Prove that $\lim_{n \to \infty}\left(\int_{a}^{b}|f(x)|^ndx\right)^{\frac{1}{n}} = \sup_{x \in [a,b]} |f(x)|$ $f:[a,b] \rightarrow \mathbb{R}$ is continuous.Prove that $$\lim_{n \to \infty}\left(\int_{a}^{b}|f(x)|^n dx\right)^{\frac{1}{n}} = \sup_{x \in [a,b]} |f(x)|$$
I was thinking of Holder's inequality for integrals, but I got: $(\int_{a}^{b}|f(x)|^n dx)^{\frac{1}{n}} \ge (\frac{1}{b-a})^{1-\frac{1}{n}}\int_{a}^{b}|f(x)dx|$ but how to bring equality with supremum? 
 A: Put $M = \sup_{x \in [a,b]} \lvert f(x) \rvert$. We see $$\left( \int_a^b \lvert f(x) \rvert^n dx \right)^{1/n} \le \left( \int_a^b M^n dx \right)^{1/n}  = M(b-a)^{1/n}, \,\,\,\,\, \text{ for all } n \in \mathbb N.$$ Sending $n \to \infty$ gives $$\lim_{n\to\infty}\left( \int_a^b \lvert f(x) \rvert^n dx \right)^{1/n} \le M.$$ On the other hand, there is some $x^*$ which achieves the supremum so by continuity, for any $\epsilon > 0$, we can find $\delta > 0$ such that $\lvert f(x) \rvert \ge M - \epsilon$ when $x \in (x^* -\delta, x^*+\delta)$ (assuming that $x^*$ is not on the boundary; the proof can easily be adjusted if $x^* =a$ or $x^* = b$). Then we see $$\left( \int_a^b \lvert f(x) \rvert^n dx \right)^{1/n} \ge \left( \int_{x^*-\delta}^{x^*+\delta} \lvert f(x) \rvert^n dx \right)^{1/n} \ge \left( \int_{x^*-\delta}^{x^*+\delta} (M-\epsilon)^n dx \right)^{1/n} = (2\delta)^{1/n}(M-\epsilon)$$ for all $n \in \mathbb N$. Sending $n\to\infty$ gives $$ \lim_{n\to\infty} \left( \int_a^b \lvert f(x) \rvert^n dx \right)^{1/n} \ge M-\epsilon.$$ But $\epsilon > 0$ is arbitrary, so we can send it to zero and see $$\lim_{n\to\infty}\left( \int_a^b \lvert f(x) \rvert^n dx \right)^{1/n} \ge M.$$ Thus $$\lim_{n\to\infty}\left( \int_a^b \lvert f(x) \rvert^n dx \right)^{1/n} = M.$$ NOTE: this isn't technically correct. Of course a priori we don't actually know that $\lim_{n\to\infty}\left( \int_a^b \lvert f(x) \rvert^n dx \right)^{1/n}$ exists, so the first time we take the limit above, we should really use $\limsup$ and the second time, we should really use $\liminf$. Then we actually proved $$M \le \liminf_{n\to\infty}\left( \int_a^b \lvert f(x) \rvert^n dx \right)^{1/n} \le \limsup_{n\to\infty}\left( \int_a^b \lvert f(x) \rvert^n dx \right)^{1/n} \le M$$ which of course implies that the limit exists and is $M$.
