Timeline for A possible trace (inequality) defined under negative Sobolev scale
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Jul 25, 2015 at 6:47 | comment | added | Johannes Hahn | The brackets are evaluations of distributions of course! That's how Sobolev spaces with negative orders are defined! | |
| Jul 24, 2015 at 17:23 | comment | added | BLM | In order that the RHS of OP's suggested definition defines a trace, the two evaluations, appearing in the RHS, must be defined independently of the LHS. In the original setting of the standard trace theorem, we have two inner products both with only functions in the Hilbert space $L^2(\Omega)$. If you want to relax the constraints on $g$ you should define how to evaluate the $(\nabla f, \nabla g)$ term appearing in the definition for an arbitrary $g$ of the new class. | |
| Jul 22, 2015 at 20:39 | comment | added | Johannes Hahn | ... and then take a limit $\epsilon\to 0$ in some suitable space of distributions on $\partial\Omega$. It seems perfectly reasonable to ask if such a limit exists and I do not see how the negativity of the exponent should cause any trouble. So the result will be a distribution, not a proper function. So what? | |
| Jul 22, 2015 at 20:38 | comment | added | Johannes Hahn | I do not speculate why the OP hasn't found any satisfying theorems since functional analysis is not my speciality. Maybe he just had bad luck with his search terms... In any case, it is still a reasonable question. The RHS of OP's equation is well-defined and linear on any $H^t(\Omega)$ and if one can verify the necessary continuity and density arguments, it can be used to define a trace operator. There are other possible avenues: Consider for example some open neighborhoods of $\partial\Omega$ of width $\epsilon$ (positive measure, so well-defined restrictions exist !) ... | |
| Jul 22, 2015 at 17:25 | comment | added | BLM | @Johannes, I hope we agree that the original poster assumes $s$ to be positive and wants to take $g\in H^{-s}(\Omega)$ (not in $H^{-s}(\partial\Omega)$). Are you suggesting that the expression OP proposes correctly defines a boundary trace $Tr(g)\in H^t(\partial\Omega)$ for some negative $t$? In my opinion, OP's question was why he cannot find trace theorems like such in the literature. | |
| Jul 18, 2015 at 16:54 | review | Late answers | |||
| Jul 18, 2015 at 16:55 | |||||
| Jul 18, 2015 at 16:48 | comment | added | Johannes Hahn | It isn't necessary to evaluate functions pointwise to make sense of boundary values. This is the whole point of the trace operator $H^1(\Omega)\to H^{1/2}(\partial\Omega)$ that it can be sensibly defined without point evaluations. | |
| Jul 18, 2015 at 16:32 | review | First posts | |||
| Jul 18, 2015 at 16:34 | |||||
| Jul 18, 2015 at 16:27 | history | answered | BLM | CC BY-SA 3.0 |