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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?

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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.

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  • $\begingroup$ So you're saying $\eta(E'', E')$ is a compatible topology. Interesting. $\endgroup$
    – user920957
    Commented Jun 4, 2023 at 12:33
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    $\begingroup$ 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. $\endgroup$
    – user920957
    Commented Jun 4, 2023 at 12:51
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    $\begingroup$ Köthe gives $c_0$ as an example where the two Mackey topologies are distinct. $\endgroup$
    – user920957
    Commented Jun 4, 2023 at 12:53
  • $\begingroup$ 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. $\endgroup$
    – Jochen
    Commented Jun 4, 2023 at 18:44

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