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 $$\lim_{n\to \infty}\left( \int_a^b f(x)^n dx\right)^{1/n} = f(x_0).$$

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 $f$.
