Timeline for Continuity + $H^1$ + Laplacian control $ \implies$ local Lipschitz property
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 28, 2016 at 21:28 | comment | added | student | @MichaelRenardy Further edited. | |
| Jan 28, 2016 at 21:28 | comment | added | student | @ConnorMooney Further edited. | |
| Jan 28, 2016 at 21:28 | comment | added | student | @DenisSerre Sorry for changing the question on you, I have added a note and edited it further. | |
| Jan 28, 2016 at 21:03 | comment | added | Denis Serre | When I answered the question, the assumption was pointwise, not away from a hypersurface. | |
| Jan 28, 2016 at 20:24 | comment | added | Connor Mooney | The function $r^{1/2}sin(\theta/2)$ on $\mathbb{R}^2$ satisfies the desired conditions (it's harmonic away from the positive $x$-axis, continuous and $H^1$) and is not Lipschitz. In this example the gradient is discontinuous on $\Sigma$ so (as Michael Renardy's comment points out) the Laplace concentrates there. | |
| Jan 28, 2016 at 19:51 | comment | added | Michael Renardy | You do not know that $\Delta u\in L^2$. $\Delta u$ might include a surface delta function if $\nabla u$ is discontinuous across $\Sigma$. | |
| Jan 28, 2016 at 18:11 | comment | added | student | Actually I wrote the question a bit wrong. Of course, statements like $|\Delta u|^2 \leq c|\nabla u|^2$ make no sense pointwise if $u$ is just continuous and $H^1$. It turns out that $u$ satisfies the above inequality pointwise everywhere except on an embedded hypersurface. But I think what you wrote still goes through? | |
| Jan 28, 2016 at 16:59 | history | answered | Denis Serre | CC BY-SA 3.0 |