Timeline for Is the kernel of a Fredholm operator stable under perturbation?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Aug 26, 2018 at 9:23 | vote | accept | Asaf Shachar | ||
| Aug 22, 2018 at 9:29 | answer | added | M.González | timeline score: 2 | |
| Aug 22, 2018 at 8:28 | history | edited | Asaf Shachar | CC BY-SA 4.0 |
Added the assumption the operators are Fredholm of index $0$.
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| Aug 22, 2018 at 8:21 | comment | added | Asaf Shachar | @JochenGlueck Thanks, in my intended application the operator is the Laplacian on differential $k$-forms (on a Riemannian manifold), which is indeed Fredholm with index $0$. | |
| Aug 22, 2018 at 7:35 | comment | added | Jochen Glueck | Just a small remark: If a bounded operator has finite dimensional kernel and closed range, then it is a so-called upper semi-Fredholm operator. If for some reason, say in a concrete application, you know that $0$ is in the topological boundary of the spectrum, then it follows that the operator is even a Fredholm operator with Fredholm index $0$; maybe this is helpful to prove the desired stability result in this special case. | |
| Aug 22, 2018 at 7:17 | history | edited | Asaf Shachar | CC BY-SA 4.0 |
I mentioned explicitly that the kernels have positive dimension.
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| Aug 22, 2018 at 7:16 | comment | added | Hannes | Yes, sure, I just wanted to point out that the further setup doesn't fit to that case before someone else does it ;-) | |
| Aug 22, 2018 at 7:12 | comment | added | Robert Israel | @Hannes All the kernels are assumed to have the same dimension, so if that dimension is $0$ the kernels are all $\{0\}$ and the answer is trivially yes. | |
| Aug 22, 2018 at 7:12 | comment | added | Asaf Shachar | Yes, of course. I will mention this explicitly. | |
| Aug 22, 2018 at 7:10 | comment | added | Hannes | Probably you want to exclude the kernel $\{0\}$ from the setup? | |
| Aug 22, 2018 at 6:36 | history | asked | Asaf Shachar | CC BY-SA 4.0 |