Let $E$ be a reflexive Banach space and let $B(E)$ be the space of bounded operators on $E$ endowed with the weak operator topology. In particular, the unit ball of $B(E)$ is then WOT-compact. $(B(E), {\rm WOT})$ is a fairly nice locally convex space. I am interested in the weak topology of this space.

1) Does $\sigma\big( (B(E), {\rm WOT} ), (B(E), {\rm WOT} )^*)={\rm WOT}$?

2) Is $(B(E), {\rm WOT} )$ a barelled space?

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    $\begingroup$ Since the WOT is a weak topology, the dual of $B(E)$ under this topology is (the linear span of) the functionals used to define it. $\endgroup$ – Gerald Edgar Oct 16 '12 at 12:04

As Gerald suggests, the answer to question 1) is yes. The answer to the second question is negative (if $E$ is infinite-dimensional) because the operator norm makes $B(E)$ a Banach space and, on any Banach space $X$, there is no strictly coarser locally convex Hausdorff topology $\tau$ which is barrelled.

Otherwise, the closure of the unit ball in the coarser topology would be a $\tau$-neighborhood of $0$, that is, the identity $(X,\|\cdot\|) \to (X,\tau)$ is almost open and thus open by the open mapping theorem (in a version which misses in most textbooks although the standard proof of the OMT gives this result).


One can say a little more about this situation. Firstly, even for a general $E$, i.e., not necessarily reflexive, one has an explicit description of the above dual space. It is the algebraic tensor product of $E'$ and $E$ and this is naturally identifiable with the space of finite rank operators on $E$. If I might make a suggestion, working on the assumption that the reason behind the question is the search for a good locally convex topology on $B(E)$ which retains some of the properties of WOT, then I would suggest the finest locally convex topology on $B(E)$ which agrees with WOT on the unit ball as a suitable candidate. This topology has the same convergent sequences and compact subsets as WOT but has some additional properties. Thus it is complete in the case of interest, i.e. where $E$ is reflexive, and its unit ball is then compact. This means that it is the dual of a Banach space---the space of all linear functions on $B(E)$ which are continuous for WOT when confined to the unit ball. It can be identified with the closure of the finite rank operators under the dual norm to the operator norm on $B(E)$. In the special case where $E$ is a Hilbert space, it is naturally identifiable with the space of nuclear soperators on $H$ (things are simpler here since we can identify $E$ with its dual space in the usual way). Presumably, it can be identified with a space of nuclear operators in the general case, but as far as I know, nobody has looked at it in detail.


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