Since I am reading some stuff about weak convergence of probability measures, I started to wonder what is the dual space of the space consisting of all the finite (signed) measures (which is well known to be a Banach space with the norm being total variation). Is there any characterization of it? We may impose extra assumptions on the underlying space if necessary.
Dual space of the space of finite measures
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3Dunford and Schwartz, *[Linear Operators I](http://books.google.com/books?id=MpqESQAACAAJ)* doesn't contain a description of that space. [In a footnote](http://i.stack.imgur.com/76dSu.png) on p.374 they remark that no satisfactory descriptions were known at the time of writing. This suggests that there is no easy characterization as you're asking for. – 2011-10-22
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0@t.b.: Thanks for sharing. Perhaps the one GEdgar gives below is one the "various sorts of representations" they mentioned in the footnote? – 2011-10-23
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0yes, I think so. It is unsatisfactory in that you can't say what you really get. Something humungous, in any case. – 2011-10-23
2 Answers
Well, your space of measures is isometric to $L^1(\mu)$ for some (probably very big, non-sigma-finite) measure $\mu$. So it is enough to know what is the dual of an $L^1$ space.
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1Thanks, could you sketch how the "big space" is obtained? Or could you give some reference? – 2011-10-22
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0Take a maximal family $\mathcal A$ of mutually singular probability measures. (Use Zorn's Lemma.) The space of measures is isometrically the $l_1$-sum of $L^1(\nu)$ as $\nu$ ranges over the family $\mathcal A$. – 2011-10-22
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0So there is no canonical representation for it? What's dual of $L^1$ when the space is not $\sigma$-finite? – 2011-10-23
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1@Xianghong: Note that you have an $l_1$-sum of $L^1(\nu)$ where $\nu$ are probabilities. Its dual space is the $l_\infty$-product of the corresponding $L^\infty(\nu)$-spaces. – 2011-10-23
In the case of measures on a compact space, you are talking about the bidual of $C(K)$. This space was investigated in detail by S. Kaplan who wrote a series of long papers on it in the Transactions---easily available online. He also produced a book summarising his results. The natural extension for completely regular spaces would be the bidual of the space of bounded, continuous functions thereon, with the strict topology. This is certainly an interesting space and many of Kaplan's results carry over in suitably modified form but nobody has written this up to my knowledge.
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3It would help a lot if you told us the title of the book (and of those articles) - Kaplan is quite a common name. – 2017-08-07