Let \((X, \mathcal{M}, \mu)\) be a measure space, and let \(f \in L^{1}(\mu) \cap L^{\infty}(\mu)\). Prove that: \[ \lim_{p \to \infty} \|f\|_p = \|f\|_{\infty}. \]
Solution
If \(\norm f_\infty=0\) then \(f=0\) almost everywhere and every \(\norm f_p\) is \(0\). Assume \(\norm f_\infty>0\), so that \(0<\norm f_1<\infty\).
1. \(\limsup_{p\to\infty}\norm f_p\le\norm f_\infty\).
For \(p>1\), \(\int\abs f^p\le\norm f_\infty^{p-1}\norm f_1\), so \(\norm f_p\le\norm f_\infty^{(p-1)/p}\norm f_1^{1/p}\), and the right side tends to \(\norm f_\infty\).
2. \(\liminf_{p\to\infty}\norm f_p\ge\norm f_\infty\).
Let \(0<\eps<\norm f_\infty\) and \(E=\{\abs f>\norm f_\infty-\eps\}\). By definition of the essential supremum \(\mu(E)>0\), and \(\mu(E)\le\norm f_1/(\norm f_\infty-\eps)<\infty\). Then \(\norm f_p\ge(\norm f_\infty-\eps)\mu(E)^{1/p}\), whose limit is \(\norm f_\infty-\eps\). Let \(\eps\to0\).
3. Q.E.D.
Steps 1 and 2 give \(\lim_{p\to\infty}\norm f_p=\norm f_\infty\).