Pointwise equivalence of maximal operators Let $M$ denotes the Hardy-Littlewood maximal operator and $f$ is a locally integrable on $\mathbb{R}^n.$
$Mf(x)=sup_{r>0}\frac{1}{r^n}\int_{|y|\leq r}{|f(x-y)|dy}$
$M'f(x)= sup_{r>0}\frac{1}{|Q(x,r)|}\int_{Q(x,r)}{|f(y)|dy}$
$M''f(x)=sup_{x \in Q}\frac{1}{|Q|}\int_{Q}{|f(y)|dy}$
where $Q(x,r) $ denotes the cube with the center at x and with side r and its sides parallel to the coordinate axes and $|Q|, |Q(x,r)|$ denotes the Lebesgue measure.
I want to show that
there exist constants $C_{i}$ depending only on the dimension $n$ such that
$C_{0}Mf(x)\leq C_{1}M'f(x) \leq C_{2}M''f(x) \leq C_{3}Mf(x)$
$i=0,1,2,3$
I have tried;
$ \frac{1}{r^n} \int_{|y| \leq r}{|f(x-y)|dy} = \frac{1}{r^n}\int_{-r \leq y \leq r}{|f(x-y)|dy} =\frac{1}{r^n}\int_{-r}^{r}{|f(x-y)|dy} = \frac{1}{r^n} \int_{x-r}^{x+r}{|f(y)|dy} = \frac{1}{r^n} \int_{[x-r,x+r]}{|f(y)|dy} $
but im stuck in here! (i will take supremum at the end)

Here $x,y$ are $n$-dimensional vectors.
$\frac{1}{r^n}$ is the volume of the cubes. After translation and reflection, the integration in the first function is exactly the integration of $|f|$ over a cube centred at $x$ with radius side length $2r$. After scaling, $M$ is comparable to $M'$ up to a constant.
$M'$ is a supremum of a subset of what's considered in $M''$, thus must be smaller.
Again, after translating the region to integrate over, you can always shift to a cube centred at $0$, so you have the third and the first maximal functions equivalent.
