I am trying to show that if a set C is closed in the dual of E, E*, then it must be closed in F^E (where F = complex numbers or reals)

So far I've figured that E*is a subset of F^E since E* is the collection of continuous (bounded) linear functionals on E and F^E is all the linear functionals on E.

Thus, I have to show that E* is closed in F^E.

It is not true that E* is closed in K^E. In fact, the closure of E* in K^E is the set of

*all* (not necessarily continuous) linear functionals on E. (I'm calling the scalar field K because I already have too many Fs – see below.)

Just to make sure that we are talking about the same thing, I'm assuming that the topology on E* is the weak* topology that it has as a dual space, and that the topology on K^E is the product topology. Then the subspace topology that E* inherits from K^E is the same as its own (weak*) topology.

Let f be a (discontinuous) linear functional on E, and denote by \(\displaystyle \mathcal{F}\) the set of all finite-dimensional subspaces of E, directed by inclusion. For \(\displaystyle F\in\mathcal{F}\), the restriction of f to F is a bounded linear functional on F, so by the Hahn–Banach theorem it has a bounded extension to an element of E*. Call this extension \(\displaystyle f_F\). (Sorry about all the different f's here, not a very good choice of notation.) Then \(\displaystyle (f_F)_{F\in\mathcal{F}}\) is a directed net in E*, which fairly obviously has the limit f (in the product topology on K^E).

It's easy to check that the set E† of all linear functionals on E is closed in K^E. It follows from the previous paragraph that E† is the closure of E* in K^E.

Going back to the original question of whether a closed subset C of E* is closed in K^E, that will certainly be the case if C is bounded (because then C will be weak*-compact and therefore closed in any space that contains it). But in general C will not be closed in K^E.