Hardy-Littlewood maximal operator 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!

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 $Mf(x) \le 2^n \mathcal Mf(x)$.
