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Regularization method for Noor’s variational inequality problem induced by a hemicontinuous monotone operator
Fixed Point Theory and Applications volume 2012, Article number: 169 (2012)
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 wellknown 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 $\u3008\cdot ,\cdot \u3009$ 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
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 oddorder 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
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 λinverse strongly monotone if there exists $\lambda >0$ such that
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 1inverse strongly monotone mapping is a nonexpansive mapping. Moreover, if T is a nonexpansive mapping, then it is well known that $(IT)$ 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 $x\in {S}_{(IT)}$. 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
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 $u,v\in K$,
$$F(u,v)+F(v,u)\le \tau {\parallel uv\parallel}^{2};$$ 
(3)
hemicontinuous in the first variable u if, for any fixed $v\in K$,
$$\underset{t\to +0}{lim}F(u+t(zu),v)=F(u,v)$$
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
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
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 kstrictly pseudocontrative if there exists a constant $k\in [0,1)$ such that
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 kstrictly pseudocontrative mapping, then the mapping $A:=IT$ is a $(\frac{1k}{2})$inverse strongly monotone. Conversely, if $A:K\to H$ is a λinverse strongly monotone with $\lambda \in (0,\frac{1}{2}]$, then $T:=IA$ is $(12\lambda )$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 kstrictly pseudocontractive mapping. Then $IT$ 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 $\{(IT)({x}_{n})\}$ converges strongly to 0, we have $(IT)(x)=0$.
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
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 gmonotone 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
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
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 ξexpanding 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
Hence, we have
for all $y\in {GVI}_{K}(A,g,S)$. This implies
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
that is,
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
for each $k\in \mathcal{N}$. Letting $k\to \mathrm{\infty}$ in (2.9), we obtain
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
for all $v\in {g}^{1}(K)$. This gives
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 semicontinuity of the norm, we know that
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
Now, we show that
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 semicontinuous 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
$$({\alpha}^{\mu}{\beta}^{\mu})\sum _{i=1}^{N}{F}_{i}({u}_{\alpha},{u}_{\beta})+\alpha \u3008g({u}_{\alpha}),g({u}_{\beta})g({u}_{\alpha})\u3009+\beta \u3008g({u}_{\beta}),g({u}_{\alpha})g({u}_{\beta})\u3009\ge 0,$$
that is,
Notice that
since $0<\alpha <\beta $. Using this one together with (2.14), we have
where $\theta =sup\{\parallel g({u}_{\alpha})\parallel :\alpha \in (0,1)\}$. Consequently, by using the boundedness of ${F}_{i}$, we have
for some ${M}_{2}>0$.
Finally, by applying Lagrange’s meanvalue theorem to a function which is defined on $[1,+\mathrm{\infty})$ by $t\mapsto {t}^{\mu}$, we see that
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:
for all $v\in H$, $g(v)\in K$, where $\{{c}_{n}\}$ and $\{{\alpha}_{n}\}$ are sequences of positive real numbers.
The welldefinedness 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
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 ξexpanding mapping. Then, for any $u,v\in {g}^{1}(K)$, we see that
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
that is,
Thus we have
and so
Hence we have
for any $v\in H$ with $g(v)\in K$ and $n\in \mathcal{N}$, where
By the similar argument, it follows from (2.1) that
and so
when ${u}_{n}$ is the solution of (2.1) and α is replaced by ${\alpha}_{n}$. Thus we have
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
and, by setting $v={z}_{n+1}$ in (3.3), we have
Thus, by adding two inequalities above, we have
Notice that since A is a gmonotone mapping and each ${A}_{i}$ for each $i\in \mathcal{I}$ is a ${\lambda}_{i}$inverse strongly monotone, we have
and
for each $i\in \mathcal{I}$. Thus, from the above fact, we obtain
This gives
for each $n\in \mathcal{N}$. Thus, using Theorem 2.1(3), we have
where
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 wellknown 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.
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Acknowledgements
The authors wish to express their gratitude to the referees for a careful reading of the manuscript and helpful suggestions. This work is supported by the Centre of Excellence in Mathematics, the commission on Higher Education, Thailand and by the Basic Science Research Program through the National Research Foundation of Korea funded by the Ministry of Education, Science and Technology (Grant Number: 20100021821)
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Cho, Y.J., Petrot, N. Regularization method for Noor’s variational inequality problem induced by a hemicontinuous monotone operator. Fixed Point Theory Appl 2012, 169 (2012). https://doi.org/10.1186/168718122012169
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Keywords
 general variational inequality problem
 regularization
 hemicontinuous mapping
 monotone mapping
 inverse strongly monotone mapping