Limit of p-norms of functions $f$ is a continuous function on the closed interval $[a,b]$ such that $f \geq 0$ on $[a,b]$. Let
$$
  M_n = \left( \int_a^b f (x)^n dx \right)^{1/n}.
$$
Show $\lim_n M_n = \sup \{ f(x) \mid x \in [a,b] \}$.
 A: Let $M=\sup\{f(x):x\in [a,b]\}$. We know that
$$ M_{n}\leq \left(\int_{a}^{b}M^{n}\right)^{1/n}=(M^{n}(b-a))^{1/n}=M(b-a)^{1/n}.$$
Let $n\rightarrow\infty$, we have $\lim M_{n}\leq M$.
On the other hand, let $y\in [a,b]$ such that $f(y)=M$. Let $\epsilon>0$ be given. Then there exists $\delta>0$ such that if $|x-y|\leq\delta$, we have $f(x)\geq M-\epsilon$. Then 
$$M_{n}=\left(\int_{a}^{b}f^{n}\right)^{1/n}\geq \left(\int_{[y-\delta,y+\delta]}f^{n}\right)^{1/n}\geq \left(\int_{[y-\delta,y+\delta]}(M-\epsilon)^{n}\right)^{1/n}=(M-\epsilon)(2\delta)^{1/n}.$$
As $n\rightarrow \infty$, then $\lim M_{n}\geq M-\epsilon$. Since $\epsilon$ is arbitrary, $\lim M_{n}\geq M$.
A: Let $s=\sup\limits_{[a,b]}\{f(x):x\in[a,b]\}$, then
$$
\begin{align}
\left(\int_a^bf(x)^n\,\mathrm{d}x\right)^{1/n}
&\le \left(\int_a^bs^n\,\mathrm{d}x\right)^{1/n}\\
&=s(b-a)^{1/n}
\end{align}
$$
If $c\lt s$, then
$$
\begin{align}
\left(\int_a^bf(x)^n\,\mathrm{d}x\right)^{1/n}
&\ge c\,|\{x:f(x)\ge c\}|^{1/n}\\
\end{align}
$$
By the Squeeze theorem
$$
c\le\lim_{n\to\infty}\left(\int_a^bf(x)^n\,\mathrm{d}x\right)^{1/n}\le s
$$
Since $c\lt s$ was arbitrary, we have
$$
\lim_{n\to\infty}\left(\int_a^bf(x)^n\,\mathrm{d}x\right)^{1/n}=\sup_{[a,b]}\{f(x):x\in[a,b]\}
$$
