Timeline for How to prove $\operatorname{Id}-K$ is a proper map when $K$ is a $C^1$ compact operator?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Dec 20, 2023 at 2:01 | comment | added | boundary | Thank you very much! @MartinVäth | |
| Dec 19, 2023 at 19:29 | comment | added | Martin Väth | Another often-used sufficient condition for unbounded $\Omega$ is that $\lVert K(x)\rVert/\lVert x\rVert\to\infty$ as $\lVert x\rVert\to\infty$ (which implies that $(I-K)^{-1}(B)$ is bounded for bounded $B$). | |
| Dec 19, 2023 at 19:25 | comment | added | Martin Väth | A sufficient condition is of course that $K(\Omega)$ is relatively compact. If you use the usual definition of compact map (maps bounded sets into relatively compact sets), it is thus sufficient that $\Omega$ is bounded, | |
| Dec 19, 2023 at 3:48 | comment | added | boundary | Is there any sufficient condition？ | |
| Dec 19, 2023 at 3:47 | comment | added | boundary | yeah, I agree with you. | |
| Dec 19, 2023 at 3:35 | comment | added | Willie Wong | In the case $X = \Omega = \mathbb{R}$, the mapping $K = Id$ is compact, but $Id - K$ is not proper. | |
| Dec 19, 2023 at 2:48 | history | edited | boundary |
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| S Dec 19, 2023 at 2:21 | review | First questions | |||
| Dec 19, 2023 at 10:28 | |||||
| S Dec 19, 2023 at 2:21 | history | asked | boundary | CC BY-SA 4.0 |