Certain compact subset of $L_1$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T17:46:10Z http://mathoverflow.net/feeds/question/108905 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/108905/certain-compact-subset-of-l-1 Certain compact subset of $L_1$ Rabee Tourky 2012-10-05T11:59:44Z 2012-10-16T19:45:16Z <p>Let $(\Omega,\Sigma, \mu)$ be a probability measure and $X$ a Banach space. I am interested in subsets $F\subseteq L_\infty (\mu,X)$ that satisfy these two compactness conditions: </p> <ol> <li>$F$ is a norm-compact subset of $L_1(\mu, X)$; and</li> <li>For any sequence $f_n$ in $F$ there exists a set $E\in \Sigma$, $\mu(E)=1$, such that each $L_1$-norm convergent subsequence of $f_n$ converges pointwise on $E$. </li> </ol> <p>An example of such a set is any compact subset $F$ of $L_\infty (\mu,X)$. To see this suppose that $f_n$ is a sequence in $F$. There is $E\in \Sigma$, $\mu(E)=1$, such that for any $m,n$ we have $\|f_m(\omega) - f_n(\omega)\|\leq \|f_m-f_n\|_\infty$ for all $\omega\in E$. Now if $f_m$ is an $L_1$-convergent subsequence of $f_n$, then it must be converging in $L_\infty$, and converging pointwise on $E$. </p> <p>More generally, the compactness condition is satisfied by </p> <ul> <li>$\star$ $F\subseteq L_\infty (\mu,X)$ and for every $\epsilon>0$ there exists $E\in \Sigma$, $\mu(E)>1-\epsilon$ with $$\{f\chi_E: f\in F\}$$ is compact in $L_\infty (\mu,X)$. </li> </ul> <p>An example of an $L_1$-compact set that does not satisfy the compactness conditions 2 or $\star$ is the set of monotone step functions $f\colon [0,1]\to \{0,1\}$. </p> <p>Now for my question: Does there exist a set satisfying 1 and 2 but not satisfying the $L_\infty$ compactness condition $\star$? </p> <p>I don't have an answer to this for the case $X=R$ and $\Omega=[0,1]$.</p> http://mathoverflow.net/questions/108905/certain-compact-subset-of-l-1/109767#109767 Answer by Rabee Tourky for Certain compact subset of $L_1$ Rabee Tourky 2012-10-15T22:47:37Z 2012-10-16T19:45:16Z <p>The answer is that (1) and (2) implies $\star$ (this resolves a nice problem in a game theory paper I'm working on though the final unresolved problem has to do with decomposable Banach spaces, which I'll ask in a different question). </p> <p>Assume (1) and (2). Let $P=\{f_n\}$ be a sequence (actual versions of equivalence classes) in $X$ such that each $f\in X$ is the $L_1$ limit of some subsequence of $f_n$. Let $E$ be the measurable set in (2). </p> <p>By (2), $P$ has compact closure $\overline{P}^p$ in the product topology $X^{E}$. Notice that $X= \overline{P}^{L_1}=\overline{P}^p$ also by (2) taken as equivalence classes. Thus, we can now quickly show by means of Egorov's theorem that for every $\epsilon>0$ there is $F\in \Sigma$, $\mu(F)>1-\epsilon$ such that $\overline{P}^p$ is compact in the topology of uniform convergence (on $F$). We have the result. </p>