Modified Mann Subgradient-like Extragradient Rules for Variational Inequalities and Common Fixed Points Involving Asymptotically Nonexpansive Mappings

Abstract

In a real Hilbert space, we aim to investigate two modified Mann subgradient-like methods to find a solution to pseudo-monotone variational inequalities, which is also a common fixed point of a finite family of nonexpansive mappings and an asymptotically nonexpansive mapping. We obtain strong convergence results for the sequences constructed by these proposed rules. We give some examples to illustrate our analysis.

1. Introduction

Let the $\langle ·,·\rangle$ and $\parallel ·\parallel$ represent the inner product and induced norm in a real Hilbert space H, respectively. We denote by $P_C$ the nearest point projection from H onto C, where $\varnothing \ne C\subset H$ and C is convex and closed. Given $T:C\to H$ a nonlinear mapping, we denote by $\mathrm{Fix}\left(T\right)$ the fixed point set of T, i.e., $\mathrm{Fix}\left(T\right)=\{x\in C:x=Tx\}$. Let the $\mathbf{R},\to$ and ⇀ indicate the set of all real numbers, the strong convergence, and the weak convergence, respectively. A self-mapping $T:C\to C$ is referred to as being asymptotically nonexpansive if $\exists \left\{{\psi }_n\right\}\subset [0,+\infty )$ s.t. ${lim}_{n\to \infty }{\psi }_n=0$ and

$$\parallel T^{n}x-T^{n}y\parallel \le \parallel x-y\parallel +{\psi }_n\parallel x-y\parallel \forall n\ge 1,x,y\in C$$

(1)

and T is nonexpansive when ${\psi }_n=0$.

Given a continuous mapping $A:H\to H$, a variational inequality problem (denoted by (VIP)) is:

$$\mathrm{find}{z}^{*}\in C\mathrm{such}\mathrm{that}\langle A{z}^{*},z-z^{*}\rangle \ge 0\forall z\in C.$$

Let us denote the set of the solution VIP by VI($C,A$). In 1976, Korpelevich [1] put forth the extragradient method, which has been one of the most effective approaches for solving the VIP:

$$\left\{\begin{array}{cc}& y_n=P_C(x_n-\zeta A{x}_n),\hfill \\ & x_{n+1}=P_C(x_n-\zeta A{y}_n)\forall n\ge 0,\hfill \end{array}\right.$$

(2)

for $\zeta \in (0,\frac{1}{L})$ with L being the Lipschitz constant of A. Weak convergence results of (2) have been obtained in studies [2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22] and references therein.

The extragradient method (2) involves solving a minimization problem over C at each iteration when $P_C$ has no closed-form solution. This could make the extragradient method (2) computationally expensive. In study [6], Censor et al. modified (2) and introduced the subgradient extragradient:

$$\left\{\begin{array}{cc}& y_n=P_C(x_n-\zeta A{x}_n),\hfill \\ & D_n=\{v\in H:\langle x_n-\zeta A{x}_n-y_n,v-y_n\rangle \le 0\},\hfill \\ & x_{n+1}=P_{D_n}(x_n-\zeta A{y}_n),\hfill \end{array}\right.$$

(3)

for $\zeta \in (0,\frac{1}{L})$ with L being the Lipschitz constant of A. Thong and Hieu [19] added an inertial extrapolation step to (3): $x_0,x_1\in H$,

$$\left\{\begin{array}{cc}& v_n=x_n+{\alpha }_n(x_n-x_{n-1}),\hfill \\ & y_n=P_C(v_n-\zeta A{w}_n),\hfill \\ & D_n=\{v\in H:\langle v_n-\zeta A{w}_n-y_n,v-y_n\rangle \le 0\},\hfill \\ & x_{n+1}=P_{C_n}(v_n-\zeta A{y}_n),\hfill \end{array}\right.$$

(4)

for $\zeta \in (0,\frac{1}{L})$ with L being the Lipschitz constant of A, and the weak convergence being obtained. In study [22], Reich et al. suggested the modified projection-type method for solving the VIP with the pseudo-monotone and uniformly continuous mapping A, given a sequence $\left\{{\alpha }_n\right\}\subset (0,1)$ and a contraction $f:C\to C$ with constant $\varrho \in [0,1)$. For any initial $x_1\in C$, the sequence $\left\{x_n\right\}$ is constructed below.

Furthermore, it was proven in study [22] that the sequence $\left\{x_n\right\}$ generated by Algorithm 1 converges strongly. Subsequently, Ceng, Yao and Shehu [21] proposed a Mann-type method of (2) to solve pseudo-monotone variational inequalities and the common fixed point problem of many finitely nonexpansive self-mappings ${\left\{T_i\right\}}_{i=1}^N$ on C and an asymptotically nonexpansive self-mapping $T_0:=T$ on C. Given a contraction $f:C\to C$ with constant $\varrho \in [0,1)$, let $\left\{{\sigma }_n\right\}\subset [0,1]$ and $\left\{{\alpha }_n\right\},\left\{{\beta }_n\right\},\left\{{\gamma }_n\right\}\subset (0,1)$ with ${\alpha }_n+{\beta }_n+{\gamma }_n=1\forall n\ge 1$ and $T_n:=T_{n\mathrm{mod}N}$. For any initial $x_1\in C$, the sequence $\left\{x_n\right\}$ is constructed below (Algorithm 2).

Algorithm 1 (see study [22]). Initialization: Given $\mu >0,l\in (0,1),\lambda \in (0,\frac{1}{\mu })$.

Iterative Steps: Given the current iterate $x_n$, calculate $x_{n+1}$ as follows: Step 1. Compute $y_n=P_C(x_n-\lambda A{x}_n)$ and $r_{\lambda }\left(x_n\right):=x_n-y_n$. If $r_{\lambda }\left(x_n\right)=0$, then stop. $x_n$ is a solution of $\mathrm{VI}(C,A)$. Otherwise; Step 2. Compute $w_n=x_n-{\zeta }_n{r}_{\lambda }\left(x_n\right)$, where ${\zeta }_n:=l^{j_n}$ and $j_n$ is the smallest nonnegative integer j, satisfying $\langle A{x}_n-A(x_n-l^j{r}_{\lambda }\left(x_n\right)),r_{\lambda }\left(x_n\right)\rangle \le \frac{\mu }{2}{\parallel r_{\lambda }\left(x_n\right)\parallel }^2$; Step 3. Compute $x_{n+1}={\alpha }_{n}f\left(x_n\right)+(1-{\alpha }_n)P_{C_n}\left(x_n\right)$, where $C_n:=\{x\in C:h_n\left(x_n\right)\le 0\}$ and $h_n\left(x\right)=\langle A{w}_n,x-x_n\rangle +\frac{{\zeta }_n}{2\lambda }{\parallel r_{\lambda }\left(x_n\right)\parallel }^2$.

Algorithm 2 (see study [21]). Initialization: Given $\mu >0,l\in (0,1),\lambda \in (0,\frac{1}{\mu })$.

Iterative Steps: Given $x_n$, compute Step 1. Set $w_n=(1-{\sigma }_n)x_n+{\sigma }_n{T}_n{x}_n$, and compute $y_n=P_C(w_n-\lambda A{w}_n)$ and $r_{\lambda }\left(w_n\right):=w_n-y_n$; Step 2. Compute $t_n=w_n-{\zeta }_n{r}_{\lambda }\left(w_n\right)$, where ${\zeta }_n:=l^{j_n}$ and $j_n$ is the smallest nonnegative integer j, satisfying $\langle A{w}_n-A(w_n-l^j{r}_{\lambda }\left(w_n\right)),w_n-y_n\rangle \le \frac{\mu }{2}{\parallel r_{\lambda }\left(w_n\right)\parallel }^2$; Step 3. Compute $z_n=P_{C_n}\left(w_n\right)$ and $x_{n+1}={\alpha }_{n}f\left(x_n\right)+{\beta }_n{x}_n+{\gamma }_n{T}^n{z}_n,$ where $C_n:=\{x\in C:h_n\left(x\right)\le 0\}$ and $h_n\left(x\right)=\langle A{t}_n,x-w_n\rangle +\frac{{\zeta }_n}{2\lambda }{\parallel r_{\lambda }\left(w_n\right)\parallel }^2$. Again set $n:=n+1$ and go to Step 1.

  Under suitable conditions, it was proven in study [21] that the sequence $\left\{x_n\right\}$ converges strongly to $x^{*}\in \mathsf{\Omega }={\bigcap }_{i=0}^N\mathrm{Fix}\left(T_i\right)\cap \mathrm{VI}(C,A)$ if and only if ${lim}_{n\to \infty }(\parallel x_n-x_{n+1}\parallel +\parallel x_n-y_n\parallel )=0$ provided $T^n{z}_n-T^{n+1}{z}_n\to 0$, where $x^{*}=P_{\mathsf{\Omega }}(I-\rho F+f)x^{*}$.

  In a real Hilbert space H, let the VIP and CFPP represent the pseudo-monotone variational inequality problem with uniformly continuous mapping A, the common fixed point problem of a finite family of nonexpansive mappings ${\left\{T_i\right\}}_{i=1}^N$, and an asymptotically nonexpansive mapping $T_0:=T$, respectively. Inspired by the above research works, we propose and analyze two modified Mann subgradient-like extragradient algorithms with the line-search process for solving the VIP and CFPP. The proposed algorithms are based on the Mann iteration method, (3) method with the line-search process, and the viscosity approximation method. Under some conditions, we establish some strong convergence results for the sequences constructed by these proposed rules. Finally, our main results are applied to handle the VIP and CFPP in an illustrated example.

The structure of the article is specified below. In Section 2, we first recall some concepts and basic results. Section 3 explores the strong convergence analysis of our proposed methods. Finally, in Section 4, an illustrated example is given. Our results complement related results by Ceng, Yao, and Shehu [21]; Reich et al. [22]; and Ceng and Shang [9]. Indeed, it is worth emphasizing that our problem of finding an element $x^{*}\in \mathsf{\Omega }={\bigcap }_{i=0}^N\mathrm{Fix}\left(T_i\right)\cap \mathrm{VI}(C,A)$ is more general and more interesting than the corresponding problem of finding an element $x^{*}\in \mathrm{VI}(C,A)$ in study [22]. Moreover, our strong convergence theorems are more advantageous and more clever than the corresponding strong convergence ones in studies [9,21] because the conclusion $x_n\to x^{*}\in \mathsf{\Omega }\iff \parallel x_n-x_{n+1}\parallel +\parallel x_n-y_n\parallel \to 0(n\to \infty )$ in the corresponding strong convergence theorems [9,21] is updated by our conclusion $x_n\to x^{*}\in \mathsf{\Omega }$. Without question, the strong convergence criteria for the sequence $\left\{x_n\right\}$ in this paper are more convenient and more beneficial in comparison with those of studies [9,21].

2. Preliminaries

We say that $T:C\to H$ is

(a) L-Lipschitz continuous (or L-Lipschitzian) if $\exists L>0$ such that $\parallel Tu-Tv\parallel \le L\parallel u-v\parallel \forall u,v\in C$;

(b) Monotone if $\langle Tu-Tv,u-v\rangle \ge 0\forall u,v\in C$;

(c) pseudo-monotone if $\langle Tu,v-u\rangle \ge 0\Rightarrow \langle Tv,v-u\rangle \ge 0\forall u,v\in C$;

(d) $\varpi$-strongly monotone if $\exists \varpi >0$ such that $\langle Tu-Tv,u-v\rangle \ge {\varpi \parallel u-v\parallel }^2\forall u,v\in C$;

(e) Sequentially weakly continuous if $\forall \left\{u_n\right\}\subset C$, we have $u_n\rightharpoonup u\Rightarrow T{u}_n\rightharpoonup Tu$.  

One can see that (b) implies (c) but the converse fails. Given $v\in H$, there exists a unique nearest point in C, denoted by $P_{C}v$ ($P_C$ is called a metric projection of H onto C), such that $\parallel v-P_{C}v\parallel \le \parallel v-z\parallel \forall z\in C$. According to reference [17], we know that the following hold:  

(a) $\langle u-v,P_{C}u-P_{C}v\rangle \ge {\parallel P_{C}u-P_{C}v\parallel }^2\forall u,v\in H$;

(b) $\langle u-P_{C}u,v-P_{C}u\rangle \le 0\forall u\in H,v\in C$;

(c) ${\parallel u-v\parallel }^2\ge \parallel u-P_C{u\parallel }^2+{\parallel v-P_{C}u\parallel }^2\forall u\in H,v\in C$;

(d) ${\parallel u-v\parallel }^2={\parallel u\parallel }^2-{\parallel v\parallel }^2-2\langle u-v,v\rangle \forall u,v\in H$;

(e) ${\parallel \lambda u+\mu v\parallel }^2={\lambda \parallel u\parallel }^2+{\mu \parallel v\parallel }^2-\lambda \mu {\parallel u-v\parallel }^2\forall u,v\in H,\forall \lambda ,\mu \in \mathbb{R}$ with $\lambda +\mu =1$.

Lemma 1

(see reference [4]). Let $H_1$ and $H_2$ be two real Hilbert spaces. Suppose that $A:H_1\to H_2$ is uniformly continuous on bounded subsets of $H_1$ and M is a bounded subset of $H_1$. Then, $A\left(M\right)$ is bounded.

It is easy from the subdifferential inequality of $\frac{1}{2}{\parallel ·\parallel }^2$:

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

Lemma 2

(see reference [23]). Let h be a real-valued function on H and define $K:=\{x\in C:h(x)\le 0\}$. If K is nonempty and h is Lipschitz continuous on C with modulus $\psi >0$, then $\mathrm{dist}(x,K)\ge {\psi }^{-1}max\{h\left(x\right),0\}\forall x\in C$, where $\mathrm{dist}(x,K)$ denotes the distance of x to K.

Lemma 3

(see reference [6], Lemma 2.1). Assume that $A:C\to H$ is pseudo-monotone and continuous. Then $u^{*}\in C$ is a solution to the VIP $\langle A{u}^{*},u-u^{*}\rangle \ge 0\forall u\in C$, if and only if $\langle Au,u-u^{*}\rangle \ge 0\forall u\in C$.

Lemma 4

(see reference [24]). Let $\left\{b_n\right\}$ be a sequence of nonnegative numbers satisfying: $b_{n+1}\le (1-{\varsigma }_n)b_n+{\varsigma }_n{\gamma }_n\forall n\ge 1$, where $\left\{{\varsigma }_n\right\}$ and $\left\{{\gamma }_n\right\}$ are sequences of real numbers such that (i) $\left\{{\varsigma }_n\right\}\subset [0,1]$ and ${\sum }_{n=1}^{\infty }{\varsigma }_n=\infty$, and (ii) ${lim\; sup}_{n\to \infty }{\gamma }_n\le 0$ or ${\sum }_{n=1}^{\infty }\left|{\varsigma }_n{\gamma }_n\right|<\infty$. Then ${lim}_{n\to \infty }{b}_n=0$.

Lemma 5

(see reference [25]). Let X be a Banach space which admits a weakly continuous duality mapping, C be a nonempty closed convex subset of X, and $T:C\to C$ be an asymptotically nonexpansive mapping with $\mathrm{Fix}\left(T\right)\ne \varnothing$. Then $I-T$ is demiclosed at zero, i.e., if $\left\{x_n\right\}$ is a sequence in C such that $x_n\rightharpoonup x\in C$ and $(I-T)x_n\to 0$, then $(I-T)x=0$, where I is the identity mapping of X.

Lemma 6

(see reference [26]). Let $\left\{{\mathrm{\Gamma }}_n\right\}$ be a sequence of real numbers such that there exists $\left\{{\mathrm{\Gamma }}_{n_k}\right\}$ of $\left\{{\mathrm{\Gamma }}_n\right\}$, satisfying ${\mathrm{\Gamma }}_{n_k}<{\mathrm{\Gamma }}_{n_k+1}$ for each integer $k\ge 1$. Define

$$\eta \left(n\right)=max\{k\le n:{\mathrm{\Gamma }}_k<{\mathrm{\Gamma }}_{k+1}\},$$

where integer $n_0\ge 1$ and $\{k\le n_0:{\mathrm{\Gamma }}_k<{\mathrm{\Gamma }}_{k+1}\}\ne \varnothing$. Then (i) $\eta \left(n_0\right)\le \eta (n_0+1)\le \cdots$ and $\eta \left(n\right)\to \infty$; (ii) ${\mathrm{\Gamma }}_{\eta \left(n\right)}\le {\mathrm{\Gamma }}_{\eta \left(n\right)+1}$ and ${\mathrm{\Gamma }}_n\le {\mathrm{\Gamma }}_{\eta \left(n\right)+1}\forall n\ge n_0$.

3. Our Contributions

Assume that

$T:C\to C$ is an asymptotically nonexpansive mapping and $T_i:C\to C$ is a nonexpansive mapping for $i=1,...,N$ such that the sequence ${\left\{T_n\right\}}_{n=1}^{\infty }$ is defined as in Algorithm 1.

$A:H\to H$ is pseudo-monotone and uniformly continuous on C, s.t. $\parallel Az\parallel \le {lim\; inf}_{n\to \infty }\parallel A{x}_n\parallel$ for each $\left\{x_n\right\}\subset C$ with $x_n\rightharpoonup z$.

$f:C\to C$ is a contraction with constant $\varrho \in [0,1)$, and $\mathsf{\Omega }={\bigcap }_{i=0}^N\mathrm{Fix}\left(T_i\right)\cap \mathrm{VI}(C,A)\ne \varnothing$ with $T_0:=T$.

$\left\{{\alpha }_n\right\},\left\{{\beta }_n\right\},\left\{{\gamma }_n\right\}\subset (0,1)$ and $\left\{{\sigma }_n\right\}\subset [0,1]$ such that

(i) ${\alpha }_n+{\beta }_n+{\gamma }_n=1\forall n\ge 1$ and ${\sum }_{n=1}^{\infty }{\alpha }_n=\infty$;

(ii) ${lim}_{n\to \infty }{\alpha }_n=0$ and ${lim}_{n\to \infty }\frac{{\psi }_n}{{\alpha }_n}=0$;

(iii) $0<{lim\; inf}_{n\to \infty }{\gamma }_n\le {lim\; sup}_{n\to \infty }{\gamma }_n<1$;

(iv) $0<{lim\; inf}_{n\to \infty }{\sigma }_n\le {lim\; sup}_{n\to \infty }{\sigma }_n<1$.

Lemma 7.

The Armijo-type search rule (5) is well defined, and consequently $\frac{\parallel r_{\lambda }\left(w_n\right){\parallel }^2}{\lambda }\le \langle A{w}_n,r_{\lambda }\left(w_n\right)\rangle$.

Proof. 

From $l\in (0,1)$ and uniform continuity of A on C, one has ${lim}_{j\to \infty }\langle A{w}_n-A(w_n-l^j{r}_{\lambda }\left(w_n\right)),r_{\lambda }\left(w_n\right)\rangle =0$. If $r_{\lambda }\left(w_n\right)=0$, then $j_n=0$. If $r_{\lambda }\left(w_n\right)\ne 0$, then ∃ (integer) $j_n\ge 0$, satisfying (3.1). By the firm nonexpansivity of $P_C$, one obtains $\parallel x-P_C{y\parallel }^2\le \langle x-y,x-P_{C}y\rangle \forall x\in C,y\in H$. Putting $y=w_n-\lambda A{w}_n$ and $x=w_n$, one gets $\parallel w_n-P_C(w_n-\lambda A{w}_n){\parallel }^2\le \lambda \langle A{w}_n,w_n-P_C(w_n-\lambda A{w}_n)\rangle$, and hence $\frac{\parallel r_{\lambda }\left(w_n\right){\parallel }^2}{\lambda }\le \langle A{w}_n,r_{\lambda }\left(w_n\right)\rangle$.    □

Lemma 8.

Let $p\in \mathsf{\Omega }$ and assume the function $h_n$ is formulated as (3.2). Then, $h_n\left(w_n\right)=\frac{{\zeta }_n}{2\lambda }{\parallel r_{\lambda }\left(w_n\right)\parallel }^2$ and $h_n\left(p\right)\le 0$. In addition, if $r_{\lambda }\left(w_n\right)\ne 0$, then $h_n\left(w_n\right)>0$.

Let $\left\{u_n\right\}$ be the sequence constructed in Algorithm 3.

Algorithm 3 Initialization: Given $\mu >0,l\in (0,1),\lambda \in (0,\frac{1}{\mu })$. Pick $u_1\in C$.

Iterative Steps: Given $u_n$, compute Step 1. Set $w_n=(1-{\sigma }_n)u_n+{\sigma }_n{T}^n{u}_n$, and compute $y_n=P_C(w_n-\lambda A{w}_n)$ and $r_{\lambda }\left(w_n\right):=w_n-y_n$; Step 2. Compute $t_n=w_n-{\zeta }_n{r}_{\lambda }\left(w_n\right)$, where ${\zeta }_n:=l^{j_n}$ and $j_n$ is the smallest nonnegative integer j, satisfying $$\langle A{w}_n-A(w_n-l^j{r}_{\lambda }\left(w_n\right)),w_n-y_n\rangle \le \frac{\mu }{2}{\parallel r_{\lambda }\left(w_n\right)\parallel }^2.$$ (5) Step 3. Compute $z_n=P_{D_n}\left(w_n\right)$ and $u_{n+1}={\alpha }_{n}f\left(u_n\right)+{\beta }_n{u}_n+{\gamma }_n{T}_n{z}_n,$ where $D_n:=\{x\in C:h_n\left(x\right)\le 0\}$ and $$h_n\left(x\right)=\langle A{t}_n,x-w_n\rangle +\frac{{\zeta }_n}{2\lambda }{\parallel r_{\lambda }\left(w_n\right)\parallel }^2.$$ (6) Again set $n:=n+1$ and go to Step 1.

Proof. 

The first claim of Lemma 8 is evident. Let us show the second claim. In fact, for $p\in \mathsf{\Omega }$, by Lemma 3 one has $\langle A{t}_n,t_n-p\rangle \ge 0$. So, one obtains that

$$h_n\left(p\right)=\langle A{t}_n,p-w_n\rangle +\frac{{\zeta }_n}{2\lambda }\parallel r_{\lambda }\left(w_n\right){\parallel }^2\le -{\zeta }_n\langle A{t}_n,r_{\lambda }\left(w_n\right)\rangle +\frac{{\zeta }_n}{2\lambda }{\parallel r_{\lambda }\left(w_n\right)\parallel }^2.$$

(7)

Furthermore, from (5) one has $\langle A{w}_n-A{t}_n,r_{\lambda }\left(w_n\right)\rangle \le \frac{\mu }{2}{\parallel r_{\lambda }\left(w_n\right)\parallel }^2$. Thus, by Lemma 7 we get

$$\langle A{t}_n,r_{\lambda }\left(w_n\right)\rangle \ge \langle A{w}_n,r_{\lambda }\left(w_n\right)\rangle -\frac{\mu }{2}\parallel r_{\lambda }\left(w_n\right){\parallel }^2\ge (\frac{1}{\lambda }-\frac{\mu }{2}){\parallel r_{\lambda }\left(w_n\right)\parallel }^2.$$

(8)

Combining (7) and (8) arrives at

$$h_n\left(p\right)\le -\frac{{\zeta }_n}{2}(\frac{1}{\lambda }-\mu ){\parallel r_{\lambda }\left(w_n\right)\parallel }^2\le 0.$$

(9)

   □

Lemma 9.

Let $\left\{u_n\right\}$ be the sequence constructed in Algorithm 3, s.t. $u_n-y_n\to 0,u_n-u_{n+1}\to 0,u_n-T^n{u}_n\to 0,u_n-T_n{u}_n\to 0$. Suppose that $T^n{u}_n-T^{n+1}{u}_n\to 0$ and $\exists \left\{u_{n_k}\right\}\subset \left\{u_n\right\}$ s.t. $u_{n_k}\rightharpoonup z$. Then $z\in \mathsf{\Omega }$.

Proof. 

Using Algorithm 3, one obtains $w_n-u_n={\sigma }_n(T^n{u}_n-u_n)\forall n\ge 1$, and hence $\parallel w_n-u_n\parallel \le \parallel T^n{u}_n-u_n\parallel$. Using the hypothesis $u_n-T^n{u}_n\to 0$, we have

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

(10)

which together with the hypothesis $u_n-y_n\to 0$, implies that

$$\parallel w_n-y_n\parallel \le \parallel w_n-u_n\parallel +\parallel u_n-y_n\parallel \to 0(n\to \infty ).$$

Besides this, combining $w_n-u_n\to 0$ and $u_{n_k}\rightharpoonup z$ yields $w_{n_k}\rightharpoonup z$.    □

Let us show that ${lim}_{n\to \infty }\parallel u_n-T_i{u}_n\parallel =0$ for $i=1,...,N$. In fact, note that for $i=1,...,N$,

$$\begin{array}{cc}\parallel u_n-T_{n+i}{u}_n\parallel \hfill & \le \parallel u_n-x_{n+i}\parallel +\parallel x_{n+i}-T_{n+i}{x}_{n+i}\parallel +\parallel T_{n+i}{x}_{n+i}-T_{n+i}{u}_n\parallel \hfill \\ & \le 2\parallel u_n-x_{n+i}\parallel +\parallel x_{n+i}-T_{n+i}{x}_{n+i}\parallel .\hfill \end{array}$$

By $u_n-u_{n+1}\to 0$ and $u_n-T_n{u}_n\to 0$, we get ${lim}_{n\to \infty }\parallel u_n-T_{n+i}{u}_n\parallel =0$ for $i=1,...,N$. This immediately arrives at

$$\underset{n\to \infty }{lim}\parallel u_n-T_i{u}_n\parallel =0\mathrm{for}i=1,...,N.$$

(11)

Moreover, we claim that $u_n-T{u}_n\to 0$ is as $n\to \infty$. In fact, combining the hypotheses $u_n-T^n{u}_n\to 0$ and $T^n{u}_n-T^{n+1}{u}_n\to 0$, guarantees that

$$\begin{array}{cc}\parallel u_n-T{u}_n\parallel \hfill & \le \parallel u_n-T^n{u}_n\parallel +\parallel T^n{u}_n-T^{n+1}{u}_n\parallel +\parallel T^{n+1}{u}_n-T{u}_n\parallel \hfill \\ & \le \parallel u_n-T^n{u}_n\parallel +\parallel T^n{u}_n-T^{n+1}{u}_n\parallel +(1+{\psi }_1)\parallel T^n{u}_n-u_n\parallel \hfill \\ & =(2+{\psi }_1)\parallel u_n-T^n{u}_n\parallel +\parallel T^n{u}_n-T^{n+1}{u}_n\parallel \to 0(n\to \infty ).\hfill \end{array}$$

(12)

Next, let us show $z\in \mathrm{VI}(C,A)$. In fact, since C is convex and closed, from $\left\{u_n\right\}\subset C$ and $u_{n_k}\rightharpoonup z$ we get $z\in C$. In what follows, we consider two cases. If $Az=0$, then it is clear that $z\in \mathrm{VI}(C,A)$ because $\langle Az,x-z\rangle \ge 0\forall x\in C$. Assume that $Az\ne 0$. Then, it follows from $w_n-y_n\to 0$ and $w_{n_k}\rightharpoonup z$ that $y_{n_k}\rightharpoonup z$. Using the assumption on A, instead of the sequentially weak continuity of A, we get $0<\parallel Az\parallel \le {lim\; inf}_{k\to \infty }\parallel A{y}_{n_k}\parallel$. So, we could suppose that $\parallel A{y}_{n_k}\parallel \ne 0\forall k\ge 1$. Furthermore, from $y_n=P_C(w_n-\lambda A{w}_n)$, we have $\langle w_n-\lambda A{w}_n-y_n,u-y_n\rangle \le 0\forall u\in C$. Thus,

$$\frac{1}{\lambda }\langle w_n-y_n,u-y_n\rangle +\langle A{w}_n,y_n-w_n\rangle \le \langle A{w}_n,u-w_n\rangle \forall u\in C.$$

(13)

According to the uniform continuity of A on C, one knows that $\left\{A{w}_{n_k}\right\}$ is bounded (due to Lemma 1). Note that $\left\{y_{n_k}\right\}$ is bounded as well. Then, by (13) we have ${lim\; inf}_{k\to \infty }\langle A{w}_{n_k},u-w_{n_k}\rangle \ge 0\forall u\in C$. To show that $z\in \mathrm{VI}(C,A)$, we pick $\left\{{\epsilon }_k\right\}\subset (0,1)$ s.t. ${\epsilon }_k\downarrow 0$ as $k\to \infty$. For each $k\ge 1$, let $m_k$ be the smallest natural number, such that

$$\langle A{w}_{n_j},u-w_{n_j}\rangle +{\epsilon }_k\ge 0\forall j\ge m_k.$$

(14)

Since $\left\{{\epsilon }_k\right\}$ is nonincreasing, it can be readily seen that $\left\{m_k\right\}$ is increasing. Noticing that $A{w}_{m_k}\ne 0\forall k\ge 1$ (due to $\left\{A{w}_{m_k}\right\}\subset \left\{A{w}_{n_k}\right\}$), we set ${\mu }_{m_k}=\frac{A{w}_{m_k}}{\parallel A{w}_{m_k}{\parallel }^2}$, and we get $\langle A{w}_{m_k},{\mu }_{m_k}\rangle =1$. By (14), it gives $\langle A{w}_{m_k},u+{\epsilon }_k{\mu }_{m_k}-w_{m_k}\rangle \ge 0$. The pseudomonotonicity of A then gives $\langle A(u+{\epsilon }_k{\mu }_{m_k}),u+{\epsilon }_k{\mu }_{m_k}-w_{m_k}\rangle \ge 0$, i.e.,

$$\langle Au,u-w_{m_k}\rangle \ge \langle Au-A(u+{\epsilon }_k{\mu }_{m_k}),x+{\epsilon }_k{\mu }_{m_k}-w_{m_k}\rangle -{\epsilon }_k\langle Au,{\mu }_{m_k}\rangle .$$

(15)

Observe that $w_{n_k}\rightharpoonup z$, $\left\{w_{m_k}\right\}\subset \left\{w_{n_k}\right\}$ and ${\epsilon }_k\downarrow 0$ as $k\to \infty$. So it follows that $0\le {lim\; sup}_{k\to \infty }\parallel {\epsilon }_k{\mu }_{m_k}\parallel ={lim\; sup}_{k\to \infty }\frac{{\epsilon }_k}{\parallel A{w}_{m_k}\parallel }\le \frac{{lim\; sup}_{k\to \infty }{\epsilon }_k}{{lim\; inf}_{k\to \infty }\parallel A{w}_{n_k}\parallel }=0$. Hence, we get ${\epsilon }_k{\mu }_{m_k}\to 0$ as $k\to \infty$.

Next, we show that $z\in \mathsf{\Omega }$. Passing to the limit as $k\to \infty$ in (15), we have $\langle Au,u-z\rangle ={lim\; inf}_{k\to \infty }\langle Au,u-w_{m_k}\rangle \ge 0\forall u\in C$. Using Lemma 3, $z\in \mathrm{VI}(C,A)$. Furthermore, for $i=1,...,N$, since Lemma 5 guarantees the demiclosedness of $I-T_i$ at zero, from $u_{n_k}\rightharpoonup z$ and $u_{n_k}-T_i{u}_{n_k}\to 0$ (due to (11)) we deduce that $z\in \mathrm{Fix}\left(T_i\right)$. Thus, $z\in {\bigcap }_{i=1}^N\mathrm{Fix}\left(T_i\right)$. Hence, from $u_{n_k}\rightharpoonup z$ and $u_{n_k}-T{u}_{n_k}\to 0$ (due to (12)), we obtain that $z\in \mathrm{Fix}\left(T\right)$. Therefore, $z\in \mathsf{\Omega }$.

Lemma 10.

Assume $\left\{w_n\right\}$ in Algorithm 3 is such that ${\zeta }_n{\parallel r_{\lambda }\left(w_n\right)\parallel }^2\to 0$ as $n\to \infty$. Then, $w_n-y_n\to 0$.

Proof. 

On the contrary, suppose that ${lim\; sup}_{n\to \infty }\parallel w_n-y_n\parallel =a>0$. Then, $\exists \left\{n_k\right\}\subset \left\{n\right\}$ s.t.

$$\underset{k\to \infty }{lim}\parallel w_{n_k}-y_{n_k}\parallel =a>0.$$

(16)

Note that ${lim}_{k\to \infty }{\zeta }_{n_k}{\parallel r_{\lambda }\left(w_{n_k}\right)\parallel }^2=0$. In what follows, we consider two cases.    □

Case 1.${lim\; inf}_{k\to \infty }{\zeta }_{n_k}>0$. In this case, we might assume that $\exists \zeta >0$ s.t. ${\zeta }_{n_k}\ge \zeta >0\forall k\ge 1$. Then, it follows that $\parallel w_{n_k}-y_{n_k}{\parallel }^2=\frac{1}{{\zeta }_{n_k}}{\zeta }_{n_k}\parallel w_{n_k}-y_{n_k}{\parallel }^2\le \frac{1}{\zeta }·{\zeta }_{n_k}{\parallel r_{\lambda }\left(w_{n_k}\right)\parallel }^2$, which hence yields

$$0<a^2=\underset{k\to \infty }{lim}{\parallel w_{n_k}-y_{n_k}\parallel }^2\le \underset{k\to \infty }{lim}\left[\frac{1}{\zeta }·{\zeta }_{n_k}{\parallel r_{\lambda }\left(w_{n_k}\right)\parallel }^2\right]=0.$$

(17)

This reaches a contradiction.

Case 2.${lim\; inf}_{k\to \infty }{\zeta }_{n_k}=0$. In this case, there exists a subsequence of $\left\{{\zeta }_{n_k}\right\}$, still denoted by $\left\{{\zeta }_{n_k}\right\}$, such that ${lim}_{k\to \infty }{\zeta }_{n_k}=0$. Putting ${\upsilon }_{n_k}=\frac{1}{l}{\zeta }_{n_k}{y}_{n_k}+(1-\frac{1}{l}{\zeta }_{n_k})w_{n_k}$, we get ${\upsilon }_{n_k}=w_{n_k}-\frac{1}{l}{\zeta }_{n_k}(w_{n_k}-y_{n_k})$. Since ${lim}_{n\to \infty }{\zeta }_n{\parallel r_{\lambda }\left(w_n\right)\parallel }^2$$=0$, we have

$$\underset{k\to \infty }{lim}\parallel {\upsilon }_{n_k}-w_{n_k}{\parallel }^2=\underset{k\to \infty }{lim}\frac{1}{l^2}{\zeta }_{n_k}·{\zeta }_{n_k}{\parallel w_{n_k}-y_{n_k}\parallel }^2=0.$$

(18)

From the step size rule (5) and the definition of ${\upsilon }_{n_k}$, it follows that

$$\langle A{w}_{n_k}-A{\upsilon }_{n_k},w_{n_k}-y_{n_k}\rangle >\frac{\mu }{2}{\parallel w_{n_k}-y_{n_k}\parallel }^2.$$

(19)

Using the uniform the continuity of A on C, from (18) we deduce that ${lim}_{k\to \infty }\parallel A{w}_{n_k}-A{\upsilon }_{n_k}\parallel =0$, which together with (19) leads to ${lim}_{k\to \infty }\parallel w_{n_k}-y_{n_k}\parallel =0$. Thus, this contradicts with (16). Consequently, ${lim}_{n\to \infty }\parallel w_n-y_n\parallel =0$.

Theorem 1.

Suppose that the sequence $\left\{u_n\right\}$ is constructed by Algorithm 3. Then, $u_n\to u^{*}\in \mathsf{\Omega }$ provided $T^n{u}_n-T^{n+1}{u}_n\to 0$, where $u^{*}\in \mathsf{\Omega }$ is the unique solution to the VIP: $\langle (I-f)u^{*},p-u^{*}\rangle \ge 0\forall p\in \mathsf{\Omega }$.

Proof. 

First of all, since $0<{lim\; inf}_{n\to \infty }{\gamma }_n\le {lim\; sup}_{n\to \infty }{\gamma }_n<1$ and ${lim}_{n\to \infty }\frac{{\psi }_n}{{\alpha }_n}=0$, we may assume, without loss of generality, that $\left\{{\gamma }_n\right\}\subset [a,b]\subset (0,1)$ and ${\psi }_n\le \frac{{\alpha }_n(1-\varrho )}{2}\forall n\ge 1$. Clearly, $P_{\mathsf{\Omega }}\circ f:C\to C$ is a contraction. Hence, there exists $u^{*}\in C$, such that $u^{*}=P_{\mathsf{\Omega }}f\left(u^{*}\right)$. Therefore, $u^{*}\in \mathsf{\Omega }={\bigcap }_{i=0}^N\mathrm{Fix}\left(T_i\right)\cap \mathrm{VI}(C,A)$ of the VIP

$$\langle (I-f)u^{*},p-u^{*}\rangle \ge 0\forall p\in \mathsf{\Omega }.$$

(20)

Next, we show the conclusion of the theorem. With this aim, we consider the following steps.

Step 1. We claim that the following inequality holds:

$$\parallel z_n{-p\parallel }^2\le {\parallel w_n-p\parallel }^2-{\mathrm{dist}}^2(w_n,D_n)\forall p\in \mathsf{\Omega }.$$

(21)

Indeed, one has

$$\begin{array}{cc}\parallel z_n{-p\parallel }^2={\parallel P_{D_n}{w}_n-p\parallel }^2\hfill & \le \parallel w_n{-p\parallel }^2-{\parallel P_{D_n}{w}_n-w_n\parallel }^2\hfill \\ & =\parallel w_n{-p\parallel }^2-{\mathrm{dist}}^2(w_n,D_n),\hfill \end{array}$$

which immediately yields

$$\parallel z_n-p\parallel \le \parallel w_n-p\parallel \forall n\ge 1.$$

(22)

Thus

$$\begin{array}{cc}\parallel w_n-p\parallel \hfill & \le (1-{\sigma }_n)\parallel u_n-p\parallel +{\sigma }_n\parallel T^n{u}_n-p\parallel \hfill \\ & \le (1-{\sigma }_n)\parallel u_n-p\parallel +{\sigma }_n(1+{\psi }_n)\parallel u_n-p\parallel \hfill \\ & \le (1+{\psi }_n)\parallel u_n-p\parallel ,\hfill \end{array}$$

which together with (22), yields

$$\parallel z_n-p\parallel \le \parallel w_n-p\parallel \le (1+{\psi }_n)\parallel u_n-p\parallel \forall n\ge 1.$$

(23)

Thus, from (23) and ${\alpha }_n+{\beta }_n+{\gamma }_n=1\forall n\ge 1$ it follows that

$$\begin{array}{cc}& \parallel u_{n+1}-p\parallel \le {\alpha }_n\parallel f\left(u_n\right)-p\parallel +{\beta }_n\parallel u_n-p\parallel +{\gamma }_n\parallel T_n{z}_n-p\parallel \hfill \\ & \le {\alpha }_n(\parallel f\left(u_n\right)-f\left(p\right)\parallel +\parallel f\left(p\right)-p\parallel )+{\beta }_n\parallel u_n-p\parallel +{\gamma }_n(1+{\psi }_n)\parallel u_n-p\parallel \hfill \\ & \le {\alpha }_n(\varrho \parallel u_n-p\parallel +\parallel f\left(p\right)-p\parallel )+{\beta }_n\parallel u_n-p\parallel +{\gamma }_n\parallel u_n-p\parallel +\frac{{\alpha }_n(1-\varrho )}{2}\parallel u_n-p\parallel \hfill \\ & =[1-\frac{{\alpha }_n(1-\varrho )}{2}]\parallel u_n-p\parallel +{\alpha }_n\parallel f\left(p\right)-p\parallel \le max\{\parallel u_n-p\parallel ,\frac{2\parallel f\left(p\right)-p\parallel }{1-\varrho }\}.\hfill \end{array}$$

Thus, $\parallel u_n-p\parallel \le max\{\parallel u_1-p\parallel ,\frac{2\parallel f\left(p\right)-p\parallel }{1-\varrho }\}\forall n\ge 1$. This $\left\{u_n\right\}$ is bounded, and so are $\left\{w_n\right\},\left\{y_n\right\},\left\{z_n\right\},\left\{f\left(u_n\right)\right\},\left\{A{t}_n\right\},\left\{T^n{u}_n\right\},\left\{T_n{z}_n\right\}$.

Step 2. Let us obtain

$$\begin{array}{cc}& {\gamma }_n\left[(1-{\sigma }_n){\sigma }_n\parallel u_n-T^n{u}_n{\parallel }^2+{\parallel z_n-w_n\parallel }^2\right]+{\beta }_n{\gamma }_n{\parallel u_n-T_n{z}_n\parallel }^2\hfill \\ & \le \parallel u_n{-p\parallel }^2-{\parallel u_{n+1}-p\parallel }^2+{\psi }_{n}K+2{\alpha }_n\langle f\left(u_n\right)-p,u_{n+1}-p\rangle \hfill \end{array}$$

for some $K>0$. To prove this, we first note that

$$\begin{array}{cc}& \parallel u_{n+1}{-p\parallel }^2={\parallel {\alpha }_n(f\left(u_n\right)-p)+{\beta }_n(u_n-p)+{\gamma }_n(T_n{z}_n-p)\parallel }^2\hfill \\ & \le \parallel {\beta }_n(u_n-p)+{\gamma }_n(T_n{z}_n-p){\parallel }^2+2{\alpha }_n\langle f\left(u_n\right)-p,u_{n+1}-p\rangle \hfill \\ & \le {\beta }_n\parallel u_n{-p\parallel }^2+{\gamma }_n\parallel z_n{-p\parallel }^2-{\beta }_n{\gamma }_n{\parallel u_n-T_n{z}_n\parallel }^2+2{\alpha }_n\langle f\left(u_n\right)-p,u_{n+1}-p\rangle .\hfill \end{array}$$

(24)

On the other hand, by Algorithm 3 one has

$$\begin{array}{cc}& \parallel z_n{-p\parallel }^2=\parallel P_{D_n}{w}_n{-p\parallel }^2\le \parallel w_n{-p\parallel }^2-{\parallel z_n-w_n\parallel }^2\hfill \\ & =(1-{\sigma }_n)\parallel u_n{-p\parallel }^2+{\sigma }_n\parallel T^n{u}_n{-p\parallel }^2-(1-{\sigma }_n){\sigma }_n\parallel u_n-T^n{u}_n{\parallel }^2-{\parallel z_n-w_n\parallel }^2\hfill \\ & \le {(1+{\psi }_n)}^2\parallel u_n{-p\parallel }^2-(1-{\sigma }_n){\sigma }_n\parallel u_n-T^n{u}_n{\parallel }^2-{\parallel z_n-w_n\parallel }^2.\hfill \end{array}$$

(25)

Substituting (25) into (24), one gets

$$\begin{array}{cc}& \parallel u_{n+1}{-p\parallel }^2\hfill \\ & \le {\beta }_n{\parallel u_n-p\parallel }^2+{\gamma }_n\left[{(1+{\psi }_n)}^2\parallel u_n{-p\parallel }^2-(1-{\sigma }_n){\sigma }_n\parallel u_n-T^n{u}_n{\parallel }^2-{\parallel z_n-w_n\parallel }^2\right]\hfill \\ & -{\beta }_n{\gamma }_n{\parallel u_n-T_n{z}_n\parallel }^2+2{\alpha }_n\langle f\left(u_n\right)-p,u_{n+1}-p\rangle \hfill \\ & \le (1-{\alpha }_n){\parallel u_n-p\parallel }^2-{\gamma }_n\left[(1-{\sigma }_n){\sigma }_n\parallel u_n-T^n{u}_n{\parallel }^2+{\parallel z_n-w_n\parallel }^2\right]\hfill \\ & +{\psi }_n(2+{\psi }_n)\parallel u_n{-p\parallel }^2-{\beta }_n{\gamma }_n{\parallel u_n-T_n{z}_n\parallel }^2+2{\alpha }_n\langle f\left(u_n\right)-p,u_{n+1}-p\rangle \hfill \\ & \le \parallel u_n{-p\parallel }^2-{\gamma }_n\left[(1-{\sigma }_n){\sigma }_n\parallel u_n-T^n{u}_n{\parallel }^2+{\parallel z_n-w_n\parallel }^2\right]-{\beta }_n{\gamma }_n{\parallel u_n-T_n{z}_n\parallel }^2\hfill \\ & +{\psi }_{n}K+2{\alpha }_n\langle f\left(u_n\right)-p,u_{n+1}-p\rangle ,\hfill \end{array}$$

where ${sup}_{n\ge 1}(2+{\psi }_n){\parallel u_n-p\parallel }^2\le K$ for some $K>0$. This immediately implies that

$$\begin{array}{cc}& {\gamma }_n\left[(1-{\sigma }_n){\sigma }_n\parallel u_n-T^n{u}_n{\parallel }^2+{\parallel z_n-w_n\parallel }^2\right]+{\beta }_n{\gamma }_n{\parallel u_n-T_n{z}_n\parallel }^2\hfill \\ & \le \parallel u_n{-p\parallel }^2-{\parallel u_{n+1}-p\parallel }^2+{\psi }_{n}K+2{\alpha }_n\langle f\left(u_n\right)-p,u_{n+1}-p\rangle .\hfill \end{array}$$

Step 3. We show that

$${\gamma }_n{\left[\frac{{\zeta }_n}{2\lambda L}{\parallel r_{\lambda }\left(w_n\right)\parallel }^2\right]}^2\le \parallel u_n{-p\parallel }^2-\parallel u_{n+1}{-p\parallel }^2+{\alpha }_n{\parallel f\left(u_n\right)-p\parallel }^2+{\psi }_{n}K.$$

Indeed, we claim that for some $L>0$,

$$\parallel z_n{-p\parallel }^2\le {\parallel w_n-p\parallel }^2-{\left[\frac{{\zeta }_n}{2\lambda L}{\parallel r_{\lambda }\left(w_n\right)\parallel }^2\right]}^2.$$

(26)

Thanks to the boundedness of $\left\{A{t}_n\right\}$, we know that $\exists L>0$ s.t. $\parallel A{t}_n\parallel \le L\forall n\ge 1$, which arrives at

$$|h_n\left(u\right)-h_n\left(v\right)|=|\langle A{t}_n,u-v\rangle |\le \parallel A{t}_n\parallel \parallel u-v\parallel \le L\parallel u-v\parallel \forall u,v\in D_n.$$

This hence ensures that $h_n(·)$ is L-Lipschitz continuous on $D_n$. By Lemmas 2 and 8, one obtains

$$\mathrm{dist}(w_n,D_n)\ge \frac{1}{L}{h}_n\left(w_n\right)=\frac{{\zeta }_n}{2\lambda L}{\parallel r_{\lambda }\left(w_n\right)\parallel }^2.$$

(27)

Combining (21) and (27) immediately yields

$$\parallel z_n{-p\parallel }^2\le {\parallel w_n-p\parallel }^2-{\left[\frac{{\zeta }_n}{2\lambda L}{\parallel r_{\lambda }\left(w_n\right)\parallel }^2\right]}^2.$$

From Algorithm 3, (23), and (26) it follows that

$$\begin{array}{cc}& \parallel u_{n+1}{-p\parallel }^2\le {\alpha }_n\parallel f\left(u_n\right){-p\parallel }^2+{\beta }_n\parallel u_n{-p\parallel }^2+{\gamma }_n{\parallel T_n{z}_n-p\parallel }^2\hfill \\ & \le {\alpha }_n\parallel f\left(u_n\right){-p\parallel }^2+{\beta }_n\parallel u_n{-p\parallel }^2+{\gamma }_n\{{(1+{\psi }_n)}^2\parallel u_n-p{\parallel }^2-{\left[\frac{{\zeta }_n}{2\lambda L}{\parallel r_{\lambda }\left(w_n\right)\parallel }^2\right]}^2\}\hfill \\ & \le {\alpha }_n\parallel f\left(u_n\right){-p\parallel }^2+(1-{\alpha }_n)\parallel u_n{-p\parallel }^2+{\psi }_n(2+{\psi }_n){\parallel u_n-p\parallel }^2-{\gamma }_n{\left[\frac{{\zeta }_n}{2\lambda L}{\parallel r_{\lambda }\left(w_n\right)\parallel }^2\right]}^2\hfill \\ & \le {\alpha }_n\parallel f\left(u_n\right){-p\parallel }^2+{\psi }_{n}K+{\parallel u_n-p\parallel }^2-{\gamma }_n{\left[\frac{{\zeta }_n}{2\lambda L}{\parallel r_{\lambda }\left(w_n\right)\parallel }^2\right]}^2.\hfill \end{array}$$

This immediately yields

$${\gamma }_n{\left[\frac{{\zeta }_n}{2\lambda L}{\parallel r_{\lambda }\left(w_n\right)\parallel }^2\right]}^2\le \parallel u_n{-p\parallel }^2-\parallel u_{n+1}{-p\parallel }^2+{\alpha }_n{\parallel f\left(u_n\right)-p\parallel }^2+{\psi }_{n}K.$$

Step 4. We show that

$$\parallel u_{n+1}{-p\parallel }^2\le (1-{\alpha }_n(1-\varrho )){\parallel u_n-p\parallel }^2+{\alpha }_n(1-\varrho )\left[\frac{2\langle f\left(p\right)-p,u_{n+1}-p\rangle }{1-\varrho }+\frac{{\psi }_n}{{\alpha }_n}·\frac{K}{1-\varrho }\right].$$

(28)

Indeed, from Algorithm 3 and (23), one has

$$\begin{array}{cc}& \parallel u_{n+1}{-p\parallel }^2={\parallel {\alpha }_n(f\left(u_n\right)-f\left(p\right))+{\beta }_n(u_n-p)+{\gamma }_n(T_n{z}_n-p)+{\alpha }_n(f\left(p\right)-p)\parallel }^2\hfill \\ & \le \parallel {\alpha }_n(f\left(u_n\right)-f\left(p\right))+{\beta }_n(u_n-p)+{\gamma }_n(T_n{z}_n-p){\parallel }^2+2{\alpha }_n\langle f\left(p\right)-p,u_{n+1}-p\rangle \hfill \\ & \le {\alpha }_n\parallel f\left(u_n\right)-f\left(p\right){\parallel }^2+{\beta }_n\parallel u_n{-p\parallel }^2+{\gamma }_n{\parallel z_n-p\parallel }^2+2{\alpha }_n\langle f\left(p\right)-p,u_{n+1}-p\rangle \hfill \\ & \le {\alpha }_n\parallel f\left(u_n\right)-f\left(p\right){\parallel }^2+{\beta }_n\parallel u_n{-p\parallel }^2+{\gamma }_n{(1+{\psi }_n)}^2{\parallel u_n-p\parallel }^2+2{\alpha }_n\langle f\left(p\right)-p,u_{n+1}-p\rangle \hfill \\ & \le \varrho {\alpha }_n\parallel u_n{-p\parallel }^2+{\beta }_n\parallel u_n{-p\parallel }^2+{\gamma }_n\parallel u_n{-p\parallel }^2+{\psi }_n(2+{\psi }_n){\parallel u_n-p\parallel }^2\hfill \\ & +2{\alpha }_n\langle f\left(p\right)-p,u_{n+1}-p\rangle \hfill \\ & \le \varrho {\alpha }_n\parallel u_n{-p\parallel }^2+{\beta }_n\parallel u_n{-p\parallel }^2+{\gamma }_n{\parallel u_n-p\parallel }^2+{\psi }_{n}K+2{\alpha }_n\langle f\left(p\right)-p,u_{n+1}-p\rangle \hfill \\ & =[1-{\alpha }_n(1-\varrho )]{\parallel u_n-p\parallel }^2+{\psi }_{n}K+2{\alpha }_n\langle f\left(p\right)-p,u_{n+1}-p\rangle \hfill \\ & =(1-{\alpha }_n(1-\varrho )){\parallel u_n-p\parallel }^2+{\alpha }_n(1-\varrho )\left[\frac{2\langle f\left(p\right)-p,u_{n+1}-p\rangle }{1-\varrho }+\frac{{\psi }_n}{{\alpha }_n}·\frac{K}{1-\varrho }\right].\hfill \end{array}$$

Step 5. We obtain the strong convergence to $u^{*}\in \mathsf{\Omega }$, satisfying (20).

Indeed, putting $p=u^{*}$, we deduce from (28) that

$$\parallel u_{n+1}-u^{*}{\parallel }^2\le (1-{\alpha }_n(1-\varrho )){\parallel u_n-u^{*}\parallel }^2+{\alpha }_n(1-\varrho )\left[\frac{2\langle f\left(u^{*}\right)-u^{*},u_{n+1}-u^{*}\rangle }{1-\varrho }+\frac{{\psi }_n}{{\alpha }_n}·\frac{K}{1-\varrho }\right].$$

(29)

Setting ${\mathrm{\Gamma }}_n={\parallel u_n-u^{*}\parallel }^2$ we show ${\mathrm{\Gamma }}_n\to 0(n\to \infty )$.    □

Case 1. Assume there exists $n_0\ge 1$ such that $\left\{{\mathrm{\Gamma }}_n\right\}$ is nonincreasing. Thus, ${lim}_{n\to \infty }{\mathrm{\Gamma }}_n=\hslash <+\infty$ and ${lim}_{n\to \infty }({\mathrm{\Gamma }}_n-{\mathrm{\Gamma }}_{n+1})=0$. Putting $p=u^{*}$, from Step 2 and $\left\{{\gamma }_n\right\}\subset [a,b]\subset (0,1)$, we obtain

$$\begin{array}{cc}& a[(1-{\sigma }_n){\sigma }_n\parallel u_n-T^n{u}_n{\parallel }^2+\parallel z_n-w_n{\parallel }^2]+(1-{\alpha }_n-b)a{\parallel u_n-T_n{z}_n\parallel }^2\hfill \\ & \le {\gamma }_n\left[(1-{\sigma }_n){\sigma }_n\parallel u_n-T^n{u}_n{\parallel }^2+{\parallel z_n-w_n\parallel }^2\right]+{\beta }_n{\gamma }_n{\parallel u_n-T_n{z}_n\parallel }^2\hfill \\ & \le \parallel u_n-u^{*}{\parallel }^2-{\parallel u_{n+1}-u^{*}\parallel }^2+{\psi }_{n}K+2{\alpha }_n\langle f\left(u_n\right)-u^{*},u_{n+1}-u^{*}\rangle \hfill \\ & \le {\mathrm{\Gamma }}_n-{\mathrm{\Gamma }}_{n+1}+{\psi }_{n}K+2{\alpha }_n\parallel f\left(u_n\right)-u^{*}\parallel \parallel u_{n+1}-u^{*}\parallel .\hfill \end{array}$$

Since $0<{lim\; inf}_{n\to \infty }{\sigma }_n\le {lim\; sup}_{n\to \infty }{\sigma }_n<1$, ${\psi }_n\to 0,{\alpha }_n\to 0$ and ${\mathrm{\Gamma }}_n-{\mathrm{\Gamma }}_{n+1}\to 0$, from the boundedness of $\left\{u_n\right\}$ one has

$$\underset{n\to \infty }{lim}\parallel u_n-T^n{u}_n\parallel =\underset{n\to \infty }{lim}\parallel u_n-T_n{z}_n\parallel =\underset{n\to \infty }{lim}\parallel w_n-z_n\parallel =0.$$

(30)

So, it follows from Algorithm 3 and (30) that

$$\parallel w_n-u_n\parallel ={\sigma }_n\parallel T^n{u}_n-u_n\parallel \le \parallel T^n{u}_n-u_n\parallel \to 0(n\to \infty ),$$

and

$$\begin{array}{cc}\parallel u_{n+1}-u_n\parallel \hfill & \le {\alpha }_n\parallel f\left(u_n\right)-u_n\parallel +{\gamma }_n\parallel T_n{z}_n-u_n\parallel \hfill \\ & \le {\alpha }_n\parallel f\left(u_n\right)-u_n\parallel +\parallel T_n{z}_n-u_n\parallel \to 0(n\to \infty ).\hfill \end{array}$$

(31)

Putting $p=u^{*}$, from Step 3 we obtain

$$\begin{array}{cc}{\gamma }_n{\left[\frac{{\zeta }_n}{2\lambda L}{\parallel r_{\lambda }\left(w_n\right)\parallel }^2\right]}^2\hfill & \le \parallel u_n-u^{*}{\parallel }^2-\parallel u_{n+1}-u^{*}{\parallel }^2+{\alpha }_n{\parallel f\left(u_n\right)-u^{*}\parallel }^2+{\psi }_{n}K\hfill \\ & ={\mathrm{\Gamma }}_n-{\mathrm{\Gamma }}_{n+1}+{\psi }_{n}K+{\alpha }_n{\parallel f\left(u_n\right)-u^{*}\parallel }^2.\hfill \end{array}$$

Since $0<{lim\; inf}_{n\to \infty }{\gamma }_n$, ${\psi }_n\to 0,{\alpha }_n\to 0$ and ${\mathrm{\Gamma }}_n-{\mathrm{\Gamma }}_{n+1}\to 0$, from the boundedness of $\left\{u_n\right\}$ one gets

$$\underset{n\to \infty }{lim}{\left[\frac{{\zeta }_n}{2\lambda L}{\parallel r_{\lambda }\left(w_n\right)\parallel }^2\right]}^2=0.$$

Hence, by Lemma 10 we deduce that

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

which immediately yields

$$\parallel u_n-y_n\parallel \le \parallel u_n-w_n\parallel +\parallel w_n-y_n\parallel \to 0(n\to \infty )$$

(32)

From the boundedness of $\left\{u_n\right\}$, it follows that there exists a subsequence $\left\{u_{n_k}\right\}$ of $\left\{u_n\right\}$ such that

$$\underset{n\to \infty }{lim\; sup}\langle f\left(u^{*}\right)-u^{*},u_n-u^{*}\rangle =\underset{k\to \infty }{lim}\langle f\left(u^{*}\right)-u^{*},u_{n_k}-u^{*}\rangle .$$

(33)

Since H is reflexive and $\left\{u_n\right\}$ is bounded, we may assume, without loss of generality, that $u_{n_k}\rightharpoonup \tilde{x}$. Thus, from (33) one gets

$$\begin{array}{cc}{\displaystyle \underset{n\to \infty }{lim\; sup}}\langle f\left(u^{*}\right)-u^{*},u_n-u^{*}\rangle \hfill & ={\displaystyle \underset{k\to \infty }{lim}}\langle f\left(u^{*}\right)-u^{*},u_{n_k}-u^{*}\rangle \hfill \\ & =\langle f\left(u^{*}\right)-u^{*},\tilde{x}-u^{*}\rangle .\hfill \end{array}$$

(34)

Furthermore, by Algorithm 3 we get $u_{n+1}-z_n={\alpha }_n(f\left(u_n\right)-z_n)+{\beta }_n(u_n-z_n)+{\gamma }_n(T_n{z}_n-z_n)$, which immediately yields

$$\begin{array}{cc}& {\gamma }_n\parallel T_n{z}_n-z_n\parallel \le \parallel u_{n+1}-z_n\parallel +{\alpha }_n(\parallel f\left(u_n\right)\parallel +\parallel z_n\parallel )+{\beta }_n\parallel u_n-z_n\parallel \hfill \\ & \le \parallel u_{n+1}-u_n\parallel +2(\parallel u_n-w_n\parallel +\parallel w_n-z_n\parallel )+{\alpha }_n(\parallel f\left(u_n\right)\parallel +\parallel z_n\parallel ).\hfill \end{array}$$

Since $u_n-u_{n+1}\to 0,w_n-u_n\to 0,w_n-z_n\to 0,{\alpha }_n\to 0,{lim\; inf}_{n\to \infty }{\gamma }_n>0$ and $\left\{u_n\right\},\left\{z_n\right\}$ are bounded, we obtain ${lim}_{n\to \infty }\parallel z_n-T_n{z}_n\parallel =0$, which together with the nonexpansivity of each $T_n$, arrives at

$$\begin{array}{cc}\parallel u_n-T_n{u}_n\parallel \hfill & \le \parallel u_n-z_n\parallel +\parallel z_n-T_n{z}_n\parallel +\parallel T_n{z}_n-T_n{u}_n\parallel \hfill \\ & \le 2\parallel u_n-z_n\parallel +\parallel z_n-T_n{z}_n\parallel \hfill \\ & \le 2(\parallel u_n-w_n\parallel +\parallel w_n-z_n\parallel )+\parallel z_n-T_n{z}_n\parallel \to 0(n\to \infty ).\hfill \end{array}$$

Since $u_n-y_n\to 0,u_n-u_{n+1}\to 0,u_n-T^n{u}_n\to 0,u_n-T_n{u}_n\to 0$ and $u_{n_k}\rightharpoonup \tilde{x}$, by Lemma 9 we infer that $\tilde{x}\in \mathsf{\Omega }$. Hence from (20) and (34) one gets

$$\underset{n\to \infty }{lim\; sup}\langle f\left(u^{*}\right)-u^{*},u_n-u^{*}\rangle =\langle f\left(u^{*}\right)-u^{*},\tilde{x}-u^{*}\rangle \le 0,$$

(35)

which immediately leads to

$$\begin{array}{cc}& {\displaystyle \underset{n\to \infty }{lim\; sup}}\langle f\left(u^{*}\right)-u^{*},u_{n+1}-u^{*}\rangle \hfill \\ & ={\displaystyle \underset{n\to \infty }{lim\; sup}}\left[\langle f\left(u^{*}\right)-u^{*},u_{n+1}-u_n\rangle +\langle f\left(u^{*}\right)-u^{*},u_n-u^{*}\rangle \right]\hfill \\ & \le {\displaystyle \underset{n\to \infty }{lim\; sup}}\left[\parallel f\left(u^{*}\right)-u^{*}\parallel \parallel u_{n+1}-u_n\parallel +\langle f\left(u^{*}\right)-u^{*},u_n-u^{*}\rangle \right]\le 0.\hfill \end{array}$$

(36)

Note that $\left\{{\alpha }_n(1-\varrho )\right\}\subset [0,1],{\sum }_{n=1}^{\infty }{\alpha }_n(1-\varrho )=\infty$, and

$$\underset{n\to \infty }{lim\; sup}\left[\frac{2\langle f\left(u^{*}\right)-u^{*},u_{n+1}-u^{*}\rangle }{1-\varrho }+\frac{{\psi }_n}{{\alpha }_n}·\frac{K}{1-\varrho }\right]\le 0.$$

Consequently, applying Lemma 4 to (29), one has ${lim}_{n\to \infty }{\parallel u_n-u^{*}\parallel }^2=0$.

Case 2. Suppose that $\exists \left\{{\mathrm{\Gamma }}_{n_k}\right\}\subset \left\{{\mathrm{\Gamma }}_n\right\}$ s.t. ${\mathrm{\Gamma }}_{n_k}<{\mathrm{\Gamma }}_{n_k+1}\forall k\in \mathcal{N}$, where $\mathcal{N}$ is the set of all positive integers. Define the mapping $\eta :\mathcal{N}\to \mathcal{N}$ by

$$\eta \left(n\right):=max\{k\le n:{\mathrm{\Gamma }}_k<{\mathrm{\Gamma }}_{k+1}\}.$$

By Lemma 6, we get

$${\mathrm{\Gamma }}_{\eta \left(n\right)}\le {\mathrm{\Gamma }}_{\eta \left(n\right)+1}\mathrm{and}{\mathrm{\Gamma }}_n\le {\mathrm{\Gamma }}_{\eta \left(n\right)+1}.$$

Putting $p=u^{*}$, from Step 2 we have

$$\begin{array}{cc}& a\left[(1-{\sigma }_{\eta \left(n\right)}){\sigma }_{\eta \left(n\right)}\parallel u_{\eta \left(n\right)}-T^{\eta \left(n\right)}{u}_{\eta \left(n\right)}{\parallel }^2+{\parallel z_{\eta \left(n\right)}-w_{\eta \left(n\right)}\parallel }^2\right]\hfill \\ & +(1-{\alpha }_{\eta \left(n\right)}-b)a{\parallel u_{\eta \left(n\right)}-T_{\eta \left(n\right)}{z}_{\eta \left(n\right)}\parallel }^2\hfill \\ & \le {\gamma }_{\eta \left(n\right)}\left[(1-{\sigma }_{\eta \left(n\right)}){\sigma }_{\eta \left(n\right)}\parallel u_{\eta \left(n\right)}-T^{\eta \left(n\right)}{u}_{\eta \left(n\right)}{\parallel }^2+{\parallel z_{\eta \left(n\right)}-w_{\eta \left(n\right)}\parallel }^2\right]\hfill \\ & +{\beta }_{\eta \left(n\right)}{\gamma }_{\eta \left(n\right)}{\parallel u_{\eta \left(n\right)}-T_{\eta \left(n\right)}{z}_{\eta \left(n\right)}\parallel }^2\hfill \\ & \le {\mathrm{\Gamma }}_{\eta \left(n\right)}-{\mathrm{\Gamma }}_{\eta \left(n\right)+1}+{\psi }_{\eta \left(n\right)}K+2{\alpha }_{\eta \left(n\right)}\langle f\left(u_{\eta \left(n\right)}\right)-u^{*},x_{\eta \left(n\right)+1}-u^{*}\rangle \hfill \\ & \le {\psi }_{\eta \left(n\right)}K+2{\alpha }_{\eta \left(n\right)}\parallel f\left(u_{\eta \left(n\right)}\right)-u^{*}\parallel \parallel x_{\eta \left(n\right)+1}-u^{*}\parallel ,\hfill \end{array}$$

which immediately yields

$$\underset{n\to \infty }{lim}\parallel u_{\eta \left(n\right)}-T^{\eta \left(n\right)}{u}_{\eta \left(n\right)}\parallel =\underset{n\to \infty }{lim}\parallel u_{\eta \left(n\right)}-T_{\eta \left(n\right)}{z}_{\eta \left(n\right)}\parallel =\underset{n\to \infty }{lim}\parallel w_{\eta \left(n\right)}-z_{\eta \left(n\right)}\parallel =0.$$

Putting $p=u^{*}$, from Step 3 we get

$$\begin{array}{cc}{\gamma }_{\eta \left(n\right)}[\frac{{\zeta }_{\eta \left(n\right)}}{2\lambda L}\parallel r_{\lambda }\left(w_{\eta \left(n\right)}\right){{\parallel }^2]}^2\hfill & \le {\mathrm{\Gamma }}_{\eta \left(n\right)}-{\mathrm{\Gamma }}_{\eta \left(n\right)+1}+{\alpha }_{\eta \left(n\right)}{\parallel f\left(u_{\eta \left(n\right)}\right)-u^{*}\parallel }^2+{\psi }_{\eta \left(n\right)}K\hfill \\ & \le {\psi }_{\eta \left(n\right)}K+{\alpha }_{\eta \left(n\right)}{\parallel f\left(u_{\eta \left(n\right)}\right)-u^{*}\parallel }^2,\hfill \end{array}$$

which hence leads to

$$\underset{n\to \infty }{lim}{\left[\frac{{\zeta }_{\eta \left(n\right)}}{2\lambda L}{\parallel r_{\lambda }\left(w_{\eta \left(n\right)}\right)\parallel }^2\right]}^2=0.$$

Utilizing the same inferences as in the proof of Case 1, we deduce that

$$\underset{n\to \infty }{lim}\parallel w_{\eta \left(n\right)}-y_{\eta \left(n\right)}\parallel =\underset{n\to \infty }{lim}\parallel w_{\eta \left(n\right)}-u_{\eta \left(n\right)}\parallel =\underset{n\to \infty }{lim}\parallel u_{\eta \left(n\right)+1}-u_{\eta \left(n\right)}\parallel =0,$$

and

$$\underset{n\to \infty }{lim\; sup}\langle f\left(u^{*}\right)-u^{*},u_{\eta \left(n\right)+1}-u^{*}\rangle \le 0.$$

On the other hand, from (29) we obtain

$$\begin{array}{cc}{\alpha }_{\eta \left(n\right)}(1-\varrho ){\mathrm{\Gamma }}_{\eta \left(n\right)}\hfill & \le {\mathrm{\Gamma }}_{\eta \left(n\right)}-{\mathrm{\Gamma }}_{\eta \left(n\right)+1}+{\alpha }_{\eta \left(n\right)}(1-\varrho )\left[\frac{2\langle f\left(u^{*}\right)-u^{*},u_{\eta \left(n\right)+1}-u^{*}\rangle }{1-\varrho }+\frac{{\psi }_{\eta \left(n\right)}}{{\alpha }_{\eta \left(n\right)}}·\frac{K}{1-\varrho }\right]\hfill \\ & \le {\alpha }_{\eta \left(n\right)}(1-\varrho )\left[\frac{2\langle f\left(u^{*}\right)-u^{*},u_{\eta \left(n\right)+1}-u^{*}\rangle }{1-\varrho }+\frac{{\psi }_{\eta \left(n\right)}}{{\alpha }_{\eta \left(n\right)}}·\frac{K}{1-\varrho }\right].\hfill \end{array}$$

which hence arrives at

$$\underset{n\to \infty }{lim\; sup}{\mathrm{\Gamma }}_{\eta \left(n\right)}\le \underset{n\to \infty }{lim\; sup}\left[\frac{2\langle f\left(u^{*}\right)-u^{*},u_{\eta \left(n\right)+1}-u^{*}\rangle }{1-\varrho }+\frac{{\psi }_{\eta \left(n\right)}}{{\alpha }_{\eta \left(n\right)}}·\frac{K}{1-\varrho }\right]\le 0.$$

Thus, ${lim}_{n\to \infty }{\parallel u_{\eta \left(n\right)}-u^{*}\parallel }^2=0$. Furthermore, note that

$$\begin{array}{cc}& \parallel u_{\eta \left(n\right)+1}-u^{*}{\parallel }^2-{\parallel u_{\eta \left(n\right)}-u^{*}\parallel }^2\hfill \\ & =2\langle u_{\eta \left(n\right)+1}-u_{\eta \left(n\right)},u_{\eta \left(n\right)}-u^{*}\rangle +{\parallel u_{\eta \left(n\right)+1}-u_{\eta \left(n\right)}\parallel }^2\hfill \\ & \le 2\parallel u_{\eta \left(n\right)+1}-u_{\eta \left(n\right)}\parallel \parallel u_{\eta \left(n\right)}-u^{*}\parallel +\parallel u_{\eta \left(n\right)+1}-u_{\eta \left(n\right)}{\parallel }^2\to 0(n\to \infty ).\hfill \end{array}$$

Thanks to ${\mathrm{\Gamma }}_n\le {\mathrm{\Gamma }}_{\eta \left(n\right)+1}$, we get

$$\begin{array}{cc}& \parallel u_n-u^{*}{\parallel }^2\le {\parallel u_{\eta \left(n\right)+1}-u^{*}\parallel }^2\hfill \\ & \le \parallel u_{\eta \left(n\right)}-u^{*}{\parallel }^2+2\parallel u_{\eta \left(n\right)+1}-u_{\eta \left(n\right)}\parallel \parallel u_{\eta \left(n\right)}-u^{*}\parallel +\parallel u_{\eta \left(n\right)+1}-u_{\eta \left(n\right)}{\parallel }^2\to 0(n\to \infty ).\hfill \end{array}$$

That is, $u_n\to u^{*}$ as $n\to \infty$.

Theorem 2.

Suppose $T:C\to C$ is nonexpansive and $\left\{u_n\right\}$ is constructed by: $u_1\in C$,

$$\left\{\begin{array}{cc}& w_n=(1-{\sigma }_n)u_n+{\sigma }_{n}T{u}_n,\hfill \\ & y_n=P_C(w_n-\lambda A{w}_n),\hfill \\ & t_n=(1-{\zeta }_n)w_n+{\zeta }_n{y}_n,\hfill \\ & z_n=P_{D_n}\left(w_n\right),\hfill \\ & u_{n+1}={\alpha }_{n}f\left(u_n\right)+{\beta }_n{u}_n+{\gamma }_n{T}_n{z}_n,\hfill \end{array}\right.$$

where for each $n\ge 1$, $D_n$ and ${\zeta }_n$ that are chosen as in Algorithm 3, then $u_n\to u^{*}\in \mathsf{\Omega }$, where $u^{*}\in \mathsf{\Omega }$ is the unique solution to the VIP:  $\langle (I-f)u^{*},p-u^{*}\rangle \ge 0\forall p\in \mathsf{\Omega }$.

Proof. 

Step 1.$\left\{u_n\right\}$ is bounded. Indeed, using the same arguments as in Step 1 of the proof of Theorem 1, we obtain the desired assertion.

Step 2.

$$\begin{array}{cc}& {\gamma }_n\left[(1-{\sigma }_n){\sigma }_n\parallel u_n-T{u}_n{\parallel }^2+{\parallel z_n-w_n\parallel }^2\right]+{\beta }_n{\gamma }_n{\parallel u_n-T_n{z}_n\parallel }^2\hfill \\ & \le \parallel u_n{-p\parallel }^2-{\parallel u_{n+1}-p\parallel }^2+2{\alpha }_n\langle f\left(u_n\right)-p,u_{n+1}-p\rangle .\hfill \end{array}$$

Indeed, using the same arguments as in Step 2 of the proof of Theorem 1, we have the result.

Step 3.

$${\gamma }_n{\left[\frac{{\zeta }_n}{2\lambda L}{\parallel r_{\lambda }\left(w_n\right)\parallel }^2\right]}^2\le \parallel u_n{-p\parallel }^2-\parallel u_{n+1}{-p\parallel }^2+{\alpha }_n{\parallel f\left(u_n\right)-p\parallel }^2.$$

The same arguments in Step 3 of the proof of Theorem 1 give the conclusion.

Step 4.

$$\parallel u_{n+1}{-p\parallel }^2\le (1-{\alpha }_n(1-\varrho )){\parallel u_n-p\parallel }^2+{\alpha }_n(1-\varrho )·\frac{2\langle f\left(p\right)-p,u_{n+1}-p\rangle }{1-\varrho }.$$

The results follow from the same arguments as in Step 4 of the proof of Theorem 1.

Step 5.$\left\{u_n\right\}$ converges strongly to $u^{*}\in \mathsf{\Omega }$, which satisfies (20), with $T_0=T$ as a nonexpansive mapping. Letting $p=u^{*}$, we deduce from Step 4 that

$$\parallel u_{n+1}-u^{*}{\parallel }^2\le (1-{\alpha }_n(1-\varrho )){\parallel u_n-u^{*}\parallel }^2+{\alpha }_n(1-\varrho )·\frac{2\langle f\left(u^{*}\right)-u^{*},u_{n+1}-u^{*}\rangle }{1-\varrho }.$$

(37)

Setting ${\mathrm{\Gamma }}_n={\parallel u_n-u^{*}\parallel }^2$, we show ${\mathrm{\Gamma }}_n\to 0(n\to \infty )$ by considering the two cases below.    □

Case 1. If there exists an integer $n_0\ge 1$ such that $\left\{{\mathrm{\Gamma }}_n\right\}$ is nonincreasing, then ${lim}_{n\to \infty }{\mathrm{\Gamma }}_n=\hslash <+\infty$ and ${lim}_{n\to \infty }({\mathrm{\Gamma }}_n-{\mathrm{\Gamma }}_{n+1})=0$. Putting $p=u^{*}$, from Step 2 and $\left\{{\gamma }_n\right\}\subset [a,b]\subset (0,1)$ we obtain

$$\begin{array}{cc}& a[(1-{\sigma }_n){\sigma }_n\parallel u_n-T{u}_n{\parallel }^2+\parallel z_n-w_n{\parallel }^2]+(1-{\alpha }_n-b)a{\parallel u_n-T_n{z}_n\parallel }^2\hfill \\ & \le {\gamma }_n\left[(1-{\sigma }_n){\sigma }_n\parallel u_n-T{u}_n{\parallel }^2+{\parallel z_n-w_n\parallel }^2\right]+{\beta }_n{\gamma }_n{\parallel u_n-T_n{z}_n\parallel }^2\hfill \\ & \le {\mathrm{\Gamma }}_n-{\mathrm{\Gamma }}_{n+1}+2{\alpha }_n\langle f\left(u_n\right)-u^{*},u_{n+1}-u^{*}\rangle \hfill \\ & \le {\mathrm{\Gamma }}_n-{\mathrm{\Gamma }}_{n+1}+2{\alpha }_n\parallel f\left(u_n\right)-u^{*}\parallel \parallel u_{n+1}-u^{*}\parallel ,\hfill \end{array}$$

which hence yields

$$\underset{n\to \infty }{lim}\parallel u_n-T{u}_n\parallel =\underset{n\to \infty }{lim}\parallel u_n-T_n{z}_n\parallel =\underset{n\to \infty }{lim}\parallel w_n-z_n\parallel =0.$$

(38)

Putting $p=u^{*}$, from Step 3 we obtain

$$\begin{array}{cc}{\gamma }_n{\left[\frac{{\zeta }_n}{2\lambda L}{\parallel r_{\lambda }\left(w_n\right)\parallel }^2\right]}^2\hfill & \le {\mathrm{\Gamma }}_n-{\mathrm{\Gamma }}_{n+1}+{\alpha }_n{\parallel f\left(u_n\right)-u^{*}\parallel }^2,\hfill \end{array}$$

which immediately leads to

$$\underset{n\to \infty }{lim}{\left[\frac{{\zeta }_n}{2\lambda L}{\parallel r_{\lambda }\left(w_n\right)\parallel }^2\right]}^2=0.$$

By inference, as in Case 1, of the proof of Theorem 1, we deduce

$$\underset{n\to \infty }{lim}\parallel w_n-y_n\parallel =\underset{n\to \infty }{lim}\parallel w_n-u_n\parallel =\underset{n\to \infty }{lim}\parallel u_{n+1}-u_n\parallel =0,$$

(39)

and

$$\underset{n\to \infty }{lim\; sup}\langle f\left(u^{*}\right)-u^{*},u_{n+1}-u^{*}\rangle \le 0.$$

(40)

Consequently, applying Lemma 4 to (37), one has ${lim}_{n\to \infty }{\parallel u_n-u^{*}\parallel }^2=0$.

Case 2. Suppose that $\exists \left\{{\mathrm{\Gamma }}_{n_k}\right\}\subset \left\{{\mathrm{\Gamma }}_n\right\}$ s.t. ${\mathrm{\Gamma }}_{n_k}<{\mathrm{\Gamma }}_{n_k+1}\forall k\in \mathcal{N}$, where $\mathcal{N}$ is the set of all positive integers. Define the mapping $\eta :\mathcal{N}\to \mathcal{N}$ by

$$\eta \left(n\right):=max\{k\le n:{\mathrm{\Gamma }}_k<{\mathrm{\Gamma }}_{k+1}\}.$$

By Lemma 6, we get

$${\mathrm{\Gamma }}_{\eta \left(n\right)}\le {\mathrm{\Gamma }}_{\eta \left(n\right)+1}\mathrm{and}{\mathrm{\Gamma }}_n\le {\mathrm{\Gamma }}_{\eta \left(n\right)+1}.$$

The conclusion follows using same arguments as in Case 2 of the proof of Theorem 1.

We introduce a viscosity extragradient-like iterative method.

We point out that Lemmas 7–10 still hold for Algorithm Section 3.

Algorithm 4 Initialization: Given $\mu >0,l\in (0,1),\lambda \in (0,\frac{1}{\mu })$. Let $u_1\in C$ be arbitrary.

Iterative Steps: Given $u_n$, calculate Step 1. Set $w_n=(1-{\sigma }_n)u_n+{\sigma }_n{T}^n{u}_n$, and compute $y_n=P_C(w_n-\lambda A{w}_n)$ and $r_{\lambda }\left(w_n\right):=w_n-y_n$. Step 2. Compute $t_n=w_n-{\zeta }_n{r}_{\lambda }\left(w_n\right)$, where ${\zeta }_n:=l^{j_n}$ and $j_n$ is the smallest nonnegative integer j, satisfying $$\langle A{w}_n-A(w_n-l^j{r}_{\lambda }\left(w_n\right)),w_n-y_n\rangle \le \frac{\mu }{2}{\parallel r_{\lambda }\left(w_n\right)\parallel }^2.$$ (41) Step 3. Compute $z_n=P_{D_n}\left(w_n\right)$ and $u_{n+1}={\alpha }_{n}f\left(u_n\right)+{\beta }_n{w}_n+{\gamma }_n{T}_n{z}_n,$ where $D_n:=\{x\in C:h_n\left(x\right)\le 0\}$ and $$h_n\left(x\right)=\langle A{t}_n,x-w_n\rangle +\frac{{\zeta }_n}{2\lambda }{\parallel r_{\lambda }\left(w_n\right)\parallel }^2.$$ (42)

Theorem 3.

Suppose $\left\{u_n\right\}$ is constructed by Algorithm 4. Then, $u_n\to u^{*}\in \mathsf{\Omega }$ provided $T^n{u}_n-T^{n+1}{u}_n\to 0$, where $u^{*}\in \mathsf{\Omega }$ is the unique solution to the VIP: $\langle (I-f)u^{*},p-u^{*}\rangle \ge 0\forall p\in \mathsf{\Omega }$.

Proof. 

By $0<{lim\; inf}_{n\to \infty }{\gamma }_n\le {lim\; sup}_{n\to \infty }{\gamma }_n<1$ and ${lim}_{n\to \infty }\frac{{\psi }_n}{{\alpha }_n}=0$, we have, without loss of generality, that $\left\{{\gamma }_n\right\}\subset [a,b]\subset (0,1)$ and ${\psi }_n\le \frac{{\alpha }_n(1-\varrho )}{2}\forall n\ge 1$. By the same arguments as in the proof of Theorem 3.1, we have $u^{*}\in \mathsf{\Omega }={\bigcap }_{i=0}^N\mathrm{Fix}\left(T_i\right)\cap \mathrm{VI}(C,A)$.

Next, we show the conclusion of the theorem. With this aim, we divide the rest of the proof into several steps.

Step 1.$\left\{u_n\right\}$ is bounded. Using the same arguments as in Step 1 of the proof of Theorem 3.1, we have inequalities (21)–(23). Thus, from (23) and ${\alpha }_n+{\beta }_n+{\gamma }_n=1\forall n\ge 1$, it follows that

$$\begin{array}{cc}& \parallel u_{n+1}-p\parallel \le {\alpha }_n(\parallel f\left(u_n\right)-f\left(p\right)\parallel +\parallel f\left(p\right)-p\parallel )+{\beta }_n\parallel w_n-p\parallel +{\gamma }_n\parallel z_n-p\parallel \hfill \\ & \le {\alpha }_n(\varrho \parallel u_n-p\parallel +\parallel f\left(p\right)-p\parallel )+{\beta }_n\parallel w_n-p\parallel +{\gamma }_n\parallel w_n-p\parallel \hfill \\ & \le {\alpha }_n(\varrho \parallel u_n-p\parallel +\parallel f\left(p\right)-p\parallel )+({\beta }_n+{\gamma }_n)(1+{\psi }_n)\parallel u_n-p\parallel \hfill \\ & \le {\alpha }_n(\varrho \parallel u_n-p\parallel +\parallel f\left(p\right)-p\parallel )+({\beta }_n+{\gamma }_n)\parallel u_n-p\parallel +\frac{{\alpha }_n(1-\varrho )}{2}\parallel u_n-p\parallel \hfill \\ & =[1-\frac{{\alpha }_n(1-\varrho )}{2}]\parallel u_n-p\parallel +\frac{{\alpha }_n(1-\varrho )}{2}·\frac{2\parallel f\left(p\right)-p\parallel }{1-\varrho }\hfill \\ & \le max\{\parallel u_n-p\parallel ,\frac{2\parallel f\left(p\right)-p\parallel }{1-\varrho }\}.\hfill \end{array}$$

Inducting, we obtain $\parallel u_n-p\parallel \le max\{\parallel u_1-p\parallel ,\frac{2\parallel f\left(p\right)-p\parallel }{1-\varrho }\}\forall n\ge 1$. Thus, $\left\{u_n\right\}$ is bounded, and so are the sequences $\left\{w_n\right\},\left\{y_n\right\},\left\{z_n\right\},\left\{f\left(u_n\right)\right\},\left\{A{t}_n\right\},\left\{T^n{u}_n\right\},\left\{T_n{z}_n\right\}$.

Step 2. We show that

$$\begin{array}{cc}& {\gamma }_n\left[(1-{\sigma }_n){\sigma }_n\parallel u_n-T^n{u}_n{\parallel }^2+{\parallel z_n-w_n\parallel }^2\right]+{\beta }_n{\gamma }_n{\parallel w_n-T_n{z}_n\parallel }^2\hfill \\ & \le \parallel u_n{-p\parallel }^2-{\parallel u_{n+1}-p\parallel }^2+{\psi }_{n}K+2{\alpha }_n\langle f\left(u_n\right)-p,u_{n+1}-p\rangle \hfill \end{array}$$

for some $K>0$. To prove this, we first note that

$$\begin{array}{cc}& \parallel u_{n+1}{-p\parallel }^2={\parallel {\alpha }_n(f\left(u_n\right)-p)+{\beta }_n(w_n-p)+{\gamma }_n(T_n{z}_n-p)\parallel }^2\hfill \\ & \le \parallel {\beta }_n(w_n-p)+{\gamma }_n(T_n{z}_n-p){\parallel }^2+2{\alpha }_n\langle f\left(u_n\right)-p,u_{n+1}-p\rangle \hfill \\ & \le {\beta }_n\parallel w_n{-p\parallel }^2+{\gamma }_n\parallel z_n{-p\parallel }^2-{\beta }_n{\gamma }_n{\parallel w_n-T_n{z}_n\parallel }^2+2{\alpha }_n\langle f\left(u_n\right)-p,u_{n+1}-p\rangle .\hfill \end{array}$$

(43)

On the other hand, using the same inferences as in (25) one has

$$\parallel z_n{-p\parallel }^2\le {(1+{\psi }_n)}^2\parallel u_n{-p\parallel }^2-(1-{\sigma }_n){\sigma }_n\parallel u_n-T^n{u}_n{\parallel }^2-{\parallel z_n-w_n\parallel }^2.$$

(44)

Substituting (44) into (43), one gets

$$\begin{array}{cc}& \parallel u_{n+1}{-p\parallel }^2\hfill \\ & \le {\beta }_n{(1+{\psi }_n)}^2\parallel u_n{-p\parallel }^2+{\gamma }_n[{(1+{\psi }_n)}^2\parallel u_n{-p\parallel }^2-(1-{\sigma }_n){\sigma }_n\parallel u_n-T^n{u}_n{\parallel }^2-\parallel z_n-w_n{\parallel }^2]\hfill \\ & -{\beta }_n{\gamma }_n{\parallel w_n-T_n{z}_n\parallel }^2+2{\alpha }_n\langle f\left(u_n\right)-p,u_{n+1}-p\rangle \hfill \\ & \le (1-{\alpha }_n)\parallel u_n{-p\parallel }^2-{\gamma }_n[(1-{\sigma }_n){\sigma }_n\parallel u_n-T^n{u}_n{\parallel }^2+\parallel z_n-w_n{\parallel }^2]\hfill \\ & +{\psi }_n(2+{\psi }_n)\parallel u_n{-p\parallel }^2-{\beta }_n{\gamma }_n{\parallel w_n-T_n{z}_n\parallel }^2+2{\alpha }_n\langle f\left(u_n\right)-p,u_{n+1}-p\rangle \hfill \\ & \le \parallel u_n{-p\parallel }^2-{\gamma }_n\left[(1-{\sigma }_n){\sigma }_n\parallel u_n-T^n{u}_n{\parallel }^2+{\parallel z_n-w_n\parallel }^2\right]-{\beta }_n{\gamma }_n{\parallel w_n-T_n{z}_n\parallel }^2\hfill \\ & +{\psi }_{n}K+2{\alpha }_n\langle f\left(u_n\right)-p,u_{n+1}-p\rangle ,\hfill \end{array}$$

where ${sup}_{n\ge 1}(2+{\psi }_n){\parallel u_n-p\parallel }^2\le K$ for some $K>0$. This immediately implies that

$$\begin{array}{cc}& {\gamma }_n[(1-{\sigma }_n){\sigma }_n\parallel u_n-T^n{u}_n{\parallel }^2+\parallel z_n-w_n{\parallel }^2]+{\beta }_n{\gamma }_n{\parallel w_n-T_n{z}_n\parallel }^2\hfill \\ & \le \parallel u_n{-p\parallel }^2-{\parallel u_{n+1}-p\parallel }^2+{\psi }_{n}K+2{\alpha }_n\langle f\left(u_n\right)-p,u_{n+1}-p\rangle .\hfill \end{array}$$

Step 3. We show that

$${\gamma }_n{\left[\frac{{\zeta }_n}{2\lambda L}{\parallel r_{\lambda }\left(w_n\right)\parallel }^2\right]}^2\le \parallel u_n{-p\parallel }^2-\parallel u_{n+1}{-p\parallel }^2+{\alpha }_n{\parallel f\left(u_n\right)-p\parallel }^2+{\psi }_{n}K.$$

Indeed, using the same argument as that of (26), we obtain that for some $L>0$,

$$\parallel z_n{-p\parallel }^2\le {\parallel w_n-p\parallel }^2-{\left[\frac{{\zeta }_n}{2\lambda L}{\parallel r_{\lambda }\left(w_n\right)\parallel }^2\right]}^2.$$

(45)

From Algorithm 4, (23), and (45) it follows that

$$\begin{array}{cc}& \parallel u_{n+1}{-p\parallel }^2\le {\alpha }_n\parallel f\left(u_n\right){-p\parallel }^2+{\beta }_n\parallel w_n{-p\parallel }^2+{\gamma }_n{\parallel z_n-p\parallel }^2\hfill \\ & \le {\alpha }_n\parallel f\left(u_n\right){-p\parallel }^2+{\beta }_n\parallel w_n{-p\parallel }^2+{\gamma }_n[\parallel w_n{-p\parallel }^2-[\frac{{\zeta }_n}{2\lambda L}\parallel r_{\lambda }\left(w_n\right){\parallel }^2{]}^2]\hfill \\ & \le {\alpha }_n\parallel f\left(u_n\right){-p\parallel }^2+{(1+{\psi }_n)}^2\parallel u_n{-p\parallel }^2-{\gamma }_n[\frac{{\zeta }_n}{2\lambda L}\parallel r_{\lambda }\left(w_n\right){{\parallel }^2]}^2\hfill \\ & \le {\alpha }_n\parallel f\left(u_n\right){-p\parallel }^2+{\parallel u_n-p\parallel }^2+{\psi }_{n}K-{\gamma }_n{\left[\frac{{\zeta }_n}{2\lambda L}{\parallel r_{\lambda }\left(w_n\right)\parallel }^2\right]}^2,\hfill \end{array}$$

which hence yields the desired assertion.

Step 4. We show that

$$\parallel u_{n+1}{-p\parallel }^2\le (1-{\alpha }_n(1-\varrho )){\parallel u_n-p\parallel }^2+{\alpha }_n(1-\varrho )\left[\frac{2\langle f\left(p\right)-p,u_{n+1}-p\rangle }{1-\varrho }+\frac{{\psi }_n}{{\alpha }_n}·\frac{K}{1-\varrho }\right].$$

Indeed, from Algorithm 4 and (3.19), one has

$$\begin{array}{cc}& \parallel u_{n+1}{-p\parallel }^2\hfill \\ & \le \parallel {\alpha }_n(f\left(u_n\right)-f\left(p\right))+{\beta }_n(w_n-p)+{\gamma }_n(T_n{z}_n-p){\parallel }^2+2{\alpha }_n\langle f\left(p\right)-p,u_{n+1}-p\rangle \hfill \\ & \le \varrho {\alpha }_n\parallel u_n{-p\parallel }^2+{\beta }_n\parallel w_n{-p\parallel }^2+{\gamma }_n{\parallel z_n-p\parallel }^2+2{\alpha }_n\langle f\left(p\right)-p,u_{n+1}-p\rangle \hfill \\ & \le \varrho {\alpha }_n\parallel u_n{-p\parallel }^2+(1-{\alpha }_n){\parallel w_n-p\parallel }^2+2{\alpha }_n\langle f\left(p\right)-p,u_{n+1}-p\rangle \hfill \\ & \le \varrho {\alpha }_n\parallel u_n{-p\parallel }^2+(1-{\alpha }_n)\parallel u_n{-p\parallel }^2+{\psi }_n(2+{\psi }_n){\parallel u_n-p\parallel }^2+2{\alpha }_n\langle f\left(p\right)-p,u_{n+1}-p\rangle \hfill \\ & \le [1-{\alpha }_n(1-\varrho )]{\parallel u_n-p\parallel }^2+{\psi }_{n}K+2{\alpha }_n\langle f\left(p\right)-p,u_{n+1}-p\rangle ,\hfill \end{array}$$

which hence leads to the desired assertion.

Step 5.$\left\{u_n\right\}$ converges strongly to the unique solution $u^{*}\in \mathsf{\Omega }$, which satisfies (20). This follows the argument in Step 5 of the proof of Theorem 1. □

Theorem 4.

Suppose $T:C\to C$ is nonexpansive and $\left\{u_n\right\}$ is constructed by: $u_1\in C$,

$$\left\{\begin{array}{cc}& w_n=(1-{\sigma }_n)u_n+{\sigma }_{n}T{u}_n,\hfill \\ & y_n=P_C(w_n-\lambda A{w}_n),\hfill \\ & t_n=(1-{\zeta }_n)w_n+{\zeta }_n{y}_n,\hfill \\ & z_n=P_{D_n}\left(w_n\right),\hfill \\ & u_{n+1}={\alpha }_{n}f\left(u_n\right)+{\beta }_n{w}_n+{\gamma }_n{T}_n{z}_n,\hfill \end{array}\right.$$

where for each $n\ge 1$, $D_n$ and ${\zeta }_n$ are chosen as in Algorithm 4, then $u_n\to u^{*}\in \mathsf{\Omega }$, where $u^{*}\in \mathsf{\Omega }$ is the unique solution to the VIP: $\langle (I-f)u^{*},p-u^{*}\rangle \ge 0\forall p\in \mathsf{\Omega }$.

Proof. 

Step 1. By Step 1 of the proof of Theorem 2, we see that $\left\{u_n\right\}$ is bounded.

Step 2. By the same arguments as in Step 2 of the proof of Theorem 2, we have

$$\begin{array}{cc}& {\gamma }_n[(1-{\sigma }_n){\sigma }_n\parallel u_n-T{u}_n{\parallel }^2+\parallel z_n-w_n{\parallel }^2]+{\beta }_n{\gamma }_n{\parallel u_n-T_n{z}_n\parallel }^2\hfill \\ & \le \parallel u_n{-p\parallel }^2-{\parallel u_{n+1}-p\parallel }^2+2{\alpha }_n\langle f\left(u_n\right)-p,u_{n+1}-p\rangle ,\hfill \end{array}$$

for some $K>0$.

Step 3. Step 3 of the proof of Theorem 2 gives

$${\gamma }_n{\left[\frac{{\zeta }_n}{2\lambda L}{\parallel r_{\lambda }\left(w_n\right)\parallel }^2\right]}^2\le \parallel u_n{-p\parallel }^2-\parallel u_{n+1}{-p\parallel }^2+{\alpha }_n{\parallel f\left(u_n\right)-p\parallel }^2.$$

Step 4. Step 4 of the proof of Theorem 2 gives

$$\parallel u_{n+1}{-p\parallel }^2\le (1-{\alpha }_n(1-\varrho )){\parallel u_n-p\parallel }^2+{\alpha }_n(1-\varrho )·\frac{2\langle f\left(p\right)-p,u_{n+1}-p\rangle }{1-\varrho }.$$

Step 5. By arguments as in Step 5 of the proof of Theorem 2, we have that $\left\{u_n\right\}$ converges strongly to the unique solution $u^{*}\in \mathsf{\Omega }$, satisfying (20). □

Remark 1.

Compared with the corresponding results in Ceng et al. [21], Reich et al. [22], and Ceng and Shang [9], our results improve and extend them in the following aspects.

(i) Although the same problem of finding an element of ${\bigcap }_{i=0}^N\mathrm{Fix}\left(T_i\right)\cap \mathrm{VI}(C,A)$ as considered in this paper was studied in reference [21], our strong convergence theorems are more advantageous and more subtle than the corresponding strong convergence ones in reference [21] because the conclusion $u_n\to u^{*}\in \mathsf{\Omega }\iff \parallel u_n-u_{n+1}\parallel +\parallel u_n-y_n\parallel \to 0(n\to \infty )$ in the corresponding strong convergence theorems [21] is updated by our conclusion $u_n\to u^{*}\in \mathsf{\Omega }$. Without doubt, the strong convergence criteria for the sequence $\left\{u_n\right\}$ in this paper are more convenient and more beneficial in comparison with those of reference [21]. In addition, to overcome the weakness of the strong convergence criteria in reference [21] (i.e., ${lim}_{n\to \infty }(\parallel u_n-u_{n+1}\parallel +\parallel u_n-y_n\parallel )=0$), we make use of Maingé’s technique (i.e., Lemma 6) to derive successfully the conclusion $u_n\to u^{*}\in \mathsf{\Omega }$.

(ii) Our results reduce to the results in reference [22] when $T_i=I$, where I is the identity mapping for $i=0,1,...,N$.

(iii) The operator A in reference [9] is extended from being Lipschitz continuous and sequentially weak in continuity mapping to A being uniformly continuous with $\parallel Az\parallel \le {lim\; inf}_{n\to \infty }\parallel A{u}_n\parallel$ for each $\left\{u_n\right\}\subset C$ with $u_n\rightharpoonup z\in C$. Furthermore, the hybrid inertial subgradient extragradient method with the line-search process in reference [9] is extended in this paper. For example, the original inertial technique $w_n=T_n{u}_n+{\alpha }_n(T_n{u}_n-T_n{x}_{n-1})$ is replaced by our Mann iteration approach $w_n=(1-{\sigma }_n)u_n+{\sigma }_n{T}^n{u}_n$, and the original iterative step $u_{n+1}={\beta }_{n}f\left(u_n\right)+{\gamma }_n{u}_n+((1-{\gamma }_n)I-{\beta }_n\rho F)T^n{z}_n$ is replaced by our simpler iterative one $u_{n+1}={\alpha }_{n}f\left(u_n\right)+{\beta }_n{u}_n+{\gamma }_n{T}_n{z}_n$. It is worth mentioning that the definition of $z_n$ in the former formulation of $u_{n+1}$ is very different from the definition of $z_n$ in the latter formulation of $u_{n+1}$.

(iv) We intend to apply the SP-iteration studied in reference [27] to the problem of finding an element of ${\bigcap }_{i=0}^N\mathrm{Fix}\left(T_i\right)\cap \mathrm{VI}(C,A)$ considered in this paper in our next project. As part of our future project, we will apply our results to the appearance of fractals using ideas given in reference [28].

4. Applications

In what follows, we give the following illustrated example. Put $\mu =l=\lambda =\frac{1}{2},{\sigma }_n=\frac{1}{3},$${\alpha }_n=\frac{1}{3(n+1)},{\beta }_n=\frac{n}{3(n+1)}$ and ${\gamma }_n=\frac{2}{3}$.

We first provide an example of Lipschitz continuous and pseudo-monotone mapping A, asymptotically nonexpansive mapping T and nonexpansive mapping $T_1$ with $\mathsf{\Omega }=\mathrm{Fix}\left(T_1\right)\cap \mathrm{Fix}\left(T\right)\cap \mathrm{VI}(C,A)\ne \varnothing$. Let $C=[-3,3]$ and $H=\mathbf{R}$ with the inner product $\langle a,b\rangle =ab$ and induced norm $\parallel ·\parallel =|·|$. The initial point $u_1$ is randomly chosen in C. Take $f\left(u\right)=\frac{1}{3}u\forall u\in C$ with $\varrho =\frac{1}{3}$. Let $A:H\to H$ and $T,T_1:C\to C$ be defined as $Au:=\frac{1}{1+|sinu|}-\frac{1}{1+\left|u\right|}$, $Tu:=\frac{2}{5}sinu$, and $T_{1}u:=sinu$ for all $u\in C$. We now claim that A is pseudo-monotone and Lipschitz continuous. Indeed, for all $u,v\in H$ we have

$$\begin{array}{cc}\parallel Au-Av\parallel \hfill & \le |\frac{\parallel v\parallel -\parallel u\parallel }{(1+\parallel u\parallel )(1+\parallel v\parallel )}|+|\frac{\parallel sinv\parallel -\parallel sinu\parallel }{(1+\parallel sinu\parallel )(1+\parallel sinv\parallel )}|\hfill \\ & \le \frac{\parallel v-u\parallel }{(1+\parallel u\parallel )(1+\parallel v\parallel )}+\frac{\parallel sinv-sinu\parallel }{(1+\parallel sinu\parallel )(1+\parallel sinv\parallel )}\hfill \\ & \le \parallel u-v\parallel +\parallel sinu-sinv\parallel \le 2\parallel u-v\parallel .\hfill \end{array}$$

This implies that A is Lipschitz continuous. Next, we show that A is pseudo-monotone. For each $u,v\in H$, it is easy to see that

$$\langle Au,v-u\rangle =(\frac{1}{1+|sinu|}-\frac{1}{1+\left|u\right|})(v-u)\ge 0\Rightarrow \langle Av,v-u\rangle =(\frac{1}{1+|sinv|}-\frac{1}{1+\left|v\right|})(v-u)\ge 0.$$

Besides, it is easy to verify that T is asymptotically nonexpansive with ${\psi }_n={\left(\frac{2}{5}\right)}^n\forall n\ge 1$, such that $\parallel T^{n+1}{z}_n-T^n{z}_n\parallel \to 0$ as $n\to \infty$. Indeed, we observe that

$$\parallel T^{n}u-T^{n}v\parallel \le \frac{2}{5}\parallel T^{n-1}u-T^{n-1}v\parallel \le \cdots \le {\left(\frac{2}{5}\right)}^n\parallel u-v\parallel \le (1+{\psi }_n)\parallel u-v\parallel ,$$

and

$$\parallel T^{n+1}{u}_n-T^n{u}_n\parallel \le {\left(\frac{2}{5}\right)}^{n-1}\parallel T^2{u}_n-T{u}_n\parallel ={\left(\frac{2}{5}\right)}^{n-1}\parallel \frac{2}{5}sin\left(T{u}_n\right)-\frac{2}{5}sin{u}_n\parallel \le 2{\left(\frac{2}{5}\right)}^n\to 0.$$

It is clear that $\mathrm{Fix}\left(T\right)=\left\{0\right\}$ and

$$\underset{n\to \infty }{lim}\frac{{\psi }_n}{{\alpha }_n}=\underset{n\to \infty }{lim}\frac{{(2/5)}^n}{1/3(n+1)}=0.$$

In addition, it is clear that $T_1$ is nonexpansive and $\mathrm{Fix}\left(T_1\right)=\left\{0\right\}$. Therefore, $\mathsf{\Omega }=\mathrm{Fix}\left(T_1\right)\cap \mathrm{Fix}\left(T\right)\cap \mathrm{VI}(C,A)=\left\{0\right\}\ne \varnothing$. In this case, Algorithm 3 can be rewritten as follows:

$$\left\{\begin{array}{cc}& w_n=\frac{2}{3}{u}_n+\frac{1}{3}{T}^n{u}_n,\hfill \\ & y_n=P_C(w_n-\frac{1}{2}A{w}_n),\hfill \\ & t_n=(1-{\zeta }_n)w_n+{\zeta }_n{y}_n,\hfill \\ & z_n=P_{D_n}\left(w_n\right),\hfill \\ & u_{n+1}=\frac{1}{3(n+1)}·\frac{1}{3}{u}_n+\frac{n}{3(n+1)}{u}_n+\frac{2}{3}{T}_1{z}_n\forall n\ge 1,\hfill \end{array}\right.$$

(46)

where for each $n\ge 1$, $D_n$ and ${\zeta }_n$ are chosen as in Algorithm 3. Then, by Theorem 1, we know that $\left\{u_n\right\}$ converges to $0\in \mathsf{\Omega }=\mathrm{Fix}\left(T_1\right)\cap \mathrm{Fix}\left(T\right)\cap \mathrm{VI}(C,A)$.

More so, since $Tu:=\frac{2}{5}sinu$ is also nonexpansive, we consider the modified version of Algorithm 3, that is,

$$\left\{\begin{array}{cc}& w_n=\frac{2}{3}{u}_n+\frac{1}{3}T{u}_n,\hfill \\ & y_n=P_C(w_n-\frac{1}{2}A{w}_n),\hfill \\ & t_n=(1-{\zeta }_n)w_n+{\zeta }_n{y}_n,\hfill \\ & z_n=P_{D_n}\left(w_n\right),\hfill \\ & u_{n+1}=\frac{1}{3(n+1)}·\frac{1}{3}{u}_n+\frac{n}{3(n+1)}{u}_n+\frac{2}{3}{T}_1{z}_n\forall n\ge 1,\hfill \end{array}\right.$$

(47)

where for each $n\ge 1$, $D_n$ and ${\zeta }_n$ are chosen as above. Then, by Theorem 2, we know that $\left\{u_n\right\}$ converges to $0\in \mathsf{\Omega }=\mathrm{Fix}\left(T_1\right)\cap \mathrm{Fix}\left(T\right)\cap \mathrm{VI}(C,A)$. In particular, we compare the performance of the new algorithm (4.2) with the Reich et al. [22] method using similar parameters as above. We choose the following initial input and take $\parallel u_{n+1}-u_n\parallel <5E^{-5}$ as the stopping criterion:

Case I: $u_1=2;$ Case II: $u_1=exp\left(\frac{150}{77}\right);$ Case III: $u_1=\frac{3}{4}\pi ;$ Case IV: $u_1=7.$

The numerical results are shown in

Table 1. Numerical results showing performance of our new method and the Reich et al. [22] method.

	New Algorithm	Riech et al. [22] alg.
	Iter. Time	Iter. Time
Case I	17 0.0082	72 0.0176
Case II	24 0.0079	168 0.0717
Case III	58 0.0131	71 0.0166
Case IV	119 0.0298	164 0.0455

and Figure 1. One can observe from the table and figures that our proposed Algorithm 1 outperforms the method proposed by Reich et al. [22] based on our test example.