Timeline for Does a weakly convergent sequence in $W^{1,p}(B_1)$ which also converges in $C^{0,\alpha}(B_1)$ converges strongly in $W^{1,p}(B_1)$?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Jul 2, 2021 at 16:36 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
Typo fixing
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| Jul 2, 2021 at 14:03 | vote | accept | Harish | ||
| Jul 2, 2021 at 13:54 | comment | added | Leo Moos | I defined them in my answer below. | |
| Jul 2, 2021 at 13:43 | answer | added | Leo Moos | timeline score: 2 | |
| Jul 2, 2021 at 13:39 | comment | added | Harish | @LeoMoos Please let me know the definition of zig zag function you are considering. | |
| Jul 2, 2021 at 13:38 | comment | added | Leo Moos | Are you sure? The Lipschitz norm is $1 + 1/n$, so that seems strange. I have the Holder norm decaying like $n^{\alpha-1}$. | |
| Jul 2, 2021 at 13:36 | comment | added | Harish | @LeoMoos Thanks for the comment, but the example you mentioned does not have a uniform bound on Holder norm. In fact the Holder norm of $f_n$ blows up as $n\rightarrow \infty$ | |
| Jul 2, 2021 at 13:22 | comment | added | Leo Moos | I don't think so. Let $f_n: [0,1] \to \mathbf{R}$ be a zig-zag function with step size $1/n$ and $\lvert f \rvert_\infty = 1/n$. Then every $f_n$ is bounded as a Lipschitz function, and the sequence converges to zero weakly in $W^{1,p}$ and strongly in $C^{0,\alpha}$ for all $\alpha \in (0,1)$. However $\lvert f_n \rvert_{1,p} \geq 1$ for all $n$ because $f_n' = 1$ almost everywhere. | |
| Jul 2, 2021 at 13:06 | history | asked | Harish | CC BY-SA 4.0 |