limit $\lim_{n\rightarrow \infty}\left(\int_0^1f^n(x)dx\right)^{\frac{1}{n}}=M$ For a continuous positive funciton $f(x)$ on $[0,1]$, with maximum value $M$, show that $$\lim_{n\rightarrow \infty}\left(\int_0^1f^n(x)dx\right)^{\frac{1}{n}}=M$$
 A: Firstly $$\left(\int_0^1f^n(x)dx\right)^{\frac{1}{n}} \leq \left(M^n\right)^{\frac{1}{n}} = M$$
Secondly, let's say $f(x_0) = M$, then for any $\epsilon > 0$, we can find $\delta > 0$ such that $|x-x_0| < \delta$ means $|f(x)- f(x_0)| < \epsilon$, so
$$\left(\int_0^1f^n(x)dx\right)^{\frac{1}{n}} \geq \left(\delta (M-\epsilon)^n\right)^{\frac{1}{n}} \to M -\epsilon$$
A: Suppose that $f$ is continuous, with maximum value $M$ occurring at $x=m$.  Then for every $\epsilon>0$ there exists $\delta>0$ such that
$$\hbox{if}\quad |x-m|<\delta\quad\hbox{then}\quad f(x)>M-\epsilon\ .$$
Therefore
$$\left(\int_0^1f^n(x)dx\right)^{\frac{1}{n}}>((M-\epsilon)^n\delta)^{1/n}\ .$$
See if you can take it from here.
A: First notice that $\left(\int f^{n}(x)dx\right)^{1/n}\leqslant M$ for all $n$.
Since $[0,1]$ is compact there is an $x$ for which $f(x)=M$. Furthermore, for all $\alpha < 1$ we can find $\varepsilon >0$ so that $f(y)>M\alpha$ whenever $|x-y|<\varepsilon$. Consequently
\begin{equation*}
\left(\int f^{n}(x)dx\right)^{1/n} \geqslant (2\varepsilon \alpha^{n}M^{n})^{1/n} \geqslant \alpha M (2\varepsilon)^{1/n}
\end{equation*}
Hence $\liminf\left(\int f^{n}(x)dx\right)^{1/n}\geqslant \alpha M$. Since $\alpha$ is arbitrary $\lim \left(\int f^{n}(x)dx\right)^{1/n} = M$.
Note that this can be rewritten $||f||_{L^{n}}\underset{n\rightarrow +\infty}{\longrightarrow} ||f||_{L^{\infty}}$.
