$\frac{\int|f|^{n+1}}{\int|f|^n} \rightarrow ||f||_{\infty}$ Let $(X, B, μ)$ be a finite measure space. Let $f \in L^\infty (μ)$. Let $α_n = \int_X |f|^n$. Show that:
$$\lim_{n\rightarrow\infty} \frac{\alpha_{n+1}}{\alpha_{n}}=||f||_{\infty}$$
I could show $\alpha_n$'s converge to a limit. I am not sure how to use the finite measure property here.
 A: Claim 1: For any $f\in L^{\infty}(\mu$), $\lim_{n\rightarrow\infty}||f||_{n}=||f||_{\infty}$.
Proof: To ease our discussion, we assume that $f\geq0$. Otherwise,
replace $f$ with $|f|$.
Clearly, the above obviously holds if $||f||_{\infty}=0$ because
in which case $f=0$ $\mu$-a.e.. Consider the case that $M:=||f||_{\infty}>0$.
Let $\alpha\in(0,M)$ be arbitrary and define $A_{\alpha}=\{x\in X\mid f(x)>\alpha\}$.
Since $\alpha$ is not an essential upper bound of $f$, $\mu(A_{\alpha})>0$.
Note that 
\begin{eqnarray*}
\int f^{n} & \geq & \int_{A}f^{n}\\
 & \geq & \mu(A_{\alpha})\alpha^{n}.
\end{eqnarray*}
Therefore $||f||_{n}\geq\alpha\mu(A_{\alpha})^{\frac{1}{n}}$ and
hence $\liminf_{n}||f||_{n}\geq\alpha$. As $\alpha\in(0,M)$ is arbitrary,
we have $\liminf_{n}||f||_{n}\geq M$. On the other hand, $\int f^{n}\leq M^{n}\mu(X)$,
so $||f||_{n}\leq M\mu(X)^{\frac{1}{n}}.$ Therefore, $\limsup_{n}||f||_{n}\leq M$.
It follows that $\lim_{n}||f||_{n}$ exists and $\lim_{n}||f||_{n}=||f||_{\infty}.$
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Now, we go back to the original question. Clearly, without loss of
generality, we may assume that $f\geq0$ (to ease our discussion,
otherwise, replace $f$ with $|f|$). Denote $M:=||f||_{\infty}$.
We assume that $M>0$. Otherwise $\alpha_{n}=0$ for all $n$ and
$\frac{\alpha_{n+1}}{\alpha_{n}}$ is undefined. Note that $\alpha_{n+1}=\int f^{n+1}\leq\int f^{n}M=M\alpha_{n}$,
so $\frac{\alpha_{n+1}}{\alpha_{n}}\leq M$. Put $p=\frac{n+1}{n}$,
$q=n+1$, then $p,q\in(1,\infty)$ with $\frac{1}{p}+\frac{1}{q}=1$.
By Holder inequality, we have 
\begin{eqnarray*}
\alpha_{n} & = & \int f^{n}\cdot1\\
 & \leq & ||f^{n}||_{p}||1||_{q}\\
 & = & \left(\alpha_{n+1}\right)^{\frac{n}{n+1}}\left(\mu(X)\right)^{\frac{1}{n+1}}.
\end{eqnarray*}
Simplifying it yields $\frac{\alpha_{n+1}}{\alpha_{n}}\geq||f||_{n}\left[\frac{1}{\mu(X)}\right]^{\frac{1}{n}}$
and hence, 
$$
||f||_{n}\left[\frac{1}{\mu(X)}\right]^{\frac{1}{n}}\leq\frac{\alpha_{n+1}}{\alpha_{n}}\leq M.
$$
Note that $||f||_{n}\rightarrow M$ by Claim 1 and $\left[\frac{1}{\mu(X)}\right]^{\frac{1}{n}}\rightarrow1$.
We conclude that $\lim_{n}\frac{\alpha_{n+1}}{\alpha_{n}}$ exists and $\lim_{n}\frac{\alpha_{n+1}}{\alpha_{n}}=M$.
