# Generic Well-Posedness for a Class of Equilibrium Problems

- Alexander J. Zaslavski
^{1}Email author

**2008**:581917

https://doi.org/10.1155/2008/581917

© Alexander J. Zaslavski. 2008

**Received: **23 December 2007

**Accepted: **6 March 2008

**Published: **17 March 2008

## 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:

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].

Clearly, is a complete metric space.

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.

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:

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 (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

## 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 .

and the following property holds.

(P4) For each satisfying the inequality holds.

Thus we have shown that the following property holds.

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

Thus we have shown that the following property holds.

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

Theorem 1.1 is proved.

## Authors’ Affiliations

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## Copyright

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