Let $M$ be a subspace of $L_1(0,1)$. If the subspace $M$ is isomorphic to $L_1(0,1)$ and complemented, then the quotient $L_1(0,1)/M$ is clearly non-reflexive if it is infinite dimensional. So as we know, there are subspaces isomorphic to $L_1(0,1)$ but not complemented. But what can we say about the reflexivity of this quotient? If the quotient is never reflexive, this would imply that if any operator $T:L_1(0,1)\rightarrow L_1(0,1)$ is an element of the boundary of the upper-semi Fredholm operators, then the residuum operator $T^{co}: L_1^{**}/L_1 \rightarrow L_1^{**}/L_1$ definied by $T^{co}(x^{**}+L_1)=T^{**}x^{**}+L_1$ is not invertible. The latter operator plays an important role in the theory of tauberian operators.
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$\begingroup$ Amir, this is the second question in a row where you ask things without giving an indication of what steps you yourself have tried in order to attack this problem. You also have not given any motivation for this series of questions. $\endgroup$– Yemon ChoiCommented Sep 2, 2013 at 18:52
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$\begingroup$ Also, the title of the question is misleading. It is trivial that there exist non-reflexive quotients of $L^1$, but you are asking for something much more specific $\endgroup$– Yemon ChoiCommented Sep 2, 2013 at 18:54
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$\begingroup$ My intention is the following Yemon: If the subspace $M$ isomorphic to $L_1(0,1)$ is complemented, then the quotient is clearly non-reflexive. So as we know, there are subspaces isomorphic to $L_1(0,1)$ but not complemented. But what can we say about the reflexivity of this quotient. I asked these questions for research about some questions with regard to tauberian operators. $\endgroup$– user39474Commented Sep 2, 2013 at 19:11
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1$\begingroup$ Please register one of your accounts and ask to merge your two accounts at mathoverflow.net/contact. $\endgroup$– François G. DoraisCommented Sep 2, 2013 at 20:09
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1$\begingroup$ @AmirBahmanNasseri: AFAIK, everything known about $L_1/L_1$ is contained in my paper "Extensions of $c_0$" and the paper of Kalton and Pelczynski on which it is based, which gives only that $L_1/L_1$ must have finite cotype and hence cannot be superreflexive. $\endgroup$– Bill JohnsonCommented Sep 7, 2013 at 4:35
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