Implicit Extragradient-Like Method for Fixed Point Problems and Variational Inclusion Problems in a Banach Space

Abstract

In a uniformly convex and q-uniformly smooth Banach space with $q\in (1,2]$, one use VIP to indicate a variational inclusion problem involving two accretive mappings and CFPP to denote the common fixed-point problem of an infinite family of strict pseudocontractions of order q. In this paper, we introduce a composite extragradient implicit method for solving a general symmetric system of variational inclusions (GSVI) with certain VI and CFPP. We then investigate its convergence analysis under some weak conditions. Finally, we consider the celebrated LASSO problem in Hilbert spaces.

1. Introduction-Preliminaries

Throughout this article, one always supposes that H is a real infinite dimensional Hilbert space endowed with norm $\parallel ·\parallel$ and inner product $\langle ·,·\rangle$. Let the nonempty subset $C\subset H$ be convex and closed, and let the mapping $P_C$ be the nearest point (metric) projection of H onto C. Let $A:C\to H$ be a nonself mapping. The celebrated variational inequality problem (VIP) is to find an element $x^{*}\in C$ s.t. $\langle y-x^{*},A{x}^{*}\rangle \ge 0$, for all $y\in C$. One denotes by VI($C,A$) the set of all solutions of the VIP. The VIP acts as a unified framework for lots of real industrial and applied problems, such as, machine learning, transportation, image processing and economics; see, e.g., [1,2,3,4,5,6,7]. One knows that Korpelevich’s extragradient algorithm [8] is now one of the most popular algorithms to numerically solve the VIP. It reads as follows:

$$\left\{\begin{array}{c}{x}_{j+1}=P_C(x_j-\ell A{y}_j)\forall j\ge 0,\hfill \\ y_j=P_C(x_j-\ell A{x}_j),\hfill \end{array}\right.$$

with $\ell \in (0,\frac{1}{L})$, where L is the Lipschitz constant of the mapping A. If its solution set is not empty, it was obtained that sequence $\{x_j\}$ is weakly convergent. The price for the weak convergence is that A must be a Lipschitz and monotone mapping. To date, now, the extragradient method and relaxed extragradient methods have received much attention and has been studied extensively; see, e.g., [9,10,11,12,13]. Next, one assumes that B is monotone set-valued mapping defined on H and A is a monotone single-valued mapping defined on H, respectively. The so-called variational inclusion problem is to find an element $x^{*}\in H$ s.t. $0\in (A+B)x^{*}$ and it has recently been studied by many authors based on splitting-based approximation methods; see, e.g., [14,15,16,17]. This model provides a unified framework for a lot of theoretical and practical problems and was investigated via different methods [18,19,20,21,22]. To the best of the authors’ knowledge, there are no few associated results obtained in Banach spaces. Next, one will turn our attention to Banach spaces.

Next, $E^{*}$ will be used to present be the dual space of Banach space E. Let $\varnothing \ne C\subset E$ be a convex and closed set. Given a nonlinear mapping T on C, one uses the symbol $\mathrm{Fix}(T)$ to denote the set of all the fixed points of T. Recall that the mapping T is said to be a Lipschitz mapping if and only if, $\forall x,y\in C$, $\parallel Tx-Ty\parallel \le L\parallel x-y\parallel$. If the Lipschitz constant L is just one, one say that T is a nonexpansive mapping whose complementary mappings are monotone.

Recall that the duality mapping $J_q:E\to 2^{E^{*}}$ is defined by

$$J_q(x)=\{\varphi \in E^{*}:\langle x,\varphi \rangle ={\parallel x\parallel }^q,\parallel \varphi \parallel ={\parallel x\parallel }^{q-1}\}.$$

For a particular case, one uses J to stand for $J_2$, which is commonly called the normalized duality. Recall that a mapping T defined on C is said to be a strict pseudocontraction of order q if, $\forall x,y\in C$, there is $j_q(x-y)\in J_q(x-y)$ such that the following inequality holds $\langle Tx-Ty,j_q(x-y)\rangle \le {\parallel x-y\parallel }^q-\varsigma {\parallel (I-T)x-(I-T)y\parallel }^q$ for some $\varsigma >0.$

The convexity modulus of space E, ${\delta }_E$, which maps the interval $(0,2]$ to the interval $[0,1]$, is defined as follows

$${\delta }_E(ϵ)=inf\{\frac{2-\parallel x+x^{\prime }\parallel }{2}:x^{\prime },x\in E,ϵ\le \parallel x^{\prime }-x\parallel ,\parallel x^{\prime }\parallel =1=\parallel x\parallel \}.$$

The smoothness modulus of space E, ${\rho }_E$, which maps the interval $[0,\infty )$ to the interval $[0,\infty )$, is defined as follows

$${\rho }_E(\tau )=sup\{\frac{\parallel \tau x^{\prime }+x\parallel +\parallel \tau x^{\prime }-x\parallel -2}{2}:x^{\prime },x\in E,\parallel x^{\prime }\parallel =\parallel x\parallel =1\}.$$

Recall that a space E is said to be uniformly convex if ${\delta }_E(e)>0$, $\forall e\in (0,2]$. Recall that a space E is said to be uniformly smooth if ${lim}_{\tau \to 0^{+}}\frac{{\rho }_E(\tau )}{\tau }=0$. Further it is said to be q-uniformly ($q>1$) smooth if $\exists c>0$ s.t. $c\ge s^{-q}{\rho }_E(s)$, $\forall s>0$. In q-uniformly ($q>1$) smooth spaces, one has the following celebrated inequality, for some $j_q(x+y)\in J_q(x+y)$,

$${\parallel x+y\parallel }^q-{\parallel x\parallel }^q\le q\langle y,j_q(x+y)\rangle \forall x,y\in E.$$

(1)

One says that an operator $\Pi :C\to D$, where D is convex and closed subset of C, is said to be a sunny mapping if, $(1-\xi )\Pi (x)+\xi x\in C$ for $\xi \ge 0$ and $x\in C$, $\Pi (x)=\Pi [\Pi (x)+\xi (x-\Pi (x))]$. If $\Pi$ is both nonexpansive and sunny, then $\langle \overline{x}-x^{\prime },J(\Pi (\overline{x})-\Pi (x^{\prime }))\rangle \ge {\parallel \Pi (\overline{x})-\Pi (x^{\prime })\parallel }^2$, $\forall \overline{x},x^{\prime }\in C$.

In the setting of q-uniformly smooth spaces, we recall that an operator B is said to be accretive if, for some $j_q(x^{\prime }-\overline{x})\in J_q(x^{\prime }-\overline{x})$, $\langle u-v,j_q(x^{\prime }-\overline{x})\rangle \ge 0,$ for all $u\in B{x}^{\prime },$ $v\in B\overline{x}$. An accretive operator B is said to be inverse-strongly of order q if, for each $x^{\prime },\overline{x}\in C$, $\exists j_q(x^{\prime }-\overline{x})\in J_q(x^{\prime }-\overline{x})$ s.t. $\langle u-v,j_q(x^{\prime }-\overline{x})\rangle \ge \alpha {\parallel u-v\parallel }^q,$ for all $u\in B{x}^{\prime },$$v\in B\overline{x}$ for some $\alpha >0$. An accretive operator B is said to be m-accretive if $(I+\lambda B)C=E$ for all $\lambda >0$. Furthermore, one can define a mapping $J_{\lambda }^B:(I+\lambda B)C\to C$ by $J_{\lambda }^B={(I+\lambda B)}^{-1}$ for each $\lambda >0$. Such $J_{\lambda }^B$ is called the resolvent of B for $\lambda >0$. In the sequel, we will use the notation $T_{\lambda }:=J_{\lambda }^B(I-\lambda A)={(I+\lambda B)}^{-1}(I-\lambda A)\forall \lambda >0$. From [9], we have $\mathrm{Fix}(T_{\lambda })={(A+B)}^{-1}0\forall \lambda >0$ and $\parallel y-T_{\lambda }y\parallel \le 2\parallel y-T_{r}y\parallel$ for $0<\lambda \le r$ and $y\in C$. From [23], one has $J_{\lambda }^{B}x=J_{\mu }^B(\frac{\mu }{\lambda }x+(1-\frac{\mu }{\lambda })J_{\lambda }^{B}x)$ for all $\lambda ,\mu >0,x\in E$ and $\mathrm{Fix}(J_{\lambda }^B)=B^{-1}0$, where $B^{-1}0=\{x\in C:0\in Bx\}$.

Let $A_1,A_2:C\to E$ and $B_1,B_2:C\to 2^E$ be nonlinear mappings with $B_{k}x\ne \varnothing \forall x\in C,k=1,2$. Consider the symmetrical system of variational inclusions, which consists of finding the pair $(x^{*},y^{*})$ in $C\times C$ s.t.

$$\left\{\begin{array}{c}0\in {\zeta }_1(A_1{y}^{*}+B_1{x}^{*})+x^{*}-y^{*},\hfill \\ 0\in {\zeta }_2(A_2{x}^{*}+B_2{y}^{*})+y^{*}-x^{*},\hfill \end{array}\right.$$

(2)

where ${\zeta }_k$ is a positive constant for $k=1,2$. Ceng, Postolache and Yao [24] obtain the fact that problem (2) is equivalent to a fixed point problem.

Based on the equivalent relation, Ceng et al. [24] suggested a composite viscosity implicit rule for solving the GSVI (2) as follows:

$$\left\{\begin{array}{c}{y}_j=J_{{\zeta }_2}^{B_2}(x_j-{\zeta }_2{A}_2{x}_j),\hfill \\ x_j={\alpha }_{j}f(x_{j-1})+{\delta }_j{x}_{j-1}+{\beta }_{j}V{x}_{j-1}+{\gamma }_j[\mu S{x}_j+(1-\mu )J_{{\zeta }_1}^{B_1}(y_j-{\zeta }_1{A}_1{y}_j)]\forall j\ge 1,\hfill \end{array}\right.$$

where $\mu \in (0,1)$, $S:=(1-\alpha )I+\alpha T$ with $0<\alpha <min\{1,\frac{2\lambda }{{\kappa }_2}\}$, and the sequences $\{{\alpha }_j\},\{{\delta }_j\},\{{\beta }_j\},\{{\gamma }_j\}\subset (0,1)$ are such that (i) ${\alpha }_j+{\delta }_j+{\beta }_j+{\gamma }_j=1\forall j\ge 1$; (ii) ${lim}_{j\to \infty }{\alpha }_j=0$, ${lim}_{j\to \infty }\frac{{\beta }_j}{{\alpha }_j}=0$; (iii) ${lim}_{j\to \infty }{\gamma }_j=1$; (iv) ${\sum }_{j=0}^{\infty }{\alpha }_j=\infty$. They proved that $\{x_j\}$ converges strongly to a point of $\mathrm{Fix}(G)\cap \mathrm{Fix}(T)$, which solves a certain VIP.

In this article, we introduce and investigate an implicitly composite solution method to solve the GSVI (2) with certain VIP and CFPP constraints. We then analyze convergence of the suggested method in the setting of real Hilbert spaces under some mild conditions. An application is also considered.

From now on, one always uses ${\kappa }_q>0$ to denote the smoothness coefficient; see [25,26]. One also lists some essential lemmas for the strong convergence theorem next.

Lemma 1

([27]). Let E be q-uniformly smooth with $q\in (1,2]$, and $\varnothing \ne C\subset E$ a closed convex set. Let $T:C\to C$ be a ς-strict pseudocontraction of order q. Given $\alpha \in (0,1)$. Define a single-valued nonlinear mapping $T_{\alpha }:C\to C$ by $T_{\alpha }x=(1-\alpha )x+\alpha Tx$. Then $T_{\alpha }$ is nonexpansive with $\mathrm{Fix}(T_{\alpha })=\mathrm{Fix}(T)$ provided $0<\alpha \le {(\frac{\varsigma q}{{\kappa }_q})}^{\frac{1}{q-1}}$.

Lemma 2

([28]). Let E be q-uniformly smooth with $q\in (1,2]$. Suppose that $A:C\to E$ is an α-inverse-strongly accretive mapping of order q. Then, for any given $\lambda \ge 0$,

$$\lambda (\alpha q-{\kappa }_q{\lambda }^{q-1}){\parallel Ax-Ay\parallel }^q+\parallel (I-\lambda A)x-(I-\lambda A){y\parallel }^q\le {\parallel x-y\parallel }^q,\forall x,y\in C.$$

Lemma 3

([26,28]). Let $q\in (1,2]$ be a fixed real number and let E be a q-uniformly smooth Banach space. Then ${\parallel x+y\parallel }^q\le q\langle y,J_q(x)\rangle +{\parallel x\parallel }^q+{\kappa }_q{\parallel y\parallel }^q,$ $\forall x,y\in E$. Let $B_1,B_2:C\to 2^E$ be two m-accretive operators. Let $A_i:C\to E(i=1,2)$ be ${\sigma }_i$-inverse-strongly accretive mapping of order q for each $i=1,2$. Define an operator $G:C\to C$ by

$$G:=J_{{\zeta }_1}^{B_1}(I-{\zeta }_1{A}_1)J_{{\zeta }_2}^{B_2}(I-{\zeta }_2{A}_2).$$

If $0\le {\zeta }_i\le {(\frac{{\sigma }_{i}q}{{\kappa }_q})}^{\frac{1}{q-1}}(i=1,2)$, then G is a nonexpansive mapping.

The following lemmas was proved in [29], which is a simple extension of the inequality established in [26].

Lemma 4.

In a uniformly convex real Banach space $E,$ there is a convex, strictly increasing, continuous function g, which maps $[0,+\infty )$ to $[0,+\infty )$ with $g(0)=0$ such that

$$\parallel \lambda x^{\prime }+\mu \overline{x}+\nu \tilde{x}{\parallel }^q\le \lambda \parallel x^{\prime }{\parallel }^q+\mu \parallel \overline{x}{\parallel }^q+\nu \parallel \tilde{x}{\parallel }^q-W_{q}g(\parallel \overline{x}-\tilde{x}\parallel ),$$

where $q>1$ is any real number and $W_q$ is a real function associated with μ and ν, for all $x^{\prime },\overline{x},\tilde{x}\in \{x\in E:\parallel x\parallel \le r\}$ (r is some real number) and $\lambda ,\mu ,\nu \in [0,1]$ such that $\lambda +\nu +\mu =1$.

Lemma 5

([26]). If E be a uniformly convex Banach space, then there exist a convex, strictly increasing, continuous function h, which maps $[0,+\infty )$ to $[0,+\infty )$ with $h(0)=0$ such that

$$h(\parallel x-y\parallel )\le {\parallel x\parallel }^q-q\langle x,j_q(y)\rangle +(q-1){\parallel y\parallel }^q,$$

where x and y are in some bounded subset of E and $j_q(y)\in J_q(y)$.

Lemma 6

([30]). Let ${\{S_n\}}_{n=0}^{\infty }$ be a sequence of self-mappings on C such that

$$\sum _{n=1}^{\infty }\underset{x\in C}{sup}\parallel S_{n}x-S_{n-1}x\parallel <\infty .$$

Then, for each $y\in C$, $\{S_{n}y\}$ converges strongly to some point of C. Moreover, let S be the self-mapping on C defined by $Sy={lim}_{n\to \infty }{S}_{n}y$ for all $y\in C$. Then ${lim}_{n\to \infty }{sup}_{x\in C}\parallel S_{n}x-Sx\parallel =0$.

Lemma 7

([31]). Let E be a strictly convex Banach space. Let $T_n$ be a nonexpansive mapping defined on a convex and closed subset C of E for each $n\ge 1$. Let ${\bigcap }_{n=0}^{\infty }\mathrm{Fix}(T_n)$ be nonempty. Let $\{{\lambda }_n\}$ be a positive sequence such that ${\sum }_{n=0}^{\infty }{\lambda }_n=1$. Then a mapping S on C defined by $Sx={\sum }_{n=0}^{\infty }{\lambda }_n{T}_{n}x\forall x\in C$ is well defined, nonexpansive and $\mathrm{Fix}(S)={\bigcap }_{n=0}^{\infty }\mathrm{Fix}(T_n)$ holds.

Lemma 8

([32]). Let E be a smooth Banach space. Let $T:C\to C$, where C is convex and closed set in E, be a self-nonexpansive mapping. Suppose that λ is constant in the interval $(0,1)$. Then $\{z^{\lambda }\}$, where $z^{\lambda }=(1-\lambda )T{z}^{\lambda }+\lambda u$, converges strongly to a fixed point $x^{*}\in \mathrm{Fix}(T)$, which solves $\langle u-x^{*},J(x^{*}-x)\rangle \ge 0$ for all $x\in \mathrm{Fix}(T)$.

Lemma 9

([33]). Let $\{a_n\}$ be a sequence in $[0,\infty )$ such that $a_{n+1}\le (1-s_n)a_n+s_n{\nu }_n\forall n\ge 0$, where $\{s_n\}$ and $\{{\nu }_n\}$ satisfy the conditions: (i) $\{s_n\}\subset [0,1]$, ${\sum }_{n=0}^{\infty }{s}_n=\infty$; (ii) ${lim\; sup}_{n\to \infty }{\nu }_n\le 0$ or ${\sum }_{n=0}^{\infty }\left|s_n{\nu }_n\right|<\infty$. Then ${lim}_{n\to \infty }{a}_n=0$.

Lemma 10

([34]). Let $\{{\Gamma }_n\}$ be a real sequence such that there exists a real sequence $\{{\Gamma }_{n_i}\}$, which is a subsequence of $\{{\Gamma }_n\}$ such that ${\Gamma }_{n_i+1}>1{\Gamma }_{n_i}$ for each integer $i\ge 1$. Let ${\{\tau (n)\}}_{n\ge n_0}$ be a integer Define the sequence defined by $\tau (n)=max\{k\le n:{\Gamma }_{k+1}>{\Gamma }_k\},$ where the integer $n_0\ge 1$ is chosen in such a way that $\{k\le n_0:{\Gamma }_{k+1}\}>{\Gamma }_k\ne \varnothing$. (i) ${\Gamma }_{\tau (n)+1}\ge {\Gamma }_{\tau (n)}$ and ${\Gamma }_{\tau (n)+1}\ge {\Gamma }_n$ for each $n\ge n_0$; (ii) $\tau (n_0)\le \tau (n_0+1)\le \cdots$ and $\tau (n)\to \infty$.

2. Results

Throughout this section, suppose that C is a convex closed set in a Banach space E, which is both uniformly convex and q-uniformly smooth with $q\in (1,2]$. Let both $B_1:C\to 2^E$ and $B_2:C\to 2^E$ be m-accretive operators. Let $A_k:C\to E$ be single-valued ${\sigma }_k$-inverse strongly accretive mapping for each $k=1,2$. Further, one assumes that $G:C\to C$ is a self mapping defined as $G:=J_{{\zeta }_1}^{B_1}(I-{\zeta }_1{A}_1)J_{{\zeta }_2}^{B_2}(I-{\zeta }_2{A}_2)$ with constants ${\zeta }_1,{\zeta }_2>0$. Let $S_n$ be a $\varsigma$-uniformly strictly pseudocontractive mapping for each $n\ge 1$. Let $A:C\to E$ and $B:C\to 2^E$ be a $\sigma$-inverse-strongly accretive mapping of order q and an m-accretive operator, respectively. Assume that the feasibility set $\Omega :={\bigcap }_{n=0}^{\infty }\mathrm{Fix}(S_n)\cap {(A+B)}^{-1}0\cap \mathrm{Fix}(G)$ is nonempty.

Algorithm 1: Composite extragradient implicit method for the GSVI (2) with VIP and CFPP constraints.

Initial Step. Given $\xi \in (0,1),\alpha \in (0,min\{1,{(\frac{\varsigma q}{{\kappa }_q})}^{\frac{1}{q-1}}\})$. Let $x_0\in C$ be an arbitrary initial.  Iteration Steps. Compute $x_{n+1}$ from the current $x_n$ as follows:  Step 1. Calculate $$\left\{\begin{array}{c}{w}_n=s_n((1-\xi )x_n+\xi G{w}_n)+(1-s_n)((1-\alpha )x_n+\alpha S_n{x}_n),\hfill \\ u_n=J_{{\zeta }_1}^{B_1}(I-{\zeta }_1{A}_1)J_{{\zeta }_2}^{B_2}(w_n-{\zeta }_2{A}_2{w}_n);\hfill \end{array}\right.$$  Step 2. Calculate $y_n=J_{{\lambda }_n}^B(u_n-{\lambda }_{n}A{u}_n)$;  Step 3. Calculate $z_n=J_{{\lambda }_n}^B(u_n-{\lambda }_{n}A{y}_n+r_n(y_n-u_n))$;  Step 4. Calculate $x_{n+1}={\alpha }_{n}u+{\beta }_n{u}_n+{\gamma }_n((1-\alpha )z_n+\alpha S_n{z}_n)$, where u is a fixed element in C, $\{r_n\},\{s_n\},\{{\alpha }_n\}$, $\{{\beta }_n\},\{{\gamma }_n\}\subset (0,1]$ with $\{{\lambda }_n\}\subset (0,\infty )$ and ${\alpha }_n+{\beta }_n+{\gamma }_n=1$.  Set $n:=n+1$ and go to Step 1.

Lemma 11.

Let the vector sequence $\{x_n\}$ be constructed by Algorithm 1. One has that the sequence $\{x_n\}$ is bounded.

Proof. 

Putting $v_n=J_{{\zeta }_2}^{B_2}(w_n-{\zeta }_2{A}_2{w}_n)$ and $S_{\alpha ,n}:=(1-\alpha )I+\alpha S_n\forall n\ge 0$, we know from Lemma 1 that each $S_{\alpha ,n}:C\to C$ is nonexpansive with $\mathrm{Fix}(S_{\alpha ,n})=\mathrm{Fix}(S_n)$. Let $p\in \Omega :={\bigcap }_{n=0}^{\infty }\mathrm{Fix}(S_n)\cap \mathrm{Fix}(G)\cap {(A+B)}^{-1}0$. Then we observe that $p=Gp=S_{n}p=J_{{\lambda }_n}^B((1-r_n)p+r_n(p-\frac{{\lambda }_n}{r_n}Ap)).$ By use of Lemmas 2 and 3, we deduce that $I-{\zeta }_1{A}_1,I-{\zeta }_2{A}_2$ and $G:=J_{{\zeta }_1}^{B_1}(I-{\zeta }_1{A}_1)J_{{\zeta }_2}^{B_2}(I-{\zeta }_2{A}_2)$ are nonexpansive mappings. It is clear that there is only one element $w_n\in C$ satisfying

$$w_n=s_n((1-\xi )x_n+\xi G{w}_n)+(1-s_n)(\alpha S_n{x}_n+(1-\alpha )x_n).$$

Since G and $S_{\alpha ,n}$ are both nonexpansive mappings, we get

$$\begin{array}{cc}\parallel w_n-p\parallel \hfill & \le s_n((1-\xi )\parallel x_n-p\parallel +\xi \parallel G{w}_n-p\parallel )+(1-s_n)\parallel S_{\alpha ,n}{x}_n-p\parallel \hfill \\ & \le s_n(\xi \parallel w_n-p\parallel +(1-\xi )\parallel x_n-p\parallel )+(1-s_n)\parallel x_n-p\parallel ,\hfill \end{array}$$

and hence $\parallel w_n-p\parallel \le \parallel x_n-p\parallel$$\forall n\ge 0.$ Using the nonexpansivity of G again, we deduce from $u_n=G{w}_n$ that $\parallel u_n-p\parallel \le \parallel w_n-p\parallel \le \parallel x_n-p\parallel$. By use of Lemmas 2 and 4, we have

$$\begin{array}{cc}\parallel y_n{-p\parallel }^q\hfill & \le \parallel (I-{\lambda }_{n}A)p-(I-{\lambda }_{n}A)u_n{\parallel }^q\hfill \\ & \le \parallel u_n{-p\parallel }^q-{\lambda }_n(\sigma q-{\kappa }_q{\lambda }_n^{q-1}){\parallel A{u}_n-Ap\parallel }^q,\hfill \end{array}$$

which leads to $\parallel y_n-p\parallel \le \parallel u_n-p\parallel$. On the other hand,

$$\begin{array}{cc}\parallel z_n{-p\parallel }^q\hfill & =\parallel J_{{\lambda }_n}^B((1-r_n)u_n+r_n(y_n-\frac{{\lambda }_n}{r_n}A{y}_n))-J_{{\lambda }_n}^B((1-r_n)p+r_n(p-\frac{{\lambda }_n}{r_n}Ap)){\parallel }^q\hfill \\ & \le r_n\parallel (I-\frac{{\lambda }_n}{r_n}A)y_n-(I-\frac{{\lambda }_n}{r_n}A){p\parallel }^q+(1-r_n){\parallel u_n-p\parallel }^q\hfill \\ & \le r_n[\parallel y_n{-p\parallel }^q-\frac{{\lambda }_n}{r_n}(\sigma q-\frac{{\kappa }_q{\lambda }_n^{q-1}}{r_n^{q-1}})\parallel A{y}_n{-Ap\parallel }^q]+(1-r_n){\parallel u_n-p\parallel }^q\hfill \\ & \le {\lambda }_n(\frac{{\kappa }_q{\lambda }_n^{q-1}}{r_n^{q-1}}-\sigma q)\parallel A{y}_n{-Ap\parallel }^q+r_n{\lambda }_n({\kappa }_q{\lambda }_n^{q-1}-\sigma q)\parallel A{u}_n{-Ap\parallel }^q+{\parallel u_n-p\parallel }^q.\hfill \end{array}$$

This ensures that $\parallel z_n-p\parallel \le \parallel u_n-p\parallel$. So it follows from (3.2) that

$$\begin{array}{cc}\parallel x_{n+1}-p\parallel \hfill & \le {\alpha }_n\parallel p-u\parallel +{\beta }_n\parallel u_n-p\parallel +{\gamma }_n\parallel S_{\alpha ,n}{z}_n-p\parallel \hfill \\ & \le {\alpha }_n\parallel p-u\parallel +(1-{\alpha }_n)\parallel x_n-p\parallel ,\hfill \end{array}$$

which leads to $\parallel x_n-p\parallel \le max\{\parallel p-u\parallel ,\parallel p-x_0\parallel \}$. □

Theorem 1.

Suppose that $\{x_n\}$ is the vector sequence generated/defined by Iterative Algorithm 1. Assume that the parameter sequences satisfy ${\sum }_{n=0}^{\infty }{\alpha }_n=\infty$ and ${lim}_{n\to \infty }{\alpha }_n=0$; $0<a\le {\beta }_n\le b<1$ and $0<c\le s_n\le d<1$; $0<r\le r_n<1$ and $0<\lambda \le {\lambda }_n<\frac{{\lambda }_n}{r_n}\le \mu <{(\frac{\sigma q}{{\kappa }_q})}^{\frac{1}{q-1}}$; $0<{\zeta }_i<{(\frac{{\sigma }_{i}q}{{\kappa }_q})}^{\frac{1}{q-1}}$, $i=1,2$. Suppose ${\sum }_{n=0}^{\infty }{sup}_{x\in D}\parallel S_{n+1}x-S_{n}x\parallel <\infty$, where D is a bounded set in set C. Define the mapping S by $Sx={lim}_{n\to \infty }{S}_{n}x$ for all $x\in C$. Then the sequence $\{x_n\}$ converges to $x^{*}\in \Omega$ strongly. The solution also uniquely solves $\langle x^{*}-u,J(x^{*}-\overline{x})\rangle \le 0$ for all $\overline{x}\in \Omega$.

Proof. 

First, we set $v_n=J_{{\zeta }_2}^{B_2}(w_n-{\zeta }_2{A}_2{w}_n)$, $x^{*}\in \Omega$ and $y^{*}=J_{{\zeta }_2}^{B_2}(x^{*}-{\zeta }_2{A}_2{x}^{*})$. Since $u_n=J_{{\zeta }_1}^{B_1}(I-{\zeta }_1{A}_1)v_n$ and $v_n=J_{{\zeta }_2}^{B_2}(I-{\zeta }_2{A}_2)w_n$, we have $u_n=G{w}_n$. Using Lemma 2 yields that

$$\parallel v_n-y^{*}{\parallel }^q\le {\zeta }_2({\kappa }_q{\zeta }_2^{q-1}-{\sigma }_{2}q)\parallel A_2{w}_n-A_2{x}^{*}{\parallel }^q+{\parallel w_n-x^{*}\parallel }^q$$

and

$$\parallel u_n-x^{*}{\parallel }^q\le {\zeta }_1({\kappa }_q{\zeta }_1^{q-1}-{\sigma }_{1}q)\parallel A_1{v}_n-A_1{y}^{*}{\parallel }^q+{\parallel v_n-y^{*}\parallel }^q.$$

Combining the last two inequalities, we have

$$\parallel u_n-x^{*}{\parallel }^q\le {\zeta }_2({\kappa }_q{\zeta }_2^{q-1}-{\sigma }_{2}q)\parallel A_2{w}_n-A_2{x}^{*}{\parallel }^q+{\zeta }_1({\kappa }_q{\zeta }_1^{q-1}-{\sigma }_{1}q)\parallel A_1{v}_n-A_1{y}^{*}{\parallel }^q+{\parallel w_n-x^{*}\parallel }^q.$$

Using (1) and Lemma 4, we obtain that

$$\begin{array}{c}\parallel x_{n+1}-x^{*}{\parallel }^q\hfill \\ \le q{\alpha }_n\langle u-x^{*},J_q(x_{n+1}-x^{*})\rangle +{\parallel {\beta }_n(u_n-x^{*})+{\gamma }_n(S_{\alpha ,n}{z}_n-x^{*})\parallel }^q\hfill \\ \le {\beta }_n\parallel x^{*}-u_n{\parallel }^q+{\gamma }_n[\parallel u_n-x^{*}{\parallel }^q-r_n{\lambda }_n(\sigma q-{\kappa }_q{\lambda }_n^{q-1}){\parallel A{u}_n-A{x}^{*}\parallel }^q\hfill \\ -{\lambda }_n(\sigma q-\frac{{\kappa }_q{\lambda }_n^{q-1}}{r_n^{q-1}})\parallel A{y}_n-A{x}^{*}{\parallel }^q]-W_{q}g(\parallel u_n-S_{\alpha ,n}{z}_n\parallel )+q{\alpha }_n\langle u-x^{*},J_q(x_{n+1}-x^{*})\rangle \hfill \\ \le (1-{\alpha }_n)\parallel x_n-x^{*}{\parallel }^q-{\gamma }_n[{\zeta }_2({\sigma }_{2}q-{\kappa }_q{\zeta }_2^{q-1}){\parallel A_2{w}_n-A_2{x}^{*}\parallel }^q\hfill \\ +{\zeta }_1({\sigma }_{1}q-{\kappa }_q{\zeta }_1^{q-1})\parallel A_1{v}_n-A_1{y}^{*}{\parallel }^q+r_n{\lambda }_n(\sigma q-{\kappa }_q{\lambda }_n^{q-1}){\parallel A{u}_n-A{x}^{*}\parallel }^q\hfill \\ +{\lambda }_n(\sigma q-\frac{{\kappa }_q{\lambda }_n^{q-1}}{r_n^{q-1}})\parallel A{x}^{*}-A{y}_n{\parallel }^q]-W_{q}g(\parallel u_n-S_{\alpha ,n}{z}_n\parallel )+{\alpha }_{n}q\langle u-x^{*},J_q(x_{n+1}-x^{*})\rangle .\hfill \end{array}$$

(3)

Put

$$\begin{array}{c}{\Gamma }_n={\parallel x_n-x^{*}\parallel }^q,\hfill \\ {\eta }_n={\gamma }_n[{\zeta }_2({\sigma }_{2}q-{\kappa }_q{\zeta }_2^{q-1})\parallel A_2{w}_n-A_2{x}^{*}{\parallel }^q+{\zeta }_1({\sigma }_{1}q-{\kappa }_q{\zeta }_1^{q-1}){\parallel A_1{v}_n-A_1{y}^{*}\parallel }^q\hfill \\ -{\lambda }_n{r}_n({\kappa }_q{\lambda }_n^{q-1}-q\sigma )\parallel A{u}_n-A{x}^{*}{\parallel }^q+(\sigma q-\frac{{\kappa }_q{\lambda }_n^{q-1}}{r_n^{q-1}}){\lambda }_n\parallel A{y}_n-A{x}^{*}{\parallel }^q]\hfill \\ -W_{q}g(\parallel u_n-S_{\alpha ,n}{z}_n\parallel )\hfill \\ {\delta }_n=\langle u-x^{*},J_q(x_{n+1}-x^{*})\rangle q{\alpha }_n.\hfill \end{array}$$

Then (3) can be rewritten as the following formula:

$${\Gamma }_{n+1}+{\eta }_n\le (1-{\alpha }_n){\Gamma }_n+{\delta }_n.$$

(4)

 □

We next give two possible cases.

Case 1. We assume that there is an integer $n_0\ge 1$ with the restriction that $\{{\Gamma }_n\}$ is non-increasing. From (4), we get ${\eta }_n\le {\Gamma }_n-{\alpha }_n{\Gamma }_n-{\Gamma }_{n+1}+{\delta }_n.$ From ${lim}_{n\to \infty }{\alpha }_n={lim}_{n\to \infty }{\delta }_n=0$, one sees that ${lim}_{n\to \infty }{\eta }_n=0$. It is easy to see from Lemma 4 that ${lim}_{n\to \infty }g(\parallel u_n-S_{\alpha ,n}{z}_n\parallel )=0$,

$$\underset{n\to \infty }{lim}\parallel A_2{w}_n-A_2{x}^{*}\parallel =\underset{n\to \infty }{lim}\parallel A_1{v}_n-A_1{y}^{*}\parallel =0$$

(5)

and

$$\underset{n\to \infty }{lim}\parallel A{u}_n-A{x}^{*}\parallel =\underset{n\to \infty }{lim}\parallel A{y}_n-A{x}^{*}\parallel =0.$$

(6)

From the fact that g is a strictly increasing, continuous and convex function with $g(0)=0$, one has

$$\underset{n\to \infty }{lim}\parallel u_n-S_{\alpha ,n}{z}_n\parallel =0.$$

(7)

By use of Lemma 5, we get

$$\begin{array}{cc}\parallel v_n-y^{*}{\parallel }^q\hfill & \le \langle w_n-{\zeta }_2{A}_2{w}_n-(x^{*}-{\zeta }_2{A}_2{x}^{*}),J_q(v_n-y^{*})\rangle \hfill \\ & \le \frac{1}{q}[\parallel w_n-x^{*}{\parallel }^q+(q-1)\parallel v_n-y^{*}{\parallel }^q-{\tilde{h}}_1(\parallel w_n-x^{*}-v_n+y^{*}\parallel )]\hfill \\ & +{\zeta }_2\langle A_2{x}^{*}-A_2{w}_n,J_q(v_n-y^{*})\rangle ,\hfill \end{array}$$

where ${\tilde{h}}_1$ is a convex, strictly increasing, continuous function as in Lemma 5. This hence entails

$$\parallel v_n-y^{*}{\parallel }^q\le \parallel w_n-x^{*}{\parallel }^q-{\tilde{h}}_1(\parallel w_n-v_n-x^{*}+y^{*}\parallel )+q{\zeta }_2\parallel A_2{x}^{*}-A_2{w}_n\parallel \parallel v_n-y^{*}{\parallel }^{q-1}.$$

In a similar way, one concludes

$$\begin{array}{cc}\parallel u_n-x^{*}{\parallel }^q\hfill & \le \langle v_n-{\zeta }_1{A}_1{v}_n-(y^{*}-{\zeta }_1{A}_1{y}^{*}),J_q(u_n-x^{*})\rangle \hfill \\ & \le \frac{1}{q}[\parallel v_n-y^{*}{\parallel }^q+(q-1)\parallel u_n-x^{*}{\parallel }^q-{\tilde{h}}_2(\parallel v_n-y^{*}-u_n+x^{*}\parallel )]\hfill \\ & +{\zeta }_1\langle A_1{y}^{*}-A_1{v}_n,J_q(u_n-x^{*})\rangle ,\hfill \end{array}$$

which hence entails

$$\begin{array}{cc}\hfill & \parallel u_n-x^{*}{\parallel }^q\hfill \\ \hfill & \le \parallel v_n-y^{*}{\parallel }^q-{\tilde{h}}_2(\parallel v_n-y^{*}-u_n+x^{*}\parallel )+q{\zeta }_1\parallel A_1{y}^{*}-A_1{v}_n\parallel \parallel u_n-x^{*}{\parallel }^{q-1}\hfill \\ \hfill & \le \parallel x_n-x^{*}{\parallel }^q-{\tilde{h}}_1(\parallel w_n-v_n-x^{*}+y^{*}\parallel )+q{\zeta }_2\parallel A_2{x}^{*}-A_2{w}_n\parallel \parallel v_n-y^{*}{\parallel }^{q-1}\hfill \\ \hfill & -{\tilde{h}}_2(\parallel v_n-u_n+x^{*}-y^{*}\parallel )+q{\zeta }_1\parallel A_1{y}^{*}-A_1{v}_n\parallel \parallel u_n-x^{*}{\parallel }^{q-1}.\hfill \end{array}$$

(8)

Note that

$$\begin{array}{cc}q\parallel y_n-x^{*}{\parallel }^q\hfill & \le q\langle (u_n-{\lambda }_{n}A{u}_n)-x^{*}+{\lambda }_{n}A{x}^{*}),J_q(y_n-x^{*})\rangle \hfill \\ & \le \parallel (u_n-{\lambda }_{n}A{u}_n)-x^{*}+{\lambda }_{n}A{x}^{*}{)\parallel }^q-h_1(\parallel u_n-y_n-{\lambda }_n(A{u}_n-A{x}^{*})\parallel )\hfill \\ & +(q-1)\parallel y_n-x^{*}{\parallel }^q,\hfill \end{array}$$

which leads us to

$$\begin{array}{cc}\parallel y_n-x^{*}{\parallel }^q\hfill & \le \parallel u_n-x^{*}{\parallel }^q-h_1(\parallel u_n-y_n-{\lambda }_n(A{u}_n-A{x}^{*})\parallel ).\hfill \end{array}$$

From (8), one has

$$\begin{array}{c}\parallel x_{n+1}-x^{*}{\parallel }^q\hfill \\ \le {\gamma }_n\{(1-r_n)\parallel u_n-x^{*}{\parallel }^q+r_n[\parallel u_n-x^{*}{\parallel }^q+{\alpha }_n{\parallel x^{*}-u\parallel }^q\hfill \\ -h_1(\parallel u_n-{\lambda }_n(A{u}_n-A{x}^{*})-y_n\parallel )]\}+{\beta }_n{\parallel u_n-x^{*}\parallel }^q\hfill \\ \le {\beta }_n\parallel x_n-x^{*}{\parallel }^q+{\alpha }_n\parallel x^{*}{-u\parallel }^q+{\gamma }_n\{\parallel x_n-x^{*}{\parallel }^q-{\tilde{h}}_1(\parallel w_n-v_n-x^{*}+y^{*}\parallel )\hfill \\ -{\tilde{h}}_2(\parallel v_n-u_n+x^{*}-y^{*}\parallel )+q{\zeta }_1\parallel A_1{y}^{*}-A_1{v}_n\parallel \parallel u_n-x^{*}{\parallel }^{q-1}\hfill \\ +q{\zeta }_2\parallel A_2{x}^{*}-A_2{w}_n\parallel \parallel v_n-y^{*}{\parallel }^{q-1}-r_n{h}_1(\parallel u_n-{\lambda }_n(A{u}_n-A{x}^{*})-y_n\parallel )\}\hfill \\ \le \parallel x_n-x^{*}{\parallel }^q+{\alpha }_n\parallel x^{*}{-u\parallel }^q-{\gamma }_n\{{\tilde{h}}_1(\parallel w_n-v_n-x^{*}+y^{*}\parallel )\hfill \\ +{\tilde{h}}_2(\parallel v_n-u_n+x^{*}-y^{*}\parallel )+r_n{h}_1(\parallel u_n-{\lambda }_n(A{u}_n-A{x}^{*})-y_n\parallel )\}\hfill \\ +q{\zeta }_1\parallel A_1{y}^{*}-A_1{v}_n\parallel \parallel u_n-x^{*}{\parallel }^{q-1}+q{\zeta }_2\parallel A_2{x}^{*}-A_2{w}_n\parallel \parallel v_n-y^{*}{\parallel }^{q-1},\hfill \end{array}$$

which immediately yields that

$$\begin{array}{c}{\gamma }_n\{{\tilde{h}}_1(\parallel w_n-v_n-x^{*}+y^{*}\parallel )+{\tilde{h}}_2(\parallel v_n-u_n+x^{*}-y^{*}\parallel )\hfill \\ +r_n{h}_1(\parallel u_n-{\lambda }_n(A{u}_n-A{x}^{*})-y_n\parallel )\}\hfill \\ \le {\alpha }_n\parallel u-x^{*}{\parallel }^q+{\Gamma }_n-{\Gamma }_{n+1}+q{\zeta }_1\parallel A_1{y}^{*}-A_1{v}_n\parallel \parallel u_n-x^{*}{\parallel }^{q-1}\hfill \\ +q{\zeta }_2\parallel A_2{x}^{*}-A_2{w}_n\parallel \parallel v_n-y^{*}{\parallel }^{q-1}.\hfill \end{array}$$

Since ${\tilde{h}}_1(0)={\tilde{h}}_2(0)=h_1(0)=0$ and the fact that ${\tilde{h}}_1,{\tilde{h}}_2$ and $h_1$ are strictly increasing, continuous and convex functions, one concludes that ${lim}_{n\to \infty }\parallel w_n-v_n-x^{*}+y^{*}\parallel ={lim}_{n\to \infty }\parallel v_n-u_n+x^{*}-y^{*}\parallel ={lim}_{n\to \infty }\parallel u_n-y_n\parallel$. This immediately implies that

$$\underset{n\to \infty }{lim}\parallel w_n-u_n\parallel =\underset{n\to \infty }{lim}\parallel u_n-y_n\parallel =0.$$

(9)

Furthermore, borrowing $w_n=s_n((1-\xi )x_n+\xi G{w}_n)+(1-s_n)S_{\alpha ,n}{x}_n$ defined as the first step in the iteration procedure, we obtain that

$$\begin{array}{c}\parallel w_n-x^{*}{\parallel }^q=\langle s_n((1-\xi )x_n+\xi G{w}_n)+(1-s_n)S_{\alpha ,n}{x}_n-x^{*},J_q(w_n-x^{*})\rangle \hfill \\ =s_n[(1-\xi )\langle x_n-x^{*},J_q(w_n-x^{*})\rangle +\xi \langle G{w}_n-x^{*},J_q(w_n-x^{*})\rangle ]\hfill \\ +(1-s_n)\langle S_{\alpha ,n}{x}_n-x^{*},J_q(w_n-x^{*})\rangle \hfill \\ \le s_n[(1-\xi )\langle x_n-x^{*},J_q(w_n-x^{*})\rangle +\xi \parallel w_n-x^{*}{\parallel }^q]+(1-s_n)\langle S_{\alpha ,n}{x}_n-x^{*},J_q(w_n-x^{*})\rangle ,\hfill \end{array}$$

which yields

$$\begin{array}{cc}\parallel w_n-x^{*}{\parallel }^q\hfill & \le \frac{1}{1-s_n\xi }[s_n(1-\xi )\langle x_n-x^{*},J_q(w_n-x^{*})\rangle +(1-s_n)\langle S_{\alpha ,n}{x}_n-x^{*},J_q(w_n-x^{*})\rangle ]\hfill \\ & \le \frac{1}{q}[\parallel x_n-x^{*}{\parallel }^q+(q-1)\parallel w_n-x^{*}{\parallel }^q]-\frac{1}{1-s_n\xi }[\frac{s_n(1-\xi )}{q}{h}_3(\parallel x_n-w_n\parallel )\hfill \\ & +\frac{1-s_n}{q}{\tilde{h}}_3(\parallel S_{\alpha ,n}{x}_n-w_n\parallel )].\hfill \end{array}$$

This further implies that

$$\begin{array}{cc}\parallel u_n-x^{*}{\parallel }^q\hfill & \le \parallel w_n-x^{*}{\parallel }^q\hfill \\ & \le \parallel x_n-x^{*}{\parallel }^q-\frac{1}{1-s_n\xi }[s_n(1-\xi )h_3(\parallel x_n-w_n\parallel )+(1-s_n){\tilde{h}}_3(\parallel S_{\alpha ,n}{x}_n-w_n\parallel )].\hfill \end{array}$$

(10)

In a similar way, one further concludes

$$\begin{array}{cc}\parallel z_n-x^{*}{\parallel }^q\hfill & \le \parallel (x^{*}-{\lambda }_{n}A{x}^{*})-(u_n-{\lambda }_{n}A{y}_n+r_n(y_n-u_n)){\parallel }^q\hfill \\ & -h_2(\parallel u_n+(r_n{y}_n-r_n{u}_n)-({\lambda }_{n}A{y}_n-{\lambda }_{n}A{x}^{*})-z_n\parallel )\hfill \\ & \le \parallel u_n-x^{*}{\parallel }^q-h_2(\parallel u_n+(r_n{y}_n-r_n{u}_n)-({\lambda }_{n}A{y}_n-{\lambda }_{n}A{x}^{*})-z_n\parallel ).\hfill \end{array}$$

Using (10) leads us to

$$\begin{array}{cc}\parallel x_{n+1}-x^{*}{\parallel }^q\hfill & \le {\gamma }_n[\parallel u_n-x^{*}{\parallel }^q+{\beta }_n{\parallel u_n-x^{*}\parallel }^q\hfill \\ & -h_2(\parallel u_n+(r_n{y}_n-r_n{u}_n)-({\lambda }_{n}A{y}_n-{\lambda }_{n}A{x}^{*})-z_n\parallel )]+{\alpha }_n{\parallel x^{*}-u\parallel }^q\hfill \\ & \le \parallel x_n-x^{*}{\parallel }^q-{\gamma }_n\{\frac{1}{1-s_n\xi }[s_n(1-\xi )h_3(\parallel x_n-w_n\parallel )\hfill \\ & +(1-s_n){\tilde{h}}_3(\parallel S_{\alpha ,n}{x}_n-w_n\parallel )]+h_2(\parallel u_n+(r_n{y}_n-r_n{u}_n)-({\lambda }_{n}A{y}_n-{\lambda }_{n}A{x}^{*})-z_n\parallel )\}\hfill \\ & +{\alpha }_n{\parallel x^{*}-u\parallel }^q.\hfill \end{array}$$

Hence,

$$\begin{array}{c}{\gamma }_n\{\frac{1}{1-s_n\xi }[s_n(1-\xi )h_3(\parallel x_n-w_n\parallel )+(1-s_n){\tilde{h}}_3(\parallel S_{\alpha ,n}{x}_n-w_n\parallel )]\hfill \\ +h_2(\parallel u_n+(r_n{y}_n-r_n{u}_n)-({\lambda }_{n}A{y}_n-{\lambda }_{n}A{x}^{*})-z_n\parallel )\}\hfill \\ \le {\Gamma }_n+{\alpha }_n{\parallel u-x^{*}\parallel }^q-{\Gamma }_{n+1}.\hfill \end{array}$$

Note that $h_2(0)=h_3(0)={\tilde{h}}_3(0)=0$ and the fact that $h_2,h_3$ and ${\tilde{h}}_3$ are strictly increasing, continuous and convex functions. From (6) and (9) we have

$$\underset{n\to \infty }{lim}\parallel x_n-w_n\parallel =\underset{n\to \infty }{lim}\parallel S_{\alpha ,n}{x}_n-w_n\parallel =\underset{n\to \infty }{lim}\parallel u_n-z_n\parallel =0.$$

(11)

By using (9) and (11), one concludes ${lim}_{n\to \infty }\parallel x_n-u_n\parallel ={lim}_{n\to \infty }\parallel x_n-z_n\parallel =0.$ It follows that

$$\begin{array}{cc}\parallel x_n-G{x}_n\parallel \hfill & \le \parallel x_n-u_n\parallel +\parallel u_n-G{x}_n\parallel \hfill \\ \hfill & \le \parallel x_n-u_n\parallel +\parallel w_n-x_n\parallel \to 0(n\to \infty ).\hfill \end{array}$$

(12)

Thanks to (11), we get ${lim}_{n\to \infty }\parallel S_{\alpha ,n}{x}_n-x_n\parallel =0,$ which, together with $S_{\alpha ,n}{x}_n-x_n=\alpha (S_n{x}_n-x_n)$, leads to

$$\parallel S_n{x}_n-x_n\parallel =\frac{1}{\alpha }\parallel S_{\alpha ,n}{x}_n-x_n\parallel \to 0(n\to \infty ).$$

(13)

From the boundedness of $\{x_n\}$ and setting $D=\overline{\mathrm{conv}}\{x_n:n\ge 0\}$, we have ${\sum }_{n=1}^{\infty }{sup}_{x\in D}\parallel S_{n}x-S_{n-1}x\parallel <\infty$. Lemma 6 yields that ${lim}_{n\to \infty }{sup}_{x\in D}\parallel S_{n}x-Sx\parallel =0$. So, ${lim}_{n\to \infty }\parallel S_n{x}_n-S{x}_n\parallel =0.$ Further, from (13), we have

$$\parallel x_n-S{x}_n\parallel \le \parallel x_n-S_n{x}_n\parallel +\parallel S_n{x}_n-S{x}_n\parallel \to 0(n\to \infty ).$$

(14)

Letting $S_{\alpha }x:=(1-\alpha )x+\alpha Sx\forall x\in C$, we deduce from Lemma 1 that $S_{\alpha }:C\to C$ is a nonexpansive mapping. It is easy to see from (14) that ${lim}_{n\to \infty }\parallel x_n-S_{\alpha }{x}_n\parallel =0.$ For each $n\ge 0$, set $T_{{\lambda }_n}:=J_{{\lambda }_n}^B(I-{\lambda }_{n}A)$. It follows that

$$\begin{array}{cc}\parallel x_n-T_{{\lambda }_n}{x}_n\parallel \hfill & \le \parallel x_n-u_n\parallel +\parallel u_n-T_{{\lambda }_n}{u}_n\parallel +\parallel T_{{\lambda }_n}{u}_n-T_{{\lambda }_n}{x}_n\parallel \hfill \\ & \le 2\parallel x_n-u_n\parallel +\parallel u_n-y_n\parallel \to 0(n\to \infty ).\hfill \end{array}$$

In light of $0<\lambda \le {\lambda }_n$ for all $n\ge 0$, we obtain

$$\parallel T_{\lambda }{x}_n-x_n\parallel \le 2\parallel T_{{\lambda }_n}{x}_n-x_n\parallel \to 0(n\to \infty ).$$

(15)

We define a mapping $\Psi :C\to C$ by $\Psi x:={\theta }_1{S}_{\alpha }x+{\theta }_{2}Gx+(1-{\theta }_1-{\theta }_2)T_{\lambda }x$ $\forall x\in C$ with ${\theta }_1+{\theta }_2<1$ for constants ${\theta }_1,{\theta }_2\in (0,1)$. Lemma 7 guarantees that $\Psi$ is nonexpansive and

$$\mathrm{Fix}(\Psi )=\mathrm{Fix}(S_{\alpha })\cap \mathrm{Fix}(G)\cap \mathrm{Fix}(T_{\lambda })=\bigcap _{n=0}^{\infty }\mathrm{Fix}(S_n)\cap \mathrm{Fix}(G)\cap {(A+B)}^{-1}0(=:\Omega ).$$

Taking into account that

$$\parallel \Psi x_n-x_n\parallel \le {\theta }_1\parallel S_{\alpha }{x}_n-x_n\parallel +{\theta }_2\parallel G{x}_n-x_n\parallel +(1-{\theta }_1-{\theta }_2)\parallel T_{\lambda }{x}_n-x_n\parallel ,$$

we deduce from (12) and (15) that

$$\underset{n\to \infty }{lim}\parallel \Psi x_n-x_n\parallel =0.$$

(16)

Let $z^{\lambda }=\lambda u+(1-\lambda )\Psi z^{\lambda },$ $\forall \lambda \in (0,1)$. Lemma 8 guarantees that $\{z^{\lambda }\}$ converges to a point $x^{*}\in \mathrm{Fix}(\Psi )=\Omega$ in norm, and $x^{*}$ further solves the VIP: $\langle x^{*}-u,J(x^{*}-p)\rangle \le 0$, $\forall p\in \Omega .$ From (1), we have

$$\begin{array}{cc}\parallel z^{\lambda }-x_n{\parallel }^q\hfill & \le \lambda q\parallel z^{\lambda }-x_n{\parallel }^q+{(1-\lambda )}^q(\parallel \Psi z^{\lambda }-\Psi x_n\parallel +\parallel \Psi x_n-x_n{\parallel )}^q+\lambda q\langle u-z^{\lambda },J_q(z^{\lambda }-x_n)\rangle \hfill \\ & \le \lambda q\parallel z^{\lambda }-x_n{\parallel }^q+{(1-\lambda )}^q(\parallel \Psi x_n-x_n\parallel +\parallel z^{\lambda }-x_n{\parallel )}^q+\lambda q\langle u-z^{\lambda },J_q(z^{\lambda }-x_n)\rangle .\hfill \end{array}$$

Further, from (16), one has

$$\underset{n\to \infty }{lim\; sup}\langle u-z_t,J_q(x_n-z_t)\rangle \le M\frac{(qt-1)+{(1-t)}^q}{qt},$$

where M is a constant such that $\parallel z_t-x_n{\parallel }^q\le M$ for all $n\ge 0$ and $t\in (0,1)$. From the properties of $J_q$ and the fact that $z_t\to x^{*}$ as $t\to 0$, one gets ${lim}_{t\to 0}\parallel J_q(x_n-x^{*})-J_q(x_n-z_t)\parallel =0.$ A simple calculation indicates that

$$\underset{n\to \infty }{lim\; sup}\langle u-x^{*},J_q(x_n-x^{*})\rangle \le 0$$

(17)

and then

$$\begin{array}{cc}\parallel x_{n+1}-x_n\parallel \hfill & \le {\alpha }_n\parallel u-x_n\parallel +{\beta }_n\parallel u_n-x_n\parallel +{\gamma }_n(\parallel S_{\alpha ,n}{z}_n-u_n\parallel +\parallel u_n-x_n\parallel )\hfill \\ & \le {\alpha }_n\parallel u-x_n\parallel +\parallel u_n-x_n\parallel +\parallel S_{\alpha ,n}{z}_n-u_n\parallel \to 0(n\to \infty ).\hfill \end{array}$$

Using (17), we have ${lim\; sup}_{n\to \infty }\langle u-x^{*},J_q(x_{n+1}-x^{*})\rangle \le 0.$ An application of Lemma 9 yields that ${\Gamma }_n\to 0$ as $n\to \infty$. Thus, $x_n\to x^{*}$ as $n\to \infty$.

Case 2. We assume that there is $\{{\Gamma }_{k_i}\}\subset \{{\Gamma }_k\}$ s.t. ${\Gamma }_{k_i}<{\Gamma }_{k_i+1}\forall i\in N$, where N is the set of all positive integers. We now give a new mapping $\tau :N\to N$ by $\tau (k):=max\{i\le k:{\Gamma }_i<{\Gamma }_{i+1}\}.$ Using Lemma 10, one concludes

$${\Gamma }_{\tau (k)+1}\ge {\Gamma }_{\tau (k)}\mathrm{and}{\Gamma }_{\tau (k)+1}\ge {\Gamma }_k.$$

Putting ${\Gamma }_k={\parallel x_k-x^{*}\parallel }^q\forall k\in \mathbf{N}$ and using the same reasoning as in Case 1 we can obtain

$$\underset{k\to \infty }{lim}\parallel x_{\tau (k)}-x_{\tau (k)+1}\parallel =0$$

(18)

$$\underset{k\to \infty }{lim\; sup}\langle u-x^{*},J_q(x_{\tau (k)+1}-x^{*})\rangle \le 0.$$

(19)

In view of ${\alpha }_{\tau (k)}>0$ and ${\Gamma }_{\tau (k)+1}\ge {\Gamma }_{\tau (k)}$, we conclude that

$$\frac{q}{1-\delta }\langle u-x^{*},J_q(x_{\tau (k)+1}-x^{*})\rangle \ge {\parallel x_{\tau (k)}-x^{*}\parallel }^q.$$

Consequently, ${lim}_{k\to \infty }{\parallel x_{\tau (k)}-x^{*}\parallel }^q=0.$ Using Lemma 3, we have that

$$\begin{array}{c}\parallel x_{\tau (k)+1}-x^{*}{\parallel }^q-{\parallel x^{*}-x_{\tau (k)}\parallel }^q\hfill \\ \le q\langle x_{\tau (k)+1}-x_{\tau (k)},J_q(x_{\tau (k)}-x^{*})\rangle +{\kappa }_q{\parallel x_{\tau (k)+1}-x_{\tau (k)}\parallel }^q\hfill \\ \le q\parallel x_{\tau (k)}-x^{*}{\parallel }^{q-1}\parallel x_{\tau (k)+1}-x_{\tau (k)}\parallel +{\kappa }_q{\parallel x_{\tau (k)+1}-x_{\tau (k)}\parallel }^q\to 0(k\to \infty ).\hfill \end{array}$$

Thanks to ${\Gamma }_k\le {\Gamma }_{\tau (k)+1}$, we get

$$\parallel x_k-x^{*}{\parallel }^q\le \parallel x_{\tau (k)}-x^{*}{\parallel }^q+q\parallel x_{\tau (k)+1}-x_{\tau (k)}\parallel \parallel x_{\tau (k)}-x^{*}{\parallel }^{q-1}+{\kappa }_q{\parallel x_{\tau (k)+1}-x_{\tau (k)}\parallel }^q.$$

It is easy to see from (18) that $x_k\to x^{*}$ as $k\to \infty$. This completes the proof.

It is well known that ${\kappa }_2=1$ in Hilbert spaces. From Theorem 1, we derive the following conclusion.

Corollary 1.

Let $\varnothing \ne C\subset H$ be a closed convex set. Let ${\{S_n\}}_{n=0}^{\infty }$ be a family of ς-uniformly strict pseudocontraction mappings defined on C. Suppose that $B_1,B_2:C\to 2^H$ are both maximal monotone operators and $A_k:C\to H$ is ${\sigma }_k$-inverse-strongly monotone mapping for $k=1,2$. Define the mapping $G:C\to C$ by $G:=J_{{\zeta }_1}^{B_1}(I-{\zeta }_1{A}_1)J_{{\zeta }_2}^{B_2}(I-{\zeta }_2{A}_2)$ for constants ${\zeta }_1,{\zeta }_2>0$. Let $A:C\to H$ and $B:C\to 2^H$ be a σ-inverse-strongly monotone mapping and a maximal monotone operator, respectively. For any given $x_0\in C,\xi \in (0,1)$ and $\alpha \in (0,min\{1,2\varsigma \})$, let ${\{x_n\}}_{n=0}^{\infty }$ be the sequence generated by

$$\left\{\begin{array}{c}{w}_n=s_n((1-\xi )x_n+\xi G{w}_n)+(1-s_n)((1-\alpha )x_n+\alpha S_n{x}_n),\hfill \\ u_n=J_{{\zeta }_1}^{B_1}(I-{\zeta }_1{A}_1)J_{{\zeta }_2}^{B_2}(w_n-{\zeta }_2{A}_2{w}_n),\hfill \\ y_n=J_{{\lambda }_n}^B(u_n-{\lambda }_{n}A{u}_n),\hfill \\ z_n=J_{{\lambda }_n}^B(u_n-{\lambda }_{n}A{y}_n+r_n(y_n-u_n)),\hfill \\ x_{n+1}={\alpha }_{n}u+{\beta }_n{u}_n+{\gamma }_n((1-\alpha )z_n+\alpha S_n{z}_n)\forall n\ge 0,\hfill \end{array}\right.$$

where the sequences $\{r_n\},\{s_n\},\{{\alpha }_n\},\{{\beta }_n\},\{{\gamma }_n\}$ are sequences in $(0,1]$ with the additional restrictions ${\alpha }_n+{\beta }_n+{\gamma }_n=1$ and $\{{\lambda }_n\}\subset (0,\infty )$ are such that ${\sum }_{n=0}^{\infty }{\alpha }_n=\infty$, ${lim}_{n\to \infty }{\alpha }_n=0$ as $n\to \infty$; $0<a\le {\beta }_n\le b<1$ and $0<c\le s_n\le d<1$; $0<r\le r_n<1$ and $0<\lambda \le {\lambda }_n<\frac{{\lambda }_n}{r_n}\le \mu <2\sigma$; $0<{\zeta }_k<2{\sigma }_k$ for $k=1,2$. Assume that ${\sum }_{n=0}^{\infty }{sup}_{x\in D}\parallel S_{n+1}x-S_{n}x\parallel <\infty$, where D is a bounded subset of C. Define a self mapping S by $Sx={lim}_{n\to \infty }{S}_{n}x\forall x\in C$, and further assume that ${\bigcap }_{n=0}^{\infty }\mathrm{Fix}(S_n)=\mathrm{Fix}(S)$. If $\Omega :={\bigcap }_{n=0}^{\infty }\mathrm{Fix}(S_n)\cap \mathrm{Fix}(G)\cap {(A+B)}^{-1}0\ne \varnothing$, then ${lim}_{n\to \infty }\parallel x_n-x^{*}\parallel =0$, where $x^{*}\in \Omega$ uniquely solves $\langle x^{*}-u,p-x^{*}\rangle \ge 0\forall p\in \Omega$.

Next, we recall the least absolute shrinkage and selection operator (LASSO) [35], which can be formulated as a convex constrained optimization problem:

$$\underset{y\in H}{min}\frac{1}{2}{\parallel Ty-b\parallel }_2^2\mathrm{subject}\mathrm{to}{\parallel y\parallel }_1\le s,$$

(20)

where T is a bounded operator on H, b is a fixed vector in H, and $s>0$ is a real number. In this section, $\Lambda$ is employed to denote the set of solutions of LASSO (20). LASSO, which acts as a unified model for a number of real problems, has been investigated in different settings. Ones know that a solution to (20) is a minimizer to the following minimization problem: ${min}_{y\in H}g(y)+h(y),$ where $g(y):=\frac{1}{2}{\parallel Ty-b\parallel }_2^2,h(y):=\lambda {\parallel y\parallel }_1$. It is known that $\nabla g(y)=T^{*}(Ty-b)$ is $\frac{1}{\parallel T^{*}T\parallel }$-inverse-strongly monotone. Hence, we have that z solves the LASSO iff z solves the problem, which consists of finding $z\in H$ s.t.

$$\begin{array}{cc}0\in \partial h(z)+\nabla g(z)\hfill & \iff z-\lambda \nabla g(z)\in z+\lambda \partial h(z)\hfill \\ & \iff z={\mathrm{prox}}_h(z-\lambda \nabla g(z)),\hfill \end{array}$$

where $\lambda >0$ is real, and ${\mathrm{prox}}_h(y)$ is the proximal of $h(y):={\lambda \parallel y\parallel }_1$ defined as follows

$${\mathrm{prox}}_h(y)={\mathrm{argmin}}_{u\in H}{\{\lambda \parallel u\parallel }_1+\frac{1}{2}{\parallel u-y\parallel }_2^2\}\forall y\in H.$$

This is separable in indices. So, $y\in H$, for $i=1,2,\dots ,n$, ${\mathrm{prox}}_h(y)={\mathrm{prox}}_{{\lambda \parallel ·\parallel }_1}(y)=({\mathrm{prox}}_{\lambda |·|}(y_1),$ ${\mathrm{prox}}_{\lambda |·|}(y_2),$…$,{\mathrm{prox}}_{\lambda |·|}(y_n)),$ with ${\mathrm{prox}}_{\lambda |·|}(y_i)=\mathrm{sgn}(y_i)max\{|y_i|-\lambda ,0\}$.

By putting $C=H,A=\nabla g,B=\partial h$ and $\sigma =\frac{1}{\parallel T^{*}T\parallel }$ in Corollary 1, we obtain the following result immediately.

Corollary 2.

Let $A_k,B_k(k=1,2)$ and ${\{S_n\}}_{n=0}^{\infty }$ be the same as in Corollary 1 with $C=H$. Assume that $\Omega :={\bigcap }_{n=0}^{\infty }\mathrm{Fix}(S_n)\cap \mathrm{Fix}(G)\cap \Lambda \ne \varnothing$. For any given $x_0\in H,\xi \in (0,1)$ and $\alpha \in (0,min\{1,2\varsigma \})$, let ${\{x_n\}}_{n=0}^{\infty }$ be the sequence generated by

$$\left\{\begin{array}{c}{w}_n=s_n((1-\xi )x_n+\xi G{w}_n)+(1-s_n)((1-\alpha )x_n+\alpha S_n{x}_n),\hfill \\ u_n=J_{{\zeta }_1}^{B_1}(I-{\zeta }_1{A}_1)J_{{\zeta }_2}^{B_2}(w_n-{\zeta }_2{A}_2{w}_n),\hfill \\ y_n={\mathrm{prox}}_h(u_n-{\lambda }_n{T}^{*}(T{u}_n-b)),\hfill \\ z_n={\mathrm{prox}}_h(u_n-{\lambda }_n{T}^{*}(T{y}_n-b)+(r_n{y}_n-r_n{u}_n)),\hfill \\ x_{n+1}={\alpha }_{n}u+(1-\alpha ){\gamma }_n{z}_n+\alpha {\gamma }_n{S}_n{z}_n+{\beta }_n{u}_n\forall n\ge 0,\hfill \end{array}\right.$$

where the sequences $\{r_n\},\{s_n\},\{{\alpha }_n\},\{{\beta }_n\},\{{\gamma }_n\}\subset (0,1]$, ${\alpha }_n+{\beta }_n+{\gamma }_n=1$ and $\{{\lambda }_n\}\subset (0,\infty )$ are such that the conditions presented in Corollary 3.1 hold where $\sigma =\frac{1}{\parallel T^{*}T\parallel }$. Then $x_n\to x^{*}\in \Omega$ as $n\to \infty$.