Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
Still a bad question: You want all $C_n \ne \varnothing$ and $C_n \supseteq C_{n+1}$. Then it would be a good question. –  GEdgar Jun 16 '11 at 19:11
    
@girdav: notice "for each" –  GEdgar Jun 16 '11 at 19:33
    
@GEdgar Thank you for your comment. That's what I want –  Jack Smith Jun 16 '11 at 19:40

3 Answers 3

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 $l=0$).

share|improve this answer
    
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

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.

share|improve this answer
    
Ok, I changed the question the original one doesn't make sense. –  Jack Smith Jun 16 '11 at 19:05

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.