Let \(f\in L^1(\RR^n)\) with \(f\neq 0\).
- Prove that there exists a \(c>0\) such that \[ Hf(x) \geq {c \over (1 + \abs x)^n } .\]
Solution
Here \(Hf(x) = \sup_{s > 0}\frac{1}{m(B(x,s))}\int_{B(x,s)}|f|\) is the Hardy–Littlewood maximal function, and \(m(B(x,s)) = \omega_n s^n\) with \(\omega_n\) the volume of the unit ball.
1. There is \(r > 0\) with \(\alpha \coloneqq \int_{B(0,r)}|f| > 0\).
\(\int_{B(0,k)}|f| \to \int_{\RR^n}|f| > 0\) as \(k \to \infty\) by monotone convergence, and \(\int |f| > 0\) because \(f \neq 0\) in \(L^1\).
2. \(Hf(x) \ge \frac{\alpha}{\omega_n(|x| + r)^n}\) for every \(x\).
For \(z \in B(0,r)\), \(|z - x| \le |z| + |x| < r + |x|\), so \(B(0,r) \subseteq B(x, |x| + r)\). Taking \(s = |x| + r\) in the definition of \(Hf\) gives \(Hf(x) \ge \frac{1}{\omega_n(|x| + r)^n}\int_{B(0,r)}|f|\).
3. Q.E.D.
\(|x| + r \le (1 + r)(1 + |x|)\), so step 2 gives \(Hf(x) \ge \frac{c}{(1+|x|)^n}\) with \(c = \frac{\alpha}{\omega_n(1 + r)^n} > 0\).