Reflexive Banach spaces and the Cantor Intersection property

Question

Let $(E, \lvert \cdot \rvert )$ be a reflexive Banach space. As a consequence of the Banach-Alaoglu Theorem and Mazur's Lemma we have that every nested non-increasing sequence $ C_1 \supset C_2 \supset \dotsm $ of non-empty bounded closed convex subsets of $E$ satisfies $ \bigcap_{k\geq 1} C_k \neq \varnothing $. Is it true that the converse implication holds? Does anyone know a reference for this?

Answer 1

The converse implication holds by James' theorem, which states that a Banach space $E$ is reflexive if and only if every continuous linear functional on $E$ attains its maximum on the closed unit ball of $E$.

If a normed space $E$ has the property that every nested sequence of non-empty bounded closed convex sets has non-empty intersection, then

1. $E$ is a Banach space, and
2. every continuous linear functional on $E$ attains its maximum on the closed unit ball of $E$.

Proof of 1.: Suppose $E$ were not complete, and let $\tilde{E}$ be its completion. Choose $x\in \tilde{E}\setminus E$ with $\lVert x\rVert \leqslant 1$. Let

$$C_n = E \cap \left\{y\in \tilde{E} : \lVert y\rVert \leqslant 1 \land \lVert y-x\rVert \leqslant \tfrac{1}{n}\right\}.$$

Then $\bigl\{ C_n : n\in \mathbb{N}\setminus \{0\}\bigr\}$ is a nested sequence of non-empty bounded closed convex sets in $E$ with empty intersection, contradicting the premise. Hence $E$ must be complete.

Proof of 2.: For $\lambda \in E'$ (or $E^\ast$, whichever your notation), let

$$C_n = \left\{ x\in E : \lVert x\rVert \leqslant 1 \land \lambda(x) \geqslant \tfrac{n-1}{n}\lVert\lambda\rVert\right\}.$$

Then $\bigl\{C_n : n \in \mathbb{N}\setminus\{0\}\bigr\}$ is a nested sequence of non-empty bounded closed convex sets, and the intersection is just the set of points with $\lVert x\rVert \leqslant 1$ and $\lambda(x) = \lVert\lambda\rVert$. As it is not empty, the maximum is attained. [$E$ is assumed real, for complex $E$ take $\operatorname{Re}\lambda$.]