The reduction of the Ѩ-problem on a Frechet nuclear space to the study of the Ѩ-operator on a Hilbert space produces a global solution u when the second member w factors globally through this Hilbert space. Easy counterexamples show that this global factorization is not in general possible and hence
Bounded factorization property for Fréchet spaces
✍ Scribed by Tosun Terzioğlu; Vyacheslav Zahariuta
- Publisher
- John Wiley and Sons
- Year
- 2003
- Tongue
- English
- Weight
- 158 KB
- Volume
- 253
- Category
- Article
- ISSN
- 0025-584X
No coin nor oath required. For personal study only.
✦ Synopsis
Abstract
An operator T ∈ L(E, F) factors over G if T = RS for some S ∈ L(E, G) and R ∈ L(G, F); the set of such operators is denoted by L^G^(E, F). A triple (E, G, F) satisfies bounded factorization property (shortly, (E, G, F) ∈ ℬ︁ℱ) if L^G^(E, F) ⊂ LB(E, F), where LB(E, F) is the set of all bounded linear operators from E to F. The relationship (E, G, F) ∈ ℬ︁ℱ is characterized in the spirit of Vogt's characterisation of the relationship L(E, F) = LB(E, F) [23]. For triples of K�othe spaces the property ℬ︁ℱ is characterized in terms of their K�othe matrices.
As an application we prove that in certain cases the relations L(E, G~1~) = LB(E, G~1~) and L(G~2~, F) = LB(G~2~, F) imply (E, G, F) ∈ ℬ︁ℱ where G is a tensor product of G~1~ and G~2~.
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