Many conditions necessary for reflexivity of a Banach space turn out to be sufficient as well, for example, compactness of the closed unit ball in the weak topology. I am wondering if there is any sort of converse to the fact that every closed subspace of a Banach space is reflexive.
Namely, let $X$ be a Banach space such that every proper closed subspace of $X$ is reflexive. Does this imply that $X$ itself is reflexive?
Best Answer
Making my comments into an answer: No there are no such Banach spaces.
Assume that every proper subspace of $X$ is reflexive. Take a non-zero continuous linear functional $\varphi: X \to \mathbb{R}$. Let $Y = \operatorname{Ker}{\varphi}$ and choose $x_0 \in X$ with $\varphi(x_0) = 1$. By continuity of $\varphi$ the space $Y$ is a closed subspace. The map $Y \oplus \mathbb{R} \to X$ given by $(y,t) \mapsto y + t x_0$ is continuous with continuous inverse $x \mapsto (x - \varphi(x)\cdot x_0,\, \varphi(x))$, hence $X \cong Y \oplus \mathbb{R}$. By hypothesis $Y$ is reflexive, hence so is $Y \oplus \mathbb{R}$ and thus $X$ is reflexive, too. Replacing $\mathbb{R}$ by $\mathbb{C}$ gives the same for complex Banach spaces.
Weakening the hypotheses as Robert suggested makes it a bit more subtle, but still manageable:
Passing to the contrapositive, if every closed subspace of infinite codimension is reflexive, then $X$ must be reflexive.
The following is a slightly expanded version of the hint given by Bourbaki:
By the Eberlein–Šmulian theorem, a non-reflexive Banach space $X$ contains a bounded sequence $(x_n)_{n=1}^{\infty}$ without weak accumulation point. Note that this implies that the $x_n$ must span an infinite-dimensional subspace of $X$. Using the Riesz-lemma on almost orthogonal vectors to closed subspaces, it is not difficult to extract a subsequence $(x_{n_k})_{k=1}^{\infty}$ and a topologically independent sequence $(y_k)_{k=1}^{\infty}$ such that $\|x_{n_k} - y_{k}\| \leq \frac{1}{k}$ [topologically independent means that no $y_k$ is in the closed linear span of $\{y_n\}_{n \neq k}$ ]. This yields that every weak accumulation point of the $y_k$ is also a weak accumulation point of the $x_n$. The closed subspace $Y$ generated by the $\{y_{2k}\}$ is thus a non-reflexive subspace of $X$ (by Eberlein–Šmulian again) and it has infinite codimension by construction.