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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 $|f|$?

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    See [$h$ere](http://math.stackexchange.com/questions/242779/limit-of-lp-norm?rq=1).2012-12-12

2 Answers 2

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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.

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    See the m.se question posted by @David Mitra above.2012-12-12
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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-

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    Note that the limit in question works only above a domain of finite measure: think about the constants.2012-12-12