TITLE: Why is it true that $\lim_{n\to\infty} \left(\int_a^b \left|f(x)\right|^ndx\right)^{1/n} = \text{sup}\{\left|f(x)\right| : x \in [a,b]\}$?
QUESTION [7 upvotes]: Can someone explain the intuition behind this/ what the proof might look for this?
If $f$ is a continuous, real-valued function on $[a,b]$, then
\begin{equation}
\lim_{n\to\infty} \left(\int_a^b \left|f(x)\right|^ndx\right)^{1/n} = \text{sup}\{\left|f(x)\right| : x \in [a,b]\}
\end{equation}
It seems like it's a way of averaging $\left|f(x)\right|$, similar to how the Lyapunov number is an average of all the slopes of an orbit in a dynamical system.

REPLY [8 votes]: Let $M = \sup \{|f(x)|: x \in [a,b]\}$, and suppose $M > 0$.  For  $M > \epsilon > 0$, if $\delta > 0$ is small enough there is an interval of length $\le \delta$ on which $|f(x)| > M - \epsilon$.  Therefore
$$ M^n (b-a) \ge \int_a^b |f(x)|^n\; dx \ge \delta (M-\epsilon)^n $$
Take the $1/n$ power, and then limit as $n \to \infty$, and use the Squeeze Theorem.