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Possible Duplicate:
Understanding proof of completeness of $L^{\infty}$

Most of the materials I have in Real Analysis consider this statement as a trivial one: "The normed space $L^{\infty}$ equipped with $\lVert\cdot\rVert_{\infty}$ is complete". But to my suprise I can't see the triviality. I am searching for it now...

Anybody with hint?

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Let $\left(f_n\right)_{n\geqslant 1}$ be a Cauchy sequence in $L^{\infty}$ endowed with the natural norm. For $n,m$, let $N_{n,m}$ a set of measure $0$ such that $|f_n(x)-f_m(x)|\leq \lVert f_n-f_m\rVert_{\infty}$ for each $x\notin N_{n,m}$. Define $N:=\bigcup_{n,m}N_{n,m}$. Then $N$ is of measure $0$ as a countable union of such sets and the sequence of functions $\left(\widetilde f_n\right)_{n\geqslant 1}$ restricted to $N^c$ is uniformly convergent. Then you can find an uniform limit $f$ (i.e. such that $\lVert f-\widetilde f_n\rVert_{\infty}\to 0$. Then just define $f$ by $0$ on $N$ to get a limit in $L^{\infty}$ (more precisely the limit will be the equivalence class of this function).

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  • $\begingroup$ Dear @Davide, I'm a beginner in measure theory and I'd like you to explain the existence of the sets $N_{n,m}$, please. I noticed Bartle uses the same argument, but I couldn't get it. Since $||f||_\infty$ is an $\inf$, then the reverse inequality $(\geq)$ seems more natural to me... Thanks in advance! $\endgroup$ – rgm Apr 7 '18 at 21:00
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    $\begingroup$ For each function $f$, the inequality $\left\lvert f(x)\right\rvert\leqslant\left\lVert f\right\Vert_\infty$ holds for almost every $x$. $\endgroup$ – Davide Giraudo Apr 7 '18 at 21:15
  • $\begingroup$ Dear @Davide, Why $\left(\widetilde f_n\right)_{n\geqslant 1}$ is uniformly convergent? $\endgroup$ – g.a.l.l.e.t.a Jun 25 '19 at 21:41
  • $\begingroup$ @g.a.l.l.e.t.a Note that $|f_n(x) - f_m(x)| \leq \epsilon$ since $(f_n)$ is Cauchy in $L^{\infty}$ $\endgroup$ – Sayan Chattopadhyay Sep 16 '19 at 19:08
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Edit: I have (hopefully) provided a non-formal way of vizualising what is going on here. Well $l^\infty = (C[a,b], ||.||_\infty)$ is incomplete because there are Cauchy sequences in $l^\infty$ which do not converge in $l^\infty$ (can you think of any?). So we want to "fill up" the space $l^\infty$ with functions so that all the Cauchy sequences converge. This "filled up" space is called $L^\infty$. The new vector space has all the functions in the space $l^\infty$, and also has discontinuous "jump" functions, but it still has the $\sup$ norm $\Vert\cdot\Vert_\infty$. We have "completed" the space $l^\infty$ and ended up with $L^\infty$ (and so we see that the space $L^\infty$ is complete). This is basically the definition of what $L^\infty$ is.

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    $\begingroup$ This is very hard to parse. Also, $C[a,b]$ is a closed subspace of $L^\infty[a,b]$, so the latter is not the completion of the former (the ess-sup norm is the same as the sup-norm for continuous functions). Moreover, writing $l^\infty$ for a space of continuous functions on an interval is very confusing, as $l^\infty$ usually stands for something completely different: the space of bounded scalar sequences with the sup-norm. $\endgroup$ – t.b. May 27 '12 at 9:34
  • $\begingroup$ Ok, well I thought L^infinity was the completion of l^infinity in the same way that L^2 is the completion of l^2. And the notation is not that confusing for me because I am used to using l^p to mean either of the two spaces you mention (and l^infinity). What space we are talking about should be obvious by the context. $\endgroup$ – Adam Rubinson May 27 '12 at 10:08
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    $\begingroup$ No, as I pointed out, $L^\infty[a,b]$ isn't the completion of $C[a,b]$ with respect to the ess-sup norm. What notation you use in private is of course up to you and it's good to know that you aren't confused by it. However, I don't remember having seen $l^p$ used this way even once in a functional analysis or a measure theory texts (an I've read dozens of them). This is an abuse of notation of a similar nature as writing $\mathbb{Q}$ for the set of integers... $\endgroup$ – t.b. May 27 '12 at 10:14
  • $\begingroup$ Actually, I am getting confused. I am thinking of (curly l)^p and these are actually the spaces of sequences. That aside, my notes define L^2[a,b] as "the smallest complete IPS containing C[a,b]", and says that for the purposes of the module, it is not important to go into the details. Isn't this the same as saying that L^2[a,b] is the completion of C[a,b]? $\endgroup$ – Adam Rubinson May 27 '12 at 10:32
  • $\begingroup$ It is true that $C[a,b]$ is dense in $L^p[a,b]$ for $1 \leq p \lt \infty$. However, $L^\infty[a,b]$ is not separable (the set of characteristic functions $[a,t]$ with $a \lt t \leq b$ is uncountable and discrete) while $C[a,b]$ is separable, so $C[a,b]$ can't be dense in $L^\infty[a,b]$. Yes, $\ell^p = l^p$ are spaces of sequences (it is about as common to use curly and non-curly letters for it). $\endgroup$ – t.b. May 27 '12 at 10:44

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