- In parts:
- Given a definition of \(L^\infty(\RR^n)\).
- Verify that \(\norm{\wait}_\infty\) defines a norm on \(L^\infty(\RR^n)\).
- Carefully proved that \((L^\infty(\RR^n), \norm{\wait}_\infty)\) is a Banach space.
- Prove that for any measurable \(f:\RR^n \to \CC\), \[ L^1(\RR^n) \intersect L^\infty(\RR^n) \subset L^2(\RR^n) \qtext{and} \norm{f}_2 \leq \norm{f}_1^{1\over 2} \cdot \norm{f}_\infty^{1\over 2} .\]
Solution
\(L^\infty(\RR^n)\) is the space of a.e.-equivalence classes of measurable \(f\colon \RR^n \to \CC\) with \(\|f\|_\infty \coloneqq \inf\theset{M \ge 0 : |f| \le M \text{ a.e.}} < \infty\).
1. \(|f| \le \|f\|_\infty\) a.e.
\(\theset{|f| > \|f\|_\infty} = \bigcup_k \theset{|f| > \|f\|_\infty + 1/k}\) is a countable union of null sets.
2. \(\|\cdot\|_\infty\) is a norm on \(L^\infty\).
By step 1, \(\|f\|_\infty = 0\) if and only if \(f = 0\) a.e. \(|\lambda f| \le M\) a.e. if and only if \(|f| \le M/|\lambda|\) a.e., for \(\lambda \neq 0\), so \(\|\lambda f\|_\infty = |\lambda|\,\|f\|_\infty\). By step 1, \(|f + g| \le \|f\|_\infty + \|g\|_\infty\) off the union of two null sets, so \(\|f + g\|_\infty \le \|f\|_\infty + \|g\|_\infty\).
3. \(L^\infty\) is complete.
Let \((f_k)\) be Cauchy in \(L^\infty\). By step 1 there is a null set \(N_{k,l}\) off which \(|f_k - f_l| \le \|f_k - f_l\|_\infty\); let \(N = \bigcup_{k,l}N_{k,l}\), a null set. Off \(N\), \((f_k)\) is uniformly Cauchy, so it converges uniformly there to a measurable \(f\); put \(f = 0\) on \(N\). Given \(\eps > 0\) choose \(K\) with \(\|f_k - f_l\|_\infty < \eps\) for \(k, l \ge K\). Letting \(l \to \infty\) gives \(|f_k - f| \le \eps\) off \(N\) for \(k \geq K\). So \(f = (f - f_K) + f_K \in L^\infty\) and \(\|f_k - f\|_\infty \le \eps\) for \(k \ge K\).
4. If \(f \in L^1 \cap L^\infty\), then \(\|f\|_2 \le \|f\|_1^{1/2}\|f\|_\infty^{1/2}\); in particular \(f \in L^2\).
By step 1, \(|f|^2 \le \|f\|_\infty|f|\) a.e., so \(\int|f|^2 \le \|f\|_\infty\|f\|_1\).