Timeline for Is an $H_0^1$ function continuous to the boundary if it is continuous in the interior?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Feb 3, 2021 at 19:43 | vote | accept | Ben Ciotti | ||
| Feb 3, 2021 at 19:43 | vote | accept | Ben Ciotti | ||
| Feb 3, 2021 at 19:43 | |||||
| Feb 3, 2021 at 5:33 | comment | added | Connor Mooney | Unfortunately not- the modification $u(x) = (1-|x|^2)\frac{x_3}{|x|^{5/4}}$ is unbounded near the origin but still in $H^1_0(\Omega) \cap C^{\infty}(\Omega)$ by the same argument as above. | |
| Feb 3, 2021 at 5:26 | comment | added | Ben Ciotti | Very nice, thank you. No wonder I was having trouble proving it. I do have a follow-up which I shall also addend to my original post: Can I at least get that $u$ is bounded and/or attains its maximum on $\overline\Omega$? | |
| Feb 3, 2021 at 4:53 | comment | added | Connor Mooney | One can show that $u \in H^1(B_1)$ by direct calculation. To show that $u \in H^1_0(\Omega)$, let $\phi \in C^{\infty}_0(B_1)$ with $\phi = 1$ in $B_{1/2}$, and argue that $(1-\phi(R\cdot))u$ (which are smooth, hence obviously in $H^1_0(\Omega)$) tend to $u$ in $H^1(\Omega)$ as $R \rightarrow \infty$. | |
| Feb 3, 2021 at 4:21 | history | edited | Connor Mooney | CC BY-SA 4.0 |
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| Feb 3, 2021 at 4:03 | comment | added | Ben Ciotti | Thanks Connor. How do you know your $u(x)$ is in $H_0^1$? | |
| Feb 3, 2021 at 3:31 | history | edited | Connor Mooney | CC BY-SA 4.0 |
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| Feb 3, 2021 at 3:10 | history | answered | Connor Mooney | CC BY-SA 4.0 |