Let $$f(x)=\frac 1{(1+|x|)^2},$$ Then what's the maximal function of $f$ ? By definition $$Mf(x)=\sup_{r>0}\frac 1{|B_r(x)|}\int_{B_r(x)}\frac 1{(1+|y|)^2}dy,$$ If one can prove that the average value $\frac 1{|B_r(x)|}\int_{B_r(x)}\frac 1{(1+|y|)^2}dy$ is monotonically decreasing with respect to $r$, for any fixed $x$, then one has $$Mf(x)=\lim_{r\to 0^+}\frac 1{|B_r(x)|}\int_{B_r(x)}\frac 1{(1+|y|)^2}dy=\frac 1{(1+|x|)^2}$$ by Lebesgue point theorem.
However, I fail to prove the monotonicity, it seems that I can only prove the it only for certain relations between $x$ and $r$.
Any help would be appreciated.
In hindsight I don't believe this is right. To be more clear, just take 1-D case as an example.
$f(x)=\frac 1{(1+|x|)^2}$. Then be definition, $$Mf(x)=\sup_{r>0}\frac 1{2r}\int_{x-r}^{x+r}\frac 1{(1+|y|)^2}dy.$$ Let's just calculate the case $x-r>0$, $$\frac 1{2r}\int_{x-r}^{x+r}\frac 1{(1+y)^2}dy=\frac 1{2r}(-\frac 1{1+y})|_{y=x-r}^{y=x+r}=\frac {1}{(x+1)^2-r^2},$$ this quantity increases as $r$ increases to $x$, therefore $$\sup_{0<r<x}\frac 1{2r}\int_{x-r}^{x+r}\frac 1{(1+|y|)^2}dy=\frac 1{2x+1}.$$