I'm trying to understand a comment regarding the maximal function in the first chapter of Stein's Singular Integrals and Differentiability Properties of Functions (emphases mine):
In contrast with the case $p>1$, when $p=1$ the mapping $f\mapsto M(f)$ is not bounded on $L^1(\mathbb{R}^n)$. Thus if $f$ is not identically zero $Mf$ is never integrable on all of $\mathbb{R}^n$. This can be seen by making the simple observation that $Mf(x)\ge c|x|^{-n}$, for $|x|\ge 1$.
Here the maximal function is defined as follows: $$ M(f)(x)=\sup_{r>0}\frac{1}{m(B(x,r))}\int_{B(x,r)}|f(y)|\,dy\ $$ [Added: where $B(x,r)$ is the ball of radius $r$, centered at $x$ and $m(B(x,r))$ denotes its Lebesgue measure.]
Could anyone show that how the "simple" observation is true? (I don't see why it is true even when $n=1$.)
The observation is simple ,but simple for Stein..:p
Ipressume that $m$ denotes the Lebesgue measure.
Now if $f$ is not zero a.e then $m(\{|f|>0\})>0$
So exists $m \in \Bbb{N}$ such that $m(\{|f| \geq \frac{1}{m}\})>0$
By inner regularity of the Lebesgue measure, for $E_m=\{|f| \geq \frac{1}{m}\}$ exists a compact set $K \subseteq E_m$ such that $m(K)>0$
Now note that $K \subseteq B(0,M)$ for some $M>0$
So $\forall |x| \geq M$ we have that $K \subseteq B(x,|x|+M)$ thus $$M(f)(x) \geq \frac{1}{(|x|+M)^d} \int_K|f(y)|dy \geq \frac{1}{(2|x|)^d} \frac{1}{m}m(K)$$
You can adjust this idea to prove the statement for $\forall |x| \geq 1$