Let $(X,M,\mu)$ be a measure space , with $\mu (X)<\infty$, and let $f:X \to R$ be a M-measurable function such that $0<||f||_\infty <\infty$
Define $\alpha_n:= ||f||_n ^ n$
Prove that $\lim_{n\rightarrow\infty} \frac {\alpha_{n+1}}{\alpha_n }= ||f||_\infty$
Sketch: WLOG, $\mu(X) = 1, f\ge 0.$ Then for any $n,$
$$\frac{\int_X f^{n+1} \, d\mu}{\int_X f^{n} \, d\mu}\le \frac{ \|f\|_\infty\int_X f^{n} \, d\mu}{\int_X f^{n} \, d\mu} = \|f\|_\infty.$$
In the other direction, note that Jensen implies
$$(\int_X f^{n}\, d\mu)^{(n+1)/n}\le \int_X f^{n+1}\, d\mu .$$
Thus
$$\|f\|_n = \frac{(\int_X f^{n}\, d\mu)^{(n+1)/n}}{\int_X f^{n} \, d\mu} \le \frac{\int_X f^{n+1} \, d\mu}{\int_X f^{n} \, d\mu}.$$
Now use the well known limit $\|f\|_n \to \|f\|_\infty.$
