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?

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.

Let $M$ be a subspace isomorphic to $L_1(0,1)$. Suppose that the Quotient $L_1(0,1)/M$ is infinite dimensional. Can the Quotient be reflexive?

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?