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If such criterion exists, since $C(\Omega)$ is closed in $L^\infty(\Omega)$, and if $\Omega$ is bounded and closed, the Ascoli-Arzela theorem has given a sufficient and necessary condition，means this criterion needs more requirements than Ascoli-Arzela theorem. So my questions:

  1. How to generalize the compactness condition in $L^\infty(\Omega)$($\Omega$ is compact in $R^N$)?
  2. If $\Omega$ is not compact, or just $R^N$, what is the compactness criterion?

Thank you!

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    $\begingroup$ Compactness of $\Omega$ can't be relevant here - $L^\infty(\Omega)$ only depends on the measure algebra of $\Omega$, and that's the same for every positive-measure set $\Omega$, so all $L^\infty(\Omega)$ are isomorphic (except the trivial ones where it is zero-dimensional). $\endgroup$
    – Nate Eldredge
    Commented Sep 13, 2017 at 4:49
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    $\begingroup$ See math.stackexchange.com/questions/955394/…. Surprise surprise: the place to look is Dunford and Schwartz. $\endgroup$
    – Nate Eldredge
    Commented Sep 13, 2017 at 4:52
  • $\begingroup$ Thanks, but why all $L^\infty(\Omega)$ are isomorphic? if $\Omega_1$ and $\Omega_2$ are topologically isomorphic, the isomorphism between $L^\infty(\Omega_1)$ and $L^\infty(\Omega_2)$ is trivial. P.s. the criterion in Dunford and Schwartz's book seems very hard to apply, sad... $\endgroup$
    – Ye Changqing
    Commented Sep 13, 2017 at 13:26
  • $\begingroup$ Topological isomorphism is neither necessary or sufficient. What you need is a bimeasurable map that takes null sets to null sets. If $\Omega$ is the standard measure-zero Cantor set and $\Omega'$ is a positive-measure "fat" Cantor set, then they are homeomorphic but $L^\infty(\Omega) = 0$ while $L^\infty(\Omega')$ is infinite dimensional. Conversely, $\Omega = [0,1]$ and $\Omega' = [0,1] \cup [2,3]$ are not homeomorphic but they have the same $L^\infty$. $\endgroup$
    – Nate Eldredge
    Commented Sep 13, 2017 at 13:36
  • $\begingroup$ Thanks again, but how to construct a bi-measurable function which takes null sets to null sets? $\endgroup$
    – Ye Changqing
    Commented Sep 13, 2017 at 14:04

1 Answer 1

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There is simple, elementary characterisation as follows: firstly, as mentioned in comments, a topology on $\Omega$ is arguably a red herring so I will consider a space with a $\sigma$-algebra and a $\sigma$-finite measure. Then a closed, bounded subset $B$ is compact in the corresponding $L^\infty$ space if and only if for each positive $\epsilon$ there is a finite partition $(A_i)$ of $\Omega$ into disjoint, measurable sets so that the variation of each $f$ in $B$ is less than $\epsilon$ (i.e., $|f(x)-f(y)| \leq \epsilon$ a.e. on each $A_i$).

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  • $\begingroup$ I believe we can prove it using the density of simple functions in $L_\infty$. Since it holds for all measure spaces, why is it that we require a $\sigma$-finite measure space? $\endgroup$
    – Smooth Alpert Frame
    Commented Aug 12, 2020 at 17:41
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    $\begingroup$ In case that anybody is looking for a reference for this result: It is proved (in a slightly different formulation though) in the article "B. M. Cherkas: Compactness in $L^\infty$ spaces (1970)" (link to zbMATH). $\endgroup$
    – Jochen Glueck
    Commented Sep 8, 2021 at 13:48

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