Consider the function \[ f(x) \da \begin{cases} {1\over \abs{x} \qty{ \log\qty{1\over x}}^2 } & \abs{x} \leq {1\over 2} \\ 0 & \text{else}. \end{cases} \]
Show that \(f \in L^1(\RR)\).
Show that there exists a \(c>0\) such that for all \(\abs{x} \leq 1/2\), \[ Hf(x) \geq {c \over \abs{x} \log\qty{1\over \abs x} } .\] Conclude that \(Hf\) is not locally integrable.
For \(x < 0\) the formula is read with \(\log(1/|x|)\) in place of \(\log(1/x)\), so that \(f\) is even.
Solution
Here \(Hf(x) = \sup_{r>0}\frac{1}{2r}\int_{x-r}^{x+r}|f(t)|\,dt\) is the Hardy–Littlewood maximal function, and \(f(x) = \frac{1}{|x|\log^2(1/|x|)}\) for \(0 < |x| \le 1/2\).
1. For \(0 < s \le 1/2\), \(\int_{-s}^{s} f = \frac{2}{\log(1/s)}\). In particular \(\int_\RR f = \frac{2}{\log 2}\), so \(f \in L^1(\RR)\).
\(f\) is even, and the substitution \(u = \log(1/t)\), \(du = -dt/t\), gives \(\int_0^{s}\frac{dt}{t\log^2(1/t)} = \int_{\log(1/s)}^\infty u^{-2}\,du = \frac{1}{\log(1/s)}\).
2. For \(0 < |x| \le 1/2\), \(Hf(x) \ge \frac{1}{2|x|\log(1/|x|)}\).
Take \(r = 2|x|\). The interval \((x - r, x + r)\) contains \((-|x|, |x|)\) and \(f \ge 0\), so by step 1 \[ Hf(x) \ge \frac{1}{4|x|}\int_{-|x|}^{|x|} f = \frac{1}{4|x|}\cdot\frac{2}{\log(1/|x|)}. \] So part (b) holds with \(c = 1/2\).
3. \(Hf\) is not integrable on any neighborhood of \(0\).
For \(0 < \delta \le 1/2\), step 2 and the substitution \(u = \log(1/x)\) give \[ \int_{-\delta}^{\delta} Hf \ge 2 \cdot \frac12\int_0^{\delta}\frac{dx}{x\log(1/x)} = \int_{\log(1/\delta)}^\infty \frac{du}{u} = \infty. \]