Suppose $h: \Bbb R \to : \Bbb R$; Hardy Littlewood maximal function $h^*: \Bbb R \to [0, \infty]$ defined by $h^*(x)=\sup_{t>0}\frac 1{2t}\int_{x-t}^{x+t}|h|$.
Now to prove $|h(x)|\leq|h^*(x)|$ for almost every $x \in \Bbb R$
I am thinking that I have to use Lebesgue differentiation theorem, first version that states "Suppose $f\in L^{1}\Bbb R$. Then
$\lim_{ t \to 0}\frac 1{2t}\int_{x-t}^{x+t}|f-f(b)|=0$ for all most all $b \in \Bbb R$.
My attempt: $||h(x)|-|h^*(x)||\leq |h(x)-h^*(x)|=|h(x)-\sup_{t>0}\frac 1{2t}\int_{x-t}^{x+t}|h(y)|dy|\leq sup_{t>0}\frac 1{2t}\int_{x-t}^{x+t}|h(x)-h(y)|dy=0$ for almost every $x$. So, we get $|h(x)|\leq h^*(x)$ for allmost every $x$.
Is my attempt coorect or is there any other way to approach? Please help
Best Answer
This version of Lebesgue differentiation, which correctly stated is $$\lim_{ t \to 0}\frac 1{2t}\int_{x-t}^{x+t}|f-f(x)|=0 \quad(\text{a.e.}\ x)$$ immediately implies the weaker one ($\frac1{2t}\int_{x-t}^{x+t} f \to f(x)$ a.e.) via triangle inequality $$ \left| \frac1{2t} \int_{x-t}^{x+t} f(y) dy - f(x)\right|\le \lim_{ t \to 0}\frac 1{2t}\int_{x-t}^{x+t}|f-f(x)|\to 0$$ Therefore, for almost every $x$, $$ |f(x)| = \lim_{t\to 0}\frac1{2t} \int_{x-t}^{x+t } |f| \le \sup_{t> 0} \frac1{2t} \int_{x-t}^{x+t } |f| = f^*(x)$$