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It is known that continuous linear functionals on the space $C_0({\mathbb{R}^n})$ are bounded Radon measures ${\cal M}({\mathbb{R}^n})$ where $C_0({\mathbb{R}^n})$ is uniform closure of the space of continuous functions with compact support $C_c({\mathbb{R}^n})$ (for reference e.g. Folland's book on real analysis chapter 7).

I am wondering if there is a similar more general result, namely: what would be a continuous linear function on $C_0({\mathbb{R}^n})$ with values in some Banach space $E$? In analogy with the previous case where we had functionals, I assume it might be in the space of vector-valued bounded Radon measures ${\cal M}({\mathbb{R}^n};E)$?

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  • $\begingroup$ I suggest to look at the Radon-Nikodym property. $\endgroup$
    – Jochen Wengenroth
    Commented Dec 2, 2013 at 15:25
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    $\begingroup$ If your functions have values in $E$, then you expect the measures to have values in $E^\ast$ at least. If $E^\ast$ is separable, then it has the Radon-Nikodym Property. $\endgroup$
    – Gerald Edgar
    Commented Dec 2, 2013 at 18:55

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The formulation is not quite clear---do you mean linear functionals on a space of Banach space valued functions or linear operators on scalar-valued functions with valued in a Banach space? In the former case, the representing measure takes its value in the dual and is regular with respect to the topology of uniform convergence on compacta. In the second case, the operator is represented by a measure with values in the bidual which is regulär with respect to the weak star topology. If you want a measure with values in the original you should suppose that the operator is weakly compact. A good reference is the classic on vector measures by Diestel and Uhl.

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  • $\begingroup$ I have had $( C_0(\mathbb{R}^n;E) )'$ on my mind: is it equal to ${\cal M}(\mathbb{R}^n;E)$? From your post I conclude it should be: $( C_0(\mathbb{R}^n;E) )'={\cal M}(\mathbb{R}^n;E')$. Thank you for your answer and the given reference book, I'll check it. $\endgroup$
    – Robert Vu
    Commented Dec 3, 2013 at 13:00
  • $\begingroup$ Yes,but note that the measures are not defined in terms of the Banach space dual, but the latter as a complete locally convex space with the so-called bounded weak star topology. Usually they are stated for functions on compact spaces but you can reduce to this case by using the one-point compactification. $\endgroup$
    – User4891
    Commented Dec 3, 2013 at 13:47

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