# Hardy Littlewood maximal function bounds

Let $$u \in W^{1,p}(\mathbb{R}^n) \cap L^{\infty}(\mathbb{R}^n)$$ be a given function for some $$1 and let $$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.

• The $\in$ in $W^{1,p}\in L^\infty$ is certainly a typo, but should it be $\subseteq$ or $\cap$? – YCor Oct 2 at 13:51
• And what does $\{B_R(x) : u \leq k\}$ really mean? – Mateusz Kwaśnicki Oct 2 at 22:31
• 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. – Piotr Hajlasz Oct 3 at 0:00
• Sorry, it is supposed to be intersection. I've fixed the typo. – Adi Oct 3 at 14:02