Prove that $\lim_{n \to \infty} (\int ^b_af(x)^n dx)^\frac{1}{n}=\sup _{x \in [a,b]} f(x)$

Question

Let $f\in C[a,b]$. prove that $$\lim_{n \to \infty} \left(\int ^b_af(x)^n dx\right)^\frac{1}{n}=\sup _{x \in [a,b]} |f(x)|$$

i really have no idea where to start

Answer 1

I assume $f$ is positive (as stated there is a problem with the left term).

Let $C := \sup_{[a,b]} f(x)$

$\left(\int ^b_af(x)^n dx\right)^\frac{1}{n} ≤ \left(\int ^b_aC^n dx\right)^\frac{1}{n} = C(b-a)^{1/n}$

Now, fix $\epsilon$ and consider $[c,d], d >c$ on which $f(x) > C - \epsilon$

$\left(\int ^b_af(x)^n dx\right)^\frac{1}{n} ≥ \left(\int ^d_cf(x)^n dx\right)^\frac{1}{n} ≥ (C- \epsilon)(d-c)^{1/n}$

To conclude, use $y^{1/n} \to 1$, for all $y > 0$.