Timeline for Reference request: Baire's theorem for operator ranges
Current License: CC BY-SA 4.0
11 events
when toggle format | what | by | license | comment | |
---|---|---|---|---|---|
Jun 1, 2021 at 12:14 | comment | added | Jochen Glueck | @Guest: Thanks for your comment. I've answered your question on Mathematics Stackexchange. | |
Jun 1, 2021 at 11:29 | comment | added | Guest | Why is pre-image also an operator range? I think it relates to my question math.stackexchange.com/questions/4155444/… | |
May 21, 2021 at 13:14 | comment | added | Jochen Glueck | Thanks for the additional information, @bathalf15320 and NateEldregde! | |
May 21, 2021 at 12:58 | vote | accept | Jochen Glueck | ||
May 21, 2021 at 4:22 | comment | added | bathalf15320 | As a matter of historical reference, the results you require are written up in Buchwalter's thesis which is available online. The chapter of interest to you ("Espaces vectoriels bornologiques") is in Publ. Dép. Math. Soc. 6 (1965) 2-53--particularly relevant are the theorems 2,3.9, 2.7.3 and 2.7.7. This is couched in a different language (not referring to French rather than English!) and the central concept is that of a "disque complétant" ("Banach disk" in the answer of Jochen Wengenroth) which goes back to Waelbroeck (1961). | |
May 20, 2021 at 20:21 | answer | added | Jochen Wengenroth | timeline score: 3 | |
May 20, 2021 at 19:27 | comment | added | Nate Eldredge | Correction: the "pretty easy to see" fact is called Pettis' lemma and while the proof is short and elementary, it's maybe not exactly obvious except in retrospect. | |
May 20, 2021 at 19:12 | comment | added | Nate Eldredge | Possibly relevant remark: if we restrict to separable Banach spaces, then operator ranges are analytic sets, and by a standard but nontrivial theorem of descriptive set theory, analytic sets have the property of Baire (BP). And using the Baire category theorem, it's pretty easy to see that a nonmeager linear subspace having the BP must have nonempty interior, and therefore equal all of $F$. This is probably overkill for you, though. | |
May 20, 2021 at 16:41 | comment | added | Jochen Glueck | @DirkWerner: Thank you for your comment! Yepp, I've check this paper before I posted the question (the paper was kindly mentioned by Mikhail Ostrovskii as an answer to an earlier question of mine; there he also confirms that there is no part II). I couldn't find (4) or (5) in this paper, though, and the focus of the article seems to be a bit different. | |
May 20, 2021 at 16:36 | comment | added | Dirk Werner | Have you checked this paper? Cross, R. W.; Ostrovskij, M. I.; Shevchik, V. V., Operator ranges in Banach spaces. I. (English) Zbl 0834.47001, Math. Nachr. 173, 91-114 (1995). (Part II does not seem to exist.) | |
May 20, 2021 at 16:23 | history | asked | Jochen Glueck | CC BY-SA 4.0 |