# Why is it true that $\lim_{n\to\infty} \left(\int_a^b \left|f(x)\right|^ndx\right)^{1/n} = \text{sup}\{\left|f(x)\right| : x \in [a,b]\}$? [duplicate]

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 $$\lim_{n\to\infty} \left(\int_a^b \left|f(x)\right|^ndx\right)^{1/n} = \text{sup}\{\left|f(x)\right| : x \in [a,b]\}$$

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.

## marked as duplicate by Jack, Community♦Mar 17 '17 at 22:55

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.
• Can I get clarification on why that inequality holds? Is $\delta(M-\epsilon)^n$ the maximum of the integral of $|f|^n$ on just the interval of length $\leq \delta$? So we know $\int_a^b |f(x)|^ndx$ is greater because that's just a portion. Then $M^n(b-a)$ is the maximum of the entire integral because $M^n$ is the supremum of $|f|^n$ so that is just the area of a rectangle with height $M^n$ and width $b-a$. – user20354139 Mar 17 '17 at 22:10
• On an interval of length $\delta$, $|f(x)|^n > |M - \epsilon|^n$, so the integral over that interval is at least $\delta |M-\epsilon|^n$, and since the integrand is nonnegative the integral over $[a,b]$ is at least that. – Robert Israel Mar 18 '17 at 0:11