Attempted Banachification of a space

Question

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.

Answer 1

At least in the locally convex world, $\tau'$ is always finer than $\tau$: Given a continuous seminorm $p$ of $(X,\tau)$ the so-called local Banach space $X_p$ is the completion of $X/p^{-1}(\lbrace 0 \rbrace)$ (endowed with the norm induced by $p$). The canonical map $\varrho:X\to X_p$ (the quotient map followed by the inclusion) is continuous and thus has closed graph, hence it is continuous with respect to $\tau'$. But the preimage of the unit ball of $X_p$ is $\lbrace x\in X: p(x)\le 1\rbrace$ which is thus a $\tau'$-neighbourhood of $0$.

A simple (even trivial) reason for $\tau'$ failing to be a Banach topology is $\tau=\tau'$ and $(X,\tau)$ not Banach. $\tau=\tau'$ holds for all ultrabornological locally convex spaces which means that a linear map $X\to Y$ into an arbitrary locally convex (or, equivalently, Banach) space is $\tau$-continuous if its composition with every $\tau$-continuous linear map from Banach spaces into $X$ is continuous (equivalently, every absolutely convex set which absorbs every Banach disc is a $\tau$-neighbourhood of $0$, equivalently, $X$ is a locally convex inductive limit of Banach spaces). Most importantly, all Frechet spaces are ultrabornological (duals of Frechet spaces endowed with the topology of uniform convergence on bounded sets may however fail to be ultrabornological, this was thoroughly analyzed by Grothendieck).

A simple criterion for an infinite dimensional locally convex space failing to be Banach is that all its bounded sets are precompact (Montel spaces). Many spaces of functions or distributions appearing in analysis have this property.

As suggested by Sergei Akbarov's comment, it could be that $\tau'$ is always the so-called associated ultrabornological topology of $\tau$.

EDIT. The problem to prove something about $\tau'$ is that "closed graph maps" do not fit well with completions. This is perhaps another good example for Examples of common false beliefs in mathematics : The "obvious" statement If $T:X\to Y$ has closed graph and $i:Y\to \tilde Y$ is the canonical map into the Hausdorff completion, then $i\circ T$ has closed graph has the following counter-example: Take a Banach space $X$, (using the axiom of choice) construct a strictly stronger norm $p$, and set $Y=(X,p)$. The identity $X\to Y$ has closed graph (because its inverse is continuous) but $i\circ id$ does not have closed graph because of the closed graph theorem.

EDIT II. My conjecture that $\tau'$ could always be the associated ultrabornological topology is not true. The reason for this is that the closed graph theorem holds for barrelled domain spaces and range spaces which are Frechet (I learned this from the book Barrelled Locally Convex Spaces of Bonet and Perez-Carreras, theorem 7.1.12). Therefore, $\tau=\tau'$ holds for all barrelled spaces. On the other hand, there are barrelled spaces (even metrizable ones) which fail to be ultrabornological (loc. cit. example 6.4.12). It could be true that $\tau'$ is the associated barrelled topology.

Answer 2

This is really a comment (but too long for that) since I'm not sure if it answers your question but I feel that it offers a perspective that might be of interest. Consider the family of all absolutely convex, bounded subsets $B$ of your tvs which are such that their linear span $E_B$ with $B$ as unit ball is a Banach space. There are plenty of these---just look at the finite dimensional subsets. The family of all such spaces is an inductive system and so we can form its locally convex inductive limit which is automatically hausdorff. Then the condition of the question is equivalent to the associated topology being banachisable (if that is a word).