This is Exercise 7.22 (a) from Rudin's RCA.
Assume that both $f$ and its maximal function $Mf$ are in $L^1(R^k)$. Prove that then $f(x)=0$ a.e.
Hint: To every other $f \in L^1(R^k)$ corresponds a constant $c = c(f)>0$ such that $$(Mf)(x) \ge c|x|^{-k}$$ whenever $|x|$ is sufficiently large.
I am stuck with this problem. Indeed how can I prove the hint to begin with? I would greatly appreciate any help.
For the hint, first notice that by dominated convergence:
$$ \|f\|_{L^1} = \int_\mathbb{R^k} |f(y)|\mathrm{d}y = \lim_{r \to \infty} \int_{B_r} |f(y)| \mathrm{d}y $$ for any $f \in L^1 (\mathbb{R}^k)$. Note also that $\|f\|_{L^!} \neq 0$ thanks to the assumption that $f$ is not a.s. 0. Now, the maximal function is given by: $$ (Mf)(x) = \sup_{B_r \ni x} \frac{1}{C(k)|r|^k}\int_{B_r(z)}|f(y)|\mathrm{d}y $$ where $B_r(z)$ is a ball centered around $z$ with radius $r$ that contains our point $x$. By the above, we may fix $\epsilon>0$ such that for $r$ sufficiently large, (i.e for $r \geq R_0$), we have: $$ \|f\|_{L^{1}} - \int_{B_r(z)}|f(y)|\mathrm{d}y <\varepsilon $$ Fix $x_0$ such that $|x| \geq R_0$. Then, by definition of $(Mf)(x)$, we have:
$$ (Mf)(x) = \sup_{B_r \ni x} \frac{1}{C(k)|r|^k}\int_{B_r(z)}|f(y)|\mathrm{d}y \geq \frac{1}{C(k)|x_0|^k }\int_{B_{|x_0|}(z)}|f(y)|\mathrm{d}y $$ For $x \geq x_0$, we may apply the above $\varepsilon$ inequality, and conclude that: $$ \frac{1}{C(k)|x_0|^k }\int_{B_|x_0|(z)}|f(y)|\mathrm{d}y \geq \frac{1}{C(k)|x_0|^k } (\|f\|_{L^1} -\epsilon) \geq \frac{1}{C(k)|x|^k}(\|f\|_{L^1} - \epsilon) $$ Multiplying all constants through on the RHS gives us a constant depending only on $f$ and dimension.