Let \(f\in L^\infty([0,1])\) and assume \(f\) is not zero almost everywhere. Define \[ a_n=\int_0^1|f(x)|^n\,dx. \] Prove that \[ \lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\|f\|_\infty. \]
Solution
Set \[ M=\|f\|_\infty>0. \]
1. The ratios are nondecreasing and bounded above by \(M\).
By the Cauchy--Schwarz inequality in $L^2$, \[ a_{n+1}^2 =\left(\int |f|^{n/2}|f|^{(n+2)/2}\right)^2 \le a_n a_{n+2}. \] Since \(a_n>0\), \[ \frac{a_{n+1}}{a_n}\le\frac{a_{n+2}}{a_{n+1}}. \] Thus the ratios are nondecreasing. Also \(|f|\le M\) almost everywhere, so \[ a_{n+1}\le M a_n, \] and hence \[ \frac{a_{n+1}}{a_n}\le M. \] Therefore the ratios converge to some \(L\le M\).
2. The limit cannot be smaller than \(M\).
Fix \(0<\varepsilon<M\). By the definition of essential supremum, the set \[ E_\varepsilon=\{x:|f(x)|>M-\varepsilon\} \] has positive measure. Hence \[ a_n\ge m(E_\varepsilon)(M-\varepsilon)^n. \] On the other hand, since the ratios increase to \(L\), each ratio is at most \(L\), so for \(n\ge1\), \[ a_n=a_1\prod_{k=1}^{n-1}\frac{a_{k+1}}{a_k}\le a_1L^{n-1}. \] Combining the two estimates and taking \(n\)th roots gives \[ M-\varepsilon \le m(E_\varepsilon)^{-1/n}a_1^{1/n}L^{(n-1)/n}. \] Letting \(n\to\infty\) yields \[ M-\varepsilon\le L. \] Since \(\varepsilon>0\) is arbitrary, \(M\le L\). Together with \(L\le M\), this gives \[ L=M=\|f\|_\infty. \]
3. Therefore\[ \boxed{ \lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\|f\|_\infty. } \]
Step 1 gives existence of the ratio limit \(L\), and step 2 proves \(L=\|f\|_\infty\).
4. Q.E.D.
Step 3 is the required limit.