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Hardy littlewood maximal function of a function supported in the unit ball

This question comes from Stein’s Note on the class$L\log L$. In the proof for Theorem 1, which establishes for an integrable function $f\in L^1(\mathbb R^n)$ supported on a finite ball $B$, that the (centered) Hardy-Littlewood maximal function $Mf$ is integrable implies $f\in L\log L$, i.e. $\int_B |f|\log^+|f|dx\lt\infty$, the following is stated:

We observe that if $f$ is supported in $B$ and$Mf$ is integrable over $B$, then $Mf$ is integrable over $B’$. In fact, suppose for simplicity that $B=\{x:|x|\le 1\}$, and $B’=\{x:|x|\le 2\}$. Then if $x\in B’-B$, we can easily verify that $Mf(x)\le cMf(\bar x)$, where $\bar x=x/|x|^2$.

Here $c=n^{n/2}\omega_n$, where $n$ is the dimension of the space and $\omega_n$ is the volume of the unit $n$-dimensional ball.

However, I don’t find this obvious. It seems rather counterintuitive how the factor $n^{n/2}$ comes in, as that was the factor between the volume of a cube and a ball with the same (or proportional) diameter. This is how $c$ appeared in the paper at first. For reference, please see this other post. But in the part quoted above, there didn’t seem to be any obvious cubes involved.

Some observations:

$Mf(\bar x)$ is achieved at $B(x,r)$ with $r=\min(|\bar x|,1-|\bar x|)$, since the argument inside the supremum (in the definition of $Mf$) could be thought of as a kind of concentration, and if the ball contains a region outside of the support of $f$, this concentration will be diluted.

If $n=1, \dfrac{1}{|B(x,r)|}\int_{B(x,r)\cap [-1,1]}|f|$ increases with $r$ for $x\in B’-B$ if $B(x,r)\cap [-1,1]$ is a nonempty proper subset of $[-1,1]$. $Mf(x)$ is achieved when $B(x,r)$ covers all of $[-1,1]$. It seems this is generalizable to higher dimensions, although I haven’t yet been successful in rigorously showing that.

$|x-\bar x|=|x|-1/|x|$. If $r\gt |x-\bar x|$, then $B(\bar x,r)\supset B(x,r-|x-\bar x|$.

Either a rigorous proof of the quoted part or an intuitive argument would be appreciated. In the meantime I might add more stuff if I do find more related to this part.


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