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Let $u \in W^{1,p}(\mathbb{R}^n) \cap L^{\infty}(\mathbb{R}^n)$ be a given function for some $1<p< \infty$ and let $k \in \mathbb{R}$ be any number and consider the following maximal function $$ \mathcal{M}_{\leq k}(|\nabla u|)(x) = \sup_{R>0}\frac{1}{|B_R(x) \cap \{u \leq k\}|} \int_{B_R(x)\cap \{u \leq k\}} |\nabla u| \ dx. $$

Question: Is it true that $\mathcal{M}_{\leq k}(|\nabla u|)(x)$ is finite almost everywhere?

Question 2: if the answer to the above question is true, then can the measure zero set where the maximal function is infinite be independent of k?
More specifically, can one find a measure zero set $E$ such that for all $k \in [a,b]$ in some closed interval, the above maximal function is finite outside $E$?

Question 3: in question 2, can I ask for a weaker conclusion, in that can $E$ be found such that $\mathbb{R}^n \setminus E$ is dense.

Analogous question regarding $\mathcal{M}_{\geq k}(|\nabla u|)(x)$ can also be asked.

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    $\begingroup$ The $\in$ in $W^{1,p}\in L^\infty$ is certainly a typo, but should it be $\subseteq$ or $\cap$? $\endgroup$ – YCor Oct 2 at 13:51
  • $\begingroup$ And what does $\{B_R(x) : u \leq k\}$ really mean? $\endgroup$ – Mateusz Kwaśnicki Oct 2 at 22:31
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    $\begingroup$ If you ask a question you need to be precise and correct. You questions is not because $W^{1,p}$ is not a subset of $L^\infty$ in general. I think I know the answer to your question, but if you do not make it right I will vote it down and ask to close. $\endgroup$ – Piotr Hajlasz Oct 3 at 0:00
  • $\begingroup$ Sorry, it is supposed to be intersection. I've fixed the typo. $\endgroup$ – Adi Oct 3 at 14:02

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