Timeline for Characterization of Besov space with Lp-modulus of continuity
Current License: CC BY-SA 4.0
12 events
when toggle format | what | by | license | comment | |
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Jul 21, 2019 at 2:25 | vote | accept | JohnLee | ||
Jul 20, 2019 at 6:09 | answer | added | Paata Ivanishvili | timeline score: 2 | |
Jul 20, 2019 at 5:55 | comment | added | Paata Ivanishvili | This was one way but it works for $p>1$. Another way is to use do what you did plus Fubini: $\int_{\mathbb{R}}\int_{x-h}^{h}|\nabla w|(s) ds dx = |h| \int_{\mathbb{R}} |\nabla w|(x) dx$ and this works just for all $p\geq 1$. | |
Jul 20, 2019 at 5:46 | comment | added | Paata Ivanishvili | Let $\tilde{M}$ be the uncentered maximal function then $|u(x+h)-u(x)|\leq \int_{x}^{x+h}|\nabla u| \leq |h| \tilde{M}|\nabla u| (x)$. Now take $p$'th, integrate in $x$ and apply Hardy--Littlewood. | |
Jul 20, 2019 at 1:52 | comment | added | Paata Ivanishvili | You can put maximal function at the very beginning (before you apply holder). Eventually you will not need to use holder at all. | |
Jul 20, 2019 at 0:46 | comment | added | JohnLee | If $|h|^{1/p}(\int_\mathbb{R}\mathrm{M}(|gradw|^p)(x)dx)^{1/p}$ is $|h|^{1/p}(\int_\mathbb{R}(\mathrm{M}(|gradw|)(x))^pdx)^{1/p}$, everything is okay, But it is not... | |
Jul 19, 2019 at 15:30 | comment | added | Paata Ivanishvili | Yes. For $p>1$ you can use the inequality $\|M(|\nabla f|)\|_{p}| \leq C_{p} \| \nabla f\|_{p}$. | |
Jul 19, 2019 at 14:54 | comment | added | JohnLee | I think I can use it. But is $\mathrm{M}(|gradw|^p)(x)\leq\mathrm{M}(|gradw|)^p(x)$ right? If it holds, I think the proof is complete. | |
Jul 19, 2019 at 14:27 | comment | added | Paata Ivanishvili | Why you cannot use maximal function inequality (the one that you wrote at the end)? | |
Jul 19, 2019 at 10:48 | history | edited | JohnLee | CC BY-SA 4.0 |
added 10 characters in body
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Jul 19, 2019 at 8:55 | review | First posts | |||
Jul 19, 2019 at 9:17 | |||||
Jul 19, 2019 at 8:52 | history | asked | JohnLee | CC BY-SA 4.0 |