Let $M$ denote the centered Hardy-Littlewood maximal function using balls, and let $M_{c}$ denote the centered Hardy-Littlewood maximal function using cubes. Exercise 2.13 in [L. Grafakos, Classical Fourier Analysis (Second Edition)] asks the reader to establish the pointwise inequality

$$v_{n}(n/2)^{n/2}\leq\dfrac{M(f)}{M_{c}(f)}\leq 2^{n}/v_{n},$$

where $v_{n}$ denotes the volume of the unit ball in $\mathbb{R}^{n}$.

I can obtain the upper bound, but I cannot figure out how to get the lower bound. I instead get $2^{n}/(v_{n}n^{n/2})$ as shown below. The errata on the author's webpage does not inclde this inequality, so what am I doing wrong?

My Solution: 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 conclude

$$\dfrac{2^{n}}{v_{n}n^{n/2}}M_{c}f(0)\leq Mf(0)$$


Your computation is correct. Let $B$ be the unit ball, $Q_i$ the cube inscribed in it, and $Q_s$ the circumscribed cube. Then the inequality should say $$ \frac{\operatorname{vol} Q_i}{\operatorname{vol} B} \le \frac{M(f)}{M_c(f)}\le \frac{\operatorname{vol} Q_s}{\operatorname{vol} B} $$ (The sharpness of both is demonstrated by placing an approximation to delta-function either at a vertex of $Q_i$, or at the center of a face of $Q_s$.)

Since $Q_s$ has sidelength $2$, $$ \frac{\operatorname{vol} Q_s}{\operatorname{vol} B} = \frac{2^n}{v_n} $$ Since $Q_s$ has sidelength $2n^{-1/2}$, $$ \frac{\operatorname{vol} Q_i}{\operatorname{vol} B} = \frac{2^n n^{-n/2}}{v_n} $$

  • $\begingroup$ Thank you! I learned today that a third edition of the book has been released and the author has corrected this erratum in it. $\endgroup$ – Matt Rosenzweig May 6 '15 at 14:37

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