Let $M$ be a subspace of $L_1[0,1]$ which is also isomorphic to $L_1[0,1]$. Is it true that $M^{**}$ is complemented in $(L_1[0,1])^{**}$?
$\begingroup$
$\endgroup$
3
-
1$\begingroup$ Did you read Bourgain's paper (at least the first page of it)? mathoverflow.net/questions/140557/quotients-of-linfty $\endgroup$– Narutaka OZAWACommented Sep 1, 2013 at 5:33
-
$\begingroup$ I echo @NarutakaOZAWA's comment. See also Bill Johnson's remarks about localization in his comments to the answer of your previous question $\endgroup$– Yemon ChoiCommented Sep 1, 2013 at 17:08
-
$\begingroup$ If you are happy with @M.G.'s answer then you should click the tick mark to "accept it". $\endgroup$– Yemon ChoiCommented Sep 3, 2013 at 4:01
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
The asnwer is no. First, $M$ is not necessarily complemented in $L_1[0,1]$ (Bourgain). In fact, $M$ is complemented in $L_1[0,1]$ iff $L_1/M$ is a $\mathcal{L}_1$-space (Lindenstrauss lifting argument). Second, $L_1/M$ is a $\mathcal{L}_1$-space iff so is its bidual $L_1[0,1]^{**}/M^{\perp\perp}$. Since Lindenstrauss argument applies also to $L_1[0,1]^{**}$, we conclude that $M^{\perp\perp}\equiv M^{**}$ is complemented in $L_1[0,1]^{**}$ iff $M$ is complemented in $L_1[0,1]$.
answered Sep 1, 2013 at 16:42