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Regularization and iterative method for general variational inequality problem in hilbert spaces
Journal of Inequalities and Applications volume 2011, Article number: 21 (2011)
Abstract
Without the strong monotonicity assumption of the mapping, we provide a regularization method for general variational inequality problem, when its solution set is related to a solution set of an inverse strongly monotone mapping. Consequently, an iterative algorithm for finding such a solution is constructed, and convergent theorem of the such algorithm is proved. It is worth pointing out that, since we do not assume strong monotonicity of general variational inequality problem, our results improve and extend some wellknown results in the literature.
1. Introduction
It is well known that the ideas and techniques of the variational inequalities are being applied in a variety of diverse fields of pure and applied sciences and proven to be productive and innovative. It has been shown that this theory provides the most natural, direct, simple, unified, and efficient framework for a general treatment of a wide class of linear and nonlinear problems. The development of variational inequality theory can be viewed as the simultaneous pursuit of two different lines of research. On the one hand, it reveals the fundamental facts on the qualitative aspects of the solutions to important classes of problems. On the other hand, it also enables us to develop highly efficient and powerful new numerical methods for solving, for example, obstacle, unilateral, free, moving, and complex equilibrium problems.
In 1988, Noor [1] introduced and studied a class of variational inequalities, which is known as general variational inequality, GVI_{ K } (A, g), is as follows: Find u* ∈ H, g(u*) ∈ K such that
where K is a nonempty closed convex subset of a real Hilbert space H with inner product 〈·, ·〉, and T, g: H → H be mappings. It is known that a class of nonsymmetric and oddorder obstacle, unilateral, and moving boundary value problems arising in pure and applied can be studied in the unified framework of general variational inequalities (e.g., [2] and the references therein). Observe that to guarantee the existence and uniqueness of a solution of the problem (1.1), one has to impose conditions on the operator A and g, see [3] for example in a more general case. By the way, it is worth noting that, if A fails to be Lipschitz continuous or strongly monotone, then the solution set of the problem (1.1) may be an empty one.
Related to the variational inequalities, we have the problem of finding the fixed points of the nonlinear mappings, which is the subject of current interest in functional analysis. It is natural to consider a unified approach to these two different problems (e.g., [3–8]). Motivated and inspired by the research going in this direction, in this article, we present a method for finding a solution of the problem (1.1), which is related to the solution set of an inverse strongly monotone mapping and is as follows: Find u* ∈ H, g(u*) ∈ S(T) such that
when A is a generalized monotone mapping, T: K → H is an inverse strongly monotone mapping, and S(T) = {x ∈ K: T(x) = 0}. We will denote by GVI_{ K } (A, g, T) for a set of solution to the problem (1.2). Observe that, if T =: 0, the zero operator, then the problem (1.2) reduces to (1.1). Moreover, we would also like to notice that although many authors have proven results for finding the solution of the variational inequality problem and the solution set of inverse strongly monotone mapping (e.g., [4, 8, 9]), it is clear that it cannot be directly applied to the problem GV I_{ K } (A, g, T) due to the presence of g.
2. Preliminaries
Let H be a real Hilbert space whose inner product and norm are denoted by 〈·, ·〉 and  · , respectively. Let K be a nonempty closed convex subset of H. In this section, we will recall some wellknown results and definitions.
Definition2.1. Let A: H → H be a mapping and K ⊂ H. Then, A is said to be semicontinuous at a point x in K if
Definition2.2. A mapping T: K → H is said to be λinverse strongly monotone, if there exists a λ > 0 such that
Recall that a mapping B: K → H is said to be kstrictly pseudocontractive if there exists a constant k ∈ [0, 1) such that
Let I be the identity operator on K. It is well known that if B: K → H is a kstrictly pseudocontrative mapping, then the mapping T := I  B is a inverse strongly monotone, see [4]. Conversely, if T: K → H is a λinverse strongly monotone with , then B := I  T is (1  2λ)strictly pseudocontrative mapping. Actually, for all x, y ∈ K, we have
On the other hand, since H is a real Hilbert space, we have
Hence,
Moreover, we have the following result:
Lemma 2.3. [10]Let K be a nonempty closed convex subset of a Hilbert space H and B: K → H a kstrictly pseudocontractive mapping. Then, I  B is demiclosed at zero, that is, whenever {x_{ n } } is a sequence in K such that {x_{ n } } converges weakly to x ∈ K and {(I  B)(x_{ n } )} converges strongly to 0, we must have (I  B)(x) = 0.
Definition2.4. Let A, g: H → H. Then A is said to be gmonotone if
For g = I, the identity operator, Definition 2.4 reduces to the wellknown definition of monotonicity. However, the converse is not true.
Now we show an example in proof of our main problem (1.2).
Example 2.5. Let a, b be strictly positive real numbers. Put H = {(x_{1}, x_{2}) a ≤ x_{1} ≤ a, b ≤ x_{2} ≤ b} with the usual inner product and norm. Let K = {(x_{1}, x_{2}) ∈ H: 0 ≤ x_{1} ≤ x_{2}} be a closed convex subset of H. Let T: K → H be a mapping defined by T(x) = (I  P_{Δ})(x), where Δ = {x := (x_{1}, x_{2}) ∈ H: x_{1} = x_{2}} is a closed convex subset of H, and P_{Δ} is a projection mapping from K onto Δ. Clearly, T is inverse strongly monotone, and S(T) = Δ ∩ K. Now, if is a considered matrix operator and g = I, where I is the 2 × 2 identity matrix. Then, we can verify that A is a gmonotone operator. Indeed, for each x := (x_{1}, x_{2}), y := (y_{1}, y_{2}) ∈ H, we have
Moreover, if , then we must have 〈A(u*), g(y)  g(u*)〉 ≥ 0, for all y = (y_{1}, y_{2}) ∈ H, g(y) ∈ K. This equivalence becomes
for all y = (y_{1}, y_{2}) ∈ H, g(y) ∈ K. Notice that g^{1}(K) = {(y_{1}, y_{2}) ∈ Hy_{1} ≥ y_{2}}. Thus, in view of (2.1), it follows that {x = (x_{1}, x_{2}) ∈ Hx_{1} = x_{2}} ⊂ GVI_{ K } (A, g). Hence, GVI_{ K } (A, g, T) ≠ ∅.
Remark 2.6. In Example 2.5, the operator A is not a monotone mapping on H.
We need the following concepts to prove our results.
Let stand for the set of real numbers. Let be an equilibrium bifunction, that is, F(u, u) = 0 for every u ∈ K.
Definition2.7. The equilibrium bifunction is said to be
(i) monotone, if for all u, v ∈ K, then we have
(ii) strongly monotone with constant τ; if for all u, v ∈ K, then we have
(iii) hemicontinuous in the first variable u; if for each fixed v, then we have
Recall that the equilibrium problem for is to find u* ∈ K such that
Concerning to the problem (2.5), the following facts are very useful.
Lemma 2.8. [11]Letbe such that F(u, v) is convex and lower semicontinuous in the variable v for each fixed u ∈ K. Then,

(1)
if F(u, v) is hemicontinuous in the first variable and has the monotonic property, then U* = V*, where U* is the solution set of (2.5), and V* is the solution set of F(u, v*) ≤ 0 for all u ∈ K. Moreover, in this case, they are closed and convex;

(2)
if F(u, v) is hemicontinuous in the first variable for each v ∈ K and F is strongly monotone, then U* is a nonempty singleton. In addition, if F is a strongly monotone mapping, then U* = V* is a singleton set.
The following basic results are also needed.
Lemma 2.9. Let {x_{ n } } be a sequence in H. If x_{ n } → x wealky and x_{ n }  → x, then x_{ n } → x strongly.
Lemma 2.10. [12]. Let a_{ n }, b_{ n }, c_{ n } be the sequences of positive real numbers satisfying the following conditions.
(i) a_{n+1}≤ (1  b_{ n } )a_{ n } + c_{ n } , b_{ n } < 1,
(ii) .
Then, lim_{n→+∞}a_{ n } = 0.
3. Regularization
Let α ∈ (0, 1) be a fixed positive real number. We now construct a regularization solution u_{ α }for (1.2), by solving the following general variational inequality problem: find u_{ α } ∈ H, g(u_{ α } ) ∈ K such that
Theorem 3.1. Let K be a closed convex subset of a Hilbert space H and g: H → H be a mapping such that K ⊂ g(H). Let A: H → H be a hemicontinuous on K and gmonotone mapping, T: K → H be λinverse strongly monotone mapping. If g is an expanding affine continuous mapping and GVI_{ K } (A, g, T) ≠ ∅, then the following conclusions are true.

(a)
For each α ∈ (0, 1), the problem (3.1) has the unique solution u_{ α }:

(b)
If α ↓ 0, then {g(u_{ α } )} converges. Moreover, for some u* ∈ GVI_{ K } (A, g, T).

(c)
There exists a positive constant M such that
(3.2)
when 0 < α < β < 1.
Proof. First, in view of the definition 2.2, we will always assume that . Now, related to mappings A, T, and g, we define functions by
for all (u, v) ∈ g^{1}(K) × g^{1}(K). Note that, F_{ A } , F_{ T } are equilibrium monotone bifunctions, and g^{1}(K) is a closed convex subset of H.
Now, let α ∈ (0, 1) be a given positive real number. We construct a function by
for all (u, v) ∈ g^{1}(K) × g^{1}(K).

(a)
Observe that, the problem (3.1) is equivalent to the problem of finding u_{ α } ∈ g ^{1}(K) such that
(3.4)
Moreover, one can easily check that F_{ α } (u, v) is a monotone hemicontinuous in the variable u for each fixed v ∈ g^{1}(K). Indeed, it is strongly monotone with constant αξ > 0, where g is an ξexpansive. Thus, by Lemma 2.8(ii), the problem (3.4) has a unique solution u_{ α } ∈ g^{1}(K) for each α > 0. This prove (a).

(b)
Note that for each y ∈ GVI_{ K } (A, g, T) we have [F_{ A } + α^{μ}F_{ T } ](y, u_{ α } ) ≥ 0. Consequently, by (3.4), we have
This means
Consequently,
that is, g(u_{ α } ) ≤ g(y) for all y ∈ GVI_{ K } (A, g, T). Thus, {g(u_{ α } )} is a bounded subset of K. Consequently, the set of weak limit points as α → 0 of the net (g(u_{ α } )) denoted by ω_{ w } (g(u_{ α } )) is nonempty. Pick z ∈ ω_{ w } (g(u_{ α } )) and a null sequence {α_{ k } } in the interval (0, 1) such that weakly converges to z as k → ∞. Since K is closed and convex, we know that K is weakly closed, and it follows that z ∈ K. Now, since K ⊂ g(H), we let u* ∈ H be such that z = g(u*) and claim that u* ∈ GVI_{ K } (A, g, T).
To prove such a claim, we will first show that g(u*) ∈ S(T). To do so, let us pick a fixed y ∈ GVI_{ K } (A, g, T). By (3.3) and the monotonicity of F_{ A } , we have
equivalently,
for each k ∈ ℕ. Using the above together with the assumption that T is an λinverse strongly monotone mapping, we have
for each k ∈ ℕ. Letting k → +∞, we obtain
On the other hand, we know that the mapping I  T is a strictly pseudocontractive, thus by lemma 2.3, we have T demiclosed at zero. Consequently, since weakly converges to g(u*), we obtain T(g(u*)) = T(g(y)) = 0. This proves g(u*) ∈ S(T), as required.
Now, we will show that u* ∈ GVI_{ K } (A, g, T). Notice that, from the monotonic property of F_{ α } and (3.4), we have
for all v ∈ g^{1}(K). That is,
for all v ∈ g^{1}(K). Since α_{ k } ↓ 0 as k → ∞, we see that (3.6) implies F_{ A } (v, u*) ≤ 0 for any v ∈ H, g(v) ∈ K. Consequently, in view of Lemma 2.8(1), we obtain our claim immediately.
Next we observe that the sequence actually converges to g(u*) strongly. In fact, by using a lower semicontinuous of norm and (3.5), we see that
since u* ∈ GVI_{ K } (A, g, T). That is, as k → ∞. Now, it is straightforward from Lemma 2.9, that the weak convergence to g(u*) of implies strong convergence to g(u*) of . Further, in view of (3.5), we see that
Next, we let , where {α_{ j } } be any null sequence in the interval (0, 1). By following the lines of proof as above, and passing to a subsequence if necessary, we know that there is such that as j → ∞. Moreover, in view of (3.5) and (3.7), we have . Consequently, since the function g(·) is a lower semicontinuous function and GVI_{ K } (A, g, T) is a closed convex set, we see that (3.7) gives . This has shown that g(u*) is the strong limit of the net (g(u_{ α } )) as α ↓ 0.

(c)
Let 0 < α < β < 1 and u_{ α } , u_{ β } are solutions of the problem (3.1). Thus, since F_{ A } and F_{ T } are monotone mappings, by (3.4), we have
that is,
Notice that,
since 0 < α < β. Using the above, by (3.8), we have
where θ = sup{g(u_{ α } ): α ∈ (0, 1)}. Moreover, since F_{ T } is a Lipschit continuous mapping (with Lipschitz constant ), it follows that
for some M_{1} > 0. Further, by applying the Lagranges meanvalue theorem to a continuous function h(t) = t^{μ}on [1, +∞), we know that
for some M > 0. This completes the proof. □
Remark 3.2. If g =: I, the identity operator on H, then we see that Theorem 3.1 reduces to a result presented by Kim and Buong [9].
4. Iterative Method
Now, we consider the regularization inertial proximal point algorithm:
The well definedness of (4.1) is guaranteed by the following result.
Proposition 4.1. Assume that all hypothesis of the Theorem 3.1 are satisfied. Let z ∈ g^{1}(K) be a fixed element. Define a bifunction F_{ z } : g^{1}(K) × g^{1}(K) → ℝ by
where c, α are positive real numbers. Then, there exists the unique element u* ∈ g^{1}(K) such that F_{ z } (u*, v) ≥ 0 for all v ∈ g^{1}(K).
Proof. Assume that g is an ξ expanding mapping. Then, for each u, v ∈ g^{1}(K), we see that
This means F is ξ(1 + cα)strongly monotone. Consequently, by Lemma 2.8, the proof is completed. □
The result of the next theorem shows some sufficient conditions for the convergent of regularization inertial proximal point algorithm (4.1).
Theorem 4.2. Assume that all the hypotheses of the Theorem 3.1 are satisfied. If the parameters c_{ n } and α_{ n } are chosen as positive real numbers such that
(C1) ,
(C2) ,
(C3) ,
then the sequence {g(z_{ n } )} defined by (4.1) converges strongly to the element g(u*) as n → +∞, where u* ∈ GVI_{ K } (A, g, T).
Proof. From (4.1) we have
that is
or equivalently,
so
Hence
where
On the other hand, by Theorem 3.1, there is u_{ n } ∈ g^{1}(K) such that
for all n ∈ ℕ. This implies
and so
Thus,
By setting v = u_{ n } in (4.2) we have
and v = z_{n+1}in (4.4) we have
and adding one obtained result to the other, we get
Notice that, since A is a gmonotone mapping, and T is a λinverse strongly monotone, we have
and
Thus, by (4.5), we obtain
that is,
Consequently,
which implies that
Using the above Equation 4.6 and (3.2), we know that
where
Consequently, by the condition (C3), we have . Meanwhile, the conditions (C2) and (C3) imply that . Thus, all the conditions of Lemma 2.10 are satisfied, then it follows that g(z_{n+1})  g(u_{n+1}) → 0 as n → ∞. Moreover, by (C1) and Theorem 3.1, we know that there exists u* ∈ GVI_{ K } (A, g, T) such that g(u_{ n } ) converges strongly to g(u*). Consequently, we obtain that g(z_{ n } ) converges strongly to g(u*) as n → +∞. This completes the proof. □
Remark 4.3. The sequences {α_{ n } } and {c_{ n } } which are defined by
satisfy all the conditions in Theorem 4.2.
Remark 4.4. It is worth noting that, because of condition (C2) of Theorem 4.2, the important natural choice {1/n} does not include in the class of parameters {α_{ n } }. This leads to a question: Can we find another regularization inertial proximal point algorithm for the problem (1.2) that includes a natural parameter choice {1/n}?
Remark 4.5. If F is a nonexpansive mapping, then I  F is an inverse strongly monotone mapping, and the fixed points set of mapping F and the solution set S(I  F) are equal. This means that our results contain the study of finding a common element of (general) variational inequalities problems and fixed points set of nonexpansive mapping, which were studied in [4–8] as special cases.
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Acknowledgements
YJC was supported by the Korea Research Foundation Grant funded by the Korean Government (KRF2008313C00050). NP was supported by Faculty of Science, Naresuan University (Project No. R2553C222), and the Commission on Higher Education and the Thailand Research Fund (Project No. MRG5380247).
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Cho, Y.J., Petrot, N. Regularization and iterative method for general variational inequality problem in hilbert spaces. J Inequal Appl 2011, 21 (2011). https://doi.org/10.1186/1029242X201121
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DOI: https://doi.org/10.1186/1029242X201121