Maximal function of $\frac{1}{|x|^a}, x \in \Bbb{R}^d$ for $0For $f \in L^1_{loc}(\Bbb{R}^d)$,we denote $$\mathcal{M}(f)(x)=\sup_{r>0} \frac{1}{|B(x,r)|}\int_{B(x,r)}|f(y)|dy$$ the Hardy-Littlewood maximal function for $f$.

Prove that if $f(x)=\frac{1}{|x|^a}, 0<a<d$ then $\exists C_{a,d}>0$ (which depends only on $a,d$) such that $$\mathcal{M}(\frac{1}{|\cdot|^a})(x) \leq C_{a,d}\frac{1}{|x|^a}$$.

For $x\in \Bbb{R}^d$ we can examine all all $r>0$ such that $$1.) r>|x|$$ $$2.) r<|x|$$
I managed to prove that $$\sup_{r>|x|}\frac{1}{|B(x,r)|}\int_{B(x,r)}\frac{1}{|y|^a}dy \leq C_{d,a}\frac{1}{|x|^a}$$ for some $C_{a,d}>0$
How can we prove a similar result examining the   case of $\sup_{|x|>r}(...)$ ?
I had a difficulty and maybe i am missing something obvious.
Can someone help me with this case?
Thank you in advance.
 A: I checked your reasoning and your calculations and I didn't find any mistake or anything I was fuzzy about. That is, you can divide the supremum in the two cases (A) and (B) (as you did) and only get bigger. Each case also makes sense to me. Anything you are not too sure about?
A: I post here an answer for my question after a week ,and it would be highly appreciated if someone can tell if it is correct.Thank you in advance.
Let  $x \in \Bbb{R}^d \setminus \{0\}$ and $R>0.$
$\textbf{(A)}$ If $R< \frac{|x|}{2}$, then for $y \in B(x,R)$ we have that $$|y| \geq |x|-|y-x| \geq \frac{|x|}{2}$$ thus $$\frac{1}{|B(x,R)|} \int_{B(x,R)} \frac{1}{|y|^a}dy \leq 2^a \frac{1}{|x|^a}$$
$\textbf{(B)}$ If $R\geq \frac{|x|}{2}$ then $$\frac{1}{|B(x,R)|} \int_{B(x,R)} \frac{1}{|y|^a}dy$$ $$\leq \frac{1}{|B(x,R)|} \int_{B(0,3R)} \frac{1}{|y|^a}dy$$ $$=\frac{1}{R^d}\int_0^{3R}r^{d-1-a}dr=\frac{3^{d-a}}{d-a}\frac{1}{R^a} \leq \frac{2^a3^{d-a}}{(d-a)|x|^a}$$
For $C_{a,d}=\frac{2^a3^{d-a}}{d-a}$ we have the desired inequality.
