How to compare the Hardy-Littlewood maximal function for balls and cubes? I am currently working through a set of notes I found on the internet at: http://math.msu.edu/~charlesb/Notes/DuoChapter2.pdf
I am up to page 8, and the Hardy-Littlewood maximal function for balls has just been introduced. Then it says that we can also define maximal functions over cubes centred at $x$. Then there is the phrase:
"Furthermore, since the $n$-dimensional volumes of the unit cube and unit ball are equal up to a multiplicative
constant depending only on $n$, it is immediate that $Mf$ and $M'f$ are comparable in the sense that $c_nM'f(x)\leq Mf(x)\leq C_nM'f(x)$
for constants $c_n$ and $C_N$ only depending on $n$."
It may be immediate to the author but it is not at all to me! I cannot understand why this is true. Does anyone have a proof? And is there a formula for $c_n$ and $C_n$?
All I can think of is maybe it's possible to come up with some sort of comparison between the size of a ball and the size of a cube both using the same $r$, but then the integrals may not be equal in order to compare the entire maximal funnction...
 A: 
Theorem. Let $\mathcal{M}$ and $M$ denote the uncentered and centered Hardy-Littlewood maximal function using
  balls, and let $\mathcal{M}_{c}$ and $M_{c}$ denote the uncentered and centered Hardy-Littlewood maximal
  function using cubes. For $f\in L_{loc}^{1}(\mathbb{R}^{n})$,
$$\dfrac{2^{n}}{(v_{n}n^{n/2})}\leq\dfrac{M(f)}{M_{c}(f)}\leq\dfrac{2^{n}}{v_{n}}, \quad \dfrac{2^{n}}{(v_{n}n^{n/2})}\leq\dfrac{\mathcal{M}(f)}{\mathcal{M}_{c}(f)}\leq\dfrac{2^{n}}{v_{n}}\qquad\text{a.e.},$$
  where $v_{n}$ denotes the volume of the unit ball in $\mathbb{R}^{n}$.

Proof. In what follows, $r>0$. Replacing $f$ by a translate, it suffices to establish the inequality at $x=0$. Observe that the cube $[-r,r]^{n}$ is almost everywhere contained in the open ball $B(0,n^{1/2}r)$. Whence,
$$\dfrac{1}{v_{n}n^{n/2}r^{n}}\int_{[-r,r]^{n}}\left|f\right|\leq\dfrac{1}{v_{n}n^{n/2}r^{n}}\int_{B(0,n^{1/2}r)}\left|f\right|=\dfrac{1}{\left|B(0,n^{1/2}r)\right|}\int_{B(0,n^{1/2}r)}\left|f\right|\leq Mf(0)$$
Multiplying by $1=2^{n}/2^{n}$ and taking the supremum over $r>0$ of the LHS, we obtain that
$$\dfrac{2^{n}}{v_{n}n^{n/2}}M_{c}f(0)\leq Mf(0)$$
Similarly, observe that the open ball $B(0,r)$ is contained in the cube $[-r,r]^{n}$. Whence,
$$\dfrac{1}{(2r)^{n}}\int_{B(0,r)}\left|f\right|\leq\dfrac{1}{(2r)^{n}}\int_{[-r,r]^{n}}\left|f\right|\leq M_{c}f(0)$$
Multiplying by $1=v_{n}/v_{n}$ and taking the supremum over $r>0$ of the LHS, we obtain that
$$\dfrac{v_{n}}{2^{n}}Mf(0)\leq M_{c}f(0)$$
A completely analogous argument establishes the inequality for the uncentered maximal functions. $\Box$
It's worth mentioning that these inequalities show that $\mathcal{M}_{c},\mathcal{M}$ are weak-type (1,1) operators and therefore are bounded operators $L^{p}(\mathbb{R}^{n})\rightarrow L^{p}(\mathbb{R}^{n})$ for $1<p\leq\infty$.
