Timeline for How to show that there's a continuous function separating convex sets of Radon measures?
Current License: CC BY-SA 3.0
9 events
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| Feb 10, 2016 at 21:15 | comment | added | dalry | If $X$ is not compact but, say, completely regular then one can consider the space of bounded, continuous functions with the so-called strict topology, which was introduced by R.C. Buck in the locally compact space and extended to the general case by several authors. Its dual is the space of bounded tight (or Radon) measures. Many features of the compact case carry over to this situation, but at the cost of replacing the concept of a Banach space by that of a Saks space. For references, google the latter phrase. | |
| Feb 10, 2016 at 20:41 | comment | added | Yun | Thank you for pointing out this. I realize that my original thought was with strong topology. Do you know how general this conclusion (with weak topology) is? I mean, how about if $X$ is noncompact? Do you have any reference explaining it? Thanks! | |
| Feb 10, 2016 at 19:48 | comment | added | dalry | We seem to be talking at cross purposes. The dual of $M(K)$ with the weak star topology is $C(K)$, just as the dual of any dual Banach space with the weak star topology coincides with the original space. | |
| Feb 10, 2016 at 18:39 | comment | added | Yun | Agree. But my comment still holds without the last sentence "i.e., by specifying the exact topology". | |
| Feb 10, 2016 at 18:36 | history | edited | Yun | CC BY-SA 3.0 |
added 3 characters in body
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| Feb 10, 2016 at 17:25 | comment | added | dalry | The exact topology IS specified---see the third line of the answer referred to. | |
| Feb 10, 2016 at 17:02 | review | Late answers | |||
| Feb 10, 2016 at 17:18 | |||||
| Feb 10, 2016 at 16:47 | review | First posts | |||
| Feb 10, 2016 at 16:47 | |||||
| Feb 10, 2016 at 16:43 | history | answered | Yun | CC BY-SA 3.0 |