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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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marked as duplicate by Nate Eldredge, sdcvvc, Micah, Davide Giraudo, David Mitra Dec 12 '12 at 20:10

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I don't see how to parse your question. – user108903 Dec 12 '12 at 18:50
You cannot prove a definition. You can prove theorems. – Hagen von Eitzen Dec 12 '12 at 18:53
Oh, you absolutely can prove that the sup is the right definition to fit in the framework of the $L_p$s. – Phira Dec 12 '12 at 18:54
See here. – David Mitra Dec 12 '12 at 19:13

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up vote 1 down vote accepted

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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I see, can you help me with my edited questions? – Zhixia Zhang Dec 12 '12 at 19:07
See the question posted by @David Mitra above. – icurays1 Dec 12 '12 at 19:20

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. – Berci Dec 12 '12 at 19:31

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