A note on well-posed null and fixed point problems
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We establish generic well-posedness of certain null and fixed point problems for ordered Banach space-valued continuous mappings. The notion of well-posedness is of great importance in many areas of mathematics and its applications. In this note, we consider two complete metric spaces of continuous mappings and establish generic well-posedness of certain null and fixed point problems (Theorems 1 and 2, resp.). Our results are a consequence of the variational principle established in [2]. For other recent results concerning the well-posedness of fixed point problems, see [1, 3]. Let (X, · , ≥) be a Banach space ordered by a closed convex cone X+ = {x ∈ X : x ≥ 0} such that x ≤ y for each pair of points x, y ∈ X+ satisfying x ≤ y. Let (K,ρ) be a complete metric space. Denote by M the set of all continuous mappings A : K → X. We equip the set M with the uniformity determined by the following base:
E() = (A,B) ∈ M × M : Ax − Bx ≤ ∀x ∈ K ,
(1)
where > 0. It is not difficult to see that this uniform space is metrizable (by a metric d) and complete. Denote by M p the set of all A ∈ M such that Ax ∈ X+
∀x ∈ K, inf Ax : x ∈ K = 0.
(2)
It is not difficult to see that M p is a closed subset of (M,d). We can now state and prove our first result. Theorem 1. There exists an everywhere dense Gδ subset Ᏺ ⊂ M p such that for each A ∈ Ᏺ, the following properties hold. (1) There is a unique x¯ ∈ K such that A¯x = 0. (2) For any > 0, there exist δ > 0 and a neighborhood U of A in M p such that if B ∈ U ¯ ≤ . and if x ∈ K satisfies Bx ≤ δ, then ρ(x, x) Copyright © 2005 Hindawi Publishing Corporation Fixed Point Theory and Applications 2005:2 (2005) 207–211 DOI: 10.1155/FPTA.2005.207
208
Well-posed problems
Proof. We obtain this theorem as a realization of the variational principle established in [2, Theorem 2.1] with fA (x) = Ax, x ∈ K. In order to prove our theorem by using this variational principle, we need to prove the following assertion. (A) For each A ∈ M p and each > 0, there are A¯ ∈ M p , δ > 0, x¯ ∈ K, and a neighborhood W of A¯ in M p such that ¯ ∈ E(), (A, A)
(3)
and if B ∈ W and z ∈ K satisfy Bz ≤ δ, then ¯ ≤ . ρ(z, x)
(4)
Let A ∈ M p and > 0. Choose u¯ ∈ X+ such that u¯ =
4
,
(5)
and x¯ ∈ K such that A¯ x ≤
8
.
(6)
Since A is continuous, there is a positive number r such that
r < min 1, Ax − A¯ x ≤
16
,
(7)
¯ ≤ 4r. for each x ∈ K satisfying ρ(x, x)
8
(8)
By Urysohn’s theorem, there is a continuous function φ : K → [0,1] such that ¯ ≤ r, φ(x) = 1 for each x ∈ K satisfying ρ(x, x)
(9)
¯ ≥ 2r. φ(x) = 0 for each x ∈ K satisfying ρ(x, x)
(10)
Define
¯ = 1 − φ(x) (Ax + u), ¯ Ax
x ∈ K.
(11)
It is clear that A¯ : K → X is continuous. Now (9), (10), and (11) imply that ¯ = 0 for each x ∈ K satisfying ρ(x, x) ¯ ≤ r, Ax ¯ ≥ u¯ for each x ∈ K satisfying ρ(x, x) ¯ ≥ 2r. Ax ¯ ∈ E(). It is not difficult to see that A¯ ∈ M p . We claim that (A, A)
(12) (13)
S. Reich and A. J. Zaslavski 209 Let x ∈ K. There are two cases: either ¯ ≥
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