Let \(f\in L^\infty([0,1])\). Prove that \[ \lim_{p\to\infty}\left(\int_0^1|f|^p\,dx\right)^{1/p}=\|f\|_\infty. \]
Give an example showing that the corresponding statement fails if \([0,1]\) is replaced by \(\mathbb R\).
Solution
1. The limit on \([0,1]\) equals the essential supremum.
Let \[ M=\|f\|_\infty. \] Since \([0,1]\) has measure \(1\), \[ \|f\|_p\le M \qquad(1\le p<\infty). \] Thus \[ \limsup_{p\to\infty}\|f\|_p\le M. \]
Fix \(\varepsilon>0\). By the definition of essential supremum, the set \[ E_\varepsilon=\{x\in[0,1]:|f(x)|>M-\varepsilon\} \] has positive measure. Therefore \[ \|f\|_p^p \ge \int_{E_\varepsilon}|f|^p \ge (M-\varepsilon)^p m(E_\varepsilon), \] so \[ \|f\|_p\ge (M-\varepsilon)m(E_\varepsilon)^{1/p}. \] Letting \(p\to\infty\) gives \[ \liminf_{p\to\infty}\|f\|_p\ge M-\varepsilon. \] Since \(\varepsilon>0\) is arbitrary, \[ \liminf_{p\to\infty}\|f\|_p\ge M. \] Combining the two bounds yields \[ \lim_{p\to\infty}\|f\|_p=M=\|f\|_\infty. \]
2. The statement fails on \(\mathbb R\) for arbitrary \(L^\infty\) functions.
Take \[ f(x)\equiv1. \] Then \(f\in L^\infty(\mathbb R)\) and \[ \|f\|_\infty=1, \] but for every finite \(p\), \[ \int_{\mathbb R}|f(x)|^p\,dx=\infty. \] Thus the finite-\(p\) norms are infinite, so they do not converge to \(\|f\|_\infty=1\).