Timeline for About a completion of a Sobolev space
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 16, 2015 at 17:52 | vote | accept | jamesC | ||
| Jan 22, 2015 at 14:42 | comment | added | jamesC | ..actually maybe the sequence is Cauchy (see the other answer).. | |
| Jan 22, 2015 at 12:47 | comment | added | jamesC | Thanks, let me think about it a bit. It seems a bit circular to me, extending the trace map to a space which is defined in terms of the trace map to be extended. Btw I made a mistake: the sequence I claimed is Cauchy is not Cauchy, so that one won't do. | |
| Jan 21, 2015 at 12:01 | history | edited | Joonas Ilmavirta | CC BY-SA 3.0 |
added 165 characters in body
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| Jan 21, 2015 at 11:59 | comment | added | Joonas Ilmavirta | @jamesC, there are many possible ways to make things rigorous. One is to redefine the trace map to $H^\epsilon(\mathcal C)\to L^2(\Omega)$ (using a cut-off $\phi$ and showing independence of the choice of the cut-off, or try to imitate the original definition of the trace map). Another option is to use the norm in your second comment (with the invariance of $\phi$). My idea was that the norm in $H^\epsilon$ will be given by the natural expression once you have extended the trace operator (which is also very natural). | |
| Jan 21, 2015 at 11:20 | comment | added | jamesC | To make this rigourous, should I not instead use as a norm $$\int\int_{\mathcal{C}}\epsilon |\nabla_x v|^2 + v_y^2 + \int_\Omega (tr_\Omega (\phi v))^2$$ and then show that this norm is independent of $\phi$ etc. Or is it the case that, upon completion wrt. the norm as it was, the norm in the completed space $H^\epsilon$ automatically will be taken care of? | |
| Jan 21, 2015 at 11:20 | comment | added | jamesC | Thanks for the answer. I understand that $tr_\Omega(\phi u) = tr_\Omega(u)$ for $u \in H^1(\mathcal{C})$. You say that the norm in $H^\epsilon(\mathcal{C})$ will be given by the natural extension of the expression $$\int\int_{\mathcal{C}}\epsilon |\nabla_x v|^2 + v_y^2 + \int_\Omega (tr_\Omega v)^2.$$ But, strictly speaking, although I do understand that the trace depends only on a neighbourhood around $y=0$, we still have $tr_\Omega: H^1(\mathcal{C}) \to H^{\frac 12}(\Omega)$. | |
| Jan 20, 2015 at 19:58 | history | answered | Joonas Ilmavirta | CC BY-SA 3.0 |