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Suppose $X\subseteq L_\infty$ is a compact subset of $L_p$ for all $1\leq p<\infty$. Does this mean that for every $\epsilon>0$ there exists a measurable set $E\subseteq [0,1]$ with $\lambda(E)> 1- \epsilon$ such that $\{f\chi_{E}\in X : f\in X\}$ is compact in $L_\infty$?

The converse is obvious.

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The answer is obviously no. Sorry. rabee –  Rabee Tourky Sep 8 '12 at 9:34
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Let $X$ be the set of all monotone functions from from $[0,1]$ to $\{0,1\}$. This set is compact in every $L_p$ and is in $L_\infty$. –  Rabee Tourky Sep 8 '12 at 9:57
    
I cannot answer it because I am new. However, here is what I would have posted---The answer is no. The set of all monotone functions from from $[0,1]$ to $\{0,1\}$ is a counter example. It is compact in each $L_p$, is in $L_\infty$ and for every measurable set $E\subseteq [0,1]$, $\lambda(E)>0$, the set $\{f\chi_E:X f\in X\}$ is nor compact in $L_\infty$.A related question remains open. If $X$ is a set of functions that are continuous (in $C([0,1]$) is the answer affirmative? –  Rabee Tourky Sep 8 '12 at 10:19
    
@RabeeTourky I don't see how being new prevents you from posting an answer to the question... You should see a button Answer Your Question somewhere in the bottom left corner. –  user31373 Sep 9 '12 at 1:23
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up vote 3 down vote accepted

The answer is no. The counterexample came to me a few seconds after hitting the post button (I'm sorry). However, this that the counter example is interesting and seems to be closely related to the non-topology of almost everywhere convergence in $L_1$.

The set of all monotone functions from $[0,1]$ to $\{0,1\}$ is a counter example. It is a compact (sublattice) in each $L_p$, is (a closed sublattice) in $L_∞$ and for every measurable set $E\subseteq [0,1]$, $λ(E)>0$, the set $X_E=\{f\chi_E: f∈X\}$ is not compact in $L_\infty$.

To see this notice for each $g\in\{f\chi_E: f∈X\}$ is separated from $\{f\chi_E: f∈X\}$ by an $L_\infty$-neighbourhood. Thus, in the relative topology of $L_\infty$ the singleton sets in $X_E$ are both open and closed. But $X_E$ has uncountably many elements. So it is not compact in $L_\infty$.

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