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...

• The point is that any ball will both contain a cube and itself be contained in a larger cube. For the maximal function you are integrating the norm of the function, so something positive - not much to worry about. – Tyr Curtis Apr 30 '15 at 16:53
• You can embed a cube in a ball embedded in a cube. Think about squares. – Emily Apr 30 '15 at 16:53
• @TyrCurtis In $R^2$, a circle of radius $r$ can fit inside a square with sides $2r$. Here the size of the ball is $\pi r^2$ and the size of the cube $4r^2$. Then a square of sides $\sqrt{2}r$ can fit inside the same circle. I can see how the size of the ball and cube can be compared. But how does the integral of the function over the ball/cube figure into it? Doesn't the function $f$ determine what the $c_n$ and $C_n$ values would be? How does it only depend on $n$? – BenJames Apr 30 '15 at 17:04
• @Arkamis See previous comment. – BenJames Apr 30 '15 at 17:05
• @TyrCurtis So for R^2, the ball inside the cube, $|B|=\pi r^2$ and $|Q|=4r^2$ so $|B|\leq \frac{\pi}{4}|Q|$? How does the integral change this? – BenJames Apr 30 '15 at 17:19

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 $1<p\leq\infty$.