Timeline for If $u_m \rightharpoonup u$, how to show using monotonicity that $f(u_m) \rightharpoonup f(u)$?
Current License: CC BY-SA 3.0
22 events
| when toggle format | what | by | license | comment | |
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| Dec 2, 2014 at 3:24 | answer | added | Michael Renardy | timeline score: 1 | |
| Mar 6, 2014 at 17:52 | answer | added | Liviu Nicolaescu | timeline score: 0 | |
| Mar 6, 2014 at 16:51 | comment | added | maximumtag | @LiviuNicolaescu I checked these books. I don't think my desired result is true. Even if $f:L^2(0,T;H^1) \to L^2(0,T;H^1)$ were maximal monotone it does not seem ot be enough. | |
| Feb 24, 2014 at 10:31 | comment | added | Liviu Nicolaescu | Viorel Barbu has two books on monotone operators and evolution equations. Try any of them. The older one, in the 70s is still THE best reference on the subject. Brezis also has a good book (in French) on this subject, also published published in the 70s. | |
| Feb 23, 2014 at 18:24 | comment | added | username | If $f$ is nicer, i.e. you have a compact embedding into $L^2((0,T)\times\Omega)$ then of course you are done, because the convergence for $u$ is strong. | |
| Feb 23, 2014 at 18:18 | comment | added | maximumtag | @AthanagorWurlitzer Oh I see. Well, my $f$ is a bit nicer than being a monotone operator so that may help. Liviu says the result holds so I guess I am missing something special about $f$ | |
| Feb 23, 2014 at 18:14 | comment | added | username | In $L^2(\Omega)$ yes, but not in $L^2(0,T)$ that's my point: if the result does not hold for constant in space functions, you are in trouble. | |
| Feb 23, 2014 at 18:10 | comment | added | maximumtag | @AthanagorWurlitzer I don't think you can, unless the convergence is in $H^1$ so we obtain strong convergence in $L^2$ via compact embedding. No monotonicity trick I know applies where both convergences are weak. | |
| Feb 23, 2014 at 17:58 | comment | added | username | My previous comment was silly. Can you solve the problem if you consider functions of time on $(0,T)$? I don't see what the space variable part changes. | |
| Feb 23, 2014 at 17:07 | comment | added | maximumtag | @LiviuNicolaescu Please can you give more details? I searched Showalter, Zeidler etc. and can find nothing appropriate to use. | |
| Feb 23, 2014 at 16:56 | comment | added | maximumtag | @AthanagorWurlitzer Sorry, why does the second convergence hold? | |
| Feb 23, 2014 at 15:11 | comment | added | Liviu Nicolaescu | There is a general result about maximal monotone operators between Hilbert spaces that implies this. | |
| Feb 23, 2014 at 3:02 | comment | added | Kelei Wang | Is this related to Yougn measure? I suspect the answer is no. Perhaps you can try some oscillating functions, which converges weakly to $0$, but after composition with $f$, the norm appears. | |
| Feb 22, 2014 at 20:49 | comment | added | maximumtag | @RicardoAndrade I agree, but at least the downvoter should say that it it is trivial. | |
| Feb 22, 2014 at 19:53 | comment | added | Ricardo Andrade | Dear @maximumtag, please understand that if the question is trivial, it is likely to be off-topic at MathOverflow. | |
| Feb 22, 2014 at 18:25 | history | edited | maximumtag | CC BY-SA 3.0 |
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| Feb 22, 2014 at 17:32 | review | Close votes | |||
| Feb 24, 2014 at 21:11 | |||||
| Feb 22, 2014 at 17:17 | comment | added | maximumtag | Not sure why this got downvoted. If the question is trivial please post the answer! | |
| Feb 22, 2014 at 17:13 | history | edited | maximumtag | CC BY-SA 3.0 |
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| Feb 22, 2014 at 17:12 | comment | added | maximumtag | @AlexDegtyarev Apologies, $X$ is a Hilbert space. All convergences are weak convergences, not strong. | |
| Feb 22, 2014 at 16:48 | comment | added | Alex Degtyarev | What is $X$? What is $(,)$? If $f$ it continuous, $f(u_m)\to u$ by definition. Thus, $f(u)=f(v)$. If $f$ is also invertible, you've got to have $u=v$. | |
| Feb 22, 2014 at 16:29 | history | asked | maximumtag | CC BY-SA 3.0 |