Iterative Algorithms for Variational Inequalities Governed by Boundedly Lipschitzian and Strongly Monotone Operators

Consider the variational inequality VI(C, F) of finding a point x∗ ∈ C satisfying the property ⟨Fx∗, x −x∗⟩ ≥ 0 for all x ∈ C, where C is a level set of a convex function defined on a real Hilbert spaceH and F : H → H is a boundedly Lipschitzian (i.e., Lipschitzian on bounded subsets ofH) and strongly monotone operator. He and Xu proved that this variational inequality has a unique solution and devised iterative algorithms to approximate this solution (seeHe and Xu, 2009). In this paper, relaxed and self-adaptive iterative algorithms are proposed for computing this unique solution. Since our algorithms avoid calculating the projection P C (calculating


Introduction
Let  be a real Hilbert space with inner product ⟨⋅, ⋅⟩ and norm ‖ ⋅ ‖, let  be a nonempty closed convex subset of , and let  :  →  be a nonlinear operator.We consider the problem of finding a point  * ∈  with the property ⟨ * ,  −  * ⟩ ≥ 0, ∀ ∈ . ( This is known as the variational inequality problem VI(, ), initially introduced and studied by Stampacchia [1] in 1964.
In recent years, variational inequality problems have been extended to study a large variety of problems arising in structural analysis, economics, optimization, operations research, and engineering sciences; see  and the references therein.Using the projection technique, one can easily show that VI(, ) is equivalent to a fixed-point problem (see, e.g., [15]).
Lemma 1.  * ∈  is a solution of (, ) if and only if  * ∈  satisfies the fixed-point relation: where  > 0 is an arbitrary constant,   is the orthogonal projection onto , and  is the identity operator on .
Moreover, a monotone operator  is called strictly monotone if the equality "=" holds only when  =  in the last relation.
In this case,  is also called a -Lipschitzian and -strongly monotone operator.It is not difficult to show the following result.
Then (, ) has a unique solution.Moreover, for any 0 <  < 2/ 2 , the sequence {  } with initial guess  0 ∈  and defined recursively by converges strongly to the unique solution of (, ).
Attempts are worth making to weaken the Lipschitz condition (4) or the strong monotonicity condition (5) so that existence of solutions of variational inequality (1) is still guaranteed.In 2009, He and Xu [14] He and Xu [14] not only proved existence and uniqueness of solutions of VI(, ) under conditions ( 5) and (7) but also estimated the range of this unique solution.
Theorem 4 (see [14]).Assume that  :  →  is boundedly Lipschitzian on  (i.e., for each bounded subset  of ,  is Lipschitzian on ).Assume also that  is -strongly monotone on .Then variational inequality (1) has a unique solution where  ∈  is an arbitrary fixed element.
Similarly, we can also introduce bounded strong monotonicity of an operator.An operator  :  →  is called boundedly strong monotone on , if, for arbitrary bounded subset  of , there exists a positive constant   depending only on the set  such that So a natural question gives rise to this: is it possible also to replace the strong monotonicity of  by bounded strong monotonicity so that the result of Theorem 4 is still guaranteed?Unfortunately, a simple example [14] gives us a negative answer.
He and Xu [14] also consider the iterative algorithms for solving VI(, ), where  :  →  is boundedly Lipschitzian and -strongly monotone on .Denote by  an arbitrary fixed element in  and denote by  a positive fixed constant such that Set   = (, ) ∩  ((, ) is a closed ball of , i.e., (, ) = { ∈  : ‖−‖ ≤ }) and denote by   the Lipschitz constant of  on the bounded closed convex subset   .
Using Theorem 4, it is easy to see that VI(, ) and VI(  , ) have the same solution.Thus one can devise iterative methods for VI(  , ) and get the unique solution of VI(, ).
However, algorithms ( 6) and (11) all have two evident weaknesses.On the one hand, they involve calculating the projections   and    , respectively, while the computation of a projection onto a closed convex subset is generally difficult.Particularly, the computation of    is maybe more difficult since the structure of   is more complicated.On the other hand, the determination of the stepsize  depends on the constants  (or   ) and .This means that, in order to implement algorithm (6) (or algorithm (11)), one has first to compute (or estimate) the constants  (or   ) and , which is sometimes not an easy work in practice.
He and Yang [22] proposed relaxed and self-adaptive algorithms in order to overcome the above weaknesses of algorithm (6) and proved strong convergence theorems.
In order to overcome the above weaknesses of algorithm (11), new relaxed and self-adaptive algorithms are proposed in this paper to solve VI(, ), where  is a level set of a convex function defined on  and  :  →  is a boundedly Lipschitzian and -strongly monotone operator.Our methods calculate    by computing  (,) (the computation of  (,) is very easy) and a sequence of projections onto half-spaces containing the original level set  and select the stepsizes through a self-adaptive way.The implementations of our algorithms are very easy since they avoid computing    directly and have no need to know any information about   (but  is assumed to be known, so our methods partly overcome the second weakness above).The algorithms in this paper improve and extend the above corresponding result of He and Xu.
The rest of this paper is organized as follows.Some useful lemmas are listed in the next section.In the last section, a relaxed algorithm for the case where  and   are all known and a relaxed self-adaptive algorithm for the case where  is known but   is unknown are proposed, respectively.The strong convergence theorems of our algorithms are proved.

Preliminaries
Throughout the rest of this paper, we denote by  a real Hilbert space and by  the identity operator on .If  :  → R is a differentiable functional, then we denote by ∇ the gradient of .We will also use the following notations.
Recall a trivial inequality, which is well known and in common use.Lemma 6.For all ,  ∈ , there holds the relation Recall that a mapping  :  →  is said to be nonexpansive if :  →  is said to be firmly nonexpansive if, for ,  ∈ , The following are characterizations of firmly nonexpansive mappings (see [7] or [23]).
Lemma 7. Let  :  →  be an operator.The following statements are equivalent.
We know that the orthogonal projection   from  onto a nonempty closed convex subset  ⊂  is a typical example of a firmly nonexpansive mapping, which is defined by It is well known that    is characterized by the inequality (for  ∈ ) The following recent result [22] is likely to become a new fundamental tool for proving strong convergence of some algorithms.Its key effect on the proofs of our main results will be illustrated in the next section.
Lemma 8 (see [22]).Assume   is a sequence of nonnegative real numbers such that where (  ) is a sequence in (0, 1), (  ) is a sequence of nonnegative real numbers, and (  ) and (  ) are two sequences in R such that Then lim  → ∞   = 0.
Recall that a function  :  → R is called convex if A differentiable function  is convex if and only if there holds the relation Recall that an element  ∈  is said to be a subgradient of A function  :  → R is said to be subdifferentiable at , if it has at least one subgradient at .The set of subgradients of  at the point  is called the subdifferential of  at  and is denoted by ().The last relation above is called the subdifferential inequality of  at .A function  is called subdifferentiable, if it is subdifferentiable at all  ∈ .If a function  is differentiable and convex, then its gradient and subgradient coincide.
Recall that a function  :  → R is said to be weakly lower semicontinuous

Iterative Algorithms
In this section, we consider the iterative algorithms for solving a particular kind of variational inequality (1) in which the closed convex subset  is of the particular structure, that is, the level set of a convex function given as follows: where  :  → R is a convex function.We always assume that  is subdifferentiable on  and  is bounded operator (i.e., bounded on bounded sets).We also assume that  :  →  is a boundedly Lipschitzian and -strongly monotone operator.Using Theorem 4, we assert that in this case VI(, ) has a unique solution, henceforth, which is denoted by  * .The computation of a projection onto a closed convex subset is generally difficult.To overcome this difficulty, Fukushima [21] suggested a way to calculate the projection onto a level set of a convex function by computing a sequence of projections onto half-spaces containing the original level set.This idea is followed by Yang [24] and Lopez et al. and so forth [25], respectively, who introduced the relaxed  algorithms for solving the split feasibility problem in a finitedimensional and infinite-dimensional Hilbert space, respectively.He and Yang [22] also used this idea to devise iterative algorithms for solving variational inequalities governed by Lipschitzian and strongly monotone operators.
In the sequel, we always assume that  is known and denote by  0 ∈  a selected arbitrarily fixed element.Using Theorem 4, the unique solution  * of VI(, ) belongs to a closed ball ( 0 , ), where  is a fixed positive constant such that  > ‖ 0 ‖/.We also always denote by   the Lipschitz constant of  on ( 0 , ).
Based on Theorem 4, we are now in a position to introduce a relaxed algorithm for computing the unique solution  * of VI(, ), where  is given as in (23).This scheme applies to the case where   is easy to be determined.Algorithm 9. Choose an arbitrary initial guess  0 ∈  and the sequence (  ) is constructed via the formula where where   ∈ (  ), the sequence (  ) in (0, 1), and   is a constant such that   ∈ (0, 2/ 2  ).
It is worth mentioning that if   is easy to be calculated, then    in Algorithm 9 can be replaced with   and it is easy to see that the whole proof of Theorem 10 is valid for this case.Therefore, if  = , VI(, ) reduces to the operator equation problem: finding  * ∈  such that  * = 0, and the following result holds.
Sometimes, the constant   is difficult to be obtained or estimated in practice (but we assume that  has been obtained).In this case, Algorithm 9 is indeed not fit for solving VI(  , ) (i.e., VI(, )).Then we now turn to introducing a relaxed and self-adaptive algorithm for the case where constant   is unknown.
Algorithm 12. Choose an arbitrary initial guess  0 ∈  and an arbitrary element  1 ∈ ( 0 , ) such that  1 ̸ =  0 .Assume that the th iterate   ( ≥ 1) has been constructed.Continue and calculate the ( + 1)th iterate  +1 via the formula where   is given as in (25), the sequence (  ) is in (0, 1),  is a constant such that  > ‖ 0 ‖/, and the sequence (  ) is determined via the relation Firstly, we show that the sequence (  ) is well defined.Noting strong monotonicity of ,  1 ̸ =  0 implies that  1 ̸ =  0 and  1 is well defined via the first formula of (40).Consequently,   ( ≥ 2) is well defined inductively according to (40) and thus the sequence (  ) is also well defined using (39). Next then ( 46) and (47) can be rewritten as the following forms, respectively: Clearly,   → 0 and ∑ ∞ =1   = ∞, together with (43)-(45), imply that   → 0 and ∑ ∞ =1     = ∞.By an argument very similar to the proof of Theorem 10, it is not difficult to verify that lim for any subsequence (  ) ⊂ ().Thus we can complete the proof by using Lemma 8.
Similar to Algorithm 9, if   is easy to be calculated, then    in Algorithm 12 can also be replaced with   and it is easy to see that the whole proof of Theorem 13 is valid for this case.The following result similar to Corollary 11 also holds.where   is given as in (40), converges strongly to the unique solution  * of the operator equation  = 0.
Finally, we give an iterative algorithm for solving a class VI(, ), in which the closed convex subset  is the intersection of finite level sets of convex functions given as follows: where  is a positive integer and   :  → R ( = 1, . . ., ) is a convex function.We always assume that   ( = 1, . . ., ) is subdifferentiable on  and   ( = 1, . . ., ) is bounded operator (i.e., bounded on bounded sets).Without loss of generality, we will consider only the case  = 2; that is,  =  1 ∩  (57) By an argument similar to the proof of Theorem 13 (together with the proof of Theorem 3.4 of [22]), we have the following result.
weakened the Lipschitz condition (4) successfully to the bounded Lipschitz condition.A mapping  :  →  is boundedly Lipschitzian on  if it is Lipschitzian on each bounded subset of ; namely, for each nonempty bounded subset  of , there exists a positive constant   depending only on the set  such that −      , ∀, ∈ .