Strong maximal function

Remarks

Each ball of radius $r$ lies in a cube of side $2r$, which is a rectangle of comparable volume. So $Mf \le C_n M_S f$ for the Hardy–Littlewood maximal function $Mf$, with $C_n$ depending only on $n$. Let $M_k$ be the one-dimensional maximal function in the $k$-th variable, with the other variables fixed. Then $M_S f \le M_1 M_2 \cdots M_n f$ almost everywhere, and the one-dimensional bounds give $\norm{M_S f}_p \le C \norm{f}_p$ for $1 < p \le \infty$. For $n \ge 2$ the operator $M_S$ is not of weak type $(1, 1)$. It satisfies the weaker bound $\abs{\set{M_S f > \lambda}} \le C \int \frac{\abs{f}}{\lambda} \left( 1 + \log^+ \frac{\abs{f}}{\lambda} \right)^{n - 1}$, due to Jessen, Marcinkiewicz and Zygmund.

Example: let $n = 2$, $f = \mathbf{1}_{[0, 1]^2}$ and $x = (a, 1/2)$ with $a > 1$. The rectangle $R = [0, a] \times [0, 1]$ contains $x$ and has $\frac{1}{\abs{R}} \int_R f = \frac{1}{a}$. Any rectangle $R' = [s, t] \times J$ containing $x$ and meeting the square has $s \le 1 < a \le t$. Its average of $f$ is at most $(1 - \max(s, 0))/(t - s)$, which is at most $1/a$. So $M_S f(x) = 1/a$.

Non-example: a square centred at $x$ and rotated by $45$ degrees is not a rectangle in the sense above. Its sides are not parallel to the axes. Allowing rectangles in all directions gives a different operator, which is unbounded on $L^p(\R^2)$ for every $p < \infty$, because of Besicovitch sets.