Let $\langle X, Y \rangle$ be a pair of vector spaces put in duality by a non-degenerate bilinear form $\langle \cdot, \cdot \rangle: X \times Y \to \mathbb{R}$. A topology $\tau$ on $X$ is called compatible with the dual pair if the dual of $X$ with respect to the topology $\tau$ equals $Y$ (where $Y$ is seen as a subspace of the algebraic dual $X^*$ by the injective map $y \mapsto \langle \cdot, y \rangle$).

The important theorem of Mackey-Arens characterizes all compatible topologies as polar topologies lying between the weak and the Mackey topology. Does there exist examples where a (or all) compatible topology can be described rather concretely? I'm especially interested in the duality of $C^\infty_c(\mathbb{R})$ and $C^\infty(\mathbb{R})$ given by integration. What can we say in this case?

  • $\begingroup$ You mean $\langle\varphi,f\rangle = \int \varphi(x)f(x)dx$, right? $\endgroup$ Oct 28, 2014 at 16:12
  • $\begingroup$ The answer to your question depends very much on what you mean by concretely. If the strong topology $\tau$ on $\mathscr E'$ (Schwartz' notation for $C^\infty(\mathbb R)'$) is concrete enough you can restrict it to $\mathscr D= C^\infty_c(\mathbb R)$ to obtain a compatible topology for the dual pair. $\endgroup$ Oct 28, 2014 at 16:17

1 Answer 1


Usha Kiran showed in An uncountable number of polar topologies and non-convex topologies for a dual pair that if the Mackey topology is distinct from the weak topology, then there are at least continuum-many distinct compatible topologies for the dual pair. The construction is based on unbounded subsets of the naturals, with equivalence given by having unbounded symmetric difference. Depending on your personal sensibilities, this means that they are not reasonably classifiable.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.