The strong maximal function is not weak type (1,1)

Question

Let $M_s(f)$ be the supremum of the averages of $|f|$ over all rectangles with sides parallel to the axes containing $x$. I want to show that $M_s(f)$ is not weak (1,1), but I can’t find any examples...

Answer 1

This is a counterexample in $\mathbb R^2$ for $M_s$.

Let $f= \chi_R(x)$, where $R=\{ (x_1,x_2): x_1\in (0,n), x_2\in (0,\frac{1}{n^2})\}$. It is easy to verify that $\|f\|_{L^1}=\frac{1}{n}$. Consider the function $x_2=g(x_1)$ as
$$ x_2=\begin{cases} \cfrac{1}{n},& 0<x_1<n,\\ \cfrac{1}{x_1},& n\le x_1\le n^2. \end{cases} $$

It is easy to verify that $$\{(x_1,x_2): x_2\le g(x_1), 0<x_1<n^2\} \subset \{M_sf(x)>\frac{1}{n}\}.$$

Thus we have $$ \cfrac{1}{n}d_{M_sf}(1/n)\ge \cfrac{1}{n}\left(\cfrac 1n \cdot n +\int_n^{n^2} x^{-1}dx\right)=\cfrac{1+\ln n}n. $$

Now suppose $M_s$ is of weak type $(1,1)$ with bound at most $C$, then we have $$ \cfrac{1+\ln n}{n}\le \cfrac{1}{n}d_{M_sf}(1/n)\le C\|f\|_{L^1}=\cfrac{C}{n} $$ but this is wrong for sufficiently large $n$.