Let $X$ be a locally convex space and $\left< X, X^{\prime} \right>$ stands for the dual pair. The bidual of $X$ is denoted by $X^{\prime \prime}$ and this is a dual of $X^{\prime}$ with a strong topology $\beta(X^{\prime}, X)$, i.e. a topology on $X^{\prime}$ of uniform convergence on all bounded subsets of $X$ (determined by a family of seminorms $p_B(f)= \sup_{x \in B} |f(x)|$, where B is a closed bounded absolutely convex subset of $X$). We can consider the natural topology on the bidual $X^{\prime \prime}$, i.e. the topology of uniform convergence on the equicontinuous subsets of $X^{\prime}$, denoted by $\tau_N(X^{\prime \prime}, X^{\prime})$. The natural topology on $X$ concides with the original topology on $X$.
Question For locally convex spaces $X$ and $Y$, let $A: (X^{\prime \prime}, \beta(X^{\prime \prime}, X^{\prime})) \rightarrow (Y^{\prime \prime}, \beta(Y^{\prime \prime}, Y^{\prime})) $ be a linear continuous map. Does it imply that $A : (X^{\prime \prime}, \tau_N(X^{\prime \prime}, X^{\prime})) \rightarrow (Y^{\prime \prime}, \tau_N(Y^{\prime \prime}, Y^{\prime}))$ is also continuous?
Since this strong topology is finer than the natural topology we have that $A : (X^{\prime \prime}, \beta(X^{\prime \prime}, X^{\prime})) \rightarrow (Y^{\prime \prime}, \tau_N(Y^{\prime \prime}, Y^{\prime}))$ is clearly continuous, but do we have something more? I don't know.
