Timeline for Existence of a bounded right inverse to a linear closed surjective operator
Current License: CC BY-SA 4.0
17 events
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| Jun 23, 2020 at 10:46 | history | edited | an_ordinary_mathematician | CC BY-SA 4.0 |
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| Jun 20, 2020 at 18:57 | vote | accept | an_ordinary_mathematician | ||
| Jun 20, 2020 at 18:57 | vote | accept | an_ordinary_mathematician | ||
| Jun 20, 2020 at 18:57 | |||||
| Jun 20, 2020 at 12:27 | answer | added | Jochen Wengenroth | timeline score: 5 | |
| Jun 20, 2020 at 8:18 | history | edited | an_ordinary_mathematician | CC BY-SA 4.0 |
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| Jun 20, 2020 at 8:05 | vote | accept | an_ordinary_mathematician | ||
| Jun 20, 2020 at 18:57 | |||||
| Jun 20, 2020 at 7:09 | comment | added | Giorgio Metafune | @Nik Weaver Yes, I am only dealing with closed subspaces in general Banach spaces. In my situation $jq(X)$ is closed since coincides with the kernel of $I-jq$. | |
| Jun 19, 2020 at 23:33 | comment | added | Nik Weaver | @GiorgioMetafune that depends on what you mean by "uncomplemented". The image of this projection need not be a closed subspace, but every subspace of a pre-Hilbert space has an unclosed algebraic complement. | |
| Jun 19, 2020 at 23:30 | answer | added | Nik Weaver | timeline score: 4 | |
| Jun 19, 2020 at 20:57 | comment | added | Giorgio Metafune | If the quotient map $q:X \to X/Y$ has a right inverse $j:X/Y \to X$ then $jq:X \to X$ is a projection whose kernel is $Y$. Such a $j$ does not exist if $Y$ is uncomplemented in $X$. | |
| Jun 19, 2020 at 14:30 | comment | added | Michael Renardy | Differentiation on $L^2(0,\infty)$ is not surjective. | |
| Jun 19, 2020 at 14:09 | comment | added | Mateusz Wasilewski | Maybe differentiation on $L^2(0,\infty)$ would work as a counterexample? The right inverse would essentially have to be the antiderivative and unboundedness of the domain should show that it is unbounded. I don't have time to check the details now, sorry. | |
| Jun 19, 2020 at 13:22 | comment | added | Nik Weaver | The kernel is closed but I don't think it has to be complemented in $D_A$. I smell a counterexample (at least if you want $R$ to be bounded) but I don't have time to think about it now ... | |
| Jun 19, 2020 at 13:12 | comment | added | an_ordinary_mathematician | Ok now I see even if the operator is not bounded the kernel is closed, right? | |
| Jun 19, 2020 at 13:09 | comment | added | an_ordinary_mathematician | But then how do you know that the quotient operator is still closed? | |
| Jun 19, 2020 at 12:37 | comment | added | Mateusz Wasilewski | Does something go wrong if you simply quotient out the kernel? Then the induced operator is bijective. | |
| Jun 19, 2020 at 11:38 | history | asked | an_ordinary_mathematician | CC BY-SA 4.0 |