In a recent blog post, Terry Tao mentioned the question of how to tell if a Hausdorff topological vector space admits a finer topological structure which happens to be the topology of a Banach space (after noting that if it does admit such a structure, then it is unique). In thinking about this question, I was led to the following definiton:

Given a Hausdorff topological vector space $(V,\tau)$, define the Banachification topology on $V$ to be the coarsest topology on $V$ such that any linear map $f: V \to W$ with $W$ a Banach space whose graph is closed with respect to $\tau$ is continuous with respect to $\tau'$.

If there is some Banach space topology $\tau^*$ on $V$ that is finer than $\tau$, then $\tau'=\tau^*$. Indeed $\tau'$ is at least as fine as $\tau^*$ because we may take $W= (V, \tau^*)$, and $\tau$ is at least as coarse as $\tau^*$ by the closed graph theorem.

So to check whether there exists a Banach space topology on $V$ that is finer than $\tau$, it suffices to check whether $\tau'$ is a Banach space topology finer than $\tau$.

This may not actually be the best way to answer the question, because $\tau$ seems hard to compute. Regardless, I ask:

How can we compute $\tau'$?

Is it possible to give a reasonable witness that proves a sequence does not converge to a certain limit in $\tau'$? (A witness that a given sequence does converge is a single Banach space $W$ and function $f$ with closed graph such that $f$ applied to that sequence converges.)

Is it possible that $\tau'$ is coarser than $\tau$, and is there a simple topological criterion for when this happens?

How many different ways can $\tau'$ fail to be a Banach space, and are there nice topological criteria for when it fails in those ways?

It's possible I should do all these with Frechet spaces or F-spaces instead of Banach spaces.