How to prove this theorem about limit of integral? I am working on this problem.
I know that I need to find $x_0$ that is the supremum of a function $f$.
If $f$ is non negative continuous function on the interval $[a,b]$ then there exists $x_0$ such that $$\lim_{n\to \infty}\left( \int_a^b f(x)^n dx\right)^{1/n} = f(x_0).$$
 A: Supposing the limit exists:
Let $M = \max_{x\in[a,b]}f(x)$ and $m = \min_{x\in[a,b]}f(x)$. Then
$$\left(\int_a^b f(x)^ndx\right)^{\frac{1}{n}}\le\left(\int_a^b M^ndx\right)^{\frac{1}{n}} = (b-a)^{\frac{1}{n}}M\xrightarrow{n\to\infty}M$$
and, similarly,
$$\left(\int_a^b f(x)^ndx\right)^{\frac{1}{n}}\ge\left(\int_a^b m^ndx\right)^{\frac{1}{n}} = (b-a)^{\frac{1}{n}}m\xrightarrow{n\to\infty}m\ .$$
Therefore,
$$\lim_{x\to\infty}\left(\int_a^b f(x)^ndx\right)^{\frac{1}{n}}\in[m,M]$$
and the result forllows by continuity of the function $f$.
A: Let $M$ be the maximum of $f(x)$ over $[a,b]$.
First,
$$
\left(\int_a^bf(x)^ndx\right)^{1/n}\le\left(\int_a^bM^ndx\right)^{1/n}=M(b-a)^{1/n}\to M\quad(n\to\infty).
$$
Let $x_0$ be the point where $f(x_0)=M$. Fix arbitrary small $\epsilon>0$. There exists $\delta>0$ such that for $x\in[x_0-\delta/2,x_0+\delta/2]$ we have $f(x)\ge M-\epsilon$ (or if $x_0=a$, or $b$ we can take one side $\delta$-neighborhood). This means that
$$
\left(\int_a^bf(x)^ndx\right)^{1/n}\ge\left(\int_{x_0-\delta}^{x_0+\delta}(M-\epsilon)^ndx\right)^{1/n}=(M-\epsilon)\delta^{1/n}\to M-\epsilon\quad(n\to\infty).
$$
This means that for sufficiently large $n$ both inequalities can be written as
$$
M-2\epsilon\le\left(\int_a^bf(x)^ndx\right)^{1/n}\le M+\epsilon,
$$
which make the limit existing and equal to $M$.
