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 
$I_n=\lim_{n\to \infty}\sqrt [n]{\lim _{k\to \infty}\sum_{i=1}^k f^n(x_i^*) dx}$
 A: 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 $\epsilon$, $$
\lim \left(\int_a^b f^n(x) dx\right)^{1/n}=\sup f
$$
A: Obviously, we have
$$
\begin{align}
\limsup_{n\to\infty}\left(\int_a^bf^n(x)\,\mathrm{d}x\right)^{1/n}
&\le\lim_{n\to\infty}(b-a)^{1/n}\left(\max_{[a,b]}f^n\right)^{1/n}\\
&=\max_{[a,b]}f\tag{1}
\end{align}
$$
Suppose that $f(x)\ge m$ on an interval of length $\delta\gt0$. Then, since $\frac{f(x)}{m}\ge1$ on this interval,
$$
\begin{align}
\liminf_{n\to\infty}\left(\int_a^bf^n(x)\,\mathrm{d}x\right)^{1/n}
&=m\liminf_{n\to\infty}\left(\int_a^b\frac{f^n(x)}{m^n}\,\mathrm{d}x\right)^{1/n}\\
&\ge m\lim_{n\to\infty}\delta^{1/n}\\
&=m\tag{2}
\end{align}
$$
If $f$ is continuous, then for any $\epsilon\gt0$, $f(x)\ge m=\max\limits_{[a,b]}f-\epsilon$ on some interval of positive length. Therefore, for any $\epsilon\gt0$,
$$
\small\max_{[a,b]}f-\epsilon\le\liminf_{n\to\infty}\left(\int_a^bf^n(x)\,\mathrm{d}x\right)^{1/n}\le\limsup_{n\to\infty}\left(\int_a^bf^n(x)\,\mathrm{d}x\right)^{1/n}\le\max_{[a,b]}f\tag{3}
$$
Since $(3)$ is true for any $\epsilon\gt0$, we have
$$
\lim_{n\to\infty}\left(\int_a^bf^n(x)\,\mathrm{d}x\right)^{1/n}=\max_{[a,b]}f\tag{4}
$$
A: Elementary tools (high school ones) reveal that
$$\lim_
{n\to\infty} \sqrt [n]{\int _a^b f^n(x) dx}=\operatorname{max}(f(x))$$
A: Okay, let's actually do this as a highschooler would or might be expected to do.
First, use the fact for positive $a_i$ that
$$\lim_{n \to \infty} (a_1^n + \cdots + a_k^n)^{1/n} = \max_{1\leq i \leq k} a_i.$$
How to show this? "Pull out" the largest $a_i$, say when $i=m$, and note that $(a_i/a_m)^n \to 0$.
So then what's the idea? The integral $\int_0^1 f(x)^n dx$ is basically the Riemann sum for large N plus an error term $\epsilon$:
$$\int_0^1 f(x)^n dx = \epsilon + \frac{1}{N} \sum_{i=1}^N f(i/N)^n.$$
Ignore the epsilon for now and take the $n$th root and use the previous fact mentioned to get that the limit is $\max f(i/N)$. Well, if $N$ is large enough, we can basically get the max of $f$ over $[0,1]$ as close as we like. So this tells us what the answer "should" be.
Now really, the error term $\epsilon = \epsilon(n,N)$. However, we can choose $N$ large enough so that $|f(i/N)-f(i+1/N)|<1$ and hence $\epsilon(n,N) < \epsilon(1, N)$. But using this puts a bound on $\epsilon$ independent of $n$, and if we include that term as we should and then take the limit, we can still show that the limit is really close to the max of $f$ on $[0,1]$ if $\max f(i/N)>1$. To make sure that happens we can just ignore all functions when it's not, and then do some scaling to make it work otherwise.
Is this a very rigorous argument? No. But I would hope a typical highschooler would follow it.
