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Riesz-Representation-Theorem states that every positive linear functional $F$ for any finite continuous $f$ on a local compact space S one can find a unique borel measure, such that $$F(f)=\int f d\mu$$.

Question:

  1. In the dual space there are not only positive linear functional. The dual space includes all linear functional. So according to the version of Riesz-representation theorem that I stated above, the set of finite borel-measure is only a subset of the dual space. What is the sufficient condition, that every element in the dual space can be written in the form of Riesz-Representation? Will the continuity of the functional help if we introduce the weak topology on the space?
  2. What about if I take a separable metrizable space S instead of local compact space S? Can we still apply the theorem?
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  • $\begingroup$ Actually, we should note that Riesz, Markov, and Kakutani all independently proved this theorem and related, at a time when there was not good communication among their countries of residence (not to mention lack of internet). $\endgroup$ – paul garrett Jul 14 '17 at 21:43
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Yes, you're right, there are two forms of the Riesz-Markov-Kakutani theorem, namely, the case of "positive functionals" on $C^o(X)$ with locally compact, Hausdorff, maybe separable $X$, and the case of continuous functionals on that space. The former is perhaps more popular, because the (LF-space, that is, strict colimit/inductive limit) topology on $C^o_c(X)$ need not be explained. But, yes, this has the deficiency that it does not produce a vector space of functionals. It has the merit that it allows $+\infty$ as values. The other version, that accommodates signed measures or complex-valued measures, does produce a vector space.

EDIT: in response to a comment, the space of positive (regular... Borel...) measures is not a vector space over $\mathbb R$, much less over $\mathbb C$, because it is not closed under scalar multiplication... Nothing too subtle here.

And, yes, we only want continuous functionals (here, integrals made from measures) on $C^o_c(X)$, which must refer to the natural LF-space topology on this space of functions.

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  • $\begingroup$ Thank you for the answer. Could you explain why the first version does not produce a vector space of functionals? And If we are talking about Borel measure as dual space, then we will apply the second version? That would mean, that we can only take the space of Borel-measure as dual space if we are using the proper topology such that all functional are continuous? $\endgroup$ – quallenjäger Jul 14 '17 at 22:01

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