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?

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    $\begingroup$ No there is no such example as in the title. Yes, to your last question: Every non-zero continuous linear functional gives you a decomposition $X \cong Y \oplus \mathbb{R}$ with $Y = \operatorname{Ker}\varphi$ reflexive by hypothesis. Direct sums of reflexive spaces are reflexive and reflexivity is preserved under isomorphism of Banach spaces. $\endgroup$ – t.b. Oct 17 '11 at 18:21
  • $\begingroup$ @t.b.: of course you require the Axiom of Choice for that! $\endgroup$ – GEdgar Oct 17 '11 at 18:41
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    $\begingroup$ @Davide: well, that's also nice, but it makes me think of sparrows and cannons... :) $\endgroup$ – t.b. Oct 17 '11 at 18:43
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    $\begingroup$ Allowing hyperplanes makes it too easy. What about if we just require that every closed subspace of infinite codimension is reflexive? $\endgroup$ – Robert Israel Oct 17 '11 at 19:16
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    $\begingroup$ @Robert: According to an exercise in Bourbaki (TVS I, Ch. IV §5, Exer. 12, p. IV.69) every non-reflexive Banach space contains a non-reflexive subspace of infinite codimension. $\endgroup$ – t.b. Oct 17 '11 at 20:01

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:

Every non-reflexive Banach space contains a non-reflexive closed subspace of infinite codimension.

(Bourbaki, Topological vector spaces 1, Exercise 12 to Chapter IV, §5, page IV.69.)

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.


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