Timeline for Is the kernel of a Fredholm operator stable under perturbation?
Current License: CC BY-SA 4.0
9 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 10:55 | history | edited | M.González | CC BY-SA 4.0 |
clarifying a point
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| Aug 22, 2018 at 10:52 | comment | added | M.González | I did not carefully read your question. If you assume $dim (\ker S_0)=dim (\ker S_t)<\infty$, closed range for all the operators, and $\|S_t-S_0\|\to 0$, then the distance from $\ker S_t$ to $\ker S_0$ tends to $0$. This is a consequence of the result I mentioned plus finite dimensión. | |
| Aug 22, 2018 at 10:17 | comment | added | Asaf Shachar | I am not sure I follow: In both your examples, $\dim(\ker S_0) \neq \dim(\ker S_t)$ for $t>0$. (while I assumed the dimension is independent of $t$). Anyway, I think that there is a chance that proposition 3.1 would suffice for my purposes (though I need to think some more about this). Thanks again for your help. | |
| Aug 22, 2018 at 10:11 | comment | added | M.González | $S_t:X_1\times X_2\times Y\to X_1\times X_2\times Y$ given by $S_t(x_1,x_2,y)=(0,tx_2,y)$, $X_1, X_2$ finite dimensional. | |
| Aug 22, 2018 at 10:05 | comment | added | M.González | You can modify $S_t$ $(t\neq 0$) so that all their kernels have the same positive dimensión. | |
| Aug 22, 2018 at 9:53 | comment | added | Asaf Shachar | Thanks for the reference. Regarding your counter-example, I assumed that all the kernels have the same (positive) dimension. | |
| Aug 22, 2018 at 9:50 | history | edited | M.González | CC BY-SA 4.0 |
deleted 18 characters in body
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| Aug 22, 2018 at 9:29 | history | answered | M.González | CC BY-SA 4.0 |