Analysis of Subgradient Extragradient Method for Variational Inequality Problems and Null Point Problems

Abstract

In this paper, we introduce a new numerical method for finding a common solution to variational inequality problems involving monotone mappings and null point problems involving a finite family of inverse-strongly monotone mappings. The method is inspired by the subgradient extragradient method and the regularization method. Strong convergence results of the proposed algorithms have been obtained under some suitable conditions.

1. Introduction

Let C be a nonempty, convex and closed subset of a real Hilbert space $\mathcal{H}$ with generated norm $∥·∥$. Let $\mathcal{A}:C\to \mathcal{H}$ be a nonlinear operator. The variational inequality problem (VIP) is to obtain a point $v\in C,$ such that

$$\langle \mathcal{A}v,u-v\rangle \ge 0,\forall u\in C,$$

(1)

whose solutions set is denoted by $VI(C,\mathcal{A}).$

The VIP was introduced by Hartman and Stampacchia in the 1960s [1] as an important tool in studying obstacle, unilateral, and equilibrium problems arising in several branches of pure and applied sciences. This problem has been studied by many authors, since it includes diverse theoretical disciplines such as optimization, optimal control, symmetric boundary value problems, fixed point problems, etc.; see [2,3,4,5,6,7,8,9,10,11,12,13,14,15,16].

Definition 1.

A mapping $\mathcal{A}:C\to \mathcal{H}$ is said to be:

(1) 

monotone, if

$$\langle \mathcal{A}x-\mathcal{A}y,x-y\rangle \ge 0,\forall x,y\in C;$$

(2) 

η-strongly monotone, if there exists a number $\eta >0$ such that

$$\langle \mathcal{A}x-\mathcal{A}y,x-y\rangle \ge \eta {∥x-y∥}^2,\forall x,y\in C;$$

(3) 

λ-inverse strongly monotone, if there exists a positive number λ such that

$$\langle \mathcal{A}x-\mathcal{A}y,x-y\rangle \ge \lambda {∥\mathcal{A}x-\mathcal{A}y∥}^2,\forall x,y\in C;$$

(4) 

L-Lipschitz continuous, if there exists $L>0$ such that

$$∥\mathcal{A}x-\mathcal{A}y∥\le L∥x-y∥,\forall x,y\in C;$$

(5) 

nonexpansive, if

$$∥\mathcal{A}x-\mathcal{A}y∥\le ∥x-y∥,\forall x,y\in C.$$

Remark 1.

A relation R is symmetric if and only if for all $x,y$: if $R(x,y)$, then $R(y,x)$. It is easy to see that the relations in Definition 1 are all symmetric. As we know, every λ-inverse-strongly monotone mapping is $\frac{1}{\lambda }$-Lipschitz continuous. Additionally, if $\mathcal{A}$ is η-strongly monotone and L-Lipschitz continuous, then $\mathcal{A}$ is $\frac{\eta }{L^2}$-inverse-strongly monotone. Furthermore, both strongly monotone and inverse strongly monotone mappings are monotone. It is also known that if the set of fixed points of a nonexpansive mapping $\mathcal{A}$ exists, then it is closed and convex.

The problem which will be studied in this paper is to find a point $u^{*}\in VI(C,\mathcal{A})\bigcap$${\bigcap }_{i=1}^N{\mathcal{A}}_i^{-1}0,$ and

$$∥u^{*}∥=inf\{∥p∥:p\in VI(C,\mathcal{A})\bigcap \bigcap _{i=1}^N{\mathcal{A}}_i^{-1}0\},$$

(2)

where ${\left\{{\mathcal{A}}_i\right\}}_{i=1}^N:C\to \mathcal{H}$ is a finite family of inverse-strongly monotone mappings and $\mathcal{A}:C\to \mathcal{H}$ is a monotone mapping.

Recently, in order to solve Lipschitz and monotone VIPs in Hilbert spaces, Korpelevich [5] introduced the extragradient method (EGM):

$$\left\{\begin{array}{c}{y}_n=P_C(x_n-\tau \mathcal{A}{x}_n),\hfill \\ x_{n+1}=P_C(x_n-\tau \mathcal{A}{y}_n),\hfill \end{array}\right.$$

(3)

where $\lambda \in (0,1/L)$ and $P_C$ is denoted by the metric projection from $\mathcal{H}$ onto C. If $VI(C,\mathcal{A})\ne \varnothing$, then the sequence $\left\{x_n\right\}$ generated by process (3) converges weakly to an element in $VI(C,\mathcal{A})$. The EGM needs to calculate two values of $\mathcal{A}$ and two projections onto C at each iteration. This can be expensive if the structures of $\mathcal{A}$ and the feasible set C are complicated, hence the efficiency might be seriously affected by using this method.

Very recently, inspired by the EGM, Censor et al. [17] introduced the following subgradient extragradient method algorithm (SEGM) for solving the VIP.

They proved that $\left\{x_n\right\}$ and $\left\{y_n\right\}$ generated by Algorithm 1 converge weakly to $u^{*}$, where $u^{*}\in VI(C,\mathcal{A})$. In the SEGM, the second projection in the EGM is replaced by a projection onto a specially constructed half-space. The projection on a half-space is inherently explicit, hence the SEGM can be considered as an improvement of the EGM over each computational step.

Algorithm 1: The subgradient extragradient algorithm (SEGM).

Initialization: Set $\tau >0$ and let $x_0\in \mathcal{H}$ be arbitrary. Step 1. Given $x_n(n\ge 0)$, compute $$y_n=P_C(x_n-\tau \mathcal{A}{x}_n),$$ construct the half-space $T_n$ the bounding hyperplane of which supports C at $y_n$, $$T_n:=\{w\in \mathcal{H}\mid \langle (x_n-\tau \mathcal{A}{x}_n)-y_n,w-y_n\rangle \le 0\},$$ and calculate the next iterate $$x_{n+1}=P_{T_n}(x_n-\tau \mathcal{A}{y}_n).$$ Step 2. If $x_n=y_n$ then stop. Otherwise, set $n:=n+1$ and return to Step 1.

Following the ideas of the EGM and the SEGM, some notable methods can be recalled here such as the projected reflected gradient method [18], the golden ratio method [19], the modified extragradient method [20], and the forward–backward–forward splitting method [21]. We note that the aforementioned methods only provide, in general, weak convergence in infinite dimensional Hilbert spaces. In order to obtain the strong convergence, which is more useful, the aforementioned methods are often combined with one or more different methods such as the Halpern iteration, the viscosity method, and the regularization method.

Most recently, inspired by the EGM and the regularization method, Hieu et al. [22] introduced the following double projection method (DPM) which has a more simple structure.

Under some suitable conditions, they proved that the iterative sequence $\left\{x_n\right\}$ constructed by Algorithm 2 converges strongly to a solution of the VIP (1).

Algorithm 2: The double-projection method.

Initialization: Set $\left\{{\tau }_n\right\}\subset (0,+\infty )$, $\left\{{\alpha }_n\right\}\subset (0,+\infty )$ and let $x_0\in C$ be arbitrary. Iterative Steps: $$\left\{\begin{array}{c}{x}_{n+1}=P_C(x_n-{\tau }_n(\mathcal{A}{y}_n+{\alpha }_n{x}_n),\hfill \\ y_n=P_C(x_n-{\tau }_n(\mathcal{A}{x}_n+{\alpha }_n{x}_n).\hfill \end{array}\right.$$

Motivated by Korpelevich [5], Censor et al. [17], Malitsky [18], Hieu et al. [19], Hieu et al. [20], Tseng [21], and Hieu et al. [22], we introduce a new numerical algorithm for finding a common solution to variational inequality problems involving monotone mappings and null point problems involving a finite family of inverse-strongly monotone mappings. The strong convergence analysis is mainly based on the extragradient method and the regularization method.

2. Preliminaries

In this work, we denote the strong and weak convergence of $\left\{x_n\right\}$ by $x_n\to x$ and $x_n\rightharpoonup x$, respectively, and denote by $\mathrm{Fix}\left(T\right)$ the set of fixed points of a mapping $T:C\to \mathcal{H}$. Namely, $\mathrm{Fix}\left(T\right)=\{u\in C:Tu=u\}$.

Lemma 1

([23]). Let $\mathcal{H}$ be a real Hilbert space and let C be a nonempty convex closed subset of $\mathcal{H}$. For $u\in H$ and $v\in C$, then $v=P_{C}u$ if and only if

$$\langle u-v,w-v\rangle \le 0,\forall w\in C.$$

Lemma 2

([24]). (Demiclosedness principle) Let $\mathcal{H}$ be a real Hilbert space and let C be a nonempty convex closed subset of $\mathcal{H}$. Let $T:C\to \mathcal{H}$ be a nonexpansive mapping. Then, the mapping $I-T$ is demiclosed, i.e., if $\left\{x_n\right\}$ is a sequence in C such that $x_n\rightharpoonup x$ and $(I-T)x_n\to y$, then $(I-T)x=y$.

Lemma 3

([24]). Let $\left\{x_n\right\}$ be a sequence in $\mathcal{H}$. If $x_n\rightharpoonup x$ and $∥x_n∥\rightharpoonup ∥x∥$, then $x_n\rightharpoonup x$.

Lemma 4

([25]). Suppose $\mathcal{A}:C\to \mathcal{H}$ is a (pseudo) monotone and hemicontinuous operator, then $x^{*}$ is a solution of the VIP if and only if $x^{*}$ is a solution of the problem:

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

Lemma 5

([26,27]). Assume $\left\{{\alpha }_n\right\}$ is a sequence of nonnegative numbers satisfying the following inequality:

$${\alpha }_{n+1}\le (1-{\beta }_n){\alpha }_n+b_n+{\beta }_n{c}_n,n\in \mathbb{N},$$

where $\left\{{\beta }_n\right\},\left\{b_n\right\},\left\{c_n\right\}$ satisfy the restrictions:

(i) 

${\sum }_{n=1}^{\infty }{\beta }_n=\infty ,{lim}_{n\to \infty }{\beta }_n=0$,

(ii) 

$b_n\ge 0,{\sum }_{n=1}^{\infty }{b}_n<\infty$,

(iii) 

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

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

Lemma 6.

Assume $\mathcal{H}$ is a Hilbert space and C is a nonempty closed convex subset of $\mathcal{H}$. Suppose ${\left\{{\mathcal{A}}_i\right\}}_{i=1}^N:C\to \mathcal{H}$ is a finite family of ${\rho }_i$-inverse-strongly monotone operators with $\rho =min\{{\rho }_1,{\rho }_2,\cdots ,{\rho }_N\}$. Assume that ${\left\{{\lambda }_i\right\}}_{i=1}^N$ is a positive sequence such that ${\sum }_{i=1}^N{\lambda }_i=1$, and ${\bigcap }_{i=1}^N{\mathcal{A}}_i^{-1}0\ne \varnothing$. Then, it holds that

(1) 

${\sum }_{i=1}^N{\lambda }_i{\mathcal{A}}_i:C\to \mathcal{H}$ is ρ-inverse-strongly monotone;

(2) 

${\left({\sum }_{i=1}^N{\lambda }_i{\mathcal{A}}_i\right)}^{-1}0={\bigcap }_{i=1}^N{\mathcal{A}}_i^{-1}0.$

Proof. 

(1) Taking any $x,y\in C,$ we find that

$$\begin{array}{ccc}\hfill \langle \sum _{i=1}^N{\lambda }_i{\mathcal{A}}_{i}x-\sum _{i=1}^N{\lambda }_i{\mathcal{A}}_{i}y,x-y\rangle & =& \sum _{i=1}^N{\lambda }_i\langle {\mathcal{A}}_{i}x-{\mathcal{A}}_{i}y,x-y\rangle \hfill \\ & \ge & \sum _{i=1}^N{\lambda }_i{\rho }_i{∥{\mathcal{A}}_{i}x-{\mathcal{A}}_{i}y∥}^2\hfill \\ & \ge & \rho \sum _{i=1}^N{\lambda }_i{∥{\mathcal{A}}_{i}x-{\mathcal{A}}_{i}y∥}^2\hfill \\ & \ge & \rho {∥\sum _{i=1}^N{\lambda }_i{\mathcal{A}}_{i}x-\sum _{i=1}^N{\lambda }_i{\mathcal{A}}_{i}y∥}^2.\hfill \end{array}$$

Hence, ${\sum }_{i=1}^N{\lambda }_i{\mathcal{A}}_i:C\to \mathcal{H}$ is ρ-inverse-strongly monotone.

(2) It is obvious that ${\bigcap }_{i=1}^N{\mathcal{A}}_i^{-1}0\subseteq {\left({\sum }_{i=1}^N{\lambda }_i{\mathcal{A}}_i\right)}^{-1}0.$ Next, we prove ${\left({\sum }_{i=1}^N{\lambda }_i{\mathcal{A}}_i\right)}^{-1}0\subseteq {\bigcap }_{i=1}^N{\mathcal{A}}_i^{-1}0.$ Assume that $p\in {\left({\sum }_{i=1}^N{\lambda }_i{\mathcal{A}}_i\right)}^{-1}0$ and $q\in {\bigcap }_{i=1}^N{\mathcal{A}}_i^{-1}0.$ Then, we have

$$\begin{array}{ccc}& & {∥p-q∥}^2\hfill \\ & =& {∥(I-r\sum _{i=1}^N{\lambda }_i{\mathcal{A}}_i)p-(I-r\sum _{i=1}^N{\lambda }_i{\mathcal{A}}_i)q∥}^2\rangle \hfill \\ & =& {∥p-q∥}^2-2r\sum _{i=1}^N{\lambda }_i\langle p-q,{\mathcal{A}}_{i}p-{\mathcal{A}}_{i}q\rangle +r^2{∥\sum _{i=1}^N{\lambda }_i{\mathcal{A}}_{i}p-\sum _{i=1}^N{\lambda }_i{\mathcal{A}}_{i}q∥}^2\rangle \hfill \\ & \le & {∥p-q∥}^2-2r\sum _{i=1}^N{\lambda }_i{\rho }_i{∥{\mathcal{A}}_{i}p-{\mathcal{A}}_{i}q∥}^2+r^2\sum _{i=1}^N{\lambda }_i{∥{\mathcal{A}}_{i}p-{\mathcal{A}}_{i}q∥}^2\rangle \hfill \\ & \le & {∥p-q∥}^2-r\sum _{i=1}^N{\lambda }_i(2\rho -r){∥{\mathcal{A}}_{i}p-{\mathcal{A}}_{i}q∥}^2\rangle .\hfill \end{array}$$

(4)

Without loss of generality, we assume $r\in (0,2\rho )$. We then deduce from (4) that

$${\mathcal{A}}_{i}p={\mathcal{A}}_{i}q,i=1,2,\cdots ,N.$$

In view of $q\in {\bigcap }_{i=1}^N{\mathcal{A}}_i^{-1}0$, we find that

$${\mathcal{A}}_{i}p={\mathcal{A}}_{i}q=0,i=1,2,\cdots ,N,$$

which implies that

$$p\in \bigcap _{i=1}^N{\mathcal{A}}_i^{-1}0.$$

Hence, we deduce ${\left({\sum }_{i=1}^N{\lambda }_i{\mathcal{A}}_i\right)}^{-1}0\subseteq {\bigcap }_{i=1}^N{\mathcal{A}}_i^{-1}0.$    □

3. Main Results

In the following, we assume C is a nonempty closed convex subset of $\mathcal{H}$, ${\left\{{\mathcal{A}}_i\right\}}_{i=1}^N:C\to \mathcal{H}$ is a finite family of ${\rho }_i$-inverse-strongly monotone operators with $\rho =min\{{\rho }_1,{\rho }_2,\cdots ,{\rho }_N\}$, and $\tilde{\mathcal{A}}:={\sum }_{i=1}^N{\lambda }_i{\mathcal{A}}_i$ such that ${\sum }_{i=1}^N{\lambda }_i=1$ for all $x\in C$. Let $\mathcal{A}:C\to \mathcal{H}$ be monotone and L-Lipschitz continuous mapping such that $\mathsf{\Omega }:=VI(C,\mathcal{A})\bigcap {\bigcap }_{i=1}^N{\mathcal{A}}_i^{-1}0\ne \varnothing .$ It is easily seen that the mapping $I-\kappa \tilde{A}$ is nonexpansive for any $\kappa \in (0,2\rho ]$ (also, see [28]). We deduce from Remark 1 that $\mathrm{Fix}(I-\kappa \tilde{A})(={\tilde{\mathcal{A}}}^{-1}0)$ is convex and closed. It is known that $VI(C,\mathcal{A})$ is closed and convex, then $\mathsf{\Omega }:=VI(C,\mathcal{A})\bigcap {\bigcap }_{i=1}^N{\mathcal{A}}_i^{-1}0$ is also closed and convex. Therefore, there exists a unique solution $u^{†}$ in $\mathsf{\Omega }$ having a minimal norm. In order to find the smallest norm solution, we firstly introduce the following generalized regularized variational inequality problem: find a point $u_{\alpha }\in C$, such that

$$\langle \mathcal{A}{u}_{\alpha }+{\alpha }^{\mu }\sum _{i=1}^N{\lambda }_i{\mathcal{A}}_i{u}_{\alpha }+\alpha u_{\alpha },v-u_{\alpha }\rangle \ge 0,\forall v\in C,\mu \in (0,1).$$

(5)

We give the following lemma for describing the relationship between $u_{\alpha }$ and $u^{†}$.

Lemma 7.

(i) Problem (5) has a unique solution $u_{\alpha }$ for each $\alpha >0$;

(ii) ${lim}_{\alpha \to 0^{+}}∥u_{\alpha }-u^{†}∥=0$, and $∥u_{\alpha }∥\le ∥u^{†}∥$, $\forall \alpha >0$;

(iii) $∥u_{\alpha }-u_{\beta }∥\le \frac{\alpha -\beta }{\alpha }M$, $\forall \alpha ,\beta >0,$ where M is a positive constant.

Proof. 

(i) Letting $\tilde{\mathcal{A}}:={\sum }_{i=1}^N{\lambda }_i{\mathcal{A}}_i$ for all $x\in C$, from Remark 1 and Lemma 6, one can derive that $\tilde{\mathcal{A}}$ is Lipschitz continuous and monotone. Hence, we deduce that $\mathcal{A}+{\alpha }^{\mu }\tilde{\mathcal{A}}+\alpha I$ is Lipschitz continuous and strongly monotone. Therefore, Problem (5) has a unique solution $u_{\alpha }$ for each $\alpha >0.$

(ii) We show that

$$∥u_{\alpha }∥\le ∥p∥,\forall p\in \mathsf{\Omega }.$$

(6)

Taking any $p\in \mathsf{\Omega }$, we have $\langle \mathcal{A}p,u_{\alpha }-p\rangle \ge 0$ and $\tilde{\mathcal{A}}p=0$. From (5), the monotone property of $\mathcal{A}$ and $\tilde{A}$, we deduce that

$$\begin{array}{ccc}\hfill \langle u_{\alpha },p-u_{\alpha }\rangle & \ge & {\alpha }^{-1}\langle \mathcal{A}{u}_{\alpha }+{\alpha }^{\mu }\tilde{\mathcal{A}}{u}_{\alpha },u_{\alpha }-p\rangle \hfill \\ & =& {\alpha }^{-1}\langle \mathcal{A}{u}_{\alpha },u_{\alpha }-p\rangle +{\alpha }^{\mu -1}\langle \tilde{\mathcal{A}}{u}_{\alpha },u_{\alpha }-p\rangle \hfill \\ & \ge & {\alpha }^{-1}\langle \mathcal{A}p,u_{\alpha }-p\rangle +{\alpha }^{\mu -1}\langle \tilde{\mathcal{A}}p,u_{\alpha }-p\rangle \hfill \\ & \ge & 0,\hfill \end{array}$$

which implies

$$\begin{array}{c}\hfill ∥u_{\alpha }∥\le ∥p∥,\forall p\in \mathsf{\Omega }.\end{array}$$

Especially, one also obtains $∥u_{\alpha }∥\le ∥u^{†}∥.$ Therefore, $\left\{u_{\alpha }\right\}$ is a bounded subset of C. As C is convex and closed, C is weakly closed. Hence, there is a subsequence $\left\{u_{{\alpha }_j}\right\}$ of $\left\{u_{\alpha }\right\}$ and some point $u^{*}\in C$ such that $u_{{\alpha }_j}\rightharpoonup u^{*}(\mathrm{as}j\to \infty ).$ By the inverse-strongly monotone property of $\tilde{\mathcal{A}}$ (notice Lemma 6), the monotone property of $\mathcal{A}$, and noting (5), we deduce

$$\begin{array}{ccc}\hfill {\alpha }_j^{\mu }\langle \tilde{\mathcal{A}}{u}_{{\alpha }_j},u_{{\alpha }_j}-p\rangle & \le & \langle \mathcal{A}{u}_{{\alpha }_j},p-u_{{\alpha }_j}\rangle +\langle {\alpha }_j{u}_{{\alpha }_j},p-u_{{\alpha }_j}\rangle \hfill \\ & \le & \langle \mathcal{A}p,p-u_{{\alpha }_j}\rangle +\langle {\alpha }_j{u}_{{\alpha }_j},p-u_{{\alpha }_j}\rangle \hfill \\ & \le & {\alpha }_j\langle u_{{\alpha }_j},p-u_{{\alpha }_j}\rangle ,\hfill \end{array}$$

which implies

$$\begin{array}{c}\hfill \langle \tilde{\mathcal{A}}{u}_{{\alpha }_j},u_{{\alpha }_j}-p\rangle \le {\alpha }_j^{1-\mu }\langle u_{{\alpha }_j},p-u_{{\alpha }_j}\rangle \to 0(\mathrm{as}j\to \infty ).\end{array}$$

(7)

Using the inverse-strongly monotone property of $\tilde{A}$, noticing $\tilde{\mathcal{A}}p=0$ and (7), we find

$$\begin{array}{ccc}\hfill \mu ∥\tilde{\mathcal{A}}{u}_{{\alpha }_j}∥& =& \mu ∥\tilde{\mathcal{A}}{u}_{{\alpha }_j}-\tilde{\mathcal{A}}p∥\hfill \\ & \le & \langle \tilde{\mathcal{A}}{u}_{{\alpha }_j}-\tilde{\mathcal{A}}p,u_{{\alpha }_j}-p\rangle \hfill \\ & =& \langle \tilde{\mathcal{A}}{u}_{{\alpha }_j},u_{{\alpha }_j}-p\rangle \hfill \\ & \le & {\alpha }_j^{1-\mu }\langle p,p-u_{{\alpha }_j}\rangle \to 0(\mathrm{as}j\to \infty ).\hfill \end{array}$$

Hence,

$$\underset{j\to \infty }{lim}\tilde{\mathcal{A}}{u}_{{\alpha }_j}=0.$$

On the other hand, since the mapping $T_{\kappa }:=I-\kappa \tilde{A}$ is nonexpansive for any $\kappa \in (0,2\rho ]$ (see [28]), then by Lemma 2, we have $T_{\kappa }$ demiclosed at zero. Hence, we can derive that $\tilde{\mathcal{A}}{u}^{*}=0,$ which implies $u^{*}\in {\tilde{\mathcal{A}}}^{-1}0={\bigcap }_{i=1}^N{\mathcal{A}}_i^{-1}0$.

Now, we show that $u^{*}\in VI(C,\mathcal{A})$. By the monotonic property of $\mathcal{A}$, $\tilde{\mathcal{A}}$, and noting (5), we have

$$\langle \mathcal{A}v+{\alpha }_j^{\mu }\tilde{\mathcal{A}}v+{\alpha }_{j}v,v-u_{{\alpha }_j}\rangle \ge 0,\forall v\in C,\mu \in (0,1).$$

It follows that

$$\langle \mathcal{A}v,u_{{\alpha }_j}-v\rangle +{\alpha }_j^{\mu }\langle \tilde{\mathcal{A}}v,u_{{\alpha }_j}-v\rangle +{\alpha }_j\langle v,u_{{\alpha }_j}-v\rangle \le 0,\forall v\in C,\mu \in (0,1).$$

(8)

Since ${\alpha }_j\to 0$ as $j\to \infty$, we see that (8) implies

$$\langle \mathcal{A}v,u^{*}-v\rangle \le 0,\forall v\in C.$$

Therefore, in view of Lemma 4, we obtain that $u^{*}\in VI(C,\mathcal{A})$. This in turn implies

$$u^{*}\in \mathsf{\Omega }.$$

(9)

Next, we show $u_{\alpha }\to u^{*}$ as $\alpha \to 0.$ Indeed, noticing the lower semicontinuous of norm and (6), we obtain

$$∥u^{*}∥\le \underset{j\to \infty }{lim\; inf}\parallel u_{{\alpha }_j}\parallel \le \underset{j\to \infty }{lim\; sup}\parallel u_{{\alpha }_j}\parallel \le ∥u^{*}∥.$$

(10)

Namely,

$$\underset{j\to \infty }{lim}\parallel u_{{\alpha }_j}\parallel =∥u^{*}∥.$$

(11)

In view of Lemma 3 and the fact that $u_{{\alpha }_j}\rightharpoonup u^{*}$, we derive that ${lim}_{j\to \infty }{u}_{{\alpha }_j}=u^{*}.$ Further, in view of (6), (9), and (11), we see that

$$∥u^{*}∥=inf\{∥p∥:p\in \mathsf{\Omega }\}=\parallel u^{†}\parallel .$$

(12)

Since $u^{†}$ is a unique element which has the smallest norm in $\mathsf{\Omega }$, then $u^{*}=u^{†}.$ We show the ${lim}_{\alpha \to 0}∥u_{\alpha }-u^{†}∥=0$, taking another subsequence $\left\{u_{{\alpha }_k}\right\}$ of $\left\{u_{\alpha }\right\}$ and some point $\widehat{u}$ such that $u_{{\alpha }_k}\rightharpoonup \widehat{u}.$ By following the lines of proof as above, we deduce that ${lim}_{k\to \infty }{u}_{{\alpha }_k}=\widehat{u}\in \mathsf{\Omega }$ and $\widehat{u}=u^{†}.$ Hence, we derive that ${lim}_{\alpha \to 0}∥u_{\alpha }-u^{†}∥=0$.

(iii) Assume that $0<\alpha <\beta <1$ and $u_{\alpha }$, $u_{\beta }$ are the solutions of the problem (5). We find that

$$\langle \mathcal{A}{u}_{\alpha }+{\alpha }^{\mu }\tilde{\mathcal{A}}{u}_{\alpha }+\alpha u_{\alpha },u_{\beta }-u_{\alpha }\rangle \ge 0$$

and

$$\langle \mathcal{A}{u}_{\beta }+{\beta }^{\mu }\tilde{\mathcal{A}}{u}_{\beta }+\beta u_{\beta },u_{\alpha }-u_{\beta }\rangle \ge 0.$$

From above two inequalities, and noticing the monotonic property of $\mathcal{A}$ and $\tilde{\mathcal{A}}$, we obtain

$$({\beta }^{\mu }-{\alpha }^{\mu })\langle \tilde{A}{u}_{\beta },u_{\alpha }-u_{\beta }\rangle +\alpha \langle u_{\alpha },u_{\beta }-u_{\alpha }\rangle +\beta \langle u_{\beta },u_{\alpha }-u_{\beta }\rangle \ge 0,\forall \mu \in (0,1),$$

which implies

$$∥u_{\alpha }-u_{\beta }∥\le \frac{\left|\alpha -\beta \right|}{\alpha }∥u_{\beta }∥+\frac{|{\beta }^{\mu }-{\alpha }^{\mu }|}{\alpha }∥\tilde{\mathcal{A}}{u}_{\beta }∥.$$

The mapping $\tilde{\mathcal{A}}$ is bounded due to the fact that the operator $\tilde{\mathcal{A}}$ is Lipschitz. Moreover, using (6) and the Lagrange’s mean-value theorem to a continuous function $f\left(x\right)=x^{-\mu }$ on $[0,+\infty )$, we obtain the conclusion (iii).    □

We now introduce the following algorithm to find the smallest norm solution $u^{†}$ of problem (2).

Assumption 1. 

(C1) $0<a\le {\tau }_n\le b<{(L+{\rho }^{-1})}^{-1}$;

(C2) ${lim}_{n\to \infty }{\alpha }_n=0$ and ${\sum }_{n=0}^{\infty }{\alpha }_n=\infty$;

(C3) ${lim}_{n\to \infty }\frac{{\alpha }_n-{\alpha }_{n+1}}{{\alpha }_n^2}=0$;

(C4) $\mathsf{\Omega }:=VI(C,\mathcal{A})\bigcap {\bigcap }_{i=1}^N{\mathcal{A}}_i^{-1}0\ne \varnothing$.

Theorem 1.

The sequence $\left\{x_n\right\}$ generated by Algorithm 3 converges strongly to the minimal norm solution $u^{†}$ of problem (2) under Assumption 1 (C1)–(C4).

Algorithm 3: The subgradient extragradient method with regularization (SEGMR).

Initialization: Set$\left\{{\tau }_n\right\}\subset (0,+\infty )$, $\left\{{\alpha }_n\right\}\subset (0,+\infty )$, $\mu \in (0,1)$and let$x_0\in \mathcal{H}$be arbitrary. Step 1. Given $x_n(n\ge 0)$, compute $$y_n=P_C\left(x_n-{\tau }_n(\mathcal{A}{x}_n+{\alpha }_n^{\mu }\sum _{i=1}^N{\lambda }_i{\mathcal{A}}_i{x}_n+{\alpha }_n{x}_n)\right),$$ construct the half-space $T_n$ the bounding hyperplane of which supports C at $y_n$, $$T_n:=\{w\in \mathcal{H}\mid \langle x_n-{\tau }_n(\mathcal{A}{x}_n+{\alpha }_n^{\mu }\sum _{i=1}^N{\lambda }_i{\mathcal{A}}_i{x}_n+{\alpha }_n{x}_n)-y_n,w-y_n\rangle \le 0\},$$ and calculate the next iterate $$x_{n+1}=P_{T_n}\left(x_n-{\tau }_n(\mathcal{A}{y}_n+{\alpha }_n^{\mu }\sum _{i=1}^N{\lambda }_i{\mathcal{A}}_i{y}_n+{\alpha }_n{x}_n)\right).$$ Step 2.Set $n:=n+1$ and return to Step 1.

Proof. 

Let $\tilde{\mathcal{A}}:={\sum }_{i=1}^N{\lambda }_i{\mathcal{A}}_i$ for all $x\in C$ and denote by $x_{{\alpha }_n}$ the solution of the problem (5) with $\alpha ={\alpha }_n$. From (C1) and Lemma 7, we have $x_{{\alpha }_n}\to u^{†}$ as $n\to \infty$. Hence, it is sufficient to prove that

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim}∥x_n-x_{{\alpha }_n}∥=0.\end{array}$$

Denoting $v_n=x_n-{\tau }_n(\mathcal{A}{y}_n+{\alpha }_n^{\mu }\tilde{\mathcal{A}}{y}_n+{\alpha }_n{x}_n)$ and noticing $x_{{\alpha }_n}\in C\subset T_n$ for any $n\in \mathbb{N}$, we deduce that

$$\begin{array}{ccc}{∥x_{n+1}-x_{{\alpha }_n}∥}^2& =& {∥P_{T_n}{v}_n-x_{{\alpha }_n}∥}^2\hfill \\ & =& \langle P_{T_n}{v}_n-x_{{\alpha }_n},P_{T_n}{v}_n-x_{{\alpha }_n}\rangle \hfill \\ & =& \langle P_{T_n}{v}_n-v_n+v_n-x_{{\alpha }_n},P_{T_n}{v}_n-v_n+v_n-x_{{\alpha }_n}\rangle \hfill \\ & =& {∥P_{T_n}{v}_n-v_n∥}^2+{∥v_n-x_{{\alpha }_n}∥}^2+2\langle P_{T_n}{v}_n-v_n,v_n-x_{{\alpha }_n}\rangle .\hfill \end{array}$$

(13)

We obtain from Lemma 1 that

$$\begin{array}{c}\hfill 2{∥P_{T_n}{v}_n-v_n∥}^2+2\langle P_{T_n}{v}_n-v_n,v_n-x_{{\alpha }_n}\rangle =2\langle P_{T_n}{v}_n-v_n,P_{T_n}{v}_n-x_{{\alpha }_n}\rangle \le 0.\end{array}$$

This together with

$$\begin{array}{ccc}& & {∥x_{n+1}-x_{{\alpha }_n}∥}^2\hfill \\ & \le & {∥v_n-x_{{\alpha }_n}∥}^2-{∥P_{T_n}{v}_n-v_n∥}^2\rangle \hfill \\ & =& {∥x_n-{\tau }_n(\mathcal{A}{y}_n+{\alpha }_n^{\mu }\tilde{\mathcal{A}}{y}_n+{\alpha }_n{x}_n)-x_{{\alpha }_n}∥}^2\rangle \hfill \\ & & -{∥x_n-{\tau }_n(\mathcal{A}{y}_n+{\alpha }_n^{\mu }\tilde{\mathcal{A}}{y}_n+{\alpha }_n{x}_n)-x_{n+1}∥}^2\rangle \hfill \\ & =& {∥(x_n-x_{{\alpha }_n})-{\tau }_n(\mathcal{A}{y}_n+{\alpha }_n^{\mu }\tilde{\mathcal{A}}{y}_n+{\alpha }_n{x}_n)∥}^2\rangle \hfill \\ & & -{∥(x_n-x_{n+1})-{\tau }_n(\mathcal{A}{y}_n+{\alpha }_n^{\mu }\tilde{\mathcal{A}}{y}_n+{\alpha }_n{x}_n)∥}^2\rangle \hfill \\ & =& {∥x_n-x_{{\alpha }_n}∥}^2-{∥x_n-x_{n+1}∥}^2+2{\tau }_n\langle \mathcal{A}{y}_n+{\alpha }_n^{\mu }\tilde{\mathcal{A}}{y}_n+{\alpha }_n{x}_n,x_{{\alpha }_n}-x_{n+1}\rangle \hfill \\ & =& {∥x_n-x_{{\alpha }_n}∥}^2-{∥x_n-x_{n+1}∥}^2+2\langle x_n-y_n,y_n-x_{n+1}\rangle \hfill \\ & & +2{\tau }_n\langle \mathcal{A}{y}_n+{\alpha }_n^{\mu }\tilde{\mathcal{A}}{y}_n+{\alpha }_n{x}_n,x_{{\alpha }_n}-y_n\rangle \hfill \\ & & +2{\tau }_n\langle (\mathcal{A}{y}_n+{\alpha }_n^{\mu }\tilde{\mathcal{A}}{y}_n)-(\mathcal{A}{x}_n+{\alpha }_n^{\mu }\tilde{\mathcal{A}}{x}_n),y_n-x_{n+1}\rangle \hfill \\ & & +2\langle x_n-{\tau }_n\mathcal{A}{x}_n-{\tau }_n{\alpha }_n^{\mu }\tilde{\mathcal{A}}{x}_n-{\tau }_n{\alpha }_n{x}_n-y_n,x_{n+1}-y_n\rangle .\hfill \end{array}$$

(14)

Noticing the definition of $T_n$ and $x_{n+1}$, we have

$$\langle x_n-{\tau }_n\mathcal{A}{x}_n-{\tau }_n{\alpha }_n^{\mu }\tilde{\mathcal{A}}{x}_n-{\tau }_n{\alpha }_n{x}_n-y_n,x_{n+1}-y_n\rangle \le 0.$$

(15)

Moreover, we have

$${∥x_n-x_{n+1}∥}^2-{∥x_n-y_n∥}^2-{∥y_n-x_{n+1}∥}^2=2\langle x_n-y_n,y_n-x_{n+1}\rangle .$$

(16)

In view of (14)–(16), we obtain

$$\begin{array}{ccc}{∥x_{n+1}-x_{{\alpha }_n}∥}^2& \le & {∥x_n-x_{{\alpha }_n}∥}^2-{∥x_n-y_n∥}^2-{∥y_n-x_{n+1}∥}^2\hfill \\ & & +2{\tau }_n\langle \mathcal{A}{y}_n+{\alpha }_n^{\mu }\tilde{\mathcal{A}}{y}_n+{\alpha }_n{x}_n,x_{{\alpha }_n}-y_n\rangle \hfill \\ & & +2{\tau }_n\langle (\mathcal{A}{y}_n+{\alpha }_n^{\mu }\tilde{\mathcal{A}}{y}_n)-(\mathcal{A}{x}_n+{\alpha }_n^{\mu }\tilde{\mathcal{A}}{x}_n),y_n-x_{n+1}\rangle .\hfill \end{array}$$

(17)

It follows that

$$\begin{array}{ccc}& & 2{\tau }_n\langle (\mathcal{A}{y}_n+{\alpha }_n^{\mu }\tilde{\mathcal{A}}{y}_n)-(\mathcal{A}{x}_n+{\alpha }_n^{\mu }\tilde{\mathcal{A}}{x}_n),y_n-x_{n+1}\rangle \hfill \\ & \le & 2{\tau }_n(∥\mathcal{A}{y}_n-\mathcal{A}{x}_n∥+∥\tilde{\mathcal{A}}{y}_n-\tilde{\mathcal{A}}{x}_n∥)∥y_n-x_{n+1}∥\hfill \\ & \le & 2{\tau }_n\overline{L}∥y_n-x_n∥∥y_n-x_{n+1}∥\hfill \\ & \le & {\tau }_n^2{\overline{L}}^2∥y_n-x_n∥+{∥y_n-x_{n+1}∥}^2,\hfill \end{array}$$

where $\overline{L}=L+{\rho }^{-1}$. This together with (17) implies that

$$\begin{array}{ccc}& & {∥x_{n+1}-x_{{\alpha }_n}∥}^2\hfill \\ & \le & {∥x_n-x_{{\alpha }_n}∥}^2-(1-{\tau }_n^2{\overline{L}}^2){∥x_n-y_n∥}^2+2{\tau }_n\langle A{y}_n+{\alpha }_n^{\mu }\tilde{A}{y}_n+{\alpha }_n{x}_n,x_{{\alpha }_n}-y_n\rangle .\hfill \end{array}$$

(18)

Note that

$$\begin{array}{ccc}& & 2{\tau }_n\langle \mathcal{A}{y}_n+{\alpha }_n^{\mu }\tilde{\mathcal{A}}{y}_n+{\alpha }_n{x}_n,x_{{\alpha }_n}-y_n\rangle \hfill \\ & =& 2{\tau }_n\langle \mathcal{A}{y}_n-\mathcal{A}{x}_{{\alpha }_n},x_{{\alpha }_n}-y_n\rangle +2{\tau }_n\langle \mathcal{A}{x}_{{\alpha }_n}+{\alpha }_n^{\mu }\tilde{\mathcal{A}}{y}_n+{\alpha }_n{x}_n,x_{{\alpha }_n}-y_n\rangle \hfill \\ & =& 2{\tau }_n\langle \mathcal{A}{y}_n-\mathcal{A}{x}_{{\alpha }_n},x_{{\alpha }_n}-y_n\rangle +2{\tau }_n\langle \mathcal{A}{x}_{{\alpha }_n}+{\alpha }_{{\alpha }_n}^{\mu }\tilde{\mathcal{A}}{x}_{{\alpha }_n}+{\alpha }_n{x}_{{\alpha }_n},x_{{\alpha }_n}-y_n\rangle \hfill \\ & & +2{\tau }_n{\alpha }_n^{\mu }\langle \tilde{\mathcal{A}}{y}_n-\tilde{\mathcal{A}}{x}_{{\alpha }_n},x_{{\alpha }_n}-y_n\rangle +2{\tau }_n{\alpha }_n\langle x_n-x_{{\alpha }_n},x_{{\alpha }_n}-y_n\rangle .\hfill \end{array}$$

(19)

As $\tilde{\mathcal{A}},\mathcal{A}$ are monotone on C and $x_{{\alpha }_n}$ is the solution of Problem (5) with $\alpha ={\alpha }_n,$ we obtain

$$\langle \mathcal{A}{y}_n-\mathcal{A}{x}_{{\alpha }_n},x_{{\alpha }_n}-y_n\rangle \le 0,\langle \tilde{\mathcal{A}}{y}_n-\tilde{\mathcal{A}}{x}_{{\alpha }_n},x_{{\alpha }_n}-y_n\rangle \le 0,$$

and $\langle \mathcal{A}{x}_{{\alpha }_n}+{\alpha }_{{\alpha }_n}^{\mu }\tilde{\mathcal{A}}{x}_{{\alpha }_n}+{\alpha }_n{x}_{{\alpha }_n},x_{{\alpha }_n}-y_n\rangle \le 0,$ which imply (by (19)) that

$$\begin{array}{ccc}& & 2{\tau }_n\langle \mathcal{A}{y}_n+{\alpha }_n^{\mu }\tilde{\mathcal{A}}{y}_n+{\alpha }_n{x}_n,x_{{\alpha }_n}-y_n\rangle \hfill \\ & \le & 2{\tau }_n{\alpha }_n\langle x_n-x_{{\alpha }_n},x_{{\alpha }_n}-y_n\rangle \hfill \\ & \le & 2{\tau }_n{\alpha }_n\langle x_n-x_{{\alpha }_n},x_{{\alpha }_n}-x_n\rangle +2{\tau }_n{\alpha }_n\langle x_n-x_{{\alpha }_n},x_n-y_n\rangle \hfill \\ & \le & -2{\tau }_n{\alpha }_n{∥x_n-x_{{\alpha }_n}∥}^2+2{\tau }_n{\alpha }_n∥x_n-x_{{\alpha }_n}∥∥x_n-y_n∥.\hfill \end{array}$$

Hence, due to the fact that $2∥x_n-x_{{\alpha }_n}∥∥x_n-y_n∥\le {∥x_n-x_{{\alpha }_n}∥}^2+{∥x_n-y_n∥}^2$, we obtain

$$\begin{array}{ccc}& & 2{\tau }_n\langle \mathcal{A}{y}_n+{\alpha }_n^{\mu }\tilde{\mathcal{A}}{y}_n+{\alpha }_n{x}_n,x_{{\alpha }_n}-y_n\rangle \hfill \\ & \le & -2{\tau }_n{\alpha }_n{∥x_n-x_{{\alpha }_n}∥}^2+{\tau }_n{\alpha }_n({∥x_n-x_{{\alpha }_n}∥}^2+{∥x_n-y_n∥}^2)\hfill \\ & \le & -{\tau }_n{\alpha }_n{∥x_n-x_{{\alpha }_n}∥}^2+{\tau }_n{\alpha }_n{∥x_n-y_n∥}^2.\hfill \end{array}$$

(20)

From (18) and (20), we have

$${∥x_{n+1}-x_{{\alpha }_n}∥}^2\le (1-{\tau }_n{\alpha }_n){∥x_n-x_{{\alpha }_n}∥}^2-(1-{\tau }_n^2{\overline{L}}^2-{\tau }_n{\alpha }_n){∥x_n-y_n∥}^2.$$

(21)

As $0<a\le {\tau }_n\le b<{\overline{L}}^{-1}$ and ${\alpha }_n\to 0$, there exists $n_0\ge 1$ such that

$$1-{\tau }_n^2{\overline{L}}^2-{\tau }_n{\alpha }_n>1-b^2{\overline{L}}^2>0,{\tau }_n{\alpha }_n\in (0,2),\forall n\ge n_0.$$

(22)

Therefore, we derive from (21) that

$${∥x_{n+1}-x_{{\alpha }_n}∥}^2\le (1-{\tau }_n{\alpha }_n){∥x_n-x_{{\alpha }_n}∥}^2.$$

(23)

Notice that

$$2∥x_{{\alpha }_{n+1}}-x_{{\alpha }_n}∥∥x_{{\alpha }_{n+1}}-x_{n+1}∥\le \frac{2}{{\tau }_n{\alpha }_n}{∥x_{{\alpha }_{n+1}}-x_{{\alpha }_n}∥}^2+\frac{{\tau }_n{\alpha }_n}{2}{∥x_{{\alpha }_{n+1}}-x_{n+1}∥}^2.$$

(24)

Hence, for each $n\ge n_0$, from (22), (24), and Lemma 7 (iii), we have

$$\begin{array}{ccc}& & {∥x_{{\alpha }_n}-x_{n+1}∥}^2\hfill \\ & =& {∥x_{{\alpha }_{n+1}}-x_{{\alpha }_n}∥}^2+{∥x_{{\alpha }_{n+1}}-x_{n+1}∥}^2-2\langle x_{{\alpha }_{n+1}}-x_{{\alpha }_n},x_{{\alpha }_{n+1}}-x_{n+1}\rangle \hfill \\ & \ge & {∥x_{{\alpha }_{n+1}}-x_{{\alpha }_n}∥}^2+{∥x_{{\alpha }_{n+1}}-x_{n+1}∥}^2-2∥x_{{\alpha }_{n+1}}-x_{{\alpha }_n}∥∥x_{{\alpha }_{n+1}}-x_{n+1}∥\hfill \\ & \ge & {∥x_{{\alpha }_{n+1}}-x_{{\alpha }_n}∥}^2+{∥x_{{\alpha }_{n+1}}-x_{n+1}∥}^2-\frac{2}{{\tau }_n{\alpha }_n}{∥x_{{\alpha }_{n+1}}-x_{{\alpha }_n}∥}^2-\frac{{\tau }_n{\alpha }_n}{2}{∥x_{{\alpha }_{n+1}}-x_{n+1}∥}^2\hfill \\ & \ge & \frac{2-{\tau }_n{\alpha }_n}{2}{∥x_{{\alpha }_{n+1}}-x_{n+1}∥}^2-\frac{2-{\tau }_n{\alpha }_n}{{\tau }_n{\alpha }_n}{∥x_{{\alpha }_{n+1}}-x_{{\alpha }_n}∥}^2\hfill \\ & \ge & \frac{2-{\tau }_n{\alpha }_n}{2}{∥x_{{\alpha }_{n+1}}-x_{n+1}∥}^2-\frac{2-{\tau }_n{\alpha }_n}{{\tau }_n{\alpha }_n}{\left(\frac{{\alpha }_{n+1}-{\alpha }_n}{{\alpha }_n}\right)}^2{M}^2\hfill \\ & =& \frac{2-{\tau }_n{\alpha }_n}{2}{∥x_{{\alpha }_{n+1}}-x_{n+1}∥}^2-\frac{(2-{\tau }_n{\alpha }_n){({\alpha }_{n+1}-{\alpha }_n)}^2}{{\tau }_n{\alpha }_n^3}{M}^2,\hfill \end{array}$$

which implies

$${∥x_{{\alpha }_{n+1}}-x_{n+1}∥}^2\le \frac{2}{2-{\tau }_n{\alpha }_n}{∥x_{{\alpha }_{n+1}}-x_{{\alpha }_n}∥}^2+\frac{2{({\alpha }_{n+1}-{\alpha }_n)}^2}{{\tau }_n{\alpha }_n^3}{M}^2.$$

(25)

Therefore, it follows from (23) and (25) that

$$\begin{array}{ccc}{∥x_{{\alpha }_{n+1}}-x_{n+1}∥}^2& \le & \frac{2-2{\tau }_n{\alpha }_n}{2-{\tau }_n{\alpha }_n}{∥x_{{\alpha }_{n+1}}-x_{{\alpha }_n}∥}^2+\frac{2{({\alpha }_{n+1}-{\alpha }_n)}^2}{{\tau }_n{\alpha }_n^3}{M}^2\hfill \\ & \le & \left(1-\frac{{\tau }_n{\alpha }_n}{2-{\tau }_n{\alpha }_n}\right){∥x_n-x_{{\alpha }_n}∥}^2+\frac{2{({\alpha }_{n+1}-{\alpha }_n)}^2}{{\tau }_n{\alpha }_n^3}{M}^2.\hfill \end{array}$$

(26)

Thus, Lemma 5, (C1), (C2), and (26) ensure that ${lim}_{n\to \infty }∥x_n-x_{{\alpha }_n}∥=0,$ which implies the ${lim}_{n\to \infty }∥x_n-u^{†}∥=0$. This finishes the proof.    □

Algorithm 3 (SEGMR) behaves better than Algorithm (3) (EGM) and Algorithm 1 (SEGM).

Remark 2.

Compared with Algorithm (3) ($\mathrm{EGM}$) and Algorithm 1($\mathrm{SEGM}$), our Algorithm 3 ($\mathrm{SEGMR}$) is very different from those in the following aspects.

• In Algorithm 3, the second projection in Algorithm (3) is replaced by a projection onto a specially constructed half-space. The projection on a half-space is inherently explicit, hence Algorithm 3 can be considered as an improvement of Algorithm (3).
• Algorithm 3 is close to Algorithm 1. However, the advantage of Algorithm 3 is its strong convergence which is more useful than the weak convergence resulting from Algorithm 1 in Hilbert spaces.
• The strong convergence of Algorithm 3 is obtained from the regularization method which is different to the viscosity-like method.

From the definition of $T_n$, we see that $C\subset T_n$. If the structures of $\mathcal{A}$ and the feasible set C are simple, then we have the following result.

Corollary 1. 

(Theorem 1 [23]) The sequence $\left\{x_n\right\}$ generated by Algorithm 2 converges strongly to the minimal norm solution $u^{†}$ of the VIP (1) under the following assumptions:

(C1) 

$0<a\le {\lambda }_n\le b<L^{-1}$;

(C2) 

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

(C3) 

${lim}_{n\to \infty }\frac{{\alpha }_n-{\alpha }_{n+1}}{{\alpha }_n^2}=0$;

(C4) 

$VI(C,\mathcal{A})\ne \varnothing$.

4. Application to Split Minimization Problems

Assume $\mathcal{H}$ is a Hilbert space and C is a nonempty closed convex subset of $\mathcal{H}$. The constrained convex minimization problem is to find a point $v^{*}\in C$ such that

$$\phi \left(v^{*}\right)=\underset{v\in C}{min}\phi \left(v\right),$$

(27)

where $\phi :C\to \mathbb{R}$ is a continuous differentiable function. We need the following lemma to prove Theorem 2.

Lemma 8

([29]). A necessary condition of optimality for a point $v^{*}\in C$ to be a solution of the minimization problem (27) is that $v^{*}$ solves the inequality

$$\begin{array}{c}\hfill \langle \nabla \phi \left(v^{*}\right),u-v^{*}\rangle \ge 0,\forall u\in C.\end{array}$$

(28)

Equivalently, $v^{*}\in C$ solves the fixed point equation

$$\begin{array}{c}\hfill v^{*}=P_C(v^{*}-\tau \nabla \phi \left(v^{*}\right))\end{array}$$

for every constant $\tau >0$. If, in addition, φ is convex, then the optimality condition (28) is also sufficient.

Theorem 2.

Let $\psi :C\to \mathcal{H}$ be a differentiable and convex function whose gradient ψ is L-Lipschitz continuous. Suppose that the optimization problem ${min}_{x\in C}\phi \left(x\right)$ is consistent, i.e., its solution set is nonempty. Then, $\left\{x_n\right\}$ defined by Algorithm 4 converges strongly to the minimal norm solution $u^{†}$ of Problem (27) under Assumption 1 (C1)–(C4).

Algorithm 4: The subgradient extragradient method with regularization for minimization problems.

Initialization: Set $\left\{{\tau }_n\right\}\subset (0,+\infty )$, $\left\{{\alpha }_n\right\}\subset (0,+\infty )$ and let $x_0\in \mathcal{H}$ be arbitrary. Step 1. Given $x_n(n\ge 0)$, compute $$y_n=P_C\left(x_n-{\tau }_n(\nabla \phi x_n+{\alpha }_n{x}_n)\right),$$ construct the half-space $T_n$ the bounding hyperplane of which supports C at $y_n$, $$T_n:=\{w\in \mathcal{H}\mid \langle x_n-{\tau }_n(\nabla \phi x_n+{\alpha }_n{x}_n)-y_n,w-y_n\rangle \le 0\},$$ and calculate the next iterate $$x_{n+1}=P_{T_n}\left(x_n-{\tau }_n(\nabla \phi y_n+{\alpha }_n{x}_n)\right).$$ Step 2. Set $n:=n+1$ and return to Step 1.

5. Numerical Examples

This section presents a experiment to show the numerical behavior of the proposed method in comparison with other methods in the literature. We compare Algorithm 3 (SEGMR) with Algorithm (3) (EGM) and Algorithm 1 (SEGM).

We first give two projection calculation formulas which need to be used in numerical experiments.

(1)

The projection of p onto a half-space $\mathcal{H}=\{x:\langle v,x-u\rangle \le 0\}$ is computed by

$$P_{\mathcal{H}}\left(p\right)=p-max\{\langle v,p-u\rangle /{∥v∥}^2,0\}v.$$

(2)

The projection of p onto a ball $\mathcal{B}(u,r)=\{x:∥x-u∥\le r\}$ is computed by

$$P_{\mathcal{B}(u,r)}=u+\frac{r}{max\{∥p-u∥,r\}}(p-u).$$

Example 3.

Let $\mathcal{H}={\mathbb{R}}^2$ and $C=\{x\in {\mathbb{R}}^2:∥x∥\le 2\}$. Let $A:C\to \mathcal{H}$ be given by $A\mathbf{x}=\frac{1}{2}\mathbf{x}$, and $A_i:C\to \mathcal{H}$ be given by $A_i\mathbf{x}=\frac{1}{2}\mathbf{x}$ for all $i=1,2\cdots ,N$. It is not difficult to check that $\mathbf{0}$ is the smallest norm solution of problem (2) in $\mathsf{\Omega }.$ Let us choose ${\alpha }_n=\frac{1}{\sqrt{n+1}}$, ${\tau }_n=\mu =\frac{1}{3}$. We test our Algorithm 3 for different values of ${\mathbf{x}}_{\mathbf{0}}$ as following figures.

As shown in Figure 1 and Figure 2, one finds that Algorithm 3 (SEGMR) behaves better than Algorithm (3) (EGM) and Algorithm 1 (SEGM).

6. Conclusions

In this paper, we proposed a new numerical method for obtaining a common solution to variational inequality problems involving monotone mappings and null point problems involving a finite family of inverse-strongly monotone mappings. This method is proposed on the basis of the the regularization method and subgradient extragradient method. Strong convergence theorems are proved for sequences generated by the method under appropriate conditions. A numerical experiment is also performed to support the efficiency of the proposed method over some existing methods.