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:
- 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?
- What about if I take a separable metrizable space S instead of local compact space S? Can we still apply the theorem?
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.