Limit of the nth root of integral of $f$ is $\sup f(x)$ Let $f:[0,1]\to\Bbb{R}$ be continuous.
How do I show that $\lim_{n\to\infty}\left(\int^1_0{f(x)^n}\right)^{1/n}=\sup_{0\leq x\leq 1}f(x)$?
I tried writing $\left(\int^1_0{f(x)^n}\right)^{1/n}=\left(f(x)^{n+1}/(n+1)\right)^{1/n}$. But I don't know what to do next. 
 A: Assume $f$ is nonnegative.
Let $M:=\sup_{x \in [0,1]} f(x)$. Then notice that for every $n$, we have that $$\bigg(\int_0^1 f(x)^ndx\bigg)^{\frac{1}{n}} \leq \bigg(\int_0^1 M^n dx\bigg)^{\frac{1}{n}} = M$$ Therefore $\limsup_{n \to \infty} \big(\int_0^1 f(x)^ndx\big)^{\frac{1}{n}} \leq M$.
On the other hand, let $\alpha$ be any nonnegative real number strictly less than $M$. Then $f(x_0)=M$ for some $x_0$ in $[0,1]$ by continuity of $f$ hence, again by continuity of $f$, we can find an interval $(c,d) \subset [0,1]$ such that $f(x)> \alpha$ for all $x \in (c,d)$. Then for every $n$, we have that $$ \bigg(\int_0^1 f(x)^ndx\bigg)^{\frac{1}{n}} \geq  \bigg(\int_c^d f(x)^ndx\bigg)^{\frac{1}{n}} >  \bigg(\int_c^d \alpha^n dx\bigg)^{\frac{1}{n}}= \alpha \; (d-c)^{\frac{1}{n}} $$ Taking limits, we get that $\liminf_{n \to \infty} \big(\int_0^1 f(x)^ndx\big)^{\frac{1}{n}} \geq \lim_{n \to \infty} \alpha \; (d-c)^{\frac{1}{n}} = \alpha$
Since $\alpha$ was an arbitrary real number which was strictly less than $M$, the above calculation implies that $\liminf_{n \to \infty} \big(\int_0^1 f(x)^ndx\big)^{\frac{1}{n}} \geq M$.
We showed that $$M \leq \liminf_{n \to \infty} \bigg(\int_0^1 f(x)^ndx\bigg)^{\frac{1}{n}} \leq \limsup_{n \to \infty} \bigg(\int_0^1 f(x)^ndx\bigg)^{\frac{1}{n}} \leq M$$ which implies that $\lim_{n \to \infty} \big(\int_0^1 f(x)^ndx\big)^{\frac{1}{n}} =M$.
