FIXED POINT PROBLEMS
SIMEON REICH AND ALEXANDER J. ZASLAVSKI Received 16 October 2004
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 (Theorems1 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 coneX+= {x∈X:x≥ 0}such thatx ≤ yfor each pair of pointsx,y∈X+satisfyingx≤y. Let (K,ρ) be a complete metric space. Denote byMthe set of all continuous mappingsA:K→X. We equip the setMwith 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 metricd) and complete.
Denote byMpthe set of allA∈Msuch that Ax∈X+ ∀x∈K,
infAx:x∈K=0. (2)
It is not difficult to see thatMpis a closed subset of (M,d).
We can now state and prove our first result.
Theorem1. There exists an everywhere denseGδsubsetᏲ⊂Mpsuch that for eachA∈Ᏺ, the following properties hold.
(1) There is a uniquex¯∈Ksuch thatAx¯=0.
(2) For any>0, there existδ >0and a neighborhoodUofAinMpsuch that ifB∈U and ifx∈KsatisfiesBx ≤δ, thenρ(x, ¯x)≤.
Copyright©2005 Hindawi Publishing Corporation Fixed Point Theory and Applications 2005:2 (2005) 207–211 DOI:10.1155/FPTA.2005.207
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 eachA∈Mpand each>0, there are ¯A∈Mp,δ >0, ¯x∈K, and a neighbor- hoodWof ¯AinMpsuch that
(A, ¯A)∈E(), (3)
and ifB∈Wandz∈KsatisfyBz ≤δ, then
ρ(z, ¯x)≤. (4)
LetA∈Mpand>0. Choose ¯u∈X+such that u¯ =
4, (5)
and ¯x∈Ksuch that
Ax¯ ≤
8. (6)
SinceAis continuous, there is a positive numberrsuch that r <min
1,
16
, (7)
Ax−A¯x ≤
8 for eachx∈Ksatisfyingρ(x, ¯x)≤4r. (8) By Urysohn’s theorem, there is a continuous functionφ:K→[0, 1] such that
φ(x)=1 for eachx∈Ksatisfyingρ(x, ¯x)≤r, (9) φ(x)=0 for eachx∈Ksatisfyingρ(x, ¯x)≥2r. (10) Define
Ax¯ =
1−φ(x)(Ax+ ¯u), x∈K. (11)
It is clear that ¯A:K→Xis continuous. Now (9), (10), and (11) imply that
Ax¯ =0 for eachx∈Ksatisfyingρ(x, ¯x)≤r, (12) Ax¯ ≥u¯ for eachx∈Ksatisfyingρ(x, ¯x)≥2r. (13) It is not difficult to see that ¯A∈Mp. We claim that (A, ¯A)∈E().
Letx∈K. There are two cases: either
ρ(x, ¯x)≥2r (14)
or
ρ(x, ¯x)<2r. (15)
Assume first that (14) holds. Then it follows from (14), (10), (11), and (5) that Ax−Ax¯ = u¯ =
4. (16)
Now assume that (15) holds. Then by (15), (11), and (5), Ax¯ −Ax =1−φ(x)(Ax+ ¯u)−Ax
≤ u¯+Ax ≤
4+Ax. (17)
It follows from this inequality, (15), (8), and (6) that Ax¯ −Ax ≤
4+Ax<
2. (18)
Therefore, in both cases,Ax¯ −Ax ≤/2. Since this inequality holds for anyx∈K, we conclude that
(A, ¯A)∈E(). (19)
Consider now an open neighborhoodUof ¯AinMpsuch that U⊂
B∈Mp: ( ¯A,B)∈E
16 . (20)
Let
B∈U, z∈K, (21)
Bz ≤
16. (22)
Relations (22), (21), (20), and (1) imply that
Az¯ ≤ Bz+Az¯ −Bz ≤ 16+
16. (23)
We claim that
ρ(z, ¯x)≤. (24)
We assume the converse. Then by (7),
ρ(z, ¯x)>≥2r. (25)
When combined with (13), this implies that
Az¯ ≥u.¯ (26)
It follows from this inequality, the monotonicity of the norm, (21), (20), (1), and (5) that Bz ≥ Az¯ −
16≥ u¯ − 16
= 4−
16=
3 16.
(27)
This, however, contradicts (22). The contradiction we have reached proves (24) and
Theorem 1itself.
Now assume that the setKis a subset ofXand
ρ(x,y)= x−y, x,y∈K. (28) Denote byMnthe set of all mappingsA∈Msuch that
Ax≥x ∀x∈K,
infAx−x:x∈K=0. (29)
Clearly,Mnis a closed subset of (M,d). Define a mapJ:Mn→Mpby
J(A)x=Ax−x ∀x∈K (30)
and allA∈Mn. Clearly, there existsJ−1:Mp→Mn, and bothJ and its inverseJ−1are continuous. ThereforeTheorem 1implies the following result regarding the generic well- posedness of the fixed point problem forA∈Mn.
Theorem2. There exists an everywhere denseGδsubsetᏲ⊂Mnsuch that for eachA∈Ᏺ, the following properties hold.
(1) There is a uniquex¯∈Ksuch thatAx¯=x.¯
(2) For any>0, there existδ >0and a neighborhoodUofAinMnsuch that ifB∈U and ifx∈KsatisfiesBx−x ≤δ, thenx−x¯ ≤.
Acknowledgments
The work of the first author was partially supported by the Israel Science Foundation founded by the Israel Academy of Sciences and Humanities (Grant 592/00), by the Fund for the Promotion of Research at the Technion, and by the Technion VPR Fund.
References
[1] F. S. De Blasi and J. Myjak,Sur la porosit´e de l’ensemble des contractions sans point fixe[On the porosity of the set of contractions without fixed points], C. R. Acad. Sci. Paris S´er. I Math.308 (1989), no. 2, 51–54 (French).
[2] A. D. Ioffe and A. J. Zaslavski,Variational principles and well-posedness in optimization and calculus of variations, SIAM J. Control Optim.38(2000), no. 2, 566–581.
[3] S. Reich and A. J. Zaslavski,Well-posedness of fixed point problems, Far East J. Math. Sci. (FJMS), (2001), Special Volume (Functional Analysis and Its Applications), Part III, 393–401.
Simeon Reich: Department of Mathematical and Computing Sciences, Tokyo Institute of Technol- ogy, 2-12-1 O-okayama, Meguro-ku, Tokyo 152-8552, Japan
E-mail address:[email protected]
Alexander J. Zaslavski: Department of Mathematics, Technion – Israel Institute of Technology, 32000 Haifa, Israel
E-mail address:[email protected]
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