convergence of $L^p $ norm 
Possible Duplicate:
Limit of $L^p$ norm 

If I define $|f|_{L^\infty}= \lim_{n\to \infty} |f|_{L^n}$. How can I prove that this limit is esssup $|f|$?
 A: The main reason to choose $\text{ess}\sup\vert f\vert$ over $\sup \vert f\vert$ is that "functions" in $L^p$ are in fact equivalence classes of functions: $f\sim g$ if $\{x:f(x)\neq g(x)\}$ has measure zero.  By construction of the Lebesgue integral, for all $1\leq p<\infty$ we have $\|f\|_p=\|g\|_p$ if $f\sim g$; we would like $\|f\|_\infty$ to have the same property.  $\sup\vert f\vert$ won't work because we can have $f\sim g$ but $\sup\vert f\vert \neq \sup\vert g\vert$, i.e. two functions in the same equivalence class will have different norm.  Since $\text{ess}\sup\vert f\vert$ "ignores" sets of measure zero, we will have $\text{ess}\sup\vert f\vert=\text{ess}\sup\vert g\vert $ if $f\sim g$ and hence the norm $\|\cdot\|_\infty$ will be well defined on our equivalence classes.
Edit: I guess the question changed as I was writing this.  This is more the reason for $\text{ess}\sup$ rather than the proof requested. 
A: First, for a finite set of positive elements $a_i$, it is clear that 
$$\sqrt[p]{\sum_i {a_i}^p}\to \max a_i \quad\text{ if }p\to\infty.$$
Similar proof can work for a bounded continuous function $f$ with domain $D$ of finite measure. 
Let $m:=\sup f$, and by dividing by $m$, we can assume $m=1$, hence $|f|\le 1$. Then for all $\epsilon>0$, we have
$$\begin{align} \left(\int_{D} |f|^p\right)^{1/p} &=
\left(\int_{(|f|>1-\epsilon)} |f|^p + \int_{(|f|\le 1-\epsilon)} |f|^p\right)^{1/p}
\\  &\ge \big(\mu(|f|>1-\epsilon)\cdot (1-\epsilon)^p+0\big)^{1/p} \\
&= \mu(|f|>1-\epsilon)^{1/p}\cdot (1-\epsilon)\to (1-\epsilon)
\end{align}$$
We also have, by $|f|\le 1$, that $\left(\int_{D} |f|^p\right)^{1/p}\le (\mu(D))^{1/p}\to 1$, 
so $||f||_p \to 1$ as $p\to\infty$. -QED-
