Let $f$ be continuous on $[a,b]$ with $f(x) \geq 0$ for all $x \in [a,b]$. Prove
$$\lim_{n \to \infty} \left( \int_{a}^{b} (f(x))^n \, dx\right)^{\frac{1}{n}} = \max_{x \in [a,b]} \{f(x)\}.$$
Attempted Solution:
Note first that it suffices to prove the above for $\max_{x \in [a,b]} \{f(x)\} = 1$, since we may scale accordingly. Let $f$ attain it's maximum at a point $c$. Let $\epsilon >0$. Since $f$ is uniformly continuous, there exists $\delta > 0$ so that $f(x) \in (1-\epsilon,1]$ whenever $x \in (c-\delta,c+\delta)$. If there exist multiple points at which $f(x) = 1$, let $$D = \bigcup_{i \in I} B_\delta(c_i),$$ an open set with finite measure. Now, partition $D$ into disjoint subsets, $D_i$, each having finite measure. We have on each $D_i$, $(f(x))^n \in ((1-\epsilon)^n,1]$. Now, $$(1-\epsilon)(\mu(D))^{\frac{1}{n}} \leq \left( \int_{a}^{b} (f(x))^n \, dx \right)^\frac{1}{n} \leq (b-a)^\frac{1}{n}$$ and since $\lim \limits_{n \to \infty} x^\frac{1}{n} = 1$,
$$(1-\epsilon) \leq \lim_{n \to \infty} \left( \int_{a}^{b} (f(x))^n \, dx \right)^{\frac{1}{n}} \leq 1$$
as desired.
Any issues here? Anything to make it more succinct? Moreover, where does continuity come into play to make this an essentially simpler proof than the general $L^p$ norm limit proof I've just found here?
