This is an exercise from Rudin's Real and Complex analysis, which I'm solving. The question asks me to show that if $a_n=\int_{X}|f|^n d\mu$, then the limit of $\frac{a_{n+1}}{a_n}$ goes to $||f||_{\infty}$ as $n\rightarrow\infty$. I've easily shown the direction that the limit is $\le$ the infinity norm. I can't quite show the other direction, and would appreciate some help.
I've tried splitting the integral in $a_n$ into two pieces, one over the set when $|f|\leq c$ for any $0<c<||f||_\infty$ and another over its complement. This is basically trying to imitate the proof of the fact that the infinity norm is the limit of $p$ norms. However, I can't manipulate this to show that the $lim $ $inf$ of the quotient is $\geq$ the infinity norm.
I would really appreciate some help on this. Thank you.
First, since $$ \alpha_{n+1}=\int \lvert f\rvert\cdot \lvert f\rvert^nd\mu\le \lVert f \rVert_{\infty}\int \lvert f\rvert^nd\mu=\lVert f \rVert_{\infty}\alpha_n, $$ we have (note that $\alpha_n>0$), $$ \limsup_{n}\frac{\alpha_{n+1}}{\alpha_n}\le \lVert f \rVert_{\infty}. $$ As for the second inequality, consider the set $A=\{\lvert f\rvert\le c\}$ for some $0<c\le \lVert f \rVert_{\infty}$. Then $$ \liminf_n \frac{\alpha_{n+1}}{\alpha_n}\cdot\frac{1}{c}=\liminf_n \frac{\int_{A^c}\lvert f/c\rvert^{n+1}d\mu}{\int_{A^c}\lvert f/c\rvert^{n}d\mu}\ge 1, $$ where the first equality follows from the fact that $\lim_n \int_A |f/c|^nd\mu=0$.