Hardy-Littlewood maximal operator

Question

Consider the centered Hardy_littlewood maximal operator

$$ \mathcal{M}f(x):= \sup_{r>0} \frac{1}{|B(x,r)|} \int_{B(x,r)} |f(y)| \text{d}y $$

and the uncentered

$$ Mf(x):= \sup_{r>0, |y-x|<r} \frac{1}{|B(y,r)|} \int_{B(y,r)} |f(t)| \text{d} t $$

In Grafakos book ("Classical Fourier Analysis") page 79, it is written that

$$ Mf(x) \leq 2^n \mathcal{M}f(x)$$

How can one prove this? Any help?

Thanking in advance!

Answer 1

It boils down to the fact that for any $y$ such that $|y-x| < r$, we have $B(y,r) \subset B(x,2r)$. Thus, for such $y$: \begin{align} \frac{1}{|B(y,r)|} \int_{B(y,r)} |f(t)| \, dt &\le \frac{1}{|B(y,r)|} \int_{B(x,2r)} |f(t)| \, dt \\ &= \frac{1}{|B(x,r)|} \int_{B(x,2r)} |f(t)| \, dt \\ &= \frac{2^n}{|B(x,2r)|} \int_{B(x,2r)} |f(t)| \, dt. \end{align} Taking the sup over $r$ yields $Mf(x) \le 2^n \mathcal Mf(x)$.