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Jul 11, 2011 at 21:22 vote accept CommunityBot
Jul 11, 2011 at 21:22 history bounty ended RadonNikodym
Jul 7, 2011 at 20:28 answer added Helge timeline score: 0
Jul 7, 2011 at 15:12 comment added Helge Of course not. $Lf$ is not defined.
Jul 7, 2011 at 14:13 comment added RadonNikodym The question is: Given f with the properties specified can I write $(-Lf,f)$ as $\int \nabla f \cdot A(x) \nabla f \pi(dx)$
Jul 7, 2011 at 3:46 comment added Helge What is the question? I just read it three times, and am not sure what the question is. My best guess is: "What does $f\in D((-L)^(1/2))$ mean?"
Jul 6, 2011 at 17:09 answer added Jeff Schenker timeline score: 1
Jul 4, 2011 at 21:37 answer added paul garrett timeline score: 4
Jul 4, 2011 at 20:24 history bounty started RadonNikodym
Jun 30, 2011 at 15:19 comment added RadonNikodym @Florian. Yes I would to know of results where $\pi$ is as general as possible (for example the underlying $\sigma$-algebra$ not necessarily countably generated, etc)
Jun 30, 2011 at 15:16 history edited RadonNikodym CC BY-SA 3.0
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Jun 30, 2011 at 15:14 comment added RadonNikodym Yes, actually I should have mentioned that assumption also. Thanks for pointing this out.
Jun 30, 2011 at 15:11 comment added Florian Am I right that you want to study this problem in the abstract (i.e. $\pi$ is a measure on some unspecified measurable space etc.)? Then your assumption that the operators $D_i$ are only skew-Hermitian seems too weak to me. A more natural assumption would be that the $D_i$ are skew-adjoint, that is, the domain of the adjoint $D_i^*$ equals that of $D_i$ (in addition to the skew-symmetry).
Jun 30, 2011 at 12:11 history asked RadonNikodym CC BY-SA 3.0