Timeline for A property of one-parameter groups of operators
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Feb 8, 2020 at 11:39 | history | edited | YCor |
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| Feb 7, 2020 at 17:01 | history | edited | YCor | CC BY-SA 4.0 |
retagged, minor formatting, slightly renamed title
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| Aug 13, 2014 at 12:17 | history | edited | user144542 | CC BY-SA 3.0 |
edited title
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| May 8, 2014 at 12:23 | vote | accept | user144542 | ||
| May 8, 2014 at 9:50 | answer | added | gsa | timeline score: 3 | |
| May 8, 2014 at 7:22 | comment | added | András Bátkai | My first comment was on the semigroup case, where you can have many weird things happening. For bounded $A$ you can modify the finite dimensional case. I will return to your question of the $C_0$-group case soon. | |
| May 8, 2014 at 0:26 | comment | added | user144542 | @Andr My problem is I want a $C_0$ group $T(t)$ or a function $e^{tA}$ and a vector $x \in X$ such that $T(t)x$ is bounded on $\mathbb{R}$, nontrivial and $\inf_{t \in \mathbb{R} }|T(t)x|=0$. It seems like I cannot modify the nilpotent translation semigroup you gave to make it a group. I am interest in the solutions on the whole real line, since $\inf_{t \geq 0 }|T(t)x|=0$ is of course simply verified by exponentially stable orbits. Thanks ! | |
| May 7, 2014 at 19:07 | comment | added | András Bátkai | The two answers to this question: mathoverflow.net/q/164182/12898 will give you full and detailed answer. | |
| May 7, 2014 at 18:34 | comment | added | Loïc Teyssier | Have you tried to reproduce the proof for finite dimesional spaces in the case of a general Banach space ? The arguments may differ but you should either be able to adapt it or to pinpoint crucial steps that may help you find a counter-example. | |
| May 7, 2014 at 15:49 | review | First posts | |||
| May 7, 2014 at 15:54 | |||||
| May 7, 2014 at 15:46 | comment | added | Christian Remling | The question is not really meaningful for a semigroup when you can only consider $t\ge 0$, and of course $\inf_{t\ge 0} \|x(t)\|=0$ wouldn't be unexpected. | |
| May 7, 2014 at 15:33 | history | asked | user144542 | CC BY-SA 3.0 |