Timeline for On the weak derivative of $|u|^{(p-2)/2}u$
Current License: CC BY-SA 4.0
24 events
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| Apr 14 at 16:23 | comment | added | Ayman Moussa | I still believe that the boundary values are irrelevant here, since you can replace $u$ by $gu$ where $g$ is some smooth bump function, and yet the question remains the same (you can choose $g$ to be equal $1$ on a small neighborhood of your choice). Of course you manage to prove that (2) and (3) $=u^{(p-2)/2}u\in H^1_0(G)$ are compatible statements, but this does not mean that they are equivalent. My point is precisely to question the implication (3) $\Rightarrow$ (2), and even (3) $\Rightarrow$ (1) as I have no way to understand (1) without assuming Sobolev regularity on $u$. | |
| Apr 14 at 15:28 | comment | added | Perelman | There seem to be a mistake also in the given counterexample. Namely boundary values certainly play a role. If you assume $u(x)=x^{1/\alpha}$ then $u^\alpha=x$ but $u^\alpha(1)=1\neq 0$. So funilly $u^\alpha$ cant be contained in $H^1_0(0,1)$. In the mean time I could prove that if $G$ is bounded and $u$ satisfies $(2)$ then necessarily $u^{(p-2)/2}u \in H_0^1(G)$. So this two things can occure at the same time and the derivative is given as in $(1)$. | |
| Mar 31 at 19:33 | comment | added | Perelman | @AymanMoussa absolutely, but I hope that its true because it would have significant impact on other results, Im interested in, that rely on that lemma. But offcourse anything could be, it could even be wrong, so I appreciate any sort of clarification. | |
| Mar 31 at 19:20 | comment | added | Ayman Moussa | The number of citation of the paper and the reputation of Pierre-Arnaud are a bit off topic, don't you think ? :) We're speaking of a tiny technical lemma, which apparently misses an assumption, this does not mean that the remaining (interesting) part of the article is wrong ... | |
| Mar 31 at 18:25 | comment | added | Perelman | @AymanMoussa Thank you very much for your efforts. But keep in mind that this article link.springer.com/article/10.1007/BF00281422 has been around for a while now and in google scholar alone, it was cited by 44 others. I would be surprised if the statement is completely wrong. Also Raviart is a distinguished mathematician. | |
| Mar 31 at 17:34 | history | edited | Ayman Moussa | CC BY-SA 4.0 |
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| Mar 31 at 17:06 | comment | added | Ayman Moussa | @Perelman I thought that something was wrong and erased the edit, I'll try to rectify it and post it tonight ! | |
| Mar 31 at 16:50 | comment | added | Perelman | @AymanMoussa, again a beautiful answer, thank you. It also provided me with valuable insights and allowed me to review Lebesgue theory. However, it is not necessary for $u$ or its derivative to be integrable over the entire domain $(0,1)$. It's only a matter of whether it's appropriate to write $D(u)$, for which $u$ and $D(u)$ need only be locally integrable. If you integrate $\sin(1/x)/x$ over $[a,b]\subseteq (0,1)$, it is Riemann integrable and hence Lebesgue integrable over $[a,b]$. So, (1) is not wrong at all. | |
| Mar 31 at 16:47 | history | edited | Ayman Moussa | CC BY-SA 4.0 |
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| Mar 31 at 15:48 | history | edited | Ayman Moussa | CC BY-SA 4.0 |
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| Mar 31 at 15:36 | history | edited | Ayman Moussa | CC BY-SA 4.0 |
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| Mar 31 at 15:22 | comment | added | Ayman Moussa | Wow. That escalated quickly. Thanks Iosif and Willie for your messages that I understood as a careful concern for MO's community. For Perelman, I reassure you : I did not waste that much of a time ... I'll try to see if I get a complete answer :D | |
| Mar 31 at 14:59 | comment | added | Iosif Pinelis | @Perelman : Your post again contains two questions, which is against the sense of the previously mentioned guidelines. Also, keeping two questions still undermines, unfairly, Ayman Moussa's answer. | |
| Mar 31 at 14:51 | comment | added | Iosif Pinelis | @Perelman : I am glad to see that you did the right thing and accepted the answer. However, I don't see any evidence that "Ayman Moussa was very aware that this wasn't an answer to [your] question", "an answer at all". To the contrary, he wrote "Only a partial answer : (2) seems strange". Indeed, he provided an answer to your question (2). I agree with you that we should always try to use common sense. Also, we should carefully recheck our questions before posting them, to avoid posting "ill-posed [questions that] needed correction" and thus avoid wasting other people's time. | |
| Mar 31 at 14:11 | vote | accept | Perelman | ||
| Mar 31 at 14:11 | comment | added | Perelman | @IosifPinelis I'm sure Ayman Moussa was very aware that this wasn't an answer to my question, as he refers to it as a partial answer. Actually, it isn't an answer at all; it just clarified that the question was ill-posed and needed correction. I suggest that you don't follow guidelines blindly but rather use your common sense. Nevertheless I will accept his answer because he invested time to provide help and his post helped me still to takle the actual problem. | |
| Mar 31 at 13:27 | comment | added | Iosif Pinelis | Previous comment continued: So, the best way to proceed here is to accept Ayman Moussa's answer, and then perhaps post the remaining question separately. That would also be in accordance with the spirit, if not the letter, of these other guidlenes. | |
| Mar 31 at 13:27 | comment | added | Iosif Pinelis | @Perelman : In addition to Willie Wong 's comment: According to these MathOverflow guidelines, users should "avoid trying to answer questions which [...] request answers to multiple questions". You had two questions, without saying which one you considered "[t]he main question". Ayman Moussa provided what you called "[n]ice counter example" to your second question. | |
| Mar 31 at 8:46 | comment | added | Perelman | @WillieWong Nevertheless, thank you for raising awareness. | |
| Mar 31 at 8:43 | comment | added | Perelman | @WillieWong The old version of the question is essentially similar to the current one. The old one was stated in a way that didn't make sense. The main question didn't change at all. Furthermore, the answer from AymanMoussa, which was provided here, didn't answer the actual question but helped to realize that the post must be edited. I gave him credit for that. If others now worked out a correct answer to the old version, it would imply that it is also an answer to the new version. Therefore, I will not make a new post. | |
| Mar 30 at 23:42 | comment | added | Willie Wong | @Perelman: on MathOverflow it is generally considered bad practice to edit the question to invalidate an answer after others have taken the time to help you. I would suggest reverting the question to the original form, and then asking a new question that is of the current form. | |
| Mar 30 at 20:16 | comment | added | Perelman | Nice counter example. It seems that (2) can not hold with these assumptions, thank you. I edited the question accordingly. | |
| Mar 30 at 19:09 | history | edited | Ayman Moussa | CC BY-SA 4.0 |
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| Mar 30 at 18:27 | history | answered | Ayman Moussa | CC BY-SA 4.0 |