Timeline for Bounding trace operator from below
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 23, 2021 at 4:22 | comment | added | Willie Wong | Though, perchance you made a typo and meant the quotient $H^1(\Omega) / \mathrm{ker}(\gamma_0)$ and not the set difference? | |
| Nov 23, 2021 at 4:20 | comment | added | Willie Wong | Just think in terms of finite dimensional inner product spaces. Given a space $V$ and a subspace $W$, there is a huge difference between $V\setminus W$ and $W^\perp$. The former, for example, is usually not a vector space. | |
| Oct 27, 2021 at 4:14 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
(Very9 Minor Math Jaxing (used $\|\cdot\|$ instead of $||\cdot||$)
|
| Oct 26, 2021 at 22:20 | comment | added | bobinthebox | Btw, in my case $\ker(\gamma_0)=H_0^1(\Omega)$, so I think that writing $u\in H^1(\Omega) \setminus \ker(\gamma_0)$ as I did makes sense: I am indeed excluding all the trace-zero functions and my $\gamma_0$ is still injective and surjective. Why do I need to write $(\ker \gamma_0)^\perp$? @LiviuNicolaescu | |
| Oct 26, 2021 at 22:11 | comment | added | bobinthebox | @LiviuNicolaescu Thanks so much, now I see the implications. The fact that $(\ker \gamma_0)^\perp\to H^{1/2}$ is bijective and continuous implies that the inverse is bounded, which is equivalent to say that it's bounded from below. | |
| Oct 26, 2021 at 21:58 | comment | added | Liviu Nicolaescu | The map from $(\ker \gamma_0)^\perp\to H^{1/2}$ is bijective and continuous | |
| Oct 26, 2021 at 19:02 | comment | added | bobinthebox | I mean, to have a bound from below, I need $\gamma_0$ to be injective and to have a closed range, and I don't see why in this case the range of $\gamma_0$ is closed @LiviuNicolaescu | |
| Oct 26, 2021 at 18:29 | comment | added | bobinthebox | @LiviuNicolaescu Thanks, I need a last clarification. How do you use the open mapping to show that bound? I really cannot see how | |
| Oct 26, 2021 at 17:40 | comment | added | Liviu Nicolaescu | You may want to assume $u\in (\ker \gamma_0)^\perp$. In this case the inequality follows from the open mapping theorem since $\gamma_0: (\ker \gamma_0)^\perp\to H^{1/2}$ is onto. Here $\perp$ refers to the orthogonal complement in the Hilbert space $H^1(\Omega)$ | |
| Oct 26, 2021 at 17:23 | comment | added | Christian Remling | Piero's example is convincing, of course, but the inequality also doesn't make sense on general grounds, because we can always make the RHS large by giving $u$ extra oscillations in the interior of our domain. | |
| Oct 26, 2021 at 16:32 | comment | added | Piero D'Ancona | I have doubts. Take a $u\in H^1_0(\Omega)$ and consider the sequence $u_\epsilon=u+\epsilon$ | |
| Oct 26, 2021 at 16:22 | history | asked | bobinthebox | CC BY-SA 4.0 |