I am trying to prove that if $Mf(x) \in L^1(R^n)$ then f = 0. Where Mf(x) is the hardy littlewood maximal function defined in the following way
$Mf(x) = sup_{0<r<\infty} 1/(\lambda(B(x,r))* \int_{B(x,r)} |f(y)|dy$
My proof is as follows -
Let $|x|>a $, for any arbitrary a
$Mf(x)\geq 1/(\lambda(B(x,2|x|))\int_{B(x,r)} |f(y)|dy$
$Mf(x) \geq 1/(\pi*(2x)^n)\int_{B(x,a)} |f(y)|dy$
Now $\int Mf(x)dx \geq \int(1/(\pi*(2x)^n)\int_{B(x,a)} |f(y)|dy)$
Let $\int_{B(x,a)} |f(y)|dy)= constant $
$\int Mf(x)dx \geq \int(1/(\pi*(2x)^n)*C)$
and we know that $1/(x^n)$ isn't integrable for $n\geq 1 $. Hence proved .
My main doubt in this proof is that can I assume that $\int_{B(x,r)} |f(y)|dy)= constant $ .
Please let me know if I am missing something at any other step too .