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Tykhonov well-posedness for lexicographic equilibrium problems

Lam Quoc Anh & Tran Quoc Duy

To cite this article: Lam Quoc Anh & Tran Quoc Duy (2016): Tykhonov well-posedness for

lexicographic equilibrium problems, Optimization, DOI: 10.1080/02331934.2016.1209673

To link to this article: http://dx.doi.org/10.1080/02331934.2016.1209673

Published online: 18 Jul 2016

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Tykhonov well-posedness for lexicographic equilibrium problems

Lam Quoc Anhaand Tran Quoc Duyb,c

aDepartment of Mathematics, Teacher College, Cantho University, Cantho, Vietnam;bDepartment of Mathematics,University of Science, Vietnam National University, Hochiminh City, Vietnam;cDepartment of Mathematics, CanthoTechnical Economic College, Cantho, Vietnam

ABSTRACT

In this paper, we consider the vector equilibrium problems involving

lexicographic cone in Banach spaces We introduce the new concepts of the

Tykhonov well-posedness for such problems The corresponding concepts

of the Tykhonov well-posedness in the generalized sense are also proposed

and studied Some metric characterizations of well-posedness for such

problems are given As an application of the main results, several results

on well-posedness for the class of lexicographic variational inequalities are

derived

ARTICLE HISTORY

Received 15 November 2015 Accepted 27 June 2016

KEYWORDS

Lexicographic order; equilibrium problems; variational inequalities; Tykhonov well-posedness

AMS SUBJECT CLASSIFICATIONS

49K40; 90C31; 91B50

1 Introduction

Well-posedness plays an important role in both theory and numerical methods for optimizationtheory This fact has been motivated and inspired many mathematicians to study the well-posednessfor problems related to optimization In 1966, Tykhonov introduced the concept of well-posednessfor unconstrained optimization problems, which has become known as Tikhonov well-posedness Aminimization problem is said to be Tykhonov well-posed if it has a unique solution toward whichevery minimizing sequence of the problem converges (see [1]) Since then, the study of Tykhonov well-posedness and its extensions has been among the very interesting and important topics in the stabilityfor optimization theory A generalization of Tykhonov well-posedness strengthened for this concepthas been discussed for constrained optimization problems and sequence optimization problems (see,e.g [2–7]) Another generalization of the concept given for optimization problems with more thanone solution requires the existence and convergence of a subsequence of each minimizing sequencetowards a solution.[8] The other fundamental generalization of Tykhonov well-posedness, whichwas first introduced for scalar optimization problem by Zolezzi [9,10], is the well-posedness underperturbations The idea of this generalized concept is embedding the Tykhonov well-posedness andthe continuous dependence of the solution on the data The study of Tykhonov well-posedness and itsextensions for problems related to equilibrium problems, such as optimization problem, variationalinequality, Nash equilibria and equilibrium problem, is a theme of great importance and has receivedincreasing attention by many researchers recently The sufficient and necessary conditions and metriccharacterizations of the well-posedness for such problems were considered For more details, we referthe reader to [11–16] and the references therein

It is well known that the class of partially ordered spaces plays an important role in vectoroptimization theory The vector problems related to optimization are usually based on partial ordersinduced by convex closed cones; i.e they base on various extensions of the Pareto order From the

CONTACT Lam Quoc Anh quocanh@ctu.edu.vn

theory of vector optimization, however, this setting leads to an optimal solution set that is usuallytoo large (see, e.g [17–20] and the references therein) Hence, reducing the optimal solution set

is the aim of many works One of the efficient approaches is to use the lexicographic cone In[21], for a fixed orthogonal base, the authors constructed a total ordering cone inRnand showedthat the lexicographic order was a unique total order in the sense that any total order onRnwasequivalent to lexicographic order Furthermore, lexicographic cone also plays a vital role in manypractical problems, such as choosing products, ranking medal table in Olympic Games; see, e.g.[22–25] Therefore, vector problems related to optimization involving lexicographic cone have beenintensively studied recently; see, e.g for variational inequalities,[24,26] optimization problems,[22,

27] equilibrium problems [19,20,25] and the references therein As far as we know, well-posednessfor the lexicographic vector equilibrium problems was discussed in only two papers [28] and [29] Inthese papers, this property was obtained under the lower semicontinuity of an auxiliary set-valuedmapping corresponding to objective function However, this assumption is difficultly checked andhard applied to practical situations since it requires the information of a solution set of the equation

In this paper, motivated and inspired by the above observations, we aim to suggest the new concepts

of the Tykhonov well-posedness and its extension to the lexicographic equilibrium problems Thecorresponding concepts of the Tykhonov well-posedness in the generalized sense are also introducedand investigated Furthermore, we also study some metric characterizations of these properties via theKuratowski measure of noncompactness and diameter of approximate solution sets of such problems.The layout of the paper is as follows: In Section2, we state the lexicographic equilibrium problemsand recall some preliminary results which are needed in the succeeding sections Section3is devoted

to the (generalized) Tykhonov well-posedness for the lexicographic equilibrium problems In Section

4, we study sufficient conditions of the (generalized) Tykhonov well-posedness under perturbations

by a sequence of approximating problems for these problems In the last section, as an application,several results on these types of well-posedness for lexicographic variational inequalities are derivedfrom the main results

2 Preliminaries

We first recall the notion of lexicographic cone in finite-dimensional spaces and the setting ofequilibrium problems involving this cone

The lexicographic cone ofRn , denoted by Clex, is the collection of zero and all vectors x ∈ Rn

which the first nonzero coordinate of x is positive, i.e.

Clex := {0} ∪ {x ∈ R n | x1= · · · = x k = 0, x k+1> 0, for some k, 0 ≤ k < n}.

For any x and y inRn, the lexicographic order is defined as follows:

x≥lexy ⇐⇒ x − y ∈ Clex.

Since Clex∪ ( − Clex) = R n , the lexicographic order is a total order Moreover, let C1 := {x ∈ R n|

x1≥ 0}, then int C1  Clex C1and

int Clex = int C1 and cl Clex= C1.

Hence, the lexicographic cone is neither closed nor open

Let E be a real Banach space and X be a nonempty closed subset of E Let E∗be the dual space of

that f = (f1, f2, , f n ) : X ×X → R n is a vector-valued function, where f iis an equilibrium function

for each i ∈ I n := {1, 2, , n}, i.e f i (x, x) = 0 for all x ∈ X We consider the following lexicographic

equilibrium problem:

(LEP) find ¯x ∈ X such that for all y ∈ X,

f (¯x, y) ≥lex0.

The following notions are employed in the sequel

Definition 1: Let Q : X ⇒ Y be a set-valued mapping between two Banach spaces.

(i) Q is said to be upper semicontinuous (usc, in short) at x0, if for any open subset U of Y with

Q(x0) ⊂ U, there is a neighborhood N of x0such that Q (N) ⊂ U.

(ii) Q is said to be lower semicontinuous (lsc, in short) at x0, if for any open subset U of Y with

Q(x0) ∩ U = ∅, there is a neighborhood N of x0such that Q (x) ∩ U = ∅, for all x ∈ N.

Q is said to be continuous at x0, if it is both usc and lsc at x0.

The following well-known assertions play an important role in our analysis

Lemma 2.1: (see, e.g [30])

(i) If Q(x0) is compact, then Q is usc at x0 if and only if for any sequence {x n } converging to x0, every sequence {y n } with y n ∈ Q(x n ) has a subsequence converging to some point in Q(x0) If,

in addition, Q(x0) = {y0} is a singleton, then such a sequence {y n } must converge to y0.

(ii) Q is lsc at x0if and only if for any sequence {x n } converging to x0and any point y ∈ Q(x0), there exists a sequence {y n } with y n ∈ Q(x n ) converging to y.

Definition 2: (see, e.g [28]) Letε be a real number An extended real-valued function g : X →

R ∪ {+∞} is said to be

(i) upper ε-level closed at ¯x ∈ X, if for any sequence {x n }, x n → ¯x,



g (x n ) ≥ ε, ∀n⇒g (¯x) ≥ ε;(ii) strongly upper ε-level closed at ¯x ∈ X, if for any sequences {x n }, x n → ¯x and {μ n } ⊂ [0; ∞),

g(x n ) + μ n ≥ ε, ∀n⇒g(¯x) ≥ ε;(iii) upper semicontinuous at ¯x ∈ X, if for any sequence {x n }, x n → ¯x, it holds that

g(¯x) ≥ lim sup

n→∞ g (x n ).

We say that F satisfies a certain property in a subset A of X if F satisfies it at each x ∈ A If A ≡ X

we omit the term ‘in X’ in the statement.

Now we recall the concepts of the Kuratowski measure of noncompactness and the Hausdorffdistance

Definition 3: (see, e.g [31]) Let M be a nonempty subset of E The Kuratowski measure of

where diam M i is the diameter of M i

Definition 4: Let A, B be nonempty subsets of E The Hausdorff distance between A and B is defined

by

H(A, B) = maxH∗(A, B), H∗(B, A),

where H∗(A, B) = sup a ∈A d(a, B) with d(a, B) = inf b ∈B d(a, b).

Lemma 2.2: (see, e.g [31]) The following assertions are true:

(iii) If {M n } is a sequence of closed subsets in E satisfying M n+1 ⊂ M n for every n ∈ N and

H(M n , K ) = 0.

3 Tykhonov well-posedness for lexicographic equilibrium problems

In this section, we study sufficient conditions for (LEP) to be Tykhonov well-posed To start our

analysis, we consider lexicographic equilibrium problems for the case n = 2, namely, f = (f1, f2) :

X ×X → R2, since the general case is similar Then, we can rewrite (LEP) in the following equivalent

way: find¯x ∈ X such that

where Z : X ⇒ X is defined by

Z(x) = {z ∈ X | f1(x, z) = 0}.

The solution set of (LEP) is denoted by S.

Let e ∈ Clex \ {0} For each ε ∈ [0, +∞), we consider the following approximate problem corresponding to e:

(LEPe,ε) find¯x ∈ X such that

f (x n , y ) + ε n e≥lex0, ∀y ∈ X.

The solution set of this approximate problem (LEPe,ε) is denoted by

S e(ε) := {x ∈ X | f (x, y) + εe ≥lex0, ∀y ∈ X}.

Definition 5: A sequence {x n } is said to be an approximating sequence for (LEP) corresponding to

e, if there exists a sequence{ε n} ⊂ R+withε n → 0 such that x n ∈ S e(ε n ) for all n.

Definition 6: The problem (LEP) is said to be

sequence{x n } for (LEP) corresponding to e, there exists a subsequence {x n i } of {x n} converging

(b) For a = (a1, a2) ∈ Clex\ {0} and b = (b1, b2) ∈ Clex\ {0}, we define a relation ∼ on Clex\ {0}

as follows:

a ∼ b ⇐⇒ there exist k, l > 0 such that a1= kb1and a2 = lb2.

One can check that∼ is an equivalence relation on Clex\ {0} Denote a be the equivalence class determined by a Then

The following example illustrates the above statement

Example 1: Let E = R2, X = {x = (x1, x2) ∈ R2 | 0 ≤ x k ≤ 1, k = 1, 2} and f = (f1, f2) : X2 →

Therefore, S e1(ε)  Se2(ε)  Se3(ε)  S e4(ε).

Proposition 3.1: Suppose that a, b ∈ Clex\ {0} and a ∼ b Then (LEP) is (generalized) Tykhonov

well-posed wrt a if and only if it is (generalized, respectively) Tykhonov well-posed wrt b.

Proof: By the similarity we verify only the case a ∈ e2as an example In this case, we need only toshow that{x n } is an approximating sequence for (LEP) corresponding to a = (a1,−a2), a1, a2 > 0

if and only if it is an approximating sequence for (LEP) corresponding to e2 Indeed, if{x n} is anapproximating sequence for (LEP) corresponding to a Then there exists a sequence {ε n} ⊂ R+,ε n→

0 such that

f (x n , y ) + ε n (a1,−a2) ≥lex 0, ∀y ∈ X.

For each n, let δ n = (a1+ a2)ε n Then{δ n} ⊂ R+,δ n→ 0 and

f1(x n , y ) + δ n > f1(x n , y ) + a1ε n ≥ 0, ∀y ∈ X,

which implies that

f (x n , y ) + δ n (1, −1) ≥lex0, ∀y ∈ X.

Hence,{x n } is an approximating sequence for (LEP) corresponding to e2

Conversely, if{x n } is an approximating sequence for (LEP) corresponding to e2 Then there exists

a sequence{ε n} ⊂ R+,ε n→ 0 such that

f (x n , y ) + δ n (a1,−a2) ≥lex 0, ∀y ∈ X,

since f1(x n , y ) + δ n a1 > f1(x n , y ) + ε n ≥ 0, ∀y ∈ X Thus, {x n} is an approximating sequence for

Motivated and inspired by the above observations, in the sequel, we choose e = e1 Then, (LEP)

is (generalized) Tykhonov well-posed wrt e if it is (generalized) Tykhonov well-posed wrt c, for all

c ∈ Clex\ {0} The following two examples illustrate that the converse is not true.

Example 2: Let E = R, X = [0, +∞) and f = (f1, f2) : X2→ R2be defined by f (x, y) = (0, y −x).

It is clear that S = {0} For each ε > 0, f (x, y) + εe ≥lex 0, ∀y ∈ X if and only if x ≤ ε; i.e.

S e(ε) = [0, ε] Hence, (LEP) is Tykhonov well-posed wrt e However, for all c = (c1, c2) ∈ intClex,one hasSc(ε) = X, so (LEP) is not Tykhonov well-posed wrt c Indeed, let x n = (n+1)/n Then, {x n} is

an approximating sequence for (LEP) corresponding to c since f (x n , y )+εc = (εc1, x n −y+εc2) >lex

By simple computations, we have S = {x ∈ X | x = (1, 1, x3 , x m ), x k ∈ [0, 1], k ∈ {3, , m}}.

For eachε > 0, one has

To simplify the presentation, in the sequel, if (LEP) is (generalized) Tykhonov well-posed with

respect to e, we omit the term ‘with respect to e’ in the statement For each ε > 0, the approximate

solution set of (LEP) corresponding to e is denoted by:

S(ε) :=S e(ε) = {x ∈ X | f1(x, y) ≥ 0, ∀y ∈ X and f2(x, z) + ε ≥ 0, ∀z ∈ Z(x)}.

The following theorem provides sufficient conditions of the Tykhonov well-posedness for (LEP).

Theorem 3.2: Assume that X is compact and

(i) f1is continuous; the Fréchet derivative D2f1of f1with respect to the second argument exists and

D2f1(x, y) is surjective for all x, y ∈ X, x = y;

(ii) f2is strongly upper 0-level closed.

Then, ( LEP) is generalized Tykhonov well-posed Furthermore, it is Tykhonov well-posed if S is a

singleton.

Proof: Let {x n} be an arbitrary approximating sequence for (LEP) Then, there exists a sequence

{ε n} ⊂ R+withε n→ 0 such that

By the compactness of X, there is a subsequence (still denoted by {x n }) converging to some ¯x in X Combining the continuity of f1 and (3), we conclude that f1(¯x, y) ≥ 0, for all y ∈ X To complete

the first conclusion of the theorem, we only need to show that¯x ∈ S Suppose, on the contrary, that

¯x /∈ S, then there is a point ¯z ∈ Z(¯x) \ {¯x}, such that

We prove that for each neighborhood V of ¯z, V ⊂ X, there exist a neighborhood U of ¯x and a mapping

s : U → V such that s(x) ∈ Z(x) for all x ∈ U Let m =D2f1(¯x, ¯z)−1 Since D2f1is surjective,Theorem 5A.1 in [32] implies that m = 0 Let α be a positive real number such that B α (¯z) ⊂ V,

where B α (¯z) is the closed ball with center at ¯z and radius α Since f1is Fréchet differentiable withrespect to the second argument, one can choose a real numberβ, with 0 < β ≤ α satisfying

2m , for all x ∈ B γ (¯x), and z ∈ B β (¯z).

For each x ∈ B γ (¯x), we construct the function ξ x : B β (¯z) → X defined by

This means thatξ x maps B β (¯z) into itself Since B β (¯z) is compact and convex, the Brouwer’s

fixed-point theorem implies that, for each x ∈ B γ (¯x), there exists a point denoted by s(x) ∈ B β (¯z) ⊂ V

such thatξ x (s(x)) = s(x), i.e.

s(x) = D2f1(¯x, ¯z)−1D2f1(¯x, ¯z), s(x)− f1(x, s(x)) ,

or equivalently, f1(x, s(x)) = 0 Thus, s(x) ∈ Z(x) This argument ensures the existence of a sequence

{z n } with z n ∈ Z(x n ), z n → ¯z It follows from (4) that f2(x n , z n ) + ε n ≥ 0, for all n Since f2is stronglyupper 0-level closed at(¯x, ¯z), we have f2(¯x, ¯z) ≥ 0, which contradicts (5) Therefore,¯x ∈ S, i.e (LEP)

is generalized Tykhonov well-posed The second conclusion of the theorem follows directly from

Going back to Example3, we immediately check that all assumptions in Theorem3.2are satisfied.Hence, (LEP) is generalized Tykhonov well-posed The following examples show that the assumptions

of Theorem3.2are essential

Example 4: (Compactness of X cannot be dispensed) Let X = E = R, and

f (x, y) = ((x − y)2, x − y).

Obviously, assumptions (i) and (ii) of Theorem3.2are satisfied and S = R However, (LEP) is not

generalized Tykhonov well-posed since the approximating sequence{x n }, x n = n, for (LEP) has no

any convergent subsequence The reason is that X is not compact.

Example 5: (Surjectivity of D2f1is essential) Let E = R, X = [0, 1] and

f (x, y) = ((1 − x)(x − y)2, y − x).

Then, assumption (ii) is satisfied by the continuity of f2 By direct computations, we see that S = [0, 1) Let x n = (n − 1)/n We can easily verify that {x n } is an approximating sequence for (LEP) but {x n}converges to 1 /∈ S Hence, (LEP) is not generalized Tykhonov well-posed The reason is that

assumption (i) is violated Indeed, for all y ∈ X, D2f1(1, y) = 0.

Example 6: (Assumption (ii) is essential) Let E = R2, X = [0, 1] × [0, 1] and f (x, y) = (f1(x, y),



.

Thus, assumption (i) is fulfilled By direct computations, we have S=(1, x2) | x2∈ {0} ∪1

2, 1and S(ε) =(1, x2) | x2∈ [0, ε] ∪1

2, 1

Let x n = (1, n+1

2n ), then x n → (1,1

2) /∈ S Hence, (LEP)

is not generalized Tykhonov well-posed The reason is that the strong upper 0-level closedness

of f2 is violated Indeed, taking x n = (1, (n + 1)/2n), y n = (1, 1/2n) and ε n = 1/n, we have

x n → x = (1, 1/2), y n → y = (1, 0), ε n → 0 Then, f2(x n , y n ) + ε n > 0, but f2(x, y) = −1/2 < 0.

Using the diameter of the approximate solution sets, we obtain a metric characterization of theTykhonov well-posedness for (LEP).

Theorem 3.3: Assume that

(i) f1 is continuous; the Fréchet derivative of f1 with respect to the second argument exists and

D2f1(x, y) is surjective for all x, y ∈ X, x = y;

(ii) f2is strongly upper 0-level closed.

Then, ( LEP) is Tykhonov well-posed if and only if

S(ε) = ∅, ∀ε > 0 and lim

ε→0diam(S(ε)) = 0.

Proof: Suppose that (LEP) is Tykhonov well-posed Then, (LEP) has a unique solution ¯x, and hence

S(ε) = ∅ for all ε > 0 If lim ε→0diam(S(ε)) = 0, then there exist r > 0, ε n > 0 with ε n → 0 and

x n , u n ∈ S(ε n ) such that

d(x n , u n ) ≥ r

Then,{x n } and {u n} are approximating sequences for (LEP) By the Tykhonov well-posedness of

(LEP), they must converge to the unique solution ¯x of (LEP) Thus, lim n→∞d(x n , u n ) = 0, which

contradicts (6)

Conversely, suppose that {x n} is an approximating sequence for (LEP) Then, there exists a

sequence{ε n }, ε n → 0 such that x n ∈ S(ε n ) for all n By taking a subsequence, if necessary,

we can assume that {ε n } is non-increasing, and hence S(ε n ) ⊂ S(ε m ) whenever m ≤ n Since

limn→∞diam(S(ε n )) = 0, {x n } is a Cauchy sequence and converges to ¯x, ¯x ∈ X Using the same

arguments as for Theorem3.2, we conclude that ¯x ∈ S To complete the proof, we show that (LEP)

has a unique solution Suppose, by contradiction, that S contains another point u with u = ¯x It is

obvious that¯x and u belong toS(ε) for any ε > 0 It follows that

0 < d (¯x, u) ≤ diam (S(ε)),

Using the Kuratowski measure of noncompactness of approximate solution sets, we establish ametric characterization of the generalized Tykhonov well-posedness for (LEP).

Theorem 3.4: Assume that

(i) f1 is continuous; the Fréchet derivative of f1 with respect to the second argument exists and

D2f1(x, y) is surjective for all x, y ∈ X, x = y;

(ii) f2is upper δ-level closed for all δ < 0.

Then, ( LEP) is generalized Tykhonov well-posed if and only if

S(ε) = ∅, ∀ε > 0 and lim

ε→0 μ(S(ε)) = 0.

Proof: Suppose that (LEP) is generalized Tykhonov well-posed We first prove that S is compact.

Indeed, let{x n } be an arbitrary sequence in S Obviously, it is also an approximating sequence for

(LEP), and hence it has a subsequence converging to some point in S Thus, S is compact Take any

Now we show that H (S(ε), S) → 0 as ε → 0 Since S ⊂ S(ε), we get H∗(S,S(ε)) = 0, and hence

we only need to prove that H∗(S(ε), S) → 0 as ε → 0 Assume, by contradiction, that there exist a

real number r > 0, a sequence {ε n }, ε n → 0, and x n ∈ S(ε n ) such that d(x n , S ) ≥ r, for all n Since

{x n } is an approximating sequence for (LEP), it has a subsequence {x n } converging to some point

Theorem 3.3: Assume that

(i) f1... characterization of the generalized Tykhonov well-posedness for (LEP).

Theorem 3.4: Assume that

(i) f1 is continuous; the Fréchet derivative...

obvious that¯x and u belong toS(ε) for any ε > It follows that

0 < d (¯x, u) ≤ diam (S(ε)),

Using the Kuratowski measure of noncompactness of approximate solution