# strong topology = inductive limit topology on duals of projective limits

I've been bothering with this for some time now, and can't find any source with an actual proof, the statement simply appears to be "well-known". If you know (a source with) a proof, I'd be happy :)

Let $E$ be a countably Hilbert space as in Gelfand/Shilov "Generalized Functions vol 2", i.e. $E=\bigcap E_n$ for a sequence of separable decreasing Hilbert spaces $E_{n+1}\subset E_n$ with dense and continuous inclusions, such that $E$ is given the projective topology, that is the weakest one which turns all the inclusions $E\subset E_n$ continuous. Then $E$ is metrizable with a complete metric ($E$ is a Frechet space) and a sequence converges in $E$ if and only if it converges in every $E_n$. One can show $E'=\bigcup E_n'$ for the duals.

There is the strong topology on $E'$, that is the one in which the semi-norms $\|x\|_A:=\sup_{y\in A}|x(y)|$, $x\in E'$, generate the topology, where $A\subset E$ is bounded.

There also is the inductive limit topology on $E'$, that is the finest locally convex topology such that every $E_n'\subset E'$ is continuous.

It seems to be well-known that these topologies coincide, but I can only prove that the inductive limit topology has more open sets than the strong one. Does anyone know where the inverse inclusion is proved?

The part of the proof relevant for me is quite quick. Basically it says "if $V$ is a base neighborhood of $0$ for the inductive topology, then $V^{\circ\circ}$ is a neighborhood of $0$ for the strong topology and $V=V^{\circ\circ}$.
• Why is $V=V^{\circ\circ}$? It seems to be known (bipolar theorem) that $V^{\circ\circ}$ is the weak closure of $V$, but in the proof it is only assumed that $V$ is closed with respect to the inductive topology (which has more closed sets).
• Why does one need that $X$ is the dual of $Y$ to get $V=V^{\circ\circ}$?