If $I_n=\sqrt[n]{\int _a^b f^n(x) dx}$,for $n\ge 1$.Find with proof $\lim _{n \to \infty}I_n$ Let $a,b \in \mathbb R ,a<b$ and let  $f:[a,b] \to [0,\infty) $ a continuous and non constant.

attempt,using Reimann series

$I_n=\sqrt [n]{\int _a^b f^n(x) dx}$
$I_n=\sqrt [n]{\lim _{k \to \infty}\sum_{i=1}^k f^n(x_i^*) dx}$
for each $i=1,2,3...n$
$x_i\in [x_{i-1},x_i]$
so our question is what is 
$I_n=\lim_{n\to \infty}\sqrt [n]{\lim _{k\to \infty}\sum_{i=1}^k f^n(x_i^*) dx}$

Let $\epsilon > 0$.
 1. As $f(x)\le \sup f$, $$\int_a^b  f^n(x) dx \le (b-a)(\sup f)^n
\\ \limsup \left(\int_a^b f^n(x) dx\right)^{1/n} \le \limsup (b-a)^{1/n}\sup f = \sup f
$$
 2. There are $a'<b'$ such as $a'<y<b' \implies f(y) > \sup f-\epsilon$, because $f$ is continuous.
$$
\liminf \left(\int_a^b f^n(x) dx\right)^{1/n}
\ge \liminf \left(\int_{a'}^{b'} (\sup f
-\epsilon)^n dx\right)^{1/n} = \sup f
-\epsilon
$$As it is true for every $\epsilon$, $$
\lim \left(\int_a^b f^n(x) dx\right)^{1/n}=\sup f
$$
