If $0<|E|<\infty$, $0<||f||_{\infty}<\infty$ and $a_k=||f||_k^k$ then $a_{k+1}/a_k\to ||f||_{\infty}$ 
Let $E\subset \mathbb{R}^n$ be a lebesgue measurable set such that $0<|E|<\infty$ abd let $f\in L^{\infty}(E)$ such that $0<||f||_{\infty}$. Define for $k\in \mathbb{N}$, $a_k=||f||_k^k=\int_{E}|f(x)|^kdx$. Prove that $a_{k+1}/a_k\to ||f||_{\infty}$ when $k\to \infty$.

It is easy to prove that $\sqrt[k]{a_k}\to ||f||_{\infty}$. Moreover, I could prove that $a_{k+1}\leq a_k+||f||_{\infty}^{k+1}|E|$ and $a_k\leq \left (|E|+a_{k+1}\right )^k$, but I cannot conclude the problem.
 A: Let us define the function 
$$
\varphi(p) := \int_E |f|^p.
$$
The function $\psi:=\log\varphi$ is convex in $[1, +\infty)$ (see [1] below).
In particular, this implies that
$$
\mathbb{N} \ni k \mapsto \psi(k+1) - \psi(k)
$$
is non-decreasing, hence the sequence
$$
\frac{a_{k+1}}{a_k} = e^{\psi(k+1)-\psi(k)} 
$$
admits a limit $l \in (0, +\infty]$.
But then this limit coincides with $\lim_n \sqrt[n]{a_n} = \|f\|_\infty$ (see Baby Rudin Thm. 3.37). 
[1] Convexity of $\psi$: if $r,s\geq 1$ and $p = \lambda r + (1-\lambda) s$ for some $\lambda\in (0,1)$:
$$
\int |f|^p = \int |f|^{\lambda r + (1-\lambda) s} \leq
\left(\int|f|^{\lambda r/\lambda}\right)^\lambda
\left(\int |f|^{(1-\lambda) s / (1-\lambda)}\right)^{1-\lambda}
= \|f\|_r^{r\lambda} \|f\|_s^{s(1-\lambda)},
$$
hence
$$
\varphi(\lambda r + (1-\lambda) s) \leq \varphi(r)^\lambda \varphi(s)^{1-\lambda}.
$$
A: Let me treat a simplified case, from which the full result follows: Assume $f\ge 0, |E|=1, \|f\|_\infty = 1.$ All integrals are over $E.$ We have
$$\tag 1\left (\int f^k\right )^{1/k} = \frac{(\,\int f^k\,)^{(k+1)/k} }{\int f^k} \le \frac{\int (f^k)^{(k+1)/k}}{\int f^k}= \frac{\int f^{k+1}}{\int f^k}\le 1.$$
The first $\le$ sign follows from Jensen's inequality, since the function $x\to x^{(k+1)/k}$ is convex and we're on a set of measure $1.$ As you noted, the left side of $(1)$ converges to $\|f\|_\infty =1.$ Thus we're done by the squeeze theorem.
