*Question: Let $E$ be a Hilbert space. Can there exist a strictly finer bornological topology on $E$?*


The background to my question is as follows. I am looking at locally complete, locally convex spaces $E$ with bounded inner products $\gamma$. Then the topology induced by $\gamma$ is weaker than the bornological topology. I am interested in ways to recognise that $(E,\gamma)$ is a Hilbert space. Consider the following three properties


1. $\gamma$ induces the bornological topology on $E$.
2. $\check \gamma : E \to E'$ is surjective; $E'$ are the bounded linear functionals on $E$.
3. $(E,\gamma)$ is a Hilbert space.


I am able to show that (1) and (2) are equivalent and that either one implies (3). If I additionally assume that $E$ is webbed, then I can show that (3) implies (1) using a closed graph theorem. The closed graph theorem in question is from (13.3.4, Jarchow, 1981): if $E$ is ultrabornological and $F$ is webbed, then every closed linear map $E \to F$ is continuous.


*Is the assumption, that $E$ is webbed, necessary? If yes, are there explicit counterexamples satisfying (3), but not (1)?*



Strictly finer bornological topology on Hilbert space
Every (real or complex) vector space $E$ can be endowed with its *finest locally convex topology* $\tau\_{flc}$ where every seminorm is continuous and (equivalently) every absolutely convex absorbing set is a $0$-neighborhood.


This topology can be described as the locally convex inductive limit of all finite dimensional subspaces (with their unique Hausdorff locally convex topologies). Since bornologicity is stable with respect to inductive limits the finest locally convex topology is thus bornological (also ultrabornological). Using e.g. a Hamel basis of $E$ you find linear functionals which are discontinuous with respect to the Hilbert space topology. As they are clearly continuous with respect to $\tau\_{flc}$ we get that the latter is strictly finer than the Hilbert topology. 


The identity $(E,\|\cdot\|) \to (E,\tau\_{flc})$ has closed graph (since it is continuous in the other direction) but is not continuous. This shows that being webbed cannot be dropped from the closed graph theorem and that $(E,\tau\_{lfc})$ is not webbed.


