Let $E$ be a locally convex space, let $E^{\prime}$ be its continuous dual space and let $F$ be a subspace of $E^{\prime}$ which is dense with respect to the strong topology on $E^{\prime}$ (i.e. the topology of uniform convergence on bounded subsets on $E$). Under what assumptions on $E$ is it possible to conclude that the weak topology induced by $F$ equals the one induced by $E^{\prime}$, i.e. when does $\sigma(E,F) = \sigma(E,E^{\prime})$ hold?
I think that the statement is false in general, but I don't know a counterexample at the moment.
As a condition on $E$, you need that $E$ is finite-dimensional (or, if Hausdorffness is not assumed, that $E/\overline{\{0\}}$ is finite-dimensional).
Since $F$ is dense in $b(E',E)$, it separates points on $E$, and thus $\sigma(E,F)$ is a locally convex Hausdorff topology, and we have
$$\sigma(E,F)' = F; \quad \sigma(E,E')' = E',$$
so if $\sigma(E,F) = \sigma(E,E')$, it follows that $F = E'$.
That means to have $\overline{F} = E' \Rightarrow \sigma(E,F) = \sigma(E,E')$ you need a condition that implies $\overline{F} = E' \Rightarrow F = E'$. The only such condition, as far as I'm aware, is finite-dimensionality.