Let $\substack{\text{lim} \\ \leftarrow i} H_i$ be a nuclear space, considered as the limit of the codirected diagram $$... \to H_2 \to H_1 \to H_0,$$ with $f_{ji}:H_i \to H_j$ being the trace class operators in the diagram. I'm hoping that $$(\substack{\text{lim} \\ \leftarrow i} H_i)' \cong \substack{\text{colim} \\ i \rightarrow} H_i$$ holds as topological vector spaces, so is this maybe a known relation? Here, the operators in the directed diagram are just the adjoints $f_{ij}^*:H_i \to H_j$. Maybe there is some literature about the connection between projective and inductive limits of topological vector spaces?

  • 1
    $\begingroup$ Did you mean to write the colimit of the duals, rather than just the $H_i$ themselves? If so, then this would be correct, because a continuous linear map to a normed TVS (such as the scalars) from a projective limit of Banach spaces factors through some limitand, for nearly definitional, easy reasons. Are $H_i$'s Banach? Hilbert? It would not be good to identify every Hilbert space with its dual. $\endgroup$ May 7, 2015 at 13:14
  • $\begingroup$ Thanks! The $H_i$'s are Hilbert spaces, i wouldn't have thought that it makes a difference to take the colimit of them rather than of their duals, but this is still nice, though! $\endgroup$ May 7, 2015 at 13:23
  • $\begingroup$ so maybe i'm stupid, but why should a functional should factor through a well-defined map $H_i \to k$? I understand that if it is well-defined it must be continuous, but not why it should exist at all (even if the projections are surjective) $\endgroup$ May 8, 2015 at 11:46
  • $\begingroup$ @JochenWengenroth's edit/addition explains why these maps factor through limitands... $\endgroup$ May 11, 2015 at 21:59

1 Answer 1


The algebraic equality is clear by the remark of Paul Garrett. There is an issue about the usual locally convex topologies: The dual of $H_\infty=\lim\limits_{\leftarrow i}H_i$ is usually endowed with the so-called strong topology of uniform convergence on bounded sets and the colimit has its colimit (or, as it is usually called, inductive) topology which is the finest locally convex topology making all maps $H_i^* \to H_\infty^*$ continuous. Spaces $H_\infty$ for which these topologies coincide were called distinguished and Grothendieck proved that e.g. reflexive Frechet spaces are distinguished. For this it would be enough to have a (projective) limit of arbitrary reflexive Banach spaces (neither Hilbert nor trace class connecting maps are needed).

EDIT (concerning the algebraic equality). That $H_\infty$ is topologically the (projective) limit of the spectrum $(H_i,f_{j,i})$ means that $H_\infty$ has the initial topology with respect to the canonical maps $f_{j,\infty}:H_\infty \to H_j$. The $0$-neighborhood filter in $H_\infty$ therefore has a basis $\lbrace f_{j,\infty}^{-1}(V): V$ $0$-neighborhood in $H_j\rbrace$. A continuous linear functional is bounded on some $0$-neighborhood and you can then well-define a continuous linear functional first on the range of $f_{j,\infty}$ and then extend it by Hahn-Banach to $H_j$. In this sense, every $\phi \in H_\infty'$ comes from a continuous linear functional $\phi_j$ on some $H_j$, i.e., $\phi=\phi_j \circ f_{j,\infty}$.


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.