The exercise reads as follows:
22. If $f\in L^1(\mathbb{R}^n),\ f\neq0$, there exist $C,R>0$ such that $Hf(x)\geq C|x|^{-n}$ for $|x|>R$.
My confusion: What are the initial conditions exactly? As far as I can tell, the conditions on function $f$ are that it is in $L^1$ and not equal to zero...however, if these are the correct initial conditions, how would function such as $\chi_{\{a\}}$ not serve as a contradiction? In particular, we know that $H\chi_{\{a\}}(x)=\sup_{r>0}(A_r\chi_{\{a\}}(x))=0$ for all $x\neq a$.
(p.s., please don't give an answer to the exercise. I would prefer not to be tempted. Thank you!)
Edit: $Hf(x)$ refers to Hardy-Littlewood's maximal function. I have replied with the definition of such function according to Folland.
Observe that $\chi_{\{a\}}$ is equal to zero! Because $L^1$ is actually a set of equivalence classes of functions up to almost-everywhere equality and $\chi_{\{a\}}$ is almost everywhere zero. We frequently commit the cardinal sin of abusing functions with their equivalence classes in $L^1$ but we should always have in mind what we are doing. A different way of saying the same thing is: $f$ is an integrable measurable function $\Bbb R^n\to\Bbb R$ and is not a.e. equal to zero. But $f\in L^1\setminus\{0\}$ is easier to say.