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)

What you wrote seems to be focused on the case $n=1$.
Because Lebesgue measure is translation invariant,
$$
Mf(x)=\sup_{r>0}\frac{c_1}{|B_r(x)|}\int_{B_r(x)} f(y)\,dy,
$$
where $B_r(x)$ is the ball of radius $r$ centered at $x$ and $|B_r(x)|=c_1\,r^n$. You always have
$$
B_r(x)\subset Q(x,r)\subset B_{\sqrt n\,r}(x).
$$
Hence
$$
|B_r(x)|\leq |Q(x,r)|\leq n^{n/2} |B_{r}(x)|.
$$
\begin{align}\tag1
\frac{c_1}{|B_r(x)|}\int_{B_r(x)} f(y)\,dy
&\leq\frac{c_1\,n^{n/2}}{|Q(x,r)|}\int_{Q(x,r)} f(y)\,dy
\end{align}
Taking sup this gives $Mf(x)\leq n^{n/2} Mf'(x)$. Since some cubes are used in $Mf'(x)$, you automatically get that $Mf'(x)\leq Mf''(x)$.
Now if $Q$ is a cube with side $r$ and $x\in Q$, then $Q\subset B_{\sqrt n\,r}(x)$. You have $|B_{\sqrt n\,r}(x)|=c_1\,n^{n/2}\,r^n=c_1\,n^{n/2}\,|Q|$.
Thus
$$
\frac1{|Q|}\int_Q|f(y)|\,dy\leq\frac1{r^n}\int_{B_{\sqrt n\,r}(x)}|f(y)|\,dy
=\frac{n^{n/2}}{(\sqrt n\,r)^n}\int_{B_{\sqrt n\,r}(x)}|f(y)|\,dy.
$$
Taking supremum,
$$
Mf''(x)\leq n^{n/2}\,Mf(x).
$$
