Suppose $X$ is a Banach space and $Y$ is a closed subspace. If $Y$ and $X{/}Y$ are reflexive, then is $X$ also reflexive? My intuition is that the answer is affirmative. I have shown that a Banach space $X$ is reflexive iff the closed unit ball of $X$ is weakly compact. I was trying to use this, but it hasn't been a success so far.
 A: Yes. More generally: If $X$ is a Banach space and $M$ is a closed subspace, then, if two out of three of $X,M,X/M$ are reflexive, then all three are reflexive. This is a result of Krein and Smulian from the 1940's, it is not trivial.
We need two lemmas that I leave as exercises, they must be easy to you:
(1) If $X$ is a Banach space and $M$ is a closed subspace, then define $M^\bot=\{\varphi\in X^*: \varphi\vert_{M}=0\}$. Then $M^\bot$ is a closed subspace of $X^*$ and $X^*/M^\bot\cong M^*$ (via the map $R$ acting as $\varphi+M^\bot\mapsto\varphi\vert_M$).
(2) If $X$ is a Banach space and $M$ is a closed subspace, then $(X/M)^*\cong M^\bot$ (via the map $S$ acting as $\varphi\mapsto \varphi\circ q$, $q:X\to X/M$ is the quotient map). Obviously, the inverse of $S$ is the map $\tau\mapsto S^{-1}(\tau)$ defined as $S^{-1}(\tau)(x+M)=\tau(x)$ for all $x\in X$.
Now suppose that $M$ and $X/M$ are reflexive spaces. Let $\chi\in X^{**}$. We want to show that this is the evaluation at some $x\in X$. Let's do it:
Define $\xi:(X/M)^*\to\mathbb{C}$ by $\xi(\varphi)=\chi(\varphi\circ q)$. As we said $\varphi\circ q\in M^\bot\subset X^*$, so our definition makes sense and actualy $\xi$ is bounded. By $X/M$'s reflexivity we may find $x_0+M\in X/M$ such that $\xi(\varphi)=\varphi(x_0+M)$ for all $\varphi\in (X/M)^*$. Note that this means that if $\tau\in M^\bot\subset X^{*}$ then $$\chi(\tau)=\chi(SS^{-1}(\tau))=\chi(S^{-1}(\tau)\circ q)=S^{-1}(\tau)(x_0+M)=\tau(x_0).$$
Now setting $\omega:X^*\to\mathbb{C}$ by $\omega(\tau)=\chi(\tau)-\tau(x_0)$ we have that $\omega\vert_{M^\bot}=0$. Thus $\omega$ induces a functional $\Omega:X^*/M^\bot\to\mathbb{C}$ by $\Omega(\tau+M^\bot)=\omega(\tau)$. This is easily seen to be bounded, so $\Omega\in (X^*/M^\bot)^*$. By (1), we have that $(X^*/M^\bot)^*\cong M^{**}$ via the map $$(X^*/M^\bot)^*\ni\varphi\mapsto\varphi\circ R^{-1}\in M^{**}$$
but $M$ is reflexive. Thus we have a point $x_1\in M$ such that $\Omega\circ R^{-1}=\text{ev}_{x_1}$. Therefores $\Omega=\text{ev}_{x_1}\circ R$, so
$$\Omega(\tau+M^\bot)=R(\tau+M^\bot)(x_1)$$
so
$$\omega(\tau)=\tau\vert_{M}(x_1)=\tau(x_1)$$
so
$$\chi(\tau)-\tau(x_0)=\tau(x_1)$$
so
$$\chi(\tau)=\tau(x_1+x_0)$$
and this is true for all $\tau\in X^*$, so we are done.
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