Show that if \(E \subseteq \mathbb{R}^n\) is a measurable set with finite measure \(\mu(E) < \infty\), then for any measurable function \(f: E \to \mathbb{C}\), \[ \lim_{p \to \infty} \|f\|_{L^p(E)} = \|f\|_{L^\infty(E)}. \]
Solution
1. Upper bound: \(\limsup_{p \to \infty} \|f\|_{L^p(E)} \le \|f\|_{L^\infty(E)}\).
1.1. If \(\|f\|_{L^\infty(E)} = \infty\), the upper bound holds vacuously.
1.2. If \(\mu(E) = 0\), then \(\|f\|_{L^p(E)} = 0\) for all \(p\) and \(\|f\|_{L^\infty(E)} = 0\), so the limit is \(0\).
1.3. Assume \(\|f\|_{L^\infty(E)} < \infty\) and \(0 < \mu(E) < \infty\).
1.4. For almost every \(x \in E\), \(|f(x)| \le \|f\|_{L^\infty(E)}\).
1.5. Integrating over \(E\):\[\|f\|_{L^p(E)} = \left( \int_E |f(x)|^p \, d\mu \right)^{1/p} \le \left( \int_E \|f\|_{L^\infty(E)}^p \, d\mu \right)^{1/p} = \|f\|_{L^\infty(E)} (\mu(E))^{1/p}.\]
1.6. Since \(\mu(E) \in (0, \infty)\), \(\lim_{p \to \infty} (\mu(E))^{1/p} = 1\).
1.7. Taking the limit superior as \(p \to \infty\):\[\limsup_{p \to \infty} \|f\|_{L^p(E)} \le \|f\|_{L^\infty(E)} \lim_{p \to \infty} (\mu(E))^{1/p} = \|f\|_{L^\infty(E)}.\]
2. Lower bound: \(\liminf_{p \to \infty} \|f\|_{L^p(E)} \ge \|f\|_{L^\infty(E)}\).
2.1. If \(\|f\|_{L^\infty(E)} = 0\), then \(f = 0\) almost everywhere, so \(\|f\|_{L^p(E)} = 0\) for all \(p \ge 1\) and the inequality holds.
2.2. Assume \(\|f\|_{L^\infty(E)} > 0\). For any real number \(M\) with \(0 < M < \|f\|_{L^\infty(E)}\), define the superlevel set\[A_M = \{x \in E : |f(x)| > M\}.\]
2.3. By definition of the essential supremum, \(\mu(A_M) > 0\).
2.4. Restricting the integral to \(A_M\):\[\|f\|_{L^p(E)} = \left( \int_E |f(x)|^p \, d\mu \right)^{1/p} \ge \left( \int_{A_M} |f(x)|^p \, d\mu \right)^{1/p} \ge \left( \int_{A_M} M^p \, d\mu \right)^{1/p} = M (\mu(A_M))^{1/p}.\]
2.5. Since \(\mu(A_M) \in (0, \infty)\), \(\lim_{p \to \infty} (\mu(A_M))^{1/p} = 1\).
2.6. Taking the limit inferior as \(p \to \infty\):\[\liminf_{p \to \infty} \|f\|_{L^p(E)} \ge M \lim_{p \to \infty} (\mu(A_M))^{1/p} = M.\]
2.7. If \(\|f\|_{L^\infty(E)} < \infty\), taking the supremum over all \(M < \|f\|_{L^\infty(E)}\) yields \(\liminf_{p \to \infty} \|f\|_{L^p(E)} \ge \|f\|_{L^\infty(E)}\).
2.8. If \(\|f\|_{L^\infty(E)} = \infty\), the bound holds for all \(M > 0\), so \(\lim_{p \to \infty} \|f\|_{L^p(E)} = \infty = \|f\|_{L^\infty(E)}\).
3. Q.E.D.
Combining steps 1 and 2 gives \(\lim_{p \to \infty} \|f\|_{L^p(E)} = \|f\|_{L^\infty(E)}\).