- Show that if \(E\subseteq \RR^n\) is measurable with \(\mu(E) < \infty\) and \(f\in L^p(X)\) then \[\norm{f}_{L^p(X)} \converges{p\to\infty}\to \norm{f}_\infty.\]
- Is it true that the converse to the DCT holds? I.e. if \(\int f_n \to \int f\), is there a \(g\in L^p\) such that \(f_n < g\) a.e. for every \(n\)?
- Prove continuity in \(L^p\): If \(f\) is uniformly continuous then for all \(p\), \[\norm{\tau_h f - f}_p \converges{h\to 0}\to 0.\]
- Prove the following inclusions of \(L^p\) spaces for \(m(X) < \infty\): \[ L^\infty(X) &\subset L^2(X) \subset L^1(X) \\ \ell^2(\ZZ) &\subset \ell^1(\ZZ) \subset \ell^\infty(\ZZ) .\]
Erratum: the second line of the displayed inclusions should read \(\ell^1(\ZZ) \subset \ell^2(\ZZ) \subset \ell^\infty(\ZZ)\). In the first part, \(X\) is the set \(E\).
Solution
1. If \(0 < \mu(E) < \infty\) and \(f\) is measurable on \(E\), then \(\norm{f}_p \to \norm{f}_\infty\) as \(p \to \infty\).
\(\norm{f}_p \le \norm{f}_\infty\,\mu(E)^{1/p}\), so \(\limsup_p \norm{f}_p \le \norm{f}_\infty\). If \(0 < M < \norm{f}_\infty\), then \(\mu\theset{|f| > M} > 0\) and \(\norm{f}_p \ge M\,\mu\theset{|f|>M}^{1/p} \to M\), so \(\liminf_p \norm{f}_p \ge M\). See E-TVBUC.
2. The converse of the dominated convergence theorem fails.
On \([0,1]\), \(f_n = n\chi_{[1/(n+1),\,1/n)}\) satisfies \(f_n \to 0\) pointwise and \(\int f_n = \frac{1}{n+1} \to 0\). Any \(g\) with \(f_n \le g\) a.e. for all \(n\) has \(g \ge \sup_n f_n\), and \(\int (\sup_n f_n)^p = \sum_n \frac{n^{p-1}}{n+1} = \infty\) for \(1 \le p < \infty\). See E-PUNDP.
3. If \(f\colon \RR^n \to \CC\) is uniformly continuous, then \(\norm{\tau_h f - f}_\infty \to 0\), and if moreover \(f \in L^p\) with \(1 \le p < \infty\), then \(\norm{\tau_h f - f}_p \to 0\).
Let \(\tau_h f(x) = f(x - h)\) and \(\omega(r) = \sup\theset{|f(x) - f(y)| : |x - y| \le r}\), which tends to \(0\) with \(r\). Then \(\norm{\tau_h f - f}_\infty \le \omega(|h|)\). For \(p < \infty\) and \(\eps > 0\), choose \(R\) with \(\int_{|x| > R}|f|^p < \eps\); for \(|h| \le 1\), the integral of \(|\tau_h f - f|^p\) over \(|x| > R + 1\) is less than \(2^p\eps\), and over \(|x| \le R + 1\) it is at most \(\omega(|h|)^p\, m(B(0, R+1))\). See E-KO5PK.
4. If \(\mu(X) < \infty\), then \(L^\infty(X) \subseteq L^2(X) \subseteq L^1(X)\); for counting measure on \(\ZZ\), \(\ell^1(\ZZ) \subseteq \ell^2(\ZZ) \subseteq \ell^\infty(\ZZ)\).
The Cauchy–Schwarz inequality gives \(\norm{f}_1 \le \mu(X)^{1/2}\norm{f}_2\), and \(\norm{f}_2 \le \mu(X)^{1/2}\norm{f}_\infty\). For sequences, \(|a_n| \le \norm{a}_{\ell^2}\) for each \(n\), and \(\sum|a_n|^2 \le \left(\sup_n|a_n|\right)\sum_n|a_n| \le \norm{a}_{\ell^1}^2\). The sequence \(a_n = 1/n\) for \(n \ge 1\), \(a_n = 0\) otherwise, lies in \(\ell^2 \setminus \ell^1\). See E-SDQ4U.