Skip to main content
12 events
when toggle format what by license comment
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
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