Lower bound for the Hardy-Littlewood maximal function implies it is not integrable

I am working on the following problem from Stein and Shakarchi:

Let $f$ be an integral function on $\mathbb{R}^d$ such that $\|f\|_{L^1} = 1$ and let $f^*$ by the Hardy-Littlewood maximal function corresponding of $f$.

• Prove that if $f$ is integrable on $\mathbb{R}^d$, and $f$ is not identically zero, then $f^*(x) \geq c/|x|^d $ for some $c > 0$ and all $|x| > 1$.
• Conclude that $f^*$ is not integrable on $\mathbb{R}^d$.
• Then, show that the weak type estimate $m(\{x:f^*(x) > \alpha \}) \geq c/\alpha$ for all $\alpha > 0$ whenever $\int|f| = 1$, is best possible in the following sense: if $f$ is supported in the unit ball with $\int|f| = 1$, then $m(\{x:f^*(x) > \alpha \}) \geq c'/\alpha$ for some $c' > 0$ and all sufficiently small $\alpha$.

[Hint: For the first part, use the fact that $\int_B|f| > 0$ for some ball $B$.]

I am not really sure where to begin with this. Any help would be greatly appreciated.

Solution 1:

Wlog, take $f$ nonnegative. Since $f \neq 0$, find an $R > 0$ large enough so that $\int_{B(0;R)} f > 0$. Fix any $x \in \mathbb{R}^d$, and take $S$ large enough so that $B(0;R) \subset B(x;S)$ (say $S = R + |x|$). So $f^*(x) \ge \frac{c}{S^d} \int_{B(x;s)} f \ge \frac{c}{(R+|x|)^d} \int_{B(0;R)} f$, where $c^{-1}$ is the volume of the unit ball in $\mathbb{R}^d$. It should be easy to obtain the last part of the estimate from here.