Let \[ \omega(\alpha)=m\{x\in\RR^n:\abs{f(x)}>\alpha\},\qquad \alpha>0, \] be the distribution function of \(f\in L^p(\RR^n)\), where \(p>0\). Does \(\alpha^p\omega(\alpha)\) tend to a limit as \(\alpha\to0^+\)? Give a proof or counterexample.
Solution
1. Use the layer-cake identity.
For the nonnegative function \(\abs{f}^p\), \[ \norm{f}_p^p =p\int_0^\infty \alpha^{p-1}\omega(\alpha)\,d\alpha<\infty. \] Thus \[ \int_0^a \alpha^{p-1}\omega(\alpha)\,d\alpha\longrightarrow0 \qquad(a\downarrow0). \]
2. The limit is\[ \boxed{\lim_{\alpha\to0^+}\alpha^p\omega(\alpha)=0}. \]
The distribution function \(\omega\) is decreasing. Hence for \(t\in[\alpha/2,\alpha]\), \[ \omega(t)\ge\omega(\alpha). \] Therefore \[ \begin{aligned} \int_{\alpha/2}^{\alpha}t^{p-1}\omega(t)\,dt &\ge \omega(\alpha)\int_{\alpha/2}^{\alpha}t^{p-1}\,dt\\ &=\frac{1-2^{-p}}p\,\alpha^p\omega(\alpha). \end{aligned} \] Consequently \[ 0\le \alpha^p\omega(\alpha) \le \frac{p}{1-2^{-p}} \int_0^\alpha t^{p-1}\omega(t)\,dt \longrightarrow0. \] The last limit follows from step 1.
3. Q.E.D.
Step 2 gives the requested limit.