# What is the (strong) dual of a projective/inductive limit?

For a family of locally convex TVS $$(E_\alpha)_{\alpha \in A}$$ it is a well known fact that the dual of the product is the sum of the individual duals and vice versa, i.e. $$\left(\prod_{\alpha \in A} E_\alpha\right)^{\prime} = \bigoplus_{\alpha \in A} E_\alpha^\prime \quad\text{and}\quad \left(\bigoplus_{\alpha \in A} E_\alpha\right)^{\prime} = \prod_{\alpha \in A} E_\alpha^\prime.$$

What I'd like to know is if there is a similar relationship between the duals of projective and inductive limits. So does, for example, hold that $$\left(\varprojlim E_\alpha\right)^\prime = \varinjlim E_\alpha^\prime \quad\text{and}\quad \left(\varinjlim E_\alpha\right)^\prime = \varprojlim E_\alpha^\prime?$$

And does the strong topology on the dual coincide with the inductive/projective topology?

Where I'm trying to get to is: how does one find the "right" topology on $$\mathcal{D}(\mathbb{R}^n)^\prime$$ and describe the space "nicely" starting from the definition of $$\mathcal{D}(\mathbb{R}^n)$$ as the inductive limit of $$\mathcal{D}(K_m)$$, where $$\bigcup_{m \in \mathbb{N}}K_m = \mathbb{R}^n$$ is a compact exhaustion. (Of course $$\mathcal{D}(K_m)$$ shall be the projective limit of the $$k$$-times differentiable functions with compact support in $$K_m$$.)

Algebraically, there are canonical isomorphisms between the duals. However, for the strong topologies, things become very subtle. A Frechet space $$E=\lim\limits_{\leftarrow} E_n$$ with Banach spaces $$E_n$$ (and, of course, continuous linear connecting maps $$\rho_{n+1}^n: E_{n+1}\to E_n$$) is called distinguished if its strong dual is barrelled, and Grothendieck showed that then one has a topological isomorphism between $$E'$$ with its strong topology and $$\lim\limits_{\to} E_n'$$. There are however non-distinguished Frechet spaces (first examples of Köthe and Grothendieck, the strikingly simple example $$C(\mathbb R) \cap L^1(\mathbb R)$$ with its natural Frechet topology was found by Taskinen).
The "dual'' situation $$(\lim\limits_\to E_n)'= \lim\limits_{\leftarrow} E_n'$$ is slightly better: It holds topologically for all countable inductive limits of Banach spaces (Grothendieck), but for Frechet $$E_n$$ this may fail again.
In the particular case of $$\mathscr D(\mathbb R^m)=\lim\limits_\to \mathscr D(K_n)$$ you get topological equality of the dual $$\mathscr D'(\mathbb R^m)$$ and the projective limit of the duals because the inductive limit is strict and hence every bounded subset of the limit is contained and bounded in some "step" (which is very helpful to calculate the strong topologies).
• Thanks for you answer! I've one question about it: are you including unbounded continuous functions in $C(\mathbb{R})$? since otherwise I'd be lead to think $C(\mathbb{R}) \cap L^1(\mathbb{R})$ is also a Banach-space, or am I missing something else? – Joseph Adams Sep 27 '18 at 9:40
• Yes, $C(\mathbb R)$ is the space of all continuous functions. The seminorms on the intersection are $\int |f(x|dx + \sup\limits_{|x|\le n} |f(x)|$. – Jochen Sep 27 '18 at 10:38