\|\rightarrow 0 and xn+1wnαnznwn+βnwn0\|x_{n+1}-w_{n}\|\leq\alpha_{n}\|z_{n}-w_{n}\|+\beta_{n}\|w_{n}\|\to 0 as nn\rightarrow\infty, and then

xn+1xnxn+1wn+wnxn0,n\|x_{n+1}-x_{n}\|\leq\|x_{n+1}-w_{n}\|+\|w_{n}-x_{n}\|\to 0,n\rightarrow\infty

This proves the asymptotic regularity of {xn}\{x_{n}\}. By repeating the relevant part of the proof of Theorem 3.1, we get ωw(xn)Ω\omega_{w}(x_{n})\subset\Omega. It is now at the position to prove the sequence {xn}\{x_{n}\} strongly converges to z=PΩ(0)z=P_{\Omega}(0). To this end, we rewrite (3.5) as xn+1=(1βn)vn+βnαn(znwn)x_{n+1}=(1-\beta_{n})v_{n}+\beta_{n}\alpha_{n}(z_{n}-w_{n}), where vn=(1αn)wn+αnznv_{n}=(1-\alpha_{n})w_{n}+\alpha_{n}z_{n}, and making use of the inequality u+v2u2+2v,u+v\|u+v\|^{2}\leq\|u\|^{2}+2\langle v,u+v\rangle which holds for all u,vHu,v\in H, we get

xn+1z2\displaystyle\|x_{n+1}-z\|^{2} =\displaystyle= (1βn)(vnz)+βn(αn(znwn)z)2\displaystyle\|(1-\beta_{n})(v_{n}-z)+\beta_{n}(\alpha_{n}(z_{n}-w_{n})-z)\|^{2}
\displaystyle\leq (1βn)2vnz2+2βnαn(znwn)z,xn+1z.\displaystyle(1-\beta_{n})^{2}\|v_{n}-z\|^{2}+2\beta_{n}\Big{\langle}\alpha_{n}(z_{n}-w_{n})-z,x_{n+1}-z\Big{\rangle}.

It follows from (2) that

vnz2=(1αn)wnz2+αnznz2αn(1αn)znwn2,\displaystyle\|v_{n}-z\|^{2}=(1-\alpha_{n})\|w_{n}-z\|^{2}+\alpha_{n}\|z_{n}-z\|^{2}-\alpha_{n}(1-\alpha_{n})\|z_{n}-w_{n}\|^{2},

and then

xn+1z2\displaystyle\|x_{n+1}-z\|^{2} \displaystyle\leq (1βn)2[(1αn)wnz2+αnznz2αn(1αn)znwn2]\displaystyle(1-\beta_{n})^{2}[(1-\alpha_{n})\|w_{n}-z\|^{2}+\alpha_{n}\|z_{n}-z\|^{2}-\alpha_{n}(1-\alpha_{n})\|z_{n}-w_{n}\|^{2}]
+2βnαn(znwn)z,xn+1z.\displaystyle+2\beta_{n}\Big{\langle}\alpha_{n}(z_{n}-w_{n})-z,x_{n+1}-z\Big{\rangle}.

Notice that znz2wnz2\|z_{n}-z\|^{2}\leq\|w_{n}-z\|^{2} from (3.10), and notice (1βn)2<1βn(1-\beta_{n})^{2}<1-\beta_{n}, hence we obtain

xn+1z2\displaystyle\|x_{n+1}-z\|^{2} \displaystyle\leq (1βn)wnz2αn(1αn)(1βn)2znwn2\displaystyle(1-\beta_{n})\|w_{n}-z\|^{2}-\alpha_{n}(1-\alpha_{n})(1-\beta_{n})^{2}\|z_{n}-w_{n}\|^{2}
+2βnαn(znwn)z,xn+1z\displaystyle+2\beta_{n}\Big{\langle}\alpha_{n}(z_{n}-w_{n})-z,x_{n+1}-z\Big{\rangle}

Submitting (3.6) into the above inequality, we have

xn+1z2\displaystyle\|x_{n+1}-z\|^{2} (1βn)[θnxnz2+(1θn)xn1z2θn(1θn)xnxn12]\displaystyle\leq(1-\beta_{n})[\theta_{n}\|x_{n}-z\|^{2}+(1-\theta_{n})\|x_{n-1}-z\|^{2}-\theta_{n}(1-\theta_{n})\|x_{n}-x_{n-1}\|^{2}] (3.13)
αn(1αn)(1βn)2znwn2+2βnαn(znwn)z,xn+1z,\displaystyle-\alpha_{n}(1-\alpha_{n})(1-\beta_{n})^{2}\|z_{n}-w_{n}\|^{2}+2\beta_{n}\Big{\langle}\alpha_{n}(z_{n}-w_{n})-z,x_{n+1}-z\Big{\rangle},

and then noticing the xn+1zxnz\|x_{n+1}-z\|\leq\|x_{n}-z\| for all nN0+1n\geq N_{0}+1, we have

xn+1z2\displaystyle\|x_{n+1}-z\|^{2} \displaystyle\leq (1βn)xn1z2θn(1θn)(1βn)xnxn12\displaystyle(1-\beta_{n})\|x_{n-1}-z\|^{2}-\theta_{n}(1-\theta_{n})(1-\beta_{n})\|x_{n}-x_{n-1}\|^{2} (3.14)
αn(1αn)(1βn)2znwn2+2βnαn(znwn)z,xn+1z\displaystyle-\alpha_{n}(1-\alpha_{n})(1-\beta_{n})^{2}\|z_{n}-w_{n}\|^{2}+2\beta_{n}\Big{\langle}\alpha_{n}(z_{n}-w_{n})-z,x_{n+1}-z\Big{\rangle}
\displaystyle\leq (1βn)xn1z2+2βnαn(znwn)z,xn+1z\displaystyle(1-\beta_{n})\|x_{n-1}-z\|^{2}+2\beta_{n}\Big{\langle}\alpha_{n}(z_{n}-w_{n})-z,x_{n+1}-z\Big{\rangle}
=\displaystyle= (1βn)xnz2(1βn)xnz2+(1βn)xn1z2\displaystyle(1-\beta_{n})\|x_{n}-z\|^{2}-(1-\beta_{n})\|x_{n}-z\|^{2}+(1-\beta_{n})\|x_{n-1}-z\|^{2}
+2βnαn(znwn)z,xn+1z\displaystyle+2\beta_{n}\Big{\langle}\alpha_{n}(z_{n}-w_{n})-z,x_{n+1}-z\Big{\rangle}
=\displaystyle= (1βn)xnz2+(1βn)(xn1z2xnz2)\displaystyle(1-\beta_{n})\|x_{n}-z\|^{2}+(1-\beta_{n})(\|x_{n-1}-z\|^{2}-\|x_{n}-z\|^{2})
+2βnαn(znwn)z,xn+1z\displaystyle+2\beta_{n}\Big{\langle}\alpha_{n}(z_{n}-w_{n})-z,x_{n+1}-z\Big{\rangle}
=\displaystyle= (1βn)an+βnbn+cn,\displaystyle(1-\beta_{n})a_{n}+\beta_{n}b_{n}+c_{n},

where an=xnz2,bn=2[αnznwn,xn+1z+z,xn+1z]a_{n}=\|x_{n}-z\|^{2},\quad b_{n}=2[\alpha_{n}\langle z_{n}-w_{n},x_{n+1}-z\rangle+\langle-z,x_{n+1}-z\rangle\,] and cn=(1βn)(xn1z2xnz2)c_{n}=(1-\beta_{n})(\|x_{n-1}-z\|^{2}-\|x_{n}-z\|^{2}). Since ωw(xn)Ω\omega_{w}(x_{n})\subset\Omega and z=PΩ(0)z=P_{\Omega}(0), which implies from (2.2) that z,qz0\langle-z,q-z\rangle\leq 0 for all qΩq\in\Omega, we deduce that

lim supnz,xn+1z=maxqωw(xn)z,qz0.\limsup_{n\rightarrow\infty}\langle-z,x_{n+1}-z\rangle=\max_{q\in\omega_{w}(x_{n})}\langle-z,q-z\rangle\leq 0. (3.15)

Since wnzn0\|w_{n}-z_{n}\|\rightarrow 0, lim¯nαn<1\overline{\lim}_{n\rightarrow\infty}\alpha_{n}<1 and xn+1z\|x_{n+1}-z\| is bounded, combining (3.15) implies that

lim supnbn\displaystyle\limsup_{n\to\infty}b_{n} =lim supn2{αnznwn,xn+1z+z,xn+1z}\displaystyle=\limsup_{n\to\infty}2\{\alpha_{n}\langle z_{n}-w_{n},x_{n+1}-z\rangle+\langle-z,x_{n+1}-z\rangle\}
=lim supn2z,xn+1z0.\displaystyle=\limsup_{n\to\infty}2\langle-z,x_{n+1}-z\rangle\leq 0.

In addition, we have

n=1cnn=1(xn1z2xnz2)<.\sum_{n=1}^{\infty}c_{n}\leq\sum_{n=1}^{\infty}(\|x_{n-1}-z\|^{2}-\|x_{n}-z\|^{2})<\infty.

These enable us to apply Lemma 2.4 to (3.14) to obtain that an0a_{n}\to 0. Namely, xnzx_{n}\to z in norm and the proof of Case 1 is complete.

Case 2. If the sequence {xnz}\{\|x_{n}-z\|\} does not decrease at infinity in the sense that there exists a sub-sequence {nk}\{n_{k}\} of {n}\{n\} such that xnkzxnk+1z\|x_{n_{k}}-z\|\leq\|x_{{n_{k}}+1}-z\| for all k0.k\geq 0. Furthermore, by Lemma 2.6, there exists an integer, non-decreasing sequence σ(n)\sigma(n) for nN1n\geq N_{1} (for some N1N_{1} large enough) such that σ(n)\sigma(n)\rightarrow\infty as nn\rightarrow\infty,

xσ(n)zxσ(n)+1z,xnzxσ(n)+1z\|x_{\sigma(n)}-z\|\leq\|x_{{\sigma(n)}+1}-z\|,\hskip 2.84544pt\|x_{n}-z\|\leq\|x_{{\sigma(n)}+1}-z\|

for each n0n\geq 0. Notice the boundedness of the sequence {xnz}\{\|x_{n}-z\|\}, which implies that there exists the limit of the sequence {xσ(n)z}\{\|x_{\sigma(n)}-z\|\} and hence we conclude that

limn(xσ(n)+1z2xσ(n)z2)=0.\displaystyle\lim_{n\rightarrow\infty}(\|x_{\sigma(n)+1}-z\|^{2}-\|x_{\sigma(n)}-z\|^{2})=0.

As a matter of fact, observe that (3.12) holds for each nn, so from (3.12) with nn replaced by σ(n)\sigma(n) and using the relation xσ(n)z2xσ(n)+1z2\|x_{\sigma(n)}-z\|^{2}\leq\|x_{\sigma(n)+1}-z\|^{2}, we have

θσ(n)(1θσ(n))(1βσ(n))xσ(n)xσ(n)12+ασ(n)(2τσ(n)2γασ(n)βσ(n))wσ(n)zσ(n)2\displaystyle\theta_{\sigma(n)}(1-\theta_{\sigma(n)})(1-\beta_{\sigma(n)})\|x_{\sigma(n)}-x_{{\sigma(n)}-1}\|^{2}+\alpha_{\sigma(n)}(2-\frac{\tau_{\sigma(n)}}{2\gamma}-\alpha_{\sigma(n)}-\beta_{\sigma(n)})\|w_{\sigma(n)}-z_{\sigma(n)}\|^{2}
+2γτσ(n)ασ(n)Bwσ(n)Bzwσ(n)zσ(n)2γ2\displaystyle+2\gamma\tau_{\sigma(n)}\alpha_{\sigma(n)}\Big{\|}Bw_{\sigma(n)}-Bz-\frac{w_{\sigma(n)}-z_{\sigma(n)}}{2\gamma}\Big{\|}^{2}
\displaystyle\leq (1βσ(n))[θσ(n)xσ(n)z2+(1θσ(n))xσ(n)1z2]xσ(n)+1z2+βσ(n)z2\displaystyle(1-\beta_{\sigma(n)})[\theta_{\sigma(n)}\|x_{\sigma(n)}-z\|^{2}+(1-\theta_{\sigma(n)})\|x_{{\sigma(n)}-1}-z\|^{2}]-\|x_{{\sigma(n)}+1}-z\|^{2}+\beta_{\sigma(n)}\|z\|^{2}
\displaystyle\leq (1βσ(n))xσ(n)z2xσ(n)+1z2+βσ(n)z2\displaystyle(1-\beta_{\sigma(n)})\|x_{\sigma(n)}-z\|^{2}-\|x_{{\sigma(n)}+1}-z\|^{2}+\beta_{\sigma(n)}\|z\|^{2}
=\displaystyle= xσ(n)z2xσ(n)+1z2βσ(n)(xσ(n)z2z2)\displaystyle\|x_{\sigma(n)}-z\|^{2}-\|x_{{\sigma(n)}+1}-z\|^{2}-\beta_{\sigma(n)}(\|x_{\sigma(n)}-z\|^{2}-\|z\|^{2})

Notice the assumptions on the parameters θσ(n),βσ(n),τσ(n)\theta_{\sigma(n)},\beta_{\sigma(n)},\tau_{\sigma(n)} and ασ(n)\alpha_{\sigma(n)}, Taking the limit by letting nn\to\infty yields

limnwσ(n)zσ(n)\displaystyle\lim_{n\to\infty}\|w_{\sigma(n)}-z_{\sigma(n)}\| =0;\displaystyle=0; (3.16)
limnxσ(n)xσ(n)12\displaystyle\lim_{n\to\infty}\|x_{\sigma(n)}-x_{\sigma(n)-1}\|^{2} =0;\displaystyle=0; (3.17)
limnBwσ(n)Bzwσ(n)zσ(n)2γ2\displaystyle\lim_{n\to\infty}\Big{\|}Bw_{\sigma(n)}-Bz-\frac{w_{\sigma(n)}-z_{\sigma(n)}}{2\gamma}\Big{\|}^{2} =0.\displaystyle=0. (3.18)

Note that we still have xσ(n)+1xσ(n)0\|x_{\sigma(n)+1}-x_{\sigma(n)}\|\to 0 and then that the relations (3.16)-(3.18) are sufficient to guarantee that ωw(xσ(n))Ω\omega_{w}(x_{\sigma(n)})\subset\Omega. Next we prove xσ(n)zx_{\sigma(n)}\to z. Observe that (3.13) holds for each nn. So replacing nn with σ(n)\sigma(n) in (3.13) and using the relation xσ(n)z2xσ(n)+1z2\|x_{\sigma(n)}-z\|^{2}\leq\|x_{\sigma(n)+1}-z\|^{2} again for nN1n\geq N_{1}, we obtain

xσ(n)+1z2\displaystyle\|x_{\sigma(n)+1}-z\|^{2} =\displaystyle= (1βσ(n))[θσ(n)xσ(n)z2+(1θσ(n))xσ(n)1z2\displaystyle(1-\beta_{\sigma(n)})[\theta_{\sigma(n)}\|x_{\sigma(n)}-z\|^{2}+(1-\theta_{\sigma(n)})\|x_{\sigma(n)-1}-z\|^{2}
θσ(n)(1θσ(n))xσ(n)xσ(n)12]\displaystyle-\theta_{\sigma(n)}(1-\theta_{\sigma(n)})\|x_{\sigma(n)}-x_{\sigma(n)-1}\|^{2}]
ασ(n)(1ασ(n))(1βσ(n))2zσ(n)wσ(n)2\displaystyle-\alpha_{\sigma(n)}(1-\alpha_{\sigma(n)})(1-\beta_{\sigma(n)})^{2}\|z_{\sigma(n)}-w_{\sigma(n)}\|^{2}
+2βσ(n)ασ(n)(zσ(n)wσ(n))z,xσ(n)+1z\displaystyle+2\beta_{\sigma(n)}\Big{\langle}\alpha_{\sigma(n)}(z_{\sigma(n)}-w_{\sigma(n)})-z,x_{\sigma(n)+1}-z\Big{\rangle}
\displaystyle\leq (1βσ(n))xσ(n)z2+2βσ(n)ασ(n)(zσ(n)wσ(n))z,xσ(n)+1z,\displaystyle(1-\beta_{\sigma(n)})\|x_{\sigma(n)}-z\|^{2}+2\beta_{\sigma(n)}\Big{\langle}\alpha_{\sigma(n)}(z_{\sigma(n)}-w_{\sigma(n)})-z,x_{\sigma(n)+1}-z\Big{\rangle},

and then we obtain

βσ(n))xσ(n)z2\displaystyle\beta_{\sigma(n)})\|x_{\sigma(n)}-z\|^{2} \displaystyle\leq xσ(n)z2xσ(n)+1z2+2βσ(n)ασ(n)(zσ(n)wσ(n))z,xσ(n)+1z\displaystyle\|x_{\sigma(n)}-z\|^{2}-\|x_{\sigma(n)+1}-z\|^{2}+2\beta_{\sigma(n)}\Big{\langle}\alpha_{\sigma(n)}(z_{\sigma(n)}-w_{\sigma(n)})-z,x_{\sigma(n)+1}-z\Big{\rangle}
\displaystyle\leq 2βσ(n)ασ(n)(zσ(n)wσ(n))z,xσ(n)+1z,\displaystyle 2\beta_{\sigma(n)}\Big{\langle}\alpha_{\sigma(n)}(z_{\sigma(n)}-w_{\sigma(n)})-z,x_{\sigma(n)+1}-z\Big{\rangle},

which means that there exists a constant MM such that M2xnzM\geq 2\|x_{n}-z\| for all nn and

xσ(n)z2\displaystyle\|x_{\sigma(n)}-z\|^{2} \displaystyle\leq 2ασ(n)(zσ(n)wσ(n))z,xσ(n)+1z\displaystyle 2\Big{\langle}\alpha_{\sigma(n)}(z_{\sigma(n)}-w_{\sigma(n)})-z,x_{\sigma(n)+1}-z\Big{\rangle} (3.19)
\displaystyle\leq Mzσ(n)wσ(n)+2z,xσ(n)+1z.\displaystyle M\|z_{\sigma(n)}-w_{\sigma(n)}\|+2\Big{\langle}-z,x_{{\sigma(n)}+1}-z\Big{\rangle}.

Now since xσ(n)+1xσ(n)0\|x_{\sigma(n)+1}-x_{\sigma(n)}\|\to 0, we have

lim supnz,xσ(n)+1z\displaystyle\limsup_{n\to\infty}\langle-z,x_{\sigma(n)+1}-z\rangle =lim supnz,xσ(n)z\displaystyle=\limsup_{n\to\infty}\langle-z,x_{\sigma(n)}-z\rangle
=maxqωw(xσ(n))z,qz0\displaystyle=\max_{q\in\omega_{w}(x_{\sigma(n)})}\langle-z,q-z\rangle\leq 0

by virtue of the facts z=PΩ(0)z=P_{\Omega}(0) and ω(xσ(n))Ω\omega(x_{\sigma(n)})\subset\Omega. Consequently, following from zσ(n)wσ(n)0\|z_{\sigma(n)}-w_{\sigma(n)}\|\to 0, the relation (3.19) assures that xσ(n)zx_{\sigma(n)}\to z, which further implies from Lemma 2.6 that

xnzxσ(n)+1zxσ(n)+1xσ(n)+xσ(n)z0.\|x_{n}-z\|\leq\|x_{\sigma(n)+1}-z\|\leq\|x_{\sigma(n)+1}-x_{\sigma(n)}\|+\|x_{\sigma(n)}-z\|\to 0.

Namely, xnzx_{n}\to z in norm, and the proof of Case 2 is complete.

(II). Now, it is time to show the strong convergence when θn0\theta_{n}\equiv 0. Clearly, if θn=0\theta_{n}=0, then wn=xn1w_{n}=x_{n-1} and xn+1=(1αnβn)xn1+αnzn1x_{n+1}=(1-\alpha_{n}-\beta_{n})x_{n-1}+\alpha_{n}z_{n-1}, where zn1=JτnA(IτnB)xn1z_{n-1}=J_{\tau_{n}}^{A}(I-\tau_{n}B)x_{n-1}. Repeating the steps from (3.10)-(3.12), we have the following similar inequality:

xn+1z2\displaystyle\|x_{n+1}-z\|^{2} \displaystyle\leq (1βn)xn1z2+βnz2+αn(τn2γ2+αn+βn)xn1zn12\displaystyle(1-\beta_{n})\|x_{n-1}-z\|^{2}+\beta_{n}\|z\|^{2}+\alpha_{n}(\frac{\tau_{n}}{2\gamma}-2+\alpha_{n}+\beta_{n})\|x_{n-1}-z_{n-1}\|^{2} (3.20)
2γτnαnBxn1Bzxn1zn12γ2.\displaystyle-2\gamma\tau_{n}\alpha_{n}\Big{\|}Bx_{n-1}-Bz-\frac{x_{n-1}-z_{n-1}}{2\gamma}\Big{\|}^{2}.

Two possible cases will be shown as follows. Case 1. There exists an integer N00N_{0}\geq 0 such that xn+1zxnz\|x_{n+1}-z\|\leq\|x_{n}-z\| for all nN0n\geq N_{0}. Then there exists the limit of the sequence {xnz}\{\|x_{n}-z\|\}, denoted by l=limnxnz2l=\lim_{n\rightarrow}\|x_{n}-z\|^{2}, and so

limn(xnz2xn+1z2)=0.\lim_{n\rightarrow\infty}(\|x_{n}-z\|^{2}-\|x_{n+1}-z\|^{2})=0.

In addition, we have

n=0(xn+1z2xnz2)=limn(xn+1z2x0z2)<\displaystyle\sum_{n=0}^{\infty}(\|x_{n+1}-z\|^{2}-\|x_{n}-z\|^{2})=\lim_{n\rightarrow\infty}(\|x_{n+1}-z\|^{2}-\|x_{0}-z\|^{2})<\infty

and therefore from (3.20) we obtain

αn(2τn2γαnβn)xn1zn12+2γτnαnBxn1Bzxn1zn12γ2\displaystyle\alpha_{n}(2-\frac{\tau_{n}}{2\gamma}-\alpha_{n}-\beta_{n})\|x_{n-1}-z_{n-1}\|^{2}+2\gamma\tau_{n}\alpha_{n}\Big{\|}Bx_{n-1}-Bz-\frac{x_{n-1}-z_{n-1}}{2\gamma}\Big{\|}^{2}
\displaystyle\leq (1βn)xn1z2+βnz2xn+1z2\displaystyle(1-\beta_{n})\|x_{n-1}-z\|^{2}+\beta_{n}\|z\|^{2}-\|x_{n+1}-z\|^{2}
=\displaystyle= (xn1z2xnz2)+(xnz2xn+1z2)+βn(z2xn1z2).\displaystyle(\|x_{n-1}-z\|^{2}-\|x_{n}-z\|^{2})+(\|x_{n}-z\|^{2}-\|x_{n+1}-z\|^{2})+\beta_{n}(\|z\|^{2}-\|x_{n-1}-z\|^{2}).

Notice the assumptions on the parameters αn\alpha_{n}, βn\beta_{n} and τn\tau_{n}, we have

limnxn1zn12=0;limnBxn1Bzxn1zn12γ2=0,\displaystyle\lim_{n\rightarrow\infty}\|x_{n-1}-z_{n-1}\|^{2}=0;\lim_{n\rightarrow\infty}\Big{\|}Bx_{n-1}-Bz-\frac{x_{n-1}-z_{n-1}}{2\gamma}\Big{\|}^{2}=0,

which imply that xn+1xn1αnzn1xn1+βnxn10\|x_{n+1}-x_{n-1}\|\leq\alpha_{n}\|z_{n-1}-x_{n-1}\|+\beta_{n}\|x_{n-1}\|\to 0 as nn\rightarrow\infty. Next we show the asymptotic regularity of {xn}\{x_{n}\}. Indeed, it follows from the relation between the norm and inner product that

xn+1xn2\displaystyle\|x_{n+1}-x_{n}\|^{2} =\displaystyle= xn+1z+zxn2\displaystyle\|x_{n+1}-z+z-x_{n}\|^{2}
=\displaystyle= xn+1z2+zxn2+2xn+1z,zxn\displaystyle\|x_{n+1}-z\|^{2}+\|z-x_{n}\|^{2}+2\langle x_{n+1}-z,z-x_{n}\rangle
=\displaystyle= xn+1z2+zxn2+2xn+1z,zxn+1+xn+1xn\displaystyle\|x_{n+1}-z\|^{2}+\|z-x_{n}\|^{2}+2\langle x_{n+1}-z,z-x_{n+1}+x_{n+1}-x_{n}\rangle
\displaystyle\leq xnz2xn+1z2+2xn+1zxn+1xn\displaystyle\|x_{n}-z\|^{2}-\|x_{n+1}-z\|^{2}+2\|x_{n+1}-z\|\cdot\|x_{n+1}-x_{n}\|
\displaystyle\leq xnz2xn+1z2+2(M+m)xn+1xn,\displaystyle\|x_{n}-z\|^{2}-\|x_{n+1}-z\|^{2}+2(M+m)\cdot\|x_{n+1}-x_{n}\|,

where MM is a constant such that MxnzM\geq\|x_{n}-z\| for all nn and m>0m>0 is a given constant, which means that

xn+1xn22(M+m)xn+1xnxnz2xn+1z2,\displaystyle\|x_{n+1}-x_{n}\|^{2}-2(M+m)\cdot\|x_{n+1}-x_{n}\|\leq\|x_{n}-z\|^{2}-\|x_{n+1}-z\|^{2},

and then

n=0[xn+1xn2(M+m)]xn+1xnn=0(xnz2xn+1z2)<.\displaystyle\sum_{n=0}^{\infty}[\|x_{n+1}-x_{n}\|-2(M+m)]\cdot\|x_{n+1}-x_{n}\|\leq\sum_{n=0}^{\infty}(\|x_{n}-z\|^{2}-\|x_{n+1}-z\|^{2})<\infty.

Therefore we obtain [xn+1xn2(M+m)]xn+1xn0[\|x_{n+1}-x_{n}\|-2(M+m)]\cdot\|x_{n+1}-x_{n}\|\rightarrow 0 as nn\rightarrow\infty.

Since xn+1xnxn+1z+xnz2M\|x_{n+1}-x_{n}\|\leq\|x_{n+1}-z\|+\|x_{n}-z\|\leq 2M, we have xn+1xn0\|x_{n+1}-x_{n}\|\to 0. This proves the asymptotic regularity of {xn}\{x_{n}\}. By repeating the relevant part of the proof of Case 1 in (I), we get ωw(xn)Ω\omega_{w}(x_{n})\subset\Omega and

xn+1z2\displaystyle\|x_{n+1}-z\|^{2} \displaystyle\leq (1βn)xn1z2αn(1αn)(1βn)2zn1xn12\displaystyle(1-\beta_{n})\|x_{n-1}-z\|^{2}-\alpha_{n}(1-\alpha_{n})(1-\beta_{n})^{2}\|z_{n-1}-x_{n-1}\|^{2} (3.21)
+2βnαn(zn1xn1)z,xn+1z\displaystyle+2\beta_{n}\Big{\langle}\alpha_{n}(z_{n-1}-x_{n-1})-z,x_{n+1}-z\Big{\rangle}
\displaystyle\leq (1βn)xnz2+(1βn)(xn1z2xnz2)\displaystyle(1-\beta_{n})\|x_{n}-z\|^{2}+(1-\beta_{n})(\|x_{n-1}-z\|^{2}-\|x_{n}-z\|^{2})
+2βnαn(zn1xn1)z,xn+1z\displaystyle+2\beta_{n}\Big{\langle}\alpha_{n}(z_{n-1}-x_{n-1})-z,x_{n+1}-z\Big{\rangle}
=\displaystyle= (1βn)an+βnbn+cn,\displaystyle(1-\beta_{n})a_{n}+\beta_{n}b_{n}+c_{n},

where an=xnz2,bn=2[αnzn1xn1,xn+1z+z,xn+1z]a_{n}=\|x_{n}-z\|^{2},\quad b_{n}=2[\alpha_{n}\langle z_{n-1}-x_{n-1},x_{n+1}-z\rangle+\langle-z,x_{n+1}-z\rangle\,] and cn=(1βn)(xn1z2xnz2)c_{n}=(1-\beta_{n})(\|x_{n-1}-z\|^{2}-\|x_{n}-z\|^{2}). The rest of proof is consistent with Case 1 in (I). So we get xnzx_{n}\to z in norm. Case 2. If the sequence {xnz}\{\|x_{n}-z\|\} does not decrease at infinity in the sense that there exists a sub-sequence {nk}\{n_{k}\} of {n}\{n\} such that xnkzxnk+1z\|x_{n_{k}}-z\|\leq\|x_{{n_{k}}+1}-z\| for all k0.k\geq 0. Furthermore, by Lemma 2.6, there exists an integer, non-decreasing sequence σ(n)\sigma(n) for nN1n\geq N_{1} (for some N1N_{1} large enough) such that σ(n)\sigma(n)\rightarrow\infty as nn\rightarrow\infty,

xσ(n)zxσ(n)+1z,xnzxσ(n)+1z\|x_{\sigma(n)}-z\|\leq\|x_{{\sigma(n)}+1}-z\|,\hskip 2.84544pt\|x_{n}-z\|\leq\|x_{{\sigma(n)}+1}-z\|

for each n0n\geq 0. Note that we still have xσ(n)+1xσ(n)0\|x_{\sigma(n)+1}-x_{\sigma(n)}\|\to 0. By repeating the relevant part of the proof of Case 2 in (I), for nN1n\geq N_{1}, we have

ασ(n)(2τσ(n)2γασ(n)βσ(n))xσ(n)1zσ(n)12\displaystyle\alpha_{\sigma(n)}(2-\frac{\tau_{\sigma(n)}}{2\gamma}-\alpha_{\sigma(n)}-\beta_{\sigma(n)})\|x_{\sigma(n)-1}-z_{\sigma(n)-1}\|^{2}
+2γτσ(n)ασ(n)Bxσ(n)1Bzxσ(n)1zσ(n)12γ2\displaystyle+2\gamma\tau_{\sigma(n)}\alpha_{\sigma(n)}\Big{\|}Bx_{\sigma(n)-1}-Bz-\frac{x_{\sigma(n)-1}-z_{\sigma(n)-1}}{2\gamma}\Big{\|}^{2}
\displaystyle\leq (1βσ(n))xσ(n)1z2xσ(n)+1z2+βσ(n)z2\displaystyle(1-\beta_{\sigma(n)})\|x_{{\sigma(n)}-1}-z\|^{2}-\|x_{{\sigma(n)}+1}-z\|^{2}+\beta_{\sigma(n)}\|z\|^{2}
\displaystyle\leq (1βσ(n))xσ(n)z2xσ(n)+1z2+βσ(n)z2\displaystyle(1-\beta_{\sigma(n)})\|x_{\sigma(n)}-z\|^{2}-\|x_{{\sigma(n)}+1}-z\|^{2}+\beta_{\sigma(n)}\|z\|^{2}
=\displaystyle= xσ(n)z2xσ(n)+1z2βσ(n)(xσ(n)z2z2)\displaystyle\|x_{\sigma(n)}-z\|^{2}-\|x_{{\sigma(n)}+1}-z\|^{2}-\beta_{\sigma(n)}(\|x_{\sigma(n)}-z\|^{2}-\|z\|^{2})

Notice the assumptions on the parameters ασ(n)\alpha_{\sigma(n)}, βσ(n)\beta_{\sigma(n)} and τσ(n)\tau_{\sigma(n)}, taking the limit by letting nn\to\infty yields

limnxσ(n)1zσ(n)1\displaystyle\lim_{n\to\infty}\|x_{\sigma(n)-1}-z_{\sigma(n)-1}\| =0;\displaystyle=0;
limnBxσ(n)1Bzxσ(n)1zσ(n)12γ2\displaystyle\lim_{n\to\infty}\Big{\|}Bx_{\sigma(n)-1}-Bz-\frac{x_{\sigma(n)-1}-z_{\sigma(n)-1}}{2\gamma}\Big{\|}^{2} =0,\displaystyle=0,

and that these relations are sufficient to guarantee that ωw(xσ(n))Ω\omega_{w}(x_{\sigma(n)})\subset\Omega. Observe that (3.21) holds for each nn. So replacing nn with σ(n)\sigma(n) in (3.21) and using the relation xσ(n)z2<xσ(n)+1z2\|x_{\sigma(n)}-z\|^{2}<\|x_{\sigma(n)+1}-z\|^{2} again for nN1n\geq N_{1}, we obtain

xσ(n)+1z2\displaystyle\|x_{\sigma(n)+1}-z\|^{2} \displaystyle\leq (1βσ(n))xσ(n)1z2ασ(n)(1ασ(n))(1βσ(n))2zσ(n)1xσ(n)12\displaystyle(1-\beta_{\sigma(n)})\|x_{\sigma(n)-1}-z\|^{2}-\alpha_{\sigma(n)}(1-\alpha_{\sigma(n)})(1-\beta_{\sigma(n)})^{2}\|z_{\sigma(n)-1}-x_{\sigma(n)-1}\|^{2}
+2βσ(n)ασ(n)(zσ(n)1wσ(n)1)z,xσ(n)+1z\displaystyle+2\beta_{\sigma(n)}\langle\alpha_{\sigma(n)}(z_{\sigma(n)-1}-w_{\sigma(n)-1})-z,x_{\sigma(n)+1}-z\rangle
\displaystyle\leq (1βσ(n))xσ(n)z2+2βσ(n)ασ(n)(zσ(n)1xσ(n)1)z,xσ(n)+1z,\displaystyle(1-\beta_{\sigma(n)})\|x_{\sigma(n)}-z\|^{2}+2\beta_{\sigma(n)}\langle\alpha_{\sigma(n)}(z_{\sigma(n)-1}-x_{\sigma(n)-1})-z,x_{\sigma(n)+1}-z\rangle,

The rest of proof is consistent with Case 2 in (I). So we get xnzx_{n}\to z in norm. (III). Finally we consider the situation θn1\theta_{n}\equiv 1. It is obvious that if θn=1\theta_{n}=1, then wn=xnw_{n}=x_{n} and xn+1=(1αnβn)xn+αnznx_{n+1}=(1-\alpha_{n}-\beta_{n})x_{n}+\alpha_{n}z_{n}, where zn=JτnA(IτnB)xnz_{n}=J_{\tau_{n}}^{A}(I-\tau_{n}B)x_{n}. Repeating the steps from (3.10)-(3.12), we have the following similar inequality

xn+1z2\displaystyle\|x_{n+1}-z\|^{2} \displaystyle\leq (1βn)xnz2+βnz2+αn(τn2γ2+αn+βn)xnzn2\displaystyle(1-\beta_{n})\|x_{n}-z\|^{2}+\beta_{n}\|z\|^{2}+\alpha_{n}(\frac{\tau_{n}}{2\gamma}-2+\alpha_{n}+\beta_{n})\|x_{n}-z_{n}\|^{2}
2γτnαnBxnBzxnzn2γ2\displaystyle-2\gamma\tau_{n}\alpha_{n}\Big{\|}Bx_{n}-Bz-\frac{x_{n}-z_{n}}{2\gamma}\Big{\|}^{2}
\displaystyle\leq (1βn)xnz2+βnz2\displaystyle(1-\beta_{n})\|x_{n}-z\|^{2}+\beta_{n}\|z\|^{2}
\displaystyle\leq max{xnz2,z2},\displaystyle\max\{\|x_{n}-z\|^{2},\|z\|^{2}\},

which means that xnz2\|x_{n}-z\|^{2} is bounded. Similar to the proof of the above situations, two possible cases will be shown as follows. Two possible cases will be shown as follows. Case 1. There exists an integer N00N_{0}\geq 0 such that xn+1zxnz\|x_{n+1}-z\|\leq\|x_{n}-z\| for all nN0n\geq N_{0}. Then there exists the limit of the sequence {xnz}\{\|x_{n}-z\|\}, denoted by l=limnxnz2l=\lim_{n\rightarrow}\|x_{n}-z\|^{2}, and so

limn(xnz2xn+1z2)=0.\lim_{n\rightarrow\infty}(\|x_{n}-z\|^{2}-\|x_{n+1}-z\|^{2})=0.

In addition, we have

n=0(xn+1z2xnz2)=limn(xn+1zx0z)<.\displaystyle\sum_{n=0}^{\infty}(\|x_{n+1}-z\|^{2}-\|x_{n}-z\|^{2})=\lim_{n\rightarrow\infty}(\|x_{n+1}-z\|-\|x_{0}-z\|)<\infty.

By repeating the relevant part of the proof of Case 1 in (I), we have

limnxnzn2=0;limnBxnBzxnzn2γ2=0,\displaystyle\lim_{n\rightarrow\infty}\|x_{n}-z_{n}\|^{2}=0;\lim_{n\rightarrow\infty}\Big{\|}Bx_{n}-Bz-\frac{x_{n}-z_{n}}{2\gamma}\Big{\|}^{2}=0,

which imply that xn+1xnαnznxn+βnxn0\|x_{n+1}-x_{n}\|\leq\alpha_{n}\|z_{n}-x_{n}\|+\beta_{n}\|x_{n}\|\to 0 as nn\rightarrow\infty. The rest of proof is consistent with Case 1 in (I). So we get xnzx_{n}\to z in norm. Case 2. If the sequence {xnz}\{\|x_{n}-z\|\} does not decrease at infinity in the sense that there exists a sub-sequence {nk}\{n_{k}\} of {n}\{n\} such that xnkzxnk+1z\|x_{n_{k}}-z\|\leq\|x_{{n_{k}}+1}-z\| for all k0.k\geq 0. Furthermore, by Lemma 2.6, there exists an integer, non-decreasing sequence σ(n)\sigma(n) for nN1n\geq N_{1} (for some N1N_{1} large enough) such that σ(n)\sigma(n)\rightarrow\infty as nn\rightarrow\infty,

xσ(n)zxσ(n)+1z,xnzxσ(n)+1z\|x_{\sigma(n)}-z\|\leq\|x_{{\sigma(n)}+1}-z\|,\hskip 2.84544pt\|x_{n}-z\|\leq\|x_{{\sigma(n)}+1}-z\|

for each n0n\geq 0.

Note that we still have xσ(n)+1xσ(n)0\|x_{\sigma(n)+1}-x_{\sigma(n)}\|\to 0. The rest of proof is consistent with Case 1 in (I). So we get xnzx_{n}\to z in norm.

This completes the proof. ∎

Remark 3.3.

It is easy to see that the convergence of our algorithms still holds even without the following condition:

n=1θnxnxn12<.\sum_{n=1}^{\infty}\theta_{n}\|x_{n}-x_{n-1}\|^{2}<\infty.

The above assumption is not necessary at all in our cases. To some extents, our inertial-like algorithms seem to have two merits:(1) Compared with the general inertial proximal algorithms, we do not need to calculate the values of xnxn1\|x_{n}-x_{n-1}\| before choosing the parameters θn\theta_{n} in numerical simulations, which make the algorithms convenient and use-friendly. (2) Compared with the general inertial algorithms, the inertial factors θn\theta_{n} are chosen in [0,1][0,1] with θn=1\theta_{n}=1 possible, which are new, natural and interesting algorithms in some ways. In particular, under more mild assumptions, our proofs are simpler and different from the others.

4 Applications

4.1 Convex Optimization

Let CC be a nonempty closed and convex subset of HH and f,gf,g be two proper, convex and lower semi-continuous functions. Moreover, assume that gg is differentiable with a 1/γ1/\gamma-Lipschitz gradient. With this data, consider the following convex minimizing problem :

minxC{f(x)+g(x)}.\displaystyle\min_{x\in C}\{f(x)+g(x)\}. (4.1)

Denoted by Ω={x:minxC{f(x)+g(x)}\Omega=\{x:\min_{x\in C}\{f(x)+g(x)\}. Recall that the subdifferential of ff at xx, denote by f\partial f:

f(x):={xC:f(y)f(x)yx,x,yH}.\displaystyle\partial f(x):=\{x^{*}\in C:f(y)-f(x)\geq\langle y-x,x^{*}\rangle,\,\forall y\in H\}.

So, by taking A=fA=\partial f and B=gB=\nabla g(: the gradient of gg) in (3.2), where AA and BB are maximal monotone operators and BB is γ\gamma-cocoercive, we can apply Theorem 3.1 and Theorem 3.2 and obtain the following results:

Theorem 4.1.

Let ff and gg be two proper, convex and lower semi-continuous functions such that g\nabla g and f\partial f are maximal monotone operators and g\nabla g also γ\gamma-cocoercive. Assume that Ω\Omega\neq\emptyset. Let {xn}\{x_{n}\} and {yn}\{y_{n}\} be any two sequences generated by the following scheme (see, e.g. ):

{wn=xn1+θn(xnxn1),xn+1=Jτnf(wnτngwn),\begin{cases}w_{n}=x_{n-1}+\theta_{n}(x_{n}-x_{n-1}),\\ x_{n+1}=J^{\partial f}_{\tau_{n}}(w_{n}-\tau_{n}\nabla gw_{n}),\\ \end{cases}

If τn(ϵ,2γϵ)\tau_{n}\in(\epsilon,2\gamma-\epsilon) for some given ϵ>0\epsilon>0 small enough, and θn[0,1]\theta_{n}\in[0,1], then the sequences {xn}\{x_{n}\} converges weakly to a point of Ω.{\Omega.}

Theorem 4.2.

Let ff and gg be two proper, convex and lower semi-continuous functions such that g\nabla g and f\partial f are maximal monotone operators and g\nabla g also γ\gamma-cocoercive. Assume that Ω\Omega\neq\emptyset. Let {xn}\{x_{n}\} and {yn}\{y_{n}\} be any two sequences generated by the following scheme:

{wn=xn1+θn(xnxn1),xn+1=(1αnβn)wn+αnJτnf(wnτngwn),\begin{cases}w_{n}=x_{n-1}+\theta_{n}(x_{n}-x_{n-1}),\\ x_{n+1}=(1-\alpha_{n}-\beta_{n})w_{n}+\alpha_{n}J^{\partial f}_{\tau_{n}}(w_{n}-\tau_{n}\nabla gw_{n}),\\ \end{cases}

where αn,βn,θn\alpha_{n},\beta_{n},\theta_{n} are satisfying the selection criteria in Algorithm3.2. If τn(ϵ,2γϵ)\tau_{n}\in(\epsilon,2\gamma-\epsilon) for some given ϵ>0\epsilon>0 small enough, then the sequences {xn}\{x_{n}\} converges strongly to z=PΩ(0)z=P_{\Omega}(0).

4.2 Variational Inequality Problem

Let CC be a nonempty closed and convex subset of HH and B:HHB:H\rightarrow H a maximal and γ\gamma-coercive operator.

Consider the classical variational inequality (VI) problem of finding a point xCx^{*}\in C such that

Bx,zx0,zC.\displaystyle\langle Bx^{*},z-x^{*}\rangle\geq 0,\,\,\,\forall z\in C. (4.2)

Denote by Ω\Omega the solution set of the problem (VI) (4.2).

So, by taking Ax:={wH:w,zx0,zC}Ax:=\{w\in H:\langle w,z-x\rangle\leq 0,\,\forall z\in C\} (: the normal cone of the set CC), the problem (VI) (4.2) is equivalent to finding zeroes of A+BA+B (see e.g., Peypouquet[24]). We apply our algorithms to the variational inequality (VI) problem and have the following theorems.

Theorem 4.3.

Let CC be a nonempty closed and convex subset of HH, B:HHB:H\rightarrow H a maximal and γ\gamma-coercive operator. Choose τn(ϵ,2γϵ)\tau_{n}\in(\epsilon,2\gamma-\epsilon) for some given ϵ>0\epsilon>0 small enough and assume that Ω\Omega\neq\emptyset. Construct the sequences {xn}\{x_{n}\} and {wn}\{w_{n}\} as follows:

{wn=xn1+θn(xnxn1),xn+1=JτnA(IτnB)wn.\begin{cases}w_{n}=x_{n-1}+\theta_{n}(x_{n}-x_{n-1}),\\ x_{n+1}=J^{A}_{\tau_{n}}(I-\tau_{n}B)w_{n}.\end{cases}

If θn[0,1]\theta_{n}\in[0,1], then the sequences {xn}\{x_{n}\} converges weakly to a point of Ω.{\Omega.}

Theorem 4.4.

Let CC be a nonempty closed and convex subset of HH, B:HHB:H\rightarrow H be a maximal and γ\gamma-cocoercive operator. Choose τn(ϵ,2γϵ)\tau_{n}\in(\epsilon,2\gamma-\epsilon) for some given ϵ>0\epsilon>0 small enough and assume that Ω\Omega\neq\emptyset. Construct the sequences {xn}\{x_{n}\}, {wn}\{w_{n}\} as follows.

{wn=xn1+θn(xnxn1),xn+1=(1αnβn)wn+αnJτnA(IτnB)wn,\begin{cases}w_{n}=x_{n-1}+\theta_{n}(x_{n}-x_{n-1}),\\ x_{n+1}=(1-\alpha_{n}-\beta_{n})w_{n}+\alpha_{n}J^{A}_{\tau_{n}}(I-\tau_{n}B)w_{n},\end{cases}

where αn,βn,θn\alpha_{n},\beta_{n},\theta_{n} are satisfying the selection criteria in Algorithm 3.2. Then the sequence {xn}\{x_{n}\} converges strongly to z=PΩ(0)z=P_{\Omega}(0).


5 Numerical Examples

In this section, we first present two numerical examples in infinite and finite-dimensional Hilbert spaces to illustrate the applicability, efficiency and stability of Algorithm 3.1 and Algorithm 3.2, and then consider sparse signal recovery from real-world life in finite dimensional spaces. In addition, the comparison results with other algorithms are also described. All the codes are written in Matlab R2016b and are preformed on an LG dual core personal computer. Example 5.1.In this example, we take H=R3H=R^{3} with Euclidean norm. Let A:R3R3A:R^{3}\rightarrow R^{3} be defined by Ax=3xAx=3x and let B:R3R3B:R^{3}\rightarrow R^{3} be defined by Bx=x3+(1,2,0)Bx=\frac{x}{3}+(-1,2,0), xR3\forall x\in R^{3}. We can see that AA and BB are maximal monotone mappings and BB with γ\gamma-cocoercive(0<γ30<\gamma\leq 3), respectively. Indeed, let x,yR3x,y\in R^{3}, then

BxBy,xy=x3y3,xy=13xy2,\langle Bx-By,x-y\rangle=\langle\frac{x}{3}-\frac{y}{3},x-y\rangle=\frac{1}{3}\|x-y\|^{2},

while

BxBy2=19xy2,\|Bx-By\|^{2}=\frac{1}{9}\|x-y\|^{2},

which means BB is 13\frac{1}{3}-cocoercive. It is not hard to check that (A+B)1(0)=(3/10,6/10,0)(A+B)^{-1}(0)=(3/10,-6/10,0). In the simulation process, we choose x0=[0.1;0.2;0.1],x1=[0.2;0.1;0.3]x_{0}=[0.1;-0.2;0.1],x_{1}=[0.2;0.1;-0.3] as two arbitrary initial points. In order to investigate the change and tendency of {xn}\{x_{n}\} more clearly, we denote by z=(3/10,6/10,0)z=(3/10,-6/10,0) and let xn+1z105\|x_{n+1}-z\|\leq 10^{-5} be the stopping criterion. The experimental results are shown in Figs.1-2, where z-axis represents the logarithm of the third coordinate value for each point.

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Figure 1: Behaviors of Algorithm 3.1 for θn=0.51/(n+1)5,θn=0,θn=1.\theta_{n}=0.5-1/(n+1)^{5},\theta_{n}=0,\theta_{n}=1.
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Figure 2: Behaviors of Algorithm 3.2 for θn=0.51/(n+1)5,θn=0,θn=1.\theta_{n}=0.5-1/(n+1)^{5},\theta_{n}=0,\theta_{n}=1.

Indeed, if θn1\theta_{n}\equiv 1, then Algorithm 3.1 coincides with the result of Moudafi[19]. In addition, from Figs.1-2, we can see that when θn=0\theta_{n}=0, two families of points alternate, and finally infinitely close to the exact null point z=(3/10,6/10,0)z=(3/10,-6/10,0).

Example 5.2. In this example, we show the behaviors of our algorithms in H=L2([0,1])H=L^{2}([0,1]). In addition, the compared results with Dadashi and Postolach[8] are considered. In the simulation, we define two mappings AA and BB by Bx(t)=x(t)2Bx(t)=\frac{x(t)}{2} and Ax(t)=3x(t)4Ax(t)=\frac{3x(t)}{4} for all x(t)L2([0,1])x(t)\in L^{2}([0,1]). Then it can be shown that BB is 12\frac{1}{2}- cocoercive. In the numerical experiment, the parameters are chosen as αn=0.51/(10n+2)\alpha_{n}=0.5-1/(10n+2), βn=1n+1\beta_{n}=\frac{1}{n+1} in Algorithm 3.2 for all n1n\geq 1. In addition, xn+1xn<104\|x_{n+1}-x_{n}\|<10^{-4} is used as stopping criterion and the following three different choices of initial functions x0(t),x1(t)x_{0}(t),x_{1}(t) are chosen: Case 1:x0(t)=sin(3t)+cos(5t)2x_{0}(t)=\frac{\sin(-3t)+\cos(-5t)}{2} and x1(t)=2sin(5t)x_{1}(t)=2sin(5t); Case 2:x0(t)=2tsin(3t)e5t200x_{0}(t)=\frac{2t\sin(3t)e^{-5t}}{200} and x1(t)=t2e2tx_{1}(t)=t^{2}-e^{-2t};

Case 3:x0(t)=2t3e5tx_{0}(t)=2t^{3}e^{5t} and x1(t)=etsin(3t)100x_{1}(t)=\frac{e^{t}\sin(3t)}{100}.

Figs.3-5 represent the numerical results for θn\theta_{n} neither 0 nor 1, Fig.6 shows the numerical results for θn=0\theta_{n}=0 and θn=1\theta_{n}=1, respectively. Fig.7 shows the comparising results with Dadashi and Postolach[8] for the initial points x0(t)=etsin(3t)100x_{0}(t)=\frac{e^{t}\sin(3t)}{100} and x0(t)=t2e2tx_{0}(t)=t^{2}-e^{-2t}, respectively. Table1 shows the comparisons with Dadashi and Postolach[8] for the initial point x0(t)=t2e2tx_{0}(t)=t^{2}-e^{-2t}.

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Figure 3: Behavior of Algorithm 3.1 and Algorithm 3.2 for Case 1 in Exp.5.2.
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Figure 4: Behavior of Algorithm 3.1 and Algorithm 3.2 for Case 2 in Exp.5.2.
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Figure 5: Behavior of Algorithm 3.1 and Algorithm 3.2 for Case 3 in Exp.5.2.
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Figure 6: Behavior of Algorithms 3.1 and 3.2 for θn=0\theta_{n}=0 and θn=1\theta_{n}=1 in Exp.5.2.
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Figure 7: Comparisons of Algorithm 3.2 with Dadashi and Postolach[8] in Exp.5.2.
Table 1: Comparisons of Algorithm 3.1, Algorithm 3.2 and Dadashi and Postolach[8]
Error Iter. CPU in second
Alg.1 Alg.2 Alg. in Dadashi et al.[31] Alg.1 Alg.2 Alg. in Dadashi et al.[31]
10310^{-3} 41 21 26 6.2813 3.4688 3.8438
10410^{-4} 49 33 52 10.0445 7.6542 7.8461
10510^{-5} 57 45 110 10.2969 8.6768 24.3212

It is clear that our algorithms, especially Algorithm 3.2, is faster, more efficient, more stable. Example 5.3. (Compressed Sensing) In this example, to show the effectiveness and applicability of our algorithms in the real world, we consider the recovery of a sparse and noisy signal from a limited number of sampling which is a problem from the field of compressed sensing. The sampling matrix TRmnT\in R^{m*n}, m<nm<n, is stimulated by standard Gaussian distribution and vector b=Tx+ϵb=Tx+\epsilon, where ϵ\epsilon is additive noise. When ϵ=0\epsilon=0, it means that there is no noise to the observed data. For further explanations the reader can consult Nguyen and Shin[14]. Let x0Rnx_{0}\in R^{n} be the KK-sparse signal, where K<<nK<<n. Our task is to recover the signal x0x_{0} from the data bb. To this end, we transform it into finding the solution of LASSO problem:

minxRnTxb2\displaystyle\min_{x\in R^{n}}\|Tx-b\|^{2}
s.t.x1t,\displaystyle s.t.\,\,\|x\|_{1}\leq t,

where tt is a given positive constant. If we define

B1(x)={{u:supx1txy,u0},ifyRn,,otherwise,B2(x)={Rm,ifx=b,,otherwise,B_{1}(x)=\begin{cases}\{u:\sup_{\|x\|_{1}\leq t}\langle x-y,u\rangle\leq 0\},&\hbox{if}\,\,y\in R^{n},\\ \emptyset,&\hbox{otherwise},\end{cases}\quad B_{2}(x)=\begin{cases}R^{m},&\hbox{if}\,\,x=b,\\ \emptyset,&\hbox{otherwise},\end{cases}

then one can see that the LASSO problem coincides with the problem of finding xRnx^{*}\in R^{n} such that

0B1(x)and   0B2(Tx),0\in B_{1}(x^{*})\,\,\,\hbox{and}\,\,\,0\in B_{2}(Tx^{*}),

which associates with the the problem for finding the solution of the problem:

0(A+B)x,0\in(A+B)x^{*},

where A=B1A=B_{1} and B=T(IJλB2)TB=T^{*}(I-J_{\lambda}B_{2})T with γ\gamma- cocoercive, γ=1TT\gamma=\frac{1}{\|T^{*}T\|}. In addition to showing the behavior of our algorithms, the results of Sitthithakerngkiet et al. [27], Kazimi and Riviz [12] without inertial process and Tang[28] with general inertial method are compared. For the experiment setting, we choose the following parameters: TRmnT\in R^{m*n} is generated randomly with m=26,27m=2^{6},2^{7}, n=28,29n=2^{8},2^{9}, x0Rnx_{0}\in R^{n} is KK-spikes (K=40,60)(K=40,60) with amplitude ±1\pm{1} distributed in whole domain randomly. For simplicity, we define the nonexpansive mappings Sn:33S_{n}:\mathbb{R}^{3}\rightarrow\mathbb{R}^{3} as Sn=IS_{n}=I for Algorithm 3.1 in Sitthithakerngkiet et al. [27], and the strongly positive bounded linear operator D=ID=I, the constant ξ=0.5\xi=0.5 and we give fixed point u=0.1u=0.1. Moreover, we take αn=0.51/(10n+2)\alpha_{n}=0.5-1/(10*n+2), βn=103/(n+1)\beta_{n}=10^{-3}/(n+1) in all compared algorithms. Let t=K0.001t=K-0.001 and xn+1xn104\|x_{n+1}-x_{n}\|\leq 10^{-4} be the stopping criterion. All the numerical results are presented in Table 2 and Fig.8.

Table 2: The comparisons of Algorithm 3.1, Algorithm 3.2, Sitthithakerngkiet et al. [27], Kazmi et al. [12], Tang[28]
Method K=40K=40,m=26m=2^{6},n=28n=2^{8} K=60K=60,m=27m=2^{7},n=29n=2^{9}
Iter.          Sec. Iter.          Sec.
Algorithm 3.1 43           0.0992 83            0.4431
Algorithm 3.2 63           0.1635 79            0.6368
Sitthithakerngkiet et al. [27] 91           0.12 122           0.6942
Kazmi et al. [12] 54          0.0771 39          0.1981
Algo.3.2-Tang[28] 54            2.2632 67          3.2541
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Figure 8: Numerical results for m=26m=2^{6}, m=28m=2^{8} and K=40K=40

6 Conclusion

We have provided two new Inertial-Like Proximal iterative algorithms (Algorithms 3.1 and 3.2) for the null point problem . We have proved that, under some mild conditions, Algorithms 3.1 converges weakly and Algorithms 3.2 strongly to a solution of the null point problem. Thanks to the novel structure of the inertial-like technique, our Algorithms 3.1 and 3.2 seem to have the following merits: (i) Theoretically, they do not need verify the traditional and difficult checking condition n=1θnxnxn12<\sum_{n=1}^{\infty}\theta_{n}\|x_{n}-x_{n-1}\|^{2}<\infty, namely, our convergence theorems still hold even without this condition.

(ii) Practically, they do not involve the computations of the norm of the difference between xnx_{n} and xn1x_{n-1} before choosing the inertial parameter θn\theta_{n}(hence less computation cost), as opposed to almost all previous inertial algorithms in the existing literature, that is, the constraints on inertial parameter θn\theta_{n} are looser and more natural, so they are extremely attractive and friendly for users.

(iii) Different from the general inertial algorithms, the inertial factors θn\theta_{n} in our Inertial-Like Proximal algorithms are chosen in [0,1][0,1] with θn=1\theta_{n}=1 possible, which are new, natural and interesting algorithms in some ways. In particular, under more mild assumptions, our proofs are simpler and different from the others. We have included several numerical examples which show the efficiency and reliability of Algorithm 3.1 and Algorithm 3.2. We have also made comparisons of Algorithm 3.1 and Algorithm 3.2 with other four algorithms in Sitthithakerngkiet et al. [27], Kazimi and Riviz [12], Tang[28], Dadashi and Postolach[8] confirming some advantages of our novel inertial algorithms.

Acknowledgements. This article was funded by the Science Foundation of China (12071316), Natural Science Foundation of Chongqing (CSTC2019JCYJ-msxmX0661), Science and Technology Research Project of Chongqing Municipal Education Commission (KJQN 201900804) and the Research Project of Chongqing Technology and Business University (KFJJ1952007).

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