# Closed convex bounded sets are weakly compact for which spaces?

It is known that for all reflexive Banach spaces, closed convex bounded sets are weakly compact (compact for the weak topology).

What is the general class of topological vector spaces for which this is true ? For example, is it true that for all reflexive complete locally convex topological vector spaces, closed convex bounded sets are weakly compact ?

• There is a beaiful theorem by Davis-Johnson-Pelczynski in" Factoring weakly compact operators", which is available freely online in sciencedirect.com/science/article/pii/0022123674900445 which states that every weakly compact operator between Banach spaces factors through a refelxive space. Oct 10, 2016 at 6:31
• You should read remark 2 in this paper, and the following discussion. It provides a positive answer for Frechet spaces, I believe. Oct 10, 2016 at 6:32

It is well-known that a Hausdorff locally convex space is semi-reflexive (i.e., the canonical map into its bidual is surjective) if and only if every weakly closed bounded set is weakly compact. This is proposition 23.18 in Introduction to Functional Analysis of Meise and Vogt.

(It is mainly a consequence of Alaoglu's theorem.)

• The statement itself has nothing to with convexity which, however, is heavily used in the proof: The bipolar of a bounded set is $\sigma(X'',X')$ compact à la Alaoglu If $X$ is semi-reflexive, this bipolar is weakly compact in $X$ and hence also the set itself is weakly compact as a closed subset. On the other hand, every element of the bidual is contained in the bipolar of some bounded set. If the absolutely convex hull is weakly compact, it coincides with the bipolar by the bipolar theorem. Mar 30 at 20:22

The following is from Shaefer's "Topological Vector Spaces", sections 5.5 and 5.6.

For a locally convex (Hausdorff) $E$, the injection into its bidual is surjective iff every bounded set in $E$ is weakly-relatively-compact. In this case $E$ is said to be semi-refelexive. The map into the bidual need not be continuous. This bijection is continuous iff $E$ is also barreled. In this case this bijection is in fact an isomorphism of topological vector spaces and $E$ is said to be reflexive.

A general topological vector space which injects into its bidual, when taken with the weak topology, is clearly locally convex Hausdorff, so we can safely restrict ourselves to this class.

So, the answer to the first question is "the general class of locally convex Hausdorff topological vector spaces for which this is true is called semi-reflexive spaces" and the answer to the second is "yes, in particular it is true for reflexive spaces".

• @JochenWengenroth: What is the difference between "precompact" and "relatively compact"? I always thought they were synonyms, both meaning "the closure is compact", and Wikipedia thinks so too. Oct 10, 2016 at 13:07
• @NateEldredge Usually I use these as synonym as well and nothing bad happens, but they have different meanings for peopel in TVS. "Relatively compact" is a property of a subset in a topological space saying that its closure is compact. "Precompact" is a property of a uniform space saying that its completion is compact. For a subset of a complete uniform space (eg a TVS) they coinside. Oct 10, 2016 at 14:24
• However, note that for a Banach space (say) the weak topology is not complete in general. In fact, its completion with respect to the weak topology is its bidual endowed with its weak* topology (I think). Thus, it is complete iff it is refelexive. Oct 10, 2016 at 14:24
• Precompactness in a locally convex space means that, for every $0$-neighbourhood $U$, the set is covered by finitely many translates $x+U$. Relative compactness means that it is contained in a compact set. As Uri said, precompactness is equivalent to relative compactness in the completion. However, the weak topology of an infinite dimensional Banach space is not complete, the best you can hope for is quasi-completeness meaning that closed bounded sets are complete. A locally convex space is semi-reflexive if and only if it is quasi-complete for its weak topology. Oct 10, 2016 at 14:36
• Thanks @JochenWengenroth. So what I wrote two steps above, the sentence ending by (I think) and the sentence after, should have regarded the unit ball in a Banach space, not the whole space, right? Oct 10, 2016 at 14:40