# Are the weakly bounded subsets of the double dual bounded for the natural topology?

I'm trying to get a better understanding of the topologies on the double dual of a Hausdorff locally convex space (l.s.c.). The following question has then come up, and I'm unable to find an answer.

Let $$E$$ be a Hausdorff l.c.s., $$E'$$ the dual space, and $$E'' = (E'_\beta)'$$ the double dual. Here I'm using the notation that $$E'_\beta$$ is $$E'$$ with the topology $$\beta(E', E) =$$ the topology of uniform convergence on the bounded subsets of $$E$$. Let $$\eta$$ denote the set of equicontinuous subsets of $$E'$$ (i.e. equicontinuous on E with its original topology). Now consider the dual pair $$(E'', E')$$. From this we can consider two topologies on $$E''$$, the natural topology $$= \eta(E'', E') =$$ the topology of uniform convergence on $$\eta$$, and the weak topology $$= \sigma(E'', E') =$$ the topology of pointwise convergence on $$E'$$.

Now the question is the following: Are the weakly bounded subsets of $$E''$$ bounded for the natural topology?

Yes, this is a consequence of the Banach-Steinhaus theorem: Let $$\mathcal B$$ be a $$\sigma(E'',E')$$-bounded subset of $$E''$$ and $$A$$ an equicontinuous subset of $$E'$$ which is contained in the (absolute) polar $$U^\circ$$ of some $$0$$-neighbourhood $$U$$ in $$E$$. Alaoglu's theorem says that $$U^\circ$$ is weak$$^*$$-compact and this implies that the linear span $$X=[U^\circ]$$ with the norm having $$U^\circ$$ as its unit ball is a Banach space. The set $$\mathcal B|_X=\{\Phi|_X:\Phi\in\mathcal B\}$$ is contained in $$X'$$ (because each $$\Phi\in E''$$ is bounded on the polar $$B^\circ$$ of some $$E$$-bounded set $$B$$ and for $$t>0$$ with $$tB\subseteq U$$ we get $$tU^\circ\subseteq B^\circ$$), and pointwise bounded by the $$\sigma(E'',E')$$-bondedness of $$\mathcal B$$. Banach-Steinhaus implies that $$\mathcal B|_X$$ is uniformly bounded on the unit ball $$U^\circ$$ of $$X$$, i.e., $$\mathcal B\subseteq cU^{\circ\circ}$$ for some $$c>0$$. As these bipolars form a basis of $$\eta(E'',E')$$ neighbourhoods of $$0$$, we have proved the $$\eta(E'',E')$$-boundedness of $$\mathcal B$$.

This is esentially the same proof as for the $$E$$-boundedness of every $$\sigma(E,E')$$-bounded set. There might be a trick to apply this very classical result instead of repeating its proof.

Edit. Indeed, the trick is to show $$(E'',\eta(E'',E'))'=E'$$ by using again Alaoglu's theorem.

Second Edit. This trick does not work by the reasons discussed in the comments.

• So you're saying $\eta(E'', E')$ is a compatible topology. Interesting. Commented Jun 4, 2023 at 12:33
• I'm not sure i believe that though, as in general the Mackey topology $\tau(E'', E')$ restricted to $E$ is coarser than the Mackey topology $\tau(E, E')$. However, if your claim is true (that $\eta(E'', E')$ is compatible), then the two Mackey topologies would always agree, which isn't true. Commented Jun 4, 2023 at 12:51
• Köthe gives $c_0$ as an example where the two Mackey topologies are distinct. Commented Jun 4, 2023 at 12:53
• I see. My idea was, that polars $U^\circ$ are $\sigma(E',E)$-compact and should thus coincide with their third polar. This is wrong since the inclusion of $(E',\sigma(E',E)$ into the weak$^*$-dual of $E''$ is not continuous. Commented Jun 4, 2023 at 18:44