Let $X$ be a completely regular space and let $C_k(X)$ be the space of all continuous functions with the compact-open topology. If $X$ is completely metrizable, is the strong dual $C(X)^*$ the strong projective limit of the spaces $C(K)$, where $K$ is a compact subset of $K$?
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2$\begingroup$ Related questions: mathoverflow.net/questions/145215/… and mathoverflow.net/questions/105147/… $\endgroup$– András BátkaiCommented Jun 20, 2018 at 11:09
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1$\begingroup$ No. By the references mentioned by András Bátkai, $\mathcal{C}(X)^*$ can be identified with the space of (Radon) measures on $X$ with compact support. This space maps by restriction to each $\mathcal{C}(K)^*$, hence to their projective limit, but the latter map is not surjective: the family of Lebesgue measures on compacts of $\mathbb{R}$ does not come from a compactly supported measure on $\mathbb{R}$. $\endgroup$– abxCommented Jun 20, 2018 at 13:44
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$\begingroup$ @abx, but if you take such a measure $\mu$ (supported on the whole real line), how do you guarantee that $\int f d \mu$ is finite for every (unbounded) continuous function? $\endgroup$– user125821Commented Jun 20, 2018 at 15:08
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1$\begingroup$ I don't. This is why the answer is negative. $\endgroup$– abxCommented Jun 20, 2018 at 16:59
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1$\begingroup$ Take for instance the discrete space $X:=\mathbb{N}$. Then $C(X)$ is the projective limit $\displaystyle{\lim_\leftarrow } $ of the embeddings $\mathbb{R}^n\to\mathbb{R}^{n+1}$, that is the Fréchet space $\mathbb{R}^\mathbb{N}$ with the product topology, and this is also the projective limit of $C(K)^*$ for $K\Subset\mathbb{N}$. On the other side, $C(X)^*$ is the space $c_c$ of compactly supported sequences, that is the inductive limit $\displaystyle{\lim_\rightarrow } $ of the $\mathbb{R}^n\to\mathbb{R}^{n+1}$, topologized by all seminorms on it, a complete, non metrizable LCTVS. $\endgroup$– Pietro MajerCommented Jun 20, 2018 at 18:30
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