Timeline for Is the "closedness of the image of operator" needed in the defintion of Fredholm operators?
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
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| S Mar 2, 2019 at 12:53 | history | suggested | Ali Taghavi |
I added a tag "operator theory"
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| Mar 2, 2019 at 12:14 | review | Suggested edits | |||
| S Mar 2, 2019 at 12:53 | |||||
| Jul 16, 2016 at 12:36 | answer | added | M.González | timeline score: 4 | |
| Jul 15, 2016 at 16:41 | answer | added | Peter | timeline score: 1 | |
| Feb 11, 2011 at 1:19 | comment | added | Theo Buehler | Yes, the cokernel has to be taken in the "algebraic" sense 1, not in the "functional analytic" sense 2. To see this, take e.g. the inclusion of the Sobolev space $H^1$ in $L^2$, which is dense, so it has zero cokernel in the second sense. | |
| Feb 11, 2011 at 1:10 | history | edited | madmath | CC BY-SA 2.5 |
added 507 characters in body; edited title; added 8 characters in body
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| Feb 10, 2011 at 5:38 | comment | added | Yemon Choi | You might also look at Theorem 1.4.7 of Murphy's "C*-algebras and operator theory", which says more or less what Chad Groft's answer does | |
| Feb 10, 2011 at 3:57 | answer | added | Chad Groft | timeline score: 9 | |
| Feb 10, 2011 at 3:26 | history | asked | madmath | CC BY-SA 2.5 |