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Let $T \subseteq \mathbb R$ be a closed set of real numbers. Let $X := C(T, \mathbb R)$ denote the Fréchet space of continuous real-valued functions on $T$. The topology on $X$ is generated by seminorms $\|x\|_K := \sup_{t \in K} |x(t)|$ for any compact $K \subseteq T$.

Let $X^*$ denote the dual space of continuous linear functionals on $X$. Is there a nice characterization of the dual space using the Riesz representation theorem?

Let $e_t : X \to \mathbb R$ denote the evaluation functional, defined by $e_t[x] := x(t)$ for all $t \in T$. Are the evaluation functionals dense in $X^*$?

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    $\begingroup$ Evaluation functionals are certainly not dense. At least you have to take linear combinations of them. $\endgroup$ Oct 18, 2013 at 20:20
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    $\begingroup$ Related question: mathoverflow.net/q/105147/2622 $\endgroup$ Oct 18, 2013 at 20:33
  • $\begingroup$ @GeraldEdgar: ha! You're absolutely right. $\endgroup$ Oct 18, 2013 at 22:02

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It is the space of compactly supported Radon measures. See Nicolas Bourbaki, Intégration, chapter 4, page 156 in Springer’s 2007-edition.

It seems to me that the space spanned by evaluation functionals is dense in the weak-*-topology (given any finite set of continuous functions choose—using compactness of the support—a finite open cover of the support such that in each of these open sets the given functions do not vary a lot, then use an evaluation functional for each of the open sets, at a point in the respective set), but not in the usual norm topology.

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  • $\begingroup$ I don't have the book handy. Is the statement true for an arbitrary locally compact space $T$? i.e., that $C(T,\mathbb R)$ is isomorphic to the space of compactly supported Radon measures. $\endgroup$ Oct 18, 2013 at 22:13
  • $\begingroup$ Yes, it is. (Notice that you will not find it in the book in exactly that formulation, since Bourbaki defines measures as functionals, so it is written in Bourbaki language) $\endgroup$
    – The User
    Oct 18, 2013 at 22:40

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