A question about the maximal function

Question

Let $n>4$, $f\in C^{\infty}(\mathbb{R}^{n})$ and 0 denote the origin of $\mathbb{R}^{n}$. We define a weighted maximal function by $$Mf(x)=\sup_{0<r<1}r^{4-n}\int_{B_{r}(x)}|f|$$ which is multiplied by an extra $r^{4}$ compared to the standard Hardy-Littlewood maximal function. Set $A=\{ y\in B_{1}(0):Mf(y)\geqslant1 \} $. It is a classical result for standard Hardy-Littlewood maximal function that $\mathcal{H}^{n}(A)\leqslant C(n)\int_{B_{6}(0)}|f|$ (using covering lemma).

However, since we have an extra $r^{4}$, can we expect something different or stronger? One thing I know is that by the same argument using covering lemma, one may obtain that the $(n-4)$-Hausdorff content (see https://en.wikipedia.org/wiki/Hausdorff_dimension#Hausdorff_content for the definition of Hausdorff content) is controlled by $\int_{B_{6}(0)}|f|$. But this is not what I want. My question is:

Does there exists a small $\delta=\delta(n)>0$, such that if $\int_{B_{6}(0)}|f|<\delta$, then $G=B_{1}(0)\setminus A$ is dense in $B_{1}(0)$? Alternatively, are there any stronger conclusions, such as $\mathcal{H}^{n}(A)=0$?

If this does not hold, can anyone give a counterexample? Any ideas or comments are welcomed.

Answer 1

The answer is no. Let $m$ be a very large positive constant. You can find smooth $f$ that equals $m$ on a small ball $B_R(0)$ and still satisfy $\int_{B_6(0)}|f|<\delta$. You can do it with $R$ such that $R^n= \delta 2^{-1}m^{-1}\omega_n^{-1}$, where $\omega_n$ is the volume of the unit ball. Indeed, you approximate the $m\chi_{B_R(0)}$ by a smooth function.

Then $$ Mf(0)\geq R^{4-n}\int_{B_R(0)}f=C(n)\delta^{\frac{4}{n}}m^{1-\frac{4}{n}}. $$ Note that $m$ is independent of $\delta$ and it can be arbitrrily large, meaning that $Mf(0)$ can be arbitrarily large and so $Mf$ will be arbitrarily large in a neighborhood of $0$. That also means that a neighborhood of $0$ will be in the set $A$. Hence $G$ will not be dense and $\mathcal{H}^n(A)>0$.