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