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.

  • $\begingroup$ I think in general it cannot be true, but I cannot find a counterexample. I've got an idea how to prove that this is false, but I don't want to suggest any solution which is not verified yet. $\endgroup$
    – Edvin Goey
    Commented Sep 6, 2011 at 1:46

1 Answer 1


The answer is "no". Here is an example:

Let $E$ be an infinite-dimensional reflexive Banach space, and let $X:=E$ with the weak topology $\sigma(E,E')$ and $Y:=E$ with the norm topology. Then both $X''=E=Y''$, and the strong topologies are the same. However, the natural topology on $X''(=E)$ is the weak topology, whereas the natural topology on $Y''(=E)$ is the norm topology. And the mapping $\mathit{id}\colon(E,\sigma(E,E')) \to (E,\text{norm-topol})$ is not continuous.

Let me ask, please, where you found the notion ‘natural topology’!


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .