Let \(g \in L^\infty([0, 1])\).
Prove that \[ \lim_{p \to \infty} \|g\|_{L^p([0, 1])} = \|g\|_{L^\infty([0, 1])}. \]
Prove that the map \[ \Lambda_g: L^1([0, 1]) \to \mathbb{C}, \qquad f \mapsto \int_0^1 f(x) g(x) \, dx \] defines a continuous linear functional in \((L^1([0, 1]))^*\) with \(\|\Lambda_g\|_{(L^1)^*} = \|g\|_{L^\infty([0, 1])}\).
Solution
Goal: Prove the \(L^p \to L^\infty\) norm limit as \(p \to \infty\) on \([0, 1]\) in (a), and compute the operator norm \(\|\Lambda_g\|_{(L^1)^*} = \|g\|_{L^\infty}\) in (b).
1. Part (a): Upper bound \(\limsup_{p \to \infty} \|g\|_{L^p} \le \|g\|_{L^\infty}\).
1.1. For almost every \(x \in [0, 1]\), \(|g(x)| \le \|g\|_{L^\infty}\).
1.2. Integrating over \([0, 1]\) for any \(p \ge 1\):\[\|g\|_{L^p} = \left( \int_0^1 |g(x)|^p \, dx \right)^{1/p} \le \left( \int_0^1 \|g\|_{L^\infty}^p \, dx \right)^{1/p} = \|g\|_{L^\infty} (m([0, 1]))^{1/p} = \|g\|_{L^\infty} (1)^{1/p} = \|g\|_{L^\infty}.\]
1.3. Taking the limit superior as \(p \to \infty\):\[\limsup_{p \to \infty} \|g\|_{L^p} \le \|g\|_{L^\infty}.\]
2. Part (a): Lower bound \(\liminf_{p \to \infty} \|g\|_{L^p} \ge \|g\|_{L^\infty}\).
2.1. If \(\|g\|_{L^\infty} = 0\), then \(g = 0\) almost everywhere, so \(\|g\|_{L^p} = 0\) for all \(p\) and the limit is \(0\).
2.2. Assume \(\|g\|_{L^\infty} > 0\). For any \(0 < M < \|g\|_{L^\infty}\), define the superlevel set \(E_M = \{x \in [0, 1] : |g(x)| > M\}\).
2.3. By definition of the essential supremum, \(m(E_M) > 0\).
2.4. Restricting the \(L^p\) integral to \(E_M\):\[\|g\|_{L^p} = \left( \int_0^1 |g(x)|^p \, dx \right)^{1/p} \ge \left( \int_{E_M} |g(x)|^p \, dx \right)^{1/p} \ge \left( \int_{E_M} M^p \, dx \right)^{1/p} = M (m(E_M))^{1/p}.\]
2.5. Since \(m(E_M) \in (0, 1]\), \((m(E_M))^{1/p} \to 1\) as \(p \to \infty\).
2.6. Thus \(\liminf_{p \to \infty} \|g\|_{L^p} \ge M \lim_{p \to \infty} (m(E_M))^{1/p} = M\).
2.7. Taking the supremum over all \(M < \|g\|_{L^\infty}\) yields \(\liminf_{p \to \infty} \|g\|_{L^p} \ge \|g\|_{L^\infty}\).
3. Part (a): Conclusion of the limit.
Combining 1 and 2 gives \(\lim_{p \to \infty} \|g\|_{L^p} = \|g\|_{L^\infty}\).
4. Part (b): Well-definedness and upper bound \(\|\Lambda_g\| \le \|g\|_{L^\infty}\).
4.1. Linearity of \(\Lambda_g\) follows directly from linearity of the Lebesgue integral.
4.2. For any \(f \in L^1([0, 1])\):\[|\Lambda_g(f)| = \left| \int_0^1 f(x) g(x) \, dx \right| \le \int_0^1 |f(x)| |g(x)| \, dx \le \|g\|_{L^\infty} \int_0^1 |f(x)| \, dx = \|g\|_{L^\infty} \|f\|_{L^1}.\]
4.3. Therefore \(\Lambda_g\) is a bounded linear functional on \(L^1([0, 1])\) with operator norm\[\|\Lambda_g\|_{(L^1)^*} = \sup_{f \neq 0} \frac{|\Lambda_g(f)|}{\|f\|_{L^1}} \le \|g\|_{L^\infty}.\]
5. Part (b): Lower bound \(\|\Lambda_g\| \ge \|g\|_{L^\infty}\).
5.1. If \(\|g\|_{L^\infty} = 0\), then \(\|\Lambda_g\| = 0 = \|g\|_{L^\infty}\).
5.2. Assume \(\|g\|_{L^\infty} > 0\). Fix \(0 < M < \|g\|_{L^\infty}\) and let \(E = \{x \in [0, 1] : |g(x)| \ge M\}\), so \(m(E) > 0\).
5.3. Define test function \(f(x) \in L^1([0, 1])\) by:\[f(x) = \frac{\overline{\operatorname{sgn}(g(x))}}{m(E)} \mathbf{1}_E(x) = \begin{cases} \frac{\overline{g(x)}}{|g(x)| m(E)} & x \in E \text{ and } g(x) \neq 0, \\ 0 & \text{otherwise.} \end{cases}\]
5.4. Compute the \(L^1\) norm of \(f\):\[\|f\|_{L^1} = \int_E \frac{1}{m(E)} \, dx = \frac{m(E)}{m(E)} = 1.\]
5.5. Evaluate \(\Lambda_g(f)\):\[\Lambda_g(f) = \int_0^1 f(x) g(x) \, dx = \frac{1}{m(E)} \int_E \frac{\overline{g(x)} g(x)}{|g(x)|} \, dx = \frac{1}{m(E)} \int_E |g(x)| \, dx \ge \frac{1}{m(E)} \int_E M \, dx = M.\]
5.6. Since \(\|f\|_{L^1} = 1\), the operator norm satisfies \(\|\Lambda_g\| \ge |\Lambda_g(f)| \ge M\).
5.7. Since this holds for all \(M < \|g\|_{L^\infty}\), we conclude \(\|\Lambda_g\|_{(L^1)^*} \ge \|g\|_{L^\infty}\).
6. Conclusion:
Combining 4 and 5 yields \(\|\Lambda_g\|_{(L^1)^*} = \|g\|_{L^\infty}\).