Limit Integral equal to a supremun: $\lim_{n\to \infty} \left[\int_a^b (f(x))^n dx\right]^{{1}/{n}}=\sup_{x\in [a,b]} f(x).$

Question

I'd like a hint to solve the following problem:

Given a $f$ continuos and positive on $[a,b]$ prove that $$\lim_{n\to \infty} \left[\int_a^b (f(x))^n dx\right]^{{1}/{n}}=\sup_{x\in [a,b]} f(x).$$

Thanks a lot!

You are missing $\lim_{n \to \infty}$. – Umberto P. Jun 05 '14 at 15:07 — Umberto P., Jun 05 '14 at 15:07
Also, the $]^{\frac 1n}$ is on the wrong side of the $dx$. – Umberto P. Jun 05 '14 at 15:15 — Umberto P., Jun 05 '14 at 15:15

score 4 · Accepted Answer · answered Jun 05 '14 at 15:14

Let $M = \sup_{x \in [a,b]} f(x)$. If $M = 0$ then $f$ is identically zero so there isn't much to prove. So assume $M > 0$. Then $$\int_a^b f(x)^n \, dx \le (b-a) M^n$$ so that $$\left( \int_a^b f(x)^n \, dx \right)^{1/n} \le (b-a)^{1/n}M.$$

On the other hand, if $0 < A < M$ then $f(x) > A$ for all $x$ in some subinterval $[c,d] \subset [a,b]$ (here we use continuity of $f$). Thus $$ \int_a^b f(x)^n \, dx \ge \int_c^d f(x)^n \, dx \ge (d-c)A^n$$ so that $$\left( \int_a^b f(x)^n \, dx \right)^{1/n} \ge (d-c)^{1/n} A.$$

Can you take it from here?

Limit Integral equal to a supremun: $\lim_{n\to \infty} \left[\int_a^b (f(x))^n dx\right]^{{1}/{n}}=\sup_{x\in [a,b]} f(x).$

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