JAMJournal of Applied Mathematics1687-00421110-757XHindawi Publishing Corporation64645210.1155/2011/646452646452Research ArticleOn Penalty and Gap Function Methods for Bilevel Equilibrium ProblemsVan DinhBui1MuuLe Dung2FangYa Ping1Faculty of Information TechnologyLe Quy Don UniversityHanoiVietnam2Control and Optimization DepartmentInstitute of Mathematics, VAST18 Hoang Quoc Viet, Cau GiayHanoi 10307Vietnam20113112011201109062011150820112011Copyright © 2011 Bui Van Dinh and Le Dung Muu.This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We consider bilevel pseudomonotone equilibrium problems. We use a penalty function to convert a bilevel problem into one-level ones. We generalize a pseudo--monotonicity concept from -monotonicity and prove that under pseudo--monotonicity property any stationary point of a regularized gap function is a solution of the penalized equilibrium problem. As an application, we discuss a special case that arises from the Tikhonov regularization method for pseudomonotone equilibrium problems.

1. Introduction

Let C be a nonempty closed-convex subset in n, and let f,g:C×C be two bifunctions satisfying f(x,x)=g(x,x)=0 for every xC. Such a bifunction is called an equilibrium bifunction. We consider the following bilevel equilibrium problem (BEP for short):find  x¯Sg  such  that  f(x¯,y)0,ySg, where Sg={uC:g(u,y)0,  yC}, that is, Sg is the solution set of the equilibrium problemsfind  uC  such  that  g(u,y)0,yC. As usual, we call problem (1.1) the upper problem and (1.2) the lower one. BEPs are special cases of mathematical programs with equilibrium constraints. Sources for such problems can be found in . Bilevel monotone variational inequality, which is a special case of problem (1.1), was considered in [4, 5]. Moudafi in  suggested the use of the proximal point method for monotone BEPs. Recently, Ding in  used the auxiliary problem principle to BEPs. In both papers, the bifunctions f and g are required to be monotone on C. It should be noticed that under the pseudomonotonicity assumption on g the solution-set Sg of the lower problem (1.2) is a closed-convex set whenever g(x,·) is lower semicontinuous and convex on C for each x. However, the main difficulty is that, even the constrained set Sg is convex, it is not given explicitly as in a standard mathematical programming problem, and therefore the available methods (see, e.g.,  and the references therein) cannot be applied directly.

In this paper, first, we propose a penalty function method for problem (1.1). Next, we use a regularized gap function for solving the penalized problems. Under certain pseudo--monotonicity properties of the regularized bifunction, we show that any stationary point of the gap function on the convex set C is a solution to the penalized subproblem. Finally, we apply the proposed method to the Tikhonov regularization method for pseudomonotone equilibrium problems.

2. A Penalty Function Method

Penalty function method is a fundamental tool widely used in optimization to convert a constrained problem into unconstrained (or easier constrained) ones. This method was used to monotone variational inequalities in  and equilibrium problems in . In this section, we use the penalty function method in the bilevel problem (1.1). First, let us recall some well-known concepts on monotonicity and continuity (see, e.g., ) that will be used in the sequel.

Definition 2.1.

The bifunction ϕ:C×C is said to be as follows:

strongly monotone on C with modulus β>0 if     ϕ(x,y)+ϕ(y,x)-βx-y2,x,yC,

monotone on C if ϕ(x,y)+ϕ(y,x)0,x,yC,

pseudomonotone on C if x,yC:    ϕ(x,y)0ϕ(y,x)0,

upper semicontinuous at x with respect to the first argument on C if limzx̅  ϕ(z,y)ϕ(x,y),yC,

lower semicontinuous at y with respect to the second argument on C if limwy̅ϕ(x,w)ϕ(x,y),xC.

Clearly, (a)(b)(c).

Definition 2.2 (see [<xref ref-type="bibr" rid="B6">17</xref>]).

The bifunction ϕ:C×C is said to be coercive on C if there exists a compact subset Bn and a vector y0BC such that ϕ(x,y0)<0,xCB.

Theorem 2.3 (see [<xref ref-type="bibr" rid="B12">18</xref>, Proposition 2.1.14]).

Let ϕ:C×C be a equilibrium bifunction such that ϕ(·,y) is upper semicontinuous on C for each yC and ϕ(x,·) is lower semicontnous, convex on C for each xC. Suppose that C is compact or ϕ is coercive on C, then there exists at least one x*C such that ϕ(x*,y)0 for every yC.

The following proposition tells us about a relationship between the coercivity and the strong monotonicity.

Proposition 2.4.

Suppose that the equilibrium bifunction ϕ is strongly monotone on C, and ϕ(x,·) is convex, lower semicontinuous with respect to the second argument for all xC, then for each yC, there exists a compact set B such that yB and ϕ(x,y)<0  for  all  xCB.

Proof.

Suppose by contradiction that the conclusion does not hold, then there exists an element y0C such that for every compact set B there is an element xBCB such that ϕ(xB,y0)0. Take B:=Br as the closed ball centered at y0 with radius r>1. Then there exists xrCBr such that ϕ(xr,y0)0. Let x be the intersection of the line segment [y0,xr] with the unit sphere S(y0;1) centered at y0 and radius 1. Hence, xr=y0+t(r)(x-y0), where t(r)>r. By the strong monotonicity of ϕ, we have ϕ(y0,xr)-ϕ(xr,y0)-βxr-y02-ϕ(xr,y0)-βt(r)2x-y02. Since ϕ(y0,·) is convex on C, it follows that ϕ(y0,x)1t(r)ϕ(y0,xr)+t(r)-1t(r)ϕ(y0,y0), which implies that ϕ(y0,x)-βt(r)x-y02-βr. Thus, ϕ(y0,x)-as  r. However, since ϕ(y0,·) is lower semicontinuous on C, by the well-known Weierstrass Theorem, ϕ(y0,·) attains its minimum on the compact set S(y0;1)C. This fact contradicts (2.9).

From this proposition, we can derive the following corollaries.

Corollary 2.5 (see [<xref ref-type="bibr" rid="B12">18</xref>]).

If the bifunction ϕ is strongly monotone on C, and ϕ(x,·) is convex, lower semicontinuous with respect to the second argument for all xC, then ϕ is coercive on C.

Corollary 2.6.

Suppose that the bifunction f is strongly monotone on C, and f(x,·) is convex, lower semicontinuous with respect to the second argument for all xC. If the bifunction g is coercive on C then, for every ϵ>0, the bifunction g+ϵf is uniformly coercive on C, for example, there exists a point y0C and a compact set B both independent of ϵ such that g(x,y0)+ϵf(x,y0)<0,xCB.

Proof.

From the coercivity of g, we conclude that there exists a compact B1 and y0C such that g(x,y0)<0  for  all  xCB1. Since f is strongly monotone, convex, lower semicontinuous on C, by choosing y=y0, from Proposition 2.4, there exists a compact B2 such that f(x,y0)<0  for  all  xCB2. Set B=B1B2, then B is compact and g(x,y0)+ϵf(x,y0)<0  for  all  xCB.

Remark 2.7.

It is worth to note that if both f, g are coercive and pseudomonotone on C, then the function f+g is not necessary coercive or pseudomonotone on C.

To see this, let us consider the following bifunctions.

Example 2.8.

Let f(x,y):=(x1y2-x2y1)ex1, g(x,y):=(x2y1-x1y2)ex2, and C={(x1,x2):x1-1,(1/10)(x1-9)x210x1+9} then we have

f(x,y),g(x,y) are pseudomonotone and coercive on C,

for all  ϵ>0 the bifunctions fϵ(x,y)=g(x,y)+ϵf(x,y) are neither pseudomonotone nor coercive on C.

Indeed,

if f(x,y)0, then f(y,x)0, thus f is pseudomonotone on C. By choosing y0=(y10,0),(0<y101) and B={(x1,x2):x12+x22r}  (r>1), we have f(x,y0)=-x2y10ex1<0  for  all  yCB, which means that f is coercive on C. Similarly, we can see that g is coercive on C,

by definition of f, we have that

fϵ(x,y)=(x2y1-x1y2)(ex2-ϵex1),ϵ>0.Take x(t)=(t,2t),  for  all  y(t)=(2t,t), then fϵ(x(t),y(t))=3t2(e2t-ϵet)>0, whereas fϵ(y(t),x(t))=-3t2(et-ϵe2t)>0 for t is sufficiently large. So fϵ is not pseudomonotone on C.

Now, we show that the bifunction fϵ(x,y)=(x2y1-x1y2)(ex2-ϵex1) is not coercive on C. Suppose, by contradiction, that there exist a compact set B and y0=(y10,y20)BC such that fϵ(x,y0)<0  for  all  xCB, then, by coercivity of fϵ, it follows, y10,y20>0 and y10y02. With x(t)=(t,kt),(t>0), we have fϵ(x(t),y0)=t(ky10-y20)(ekt-ϵet). However

if y10>y20, then from 1<k<10 it follows that x(t)C and fϵ(x(t),y0)>0 for t is sufficiently large, which contradicts with coercivity,

if y10<y20, then, by choosing 1/10<k<1, we obtain x(t)C and fϵ(x(t),y0)>0 for t is large enough. But this cannot happen because of the coercivity of fϵ.

Now, for each fixed ϵ>0, we consider the penalized equilibrium problem PEP(C,fϵ) defined as find  x¯ϵC  such  that  fϵ(x¯ϵ,y):=g(x¯ϵ,y)+ϵf(x¯ϵ,y)0,yC. By SOL(C,fϵ), we denote the solution set of PEP(C,fϵ).

Theorem 2.9.

Suppose that the equilibrium bifunctions f,g are pseudomonotone, upper semicontinuous with respect to the first argument and lower semicontinuous, convex with respect to the second argument on C, then any cluster point of the sequence {xk} with xk SOL(C,fϵk),  ϵk0 is a solution to the original bilevel problem (1.1). In addition, if f is strongly monotone and g is coercive on C, then for each ϵk>0 the penalized problem PEP(C,fϵk) is solvable, and any sequence {xk} with xkSOL(C,fϵk) converges to the unique solution of the bilevel problem (1.1) as k.

Proof.

Let {xk} be any sequence with xkSOL(C,fϵk), and let x¯ be any of its cluster points. Without lost of generality, we may assume that xkx¯  as  k. Since xkSOL(C,fϵk), one has g(xk,y)+ϵkf(xk,y)0,yC. For any zSg, we have g(z,y)0,  for  all  yC and in particular, g(z,xk)0. Then, by the pseudomonotonicity of g, we have g(xk,z)0. Replacing y by z in (2.13), we obtain g(xk,z)+ϵkf(xk,z)0, which implies that ϵkf(xk,z)-g(xk,z)0f(xk,z)0. Let k, by upper semicontinuity of f, we have f(x¯,z)0  for  all  zSg.

To complete the proof, we need only to show that x¯Sg. Indeed, for any yC, we have g(xk,y)+ϵkf(xk,y)0,yC. Again, by upper semicontinuity of f and g, we obtain in the limit, as ϵk0, that g(x¯,y)0  for  all  yC. Hence, x¯Sg.

Now suppose, in addition, that f is strongly monotone on C. By Corollary 2.6, fϵk is uniformly coercive on C. Thus, problem PEP(C,  fϵk) is solvable and, for all ϵk>0, the solution sets of these problems are contained in a compact set B. So any infinite sequence {xk} of the solutions has a cluster point, say, x¯. By the first part, x¯ is a solution of (1.1). Note that, from the assumption on g, the solution set Sg of the lower equilibrium (EP(C,g)) is a closed, convex, compact set. Since f is lower semicontinuous and convex with respect to the second argument and is strongly monotone on C, the upper equilibrium problem EP(Sg,f) has a unique solution. Using again the first part of the theorem, we can see that xkx¯  as  k

Remark 2.10.

In a special case considered in , where both f and g are monotone, the penalized problem (PEP) is monotone too. In this case, (PEP) can be solved by some existing methods (see, e.g., [6, 1114, 19]) and the references therein. However, when one of these two bifunctions is pseudomonotone, the penalized problem (PEP), in general, does not inherit any monotonicity property from f and g. In this case, problem (PEP) cannot be solved by the above-mentioned existing methods.

3. Gap Function and Descent Direction

A well-known tool for solving equilibrium problem is the gap function. The regularized gap function has been introduced by Taji and Fukushima in  for variational inequalities, and extended by Mastroeni in  to equilibrium problems. In this section, we use the regularized gap function for the penalized equilibrium problem (PEP). As we have mentioned above, this problem, even when g is pseudomonotone and f is strongly monotone, is still difficult to solve.

Throughout this section, we suppose that both f and g are lower semicontinuous, convex on C with respect to the second argument. First, we recall (see, e.g., ) the definition of a gap function for the equilibrium problem.

Definition 3.1.

A function φ:C{+} is said to be a gap function for (PEP) if

φ(x)0,  for  all  xC,

φ(x¯)=0 if and only if x¯ is a solution for (PEP).

A gap function for (PEP) is φ(x)=-minyCfϵ(x,y). This gap function may not be finite and, in general, is not differentiable. To obtain a finite, differentiable gap function, we use the regularized gap function introduced in  and recently used by Mastroeni in  to equilibrium problems. From Proposition 2.2 and Theorem 2.1 in , the following proposition is immediate.

Proposition 3.2.

Suppose that l:C×C is a nonnegative differentiable, strongly convex bifunction on C with respect to the second argument and satisfies

l(x,x)=0  for  all  xC,

yl(x,x)=0  for  all  xC.

Then the function φϵ(x)=-minyC[g(x,y)+ϵ[f(x,y)+l(x,y)]] is a finite gap function for (PEP). In addition, if f and g are differentiable with respect to the first argument and xf(x,y),xg(x,y) are continuous on C, then φϵ(x) is continuously differentiable on C and φϵ(x)=-xg(x,yϵ(x))-ϵx[f(x,yϵ(x))+l(x,yϵ(x))]=-xgϵ(x,yϵ(x)), where   gϵ(x,y)=g(x,y)+ϵ[f(x,y)+l(x,y)],yϵ(x)=argminyC{gϵ(x,y)}.

Note that the function l(x,y):=(1/2)M(y-x),y-x, where M is a symmetric positive definite matrix of order n that satisfies the assumptions on l.

We need some definitions on -monotonicity.

Definition 3.3.

A differentiable bifunction h:C×C is called as follows:

strongly -monotone on C if there exists a constant τ>0 such that, xh(x,y)+yh(x,y),y-xτy-x2,x,yC,

strictly -monotone on C if xh(x,y)+yh(x,y),y-x>0,x,yC,xy,

-monotone on C if xh(x,y)+yh(x,y),y-x0,  x,yC,

strictly pseudo--monotone on C if xh(x,y),y-x0yh(x,y),y-x>0,x,yC,xy,

pseudo--monotone on C if xh(x,y),y-x0yh(x,y),y-x0,x,yC.

Remark 3.4.

The definitions (a), (b), and (c) can be found, for example, in [8, 11]. The definitions (d) and (e), to our best knowledge, are not used before. From the definitions, we have (a)(b)(c)(e),  (a)(b)(d)(e). However, (c) may not imply (d) and vice versa as shown by the following simple examples.

Example 3.5.

Consider the bifunction h(x,y)=ex2(y2-x2) defined on C×C with C=. This bifunction is not -monotone on C, because xh(x,y)+yh(x,y),y-x=2ex2(y-x)2(x2+xy+1)   is negative for x=-1,    y=3. However, h(x,y) is strictly pseudo- -monotone. Indeed, we have xh(x,y),y-x=2xex2(y2-x2-1)(y-x)0x(y2-x2-1)(y-x)0,yh(x,y),y-x=2yex2(y-x)>0y(y-x)>0. It is not difficult to verify that x(y2-x2-1)(y-x)0y(y-x)>0,  as  xy.

Hence this function is strictly pseudo- -monotone but is not -monotone.

Vice versa, considering the bifunction h(x,y)=(y-x)TM(y-x) defined on n×n, where M is a matrix of order n×n, we have the following:

h is -monotone, because xh(x,y)+yh(x,y),y-x=-(y-x)T(M+MT)+(y-x)T(M+MT),y-x=0,x,y. Clearly, h is not strictly--monotone,

h is strictly pseudo -monotone if and only if xh(x,y),y-x=-(y-x)T(M+MT),y-x0 implies

yh(x,y),y-x=(y-x)T(M+MT),y-x>0,x,y,xy. The latter inequality equivalent to M+MT is a positive definite matrix of order n×n.

Remark 3.6.

As shown in  when h(x,y)=T(x),y-x with T a differentiable monotone operator on C, h is monotone on C if and only if T is monotone on C, and in this case, monotonicity of h on C coincides with -monotonicity of h on C.

The following example shows that pseudomonotonicity may not imply pseudo--monotonicity.

Example 3.7.

Let h(x,y)=-ax(y-x), defined on +×+, (a>0). It is easy to see that h(x,y)0h(y,x)0,x,y0. Thus, h is pseudomonotone on +.

We have xh(x,y),y-x=-a(y-x)(y-2x)<0,y>2x>0.

But yh(x,y),y-x=-ax(y-x)<0,y>2x>0.

So h is not pseudo--monotone on +.

From the definition of the gap function φϵ, a global minimal point of this function over C is a solution to problem (PEP). Since φϵ is not convex, its global minimum is extremely difficult to compute. In , the authors have shown that under the strict -monotonicity a stationary point is also a global minimum of gap function. By a counterexample, the authors in  also pointed out that the strict -monotonicity assumption cannot be relaxed to -monotonicity. The following theorem shows that the stationary property is still guaranteed under the strict pseudo--monotonicity.

Theorem 3.8.

Suppose that gϵ is strictly pseudo- -monotone on C. If x¯ is a stationary point of φϵ over C, that is, φϵ(x¯),y-x¯0,yC. then x¯ solves (PEP).

Proof.

Suppose that x¯ does not solve (PEP), then yϵ(x¯)x¯.

Since x¯ is a stationary point of φϵ on C, from the definition of φϵ, we have φϵ(x¯),y-x¯=-xgϵ(x,yϵ(x)),yϵ(x)-x0. By strict pseudo--monotonicity of gϵ, it follows that ygϵ(x¯,yϵ(x¯)),yϵ(x¯)-x¯>0. On the other hand, since yϵ(x¯) minimizes gϵ(x,·) over C, we have ygϵ(x¯,yϵ(x¯)),yϵ(x¯)-x¯0, which is in contradiction with (3.21).

To compute a stationary point of a differentiable function over a closed-convex set, we can use the existing descent direction algorithms in mathematical programming (see, e.g., [8, 21]). The next proposition shows that if y(x) is a solution of the problem minyCgϵ(x,y), then y(x)-x is a descent direction on C of φϵ at x. Namely, we have the following proposition.

Proposition 3.9.

Suppose that gϵ is strictly pseudo--monotone on C and x is not a solution to Problem (PEP), then φϵ(x),yϵ(x)-x<0.

Proof.

Let dϵ(x)=yϵ(x)-x. Since x is not a solution to (PEP), then dϵ(x)0. Suppose that, by contradiction, dϵ(x) is not a descent direction on C of φϵ at x, then φϵ(x),yϵ(x)-x0-xgϵ(x,yϵ(x)),yϵ(x)-x0, which, by strict pseudo--monotonicity of gϵ, implies ygϵ(x,yϵ(x)),yϵ(x)-x>0. On the other hand, since yϵ(x) minimizes gϵ(x,·) over C, by the well-known optimality condition, we have ygϵ(x,yϵ(x)),yϵ(x)-x0, which contradicts (3.25).

Proposition 3.10.

Suppose that g(x,·) is strictly convex on C for every xC and g is strictly pseudo--monotone on C. If xC is not a solution of (PEP), then there exists ϵ¯>0 such that yϵ(x)-x is a descent direction of φϵ on C at x for all 0<ϵϵ¯.

Proof.

By contradiction, suppose that the statement of the proposition does not hold, then there exist ϵk0 and xC such that φϵk(x),yϵk(x)-x0-xgϵk(x,yϵk(x)),yϵk(x)-x0. Since gϵ(x,·) is strictly convex differentiable on C, by Theorem 2.1 in , the function ϵyϵ(x) is continuous with respect to ϵ, thus yϵk(x) tends to y0(x) as ϵk0, where y0(x)=argminyCg(x,y). Since gϵk(x,y)=g(x,y)+ϵkf(x,y) is continuously differentiable, letting ϵk0 in (3.27), we obtain -xg(x,y0(x)),y0(x)-x0. By strict pseudo--monotonicity of g, it follows that yg(x,y0(x)),y0(x)-x>0. On the other hand, since yϵk(x) minimizes gϵk(x,·) over C, we have ygϵk(x,yϵk(x)),  yϵk(x)-x0. Taking the limit, we obtain yg(x,y0(x)),y0(x)-x0, which contradicts (3.29).

To illustrate Theorem 3.8, let us consider the following examples.

Example 3.11.

Consider the bifunctions g(x,y)=ex2(y2-x2) and f(x,y)=10x2(y2-x2) defined on ×. It is not hard to verify that,

g(x,y),f(x,y) are monotone, strictly pseudo--monotone on ,

for all ϵ>0 the bifunction g(x,y)+ϵf(x,y) is monotone and strictly pseudo--monotone on and satisfying all of the assumptions of Theorem 3.8.

Example 3.12.

Let f(x,y)=-x2-xy+2y2 and g(x,y)=-3x2y+xy2+2y3 defined on +×+ it is easy to see that,

g,f are pseudomonotone, strictly -monotone on +,

for all ϵ>0 the bifunction g(x,y)+ϵf(x,y) is pseudomonotone and strictly -monotosne on + and satisfying all of the assumptions of Theorem 3.8.

4. Application to the Tikhonov Regularization Method

The Tikhonov method  is commonly used for handling ill-posed problems. Recently, in  the Tikhonov method has been extended to the pseudomonotone equilibrium problem Find  x*C  such  that  g(x*,y)0,yC, where, as before, C is a closed-convex set in n and g:C is a pseudomonotone bifunction satisfying g(x,x)=0 for every xC.

In the Tikhonov regularization method considered in , problem (EP(C,g)) is regularized by the problems find  x*C  such  that  gϵ(x*,y):=g(x*,y)+ϵf(x*,y)0,  yC, where f is an equilibrium bifunction on C and ϵ>0 and play the role of the regularization bifunction and regularization parameter, respectively.

In , the following theorem has been proved.

Theorem 4.1.

Suppose that f(·,y),  g(·,y) are upper semicontinuous and f(x,·),  g(x,·) are lower semicontinuous convex on C for each x,yC and that g is pseudomonotone on C. Suppose further that f is strongly monotone on C satisfying the condition δ>0:   |f(x,y)|δx-xgy-x,x,yC, where xgC (plays the role of a guess solution) is given.

Then the following three statements are equivalent:

the solution set of (EP(C,gϵ)) is nonempty for each ɛ>0 and limɛ0+x(ɛ) exists, where x(ɛ) is arbitrarily chosen in the solution set of (EP(C,gϵ)),

the solution set of (EP(C,gϵ)) is nonempty for each ɛ>0 and limɛ0+supx(ɛ)<, where x(ɛ) is arbitrarily chosen in the solution set of (EP(C,gϵ)),

the solution set of (EP(C,g)) is nonempty.

Moreover, if any one of these statements holds, then limɛ0+x(ɛ) is equal to the unique solution of the strongly monotone equilibrium problem EP(Sg,f), where Sg denotes the solution set of the original problem (EP(C,g)).

Note that, when g is monotone on C, the regularized subproblems are strongly monotone and therefore, they can be solved by some existing methods. When g is pseudomonotone, the subproblems, in general, are no longer strongly monotone, even not pseudomonotone. So solving them becomes a difficult task. However, the problem of finding the limit point of the sequences of iterates leads to the unique solution of problem EP(Sg,f).

In order to apply the penalty and gap function methods described in the preceding sections, let us take, for instant, f(x,y)=x-xg,y-x. Clearly, f is both strongly monotone and strongly -monotone with the same modulus 1. Moreover, f satisfies the condition (4.1). Therefore, the problem of finding the limit point in the above Tikhonov regularization method can be formulated as the bilevel equilibrium problemfind  xSg  such  that  f(x*,y)0,ySg, which is of the form (1.1). Now, for each fixed ϵk>0, we consider the penalized equilibrium problem PEP(C,fϵk) defined asfind  x¯kC  such  that  fϵk(x¯k,y):=g(x¯k,y)+ϵkf(x¯k,y)0,yC. As before, by SOL(C,fϵk), we denote the solution set of PEP(C,fϵk).

Applying Theorems 2.9 and 3.8, we obtain the following result.

Theorem 4.2.

Suppose that the bifunction g satisfies the following conditions:

g(x,·) is convex, lower semicontinuous for all xC,

g is pseudomonotone and coercive on C. Then for any ϵk>0, the penalized problem PEP(C,fϵk) is solvable, and any sequence {xk} with xkSOL(C,fϵk) for all k converges to the unique solution of the problem (4.3) as k.

In addition, if  g(x,y)+ϵkf(x,y) is strictly pseudo- -monotone on  C (in particular,  g(x,y) is  -monotone), and  x¯k is any stationary point of the mathematical program  minxCφk(x) with φk(x):=minyC{g(x,y)+ϵkf(x,y)}, then  {x¯k} converges to the unique solution of the problem (4.3) as  k.

5. Conclusion

We have considered a class of bilevel pseudomonotone equilibrium problems. The main difficulty of this problem is that its feasible domain is not given explicitly as in a standard mathematical programming problem. We have proposed a penalty function method to convert the bilevel problem into one-level ones. Then we have applied the regularized gap function method to solve the penalized equilibrium subproblems. We have generalized the pseudo--monotonicity concept from -monotonicity. Under the pseudo--monotonicity property, we have proved that any stationary point of the gap function is a solution to the original bilevel problem. As an application, we have shown how to apply the proposed method to the Tikhonov regularization method for pseudomonotone equilibrium problems.

Acknowledgment

This work is supported by the National Foundation for Science Technology Development of Vietnam (NAFOSTED).

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