Regularization method for Noor’s variational inequality problem induced by a hemicontinuous monotone operator

Abstract

In this paper, we provide a regularization method for finding a solution of Noor’s variational inequality problem induced by a hemicontinuous monotone operator. Moreover, such a solution is related to the set of zero of inverse strongly monotone mappings. Consequently, since we do not assume the strong monotonicity of the considered operator, our results are general and extend some well-known results in the literature.

1 Introduction and preliminaries

It is well known that the variational inequalities theory, which was introduced by Hartman and Stampacchia [1] in early 1960s, provides the most natural, direct, simple, unified and efficient framework for a general treatment of a wide class of linear and nonlinear problems. Of course, it has been extended and generalized in several directions.

Let H be a real Hilbert space with inner product $〈\cdot ,\cdot 〉$ and norm $\parallel \cdot \parallel$. An important and useful generalization of variational inequality problem is called the general variational inequality introduced by Noor [2] in 1988, which is a problem of finding $u^{\ast }\in H$ with $g(u^{\ast })\in K$ such that

$$〈A\left(u^{\ast }\right),g(v)-g\left(u^{\ast }\right)〉\ge 0$$

(1.1)

for all $v\in H$ with $g(v)\in K$, where K is a nonempty closed convex subset of H and $A,g:H\to H$ are mappings.

It is known that a class of nonsymmetric and odd-order obstacle, unilateral and moving boundary value problems arising in pure and applied sciences can be studied in the unified framework of general variational inequality (1.1), see [3–12] and the references therein. However, it is worth knowing that to guarantee the existence and uniqueness of a solution of the problem (1.1), one has to impose some conditions on the operator A and g (see [13] for an example in a more general case). Indeed, it has been shown that if A fails to be Lipschitz continuous or strongly monotone, then the solution set of the problem (1.1), which will be denoted by ${GVI}_K(A,g)$, may be empty.

On the other hand, for a nonempty subset K of H, recall that a mapping $T:K\to H$ is said to be nonexpansive if

$$\parallel Tx-Ty\parallel \le \parallel x-y\parallel$$

for all $x,y\in K$. For a mapping $T:K\to H$, an element $x\in K$ is called a fixed point of T if $x=Tx$. Here, we will denote the set of all fixed points of the mapping T by $F(T)$. It is well known that the problem of finding the fixed point sets for the class of nonexpansive mappings is the subject of current interest in nonlinear functional analysis. A large number of mathematicians are interested in this field, one may see some important results in [14–17]. Moreover, it is well known that there are some relations between the fixed point problems and many types of variational inequality problems. Hence, of course, it is natural to consider a unified approach to these two different problems (see [13, 18–21] for examples).

A mapping $A:K\to H$ is said to be λ-inverse strongly monotone if there exists $\lambda >0$ such that

$$〈A(x)-A(y),x-y〉\ge \lambda {\parallel A(x)-A(y)\parallel }^2$$

for all $x,y\in K$.

For a mapping $A:K\to H$, we put $S_A=\{x\in K:A(x)=0\}$. Further, if $x\in S_A$, then x is called a zero of A.

Remark 1.1 It is easy to see that any 1-inverse strongly monotone mapping is a nonexpansive mapping. Moreover, if T is a nonexpansive mapping, then it is well known that $(I-T)$ is $\frac{1}{2}$-inverse strongly monotone mapping (see [22, 23]). Also, let us notice that the problem of finding an element of $F(T)$ is equivalent to the problem of finding an element of $x\in S_{(I-T)}$. These mean that the problem of finding zeros of the inverse strongly monotone mappings contains the problem of finding fixed points of nonexpansive mappings as a special case.

Let $A:H\to H$ be a mapping and $A_i:K\to H$ be ${\lambda }_i$-inverse strongly monotone mapping for each $i=1,2,\dots ,N$. In this paper, we present a method for finding a solution of the problem (1.1), which is related to the solution sets of those inverse strongly monotone mappings as follows: find $u^{\ast }\in H$ with $g(u^{\ast })\in S$ such that

$$〈A\left(u^{\ast }\right),g(v)-g\left(u^{\ast }\right)〉\ge 0$$

(1.2)

for all $v\in H$ with $g(v)\in K$, where $S={\bigcap }_{i=1}^N{S}_{A_i}$. Further, ${GVI}_K(A,g,S)$ denotes a set of solutions of the problem (1.2).

Remark 1.2

• (1) If $A_i=:0$ (: the zero operator) for all $i=1,2,\dots ,N$, then the problem (1.2) reduces to (1.1).

• (2) If $g=:I$ (: the identity operator), then the problem (1.2) is reduced to the problem of finding the common element of the solution of classical variational inequality problem and the set S, which has been studied by many authors (see [23–25] for examples).

We need the following concepts in order to prove our results.

Throughout this paper, we let $\mathcal{R}$ and $\mathcal{N}$ stand for the set of real numbers and the set of natural numbers, respectively. Let $F:K\times K\to \mathcal{R}$ be a bifunction with $F(u,u)=0$ for all $u\in K$.

Definition 1.3 A bifunction $F:K\times K\to \mathcal{R}$ is said to be:

• (1) monotone if, for all $u,v\in K$,

  $$F(u,v)+F(v,u)\le 0;$$

• (2) strongly monotone with constant τ if, for all $u,v\in K$,

  $$F(u,v)+F(v,u)\le -\tau {\parallel u-v\parallel }^2;$$

• (3) hemicontinuous in the first variable u if, for any fixed $v\in K$,

  $$\underset{t\to +0}{lim}F(u+t(z-u),v)=F(u,v)$$

for all $(u,z)\in K\times K$.

Recall that the equilibrium problem for a bifunction $F:K\times K\to \mathcal{R}$ is to find $u^{\ast }\in K$ such that

$$F(u^{\ast },v)\ge 0$$

(1.3)

for all $v\in K$.

Considering the problem (1.3), the following are very useful.

Lemma 1.4 [26]

Let $F:K\times K\to \mathcal{R}$ be a bifunction. If $F(u,v)$ is convex and lower semicontinuous in the variable v for any fixed $u\in K$, then we have the following:

• (1) If $F(u,v)$ is hemicontinuous in the first variable and has the monotonic property, then $U^{\ast }=V^{\ast }$, where $U^{\ast }$ is the solution set of (1.3), $V^{\ast }$ is the solution set of $F(u,v^{\ast })\le 0$ for all $u\in K$. Moreover, in this case, they are closed and convex;

• (2) If $F(u,v)$ is hemicontinuous in the first variable for each $v\in K$ and F is strongly monotone, then $U^{\ast }$ is a nonempty singleton. In addition, if F is a strongly monotone mapping, then $U^{\ast }=V^{\ast }$ is a singleton set.

In order to prove our results, we also need the following concepts and facts.

Definition 1.5 A mapping $A:H\to H$ is said to be hemicontinuous at a point $x\in H$ if

$$\underset{t\to 0}{lim}〈A(x+th),y〉=〈A(x),y〉$$

for all $h,y\in H$.

Definition 1.6 Let H be a real Hilbert space and K be a subset of H. A mapping $T:K\to H$ is said to be k-strictly pseudocontrative if there exists a constant $k\in [0,1)$ such that

$${\parallel Tx-Ty\parallel }^2\le {\parallel x-y\parallel }^2+k{\parallel (I-T)(x)-(I-T)(y)\parallel }^2$$

for all $x,y\in K$, where I is the identity operator on K.

Remark 1.7 It is well known that if $T:K\to H$ is a k-strictly pseudocontrative mapping, then the mapping $A:=I-T$ is a $(\frac{1-k}{2})$-inverse strongly monotone. Conversely, if $A:K\to H$ is a λ-inverse strongly monotone with $\lambda \in (0,\frac{1}{2}]$, then $T:=I-A$ is $(1-2\lambda )$-strictly pseudocontractive mapping (see [23]).

Lemma 1.8 [27]

Let K be a nonempty closed convex subset of a Hilbert space H and $T:K\to H$ be a k-strictly pseudocontractive mapping. Then $I-T$ is demiclosed at zero, that is, whenever $\{x_n\}$ is a sequence in K such that $\{x_n\}$ converges weakly to $x\in K$ and $\{(I-T)(x_n)\}$ converges strongly to 0, we have $(I-T)(x)=0$.

From now on, the symbols ‘⇀’ and ‘→’ stand for ‘weak convergence’ and ‘strong convergence’, respectively.

Lemma 1.9 Let H be a real Hilbert space and $\{x_n\}$ be a sequence of H. If $x_n\rightharpoonup x_0$ and $\parallel x_n\parallel \to \parallel x_0\parallel$, then $x_n\to x_0$.

Lemma 1.10 [28]

Let $\{a_n\}$, $\{b_n\}$, $\{c_n\}$ be the sequences of positive numbers satisfying the conditions:

• (a) $a_{n+1}\le (1-b_n)a_n+c_n$, where $b_n<1$;

• (b) ${\sum }_{n=0}^{\mathrm{\infty }}{b}_n=+\mathrm{\infty }$ and ${lim}_{n\to +\mathrm{\infty }}\frac{c_n}{b_n}=0$.

Then $a_n\to 0$ as $n\to \mathrm{\infty }$.

2 Regularization

Through this paper, we will denote by $\mathcal{R}$ and $\mathcal{N}$ the set of all real numbers and the set of natural numbers, respectively. Let H be a real Hilbert space and K be a closed convex subset of H. Let $A:H\to H$ and $A_i:K\to H$ be mappings, where $i=1,2,\dots ,N$. For each $\alpha \in (0,1)$, we now construct a regularization solution $u_{\alpha }$ for the problem (1.2). In fact, we have to solve the following general variational inequality problem: find $u_{\alpha }\in H$ with $g(u_{\alpha })\in K$ such that

$$〈A(u_{\alpha })+{\alpha }^{\mu }\sum _{i=1}^N(A_i\circ g)(u_{\alpha })+\alpha g(u_{\alpha }),g(v)-g(u_{\alpha })〉\ge 0$$

(2.1)

for all $v\in H$ with $g(v)\in K$, where $\mu \in (0,1)$ is a fixed real number.

Theorem 2.1 Let K be a closed convex subset of a real Hilbert space H and $g:H\to H$ be a mapping such that $K\subset g(H)$. Let $A:H\to H$ be a hemicontinuous and g-monotone mapping. Let $A_i$ be a ${\lambda }_i$-inverse strongly monotone mapping of K into H for each $i=1,2,\dots ,N$. If g is an expanding affine continuous mapping and ${GVI}_K(A,g,S)\ne \mathrm{\varnothing }$, then the following conclusions are true:

• (1) For each $\alpha \in (0,1)$, the problem (2.1) has the unique solution $u_{\alpha }$;

• (2) ${lim}_{\alpha \to 0^{+}}g(u_{\alpha })=g(u^{\ast })$ for some $u^{\ast }\in {GVI}_K(A,g,S)$;

• (3) There exists a positive constant M such that

  $${\parallel g(u_{\alpha })-g(u_{\beta })\parallel }^2\le \frac{M|\alpha -\beta |}{{\alpha }^2}$$

  (2.2)

for all $\alpha ,\beta \in (0,1)$.

Proof Put $\mathcal{I}=\{1,2,\dots ,N\}$. Firstly, due to the definition of an inverse strongly monotone mapping, we may always assume that ${\lambda }_i\in (0,\frac{1}{2}]$ for each $i\in \mathcal{I}$. Now, for each $i\in \mathcal{I}$, we define functions $F_0,F_i:g^{-1}(K)\times g^{-1}(K)\to \mathcal{R}$ by

$$F_0(u,v)=〈A(u),g(v)-g(u)〉,F_i(u,v)=〈(A_i\circ g)(u),g(v)-g(u)〉$$

for all $(u,v)\in g^{-1}(K)\times g^{-1}(K)$. Note that since g is an affine continuous mapping, we have $g^{-1}(K)$ is a closed convex subset of H. Moreover, it is easy to see that $F_0$ and $F_i$ are monotone bifunctions on $g^{-1}(K)$ for each $i\in \mathcal{I}$.

Now, let $\alpha \in (0,1)$ be a given real number. We construct a function $F_{\alpha }:g^{-1}(K)\times g^{-1}(K)\to \mathcal{R}$ by

$$F_{\alpha }(u,v)=F_0(u,v)+{\alpha }^{\mu }\sum _{i=1}^N{F}_i(u,v)+\alpha 〈g(u),g(v)-g(u)〉$$

(2.3)

for all $(u,v)\in g^{-1}(K)\times g^{-1}(K)$.

• (1) Observe that the problem (2.1) is equivalent to the problem of finding $u_{\alpha }\in g^{-1}(K)$ such that

  $$F_{\alpha }(u_{\alpha },v)\ge 0$$

  (2.4)

for all $v\in g^{-1}(K)$. Moreover, one can easily check that $F_{\alpha }(u,v)$ is a monotone hemicontinuous in the variable u for each fixed $v\in g^{-1}(K)$. Indeed, if g is an ξ-expanding mapping, then we have $F_{\alpha }(u,v)$ is a strongly monotone mapping with constant $\alpha \xi >0$. Thus, by Lemma 1.4(2), the problem (2.4) has the unique solution $u_{\alpha }\in g^{-1}(K)$ for each $\alpha >0$. This proves (1).

• (2) Observe that for each $y\in {GVI}_K(A,g,S)$, we have $F_0(y,u_{\alpha })\ge 0$ and $F_i(y,u_{\alpha })=0$ for each $i\in \mathcal{I}$. This implies that

  $$F_0(y,u_{\alpha })+{\alpha }^{\mu }\sum _{i=1}^N{F}_i(y,u_{\alpha })\ge 0.$$

  (2.5)

Consequently, by using the monotonicity of $F_0$ and $F_i$, we see that (2.3), (2.4) and (2.5) imply

$$\begin{array}{rcl}0& \ge & -F_{\alpha }(u_{\alpha },y)\\ =& -[F_0(u_{\alpha },y)+{\alpha }^{\mu }\sum _{i=1}^N{F}_i(u_{\alpha },y)+\alpha 〈g(u_{\alpha }),g(y)-g(u_{\alpha })〉]\\ \ge & -F_0(u_{\alpha },y)-{\alpha }^{\mu }\sum _{i=1}^N{F}_i(u_{\alpha },y)-\alpha 〈g(u_{\alpha }),g(y)-g(u_{\alpha })〉\\ -[F_0(y,u_{\alpha })+{\alpha }^{\mu }\sum _{i=1}^N{F}_i(y,u_{\alpha })]\\ =& -[F_0(u_{\alpha },y)+F_0(y,u_{\alpha })]-{\alpha }^{\mu }\sum _{i=1}^N[F_i(u_{\alpha },y)+F_i(y,u_{\alpha })]-\alpha 〈g(u_{\alpha }),g(y)-g(u_{\alpha })〉\\ \ge & \alpha 〈g(u_{\alpha }),g(u_{\alpha })-g(y)〉.\end{array}$$

(2.6)

Hence, we have

$$〈g(u_{\alpha }),g(y)-g(u_{\alpha })〉\ge 0$$

for all $y\in {GVI}_K(A,g,S)$. This implies

$$\parallel g(u_{\alpha })\parallel \le \parallel g(y)\parallel$$

(2.7)

for all $y\in {GVI}_K(A,g,S)$. Thus, $\{g(u_{\alpha })\}$ is a bounded subset of K. Consequently, the set of weak limit points as $\alpha \to 0$ of the net $(g(u_{\alpha }))$, denoted by ${\omega }_w(g(u_{\alpha }))$, is nonempty. This allows us to pick $z\in {\omega }_w(g(u_{\alpha }))$ and a null sequence $\{{\alpha }_k\}$ in the interval $(0,1)$ such that $\{g(u_{{\alpha }_k})\}$ weakly converges to z as $k\to \mathrm{\infty }$. Notice that since K is closed and convex, we know that K is weakly closed, and so $z\in K$. Consequently, since $K\subset g(H)$, we let $u^{\ast }\in H$ be such that $z=g(u^{\ast })$.

Now, we claim that $u^{\ast }\in {GVI}_K(A,g,S)$. To prove this claim, we divide the proof into two steps:

Step 1: We show that $g(u^{\ast })\in S$. Observe that $F_0(y,u_{{\alpha }_k})\ge 0$ for each $y\in {GVI}_K(A,g,S)$ and $k\in \mathcal{N}$. Thus, by (2.3) and the monotonicity of $F_0$, we see that

$${\alpha }_k^{\mu }\sum _{i=1}^N{F}_i(u_{{\alpha }_k},y)+{\alpha }_k〈g(u_{{\alpha }_k}),g(y)-g(u_{{\alpha }_k})〉\ge -F_0(u_{{\alpha }_k},y)\ge F_0(y,u_{{\alpha }_k})\ge 0,$$

that is,

$$\sum _{i=1}^N{F}_i(u_{{\alpha }_k},y)+{\alpha }_k^{1-\mu }〈g(u_{{\alpha }_k}),g(y)-g(u_{{\alpha }_k})〉\ge 0$$

(2.8)

for any $y\in {GVI}_K(A,g,S)$ and $k\in \mathcal{N}$. Let $j\in \mathcal{I}$ be given. Pick $y\in {GVI}_K(A,g,S)$. Since $A_i$ is a ${\lambda }_i$-inverse strongly monotone mapping for each $i\in \mathcal{I}$, in view of (2.7), we have

$$\begin{array}{rcl}{\lambda }_j{\parallel A_j\circ g(u_{{\alpha }_k})-A_j\circ g(y)\parallel }^2& \le & 〈A_j(g(u_{{\alpha }_k}))-A_j(g(y)),g(u_{{\alpha }_k})-g(y)〉\\ =& 〈A_j(g(u_{{\alpha }_k})),g(u_{{\alpha }_k})-g(y)〉\\ \le & \sum _{i=1}^N〈A_i(g(u_{{\alpha }_k})),g(u_{{\alpha }_k})-g(y)〉\\ =& -\sum _{i=1}^N{F}_i(u_{{\alpha }_k},y)\\ \le & {\alpha }_k^{1-\mu }〈g(u_{{\alpha }_k}),g(y)-g(u_{{\alpha }_k})〉\\ \le & {\alpha }_k^{1-\mu }[\parallel g(u_{{\alpha }_k})\parallel \parallel g(y)\parallel -{\parallel g(u_{{\alpha }_k})\parallel }^2]\\ \le & {\alpha }_k^{1-\mu }{\parallel g(y)\parallel }^2\end{array}$$

(2.9)

for each $k\in \mathcal{N}$. Letting $k\to \mathrm{\infty }$ in (2.9), we obtain

$$\underset{k\to \mathrm{\infty }}{lim}\parallel A_j\circ g(u_{{\alpha }_k})\parallel =\underset{k\to \mathrm{\infty }}{lim}\parallel A_j\circ g(u_{{\alpha }_k})-A_j\circ g(y)\parallel =0.$$

(2.10)

On the other hand, we know that the mapping $T_j:=I-A_j$ is strictly pseudocontractive for each $j\in \mathcal{I}$, and hence by Lemma 1.8, we have $A_j=I-T_j$ is demiclosed at zero. It follows that $A_j(g(u^{\ast }))=0$. This means $g(u^{\ast })\in S_j$. Consequently, since $j\in \mathcal{I}$ is arbitrary chosen, we conclude that $g(u^{\ast })\in {\bigcap }_{i=1}^N{S}_i$. This proves Step 1.

Step 2: We show that $u^{\ast }\in {GVI}_K(A,g)$. From the monotonic property of $F_{\alpha }$ and (2.4), we have

$$F_0(v,u_{{\alpha }_k})+{\alpha }_k^{\mu }\sum _{i=1}^N{F}_i(v,u_{{\alpha }_k})+{\alpha }_k〈g(v),g(u_{{\alpha }_k})-g(v)〉=F_{\alpha }(v,u_{{\alpha }_k})\le -F_{\alpha }(u_{{\alpha }_k},v)\le 0$$

for all $v\in g^{-1}(K)$. This gives

$$F_0(v,u_{{\alpha }_k})+{\alpha }_k^{\mu }\sum _{i=1}^N{F}_i(v,u_{{\alpha }_k})\le {\alpha }_k〈g(v),g(v)-g(u_{{\alpha }_k})〉$$

(2.11)

for all $v\in g^{-1}(K)$. Since the operator $A_i$ is a Lipschitzian mapping for each $i\in \mathcal{I}$, it follows that $F_i$ is a bounded mapping for each $i\in \mathcal{I}$. Thus, letting $k\to \mathrm{\infty }$, since ${\alpha }_k\to 0^{+}$ and $g(u_{{\alpha }_k})\rightharpoonup g(u^{\ast })$ as $k\to \mathrm{\infty }$, it follows from (2.11) that $F_0(v,u^{\ast })\le 0$ for any $v\in H$, $g(v)\in K$. Consequently, in view of Lemma 1.4(1), Step 2 is proved.

Hence, from Steps 1 and 2, we conclude that $u^{\ast }\in {GVI}_K(A,g,S)$ as required.

Next, we observe that the sequence $\{g(u_{{\alpha }_k})\}$ actually converges strongly to $g(u^{\ast })$. In fact, by using the lower semi-continuity of the norm, we know that

$$\parallel g\left(u^{\ast }\right)\parallel \le \underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel g(u_{{\alpha }_k})\parallel .$$

(2.12)

Subsequently, since $u^{\ast }\in {GVI}_K(A,g,S)$, we see that (2.7) and (2.12) imply $\parallel g(u_{{\alpha }_k})\parallel \to \parallel g(u^{\ast })\parallel$ as $k\to \mathrm{\infty }$. Then it is straightforward from Lemma 1.9 that the weak convergence to $g(u^{\ast })$ of $\{g(u_{{\alpha }_k})\}$ implies the strong convergence to $g(u^{\ast })$ of $\{g(u_{{\alpha }_k})\}$. Moreover, in view of (2.7), we see that

$$\parallel g\left(u^{\ast }\right)\parallel =inf\{\parallel g(y)\parallel :y\in {GVI}_K(A,g,S)\}.$$

(2.13)

Now, we show that

$$\underset{\alpha \to 0^{+}}{lim}g(u_{\alpha })=g(u^{\ast }).$$

To do so, let $\{g(u_{{\alpha }_j})\}\subset (g(u_{\alpha }))$, where $\{{\alpha }_j\}$, be any null sequence in the interval $(0,1)$. By following the lines of proof as above (passing to a subsequence if necessary), we know that there exists $\tilde{u}\in {GVI}_K(A,g,S)$ such that $g(u_{{\alpha }_j})\to g(\tilde{u})$ as $j\to \mathrm{\infty }$. Moreover, from (2.7) and (2.13), we have $\parallel g(\tilde{u})\parallel =\parallel g(u^{\ast })\parallel$. Consequently, since the function $\parallel g(\cdot )\parallel$ is a lower semi-continuous function and ${GVI}_K(A,g,S)$ is a closed convex subset of H, it follows from (2.13) that $u^{\ast }=\tilde{u}$. This implies that $g(u^{\ast })$ is the strong limit of the net $(g(u_{\alpha }))$ as $\alpha \to 0^{+}$.

• (3) Let $\alpha ,\beta \in (0,1)$ and $u_{\alpha }$, $u_{\beta }$ be the solutions of the problem (2.1) associated with α and β, respectively. Without loss of generality, we assume that $\alpha <\beta$. For each $i\in \mathcal{I}$, we know that $F_0$ and $F_i$ are monotone mappings. Then by applying (2.4), we have

  $$({\alpha }^{\mu }-{\beta }^{\mu })\sum _{i=1}^N{F}_i(u_{\alpha },u_{\beta })+\alpha 〈g(u_{\alpha }),g(u_{\beta })-g(u_{\alpha })〉+\beta 〈g(u_{\beta }),g(u_{\alpha })-g(u_{\beta })〉\ge 0,$$

that is,

$$〈g(u_{\alpha })-\frac{\beta }{\alpha }g(u_{\beta }),g(u_{\alpha })-g(u_{\beta })〉\le \frac{{\beta }^{\mu }-{\alpha }^{\mu }}{\alpha }\sum _{i=1}^N|F_i(u_{\alpha },u_{\beta })|.$$

(2.14)

Notice that

since $0<\alpha <\beta$. Using this one together with (2.14), we have

$${\parallel g(u_{\alpha })-g(u_{\beta })\parallel }^2\le \frac{\beta -\alpha }{\alpha }{\theta }^2+\frac{{\beta }^{\mu }-{\alpha }^{\mu }}{\alpha }\sum _{i=1}^N|F_i(u_{\alpha },u_{\beta })|,$$

(2.15)

where $\theta =sup\{\parallel g(u_{\alpha })\parallel :\alpha \in (0,1)\}$. Consequently, by using the boundedness of $F_i$, we have

$${\parallel g(u_{\alpha })-g(u_{\beta })\parallel }^2\le \frac{\beta -\alpha }{\alpha }{\theta }^2+\frac{{\beta }^{\mu }-{\alpha }^{\mu }}{\alpha }{M}_2$$

for some $M_2>0$.

Finally, by applying Lagrange’s mean-value theorem to a function which is defined on $[1,+\mathrm{\infty })$ by $t\mapsto t^{-\mu }$, we see that

$${\parallel g(u_{\alpha })-g(u_{\beta })\parallel }^2\le \frac{\beta -\alpha }{{\alpha }^2}{\theta }^2+\frac{\beta -\alpha }{{\alpha }^2}\cdot \mu M_2\le \frac{\beta -\alpha }{{\alpha }^2}M,$$

where $M=max\{{\theta }^2,\mu M_2\}$. This completes the proof. □

Remark 2.2 As a special case of Theorem 2.1, if g is the identity operator on H, then we recover a recent result presented by Kim and Buong [24].

3 Main results

In this section, we provide a regularization inertial proximal point algorithm for finding a solution of the problem (1.2). In fact, starting with an element $z_1\in H$ such that $g(z_1)\in K$, we consider the following processes:

$$〈c_n[A(z_{n+1})+{\alpha }_n^{\mu }\sum _{i=1}^N(A_i\circ g)(z_{n+1})+{\alpha }_{n}g(z_{n+1})]+g(z_{n+1})-g(z_n),g(v)-g(z_{n+1})〉\ge 0$$

(3.1)

for all $v\in H$, $g(v)\in K$, where $\{c_n\}$ and $\{{\alpha }_n\}$ are sequences of positive real numbers.

The well-definedness of algorithm (3.1) is guaranteed by the following proposition:

Proposition 3.1 Assume that all hypotheses of Theorem 2.1 are satisfied. Let $z\in g^{-1}(K)$ be a given element and c, α be positive real numbers. Define a bifunction $F_{(z,c,\alpha )}:g^{-1}(K)\times g^{-1}(K)\to \mathcal{R}$ by

$$F_{(z,c,\alpha )}(u,v):=〈c[A(u)+{\alpha }^{\mu }\sum _{i=1}^N(A_i\circ g)(u)+\alpha g(u)]+g(u)-g(z),g(v)-g(u)〉$$

for all $u,v\in g^{-1}(K)$. Then there exists a unique element $u^{\ast }\in g^{-1}(K)$ such that $F_{(z,c,\alpha )}(u^{\ast },v)\ge 0$ for all $v\in g^{-1}(K)$.

Proof Assume that g is an ξ-expanding mapping. Then, for any $u,v\in g^{-1}(K)$, we see that

$$\begin{array}{rl}{F}_{(z,c,\alpha )}(u,v)+F_{(z,c,\alpha )}(v,u)& \le (1+c\alpha )〈g(u)-g(v),g(v)-g(u)〉\\ =-(1+c\alpha ){\parallel g(u)-g(v)\parallel }^2\\ \le -\xi (1+c\alpha ){\parallel u-v\parallel }^2.\end{array}$$

This means $F_{(z,c,\alpha )}$ is $\xi (1+c\alpha )$-strongly monotone. Consequently, by Lemma 1.4(2), the proof is completed. □

Now, we provide some sufficient conditions for the convergence of the regularization inertial proximal point algorithm (3.1).

Theorem 3.2 Assume that all hypotheses of Theorem 2.1 are satisfied. If the parameters $c_n$ and ${\alpha }_n$ are positive real numbers such that

• (a) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$;

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

• (c) ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{c}_n{\alpha }_n>0$,

then the sequence $\{g(z_n)\}$ defined by (3.1) converges strongly to an element $g(u^{\ast })\in H$ for some $u^{\ast }\in {GVI}_K(A,g,S)$.

Proof Let $v\in H$ such that $g(v)\in K$. From (3.1), for each $n\in \mathcal{N}$, we see that

$$〈c_n[A(z_{n+1})+{\alpha }_n^{\mu }\sum _{i=1}^N(A_i\circ g)(z_{n+1})]+(1+c_n{\alpha }_n)g(z_{n+1})-g(z_n),g(v)-g(z_{n+1})〉\ge 0,$$

that is,

Thus we have

and so

Hence we have

$$〈{\gamma }_n[A(z_{n+1})+{\alpha }_n^{\mu }\sum _{i=1}^N(A_i\circ g)(z_{n+1})]+g(z_{n+1}),g(v)-g(z_{n+1})〉\ge {\beta }_n〈g(z_n),g(v)-g(z_{n+1})〉$$

for any $v\in H$ with $g(v)\in K$ and $n\in \mathcal{N}$, where

$${\beta }_n=\frac{1}{(1+c_n{\alpha }_n)},{\gamma }_n=c_n{\beta }_n.$$

(3.2)

By the similar argument, it follows from (2.1) that

$$〈c_{n}A(u_n)+{\alpha }_n^{\mu }\sum _{i=1}^N(A_i\circ g)(u_n)+{\alpha }_{n}g(u_n)+g(u_n)-g(u_n),g(v)-g(u_n)〉\ge 0$$

and so

when $u_n$ is the solution of (2.1) and α is replaced by ${\alpha }_n$. Thus we have

$$〈{\gamma }_n[A(u_n)+{\alpha }_n^{\mu }\sum _{i=1}^N(A_i\circ g)(u_n)]+g(u_n),g(v)-g(u_n)〉\ge {\beta }_n〈g(u_n),g(v)-g(u_n)〉,$$

(3.3)

for any $v\in H$ with $g(v)\in K$ and $n\in \mathcal{N}$. By setting $v=u_n$ in (3.2), we have

$$〈{\gamma }_n[A(z_{n+1})+{\alpha }_n^{\mu }\sum _{i=1}^N(A_i\circ g)(z_{n+1})]+g(z_{n+1}),g(u_n)-g(z_{n+1})〉\ge {\beta }_n〈g(z_n),g(u_n)-g(z_{n+1})〉$$

and, by setting $v=z_{n+1}$ in (3.3), we have

$$〈{\gamma }_n[A(u_n)+{\alpha }_n^{\mu }\sum _{i=1}^N(A_i\circ g)(u_n)]+g(u_n),g(z_{n+1})-g(u_n)〉\ge {\beta }_n〈g(u_n),g(z_{n+1})-g(u_n)〉.$$

Thus, by adding two inequalities above, we have

(3.4)

Notice that since A is a g-monotone mapping and each $A_i$ for each $i\in \mathcal{I}$ is a ${\lambda }_i$-inverse strongly monotone, we have

$$〈A(z_{n+1})-A(u_n),g(u_n)-g(z_{n+1})〉\le 0$$

and

$$〈A_i\circ g(z_{n+1})-A_i\circ g(u_n),g(u_n)-g(z_{n+1})〉\le 0$$

for each $i\in \mathcal{I}$. Thus, from the above fact, we obtain

$$〈g(z_{n+1})-g(u_n),g(u_n)-g(z_{n+1})〉\ge {\beta }_n〈g(z_n)-g(u_n),g(u_n)-g(z_{n+1})〉.$$

This gives

$$\parallel g(z_{n+1})-g(u_n)\parallel \le {\beta }_n\parallel g(z_n)-g(u_n)\parallel$$

(3.5)

for each $n\in \mathcal{N}$. Thus, using Theorem 2.1(3), we have

$$\begin{array}{rcl}\parallel g(z_{n+1})-g(u_{n+1})\parallel & \le & \parallel g(z_{n+1})-g(u_n)\parallel +\parallel g(u_n)-g(u_{n+1})\parallel \\ \le & {\beta }_n\parallel g(z_n)-g(u_n)\parallel +\sqrt{\frac{M({\alpha }_n-{\alpha }_{n+1})}{{\alpha }_{n+1}^2}}\\ \le & (1-b_n)\parallel g(z_n)-g(u_n)\parallel +d_n,\end{array}$$

where

$$b_n=\frac{c_n{\alpha }_n}{(1+c_n{\alpha }_n)},d_n=\sqrt{\frac{M({\alpha }_n-{\alpha }_{n+1})}{{\alpha }_{n+1}^2}}.$$

Consequently, by the condition (c), we have ${\sum }_{n=1}^{\mathrm{\infty }}{b}_n=\mathrm{\infty }$. Meanwhile, the conditions (b) and (c) imply ${lim}_{n\to \mathrm{\infty }}\frac{d_n}{b_n}=0$. Thus, all the conditions of Lemma 1.10 are satisfied, and so $\parallel g(z_{n+1})-g(u_{n+1})\parallel \to 0$ as $n\to \mathrm{\infty }$. Moreover, by the condition (a) and Theorem 2.1(2), we know that there exists $u^{\ast }\in {GVI}_K(A,g,S)$ such that $\{g(u_n)\}$ converges strongly to $g(u^{\ast })$. This implies that $\{g(z_n)\}$ converges strongly to $g(u^{\ast })$ as $n\to +\mathrm{\infty }$. This completes the proof. □

Example 3.3 Let $p\in (0,1)$ be a fixed real number. Let ${\alpha }_n={(\frac{1}{n})}^p$ and $c_n=\frac{1}{{\alpha }_n}$ for each $n\in \mathcal{N}$. Then we can check that the sequences $\{{\alpha }_n\}$ and $\{c_n\}$ satisfy all the conditions in Theorem 3.2.

Remark 3.4 Note that because of the condition (b) of Theorem 3.2, the choice $\{\frac{1}{n}\}$ is not included in the class of the parameters $\{{\alpha }_n\}$. This may lead to an important further research work, that is, finding other (regularization inertial proximal point) algorithms for the problem (1.2) including the natural parameter choice $\{\frac{1}{n}\}$.

4 Conclusion

In this paper, we provide a regularization method for general monotone variational inequality, where the regularizer is a Lipschitz continuous and strongly monotone operator. Also, an iterative method as discretization of regularization method is introduced. We would like to point out, as we have mentioned in Remark 1.1 and Remark 1.2, that our results are general and extend some well-known results which have been investigated in literature; for more examples, see [29–34]. Further, let us notice that although many authors have proved results for finding the solution of the variational inequality problem and the solution set of a finite inverse strongly monotone mapping, it is clear that it cannot be directly applied to our main considered problem (1.2) due to the presence of g.