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I have a set $K$ which is weakly compact in a Banach space $E$. Also, its span is norm-dense in $E$. ($E$ is weakly compactly generated. But this doesn't enter the question anywhere.)

I need to construct a reflexive Banach space, so I need its unit ball to be compact. Since $K$ is compact, I thought I might try to use the space $\operatorname{span}(K)$. I want its unit ball to still be weakly compact, is this the case?

That is, if $K$ is weakly compact in $E$, then is $B_{Y}$ weakly compact in $Y$, where $Y = \operatorname{span}(K)$?

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Note that $Y$ will usually not be complete. –  Nate Eldredge Feb 25 '12 at 23:18

2 Answers 2

up vote 5 down vote accepted

I think the answer is no. Let $E=\ell^1(\mathbb{N})$ and let $\{e_n\}_{n\in\mathbb{N}}$ be the canonical basis. Then put $$ K=\left\{\frac1n\,e_n: \ n\in\mathbb{N}\right\}\cup\{0\}. $$ Then $K$ is even norm-compact (because it is itself a norm-convergent sequence), but the closure of the span of $K$ is all of $\ell^1(\mathbb{N})$.

The same trick can be used on any separable Banach space, many of which are not reflexive and thus have non-weak-compact balls.

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Good point! I'll add it right now. As for the second part of the comment, I intended to say that, and I forgot. I'll add it too. –  Martin Argerami Feb 25 '12 at 23:11
    
The norm closure of the span of $K$ is all of $\ell^{1}$, so that means it can't be compact. But I don't see how this denies weak compactness? –  Kyle Schlitt Feb 25 '12 at 23:30
    
The unit ball of $\ell^1$ not weakly compact. This is because the unit ball is weakly compact if and only if the space is reflexive. Alternatively, you can replace $\ell^1$ with your favourite space where the unit ball is not weakly compact, and do the same trick on any countable subset of unit vectors that spans the whole space. –  Martin Argerami Feb 26 '12 at 0:06
    
But why would that unit ball be weakly compact? In the case of your example, $B_{Y}$, where $Y:= span(K)$, is norm dense in $B_\ell^{1}$ but not equal to it necessarily. –  Kyle Schlitt Feb 26 '12 at 1:00
    
I see it now. The norm closure is contained in the weak closure. –  Kyle Schlitt Feb 26 '12 at 1:11

I am going to suggest a different answer to the accepted one, but I hope it may be of interest to someone. In particular, Corollary $3$ below may be noteworthy. In this answer we shall need to use the Krein-Smulian Weak Compactness Theorem, which says the following: the closed convex hull of a weakly compact subset of a Banach space is itself weakly compact (the original reference for this result is M. Krein and V. Smulian, On regularly convex sets in the space conjugate to a Banach space, Ann. Math. (second series) 41 (1940) pp.556-583, but proofs abound.)

The idea will be to take the absolutely convex hull of $K$ (that is, the convex hull of the weakly compact set $K\cup (-K)$), and form a Banach space from this set; this idea comes from the paper Factoring Weakly Compact Operators, J. Funct. Anal. 17 (1974) pp.311–327 by Davis, Figiel, Johnson and Pelczynski. I won't give all the details from their paper, but I will quote below several parts of the text of their paper:

Let $W$ be a bounded, convex, symmetric subset of a Banach space $(X,\,\Vert \cdot\Vert)$. For each $n=1, 2,\ldots$ the gauge $\Vert\cdot\Vert_n$ of the set $U_n=2^nW+2^{-n}B_X$ is a norm equivalent to $\Vert\cdot\Vert$. Define, for each $x\in X$, $\vert\vert\vert\, x \, \vert\vert\vert= (\sum_{n=1}^\infty\Vert x\Vert_n^2)^{1/2}$, let $Y= \{ x\in X \mid \vert\vert\vert\, x \, \vert\vert\vert<\infty\}$ and $C=B_Y = \{ x\in X \mid \vert\vert\vert\, x \, \vert\vert\vert\leq 1\}$. Finally, let $j$ denote the identity embedding of $Y$ into $X$.

$\vdots$

Here are some properties of this construction:

Lemma 1.

(i) $W\subseteq C$.

(ii) $(Y,\, \vert\vert\vert\cdot\vert\vert\vert)$ is a Banach space and $j$ is continuous.

(iii) $j^{\ast\ast}:Y^{\ast\ast}\longrightarrow X^{\ast\ast}$ is one-to-one and $(j^{\ast\ast})^{-1}(X)=Y$.

(iv) $Y$ is reflexive if and only if $W$ is weakly relatively compact.

Note that in part (iii) of Lemma $1$ above, $X$ and $Y$ are considered as subsets of their respective biduals in the usual way. I will not reproduce the proof of Lemma $1$ here, but proofs of the following three corollaries of Lemma $1$ are brief enough that it is easy for me to reproduce them here (I shall add a little information in the proof of the third corollary.)

Corollary 1. Weakly compact operators factor through reflexive spaces.

Proof. Let $T:Z\longrightarrow X$ be weakly compact and let $W$ of Lemma $1$ be $T(B_Z)$. The operators $j^{-1}\circ T:Z\longrightarrow Y$ and $j: Y\longrightarrow X$ provide the required factorization. $\square$

Corollary 2. Every weakly compact subset $K$ of a Banach space $X$ is affinity homeomorphic (in the respective weak topologies) to a subset of a reflexive Banach space.

Proof. In Lemma $1$ let $W$ be the convex hull of $K\cup (-K)$. Then, by the Krein-Smulian Weak Compactness Theorem, $W$ is weakly relatively compact, hence, by (iv) above, $Y$ is reflexive. Therefore $K'=j^{-1}(K)$ is weakly relatively compact (being weakly closed and bounded, by (i).) Note that $j|_{K'}$ is the homeomorphism we need. $\square$

Corollary 3. A Banach space $X$ is weakly compact generated if and only if there is a reflexive Banach space $R$ and a one-to-one operator $T:R\longrightarrow X$ with $T(R)$ dense in $X$.

Proof. The sufficiency is obvious. For the necessity, let $K$ be a weakly compact subset of $X$ whose linear span is norm dense in $X$ and in Lemma $1$ let $W$ be the convex hull of $K\cup (-K)$, so that $Y$ is reflexive and $j$ gives the desired one-to-one operator. $\square$

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Very nice summary! Thanks for the effort and the links. –  t.b. Mar 7 '12 at 14:25

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