Generic Well-Posedness for a Class of Equilibrium Problems

Abstract

We study a class of equilibrium problems which is identified with a complete metric space of functions. For most elements of this space of functions (in the sense of Baire category), we establish that the corresponding equilibrium problem possesses a unique solution and is well-posed.

1. Introduction

The study of equilibrium problems has recently been a rapidly growing area of research. See, for example, [1–3] and the references mentioned therein.

Let be a complete metric space. In this paper, we consider the following equilibrium problem:

(P)

where belongs to a complete metric space of functions defined below. In this paper, we show that for most elements of this space of functions (in the sense of Baire category) the equilibrium problem (P) possesses a unique solution. In other words, the problem (P) possesses a unique solution for a generic (typical) element of [4–6].

Set

(1.1)

Clearly, is a complete metric space.

Denote by the set of all continuous functions such that

(1.2)

We equip the set with the uniformity determined by the base

(1.3)

where . It is clear that the space with this uniformity is metrizable (by a metric ) and complete.

Denote by the set of all for which the following properties hold.

(P1) For each , there exists such that for all .

(P2) For each , there exists such that for all satisfying .

Clearly, is a closed subset of . We equip the space with the metric and consider the topological subspace with the relative topology.

For each and each subset , put

(1.4)

For each and each , set

(1.5)

Assume that the following property holds.

(P3) There exists a positive number such that for each and each pair of real numbers satisfying , there is such that

In this paper, we will establish the following result.

Theorem 1.1.

There exists a set which is a countable intersection of open everywhere dense subsets of such that for each , the following properties hold:

• (i) there exists a unique such that

  

  (1.6)

• (ii) for each , there are and a neighborhood of in such that for each and each satisfying , the inequality holds.

In other words, for a generic (typical) , the problem (P) is well-posed [7–9].

2. An Auxiliary Density Result

Lemma 2.1.

Let and . Then there exist and such that and for all .

Proof.

By (P1) there is such that

(2.1)

Set

(2.2)

For each , there is such that

(2.3)

For each , there is such that

(2.4)

For each , there is such that

(2.5)

For each , set

(2.6)

For any , put

(2.7)

and for any , put

(2.8)

Clearly, is an open covering of . Since any metric space is paracompact, there is a continuous locally finite partition of unity subordinated to the covering . Namely, for any , is a continuous function and there exist such that and that

(2.9)

Define

(2.10)

Clearly, is well defined, continuous, and satisfies

(2.11)

Let . Then

(2.12)

Assume that and that . Then

(2.13)

If , then in view of (2.5), (2.6), and (2.13), , a contradiction (see (2.12)). Then , and by (2.7),

(2.14)

Since this equality holds for any satisfying , it follows from (2.10) that

(2.15)

for all .

Relations (2.1), (2.2), and (2.15) imply that

(2.16)

By (1.2), (2.7), (2.8), and (2.10)

(2.17)

Assume that

(2.18)

Then in view of (2.2) and (2.18), Together with (2.7) and (2.10), this implies that

(2.19)

Combined with (2.11), this implies that

(2.20)

for all

Let

(2.21)

and assume that

(2.22)

Then in view of (2.22),

(2.23)

By (2.23) and the choice of (see (2.3)–(2.6)), and by (2.4), (2.6), (2.7), and (2.8),

(2.24)

Since the inequality above holds for any satisfying (2.22), the relation (2.10) implies that

(2.25)

Together with (2.11), (2.12), and (2.15), this implies that for all

(2.26)

By (2.17), . In view of (2.16), possesses (P1). Since possesses (P2), it follows from (2.7), (2.8), and (2.10) that possesses (P2). Therefore and Lemma 2.1 now follows from (2.16) and (2.26).

3. A Perturbation Lemma

Lemma 3.1.

Let , , and let satisfy

(3.1)

Then there exist and such that

(3.2)

and if satisfies then .

Proof.

By (P2) there is a positive number

(3.3)

such that

(3.4)

Set

(3.5)

Define

(3.6)

(3.7)

Clearly, is continuous and

(3.8)

By (3.6) and (3.7),

(3.9)

Let . We estimate . If , then by (3.6) and (3.7),

(3.10)

Assume that

(3.11)

By (3.3) and (3.11),

(3.12)

By (3.5), (3.6), (3.7), (3.11), and (3.12),

(3.13)

Together with (3.10) this implies that

(3.14)

Assume that . In view of (P3) and (3.3), there is such that

(3.15)

It follows from (3.15) and (3.9) that

(3.16)

(3.17)

for all . Set

(3.18)

Clearly, the function is continuous and

(3.19)

In view of (3.1), (3.18), and (3.6),

(3.20)

Since the function possesses (P2), it follows from (3.9), (3.20), and (3.18) that possesses the property (P2). Thus .

By (3.6), (3.14), and (3.18) for all

(3.21)

Assume that

(3.22)

If , then by (3.6) and (3.18),

(3.23)

and together with (3.17), this implies that

(3.24)

This inequality contradicts (3.22). The contradiction we have reached proves that

(3.25)

This completes the proof of the lemma.

4. Proof of Theorem 1.1

Denote by the set of all for which there exists such that for all . By Lemma 2.1, is an everywhere dense subset of .

Let and be a natural number. There exists such that

(4.1)

By Lemma 3.1, there exist and such that

(4.2)

and the following property holds.

(P4) For each satisfying the inequality holds.

Denote by the open neighborhood of in such that

(4.3)

Assume that

(4.4)

By (1.3), (4.3), and (4.4),

(4.5)

In view of (4.5) and (P4),

(4.6)

Thus we have shown that the following property holds.

(P5) For each and each satisfying (4.4), the inequality holds.

Set

(4.7)

Clearly, is a countable intersection of open everywhere dense subset of . Let

(4.8)

Choose a natural number . There exist and an integer such that

(4.9)

The property (P4), (4.3), and (4.9) imply that for each satisfying

(4.10)

we have

(4.11)

Thus we have shown that the following property holds.

(P6) For each satisfying (4.10), the inequality holds.

By (P1) there is a sequence such that

(4.12)

In view of (4.12) and (P6) for all large enough natural numbers , we have

(4.13)

Since is any positive number, we conclude that is a Cauchy sequence and there exists

(4.14)

Relations (4.12) and (4.14) imply that for all

(4.15)

We have also shown that any sequence satisfying (4.12) converges. This implies that if satisfies for all , then . By (P6) and (4.15),

(4.16)

Let and satisfy (4.4). By (P5), . Together with (4.16), this implies that

(4.17)

Theorem 1.1 is proved.