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Dec 3, 2013 at 13:47 comment added User4891 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.
Dec 3, 2013 at 13:04 vote accept Robert Vu
Dec 3, 2013 at 13:00 comment added Robert Vu 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.
Dec 2, 2013 at 23:28 review First posts
Dec 2, 2013 at 23:31
Dec 2, 2013 at 23:10 history answered User4891 CC BY-SA 3.0