Timeline for A non-condensing operator with a power condensing
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Sep 26, 2019 at 19:02 | history | bounty ended | CommunityBot | ||
| Sep 23, 2019 at 15:28 | comment | added | Motaka | Ok I get it!Many Thanks.. | |
| Sep 23, 2019 at 15:05 | comment | added | Jochen Glueck | @Motaka: My motivation was as follows: You said that you find Example 1 too trivial for your purposes. However, Example 2 is, in a sense, just a product of two trivial examples (a nilpotent mapping and a compact mapping), and thus somewhat trivial on its own. The point about Example 3 is that we cannot simply split off a nilpotent part here because the images of all powers of $A$ are dense. | |
| Sep 23, 2019 at 14:38 | comment | added | Motaka | I mean why we should be interested in this property? | |
| Sep 23, 2019 at 13:17 | comment | added | Jochen Glueck | @Motaka: Just to make sure I understand your question correctly: Do you want to know why $A^n(E,E)$ is dense in $E$ in Example 3, or do you want to know why we should be interested in this property? | |
| Sep 23, 2019 at 10:59 | vote | accept | Motaka | ||
| Sep 23, 2019 at 10:58 | comment | added | Motaka | Great! Thanks Jochen ..I have a little question, why do we need $A^n(E,E)$ to be dense in $E$ in the third example? | |
| Sep 19, 2019 at 20:10 | comment | added | Jochen Glueck | @Motaka: Well, what is trivial and what is not depends on the perspective. I've added two more examples with increasing level of complexity, so that you can choose the level of "non-triviality" that is required for your purposes. ;-) | |
| Sep 19, 2019 at 20:08 | history | edited | Jochen Glueck | CC BY-SA 4.0 |
Added two further examples in response to a comment of the OP.
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| Sep 19, 2019 at 16:42 | comment | added | Motaka | If we can construct an operator such that one of its power is compact, it xill be great | |
| Sep 19, 2019 at 16:40 | comment | added | Motaka | Thank you for your reply. It's a good idea, and what you have written is absolutely true. But the nilpotant case is somehow "trivial": the fact that $A$ is nilpotant makes the condition $A^n$ is a condensing operator that we look for trivial . So It's not the answer that I look for, unfortunately! | |
| Sep 18, 2019 at 18:51 | history | answered | Jochen Glueck | CC BY-SA 4.0 |