Suppose that \(f:[a,b]\to\mathbb{R}\) is continuous, \(f\geq 0\) on \([a,b]\), and put \(M=\sup\{f(x):x\in[a,b]\}\). Prove that \[\lim_{p\to\infty}\left(\int_a^b f(x)^p\,dx\right)^{1/p}=M.\]
Solution
1. Upper bound: \((\int_a^b f^p)^{1/p} \le M\,(b-a)^{1/p}\).
\(0 \le f \le M\) on \([a,b]\), so \(\int_a^b f^p \le M^p (b-a)\), and \((b-a)^{1/p} \to 1\).
2. Hence \(\limsup_{p\to\infty}(\int_a^b f^p)^{1/p} \le M\).
1 and \((b-a)^{1/p} \to 1\).
3. Lower bound: for each \(0 < \eps < M\), \(\liminf_{p\to\infty}(\int_a^b f^p)^{1/p} \ge M - \eps\).
since \(f\) is continuous, it attains \(M\) at some \(x_0 \in [a,b]\).
4. \(\lim_{p\to\infty}(\int_a^b f^p)^{1/p} = M\).
5. Q.E.D.