# p-norm and the sup norm

$$f$$ 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

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.