$f$ is a positive continuous function on a compact interval $[a,b]$. Determine the limit
$\lim_{n \to \infty}[\int_{a}^{b}f(x)^ndx]^{1/n}$.
For this question, isn't the limit just the sup norm of $f$? If it is, how to show it formally? (and why does $f$ have to be positive?)
Thanks
On
It is clear that $(\int_a^{b} f(x)^{n}dx)^{1/n} \leq M$ where $M =\sup \{|f(x)|: 0\leq x \leq 1\}$. Not let $\epsilon >0$ and choose a point $x$ such that $f(x) >M-\epsilon$. There exists $\delta >0$ such that $f(y) >M-\epsilon$ for all $y \in (x-\delta,x+\delta)$. Now $(\int_a^{b} f(x)^{n}dx)^{1/n} \geq (\int_{x-\delta}^{x+\delta} f(x)^{n}dx)^{1/n} \geq (M-\epsilon) (2\delta)^{1/n}$. Since $(2\delta)^{1/n} \to 1$ we are done.
Positivity of $f$ is not required provided $f(x)$ is replaced by $|f(x)|$.
It needs to be posetive to ensure the limit exists. For example, let $f(x) = -1$ and $[a,b] = [0,1]$ we then have that $$ \int_a^b f(x)^n dx = \begin{cases} 1, & \text{for n even} \\ -1, & \text{for n odd} \end{cases} $$ which means it does not converge for $n\rightarrow \infty$ and you will have to take roots from negative numbers. For a formal proof of your question you can check this older question: Limit of $L^p$ norm
By the way if you are checking if the norm converges then your integral should look like this: $$ \left[ \int_a^b \vert f(x)\vert ^n dx\right] ^\frac{1}{n} $$ and in this case $f$ does not need to be positive.