Hardy-Littlewood maximal function $Mf$ is greater than $f$

Question

Suppose that $f\in L^{1}(\mathbb R)$ and $x\in \mathbb R.$ We denote $B_{x}$ by the ball in $\mathbb R$ with centred $x,$ and $|B_{x}|=$ length(Lebsgue measure) of $B_{x}.$

Put, $Mf(x)=\sup \{|B_{x}|^{-1}\int_{B_{x}} |f(y)| dy: B_{x}\subset \mathbb R\}.$

My Question: (1) Why $Mf$ is known as maximal function of $f$ ? (2) Why $|f(x)|\leq Mf(x)$ for all most all $x$ ? (3) Bit rough question: Why this function got some attention by analyst ? (any application ?)

Thanks,

Answer 1

1. Because there is a maximum (well, supremum) in its definition.

2. Answered here: I want to show that $|f(x)|\le(Mf)(x)|$ at every Lebesgue point of $f$ if $f\in L^1(R^k)$

3. The important properties are: $M$ is a bounded (nonlinear) operator on $L^p$ for $1<p<\infty$, and also from $L^1$ to weak $L^1$. This fact offers more control over the local behavior of $L^p$ functions than the finiteness of $L^p$ norm itself. Applications are listed on Wikipedia.