We're talking about weak topologies in my FA courses and I thought of the following question, that I don't know the answer to.
Say we have a normed vector space $X$ which we view as a TVS with its weak topology. I'm wondering if this is complete as a TVS. Surely we need $X$ to be at least Banach for this to hold, but it doesn't seem enough.
I thought of the following: take the canonical embedding $\iota : X \to X^{**}.$ If we endow $X$ with its weak topology, say $\tau,$ and $X^{**}$ with its weak* topology, say $\tau^{*}$, then $\iota$ implements a homeomorphic embedding between these two spaces. But we know $\iota(X)$ is dense in $(X^{**},\tau^{*})$ so I think this shows unless $X$ is reflexive, $(X,\tau)$ cannot be complete as a TVS as $\tau(X)$ is not closed in $(X^{**},\tau^{*})$.
Then the question is, is it true that if $X$ is reflexive $(X, \tau)$ is complete as a TVS? Also, if $(X, \tau)$ is not complete, what can we say about its (Hausdorff) completion? For example, do we have an explicit description of it? Can it be the case that when $(X, \tau)$ is not complete its completion can still be realised by putting the weak topology on some other normed vector space?
