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Let $X$ be a Banach space and $B_X$ be the unit ball.

Suppose that for each $\lbrace C_n\rbrace_{n=1}^\infty\subset B_X$ satisfying $C_n$ are closed convex and $C_n\supset C_{n+1}$ has a nonempty intersection.

Is it true that $X$ is reflexive?

Jack Smith
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3 Answers3

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We will admit the following result:

A Banach space $X$ is reflexive if and only if for all $l:X\rightarrow \mathbb R$ linear and continuous we can find $x_0$ such that $\lVert x_0\rVert =\lVert l\rVert = \sup_{x\neq 0}\frac{l(x)}{\lVert x\rVert}$.

Let $l$ such a map. For all $n\in\mathbb{N}^*$, we can find $x_n$ with $\lVert x_n\rVert =1$ and $\langle l,x_n\rangle \geq \lVert l\rVert -\frac 1n$. Let $C_n$ the closed convex hull of the set $\left\{x_k,k\geq n\right\}$. $\left\{C_n\right\}$ is a decreasing sequence of closed convex non empty subsets of $B_X$. Let $x\in \bigcap_{n\in \mathbb{N}^*}C_n$. Let $\varepsilon>0$ and $k_0\in\mathbb N^*$. We can find $N\in\mathbb N$ and $k_1<\cdots<k_N$ integers with $k_1\geq k_0$ and $(\alpha_j)_{1\leq j\leq N}\in \left[0,1\right]$ such that $\lVert x-x'\rVert <\varepsilon$ with $x'=\sum_{j=1}^N\alpha_jx_{k_j}$ and $\sum_{j=1}^N\alpha_j=1$. Since each $k_j$ is greater than $k_0$ we have \begin{align*} \langle l,x\rangle& =\langle l,x-x'\rangle+\langle l,x'\rangle\\ &= \langle l,x-x'\rangle +\sum_{j=1}^N\alpha_j \langle l,x_{k_j}\rangle\\ &\geq -\varepsilon\lVert l\rVert+\sum_{j=1}^N\alpha_j\left(\lVert l\rVert -\frac 1{k_j}\right)\\ &=\lVert l\rVert (1-\varepsilon)-\sum_{j=1}^N\frac{\alpha_j}{k_j}\\ \langle l,x\rangle&\geq \lVert l\rVert (1-\varepsilon)-\frac 1{k_0}. \end{align*} Since the last inequality is true for all $\varepsilon >0$ and $k_0>0$ we get that $\lVert l\rVert\lVert x\rVert \leq \langle l,x\rangle$ and it's an equality (we notice that $x\neq 0$, exect in the case in which $l=0$).

Davide Giraudo
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  • Very nice proof. My thoughts went exactly in the same direction. I remembered seeing the result about reflexivity in the book "Functional Analysis, Sobolev Spaces and Partial Differential Equations" by Haim Brezis, 2011 edition. It is a remark on page 4, and has some nice references for the proof of this result. – Beni Bogosel Jun 16 '11 at 21:42
  • I have already red this result in the Haim Brezis' book. A reference given is "Geometry of Banach Spaces: selected topics" by Diestel. – Davide Giraudo Jun 16 '11 at 21:47
  • Do you think the result you admitted is easier than the result to be proved? – GEdgar Jun 16 '11 at 21:59
  • @GEdgar: Indeed, this is the main problem of this proof. Since I don't have any other idea, I don't know if the result asked by @Jack Smith is easier. I will look at the reference given by @Zakk. – Davide Giraudo Jun 16 '11 at 22:13
  • I didn't find Zakk's reference, but we answered Jack Smith's question. – Davide Giraudo Jun 17 '11 at 11:04
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    @girdav: here's the original paper of Smulyan. It's in Russian but it contains an English summary at the end. Dieudonné has an article on (generalizations of) this work here. I think Holmes's book treats these results nicely. Essentially it's a criterion for weak compactness of bounded sets. Keywords: Smulyan, Eberlein-Smulyan. – t.b. Jun 21 '11 at 17:34
  • @Theo Buehler Thanks for these references. – Davide Giraudo Jun 21 '11 at 18:22
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Such $C_n$ exist in any Banach space. For example take $C_n$ to be the closed ball of radius $1-1/n$ centred at the origin.

Chris Eagle
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It's true This is an old (1939) theorem due to Smulyan Look at any Banach space theory book for the proof

Zakk
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