Relation between dual of nuclear space $(\substack{\text{lim} \\ \leftarrow i} H_i)'$ and $\substack{\text{colim} \\ i \rightarrow } H_i$ 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?
 A: 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., $\phi=\phi_j \circ f_{j,\infty}$.
