Timeline for Monotonically increasing and bounded function is in $BV_{loc}$?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Apr 16, 2021 at 10:49 | comment | added | user99432 | Thank you very much! | |
| Apr 16, 2021 at 10:42 | comment | added | leo monsaingeon | In the future if you crosspost to math.MO a question initially posted on math.SE that never received any answer then DO mention it in the MO post, and include the link. | |
| Apr 16, 2021 at 10:41 | review | Close votes | |||
| Apr 30, 2021 at 20:44 | |||||
| Apr 16, 2021 at 10:40 | comment | added | leo monsaingeon | yes, exactly, you tried to compute the ac part. But proving that the a.e.-well-defined pointwise derivative coincides with the absolutely continuous part of the distributional derivative (the measure) in the Lebesgue decomposition is not immediate. and indeed $\|f_n'\|_{TV}\leq 1$ gives that $f_n$ is BV, since by definition $\|f_n\|_{BV(K)}=\|f_n\|_{L^1(K)}+\|f_n'\|_{TV(K)}$ on any compact set $K$ and my "argument" proves that this quantity is finite | |
| Apr 16, 2021 at 10:35 | comment | added | user99432 | Thanks for your answer. I did ask it on math SE. Having $\|f_n'\|=1$ for all $n\in \mathbb{N}$ can I conclude that $f_n$ is bounded in $BV$ ? Maybe like this: Since $f_n'$ defines a measure, I can decompose it by Lebesgue's decomposition into two measures, the absolutely continuous part and the singular part. What I tried to compute is then just the absolutly continuous part ? | |
| Apr 16, 2021 at 10:19 | comment | added | leo monsaingeon | This is more suited for math.SE, please consider cross-posting there. As far as your question is concerned: your first integration by parts is not legitimate to start with, since the distributional derivative $f_n'$ can contain singular parts. But because $f_n'\geq 0$ in the sense of distributions you know it must be a measure, with moreover $\|f_n'\|_{TV}=1-0$ due to your "boundary conditions" at infinity. In particular your first integration by parts should read accordingly $\int \phi(x) df_n'(x)$, instead of using the absolutely continuous measure $f_n'(x) dx$ as you did | |
| Apr 16, 2021 at 9:47 | review | First posts | |||
| Apr 16, 2021 at 9:59 | |||||
| Apr 16, 2021 at 9:31 | history | asked | user99432 | CC BY-SA 4.0 |